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Calculating the square root of 2
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Since 2–√2 is irrational, is there a way to compute the first 20 digits of it?
What I have done so far
I started the first digit decimal of the 2–√2 by calculating iteratively so that it would not go to 3 so fast. It looks like this:
2–√2–√2–√=1.4 2≡1.96=1.41 2≡1.9881=1.414 2≡1.999396…2=1.4 2≡1.96 2=1.41 2≡1.9881 2=1.414 2≡1.999396…
First I tell whether it passes such that 1.x 2 1.x 2 would be not greater than 3.
If that passes, I will add a new decimal to it. Let's say y.y.1.x y 2 1.x y 2
If that y fails, I increment y y by 1 and square it again.
The process will keep repeating. Unfortunately, the process takes so much time.
approximation
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edited Sep 15, 2018 at 11:36
amWhy
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asked Sep 14, 2018 at 12:10
MMJMMMJM
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4 You can go on trying to compute the square of 1.414 x 1.414 x, where x x is a number between 0 0 and 9 9. The greatest number between 1.4140 1.4140 and 1.4149 1.4149 such that its square is less then 2 2 is your next candidate to repeat the process. –Gibbs Commented Sep 14, 2018 at 12:14
2 See en.wikipedia.org/wiki/Methods_of_computing_square_roots –lhf Commented Sep 14, 2018 at 12:17
@Gibbs I tried that so far. But the reason is that it takes more time to compute it. –MMJM Commented Sep 14, 2018 at 12:21
2 Possible duplicate of 1. calculate-more-digits-of-square-root-of-22. is-there-any-simple-method-to-calculate-sqrt-x-without-using-logarithm3. –user202729 Commented Sep 14, 2018 at 15:22
1 @Gibbs Please don't post answers as comments. –David Richerby Commented Sep 14, 2018 at 18:44
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Calculating the square root of a number is one of the first problems tackled with numerical methods, known I think to the ancient Babylonians. The observation is that if x,y>0 x,y>0 and y≠x−−√y≠x then y,x/y y,x/y will be on opposite sides of x−−√x, and we could try averaging them. So try y 0=1,y n+1=1 2(y n+x y n)y 0=1,y n+1=1 2(y n+x y n). This is actually the Newton-Raphson method 5xum mentioned. The number of correct decimal places approximately doubles at each stage, i.e. you probably only have to go as far as y 5 y 5 or so.
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edited May 16, 2019 at 13:44
answered Sep 14, 2018 at 12:17
J.G.J.G.
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18 Definitely one of the fastest methods: y 0=1.0;y 1=1.5;y 2=1.41 666666666666666666666666666...;y 3=1.41421 568627450980392156862745...;y 4=1.41421356237 468991062629557889...;y 5=1.41421356237309504880168 962350...;⋯y 0=1.0;y 1=1.5;y 2=1.41 666666666666666666666666666...;y 3=1.41421 568627450980392156862745...;y 4=1.41421356237 468991062629557889...;y 5=1.41421356237309504880168 962350...;⋯ –Oleg567 Commented Sep 14, 2018 at 12:26
6 @Oleg567 We could go even faster with post-Newton Householder methods, but the individual steps become more computationally complex. BTW the calculator you used to check that probably also used Newton-Raphson for the division. –J.G. Commented Sep 14, 2018 at 12:30
2 The beauty of this method is that the initial estimate can be way off and the method will converge quickly anyway. of course, making an educated guess to pick the initial estimate helps to reduce the number of iterations. –Vasili Commented Sep 14, 2018 at 12:32
4 Love the intuitive explanation for it! –Sort of Damocles Commented Sep 14, 2018 at 12:52
1 @Paul Since it's Newton-Raphson it'll be about 2 n 2 n of them, but a more detailed answer than that can't be obtained without careful analysis of the specifics of the problem. However, if you look at how which digits have "gotten stuck", you can be confident from the shrinking error terms that they won't change. See the black digits in Oleg567's comment for an example. –J.G. Commented Sep 14, 2018 at 22:04
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Here's the way I learnt to obtain decimal digit after decimal digit when I began middle school:
2 1 00−96−0 4 00−2 81−0 119 00 0−112 96 00 604 00(1.414 2…24×4=96<100 25×5=125>100 281×1<400 282×2>400 2824×4<11900 2825×5>11900 28282×2<60400 28283×3>60400 2(1.414 2…1 00 24×4=96<100−96 25×5=125>100−0 4 00 281×1<400−2 81 282×2>400−0 119 00 2824×4<11900 0−112 96 2825×5>11900 00 604 00 28282×2<60400 28283×3>60400
&c.
Let me explain the procedure on the first two steps. It relies on a clever use of the identity (x+y)2=x 2+2 x y+y 2(x+y)2=x 2+2 x y+y 2. Suppose more generally we want to find the square root of a number a a.
We first find the greatest natural number n n such that n 2≤a n 2≤a.
If a a is not a perfect square, i.e. if n 2<a n 2<a, let d d be the first decimal digit of the square root. This is the greatest digit such that (n+d 10)2≤a(n+d 10)2≤a. We'll transform this inequality into a more easy-to-use test: (n+d 10)2≤a⟺2 n 10 d+d 2 100<a−n 2⟺(10×2 n+d)×d≤(a−n 2)×100(n+d 10)2≤a⟺2 n 10 d+d 2 100<a−n 2⟺(10×2 n+d)×d≤(a−n 2)×100 In practice, this means, we calculate the difference a−n 2 a−n 2 and add two 0s. Then we double n n, add a digit d (this is the result of calculating 10×2 n+d 10×2 n+d) and multiply what we obtain by this digit. Last, we test whether the result is less than 100(a−n 2)100(a−n 2), and retain the largest possible digit.
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edited Sep 15, 2018 at 9:43
answered Sep 14, 2018 at 12:55
BernardBernard
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4 Looks interesting, can you talk us through it a bit? I don't really get it. e.g. where does 100 come from? –goblin GONE Commented Sep 15, 2018 at 1:57
@goblin, There are some references for this method at math.stackexchange.com/a/538055/117057 and math.stackexchange.com/q/376365/117057 –shoover Commented Sep 15, 2018 at 5:21
@goblin: I've added an explanation for the first two steps. The following stepsruns along te same lines, only the first step is different. Hope this will make it clear. –Bernard Commented Sep 15, 2018 at 9:24
@Bernard, thanks. –goblin GONE Commented Sep 15, 2018 at 9:39
@goblin You start off with 1 because 1 is the largest integer whoose square is less than 2. Then extend 1 by the next two digits, 00, to get 100. Now double the 1 just obtained and find the largest digit such that 2x times x is less than 100. –Paul Evans Commented Sep 15, 2018 at 9:41
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On a similar note to the answer by R. Romero: in the special case of taking the square root of an integer N N, it is fairly straightforward to calculate the continued fraction representation of N−−√N.
In the particular case N=2 N=2, we have:
2–√=1+1 2+1 2+1 2+⋱.2=1+1 2+1 2+1 2+⋱.
(This follows from the fact that if x=2–√−1 x=2−1, then x=2–√−1=1 2√+1=1 2+x x=2−1=1 2+1=1 2+x.)
Now, from this we can calculate subsequent rational approximations to 2–√2:
0 1 1 0 1 1 1 2 3 2 2 7 5 2 17 12 2 41 29 2 99 70⋯⋯⋯1 2 2 2 2 2⋯0 1 1 3 7 17 41 99⋯1 0 1 2 5 12 29 70⋯
So, for example 99 70≈1.4142857 99 70≈1.4142857 whereas 2–√≈1.4142136 2≈1.4142136.
(It also happens that this procedure generates solutions to Pell's equation a 2−2 b 2=±1 a 2−2 b 2=±1; for example, 99 2−2⋅70 2=1 99 2−2⋅70 2=1. The connection is: if a 2−2 b 2=±1 a 2−2 b 2=±1 then a−b 2–√=±1 a+b 2√a−b 2=±1 a+b 2; so if a a and b b are large positive integers satisfying Pell's equation, then a−b 2–√≈±1 2 a a−b 2≈±1 2 a which implies a b−2–√≈±1 2 a b≈±1 a 2 2√a b−2≈±1 2 a b≈±1 a 2 2.)
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answered Sep 14, 2018 at 22:43
Daniel ScheplerDaniel Schepler
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2 Is there somewhere I can read more about this, especially the connection between continued fractions and Pell's equation? –goblin GONE Commented Sep 15, 2018 at 2:00
1 Once you see the first few rational approximations it's easy to guess and prove the recursion for p/q p/q, namely, p n=p n−1+2 q n−1 p n=p n−1+2 q n−1, q n=p n−1+q n−1 q n=p n−1+q n−1..See en.wikipedia.org/wiki/…, en.wikipedia.org/wiki/Pell%27s_equation –Ethan Bolker Commented Sep 15, 2018 at 13:13
There's also a simple two step recursion, which is identical for the numerator & denominator sequences. Using Ethan's notation, p n+1=2 p n+p n−1 p n+1=2 p n+p n−1 and q n+1=2 q n+q n−1 q n+1=2 q n+q n−1. Also p n=q n+1−q n p n=q n+1−q n and q n=(p n+p n−1)/2 q n=(p n+p n−1)/2. –PM 2Ring Commented May 26, 2021 at 10:19
can someone please explain that table where "subsequent rational approximations" appear? how does one get that from a known continued fraction? –Noone AtAll Commented Aug 15, 2021 at 12:46
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The number 2–√2 is the solution to the equation x 2−2=0 x 2−2=0, so any method for numerically approximating the roots of an equation (such as the Newton method) will be able to approximate 2–√2.
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answered Sep 14, 2018 at 12:13
5xum5xum
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9 I don't see how this qualifies as an answer. It is just a general statement. –M. Wind Commented Sep 16, 2018 at 5:29
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Okay, I searched through the answers, but none seems to mention this one: long quadratic root calculation.
From the name it is obvious that it resembles long division, like this:
2.00 00 00 00..−−−−−−−−−−−−√2.00 00 00 00..
Notice how they are grouped into tuples. Now estimate the first digit, namely 1 1:
1 1.2.00 00 00 00..−−−−−−−−−−−−√1 1 00¯¯¯¯¯¯¯¯1.1 2.00 00 00 00..1 1 00¯
We calculate 1×1=1 1×1=1, write it down, and calculate the "remainder", just like divisions. Notice that we append 2 digits behind instead of 1.
Next, double the number on the top, and write it on the left of 1 00 1 00:
1 2∗1.∗2.00 00 00 00..−−−−−−−−−−−−√1|1 00¯¯¯¯¯¯¯¯1.∗1 2.00 00 00 00..1 2∗|1 00¯
Now we estimate the next digit, . It is written both on the top and to the left. Of course, we know that it is 4, so:
1 24 1.4∗2.00 00 00 00..−−−−−−−−−−−−√1|1 00¯¯¯¯¯¯¯¯|96 2 8∗|4 00¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯1.4∗1 2.00 00 00 00..1 24|1 00¯|96 2 8∗|4 00¯
We double the numbers on the top again to get 28∗28∗, and repeat the process:
1 24 1.4 1 2.00 00 00 00..−−−−−−−−−−−−√1|1 00¯¯¯¯¯¯¯¯|96 2 8 1|4 00¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯|2 81 1.4 1 1 2.00 00 00 00..1 24|1 00¯|96 2 8 1|4 00¯|2 81
I found a picture, but not of 2–√2:
This is extremely inefficient for computers, but great for manual calculation. After all, we don't do multiplication through fast Fourier transforms!
Also, this method is developed in ancient China.
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answered Sep 17, 2018 at 7:44
TreborTrebor
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Suppose you want to find the square root of p p and suppose your initial guess is x/y x/y:
Let M=[1 1 p 1]M=[1 p 1 1] and q=(x y)q=(x y) Then M M M...q M M M...q gives a numerator and denominator the ratio of which converges to the square root of p p. This gives an approximation to the square root of 2 2 as fast as the other methods but with no floating point arithmetic until the final division.
Performs well for calculation tools optimized for Matrix arithmetic. This also gives you solutions for Pell's equation for p=2 p=2 as mentioned by Daniel Schepler.
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edited Sep 15, 2018 at 3:29
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In this answer, there is a method using continued fraction approximations for 2–√2 and the generating function for the central binomial coefficients to get some very quickly convergent series for 2–√2. For example,
2–√=7 5∑k=0∞(2 k k)1 200 k(1)(1)2=7 5∑k=0∞(2 k k)1 200 k
and
2–√=239 169∑k=0∞(2 k k)1 228488 k(2)(2)2=239 169∑k=0∞(2 k k)1 228488 k
For example, summing to k=4 k=4 in (2)(2) gives
2–√=1.414213562373095048801688 2=1.414213562373095048801688
which is accurate to 23 23 places.
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edited Sep 15, 2018 at 23:39
answered Sep 15, 2018 at 13:50
robjohn♦robjohn
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(+) This method could be used to calculate millions of 2–√2 digits (especially when notice that the series has rational terms, and apply en.wikipedia.org/wiki/Binary_splitting technique). –Oleg567 Commented Sep 24, 2018 at 4:44
Good description of the elementary school method. You are not likely to succeed unless you do at least one square root every day. After age 75, you have to do more than one every day. That is why there are calculators. –richard1941 Commented Sep 24, 2018 at 23:53
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Binary search for it.
Since 1<2<4 1<2<4, we must have 1–√<2–√<4–√1<2<4, so 2–√∈(1,2)2∈(1,2). Now repeatedly: find the midpoint, m m, of the current interval, (a,b)(a,b), square m m and compare with 2 2, and if 2=m 2 2=m 2 declare that m=2–√m=2, or if 2<m 2 2<m 2, make the new interval (a,m)(a,m), otherwise make the new interval (m,b)(m,b). This process halves the size of the interval on each step. Since log 2(10−20)=−66.438…log 2(10−20)=−66.438…, after 67 doublings, the error in taking any value from the interval is <10−20<10−20 (but, if the interval straddles a digit change, you may have to perform additional steps to find out on which side of the change is 2–√2).
This process is shown in the table below. Each decimal number is computed to 21 21 digits and has trailing zeroes stripped. If there are still 21 21 digits, a space is inserted between the 20 th 20 th and 21 st 21 st.
step 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 interval(1.,2.)(1.,1.5)(1.25,1.5)(1.375,1.5)(1.375,1.4375)(1.40625,1.4375)(1.40625,1.421875)(1.4140625,1.421875)(1.4140625,1.41796875)(1.4140625,1.416015625)(1.4140625,1.4150390625)(1.4140625,1.41455078125)(1.4140625,1.414306640625)(1.4141845703125,1.414306640625)(1.4141845703125,1.41424560546875)(1.4141845703125,1.414215087890625)(1.4141998291015625,1.414215087890625)(1.41420745849609375,1.414215087890625)(1.414211273193359375,1.414215087890625)(1.4142131805419921875,1.414215087890625)m 1.5 1.25 1.375 1.4375 1.40625 1.421875 1.4140625 1.41796875 1.416015625 1.4150390625 1.41455078125 1.414306640625 1.4141845703125 1.41424560546875 1.414215087890625 1.4141998291015625 1.41420745849609375 1.414211273193359375 1.4142131805419921875 1.41421413421630859375 m 2 2<2.25 1.5625<2 1.890625<2 2<2.06640625 1.9775390625<2 2<2.021728515625 1.99957275390625<2 2<2.0106353759765625 2<2.005100250244140625 2<2.00233554840087890625 2<2.00095391273498535156 3 2<2.00026327371597290039 1.99991799890995025634 8<2 2<2.00009063258767127990 7 2<2.00000431481748819351 2 1.99996115663088858127 6<2 1.99998273566598072648<2 1.99999352522718254476 8<2 1.99999892001869739033 3<2 2<2.00000161741718329722 step interval m m 2 1(1.,2.)1.5 2<2.25 2(1.,1.5)1.25 1.5625<2 3(1.25,1.5)1.375 1.890625<2 4(1.375,1.5)1.4375 2<2.06640625 5(1.375,1.4375)1.40625 1.9775390625<2 6(1.40625,1.4375)1.421875 2<2.021728515625 7(1.40625,1.421875)1.4140625 1.99957275390625<2 8(1.4140625,1.421875)1.41796875 2<2.0106353759765625 9(1.4140625,1.41796875)1.416015625 2<2.005100250244140625 10(1.4140625,1.416015625)1.4150390625 2<2.00233554840087890625 11(1.4140625,1.4150390625)1.41455078125 2<2.00095391273498535156 3 12(1.4140625,1.41455078125)1.414306640625 2<2.00026327371597290039 13(1.4140625,1.414306640625)1.4141845703125 1.99991799890995025634 8<2 14(1.4141845703125,1.414306640625)1.41424560546875 2<2.00009063258767127990 7 15(1.4141845703125,1.41424560546875)1.414215087890625 2<2.00000431481748819351 2 16(1.4141845703125,1.414215087890625)1.4141998291015625 1.99996115663088858127 6<2 17(1.4141998291015625,1.414215087890625)1.41420745849609375 1.99998273566598072648<2 18(1.41420745849609375,1.414215087890625)1.414211273193359375 1.99999352522718254476 8<2 19(1.414211273193359375,1.414215087890625)1.4142131805419921875 1.99999892001869739033 3<2 20(1.4142131805419921875,1.414215087890625)1.41421413421630859375 2<2.00000161741718329722
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35(1.4142131805419921875,1.41421413421630859375)(1.4142131805419921875,1.41421365737915039062 5)(1.41421341896057128906 2,1.41421365737915039062 5)(1.41421353816986083984 4,1.41421365737915039062 5)(1.41421353816986083984 4,1.41421359777450561523 4)(1.41421353816986083984 4,1.41421356797218322753 9)(1.41421355307102203369 1,1.41421356797218322753 9)(1.41421356052160263061 5,1.41421356797218322753 9)(1.41421356052160263061 5,1.41421356424689292907 7)(1.41421356052160263061 5,1.41421356238424777984 6)(1.41421356145292520523,1.41421356238424777984 6)(1.41421356191858649253 8,1.41421356238424777984 6)(1.41421356215141713619 2,1.41421356238424777984 6)(1.41421356226783245801 9,1.41421356238424777984 6)(1.41421356232604011893 3,1.41421356238424777984 6)1.41421365737915039062 5 1.41421341896057128906 2 1.41421353816986083984 4 1.41421359777450561523 4 1.41421356797218322753 9 1.41421355307102203369 1 1.41421356052160263061 5 1.41421356424689292907 7 1.41421356238424777984 6 1.41421356145292520523 1.41421356191858649253 8 1.41421356215141713619 2 1.41421356226783245801 9 1.41421356232604011893 3 1.41421356235514394938 9 2<2.00000026871771297010 1 1.99999959436814833679 8<2 1.99999993154291644259 5<2 2<2.00000010013031115363 4 2<2.00000001583661290993 6 1.99999997368976445422 1<2 1.99999999476318862656 8<2 2<2.00000000529990075437 4 2<2.00000000003154468700 1 1.99999999739736665591 7<2 1.99999999871445567124 2<2 1.99999999937300017906 8<2 1.99999999970227243302 1<2 1.99999999986690856000 8<2 1.99999999994922662350 4<2 21(1.4142131805419921875,1.41421413421630859375)1.41421365737915039062 5 2<2.00000026871771297010 1 22(1.4142131805419921875,1.41421365737915039062 5)1.41421341896057128906 2 1.99999959436814833679 8<2 23(1.41421341896057128906 2,1.41421365737915039062 5)1.41421353816986083984 4 1.99999993154291644259 5<2 24(1.41421353816986083984 4,1.41421365737915039062 5)1.41421359777450561523 4 2<2.00000010013031115363 4 25(1.41421353816986083984 4,1.41421359777450561523 4)1.41421356797218322753 9 2<2.00000001583661290993 6 26(1.41421353816986083984 4,1.41421356797218322753 9)1.41421355307102203369 1 1.99999997368976445422 1<2 27(1.41421355307102203369 1,1.41421356797218322753 9)1.41421356052160263061 5 1.99999999476318862656 8<2 28(1.41421356052160263061 5,1.41421356797218322753 9)1.41421356424689292907 7 2<2.00000000529990075437 4 29(1.41421356052160263061 5,1.41421356424689292907 7)1.41421356238424777984 6 2<2.00000000003154468700 1 30(1.41421356052160263061 5,1.41421356238424777984 6)1.41421356145292520523 1.99999999739736665591 7<2 31(1.41421356145292520523,1.41421356238424777984 6)1.41421356191858649253 8 1.99999999871445567124 2<2 32(1.41421356191858649253 8,1.41421356238424777984 6)1.41421356215141713619 2 1.99999999937300017906 8<2 33(1.41421356215141713619 2,1.41421356238424777984 6)1.41421356226783245801 9 1.99999999970227243302 1<2 34(1.41421356226783245801 9,1.41421356238424777984 6)1.41421356232604011893 3 1.99999999986690856000 8<2 35(1.41421356232604011893 3,1.41421356238424777984 6)1.41421356235514394938 9 1.99999999994922662350 4<2
36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69(1.41421356235514394938 9,1.41421356238424777984 6)(1.41421356236969586461 8,1.41421356238424777984 6)(1.41421356236969586461 8,1.41421356237697182223 2)(1.41421356236969586461 8,1.41421356237333384342 5)(1.41421356237151485402 1,1.41421356237333384342 5)(1.41421356237242434872 3,1.41421356237333384342 5)(1.41421356237287909607 4,1.41421356237333384342 5)(1.41421356237287909607 4,1.41421356237310646974 9)(1.41421356237299278291 2,1.41421356237310646974 9)(1.41421356237304962633,1.41421356237310646974 9)(1.41421356237307804804,1.41421356237310646974 9)(1.41421356237309225889 5,1.41421356237310646974 9)(1.41421356237309225889 5,1.41421356237309936432 2)(1.41421356237309225889 5,1.41421356237309581160 8)(1.41421356237309403525 2,1.41421356237309581160 8)(1.41421356237309492343,1.41421356237309581160 8)(1.41421356237309492343,1.41421356237309536751 9)(1.41421356237309492343,1.41421356237309514547 5)(1.41421356237309503445 2,1.41421356237309514547 5)(1.41421356237309503445 2,1.41421356237309508996 3)(1.41421356237309503445 2,1.41421356237309506220 8)(1.41421356237309504833,1.41421356237309506220 8)(1.41421356237309504833,1.41421356237309505526 9)(1.41421356237309504833,1.41421356237309505180 0)(1.41421356237309504833,1.41421356237309505006 5)(1.41421356237309504833,1.41421356237309504919 7)(1.41421356237309504876 4,1.41421356237309504919 7)(1.41421356237309504876 4,1.41421356237309504898)(1.41421356237309504876 4,1.41421356237309504887 2)(1.41421356237309504876 4,1.41421356237309504881 8)(1.41421356237309504879,1.41421356237309504881 8)(1.41421356237309504879,1.41421356237309504880 4)(1.41421356237309504879 8,1.41421356237309504880 4)(1.41421356237309504880 1,1.41421356237309504880 4)1.41421356236969586461 8 1.41421356237697182223 2 1.41421356237333384342 5 1.41421356237151485402 1 1.41421356237242434872 3 1.41421356237287909607 4 1.41421356237310646974 9 1.41421356237299278291 2 1.41421356237304962633 1.41421356237307804804 1.41421356237309225889 5 1.41421356237309936432 2 1.41421356237309581160 8 1.41421356237309403525 2 1.41421356237309492343 1.41421356237309536751 9 1.41421356237309514547 5 1.41421356237309503445 2 1.41421356237309508996 3 1.41421356237309506220 8 1.41421356237309504833 1.41421356237309505526 9 1.41421356237309505180 0 1.41421356237309505006 5 1.41421356237309504919 7 1.41421356237309504876 4 1.41421356237309504898 1.41421356237309504887 2 1.41421356237309504881 8 1.41421356237309504879 1.41421356237309504880 4 1.41421356237309504879 8 1.41421356237309504880 1 1.41421356237309504880 3 1.99999999999038565525 2<2 2<2.00000000001096517112 7 2<2.00000000000067541319 0 1.99999999999553053422 1<2 1.99999999999810297370 5<2 1.99999999999938919344 7<2 2<2.00000000000003230331 9 1.99999999999971074838 3<2 1.99999999999987152585<2 1.99999999999995191458 5<2 1.99999999999999210895 2<2 2<2.00000000000001220613 5 2<2.00000000000000215754 3 1.99999999999999713324 7<2 1.99999999999999964539 5<2 2<2.00000000000000090146 9 2<2.00000000000000027343 2 1.99999999999999995941 4<2 2<2.00000000000000011642 3 2<2.00000000000000003791 8 1.99999999999999999866 6<2 2<2.00000000000000001829 2 2<2.00000000000000000847 9 2<2.00000000000000000357 3 2<2.00000000000000000111 9 1.99999999999999999989 3<2 2<2.00000000000000000050 6 2<2.00000000000000000019 9 2<2.00000000000000000004 6 1.99999999999999999996 9<2 2<2.00000000000000000000 8 1.99999999999999999998 9<2 1.99999999999999999999 8<2 2<2.00000000000000000000 3 36(1.41421356235514394938 9,1.41421356238424777984 6)1.41421356236969586461 8 1.99999999999038565525 2<2 37(1.41421356236969586461 8,1.41421356238424777984 6)1.41421356237697182223 2 2<2.00000000001096517112 7 38(1.41421356236969586461 8,1.41421356237697182223 2)1.41421356237333384342 5 2<2.00000000000067541319 0 39(1.41421356236969586461 8,1.41421356237333384342 5)1.41421356237151485402 1 1.99999999999553053422 1<2 40(1.41421356237151485402 1,1.41421356237333384342 5)1.41421356237242434872 3 1.99999999999810297370 5<2 41(1.41421356237242434872 3,1.41421356237333384342 5)1.41421356237287909607 4 1.99999999999938919344 7<2 42(1.41421356237287909607 4,1.41421356237333384342 5)1.41421356237310646974 9 2<2.00000000000003230331 9 43(1.41421356237287909607 4,1.41421356237310646974 9)1.41421356237299278291 2 1.99999999999971074838 3<2 44(1.41421356237299278291 2,1.41421356237310646974 9)1.41421356237304962633 1.99999999999987152585<2 45(1.41421356237304962633,1.41421356237310646974 9)1.41421356237307804804 1.99999999999995191458 5<2 46(1.41421356237307804804,1.41421356237310646974 9)1.41421356237309225889 5 1.99999999999999210895 2<2 47(1.41421356237309225889 5,1.41421356237310646974 9)1.41421356237309936432 2 2<2.00000000000001220613 5 48(1.41421356237309225889 5,1.41421356237309936432 2)1.41421356237309581160 8 2<2.00000000000000215754 3 49(1.41421356237309225889 5,1.41421356237309581160 8)1.41421356237309403525 2 1.99999999999999713324 7<2 50(1.41421356237309403525 2,1.41421356237309581160 8)1.41421356237309492343 1.99999999999999964539 5<2 51(1.41421356237309492343,1.41421356237309581160 8)1.41421356237309536751 9 2<2.00000000000000090146 9 52(1.41421356237309492343,1.41421356237309536751 9)1.41421356237309514547 5 2<2.00000000000000027343 2 53(1.41421356237309492343,1.41421356237309514547 5)1.41421356237309503445 2 1.99999999999999995941 4<2 54(1.41421356237309503445 2,1.41421356237309514547 5)1.41421356237309508996 3 2<2.00000000000000011642 3 55(1.41421356237309503445 2,1.41421356237309508996 3)1.41421356237309506220 8 2<2.00000000000000003791 8 56(1.41421356237309503445 2,1.41421356237309506220 8)1.41421356237309504833 1.99999999999999999866 6<2 57(1.41421356237309504833,1.41421356237309506220 8)1.41421356237309505526 9 2<2.00000000000000001829 2 58(1.41421356237309504833,1.41421356237309505526 9)1.41421356237309505180 0 2<2.00000000000000000847 9 59(1.41421356237309504833,1.41421356237309505180 0)1.41421356237309505006 5 2<2.00000000000000000357 3 60(1.41421356237309504833,1.41421356237309505006 5)1.41421356237309504919 7 2<2.00000000000000000111 9 61(1.41421356237309504833,1.41421356237309504919 7)1.41421356237309504876 4 1.99999999999999999989 3<2 62(1.41421356237309504876 4,1.41421356237309504919 7)1.41421356237309504898 2<2.00000000000000000050 6 63(1.41421356237309504876 4,1.41421356237309504898)1.41421356237309504887 2 2<2.00000000000000000019 9 64(1.41421356237309504876 4,1.41421356237309504887 2)1.41421356237309504881 8 2<2.00000000000000000004 6 65(1.41421356237309504876 4,1.41421356237309504881 8)1.41421356237309504879 1.99999999999999999996 9<2 66(1.41421356237309504879,1.41421356237309504881 8)1.41421356237309504880 4 2<2.00000000000000000000 8 67(1.41421356237309504879,1.41421356237309504880 4)1.41421356237309504879 8 1.99999999999999999998 9<2 68(1.41421356237309504879 8,1.41421356237309504880 4)1.41421356237309504880 1 1.99999999999999999999 8<2 69(1.41421356237309504880 1,1.41421356237309504880 4)1.41421356237309504880 3 2<2.00000000000000000000 3
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answered Sep 16, 2018 at 5:59
Eric TowersEric Towers
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Using the fact that sin π 4=2√2 sinπ 4=2 2, then we have to find 2 sin π 4 2 sinπ 4.
We can approximate sin x sinx using the Taylor series to three terms:
sin x=x−x 3 3!+x 5 5!+O(x 6),sinx=x−x 3 3!+x 5 5!+O(x 6),
so we have:
sin π 4≈π 4−(π/4)3 3!+(π/4)5 5!.sinπ 4≈π 4−(π/4)3 3!+(π/4)5 5!.
If we approximate π π as 22 7 22 7, then we have π 4=11 14 π 4=11 14, then we have:
sin π 4≈11 14−(11/14)3 3!+(11/14)5 5!,sinπ 4≈11 14−(11/14)3 3!+(11/14)5 5!,
which when you multiply by 2 2 to get 2–√2, gives 1.4147 1.4147, while the actual value is 1.4142 1.4142.
If we expand the Taylor series to more terms, or improve the approximation of π π (such as 355 113 355 113), then we can get to 20 20 correct digits.
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answered Sep 14, 2018 at 13:49
Toby MakToby Mak
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2 Don’t you need pi to nearly 20 digits for this to work? –JTP - Apologise to Monica Commented Sep 15, 2018 at 2:10
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There's a general method that converges about as quickly as Newton-Raphson but is somewhat more general. It's based off of Continued Fractions:
Suppose you want to find the square root of N N. Let a+b=N a+b=N where b b has an easy to calculate square root.
let y n+1=b√+a b√+y n y n+1=b+a b+y n
y n+1 y n+1 converges to N−−√N.
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edited Sep 14, 2018 at 22:33
asky
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answered Sep 14, 2018 at 20:35
TurlocTheRedTurlocTheRed
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Start with an initial guess x x for the square root of 2 2. Then add a correction term y y. Write down (x+y)2−2=0(x+y)2−2=0. Solve this equation for y y by expanding it up to third order in the difference (2−x 2)(2−x 2). This is a straightforward calculation. Combining all contributions, the result is elegant:
x+y=(x 4+12 x 2+4)/(4 x 3+8 x)x+y=(x 4+12 x 2+4)/(4 x 3+8 x)
For a rational initial guess x x the result (x+y)(x+y) is also rational, but much closer to the desired value.
For example if we take x=3/2 x=3/2, then (x+y)=577/408(x+y)=577/408, which differs from the square root of 2 by a factor 1.0000015. If we start with x=7/5 x=7/5, the result is 19601/13860 19601/13860, which differs from the square of root of 2 2 by a factor 1.0000000013 1.0000000013
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edited Sep 16, 2018 at 16:42
answered Sep 15, 2018 at 5:36
M. WindM. Wind
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Please show what happens with 140/99. I find the error to be 1.2 10^-18 on my WP-34s iPhone emulator in double precision mode (good to at least 30 digits). If you recycle 577/408, you get an error 9.0 10^-25. That meets to goal of 20 digits. Recycling 19601/13860 gives an error of absolute zero (on the calculator). –richard1941 Commented Dec 21, 2018 at 17:50
Thanks! The initial values 99/70 99/70 and 140/99 140/99 both result in 768398401/543339720 768398401/543339720. –M. Wind Commented Dec 23, 2018 at 19:39
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You can compute it manually using the algorithm:
p=0 p=0, r=0 r=0, i=0 i=0
Split the number into sections of two digits
Take i'th section n i n i, let k=100 t+n i k=100 t+n i
Find the greatest number x x, such that y=x(20 p+x)≤k y=x(20 p+x)≤k
Assign p=10 p+x p=10 p+x, i=i+1 i=i+1, if the accyracy of the result is not satisfied, then return to 3.
Example:
02.00 00 00 00 00
n 0=2 n 0=2, k=2 k=2, therefore for x=1 x=1: y=1 y=1 and p=1 p=1
n 1=0 n 1=0, k=100 k=100, so for x=4 x=4: y=24∗4=96<100 y=24∗4=96<100 and p=14 p=14
n 2=0 n 2=0, k=400 k=400, so for x=1 x=1, y=281∗1=281<400 y=281∗1=281<400 and p=141 p=141
n 3=0 n 3=0, k=11900 k=11900, so for x=4 x=4, y=2824∗4=11296<11900 y=2824∗4=11296<11900 and p=1414 p=1414
n 4=0 n 4=0, k=60400 k=60400, so for x=2 x=2, y=28282∗2=56564<60400 y=28282∗2=56564<60400 and p=14142 p=14142
n 5=0 n 5=0, k=383600 k=383600, so for x=1 x=1, y=282841∗1=282841<383600 y=282841∗1=282841<383600 and p=141421 p=141421
...
After all just remember to point the comma in place, where it should be, ie. after first number (it depends how many sections were there on the left side of our number), so you'll have:
2–√≈1.41421 2≈1.41421
To obtain accuracy of 20 numbers after the comma, you should append 20 sections of 00 in the step 2. , ie.:
02.00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
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answered Sep 14, 2018 at 13:05
Jaroslaw MatlakJaroslaw Matlak
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Newton-Rhapson is a good idea because of the convergence rate. However, I am more of a fan of using Taylor's expansions here since it is super easy to derive on the go to give fairly ok estimates in quite a reasonable time. So, the way to go to find x−−√x is to find first the closest integer which approximates x−−√x and call this a a, then apply Taylor to a 2 a 2. Then Taylor says
x−−√≈a+(x−a 2)⋅1 2 a−(x−a 2)2/2⋅1 4 a 3+⋯.x≈a+(x−a 2)⋅1 2 a−(x−a 2)2/2⋅1 4 a 3+⋯.
The thing that is nice here is that you also get bounds on the error you make. So, denote f(x)=x−−√f(x)=x, then the error of a n n th order approximation (i.e., going as far as (x−a 2)n/n!⋅f(n)(a 2)(x−a 2)n/n!⋅f(n)(a 2) in the approximation above) is given by
(x−a)n+1/(n+1)!⋅f(n+1)(ξ)(x−a)n+1/(n+1)!⋅f(n+1)(ξ)
for a certain ξ ξ between a 2 a 2 and x x. This can be estimated quite easily since this f(n+1)f(n+1) is monotone around x x. Thus look at the boundaries of the domain of ξ ξ and find the 'best' maximal value which you can calculate without a calculator.
Example for x=2 x=2. Apparently 1 1 is the closest integer to 2–√2 and thus we will take a=1 a=1. Then, let's take a second order approximation
2–√≈1+(2−1)⋅1 2−(2−1)2/2⋅1 4=1+0.5−0.125=1.375 2≈1+(2−1)⋅1 2−(2−1)2/2⋅1 4=1+0.5−0.125=1.375
and the absolute error is given by
E=∣∣∣(2−1)3/3!⋅3 8⋅ξ 2 ξ√∣∣∣=1 16⋅1|ξ 2 ξ√|E=|(2−1)3/3!⋅3 8⋅ξ 2 ξ|=1 16⋅1|ξ 2 ξ|
for a certain ξ ξ between 1 1 and 2 2. Since this is a decreasing function on (1,2)(1,2). The maximum is attained at 1 1 and hence the error is bounded by
E≤1 16 E≤1 16
which seems to be a good estimate since E=0.039…E=0.039… and 1/16=0.0625 1/16=0.0625.
Edit As some of you noted this method 'looks' more difficult than Newton-Rhapson and the convergence is slower. The last part is obviously true and I would answer this question with: How quick do you need it to be and do you want to calculate it in your head or do you have a computer? Do you need to have a quick guess which is approximately equal to the value of 2–√2 or do you need a precise estimate. If you don't have a computer but pen and paper, the best method is Newton-Rhapson.
I would argue that my method is better if you don't have pen and paper or a computer and you are asked to give an estimation of 10−−√10 on the go (especially for x−−√x with x x big, the Taylor approximation is better since the ∙–√∙ function becomes more linear as x x grows).
I agree that my method looks way more difficult but it isn't if you get more familiar with it. Also, this method is super quick in terms of calculation time in your head and if you practice a little with it, it becomes way easier. Also, this method works particularly nice for x−−√x where x x differs one from a perfect square because then the (x−a 2)n(x−a 2)n term will always be one.
Let's look at an example here. Suppose you need to calculate 122−−−√122, then first order approximation of my method gives
122−−−√≈11+1 2⋅11.122≈11+1 2⋅11.
It took me less than one second to find this approximation and the second order approximation works almost as quick here. You just need to add −1 8⋅11 3−1 8⋅11 3. Please note that the error of the first order approximation here is approximately equal to 10−4 10−4.
If you apply Newton-Rhapson here you get the same approximation after one step if you choose x 0=11 x 0=11. The only thing is that I always forget what the exact form is of Newton-Rhapson. So when I want to apply it, I have to think about it where I could have immediately applied Taylor but I would say that is just my particular preference.
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edited Sep 15, 2018 at 11:09
answered Sep 14, 2018 at 12:39
Stan TendijckStan Tendijck
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2 I'd say this is more difficult, less precise, and not as generally applicable as Newton-Raphson. –leftaroundabout Commented Sep 14, 2018 at 14:48
I would say it is less difficult since when you apply Newton-Rhapson you always have to find the exact algorithm and this method can be applied to find 2.243−−−−√2.243 also quite quickly. –Stan Tendijck Commented Sep 14, 2018 at 15:38
I agree with @leftaroundabout, but perhaps if you edit into your post an illustration of how this method could be used by hand to compute rad 2 to high accuracy, it would appear simpler. Right now, it looks much more difficult. –Wildcard Commented Sep 14, 2018 at 18:17
3 Taylor's converges much more slowly than Newton Raphson. Note the second order term starting with initial guess 1 is 1.4166.... already correct to two digits behind the decimal. You might get an additional correct digit at each step of the calculation heavy Taylor series. The accuracy doubles per step for Newton Raphson without the difficulty of calculating the Taylor coefficients. There might be ways to patch it up. There's an alternative series to the Taylor series for arctan that converges much faster than Taylor. –TurlocTheRed Commented Sep 15, 2018 at 2:09
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I came up with an interesting, but terribly inefficient method.
Consider the sequence {x n x n}: 1,1 2,1 2,1 3,1 3,1 3,1 4,1 4,1 4,1 4 1,1 2,1 2,1 3,1 3,1 3,1 4,1 4,1 4,1 4, ...
Suppose you want k digits of the square root of 2. Then add up the first 100 k 100 k terms and then divide the sum by 10 k 10 k.
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answered Sep 28, 2018 at 1:09
TurlocTheRedTurlocTheRed
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I know an easy way to calculate the binary digits of 2–√2. Take the ordered pair (1, 2) 1 2 1 2 is less than 2 and 2 2 2 2 is more than 2. Calculate the square of the average 1.5 2 1.5 2 in base 2. The square of the average is just the average of the squares minus 1 4 1 4. The result expressed in binary is 10.01 so the first binary digit after the decimal is 0. Take the next ordered pair to be (1, 1.5) and calculate the square of its average which is the average of its squares minus 1 16 1 16. The result expressed in binary is 1.1001 so the next binary digit is 1.
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answered Oct 24, 2018 at 3:56
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Towers' bisection method above is similar to your own approach, but more efficient. Another method that is not as good as binary search, but is better than your own method, is to increment the last digit in bigger steps. I would try incrementing by 3. The worst case is that you reach the correct digit in 5 steps instead of 9.
My favorite method for mental approximation is to find the next lowest square, determine the error, and add to its square root the error divided by double the guess. For sqrt(200), the lowest square is 196. The error is 4, so my mental estimate is 14 + 4/14 = 14.142857...
I apologize for off-topic, but note that square roots can be used to calculate logarithms by a process similar to bisection. I suspect that is how it was done in the late 16th century, as they did not yet have calculus. In our times, there are extremely accurate formulas for logarithm that still require square roots. This exercise should make you appreciate the power of a square root button on a calculator, even if you have no "scientific" functions.
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answered Dec 21, 2018 at 17:26
richard1941richard1941
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Project MUSE - The Concept of Will in Early Latin Philosophy
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Journal of the History of Philosophy
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The Concept of Will in Early Latin Philosophy
Neal Ward Gilbert
Journal of the History of Philosophy
Johns Hopkins University Press
Volume 1, Number 1, October 1963
pp. 17-35
10.1353/hph.2008.1582
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The Concept of Will in EarlyLatin Philosophy NEAL W. GILBERT AN HISTORICALDISCUSSIONOf the concept of will is best begun with an analysis of the use of voluntas in Latin philosophy, from its earliest occurrences in Lucretius and Cicero on down to Augustine and medieval times. This development can be traced without much controversy because the line of transmission and development is more or less unbroken. But the correlating of Latin psychological terms with their Greek originals presents problems. Greek philosophy was indeed the source from which Latin philosophical psychology developed: Lucretius reflects Epicurean doctrine, while Cicero and Seneca were strongly influenced by Stoic philosophy when they discussed moral action. However, many modern scholars are convinced that the Greeks had no word that corresponds to our "will" at all, and so we seem to be left with no choice but to regard the will as an original creation of Latin philosophy . But this would be to overlook the fact that ancient Latin writers sometimes specifically equated voluntas with certain Greek terms, and such evidence ought not to go unexamined. Voluntas was well established in Latin usage before Roman writers began to concern themselves with philosophical problems, but it did not have a technical sense. It meant simply "good will," or "favor," or concretely, a "will, or testament. ''I Clustered around it were other derivatives of "volo," such as "benevolentia" and "malevolentia," "well-wishing" and "ill-wishing." When Roman writers began to deal with philosophical problems, there were a number of different contexts in which voluntas could play a leading role. Chief among them, naturally, was the "freedom of the will" which, then as now, intrigued the ordinary man or the moralist. However, discussions revolving around the determinist-libertarian issue often fail to clarify the concept of the will itself. Major philosophers (Chrysippus, Augustine, or Hume, for example) offer fine distinctions within moral action when considering the determinist-libertarian dispute, but lesser figures are apt to "C'est seulement lots de la creation du vocabulairephilosophiqueque voluntas a pris le sens abstrait et technique de 'volont6'." Ernout, A., and A. Meillet,Dictionnaire Etymologique de la langue latine, 4th ed. (Paris, 1959),sub "volo." 18 HISTORY OF PHILOSOPHY neglect the detailed analysis of moral action in favor of sweeping metaphysical pronouncements. When a Latin writer wished to say that a person did something of his own accord, the locution most natural to him was "sua sponte," an ablative absolute derived from the same root as our word "spontaneous." The nominative case of "sponte" ("spons") was almost never used, so that dictionaries are perhaps somewhat misleading in suggesting that the word "spons" means "free will": it would be more correct to say that the phrase "sua sponte" means "of his own free will." Latin writers also indicate that someone did something willingly by saying that he did it "non invitus" ("not unwillingly "). Finally, they could apply the adjective "voluntarius" to such an action, or say that it was done "ex voluntate." When the latter term begins to be used as a designation for a separate faculty of the soul distinct from reason or intelligence, it begins to resemble the modern concept of will. Augustine is usually given credit for introducing the concept of will into philosophy, but even the earliest Latin treatment of the free-will problem framed the issue in terms of voluntas. Nevertheless, I think that the received view is substantially correct, although it would be more proper to say that what Augustine introduced into philosophy was not the concept of will in general but the concept of the evil will. What Augustine called the "good" will ("bona voluntas") was closely related to the "reasonable desire" of the Stoic sage, rendered into Latin by Cicero as voluntas, without any adjective. The addition of the evil will altered the outlines of moral analysis considerably , involving among other things the major shift from a Socratic or Platonic psychology and ethics (usually labeled "intellectualistic") to a Christian psychology and ethics (usually labeled "voluntaristic"). Such a characterization is likely to be drawn too sharply, and to overlook anticipations of voluntarism in Greek thought or reminiscences of Greek intellectualism in Christian thought. Hence it may be useful...
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Additional Information | ISSN | 1538-4586 |
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| Print ISSN | 0022-5053 |
| Pages | pp. 17-35 |
| Launched on MUSE | 2008-01-01 |
| Open Access | No |
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Covering a complete graph with as few complete bipartite subgraphs as possible
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The biclique covering number b c(G)b c(G) of a graph G G is the smallest number of bicliques (complete bipartite subgraphs) of G G such that every edge of G G belongs to at least one of these bicliques.
Quoted from the paper "On covering graphs by complete bipartite subgraphs" by S. Jukna and A.S. Kulikov, 2009:
We have b c(K n)≤⌈log 2 n⌉b c(K n)≤⌈log 2n⌉: just encode the vertices of K n K n (complete graph with n n vertices) by binary vectors of length m=⌈log 2 n⌉m=⌈log 2n⌉ and define, for each i=1,…,m i=1,…,m, a biclique containing all edges, the codes of whose endpoints differ in the i i th coordinate.
For example, to cover K 6 K 6, we write
1=001 2=010 3=011 4=100 5=101 6=110 1=001 2=010 3=011 4=100 5=101 6=110
The three bicliques are thus K{1,2,3},{4,5,6},K{1,4,5},{2,3,6},and K{1,3,5},{2,4,6}K{1,2,3},{4,5,6},K{1,4,5},{2,3,6},and K{1,3,5},{2,4,6}, corresponding to the leftmost, the middle, and the rightmost bits, respectively.
Note that the result above is b c(K n)≤⌈log 2 n⌉b c(K n)≤⌈log 2n⌉. My question is what is the exact value of b c(K n)b c(K n) and how to prove it? In other words, what is the smallest number of bicliques to cover K n K n?
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edited Jun 5 at 4:56
The Amplitwist
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asked Mar 31, 2017 at 14:04
hengxinhengxin
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This is tight: we have b c(K n)=⌈log 2 n⌉b c(K n)=⌈log 2n⌉. This proof of the lower bound can be found in Fishburn and Hammer's Bipartite dimensions and bipartite degrees of graphs.
We prove that b c(K n)≥log 2 n b c(K n)≥log 2n by strong induction on n n. To cover K n K n, suppose that K p,q K p,q is the first biclique we put down. We may assume p+q=n p+q=n, since if we didn't include a vertex at all, we could add it to either side of the biclique and only increase the set of edges covered.
The vertices on either side of the first biclique induce a K p K p and a K q K q respectively that we haven't touched at all yet. Since p+q=n p+q=n, either p≥n 2 p≥n 2 or q≥n 2 q≥n 2, so we have a clique of size at least n 2 n 2 left to cover. By the inductive hypothesis, this needs at least log 2 n 2=(log 2 n)−1 log 2n 2=(log 2n)−1 more bicliques.
Together with the first biclique we put down, that brings us to a total of log 2 n log 2n bicliques needed to cover K n K n, which proves the result by induction on n n.
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edited Mar 31, 2017 at 18:04
answered Mar 31, 2017 at 14:41
Misha LavrovMisha Lavrov
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Thanks for the correction. Deleting my answer. –bof Commented Mar 31, 2017 at 22:00
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Published Time: Fri, 20 Jan 2023 21:58:48 GMT
arXiv:1511.00266v1 [math.GN] 1 Nov 2015
GENERALIZED INVERSE LIMITS INDEXED BY TOTALLY ORDERED SETS
SCOTT VARAGONA
Abstract. Although inverse limits with factor spaces indexed by the positive integers are most commonly studied, Ingram and Mahavier have defined inverse limits with set-valued functions broadly enough for any directed index set to be used. In this paper, we investigate generalized inverse limits whose factor spaces are indexed by totally ordered sets. Using information about the projections of such inverse limits onto finitely many coordinates, we generalize various well-known theorems on connectedness in inverse limits. Moreover, numerous theorems and examples are given addressing the special case of an inverse limit with a single idempotent surjective u.s.c. bonding function.
Introduction
In recent years, the vast majority of work on generalized inverse limits has fo-cused only on the case where the factor spaces are indexed by the positive integers. However, in , W. T. Ingram and William S. Mahavier define generalized inverse limits much more broadly, so that any directed index set could be used—for exam-ple, an uncountable totally ordered set such as the limit ordinal ω1. In the case of traditional inverse limits with continuous bonding functions, such “long” inverse limits have proven to be fruitful for continuum theory; for example, Michel Smith and later David Bellamy used long traditional inverse limits to construct non-metric indecomposable continua with remarkable properties. It is therefore natural for us to investigate long generalized inverse limits as well. After the groundbreaking work of Ingram and Mahavier, some other researchers have worked with generalized inverse limits using alternate index sets. Inverse limits indexed by the set of all integers have been investigated by the author and, to a more significant degree, by Patrick Vernon ; theorems about Mahavier Products indexed by totally ordered sets have been proven by Wlodzimierz Charatonik and Robert Roe in . However, in general, relatively little is known about generalized inverse limits indexed by any set aside from the positive integers. The goal of this paper is to expand the theory of generalized inverse limits indexed by totally ordered sets. To that end, we present a variety of theorems and examples that should be a good starting point for future research. After giving basic definitions and background information in Section 2, in Section 3 we discuss
2010 Mathematics Subject Classification. Primary 54F15, 54H20.
Key words and phrases. inverse limits, upper semi-continuous inverse limits, connected, set valued function, totally ordered sets, Continuum Theory. The author is indebted to W. T. Ingram and William Mahavier for their groundbreaking work on this topic (in ), to Michel Smith, who first introduced the author to “long” inverse limits, and to Van Nall, Sina Greenwood, and Steven Clontz for their feedback on the presentation this paper is based upon.
12SCOTT VARAGONA
the behavior of the projections of inverse limits indexed by totally ordered sets. In Section 4, we study the properties of idempotent u.s.c. functions; such functions, it seems, will be important to this type of inverse limit. Next, in Section 5, we further generalize some of the connectedness results by Ingram and Mahavier in . We also extend some of the connectedness results by Van Nall in to the case of generalized inverse limits indexed by totally ordered sets, at least in the case of inverse limits with a single idempotent u.s.c. bonding function f . Then, in Section 6, we apply our results to study examples of inverse limits indexed by some limit ordinal γ ≥ ω. Just as inverse limits indexed by the positive integers have been useful for representing complicated metric continua in a straightforward way, inverse limits indexed by “long” initial segments of the ordinals can be useful in the same way for certain non-metric continua. Finally, we conclude by stating some questions for future investigation in Section 7. 2. Definitions and Background Theorems
If X is a non-empty compact Hausdorff space, let 2X denote the set of non-empty compact subsets of X. C(X) denotes the set of connected members of 2X . Suppose both X and Y are non-empty compact Hausdorff spaces and f : X → 2Y is a set-valued function; then we say f is upper semi-continuous (u.s.c.) if, for all x ∈ X,whenever V is open in Y containing f (x), there exists an open U in X containing
x so that f (u) ⊆ V for each u ∈ U . The graph of f , denoted in this paper by
Graph (f ), is the set {(x, y ) ∈ X × Y | y ∈ f (x)}; it is well-known that f is u.s.c. iff Graph (f ) is a closed subset of X × Y . If f (x) = {y} for some x ∈ X, y ∈ Y , we simply write f (x) = y. Given some u.s.c. function f : X → 2Y , the preimage via f
of a given y ∈ Y is f −1(y) = {x ∈ X | y ∈ f (x)}. (It is not hard to see that f −1(y)
is a compact subset of X.) If A ⊆ X, then f (A) = ⋃
a∈A
f (a). Similarly, if B ⊆ Y ,
f −1(B) = ⋃
b∈B
f −1(b). We say f is surjective if, for all y ∈ Y , f −1(y) is non-empty. If f : X → 2Y is surjective, then the inverse of f , denoted f −1, is the u.s.c. function f −1 : Y → 2X whose graph is Graph (f −1) = {(y, x ) | (x, y ) ∈ Graph (f )}.Given non-empty compact Hausdorff spaces X, Y and Z and u.s.c. functions
f : X → 2Y and g : Y → 2Z , the composition g ◦ f : X → 2Z is the u.s.c. function given by (g ◦ f )( x) = {z ∈ Z | ∃ y ∈ Y such that y ∈ f (x) and z ∈ g(y)}. In the special case where f : X → 2X is u.s.c., the composition f ◦ f is denoted f 2. When
f 2 = f , we say f is idempotent .If Xα is a compact Hausdorff space for each α in an index set A, then ∏
α∈A
Xα
denotes the product space with the usual product topology. If B ⊆ A, then πB
denotes the projection map from ∏
α∈A
Xα into ∏
α∈B
Xα. If β ∈ A, we will write
π{β} as πβ . Let us use a boldface x to denote an element (xα)α∈A of the product ∏
α∈A
Xα. (So, if β ∈ A, πβ (x) = xβ .) A directed set (D, ) is a set D together with a relation on D that is reflexive, transitive, and has the property that whenever α, β ∈ D, there exists some η ∈ D
such that α η and β η. A directed set (D, ) will often be denoted simply by D when the relation is understood. A directed set D is totally ordered if,
∀α, β ∈ D, either 1) α β and β 6 α, 2) β α and α 6 β, or 3) α β and β α,in which case α = β. Whenever we write a finite subset {β1, β 2, . . . , β n} of some totally ordered set D, we tacitly assume that βi βj in the ordering on D iff i ≤ j
in the standard ordering on the natural numbers. Also, if α β but α 6 = β, then we write α ≺ β.GEN. INVERSE LIMITS INDEXED BY TOTALLY ORDERED SETS 3
The following notation, and indeed, most of the notation used in this paper, is intended to coordinate with the notation used in . Suppose that, for each element
α of a directed set D, Xα is a non-empty compact Hausdorff space. Moreover, if
α, β ∈ D with α β, let fαβ : Xβ → 2Xα be u.s.c. (where fαα always denotes the identity on Xα). If, for all α β η in D, fαβ ◦ fβη = fαη , then the collection
{fαβ | α β ∈ D} is called exact . Assuming Xα is compact Hausdorff ∀α ∈ D,
fαβ : Xβ → 2Xα is u.s.c. for all α β ∈ D, and f = {fαβ | α β ∈ D} is exact, we say {Xα, f αβ , D } is an inverse limit system (or simply, a system ). The inverse limit , lim
←− f, of this system is given by {x ∈ ∏
α∈D
Xα | xα ∈ fαβ (xβ ) ∀α β ∈ D}.The spaces Xα are called the factor spaces of the inverse limit, and the u.s.c. functions fαβ are called the bonding functions . In the special case where X is compact Hausdorff and f : X → 2X is an idempotent u.s.c. function, if Xα = X
for all α ∈ D and fαβ = f for all α ≺ β ∈ D, then {Xα, f αβ , D } = {X, f, D } is a system and lim
←− f is an inverse limit with a single idempotent bonding function f .(Note that, in this special case, the collection f is automatically exact because f
is idempotent.) If {Xα, f αβ , D } is a system so that, for each η ∈ D and for each
p ∈ Xη, there exists x ∈ ∏
α∈D
Xα with xα ∈ fαβ (xβ ) for all α β η and xη = p,then we say the system is consistent .As we will see, certain sets will be helpful for our study of inverse limits. Suppose
{Xα, f αβ , D } is a system. If η ∈ D, we define Gη = {x ∈ ∏
α∈D
Xα | xα ∈
fαβ (xβ ) for all α β η}. If {β1, β 2, . . . , β n} is a finite subset of D, then let us say G(β1, β 2, . . . , β n) = {(xβ1 , x β2 , . . . , x βn ) ∈ ∏
1≤i≤n
Xβi | xβi ∈ fβiβj (xβj )
for 1 ≤ i ≤ j ≤ n}. (In the case where two different inverse limits, e.g., lim
←− f
and lim
←− g, with the same index set D are being discussed at the same time, we may use subscripts Gf (β1, β 2, . . . , β n) and Gg (β1, β 2, . . . , β n) to help distinguish between the two corresponding sets.) We also define G′(β1, β 2, . . . , β n) = {x ∈∏
α∈D
Xα | xβi ∈ fβiβj (xβj ) for 1 ≤ i ≤ j ≤ n}. Note that G(β1, β 2, . . . , β n)
is a subset of ∏
1≤i≤n
Xβi , whereas G′(β1, β 2, . . . , β n) is the analogous subset of
∏
α∈D
Xα.Finally, let us say K is a Hausdorff continuum if K is a non-empty, compact and connected subset of a Hausdorff space. If K happens to be metrizable, we call
K a metric continuum or simply a continuum . If H is a set and K is a Hausdorff continuum that is a subset of H, we call K a subcontinuum of H. If X and Y are compact Hausdorff spaces and the u.s.c. function f : X → 2Y has the property that f (x) is connected for each x ∈ X, then we say f is Hausdorff continuum-valued
and we write f : X → C(X).A great deal of the initial work on generalized inverse limits indexed by totally ordered sets was done by Ingram and Mahavier in . Although, in general, inverse limits indexed by arbitrary directed sets may fail to be non-empty, Ingram and Ma-havier showed that the inverse limit of a consistent system with non-empty compact Hausdorff spaces indexed by a directed set D is non-empty and compact (Theorem 111, ). Charatonik and Roe recently showed that any system indexed by a totally ordered set D is automatically consistent. When combining this result with the original theorems of Ingram and Mahavier, we obtain an important background theorem:
Theorem 2.1. Let {Xα, f αβ , D } be a system with non-empty compact Hausdorff factor spaces, u.s.c. bonding functions, and a totally ordered index set D. Then 1) for each η ∈ D, Gη is non-empty and compact; 2) for each finite subset {β1, β 2, . . . , β n}4 SCOTT VARAGONA
of D, G(β1, β 2, . . . , β n) and G′(β1, β 2, . . . , β n) are non-empty and compact; 3) lim
←− f
is non-empty and compact.
Since lim
←− f is always non-empty and compact in the context we have described, it is natural to look for conditions under which such an inverse limit would also be connected. The following result (which combines the work of Ingram, Mahavier, Charatonik and Roe) is given as Corollary 2.3 in (although, in that paper the result is stated in the wider context of Mahavier Products).
Theorem 2.2. Let {Xα, f αβ , D } be a system for which each factor space is a Hausdorff continuum, each bonding function is u.s.c., and D is totally ordered. Suppose that, for each α, β ∈ D with α β, either fαβ is Hausdorff continuum-valued or fαβ (Xβ ) is connected with f −1
αβ
(y) a Hausdorff continuum for each y ∈
fαβ (Xβ ). Then lim
←− f is a Hausdorff continuum.
The above theorem generalizes two important theorems on connectedness in inverse limits given in . Later in this paper, we will generalize other well-known results as well. Whenever possible, we will prove theorems in the broad setting of inverse limit systems {Xα, f αβ , D } where each Xα is compact Hausdorff, each fαβ
is u.s.c., and D is a totally ordered set. However, we will also give a number of theorems that are particular to the case of an inverse limit with a single idempotent, surjective, u.s.c. bonding function f .3. Projections of Inverse Limits Onto Finitely Many Coordinates
To prepare for the main results in the later sections, we first need to discuss the behavior of the projections of an inverse limit. Unfortunately, for a general inverse limit system {Xα, f αβ , D }, if H = {β1, β 2, . . . , β n} is a finite subset of the index set D, then πH (lim
←− f ) is not necessarily equal to G(β1, β 2, . . . , β n). For example, Ingram and Mahavier give an example (, Example 106) of a system with compact factor spaces, surjective bonding functions, and a (non-totally ordered) directed index set D where lim
←− f is the empty set, but G(α, β ) is non-empty for every α β ∈ D. On the other hand, if f : [0 , 1] → 2[0 ,1] is given by f (x) = 0
for 0 ≤ x ≤ 1, then the inverse limit lim
←− f of the system {[0 , 1] , f, ω } (with the single bonding function f ) is the singleton {(0 , 0, 0, . . . )}. Therefore, in this case,
π{0,1}(lim
←− f) = {(0 , 0) }, whereas G(0 , 1) , being homeomorphic to the graph of f , is an arc. Let us say an inverse limit lim
←− f is cordial if, for each finite subset {β1, β 2, . . . , β n}
of the index set D, π{β1,β 2,...,β n}(lim
←− f) = G(β1, β 2, . . . , β n). The goal of this section is to show that an inverse limit (with compact Hausdorff factor spaces indexed by a totally ordered set) is cordial if and only if its bonding functions are all surjective.
Lemma 3.1. Suppose {Xα, f αβ , D } is a system with non-empty compact Hausdorff factor spaces, surjective u.s.c. bonding functions and a totally ordered index set D.Then lim
←− f is cordial. Proof. Let a finite subset H = {β1, β 2, . . . , β n} of D be given. We aim to show that πH (lim
←− f) = G(β1, β 2, . . . , β n). Clearly πH (lim
←− f) ⊆ G(β1, β 2, . . . , β n), so let
p = ( pβ1 , p β2 , . . . , p βn ) ∈ G(β1, β 2, . . . , β n). We need to show that there exists
y ∈ lim
←− f such that yβi = pβi for 1 ≤ i ≤ n.GEN. INVERSE LIMITS INDEXED BY TOTALLY ORDERED SETS 5
If M = {η1, η 2, . . . , η k} is a finite subset of D with H ⊆ M , let us define
p∗(M ) = {x ∈ ∏
α∈D
Xα | xβi = pβi for 1 ≤ i ≤ n and xσ ∈ fστ (xτ ) for all
σ, τ ∈ M with σ τ }. We intend to show that p∗(M ) is non-empty and compact. Let us construct an element x of p∗(M ) as follows. First, let xβi = pβi for
1 ≤ i ≤ n. If {η1, η 2, . . . , η m} is the set of all members of M with η1, η 2, . . . , η m ≺
β1, then let xηm ∈ fηmβ1 (xβ1 ), and then continue inductively: for each integer
j with m − 1 ≥ j ≥ 1, once xηj+1 has been defined, let xηj ∈ fηj ηj+1 (xηj+1 ).Similarly, for a given integer i with 1 ≤ i ≤ n − 1, suppose ηr , η r+1 , . . . , η s are all the members of M lying strictly between βi and βi+1 in the ordering on D.Then since xβi ∈ fβiβi+1 (xβi+1 ) and f is exact, there must exist some xηs ∈ Xηs
so that xηs ∈ fηsβi+1 (xβi+1 ) and xβi ∈ fβiηs (xηs ). Using the same argument, we may continue in this way inductively: for each integer j with s − 1 ≥ j ≥ r,once xηj+1 has been defined, we select xηj ∈ Xηj so that xηj ∈ fηj ηj+1 (xηj+1 ) and
xβi ∈ fβiηj (xηj ). Finally, if ηt, η t+1 , . . . , η u are all the members of M that are strictly greater than βn in the ordering on D, then because each bonding function is surjective, we can choose some xηt ∈ f −1
βnηt
(xβn ) and then proceed inductively: for each integer j with t + 1 ≤ j ≤ u, once xηj−1 has been defined, choose xηj ∈
f −1
ηj−1ηj
(xηj−1 ). For each α ∈ D − M , we let xα ∈ Xα be chosen arbitrarily. Then
x satisfies xβi = pβi for 1 ≤ i ≤ n, and (by construction) whenever σ, τ ∈ M
with σ τ , we have xσ ∈ fστ (xτ ). So x ∈ p∗(M ). To see that p∗(M ) is closed, for 1 ≤ i ≤ n, let Oβi = π−1
βi
(Xβi − { pβi }). Each Oβi is open in ∏
α∈D
Xα,so G′(η1, η 2, . . . , η k) − ⋃
1≤i≤n
Oβi is closed. However, it is not hard to see that
p∗(M ) = G′(η1, η 2, . . . , η k) − ⋃
1≤i≤n
Oβi , so p∗(M ) is compact. Note that, whenever M and N are two finite subsets of D with H ⊆ M and H ⊆
N , then p∗(M ∪ N ) ⊆ p∗(M ) ∩ p∗(N ). This means the collection Λ = {p∗(M ) | M
is a finite subset of D with H ⊆ M } has the finite intersection property, and ⋂ Λ
is non-empty. Let y ∈ ⋂ Λ; we claim that y ∈ lim
←− f. Let α1, α 2 ∈ D with α1 α2.Then since y ∈ p∗({α1, α 2} ∪ H), yα1 ∈ fα1 α2 (yα2 ). So indeed, y ∈ lim
←− f, which means that there is an element y ∈ lim
←− f with yβi = pβi for 1 ≤ i ≤ n. We conclude that G(β1, β 2, . . . , β n) = πH (lim
←− f).
Theorem 3.2. Suppose {Xα, f αβ , D } is a system with non-empty compact Haus-dorff factor spaces, u.s.c. bonding functions and a totally ordered index set D. Then
lim
←− f is cordial if and only if, for all α, β ∈ D with α β, fαβ is surjective. Proof. By Lemma 3.1, if each fαβ is surjective, lim
←− f is cordial. On the other hand, suppose lim
←− f is cordial and let α, β ∈ D with α β. Let y ∈ Xα; we intend to show that f −1
αβ
(y) is non-empty. Because lim
←− f is cordial, G(α) = πα(lim
←− f). But
G(α) = Xα. So, there exists x ∈ lim
←− f such that xα = y, and therefore xβ ∈ f −1
αβ
(y).This means fαβ is surjective.
It may also be interesting to note that, since G(α) = Xα for each α in a directed index set D, when lim
←− f is cordial the corresponding system {Xα, f αβ , D } must be consistent. However, as the previous example (where f was the constant map 0)shows, consistent systems do not always produce cordial inverse limits. 4. Idempotent u.s.c. Functions
The main obstacle to constructing concrete examples of inverse limits indexed by some “long” totally ordered set (e.g., an uncountable limit ordinal such as ω1) is 6 SCOTT VARAGONA
finding a large collection f of bonding functions that is exact. One way around this difficulty is to consider generalized inverse limits with a single idempotent u.s.c. bonding function f ; in this context, because f 2 = f , the collection of bonding functions is automatically exact. So, let us suppose X is a non-empty compact Hausdorff space, f : X → 2X is an idempotent u.s.c. bonding function, and D is a totally ordered set. Then, as we have seen, {X, f, D } is a system and the inverse limit lim
←− f with the single idempotent bonding function f is defined. Generalized inverse limits indexed by the positive integers with a single bonding function f = fi i +1 for each i ∈ Z+ (and fij = fi i +1 ◦fi+1 i+2 ◦· · ·◦ fj−1 j whenever
i < j ) are commonly studied, and have given rise to many interesting problems and theorems. Therefore, we believe generalized inverse limits indexed by a totally ordered set D with a single idempotent bonding function f should be a natural (and, hopefully, fruitful) next step for the theory. In this section, we discuss some basic properties of idempotent u.s.c. functions and show how to construct some simple examples of such functions. The lemmas in this section will also be needed for some of the theorems and examples seen later in this paper.
Lemma 4.1. Suppose X is a compact Hausdorff space and f : X → 2X is surjec-tive. Then f is idempotent if and only if f −1 is idempotent. Proof. Let f be idempotent. (y, x ) ∈ Graph (f −1) ⇔ (x, y ) ∈ Graph (f ) = Graph (f ◦
f ) ⇔ ∃ z ∈ X such that z ∈ f (x) and y ∈ f (z) ⇔ ∃ z ∈ X such that x ∈ f −1(z) and
z ∈ f −1(y) ⇔ (y, x ) ∈ Graph (f −1 ◦ f −1). Thus, f −1 is idempotent. The argument can easily be reversed by changing the roles of f and f −1.
Lemma 4.2. Let X be a compact Hausdorff space and let f : X → 2X be u.s.c. Then: 1. f 2 = f if and only if, for each A ⊆ X satisfying f (x) = A for some x ∈ X,
f (A) = A.2. Suppose f 2 = f . If f (x) = y for some x, y ∈ X, then f (y) = y.3. If there is some B ⊆ X so that, for all x ∈ X, either f (x) = x or f (x) = B,then f 2 = f .4. Suppose that, for some A, B ⊆ X where A ∩ B = ∅, we have f (a) = X for each a ∈ A and f (x) = B whenever x 6 ∈ A. Then f 2 = f .Proof. To prove statement 1, first note that, if f 2 = f and f (x) = A, then f (A) =
f (f (x)) = f (x) = A. To prove the other direction of the equivalence, we let
x ∈ X, so f (x) = A for some A ⊆ X. However, f (A) = A, so that f 2(x) =
f (f (x)) = f (A) = A = f (x). Applying the forward implication in statement 1 in the case where A = {y} gives us statement 2. As for statement 3, if f satisfies the given conditions, then whenever x ∈ X, either f (x) = x (in which case, clearly
f 2(x) = x) or f (x) = B (in which case, for each b ∈ B, either f (b) = b or
f (b) = B, so that f (B) = ⋃
b∈B
f (b) = B, and therefore f 2(x) = B). Thus,
f 2 = f . Finally, addressing statement 4, we note that when x ∈ A, f (x) = X (so of course f 2(x) = X), and if x 6 ∈ A, then f (x) = B, so that f 2(x) = f (B). However, if b ∈ B, since b 6 ∈ A, f (b) = B. Thus, f (B) = B, so that f 2(x) = B, and we conclude that f 2 = f .
In the special case where X = [0 , 1] , we also have the following. These statements are easy to verify directly and so we omit the proof. (Statement 3 below has been observed before by Ingram in .) GEN. INVERSE LIMITS INDEXED BY TOTALLY ORDERED SETS 7
Lemma 4.3. Let f : [0 , 1] → 2[0 ,1] be u.s.c. 1. If, for all x ∈ [0 , 1] , either f (x) = x or f (x) = [0 , x ], then f 2 = f .2. If, for all x ∈ [0 , 1] , either f (x) = x or f (x) = [ x, 1] , then f 2 = f .3. If, for all x ∈ [0 , 1] , f (x) = {x, 1 − x}, then f 2 = f .
Despite how restrictive the condition f 2 = f may seem, the following lemma shows that one often has more freedom when constructing u.s.c. idempotent func-tions than it may first appear.
Lemma 4.4. Let a ∈ (0 , 1) and let K be any closed subset of [0 , 1] 2 such that
K ⊆ ([0 , a ) × (a, 1]) ∪ { (a, a )}. Let ∆ = {(x, x ) ∈ [0 , 1] 2 | x ∈ [0 , 1] }. Then K ∪ ∆
is the graph of an idempotent surjective u.s.c. function f : [0 , 1] → 2[0 ,1] .Proof. K ∪ ∆ is a closed subset of [0 , 1] 2 whose projections map onto both coordi-nates, so K ∪ ∆ is the graph of a surjective u.s.c. function f : [0 , 1] → 2[0 ,1] ; it re-mains to check that f is idempotent. If a ≤ x ≤ 1, then f (x) = x and so f 2(x) = x
also. If 0 ≤ x < a , then f (x) = {x} ∪ Hx, where Hx = π2(K ∩ ({x} × (a, 1])) .In particular, Hx ⊆ (a, 1] , and since f |[a, 1] is the identity, f (Hx) = Hx. Thus,
f 2(x) = f (x) ∪ f (Hx) = ( {x} ∪ Hx) ∪ Hx = {x} ∪ Hx = f (x). So, f 2 = f .
We close this section with a remark: if f : X → X is continuous, idempotent and surjective, then by Lemma 4.2 part 2, f can only be the identity on X. However, a multitude of distinct idempotent, surjective, u.s.c. functions exist, which is an unexpected advantage to working in the general setting of u.s.c. functions. 5. Connectedness Results
A considerable subset of the literature on generalized inverse limits has been devoted to finding necessary and/or sufficient conditions for lim
←− f to be connected. The difficulty of this problem is very well-known; still, researchers have produced various helpful results. In this section, we intend to reformulate some of these results in the context of generalized inverse limits indexed by a totally ordered set. The key to proving the main theorems of this section will be the various “ G-sets” introduced in Section 2 of this paper. So, the following lemma (which was originally proved by Ingram and Mahavier amid the proof of Theorem 124 in ) will be useful:
Lemma 5.1. (Ingram & Mahavier) Let {Xα, f αβ , D } be a system for which each factor space is non-empty compact Hausdorff, each bonding function is u.s.c., and D
is totally ordered. Let η ∈ D be fixed, and let Γ = {G′(β1, β 2, . . . , β n) | { β1, β 2, . . . , β n}
is a finite subset of D with βi η for 1 ≤ i ≤ n}. Then ⋂ Γ = Gη .Proof. Let x ∈ Gη and let G′(β1, β 2, . . . , β n) be an arbitrary member of Γ. Then because x ∈ Gη, we have xβi ∈ fβiβj (xβj ) for 1 ≤ i ≤ j ≤ n. Thus, x ∈
G′(β1, β 2, . . . , β n), which implies x ∈ ⋂ Γ. On the other hand, suppose x ∈ ⋂ Γ.Then for any α β η, because x ∈ G′(α, β ), xα ∈ fαβ (xβ ). Thus, x ∈ Gη .
The first main theorem of this section is an easy generalization of background Theorem 2.2. Note that the original theorem required fαβ to be Hausdorff continuum-valued (or fαβ (Xβ ) to be connected and f −1
αβ
to be Hausdorff continuum-valued) for each α β in the index set D. This condition was imposed in order to force each G(β1, β 2, . . . , β n) to be connected, which was the key to the proof; however, there are more general circumstances in which G(β1, β 2, . . . , β n) also turns out to 8 SCOTT VARAGONA
be connected, so the following theorem will be helpful. The proof requires only a minor adjustment to the original proofs of Theorems 124 and 125 given by Ingram and Mahavier in .
Theorem 5.2. Let {Xα, f αβ , D } be a system for which each factor space is a Hausdorff continuum, each bonding function is u.s.c., and D is a totally ordered set. Suppose that, for each finite subset {β1, β 2, . . . , β n} of D, G(β1, β 2, . . . , β n) is connected. Then lim
←− f is a Hausdorff continuum. Proof. Let η ∈ D be fixed; we intend to show that Gη is a Hausdorff continuum. Let
H = {β1, β 2, . . . , β n} be a finite subset of D. Then G′(β1, β 2, . . . , β n) is homeomor-phic to G(β1, β 2, . . . , β n)×∏
α∈D−H
Xα, a product of Hausdorff continua; therefore,
G′(β1, β 2, . . . , β n) is a Hausdorff continuum. Let Γ = {G′(β1, β 2, . . . , β n) | { β1, β 2, . . . , β n}
is a finite subset of D with βi η for 1 ≤ i ≤ n}. Next, note that whenever
{β1, β 2, . . . , β n} and {σ1, σ 2, . . . , σ s} are finite subsets of D, if we let {τ1, τ 2, . . . , τ r } =
{β1, β 2, . . . , β n} ∪ { σ1, σ 2, . . . , σ s}, then we may conclude that G′(τ1, τ 2, . . . , τ r ) ⊆
G′(β1, β 2, . . . , β n) ∩ G′(σ1, σ 2, . . . , σ s). Thus, Γ is a collection of Hausdorff continua with the property that the intersection of any finite subcollection of members of
Γ contains another member of Γ. It follows that ⋂ Γ is a Hausdorff continuum. However, by Lemma 5.1, ⋂ Γ = Gη . Gη is therefore a Hausdorff continuum for each η ∈ D.Finally, it is easy to see that lim
←− f = ⋂
η∈D
Gη , but {Gη | η ∈ D} is a nested collection of Hausdorff continua, so lim
←− f is also a Hausdorff continuum.
We may apply the above theorem to obtain a characterization of connectedness in inverse limits with surjective bonding functions:
Theorem 5.3. Let {Xα, f αβ , D } be a system for which each factor space is a Hausdorff continuum, each bonding function is u.s.c. and surjective, and D is a totally ordered set. Then lim
←− f is a Hausdorff continuum iff for each finite subset
{β1, β 2, . . . , β n} of D, G(β1, β 2, . . . , β n) is connected. Proof. Suppose that, for each finite subset {β1, β 2, . . . , β n} of D, the set G(β1, β 2, . . . , β n)
is connected. By Theorem 5.2, we may conclude that lim
←− f is a Hausdorff con-tinuum. On the other hand, suppose lim
←− f is a Hausdorff continuum. Since the bonding functions are surjective, lim
←− f is cordial, and therefore for any finite sub-set H = {β1, β 2, . . . , β n} of D, πH (lim
←− f) = G(β1, β 2, . . . , β n), which implies that
G(β1, β 2, . . . , β n) is connected.
A similar characterization, this time in terms of Gη, is also worth noting. (This result was inspired by Theorem 2.1 in .)
Theorem 5.4. Let {Xα, f αβ , D } be a system for which each factor space is a Hausdorff continuum, each bonding function is u.s.c. and surjective, and D is a totally ordered set. Then lim
←− f is a Hausdorff continuum iff Gη is connected for each η ∈ D.Proof. Since lim
←− f = ⋂
η∈D
Gη , if Gη is connected for each η ∈ D, then lim
←− f is the intersection of a nested collection of Hausdorff continua and is therefore a Hausdorff continuum. On the other hand, suppose lim
←− f is a Hausdorff continuum and η ∈ D.By Theorem 5.3, when {β1, β 2, . . . , β n} is a finite subset of D, G(β1, β 2, . . . , β n) is a Hausdorff continuum, which would mean that G′(β1, β 2, . . . , β n) is also a Hausdorff GEN. INVERSE LIMITS INDEXED BY TOTALLY ORDERED SETS 9
continuum. So, if Γ = {G′(β1, β 2, . . . , β n) | { β1, β 2, . . . , β n} is a finite subset of
D with βi η for 1 ≤ i ≤ n}, then by the same argument given in the proof of Theorem 5.2, ⋂ Γ is a Hausdorff continuum. By Lemma 5.1, ⋂ Γ = Gη.
In the last two theorems, the hypothesis that each bonding function is surjective is necessary. To see this, consider the following example (which can also be found as Example 1.8 in ). Let f : [0 , 1] → 2[0 ,1] be given by f (x) = 0 for 0 ≤ x < 1
and f (1) = {0, 1/2}, let the index set be the positive integers, and let fi i +1 = f for each positive integer i. Then the inverse limit lim
←− f is the singleton {(0 , 0, 0, . . . )},whereas G(1 , 2) , being homeomorphic to the graph of f , is not connected. G2 is not connected for similar reasons. The source of the trouble here is that, because the bonding functions are not surjective, lim
←− f is not cordial, and so, G(1 , 2) need not equal π{1,2}(lim
←− f). This example also shows that, in Theorem 2.1 of , the bonding functions should have been assumed to be surjective in order for lim
←− f
being a continuum to imply that each Gn is connected. (The converse, which is more commonly used, is true as it stands, without having to assume surjectivity.) Ingram has asked the author to make the readers of this paper aware of the error. Next, we give a simple sufficient condition for non-connectedness in an inverse limit with surjective bonding functions. (This result generalizes an observation by Nall in .) Once again, the previous example shows that the bonding functions in this theorem must be surjective.
Theorem 5.5. Let {Xα, f αβ , D } be a system for which each factor space is non-empty, compact and Hausdorff, each bonding function is u.s.c. and surjective, and
D is totally ordered. If, for some α β ∈ D, the graph of fαβ is not connected, then lim
←− f is not connected. Proof. Suppose lim
←− f is connected. Then because it is cordial (by Lemma 3.1),
π{α,β }(lim
←− f) = G(α, β ) is connected for every α β ∈ D. Since, for each α β ∈
D, G(α, β ) is homeomorphic to Graph (fαβ ), the proof is complete.
The reader should be warned that the converse of the previous theorem is not true, as illustrated by an example from Jonathan Meddaugh that is given in . For the remainder of this section, we will focus on generalized inverse limits with a single idempotent surjective u.s.c. bonding function. In , Van Nall presents a variety of theorems that give information about the connectedness of a generalized inverse limit (indexed by the positive integers) where there is some u.s.c. function
f so that fi i +1 = f for each i > 0. Using techniques from the original proofs by Nall, we will reformulate two of those theorems in the setting of an inverse limit (indexed by some totally ordered set D) with a single idempotent u.s.c. bonding function f .To prepare for the first of these theorems, we give a lemma that restates a key detail from Nall’s original Theorem 3.1 in , but in the context of this paper. (For the proof of this lemma, we refer the reader to Nall’s paper.) Suppose X
is compact Hausdorff, f : X → 2X is u.s.c., {β1, β 2, . . . , β n} (with n ≥ 2) is a finite subset of a totally ordered set D, and Xβi = X for 1 ≤ i ≤ n. Then let us define K(β1, β 2, . . . , β n) = {(xβ1 , x β2 , . . . , x βn ) ∈ ∏
1≤i≤n
Xβi | xβi ∈ f (xβi+1 ) for
1 ≤ i ≤ n − 1}.
Lemma 5.6. (Nall) Suppose X is a metric continuum and f : X → 2X is a surjective u.s.c. function whose graph is connected and is the union of a collection 10 SCOTT VARAGONA
of u.s.c. functions, each of which has domain X and maps into C(X). Suppose also that {β1, β 2, . . . , β n} (with n ≥ 2) is a finite subset of a totally ordered set D, and
Xβi = X for 1 ≤ i ≤ n. Then K(β1, β 2, . . . , β n) is a continuum for each integer
n ≥ 2.
Nall proved the above result in the case of K(1 , 2, . . . , n ), but of course the same result holds if we replace the natural numbers 1, 2, . . . , n with other sym-bols β1, β 2, . . . , β n from a totally ordered set D. Also, not surprisingly, the set
K(β1, β 2, . . . , β n) can be rewritten using the “ G” notation of this paper:
Lemma 5.7. Let X be a non-empty compact Hausdorff space, and let f : X → 2X
be an idempotent surjective u.s.c. function. Suppose {X, f, D } is a system with the single idempotent bonding function f and a totally ordered index set D. If
{β1, β 2, . . . , β n} (with n ≥ 2) is a finite subset of D, then K(β1, β 2, . . . , β n) =
G(β1, β 2, . . . , β n).Proof. If (xβ1 , x β2 , . . . , x βn ) ∈ G(β1, β 2, . . . , β n), then because we have xβi ∈ fβiβi+1 (xβi+1 ) =
f (xβi+1 ) for 1 ≤ i ≤ n − 1, it follows that (xβ1 , x β2 , . . . , x βn ) ∈ K(β1, β 2, . . . , β n).Now let (xβ1 , x β2 , . . . , x βn ) ∈ K(β1, β 2, . . . , β n). Clearly xβi ∈ fβiβi+1 (xβi+1 ) for
1 ≤ i < n . So now, for a fixed i with 1 ≤ i < n , we will show inductively that
xβi ∈ fβiβj (xβj ) when i < j ≤ n. Suppose we have shown xβi ∈ fβiβk (xβk )
for some k with i + 1 ≤ k < n . Then since xβk ∈ fβk βk+1 (xβk+1 ), we know
xβi ∈ fβiβk ◦ fβk βk+1 (xβk+1 ). However, fβiβk ◦ fβkβk+1 = f ◦ f = f = fβiβk+1 , so we have xβi ∈ fβiβk+1 (xβk+1 ). Thus, xβi ∈ fβiβj (xβj ) for all 1 ≤ i ≤ j ≤ n, and
(xβ1 , x β2 , . . . , x βn ) ∈ G(β1, β 2, . . . , β n).
Theorem 5.8. Suppose X is a metric continuum and f : X → 2X is an idempotent surjective u.s.c. function whose graph is connected and is the union of a collection of u.s.c. functions, each of which has domain X and maps into C(X). Let {X, f, D }
be a system with the single idempotent bonding function f and a totally ordered index set D. Then the inverse limit lim
←− f of the system {X, f, D } is a Hausdorff continuum. Proof. By Theorem 5.2, it suffices to show that for each finite subset {β1, β 2, . . . , β n}
of D, G(β1, β 2, . . . , β n) is connected. Of course, G(β1) = X is connected, so assume {β1, β 2, . . . , β n} is a finite subset of D with n ≥ 2. By Lemma 5.7,
G(β1, β 2, . . . , β n) = K(β1, β 2, . . . , β n). However, by Lemma 5.6, K(β1, β 2, . . . , β n)
is a continuum, so the proof is complete.
Lemma 4.1 stated that if f is u.s.c., idempotent and surjective, then so is f −1.Thus, assuming X is a non-empty compact Hausdorff space and f : X → 2X is u.s.c., idempotent, and surjective, then if D is totally ordered, not only is the inverse limit lim
←− f of the system {X, f, D } defined, but also the inverse limit lim
←− f–1 of the system {X, f −1, D }. This is the basis for the last theorem of this section, which retains the flavor of Theorem 3.3 in .
Theorem 5.9. Suppose X is a Hausdorff continuum, f : X → 2X is an idem-potent surjective u.s.c. function, and D is a totally ordered set. Let lim
←− f be the inverse limit of the system {X, f, D }, and let lim
←− f–1 be the inverse limit of the system {X, f −1, D }. Then lim
←− f is a Hausdorff continuum if and only if lim
←− f–1 is a Hausdorff continuum. GEN. INVERSE LIMITS INDEXED BY TOTALLY ORDERED SETS 11
Proof. We employ a similar proof technique as the one used by Nall. Let {β1, β 2, . . . , β n}
be a finite subset of D. If lim
←− f is a Hausdorff continuum, then because it is cor-dial, we may conclude that the set Gf (β1, β 2, . . . , β n) is a Hausdorff continuum. Also, (xβ1 , x β2 , . . . , x βn ) ∈ Gf (β1, β 2, . . . , β n) if and only if (xβn , x βn−1 , . . . , x β1 ) ∈
Gf −1 (β1, β 2, . . . , β n). (Justification: xβi ∈ f (xβj ) for each 1 ≤ i < j ≤ n if and only if xβj ∈ f −1(xβi ) for each 1 ≤ i < j ≤ n.) Thus, the map that reverses the order of the sequence (xβ1 , x β2 , . . . , x βn ) is a homeomorphism from Gf (β1, β 2, . . . , β n) to
Gf −1 (β1, β 2, . . . , β n), and therefore, Gf −1 (β1, β 2, . . . , β n) is a Hausdorff continuum. We apply Theorem 5.2 to conclude that lim
←− f–1 is a Hausdorff continuum. The argument is easily reversed to obtain the other direction of the equivalence.
Examples
As we will show, in each of the following examples, the given surjective u.s.c. function f : [0 , 1] → 2[0 ,1] is idempotent. Thus, if D is a totally ordered set, the inverse limit lim
←− f of the system {X, f, D } with the single bonding function f is defined. We will use a limit ordinal γ ≥ ω for our index set D in these examples. (Recall that an ordinal γ is equal to the set of its predecessors. Although the index sets of inverse limits typically start at 1 because the positive integers are so often used, when we use ordinals our index set will start with 0. For more background material on ordinals, see, e.g., .) Of course, if γ = ω, then lim
←− f is homeomorphic to the usual generalized inverse limit indexed by the positive integers. Versions of the first four examples have been studied previously by others (e.g., Ingram in ) in that setting; however, as we shall see below, some striking changes can occur when larger limit ordinals are chosen for the index set—especially γ ≥ ω1.
Example 6.1. Let f : [0 , 1] → C([0 , 1]) be given by f (0) = [0 , 1] and f (x) = x for each x 6 = 0 . (f is idempotent by Lemma 4.2, part 3.) The inverse limit lim
←− f of the system {[0 , 1] , f, γ } is a fan that is the union of one arc for each ordinal β with
1 ≤ β ≤ γ, all intersecting at the vertex (0 , 0, 0, . . . ).Proof. For 1 ≤ β ≤ γ, let Aβ = {x ∈ ∏
α<γ
[0 , 1] | xα = x0 if α < β and xα = 0 if
β ≤ α < γ }. Then for 1 ≤ β ≤ γ, Aβ is an arc and lim
←− f = ⋃
1≤β≤γ
Aβ . Note that
⋂
1≤β≤γ
Aβ = {(0 , 0, 0, . . . )}.
Example 6.2. Let f : [0 , 1] → C([0 , 1]) be given by f (0) = [0 , 1] and f (x) = 1 for each x 6 = 0 . (f is idempotent by Lemma 4.2, part 4.) The inverse limit lim
←− f of the system {[0 , 1] , f, γ } is homeomorphic to the set (γ × [0 , 1)) ∪ { (γ, 0) } with the lexicographic order topology. (If γ = ω1, this space is the traditional compactified long line.) Proof. Let Y be the space (γ×[0 , 1)) ∪{ (γ, 0) } with the lexicographic order topology, and denote the points (0 , 0, 0, . . . ) and (1 , 1, 1, . . . ) of lim
←− f by 0 and 1, respectively. Let h : lim
←− f → Y be given as follows: suppose x = ( xα)α<γ ∈ lim
←− f. If x = 1, then
h(x) = ( γ, 0) ; if x 6 = 1, then h(x) = ( β, x β ), where β is the least ordinal < γ such that xβ 6 = 1 . We intend to show that h is a homeomorphism. Let x, y ∈ lim
←− f with x 6 = y. If exactly one of x or y is 1, then clearly h(x) 6 =
h(y). So, suppose both x and y are not 1. Then h(x) = ( β1, x β1 ) and h(y) = (β2, x β2 ) for some β1, β 2 < γ . If β1 6 = β2, then of course h(x) 6 = h(y). If β1 = β2,then by the way f and h were defined, xα = yα = 1 for all α < β 1, and xα =
yα = 0 for all α > β 1. However, x 6 = y, and that forces xβ1 6 = yβ1 , which means 12 SCOTT VARAGONA
h(x) 6 = h(y). So, h is one-to-one. To show h is onto, we recall that h(1) = ( γ, 0) ,so let (β, t ) ∈ Y − { (γ, 0) }. Then if x = ( xα)α<γ is the element of the inverse limit that satisfies xα = 1 for all α < β , xβ = t, and xα = 0 for all α > β , we have
h(x) = ( β, t ). So, h is indeed onto. Finally, to show h is continuous, we let x ∈ lim
←− f and let V be an open set in
Y containing h(x). We consider the subcase where h(x) = ( β, x β ), where β < γ
and 0 < x β < 1. Then V contains an open interval of form (( β, s ), (β, t )) where
0 < s < x β < t < 1. Let U = ( ∏
α<γ
Uα) ∩ lim
←− f, where Uα = [0 , 1] for all α 6 = β
and Uβ = ( s, t ). Then U is open in lim
←− f and x ∈ U . To see that h(U ) ⊆ V , let
y ∈ U and note that yβ ∈ (s, t ). Since 0 < s < y β < t < 1, we may conclude yα = 1
for all α < β , and that means h(y) = ( β, y β ), which lies in (( β, s ), (β, t )) ⊆ V . The remaining subcases (i.e., when h(x) = ( β, 0) for some 0 ≤ β ≤ γ), though slightly more complicated, are similar and will be left to the reader. Thus, h is one-to-one, onto, and continuous, so h−1 is also continuous and h is a homeomorphism.
A similar argument can be used for the following example, so we omit the proof.
Example 6.3. Let f : [0 , 1] → C([0 , 1]) be given by f (0) = [0 , 1/2] , f (1) = [1 /2, 1] ,and f (x) = 1 /2 whenever 0 < x < 1. ( f can be shown to be idempotent directly, or by applying Lemma 4.2, part 1.) The inverse limit lim
←− f of the system {[0 , 1] , f, γ }
is an arc homeomorphic to the union of two copies of the space produced in Example 6.2 intersecting at the common compactification point (1 /2, 1/2, 1/2, . . . ).
It may be interesting to note that in the previous two examples, if γ is chosen to be any limit ordinal < ω 1, then the inverse limit is simply a metric arc. Only once γ = ω1 is chosen do we get a non-metric arc.
Example 6.4. Let f : [0 , 1] → 2[0 ,1] be given by f (x) = {x, 1−x} for each x ∈ [0 , 1] .(f is idempotent by Lemma 4.3, part 3.) The inverse limit lim
←− f of the system
{[0 , 1] , f, γ } is a cone over the set {0, 1}γ with vertex (1 /2, 1/2, 1/2, . . . ).Proof. For a given y ∈ { t, 1 − t}γ and a ∈ [0 , 1] , let y(a) denote the sequence y
with each t replaced by a. For a given y ∈ { t, 1 − t}γ , let Ay = {y(a) | a ∈ [0 , 1] }.Then lim
←− f = ⋃
y∈{ t, 1−t}γ
Ay. Note that, for each y, Ay is an arc from y(0) to y(1)
and whenever y, z ∈ { t, 1 − t}γ with y 6 = z, Ay ∩ Az = {(1 /2, 1/2, 1/2, . . . )}.
Example 6.5. Let f : [0 , 1] → 2[0 ,1] satisfy the hypothesis of Lemma 4.4 (so f is idempotent), with the additional requirement that K is a continuum containing the point (a, a ). Then the inverse limit lim
←− f of the system {[0 , 1] , f, γ } is a Hausdorff continuum. (This continuum contains a fan of copies of K, with one copy of K for each ordinal 1 ≤ β < γ .) Proof. For 1 ≤ β < γ , let Aβ = {x ∈ lim
←− f | ∃ (s, t ) ∈ K with xα = t ∀α < β and
xα = s ∀α ≥ β}. Let B = {x ∈ lim
←− f | ∃ t ∈ [0 , 1] such that xα = t ∀α < γ }.Then lim
←− f = ( ⋃
1≤β<γ
Aβ ) ∪ B. Each Aβ can be seen to be homeomorphic to K, a continuum, and B is an arc. Since the sequence a = ( a)α<γ is an element of each
Aβ as well as B, lim
←− f is connected.
Conclusion
Virtually any question that has been stated for inverse limits indexed by the positive integers has an analogue for inverse limits indexed by totally ordered sets, GEN. INVERSE LIMITS INDEXED BY TOTALLY ORDERED SETS 13
so there are ample opportunities for further research. A number of problems (stated mainly for inverse limits indexed by the positive integers) can be found in and in Chapter 6 of . One question of interest would be the following:
Question 7.1. Choose some totally ordered set D. What continua are homeomor-phic to an inverse limit with a single u.s.c. idempotent surjective bonding function
f : [0 , 1] → 2[0 ,1] with index set D?
Choosing index sets other than the positive integers can have surprising effects on this problem. For example, Patrick Vernon used the integers as his index set to produce a 2-cell in , whereas Van Nall proved (in ) that a 2-cell cannot be produced if the index set consists only of the positive integers. (The bonding function used in Vernon’s example was not idempotent, however.) Let us also state a much more open-ended question:
Question 7.2. If lim
←− f is the inverse limit of a system {[0 , 1] , f, D } with a single u.s.c. idempotent surjective bonding function f : [0 , 1] → 2[0 ,1] and totally ordered index set D, what can be said of lim
←− f?
When the index set D is large, finding collections of u.s.c. functions f that are exact (without being trivial, e.g., by making almost all of the bonding functions be the identity) remains a difficult problem. Using a single idempotent bonding function is only one possible solution. The collection of continuous bonding func-tions used by Michel Smith in is exact, although the factor spaces become increasingly complicated as one moves deeper into the index set D.
Question 7.3. What other techniques are there for generating non-trivial collec-tions of u.s.c. (surjective) functions that are exact?
The examples section showed how strongly the choice of index set D can affect the properties of the inverse limit space. So, we close with another open-ended question:
Question 7.4. If the index set D has a given property P , under what conditions (and to what degree) does that impact the properties of lim
←− f?
References
David Bellamy, Indecomposable continua with one and two composants , Fund. Math. 101 (1978), no. 2, 129-134. Wlodzimierz J. Charatonik and Robert P. Roe, On Mahavier Products , Topology and its Applications, 166, (2014), 92-97. W. T. Ingram, William S. Mahavier, Inverse Limits: from Continua to Chaos , Springer, Developments in Mathematics (vol. 25), 2012. W. T. Ingram, William S. Mahavier, Inverse limits of upper semi-continuous set valued functions , Houston Journal of Mathematics, vol. 32 (2006) no. 1, 119-130. W. T. Ingram, Concerning Nonconnected Inverse Limits with Upper Semi-Continous Set-Valued Functions , Topology Proceedings vol. 40 (2012), 203-214. W. T. Ingram, An introduction to inverse limits with set-valued functions , Springer Briefs in Mathematics, 2012. W. T. Ingram, Inverse Limits with Upper Semi-Continuous Bonding Functions: Problems and Some Partial Solutions , Topology Proceedings, vol. 34 (1) (2010), 353-373. Kenneth Kunen, Set Theory: An Introduction to Independence Proofs , Elsevier B.V., 1980. 14 SCOTT VARAGONA
Van Nall, Inverse limits with set valued functions , Houston Journal of Mathematics, 37 (2011), no. 4, 1323-1332. Van Nall, Connected inverse limits with a set-valued function , Topology Proc. 40 (2012), 167-177. Michel Smith, Generating Large Indecomposable Continua , Pacific Journal, vol. 62, No.2 (1976), 587-593. Scott Varagona, Simple techniques for detecting decomposability or indecomposability of gen-eralized inverse limits , Ph.D. dissertation (Auburn University), 2012. Patrick Vernon, Inverse limits of set-valued functions indexed by the integers , Topology Ap-plications 171 (2014), 35-40.
Department of Biology, Chemistry & Mathematics; University of Montevallo; Montevallo, Alabama 35115
E-mail address : [email protected]
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Forced Stalemates
I was thinking, is it possible to be forced into a Stalemate? If so provide examples.
Here in a winning position Reshevsky blunders with Qxg3? This allows Evans to sacrifice his queen and force a draw. If the rook is captured it will be stalemate and if the king moves it will be perpetual check.
It's quite hard.
Even in my this "lost game", my opponent made an unforced error and allowed me to escape with a draw, I would not classify this as a forced perpetual check.
Forced stalemates are much harder, and so far all of my games that ended in stalemate (both the situations where I was stalemating the opponent and where I was stalemated) appear due to something unforced.
Depends on how you define forced stalemate, but I doubt this example counts.
Here in a winning position Reshevsky blunders with Qxg3? This allows Evans to sacrifice his queen and force a draw. If the rook is captured it will be stalemate and if the king moves it will be perpetual check.
This one is not strictly a forced stalemate, because Reshevsky does have the choice between capturing the rook (draw by stalemate) or moving the king (perpetual check eventually forcing threefold repetition). I suppose there might be desperado positions where that choice does not exist and the only legal move to get out of check is the capture that creates stalemate - but cannot provide examples.
Turns out it is impossible to be legally forced into stalemate since 1997:
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| 5.2.2 | The game is drawn when a position has arisen in which neither player can checkmate the opponent’s king with any series of legal moves. The game is said to end in a ‘dead position’. This immediately ends the game, provided that the move producing the position was in accordance with Article 3 and Articles 4.2 – 4.7. |
Therefore, as soon as all moves that otherwise would be legal result in a stalemate, the game already ends in a dead position - being forced into stalemate is impossible because the position forcing stalemate is a dead position. Example position:
White is in check and would otherwise have two legal moves to get out of check: Kxg5 or hxg5. However, both would create stalemate. Therefore the position already is a dead position and the game has already ended.
It is more common, I think, to force a draw by repetition where stalemate is a threat. That's what happened here. Black to move in all three diagrams.
But, stalemate can be forced from many positions.
My opponent forced stalemate from:
And I did from:
Pfft! Who would want to force stalemate? If you're determined not to win you might as well resign.
Yeah cuz I'm sure if you were in a tournament you would throw away the half a point.
It is more common, I think, to force a draw by repetition where stalemate is a threat. That's what happened here. Black to move in all three diagrams.
But, stalemate can be forced from many positions.
My opponent forced stalemate from:
As I pointed out, no, under 5.2.2 stalemate can never be forced:
White would otherwise have one legal move to get out of check, but this would produce stalemate. Since all legal moves draw, the position on board is a dead position, the game already is ended and it is illegal for White to actually capture the rook and produce stalemate.
Choice between stalemate and threefold repetition is not a dead position. In case of perpetual check, the defender needs to actually keep up checking until third repetition and then actually claim. The defender might legally blunder into making a different move and getting checkmated after all. Therefore the game does not end in dead position before the attacker actually plays stalemate or either player claims threefold repetition (or a fivefold repetition happens).
Pfft! Who would want to force stalemate? If you're determined not to win you might as well resign.
If I have an inferior position, I would be thrilled to draw!
Half a point beats no points all day and twice on Sunday.
Only moronic clowns don't appreciate the draw in the game of chess and the many ways to achieve it!
Turns out it is impossible to be legally forced into stalemate since 1997:
Therefore, as soon as all moves that otherwise would be legal result in a stalemate, the game already ends in a dead position - being forced into stalemate is impossible because the position forcing stalemate is a dead position. Example position:
White is in check and would otherwise have two legal moves to get out of check: Kxg5 or hxg5. However, both would create stalemate. Therefore the position already is a dead position and the game has already ended.
Actually, forced stalemate is possible based on the rules given.
WKc8, WQh8, WPb5, BKa7, BQg3, BPb6. Black to Move.
The game is not over as Black can legally play one of many legal moves, and many lose, but he has a forced stalemate with 1...Qc7 if he wants it. Of course, instead of the losing moves or 1...Qc7, I would advise Black to play 1...Qb8 and win!
As I pointed out, no, under 5.2.2 stalemate can never be forced:
Chess.com does not use FIDE rules. In the US, very few events follow FIDE rules. Under USCF rules, stalemate is a legal move.
I saw your post before I created mine. It is useful, but does not answer the OP's question.
What about the "crazy rock"? Where you have only one rock beside your king and this one can't move at all (or there are also other materials that can't move anymore).
Therefore, you just attack the opposent's king with your rock at each move; either he moves and you go on with attacks, or he takes the rock and lt's draw.
Whatever the circomstances, the goal might be for the team's score, or when the victory isn't possible anymore (or you are in a hurry to eat!!).
What about the "crazy rock"? Where you have only one rock beside your king and this one can't move at all (or there are also other materials that can't move anymore).
Therefore, you just attack the opposent's king with your rock at each move; either he moves and you go on with attacks, or he takes the rock and lt's draw.
Yes, that´s the situation in all the diagrams.
But my point is that under the Fide rules since 1997, none of them is a "forced stalemate", and there cannot be a "forced stalemate". Either the king has a choice between capturing the rook (and drawing by stalemate) and moving (and being chased by perpetual check into threefold repetition) and then the stalemate is not specifically forced because the king can choose to draw by repetition instead of stalemate. Or else the king has nowhere to move, but then it is a dead position and the game should end in a draw then, without actually capturing the rook.
What about the "crazy rock"? Where you have only one rock beside your king and this one can't move at all (or there are also other materials that can't move anymore).
Therefore, you just attack the opposent's king with your rock at each move; either he moves and you go on with attacks, or he takes the rock and lt's draw.
Yes, that´s the situation in all the diagrams.
But my point is that under the Fide rules since 1997, none of them is a "forced stalemate", and there cannot be a "forced stalemate". Either the king has a choice between capturing the rook (and drawing by stalemate) and moving (and being chased by perpetual check into threefold repetition) and then the stalemate is not specifically forced because the king can choose to draw by repetition instead of stalemate. Or else the king has nowhere to move, but then it is a dead position and the game should end in a draw then, without actually capturing the rook.
Please bear in mind that FIDE rules apply to FIDE sanctioned competition, and only FIDE sanctioned competition.
It does not apply to competition sanctioned by other bodies, nor to casual chess, nor to the vast majority of websites (and those claiming to employ FIDE rules likely do not encode the rule as you have explained it).
Hence, your post is interesting, but barely relevant.
king and pawn vs king endgame where the lone king has the opposition is an example.
force a stalemate with white
force a stalemate with white
Forced stalemate from black
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ANNALES DE L’I. H. P., SECTION B PHILIP S. GRIFFIN TERRY R. MCCONNELL GREGORY VERCHOTA Conditioned brownian motion in simply connected planar domains Annales de l’I. H. P., section B, tome 29, no 2 (1993), p. 229-249 © Gauthier-Villars, 1993, tous droits réservés.
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques 229.
Conditioned Brownian motion in simply connected planar domains Philip S. GRIFFIN (), Terry R. McCONNELL () and Gregory VERCHOTA () Department of Mathematics, Syracuse University, Syracuse, NY 13244-1150, U.S.A Ann. Inst. Henri Poincaré, Vol. 29, n° 2, 1993, p.
249.
Probabilités et Statistiques ABSTRACT. - The purpose of this paper is to study Doob’s conditioned Brownian motions in simply connected domains in [R2. We obtain the precise value of the best constant in the lifetime inequality of Cranston and McConnell and prove a related maximum principle. We also exhibit a connection between these processes and the classical isoperimetric inequalities.
Key words : Conditioned Brownian motion, expected lifetime, positive harmonic function, isoperimetric inequality.
1. INTRODUCTION Let D be a domain in !R" and p (t, a, P) the transition densities of Brownian motion killed on exiting D. Let positive harmonic function in D}.
Classification A.M.S. : (1980) Subject classification (1985 Revision) : 60 J 45, 60 J 65.
() Research supported in part by N.S.F. Grant DMS-8700928.
() Research supported in part by N.S.F. Grant DMS-8900503.
() Research supported in part by N.S.F. Grant DMS-8915413.
Annales de l’lnstitut Henri Poincaré - Probabilites et Statistiques - 0246-0203 Vol. 29/93/02/229/21/$ 4,10/ © Gauthier-Villars 230 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA For h E H + D set Let P a denote the measure on path space induced by these transition densities and Ea expectation with respect to Pa. The canonical process, which we denote by Zt [or sometimes Z (t)], is then a Doob conditioned Brownian motion or h-process. Its lifetime is given by If no confusion can arise we will often drop the subscript D and simply write T for the lifetime.
In 1983, Cranston and McConnell proved that for any domain D ~ 1R2, there exists a constant cD such that and furthermore Note that ( 1.1 ) is only of interest when D has finite area. See ] for references to extensions, applications and related results.
To translate this result into analytic terms, we only need observe that where G is the (probabilist’s) Green function for D, i. e., minus two times the analyst’s Green function and m is Lebesgue measure.
One of the main results of this paper is an evaluation of the best possible constant among all simply connected domains, i. e.
c = sup { CD : D is simply connected}. _ We will prove that c = 1 and furthermore that the su p remum can not be attained. " Examples of domains for which c D approaches - are given in section 3 7T (see Theorem 3 . 4) and include long thin rectangles. The opposite extreme to these domains is the disc and here we explicitly compute CD to be (4 log 2 - 2)/~c ^-~ . .7726jn.
If h is a Martin kernel (i. e. minimal) function with pole at where aM D = minimal Martin boundary of D, then we write P a and E~ for P a and Ea respectively. In this case we prove that, as a function of a E D, satisfies a maximum principle; see section 2. It is not clear Annales de l’Institut Henri Poincaré - Probabilités et Statistiques 231 CONDITIONED BROWNIAN MOTION how important it is that h be minimal in this result, but we should point out that does not satisfy the maximum principle for all hEH+ D, for example when h is constant.
In the final section we extend the best constant result to the case of superharmonic h and discuss various topics including the connection with the isoperimetric inequality; see Remark 5.4. -The authors would like to thank Tadeusz Iwaniec for informing us of the work of Carleman, Gerry Cargo for a simplification in the proof of Theorem 2.1 and Eugene Poletsky for several enlightening discussions.
Finally we would like to thank the referee for pointing out an oversight on our part in the statement of the strong maximum principle in section 2, and for the observation that 0~4/71: for any planar domain. This bound can be bound in Banuelos .
2. A MAXIMUM PRINCIPLE In this section we will prove a strong maximum principle for as a function of a E D whenever D c 1R2 is simply connected and I D oo .
Here z) and K(3 is any kernel function for D with pole at By replacing K(3 with the constant function we see that such a principle cannot hold in general for positive harmonic functions. We will also prove a maximum principle for simply connected domains of infinite area in 1R2.
Before giving the strong maximum principle we make a few simple observations. Let B denote the unit disc and aB its euclidean boundary.
Fix a E B, b E aB and let 03A6: B ~ D be the 1-1, conformal map of B onto D such Here and elsewhere below we assume that 03A6 has been extended to a homeomorphism of the Martin closures, and we identify OM B with aB. Then by conformal invariance of Green functions and kernel functions where, Vol. 29, n° 2-1993.
232 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA is the Green function for B with pole at a and is a kernel function for B with pole at b E aB. We can always find such a conformal map for which a = 0 and b =1. In this case 0 maps the interval [ -1, 1] onto the hyperbolic geodesic through a and P. Thus the strong maximum principle is a consequence of the following stronger result: THEOREM 2. 1. - Let F not identically zero be holomorphic in the disc B and FeL2(B). let Then L’ (0) O. ’ ’ Remark. - This result is stronger than the strong maximum principle in two ways. First it allows us to replace C’ in (2.2) with an arbitrary holomorphic function Fe L 2 (B), and secondly it shows that if uo is on the hyperbolic geodesic connecting a point ex E D = D U OM D to P, iD increases as ao moves along the geodesic away from P.
Proof. - One easily sees that the derivative of the numerator may be taken inside the integral. Evaluating the derivative one gets Let akzk. In polar coordinates dm (z) = r dr d8 so that by orthogonality while Thus integrating in 8 first and then in r, the theorem follows if the inequality holds.
Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques 233 CONDITIONED BROWNIAN MOTION By taking absolute values, it suffices to prove (2. 3) for nonnegative ak.
Let aM be the first nonzero coefficient. Then by the geometric-arithmetic mean inequality This shows the inequality in (2. 3) is strict provided all quantities in (2.4) 00 k are finite. To see this observe that if then b~. Thus j=O ;=o by the Cauchy-Schwarz inequality, and the last quantity is finite since F E L2 (B)..
. COROLLARY 2. 2. - Let D be a nonempty simply connected plane domain of finite area. Let 03B1, 03B2 belong to OM D and exo lie on the hyperbolic geodesic r joining ex and 03B2. Then where the kernel functions are normalized to be 1 at,some point, call it ç, on h.
Proof. - We begin by showing that if a" converges to a in the Martin topology then To see this, map the unit disk 1-1 and conformally onto D so that -1, 0, and 1 are mapped to cx, ç, and P respectively. Then (2 . 6) follows from its counterpart in the unit disk, which reads This is a straightforward computation.
Vol. 29, n° 2-1993.
234 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA Now let I> denote the conformal map described above. Choose a strictly decreasing sequence wn of negative real nurnbers such that lim wn = -1.
n ~ o0 It follows from Theorem 2 .1 that is strictly increasing. Thus By [ 11 ], Corollary 1. 2, the integrands on the right-hand side of the last inequality are uniformly integrable. Thus we obtain (2. 5) from (2.6) by passage to the limit as n - oo. M Remark 2.3. - Essentially the same argument shows that if an is any sequence converging to a in the Martin topology, then where the kernel functions are normalized at some point ~ lying on the hyperbolic geodesic joining cx and ~. The right-hand side is the expected lifetime of Brownian motion in D started at the entrance boundary point cx and conditioned to die at the boundary point ~. (See .) Accordingly, we denote it by Ea ’to COROLLARY 2.4. - Let D be a nonempty simply connected plane domain with a Green function. Let belong to aM D and CXo lie on the hyperbolic geodesic r joining oc and ~. Then Proof. - By conformal invariance it is enough to prove for any conformal map 03A6 defined in B.
By monotone convergence the right hand side of (2. 8) may be written as Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques 235 CONDITIONED BROWNIAN MOTION The left hand side may be written The first integral converges by monotone convergence and the second by dominated convergence. For the third the functions may be extended to {z: l/2~z 1 and Rez>0} so that con-verge uniformly there to /(z) = ’20142014L log -2014-. This function is uni-formly bounded away from zero. As has already been noted the functions gr (z) = (r2-|z|2)2 |r2-z2|2 for {z:|z| r}, extended to be zero outside { z: |z|r}, increase monotonically on [z: z| 1} to g (z) = -2014’ . . Thus given any E > 0, by Fatou’s lemma, uniform convergence to f and monotone convergence, we have If the same is true of since f is bounded away from zero. Thus the left side of (2. 8) is Now by Corollary 2. 2 for each r > 0 the integrals in (2. 9) are strictly greater than those in (2 .10) and (2. 8) follows..
Remark 2 . 5. - If D 1 = 00 then it is possible that ’t = 00 for CXo E D so that the above proof can not give us a strict inequality in (2.7).
However it would be interesting to know whether or not the strict maximum principle holds in the case where the maximum expected life-time in D is known to be finite but D has infinite area. -Vol. 29, n° 2-1993.
236 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA 3. THE BEST CONSTANT In this section we show that - is the best constant in the lifetime 7T inequality when the terminal point lies on the boundary, or, more generally, when the conditioning function is positive harmonic. The ine-quality continues to hold with the same constant even when the terminal point lies in the interior. The more difficult proof of that fact is deferred to section 5. The precise result to be proved here is the following THEOREM 3.1. 2014 Let D be a nonempty simply connected plane domain of finite area, u e D, and /!>0 harmonic on D. Then and the constant 1 is best possible.
It suffices to prove (3 .1 ) when h is minimal, say h = K03B2, 03B2~~MD.
Moreover, by the results of Section 2 we may assume Recall (see Corollary 2. 2 and Remark 2. 3) that if ç lies on the hyperbolic geodesic joining a and P, and if K" and K~ are normalized so that K" (~) == K~) = 1, then If O is the unique 1-1 conformal map of the unit disc B onto D such that c~ ( -1 ) = cx, d~ (0) =~, and 0 ( 1 ) = P, then The unit disc here may be replaced by any other model simply connected domain, . and it is more convenient for our purposes in this section to replace it by the infinite strip An easy computation using e. g., the fact that log ( 20142014) maps B to S, shows that 1 - z Annales de l’lnstitut Henri Poincare - Probabilités et Statistiques 237 CONDITIONED BROWNIAN MOTION where ’P maps S 1-1 and conformally onto D with 03A8 ( - oo) = a, W (0) = ç, and (This formula has been noted independently by R. Banuelos and T. Carrol .) By the half-angle formula, where The proof of (3 . 1) thus reduces to showing that the second term on the right-hand side of (3 . 3) is strictly negative, which, in turn, is an immediate consequence of the following two results.
LEMMA 3. 2. - Let f be a strictly increasing real-valued function and g a strictly decreasing function on a finite interval [a, b]. Then LEMMA 3. 3. - The function H defined in (3 . 4) above is strictly increasing To complete the proof of (3.1), apply Lemma 3.2 with [a, b] = 0,- , f=H, and g(y)=cos(2y).. Proof of Lemma 3 . 2. - Let 1=20142014 f(y)dy and choose such that f~I on [a, y] andf>I on (y, b]. Let/=/-!. Then We should remark that this result is a special case of an inequality of Chebyshev. See, e. g., Theorem 43 (2.17.1) in .
Proof of Lemma 3. 3. - We begin by showing that H is bounded on [0, A] for each 0A03C0 2. Fix B such that AB03C0 2. Since Vol. 29, n° 2-1993.
238 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA it follows from Fubini’s Theorem that we may further assume B is chosen so that H (B) 00. By similar reasoning applied to integrals over vertical line segments we may choose Xn -+ oo and a constant C independent of n such that Let z0=x0+iy0 satisfy and let rn denote the positively oriented boundary of the rectangle [- x~, xn] X [ - B, B]. By the Cauchy integral formula we have for xn> 1 Xo I It follows from the Schwarz inequality and (3. 5) that the contribution to this integral from the vertical ends of rn vanishes in the limit as n - oo .
Thus we obtain the representation valid for all z~ = xo + iyo satisfying A.
Since H (B) 00, one may apply results from the theory of HP-spaces with p = 2 (e. g., the corollary on p. 172 of ) to each of the two integrals above to conclude that Next, by a well-known Paley-Wiener Theorem (see, e. g. , p. 174) is the Fourier transform of a nonzero function cp such that ey|03BB|03C6(03BB)~L2(R) for each |y|03C0 2. Thus, by Plancherel’s Theorem the function H (y) is a positive multiple of cosh(2y03BB)|03C6(03BB)|2d03BB, which is clearly strictly increasing in y..
To show that 1 is best-possible, let Dp denote the image of B under the 7T 0pl. Then a direct, but lengthy, com-putation shows that .
Annales de l’Institut Henri Poincaré - Probabilités et Statistiques 239 CONDITIONED BROWNIAN MOTION Alternatively, one may apply Theorem 3.4 below (see esp. example 3.6) but perhaps the easiest way to see this is by means of a probabilistic argument, which we sketch at the end of this section.
Since strict inequality always holds in (3 . 1) there are no extremal domains, but domains such as DP above for p close to 1 may be considered "near extremal". The following result shows that there are many near extremal domains and provides some additional insight into how the geometry of the domain influences the expected lifetime.
THEOREM 3 4. - Let D be a bounded convex domain which is symmetric with respect to one of its diameters. Let R = R (D) denote the supremum of radii of all open discs contained in D, P = P (D) the perimeter, and 0394 = 0394 (D) the diameter of D. Then there exist points cx E aD and 03B2 E aD such that Remark 3 . 5. - J. Xu [1 5] has shown that there is a positive constant y so that for all convex plane domains.
Proof of Theorem 3 4. - We may assume that D is symmetric with respect to the x-axis and that there are points a and 03B2 on the x-axis so that [a, P] is a diameter of D. For technical reasons it is convenient to replace D with a slightly smaller convex domain having smooth boundary.
Thus, let D be a 1-1 conformal map of B onto D such D(1)==P. Let for 0pl, and let Dp be the image of B under Op. Then Dp is also convex by Study’s Theorem (see, e. g. , p. 224) and symmetric with respect to the x-axis.
- . Note that ’Pp maps S conformally onto Dp, is real on the x-axis and maps the upper boundary of S to the upper boundary of Dp. It is easy to check that Dp has a smooth boundary and also that is analytic in an open set containing the euclidean closure of S, (3 . 7) ) lP§ I is bounded on the euclidean closure of S, (3 . 8) Vol. 29, n° 2-1993.
240 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA and Since the left-hand side of (3 .10) equals Ea i and the right-hand side represents the limit of the analogous quantities for the Dp, it is sufficient to prove (3. 6) for the domains DP, with the quantity A (Dp) replaced by the length of the intersection of Dp with the x-axis. To simplify the notation, we shall assume for the remainder of the proof that D = Dp for some 0 p 1, and drop the "p" throughout.
Our method now is to find a lower bound for the second term on the right-hand side of (3 . 3), i. e., Let u, v real, and note that, by (3 . 7) and the Cauchy-Riemann equations, Here all expressions are evaluated at O~~~Tr/2, and K is the curvature of the curve ~ -+ ’P (t, y). Thus, since D is convex we have ~ ~y(|03A8’|2~0 for y=03C0 2; since ’P maps the x-axis to the x-axis we have ~(|03A8’|2)=0 for y=0. On the other hand, ly =Re [.q"J 1- so that the =2K ~ is 12 2 ’ ’ harmonic on S, and bounded on S by (3.8). By the maximum principle we may conlude for 03C0 2, -~x~.
Recalling that the expression in (3.11) is negative and integrating by parts on y, we have Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques 241 CONDITIONED BROWNIAN MOTION By Koebe’s distortion theorem (See e. g., , p. 147, where the result is stated for the unit disc. The version we use is easily obtained by conformal mapping.) Thus, which completes the proof..
Example 3.6. - Take for D the rectangle C- ~ ~ 21 x C 2’ 21’ Then by the proof of Theorem 3 . 4 ~ ~ ~ ~ We conclude this section by sketching a simple probabilistic argument to show that long thin rectangles are near-extremal. For given E > 0 let Dn denote the rectangle [ - En, n] [-03C0 2,03C02]. Let h (z) = ex cos y, so that h is positive harmonic on Dn. Finally, let tn = We will show that E© tn ’" n as n - ~, hence E© Dn 1-+ 1 as n - 00.
n(l+E) The h-process Zt = Yt) started from 0 satisfies the stochastic differen-tial equation with Zo=0. Here Wt is a standard 1R2-valued Brownian motion started from 0. Let 03B2t denote the x-component of Wt. Since the drift is given by V/x ’ 2014(z)=(l, - tan y), we have that and moreover, , h ( ) ( ~ .v)~ t at > > (Note that Zt cannot leave Dn through the top or bottom boundaries since h vanishes there.) Since both points - E n and n are accessible for the 1-dimensional diffusion fl~ + t, it follows that oo . Thus, by Wald’s Vol. 29, n° 2-1993.
242 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA identity, E ~3~n = o. We conclude that The desired result follows if we can show pð = n) - 1 as n ~ ~. One way to see this is to note that -+ 0, a. s., as t - 00 by the strong law, hence ~3t + t has sample paths which are almost surely bounded below.
4. THE DISC As we have already indicated, the extremal domains in the best constant problem are long thin rectangles. The opposite extreme to these domains is the disc. We will explicitly compute the best constant in this case. By the maximum principle of section 2, the worst case occurs on the boundary, i. e. Brownian motion conditioned to go from one boundary point to another. The first step is to show that the boundary points are diametri-cally opposite. We will do this probabilistically, by a coupling argument.
We begin by recalling some basic facts about h-processes. Let D 1 and D2 be two domains and C a 1-1 conformal map of D 1 onto D~. We will write where z = x + iy and w = u + iv. If h2 is a positive harmonic function in D2, then is a positive harmonic function in Di. If Zt is an hi-process in D 1 then 03A6(Zt) is a time change of an h2-process in D2; to be precise, there exists an h2-process Wt such that As an immediate consequence, we have that If we apply this with D1=B, D2 = ~ w : u > 0 ~, h2 (w) = u and ~ (z) _ ( 1 + z) ( 1- z) -1, we see that Furthermore the h2-process W = (U, V) is very simple; U is a Bessel process of index 3 and V is an independent 1-dimensional Brownian motion. If we let be the inverse of ~, and apply the above in reverse, we can conclude that where H is the half space M>0. and ’tH = 00 a. s., to show that iB is maximized when Z starts at z = -1 is equivalent to showing Annales de l’Institut Henri Poincaré - Probabilités et Statistiques 243 CONDITIONED BROWNIAN MOTION is maximized when W starts at w = O. Since we already know the maximum occurs on the boundary, we only need consider starting points on the imaginary axis u = O.
PROPOSITION 4 . 1. - Let w~ = iv~, j =1, 2, be two points on the boundary of H such that I v21. Then on the same probability space, one can define two h2-processes W and W 2 starting at w l and w2 respectively, such that Proof - Let W1 = (Ul, be an h2-process started at w 1 defined on some probability space. Let Define W~=(U~, V2) by U2 = U1 and By the reflection principle V; is a Brownian motion starting at v2 indepen-dent of U2. Thus W2 is an h2-process started at w~ and by construction V; = V; and I V; I ~ for all t. Since it follows that for all s, from which the result is immediate..
As a consequence of the previous result we have that Thus by the Poisson representation of positive harmonic functions in B the best possible constant c B for the unit disk is 1 03C0E1 1 which we now compute. " Vol. 29, n° 2-1993.
244 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA PROPOSITION 4 . 2. - is = 4 log 2 - 2 ~ . 7726.
Proof. - Recall that by (3 . 2) Among convex domains the disc is at the opposite extreme to the long thin rectangles, i. e., it minimizes the perimeter for a given area. Thus it seems reasonable to ask Open Question. - Amongst all convex domains D of area x, is minimized when D = B?
If we remove the convexity assumption then the result is false. Indeed by an example of Xu [ 15] there are simply connected domains of infinite area having sup E~ r as small as desired.
a, fi e D It is interesting to note that in the case of unconditioned Brownian motion (i. e., A= 1), the disc is in fact the worst case, i. e., and in the simply connected case, equality holds iff D is a disc. This is a consequence of classical isoperimetric theory which says that the distribu-tion function of the Green function is pointwise maximized over domains of equal area only in the ball; see .
5. THE SUPERHARMONIC CASE In this section we wish to establish (3 .1 ) in simply connected domains with finite area in the plane when h is superharmonic with nonnegative Annales de l’Institut Henri Poincaré - Probabilités et Statistiques 245 CONDITIONED BROWNIAN MOTION boundary values. See . To do this it suffices by conformal mapping and the Riesz Representation Theorem to prove the following result.
THEOREM 5.1. - Let F not identically zero be holomorphic in the unit disc B with F e L2 (B). Then for all b e B Our proof is based on a lemma that has a very close relationship with Carleman’s generalization of the isoperimetric inequality . Denote rd 9 by dc and LEMMA 5.2. - Let F1, ..., Fn+l’ n =1, 2, ..., not identically zero be holomorphic in B with F1F2...Fn+1 ~ L2(B). Then Proof. - Let Fj(z)= 03A3 a(j)k zk, 1 - j - n + 1. The square of the modulus k=0 of the jo-th coefficient of F 1 (z)... (z) is by the Schwarz inequality.
Let I denote the left side of (5 . 2). Integrating in 8 and using (5.3) Vol. 29, n° 2-1993.
246 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA By an induction on n > 1, for all J > O. Thus Proof of Theorem 5 . 1. - It suffices to take 0 b 1. Since it follows that Thus Apply the lemma for each ~1 with F~=F~= ...F~+i= 201420142014.
Using the harmonicity of -2014’2014L m B ~l-&z~ Thus (5.4) is less than Annales de l’Institut Henri Poincaré - Probabilités et Statistiques 247 CONDITIONED BROWNIAN MOTION Now use the identity valid for all 0~1. N Remark 5. 3. - An alternative approach to Theorem 5. 1 would be to obtain a maximum principle similar to the one in section 2 but with both points lying in the interior. Unfortunately, we have been unable to prove this.
Remark 5.4. - The proof of Lemma 5.2 is a generalization of Carle-man’s proof when n =1 and the log and the weights (1- z|2)n are not present. Many different lemmas of this type may be stated. This one is devised for Theorem 5.1. In Beckenbach and Rado generalized Carleman’s result to deal with integrands that are logarithmically sub-harmonic.
Carleman’s result can be used to give an alternative proof of Theo-rem 3 .1, which we very briefly sketch below leaving the details to the interested reader. Using Carleman’s isoperimetric inequality, the Cauchy-Schwarz inequality, and the coarea formula, we have The second inequality is strict except when F is a constant multiple of 201420142014, i, e., the derivative of the conformal map taking B to an infinite strip. This is consistent with the observation that long thin domains are extremal for the expected lifetime. Furthermore in this case is the Jacobian of a linear fractional transformation. The isoperimetric inequality is then also sharp since the t ~ are discs.
Vol. 29, n° 2-1993.
248 P. S. GRIFFIN, T. R. McCONNELL AND G. VERCHOTA Remark 5. 5. - A simple computation shows that K -1 is logarithmi-cally harmonic. Hence, by conformal mapping, given any two kernel functions K°‘, KP for a simply connected domain D c R2, is loga-rithmically harmonic. Thus for all zeD. More generally, given any kernel function K and any positive harmonic function h for D, h/K is logarithmically subharmonic. This is because when transferred to the disc, by the Poisson integral representa-tion, h/K has the representation I(z)= f" |F03B8(z)|2 d (03B8) where Fo (z) is holomorphic in B for each 8 and dJl is a positive Borel measure. Taking the log and then applying the Laplacian yields which is positive by the Schwarz inequality. As a consequence Given an h-process the quantity ~h is called the drft. We have thus h proved the following.
PROPOSITION 5. 6. - In a simply connected domain D c [R2 the magnitude of the drift at a point z e D is maximized over all positive harmonic functions h in D precisely by the kernel functions, and this maximum is attained by the drift associated with every kernel function for D.
Remark 5. 7. - The quantity G03B1(z)G03B2(z) G03B1(03B2) dm (z) can be obtained as the limit of quantities D G« ( ~) where BE is a small disc shrinking to a point and 03B2 e aBE. Thus 1 /03C0 is the best constant in the lifetime inequality with terminal point on the boundary in a doubly connected domain when the hole is small enough. Saying anything more than this about the best constant in multiply connected domains’ is an open problem.
Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques 249 CONDITIONED BROWNIAN MOTION REFERENCES C. BANDLE, Isoperimetric inequalities and applications, Pitman Publishing, Boston, 1980.
R. BAÑUELOS, On the estimate of Cranston and McConnell for elliptic diffusions in uniform domains, Prob. Th. Rel. Fields, Vol. 76, 1987, pp. 311-323.
R. BAÑUELOS and T. CARROL, Conditional Brownian motion and hyperbolic geodesics in simply connected domains, (Preprint).
E. F. BECKENBACH and T. RADO, Subharmonic functions and surfaces of negative curvature, Trans. Amer. Math. Soc., Vol. 35, 1933, pp. 662-674.
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S. SAKS and A. ZYGMUND, Analytic Functions, 3rd Ed., American Elsevier Publishing Co., New York, 1971.
T. SALISBURY, A Martin boundary in the plane, Trans. Amer. Math. Soc., Vol. 293, 1986, pp. 623-642.
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(Manuscript received November 22, 1991; revised September 9, 1992.) Vol. 29, n° 2-1993.
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