Factorial of fractional numbers in an ARFIMA (long-memory time series analysis) context?
409602
Search for "Gamma function" in/at/on Google.\n\nThe factorial function has 0! = 1 "the empty product" and for integers n > 0 n! = 1 times 2 times ... times n.\n\nI believe it was Euler who noticed that if you compute the following definite integral -- let's call it Gamma(n) for short --\n\nintegral from 0 to positive infinite\nof x^n e^(-x) dx\n\n-- that you get:\n\nGamma(0) =\nintegral from 0 to positive infinite\nof x^n e^(-x) dx\n\n= 1\n\nand for integers n > 0\n\nIntegrate by parts: u = x^n, dv = e^(-x) dx\nintegral of u dv = uv - (integral of v du)\nGiven u and dv, we find that du = nx^(n-1) and\nv = - e^(-x)\n\nSo the integral above is equal to\n(- x^n e^(-x) from 0 to positive infinity)\nminus\n(the integral from 0 to positive infinity\nof - n x^(n-1) e^(-x) dx).\n\nx^k e^(-x) goes to zero as x goes to\npositive infinity for any k, so the\nresult is that, and at the lower limit\nof 0 you have x = 0.  So you get:\n\nGamma(0) = 1\nGamma(n) = n * Gamma(n-1)\n\nLook familiar?\n\nGamma(5) = 4 * 3 * 2 * 1 * 1 = 4!\nand in general n! = Gamma(n + 1).\n\nSo this gives a way to compute x! when\nx is not a nonnegative integer.  Compute\nGamma(x + 1) and call it x!.  In this\nmanner Euler extended the definition\nof the factorial function to include\nnon-integers.
885409
A number that can be written as a fraction.
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
515668
F = (m dot * V)e - (m dot * V)0 + (pe - p0) * Ae
906131
Having researched this topic as far as possible online, I found very few helpful resources.\n\nWhile the result you mention, using the subsequence 6(5)^n, is intriguing, I suspect that other sequences will yield vastly different bounds on the index.  In particular, is it not true that any prime number p always has a corresponding Fibonacci number F(p+1) of which it is a factor?\n\nGood luck with your own research!
447322
Try this site:\n\nhttp://www.mathsisfun.com/long_division2.html
1080867
This is the Gaussian binomial coefficient [n,1] for q=2. \n\nGreetings from The On-Line Encyclopedia of Integer Sequences!
917477
where else?\n\nhttp://www.factoring-polynomials.com/
594715
are these sociable numbers? ie each is the sum of the previous term's factors?
958614
#1.)-----------------------------------\n  \n25b^ -20by + 4y^\n\nMultiply 1st and 3rd term together to get 100b^y^.\n\nFactor 100b^y^ into two factors whose sum = the middle term of -20by.\n\n1by   X 100by\n2by   X 50by\n4by   X 25by\n5by   X 20by\n10by X 10 by   ....................BINGO!\n\n-10by + (-10by) = - 20by   &   (-10by)(-10by) = 100by\n\nThis is a really kewl factor by grouping trick, I LOVE it!\n\nBring down the first term of: (25b^)\n\nNext, write the SMALLER of the two factors 1st.  Followed by the LARGER of the two factors.  In this case, they're both the same, so it doesn't matter which one goes first: (-10by)\n\nFinally, bring down the third term of: (+4y^)\n\n25b^ -10by -10by + 4y^\n\nNow, lets factor by grouping! \nFactor out the GCFs.\nGCF#1 = 5b\nGCF#2 = -2y\n\n(25b^ -10by) + (-10by +4y^)\n5b(5b - 2y) -2y(5b - 2y)\n\nGCF#3 = (5b - 2y)\n\n(5b - 2y)(5b - 2y)\n\nSimplifies further as: (5b - 2y)^   ....Whatdayaknow?! \n.......................................................................................................It's a Perfect Square Trinomial.\n\n#2.)------------------------------------------------------------------\n\n100 - 9b^\n\nAh, the difference of two perfect squares.\n\n---------------------------------------------\n\nFORMULA:\na^ - b^ = (a-b)(a+b)\n\n---------------------------------\n\nSOLUTION:\n100 - 9b^\n(10)^ - (3b)^\n\n-----------------------\n\nANSWER:\n(10-3b)(10+3b)
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
735909
If you want an analytical method, use a Taylor series.\nIf you want an iterative method (successive approximation, to arbitrary precision), use the CORDIC algorithm.\nThere are surely heaps of other methods.\n\nTaylor series:\narcsin z = z + (1/2) z^3/3 + (1*2)/(3*4) z^5/5 + (1*3*5)/(2*4*6) z^7/7 + ...\nSee http://en.wikipedia.org/wiki/Arcsin#Infinite_series\n\nCORDIC algorithm:\nCORDIC is an iterative (not analytical) method involving shift-and-add/subtract, and some table lookup.\nIt doesn't even use multiplication! It's very efficient for both hardware and software implementations.\nhttp://en.wikipedia.org/wiki/CORDIC\n\nAs to gp4rts's comment, by "division" I presume you mean the hard kind, i.e. floating-point division esp. by a variable not an integer constant. 1/(2*4*6) is pretty trivial stuff.\nFor what it's worth both logarithms and square-rooting require either division and/or iterative methods (http://en.wikipedia.org/wiki/Division_%28electronics%29).\n\nSo the bottom line is if you want to avoid *all* division, even the easy integer stuff, use CORDIC. CORDIC is commonly used in chips today, it's a very compact implementation whose speed is good-enough for many applications, and can be pipelined.
1002263
hope this helps\n\nhttp://www.thea.nesinc.com/ptmath.htm
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
520825
Sorry but the model TI-83 is an outdated tool, that was used by the Aertms generation, six centuries ago.  This model could only perform 2,000 logic calculations per second.  In this age that is pathetic.  I am sorry all knowledge that is more then 2 centuries old has been deleted from my memory banks.
1140302
Here is an interactive page that discusses one way to do it:\n\nhttp://dilc.upd.edu.ph/loDeploy/math/math10a.html\n\nFor large numbers, bequalmings technique using logs is an easier way to go.
584781
The Ring of polynoms over the Field lets say G2 (just for denoting) is the right algebraic model to use if we want to study CRCs. \n\nA sequence of bit like 10011101 is modeled by 1x7 + 0x6 + 0x5 + 1x4 + 1x3 + 1x2 + 0x1 + 1x0 which we can write shorter as x7 + x4 + x3 + x2 + 1.\n\nI hope I helped...