Euler's Method Question?
822820
The true solution is y=ln(x)+1. Here is the approximation using Euler's method:\n\ny(1) = 1\ny(2)=1.000000 + 1*(1/2)=1.500000\ny(3)=1.500000 + 1*(1/3)=1.833333\ny(4)=1.833333 + 1*(1/4)=2.083333\ny(5)=2.083333 + 1*(1/5)=2.283333\ny(6)=2.283333 + 1*(1/6)=2.450000\ny(7)=2.450000 + 1*(1/7)=2.592857\ny(8)=2.592857 + 1*(1/8)=2.717857\ny(9)=2.717857 + 1*(1/9)=2.828968\ny(10)=2.828968 + 1*(1/10)=2.928968\n\nNow to compare, y(10) - (ln(10)+1) = -0.3736
106376
Also, if you agree that the formula for the number of combinations of n objects taken r at a time is\n\nn!/(r!(n-r)!)\n\nand that there is only one way to get a combination of n objects taking all n at a time, you'd get\n\nn!/(n!(n-n)!)  or n!/(n!0!), or canceling, 1/0! which must equal 1.\n\nSo 0! must equal 1.  More of a demonstration than a proof but it's easy to understand.
508252
im going to assume you have a copy of the binomial theorem already, so ill just use your example problem (it looks like a lot, but stick with me here . . .)(also im sorry if i sound like im babying you, i just want to be clear is all)\n\nmaybe it will help to see the complete binomial explansion first.  try to look at the pattern with the exponents-\n(its hard to type math on the computer, so for example read the first term like this: 1 times (2a to the 7th power) times (-b to the 0th power)\n\n1 (2a)^7 (-b)^0 + 7 (2a)^6 (-b)^1 + 21 (2a)^5 (-b)^2 + \n35 (2a)^4 (-b)^3 + 35 (2a)^3 (-b)^2 + 21 (2a)^2 (-b)^3 +\n7 (2a)^1 (-b)^6 + 1 (2a)^0 (-b)^7\n\nnow, thats what it looks like before you simplify.  note that things to the 0th power will equal one and are irrelevent to the term.\n\neach term has three main parts: a coefficient, an a part, and a b part.\n\nthe a part starts at the highest power, which in this case is 7, while the b part starts at 0.  with each following term, the power of the a part goes down one and the power of the b part goes up one.  the last term has the a part at 0 and the b part at 7.\n\nnow for the coefficients.  the easiest way to understand what coeffiecient is using combinations, which i think you have probably learned already.  (ill use the notation 7-C-2 to mean "7 choose 2").  each coefficient will be 7-C-[ ] and the [ ] depends on the power of that term.  you can make it match the a power (and start with 7-C-7) or the b power (and make it 7-C-0).  this is because, as you probably know, 7-C-3 is the same as 7-C-4 (because 3+4=7).\n\nnow look at the coeffiecients in the expansion i wrote out:\n \nthe first one is 7-C-7 = 1\nnext is 7-C-6 = 7\nnext is 7-C-5 = 21\netc. . .\n\n\n\nso now the question is, how do you go straight to the 4th term without writing the whole thing?\n\nstart by counting backwards from 7 to determine the power of a.  the first term will be 7, the second will be 6, the third will be 5, the fourth will be 4.  ok good, so write down what you know:\n\n(2a)^4\n\nnext think that the powers of a and b must add up to 7, so the b power has to be 7-4=3.  now add that\n\n(2a)^4 (-b)^3\n\nnow stick the coeffiecient n front, knowing that it has to  be 7-C-4 or 7-C-3, depending on how you want to look at it.\n\n35 (2a)^4 (-b)^3\n\nnow just simplify it (and dont forget to note the negatives)\n35 * 16a^4 * -b^3  ----->  -560 a^4 b^3\n\nand thats it! the end! i hope that helped. contact me if you have a question about it
755203
Dear friend,\n\nYou're proposing us the famous Basel Problem, solved by Leonhard Euler in 1735. You should try Euler's approach to the problem, you can see it in the following link:\n\nhttp://en.wikipedia.org/wiki/Basel_problem#Euler_attacks_the_problem\n\nHope I've helped on something, and wish you a nice day!
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
358449
take the rise over the run.\n\nstart at (0,2)\nto get to (4,3) you move 1 up and 4 to the right.\nSo rise=1 run=4\n\nslope is 1/4
487838
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
136546
here's an output step by step:\n\nthis is the initial state of deque D\nD :: first -> <- last\n\naddFirst(3)\nD :: first -> 3 <- last\n\naddLast(8)\nD :: first -> 3 8 <- last\n\naddLast(9)\nD :: first -> 3 8 9 <- last\n\naddFirst(5)\nD :: first -> 5 3 8 9 <- last\n\nremoveFirst()\nD :: first -> 3 8 9 <- last\n\nremoveLast()\nD :: first -> 3 8 <- last\n\nfirst()\nfirst = 3\n\naddLast()\nD :: first -> 3 8 <- last\n\nremoveFirst()\nD :: first -> 8 <- last\n\nlast()\nlast = 8\n\nremoveLast()\nD :: first -> <- last\n\nhope this helps...
845874
1.  This is equal to\nk*(lnx)/x, and since \n(lnx)/x --> 0, so does this expression.\n\n2. Sorry, blank on this and 3.\n\n4. (n^k)/2n = (n^(k-1))/2, and I guess as long as \nk > or = 1\nthis will approach 0.\n\nOf course, if k < 1 then it has no limit, or "approaches infinity"
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
618855
Use the sigma function:\n\n1/16^n * ( (4/(8n+1)) - (2/(8n+4)) - (1/(8n+5)) - (1/(8n+6)) )\n\nfor n = 0 to infinity. You should get pi. This isn't really the easiest way, but it's a start!
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
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)
528695
x to the n + y to the n = z to the n where n is an integer? ... damn it's been so long I can't remember for sure. Some Brit solved the proof already.
500014
a. (x+9)(x-9)\nb.(t-2)(t+2)\nc. (b-12)(b+12)
497395
1.\nBecause e to the power minus infinity is 0.
1065017
I'm writing up a proof right now that this is not possible.\n\nCheck out the pdf at:\nhttp://www.geocities.com/euleratuo/