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 December 22nd, 2010, 11:43 AM #1 Newbie   Joined: Dec 2010 Posts: 5 Thanks: 0 binomial series precision Hello, I'm new in this forum so I'm not sure whether this is the right place to post my question here. I'm doing some computing with binomial series: http://en.wikipedia.org/wiki/Binomial_series. My problem is that I don't know how many elements I should include in to my sum that it would match certain precision (eg. 0.0001). Thanks for the answers.
 December 22nd, 2010, 05:56 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,030 Thanks: 2260 Any series in particular?
 December 23rd, 2010, 03:00 AM #3 Newbie   Joined: Dec 2010 Posts: 5 Thanks: 0 Re: binomial series precision ammm, nope just (1 + x)^k
 December 23rd, 2010, 01:19 PM #4 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 Re: binomial series precision The number of terms you need for a specific precision depends on x and k.
 December 24th, 2010, 12:49 AM #5 Newbie   Joined: Dec 2010 Posts: 5 Thanks: 0 Re: binomial series precision I know that, and in specific cases I've got x and k. So I want to know how using x and k I can find the number of terms for specific precision. Thanks.
 December 24th, 2010, 01:12 PM #6 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 Re: binomial series precision Dumb question: Why not simply calculate (1+x)^k directly?
 December 25th, 2010, 12:07 AM #7 Newbie   Joined: Dec 2010 Posts: 5 Thanks: 0 Re: binomial series precision well the requirement is to write it with taylor series I know that there exists some forms of calculating the remainder, eg. Langrange form (http://en.wikipedia.org/wiki/Taylor's_theorem) but when i calculate remainder using this form it's growing really fast which is nonsense i guess because the remainder should keep getting smaller. This happens when i use x between [-1;0]. Any ideas?
 December 25th, 2010, 09:40 AM #8 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: binomial series precision There are indeed several methods to determine the remainder. The problem with the Lagrange form is that you use a bounding estimate, and so if you wanted to find the remainder for x=-1.5, say, you would see the remainder is simply not bounded. This is true also for x = -1. But if we think back to Taylor's theorem, we might recall that is requires the function to be differentiable at x. Well, the function is not smooth at x = -1, so we do not use x = -1. But this allows us to bound our error for any particular x, although it still depends on the x.
 December 25th, 2010, 10:49 AM #9 Newbie   Joined: Dec 2010 Posts: 5 Thanks: 0 Re: binomial series precision Well yeah I've notice that it depends on the x. I get correct calculations using Lagrange form with x [0..1], so what should I do with x (-1..0) ? I've read about the Cauchy's form, would it help?

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