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-   -   binomial series precision (http://mymathforum.com/pre-calculus/16558-binomial-series-precision.html)

povilasb December 22nd, 2010 12:43 PM

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.

skipjack December 22nd, 2010 06:56 PM

Any series in particular?

povilasb December 23rd, 2010 04:00 AM

Re: binomial series precision
 
ammm, nope just (1 + x)^k

mathman December 23rd, 2010 02:19 PM

Re: binomial series precision
 
The number of terms you need for a specific precision depends on x and k.

povilasb December 24th, 2010 01:49 AM

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.

mathman December 24th, 2010 02:12 PM

Re: binomial series precision
 
Dumb question: Why not simply calculate (1+x)^k directly?

povilasb December 25th, 2010 01:07 AM

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?

DLowry December 25th, 2010 10:40 AM

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.

povilasb December 25th, 2010 11:49 AM

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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