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binomial series precisionHello, 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. |

Any series in particular? |

Re: binomial series precisionammm, nope just (1 + x)^k |

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

Re: binomial series precisionI 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. |

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

Re: binomial series precisionwell 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? |

Re: binomial series precisionThere 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. |

Re: binomial series precisionWell 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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