- #1

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I've being using sum of a binomial times something, but was wondering if there is anything more simple.

Thanks!

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- Thread starter kaleidoscope
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- #1

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I've being using sum of a binomial times something, but was wondering if there is anything more simple.

Thanks!

- #2

Stephen Tashi

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What do you mean by "to pick"? Can you state your question mathematically?What function would you use to pick only the last N terms?

- #3

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Here:

where

for example:

where

for example:

- #4

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^ I think he might just mean an (ordered) set of terms from the polynomial. The order is established by listing the coefficients in descending order of their degree.

I think what you want isn't so complicated. It's as follows: if [itex]p \in \mathbb{R}\left[x\right][/itex] is a polynomial with degree n, i.e. [itex]p(x) = {p_n}{x^n} + {p_{n - 1}}{x^{n - 1}} + ... + {p_1}x + p_0[/itex], define the function [itex]F_k: \mathbb{R}\left[x\right] \rightarrow \mathbb{N}^k[/itex] by [itex]{F_k}(p) = (p_n, p_{n - 1}, ..., p_{n - k})[/itex]. This "gets" the first (or if you want, last) k coefficients, which is really all you need to define a polynomial.

I think what you want isn't so complicated. It's as follows: if [itex]p \in \mathbb{R}\left[x\right][/itex] is a polynomial with degree n, i.e. [itex]p(x) = {p_n}{x^n} + {p_{n - 1}}{x^{n - 1}} + ... + {p_1}x + p_0[/itex], define the function [itex]F_k: \mathbb{R}\left[x\right] \rightarrow \mathbb{N}^k[/itex] by [itex]{F_k}(p) = (p_n, p_{n - 1}, ..., p_{n - k})[/itex]. This "gets" the first (or if you want, last) k coefficients, which is really all you need to define a polynomial.

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- #5

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Here:

where

for example:

Exactly! This is the function I've being using. What I wonder is if there is a more simple version of it.

Thank you very much anyways!

- #6

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I think it's the simplest way. May be, if you don't want to deal with combinatorics, use "[URL [Broken] triangle[/URL]

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- #7

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That only works for polynomials which can be expressed in the form (1 + x)^{n}, though ...

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