What is the simplest way of selecting the last N terms of a polynomial?

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The simplest way to select the last N terms of a polynomial, such as (1+x)^6, is to define a function F_k that extracts the last k coefficients from the polynomial's expression. This function is mathematically represented as F_k(p) = (p_n, p_{n - 1}, ..., p_{n - k}), where p is a polynomial of degree n. While some users have suggested using combinatorial methods, the direct extraction of coefficients is deemed the most straightforward approach for polynomials in the form of (1 + x)^n.

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If you have a polynomial like (1+x)^6 = x^6+6 x^5+15 x^4+20 x^3+15 x^2+6 x+1, What function would you use to pick only the last N terms? For instance, for N=3 pick x^6+6 x^5+15 x^4

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

Thanks!
 
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kaleidoscope said:
What function would you use to pick only the last N terms?
What do you mean by "to pick"? Can you state your question mathematically?
 
Here:

gif.latex?\dpi{120}%20(x+y)^n=\sum_{\mu=0}^{n}\binom{n}{\mu}x^ny^{n-\mu}.gif

where

gif.latex?\dpi{120}%20\binom{n}{\mu}=\frac{n!}{\mu!(n-\mu)!}.gif


for example:
tex?\dpi{150}%20(x+1)^3=\binom{3}{0}+\binom{3}{1}x+\binom{3}{2}x^2+\binom{3}{3}x^3=1+3x+3x^2+x^3.gif
 
^ 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 p \in \mathbb{R}\left[x\right] is a polynomial with degree n, i.e. p(x) = {p_n}{x^n} + {p_{n - 1}}{x^{n - 1}} + ... + {p_1}x + p_0, define the function F_k: \mathbb{R}\left[x\right] \rightarrow \mathbb{N}^k by {F_k}(p) = (p_n, p_{n - 1}, ..., p_{n - k}). This "gets" the first (or if you want, last) k coefficients, which is really all you need to define a polynomial.
 
Last edited:
Black Integra said:
Here:

gif.latex?\dpi{120}%20(x+y)^n=\sum_{\mu=0}^{n}\binom{n}{\mu}x^ny^{n-\mu}.gif

where

gif.latex?\dpi{120}%20\binom{n}{\mu}=\frac{n!}{\mu!(n-\mu)!}.gif


for example:
tex?\dpi{150}%20(x+1)^3=\binom{3}{0}+\binom{3}{1}x+\binom{3}{2}x^2+\binom{3}{3}x^3=1+3x+3x^2+x^3.gif

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!
 
I think it's the simplest way. May be, if you don't want to deal with combinatorics, use "[URL triangle[/URL]
 
Last edited by a moderator:
That only works for polynomials which can be expressed in the form (1 + x)n, though ...
 

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