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

  • Thread starter Thread starter kaleidoscope
  • Start date Start date
  • Tags Tags
    Polynomial Terms
AI Thread Summary
To select the last N terms of a polynomial, one can define a function that retrieves the coefficients of the polynomial in descending order of their degree. For a polynomial p of degree n, the function F_k can be used to extract the last k coefficients, which represent the desired terms. While some users have suggested using binomial coefficients or combinatorial methods, the simplest approach remains the direct extraction of coefficients. This method is effective for polynomials expressed in the form (1 + x)^n. Overall, the discussion emphasizes the utility of straightforward coefficient extraction for polynomial term selection.
kaleidoscope
Messages
66
Reaction score
0
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!
 
Mathematics news on Phys.org
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 ...
 
Back
Top