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I got stuck on this problem and would appreciate if someone could give me a hint or, maybe even better, a complete solution of it, and it goes like this:

While doing some sums I arrived at the polynomial of degree $n+1$ which is of the following form:

$\sum_{k=0}^{[(n+1)/2]}(w_0)^k{n+1 \choose 2k}(x+a)^{n+1-2k}$, where $[(n+1)/2]$ is the integer part of $(n+1)/2$, and $w_0$ and $a$ are constants.

Now, I would like to write it in its "standard form", i.e. in the form $p_{n+1}(x)=\sum_{i=0}^{n+1}b_ix^i$, so the problem is to find coefficients $b_i$, $i=0,1,...,n+1$ as a function of $n,w_0,a$, in other words, is there any known general method for calculating $b_i=f_i(n,w_0,a)$?

EDIT: I forgot to mention that in this problem I have $w_0<0$ (because somehow this fact doesn`t seem crucial for this problem, at least for me).

2 Answers2

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Hint: Use the Binomial theorem. What do you get when you expand

$$[\sqrt{w_0} + (x+a) ]^{n+1}?$$

How about

$$ [\sqrt{w_0} - (x+a) ]^{n+1}? $$

How about

$$ [\sqrt{w_0} + (x+a) ]^{n+1} + [\sqrt{w_0} - (x+a) ]^{n+1}? $$ $$ [\sqrt{w_0} + (x+a) ]^{n+1} - [\sqrt{w_0} - (x+a) ]^{n+1}? $$

Calvin Lin
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  • What do I get? If I can get the coefficients $b_i$ in that way please explain how? –  Aug 16 '13 at 16:24
  • @Amateur Do you know what the binomial theorem is? Write out the expansion, and compare. – Calvin Lin Aug 16 '13 at 16:39
  • I know about binomial theorem but cannot see how your $[w_0 - (x+a) ]^{n+1}?$ is equal to mine $\sum_{k=0}^{[(n+1)/2]}(w_0)^k{n+1 \choose 2k}(x+a)^{n+1-2k}$? –  Aug 16 '13 at 16:41
  • @Amateur It is a hint. I've added more equations to guide you through. – Calvin Lin Aug 16 '13 at 16:42
  • I cannot take the square root of $w_0$ because in my problem $w_0<0$, maybe I should mention that, I didn`t knew that it would be important. –  Aug 16 '13 at 16:44
  • @Amateur It's not important/relevnat. You can work in complex numbers, and have the answer evaluate to be a real value. E.g. $(1+i) + (1-i) = 2 $. – Calvin Lin Aug 16 '13 at 17:29
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Expand the polynomial in a (finite) Maclaurin series. The coefficients you want are then the ith derivative of your polynomial evaluated at zero and divided by i factorial. It is easy to develop a closed form version for these derivatives.

Starnes
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  • I totally forgot about Maclaurin series, it surely looks as a nice and elegant way to solve this problem, thank you! –  Aug 16 '13 at 17:12