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My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.

Page 57. Exercise 12. Show that the following are polynomials by converting them to the form $\sum_{k=0}^{m}a_kx^k$ for a suitable $m$. In each case $n$ is a positive integer.

$a)$ $(1+x)^{2n}.$

$b)$ $\frac{1-x^{n+1}}{1-x}, x\not=1.$

$c)$ $\prod_{k=0}^{n}(1+x^{2^k}).$

The attempt at a solution: a) part of the problem is pretty easy I guess, it is example of binomial theorem, so the answer would be $(1+x)^{2n}=\sum_{k=0}^{2n}(^{2n}_{k})x^k.$ Answer to part b) would be the following: $$\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2+\cdots+x^n)}{1-x}=1+x+x^2+\cdots+x^n=‌​\sum_{k=0}^{n}x^k.$$thanks to @DiegoMath's hint.

As for part c), we have $$\prod_{k=0}^{n}(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})$$ and I have trouble "converting" this to sum which would be of a form of polynomial.

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George Apriashvili
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1 Answers1

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I think I understand all questions from this problem and I have the solutions, so since no one is posting good answer I will, hope its ok.

Answer to $a)$ part of the problem I already listed but here it is anyway.

$a)$ Using binomial theorem, we have $$(1+x)^{2n}=\sum_{k=0}^{2n}(^{2n}_{k})x^k.$$

For $b)$, @DiegoMath's hint helped me, and answer would be:

$b)$ $$\frac{1-x^{n+1}}{1-x}=\frac{(1-x)(1+x+x^2+\cdots+x^n)}{1-x}=1+x+x^2+\cdots+x^n=‌​\sum_{k=0}^{n}x^k.$$

$c)$ We have $$\prod_{k=0}^{n}(1+x^{2^k})=(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n}),$$and$$(1+x)(1+x^2)\cdots(1+x^{2^n})=1+x+x^2+\cdots+x^{2^{n+1}-1},$$therefore$$\prod_{k=0}^{n}(1+x^{2^k})=\sum_{k=0}^{2^{n+1}-1}x^k.$$

This is my solutions, please post if you have better solutions or if you think my reasoning is wrong.

George Apriashvili
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