Discussion between Mathy Person and ir7

11:42 PM

A: How can I write this power series as a power series representation?

Hint: using $y=x^2$ and derivative in $y$: $$(1+x)(1+2x^2+3x^4+\ldots) $$ $$ =(1+x)(1+2y+3y^2+4y^3 +\ldots)$$ $$= (1+x)(y+y^2+y^3+y^4+\ldots)'$$ $$ = (1+x)\left( \frac{y}{1-y}\right)'$$

Mathy Person

I believe $(y+y^2+y^3+y^4+....)$ = $\frac{1}{1-y}$, if I am not mistaken. Then is it: $\frac{1+x}{1-y}$?

Oh wait, I see that you had $\frac{y}{1-y}$ instead. Is it: $\frac{(1+x)(y)}{1-y}$?

And $\frac{(1+x)(y)}{1-y}$, and $y=x^2$, then $\frac{(1+x)(x^2)}{1-x^2}$?

Simplifying would result in: $\frac{x^2}{1-x}$?

ir7

Just take derivative in y of $y/(1-y)$, then replace $y$ with $x^2$.

Mathy Person

yep, i did. i got $\frac{x^2}{1-x}$, but I see that SBareS had $\frac{1+x}{1-x}$ instead. Which one is correct?

ir7

you should get $(1+x)/(1-x^2)^2$

confirmation is here wolframalpha.com/input/…

Mathy Person

Can I ask you a further question? I will move the discussion to chat.

ir7

11:43 PM

I'm here.

Mathy Person

I am supposed to find $\frac{1}{\frac({1+x}{(1-x^2)^2})}$, and then I got $\frac{(1-x^2)^2}{1+x}. What would that look like as a polynominal in expanded form?

I tried this: wolframalpha.com/input/… but it seemed to be infinite? @ir7

ir7

I edit it, via derivative of $y/(1-y)$ which is $1/(1-y)^2$.

Mathy Person

@ir7: where y = x^2?

ir7

try the inverse: power series expansion of \frac{1+x}{(1-x^2)^2}

after derivative, we get: $(1+x)\frac{1}{(1-y)^2}$. Then we plug in $y=x^2$.

Mathy Person

@ir7: Here is the problem I am referring to: Let $U(x)=\sum_{n=0}^{\infty}$ $u_nx^n$, where $u_n$ is the number of partitions of n into at most two parts. For example, $u_4=3$ because 4 can be partitioned into at most two parts as 4, 3+1, or 2+2. Use the convention that $u_0=1$.

Then $\frac 1{U(x)}$ is a polynomial. What polynomial is it? (Enter your answer in expanded form.)

@ir7: U(x) is the polynomial I was asking about in my original problem statement

@ir7: I was planning on finding the power series representation of U(x), find 1 / that power series representation, simplify that, and then expand it out to find the polynomial

ir7

11:52 PM

If I'm not wrong, I think this is different from the one we just solved. Why don't you post the question is separately?

Mathy Person

@ir7: I only needed help with the converting U(x) to power series representation part

@SBareS hello!

@ir7: Would U(x) not yield the power series I posted?