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I came across this question

Use a power series representation for the function $\frac{1}{1-x}$ to write a power series representation for the function $\frac{4 x^3}{(1-x)^2}$, for $-1 < x < 1$

So my question is this:

I know that it's $$4x^3\frac{1}{(1-x)^2}$$ And I know that $$\int \frac{1}{(1-x)^2} = \frac{1}{(1-x)}$$ which means $$\frac{1}{(1-x)^2} = \frac{d}{dx}\frac{1}{(1-x)}$$ and $$ \frac{1}{(1-x)} = \sum_{n=0}^\infty x^n$$

But I'm not sure mathematically how to continue without ruining it, because basically what I was thinking is maybe this works:

$$4x^3\frac{d}{dx}\frac{1}{(1-x)}$$ $$4x^3\frac{d}{dx}\sum_{n=0}^\infty x^n$$ $$4x^3\sum_{n=0}^\infty n {x^n}^{-1}$$ $$\sum_{n=0}^\infty 4n {x^n}^{+2}$$

But I'm sure I've done something wrong, please let me know what it is and how I should go about solving this.

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  • $\begingroup$ This is quite hand-wavy in some parts (you omit the indices of the summation and exchange the derivation and summation without explain why you can do that) but it is essentially correct. $\endgroup$ – Francesco Carzaniga May 8 '18 at 10:37
  • $\begingroup$ @FrancescoCarzaniga Edited! Thank you. $\endgroup$ – Kode Ch May 8 '18 at 10:43
  • $\begingroup$ The way you edited the indices is not correct, and you still did not explain why you are allowed to perform derivation before summation. $\endgroup$ – Francesco Carzaniga May 8 '18 at 10:47
  • $\begingroup$ @FrancescoCarzaniga That's mainly why I'm asking, because I'm not sure if I could do that in the first place. In terms of the indices I just copied them from the book, so I'm also not sure why they're incorrect. $\endgroup$ – Kode Ch May 8 '18 at 10:49
  • $\begingroup$ Go straight to the point, there is no need to mention an integral. $4x^3/(1-x)^2=4x^3(1/(1-x))'$. $\endgroup$ – Yves Daoust May 8 '18 at 12:56
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There is another way of doing this computation. You correctly pointed out that: $$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$

so:

$$\frac{1}{(1-x)^2} = \frac{1}{1-x}\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty}\sum_{k=0}^{n}x^kx^{n-k} = \sum_{n=0}^{\infty}(n+1)x^n$$

This series is convergent since it is the Cauchy product of two absolutely convergent series.

Now we can complete using the same reasoning you used: $$4x^3\dfrac{1}{(1-x)^2} = 4x^3\sum_{n=0}^{\infty}(n+1)x^n = \sum_{n=0}^{\infty}4(n+1)x^{n+3}$$

To answer your question in the comments, you cannot easily exchange differentiation and summation since you require uniform convergence of the derivatives, which in this case you don't have.

Edit:

Since you haven't seen the Cauchy product we can do it with the monotone convergence theorem. Call $f_m(x) = \sum_{n=0}^{m}(n+1)x^n$ and $f(x) = \sum_{n=0}^{\infty}(n+1)x^n$, then $$\int f_m(x)dx = \int \sum_{n=0}^{m}(n+1)x^n dx = \sum_{n=0}^{m}\int(n+1)x^n dx$$ since this time the sum is finite and we have no problem integrating term by term. Now if we manage to prove $$\sum_{n=0}^{\infty}\int(n+1)x^n dx = \lim_{m\to\infty} \int f_m(x) dx = \int f(x) dx = \int \sum_{n=0}^{\infty}(n+1)x^n dx $$ we will have that $$\frac{d}{dx}\sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty}(n+1)x^n$$ and then we will be able to proceed as before.

To do this we apply the dominated convergence theorem, and since clearly $f_m(x) > 0$ for all $x \in (-1, 1)$ we just need to show that $f \geq f_m$:

$$f - f_m = \sum_{n=0}^{\infty}(n+1)x^n - \sum_{n=0}^{m}(n+1)x^n = \sum_{n=m}^{\infty}(n+1)x^n$$

which is just the original series up to rearrangement, so it is also greater than $0$ and we are done.

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  • $\begingroup$ This is a fantastic explanation, unfortunately we haven't learned how to multiply power series, so I wouldn't know how to use this. $\endgroup$ – Kode Ch May 8 '18 at 12:08
  • $\begingroup$ I hope you know what dominated convergence is, otherwise I don't know how else to show what you want without knowing what tools you are allowed to use. $\endgroup$ – Francesco Carzaniga May 8 '18 at 12:55
  • $\begingroup$ I unfortunately don't...all we've learned so far is that there is possibility of differentiating or integrating the power series to solve it. $\endgroup$ – Kode Ch May 8 '18 at 13:08
  • $\begingroup$ Then you are probably expected to just differentiate the power series without thinking too much about why you're allowed to do it. $\endgroup$ – Francesco Carzaniga May 8 '18 at 13:25
  • $\begingroup$ Just to make sure in the end, my answer was correct with all the current information I have? $\endgroup$ – Kode Ch May 8 '18 at 13:55

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