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.

  • $\begingroup$ 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

There is another way of doing this computation. You correctly pointed out that: $$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$


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

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.


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.

  • $\begingroup$ 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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