0
$\begingroup$

I came across this question:

What's the domain of $f(x)=\int (x+2x^2+3x^3+\dots) dx$?

I think it'd be a polynomial, something like $\frac{x^2}{2}+\frac{2x^3}{3}+\dots$, which domain is all real numbers ($\mathbb R$) but it's not correct according to the book. I'm stuck.

EDIT:

The answer should be one of the followings:

  1. $(-1,1)$
  2. $(-1,1]$
  3. $[-1,1)$
  4. $[-1,1]$
$\endgroup$
6
  • 3
    $\begingroup$ May be another way of asking the radius of convergence of the series you obtained on integrating? $\endgroup$ – user21436 Jan 31 '12 at 21:41
  • 2
    $\begingroup$ Let's first recognize that $x + 2x^2 + 3x^3 + \cdots$ is the derivative of the geometric series $1 + x + x^2 + \cdots$ multiplied by $x$. What's the radius of convergence for that? [It also seems like it would be better to write $f(x) = \int_0^x (t + 2t^2 + \cdots)\,dt$ but this seems like a common abuse.] $\endgroup$ – Dylan Moreland Jan 31 '12 at 21:43
  • $\begingroup$ @DylanMoreland: $\frac {1}{1-x}$? It's written like that in my book! $\endgroup$ – Gigili Jan 31 '12 at 22:07
  • 2
    $\begingroup$ I hope your book says something more than just $\frac{1}{1-x}$: it also says the series converges for $|x|<1$ and diverges otherwise. The same is true for $x + 2 x^2 + 3 x^3 + \ldots$. $\endgroup$ – Robert Israel Jan 31 '12 at 23:01
  • $\begingroup$ What does the book say the answer is? And why have none of those who have commented asked that question?? $\endgroup$ – Michael Hardy Feb 1 '12 at 0:52
3
$\begingroup$

Given the answer choices, it's probably intended for you to integrate term-wise and then investigate the convergence of the resulting power series. (Also: polynomials have finite degree, but these don't and hence are not polynomials.) Thus

$$\int(x+2x^2+3x^3+\cdots)dx=\frac{1}{2}x^2+\frac{2}{3}x^3+\frac{3}{4}x^4+\cdots.$$

The ratio test tells us that this does converge when $|x|<1$ but not $|x|>1$. Similarly, plugging in either of $x=\pm1$ results in a divergent series, so the domain is $(-1,1)$.

$\endgroup$
3
$\begingroup$

Lets write your function this way

$x+2x^2+3x^3+\cdots = x \frac{d}{dx} (x+x^2+x^3+x^4+\cdots)$

It is elementary that $x+x^2+x^3+x^4 + \cdots = \frac{x}{1-x} $

Thus we have $x+2x^2+3x^3+\cdots = x \frac{d}{dx} \left(\frac{x}{1-x}\right)$

This produces $x+2x^2+3x^3+\cdots = \frac{x}{(1-x)^2}$

Remember we always have to consider $ |x| < 1 $ to make sure the sum converges.

Returning to you problem, we can integrate to get:

$ \int {\frac{x}{(1-x)^2} dx} = \frac{1}{1-x} + \log(1-x) + C $

EDIT: Rigorously, the series will only converge for $|x|<1$ and you'll need $x \neq 1 $ for the $\log$ to be defined. So $\mathbb{D} = (-1,1)$.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer, but this is not correct. $\endgroup$ – Gigili Feb 1 '12 at 6:08
  • $\begingroup$ I think your solution is correct except the last part, when you said $|x| < 1$ so the domain should be that, am I right? $\endgroup$ – Gigili Feb 1 '12 at 6:11
  • $\begingroup$ I accepted the other answer because it was correct according to my book, but I guess you missed a part of your own explanation, otherwise yours is correct as well. I'll accept your answer when you edit it. Thank you again. $\endgroup$ – Gigili Feb 1 '12 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.