# Thread: Taylor series, is this correct?

1. ## Taylor series, is this correct?

Find Taylor series of f around x=0 where

$f(x)=\int_0^{x^2}\arctan(t)dt$.

Well first of I derived the series expansion of $arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}$.

Using that:

$f(x)=\int_0^{x^2}\arctan(t)dt=\int_0^{x^2}\left(\s um_{n=0}^{\infty}\frac{(-1)^n}{2n+1}t^{2n+1}\right)dt=\sum_{n=0}^{\infty}\f rac{(-1)^n}{2n+1}\int_0^{x^2}t^{2n+1}dt$

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

Is this correct? I could not make wolfram alpha show me the series expansion of f(x).

2. I can't see anything wrong here. If you want to be really certain, here's some options.

Integrating argtan and plugging in the appropriate limits we get:
$f(x) = x^2 arctan(x^2) - \frac{1}{2} ln(x^4 + 1)$.
You could at this point just pick some representative values of x, plug them into f(x) above, and plug them into the series you obtained (evaluated over, say, the first few thousand terms) to see if you get roughly the same values. Even better, you can plot f(x) and plot your series (again, really a truncated version of your series) to see if the plots line up.

Alternatively, we know that $arctan(x^2) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1} x^{4n+2}$ and $ln(x^4+1) = \sum_{n = 1}^{\infty} \frac{(-1)^{n+1}}{n}x^{4n} = \sum_{n = 0}^{\infty} \frac{(-1)^n}{n+1}x^{4n+4}$. Plug these into the f(x) above and I think you'll arrive at the same answer you got.

3. Question does this not work for other methods, such as sin x, cos x, ln x? where you have the basic series expansion of x and you can substitute different types of x, such as x^2 assuming the series expansion you use was of a positive power not ^-x or such, ?

4. Yes, and you can replace x with anything you want, the only caveat being that you must update the region of convergence as well. For example:
$arctan(x) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1}x^{2n+1},|x| \le 1$
To get the expansion for arctan(5x), for example, replace x with 5x everywhere, including the ROC:
$arctan(5x) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1}x^{10n+5}, |5x| \le 1$.
So the new series converges only for $|x| \le 1/5$.

Originally Posted by RockHard
Question does this not work for other methods, such as sin x, cos x, ln x? where you have the basic series expansion of x and you can substitute different types of x, such as x^2 assuming the series expansion you use was of a positive power not ^-x or such, ?