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I'm stumped here. I''m supposed to find the Maclaurin series of $\frac1{1+x^2}$, but I'm not sure what to do. I know the general idea: find $\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$. What I want to do is find some derivatives and try to see if there's a pattern to their value at $0$. But after the second derivative or so, it becomes pretty difficult to continue. I know this:

$$f(0) = 1$$

$$f'(0) = 0$$

$$f''(0) = -2$$

$$f^{(3)}(0) = 0$$

$$f^{(4)}(0) = 0$$

But when trying to calculate the fifth derivative, I sort of gave up, because it was becoming too unwieldly, and I didn't even know if I was going somewhere with this, not to mention the high probability of making a mistake while differentiating.

Is there a better of way of doing this? Differentiaing many times and then trying to find a pattern doesn't seem to be working.

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up vote 8 down vote accepted

$1$) Write down the series for $\frac{1}{1-t}$. You have probably have already seen this one. If not, it can be computed by the method you were using on $\frac{1}{1+x^2}$. The derivatives are a lot easier to get a handle on than the derivatives of $\frac{1}{1+x^2}$.

$2$) Substitute $-x^2$ for $t$, and simplify.

Comment: It can be quite difficult to find an expression for the $n$-th derivative of a function. In many cases, we obtain the power series for a function by "recycling" known results. In particular, we often get new series by adding known ones, or by differentiating or integrating known ones term by term. Occasionally, substitution is useful.

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It is worth taking home the explicit lesson that successive differentiation is not usually a practical way to find power series (except in special lucky cases such as $e^x$). It is more of theoretical importance (such as in proving that the power series for a function is unique if it exists). – Henning Makholm Oct 28 '11 at 18:26
Thanks, I get it. I left a comment but deleted it because I had made a mistake. I now get $\displaystyle \sum_{n=0}^\infty (-1)^n x^{2n}$. Is that right? – Javier Oct 28 '11 at 18:29
Yes, that's right. I had left a comment giving the $(-1)^n$ stuff, but deleted it when you deleted your question, since I decided you had figured it out. By the way, soon you will be calculating new series by differentiating or integrating known ones term by term. – André Nicolas Oct 28 '11 at 18:37
Great, thank you. – Javier Oct 28 '11 at 18:39

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