# Maclaurin series of a function

1. Nov 15, 2012

### Mangoes

1. The problem statement, all variables and given/known data

Find the maclaurin series of:

$$f(x) = \int_{0}^{x}(e^{-t^2}-1) dt$$

3. The attempt at a solution

I know $$e^t = \sum_{n=0}^{∞} \frac{t^n}{n!}$$

Simple substitution gives me:

$$e^{-t^2} = \sum_{n=0}^{∞}\frac{(-t^2)^n}{n!}$$

Which I rewrote as

$$e^{-t^2} = \sum_{n=0}^∞\frac{(-1)^n(t^{2n})}{n!}$$

Since I notice that when n = 0, the term simplifies to 1,

$$e^{-t^2} - 1 = \sum_{n=1}^∞ \frac{(-1)^n(t^{2n})}{n!}$$

Going back to the original expression and substituting the infinite series

$$\int_0^x\sum_{n=1}^∞ \frac{(-1)^n(t^{2n})}{n!} dt$$

Now, I'm integrating with respect to t, so after using the power rule and applying the upper limit (lower limit simplifies to 0)

$$\sum_{n=1}^∞\frac{(-1)^nx^{2n+1}}{n!(2n+1)}$$

And this is what I would think is the maclaurin representation of the function, but it's wrong and I have no idea why. Would anyone please point to where I'm going wrong with this?

Last edited: Nov 15, 2012
2. Nov 15, 2012

### Dick

How did your lower limit n=0 appear in the appear in the answer? It was n=1 in the step before.

3. Nov 15, 2012

### Mangoes

Sorry about that, typo from copying and pasting.

4. Nov 15, 2012

### SammyS

Staff Emeritus
How do you know it's wrong?

You can change the sum to start at n=0, if that's the problem.

5. Nov 15, 2012

### Dick

Well, then I'm having trouble finding anything wrong with it. But does the answer you are checking against expect you to shift the lower limit to n=0?

6. Nov 15, 2012

### Mangoes

$$\sum_{n=0}^∞\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)(n+1)!}$$

They shifted the lower bound to start at 0 and compensated by replacing n by (n+1).

Pretty frustrating considering I've been hitting my head against the wall thinking I was doing something wrong all this time...

Thanks a lot for the help to both of you. Just learned about power series about a week ago so still not too used to working with them.

7. Nov 15, 2012

### rbj

there isn't anything wrong with it. this

$$f(x) = \int_{0}^{x}(e^{-t^2}-1) dt = \sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{n!(2n+1)}$$

is in fact correct.