Maclaurin series of a function

[Mangoes]

Homework Statement

Find the maclaurin series of:

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

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?

 

[Dick]

Homework Statement

Find the maclaurin series of:

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

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)

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

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?

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

 

[Mangoes]

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

Sorry about that, typo from copying and pasting.

 

[SammyS]

...

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?

How do you know it's wrong?

If you have the correct answer, please give it.

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

 

[Dick]

Sorry about that, typo from copying and pasting.

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?

 

[Mangoes]

Your replies made me look at the answer list a little harder and I just noticed why my answer isn't matching up.

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

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.

 

[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.