Maclaurin Series Problem Correct?

Riemann sum of n = 0 to infinity, (-1)^n * x^(2n) / n! for all real x, then f''(0) = ?

I changed this to (-x^2)^n / n! in order to make it similar to the e^x maclaurin series. I then used e^(-x^2) as my function for the series and took the second derivative which I got to be -2 + 6x^2 - 5x^4 +... I plugged in 0 for x and thus the answer is -2. Is this logic correct?

Re: Maclaurin Series Problem Correct?

Quote:

Originally Posted by

**achiu17** Riemann sum of n = 0 to infinity, (-1)^n * x^(2n) / n! for all real x, then f''(0) = ?

I changed this to (-x^2)^n / n! in order to make it similar to the e^x maclaurin series. I then used e^(-x^2) as my function for the series and took the second derivative which I got to be -2 + 6x^2 - 5x^4 +... I plugged in 0 for x and thus the answer is -2. Is this logic correct?

In my opinion your solution is correct. I've used an other method:

The Maclaurin serie of a function $\displaystyle f(x)$ is given by

$\displaystyle f(x)= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0)+f'(0)x+\frac{x^2}{2}f''(0)+\ldots$

In this case we have the following sum

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

If we compare both series the we notice that $\displaystyle \frac{f''(0)}{2} = -1 \Rightarrow f''(0)=-2$