The Maclaurin series of $f(x)$ is
$$ \sum_{n=0}^\infty \frac{1}{2n-1}x^{2n-1} $$
Find the fifth derivative of $ f(x) $ at $ x = 0$.
A. 3!
B. 4!
C. 5!
D. 6!
E. 7!
The correct answer should be B, but I'm stumped on this one. Can someone lend me a hand? The farthest I got was noticing that the inner equation is obviously the integral of $ x^{2n} $, but I don't really know where to go from there.
On
The $k$th derivative of a Maclaurin series is $k!$ times the coefficient of $x^k$ (i.e. for $ f(x) = \sum_{n=0}^{\infty} a_n x^n$, $f^{(k)}(0) = k!a_k$). Here, $k=5$, and the power of $x$, is $2n-1$. $2n-1=5$ when $n=3$. The coefficient of $x^5 = x^{2 \times 3-1}$ is then $1/(2 \times 3-1)=1/5$, so the answer is $$ 5! \times \frac{1}{5} = 4!. $$
The coefficient on the $n$'th term of a taylor series around $a$ is given by $$ \frac{f^{k}(a)}{k!}=a_k $$ Setting $a=0$ since it is a maclaurin series, $k=2n-1=5$, we have $$ \frac{f^{5}(0)}{5!}=a_5=\frac{1}{5}\implies f^{5}(0)=\frac{5!}{5}=4! $$