Using Taylor series for $arctan(x)$, find $f^{(5)}(0)$ and $f^{(6)}(0)$ for $f(x)=arctan(x)$
I figure for this problem I compare the general Taylor series formula to the Taylor series for $arctan(x)$ and extract the derivatives form there.
$arctan(x) = 0 + x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}$
General Taylor series = $f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\frac{f^{(5)}(0)}{5!}x^5+\frac{f^{(6)}(0)}{6!}x^6$
*Both series centered around $0$
5th term of $arctan(x)$ is $\frac{x^9}{9}$
5th term of the Taylor series is $\frac{f^5(0)}{5!}$
Set equal and solve for $f^5(0)$
After solving I get $f^5(0) = \frac{x^4*5!}{9}$ which equals $0$ when x equals $0$. The actual derivative at $x=0$ is 24 so I am not sure where I am going wrong.
Best Answer
Your expression for the Taylor series is wrong, it should be $$f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \ldots$$ What you've written isn't a power series, since each term after the second are constants rather than constants times powers of $x$. (Also, you should be careful to not conflate finite Taylor polynomials with the infinite Taylor series.)
Equating the $5$-th term coefficients should instead yield $\frac{f^{(5)}(0)}{5!}x^5 = \frac{x^5}{5}$, and thus $\frac{f^{(5)}(0)}{5!} = \frac{1}{5}$.