How do you use taylor's theorem to solve this approximation? I'm supposed to use taylor theorem to find an approximation to $\frac{\pi}{4}$ to the 12th decimal place using $tan(x)$ but I can't get a way to even start.  I saw the equation for taylor theorem and it requires a differentiation of the riemann-zeta function, but I don't think it should be that complicated.
Is there any insight to this?
 A: As said in comment, you need to use
$$\tan^{-1}(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$$ Since it is an alternating series, if you want the result to be correct up to $p$ decimal places, since $x=1$, you need $$\frac 1{2n+1} < 10^{-p}\implies n> \frac {10^p}2$$ So, be very patient if you need $p=12$ !
A: You can use the half-angle relations to quite easily find that
$$
\frac\pi8=\arctan(\sqrt2-1)
$$
which requires less than 20 terms of the arcus tangent series to get to your target accuracy.

You can use the Euler-Machin formulas like the classical
$$
(2+i)(3+i)=5(1+i)\implies \frac\pi4=\arctan\frac12+\arctan\frac13
$$
which requires 20 and 14 terms to reach target accuracy, or
$$
(1-i)(5+i)^4=4(239+i)\implies\frac\pi4 =4\arctan\frac15 - \arctan\frac1{239}
$$
which uses even less terms.

You can also accelerate the $\arctan(1)$ series by comparing it with series of similar convergence speed but known sum
$$
\frac\pi4=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}
=1-2\sum_{n=1}^\infty\frac{1}{(4n)^2-1}
$$
where one finds the similar series
$$
\sum_{n=1}^\infty\frac{1}{(4n)^2-4}
=\frac14\sum_{n=1}^\infty\frac{1}{(2n-1)(2n+1)}=\frac18
$$
Thus inserting to cancel the $O(\frac1n)$ convergence one gets
$$
\frac\pi4=\frac34-2\sum_{n=1}^\infty\left[\frac{1}{(4n)^2-1}-\frac{1}{(4n)^2-4}\right]=\frac34+\frac32\sum_{n=1}^\infty\frac1{((4n)^2-1)((2n)^2-1)}
$$
which now has convergence $O(n^{-3}$ so that you get the desired precision with "only" about $10000$ terms.
Then there is Richardson extrapolation which reaches target accuracy with 1000 terms in the third recursion, Wynn's epsilon algorithm which might be even faster,...
