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?
On
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,...
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 $p=12$ !