approximating π

$\frac{\pi }{4}=4ta{n}^{-1}(\frac{1}{5})-ta{n}^{-1}(\frac{1}{x})$
$ta{n}^{-1}(\frac{1}{x})=-\frac{\pi }{4}+4ta{n}^{-1}(\frac{1}{5})$
$\frac{1}{x}=tan(-\frac{\pi }{4}+4ta{n}^{-1}(\frac{1}{5}))$
$x=239$

$\frac{\pi }{4}=4ta{n}^{-1}(\frac{1}{5})-ta{n}^{-1}(\frac{1}{239})$
$\pi =16ta{n}^{-1}(\frac{1}{5})-4ta{n}^{-1}(\frac{1}{239})$

$f(a)+f^{\prime }(a)(x-a)+\frac{f^{\prime \prime }(a)}{2!}(x-a{)}^2+\cdots +\frac{f^n(a)}{n!}(x-a{)}^n$
$g(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots +(-1{)}^n\frac{x^{2n+1}}{2n+1}$

$\pi =16g(\frac{1}{5})-4g(\frac{1}{239})$
err of $g$ needs to be $<5\times 10^{-7}$ for proper 6 places

for some $n$ terms, error bound is $|ta{n}^{-1}(x)-g_n(x)|\le |\frac{x^{2(n+1)+1}}{2(n+1)+1}|$, since next term is omitted and term values are decreasing because $|x|<1$

need $\sum k\times |\frac{x^{2(n+1)+1}}{2(n+1)+1}|<5\times 10^{-7}$ for $k=16,x=\frac{1}{5}$ and $k=4,x=\frac{1}{239}$
$16|\frac{{\frac{1}{5}}^{2(n+1)+1}}{2(n+1)+1}|+4|\frac{{\frac{1}{239}}^{2(n+1)+1}}{2(n+1)+1}|<5\times 10^{-7}$

solving, $n=3.1743$
so choose terms up to $k=4$, $g(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}$