Approximate $\arctan\frac89$ using $\arctan 1$ with Taylor of first order. How many digits can I guarantee?
My apporximation is $\frac\pi4-\frac1{18}\sim0.7298$.
$f''(x)=\frac{-2x}{(x^2+1)^2}$
We know that: $f(x)=f(a)+f'(a)(x-a)+\frac12f''(z)(x-a),$ z between $a$ and $x$
Let's say the error is $\epsilon\left(\frac89\right)=-\frac{1}{81}\frac{z}{(z^2+1)^2}$
As $\frac89<z<1 \rightarrow 0<\frac{z}{(z^2+1)^2}<1 \rightarrow-\frac1{81}<\epsilon\left(\frac89\right)<0$.
We have that
$$\frac\pi4-\frac1{18}>\arctan\left(\frac89\right)>\frac\pi4-\frac1{18}-\frac1{81}$$ $$0.7298>\arctan\left(\frac89\right)>0.7174$$
More precisely (with a calculator) I know that $\arctan\left(\frac89\right)\sim0.726642$
I can guarantee just 1 digit. But I see that I really can guarantee 2, did I make a mistake?
You could do better using $$\tan ^{-1}(a)-\tan ^{-1}(b)=\tan ^{-1}\left(\frac{a-b}{1+a b}\right)$$ So $$\tan ^{-1}(1)-\tan ^{-1}\left(\frac{8}{9}\right)=\tan ^{-1}\left(\frac{1}{17}\right)\implies \tan ^{-1}\left(\frac{8}{9}\right)=\frac \pi 4-\tan ^{-1}\left(\frac{1}{17}\right)$$ $$\frac{1}{17}-\frac{1}{3(17)^3} <\tan ^{-1}\left(\frac{1}{17}\right) <\frac{1}{17}$$ So, converting to decimal, $$0.7265746<\tan ^{-1}\left(\frac{8}{9}\right)<0.7266425$$ while the exact value is $0.7266423$.