Approximation of $\arctan(x)$ with $\pi/2 + a/x + b/x^3$

Question

I have to find numbers $a, b \in \mathbf{R}$ that function $f(x) = \pi/2 + a/x + b/x^3$ is a very good approximation to the function $g(x) = \arctan(x)$ for big $x$ in terms of that the error goes towards $0$ faster than $1/x^4$, where $x$ goes towards $\infty$.

So I calculate the derivative of functions $f$ and $g$ and equate them. When I all calculate then I got $(a +1) x^4 + (a + b) x^2 + b = 0$

Is this the right path for this exercise or not?

Answer 1

Hint: Use the fact that for $x > 0$, we have $\arctan x = \frac{\pi}{2} - \arctan(\frac{1}{x})$. Then, apply the Taylor series $$\arctan t = \sum_{n=0}^\infty \frac{(-1)^{2n+1}}{2n+1} t^{2n+1}.$$

Answer 2

Another approach is to use

$$\arctan x=\int_0^x \frac{dt}{1+t^2}$$ $$ = \frac{\pi}{2} - \int_x^\infty\frac{dt}{1+t^2} = \frac{\pi}{2}-\int_x^\infty\frac{1}{t^2}\frac{1}{1+1/t^2}\, dt.$$

To deal with the last integral, recall that $\dfrac{1}{1+u} = 1-u+u^2 - \cdots$ for $|u|<1.$