By expanding the derivative f'(x) in a power series in x
By expanding the derivative $f'(x)$ in a power series in $x$ and integrating term by term, write the power series expansion in $x$ for $f(x)=\arctan x$.
Explanation of Method
The derivative of $\arctan x$ is the simple rational function $1/(1+x^2)$, which is itself a geometric series; integrating that series term by term and fixing the constant via $f(0)=0$ recovers the classical arctangent series.
- Differentiate to get $f'(x)=1/(1+x^2)$.
- Expand as a geometric series and integrate term by term.
- Fix the constant using $f(0)=0$ and check both endpoints.
Step 1 · Differentiate
$$ f'(x) = \frac1{1+x^2}. $$
Step 2 · Expand and integrate
$$ f'(x) = \sum_{n=0}^{\infty}(-1)^nx^{2n}, \qquad |x|<1. $$ Integrating term by term: $$ f(x) = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1} + C. $$ Since $f(0)=\arctan0=0$, $C=0$.
Step 3 · Check the endpoints
At $x=1$: $\sum(-1)^n/(2n+1)=1-\frac13+\frac15-\cdots$ converges by the Leibniz test, and equals $\arctan1=\frac\pi4$ (the Leibniz formula for $\pi$). By symmetry, $x=-1$ also converges. So $$ \boxed{\arctan x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}}, \qquad |x|\le1. $$
Summary Table
| Property | Value |
|---|---|
| Series | $\sum(-1)^nx^{2n+1}/(2n+1)$ |
| At $x=1$ | $\pi/4$ (Leibniz formula) |
| Region of validity | $|x|\le1$ |
Additional Notes
- Setting $x=1$ gives the celebrated Leibniz formula $\frac\pi4=1-\frac13+\frac15-\frac17+\cdots$, one of the earliest infinite series ever discovered for $\pi$ (though it converges extremely slowly).
- This derivative-then-integrate technique is the same one used in Problem 268 for $\arcsin x$ and Problem 269 for $\sinh^{-1}x$ — all three inverse-function series arise from a rational or algebraic derivative expanded by geometric or binomial series.
- Problem 273's $(1+x^2)\arctan x$ builds directly on this series by multiplying it by $(1+x^2)$ and reindexing.