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 $f(x)=\arccos(1-2x^2)$.
Explanation of Method
Differentiating with the chain rule and simplifying the resulting square root collapses the derivative to exactly twice the derivative of $\arcsin x$ — so the whole function turns out to equal $2\arcsin x$ in disguise.
- Differentiate and simplify $\sqrt{1-(1-2x^2)^2}$.
- Expand the simplified derivative via the binomial series.
- Integrate and recognize the closed form $2\arcsin x$.
Step 1 · Differentiate and simplify
$$ f'(x) = \frac{4x}{\sqrt{1-(1-2x^2)^2}}. $$ Since $1-(1-2x^2)^2=4x^2(1-x^2)$: $$ f'(x) = \frac{4x}{2|x|\sqrt{1-x^2}} = \frac2{\sqrt{1-x^2}} = 2(1-x^2)^{-1/2}. $$
Step 2 · Binomial expansion
$$ f'(x) = 2\sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^{2n} = 2+x^2+\frac{3x^4}4+\frac{5x^6}8+\cdots $$
Step 3 · Integrate and identify the closed form
$$ f(x) = \sum_{n=0}^{\infty}\frac{2(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1} + C, \qquad f(0)=\arccos1=0\implies C=0. $$ Comparing with Problem 268's $\arcsin x=\sum\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}$, this series is exactly twice it: $$ \arccos(1-2x^2)=2\arcsin x. $$ (Consistent with $1-2\sin^2\theta=\cos2\theta$: setting $x=\sin\theta$ gives $\arccos(1-2x^2)=2\theta=2\arcsin x$.)
Summary Table
| Property | Value |
|---|---|
| Series | $\sum\frac{2(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}$ |
| Closed form | $2\arcsin x$ |
| Region of validity | $|x|\le1$ |
Additional Notes
- The identity $\arccos(1-2x^2)=2\arcsin x$ is the double-angle formula $\cos2\theta=1-2\sin^2\theta$ read backwards, and it means this series carries no new information beyond Problem 268 — just a factor of $2$.
- The simplification $\sqrt{4x^2(1-x^2)}=2|x|\sqrt{1-x^2}$ silently assumes $x\ge0$; the final closed form $2\arcsin x$ is odd and matches the function correctly on both sides once verified directly.
- This is a useful illustration that composing a function with a polynomial substitution (here $1-2x^2$ inside $\arccos$) can produce a series identical in structure to a known one, rescaled — a pattern also seen when $e^x$ absorbed a linear shift in Problem 249.