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 $f(x)=\ln\bigl(x+\sqrt{1+x^2}\bigr)$.
Explanation of Method
This function is the inverse hyperbolic sine, $\sinh^{-1}x$. Its derivative simplifies remarkably to $(1+x^2)^{-1/2}$, which — like Problem 268's $\arcsin$ derivative — is a binomial series, but here with alternating signs.
- Differentiate $f(x)$ and simplify.
- Expand $(1+x^2)^{-1/2}$ via the binomial series.
- Integrate term by term and compare with $\arcsin x$.
Step 1 · Differentiate
$$ f'(x) = \frac1{x+\sqrt{1+x^2}}\left(1+\frac x{\sqrt{1+x^2}}\right) = \frac1{x+\sqrt{1+x^2}}\cdot\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}} = \frac1{\sqrt{1+x^2}}. $$
Step 2 · Binomial expansion
With $t=x^2$ in the binomial series ($\alpha=-\frac12$): $$ (1+x^2)^{-1/2} = \sum_{n=0}^{\infty}\binom{-1/2}nx^{2n} = \sum_{n=0}^{\infty}\frac{(-1)^n(2n)!}{4^n(n!)^2}x^{2n} = 1-\frac{x^2}2+\frac{3x^4}8-\frac{5x^6}{16}+\cdots $$
Step 3 · Integrate and compare to $\arcsin x$
$$ f(x) = \sum_{n=0}^{\infty}\frac{(-1)^n(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1} + C, \qquad f(0)=\ln1=0\implies C=0. $$ This has exactly the same coefficient magnitudes as Problem 268's $\arcsin x$ series, but with alternating signs — consistent with $\sinh^{-1}x=-i\arcsin(ix)$. Both endpoints converge, with $f(1)=\ln(1+\sqrt2)\approx0.8814$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum\frac{(-1)^n(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}$ |
| Identity | $f(x)=\sinh^{-1}x$ |
| At $x=1$ | $\ln(1+\sqrt2)$ |
| Region of validity | $|x|\le1$ |
Additional Notes
- Same central-binomial-coefficient magnitudes as $\arcsin x$ (Problem 268), but alternating in sign — a direct consequence of the identity $\sinh^{-1}(x)=-i\sin^{-1}(ix)$.
- This series matches the one built earlier from the binomial expansion of $\sqrt{1+x^2}$ itself in Problem 258, since $\ln(x+\sqrt{1+x^2})$ is precisely the antiderivative of $1/\sqrt{1+x^2}$ used to define $\sinh^{-1}x$.
- Both endpoints $x=\pm1$ converge because the coefficients decay like $1/n^{3/2}$ (by Stirling's approximation on the central binomial coefficient), which is fast enough for conditional (in fact absolute, up to sign) convergence.