Find the Taylor series of f(x) = ln(sqrt(x^2 + 1) + x) about x = 0. Give the first few nonzero terms, a formula for the general term, and the interval of convergence.

Step-by-step solution

1. 1

   Recognise the function before differentiating anything. The combination $\ln\!\left(x+\sqrt{x^2+1}\right)$ is the standard logarithmic form of the inverse hyperbolic sine,

   $\operatorname{arsinh}x=\ln\!\left(x+\sqrt{x^2+1}\right),$

   so $f(x)=\operatorname{arsinh}x$. This matters because computing $f^{(n)}(0)$ directly from the logarithm is brutal, while $\operatorname{arsinh}$ has a very simple derivative. Whenever a Maclaurin expansion looks unpleasant, check whether the derivative is the friendly object.

2. 2

   Differentiate once and watch the logarithm collapse. By the chain rule,

   $f'(x)=\frac{1}{\sqrt{x^2+1}+x}\left(\frac{x}{\sqrt{x^2+1}}+1\right)=\frac{1}{\sqrt{x^2+1}+x}\cdot\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}.$

   The bracket $x+\sqrt{x^2+1}$ cancels the denominator exactly, leaving

   $f'(x)=\frac{1}{\sqrt{1+x^2}}=(1+x^2)^{-1/2}.$

   That cancellation is the whole reason this problem is easy.

3. 3

   Expand the derivative with the binomial series. For $|t|<1$,

   $(1+t)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}t^{n}.$

   Put $t=x^{2}$ and $\alpha=-\tfrac12$:

   $(1+x^2)^{-1/2}=1-\frac12x^2+\frac38x^4-\frac{5}{16}x^6+\frac{35}{128}x^8-\cdots$

   Only even powers appear, which already tells you $f$ will have only odd powers — consistent with $f$ being an odd function.

4. 4

   Integrate term by term using $f(0)=0$. Since $f(0)=\ln 1=0$, we have $f(x)=\int_0^x f'(t)\,dt$, and a power series may be integrated term by term inside its radius of convergence:

   $f(x)=x-\frac12\cdot\frac{x^3}{3}+\frac38\cdot\frac{x^5}{5}-\frac{5}{16}\cdot\frac{x^7}{7}+\frac{35}{128}\cdot\frac{x^9}{9}-\cdots$

   $f(x)=x-\frac{x^{3}}{6}+\frac{3x^{5}}{40}-\frac{5x^{7}}{112}+\frac{35x^{9}}{1152}-\cdots$

5. 5

   Write the general term. Integrating $\binom{-1/2}{n}x^{2n}$ gives $\binom{-1/2}{n}\frac{x^{2n+1}}{2n+1}$, and $\binom{-1/2}{n}=(-1)^n\frac{(2n)!}{4^n (n!)^2}$, so

   $\ln\!\left(\sqrt{x^2+1}+x\right)=\sum_{n=0}^{\infty}(-1)^{n}\frac{(2n)!}{4^{n}(n!)^{2}(2n+1)}\,x^{2n+1}.$

   Check the first two coefficients: $n=0$ gives $x$, and $n=1$ gives $-\frac{2}{4\cdot 1\cdot 3}x^3=-\frac{x^3}{6}$. ✓

6. 6

   Fix the interval of convergence and test the answer numerically. The binomial series for $(1+x^2)^{-1/2}$ needs $|x^2|<1$, i.e. $|x|<1$; at $x=\pm1$ the integrated series has terms of size $\sim n^{-3/2}$ and converges absolutely, so the interval is $[-1,1]$. A numeric check at $x=0.2$: the five printed terms give $0.198690110\ldots$ against $\operatorname{arsinh}(0.2)=0.198690110\ldots$ ✓