Find the Taylor series of sqrt(x^2 + 1) about x = 0, give the general term, and state the interval of convergence.

Step-by-step solution

1. 1

   Apply the binomial series with $t=x^{2}$. The generalised binomial theorem gives

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

   and here $\alpha=\tfrac12$, $t=x^{2}$. Because $t$ is itself a square, only even powers of $x$ can appear — which is exactly right, since $\sqrt{x^{2}+1}$ is an even function.

2. 2

   Compute the coefficients one at a time. With $\binom{1/2}{n}=\frac{\tfrac12\left(\tfrac12-1\right)\cdots\left(\tfrac12-n+1\right)}{n!}$:

   $n=1:\ \tfrac12x^{2};\qquad n=2:\ \frac{\tfrac12\left(-\tfrac12\right)}{2}x^{4}=-\tfrac18x^{4};$

   $n=3:\ \frac{\tfrac12\left(-\tfrac12\right)\left(-\tfrac32\right)}{6}x^{6}=\tfrac{1}{16}x^{6};\qquad n=4:\ \frac{\tfrac12\left(-\tfrac12\right)\left(-\tfrac32\right)\left(-\tfrac52\right)}{24}x^{8}=-\tfrac{5}{128}x^{8}.$

   Unlike the expansion of $\sqrt{1+2x}$, there is no $2^{n}$ factor to carry here — the substitution supplies powers of $x$, not a constant.

3. 3

   Write the expansion. Collecting terms,

   $\sqrt{1+x^{2}}=1+\frac{x^{2}}{2}-\frac{x^{4}}{8}+\frac{x^{6}}{16}-\frac{5x^{8}}{128}+\cdots$

   The signs alternate from the $x^{4}$ term onward, and the coefficients shrink quickly, so a handful of terms is very accurate for small $x$.

4. 4

   State the general term. In closed form,

   $\sqrt{1+x^{2}}=\sum_{n=0}^{\infty}\binom{1/2}{n}x^{2n},\qquad \binom{1/2}{n}=\frac{(-1)^{n-1}}{4^{n}(2n-1)}\binom{2n}{n}\ (n\ge1).$

   Check $n=2$: $\frac{-1}{16\cdot 3}\cdot 6=-\frac{6}{48}=-\frac18$ ✓, matching the $x^{4}$ coefficient.

5. 5

   Determine the interval of convergence. The binomial series requires $|t|=|x^{2}|<1$, that is $|x|<1$. At $x=\pm1$ the terms behave like $n^{-3/2}$, so the series converges absolutely there as well, giving the closed interval $[-1,1]$. At $x=1$ it converges to $\sqrt2=1.4142136$, but only slowly: truncating after the $x^{8}$ term gives just $1.3984375$, ten terms give $1.40993$ and forty terms still only $1.41366$. Fast convergence is a feature of the interior, not of the endpoints.