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

Step-by-step solution

1. 1

   Use the binomial series rather than repeated differentiation. The generalised binomial theorem states

   $(1+t)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}t^{n},\qquad \binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!},$

   valid for $|t|<1$ and any real $\alpha$. Matching the problem: $\alpha=\tfrac12$ and $t=2x$. Computing $f^{(n)}(0)$ by hand would take four rounds of the chain rule to reach the same place.

2. 2

   Compute the first few coefficients, keeping the powers of 2. Each term is $\binom{1/2}{n}(2x)^{n}$, so the $2^{n}$ must be carried:

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

   $n=3:\ \frac{\tfrac12\left(-\tfrac12\right)\left(-\tfrac32\right)}{6}(8x^{3})=\tfrac12x^{3};\qquad n=4:\ \frac{\tfrac12\left(-\tfrac12\right)\left(-\tfrac32\right)\left(-\tfrac52\right)}{24}(16x^{4})=-\tfrac58x^{4}.$

   Forgetting the $2^{n}$ is the single most common error and would give $\tfrac18x^{2}$ instead of $-\tfrac12x^{2}$.

3. 3

   Write out the expansion. Collecting the terms:

   $\sqrt{1+2x}=1+x-\frac{x^{2}}{2}+\frac{x^{3}}{2}-\frac{5x^{4}}{8}+\cdots$

   Note the signs: after the first two terms they alternate, because every factor $\left(\tfrac12-k\right)$ with $k\ge1$ is negative.

4. 4

   State the general term. In closed form,

   $\sqrt{1+2x}=\sum_{n=0}^{\infty}\binom{1/2}{n}2^{n}x^{n}=1+x+\sum_{n=2}^{\infty}\frac{(-1)^{n-1}(2n-3)!!}{n!}\,x^{n},$

   where $(2n-3)!!=1\cdot3\cdots(2n-3)$. Checking $n=4$: $\frac{-1\cdot(1\cdot3\cdot5)}{24}=-\frac{15}{24}=-\frac58$ ✓.

5. 5

   Find the interval of convergence and test numerically. The binomial series needs $|t|=|2x|<1$, i.e. $|x|<\tfrac12$; since $\alpha=\tfrac12>0$ the series also converges absolutely at both endpoints, giving $\left[-\tfrac12,\tfrac12\right]$. At $x=0.05$ the four printed terms give $1.0488086$ against $\sqrt{1.1}=1.0488088$ ✓; at $x=0.1$ they give $1.0954375$ against $1.0954451$ ✓ — the error grows as $x$ approaches the endpoint, exactly as expected.