How can I arrive at a series expansion of $ (1+x^{\frac{1}{2}})^{\frac{1}{2}}$

Question

How can I arrive at a series expansion of $ (1+x^{\frac{1}{2}})^{\frac{1}{2}}$

Is this a power series? I've tried using Maclaurin series, but I don't seem to be getting anywhere.

Please briefly explain to me how I can obtain a sensible solution.

Answer 1

The result is not going to be a Taylor (MacLaurin) series, since such a series is of the form $f(x)=\sum_{n=0}^\infty a_n x^n$ for $|x|< R$ (where $R$ is the convergence radius), i.e., it only has integer powers.

However, we can derive a similar series, with fractional powers. The reasion is that your function $f$ satisfies, for every $x>0$, $$ f(x) = g(\sqrt{x})\tag{1} $$ where $g(u) = (1+u)^{1/2}$ does have a MacLaurin expansion: $$ g(u) = \sum_{n=0}^\infty \binom{1/2}{n} u^n= 1+\frac{u}{2}-\frac{u^2}{8} + \dots\,,\qquad |u|<1. \tag{2} $$ Consequently, we have that for every $0\leq x< 1$, since $\sqrt{x}< 1$, $$ f(x) = g(\sqrt{x}) = \sum_{n=0}^\infty \binom{1/2}{n} x^{n/2} = 1+\frac{x^{1/2}}{2}-\frac{x}{8} + \dots\,. \tag{3} $$

Answer 2

Hint: The Generalized Binomial Theorem will be useful$$(1+x)^n=1+nx+\frac {n(n-1)}{2!}x^2+\frac {n(n-1)(n-2)}{3!}x^3+\cdots$$Now set $n=\tfrac 12$ and replace $x$ with $\sqrt x$.