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

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.

• Use binomial expansion Sep 9 '18 at 2:56
• You cannot expect a Maclaurin expansion because it's derivative at $0$ is infinite. But you can get something very similar to a Maclaurin series using binomial formula. Sep 9 '18 at 2:59
• Or you can use the MacLaurin expansion for $(1+u)^{1/2}$, then replace $u$ in the result by $x^{1/2}$. Sep 9 '18 at 3:08
• @Eclipse, Isn't a binomial theorem/series derived from Maclaurin series? Thanks, I will also research more on this. Sep 9 '18 at 3:11
• @AnikBhowmick, thanks. Sep 9 '18 at 3:11

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 $\sqrt x$.
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 $\sqrt{x}< 1$, $$f(x) = g(\sqrt{x}) = \sum_{n=0}^\infty \binom{1/2}{n} x^{n/2} = 1+\frac{x^{1/2}}{2}-\frac{x}{8} + \dots\,. \tag{3}$$