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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.

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  • $\begingroup$ Use binomial expansion $\endgroup$ Sep 9 '18 at 2:56
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    $\begingroup$ 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. $\endgroup$ Sep 9 '18 at 2:59
  • $\begingroup$ Or you can use the MacLaurin expansion for $(1+u)^{1/2}$, then replace $u$ in the result by $x^{1/2}$. $\endgroup$
    – Clement C.
    Sep 9 '18 at 3:08
  • $\begingroup$ @Eclipse, Isn't a binomial theorem/series derived from Maclaurin series? Thanks, I will also research more on this. $\endgroup$
    – Lucy M
    Sep 9 '18 at 3:11
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    $\begingroup$ @AnikBhowmick, thanks. $\endgroup$
    – Lucy M
    Sep 9 '18 at 3:11
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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 $\sqrt x$.

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  • $\begingroup$ Thanks @Frank W $\endgroup$
    – Lucy M
    Sep 9 '18 at 3:13
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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 $\sqrt{x}< 1$, $$ f(x) = g(\sqrt{x}) = \sum_{n=0}^\infty \binom{1/2}{n} x^{n/2} = 1+\frac{x^{1/2}}{2}-\frac{x}{8} + \dots\,. \tag{3} $$

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  • $\begingroup$ Thanks @Clement C $\endgroup$
    – Lucy M
    Sep 9 '18 at 3:51
  • $\begingroup$ @LucyM You're welcome! $\endgroup$
    – Clement C.
    Sep 9 '18 at 4:02

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