Find complicated Taylor Series

According to some software, the power series of the expression, $$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}}$$ around $x=0$ is

$$\sqrt{x}-x^{3/2}+\mathcal{O}(x^{5/2}).$$ When I try to do it I find that I can't calculate Taylor because there are divisions by zero. Also I do not understand how Taylor could give non integer powers.

Does anybody know how this expression is calculated?

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It's not a Taylor Series per se. It's one of these: en.wikipedia.org/wiki/Puiseux_series –  Antonio Vargas Mar 23 '14 at 14:58

Close to $x=0$,$$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}} \simeq \frac{1}{2} \sqrt{-1+ (1+4x)}\simeq \sqrt x$$ So, this must be the start of the development (in order that, locally, your expansion looks like the formula)

If you start using $$\sqrt{1+8 x} \simeq 1+4 x-8 x^2+32 x^3+O\left(x^4\right)$$ then $$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}} \simeq \frac{1}{2}\sqrt {4 x-8 x^2+32 x^3+O\left(x^4\right)}=\sqrt {x-2 x^2+8 x^3+O\left(x^4\right)}$$ Now, extract $\sqrt x$ and you get $$\sqrt x \sqrt {1-2x+8 x^2}$$ Develop the second square root and you arrive to your wanted result.

Suppose we change the problem to $$\frac{1}{2} \sqrt{-1+\sqrt{1+8 \sqrt{x}}}$$ For the same reasons, the first term should be $x^{1/4}$ and the development would be $$x^{1/4}-x^{3/4}+\frac{7 x^{5/4}}{2}-\frac{33 x^{7/4}}{2}+O\left(x^{9/4}\right)$$

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I'm having trouble with the constants. Anyway, here's what I thought, I hope it helps (at least it shows that non-integer powers are possible):

You do know that $(1+x)^{1/2} = 1 + \frac{x}{2} - \frac{1}{8}x^2 + o(x^3)$

then $\frac{1}{2}\sqrt{-1+\sqrt{1+8x}}=\frac{1}{2}\sqrt{4x-8x^2+o(x^3)}=\sqrt{x}\sqrt{1-2{x}+o(x^2)}={\sqrt{x}}(1 -x + o(x^2)) = (\sqrt x-x^{3/2}+o(x^{5/2}))$

edit: now it works

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I fixed it up! Thank you :) –  nelv Mar 23 '14 at 14:43

$$f(x) =\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}}$$

The argument of the square root is 0, hence $f$ has no derivative in 0. That is why you get non integer powers.

First, get the equivalent:

$$\sqrt{1+8 x}-1 \sim 4x\\ f(x) = \frac 12\sqrt{-1+\sqrt{1+8 x}}\sim \frac 12\sqrt{4x} = \sqrt{x}$$ and then: $$f(x) -\sqrt{x} = \frac 12\sqrt{-1+\sqrt{1+8 x}} - \frac 12\sqrt{4x} \\= \frac 12 \frac{-1+\sqrt{1+8 x} +4x} {\sqrt{-1+\sqrt{1+8 x}} - \sqrt{4x}} \sim \frac 12 \frac{-\frac 18 (8x)^2}{4\sqrt{x}}=-x\sqrt{x}$$

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