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Compute the first five terms in the Taylor series about $0$ of the principal branch of $\arcsin{z}$ with a real part between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

Hint: Consider the branch of the function $\sqrt{1-z^2}$ with a positive real part. What is the relation between $\arcsin{z}$ and $\sqrt{1-z^2}?$

Thank you for your help.

I read this link Clever derivation of $\arcsin(x)$ Taylor series (This is another way and longer than this way I think)

Ross
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1 Answers1

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Near $w=0$, $z=\sin{w}$ satisfies $dz/dw = \cos{w} = \sqrt{1-z^2}$. Therefore by the inverse function theorem the local inverse $w=\arcsin{z}$ satisfies $$ \frac{dw}{dz} = \frac{1}{\sqrt{1-z^2}} $$ near $z(0)=0$. Hence the Taylor series can be found by integrating that of $(1-z^2)^{-1/2}$ term-by-term.

Chappers
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  • So the answer would be $f(z)= 0+z-\frac{1}{2}z^2 +\frac{3}{4}z^3-\frac{15}{8}z^4+...$. Is it true? I just used Taylor expansion, I did not integrate it term-by-term. – Ross Oct 14 '17 at 06:39
  • Sine is odd, so its inverse function should also be odd. (Equally, the derivative of the inverse is even and $\arcsin{0}=0$.) – Chappers Oct 14 '17 at 12:46