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Calculate Maclaurin series of $\sinh^{-1}(x)$

Using the formula $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3...$$

The derivatives are as follows: $$f'(x) = \frac{1}{\sqrt{1-x^2}}, f''(x) = \frac{(3 x^2)}{(1 - x^2)^{5/2}} + \frac{1}{(1 - x^2)^{3/2}}, f'''(x) = \frac{(9 x)}{(1 - x^2)^{5/2}} + \frac{(15 x^3)}{(1 - x^2)^{7/2}}$$

Plugging in $0$ I get the following: $f(0)=0, f'(0) = 1, f''(0)=1, f'''(0) = 0$

$$\implies f(0) = 0 + x + \frac{x^2}{2!}+0 \cdots$$

However when I use the binomial expansion and integrate It produces different results. How do I get this by the Maclaurin series?

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    First few terms:$$\sinh^{-1}(x) = x - \frac{1}{6} x^3 + \frac{3}{40}x^5 - \frac{5}{112} x^7+\dots$$ Use $\sinh^{-1}(x) = \int\frac{dx}{\sqrt{1+x^2}}$ where you know the series for $\frac{1}{\sqrt{1+x^2}}$ – GEdgar Nov 29 '21 at 00:09
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    $\operatorname{arcsinh}'(x)=\frac1{\sqrt{1+x^2}}$. –  Nov 29 '21 at 00:11
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    @GEdgar I too am curious, as how would you get the series for that equation? I imagined we just square root both sides like $\sqrt{\sum_{n=0}^{\infty}(-x^2)^n}$, is there a cleaner expression than this? I thought to instead use the macluarin expansion and then integrate – me.limes Dec 12 '21 at 19:13
  • @me.limes The series for $\frac{1}{\sqrt{1+x^2}}$ is done by writing it as $(1+x^2)^{-1/2}$ and using Newton's binomial series for exponent $-1/2$. – GEdgar Dec 12 '21 at 22:00
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    This question is similar to: Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Integreek Nov 09 '24 at 11:10

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