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I'm trying to find the first five terms of the Maclaurin expansion of $\arcsin x$, possibly using the fact that

$$\arcsin x = \int_0^x \frac{dt}{(1-t^2)^{1/2}}.$$

I can only see that I can interchange differentiation and integration but not sure how to go about this. Thanks!

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up vote 1 down vote accepted

If I was doing this I would start with one of the very common series like $\sin(x)$, $e^x$, etc. and then use substitution. The one that fits best here in my opinion is: $$ \sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{5x^4}{128} + \dotsc $$ Through substitution, we can obtain: $$ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} + \frac{35x^8}{128} + \dotsc $$ Then by integration: $$ \arcsin(x) = \int^x_0 \frac{1}{\sqrt{1-t^2}}\,dt = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} + \dotsc $$

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you want $-x^2$ or $-t^2$ in the denominators, and the signs on the right hand side need to be fixed too. – Maesumi Feb 27 '14 at 10:25

Hint: Use the following binomial series $$ (1+z)^\alpha=\sum\limits_{k=0}^{+\infty}\frac{\alpha(\alpha-1)\ldots(\alpha-k+1)}{k!}z^k $$ with $z=-t^2$, $\alpha=-1/2$. Then integrate.

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Change variables $t = x u$: $$ \arcsin(x) = \int_0^1 \frac{ x}{(1- x^2 u^2)^{1/2}} \mathrm{d} u $$ Now using Taylor series expansion of the integrand: $$ \frac{ x}{(1- x^2 u^2)^{1/2}} = \sum_{n=0}^\infty \frac{1}{2^{2n}}\binom{2n}{n} x^{2n+1} u^{2n} $$ Integrating term-wise: $$ \arcsin(x) = \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ x^{2n+1}}{2n+1} $$

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@Chris'ssister I am now:) – Sasha Sep 17 '12 at 14:56
Just type this in Limit[Sum[Cos[(i + j)/n]/n^2, {j, 2, n}, {i, 1, j - 1}],n->Infinity]. It promptly comes back with $2 \cos(1) \sin^2\left(\frac{1}{2}\right)$. – Sasha Sep 17 '12 at 15:20
thank you very much! :-) – user 1618033 Sep 17 '12 at 15:21

If all you want is the terms of $\arcsin x$ out to $x^5$, it’s also easy enough to invert (“revert”) the series $x-x^3/6+x^5/120$ by hand. You’re looking for a series $A(x)=x + ax^3+bx^5$ so that $\sin(A(x))=x$, but you’re willing to do it modulo terms of degree six or more. Just working modulo $(x^4)$, you see immediately that $a=+1/6$, so the only work is to find $b$. When you calculate $\sin(A(x))$ out to degree five, always ignoring any term in $x^n$ for $n>5$, you get $x+(b-1/12+1/120)x^5$, and you want that last part to be zero, so $b=3/40$.

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The following Python + SymPy script:

from sympy import *

# symbolic variable
x = Symbol('x')

# compute and print Maclaurin series
print series(asin(x),x,0,11,'+')

produces the following output:

x + x**3/6 + 3*x**5/40 + 5*x**7/112 + 35*x**9/1152 + O(x**11)

Therefore, assuming that the implementation is correct, the answer to your question is:

$$\displaystyle\arcsin(x) \approx x + \frac{1}{6} x^3 + \frac{3}{40} x^5 + \frac{5}{112} x^7 + \frac{35}{1152} x^9$$

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