Exercise 8.3.11

Question

Assuming that the derivative of $\arcsin(x)$ is indeed $1 / \sqrt{1-x^2}$, supply the justification that allows us to conclude
$$\arcsin(x) = \sum^\infty_{n=0} \frac{c_n}{2n+1} x^{2n+1} \text{ for all } |x| < 1$$

Ulisse Mini · Accepted Answer

Term-by-term antidifferentiation is a valid operation with the expected results, as proven in Exercise 6.5.4.

Exercise 8.3.11

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Exercise 8.3.11

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