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I have a function $F(t)=\int_0^t \sqrt{1-x^8} dx.$ I have to find the first three non-zero terms of a Taylor series of $F$ around the point $a=0.$

Since I want the Taylor series I started with the definition formula and I got $T(t)=\sum_{n=0}^{\infty} \frac{F^{(n)}(0)}{n!}t^n$.

I know that $F'(t) = \sqrt{1-t^8}.$ I have trouble finding the formula for n-th derivative. Because of how complicated the derivatives get I don't really think that this is the right way to solve this problem.

Thanks for your answers

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    $\begingroup$ Expand $\sqrt{1-y}$ into a series, replace $y$ with $x^8$, and integrate. $\endgroup$ – Daniel Fischer Jun 20 '15 at 23:40
  • $\begingroup$ The problem as phrased doesn't ask for $F^{(n)}(t)$, so you could just do it by checking for $n = 0, 1, \ldots$ until you have $3$ terms for which $F^{(n)}(0) \neq 0$. $\endgroup$ – AJY Jun 20 '15 at 23:48
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Hint: You can ignore almost all of the terms with a $t^k$ term in front of the radical; I don't think you'll have much luck finding a general formula for $F^{(n)}(t)$.

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