Exercise 6.7.4

Question

Show that $f(x)=\sqrt{1-x}$ has Taylor series coefficients $a_{n}$ where $a_{0}=1$ and
$$
a_{n}=\frac{-1 \cdot 3 \cdot 5 \cdots(2 n-3)}{2 \cdot 4 \cdot 6 \cdots 2 n}
$$
for $n \geq 1$.

Ulisse Mini · Accepted Answer

For $n \geq 1$:
$$f^{(n)} = -\frac{\prod^{N-1}_{i=1} (2i - 1)}{2^n} (1-x)^{-\frac{2n-1}{2}}$$
$$a_n = \frac{f^{(n)}(0)}{n!}
= -\frac{ \prod^{N-1}_{i=1} (2i - 1) }{ \left(\prod^{N}_{i=1}2\right) \left(\prod^N_{i=1}i\right) }
=\frac{-1 \cdot 3 \cdot 5 \cdots(2 n-3)}{2 \cdot 4 \cdot 6 \cdots 2 n}
$$

Exercise 6.7.4

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Exercise 6.7.4

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