Exercise 8.3.6

Question

Show that $1 / \sqrt{1-x}$ has Taylor expansion $\sum^\infty_{n=0} c_n x^n$, where $c_0 = 1$ and
$$c_n = \frac{(2n)!}{2^{2n} (n!)^2} = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{ 2 \cdot 4 \cdot 6 \cdots 2n}$$
for $n \geq 1$.

Ulisse Mini · Accepted Answer

$c_0 = 1 / \sqrt{1-0} = 1$, and we computed in Exercise 6.6.10 (a) that
$$c_n = \frac{\prod^n_{i=1} (2i - 1)}{2^n n!} = \frac{(2n-1)!}{\left(2^n n!\right)\left( 2^{n-1} (n-1)!\right)} =\frac{(2n)!}{2^{2n} (n!)^2} $$

Exercise 8.3.6

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Exercise 8.3.6

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