Question

How do you find the `n`-th derivative of a power series?

Answer 1

If `f(x)=sum_{k=0}^infty c_kx^k`, then
`f^{(n)}(x)=sum_{k=n}^infty k(k-1)(k-2)cdots(k-n+1)c_kx^{k-n}`

By taking the derivative term by term,
`f'(x)=sum_{k=1}^infty kc_kx^{k-1}`
`f''(x)=sum_{k=2}^infty k(k-1)c_kx^{k-2}`
`f'''(x)=sum_{k=3}^infty k(k-1)(k-2)c_kx^{k-3}`
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`f^{(n)}(x)=sum_{k=n}^infty k(k-1)(k-2)cdots(k-n+1)c_kx^{k-n}`