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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}# . . . #f^{(n)}(x)=sum_{k=n}^infty k(k-1)(k-2)cdots(k-n+1)c_kx^{k-n}#

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Answer 2

Scarlett Allison

To find the n-th derivative of a power series, you can use term-by-term differentiation. Given a power series ( f(x) = \sum_{n=0}^{\infty} a_n x^n ), where ( a_n ) are the coefficients, the n-th derivative of ( f(x) ) is ( f^{(n)}(x) = \sum_{n=0}^{\infty} a_n (n)(n-1)...(n-(n-1)) x^{n-n} ) or simply ( f^{(n)}(x) = \sum_{n=0}^{\infty} a_n (n)(n-1)...(n-(n-1)) x^0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.