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bcjochim07

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## Homework Statement

Is there a way to check if a power series representation of a function is correct?

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bcjochim07

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Is there a way to check if a power series representation of a function is correct?

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Dick

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bcjochim07

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bcjochim07

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Ok f(0) = the value of the power series at x=0

The first derivative of the function is (-x^2 + 1)/(1-x)^4 so f'(0)=1

The first derivative of the power series is sum from n=1 to infinity of n^2*x^(n-1) and if I plug in x, I will get zero. What am I not understanding?

The first derivative of the function is (-x^2 + 1)/(1-x)^4 so f'(0)=1

The first derivative of the power series is sum from n=1 to infinity of n^2*x^(n-1) and if I plug in x, I will get zero. What am I not understanding?

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Dick

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Ok f(0) = the value of the power series at x=0

The first derivative of the function is (-x^2 + 1)/(1-x)^4 so f'(0)=1

The first derivative of the power series is sum from n=1 to infinity of n^2*x^(n-1) and if I plug in x, I will get zero. What am I not understanding?

The power series is x+2x^2+3x^3+... The derivative of that at x=0 is most certainly 1. What went wrong with your reasoning above?

- #6

bcjochim07

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Oh, ooops I wasn't thinking right

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bcjochim07

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I forgot the first x term

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HallsofIvy

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However, in general checking that derivatives at a specific point are the same does NOT prove that a power series is equal to a function. That is only true if the function in question is "analytic" at the point which means, by definition, that it is equal to its Taylor Series expansion about the point.

For example, it is easy to show that the function

[tex]f(x)= \begin{array}{c}e^{-\frac{1}{x^2}} x\ne 0 \\ 0 x= 0\end{array}[/tex]

is infinitely differentiable, has all derivatives at 0 equal to 0 and so has Taylor series expansion about 0 [itex]\Sum 0\cdot x^n[/itex] but is not equal to that anywhere except at x= 0.

For example, it is easy to show that the function

[tex]f(x)= \begin{array}{c}e^{-\frac{1}{x^2}} x\ne 0 \\ 0 x= 0\end{array}[/tex]

is infinitely differentiable, has all derivatives at 0 equal to 0 and so has Taylor series expansion about 0 [itex]\Sum 0\cdot x^n[/itex] but is not equal to that anywhere except at x= 0.

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- #9

bcjochim07

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So then is there a way to prove that a power series is equal to a function?

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bcjochim07

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Anyone have thoughts on this?

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