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what does the power series for f(2x) look like?

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- #1

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what does the power series for f(2x) look like?

- #2

HallsofIvy

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You have a problem with this? If "f(x) has power series a_n(x-a)^n" (by which I assume you mean

what does the power series for f(2x) look like?

[tex]\Sigma a_n (x-a)^n[/tex]

Doesn't it follow that:

[tex]f(2x)= \Sigma a_n (2x-a)^n[/tex]?

- #3

arildno

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I would think he is after something like this:

[tex]f(2x)=\sum_{n=0}^{\infty}a_{n}(2x-a)^{n}=\sum_{n=0}^{\infty}{a_{n}}{2^{n}}\sum_{l=0}^{n}\binom{n}{l}(\frac{a}{2})^{n-l}(x-a)^{l}[/tex]

and then reorganize this double sum into some expression:

[tex]f(x)=\sum_{n=0}^{\infty}b_{n}(x-a)^{n}[/tex]

with a known sequence b_n

- #4

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[tex]f(2x) = \sum_{n=0}^{\infty}a_n(2x-a)^n = \sum_{n=0}^{\infty}a_n2^n\left(x-\frac a 2\right)^n =\sum_{n=0}^{\infty}b_n\left(x-\frac a 2\right)^n[/tex]

where [itex]b_n=a_n2^n[/itex]

- #5

JasonRox

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You did all the work!

[tex]f(2x) = \sum_{n=0}^{\infty}a_n(2x-a)^n = \sum_{n=0}^{\infty}a_n2^n\left(x-\frac a 2\right)^n =\sum_{n=0}^{\infty}b_n\left(x-\frac a 2\right)^n[/tex]

where [itex]b_n=a_n2^n[/itex]

Oh well, I hope he sees how it came around to this.

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