Power Series

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If f(x) has a power series: a_n(x-a)^n (centered at a)

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

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  • #2
HallsofIvy
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If f(x) has a power series: a_n(x-a)^n (centered at a)

what does the power series for f(2x) look like?
You have a problem with this? If "f(x) has power series a_n(x-a)^n" (by which I assume you mean
[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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Hmm..
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
D H
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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_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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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_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]
You did all the work!

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

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