Power Series for f(2x): Exploring the Expansion

[moo5003]

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?

 

[HallsofIvy]

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
$$\Sigma a_n (x-a)^n$$
Doesn't it follow that:
$$f(2x)= \Sigma a_n (2x-a)^n$$?

 

[arildno]

Hmm..
I would think he is after something like this:
$$f(2x)=\sum_{n=0}^{\infty}a_{n}(2x-a)^{n}=\sum_{n=0}^{\infty}{a_{n}}{2^{n}}\sum_{l=0}^{n}\binom{n}{l}(\frac{a}{2})^{n-l}(x-a)^{l}$$
and then reorganize this double sum into some expression:
$$f(x)=\sum_{n=0}^{\infty}b_{n}(x-a)^{n}$$
with a known sequence b_n

 

[D H]

I would think he is after something like this:

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

where $b_n=a_n2^n$

 

[JasonRox]

I would think he is after something like this:

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

where $b_n=a_n2^n$

You did all the work!

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