Power Series Expansions

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  • #1
STEMucator
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I had a quick question about an expansion. Wolfram and maple have not been very useful in verifying the series.

Could I do these:

Centered about a=1:

##f(x) = e^{-3x^2} = e^{-3}e^{-3(x^2-1)} = \sum_{n=0}^{∞} \frac{e^{-3}(-3)^n}{n!} (x^2-1)^n##

Centered about a=2:

##f(x) = e^{-3x^2} = e^{-12}e^{-6(\frac{x^2}{2}-2)} = \sum_{n=0}^{∞} \frac{e^{-12}(-6)^n}{n!} (\frac{x^2}{2}-2)^n##

Centered about a=3:

##f(x) = e^{-3x^2} = e^{-27}e^{-9(\frac{x^2}{3}-3)} = \sum_{n=0}^{∞} \frac{e^{-27}(-9)^n}{n!} (\frac{x^2}{3}-3)^n##

My overall hunch from this pattern centered about a=k:

##f(x) = e^{-3x^2} = e^{-3k^2}e^{-3k(\frac{x^2}{k}-k)} = \sum_{n=0}^{∞} \frac{e^{-3k^2}(-3k)^n}{n!} (\frac{x^2}{k}-k)^n##

and for any ##\alpha## centered about a=k:

##f(x) = e^{-\alpha x^2} = e^{-\alpha k^2}e^{-\alpha k(\frac{x^2}{k}-k)} = \sum_{n=0}^{∞} \frac{e^{-\alpha k^2}(-\alpha k)^n}{n!} (\frac{x^2}{k}-k)^n##

EDIT: Also one more, if I replace ##x^2## by ##x^n## in the above:

##f(x) = e^{-\alpha x^n} = e^{-\alpha k^n}e^{-\alpha k(\frac{x^n}{k}-k)} = \sum_{n=0}^{∞} \frac{e^{-\alpha k^n}(-\alpha k)^n}{n!} (\frac{x^n}{k}-k)^n##
 

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  • #2
jbunniii
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I'm not sure exactly what you're trying to do, so my comment may be off base. But the series
$$\sum_{n=0}^{∞} \frac{e^{-3}(-3)^n}{n!} (x^2-1)^n$$
is not centered at ##x=1##, at least not in the usual sense of that phrase, because the polynomials ##(x^2 - 1)^n## are not centered at ##x=1##. They are centered at (symmetric around) ##x=0##.

A series which is centered about ##x=1## would be of the form
$$\sum_{n=0}^\infty a_n (x-1)^n$$
 
  • #3
STEMucator
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I'm not sure exactly what you're trying to do, so my comment may be off base. But the series
$$\sum_{n=0}^{∞} \frac{e^{-3}(-3)^n}{n!} (x^2-1)^n$$
is not centered at ##x=1##, at least not in the usual sense of that phrase, because the polynomials ##(x^2 - 1)^n## are not centered at ##x=1##. They are centered at (symmetric around) ##x=0##.

A series which is centered about ##x=1## would be of the form
$$\sum_{n=0}^\infty a_n (x-1)^n$$

I was trying to find a general expression without having to find ##f^{(n)}(a)##. It seemed to work for (center a=1):

##f(x) = e^{-3x} = e^{-3} e^{-3(x-1)} = \sum_{n=0}^{∞} \frac{e^{-3}(-3)^n}{n!} (x-1)^n##

I see that won't work now though for higher powers of ##x## as I would produce series that are not centered around the desired point.

Is there a way I could adapt this little trick to higher powers?
 

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