Power Series

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  • #1
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[tex]\mbox {Suppose I have: } \sum_{n=1}^\infty (\frac {x} {3})^{2n} [/tex]

[tex]\mbox{Can I define } y= \frac {x} {3} [/tex]

[tex]
a_k(y) = \left\{
\begin{array}{c l}
(y)^k, & \mbox{if } k= 2n\\
\\
(0)^k, & \mbox{otherwise}
\end{array}
\right.
[/tex]

[tex] \mbox {And then use all the cool properties of power series on } \sum_{k=1}^\infty a_k(y)[/tex]

[I edited my question]
 
Last edited:

Answers and Replies

  • #2
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5
Of course, but the convergence properties you know are given in terms of [itex]y[/itex] then. You need to translate them back in terms of [itex]x[/itex]. Essentially, substitute [itex]y = (x/3)^{2}[/itex] everywhere.
 
Last edited:
  • #3
HallsofIvy
Science Advisor
Homework Helper
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966
[tex]\mbox {Suppose I have: } \sum_{n=1}^\infty (\frac {x} {3})^{2n} [/tex]

[tex]\mbox{Can I define } y= \frac {x} {3} [/tex]

[tex]
a_k(y) = \left\{
\begin{array}{c l}
(y)^k, & \mbox{if } k= 2n\\
\\
(0)^k, & \mbox{otherwise}
\end{array}
\right.
[/tex]

[tex] \mbox {And then use all the cool properties of power series on } \sum_{k=1}^\infty a_k(y)[/tex]

[I edited my question]
The simpler thing to do is write this as
[tex]\sum_{n=1}^\infty \left(\frac{x^2}{9}\right)^n[/tex]
so it is a geometric series with "common ratio" of [itex]x^2/9[/itex].
 
  • #4
Gib Z
Homework Helper
3,346
6
[tex]\mbox {Suppose I have: } \sum_{n=1}^\infty (\frac {x} {3})^{2n} [/tex]

[tex]\mbox{Can I define } y= \frac {x} {3} [/tex]

[tex]
a_k(y) = \left\{
\begin{array}{c l}
(y)^k, & \mbox{if } k= 2n\\
\\
(0)^k, & \mbox{otherwise}
\end{array}
\right.
[/tex]

[tex] \mbox {And then use all the cool properties of power series on } \sum_{k=1}^\infty a_k(y)[/tex]

[I edited my question]

Indeed you can, and that's how you can easily do problems with out getting fooled like

[tex]\sum_{n=1}^{\infty} n^n z^{n^n}[/tex]
 
  • #5
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0
Thank you guys!
 

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