Can I Define y and Explore Power Series Properties with This Sum?

Click For Summary

Discussion Overview

The discussion centers around the use of power series and their properties, specifically in relation to the sum \(\sum_{n=1}^\infty \left(\frac{x}{3}\right)^{2n}\). Participants explore whether defining a variable \(y = \frac{x}{3}\) can facilitate the application of power series properties.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant proposes defining \(y = \frac{x}{3}\) and using the series \(\sum_{k=1}^\infty a_k(y)\) where \(a_k(y)\) is defined piecewise based on the parity of \(k\).
  • Another participant suggests that while this approach is valid, the convergence properties must be translated back to \(x\) by substituting \(y = \left(\frac{x}{3}\right)^{2}\).
  • A different participant offers a simplification of the original series to a geometric series, \(\sum_{n=1}^\infty \left(\frac{x^2}{9}\right)^n\), indicating it has a common ratio of \(\frac{x^2}{9}\).
  • Another participant agrees with the initial proposal and emphasizes its utility in solving problems without confusion.

Areas of Agreement / Disagreement

Participants generally agree on the validity of defining \(y\) and using power series properties, but there are differing views on the necessity of translating convergence properties back to \(x\) and the approach to simplifying the series.

Contextual Notes

Some assumptions about convergence and the definitions of the series are not fully explored, and the implications of the simplifications are not resolved.

estro
Messages
239
Reaction score
0
[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\{<br /> \begin{array}{c l}<br /> (y)^k, & \mbox{if } k= 2n\\<br /> \\<br /> (0)^k, & \mbox{otherwise}<br /> \end{array}<br /> \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:
Physics news on Phys.org
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:
estro said:
[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\{<br /> \begin{array}{c l}<br /> (y)^k, & \mbox{if } k= 2n\\<br /> \\<br /> (0)^k, & \mbox{otherwise}<br /> \end{array}<br /> \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].
 
estro said:
[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\{<br /> \begin{array}{c l}<br /> (y)^k, & \mbox{if } k= 2n\\<br /> \\<br /> (0)^k, & \mbox{otherwise}<br /> \end{array}<br /> \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 without getting fooled like

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

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K