Simplifying a Sum of Squared Terms: A Geometric Series Approach

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Homework Help Overview

The original poster seeks to simplify the infinite sum of squared terms represented as the series from j=0 to infinity of x^(2j), which relates to geometric series concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants identify the series as resembling a geometric series and discuss its convergence criteria based on the absolute value of x. Others suggest examining partial sums and limits to understand the series' behavior. There is also a mention of manipulating the terms to relate them to a standard geometric series form.

Discussion Status

The discussion includes various approaches to understanding the series, with participants providing insights into convergence and manipulation of terms. There is no explicit consensus, but several productive lines of reasoning have been suggested.

Contextual Notes

Participants note the convergence conditions for the series, specifically that it diverges for |x| > 1 and converges for 0 < |x| < 1. There is also a reference to the need for a formula related to factoring differences of higher powers.

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Homework Statement




How can I simplify sum from j=0 to infinite of x^(2j) ?


Homework Equations





The Attempt at a Solution


THis is close to the geometric series but I'd have to square each individual term
 
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well this looks like a gjeometric series, it will diverge for IxI>1, and it will converge for 0<IxI<1

what else are u looking for?
 
Look at its partial sums, take the limit and you will get the result, i mean where it converges to for 0<IxI<1
 
Ignore what i just said, in my posts #2,3
EDIT: Well don't ignore them, they seem to be right. Can you go from there?
 
Last edited:
You might also observe that x^{2j} = (x^2)^j so start with the problem \sum_{j=0}^\infty a^j
and later set a = x^2

while you're at it recall how to factor differences of higher powers, e.g. a^5 - 1 =? There's a key formula you'll need.
 

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