Convergence of Sin(x) Series: Understanding the Pattern and Proving Convergence

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Homework Help Overview

The discussion revolves around the convergence of the series involving sin(x), specifically the expression sin(x) - 1/2sin^2(x) + 1/4sin^3(x) - 1/8 sin^4(x) and its equivalence to 2sin(x)/(2+sin(x)). Participants are exploring the conditions under which this series converges for all real values of x.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of standard series identities and the conditions for convergence, particularly focusing on the substitution that could simplify the series. There is also a suggestion to analyze specific values of x, such as π/2 and -π/2, to understand the behavior of the series.

Discussion Status

The discussion is ongoing, with participants offering hints and suggestions for approaching the problem. Some guidance has been provided regarding the use of series identities, but there is no explicit consensus on the convergence of the series or a complete solution yet.

Contextual Notes

Participants are navigating the complexities of the series and its convergence, with some expressing uncertainty about how to begin the problem. There may be assumptions about the behavior of sin(x) that are being questioned.

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Homework Statement



Why does sin(x) - 1/2sin^2(x) + 1/4sin^3(x) - 1/8 sin^4(x) + ... = 2sin(x)/2+sin(x)

How do you know for certain the series converges for all real values of x?

Homework Equations





The Attempt at a Solution



Have no clue where to even start...

Thanks for any help...
 
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You should use the standard series identity

[tex]\sum_{n=0}^{\infty} y^n = \frac{1}{1-y}[/tex]

The series converges for |y|<1. With an appropriate substitution you can make this series appear in your problem.
 
You might want to consider, separately, what happens when x= [itex]\pi/2[/itex]or x= [itex]-\pi/2[/itex].
 
To make dhris's hint more obvious, just show that

[tex]\sum_{n=0}^{\infty} (\frac{- \sin x}{2})^n = \frac{2}{2+ \sin x}[/tex]
then multiply through out by sin x.
 

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