Steps for Solving an Inequality Proof

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Homework Help Overview

The discussion revolves around proving an inequality related to a series expansion, specifically involving the sine function and its Taylor series representation. Participants are exploring the mathematical reasoning behind the inequality and its components.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Some participants discuss the relationship between the series expansion of sine and the terms involved in the inequality. Others inquire about the specific steps needed to approach the proof, indicating a desire for clarification on the methodology.

Discussion Status

Participants are actively engaging with the problem, with some offering insights into the properties of the series involved. There is a focus on understanding the implications of the alternating series and its convergence, though no consensus on a specific solution has been reached.

Contextual Notes

There are references to external links for additional context, suggesting that participants may be working with incomplete information or seeking further clarification on the problem setup.

transgalactic
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the question and how i tried to solve it are in this link

http://img155.imageshack.us/img155/4710/55108660qi1.gif
 
Last edited by a moderator:
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You need a little bit more. It is true that
[tex]x- \frac{x^3}{3!}+ \frac{x^5}{5!}> x- \frac{x^3}{3!}[/tex]
but you have to fit
[tex]sin(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}[/tex]
into that. Notice that because this is an alternating series with decreasing terms, the entire "tail" (all terms past "n") is less than the difference between the n-1 and n terms.
 
what is the steps for the solution?
 

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