Taylor series with summation notation

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Homework Help Overview

The discussion revolves around the function f(x) = (1 - cos(x^2)) / x^3 and the application of Taylor series, particularly in relation to trigonometric functions. Participants are exploring how to manipulate the expression and utilize series expansions effectively.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster inquires about which identity to use and seeks general tips for similar problems. Some participants suggest considering the Taylor series for cos(x) and discuss the implications of the x^3 term in the denominator. Others mention that the summation can be manipulated by factoring out x.

Discussion Status

The discussion is active, with participants providing insights into the use of Taylor series and discussing the structure of the expression. There is an exchange of ideas regarding how to approach the problem, but no consensus has been reached on a specific method or solution.

Contextual Notes

Participants are navigating the complexities of the function's form and the implications of the Taylor series, particularly in relation to the presence of x in the denominator. There is an emphasis on understanding the series expansion without a clear resolution of the problem at this stage.

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Homework Statement



f(x) = [tex]\frac{1-cos(X^2)}{x^3}[/tex]

which identity shoud i use?
and tips on this type of questions? once i can separate them, then i'll be good


thanks!
 
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Do you know a Taylor series for [itex]\cos x[/itex]?
 
benorin said:
Do you know a Taylor series for [itex]\cos x[/itex]?

yeah, but there's a x^3 on the bottom...
 
Sure, but the summation isn't over x so you can put the x in the sum or outside the sum.
 
Example:

[tex]\frac{1-\sin 2x^3}{x}=\frac{1-\sum_{k=0}^{\infty}\frac{\left( 2x^3\right)^{2k+1}}{(2k+1)!}}{x} = {\scriptstyle \frac{1}{x}}-{\scriptstyle \frac{1}{x}}\sum_{k=0}^{\infty}\frac{2^{2k-1}x^{6k+3}}{(2k+1)!}[/tex]
= {\scriptstyle \frac{1}{x}}-\sum_{k=0}^{\infty}\frac{2^{2k-1}x^{6k+2}}{(2k+1)!}[/tex]​
 

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