Maclaurin series help

  • Thread starter Muzly
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  • #1
Muzly
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I've looked at a number of explanations for the Maclaurin series of cos(x) yet none have given an easily understood answer, i was wondering if anyone has a way of explaining it when it is used as only a part of a function

eg. use that Maclaurin series of cos(x) to obtain the Macalurin series for

h(x)=(cos(3x)-1)/x^2
 

Answers and Replies

  • #2
lurflurf
Homework Helper
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The formula for Maclaurin series is
[tex]f(x)=\sum_{k=0}^\infty \frac{x^k}{k!}f^{(k)}(x)[/tex]
as can be easily derived several ways
-integration by parts
-formal operator manipulation
-matching derivatives at zero
-complex contour integration
This can be applied to cos(x) as
[tex]\cos(x)=\sum_{k=0}^\infty \frac{x^k}{k!}\cos^{(k)}(0)=\sum_{k=0}^\infty \frac{x^k}{k!}\cos(k\pi/2)=\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}(-1)^k[/tex]
where the index is changed in the final step to indicate odd terms are 0
Often a known Maclaurin series can be used to find a closely related one.
Just apply the known series
h(x)=(cos(3x)-1)/x^2
[tex]h(x)=\frac{1}{x^2}\left(\sum_{k=0}^\infty \frac{(3x)^{2k}}{(2k)!}(-1)^k-1\right)[/tex]
simplify that into standard form.
 

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