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

h(x)=(cos(3x)-1)/x^2

- Thread starter Muzly
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- #1

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

h(x)=(cos(3x)-1)/x^2

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lurflurf

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[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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