Understanding the Maclaurin Series of cos(x) for Use in Functions

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SUMMARY

The discussion focuses on the Maclaurin series of cos(x) and its application in deriving the series for the function h(x) = (cos(3x) - 1)/x^2. The formula for the Maclaurin series is f(x) = ∑(k=0 to ∞) (x^k/k!) f^(k)(0). The series for cos(x) is expressed as cos(x) = ∑(k=0 to ∞) (x^(2k)/(2k)!)(-1)^k, where odd terms are eliminated. The transformation of the known series into the form of h(x) is essential for simplification and analysis.

PREREQUISITES
  • Understanding of Maclaurin series and Taylor series expansions
  • Familiarity with calculus concepts such as derivatives and limits
  • Knowledge of trigonometric functions and their series representations
  • Basic skills in algebraic manipulation and simplification of series
NEXT STEPS
  • Study the derivation of the Maclaurin series for sin(x) and its applications
  • Explore the use of Taylor series in approximating functions beyond trigonometric functions
  • Learn about convergence criteria for power series and their implications
  • Investigate the application of complex contour integration in deriving series
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Mathematicians, students of calculus, and anyone interested in advanced series expansions and their applications in function analysis.

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
 
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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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