Summing Maclaurin Series for x^2

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SUMMARY

The sum of the Maclaurin series for \( \sum \frac{x^{2k}}{k!} \) can be derived by substituting \( x^2 \) into the Taylor series expansion of \( e^x \). The original attempt to relate this series to known Taylor series such as \( \sin x \) and \( e^x \) was unsuccessful due to differences in factorial terms and the nature of the series' signs. Ultimately, the correct approach simplifies the problem significantly by leveraging the properties of the exponential function.

PREREQUISITES
  • Understanding of Maclaurin series
  • Familiarity with Taylor series expansions
  • Knowledge of factorial notation and its applications
  • Basic concepts of exponential functions
NEXT STEPS
  • Study the derivation of the Taylor series for \( e^x \)
  • Explore the properties of Maclaurin series
  • Investigate the relationship between factorials and series expansions
  • Learn about convergence of series and their applications in calculus
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Students studying calculus, mathematicians interested in series expansions, and educators teaching series convergence and function approximations.

JakeD
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Homework Statement


How do I find the sum of \sum\frac{x^{2k}}{k!}?

The Attempt at a Solution


I tried transforming various known Taylor series, such as sin x, e^x, and so on, but they didn't fit for 2 reasons:
1. In all of them, the degree of the factor equals the power of x. i.e. if you have x^2k in the nominator, then you have (2k)! in the denominator, whereas here, you have x^2k in the nominator, while having k! (not (2k!)) in the denominator.

2. In sin x, you have alternating pluses and minuses, while in the required sum, they are all pluses.Any help will be appreciated
 
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OK, found the solution.

By replacing x with x^2 in the Taylor series of e^x, I get the desired sum.
 

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