Quicker Method to Find Taylor Series of sinz - sinhz?

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SUMMARY

The discussion focuses on finding the first three non-zero terms of the Taylor series for the functions sin(z) and sinh(z). The user initially calculated the terms as -z³/3 - z⁷/2520 - z¹¹/19958400 using a brute force method. However, they discovered that utilizing the generalized form of the Taylor series for sin and sinh significantly accelerates the process. The conversation highlights the efficiency gained by applying the correct mathematical framework.

PREREQUISITES
  • Understanding of Taylor series expansion
  • Familiarity with the functions sin(z) and sinh(z)
  • Basic knowledge of factorial notation (n!)
  • Ability to perform algebraic manipulations
NEXT STEPS
  • Study the generalized Taylor series for sin(z) and sinh(z)
  • Learn about convergence and error analysis in Taylor series
  • Explore computational tools for symbolic mathematics, such as SymPy
  • Practice deriving Taylor series for other functions
USEFUL FOR

Mathematicians, students studying calculus, and anyone interested in improving their efficiency in calculating Taylor series expansions.

connor415
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I have to find the first three non zero terms of this series by hand. I know the answer and it is

-(z^3/3) - z^7/2520 - z^11/19958400

Which will take ages to get to by brute force. Is there a quicker way?
 
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hi connor415! :smile:

but that's just (n!)/2 …

what's brutish about that? :confused:
 
The method I used was brutish. I didnt use the generalized form of the series for sin and sinh that I think youre referring to. I now am, and its faster. Thanks
 

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