Complex Analysis: Taylor's Theorem

Click For Summary
SUMMARY

The discussion focuses on finding the Maclaurin series representation of the function f(z) defined as sinh(z)/z for z ≠ 0 and 0 for z = 0. The solution identifies that the Maclaurin series for sinh(z) is expressed as the sum of z^(2n+1)/(2n+1)!. By dividing this series by z, the resulting series representation for f(z) is confirmed as the sum of z^(2n)/(2n+1)!. The conclusion emphasizes that this series representation holds for the entire domain, including at z = 0, where it evaluates to 0.

PREREQUISITES
  • Understanding of Maclaurin series and Taylor's Theorem
  • Familiarity with hyperbolic functions, specifically sinh(z)
  • Basic knowledge of limits and continuity in complex analysis
  • Experience with series convergence and manipulation
NEXT STEPS
  • Study the derivation of Taylor series for various functions
  • Explore the properties and applications of hyperbolic functions
  • Learn about series convergence tests in complex analysis
  • Investigate the implications of singularities in function representations
USEFUL FOR

Students and educators in mathematics, particularly those focusing on complex analysis, as well as anyone interested in the applications of Taylor's Theorem and series expansions.

tylerc1991
Messages
158
Reaction score
0

Homework Statement



Find the Maclaurin series representation of:

f(z) = {sinh(z)/z for z =/= 0 }
{0 for z = 0 }

Note: wherever it says 'sum', I am noting the sum from n=0 to infinity.

The Attempt at a Solution



sinh(z) = sum [z^(2n+1)/(2n+1)!]

=> sinh(z)/z = sum [z^(2n)/(2n+1)!] (referenced as (1))

=> the Maclaurin series representation for f(z) is (1) when z =/= 0 and 0 when z=0

Thank you for the help, I hope the text is not confusing.
 
Physics news on Phys.org
Correct, but

tylerc1991 said:
the Maclaurin series representation for f(z) is (1) when z =/= 0 and 0 when z=0

The distinction is not necessary in this case. If z=0, then (1) already yields 0. Thus (1) is the series representation on the entire domain!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K