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Trivial solution for cosh(x)=0 and sinh(x)=0

• Context: Undergrad
• Thread starter Thread starter ksukhin
• Start date Start date Apr 12, 2015

Click For Summary SUMMARY

The discussion focuses on finding trivial solutions for the hyperbolic functions cosh(x) and sinh(x) in the context of Sturm-Liouville problems. It is established that cosh(x) is never zero, while sinh(x) equals zero at x=0. The user seeks clarification on these functions to determine eigenvalues for λ in their mathematical work.

PREREQUISITES

• Understanding of hyperbolic functions, specifically sinh(x) and cosh(x)
• Familiarity with Sturm-Liouville theory and eigenvalue problems
• Basic knowledge of exponential functions and their properties
• Mathematical notation for trigonometric and hyperbolic equations

NEXT STEPS

• Study the properties of hyperbolic functions in detail
• Explore Sturm-Liouville theory and its applications in differential equations
• Learn about eigenvalues and eigenfunctions in the context of boundary value problems
• Investigate the implications of hyperbolic identities in mathematical modeling

USEFUL FOR

Mathematicians, physics students, and engineers working on differential equations, particularly those involved in Sturm-Liouville problems and eigenvalue analysis.

ksukhin Messages 33 Reaction score 1 I'm doing Sturm-Loiuville problems and I need to find the eigenvalues for λ I'm having difficulty understanding the trivial solutions for the hyperbolic sin and cos when they equal 0. I know that cos(x)=0 when ##x= π/2 + πn = (2n+1)π/2## sin(x) = 0 when ##x=πn## What about cosh(x) and sinh(x)? Please help Physics news on Phys.org

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Mark44 Mentor Insights Author Messages 38,095 Reaction score 10,646

ksukhin said: I'm doing Sturm-Loiuville problems and I need to find the eigenvalues for λ I'm having difficulty understanding the trivial solutions for the hyperbolic sin and cos when they equal 0. I know that cos(x)=0 when ##x= π/2 + πn = (2n+1)π/2## sin(x) = 0 when ##x=πn## What about cosh(x) and sinh(x)? Please help

##sinh(x) = \frac{e^x - e^{-x}}{2}## ##cosh(x) = \frac{e^x + e^{-x}}{2}## Clearly cosh(x) is never zero. It's pretty easy to find the zeroes of sinh(x). mathman Science Advisor Homework Helper Messages 8,130 Reaction score 575 sinh(x)=0 when x=0.

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