Orthogonality Property of Hyperbolic functions ?

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The discussion centers on the orthogonality property of hyperbolic functions, specifically sinh and cosh, in comparison to trigonometric functions. It is noted that while trigonometric functions exhibit orthogonality, hyperbolic functions may not share this property due to their derivation from Sturm–Liouville eigenvalue equations, where solutions differ based on the sign of the eigenvalue. The integral of the inner product for hyperbolic functions does not yield zero, suggesting a lack of orthogonality. The conversation also touches on the challenges of extracting coefficients from series solutions of Laplace's equation when hyperbolic functions are involved. Ultimately, it concludes that hyperbolic functions do not possess an orthogonal property akin to that of trigonometric functions.
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Orthogonality Property of Hyperbolic functions ?

Hi all,
I have seen Orthogonal property for trigonomeric functions but I am unsure if there is something similar for sinh() , cosh() ? . I know that the integral of inner product of the two functions should be zero for them to be called as orthogonal.

I am asking this question because if one applies Variable separable form of F(x)G(y) to solve a laplace equation, then one always gets one of F(x) or G(y) to be hyperbolic. With this being the case ,doesn anybody know how to extract coefficients from series form of solution
of laplace equation when (say) only sinh(kx) is remaining :


for e.g Summation {Pn cosh(kx) } =phi . How to extract Pn ?


Had it been Summation {Pn cos(kx) } =phi ,then one can use orthogonal property of trigonometric function .

I hope ,I am clear.

Thanks ,
A.S
 
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I think most of the orthogonal functions are obtained from the solutions of the Sturm–Liouville eigenvalue equations

(py')'+(q+\lambda r)y=0

where \lambda is a constant.

The trigonometric and hyperbolic come from solving the following equation
y''+\lambda y=0.
As a general solution, we obtain trigometric solution if \lambda > 0 and hyperbolic solution if \lambda < 0. But if we further impose the boundary conditions then for \lambda < 0 the solution is trivial y=0.

So I suspect that there is no such thing as orthogonal property for hyperbolic functions.
 


matematikawan said:
I think most of the orthogonal functions are obtained from the solutions of the Sturm–Liouville eigenvalue equations

(py')'+(q+\lambda r)y=0

where \lambda is a constant.

The trigonometric and hyperbolic come from solving the following equation
y''+\lambda y=0.
As a general solution, we obtain trigometric solution if \lambda > 0 and hyperbolic solution if \lambda < 0. But if we further impose the boundary conditions then for \lambda < 0 the solution is trivial y=0.

So I suspect that there is no such thing as orthogonal property for hyperbolic functions.

This is true.
 

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