MHB Are Equal: $$\sin \alpha_nx - \sinh \alpha_nx$$

[Dustinsfl]

$$
\sin \alpha_nx - \sinh \alpha_nx - \frac{\sin \alpha_n + \sinh \alpha_n}{\cos\alpha_n + \cosh\alpha_n}(\cos \alpha_nx -
\cosh \alpha_nx) = \sin \alpha_nx - \sinh \alpha_nx - \frac{\sin \alpha_n - \sinh \alpha_n}{\cos\alpha_n - \cosh\alpha_n}(\cos \alpha_nx -
\cosh \alpha_nx)
$$

When I plotted them, they were the same but I was only looking at a finite range since I am considering a beam of length 1.

In the range [0,1], they converge at every eigenvalue which are defined by $\sin x\cosh x = \cos x\sinh x$

 

[Sudharaka]

$$
\sin \alpha_nx - \sinh \alpha_nx - \frac{\sin \alpha_n + \sinh \alpha_n}{\cos\alpha_n + \cosh\alpha_n}(\cos \alpha_nx -
\cosh \alpha_nx) = \sin \alpha_nx - \sinh \alpha_nx - \frac{\sin \alpha_n - \sinh \alpha_n}{\cos\alpha_n - \cosh\alpha_n}(\cos \alpha_nx -
\cosh \alpha_nx)
$$

When I plotted them, they were the same but I was only looking at a finite range since I am considering a beam of length 1.

In the range [0,1], they converge at every eigenvalue which are defined by $\sin x\cosh x = \cos x\sinh x$

Hi dwsmith, :)

You can show that,

\[\sin \alpha_n\cosh \alpha_n = \cos \alpha_n\sinh \alpha_n\]

\[\Rightarrow\frac{\sin \alpha_n - \sinh \alpha_n}{\cos\alpha_n - \cosh\alpha_n}=\frac{\sin \alpha_n + \sinh \alpha_n}{\cos\alpha_n + \cosh\alpha_n}\]

So both sides are equal provided, \(\sin \alpha_n\cosh \alpha_n = \cos \alpha_n\sinh \alpha_n\).

Kind Regards,
Sudharaka.