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- #1

$$

\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 $\sin x\cosh x = \cos x\sinh x$

\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 $\sin x\cosh x = \cos x\sinh x$

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