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Question Number 199382 by mokys last updated on 02/Nov/23

$$ifA=\left[\begin{array}{c}cosh\left(x\right)sinh\left(x\right)\\ sinh\left(x\right)cosh\left(x\right)\end{array}\right]find{A}^k?$$

Answered by manxsol last updated on 02/Nov/23

$$A^2={cosh}^2-{sinh}^2=1$$$$A^3=A$$$$A^4=1$$$$A^k=\{\begin{array}{l}Ak=2m-1\\ 1k=2m\end{array}$$$$$$$$$$

Answered by witcher3 last updated on 02/Nov/23

$$\mathrm{A}=\left(\begin{array}{c}\mathrm{ch}\left(\mathrm{x}\right)\mathrm{sh}\left(\mathrm{x}\right)\\ \mathrm{sh}\left(\mathrm{x}\right)\mathrm{ch}\left(\mathrm{x}\right)\end{array}\right)$$$${\mathrm{A}}^2=\left(\begin{array}{c}{\mathrm{sh}}^2\left(\mathrm{x}\right)+{\mathrm{sh}}^2\left(\mathrm{x}\right)2\mathrm{sh}\left(\mathrm{x}\right)\mathrm{ch}\left(\mathrm{x}\right)\\ 2\mathrm{sh}\left(\mathrm{x}\right)\mathrm{ch}\left(\mathrm{x}\right){\mathrm{sh}}^2\left(\mathrm{x}\right)+{\mathrm{ch}}^2\left(\mathrm{x}\right)\end{array}\right)$$$${\mathrm{A}}^2=\left(\begin{array}{c}\mathrm{ch}\left(2\mathrm{x}\right)\mathrm{sh}\left(2\mathrm{x}\right)\\ \mathrm{sh}\left(2\mathrm{x}\right)\mathrm{ch}\left(2\mathrm{x}\right)\end{array}\right)$$$$\mathrm{claim}{\mathrm{A}}^{\mathrm{n}}=\left(\begin{array}{c}\mathrm{ch}\left(\mathrm{nx}\right)\mathrm{sh}\left(\mathrm{nx}\right)\\ \mathrm{sh}\left(\mathrm{nx}\right)\mathrm{ch}\left(\mathrm{nx}\right)\end{array}\right)$$$${\mathrm{A}}^{\mathrm{n}+1}={\mathrm{A}}^{\mathrm{n}}.\mathrm{A}=\left(\begin{array}{c}\mathrm{ch}\left(\mathrm{nx}\right)\mathrm{sh}\left(\mathrm{nx}\right)\\ \mathrm{sh}\left(\mathrm{nx}\right)\mathrm{ch}\left(\mathrm{nx}\right)\end{array}\right)\left(\begin{array}{c}\mathrm{ch}\left(\mathrm{x}\right)\mathrm{sh}\left(\mathrm{x}\right)\\ \mathrm{sh}\left(\mathrm{x}\right)\mathrm{ch}\left(\mathrm{x}\right)\end{array}\right)$$$$\mathrm{ch}(\mathrm{n}+1)\mathrm{x}=\mathrm{ch}\left(\mathrm{nx}\right)\mathrm{ch}\left(\mathrm{x}\right)+\mathrm{sh}\left(\mathrm{nx}\right)\mathrm{sh}\left(\mathrm{x}\right)$$$$\mathrm{sh}(\mathrm{n}+1)\mathrm{x}=\mathrm{sh}\left(\mathrm{nx}\right)\mathrm{ch}\left(\mathrm{x}\right)+\mathrm{ch}\left(\mathrm{nx}\right)\mathrm{sh}\left(\mathrm{x}\right)...$$$$=\left(\begin{array}{c}\mathrm{ch}(\mathrm{n}+1)\mathrm{x}\mathrm{sh}(\mathrm{n}+1)\mathrm{x}\\ \mathrm{sh}(\mathrm{n}+1)\mathrm{x}\mathrm{ch}(\mathrm{n}+1)\mathrm{x}\end{array}\right)..\mathrm{worck}$$$$\mathrm{\forall }\mathrm{n}⩾1{\mathrm{A}}^{\mathrm{n}}=\left(\begin{array}{c}\mathrm{ch}\left(\mathrm{nx}\right)\mathrm{sh}\left(\mathrm{nx}\right)\\ \mathrm{sh}\left(\mathrm{nx}\right)\mathrm{ch}\left(\mathrm{nx}\right)\end{array}\right),\mathrm{\forall }\mathrm{x}\in \mathbb{R}$$

Commented by MathematicalUser2357 last updated on 03/Nov/23

Hey Is sh hyperbolic sine and ch is hyperbolic cosine?

Commented by witcher3 last updated on 03/Nov/23

$$\mathrm{yes}\mathrm{ch}\left(\mathrm{x}\right)=\cosh \left(\mathrm{x}\right)$$$$\mathrm{sh}\left(\mathrm{x}\right)=\sinh \left(\mathrm{x}\right)\mathrm{in}\mathrm{france}\mathrm{we}\mathrm{use}\mathrm{this}\mathrm{notation}$$

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