if-A-cosh-x-sinh-x-sinh-x-cosh-x-find-A-k-

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

$${if}\:{A}\:=\:\begin{bmatrix}{{cosh}\left({x}\right)\:\:\:\:\:\:{sinh}\left({x}\right)\:}\\{{sinh}\left({x}\right)\:\:\:\:\:\:{cosh}\left({x}\right)}\end{bmatrix}{find}\:{A}^{{k}} \:? \\ $$

Answered by manxsol last updated on 02/Nov/23

A^2 =cosh^2 −sinh^2 =1 A^3 =A A^4 =1 A^k = { ((A k=2m−1)),((1 k=2m)) :}

$${A}^{\mathrm{2}} ={cosh}^{\mathrm{2}} −{sinh}^{\mathrm{2}} =\mathrm{1} \\ $$$${A}^{\mathrm{3}} ={A} \\ $$$${A}^{\mathrm{4}} =\mathrm{1} \\ $$$${A}^{{k}} =\begin{cases}{{A}\:\:\:\:{k}=\mathrm{2}{m}−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:{k}=\mathrm{2}{m}}\end{cases} \\ $$$$ \\ $$$$ \\ $$

Answered by witcher3 last updated on 02/Nov/23

A= (((ch(x) sh(x))),((sh(x) ch(x))) ) A^2 = (((sh^2 (x)+sh^2 (x) 2sh(x)ch(x))),((2sh(x)ch(x) sh^2 (x)+ch^2 (x))) ) A^2 = (((ch(2x) sh(2x))),((sh(2x) ch(2x))) ) claim A^n = (((ch(nx) sh(nx))),((sh(nx) ch(nx))) ) A^(n+1) =A^n .A= (((ch(nx) sh(nx))),((sh(nx) ch(nx))) ) (((ch(x) sh(x))),((sh(x) ch(x))) ) ch(n+1)x=ch(nx)ch(x)+sh(nx)sh(x) sh(n+1)x=sh(nx)ch(x)+ch(nx)sh(x)... = (((ch(n+1)x sh(n+1)x)),((sh(n+1)x ch(n+1)x)) )..worck ∀n≥1 A^n = (((ch(nx) sh(nx))),((sh(nx) ch(nx))) ),∀x∈R

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

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

yes ch(x)=cosh(x) sh(x)=sinh(x) in france we use this notation

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

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