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Question Number 27343 by abdo imad last updated on 05/Jan/18

let give A=(_(2 2) ^(1 2) ) find A^n and e^A and e^(tA) . we remind that e^A = Σ_ (A^n /(n!))

$${let}\:{give}\:{A}=\left(_{\mathrm{2}\:\:\:\:\:\:\:\mathrm{2}} ^{\mathrm{1}\:\:\:\:\:\:\mathrm{2}} \right)\:\:\:{find}\:\:{A}^{{n}} \:\:\:{and}\:\:{e}^{{A}} \\ $$$${and}\:\:{e}^{{tA}} \:\:\:\:\:\:.\:{we}\:{remind}\:{that}\:\:{e}^{{A}} =\:\sum_{} \:\frac{{A}^{{n}} }{{n}!} \\ $$

Commented by Rasheed.Sindhi last updated on 05/Jan/18

What′s n in e^A = Σ_ (A^n /(n!)) ?

$$\mathrm{What}'\mathrm{s}\:{n}\:\mathrm{in}\:{e}^{{A}} =\:\sum_{} \:\frac{{A}^{{n}} }{{n}!}\:\:? \\ $$

Commented by abdo imad last updated on 05/Jan/18

n integer.

$${n}\:{integer}. \\ $$

Commented by Rasheed.Sindhi last updated on 06/Jan/18

My question is actually this that why n is involved in e^A ?

$$\mathrm{My}\:\mathrm{question}\:\mathrm{is}\:\mathrm{actually}\:\mathrm{this}\:\mathrm{that} \\ $$$$\mathrm{why}\:\mathrm{n}\:\mathrm{is}\:\mathrm{involved}\:\mathrm{in}\:\mathrm{e}^{\mathrm{A}} ? \\ $$

Commented by Rasheed.Sindhi last updated on 07/Jan/18

thanks sir. I misunderstood.

$${thanks}\:{sir}.\:{I}\:{misunderstood}. \\ $$

Commented by prakash jain last updated on 06/Jan/18

e^x =Σ_(i=0) ^∞ (x^i /(i!))

$${e}^{{x}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{i}} }{{i}!} \\ $$

Commented by Rasheed.Sindhi last updated on 06/Jan/18

Is e^A = Σ_ (A^n /(n!)) correct?

$$\:\mathrm{Is}\:\:{e}^{{A}} =\:\sum_{} \:\frac{{A}^{{n}} }{{n}!}\:\mathrm{correct}? \\ $$

Commented by abdo imad last updated on 06/Jan/18

e^A is also a matrise defined by e^A = Σ_(n≥0) (A^n /(n!)) like e^x = Σ (x^n /(n!))

$${e}^{{A}} \:{is}\:{also}\:{a}\:{matrise}\:{defined}\:{by}\: \\ $$$${e}^{{A}} =\:\sum_{{n}\geqslant\mathrm{0}} \:\:\frac{{A}^{{n}} }{{n}!}\:\:{like}\:\:{e}^{{x}} =\:\Sigma\:\frac{{x}^{{n}} }{{n}!}\: \\ $$

Answered by prakash jain last updated on 07/Jan/18

A^n = ((a_(n−1) ,b_(n−1) ),(c_(n−1) ,d_(n−1) ) ) ∙ ((1,2),(2,2) ) a_n =a_(n−1) +2b_(n−1) b_n =2a_(n−1) +2b_(n−1) c_n =2c_(n−1) +2d_(n−1) c_n =2c_(n−1) +2d_(n−1) b_(n−1) =((a_n −a_(n−1) )/2) b_n =2a_(n−1) +2b_(n−1) =a_n +a_(n−1) a_n =a_(n−1) +2b_(n−1) =a_(n−1) +2a_(n−1) +2a_(n−2) a_n −3a_(n−1) −2a_(n−2) =0 x^2 −3x−2=0⇒x=((3±(√(17)))/2) a_n =c_1 (((3+(√(17)))/2))^n +c_2 (((3−(√(17)))/2))^n a_1 =1⇒c_1 (3+(√(17)))+c_2 (3−(√(17)))=1 a_2 =5⇒c_1 (3+(√(17)))^2 +c_2 (3−(√(17)))^2 =20 c_1 =((20−(3−(√(17))))/((3+(√(17)))^2 −(3+(√(17)))(3−(√(17))))) =((17+(√(17)))/(9+17+6(√(17))−9+17))=((17+(√(17)))/(34+6(√(17)))) =((1+(√(17)))/(2((√(17))+3))) c_2 =((20−(3+(√(17))))/((3−(√(17)))^2 −(3+(√(17)))(3−(√(17)))))=((17−(√(17)))/(34−6(√(17)))) c_2 =((1−(√(17)))/(2(3−(√(17))))) Similarly we can find b_n ,c_n and d_n .

$${A}^{{n}} =\begin{pmatrix}{{a}_{{n}−\mathrm{1}} }&{{b}_{{n}−\mathrm{1}} }\\{{c}_{{n}−\mathrm{1}} }&{{d}_{{n}−\mathrm{1}} }\end{pmatrix}\:\centerdot\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{2}}&{\mathrm{2}}\end{pmatrix} \\ $$$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{b}_{{n}−\mathrm{1}} \\ $$$${b}_{{n}} =\mathrm{2}{a}_{{n}−\mathrm{1}} +\mathrm{2}{b}_{{n}−\mathrm{1}} \\ $$$${c}_{{n}} =\mathrm{2}{c}_{{n}−\mathrm{1}} +\mathrm{2}{d}_{{n}−\mathrm{1}} \\ $$$${c}_{{n}} =\mathrm{2}{c}_{{n}−\mathrm{1}} +\mathrm{2}{d}_{{n}−\mathrm{1}} \\ $$$${b}_{{n}−\mathrm{1}} =\frac{{a}_{{n}} −{a}_{{n}−\mathrm{1}} }{\mathrm{2}} \\ $$$${b}_{{n}} =\mathrm{2}{a}_{{n}−\mathrm{1}} +\mathrm{2}{b}_{{n}−\mathrm{1}} ={a}_{{n}} +{a}_{{n}−\mathrm{1}} \\ $$$${a}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{b}_{{n}−\mathrm{1}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{a}_{{n}−\mathrm{1}} +\mathrm{2}{a}_{{n}−\mathrm{2}} \\ $$$${a}_{{n}} −\mathrm{3}{a}_{{n}−\mathrm{1}} −\mathrm{2}{a}_{{n}−\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{2}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{3}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${a}_{{n}} ={c}_{\mathrm{1}} \left(\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{{n}} +{c}_{\mathrm{2}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{{n}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1}\Rightarrow{c}_{\mathrm{1}} \left(\mathrm{3}+\sqrt{\mathrm{17}}\right)+{c}_{\mathrm{2}} \left(\mathrm{3}−\sqrt{\mathrm{17}}\right)=\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{5}\Rightarrow{c}_{\mathrm{1}} \left(\mathrm{3}+\sqrt{\mathrm{17}}\right)^{\mathrm{2}} +{c}_{\mathrm{2}} \left(\mathrm{3}−\sqrt{\mathrm{17}}\right)^{\mathrm{2}} =\mathrm{20} \\ $$$${c}_{\mathrm{1}} =\frac{\mathrm{20}−\left(\mathrm{3}−\sqrt{\mathrm{17}}\right)}{\left(\mathrm{3}+\sqrt{\mathrm{17}}\right)^{\mathrm{2}} −\left(\mathrm{3}+\sqrt{\mathrm{17}}\right)\left(\mathrm{3}−\sqrt{\mathrm{17}}\right)} \\ $$$$=\frac{\mathrm{17}+\sqrt{\mathrm{17}}}{\mathrm{9}+\mathrm{17}+\mathrm{6}\sqrt{\mathrm{17}}−\mathrm{9}+\mathrm{17}}=\frac{\mathrm{17}+\sqrt{\mathrm{17}}}{\mathrm{34}+\mathrm{6}\sqrt{\mathrm{17}}} \\ $$$$=\frac{\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{2}\left(\sqrt{\mathrm{17}}+\mathrm{3}\right)} \\ $$$${c}_{\mathrm{2}} =\frac{\mathrm{20}−\left(\mathrm{3}+\sqrt{\mathrm{17}}\right)}{\left(\mathrm{3}−\sqrt{\mathrm{17}}\right)^{\mathrm{2}} −\left(\mathrm{3}+\sqrt{\mathrm{17}}\right)\left(\mathrm{3}−\sqrt{\mathrm{17}}\right)}=\frac{\mathrm{17}−\sqrt{\mathrm{17}}}{\mathrm{34}−\mathrm{6}\sqrt{\mathrm{17}}} \\ $$$${c}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{17}}}{\mathrm{2}\left(\mathrm{3}−\sqrt{\mathrm{17}}\right)} \\ $$$$\mathrm{Similarly}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:{b}_{{n}} ,{c}_{{n}} \:\mathrm{and}\:{d}_{{n}} . \\ $$

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