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Question Number 28533    Answers: 0   Comments: 1

let give the matrice A= (((1 2 )),((2 1)) ) calculate A^n then find e^A .

$${let}\:{give}\:{the}\:{matrice}\:\:{A}=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${calculate}\:\:{A}^{{n}} \:\:{then}\:{find}\:\:{e}^{{A}} \:. \\ $$

Question Number 28532    Answers: 0   Comments: 1

let give A_n = ( C_n ^0 .C_n ^1 ....C_n ^n )^(1/(n+1)) find^n (√A) _n .

$${let}\:{give}\:\:{A}_{{n}} =\:\left(\:{C}_{{n}} ^{\mathrm{0}} \:.{C}_{{n}} ^{\mathrm{1}} \:....{C}_{{n}} ^{{n}} \right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} \:\:\:{find}\:^{{n}} \sqrt{{A}}\:_{{n}} . \\ $$

Question Number 28531    Answers: 0   Comments: 0

let give S_n = Σ_(k=1) ^n k^p , p∈N and n≥1 find the radius of convergence of the serie Σ_(n≥1) (x^n /S_n ) .

$${let}\:{give}\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{{p}} \:\:\:\:,\:{p}\in{N}\:\:{and}\:{n}\geqslant\mathrm{1} \\ $$$${find}\:{the}\:{radius}\:{of}\:{convergence}\:{of}\:{the}\:{serie}\:\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{{x}^{{n}} }{{S}_{{n}} }\:. \\ $$

Question Number 28530    Answers: 0   Comments: 0

prove that ∣sinx∣= (8/π) Σ_(n=1) ^(+∞) ((sin^2 (nx))/(4n^2 −1)) .

$${prove}\:{that}\:\mid{sinx}\mid=\:\frac{\mathrm{8}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\frac{{sin}^{\mathrm{2}} \left({nx}\right)}{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}\:. \\ $$

Question Number 28529    Answers: 0   Comments: 2

solve the d.e. (x^2 −1)y^′ +xy= x^2 −e^x .

$${solve}\:{the}\:{d}.{e}.\:\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{'} +{xy}=\:{x}^{\mathrm{2}} −{e}^{{x}} \:. \\ $$

Question Number 28540    Answers: 0   Comments: 0

find F( e^(−ax^2 ) ) where F mean fourier transform. a>0

$${find}\:\boldsymbol{{F}}\left(\:{e}^{−{ax}^{\mathrm{2}} } \right)\:\:\:{where}\:\:\boldsymbol{{F}}\:\:{mean}\:{fourier}\:{transform}. \\ $$$${a}>\mathrm{0} \\ $$

Question Number 28518    Answers: 1   Comments: 2

Question Number 28516    Answers: 2   Comments: 1

Question Number 28515    Answers: 1   Comments: 0

Question Number 28512    Answers: 1   Comments: 1

Question Number 28508    Answers: 1   Comments: 1

Question Number 28506    Answers: 0   Comments: 3

Question Number 28505    Answers: 2   Comments: 0

Question Number 28501    Answers: 2   Comments: 1

Question Number 28504    Answers: 0   Comments: 0

Find the direction cosines of two lines which are connected by relation l+m+n=0 mn−2nl−2lm=0 my solution l=−(m+n) mn+2n(m+n)+2(n+m)n=0 2m^2 +5mn+2n^2 =0 (2m+n)(m+2n)=0 m=−2n, or m=−(1/2)n case 1: m=−2n l=−(m+n)=−n l^2 +m^2 +n^2 =1 6n^2 =1⇒n=±(1/(√6)) (l,m,n)=±(−(1/(√6)),−(2/(√6)),(1/(√6))) case 2:m=−(1/2)n l=−(1/2)n l^2 +m^2 +n^2 =1 (n^2 /4)+(n^2 /4)+n^2 =1⇒n=±(2/(√6)) (l,m,n)=±(−(1/(√6)),−(1/(√6)),(2/(√6))) so i get a total of 4 solution. Books answer is 2 lines ((1/(√6)),(1/(√6)),−(2/(√6))) and ((1/(√6)),−(2/(√6)),(1/(√6))) please help.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{cosines}\:\mathrm{of}\:\mathrm{two} \\ $$$$\mathrm{lines}\:\mathrm{which}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{relation} \\ $$$${l}+{m}+{n}=\mathrm{0} \\ $$$${mn}−\mathrm{2}{nl}−\mathrm{2}{lm}=\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{solution}} \\ $$$${l}=−\left({m}+{n}\right) \\ $$$${mn}+\mathrm{2}{n}\left({m}+{n}\right)+\mathrm{2}\left({n}+{m}\right){n}=\mathrm{0} \\ $$$$\mathrm{2}{m}^{\mathrm{2}} +\mathrm{5}{mn}+\mathrm{2}{n}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{m}+{n}\right)\left({m}+\mathrm{2}{n}\right)=\mathrm{0} \\ $$$${m}=−\mathrm{2}{n},\:{or}\:{m}=−\left(\mathrm{1}/\mathrm{2}\right){n} \\ $$$${case}\:\mathrm{1}:\:{m}=−\mathrm{2}{n} \\ $$$${l}=−\left({m}+{n}\right)=−{n} \\ $$$${l}^{\mathrm{2}} +{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{6}{n}^{\mathrm{2}} =\mathrm{1}\Rightarrow{n}=\pm\frac{\mathrm{1}}{\sqrt{\mathrm{6}}} \\ $$$$\left({l},{m},{n}\right)=\pm\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}\right) \\ $$$${case}\:\mathrm{2}:{m}=−\left(\mathrm{1}/\mathrm{2}\right){n} \\ $$$${l}=−\frac{\mathrm{1}}{\mathrm{2}}{n} \\ $$$${l}^{\mathrm{2}} +{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{{n}^{\mathrm{2}} }{\mathrm{4}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}+{n}^{\mathrm{2}} =\mathrm{1}\Rightarrow{n}=\pm\frac{\mathrm{2}}{\sqrt{\mathrm{6}}} \\ $$$$\left({l},{m},{n}\right)=\pm\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},\frac{\mathrm{2}}{\sqrt{\mathrm{6}}}\right) \\ $$$$\mathrm{so}\:\mathrm{i}\:\mathrm{get}\:\mathrm{a}\:\mathrm{total}\:\mathrm{of}\:\mathrm{4}\:\mathrm{solution}. \\ $$$$\mathrm{Books}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2}\:\mathrm{lines} \\ $$$$\left(\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}}\right)\:\mathrm{and}\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}\right) \\ $$$$\mathrm{please}\:\mathrm{help}. \\ $$

Question Number 28490    Answers: 2   Comments: 0

lim_(x → a) (((cos(x) − cos(a))/(x − a))) Don′t use L′ho^^ spital rule.

$$\underset{\mathrm{x}\:\rightarrow\:\mathrm{a}} {\mathrm{lim}}\:\left(\frac{\mathrm{cos}\left(\mathrm{x}\right)\:−\:\mathrm{cos}\left(\mathrm{a}\right)}{\mathrm{x}\:−\:\mathrm{a}}\right) \\ $$$$ \\ $$$$\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{L}'\mathrm{h}\bar {\mathrm{o}spital}\:\mathrm{rule}. \\ $$

Question Number 28483    Answers: 0   Comments: 0

Question Number 28481    Answers: 0   Comments: 1

Question Number 28480    Answers: 1   Comments: 1

Question Number 28479    Answers: 1   Comments: 0

Question Number 28475    Answers: 0   Comments: 0

Question Number 28468    Answers: 0   Comments: 0

Prove that for all values of u, the point [a cosh(u) , b sinh(u)] lies on the hyperbola whose equation is. (x^2 /a^2 ) − (y^2 /b^2 ) = 1 And that the tangent at that point is : (x/a) cosh(u) − (y/b) sinh(u)

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:\mathrm{u},\:\mathrm{the}\:\mathrm{point}\:\left[\mathrm{a}\:\mathrm{cosh}\left(\mathrm{u}\right)\:,\:\:\mathrm{b}\:\mathrm{sinh}\left(\mathrm{u}\right)\right]\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{hyperbola}\:\mathrm{whose}\:\mathrm{equation}\:\mathrm{is}.\:\:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\mathrm{And}\:\mathrm{that}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}\:\mathrm{is}\::\:\:\:\frac{\mathrm{x}}{\mathrm{a}}\:\mathrm{cosh}\left(\mathrm{u}\right)\:\:−\:\:\frac{\mathrm{y}}{\mathrm{b}}\:\mathrm{sinh}\left(\mathrm{u}\right) \\ $$

Question Number 28467    Answers: 2   Comments: 1

Question Number 28464    Answers: 1   Comments: 0

Question Number 28461    Answers: 0   Comments: 2

Question Number 28456    Answers: 1   Comments: 1

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