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Question Number 28543    Answers: 0   Comments: 0

find the value of ∫_(−∞) ^(+∞) (dt/((t^2 +t +h^2 )^2 +h^2 )) .

$${find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} +{t}\:+{h}^{\mathrm{2}} \right)^{\mathrm{2}} \:+{h}^{\mathrm{2}} }\:\:\:. \\ $$

Question Number 28542    Answers: 0   Comments: 0

if (1/(1−x−y−xy)) = Σ_(n=0) ^∞ g_n (y) x^n find g_n .

$${if}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{x}−{y}−{xy}}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{g}_{{n}} \left({y}\right)\:{x}^{{n}} \:\:{find}\:{g}_{{n}} \:\:. \\ $$

Question Number 28541    Answers: 0   Comments: 0

prove that ∫_0 ^∞ ((sinx)/(e^(ax) −1))dx= Σ_(p=1) ^∞ (1/(1+p^2 a^2 )) with a>0

$${prove}\:{that}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{e}^{{ax}} −\mathrm{1}}{dx}=\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{p}^{\mathrm{2}} {a}^{\mathrm{2}} }\:\:\:\:\:{with}\:{a}>\mathrm{0} \\ $$

Question Number 28539    Answers: 0   Comments: 0

find the value of ∫_0 ^∞ ((1−cos(xt))/t^2 ) e^(−t) dt .

$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}^{\mathrm{2}} }\:{e}^{−{t}} {dt}\:. \\ $$

Question Number 28538    Answers: 0   Comments: 0

study the convergence of u_(n ) =Σ_(k=0) ^n (1/(√(1+k^2 ))) −argsh(n) and v_n = Σ_(k=0) ^n (1/(√(1+k^2 ))) .

$${study}\:{the}\:{convergence}\:{of} \\ $$$${u}_{{n}\:} \:\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}\:\:−{argsh}\left({n}\right)\:\:{and} \\ $$$${v}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}\:\:. \\ $$$$ \\ $$

Question Number 28537    Answers: 0   Comments: 0

prove that ∀ n∈ N (1/(√(1+(n+1)^2 ))) ≤ argsh(n+1) −argsh(n)≤ (1/(√(1+n^2 ))) .

$${prove}\:{that}\:\:\:\forall\:{n}\in\:\mathbb{N} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\left({n}+\mathrm{1}\right)^{\mathrm{2}} }}\:\leqslant\:{argsh}\left({n}+\mathrm{1}\right)\:−{argsh}\left({n}\right)\leqslant\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }}\:. \\ $$

Question Number 28536    Answers: 0   Comments: 0

study the convergence of V_n = Π_(1≤p≤n) (1 +(i/p)) i∈ C and i^2 =−1 .

$${study}\:{the}\:{convergence}\:{of}\:\:{V}_{{n}} =\:\prod_{\mathrm{1}\leqslant{p}\leqslant{n}} \left(\mathrm{1}\:+\frac{{i}}{{p}}\right) \\ $$$${i}\in\:{C}\:{and}\:{i}^{\mathrm{2}} =−\mathrm{1}\:. \\ $$

Question Number 28535    Answers: 0   Comments: 0

study the convergence of U_n = (1+(z/n))^n with z∈C .

$${study}\:{the}\:{convergence}\:{of}\:\:{U}_{{n}} =\:\left(\mathrm{1}+\frac{{z}}{{n}}\right)^{{n}} \:{with}\:{z}\in{C}\:. \\ $$

Question Number 28534    Answers: 0   Comments: 1

find n from N in ordre tohave x^2 +x+1 divide (x+1)^n −x^n −1.

$${find}\:{n}\:{from}\:{N}\:\:{in}\:{ordre}\:{tohave}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:{divide} \\ $$$$\left({x}+\mathrm{1}\right)^{{n}} −{x}^{{n}} −\mathrm{1}. \\ $$

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

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