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Question Number 67525 Answers: 0 Comments: 3

let a>b>0 calculate ∫_0 ^(2π) (dx/((a+bsinx)^2 ))

$${let}\:{a}>{b}>\mathrm{0}\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\left({a}+{bsinx}\right)^{\mathrm{2}} } \\ $$

Question Number 67524 Answers: 0 Comments: 1

prove that ∀z ∈C we have sinz =z Π_(n=1) ^∞ (1−(z^2 /(n^2 π^2 )))

$${prove}\:{that}\:\forall{z}\:\in{C}\:\:{we}\:{have} \\ $$$${sinz}\:={z}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right) \\ $$

Question Number 67522 Answers: 0 Comments: 0

let z from C−Z prove that (π/(sin(πz))) =(1/z) +Σ_(n=1) ^∞ (((−1)^n 2z)/(z^2 −n^2 )) and ((πcos(πz))/(sin(πz))) =(1/z) +Σ_(n=1) ^∞ ((2z)/(z^2 −n^2 ))

$${let}\:{z}\:{from}\:{C}−{Z}\:\:\:\:\:{prove}\:{that} \\ $$$$\frac{\pi}{{sin}\left(\pi{z}\right)}\:=\frac{\mathrm{1}}{{z}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{2}{z}}{{z}^{\mathrm{2}} −{n}^{\mathrm{2}} }\:\:{and} \\ $$$$\frac{\pi{cos}\left(\pi{z}\right)}{{sin}\left(\pi{z}\right)}\:=\frac{\mathrm{1}}{{z}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}{z}}{{z}^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$

Question Number 67521 Answers: 0 Comments: 0

calculate A(x) =Σ_(n=1) ^∞ (((−1)^n cos(nx))/n) and B(x) =Σ_(n=1) ^∞ (((−1)^n sin(nx))/n)

$${calculate}\:\:\:{A}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{cos}\left({nx}\right)}{{n}} \\ $$$${and}\:{B}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{sin}\left({nx}\right)}{{n}} \\ $$

Question Number 67520 Answers: 0 Comments: 0

let f(x,z) =((z e^(xz) )/(e^z −1)) (x and z from C) 1) prove that f(x,z) =Σ_(n=0) ^∞ B_n (x)(z^n /(n!)) with B_n (x) is a unitaire polynome with degre n determine B_n (x) interms of B_n (number of bernoulli) 2)prove that B _n^′ (x)=nB_(n−1) (x) B_n (x+1)−B_n (x) =nx^(n−1) prove that f(x,z)=f(1−x,−z) and B_n (1−x) =(−1)^n B_n (x)

$${let}\:{f}\left({x},{z}\right)\:=\frac{{z}\:{e}^{{xz}} }{{e}^{{z}} −\mathrm{1}}\:\:\:\:\:\:\left({x}\:{and}\:{z}\:{from}\:{C}\right) \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}\left({x},{z}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{B}_{{n}} \left({x}\right)\frac{{z}^{{n}} }{{n}!} \\ $$$${with}\:{B}_{{n}} \left({x}\right)\:{is}\:{a}\:{unitaire}\:{polynome}\:{with}\:{degre}\:{n} \\ $$$${determine}\:{B}_{{n}} \left({x}\right)\:{interms}\:{of}\:{B}_{{n}} \left({number}\:{of}\:{bernoulli}\right) \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:{B}\:_{{n}} ^{'} \left({x}\right)={nB}_{{n}−\mathrm{1}} \left({x}\right) \\ $$$${B}_{{n}} \left({x}+\mathrm{1}\right)−{B}_{{n}} \left({x}\right)\:={nx}^{{n}−\mathrm{1}} \\ $$$${prove}\:{that}\:{f}\left({x},{z}\right)={f}\left(\mathrm{1}−{x},−{z}\right)\:\:{and}\:{B}_{{n}} \left(\mathrm{1}−{x}\right)\:=\left(−\mathrm{1}\right)^{{n}} \:{B}_{{n}} \left({x}\right) \\ $$

Question Number 67519 Answers: 0 Comments: 0

if (z/(e^z −1)) =Σ_(n=0) ^∞ B_n (z^n /(n!)) 1) calculate B_0 ,B_1 ,B_2 ,B_3 ,B_4 2)prove that z→(1/(e^z −1))+(1/2) is a odd function conclude that B_(2n+1) =0 for n≥1

$${if}\:\frac{{z}}{{e}^{{z}} −\mathrm{1}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{B}_{{n}} \:\frac{{z}^{{n}} }{{n}!} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{B}_{\mathrm{0}} ,{B}_{\mathrm{1}} ,{B}_{\mathrm{2}} ,{B}_{\mathrm{3}} ,{B}_{\mathrm{4}} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:{z}\rightarrow\frac{\mathrm{1}}{{e}^{{z}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\:{is}\:{a}\:{odd}\:{function}\:\:{conclude}\:{that} \\ $$$${B}_{\mathrm{2}{n}+\mathrm{1}} =\mathrm{0}\:\:{for}\:{n}\geqslant\mathrm{1} \\ $$

Question Number 67518 Answers: 1 Comments: 1

if z =x+iy find lnz interms of x and y

$${if}\:{z}\:={x}+{iy}\:\:\:{find}\:\:{lnz}\:\:{interms}\:{of}\:{x}\:{and}\:{y} \\ $$$$ \\ $$

Question Number 67517 Answers: 0 Comments: 0

let z ∈C and ∣z∣<1 prove that (z/(1−z^2 )) +(z^2 /(1−z^4 )) +.....+(z^2^n /(1−z^2^(n+1) ))+...=(z/(1−z)) (z/(1+z)) +((2z^2 )/(1+z^2 )) +....+((2^n z^2^n )/(1+z^2^n )) +....=(z/(1−z))

$${let}\:{z}\:\in{C}\:{and}\:\mid{z}\mid<\mathrm{1}\:\:{prove}\:{that} \\ $$$$\frac{{z}}{\mathrm{1}−{z}^{\mathrm{2}} }\:+\frac{{z}^{\mathrm{2}} }{\mathrm{1}−{z}^{\mathrm{4}} }\:+.....+\frac{{z}^{\mathrm{2}^{{n}} } }{\mathrm{1}−{z}^{\mathrm{2}^{{n}+\mathrm{1}} } }+...=\frac{{z}}{\mathrm{1}−{z}} \\ $$$$\frac{{z}}{\mathrm{1}+{z}}\:+\frac{\mathrm{2}{z}^{\mathrm{2}} }{\mathrm{1}+{z}^{\mathrm{2}} }\:+....+\frac{\mathrm{2}^{{n}} \:{z}^{\mathrm{2}^{{n}} } }{\mathrm{1}+\mathrm{z}^{\mathrm{2}^{\mathrm{n}} } }\:+....=\frac{\mathrm{z}}{\mathrm{1}−\mathrm{z}} \\ $$

Question Number 67516 Answers: 2 Comments: 2