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Question Number 67525 Answers: 0 Comments: 3

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

$l e t a > b > 0 c a l c u l a t e \int_{0}^{2 π} \frac{d x}{{(a + b s i n x)}^{2}}$

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 )))

$p r o v e t h a t \forall z \in C w e h a v e$ $s i n z = z \prod_{n = 1}^{\infty} (1 - \frac{z^{2}}{n^{2} π^{2}})$

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 ))

$l e t z f r o m C - Z p r o v e t h a t$ $\frac{π}{s i n (π z)} = \frac{1}{z} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 2 z}{z^{2} - n^{2}} a n d$ $\frac{π c o s (π z)}{s i n (π z)} = \frac{1}{z} + \sum_{n = 1}^{\infty} \frac{2 z}{z^{2} - n^{2}}$

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)

$c a l c u l a t e A (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} c o s (n x)}{n}$ $a n d B (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} s i n (n x)}{n}$

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)

$l e t f (x, z) = \frac{z e^{x z}}{e^{z} - 1} (x a n d z f r o m C)$ $1) p r o v e t h a t f (x, z) = \sum_{n = 0}^{\infty} B_{n} (x) \frac{z^{n}}{n!}$ $w i t h B_{n} (x) i s a u n i t a i r e p o l y n o m e w i t h d e g r e n$ $d e t e r m i n e B_{n} (x) i n t e r m s o f B_{n} (n u m b e r o f b e r n o u l l i)$ $2) p r o v e t h a t B_{n}^{^{'}} (x) = {n B}_{n - 1} (x)$ $B_{n} (x + 1) - B_{n} (x) = {n x}^{n - 1}$ $p r o v e t h a t f (x, z) = f (1 - x, - z) a n d B_{n} (1 - x) = {(- 1)}^{n} B_{n} (x)$

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

$i f \frac{z}{e^{z} - 1} = \sum_{n = 0}^{\infty} B_{n} \frac{z^{n}}{n!}$ $1) c a l c u l a t e B_{0}, B_{1}, B_{2}, B_{3}, B_{4}$ $2) p r o v e t h a t z \to \frac{1}{e^{z} - 1} + \frac{1}{2} i s a o d d f u n c t i o n c o n c l u d e t h a t$ $B_{2 n + 1} = 0 f o r n ⩾ 1$

Question Number 67518 Answers: 1 Comments: 1

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

$i f z = x + i y f i n d l n z i n t e r m s o f x a n d y$

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))

$l e t z \in C a n d ∣ z ∣< 1 p r o v e t h a t$ $\frac{z}{1 - z^{2}} + \frac{z^{2}}{1 - z^{4}} + . . . . . + \frac{z^{2^{n}}}{1 - z^{2^{n + 1}}} + . . . = \frac{z}{1 - z}$ $\frac{z}{1 + z} + \frac{2 z^{2}}{1 + z^{2}} + . . . . + \frac{2^{n} z^{2^{n}}}{1 + z^{2^{n}}} + . . . . = \frac{z}{1 - z}$