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Let consider (a_n )_n and (u_n )_n two reals sequence defined such as a_0 =1 , ∀ n>1 a_(n+1) =Σ_(p=0) ^n a_p a_(n−p) and Σ_(p=0) ^n a_p u_(n−p) =0 Part1 1)Express ∀ n >1 a_n in terms of n 2) Find the largest domain of convergence of the integer serie {a_n x^n } 3)Determinate ∀ x∈D the sum f(x) of {a_n x^n } 4)Find the radius of convergence of the serie {u_n x^n } 5) Give the relation that between the sum S(x) of the second serie and (x/(f(x))) 6) Can you developp in integer serie g(x)=((πx)/(tan(πx))) Part2 Now do the part 1 but in the order 2)−1)−3)−4)−5)−6)

$L e t c o n s i d e r {(a_{n})}_{n} a n d {(u_{n})}_{n} t w o r e a l s s e q u e n c e$ $d e f i n e d s u c h a s a_{0} = 1, \forall n > 1 a_{n + 1} = \underset{p = 0}{\sum^{n}} a_{p} a_{n - p} a n d \underset{p = 0}{\sum^{n}} a_{p} u_{n - p} = 0$ $P a r t 1$ $1) E x p r e s s \forall n > 1 a_{n} i n t e r m s o f n$ $2) F i n d t h e l a r g e s t d o m a i n o f c o n v e r g e n c e o f t h e i n t e g e r s e r i e {a_{n} x^{n}}$ $3) D e t e r m i n a t e \forall x \in D t h e s u m f (x) o f {a_{n} x^{n}}$ $4) F i n d t h e r a d i u s o f c o n v e r g e n c e o f t h e s e r i e {u_{n} x^{n}}$ $5) G i v e t h e r e l a t i o n t h a t b e t w e e n t h e s u m S (x) o f t h e s e c o n d s e r i e a n d \frac{x}{f (x)}$ $6) C a n y o u d e v e l o p p i n i n t e g e r s e r i e g (x) = \frac{π x}{t a n (π x)}$ $P a r t 2$ $N o w d o t h e p a r t 1 b u t i n t h e o r d e r 2) - 1) - 3) - 4) - 5) - 6)$

Question Number 68219 Answers: 1 Comments: 0

Question Number 68149 Answers: 0 Comments: 2

Explicit f(a)=Σ_(n=1) ^∞ (((−1)^n )/(n(an+1)))

$E x p l i c i t f (a) = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n (a n + 1)}$

Question Number 68145 Answers: 0 Comments: 5

Find the arc length, given the curve x(t) = sin (πt), y(t) = t , 0 ≤ t ≤ 1

$Find the arc length, given the curve$ $x (t) = \sin (π t), y (t) = t, 0 ⩽ t ⩽ 1$

Question Number 68141 Answers: 0 Comments: 1

the 2 formulas for solving ∫(dx/(x^3 +px+q)) with “nasty” solutions of x^3 +px+q=0 with p, q ∈R case 1 D=(p^3 /(27))+(q^2 /4)>0 ⇒ x^3 +px+q=0 has got 1 real and 2 conjugated complex solutions u=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) ∧v=((−(q/2)−p(√((p^3 /(27))+(q^2 /4)))))^(1/3) x_1 =u+v x_2 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v x_3 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v α=u+v∧β=((√3)/2)(u−v) ⇔ u=(α/2)+(β/(√3))∧v=(α/2)−(β/(√3)) x_1 =α x_2 =−(α/2)+βi x_3 =−(α/2)−βi ∫(dx/(x^3 +px+q))=∫(dx/((x−α)(x^2 +αx+((α^2 +4β^2 )/4))))= =(1/(9α^2 +4β^2 ))(∫(dx/((x−α)))−∫((x+2α)/(x^2 +αx+((α^2 +4β^2 )/4)))dx)= =(1/(9α^2 +4β^2 ))(ln ∣x−α∣ −(1/2)ln ((2x+α)^2 +4β^2 ) −((3α)/(2β))arctan ((2x+α)/(2β))) +C ...now calculate the constants case 2 D=(p^3 /(27))+(q^2 /4)<0 ⇒ x^3 +px+q=0 has got 3 real solutions x_k =(2/3)(√(−3p)) sin (((2π)/3)k+(1/3)arcsin ((3(√3)q)/(2(√(−p^3 ))))) with k=0, 1, 2 let x_1 =α, x_2 =β, x_3 =γ ∫(dx/(x^3 +px+q))=∫(dx/((x−α)(x−β)(x−γ)))= =(1/((α−β)(α−γ)))∫(dx/(x−α))+(1/((β−α)(β−γ)))∫(dx/(x−β))+(1/((γ−α)(γ−β)))∫(dx/(x−γ))= =((ln ∣x−α∣)/((α−β)(α−γ)))+((ln ∣x−β∣)/((β−α)(β−γ)))+((ln ∣x−γ∣)/((γ−α)(γ−β)))+C ...now calculate the constants

$the 2 formulas for solving \int \frac{d x}{x^{3} + p x + q} with$ $‘ ‘ {nasty}^{″} solutions of x^{3} + p x + q = 0 with p, q \in R$ $case 1$ $D = \frac{p^{3}}{27} + \frac{q^{2}}{4} > 0 \Rightarrow x^{3} + p x + q = 0 has got 1 real$ $and 2 conjugated complex solutions$ $u = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} \land v = \sqrt[3]{- \frac{q}{2} - p \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}}$ $x_{1} = u + v$ $x_{2} = (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) v$ $x_{3} = (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) v$ $α = u + v \land β = \frac{\sqrt{3}}{2} (u - v) \Leftrightarrow u = \frac{α}{2} + \frac{β}{\sqrt{3}} \land v = \frac{α}{2} - \frac{β}{\sqrt{3}}$ $x_{1} = α$ $x_{2} = - \frac{α}{2} + β i$ $x_{3} = - \frac{α}{2} - β i$ $\int \frac{d x}{x^{3} + p x + q} = \int \frac{d x}{(x - α) (x^{2} + α x + \frac{α^{2} + 4 β^{2}}{4})} =$ $= \frac{1}{9 α^{2} + 4 β^{2}} (\int \frac{d x}{(x - α)} - \int \frac{x + 2 α}{x^{2} + α x + \frac{α^{2} + 4 β^{2}}{4}} d x) =$ $= \frac{1}{9 α^{2} + 4 β^{2}} (\ln ∣ x - α ∣ - \frac{1}{2} \ln ({(2 x + α)}^{2} + 4 β^{2}) - \frac{3 α}{2 β} \arctan \frac{2 x + α}{2 β}) + C$ $. . . now calculate the constants$ $case 2$ $D = \frac{p^{3}}{27} + \frac{q^{2}}{4} < 0 \Rightarrow x^{3} + p x + q = 0 has got 3 real solutions$ $x_{k} = \frac{2}{3} \sqrt{- 3 p} \sin (\frac{2 π}{3} k + \frac{1}{3} \arcsin \frac{3 \sqrt{3} q}{2 \sqrt{- p^{3}}}) with k = 0, 1, 2$ $let x_{1} = α, x_{2} = β, x_{3} = γ$ $\int \frac{d x}{x^{3} + p x + q} = \int \frac{d x}{(x - α) (x - β) (x - γ)} =$ $= \frac{1}{(α - β) (α - γ)} \int \frac{d x}{x - α} + \frac{1}{(β - α) (β - γ)} \int \frac{d x}{x - β} + \frac{1}{(γ - α) (γ - β)} \int \frac{d x}{x - γ} =$ $= \frac{\ln ∣ x - α ∣}{(α - β) (α - γ)} + \frac{\ln ∣ x - β ∣}{(β - α) (β - γ)} + \frac{\ln ∣ x - γ ∣}{(γ - α) (γ - β)} + C$ $. . . now calculate the constants$