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Question Number 175402 by Linton last updated on 29/Aug/22
solve for x x_(n+1) =rx_n (1−x_n )
solveforxxn+1=rxn(1−xn)
Commented by a.lgnaoui last updated on 31/Aug/22
x_1 =rx_0 (1−x_0 ) x_2 =rx_1 (1−x_1 ) x_3 =rx_2 (1−x_2 ) ....... x_(n+1) =rx_n (1−x_n ) −−−−−−−− Π _(i=1)^n (x_i )=r^n Πx_i (1−x_i )=x_1 .x_2 .x_3 ......x_n 1=r^n x_0 (1−x_n ) x_n =1−(1/(x_0 r^n )) (1) x_1 =rx_0 (1−x_0 ) x_2 =rx_1 (1−x_1 )=r[rx_0 (1−x_0 )][1−rx_0 (1−x_0 )] =r^2 x_0 (1−x_0 )−r^3 x_0 ^2 (1−x_0 )^2 x_3 =rx_2 (1−x_2 )=[r(rx_1 (1−x_1 ))][1−rx_1 (1−x_1 )] =r^2 x_1 (1−x_1 )−r^3 x_1 ^2 (1−x_1 )^2 x_4 =r^2 x_2 (1−x_2 )−r^3 x_2 ^2 (1−x_2 )^2 ........ x_n =r^2 x_(n−2) (1−x_(n−2) )−r^3 (1−x_(n−2) )^2 Σx_i =r[(r(x_0 (1−x_0 )+rx_1 (1−x_1 )+rx_2 (1−x_2 )+.....rx_(n−2) (1−x_(n−2) )−r×[r^2 [x_0 (1−x_0 )^2 +x_1 ^2 (1−x_1 )^2 +x_3 ^2 (1−x_3 )^2 +.....x_(n−2) ^2 (1−x_(n−2) )^2 Σx_i =r[x_1 +x_2 +x_3 +....x_(n−1) ]+r[x_1 ^2 +x_2 ^2 +x_3 ^2 +...x_(n−1) ^2 ] ? (1−r)Σx_i =r(Σx_i ^2 ) Σx_i =((rΣx_i ^2 )/(1−r)) ? .............
x1=rx0(1−x0)x2=rx1(1−x1)x3=rx2(1−x2)…….xn+1=rxn(1−xn)−−−−−−−−Πi=1n(xi)=rnΠxi(1−xi)=x1.x2.x3……xn1=rnx0(1−xn)xn=1−1x0rn(1)x1=rx0(1−x0)x2=rx1(1−x1)=r[rx0(1−x0)][1−rx0(1−x0)]=r2x0(1−x0)−r3x02(1−x0)2x3=rx2(1−x2)=[r(rx1(1−x1))][1−rx1(1−x1)]=r2x1(1−x1)−r3x12(1−x1)2x4=r2x2(1−x2)−r3x22(1−x2)2……..xn=r2xn−2(1−xn−2)−r3(1−xn−2)2Σxi=r[(r(x0(1−x0)+rx1(1−x1)+rx2(1−x2)+…..rxn−2(1−xn−2)−r×[r2[x0(1−x0)2+x12(1−x1)2+x32(1−x3)2+…..xn−22(1−xn−2)2Σxi=r[x1+x2+x3+….xn−1]+r[x12+x22+x32+…xn−12]?(1−r)Σxi=r(Σxi2)Σxi=rΣxi21−r?………….

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