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solve-for-x-x-n-1-rx-n-1-x-n-




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(1xn)
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(1x0)x2=rx1(1x1)x3=rx2(1x2).xn+1=rxn(1xn)Πi=1n(xi)=rnΠxi(1xi)=x1.x2.x3xn1=rnx0(1xn)xn=11x0rn(1)x1=rx0(1x0)x2=rx1(1x1)=r[rx0(1x0)][1rx0(1x0)]=r2x0(1x0)r3x02(1x0)2x3=rx2(1x2)=[r(rx1(1x1))][1rx1(1x1)]=r2x1(1x1)r3x12(1x1)2x4=r2x2(1x2)r3x22(1x2)2..xn=r2xn2(1xn2)r3(1xn2)2Σxi=r[(r(x0(1x0)+rx1(1x1)+rx2(1x2)+..rxn2(1xn2)r×[r2[x0(1x0)2+x12(1x1)2+x32(1x3)2+..xn22(1xn2)2Σxi=r[x1+x2+x3+.xn1]+r[x12+x22+x32+xn12]?(1r)Σxi=r(Σxi2)Σxi=rΣxi21r?.

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