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Question Number 209232 by alcohol last updated on 04/Jul/24

u_0  = a, u_(n+1)  = (√(u_n v_n ))  v_0  = b ∈ ]0,1[ , v_(n+1)  = (1/(2(u_n +v_n )))  • show that a≤u_n ≤u_(n+1) ≤v_n ≤v_(n+1) ≤b  • show that v_n  − u_n  ≤ ((a+b)/2^n )

u0=a,un+1=unvnv0=b]0,1[,vn+1=12(un+vn)showthataunun+1vnvn+1bshowthatvnuna+b2n

Answered by Berbere last updated on 04/Jul/24

u_0 =a≥0   if a<0 U_1 Will not defined  v_0 =b>a  we will show  u_n >0 and v_n >0   ....P(n)  u_0 =a>0 & v_0 =b>0     p(n+1)..{u_(n+1) (√(u_n v_n ))>0;v_(n+1) =(1/(2(u_n +v_n )))>0}  a=(8/(10));b=(9/(10))  u_1 ≤^? v_1 ⇒(√((8.9)/(100)))≤(1/(2(((17)/(10)))))=(5/(17))  ⇒8<(√(72))≤((50)/(17))<3 false! tchek Quation sir!

u0=a0ifa<0U1Willnotdefinedv0=b>awewillshowun>0andvn>0....P(n)u0=a>0&v0=b>0p(n+1)..{un+1unvn>0;vn+1=12(un+vn)>0}a=810;b=910u1?v18.910012(1710)=5178<725017<3false!tchekQuationsir!

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