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Let pn = (an+bn)/2 , p = limn→∞pn, and en = p−pn. Here [an,bn], with n...

Let pn = (an+bn)/2 , p = limn→∞pn, and en = p−pn. Here [an,bn], with n ≥ 1, denotes the successive intervals that arise in the Bisection method when it is applied to a continuous function f.

Show that |pn −pn+1| = .2^{-n-1}(b1-a1)

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Prantben, keranston ant t I bisection method, Either, [ann, bron] = [an, Pn] or, [and, but] = [en, bu] Case I [anos, buri] =11Pn-Pasil = Ibn-an) 4 Case I [ant, bad]=[Pn, br] > ant=Pn= anton e bur= bn • Ano-an-an-bn, bir-bn ao an- anos + on-tomon lan2 So in both cases 6 & Tipo - Pao) = 16.2 mm - We also know that in binction method, lant-Bonoll - I An-bnl 1an-1-bm-11 lan-blan-bal= 18,-a, 1 substitute in Ibn-Pnel = 10,-a, 1 4.27 » IPn-PnHl= 2h-10/6-911

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