Need help with this question. Thank you :) When f=C¹ (R,R), and X*ER is a point with f(x*)=0 and f'(x*) ±0, then we know that there existssome 8>0 such that for every xBg(x*), the parallel chord method converges to x*. What isthe deeper reason why this is true?(You should know this reason, and you are simply asked to recognise it, so there is no needfor proofs or counterexamples in this question.)a. For sufficiently small 8>0, the parallel chord method g(x)=x-f(x)/f'(x。) is a contractionon B8(x*).O b. For sufficiently small 8>0 the parallel chord method g(x)=x-f(x)/f'(x。) is a contractionfrom R into Bg(x*), i.e. for every XER, there exists N≥0 such that g(x)=Bs(x*) for allk₂N.O c.I think that my friend thinks that the tutor thinks that the lecturer thinks that thisstatement is true. So my lecturer thinks that my tutor thinks that my friend thinksthat I should tick this box. So this must be the box to tick.Od. The parallel chord method g(x)=x-f(x)/f'(x。) has the Newton property.

From the given expression option "b" is true

When feC¹ (R,R), and X*ER is a point with f(x*)=0 and f'(x*) ±0, then we know that there exists some 8>0 such that for every x₁=BS(x*), the parallel chord method converges to x*. What is the deeper reason why this is true? (You should know this reason, and you are simply asked to recognise it, so there is no need for proofs or counterexamples in this question.) a. For sufficiently small 8>0, the parallel chord method g(x)=x-f(x)/f'(x) is a contraction on B8(x*). b. For sufficiently small 8>0 the parallel chord method g(x)=x-f(x)/f'(xò) is a contraction from R into Bg(x*), i.e. for every XÃER, there exists N≥0 such that g(x)=Bg(x*) for all k≥N. C. I think that my friend thinks that the tutor thinks that the lecturer thinks that this statement is true. So my lecturer thinks that my tutor thinks that my friend thinks that I should tick this box. So this must be the box to tick. d. The parallel chord method g(x)=x-f(x)/f'(x) has the Newton property.

When feC¹ (R,R), and XER is a point with f(x)=0 and f'(x) ±0, then we know that there exists some 8>0 such that for every x₁=BS(x), the parallel chord method converges to x. What is the deeper reason why this is true? (You should know this reason, and you are simply asked to recognise it, so there is no need for proofs or counterexamples in this question.) a. For sufficiently small 8>0, the parallel chord method g(x)=x-f(x)/f'(x) is a contraction on B8(x). b. For sufficiently small 8>0 the parallel chord method g(x)=x-f(x)/f'(xò) is a contraction from R into Bg(x), i.e. for every XÃER, there exists N≥0 such that g(x)=Bg(x) for all k≥N. C. I think that my friend thinks that the tutor thinks that the lecturer thinks that this statement is true. So my lecturer thinks that my tutor thinks that my friend thinks that I should tick this box. So this must be the box to tick. d. The parallel chord method g(x)=x-f(x)/f'(x) has the Newton property.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.6: Permutations

Problem 47E

See similar textbooks