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Precalculus Mathematics Homework Help
Iterative root finding for the cube root of 17
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[QUOTE="AN630078, post: 6416504, member: 680923"] [B]Homework Statement:[/B] Hello, I have found this problem below but I am not sure how to approach it, perhaps because I am confused by what I am being asked. I have attached the question below but I will type it out also (apologies for not writing in LaTEX). The formula xr+1 =1/2(xr+N/xr^2) can be used to find an approximate value for the cube root of N. Starting with x0=2 find the value of the cube root 17 to 3.s.f. [B]Relevant Equations:[/B] xr+1 =1/2(xr+N/xr^2) Firstly, the cube root of 17 is 2.571281591 which is 2.57 to 3.s.f. Initially, I thought about just approaching this problem using the Newton-Raphson Method when x0=2. In which case; x^3=17 x^3-17=0 Using the Newton-Raphson iterative formula xn+1=xr-f(xn)/f’(xn) f(x)=x^3-17 f’(x)=3x^2 x0+1=2-(2^3-17)/(3(2)^2) x1=2.75 x1+1=2.75-(2.75^3-17)/(3(2.75)^2) x2=2.58264... x2+1= 2.58264-(2.58264^3-17)/3(2.58264)^2 x3=2.571331512 x4=2.571281592 x5=2.571281591 x6= 2.571281591 With 5 iterations the sequence converges to the the cube root of 17 which is equal to 2.571281591 Should I actually be using the formula from the question, taking N = 17? In which case; xr+1 =1/2(xr+N/xr^2) x0+1 =1/2(2+17/2^2) x1=3.125 x1+1=1/2(3.125+17/3.125^2) x2=2.4329 x3=2.652502803 x4=2.53436347 x5=2.590551248 x6=2.561861233 x7=2.576043793 x8=2.568913687 x9=2.572468818= 2.57 to 3.s.f Therefore, x9 is the first iterate that equals the cube root of 17 when both are given to three significant figures. Would this be correct? I do not know whether my method is appropriate or if my workings/solution satisfy the question fully? I would be very grateful for any advice 👍 [/QUOTE]
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Precalculus Mathematics Homework Help
Iterative root finding for the cube root of 17
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