 #1
AN630078
 242
 25
 Homework Statement

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.
 Relevant Equations
 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 NewtonRaphson Method when x0=2. In which case; x^3=17
x^317=0
Using the NewtonRaphson iterative formula xn+1=xrf(xn)/f’(xn)
f(x)=x^317
f’(x)=3x^2
x0+1=2(2^317)/(3(2)^2)
x1=2.75
x1+1=2.75(2.75^317)/(3(2.75)^2)
x2=2.58264...
x2+1= 2.58264(2.58264^317)/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 👍
Initially, I thought about just approaching this problem using the NewtonRaphson Method when x0=2. In which case; x^3=17
x^317=0
Using the NewtonRaphson iterative formula xn+1=xrf(xn)/f’(xn)
f(x)=x^317
f’(x)=3x^2
x0+1=2(2^317)/(3(2)^2)
x1=2.75
x1+1=2.75(2.75^317)/(3(2.75)^2)
x2=2.58264...
x2+1= 2.58264(2.58264^317)/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 👍