- #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 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

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