# Iterative root finding for the cube root of 17

• AN630078
In summary: The only thing that I noticed is that you stopped too early on the Newton-Raphson method. You should have kept going until your result was the same to three significant figures as the last one you tried. So in that case your answer should have been ## x_{10}=2.571285527 ##. The answer you got (x_9) is a very good approximation to the correct answer, accurate to 5 significant figures. I hope this helps.In summary, the cube root of 17 can be found using the Newton-Raphson method or the formula given in the question. Both methods result in a value of 2.57 to 3 significant figures, with the Newton-Raphson method
AN630078
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

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I'm going to write your formula with Latex:
##x_{r+1}=\frac{1}{2}(x_r+\frac{N}{x_r^2}) ##.
This makes sense because ##x=\frac{N}{x^2} ##.It would also make sense because we also have ## x =\frac{1}{2}(x+\frac{N}{x^2}) ##.
I'll leave it for others to comment on this, now that it should be more readable.
One simple comment is you will find different iterative expressions converge at different rates. The Newton-Raphson method converges much quicker.

Last edited:
Delta2 and sysprog
additional comment: Even though the Newton-Raphson method comes from derivatives, you can also see how the formula works by looking at the difference of cubes: ## x_{no}^3-x_n^3=(x_{no}-x_n)(x_{no}^2+x_{no}x_n+x_n^2) \approx (x_{no}-x_n)(3 x_n^2) ##. This then gives ##x_{no}=x_n-\frac{x_n^3-x_{no}^3}{3 x_n^2} ##, where ## x_{no} ## is the exact answer. Note ## x_{no}^3=N ##.

Delta2 and sysprog
One additional comment: It is a very good job that you did of solving the problem. You compared two different iterative methods, which is perhaps even worthy of extra credit. Your results look very accurate.

## 1. What is the concept of iterative root finding for the cube root of 17?

Iterative root finding is a method used to approximate the cube root of a number, in this case 17. It involves repeatedly guessing and refining the guess until the desired accuracy is achieved.

## 2. Why is iterative root finding used for finding the cube root of 17?

Iterative root finding is used because there is no simple algebraic solution for finding the cube root of 17. This method allows for a close approximation to be found.

## 3. How does iterative root finding work for finding the cube root of 17?

The process involves starting with a guess for the cube root of 17, and then using that guess to refine the next guess. This is done by using a formula that takes into account the previous guess and the original number. The process is repeated until the desired level of accuracy is reached.

## 4. What is the formula used for iterative root finding of the cube root of 17?

The formula used is: xn+1 = (2xn + 17/(xn2))/3. This formula is used to refine the guess for the cube root of 17 in each iteration.

## 5. How accurate is iterative root finding for finding the cube root of 17?

The accuracy of the method depends on the number of iterations performed. The more iterations, the closer the approximation will be to the actual cube root of 17. However, it is important to note that iterative root finding will always give an approximation and not the exact value.

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