We are going to compute the cube root of 9 to 6 significant digits using Newton's method.
Choose a function of the form
𝑓(𝑥)=(𝑥^𝑚)−𝐶
Give the formula for 𝑥_(𝑛+1) in terms of the previous value 𝑥_(𝑛) .
you want y = x^3 - 9
Now just apply Newton's method, which I'm sure you can find
(a) in your text
(b) online.
It doesn't even really matter how close your first guess is, since it converges quite rapidly (if at all, but x^3 is a nice well-behaved function).
Here is my simplistic GWBASIC program (from the early 80's)
10 x = 2
20 xnew = x - (x^3 - 9)/(3*x^2)
30 if ABS(xnew - x) < .0000001 THEN END
40 PRINT XNEW
50 X = XNEW
60 GOTO 20
2.083333
2.080089
2.080084
Well, here's a formula that should do the trick:
𝑥_(𝑛+1) = (2/3)*(𝑥_n) + (2/(3*(𝑥_n)^2)) + (𝑥_n/3)
It's like a mathematical dance move - two steps forward, one step back, and a little shimmy at the end!
In Newton's method, the formula for the next value, 𝑥_(𝑛+1), is given by:
𝑥_(𝑛+1) = 𝑥_𝑛 - 𝑓(𝑥_𝑛)/𝑓'(𝑥_𝑛)
Where 𝑓(𝑥_𝑛) represents the function value at the current estimate 𝑥_𝑛, and 𝑓'(𝑥_𝑛) represents the derivative of the function at 𝑥_𝑛.
In this case, we want to compute the cube root of 9, so we can choose the function 𝑓(𝑥) = 𝑥^3 - 9.
The derivative of 𝑓(𝑥) is 𝑓'(𝑥) = 3𝑥^2.
Therefore, the formula for 𝑥_(𝑛+1) in terms of 𝑥_𝑛 becomes:
𝑥_(𝑛+1) = 𝑥_𝑛 - (𝑥_𝑛^3 - 9)/(3𝑥_𝑛^2)
To compute the cube root of 9 using Newton's method, we can choose a function of the form:
𝑓(𝑥) = 𝑥^3 - 9
Newton's method involves iteratively improving an initial guess to find a more accurate root. For each iteration, we can use the formula:
𝑥_(𝑛+1) = 𝑥_𝑛 - 𝑓(𝑥_𝑛)/𝑓'(𝑥_𝑛)
where 𝑥_𝑛 represents the previous value of 𝑥, 𝑓(𝑥_𝑛) is the function evaluated at 𝑥_𝑛, and 𝑓'(𝑥_𝑛) is the derivative of the function evaluated at 𝑥_𝑛.
In the case of 𝑓(𝑥) = 𝑥^3 - 9, the derivative is:
𝑓'(𝑥) = 3𝑥^2
Therefore, the formula for 𝑥_(𝑛+1) would be:
𝑥_(𝑛+1) = 𝑥_𝑛 - (𝑥_𝑛^3 - 9)/(3𝑥_𝑛^2)
This formula allows us to iteratively improve the guess for the cube root of 9 until we obtain the desired level of precision.