Estimate #sqrt(12)# using Newton-Rhapson?
3 Answers
Newton's method says that
We know that
1st iteration:
Now we iterate with
2nd iteration:
3rd iteration:
4th iteration:
Explanation:
This link here lists several methods to compute square roots: https://en.wikipedia.org/wiki/Methods_of_computing_square_roots
The digit-by-digit method is an extremely useful method to find exact answers but is extremely slow.
A faster method is to use Newton's method, but it can only be used to approximate square roots.
Newton's method states that the zeros of a function
This means that, if we have a guess
Take note of three things. First, our new guesses will always be slightly larger than the actual answer (except when the guess is exactly the answer, then the new guess would be equal to the previous guess). We can round our final answer down to account for this. Second, some guesses may not work (for example, if our first guess is
Let's first try finding the square root of
Now, find the square root of
The following list shows how the formula produces more accurate answers each iteration:
#x_0=3# (This is our initial guess)#x_1=3.5# #x_2=3.4642857142857144...# #x_3=3.4641016200294548...# #x_4=3.4641016151377544...#
After
The actual answer is
# sqrt(12) = 3.4641# to 4 dp via Newton-Rhapson
Explanation:
We want to find
# x = sqrt(12) => x^2=12 #
# :. x^2-12 = 0 #
Let:
# f(x) = x^2-12 #
Our aim is to solve
# { (x_1,=x_0), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :} #
Therefore we need the derivative:
# \ \ \ \ \ \ \f(x) = x^2-12 #
# :. f'(x) = 2x #
So our iterative formula is:
# { (x_1,=x_0), ( x_(n+1), = x_n - (x^2-12)/(2x) ) :} #
Then using excel working to 8dp with
We get convergence to
We could equally use a modern scientific graphing calculator as most new calculators have an "Ans" button that allows the last calculated result to be used as the input of an iterated expression.
And we conclude that the solution is (to 4dp):
# x = 3.4641#
The actual value via a calculator is:
# sqrt(12) =3.46410161 ... #