Question

question

Answer 1

Use newtons method to approximate

\[\sqrt{16.2}\]

Answer 2

@mathmate

Answer 3

Newtons methos is used to find the solution of

x^2 - S = 0 where in this case S = 16.2

Answer 4

i'm a little unsure. how did you get that? don't we have to approximate the value of (16.2)^0.5?

Answer 5

we solve this for x by starting with an approximation and performing iterative steps

here it reduces to


x (n + 1) = 1/2 (x (n) + 16.2/x(n) )

Answer 6

suppose we take 4.1 as a first approximation

x n+1 = 1/2 * ( 4.1 * 16.2/4.1) = 4.0256

now replace x(n) by 4.256 and repeat until you get the required accuracy

Answer 7

finding the square root of S is the same as solving x^2 - S = 0

Answer 8

This is what I have in my book

\[f(x+\Delta~x) ~~~ f(x)+f'(x)*\Delta~X\]

Answer 9

I'm wondering if someone could explain this? would delta x be f'(x)

Answer 10

@mathmate

Answer 11

delta x means a small increment of x

Answer 12

If you go to the basis of the method, it's basically derived from a truncated Taylor's expansion,
so
\(y=f(x_n)+f'(x_n)(x_{n+1}-x_n)\)
where \(x_n\) is the current approximation, and \(x_{n+1}\) the next one.
now set y=0 (solution of f(x)=0)
\(0=f(x_n)+f'(x_n)(x-x_n)\)
transpose, and solve for \(x_{n+1}\) to get
\(x_{n+1}=x_n-f(x_n)/f'(x_n)\)
Nice thing about Newton's method is that it converges quadratically, i.e. the number of accurate digits doubles after each iteration.
Now the drawback:
1. It does not always converge, because there are criteria for convergence.
2. The starting point is critical because it affects \(which|) solution we're going to get. Graphing will give us a good hint as to where to start, if we start with a point close to a solution.

Answer 13

Try square-root of 5, i.e. solving
f(x)=x^2-5=0
we will choose initial value as x0=2 (close enough), and calculate f'(x)=2x.
So we will "iterate" using the equation
xn+1=xn-f(xn)/f'(xn)
x1=2-(2^2-5)/(2(2))=2.25
x2=2.25-(2.25^2-5)/(2(2.25))=2.236111111111111
x3=2.236067977915804
x4=2.23606797749979
(sqrt(5)=2.23606797749979, accurate to the last digit)
as you can see, the accuracy improves very quickly.

Answer 14

Another drawback of Newton's method is that the function must be differentiable in the vicinity of the solution. Extra work is required to calculate the derivative as well.

Answer 15

Converging means? what like approaching a certain number?