Question

question

Answer 1

Newtons method

Using the following:

\[x_{n+1} = x_{0}-\frac{ f(x_{0}) }{ f'(x_{0}) }\]

approximate \[\sqrt{3}\]

Answer 2

using

\[f(x) = x^{2}-3 ; f'(x) = 2x \]

Answer 3

\[\frac{ x_{0}-f(x_{0}) }{ f'(x_{0}) } = x_{n+1}\]

Answer 4

@Preetha @mathmale I had another question too. it says for newtons method the approximations get more accurate, why is this so?

Answer 5

so that's newton's method for approximating roots of a poynomial..

so let

\[x = \sqrt{3}\]

now make an quadratic equation... out of it

Answer 6

and you need an initial estimate of the root for it to work..

Answer 7

so the root is (3)^0.5

Answer 8

so the value is between 1 and 4 so pick an initial value... and substitute it into the function and the derivative

Answer 9

okay so let's chose 1

Answer 10

actually the value is between 1 and 2 oops my mistake to choose 1.5

\[x_{1} = 1.5 - \frac{f(1.5)}{f'(1.5)}\]

Answer 11

so we're approximating the value by creating a range that is choosing a value that's between the number given

Answer 12

well in Newton's method you need to pick an inityal value

so pick values that the root will be between and check

so its between 1 and 2

so find f(1) and f(2) you should see a change in sign, indicating an root

f(1) = -2 f(2) = 1

so the root is closer to 2 than it is to 1...

a simple method is the pick the halfway value...

if you look at f(1) and f(2) ... then maybe x = 1.6 could be a better approximation

Answer 13

hope it makes sense

Answer 14

Like this?

\[x_{0}-\frac{ x^{2}-3 }{ 2x } => \frac{ \frac{ 3 }{ 2 }-(\frac{ 3 }{ 2 })^{2}-3 }{ 2*(\frac{ 3 }{ 2 }) }\]

Answer 15

It's probably more believable if you look at a graph of successive iterations maybe

Answer 16

that's correct... and I think that is about 1.75

if you use 1.6

\[x_{1} = 1.6 - \frac{f(1.6)}{f'(1.6)}\]

which is probably a better approximation after 1 interation

Answer 17

so there is a little arithmetic and a little intuition in getting the initial value for the estimate if it isn't given to you

Answer 18

don't know why i'm getting 3/4

Answer 19

this is a high school type question where they only normally ask for 1 or 2 applications... so don't worry about graphing... just make some initial choices... test for a change in sign then get an initial guestimate

Answer 20

the numerator is -3/4 and the denominator is 3

\[x_{1} = 1.5 - \frac{(1.5)^2 - 3}{3} = 1.5 - \frac{-0.75}{3}\]

Answer 21

Like I get the fact that I'm using this to get an estimate of the value of the function

Answer 22

the numerator is

\[\frac{9}{4} - 3 = -\frac{3}{4}\]

Answer 23

I see, so what would determine whether the approximations are more accurate?

Answer 24

would we pick a smaller interval?

Answer 25

just substitution into the initial function

\[f(x) = x^2 - 3\]

so I know the root is between 1 and 2

and when you substitute x = 1 you get f(1) = -2

and substituting x = 2 f(2) = 1

so which value of f(x) is closer to zero...f(2)

which to me suggests the root is closer to x = 2 than x = 1

so pick a value closer to 2 maybe x = 1.75

this part is where you need to think about the answers that the substitutions give...

so if you did another iteration using x = 1.75

the solution from the 1st iteration you would find you are starting to get very close to the answer

Answer 26

hope it makes sense

Answer 27

@photon336

It seems that you have the following questions:
1. How do we choose the initial approximation so that it will converge to the answer we want, i.e. how to find the interval of convergence.
2. How do we know if we chose the right initial approximation (i.e. Newton's method will work).

The first question actually answers the second, but the answer to (1) is a little more involved (see answer later).

There is an easier way to tell if you have made the right choice, i.e. answer to question (2).

If you have read my previous post to one of your questions on the subject, you'd recall from the example that when it works, the method converges very fast, doubling the accuracy after each iteration.

However, when the initial approximation does not converge to the expected answer, two things can happen:
1. it quickly converges, but to a wrong number. If you know what answer to expect, you would know this is happening.
2. It diverges, i.e. numbers are flying around without converging. This is also easy to spot.

If you are in university doing a numerical analysis course, you will \(need\) the answer to the first question. I'll be glad to send you a document that gives the answer (requires elementary university mathematics to follow). If you are learning a new technique to solve non-linear equations of a single unknown, use graphing to find an approximate solution, and use Newton's to refine it. Or even make a guess to the answer and try it out.

For the case of square-root problem, usually bracketing between two integers should work very well.
Example: to find sqrt(5), we know 2^2=4<5, 3^2=9>5, so try 2.5 to start.