How to find initial guess for Newton's Method?

Much depends on what you are doing, on the specific problem. If you have a choice, a good method is often a graphical method. Plot the function. Look for where it crosses the x axis. I'd bet that gives you a good starting value.

On other problems you won't be given such an option. Finding an automatic good choice of starting value is a terribly difficult thing for a completely general function. On one variable problems, a good solution is to sample the function at a sequence of points. Look for a pair of points that bracket a root. That is, find x_1 and x_2 such that f(x_1) and f(x_2) have different signs. This will often provide a starting point, although it does not ensure that fact. For example, find a root of f(x) = 1/x. But while f(-1) and f(1) have opposite signs, there is no root that lies in the interval [-1,1].

Newton's method will converge on the particular problem you have posed for almost any guess (except zero.) So there may be little reason to find a good stating point, as it will converge rapidly almost no matter what. On other problems, the iterative scheme will diverge for bad starting values. So an important thing to understand is the basin of convergence. This is the set of points that will converge to a given solution, given that they are used as starting values for a given solver. What is the basin of convergence for Newton's method on this problem? There are two roots (solutions) to this problem. See that Newton will converge to sqrt(2) for any finite value in the interval (0,inf), and converge to -sqrt(2) for any finite starting point in the interval (-inf,0).

Ok, so the above may not really help us much, since the objective is to choose an intelligent guess. Sometimes you may be able to find an approximation for your function, one that is easily inverted. This may be based on a low order (truncated) series for your function.

Finally, lets look more carefully at the specific problem posed. Find a good starting point to solve for a root of:

f(x) = x^2 - 2 =0.

Note that for small x (any starting point that is less than sqrt(2)), the first iteration will drive out a long way, and then be forced to move inwards.

Writing out the iterative scheme for Newton's method, we see it looks like

x(n+1) = x(n) - f(x(n))/f'(x(n) = x(n) - (x(n)^2 - 2)/(2*x(n))

I'll do this in MATLAB. Starting at x = 0.1, we see the iteration immediately jump far out, then

>> x = 0.1;

>> x = x - (x^2 - 2)/(2*x)

x =

10.05

>> x = x - (x^2 - 2)/(2*x)

x =

5.12450248756219

>> x = x - (x^2 - 2)/(2*x)

x =

2.75739213841957

>> x = x - (x^2 - 2)/(2*x)

x =

1.74135758044959

>> x = x - (x^2 - 2)/(2*x)

x =

1.44494338195892

>> x = x - (x^2 - 2)/(2*x)

x =

1.41454033012869

>> x = x - (x^2 - 2)/(2*x)

x =

1.4142136001158

>> x = x - (x^2 - 2)/(2*x)

x =

1.4142135623731

>> x = x - (x^2 - 2)/(2*x)

x =

1.4142135623731

Newton's method here is quadratically convergent. In fact, it has converged here to sqrt(2) as we should expect. All of the digits displayed are in agreement with the known solution.

>> sqrt(2)

ans =

1.4142135623731

In fact, see that for any starting point greater than the root here, Newton's method will converge quickly. So while the first point was a terrible choice, it moved immediately into a realm where the iteration became one that is monotonically convergent.

Finally, you need to understand that on some problems, Newton's method is not so quickly convergent. For example, find a root for the function f(x) = x^4 using Newton iterations. Newton's method iterations will reduce to

x(n+1) = 3/4*x(n)

While these iterations will converge to x = 0, it is not a rapid convergence.

The point of this long discussion is to recognize that every problem is different, and that you will gain greatly by understanding your problem, as well as the numerical method employed for its solution.

Newton's method for what? Finding zeros? As humans, the best way is probably a combination of graphical and theoretical analysis. You just have to make some inferences about each function to decide how many roots there are and where they should be, roughly. If you're writing a computer program, you need something automatic, so you'd have to come up with some sort of search algorithm. With x^2 - 7, it doesn't matter where you start. It will converge from anywhere. This is not true in general so you do have to try to get "close". For square-root finding, I've done things like start with n/2 (you want the square root of 7, so start with 3.5). For x^3 - cos x, I'd go through the following theoretical analysis: the cosine term is always between -1 and +1. So for all x large enough (>1 actually), I know f(x) will be >0, and for all x < -1, f(x) will be < 0. So any roots have to lie in [-1, 1]. But if you plot the function, you'll see exactly how many roots there are and where they should be (roughly).

Hi :

It one of of my most favorite time waster way of sloving a problem but it does work

the way I do this is find the the closest number to it with out going over

Make a guess number that is the closest number to it with out going over

Step 1 :

solve the equation and see if your too high

if you are not

than add one unit to the guess number

Repeat Step1

- than take it down one unit

add 5/(10^next decimal place number)

Is it to how many decimal places to where you want to go to?

if yes - Stop

Repeat Step 1

Your problem for example

closest root to 2 with going over 1 so

1^2 - 2 = 0

1 -2 = 0

-1 = 0 - too low

try 2

2^2- 2 = 0

4 - 2 = 0 too high

back to 1

add .5

Now try 1 .5

1.5^2 - 2 = 0

2.25 - 2 = 0

.25 = 0 - too high but closer to it

so your answer must be somewhere between 1 and 1.5

try 1.2

1.2^2 - 2 = 0

1.44 - 2 = 0

-1.56 = 0 - too low

try

1.4

1.4^2

1.96 - 2 = ..04 too high but really close

so the answer must be between 1.4 and 1.5

add .05 to 1.4

1.45^-2 = 0

2.1025- 2 = 0.1025 - too high

try 1.42^2 - 2 = 0

2.0164- 2 = .0164 still too high but closer to it

try 1.41

1.9881 - 2 = -0.0119 way too low but since 1.42 is too high

the answer must be 1.41 accurate to 2 decimal place

and .005 to

and keep doing it

and if you keep doing it

you will get 1.4142135623730950488016887242097 and -1.4142135623730950488016887242097 for your answer

Really, you just avoid the minimum, and close to where you think the answer is.

Don't use 0, and try around 1.4-1.5