Calculus – Why Does the Newton-Raphson Method Not Converge for Some Functions?

calculusnumerical methods

$f(x)=2x^2-x^3-2$. This is a cubic type graph as shown. The real root of this graph is $(-0.839,0)$.

So, the question is to use Newton's approximation method twice to approximate a solution to this $f(x)$.

I use an initial starting value of $x_0=1$.
My first approximation is $x_1=2$, and my second one is $x_2=1.5$.
I seem to not move any closer to the real solution as I keep iterating through the method.

Am I misunderstanding how to use this approximation? Is the issue that my first guess was too far from the actual solution?

Best Answer

Newton's method does not always converge. Its convergence theory is for "local" convergence which means you should start close to the root, where "close" is relative to the function you're dealing with. Far away from the root you can have highly nontrivial dynamics.

One qualitative property is that, in the 1D case, you should not have an extremum between the root you want and your initial guess. If you have an odd number of extrema in the way, then you will start going away from the root you want, as you see here. If you have an even number of extrema in the way, then you will start going the right way, but you may later find yourself in a spot with an odd number of extrema in the way, leading to problems later.

Of course you may eventually find an occasion where there are an even number of extrema in the way, and then you manage to skip over all of them and get to the right side. At that point things will usually work out (not always, though). In this problem with your initial guess, that eventually happens, because the system eventually finds its way just slightly to the right of the extremum on the right, which sends it far off to the left.

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[Math] Trouble Understanding Newton-Raphson Iterative Method

Suppose you want to compute $\sqrt{2}$, so you use $f(x) = x^2-2$ and start with $x_1=1$. Say you want 50 places. Here we go:

x[ 1] = 1.00000000000000000000000000000000000000000000000000 
x[ 2] = 1.50000000000000000000000000000000000000000000000000 
x[ 3] = 1.41666666666666666666666666666666666666666666666667 
x[ 4] = 1.41421568627450980392156862745098039215686274509804 
x[ 5] = 1.41421356237468991062629557889013491011655962211574 
x[ 6] = 1.41421356237309504880168962350253024361498192577620 
x[ 7] = 1.41421356237309504880168872420969807856967187537723 
x[ 8] = 1.41421356237309504880168872420969807856967187537695 
x[ 9] = 1.41421356237309504880168872420969807856967187537695 
x[10] = 1.41421356237309504880168872420969807856967187537695 
x[11] = 1.41421356237309504880168872420969807856967187537695 
x[12] = 1.41421356237309504880168872420969807856967187537695 
x[13] = 1.41421356237309504880168872420969807856967187537695 
x[14] = 1.41421356237309504880168872420969807856967187537695 
x[15] = 1.41421356237309504880168872420969807856967187537695

So presumably we can stop at $x_8$.

[Math] How to correctly apply Newton-Raphson method to Backward Euler method

Your aim is to solve the following equation for $y_{k+1}$: $$ g(y_{k+1}):=y_{k+1}-y_k-hf(t_{k+1},y_{k+1})=0,$$ where $f$ is a known function and $y_k, t_{k+1}$ and $h$ are known values. This gives you the $g$ for a Newton-Raphson method. As an initial guess, I'd suggest that you use a 1-step forward Euler method to explicitly calculate $$ \hat{y}_{k+1} := y_k+hf(t_{k+1},y_k),$$ and use $\hat{y}_{k+1}$ as your initial guess for $y_{k+1}$.

Best Answer

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[Math] Trouble Understanding Newton-Raphson Iterative Method

[Math] How to correctly apply Newton-Raphson method to Backward Euler method

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