MHB What conditions guarantee convergence of Newton's method for approximating pi/2?

[mathmari]

Hey!

I have calculated an approximation to $\frac{\pi}{2}$ using Newton's method on $f(x)=\cos (x)$ with starting value $1$. After 2 iterations we get $1,5707$.
Which conditions does the starting point has to satisfy so that the convergence of the sequence of the Newton iterations to $\frac{\pi}{2}$ is guaranteed? :unsure:

 

[I like Serena]

Suppose we start at a point $x_0$ with $0<x_0<\frac\pi 2$. Then the first iteration $x_1$ will be greater than $\frac\pi 2$.
We need that $x_1$ is closer to $\frac\pi 2$ than $x_0$. And if it is, it will be convergent, won't it?

 

[mathmari]

Suppose we start at a point $x_0$ with $0<x_0<\frac\pi 2$. Then the first iteration $x_1$ will be greater than $\frac\pi 2$.

Why will $x_1$ be greater than $\frac\pi 2$ ? :unsure:

We need that $x_1$ is closer to $\frac\pi 2$ than $x_0$. And if it is, it will be convergent, won't it?

We need that because that would mean that the sequence $(x_i)$ converges to $\frac{\pi}{2}$, or not? :unsure:

 

[I like Serena]

Why will $x_1$ be greater than $\frac\pi 2$ ?

Geometrically, the method takes the intersection of the tangent line with the x-axis.
Since $\cos$ has a decreasing slope on $(0,\frac\pi 2)$, that intersection point must be on the other side of $\frac\pi 2$.

Alternatively, we can look at it algebraically.
wiki explains that we can write the successive errors as:
$$\varepsilon_{n+1} = \frac {- f'' (\xi_n)}{2 f'(x_n)} \cdot {\varepsilon_n}^2$$
where $x_n$ are the successive approximations of a root $a$, $\varepsilon_n=a-x_n$ is the error in that approximation, and $\xi_n$ is a point between $x_n$ and the root $a$.

So if $\varepsilon_1$ is negative, we have that $x_1$ is on the right side of the root. Can we deduce that?

We need that because that would mean that the sequence $(x_i)$ converges to $\frac{\pi}{2}$, or not?

We're trying to find which starting value we need so that the sequence converges.
We get that if the remaining error converges to zero, don't we?

 

[mathmari]

Alternatively, we can look at it algebraically.
wiki explains that we can write the succesive errors as:
$$\varepsilon_{n+1} = \frac {- f'' (\xi_n)}{2 f'(x_n)} \cdot {\varepsilon_n}^2$$
where $x_n$ are the successive approximations of a root $a$, $\varepsilon_n=a-x_n$ is the error in that approximation, and $\xi_n$ is a point between $x_n$ and the root $a$.

So if $\varepsilon_1$ is negative, we have that $x_1$ is on the right side of the root. Can we deduce that?

We have that
$$\varepsilon_{1} = \frac {- f'' (\xi_0)}{2 f'(x_0)} \cdot {\varepsilon_0}^2= \frac {\cos (\xi_0)}{-2 \sin (1)} \cdot \left (\frac{\pi}{2}-1\right )^2 \approx \frac {\cos (\xi_0)}{-1.6829} \cdot 0.3258 = -0.1936 \cdot \cos (\xi_0)$$
So it is negative when $ \cos (\xi_0)$ is positive, right? :unsure:

 

[I like Serena]

We have that
$$\varepsilon_{1} = \frac {- f'' (\xi_0)}{2 f'(x_0)} \cdot {\varepsilon_0}^2= \frac {\cos (\xi_0)}{-2 \sin (1)} \cdot \left (\frac{\pi}{2}-1\right )^2 \approx \frac {\cos (\xi_0)}{-1.6829} \cdot 0.3258 = -0.1936 \cdot \cos (\xi_0)$$
So it is negative when $ \cos (\xi_0)$ is positive, right?

You have picked $x_0=1$ in which case $\xi_0$ must be between $1$ and $\frac\pi 2$. So for each possible value of $\xi_0$ we have indeed that $\cos (\xi_0)$ is positive.
But aren't we trying the find the $x_0$ that is 'just' sufficient to get convergence?
We have already found that for $x_0=1$ we get convergence. So the $x_0$ that we are trying to find should be smaller, shouldn't it?

 

[mathmari]

You have picked $x_0=1$ in which case $\xi_0$ must be between $1$ and $\frac\pi 2$. So for each possible value of $\xi_0$ we have indeed that $\cos (\xi_0)$ is positive.
But aren't we trying the find the $x_0$ that is 'just' sufficient to get convergence?
We have already found that for $x_0=1$ we get convergence. So the $x_0$ that we are trying to find should be smaller, shouldn't it?

So we want to find the boundary ofall possible $x_0$'s to get convergence, right?

Do we have to use for that the formula for the remaining error?

:unsure:

 

[I like Serena]

That's how I interpret the OP.

The formula is for the general case. In our case it should suffice if we solve for the case that after the first iteration we have the same error.

 

[mathmari]

The formula is for the general case. In our case it should suffice if we solve for the case that after the first iteration we have the same error.

I got stuck right now.Why does the error have to be the same as the previous error? :unsure:

 

[I like Serena]

I got stuck right now.Why does the error have to be the same as the previous error?

Then we should have the critical initial starting value.
A bigger starting value (that is less than $\frac\pi 2$) should guarantee convergence shouldn't it?

 

[mathmari]

A bigger starting value (that is less than $\frac\pi 2$) should guarantee convergence shouldn't it?

Yes. The nearer the starting value is to $\frac{\pi}{2}$ the faster it converges, right? :unsure: