0

Consider the function $f(x)=\arctan(x)$ defined on $\mathbb{R}$. Let $x_0\neq0$ be an initial guess for the root of $f$ and apply Newton's Method with $x_0$. Further assume that the iterations produce an oscillating sequence $(x_n)_n=(x_0,x_1,x_0,x_1,...)$ with period $2$ and $x_0,x_1$ as repeating terms.

Problem: How to prove that $x_0$ and $x_1$ are roots of the function $g(x):=\frac{2x}{x^2+1}-f(x)$ ?

Raffaele
  • 26,371
Silver Pine
  • 113

1 Answers1

0

As $g(x)$ is an odd function, if $x_0$ is a root, then $x_1=-x_0$ is a root as well.

The periodic sequence is

$(x_n)=x_0,-x_0,x_0,\ldots$

Apply Newton method to find the root of $f(x)=\arctan x$

$x_1=x_{0}-\frac{f(x_{0})}{f'(x_{0}}= -x_0$

$x_1=-x_0$ so we get

$-x_0=x_0-\frac{\arctan x_0}{\frac{1}{x_0^2+1}}$

expand

$2 x_0=\left(x_0^2+1\right) \arctan x_0$

Which proves that $\pm x_0$ are roots of $g(x)=\frac{2 x}{x^2+1}-\arctan x$

$x_0\approx 1.39175$

Hope this helps

Edit

They say

$x_0,x_1$ form a periodic sequence when used in Newton method to find roots of $\arctan x$. Show that exactly those two values are roots of $g(x)$.

I say: if they are roots of $g(x)$ then $x_1=-x_0$ because $g(x)$ is odd (as shown in the image below). Then I prove that those values are exactly the values found in the first part.

$$...$$

enter image description here

Raffaele
  • 26,371
  • How do I know that $x_1=-x_0$? – Silver Pine Jan 14 '21 at 10:18
  • As $\frac{2 x}{x^2+1}-\arctan x$ is odd then must be $x_1=-x_0$. @SilverPine – Raffaele Jan 14 '21 at 10:30
  • Yes I agree that the function is odd, but how does that imply $x_1=-x_0$? I mean I only know that the sequence is oscillating between the two points $x_0$ and $x_1$, but how to show that $x_1=-x_0$? – Silver Pine Jan 14 '21 at 10:44
  • Because they say that $x_0$ and $x_1$ are both roots of $g(x)$ @SilverPine – Raffaele Jan 14 '21 at 12:45
  • Well, that is what I want to prove, so it's not given yet... – Silver Pine Jan 14 '21 at 12:49
  • @SilverPine I edited my answer adding some details. You know as hypothesis that we are looking for the roots of $g(x)$. If we were speaking of another function, then the claim would be false. – Raffaele Jan 14 '21 at 12:54
  • I see. Well your argumentation still assumes $x_1=-x_0$ at the very beginning. Now my question is, why? I mean, surely, if $x_1=-x_0$, then your proof works good, but how do I prove that $x_1=-x_0$, given the assumptions I presented in the problem? – Silver Pine Jan 14 '21 at 13:07
  • @Silverpine Isn't it strange that I am the ONLY one who answered and commented on this question? Viewed 29 times but no one said a word in four hours: this is unusual AF! – Raffaele Jan 14 '21 at 13:12
  • Yes I agree! I really appreciate your effort in trying to help me, but somehow there seems to be a missing block in the argumentation. Maybe my problem/question actually doesn't make sense, but currently I see it as sensible. Do you understand my issue? – Silver Pine Jan 14 '21 at 13:17
  • @SilverPine I can't help you more than this. I hope that someone with more reputation than me will comment on this question. What is the origin? Is it a homework or what? – Raffaele Jan 14 '21 at 13:25