# Convergence of a Recursive Sequence

## Homework Statement

The following sequence comes from the recursion formula for Newton's Method.
x0= 1 , xn+1=xn-(tanxn-1)/sec2xn
Show if the sequence converges or diverge.

## The Attempt at a Solution

I don't really know where to start on this problem, I have tried to use some trig identities to no avail, but I don't know what else to do. Any help is appreciated, thanks!

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What's the question?

oh sorry your supposed to show if it converges or not.

If the x_n converges to a limit L what happens to x_n+1 if n->inf?

I'm not sure, does it also approach L?

Ray Vickson
Homework Helper
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## Homework Statement

The following sequence comes from the recursion formula for Newton's Method.
x0= 1 , xn+1=xn-(tanxn-1)/sec2xn

## The Attempt at a Solution

I don't really know where to start on this problem, I have tried to use some trig identities to no avail, but I don't know what else to do. Any help is appreciated, thanks!
What is stopping you from computing a few of the initial x_n values, say for n = 1, 2, 3,... ? This might offer some insight.

One possible problem I can see is that your initial definition
$$f(x) = x - \frac{\tan(x)-1}{\sec^2(x)}$$
is meaningless at ##x = \pi/2, 3\pi/2, \ldots## where both the numerator and the denominator are infinite, but the algebraically equivalent form
$$f_1(x) = x - \sin(x) \cos(x) + \cos^2(x)$$
has no problems at any values of x. In other words, while your original f(x) is not defined at some x-values, we have ##f(x) = f_1(x)## at ##x \neq \pi/2, 3\pi/2, \ldots## and
$$\lim_{x \to p} f(x) = \lim_{x \to p} f_1(x) = f_1(p)$$
holds for ##p = \pi/2, 3\pi/2, \ldots ##. In other words, f(x) has "removable" discontinuities, and can thus be replace by f_1(x) in the optimization algorithm. Is that the source of your problems?

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Also, how could I prove that x_n converges to a limit L? Thanks for responding, I really want to get this problem, I just can't seem to get it on my own.

What are you attempting to do? What is stopping you from computing a few of the initial x_n values, say for n = 1, 2, 3,... .

One possible problem I can see is that your initial definition
$$f(x) = x - \frac{\tan(x)-1}{\sec^2(x)}$$
is meaningless at ##x = \pi/2, 3\pi/2, \ldots## where both the numerator and the denominator are infinite, but the algebraically equivalent form
$$f_1(x) = x - \sin(x) \cos(x) + \cos^2(x)$$
has no problems at any values of x. In other words, while your original f(x) is not defined at some x-values, we have ##f(x) = f_1(x)## at ##x \neq \pi/2, 3\pi/2, \ldots## and
$$\lim_{x \to p} f(x) = \lim_{x \to p} f_1(x) = f_1(p)$$
holds for ##p = \pi/2, 3\pi/2, \ldots ##. In other words, f(x) has "removable" discontinuities, and can thus be replace by f_1(x) in the optimization algorithm. Is that the source of your problems?
I need to find what it converges to, so I believe that entails taking the lim as x -> ∞ but the lim of any trig function as it goes to infinity is undefined. Is that what I'm supposed to do?

Ray Vickson
Homework Helper
Dearly Missed
I need to find what it converges to, so I believe that entails taking the lim as x -> ∞ but the lim of any trig function as it goes to infinity is undefined. Is that what I'm supposed to do?
No, that is not what you were asked about. You were asked to look at the sequence ##x_0 = 1, \: x_{n+1} = f(x_n)## for a given function f(x). You were asked if the sequence ##x_0, x_1, x_2, \ldots## is convergent.

No, that is not what you were asked about. You were asked to look at the sequence ##x_0 = 1, \: x_{n+1} = f(x_n)## for a given function f(x). You were asked if the sequence ##x_0, x_1, x_2, \ldots## is convergent.
So when I try to evaluate x1 I get x1=x0+1=x0-sin(x0)cos(x0)+cos2(x0)=(1)-sin(1)cos(1)+cos2(1). How can you evaluate this?

Ray Vickson
Homework Helper
Dearly Missed
So when I try to evaluate x1 I get x1=x0+1=x0-sin(x0)cos(x0)+cos2(x0)=(1)-sin(1)cos(1)+cos2(1). How can you evaluate this?
Use a calculator, or a computer package. You can even use an on-line facility such as Wolfram Alpha.

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Use a calculator, or a computer package. You can even use an on-line facility such as Wolfram Alpha.
Ok, I used a calculator to continue to find the next number in the sequence about 4 or 5 inputs and I start to get .7853981634... every time. This should then be the number the sequence converges to, the only problem is that the initial equation was of the form of Newton's method, which is used to find x intercepts (roots/zeros). Therefore this sequence HAS to converge to zero, but I should be able to prove this.

Sorry, that is the answer, much thanks for the help!

Ray Vickson