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Convergence of a Recursive Sequence

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations



    3. 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!
     
    Last edited: Oct 31, 2013
  2. jcsd
  3. Oct 31, 2013 #2
    What's the question?
     
  4. Oct 31, 2013 #3
    oh sorry your supposed to show if it converges or not.
     
  5. Oct 31, 2013 #4
    If the x_n converges to a limit L what happens to x_n+1 if n->inf?
     
  6. Oct 31, 2013 #5
    I'm not sure, does it also approach L?
     
  7. Oct 31, 2013 #6

    Ray Vickson

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    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
    [tex] f(x) = x - \frac{\tan(x)-1}{\sec^2(x)}[/tex]
    is meaningless at ##x = \pi/2, 3\pi/2, \ldots## where both the numerator and the denominator are infinite, but the algebraically equivalent form
    [tex] f_1(x) = x - \sin(x) \cos(x) + \cos^2(x) [/tex]
    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
    [tex] \lim_{x \to p} f(x) = \lim_{x \to p} f_1(x) = f_1(p)[/tex]
    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?
     
    Last edited: Oct 31, 2013
  8. Oct 31, 2013 #7
    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.
     
  9. Oct 31, 2013 #8
    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?
     
  10. Oct 31, 2013 #9

    Ray Vickson

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    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.
     
  11. Oct 31, 2013 #10
    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?
     
  12. Oct 31, 2013 #11

    Ray Vickson

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    Use a calculator, or a computer package. You can even use an on-line facility such as Wolfram Alpha.
     
  13. Oct 31, 2013 #12
    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.
     
  14. Oct 31, 2013 #13
    Sorry, that is the answer, much thanks for the help!
     
  15. Oct 31, 2013 #14

    Ray Vickson

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    NO, that is not the case. Newton's method seeks a solution of an equation g(x) = 0; it is g that needs to go to zero, NOT x! For example, if you want to compute √2 you can use Newton's method on the function g(x) = x^2 - 2. You will find that the x_n converge to 1.414213562... , and that g(x_n) → 0, and this is exactly what you need.

    As for your problem, I just suggested you do computations in order to get a "feel" for what is happening. Whether or not that is an acceptable solution to your problem is something your instructor will decide. Perhaps he/she wants an actual proof of convergence, in which case the numbers themselves would not be enough.
     
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