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Function f(a)=a

  1. Mar 27, 2007 #1
    I find it interesting that some functions have specific values of a where f(a)=a.
    Is there a way to find x that satisfies these following equations?:

    e^-x=x or -log(x)=x
    cos(x)=x

    They seem like they should be interesting values, and I was trying to figure out if they could be represented using pi, e, or what not. Also, sin(x)=x has only the trivial value of 0 while tan(x)=x has infinitley many solutions.
     
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  3. Mar 27, 2007 #2

    JasonRox

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    That's actually harder than it looks because you have a algebraic equation on one said and a exponential or trigonometric function on the other. Generally, they aren't solvable unless it is trivial, like you pointed out for sin(x).
     
  4. Mar 27, 2007 #3

    Integral

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    I believe you are referring to what is called a fixed point of the function. If the derivative of the function is <1 around the fixed point then you may be able to find the fixed point by iteration. Make a guess, compute the functional value, repeat with the generated value as your new x.
     
  5. Mar 28, 2007 #4
    for x=e^-x

    Defn. y=e^-x - x

    Obj. find where y(x)=0
    Use Newton Raphson's method

    x(n+1)=x(n)+(e^-x - x)/(e^-x +1)

    x(0)=1, it then goes:
    1 0.5378828427399903
    2 0.5669869914054132
    3 0.5671432859891229
    4 0.5671432904097839
    5 0.5671432904097839

    Which can be confirmed by substitution e^-x=x
     
  6. Mar 28, 2007 #5
    For cos(x)
    x(n+1)=x+(cos(x)-x)/(sin(x)+1)

    x(0)=1
    1 0.7503638678402439
    2 0.7391128909113617
    3 0.7390851333852839
    4 0.7390851332151607
    5 0.7390851332151607
    6 0.7390851332151607
     
  7. Mar 28, 2007 #6

    Integral

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    Or as a simple fixed point iteration, its slower but a better demonstration of the idea of a fixed point

    let [tex] x_{n+1} = e^{-x_{n}} [/tex]

    let [tex] x_0 = 1 [/tex]

    1.000000000
    0.367879441
    0.692200628
    0.500473501
    0.606243535
    0.545395786
    0.579612336
    0.560115461
    0.571143115
    0.564879347
    0.568428725
    0.566414733
    0.567556637
    0.566908912
    0.567276232
    0.567067898
    0.567186050
    0.567119040
    0.567157044
    0.567135490
    0.567147714


    now for the cos

    let [tex] x_{n+1} = cos(x_n) [/tex]

    [tex] x_0 = 1 [/tex]

    1
    0.540302306
    0.857553216
    0.65428979
    0.793480359
    0.701368774
    0.763959683
    0.722102425
    0.750417762
    0.731404042
    0.744237355
    0.73560474
    0.741425087
    0.737506891
    0.740147336
    0.738369204
     
  8. Mar 28, 2007 #7
    there is another way to do that. If u can sketch the graphs of two functions (using a graph software,etc.) u can find the point where the 2 curves meet.
     
  9. Mar 28, 2007 #8
    Thank you for telling me the name for this kind of math. I have now gone to my college library and checked out a book on Fixed Point Theory.
     
  10. Mar 28, 2007 #9
    The Newton-Raphson method is a special kind of fixed point iteration as well.

    Fixed points in practice are used to find the roots of a function/equation. Something like 2x+cos(x)sin(x)exp(-x/2)=0. It's easy to see that if the function g(x)=-cos(x)sin(x)exp(-x/2)/2 has a fixed point then this fixed point is the root of the original equation. Using theorems on fixed points we might prove the existence of such a point.
     
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