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Simple curve - not so simple function?

  1. May 25, 2014 #1
    What is f(x) when eg

    x = 0, 1, 2, 3, 4, 5, 6, etc
    y = 0, 8, 12, 14, 15, 15.5, 15.75 etc (the y value at x =1 is arbitrary)

    ie each successive y value adds half the difference of the preceding 2 values.

    (It is effectively the inverse of a first order exponential decay where the y value halves at constant x intervals).

    But it doesn’t fit an exponential ....! Could it be a type of hyperbola? or polynomial? or both? or something else entirely?

    I have no decent curve fitting software and struggling to find this.
    Thank you!
     
  2. jcsd
  3. May 25, 2014 #2

    adjacent

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    Gold Member

    Try drawing it on a graph paper.
     
  4. May 25, 2014 #3
    Thanks ... but how would that help, if I've already tried to fit it in Excel?
     
  5. May 25, 2014 #4

    adjacent

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    You can try finding the nth term of the sequence(That's what I do).
     
  6. May 25, 2014 #5

    HallsofIvy

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    So you are saying that [itex]x_{n+2}= x_{n+1}+ (x_n+ x_{n+1})/2= (3/2)x_{n+1}+ (1/2)x_n[/itex]

    That's a "second order difference equation" which has associated characteristic equation [itex]s^2= (3/2)s+ 1/2[/itex] or [itex]s^2- (3/2)s- 1/2= 0[/itex]. The solutions to that equation are [itex]s= \frac{3\pm \sqrt{17
    }}{4}[/itex]. That means that the general solution to the difference equation is [itex]8^{n/4}(C2^{\sqrt{17}n/4}+ D2^{-\sqrt{17}n/4})[/itex]
     
    Last edited: May 25, 2014
  7. May 25, 2014 #6

    micromass

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    So you have a sequence defined by

    [tex]x_0 = 0,~x_1 = 8,~x_{n+1} = x_n + \frac{1}{2}(x_n - x_{n-1}) = \frac{3}{2}x_n - \frac{1}{2}x_{n-1}[/tex]

    Let's find a general term. We can express this sequence as follows:

    [tex]\left(
    \begin{array}{cc}
    x_{n+1}\\ x_n
    \end{array}
    \right)
    =
    \left(
    \begin{array}{cc}
    \frac{3}{2} & -\frac{1}{2}\\ 1 & 0
    \end{array}
    \right)
    \left(
    \begin{array}{cc}
    x_{n}\\ x_{n-1}
    \end{array}
    \right)
    [/tex]

    So, we get

    [tex]\left(
    \begin{array}{cc}
    x_{n+1}\\ x_n
    \end{array}
    \right)
    =
    \left(
    \begin{array}{cc}
    \frac{3}{2} & -\frac{1}{2}\\ 1 & 0
    \end{array}
    \right)^n
    \left(
    \begin{array}{cc}
    x_1\\ x_0
    \end{array}
    \right)
    [/tex]

    By using the diagonalization theory of linear algebra, we can write this as

    [tex]\left(
    \begin{array}{cc}
    x_{n+1}\\ x_n
    \end{array}
    \right)
    =
    \left(
    \begin{array}{cc}
    \frac{1}{2} & 1\\ 1 & 1
    \end{array}
    \right)
    \left(
    \begin{array}{cc}
    \frac{1}{2^n} & 0\\ 0 & 1
    \end{array}
    \right)
    \left(
    \begin{array}{cc}
    -2 & 2\\ 2 & -1
    \end{array}
    \right)
    \left(
    \begin{array}{cc}
    x_1\\ x_0
    \end{array}
    \right)
    [/tex]

    and thus

    [tex]\left(
    \begin{array}{cc}
    x_{n+1}\\ x_n
    \end{array}
    \right)
    =
    \left(
    \begin{array}{cc}
    2 - \frac{1}{2^n} & \frac{1}{2^n}- 1\\ 2 - \frac{1}{2^{n-1}} & \frac{1}{2^{n-1}}-1
    \end{array}
    \right)
    \left(
    \begin{array}{cc}
    x_1\\ x_0
    \end{array}
    \right)
    [/tex]

    Thus we get

    [tex]x_{n+1} = 2x_1 - \frac{x_1}{2^n} + \frac{x_0}{2^n} - x_0[/tex]

    Since ##x_1 = 8## and ##x_0 = 0##, we get

    [tex]x_{n+1} = 16 - \frac{8}{2^n}[/tex]
     
  8. May 25, 2014 #7
    Thanks both! I'm not a mathematician ... just a chemist struggling to make sense of some data!

    So Hallsofivy, is a "second order difference equation" the same thing as a second order polynomial? I tried to fit the data to a polynomial and it required 6th order before R2= 1 in Excel! (And not sure how you got from the series to the associated characteristic equation)

    I see you guys have both written it as a series, but can it be expressed in terms of y=f(x)? And described as eg a power function or polynomial etc?
     
  9. May 25, 2014 #8

    PeroK

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    Homework Helper
    Gold Member

    You can manipulate Micromass's approach to get, for example:

    [tex]f(x) = 16 - 2^{4-x}[/tex]

    And, if f(1) = a, then:

    [tex]f(x) = a(2 - 2^{1-x}) = 2a(1 - \frac{1}{2^x})[/tex]
     
    Last edited: May 25, 2014
  10. May 25, 2014 #9
    .... building on micromass's reply .... something like y = 2a -(a/2^x)? where a is an arbitrary constant. Is it a hyperbolic function??! Or a hyperbolic polynomial! does that exist?
     
  11. May 25, 2014 #10
    Thanks Perok ... I was sending mine while yours was arriving - I was close! So does the function fit one of those categories?
     
  12. May 25, 2014 #11

    PeroK

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    I would say it's just a variation of a power function (powers of 1/2).
     
  13. May 25, 2014 #12
    OK - a bit weird though, that it fits precisely to polynomial order 6! Coincidence?
     
  14. May 25, 2014 #13

    PeroK

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    Any set of n values will fit a polynomial of order n-1.
     
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