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Finding a function from its series equivalent

Discussion in 'Electronic Design' started by Don Lancaster, Mar 12, 2007.

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  1. It is usually a fairly simple matter to get from a trig or exponantial
    function of one sort or another to its underlying series form.

    But how can you get from a known accurate series expression to a
    nonobvious and crucially esoteric equivalent function?

    Specifically, the "raw" power series

    [-1517.83 5094.6 821.18 -29457.7 61718.9 -61268.8 30448.6 -4770.84
    -269.684 -2892.14 3300.63 -1460.88 213.578 78.8959 -49.2164 12.3083
    -1.74731 0.149743 -0.00245142 0.103691]

    where 0.103691 is the x^1 term, -0.00245142 is x^2 etc...

    The equivalent McLauran Series (or Taylor about zero) is found by
    dividing each term by its factorial. 0.103691/1! , -0.00245142/2!...

    ... may be of extreme interest in finding a closed form expression
    that involves trig products and possibly exponantials. The range of
    interest is from 0 to 1.

    The function appears continuous and monotonic with well behaved
    derivatives. There is no zero offset.

    The trig angle of 84.0000 degrees is also expected to play a major role
    in the solution. As is the trig identity of cos(a+b) = cos(a)cos(b) -
    sin(a)sin(b). As is a magic constant of 0.104528. Everything happens in
    the first quadrant.

    Sought after is a closed form determnistic solution that accepts the 0-1
    value, the 84 degree angle, and the magic constant that evaluates to the
    above series.

    --
    Many thanks,

    Don Lancaster voice phone: (928)428-4073
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    rss: http://www.tinaja.com/whtnu.xml email:

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  2. I would think you'd find the coefficients of the Taylor series by
    multiplying the k-th term by k-factorial, not dividing. E.g.,
    the coefficients of the power series for e^x are 1, 1, 1/2, 1/6, ...,
    while the Taylor coefficients, the coefficients of (x^n)/(n!), are
    1, 1, 1, 1, ....
     
  3. Dividing plots beautifully. just checked it.
    You basically have a ramp that gets VERY steep near a=0.

    Also, multiplying -1517.83 by 20! might end up a tad on the largish side
    of humongous. In any sane function, the higher power terms should be
    quite small.

    There are apparently two or three summation terms to the real answer.
    One is an easily found and apparently fully determninistic linear ramp.
    The second is an ever increasing "bent" variable that may or may not
    need some extra help above 0.9. It may or may not have its own small
    ramp component.

    I suspect it does.


    --
    Many thanks,

    Don Lancaster voice phone: (928)428-4073
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    rss: http://www.tinaja.com/whtnu.xml email:

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  4. Question about terminology here. By the above, are you saying that
    0.103691 = a1 * x^1 and
    -0.00245142 = a2 * x^2
    etc.

    Or are you saying that 0.103691 is the coefficient of x^1, -0.00245142
    is the coefficient of x^2, etc. so

    f(x) = (0.103691 * x^1) + (-0.00245142 * x^2) + .....


    BTW, you posted a similar, but slightly different problem a half hour
    earlier with terms starting with x^0 as opposed to x^1 (above). Which is
    correct?
     
  5. Cool. But the function you're plotting isn't the one you're asking
    about. The power series a_1 x + a_2 x^2 + a_3 x^3 + ... is
    (obviously!) equivalent to the Taylor series
    a_1 x + (2 a_2) (x^2 / 2) + (6 a_3) (x^3 / 6) + ...
    and not equivalent to
    a_1 x + (a_2 / 2)(x^2 / 2) + (a_3 / 6) (x^3 / 6) + ....

    So either you're not doing what you say you're doing, or else
    the function you're plotting has nothing to do with the function
    you say interests you.
     
  6. Chip Eastham

    Chip Eastham Guest

    Is the function known to be periodic?

    A Fourier series expansion, if so, would be more
    illuminating than a power series.

    regards, chip
     
  7. Jon

    Jon Guest

     
  8. whit3rd

    whit3rd Guest

    This is the art of curve fitting. You need one additional piece of
    information, the error (tolerance) of each of the measurements,
    to do it properly.

    Since you have specified this over the range 0 to 1, the classical
    approach would be to use a set of functions that are orthogonal
    over this range. Since you already have a polynomial,
    a kind of Legendre polynomial is your starting point (I think the
    Legendre polynomials are orthogonal over (-1, +1), so there's
    some scaling and shifting involved).

    Your goal is, from that starting point, to find a simpler expression
    that holds the same data (i.e. reproduces the same function to
    the given tolerance). Sine/Cosine, polynomial, Bessel, there
    are lots of sets of orthogonal functions to choose from, and
    one of the sets might have all-but-one-term-vanishes character.

    If there is any OTHER information about the boundary conditions,
    (if it's periodic, use sines; if it doesn't blow up at infinity,
    don't use
    polynomials) that can help, you need to consider that now.
    Finally, there's a theorem, the Wiener-Hopf theorem, that
    states that a set of fit functions is optimal if its autocorrelation
    is the same as that of the data. That means that data with
    big ripples doesn't fit well with sinewaves with little tiny ripples,
    but you can find that kind of thing out the hard way, too.

    For orthogonal functions, the fit procedure is simple: there's a
    unique
    right set of coefficients, and a procedure to find it (inner product
    of your data and the function). For other kinds of functions,
    like wavelets and fractals and such, it gets ... challenging.

    Your starting point has 20 coefficients, so anything that expresses
    the data with 19 or less, and hits within the measurement tolerance,
    is some kind of success.

    Finally, I should note that the 'measurement tolerance' defines a kind
    of weight, a statistical weight function, that cannot be ignored.
    The orthogonal polynomials (Legendre polynomials) with
    constant weight are very different from the orthogonal polynomials
    (Chebyshev polynomials) with (x(1-x)) weight, and the weight
    is part of that 'inner product' step as well.

    Usually, one starts with a graph of the function, makes a guess as
    to its form, then looks at a graph of the difference-from-the-guess.
    If you get lucky with a guess, and it was a natural measurement
    you started with, you've just created a scientific theory. Kudos!
     
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