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Error on using a curve of best fit for extrapolation

  1. Feb 25, 2009 #1
    Hello everyone,

    I did a quick search but could not find this in the forums.

    I have quite a basic situation. I have been gathering data points from an experiment and was able to fit an exponential curve of best fit to it. What I want to do is approximate some values between my data points using this curve. My approach is to just use the equation of the curve and plug in the desired number. I will then add the uncertainty to it.

    This uncertainty is what I am drawing a blank with. Does anyone know what method I could use to calculate this uncertainty. I have a feeling this is really simple, but... I cant remember anything yet.

    Any help would be appreciated.
  2. jcsd
  3. Feb 25, 2009 #2

    If your curve is:


    Your extrapolation error will be roughly:

    Y=exp(sigma_c (x2-x1))+sigma_yo

    sigma_c is the standard deviation in your estimate of c.
    x1 is the point you are extrapolating from
    x2 is the point you are extrapolating to.
    sigma_yo is your initial error.
  4. Feb 25, 2009 #3
    which method are you using to get the error equation?

    I was thinking of something along the lines of:
    If I have y = A exp(Bx) as my best fit, I can take the ln of both sides:

    ln y = B lnA + B ln x

    I can now do a line of best fit and find sigma lny, sigma A and sigma B.
    These are the uncertainties in the fit which I get from the least - squares method. I can now use the error propagation formula to get the uncertainty in y.

    My question is, where do I incorporate the uncertainty in my measurments.
  5. Feb 25, 2009 #4
    Well, lets say y=<y(x)>+rv
    <y(x)> is the mean value of y at x and rv is a random variable.

    Lets assume for a sec that rv is a quasian distribution then:


    To get ln y do a change of variables on the above expression for P(y)

    u=ln(y) <=> y=exp(u)

    The term inside the integrand will be your new distribution function for ln y

    See functions of random variables:

    Keep in mind that
    <ln y>
    does not equal
    ln <y>

    Thus a least mean squared fit of ln y will be biased. Measurements where y is much smaller then the standard deviation of the random variable will be more biased. You can do the transformation before the fit but if your error is large compared to your measurement then you’ll need to use maximum likelihood. If you use least squares then weight large values of y exponentially more then small values of y.

    Also note that:

    if <y>=Aexp(BX)
    <ln y> does not equal: BX + ln(A)

    Rather it is equal to <ln (<y(x)> +rv)>

    Which we can’t really separate neatly into the form you gave.
  6. Feb 25, 2009 #5
    Ah thank you for clarifying. I knew something was too easy about my approach. I dont have time to try this now, but Im sure it will work. Thank you.
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