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Integral Error

  1. Nov 25, 2015 #1
    Consider two functions f, g that take on values at t=0, t=1, t=2.

    Then the total error between them is:

    total error = mod(f(0)-g(0)) + mod(f(1)-g(1)) + mod(f(2)-g(2))

    where mod is short for module.

    This seems reasonable enough.

    Now, consider the two functions to be continuous on [0,2].
    What is the total error now?

    My guess is that it is the integral of the absolute value of their difference divided by the length of the interval:
    total error =1/2 * integral from 0 to 2 of mod(f(x)-g(x)) dx

    Is this right?

    Or is the error evaluation done in a different way?
     
  2. jcsd
  3. Nov 25, 2015 #2
  4. Nov 25, 2015 #3

    pasmith

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    If [itex]f(x) = 10^7 \sin \pi x[/itex] and [itex]g(x) = 0[/itex] then [tex]\int_0^2 f(x) - g(x)\,dx = 0.[/tex] Does it seem reasonable to say that the error between [itex]f[/itex] and [itex]g[/itex] is zero?

    Errors should be defined in terms of norms. There are many norms one can place on the space of continuous functions on [itex][0,2][/itex], for example [tex]
    \|f\|_{\infty} = \max \{|f(x)| : x \in [0,2]\}[/tex] or [tex]
    \|f\|_p = \left( \int_0^2 |f(x)|^p\,dx \right)^{1/p},\quad p \geq 1.
    [/tex]
     
  5. Nov 25, 2015 #4
    No. I take the module. I integrate the module. I specified this in the first post.

    It is not zero if integrating the absolute value.
     
  6. Nov 25, 2015 #5

    pasmith

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    You then say "measuring the error between the curves using the area between them" - ie [itex]\int_0^2 f(x) - g(x)\,dx[/itex] - "is good enough", which of course it isn't.
     
  7. Nov 25, 2015 #6
    What if I do the following thing.

    So, consider the function f(x) that takes on values at x=0, x=1, x=2. So the space between two points on the x axis is 1.
    And I want to sum the values:
    sum = f(0)+f(1)+f(2)

    Now, I want to make the space between the points infinitesimally small. For this I write like this:
    sum = f(0) + f(0+Dx) + f(0+2Dx) where Dx = 1
    And I make Dx to go to 0. Thus:

    sum = f(0) + f(0+dx) + f(0+2dx)
    And I sum like this from x = 0 to x = 2.

    Then I do this for mod(f(x)-g(x)). So I sum the absolute value of their differences at each point dx on the x axis. Would this work? Something similar is made when going from Fourier series to Fourier transform.

    I mean, the sum would have to be infinite. But, in theory, I think it is correct.
    Or not. If it is infinite, who decides that it should stop at x=2 or x=10. For both it is sum from k=0 to inf of f(k*dx). I dont know.
     
    Last edited: Nov 25, 2015
  8. Nov 25, 2015 #7
    The absolute value of the area. It was self-implied from the first post.
     
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