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Approximation by polynomails

  1. Apr 20, 2009 #1
    I`m studying Series/Sequences/Approximation by polynomials..

    -We approximate a function f(x) by getting a polynomial (I don`t know how we get it, and I don`t know what characteristics it should have, and I`d like to know please)

    -when we need more accuracy we add a higher derivatives, but why is adding a higher derivative gives more accuracy? I tried to imagine what a second derivative represents on a graph, and I came out with a result that may be true, If a second derivative is the rate of change of a first derivative then I should draw a new graph where the independent variable is the function that we defferentiated for once, is that true? I think understanding what a second, third, etc... derivatives represent on a graph can make it clear for me to understand how a higher derivative gives more accuracy.

    So, to sum up the questions:
    1- how do we find the polynomial when we approximate a function?
    2- what does a second, third, etc... derivative represent on a graph?
    3- why adding higher derivatives add more accuracy to the "function"? -in other words: precisely, whats the relationship between taking higher derivatives and the accuracy of the result of the polynomial?-

    Thank you,
     
  2. jcsd
  3. Apr 20, 2009 #2

    alxm

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    Try http://en.wikipedia.org/wiki/Taylor_series" [Broken].

    It's in any introductory calculus book. If you want a proof, try an introductory complex analysis book, Taylor's theorem follows easily from the Laurent series.
     
    Last edited by a moderator: May 4, 2017
  4. Apr 20, 2009 #3
    A very basic version to get you started:

    1. Pick a point in the neighborhood of where you want to approximate the function.
    2. The approximation is given by the Taylor series, and the polynomial is:

    [f(a) (x-a)^0 / 0!] + [f '(a) (x-a)^1 / 1!] + [f ''(a) (x-a)^2 / 2!] + ... + [f^n(a) (x-a)^n / n!]

    Example:

    f(x) = e^x. Let's approximate around x=0. So a=0.

    f(0) = 1.
    f '(0) = 1.
    f ''(0) = 1.

    So

    p(x) = 1 + (x) + [(x)^2] / 2 is the best approximation to e^x around x=0 using a 2nd order polynomial.

    Higher derivatives represent slopes of slopes of slopes ... of slopes. In terms of the original function, higher order derivatives start losing meaning after the first few (first derivative is slope, 2nd derivative is concavity, third derivative is... what? the 78th derivative...?) However, each provides a little more information on exactly how the function behaves.

    The thing about Taylor polynomials is that by matching the derivatives of a function at a point, it becomes harder and harder to distinguish the function and your polynomial in a neighborhood around that point. Imagine using a magnifying glass around a point. If the first, second, third, etc. derivatives match, it starts looking like the same function - exactly. Of course, not all functions work like this, but many do.
     
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