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Math Help - Proofs for Properties of Taylor Series...

  1. #1
    Radders3
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    Proofs for Properties of Taylor Series...

    Hi everyone!

    I was wondering if anyone knew rigorous proofs from the following properties of Taylor series:

    (a) Addition and subtraction of series, (eg: taylor series for sinx+cosx = taylor series for sinx + taylor series for cosx )

    (b) Substitution of z for something else to form a new Taylor series for a different function, (eg: replacing x with x^3 in the taylor series for e^x to give taylor series for e^(x^3) )

    (c) Multiplication and division of series. (eg: taylor series for sinx*cosx = taylor series for sinx multiplied by taylor series for cos x )

    I am a Mathematics undergraduate and am thus very familiar with using these properties, but I have never seen a rigorous proof for any of them.

    If anyone could post any of the proofs then I would be extatic!

    Many thanks in advance.

    Radders.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Radders3 View Post
    Hi everyone!

    I was wondering if anyone knew rigorous proofs from the following properties of Taylor series:

    (a) Addition and subtraction of series, (eg: taylor series for sinx+cosx = taylor series for sinx + taylor series for cosx )
    Better to confine attaention to McLaurin series makes things a bit
    simpler, but for (a) it is a consequence of the additivity of differential
    operators:

    <br />
D^{(n)}[f(x)+g(x)]=D^{(n)}f(x) + D^{(n)}g(x)<br />

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Radders3 View Post
    (b) Substitution of z for something else to form a new Taylor series for a different function, (eg: replacing x with x^3 in the taylor series for e^x to give taylor series for e^(x^3) )
    There are a number of ways of showing that the McLaurin series for f(x^k)
    is the same as the McLaurin series for f(u) with u=x^k.

    The most obvious is to look at the corresponding terms of the series, when
    with a little work it is obvious that they are the same.

    A second idea is to use the uniqueness of McLaurin series for analytic functions.
    The two series are McLaurin series for two functions, but by construction the
    functions are equal in some neighbourhood of 0, so have the
    same McLaurin series.

    RonL
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  4. #4
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    If \sum_{k=0}^n \frac{f^{(k)}(0)x^k}{k!} and \sum_{k=0}^n \frac{g^{(k)}(0)x^k}{k!} are the Taylor polynomials for f(x),g(x) then their sum is \sum_{k=0}^n \frac{[f^{(k)}(0)+g^{(k)}(0)]x^k}{k!} is a Taylor polynomial for f(x)+g(x).
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    If \sum_{k=0}^n \frac{f^{(k)}(0)x^k}{k!} and \sum_{k=0}^n \frac{g^{(k)}(0)x^k}{k!} are the Taylor polynomials for f(x),g(x) then their sum is \sum_{k=0}^n \frac{[f^{(k)}(0)+g^{(k)}(0)]x^k}{k!} is a Taylor polynomial for f(x)+g(x).
    Once again this is for McLaurin series , and what I said in an earlier post about
    the additivity of powers of the differential operator, or uniquiness of McLaurin
    polynomials to complete.

    Not that I am complaining, its an argument that I would use.

    RonL
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