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Math Help - Deriving Taylor and Maclaurin Polynomials

  1. #1
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    Deriving Taylor and Maclaurin Polynomials

    I am just reading about Taylor and Maclaurin Polynomials. I was wondering how they were derived. Could someone please show me how in both a rigourous mathematical way and an intuitive way. I hate to just use formulas that I don't have at least some grasp of the theory behind them. As a toddler my parents gave me a set of real tools (yes they were insane) and I comenced dismantling everything. I took apart tables, telephones and finally the TV, which put an end to my tools.

    Anyway I appreciate the insight from whoever, the more the merrier.

    ManyArrows
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  2. #2
    Member Mentia's Avatar
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    Take a look at Taylor Series Expansion by mtu.edu for a derivation.


    Basically, you look at a function and say, "How could I expand this in a power series?" The Taylor Series is one way to expand a function as a power series. Expanding a function in this way can lead to some useful approximations. For instance, lets look at sin[x] about zero:

    About zero, \sin (x) = x-\frac{x^3}{6}+\frac{x^5}{120}+ ...

    So one can make approximations to whatever order they wish. For instance, to "zeroeth" order, sin(x) about zero = 0. To first order, sin(x) about zero = x. To third order, sin(x) about zero = x-\frac{x^3}{6}. Etc.

    In addition, it can be helpful in finding limits of functions:

      \lim_{x->0 } \frac {\sin (x)}{3x - (1/2)x^3} = ?

    Notice that in the limit that x goes to zero, the Taylor series to any arbitrary order is exactly equal to Sin(x):

      \lim_{x->0 } \sin (x) = \lim_{x->0 } x = \lim_{x->0 } x-\frac{x^3}{6} = \lim_{x->0 } x-\frac{x^3}{6}+\frac{x^5}{120} = ...

    So,

      \lim_{x->0 } \frac {\sin (x)}{3x - (1/2)x^3} =  \lim_{x->0 } \frac {x-\frac{x^3}{6}}{3x - (1/2)x^3} = \frac {1}{3}

    In physics we make approximations using Taylor series all the time because instruments can only measure to a certain degree of accuracy anyway.
    Last edited by Mentia; March 11th 2009 at 08:54 PM. Reason: expanded reply
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  3. #3
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    One down

    Ok, that explains the mathematical derivation. It was easy to follow, thank you.

    Now though, what is the theory behind it. Derivations are a way to prove an argument, right? So, when Mr. Taylor and Mr. Maclaurin were pondering the universe, what led them down this path.

    Also, when they say that Maclaurin polynomials are accurate close to zero, how close is close. And the same with Taylor, how close to the refernce point is close. Does this have something to do with the margin of error.

    Also, I haven't learned power series yet. For some reason my textbook has Taylor and Maclaurin in chpt 9 and power series in chpt 11.

    Thanks ManyArrows
    Last edited by manyarrows; March 11th 2009 at 08:59 PM. Reason: Further clairification
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  4. #4
    Member Mentia's Avatar
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    Well, if you were to take the Taylor or Maclaurin series out to infinite order they would be exactly equal to the function. The only caveat is that the nth derivative at the given point must be defined for the function. So a function like:

    f(x) = \sqrt[]{x} does not have a Maclaurin series, but it does have a Taylor series about non-zero points.
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  5. #5
    Member Mentia's Avatar
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    Strange about your book and the taylor/powerseries thing. Check out the wiki on power series: Power series - Wikipedia, the free encyclopedia

    Notice how the taylor and maclaurin series are really just specific types of power series.
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  6. #6
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    textbook wierdness

    the chpt on the taylor series also didn't explain why they were usefull. I read it and was like why do I need to approximate a function that I already have. Luckily the class notes explained it, which I read after the chpt. I am taking calc 2 online after taking calc 1 10 years ago. It's alot of fun self teaching math.

    ManyArrows
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