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Math Help - Taylor Polynomial, Please Help!

  1. #1
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    Taylor Polynomial, Please Help!

    I have no idea where to begin or what to do because I have no notes, etc... on this stuff. I really need this as an example to work with. Thanks.

    (a) Determine the second Taylor Polynomial of f(x) = x at x = 16.

    (b) Use the polynomial found in part (a) to estimate 16.1 to the fifth decimal place.
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  2. #2
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    Google. Best place to start if you have no clue about it at all.
    Taylor series - Wikipedia, the free encyclopedia

    As taylor polynomial is based on set of terms basic on the nth derivative of a function you are trying to find a approximation for in a series.

    The general taylor polynomial is as such, a general taylor polynomial starts at the 0th degree and you continue on till the degree which you were given

    X = A = 16
    in your case.

    P_n=\frac{f^{n}(a)}{n!}(x-a)^n

    You are taking the n'th derivative of the function you were given to a degree over the nth factorial center at x - a to the nth power.

    Here is your example, it helps to build a table of each nth derivative you take
    Your problem states to the 2nd degree, will do as such and it is centered at x = 16

    Making a table of each nth derivative as such evulated at a which is equal to x which is 16

    f^{0}(a) = (a)^{\frac{1}{2}}

    f^{1}(a) = \frac{1}{2(a)^\frac{1}{2}}


    f^{2}(a) = - \frac{1}{4(a)^\frac{3}{2}}

    Now evulate at a which equals 16 and set up your polynomial keeping in mind the nth factiorial and (x-16)^n
    Last edited by RockHard; December 16th 2009 at 08:26 PM.
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  3. #3
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    im struggling with these as well.. is there anyone here who can tell me if im anywhere close to the right track

    f(x)+f'(x)(x-16)+f''(x)/2! * (x-16)^2

    and get: 4+ 1/8 (x-16) -1/128 (x-16)^2

    i don't know how to do the 2nd part at all, but is that remotle right?
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  4. #4
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    Check out my edited post above
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  5. #5
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    is that what mine said, or no?

    i think theyre the same...
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  6. #6
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    Yes correct, but your last term in incorrect, you divided the denominator on the top by 2, when you should be multiplying the whole fraction by 1/2

    \frac{\frac{1}{256}}{2} \ne \frac{1}{128}

    If you visualize the expression as such

    \frac{\frac{1}{256}}{\frac{2}{1}}

    to divide a fraction by a fraction you simply multiply the top by the reciprocal of the bottom


    As for the second part, what you have here is an alternating series, from your polynomial form a general expression of the polynomial as a power series, and using the theorem of how to estimate an a value of the series with in a given error tolerance, in your case .00001
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  7. #7
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    does that have anything to do with "little o" notation or is that something completely different? now that i got the basics down , those are next on the chopping block

    ps. sorry to the op for highjacking his thread
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  8. #8
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    What part are you asking pertains to the "little" notation, your next step is to approx the function to 5 decimals place, aka, your tolerance

    The standard of how to approximate an sum of an alternating series with in a given tolerance is as such

    |S-S_n| \le a_{n+1} \le .00001

    Leaving you only to work with

     a_{n+1} \le .00001

    Where you solve for n, to find how many terms you must evaluate to get the approximation within 5 decimal places
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