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Math Help - question on taylor polynomials

  1. #1
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    question on taylor polynomials

    Let f(x) be the Taylor Polynomial P_7(x) of order 7 of arctan(x) about x=0. Then it follows that, if -1.5<x<1.5,
    A) f(x) = arctan(x)
    B) f(x) </= arctan(x)
    C) f(x) >/= arctan(x)
    D) f(x) > arctan(x) if x < 0, but f(x) < arctan(x) if x > 0
    E) f(x) < arctan(x) if x < 0, but f(x) > arctan(x) if x > 0

    my book says the answer is D) but i have no idea how they figured out whether or not f(x) is greater than arctan(x). please help.
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  2. #2
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    Quote Originally Posted by oblixps View Post
    Let f(x) be the Taylor Polynomial P_7(x) of order 7 of arctan(x) about x=0. Then it follows that, if -1.5<x<1.5,
    A) f(x) = arctan(x)
    B) f(x) </= arctan(x)
    C) f(x) >/= arctan(x)
    D) f(x) > arctan(x) if x < 0, but f(x) < arctan(x) if x > 0
    E) f(x) < arctan(x) if x < 0, but f(x) > arctan(x) if x > 0

    my book says the answer is D) but i have no idea how they figured out whether or not f(x) is greater than arctan(x). please help.
    What is the Taylor's polynomial for arctan(x)? In particular is the term involving x^8, the first term not in the 7 th order polynomial, have positive or negative coefficient?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    What is the Taylor's polynomial for arctan(x)? In particular is the term involving x^8, the first term not in the 7 th order polynomial, have positive or negative coefficient?
    the taylor polynomial for arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + (x^9)/9 +...

    the term involving x^8 is 0 but the coefficient of the term involving x^9 is positive. but i'm still unclear about how to know whether or not f(x) is greater than or less than arctan(x) and why D is the correct answer.
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  4. #4
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    Quote Originally Posted by oblixps View Post
    the taylor polynomial for arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + (x^9)/9 +...

    the term involving x^8 is 0 but the coefficient of the term involving x^9 is positive. but i'm still unclear about how to know whether or not f(x) is greater than or less than arctan(x) and why D is the correct answer.
    For any positive x the error in the seventh orther Taylor polynomial is positive (that is f(x)>P_7(x)), and for negative x the error is -ve (look at the Lagrange form of the remainder, this may be more convieniently done with the 8th order polynomial). The observation need only be: for any positive x the error in the seventh orther Taylor polynomial is positive, since the other follows from the polynomial and the function both being odd functions.

    CB
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  5. #5
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    the solution in my book reads:

    Note that the Taylor seires for arctan(x) satisfies the Alternating Series Test and that f(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 = P_7(x). If x<0, then the first omitted term, (x^9)/9 is negative. Hence P_7(x) exceeds arctan(x).

    But that doesn't make sense since if the error of the seventh order polynomial is negative, how can P_7(x) be greater than arctan(x)?
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  6. #6
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    Quote Originally Posted by oblixps View Post
    the solution in my book reads:

    Note that the Taylor seires for arctan(x) satisfies the Alternating Series Test and that f(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 = P_7(x). If x<0, then the first omitted term, (x^9)/9 is negative. Hence P_7(x) exceeds arctan(x).

    But that doesn't make sense since if the error of the seventh order polynomial is negative, how can P_7(x) be greater than arctan(x)?
    You have to subtract a positive quantity (add a negative) from p_7(x) to get f(x); hence f(x) is less than p_7(x).

    CB
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