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Math Help - Taylor's Series.

  1. #1
    Junior Member pearlyc's Avatar
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    Taylor's Series.

    Given f(x) = sinh(2x)

    (i) Find the Taylor series associated with the function f(x) at x=0.
    I found the Taylor polynomials for this function but how do I convert it do Taylor series?

    (ii) Use the ratio test to determine for reach values of x this Taylor series is a convergent series.

    (iii) By analysing Lagrange's remainder term, show that the Taylor series represent f(x) exactly in the interval you found in the part (b) above.
    Do we do this by proving the remainder is equivalent to 0, hence representing f(x)?

    Thank you.
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  2. #2
    Junior Member pearlyc's Avatar
    Joined
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    Okay, I have been trying and I am not sure if I am on the right track, somebody please kindly enlighten me. Thank you.

    This is what I got so far, I converted sinh 2x back to its basic definition,

     sinh 2x = \frac{e^{2x} - e^{-2x}}{2}

    Then I differentiated the basic definition instead.

    f'(x) = 2 . \frac{e^{2x}}{2} - (-1)(2)\frac{e^{-2x}}{2}
    f''(x) = 2 . 2 \frac{e^{2x}}{2} - (-1)(-1)(2)(2)\frac{e^{-2x}}{2}
    f'''(x) = 2. 2 . 2 \frac{e^{2x}}{2} - (-1)(-1)(-1)(2)(2)(2)\frac{e^{-2x}}{2}
    .
    .
    .
    .
    f^{(n)}(x) = 2^n\frac{e^{2x}}{2} - (-1)^{n}2^{n}\frac{e^{-2x}}{2}

    Further simplifying,
    f^{(n)}(x) = 2^{n-1}e^{2x} - (-1)^{n}2^{n-1}e^{-2x}
    f^{(n)}(0) = 2^{n-1} - (-1)^{n}2^{n-1}
    f^{(n)}(0) = 2^{n-1}[1 - (-1)^{n}]

    Therefore, the Taylor series associated with the function f(x) is
    \sum^\infty_{n=0} \frac{2^{n-1}[1- (-1)^{n}]}{n!}x^n.

    For part (b), to prove with ratio test that this series is convergent means \lim_{n\rightarrow\infty} r_n = \lim_{n\rightarrow\infty} | \frac{a_{n+1}}{a_n} | < 1

    \lim_{n\rightarrow\infty} | \frac{a_{n+1}}{a_n} | = \frac{2x [ 1 - (-1)^{n+1} ]}{n! [1 - (-1)^n ]}

    How should I proceed from here?

    For part (c), to show that the Taylor series represents f(x) exactly in the interval in part (b) above means that I'll have to prove that the remainder is equivalent to zero as n \rightarrow \infty right?

    R_n(x) = \frac{2^ne^{2c} - (-1)^{n+1}2^{n}e^{-2c}}{(n+1)!}{x^{n+1}}

    I don't know how to proceed from here either.

    Thank you for your time!!
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