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Math Help - Power Series

  1. #1
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    Power Series

    Recall that the power series of arctan x is the Sum from 0 to infinity = 0(−1)^n (x^(2n+1) / (2n+1)) , with a radius of convergence of 1.

    a) What is the value of the sixteenth derivative of arctan x at x = 0? What is the value of the seventeenth derivative of arctan x at x = 0? (Hint: look at the term of that degree in the power series.)

    b) What is limx−>0 ((arctan(x^2)−(arctan (x))^2)/x4) ? (Do not use líHospital! - it would be a waste of time.)

    This is a homework problem and i have no clue how to even go about starting it. Can someone help me out. TIA!!!!
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  2. #2
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    Quote Originally Posted by TreeMoney View Post
    Recall that the power series of arctan x is the Sum from 0 to infinity = 0(−1)^n (x^(2n+1) / (2n+1)) , with a radius of convergence of 1.

    a) What is the value of the sixteenth derivative of arctan x at x = 0? What is the value of the seventeenth derivative of arctan x at x = 0? (Hint: look at the term of that degree in the power series.)
    If:

    <br />
f(x)=\sum_0^{\infty} a_n x^n<br />

    is sufficiently well behaved then:

    <br />
f'(x)=\sum_1^{\infty} a_n n x^{n-1}<br />

    and in general:

    <br />
f^{(k)}(x)=\sum_k^{\infty} a_n \frac{n!}{(n-k)!} x^{n-k}<br />

    so:

    <br />
f^{(k)}(0)=a_k \, {k!}<br />

    Which should be sufficient to do part a).

    (In this case sufficiently well behaved is satisfied by the series being absolutely
    convergent in some neighbourhood of x=0)

    RonL
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  3. #3
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    b) What is limx−>0 ((arctan(x^2)−(arctan (x))^2)/x4) ? (Do not use líHospital! - it would be a waste of time.)
    It turns into a monstrisity with L'Hopital, though, it can be done.

    What they're getting at is Taylor series.

    Expand the Taylor series for tan^{-1}(x^{2})-(tan^{-1}(x))^{2} and you get an end term of \frac{2}{3}x^{4}

    When you divide by x^{4}, you get \frac{2}{3}

    Which is the limit as x approaches 0.
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  4. #4
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    Still don't get it

    I tried doing part A and am still clueless. Can someone help me out. TIA
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