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Math Help - Help with computing a series

  1. #1
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    Help with computing a series

    imgur: the simple image sharer

    Would someone mind explaining how they get

    Icos(theta) + Asin(theta)


    Additionally, or a more urgent matter I don't understand what's happened here:
    imgur: the simple image sharer
    Last edited by 99.95; November 5th 2012 at 05:18 AM.
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  2. #2
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    Re: Help with computing a series

    They are using the power series expansion of e^x:
    e^x= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot= \sum_{n=0}^\infty \frac{x^n}{n!}

    The given sum is 1+ \theta A+ \frac{\theta^2}{2!}A^2+ \frac{\theta^3}{3!}A^3+ \cdot\cdot\cdot.
    We can combine the " \theta" and " A" terms to get
    1+ (\theta A)+ \frac{(\theta A)^2}{2!}+ \frac{(\theta A)^3}{3!}+ \cdot\cdot\cdot= e^{\theta A}.

    However, the answer to your question depends upon what A is. The problem you gives says "Using your result from part (a)" but you don't tell us what "part (a)" is or what your result was. Without that information, it is impossible to go further.

    In your second question, you are given I(n)= \int_0^\infty x^ne^{-x}dx. Integrating by parts, letting u= x^n and dv= e^{-x}dx, we have du= nx^{n-1}dx and v= -e^{-x} so that the integral becomes \int_a^b udv= \left[uv\right]_a^b- \int_a^b vdu
    \left[-x^ne^{-x}\right]_0^\infty- \int_0^\infty (nx^{n-1})(-e^{-x})dx

    The "suitable rigor" (did they really mispell "suitable" in your book?) referred to means take the limits to be "0" and "A" and then taking the limit as A goes to infinity: \left[-x^ne^{-x}\right_0^\infty= \lim_{A\to\infty}\left[-x^ne^{-x}\right]_0^\infty = \lim_{A\to\infty}(-A^ne^{-A}- 0)= 0 for all n> 0. That makes the integral I(n)= \int_0^\infty x^n e^{-x}dx= n\int x^{n-1}e^{-x}dx= nI(n-1).

    (In what you have they have used " \epsilon" instead of my "A".)
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  3. #3
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    Re: Help with computing a series

    Thanks a heap!

    I understand the second question now,
    my apologies for not providing the previous answer for question 1.

    Here it is: imgur: the simple image sharer

    Hopefully that is sufficient, if you could please get back to me on that one .

    Thanks!
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