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Math Help - question on series

  1. #1
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    question on series

    hi
    well looks like i found a good site to help me with my calc hw.
    have a question on power series

    the first one i wanna know if i did it right

    1) find power series for f(x)=ln(6+x)
    ln(6+x) = 1/(6+x) = (1/6) /(1-(-x/6)) = a/1-r
    ln(6+x) = sum ar^n = sum (1/6)(-x/6)^n
    (-x/6)<1 = -6<x<1

    2) Use power series to find the integral of 3/(4-x^2) and then approximate the integral from x=0 to x=.9

    kinda stuck on this one there are no examples like this in our book im guessing just do the same thing i did in question 1 right? could someone work this one out.

    thanks

    oh how do you enter in math formulas in on here?
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  2. #2
    MHF Contributor matheagle's Avatar
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    You let out the integral in (1)

    Quote Originally Posted by amoochio46 View Post
    hi
    well looks like i found a good site to help me with my calc hw.
    have a question on power series

    the first one i wanna know if i did it right

    1) find power series for f(x)=ln(6+x)
    ln(6+x) = INTEGRAL 1/(6+x) = (1/6)INTEGRAL /(1-(-x/6)) = a/1-r
    ln(6+x) = sum ar^n = sum (1/6)(-x/6)^n
    (-x/6)<1 = -6<x<1

    2) Use power series to find the integral of 3/(4-x^2) and then approximate the integral from x=0 to x=.9

    kinda stuck on this one there are no examples like this in our book im guessing just do the same thing i did in question 1 right? could someone work this one out.

    thanks

    oh how do you enter in math formulas in on here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor matheagle's Avatar
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    (2) You could use partial fractions and that may work out better, but I'll do it directly...

     \int_0^{.9} {3dx\over 4-x^2} = (3/4) \int_0^{.9} {dx\over 1-(x^2/4)}=(3/4) \int_0^{.9}\sum_{k=0}^{\infty} (x^2/4)^k  .

    Now use the geometric series with r=x^2/4

    {1\over 1-r} =\sum_{k=0}^{\infty} r^k since -1<r<1.
    Last edited by matheagle; March 29th 2009 at 06:45 PM.
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  4. #4
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    Quote Originally Posted by amoochio46 View Post
    [snip]
    oh how do you enter in math formulas in on here?
    Read this: http://www.mathhelpforum.com/math-he...-tutorial.html
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  5. #5
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    Quote Originally Posted by matheagle View Post
    You let out the integral in (1)
    not sure i get what your saying. what integral did i leave out?
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  6. #6
    MHF Contributor matheagle's Avatar
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    ln(6+x) = INTEGRAL 1/(6+x) = (1/6)INTEGRAL /(1-(-x/6)) = a/1-r


    You left out the integral where I wrote INTEGRAL


    ln(6+x) does NOT equal 1/(6+x)


    ln(6+x) = INTEGRAL 1/(6+x)
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