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Math Help - expand (e^2x)/(2x+1) up to x^3

  1. #1
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    expand (e^2x)/(2x+1) up to x^3

    I want to expand (e^2x)/(2x+1) up to x^3. I can expand e^2x but I am then lost.
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  2. #2
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    Re: expand (e^2x)/(2x+1) up to x^3

    Quote Originally Posted by Stuck Man View Post
    I want to expand (e^2x)/(2x+1) up to x^3. I can expand e^2x but I am then lost.
    just expand e^{2x} out as you have and do the polynomial division. Keep going until you have the x^3 term.
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  3. #3
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    Re: expand (e^2x)/(2x+1) up to x^3

    Every term that I expand to e.g 2x^6 from e^2x will add something to the low power terms. If the question said divide by 2x it would be easy.
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    Re: expand (e^2x)/(2x+1) up to x^3

    I have started to do this question using MacLaurin's series. The question was not appropriate for the book its in.
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    Re: expand (e^2x)/(2x+1) up to x^3

    Quote Originally Posted by Stuck Man View Post
    I have started to do this question using MacLaurin's series. The question was not appropriate for the book its in.
    what you could do is let

    u=2x+1 \Rightarrow 2x=u-1

    then

    e^{2x}=e^{u-1}=\frac{1}{e}e^u \Rightarrow

    \frac{e^{2x}}{2x+1}=\frac{\frac{1}{e}e^u}{u}

    \frac{\frac{1}{e}e^u}{u}=\frac{1}{e}\left(\frac{1+  u+\frac{u^2}{2}+\frac{u^3}{6}+\frac{u^4}{24}...}{u  }\right)=

    \frac{1}{e}\left(\frac{1}{u}+1+\frac{u}{2}+\frac{u  ^2}{6}+\frac{u^3}{24}...}\right)

    and just substitute u=2x+1 in the expression above.
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  6. #6
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    Re: expand (e^2x)/(2x+1) up to x^3

    Here is another method:

    \dfrac{e^{2x}}{2x+1} = e^{2x}\dfrac{1}{1-(-2x)} = \left( \sum_{n\ge 0} \dfrac{(2x)^n}{n!} \right) \left( \sum_{n\ge 0}(-2x)^n \right)

    So, you have the product of two infinite series. You want to multiply out:

    \begin{align*} & (2x)^0\cdot (-2x)^0 + (2x)^0\cdot (-2x)^1 + (2x)^0\cdot (-2x)^2 + (2x)^0\cdot (-2x)^3 + \cdots \\ + & (2x)^1\cdot (-2x)^0 + (2x)^1\cdot (-2x)^1 + (2x)^1\cdot (-2x)^2 + \cdots \\ + & \dfrac{(2x)^2}{2}(-2x)^0 + \dfrac{(2x)^2}{2}(-2x)^1 + \cdots \\ + & \dfrac{(2x)^3}{3!}(-2x)^0 + \cdots \end{align*}

    Those are all of the terms with the power of x less than or equal to 3.

    Another way to write this:

    \sum_{n\ge 0} \left( \sum_{k = 0}^n \dfrac{(-1)^{n-k}}{k!} \right) (2x)^n
    Last edited by SlipEternal; January 27th 2014 at 03:01 PM.
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  7. #7
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    Re: expand (e^2x)/(2x+1) up to x^3

    Oh, thinking about it, \sum_{k=0}^n \dfrac{(-1)^{n-k}}{k!} = (-1)^n\dfrac{!n}{n!} where !n is the number of derangements of an n element set. So, you can write it:

    \dfrac{e^{2x}}{2x+1} = \sum_{n\ge 0}(-2)^n\dfrac{!n}{n!}x^n
    Last edited by SlipEternal; January 27th 2014 at 04:03 PM.
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    Re: expand (e^2x)/(2x+1) up to x^3

    first time I've ever seen !n... learn something new every day
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    Re: expand (e^2x)/(2x+1) up to x^3

    Quote Originally Posted by romsek View Post
    first time I've ever seen !n... learn something new every day
    It is also called subfactorial (link).
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  10. #10
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    Re: expand (e^2x)/(2x+1) up to x^3

    The substitution method doesn't work for the same reason as with what I first tried. I have used the method given by SlipEternal but the combined sigma notation at the end I don't know how to obtain. I haven't seen subfactorial before so I won't be looking at the last one.
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  11. #11
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    Re: expand (e^2x)/(2x+1) up to x^3

    Quote Originally Posted by Stuck Man View Post
    The substitution method doesn't work for the same reason as with what I first tried. I have used the method given by SlipEternal but the combined sigma notation at the end I don't know how to obtain. I haven't seen subfactorial before so I won't be looking at the last one.
    All I did was apply the multinomial theorem. If you are working with the product of two infinite series, this should be well-known:

    \left(\sum_{n\ge 0}a_n x^n \right) \left(\sum_{n\ge 0} b_n x^n \right) = \sum_{n\ge 0} \left(\sum_{k=0}^n a_k b_{n-k} \right) x^n

    So, applying that to the product of infinite series I gave you, you get the solution I provided (the double summation at the end of post #6).

    Btw, Wolframalpha can compute the subfactorial, so if you want to check your answer: link. In the link I offered for subfactorial, it actually states that the exponential generating function is \dfrac{e^{-x}}{1-x}, so plugging in -2x, you get the same expansion I offered, so this is a correct solution.
    Last edited by SlipEternal; January 28th 2014 at 05:49 AM.
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  12. #12
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    Re: expand (e^2x)/(2x+1) up to x^3

    Quote Originally Posted by Stuck Man View Post
    The substitution method doesn't work for the same reason as with what I first tried. I have used the method given by SlipEternal but the combined sigma notation at the end I don't know how to obtain. I haven't seen subfactorial before so I won't be looking at the last one.
    well... if all else fails you can always find the Taylor series about 0.

    If you take the derivatives of f(x)=\frac{e^{2x}}{2x+1} and let x=0

    and form f(x)=\frac{e^{2x}}{2x+1} \approx \displaystyle{\sum_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k

    you find

    \frac{e^{2x}}{2x+1}\approx 1+2 x^2 - \frac{8}{3}x^3 + ...

    This is identical to it's MacLaurin series.
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