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

Printable View

- Jan 27th 2014, 05:42 AMStuck Manexpand (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.

- Jan 27th 2014, 06:01 AMromsekRe: expand (e^2x)/(2x+1) up to x^3
- Jan 27th 2014, 06:25 AMStuck ManRe: 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.

- Jan 27th 2014, 10:47 AMStuck ManRe: 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.

- Jan 27th 2014, 02:10 PMromsekRe: expand (e^2x)/(2x+1) up to x^3
what you could do is let

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

then

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

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

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

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

and just substitute $\displaystyle u=2x+1$ in the expression above. - Jan 27th 2014, 02:48 PMSlipEternalRe: expand (e^2x)/(2x+1) up to x^3
Here is another method:

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

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

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

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

Another way to write this:

$\displaystyle \sum_{n\ge 0} \left( \sum_{k = 0}^n \dfrac{(-1)^{n-k}}{k!} \right) (2x)^n$ - Jan 27th 2014, 03:53 PMSlipEternalRe: expand (e^2x)/(2x+1) up to x^3
Oh, thinking about it, $\displaystyle \sum_{k=0}^n \dfrac{(-1)^{n-k}}{k!} = (-1)^n\dfrac{!n}{n!}$ where $\displaystyle !n$ is the number of derangements of an $\displaystyle n$ element set. So, you can write it:

$\displaystyle \dfrac{e^{2x}}{2x+1} = \sum_{n\ge 0}(-2)^n\dfrac{!n}{n!}x^n$ - Jan 27th 2014, 03:59 PMromsekRe: expand (e^2x)/(2x+1) up to x^3
first time I've ever seen !n... learn something new every day

- Jan 27th 2014, 04:06 PMSlipEternalRe: expand (e^2x)/(2x+1) up to x^3
It is also called subfactorial (link).

- Jan 28th 2014, 05:26 AMStuck ManRe: 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.

- Jan 28th 2014, 05:36 AMSlipEternalRe: expand (e^2x)/(2x+1) up to x^3
All I did was apply the multinomial theorem. If you are working with the product of two infinite series, this should be well-known:

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

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 $\displaystyle \dfrac{e^{-x}}{1-x}$, so plugging in $\displaystyle -2x$, you get the same expansion I offered, so this is a correct solution. - Jan 28th 2014, 05:46 AMromsekRe: expand (e^2x)/(2x+1) up to x^3
well... if all else fails you can always find the Taylor series about 0.

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

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

you find

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

This is identical to it's MacLaurin series.