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
- Jan 27th 2014, 02:48 PMSlipEternalRe: expand (e^2x)/(2x+1) up to x^3
Here is another method:

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

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

Another way to write this:

- Jan 27th 2014, 03:53 PMSlipEternalRe: expand (e^2x)/(2x+1) up to x^3
Oh, thinking about it, where is the number of derangements of an element set. So, you can write it:

- 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:

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 , so plugging in , 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