in Combinatory edited by
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Find the sequence with each of these functions as its exponential generating function
g(x) = $e^{-2x} - \frac{1}{1-x}$

so I expanded it like
$\sum_{r=0}^{\infty}\frac{(-2^r)x^r}{r!} \,-\sum_{r=0}^{\infty}x^r$
and finally after resolving it i got expression below

$\left\{ (\frac{-2-1!}{1!}).x + (\frac{2^2-2!}{2!}).x^2  + (\frac{-2^3-3!}{3!}).x^3 + (\frac{2^4-4!}{4!}).x^4+.......\right\}$
so This is my exponential series that I got and it's coefficient term can be generalized as
$a_n=(-2)^n -n!\, , for\, n\,\geq1$ and $a_0=0$
but rosen's key says it's $a_n=(-2)^n+n!$ and they even have not given any condition which implies it is valid for $n\geq0$. So, according to rosen's answer $a_0$ should exist but in my answer $a_0$ is 0. Please let me know where I am missing something.
in Combinatory edited by
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4 Comments

Anyway, Your solution is correct... And Why do you require to specify $a_0 = 0$ separately ? It is implied from $(-2)^n - n!$.
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Itna kuch krne ki jarurat nhi

See my solution
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Is there anything wrong in solution?
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1 Answer

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$G\left ( x \right )=e^{-2x}-\frac{1}{1-x}$

$G\left ( x \right )=\sum \frac{\left ( -2 \right )^{n}.x^{n}}{n!}-\sum x^{n}$

                  $=\sum \left ( \frac{\left ( -2 \right )^{n}.}{n!}- 1 \right ).x^{n}$

$a_{n}=\left ( \frac{\left ( -2 \right )^{n}.}{n!}- 1 \right )$

3 Comments

@sreshtha-it has to expressed in terms of exponential generating function and not the regular case of $x^r$
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It is exponential generating function only

where u got it is not exponential generating??

G(x) always holds that property
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@Sreshtha-Yours and deepak's answers are correct.

It was given in rosen

The exponential generating function for the sequence $a_n$ is the series

$\sum_{n=0}^{\infty}\frac{a_n}{n!}x^n$

so for $e^x$ this $a_n$ is 1.

We had to express $a_n$ as given definition of above exponential generating function disregarding the n! the term in the denominator.

Well, you and Deepak have taken a general view of the GF as $\sum_{n=0}^{\infty}a_n x^n$ and in that way you expressed your $a_n$ and that is also correct. I was getting confused between the two and troubled up you guys :P
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