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Jan
7
accepted How $(r-k){r \choose r-k}$ becomes $r{r-1 \choose r-k-1}$?
Jan
7
comment Why the generating function of $a_r=r$ is $xf'(x)$ instead of $f'(x)$?
Thanks. There's something I'm still on doubt, when I posted the question, I didn't fully understand $(iv)$. Does it mean that to obtain the generating function of an arbitrary sequence, say $a_r=\sin r$, I just need to derive $\sum_k a_k n^k$ and multiply it by $\sin r$?
Jan
7
asked Why the generating function of $a_r=r$ is $xf'(x)$ instead of $f'(x)$?
Jan
6
comment Why the sequence generated by $x+e^x$ is $(1,x,\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!}\ldots)$?
Yes. My mistake was to think that $x$ was a series like $e^x$, it's only a monomial, I guess.
Jan
6
comment Why the sequence generated by $x+e^x$ is $(1,x,\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!}\ldots)$?
Oh, now I get it. While $e^x$ is a series, $x$ is only a monomial. I thought it was a series too.
Jan
6
comment Why the sequence generated by $x+e^x$ is $(1,x,\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!}\ldots)$?
I don't understand: Why $x$ generates the sequence $(0,1,0,0,\ldots)$? I was thinking that it would generate the sequence $(0,1,2,3,4,\ldots)$.
Jan
6
comment Why the sequence generated by $x+e^x$ is $(1,x,\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!}\ldots)$?
Why the sequence for $x$ is just $(0,1,0,0,\ldots)$?
Jan
6
comment Why the sequence generated by $x+e^x$ is $(1,x,\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!}\ldots)$?
Isn't it needed to use $\frac{1}{1-x}$? There is a part in the book where they derive it via $\frac{1}{1-x}$, I also questioned myself why all that was needed, but perhaps it was to enforce that the generating function is on the form $\sum_k a_kn^k$.
Jan
6
asked Why the sequence generated by $x+e^x$ is $(1,x,\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!}\ldots)$?
Jan
6
accepted Differentiating $1/(1-x)$ using the fact that $[x^n]'=nx^{n-1}$?
Jan
6
revised Differentiating $1/(1-x)$ using the fact that $[x^n]'=nx^{n-1}$?
added 2 characters in body
Jan
6
comment Differentiating $1/(1-x)$ using the fact that $[x^n]'=nx^{n-1}$?
Yes. But somehow, I thought that I could do using only that fact.
Jan
6
comment Differentiating $1/(1-x)$ using the fact that $[x^n]'=nx^{n-1}$?
Oh, thanks. I really thought I could do it using only that fact.
Jan
6
asked Differentiating $1/(1-x)$ using the fact that $[x^n]'=nx^{n-1}$?
Jan
6
awarded  Nice Question
Jan
6
asked What is distinguishble and indistinguishble in generating functions?
Jan
6
asked Why is it important to have the closed form of a generating function?
Jan
5
comment Why the generating function of $\frac{2^K}{k!}$ is $e^{2x}$ instead of $e^{2^{x}}$?
Winther also pointed something very interesting in the comments. Take a look.
Jan
5
comment Why the generating function of $\frac{2^K}{k!}$ is $e^{2x}$ instead of $e^{2^{x}}$?
@Winther That was exactly what I was looking for. You should put it as an answer instead. In my course, we still didn't have a good training in sums, I'll have to study it alone in Knuth's Concrete Mathematics. That's the maing reason I couldn't guess that an equality of sums would solve it.
Jan
5
comment Why the generating function of $\frac{2^K}{k!}$ is $e^{2x}$ instead of $e^{2^{x}}$?
@Winther Yes, but instead of $2x$, it could be other thing that would still give me a power series. I'm curious about some sistematic way to show that it needs to be $2x$. Using $2x$ is just like a lucky guess, I'm curious about a method that shows that only $2x$ would work.