5,906 reputation
62584
bio website
location
age
visits member for 2 years, 11 months
seen yesterday

2d
comment Checking: finding extremals for a functional
DISTRACTED FOR WHAT?!
2d
revised Confusion about Hajós construction?
added 63 characters in body
2d
asked Confusion about Hajós construction?
Jan
18
comment Why don't we use Presburger's arithmetic instead of Peano's arithmetic?
@AndréNicolas Why can't we define it?
Jan
18
comment Why don't we use Presburger's arithmetic instead of Peano's arithmetic?
@AsafKaragila I'm not sure what is the meaning of write something in theory X. But I guess that it means to write something in terms of the given axioms. I'm also not really sure how I should write that proposition with Peano arithmetic. Could you expand a little or provide me with some reference on how to do it?
Jan
18
comment Why don't we use Presburger's arithmetic instead of Peano's arithmetic?
@AndréNicolas I don't understand. Could you expand?
Jan
18
asked Why don't we use Presburger's arithmetic instead of Peano's arithmetic?
Jan
18
comment Why is it differential equations exist on an interval instead of a domain?
@MarkMcClure I guess he's speaking about domains in the sense of $\mathbb{R},\mathbb{Q},\mathbb{C}$, etc. How are these domains called? In order to differentiate them from having an interval as a domain.
Jan
17
comment Why the generating function of $x^2(1-3x)^{-1}$ is $(0,0,1,3,3^2,3^3)$ instead of $(0,0,1,-3,3^2,-3^3)$?
Oh, got it. I was just expanding with the formula. Sum[Binomial[-1, n] (3 x)^n, {n, 0, 6}]. But what's not clear to me is why what I made doesn't work. It gives almost the same numbers modulo-wise.
Jan
17
comment Why the generating function of $x^2(1-3x)^{-1}$ is $(0,0,1,3,3^2,3^3)$ instead of $(0,0,1,-3,3^2,-3^3)$?
Oh, got it. I was just expanding with the formula. Sum[Binomial[-1, n] (3 x)^n, {n, 0, 6}]. But what's not clear to me is why what I made doesn't work. It gives almost the same numbers modulo-wise.
Jan
17
asked Why the generating function of $x^2(1-3x)^{-1}$ is $(0,0,1,3,3^2,3^3)$ instead of $(0,0,1,-3,3^2,-3^3)$?
Jan
17
awarded  Nice Question
Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
What is the meaning of $\displaystyle \left( {n \choose k}\right)$?
Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
@GrigoryM Nevermind. Did you see what I suggested? That's what's leaving me a little bit confused.
Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
@GrigoryM See here. Excerpt from Knuth/Patashnik/Graham's Concrete Mathematics, page 154.
Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
@graydad Yes. I am the one who made the question pointed by GrigroyM, but I am trying to look at it with a different intention.
Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
@GrigoryM Not, it is not. It's a completely different question, read both and check it. At that question, I asked how to use the definition to check that, in this question, I'm askinf if I could assume that $\frac{(-1)!}{(-1)!}=1$.
Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
@NajibIdrissi Yes, it appeared in a formula. But the propositions in the books I've read work with it perfectly. I'm reading a brazilian book on combinatorics and I also read a bit of Knuth/Patashnik/Graham's Concrete Mathematics. Im both books, they appear to make sense, the latter book even discusses about it.
Jan
16
comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?
@graydad I still don't know about the gamma function.
Jan
16
asked On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$?