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21h
comment Checking: finding extremals for a functional
DISTRACTED FOR WHAT?!
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
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
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
13
comment How to identify which terms are infinite sequences?
@ThomasAndrews Could you suggest something better? I'm just speaking about it as I looked in the book.
Jan
7
comment Why the generating function of $a_r=r$ is $xf'(x)$ instead of $f'(x)$?
Got it. Then the only allowable product are the products that leave it as a power series?
Jan
7
comment Why the generating function of $a_r=r$ is $xf'(x)$ instead of $f'(x)$?
Oh, good. But does that work for arbitrary sequences? My guess is yes, but I'm really not sure. For example, would it be possible to do as the example I gave earlier [$\sin x f'(x)$]?
Jan
7
comment Why the generating function of $a_r=r$ is $xf'(x)$ instead of $f'(x)$?
But then, the $x$ in this case could be other sequences too? Is it possible to make - for example - $2xf'(x)$?
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
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.