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Nov
30 |
comment |
Summing a function using modulus.
If I need ask things here and just here (what's is annoying thing) tell me that I'll remember. But, but you may point in faq where is it? Or put there (not in meta) that's not a polite thing. Maybe I had been some agressive, but I think all community (even me, a beginner here) should opine to have a better site. |
Nov
30 |
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Summing a function using modulus.
Well it has been very difficult join to this community. I think I have good questions, with Mathematics interest (like said in faq). The truth is, I didn't asked here because I didn't get ansser in Math.SE, I asked here because I wanted too, I would like another different opinions. I had seen questions in Math.SE with over 30 votes with no answer and answered here. So, what's the problem? If I ask in Math.SE and write in the question, you close my question. I I ask there too and write in the question, you close my question. Do you hate the Math.SE, didn't like no cross platafform? |
Nov
28 |
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Summing a function using modulus.
@AH, What about if $f$ a polynomial or a analytic function. |
Nov
28 |
revised |
Summing a function using modulus.
added 514 characters in body |
Nov
28 |
revised |
Summing a function using modulus.
added 2 characters in body |
Nov
28 |
awarded | Supporter |
Nov
28 |
revised |
Summing a function using modulus.
added 227 characters in body |
Nov
28 |
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Summing a function using modulus.
@JacquesCarette. I think you're right about the $S_f$ and $MS_f$. Your formula using $x_0=0$ and $m=2$ shows what I need. And you're right again when you say I'm looking for something like $MS_f(x,x_0,m)=\displaystyle\sum_{i=0}^{m-1}a_iS_f(w^ix)$. I already know this A=B book and it's really a good one, about the other I'm looking for. But isn't clear to me how to find this $a_i$'s, can you point something or a algorithm on the books to treat this? |
Nov
28 |
revised |
Summing a function using modulus.
edited title |
Nov
28 |
asked | Summing a function using modulus. |
Nov
7 |
awarded | Commentator |
Nov
7 |
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Matrices. $XX^t=A$. $X=?$
@BrendanMcKay I will try to get this article and take a look and return... But I think this will not can help me so much, I need find the other solutions of this problem, because this solution, by adjacency matrix I already know but I need the others, if exists. |
Nov
7 |
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Matrices. $XX^t=A$. $X=?$
@BrendanMcKay says, I'm investigating graphs but by linear algebra. And I think, but don't know why yet, it's possible find a good method to find the solutions because in that link of the integers matrices, he claims to find all integers matrices, so, binary matrices should be more easy to find but that process it's a bit fuzzy to me. |
Nov
6 |
asked | Matrices. $XX^t=A$. $X=?$ |
Nov
2 |
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How to isolate $f(x)$ in $f(x+a)=f(x)+a\times g(x)$?
Didn't understand very well. Can you give an example? For example, using $a=1$ ang $g(x)=\frac{-1}{x(x-1)}$ who mathematica don't solves. |
Nov
1 |
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How to isolate $f(x)$ in $f(x+a)=f(x)+a\times g(x)$?
I think use DiracDelta is special functions no? And using Fourier Transforms, for example, DiracDelta appears frequently. |
Nov
1 |
revised |
How to isolate $f(x)$ in $f(x+a)=f(x)+a\times g(x)$?
edited tags; added 119 characters in body |
Nov
1 |
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How to isolate $f(x)$ in $f(x+a)=f(x)+a\times g(x)$?
@shrdlu Yes I agree, but if $f$ is unknown, how to find, at least, one solution to $f$? |
Nov
1 |
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How to isolate $f(x)$ in $f(x+a)=f(x)+a\times g(x)$?
@BR , Are you saying it's possible find $f(x)$ in my exampling using Inverse Fourier Transform, but Mathematica, don't do it? |
Nov
1 |
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How to isolate $f(x)$ in $f(x+a)=f(x)+a\times g(x)$?
@AliBleybel I can't embrace you answer as solution how it presents now. When I put the FourierTransform link and that formulas, I tried it before. I tried isolate $f(x)$ using that. And tried using ZTransform too. It's not a easy question. Maybe we can use ZTransform to several things and FourierTransform to many others and other transform and find all solutions. Don't know, but if $a=1$ or $a\implies 0$, I think there limitated solutions. And I'm searching a way to find them. |