Wikipedia:Reference desk/Archives/Mathematics/2018 March 7

= March 7 =

Functional Completeness-Related Question
A is the set of formulae that are conjunction of the following clauses:
 * clauses of the form $$\neg a$$ or $$\neg a\vee\neg b$$
 * clauses with the connector $$\vee$$ only (without $$\neg$$)
 * clauses with the connectors $$\wedge, \to$$ (without $$\neg$$), such that $$\to$$ appears in the clause exactly once

B is set of formulae defined just like A, but in the last kind of clauses they have also the connector $$\vee$$.

Is that true that $$\forall\psi \in B\exists \varphi\in A: \forall x\in\{0,1\}^n(\exists y\in\{0,1\}^m. \varphi(x,y)\leftrightarrow \exists z\in\{0,1\}^k. \psi(x,z))$$?

I think I can solve this somehow with a technique similar to the one used for functional completeness proofs, but I can't figure out how exactly. עברית (talk) 19:53, 7 March 2018 (UTC)
 * Pretty sure I'm not understanding something, but it looks to me like A⊋B, so you could just take φ to be ψ. Maybe explain the context more? --RDBury (talk) 14:14, 8 March 2018 (UTC)
 * It's the opposite: B⊋A, since B has the extra connector. Is my question clear now? עברית (talk) 16:44, 8 March 2018 (UTC)

However, I found a better formulation of what I need:

We say that a formula $$\varphi\in C_{n,k}$$ is 3-symmetric if $$\forall x\in\{0,1\}^n, y\in\{0,1\}^k, \sigma_1,\sigma_2,\sigma_3\in S_{k/3}:\varphi(x,y)=\varphi(x,\sigma_1(y_1),\sigma_2(y_2),\sigma_3(y_3))$$ (where $$S_n$$ is the symmetric group)

$$A_k$$ is the set of k-variable formulae that are conjunction of expressions (or, clauses) that satisfy the following property: each expression uses the connectors $$\wedge, \to$$ (but not $$\neg,\vee$$), such that $$\to$$ appears in the clause exactly once

$$B_k$$ is defined just like $$A_k$$, but in that case using the connector $$\vee$$ is also allowed.

$$C_k$$ is the set of k-variable formulae that satisfy the 3-symmetry.

Is the following statement true?

$$\forall n, m \exists d\exists k\leq (n+m)^d\forall \varphi\in C_{n,\max\{m,k\}} \forall\psi_b\in B_m \exists\psi_a\in A_k \forall x\in\{0,1\}^n $$ $$(\exists y\in\{0,1\}^k. \varphi(x,y)\wedge\psi_a(y)\leftrightarrow \exists z\in\{0,1\}^m. \varphi(x,z)\wedge\psi_b(z))$$

Motivation: This is a definition of equivalence between two formulae (of course, it's different from the usual definition). I would like to define them as equal if either both have a "true" in their truth table or both don't have. But this definition is too weak, so I strengthened the definition: Even after "conjuncting" with any other formula, both have or both don't have a "true" in their truth tables. Finally, the question is whether the formulae in A are equivalent to the formulae in B. That is, whether adding the connector $$\vee$$ enlarges the group also under this definition of equivalence. עברית (talk) 18:39, 8 March 2018 (UTC)
 * I misinterpreted what you said before, but I'm still not entirely sure I understand the difference between A and B or what difference "conjuncting" with other expressions makes. In any case, maybe you should start with the simplest non-trivial case. I think here it would be m=3, n=0, φ(p, q, r) = p→(q∨r), ψb = True. What is the A-equivalent form of φ in this case? Once you have a few examples to go on, you will need to construct some kind of inductive argument, presumably based on the length of φ. This seems like a lot of work, but it would help to define A and B recursively as they do for formal languages. --RDBury (talk) 08:21, 9 March 2018 (UTC)


 * Could you clean up the statement you're asking about? You've clearly made mistakes in stating it.  For example, phi has arity n, but you've fed it a tuple of length n + k + m.2601:245:C500:754:4831:C848:4E82:B552 (talk) 12:48, 9 March 2018 (UTC)


 * Thank you. I fixed this, and cleaned up my statement.
 * I thought about some solution: we can just copy all of the variables in the disjunction to separated clauses: for example $$\psi_b=(a\wedge b)\to (c\vee (d\wedge e))$$ goes to $$\psi_a=[(a_1\wedge b_1)\to c]\wedge [(a_2\wedge b_2)\to (d\wedge e)]$$.
 * However, in such a way the number of variables grows exponentially, so I added a new requirement that k is polynomial in n,m. עברית (talk) 07:06, 10 March 2018 (UTC)

Associated Legendre polynomials
I've done a few experiments with Mathematica, but have not found the general result.

Am I correct that the set $$\frac{d}{dx} P_\ell^{2}(x), n \geq 2$$ form an orthogonal set of polynomials under $$ \int^1_{-1} \frac{d}{dx} P_\ell^{2}(x)\frac{d}{dx} P_k^{2}(x) dx $$? If so, what is the result for $$ \int^1_{-1} \left(\frac{d}{dx} P_\ell^{2}(x) \right)^2 dx $$?--Leon (talk) 21:19, 7 March 2018 (UTC)


 * I get


 * $$\begin{align}\int _{- 1}^{1}\left( \frac{\partial}{\partial x} {P}_2^2(x)\right) \left( \frac{\partial}{\partial x} {P}_4^2(x)\right) \partial x & = \int _{- 1}^{1}\left( \frac{\partial}{\partial x} (3 (1 - x^{2}))\right) \left( \frac{\partial}{\partial x} \left( \frac{15}{2} (7 x^{2} - 1) (1 - x^{2})\right)\right) \partial x \\

& = \int _{- 1}^{1}180 (7 x^{4} - 4 x^{2}) \partial x \\ & = \left[ 180 \left( 7 \frac{x^{5}}{5} - 4 \frac{x^{3}}{3}\right)\right]^{x = 1}_{x = - 1} \\ & = 24 \end{align}$$


 * which is non-zero. The integral is zero if one of $$k$$ and $$l$$ is odd and one even, since the integrand is then odd in $$x$$. --catslash (talk) 12:38, 8 March 2018 (UTC)
 * I am not sure why you think that the first derivatives of associated Legendre polynomials form an orthogonal set? And what is n?Ruslik_ Zero 20:25, 8 March 2018 (UTC)
 * n is the degree, as is clear from the condition that it be no less than the order - so $$\frac{d}{dx} P_n^{2}(x), n \geq 2$$ was intended. Orthogonality is not an entirely daft supposition, and could still be the case with the introduction of a suitable weight function. --catslash (talk) 13:15, 9 March 2018 (UTC)


 * My goals is a "complete" set of polynomials with the following properties.


 * 1) $$P_n(-1) = P_n(1) = 0$$.
 * 2) $$\int^1_{-1} P_n(x) P_m(x) dx = 0$$ if and only if $$n \neq m$$.
 * 3) $$\int^1_{-1} P'_n(x) P'_m(x) dx = 0$$ if and only if $$n \neq m$$.
 * When I says "complete", I mean being able to express all polynomials for which $$P_n(-1) = P_n(1) = 0$$ is true, not all polynomials full stop.
 * I tried something with sines and cosines instead if polynomials, but whilst it "worked", it lacked other properties that I wanted.--Leon (talk) 13:33, 9 March 2018 (UTC)
 * You can read this or this. Ruslik_ Zero 19:28, 9 March 2018 (UTC)