Legendre Polynomial Identity

I was looking for a method to do the following integral:

I know there should be an explicit representation for the result but I am struggling to work it out.

linear algebra – Problems relates to minimal polynomial and $AB, BA$

Let $A$ and $B$ be complex $3times 3$ matrices. The minimal polynomial of $A$ is $x^3-1$ and the minimal polynomial of $B$ is $(x-1)^3$. At this time show that $ABneq BA$

I know that the characteristic and minimal polynomials of $AB$ and $BA$ are the same at least one of $A$,$B$ is invertible. But in general, $AB$ and $BA$ do not have the same minimal polynomial. Am I right?

How do we answer this question? Any hints would be appreciated.

complexity theory – Polynomial kernelization for “Set coloring” problem

Given an universum $U$ and a system of its subsets $Fsubseteq mathcal{P}(U)$, given a $kin mathbb{Z}_{geq 0}$. The decision problem is to say whether there exists a $2$-coloring of $U$ s.t. atleast $k$ sets from $F$ contain elements of both colors.

I wish to show that there exists kernel with $2k$ sets and $|U|in O(k^2)$. So far I have these reduction rules:

Rule 1: If there is $xin U$ s.t. there’s no $Win F$ s.t. $xin W$, then we can delete $x$ from U and work with a new instance. $(F,Usetminus{x},k)$.

Rule 2: If $Win F$ is a singleton, then we can remove it and work with $(Fsetminus {W},U,K)$ because such $W$ will never contain two elements of different colors.

Not really sure how to proceed from here. Any hints appreciated.

ag.algebraic geometry – Under what conditions is the polynomial of degree $6$ irreducible?

Let $k$ be a perfect field of characteristic $p neq 2,3$ such that $omega := sqrt(3){1} in k$, where $omega neq 1$. Consider an absolutely irreducible (not necessarily homogenous) quadratic polynomial $Q in k(s_1, s_2)$ in two variables $s_1, s_2$. Under what conditions is the polynomial $Q^prime(t_1,t_2) := Q(t_1^3, t_2^3)$ (of degree $6$) absolutely irreducible (or at least irreducible over $k$) ?

Thank you in advance.

complexity theory – Is there a non-deterministic polynomial by time Turing machine such that: $L(M)in NPC$ and $L(overline{M})in P$

When $overline{M}$ is a non-deterministic polynomial by time Turing machine that final states switched: accept to reject and vice versa.
I’m thinking that this equal to $P=NP$, but I saw a solution (an example) that I disagree with:
$M$ is a non-deterministic polynomial by time Turing machine that decide $SAT$, if all that paths are rejected then $L(overline{M})=Sigma^*in P$

Is it a valid solution, or as I’m thinking $L(M)in NPC$ and $L(overline{M})in P Leftrightarrow P=NP$

algebraic geometry – Reducible polynomial & radicals of ideals..

I was at a seminar about algebraic geometry recently and the first point made was to find out if this object was reducible, thus I want to improve my intuition about reducibility of a polynomial generating an ideal by comparing it against intuition about the capabilities of that ideal having a radical:

If a polynomial is reducible and the factors are equal to each other then a radical of the ideal could exist, such as $x^{2}+2x+1 = (x+1)(x+1)$, $(x+1)^{2}$ is obviously in that ideal.

If the polynomials reducible and the factors don’t equal eachother then a radical could not be different such as $(x+1)(x-1) = x^{2} -1$, neither of the factors raised to a power could at least equal multiplying by the other factor.

If the polynomials irreducible then a radical could not exist the same fashion as a reducible polynomial without equal factors.

So that means in terms of Hilbert Nullstellensatz theorem whichs that $mathbb{I}(mathbb{V}(I)) = sqrt{I}$, the ideal of functions vanishing on the zero set of an ideal is the radical of that ideal, that if the ideal is the radical in the first place – it is simply an ideal thats obtained by a reducible polynomial without equal factors or an irreducible polynomial.

reference request – Interpolation estimate for trigonometric polynomial

J^s u := sum_{m in mathbb{N}} (1+vert m vert^2)^{s/2}hat u_m e^{imt}

Vert varphi Vert_{W^{s,p}} := Vert J^s u Vert_{L^p}

P_n u = sum^{n-1}_{m=-n} hat u_m e^{imt}

Q_n u = sum^{2n-1}_{k=0} u(t_k) L_k, quad t_k = frac{k}{n} pi, k = 0,1,cdots, 2n-1

L_k(t) := frac{1}{2n} sum ^{n-1}_{m=-n} e^{im(t – t_k)}, quad k = 0,1,cdots, 2n-1

The estimate
Vert varphi – P_n varphi Vert_{W^{s,p}} leq C frac{1}{n^{r-s}} Vert varphi Vert_{W^{r,p}}

holds for $ p = 2 $ and $ 1 <p < infty, p neq 2 $ with any $ -infty < s < r <infty $(see (McLean86)).

Replacing orthogonal projection $ P_n $ with interpolation operator $ Pi_n $, the above estimate of case $ p= 2 $ can be found in (Kirsch96) or (Kress14), that is,
Vert varphi – Q_n varphi Vert_{W^{s,p}} leq C frac{1}{n^{r-s}} Vert varphi Vert_{W^{r,p}}

for any $ 0 leq s leq r $ and $ r > frac{1}{2} $.

Question: Can the latter estimate also be extended to the case $ 1 <p < infty, p neq 2 $?


(McLean86) A Spectral Galerkin Method for a Boundary Integral Equation Mathematics of Computation Volume 47, Number 176 October 1986. Pages 597-607.

(Kirsch96) An Introduction to the Mathematical Theory of Inverse Problems. Springer, New York, 1996.

(Kress14) Linear integral equations Third edition. Springer-verlag, New York, 2014.

analysis – Taylor polynomial of degree $3$ around the point $0$ using the series expansion

I want to calculate the Taylor polynomial of degree $3$ around the point $0$ for the function $f(x)=tan (x)cdot log (1+2x)$ using the series expansion.

We have that $tan (x)=x+frac{x^3}{3}+frac{2x^5}{15}+ldots $ and $log (1+2x)=2x-2x^2+frac{8x^3}{3}-4x^4+ldots$.

So is the Taylor polynomial of degree $3$ for $f(x)$ equal to $xcdot (2x-2x^2)=2x^2-2x^3$ ?

Is there any proof that says "For each problem in NP there is a randomized algorithm that solves that problem in expected polynomial time."

Is it known that "For each problem in NP there is a randomized algorithm that solves it in polynomial time"? If not true then is there any proof of that. Or does it belongs to the unknown domain?

equation solving – NSolve – Problem with univariate long polynomial

Below I am trying to find the steady state of x1 (x2 is a func of x1) in terms of a variable q1. x1stst. Then I wish to compute q1 values giving “trace = 0” in equq1 (equq1 includes x1 steady states as a function of q1). Unfortunately NSolve doesn’t work due to the hideous polynomial of equq1

(*Find q1*)
q2 = 46.2857;
q3 = 0.111;
q4 = 5.714285714285714`;
q5 = 0.14285714285714288`;
NSolve(q1 q2  (q3 + x1)/(1 + x1) x2 - (q4 x1)/(q5 + x1) == 0 && 
   1/(1 + x1^2) - x2 == 0, {x1, x2});
x1stst = NSolve(
   q1 q2 (q3 + x1)/(1 + x1) 1/(1 + x1^2) - (q4 x1)/(q5 + x1) == 0, x1);
equq1 = -1 + (5.714285714285714` x1)/(0.14285714285714288` + x1)^2 - 
   5.714285714285714`/(0.14285714285714288` + x1) - (
   46.285714285714285` (0.111` + x1) x2 q1)/(1 + x1)^2 + (
   46.285714285714285` x2 q1)/(1 + x1);
NSolve(equq1 == 0 /. 
   x1stst((1)) /. {x2 -> 1/(1 + x1stst((1))^2)}, q1, Reals)

I tried Together Collect etc. non of them worked. Looked similar questions but again couldn’t find anything.

Any advice would be appreciated. Thank you.