Suppose $A=(a_{ij})$ is a n×n matrix by $a_{ij}=\sqrt{i^2+j^2}$. I have tried to check its sign by matlab. l find that the determinant is positive when n is odd and negative when n is even. How to prove it？

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Jimmy

Question: How to show the relation $$ J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\frac 1{64}\pi^4 $$ (using a "minimal industry" of relations, possibly remaining inside the real analysis)?

So i have found a solution to the problem, it is part of my solution for math.stackexchange.com - questions - 3854736, but not a satisfactory solution. "There should be more", explaining why there is a "clean result" for the integral.

Here, i am not strictly interested in a computational approach. I just want to share this with the community in these days of isolation. Any idea to attack this, or a related integral involving "three log factors" is welcome. (Well, the $\arctan$ is a sort of $\log$ in a sense that i don't want to define closer, see below.) Computations may be safely done "modulo integrals involving two or one log factor". But an illuminating, short way to show the above formula for $J$ would be wonderful.

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Motivation: The above relation appeared as i tried to solve the integral posted at the above link:

Calculate $\displaystyle\int_0^{2\pi} x^2\; \cos x \cdot\operatorname{Li}_2(\cos x)\; dx$ .

After several simplifications and substitutions, it turns out that the above integral is related to integrals of the shape

• $\int_0^1\log t\; R(t)\; dt$ , and
• $\int_0^1\arctan t\cdot \log t\; R(t)\; dt$ , and
• $\int_0^1\arctan^2 t\cdot \log t\; R(t)\; dt$ ,
• and "similar" expressions.

Here $R$ is in each case a (rather simple) rational function. (The more log and/or arctangent factors, the higher the computational complexity.)

I could compute more or less algorithmically most of the the needed integrals to solve the linked problem, all of them but the integral $$ K=\int_0^1\arctan^2 t\cdot\log t\cdot\frac2{1-t^2}\; dt\ , $$ which turned out to be very hard to attack with the methods of real analysis. Computing this integral is more or less equivalent to computing $J$, and the question wants $J$ instead, since we have a "clean formula", so that some speculation about a "clever substitution" may be accepted.

My solution (for $K$) works in complex analysis, the first step is to write $$ \int_0^1 =\int_0^i+\int_i^1\ , $$ then parametrize the first integral using a linear path, the second one using a path on the unit circle.

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Some comments: I will say some more words, because the situation is rich in coincidences. Since a numerical evidence is the simplest and shortes way to present (instead of showing how to show), i will use this method to at least list the coincidences. Many equalities below are "equivalent" (modulo computation of integrals of lower complexity) to the formula for $J$.

• First of all, a numerical experiment using pari/gp delivers some connection between $K$ and a "cousin" of $J$:

    ? 2 * intnum( t=0, 1, atan(t)^2 * log(t) / (1-t^2) )
    %88 = -0.357038604620289042902893412499686912781214141574556097366337
    ? real(intnum( t=0, I, (pi/4 - atan(t))^2 * log(t) / (1-t^2) ))
    %89 = -0.357038604620289042902893412499686912781214141574556097366337
    ? intnum( t=0, 1, (pi/4 - atan(t))^2 * log(t) / (1-t^2) )
    %90 = -0.357038604620289042902893412499686912781214141574556097366337
  

In words: $$ \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &= \Re \int_0^i\left(\frac \pi 4-\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \\ &= \int_0^1\left(\frac \pi 4-\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \ . \end{aligned} $$ Note the integration margins. What happens if we take the integral on $[0,i]$ instead of $[0,1]$ in the $K$-integral? Numerically:

    ? 2 * real(intnum( t=0, i, atan(t)^2 * log(t) / (1-t^2) ))
    %98 = 1.52201704740628808181938019826101736327699352613570971392919
    ? pi^4/64
    %99 = 1.52201704740628808181938019826101736327699352613570971392919

In words: $$ \begin{aligned} K^* &:= \Re\int_0^i\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &=\frac 1{64}\pi^4 \\ &= -\int_0^1\left(\frac \pi 4+\arctan t\right)^2 \cdot\log t\;\frac 1{1-t^2}\; dt \\ &=-J\ . \end{aligned} $$

(These observations were leading to the formula for $K$ in loc. cit. .)

• One idea is to use partial integration in $J$ or $K$. Well, we have for $K$: $$ \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\left(-\log\frac {1-t}{1+t}\right)'\; dt \\ &= \underbrace{\int_0^1\arctan^2 t\cdot\frac 1t\cdot \log\frac {1-t}{1+t}\; dt}_{=2K\text{ (why?)}} \\ &\qquad\qquad+ \underbrace{\int_0^12\arctan^2 t\cdot\frac 1{t+t^2}\cdot \log t\cdot \log\frac {1-t}{1+t}\; dt}_{=-K\text{ (why?)}} \ . \end{aligned} $$

• Note that $\arctan$ is related to the logarithm (over $\Bbb C$), we have the relation (around $0$) $$ \arctan t=\frac 1{2i}\log\frac {1+it}{1-it}\ . $$ The substitution $t=\frac{1-u}{1+u}$ and the formula for $\tan(\arctan 1-\arctan u)$ are giving: $$ \begin{aligned} K &= \int_0^1\arctan^2 t\cdot\log t\;\frac{2}{1-t^2}\; dt \\ &=\int_0^1 \left(\frac\pi2-\arctan u\right)^2\cdot\log\frac {1-u}{1+u}\cdot \frac {du}u\ . \\ &=\int_1^\infty \left(\frac\pi2-\arctan u\right)^2\cdot\log\frac {u-1}{1+u}\cdot \frac {du}u\ . \end{aligned} $$ (Write $\log t=\frac 12\log t^2$ to have the same expression under the integral on $(0,1)$ and on $(1,\infty)$.)

• Note the fact that the factor $\frac 2{1-t^2}$ is not "random". It is the right one to make things feasible. It is the derivative of $\displaystyle -\log\frac{1-t}{1+t}$, and plugging in $t=iu$ into $\displaystyle \log\frac{1-t}{1+t}$ leads to an expression related to $\arctan u$. And conversely, $\arctan(iu)$ is related to such a logarithmic expression in $u$.

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dan_fulea

4 Jobs You Can Do With a Major in Software Engineering

from Math Blog https://ift.tt/3eWcxRh

In ordinary discourse, when we say that $A$ implies $B$, we shall formalize it by writing the following: $$A\rightarrow B$$ But when we say that $A$ does not imply $B$, we cannot formalize it as the following: $$\neg(A\rightarrow B)$$ Because by material implication, we have the following equivalences for the second formula: $$\neg(A\rightarrow B)\Longleftrightarrow\neg(\neg A\vee B)\Longleftrightarrow A\wedge\neg B$$ But the ordinary meaning of the sentence "$A$ does not imply $B$" is that $B$ does not follow from $A$: if $A$ is true, $B$ is either false or undecidable. I wonder how "$A$ does not imply $B$" is formally expressed in the object language (not at the meta-level). Thanks!

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Fred

Given the following recurrence relation :

$f(n) = 5f(n-1) - 2f(n-2)$ where $f(0) = 0, f(1) = 1$

I need to find out if an integer $F_n$ is present in the sequence in $O(1)$ time and space.

Solving the equation, there are two distinct real roots.

$\phi = \frac{5 + \sqrt17}2$

$\psi = \frac{5 - \sqrt17}2$

Therefore, $F_n = \frac{\phi^n - \psi^n}{\sqrt17}$

Similar to Binet's rearranged formula, I want to solve for $n$ in terms of $F_n$.

Since, $\psi = \frac{2}{\phi}$

$\sqrt17F_n = \phi^n - \frac{2^n}{\phi^n}$

$Or,$

$\phi^{2n} - \sqrt17F_n\phi^n-2^n = 0$

Here I'm not able to find out a solution to express $n$ purely in terms of $F_n$ so that I can calculate the perfect square just like in Binet's formula.

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Rohit Roy Chowdhury

I was looking into the proof that there are only five Platonic solids in Basic Concepts of Algebraic Topology by F.H. Croom at page 29, Theorem 2.7. To clarify,

• We define a Platonic solid as a simple, regular polyhedron.
• We define a simple polyhedron to be a polyhedron homeomorphic to the $2$-sphere.
• We define a regular polyhedron to be a polyhedron whose faces are regular polygons all congruent to each other and whose local regions near the vertices are all congruent to each other.

Using homology theory, one can prove that the Euler formula $V-E+F=2$ must hold for Platonic solids. Then by using Euler's formula and invoking a counting argument, we find that there are five possible tuples $(V, E, F)$. This is a beautiful proof, but I am unsatisfied with a question: How do we know there can't be two non-similar Platonic solids that have the same $(V, E, F)$-tuple?

Almost all sources I've looked at seem to assume it is obvious that two Platonic solids with the same $(V, E, F)$-tuple are similar, and it is not obvious to me.

Does anyone have any suggestions for how to prove this? Alternatively, does anyone know of a reference where this is proved rigorously?

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Edit 1: It's not completely clear, but it seems like the definition I used for "regular polyhedra" is different than the one commonly used. Note that I am not assuming any global symmetry, so if any global symmetry is to be invoked, it needs to be proven.

Edit 2: I've been made aware of Cauchy's rigidity theorem, which is proven in, e.g., Proofs From the BOOK by Aigner & Zeigler. One can show that any two Platonic solids that have the same $(V, E, F)$-tuple must be combinatorically equivalent. However, in order for the theorem to apply, we need to show that our Platonic solids are convex. I can't seem to think of any rigorous argument for why the Platonic solids have to be convex.

And actually, you don't need to show that the entire polyhedron is convex. If I'm not mistaken, the proof for Cauchy's rigidity theorem only relies on the vertices of the polyhedron being locally convex. So really it suffices to show the vertices are convex.

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Maximal Ideal