- Evaluating a definite integral by introducing change of variablesby JD_PM on March 5, 2021 at 9:36 am
In a Physics class, I encountered the following integral: $$\int_{0}^{\infty} \left(\frac{x^2}{\exp\left(x-a\right)+1}-\frac{x^2}{\exp\left(x+a\right)+1}\right)\,\mathrm{d}x\text{,}$$ where $a$ is a constant. It is stated to yield: $$\int_{0}^{\infty}\left(\frac{x^2}{\exp\left(x-a\right)+1}-\frac{x^2}{\exp\left(x+a\right)+1}\right)\,\mathrm{d}x=\frac{1}{3}\left[\pi^2a+a^3\right]$$ I first introduced a change of variables: $$y:=x-a$$ $$z:=x^\prime+a$$ This gives: $$\int_{0}^{\infty}\frac{x^2}{\exp\left(x-a\right)+1}\,\mathrm{d}x-\int_{0}^{\infty}\frac{\left(x^\prime\right)^2}{\exp\left(x^\prime+a\right)+1}\,\mathrm{d}x^\prime=\int_{-a}^{\infty}\frac{\left(y+a\right)^2}{e^y + 1}\,\mathrm{d}y-\int_{a}^{\infty}\frac{\left(z-a\right)^2}{e^z+1}\,\mathrm{d}z$$ However, when introduced to a computer, the definite integrals are stated to have no solution. I actually expected to get integrals of the form: $$\int_{0}^{\infty}\frac{y}{e^y+1}\,\mathrm{d}y=\frac{\pi^2}{12}$$ What am I missing here?

- Set with a propertyby brudder on March 5, 2021 at 1:17 am
This was a mock math olympiad problem sent to me by a friend. Let $P$ be a set of prime numbers. Then, create a set $S$ of positive integers that satisfies the following property: For every element $p \in P$, $p$ is a factor of at least three elements in $S$. Prove that for all sets $P$ and $S$, it is possible to divide $S$ into 4 nonempty subsets such that each $p$ is a factor of at least three integers. I tried to consider the element $a \in P$ that was a factor of the least number of elements in $S$ and see what I could discover, but got nowhere. I also thought that perhaps some sort of algorithm would work too, but I didn’t make any progress. Overall, I’m just not quite sure how to even start.

- Looking for an other method to compare $100!$ and $50^{100}$by Khosrotash on March 4, 2021 at 7:46 pm
There is a question asked me to compare $100! \ and \ 50^{100}$. I think more than 2 hours to solve it. I show my work below, but I looking for other Ideas to prove the inequality. Thanks in advance for any hints or new Ideas. my work : $$50 \times 50 > 1 \times 99 \times 2\\ 50 \times 50 > \ 2 \times 98\\ 50 \times 50 > \ 3 \times 97\\ 50 \times 50 > \ 4 \times 96\\ \vdots \\ 50 \times 50 > \ 48 \times 52=(50+2)(50-2)=50^2-4\\ 50 \times 50 > \ 49 \times 51=(50+1)(50-1)=50^2-1\\ 50 \times 50 \geq 50 \times 50 $$ then multiply them $$50^{49}\times 50^{49}\times 50^2 >1.2.3…50….99.(2.50)\\50^{100}>100!$$ Remark: I use numerical approximation for $50^{100}\sim 7.8886\times 10^{169} $ and $100!=9.33236\times 10^{157}$ But is not interesting( Ithink).

- Does this discrete curve have a name?by B Williams on March 4, 2021 at 4:47 pm
Generated using the following logic. Begin at a center starting point and draw two one-unit line segments from the center point in opposite directions. Draw two one-unit line segments that begin at the end of each of the previous step’s endpoint, perpendicular to the previous segment and opposite one another. Continue doing this while a given path does not terminate. A path terminates if both new segments would touch another path from this or a previous step. Here is a visual of the first few generations, center point and terminal points marked.

- Are there nontrivial rational solutions to $\sqrt{1-x^2} + \sqrt{1-y^2} = \sqrt{1-z^2}$?by Elmusfire on March 4, 2021 at 11:56 am
Obviously $(a,0,a)$,$(-a,0,a)$,$(0,a,a)$,$(0,-a,a)$ are solutions. I tried finding solutions brute-forcing this problem, but I discovered there are no solutions with numerator and denominator smaller than 100’000. I can prove that the question is equivalent to finding a bunch of natural numbers that satisfy: $x’^2+ka^2 = w^2\\y’^2+kb^2 = w^2\\z’^2+kc^2 = w^2\\a+b=c$ (here is $x=x’/w$, $y=y’/w$, $z=z’/w$) It can be proven they are equivalent as follows: To go from the title to the 4 equations: Call $w$ the lcm of the denominators of x, y, z Define $x’=xw$, $y’=yw$, $z’=zw$ The equation is now $\sqrt{w^2-x^2} + \sqrt{w^2-y^2} = \sqrt{w^2-z^2}$ This equation can only hold if all sqrt are natural numbers, up to a common factor $k$ by this theorem Now call $a=\sqrt{w^2-x^2}/\sqrt{k}$, $b=\sqrt{w^2-y^2}/\sqrt{k}$, $c=\sqrt{w^2-z^2}/\sqrt{k}$ The original equation becomes $a+b=c$, the other 3 can be calculated by the definitions of the previous step. To go from the 4 equations to the title: Rewrite the first 3 equations to $a=\sqrt{w^2-x^2}/\sqrt{k}$, $b=\sqrt{w^2-y^2}/\sqrt{k}$, $c=\sqrt{w^2-z^2}/\sqrt{k}$ Plug in in equation 4. Substitude $x=x’/w$, $y=y’/w$, $z=z’/w$ But apparently being able to split a number into two squares in at least three different ways, is quite a rare property, so I would suspect that finding three with the extra property that $a+b=c$ is probably impossible, which makes me believe that there are no solutions, or that solutions are very rare. But is there a way to find a solution, or prove that non exists except for the trivial case? Edit: Added proof of equivalence.

- $\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$by DatBoi on March 4, 2021 at 6:45 am
Evaluate $$S=\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$$ Its quite simple to conclude that this limit does not exist. Now here comes the interesting part. Consider the limit $$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{1-\frac{\sin x}x}{1+\frac{\cos x}x}=1$$ But this limit is of the form $\frac \infty\infty$. So from L’Hopital’s $$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{(x-\sin x)’}{(x+\cos x)’}=S$$ $$\implies S=1$$ Is this reverse reasoning correct? Is there any faulty assumption that is made here?

- Yoneda’s lemma: group morphisms give Hopf-algebra morphismsby user839372 on March 3, 2021 at 3:35 pm
Let $k$ be a commutative ring. Let $\text{Alg}$ be the category of commutative $k$-algebras and $\text{CHopf}$ the category of commutative Hopf-algebras. Let us also write $[\text{Alg}, \text{Grp}]$ for the category of representable functors $\text{Alg} \to \text{Grp}$ (the morphisms are natural transformations between such functors). We have an obvious functor $$\text{CHopf}^{op} \to [\text{Alg}, \text{Grp}]: H \mapsto \text{Hom}_{\text{Alg}}(H,-)$$ since the sets $\text{Hom}_{\text{Alg}}(H,A)$ obtain a group structure via convolution for every algebra $A$. Concretely, the multiplication is defined by $$f \star g := \mu(f \otimes g)\Delta$$ where $\mu$ is the multiplication on $A$ and $\Delta$ is the comultiplication on $H$. On the level of morphisms, the above functor is defined using the Yoneda lemma. I’m trying to show that this functor is fully faithful. For this, I need to show that if $\lambda: \text{Hom}_{\text{Alg}}(H,-) \to \text{Hom}_{\text{Alg}}(H’,-)$ is a natural transformation that consists of group morphisms and where $H,H’$ are commutative Hopf-algebras, then the morphism $$\lambda_H(\text{id}_H) \in \text{Hom}_{\text{Alg}}(H’,H)$$ is a Hopf-algebra morphism. In particular, I try to check that $$(\lambda_H(\text{id}_H) \otimes \lambda_H(\text{id}_H))\Delta_{H’}= \Delta_H \lambda_H(\text{id}_H).$$ First of all, is this true? If it is true, how can I show it? I have tried to exploit the fact that $\lambda_H$ is a group morphism but could not conclude.

- If a forgetful functor between varieties preserves coproducts, does it have a right adjoint?by Fawzy Hegab on March 3, 2021 at 5:26 am
Suppose that $C,D$ are two varieties (equational classes) and that $U:C\rightarrow D$ is the forgetful functor. Is it true that if this fuctor preserves coproducts and initial objects then it has a right adjoint? I remember someone once told me this fact but I do not remember whom it was to ask for details. I just remember that I have been told that it follows from Freyd’s criterion and some properties of varieties, but whenever I look at Freyd’s criterion, it only talks about the existence of left adjoints and not right adjoints. Maybe this is easy, but I did not study the subject properly but need to use this result but I want to make sure it is true and understand some of its backgrounds. I would appreciate any clarification/reference where I can find proof for this fact.

- What subset of the reals have unique prime factorizations if you allow rational exponents?by Daniel H on March 3, 2021 at 1:07 am
By the fundamental theorem of arithmetic we know that all positive integers have a unique prime factorization. So if $n$ is some positive integer, then $$n = \prod_{i\in\mathbb{N}}{p_i^{e_i}}$$ where $p_i$ is the $i$th prime number and $\{e_i\}$ is a sequence of natural (I’m including $0$) numbers, all but finitely many of which are $0$. If we allow the elements of $\{e_i\}$ to be integers and not just naturals, then the subset of $\mathbb{R}$ with “prime factorizations” is $\mathbb{Q}$, but what happens when we allow the exponents to be rationals? What reals have a unique “prime factorization” then? It’s some subset of the algebraic numbers, but I don’t know if there’s any other characterization of this set. Also, in the case of integer or natural number exponents, it was necessary for all but finitely many exponents to be 0 for the product to converge, but given that rationals are dense in the reals, it seems like you could have products of the form $$\prod_{i\in\mathbb{N}}{p_i^{e_i}}$$ that converge without needing only finitely many non-zero exponents so long as they converged to $0$ quickly enough. What subset of the reals have unique “factorizations” in this sense? It seems plausible to me that these sorts of factorizations might not be unique too.

- Squarefree totient sumby Unit on March 3, 2021 at 12:40 am
Does anybody have a reference/proof for the asymptotic growth rate of $$A(x) = \!\!\!\!\!\!\sum_{\substack{n \leqslant x \\ n \ \text{squarefree}}} \!\!\!\!\!\! \varphi(n)$$ as $x \to \infty$? Here $\varphi(n)$ denotes Euler’s totient function. Numerical evidence suggests it grows like $cx^2$ where $c \approx 0.21$. I tried using the identity $$ \varphi(n) = n \sum_{d \mid n} \frac{\mu(d)}{d}$$ along with the fact that $\mu(n)^2 = 1$ iff $n$ is squarefree, and I got the expression $$A(x) = \sum_{d \le x} \mu(d) \sum_{m \le x/d} m \mu(md)^2.$$ I realize that $md$ is squarefree iff $m$ and $d$ are both squarefree and coprime, but it feels like that just makes the problem harder. Any help would be appreciated!

- Does every vector bundle have a ‘tensor inverse’?by Michael Albanese on March 3, 2021 at 12:14 am
For any vector bundle $E$ over a finite-dimensional CW complex, there is a vector bundle $E’$ such that $E\oplus E’$ is trivial. For a compact Hausdorff base, this is Proposition 1.4 of Hatcher’s Vector Bundles and K Theory, while the general case can be deduced from Tom Goodwillie’s comments here. Finite-dimensionality of the base is a necessary hypothesis as the tautological line bundle $\gamma^1 \to \mathbb{RP}^{\infty}$ demonstrates (see Example 3.6 of Hatcher). While these references only deal with real vector bundles, the analogous statements are true for complex vector bundles too (the arguments carry over easily). My question is whether there is an analogue of the above result when direct sum is replaced by tensor product. Question: Let $E$ be a vector bundle over a finite-dimensional CW complex. Is there a vector bundle $E” \neq 0$ such that $E\otimes E”$ is trivial? One case where such a bundle exists is when $E$ is a line bundle: we can take $E” = E^*$ as $E\otimes E^* \cong \operatorname{End}(E)$ which is a rank one bundle with a nowhere-zero section, namely $\operatorname{id}_E$. Stiefel-Whitney classes do not provide any obstructions to the existence of such a bundle. More precisely, for any bundle $E$ over a CW complex of dimension $d$, there is a bundle $E” \neq 0$ such that $w(E\otimes E”) = 1$. To see this, first choose an integer $j$ such that $2^j > d$. Now note that \begin{align*} w(E\otimes\varepsilon^{2^j}) &= w(E^{\oplus 2^j})\\ &= w(E)^{2^j}\\ &= (1 + w_1(E) + \dots + w_d(E))^{2^j}\\ &= 1 + w_1(E)^{2^j} + \dots + w_d(E)^{2^j}\\ &= 1 \end{align*} as $\deg w_i(E)^{2^j} = i2^j > d$ for $i > 0$. Although one can compute Pontryagin classes of tensor products, see here, I haven’t been able to establish the existence of a bundle $E” \neq 0$ with $p(E\otimes E”) = 1$ (or even just $p_1(E\otimes E”) = 0$). You can use Pontryagin classes to show that $E\otimes\varepsilon^{2^j}$ can be non-trivial, i.e. the above trick to kill Stiefel-Whitney classes won’t necessarily give rise to a trivial bundle. For example, if $E = T\mathbb{CP}^2$, the bundle $E\otimes\varepsilon^{2^j}$ is non-trivial for all $j \geq 0$ as $p_1(E^{\oplus 2^j}) = 2^jp_1(E) = 2^jp_1(T\mathbb{CP}^2) \neq 0$. One thing which makes the direct sum analogue easier to attack is the fact that the existence of a bundle $E’$ with $E\oplus E’$ trivial is equivalent to $E$ being a subbundle of a trivial bundle. I am not aware of a condition on $E$ that is equivalent to the existence of a bundle $E” \neq 0$ with $E\otimes E”$ trivial. I was hoping that the existence of such a bundle could be interpreted in terms of K theory, but I don’t think this is possible as the triviality of $E\otimes E”$ doesn’t imply the triviality of $(E\oplus\varepsilon^k)\otimes E”$.

- Music and Maths – Is there a way to prove that there are only $7$ Modes of Limited Transposition?by DPJDPJ on March 2, 2021 at 1:30 pm
In music, ‘Modes of limited transposition’ are modes that have a limited availability of transpositions. Unlike a major scale that has $12$ possible unique transpositions, the seven modes of limited transposition have fewer than $12$ possibilities of transposition. French Composer Olivier Messiaen coined this idea, and he wrote, “Their series is closed, it is mathematically impossible to find others, at least in our tempered system of $12$ semitones.” I want to try and figure out why. Might anyone have any idea how I could use mathematics to prove this statement? Edit: See Link in comments for ideas surrounding Group Theory

- A dig at Ramanujan’s: $\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,~ p>0$by Z Ahmed on March 2, 2021 at 7:19 am
Ramanujan’s claim on page 98, in the book (`Ramanujan’s note book part 1′ by Bruce C. Berndt) is that $$\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,\qquad p>0\tag{1}\label{theclaim}$$ The proposed proofs for $p=1,2$ are reported to be incorrect. For $p>2$ the claim \eqref{theclaim} has been disproved. In the year 1996, my proofs for $p=1,2$ were evaluated to be correct by American Math. Monthly, however, similar proofs were told to have been published in the year 1995, somewhere. The result (1) being asymptotic $x$ needs to be positive and large. The question here is: What is the latest about this result when $p\in (0,2]$ any information or proof is welcome.

- On the parity of $\left\lfloor{\frac{3^n}{2^n}}\right\rfloor$by lsr314 on March 2, 2021 at 7:15 am
Let $a_n=(-1)^{\left\lfloor{\frac{3^n}{2^n}}\right\rfloor}$ and $$s_n=\sum_{k=1}^na_k.$$ Is it true that $s_n\le 0$ for all $n\geq 1$ ? (This is true for $n\le 100000$.) In other words, odd numbers are always more than even numbers on the sequence $\left\lfloor{\frac{3^n}{2^n}}\right\rfloor$. This is unexpected, I think they should be roughly equal, and even numbers will exceed odd numbers sometimes.

- Why does ${x}^{x^{x^{x^{\,.^{\,.^{\,.}}}}}}$ bifurcate below $\sim0.065$?by Kevin Ndayishimiye on March 2, 2021 at 4:46 am
When you calculate what ${x}^{x^{x^{x\cdots }}}$ converges to between 0 and 1, before approximately 0.065 the graph bifurcates. Why does this happen and is there a reason for it happens at that number?

- How unique is the connecting morphism in the snake lemma?by Thorgott on March 2, 2021 at 1:03 am
This question was inspired by this question. The question whether the connecting morphism in the snake lemma is unique needs to be formulated precisely. Of course, in any specific exact sequence coming from the snake lemma, there may be a whole lot of other morphisms making the sequence exact, but this isn’t very satisfying. Hagen von Eitzen gives the much more interesting example $-\delta$ in the linked question, which happens to be natural as well. My question essentially is how unique the connecting morphism from the snake lemma becomes when imposing this naturality condition. To set the stage, fix some abelian category. Consider the category of diagrams with exact rows in this abelian category of the form $$\require{AMScd} \begin{CD} @. A @>{f}>> B @>{g}>> C @>>> 0\\ @. @VV{a}V @VV{b}V @VV{c}V \\ 0@>>> A’ @>>{f’}> B’ @>>{g’}> C’ \end{CD}$$ The morphisms in this category are morphisms of diagrams. The snake lemma construction in fact gives us a functor from this category to the category of six-term exact sequences in the abelian category (morphisms again being morphisms of diagrams). Explicitly, it maps such a diagram to the exact sequence $$\ker a\rightarrow\ker b\rightarrow\ker c\stackrel{\delta}{\rightarrow}\operatorname{coker}a\rightarrow\operatorname{coker}b\rightarrow\operatorname{coker}c.$$ The unlabeled morphisms are the ones induced by the morphisms in the diagrams. Similarly, a morphism between these diagrams induces a morphism between the associated six-term sequences whose components are induced by the components of the original morphism. Now we can make the question precise: how many such functors are there acting the same way on objects and morphisms except for possibly associating a different connecting morphism $\delta^{\prime}\colon\ker c\rightarrow\operatorname{coker}a$ to each diagram? As a preliminary observation, note that if $\psi$ is an automorphism of the identity functor of the abelian category, i.e. a choice of automorphism $\psi_A\colon A\rightarrow A$ for every object $A$ such that $\psi_B\circ f=f\circ\psi_A$ for any morphism $f\colon A\rightarrow B$, we can consequently replace $\delta$ by $\psi_{\operatorname{coker}a}\circ\delta=\delta\circ\psi_{\ker c}$ and obtain another such functor. In the category of abelian groups, additive inversion provides such an automorphism and this recovers Hagen von Eitzens example mentioned earlier. Conversely, if we have such a functor, then to the diagram $$\require{AMScd} \begin{CD} @. 0 @>>> A @= A @>>> 0\\ @. @VVV @| @VVV \\ 0@>>> A @= A @>>> 0 \end{CD}.$$ there is the associated exact sequence $$0\rightarrow0\rightarrow A\stackrel{\psi_A}{\rightarrow}A\rightarrow0\rightarrow0.$$ Functoriality implies naturality of the map $\psi_A$ and it is an automorphism by exactness. Thus, we see that each such functor induces an automorphism of the identity functor. If we start with an automorphism of the identity functor, pass to the functor and then pass back to an automorphism of the identity functor via these constructions, we end up where we started again. If we carry out the two constructions in the reverse order, it is not as clear what happens. That is, is such a functor determined by how it acts on diagrams of the specific shape from the last paragraph? Are there any such functors not arising from the construction described two paragraphs ago? And, ultimately, how many such functors are there/can we give a meaningful characterization of them?

- Sufficient & necessary conditions for comparing sums of square roots: $\sum_{i=0}^m \sqrt{x_i} \ge \sum_{j=0}^n \sqrt{y_j}$by Azat Ibrakov on March 1, 2021 at 6:50 pm
I’m trying to implement an algorithm for comparing sums of square roots, i.e. to find out whether one sum is greater than/equal to the other $$ \sum\limits_{i=1}^{m} \sqrt{x_i} \ge \sum\limits_{j=1}^{n} \sqrt{y_j} $$ where $m$ may be not equal to $n$. The problem is that we can’t in general calculate square root “exactly” and cumulative error (even with increased precision) causes problems downstream. I have no problems for $m + n < 5$, because it can be solved using consecutive squaring, but it’s not always the case. So I want to know if there exists some sufficient condition which doesn’t involve square roots or some iterative process that minimises them. Knowing about Jensen’s inequality and the fact that square root is concave we can show that $$ \sqrt{m \sum\limits_{i=1}^{m} x_i} \ge \sum\limits_{i=1}^{m} \sqrt{x_i} $$ also following this answer we have $$ \sum\limits_{i=1}^{m} \sqrt{x_i} \ge \sqrt{\sum\limits_{i=1}^{m} x_i} $$ hence something like $$ m \sum\limits_{i=1}^{m} x_i \ge \sum\limits_{j=1}^{n} y_j $$ is necessary condition for original inequality and $$ \sum\limits_{i=1}^{m} x_i \ge n \sum\limits_{j=1}^{n} y_j $$ is sufficient one, which is great because doesn’t involve any square roots, but I hope this can be improved. So I want to know if there are “more precise” ones. Ideally something like $$ \sum\limits_{i=1}^{m} x_i \ge \sum\limits_{j=1}^{n} y_j \implies \sum\limits_{i=1}^{m} \sqrt{x_i} \ge \sum\limits_{j=1}^{n} \sqrt{y_j} $$ will be awesome.

- What is the largest volume of a polyhedron whose skeleton has total length 1? Is it the regular triangular prism?by RavenclawPrefect on March 1, 2021 at 5:12 pm
Say that the perimeter of a polyhedron is the sum of its edge lengths. What is the maximum volume of a polyhedron with unit perimeter? A reasonable first guess would be the regular tetrahedron of side length $1/6$, with volume $\left(\frac16\right)^3\cdot\frac1{6\sqrt{2}}=\frac{\sqrt{2}}{2592}\approx 0.0005456$. However, the cube fares slightly better, at $\frac{1}{1728}\approx 0.0005787$. After some experimentation, it seems that the triangular prism with all edges of length $1/9$ is optimal, at a volume of $\frac{\sqrt{3}}{2916}\approx0.00059398$. I can prove that this is optimal among all prisms (the Cartesian product of any polygon with an interval), and that there is no way to cut off a small corner from the shape and improve it. Is the triangular prism the largest polyhedron with a fixed perimeter? I can prove a weak upper bound of $\frac1{12\pi^2\sqrt{2}}\approx 0.00597$ on the volume of such a polyhedron, by combining the isoperimetric inequalities in both $2$ and $3$ dimensions (i.e., the fact that polygons cannot enclose more surface area than a circle and that a polyhedron cannot enclose more volume that a sphere of the same surface area) along with the observation that a single face of a polyhedron cannot take up the majority of its surface area. Note the number of leading zeros – this upper bound is a bit over $10$ times my lower bound!

- Find $ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta$by Andrew John on March 1, 2021 at 5:06 pm
Evaluate $$ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left[\frac{1+2r^2}{1-r^2\cos\left(2\theta\right)}\right]^{1/3}{\rm d}\theta $$ Question- Can I take the limit inside the integral? My try- $$I= \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left(\frac{1+2r^2}{1-r^2\cos2\theta} \right)^{1/3}\, d\theta $$ $$ I= \int_{-\pi}^{\pi} \lim_{r \to 1^{-}} (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta $$ $$ I=3^{1/3} \int_{-\pi}^{\pi} \frac{1}{(1-\cos2\theta)^{1/3}}d\theta $$ $$ I= 3^{1/3}2\int_{0}^{\pi} \frac{1}{(1-\cos2\theta)^{1/3}} d\theta $$ $$ I= 3^{1/3}4\int_{0}^{\pi/2} \frac{1}{(1-\cos2\theta)^{1/3}} d\theta $$ $$ I= (3/2)^{1/3}4\int_0^{\pi/2} \sin^{-2/3}\theta \ d\theta $$ $$I= 4(3/2)^{1/3} \frac{\Gamma(1/6)\Gamma(1/2) }{\Gamma(2/3)}. $$

- How many arrangements of red and blue balls are there so that, the number of red balls with: the ball immediately to the right is also red, is $9$.by Adam Rubinson on March 1, 2021 at 2:14 pm
The question is too long to fit in the title, but I tried. $50$ balls: $23$ indistinguishable red balls; $27$ indistinguishable blue balls. The balls are arranged in a line. How many distinct arrangements of the balls are there so that the following property holds: The number of red balls such that the ball immediately to the right of it is also red, is $9.\quad (1)$ ? Can I please get my following solution verified. Also, feel free to post alternative solution of course. Let the number of red balls with property $(1)$ be $k$. If we start by placing all $23$ red balls next to each other, then $k=22.\ $ Now between each red ball, we imagine there is a bin. We don’t yet have a bin to the left of the first red or a bin to the right of the last red. There are $22$ bins between the first and last red ball. Each time you put a blue ball into one of these bins, you reduce the number $k$ by $1$. So what we can first do is find the number of distinct ways to put $(22-9=)\ 13$ blue balls into $22$ bins so that no two balls are in the same bin. This is just $\binom{22}{13}.$ Now we have placed $13$ blue marbles into bins such that no two blue marbles are in the same bin, and we did this so that $k=9$, and so property $(1)$ holds. We now have $27-13=14$ blue marbles remaining to put into bins. Ah, but not $22$ bins because that would ruin property $(1)$, and not just $13$ bins: the ones that we have just put the blue balls into, but… Now we introduce a bin to the left of the first red and another bin to the right of the last red. We put the remaining $(27-13=)\ 14 $ blue balls into $(1+13+1=)\ 15$ bins! How many ways are there of doing this? Well, it is a classic stars and bars problem, in particular see Theorem 2 on wikipedia. The number of ways of putting the remaining $14$ blue balls into $15$ bins is $\binom{14+15-1}{14} =\binom{28}{14}.$ I believe all possible arrangements can be done like this, and there are no repeats in this method. Therefore the final answer is: $$ \binom{22}{13} \times \binom{28}{14} = 497420 \times 40116600 = 19954799172000. $$

- Given a set of unitary matrices, can one find a vector whose images under these unitary matrices span the underlying Hilbert space?by mathwizard on March 1, 2021 at 3:00 am
Given a set of (linearly independent) $d\times d$ complex unitary matrices $\{U_i\}_{i=1}^n \subseteq M_d$ with $n\geq d$, does there exist a vector $v\in \mathbb{C}^d$ such that $$\text{span} \{U_1v, U_2v, \ldots , U_nv\} = \mathbb{C}^d ?$$ The motivation for this question comes from the theory of mixed unitary quantum channels. A quantum channel $\Phi: M_d \rightarrow M_d$ is a completely positive and trace preserving linear map. Any such map admits a Kraus representation of the form $\Phi (X) = \sum_{i=1}^k A_i X A_i^*$, where $\{A_i \}_{i=1}^k \subseteq M_d$ and $\sum_{i=1}^k A_i^* A_i = \mathbb{I}_d$. We say that a quantum channel is mixed unitary if it can be expressed as a convex combination of unitary conjugations: $\Phi (X) = \sum_{i=1}^n p_i U_i X U_i^*$. Our aim then is to look for a rank one input projector $X = vv^*$ for some $v\in \mathbb{C}^d$ such that the output $\Phi (vv^*) = \sum_{i=1}^n p_i (U_i v) (U_i v)^*$ has full rank. This is possible only if $n\geq d$. To avoid trivial counterexamples, we can also assume that $\{ U_i\}_{i=1}^n \subseteq M_d$ is linearly independent. Follow-up question: Since it has been shown below that the question can be answered in the negative for $d\geq 4$, the natural way of progression would be to ask if one can provide a classification of all the sets of (linearly independent) unitary matrices $\{U_i\}_{i=1}^n \subseteq M_d$ which allow for the existence of $v\in \mathbb{C}^d$ such that $$\text{span}\{U_1v, U_2v, \ldots ,U_nv\}=\mathbb{C}^d.$$ One can also try to answer this question for (linearly independent) sets of arbitrary complex matrices: $\{ A_i \}_{i=1}^k \subseteq M_d$.

- Evaluate alternatively $\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\pm\sin^4x\right)dx$by Naren on February 28, 2021 at 7:57 pm
In my recent article here at Researchgate on page no 9, I was able to deduce the following two identities from the general result (which I treat it as Theorem 3.2). Those two identities are $$\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\sin^4x\right)dx =\frac{7\pi^2}{12}-\frac{25}{4}\ln^2(2)-4\operatorname{Li}_2\left(\frac{1}{\sqrt 2}\right)+2\operatorname{Li}_2\left(\frac{2-\sqrt 2}{4}\right)-\operatorname{arcsinh}^2(1)+3\ln(2)\operatorname{arcsinh}(1)\approx 0.581222\cdots$$ also $$\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(-\sin^4x\right)dx=\operatorname{Li}_2\left(\frac{1-\sqrt 2}{2}\right)+4\operatorname{Li}_2\left(-\sqrt{\frac{1+\sqrt 2}{2}}\right)+\frac{\pi^2}{3}-\frac{19}{4}\ln^2(2)+2\operatorname{Li}_2\left(\frac{\sqrt 2-\sqrt{1+\sqrt 2}}{2\sqrt 2}\right)-\frac{\operatorname{arcsinh}^2(1)}{2}-\ln^2\left(\sqrt {2}+\sqrt{1+\sqrt{2}}\right)-3\ln(2)\ln\left(\sqrt{1+\sqrt 2}-\sqrt {2}\right)\approx -0.32379\cdots$$ These two beautiful identities are the special cases of general result $$\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\pm16v\sin^4x\right)=\text{Result of theorem 3.2}$$for all $v\leq 1/16$. However, my interest is to seek an alternative path for both general result as well as those two special cases mentioned above other than technique used in the article. If there are other ways I would be glad to known. Using the same technique one can easily find the closed form for the integral of type $$\displaystyle \frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\pm256w\sin^8x\right)dx , \;\; |w|\leq 1/256$$

- Teacher claims this proof for $\frac{\csc\theta-1}{\cot\theta}=\frac{\cot\theta}{\csc\theta+1}$ is wrong. Why?by Ramblin Wreck on February 28, 2021 at 5:48 pm
My son’s high school teacher says his solution to this proof is wrong because it is not “the right way” and that you have to “start with one side of the equation and prove it is equal to the other”. After reviewing it, I disagree. I believe his solution is correct, even if not “the right way”, whatever that means. I asked my son how he did it: he cross-multiplied the given identity, simplified it to a known/obvious equality, and then reversed the steps for the proof. This was a graded exam, and the teacher gave him a zero for this problem. What do you think about my son’s solution? Thanks! Problem: prove the following trigonometric identity \begin{align*} \frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1}\ .\\ \end{align*} Solution: for all real $\theta$ not equal to an integer multiple of $\pi/2$, we have \begin{align*} \cot^2(\theta)&=\cot^2(\theta)\\[8pt] \frac{\cos^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt] \frac{1-\sin^2(\theta)}{\sin^2(\theta)}&=\cot^2(\theta)\\[8pt] \csc^2(\theta)-1&=\cot^2(\theta)\\[8pt] \frac{\csc^2(\theta)-1}{\cot(\theta)}&=\cot(\theta)\\[8pt] \frac{\csc(\theta)-1}{\cot(\theta)}&=\frac{\cot(\theta)}{\csc(\theta)+1} \end{align*}

- Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$by xldd on February 28, 2021 at 12:56 pm
Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?

- Mistake in OEIS A103904?by Marie on February 28, 2021 at 12:12 pm
The sequence OEIS A103904 is described as Number of perfect matchings of an $n \times (n+1)$ Aztec rectangle with the third vertex in the topmost row removed. Definition of $M \times N $ Aztec rectangle (one can see such a definition by C. Krattenthaler): Consider a $(2M+1) \times (2N+1)$ rectangular chessboard and suppose that the corners are black. Then an $M \times N$ Aztec rectangle is the graph whose vertices are the white squares and whose edges connect precisely those pairs of white squares that are diagonally adjacent. So an $n \times (n+1)$ Aztec rectangle is when $M=n$ and $N=n+1$. see the following figures for $n \times (n+1)$ Aztec rectangles as examples: When $n=2$, I try to calculate the number of perfect matchings of $2 \times (2+1)$ Aztec rectangle with the third vertex in the topmost row removed: The number of perfect matching is actually $8$, which is not in A103904. Is there any mistake in this sequence or the definition is wrong? Or maybe I misunderstand it. I am very happy to hear from anyone and thank you very much in advance!

- Why does an exponential function eventually get bigger than a quadraticby John L on February 27, 2021 at 7:30 pm
I see this answer to this question and this one. My $7$th grade son has this question on his homework: How do you know an exponential expression will eventually be larger than any quadratic expression? I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L’Hopital’s rule, Limits, etc, Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase “eventually” makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don’t like the way the question is written.

- Another Topology on the Prime Spectrum of a Ringby Brian Shin on February 27, 2021 at 4:07 pm
Let’s fix a commutative ring $R$. We’ll write $X = \operatorname{Spec}R$ for the set of prime ideals of $R$. Finally, let’s write $\beta X$ for the Stone-Čech compactification of $X$. We can define a function $\lim : \beta X \to X$ using the following recipe: $$ f \in \lim \mu \iff \mu(\{p \in \operatorname{Spec} R : f \in p \}) = 1$$ Indeed, $\lim \mu$ is prime by Łoś’s theorem, or an easy direct verification. I believe this function actually endows $X$ the structure of a $\beta$-algebra. In other words, this endows $X$ with a compact Hausdorff topology. In particular, it must be distinct from the Zariski topology. My questions: does this topology have a name? Is there somewhere I could read about this?

- Showing the Fibonacci inequality $f_1^{f_1}f_2^{f_2}f_3^{f_3}\cdots f_n^{f_n}\leq f_1!f_2!f_3!\cdots f_n!\;e^{({f_{n+2}-n-1)}}$ without induction.by user1055 on February 27, 2021 at 8:42 am
Today I saw the following 2 beautiful inequalities on a facebook page. $$1^12^23^34^4\cdots n^n\leq 1!2!3!4!\cdots n!e^{\frac{n(n-1)}{2}}$$ $$f_{1}^{f_{1}}f_{2}^{f_{2}}f_{3}^{f_{3}}\cdots f_{n}^{f_{n}}\leq f_{1}!f_{2}!f_{3}!\cdots f_{n}!e^{({f_{n+2}-n-1)}}$$ Here $f_{n}$ denotes the fibonacci numbers $f_{1}=f_{2}=1$ and $f_{n+2}=f_{n+1}+f_{n}$ for $n\in N$ Here is How I proved the first one. Method 1:- $$1^12^23^34^4…n^n\leq 1!2!3!4!…n!e^{\frac{n(n-1)}{2}}$$ $$\implies\bigg(\frac{1^1}{1!}\bigg)\bigg(\frac{2^2}{2!}\bigg)\bigg(\frac{3^3}{3!}\bigg)…\bigg(\frac{n^n}{n!}\bigg)\leq e^{\frac{n(n-1)}{2}}$$ $\implies \displaystyle\prod_{x=1}^{n} \frac{x^x}{x!}\leq e^{\frac{n(n-1)}{2}}$ Now taking logarithm on both sides of inequality. $\implies \displaystyle\sum_{x=1}^{n} ln\bigg(\frac{x^x}{x!}\bigg)\leq {\frac{n(n-1)}{2}}$ $\implies \displaystyle\sum_{x=1}^{n} ln\bigg(\frac{x^x}{x!}\bigg)\leq {\frac{n(n+1)}{2}}-n$ $\implies \displaystyle\sum_{x=1}^{n} ln\bigg(\frac{x^x}{x!}\bigg)\leq \bigg(\sum_{x=1}^{n} x \bigg)-n$ $\implies \displaystyle \ \sum_{x=1}^{n} x -\displaystyle\sum_{x=1}^{n} ln\bigg(\frac{x^x}{x!}\bigg)\geq n$ $\implies \displaystyle \ \sum_{x=1}^{n} x + ln\bigg(\frac{x!}{x^x}\bigg)\geq n$ $$\implies 1+1.30+1.49+1.63+1.74+1.82+1.90+1.97+2.02+…\geq n$$ $\implies 1+(1+0.30)+(1+0.49)+(1+0.63)+(1+0.74)+…\geq {\smash[b]{1+\! \underbrace{1+1+\cdots1\,}_\text{$n$ times}}}$ This proves our first inequality. Method 2:- Using $\frac{n^n}{n!}\leq e^{n-1}$ for $n\in N$. $\frac{1^1}{1!}\leq e^{0}$ $\frac{2^2}{2!}\leq e^{1}$ $\frac{3^3}{3!}\leq e^{2}$ $\vdots\\$ $\frac{n^n}{n!}\leq e^{n-1}$ Now multiply all above inequalities we get, $\implies \displaystyle\prod_{x=1}^{n} \frac{x^x}{x!}\leq e^{\frac{n(n-1)}{2}}$ Using the same approach for $2^{nd}$ inequality we get, $$\sum_{x=1}^{n} ln\bigg(\frac{{f_{x}}^{f_{x}}}{f_{x}!}\bigg)\leq f_{n+2}-n-1$$ Now I got stuck on this step. How can we prove the $2^{nd}$ inequality. Moreover can we show without numerically calculating(as I do above) that $\displaystyle \ \sum_{x=1}^{n} x + ln\bigg(\frac{x!}{x^x}\bigg)\geq n$

- Does a semicircle have 2 right angles?by Jack Freeth on February 27, 2021 at 8:26 am
If you have a line passing through the middle of a circle, does it create a right angle at the intersection of the line and curve? More generally, is it valid to define an angle created between a line and a curve? Is the tangent to the curve at the point of intersection a valid interpretation (I.e a semi circle has 2 right angles) I saw it in a debate thread and it got me curious now.

- What is the deal with the three isomorphism theorems?by Billy Rubina on February 27, 2021 at 3:53 am
I’ve been having lectures in group theory with Hungerford’s book. We were presented with the following theorem: And then with: Previous to the lectures about them, I was understanding most of the stuff, that is: I kinda could figure out the motivation for things. But when I came into these theorems, things got pretty weird for me. I don’t understand the motivation for them. I have the following guesses: Theorem 5.6 is used to prove Corollary 5.7 but I think it has a utility on its own: If given an homomorphism $f: G \to H$ with $N\lhd G$, we have a unique homomorphism $\overline{f} : G/N\to H$, then we could use this to know what homomorphisms could exist betwen $G$ and $H$ and this depends only on the normal subgroups of $G$. Does that make sense? For the isomorphism theorems, I noticed they are corollaries so they must be very important. One of the reasons I conjectured for them is that we can construct very interesting isomorphisms. But why are those isomorphisms interesting? It’s not clear to me what interesting stuff one could do with them. I know my guesses are maybe too obvious or wrong, but I can’t figure out on my own the answer to that question. Can you help me?