Sunday, March 28, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

What is the limit of this function as x tends to 0

Posted: 28 Mar 2021 08:46 PM PDT

I am trying to evaluate the limit of

$x[1/x]$ as x tends to zero, where $[.]$ is greatest integer function. I know this is dumb question but can I write [1/x] as 1/[x]? Thanks in advance.

All triangles are isosceles Theorem

Posted: 28 Mar 2021 08:43 PM PDT

Let ABC be any triangle. Bisect BC at D, and from D draw DE at right angles to BC. Bisect the angle BAC.

(1) If the bisector does not meet DE, they are parallel. Therefore the bisector is at right angles to BC. Therefore AB = AC, i.e., ABC is isosceles.

(2) If the bisector meets DE, let them meet at F. JOin FB, FC, and from F draw FG, FH, at right angles to AC, AB.

Then the triangles AFC, AFH are equal, because they have the side AF common, and the angles FAG, AGF equal to the angles FAH, AHF. Therefore AH = AG, and FH = FG.

Again, the triangles BDF, CDF are equal, because BD = DC, DF is common, and the angles at D are equal. Therefore FB = FC.

Again, the triangles FHB, FGC are right-angled. Therefore the square on FB = the squares on FH, HB; and the square on FC = the squares on FG, GC. But FB = FC, and FH = FG. Therefore the square on HB = the square on GC. Therefore HB = GC. Also, AH has been proved = to AG. Therefore AB = AC; i.e., ABC is isosceles. Therefore the triangle ABC is always isosceles.

Questions :

  1. Draw pictures for all cases of the proof described above.
  2. Explain the flaws in the above reasoning.

Commutativity of a group which can be expressed as internal direct product

Posted: 28 Mar 2021 08:42 PM PDT

If G is a group and it is an internal direct product of the subgroups H and K then, hk=kh, for all h€H, k€K. Does that imply that G is abelian? If not can you provide a counter example?

A smooth action of a Lie group on a smooth manifold. Circle Example

Posted: 28 Mar 2021 08:33 PM PDT

I want to check that I understand this definition correctly and that my example makes sense.

Let $\varphi:G\to Diff(M)$ be a action of a Lie group $G$ on a smooth manifold $M$. We say that $\varphi$ is a smooth action if the evaluation map, i.e. $$ev:G\times M\to M$$ given by $ev(g,m)=\varphi(g)(m)$, is a smooth map.

Example:

Let's show that action $\varphi$ of $S^1$ on $\mathbb{R}^2$, given by $\varphi:S^1\to Diff(\mathbb{R}^2)$, $\varphi(\theta)(x,y)=(x+\cos(\theta),y+\sin(\theta))$, is a smooth action. It's enough to show that the evaluation map $$ev:S^1\times \mathbb{R}^2\to \mathbb{R}^2\text{ is smooth}.$$ Indeed, consider the open subset $U=S^2-\{N\}$, a homeomorphism $\psi:U\to\mathbb{R}$ with $\psi(x,y)=\frac{x}{1-y}$, and its inverse $\psi^{-1}:\mathbb{R}\to U$ with $\psi^{-1}(t)=(\frac{2t}{1+t^2},\frac{t^2-1}{t^2+1})$.

Then we have a map $$(ev\circ\psi^{-1}(t))(t,(x,y))=ev((\frac{2t}{1+t^2},\frac{t^2-1}{t^2+1}),(x,y))=(x+\frac{2t}{1+t^2},y+\frac{t^2-1}{t^2+1})$$ which is obviously a smooth map. A similar map we can obtain by considering another open subset of $S^1$. Therefore, the action $\varphi$ is smooth.

OEIS sequence A308092: run lengths of bits and run lengths of an auxillary sequence.

Posted: 28 Mar 2021 08:32 PM PDT

In February 2018, when the On-Line Encyclopedia of Integer Sequences (OEIS) was approaching it's 300,000th sequence, Neil Sloane sent an email out to the SeqFan mailing list announcing hand-picked sequences for A300000, A300001, A300002, and A300003.

In May of the following year, I submitted sequence A308092, which is the base-2 analog of A300000:

The sum of the first $n$ terms of the sequence is the concatenation of the first $n$ bits of the sequence read as binary, with $a(1) = 1$.

In decimal, the sequence begins $$ 1, 2, 3, 7, 14, 28, 56, 112, 224, 448, 896, 1791, 3583, 7166, 14332, 28663, 57326, 114653, 229306, \dots. $$

but it is more natural to look at the sequence in binary $$ 1_2, 10_2, 11_2, 111_2, 1110_2, 11100_2, 111000_2, 1110000_2, 11100000_2, 111000000_2, \dots $$

For example, $a(4) = 7$ because $7$ is the right choice so that the first four bits of the sequence ($1101_2$) is equal to the sum of the first four terms: $1+2+3+7=13=1101_2$.

One thing that I noticed about this sequence—and that Matthew Scroggs proved in June 2019—is that the number of $1$s in each term seems to be weakly increasing: $$ \underbrace{1_2}_{1}, \underbrace{10_2}_{1}, \underbrace{11_2}_{2}, \underbrace{111_2}_{3}, \underbrace{1110_2}_{3}, \underbrace{11100_2}_{3}, \underbrace{111000_2}_{3}, \underbrace{1110000_2}_{3}, \underbrace{11100000_2}_{3}, \underbrace{111000000_2}_{3}, \dots, $$ which gives the auxiliary sequence A324608, "Number of 1's in binary expansion of $A308092(n)$." $$ 1, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 10, 11, 11, 11, 13, 13, 14, 14, 14, 16, 16, 16, 17, 17, \dots. $$

The question

That was a lot of background, but now we're ready for the question!

The run-lengths on the bits of A308092 (the first sequence), $$ \underbrace{1_2, 1}_2\underbrace{0_2}_1, \underbrace{11_2, 111_2, 111}_8\underbrace{0_2}_1, \underbrace{111}_3\underbrace{00_2}_2, \underbrace{111}_3\underbrace{000_2}_3, \underbrace{111}_3\underbrace{0000_2}_4, \underbrace{111}_3\underbrace{00000_2}_5, \underbrace{111}_3\underbrace{000000_2}_{6}, \dots, $$ appear to be equal to the run-lengths of the auxiliary sequence A324608, $$ \underbrace{1, 1}_2, \underbrace{2}_1, \underbrace{3, 3, 3, 3, 3, 3, 3, 3}_8, \underbrace{10}_1, \underbrace{11, 11, 11}_3, \underbrace{13, 13}_2, \underbrace{14, 14, 14}_3, \underbrace{16, 16, 16}_3, \underbrace{17, 17, 17}_3, \dots. $$

It feels like there should be some obvious, elementary reason for why this is, but I'm missing it. Any help?

Non-negative martingale $X_n \rightarrow 0$ a.s. prove that $P[X^* \geq x | \mathcal{F}_0]= 1 \wedge X_0 / x$

Posted: 28 Mar 2021 08:25 PM PDT

I need to prove the following statement.

Let $X$ be a non negative martingale such that $X_n\rightarrow 0$ a.s. when $n\rightarrow \infty$. Define $X^*=supX_n$. Prove that for all $x>0$

$$P[X^* \geq x | \mathcal{F}_0]= 1 \wedge X_0 / x$$

I think I've got the easy case

  • if $x\leq X_0$

Then necessarily $x\leq X^*$ for the sup property. Then it follows that for $1\leq X_0 /x$ we have that $P[X^* \geq x | \mathcal{F}_0]= 1$. But I can't figure out the other case.

uniform convergence of the function sequence

Posted: 28 Mar 2021 08:28 PM PDT

Investigate uniform convergence of the function sequence :

  1. $f_n(x) = \frac{nx^2}{1+n+2x}, 0 \le x \lt +\infty $
  2. $f_n(x) = n(1-x)x^n, 0 \le x \le 1$

Permutations of a Graph Under Isomorphism

Posted: 28 Mar 2021 08:11 PM PDT

Let's say we have the graph $G$ on $4n$ vertices that forms a cycle. Furthermore, connect one vertex with it's antipode (so that we have a total of $4n+1$ edges in the graph).

How many different isomorphic permutations are there of the vertex set?

My Progress

Label the vertices $1, 2, \cdots 4n$ and assume that vertex $1$ is connected to its antipode, $2n+1$. Let the isomorphism be $\phi$. Then, vertices $1$ and $2n+1$ are the only vertices in the graph with degree $3$. Hence, any isomorphism must map $1$ and $2n+1$ either to themselves or to each other.

After that, we can map the two other neighbors of $1$, $2$ and $4n$, to the two sides of $\phi(1)$. Afterwards, all the other numbers must be fixed (there is only one place to map $3$ which gives only one place to map $4$, and so on so forth).

Similarly, we can choose how to map $2n$ and $2n+2$ on either side of $\phi(2n+1)$.

All in all, this gives me $8$ isomorphisms - are there any more?

generators of $SL(2,8)$

Posted: 28 Mar 2021 08:10 PM PDT

I am trying to find the generators of the special linear group $SL(2,8)$. I have read some posts about the generators of $SL(2,p^r)$ but they are supposing that $p$ is odd. I am not sure if there is a theorem works with this case..

Verification of Proof: "$K$ is compact $\implies$ $\exists \inf K \in K$ and $\exists \sup K \in K$"

Posted: 28 Mar 2021 08:07 PM PDT

I hope life is going okay for everyone; I am hoping someone here can comment on aspects of my proof's accuracy, coherence, simplicity, readability, etc.. Anything really goes, I am trying to improve all fronts of mathematical intuition - please criticize as much of my work as you can. Note that the problem below is an exercise (3.3.1.) in Understanding Analysis ed.2 by S. Abbott.

Problem

Show that if $K$ is compact and nonempty, then $\sup K$ and $\inf K$ both exist and are elements of $K$.

Definitions

Compact Set: A set $K \subseteq \mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a limit that is also in $K$.

Closed Set: A set $F \subseteq R$ is closed if it contains its limit points.

Bounded Set: A set $A \subseteq \mathbb{R}$ is bounded if there exists $M > 0$ such that $|a|\leq M$ for all $a \in A$.

Theorem 2.5.2. - Subsequences of a convergent sequence converge to the same limit as the original sequence.

Theorem 3.3.4. - A set $K \subseteq \mathbb{R}$ is compact if and only if it is closed and bounded.

Lemma 1.3.8.

  1. Assume $s \in R$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then, $s = \sup A$ if and only if, for every choice of $\epsilon > 0$, there exists an element $a \in A$ satisfying $s - \epsilon < a$.
  2. Assume $i \in R$ is a lower bound for a set $A \subseteq \mathbb{R}$. Then, $i = \inf A$ if and only if, for every choice of $\epsilon > 0$, there exists an element $a \in A$ satisfying $a < i + \epsilon $.

Axiom of Completeness: Every nonempty set of real numbers that is bounded above has a least upper bound and every nonempty set of real numbers that is bounded below has a greatest lower bound.

Limit Point: A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $\mathcal{B}_{\epsilon}(x)$ of $x$ intersects the set $A$ at some point other than $x$.

My Proof

$\text{ }$ Suppose that a set $K \subseteq \mathbb{R}$ is compact and nonempty. Let $\alpha = \sup K$ for some $\alpha \in \mathbb{R}$. First we want to show $\alpha$ exists and $\alpha \in K$. By Theorem 3.3.4., $K$ is bounded, i.e. there exists $M > 0$ such that $-M \leq k \leq M$ for all $k \in K$. This implies that $\alpha$ exists by the Axiom of Completeness. Now let $\epsilon > 0$. By Lemma 1.3.8. we know that there exists an element $k \in K$ satisfying $\alpha - \epsilon < k$. In particular, we have $\alpha - \epsilon < k < \alpha + \epsilon$, which means that $k \in \mathcal{B}_{\epsilon}(\alpha)$. Moreover, $\alpha$ is a limit point of $K$ since $k \in \mathcal{B}_{\epsilon}(\alpha) \cap K$. Since $K$ is closed by Theorem 3.3.4., then we have $\alpha \in K$, as desired.

$\text{ }$ Let $\beta = \inf K$ for some $\beta \in \mathbb{R}$. What remains to be shown is that $\beta$ exists and $\beta \in K$. Since $K$ is bounded by Theorem 3.3.4., the Axiom of Completeness implies that $\beta$ exists. Let $\epsilon > 0$. By Lemma 1.3.8. we know that there exists an element $k \in K$ satisfying $\beta + \epsilon > k$. In particular, we have $\beta + \epsilon > k > \beta - \epsilon$, which means that $k \in \mathcal{B}_{\epsilon}(\beta)$. Since $k \in \mathcal{B}_{\epsilon}(\beta) \cap K$, we have that $\beta$ is a limit point of $K$. Thus, $\beta \in K$ since $K$ is closed.

zeros of a bijective holomorphic function

Posted: 28 Mar 2021 08:07 PM PDT

Let $V,W \subset \mathbb{C}$ be open sets. Let $f\colon V \to W$ be a bijective holmorphic map. Then, the set $Z:= \left\{v\in V \colon f'(v)=0\right\}$ is empty.

I know that $f$ is a homeomorphism and that $Z$ is a discrete closed set in $W$. but I'm not sure if $Z$ is empty or not. Any help?

Is it true that range of $(I + T)$ closed whenever $T$ is compact?

Posted: 28 Mar 2021 08:29 PM PDT

Let $\mathcal H$ be a Hilbert space and $T \in \mathcal L (\mathcal H)$ be a compact operator. Then is it true that range of $(I + T)$ is closed?

If the answer is in affirmative sense then how to prove it? Any help in this regard will be warmly appreciated.

Thanks for your time.

Checking the a section $\sigma$ of a vector bundle $f:E\to M$ is smooth

Posted: 28 Mar 2021 08:34 PM PDT

I have a clarifying question about sections of vector bundles. Let $\pi:E\to M$ be a smooth vector bundle of rank $n$, and $\sigma:M\to E$ be some map. I want to show that $\sigma$ is a section of $E$.

First, I want to show that $(\pi\circ\sigma)(p)=p$ for all $p\in M$.

Second, I want to show that $\sigma$ is a smooth map. Since $E$ is a smooth vector bundle over $M$ of rank $n$, then there exists an open cover $\{U_i\}$ of $M$ s.t. for all i, there is a diffeomorphism $$\varphi_i:\pi^{-1}(U_i)\to U_i\times\mathbb{R}^n$$ So, in order to show that $\sigma$ is smooth, I want to show that $\sigma|_{U_i}$ is smooth for all $i$ where $$\sigma|_{U_i}:=\varphi_i\circ\sigma:U_i\to U_i\times\mathbb{R}^n.$$ To show that $\sigma|_{U_i}$ is smooth, it's enough to show that $\sigma|_{U_i}$ is smooth on open charts of $M$ i.e. $$(\psi_i\times id)\circ \sigma|_{U_i}\circ \psi_i^{-1}:\hat{U_i}\to \hat{U_i}\times\mathbb{R}^n\text{ is smooth for all }i$$ where, assuming that $\dim(M)=k$, $\psi_i:U\to \hat{U}\subset \mathbb{R}^k$ is a homeomorphism. If the original $U_i$ is big enough, we always can make it smaller to obtain $\psi_i$.

For example:

Consider a tangent bundle of a circle, i.e. $TS^1$, and a map $\sigma:S^1\to TS^1$ where $\sigma(\theta)=(\theta,\frac{\partial}{\partial\theta})$. So, if I want to show that $\sigma$ is a smooth section, then I need to choose an open cover of $S^1$ which will give me a local trivialization of a tangent bundle and write that map in local coordinates.

But, we can use another approach where I will start with some opens $U_i\times\mathbb{R}$ and smooth maps $g_{ij}:U_{i}\cap U_j\to GL(\mathbb{R})$, i.e. so I can use them to glue my tangent bundle. Then if we know each $$s_i:U_i\to U_i\times\mathbb{R}$$ where $s_i(p)=(p,\phi_i(p))$ and that $\phi_j(p)=g_{ij}(p)\phi_i(p)$, then it's enough to check that each $s_i$ is smooth map to obtain that $s$, as a section over tangent bundle, is smooth.

So, my question is basically comparing this two approaches and a general idea how to show that some map is a section.

Solving a cubic equation... [duplicate]

Posted: 28 Mar 2021 08:44 PM PDT

I've been wondering about solving a cubic equation. Sure, I know how to do a linear equation, and even a quadratic, but how do I do a cubic? It's really hard for me to factor, and the formula is really big. I searched up the process for solving, where you 'depress the cubic' and then solve it like that. But I don't really understand how it works. Can someone explain this process to me?

convex function $f(x)$ bounded by $\|x\|_0$ must be bounded by $\|x\|_1$?

Posted: 28 Mar 2021 08:23 PM PDT

One of my professors stated the following result without proof: Suppose that $f \colon \mathbb{R}^p \to \mathbb{R}$ is convex such that $f({\bf x}) \leq \|{\bf x}\|_0$, then $f({\bf x}) \leq \|{\bf x}\|_1$. (Recall that $\|{\bf x}\|_0$ counts the number of non-zero components of ${\bf x}$). The argument for $p = 1$ is fairly easy, but I did not see a way to prove this result in higher dimensions. Thank you very much!

Edit: we might need $f$ to be defined only on a bounded (convex) subset of $\mathbb{R}^p$ in order to avoid triviality.

Infinitely Many Prime Divisors $k\pmod n$

Posted: 28 Mar 2021 08:24 PM PDT

Fix $f(x)\in\mathbb Z[x]$ a polynomial. For a given prime $p,$ we say $p$ is a "prime divisor" of $f$ if and only if $p$ divides $f(k)$ for some $k\in\mathbb Z.$ It is known that nonconstant $f$ have infinitely many prime divisors, and one can even exhibit polynomials $f$ with fun properties; for example $$p\;\;\text{divides}\;\;\Phi_n(nx)\implies p\equiv1\pmod n,$$ where $\Phi_n(x)\in\mathbb Z[x]$ is the $n$th cyclotomic polynomial. One can even show, using this, that every nonconstant polynomial $f$ have infinitely many $1\pmod n$ prime divisors. (See discussion after Theorem 3 here.) Explicitly, we can use Dedekind-Kummer to take all but finitely many of the primes which split completely in the closure of $\mathbb Q(\zeta_n)$ and $\mathbb Q(\alpha)$ where $\alpha\in\mathbb C$ is a root of $f.$

Showing the general case, however, looks harder. The question is as follows.

For nonconstant polynomials $f(x)\in\mathbb Z[x],$ are there infinitely many $k\pmod n$ prime divisors, for any $n,k\in\mathbb N$ with $\gcd(k,n)=1$?

In particular, I suspect that purely "algebraic" techniques (even allowing, say, Dedekind-Kummer) are insufficient here, but I have no strong reason to believe this.

Actually, the question is false for somewhat easy reasons, as was pointed out to me: the polynomial $\Phi_n(nx)$ itself fails to hit every single$\pmod n$ class except for $1\pmod n$ and so provides a counterexample. The revised question, then, is for a classification of which modular classes of primes are hit infinitely often.

For nonconstant polynomials $f(x)\in\mathbb Z[x],$ when are there infinitely many $k\pmod n$ prime divisors, for given $n,k\in\mathbb N$ with $\gcd(k,n)=1$?

2D integral of a 1D delta function, and changing coordinate systems

Posted: 28 Mar 2021 08:21 PM PDT

I'd like to evaluate an integral over a 2D area with a 1D delta function $$ \operatorname{I}\left(\vec{q},\mu\right) = \int_{-\pi}^{\pi}{\rm d}x\int_{-\pi}^{\pi}{\rm d}y\,\, \delta\left(\mu - \operatorname{f}\left(x,y\right)\right) \left[\vec{q}\cdot\nabla f(x,y)\right]^{2} $$ What is the proper way to do this? For simplicity, take $\vec{q}=q_x\hat{x}$. I'm accustomed to using the change of variables to change this as $$ I(q_x,\mu) = q_x^2 \int_{-\pi}^\pi dy \,\, \frac{1}{\left|\frac{\partial}{\partial x}f\right|}\left(\frac{\partial}{\partial x}f\right)^2\Bigg|_\left(f(x,y)=\mu\right) = q_x^2 \int_{-\pi}^\pi dy \,\,\left|\frac{\partial}{\partial x}f\right|\Bigg|_\left(f(x,y)=\mu\right) $$ where the function is now integrated over the level set of $x,y$ such that $f(x,y)=\mu$, but my gut tells me that the factor of $\left|\frac{\partial}{\partial x}f\right|$ is not right.

I'm sure there should be a nice result for this. I'd imagine there is a way to change coordinate system to one defined by level sets of $f$, and do the calculation nicely in those coordinates, but I am not sure how to go about that. For example, clearly I could define $\chi=f(x,y)$ as one of my coordinates, but how would I define the other coordinate? I would want to define it in such a way that its unit vector is orthogonal to the level sets of $f$.

Calculators cant process this integral

Posted: 28 Mar 2021 08:29 PM PDT

Calculators can't seem to process this equation, and I don't know why. Is anyone able to explain why this happens

enter image description here $$ \frac{1}{2}\int_{0.15859433}^{3.14619322} \left[2\ln\left(x\right)-2x+4\right]^{\, 2}\,{\rm d}x $$

Which groups with order 2ⁿ, n≤128 have small commuting probability and still high element order?

Posted: 28 Mar 2021 08:33 PM PDT

I have found some formulae to calculate/estimate the commuting probability $P_G$ of some finite non Abelian groups in this thesis. For example: $S_n$, $D_n$, $Q_8$ among others. The options from the list seem to be either of prime/odd order, or with $P_G \approx \frac{1}{4}$. That makes me wonder whether there is some power-of-two group that stands not far behind from the "anti-abelianess" of $S_n$. And, I am also asking my self whether $S_n$ stablishes some kind of lower limit for $P_G$, even considering groups with order different from powers of two.

An example of power-of-two group with $P_G \ll \frac{1}{4}$ is the multiplication over unitriangular matrices (modulo). However, it is still much higher than that of $S_n$ for similar set sizes. I tried also direct product of dihedral groups and achieved even higher $P_G$ than with the matrices.

I have been putting some code and results here, in the case someone is interested in further investigation; see sections Abstract algebra module, Commutativity degree of groups and Tendency of commutativity on Mn.

I am looking for a group that is algorithmically feasible to be implemented, like $S_n$, $D_n$, $Z_n$ etc. which are straight forward to code and fast to calculate.

UPDATE:

It seems that groups of order $2^n$ are fated to be almost Abelian because prime power order implies nilpotent.

UPDATE 2:

My best attempt was a direct product of $D_{34}$ and $Z_{2^{128}-159}$ (I doubled the order to avoid half bits when importing 128 bits from legacy systems, among other benefits [and disadvantages]). Unfortunately, another attempt was to use the matrices mentioned above in the place of S_{34}, but, in a small test with 45 bits the matrices already commuted with p = ~5e-06 after 40 million pairs. The direct product of dihedral groups give formula values worse than the matrices.

The usefulness of all this is to combine identifiers that can be nicely operated within certain guarantees/properties.

PS. Thanks to the help of a couple mathematicians, almost a year by now, I am almost not feeling in completely strange waters. I even learned what is a happy family - pun intended. [and monster group]

Compute integral of general form $ \int_0^\infty \left(\frac{x}{\sinh x}\right)^n d x $

Posted: 28 Mar 2021 08:08 PM PDT

I encountered the following integral and can't compute its value in general: $$ I_n = \int_0^\infty \left(\frac{x}{\sinh x}\right)^n d x $$ where $n\in \mathbb N$.

From integral tables/mathematica, I know that

$$I_1 = \frac{\pi^2} 4,\>\>\>\>\>I_2 = \frac{\pi^2}6$$

But, for general $n\in\mathbb N$, I don't know the result though. Any ideas?

Sequences in a subset of $\mathbb R^4$

Posted: 28 Mar 2021 08:37 PM PDT

I'm having some problems with the following question:

Consider $X=\{x \in \mathbb R^4 \mid x_1x_4 - x_2x_3\neq 0\}$. Show that given $a \in \mathbb R^4$ exists a sequence $(x_n)_{n \in \mathbb N}$ with elements in $X$, such that $x_n \rightarrow a$.

My attempt was to show that $X$ is dense at $\mathbb R^4$ (it seemed more natural to me). So, the idea was to take an arbitrary $x$ in $\mathbb R^4$ and an open ball $B_x$ centered in $x$ with radius $\varepsilon$. Soon after I was trying to show that $B_x \cap X \neq \emptyset$. However, the solution did not come out, I was unable to define a general element that belongs to $X$ and $B_x$. Could someone give me a tip?

Thank you very much in advance.

Prove bounded maximum is primitive recursive [closed]

Posted: 28 Mar 2021 08:44 PM PDT

Let T(x, y) be a primitive recursive relation. Define f(x, y) as the greatest z < y such that T holds (if such z exists). Define f as y if no such z exists. Prove that f (bounded maximum) is primitive recursive

Note: This is for a beginning course in computability & complexity

My general instinct would be to break f down into primitive recursive "chunks." For example, z < y can be rewritten as x > z and that relation is primitive recursive (from class examples). We already know that T is primitive recursive. However I feel I may be oversimplifying things.

Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$

Posted: 28 Mar 2021 08:19 PM PDT

Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.

Attempt

First attempt: I was trying see the geometric meaning, but I´m fall.

Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$ and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.

Third attempt:comparing $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.

Any hint or advice of how I should think the problem was very useful.

Homework hint: use dot product to show inscribed angle of a semicircle is a right angle

Posted: 28 Mar 2021 08:26 PM PDT

I'm currently stuck on a homework problem from MIT OCW 18.02 multivariate calculus.

The problem is in a section on dot products, and asks to use vector methods (without components) to show that the inscribed angle of a semicircle is a right angle.

I have been toying around with it the past couple of days. The first thing I did was sketch it out, and I quickly came to a solution by summing up the angles in my sketch. (I drew two triangles and determined the relationship between the angles, which led me to deduce the inscribed angle was $\frac{\pi}{2}$)

But that solution doesn't feel in keeping with the rest of the problems, which makes me wonder what I missed.

I know I could just peek the answer, but I also know that wouldn't really help. I would prefer a gentle nudge in the right direction instead. Can you provide a hint that may point me in the direction I need to think?

derivative of product matrix

Posted: 28 Mar 2021 08:35 PM PDT

Let $$f(\alpha)= \alpha^{3} A(B+\alpha I)^{-3}C,$$ where A , B and C are square real matrices that do not depend on the scalar $\alpha$ and I is the identity matrix. I need help with computing the derivative of f with respect tot $\alpha$.

$(C^1(\mathbb T),\|\cdot\|) , \|f\|:=|f(0)|+\|f′\|_{L^2(\mathbb T)}$ is not a Banach space.

Posted: 28 Mar 2021 08:46 PM PDT

Show that the normed space $\left(C^1(\mathbb T),\lVert \text{.}\rVert\right)$ where $C^1$ is the space of continuously differentiable functions, and $ \lVert f\rVert:=\lvert f(0)\rvert+\lVert f'\rVert_{L^2(\mathbb T)}$ is not a Banach space

  • We define $\mathbb T:=\mathbb R/2π\mathbb Z$, the topological space of $2π$-periodic functions, in other words $\mathbb T$ can be seen as the interval $[0,2π]$ but the points $0$ and $2π$ are the same if we move leftwords to $2π.$

Edit: I've solved it in the answers.

Finding the fractal dimension of the Mandelbrot set using the box counting method

Posted: 28 Mar 2021 08:35 PM PDT

So I'm trying to calculate the fractal dimension of the perimeter of the mandelbrot set using the box-counting or Minkowski–Bouligand definition of fractal dimension. According to this definition, my results should be greater than 2, but for some reason, I keep getting around 1.36 as my dimension value.

I was wondering if what I'm doing is incorrect, even though it seems to be the proper method.

Box side length 2, 45 boxes Box side length 2, 45 boxes

Box side length 1, 122 boxes Box side length 1, 122 boxes

Box side length 0.5, 314 boxes Box side length 0.5, 314 boxes

As far as I understand it, with the box side length increasing in size by 2x, the number of perimeter boxes should be divided by 2^d, where although d should be 2, I'm getting a value of about 1.36. Any help?

How big is the equilateral triangle inscribed in a triangle?

Posted: 28 Mar 2021 08:42 PM PDT

Given a scalene triangle $ABC$ and an inscribed equilateral triangle whose vertices lie on different sides of $\triangle ABC$, what is the maximal ratio of the area of the equilateral triangle to that of the original triangle?

I would expect an answer as a ratio of polynomials in sines and cosines of the angles of $\triangle ABC$. I got such an expression via a clunky unsymmetrical method, but it but it was so messy that I gave up trying to simplify it. However, a more intelligent method might well yield a formula of reasonable length.

quadratic equation form maximum solutions

Posted: 28 Mar 2021 08:46 PM PDT

My Pearson intermediate algebra book has a "concept check" question in its section on solving equations by using quadratic methods. These questions are supposed to highlight fundamental concepts that indicate full or poor understanding of the subject. The question asks:

a. True or false? The maximum number of solutions that a quadratic equation can have is 2. b. True or false? The maximum number of solutions that an equation in quadratic form can have is 2.

The answers are listed as a. true and b. false.

I'm having difficulty searching for information on this point because search results yield explanations of how to determine the number of solutions based on the discriminant, but don't seem to get into why an equation in quadratic form is not necessarily a quadratic equation, or why it wouldn't have the same of maximum number of solutions as a quadratic equation. I'm also not finding an explanation anywhere in the text, which is mostly examples and the phrase "equation in quadratic form" is nowhere to be found.

Surprising identities / equations

Posted: 28 Mar 2021 08:09 PM PDT

What are some surprising equations/identities that you have seen, which you would not have expected?

This could be complex numbers, trigonometric identities, combinatorial results, algebraic results, etc.

I'd request to avoid 'standard' / well-known results like $ e^{i \pi} + 1 = 0$.

Please write a single identity (or group of identities) in each answer.

I found this list of Funny identities, in which there is some overlap.
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