Sunday, November 14, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

What makes every strong monad on a certain category be a monoidal functor?

Posted: 14 Nov 2021 10:29 PM PST

A concept named Monad is used a lot in functional programming. And in spite their definition is not completely same with the definition of monad in category theory, as I know, Monad on a programming language is equivalent to certain category theoretical monad. The category has data types as objects and pure functions between them as morphisms. Since data types can be seen as a set of values, it's simply a subcategory of Set.

I found that in functional programming, a Monad is always an Applicative Functor, which is said to be 'programming equivalent of a lax monoidal functor with tensorial strength in category theory.'

Exactly how is this possible? Specifically, What makes monad on Set always be a monoidal functor?

I tried to prove this from that every monad in Set are strong monad, but failed. The only thing I could find was that strong monads are monoidal if they are commutative. The problem is that even in the category of sets, monads are not commmutative in general. There are many monad which are not commutative in Set, including free monoid monad.

Just proving that every monad in Set are monoidal functor perhaps not be that hard. I rather want to know more fundamental, or generalizable reason behind that. (If exist.) What makes every strong monad in certain category be always monoidal functor, even when they are not commutative?

Is the operator the operator $T : C[0, 1] → C[0, 1]$ defined by $(T f)(x) = x^2f(x)$ for x ∈ [0, 1] continuous?

Posted: 14 Nov 2021 10:14 PM PST

Let the operator $T : C[0, 1] → C[0, 1]$ be defined by $$(T f)(x) = x^2f(x)$$ for x ∈ [0, 1]. If we equip C[0, 1] with the norm $$||f||_{L^p[0,1]} =(\int^ 1_0|f(x)|^p dx)^{1/p}$$ for $1 ≤ p < ∞$, show that $T$ is linear, bounded, and find its operator norm $||T||$.

Proof idea: Linearity follows simply from the definition of Tf. (no questions here)

Boundedness (thus continuity):

$||Tf||^p = ||x^2 f(x)||^p = \int_0^1 |x^2f(x)|^p dx =\int_0^1 |x^{2p}||f(x)|^p dx\leq \int_0^1 |f(x)|^p$ since $x^{2p}\in [0,1]$.

So $||Tf||^p = 1 ||f||^p_{L^p[0,1]}<\infty$ so $||Tf||$ is bounded thus continuous.

$||T||=sup\frac{||Tf||}{||f||}\leq sup\frac {||f||}{||f||}=1$.

Let $f_1=1 \in C[0,1]$ then $||f_1||=1$ and $||T||=sup\frac{||Tf||}{||f||}\geq \frac {||Tf_1||}{||f_1||}=1/1=1$. So $||T||=1$.

I am uncertain about it. Is this the norm? also is my proof of boundedness correct? Thanks and regards,

Minimizing trace equivalence

Posted: 14 Nov 2021 09:55 PM PST

In this paper: https://web.stanford.edu/~boyd/papers/pdf/rank_min_heur_hankel.pdf

They provide an interative algorithm such as this one:

$$X_{k+1}=\underset{X \in \mathcal{C}}{\operatorname{argmin}} \operatorname{Tr}\left(X_{k}+\delta I\right)^{-1} X$$

Where $\delta > 0 $ is a small regularization constant. They further say:

If we choose $X_0 = I$, the first iteration is equivalent to minimizing the trace of $X$.

My question is then how does this hold for all positive $\delta$. If $\delta$ is small enough $I + \delta I \approx I$ and then the equivalence obviously holds. However how would it hold if $\delta$ is not small enough?

how to find the variance on this problem?

Posted: 14 Nov 2021 09:51 PM PST

I'm having problems solving this problem "Let X be a uniform r.v. over (−1, 1). Let $Y = X^n.$ Calculate: the covariance of X and Y and the correlation coefficient of X and Y.

first we'll be finding the covariance (it looks like this:) $$cov(x,y) = \frac{1}{n+2} \space\forall n -\space odd $$ $$cov(x,y) = 0\space \forall n -\space even $$

after this it is clear to me that i should use this formula: $$\frac{cov(x,y)}{\sqrt{var x} \sqrt{var y}}$$

finding the var of x should look something like this: $$\frac{(b-a)^2}{12} = \frac{1-(-1)^2}{12} = \frac{1}{3}$$ but the problem comes for var y, it is not clear to me how am i supposed to find that parameter, i know it has something to do with this formula:

$$ 𝖵𝖺𝗋(Y)=𝖤(Y^2)−(𝖤(Y))^2.$$ but i don't understand how to use it with X^n, anyone can help?

Gaussian integral $\int_{0}^{\infty} dx \, x \, e^{-a^2 x^2} \left( \sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2} \right)$

Posted: 14 Nov 2021 09:51 PM PST

I've been having trouble evaluating the Gaussian integral of the form

$$\int_{0}^{\infty} dx \, x \, e^{-a^2 x^2} \left( \sqrt{(x - c)^2 + b^2} - \sqrt{(x + c)^2 + b^2} \right) \, ,$$

which can be rearranged into integrals of the form

$$\int_{0}^{\infty} dx \, x^k \sqrt{x^2 + b^2} \left( e^{-a^2 (x + c)^2} \pm e^{-a^2 (x - c)^2} \right) \, ,$$

for $a, \, b, \, c > 0$ and $k \geq 0$. Does anyone know how to evaluate these expressions analytically? Or know any good integral tables on Gaussians I should refer to?

How do I prove that a vector space endomorphism is a commutative ring if and only if the vector space has dimension 1?

Posted: 14 Nov 2021 09:49 PM PST

Let $V$ be a vector space over $k$. Show that $\mathrm{Hom}(V,V)$ is a commutative ring (with respect to addition and composition of functions) if and only if $V$ has dimension 1.

We know that $\mathrm{Hom}(V,V)$ is a ring (with respect to addition and composition of functions) and would like to prove that it is commutative.

If $V$ has dimension 1, then every $v\in V$ is a multiple of some $v_0\in V$, and every $T\in\mathrm{Hom}(V,V)$ is a scalar multiplication. By commutativity of scalar multiplication, $\mathrm{Hom}(V,V)$ is a commutative ring.

Now, given that $T_2\circ T_1=T_1\circ T_2$ for every $T_1,T_2\in\mathrm{Hom}(V,V)$, how do we show that $V$ has dimension 1?

Why is the number of heads and the number of tails independent in $n$ coin flips?

Posted: 14 Nov 2021 09:44 PM PST

In probability, in different manifestations, I have noticed that the sum of random variables $X$ and $Y$ may be given as $n$, and yet $X$ and $Y$ may still be independent. I find this very counter intuitive. If $Y = n - X$, how can it be independent of $X$?

For example, if I flip a coin 10 times, are the number of heads and the number of tails independent? It doesn't quite make sense for them to be to me.

Thanks in advance

Tuning Beta model: Bayes Rules! Book exercise

Posted: 14 Nov 2021 09:43 PM PST

I'd like to verify my solution and receive wider clarification where it's possible. This is an exercise from Bayes Rules! book.

Exercise 3.1 (Tune your Beta prior: Take I) In each situation below, tune a Beta(α,β ) model that accurately reflects the given prior information. In many cases, there's no single "right" answer, but rather multiple "reasonable" answers.

a. Your friend applied to a job and tells you: "I think I have a 40% chance of getting the job, but I'm pretty unsure." When pressed further, they put their chances between 20% and 60%.

First of all, I think that "a 40% chance of getting the job" should be treated as mode of the beta distribution, because according to the context 40% is the most probable value where pdf respectively reaches its maximum. Playing with numbers around using the formula $$\frac{\alpha-1}{\alpha+\beta-2}$$

I found reasonable $\alpha=11$ and $\beta=16$

enter image description here

b. A scientist has created a new test for a rare disease. They expect that the test is accurate 80% of the time with a variance of 0.05.

In this case I think that 80% must be treated as mean of the beta. "A variance of 0.05" is actually a variability and therefore must be treated as standard deviation of the beta. In fact I raised an issue about the misleading formulation https://github.com/bayes-rules/bayesrules/issues/87. Using the formula for the mean of the beta distribution $$\frac{\alpha}{\alpha+\beta}$$ it's obvious that $\beta=\frac{1}{4}\alpha$

Trying various numbers, I stopped on $\alpha=48$ and $beta=12$. With these parameters $SD(x) = 0.05121475$ which is close to "a variance of 0.05". enter image description here

c. Your Aunt Jo is a successful mushroom hunter. She boasts: "I expect to find enough mushrooms to feed myself and my co-workers at the auto-repair shop 90% of the time, but if I had to give you a likely range it would be between 85% and 100% of the time."

"90% of the time" is mean of the beta. So $\beta=\frac{1}{9}\alpha$. I found the reasonable values of $\alpha=90$ and $beta=10$. The beta cdf(0.85) with these parameters gives us 0.05952991 which I consider fitting for "likely range it would be between 85% and 100% of the time".

enter image description here

d. Sal (who is a touch hyperbolic) just interviewed for a job, and doesn't know how to describe their chances of getting an offer. They say "I couldn't read my interviewer's expression! I either really impressed them and they are absolutely going to hire me, or I made a terrible impression and they are burning my resumé as we speak."

This is the most interesting case in my opinion. As all the events are equally possible then it leads to the uniform probability distribution with $\alpha=1$ and $\beta=1$.

enter image description here

How can one find a polynomial with a given solution?

Posted: 14 Nov 2021 09:53 PM PST

I was just looking at "algebraic" or non-transcendental numbers and apparently every number that can be created with multiplication, addition, subtraction, reciprocation, and radicals using a finite number of terms is algebraic and the solution to a polynomial with integer coefficients. Given a certain algebraic number, how can one find the corresponding polynomial with integer coefficients? Is it easier or harder than solving for the solutions of a degree 5+ polynomial?

Concave Lagrange dual function

Posted: 14 Nov 2021 09:38 PM PST

It can be proved that the Lagrange dual function is concave because the point-wise infimum of affine functions is concave.

Consider a simple example to minimize $f(x)=-\max(x, 0)$ under the equality constraint $|x|-3=0$, the Lagrangian is $L(x, \gamma) = f(x) + \gamma (|x|-3)$. It is not difficult to work out the Lagrange dual $g(\gamma)=\inf_x L(x, \gamma)$ to be $g(\gamma)=-3\gamma$ if $\gamma \geq 1$ and $g(\gamma)=-\infty$ if $\gamma < 1$.

Now I am very confused because this $g(\gamma)$ does not look like a concave function. For example, consider the interval $\gamma \in [0, 2]$, $g(\gamma)$ does lie below the straight line joining $(0, g(0))$ and $(2, g(2))$. What goes wrong?

Let $n \in N$, $(X, A, \mu)$ be a measure space and $E_1 ,. . . , E_n \in A$.

Posted: 14 Nov 2021 09:54 PM PST

Let $n \in N$, $(X, A, \mu)$ be a measure space and $E_1 ,. . . , E_n \in A$. For each $j \in \{1 ,. . . , n\}$ define $$C_j∶= \{x \in X | x \in E_k \quad \text{for exactly $j$ indices}, k ∈ \{1. . . , n\}\}.$$ I want to find a general expression for all the $C_j$ since I need to prove that each $C_j$ is a measurable set, I made an example and it is as follows

With $n = 3$

$$C_1 = (E_1\cap E_2^c \cap E_3^c)\cup (E_1^c\cap E_2 \cap E_3^c)\cup(E_1^c\cap E_2^c \cap E_3)$$ $$C_2 =(E_1\cap E_2 \cap E_3^c)\cup (E_1\cap E_2^c \cap E_3)\cup (E_1^c\cap E_2^c \cap E_3)$$ $$C_3 = \bigcap_{k=1}^3 E_k$$ So for $C_n$ we already have a general expression: $$C_n = \bigcap_{k=1}^n E_k$$ for $C_1$: $$C_1=\bigcup_{j=1}^n \left(E_j \cap \bigcap_{k\neq j}^n E_k^c\right) $$ I was thinking of introducing permutations but actually I don't know if it works, someone can guide me a bit to express the $ C_j $ sets in a manageable way

Proving Continuity in Higher Dimensions

Posted: 14 Nov 2021 09:40 PM PST

Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $f(x,y) = (x^2-y^2, 2xy)$. Show that $f$ is continuous.

In one dimension ($\mathbb{R}$), using epsilon-delta proofs is a very straightforward process. You start with $|f(x) - f(x_0)| < \varepsilon$ and you manipulate the inequality algebraically until you can obtain $|x-x_0| <$ [expression]. Then you let $\delta = $ [expression]. And doing the algebra in reverse gives the conclusion.

Now, in more than one dimension (i.e. when the codomain is multi-dimensional), it becomes less obvious how to use this same process to obtain a $\delta$ in terms of $\varepsilon$. I show my work below.

Proof Attempt.

In order to show that $f$ is continuous, we show that

$$ \forall \, \varepsilon > 0, \exists \, \delta > 0 \, \forall \, j_0 \in \mathbb{R}^2, \|j - j_0 \| < \delta \implies \|f(j) - f(j_0) \| < \varepsilon$$

where $j_0 \in \mathbb{R}^2$ and for fixed $j \in \mathbb{R}^2$.

So we start with $\|f(j) - f(j_0)\| < \varepsilon$.

$$ \| (x^2-y^2, 2xy) - (x_0^2 - y_0^2, \, 2x_0y_0) \| < \varepsilon$$

$$\| (x^2-y^2 - ( x_0^2 - y_0^2), 2xy - 2x_0y_0 ) \| < \varepsilon$$

$$\sqrt{\Big[x^2-y^2 - ( x_0^2 - y_0^2) \Big]^2 + \Big[ 2xy - 2x_0y_0 \Big]^2} < \varepsilon$$

$$\Big[x^2-y^2 - ( x_0^2 - y_0^2) \Big]^2 + \Big[ 2xy - 2x_0y_0 \Big]^2 < \varepsilon^2$$

$$\vdots$$

$$\|(x-x_0, y- y_0) \| < \delta$$

Am I on the right track here? The \vdots are the remaining steps I need to obtain my $\delta$ but it doesn't appear to me that I can end up with $|(x-x_0 , y-y_0) | <$ [expression] without lumping variables over on the righthand side.

Any advice or suggestions are greatly appreciated.

A few questions on Linear Algebra

Posted: 14 Nov 2021 10:20 PM PST

I actually posted this question a few weeks ago where I wanted my solutions to a few Linear Algebra questions checked. Now thanks to useful links provided by @GerryMyerson I can ask my questions(and verify my answers).

The first three questions I just want solution verifications and suggestions for methods which are faster than the ones I used(Also no answers were provided by the creator, so i'm not even sure of the correct answer).

Question 1:

Show that the equation:

$x^2 + y^2 + z^2 + 8x -6y + 2x + 17 = 0 $

Represents a sphere, and find it's centre and radius.

My Solution:

$x^2 + y^2 + z^2 + 8x -6y + 2x + 17 = 0 $

By completing the square we have that:

$((x+4)^2 -16) + ((y-3)^2 -9) + ((z+1)^2 -1) + 17 = 0 $

Simplifying and collecting we have that:

$(x+4)^2 + (y-3)^2 + (z+1)^2 = 9 $.

Since a sphere is an equation of the form:

$(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = r^2$ where $x_0,y_0,z_0 = (-4,3,-1)$ are the coordinates of the centre of the sphere. Since $r^2 = 9$ the radius is $3$. Our expression can be rewritten as:

$(x-(-4))^2 + (y-(3))^2 + (z-(-1))^2 = 0$, which is indeed the representation of a sphere(I think).

Question 2:

Find the area of the triangle with vertices $P(1,-2,3), Q(0,3,1)$ and $R(-1,1,0)$.

My Solution:

Since $\vec{PQ} = \langle -1,5,-2 \rangle$ and $\vec{PQ} = \langle -2,3,-3 \rangle$ we take $\vec{PQ} \cdot \vec{QR}$ and have that:

\begin{vmatrix} i & j & k\\ -1 & 5 & -2 \\ -2 & 3 & -3 \end{vmatrix}

Simplifying we get: $-9i + j + 7k$ therefore $\vec{PQ} \cdot \vec{QR} = \langle -9, 1, 7\rangle$.

$\Vert \langle \vec{PQ} \cdot \vec{QR} \rangle \Vert = \sqrt{(-9)^2 + (1)^2 + (7)^2} = \sqrt{131}$. Therefore using the formula $A = \frac{1}{2}\Vert \vec{v} \cdot \vec{u} \Vert$ we get:

$A = \frac{1}{2} \cdot \sqrt{131}$
$A = 0.5 \cdot 11.4455$
$A = 5.723$ unites squared

Question 3:

Find the volume of the paralelpiped spanned by the vectors $a = \langle 1,2,3 \rangle$, $b = \langle 0,1,-1 \rangle$ and $c = \textbf{i} + \textbf{j}$

My Solution:

Vector $\textbf{c}$ can be rewritten as $\langle 1,1,0 \rangle$. Taking the $3$x$3$ matrix we get that:

$\begin{vmatrix} 1 & 2 & 3\\ 0 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix} = 1 \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} - 2 \begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix} + 3 \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}$

Simplifying the determinant we get that the volume is $2$ units cubed.

Those were the three question's that I need method/solution verification and improvements(if needed) for. I am still a beginner so if you do have any advice could you please explain it in a more fundamental way. Thanks!

The Laplace transform $\mathcal L$ is a self-adjoint operator on $L^2(\mathbb R_+)$

Posted: 14 Nov 2021 09:52 PM PST

Let $\mathbb R_+ = [0, \infty)$ and consider the kernel $K(x,y) = e^{-xy}$ on $\mathbb R_+ \times \mathbb R_+$. The associated integral operator on $L^2(\mathbb R_+)$ is called the Laplace transform $\mathcal L$, i.e. $$(\mathcal L f)(x) = \int_0^\infty K(x,y) f(y)\, dy$$ Show that the Laplace transform $\mathcal L$ is a self-adjoint operator on $L^2(\mathbb R_+)$.

This question is similar to another question that I'd asked a few days ago, but the same approach doesn't work because Fubini's theorem may not be applicable here (for the kernel is not square-integrable) - at least similar bounds do not work.

My work:
We want to show $\langle \mathcal Lf, g\rangle = \langle f, \mathcal Lg\rangle$ for all $f,g\in L^2(\mathbb R_+)$. We have $$\begin{align*} \langle \mathcal Lf,g\rangle &= \int_0^\infty \mathcal Lf(x) \overline{g(x)} \, dx\\ &= \int_0^\infty \int_0^\infty e^{-xy} f(y) \overline{g(x)}\, dy\, dx \end{align*}$$ and $$\begin{align*} \langle f, \mathcal L g\rangle &= \int_0^\infty f(x) \overline{\mathcal Lg(x)}\, dx\\ &= \int_0^\infty \int_0^\infty e^{-xy} f(x) \overline{g(y)} \, dy\, dx\\&= \int_0^\infty \int_0^\infty e^{-xy} f(y) \overline{g(x)} \, dx\, dy \end{align*}$$ So, we must show $$\int_0^\infty \int_0^\infty e^{-xy} f(y) \overline{g(x)}\, dy\, dx = \int_0^\infty \int_0^\infty e^{-xy} f(y) \overline{g(x)} \, dx\, dy $$ This can be done easily if Fubini's theorem is applicable. For Fubini's theorem, we require $$\int_0^\infty \int_0^\infty |e^{-xy}| |f(y)| |g(x)| \, dx\, dy < \infty$$ By applying Cauchy-Schwarz inequality twice (as in the linked post's answer), I was able to get $$\int_0^\infty \int_0^\infty |e^{-xy}| |f(y)| |g(x)| \, dx\, dy \le \|f\|_2 \|g\|_2 \left(\int_0^\infty \int_0^\infty e^{-2xy}\, dx\, dy \right)^{1/2}$$ Unfortunately, $$\int_0^\infty \int_0^\infty e^{-2xy}\, dx\, dy$$ diverges, so this bound is useless.

Two possibilities arise: (i) either Fubini's theorem is applicable, and we must find a stronger bound to show $\int_0^\infty \int_0^\infty |e^{-xy}| |f(y)| |g(x)| \, dx\, dy < \infty$, or (ii) we must show that the two integrals are equal using some technique other than Fubini's theorem. To me, (ii) seems less likely.

I'd appreciate any help in completing the proof. Thank you!

Note:

  1. $\mathcal L$ is a bounded operator on $L^2(\mathbb R_+)$, in fact, $\| \mathcal L\| = \sqrt\pi$. $\mathcal L$ is also bijective. The explicit calculations can be found in Rajendra Bhatia's Notes on Functional Analysis, Pg. $26-27$, and also in Setterqvist E., Unitary Equivalence: A New Approach to the Laplace transform and the Hardy operator. Master's Thesis, Department of Mathematics Lule ̊a University of Technology, 2005:329 CIV here. Further, see Post 1, Post 2 and Post 3.

to determine the integration of $ \int_\limits{0}^{\infty} \exp-\bigg(\frac{ax^2+bx+c}{gx+h}\bigg) dx$.

Posted: 14 Nov 2021 09:29 PM PST

What is the integration of the following function- $$ \int_\limits{0}^{\infty} \exp-\bigg(\frac{ax^2+bx+c}{gx+h}\bigg) dx$$

What I have done is as follows-

enter image description here

Here, $\kappa=c-\bigg(\frac{bg-ah}{g^2}\bigg)h$. $$\implies\int_\limits{0}^{\infty} \exp-\bigg(\frac{ax^2+bx+c}{gx+h}\bigg) dx =\int_\limits{0}^{\infty}\exp-\bigg(\frac{a}{g}x+\frac{b.g-a.h}{g^2}\bigg).\exp\bigg(-\frac{\kappa}{gx+h}\bigg)dx$$ I am finding it difficult to proceed further from here. Is this approach correct or is there any other intuitive way to solve this problem? Thanks in advance for the help.

About the motivation behind a proof

Posted: 14 Nov 2021 09:36 PM PST

I'm wondering about the motivation for proving the limit.

$$\displaystyle \lim_{x\to 1} \sqrt[n] {x} =1 $$

I saw the solution, which uses the squeeze theorem.

Solution. a) Fix $n \in \mathbb{N}$. Let's show first that $\lim_{x \to 1} \sqrt[n]{x}=1$. Denote by $\sqrt[n]{x} = y$ then $y^n=x$ and using standard identity we have $$|1 - x|=|1-y^n|=|1-y|\cdot|1 + y + y^2 + \cdots + y^{n-1}| \geq |1 -y|$$ since $|1+y+y^2+\cdots y^{n-1}| \geq 1$ for $y \geq 0$. Thus, we have $$0 \leq |1 - \sqrt[n]{x}| \leq |1-x|.$$ Using the squeeze theorem and the fact that $\lim_{x\to 1}(1 - x) = 0$, we prove a) in the case of $a = 1$.

Even though I understand this, I don't know how to come up with this idea.

When proving this, I thought about using the squeeze theorem, but I failed.

When trying using squeeze theorem, I attempt to find inequality like : $g(x)$<$\;\sqrt[n] {x}\;$<$\;f(x)$ and inequality like :$g(x)$<$\;\sqrt[n] {x}-1\;$<$\;f(x)$.

However, I finally give up. Then I see the solution, and it uses the absolute value to find the inequality. Even though the solution is correct, I think this is slightly unreasonable and logically disconnected, because I don't see any hints for using absolute value.

Therefore, I want to ask that is there some clues in this question that motivate us to use absolute value or is there a better and more "natural" way to use squeeze theorem for this?

What does The line is reflected in the line mean?

Posted: 14 Nov 2021 09:37 PM PST

I am doing the $2014$ CIMC (Canadian Intermediate Mathematics Contest) contest and I am kind confused with the question A6 meaning.

I just wonder what does the line is reflected in the line mean?

The line is reflected in the line with equation $x + y = 1$.

Here is the whole question: A line has equation $y=kx$,where $k$ not equal $0$ and $k$ not equal $−1$. The line is reflected in the line with equation $x + y = 1$. Determine the slope and the y-intercept of the resulting line, in terms of $k$.

From CIMC $2014$ part a 6.

9-(-3^2) and 9+(-3^2) - Order of Operation. Exponent (^) and Minus (-) sign | Spreadsheet and Calculator give opposite answers. Why?

Posted: 14 Nov 2021 10:12 PM PST

=9-(-3^2) isn't this the same as =9-(-3*-3) (A?) Google spreadsheet result: 0. Google scientific calculator result: 18

Note: =9 - (-(3^2)) - The spreadsheet automatically adds in two spaces into the input [column A and B only]. Why? (D?)Edit: Seems to have resolved itself

=9+(-3^2) isn't this the same as =9+(-3*-3) (B?) Google spreadsheet result: 18. Google scientific calculator result: 0

Google scientific calculator interprets it as 9-(-(3^2)) and 9+(-(3^2)) rather than 9-((-3)^2) and 9+((-3)^2)

Therefore never accepting -3 by itself as an integer/real number without brackets when dealing with exponents. (C?)

Why is this? (A, B, C)

Definite integral of a power times a cosine

Posted: 14 Nov 2021 10:38 PM PST

The following integral appears in Gradshteyn & Rhyzik, page 421 3.769 $$\int_0^{\infty } x^{\mu -1} \cos (a x) \, dx=\frac{\Gamma (\mu ) \cos \left(\frac{\mu \pi }{2}\right)}{a^{\mu }}$$ (a>0, $0<\Re(\mu )<1$)

I am looking for a derivation

Show that if $0 \leq a \leq b$ then $f_m(a) \leq f_n(b)$ for $m \leq n.$

Posted: 14 Nov 2021 09:53 PM PST

Let $n \in \mathbb N$ and let $f_n$ be the function $t \mapsto t \left (t + \frac {1} {n} \right )^{-1}$ on $\mathbb R_{+}.$ Let $a$ and $b$ be two positive elements in a $C^{\ast}$-algebra $A$ such that $a \leq b$ and let $m \leq n.$ Then $f_m (a) \leq f_n (b),$ where $f_m(a)$ and $f_n(b)$ are the images of $f_m$ and $f_n$ under the continuous functional calculus for $a$ and $b$ respectively.

My Attempt $:$ Here $$f_m(a) = a \left (a + \frac {1} {m} \right )^{-1} = \left (a + \frac {1} {m} \right )^{-1} a\ \ \text{and}\ \ f_n (b) = b \left (b + \frac {1} {n} \right )^{-1} = \left (b + \frac {1} {n} \right )^{-1} b.$$

From here I couldn't quite able to show the required inequality.

Since $m \leq n$ we have $0 \leq a + \frac {1} {n} \leq a + \frac {1} {m}.$ Hence $0 \leq \left (a + \frac {1} {m} \right )^{-1} \leq \left (a + \frac {1} {n} \right )^{-1}.$ So $$\begin{align*} a \left (a + \frac {1} {m} \right )^{-1} & = a^{\frac {1} {2}} \left (a + \frac {1} {m} \right )^{-1} a^{\frac {1} {2}} \\ & \leq a^{\frac {1} {2}} \left (a + \frac {1} {n} \right )^{-1} a^{\frac {1} {2}} \\ & = a \left (a + \frac {1} {n} \right )^{-1} \\ & = \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} a \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} \\ & \leq \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} b \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} \\ & = b \left (a + \frac {1} {n} \right )^{-1} \\ & = b^{\frac {1} {2}} \left (a + \frac {1} {n} \right )^{-1} b^{\frac {1} {2}} \end{align*}$$ At this stage I got stuck. I know that $0 \leq a + \frac {1} {n} \leq b + \frac {1} {n}.$ So we have $0 \leq \left (b + \frac {1} {n} \right )^{-1} \leq \left (a + \frac {1} {n} \right )^{-1},$ which is not what I wanted to have. Could anyone give me some suggestion regarding this?

Thanks for your time.

EDIT $:$ I think the last two equalities doesn't hold in general unless $a$ and $b$ commute.

Finding a closed formula for calculating $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$

Posted: 14 Nov 2021 10:16 PM PST

For reasons I can't even remember, the other day I wanted to find out if there was a closed formula for calculating the $n$-th derivative $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$ for the function $f\left(x\right)=e^{x^2}$. Where I ended up after some trial and error is the formula $$\frac{d^n}{{dx}^n}e^{x^2}=c_n\left(\sum_{0 \leq i \leq \lfloor\frac{n}{2}\rfloor} {p_i x^{n-2i}}\right)e^{x^2}=c_n\left(c_{n-1}x^n + \sum_{1 \leq i \leq \lfloor\frac{n}{2}\rfloor} {p_i x^{n-2i}}\right)e^{x^2},$$ with $c_n=2^{n-\lfloor\frac{n}{2}\rfloor}$.

The $p_{i \geq1}$ turn out to be as follows:

$n$ $p_{i=1}$ $p_{i=2}$ $p_{i=3}$ $p_{i=4}$ $...$
0 – – – – $...$
1 – – – – $...$
2 1 – – – $...$
3 3 – – – $...$
4 12 3 – – $...$
5 20 15 – – $...$
6 60 90 15 – $...$
7 84 210 105 – $...$
8 224 840 840 105 $...$
9 288 1512 2520 945 $...$
$...$ $...$ $...$ $...$ $...$ $...$

I have not yet understood the rule behind the $p_i$ sequences $$p_{i=1}:\left(1,3,12,20,60,84,224,288,...\right),$$

$$p_{i=2}:\left(3,15,90,210,840,1512,...\right),$$

$$p_{i=3}:\left(15,105,840,2520,...\right),$$

$$p_{i=4}:\left(105,945,...\right),$$

$$...$$ I suspect it has something to do with binomial coefficients, since the coefficients $p_i$ arise from multiplying binomials during the derivation. One regularity I've noticed so far is that starting at $p_{i=2}$, the first values always correspond to the second ones of the previous $p$ sequence.

Do any of you have an idea how I can formalize the coefficients $p_i$ and the timing of their occurrence and integrate them into the above closed formula? Or do you know if there even already exists a known solution to the problem, namely finding a closed formula to calculate the $n$-th derivative $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$?

Thank you and best regards!

What is the probability that none of the friends was born on the same day of the month as the host?

Posted: 14 Nov 2021 10:34 PM PST

A man invited five friends. He was born in April as also all the invited friends. What is the probability that none of the friends was born on the same day of the month as the host?

The way I approached it was $(30\times 29^5)/(30^6)$. However, there is yet another equally convincing way i.e. Probability that a friend's birthday is on the same day as the host is $1/30$. So if this goes for all friends then we have $(1/30)^5$. And we want the negation of it so $1-(1/30)^5$.

Which one is correct?

Sets order isomorphic to $[\omega_0, \omega_1)$

Posted: 14 Nov 2021 10:32 PM PST

Let $\omega_0$ be the first countable ordinal and $\omega_1$ be the first uncountable ordinal. Use $[0, \omega_1)$ to denote the set of all countable ordinals and hence $\omega_1 = [0, \omega_1)$. Could someone construct a set (other than the set of ordinals) that is order isomorphic to the set $[\omega_0, \omega_1)$. In this post, Arthur (in the top post) provided an example, which is the following:

For each $\alpha\in[0, \omega_1)$, let $X_{\alpha}$ be the set of subsets of $\mathbb{Q}$ , each of which is order isomorphic to $\alpha$, and define $X_{\alpha} < X_{\beta}$ iff $\alpha < \beta$. Then it is claimed that the following set $A$, equipped with the order "<"

$$A = \{X_{\alpha}\,\vert\,\alpha\in[0, \omega_1)\}\hspace{1cm}(\subseteq\mathcal{P}[\mathcal{P}(\mathbb{Q})])$$

is order isomorphic to $[0, \omega_1)$. If we can find a uncountable family of sets (ideally not subsets of the set of ordinals) so that the supremum is not in the family and no countable union of elements will be equal to the supremum, will that work for this question?

Reindexing double sum where lower limit of inner sum is dependent on lower limit of outer sum

Posted: 14 Nov 2021 10:27 PM PST

I am new to the math world and I am currently struggling with seeing how the reindexing for the following identity works: $$ \sum_{k=0}^{n}{\sum_{i=k}^{n}{\binom{n}{i}\binom{i}{k}}} = \sum_{i=0}^{n}{\sum_{k=0}^{i}{\binom{n}{i}\binom{i}{k}}} $$ I see how one of the last steps, changing the order of the sums with $$ \sum_{k=0}^{i}{\sum_{i=0}^{n}{\binom{n}{i}\binom{i}{k}}} = \sum_{i=0}^{n}{\sum_{k=0}^{i}{\binom{n}{i}\binom{i}{k}}} $$ works but I am not able to even get to this point. Could somebody please point me into the right direction or even be so kind to explain it to me?

Thanks a lot in advance

What arangement of $n$ points in the plane minimizes the dispersion of the distances between them?

Posted: 14 Nov 2021 09:57 PM PST

If $n=3$ the solution is an equilateral triangle where all sides have the same length. In general, there are $N = \frac{n}2(n-1)$ distances between the points. Use the empirical coefficient of variation as a relative measure of dispersion of the distances $x_i$

$$v = \frac1{\bar{x}} \sqrt{\frac1N \sum_{i=1}^N (x_i - \bar{x})^2} \qquad \bar{x} = \frac1N \sum_{i=1}^N x_i$$

Below are some arangements of four points and their values of $v$ (distances with the same length are coloured). I have not found an arangement for $n=4$ with a smaller variation than the square. It is possible to show that for two points at $(\pm 1,0)$ and the two other ones at $(0,\pm y)$ the minimum of $v$ occurs at $y = 1$.

$\hspace{4cm}$

For $n < 8$ the regular polygons seem to minimize the variation $v = P(n)$. After that it is better to place one point at the center and arange the others evenly around it like in the arangement on the right. In that case, call $v=Q(n)$

$$P(n) = \tan{\left( \frac\pi{2n} \right)} \sqrt{N - \cot^2{\left( \frac\pi{2n} \right)}} \qquad Q(n) = \frac{\sqrt{\frac{n^2}2 - \left( \cot{\left( \frac\pi{2(n-1)} \right)} + 1 \right)^2}}{\cot{\left( \frac\pi{2(n-1)} \right)} + 1}$$

$$\begin{array}{c|cccc|ccc} n & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline P(n) & 0.172 & 0.236 & 0.277 & 0.3066 & 0.328 & 0.345 & 0.359 \\ Q(n) & 0.268 & 0.269 & 0.287 & 0.3068 & 0.324 & 0.339 & 0.351 \end{array}$$

Are there better strategies for $n \to \infty$? And is there a way to prove an arangement is optimal for a given $n$?

Expected waiting time in M/M/1 queue

Posted: 14 Nov 2021 09:43 PM PST

What is the expected waiting time in an $M/M/1$ queue where order service is last-in-first-out? On service completion, the next customer served is the most recent arrived. Suppose we do not know the order of service (think of a busy retail shop that does not have a "take a number" system).

I tried many things like using $L = \lambda w$ but I am not able to make progress with this exercise. I am new to queueing theory and will appreciate some help. I can't find very much information online about this scenario either.

I found this online: https://people.maths.bris.ac.uk/~maajg/teaching/iqn/queues.pdf

It helped me understand more.

But it does help me with this problem.

I think that the expected waiting time (time waiting in queue plus service time) in LIFO is the same as FIFO. But I am not completely sure. I remember reading this somewhere. Maybe this can help?

Boundedness of the Hilbert-Hankel operator on $L^p(\mathbb{R^+})$

Posted: 14 Nov 2021 10:39 PM PST

This is an exercise in Lax.

The Hilbert-Hankel operator is defined to be $f\mapsto g(r)=\int^\infty_0 \frac{f(t)}{t+r}dt$. The question is to show the operator is a bounded map of $L^p(\mathbb{R^+})\rightarrow L^p(\mathbb{R^+})$.

Here is what I have so far:

I need to show that $\frac{||g||_p}{||f||_p}\le C$ for all $f\in L^p(\mathbb{R^+})$. Assume $f\ge 0$, by Minkowski's inequality for integrals, $||g||_p\le c\int^\infty_0 f(t)t^{1/p-1}dt=cI$.

Now I have no idea to show the bound of $I/||f||_p$, or equivalently the bound of $I^p/||f||_p^p$. Or do it need to divide the integral for different treatments?

Finding the length of a side of an equilateral triangle

Posted: 14 Nov 2021 10:01 PM PST

There is a large right isosceles triangle with a hypotenuse length of $24$. Inside the triangle is an equilateral triangle with a vertex on the midpoint of the hypotenuse. If the length of each side of the equilateral triangle is $k(\sqrt{3}-1)$, find $k$. 

I know that if $y$ is the length of one side of the equilateral triangle:

  1. $\triangle BDE$ is a 30-60-90 triangle so $BC= \frac{y\sqrt{3}}{2}$.
  2. $\triangle ABG$ is an isosceles triangle where $AB \cong BG$, so $AB=12$.
  3. $AC= \frac{y}{2}$
  4. $AC= AB+BC$, or $12= \frac{y\sqrt{3}}{2}+\frac{y}{2}$.

I'm just a bit confused as to reasoning behind the above/why it works.

  1. I thought $BC= y\sqrt{3}$, not $\frac{y\sqrt{3}}{2}$. Why is it divided by $2$?
  2. How do we know $\triangle ABG$ is isosceles?
  3. I don't understand how $AC= \frac{y}{2}$.
  4. I can do step 4.

enter image description here

Let $X$ and $Y$ be i.i.d. $\operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$

Posted: 14 Nov 2021 09:53 PM PST

Let $X$ and $Y$ be i.i.d. $\operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$.

I have generally difficulties with such problems, as I get easily confused. Below I detailed my (most probability incorrect) approach. Besides the correct approach, I also would like to have some more general advice on how to think about and tackle such problems.

The random variables $X$ and $Y$ have geometric distributions, so $P(X=x)=(1-p)^xp$ and $P(Y=y)=(1-p)^yp$. I also know that the sum of two i.i.d. geometric random variables has a negative binomial distribution, thus $P(N=n) = \binom{n+r-1}{n} (1-p)^n p^r$, where $r$ is the number of success. In our case $r=2$, as we are dealing with the sum of two random variables. Hence, $P(N=n) = (n+1)(1-p)^n p^2$.

I'm asked to find $P(X=x, Y=y, N=n)$. I don't know how to write that down, so I condition on $N=n$ and write

\begin{align} P(X=x, Y=y, N=n) &= P(X=x, Y=y \mid N=n)P(N=n) \\ &= P(X=x, Y=y \mid N=x+y)P(N=n), \end{align} because I know that $n=x+y$. But conditional on $N=x+y$, $X=x$ and $Y=y$ are the same event. Thus,

\begin{align} P(X=x, Y=y, N=n) &= P(X=x \mid N=x+y)P(N=n) \end{align}

I can assume that $X$ is the number of failures till the first of both successes, otherwise I could just consider $P(Y=y \mid N=x+y)$. The probability that there are $X=x$ failures till the frist success, given that there are in total $N=x+y$ failures is the same is having the first success between the $x^{th}$ and $(x+1)^{th}$ failure. As both $X$ and $Y$ have the same probability of success, the probability of this to happen is $\frac{1}{x+y+1}$. Hence,

\begin{align} P(X=x, Y=y, N=n) &= \frac{1}{x+y+1} (n+1)(1-p)^n p^r \\ &= (1-p)^n p^2. \end{align}

But this looks like nonsense.

Correspondence theorem for rings.

Posted: 14 Nov 2021 09:38 PM PST

Could someone provide a reference that includes a full and honest proof of the Correspondence Theorem for rings?

Let $A$ be a multiplicative ring with identity and $I$ an ideal of $A$. There is a one-to-one correspondence between the ideals of $A$ that contain $I$ and the ideals of the quotient ring $A/I$.
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