Tuesday, February 8, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

Trace inequality on positive definite matrices

Posted: 08 Feb 2022 12:08 PM PST

My question is the following:

Suppose that $A, B$ are symmetric positive definite real matrices. Is it true that $$ \mathrm{trace}\big((A+B)^{-1}\big)\geq \frac{1}{2} \min\Big\{\mathrm{trace}(A^{-1}), \mathrm{trace}(B^{-1})\Big\} $$

Some comments:

  • This is clearly true in the scalar case: $$ \frac{1}{a + b} \geq \frac{1}{2 \max\{a, b\}} = \frac{1}{2} \min\{a^{-1}, b^{-1}\}. $$

  • What I could show so far is that, using the inequality above, $$ \mathrm{trace}((A+B)^{-1}) \geq \frac{1}{2} \sum_{i=1}^d \min\{\lambda_i(A^{-1}), \lambda_i(B^{-1})\}, $$ Above, $d$ is the dimension of the matrices and $\lambda_i$ denotes the $i$th largest eigenvalue. This isn't quite strong enough, though.

Explanation of Frobenius endomorphism on elliptic curves

Posted: 08 Feb 2022 12:06 PM PST

I'm trying to understand how to calculate the Frobenius endomorphism for an elliptic curve.

Specifically, If $E$ is defined over $\mathbb{F}_q$, then the Frobenius endomorphism $\pi$ is defined as

$\pi : E \rightarrow E,\,\,\,\,\,\,\,\,\,$ $(x,y) \rightarrow (x^q,y^q)$

This definition doesn't make sense to me b/c the Frobenius endomorphism is defined as carrying an element, $a$ to $a^p$ where $p$ is the characteristic of the ring (and is prime). So it seems like the definition should be $(x, y) \rightarrow (x, y)^q$ = $(x, y) + (x, y) + ... (x, y)$, $q$ times since elliptic curve addition is between 2 points. Why are the $x$ and $y$ values calculated separately?

Identify the parameter $\alpha$ that forces a row exchange

Posted: 08 Feb 2022 12:05 PM PST

Problem. Consider the system:

$$2x + 6y + z = 1$$ $$3x + \alpha y + z = 6$$ $$0x + 1y - 1z = 3$$

What number $\alpha$ forces a row exchange, and what is the solution for that $\alpha$? What $\alpha$ makes the system singular?

I'm taking an introductory linear algebra course, and I'm stuck on the above problem. The instructor has avoided further clarification.

It isn't clear to me whether there exists a single $\alpha$ that forces a row change (when attempting to solve the system by putting it in RREF form, then performing regression) or, if so, how such a number may be determined.

$\alpha = 6$, for instance, might "force a row change":

$$\begin{bmatrix} 2 & 6 & 1 &|&1\\ 3 & 6 & 1 &|&6\\ 0&1&-1&|&3 \end{bmatrix} \to \text{subtract (1) from (2)} \to \begin{bmatrix} 2 & 6 & 1 &|&1\\ 1 & 0 & 0 &|&5\\ 0&1&-1&|&3 \end{bmatrix}$$

...we'd then perhaps want to swap rows (2) and (1). Progressing further, however, it doesn't seem that this system is solvable. Is this "the $\alpha$ that forces a row change"?

Similarly, it isn't clear to me how to identify the $\alpha$ that makes the system singular, outside of simply guessing.

What kind of probability generating function is this?

Posted: 08 Feb 2022 12:12 PM PST

What kind of probability generating function is this? $$\Pi (s,t) = \bigg( {\frac{q+p(1-s)\cdot e^{-vt}}{1-ps}} \bigg)^{\frac{\lambda}{v}}$$

Is the Hausdorff Dimension of any space-filling curve 2?

Posted: 08 Feb 2022 12:02 PM PST

I've been looking into space-filling curves lately for a project I am working on and was wondering if the Hausdorff dimension is always 2? Intuitively this makes sense since their image is the entire plane. Would there be a way to formally prove this? Do you have any references to papers which prove this? Thank you

Find the IDFT of a sequence with eight elements

Posted: 08 Feb 2022 11:59 AM PST

I'm trying to solve the following question:

Let $X(k)$ be a DFT of $x(n)$ series so: $$ X(k)=\begin{cases} 1 & k=2\\ 1 & k=6\\ 0 & 0,1,3,4,5,7 \end{cases} $$ Find $x(n)$.

What I did: $$ x(n)=\frac{1}{N}\sum_{k=0}^{N-1}X(k)e^{i2\pi k\frac{n}{N}}=\frac{1}{8}\left(e^{i2\pi2\frac{n}{8}}+e^{i2\pi6\frac{n}{8}}\right) $$ At now I don't see if that's it or if I can continue from here somehow. If it is, the question feels really "empty" (I just did two steps). Otherwise, how can I continue from here? Also, did I use the right formula? In some places I see that they multiply by $\frac{1}{N}$ and in other places they don't. What is the correct way?

because $f$ satisfies $f''+f^p \geq 3 \varepsilon f'$?

Posted: 08 Feb 2022 11:57 AM PST

I would like to understand how the author makes this estimate $f''+f^p \geq 3 \varepsilon f'$ for $s \leq 1/2$. Is there a way to tell if $\epsilon$ is small or large? Is there any formality here? It just seems that he compares the degree of polynomials without much formality. Here $p>1$.

enter image description here

If $f$ is in a normed linear space and $\psi_f\in X^{**}$ defined by $\psi_f(\varphi)=\varphi(f)$ then $\|\psi_f\|_{**} = \|f\|$.

Posted: 08 Feb 2022 12:10 PM PST

Let $(X,\|\cdot\|)$ be a normed linear space and define $\psi_f\in X^{**}$ by the equation $\psi_f(\varphi)=\varphi(f)$ for every $\varphi\in X^*$. Claim: $\|\psi_f\|_{**}=\|f\|$.

Proof: $\|\psi_f\|_{**}\le \|f\|$ is trivial from the definition.

For the reverse I tried thinking of a bounded linear functional $\varphi$ such that $|\psi_f(\varphi)|=|\varphi(f)|=\|f\|$. But for a general space, I can't think of many interesting bounded linear functionals that are always in the dual space. There's the zero functional, but that doesn't help here. I thought about the norm itself, since this is a functional, but it's not necessarily linear.

Is there a linear functional such that $\varphi(f)=\|f\|$ and we somehow linearly extend this to all other points in $X$? Obviously we could define $\varphi(af) = a\|f\|$ for every $a\in\Bbb R$. It would be nice to have an inner product in order to decide which vectors are orthogonal to $f$ and map them all to zero, and then use that to define $\varphi$ on a general vector. But if we only have a norm, I don't know of a way to obtain such an inner-product. I'm only familiar with going the other way, using the inner-product to say what the norm is.

How do I prove this measure of entropy is constant?

Posted: 08 Feb 2022 12:15 PM PST

This actually came about from a different question: "A random natural number $0 \leq n \leq 999$ is chosen and written down in base 10 by person A (including leading zeros, if $n<100$). Person B then writes down a three-digit number $m$ (also including any necessary leading zeroes) and sees how many, if any, of the digits match $n$. What is the associated entropy of the set of matches?"

We can calculate these probabilities fairly easily. There is a .001 chance that $m=n$. There is then a .009 chance that the first two digits match but the third does not, a .009 chance for the second two match but the first does not, and a .009 chance for the first and last to match but not the middle; in total, there is a .027 chance that two digits match. We can use a similar argument to see that there is a .243 probability that one digit matches, and a .729 probability that no digits match. The entropy function in its general form (when measured in bits) is equal to $H=\Sigma_i^np_i\log_2 (\frac{1}{p_i})$, where $p_i$ is a the probability that a certain outcome occurs. Thus, our entropy is

  • $\frac{1}{1000}\log_2{(1000)}+\frac{27}{1000}\log_2{(\frac{1000}{9})}+\frac{243}{1000}\log_2{(\frac{1000}{81})}+\frac{729}{1000}\log_2{(\frac{1000}{243})}\approx 1.407$

One quickly sees that this can be generalized for $10^b$ with the following formula

$$H(X_b)=\Sigma_{n=0}^b {b \choose n}\left(\frac{9^n}{10^b}\right)\log_2{\left(\frac{10^b}{9^n}\right)}$$

Here's where things get interesting: I was looking at this through Desmos, and I found the following relation.

$\frac{\log_2{(10^b)}}{H(X_b)} = 7.083068882...$

This seems to hold for natural $b$ - there is some small variance from this value when $b\in \{\mathbb{R^+}\}/\{\mathbb{N}\}$, but I am more astonished at this constant appearing out of the blue. I can't find anything on OEIS suggesting that this is a known constant. Where did it come from, and what is its exact form?

Use the isomorphism theorem to determine the group $GL_2(\mathbb{R})/SL_2(\mathbb{R})$.

Posted: 08 Feb 2022 12:13 PM PST

Use the isomorphism theorem to determine the group $GL_2(\mathbb{R})/SL_2(\mathbb{R})$. Here $GL_2(\mathbb{R})$ is the group of $2\times 2$ matrices with determinant not equal to $0$, and $SL_2(\mathbb{R})$ is the group of $2\times 2$ matrices with determinant $1$. In the first part of the problem, I proved that $SL_2(\mathbb{R})$ is a normal subgroup of $GL_2(\mathbb{R})$. Now it wants me to use the isomorphism theorem. I tried using $$|GL_2(\mathbb{R})/SL_2(\mathbb{R})|=|GL_2(\mathbb{R})|/|SL_2(\mathbb{R})|,$$

but since both groups have infinite order, I don't think I can use this here.

Need help understanding this moderately basic rings/ideals proof

Posted: 08 Feb 2022 12:02 PM PST

I was doing some practice abstract algebra questions off the internet since I have a quiz coming up soon. However, I am not very skilled at abstract algebra. In fact, I did very average in my group theory class, so I am struggling in my ring theory one. Can someone please help explain what is happening in this proof? I'm very sorry if it's extremely straightforward, I just think I need some time to get used to the way of thinking that's required to solve these questions.

Let $R_1$ and $R_2$ be commutative rings with identities and let $R = R_1 × R_2$. The question asks to show that every ideal $I$ of $R$ is of the form $I = I_1 × I_2$ with $I_1$ an ideal of $R_1$ and $I_2$ an ideal of $R_2$.

The proof/solution given goes like this:

Let $I$ be an ideal of $R$, and let $(a, b) ∈ I$. Then $(a, b)·(1, 0) = (a, 0)$, and $(a, b)·(0, 1) = (0, b)$, so $I = I_1 × I_2$, where $I_1$ is the set of $x ∈ R_1$ such that $(x, y) ∈ I$ for some $y$. Similar for $I_2$.Then it is easy to show that these sets are ideals.

I don't completely understand this though. Not even just a specific part, but I guess how $I = I_1 \times I_2$ is implied from what comes before it. In my head, I keep (kind of) feeling like this means that $a \in I_1$ and $b \in I_2$, but I don't get how/why? Or if I'm wrong, could someone correct that too? :/

Again, really sorry if this is too basic. I genuinely feel dumb in this class sitting with people I can't compete with, so I can't even ask them for help, and I couldn't think of anywhere else I could find an explanation :/

How do I prove that this ideal is not a prime ideal?

Posted: 08 Feb 2022 12:00 PM PST

I have the following problem:

Let $K$ be a field and $R=K[X,Y]/(XY)$. For $P\in K[X,Y]$ we denote $[P]$ its class in $R$. Show that the Ideal $(XY)$ is not a prime ideal.

My Idea was the following:

Let us assume $(XY)$ is a prime ideal, this means that $$PQ\in (XY)\Rightarrow Q\in (XY)~~~or~~~P\in (XY)$$ Now take $PQ=XY$, then $X\in (XY)$ but this means that $X=P(X,Y)\cdot XY$ where $P(X,Y)\in K[X,Y]$. Now in the solution they took $Y=0$ then clearly $X=0$. But then they said that this is a contradiction. But I don't see why $X=0$ is a contradiction.

Thanks for your help

Is this set open in the Euclidean topology on $\mathbb{R}^n$, and if so, how can it be represented as a union of open balls?

Posted: 08 Feb 2022 12:17 PM PST

I am working on an exercise in a real analysis book and have come to a point where I'm not sure what to do. In my proof, I have defined sets in $\mathbb{R}^n$ (where $n\in\mathbb{N}$) that have the following form: \begin{equation} E(x,\hat{r}) = \prod_{i=1}^{n} \hat{B}(x_{i},\hat{r}_{i}) \end{equation} Here $\hat{B}(x_{i},\hat{r}_{i})$ is an open ball (i.e. open interval) in $\mathbb{R}$ with center $x_{i}$ and radius $\hat{r}_{i}$. Here $\hat{r}_{i}$ can equal $\infty$ so that $\hat{B}(x_{i},\hat{r}_{i})$ can be equal to $\mathbb{R}$.

Is it true that every such $E(x,\hat{r})$ is an open set in $\mathbb{R}^{n}$ ? If so, how do you write a given $E(x,\hat{r})$ as the union of a set of open balls in $\mathbb{R}^{n}$ ? Since the set of open balls in $\mathbb{R}^{n}$ is a basis for the Eucldean topology, if $E(x,\hat{r})$ is open shouldn't there be a way to do it ?

Equivalence between sections of the pullback bundle and lifts in the corresponding commutative diagram

Posted: 08 Feb 2022 12:06 PM PST

Let $\pi: P \to B$ denote a principal $G$-bundle over base $B$, and let $f: B' \to B$ be a continuous map from another space $B'$ to $B$.

I've been reading Stephen Mitchell's Notes on Principal Bundles and Classifying Spaces. On page 3, he says that sections of the pullback bundle $f^{*}P \to B'$ are in bijective correspondence with lifts in the diagram

enter image description here

Now, I understand that this fits into a bigger commutative diagram, namely

enter image description here

where $\text{pr}_2$ is the projection onto the second factor (thinking, as usual, of the pullback bundle as $f^{*}P = \{ (b', p) \in B' \times P ~|~ f(b') = \pi(p) \}$).

This suggests that given a section $s' : B' \to f^{*}P$, one can construct the lift by defining $\widetilde{f} = \text{pr}_2 \circ s'$.

Question: What about the converse? That is, given a lift $\widetilde{f}: B' \to P$ that fits into the above commutative diagram, how does one construct a (local) section of $\pi' : f^{*}P \to B'$?

In the textbook, ``Lecture Notes on Algebraic Topology,'' by Davis and Kirk [cf. pg. 168], it is stated that the cross-section problem for pullback bundles is equivalent to the so-called relative lifting problem. On the one hand, the existence of a commutative square is equivalent to the object in the top left corner being isomorphic to the pullback bundle (this is also mentioned in Mitchell's notes referred to above). However, without resorting to a local trivialization, is there a way to understand the claimed bijective correspondence?

$B = \{\{x\}|x\in X\}$ Prove B is a basis for a topology on $X$

Posted: 08 Feb 2022 12:08 PM PST

Say $X$ is a set and we know $\mathcal{B} = \{\{x\}|x\in X\}$.

I'm trying to show that:

(a) $\mathcal{B}$ is a basis for a topology on $X$

(b) that $\mathcal{B}$ generates the discrete topology.

The criteria for a basis are:

For each $x\in X$ there is some $B\in \mathcal{B}$ such that $x\in B$

For any $B_1$,$B_2 \in \mathcal{B}$ and any $x \in B_1 \cap B_2$, there is some $B \in \mathcal{B}$ such that $x \in B⊆B_1\cap B_2$.

For part (a), I'm honestly somewhat lost. $\mathcal{B}$ is the set of points, so there must be subsets that have fewer points than $\mathcal{B}$, but how do I know if the second criterion for a basis applies without knowing the size of a subset?

For part (b), I know that the discrete topology is the largest topology containing all subsets as open sets. Assuming that I know part (a) has given me the basis for a topology:

We have a collection of all one-point subsets on $X$, which is the basis for the discrete topology by definition.

Proving if $X$ is discrete then every set of one element is open is easy. If $X$ is discrete, we know this by definition.

I'm less sure how to prove $X$ is discrete because all of the one-point subsets are open. I would guess that you would take some arbitrary set in $X$, something like $U$, and say that $U = \bigcup_{u \in U}\{u\}$. Then you know it's a union of open sets. Because it was arbitrary, $X$ only contains open sets and must be discrete.

Group action and orbit question

Posted: 08 Feb 2022 11:58 AM PST

I have the following group action question that I would like some advice on how should i proceed.

Let $X$ be the group $\mathbb{Z}/n\mathbb{Z}$ and $G$ be the group $(\mathbb{Z}/n\mathbb{Z})^*$. Let $G$ act on $X$ via

$$g\cdot x=gx$$

Given $m\in X$, show that the orbit containing $m$ is the same as the orbit containing $d$ where $(m,n)=d$.

I have tried using the following to solve it to no avail.

  1. Bezout's identity: There exists $a,b\in \mathbb{Z}$ such that $am+bn=d$
  2. Let $a,b>0$ be integers and $d = (a, b)$. Suppose $d|N$. Then the linear congruence

$$ax \equiv N (\text{mod }b)$$

$\space\space\space\space\space\space\space$ has $d$ mutually incongruent solutions modulo $b$.

Does anybody have any advice on how I should solve this?

If x and y are real numbers such that $4x^2+2xy+9y^2=100$ then what are all possible values of $x^2+2xy+3y^2$

Posted: 08 Feb 2022 12:02 PM PST

If $x$ and $y$ are real numbers and $4x^2+2xy+9y^2=100$ then what are all possible values of $x^2+2xy+3y^2$.

The first thing I did with this equation was to try and factor the first equation, but that didn't work so I tried to simplify it into a standard elliptical equation

$$\left(\frac{x}{5}\right)^2+\frac{xy}{50}+\left(\frac{3y}{10}\right)^2=1$$

Then I noticed that the numerator of the $x$ coefficient and the coefficient lined up but i couldn't get the denominators to cancel out.

Now I'm stuck and don't know how I should progress. Any help would be appreciated.

How to bridge the gap between finite and infinite index sets in the proof of sigma subadditivity?

Posted: 08 Feb 2022 12:05 PM PST

From a standard probability theory course, there is an exercise to prove that $$P\bigg(\bigcup_{k=1}^\infty A_k\bigg) \leq \sum_{k=1}^\infty P(A_k)$$

for some probability measure $P$, and countable (not necessarily disjoint) family of events $(A_k)$.

I assumed it should be relatively simple: When I first tried proving this I had something along the lines of $$P\bigg(\bigcup_{k=1}^N A_k\bigg) = \sum_{k=1}^N P(A_k) - \sum_{k=1}^{N-1} P\bigg(\bigcup_{k=1}^{N-j} A_k \bigcap A_{N-j+1}\bigg)$$

where the formula above is true for any $N$ (if I have written it correctly, but the crux of the question is less about the exact formula), which can be proved by induction.

Back then I assumed I could simply go from the above to concluding sigma subadditivity. But revisiting this I realise this is wrong; the claim I proved via induction is for every finite index set, but it is NOT for a single countably infinite set.

Is there any way to salvage this approach?

What's wrong with this tensor product approach to diagonalize a 4x4 matrix?

Posted: 08 Feb 2022 12:05 PM PST

I have a matrix that looks like: \begin{align} H &= \begin{bmatrix} 0 & 2\sin(2K_y) & 2\sin(2K_x) & 0\\ 2\sin(2K_y) & 0 & 0 & \gamma\\ 2\sin(2K_x) & 0 & 0 & 2\sin{2K_y}\\ 0 & \gamma & 2\sin{2K_y} & 0\\ \end{bmatrix} \end{align}

where $\gamma = 2\sin(2K_x + 2\pi/N)$, and $N = 2,3$. I write this in the tensor product form: \begin{align} H = \sigma^x \otimes \Bigg[(\frac{I_{2x2}+\sigma^z}{2})2\sin (2Kx) + (\frac{I_{2x2}-\sigma^z}{2}) \gamma\Bigg] + 2\sin(2K_y) I_{2x2} \otimes \sigma^x \end{align} Let's say our solution looks like $\psi_x \otimes \psi_y$. And then suppose that we have a sub equation $\sigma^x \psi_x = l \psi_x$, then we can solve the eigenvalues of H by subbing this equation in and getting: \begin{align} \Bigg[l \Big[(\frac{I_{2x2}+\sigma^z}{2})2\sin (2Kx) + (\frac{I_{2x2}-\sigma^z}{2}) \gamma\Big] + 2\sin(2K_y) \sigma^x \Bigg] \psi_y = \lambda \psi_y \end{align}

Of course, we have to make sure that the $\psi_x$ eigenequation has a solution, and it does for $l=\pm 1$.

We then solve the vector equation for $\psi_y$ to get the following eigenvalues: \begin{align} \lambda_{\pm} = \frac{l (2x + \gamma)}{2} \pm \frac{\sqrt{(2x + \gamma)^2 + 16y^2 - 8\gamma x}}{2} \end{align} Where I've used the short form $x = \sin (2K_x), y = \sin(2K_y)$.

Which gives me the following graph for N=2 (makes sense, as $\gamma \rightarrow -2\sin(2K_x)$ for N=2, thus the solution becomes $\lambda = \pm 2\sqrt{\sin^2 x + \sin^2 y}$): N=2 And the following graph for N=3: N=3 To check this, we can diagonalize the 4x4 matrix directly, without reducing it down to the tensor product form. It agrees with the tensor solution for N=2, but gives me a completely different graph for N=3! (It doesn't seem very apparent in this figure but even here we can see that they have different answers at the edges. And this has a depression that the tensor solution does not) N=3, straight diagonalization

I trust the straight diagonalization more than my tensor solution, but I can't place any obvious mistakes in my calculations. Is my tensor decomposition of H at fault? Is the agreement of the 2 graphs for the case of N=2 just a fluke?

How to show the basic formula of the Mean Residual Life Function in survival analysis?

Posted: 08 Feb 2022 12:09 PM PST

The basic quantity employed to describe time-to-event phenomena is the survival function, the probability of an individual surviving beyond time x (experiencing the event after time x). If $T$ is time of death, survival at time $x$ is is defined as $S(x) = Pr (T > x)$.

Also, the survival function is the integral of the probability density function, $f(x)$, that is,

$$ S(x)=Pr[T>x]=\int^\infty_x f(t)\,dt$$

The mean residual life function is defined as $$\operatorname{mrl}(x):=E[T-x|T>x]$$

Can anyone tell me how to get the formula below? $$ \operatorname{mrl}(x) =\frac{\int^\infty_x (t-x)f(t)\,dt}{S(x)}=\frac{\int^\infty_x S(t)\,dt}{S(x)} $$

Combined mean of gamma variates [closed]

Posted: 08 Feb 2022 12:06 PM PST

Would someone please help on the following.

Suppose $X_i$ follows Gamma(a,b),where a and b are shape and scale parameters of gamma distribution, respectively. I know the distribution of $\bar{X}$ is also gamma with parameters na and nb. What are the distributions of the following?

  1. $(n_1\bar{X_1}+n_2\bar{X_2})/(n_1+n_2)$

  2. $(n_1\bar{X_1}+n_2\bar{X_2}+n_3\bar{X_3})/(n_1+n_2+n_3)$

Would someone please help to answer with some mathematics? Thank you

Integral extension of a DVR

Posted: 08 Feb 2022 12:00 PM PST

Let $A$ be a discrete valuation ring, with maximal ideal $\mathfrak m=(\pi)$ for some $\pi\in A$; show that $B:=A[\sqrt{\pi }]$ is a discrete valuation ring. I know that this question has already been raised on M.SE., but the argument that I used is different.

The fact that $B$ is isomorphic to $A[X]/(X^2-\pi)$ shows at once that it is a Noetherian domain. Moreover $A\subset B$ is an integral extension and $\operatorname {dim}A=1$, so also $\operatorname {dim}B=1$ (basically by the going-up theorem).

Hence is sufficient to prove that the ideal $(\sqrt\pi)\subset B$ is the only maximal one. Observe that $A\cap(\sqrt\pi)=\mathfrak m$: indeed $\pi\in A\cap(\sqrt\pi)$, so $\mathfrak m\subseteq A\cap(\sqrt\pi)$, and the maximality of $\mathfrak m$ implies the equality. Thus the following isomorphisms show that $(\sqrt\pi)$ is a maximal ideal: $$\frac{A[\sqrt\pi]}{(\sqrt\pi)}\cong \frac {A}{A\cap(\sqrt\pi)}\cong \frac A{\mathfrak {m}}.$$ Plus $\mathfrak M\cap A$ must be a maximal ideal for any maximal ideal $\mathfrak M\subset B$, as $A\subset B$ is integral, so $\mathfrak M\cap A=\mathfrak m$; since $\pi\in \mathfrak M$ implies that $\sqrt\pi\in\mathfrak M$, we should be done.

Probability of a sum of two die rolls being 5 and not 7 in a game

Posted: 08 Feb 2022 12:19 PM PST

I've been working on this problem for a while and I'm not sure how to approach it. This was given in a class on Stochastic Processes, and it's meant to be solved using a Markov chain:

A single die is rolled repeatedly. The game stops the first time that the sum of two successive rolls is either 5 or 7. What is the probability that the game stops at a sum of 5?

What's the correct way to tackle this problem?

Hyperelliptic integral involving square root of sextic polynomial

Posted: 08 Feb 2022 12:20 PM PST

I am interesting in finding the integral of $$ I = \int \frac{{\rm d}x}{-\sqrt{x^6+x^2-c}}$$ Where $c \leq x^6 + x^2$ such that the integral is always real. I was trying to look at a reduction of 579.00 from the Handbook of Elliptic integrals for scientists and engineers by Byrd and Friedman, i.e., $$ \int\frac{{\rm d}x}{a_0x^6 + a_1x^5 + a_2+x^4 + a_3x^3 + a_2x^2 + a_1x + a_0}, $$ But in my case $c \neq -1$ in general. I know Wolfram can solve this equation, but I am interested in learning how to apply the reduction and subsequent integration. Moreover, I am unsure on how to check that the roots of the polynomial form three pairs of points of an involution.

Finite differences and Neumann/Periodic BC of a fourth order PDE

Posted: 08 Feb 2022 12:15 PM PST

Before I start, I have already read this similar thread and didn't get anywhere unfortunately.

I have a PDE of the form

$$ \frac{\partial u}{\partial t} = \nabla^2[2u(u-1)(2u-1) - \gamma \nabla^2 u] $$ where $u = u(x,y,t)$.

The fourth derivative in a central difference scheme is written as: $$ \left(\frac{\partial^4 u}{\partial x^4}\right)_{ij} \approx \frac{1}{\Delta x^4} (u_{i+2, j} - 4 u_{i+1,j} + 6u_{i,j} -4u_{i-j,j} + u_{i-2, j}) $$

and similarly for the $y$ derivative. Thus, at the boundary point $i=N$, I will need ghost points $i=N+1$ and $i=N+2$. I know how to deal with $N+1$ via the Neumann boundary condition I am interested in

$$ \frac{\partial u}{\partial x} = 0 \implies u(x_i = u_{N-1} $$ in the central difference scheme. But I don't know how to deal with the second ghost point.

For periodic BC, I assume for i = N, i+1 wraps around to 1 and i+2 wraps around to 2, and similarly for the other boundaries. But do the coefficients change or do they stay the same? How do I write this in matrix form?

Prove or disprove a partial refinement of : $x^a+a^x>1$ for $a,x>0$

Posted: 08 Feb 2022 12:11 PM PST

Problem:

Let $a,x>0$ then prove or disprove that : $$a^{x}+\left(\frac{1}{a^{x}}\right)^{\left(1-e^{-1}\right)}+x^{a}+\left(\frac{1}{x^{a}}\right)^{\left(1-e^{-1}\right)}> 3-e^{-1}+\sqrt{\frac{1}{1-e^{-1}}}$$

My attempt

From here ($x^y+y^x>1$ for all $(x, y)\in \mathbb{R_+^2}$) we have $a^x+x^a>1$.

But it's not enough to show the proposed problem.

As second attempt I have tried AM-GM (for two variables) which is also insufficient.

Edit 01/02/2022:

Case $a\geq 1$ :

It's not hard to check that :

$$f(x)=a^{x}+x^{a}-\left(3-e^{-1}+\sqrt{\frac{1}{1-e^{-1}}}\right)$$

Is convex $(0,\infty)$

On the other hand the function $g(x)$ :

$$g(x)=-\left(\left(\frac{1}{x^{a}}\right)^{\left(1-e^{-1}\right)}+\left(\frac{1}{a^{x}}\right)^{\left(1-e^{-1}\right)}\right)$$

Is concave negative on the same interval. We deduce that $h(x)=f(x)-g(x)$ is convex so we have the inequality :

$$h(x)\geq h'(1)(x-1)+h(1)$$

I conjecture that $x_r$ wich check $h'(1)(x_r-1)+h(1)=0$ is less that $x_{min}$ wich check $h'(x_{min})=0$.If true it solves the problem in this case .

Remains to show the hardest case I mean $0<a\leq 1$ .

Edit 02/02/2022 :

As $h(x)$ is convex on $(0,\infty)$ the first derivative is increasing one can show that $h'(1)>0$ so the function $h(x)$ is increasing remains to show the positiveness of $h(1)$ .Now we can conclude in the case $a,x\geq 1$

How to (dis)prove it ?

Thanks !

Truncation Error in Mixed Derivative

Posted: 08 Feb 2022 12:07 PM PST

In Pattern Recognition and Machine Learning Ch 5.4.4 Finite Differences equation 5.90 gives a finite difference approximation to a mixed partial derivative:

$$\frac{\partial^2E}{\partial w_{ji}\partial w_{lk}}=\frac{1}{4\epsilon^2}[E(w_{ji}+\epsilon,w_{lk}+\epsilon)-E(w_{ji}+\epsilon,w_{lk}-\epsilon)-E(w_{ji}-\epsilon,w_{lk}+\epsilon)+E(w_{ji}-\epsilon,w_{lk}-\epsilon))] + O(\epsilon^2)$$

I tried to derive this myself and it seems like to derive it in the general case you need to treat the perturbations $\epsilon_1$ and $\epsilon_2$ of $w_{ji}$ and $w_{lk}$ independently. Then you use the second order Taylor approximations and in each expansion you have a remainder term $O(\epsilon_1^3,\epsilon_2^3)$ since they are second order approximations. In order to solve for the mixed partial derivative above I believe based on my own calculations and this answer here that you need to divide the remainder term $O(\epsilon_1^3,\epsilon_2^3)$ by $\epsilon_1\epsilon_2$ and the result according to Pattern Recognition and Machine Learning should be $O(\epsilon^2)$ where $\epsilon_1=\epsilon_2=\epsilon$. I'm somewhat familiar with big O notation but don't know much about truncation error and I'm wondering how you justify

$$O(\epsilon_1^3,\epsilon_2^3)/(\epsilon_1\epsilon_2)=O(\epsilon^2)$$

when $\epsilon_1=\epsilon_2=\epsilon$. It seems like dividing by the product $\epsilon_1\epsilon_2$ you first divide the $\epsilon_1^3$ by the $\epsilon_1$ and you get $O(\epsilon_1^2,\epsilon_2^3)$ and then you divide by the $\epsilon_2$ and you get $O(\epsilon_1^2,\epsilon_2^2)$ which if $\epsilon_1=\epsilon_2=\epsilon$ equals $O(\epsilon^2)$.

Is my intuition about this remainder correct and if so what justifies the fact that you can split the division component wise?

EDIT: It looks like the text here does not use an $O(\epsilon_1^3,\epsilon_2^3)$ term but instead uses two terms which are $O(\epsilon_1^2)$ and $O(\epsilon_2^2)$ but I haven't had the time to read this text in detail yet, I suspect the method of derivation is different to mine.

Order of growth rate in increasing order

Posted: 08 Feb 2022 12:01 PM PST

This question is related to maths, so I post here. Actually it's a computer science question and I am facing this type of question while learning Design and Analysis of Algorithms, but we all know that computer science has complete relation with maths.

Arrange the following functions in increasing order of growth rate (with $g(n)$ following $f(n)$ in your list if and only if $f(n)=O(g(n)))$.

a) $2^{log(n)}$
b) $2^{2log(n)}$
c) $n^5/2$
d) $2^{n^2}$
e) $n^2 log(n)$

So i think the answer, in increasing order, is CEDAB.

Is it correct? I have confusion in option A and B. I think option A should be first... the one with the lower growth rate I mean. Please help me solve this. I faced this question in algorithm course part 1 assignment (Coursera).

What does the semicolon ; mean in a function definition

Posted: 08 Feb 2022 12:20 PM PST

Cauchy's Hypothesis or Noll's theorem states that $\vec{t}(\vec{X},t;\partial \Omega) = \vec{t}(\vec{X},t;\vec{N})$ where $\vec{N}$ is the outward unity normal to the positively oriented surface $\partial \Omega$. This translates to words as the dependence of the surface interaction vector on the surface on which it acts is only through the normal $\vec{N}$. My question is what is the significance of the semicolon (;)? How does it differ from the comma (,) used to separated the function's first two arguments?
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