text
stringlengths 83
79.5k
|
|---|
H: Is $\mathbb{Q} [X,Y]/[x^{20},y^{20}]$ a local ring?
Is $\mathbb{Q}[X,Y]/[x^{20},y^{20}]$ is a local ring?
My approach is to look for a maximal ideal, but got stuck how to find a single maximal ideal. Any feedback would be appreciated.
AI: Well, $\Bbb Q[X,Y]/\langle X^{20},Y^{20}\rangle$ is a local ring.
The ideal $\langle X,Y\rangle$ is maximal in $\Bbb Q[X,Y]$, and as
$\langle X,Y\rangle^{40}\subseteq\langle X^{20},Y^{20}\rangle
\subseteq\langle X,Y\rangle$ the radical of $\langle X^{20},Y^{20}\rangle$
is $\langle X,Y\rangle$. This means that the image of $\langle X,Y\rangle$
in $\Bbb Q[X,Y]/\langle X^{20},Y^{20}\rangle$ is the unique maximal ideal.
|
H: Series $\sum_0^{\infty} a_n$ and $\sum_0^{\infty} b_n$ are convergent series then so is $\sum_0^{\infty} c_k$ where $c_k= \sum_{l=0}^{k}a_lb_{k-l}$
Series $\sum\limits_0^{\infty} a_n$ and $\sum\limits_0^{\infty} b_n$ are convergent series then so is $\sum\limits_0^{\infty} c_k$ where $c_k= \sum\limits_{l=0}^{k}a_lb_{k-l}$
Solution I tried: I tried to prove this using a counterexample like $$a_n\frac{(-1)^n}{(n+1)^{\frac{1}{3}}}$$ and $$b_n=\frac{(-1)^n}{(n+2)^{\frac{2}{3}}}$$ then after that we get $$c_k=\frac{(-1)^{2n}}{(n+1)}$$ which is divergent, but I contrardict this with an example, is there any way we can prove this divergent or convergent with proper definition?
Please help
Thank you
AI: Consider
$$a_n = b_n = \frac{(-1)^n}{\sqrt{n+1}}.$$
Then
$$c_n = (-1)^n \sum_{k=0}^n \frac{1}{\sqrt{(k+1)(n-k-1)}},$$
and $k+1, n-k-1$ are both less than $n+1$, so we have
$$|c_n| \geq \sum_{k=0}^n \frac{1}{n+1} = 1,$$
and so $c_n$ does not go to $0$. Hence $\sum c_n$ diverges.
|
H: best approximation of $\sqrt[3]{8.1}$ using cubic function
Approximate value of $\sqrt[3]{8.1}$ using linear approximation of $g(x)=(x+1)^{\frac{1}{3}}$
What i try:
Linear approximation of $f(x)$ at $=a$ is given by
$$L(x)=g(a)+g'(a)(x-a)$$
$$L(x)=\sqrt[3]{a+1}+\sqrt[3]{(a+1)^{-2}}(x-a)$$
I did not understand what should i take $a$ so that i can approximate $\sqrt[3]{8.1}.$
Help me please
AI: Take $a=0$. If $f(x) = \sqrt[3]{1+x}$, then $f'(0) =1/3$ so that $f(x) = 1 + \frac{1}{3}x + \mathcal{O}(x^2)$ as $x\to 0$. The best linear approxiamtion to $\sqrt[3]{8.1} = 2\sqrt[3]{1+0.1/8}$ is then
$$2+\frac{2}{3}\cdot0.0125=2.0083...$$
|
H: Probability problem socks
A drawer has 7 different pairs of socks. Each sock is taken randomly and without replacement until the first pair is found. What is the probability of having the first pair until the 8th extraction?
My answer is $\frac{2^7}{\binom{14}{7}}=0.037$, is my answer right?
AI: The probability is
$$\frac{14}{14} \times \frac{12}{13} \times \frac {10}{12} \cdots \times \frac{2}{8},$$
where there are seven fractions, each of which refers to the probability of picking a sock from an un-picked pair at each step.
This does indeed rearrange to $\frac{2^7}{\binom{14}{7}}$.
|
H: Is it true that for any natural number n, that nth order Taylor polynomial of any polynomial at $\vec0$ is just the polynomial itself?
Is it true that for any natural number $n$, that $n$-th order Taylor polynomial of any polynomial at $\vec0$ is just itself ? (If $n\ge$degree of the polynomial)
For example if $f(x_1,\dots,x_m)$ is a polynomial, then $n$-th order Taylor polynomial of $f$ at $\vec 0$ is just $f$ itself ? i.e. $f(x_1,\dots,x_m)=4x_1^2x_2+5x_3^3x_5+7x_1^3+2x_4,n\ge4$ etc.
AI: JUst repeating the comments: Yes, it's true, but as I recall, the proof is pretty tedious. In the one-dimensional case, I know that Spivak's Calculus has a proof (somewhere in Chapter 20, on Taylor polynomials). In the multivariable case, I know that Henri Cartan's Differential Calculus text (Chapter 7) contains a proof. Anyway, just try it out in the 2-variable case by writing out the taylor polynomial for small n. You should see how to generalize.
|
H: Number of $h$ hop paths between two vertices with shortest path $s$ on an $n$ sided polygon.
We have a regular polygon with $n$ vertices. We start at one vertex and want to get to another. The least number of hops we need to get from the starting to target vertex is $s$. How many ways are there to go from the starting to the target vertex in exactly $h$ hops. This becomes a function $f(n,s,h)$ ($s \in [0,\left[\frac{n}{2}\right]]$). Is there a closed-form or recurrence for this function?
Example: Let $n=3$ (triangle), $s=1$. We get:
$$f(3,1,h)=f(3,0,h-1)+f(3,1,h-1)$$
$$=f(3,1,h-1)+2 f(3,1,h-2)$$.
This recurrence yields the Jacobsthal numbers.
AI: Let us define $f(s, h)$ as the number of walks to get to the target vertex in $h$ hops, provided we're currently $s$ hops away (oriented, so $s$ can be negative), but instead of a polygon we have a line (this is analogous to $n = \infty$). Then, it is not hard to see that
$$
f(n, s, h) = \sum_{k=-\infty}^{\infty} f(s + kn, h).
$$
(Only finitely many of the summands will be nonzero, assuming $n > 0$.) This is just accounting for walks which go around the polygon a certain number of times. Now we will focus our attention towards finding an explicit formula for $f(s, h)$. We have the recurrence
$$
f(s, h) =
\begin{cases}
1 & h = 0, s = 0 \\
0 & h = 0, s \neq 0 \\
f(s+1, h-1) + f(s-1, h-1) & h > 0.
\end{cases}
$$
This recurrence has solution
$$
f(s, h) = \begin{cases}
\binom{h}{(h+s)/2} & h = s \bmod{2} \\
0 & h \neq s \bmod{2},
\end{cases}
$$
where $\binom{n}{r}$ is defined to be $0$ for $r < 0$ or $r > n$.
Combining this with the summation from before, we have a couple cases for $n$ even or odd and $h - s$ even or odd. In all of these cases, we want to compute a summation of the form
$$
\sum_{k=-\infty}^{\infty} \binom{h}{x+ky},
$$
where $x$ is either $\frac{h+s}{2}$ or $\frac{h+s+n}{2}$ and $y$ is either $\frac{n}{2}$ or $n$ depending on the parities mentioned above. We can compute this summation using any of a number of strategies, such as a root of unity filter; in particular, if we let $\omega$ be a primitive $y$-th root of unity, then the solution can be expressed in the form
$$
\sum_{j=0}^{y-1} \omega^{-jx} (1 + \omega^j)^h.
$$
This can't be reduced further easily, but it's closer to a closed form than nothing.
|
H: Unable to explain flow of steps in this basic modular expression?
Consider the expresion $(13 + 11)· 18 (\mod 7)$:
$(13+11)· 18 ≡ (6+4)· 4 (\mod 7)$ Note the transition from $(13+11)·
> 18$$\implies$ $(6+4)· 4$
$≡ 10 · 4 (\mod 7)$
$≡ 3 · 4 (\mod 7)$ Note the transition from $10 · 4$$\implies$ $3 · 4$
$≡ 12 (\mod 7)$
$≡ 5 (\mod 7)$
$≡ 5$
These 2 transitions involve subtracting 7, but in each case they were either the factor( the $10$ going to $7$ in the 2nd transition) or a component of a factor( the $13$ and $11$ going to $6$ and $4$ in first transition).
I would have understood if they subtracted 7 from the product itself( like the $12$ going to $7$ in the final step) because I can intutively understand that the equivalence that both sides have the same remainder when divided by 7 still holds.
I didn't get how this was possible( there wasn't any law/theorem stating you could do that). Few pages down , I saw this corollary:
$ab ≡ [(a \mod n)(b \mod n)](\mod n)$
Is the transitions some result of the corollary or is there some knowledge I'm lacking entirely to explain those transitions?
AI: The corollary is indeed the key! In case we have:
$$
a\pmod n=x\\
b\pmod n=y
$$
we can write
$$
a=pn+x\\
b=qn+y
$$
and therefore
$$
a\cdot b=pq\cdot n+(py+qx)n+x\cdot y
$$
showing that
$$
a\cdot b\equiv x\cdot y\pmod n
$$
so this establishes the result, which could also be formulated as
When you do modular arithmetics on products, you can reduce each factor first.
To clarify why partial reductions will also work, let us prove a slightly different result that
$$
a\cdot b\equiv(a+kn)\cdot b\pmod n
$$
This can be seen from
$$
(a+kn)\cdot b=a\cdot b+bk\cdot n
$$
So when $a$ is changed by $kn$ the product is changed by $bk\cdot n$
Changing one factor of a product by a multiple of $n$ changes the product by a multiple of $n$
|
H: find $\sum_{n=0}^{\infty}n^2 x^n$
I'm trying to find $\sum_{n=0}^{\infty}n^2 x^n$
What I tried doing is:
$\sum_{n=0}^{\infty}n^2 x^n$ = $\sum_{n=0}^{\infty}n(n-1) x^n + \sum_{n=0}^{\infty}n x^n$
= $x^2\sum_{n=0}^{\infty}n(n-1) x^{n-2}+x \sum_{n=0}^{\infty}n x^{n-1} =x^2(\frac{1}{1-x})'' + x(\frac{1}{1-x})'$
But anyway I continute gets me a wrong answer so I wonder if there's some mistake here?
The answer should be $ -\frac{x(x+1)}{(x-1)^3} $
AI: Your solution is mostly correct. Your lower bounds for the summation are wrong. But that is irrelevant when it comes to your argument simplification which you must not have done correctly because you will get the right answer if you continue down the path you are currently going on.
|
H: Calculate $\iiiint_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}\,dx\,dy\,du\,dv$
$$\iiiint_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}\,dx\,dy\,du\,dv$$
So we just learned substitution and i thought maybe for this integral doing 2 polar subs for x,y and for u,v but i'm not sure this is the right way for this.
Any hints will be welcome
AI: Nothing wrong with what you said, we'll just have to do an additional substitution afterwards. Let
$$\begin{cases} x = r\cos\theta \\ y = r\sin\theta \\ u = s\cos \phi \\ v = s\sin\phi \\ \end{cases} \implies J = rs$$
which gives us the integral
$$\int_{r^2+s^2\leq 1 \:\cap\:(\theta,\phi)\in[0,2\pi]^2} rse^{r^2-s^2}\:dr\:ds\:d\theta\:d\phi = 4\pi^2\int_{r^2+s^2\leq 1} rs e^{r^2-s^2}\:dr\:ds$$
Now let
$$\begin{cases}r = \rho \cos\gamma \\ s = \rho \sin\gamma \\ \end{cases} \implies J = \rho$$
giving us the integral
$$4\pi^2 \int_0^1 \int_0^{\frac{\pi}{2}} \rho^3\sin\gamma\cos\gamma \: e^{\rho^2(\cos^2\gamma - \sin^2\gamma)}\:d\gamma\:d\rho = \pi^2 \int_0^1 \int_0^{\frac{\pi}{2}} 2\rho^3\sin2\gamma \:e^{\rho^2\cos2\gamma} \:d\gamma\:d\rho$$
since $r,s>0$ by definition. Integrating gives us
$$\pi^2 \int_0^1-\rho \:e^{\rho^2\cos2\gamma}\Bigr|_0^{\frac{\pi}{2}}\:d\rho = \pi^2 \int_0^1 2\rho\:\sinh(\rho^2) \:d\rho = \pi^2(\cosh 1 - 1)$$
|
H: how many ways to arrange $3$ people to a circular table which has $6$ seats?
As the title, how can I solve this problem?
Should I use combination with repitition or something else different?
Any help would be appreciated
3 people are distinguish. Below here is an example of symmetry situation which only consider as 1 way
but this situation is 2 ways
AI: Due to the rotation invariance, person $p_1$ can sit anywhere. Then $p_2$ can be one seat away from $p_1$, or two or three. Due to symmetry invariance, we can suppose that it is to the right of $p_1$. For each of these 3 options, person $p_3$ has 4 possible seats, but in the case where $p_2$ is 3 seats away from $p_1$, the positions of $p_3$ come in symmetric pairs. Hence there are $2\times 4 + 2 = 10$ ways to arrange the 3 people.
|
H: Find $\int(\frac{\sin{x}+\cos{x}}{x^2}+\frac{\cos{x}-\sin{x}}{x})dx$
Problem: Find
$$\int\left(\frac{\sin{x}+\cos{x}}{x^2}+\frac{\cos{x}-\sin{x}}{x}\right)dx$$
The possible methods I thought were:
Substitute $t=\tan{\left({x\over2}\right)}$: failed because of polynomials in the denominators
Put $t={\pi\over2}-x$: failed because of polynomials in the denominators
Substitute $t=\frac{\cos{x}-\sin{x}}{x}$: failed - it is difficult to express $\frac{\sin{x}+\cos{x}}{x^2}$ in terms of $t$.
Partial integration: but I couldn't find the way to split into $f'$ and $g$.
Using $\sin{x}+\cos{x}=\sqrt{2}\sin\left({x+{\pi\over4}}\right)$ and $\cos{x}-\sin{x}=\sqrt{2}\sin\left({x-{\pi\over4}}\right)$, but failed since I also don't know how to integral $\int\frac{\sin\left({x+{\pi\over4}}\right)}{x^2}$ and $\int\frac{\sin\left({x-{\pi\over4}}\right)}{x}$.
I wanted to focus on the fact that
$$\cos{x}-\sin{x}=\left(\sin{x}+\cos{x}\right)'$$
but it also did not work well.
Is there a special idea to solve these types of integrals - when the fraction includes polynomials in the denominator and trigonometry functions in the numerator? Thanks.
AI: Unfortunately, this integral requires nonelementary functions, such as $\text{Si}(x)$; there is no elementary way to describe it.
In case you accidentally missed a minus sign when transcribing your integral, observe that
$$\frac{d}{dx} \frac{\sin x + \cos x}{x} = -\frac{\sin x + \cos x}{x^2} + \frac{\cos x - \sin x}{x}.$$
|
H: Why $\lim\limits_{n\to\infty}\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=0$?
Consider the following theorem regarding convergence in distribution
THEOREM 1: Let $(X_n)_{n\geq1}$, $X$ be $\mathbb{R}^d$-valued random variables. Then $X_n$ converges to $X$ in distribution if and only if $\lim\limits_{n\to\infty}E\{f(X_n)\}=E\{f(X)\}$ for all continuous, bounded functions $f$ on $\mathbb{R}^d$.
Then consider the following theorem and its proof
THEOREM 2 Let $(X_n)_{n\geq1}$, $X$ be defined on a given fixed probability space $(\Omega\text{, }\mathcal{A}\text{, }\mathbb{P})$. If $X_n$ converges to $X$ in distribution, and if $X$ is a r.v. a.s. equal to a constant, then $X_n$ converges to $X$ in probability as well.
PROOF: Suppose that $X$ is a.s. equal to a constant $a$ (that is $\mathbb{P}(X=a)=1)$. The function $f(x)=\frac{|x-a|}{1+|x-a|}$ is bounded and continous. Therefore, $\lim\limits_{n\to\infty}E\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=0$, hence $X_n$ converges to $a$ in probability by a result assuring that.
I have a doubt about the proof. In particular, since $X_n$ converges to $X$ in distribution, given THEOREM 2, $\lim\limits_{n\to\infty}E\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=E\Big\{\lim\limits_{n\to\infty}\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}\Big\}$. At this point, since in the proof it is directly stated that "$\lim\limits_{n\to\infty}E\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=0$", I guess it holds that $\lim\limits_{n\to\infty}\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=0$, but I cannot figure out why this is true.
So, the question is: how can I show that $$\lim\limits_{n\to\infty}\Bigg\{\frac{|X_n-a|}{1+|X_n-a|}\Bigg\}=0$$?
AI: We have $X_n \stackrel{d}\to a$ (convergence in distribution). Thus
$$\lim_n\mathbb{E}(f(X_n)) = \mathbb{E}(f(a))= f(a)$$
for every bounded continuous function $f$.
With your choice of $f$, the result immediately follows.
The following result/definition was used:
$$X_n \stackrel{d}\to X$$
$$\iff$$
$$\mathbb{P}_{X_n} \stackrel{w}\to \mathbb{P}_X$$
$$\iff$$
$$\int fd\mathbb{P}_{X_n }= \int f(X_n) d \mathbb{P}=\mathbb{E}(f(X_n))\to \mathbb{E}(f(X)) = \int f(X) d \mathbb{P} = \int f d\mathbb{P}_X$$
for all $f \in C_b$.
Your claim that $$\lim_n \frac{|X_n-a|}{1+|X_n-a|}=0$$
need not be true.
|
H: Ideal $I=\langle x^2+1 \rangle$ in $R=C[0,1]$
Let $R$ be the ring all continous functions from $[0,1]$ to $\mathbb{R}$. Let $I$ be the ideal generated by $f(x)$ where $f(x)=x^2+1$ .Then
$(a) I$ is maximal ideal of $R$
$(b) I$ is prime ideal of $R$.
$(c) \frac RI $ is finite
$(d)char\frac RI$ is a prime number.
My attempt:-
For $(a)$
$I$ is not a maximal ideal since every maximal ideal of $R$ is of the form $\{f\in R: f(c)=0\}$, for some $c\in \mathbb{R}$, and the given polynomial doesn't vanish anywhere.
But I have one question here: $R$ is a ring with identity ,namely $f(x)=1, \forall x\in [0,1]$, so every proper ideal is contained in a maximal ideal.
But how can a non-vanishing function be in the set of functions vanishing at a particular point?[or if the question is simply put , in which maximal ideal is $I$ contained?]
For $(c)$
We have $\{ax+b+\langle x^2+1\rangle : a,b \in \mathbb{R}\}\subset\frac RI$ thus $\frac RI$ is not finite.
I have no idea about other options
Am I thinking properly? Please help me clear my doubts and answer rest of the parts.
Thanks for your valuable time.
AI: You actually have $I = R$ since the generator is invertible with inverse $x \mapsto \dfrac{1}{x^2+1}$.
(a) Your observation that it not a maximal ideal was correct. However, in this case, the reason is that it is not even a proper ideal!
This is precisely why you struggled with the last part of the doubt. As you noted, every proper ideal is part of a maximal ideal.
(c) We clearly know that $R/I$ is finite now. The error in your reasoning is that you were probably thinking that you are in the polynomial ring.
In that case, $$ax + b + \langle x^2 + 1\rangle \neq a'x + b' + \langle x^2 + 1\rangle$$ iff $(a, b) \neq (a', b')$.
However, here you don't have that anymore.
(b) and (d) should be answered now based on the knowledge that $I = R$.
But just for completeness -
(b) $I$ is not prime since it is not proper.
(d) ${\rm char}\ R/I = 1$, not a prime.
|
H: Dieudonné: Treatise on Analysis, chapter 16, differentiable manifolds: Common principle used for proofs
In studying Dieudonné's Treatise on Analysis, vol. 3, chapter 16, differentialbe manifolds, I stumbled over a principle which he uses frequently for his proofs. I encounterd it for the first time in the proof of proposition 16.3.1, p.12, following the definition of a morphism of differentiable manifolds:
Here is a hand-drawn picture illustrating the situation for the proof of sufficiency of 16.3.1.
The point which I don't understand is why, at the end of the proof of sufficiency of condition 16.3.1, we can limit the discussion to the intersection $ U \cap U' $. It seems obvious to me that we can do this with respect to the chart of U, because the condition which holds for the chart of U also holds for the restricted chart of $ U \cap U' $. But then what we derive for $ U \cap U' $ as being the restricted chart of U', why does it hold for a chart on all of U'? I see no indication in all of the preceeding text that we are allowed to draw such a conclusion.
This principle of proving, limiting the discussion of what is valid for different charts to their intersection, thus always assuming that their domain be equal, appears again and again in the following, for example in 16.5.1.1. So it would be very helpful for me to properly understand this point.
PS: Of course I am aware that compatibility between charts is a question which is limited to the intersection of their domains. But this does not explain to me the above question.
Thanks for any help from someone maybe familiar with Dieudonné !
AI: Differentiability is a purely local property, i.e. the question whether $f:X\rightarrow Y$ is differentiable at $x_0\in X$ can be decided on any neighbourhood $U $ of $x_0$.
So in the proof, you want to show differentiability in $U^\prime$. To this end you choose an arbitrary point $x_0$, and show differentiability in this point (by restricting to a smaller neighbourhood, if necessary, using the fact that differentiability is a local property). Since $x_0$ has been chosen arbitrarily, the conclusion will then hold in all of $U^\prime$.
(This is not a reasoning which is used only by Dieudonné)
|
H: Calculating matrix exponential for $n \times n$ Jordan block
I want to calculate exponential of the matrix which on diagonal has some $a \in \mathbb{R}$ and ones above. The $n\times n$ matrix looks like following
$$ A = \left( \begin{matrix} a & 1 & 0 & 0 & \cdots & 0 \\ 0 & a & 1 & 0 & \cdots & 0 \\ 0 & 0 & a & 1 & 0 & 0 \\ 0 & 0 & 0 & a & \ddots & 0 \\ 0 & 0 & 0 & 0 & \ddots & 1 \\ 0 & 0 & 0 & 0 & 0 & a \end{matrix} \right) $$
I tried to do it by counting determinant of matrix $A-\lambda I$ by the following algorithm :
Divide last row by $a-\lambda$, so the $n$-th row is just $0$ and $1$ in the $n$-th column.
Subtract $(n-1)$-th row by $n$-th row. Then the $1$ in the $n$-th column and $n-1$ row disappears.
Divide $n-1$ row by $a-\lambda$, so the $(n-1)-$ th row is just $0$ and $1$ in the $(n-1)$-th column.
And so on so on. By algorithm above we get matrix with only ones at diagonal, so the determinant of that matrix is just $(a - \lambda)^n$. So we have $n$-th fold eigenvalue equals to $\lambda$. I now I have a problem with derivation of eigenvectors of matrix $A$. Can you give me some advice? Is there any simplest way to calculate that?
AI: Actually this matrix only has a one dimensional eigenspace, spanned by the first unit vector. This is an example of a 'Jordan'-block, you will find lots of theory if you google it.
Maybe a simpler way to calculate the matrix exponential is to write
\begin{align}
A=a\mathbb{I}+B
\end{align}
where $\mathbb{I}$ is the unit matrix and $B$ has only one's above the diagonal. Since $\mathbb{I}$ and $B$ commute, you get
\begin{align}
\exp(A)=\exp(a\mathbb{I})\cdot \exp(B)
\end{align}
Calculating the exponential of $a\mathbb{I}$ is straightforward and calculating the exponential of $B$ is also not too difficult, since $B$ is nilpotent, i.e $B^k=0$ for some appropriate $k$.
|
H: Integral of $\int^{\infty}_0 \frac{x^n}{x^s+1}dx$
$$R(s;n)= \int^{\infty}_0 \frac{x^n}{x^s+1}dx$$
From a previously asked question, I know:
$$R(s;0)=\frac{1}{s} \varGamma\left(\frac{1}{s}\right) \varGamma\left(1-\frac{1}{s}\right)$$
The obvious approach is to do integration by parts but I did not manage to find it using that approach, can any of you provide hints or solutions?
AI: Set $y = x^{n+1}$, then $dy = (n+1)x^n dx$, and $ x^s = y^{ s/({n+1})}$, so
$$R(s;n)=\frac1{n+1}R\left(\frac{s}{n+1};0\right)$$
|
H: Calculate $\iint\frac{dxdy}{(1+x^2+y^2)^2}$ over a triangle
Calculate
$$\iint\frac{dxdy}{(1+x^2+y^2)^2}$$ over the triangle $(0,0)$, $(2,0)$, $(1,\sqrt{3})$.
So I tried changing to polar coordinates and I know that the angle is between $0$ and $\frac{\pi}{3}$ but I couldn't figure how to set the radius because it depends on the angle.
AI: Yes, using polar coordinates is a good idea. We find
$$\iint_T\frac{dxdy}{(1+x^2+y^2)^2}=\int_{\theta=0}^{\pi/3}d\theta\int_{\rho=0}^{f(\theta)}\frac{\rho d\rho}{(1+\rho^2)^2}
=-\frac{1}{2}\int_{\theta=0}^{\pi/3}\left[\frac{1}{1+\rho^2 }\right]_{\rho=0}^{f(\theta)}\,d\theta$$
where the upperbound $\rho=f(\theta)$ can be obtained from the line joining the points $(1,\sqrt{3})$ and $(2,0)$,
$$\rho\sin(\theta)=y=\sqrt{3}(2-x)=\sqrt{3}(2-\rho\cos(\theta))$$
and therefore
$$\rho=f(\theta)=\frac{2\sqrt{3}}{\sin(\theta)+\sqrt{3}\cos(\theta)}
=\frac{\sqrt{3}}{\sin(\theta+\pi/3)}.$$
Can you take it from here?
|
H: Integral $\int \frac{x\,dx}{(a-bx^2)^2} $
How can I integrate
$$
\int \frac{x\,dx}{(a-bx^2)^2}
$$ I've tried to use partial fraction decomposition, but I'm getting six equations for four variables, and they don't give uniform answers.
AI: Substitute $u=a-bx^2$ so your integral is $-\frac{1}{2b}\int\frac{du}{u^2}=\frac{1}{2bu}+C$.
|
H: Find $E(X_1X_2 \mid X_{(1)})$ where $X_i$'s are i.i.d Exponential. Is my solution correct?
Let $X_1, ... , X_n$, ($n\ge 2$) be a sample from exponential distribution with parameter $\lambda=1$ and $X_{(1)}={\rm{min}}(X_1,...,X_n)$. Then we need to find the conditional expectation $E(X_1X_2\mid X_{(1)})$.
Is this correct:
\begin{align}
&E(X_1X_2\mid X_{(1)}=a)\cdot P(X_{(1)}=a)\\&=a\cdot\int_a^\infty f_{X_2}(x_2)x_2dx_2\cdot \prod_{i=3}^n \int_a^\infty f_{X_i}(x_i)dx_i\\
&+\int_a^\infty f_{X_1}(x_1)x_1dx_1\cdot a \cdot \prod_{i=3}^n \int_a^\infty f_{X_i}(x_i)dx_i\\
&+\sum_{k=3}^n \int_a^\infty f_{X_1}(x_1)x_1dx_1\cdot\int_a^\infty f_{X_2}(x_2)x_2dx_2\cdot \prod_{\substack{i=3\\ i\neq k}}^n \int_a^\infty f_{X_i}(x_i)dx_i,
\end{align}
where $P(X_{(1)})=ne^{-(n-1)a}$.
The formula above is self-explanatory: we consider $n$ cases where $X_{(1)}=X_k$, $k=1,2,...,n$ and take the sum of expectations in each case.
This seems intuitively correct to me, but I'm not sure if this is so. Especially I can't understand how do we use known properties of conditional mean to prove this?
By calculating this expression with $f_X(x)=e^{-x}, x>0$ I find
$$
E(X_1X_2\mid X_{(1)}=a)=(1+a)\left(a+1-\frac2n\right).
$$
EDIT. Whoever can give a definitive answer to this question will be given 200 pt bonus. Thanks.
AI: It is necessary to suppose that the n exponential rv are independent so $\mathbb{E}[X_1 X_2]=\mathbb{E}[X_1]\mathbb{E}[X_2]=1$
Now, reminding the memoryless property of Exp Neg Law, when the minimum $X_{(1)}=a$ means that in $a$ the n rv's are "good as new" so their conditional expected value is $a+1$
EDIT: the previous solution is NOT CORRECT. Finally I found
$\mathbb{E}[X_1 X_2 |X_{(1)}=a]=\frac{2}{n}a(a+1)+\frac{n-2}{n}(a+1)^2$
that is equivalent to the solution found by the OP
I explain my brainstorming:
First of all, observe that $f(x|x>a)=e^a e^{-x}\mathbb{1}_{[a;+\infty)}(x)$
and so $$\mathbb{E}[X|X>a]=e^a \int_a^{+\infty}xe^{-x}dx=a+1$$
Now, if the minimun is NOT $X_1$ or $X_2$ the requested probability is like the following
$\mathbb{E}[X|X>a]\mathbb{E}[Y|Y>a]=(a+1)^2$
For similar reason, if $X_1$ or $X_2$ is the minimum, the expected value will be $a(a+1)$
As the probability of one in $n$ independent rv's to be the minimum is constant $=\frac{1}{n}$ the solution is what I showed
|
H: Find the convergence radius for $\sum_{n=1}^{\infty}\frac{2^{n-1}x^{2n-1}}{(4n-3)^2}$
Find the convergence radius for $\sum_{n=1}^{\infty}\frac{2^{n-1}x^{2n-1}}{(4n-3)^2}$
I'm trying to understand why this solution is wrong:
We can look at this as a power series with $a_n = \begin{cases}
& \frac{2^{n-1}}{(4n-3)^2} \text{ if power of } x \text{ is odd} \\
& 0 \text{ if power of } x \text{ is even}
\end{cases}$
And then using Cauchy–Hadamard we would get $R=\frac{1}{2}$.
On the other hand, if we substitute $t=x^2$ we get that the radius is $R=\frac{1}{2}$ for t, meaning $R=\frac{1}{\sqrt{2}}$ for x. Which is actually the correct solution.
I don't get why the first solution is wrong, explanation will be appreciated.
AI: In the first method your formula for $a_n$ is not correct. To find $a_n$, the coefficient of $x^{n}$, where $n$ is odd, write $n=2m-1$ or $m=\frac {n+1} 2$. Then you get $a_n=\frac {2^{m-1}} {(4m-3)^{2}}=\frac {2^{(n-1)/2}}{ (2n-1)^{2}}$
|
H: How do I obtain the following upper bound for the number of heads for coin tosses?
I have $n$ coins, each with probability
$p$ of being head and coin flips are independent of each other, the probability of at least $n/2$ heads is at most
${n\choose n/2} p^{n/2}$. Why is this so?
In other words, I am asking the proof of why \begin{align*} P[\text{#heads}\geq n/2]\leq{n\choose n/2}p^{n/2}\end{align*}
AI: If the number of heads is $\geq n/2$, then there is some subset $A \subset \{1,2,\ldots,n\}$ with $|A| = n/2$ such that $i \in A$ implies flip $i$ was a head.
That is, $$\{\#\text{heads} \geq n/2\} \subseteq \bigcup_{|A| = n/2} \{\forall i \in A, \text{ flip } i \text{ is a head}\}$$
By monotonicity and subadditivity of $P$, we have
$$P[\#\text{heads} \geq n/2] \leq \sum_{|A|=n/2}P[\forall i \in A, \text{ flip } i \text{ is a head}] = \sum_{|A|=n/2} p^{n/2} = \binom{n}{n/2}p^{n/2}$$
|
H: Atiyah-Macdonald Q1.27 Ring of Polynomial Equations on X.
I do not understand his last statement where he mentions that $P(X)$ is the ring of polynomial functions on $X$. What does he mean by that? Is it the ring of polynomials with coefficients in $X$? If so how is $X$ a ring in the first place.
AI: To avoid confusion with the indeterminate $X$ commonly used in polynomials let me write $V$ instead of $X$.
With $V$ being a subset of $K^n$ we can consider polynomial functions $p:V\rightarrow K$, meaning functions which are given by evaluating a polynomial $p‘\in K[T_1,...,T_n]$ at points in $V$. Now two polynomials $p‘,p‘‘$ yield the same polynomial function on $V$, if they agree at each point, that is if $p‘-p‘‘=0$ on $X$. But this is exactly the defining property of $I(V)$, so the set of polynomial functions on $X$ is precisely the set $K[T_1,...,T_n]/I(V) =: K[V]$.
|
H: Compositeness test using $S_i=6S_{i-1}-11S_{i-2}+6S_{i-3}$ recurrence relation
Can you prove or disprove the following claim:
Let $S_i=6S_{i-1}-11S_{i-2}+6S_{i-3}$ with $S_0=0$ , $S_1=1$ , $S_2=1$ . Let $n$ be a natural number greater than $3$, then:
$$\text{If } n \text{ is a prime number then } S_{n-1} \equiv 0 \pmod{n}$$
You can run this test here. I have verified this claim for all $n$ up to $100000$ .
I was searching for counterexample using the following PARI/GP code:
rec(m,P,Q,R)={s0=0;s1=1;s2=1;l=3;while(l<=m,s=P*s2+Q*s1+R*s0;s0=s1;s1=s2;s2=s;l++);return(s);}
RPT(n1,n2)={forprime(n=n1,n2,if(Mod(rec(n-1,6,-11,6),n)!=0,print(n);break))}
AI: Note that the roots of
$$x^3 - 6x^2 + 11x - 6 = 0$$
are $1, 2, 3$.
Thus, the general term can be written as
$$S_n = A\cdot1^n + B\cdot2^n + C\cdot3^n$$
for some constants $A, B, C$.
(See this for details.)
Using $S_0, S_1, S_2$, we can determine $(A, B, C)$. It turns out to be $(-2, 3, -1)$. That is,
$$S_n = -2 + 3\cdot2^n - 3^n.$$
Now, since $n$ is a prime number coprime to $2$ and $3$, we have
$$2^{n-1} \equiv 1 \pmod n$$
and
$$3^{n-1} \equiv 1 \pmod n.$$
This follows from Fermat's little theorem.
In this case, we have
$$S_{n-1} \equiv -2 + 3 - 1 \equiv 0 \pmod n,$$
as desired.
EDIT: Additional observation! We didn't even have to calculate $(A, B, C).$
Once we know that $S_n = A + B\cdot2^n + C\cdot3^n$, Fermat's little theorem would directly give us
$$S_{n-1} \equiv A + B +C \equiv S_0 \pmod n.$$
We already have that $S_0 = 0 \equiv 0\pmod n$ and thus, we would be done.
|
H: Assign 25 identical items to 8 different people
In how many different ways $24$ identical gifts can be shared between $8$ different people such that
One of them doesn't accept more than $2$ gifts , and
The other people accept maximum of $5$ gifts.
(Similar problems are in Combinatorics a guided tour pdf book in the chapter of Algebraic, generating functions).
AI: We have $N\in\{24,25\}$. The generating function is
$$g(x):=(1+x+x^2)(1+x+\ldots+x^5)^7={(1-x^3)(1-x^6)^7\over(1-x)^8}\ .$$
Now collect all terms of degree $\leq N$ of the polynomial $g_1(x):=(1-x^3)(1-x^6)^7$, and note that
$$g_2(x):={1\over(1-x)^8}=\sum_{k=0}^\infty{7+k\choose k}x^k\ .$$
Now find the coefficient of $x^N$ in the product $g_1(x)g_2(x)$.
|
H: Inequality question involving maximum value and three variables
If $x, y, z > 0$ and $x + y + z = 1$, prove that
(a) $x^2+y^2+z^2 \geq \frac{1}{3}$
(b) $x^2yz \leq \frac{1}{64}$
For the first part, since
$\begin{align} x+y+z=1 &\geq 3\cdot \sqrt[3]{xyz} \\
\Rightarrow \frac{1}{9} &\geq \sqrt[3]{x^2y^2z^2} \end{align}$
and thus
$\begin{align} x^2+y^2+z^2 &\geq 3 \cdot \sqrt[3]{x^2y^2z^2} \\
\therefore x^2+y^2+z^2 &\geq \frac{1}{3} \blacksquare. \end{align}$
I dont understand how i should proceed for the second part tho. Here is my attempt at it with weighted AM GM.
$\begin{align} 2x+y+z &\geq 4 \cdot \sqrt[4]{x^2yz} \\
\frac{1+x}{4} &\geq \sqrt[4]{x^2yz} \\
\frac{x^4+4x^3+6x^2+4x+1}{256} &\geq x^2yz \\ \end{align}$
But i don't get how i should proceed from here. Please check the solutions and help with the second part. Thanks
AI: For (b) apply AM/GM to $x$, $x$, $2y$ and $2z$. We get
$$\frac{2(x+y+z)}{4}\ge\sqrt[4]{4x^2yz}$$
etc.
|
H: Weak-* convergence in $L^{\infty}$ implies weak convergence in $L^p$ on bounded set
In a lecture I found the following result: "We remark that when $\Omega$ is bounded the weak-* convergence of $u_{n}$ in $L^{\infty}(\Omega)$ to some $u \in L^{\infty}(\Omega)$ implies weak convergence of $u_{n}$ to $u$ in any $L^p$, $1 \leq p < \infty$."
Does anyone know a book or some other source where this result is proved?
AI: I don't know of a reference but that might be because the result isn't so hard to see. Since $\Omega$ is bounded, we have that $L^q(\Omega) \subseteq L^{1}(\Omega)$ for every $1 \leq q \leq \infty$.
Now weak-$*$ convergence in $L^\infty(\Omega)$ means that for every $v \in L^1(\Omega)$,
$$\int u_n v dx \to \int u v dx$$
as $n \to \infty$. To prove the weak convergence in $L^p(\Omega)$ we need to prove that if $v \in L^q(\Omega)$ where $p^{-1} + q^{-1} = 1$, then $\int u_n v \to \int u v$. But since $L^q(\Omega) \subseteq L^1(\Omega)$, this is immediate by the weak-$*$ convergence in $L^\infty$.
|
H: Question about Zariski density and polynomials with full Galois group
Let $A_3\subseteq {\mathbb Q}^4$ be the sets of all $q=(q_3,q_2,q_1,q_0)$ such that $P_q=X^4+q_3X^3+q_2X^2+q_1X+q_0$ has no rational root. Let $A_2 \subseteq A_3$ be the subset of all $q$'s such that $P$ is irreducible over $\mathbb Q$. Let $A_1 \subseteq A_2$ be the sub-subset of all $q$'s such that $P$ has Galois group $S_4$ over $\mathbb Q$.
My questions : Is $A_1$ Zariski-dense in $A_2$ ? Is $A_2$ Zariski-dense in $A_3$ ?
My thoughts : $B=\cap_{r\in {\mathbb Q}} \lbrace q \ | \ P_q(r)\neq 0 \rbrace$ is a countable intersection of open sets, so it is probably not open or closed.
AI: The Hilbert irreducibility theorem says that $A_1$ is already Zariski dense in all of $\mathbb Q^4$, so the answer to both of your questions is yes.
The Hilbert irreducibility theorem says that $A_1$ is already Zariski dense in all of $\mathbb Q^4$, so the answer to both of your questions is yes.
First, one can see that the Galois group of your polynomial over the field $K=\mathbb Q(q_0,q_1,q_2,q_3)$ is $S_4$. A lazy way to see this is to observe that the Galois group is at most that large and that there are specializations which have Galois group $S_4$ (of course this is ``historically'' a backward approach).
Next, let $\lambda$ be a primitive element for the splitting field of $f(x)$ over $K$, which we may take to be a $K$-linear combination of the roots of $f(x)$, and $g(x)$ the minimal polynomial of $\lambda$ over $K$. Tweaking the linear combination if necessary, we may take $\lambda$ to be integral over $\mathbb Q[q_0,q_1,q_2,q_3]$ so that $g(x)$ is in $\mathbb Q[q_0,q_1,q_2,q_3][x]$, making it a polynomial. By construction, $g$ is irreducible in $K[x]$. By the HIT the set of specializations which preserve the irreducibility of $g$ is Zariski dense. All we need to do is check that these also give rise to specializations of $f(x)$ which have Galois group $S_4$.
This is easy: the Galois group of $f(x)$ (specialized somewhere) embeds in $S_4$. On the other hand, the specialization of $\lambda$ is still a linear combination of roots of $f(x)$ and so the splitting field of $f(x)$ contains it. But $\lambda$ has degree $4!$ over $\mathbb Q$ by our choice of specialization and so the degree of the splitting field must be as large as possible, meaning we have the desired Galois group.
I suppose to be more a little careful with the specialization (and how it extends to $\lambda$) one would want to talk about a prime of the integral closure of $\mathbb Q[q_0,q_1,q_3,q_4]$ in $K(\lambda)$ lying over the ideal $(q_0-a_0,q_1-a_1,q_2-a_2,q_3-a_3)$. The HIT is a statement about decomposition groups in this setting. If $\lambda$ were not chosen to be integral one would have to localize to resolve that issue, and then use the version of HIT which gives specializations which don't vanish on specified polynomials in order to avoid trouble with the denominators.
|
H: Triangle with altitudes and projections.
Let $ABC$ be triangle with $AA_1$ , $BB_1$ and $CC_1$ as its altitude. $M$ and $N$ are the projection of $C_1$ to $AC$ , $BC$ respectively. Let $MN$ intersects $B_1C_1$ at $P$. Show that $P$ is midpoint of $B_1C_1$.
I have tried using Menelaus’ Theorem and similar triangles. Then , I use trigonometry and complex bashing. But , still can’t prove it. Should I draw some more line(s) or use any theorem? Can anyone give me some hints (or solution) please. Thank you!
AI: Let $MN$ cut $BB_1$ at X.
$\angle CMC_1 = \angle CNC_1 = 90^0$ means $CMC_1N$ is cyclic. Then, $\angle NMC_1 = \angle NCC_1$.
Note also that $\angle C_1B_1 B = \angle NCC_1$ because of the properties of orthic triangle.
This means $MB_1XC_1$ is cyclic and is a rectangle. Result follows.
|
H: A positive Ricci curvature problem from Peter Petersen's book
The book "Riemannian Geoemetry, the third edition" by Peter Petersen says the following on page 304: $S^k \times S^1$ does not admit any Ricci flat metrics when $k=2, 3.$ My question is whether $S^k \times S^1$ admits metrics of positive Ricci curvature when $k=2, 3.$ If it is correct, how to prove it?
AI: No. If a manifold $M$ has positive Ricci curvture, then it follows from Myers's theorem that its fundamental group $\pi_1(M)$ is finite. But the fundamental group of $S^k\times S^1$ is $\mathbb{Z}$ for any $k\geq 2$.
|
H: Eigenvalues and reducible polynomials
Let $n \geq 3$. Is it true that an $n \times n$ matrix $A$ with entries from a field has an eigenvalue if and only if every monic polynomial of degree $n$ is reducible? I believe it is true, but I couldn't come up with a proof. Is this known?
AI: Apologies if I'm misunderstanding the question, but I think my counterexample from the comments can be modified to work for all $n$. In fact we can show that neither implication is true.
First Implication
We claim that the existence of eigenvalues does not require every polynomial of degree $n$ to be reducible.
Take our field to be $\mathbb{Q}$, and let $A = I_n$ be the $n\times n$ identity matrix. Then certainly $A$ has an eigenvalue (namely $1$).
However, we claim that there is always an irreducible polynomial in $\mathbb{Q}[x]$ of degree $n$. It suffices to show that $x^n - 2$ is irreducible, which is a trivial consequence of Eisenstein's criterion, since it is irreducible over $\mathbb{Z}$, hence over $\mathbb{Q}$, by a result sometimes called Gauss's Lemma. Alternatively, we can note that any nontrivial, proper factor of $x^n - 2$ would have constant coefficient $\pm \omega2^{a/n}$ for some $1 \leq a \leq n-1$ and some root of unity $\omega$, and no number of this form is ever rational.
Second Implication
We claim that if every polynomial of degree $n$ is reducible, it is possible for find an $n\times n$ matrix with no Eigenvalues.
Take our field to be $\mathbb{R}$, since every real polynomial of degree at least $3$ is reducible, and consider the $4\times 4$ matrix
$$
A = \Bigg(\begin{matrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{matrix}\Bigg)
$$
This matrix has characteristic polynomial $\chi_A(x) = (x^2+1)^2$, which has no roots, hence $A$ has no Eigenvalues.
|
H: If $f:A \to B$ then prove that $|A| \geqslant |f(A)| $
So, $A$ and $B$ are non empty finite sets and there is a function $f:A \to B$ and I need to prove that $|A| \geqslant |f(A)| $. So, $|A|$ is the cardinality of set $A$ and $ |f(A)| $ is the cardinality of the co-domain of the function. So, here is my thinking on this. Since $f$ is a function, each element in $A$ is mapped to some element in $B$. In the worst case, all elements of $A$ are mapped to a single element of $B$. And since $A$ is non-empty, there is at least one element in $A$. So, if there is only one element in $A$, then we have $|A| = |f(A)| $. Another case is when $|A| > 1$ and $ |f(A)| = 1$. Here we would have $|A| > |f(A)| $. And another case is when the function is one to one. So, every element of $A$ must be mapped to a different element of $B$. So, we must have $|A| = |f(A)|$. So, in any case, we have $|A| \geqslant |f(A)| $. Now, would this be considered a valid proof ? I don't know if this proof is without any loopholes.
Thanks
AI: For each $b\in B$, let $N(b)$ be the number of elements in $A$ that are mapped to $b$ by $f$;
$$
N(b) = | \{ a \in A : f(a)=b \} |
$$
Then $\sum_{b\in B} N(b) = |A|$ because every element of $A$ is mapped to some element of $B$. (This is the statement that the fibers of $f$ partition $A$.)
Therefore
$$
|A| = \sum_{b\in B} N(b) = \sum_{b\in f(A)} N(b) \ge \sum_{b\in f(A)} 1 = |f(A)|
$$
|
H: $|X|=|\mathbb{N}|$ for $X$ finite?
Consider a countable set $X$, which may or may not be finite (note: we define a countable set to be either finite or infinite). Is it necessarily true that a bijection exists from $X$ to $\mathbb{N}$ (i.e. is it true that $|X|=|\mathbb{N}|$) for finite $X$?
I know from definitions that if $X$ is countably infinite (i.e. countable and infinite), then $|X|=|\mathbb{N}|$. But I was also wondering if the same is true if $X$ is a finite set. My intuition thinks this is true, but I am looking for some clarity.
AI: No, it is not true that a bijection exists from finite $X$ to $\mathbb N$. If you tried to construct one, you'd inevitably find that there are still infinitely many elements of $\mathbb N$ left after you mapped all the elements of $X$.
However if $X$ is countable, including finite, there always exists an injection to $\mathbb N$. That is, you can map each element to a different natural number. Or in other words, you can count the elements of $X$ (because counting is nothing else but assigning natural numbers). That's where the term countable comes from.
|
H: Why can the average/midpoint of two numbers be described as the sum of the numbers divided by two?
Say I have two numbers, A and B. The "average" or the midpoint of the two numbers is given by $$\frac{A+B}{2}$$
My question is, why does this formula work?
Intuitively, I can derive the equation as follows. The "midpoint" of the two numbers can be given by:
$$A+\frac{B-A}{2}$$
$$=\frac{2A}{2}+\frac{B-A}{2}$$
$$=\frac{A+B}{2}$$
But why? Why is it when you add two numbers and divided it by two you get its "midpoint"? I can't seem to find a way to intuitively visualize this. Please help, thank you!
AI: Let's have this visualization:
$\mathtt{|----|------+------|----|}\\
O\hphantom{----}A\hphantom{------}\ \ I\hphantom{------}\ \ \,B\hphantom{---}A+B$
$I$ is the midpoint of $[A,B]$
but $\operatorname{dist}(O,A)=\operatorname{dist}(B,A+B)=a$
Thus $I$ is also the midpoint of $[O,A+B]$.
And since you agree it is $I=O+\dfrac{(A+B)-O}2$ and that $O$ can be identified to the zero of point addition, then $I=\dfrac{A+B}{2}$.
|
H: Integral $\int^{\infty}_{0} \int^{x}_{0}x.e^{-x^2/y} \, dy\,dx$
This question was asked a couple of times in my Engineering Maths exam in last 5 years but strangely I couldn't find any solution of it on the Internet by Googling, as far as I have tried.
Solve by changing the order of integration:
$$\int^{\infty}_{0} \int^{x}_{0}x.e^{-x^2/y} \, dy\,dx$$
Region of area on the graph should look like following I think.
If I am not wrong the this should be the new order: $$\int^{y}_{0} \int^{x}_{0}x.e^{-x^2/y} \, dy\,dx$$
Upon trying, I am encountering endless integration. Could anyone help how we will reach the solution. Online calculators aren't giving any solution (error or not solvable kind of thing)
AI: $\int_0^{\infty} \int_0^{x} xe^{-x^{2}/y} dydx=\int_0^{\infty} \int_y^{\infty} xe^{-x^{2}/y}dx dy$. This becomes $\int_0^{\infty} -\frac y 2 e^{-x^{2}/y}|_y^{\infty} dx=\frac 1 2\int_0^{\infty} ye^{-y}dy=\frac 1 2$.
|
H: volume of solid generated by revolving the triangle about $y$ axis
Volume of solid generated by revolving the region formed by triangle $(5,0),(5,2),(7,2)$ about $y$ axis is
What i Try: Let $3$ points be $A(5,0)$ and $B(5,2)$ and $C(7,2)$
When we rotate a $\triangle ABC$ about $y$ axis. We get a cone
Whose outer radius $r_{1}=y=f(x)=x-5.$
i.e equation of line $AC$ is $y=x-5$
And here inner radius $r_{2}=5$
So volume is $$\int^{7}_{5}\pi\bigg(r^2_{1}-r^2_{2}\bigg)dx$$
$$\int^{7}_{5}\bigg((x-5)^2-5^2\bigg)dx=-\frac{142\pi}{3}$$
I did not understand where is my solution wrong and why I am getting negative answer. Please, help me.
AI: Hints:
$$V_y=\pi\int\limits_0^2((y+5)^2 - 5^2)dy$$
|
H: How to prove it is a continuous function?
Assume that $f\in L^1(\mathbb{R}^n)$. Let $B(x,r)$ be a ball centered at $x$ with radius $r$. By Lebesgue Differentiation Theorem,
$$ \lim_{r\to 0} \dfrac{1}{m(B(x,r))}\int_{B(x,r)}f(y)dy=f(x)$$
for $x\in\mathbb{R}^n$ almost everywhere. Assume that $x_0\in\mathbb{R}^n$ is a point that such that $f(x_0)$ is finite and such that the equality above holds. Define
$$F(r):=\dfrac{1}{m(B(x_0,r))}\int_{B(x_0,r)}f(y)dy.$$ Is $F(r)$ a continuous function of $r$ on $(0,+\infty)$? I think so but I’m not sure how to prove it. I tried to prove it by writing the definitions but feel completely stuck. Here’s some of my attempts.
By Lebesgue Differentiation Theorem,
$$\lim_{r\to 0^+} F(r)=f(x_0)<\infty.$$
And
$$\lim_{r\to +\infty} F(r)= \lim_{r\to +\infty} \dfrac{1}{m(B(x_0,r))}\int_{B(x_0,r)}f(y)dy\le \lim_{r\to +\infty} \dfrac{1}{m(B(x_0,r))}\int_{\mathbb{R}^n} f(y)dy
= \lim_{r\to +\infty} \dfrac{1}{m(B(x_0,r))}\cdot \Vert f\Vert_1=0.$$
Could you give me some ideas? Thanks!
AI: It is a ratio of two continuous functions. Note that $m(B(x,r))=r^{n} m(B(0,1))$. To show that the numerator is continuous use the following facts:
For any integrable function $f$ , $\int_E f(x)dx \to 0$ as $m(E) \to 0$
$|\int_E f(x)dx - \int_F f(x)dx | \leq \int_{E\Delta F} |f(x)| dx$.
$m(B(x,r+h)\Delta B(x,r))=|(r+h)^{n} -r^{n}| m(B(0,1)) \to 0$ as $ h \to 0$.
Notation: $A \Delta B=(A \setminus B) \cup (B \setminus A)$
|
H: getting a number correct on 3 successive dice rolls
I'm working through some practice problems. If we guess a number between $1..6$ and then roll 3 regular six sided die, what is the probability that our guess will appear on any of the dice?
I'm new to probability and I'm struggling a bit to understand this....
I know it's not 1/6 * 3 or $(1/6)^3$.
Thanks for your help
AI: What is the possible number of total outcomes? There are three dices each may give $6$ numbers that is:
$6^3$
What is the possible number of desired outcomes?
Notice that you said that you want $6$ to appear anywhere, so we might just go about and find the number of not desired outcomes. A common practice, btw. Well, that is the same as if we would not want $6$ to appear anywhere, which is the same as the number of outcome on $3$ $5$-side dices, with $6$'s removed, and if so these now have
$5^3$ outcomes.
That leaves $6^3$ - $5^3$ and the probability is then the number of desired outcomes divided by the number of total outcomes.
$$ \frac{6^3 - 5^3}{6^3} $$
You do the math. ;)
|
H: Does convergence to zero follow from $x\cdot a_n \to 0,\ x\in H$?
Let $H$ be an arbitrary Hilbert space. Suppose that for some sequence $\{a_n\} \subseteq H$ we have
$$\forall x\in H,\quad x\cdot a_n \to 0\in\mathbb K.$$
Does it follow that $a_n\to 0$?
I have tried various manipulations of the expression $a_n\cdot a_n$ to no avail. We could also take an orthonormed basis $E\subseteq H$, then $H \cong \ell _2(E)$. But the assumption would only justify coordinate-wise convergence to zero, which I don't think is equivalent to convergence, in general.
AI: No. If $(e_n)$ is an orthonormal basis for $H$ then $x.e_n \to 0$ for every $x$ but $\|e_n\|=1$ so $e_n$ does not tend to $0$.
|
H: How to find $a, b$ and $c$?
I was given the following two equations and told to find $a, b$ and $c$:
$$\left\{\begin{array}{c}
a+b-c=1\\
a^2+b^2-c^2=-1\end{array}\right.$$
I tried to form a matrix to solve it and I made this:
$$\left[\begin{array}{ccc|c}
1 & 1 & -1 & 1\\
a & b & -c & -1
\end{array}
\right]$$
Then I performed elimination and got this:
$$\left[\begin{array}{ccc|c}
1&0&\frac{-b+c}{-a+b}&\frac{b+1}{-a+b}\\
0&1&\frac{a-c}{-a+b}&-\frac{a+1}{-a+b}
\end{array}\right]$$
After this I tried to find a specific solution when $c=0$ and a made this:
$$\left[\begin{matrix}
\frac{1\pm i\sqrt 3}{2}\\
\frac{1\mp i\sqrt 3}{2}\\
0
\end{matrix}\right]$$
Then, to find all the solutions I tried to find the null space while making $c=1$, then I found that all values are of the form:
$$\left[\begin{matrix}
\frac{1\pm i\sqrt 3}{2}\\
\frac{1\mp i\sqrt 3}{2}\\
0
\end{matrix}\right]
%
+C\left[\begin{matrix}
0/1\\
1/0\\
1
\end{matrix}\right]$$
Note: ‘/’ denotes $“or”$ and $C$ can be any integer
But when tried to plug this back into my question taking $C=1$ it didn’t work, but when $C=0$, it works, please help me.
AI: There are only $2$ equations for $3$ variables so you can try to find two of them in function of the third one.
Since $a,b$ play a symmetrical role, we can choose between $b$ or $c$ as free variable. The choice of $c$ leads to solve quadratic equations, so we will choose $b$ instead.
First let $b=1$ then the system becomes $\begin{cases}a=c\\a^2-c^2=-2\iff 0=-2\end{cases}\quad$ the system has no solution.
So now we can exploit $a^2-c^2=\underbrace{(a-c)}_{1-b}(a+c)=-1-b^2\quad$ and solve
$\begin{cases}a=\frac b{b-1}\\c=\frac{b^2-b+1}{b-1}\end{cases}$
|
H: Built a sequence
They ask me to create a sequence $\{A_n\}$ that fulfill:
$P( \limsup A_n) = \alpha$ with $\alpha\in[0,1]$
$\sum_{n \geq 1}{P(A_n)} = \infty $ and $P( \limsup A_n) = 0$
For the first case I don't know how to start and do it, for the second my idea is something that the probabilities are equiprobable like $\frac{1}{n}$ because it doesn't converge but I don't know what $\Omega$ and probability take.
AI: 1)
Choose some event $A$ and let $A_n=A$ for every $n$.
Then $\limsup A_n=A$ so that $P(\limsup A_n)=P(A)\in[0,1]$.
If $\alpha$ some fixed number then you can go for $\Omega=\{0,1\}$, $A=\{0\}$, $P(\{0\})=\alpha$ and $P(\{1\})=1-\alpha$.
2)
You can take $\Omega=\mathbb N_+$ together with $P(\{n\})=\frac{1}{n}-\frac1{n+1}$ for $n=1,2,\dots$
Then for $A_n=\{n,n+1,n+2,\dots\}$ we have $P(A_n)=\frac1n$ and $\limsup A_n=\varnothing$ so that the conditions mentioned under 2) are satisfied.
|
H: find the value of $\sec^4(\pi/9) + \sec^4( 2\pi/9) + \sec^4 (4\pi/9)$
find the value of $\sec^4(\pi/9) + \sec^4( 2\pi/9) + \sec^4 (4\pi/9)$
I tried converting $\sec^2(\theta)$ to $\tan^2(\theta)$ but for no vain.
AI: Like Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$
$$\cos3t=4\cos^3t-3\cos t$$
Observe that $\sec\left(3\cdot\dfrac{\pi}9\right)=2,\sec\left(3\cdot\dfrac{2\pi}9\right)=-2, \sec\left(3\cdot\dfrac{4\pi}9\right)=-2$
Put $3t=\dfrac\pi3$
So, the roots of $4c^3-3c-\dfrac12=0\iff8c^3-6c-1=0$ are $\cos\dfrac{(6n+1)\pi}9$ where $n=0,1,2$
As $\cos(\pi\pm y)=-\cos y$
$\cos\dfrac{(6\cdot1+1)\pi}9=\cdots=-\cos\dfrac{2\pi}9$ and $\cos\dfrac{(6\cdot2+1)\pi}9=\cdots=-\cos\dfrac{4\pi}9$
Set $\dfrac1c=s$ to find the roots of $s^3+6s^2-8=0$ are $$ \sec\dfrac\pi9, -\sec\dfrac{2\pi}9, -\sec\dfrac{4\pi}9$$
Square both sides of $s^3=8-6s^2$ and replace $s^2=t$ to find the roots of $$t^3-36t^2+96t-64=0$$ are $$a=\sec^2\dfrac\pi9,b=\sec^2\dfrac{2\pi}9,c=\sec^2\dfrac{4\pi}9$$
We need $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$$
|
H: Area of parametric curve
I have a parametric curve (shapes like a heart):
$r(t) = (16sin^3(t), 13cos(t) - 5cos(2t) - 2cos(3t) - cos(4t))$
$0\le t \le 2\pi$
And I'm going to find the area limited by the curve.
I have to find
$\int_0^{2\pi} y*x' dt$
$\int_0^{2\pi} (13cos(t) - 5cos(2t) - 2cos(3t) - cos(4t))48sin^2(t)cos(t) dt$, which is pretty ugly.
But I'm given a hint:
$\int_0^{2\pi} sin(nx)sin(mx) dx = \int_0^{2\pi} cos(nx)cos(mx) dx = \begin{cases}
\pi & \text{for m = n} \\[2ex]
0 & \text{otherwise}
\end{cases}$
But I'dont know how to use this identity in this case.
AI: hint
In your ugly integral, Replace $ \; \sin^2(t) \; $ by $ \;(1-\cos^2(t)) $
and linearise by$$\cos^3(t)=\Bigl(\frac{e^{it}+e^{-it}}{2}\Bigr)^3$$
to get $$\cos^3(t)=\frac 14\Bigl(\cos(3t)+3\cos(t)\Bigr)$$
then you can use $$\int_0^{2\pi}\cos(nt)\cos(mt)dt=\pi \; if \; m=n \; and \; 0 \; otherwise$$
Your area will be about
$$180\pi$$
|
H: The number of letters between $z$ and $y$ is less than that between $y$ and $x$
QUESTION: Consider all the permutations of the $26$ English alphabets that start with $z$. In how many of these permutations the number of letters between $z$ and $y$ is less than that between $y$ and $x$?
Source: ISI BMath 2017 UGA
MY APPROACH: This is what I have done-
The $13^{th}$ place is the mid point of the $26$ places we have for the $26$ letters. Now, the first place is obviously occupied by $z$. Observe that,
If $y$ sits somewhere right to the $13^{th}$ place then $x$ cannot be placed to the right of $y$, and therefore, $x$ will only occupy any position to the left of $y$. So, that's not required.
Note From here on we will not mention that $x$ can always be placed at the left of $y$ (since that's unnecessary). We will only bother with the cases where $x$ is to the right of $y$.
If $y$ occupies the $13^{th}$ place then $x$ can occupy the $26^{th}$ place only, therefore we get only one way for $x$ here.
If $y$ occupies the $12^{th}$ place then $x$ can be placed at $26^{th}$,$25^{th}$ or $24^{th}$ place, thus $3$ ways.
Following this pattern, we get the series (as $y$ moves from $12^{th}$ to the $2^{nd}$ place) $$1+3+5+...+23$$
So, we are done with all possible cases where $x$ is to the right of $y$.
And therefore,, our answer must be $$[144].23!$$
Am I correct?
If not, where is the mistake that I have made?
Can you please provide me with some alternate solution? This seems a bit lengthy..
Thank you so much.
AI: That’s definitely bad — if x is to the left of y, the number of letters between x and y is certainly less than that between z and y, which is the wrong way around!
(this is because the sequence must start with z)
On the other hand, $144\times 23!$ looks right to me, if you disinclude those cases.
|
H: Cone over $X$, equivalence relation
I have a question about the definition of the cone over $X$.
If $X$ is a space, define an equivalence relation $X\times [0,1]$ by $(x,t)\sim (x',t')$ if $t=t'=1$. Denote the equivalence class of $(x,t)$ by $[x,t]$. The cone over $X$, denoted by $CX$, is the quotient space $X\times [0,1]/\sim$.
I acutally do not get this definition of the equivalence relation....
Sure, two points $(x,t), (x',t')$ are equivalent if $t=t'=1$.
But which points are equivalent to (for example) $(x,\tfrac12)$? The relation does not tell anything about the cases when $t\neq 1$ or $t\neq t'$, which feels incomplete. But I think I am doing a horrible mistake here.
How does this relation include every pair $(x,t)\in X\times [0,1]$, when the relation is only defined for $t=1$?
I am currently studying "Introduction to algebraic topology" by Joseph J. Rotman.
An exercise goes as follows:
For fixed $t$ with $0\leq t<1$, prove that $x\mapsto [x,t]$ defines a homeomorphism from a space $X$ to a subspace of $CX$.
Which revealed my misunderstanding.
So I am not understanding which equivalence classes there are.
For every point $(x,t)$ with $t\neq 0$, the equivalcence class should just contain this one point.
Can you elaborate more?
Thanks in advance.
AI: When discussing equivalence relations, one often uses the following fact of set theory:
For every relation $R \subset X \times X$ on a set $X$ there is a unique smallest equivalence relation $E \subset X \times X$ on the set $X$ such that $xRy \implies x E y$. This equivalence relation $E$ is the intersection (in $X \times X$) of all equivalence relations $S$ which have the property $x R y \implies x S y$. We say that $E$ is the equivalence relation generated by $R$.
You can construct $E$ rather concretely. First take the reflexive closure, by adding all pairs $(x,x) \in X \times X$ to the relation $R$. Then take the symmetric closure, by adding all pairs $(y,x)$ for which $(x,y)$ is already in the relation. Finally take the transitive closure: for all sequences $x_0,x_1,x_2,...,x_n$ such that each of the pairs $(x_0,x_1), (x_1,x_2), ..., (x_{n-1},x_n)$ is already in the relation, add $(x_0,x_n)$ to the relation.
So, when you get an equivalence relation on a set (such as $X \times [0,1]$) which appears to only be partially defined, what you are supposed to do is to use that partial definition to define a relation $R$ on the set, and then you should take the equivalence relation generated by $R$.
In particular, since the point $(x,1/2) \in X \times [0,1]$ has not even been mentioned in the definition of $R$, it will follow that point $(x,1/2)$ is the only point in its equivalence class, i.e. its equivalence class is $\{(x,1/2)\}$.
|
H: Integration of some two-variables differential (quadratic polynomial) - apparent absurdum
Let us consider some function $f(x,y):\,\mathbb R^2\to\mathbb R$, such that
$$\frac{\partial f(x,y)}{\partial x}=a\,x+b\,y\qquad\qquad\frac{\partial f(x,y)}{\partial y}=b\,x+c\,y\qquad\qquad f(0,0)=0$$
where $a,b,c\in\mathbb R$.
From this knowledge, i would like to find $f(x,y)$
I proceeded as follows
$$df=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy=(a\,x+b\,y)\,dx+(b\,x+c\,y)\,dy$$
Then
$$f(x,y)=\int_{f(0,0)}^{f(x,y)}df'=\int_{0}^x(a\,x'+b\,y)\,dx'+\int_{0}^y(b\,x+c\,y')\,dy'=\frac{1}{2}\,a\,x^2+2\,b\,x\,y+\frac{1}{2}\,c\,y^2$$
where $\square'$ indicates the variable of integration
But then, if i differentiate the expression for $f$ found, partial derivatives are NOT the ones i started with
$$\frac{\partial f(x,y)}{\partial x}=a\,x+2\,b\,y\neq a\,x+b\,y\qquad\qquad\frac{\partial f(x,y)}{\partial y}=2\,b\,x+c\,y\neq b\,x+c\,y$$
Where am i wrong? How to reach the correct expression for $f$, which by 'guessing' should be $f=\frac{1}{2}\,a\,x^2+\,b\,x\,y+\frac{1}{2}\,c\,y^2$?
AI: You've gotten the formula for computing $f(x,y)$ wrong; you need to set $y$ to be $0$ in your integral using $dx'$.
The reason for this is that you're finding $f(x,y)$ by integrating along two paths; first from $(0,0)$ to $(x,0)$, and then from $(x,0)$ to $(x,y)$. The first path has $y=0$ at all times.
|
H: Do these two commutative diagrams represent the same homomorphism?
Consider the two many-sorted algebras $\mathtt{A}$, $\mathtt{B}$ and the homomorphism $h:\mid \mathtt{A}\mid \to \mid \mathtt{B}\mid$ below. Do the commutative diagrams in Figure 1 and Figure 2 both adequately represent $h$? Can $h_2$ be dropped from Figure 1 to give Figure 2? As far as I know identity and composition are not labelled on commutative diagrams.
$\mathtt{A}$, $\mathtt{B}$ and $h$ are based on this paper.
In algebra $\mathtt{A}$ a point is represented by a triple consisting of a name and integer coordinates. Points are considered equal if they have the same name.
\begin{aligned}
& \mathtt{Point_A} = \{ \langle name,x,y \rangle \mid name \in \mathtt{String}, x,y \in \mathbb{Z} \} \\
& \mathtt{equal_A} : \mathtt{Point_A} \times \mathtt{Point_A} \to \mathbb{B}\\
& \mathtt{equal_A}(\langle name_1,x_1,y_1 \rangle , \langle name_2,x_2,y_2 \rangle) \triangleq (name_1 =_{\mathtt{String}} name_2)
\end{aligned}
Algebra $\mathtt{B}$ represents a point as a coordinate 2-tuple with point equality based the integer value of each coordinate component.
\begin{aligned}
& \mathtt{Point_B} =\{ \langle x,y\rangle \mid x,y \in \mathbb{Z}\} \\
& \mathtt{equal_B} : \mathtt{Point_B} \times \mathtt{Point_B} \to \mathbb{B}\\
&\mathtt{equal_B}(\langle x_1,y_1 \rangle ,\langle x_2,y_2 \rangle ) \triangleq ((x_1 =_\mathbb{Z} x_2) \land (y_1 =_\mathbb{Z} y_2))
\end{aligned}
The homomorphism $h:\mid \mathtt{A}\mid \to \mid \mathtt{B}\mid$ is defined as follows:
\begin{aligned}
& h_1 : \mathtt{Point_A} \to \mathtt{Point_B}\\
& h_1(\langle name,x,y \rangle) = \langle x,y \rangle \\
& h_2 : \mathbb{B} \to \mathbb{B}\\
& h_2(x) = x\\
\end{aligned}
Where $h_1$ forgets the name and $h_2$ is an identity mapping.
The commutative diagram in Figure 1 represents homomorphism: $h_2(\mathtt{equal_A}(a,b)) = \mathtt{equal_B}(h_1(a), h_1(b))$?
The commutative diagram in Figure 2 represents homomorphism: $\mathtt{equal_A}(a,b) =\mathtt{equal_B}(h_1(a), h_1(b))$.
AI: According to your definition of $h_2$ as the identity on $\mathbb{B}$, Figure 1 and 2 are equivalent, i.e. $h_2$ can be dropped from Figure 1 to give Figure 2, which has the same meaning as Figure 1.
The problem is that the diagrams in Figure 1 and 2 do not commute! Take in $|\mathtt{A}|$
\begin{align}
a_1 &= \langle \text{Rome}, 0, 1 \rangle & a_2 &= \langle \text{Rome}, 0, 2 \rangle
\end{align}
Then, $\mathtt{equal_A}(a_1, a_2)$ is true because $\text{Rome} =_\mathtt{String} \text{Rome}$, but $h_1(a_1) = \langle 0,1 \rangle$ and $h_1(a_2) = \langle 0, 2\rangle$, hence $\mathtt{equal_B}(h_1(a_1), h_1(a_2)) = \mathtt{equal_B}(\langle 0,1\rangle, \langle 0 , 2\rangle)$ is false because $1 \neq_\mathbb{Z} 2$.
Of course, the two diagrams commute if you assume that there are no different places with the same name.
|
H: Prove sum of two convex funtions is convex using first and second derivatives
Consider if I have two functions f(x) and g(x) which are both convex. How do I prove that their sum is also convex using the first and second derivatives?. I know if the second derivative of a function is positive then it is convex but I'm not sure how to prove this for a general case like this.
AI: Since you're talking about second derivatives, I'll assume $f$ and $g$ are twice differentiable.
Then $f, g$ are convex if their second derivatives are nonnegative. But then the second derivative of $f+g$ is $f''+g'' \geq 0$ since $f'' \geq 0, g'' \geq 0$.
So $f+g$ is convex, since it has nonnegative second derivative.
|
H: Probability chain rule given some event.
I'm dealing with a probability of the form $\mathbb{P}(\cap_{l \in S} E_l | X)$. Using the conditional probability formula and the probability chain rule would give me a product over the set $\{E_l: l \in S\} \cup \{X\}$. What if I just want to write the product over the set $\{E_l: l \in S\}$?
I am actually trying to understand the following statement, if that helps:
AI: I will try to explain the equation in black, since this is what you are trying to understand. So this is just Bayes applied over and over. Note that:
$$\mathbb P(E_1 \cap E_2 \mid \cap_{i\in S_2} A_i) =
\mathbb P(E_1 \mid \cap_{i\in S_2} A_i)\mathbb P(E_2 \mid (E_1) \cap (\cap_{i\in S_2} A_i))$$
Now, just use this for your case with $\mathbb P(\cap_{i \in S_1}E_i \mid \cap_{i\in S_2} A_i) $.
|
H: Proving that if $H$ and $K$ are subgroups of a finite group G, then $|HK|=\frac{|H||K|}{|H \cap K|}\le |G|$.
Notations: $|C|$= no. of distinct elements in $C$= order of group $C$, if $C$ denotes a group.
$HK=\{hk : \forall \;\;h\in H, g \in G\}.$
The problem lies here: $HK$ may not be a subgroup of $G$. If $HK$ is a subgroup of $G$ then indeed we must have $|HK|\le |G|$, but if $HK$ is not a subgroup of $G$, then there is no reason to believe that $|HK|\le |G| $. Can it be proven that nonetheless, $|HK|\le |G| $? $\tag{A}$
This doubt comes from while proving that every group of order $2p$ (where $p$ is a prime $\gt 2$) is isomorphic to either $\mathbb Z_{2p}$ or $D_p$ (dihedral group of order $2p$).
Following is how the proof goes:
If $G$ contains an element of order $2p$ then $G$ is cyclic of order $2p$ and hence isomorphic to $Z_{2p}$.
So suppose that $G$ does not have any element of order $2p$. By Lagrange's theorem, possible orders of non -identity elements of $G$ are $2$ and $p$. Assuming that all elements of $G$ have order $2$ gives a contradiction as suppose that $a,b \in G$ then $\{e,a,b,ab\}$ is a subgroup of order $4$, which contradicts Lagrange's theorem. Hence there must exist atleast one element of order $p$ in $G$. Let it be $a$.
Suppose that $b\in G$ such that $b \notin \langle a\rangle$, $|b|$ can be either $2$ or $p$. $|\langle a \rangle \cap\langle b \rangle|=1 $ since $\langle a \rangle \cap\langle b \rangle$ is a subgroup of $\langle a \rangle$, and$\langle a \rangle \ne \langle b \rangle$. Now $|b|$ must be $2$ because if $|b|=p$, then
$$|\langle a \rangle \langle b \rangle|=\frac{|\langle a \rangle||\langle b \rangle|}{|\langle a \rangle \cap\langle b \rangle|}=p^2\gt 2p \tag {1}$$
$(1)$ is a contradiction. Due to the problem mentioned above in $(A)$, I don't understand how $(1)$ can be a contradiction?
If $(1)$ is a contradiction, then it will follow that $|b|=2$. In particular, for all elements of $G$ not in $\langle a \rangle $, order $=2$. Therefore since $ab \notin \langle a \rangle $, we have $|ab|=2 \implies ab =ba^{-1}$. Writing Cayley Table, we observe that $G$ is isomorphic to $D_{2p}$.
Can you please help me understand $(A)$ and $(1)$? Thanks for your time.
AI: For $(A)$:
$HK$ is a subset of $G$, hence $|HK|\le|G|$ (or if you prefer, the function $f:HK\to G$ defined by $x\mapsto x$ is injective so by definition $|HK|\le|G|$)
Why is it a subset of $G$?
Well let $g\in HK$, then there exists $h\in H, k\in K$ such that $g=hk$. $H$ and $K$ are subsets of $G$, hence $h\in G, k\in G$, and since $G$ is a group: $g=hk\in G$.
This is true for all $g\in HK$, hence $HK\subseteq G$
For the actual proof that $|HK|=\frac{|H||K|}{|H\cap K|}$, you can see the several proofs in the answers to this question.
|
H: Binomial distribution as a function of $n$ and $k$
As we know, the probability that we get exactly $k$ successes after $n$ experiments is: $\mathbb{P}(S_n = k) = C_n^k\cdot p^k(1-p)^{n-k}$, where $p$ is the probability to get a success in $i^{th}$ experiment.
Let us look at $\mathbb{P}(S_n = k)$ as a function of two variables; $n$ and $k$. Then:
1. If we fix $k$, for what values of $n$, the probability is maximal?
2. If we fix $n$, for what values of $k$, the probability is maximal?
Logically, the answer to the second one must be $\left\lceil\frac{n}{2}\right\rceil$ if $p>1/2$ and $\left\lfloor\frac{n}{2}\right\rfloor$ if $p<1/2$.
I can't come up with a proof. Obviously differentiation is not an option. I tried using Stirling's formula or taking $\ln$ of both sides, but still couldn't get anything.
AI: Your answer to the second part is wrong, it’s either the floor or ceiling of $pn$. The proof of this is to look at successive ratios between terms. Notice that it increases up to $pn$, then decreases afterwards, and hence the maximum must be around there.
On the other hand, if you fix $k$ and want to maximise it for $n$, the method is exactly the same.
|
H: $P$ and $Q$ are the two vertices of a regular polygon having $12$ sides such that $PQ$ is a diameter of the circle circumscribing the polygon. Then...
QUESTION: Let $P=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ and $Q=(−\frac{1}{\sqrt{2}},−\frac{1}{\sqrt{2}})$ be two vertices of a regular polygon having $12$ sides such that $PQ$ is a diameter of the circle circumscribing the polygon. Which of the following points is not a vertex of this polygon?
(A) $(\frac{\sqrt{3}-1}{2\sqrt{2}},\frac{\sqrt{3}+1}{2\sqrt{2}})$
(B) $(\frac{\sqrt{3}+1}{2\sqrt{2}},\frac{\sqrt{3}-1}{2\sqrt{2}})$
(C) $(\frac{\sqrt{3}+1}{2\sqrt{2}},\frac{1-\sqrt{3}}{2\sqrt{2}})$
(D) $({-\frac{1}2},\frac{\sqrt{3}}{2})$
MY APPROACH: If $P$ and $Q$ are the end points of the diameter, it is quite clear that the equation of the circle must be $$x^2+y^2=1$$ Therefore, all the vertices must lie on this circle. Now, checking from the options, we find that every point given in the options satisfies the above equation. Now I am stuck.
How else should I tackle the sum? Thank you in advance .
AI: D)
But how did I get that answer? Well, firstly have you drawn a picture of the diagram, with angles labelled? If so, then you’ll notice that the arg (angle from the x axis) of any point on the polygon must be 15 mod 30. The last option doesn’t satisfy this property (the angle it forms is 120 degrees).
However, another approach is to notice — again from a well drawn diagram — that the figure is symmetric in the x axis, the y axis, and the x=y line. Since the first three options are just reflections of each other in these lines, the answer must be D. Meta analysis is useful in multiple choice!
|
H: Prove that $\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$
For $a,b,c \in (0,1)$ such that $ab+bc+ca=1$ Prove that $\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$
I tried repleace $1$ by $ab+bc+ca$ so we need to prove:
$\frac{a^2+b^2}{(ab+bc+ca-a^2)(ab+bc+ca-b^2)} + \frac{b^2+c^2}{(ab+bc+ca-b^2)(ab+bc+ca-c^2)}+\frac{c^2+a^2}{(ab+bc+ca-c^2)(ab+bc+ca-a^2)} \geq \frac{9}{2}$
Then i try to factor the denominator. This didn't seem to help me.
AI: Another way:
$$\sum_{cyc}\frac{a^2+b^2}{(1-a)^2(1-b^2)}\geq\sum_{cyc}\frac{\frac{1}{2}(a+b)^2}{(1-ab)^2}=\sum_{cyc}\frac{\frac{1}{2}(a+b)^2}{c^2(a+b)^2}=$$
$$=\frac{1}{2}\sum_{cyc}\frac{1}{a^2}\geq \frac{1}{2}\sum_{cyc}\frac{1}{ab} =\frac{1}{2}\sum_{cyc}ab\sum_{cyc}\frac{1}{ab}\geq\frac{9}{2}.$$
|
H: How this integral is derived?
There is an interesting definite integral
$$\int_0^1{x^m(1 - x)^n} dx = \frac{n!m!}{(m + n + 1)!}.$$
How to derive this integral?
AI: Let
$$I(m,n) = \int_0^1 x^m (1-x)^n \, dx.$$
It's straightforward to prove that $I(n,0)=I(0,n)=\frac{1}{n+1}$, which satisfies the given formula.
Now note that
$$x^m(1-x)^n = x^m(1-x)^{n-1}(1-x) = x^m(1-x)^{n-1} - x^{m+1}(1-x)^{n-1}.$$
So
$$I(m,n) = I(m,n-1) - I(m+1, n-1),$$
which gives
$$I(m+1,n-1) = I(m, n-1) - I(m,n).$$
Now we induct on $m$; if it is true for some $m$, and all $n$, then we have
\begin{align*}
I(m+1, n-1) &= \frac{m!(n-1)!}{(m+n)!} - \frac{m!n!}{(m+n+1)!}\\
&= \frac{m!(n-1)!}{(m+n+1)!}\left(m+n+1-n\right)\\
&= \frac{m!(n-1)!}{(m+n+1)!} \cdot (m+1)\\
&= \frac{(m+1)!(n-1)!}{(m+n+1)!}.
\end{align*}
So the statement is now true for $m+1$ and all $n$, and we are done.
|
H: How to prove elementary set theory problem: $(A \cap B) \cup (A \cap B^c)=A$
I am trying to prove that $A \cap B$ and $A \cap B^c$ are disjoint, with $(A \cap B) \cup (A \cap B^c)=A$, but I do not know how to approach it. Intuitively expressed in Venn Diagrams, the above statement is correct.
AI: Disjoint:
$$(A\cap B)\cap(A\cap B^c) = A\cap A\cap B\cap B^c = A\cap\emptyset = \emptyset$$
Union is $A$:
$$(A\cap B)\cup(A\cap B^c) = A\cap(B\cup B^c) = A$$
|
H: Linear Factorization of $X^n+1$ in $\mathbb{F}p$
I was just playing around with SageMath and noticed that I can find for each $n < 200$ (at least) a prime $p$ such that the polynomial $X^n+1$ factorizes into polynomials of degree one in $\mathbb{F}_p[X]$. I have a feeling that this statement is in general true, but is there an easy way to see this?
AI: You are essentially asking that the field contain the $(2n)^{th}$ roots of unity. In order for that to happen we need a prime $p$ such that $2n\,|\,(p-1)$. Equivalently, we want a prime $p\equiv 1 \pmod {2n}$. Since $\gcd(1, 2n)=1$, Dirichlet assures us that infinitely many such primes exist.
Example: $n=17$. We want $p$ such that $p\equiv 1 \pmod {34}$. The least such is $p=103$ and we get $$x^{17}+1\equiv (x + 1) (x + 8) (x + 9) (x + 13) (x + 14) (x + 23) (x + 30) (x + 34) (x + 61)\cdots $$ $$\cdots (x + 64) (x + 66) (x + 72) (x + 76) (x + 79) (x + 81) (x + 93) (x + 100) \pmod {103}$$
|
H: How is ($\neg P \lor R$ ) $\land$ ($\neg Q \lor R$) equivalent to ($\neg P \land \neg Q) \lor R $
How is ($\neg P \lor R$ ) $\land$ ($\neg Q \lor R$) equivalent to ($\neg P \land \neg Q) \lor R $ ?
I know this by making truth table but how to solve this using rules of logic ? Thank you
AI: Just use distributivity of $\wedge$ with respect to $\vee$.
|
H: If $X,Y,Z \sim $ Poisson($\lambda$) are independent, how do I calculate $P(X+Y=2,X+Z=3)$?
I tried only with some algebric manipulation like
$$P(X+Y=2,X+Z=3)=P(X=2-Y,Z=1+Y)=P(X=2-Y)P(Z=1+Y)$$
But it seems to me that this is not going anywhere. Maybe can be useful the fact that $X+Y,X+Z \sim $ Poisson($2\lambda$), but I don't know how to use it.
AI: Note that $X+Y=2, X+Z=3$ requires that $X\leq 2.$ You can now use the following:
$$P(X+Y =2, X+Z = 3) = \sum_{x=0}^2 P(X+Y = 2, X+Z=3|X=x)P(X=x) = \sum_{x=0}^2 P(Y=2-x)\cdot P(Z=3-x)\cdot P(X=x),$$
using independence of $X,Y,Z.$ This is ugly, but it is something that you can just calculate by hand, in principle.
|
H: Size of quotient groups
This question will be really easy to answer I think, I know that there are other questions on roughly the same topic but honestly I could not find what i was looking for in an answer.
I am a physics student trying to learn group theory, I am studying on Pierre Ramond's book "Group theory - a physicist's survey".
I am just really confused about a pretty elementary fact.
When talking about finite size groups the book introduces the notion of quotient groups and explains their construction. Given a group $G$ and a normal subgroup $\mathcal{H}$ then the cosets $g_i\mathcal{H}$ can be given a group structure using the normality properties of $\mathcal{H}$, .
The book also says that the size of this group of cosets $G/\mathcal{H}$ is $\dfrac{n_G}{n_\mathcal{H}}$ with $n_G$ being the size of $G$ and $n_\mathcal{H}$ being the size of the subgroup $\mathcal{H}$ the book does not give any proof of this statement about the size of the quotient group treating it as a trivial result but the counting just doesn't add up to me.
I mean, in $G$ we have $n_G-n_\mathcal{H}$ elements that are not in $\mathcal{H}$, in fact we could write sumeting like $G = \{g_1, g_2, g_3, ..., g_{n_G-n_\mathcal{H}}\} \bigcup \mathcal{H}$. This would imply that the cosets I can build are:
$$
G/\mathcal{H} = \{\mathcal{H}, g_1 \mathcal{H}, g_2 \mathcal{H},..., g_{n_G-n_\mathcal{H}}\mathcal{H}\}
$$
Giving it a size of $n_G - n_\mathcal{H} + 1$.
I know i must be at fault on this thing since the fact that the size of the quotient group is $\dfrac{n_G}{n_\mathcal{H}}$ is stated pretty much everywhere in group theory books but i just can't seem to find where I am wrong about this.
Can you give me a simple proof that the size of $G/\mathcal{H}$ is $\dfrac{n_G}{n_\mathcal{H}}$?
AI: While it is true that $G=\{g_1,\ldots, g_{n_G-n_H}\}\cup H$, many of the elements $g_i$ give the same coset. In fact, you should prove that $g_iH=g_jH$ if and only if $g_j^{-1}g_i\in H$. Here's an outline for the proof of the fact that$|G/H|=n_G/n_H$:
First, one shows that all the cosets $gH$ have the same size. In particular, they all have size $|1\cdot H|=|H|=n_H$. Now show that $G$ is the disjoint union of the distinct cosets, so you get $$n_G=|G|=\left|\bigcup g_iH\right|.$$ But there are exactly $|G/H|$ cosets, so the right hand side is $|G/H|\cdot n_H$. Thus $n_G=|G/H|\cdot n_H$ which is what you're looking for.
|
H: Is there a real-analytic monotone function $f:(0,\infty) \to \mathbb{R}$ which vanishes at infinity, but whose derivative admits no limit?
A function $f:\mathbb{R} \to \mathbb{R}$ is called real-analytic if for each $x_0 \in \mathbb{R}$ there exists a neighbourhood of $x_0$ where $f$ is given by a convergent power series centred at $x_0$.
Problem: Is there a real-analytic monotone function $f:(0,\infty) \to \mathbb{R}$ which vanishes at infinity, but whose derivative admits no limit as $x \to \infty$?
We can note some weaker, but related, results. The (non-monotone) function $f(x)=x^{-1} \sin x^2$ is a real-analytic function on $(0, +\infty)$ and has the property that $\lim_{x \to +\infty} f(x) = 0$ but $\lim_{x \to + \infty} f'(x)$ fails to exist. It's not difficult to construct monotone examples if real-analyticity is weakened to merely being infinitely differentiable. The basic construction is straightforward. For each integer $n \geq 2$, and on each interval $[n, n+1-1/n^3]$, set $f(x)=1/n$, and on intervals $[ n+1-1/n^3, n+1]$ the function is linear, and decreasing from $\frac{1}{n}$ to $\frac{1}{n+1}$. This function is piecewise linear, and not smooth at the transition points, but it's trivial to smoothen this construction by utilizing appropriate variants of $\exp(1/x)$, rather than a linear interpolation. By the mean value theorem, we have that $\sup_{x \in [n+1-1/n^3, n+1]} |f'(x)| \geq \left|\frac{\frac{1}{n+1} - \frac{1}{n}}{\frac{1}{n^3}}\right|=\frac{n^3}{n(n+1)} \xrightarrow{n \to + \infty} + \infty$ hence $\lim f'(x)$ fails to exist.
However, I don't think one can use these ideas to obtain a real-analytic monotone function with the desired properties, since there's no real-analytic "transition" functions.
AI: What we need is a real-analytic non-negative and integrable $g$ that has no limit at $+\infty$. Then
$$f(x) = \int_x^{+\infty} g(t)\,dt$$
fits the bill.
Consider
$$g(x) = \biggl(\frac{2 + \cos x}{3}\biggr)^{6 x^5}\,.$$
It is evident that $g$ is strictly positive, real-analytic on $(0,+\infty)$, and has no limit as $x \to +\infty$. It remains to see that $g$ is integrable. For a positive integer $n$, consider the interval of length $\pi$ with midpoint $n\pi$. In this interval, for $\lvert x - n\pi\rvert \geqslant \frac{1}{n^2}$ we have
$$\lvert \cos x\rvert \leqslant \cos \bigl(n^{-2}\bigr) \leqslant 1 - \frac{1}{2n^4} + \frac{1}{24n^8} \leqslant 1 - \frac{1}{3n^4}$$
by Taylor expansion, and hence (using $\bigl(n - \frac{1}{2}\bigr)\pi > \frac{3}{2}n$)
$$g(x) \leqslant \biggl(1 - \frac{1}{9n^4}\biggr)^{9n^5} \leqslant \exp \bigl(-n\bigr)\,.$$
Hence the integral of $g$ over that interval is bounded by
$$\frac{2}{n^2} + \pi\cdot e^{-n}\,,$$
which is a summable sequence.
|
H: Set of distinct combinations of combinations
I recently solved a task on Stackoverflow. Please see here
Out of an array with 9 values, triplets consisting of 3 distinct values shall be generated. Calculating them is simple. Result is n choose k, 9 choose 3 = 84 triplets.
Next I needed to select 3 triplets, a triplet set, out of this set of 84 triplets where all values in the triplet set are also distinct. No double value anywhere.
Overall sets of triplets could again be calculated by n choose k, 84 choose 3 triplets = 95284 triplet sets. But, with the condition that all values in the triplet set should be distinct.
As a result I got 280 triplet groups (Via brute force).
Now, I do not know, if this number 280 is correct.
How can this value be calculated?
And how can the groups be calculated without brute force?
AI: After you choose $3$ from the $9$, choose another $3$ from the remaining $6$. This splits all of the numbers into the three groups you want.
However, we have also overcounted. We have counted every permutation of $3$ triplets, so we need to divide by $3!$ to only get the combination of triplets.
$$
\frac{1}{3!}\binom{9}{3}\binom{6}{3} = 280
$$
|
H: Separated morphisms are stable under base change
Suppose that the map $f$ in the following diagram is a separated morphism (i.e. $\Delta_{X/S}:X\rightarrow X\times_{S}X$ is a closed immersion). I want to prove that $p_{2}$ is also a separated morphism.
$$\require{AMScd}$$
\begin{CD}
X\times_{S}Y @>{p_{1}}>> X\\
@VV{p_{2}}V @VV{f}V\\
Y @>{h}>> S
\end{CD}
To prove that $p_{2}$ is also separated we have to show that the diagonal morphism $\Delta_{X\times_{S}Y/Y}: X\times_{S}Y\rightarrow (X\times_{S}Y)\times_{Y}(X\times_{S}Y)$ is a closed immersion.
My strategy was to construct a cartesian diagram containing the $\Delta_{X/S}$ and $\Delta_{X\times_{S}Y/X}$ and use the fact that closed immersions are stable under base change, i.e. if we have a cartesian diagram
$$\require{AMScd}$$
\begin{CD}
Z @>>> Y\\
@VVV @VVV\\
X @>>> S
\end{CD}
such that $X\rightarrow S$ is a closed immersion, then also $Z\rightarrow Y$ is a closed immersion.
Unfortunately I couldn't find such a cartesian diagram.
AI: I guess that your right horizontal $g$ in your first diagram is your $f: X \to S$ and your $p_2: X \times_S Y \to Y$ is the pullback of $f$ along horizontal $Y \to S$ (you called it also $f$ but in your first sentence you reserved $f$ for $X \to S$. Let call the horizontal arrow $Y \to S$ $h$. Clearly the diagram below is a pullback because $X \times_{ X \times_S X} (X\times_{S}Y \times_Y X\times_{S}Y) = X \times_{ X \times_S X} X \times_S X \times_S Y=X\times_{S}Y$ (use universal property of fiber product).
\begin{CD}
X\times_{S}Y @>{p_{1}}>> X\\
@VV\Delta_{}V @VV\Delta_{X/S}V\\
X\times_{S}Y \times_Y X\times_{S}Y @>{pr \times_h pr}>> X \times_S X
\end{CD}
Now closed immersions are preserved under base change and you assumed $\Delta_{X/S}:X\rightarrow X\times_{S}X$ be closed immersion.
|
H: How to find the number of elements of order $p$.
This is a follow up to this question.
Let $G = \langle x,y,z\mid{x^{{p^2}}} = {y^p} = {z^p} = 1,{x^y} = {x^{1+p}},[x,z] = [y,z] = 1\rangle$.
By euler phi function, the number of elements of order $p$ is $p-1$.
For each generator, the number of elements of order $p$ is in the following.
generator $x$ = $p-1$.
generator $y$ = $p-1$.
generator $z$ = $p-1$.
There are $3(p-1)$ elements of order $p$ from the generators.
But, how to find the rest of the elements of order $p$? Is it there are more simple way to find the number of elements of order $p$ in a finite group?
I hope someone can give me some idea regarding this question? Thank you
AI: $\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$First of all, note that $G$ is an (internal) direct product
$$
G = \Span{x, y} \times \Span{z}.
$$
Since $\Span{z}$ is cyclic of order $p$, you have only to find the number $N$ of elements of order dividing $p$ in $H = \Span{x, y}$. Then the total number of elements of order $p$ of $G$ will be $N p - 1$.
Now $N$ is different for $p = 2$ and for $p$ odd.
For $p = 2$, the group $H$ is dihedral of order $8$, and so you should know that it has $5$ elements of order $2$, so $N = 6$ here.
For $p > 2$, a direct computation yields
$$
\tag{power}
(x^{i} y^{j})^{p}
=
x^{i p} y^{j p}
=
x^{i p},
$$
so that $x^{i} y^{j}$ has order dividing $p$ iff $p$ divides $i$. Thus $N = p^{2}$ here.
You get (power) from the standard identity, valid in a group where commutators are central
$$
(a b)^{n} = a^{n} b^{n} [b, a]^{\binom{n}{2}}.
$$
|
H: The $y(\frac{∂z}{∂x})-x(\frac{∂z}{∂y})=0$ equation
The $y(\frac{∂z}{∂x})-x(\frac{∂z}{∂y})=0$ equation,
$u=x$
$v=x^2+y^2$
Write according to the new u, v variables given.
I don't know the question at all. Can you please help for solution?
AI: Taking $z=z(x, y)=z(x(u,v),y(u,v))$ can be obtained system
$$z_{u}^{'} = z_{x}^{'}x_{u}^{'}+z_{y}^{'}y_{u}^{'}$$
$$z_{v}^{'} = z_{x}^{'}x_{v}^{'}+z_{y}^{'}y_{v}^{'}$$
Knowing formulas for $x$ and $y$ you can solve from system $z_{x}^{'}$ and $z_{y}^{'}$
So
$$z_{x}^{'} = \dfrac{z_{u}^{'}y_{v}^{'} - z_{v}^{'}y_{u}^{'}}{x_{u}^{'}y_{v}^{'} - x_{v}^{'}y_{u}^{'}}$$
$$z_{y}^{'} = \dfrac{x_{u}^{'}z_{v}^{'} - x_{v}^{'}z_{u}^{'}}{x_{u}^{'}y_{v}^{'} - x_{v}^{'}y_{u}^{'}}$$
Then we can obtain
$$z_{x}^{'} =z_{u}^{'} +2u z_{v}^{'}$$
$$z_{y}^{'}=2y z_{v}^{'}$$
Finally your equation become
$$yz_{u}^{'}+2u(1-y)z_{v}^{'}$$
where $y$ is from system.
|
H: Explicit presentation of a group
$$\langle G\mid \{abc\mid a,b,c \in G, abc = 1\}\rangle$$ is a presentation of $G$.
Why did we specifically choose the relations to be of length 3?
(Why not something like $\{ab\mid a,b \in G, ab = 1\}$ or $\{abcd\mid a,b,c,d \in G,abcd = 1 \}$?)
(This example appears in https://www.math.ucla.edu/~azhou/notes/1/210ABC/notes.pdf, page 34)
AI: I found the notation confusing at first, hence my earlier comment, but once I saw the context, now I see the point.
The authors mean that any group $G$ has a presentation in which the set of generators is the whole set $G$, and the relations correspond to all the products. That is, for all $a, b \in G$, you write a relation $a b c = 1$, where $c = (a b)^{-1}$.
|
H: How big is $\{n\in\Bbb N\mid 1\leq n\leq 2000\text{ and the digital sum of }n^2=21\}$?
well,the title of the question makes it clear,
The question is :find the number of natural numbers between 1 to 2000 such that the sum of digits of their squares is equal to 21.
my approach:
just to make the question more clear I want to illustrate with help of an example .
$$89^2 = 7921$$and sum of digits of $7921$ is $7+9+2+1=19$
now since the squares of numbers from $1$ to $2000$
can be from $1$ to $7$ digits ,I could not find any way except calculating the squares by hand and adding their digits .so,is there a better method for solving this question? any help is greatly appreciated.
AI: The number is $0$.
Note that the sum of the digits of $n$ is congruent to $n$ mod $9$.
If $n$ is divisible by $3$ then $n^2$ is divisible by $9$, but $21$ is not. If $n$ is not divisible by $3$ then neither is $n^2$, but $21$ is. And we are done.
|
H: Assign the number of ways to share 16 identical objects
In how many different ways can be shared 16 identical objects to 7 different persons such that 3 of them can accept maximum of 2 objects, 3 of them at least 2 objects and for the other person don't have restriction.
I don'n know if I'm on the right way but I started like this:
f(x)= (1+x+x^2)^3 (x^2+x^3+...+x^16)^3 (1+x+x^2+x^3+...+x^16) then the number of ways is the coefficient of x^16.
AI: You can just give two to each of the three that needs at least two, then distribute the remaining $10$ to ease the restrictions. That removes the overall factor $x^6$ from your expression. Your approach is fine, but this makes the maximum exponent $10$ which simplifies things a bit. You can also extend the sums to infinity because powers above $10$ will not matter. Then you get $(1+x+x^2)^3(\frac 1{1-x})^4$ and you are looking for the coefficient of $x^{10}$. Alpha says the answer is $3483$. You may have to click on More Terms to see it.
|
H: Let $a$ and $b$ be elements of odd order in a finite group. Show that $a^2$ and $b^2$ commute if and only if $a$ and $b$ commute.
I really don't know how to solve this problem. I just know that if $|a|=2k_1+1$ and $|b|=2k_2+1$, then,
$a^{2k_1+1}=e=a^0$ and $b^{2k_2+1}=e=b^0$. Also, if $|G|=n$, then, $2k_1+1,2k_2+1\equiv 0 \pmod n$.
AI: You got it the other way round. It's $n \equiv 0 \pmod{2k_1 + 1}$ and $n \equiv 0 \pmod{2k_2 + 1}$.
Suppose $a^2b^2 = b^2a^2$. Then:
\begin{align*}
ab &= a^{2k_1+1}(ab)b^{2k_2+1} \\
&= a^{2k_1+2}b^{2k_2+2} \\
&= (a^2)^{k_1+1}(b^2)^{k_2+1} \\
&= (b^2)^{k_2+1}(a^2)^{k_1+1} \\
&= b^{2k_2+1}(ba)a^{2k_1+1} \\
&= ba
\end{align*}
|
H: Let $A$ be a normal matrix. Prove that if $|\lambda| = 1$ for all eigenvalues $\lambda$ of $A$ then $A$ is unitary.
Let $A$ be a normal matrix. Prove that if $|\lambda| = 1$ for all
eigenvalues $\lambda$ of $A$ then $A$ is unitary.
Normal is defined as : $A^*A = AA^*$.
I am unable to find a theorem that makes $|\lambda| =1$ useful in anyway. I have been trying to use Schur's theorem to make $U^*AU = T$ where the diagonal of $T$ will be eigenvalues of $A$.
Can someone please help out with this proof?
AI: Let $A x_i = \lambda x_i$, where $\lvert \lambda_i \rvert = 1$ and $\langle x_i, x_j \rangle = \delta_{ij}$ with $1 \leq i, j \leq n$. Write every vector $x \in \mathbb{C}^n$ as $x = \sum_{i = 1}^n \alpha_i x_i$, then it follows that
$$A^\ast A x = A^\ast \left ( \sum_{i = 1}^n \alpha_i A x_i \right ) = A^\ast \left (\sum_{i =1}^n \alpha_i \lambda_i x_i \right).$$
Since $A^\ast x_i = \bar{\lambda}x_i$ holds for all $i = 1, \dots, n$, this equation becomes
$$A^\ast Ax = \sum_{i = 1}^n \alpha_i \lambda_i A^\ast x_i = \sum_{i = 1}^n \alpha_i \lvert \lambda_i \rvert^2 x_i = \sum_{i = 1}^n \alpha_i x_i = x.$$
Thus $A^\ast Ax = x$ for every $x \in \mathbb{C}^n$ and hence $A^\ast A = I$.
|
H: Graph complexity
Can someone please explain briefly or direct me to some relevant tutorials/material what we maen by complexity of a graph('graph' of graph theory) .It seems there is no one definition and that is confusing me .I would be highly obliged for any help in this regard
AI: Posting as an answer since too long for a comment
I believe different people may have different answers to this question. From this paper,
"... complexity of a graph has been measured in several different ways. For example, the complexity of a graph has been defined to be the number of its spanning trees. It has been defined to be the value of a certain formula involving the number of vertices, edges, and proper paths in a graph. It has also been defined as the number of Boolean operations, based on a pre-determined set of Boolean operators (usually union and intersection), necessary to construct the graph from a fixed generating set of graphs".
The paper then goes on to define yet another method for measuring the complexity of a graph.
According to this article,
Computational complexity of graphs is the smallest number of union and intersection operations required to generate them when starting from simplest sets of edges: stars or cliques.
Another opinion is to use Kolmogorov Complexity to determine
how long is a minimal program to produce the graph?
I am sure there are several other points of view on this as well.
|
H: Combinations series: $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$
Evaluate $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$ for $n=10$.
Attempt: I'll deal with the case n being even, as we need to evaluate for n=10.
the numerator is
$${n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots$$
$$=\sum_{r=odd} {n \choose r}(n-r)^3$$(not sure if this is a correct notation).
$$=\sum_{r=odd}{n \choose n-r}r^3=\sum_{r=odd} {n \choose r}r^3$$(parity being same as n is even and r is odd, although I don't think this matters much).
Using the identity ${n \choose r}=\frac{n}{r} {n-1 \choose r-1}$ repeatedly in following steps,
$$=n\sum_{r=even} {n-1 \choose r-1}r^2$$
$$=[n(n-1)](1+\sum_{r=odd} {n-2 \choose r-2}[(r-2+3)+\frac{1}{r-1}]$$
$$=[n(n-1)](1+(n-2)\sum_{r=even}{n-3 \choose r-3}+3\sum_{r=odd}{n-2 \choose r-2}+\frac{1}{n-1} \sum_{r=even}{n-1 \choose r-1} -1)$$
$$=[n(n-1)]((n-2)\cdot 2^{n-4} +3\cdot 2^{n-4}+\frac{2^{n-2}}{n-1}$$
This simplfies to $n \cdot 2^{n-4} (n^2+7n-4)$.
Which is incorrect. The answer for $n=10$ (numerator/denominator is given as $\frac{1}{16}$).
Where am I going wrong?
Also the hint given for this problem was "expand $\frac{(e^x+1)^n - (e^x-1)^n}{2}$ in two different ways". I didn't quite understand this approach?
Could someone please explain this approach and any other approach also?
AI: Following the given hint, we have that
$$\begin{align}\sum_{r \text{ odd}} {10 \choose r}(10-r)^3&=\frac{1}{2}\left[\left((e^x+1)^{10}-(e^x-1)^{10}\right)'''\right]_{x=0}\\
&=\left[360(e^x+1)^7e^{3x}+135(e^x+1)^8e^{2x}+5(e^x+1)^9e^x\right.\\
&\quad \left.-360(e^x-1)^7e^{3x}-135(e^x-1)^8e^{2x}-5(e^x-1)^9e^x\right]_{x=0}\\
&=360\cdot 2^7+135\cdot 2^8+5\cdot 2^9.
\end{align}.$$
The same approach works for any integer $n\geq 4$:
$$\begin{align}\sum_{r \text{ odd}} {n \choose r}(n-r)^3
&=\frac{1}{2}\left[\left((e^x+1)^{n}-(e^x-1)^{n}\right)'''\right]_{x=0}\\
&=3\binom{n}{3}2^{n-3}+3\binom{n}{2}2^{n-2}+n2^{n-2}
=\frac{n^2(n+3)2^n}{16}.
\end{align}$$
P.S. We could also use the Taylor series of $e^x$: for $n\geq 4$
$$\begin{align}\frac{1}{2}\left[\left((e^x+1)^{n}-(e^x-1)^{n}\right)'''\right]_{x=0}
&=\frac{3!}{2}[x^3]\left((e^x+1)^{n}-(e^x-1)^{n}\right)\\
&=3[x^3]\left(2+x+\frac{x^2}{2}+\frac{x^3}{6}\right)^{n}\\
&=3[x^3]n2^{n-1}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)\\
&\quad +3[x^3]\binom{n}{2}2^{n-2}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)^2\\
&\quad +3[x^3]\binom{n}{3}2^{n-3}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)^3\\
&=n2^{n-2}+3\binom{n}{2}2^{n-2}+3\binom{n}{3}2^{n-3}\\
&=\frac{n^2(n+3)2^n}{16}.
\end{align}$$
|
H: Trigonometric equation with two variables
I want to find solutions of the following equation
\begin{equation}
4\cos(x) + 5\cos (x+y)=5
\end{equation}
and the best I’ve managed so far is to express $y$ in terms of $x$
$y= \pm\arccos\biggl(1-\frac{4\cos(x)}{5}\biggr) - x + 2\pi n$ with $n\in \mathbb{Z}$
Is there a way to simplify this further?
My thinking is that since I have a single equation but two unknown variables, I cannot define a finite set of solutions. But how can I define the range for which $x$ exists?
AI: As I suggested: substitute $z=x+y$ and divide by 5:
$$1-4/5\cos x=\cos z.$$
Since $|\cos z|\le 1$, you get $\cos x\ge 0$. This gives $$-\pi/2+2\pi k\le x\le \pi/2+2\pi k $$ for every integer $k$. Each such $x$ determines a $z$ and a $y$.
|
H: Singleton in a complete metric space is a complete metric space, but has no interior point. Baire?
I am confused by the above statement with the follwing version of Baire's category theorem:
If a non-empty complete metric space $(M,d)$ is the countable union of closed sets, then one of these closed sets has non-empty interior.
A singleton in a complete metric space is a complete metric space. By Baire's theorem the singleton should have an interior point, which I believe it does not.
AI: Yes, it has an interior point. If $p\in M$, then $p$ is an interior point of $\{p\}$, if we see $\{p\}$ as a metric subspace of $M$. Of course, in $M$, the set $\{p\}$ has, in general, no interior point (it has only if $p$ is an isolated point of $M$).
|
H: How does this show that this homomorphism is onto?
How does $\varphi (x,e') = x$ for all $x\in G$ show that this homomorphism is surjective? I'm not sure how this fits the definition of surjectivity.
AI: Surjective means every point in the target of a function lies in the image of the function. Hence picking any $x \in G$ you have to show it lies in the image of $\varphi$, which it does, since $\varphi(x,e‘)=x$...
|
H: Solution of systems of equations with increasing funcitons
For functions $f,g: I \to \mathbb{R}$ increasing on interval $I$, prove that for the solution of the system of equations
$$
\begin{cases}
\begin{align}
f(x_{1}) &= g(x_{2}) \\
f(x_{2}) &= g(x_{3}) \\
&\;\;\vdots \notag \\
f(x_{n}) &= g(x_{1})
\end{align}
\end{cases}
$$
$x_1 = x_2 = · · · = x_n$ is true.
I believe there might be a lemma for this, unfortunately, I can't find anything. Is there an elementary way to prove this?
AI: Suppose that $x_2\ge x_1$. Then $f(x_2)\ge f(x_1)$. Hence $g(x_3)\ge g(x_2)$, whence $x_3\ge x_2$. Hence $x_4\ge x_3$, and so on: $x_n\ge x_{n-1}$ and $x_1\ge x_n$, but that implies $x_1=x_2-...=x_n$.
Now suppose $x_2\le x_1$. Then similarly $x_3\le x_2$,..., $x_n\le x_{n-1}$ and $x_1\le x_n$, so again $x_1=x_2=...=x_n$.
|
H: what is the probability for a green marble to be drawn?
Bag A contains 3 green marbles and 2 red marbles.
Bag B contains 4 green marbles and 5 red marbles.
choose one bag randomly and pick one marble from it.
what is the probability for a green marble to be drawn?
AI: Assuming that you are equally likely to pick Bag A as Bag B, here is how I would approach this problem. There are two cases we must consider:
You randomly select Bag A and draw a green from it
You randomly select Bag B and draw a green from it
Since these events are mutually exclusive, we can just add up the probabilities of these cases.
$$
P(\text{Drawing a green from Bag A})=\frac{1}{2} \times \frac{3}{5} = \frac{3}{10} \\
P(\text{Drawing a green from Bag B})=\frac{1}{2} \times \frac{4}{9} = \frac{4}{18}
$$
Adding these two up gives us the answer:
$$
\frac{3}{10} + \frac{4}{18} = \frac{27}{90} + \frac{20}{90} = \frac{47}{90}
$$
|
H: How can I prove that $\left (\dfrac 1f \right)''(x) = -\dfrac {f''(x)}{f(x)^2} + 2\dfrac {f'(x)^2}{f(x)^3}$
I think its need some theorem what idk. Can someone help me?
$$\left (\dfrac 1f \right)''(x) = -\dfrac {f''(x)}{f(x)^2} + 2\dfrac {f'(x)^2}{f(x)^3}$$
AI: Note that:
$$(f^n)'=nf^{n-1}f'$$
And the derivative of a product of two functions is :
$$(fg)'=f'g+fg'$$
Now you can differentiate twice
$f^{-1}(x)$.
|
H: When is $f(x)$ reducible over $F$?
Suppose $f(x)$ has a factor $(x-a)$ in the field $F[X]$. Then, is $f(x)$ reducible over $F$? I know that we can write
$f(x) = (x-a)q(x)$ for some $q(x) \in F[x]$. Now, of course, deg$[(x-a)] = 1$ and deg$[f(x)] \geq 1$. This means that it is possible that deg$[f(x)] = 1$ (which would mean that deg$[q(x)] = 0$, which is fine). So, we cannot always express $f(x)$ as a product of two polynomials whose respective degrees are strictly less than the degree of $f(x)$.
The problem is that this argument seems to produce a very counter-intuitive result for me. Is the argument even correct?
Definition of a reducible polynomial:
A nonconstant polynomial $f(x) \in F[x]$ is reducible over $F$ if $f(x)$ can be expressed as a product $g(x)h(x)$ of two polynomials $g(x)$ and $h(x)$ in $F[x]$ both of lower degree than the degree of $f(x)$.
AI: As you've seen, the result is not true if $\deg f = 1$.
Consider the polynomial $f(x) = x$. It has a factor, namely, itself but it is irreducible.
In fact, your title says
If $f(x)$ has a factor in the field $F[X]$, then $f(x)$ is reducible over $F$?
Even that is not true. To see this, pick your favourite irreducible polynomial $f$ (not necessarily of degree $1$). Then, $f$ is a factor of $f$ but it is not reducible.
For example, you could choose $f(x) = x^2 + 1 \in \Bbb R[x]$.
What you do have is that if $f(x) \in F[x]$ has a linear factor in $F[x]$, then $f$ is reducible iff $\deg f \ge 2.$
A proof of this has been given in your other question here.
|
H: Shorter notation for sine function
TL;DR: Are there shorter, established notations for the sine function and other trigonometric functions than "$ \sin(x) $" etc.?
In a some homeworks and tests, there are exercises (e.g. differential equations or matrices) which contains sine and cosine, which I have to write very often - obviously, this becomes quite annoying.
My usual "workaround" is, that I define $ \sin(x) = s $ and $ \cos(x) = c $ to safe me from writing the rest and loose time unessecarily. As far as I can see, this is totally correct and can't be marked as wrong.
Yet this reduces the readability (and probably annoys the corrector). So are there any shorter notations for the trigonometric functions, which are established or at least sometimes used?
AI: In general, there is no such other notation. Your relabelling approach is certainly fine (except maybe if the work is being marked by someone unwilling to look through your custom symbol definitions). But I'd concentrate on whether your proofs are more verbose than necessary for other reasons. Writing in terms of $e^{ix}$ would shorten many proofs, especially if it's appropriate in that context to write e.g. $z=e^{ix}$.
In contexts where multiple angles are indexed viz. $\theta_i$, it's common to use notation such as $c_i:=\cos\theta_i$ etc. See e.g. the tabulated results in terms of proper Euler angles & Tait–Bryan angles here. Credit goes to @JohnDouma for this observation.
|
H: How do I find a slice of the area of a semi-circle?
A semi-circle around the origin, with radius $r$, is given by $$f(x) = \sqrt{r^2 - x^2}$$ The area of this semi-circle can be written as $$\int_{-r}^r \sqrt{r^2 - x^2} dx = \frac{\pi r^2}{2}$$
How do I find only a slice of this area, i.e. $$\int_a^b \sqrt{r^2 - x^2} dx$$ for $-r \le a < b \le r$?
There's probably a geometric approach to this, but I'm kinda stuck. If it's easier to calculate a slice of the area of a full circle that's also fine.
AI: Take $x = r\sin\theta$ so that $r^{2}-x^{2}$ becomes $r^{2}(1-\sin^{2}\theta) = r^{2}\cos^{2}\theta$ and $\sqrt{r^{2}-x^{2}}$ becomes $r\sin\theta$. Now, with this change of variable, one have $dx \to r\cos\theta d\theta$ and, thus:
$$\int_{a}^{b}\sqrt{r^{2}-x^{2}}dx = \int_{\sin^{-1}(a/r)}^{\sin^{-1}(b/r)}r^{2}\cos^{2}\theta d\theta = r^{2}\bigg{(}\frac{1}{2}\theta+\frac{1}{4}\sin2\theta\bigg{)}\bigg{|}_{\sin^{-1}(a/r)}^{\sin^{-1}(b/r)}$$
|
H: Tricking the CLT
I am looking for a sequence of i.i.d. RVs $X_{n}$ with mean $0$ and variance $1$, a standard normal RV $X$ and a set $A \subset \mathbb{R}$ such that for all $n$ we have $P(\frac{1}{\sqrt n}\sum_{k=1}^{n}X_{k} \in A) = 1$ but
$P(X \in A)=0$.
By Portmanteau, obviously we need $P(X \in \partial{A})>0$.
So my idea was to choose $A=\mathbb{Q}$ and each $X_{n}$ with values in the rationals, but I can´t find a specific construction that also ensures independence.
AI: Let $P(X_1 = 1)=P(X_1=-1)=1/2$ (this is the Rademacher distribution). This has mean $0$ and variance $1$. Now let
$$A:= \{a/\sqrt{b}: a\in \mathbb{Z}, b\in\mathbb{N}\}.$$
Then $A$ is countable, since $|A|\leq |\mathbb{Z}\times\mathbb{N}|$ and the latter is countable. It is clear that $\sum_{k=1}^nX_k \in \mathbb{Z}$ a.s. for any $n.$ Hence for any $n$,
$$P\left(\frac{1}{\sqrt{n}}\sum_{k=1}^n X_k\in A\right) = 1.$$
However, since $A$ is countable, $P(X\in A)=0$.
|
H: Giving a counterexample to $ 2^{n-1}- 1 = n \cdot a \iff n \text{ is prime}$
Fermat's little theorem asserts: $ n \text{ is prime} \implies 2^{n-1}- 1 = n \cdot a $.
However , the converse , $ 2^{n-1}- 1 = n \cdot a \implies n$ is prime, is not true . How can we prove it , taking odd $n$ (without using a computer)?
Edit:
I know that $341$ works , but how can I prove it's a counterexample without using a computer?
AI: As commented by hardmath, you are looking for pseudoprimes base $2$, and $341$ is a counterexample. To prove it without using a computer, note that $2^{5}=32=31+1\equiv1\bmod31$ and $2^5=32=33-1\equiv-1\bmod11$, so $2^{10}\equiv1\bmod31$ and $11$ and therefore $\bmod 341$, so $2^{340}\equiv1\bmod341$.
|
H: What's a good symbol to represent many summations? Can a tensor product glyph be used?
What's a good symbol for many repeated summations? I vaguely remember seeing something like
$$
\otimes_{j=1}^N \sum_{n_j=-\infty}^\infty f(\vec{n})= \sum_{n_1=-\infty}^\infty\sum_{n_2=-\infty}^\infty\cdots\sum_{n_N=-\infty}^\infty f(\vec{n}).$$
Is this used in practice? If not, is there a commonly accepted shorthand?
AI: You essentially want to sum over all vectors of length $N$ with integer entries. If we say $\mathcal{Z} = \otimes^N_{j=1} \mathbb{Z}$, then you can write the above sum compactly as:
$$\sum_{\vec{z} \in \mathcal{Z}} f(\vec{z})$$
|
H: Finding Coefficients via Generating Functions
I need to find the coefficient of $x^5$ in $\frac{2x}{1 - 3x}.$ I have found two different solutions, but I am not sure which is correct. Is the coefficient of $x^5$ given by $3^5 \cdot 2x$ under the interpretation $2x \cdot \frac{1}{1 - 3x},$ or is it given by $2 \cdot 3^4$ under the interpretation $2 \frac{x}{1-3x}?$
AI: Consider the formal power series $f(x) = \frac 1 {1 - x} = \sum_{n = 0}^\infty x^n.$ We want to find the coefficient of $x^5$ in $g(x) = \frac{2x}{1 - 3x},$ so we can use the formal power series of $f(x)$ to find the formal power series of $g(x).$ Explicitly, we have that $g(x) = 2x f(3x).$ Can you finish the solution from here?
|
H: Bayes Formula Problem Solution Check
The occurrence of a disease is $\frac{1}{100} = P(D)$
The false negative probability is $\frac{6}{100} = P(- | D)$, and the false positive is $\frac{3}{100} = P(+ | \neg D)$
Compute $P(D | +)$
By bayes formula, $P(+) = P(+ | D) P(D) + P(+ | \neg D) P(\neg D) = \frac{97}{10000} + \frac{297}{10000} = \frac{394}{10000}$
Similarly $P(D | +) = \frac{P(+ | D) P(D)}{P(+)} = \frac{97}{394} = 0.246$
Is this correct?
AI: no
$\mathbb{P}[D|T^+]=\frac{\mathbb{P}[T^+ \cap D]}{\mathbb{P}[T^+ \cap D]+\mathbb{P}[T^+ \cap \overline{D}]}=\frac{(100-6)\times1}{(100-6)\times1+3\times(100-1)}=24.04$
Notes
I expressed the data in % to avoid lot of decimals, so the result is 24.04%
All the data expressed in formulas, e.g $(100-1)$, mean that they are calculated and not given by the text
|
H: Calculus proof: $0=1$ What is my mistake?
The quotient rule states that:
$$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{{g(x)}^2}=\frac{f'(x)}{g(x)}-\frac{f(x)g'(x)}{{g(x)}^2}$$
Integrating tells us that
$$\frac{f(x)}{g(x)}=\int\frac{f'(x)}{g(x)}dx-\int\frac{f(x)g'(x)}{{g(x)}^2}dx$$
If I let $f(x)=g(x)$,this gives us:
$$\frac{f(x)}{f(x)}=\int\frac{f'(x)}{f(x)}dx-\int\frac{f(x)f'(x)}{{f(x)}^2}dx=\int\frac{f'(x)}{f(x)}dx-\int\frac{f'(x)}{f(x)}dx$$
Simplifying both sides:
$$1=0$$
Where have I erred? Where and what is my mistake? Any help would be appreciated greatly.
AI: Well, straight ahead letting $f=g$ means you're differentiating $1$ so you get $0$,
but let's go through this again:
$$\frac{d}{dx}\frac{f(x)}{f(x)}=\frac{f'(x)f(x)-f(x)f'(x)}{{f(x)}^2}=\frac{f'(x)}{f(x)}-\frac{f'(x)}{f(x)}$$
$$\implies 1= \int(\frac{f'(x)}{f(x)}-\frac{f'(x)}{f(x)}) dx=\int0 \ dx=C$$
So no contradiction.
Update: you did it in another way:
$$ 1= \int\frac{f'(x)}{f(x)}dx-\int\frac{f'(x)}{f(x)} dx$$
Now let $y=f(x)$, we get
$$\int \frac{f'(x)}{f(x)}dx= \int \frac{1}{y}dy=\ln(y)+C=\ln(f(x))+C$$
So
$$1=\ln(f(x))+C_1 - (\ln(f(x))+C_2)=C_1-C_2$$
again, no contradiction.
|
H: Conjecture: All but 21 non-square integers are the sum of a square and a prime
Update on 6/19/2020. This discussion led to deeper and deeper results on the topic. The last findings are described in my new post (including my two answers), here.
I came up with the following conjecture. All non-square integers $z$ can be represented as $z=x^2 + y$ where $x$ is an integer and $y$ is a prime. The exceptions are
z = 10, 34, 58, 85, 91, 130, 214, 226, 370, 526, 706, 730, 771, 1255, 1351, 1414, 1906, 2986, 3676, 9634, 21679.
Note that this is deeper than Goldbach conjecture (all even numbers are the sum of two primes) because squares are far rarer than primes. Also, few numbers are the sum of two squares, such numbers (sums of two squares) are far more abundant than primes, but their natural density is also zero. But all numbers are the sum of four squares. Surprisingly, all integers can be represented as $z = \lfloor x^c \rfloor +
\lfloor y^c \rfloor$ where $x, y$ are positive integers and $c < \log_{22} 63$ is a positive constant; but this fails at $c = \log_{22} 63$ as $z=73$ becomes an exception. See section 1 in this article for details; this is also a conjecture.
Question: Can you verify if my conjecture is true up to some very large $z$? I tested it only for $0\leq z < 750000$.
Heuristics behind this conjecture
This is by no mean a proof, but rather, I explain here why I think it could be true. Let denote as $r(z)$ the number of solutions to $x^2 +y \leq z$ where $x, y$ are integers and $y$ is prime. For a fixed large $z$, we want to count the number of integer couples $(x, w)$ below the curve $z=x^2+ w\log w$, with $x, w\geq 0$, in order to approximate $r(z)$. The choice of $w \log w$ is a direct consequence of the prime number theorem, replacing primes by their approximation, for large primes. That count $r(z)$ grows faster than $O(z)$. The derivative $dr(z)/dz$ thus grows faster than $O(1)$, and it shows how the number of solutions to $z=x^2+y$ grows on average, faster than $O(1)$ as $z$ increases.
More details about the heuristic approach
Essentially, we are trying to count the number of blue points under the red curve in the plot below (in this example, $z=100$). The equation for the curve is $w \log w = z-x^2$, and $z$ is assumed to be fixed.
The equation can be re-written as $w = (z-x^2)/W(z-x^2)$ where $W$ is the Lambert function, which behaves asymptotically like the $\log$ function. Thus the number of points below the red curve is asymptotically (for large values of $z$) equal to
$$r(z) \sim \int_0^\sqrt{z} \frac{z-x^2}{W(z-x^2)}dz \sim
\int_0^\sqrt{z} \frac{z-x^2}{\log(z-x^2)}dz = \frac{1}{2}\int_0^z \frac{u}{\sqrt{z-u}\cdot\log u}du.$$
Let us denote as $\phi(z)$ the function defined by the rightmost integral. We have $r(z) \sim \phi(z)$. I computed the exact values of $r(z)$ and $\phi(z)$ for various small and large $z$, and clearly, $r(z) \rightarrow C \cdot \phi(z)$, but I am not sure if $C=1$. See WolframAlpha computations here.
The number of solutions to $z=x^2+y$ (with $y$ prime) is thus, on average, as $z$ gets larger and larger, asymptotically equivalent to $d\phi(z) / dz$. Below is a table featuring $r(z)$ and $\phi(z)$.
Good asymptotic approximations for very large $z$ are
$$\phi(z)\approx\frac{2}{3}\cdot \frac{z^{3/2}}{\log z} \mbox{ and }
\frac{d\phi(z)}{dz}\approx \frac{\sqrt{z}}{\log z}.$$
The last result is compatible with the one posted in the answer by Dietrich Burde, confirming that the approach I used here is sound. Note that the same methodology could be applied to sums of squares or sums of primes or any sums of integers. It is pretty generic.
Final comment
The number of solutions to $z = x^2 + y$ (with $y$ prime, $x$ an integer) is equal to $r(z)-r(z-1)$. In all cases, $r(z)$ grows slowly (polynomial at most) and thus $r(z)-r(z-1) \sim dr(z)/dz$. We could get deeper results with second- and third-order approximations in all the asymptotic results used in this article, rather than just first-order approximations.
Below is a chart featuring the distribution for the number of solutions to $z=x^2+y$ [that is, the distribution of $r(z)-r(z-1)$] for $700000\leq z < 740000$.
For instance, there are $441$ different $z$'s between $z = 700000$ and $z = 740000$ for which $z=x^2 + y$ has exactly $50$ solutions. Below is the same chart, but for $100000\leq z < 140000$. The two distributions are strikingly similar in shap2.
Finally, among the first 750,000 $z$'s, we have:
$z = 78754$ is the last one to admit only one decomposition as $z =
x^2+y$
$z = 101794$ is the last one to admit exactly two decompositions
$z = 339634$ is the last one to admit exactly three decompositions
$z = 438166$ is the last one to admit exactly four decompositions
$z = 383839$ is the last one to admit exactly five decompositions
The $z$'s that admit only one decomposition are listed below. I searched for this sequence to see if it had been discovered, but could not find any reference.
z = 2, 5, 8, 13, 15, 22, 24, 26, 31, 37, 40, 46, 50, 55, 61, 70, 74, 76, 82, 94, 99, 106, 115, 120, 127, 133, 136, 142, 145, 154, 159, 166, 170, 178, 184, 202, 205, 219, 221, 235, 246, 250, 253, 265, 268, 274, 295, 298, 301, 310, 316, 319, 325, 328, 334, 340, 346, 379, 391, 394, 399, 412, 424, 436, 439, 442, 445, 469, 490, 505, 511, 559, 562, 571, 574, 586, 589, 610, 616, 646, 694, 781, 793, 799, 829, 834, 835, 874, 914, 922, 946, 949, 970, 979, 991, 994, 1030, 1045, 1066, 1090, 1105, 1164, 1204, 1219, 1243, 1324, 1354, 1366, 1384, 1411, 1450, 1501, 1549, 1555, 1642, 1717, 1726, 1765, 1786, 1810, 1885, 1981, 1990, 2041, 2059, 2074, 2146, 2167, 2245, 2266, 2284, 2344, 2410, 2416, 2479, 2650, 2806, 2821, 2854, 2899, 2926, 3004, 3094, 3151, 3166, 3184, 3319, 3418, 3502, 3811, 3859, 3865, 3964, 3991, 4216, 4222, 4279, 4330, 4414, 4504, 4510, 4645, 4654, 4711, 4930, 5482, 5506, 5545, 5986, 6031, 6049, 6274, 6439, 7009, 7081, 7441, 7549, 7954, 8086, 8584, 8824, 9214, 9571, 10165, 10774, 11509, 11806, 13834, 15106, 15334, 15565, 16081, 16186, 23851, 31879, 33205, 44536, 78754
AI: This is Hardy and Littlewood's Conjecture $H$. It says that this sequence $a(n)=
10,34,58,85,\ldots$ is finite
and that the number of representations of $n$ as the sum of a prime and a square is asymptotically
$$
\frac{\sqrt{n}}{\log (n)} \cdot \prod_{p > 2}\left( 1 - \frac{(n / p)}{p - 1}\right)$$
where $(n / p)$ is the Legendre symbol.
References: Mikawa, Narkiewicz, Wang
The conjecture is tested up to $10^{11}$ so far, i.e., it is known that $a(22) > 10^{11}$, if it exists.
|
H: Is the rank of a matrix equal to the number of non-zero eigenvalues?
I have studied before that the rank of a matrix = number of non zero Eigen values. But recently i came across a problem and i dont think it is valid there. I know i am going wrong somewhere.
$$A=
\begin{bmatrix}
0 & 4 & 0 \\
0 & 0 & 4\\
0 & 0 & 0 \\
\end{bmatrix}
\quad
$$
The Rank of this matrix is 2. So there should be 2 non zero eigen values. But I only get 0 as the eigen value(λ) using $$[A-λI]=0$$
Can anybody explain? Thanks
AI: rank of a matrix = number of non zero Eigen values
is not true, as you have witnessed.
Consider that $A^3=0$, so if $A$ has an eigenvalue $\lambda$ and $v\neq0$ is a corresponding eigenvector, then
$$
0=A^3v=\lambda^3v
$$
meaning $\lambda^3=0$, so $\lambda$ must be $0$.
The rank is, however, equal to the dimension of the image. Which is to say, the size of the largest possible set of linearly independent vectors of the form $Av$.
It is also the case that nilpotency (or more specifically the fact that the image may contain elements of the kernel) is in some sense the only thing that can go wrong with your statement.
|
H: Question about the ratio of two random variables
We have a couple $(X,Y)$ of random variables and its joint density function is:
$$p_{X,Y}(x,y)=\begin{cases} \dfrac{e^{-1}}{3(x-1)!} & \mbox{if }x \in \mathbb{N}\setminus \{0\},\ y \in \{-x,0,x\} \\ 0 & \mbox{otherwise}\end{cases}$$
Find the distribution of $W=Y/X$.
The first thing I noticed is that $W \in \{-1,0,1\}$. Then I calculated the marginal distributions:
$$p_X(x)=\sum _{y \in \{-x,0,x\}} p_{X,Y}(x,y)=3\dfrac{e^{-1}}{3(x-1)!}=\dfrac{1}{e(x-1)!} \\
p_Y(y)=\sum _{x=1}^{\infty} \dfrac{e^{-1}}{3(x-1)!}=\dfrac{1}{3e}\sum_{x=1}^{\infty} \dfrac{1}{(x-1)!}=\dfrac{1}{3e}\sum_{x=0}^{\infty}\dfrac{1}{x!}=\dfrac{1}{3e}e=\dfrac{1}{3}$$
And then I proceeded to calculate $p_W$ in its three values using the disintegration formula as it follows:
$$P(W=-1)=P(Y/X=-1)=P(Y=-X)=\sum_{x=1}^{\infty}P(Y=-x,X=x)P(X=x)\\
=\sum_{x=1}^{\infty} \dfrac{e^{-1}}{3(x-1)!}\dfrac{1}{e(x-1)!}=\dfrac{1}{3e^2}\sum_{x=0}^{\infty}\left(\dfrac{1}{x!}\right)^2$$
At this point I don't know what to do, it seems reasonable to me that $W$ is uniform (Also because of the fact that $P(W=0)=P(Y=0)=1/3$), and so my thought was that I was able to obtain an $e^2$ from the sum. I'm not able to understand if I did something wrong or not.
AI: I think you meant $P(Y = -X) = \sum_{x = 1}^{\infty} P(Y = -x \mid X = x) P(X = x)$ rather than $\sum_{x=1}^{\infty} P(Y = -x, X = x) P(X = x)$. This leads to
$$ P(Y = -X) = \sum_{x = 1}^{\infty} P(Y = -x, X = x) $$
which then gives $P(W = -1) = 1/3$ as expected.
|
H: Flat modules are torsion-free
I need to prove the following assertion:
Let $A$ be an integral domain. If $M$ is a flat $A$-module, then $M$ is torsion-free.
My definition of a flat module is: an $R$-module $F$ is flat if the functor $F \otimes_R \star \colon M \mapsto F \otimes_R M$ transforms exact sequences in exact sequences.
I am confused since I cannot find the relation between the definition and to be torsion-free.
Does anyone have some recommendation for me?
Thank you.
AI: Hint : let $r\in A$ and let $f: A\to A$ be the map "multiplication by $r$".
What happens if you tensor it with $M$ ?
|
H: Show for any monic polynomial $p(x)$ and for any $k$ that there are $k$ primes $q_i$ and $k$ integers that $n_i$ such that $q_i|p(n_i)$
Problem: Let $p(x)$ be a monic polynomial with integral coefficients, I want to show using induction for any integer $k$ that there exists $k$ distinct primes $q_1,\ldots,q_k$ and $k$ integers $n_1,\ldots, n_k$ such that $q_i\big|p(n_i)$.
My solution (after struggling with this problem for a couple of hours sigh) which is based on Fermat’s little theorem and Chinese remainder theorem is as follows:
For $k=1$ it is trivial so let’s assume the inductive hypothesis that we have $k$ prime and $k$ integers with the desired property. We want to show that there’s another prime-integer pair $(q_{k+1},n_{k+1})$ such that $q_{k+1}|p(n_{k+1})$.
Now notice that $p(x) \mod q_i$ is equivalent to a $\le \deg q_i-1$ polynomial by Fermat’s little theorem, and so for one of the equivalence classes of $q_i$, $p(x)\not=0\mod q_i$ (Since the number of solutions $\mod q_i$ can not exceed $\deg p(x)\mod q_i$. Hence, by the CRT there’s a positive integer $n$ that satisfies the desired equivalence classes,i.e, $p(n)\not=0\mod q_i$. Therefore, there must be a prime $q_{k+1}$ that divides $p(n)$.
I think this work but I could be wrong.However, I am not satisfied with this answer since it utilizes polynomials over finite fields which is out of the scope of the chapter I am working on (I may be wrong here as well). Any hints to solve this problem that’s different from my approach? Additionally, the polynomial’s monicity isn’t utilized which tells me that I made a mistake.
AI: Hint. For the $k \Rightarrow k+1$ step you may consider
if $p(0)=0$, then any $q_{k+1}\notin \{q_1,...,q_k\}$ will satisfy $q_{k+1}\mid p(q_{k+1})$. In this case $n_{k+1}=q_{k+1}$. Otherwise ...
$p\left(\color{red}{p(0)}\cdot\color{blue}{\prod\limits_{j=1}^k q_j}\right)=\color{red}{p(0)}\cdot\left(\color{green}{Q}\cdot\color{blue}{\prod\limits_{j=1}^k q_j}+1\right)$ ($Q$ some integer) and $Q\cdot\prod\limits_{j=1}^k q_j+1$ will either be prime or divisible by a prime $q_{k+1} \notin \{q_1,...,q_k\}$. In this case $n_{k+1}=p(0)\cdot\prod\limits_{j=1}^k q_j$
The latter is due to
$$p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \Rightarrow \\
p(\color{red}{a_0}\cdot \color{blue}{x})=a_n(a_0\cdot x)^n+a_{n-1}(a_0\cdot x)^{n-1}+...+a_1(a_0\cdot x)+a_0=\\
\color{red}{a_0}\left(\color{blue}{x}\cdot\color{green}{\left(a_n(a_0\cdot x)^{n-1}+...+a_2(a_0\cdot x)+a_1\right)}+1\right)$$
and of course $p(0)=a_0$
|
H: Proving that CDFs of maximum and $\frac{1}{k}$ sum are equal for $X_i \sim \text{Exp}(1)$.
Let $(X_i)_{i \in \mathbb{N}}$ i.i.d random variables such that $X_i \sim \text{Exp}(1)$.
Prove that CDF of $\max_{1 \leq k \leq n} X_k$ is identical to CDF of $\sum_{k=1}^{n} \frac{1}{k} X_k$, i.e $$\mathbb{P}\Big(X_1 + \dotsc + \frac{X_{n}}{n} \leq t\Big) = \mathbb{P}\big(\max_{1 \leq k \leq n} X_k \leq t\big).$$
I've showed that $\mathbb{P}\big(\max_{1 \leq k \leq n} X_k \leq t\big) = (1-e^{-t})^n$ proved the identity for $n = 2$. Then tried to show it by induction, but struggled with the final integral
$$\begin{align*}
\mathbb{P}\bigg(X_1 + \dotsc + \frac{X_{n+1}}{n+1} \leq t\bigg) &= \int_{x_{n+1}} \mathbb{P}\bigg(X_1 + \dotsc + \frac{X_{n+1}}{n+1} \leq t, X_{n+1} = (n+1)x_{n+1}\bigg) \ \text{d}x_{n+1} \\
&= \int_{0}^{t} \mathbb{P}\bigg(X_1 + \dotsc + \frac{X_{n}}{n} \leq t - x_{n+1} \bigg) e^{-(n+1)x_{n+1}} \ \text{d}x_{n+1} \\
&= \int_{0}^{t} \Big(1-e^{x_{n+1}- t}\Big)^n e^{-(n+1)x_{n+1}} \ \text{d}x_{n+1}
\end{align*}$$
I feel I did something wrong with marginalisation of $X_{n+1}$. Any help would be great!
AI: 1) if $X\sim Exp(1)$ it is very easy to verify that $\frac{X}{k}\sim Exp(k)$
2) the density of the $max(X_i)$ is the following
$$ \bbox[5px,border:2px solid black]
{
f_{max}(z)=n e^{-z}(1-e^{-z})^{n-1}
\qquad (1)
}
$$
Now it is very easy to check the density of the various sum with the convolution
n=2
$$f_Z(z)=\int_0^z e^{-x}2e^{-2(z-x)}dx=2e^{-z}(1-e^{-z})$$
n=3
$$f_Z(z)=\int_0^z 2e^{-x}(1-e^{-x})3e^{-3(z-x)}dx=3e^{-z}(1-e^{-z})^2$$
and so on...
this prove what requested: as you can see the densities are equivalent to the general formula (1)
|
H: Proving $Ext_{\mathbb{Z}}^1(A,B)$ is a torsion-free abelian group, given that $A$ is divisible and $B$ is torsion-free
I have been trying to prove the following:
Let $A,B$ be abelian groups. If $A$ is divisible and $B$ is torsion-free, then the group $Ext_{\mathbb{Z}}^1(A,B)$ is torsion-free.
So, I study homological algebra and in my class we have come as far as to define the $Ext$ functor. I have tried proving this result using long exact sequences, since I Know that $Ext_{\mathbb{Z}}^i(A,B)=0$ for all $i \geq 2$ and all abelian groups $A,B$. After that, I am completely stumped. I should also mention that in my class we did not define the $Tor$ functor and we defined the $Ext$ using projective resolutions.
I tried looking in the following books:
Basic Homological Algebra by M. Scott Osborne
An Introduction to Homological Algebra by Joseph J. Rotman
to no avail. Of course, it is possible that the answer is in one of these books and I may have missed it.
AI: Since $A$ is divisible, the map $n:A\to A$ is surjective for any nonzero integer $n$. There is thus a short exact sequence $$0\to K\to A\stackrel{n}\to A\to 0$$ which gives a long exact sequence $$0\to \operatorname{Hom}(A,B)\stackrel{n}\to\operatorname{Hom}(A,B)\to \operatorname{Hom}(K,B)\to \operatorname{Ext}^1(A,B) \stackrel{n}\to \operatorname{Ext}^1(A,B)\to \operatorname{Ext}^1(K,B)\to 0.$$
But since $K$ is the kernel of multiplication by $n$, it is $n$-torsion, so $\operatorname{Hom}(K,B)=0$ since $B$ is torsion-free. Exactness of the sequence above now says that $n:\operatorname{Ext}^1(A,B)\to \operatorname{Ext}^1(A,B)$ is injective. Since $n$ is an arbitrary nonzero integer, this says that $\operatorname{Ext}^1(A,B)$ is torsion-free.
|
H: MLE and log MLE
I'm getting confused on likelihoods and the use of the log function.
negloglik = negative log likelihood
This statement on negloglik seems to be wrong to me:
"
negloglik is an exponential scale, therefore small changes in negloglik represent a very large change in likelihood.
"
Can anyone explain why it's true? The opposite seems to be the case if you ask me..
AI: Let $L(x)$ denote the likelihood function and $N(x) = -\ln L(x)$ be the negative log of the likelihood function.
What happens to $L(x)$ if $N(x)$ is changed to $N(x) + \epsilon$? We get $L(x) = e^{-N(x)}$ so if $N'(x) = N(x) + \epsilon$ we get
$$
L'(x)
= e^{-N'(x)} = e^{-N(x)-\epsilon}
= \frac{L(x)}{e^\epsilon}.
$$
So for example
if $\epsilon = 0.1$ then $L'(x) \approx 0.9 L(x)$,
and if $\epsilon = 0.001$ then $L'(x) \approx 0.999 L(x)$,
but if $\epsilon = 1$ then $L'(x) \approx 0.36 L(x)$.
Not sure if this is big enough or not...
UPDATE
Your specific example, if $L(x)$ goes from $0.1$ to $0.3$, then $N(x)$ goes from $2.3$ to $1.2$, more than 4 times the size of change in $L(x)$.
|
H: How to calculate the volume between two cylinders
I am asked to calculate the volume between the two cylinder : $x^2+y^2=1$ and $x^2+z^2=1$
So my assumption here that the limit of $y$ and $z$ must be equal, and we are looking on the unit circle so the integral needs to be:
$$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} 1 dx\,dy\,dz$$
but the answer is just one and this is doesn't make sense (intuitively) because the radius it self is 1 and what I imagine in my head is that the marked domain is the line between $x=1$ and $x=-1$
AI: I've attached a plot of the problem shown in the $xy$-plane. The red circle is a circular cross section of the cylinder $x^2+y^2=1$ where $z=0$, and the blue rectangle is a rectangular cross section of the cylinder $x^2+z^2=1$ where $z=0$ as well. We can see here then that we might choose to integrate from $z=-1$ to $z=1$, and we can determine the bounds of the $x$ and $y$ integrals by using the boundary conditions for the cylinders:
$$
x^2+z^2=1 \\
x=\pm \sqrt{1-z^2}
$$
and
$$
x^2+y^2=1\\
y=\pm\sqrt{1-x^2}
$$
So, the integral is:
$$
\int_{-1}^{1}\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\text{d}y\text{d}x\text{d}z
$$
|
H: Showing triangle inequality in this metric.
I'm struggling to show that the following function on $\mathbb{Z}$ is a metric: specifically, showing the triangle inequality.
Fix an odd prime $p,$ and define $d:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{R}$ by $$d(m,n)=0 {\text{ if }} m-n=0,$$ or as $$d(m,n)=\frac{1}{r+1} {\text{ if }} p^r {\text{ is the largest power of }} p {\text{ which divides }} m-n.$$ In my attempt to show $d(x,z)\leq d(x,y)+d(y,z),$ I took distinct $x,y,z\in\mathbb{Z}$ and supposed that $p^t$ was the largest power of $p$ which divided $x-z,$ so $d(x,z)=\frac{1}{t+1}.$ I tried to write $$p^t\hspace{0.1cm}|\hspace{0.1cm}x-z\implies p^t\hspace{0.1cm}|\hspace{0.1cm}(x-y)+(y-z),$$but this does not necessarily imply that $p^t\hspace{0.1cm}|\hspace{0.1cm}x-y$ and $p^t\hspace{0.1cm}|\hspace{0.1cm}y-z.$
I understand that writing $x-z=(x-y)+(y-z)$ makes the two individual differences smaller and thus a smaller power of $p$ can divide them, resulting in their distance $\frac{1}{r+1}$ to be larger, but I am having trouble making this explicit.
Am I thinking about this the wrong way? Any guidance is appreciated.
Thank you.
AI: I think you're thinking about it wrong. It's not about the size of the difference $n - m$, it's about the highest power of $p$ which divides it. You're right to consider the decomposition
$$ x - z = (x - y) + (y - z).$$
The key observation is if $p^r$ divides both $x - y$ and $y - z$ then $p^r$ divides $x - z$. In particular take $p^r$ to be the highest power of $p$ such that $p^r$ divides both $x - y$ and $y - z$. Can you show that
$$ d(x, z) \leq \frac{1}{r + 1} \leq d(x, y) + d(y, z) ? $$
|
H: Whats the joint PDF of Z=XY given the a joint pdf f(x,y)?
Let $X$ and $Y$ be random variables with joint pdf:
$$f(x,y)=x + y \quad \text{if } x \ge 0, y \le 1$$
Let $Z=XY$. Calculate the pdf of $Z$.
I'm a bit confused about solving this problem, I'm trying to get to the pdf by calculating the cdf to derive it afterwards, so I know that the cdf of $Z$ would be something like this:
$$F(XY \le z) = \iint(x+y) \,dydx.$$
But I'm not so sure how to would the limits of the definite would be...Im guessing it's:
$$F(XY \le z)= \int_0^\infty\int_{-\infty}^{z/x}(x+y) \,dy dx.$$
But this integral's result is divergent, so I know something is wrong but I'm a bit lost there. Is there a better approach on solving it? Any thoughts?
I appreciate any help!
AI: So indeed, if $Z = XY$, let $\mathbb{I}(A)$ denote the indicator of the event $A$ (i.e. $\mathbb{I}(A)=1$ if $A$ is true and is $0$ otherwise). You have
$$
\begin{split}
F_Z(z)
&= \mathbb{P}[Z \le z] \\
&= \mathbb{P}[XY \le z] \\
&= \int_{-\infty}^\infty
\int_{-\infty}^\infty
\mathbb{I}(xy \le z) f_{X,Y}(x,y)\ dxdy \\
&= \int_{x=-\infty}^{x=0}
\int_{y=z/x}^{y=\infty} f(x,y) dy dx
+ \int_{x=0}^{x=\infty}
\int_{y=-\infty}^{y=z/x} f(x,y) dy dx
\end{split}
$$
UPDATE
Sorry, I missed that you gave the definition for $f(x,y) = x+y$ for $x \ge 0$ and $y \le 1$. I don't understand how this is a valid pdf -- you must have
$$
1 = \int_{x=0}^{x=\infty} \int_{y = -\infty}^{y=1} (x+y)dxdy
$$
but the RHS integral diverges...
UPDATE 2
I think the intent is to have $f(x,y)=x+y$ for $0 \le x,y \le 1$, which means $0 \le x \le 1$ and $0 \le y \le 1$. Indeed,
$$
\begin{split}
\int_0^1 \int_0^1 (x+y)dxdy
&= \int_0^1 \left[y + \left(\int_0^1 x dx\right) \right]dy\\
&= \int_0^1 \left[y + \frac12 \right]dy \\
&= \int_0^1 y dy + \frac12 \\
&= \frac12 + \frac12 \\
&= 1.
\end{split}
$$
Then,
$$
\begin{split}
F_Z(z)
&= \int_{x=0}^{x=1}
\int_{y=0}^{y=\min\{z/x,1\}} (x+y) dy dx
\end{split}
$$
Can you now finish?
|
H: Finding the probability transition matrix
Find $P_t$ when Q =
\begin{bmatrix}
-\alpha & \beta \\
\alpha & -\beta
\end{bmatrix}
and $\alpha, \beta > 0$ satisfying the backwards equation
AI: You have to find exp(tQ) to solve this problem.
|
H: Evaluating $\int_{0}^{\infty} (\frac{\sin x}{x})^2 dx$ using complex analysis
I need to calculate $\displaystyle\int_{0}^{\infty} \left(\frac{\sin x}{x}\right)^2dx$.
I have started with defining:
$$f(z) = \frac{1-e^{2iz}}{z^2},\quad z\in\mathbb{C}\;.
$$
Then divided it into four contour integrals, just standard stuff:
$\int_{-R}^{-r} \frac{1-e^{2ix}}{x^2} dx + \int_{r}^{R} \frac{1-e^{2ix}}{x^2} dx + \int_{C_r }^{} \frac{1-e^{2iz}}{z^2} dz+\int_{C_R }^{} \frac{1-e^{2iz}}{z^2} dz$
First two integrals add up to $2\int_{r}^{R} \frac{1-\cos(2x)}{x^2} dx$
Third integral, Laurent series for $\frac{1-e^{2iz}}{z^2} = \frac{-2i}{z} +2 + \frac{4iz}{3} - \frac{2z^2}{3} +... = \frac{-2i}{z} + P(z)$ and $\int_{C_r} P(z) = 0$ as $r \to 0$, so:
$\int \frac{-2i}{z}dz=-\int_0^\pi \frac{2i^2re^({ti})}{re^{ti}}dt = 2\pi$ for $z = re^{ti}$
$\int_{C_R }^{} \frac{1-e^{2iz}}{z^2} dz \leq \frac{2}{R} +\frac{|e^{2xi}|e^{-2y}}{|z^2|} \leq \frac{2+\pi}{R}= 0$ as $R \to \infty$
$2\int_{0}^{\infty} \frac{1-\cos(2x)}{x^2} dx + 2\pi = 4\pi$ so my integral is equal to $\pi$, whereas it should be $\frac{\pi}{2}$. Where did I go wrong?
AI: $$1-\cos{2x}=2\sin^2{x}$$
You're missing the $\frac{1}{2}$ when you converted to exponential form.
I believe another mistake you made is that the bounds for the line integral of the semi-circle with little radius r should be $\pi$ to $0$ because it should be in the clockwise direction. You would get $-2\pi$ from that instead of $2 \pi$. Then, you should use Cauchy's first theorem instead of second because the semi-circle of little radius r avoids the singularity:
$$ 2\int_0^{\infty} \frac{1-\cos{2x}}{x^2} -2\pi=0$$
$$\int_0^{\infty} \frac{\sin^2{x}}{x^2} \; dx= \boxed{\frac{\pi}{2}}$$
|
H: Reasoning with congruences: does a positive integer $x$ exist with the following properties?
Question:
Find $x$ such that the following properties are true or prove that no such $x$ exists.
Let:
$x>0$ be an integer
$p_1, p_2, p_3$ be distinct odd primes
$1 \le a < p_3$ is an integer
Find $x$ with the following properties:
$x \equiv p_2 \pmod {p_3}$
$x \equiv p_3 \pmod {p_1}$
$x = p_1 + ap_2$
Here are my observations:
For any specific $p_1, p_2, p_3$, the problem is straight forward. Use the Chinese Remainder Theorem to solve for:
$x \equiv p_1 \pmod {p_2}$
$x \equiv p_2 \pmod {p_3}$
$x \equiv p_3 \pmod {p_1}$
Then see if the minimal $x$ has the form $p_1 + ap_2$. In all the cases I am testing, $x > p_1 + (p_3 - 1)p_2$ in which case no such $a$ exists.
For example:
$x \equiv 3 \pmod 5$
$x \equiv 5 \pmod 7$
$x \equiv 7 \pmod 3$
The minimal $x$ is $103 > 3 + 5\times 7 = 38$
So, in this case, no such $a$ exists.
I am having troubling proving that no solution exists and I am not able to find example where it is true.
AI: You don't mention any specific ordering requirements among the primes, so consider $p_1 = 7$, $p_2 = 5$ and $p_3 = 3$. You then get
$$x \equiv p_1 \pmod{p_2} \implies x \equiv 7 \pmod{5} \tag{1}\label{eq1A}$$
$$x \equiv p_2 \pmod{p_3} \implies x \equiv 5 \pmod{3} \tag{2}\label{eq2A}$$
$$x \equiv p_3 \pmod{p_1} \implies x \equiv 3 \pmod{7} \tag{3}\label{eq3A}$$
You can easily confirm that $x = 17$ satisfies the $3$ equations above, and it's the smallest positive integer that does. Also,
$$p_1 + (p_3 - 1)p_2 = 7 + (3 - 1)5 = 17 = x \tag{4}\label{eq4A}$$
so $a = p_3 - 1 = 2$ in this case satisfies your requirements.
I haven't checked, but I'm fairly certain counter-examples will exist even if you impose an ordering requirement on the primes, e.g., $x_1 \lt x_2 \lt x_3$.
|
H: Prove that $-X$ is measurable with respect to some sigma field.
I am reading A Second Course in Probability by Ross and Peköz. I came across the following question:
1.10.4. Show that if $X$ and $Y$ are real-valued random variables measurable with respect to some given sigma field, then so is $XY$ with respect to the same sigma field.
My attempt was to first show that:
(i) $X+Y$ is measurable
(ii) $cX$ is measurable for any $c\in\mathbb R$
(iii) $X^2$ is measurable.
Then, since $XY = \frac14[(X+Y)^2-(X-Y)^2]$, we just use the above properties.
Property (i) was proven in Chapter 1.
However, I am having a bit of trouble showing property (ii) in the case that $c < 0$. From the problem setup, we know that for any $x\in\mathbb R$, $\{X\leq x\}\equiv\{\omega\in\Omega:X(\omega)\leq x\}\in \mathcal F$.
So $\{-X\leq x\}=\{X\geq -x\} = \{X<-x\}^c$, but I don't know if $\{X<-x\}\in\mathcal F$. If I expand a little more, it seems I essentially need to show that $\{X=x\}\in\mathcal F$, but I don't see how that is necessarily true.
I have seen the question Product of two random variables, though in the answers, they assume knowledge of the fact that $cX$ is measurable, and they also seem to be working with the Borel $\sigma$-algebra.
I have also seen these notes, but they omit the case where $c < 0$.
My Question: How do we show that $\{-X\leq x\}\in\mathcal F$ if we already know that $\{X\leq x\}\in\mathcal F$? Do we need to assume something about the sigma field in order to show this for real-valued random variables (e.g. Borel sigma field)?
AI: Note that $\{X<x\}=\cup_{n=1}^\infty \{X\leq x-\frac{1}{n}\}$. Since a Sigma field is closed under countable unions this implies $\{X<x\}\in\mathcal{F}$ for every $x\in\mathbb{R}$.
A general note: If $X:\Omega\to\mathbb{R}$ is a random variable then we have $X^{-1}(B)\in\mathcal{F}$ for every Borel set $B\subseteq\mathbb{R}$. In many books this is actually the definition of a random variable. The definition you are using, that is $\{X\leq x\}\in\mathcal{F}$ for every $x\in\mathbb{R}$, is an equivalent definition. It is good to know both definitions. To prove the equivalence note that the collection of sets $\{B\subseteq\mathbb{R}: X^{-1}(B)\in\mathcal{F}\}$ is a Sigma field on $\mathbb{R}$, and from your definition it is not too hard to show it contains all open subsets of $\mathbb{R}$, hence must contain the whole Borel sigma field.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.