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H: Chain rule method doesn't result with same answer as u-sub. Why?
$\int \ln \left(2x\right)\,\mathrm{d}x$ is the integral in question.
I know how to solve it with the chain rule. $\frac{1}{2x}\times 2x = \frac{1}{x}$
But, because I know $u$-sub method, I wish to use it upon this integral.
$\int \ln \left(2x\right)\,\mathrm{d}x$
$u = 2x$
$du = 2dx \Rightarrow (1/2)du = dx$
Thus tranforming the integral into: $(1/2)\int \ln u \, \mathrm{d}u$
Which gives us $\frac{1}{2u}$ and undoing the sub results in $\frac{1}{4x}$
Now, this can not be, thus I am at fault and I do not see how. Please, help explain where I am wrong.
AI: We usually integrate something like $\int \ln(2x) \, dx$ by parts, as it is quite efficient.
We use the formula $\int u \,dv=uv-\int v\,du$.
Let
\begin{align*}
u &=\ln(2x) \quad\quad v=x \\
du &=\frac{1}{x}\, dx \,\,\,\,\,\,\,\,\,\, dv=dx
\end{align*}
Then
\begin{align*}
&uv-\int v\,du \\
=&x\ln(2x)-\int 1 \, dx \\
=&x\ln(2x)-x+c.
\end{align*}
If you were to put yourself in a situation where, after a $u$ substitution, all you had to integrate was $\ln(u)$, you would probably integrate that by parts as well (and then back substitute to the same result if done correctly), thus diminishing your efforts to demonstrate an alternate method. |
H: what is $X / \cong $ ?? where $\cong $ is given by
what is $X / \cong $ ?? Suppose $X = \mathbb{R}^2 $. and we define
$$ (x_1,y_1) \cong (x_2, y_2) \iff x_1 + y_1 =x_2 + y_2 $$
With this equivalence relation, we get that the partition is the trivial one since every point is congruent to itself and hence $X$ is partitioned trivially. So we get the same space for the quoetient space.
But if we define $$ (x_1,y_1) \cong (x_2, y_2) \iff x^2_1 + y^2_1 =x^2_2 + y^2_2 $$
then $X / \cong$ is half plane.
Are these correct? CAn someone give me some feedback. thanks
AI: Expanding on Will Jagy's answer, notice that $(x,y)~(x',y')$ iff $\frac{y'-y}{x'-x}=-1$ (since $\frac{y'-y}{x'-x}=\frac{(c-x')-(c-x)}{x'-x}=-1$ , so that two points are equivalent iff they are on the same line with slope =-1 . You can choose, as a representative of each line, its point of intersection with the $x-$ axis, and get an equivalence ( a bijection) with $\mathbb R$
Similarly, for the second case, two points are equivalent if they are on the same circle centered at the origin. Choose, e.g., the intersection of each circle with the positive x-axis, and you get a bijection with {$x:x $ in $\mathbb R: x\geq 0$} |
H: Why is there Inequality in Fatou's Lemma?
I'm studying measure theory for the first time, and I just came across Fatou's Lemma.
Why isn't it true that for any sequence of functions $\left\{ f_n \right\}$ in $L^+$ we always have that $$\int \displaystyle \liminf_{n\rightarrow \infty} f_n d\mu =\liminf_{n\rightarrow \infty} \int f_n d\mu\ ?$$
AI: consider the sequence of functions$$f_n = n 1_{(0,\frac{1}{n}]} $$
Notice
$$ \int f_n = 1 \implies \lim \int f_n = 1 \implies \liminf \int f_n = 1$$
But, $\lim f_n = 0 \implies \liminf f_n = 0 \implies \int (\liminf f_n ) = 0 $ |
H: Pigeonhole principle problem involving inequality 0 < |$\sqrt{x} - \sqrt{y}$| < 1
21 integers are selected from {1, 2, 3, ..., 400}. Prove that two of them, say x and y, satisfy 0 < |$\sqrt{x} - \sqrt{y}$| < 1.
I am confident you have to use and apply the Pigeon Hole Principle. From what I gathered, there are 400 numbers in the set and $\sqrt{400} = 20$. The minimal difference is obtained by looking at consecutive integers. I don't know where to go from here.
AI: Divide the 400 integers into 20 groups $g_1\ldots g_{20}$, where $n\in g_i$ if $i\le\sqrt n\lt i+1$. That is: $$\begin{align}
g_1 & = \{\mathbf{1}, 2, 3\} \\
g_2 & = \{\mathbf{4}, 5, 6, 7, 8\} \\
g_3 & = \{\mathbf{9}, 10, \ldots, 15\} \\
& \;\vdots \\
g_{19} & = \{\mathbf{361}, 362, \ldots, 399\} \\
g_{20} & = \{\mathbf{400}\}
\end{align}
$$
Of the 21 pigeons, two must be in the same group $g_i$. |
H: Pigeonhole Principle / Number Theory
Let $S$ be a subset of $A=\{1,2,3,...,1000\}$. Find the largest number of elements in $S$ such that for any $a, b \in S$ with $a>b$, $a-b$ does not divide $a+b$.
I've tried numerous approaches, even brute force (listing), but to no avail. Any help would be greatly appreciated :)
AI: Hint. $S$ cannot contain consecutive numbers, because if $a-b=1$ then $a-b$ divides anything. $S$ cannot contain numbers 2 units apart, because if $a-b=2$ then $a$ and $b$ have the same parity, so $a+b$ is even, hence $a-b$ divides $a+b$.
Now, can $S$ contain numbers 3 units apart? |
H: Why aren't multi valued functions invertible?
I recently learnt that functions are invertible if and only if they are bijective. But why aren't multi-valued surjective 'functions' invertible?
AI: Noone would kill you if you wrote $f(x) = x^2$ and then $f^{-1}(4) = \{2, -2\}$ (You kind-of define your inverse as a set-valued function). But, forget about its linguistic content, "Invertible" is a technical term reserved for a specific purpose to signify a bijective function. You may say hello when leaving, and goodbye while coming, but that won't help much, will it? |
H: Proving a map is an automorphism
Let $G$ be a group; for $g\in G$ define $T_g:G\to G$ by $xT_g = g^{-1}xg$ for all $x\in G.$ Prove that $T_g$ is an automorphism of $G$.
Since $T_g$ is onto then for $y\in G$ let $x = gxg^{-1}$ then $xT_g = g^{-1}(x)g = g^{-1}(gyg^{-1})g = y$. Also, one to one is easy to see but how can I show $T_g$ is a homomorphism? If I have $x,y\in G$ then $(xy)T_g = g^{-1}(xy)g \ = \ ... \ = (xT_g)(yT_g)$.
How can I complete the $... \ $?
AI: $$(xT_g)(yT_g) = \Big(g^{-1} x g\Big)\Big(g^{-1} y g\Big) = g^{-1} x g g^{-1} y g = ... ?$$ |
H: This operator is invertible
I'm thinking if the self-adjoint operators are invertible. I'm really stuck, I don't know even how to begin,I need a hint or or a counter-example if it's not true.
Thanks
AI: The constant zero operator is self-adjoint, but clearly not invertible.
EDIT:
More generally, for an operator on finite dimensional real vector space represented in some bases by a matrix A it will be self-adjoint if the matrix is symmetric (in the complex case, if the matrix is Hermitian) and it will be invertible if the matrix is invertible. It is very easy to construct symmetric non-zero matrices that are not invertible, for example diagonal matrices that have at least one zero on the diagonal. |
H: Find $ \int_{0}^{3} \left[- \frac {x^2}{4} - 8\right] \,dx$
Consider the following definite integral
$$A = \int_{0}^{3} \left[- \frac {x^2}{4} - 8\right] \,dx$$
Calculate the Riemann sum that approximates the value of A as a closed-form formula in n (i.e. remove the Sigma using the necessary formulas). Use right-handed endpoints.
Next, find the actual value of A by taking a limit of your formula.
No idea where to start.
AI: If you split your interval $[a,b]$ into $n$ even sub-intervals, then we know that the width of each sub-interval is $\Delta x = \dfrac{b-a}{n}$. Let $a=x_0<x_1<\ldots<x_n=b$ evenly partition the interval $[a,b]$ (by evenly partition, I mean $x_{i+1}-x_i=\Delta x$). Then it follows that $x_1=a+\Delta x$, $x_2=a+2\Delta x$, and hence $x_i=a+i\Delta x$ for $1\leq i\leq n$. Since we're after a right Riemann sum, we take the right endpoint to be $x_k^{\ast}=x_k=a+k\Delta$ (see remark at bottom). We can now see that the area of each rectangle is $f(x_k^{\ast})\Delta x$. Thus the definite integral is given by the following right Riemann sum:
$$\int_a^b f(x)\,dx =\lim_{n\to\infty}\sum_{k=1}^nf(x_k^{\ast})\Delta x = \lim_{n\to\infty}\frac{b-a}{n}\sum_{k=1}^n f\left(a+\frac{k}{n}(b-a)\right)$$
In the particular problem that you have, we can easily see that $f(x)=-\frac{1}{4}x^2-8$ and that we're computing the integral over the interval $[0,3]$. Hence $\Delta x=\dfrac{3}{n}$ and thus your first step is to find the closed form of the partial sum
$$S_n = \sum_{k=1}^n f(x_k^{\ast})\Delta x = \frac{3}{n}\sum_{k=1}^n\left( -\frac{1}{4}\left(\frac{3k}{n}\right)^2 - 8\right)=\ldots$$
I'll leave the rest of the simplification of this sum to you.
Once you finally have the closed form of the above partial sum, take the limit as $n\to\infty$ to get the value of the definite integral you were originally given.
Remark: Depending on the type of Riemann sum one wants to evaluate, $x_k^{\ast}$ will be one of three things:
Right endpoint RS: $x_k^{\ast}=x_k,\, 1\leq k\leq n$
Left endpoint RS: $x_k^{\ast}=x_{k-1},\, 1\leq k\leq n$
Midpoint RS: $x_k^{\ast} = \frac{1}{2}\left(x_{k-1}+x_k\right),\, 1\leq k\leq n$ |
H: Couples around a table
Find the number of ways in which $n$ men and $n$ women can sit around a round table such that every man can pair off with a woman sitting next to him (to form $n$ couples).
Anyone can help with this problem? Thanks :)
EDIT: Ignore rotations of the table. For example, when $n=1$ there should be only 1 seating arrangement.
AI: First of all, we can assign seats in two stages. First, decide which $n$ seats will be for men and which for women. Second, arrange men on male seats and women on female seats.
For the second step, there are always $(n!)^2$ possibilities. So the answer to the problem is $k \cdot (n!)^2$, where k is the number of ways to decide which $n$ seats will be for men. So what is left is to find $k$.
For convenience, let us assign numbers to seats (say, clockwise): $1, 2, \ldots, 2n$. When every male seat is paired with a nearby female seat, there are two possibilities: either the paired seats are $(1, 2), (3, 4), \ldots, (2n-1, 2n)$, or they are $(2, 3), (4, 5), \ldots, (2n-2, 2n-1), (2n, 1)$.
Let $A$ be the set of all male-female seat assignments that allow pairing of the first kind, and $B$ the set of all assignments that allow pairing of the second kind. Now $k$ is the cardinality of their union $A \cup B$.
It is easy to find $|A|$. In each pair we pick arbitrarily which seat will be male and which female. There are $2^n$ choices, so $|A| = 2^n$. In a similar fashion, $|B|=2^n$.
Now, to find $|A \cup B|$, we note that $|A \cup B| = |A| + |B| - |A \cap B|$. What is $A \cap B$? It is the set of all male-female assignments that can be split into pairs in both ways: "$(1, 2), (3, 4), \ldots$" and "$(2, 3), (4, 5), \ldots$". There are only 2 assignments like this: one is when we assign all odd seats to men and even seats to women, the other is when we do the opposite. So $|A \cap B| = 2$.
Then we see that $k = 2^n + 2^n -2$, and the answer is $(2^{n+1}-2) \cdot (n!)^2$.
UPDATE: Ok, I assumed this was clear, but maybe not. I assume that all seats are distinguishable, and so are all the $2n$ people. So, $2n$ people enter a room, and you know each of them personally. And $2n$ seats are standing at fixed places around the table, and each seat is familiar to you, and each is precious in its own way, like an old friend. Now people take their seats at the table, and for you it is important who sits where exactly. If two men swap positions - you notice, because you distinguish them. If everyone stands up and moves to the seat to the right, thus rotating the whole arrangement - you also notice, because you know and distinguish the seats as well as people.
If you decide you don't want to distinguish between different men and different women, you drop the $(n!)^2$ factor. If you decide in addition that you don't even distinguish between sexes, you divide the answer by 2.
If your understanding is that people are all distinguishable, but rotations of the table don't really change the arrangement, then simply divide my answer by $2n$. The justification is simple: given one allowed arrangement, you can get $2n$ distinct arrangements from it by rotations. I have counted each one as a separate arrangement, but for you they are all the same, so you divide the answer by $2n$ to get $(2^n - 1) \cdot n! \cdot (n-1)!$. |
H: Infinite sets with cardinality less than the natural numbers
Are there any infinite sets that have a lower cardinality than the natural numbers? Is there a proof of this?
AI: No there are none. If $A$ has cardinality of at most the natural numbers, we may assume that it is a subset of the natural numbers.
One can show that a subset of the natural numbers is either bounded and finite, or unbounded and equipotent to the natural numbers themselves. |
H: Number of Nonnegative integers = number of integers
I just came across the idea that says the number of non-negative integers is equal to the number of integers. How is this possible? Isn't it true that non-negative integers are a subset of all integers? Then, it should follow that the number of all integers is bigger, twice at least, than the number of non-negative integers?
If the passage is indeed true, what is the implication of this?
AI: The idea here is that if we can put a set into a one-to-one correspondence with the Natural Numbers, then the sets have the same cardinality $\aleph_0$.
So the number of nonnegative integers (we can also call this the cardinality of the set $\mathbb{N}$) is the same as the number of integers (we can call this the cardinality of $\mathbb{Z}$). We can show that $\mathbb{Z}$ has cardinality $\aleph_0$ because the function $f:\mathbb{Z}\rightarrow \mathbb{N}$ given by:
$$f(z) = \begin{cases} 2z + 2 \quad\quad \;if\;z\geq0\\-2z-1, \quad if\; z<0\end{cases}$$
is a bijection.[*]
This can be a little tricky to wrap your head around - if you think that it's not possible because both sets "go to infinity" and one of them seems "smaller" than the other, remember that infinity is not a number.
[*] The function described there maps from $\mathbb{Z}$ to $\mathbb{N}$, where $\mathbb{N}$ is defined as the positive integers. Given that the OP asked for the nonnegative integers, i.e. $\mathbb{N}$ indexed from $0$, the function is not perfectly applicable, but the idea remains the same, and modifying the function to map from $\mathbb{Z}$ to the nonnegative integers is easy, perhaps best left as an exercise for OP. |
H: Questions on covering space
Following is a paragraph of a paper I am reading:
But I cannot understand this image, maybe it is because I have no idea about coverings. Could anyone explain it to me?
Particularly, you could just explain to me: what does $X_{|U_\alpha}$ mean?
AI: $X|_{U_\alpha}$ should mean "the preimage of $U_\alpha$ in $X$". The rest is just a reinterpretation of the definition of covering map/covering space with a nice little diagram. |
H: Intermediate value theorem for mean
Suppose the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous. For a natural number $k$, let $x_1,x_2,...,x_k$ be points in $[a,b]$. Prove there is a point $z$ in $[a,b]$ at which $$f(z)=(f(x_1)+f(x_2)+...+f(x_k))/k$$.
So I'm thinking about applying the intermediate value theorem:
If $$a<x_1<b,a<x_2<b,...,a<x_k<b$$, then $$f(a)<f(x_1)<f(b),...,f(a)<f(x_k)<f(b)$$ or $$k.f(a)<f(x_1)+...+f(x_k)<k.f(b)$$ $$f(a)<(f(x_1)+f(x_2)+...+f(x_k))/k<f(b)$$
But I couldn't think of any way to prove that $f(a)<f(x_k)<f(b)$ or is it even true?
EDIT: Thanks everyone for your effort.
AI: Hint:
Take $z\in\{x_1,\cdots x_k\}$ such that $f(z)\le f(x_j)$ for any $x_j$.
Take $w\in\{x_1,\cdots x_k\}$ such that $f(w)\ge f(x_j)$ for any $x_j$.
Then $$f(z)\le \frac{f(x_1)+\cdots+f(x_k)}{k}\le f(w)$$
Now, use the intermediate value theorem. |
H: How to integrate $\int_{0}^{a}x^{n-1}e^{-x}dx$
We know that $$\int_{0}^{\infty }x^{n-1}e^{-x}dx = \Gamma (n)$$
But how do we integrate this?
$$\int_{0}^{a}x^{n-1}e^{-x}dx$$
AI: This integral can be viewed as a recurrance
$$\begin{align*}
I_n &= \int_0^a x^{n-1}e^{-x}dx\\
&= -\int_0^a x^{n-1}de^{-x}\\
&= -\left[x^{n-1}e^{-x}\right]_0^a + \int_0^a e^{-x}dx^{n-1}\\
&= -a^{n-1}e^{-a} + (n-1)\int_0^a x^{n-2}e^{-x}dx\\
&= -a^{n-1}e^{-a} + (n-1)I_{n-1}
\end{align*}$$
With $$I_1 = \int_0^a e^{-x}dx = -\left[e^{-x}\right]_0^a = -e^{-a}+1$$ |
H: $a^{1/n}$ - How do you explain it cannot be $0$ for $a > 0$?
How do you explain formally, $a^{1/n}$ cannot be equal to $0$ for every $a > 0$?
Thanks
AI: If $b=a^{1/n}$, we have $b^n=a$. Now, can $0^n=a>0$? |
H: How to determine the coefficient of binomial
Suppose I have $\left(x-2y+3z^{-1}\right)^4$
How to determine coefficient binom of $xyz^{-2}$?
I've tried using trinominal expansion like this:
$\displaystyle \frac{4!}{1!1!1!} (1)^1 (-2)^1 (3)^2$
AI: You can do this by elementary combinatorics, think about how products of this kind are expanded. You need to find all the possible (ordered) products of $x$, $y$ and $z^{-1}$ with 4 factors that give $xyz^{-2}$. In order to obtain $xyz^{-2}$ one of the factors needs to $x$, one needs to be $y$ and two need to be $z^{-1}$, so we are looking for all the permutations of $(x,y,z^{-1},z^{-1})$. In general, there are $4!=24$ permutations of a $4$-tuple, but since two of the entries are equal, we are left with $4!/2=12$ different permutations. Now you just multiply the coefficients of these monomials with how often the desired combination appears in the expansion and obtain
$$12\cdot x\cdot(-2)y\cdot3z^{-1}\cdot3z^{-1} = -216xyz^{-2}.$$
This is the reasoning that leads to multinomial expansion in the end, but I find it a lot easier to remember how this works, instead of shooting up the multinomial formula if I only need a single coefficient. |
H: Minimize the area of a triangle
Let $A \neq B$ be fixed points outside a fixed circle with centre
$C$. The point $D$ can be chosen freely on the circle. The goal is to minimise
the area of triangle $ABD$. Degenerate triangles (triangles that are merely line
segments) are excluded. In which configurations of $A, B, C$ and the circle does
this problem have a solution and how can one construct its solution?
I expressed the area as $A = \frac{1}{2}ab \sin(\gamma)$ and then derived this expression with respect to $\gamma$ but this gives a maximum, not a minimum. I think a minimum would occur for $\gamma \rightarrow 0$ but this would resulte in a degenerate triangle. So I'm inclined to say that a solution does not exist. Does anyone have any thoughts on this?
AI: Forget about $\frac{1}{2}ab\sin \gamma$. The area of $ABD$ equals $\frac{1}{2}|AB| \cdot h$, where $h$ is the distance from line $AB$ to point $D$. Points $A$ and $B$ are fixed, so the problem is to minimize the distance from point $D$ to the fixed line $AB$. The nondegenerate condition means that you cannot pick $D$ on line $AB$. Can you solve the problem when formulated like that? |
H: Matchmaker's problem
This is a past exam exercise I'm unable to solve.
$B$ is finite set.
$h:B\rightarrow \mathcal{P}(G)$ where $\mathcal{P}(G)$ is the powerset of $G$ with the following properties:
for every $x\in B$, $h(x)$ is a finite subset of $G$
$X\subseteq B\rightarrow |X|\leq|\cup \{h(x):x\in X\}|$
I would like to prove that there exists a monomorphism $f:B\rightarrow G$ such that
$(\forall x\in B)[f(x)\in h(x)]$.
Is this known as "the matchmaker's problem?"
Thank you very much for your time and effort.
AI: This is known as Hall's theorem. Here is a proof by induction on $|B|.$ The theorem is trivial if $B=\emptyset,$ so we assume that $|B|\ge1.$ For $X\subseteq B$ define $h(X)=\bigcup\{h(x):x\in X\}.$ Define a "critical set" to be a set $C\subseteq B\ $ such that $|C|=|h(C)|, C\ne\emptyset,$ and $C\ne B.$ We consider two cases, depending on whether or not there is a critical set.
Case I. There is no critical set. In other words, $|X|\lt|h(X)|$ whenever $\emptyset\ne X\subsetneq B\ $. Choose $b\in B$ and $g\in h(b)$. Let $B'=B\setminus\{b\}$, and define $h':B'\to\mathcal P(G)$ by setting $h'(x)=h(x)\setminus\{g\}\ $ for $x\in B'$. Then conditions 1 & 2 hold with $B$ and $h$ replaced by $B'$ and $h'$. By the induction hypothesis, there is an injection $f':B'\to G$ such that $f'(x)\in h'(x)$ for all $x\in B'.$ Now the function $f=f'\cup\{(b,g)\}$ does the job.
Case II. There is a critical set. I.e., there is a set $C$ such that $\emptyset\ne C\subsetneq B$ and $|C|=|h(C)|.$ By the induction hypothesis, there is an injection $f_C:C\to G$ such that $f_C(x)\in h(x)$ for all $x\in C$. Let $B'=B\setminus C$. Define $h':B'\to\mathcal P(G)$ by setting $h'(x)=h(x)\setminus h(C)$ for $x\in B'.$ Then conditions 1 & 2 hold with $B$ and $h$ replaced by $B'$ and $h'$. By the induction hypothesis, there is an injection $f':B'\to G$ such that $f'(x)\in h'(x)$ for all $x\in B'.$ Now the function $f=f'\cup f_C$ does the job. |
H: Is $\left\{ \frac{1}{n}: n \in \mathbb{N} \right\} \cup \left\{ 0\right\}$ closed set?
Is it true that $\left\{ \frac{1}{n}: n \in \mathbb{N} \right\} \cup \left\{ 0\right\}$ is closed set? I suppose that yes, but I have no idea how can I prove it.
AI: Hint : It would be a bit easy if you can realize :
$\big(\{ \frac{1}{n} : n\in \mathbb{N}\}\cup \{0\}\big)^C=\bigcup_{n\in \mathbb{N}}\big(\frac{1}{n+1},\frac{1}{n}\big)\text{with} (-\infty,0)\cup (1,\infty)$ |
H: Continuity of translation for p=infinite
For $f\in L^p$, $1\leq p<\infty$, the property of continuity of translation holds:
$$\lim_{|x|\to 0}\|f_x-f\|_p=0,$$
where
$$f_x(y):=f(x+y).$$
When $p=\infty$, is continuity of translation still true? If not, can you give a counterexample?
AI: No. Take $f = \chi_{[0,1]}$, then for any $x \in \mathbb{R}\setminus\{0\}$,
$$
\|f_x -f\| = \|\chi_{[-x,1-x]} - \chi_{[0,1]}\| = 1
$$ |
H: Laplace Transform using t-shift (second shift)
$$f(t) = tu(t-π)$$
I know I have to get t in terms of $$(t-π)$$ and to do that I have done
$$ t = a(t-π) + b$$
$$ t = at-aπ + b$$
$$ t = (a-π)t + b$$
$$ (a-π) = 1$$ and $$b = 0$$
Then I think I have done the write thing and being to t-shift with
$$ f(t) = (a-π)(t-π)u(t-π)$$ but when I apply the t-shift theorem I get lost. Is this the write method? have I done it right?
AI: Assuming that $*$ doesn't mean convolution in your original statement, note that
$$tu(t-\pi) = (t-\pi + \pi)u(t-\pi) = (t-\pi)u(t-\pi) + \pi u(t-\pi)$$
Thus,
$$\mathcal{L}\{tu(t-\pi)\} = \mathcal{L}\{(t-\pi)u(t-\pi)\} + \pi\mathcal{L}\{u(t-\pi)\} = \ldots$$ |
H: Number of orbits of a subgroup of the symmetric group $S_9$
Let $H=\langle (3\quad 4\quad 5),(1\quad 2\quad3)(7\quad8\quad9)\rangle \le S_9$ be a subgroup in $S_9$. Find number of orbits and their order.
First I noticed $\mathrm{orb}(6)=\{6\}$. I also think that $\mathrm{orb}(3)=\{1,2,3,4,5\}$ but I can't explain how (and if so, then also $\mathrm{orb}(7)=\{7,8,9\}$). Another problem is I don't know to explain myself why a permutation send one of $\{1,2,3,4,5\}$ to $\{7,8,9\}$ does not exist. How can I explain both problems?
AI: This is not a complete answer, but rather a hint on what you need to understand.
I assume (though you should specify) that the set upon which your $S_9$ is acting is $\{1,2,3,4,5,6,7,8,9\}$. If this is the case, then the permutation $\sigma=(3,4,5)$ takes $3$ to $4$, takes $4$ to $5$, takes $5$ to $3$, and leaves $1,2,6,7,8,9$ fixed. In the same manner you can see what the permutation $\tau=(1,2,3)(7,8,9)$ does to each element of the set $\{1,\dots,9\}$. In order to compute the orbits of this group action you need also to consider composing the two permutations that generate your subgroup, for example, $(\tau\circ\sigma)(5)=1$. Methods for doing that involve Burnside's lemma and the orbit-stabilizer theorem. |
H: Fourier transform of $\exp(-t^2)$ using contour integration.
I am calculating the Fourier transform of $\exp(-t^2)$ using contour integration.
I am left with the integral $\int_{-\infty}^\infty \exp(i\omega t)\exp(-t^2)$.
Usually I would now use the residue theorem, but I cannot find the singularities.
Can someone help me?
AI: We can write
$$\exp (i\omega t)\exp(-t^2) = \exp\left(-(t-i\omega/2)^2\right)\exp\left((i\omega/2)^2\right).$$
The factor $\exp (-\omega^2/4)$ can be pulled out of the integral, and we are left with
$$\int_{-\infty + i0}^{+\infty+i0} \exp \left(-(z-i\omega/2)^2\right)\,dz.$$
Using the Cauchy integral theorem, we can shift the integration to
$$\int_{-\infty + i\omega/2}^{+\infty+i\omega/2}\exp \left(-(z-i\omega/2)^2\right)\,dz,$$
and parametrising as $z = t + i\omega/2$, that becomes
$$\int_{-\infty}^{\infty} \exp(-t^2)\,dt.$$
The shift of the contour of integration is justified by applying the Cauchy integral theorem to a rectangle with vertices $-R,\, R,\, R+i\omega/2,\, -R+i\omega/2$ and taking the limit $R\to +\infty$. |
H: Show that the topological space ( X, $\tau$ ) is not metrizable
For the topological space ( X, $\tau$ ), with X = {0, 1} and $\tau$ = { $\emptyset$ , {0}, {0,1} } , prove that ( X, $\tau$ ) is not metrizable.
I know intuitively it can't be but don't know how to prove it. The only idea I have is that {0, 1} cannot be open as X = {0, 1} is discrete but I'm not even sure if this is right.
AI: You can use the fact that any metrizable space is Hausdorff. This follows the definition of a metric, in particular the non-negative axiom ($d$ is a metric implies that $d(x,y)\geq0$ and $d(x,y)=0$ if and only if $x=y$).
Let me know if you'd like me to flesh out the details.
Edit: A topological space $(X,\tau)$ is Hausdorff if and only if for any two points $x,y$ in $X$ you can find two sets $U$, $V$ in $\tau$ such that $x\in U$, $y\in V$ and $U\cap V=\emptyset$ (that is, if you can "house" any two points in disjoint open sets).
This is clearly not the case in your example, since you cannot find such sets for $x=0$ and $y=1$ (try!).
So we know that $(X,\tau)$ is not Hausdorff. If we can show that any metrizable space must be Hausdorff then it follows that $(X,\tau)$ is not metrizable.
Suppose that $(X,\tau)$ is metrizable, pick a metric $d$ and any two points $x,y\in X$. Let $\varepsilon=\frac{1}{2}d(x,y)>0$ (by the metric axioms).
Then $U:=B_\varepsilon(x)$ and $V:=B_\varepsilon(y)$--where $B_\varepsilon(x)$ denotes the open ball of radius $\varepsilon$ centred on $x$--are open, disjoint and $x\in U$, $y\in V$. Since, $x,y$ where arbitrary we have that $(X,\tau)$ is Hausdorff. |
H: If $[G:H]=n$, is it true that $x^n\in H$ for all $x\in G$?
Let $G$ be a group and $H$ a subgroup with $[G:H]=n$. Is it true that $x^n\in H$ for all $x\in G$?
Remarks. The answer is positive whenever $H$ is normal, e.g., for $n=2$. In general, by using the normal core of $H$, one can find an $m\ge 1$ such that $x^m\in H$ for all $x\in G$.
AI: For a counterexample, take $S_3$ and the subgroup $H = \{\rm{id},(12)\}$. This has index $3$, but $(13)^3 = (13)\not\in H$. |
H: Central Limit Theorem: asymptotic distribution
I keep reading in the documents about the asymptotic distribution of the CLT. I am learning things by myself so try to figure it out on my own and don't have a teacher to ask the question to.
So it's my understanding that the variance of the sample mean is:
$Var(\bar X) = {{ \sigma^2 } \over { n }}.$
I also know that you can write this variance as $E[(\bar X - \mu)^2]$ where $\bar X$ is the sample mean and $\mu$ is the population mean. you can definitely plot the first equation (if you know the population variance $\sigma^2$) and yes you would get a curve which is going very close to the x-axis (without even touching it); this my understanding of what they "asymptotic" word means.
I am trying to understand if this is what it means when I find in text books that the CLT has an asymptotic distribution? Could someone explained a bit more what it means when one says that "the Central Limit Theorem provides the asymptotic distribution of $\sqrt{N}(\bar X − \mu)$"?
Thank you very much.
AI: The fact that the variance of $\bar X_n$ goes to zero when $n\to\infty$ (like $\sigma^2/n$) indicates that $\bar X_n$ looks more and more like the constant random variable equal to $E[\bar X_n]=\mu$. The CLT is a second order expansion of $\bar X_n$ around $\mu$. The surprise, when one first sees this result, is that the convergence of $\bar X_n-\mu$ suitably normalized happens in distribution (but not in probability, even less almost surely).
At the end of the day, the CLT is a relatively weak statement saying, first, that the correct normalization is $\sqrt{n}$ (not a surprise if one wants the variance to converge to a positive finite limit) and, second, that $Z_n=\sqrt{n}(\bar X_n-\mu)$ converges in distribution to a normal random variable $Z$ with mean $0$ and variance $\sigma^2$. This means that, for every real number $z$,
$$
P[Z_n\leqslant z]\to P[Z\leqslant z].
$$
Since one knows the distribution of $Z$, one knows that the value of the RHS is
$$
P[Z\leqslant z]=\int_{-\infty}^{x/\sigma}\frac1{\sqrt{2\pi}}\mathrm e^{-t^2/2}\mathrm dt.
$$
Note that the almost sure behaviour of $Z_n$ is another matter, since $\limsup\limits_{n\to+\infty}Z_n=+\infty$ and $\liminf\limits_{n\to+\infty}Z_n=-\infty$, almost surely (a keyword here being law of iterated logarithm).
To sum up, the fact that the variance is $\sigma^2/n$ suggests the normalization in the CLT but the CLT itself is a much more encompassing result. For example, the fact that a unique distribution (the normal one) captures the behaviour of every $\bar X_n$ (provided square integrability) should be truly bewildering. |
H: Elliptic Curve: Multiplying points over a finite field
Let $E$ be an elliptic curve over a finite field $\mathbb{F}_q$ where $q$ is prime. Let $P$ be a point on $E$.
Consider the point $Q=(q+1)P=P+\cdots+P$, which is $P$ added to itself $q+1$ times. Due to the fact that we are in the finite field $\mathbb{F}_q$, I would expect the "scalar", in this case, $q+1$, to respect the ground field property and have $Q=P$, but this is not the case.
I am told that $E(\mathbb{F}_q)$ is either cyclic or a product of two cyclic groups, and that the order may not be $q$. Are there underlying concepts that I may be missing?
AI: You are right; you are missing some points. As of now you seem to be confusing the torsion points of elliptic curves with the (torsion elements of) multiplicative group $\mathbb F_q^*$. This is not a bad analogy at a higher level; but it is not superficially true. You will need to spend quite some time to understand everything. You can find a rigorous proof of this fact in Silverman's book on elliptic curves. If you want to understand heuristically, here is the following.
An elliptic curve over $\mathbb C$ is isomorphic to $S^1 \oplus S^1$ as a topological group, where $S^1$ is the circle group. The $m$-torsion has size $\mathbb Z/m\mathbb Z\oplus\mathbb Z/m\mathbb Z$. Due to the fact that elliptic curves are actually algebraic, and in particular, the torsion points are algebraic, these facts largely carry over to elliptic curves over algebraically closed char-$p$ fields(Here is where I am fudging things). The $\mathbb F_q$-rational points are obviously finite in number and they belong to some $m$-torsion group for the elliptic curve considered over the algebraic closure of $\mathbb F_q$. That it is either cyclic or a direct sum of two cyclic group now follows from the structure theorem of finitely generated abelian groups. |
H: A linear operator $T: V \rightarrow V$ commuting with all linear operators is a scalar multiple of the identity.
Let $\mathbb{K}$ a field, $V$ a vector space over $\mathbb{K}$. If $T:V\to V$ commutes with all other linear operators $V \to V$, then there exists $\lambda \in \mathbb{K}$ such that $T= \lambda I$, where $I$ is the identity V.
AI: First show that $Tv$ is scalar multiple of $v$ for all $v\in V$. You can do this by assuming, for the sake of contradiction, that there exists $v$ such that the vectors $v$ and $Tv=w$ are linearly independent. Form a basis $\mathcal B$ which includes both $v$ and $w$. Now, define a new linear operator $L:V\to V$ which maps every basis element to $0$ except $w$, which we map to $v$. You will get a contradiction.
Now, you must show that there is just one scalar $\lambda$ such that $Tv=\lambda v$ for all $v$. Say $Tv_1=\lambda_1v_1$ and $Tv_2=\lambda_2v_2$ where neither $v_1$ nor $v_2$ is a scalar multiple of the other--i.e. they are linearly indepdendent--notice that the case where they are linearly dependent is easily handled. Write $T(v_1+v_2)=\lambda_3(v_1+v_2)$. Then the equation $$(\lambda_3-\lambda_1)v_1+(\lambda_3-\lambda_2)v_2=0$$ implies $\lambda_1=\lambda_3=\lambda_2$ as desired. |
H: Arbitrary intersection of uncountable subfields of $\mathbb{C}$
What is about arbitrary intersection of uncountable nested subfields of $\mathbb{C}$?
Does it have to be uncountable too or it can be countable?
Edit: {\it Nested} subfields of $\mathbb{C}$ means that we are considering an infinite chain of field $F_1, F_2, \dots, F_\alpha,etc..$ of subfields of $\mathbb{C}$ such that for every two ordinals $\beta,\lambda < \kappa$ such that $\beta < \lambda$ we have $F_\beta \supset F_\lambda$ and $\kappa$ is some limit ordinal, which might be uncountable.
Of course we are assuming that every $F_\alpha$ is uncountable for $\alpha < \kappa$.
AI: The answer is no.
Let $A=\{a_\alpha\mid\alpha<\kappa\}$ be a set of generators for $\Bbb C$ over $\Bbb Q$ (that is, $\Bbb Q(A)=\Bbb C$). Now let $F_\alpha=\Bbb Q(\{a_\beta\mid\alpha<\beta\})$.
Each $F_\alpha$ is uncountable, but $\bigcap F_\alpha=\Bbb Q$. |
H: Extensions of continuous bounded functions
If $u:U \rightarrow \mathbb{R}^{n}$ is bounded and continuous can $u$ always be extended such that $u \in C(\bar{U})$? and is $u$ uniformly continuous?
AI: No, this is not always possible, let $n = 1$ and $U = (0,1) \subseteq \mathbb R$, define $u(x) = \sin(x^{-1})$. Then $u$ is bounded (by 1) and continuous, but is neither uniformly continuos nor extendable to $[0,1] = \bar U$. |
H: Upper bound of an integral on a circle segment
Say you have the integral:
$$\int_S f(z)dz = \int_S \frac{1}{x^{12}+1}dz$$
Where $S$ is a circle segment that runs from $R$ to $Re^{i\pi/6}$. An example question suggests it is possible to estimate this integral with:
$$\left| \int_S f(z)dz \right| \le \frac{\pi R}{6} \frac1{R^{12}-1}$$
This looks like the length of the circle segment times the maximum possible value of $f(Re^{i\theta})$.
I think the maximum possible value because the triangle inequality says that, for a certain $|z|$, $|z+1|$ is largest when z lies on the positive real axis. So for a given $R \gt 1$, $\frac1{R^{12}e^{i\theta}+1}$ cannot be larger than $\frac1{R^{12} -1}$.
Is this reasoning correct? Is there a formal name or formula for the above estimate?
AI: There is something called the "$ML$" inequality, which states that the magnitude of a contour integral is bounded by the product of the maximum value of the magnitude of the integrand ($M$) times the length of the contour ($L$).
In your case, yes, sub $z=R e^{i \theta}$ to get
$$i R \int_0^{\pi/6} d\theta \, e^{i \theta}\, \frac{1}{R^{12} e^{i 12 \theta}+1}$$
The max value of the magnitude of the integrand is $1/(R^{12}-1)$ because the min value of the denominator occurs when $\theta=\pi/12$, which lies within the integration interval. By the $ML$ inequality, the magnitude of the integral is bounded by
$$\frac{\pi}{6} R \frac{1}{R^{12}-1}$$ |
H: Algebra Homomorphism
This is a follow-up to a question I asked here yesterday. It's coming from a (non-examinable) exercise sheet and I really can't get my heard around how this question is posed and to be approached.
For clarity I'll re-type relevant parts of the initial post.
Suppose we have some field $K$ and non-zero elements $a,b,$ in $K$. Define $H=H(a,b)$ to be the $K$-algebra with basis $\{1,x,y,z \}$ over $K$ satisfying $$x^2=a, \\ y^2=b, \\ z=xy=-yx$$
Additional Information Now we're asked to suppose that $\sqrt{a}$ is in fact an element in the the field $K$.
Question We need to show that there is an algebra homomorphism $\phi : H \rightarrow M_2(K)$ such that
$$ \phi(x)= \left( \begin{array}{ccc}
\sqrt{a} & 0 \\
0 & -\sqrt{a} \end{array} \right) , \\
\phi(y)= \left( \begin{array}{ccc}
0 & b \\
1 & 0 \end{array} \right)
$$
Initial Thoughts It is clear that $\{\phi(x)\}^2$ is $a.\text{Id}$ and $\{\phi(y)\}^2$ is $b.\text{Id}$, which will presumably tie in nicely with the fact that $x^2=a$ and $y^2=b$.
If anyone can shed light on what this homomorphism actually is, and why it's a homomorphism, I'd be very grateful.
AI: $H$ has a basis $1, x,y,z$. To give a linear map $\phi \colon H \to M_2(K)$ it is enough to give the values on the basis. As we want $\phi$ to be an algebra homomorphism, we let
$$ \phi(1) = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$
$\phi(x)$ and $\phi(y)$ are given by the exercise. Moreover
$$ \phi(x) \phi(y) = \begin{pmatrix} 0 & b \sqrt a\\ - \sqrt{a} & 0 \end{pmatrix} $$
So we let
$$ \phi(z) = \begin{pmatrix} 0 & b \sqrt a\\ - \sqrt{a} & 0 \end{pmatrix}. $$
Alltogether we have
$$ \phi(\alpha + \beta x + \gamma y + \delta z) =
\begin{pmatrix} \alpha + \beta \sqrt a & \gamma b + \delta b\sqrt{a}\\ \gamma - \delta \sqrt a & \alpha - \beta \sqrt a \end{pmatrix}
$$
Now we have to check that $\phi$ is multiplicative, it suffices to check that
$$ \phi(x)^2 = a\,{\rm Id}, \phi(y)^2 = b\,{\rm Id}, \phi(x)\phi(y) = -\phi(y)\phi(x) = \phi(z) $$
which are all true. So $\phi$ is a homomorphism of $K$-algebras. |
H: Laplace Transform using t-shift
$$f(t)=\begin{cases}cos(πt), & 1\leq t < 4 \\ 0, &elsewhere \end{cases}$$
Okay, I attempted to write it in terms of step functions and I got
$$ f(t) = cos(πt)u(t-1)-cos(πt)u(t-4)$$
But Now I'm not sure how to get the $$cos(πt)$$ for both parts in terms of (t-1) and (t-4) so then I could use the t-shift theorem. Can someone help me out?
Thanks!
AI: Remember that $$\mathcal L(g(t)u(t-a)) = e^{-as}\mathcal L(g(t+a)),$$ where $\mathcal L$ is the Laplace transform. This is probably the version of the 't-shift' theorem you want to use here. |
H: Why is $\mathbf{Rel} \cong \mathbf{Rel}^{\mathbb{op}}$?
I haven't yet fully grasped Category Theory so I am doing the exercises in Awodey's book for the first chapter, and exercise 2a) is confusing me very much.
The question is to prove or disprove that $\mathbf{Rel} \cong \mathbf{Rel}^{\mathbf{op}}$. I get that the idea is
$$f^{\mathbf{op}}: D \longrightarrow C = \{\langle d,c \rangle \in D\times C| (c,d) \in f\}$$
But I can't write the functor because according to the definition: $$F(f:C\longrightarrow D)=F(f):F(C)\longrightarrow F(D)$$
So if there are two relations from $C$ to different objects, $F(C)$ would not be defined.
What am I doing/understanding wrong? Thank you!
edit: I should add that I say $\mathbf{Rel} \cong \mathbf{Rel}^{\mathbb{op}}$ not because I proved it but because I read a document with the solutions...
AI: One of the problems with these exercises in Awodey is that he didn't actually define "isomorphism of categories" in the first chapter. But let us take it to mean "isomorphism in $\sf Cat$", the category of (small) categories and functors.
In the preceding exercise, one is asked to show that:
$$C: {\sf Rel}^{\rm op} \to {\sf Rel}$$
given as the identity on objects, and on relations as $C(R^{\rm op}: D^{\rm op} \to C^{\rm op}) = R^c: D \to C$ (note that $R^{\rm op}:D^{\rm op} \to C^{\rm op}$ corresponds to $R: C \to D$ in $\sf Rel$) is a functor.
Let us check the composition property of $C$ on $S^{\rm op}: E^{\rm op}\to D^{\rm op}$ and $R^{\rm op}$:
\begin{align}
C(S^{\rm op}R^{\rm op}) &= C((RS)^{\rm op}) & &\text{Definition of composition in $\sf Rel^{\rm op}$}\\
&= (RS)^c & &\text{Definition of $C$}\\
&= \{(e,c) \mid (c,e) \in RS\} &&\text{Definition of $(-)^c$}\\
&= \{(e,c)\mid \exists d \in D: (c,d)\in R, (d,e)\in S\} &&\text{Definition of $RS$}\\
&= \{(e,c)\mid \exists d \in D: (d,c)\in R^c, (e,d) \in S^c\}\\
&= S^c R^c = C(S^{\rm op})C(R^{\rm op})
\end{align}
It's not hard now to verify that $C$ is indeed a functor. I'm sure you can define a functor $C^{\rm op}: {\sf Rel}\to {\sf Rel}^{\rm op}$ that together with $C$ gives ${\sf Rel} \cong {\sf Rel}^{\rm op}$.
As a general advice, I suggest that you meticulously write the $\rm op$ superscript on every object and morphism in an opposite category until you know your way around. This can help to overcome some of the initial confusion. |
H: Finite group of two generators
My question is simple :
Any finite group of two generators is cyclic, semidirect sum, or direct sum ?
AI: No. The group $A_5$ is generated by $(123)$ and $(12345)$. It isn't cyclic and cannot be represented as a nontrivial semidirect product, let alone a direct product. |
H: What do $x\in[0,1]^n$ and $x\in\left\{ 0,1\right\}^n$ mean?
$x\in[0,1]^n$
$x\in\{0,1\}^n$
Thank you in advance.
AI: If $x=(x_1,\ldots,x_n)\in [0,1]^n$, then $0\leq x_i\leq 1$ for all $i$. If instead $x=(x_1,\ldots,x_n)\in \{0,1\}^n$, then $x_i=0$ or $x_i=1$ for all $i$.
For example, if $n=2$, then $[0,1]^2$ is the (filled) square in $\mathbb{R}^2$ with corners $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$, while $\{0,1\}^2$ is just the corners. |
H: Calculate the length of AC
The diameter $AB$ of the circle is $10\,\text{cm}$. The length of $BC$ is $6\,\text{cm}$. Calculate the length of $AC$.
I'm doing a mock exam and I'm not sure how to work out the length of $AC$. Any ideas? Thanks.
AI: The Pythagorean Theorem applies: the right angle is $\angle ACB$, by Thales Theorem. So the hypotenuse is $AB = 10$.
$$\begin{align} |AB|^2 & = |AC|^2 + |BC|^2 \\ \\ \iff |AC|^2 & = |AB|^2 - |BC|^2 \\ \\ \iff |AC| & = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\end{align}$$ |
H: Nth Term of Fractions
How do I work out the Nth term of these fractions?
$$1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\frac{1}{25},\dots$$
Would I need to change them all into decimals but that would be quite complicated and then go from there? Yes I can see a pattern, the denominators are squared numbers but what do I do after? I know that answer is going to have a $^2$ in it.
AI: Do you recognize the sequence that the numbers in the denominators form? 1,4,9,16,25,.. is a well known sequence. If you've figured that out, it will be quite easy to write down a direct expression for the nth term of the sequence. |
H: What causes the change in the expected value of the product of random variables?
The following question is part of a homework exercise on portfolio theory that I have to do.
Suppose that $Y$ is a random variable representing the returns on an investment. Now, let $f$ be a continuous, concave, strictly increasing function of $Y$. Let $\xi^*$ be the amount invested in security Y that maximizes $E[f(\xi^*Y)]$. Show that $(\xi^*>0)$ iff $E[Y]>0$.
AI: The function $u$ can be seen as a utility function, hence the idea is to find where your expected marginal utility is 0, similar to normal economics with deterministic payoff functions and utilities. Since the function is integrable over the reals, you can take the derivative inside the integral of the LHS of your objective function to get the marginal expected utility function: $\frac{d}{d\xi}\int u(\xi y) f(y) dy = \int \frac{d}{d\xi}u(\xi y) f(y) dy = \int yu'(\xi y) f(y) dy = E[Yu'(\xi Y)] =$ expected marginal utility.
Setting it to zero gives you the condition you mentioned: $E[Yu'(\xi^* Y)] =0$
Now, using Jensen's inequality, we know $E[u(\xi Y)] \leq u(\xi E[Y])$ since the function is strictly increasing, convex function. Therefore, we can only hope to have a positive utility on our investment if $E[Y]>0$. There is an implicit assumption that I made here that $(x<0) \longrightarrow (u(x)<0)$, otherwise, you would benefit from a loss. |
H: partitions that contain two singleton blocks one and n.
How many partitions [n] contain at least one of the singleton blocks 1 and n? I am having trouble doing this problem using the Sieve formula. Is it possible? and if so how do i go about it?
AI: I would use inclusion-exclusion on this one. Your answer is equal to:
The number of partitions containing the singleton 1
$+$ The number of partitions containing the singleton n
$-$ The number of partitions containing both singletons.
Now, a partition containing the singleton 1 is just any partition of the $n-1$ other numbers, plus the singleton 1, so the number of partitions needed for the first line is the same as the total number of partitions of $n-1$. Can you take it from here? |
H: Prove that $n$ is prime $\iff \forall a\in\mathbb{Z}(\gcd(a,n)=1\lor n\mid a)$
I need to prove that given $n\in\mathbb{N}$ ($n>1$), $n$ is prime $\iff \forall a\in\mathbb{Z}(\gcd(a,n)=1\lor n\mid a)$.
I proved the first part, assuming that $n$ is prime and proving that for every $a$, $\gcd(a,n)=1$ or $n\mid a$, but I'm struggling with the proof of the second part.. I assumed that for every $a$, $\gcd(a,n)=1\lor n\mid a$, and then assumed that $n$ is not prime: from here we know that there exists $1<d<n$ such that $d\mid n$... And I don't know how to proceed from here.
Thanks in advance.
AI: Better to prove the contrapositive. Suppose that $n$ is not prime, and find an appropriate $a$ that breaks the other expression. This almost writes itself; if not prime then $n=cd$ for some $c,d$ that are neither $1$ nor $n$. |
H: Question on the proof that $C(\Omega)$ is a Frechet space
I am using Rudin's book on Functional Analysis. I am studying the proof that the space $C(\Omega)$ of continuous functions on an open set $\Omega \subseteq \mathbb{C}$ is a Frechet space. I encountered a detail that I can't understand and I hope you guys here can help me.
Here is a screenshot of the part:
Here, the $p_n$'s are seminorms defined by supremum of $f$ on a compact set $K_n$. I understand that the sequence $\{ f_i \}$ converges uniformly on each $K_n$ and its limit is continuous on $K_n$.
My problem is: how do we know that the limit $f$ is continuous on $\Omega$? In fact, I'm not sure why we get the same limit when we consider different $K_n$'s?
AI: If we restrict $f_{i}$ to the interior of $K_n$, then it converges uniformly there to $f\mid_{K_n}$, so $f\mid_{K_n}$ is continuous on the interior of $K_n$, and since continuity is a local property, $f$ is continuous on all of $\Omega$.
To see that you get the same limit when considering different $K_n$, so that we have a well-defined limit on all of $\Omega$, suppose that $f$ and $f'$ are the limits of $f_i$ considered on the compact sets $K$ and $K'$. We wish to show that these are compatible, so they agree on $K \cap K'$. But this is clear, for suppose that they did not agree at $x \in K \cap K'$, i.e. $f(x) \neq f'(x)$. Then the sequence $f_i$ cannot both converge to $f$ on $K$ and $f'$ on $K'$ in the appropriate metrics, because they cannot both converge pointwise at $x$. |
H: What is the best way to solve this high school exercise?
Can you share with me how would you best solve this exersise to a high school student?
Show that $f(x)=x^2-6x+2$ , $x\in(-\infty,3]$ is $1-1$ and find its inverse.
AI: Hint: To prove the function is 1-1 use the definition
$$ f(x_1)=f(x_2) \implies x_1 = x_2 .$$ |
H: Tangent vectors as curves equivalence relation
I do not understand the definition of the equivalence relation that is defined on the curves creating a tangent vector space.
Let $X$ be any manifold, a point $x \in X$, two curves $\alpha:(-a,a) \to X, \beta:(-b,b) \to X$. Then $\alpha$ is equivalent to $\beta$ at $x$ iff $\alpha(0)=x, \beta(0)=x$ and for a chart $\phi:X \to \mathbb{R}^n$ to some euclidean space $(\phi \circ \alpha)'(0)=(\phi \circ \beta)'(0)$.
The first point I do not understand is why we need a chart at all, it is a bijection so it should be true$(\phi \circ \alpha)'(0)=(\phi \circ \beta)'(0)$ iff $(\alpha)'(0)=(\beta)'(0)$.
Secondly imagine curves $\alpha(t)=(t,t), \beta(t)=(2t,2t)$ to euclidean space with their identity chart. Such curves have at every point the same direction, however their derivative is different, therefore such curves would not be equivalent at $t=0$ according to that definition.
Therefore the tangent space $T_xX$ consists of equivalence classes of curves at $x$ that do not have the same direction only, but their derivatives (vectors) have the same magnitude too. Is that right?
AI: For your second question, yes, the tangent space $T_xX$ consists of vectors, so not only direction but magnitude matters as well. Actually, $T_xX$ is isomorphic to the vector space $\Bbb R^n$, using a fixed chart.
The point is that, we a priori don't know yet what tangent vectors are in an abstract manifold $X$, though $\alpha'(0)$ should be a tangent vector, an element of $T_{\alpha(0)}X$, which we are about to define just now. That's why chart is needed, as differentiation and vectors are already defined in $\Bbb R^n$, and we can use those. |
H: free subgroups of $SL(2,\mathbb{R})$
In the example section of the wikipedia article on the the Ping Pong lemma, you can see how to construct a free subgroup of $SL(2,\mathbb{R})$ with two generators
$$ a_1 =
\begin{pmatrix}
1 & 2 \\
0 & 1
\end{pmatrix},
\ \ \ \ \ a_2 =
\begin{pmatrix}
1 & 0 \\
2 & 1
\end{pmatrix}.
$$
Is it possible to construct free subgroups of $SL(2,\mathbb{R})$ with an arbitrary number of generators in a similar way (using the Ping Pong lemma)? Do such subgroups even exist?
Sorry if this is a stupid question, I'm a noob at group theory.
AI: Subgroups of free groups are free (this is well-known theorem by Nilsen-Schreier). Free group $H$ on two generators $F_2$ contains a subgroup of countable rank (for example, its commutator subgroup). This implies that $F_2$ contains any subgroup of finite rank. Indeed, if $H=\langle a_1,a_2,a_3,\dots\rangle$, then $\langle a_1,\dots, a_n\rangle$ gives you a free subgroup in $F_2$ of rank $n$.
So the free subgroup on $2$ generators that you've found in $SL(2,\mathbb{R})$ gives you automatically free subgroups of $SL(2,\mathbb{R})$ of arbitrary countable rank. |
H: What are the chances of winning with a specific card in Spades
In the game of spades, a standard deck is shuffled then all the cards are dealt in a clockwise manner until each of the 4 players has 13 cards. The first play of the game is for each player to throw their lowest club (clubs are ordered from low to high: 2,3,4,...,Queen,King,Ace).
When all four lowest clubs are on the table, the player who threw the highest of those four cards wins the "trick" but if a player has no clubs, he or she must play a heart or a diamond, and that card has no chance of winning the trick. If a player has no clubs, no hearts, and no diamonds, then the player must play a spade, and will be guaranteed to win the trick.
I simulated this game by counting how many times a specific card won and I divided it by the number of tricks and I got a winning probability of 0 with the card 2 of Clubs, I got an approximate probability of 9.15 with 10 of Clubs, 3.59 with King of Clubs, and 11.98 with 9 Clubs but now I want to solve it mathematically. I think simulation is way too complicated because I have simulate 1,000,000 tricks or more to get closer to the true value.
AI: You haven't defined the rules completely. If all the hands are $4-3-3-3$, the middle five tricks will have no clubs or spades, so a heart or diamond must win. But I think simulation is the only reasonable approach here-there are too many different possibilities for hand work. If a card has a $10\%$ chance of winning a trick, the error in $10,000$ runs will be about $\sqrt {10,000\cdot 0.1 \cdot 0.9}=30$ or about $0.3\%$ and it doesn't seem even a million runs with error about $0.03\%$ should take very long. You can keep track of the chance for every card on one run through, so don't have to do a separate run for each card. |
H: Let $G=\langle x\rangle$ be cyclic of order $n$. Prove that $\langle x^r\rangle⊆\langle x^s\rangle$ iff r is a multiple of $s$ modulo $n$.
Let $G=\langle x\rangle$ be cyclic of order $n.$ Prove that $\langle x^r\rangle\subseteq \langle x^s\rangle \iff r$ is a multiple of $s$ modulo $n.$
I know you have to approach this from both ways, $\Longrightarrow$ and $\Longleftarrow$ but im not sure of the correlation from both ways, can someone help me?
AI: If $\subseteq$, then $x^r\in \langle x^s\rangle=\{x^{sk}:k\in\mathbb{N}\}$. Hence there is some $k\in\mathbb{N}$ such that $x^r=x^{sk}$. Since a group is cancellative, we have $1=x^{sk-r}$. Since $G$ is cyclic of order $n$, then $n|sk-r$ or $sk\equiv r\pmod{n}$.
The other direction is simple. |
H: Prove, square of quadrilateral is the sum of squares of 4 triangles
Let $A_1$, $B_1$, $C_1$, and $D_1$ - midpoints of the sides $AB$, $BC$, $CD$ and $DA$ convex quadrilateral $AВСD$. Directs $AC_1$, $ВD_1$, $CA_1$ and $DВ_1$ - divide it by $5$ quadrilaterals and $4$ triangles. Prove that the area of the central quadrilateral is the sum of squares of $4$ triangles.
I think, it should be proved in such a way: the square of quadrilateral is finding out by such formula: $S=(p-a)(p-b)(p-c)-abcd\cos^2\left(\frac{A+B}{2}\right)$. If we will prove that the sum of big triangles are the same as the square of $ABCD$, it will be automatically proved area of the central quadrilateral is the sum of areas of small triangles.
AI: Since $A_1A=A_1B$, $~C_1C=C_1D$, then
$$
S_{A_1BC} = S_{AA_1C}, \qquad S_{C_1DA} = S_{CC_1A};
$$
since $D_1D=D_1A$, $~B_1B=B_1C$, then
$$
S_{D_1AB} = S_{DD_1B}, \qquad S_{B_1CD} = S_{BB_1D};
$$
so,
$$
S_{A_1BC}+S_{B_1CD}+S_{C_1DA}+S_{D_1AB} = S_{AA_1C}+S_{BB_1D}+S_{CC_1A}+S_{DD_1B},
$$
other words,
$$
\color{green}{S_{\large{green}}} + 2\cdot \color{blue}{S_{\large{blue}}} =
\color{green}{S_{\large{green}}} + 2\cdot \color{red}{S_{\large{red}}};
$$
$$
\color{blue}{S_{\large{blue}}} =
\color{red}{S_{\large{red}}}.
$$ |
H: Denoting the set of initial segments of a binary sequence
The index is an infinite, innumerable binary sequence in $\{0,1\}$. $ I= \{f \mid f: \Bbb {N} \to \{0,1\}\} $
Is there a way to get a set $X_i$ from the infinite index number $10110\ldots$
$ X_i = \{1,10,101,1011,10110,\ldots\}$ where $i \in I$ and $i=10110\ldots$
AI: If the elements of $X_f$ are supposed to be decimal numbers, then this would work:
$$X_f = \left\{\sum_{i=0}^n f(i)10^{n-i}\,\middle\vert\, n \in \Bbb N\right\}$$
If the elements of $X_f$ are supposed to be written in binary instead, replace the $10$ by a $2$.
Added: The $\sum$ notation denotes a sum. $f(i)$ is simply the $i$th element of your sequence. In this case, $$\sum_{i=0}^n f(i)10^{n-i} = f(0)10^n + f(1)10^{n-1}+\ldots+f(n)10^0$$
So for $n=0$, we have $f(0)10^0 = 1$. For $n=1$, we have $f(0)10^1+f(1)10^0 = 10+0=10$. For $n=2$, we arrive at $101$ (try it!).
The iteration variable $i$ is called the index (variable). In this example, $\sum\limits_{i=0}^n$ means "sum the results of replacing $i$ by $0, 1,2,\ldots n$ in the expression on the right". A Google search on "sum notation" or "sigma notation" will likely help you further.
On the other hand, if you want $X_f$ to be the set of finite (non-empty) initial segments of $f$ (i.e. the sequences $(1), (1,0), (1,0,1)$ etc.) then this would work:
$$X_f = \left\{f \restriction_{\Bbb N_n}: \Bbb N_n \to \{0,1\} \,\middle\vert\, n \in \Bbb N\right\}$$
where $\Bbb N_n = \{0,1,\ldots n\}$ and $f\restriction_{\Bbb N_n}$ is the function obtained by restricting the domain of $f$ to $\Bbb N_n$. |
H: p.d.f of function of random variable
Suppose $X$ is a continuous random variable with p.d.f:
$$
f_X(x) = \begin{cases}1 & x \in [0, \frac{1}{2}) \\ 1 & x \in [1, \frac{3}{2}) \\ 0 & \text{otherwise} \end{cases}
$$
What is the p.d.f of $Y = {(X - 1)}^2$?
Let's plot $g(X) = {(X - 1)}^2$ over the interval $0 \leq x \leq \frac{3}{2}$:
$g(X)$ is one-to-one over the interval $0 \leq x < \frac{1}{2}$ and is not one-to-one over $\frac{1}{2} \leq x \leq \frac{3}{2}$
So, consider $0 \leq y < \frac{1}{4}\quad$ ($\frac{1}{2} \leq x < \frac{3}{2}$):
$$
\begin{eqnarray*}
F_Y(y)&=&P(Y \leq y)\\
&=&P({(X - 1)}^2 \leq y)\\
&=&P(\sqrt{{(X - 1)}^2} \leq \sqrt{y})\\
&=&P(1 - \sqrt{y} \leq X \leq 1 + \sqrt{y})\\
&=&F_X(1 + \sqrt{y}) - F_X(1 - \sqrt{y})\\
f_Y(y)&=&F'_Y(y)\\
&=&F'_X(1 + \sqrt{y}) - F'_X(1 - \sqrt{y})\\
&=&f_X(1 + \sqrt{y})\cdot(1 + \sqrt{y})' - f_X(1 - \sqrt{y})\cdot(1 - \sqrt{y})'\\
&=&1\cdot\frac{1}{2}y^{-\frac{1}{2}} - 0\cdot(1 - \sqrt{y})'\\
&=&\frac{1}{2}y^{-\frac{1}{2}}
\end{eqnarray*}
$$
Note that $f_X(1 - \sqrt{y}) = 0$ because $f_X(x) = 0$ over $\frac{1}{2} \leq x < 1$
Now, consider $\frac{1}{4} \leq y < 1\quad$ ($0 \leq x < \frac{1}{2}$):
$$
\begin{eqnarray*}
F_Y(y)&=&P(Y \leq y)\\
&=&P((X - 1)^2 \leq y)\\
&=&P(\sqrt{(X - 1)^2} \leq \sqrt{y})\\
&=&P(- (X - 1) \leq \sqrt{y})\\
&=&P(X \geq 1 - \sqrt{y})\\
&=&1 - P(X \leq 1 - \sqrt{y})\\
&=&1 - F_X(1 - \sqrt{y})\\
f_Y(y)&=&F'_Y(y)\\
&=&- F'_X(1 - \sqrt{y})\\
&=&- f_X(1 - \sqrt{y})\cdot(1 - \sqrt{y})'\\
&=&- 1\cdot(- \frac{1}{2}y^{-\frac{1}{2}})\\
&=&\frac{1}{2}y^{-\frac{1}{2}}
\end{eqnarray*}
$$
Note that $\sqrt{(X - 1)^2} = - (X - 1)$ because $g$ is a decreasing function of $X$ over $0 \leq x < \frac{1}{2}$
So,
$$
f_Y(y) = \begin{cases}\frac{1}{2}y^{-\frac{1}{2}} & y \in [0, \frac{1}{4}) \\ \frac{1}{2}y^{-\frac{1}{2}} & y \in [\frac{1}{4}, 1) \\ 0 & \text{otherwise} \end{cases}
$$
Is this correct?
I did check and $\int_0^1 f_Y(y)\,\mathrm{d}y = 1$, but I am interested in the reasoning too.
AI: Actually, the function is one-to-one on the relevant domain. And the image (and hence the range of $Y$) is $(1/4,1] \cup [0,1/4]=[0,1]$
Hence, you can simply apply the formula $$f_Y(y)= \frac{f_X(x)}{|g'(x)|}=\frac{1}{2 |x-1|}$$
But $|x-1|= \sqrt{y}$
Hence $$f_Y(y) = \frac{1}{2 \sqrt{y}} , \hspace{1cm} y\in (0,1]$$
Your procedure is right, of course. |
H: term for a "squared simplex"
The set of points
$$\{(x_0,...,x_n)|\forall{i}: x_i \in [0,1], \ and \ x_0+..+x_n=1\}$$
is an n-simplex.
What can I call a set of points:
$$\{(x_0,...,x_n,y_0,...,y_n)|\forall{i}: x_i,y_i \in [0,1], \ and \ x_0+...+x_n = 1 \ and \ y_0+...+y_n=1\}$$ ?
for $n=1$, a 1-simplex is a line segment, and my shape is a square. This shape is like a "square simplex" - it is a simplex in $n$ dimensions, and a simplex in the other $n$ dimensions, but not in $2n$ dimensions.
Does this shape has a standard name? If not, what could be a good name for it?
AI: What you call an"$n$-simplex" is in fact an $n$-dimensional hyperplane in $\Bbb R^{n+1}$. The second expression defines a $2n$-dimensional hyperplane in $\Bbb R^{2n+2}$.
Edit: Regarding your revised question: what you have isn't a type of simplex but a type of square (i.e. a square of a simplex). The term "simplex square" seems not to be laden with other mathematical meanings; so it would be one to consider. |
H: Solid Angle Calculation - Understanding a formula
I'm currently reading a paper and try to understand this one formula. The problem is: In an n dimensional space. A cone with half-angle $\theta$ is given (the top of the cone is in the origin). We are interested in the solid angle of the cap cut out by this cone of the unit sphere.
According to this paper, the procedure is: Sum up all contributions of ring-shaped elements of area.
The formula follwing from this procedure is
$\Omega(\theta_1) = \frac{(n-1)\pi^{(n-1)/2}}{\Gamma(\frac{n+1}{2})} \int_0^{\theta_1} (sin \theta)^{n-2} d\theta$
Obviously, the first fraction is the surface of the unit (n-1)-ball. What I don't understand are 2 points:
Why does he choose the (n-1) surface instead of the n surface? (Later, the calculates the fraction of this solid angle by dividing by the n-ball surface)
Where does this integral come from?
I hope somebody does have an idea...
If anyone is interested, the paper I'm reading is 'Probability of Error for Optimal Codes in a Gaussian Channel' by Claude Shannon
AI: The integral represents how much of the surface is cut out. Since $\theta$ is measured at the center of the sphere, and it is a unit sphere, $sin\theta$ will give the radius at the surface.
Consider n=3:
The first fraction will evaluate to $2\pi$, the circumference of the unit circle (1-sphere).
Notice that $\Omega(\theta_1) = \int_0^{\theta_1} \!2\pi r \,d\theta$.
Consider n=4:
The first fraction will evaluate to $4\pi$, the surface area of the unit 2-sphere.
Notice that $\Omega(\theta_1) = \int_0^{\theta_1} \!4\pi r^2 \, d\theta$.
So yes, you do sum up all contributions of ring-shaped elements of area. In an $n$-dimensional space, each element of area will be that of a $(n-2)$-sphere. Multiply the element of area for the unit $(n-2)$-sphere by the radius raised to the $(n-2)$th power, and integrate the product over $\theta$. |
H: Given a non-abelian group $G$ with $|G|=p^3$ and $p$ prime, how do I show that $|Z(G)|=p$?
Given a non-abelian group $G$ with $|G|=p^3$ and $p$ prime, how do I show that $|Z(G)|=p$? $Z(G)$ is as always center of $G$.
It is easy to see that $|Z(G)|\in\{p, p^2\}$, but how do I exclude $p^2$. Thanks in advance!
AI: Hint:
If $\frac{G}{\rm{Z}(G)}$ is cyclic then $G$ is abelian. |
H: Showing that function are equal almost everywhere in Sobolev Spaces
Consider the Holder space $C^{0,1-\frac{n}{p}}(\mathbb{R}^{n})$ and the Sobolev Space $W^{1,p}(\mathbb{R}^{n})$. Take $u_{m} \in C_{c}^{\infty}(\mathbb{R}^{n})$ such that Morrey's Inequality we have $||u_{m}||_{C^{0,1-\frac{n}{p}}(\mathbb{R}^{n})} \leq C||u_{m}||_{W^{1,p}(\mathbb{R}^{n})}$. We are also given that $u_{m} \rightarrow \bar{u}$ in $W^{1,p}(\mathbb{R}^{n})$ and $u_{m} \rightarrow u^{*}$ in $C^{0,1-\frac{n}{p}}(\mathbb{R}^{n})$.
How does it follow that $u^{*} = u \text{ a.e}$? What result is used?
Thanks for assistance. Let me know if anything is unclear.
AI: Since $u_m \to u^*$ in $C^{0,1-\frac np}$ you know that $u_m(x) \to u^*(x)$ for all $x$. On the other hand, since $u_m \to u$ in $W^{1,p}$ (and in particular in $L^p$) there is a subsequence with the property that $u_{m_j}(x) \to \bar u(x)$ almost everywhere. Since $u_{m_j}(x) \to u^*(x)$ for all $x$, you get a.e. equality. |
H: Negating the Definition of a Convergent Sequence to Find the Definition of a Divergent Sequence
My task is to write a precise mathematical statement that "the sequence $(a_n)$ does not converge to a number $\mathscr l$"
So, I have my definition of a convergent sequence:
"$\forall\varepsilon>0$ $\exists N\in\Bbb R$ such that $|x_n -\mathscr l|<\varepsilon$
$\forall n \in \Bbb N$ with $n>N$"
Would the correct negation of this be "$\forall\varepsilon>0$ $\exists N\in\Bbb R$ such that $|x_n -\mathscr l|>\varepsilon$
$\forall n \in \Bbb N$ with $n>N$"?
It doesn't seem that this is the answer as the next part of my task is to prove that a sequence is divergent using my formed proof, but it'd be difficult to do since it's a general proof of divergence and not just a proof that $(a_n)$ doesn't converge a specific number $\mathscr l$
Perhaps I should find a prove that $(a_n)$ tends to $\pm\infty$? This is more simple but it does not include monotone sequences such as $x_n:=(-1)^n$.
Can someone assist me with this task? All comments and answers are appreciated.
AI: This is not the correct negation. Consider $x_n = (-1)^n$ and $l = 1$. The correct negation can be expressed as
$$\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon$$ |
H: which one of the following probability mass function can define a probability distribution?
a) $f(x)=(5-x^2)/6$ for $x=0,1,2,3$
b) $f(x)=x/15$ for $x=1,2,3,4,5$
c) $f(x)=1/2^x$ for $x=0,1,2,3,4$
d) $f(x)=1/4$ for $x=2,3,4,5,6$
Can you please suggest me how to solve these questions ?
AI: Hint: The probabilities should sum to 1 and all be non-negative. Which of your alternatives satisfy these two conditions? |
H: Convert output of atan to range -1, 1,-1
I'm trying to convert the output of an arctan function from a range of -PI,0,PI to -1,1,-1 - in fact I've succeeded! But it's not very elegant:
angle = Mathf.Atan2(y, x);
angle = Mathf.Abs(angle);
angle /= Mathf.PI;
angle = angle - 1f;
angle = Mathf.Abs(angle);
angle *= 2f;
angle -= 1f;
So that does the job, but for the sake of my self-improvement, how could it be done more efficiently?
Thanks for looking & hope it's ok to post code here.
AI: How about $f(\theta)=1-2\frac {|\theta|}{\pi}$? It takes $\theta$ ranging over $[-\pi,\pi]$ and folds it so both ends go to $-1$ and the middle goes to $1$. That seems to be what you mean by $-1,1,-1$. |
H: Show that $\sqrt{n+1}-\sqrt{n}\to0$
Let $\ a_n=\sqrt{n+1}-\sqrt{n}$. I have to show that $\lim_{n\to \infty}a_{n}=0$.
How should I start? Do I have to use any theorem?
AI: Use the fact that
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}.$$ |
H: Help with Elementary Vector subspace proof
I am having trouble getting started with a proof of the following statement (I translated it from German, so bear with me):
"$V$ is a vector space and $U$ and $W$ are vector subspaces of $V$. Prove that the following two are equivalent:
a) $U \cap W = \{0\} $
b) Every $v \in U+W$ (with $U+W$ denoting the direct sum) can only be represented in one unique way as $v=u+w$ with $u \in U$ and $w \in W$."
I know that to prove equivalence (i.e. $a \Leftarrow\Rightarrow$ b), I have to show that $a \Rightarrow b$ and $b \Rightarrow a$ but I am having trouble getting anywhere. I attempted to assume $a$ is true and $b$ is false and prove by contradiction, but I couldn't get anywhere. Could you give me one or two hints so I can solve this on my own? Thank you! :)
AI: Hint:
Assume $$v=u_1+w_1=u_2+w_2$$ with $u_i\in U$ and $w_i\in W$. Then
$$0=v-v=(u_1-u_2)+(w_1-w_2)$$
Now the first summand is in $U$ the second summand lies in $W$. Can you show that if the $u_i$ and $w_i$ are distinct that the intersection $U\cap W$ can't be trivial?
For the other direction: Assume
$$v=u+w$$
and $x\in U\cap W$. Now consider
$$v=(u+x)+(w-x)$$ |
H: Covering an area equally with layers of non-tesselating polygons
A series of hexagons on an hexagonal lattice means that the every point in the entire area is covered by one polygon only.
A grid of octagons will not tesselate, leaving square holes such that 4/18 of the area is not covered.
Is it possible to stack multiple layers of identical, non-tesselating, regular polygons such that every point in the area underneath is covered by the same number of integer polygons? e.g. every point being covered by n - m polygons from n layers of k-sided polygons spaced on a specified lattice (m < n).
I have started by considering the case of the octagon grid. I don't think it is possible with this basis, as it would require being able to tesselate other squares around the square vacancies, which they cannot. Are other values of k viable?
AI: If you don't require regular octagons, you can do it. In the tessellation below, let the horizontal and vertical sides of the octagons be $1$ and the diagonal sides of the octagons be $\sqrt 2$. Then the distance between squares is $3$ and the squares cover $\frac 18$ of the plane. You can stack eight of these and cover each point seven times. You can do a similar thing with the canonical dodecagon/equilateral triangle tiling-make the sides of the dodecagons that touch each other the proper value and the triangles will fill the plane. |
H: evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n}$
I am trying to compute the sum $$\sum_{n=1}^\infty \frac{n^2}{3^n}.$$
I would prefer a nice method without differentiation, but if differentiation makes it easier, then that's fine.
Can anyone help me?
Thanks.
AI: Here's a bit of a slick trick. Let's put $$S=\sum_{n=1}^\infty\frac{n^2}{3^n}$$ and $$L=\sum_{n=1}^\infty\frac{n}{3^n}.$$ You should be able to determine (via ratio test or some such) that both series are convergent. (You'll see why I brought up the other series in a moment.)
Note that $$\begin{align}S &= \frac13+\sum_{n=2}^\infty\frac{n^2}{3^n}\\ &= \frac13+\sum_{n=1}^\infty\frac{(n+1)^2}{3^{n+1}}\\ &= \frac13+\sum_{n=1}^\infty\frac{n^2}{3^{n+1}}+2\sum_{n=1}^\infty\frac{n}{3^{n+1}}+\sum_{n=1}^\infty\frac1{3^{n+1}}\\ &= \frac13+\frac13 S+\frac13 L+\frac13\sum_{n=1}^\infty\frac1{3^n}\\ &= \frac13 S+\frac13 L+\frac13\sum_{n=0}^\infty\frac1{3^n},\end{align}$$ so $$\frac23 S=\frac13 L+\frac13\sum_{n=0}^\infty\frac1{3^n},$$ so $$S=\frac12 L+\frac12\sum_{n=0}^\infty\frac1{3^n}.\tag{$\star$}$$ Now a similar trick shows us that $$\begin{align}L &= \frac13+\sum_{n=2}^\infty\frac{n}{3^n}\\ &= \frac13+\sum_{n=1}^\infty\frac{n+1}{3^{n+1}}\\ &= \frac13L+\frac13\sum_{n=0}^\infty\frac1{3^n},\end{align}$$ so $$\frac23 L=\frac13\sum_{n=0}^\infty\frac1{3^n},$$ so $$L=\frac12\sum_{n=0}^\infty\frac1{3^n},$$ and so by $(\star),$ we have $$S=\frac34\sum_{n=0}^\infty\frac1{3^n}.$$ All that's left is to evaluate the geometric series. |
H: Approximating the modulus of a Complex Function near a point.
Let $\Omega$ be a domain in $\mathbb{C}$, and let $z_0 \in \Omega$.
Let $f$ be analytic on $\Omega$.
Let $z=z_0+re^{i\theta}$ for $r$ small.
Assume that $f(z_0) \neq 0$ and $f'(z_0) \neq 0$.
I want to show $|f(z)| = |f(z_0)|\big(1+\lambda r \cos(\theta+\phi)+O(r^2)\big)$
where $\displaystyle \frac{f'(z_0)}{f(z_0)}=\lambda e^{i\phi}, \lambda > 0.$
Now:
$f(z)=f(z_0)+f'(z_0)(z-z_0)+O((z-z_0)^2) = $
$ f(z_0)+f'(z_0)(re^{i\theta})+O(r^2) =$
$|f(z_0)|\big | 1+\lambda r e^{i(\theta+\phi)} + O(r^2) \big | =$
$|f(z_0)|\big | 1+\lambda r (\cos(\theta+\phi)+i\sin(\theta + \phi)) + O(r^2) \big |$
It is not clear to me how I finish this off...
AI: Let us for simplicity write $\psi = \theta + \phi$. When dealing with absolute moduli, it is often convenient to square them. Here, that yields
$$\begin{align}
\lvert 1 + \lambda re^{i\psi}\rvert^2 &= \lvert (1+\lambda r\cos\psi) + i\lambda r\sin\psi\rvert^2\\
&= (1+\lambda r\cos\psi)^2 + \lambda^2 r^2\sin^2\psi\\
&= 1 + 2\lambda r\cos\psi + \lambda^2r^2.
\end{align}$$
Taking the square root, using $\sqrt{1+x} = 1 + \frac{x}{2} + O(x^2)$ yields
$$\lvert 1 +\lambda re^{i\psi}\rvert = 1 + \lambda r\cos\psi + \frac12\lambda^2r^2 + O(\lambda^2r^2) = 1 + \lambda r\cos\psi + O(r^2)$$
as desired. |
H: $G=\{(x,y)\in\Bbb R^2 | 1\leq x^2 + y^2\leq16 \text{ and } 0\leq x \text{ and }0\leq y \}$ Calculate $\underset{G}{\int\int}xy^2dxdy $
I'm making some exercises for my analysis exam, and i'm having trouble with this exercise.
$G=\{(x,y)\in\Bbb R^2 | 1\leq x^2 + y^2\leq16 \text{ and } 0\leq x \text{ and }0\leq y \}$
Calculate $$\underset{G}{\int\int}xy^2dxdy $$
I think that:
$$\underset{G}{\int\int}xy^2dxdy = \int_0^{\tfrac{1}{2}\pi}\int_1^4 r^4 \cos \theta \sin \theta^2 dr d\theta $$
But I don't how to continue. Am I on the right track?
AI: As you've got:
$$
\underset{G}{\int\int}xy^2dxdy = \int_0^{\tfrac{1}{2}\pi}\int_1^4 r^4 \cos \theta \sin^2 \theta dr d\theta = \int_1^4 r^4dr\int_0^1 u^2 du=
\left. \frac{r^5}{5}\right|_{r=1}^4 \, \left.\frac{u^3}{3}\right|_{u=0}^1=
\frac{4^5-1^5}{5}\frac{1-0}{3}=\frac{1023}{15},
$$
where I used the substitution $u=\sin \theta \Rightarrow du=\cos \theta d\theta$. |
H: How to solve $\lim_{x\to-\infty}x^2e^x$
How to solve the following limit?
\begin{eqnarray}
\\\lim_{x\to-\infty}x^2e^x\\
\end{eqnarray}
According to the website WolframAlpha, L'H rule can be used here but it is $-\infty/1$ instead of $-\infty/-\infty$, $\infty/\infty$ or $0/0$. So how to solve this question?
Thank you for your attention
AI: HINT:
Putting $y=-x,$ $$\lim_{x\to-\infty}x^2e^x=\lim_{y\to\infty}y^2e^{-y}=\lim_{y\to\infty}\frac{y^2}{e^y}$$ which is of the form $\frac{\infty}{\infty}$
So, apply L'H rule twice as the exponent of $y$ in the numerator is $2$ |
H: What is the remainder when $2^{1990}$ is divided by $1990$?
I actually do not have the basic idea on how to approach these type of questions....so please tell me a generalized method about all this too.
It came in RMO, and the question is:
What is the remainder when $2^{1990}$ is divided by $1990$ ?
AI: Using Fermat's Little Theorem,
$2^4\equiv1\pmod 5\implies 2^{1990}=2^2\cdot(2^4)^{997}\equiv4\cdot1^{997}\pmod 5\equiv4\ \ \ \ (1)$
$2^{198}\equiv1\pmod {199}\implies 2^{1990}=2^{10}\cdot(2^{198})^{10}\equiv1024\cdot1^{10}\pmod{199}\equiv 29\ \ \ \ (2)$
Clearly, $2^{1990}\equiv0\pmod2\ \ \ \ (3)$
Apply the famous CRT on $(1),(2),(3)$ |
H: Is $\lambda(n) + \max\limits_{p\mid n} v_p(n)\leqslant n$?
Given an integer $n = \prod\limits_{n\mid p}p^{v_p(n)}$, is
$$\lambda(n) + \max\limits_{p\mid n} v_p(n)\leqslant n$$
where $\lambda(n)$ is the Carmichael function?
AI: For a single-prime factor, i.e. $n=p^v$, this is true, since
$$\begin{align}
\lambda(p^v) &\leqslant \phi(p^v) = p^v-p^{v-1} \stackrel{v\geqslant1}\leqslant p^v - v
\end{align}$$
(since $p^{v-1}-v$ is strictly increasing in $v$) and therefore $$\lambda(p^v)+v\leqslant\phi(p^v)+v\leqslant p^v.$$ Note how for $v\geqslant\begin{cases}3 & p=2 \\ 2 & p > 2\end{cases}$ this is even $\phi(p^v)+v < p^v$.
For $n=m\cdot p^v$ where $m\geqslant2$ and $p$ are coprime, observe that
$$\begin{align}
\lambda(n) &= \lambda(m\cdot p^v)
\\ &\leqslant \phi(m\cdot p^v) = \overbrace{\phi(m)}^{< m}\overbrace{\phi(p^v)}^{\leqslant p^v - v}
\\ &< m(p^v-v) = (mp^v) - \underbrace{\overbrace{m}^{\geqslant v_\max(m)}v}_{\geqslant vv_\max(m)\geqslant v_\max(n)}
\\ &\leqslant n - v_\max(n)
\end{align}$$
where $v_\max(m) := \max\limits_{p\mid m}v_p(m)$.
So in summary the even stronger relation
$$\phi(n) + v_\max(n) < n \quad n\in\mathbb N\backslash(\mathbb P\cup\{4\})$$
holds for all composite $n\neq4$. |
H: Why am I getting half the correct answer by using Green's Theorem?
I had this homework problem that asked me to use Green's Theorem to solve it, so I did. Unfortunately, my answer was wrong. I looked for an error in my reasoning, but did not find it. I eventually solved by way of the line integral, which is usually harder than using Green's Theorem, so now I know that the correct answer is $-64\pi$. Yet, the answer I get from trying to use Green's Theorem is $-32\pi$, exactly half. Below is my attempt, and I'd appreciate a pointer as to where my error is.
$\vec{F} = 4y\hat{i} + 5xy\hat{j}$
$C =$ circle of radius $4$ centered on the origin, oriented counter-clockwise.
$$\begin{align*}
\int_C \vec{F}\cdot\text{d}\vec{r} &= \int^{2\pi}_0\int^4_0 (5r\sin(\theta)-4) \text{ d}r\text{ d}\theta \\
&= \int^{2\pi}_0 \left.\left( \frac{5}{2}r^2\sin(\theta) - 4r \right)\right|^{r=4}_{r=0} \text{ d}\theta \\
&= \int^{2\pi}_0 (40\sin(\theta)-16) \text{ d}\theta \\
&= \left. (-40\cos(\theta)-16\theta) \right|^{\theta=2\pi}_{\theta=0} \\
&= -32\pi \\
&\ne -64\pi
\end{align*}$$
AI: When you switch to polar coordinates, you should replace $dx dy$ with $r dr d\theta$. You are missing a factor of $r$. |
H: Prove that $f$ is constant
Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be an entire function. if there exists $\delta> 0$ and $w\in \mathbb{C}$ such that
$$\left | f(z)-w \right | \geq \delta \qquad \forall z\in\mathbb C $$
Prove that $f$ is constant.
AI: Let $g(z) = \frac{1}{f(z)-w}$, then $g$ is entire and $|g(z)| \le \delta$ for all $z$. What can you say about $g$ and hence $f$? |
H: Prove that the identity map $(C[0,1],d_1) \rightarrow (C[0,1],d_\infty)$ is not continuous
$$d_\infty = \max|x_i - y_i|$$
$$d_1 = \sum_{i=1}^n |x_i - y_i|$$
The first part of this question was to prove that the identity map $$(C[0,1],d_\infty) \rightarrow (C[0,1],d_1)$$ is continuous, which I did using $\epsilon = n \delta$. I'm not sure how to prove it is not continuous in the opposite direction. I tried using my logic in reverse but it doesn't seem to be working. Can anyone offer any hints or suggestions?
AI: Is the identity map $C[0,1],d_1\to C[0,1],d_{\infty}$ bounded? $C[0,1],d_1$ is incomplete
$f_n(x)= 2nx; x\in [0,{1\over 2n}]$
$f_n(x)= 2-2nx; x\in [{1\over 2n},{1\over n}]$
$f_n(x)= 0$ else
$f_n$ converges to $0$ in $d_1$, it does not convrges to $d_{\infty}$ norm,since the value of $\sup f_n(x)=1$ always |
H: Bound for analytic function on a disk given values at 0
I am trying to prove the following problem from An Introduction to Complex Function Theory by Palka.
If $f$ is analytic in the unit disk $D(0,1)$, and if $f(0)=f'(0)=\ldots= f^{(k)}(0)=0$, and $|f^{(k)}(z)|\leq 1$ for every $z\in D(0,1)$, then show that $|f(z)|\leq |z|^{k+1}/(k+1)!$ in the disk.
I tried to use Schwarz lemma or maximum modulus principle, but what I get is something like $|f(z)|\leq |z|^{k+1}$, and I can't seem to use all the information given.
AI: Hint:
By Schwarz Lemma, $|f^{(k)}(z)|\le|z|$.
Using $g(z)-g(0)= \int_0^zg'(w)dw$ for any holomorphic $g$ defined on $D(0,1)$ or not, prove that $|f^{(k-j)}(z)|\le \frac{|z|^{j+1}}{(j+1)!}$ for $j\le k$ by induction on $j$. |
H: For sets $A,B,C$, $(A\setminus B)\subset (A\setminus C)\cup (C\setminus B)$
First of all, I am sorry for my bad english, I am from Brazil :-) I have problem with proof for some set theory task.
Here it is:
$A,B,C$ are three sets. Show that:
$$(A\setminus B) \subset (A\setminus C) \cup (C\setminus B)$$
It is clear by looking at diagrams but I do not know how I show! It is not too hard I think...
Thank you for the help,
Igor
AI: Suppose $\;x\in A\setminus B\;$ . We have only two possibilites:
$$\begin{align*}(1)&\;\;x\in C : \;\;\implies x\in C\setminus B\implies x\in (A\setminus C)\cup (C\setminus B)\\
(2)&\;\;x\notin C :\;\;\implies x\in A\setminus C\implies x\in (A\setminus C)\cup (C\setminus B)\end{align*}$$ |
H: Proving that $\frac{1}{2} \le \frac{1}{2^n+1} + \frac{1}{2^n+2} + ... + \frac{1}{2^n + 2^n}$
I'm having trouble proving that:
$$\frac{1}{2} \le \frac{1}{2^n+1} + \frac{1}{2^n+2} + ... + \frac{1}{2^n + 2^n}$$
Edit: The next step is actually a mistake. I've put up a comment to the accepted answer explaining why it's a mistake.
I've tried reducing every single fraction on the right side to $\frac{1}{2^n + 2^n}$, but then I get result:
$$\frac{1}{2} \le \frac{1-\frac{1}{2^n}}{2}$$
which is not true.
I also tried to put this harmonic sequence in Wolfram Alpha and it tells me it equals $\ln(2)$, which is greater than $\frac{1}{2}$.
AI: For any $\;1\le k\le 2^n\;$ we have that
$$\frac1{2^n+k}\ge\frac1{2^n+2^n}=\frac1{2^{n+1}}\;,\;\;\text{so:}$$
$$\underbrace{\frac1{2^n+1}+\ldots+\frac1{2^n+2^n}}_{2^n\;\text{summands}}\ge\frac{2^n}{2^{n+1}}=\frac12$$ |
H: I am trying to prove that for all $x \ge 1, (x^{\frac{1}{n}}) \to 1$.
$\textbf{Proof:}$
Let $x \ge 1$ be arbitrary and note that $\forall n \in \mathbb{N}$, $x^{\frac{1}{n}} > 1$.
So $ \forall n \in \mathbb{N},\ x^{\frac{1}{n}} - 1 \ge 0 > -1$.
Therefore I can write $x^{\frac{1}{n}} = 1 + ( x^{\frac{1}{n}} -1 )$ and use Bernoulli's inequality.
So $ \forall n \in \mathbb{N}, \left ( 1 + ( x^{\frac{1}{n}} -1) \right )^{n} \ge 1 + n( x^{\frac{1}{n}} -1)$
$\implies \> x \ge 1 + n( x^{\frac{1}{n}} -1) $
$\implies \> x - 1 \ge n( x^{\frac{1}{n}} -1) $
$\implies \> \frac{x - 1}{n} \ge ( x^{\frac{1}{n}} -1) \ge 0$
Note that as $n \to\infty, \left ( \frac{x-1}{n} \right ) \to 0 $. So by the sandwich theorem $ ( x^{\frac{1}{n}} -1) \to 0 $.
Therefore by the sum rule for convergent sequences, $(x^{\frac{1}{n}}) \to 1$.
I was wondering if my proof is correct, as it makes sense to me but I just to need to check it.
AI: For $n$ sufficiently large, $$1\le x\le n$$ $$\sqrt[n]{1}\le \sqrt[n]{x}\le \sqrt[n]{n}$$
By the squeeze theorem, as $n\to \infty$, since $\sqrt[n]{1}\to 1$ and $\sqrt[n]{n}\to 1$, also $\sqrt[n]{x}\to 1$. |
H: How to find the exact value of an upper bound for an exponential random variable
Suppose the waiting time between people entering a store can be modeled by the exponential random variable $X$ with parameter $\lambda=5$. If you use markov's inequality you can find the $P(X\ge 20)$ is $.25$. How would I find the exact value of $P(X\ge 20)$ ? Thanks for your help
AI: Use the fact that $P(X \ge 20) = 1 - P(X \le 20)$ and use the CDF of the exponential r.v. Thus your answer is $1 - F_X(20)$ where $X$ is exponential with $\lambda =5$ and $F_X$ is its CDF.
Let's do the computation:
$P(X \ge 20) = 1-F_X(20)= 1 - (1-e^{-(5)(20)})= 1 - 1 + e^{-100}= e^{-100} = 3.72 \times 10^{-44}$
Indeed, this is a very small number. Which makes sense, intuitively. If the parameter $\lambda = 5$ we have that the mean of the r.v. $X$ is $1/5 = 0.2$.
Putting this in words: the mean waiting time between two persons entering the store is 0.2 minutes. What is the probability that we wait TWENTY minutes or more for the next person to enter? (i.e. $P(X \ge 20)$)
It has to be very small. |
H: Finding the limit of $\frac{e^{x}+x-\cos(2x)}{x^2}$
How would one find the limit for the following problem.
$x\rightarrow\infty$
$\frac{e^{x}+x-\cos(2x)}{x^2}$
I did the hospital rule.
$\frac{e^x+1+2\sin(2x)}{2x}$
But now I am stuck I did this but I feel it diverges.
$e^x+1+2\sin(2x)*\frac{1}{2x}$
AI: $$\lim_{x\to\infty}\frac{e^x+x-\cos 2x}{x^2}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{e^x+1+2\sin 2x}{2x}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{e^x+4\cos 2x}2=\infty$$
The limit thus exists (in the wide sense of the word), since $\;\cos 2x\;$ is bounded. |
H: Find the domain of $g(x) = \ln \left( {\frac{x}{{x - 1}}} \right)$
I know the argument of the function has to be greater than 0, so:
$\eqalign{
& \left( {\frac{x}{{x - 1}}} \right) > 0 \cr
& x > 0 \cr} $
however in this case $x \ne 1$, $x \ne 0$ as they result in an answer which is undefined, so I think it's reasonable for me to say the domain is $x>1$, however this is not the correct answer.
AI: To find the actual intervals on which the function is defined, what we need is:
$$\Big(\underbrace{x-1 > 0 \;\text{ AND } \;x > 0}_{\large \;\rightarrow x \;\gt \;1}\Big) \;\text{ or else }\; \Big(\underbrace{x - 1 <0 \;\text{AND}\;x< 0}_{\large \rightarrow \;x\;< \;0}\Big)$$
So the the function's domain is given by $$(-\infty, 0) \cup (1,\infty)$$ |
H: I've seen "hyperbolic rotation" - from this: generalization to multisection rotation: is this possible?
This question is more in recreational mathematics area
By accident I came across the concept of "hyperbolic rotation" where we use a matrix containing $\cosh$ and $\sinh$ instead of the trigonometric $\cos$ and $\sin$ for the rotation by matrix-multiplication such that we have:
$$ \tag{trigonometric} \qquad T_t(\varphi) = \begin{bmatrix} \cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{bmatrix} $$
$$ \tag{hyperbolic} \qquad T_h(\varphi) = \begin{bmatrix} \cosh \varphi & \sinh \varphi \\ \sinh \varphi & \cosh \varphi \end{bmatrix} $$
A key-feature is surely, that both rotation-matrices have a determinant of $1$ ( because $$\small \cos^2(\varphi) + \sin^2(\varphi) = 1 \tag{trigonometric} $$ and $$\small \cosh^2(\varphi) - \sinh^2(\varphi) = 1 \tag{hyperbolic} $$ ) .
Now I toyed a bit to extend this to the case of 3-multisection series of the exponential; an example for what I mean is my older question; let's call that three functions just $f(x),g(x),h(x)$ such that $f(x)+g(x)+h(x) = \exp(x)$ and the analogon to the square-formulae which equal $1$ is:
$$ f(x)^3 + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1 \tag{3-multisection} $$
At least, such a "rotation"-matrix must have size of $3 \times 3$ but possibly even more - if it is constructable at all.
Qu1: Is such a generalization to higher multisections (here order 3) possible?
Qu2: and if, how could such a "rotation"- matrix be contructed?
[Update]: I've just found, that
$$ \tag{3-multisection} \qquad T_{3m}(\varphi) = \begin{bmatrix} f(\varphi) & h(\varphi) & g(\varphi)\\ g(\varphi)&f(\varphi)&h(\varphi)\\ h(\varphi) & g(\varphi) &f(\varphi) \end{bmatrix}$$
has determinant $1$ and could be a candidate model. But I didn't find nice properties so far. Perhaps it is even better to not to stick to the determinant 1-condition, but allow determinant $-1$ here; the "rotation"-matrix could then be a simple circulant one, like in the hyperbolic case.
The log of that matrix looks like
$$ \log(T_{3m}(\varphi))=\small \begin{bmatrix}
0 & 0 & \varphi \\
\varphi & 0 & 0 \\
0 & \varphi & 0
\end{bmatrix}$$
and I think it is a good hint, that this is equivalent to the rotation-matrices in the trigonometric/hyperbolic-cases, where the form of the matrix-log comes out to be much similar.
AI: The analogy I would suggest comes from quadratic forms. Your first two examples are are linear changes of variable that preserve a quadratic form. That is, a matrix $P$ such that $P^T A P = A.$ In the first one, $A= I,$ in the second $A$ is the diagonal matrix with diagonal entries $(1,-1.)$ This is very convenient for homogeneous degree two because of matrices; but we could write
$$ u = a x + b y, v = c x + d y, u^2 + v^2 = x^2 + y^2 ?? $$ and ask about possible $a,b,c,d.$ Same for $u^2 - v^2 = x^2 - y^2.$
So that suggests $$ u = ax+by+cz, v=dx+ey+fz, w=gx+hy+iz, $$ what about
$$ u^3 + v^3 + w^3 - 3 u v w = x^3 + y^3 + z^3 - 3 x y z ?? $$
and ask about possible $a,b,c,d,e,f,g,h,i.$ These linear variable changes make a group over the reals. Hmmm. They should, but I need to think about why the determinant of the evident 3 by 3 matrix is nonzero; obvious for the quadratic form case.
EDIT: got it, in the version I am pushing.
Given cubic form
$$ f(x,y,z) = x^3 + y^3 + z^3 - 3 x yz. $$
$$ f(x,y,z) =
\det \left( \begin{array}{ccc}
x & y & z \\
z & x & y \\
y & z & x
\end{array}
\right) . $$
$f(x,y,z) = \det(xI + y A + z A^2),$ where
$$ A = \left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}
\right) $$ Then $A^3 = I$ and $A^4 = A .$
Now, since both the characteristic and minimal polynomials for $A$ are $x^3 - 1,$
it follows that all matrices that commute with $A$ are of the form $rI + s A + t A^2.$ |
H: $C[0,1]$ endowed with integral norms
Consider the following norms on $C[0,1]$: for each $p\in [1,+\infty]$ $||x||_p=(\int_0^1|x(t)|^pdt)^{1/p}$ (this $p$-norm is induced from $L^p$ wich contains $C[0,1]$). Are they equivalent? Of course, $||x||_p$ is not equivalent to $||x||_{\infty}$ (it is easy to build a counterexample), but I also believe that $p$-norm is not equivalent to $q$-norm (when $p\ne q$). Really?
AI: Given $n\ge 1$, let $x_n(t)=t^n$, $\forall ~t\in[0,1]$. By definition, $x_n\in C[0,1]$, and for every $p\ge 1$,
$$\|x_n\|_p=\left(\frac{1}{1+np}\right)^{\frac{1}{p}}.$$
It follows that when $p>q\ge 1$,
$$\lim_{n\to\infty}\frac{\|x_n\|_p}{\|x_n\|_q}=\infty,$$
i.e. $\|\cdot\|_p$ and $\|\cdot\|_q$ are not equivalent on $C[0,1]$. |
H: How could you express the following double integral in term of a single integral?
How could I express the $$\int\int e^{(x^2+y^2)^2} dA$$ in terms of a single integral with respect to r where D is a disk with center (0,0) and radius 1
AI: Change to polar coordinates:
$$x=r\cos t\;,\;\;y=r\sin t\;,\;\;0\le t\le 2\pi \implies$$
$$\int\int\limits_D e^{(x^2+y^2)^2}dA=\int\limits_0^1\int\limits_0^{2\pi}re^{r^4} dt\,dr=2\pi\int\limits_0^1re^{r^4}dr$$ |
H: Find the $P{X < Y}$ from this joint density function $f(x, y) = 6x^2y$
I have a joint density function given as:
$$f(x, y) = 6x^2y \\
0\le x\le 1 \\
0\le y\le 1$$
Now I am asked to find the $P\{X < Y\}$, as well as the $P\{X < 2Y\}$
In order to solve this, I did the following integrals:
\begin{align}
P\{X < Y\} &= \int_0^1\int_0^Y6x^2y\,dx\,dy \\
&= 6\int_0^1y\,dy\int_0^Yx^2\,dx \\
&= 3\int_0^Yx^2\,dx \\
&= y^3
\end{align}
That seems correct to me, as plugging in a value of $1$ for $y$ results in a probability of $1$, which is intuitively correct.
And for $P\{X < 2Y\}$:
\begin{align}
P\{X < 2Y\} &= \int_0^1\int_0^{2Y} 6x^2y\,dx\,dy \\
&= 6\int_0^1y\,dy\int_0^{2Y}x^2\,dx
\end{align}
...same as above, but ending up with $\boxed{8y^3}$.
Now this intuitively seems correct as well, since plugging in a value of $0.5$ for $y$ is like asking what is the probability that $X$ is less than $2\cdot 0.5$, or 1, which should also be 1, and it is with $8\cdot(0.5)^3 = 1$
However what I'm confused about slightly now is the range. I assumed that the function range should be redefined as:
$$P\{X < 2Y\} =
\begin{cases}
8y^3, & 0 \le Y \le 1/2 \\
1, & 1/2 \le Y \le 1 \\
\end{cases}$$
...since obviously the probability cannot be greater than $1$, and the function continues up to $8$ within the original range. Does this all seem correct?
AI: You aren't entirely off track, but instead, you should have $$\int_0^1\int_0^y6x^2y\,dx\,dy.$$ We can certainly pull the $y$ out of the first integrand, but note that our limits of integration still have a $y$ in them, so we can't simply turn this into a product of integrals. Instead, evaluating the inner integral first, we have $$\int_0^1y\left(\int_0^y6x^2\,dx\right)\,dy=\int_0^1y(2y^3)\,dy,$$ and I'm sure you can take it from there. The other problem has similar issues and fixes.
The thing to keep in mind, here, is that $Y$ is the name of a function, that can take on values $y$. |
H: Does the interval (a, a) contain any real numbers, assuming the interval is on the real line?
The title pretty much describes my question. I apologize if the language is not as mathematically precise as it could be. Many thanks in advance for your thoughts.
AI: Hardly. The interval $(a, a)$ is empty; it doesn't even contain $a$.
On the otherhand, $[a, a] = \{a\}$. |
H: Inverse of Natural Projection?
May this be a silly question, but can I construct an inverse of a natural projection $p$ from a module $M$ to its quotient module $M/A$? Of course more than one element can be assigned for each coset in $M/A$, but if we limit the inverse's range to direct complement of $A$, I think such inverse could be well-defined.
As far as I know such inverse homomorphism might be useful for various situations but I have never seen any.
There were only isomorphisms from the quotient module $M/A$ itself. There might be some reason for this. Can anyone teach me? Thank you!
AI: You have an exact sequence
$$ 0 \longrightarrow A \overset{i}\longrightarrow M \overset{p}\longrightarrow M/A \longrightarrow 0.$$
By the Splitting lemma, this sequence splits, i.e. $M\cong A\oplus M/A$, exactly if there is a morphism $r:M/A\to M$ such that $p\circ r = \operatorname{id}_{M/A}$. Look at the proof of the splitting lemma to see how to construct $r$ from the isomorphism $M\overset{\sim}\rightarrow A\oplus M/A$ and how to construct this isomorphism from a given right-inverse $r$ of $p$. |
H: How to find the number of positive integral solutions for the equations $\frac1x+\frac1y=\frac1{n!}$?
I was trying to solve a question over hackerrank and the question link is EQUATION
How to approach for it?
AI: Hint :
$$ \frac{1}{x} + \frac{1}{y} = \frac{1}{n!} \Rightarrow x = \frac{n!y}{y-n!}$$
but $$ x \in \Bbb Z^+ \Rightarrow y-n! \mid n!y = n!y - n!^2 + n!^2 = n!(y-n!) + n!^2 \Rightarrow y - n! \mid n!^2$$ |
H: Finding the limit of $(1+2x)^{3\csc(2x)}$
Find the limit $\displaystyle{\lim_{x\to0}(1+2x)^{3\csc(2x)}}$
I did the following $(1+2x)^{3\csc(2x)}=e^{\ln(1+2x)3\csc(2x)}$, took $\ln$ on the lim getting
$3\frac{\ln(1+2x)}{\frac{1}{\csc(2x)}}$
I did hospital rule
and got
$\lim3\cdot2\frac{\frac{1}{2x}}{\frac{-1}{\csc(2x)cot(2x) 2}}$
then I find myself stuck on what to do.
AI: $$
\frac{\ln(1+2x)}{\frac1{\csc 2x}}=\frac{\ln(1+2x)}{\sin 2x}.
$$
Being a $0/0$ of differentiable functions, L'Hopital applies, so
$$
\lim_{x\to0}\frac{\ln(1+2x)}{\frac1{\csc 2x}}=\lim_{x\to0}\frac{\ln(1+2x)}{\sin 2x}=\lim_{x\to0}\frac{\frac2{1+2x}}{2\cos2x}=\lim_{x\to0}\frac1{(1+2x)\cos2x}=1
$$
Or, one can use Taylor:
$$
\frac{\ln(1+2x)}{\sin 2x}=\frac{2x+O(x^2)}{2x+O(x^3)}=\frac{1+O(x)}{1+O(x^2)}\to1.
$$ |
H: Finding the limit of $( 1 + a + a^2 + \ldots+ a^n)/(1 + b + b^2 + \ldots+ b^n)$
I have some problems finding the limit of $$\frac{ 1 + a + a^2 +\cdots + a^n}{1 + b + b^2 + \cdots + b^n}.$$
$0\le a,b \le +∞$
Here is what I got :
Forcefuly factorize $a^n$ and $b^n$ :
$$ \frac{a^n ( 1 + \frac1{a} + \frac1{a^2} + ... + \frac1{a^n})}{b^n( 1 + \frac1{b} + ... + \frac1{b^n})}
= \left(\frac ab\right)^n\cdot \frac{\dfrac{1-\frac{1}a^{n+1}}{1-\frac{1}a}}{\dfrac{1-\frac{1}b^{n+1}}{1-\frac{1}b}} $$
From here I'm kinda stuck since I don't know the limit of $\left(\frac ab\right)^n$ unless I take all the possible cases. Same goes for the other part.
I want to know the next step into solving this limit.
AI: The $a$ and $b$ parts of your fraction have their own limits, without taking any ratio. When $a$ and $b$ are both less than $1$, this answers the question. When $a$ or $b$ is $\geq 1$, both parts have $+\infty$ as their limit, but they are getting there at very different rates (or the same rate when $a=b$) and this solves the other case.
There should be no need to use L'Hopital rule, and there are indeed potentially $9$ cases, but most of them require no calculation. The significant cases are the ones in the previous paragraph. |
H: If every element of $R$ is irreducible or a unit, then $R$ is a field.
This is not for homework, but I seem to be stuck a would like a hint please. The question asks
If every nonzero element of an integral domain $R$ is either a unit or irreducible, then $R$ is a field.
The question looks non-threatening, and I'm surely missing something obvious. I started by choosing some nonzero $r \in R$. If $r$ is a unit then there is nothing to prove. If $r$ is irreducible, however, I don't immediately see how to conclude that $r$ is invertible. Any hints would be greatly appreciated!
AI: Hint: If $r$ is irreducible, then what can be said about $r^2$? What does this imply about $r$? |
H: A problem related to Rouche's Theorem.
In Rouche's theorem, If we replace analytic property of functions $f(z)$ and $h(z)$ with meromorphic then this theorem will not be valid anymore.
I want to illustrate this fact by producing some $f(z)$ and $h(z)$ which are meromorphic on some bounded domain D (where D have piecewise smooth boundary $\partial D$). Such that $f(z)$ and $h(z)$ have no poles on $\partial D$ and $|h(z)|<|f(z)|$, $\forall z\in\partial D$ but $f+h$ and $h$ have different number of zeros in D.
AI: (I assume you mean $f+h$ and $f$.) Take $D$ the closed unit disc, $h(z) = z^{-1}$ and $f(z) = 2$. |
H: Area below and above a curve compensate
Consider the graph below:
Don't worry about the units in the axes, it might look like a physics question, but my question is purely mathematical. Basically, the work done is the area under this force/displacement graph. For $x_0 = 10$, this work done is $100J$, which in our case also equals the kinetic energy, as there is no friction. Sorry if this sounds to physics-y, but I want to know at which displacement the kinetic energy equals 0.
So the physics behind it I've done; At $x=-10$, the KE reaches a maximum of $150J$. So I figured out that the area between the x-axis and the positive part of the line must equal the area between $x=10$ and $x=-10$, which as I've said is 150.
So now my question: How can I find at which value of $x$ the area below the line equals 150? In other words:
$$\displaystyle \int_{-q}^{-10} f(x) = 150 $$
Where $-q$ is the value we want. How can I obtain it?
AI: You can find the equation of $f(x)$ and then integrate it from $-q$ to $-10$. For the equation of $f$ you need its slope. The slope of $f$ is $-1$. So its equation is $y-y_0=m(x-x_0)$ where $m$ is the slope. Take $(x_0,y_0)=(0,-10)$, so
$$y-0=-1(x-(-10))=-x-10$$
$$\int_{-q}^{-10}(-x-10)\ dx=150$$ |
H: How to prove that $\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n)$
I know that $$\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n),$$ but I cannot find a way how to prove it. I tried induction but it did not work. On wiki they say that I should use differentiation but I do not know how to apply it to binomial coefficient.
Thanks for any response.
(Presumptive) Source: Theoretical Exercise 1.12(b), P18, A First Course in Pr, 8th Ed, by S Ross
AI: This is equivalent to proving $$\sum_{k=0}^n k^2 {n\choose k}=n(n+1)2^{n-2}.$$ Given $n$ people we can form a committee of $k$ people in ${n\choose k}$ ways. Once the committee is formed we can pick a committee leader and a committee planner. If we allow each person to hold both job titles there are $k$ ways for this to happen. If no person is allowed to have both job titles, then there are $k$ choices for the committee leader and $k-1$ choices for the committee planner for a total of $k(k-1)$ choices for a different committee leader and committee planner. The total number of choices for a committee leader and a committee planner is $k+k(k-1)=k^2$ . We now sum for all possible $k$ values from $0$ to $n$ to form every committee of $k$ people with a committee leader and a committee planner, that is, $\sum_{k=0}^n k^2 {n\choose k}$. We can count the same thing by first picking committee leaders and committee planners and then filling in the committee with the remaining people. If we allow each person to hold both job titles, then there are $n$ ways for this to be done. The remaining $n-1$ people can form the committee in $2^{n-1}$ ways since each person has two choices, either they are in the committee or they are not in the committee. The total number of committees where each person can hold both titles is $n2^{n-1}$. If no person is allowed to hold both job titles, then there are $n$ choices for a committee leader and $n-1$ choices for a committee planner. The remaining $n-2$ people can form the committee in $2^{n-2}$ ways since each person has two choices, either they are in the committee or they are not in the committee. The total number of ways to form a committee where no person can have both job titles is $n(n-1)2^{n-2}$. Thus the total number of ways to form the committee is $n2^{n-1}+n(n-1)2^{n-2}=n(n+1)2^{n-2}$. Hence $\sum_{k=0}^n k^2 {n\choose k}=n(n+1)2^{n-2}$.
Another way to do this is to consider $$(1+x)^n=\sum_{k=0}^n {n\choose k}x^k.$$ Differentiating both sides with respect to $x$ we obtain $$n(1+x)^{n-1}=\sum_{k=1}^n k{n\choose k}x^{k-1}.$$ Multiplying both sides of the equation by $x$ and differentiating with respect to $x$ we obtain $$n(1+x)^{n-1}+nx(n-1)(1+x)^{n-2}=\sum_{k=1}^nk^2{n\choose k}x^{k-1}.$$ Letting $x=1$ and simplifying the left-hand side we see that $$n(n+1)2^{n-2}=\sum_{k=1}^n k^2{n\choose k}.$$ |
H: Expectation of throwing $n$ balls into $n$ bins
Suppose we throw $n$ indistinguishable balls in $n$ bins at random. The throws are independent. What is the expected number of empty bins? What is the expected number of bins with one ball.
Using indicator random variables, expectations, some sloppy math and some questionable logic, I arrive at the conclusion that both are approximately $n/e$. I'm not able to find simple formulas. It also sounds very counter-intuitive to me.
Worse, I wrote a small Ruby program to simulate it and the results appear to be (approximately) correct.
Can anybody show me a good solution?
AI: The expected number of empties has been done endlessly. So we do the expected number with $1$ ball. Let $X_i=1$ if Bin $i$ has $1$ ball, and let $X_i=0$ otherwise.
The probability that $X_i=1$ is the probability that one of the $n$ balls falls into Bin $i$, and the others don't. This probability is $\binom{n}{1}\frac{1}{n}\left(1-\frac{1}{n}\right)^{n-1}$, which simplifies to $\left(1-\frac{1}{n}\right)^{n-1}$.
Let random variable $Y_1(n)$ be the number of bins with $1$ ball. Then $Y_1(n)=X_1+\cdots+X_n$, and therefore
by the linearity of expectation we have
$$E(Y_1(n))=n\left(1-\frac{1}{n}\right)^{n-1}.$$
And indeed
$$\lim_{n\to\infty}\frac{E(Y_1(n))}{n}=\lim_{n\to\infty} \left(1-\frac{1}{n}\right)^{n-1}=e^{-1}.$$ |
H: Proof about pointwise $\lim\limits_{x\to c} f(x)g(x)$
Let $f$ be bounded and let
$$\lim_{x\to c} g(x) = L$$
Prove:
(a) If $L=0$ then $\lim\limits_{x\to c} f(x) g(x) = 0$
(b) if $0<L<\infty$ then $\lim\limits_{x\to c} f(x) g(x)$ exists.
(c) $L=\infty$ then then $\lim\limits_{x\to c} f(x) g(x) = \infty$.
when we say exists of course this means limit is finite.
I have the thought that this may be as simple as creating what ever type of $f(x)$ and $g(x)$ I want and use $\epsilon$-$\delta$ as a proof.
AI: Let $|f| < M$ be the bound. Then we have
(a)
$$|\lim_{x\to c} f(x) g(x)| \leq |\lim_{x\to c} M g(x)| = M |\lim_{x\to c} g(x) | = 0$$
Can you do (b) and (c) in a similar way? |
H: True or False: Continuous Functions (Extreme Value Theorem)
Do my justifications seem appropriate?
a.) Every function $f:[0,1]\rightarrow \mathbb{R}$ has a maximum.
True; if $f:[0,1]\rightarrow \mathbb{R}$, $f$ wil be closed and bounded above, so it will have a maximum.
b.) Every continuous function $f:[a,b]\rightarrow \mathbb{R}$ has a maximum.
True; if $f$ is continuous on a closed, bounded interval, then it will have a min and a max.
c.) Every continuous function $f:(0,1)\rightarrow \mathbb{R}$ has a maximum.
False; $f$ is not on a closed interval.
d.) Every continuous function $f:(0,1)\rightarrow \mathbb{R}$ has a bounded image.
False; if $f$ has a vertical asymptote at the $0$ endpoint, the image will be unbounded.
e.) If the image of a continuous function $f:(0,1)\rightarrow \mathbb{R}$ is bounded below, then the function has a minimum.
False; only if the inf$f(D)$ is a functional value.
AI: For the true part(s), you may want to be more precise and refer to specific results. I can't fill in the details for you without knowing more about what class you're in (calculus 1? real analysis? topology?).
Your part (a) is wrong because $f$ is not assumed to be continuous. You can cook up as nasty a function as you want as a counterexample.
For parts (c), (d), and (e), you should probably write down counterexamples. For instance, in (c), you note: "$f$ is not on a closed interval." This is begging the question, because that's more or less what you're being asked to justify. There are many continuous functions on an open interval that are bounded, so you should explicitly produce a function on an open interval that is not bounded. |
H: an example of a simple ring which is not a division ring
I was thinking about a simple ring and I found a question that every simple ring is a division ring. I think this is not a correct theorem and I want to find an example.
AI: A matrix ring over a field $M_n(F)$ is simple but not a division ring for $n>1$.
I think another different example is furnished by infinite dimensional Clifford algebras. If you have a real, infinite dimensional vector space $V$ with a nondegenerate symmetric bilinear form $B$, I think that $C\ell(V,B)$ is a simple not-Artinian ring that isn't a domain.
Weyl algebras are simple domains which are not usually division rings. If I read this right, Weyl algebras can be thought of as Clifford algebras using symplectic forms rather than symmetric ones.
Commutative simple rings are all fields, so if that is what you proved, you are fine. |
H: $S_n$ and its subgroups
Show that $A_n$ is unique in $S_n$ with index $2$.
I'm trying to use Quotient Group and Lagrange's Theorem to approach this problem but I'm still clueless. Can anyone tell me how to do this problem? Thanks.
AI: As I comment above, for n=1,2,3,4, this can be done explicitly. Assume n $\geq 5$. We know that any index 2 subgroup must be normal (Quite easy to show). Let N be another subgroup of $S_n$ of index two. Then both $A_n$ and $N$ are normal, so the intersection $A_n \cap N$ is also normal (in $A_n, N$ and $S_n$). But, since $A_n$ is simple, this means that $A_n \cap N=\{1\}$. So, $|A_n \cup N|=n!-1$. So there is a single nonidentity element of $S_n$ not in either of our index two groups, call it $x$. Any normal subgroup is the union of conjugacy classes (of $S_n$ here). So $x$ lies in its own conjugacy class of $S_n$, and hence lies in the center of $S_n$. BUT, the center of $S_n$ is trivial. Contradiction. $x$ cannot be in the center. |
H: define the reals in a non-archimedean elementary extension of the real field.
Can it be done?
We have the real field $(\Bbb R,+,-,\times,0,1,<)$, of course $(0,1,-,<)$ are definable using the rest.
We take an elementary non-archimedean extension. Can we define the original set of reals in it?
AI: No, you cannot: You have the supremum property for bounded definable sets, since this is first-order and your extension is elementary. If $\mathbb R$ is definable, it would be bounded (by any infinite element of the field). So there would be a "least" infinite number. |
H: Center is never a maximal proper subgroup
Prove that center is not a maximal proper subgroup of group $G$.
AI: Hint: We can assume that $Z(G)\neq G$. If $x\in G\setminus Z(G)$ what can you say about $C_G(x)$? |
H: Injective function: example of injective function that is not surjective.
Does there exist an injective function that is not surjective? Could I have an example, please?
AI: There are many examples. It just all depends on how your define the range and domain.
For example $\operatorname{f} : \mathbb{R} \to \mathbb{R}$ given by $\operatorname{f}(x)=x^3$ is both injective and surjective. But then I can change the image and say that $\operatorname{f} : \mathbb{R} \to \mathbb{C}$ is given by $\operatorname{f}(x) = x^3$. Now it is still injective but fails to be surjective.
Similarily, the function $\operatorname{g} : \mathbb{R} \to \mathbb{R}$ given by $\operatorname{g}(x)=x^2$ is neither surjective nor injective. But if I change the range and domain to $\operatorname{g}: \mathbb{R}^+ \to \mathbb{R}^+$ then it is both injective and surjective.
Edit: As requested by the OP:
An example of an injective function $\mathbb{R}\to\mathbb{R}$ that is not surjective is $\operatorname{h}(x)=\operatorname{e}^x$. This "hits" all of the positive reals, but misses zero and all of the negative reals. But the key point is the the definitions of injective and surjective depend almost completely on the choice of range and domain. |
H: Does $A \Delta N = A \cup N'$ in this proof?
Please have a look at this topic:
$\sigma$- ideal
The answer says:
$A\Delta N=A\cup N'$ with $N'=N\setminus A$.
I do not see why this is correct..
Usually it is $A\Delta N=(A\setminus N)\cup (N\setminus A)$.
Unfortunately the helper does not answer and I really need a reason. I do not want to be impolite but i am in a hurry and need an answer to this little problem if possible.
AI: From what I can see, it is a false, or at best unjustified, assertion.
It holds iff $A \setminus N = A$, i.e. $A \cap N =\varnothing$. But we don't know that. |
H: $\int_0^{2\pi} \sqrt{1-\cos(x)}\,dx = 4\sqrt{2}$. Why?
According to the textbook, and Wolfram Alpha the above is correct.
Here is the step by step procedure from Wolfram Alpha for evaluating the indefinite integral:
Take the integral: $$\int\sqrt{1-\cos(x)}\,dx$$ For the integrand $\sqrt{1-\cos(x)}$, substitute $u=1-\cos(x)$ and $du=\sin(x)\,dx$: $$=\int-\frac{1}{\sqrt{2-u}}\,du$$ Factor out constants: $$=-\int\frac{1}{\sqrt{2-u}}\,du$$ For the integrand $1/\sqrt{2-u}$, substitute $s=2-u$ and $ds=-du$: $$=\int\frac{1}{\sqrt{s}}\,ds$$ The integral of $1/\sqrt{s}$ is $2\sqrt{s}$: $$=2\sqrt{s}+\text{constant}$$ Substitute back for $s=2-u$: $$=2\sqrt{2-u}+\text{constant}$$ Substitute back for $u=1-\cos(x)$: $$=2\sqrt{\cos(x)+1}+\text{constant}$$ Which is equivalent for restricted $x$ values to: $$\boxed{=-2\sqrt{1-\cos(x)}\cot\big(\frac{x}{2}\big)+\text{constant}}$$
I understand up to the below (which is a valid solution to the integral): $$2\sqrt{\cos(x)+1}+\text{constant}$$
However, if you evaluate this at $2\pi$ and $0$, you get the same thing, so the definite integral evaluates to zero.
After, you transform the above to: $$-2\sqrt{1-\cos(x)}\cot\big(\frac{x}{2}\big)+\text{constant}$$
The expression is indeterminate at $2\pi$ and $0$ of the form $0 \times \infty$. So I guess you would set up a limit and then use L'Hospital's rule to evaluate the expression at $2\pi$ and $0$ and get the answer to the definite integral?
In any case, all this seems strange. Why should the definite integral evaluated one way give $0$, and in another way give something else?
AI: $$
\int_0^{2\pi}\sqrt{1-\cos x}\,dx=\int_0^{2\pi}\sqrt{2\sin^2\frac{x}{2}}\,dx
=\sqrt{2}\int_0^{2\pi}\sin\frac{x}{2}\,dx=-2\sqrt{2}\cos\frac{x}{2}\Big|_0^{2\pi}=4\sqrt{2}.
$$ |
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