Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Derive a rotation from a 2D rotation matrix I have a rotation 2D rotation matrix. I know that this matrix will always ever only be a rotation matrix.
$$\left[
\begin{array}{@{}cc}
\cos a & -\sin a \\
\sin a & \cos a \\
\end{array}
\right]$$
How can I extract the rotation from this matrix?
The less steps, the better, since this will be done on a computer and I don't want it to constantly be doing a lot of computations!
| Pick any non-zero vector $v$ and compute the angle between $v$ and $Av$, where $A$ is the matrix above.
A simple vector is $e_1 = \binom{1}{0}$, and $Ae_1 = \binom{\cos \alpha}{\sin \alpha} = \binom{A_{11}}{A_{21}}$, hence the angle $\alpha$ can be computed from $\text{atan2}(\sin \alpha, \cos \alpha) = \text{atan2}(A_{21}, A_{11}) $. (Note that $\text{atan2}$ usually takes the $y$-component as the first argument.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How many odd numbers with distinct digits between 1000 and 9999 How many numbers with distinct digits are there between 1000 and 9999. [1]
I came up with a solution like this.
Since we can't know what numbers have been used, in the tens, hundreds and thousands we start counting at the ones.
1s: { 1, 3, 5, 7, 9 }, so 5 initial possibilities
10s: { 0, 1, ... , 9 }, so 10 initial possibilities, 1 taken: 9 left
100s: { 0, 1, ... , 9 }, so 10 initial possibilities, 2 taken: 8 left
1000s: { 1, 2, ... , 9 }, so 9 initial possibilites, 3 taken: 6 left
So then we arrive at the following: 5 * 9 * 8 * 6 = 2160 possibilities. I thought this was pretty straight forward.
Than I had a glimpse at the solution sheet... And lo an answer which really doesn't make much sense at its first glimpse.
Calculate the sum of those odd numbers with distinct digits with no 0’s, a 0 in
the tens place, or a 0 in the hundreds place. No 0’s: 5 choices for the ones place,
then 8 · 7 · 6 choices for the other three places; 0 in the tens place: 5 choices for the
ones place and 1 choice for the tens place, then 8 · 7 choices for the other two
places; 0 in the hundreds place: 5 choices for the ones place and 1 choice for the
hundreds place, then 8 · 7 choices for the other two places;
(5 · 8 · 7 · 6) + (5 · 1 · 8 · 7) + (5 · 1 · 8 · 7) = 2240;
Why are the 0's treated special? The exercise states it should be an odd number, with distinct digits. I thought I adhered to that proposition....
[1] Exercise 2.7.15 from Applied Combinatorics 2nd edition by Fred S. Roberts and Barry Tesman
| Your problem comes when you assume that there were $3$ taken from your last set. It's quite possible that the second or third choice was a $0$, so that you in fact have only eliminated $2$ options from the last set.
Instead, choose the last digit first ($5$ choices), then the first digit ($8$ choices, since we can't have picked $0$ for the last digit), then deal with the second and third digit ($8$ and $7$ choices, in some order). That will get you $5\cdot 8\cdot 8\cdot 7=2240$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Fractions in binary? How would you write a fraction in binary numbers; for example, $1/4$, which is $.25$? I know how to write binary of whole numbers such as 6, being $110$, but how would one write fractions?
| $1/4=0\cdot(1/2)^0+0\cdot(1/2)^1+1\cdot(1/2)^2=0.01$ in base $2$, you just go in reverse with powers $(1/2)^n, n=0,1,2,...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 8,
"answer_id": 1
} |
Combination / Permutation Question
There are 3 bags and 5 different marbles. In how many ways can the marbles be put into the bags?
(disclosure - the question is one of many in a teacher prep study guide. I am taking the qualification exam for teaching middle school math next month.)
The answer and explanation for the above was given as 3^5 = 243.
Earlier on, the book offered the distinction between say "3 kids of a group of ten are being chosen to play...." and the more specific "3 kids of 10 are being chosen for pitcher/catcher/first base" - the former being a lower number since it's not for specific positions.
With the marble bags, it seems the author counts all in bag one separate from all in bag two, or bag three. I'm okay with a wrong answer, but asking - given a test environment where I'm not going to be able to ask for any clarification, what in the wording above leads to that conclusion that the bags must be treated differently? Note - had the question said "3 different colored bags", I'd have been satisfied with the answer.
Edit - I've passed the exam. Remarkably, this is the most unsatisfying way I've ever had an exam reported, simply pass/fail. One would think especially for math, that the test taker would like a result with precision, even if it's not offered to prospective employers.
| Hint: Think about the balls. In how many bags can one ball be put in? How many balls are there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
showing compactness of the intersection of a compact set and disjoint open sets. Here's the problem I'm trying to solve.
"If $A\subseteq B_1\cup B_2$ where $B_1, B_2$ are disjoint open sets and $A$ is compact, show that $A\cap B_1$ is compact. Is the same true if $B_1$ and $B_2$ are not disjoint?"
Hope you can help, I can't seem to wrap my head around this one. Thanks!
| Let $\{V_{\alpha}\}_{\alpha\in J}$ be a family of open sets such that:
$$A\cap B_1\subseteq\cup_{\alpha\in J}V_{\alpha}$$Since $A\subseteq B_1\cup B_2$, it follows that $A\subseteq\ B_2\cup(\cup_{\alpha\in J}V_{\alpha})$. Since $A$ is compact , therefore there is a finite subset of $\{{V_{\alpha}}|\alpha\in J\}\cup\{B_2\}$ that covers $A$. Now try to use this subset (along with the fact that $B_1\cap B_2=\emptyset$) to make a finite subset of $\{{V_{\alpha}}|\alpha\in J\}$ that covers $A\cap B_1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof of $\displaystyle \lim_{z\to 1-i}[x+i(2x+y)]=1+i$ I am having some difficulty with the epsilon-delta proof of the limit above.
I know that $|x+i(2x+y)-(1+i)|<\epsilon$ when $|x+iy-(1-i)|<\delta$.
I tried splitting up the expressions above in this way:
$|x+i(2x+y)-(1+i)|\\
=|(x-1)+i(y+1)+i(2x+2)|\\
\le|(x-1)+i(y+1)|+|i(2x+2)|\\
<\delta+|i(2x+2)|$
Is this the correct approach? I don't know how else I can manipulate the expression bounded by $\delta$. Also, what can I do about the i's?
* $z=x+iy$
| Hint: If $z\to 1-i$ and $z=x+iy$, then $x\to 1$ and $y\to -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is mathematical research like? I'm planning on applying for a math research program over the summer, but I'm slightly nervous about it just because the name math research sounds strange to me. What does math research entail exactly? For other research like in economics, or biology one collects data and analyzes it and draws conclusions. But what do you do in math? It seems like you would sit at a desk and then just think about things that have never been thought about before. I appologize if this isn't the correct website for this question, but I think the best answers will come from here.
| For me as an independent mathematical researcher, it includes:
1) Trying to find new, more efficient algorithms.
2) Studying data sets as projected visually through different means to see if new patterns can be made visible, and how to describe them mathematically.
3) Developing new mathematical language and improving on existing language.
It is not so different from how you describe biological or economical research, only that you try to find patterns linked to mathematical laws rather than biological or economical laws.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "71",
"answer_count": 9,
"answer_id": 7
} |
Is the Euler phi function bounded below? I am working on a question for my number theory class that asks:
Prove that for every integer $n \geq 1$, $\phi(n) \geq \frac{\sqrt{n}}{\sqrt{2}}$.
However, I was searching around Google, and on various websites I have found people explaining that the phi function has a defined upper bound, but no lower bound. Am I reading these sites incorrectly, or am I missing something in the problem itself?
| EEDDIITT: this gives a proof of my main claim in my first answer, that a certain function takes its minimum value at a certain primorial. I actually put that information, with a few examples, into the wikipedia article, but it was edited out within a minute as irrelevant. No accounting for taste.
ORIGINAL: We take as given Theorem 327 on page 267 of Hardy and Wright, that for some fixed $0 < \delta < 1,$ the function
$$ g(n) = \frac{\phi(n)}{n^{1-\delta}} $$
goes to infinity as $n$ goes to infinity.
Note that $g(1) = 1$ but $g(2) < 1.$ For some $N_\delta,$ whenever $ n > N_\delta$ we get $g(n) > 1.$ It follows that, checking all $1 \leq n \leq N_\delta,$ the quantity $g(n)$ assumes a minimum which is less than 1. Perhaps it assumes this minimum at more than one point. If so, we are taking the largest such value of $n.$
Here we are going to prove that the value of $n$ at which the minimum occurs is the primorial created by taking the product of all the primes $p$ that satisfy
$$ p^{1-\delta} \geq p-1. $$
As I mentioned, in case two the minimum occurs at two different $n,$ this gives the larger of the two.
So, the major task of existence is done by Hardy and Wright. We have the minimum of $g$ at some
$$ n = p_1^{a_1} p_2^{a_2} p_3^{a_3} \cdots p_r^{a_r}, $$
with $$ p_1 < p_2 < \cdots < p_r. $$
First, ASSUME that one or more of the $a_i > 1.$ Now,
$$ \frac{ g(p_i)}{g(p_i^{a_i})} = p^{\delta - a_i \delta} = p^{\delta (1 - a_i)} < 1. $$
As a result, if we decrease that exponent to one, the value of $g$ is lowered, contradicting minimality. So
all exponents are actually 1.
Second, ASSUME that there is some gap, some prime $q < p_r $ such that $q \neq p_j$ for all $j,$ that is $q$ does not divide $n.$
Well, for real variable $x > 0,$ the function $$ \frac{x-1}{x^{1-\delta}} $$ is always increasing, as the first derivative is
$$ x^{\delta - 2} (\delta x +(1-\delta)). $$
It follows that, in the factorization of $n,$ if we replace $p_r$ by $q,$ the value of $g$ is lowered, contradicting minimality. So
the prime factors of $n$ are consecutive, beginning with 2, and $n$ is called a primorial.
Finally, what is the largest prime factor of $n?$ Beginning with 2, multiplying by any prime $p$ with
$$ \frac{p-1}{p^{1-\delta}} \leq 1 $$
shrinks the value of $g$ or keeps it the same, so in demanding the largest $n$ in case there are two attaining the minimum of $g,$ we take $n$ to be the product of all primes $p$ satisfying
$$ p - 1 \leq p^{1-\delta}, $$ or
$$ p^{1-\delta} \geq p-1 $$
as I first wrote it.
Examples are given in my first answer to this same question.
EEDDIITTTT: Jean-Louis Nicolas showed, in 1983, that the Riemann Hypothesis is true if and only if, for all primorials $P,$
$$ \frac{e^\gamma \phi(P) \log \log P}{P} < 1. $$
Alright, the exact reference is: Petites valeurs de la fonction d'Euler. Journal of Number Theory, volume 17 (1983), number 3, pages 375-388.
On the other hand, if RH is false, the inequality is true for infinitely many primorials and false for infinitely many. So, either way, it is true for infinitely many primorials (once again, these are $P = 2 \cdot 3 \cdot 5 \cdots p$ the product of consecutive primes beginning with 2).
For whatever reason, the criterion of Guy Robin, who was a student of Nicolas, got to be better known.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 4,
"answer_id": 3
} |
How to solve this integration: $\int_0^1 \frac{x^{2012}}{1+e^x}dx$ I'm having troubles to solve this integration: $\int_0^1 \frac{x^{2012}}{1+e^x}dx$
I've tried a lot using so many techniques without success. I found $\int_{-1}^1 \frac{x^{2012}}{1+e^x}dx=1/2013$, but I couldn't solve from 0 to 1.
Thanks a lot.
| You have $$\int_{-1}^{1} \frac{x^{2012}}{1+e^{x}} \ dx =\underbrace{\int_{-1}^{0}\frac{x^{2012}}{1+e^{x}}}_{I_{1}} \ dx + \int_{0}^{1}\frac{x^{2012}}{1+e^{x}} \ dx \qquad \cdots (1)$$
In $I_{1}$ put $x=-t$, then you have $dx = -dt$, and so the limits range from $t=0$ to $t=1$. So you have $$I_{1}= -\int_{1}^{0} \frac{e^{t}\cdot t^{2012}}{1+e^{t}} \ dt = \int_{0}^{1} \frac{e^{x}\cdot x^{2012}}{1+e^{x}} \ dx$$
Put this in equation $(1)$ to get the value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Definition of Tangents When we had not learnt Calculus, we met the concept of Tangent in Circle, which was defined as the line touching the circle at ONE point.
Then, after learning Calculus, we knew that a curve could intersect with its tangent at more than one point, and a line intersecting with a curve at only one point is not necessarily a tangent. Hence, we used Limit to define tangent, which involved TWO points and we considered one approached the other to obtain the tangent.
My question is:
The definition of a tangent to a curve should be more general than that to a circle, and hence, we can say that the definition of a tangent to a circle can be derived from the definition of a tangent to a curve.
However, limit uses TWO points, even though they are very close to each other. If they overlap with each other to become ONE point, then no line occurs. So, in theory, how can we proof that the two definitions (a general curve VS a circle) are consistent?
| Limit means approaching, not coincidence. So, if you take two points on the circle, line that goes through them is not a tangent, of course. But if you make one point closer to another, that line goes closer to the tangent. If you take it even more close, then line will be also closer. And here comes the limit into the business. Just like derivatives which are based on difference of two values. But to be true derivative one of your values "kind of" approaching another, although you never say that two values are the same, so you cannot take ratio because of denominator being "zero". That's why we define derivative at the point, not two points, and that's how we define tangent as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Applications of prime-number theorem in algebraic number theory? Dirichlet arithmetic progression theorem, or more generally, Chabotarev density theorem, has applications to algebraic number theory, especially in class-field theory.
Since we might think of the density theorem as an analytic theorem, and as prime number theorem is one main theorem of analytic number theory, one is led to wonder:
if there is any application of prime number theorem to algebraic number theory.
Thanks for any attention in advance.
| I don't know if this can be considered algebraic number theory or if it is more algebraic geometry; but, here goes.
Deligne uses the methods of Hadamard-de la Vallée-Poussin to prove the Weil conjectures. Even if it is not an application of the ordinary PNT as such, the exact same methods are applied elsewhere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is wrong with my proof: Pseudoinverse and SVD I was trying to prove the following:
Let $U\Sigma V$ be the $SVD$ decomposition of $A\in\mathbb R^{m\times n}$, where $\textrm{rank}(A)=k$. Show that the pseudoinverse of $A$ is given by,
$$
\displaystyle A^\dagger=\sum_{i=1}^k\sigma_i^{-1}v_iu_i^T.
$$
${\bf Proof:}$ Let us show that $AA^\dagger A=A$, $A^\dagger AA^\dagger=A^\dagger$, $(A^\dagger A)^T=A^\dagger A$ and $(AA^\dagger)^T=AA^\dagger$. Particionate $A$ as follows,
\begin{align*}
\displaystyle A=\left[\begin{array}{c|c}
U_1&U_2
\end{array}\right]\left[\begin{array}{c|c}
\tilde{\Sigma}&0\\ \hline 0&0
\end{array}\right]\left[\begin{array}{c}
V_1^T\\ V_2^T
\end{array}\right],
\end{align*}
where $\tilde{\Sigma}=\textrm{diag}(\sigma_1, \ldots, \sigma_k)$. Then, $A=U_1\tilde{\Sigma}V_1^T$ whereas $A^\dagger=V_1\tilde{\Sigma}^{-1}U_1^T$. Since $V^TV=U^TU=I$ we see $U_1^TU_1=V_1^TV_1=I$. Then,
\begin{align*}
\displaystyle (A^\dagger A)^T=(V_1\tilde{\Sigma}U_1^T)(U_1\tilde{\Sigma}^{-1}V_1^T)=I=A^\dagger A.
\end{align*}
Analogously, we see $(AA^\dagger)=I=AA^\dagger$. Using that $A^\dagger A=AA^\dagger=I$ (how can that be? $A$ is not squared) we have,
\begin{align*}
\displaystyle (A^\dagger AA^\dagger)=(A^\dagger A)A^\dagger=IA^\dagger\ \textrm{and}\ (AA^\dagger A)=A(A^\dagger A)=AI=A.
\end{align*}
${\bf Problems:}$ The main problem is the part $(A^\dagger A)=I$ for $A$ is not a squared matrix. Furthermore if that identity held the problem would be almost trivial.. What is the problem with my proof?
| You shouldn't assume that $A^\dagger A$ is equal to $I$:
\begin{align*}
A &= U_1 \widetilde{\Sigma} V_1^T,\\
A^\dagger &= V_1 \widetilde{\Sigma}^{-1} U_1^T,\\
\Rightarrow A^\dagger A &= V_1V_1^T \ \text{ is symmetric}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\Gamma (-n+x)=\frac{(-1)^n}{n!}\left [ \frac{1}{x}-\gamma +\sum_{k=1}^{n}k^{-1}+O(x) \right ]$ Prove that $\Gamma (-n+x)=\frac{(-1)^n}{n!}\left [ \frac{1}{x}-\gamma +\sum_{k=1}^{n}k^{-1}+O(x) \right ]$
I don't know how to do this ?
Note that $\gamma $ is the Euler-Mascheroni constant
| A standard trick is to use the reflection identity $$\Gamma(-n+x) \Gamma(1+n-x) = -\frac{\pi}{\sin(\pi n - \pi x)}$$
giving, under the assumption of $n\in \mathbb{Z}$
$$
\Gamma(-n+x) = (-1)^n \frac{\pi}{\sin(\pi x)} \frac{1}{\Gamma(n+1-x)} = (-1)^n \frac{\pi}{\sin(\pi x)} \frac{1}{\color\green{n!}} \frac{\color\green{\Gamma(n+1)}}{\Gamma(n+1-x)}
$$
Assuming $n \geqslant 0$,
$$ \begin{eqnarray}\frac{\Gamma(n+1)}{\Gamma(n+1-x)} &=& \frac{1}{\Gamma(1-x)} \prod_{k=1}^n \frac{1}{1-x/k} \\ &=& \left(1+\psi(1) x + \mathcal{o}(x)\right) \left(1+\sum_{k=1}^n \frac{x}{k} + \mathcal{o}(x) \right) \\ &=& 1 + \left( \psi(1) + \sum_{k=1}^n \frac{1}{k} \right) x + \mathcal{o}(x)
\end{eqnarray}$$
where $\psi(x)$ is the digamma function.
Also using
$$
\frac{\pi}{\sin(\pi x)} = \frac{1}{x} + \frac{\pi^2}{6} x + \mathcal{o}(x)
$$
and multiplying we get
$$
\Gamma(n+1) = \frac{(-1)^n}{n!} \left( \frac{1}{x} + \psi(1) + \sum_{k=1}^n \frac{1}{k} + \mathcal{O}(x) \right)
$$
Further $\psi(1) = -\gamma$, where $\gamma$ is the Euler-Mascheroni constant, arriving at your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Question about closure of the product of two sets Let $A$ be a subset of the topological space $X$ and let $B$ be a subset of the topological space $Y$. Show that in the space $X \times Y$, $\overline{(A \times B)} = \bar{A} \times \bar{B}$.
Can someone explain the proof in detail? The book I have kind of skims through the proof and I don't really get it.
| $(\subseteq)$: The product of closed sets $\overline{A} \times \overline{B}$ is closed. For every closed $C$ that contains $\overline{A} \times \overline{B}$, $A \times B \subseteq C$ so $\overline{A \times B} \subseteq \overline{\overline{A} \times \overline{B}} = \overline{A} \times \overline{B}$.
$(\supseteq)$: Choose any $(a,b) \in \overline{A} \times \overline{B}$. Notice that for every open neighborhood $W \subseteq X \times Y$ that contains $(a, b)$, $U \times V \subseteq W$ (by the definition of the product topology) for some open neighborhood $U$ of $a$ and some open neighborhood $V$ of $b$. By the definition of closure points, $U$ intersects $A$ at some $a'$. Similarly, define $b' \in V \cap B$. Hence, $(a', b') \in W \cap (A \times B)$.
To summarize, every open neighborhood $W \subseteq X \times Y$ that contains $(a,b)$ must intersect $A \times B$, therefore $(a,b) \in \overline{A\times B}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0Evaluate by complex methods
$$\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0<a<b<1$$
Sis.
| It can be done easily without complex, if we note that
$$ \frac{\sin x}{1-2a\cos x+a^2}=\sum_{n=0}^{+\infty}a^n\sin[(n+1)x]$$
Just saying.
EDIT: for proving this formula, we actually use complex method
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Vector Space with Trivial Dual How to construct a Vector Space $E$ (non trivial) such that, the only continuous linear functional in $E$ is the function $f=0$?
| In addition to what Asaf has written: There are non-trivial topological vector spaces with non-trivial topologies which have a trivial dual. I think in Rudin's "Functional Analysis" it is shown that the $L^p$-spaces with $0<p<1$ are an example of this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Finding the smallest positive integer $N$ such that there are $25$ integers $x$ with $2 \leq \frac{N}{x} \leq 5$
Find the smallest positive integer $N$ such that there are exactly $25$ integers $x$ satisfying $2 \leq \frac{N}{x} \leq 5$.
| $x$ ranges from $N/5$ through $N/2$ (ignoring the breakage) so $N/2-N/5+1=25$ so $N=80$ and a check shows $x$ goes $16$ through $40$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to read mathematical formulas? I'm coming from a programmers background, trying to learn more about physics. Immediately, I was encountered by math, but unfortunately unable to read it.
Is there a good guide available for reading mathematical notation? I know symbols like exponents, square roots, factorials, but I'm easily confused by things like sub-notation. For example, I have no idea what this is:
fn
I can easily express values using programmatic notation, ie pseudocode:
milesPerHour = 60
distanceInFeet = 100
feetPerMillisecond = ((milesPerHour * 5280) / (1e3 * 60 * 60))
durationInMilliseconds = 100 / feetPerMillisecond
However, I have no clue even where to begin when trying to express the same logic in mathematical notation.
How can I improve my ability to read and interpret mathematical formulas in notation?
| I guess the most natural anwer to "How can I improve my ability to read and interpret mathematical formulas in notation?" is: through practice. If you're trying to read physics, you're probably familiar with Calculus. I would advise then that you do a Real Analysis course, only to get used to these notations, to mathematical logic and for the fun of it :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Prove $\lim_{j\rightarrow \infty}\sum_{k=1}^{\infty}\frac{a_k}{j+k}=0$ I am only looking for a hint to start this exercise, not a full answer to the problem, please take this into consideration.
Suppose that $a_k \geq 0$ for $k$ large and that $\sum_{k=1}^\infty\frac{a_k}k$ converges. Prove that
$$\lim_{j\rightarrow \infty}\sum_{k=1}^{\infty}\frac{a_k}{j+k}=0$$
What I can see so far that may help is that, since $\sum_{k=1}^\infty a_k/k$ converges,
$\forall \epsilon > 0$, $\exists N\in \mathbb{N}$ such that $n\geq N\Rightarrow |\sum_{k=n}^\infty \frac{a_k}k|<\epsilon$, which is a result of the Cauchy Criterion. Once again, I am only looking for a hint to start this exercise, not a full proof.
| I gave the hint in my comment. For a full solution, read below:
Let $\epsilon>0$.
Choose $N>1$ so that $a_j\ge 0$ for $j\ge N$ and such that $\sum\limits_{k=N}^\infty {a_k\over k}<\epsilon/2$. Note that for $j>0$, we then have
$$\tag{1}0\le \sum\limits_{k=N}^\infty {a_k\over k+j} \le\sum\limits_{k=N}^\infty {a_k\over k}<\epsilon/2.$$
So, we can make the tails $\sum\limits_{k=N}^\infty {a_k\over k+j}$ small (independent of $j$ in fact).
Let's see how to make the remaining part of the sum, $\sum\limits_{k=1}^{N-1} {a_k\over j+k}$, small:
Let $M=\max\{|a_1|,\ldots |a_{N-1}| \}$. Choose $J> M(N-1)/(2\epsilon)$.
Then for $j\ge J$:
$$\tag{2}\Bigl| \,\sum\limits_{k=1}^{N-1} {a_k\over j+k}\,\Bigr|\le
\sum\limits_{k=1}^{N-1} {M\over J }=(N-1)M\cdot{1\over J}<\epsilon/2.
$$
Using $(1)$ and $(2)$, we have for $j>J $:
$$
\Bigl|\,\sum_{k=1}^\infty {a_k\over j+k}\,\Bigr|
\le
\Bigl|\,\sum_{k=1}^{N-1} {a_k\over j+k}\,\Bigr|+
\Bigl|\,\sum_{k=N }^\infty {a_k\over j+k}\,\Bigr|
\le{\epsilon\over2}+
{\epsilon\over2}=\epsilon.
$$
Since $\epsilon$ was arbitrary, the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/302635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Automorphisms on Punctured Disc I have to find the automorphism group of the punctured unit disc $D = \{|z| <1\}\setminus \{0\}$.
I understand that if $f$ is an automorphism on $D$, then it will have either a (i) removable singularity or (ii) a pole of order 1 at $z=0$.
If it has a removable singularity at 0, then $f$ is a rotation. I am stuck at case (ii).
Also, using this result, later I also have to find the automorphism group of $\{|z|<1\}\setminus \{1/2\}$
Can anybody please help ?
| A bounded holomorphic function does not have a pole.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/303734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Combinatorics Catastrophe How will you solve
$$\sum_{i=1}^{n}{2i \choose i}\;?$$
I tried to use Coefficient Method but couldn't get it! Also I searched for Christmas Stocking Theorem but to no use ...
| Maple gives a "closed form" involving a hypergeometric function:
$$ -1-{2\,n+2\choose n+1}\; {\mbox{$_2$F$_1$}(1,n+\frac32;\,n+2;\,4)}-\frac{i
\sqrt {3}}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/303816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Semigroup homomorphism and the relation $\mathcal{R}$ Let $S$ be a semigroup and for $a\in S$ let $$aS = \{as : s \in S \}\text{,}\;\;\;aS^1 = aS \cup \{a\}\text{.}$$The relation $\mathcal{R}$ on a semigroup $S$ is defined by the rule: $$a\;\mathcal{R}\; b \Leftrightarrow aS^1 = bS^1 \;\;\;\;\forall \;\;a,b\in S\text{.}$$
Let $S,T$ be semigroups and let $\phi : S \to T$ be a homomorphism.
Show that if $a,b \in S$ and $a\;\mathcal{R} \;b$ in $S$ then $\phi(a) \;\mathcal{R} \;\phi(b)$ in $T$.
| Use that $a\mathcal R b \iff (a=b) \lor (a\in bS\land b\in aS)$.
(For this, in direction $\Leftarrow$, in the case of $a\ne b$ we conclude $aS^1\subseteq bS$ and $bS^1\subseteq aS$.)
We can assume $a\ne b$, then $a\mathcal Rb$ means $a=bs$ for some $s$ and $b=as'$ for some $s'\in S$, so $\phi(a)=\phi(b)\phi(s)\in\phi(b)T$ and $\phi(b)=\phi(a)\phi(s')\in\phi(a)T$. -QED-
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/303960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the asymptotical bound of this recurrence relation? I have the recurrence relation, with two initial conditions
$$T(n) = T(n-1) + T(n-2) + O(1)$$
$$T(0) = 1, \qquad T(1) = 1$$
With the help of Wolfram Alpha, I managed to get the result of $O(\Phi^n)$, where $\Phi = \frac{1+\sqrt 5}{2} \approx 1.618$ is the golden ratio.
Is this correct and how can that be mathematically proven?
| You have essentially stated the Fibonacci sequence, or at least asymptotically. There are numberless references, here for instance. And your result is not correct; as you will see from the reference, the Fib sequence behaves as $\phi^n/\sqrt{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Superharmonic function and super martingale. The definition (from Durrett's "Probability: Theory and Examples"):
Superharmonic functions. The name (super martingale)
comes from the fact that if $f$ is superharmonic (i.e., f has continuous derivatives of
order $\le 2$ and $\partial^2 f /\partial^2 x_1^2 + · · · + \partial^2 f /\partial^2 x_d^2)$, then
$$
f (x) \ge \frac 1 {|B(0, r)|} \int_{B(x,r)} f(y) dy
$$
where $B(x, r) = \{y : |x − y| \le r\}$ is the ball of radius $r$, and $|B(0, r)|$ is the volume of the ball of radius $r$.
The question is
Suppose $f$ is superharmonic on $R^d$. Let $\xi_1 , \xi_2 , ...$ be i.i.d. uniform
on $B(0, 1)$, and define $S_n$ by $S_n = S_{n−1} + \xi_n$ for $n \ge 1$ and $S_0 = x$. Show that $X_n = f (S_n)$ is a supermartingale.
Here the filtration should be $\mathcal{F}_n = \sigma \{X_n, X_{n-1}...\}$.
I know to prove $X_n, n \ge 0$ is a super martingale, we only need to show
$$
E\{X_{n+1} ~|~ \mathcal{F}_n\} \leq X_n.
$$
This is easy when $n=0$. But for $n > 0$, I've no idea how to, or if it is possible, to derive the formula of the conditional expectation.
| $f(S_{n+1})=f(S_n+\xi_{n+1})$
Using Did's Hint: If $g(x)=E(f(x,Y))$ then $g(X)=E(f(X,Y)|X)$
$\Rightarrow E(f(S_n+\xi_{n+1})|\mathcal F_n)=g(S_n)$
but $\displaystyle g(x)=\int f(x+y)\nu(dy)$
and in our case $\nu(dy)=\mathbf 1_{B(0,1)}\frac{1}{|B(0,1)|}dy$ (because $\xi_i$ are uniform on $B(0,1)$)
$\displaystyle\Rightarrow g(S_n)=\int f(S_n+y)\mathbf 1_{B(0,1)}\frac{1}{|B(0,1)|}dy=\frac{1}{|B(0,1)|}\int_{B(S_n,1)}f(y)dy\le f(S_n)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Are these exactly the abelian groups? I'm thinking about the following condition on a group $G$.
$$(\forall A\subseteq G)(\forall g\in G)(\exists h\in G)\ Ag=hA.$$
Obviously every abelian group $G$ satisfies this condition. Are there any other groups that do? Can we give a familiar characterization for them? Can we give one if we confine the considerations to finite groups?
Certainly not all groups satisfy the condition. Let $G$ be the free group on $\{x,y,z\}.$ Let $A=\{x,y\}$ and $g=z.$ Then $$Ag=\{x,y\}z=\{xz,yz\}.$$ Suppose there is $h\in G$ such that $\{hx,hy\}=hA=\{xz,yz\}.$ Then either $$\begin{cases}hx=xz\\hy=yz\end{cases}$$
or $$\begin{cases}hx=yz\\hy=xz\end{cases}$$
From the first case we get $h=xzx^{-1}$ and $h=yzy^{-1}$, which is a contradiction. From the second case we get $h=yzx^{-1}$ and $h=xzy^{-1}$, which is also a contradiction.
| Assume $ab\ne ba$. Let $A=\{1,a\}$, $g=b$. Then there is $h\in G$ such that $\{h,ha\}=\{b,ab\}$. This needs $h=b\lor h=ab$.
In the first case $ha=ba\ne ab$, so this fails.
Therefore $h=ab$ and $ha=aba=b$. Similarly, $bab=a$. This implies $aa=abab=bb$. We conclude
$$a=bab=bbabb=aaaaa, $$
hence $a^4=1$ and similarly $b^4=1$.
Now take $A=\{1,a,b\}$ and $g=b$. Then there is $h\in G$ such that $\{h,ha,hb\}=\{b,ab,b^2\}$.
*
*$h=b$: Then $hb=b^2$ implies $ba=ha=ab$, contradiction
*$h=ab$: Then $ha=aba=b$ implies $hb=b^2$, i.e. $a=1$ and of course $ab=ba$, contradiciton.
*$h=b^2=a^2$: Then $ha=a^3=a^{-1}\ne b$, hence $ha=ab$, i.e. $a^2=b$ and of course $ab=ba$, contradiction
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
} |
$\epsilon$-$\delta$ proof of discontinuity How can I prove that the function defined by
$$f(x) = \begin{cases} x^{2}, & \text{if $x \in \mathbb{Q}$;} \\ -x^{2}, & \text{if $x \notin \mathbb{Q}$;} \end{cases} $$ is discontinuous?
I see that it is true by using sequences but I cannot prove using only $\epsilon$'s and $\delta$'s.
| Hint: Check if the sequence $f(1+\frac{\sqrt{2}}{n})$ becomes very close to $f(1)$ for large $n$
To do it using $\epsilon,\delta$ definition. Suppose there exists $\delta>0$ such that for $\forall y\in \mathbb{R}[|x-y|<\delta\implies|f(x)-f(y)|<0.1]$. Now choose $n$ sufficently large such that $|(1+\frac{\sqrt{2}}{n})-1|<\delta$ (Use the archemidean principle to do this). Thus $|f(1+\frac{\sqrt{2}}{n})-f(1)|<0.1$ Does this lead to a contradiction ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
What are your favorite proofs using mathematical induction? I would like to get a list going of cool proofs using mathematical induction.
Im not really interested in the standard proofs, like $1+3+5+...+(2n-1)=n^2$, that can be found in any discrete math text. I am looking for more interesting proofs.
Thanks a lot.
|
Let $a>0$ and $d\in\mathbb{N}$ and define the simplex $S_d(a)$ in $\mathbb{R}^d$ by
$$
S_d(a)=\{(x_1,\ldots,x_n)\in\mathbb{R}^d\mid x_1,\ldots,x_d\geq 0,\;\sum_{i=1}^d x_i\leq a\}.
$$
Then for every $a>0$ and $d\in\mathbb{N}$ we get the following
$$
\lambda_d(S_d(a))=\frac{a^d}{d!},\qquad (*)
$$
where $\lambda_d$ is the $d$-dimensional Lebesgue measure.
Proof: Let $d=1$. Then
$$
\lambda_1(S_1(a))=\lambda_1([0,a])=a=\frac{a^1}{1!},
$$
so this case holds. Assume $(*)$ holds for $d\in\mathbb{N}$ and let us prove that it also holds for $d+1$. Now we use that for $B\in\mathcal{B}(\mathbb{R}^{n+m})$ the following holds:
$$
\lambda_{n+m}(B)=\int_{\mathbb{R}^n}\lambda_m(B_x)\,\lambda_n(\mathrm dx),
$$
where $B_x=\{y\in \mathbb{R}^m\mid (x,y)\in B\}$. Using this we have
$$
\lambda_{d+1}(S_{d+1}(a))=\int_{\mathbb{R}}\lambda_d((S_{d+1}(a))_{x_1})\,\lambda_1(\mathrm dx_1).
$$
But
$$
\begin{align}
(S_{d+1}(a))_{x_1}&=\{(x_2,\ldots,x_{d+1})\in\mathbb{R}^d\mid (x_1,x_2,\ldots,x_{d+1})\in S_d(a)\}\\
&=\{(x_2,\ldots,x_{d+1})\in\mathbb{R}^d\mid x_1,x_2,\ldots,x_{d+1}\geq 0,\; x_2+\cdots+ x_{d+1}\leq a-x_1\}\\
&=
\begin{cases}
S_{d}(a-x_1)\quad &\text{if }0\leq x_1\leq a,\\
\emptyset &\text{otherwise}.
\end{cases}
\end{align}
$$
Thus
$$
\begin{align}
\lambda_{d+1}(S_{d+1}(a))&=\int_0^a\lambda_d(S_d(a-x_1))\,\lambda_1(\mathrm dx_1)=\int_0^a\frac{(a-x_1)^d}{d!}\,\lambda_1(\mathrm dx_1)\\
&=\frac{1}{d!}\left[-\frac{1}{d+1}(a-x_1)^{d+1}\right]_0^a=\frac{a^{d+1}}{(d+1)!}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
The set of all functions from $\mathbb{N} \to \{0, 1\}$ is uncountable? How can I prove that the set of all functions from $\mathbb{N} \to \{0, 1\}$ is uncountable?
Edit: This answer came to mind. Is it correct?
This answer just came to mind. By contradiction suppose the set is $\{f_n\}_{n \in \mathbb{N}}$. Define the function $f: \mathbb{N} \to \{0,1\}$ by $f(n) \ne f_n(n)$. Then $f \notin\{f_n\}_{n \in \mathbb{N}}$.
| Hint : use the diadic developpement of elements of $[0,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Decomposition of $C_0^{\infty}(\mathbb{R}^n)$ -function I got the following question as part of a fourier-analysis course..
Consider $\phi\in C_0^{\infty}(\mathbb{R}^n)$ with $\phi(0)=0$. Apparantly then we can write $$\phi =\sum_{j=1}^nx_j\psi_j $$ for functions $\psi_j$ in the same space, and I would like to prove this.
However I'm little stuck on this. The steps would involve that we start integrating $$\int_0^{x_1}D_1\phi(t,x_2,\cdots, x_n)dt+\phi(0,x_2,\cdots, x_n) $$
and continue doing so w.r.t. to the other variables, and change the interval of integration to [0,1]. Then get everything in $C_0^{\infty}(\mathbb{R}^n)$ by writing $$\phi = \sum \frac{x_i^2\phi(x)}{\left\|x\right\|^2
} $$
and patch everything together with a partition of unity. I just dont quite see what they mean...A partition of unity is a sequence of functions that sums up to 1 for all $x\in \mathbb{R}^n$.
| The result can be shown by induction on $n$.
For $n=1$, just write $\phi(x)=\int_0^x\phi'(t)dt=x\int_0^1\phi'(sx)ds$, and the map $x\mapsto \int_0^1\phi'(sx)ds$ is smooth, and has a compact support.
Assume the result is true for $n-1\geqslant 1$. We have
$$\phi(x)=x_n\int_0^1\partial_n\phi(x_1,\dots,x_{n-1},tx_n)dt+\varphi(x_1,\dots,x_{n-1},0).$$
As similar argument as in the case $n=1$ shows that $(x_1,\dots,x_n)\mapsto \int_0^1\partial_n\phi(x_1,\dots,x_{n-1},tx_n)dt$ is smooth with compact support. By the induction, as $(x_1,\dots,x_{n-1})\to \phi(x_1,\dots,x_{n-1},tx_n)dt+\varphi(x_1,\dots,x_{n-1},0)$ is smooth with compact support, we can write it as $\sum_{j=1}^{n-1}x_j\psi_j(x_1,\dots,x_{n-1})$.
But it's not finished yet, as $(x_1,\dots,x_n)\mapsto \psi_j(x_1,\dots,x_{n-1})$ doesn't have a compact in $\Bbb R^n$ unless $\psi_j\equiv 0$. However, we can find a smooth function with compact support $\chi$ such that $\chi(x)=1$ whenever $x\in\operatorname{supp}(\phi)$. As $\phi(x)=\phi(x)\chi(x)$, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Formal proof for $(-1) \times (-1) = 1$ Is there a formal proof for $(-1) \times (-1) = 1$? It's a fundamental formula not only in arithmetic but also in the whole of math. Is there a proof for it or is it just assumed?
| In any ring, it holds, where $1$ denotes the unit element ($1x=x=x1$ for all $x$) and $-x$ denotes the additive inverse ($x+(-x)=0$ for all $x$).
$x=1\cdot x=(1+0)\cdot x=1\cdot x+0\cdot x=x+0\cdot x$. Then, using the additive group, it follows that $0\cdot x=0$ for all $x$.
Now use distributivity for
$$0=(1+(-1))(-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "93",
"answer_count": 14,
"answer_id": 7
} |
Unions of disjoint open sets. Let $X$ be a compact metric space (hence separable) and $\mu$ a Borel
probability measure.
Given an open set $A$ and $r,\epsilon>0$ $\ $does there exist a finite set of
disjoint open balls $\left\{ B_{i}\right\} $ contained in $A$ and of radius smaller than $r$ , so that $\mu(\cup B_{i})\geq\mu(A)-\epsilon.$
| Recall
Lemma (Finite Vitali covering lemma) Let $(X,d)$ be a metric space, $\{B(a_j,r_j),j\in [K]\}$ a finite collection of open balls. We can find a subset $J$ of $[K]$ such that the balls $B(a_j,r_j),j\in J$ are disjoint and
$$\bigcup_{i\in [K]}B(a_i,r_i)\subset \bigcup_{j\in J}B(a_j,3r_j).$$
A proof is given page 41 in the book Ergodic Theory: with a view towards Number Theory, Einsiedler M., Ward T.
For each $a\in A$, fix $r_a<r/3$ such that $B(a,3r_a)\subset A$. As $X$ is separable, we can, by Lindelöf property, extract from the cover $\{B(a,r_a),a\in A\}$ of $A$ a countable subcover $\{B(a_j,r_j),j\in \Bbb N\}$. Now take $N$ such that $\mu(A)-\mu\left(\bigcup_{j=0}^NB(a_j,r_j)\right)<\varepsilon$. Then we conclude by finite Vitali covering lemma.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate a definite integral that involves $(dx)^2$? For example: $$\int_0^1(15-x)^2(\text{d}x)^2$$
| Just guessing, but maybe this came from $\frac {d^2y}{dx^2}=(15-x)^2$ The right way to see this is $\frac d{dx}\frac {dy}{dx}=(15-x)^2$. Then we can integrate both sides with respect to $x$, getting $\frac {dy}{dx}=\int (15-x)^2 dx=\int (225-30x+x^2)dx=C_1+225x-15x^2+\frac 13x^3$ and can integrate again to get $y=C_2+C_1x+\frac 12 225x^2-5x^3+\frac 1{12}x^4$ which can be evaluated at $0$ and $1$, but we need a value for $C_1$ to get a specific answer.
As I typed this I got haunted by the squares on both sides and worry that somehow it involves $\frac {dy}{dx}=15-x$, which is easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Question on the proof of a property of the rank of a matrix The task is that I have to prove this statement:
Given $(m+1)\times(n+1)$ matrix $A$ like this:
$$
A=\left[\begin{array}{c |cc}
1 & \begin{array}{ccc}0 & \cdots & 0\end{array} \\
\hline
\begin{array}{c}0\\ \vdots\\0\end{array} & {\Large B} \\
\end{array}\right]
$$
where $B$ is a $m\times n$ sub-matrix of $A$.
Show: $\operatorname{rank}(A) = r$ implies $\operatorname{rank}(B) = r - 1$.
Here are the steps that I constructed:
(1) First I already proved this claim: if $m\times n$ matrix $S$ has rank $r$, then $r\le m$ and $r\le n$. And I let $\operatorname{rank}(B) = k$.
(2) Use (1), I say the following 2 statements:
*
*for matrix $A$, $r\le m+1$ and $r\le n+1$. Thus $r-1\le m$ and $r-1 \le n$
*for matrix $B$, $k\le m$ and $k\le n$.
(3) Now, this is the part that I feel shaky about. I plan to say that using (2), I have $r-1\le k$ and $k\le r - 1$. Thus $r - 1 = k$. But I'm not certain whether this argument is valid. After noticing that $r - 1\le m$ and $k \le m$ at a same time, I come up with the above relation between $k$ and $r - 1$.
Would someone please help me check if there is anything wrong or missing in this proof?
Thank you.
| In analogy, you're arguing: "$5 \leq 7, 4 \leq 7 \Rightarrow 5 \leq 4$", which, when put this way, is clearly not true.
The way to prove this exercise depends on how your class introduced these concepts. Absent that knowledge, I would argue as follows:
When reduced to row-echelon form, you can read off the rank of a matrix by the number of non-zero rows. So reduce your matrix $B$ to $1 \leq k \leq \min(n, m)$ non-zero rows (which equals the rank of $B$), and note that the elementary matrices multiplied to the left and right of $A$, to achieve this, will not change column 1 (this is quite obvious as all but the first element of column $1$ and row $1$ are $0$). So you still have the $1$ in position $a_{11}$ (and $0$ in the other entries of the first column), and the entire matrix has $k+1$ non-zero rows in row-echelon form which equals the rank of $A$.
So $\operatorname{rank}(A) = \operatorname{rank}(B)+1,$ which was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
About an extension of Riesz' Lemma for normed spaces The Riesz' Lemma is as follows:
Let $Y$ and $Z$ be subspaces of a normed space $X$ of any dimension (finite or infinite) such that $Y$ is closed (in $X$) and is also a proper subset of $Z$. Then for every real number $\theta$ in the open interval $(0,1)$, there is a point $z$ in $Z$ such that $$||z|| = 1$$ and $$||z-y|| \geq \theta$$ for every $y$ in $Y$.
Now we want to prove the following assertion:
If $Y$ is finite-dimensional, then we can even take $\theta$ to be equal to $1$ in the statement of the Riesz' Lemma.
How to prove this?
| I would apply the following trick in case $\dim Y<\infty$:
Let $z_0\in Z\setminus Y$ arbitrary and imagine the finte dimensional normed space $U$ spanned by $\langle Y,z_0\rangle$. Then there are many ways to continue, for example, the unit ball is compact in $U$, thus applying Riesz's lemma to $\theta_n:=1-\frac1n$ and $Y\subset U$, we get a sequence $u_n$ with $||u_n||=1$ and $d(u_n,Y)>1-\frac1n$. Pick a convergent subseqence and let $z$ be its limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Open and Closed Set Problems using a Ball I am having trouble with these two questions. In particular, using a ball and choosing an $r$ to show that a set is open.
(a)
$$X = \left \{ \mathbf{x} \in \mathbb{R}^d | \: ||\mathbf{x}|| \leq 1 \right \} .$$
So, $X$ is closed if its compliment $X^c$ is open. So if I can show that $X^c$ is open, then it follows that $X$ is closed. I'll start with the definition I am using for a ball.
The ball about $\mathbf{a}$ in $\mathbb{R}^n$ of radius $r$ is the set
$$B_{r}(\mathbf{a})= \left \{ \mathbf{x} \in \mathbb{R}^n : || \mathbf{x} - \mathbf{a} || < r\right \}$$
A subset $U$ of $\mathbb{R}^n$ is open if for every $\mathbf{a} \in U$, there is some $r=r(a) > 0$ such that the ball is contained in $U$.
So, I have a set $X^{c} = \left \{ \mathbf{v} \in \mathbb{R}^d | \: ||\mathbf{v}|| > 1 \right \} $. It consists of points whose lengths are longer than 1. Let $\mathbf{a} \in X^{c}$, then $|| \mathbf{a} || > 1$. Now, using $B_{r}(\mathbf{a})$ as it is written above, how do I find an explicit formula for $r>0$? Intuitively it seems to make sense if $\mathbf{x}$ is in $X$, then $0<|| \mathbf{x} - \mathbf{a} ||$ as $||\mathbf{x}|| \neq ||\mathbf{a}||$. I am not sure how to proceed.
(b)
$$X = \mathbb{R}^2 \setminus \left \{ \mathbf{x} \in \mathbb{R}^2 | \mathbf{x}=(x,0) \: \right \}$$
I know that this set is open, but it again comes down to choosing some $r$ and using a ball. The set is $\mathbb{R}^2$ less a line across the x-axis - that is, any $x$ and any $y$ with $y \neq 0$. The line has no height, so it should be easy to show that for any $\mathbf{a} \in X$ there exists $r>0$ such that $B_r(a) \subseteq X$.
I've done lots of scratch work and diagrams etc. but I just can't seem to put this concept together. Any help and clarification would be appreciated.
| Regarding you second question: Let $a\in X$, and you want to find $r>0$ such that $B_r(a)\subseteq X$. What could "go wrong"? Well, this ball might contain points of the form $(x,0)$, and these are not in $X$. So all we have to do is to eliminate this option. Try to figure out a general way of doing it, using an example. Suppose $a=(1,8)$. If you take $r=9$ it won't work, right? But it will work for all $r<8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Construct a pentagon from the midpoints of its sides Let $p_{1},p_{2},p_{3},p_{4},p_{5}$ be five points in the euclidean plane such that no set of three of those points lie on the same line. It is easy to prove that there exists a unique pentagon such that $p_{1},p_{2},p_{3},p_{4},p_{5}$ are the midpoints of its sides (In fact there is a more general result saying that the same is true for any odd number n of points as the midpoints of the sides of an n-gon).
The proof uses $\mathbb{C}$ as a model of the euclidean plane and then proves, that the system of linear equations $$\frac{1}{2}(x_{i}+x_{i+1}) = p_{i} \space\space\space \space 1 \leq i \leq 5$$ where $x_{6} = x_{1}$, has unique solutions for $x_{1},x_{2},x_{3},x_{4},x_{5}\in\mathbb{C}$ since the corresponding 5x5 matrix is invertible.
My question is whether there is a way to construct the solution using ruler and compass. (which is possible in the case with only 3 points)
| Let $A,B,C,D,E$ be the given midpoints opposite the unknown vertices $V,W,X,Y,Z$ respectively; both sets of points in rotational order. Then quadrilateral $WXYZ$ must have the midpoints of its sides on a parallelogram, and three of those midpoints are given by $E,A,B$. To find the midpoint $M$ of $\overline{ZW}$, construct lines parallel to $\overline{AB}$ through $E$ and parallel to $\overline{EA}$ through $B$. These lines intersect at $M$.
Then $C,D,M$ are midpoints of the sides of $\triangle ZVW$, and the sides of that triangle are parallel to those of $\triangle DMC$. Construct lines parallel to $\overline{MC}$ through $D$ and parallel to $\overline{DM}$ through $C$. These lines intersect at $V$, and the lines contain the sides of the pentagon through $V$.
Once $V$ is obtained, the remaining vertices follow readily. The given midpoint $D$ is the midpoint of $W$, so $W$ is thereby obtained from $V$ and $D$. Similarly $X$ is obtained from $W$ and $E$, and so on around the pentagon.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Existence of vector fields Does there exists two vector fields $X$ and $Y$ on $\mathbb R^2$ such that the following are satisfied?
*
*$X(0)= Y(0)= 0$, where $0\in \mathbb R^2$ and for others points
$q\in \mathbb R^2$, we have $X(q)\neq 0, Y(q)\neq 0$.
*For any curve $\gamma\in \mathbb R^2$, we have $\langle
X'(t),Y(t)\rangle\geq 0$.
*For any curve $\gamma\in \mathbb R^2$, we have $[X'(t),Y(t)]=0$.
What happen if we change condition $3$ to $[X(t),Y(t)]=0$?
Clarification: As $X$ is a vector field on $\mathbb R^2$, then for any path $\gamma :[0,1]\to \mathbb R^2$, we have a map $\tilde{X}(t):[0,1]\to \mathbb R^2$ by $\widetilde{X}(t)= X(\gamma(t)$, In above question, I mean $X(t)$ by $\widetilde{X}(t)$ and $X'(t)$ is derivative of this map which can be identified with an element in $\mathbb R^2$.
| I think $ X(x,y)=(x^2+y^2)\frac{\partial}{\partial y} $ and $ Y(x,y)=(x^2+y^2)\frac{\partial}{\partial x} $ should work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding second order linear homogenous ODE from the fundamental set of solutions
Find second order linear homogeneous ODE with constant coefficients if its fundamental set of solutions is {$e^{3t},te^{3t}$}.
Attempt: Had this question in my midterm. So, since the fundamental set of solutions is $$y=y_1+y_2=c_1e^{3t}+c_2te^{3t}$$
the characteristic equation of the second order ODE has only one root. I don't know what to do next. Help please.
| The eigenvalues and eigenvectors for the coefficient matrix $A$ in the linear homogeneous system:
$Y'= AY$ are $\lambda_{1} = 3$ with $v_1 =< a; b >$ and $\lambda_2 = 3$ with $v2 =< c; d >$
The fundamental form of the solution is:
$$ Y = c_1 e^{3t}v_1 + c_2t e^{3t}v_2$$
Take the second derivative, $Y''$ for the DEQ.
Your original system will be of the form:
$$y'' - 6y' + 9 = 0$$
to give you the double eigenvalue $\lambda_{1,2} = 3$. You can actually solve this to find the corresponding eigenvectors.
Regards
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Arcs of a circle Geometry and angles If a central angle of measure $30°$ is subtended by a circular arc of length $6\,\mathrm{m}$, how many meters in length is the radius of the circle?
A. $\frac{π}{36}$
В. $\frac{1}{5}$
С. $π$
D. $\frac{36}{π}$
E. $180$
How do I find out what the radius length of the angle is? The answer is (D) by the way.
| HINT: The circumference of a circle of radius $r$ is $\pi r$. The $30^\circ$ angle is $\frac{30}{360}=\frac1{12}$ of the total angle at the centre of the circle, so $6$ metres is $\frac1{12}$ of the circumference of the circle. The whole circumference is therefore $6\cdot12=72$ metres, which, as already noted, is $2\pi r$. Therefore $r$ is ... ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
trying to verify pdf for distance between normally distributed points Math people:
I am trying to find the probability density function for the distance between two points in $\mathbb{R}^3$ selected independently according to the Gaussian pdf $F(\mathbf{z}) = \left(\frac{1}{\sqrt{2\pi}}\right)^3 \exp(-\frac{1}{2}|\mathbf{z}|^2)$. I keep getting $f(t) = \frac{t^2\exp(-\frac{t^2}{4})}{2\sqrt{\pi}}$. I found a paper (http://www.pupr.edu/hkettani/papers/HICS2008.pdf) which, if I understand correctly, states that the actual pdf is $\sqrt{\frac{2}{\pi}} t^2\exp(-\frac{t^2}{2})$ (see p. 10, second "$f(x)$"). That paper actually proves a much more general result and gives this as a special case. I am reluctant to just take their word for it, since I used only Calc 3 techniques and their arguments are much more sophisticated and difficult. Has anyone seen this problem before, perhaps as a calculus or probability exercise? Assuming that either their formula or mine is right, is there an easy way to test which is right experimentally (generating random vectors, etc.)?
Stefan (STack Exchange FAN)
| The square of the distance is $r^2=(X_1-X_2)^2+(Y_1-Y_2)^2+(Z_1-Z_2)^2$, where the $X_i$s, $Y_i$s, and $Z_i$s are independent standard normal r.v.s. So, $r^2/2$ is $\chi^2$ distributed with $3$ degrees of freedom, and as you say, the distance $r$ has density $r^2\exp(-\frac{r^2}{4})/(2\sqrt{\pi}).$ p. 10 of the paper you mention is treating a different case, where each component of the difference between the two points is standard normal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$ I am trying to show $$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$$
I tried breaking it down, and got stuck when trying to $\left( \frac{2^{n}n!}{n^{n}} \right)$ goes to 0.
| It is not true that $\left( \frac{(2n)!}{n^{2n}} \right)=\left( \frac{2^{n}n!}{n^{n}} \right)$ The left side has a factor $2n-1$ in the numerator while the right side does not. But you can use Stirling's approximation to say $$\frac {(2n)!}{n^{2n}}\approx \frac {(2n)^{2n}}{(ne)^{2n}}\sqrt{4 \pi n}$$ and the powers of $\frac 2e$ take it to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Does $\sin(x)=y$ have a solution in $\mathbb{Q}$ beside $x=y=0$ Is there a way to show, that the only solution of
$$\sin(x)=y$$
is $x=y=0$ with $x,y\in \mathbb{Q}$.
I am seaching a way to prove it with the things you learn in linear algebra and analysis 1+2 (with the knowledge of a second semester student).
| Sorry for the previous spam. I shall prove for $\cos$, cosine of any rational numbers except for 0 cannot get rational numbers.
By using polynomial argument, we shall only have to prove for integers.
Suppose that $m\in\mathbb{N}$, $\cos(m)\in\mathbb{Q}$.
For any fixed prime number $p>m$, consider polynomial $x\in(0,m)$
$$f(x) = \frac{(x-m)^{2p}(m^2-(x-m)^2)^{p-1}}{(p-1)!}$$
And
$$F(x) = \sum_{n = 0}^{2p-1} (-1)^{n+1}f^{2n}(x)$$
Which satisfies
$$(F'(x)\sin(x)-F(x)\cos(x))' = F''(x)\sin(x) +F(x)\sin(x) = f(x)\sin(x)$$
since the other terms cancelled.
$$\int_0^mf(x)\sin(x)dx = F'(m)\sin(m)-F(m)\cos(m)+F(0)$$
Consider $f$ is a polynomial of $(x-m)^2$, thus $F'(m) = 0$, and we can see that
$$f(m-x) = x^{2p}(m^2-x^2)^{p-1}/(p-1)!$$
By computing, $p|f^{(l)}(m)$ for every $l$. That means $F(m)$ is a multiple of $p$ by definition of $F$, say $pM$.
If $\cos(m) = s/t$, then
$$t\int_0^m f(x)\sin(x)dx = -spM+tN$$
is an integer,
While
$$f\le \frac{m^{4p-2}}{(p-1)!} $$
thus
$$t\int_0^mf(x)\sin(x)dx\le t\frac{m^{4p-2}}{(p-1)!}\cdot m <1 $$
when $p$ is large enough. Contradiction of it is an integer.
As a result of this, $\sin$ should also satisfy this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Property kept under base change and composition is preserved by products The following is true? Why?
Let $P$ be a property of morphisms preserved under base change and composition. Let $X\to Y$ and $X'\to Y'$ be morphisms of $S$-schemes with property $P$.
Then the unique morphism $X\times_S X' \to Y\times_S Y'$ has property $P$.
| Yes this is true.
The canonical morphism $X\times_S X'\to Y\times_S Y'$ is the composition of $X\times_S X'\to Y\times_S X'$ and $Y\times_S X'\to Y\times_S Y'$. The latter verify property P because they are obtained by base change (the first one is $X\to Y$ base changed to $Y\times_S X'$, the second one is similar). As P is stable by composition, your canonical morphism satisfies P.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Schoen Estimates (part 3) I'm referring to the article 'Estimates for stable minimal surfaces in three dimensional manifolds' of Richard Schoen
In the first paragraph of the proof of theorem 2 the author seems to assert that the universal covering of $ M $ is conformally equivalent to the unit disk (with standard metric). But i have some doubt about this thing. In fact if $ M $ is complete and non compact, its universal covering space has to be conformally equivalent to the complex plane (applying methods of Fischer Colbrie Schoen and observing that non negative ricci curvature implies non negative scalar curvature).
Thanks
| Indeed, $M$ could be simply the plane, which is its own universal cover. So yes, their statement
we may assume that $M$ is represented by a conformal immersion $f:D\to N$ [where $D=D_1$ is presumably the unit disk from page 116]
needs to be modified. But I think it suffices to replace $M$ with $B_R(P_0)$, which is a non-complete manifold covered by $D$. As other results in the paper, Theorem 2 is local: there is no statement made about the points outside of $B_R(P_0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find (or 'generate') combinatorial meaning for the given expression $\left(\dfrac{6(k-n)(k-1)}{(n-2)(n-1)}+1\right)\dfrac{30}{n(n+1)(n+2)}$
(for $n\geq 3$ and $1\leq k \leq n$)
The expression comes from question https://math.stackexchange.com/questions/304876/please-help-to-find-function-for-given-inputs-and-outputs where it is used to get answer for some unknown problem (I believe that's combinatorial, with possibility of used Monte-Carlo method). The OP didn't give any interpretation of the expression.
Can you please share your best ideas about combinatorial interpretation of the expression? Big thanks in advance.
P.S. There's tiny possibility that it was mathematical/SE joke.
| This post does not show a meaning of the expression, but explains how one might arrive at it.
I want a quadratic function with zero mean defined on $\{1,\dots,n\}$. Naturally, it should be symmetric about the midpoint of the interval. The obvious symmetric function is $k(k-n-1)$, but it does not have mean zero. The mean is
$$\frac{1}{n}\sum_{k=1}^n k(k-n-1) = -\frac{(n+1)(n+2)}{6}$$
So I subtract that, arriving at $k(k-n-1)+\frac{(n+1)(n+2)}{6}$. But maybe I also care about the second moment, which is
$$\sum_{k=1}^n \left(k(k-n-1)+\frac{(n+1)(n+2)}{6}\right)^2 = \frac{(n-2)(n-1)n(n+1)(n+2)}{180}$$
Dividing by the second moment yields
$$\frac{180}{(n-2)(n-1)n(n+1)(n+2)}\left(k(k-n-1)+\frac{(n+1)(n+2)}{6}\right) $$
which is the formula in question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A improper integral with Glaisher-Kinkelin constant Show that :
$$\int_0^\infty \frac{\text{e}^{-x}}{x^2} \left( \frac{1}{1-\text{e}^{-x}} - \frac{1}{x} - \frac{1}{2} \right)^2 \, \text{d}x = \frac{7}{36}-\ln A+\frac{\zeta \left( 3 \right)}{2\pi ^2}$$
Where $\displaystyle A$ is Glaisher-Kinkelin constant
I see Chris's question is a bit related with this Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$
| Here is an identity for log(A) that may assist.
$\displaystyle \ln(A)-\frac{1}{4}=\int_{0}^{\infty}\frac{e^{-t}}{t^{2}}\left(\frac{1}{e^{t}-1}-\frac{1}{t}+\frac{1}{2}-\frac{t}{12}\right)dt$.
I think Coffey has done work in this area. Try searching for his papers on the Stieltjes constant, log integrals, Barnes G, log Gamma, etc.
Another interesting identity is $\displaystyle 2\int_{0}^{1}\left(x^{2}-x+\frac{1}{6}\right)\log\Gamma(x)dx=\frac{\zeta(3)}{2{\pi}^{2}}$.
Just some thoughts that may help put it together.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Compound interest derivation of $e$ I'm reviewing stats and probability, including Poisson processes, and I came across:
$$e=\displaystyle \lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n$$
I'd like to understand this more fully, but so far I'm struggling. I guess what I'm trying to understand is how you prove that it converges. Can anyone point me toward (or provide) a good explanation of this?
| It's not too hard to prove, but it does rely on a few things. (In particular the validity of the taylor expansion of $\ln$ around 1 and that $\exp$ is continuous.)
Consider in general the sequence $n\ln(1+x/n)$ which is defined for all $x$, positive or negative provided $n$ is large enough. (In fact the proof that follows can also be modified slightly to work for complex $x$). Using the power series representation $\ln(1+y) = \sum_{i=1}^\infty (-1)^{i+1}\frac{x^i}{i}$ we get
$$n\ln(1+\frac{x}{n}) = x + \sum_{i=2}^\infty(-1)^{i+1}\frac{x^i}{in^{i-1}}$$
The absolute values of the terms in the series on the right hand side are bounded above by a geometric series (for n large enough) whose summation tends to 0 as $n$ tends to infinity. Hence
$$\lim_{n\rightarrow \infty} n\ln(1+\frac{x}{n}) = x$$
And so using the fact that $\exp(x)$ is a continuous function:
$$\lim_{n\rightarrow \infty}\bigg(1+\frac{x}{n}\bigg)^n = \exp\bigg(\lim_{n\rightarrow \infty}n\ln(1+\frac{x}{n})\bigg) = \exp(x)$$
In particular, set $x = 1$ to get the result you ask for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Wave Equation, Energy methods. I am reading the book of Evans, Partial differential Equations ... wave equation section 2.4; subsection 2.4.3: Energy methods. Arriving at the theorem:
Theorem 5 (Uniqueness for wave equation). There exists at most one function $u \in C^{2}(\overline{U}_{T})$ solving
$u_{tt} -\Delta u=f $ in $ U_{T}$
$u=g $ on $ \Gamma_{T}$
$u_{t}=h$ on $U \times \{t=0\}.$
Proof. If $\tilde{u}$ is another such solution, then $ w:=u-\tilde{u}$ solves
$w_{tt} -\Delta w=0 $ in $ U_{T}$
$w=0 $ on $ \Gamma_{T}$
$w_{t}=0$ on $U \times \{t=0\}.$
Define the "energy"
$e(t):=\frac{1}{2} \int_{U} w^{2}_{t}(x,t)+ \mid Dw(x,t)\mid ^{2} dx (0\leq t \leq T).$
We compute
$\dot{e}(t)=\int_{U} w_{t}w_{tt}+ Dw \cdot Dw_{t}dx (\cdot = \frac{d}{dt})$
$=\int_{U}w_{t}(w_{tt} - \Delta w)dx=0$.
There is no boundary term since $w=0$, and hence $w_{t}=0$, on $\partial U \times [0,T].$ Thus for all $0\leq t \leq T, e(t)= e(0)=0$, and so $w_{t}, Dw \equiv 0$ within $U_{T}$. Since $w \equiv 0$ on$ U \times \{t=0\}$, we conclude $w=u-\tilde {u}\equiv 0$ in $U_{T}$.
I have two questions:
1) What is the motivation for the definition of $e(t)$
2)$\int_{U} w_{t}w_{tt}+ Dw \cdot Dw_{t}dx $
$=\int_{U}w_{t}(w_{tt} - \Delta w)dx$. How to justify this equality?
Thank very much.
| This quantity, the definition of $e(t)$, can be easily recognized as the Hamiltonian (basically another name of energy) of the system by a physics student. Let's explore more details.
The PDE $w_{tt}-\Delta w=0$ is equivalent to a variation problem (under some boundary conditions, maybe), whose Lagrangian (or Lagrangian density) is
$$
\mathcal L = \frac{1}{2}(w_t^2-(\nabla w)\cdot(\nabla w))
$$
There is a standard process obtaining the Hamiltonian from the Lagrangian. First, calculate the canonical momentum $\pi$:
$$
\pi = \frac{\partial \mathcal L}{\partial w_t}=w_t.
$$
Then use the Legendre transformation:
$$
\mathcal H = \pi w_t - \mathcal L = \frac{1}{2} (w_t^2+(\nabla w)^2),
$$
which is the integrand in the definition of $e(t)$. This is actually the Hamitonian density and should be integrated over space to give the total energy.
P.S.
As a somewhat more complicated case, consider the following equation:
$$
u_{tt} - \nabla \cdot (c^2(x) \nabla u)+q(x)u = 0
$$
with some appropriate boundary conditions and initial conditions imposed. The two given functions $c,q\ge 0$ and depend on $x$ only. You can check that the Lagrangian is
$$
\mathcal L = \frac{1}{2}(u_t^2 - c^2 (\nabla u)^2-qu^2),
$$
which gives the Hamiltonian
$$
\mathcal H = \frac{1}{2} (u_t^2+c^2(\nabla u)^2+qu^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How to compute the SVD of $2\times 2$ matrices? What's an efficient algorithm to get the SVD of $2\times 2$ matrices?
I've found papers about doing SVD on $2\times 2$ triangular matrices, and I've seen the analytic formula to get the singular values of a $2\times 2$ matrix. But how to use either of these to get the SVD of an arbitrary $2\times 2$ matrix?
Are the general algorithms built on these, or are these just some special cases?
| The SVD of a $2\times 2$ matrix has a closed-form formula, which can be worked out by writing the rotation matrices in terms of a single unknown angle each, and then solving for those angles as well as the singular values.
It is worked out here, for instance.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Functions of algebra that deal with real number
If the function $f$ satisfies the equation $f(x+y)=f(x)+f(y)$ for every pair of real numbers $x$ and $y$, what are the possible values of $f(0)$?
A. Any real number
B. Any positive real number
C. $0$ and $1$ only
D. $1$ only
E. $0$ only
The answer for this problem is E. For the following problem to find the answer do you have to plug in 0 to prove the function?
| Just to add to the collection above, if the vector space is $V$:
$$f(v)=f(v+0)=f(v)+f(0)\Longrightarrow f(0)=0$$
for any $v\in V$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Exercise of complex variable, polynomials. Calculate the number of zeros in the right half-plane of the following polynomial:
$$z^4+2z^3-2z+10$$
Please, it's the last exercise that I have to do. Help TT. PD: I don't know how do it.
| Proceed like the previous problem for first quadrant.
You will find one root.
And note that roots will be conjugates.
So 2 roots in the right half-plane..
For zero in the first quadrant, consider the argument principle: if $Z$ is the number of zeroes of $f$ inside the plane region delimited by the contour $\gamma$, then $\Delta_\gamma(\textrm{arg}f)=2\pi Z$, i.e. the variation of the argument of $f$ along $\gamma$ equals $Z$ times $2\pi$.
Take a path from the origin, following the real axis to the point $M>0$, then make a quarter of circle or radius $M$, reaching the point $iM$ and then go back to the origin along the imaginary axis. Now try to determine the variation of the argument of $f(z)$ along this path for $M\to\infty$:
*
*along the real axis, the function is $f(t)=t^4-2t+2t^3+10$, therefore $f(t)$ is real for $t\geq0$ so the total change of argument along this part of the path is $0$.
*along the path $Me^{i\theta}$ for $0\leq\theta\leq \pi/2$, if $M$ is very large, the function is near to $g(\theta)=M^4e^{i4\theta}$; therefore the argument goes from $0$ to $2\pi$.
*along the imaginary axis, the function's argument doesn't change.
So, the total change of the argument is $2\pi$, implying that the function has only one zero in that quadrant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Ask for a question about independence This is the question I met while reading Shannon's channel coding theorem. Assume a random variable $X$ is transmitted through a noisy channel with transition probability $p(y|x)$. At the receiver a random variable $Y$ is obtained. Assume we have an additional random variable $X'$ which is independent of $X$. How to show that $X'$ is independent of bivariate random variable $(X,Y)$ and $X'$ is independent of $Y$? It looks obvious because $Y$ is generated only from $X$, but I just can not prove it rigorously, i.e., that $p(x,y|x')=p(x,y)$ and $p(y|x')=p(y)$. Thanks a lot for your answer!
| Two given conditions are:
(1)$X'$ is independent of $X$,
(2)$Y$ is generated only from $X$, i.e., $p(y|x)=p(y|x,x')$. This acutually means $X'\to X\to Y$ forms a Markov chain.
Now let's prove $X'$ is independent of bivariate random variable $(X,Y)$, i.e., $p(x,y|x')=p(x,y)$:
$p(x,y|x')=p(x|x')p(y|x,x')=p(x)p(y|x,x')$(from (1))$=p(x)p(y|x)$(from (2))$=p(x,y)$.
It can easily shown that if $X'$ is independent of $(X,Y)$, then $X'$ is independent of each component. I have shown $X'$ is independent of $(X,Y)$, so $X'$ is also independent of $Y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
what is the logic behind the 'UPC -A' check digit? In UPC-A barcode symbology, the 12th digit is known as check digit and it is used by the scanner to determine if it scanned the number correctly or not.
the check digit is calculated as follows:
1.Add the digits in the odd-numbered positions (first, third, fifth, etc.) together and multiply by three.
2.Add the digits in the even-numbered positions (second, fourth, sixth, etc.) to the result.
3.Find the result modulo 10 (i.e. the remainder when divided by 10.. 10 goes into 58 5 times with 8 leftover).
4.If the result is not zero, subtract the result from ten.
For example, in a UPC-A barcode "03600029145x" where x is the unknown check digit, x can be calculated by
adding the odd-numbered digits (0 + 6 + 0 + 2 + 1 + 5 = 14),
multiplying by three (14 × 3 = 42),
adding the even-numbered digits (42 + (3 + 0 + 0 + 9 + 4) = 58),
calculating modulo ten (58 mod 10 = 8),
subtracting from ten (10 − 8 = 2).
The check digit is thus 2.
What is the logic behind calculating 'check digit' in such manner?
Can't two different combination of digits produce the same check digit?
| Just a remark on your question
Can't two different combination of digits produce the same check digit?
Of course this happens, after all there are $10^{11}$ possibilities for the first $11$ digits, and only $10$ for the $12$-th one. So plenty of valid codes will share the same $12$-th digit, and a mistake that takes you from one to the other can't be caught. But as noted by @bubba, the code aims at catching two of the commonest errors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Vantage point of character theory I am not sure whether I can frame my question properly, or whether at this point my understandings permit me to comprehend the perspectives of the answers to come, but somehow I find it pretty amazing that while doing representation of finite groups over characteristic 0 fields, the character of representation plays such an important role.
If I look at it separately the trace of a matrix hardly reveals anything about the matrix except if the matrix is diagonal. Does the fact that there exist bases such that I can make $ \rho (g) $ diagonal ( for nice underlying fields at least) for any element g of the group G make a significant difference in our considerations?
Also, characters being equal implying equivalence of representations is such a stunning fact. Is there any intuitive basis for this?
And lastly what kind of information comes under the realm of character theory and what is purely representation theory's realm?
I apologize for such a vague question, and reading representation theory for the first time surely I don't understand things too well. But any motivation towards the perspective and vantage points of the subject will be very beneficial.
| I know I'm digging up an old question, but one seems to have brought up this point, although darij grinberg did briefly allude to it in the comments:
For any fixed $g \in G$, knowing the trace of $\rho(g)$ doesn't tell you much. However, the character contains the additional information of the trace of $\rho(g^k)$ for all $k$. If $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $\rho(g)$, then $$\operatorname{Tr}(\rho(g^k)) = (\lambda_1)^k + \cdots + (\lambda_n)^k.$$ By Newton's identity, knowing this for all $k$ allows one to recover the data of the multiset $\{\lambda_1, \ldots, \lambda_n\}$. Since $\rho(g)$ is diagonalizable, it follows that you know the similarity type of $\rho(g)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
Use Maple I can find $x \in \{1;ab+bc+ca\}$
| I have a partial solution, as follows:
Note that $\frac{(b-c)(1+a^2)}{x+a^2}=\frac{(b-c)\left((x+a^2)+(1-x)\right)}{x+a^2}=(b-c)+\frac{{(b-c)}(1-x)}{x+a^2}$. Likewise, $\frac{(c-a)(1+b^2)}{x+b^2}=(c-a)+\frac{(c-a)(1-x)}{x+b^2}$ and $\frac{(a-b)(1+c^2)}{x+c^2}=(c-a)+\frac{(a-b)(1-x)}{x+c^2}$.
Now, $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=\frac{(b-c)(1-x)}{x+a^2}+\frac{(c-a)(1-x)}{x+b^2}+\frac{(a-b)(1-x)}{x+c^2}$ as $(b-c)+(c-a)+(a-b)=0$.
Hence $(1-x)\left(\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}\right)=0$ and so $x=1$ or $\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Proving that $\|A\|$ is finite. Let $|v|$ be the Euclidean norm on $\mathbb{R^n} $. For $A\in \mathrm{Mat}_{n\times n}(\mathbb{R})$ we define $\displaystyle \|A\|:= \sup_{\large v\in \mathbb{R^n},\,v \neq 0}\frac{|Av|}{|v|}$. How to show that $\|A\|$ is finite for every $A$?
It would be very helpful if someone could give hints. I think I should show that $\|-\|$ is bounded,but I don't know how...
| Hint:
Let $S=\{v\in\mathbb{R}^n\;|\;|v| = 1\}, N = \{\frac{|Av|}{|v|}\;|\;v\in\mathbb{R}^n.\;v\ne 0\}, N' = \{|Av|\;|\;v\in\mathbb{R}^n.\;|v|=1\}$
Step 1: $||A|| = \sup N = \sup N'$
Step 2: Show that $x\to|Ax|$ is a continuous map. $S$ is closed and bounded in $\mathbb{R}^n$ therefore compact, |Ax| attains max on $S$.
Done
Do you happen to be in Linear Algebra II class?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Subgroups of order $p$ and $p^{n-1}$ in a group of order $p^n$. I have a group $G$ of order $p^n$ for $n \ge 1$ and $p$ a prime. I am looking for two specific subgroups within $G$: one of order $p$ and one of order $p^{n-1}$. I don't think I would use the Sylow theorems here because those seem to apply to groups with a "messier" order than simply $p^n$. Would Cauchy's Theorem allow me to generate the two requisite subgroups? I could use it to find an element of order $p$ and an element of order $p^{n-1}$ and then consider the cyclic subgroups generated by these two elements?
|
Let $P$ act on itself by conjugation. $1$ appears in an orbit of size $1$, and everything else appears in an orbit of size $p^k$ for some $k$. Since the sum of the orbit sizes is equal to $|P|$, which is congruent to $0\mod{p}$, that means there has to be at least one more orbit of size $1$. Orbits of size $1$ under conjugation contain elements which commute with everything in the group; they compose $Z(P)$. Now suppose inductively that $\exists S\unlhd P$ with $|S|=p^k$. Then by the above lemma and Cauchy's theorem $P/S$ has a central subgroup $\overline{Q}$ of order $p$. Lifting $\overline{Q}$ back to $P$, we obtain a normal subgroup of order $p^{k+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet
First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet.
$$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\
&= m^4 -2m^2n^2 + n^4 + 4m^2n^2 \\
&= m^4 + 2m^2n^2 + n^4 \\
&= 1\end{align*}$$ which shows that it respect $a^2+b^2 = c^2$
let p be a prime number, $ p|(m^2 + n^2) \text { and } p|(m^2 - n^2) $
if $gcd(m^2 + n^2, (m^2 - n^2)) = 1$
$p | (m^2 + n^2) , \text { so, } p |m^2 \text { and } p |n^2$
that means $ (m^2 + n^2) \text { and } (m^2 - n^2) $ are prime together
I'm kind of lost when I begin to show the gcd = 1... I think I know what to do, just not sure how to do it correctly.
Thanks
| To show $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ is equivalent to showing $(m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 = 0$ so \begin{align*} && (m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 \\ &=& m^4 - 2m^2n^2 + n^4 + 4m^2n^2 - m^4 - 2m^2n^2 - n^4 \\ &=& m^4 + n^4 - m^4 - n^4 \\ &=& 0\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Sine not a Rational Function Spivak This is Chapter 15 Question 31 in Spivak:
a) Show sin is not a rational function.
By definition of a rational function, a rational function cannot be $0$ at infinite points unless it is $0$ everywhere. Obviously, sin have infinite points that are 0 and infinite points that are not zero, thus not a rational function.
b) Show that there do not exist rational functions $f_0, \ldots, f_{n-1}$ such that $(\sin x)^n + f_{n-1}(x)(\sin x)^{n-1} + \ldots + f_0({x}) = 0$ for all x
First choose $x = 2k\pi$, so $f_0(x) = 0$ for $x = 2k\pi$. Since $f_0$ is rational $\implies f_0(x) = 0$ for all x.
Thus can write $\sin x[(\sin x)^{n-1} + f_{n-1}(\sin x)^{n-2} ... +f_1(x)] = 0$
Question: The second factor is $0$ for all $x \neq 2k\pi$ How does this imply that it is 0 for all x? And how does this lead to the result?
| Hint: Use continuity of $\sin$ and rational functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Stadium Seating A circular stadium consists fo 11 sections with aisles in between. With 13 tiers of concrete steps for the final section, section K. Seats are places along every concrete step, with each step 0.45 m wide. The arc AB at the fron of the first row is 14.4 m long while the arc CD at the back of the back row is 20.25m long.
1. How wide is each concrete step?
2. What is the length of the arc of the back of row 1, row 2, row 3, and so on?
| If the inner radius of row $i$ is $r_i$, then the arc length of the inner arc of that row is given by
$$L_i = \frac{2\pi r_i}{11}.$$
If the width $w$ of each step is constant, then
$$r_i = r_1 + (i-1)w$$
and the arc length of the outer arc of step $i$ is
$$M_i = \frac{2\pi (r_i+w)}{11}.$$
You now have two unknowns -- $r_1$ and $w$ -- and two equation relating the unknowns to known quantities, namely $L_1$ and $M_{13}$. Solve for $w$ and $r_1$ and then plug into the equation above to find $M_i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If the matrix of a linear map is independent from the basis, then the map is a multiple of the identity map. Let $V$ be a finite dimensional vector space over $F$, and let
$$T:V\to V$$
be a linear map.
Suppose that given any two bases $B$ and $C$ for $V$, we have that the matrix of $T$ with respect $B$ is equal to that with respect to $C$.
How can we show that this implies that there exists some $\lambda\in F$ such that $T(v)=\lambda v$, $\forall v\in V$?
| Hint: This is equivalent to say that any nonzero vector is an eigenvector of $T$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Legendre symbol- what is the proof that it is a homomorphism? I know that one property of the Legendre symbol is that it is a homomorphism. However, I have not been able to find a proof that this is the case. If someone could give me or show me to a thorough proof of this, that would be great.
I am going with the definition:
$\sigma(x) = 1$ when $ x=y^2$
$\sigma(x)= -1 $ otherwise
where $\sigma$ is a map st $\sigma: {\mathbb{Z}_p}^{\times} \rightarrow (-1,1)$
EDIT: How would we sketch a proof that the symbol is a homomorphism using that
if $(G, *)$ is a finite group and $H\subset G$ is a subgroup, and we have an equivalence relation on $G: x \sim y$ iff $\exists h \in H$ st $y= x*h$, and $P$ is an equivalence relation: then we know $\# P = \# H$ and $\# H$ divides $\# G$.
Basically how would we prove, using this fact, that the set of squares in $\mathbb{Z}_p$ is a subgroup of $\mathbb{Z}_p$?
| I am wondering if it is permissible to use the primitive roots modulo a prime $p$.
Suppose so, and then we look at the definition of a Legendre symbol, and then give a proof that it is a homomorphism. So fix a prime $p$ first.
Let $x$ be a number not divisible by $p$. Suppose that we already have at disposal a primitive root modulo $p$, denoted by $g$. Then it is easily seen that $g$ is a non-residue of $p$, and that every integer is congruent to $g^n$ for some $n$. Moreover, $x$ is a residue if and only if $x\equiv g^{2k}$ for some $k$. Then the fact that Legendre symbol is a homomorphism follows from the rules that even+even=even, even+odd=odd, and that odd+odd=even.
Suppose we are not given the existence of a primitive root modulo $p$. Then we use another approach:
Let $p$ be an odd prime. Define a homomorphism from $\mathbb Z_p^*={1, \cdot\cdot\cdot,p-1}$ to itself by squaring: we send $x$ to $x^2$. Its kernel, that is, {$x: x^2\equiv 1\pmod p$}, consists of two numbers: $1$ and $p-1$. Its image is the set of residues modulo $p$, which is also a group $\mathbb S_p$ under multiplication. Since multiplication of integers is abelian, the group $\mathbb S_p$ is a normal subgroup of $\mathbb Z_p^*$. So we can form a quotient $\mathbb Z_p^*/\mathbb S_p$. It is of order $2$, hence isomorphic with the cyclic group of order $2$, {$1,-1$}. And we send $x\in \mathbb Z_p^*$ to its image in $\mathbb Z_p^*/\mathbb S_p$: this is the Legendre symbol. By this definition, it is obvious that Legendre symbol is a homomorphism.
I wonder also if we can talk about the wonderful interpretation due to Zolotarev: it is nothing but the theory of permutations, yet is a magic observation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Tricky T/F... convergence T or F?
1) If $x_n \rightarrow 0$ and $x_n \neq 0$ for all $n$, then the sequence {$1/n$} is unbounded.
Also similarly...
2) If {$x_n$} is unbounded and $x_n \neq 0$ for all $n$, then $1/x_n \rightarrow 0$.
For the first one, I would say that is true because the limit of {$1/n$} would approach infinity, thus making it unbounded?
And the second one, also seems true by similar logic. Am I overlooking something here?
| The first one is true; but note the sequence $(1/x_n)$ need not converge to $\infty$. The sequence $(1/|x_n|)$, however, would. Consider here, for example, the sequence $(1/2,-1/3,1/4,-1/5,\ldots)$.
For the second one, consider the sequence $(1,1,2,1,3,1,4,1,5,\ldots)$. Note this sequence is unbounded, but the sequence of reciprocals does not converge to $0$.
(If you knew $(x_n)$ converged to $\infty$ (or to $-\infty$), then you could conclude the sequence of reciprocals would converge to $0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$ Find
$$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$
| Here is another sketch of proof.
Let
$$J_n = \{(j, k) : 0 \leq j, k < n \text{ and } (j, k) \neq (0, 0) \}.$$
Then for each $(j, k) \in J_n$ and $(x_0, y_0) = (j/n, k/n)$, we have
$$ \frac{x_0 + y_0}{(x_0+\frac{1}{n})^{3} + (y_0 + \frac{1}{n})^3} \leq \frac{x+y}{x^3 + y^3} \leq \frac{x_0 + y_0 + \frac{2}{n}}{x_0^3 + y_0^3} $$
for $(x, y) \in [x_0, y_0] \times [x_0 + 1/n, y_0 + 1/n]$. Thus if we let $D_n$ be the closure of the set $[0, 1]^2 - [0, 1/n]^2$, then
$$ \sum_{(j,k) \in J_n} \frac{j+k}{(j+1)^3 + (k+1)^3} \leq \int_{D_n} \frac{x+y}{x^3 + y^3} \, dxdy \leq \sum_{(j,k) \in J_n} \frac{j+k+2}{j^3 + k^3}. $$
It is not hard to establish the relation that
$$ \sum_{(j,k) \in J_n} \frac{j+k}{(j+1)^3 + (k+1)^3} = \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} + O(1) $$
and that
$$ \sum_{(j,k) \in J_n} \frac{j+k+2}{j^3 + k^3} = \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} + O(1). $$
By noting that
\begin{align*}
\int_{D_n} \frac{x+y}{x^3 + y^3} \, dxdy
&= 2\int_{\frac{1}{n}}^{1} \int_{0}^{y} \frac{x+y}{x^3 + y^3} \, dxdy \\
&= (2 \log n) \int_{0}^{1} \frac{1}{x^2 - x + 1} \, dx \\
&= \frac{4 \pi \log n}{3\sqrt{3}},
\end{align*}
we obtain the asymptotic formula
$$ \frac{1}{\log n} \sum_{j,k=1}^{n} \frac{j+k}{j^3 + k^3} = \frac{4 \pi}{3\sqrt{3}} + O\left( \frac{1}{\log n} \right) $$
and hence the answer is $\displaystyle \frac{4 \pi}{3\sqrt{3}} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Powers and the logarithm By example:
*
*$4^{\log_2(n)}$ evaluates to $n^2$
*$2^{\log_2(n)}$ evaluates to $n$
What is the rule behind this?
| Here's a method that relies more on applying an appropriate strategy than on formulas.
If you want to rewrite $4^{\log_2(n)}$ as a power of $n$, then you simply want to solve for $u$ in the following equation:
$$4^{\log_2(n)} = n^u$$
The standard way of solving for something that appears in the exponent of an exponentiated expression is to take the logarithm (to some fixed base) of both sides. Since $\log_2$ already shows up, we may as well take the logarithm-base-$2$ of both sides (otherwise, two different kinds of logarithms will show up, and we'd have to take care of this later):
$$\log_{2}\left(4^{\log_2(n)}\right) = \log_{2}\left(n^u\right)$$
$$\log_{2}(n) \cdot \log_{2}(4) = u \cdot \log_{2}(n)$$
$$\log_{2}(4) = u$$
$$2 = u$$
In the last step I used the fact that $\log_2(4) = \log_{2}(2^2) = 2$. However, if I didn't know this, or if the numbers weren't nice (i.e. we got something like $\log_2(5) = u$), we'd still have what we wanted -- a numerical expression for the exponent of $n.$ The same method allows you to (even more easily) determine that $2^{\log_2(n)} = n.$ Here's an application of the same method to some second semester calculus (in the U.S.) semi-challenging $p$-series convergence/divergence problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$E^2[X]$ vs.$E[X^2]$, what's the difference? This has come up in a homework problem, but I've never seen exponents defined in terms of random variables and expected values. I've tried googling this, but I must not be using the right words. If anyone could define both terms for me, or briefly explain the difference, it would help a lot.
Thanks in advance.
| The difference is exactly the same as with $$ a^2 + b^2 \neq (a+b)^2$$
Another thing to see is that $\rm E[X^2]$ is the $\rm L^2$-norm is $\rm X$ is centered ; $(\rm E[X])^2$ is the square of the $\rm L^1$-norm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Computing $\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$ What ways would you propose for the limit below?
$$\lim_{n\to\infty}n\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right)$$
Thanks in advance for your suggestions, hints!
Sis.
| OK, it turns out that
$$\sum_{k=1}^n\left( \frac{1}{(2k-1)^2} - \frac{3}{4k^2}\right) = \sum_{k=1}^{n-1} \frac{1}{(k+n)^2}$$
This may be shown by observing that
$$\sum_{k=1}^n \frac{1}{(2k-1)^2} = \sum_{k=1}^{2 n-1} \frac{1}{k^2} - \frac{1}{2^2} \sum_{k=1}^n \frac{1}{k^2}$$
The desired limit may then be rewritten as
$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{(1 + (k/n))^2}$$
which you may recognize as a Riemann sum, equal to
$$\int_0^1 dx \: \frac{1}{(1+x)^2} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 2
} |
Differentiation in 3d of sin root and fractions in one! -> to find the normal to a function I have to find the normal of this function at a defined point $x$ and $z$, I have done A Level maths but that was some time ago but I don't think it was covered to this level, I am now doing a CS degree. I thought the best way would be to differentiate the function to get the tangent, then get the normal.
I have no idea where to start.
Any ideas on how I should go about it?
$$y = \frac{\sin (\sqrt{x^2+z^2})}{\sqrt{x^2+z^2}}$$
| The normal to a curve at a point is perpendicular to the gradient at that point. In your case:
$$f(x,z) = \frac{\sin{\sqrt{x^2+z^2}}}{\sqrt{x^2+z^2}}$$
$$\begin{align}\nabla f &= \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial z}\right ) \\ &= \left ( \frac{x \cos
\left(\sqrt{x^2+z^2}\right)}{x^2+z^2
}-\frac{x \sin
\left(\sqrt{x^2+z^2}\right)}{\left(x
^2+z^2\right)^{3/2}}, \frac{z \cos
\left(\sqrt{x^2+z^2}\right)}{x^2+z^2
}-\frac{z \sin
\left(\sqrt{x^2+z^2}\right)}{\left(x
^2+z^2\right)^{3/2}}\right ) \end{align}$$
The unit normal to the curve is then
$$\frac{\left ( -\frac{\partial f}{\partial z}, \frac{\partial f}{\partial x}\right )}{\sqrt{\left ( \frac{\partial f}{\partial x} \right )^2+\left ( \frac{\partial f}{\partial z} \right )^2}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why isn't this a valid argument to the "proof" of the Axiom of Countable Choice? I am having a little trouble identifying the problem with this argument:
Let $\{A_1, A_2, \ldots, A_n, \ldots\}$ be a sequence of sets.
Let $X:= \{n \in \mathbb{N} : $ there is an element of the set $A_n$ associated to $n \}$
(1) $A_1$ is not empty. Therefore, there exists a $x_1 \in A_1$
(2) Given an $A_n$ (and a $x_n \in A_n$), we have that $A_{n+1}$ is non-empty and, therefore, there exists a $x_{n+1} \in A_{n+1}$
So, by induction, $X=\mathbb{N}$ and the axiom of countable choice is "proven".
Where is the error?
| To avoid Choice, you need to have a definitive way of choosing an element from each set.
For example, if each $A_n$ is a pair of shoes, you may always choose the left.
A typical useful case is when each $A_n$ has a distinguished member such as a unique minimum that you can choose. For example, Baire Category Theorem for a complete SEPARABLE metric space can be proved by a trick like this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
} |
Combination of three items with no adjacent items the same I am looking for a closed expression for calculating the number of combinations of $n = n_1 + n_2 + n_3$ objects arranged in an ordered list, where there are $n_1$ $a$, $n_2$ $b$ and $n_3$ $c$, under the constraint that $a$ may not appear adjacent to $a$, $b$ not adjacent to $b$, and $c$ not adjacent to $c$.
If $n_1 = n_2 = n_3$, the following is an upper bound on the number of combinations:
$3$ ways of picking the first, and $2$ ways of picking all subsequent objects gives
$$3 \cdot 2^{n-1}$$
A solution only exists if $n_1$ is not greater than $1+n_2+n_3$ and similarly for $n_2$ and $n_3$
Thanks to @Byron Schmuland for pointing me to a general answer; the integral that needs to be evaluated looks a bit daunting.
Now I need a solution to the integral for three items like the one listed on page 9 of the paper cited in the answer for two items.
| In addition to the formula linked above there is a nice generating function. The number of solutions with $i$ $a$'s, $j$ $b$'s and $k$ $c$'s is the coefficient of $a^ib^jc^k$ in $$\frac{1}{1 - \frac{a}{1+a} - \frac{b}{1+b} - \frac{c}{1+c}}.$$ This is a consequence of the beautiful "Carlitz-Scoville-Vaughan" theorem; see this MathOverflow question for an explanation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $G$ a group. Let $x\in G$. Assume that for every $y\in G, xyx=y^3$. Prove that $x^2=e$ and $y^8=e$ for all $y\in G$. To put this in context, this is my first week abstract algebra.
Let $G$ a group. Let $x\in G$. Assume that for every $y\in G, xyx=y^3$.
Prove that $x^2=e$ and $y^8=e$ for all $y\in G$.
A hint would be appreciated.
| $\textbf{Full answer:}$
Let $x\in G$. Suppose $(\forall y\in G)(xyx=y^3)$.
(i) Set $y=e$ to get $xyx=xex=x^2=e=e^3=y^3$. (The OP already knew this part).
(ii) The hypothesis is $(\forall y\in G)(xyx=y^3)$. Let $y\in G$ be taken arbitrarily. By the hypothesis we have $xyx=y^3$. Cubing both sides we get $(xyx)(xyx)(xyx)=y^9$, i.e., $xy(x^2)y(x^2)yx=y^9$. Therefore $xyyyx=y^9$, that is, $xy^3x=y^9$. Recalling that $y^3=xyx$ it follows $y^9=xy^3x=x(xyx)x=x^2yx^2=eye=y$. From $y^9=y$ you can conclude that $y^8=e$ by multiplying by $y^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Appearance of Formal Derivative in Algebra When studying polynomials, I know it is useful to introduce the concept of a formal derivative. For example, over a field, a polynomial has no repeated roots iff it and its formal derivative are coprime. My question is, should we be surprised to see the formal derivative here?
Is there some way we can make sense of the appearance of the derivative (which to me is an analytic object) in algebra? I suspect it might have something to do with the fact that the derivative is linear and satisfies the product rule, which makes it a useful object to consider. It would also be interesting to hear an explanation which explains this in the context of algebraic geometry.
Thanks!
| Formal derivatives appear naturally when trying to rewrite a polynomial in $ x $ ( over a characteristic $ 0 $ field ) as a polynomial in $ (x-c) $ :
Let $ f(x) = a_n x^n + \ldots + a_1 x + a_0 \in K[x] $ ( with $ a_n \neq 0 $ ). Given any $ c \in K $, we can expand $ f(x) $ as a polynomial in $ (x-c) $ to get
$ \displaystyle f(x) = \sum_{j=0}^{n} a_j x^j = \sum_{j=0}^{n} a_j (x-c+c)^j $ $ \displaystyle = \sum_{j=0}^{n} a_j \bigg( \sum_{r = 0}^{j} \binom{j}{r} (x-c)^{r} c^{j-r} \bigg) $ $ \displaystyle = \sum_{0 \leq r \leq j \leq n} (x-c)^{r} a_j \binom{j}{r}c^{j-r} $ $ \displaystyle = \sum_{r=0}^{n} (x-c)^{r} \bigg( \sum_{j=r}^{n} a_j \binom{j}{r}c^{j-r} \bigg) $
$ \big[ $ Here "$ \binom{j}{r} $" should be understood as $ 1 + \ldots + 1 $ with $ \frac{j(j-1)..(j-(r-1))}{r!} $ many $ 1$s $ \big] $
Now assuming $ \text{char}(K) = 0 $ we can write $ \binom{j}{r} = (r!)^{-1} ( j(j-1)..(j-r+1) )$ [ where again each term "$ n $" should be understood as $ 1 + \ldots + 1 $ with $n$ many $1$s ], giving us
$ f(x) \displaystyle = \sum_{r=0}^{n} \frac{(x-c)^{r}}{r!} \bigg( \sum_{j=r}^{n} a_j \: j(j-1)\ldots(j-(r-1)) \: c^{j-r} \bigg) $.
But $ \displaystyle \sum_{j=r}^{n} a_j \: j(j-1)\ldots(j-(r-1)) \: c^{j-r} $ is just $ (D^{r} f)(c) $, where $ D : K[x] \rightarrow K[x] $ is a map sending $ c_0 + c_1 x + c_2 x^2 + \ldots + c_n x^n $ $ \mapsto c_1 + 2 c_2 x + \ldots + n c_n x^{n-1} $.
To summarise : Let $ f(x) = a_0 + \ldots + a_n x^n \in K[x] $, $ c \in K $, and $ \text{char}(K) = 0 $. Then $ f(x) = f(c) + (Df)(c) \: (x-c) $ $ + \frac{(D^2 f)(c)}{2!} \: (x-c)^2 + \ldots + \frac{(D^n f)(c)}{n!} \: (x-c)^n $, where $ D : K[x] \rightarrow K[x] $ sends $ c_0 + c_1 x + \ldots + c_n x^n $ $ \mapsto c_1 + 2 c_2 x + \ldots + n c_n x^{n-1} $ [ especially if $ \alpha \in K $ is a root of $ f $, it's multiplicity is the least $ j $ for which $ (D^j f)( \alpha ) \neq 0 $ ].
Edit: [This should work in general characteristic]
For every $f \in K[x]$ there is a unique $\psi _f \in K[x,y]$ such that $f(y) - f(x) = (y-x) \psi _f (x,y).$
Explicitly, taking $f(x) = \sum _{0} ^{n} a _i x ^i,$ we get $f(y) - f(x)$ ${= \sum _{0} ^{n} a _i (y ^i - x ^i)}$ ${= (y-x) \left( \sum _{1} ^{n} a _i (y ^{i-1} + y ^{i-2} x + \ldots + x ^{i-2} y + x ^{i-1}) \right).}$
Now $K[x] \longrightarrow K[x,y]$ sending $f(x) \mapsto \psi _f (x,y)$ is linear.
$f(y) - f(x) = (y-x) \psi _f,$ and $g(y) - g(x) = (y-x) \psi _g.$ Taking ${\alpha \cdot (\text{eq}1) + \beta \cdot (\text{eq}2)}$ gives $\psi _{\alpha f + \beta g} = \alpha \psi _f + \beta \psi _g.$
Also $\psi _{fg} = \psi _f g + f \psi _g + (y-x) \psi _f \psi _g$.
$(fg)(y)$ ${= (f(x) + (y-x) \psi _f)(g(x) + (y-x) \psi _g)}$ ${ = (fg)(x) + (y-x)(\psi _f g+ f \psi _g + (y-x) \psi _f \psi _g)}$
So the map $K[x] \overset{D}{\longrightarrow} K[x]$ sending $f(x) \mapsto \psi _f (x,x)$ is linear, with "Leibnitz rule" $D(fg) = (Df)g + f(Dg).$ From the expression for $\psi _f$ we get the explicit description $D\left( \sum a _i x ^i \right) = \left(\sum i a _{i} x ^{i-1} \right).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
"answer_count": 8,
"answer_id": 4
} |
A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$ The following question comes from Some integral with sine post
$$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$
but now I'd be curious to know how to deal with it by methods of complex analysis.
Some suggestions, hints? Thanks!!!
Sis.
| A complete asymptotic expansion for large $n$ may be derived as follows. We write
$$
\int_0^{ + \infty } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t} = \int_0^\pi {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t} + \int_\pi ^{ + \infty } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t} .
$$
Notice that
$$
\int_\pi ^{ + \infty } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t} = \frac{\pi }{{\pi ^n }}\int_1^{ + \infty } {\left( {\frac{{\sin (\pi s)}}{s}} \right)^n {\rm d}s} = \mathcal{O}(\pi ^{ - n} )
$$
as $n\to+\infty$. By Laplace's method,
$$
\int_0^{ \pi } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t} = \int_0^\pi {\exp \left( { - n\log (t\csc t)} \right){\rm d}t} \sim \sqrt {\frac{{3\pi }}{{2n}}} \sum\limits_{k = 0}^\infty {\frac{{a_k }}{{n^k }}}
$$
as $n\to+\infty$, where
$$
a_k = \left( {\frac{3}{2}} \right)^k \frac{1}{{k!}}\left[ {\frac{{{\rm d}^{2k} }}{{{\rm d}t^{2k} }}\left( {\frac{{t^2 }}{{6\log (t\csc t)}}} \right)^{k + 1/2} } \right]_{t = 0} .
$$
Accordingly,
\begin{align*}
\int_0^{ + \infty } {\left( {\frac{{\sin t}}{t}} \right)^n {\rm d}t} &\sim \sqrt {\frac{{3\pi }}{{2n}}} \sum\limits_{k = 0}^\infty {\frac{{a_k }}{{n^k }}} \\ & = \sqrt {\frac{{3\pi }}{{2n}}} \left( {1 - \frac{3}{{20n}} - \frac{{13}}{{1120n^2 }} + \frac{{27}}{{3200n^3 }} + \frac{{52791}}{{3942400n^4 }} + \ldots } \right)
\end{align*}
as $n\to+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 6,
"answer_id": 5
} |
How can I prove that a function $f(x,y)= \frac{x^2}{y}$ is convex for $y \gt 0$? How can I prove that a function $f(x,y)= \frac{x^2}{y}$ is convex for $ y \gt 0$?
I take the Hessian matrix of $\displaystyle \frac{x^2}{y}$, and I got:
$$H = \displaystyle\pmatrix{\frac{2}{y} & -\frac{2x}{y^2} \\-\frac{2x}{y^2} & \frac{2x^2}{y^3}}$$
further more, we have:
$$H = \frac{2}{y^3}\displaystyle\pmatrix{y^2 & -xy \\-xy & x^2}$$
I need to prove that H is semi-define positive matrix for y>0. I think I am close to the answer. But lack some knowledge to prove it since I try to calculate the $det(H)$ and it ends up equal to 0. Anyone can help? Thanks.
| You're basically done:
A symmetric matrix $(a_{ij})_{i,j\le n}$ is positive semidefinite iff the subdeterminants $\det(a_{ij})_{i,j\in H}$ are $\ge 0$ for all $H\subseteq\{1,2,..,n\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Convergence of a particular double series For the double series
$$ \sum_{m,n=1}^{\infty} \frac{1}{(m+n)^p} , $$
I was wondering when it converges. I want to use double integrals to estimate it, but I don't know how to write the process accurately... Could you show me a detailed computation? Thanks!
| While we wait for someone to do it using double integrals, here's another way:
Given a positive integer $k$, the number of pairs $m,n$ with $m\ge1$, $n\ge1$, and $m+n=k$ is $k-1$. So your sum is $\sum_{k=2}^{\infty}(k-1)/k^p$, and the usual single-series methods apply.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Really Stuck on Partial derivatives question Ok so im really stuck on a question. It goes:
Consider $$u(x,y) = xy \frac {x^2-y^2}{x^2+y^2} $$ for $(x,y)$ $ \neq $ $(0,0)$ and $u(0,0) = 0$.
calculate $\frac{\partial u} {\partial x} (x,y)$ and $\frac{\partial u} {\partial y} (x,y)$ for all $ (x,y) \in \Bbb R^2. $ show that $ \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0) $. Check, using polar coordinates, that $ \frac {\partial u}{\partial x} and \frac {\partial u}{\partial y} $ are continuous at $(0,0)$
Any help really appreciated.
Cheers
| We are given:
$$u(x, y)=\begin{cases}
xy \frac {x^2-y^2}{x^2+y^2}, ~(x, y) \ne (0,0)\\\\
~~~0, ~~~~~~~~~~~~~~~(x, y) = (0,0)\;.
\end{cases}$$
I am going to multiply out the numerator for ease in calculations, so we have:
$$\tag 1 u(x, y)=\begin{cases}
\frac {x^3y - xy^3}{x^2+y^2}, ~(x, y) \ne (0,0)\\\\
~~~0, ~~~~~~~~~~~~~~(x, y) = (0,0)\;.
\end{cases}$$
We are asked to:
*
*(a) Find: $\displaystyle \frac{\partial u} {\partial x} (x,y) ~\forall x~ \in \Bbb R^2$
*(b) Find: $\displaystyle \frac{\partial u} {\partial y} (x,y) ~\forall x~ \in \Bbb R^2$
*(c) Show that $ \displaystyle \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$
*(d) Check, using polar coordinates, that $\displaystyle \frac {\partial u}{\partial x} \text{and} \frac {\partial u}{\partial y} $ are continuous at $(0,0)$
Using $(1)$, for part $(a)$, we get:
$\tag 2 \displaystyle \frac{\partial u} {\partial x} (x,y) = \frac{(3x^2y- y^3)(x^2+y^2) - 2x(x^3y - xy^3)}{(x^2 + y^2)^2} = \frac{x^4y + 4x^2y^3-y^5}{(x^2+y^2)^2}$
Using $(1)$, for part $(b)$, we get:
$\tag 3 \displaystyle \frac{\partial u} {\partial y} (x,y) = \frac{(x^3-3xy^2)(x^2+y^2) - 2y(x^3y - xy^3)}{(x^2+y^2)^2} = \frac{x^5 - 4x^3y^2-xy^4}{(x^2+y^2)^2}$
Next, we need mixed partials, so using $(2)$, we have:
$ \tag 4 \displaystyle \frac {\partial^2 u} {\partial x \partial y} (x, y) = \frac{(x^4+ 12x^2y^2-5y^4)(x^2+y^2)^2 - 2(x^2+y^2)(2y)(x^4y + 4x^2y^3-y^5)}{(x^2+y^2)^4} = \frac{x^6 + 9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3} = \frac {\partial^2 u} {\partial y \partial x} (x, y)$
Thus, we get:
$$\tag 5 \displaystyle \frac {\partial^2 u} {\partial x \partial y} (x, y) = \frac {\partial^2 u} {\partial y \partial x} (x, y) = \begin{cases}
\frac{x^6 + 9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}, ~(x, y) \ne (0,0)\\\\
~~~~~~~~~~~~~~~~~~0, ~~~~~~~~~~~~~~~~~~~~~~~~(x, y) = (0,0)\;.
\end{cases}$$
Now, for part $(c)$, we want to show that $ \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$, so we need to find the limits of each mixed partial.
We have:
$ \tag 6 \displaystyle \frac{\partial^2 u} {\partial x \partial y} (0,0) = \lim\limits_{h \to 0} \frac{\frac{\partial u}{\partial x} (0, h) - \frac{\partial u}{\partial x} (0, 0)}{h} = \lim\limits_{h \to 0} \frac{-h^5/h^4}{h} = \lim\limits_{h \to 0} \frac{-h}{h} = -1$, and
$ \tag 7 \displaystyle \frac{\partial^2 u} {\partial y \partial x} (0,0) = \lim\limits_{h \to 0} \frac{\frac{\partial u}{\partial y} (h, 0) - \frac{\partial u}{\partial y} (0, 0)}{h} = \lim\limits_{h \to 0} \frac{h^5/h^4}{h} = \lim\limits_{h \to 0} \frac{h}{h} = +1$
$\therefore$, for part $(c)$, we have shown that:
$$\frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$$
as desired.
Can you handle part $(d)$?
Regards
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence of a sum of rational functions I have the complex functions $f_n(z) = 1/(1+z^n)$ and I'm supposed determine where $\sum_{n=1}^\infty f_n(z)$ converges for $z \in\mathbb{C}$
Extra Info:
I was only able to determine convergence for $|z| \gt 1$. The argument applies the ratio test:
$|\frac{f_{n+1}}{f_n}| = \frac{|1 + z^n|}{|1+z^{n+1}|}$
Now note that the denominator $|1 + z^{n+1}| = |1 -(-z^{n+1})| \ge |1 - |-z^{n+1}|| = |1 - |z|^{n+1}| \ge \frac{1}{K}|z|^{n+1}$
for large $n$ and $K \gt 1$
So then the ratio can be bounded
$\frac{|1 + z^n|}{|1+z^{n+1}|} \le K\frac{1 + |z|^n}{|z|^{n+1}} = \frac{K}{|z|^{n+1}} + \frac{K}{|z|}$ which limits to $\frac{K}{|z|}$ so convergence is guaranteed by ratio test when $|z| \gt K$ and $K$ was arbitrarily greater than $1$
| For any point $z \in \bar{\mathbb D}$ you can apply the divergence test to your summation to conclude that it does not converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A question about normal subgroup $H<G,aHa^{-1}<G$. Then, $H$ is isomorphic to $aHa^{-1}$.
I want to show $aha^{-1}\in H,\forall h\in H,a\in G$, but I cannot figure it out. Any hint is appreciated.
| To show that $H\cong gHg^{-1}$, you need to show that for any $a\in G$, the conjugation $$\kappa_a : G\to G,\quad g\mapsto aga^{-1}$$ is an automorphism of $G$. The proof is easy: For $g,h\in G$ we have $$\kappa_a(gh) = agha^{-1} = ag(a^{-1}a)ha^{-1} = (aga^{-1})(aha^{-1}) = \kappa_a(g)\kappa_a(h)\text{,}$$
so $\kappa_a$ is a homomorpisms. To show that $\kappa_a$ is one-to-one, check $\kappa_a\circ\kappa_{a^{-1}} = \operatorname{id}$.
Now every subgroup $H$ is isomorphic to $\kappa_a(H) = aHa^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Multidimensional Hensel lifting I have a question about a practical application of (some) generalised form of Hensel's Lemma. I cannot find it stated in an appropriate form in Bourbaki or anywhere else, so here goes ...
Let $p$ be an odd prime: we work over the $p$-adic numbers $Q_p$ with ring of integers $Z_p$ and residue field $F_p$. I have a bunch of quadratic and quartic polynomial equations in N variables, with coefficients in $F_p$, and I have a (non-unique) solution set in $F_p^{N}$.
To simplify things assume that the number of variables N exceeds the total number of equations.
Is it possible to conclude from the non-vanishing of some sort of generalised Jacobian determinant, that my solution set lifts to $Z_p$?
| The same proof method works in higher dimension.
Let $F: \mathbb{Z}_p^m \to \mathbb{Z}_p^n$ be our system of polynomial functions.
If $x \in \mathbb{Z}_p^m$ satisfies
$$ F(x) \equiv 0 \pmod {p^e} $$
then
$$ F(x) \equiv p^e y \pmod {p^{e+1}} $$
for some vector $y$. Furthermore, the Taylor expansion about $x$ tells us
$$ F(x + p^e z) \equiv p^e y + p^e dF(x) \cdot z \pmod{p^{e+1}} $$
so as long as
$$ dF(x) \cdot z \equiv -y \pmod p$$
always has a solution for the vector $z$ (e.g. more variables than equations and $dF(x)$ has full rank), then lifts exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Roots of the equation $I_1(b x) - x I_0(b x) = 0$ I'm interested in the roots of the equation:
$I_1(bx) - x I_0(bx) = 0$
Where $I_n(x)$ is the modified Bessel function of the first kind and $b$ is real positive constant.
More specifically, I'm interested in the behaviour of the largest non-negative root for $x \geq 0$. I have solved the problem numerically in a crude fashion to see what the function looks like, but as far as I can tell there isn't a closed form.
If the value of the largest root is given by $x(b)$, then based on the numerical solution I think the following is true:
$x(b) < 0 \; $ for $\; b < 2$
$\lim_{b\to\infty} x(b) = 1$
Can anyone give me some pointers to the possible approaches I might take to approximating $x(b)$ for $b>2$ ?
Thanks!
| Notice that the equation also can be rewritten as
$$
\frac{I_1(b x)}{I_0(b x)} = x
$$
Using the asymptotic series expansion for large $b$ and some fixed $x>0$ we get:
$$
1 - \frac{1}{2 x b} - \frac{3}{8 b^2 x^2} - \frac{1}{8 b^3 x^3} + \mathcal{o}\left(b^{-1}\right) = x
$$
resulting in
$$
x(b) = 1 - \frac{1}{2 b} - \frac{3}{8 b^2} - \frac{9}{16 b^3} + \mathcal{o}\left(b^{-3}\right)
$$
Confirming $\lim_{b \to + \infty} x(b) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving {$b_n$}$_{n=1}^\infty$ converges given {$a_n$}$_{n=1}^\infty$ and {$a_n b_n$}$_{n=1}^\infty$ Suppose {$a_n$}$_{n=1}^\infty$ and {$b_n$}$_{n=1}^\infty$ are sequences such that {$a_n$}$_{n=1}^\infty$ coverges to A$\neq$0 and {$a_n b_n$}$_{n=1}^\infty$ converges. Prove that {$b_n$}$_{n=1}^\infty$ converges.
What I have so far:
$b_n = {a_n b_n \over a_n}$ $\to$ $C \over A$, $A\neq0$
|$b_n - {C \over A}$| = |${a_n b_n \over a_n} - {C \over A}$| = |${Aa_nb_n - Ca_n \over Aa_n}$| $ \leq $ |${1 \over Aa_n}||Aa_nb_n - Ca_n$|=|${1 \over Aa_n}||a_n(Ab_n - C)$|
$\leq |{1 \over Aa_n}||a_n||(Ab_n - C)$|. Note: since $a_n$ converges, there is M>0 such that |$a_n| \leq$M for all n $ \in\Bbb N$.
Thus, |${1 \over Aa_n}||a_n||(Ab_n - C)$| = |${1 \over M}||M||(Ab_n - C)$|. And this is where I get lost. Any thoughts? Or am I completely wrong to begin with?
| Hint: Let $A=\lim\limits_{n\to\infty}a_n$ and $B=\frac{\lim\limits_{n\to\infty}a_nb_n}{A}$.
Since $|A|>0$, for $n$ large enough, $|a_n-A|\le\frac{|A|}{2}$. Show that then, $|a_n|\ge\frac{|A|}{2}$.
Then note that
$$
a_n(b_n-B)=(a_nb_n-AB)-B(a_n-A)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Linear Transformations and Identity Matrix Suppose that $V$ is a finite dimensional vector space and $T$ is a linear transformation from $V$ to $V$. Prove that $T$ is a scalar multiple of the identity matrix iff $ST=TS$ for every linear transformation $S$ from $V$ to $V$.
| One direction is trivial. For the other, take $\{v_1,...,v_n\}$ to be any basis for $V$, and consider the transformations $E_{i,j}:V\to V$ induced by $$v_k\mapsto\begin{cases} v_j & \text{if }k=i,\\0 & \text{otherwise.}\end{cases}$$ What can we conclude about $T$ from the fact that $E_{i,j}T=TE_{i,j}$ for all $i,j\in\{1,...,n\}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to rewrite $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta,\cos4\theta$? I need help with writing $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta, \cos4\theta$.
My attempts so far has been unsuccessful and I constantly get developments that are way to cumbersome and not elegant at all. What is the best way to approach this problem?
I know that the answer should be:
$\sin^4 \theta =\frac{3}{8}-\frac{1}{2}\cos2\theta+\frac{1}{8}\cos4\theta$
Please explain how to do this.
Thank you!
| Write
$$\sin^4{\theta} = \left ( \frac{e^{i \theta} - e^{-i \theta}}{2 i} \right )^4$$
and use the binomial theorem.
$$\begin{align}\left ( \frac{e^{i \theta} - e^{-i \theta}}{2 i} \right )^4 &= \frac{1}{16} (e^{i 4 \theta} - 4 e^{i 2 \theta} + 6 - 4 e^{-i 2 \theta} + e^{-i 4 \theta}) \\ &= \frac{1}{8} (\cos{4 \theta} - 4 \cos{2 \theta} + 3)\end{align}$$
Item of curiosity: the Chebyshev polynomials are defined such that
$$T_n(\cos{\theta}) = \cos{n \theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $P(A \cup B \cup C) = 1$, $P(B) = 2P(A) $, $P(C) = 3P(A) $, $P(A \cap B) = P(A \cap C) = P(B \cap C) $, then $P(A) \le \frac14$ We have ($P$ is probability):
$P(A \cup B \cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A \cap B) = P(A \cap C) = P(B \cap C) $. Prove that $P(A) \le \frac{1}{4} $.
Well, I tried with the fact that $ 1 = P(A \cup B \cup C) = 6P(A) - 3P(A \cap B) + P(A \cap B \cap C) $ but I got stuck... Could anyone help me, please?
| This solution is quite possibly messier than necessary, but it’s what first occurred to me. For convenience let $x=P\big((A\cap B)\setminus C\big)$ and $y=P(A\cap B\cap C)$. Let $a=P(A)-2x-y$, $b=P(B)-2x-y$, and $c=P(C)-2x-y$; these are the probabilities of $A\setminus(B\cup C)$, $B\setminus(A\cup C)$, and $C\setminus(A\cup B)$, respectively. (A Venn diagram is helpful here.)
Now $$b+2x+y=2(a+2x+y)=2a+4x+2y\;,$$ so $b=2a+2x+y$. Similarly, $$c+2x+y=3(a+2x+y)=3a+6x+3y\;,$$ so $c=3a+4x+2y$. Then
$$\begin{align*}
1&=P(A\cup B\cup C)\\
&=a+b+c+3x+y\\
&=6a+9x+4y\\
&=4(a+2x+y)+2a+x\\
&=4P(A)+2a+x\;.
\end{align*}$$
Since $2a+x\ge 0$, we must have $P(A)\le \frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find and example of two elements $a,b$ in a finite group $G$ such that $|a| = |b| = 2, a \ne b$ and $|ab|$ is odd. Find and example of two elements $a,b$ in a finite group $G$ such that $|a| = |b| = 2, a \ne b$ and $|ab|$ is odd. Any ideas as to how I would go about finding it?
Thanks
| There is a very general example you should know about, that of dihedral groups. A dihedral group has order $2n$, for any $n \ge 2$, and it is generated by two elements of order $2$, whose product has order $n$.
Probably the simplest way to see these groups is as a group of bijective maps on $\mathbf{Z}_{n}$,
$$
a : x \mapsto -x, \qquad b: x \mapsto -x-1.
$$
(Coefficients are written as integers, but meant to be in $\mathbf{Z}_{n}$.) These are clearly elements of order $2$, while because of
$$
a \circ b(x) = (a(b(x)) = -(-x - 1) = x + 1
$$
$ab = a \circ b$ has order $n$.
(Geometrically, such a group is the group of congruences of a regular $n$-gon.)
If you do the same thing over $\mathbf{Z}$, you find that $ab$ has infinite order.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is this the way to estimate the amount of lucky twins? To estimate the amount of prime twins between $3$ and $x$ we just take $x \prod_{p}(1-2/p)$ where $p$ runs over the primes between $3$ and $\sqrt x$. Lucky numbers are similar to prime numbers. Does this imply that a good way to estimate the amount of lucky twins between $3$ and $x$ is $x \prod_{l}(1-2/l)$ where $l$ runs over the lucky numbers between $3$ and $\sqrt x$?
*EDIT:*I Think I can improve the question by stating it as follows.
$1$) If $x$ goes to infinity and $l(x)$ denotes the amount of lucky numbers between $3$ and $x$ does
$\dfrac{x \prod_{l}(1-2/l)}{l(x)}=Constant$ ?
$2$) If $x$ goes to infinity, does
$\dfrac{\prod_{p}(1-2/p)}{\prod_{l}(1-2/l)}=Constant$ ?
If no theoretical answer is possible are $1$) and $2$) supported numerically ?
| I think that to answer this question one would need to start from the paper by Bui and Keating on the random sieve (of which the lucky sequence is a particular realization) and then carefully examine the proofs that they cite of Mertens' theorems for that sieve. The issue is whether there is a correction factor like Mertens' constant or the twin prime constant in the product that you wrote down for the number of lucky twins, and if there is, is it calculated (up to $1+o(1)$ factors) by the same formula as for the primes.
http://arxiv.org/abs/math/0607196
The paper gives an asymptotic formula for random twin primes that does not have any twin prime constant, but the explanation there is not exactly about the same situation. It looks to me like some analysis of the proofs of the analogues of Mertens' theorems is needed to understand what product over random primes is the correct estimate of the number of random twin primes.
I cannot tell the answer to the question from Bui and Keating paper by itself, but would guess that up to multiplication by a constant, the product you wrote down is, for nearly all realizations of the random sieve (ie., with probability 1) asymptotic to the number of random-sieve twins, and this is the heuristic guess for what happens with the lucky numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Isomorphism with Lie algebra $\mathfrak{sl}(2)$ Let $L$ be a Lie algebra on $\mathbb{R}$. We consider $L_{\mathbb{C}}:= L \otimes_{\mathbb{R}} \mathbb{C}$ with bracket operation
$$ [x \otimes z, y \otimes w] = [x,y] \otimes zw $$
far all $x,y \in L$ and $z,w \in \mathbb{C}$. We have that $L_{\mathbb{C}}$ is a Lie algebra.
If $L= \mathbb{R}^{3}$ and for $x,y \in L$ we define $[x,y]:= x \wedge y$ (where $\wedge$ denotes the usual vectorial product). We have that $(L, \wedge)$ is a Lie algebra. I have to prove that $L \simeq \mathfrak{sl}(2)$. In order to do this I'd like to prove that $L \simeq \mathfrak{so}(3,\mathbb{R})$. Than, because $\mathfrak{so}(3,\mathbb{R}) \otimes \mathbb{C} \simeq \mathfrak{sl}(2)$ and $\mathfrak{sl}(2)$, up to isomorphism, is the unique $3$-dimetional semisimple algebra, I complete my proof. So my questions are: 1) How to prove that $(\mathbb{R}^{3}, \wedge) \simeq \mathfrak{so}(3, \mathbb{R})$ ? 2) Why $\mathfrak{so}(3,\mathbb{R}) \otimes \mathbb{C} \simeq \mathfrak{sl}(2)$ ?
| For (1):
Let
$$i:=(1,0,0),\quad j:=(0,1,0)\text{ and } k:=(0,0,1)\in\mathbb R^3$$
and
$$
A:=\left(\begin{array}{ccc}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right),\quad
B:=\left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{array}\right)
\text{ and }
C:=\left(\begin{array}{ccc}0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0\end{array}\right) \in \mathfrak{so}(3,\mathbb R).
$$
So,
$$i\wedge j = k,\quad i\wedge k = -j\text{ and }j\wedge k = i \in \mathbb R^3$$
and
$$[A,B] = C,\quad [A,C] = -B\text{ and }[B,C]=A\in \mathfrak{so}(3,\mathbb R).$$
Then the linear application $\mathbb R^3\to\mathfrak{so}(3,\mathbb R)$ given by
$$\begin{array}{rcl}
i & \to & A \\
j & \to & B \\
k & \to & C
\end{array}$$
is an isomorphism of Lie algebras.
For (2):
Let
$$
X:=\left(\begin{array}{cc}0 & i \\ i & 0\end{array}\right),\quad
Y:=\left(\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right)\text{ and }
Z:=\left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right)\in\mathfrak{sl}(2,\mathbb C)
$$
and
$$
X'=A\otimes 2,\quad
Y'=B\otimes 2\text{ and }
Z'=C\otimes 2\in\mathfrak{so}(3,\mathbb R)\otimes\mathbb C,
$$
for $A$, $B$ and $C\in\mathfrak{so}(3,\mathbb R)$ as in item (1). So,
$$[X,Y] = 2Z,\quad [X,Z] = -2Y\text{ and }[Y,Z]=2X\in \mathfrak{sl}(2,\mathbb C)$$
and
$$[X',Y'] = 2Z',\quad [X',Z'] = -2Y'\text{ and }[Y',Z']=2X'\in \mathfrak{so}(3,\mathbb R)\otimes\mathbb C.$$
Then, the linear application $\mathfrak{sl}(2,\mathbb C)\to\mathfrak{so}(3,\mathbb R)\otimes\mathbb C$ given by
$$\begin{array}{rcl}
X & \to & X' \\
Y & \to & Y' \\
Z & \to & Z'
\end{array}$$
is an isomorphism of Lie algebras.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
quotient of finitely presented module $\DeclareMathOperator\Coker{Coker}$Assume the exact sequence
$$A \xrightarrow{f} B \xrightarrow{} \Coker f \xrightarrow{} \{0\} $$ where $A$ is a finitely generated module and $B$ is a finitely presented module.
Is it true that $\Coker f$ is finitely presented module ?
In general, is the quotient of a finitely presented module to a finitely generated submodule, finitely presented ?
| This is true.
I'll denote the ring by $R$, while remaining silent about whether these are left or right modules.
Write $B = R^n/I$ for some finitely generated submodule $I \subseteq R^n$. Then the image of $f \colon A \to B$ is of the form $J/I$ for some submodule $J \subseteq R^n$, and $J/I$ is a finitely generated module. Then $J$ is finitely generated as well: taking finitely many elements of $J$ whose images generate $J/I$, along with a finite generating set for $I$, will give a finite generating set for $J$.
Thus $Coker(f) \cong R^n/J$ is finitely presented.
If you're looking for a reference, see Exercise 4.8 in T.Y. Lam's book Exercises in Modules and Rings.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
limit of the error in approximating definite integral with midpoint rule I want to calculate $\lim_{n \rightarrow \infty} n^2 |\int_{[0,1]}f(x)-I_n(x)|$ where $I_n$ is the integral approximation by midpoint rule:
$I_n=\frac{1}{n}\sum_{k=1}^nf(c_k)$ and $c_k$ is the point in the middle of $k^{th}$ interval.
My attempt:
I know the error is bounded by $\frac{max_{[0,1]} f''(x)}{24n^2}$.
Also by Lagrange theorem, we know that the remainder at point $x$ is $\frac{f^{(n+1)}(\xi)(x-x_\circ)^{n+1}}{(n+1)!}$.
Now if I want to use the Lagrange formula I'll have to integrate over the whole $[0,1]$ interval. This is what confuses me, because if $n$ is even the integral will become zero and if it odd nonzero.
Even if I can integrate, $\lim_{n \rightarrow \infty} n^2 \frac{f^{(n+1)}(\xi)(x-x_\circ)^{n+1}}{(n+1)!}$ is $\infty \times 0$ and I don't know how to deal with it.
I appreciate any help.
| Assume $f''$ is continuous. The error in a single Midpoint Rule interval of length $h$ is
$$ \int_a^{a+h} f(x)\ dx - h f(a+h/2) = -\frac{h^3}{24} f''(\xi)$$
for some $\xi \in [a,a+h]$. The error for $n$ equal subintervals of $[0,1]$ is the sum of the errors in each: if $h = 1/n$ and $x_j = j h$ we get
$$\int_0^1 f(x)\ dx - I_n = \sum_{j=0}^{n-1} \left(\int_{x_j}^{x_{j+1}} f(x)\ dx - \frac{1}{n} f(x_j + h/2)\right) = - \frac{1}{24n^3} \sum_{j=0}^{n-1} f''(\xi_j) $$
where $\xi_j \in [x_j, x_{j+1}]$.
Now notice that $\displaystyle \frac{1}{n} \sum_{j=0}^{n-1} f''(\xi_j)$ is a Riemann sum
for $\int_0^1 f''(x)\ dx = f'(1) - f'(0)$, so ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Applications of the Isomorphism theorems In my study of groups, rings, modules etc, I've seen the three isomorphism theorems stated and proved many times. I use the first one ( $G/\ker \phi \cong \operatorname{im} \phi$ ) very often, but I can't recall having ever used the other two. Can anyone give some examples where they are used in a crucial way in some proof?
For clarity, let us say that the 2nd one is : $(M/L)/(N/L) \cong M/N$ under the appropriate conditions, and the 3rd one is $(M+N)/N \cong M/(M\cap N).$
| E.g., in studying solvable groups.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 7,
"answer_id": 6
} |
Cantor set + Cantor set =$[0,2]$ I am trying to prove that
$C+C =[0,2]$ ,where $C$ is the Cantor set.
My attempt:
If $x\in C,$ then $x= \sum_{n=1}^{\infty}\frac{a_n}{3^n}$ where $a_n=0,2$
so any element of $C+C $ is of the form $$\sum_{n=1}^{\infty}\frac{a_n}{3^n} +\sum_{n=1}^{\infty}\frac{b_n}{3^n}= \sum_{n=1}^{\infty}\frac{a_n+b_n}{3^n}=2\sum_{n=1}^{\infty}\frac{(a_n+b_n)/2}{3^n}=2\sum_{n=1}^{\infty}\frac{x_n}{3^n}$$
where $x_n=0,1,2, \ \forall n\geq 1$.
Is this correct?
| Short answer for this question is "your argument is correct ". To justify the answer consider some particular $n_{0}\in \mathbb{N}$. Since $$\sum_{n=1}^{\infty}\dfrac{a_{n}}{3^{n}},\sum_{n=1}^{\infty}\dfrac{b_{n}}{3^{n}}\in C$$ we have that $ x_{n_{0}}=\dfrac{a_{n_{0}}+b_{n_{0}}}{2}\in \{0,1,2\} $. ($ x_{n_{0}}=0 $ if $ a_{n_{0}}=b_{n_{0}}=0 $. $ x_{n_{0}}=2 $ if $ a_{n_{0}}=b_{n_{0}}=2 $. Otherwise $ x_{n_{0}}=1 $. )
Then clearly $$\sum_{n=1}^{\infty}\dfrac{x_{n}}{3^{n}}\in [0,1].$$ So $$2\sum_{n=1}^{\infty}\dfrac{x_{n}}{3^{n}}=\sum_{n=1}^{\infty}\dfrac{a_{n}}{3^{n}}+\sum_{n=1}^{\infty}\dfrac{b_{n}}{3^{n}}\in [0,2].$$ Hence $ C+C\subseteq [0,2] $. To complete the proof you must show that the other direction as well. To show $ [0,2]\subseteq C+C $ it is enough to show $ [0,1]\subseteq \dfrac{1}{2}C+\dfrac{1}{2}C $.
Observe that $ b\in \dfrac{1}{2}C $ if and only if there exists $ t\in C $ such that $ b=\dfrac{1}{2}t $.
Hence $$ b\in \dfrac{1}{2}C\text{ if and only if }b=\sum\limits_{n = 1}^\infty \frac{b_n}{3^n}\text{ ; where }b_{n}=0\text{ or }1. $$ Now let $ x\in [0,1] $. Then $$ x= \sum\limits_{n = 1}^\infty \frac{x_n}{3^n}\text{ ; where }x_{n}=0,1\text{ or }2. $$
Here we need to find $ y,z\in \dfrac{1}{2}C $ such that $ x=y+z $. Let's define $ y=\sum\limits_{n = 1}^\infty \frac{y_n}{3^n} $ and $ z=\sum\limits_{n = 1}^\infty \frac{z_n}{3^n} $ as follows.
For each $n\in \mathbb{N}$, $ y_{n}=0 $ if $ x_{n}=0 $ and $ y_{n}=1 $ if $ x_{n}=1,2 $.
For each $n\in \mathbb{N}$, $ z_{n}=0 $ if $ x_{n}=0,1 $ and $ z_{n}=1 $ if $ x_{n}=2 $.
Thus $y,z\in \dfrac{1}{2}C $ and for each $n\in \mathbb{N}$, $ y_{n}+z_{n}=0 $ if $ x_{n}=0 $ , $ y_{n}+z_{n}=1 $ if $ x_{n}=1 $ and $ y_{n}+z_{n}=2 $ if $ x_{n}=2 $.
Therefore $x=y+z\in \dfrac{1}{2}C+\dfrac{1}{2}C$ and hence $[0,1] \subseteq \dfrac{1}{2}C+\dfrac{1}{2}C$. $\square $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 6,
"answer_id": 3
} |
Riesz representation theorem on dual space The Riesz representation theorem on Hilbert spaces is well known, It asserts we can represent a bounded linear function on a Hilbert space $H$ with an inner product on $H$ and vice-versa.
My question: Given an inner product in $H^*$, say $(a,b)_{H^*}$, can I write it as $$(a,b)_{H^*} = \langle a, f \rangle_{H^*, H}$$ where $f \in H$? This is the RRT applied to the Hilbert space $H^*$ with its dual $H$. I think it works but I never saw it so I should get it clarified.
| Well, for an arbitrary inner product on $X:=H^*$ it is not going to work, since then $X^*$ need not be isomorphic to $H$.
On the other hand, the Riesz representation gives a linear isomorphism $H\to H^*$, and if the inner product is defined via this isomorphism, i.e. if
$$(\langle x,-\rangle,\langle y,-\rangle)_{H^*}=\langle x,y\rangle_{H} $$
for all $x,y\in H, \ f\in H^*$, then your claim is valid:
Let $a,b\in H^*$ then $a=\langle x,-\rangle$ and $b=\langle y,-\rangle$ for some $y\in H$ by Riesz representation, and $a(y)=\langle x,y\rangle$ so we have
$$(a,b)_{H^*} = (a,\langle y,-\rangle)=a(y)=\langle a,y\rangle_{H^*,H}\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Expectation of $X$ given a Cumulative Function I've (hard) to find the expectation of $X$:
CDF (Cumulative Function) = $1 - x^{-a}$, $1 \leqslant x < \infty$
$E[X]$ = ??
| First you have to establish the proper domain for the random variable $X$. That can be done by using $F(x_{\min}) = 0$. You random variable will be supported on $[x_{\min}, \infty)$.
*
*Find the probability density function, $f_X(x)$ by differentiating the cumulative function, $f_X(x) = F_X^\prime(x)$.
*Apply the law of the unconscious statistician: $$ \mathbb{E}(X) = \int_{x_\min}^\infty x f_X(x) \mathrm{d}x $$
*For what values of parameter $a$ does the integral exist?
*Once you found the answer, you have found the mean of the Pareto distribution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$T:P_3 → P_3$ be the linear transformation such that... Let $T:P_3 \to P_3$ be the linear transformation such that $T(2 x^2)= -2 x^2 - 4 x$, $T(-0.5 x - 5)= 2 x^2 + 4 x + 3$, and $T(2 x^2 - 1) = 4 x - 4.$
Find $T(1)$, $T(x)$, $T(x^2)$, and $T(a x^2 + b x + c)$, where $a$, $b$, and $c$ are arbitrary real numbers.
| hint:T is linear transformation $T(a+bx+cx^2)=aT(1)+bT(x) +cT(x^2) $ $$T(2 x^2)= -2 x^2 - 4 x \to T( x^2)= - x^2 - 2 x$$$$T(-0.5 x - 5)= 2 x^2 + 4 x + 3\to-0.5T( x) - 5T(1)= 2 x^2 + 4 x + 3\to 16x^2+72x-46$$$$T(2 x^2 - 1) = 4 x - 4.\to T( x^2) - \frac12T(1) = 2 x - 2\to T(1)=-2 x^2 - 8x +4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Truth set of $-|x| \lt 2$? An exercise in my Algebra I book (Pearson and Allen, 1970, p. 261) asks for the graph of the truth set for $-\left|x\right| \lt 2; x \in \mathbb{R}$.
I've re-stated the inequality in the equivalent form of $\left|x\right| \gt -2$. I know that the truth set of $\left|x\right| = -2$ is $\emptyset$, but I'm not certain how to handle the inequality in conjunction with the absolute value.
I suspect the truth set is $\{x \mid x \ge 0\}$, but I am not certain whether this is correct, or how to prove it using the algebraic concepts I've learned thus far. (I suspect this is a flaw with the book as this is not the first time it assumes knowledge that hasn't yet been presented.)
Is the truth set I arrived at correct? Is there a simple proof of the solution using Algebra I concepts (i.e. the field axioms, basic order properties, etc.)?
Bibliography
Pearson, H. R and Allen, F. B., 1970. Modern Algebra - A Logical Approach (Book I). Boston: Ginn and Company.
| $$-|x|<2\stackrel{\text{multiplication by}\,\,(-1)}\Longleftrightarrow |x|>-2$$
So: for what values of (real) $\,x\,$ it is true that $\,|x|>-2\,$ ? Hint: this is a rather huge subset of the reals...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Can any mathematical relation be called an 'operator'? Mathematics authors agree that $+,-,/,\times$ are basic operators. There are also logical operators like $\text{or, and, xor}$ and the unary negation operator $\neg$.
Where there seems to be a disagreement, however, is whether certain symbols used in the composition of propositions qualify as operators. By this definition:
an operation is an action or procedure which produces a new value from
one or more input values, called "operands"
symbols that denote a relation should also be operators.
For instance, $a<b$ takes integral operands and returns a value in the Boolean domain. Even something like $x \in S$ should be an operator that takes an element and a set, and also returns a truth value. The same could be said for the equality operator $=$.
Computer scientists call these 'relational operators', but mathematicians rarely do this. Is there a reason for this discrimination?
| A serious disadvantage of treating logical connectives as operators returning a value (that is, as functions) is that in non-classical logics their "range" may fail to correspond to any two-element set $\{\top,\bot\}$, and according to the Wikipedia page Truth value, may even fail to correspond to any set whatsoever: "[n]ot all logical systems are truth-valuational in the sense that logical connectives may be interpreted as truth functions. For example, intuitionistic logic lacks a complete set of truth values...." (Note that in particular, the naive attempt to consider logical connectives as mapping into a three-element set $\{\top,\bot,\text{some third value}\}$ does not work.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that any finite set of words is regular. How many states is sufficient for a single word $w_1...w_m$? DFA
Prove that any finite set of words is regular. How many states is sufficient for a single word $w_1...w_m$?
For part 2, wouldn't it require M states if the word length is M?
| A language over the alphabet $\Sigma$ will be a regular language given that it follows the following clauses:
*
*$\epsilon$, {$a$} for $a\in\Sigma$.
*If $L_1$ and $L_2$ are regular languages, then $L_1\cup L_2$, $L_1L_2$, and $L^*$ are also regular.
*$L$ is not a regular language unless taken from those two clauses.
A finite set of words taken from some alphabet will clearly be a regular language. As for the second part, you need $m+1$ states (hint: there is a tiny element in the first clause that should give you the reason to this).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How is the set of all programs countable? I'm having a hard time seeing how the number of programs is not uncountable, since for every real number, you can create a program that's prints out that number. Doesn't that immediately establish uncountably many programs?
| All the answers to fit the question number of programs are countable use the discrete finite definition of program , using either finite memory, finite ( countable ) instruction etc.
How ever in the old analog days where voltage was considered as output it was a trivial task to construct a circuit that prdouced all the possible voltage between 0 and 1. Now of course some physics savvy people would point out that voltage is discrete therefor you really dont end up producing all the real numbers as a voltage output between 0 and 1. But that is a physical constraint.
So yes a classical program with all of it's finite/countable restriction on memory, instructions etc. can be shown to end up as a point in countable set.
But an analog machine, like the ones constructed by pullies and ropes by Mayan's can indeed produce all the real numbers between 0 and 1, rest of the real line could have been achieved by a multiplication factor ( again some type of pully and rope computation).
So the statement that set of all the programs is countable depends on what is the computation model that it is being set in, otherwise it neither true or false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 12,
"answer_id": 7
} |
Examples of categories where morphisms are not functions Can someone give examples of categories where objects are some sort of structure based on sets and morphisms are not functions?
| Let $G$ be a directed graph. Then we can think of $G$ as a category whose objects are the vertices in $G$. Given vertices $a, b \in G$ the morphisms from $a$ to $b$ are the set of paths in the graph $G$ from $a$ to $b$ with composition being concatenation of paths. Note that we allow "trivial" paths that start at a given vertex $a$, traverse no edges, and end at the starting vertex $a$. We have to do this for the category to have identities.
Also we can create categories whose morphisms are not exactly maps but instead equivalence classes of maps. For example given an appropriately nice ring $R$ we can create the stable module category whose objects are $R$-modules. Given two $R$-modules $A$ and $B$ the set of morphisms from $A$ to $B$ is the abelian group of $R$-module homomorphisms from $A$ to $B$ modulo the subgroup of homomorphisms that factor through a projective $R$-module.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 9,
"answer_id": 8
} |
Finding the standard deviation of a set of angles. My question is given a set of angles in the form of degrees minutes and seconds once finding the mean how do you find the standard deviation.
I know how to find the average or mean of a set (see below) but i'm not sure how to find the standard deviation.
For example say we have the following set of angles:
$$39^\circ 15'01''$$
$$39^\circ 14'15''$$
$$39^\circ 14'32''$$
The average is $39^\circ 14'36''$. Now, how do I find the standard deviation.
I looked at the wiki page, but can't make since of it using degrees minutes seconds instead..
| Your first number is $39 + \dfrac{15}{60} + \dfrac{1}{60^2} = 39.250277777\ldots$. Deal similarly with the others.
And remember to do as little rounding as possible at least until the last step. Rounding should always wait until the last step except when you know how much effect it will have on the bottom line. One way to do avoid it is to work with exact fractions. (Except that the seconds are of course probably rounded . . . .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.