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Locally exact form $P\;dx+Q\;dy$ , and the property $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ This is a very known result, but I don't have some proof. Someone known or has some proof of it?
Let be $\omega = P\;dx + Q\;dy$ be a $C^1$ differential form on a domain $D$. If $$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ,$$ then $\omega$ is locally exact.
| If you have a curl-free field $W = (W_1, W_2, W_3)$ in a neighborhood of the origin, it is the gradient of a function $f$ given by
$$ f(x,y,z) = \int_0^1 \; \left( \; x W_1(tx, ty,tz) + y W_2(tx, ty,tz) + z W_3(tx, ty,tz) \; \right) dt.$$
In your case, take $W_3 = 0$ and drop the dependence on $z$ from $f, \; W_1$ and $W_2.$ Note how this is set up so that $f=0$ at the origin.
There is more information at Anti-curl operator
| {
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Proving that crossing number for a graph is the lowest possible How would one go about proving that the crossing number for a graph is the lowest possible?
To be more specific, given a specific representation of a particular cubic graph $G$, how do I prove that the crossing number can not be lowered any further?
This $G$ has $|V|<20$, and $\operatorname{cr}{(G)}\geq 2$. Furthermore, various online sources say the $\operatorname{cr}{(G)}$ I have found is correct, but offer no proof of why.
Complexity for this in general is hard, supposedly NP-hard, and no general solution is known, but given that the number of vertices is small enough and graph is 3-regular, there must be a way.
| If the graph is small enough and you were willing to prove it by hand, you could do a case analysis similar to how students show a graph is non-planar ($cr(G) \geq 1$) by hand.
Take a long cycle (hopefully Hamiltonian), and place it evenly spaced on a circle, then start adding in edges. You're done if you can show by case analysis that no matter how the edges are added, more than one crossing is created.
| {
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Checking for Meeting Clashes I've been sent here from StackOverflow with my mathematical / algorithm question.
I am currently working with an organisation developing a web system, one area attempting to resolve in-house training clashes.
An example (as best as I can describe is):
What the company is attempting to do is prevent major clashes (>10 people affected) when planning training course times.
*
*100 people are attending training course A.
*75 people are attending training course B.
*25 people are attending training course C.
*5 people are attending training course D.
If 75 people attending B are all attending course A, and B were to run at the same time, there would be a total of 75 clashes.
If all 25 people from course C are attending course A and B, running any of these courses at the same time would result in at minimum of 25 clashes.
If 3 people were attending A and D, and they were to run at the same time only 3 would have an issue and therefore not be a major problem.
The system they are attempting to develop does not necessarily need to resolve the clash itself, just highlight if a certain number of clashes are likely to occur when arranging a new time.
I hope that explains the situation - I am a programmer by profession so this sort of thing is new to me, any points in the right direction would be fantastic!
| If you intend to estimate the expected number of clashes (not necessarily the unique or best measure, but perhaps the more easy to compute) you need a probabilistic model: in particular, you need to know the size of the total population ($N$) and if there is some dependency among courses attendance (i.e. if given that a particular person attends course A then it's more or less probable that he attends course B). Assumming the most simple scenario -no dependencies-, a simple computation (see this related question) shows that if $N_A$ people from a total population of $N$ attends course A, and $N_B$ attends course B, then the expected number of classhes is:
$$E(N_{AB}) = \frac{N_A N_B} {N}$$
| {
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What can be said given that $\Psi$ is a Homomorphism? Let G and H be nontrivial finite groups with relatively prime orders When $\Psi: G\to H$ be a homomorphism, what can be said about $\Psi$ ?
| If $\Psi : G \to H$ is a homomorphism then by the (first) isomorphism theorem you know $G / \ker \Psi \cong \Psi (G)$. This means that $\frac{|G|}{|\ker \Psi|} = |\Psi(G)|$ so you know that $\Psi(G)$ divides $|G|$.
Next you know that $\Psi (G) $ is a subgroup of $H$ and hence by the Lagrange theorem it divides the order of $|H|$.
Putting these two things together you now know that $|\Psi (G)|$ divides $|H|$ and $|\Psi(G)|$ divides $ |G| $.
But $\gcd(|G|, |H|) = 1$ i.e. their only common divisor is $1$ and so $|\Psi(G)| = 1$ and so $\Psi = 0$ is the trivial map.
| {
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The letters ABCDEFGH are to be used to form strings of length four
The letters ABCDEFGH are to be used to form strings of length four.
How many strings contain the letter A if repetitions are not allowed?
The answer that I have is :
$$ \frac{n!}{(n-r)!} - \frac{(n-1)!}{(n-r)!} = \frac{8!}{4!} - \frac{7!}{4!} = 8 \times 7 \times 6 \times 5 - (7 \times 6 \times 5) = 1470 $$ strings.
If you could confirm this for me or kindly guide in me the right direction, please do let me know.
| I presume I am correct. Here is a detailed proof.
First exclude 'A' and permute the rest (7P3). Which can be done in $\frac{7!}{4!}$ ways.
Then, include 'A' back into those permuted cases. $|X_1|X_2|X_3|$ and as indicated by the vertical lines can be in 4 locations. So, the answer is
$$\frac{7!}{4!} \times 4 = 840$$
| {
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reference for "compactness" coming from topology of convergence in measure I have found this sentence in a paper of F. Delbaen and W. Schachermayer with the title: A compactness principle for bounded sequences of martingales with applications. (can be found here)
On page 2, I quote: "If one passes to the case of non-reflexive Banach spaces there is—in general—no
analogue to theorem 1.2 pertaining to any bounded sequence $(x_n )_{n\ge 1} $ , the main
obstacle being that the unit ball fails to be weakly compact. But sometimes there
are Hausdorff topologies on the unit ball of a (non-reflexive) Banach space which
have some kind of compactness properties. A noteworthy example is the Banach
space $ L^1 (Ω, F, P) $ and the topology of convergence in measure."
So I'm looking for a good reference for topology of convergence in measure and this property of "compactness" for $ L^1 $ in probability spaces.
Thx
math
| So that this question has an answer: t.b.'s comment suggests that the quotes passage relates to the paper's Theorem 1.3, which states:
Theorem. Given a bounded sequence $(f_n)_{n \ge 1} \in L^1(\Omega, \mathcal{F}, \mathbb{P})$ then there are convex combinations
$$g_n \in \operatorname{conv}(f_n, f_{n+1}, \dots)$$
such that $(g_n)_{n \ge 1}$ converges in measure to some $g_0 \in L^1(\Omega, \mathcal{F}, \mathbb{P})$.
This is indeed "some kind of compactness property" as it guarantees convergence after passing to convex combinations.
| {
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Does $\mathbb{R}^\mathbb{R}$ have a basis? I'm studying linear algebra on my own and saw basis for $\mathbb{R}, \mathbb{R}[X], ...$ but there is no example of $\mathbb{R}^\mathbb{R}$ (even though it is used for many examples). What is a basis for it? Thank you
| The dimension of $\mathbb R^\mathbb R$ over $\mathbb R$ is $2^{\frak c}$. It is not even the size of the continuum. As Jeroen says, this space is not finitely generated. Not even as an algebra.
Even as an algebra it is not finitely generated. What does that mean? Algebra is a vector space which has a multiplication operator. In $\mathbb R[x]$, the vector space of polynomials, we can write any polynomial as a finite sum of scalars and $x^n$'s. This is an example for a vector space which is not finitely generated, but as an algebra it is finitely generated.
In introductory courses it is customary to deal with well understood spaces. In the early beginning it is even better to use only finitely generated spaces, which are even better understood.
The axiom of choice is an axiom which allows us to "control" infinitary processes. Assuming this axiom we can prove that every vector space has a basis, but we cannot necessarily construct such space.
| {
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What are some "natural" interpolations of the sequence $\small 0,1,1+2a,1+2a+3a^2,1+2a+3a^2+4a^3,\ldots $? (This is a spin-off of a recent question here)
In fiddling with the answer to that question I came to the set of sequences
$\qquad \small \begin{array} {llll}
A(1)=1,A(2)=1+2a,A(3)=1+2a+3a^2,A(4)=1+2a+3a^2+4a^3, \ldots \\
B(1)=1,B(2)=1+3a,B(3)=1+3a+6a^2,B(4)=1+3a+6a^2+10a^3, \ldots \\
C(1)=1,C(2)=1+4a,C(3)=1+4a+10a^2,C(4)=1+4a+10a^2+20a^3, \ldots \\
\ldots \\
\end{array} $
with some indeterminate a .
We had the discussion often here in MSE, that interpolation to fractional indexes, say A(1.5)=?? is arbitrary, considering, that an initial solution composed with any 1 -periodic function satisfies the condition. But here the embedding in a set of sequences, which are constructed from binomial-coefficients might suggest some "natural" interpolation, such as for
$\qquad \small K(1)=1, K(2)=1+a, K(3)=1+a+a^2, \ldots $
the interpolation $\small K(r) = {a^r-1 \over a-1}$ seems the most "natural" which even can smoothly be defined for a=1. This observation made me to refer to "q-analogues" $\small [r]_a $ in my answer in the initiating MSE-question, but it's not obvious how to interpolate the shown sequences of higher orders A , B , C (I think they're not related to the "q-binomial-analogues" , for instance ).
Q: So what would be some "natural" interpolation to fractional indexes for the sequences A, B, C, and possibly in general for sequences generated in the obvious generalized manner?
Agreeing mostly with Henning's ansatz I got now the general form as
$$ A_m(n) = {1 \over (1-a)^m} - \sum_{k=0}^{m-1} \binom{n+m}{k}{a^{n+m-k} \over (1-a)^{m-k} } $$
I do not yet see, whether some examples of fractional indexes agree with the solutions of all three given answers so far, for instance: given a=2.0 what is A(1.5), B(4/3), C(7/5)? With my programmed version I get now
$\qquad \small A(1.5)\sim 9.48528137424 $
$\qquad \small B(4/3) \sim 11.8791929545 $
$\qquad \small C(7/5) \sim 18.4386165488 $
(No interpolation for fractional m yet)
[update 2] the derivative-versions of Sivaram and Michael arrive at the same values so I think, all versions can be translated into each other and mutually support each other to express a "natural" interpolation.
[update 3] I had an index-error in my computation call. Corrected the numerical results.
| I'll build from Michael's work (thanks for doing the heavy lifting!) and start with
$$A_n(r)=\frac1{n!}\frac{\mathrm d^n}{\mathrm da^n} \frac{a^{r+n}-1}{a-1}$$
Let's switch back to the series representation, and swap summation and differentiation:
$$A_n(r)=\frac1{n!}\sum_{k=0}^{r+n-1} \frac{\mathrm d^na^k}{\mathrm d a^n}$$
and make a suitable replacement:
$$A_n(r)=\frac1{n!}\sum_{k=0}^{r+n-1} \frac{k!a^{k-n}}{(k-n)!}=\frac1{n!}\sum_{k=-n}^{r-1} \frac{(k+n)!a^k}{k!}=\frac1{n!}\sum_{k=0}^{r-1} \frac{(k+n)!a^k}{k!}$$
where a reindexing and removal of extraneous zero terms was done in the last two steps.
The result can then be expressed as:
$$A_n(r)=\frac{\mathrm{I}_{1-a}(n+1,r)}{(1-a)^{n+1}}=\frac{1-\mathrm{I}_a(r,n+1)}{(1-a)^{n+1}}$$
where $\mathrm{I}_z(r,n)$ is a regularized incomplete beta function.
In terms of the Gaussian hypergeometric function, we have
$$\begin{align*}
A_n(r)&=\frac{\mathrm{I}_{1-a}(n+1,r)}{(1-a)^{n+1}}\\
&=\frac{a^r}{(n+1)\mathrm{B}(n+1,r)}{}_2 F_1\left({{n+r+1, 1}\atop{n+2}}\mid1-a\right)\\
&=\binom{n+r}{n+1} {}_2 F_1\left({{1-r,n+1}\atop{n+2}}\mid1-a\right)
\end{align*}$$
where the third one is derived from the second one through the Pfaff transformation. In particular, the third one gives the representation
$$A_n(r)=r\binom{n+r}{r}\sum_{k=0}^{r-1}\binom{r-1}{k}\frac{(a-1)^k}{n+k+1}$$
which is valid for arbitrary $n$ and nonnegative integer $r$.
Other hypergeometric representations can be derived.
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Continuous functions are Riemann-Stieltjes integrable with respect to a monotone function
Let $g:[a,b] \to \mathbb{R}$ be a monotone function. Could you help me prove that $\mathcal{C}([a,b])\subseteq\mathcal{R}([a,b],g)$?
(Here $\mathcal{R}([a,b],g)$ is the set of all functions that are Riemann-Stieltjes integrable with respect to $g$.)
Definition of the Riemann-Stieltjes integral. Suppose $f,g$ are bounded on $[a,b]$. If there is an $A \in \mathbb{R}$ such that for every $\varepsilon >0$, there exists a partition $\mathcal{P}$ of $[a,b]$ such that for every refinement $\mathcal{Q}$ of $\mathcal{P}$ we have $|I(f,\mathcal{Q},X,g)-A|<\varepsilon$ (where if $\mathcal{P}=\{a=x_0<\ldots<x_n=b\}$ and $X$ is an evaluation sequence $X=\{x_1^\prime,\ldots,x_n^\prime\}$ and $I(f,\mathcal{Q},X,g)=\sum_{j=1}^n f(x_j^\prime)(g(x_j)-g(x_{j-1}))$), then $f$ is R-S integrable with respect to $g$, and the integral is $A$.
| Assume that $g$ is increasing. I suppose that you know that $f\in\mathcal{R}([a,b],g)$ iff $f$ satisfies the Riemann's condition. The Riemann's condition says:
$f$ satisfies the Riemann's condition respect to $g$ in $[a,b]$ if for every $\epsilon\gt 0$, there exist a partition $P_\epsilon$ of $[a,b]$ such that if $P$ is a refinement of $P_\epsilon$ then $$0\leq U(P,f,g)-L(P,f,g)\lt \epsilon,$$
where $U(P,f,g)$ and $L(P,f,g)$ are the upper and lower Riemann-Stieltjes sums respectively.
With this and the hint in the comments the result holds.
Perhaps the chapter 7 of Mathematical Analysis of Tom M. Apostol can be useful to you.
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How do I show that $\mathbb{Q}(\sqrt[4]{7},\sqrt{-1})$ is Galois? How do I show that $\mathbb{Q}(\sqrt[4]{7},\sqrt{-1})$ is Galois?
At first I thought it was the splitting field of $x^4-7$, but I was only able to prove that it was a subfield of the splitting field. Any ideas?
I'm trying to find all the intermediate fields in terms of their generators, but I don't understand how. I am trying to imitate Dummit and Foote on this. I am looking at the subgroups of the Galois group in terms of $\tau$ where $\tau$ is the automorphism that takes $\sqrt[4]{7}$ to itself and $i$ to $-i$, and $\sigma$ that takes $\sqrt[4]{7}$ to $i\sqrt[4]{7}$ and $i$ to itself. How do I, for example, find the subfield corresponding to $\{1, \tau\sigma^3\}$? I know I am supposed to find four elements of the galois group that $\tau\sigma^3$ fixes, but so far i can only find $-\sqrt[4]{7}^3$.
| The polynomial $f(x)=x^4-7$ factors as $(x-7^{1/4})(x+7^{1/4})(x-i7^{1/4})(x+i7^{1/4})$, and all these irreducible factors are distinct. Hence, $x^4-7$ is separable. Moreover, the field $L= \mathbb{Q}(7^{1/4},i)$ contains all its roots and is the minimal field where $x^4-7$ factors completely in. Hence, it is the minimal splitting field of $f(x)$. So the extension $K \subset L$ is both normal and separable and hence, is Galois.
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Chaos and ergodicity in hamiltonian systems EDIT : I formerly claimed something incorrect in my question. The Liouville measure needs NOT be ergodic on hypersurfaces of constant energy. Also, I found out that NO hamiltonian system can be globally ergodic.
So the new formulation of my question is now this :
Do we call chaotic any hamiltonian system that exhibits the usual chaotic properties on each hypersurface of constant energy (e.g. ergodicity, mixing, positive entropy, positive Lyapunov exponent, etc.) or do we require a "complicated" geometry of those hypersurfaces ?
For example, imagine a system whose hypersurfaces of constant energy are very simple, like planes $z$=constant, but with complicated, chaotic behaviour on each of those planes. Would you call that system chaotic ? Thank you for your thoughts !
| *
*Yep: Alfredo M. Ozorio de Almeida wrote about this:
http://books.google.co.uk/books?id=nNeNSEJUEHUC&pg=PA60&lpg=PA60&dq=hamiltonian+chaos+liouville+measure&source=bl&ots=63Wnmn-xvT&sig=Z0eRtIQxmdQvgWUcLBab7ZJ9y-U&hl=en&ei=0EXfTuvzJcOG8gP5mZjaBQ&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDcQ6AEwAw#v=onepage&q=hamiltonian%20chaos%20liouville%20measure&f=false
*What is meant by Chaos:
Laplace said, standing on Newton's shoulders, "Tell me the force and where we are, and I will predict the future!" An elusive claim, which assumes the absence of deterministic chaos:
Deterministic time evolution does not guarantee predictability, which is particularly relevant for mechanical systems whose equations are non-integrable - common in systems which have nonlinear differential equations with three of more variables.
Knowing this, we enter the realm of physics:
In the hamiltonian formulation of classical dynamics, a system is described by a pair of first-order ordinary equations for each degree of freedom, so in addition we re-impose the conditions from deterministic time evolution (the nonlinear differential equations) and given the space is constrained: Ergodicity!
Hamiltonian Chaos: http://www.phys.uri.edu/~gerhard/hamchaos.html
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Proving the existence of point $z$ s.t. $f^{(n)}(z) = 0$
Suppose that $f$ is $n$ times differentiable on an interval $I$ and there are $n + 1$ points $x_0, x_1, \ldots, x_n \in I, x_0 < x_1 < \cdots < x_n$, such that $f(x_0) = f(x_1) = \cdots = f(x_n) = 0$. Prove that there exists a point $z \in I$ such that $f^{(n)}(z) = 0$.
I am trying to solve this, but other than using then using Rolle's Theorem, I am not sure how to proceed.
| Here’s a fairly broad hint:
You know from Rolle’s theorem that it’s true when $n=1$. Try it for $n=2$; Rolle’s theorem gives you points $y_0$ and $y_1$ such that $x_0<y_0<x_1<y_1<x_2$ and $f\;'(y_0)=f\;'(y_1)=0$. Can you now apply Rolle’s theorem to $f\;'$ on some interval to get something useful?
In order to generalize the hint from $2$ to $n$, you’ll need to use mathematical induction.
| {
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Encyclopedic dictionary of Mathematics I'm looking for a complete dictionary about Mathematics, after searching a lot I found only this one http://www.amazon.com/Encyclopedic-Dictionary-Mathematics-Second-VOLUMES/dp/0262090260/ref=sr_1_1?ie=UTF8&qid=1323066833&sr=8-1 .
I'm looking for a book that can give me a big picture, well written with a clever idea in mind from the author, a big plus can be the presence of railroad diagram or diagram with logic connections between the elements.
| You might first try the updated version of the book you mention, the Encyclopedia of Mathematics which is freely available via the quoted link. Then, since this encyclopedia is very weak on applied mathematics you could have a try with Engquist (ed): Encyclopedia of Applied and Computational Mathematics. This will give you a cross-disciplinary overview of mathematics.
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Conditional probability with union evidence In the problem of cancer (C) and tests (t1, t2), or any other example,
How can I calculate: $P(C^+|(t1^+ \text{ or } t2^+)$
I think this would be the same as finding: $$P(t1^+ \text{ or } t2^+|C^+) P(C^+)\over P(t1^+ \text{ or } t2^+).$$
But is $$P(t1^+ \text{ or } t2^+|C^+) = P(t1^+|C^+)+P(t2^+|C^+)-P(t1^+ \text{ and } t2^+|C^+)?$$
On the other side, is it true in other problems that
$$P(t2^+|t1^+) ={ P(t1^+ \text{ and } t2^+)\over P(t1^+)}?$$
Thanks.
| The answer for your third (last) question is "yes"; this is just the definition of conditional probability. (I answer this first, since it is used later, here).
Your initial instinct is right. For any two events $A$ and $C$:
$$
P(A|C)={P(C\cap A)\over P(C)}={P(A\cap C)\over P(C)}={P(A)P(C| A)\over P(C)}.
$$
The answer to your second question is "yes": $$\eqalign{P(t1^+\cup t2^+|C^+)&={P(( t1^+\cup t2^+)\cap C+)\over P(C^+)}\cr & ={P( ( t1^+\cap C^+)\cup (t2^+\cap C^+))\over P(C^+)}\cr
&={P( t1^+\cap C^+)+P (t2^+\cap C^+)- P (t2^+\cap t1^+\cap C^+) \over P(C^+) }\cr
&={P( t1^+\cap C^+) \over P(C^+) }
+{P (t2^+\cap C^+) \over P(C^+) } -{ P (t2^+\cap t1^+\cap C^+) \over P(C^+) }\cr
&={P( t1^+| C^+)+P (t2^+| C^+)- P (t2^+\cap t1^+| C^+) . }
}$$
| {
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Rank of a degenerate conic This question comes from projective geometry. A degenerate conic $C$ is defined as $$C=lm^T+ml^T,$$
where $l$ and $m$ are different lines. It can be easily shown, that all points on $l$ and m lie on the $C$. Because, for example, if $x\in l$, then by definition $l^Tx=0$ and plugging it into conic equation makes it true.
Question: Find the rank of $C$.
(We can limit ourself to 3-dimensional projective space.)
P.S. I'm reading a book, where it is guessed and checked, but I would like to have a proof without guessing. I do not provide the guess, since it can distract you, but if you really need it just leave a note.
| The rank of $lm^T$ is one. The same goes for $ml^T$. In most cases, the rank of the symmetric matrix $C$ as you define it will be 2. This corresponds to a conic degenerating into two distinct lines. If the lines $l$ and $m$ should coincide, though, the rank of $C$ will be 1.
If you need a proof, you can show this assumption for specific cases without loss of generality. Of you can have a look at the corresponding dual conics and how that relates to adjoint matrices.
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Can you give an example of a complex math problem that is easy to solve? I am working on a project presentation and would like to illustrate that it is often difficult or impossible to estimate how long a task would take. I’d like to make the point by presenting three math problems (proofs, probably) that on the surface look equally challenging. But…
*
*One is simple to solve (or prove)
*One is complex to solve (or prove)
*And one is impossible
So if a mathematician can’t simply look at a problem and say, “I can solve that in a day, or a week, or a month, how can anyone else that is truly solving a problem? The very nature of problem solving is that we don’t know where the solutions lies and therefore we don’t know how long it will take to get there.
Any input or suggestions would be greatly appreciated.
| What positive integers can be written as the sum of two squares? Sum of three squares? Sum of four?
For two squares, it's all positive integers of the form $a^2b$, where $b$ isn't divisible by any prime of the form $4k+3$, and the proof is easy.
For four squares, it's all positive integers, and the proof is moderately difficult, but covered in any course on number theory.
For three squares, it's positive integers not of the form $4^a (8b+7)$ and, while not impossible, the proof is considerably more difficult than the previous two.
| {
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Trouble forming a limit equation Here's the question:
The immigration rate to the Czech republic is currently $77000$ peeople per year. Because of a low fertility rate, the population is shrinking at a continuous rate of $0.1$% per year. The current Czech population is ten million.
Assume the immigrants immediately adopt the fertility rate of their new country. If this scenario did not change, would there be a terminal population projected for the Czech republic? If so, find it.
I need to establish a limit as time would approach infinity (terminal population) but the only way Ive been able to solve it has been using $(x_{n-1} + 77000)(0.999) = x_n$ and I'm not sure exactly how to establish a limit off that (can't use differential equations, we never learned this year). Is there another way to solve for the terminal population without using differential equations?
I've tried multiple ways to get a better answer, but the prof said we need to use limits and I'm stumped.
| We will proceed in two steps: First, assuming that the limit $\lim \limits_{n \to \infty} x_n$ exists, we will find it. Of course, we need to justify our assumption. So we will come back and show the existence of the limit.
Finding the limit. Suppose $x = \lim \limits_{n \to \infty} x_n$. Then allowing $n$ to go to infinity in
$$
x_n = 0.999(x_{n-1} + 77000),
$$
we get
$$
x = 0.999 x + 76992.3 \qquad \implies x = \cdots.
$$
(Exercise: Fill in the blank!)
Showing the existence of the limit. We suspect that for large $n$, we must have $x_n$ approaching $x = \cdots$, so we will think about the "corrected" sequence $x_n - x$ instead. That is, subtracting $x$ from both sides of the equation
$$
x_n = 0.999(x_{n-1} + 77000),
$$
we get
$$
x_n - x = 0.999(x_{n-1} + 77000) - x \stackrel{\color{Red}{(!!)}}{=} 0.999(x_{n-1} - x).
$$
Be sure to check the step marked with $\color{Red}{(!!)}$.
So we end up with
$$
(x_n - x) = 0.999(x_{n-1} - x).
$$
Making the substitution $x_n - x = y_n$ (for all $n$), we get
$$
y_n = 0.999 y_{n-1}.
$$
Can you take it from here?
| {
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Function which has no fixed points Problem:
Can anyone come up with an explicit function $f \colon \mathbb R \to \mathbb R$ such that $| f(x) - f(y)| < |x-y|$ for all $x,y\in \mathbb R$ and $f$ has no fixed point?
I could prove that such a function exists like a hyperpolic function which is below the $y=x$ axis and doesn't intersect it. But, I am looking for an explicit function that satisfies that.
| $f(x)=2$ when $x\leq 1$, $f(x)=x+\frac{1}{x}$ when $x\geq 1$.
Another example:
Let $f(x)=\log(1+e^x)$. Then $f(x)>x$ for all $x$, and since $0<f'(x)<1$ for all $x$, it follows from the mean value theorem that $|f(x)-f(y)|<|x-y|$ for all $x$ and $y$.
| {
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Why does $\int \limits_{0}^{1} \frac{1}{e^t}dt $ converge? I'd like your help with see why does $$\int^1_0 \frac{1}{e^t} \; dt $$ converge?
As I can see this it is suppose to be:
$$\int^1_0 \frac{1}{e^t}\;dt=|_{0}^{1}\frac{e^{-t+1}}{-t+1}=-\frac{e^0}{0}+\frac{e}{1}=-\infty+e=-\infty$$
Thanks a lot?
| I assume you are integrating over the $t$ variable. $1/e^t$ is a continuous function, and you are integrating over a bounded interval, so the integral is well defined. An antiderivative of $1/e^t=e^{-t}$ is equal to $-e^{-t}$. So the integral equals $1-1/e$
| {
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Correspondences between Borel algebras and topological spaces Though tangentially related to another post on MathOverflow (here), the questions below are mainly out of curiosity. They may be very-well known ones with very well-known answers, but...
Suppose $\Sigma$ is a sigma-algebra over a set, $X$. For any given topology, $\tau$, on $X$ denote by $\mathfrak{B}_X(\tau)$ the Borel algebra over $X$ generated by $\tau$.
Question 1. Does there exist a topology, $\tau$, on $X$ such that $\Sigma = \mathfrak{B}_X(\tau)$?
If the answer to the previous question is affirmative, it makes sense to ask for the following too:
Question 2. Denote by ${\frak{T}}_X(\Sigma)$ the family of all topologies $\tau$ on $X$ such that $\Sigma = \mathfrak{B}_X(\tau)$ and let $\tau_X(\Sigma) := \bigcap_{\tau \in {\frak{T}}_X(\Sigma)} \tau$. Is $\Sigma = \mathfrak{B}_X({\frak{T}}_X(\Sigma))$?
Updates. Q2 was answered in the negative by Mike (here).
| I think that I can answer the second question. For each point $p \in \mathbb{R}$, let $\tau_p$ be the topology on $\mathbb{R}$ consisting of $\varnothing$ together with all the standard open neighbourhoods of $p$. Unless I've made some mistake, the Borel sigma-algebra generated by $\tau_p$ is the standard one. However, $\bigcap_{p \in \mathbb{R}} \tau_p$ is the indiscrete topology on $\mathbb{R}$.
| {
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How to prove $k^n \equiv 1 \pmod {k-1}$ (by induction)? How to prove $k^n \equiv 1 \pmod {k-1}$ (by induction)?
| Well, I'll leave the case of $n=1$ to you.
So, for a fixed $k$, suppose that $k^n\equiv 1 \mod(k-1)$ for some $n\in \mathbb{N}$.
We want to show that $k^{n+1} \equiv 1 \mod(k-1)$. Well, $k^{n+1}=k^n k$, and we know that $k^n\equiv 1 \mod(k-1)$ (since this is the induction hypothesis). So, what is $k^{n+1}$ congruent to mod $k-1$?
| {
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Arithmetic Mean Linear? Is a function that finds the arithmetic mean of a set of real numbers a linear function?
So is $\left(X_1 + X_2 + \cdots + X_n\right)/n$ linear or not?
I'm not sure because so long as the set stays the same size $n$ could be defined as a constant.
| To talk about linearity, the domain and range of a function must be vector spaces, in this case over the real numbers. So your first question should be, what vector space do you take the arithmetic mean to be defined on? It turns out you must fix the value of $n$ to get any reasonable vector space with the arithmetic mean defined. If you mix various length sequences, they cannot be added (and extending the shorter sequence by zeros in order to perform the addition is not an option, because this changes its mean value). Once you realise $n$ must be fixed, you should have no difficulty seeing that the arithmetic mean is a linear function.
| {
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probability a credit-card number has no repeated digits It seems as though every Visa or Mastercard account number I've ever had (in the United States) has had at least two consecutive digits identical. I was wondering what the probability is that a particular account number will have at least two consecutive digits identical.
Assume the account number is $16$ digits long, totally ordered, with digits called $(d_i)_{i=1}^{16}$.
Let $$a_i=\left\{\begin{array}{rl}d_i&i \textrm{ is even}\\2d_i&i \textrm{ is odd and }2d_i\lt9\\2d_i-9&i \textrm{ is odd and }2d_i\gt9\end{array}\right.$$
Assume $d_1=4$, $d_2$ through $d_{15}$ are chosen arbitrarily, and $d_{16}$ is chosen so that $$\sum_{i=1}^{16}a_i\in10\mathbb Z.$$
(Fwiw, those assumptions are not correct in the real world, but they're based on real-world facts.)
My answer so far is this:
Consecutive integers are distinct iff all the following hold:
*
*For $2\le i\le15$, $d_i\ne d_{i-1}$ (probability $\frac9{10}$ each).
*Since the ten digits all appear as values of $a_i$ (for fixed $i$, as $d_i$ varies), we might as well assume, in computing $d_{16}$, that the $d_i$ are used in the sum, i.e., that $\sum_id_i\in10\mathbb Z$. But since $d_2,\ldots,d_{15}$ are with equal probability any digit, so is $d_{16}$, so the probability it's distinct from $d_{15}$ is just $\frac9{10}$ again.
So we get $1-.9^{15}\approx .79$.
However I'm very unsure about that last bullet point. Can someone make it more rigorous or correct it?
| It looks fine to me. What you’re using is the fact that if the random variables $X$ and $Y$ are uniformly distributed in $\{0,1,\dots,n-1\}$, then so is the reduced sum $Z=(X+Y)\bmod n$, where $\bmod$ here denotes the binary operation. (In your case $n$ is of course $10$.)
To see this, you can observe that $X+Y$ itself takes values in the set $\{0,1,\dots,2n-2\}$, with the following probabilities:
$$\mathbb{P}[X+Y=k]=\begin{cases}
\frac{k+1}{n^2},&k\le n-1\\\\\\
\frac{2n-1-k}{n^2},&k\ge n-1\;.
\end{cases}$$
(Just count the number of ways that each sum can occur.)
But $Z=(X+Y)\bmod n=k$ iff $X+Y=k$ or $X+Y=k+n$, so
$$\mathbb{P}[Z=k]=\frac1{n^2}\bigg((k+1)+\big(2n-1-(k+n)\big)\bigg)=\frac1{n}\;,$$
i.e., $Z$ is uniformly distributed in $\{0,1,\dots,n-1\}$. By induction the same is true for any finite sum reduced modulo $n$.
| {
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Consequence of Cauchy Integral Formula for Several Complex Variables in Gunning's book I am reading Gunning's book Introduction to Holomorphic Functions of Several Variables, Vol. I, and I am stuck in the proof of Maximum modulus theorem: if $f$ is holomorphic in a connected open subset $D \subset \mathbb{C}^{n}$ and if there is a point $A \in D$ such that $|f(Z)| \leq |f(A)|$ for all points $Z$ in some open neighborhood of $A$, then $f(Z) = f(A)$ for all points $Z \in D$. In the proof Gunning says that for any polydisc $\Delta = \Delta(A; R) \subset D$ for which $\overline{\Delta} \subset D$ we have as a consequence of the Cauchy integral formula that
$$|\Delta| f(A) = \int_{\Delta} f(Z) dV(Z),$$
where $dV(Z)$ is the Euclidean volume element in $\mathbb{C}^{n} = \mathbb{R}^{2n}$ and $|\Delta| = \int_{\Delta} dV(Z) = \pi^{n}r_{1}^{2} \cdots r_{n}^{2}$ is the Euclidean volume of $\Delta$.
It looks very easy, but I am stuck on it for a long time. I can not see how this is a consequence of Cauchy integral formula, since the integral is on $\Delta$ and not on the product of $|\zeta_{j} - a_{j}| = r_{j}$. We can not apply Stokes, because the form is of degree $n$. Maybe the Intermediate Value Theorem for integrals solve it, but how to assure that the point giving the equality is $A$? Maybe a change of variables?
Thanks for help.
| It is an integrated form of the Cauchy formula. The single complex variable case illustrates what's going on. For example,
$$
f(0) = {1\over 2\pi i}\int_{|z|=1} {f(z)\over z}\;dz
= {1\over 2\pi} \int_0^{2\pi} {f(re^{i\theta})\over r\,e^{i\theta}}\,d(re^{i\theta})
= {1\over 2\pi} \int_0^{2\pi} f(re^{i\theta})\;i\,d\theta
$$
can be integrated (for example) $\int_0^1 \ldots r\,dr$:
$$
f(0)\cdot \int_0^11\cdot r\,dr \;=\; {1\over 2\pi}\int_0^1 \int_0^{2\pi}
f(re^{i\theta})\;r\;d\theta\,dr
$$
The obvious absolute-value estimate gives
$$
|f(0)|\;\le\; {1\over \pi} \int_{|z|\le 1} |f(z)|\,dV
$$
| {
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An application of Gronwall's lemma I've come across a creative use of Gronwall's lemma which I would like to submit to the community. I suspect that the argument, while leading to a correct conclusion, is somewhat flawed.
We have a continuous mapping $g \colon \mathbb{R}\to \mathbb{R}$ such that
$$\tag{1} \forall \varepsilon>0\ \exists \delta(\varepsilon)>0\ \text{s.t.}\ \lvert x \rvert \le \delta(\varepsilon) \Rightarrow \lvert g(x) \rvert \le \varepsilon \lvert x \rvert$$
and a continuous trajectory $x\colon [0, +\infty) \to \mathbb{R}$ such that
$$\tag{2} e^{\alpha t}\lvert x(t)\rvert \le \lvert x_0\rvert+\int_0^t e^{\alpha s}\lvert g(x(s))\rvert\, ds. $$
Here $x_0=x(0)$ is the initial datum, which we may choose small as we wish, but $\alpha >0$ is a fixed constant that we cannot alter in any way.
Now comes the point. Fix $\varepsilon>0$. The lecturer says: Suppose
we can apply (1) for all times $t \ge 0$. Then inserting (1) in (2)
we get
$$e^{\alpha t}\lvert x(t) \rvert \le \lvert x_0\rvert + \varepsilon \int_0^t e^{\alpha s} \lvert x(s)\rvert \, ds$$
and from Gronwall's lemma we infer
$$\tag{3} \lvert x(t)\rvert \le e^{(\varepsilon - \alpha)t}\lvert x_0\rvert.$$
So if $\varepsilon <\alpha$ and $\lvert x_0 \rvert < \delta(\varepsilon)$, $\lvert x(s) \rvert$ is small at all times and
our use of (1) is justified. We conclude that inequality (3) holds.
Does this argument look correct to you? I believe that the conclusion is correct, but that it requires more careful treatment.
Thank you.
| As you presented it, this is completely bogus: it is an example of the logical fallacy called "begging the question".
| {
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Prove equations in modular arithmetic Prove or disprove the following statement in modular arithmetic.
*
*If $a\equiv b \mod m$, then $ a^2\equiv b^2 \mod m$
*If $a\equiv b \mod m$, then $a^2\equiv b^2 \mod m^2$
*If $a^2\equiv b^2\mod m^2$, then $a\equiv b\mod m$
My proofs.
*
*$$ a\equiv b \mod m \implies (a-b) = mr, r\in\mathbb{Z}$$
$$ a^2-b^2 = (a+b)(a-b) = (a+b)mr = ms \text{ where } s = (a+b)\cdot r$$
So the first statement is true
*$$ a\equiv b \mod m \implies (a-b) = mr, r\in\mathbb{Z}$$
$$a^2-b^2 = (a+b)(a-b) = (a+b)mr$$
but $(a+b)\neq ms$ $\forall s\in\mathbb{Z}$ in general. So the second one is false.
*$$a^2-b^2 = m^2r, \exists r\in\mathbb{N}$$
$$a^2-b^2= (a+b)(a-b) $$
Then I kind of got stuck here. I'm not sure how to continue it. Am I missing some properties I don't know? Or there is a algebra trick that could be applied here?
| HINT $\: $ for $\rm (3),\ \ m^2\ |\ a^2 - b^2\ \Rightarrow\ m\ |\ a-b\ $ fails if $\rm\: m > 1 = a - b\:.\:$ Then $\rm\:a^2-b^2 = 2\:b+1\:$ so any odd number with a square factor $\rm\:m^2 \ne 1\:$ yields a counterexample.
| {
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Algorithm wanted: Enumerate all subsets of a set in order of increasing sums I'm looking for an algorithm but I don't quite know how to implement it. More importantly, I don't know what to google for. Even worse, I'm not sure it can be done in polynomial time.
Given a set of numbers (say, {1, 4, 5, 9}), I want to enumerate all subsets of this set (its power set, really) in a certain order: increasing sum of the elements.
For example, given {1, 4, 5, 9}, the subsets should be enumerated in this order, "smaller" sets first:
{} = 0
{1} = 1
{4} = 4
{5} = 5
{1, 4} = 5
{1, 5} = 6
{9} = 9
{4, 5} = 9
{1, 9} = 10
{1, 4, 5} = 10
{4, 9} = 13
{5, 9} = 14
{1, 4, 9} = 14
{1, 5, 9} = 15
{4, 5, 9} = 18
{1, 4, 5, 9} = 19
This feels like some unholy mix between a breadth-first search and a depth-first search, but I can't wrap my head around the proper way to mix these two search strategies.
My search space is very large ($2^{64}$ elements) so I can't precompute them all up-front and sort them. On that note, I also don't need to enumerate the entire search space -- the smallest 4,096 subsets is fine, for example.
Can anyone give any pointers or even any clues to google for? Many thanks.
| Here's an algorithm. The basic idea is that each number in the original set iterates through the list of subsets you've already found, trying to see if adding that number to the subset it's currently considering results in the smallest subset sum not yet found.
The algorithm uses four arrays (all of which are indexed starting with $0$).
*
*$N$ consists of the numbers in the original set; i.e., $N = [1, 4, 5, 9]$ in your example.
*$L$ is the list of subsets found so far.
*$A[i]$ contains the subset that $N[i]$ is currently considering.
*$S[i]$ is the sum of the elements of subset $i$ in $L$.
Algorithm:
*
*Initialize $N$ to numbers in the original set, all entries of $A$ to $0$, $L[0] = \{\}$, $S[0] = 0$. Let $j = 1$.
*For iteration $j$ find the minimum of $S[A[i]] + N[i]$ over all numbers $N[i]$ in the original set. (This finds the subset with smallest sum not yet in $L$.) Tie-breaking is done by number of elements in the set. Let $i^*$ denote the argmin.
*Let $L[j] = L[A[i^*]] \cup \{N[i^*]\}$. Let $S[j] = S[A[i^*]] + N[i^*]$. (This updates $L$ and $S$ with the new subset.)
*Increase $A[i^*]$ to the next item in $L$ that has no number larger than $N[i^*]$. If there is none, let $A[i^*] =$ NULL. (This finds the next subset in $L$ to consider for the number $N[i^*]$ just added to an existing subset in $L$ to create the subset just added to $L$.)
*If all entries in $A[i]$ are NULL, then stop, else increment $j$ and go to Step 2.
For example, here are the iterations for your example set, together with the subset in $L$ currently pointed to by each number.
Initialization:
{} 1, 4, 5, 9
Iteration 1:
{} 4, 5, 9
{1}
Iteration 2:
{} 5, 9
{1} 4
{4}
Iteration 3:
{} 9
{1} 4, 5
{4}
{5}
Iteration 4:
{} 9
{1} 5
{4}
{5}
{1,4}
Iteration 5:
{} 9
{1}
{4} 5
{5}
{1,4}
{1,5}
Iteration 6:
{}
{1} 9
{4} 5
{5}
{1,4}
{1,5}
{9}
Iteration 7:
{}
{1} 9
{4}
{5}
{1,4} 5
{1,5}
{9}
{4,5}
Iteration 8:
{}
{1}
{4} 9
{5}
{1,4} 5
{1,5}
{9}
{4,5}
{1,9}
Iteration 9:
{}
{1}
{4} 9
{5}
{1,4}
{1,5}
{9}
{4,5}
{1,9}
{1,4,5}
And the rest of the iterations just involve adding $9$ successively to each subset already constructed that doesn't include $9$.
| {
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Foreign undergraduate study possibilities for a student in Southeastern Europe In the (non-EU) country I live in, the main problem with undergraduate education is that it's awfully constrained. I have only a minimal choice in choosing my courses, I cannot take graduate courses, and I have to take many applied and computational classes. My problem is that
*
*I have already learned (or I plan to learn in the 9 months until university) a lot of the pure mathematics that I would learn at the undergraduate programme
*I am not interested in applied or computational classes
I'm interested in:
Is there any undergraduate programme, affordable to an average but dedicated student with a modest budget, which either has very pure emphasis or freedom in choosing the classes? I'm ready to learn a new language if it is required.
Thank you for any help!
| Hungary also has a very strong mathematical tradition, especially in discrete math, and has relatively cheap living standards. Many great mathematicians have studied at Eötvös Loránd University (ELTE) in Budapest. You can try looking there as well.
| {
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Trick to find multiples mentally We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.
A few months ago, I heard a simple and elegant way to find multiples of $7$:
Cut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.
Example:
$21553$
Cut digits into pairs:
$2, 15, 53$
Multiply $53$ by $1, 15$ by $2, 2$ by $4$:
$8, 30, 53$
Add:
$8+30+53=91$
As $91$ is a multiple of $7$ ($13 \cdot 7$), then $21553$ is too.
This works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 → 05 = 7 - 2, 112 → 12 = 14 - 2, \cdots)$
I figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.
Exemple with $91$:
$91$
$9, 1$
$9\cdot3, 1\cdot1$
$27, 1$
$28$
My question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$?
| One needn't memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \ $ is $\rm\: (d_2\cdot 10 + d_1)\ 10 + d_0\ \equiv\ (d_2\cdot 3 + d_1)\ 3 + d_0\ (mod\ 7)\ $ since $\rm\ 10\equiv 3\ (mod\ 7)\:.\:$ So we compute the remainder $\rm\ (mod\ 7)\ $ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\rm\:(mod\ 7)\:.\:$
For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\ d_{n-1}\ $ by $\rm\ d_n\cdot 3 + d_{n-1}\:\ (mod\ 7),\:$ namely
$\rm\qquad\phantom{\equiv} \color{red}{4\ 3}\ 2\ 1\ 1$
$\rm\qquad\equiv\phantom{4} \color{green}{1\ 2}\ 1\ 1\quad $ by $\rm\quad \color{red}4\cdot 3 + \color{red}3\ \equiv\ \color{green}1 $
$\rm\qquad\equiv\phantom{4\ 3} \color{royalblue}{5\ 1}\ 1\quad $ by $\rm\quad \color{green}1\cdot 3 + \color{green}2\ \equiv\ \color{royalblue}5 $
$\rm\qquad\equiv\phantom{4\ 3\ 5} \color{brown}{2\ 1}\quad $ by $\rm\quad \color{royalblue}5\cdot 3 + \color{royalblue}1\ \equiv\ \color{brown}2 $
$\rm\qquad\equiv\phantom{4\ 3\ 5\ 2} 0\quad $ by $\rm\quad \color{brown}2\cdot 3 + \color{brown}1\ \equiv\ 0 $
Hence $\rm\ 43211\equiv 0\:\ (mod\ 7)\:,\:$ indeed $\rm\ 43211 = 7\cdot 6173\:.\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\rm\: \pm\{0,1,2,3\}\ \:(mod\ 7)\:.$ Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9\:$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Sparsest cut is solvable on trees The problem is to prove that Sparsest cut is solvable on trees in polynomial time.
A short review, a sparsest cut is linear program
$$\min \frac{c(S,\overline{S})}{D(S,\overline{S})}$$
where $c(S,\overline{S})$ - sum of edge weights for every edge that crosses the cut $S,\overline{S}$
and $D(S,\overline{S})$ - sum of demands between $s_{i}$ and $t_{i}$ are separated by $S,\overline{S}$
The proof is based on the claim.
There exists a sparsest cut $(S,\overline{S})$, such that the graphs $G[S]$ and $G[\overline{S}]$ are connected.
Proof of the claim:
Without loss of generality, assume $G[S]$ is not connected. Say it has components $C_{1}...C_{t}$. Let the total capacity of edges from $C_{i}$ to $\overline{C}_{i}$ be $c_{i}$ and the demand be $d_{i}$. The sparsity of cut $S$ is $\frac{c_{1}+\cdots+c_{t}}{d_{1}+\cdots+d_{t}}$.
Now since all quantities $C_{i},d_{i}$ are non-negative, by simple arithmetic, there exists $i$ such that $\frac{c_{i}}{d_{i}} \leq \frac{c_{1}+\cdots+c_{t}}{d_{1}+\cdots+d_{t}}$. This implies that cut $C_{i}$ is at least as good as $S$.
Using this claim we know that the sparsest cut on trees will be exactly one edge. Therefore, the sparsest cut problem on trees becomes easy to solve in polynomial time.
The problem is I don't really understand how the claim was proved. I think it was proved by contradiction, firstly we assumed that $G[S]$ is not connected and found a better that $(S,\overline{S})$ cut for a component $C_{i}$, which is contradiction therefore $S$ is connected. Am I wrong? Secondly, how does this claim apply that sparsest cut is solvable on a tree. Just because as a component $C_{1}$ we can take a one edge of the tree? And how to show that it's solvable in polynomial time?
Thanks!
| Actually, you understand more than you think :). The proof indeed goes by contradiction, in that if the optimal cut induced disconnected components, then one of the components would give a better cut value. The rest of the proof follows from the fact that you can now parametrize the set of candidate optimal solutions by edges from the tree (since each edge removal creates two connected subgraphs). There are n-1 edges, and for each edge you can compute the cut cost in poly time.
| {
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Count the number of integer solutions to $x_1+x_2+\cdots+x_5=36$ How to count the number of integer solutions to $x_1+x_2+\cdots+x_5=36$ such that $x_1\ge 4,x_3 = 11,x_4\ge 7$
And how about $x_1\ge 4, x_3=11,x_4\ge 7,x_5\le 5$
In both cases, $x_1,x_2,x_3,x_4,x_5$ must be nonnegative integers.
Is there a general formula to calculate things like this?
| $\infty$, if you have no constraint on $x_2$ and $x_5$ other than that they are integers: note that you can always add $1$ to one of these and subtract $1$ from the other. Or did you mean nonnegative (or positive) integers?
| {
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Rules for algebraically manipulating pi-notation? I'm a bit of a novice at maths and want to learn more about algebraically manipulating likelihoods in statistics.
There are a lot of equations that involve taking the product of a set of values given a model.
I know a few rules for manipulating sigma-notation (e.g., here and here).
*
*What are the basic rules for manipulating sequences of products (i.e, $\prod_{i=1}^{I} ... $)? Is there a web page that you could direct me to?
e.g.,
*
*$\prod_{i=1}^{I} x_i$
*$\prod_{i=1}^{I} x_i y_i$
*$\prod_{i=1}^{I} a + b x_i$
*$\prod_{i=1}^{I} \exp x_i$
| This might be inappropriate for an answer but I believe you tried yourself too hard at here. The $\prod $ sign just means multiplying some elements together, with a label in the bottom to denote the beginning element and a label on the top to denote the end element. The files you provided are very good and you should get a sense of what the general situation is. You will get better in using it if you read articles proving things using this symbol. But as any other short hand symbol writing the product of elements in this way does not really simplify anything; you should be able to write out any such products explicitly if you can write it in the $\prod$ sign.
| {
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Curve defined by 3 equations Suppose that $X$ is a curve in $\mathbb{A}^3$ (in the AG sense, let's say over an algebraically closed field $k$) that contains no lines perpendicular to the $xy$ plane, and that there exist two polynomials $f,g\in k[x,y,z]$ such that $\{f=0\}\cap\{g=0\}=X\cup l_1\cup\cdots\cup l_n$, where $l_i$ are lines perpendicular to the $xy$ plane (and can possibly intersect $X$). Is it possible to find a third polynomial $h$ such that the intersection $\{f=0\}\cap\{g=0\}\cap\{h=0\}=X$?
Since $X$ is algebraic, of course given a point that does not lie on $X$, there is a polynomial that is zero on $X$ and not zero on that point. I want to see if I can "cut out" $X$ with one other equation.
| Yes. I believe this is from a Shafarevich problem?
For instance, suppose $\ell_1$ intersects the $x-y$ plane at $(x,y) = (a,b)$. Consider the homomorphism $k[x,y,z] \rightarrow k[z]$ sending $x\mapsto a$ and $y \mapsto b$. The image of $I(X)$ is some prime ideal of $k[z]$, which is principal. Now look at the pullback of a generator.
To treat the general case, use this idea along with the Chinese remainder theorem.
| {
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Find all analytic functions such that... Here is the problem: find all functions that are everywhere analytic, have a zero of order two in $z=0$, satisfy the condition $|f'(z)|\leq 6|z|$ and such that $f(i)=-2$. Any hint is welcomed.
| Here is a hint: consider $f'(z)/z$.
Since $f(z)$ has a zero of order two at $z=0$, the derivative $f'(z)$ is also holomorphic, and $f'(0)=0$. Thus, you may write $f'(z)$ as $z\cdot g(z)$, with $g(z)$ holomorphic. Then, the bound in the statement tells you that $|g(z)|$ is bounded.
| {
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A problem related to basic martingale theory In our probability theory class, we are supposed to solve the following problem:
Let $X_n$, $n \geq 1 $ be a sequence of independent random variables such that
$ \mathbb{E}[X_n] = 0, \mathbb{Var}(X_n) = \sigma_n^2 < + \infty $ and
$ | X_n | \leq K, $ for some constant $ 0 \leq K < + \infty, \ \forall n \geq 1$.
Use martingale methods to show that
$$ \sum \limits_{n = 1}^{\infty} \ X_n \ \mbox{ converges } \mathbb P-\mbox{a.s.} \ \Longrightarrow \ \sum\limits_{n = 1}^{\infty} \ \sigma_n^2 < + \infty .$$
Could anybody give me a hint?
Thanks a lot for your help!
Regards, Si
| Set $S_n:=\sum_{i=1}^n X_i$ and $S_0=0$. $S$ is a martingale (wrt the natural filtration), so $S^2$ is a sub-martingale.
Using the Doob's decomposition we can write $S^2=M+A$ where M is a martingale and A is predictable (increasing) process, both null at 0.
It turns out that $A_n=\sum_{i=1}^n \sigma_i^2$.
Define the stopping time $T_\alpha$ as $T_\alpha=\inf\{n \geq0 \ : \ |S_n|>\alpha\}$.
By OST we have $$E[S_{T_\alpha \wedge n}^2]-E[A_{T_\alpha \wedge n}]=0;\qquad \forall n.$$
Also $|S_{T_\alpha \wedge n}| \leq \alpha+K$, since $|X_i| \leq K $; then
$$E[A_{T_\alpha \wedge n}] \leq (\alpha+K)^2;\qquad \forall n.$$
Since $\sum X_i$ converges a.s., the partial sum $S_n$ are a.s. bounded, then for some $\alpha$ we have that $(T_{\alpha}=\infty)$ has positive probability. So applying this fact to the last inequality, we get $A_{\infty}=\sum\sigma_n^2 < \infty$.
P.S.: I hope it is clear; I am a new user and I want to say hi to everyone.
Also I noticed that I can not comment in some post; is there any restriction?
Thanks.
| {
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The expected value of magnitude of winning and losing when playing a game Suppose we play a game at a casino. There is a \$5 stake and three possible outcomes: with probability $1/3$ you lose your stake, with with probability $1/3$ the bank returns your stake plus \$5, and with probability $1/3$ the bank simply returns your stake.
Let $X$ denote your winning in one play. And you play 1000 times. What would you expect the magnitude of your win or loss to be approximately?
Is this question asking the same as to find the $Var(S)$ where $S=X_1+\cdots X_{1000}$, so the final answer should be $1000\cdot Var(X)$?
| If I understand correctly, you are interested in the expected value of the magnitude of the win (or) loss (and not the magnitude of the expected value of the win (or) loss). Hence, you are interested in computing $\mathbb{E}(S_n)$ of the underlying random variable where $$S_n = \left| X_1 + X_2 + \cdots + X_n \right|.$$
Note that $S_n$ can take values in the set $\{0,5,10,\ldots,5n\}$.
Now let us find out the probability that $S_n = 5m$ for some $m \in \{0,1,2,\ldots,n\}$.
This means we are interested in the event $X_1 + X_2 + \cdots + X_n = \pm 5m$.
Let us first evaluate the probability of the event $X_1 + X_2 + \cdots + X_n = 5m$.
For this event to occur, if you lose $k$ times, you need to win $m+k$ times and get back your stake the remaining $n-m-2k$ times where $k \in \left \{0,1,2,\ldots, \left \lfloor \frac{n-m}{2} \right \rfloor \right\}$.
Hence, the desired probability of the event $X_1 + X_2 + \cdots + X_n = 5m$ is given by $$\sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac1{3^n}$$
Hence, the desired probability of the event $S_n = 5m$ is given by $$P_n(m) = \sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac2{3^n}$$
Hence, the expected value of $S_n$ is given by $$\sum_{m=0}^{n} 5m P_n(m) = \sum_{m=0}^{n} \sum_{k=0}^{\left \lfloor \frac{n-m}{2} \right \rfloor} \frac{n!}{k!(m+k)!(n-m-2k)!} \frac{10m}{3^n}$$
| {
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Properties about a certain martingale I asked this question here. Unfortunately there was not a satisfying answer. So I hope here is someone who could help me.
I'm solving some exercises and I have a question about this one:
Let $(X_i)$ be a sequence of random variables in $ L^2 $ and a filtration $ (\mathcal{F}_i)$ such that $X_i$ is $\mathcal{F}_i$ measurable. Define
$$ M_n := \sum_{i=1}^n \left(X_i-E(X_i|\mathcal{F}_{i-1})\right) $$
I should show the following:
*
*$M_n $ is a martingale.
*$M_n $ is square integrable.
*$M_n $ converges a.s. to $ M^*$ if $ M_\infty := \sum_{i=1}^\infty E\left((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1}\right)<\infty$ .
*If $\sum_{i=1}^\infty E(X_i^2) <\infty \Rightarrow 3)$
I was able to show 1 and with Davide Giraudo's comment 2. is clear too. But I got stuck at 3. and 4. So I'm very thankful for any help!
hulik
| For 3, compute $E[M_n^2]$. Having done this conclude that $E[M_n^2]\le M_\infty$ for all $n$. This means that $\{M_n\}$ is an $L^2$-bounded martingale, to which the martingale convergence theorem may be applied.
For 4, the $i$th term in the sum defining $M_\infty$ is equal to
$E[(X_i-E[X_i|\mathcal{F}_{i-1}])^2]$, which in turn is equal to
$E[X_i^2] -E[(E[X_i|\mathcal{F}_{i-1}])^2]\le E[X_i^2]$.
| {
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A primitive and irreducible matrix is positive for some power $k.$ Prove it's positive for any power $k+i,$ where $i=1,\dots.$
Let $P = [p_{ij}]_{1 \leqslant i,j \leqslant m} \geqslant 0$ a primitive and irreducible matrix. And $P^k > 0$ for some $k.$ Prove that $ P^{k+i} > 0, i =1,2, \dots.$
I have used a hint suggested by N.S below and wrote this proof what do you think?
By the irreducibility of $P \geqslant 0$ there exists a permutation matrix $M$ (an identity matrix whose rows and columns have been reordered) such that $MP$ can be written in the form $ MP = \begin{pmatrix} P_{11} & 0 \\ P_{21} & P_{22} \end{pmatrix}$ where $P_{11}$ and $P_{22}$ are square matrices. This implies $ P = M^{-1}\begin{pmatrix} P_{11} & 0 \\ P_{21} & P_{22} \end{pmatrix}.$ The matrix $P$ has at least a non-zero element (which is in fact positive) in each row, thus we use this fact in the proof below
We have $P^k >0.$ At $i=1$ we have $P^{k+1} = P P^k > 0.$ We assume at $i$ that we have $P^{k+i} >0.$ Thus at $i+1$ we have $ P^{k+i+1} = PP^{k+i} > 0.$ This is correct for all $i = 1,2, \dotsm$
| Hint Prove first that any row of $P$ has a non-zero element. Then
$P^{k+i+1}=PP^{k+i}$
| {
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Need help finding limit $\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$ Facing difficulty finding limit
$$\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$$
For starters I have trouble simplifying it
Which method would help in finding this limit?
| $$
\begin{eqnarray}
\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}=\lim \limits_{x\to \infty}\left(\frac{x-1+1}{x-1}\right)^{2x+1}
=\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{2x+1}\\= \lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}}
=\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}}=e^{\lim \limits_{x\to \infty}\frac{2x+1}{x-1}}=e^2
\end{eqnarray}
$$
| {
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Simple Combinations with Repetition Question
In how many ways can we select five coins from a collection of 10 consisting of one penny, one nickel, one dime, one quarter, one half-dollar and 5 (IDENTICAL) Dollars ?
For my answer, I used the logic, how many dollars are there in the 5 we choose?
I added the case for 5 dollars, 4 dollars, 3 dollars and 2 dollars and 1 dollars and 0 dollars.
$$C(5,5) + C(5,4) + C(5,3) + C(5,2) + C(5,1) + 1 = 32$$
which is the right answer ... but there has to be a shorter way, simpler way. I tried using the repetition formula that didn't pan out.
If you could introduce me to a shorter way with explanation I appreciate it.
| Decide for each small coin whether you select that or not. Then top up with dollars until you have selected 5 coins in total.
The topping-up step does not involve any choice, so you have 5 choices to make, each with 2 options, giving $2^5=32$ combinations in all.
| {
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Interesting Property of Numbers in English I was playing with the letters in numbers written in English and I found something quite funny. I found that if you count the number of letters in the number and write this as a number and then count the number of letters in this new number and keep repeating the process, you will arrive at the number 4.
I've confirmed this (using a computer program) for all numbers up to 999999 and was wondering if there's a way to prove this or to find a counter example for which it does not hold.
Just to give an example of the above statement, let's start with thirty seven (I chose this randomly)
Thirty seven has 11 letters in it, Eleven has 6 letters in it, Six has three letters in it, Three has 5 letters in it, Five has 4 letters in it.
It may look like I just picked this number, so let me show this for another random number, say 999.
Nine hundred and ninety nine has 24 letters in it, Twenty four has 10 letters in it, Ten has 3 letters in it, Three has 5 letters in it, Five has 4 letters in it.
What are your thoughts on how to prove this?
(Just a note: I only confirmed this for numbers written in the standard British way of writing numbers - for example 101 is one hundred and one)
| Define $f: \mathbb{N} \to \mathbb{N}$ as the number of letters in a given natural number spelled out.
Four is the only fixed point under $f$, and it's not too difficult to see that $f$ is almost always strictly decreasing with the only exceptions being one, two, three and four. So the $n^{th}$ iterate of $f$ must eventually become smaller than 5, which doesn't leave very many cases to verify.
| {
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Isometries of $\mathbb{R}^3$ So I'm attempting a proof that isometries of $\mathbb{R}^3$ are the product of at most 4 reflections. Preliminarily, I needed to prove that any point in $\mathbb{R}^3$ is uniquely determined by its distances from 4 non-coplanar points, and then that an isometry sends non-coplanar points to non-coplanar points in $\mathbb{R}^3$. I've done the first preliminary step, and finished the proof assuming the second, but I can't find a simple way to prove the second...
Intuitively it makes a lot of sense that non-coplanar points be sent to non-coplanar points, but every method I've stumbled upon to prove such has been quite heavy computationally...
I know for example that any triangle chosen among the four points, A, B, C, D must be congruent to the triangles of their respective images, but what extra bit of information would allow me to say that the image of the whole configuration can't be contained in a single plane...
| Would this work? Construct the altitude from $D$ to the plane containing $A$, $B$, and $C$. Call the foot of this altitude $E$ (the point where the altitude meets the plane). Triangles $ADE$, $BDE$, and $CDE$ all have right angles at $E$ and you know that the isometry preserves angles, so triangles $A'D'E'$, $B'D'E'$, and $C'D'E'$ all have right angles at $E'$, which makes $A'$, $B'$, $C'$, and $E'$ coplanar and since $D'E'\ne 0$ and $\overline{D'E'}$ is perpendicular to that plane, $D'$ cannot be coplanar with the other points.
| {
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Why is the Kendall tau distance a metric? So I am trying to see how the Kendall $\tau$ distance is considered a metric; i.e. that it satisfies the triangle inequality.
The Kendall $\tau$ distance is defined as follows:
$$K(\tau_1,\tau_2) = |(i,j): i < j, ( \tau_1(i) < \tau_1(j) \land \tau_2(i) > \tau_2(j) ) \lor ( \tau_1(i) > \tau_1(j) \land \tau_2(i) < \tau_2(j) )|$$
Thank you in advance.
| Kendall tau rank distance is a metric only if you compare ranking of the elements.
If you perform Kendall function comparing elements you will find cases where the triangular inequality does not work.
Example:
0 0 0 10 10 10
and
5 5 5 0 0 0
scores 9 (using Kendall comparing elements)
While
0 0 0 10 10 10
and
5 3 5 7 5 2
scores 3
and
5 3 5 7 5 2
and
5 5 5 0 0 0
scores 4
so 9 > 3 + 4. So triangular inequality is not working here.
But if you operate with the sorting position of each element in it vector (aka ranking) triangular inequality is gauaranteed.
This happens beacuase of repetitions of elements within vectors.
We should call "Kendall tau ranking distance" to one of the algorithms and "Kendall tau distance" to the other
Best
Mijael
| {
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Can this be simplified any further I've been working on a formula, which I have managed to simplify to the following expression, but I wonder if anyone can spot a way to simplify it further?
$$2^{1 -\frac{1}{2}\sum_i \log_2 \frac{(a_i + c_i)^{(a_i + c_i)}}{a_i^{a_i}c_i^{c_i}}}$$
| Does this look simpler to you?
$$
2 \left( \prod_i \frac{a_i^{a_i}c_i^{c_i}}{(a_i + c_i)^{(a_i + c_i)}} \right)^\frac{1}{2}
$$
| {
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Probability of getting different results when tossing a coin Here's a question I got for homework:
In every single time unit, Jack and John are tossing two different
coins with P1 and P2 chances for heads. They
keep doing so until they get different results. Let X be the number of
tosses. Find the pmf of X (in discrete time units). What kind of
distribution is it?
Here's what I have so far:
In every round (time unit) the possible results
HH - p1p2
TT - q1q2
TH - q1p2
HT - q2p1
and so P(X=k) = ((p1p2 + q1q2)^(k-1))*(q1p2+q2p1)
Which means we're dealing with a geometric distribution.
What doesn't feel right is that the question mentions 'discrete time units'. That makes me think about a Poisson distribution, BUT - Poisson is all about number of successes in a time unit, while here we only have one round in every time unit.
If I'm not too clear its only because I'm a little confused myself. Any hint would be perfect. Thanks in advance
| You can work out the probability that they get different results on the first toss, namely $p_1 (1-p_2)+ (1-p_1)p_2 = p_1+p_2 - 2p_1 p_2$.
If they have not had different results up to the $n$th toss, then the conditional probability they get different results on the next toss is the same; this is the memoryless property and so (since the number of tosses is a positive integer, i.e. discrete) you have a geometric distribution, as you spotted.
| {
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Prove that exist $z_0 \in \mathbb C$ that satisfy $f(z_0)=0$. I would be glad to get some help with this question:
Let $f(z)$ be an entire function. Assume that there exists a
monotonous increasing and unbounded sequence $\{r_n\}$ such that $\lim\limits_{n \to \infty} \min\limits_{|z|=r_n} |f(z)|=\infty$. I want to show that there exists a $z_0 \in \mathbb C$ that satisfies $f(z_0)=0$.
I'd especially like to know how to use that fact about the sequence.
Thanks.
| Assume $f(z)\ne0$ for all $z\in\mathbb{C}$. Then $h(z)=1/f(z)$ is also an entire function. Apply the maximum modulus principle to $h$.
| {
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What function does $\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$ represent, evaluated at some number $x$? I need to know what the function $$\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$$ represents evaluated at a particular point.
For example if the series given was $$\sum \limits_{n=0}^{\infty}\frac{3^n}{n!}$$ the answer would be $e^x$ evaluated at $3$.
Yes, this is homework, but I'm not looking for any handouts, any help would be greatly appreciated.
| Take $f(x) = \displaystyle \sum_{n=1}^\infty \frac{x^n}{n}$.
Then,
$f^\prime (x) = \displaystyle \sum_{n=1}^\infty \frac{n x^{n-1}}{n} = \sum_{n=0}^\infty x^n$.
The last expression is a geometric series and, as long as $x < 1$, it can be expressed as
$f^\prime (x) = \displaystyle \frac{1}{1-x}$.
Therefore,
$f(x) = - \ln | 1 - x | + \kappa$
Where $\kappa$ is a constant. But if you take the original expression for $f(x)$, you can see that $f(0) = 0$ and, therefore, $\kappa = 0$.
So $f(x) = -\ln | 1 - x |$.
The answer to your question is just $f \left(\frac{1}{3} \right)$.
You can also obtain this result by Taylor expanding $\ln ( 1 - x )$.
| {
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A group of order $195$ has an element of order $5$ in its center Let $G$ be a group of order $195=3\cdot5\cdot13$. Show that the center of $G$ has an element of order $5$.
There are a few theorems we can use here, but I don't seem to be able to put them together quite right. I want to show that the center of $G$ is divisible by the prime number $5$. If this is the case, then we can apply Cauchy's theorem and we are done.
By Sylow's theorems we get that there are unique $3$-Sylow, $5$-Sylow, and $13$-Sylow subgroups in $G$. Since they are of prime order, they are abelian. Furthermore, their intersection is trivial (by a theorem I beleive). Does this then guarantee that $G=ABC$ and that $G$ is abelian?
| Hint: There are unique, hence normal, $5$- and $13$-Sylows. Their internal direct product is thus normal and has complementary subgroup equal to one of the $3$-Sylows, so $G$ is a semidirect product of $H_5 H_{13}$ and $H_3$, where $H_p$ denotes a $p$-Sylow (not necessarily unique). What can you say about $\varphi: H_3 \to \text{Aut}(H_5 H_{13})$?
| {
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Maxima of bivariate function [1] Is there an easy way to formally prove that,
$$
2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y\ge -x^{4} -y^{4} +2x^{3} +2y^{3} -2x^{2} -2y^{2} +x+y$$
$${0<x,y<1}$$
without resorting to checking partial derivatives of the quotient formed by the two sides, and finding local maxima?
[2] Similarly, is there an easy way for finding $$\max_{0<x,y<1} [f(x,y)]$$
where,
$$f(x,y)=2x(1+x)+2y(1+y)-8xy-4(2xy^{2} +2x^{2} y-2x^{2} y^{2} -4xy+x+y)^{2}$$
| Your first question:
With a little manipulation you get that it is equivalent to
$$x^2((1-x)^2+1)+y^2((1-y)^2+1) \ge 2xy[(1-x)(1-y)+1].$$
This can be obtained from addition of two inequalities
$$x^2(1-x)^2+y^2(1-y)^2 \ge 2xy(1-x)(1-y)$$
$$x^2+y^2\ge 2xy.$$
Both of them are special cases of $a^2+b^2\ge 2ab$, which follows from $(a-b)^2\ge 0$. (Or, if you prefer, you can consider it as a special case of AM-GM inequality.)
Note: To check the algebraic manipulations, you can simply compare the results for
2xy^2 +2x^2 y-2x^2 y^2 -4xy+x+y - ( -x^4 -y^4 +2x^3 +2y^3 -2x^2 -2y^2 +x+y)
expand x^2((1-x)^2+1)+y^2((1-y)^2+1) -2xy[(1-x)(1-y)+1]
Or simply subtract the two expressions:
2xy^2 +2x^2 y-2x^2 y^2 -4xy+x+y - ( -x^4 -y^4 +2x^3 +2y^3 -2x^2 -2y^2 +x+y) - [x^2((1-x)^2+1)+y^2((1-y)^2+1) -2xy[(1-x)(1-y)+1]]
I did not succeed in finding similar type of solution for your second problem.
| {
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formal proof challenge I am desperately trying to figure out the formal proof for this argument.
$$\begin{array}{r}
A\lor B\\
A\lor C\\
\hline
A\lor (B \land C)
\end{array}$$
I am trying to apply the backwards method here. I am trying to infer A, in order to use vIntro in the last step and introduce the final disjunction. But I got stuck finding sufficient proof for A.
Any hint will be greatly appreciated. Thank you!
|
I am trying to apply the backwards method here. I am trying to infer A, in order to use vIntro in the last step and introduce the final disjunction. But I got stuck finding sufficient proof for A.
You don't prove it; you assume it -- To be precise: you assume both cases aiming to derive the same conclusion from each. The disjunction elimination rule then discharges the assumptions. This is also known as the Proof By Cases argument. $$\dfrac{\dfrac{[A]^\star\\~~\vdots}{A\lor(B\land C)}\quad\dfrac{[B]^\star\\~~\vdots}{A\lor(B\land C)}\quad\lower{1.5ex}{A\lor B}~}{A\lor(B\land C)}{\small\lor\mathsf E^\star}$$
So, in this proof you need two $\lor$ eliminations and therefore four assumptions must be made and discharged. The trick is how you combine the assumptions of $B$ and $C$.
$$\dfrac{\dfrac{[A]^1}{A\lor(B\land C)}{\small\lor\mathsf I}\quad\dfrac{\dfrac{[A]^2}{A\lor(B\land C)}{\small\lor\mathsf I}\quad\dfrac{\dfrac{[B]^1\quad[C]^2}{B\land C}{\small\land\mathsf I}}{A\lor(B\land C)}{\small\lor\mathsf I}\quad\lower{1.5ex}{A\lor C}}{A\lor (B\land C)}{\small\lor\mathsf E^2}\quad\lower{1.5ex}{A\lor B}}{A\lor (B\land C)}{\small\lor\mathsf E^1}$$
| {
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Is my trig result unique? I recently determined that for all integers $a$ and $b$ such that $a\neq b$ and $b\neq 0$,
$$
\arctan\left(\frac{a}{b}\right) + \frac{\pi}{4} = \arctan\left(\frac{b+a}{b-a}\right)
$$
This implies that 45 degrees away from any angle with a rational value for tangent lies another angle with a rational value for tangent. The tangent values are related.
If anyone can let me know if this has been done/shown/proven before, please let me know. Thanks!
| If you differentiate the function $$f(t)=\arctan t - \arctan\frac{1 + t}{1 - t},$$ you get zero, so the function is constant in each of the two intervals $(-\infty,1)$ and $(1,+\infty)$ on which it is defined.
*
*Its value at zero is $\pi/2$, so that $f(t)=-\pi/4$ for all $t<1$, so
$$ \arctan t + \frac\pi4 = \arctan\frac{1 + t}{1 - t},\qquad\forall t<1.$$
*On the other hand, one easily shows that $\lim_{t\to+\infty}f(t)=\frac{3\pi}{4}$, so
$$ \arctan t - \frac{3\pi}4 = \arctan\frac{1 + t}{1 - t},\qquad\forall t>1.$$
If $t=a/b$ is a rational number smaller that $1$, then the first point is your identity. If it larger than $1$, we see that you have to change things a bit.
| {
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Prove that $\mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^2$ is isomorphic to $\{\pm1\}$. Prove that $\mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^2$ is isomorphic to $\{\pm1\}$, where $p$ is a prime integer.
| I take it that you mean to prove that $\mathbb{F}_p^\times/(\mathbb{F}_p^\times)^2 \cong \{\pm 1\}$, where $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$.
If so, use the fact that the map $(\mathbb{Z}/p\mathbb{Z})^\times \to \{\pm 1\}$ given by $a\bmod p\mapsto (\frac{a}{p})$ is a homomorphism of groups, where $(\frac{a}{p})$ is the Legendre symbol of $a$ over $p$.
| {
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Proving the Cantor Pairing Function Bijective How would you prove the Cantor Pairing Function bijective? I only know how to prove a bijection by showing (1) If $f(x) = f(y)$, then $x=y$ and (2) There exists an $x$ such that $f(x) = y$
How would you show that for a function like the Cantor pairing function?
| I will denote the pairing function by $f$. We will show that pairs $(x,y)$ with a particular value of the sum $x+y$ is mapped bijectively to a certain interval, and then that the intervals for different value of the sum do not overlap, and that their union is everything.
Let $m$ be a natural number and suppose $m=x+y$. The least value that $f(x,y)$ can take is $\frac{m(m+1)}{2}$ (if $x=m$) and the largest value it can take is $\frac{m(m+1)}{2}+m$ (if $y=m$). It can also take all values in between. It is thus easy to see that the $m+1$ pairs $(x,y)$ with sum $m$ are mapped bijectively to an interval.
If $x+y=m+1$ then the least possible value of $f(x,y)$ is $\frac{(m+1)(m+2)}{2}$. We can check that $\frac{(m+1)(m+2)}{2} - (\frac{m(m+1)}{2}+m)=1$ so the intervals for the different value of the sum do not overlap and it is easy to see that their union is $\mathbb{N}$.
| {
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Characteristic equation of a recurrence relation Yesterday there was a question here based on solving a recurrence relation and then when I tried to google for methods of solving recurrence relations, I found this, which gave different ways of solving simple recurrence relations.
My question is how do you justify writing the recurrence relation in its characteristic equation form and then solving for its roots to get the required answer.
For example, Fibonacci relation has a characteristic equation $s^2-s-1=0$.
How can we write it as that polynomial?
| The characteristic equation is the one that a number $\lambda$ should satisfy in order for the geometric series $(\lambda^n)_{n\in\mathbf N}$ to be a solution of the recurrence relation. Another interpretation is that if you interpret the indeterminate $s$ as a left-shift of the sequence (dropping the initial term and renumbering the renaming terms one index lower), then the characteristic equation gives the lowest degree monic polynomial that when applied to this shift operation kills all sequences satisfying the recurrence. In the case of Fibonacci recurrence, applying $s^2-s-1$ to a sequence $A=(a_i)_{i\in\mathbf N}$ gives the sequence $(a_{i+2}-a_{i+1}-a_i)_{i\in\mathbf N}$, which is by definition identically zero if (and only if) $A$ satisfies the recurrence.
A different but related polynomial that is of interest is obtained by reversing the order of the monomials (giving a polynomial starting with constant term $1$), so for Fibonacci it would be $1-X-X^2$. This polynomial $P$ has the property that the formal power series $F=\sum_{i\in\mathbf N}a_iX^i$ associated to a sequence satisfying the recurrence, when multiplied by $P$ gives a polynomial $R$ (with $\deg R<\deg P$), in other words all terms from the index $\deg P$ on are killed. This is basically the same observation as for the shift operation, but the polynomial $R$ permits describing the power series of the sequence, including its initial values, formally as $F=\frac RP$. For the Fibonacci sequence one finds $R=X$ so its formal power series is $F=\frac X{1-X-X^2}$.
| {
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How to find all rational points on the elliptic curves like $y^2=x^3-2$ Reading the book by Diophantus, one may be led to consider the curves like:
$y^2=x^3+1$, $y^2=x^3-1$, $y^2=x^3-2$,
the first two of which are easy (after calculating some eight curves to be solved under some certain conditions, one can directly derive the ranks) to be solved, while the last , although simple enough to be solved by some elementary consideration of factorization of algebraic integers, is at present beyond my ability, as my knowledge about the topic is so far limited to some reading of the book Rational Points On Elliptic Curves, by Silverman and Tate, where he did not investigate the case where the polynomial has no visible rational points.
By the theorem of Mordell, one can determine its structure of rational points, if the rank is at hand. So, according to my imagination, if some hints about how to compute ranks of elliptic curves of this kind were offered, it would certainly be appreciated.
Thanks in advance.
| Given your interest in Mordell's equation, you really ought to buy or borrow Diophantine Equations by Mordell, then the second edition of A Course in Number Theory by H. E. Rose, see AMAZON
Rose discusses the equation starting on page 286, then gives a table of $k$ with
$ -50 \leq k \leq 50$ for which there are integral solutions, a second table for which there are rational solutions. The tables are copied from J. W. S. Cassels, The rational solutions of the diophantine equation $y^2 = x^3 - D.$ Acta Arithmetica, volume 82 (1950) pages 243-273.
Other than that, you are going to need to study Silverman and Tate far more carefully than you have done so far. From what I can see, all necessary machinery is present. Still, check the four pages in the Bibliography, maybe you will prefer something else.
| {
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Generalization of manifold Is there a generalization of the concept of manifold that captures the following idea:
Consider a sphere that instead of being made of a smooth material is actually made up of a mesh of thin wire. Now for certain beings living on the sphere the world appears flat and 2D, unware that they are actually living on a mesh, but for certain other smaller beings, the world appears to be 1D most of the time (because of the wire mesh).
| One thing to look at is foliations (and laminations), which are decompositions of manifolds into lower-dimension manifolds. While there is no "mesh" because each lower-dimension manifold has another lower-dimension manifold in any neighborhood, there is still a lower-dimensionality that is something like what you seek. (When you're looking at surfaces in a $3$-manifold, you can also look at the one-dimensional transversals.) See, e.g., H. B. Lawson, Foliations, Bulletin of the AMS 80:3 (1974), 369–418, MR 0343289 (49 #8031).
| {
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Find the ordinary generating function $h(z)$ for a Gambler's Ruin variation. Assume we have a random walk starting at 1 with probability of moving left one space $q$, moving right one space $p$, and staying in the same place $r=1-p-q$. Let $T$ be the number of steps to reach 0. Find $h(z)$, the ordinary generating function.
My idea was to break $T$ into two variables $Y_1,Y_2$ where $Y_1$ describes the number of times you stay in place and $Y_2$ the number of times we move forward or backward one. Then try to find a formula for $P(T=n)=P(Y_1+Y_2=n)=r_n$, but I'm getting really confused since there are multiple probabilities possible for each $T=n$ for $n\geq 3$. Once I have $r_n$ I can then use $h_T(z)=\sum_{n=1}^\infty r_n z^n$, but I'm not sure where to go from here.
| A classical way to determine $h(z)$ is to compute $h_n(z)=\mathrm E_n(z^T)$ for every $n\geqslant0$, where $\mathrm E_n$ denotes the expectation starting from $n$, hence $h(z)=h_1(z)$.
Then $h_0(z)=1$ and, considering the first step of the random walk, one gets, for every $n\geqslant1$,
$$
h_n(z)=rzh_n(z)+pzh_{n+1}(z)+qzh_{n-1}(z),
$$
with $r=1-p-q$. Fix $z$ in $(0,1)$. Then the sequence $(x_n)_n=(h_n(z))_n$ satisfies the relations $x_0=1$, $x_n\to0$ when $n\to\infty$, and $ax_{n+1}-bx_n+cx_{n-1}=0$ for every $n\geqslant1$, for some positive $(a,b,c)$.
Hence $x_n=\alpha s_z^n+(1-\alpha)t_z^n$, where $s_z$ and $t_z$ are the roots of the polynomial $Q_z(t)=at^2-bt+c$. Since $a=pz$, $b=1-rz$ and $c=qz$, one sees that $Q_z(0)=qz\gt0$, $Q_z'(0)=-(1-rz)\lt0$ and $Q_z(1)=-(1-z)\lt0$. Thus, $0\lt s_z\lt1\lt t_z$. If $\alpha\ne1$, $|x_n|\to\infty$ when $n\to\infty$. But $(x_n)_n$ should stay bounded hence this shows that $\alpha=1$. Finally, $x_n=s_z^n$ for every $n\geqslant0$, where $Q_z(s_z)=0$ and $0\lt s_z\lt 1$.
In particular, for every $z$ in $(0,1)$, $h(z)=s_z$, that is,
$$
h(z)=\frac{1-rz-\sqrt{(1-rz)^2-4pqz^2}}{2pz}.
$$
Note that the limit of $h(z)$ when $z\to1^-$ is the probability $\mathrm P_1(T\lt\infty)$, which is $1$ if $p\leqslant q$ and $q/p\lt1$ if $p\gt q$.
The technique above is flexible enough to be valid for any random walk. If the steps are $i$ with probability $p_i$, the recursion becomes
$$
h_n(z)=z\sum\limits_ip_ih_{n+i}(z).
$$
The case at hand is $p_{-1}=q$, $p_0=r$ and $p_1=p$. When $p_{-1}$ is the only nonzero $p_i$ with $i$ negative, a shortcut is to note that one can only go from $n\geqslant1$ to $0$ by first reaching $n-1$, then reaching $n-2$ from $n-1$, and so on until one reaches $0$ starting from $1$. These $n$ hitting times are i.i.d. hence $h_n(z)=h(z)^n$ for every $n\geqslant1$, and one is left with
$$
h(z)=zp_{-1}+z\sum\limits_{i\geqslant0}p_ih(z)^{i+1}.
$$
In the present case, this reads
$$
h(z)=qz+rzh(z)+pzh(z)^2,
$$
hence the expression of $h(z)=s_z$ as solving a quadratic equation should not be surprising.
| {
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Free products of cyclic groups Given $G$, $H$, $G'$, and $H'$ are cyclic groups of orders $m$, $n$, $m'$, and $n'$ respectively.
If $G*H$ is isomorphic to $G'* H'$, I would like to show that either $m = m'$ and $n = n'$ or else $m = n'$ and $n = m'$ holds. Where * denotes the free product.
My approach:
$G*H$ has an element of order $n$, thus $G' * H'$ has one too.
But already the next step is not clear to me, should I show that there is an element of length $> 1$ which has infinite order or what would be the right approach here?
Thank you.
| Let me give you an alternative argument for the claim $m=m'$ in Arturo Magidin's answer.
Take the abelianizations of the groups $G*H$ and $G'*H'$, since $G,G,H,H'$ are abelian, their abelianizations are $G\oplus H$ and $G'\oplus H'$ respectively. Then you get $G\oplus H\cong G'\oplus H'$ and in particular, their orders $mn=m'n'$ are equal. Since we already had $n=n'$, we conclude that $m=m'$.
| {
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Proving $\mathbb{N}^k$ is countable Prove that $\mathbb{N}^k$ is countable for every $k \in \mathbb{N}$.
I am told that we can go about this inductively.
Let $P(n)$ be the statement: “$\mathbb{N}^n$ is countable” $\forall n \in \mathbb{N}$.
Base Case: $\mathbb{N}^1 = \mathbb{N}$ is countable by definition, so $\checkmark$
Inductive Step: $\mathbb{N}^{k+1}$ $“=”$ $\mathbb{N}^k \times \mathbb{N}$
We know that $(A,B)$ countable $\implies$ $A \times B$ is countable. I am stuck on the part where I have to prove the rest, but I know that, for example, $(1,2,7) \in \mathbb{N}^3 \notin \mathbb{N}^2 \times \mathbb{N}$ but instead $((1,2),7) \in \mathbb{N}^2 \times \mathbb{N}$. So how would I go about proving the statement.
| The function $f:\mathbb{N^K}\to \mathbb{N^K\times \{m \}}$ defined by $$f(a_1,a_2,\cdots, a_k)=(a_1,a_2,\cdots, a_k,m)$$ is clearly a bijection for fixed $m\in \mathbb{N}$ and we can write $\mathbb{N^{K+1}}$ as $$\mathbb{N^{K+1}}=\bigcup_{m=1}^{\infty}\{\mathbb{N^K\times \{m \}}\}$$ and this being a countable union of countable sets is countable.
| {
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Define when $y$ is a function of $x$ Hello guys I want to make sure myself in determming when is $y$ as function of $x$,so for this, let us consider following question. If the equation of circle is given by this
$$x^2+y^2=25$$
and question is find the equation of tangent of circle at point $(3,4)$,then it is clear that we need calculate the derivative and also we can express $ y$ as
$$y=\sqrt{25-x^2}$$
but, let us consider following situation
$$x^3+y^3=6\cdot x \cdot y$$
it's name is The Folium of Descartes and my question is the same : calculate the equation of the tangent of this folium at point $(3,4)$.
I know that somehow expressing $y$ as a function of $x$ is difficult but could't we do it by some even long mathematical manipulation? So what is the strict explanation when is $y$ a function of $x$ and when is it not? I need to be educated in this topic and please explain me ways of this,thanks a lot.
| The equation $$
\tag{1}x^3+y^3=6xy
$$does define $y$ as a function of $x$ locally (or, rather, it defines $y$ as a function of $x$ implicitly). Here, it is difficult to write the defining equation as $y$ in terms of $x$. But, you don't have to do that to evaluate the value of the derivative of $y$.
[edit] The point $(3,4)$ does not satisfy equation (1); so there is no tangent line at this point.
Let's, instead, consider the point $(3,3)$, which does satisfy equation (1):
To find the slope of the tangent line at $(3,3)$, you need to find $y'(3)$. To find this, first
implicitly differentiate both sides of the defining equation for $y$
(equation (1)). This gives
$$
{d\over dx} (x^3+y^3)={d\over dx} 6xy
$$
So, using the chain and product rules:
$$
3x^2+3y^2 y' =6y+6x y'.
$$
When $x=3$ and $y=3$, you have
$$
3\cdot 3^2+3\cdot 3^2 y'(3)=6\cdot 3+6\cdot 3\cdot y'(3).
$$
Solve this for $y'(3)={3\cdot 3^2-6\cdot 3\over6\cdot3 -3\cdot3^2}=-1.$
Now you can find the equation of the tangent line since you know the slope and that the point $(3,3)$ is on the line.
Generally any "nice" equation in the variables $x$ and $y$ will define $y$ as a function of $x$ in some neighborhood of a given point. Given the $x$ value, the corresponding $y$ value is the solution to the equation ("the" solution in a, perhaps small, neighborhood of the point).
Of course, sometimes it is extremely difficult (if not impossible) to find an explicit form of the function; that is, of the form $y=\Phi(x)$ for some expression $\Phi(x)$. In these cases, to find the derivative of $y$, you have to use the approach above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of irrational numbers Well, in this question it is said that $\sqrt[100]{\sqrt3 + \sqrt2} + \sqrt[100]{\sqrt3 - \sqrt2}$, and the owner asks for "alternative proofs" which do not use rational root theorem. I wrote an answer, but I just proved $\sqrt[100]{\sqrt3 + \sqrt2} \notin \mathbb{Q}$ and $\sqrt[100]{\sqrt3 - \sqrt2} \notin \mathbb{Q}$, not the sum of them. I got (fairly) downvoted, because I didn't notice that the sum of two irrational can be either rational or irrational, and I deleted my (incorrect) answer. So, I want help in proving things like $\sqrt5 + \sqrt7 \notin \mathbb{Q}$, and $(1 + \pi) - \pi \in \mathbb{Q}$, if there is any "trick" or rule to these cases of summing two (or more) known irrational numbers (without rational root theorem).
Thanks.
| Here is a useful trick, though it requires a tiny bit of field theory to understand: If $\alpha + \beta$ is a rational number, then $\mathbb{Q}(\alpha) = \mathbb{Q}(\beta)$ as fields. In particular, if $\alpha$ and $\beta$ are algebraic, then the degrees of their minimal polynomials are equal.
So, for example, we can see at a glance that $\sqrt{5} + \sqrt[3]{7}$ is irrational, because $\sqrt{5}$ and $\sqrt[3]{7}$ have algebraic degree $2$ and $3$ respectively.
Note that this trick doesn't work on your original example, because $\alpha=\sqrt[100]{\sqrt{3} + \sqrt{2}}$ and $\beta=\sqrt[100]{\sqrt{3} - \sqrt{2}}$ do have the same degree. But we can also use field theory: since $\alpha \beta = 1$, if $\alpha+\beta$ is rational then $\alpha$ and $\beta$ satisfy a rational quadratic. However, $\alpha^{100}= \sqrt{3}+\sqrt{2}$ already has degree $4$ over $\mathbb{Q}$, so $\alpha$ certainly has degree bigger than $2$.
| {
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Heads or tails probability I'm working on a maths exercise and came across this question.
The probability of a "heads" when throwing a coin twice is 2 / 3. This could be explained by the following:
• The first time is "heads". The second throw is unnecessary. The result is H;
• The first time is "tails" and twice "heads". The result is TH;
• The first time is "tails" and twice "tails". The result is TT;
The outcome: {H, TH, TT}. two of the three results include a "heads", it follows that the probability of a "heads" is 2/3
What's wrong with this reasoning?
I think the answer is 1/2, is that right?
Ps. my first language isn't english,
Thanks Jef
| The reason your result, as Shitikanth has already pointed out, is wrong, is that you've applied the principle of indifference where it doesn't apply. You can only assume that events will all be equally likely if they're all qualitatively the same and there's nothing (other than names and labels) to distinguish them from each other. The prototypical examples are the two sides of a coin and the six sides of a six-sided die.
In your case, on the other hand, the event H is qualitatively different from the two events TH and TT, so there's no reason to expect that these three will be equiprobable, and the principle of indifference doesn't apply. To apply it, you need to look at qualitatively similar events. In this case, that would be HH, HT, TH and TT. Of these four, three contain a heads, so, as Shitikanth has already stated, the probability is $3/4$.
| {
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Finding angles in a parallelogram without trigonometry
I'm wondering whether it's possible to solve for $x^{\circ}$ in terms of $a^{\circ}$ and $b^{\circ}$ given that $ABCD$ is a parallelogram. In particular, I'm wondering if it's possible to solve it using only "elementary geometry". I'm not sure what "elementary geometry" would usually imply, but I'm trying to solve this problem without trigonometry.
Is it possible? Or if it's not, is there a way to show that it isn't solvable with only "elementary geometry" techniques?
| The example by alex.jordan does finish the matter, and similar ones may be constructed. We have an angle
$$ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $$
and we wish to know whether $ x = \frac{\theta}{\pi} $ is the root of an equation with rational coefficients.
Well,
$$ e^{i \theta} = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $$
Next, $\cos 2 \theta = 2 \cos^2 \theta - 1 = \frac{11}{13}.$ So, by Corollary 3.12 on page 41 of NIVEN we know that $2 \theta$ is not a rational multiple of $\pi.$ So, neither is $\theta,$ and
$$ x = \frac{\theta}{\pi} $$
is irrational.
Now, the logarithm is multivalued in the complex plane. We may choose
$$ \log(-1) = \pi i. $$ With real $x,$ we have chosen
$$ (-1)^x = \exp(x \log(-1)) = \exp(x\pi i) = \cos \pi x + i \sin \pi x. $$
With our $ x = \frac{\theta}{\pi}, $ we have
$$ (-1)^x = e^{i \pi x} = e^{i \theta} = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $$
The right hand side is algebraic.
The Gelfond-Schneider Theorem, Niven page 134, says that if $\alpha,\beta$ are nonzero algebraic numbers, with $\alpha \neq 1$ and $\beta$ not a real rational number, then any value of $\alpha^\beta$ is transcendental.
Taking $\alpha = -1$ and $\beta = x,$ which is real but irrational. We are ASSUMING that $x$ is algebraic over $\mathbb Q.$ The assumption, together with Gelfond-Schneider, says that $ (-1)^x$ is transcendental. However, we already know that $ (-1)^x = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $ is algebraic. This contradicts the assumption. So $x = \theta / \pi$ is transcendental, with $ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $
| {
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Adding a different constant to numerator and denominator Suppose that $a$ is less than $b$ , $c$ is less than $d$.
What is the relation between $\dfrac{a}{b}$ and $\dfrac{a+c}{b+d}$? Is $\dfrac{a}{b}$ less than, greater than or equal to $\dfrac{a+c}{b+d}$?
| One nice thing to notice is that
$$
\frac{a}{b}=\frac{c}{d} \Leftrightarrow \frac{a}{b}=\frac{a+c}{b+d}
$$
no matter the values of $a$, $b$, $c$ and $d$. The $(\Rightarrow)$ is because $c=xa, d=xb$ for some $x$, so $\frac{a+c}{b+d}=\frac{a+xa}{b+xb}=\frac{a(1+x)}{b(1+x)}=\frac{a}{b}$. The other direction is similar.
The above is pretty easy to remember, and with that intuition in mind it is not hard to imagine that
$$
\frac{a}{b}<\frac{c}{d} \Leftrightarrow \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}
$$
and similar results.
| {
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Tricky Factorization How do I factor this expression: $$ 0.09e^{2t} + 0.24e^{-t} + 0.34 + 0.24e^t + 0.09e^{-2t} ? $$
By trial and error I got $$ \left(0.3e^t + 0.4 + 0.3 e^{-t}\right)^2$$ but I'd like to know how to formally arrive at it.
Thanks.
| The most striking thing about the given expression is the symmetry. For anything with that kind of symmetric structure, there is a systematic approach which is definitely not trial and error.
Let
$$z=e^t+e^{-t}.$$
Square. We obtain
$$z^2=e^{2t}+2+e^{-2t},$$
and therefore $e^{2t}+e^{-2t}=z^2-2$.
Substitute in our expression. We get
$$0.09(z^2-2)+0.24 z+0.34.\qquad\qquad(\ast)$$
This simplifies to
$$0.09z^2 +0.24z +0.16.$$
The factorization is now obvious. We recognize that we have simply $(0.3z+0.4)^2$. Now replace $z$ by $e^t+e^{-t}$.
If the numbers had been a little different (but still with the basic $e^t$, $e^{-t}$ symmetry) we would at the stage $(\ast)$ obtain some other quadratic. In general, quadratics with real roots can be factored as a product of linear terms. It is just a matter of completing the square. For example, replace the constant term $0.34$ by, say, $0.5$. We get a quadratic in $z$ that does not factor as prettily, but it does factor.
Comment: For fun we could instead make the closely related substitution $2y=e^t+e^{-t}$, that is, $y=\cosh t$. If we analyze the substitution process further, we get to useful pieces of mathematics, such as the Chebyshev polynomials.
The same idea is the standard approach to finding the roots of palindromic polynomials. For example, suppose that we want to solve the equation $x^4 +3x^3-10x^2+3x+1=0$. Divide through by $x^2$. We get the equation
$$x^2+3x-10+\frac{3}{x}+\frac{1}{x^2}=0.$$
Make the substitution $z=x+\frac{1}{x}$. Then $x^2+\frac{1}{x^2}=z^2-2$. Substitute. We get a quadratic in $z$.
| {
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$f$ uniformly continuous and $\int_a^\infty f(x)\,dx$ converges imply $\lim_{x \to \infty} f(x) = 0$ Trying to solve
$f(x)$ is uniformly continuous in the range of $[0, +\infty)$ and $\int_a^\infty f(x)dx $ converges.
I need to prove that:
$$\lim \limits_{x \to \infty} f(x) = 0$$
Would appreciate your help!
| Suppose $$\tag{1}\lim\limits_{x\rightarrow\infty}f(x)\ne 0.$$
Then we may, and do, select an $\alpha>0$ and a sequence $\{x_n\}$ so that for any $n$, $$\tag{2}x_n\ge x_{n-1}+1$$
and
$$\tag{3}|f(x_n)|>\alpha.$$
Now, since $f$ is uniformly continuous, there is a $1>\delta>0$ so that
$$\tag{4}|f(x)-f(y)|<\alpha/2,\quad\text{ whenever }\quad |x-y|<\delta.$$
Consider the contribution to the integral of the intervals $I_n=[x_n-\delta/2,x_n+\delta/2]$:
We have, by (3), and (4) that
$$\biggl|\,\int_{I_n} f(x)\, dx\,\biggr|\ge {\alpha\over2}\cdot \delta$$ for each positive integer $n$.
But, by (2), the $x_n$ tend to infinity. This implies that $\int_a^\infty f(x)\,dx$ diverges, a contradiction.
Having obtained a contradiction, our initial assumption, (1), must be incorrect. Thus, we must have $\lim\limits_{x\rightarrow\infty}f(x)= 0 $.
Take this with a grain of salt, but,
informally, the idea used above is based on the following:
For clarity, assume $f>0$, here.
If the integral $\int_a^\infty f(x)\,dx$ is convergent, then for large $x$, the graph of $f$ is close to the $x$-axis "most of the time" and, in fact, the positive $x$-axis is an asymptote of "most of" the graph of $f$.
I say "most of the time" and "most of" because is not necessarily so that a function $f$ which is merely continuous must tend to 0 when $\int_a^\infty f(x)\,dx$ converges. There may be spikes in the graph of $f$ as you go out in the positive $x$ direction. Though the height of the spikes can be large, the width of the spikes would be small enough so that the integral
converges (so, the sum of the areas of the spikes is finite).
But the graph of a uniformly continuous function that is "mostly asymptotic to the $x$-axis" does not have very tall spikes of very short widths arbitrarily far out in the $x$-axis.
| {
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Simplifying trig expression I was working through some trig exercises when I stumbled upon the following problem:
Prove that: $ \cos(A+B) \cdot \cos(A-B)=\cos^2A- \sin^2B$.
I started out by expanding it such that
$$ \cos(A+B) \cdot \cos(A-B)=(\cos A \cos B-\sin A \sin B) \cdot (\cos A \cos B+ \sin A \sin B),$$
which simplifies to:
$$ \cos^2 A \cos^2 B- \sin^2 A \sin^2 B .$$
However, I don't know how to proceed from here. Does anyone have any suggestions on how to continue.
| The identities
$$\cos(\theta) = \frac{e^{i \theta}+e^{- i \theta}}{2}$$
$$\sin(\theta) = \frac{e^{i \theta}-e^{- i \theta}}{2i}$$
can reduce a trigonometric identity to a identity of polynomials. Let's see how this works in your example:
$$\cos(A+B) \cos(A-B)=\cos(A)^2-\sin(B)^2$$
is rewritten into:
$$\frac{e^{i (A+B)}+e^{- i (A+B)}}{2} \frac{e^{i (A-B)}+e^{- i (A-B)}}{2}=\left(\frac{e^{i A}+e^{- i A}}{2}\right)^2-\left(\frac{e^{i B}-e^{- i B}}{2i}\right)^2$$
now we change it into a rational function of $X = e^{i A}$ and $Y = e^{i B}$:
$$\frac{(X Y + \frac{1}{X Y})(\frac{X}{Y} + \frac{Y}{X})}{4} = \frac{(X + \frac{1}{X})^2 + (Y - \frac{1}{Y})^2}{4}$$
and you can simply multiply out both sides to see that they are both $\frac{1}{4}\left(X^2 + \frac{1}{X^2} + Y^2 + \frac{1}{Y^2}\right)$ which proves the trigonometric equality.
| {
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Quadratic forms and prime numbers in the sieve of Atkin I'm studying the theorems used in the paper which explains how the sieve of Atkin works, but I cannot understand a point.
For example, in the paper linked above, theorem 6.2 on page 1028 says that if $n$ is prime then the cardinality of the set which contains all the norm-$n$ ideals in $\mathbf Z[(-1+\sqrt{-3})/2]$ is 2. I don't understand why, and I am not able to relate this result to the quadratic form $3x^2+y^2=n$ used in the proof.
| The main thing is that the norm of $s + t \omega$ is $s^2 + s t + t^2,$ which is a binary form that represents exactly the same numbers as $3x^2 + y^2.$
It is always true that, for an integer $k,$ the form $s^2 + s t + k t^2$ represents a superset of the numbers represented by $x^2 + (4k-1)y^2.$ For instance, with $k=2,$ the form $x^2 + 7 y^2$ does not represent any numbers $2\pmod 4,$ otherwise it and $s^2 + s t + 2 t^2$ agree.
With $k=-1,$ it turns out that $x^2 - 5 y^2$ and $s^2 + s t - t^2$ represent exactly the same integers.
Take $s^2 + s t + k t^2$ with $s = x - y, \; t = 2 y.$ You get
$$ (x-y)^2 + (x-y)(2y) + k (2y)^2 = x^2 - 2 x y + y^2 + 2 x y - 2 y^2 + 4 k y^2 = x^2 + (4k-1) y^2.$$
| {
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Finding the limit of a sequence $\lim _{n\to \infty} \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right)$ If there were a regular square root I would multiply the top by its adjacent and divide, but I've tried that with this problem and it doesn't work. Not sure what else to do have been stuck on it.
$$ \lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}-
\sqrt [3]{n} \right) .$$
| $$
\begin{align*}
\lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}-
\sqrt [3]{n} \right)
&= \lim _{n\to \infty } \sqrt [3]{n^2} \cdot \sqrt[3]{n} \left( \sqrt [3]{1+ \frac{1}{n}}-
1 \right)
\\ &= \lim _{n\to \infty } n \left( \sqrt [3]{1+ \frac{1}{n}}-
1 \right)
\\ &= \lim _{n\to \infty } \frac{\sqrt [3]{1+ \frac{1}{n}}-
1 }{\frac{1}{n}}
\\ &= \lim _{h \to 0} \frac{\sqrt [3]{1+ h}-
1 }{h}
\\ &= \left. \frac{d}{du} \sqrt[3]{u} \ \right|_{u=1}
\\ &= \cdots
\end{align*}
$$
| {
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Expected value of $XYZ$, $E(XYZ)$, is not always a $E(X)E(Y)E(Z)$, even if $X$, $Y$, $Z$ are not correlated in pairs Could you prompt me, please, is it true?
Expected value of $XYZ$, $E(XYZ)$, is not always $E(X)E(Y)E(Z)$, even if $X$, $Y$, $Z$ are not correlated in pairs, because if $X$, $Y$, $Z$ are not correlated in pairs it doesn't entail that they are uncorrelated in aggregate (it is my idea)?
| Suppose
$$
(X,Y,Z) = \begin{cases}
(1,1,0) & \text{with probability }1/4 \\
(1,0,1) & \text{with probability }1/4 \\
(0,1,1) & \text{with probability }1/4 \\
(0,0,0) & \text{with probability }1/4
\end{cases}
$$
Then $X,Y,Z$ are pairwise independent, and $E(X)E(Y)E(Z)=1/8\ne 0 = E(XYZ)$.
| {
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Proofs for an equality I was working on a little problem and came up with a nice little equality which I am not sure if it is well-known (or) easy to prove (It might end up to be a very trivial one!). I am curious about other ways to prove the equality and hence I thought I would ask here to see if anybody knows any or can think of any. I shall hold off from posting my own answer for a couple of days to invite different possible solutions.
Consider the sequence of functions:
$$
\begin{align}
g_{n+2}(x) & = g_{n}(x) - \left \lfloor \frac{g_n(x)}{g_{n+1}(x)} \right \rfloor g_{n+1}(x)
\end{align}
$$
where $x \in [0,1]$ and $g_0(x) = 1, g_1(x) = x$. Then the claim is:
$$x = \sum_{n=0}^{\infty} \left \lfloor \frac{g_n}{g_{n+1}} \right \rfloor g_{n+1}^2$$
| For whatever it is worth, below is an explanation on why I was interested in this equality. Consider a rectangle of size $x \times 1$, where $x < 1$. I was interested in covering this rectangle with squares of maximum size whenever possible (i.e. in a greedy sense).
To start off, we can have $\displaystyle \left \lfloor \frac{1}{x} \right \rfloor$ squares of size $x \times x$. Area covered by these squares is $\displaystyle \left \lfloor \frac{1}{x} \right \rfloor x^2$.
Now we will then be left with a rectangle of size $\left(1 - \left \lfloor \frac1x \right \rfloor x \right) \times x$.
We can now cover this rectangle with squares of size $\left(1 - \left \lfloor \frac1x \right \rfloor x \right) \times \left(1 - \left \lfloor \frac1x \right \rfloor x \right)$.
The number of such squares possible is $\displaystyle \left \lfloor \frac{x}{\left(1 - \left \lfloor \frac1x \right \rfloor x \right)} \right \rfloor$.
The area covered by these squares is now $\displaystyle \left \lfloor \frac{x}{\left(1 - \left \lfloor \frac1x \right \rfloor x \right)} \right \rfloor \left(1 - \left \lfloor \frac1x \right \rfloor x \right)^2$.
And so on.
Hence, at $n^{th}$ stage if the sides are given by $g_{n-1}(x)$ and $g_n(x)$ with $g_n(x) \leq g_{n-1}(x)$, the number of squares with side $g_{n}(x)$ which can be placed in the rectangle of size $g_{n-1}(x) \times g_n(x)$, is given by $\displaystyle \left \lfloor \frac{g_{n-1}(x)}{g_{n}(x)} \right \rfloor$.
These squares cover an area of $\displaystyle \left \lfloor \frac{g_{n-1}(x)}{g_{n}(x)} \right \rfloor g^2_{n}(x)$.
Hence, at the $n^{th}$ stage using squares we cover an area of $\displaystyle \left \lfloor \frac{g_{n-1}(x)}{g_{n}(x)} \right \rfloor g^2_{n}(x)$.
The rectangle at the $(n+1)^{th}$ stage is then given by $g_{n}(x) \times g_{n+1}(x)$ where $g_{n+1}(x)$ is given by $g_{n-1}(x) - \left \lfloor \frac{g_{n-1}(x)}{g_n(x)} \right \rfloor g_n(x)$.
These squares end up covering the entire rectangle and hence the area of all these squares equals the area of the rectangle.
This hence gives us $$x = x \times 1 = \sum_{n=1}^{\infty} \left \lfloor \frac{g_{n-1}(x)}{g_{n}(x)} \right \rfloor g^2_{n}(x)$$
When I posted this question, I failed to see the simple proof which Srivatsan had.
| {
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Deriving SDE(s) and Expectation from Given PDE We want to solve the PDE $u_t + \left( \frac{x^2 + y^2}{2}\right)u_{xx} + (x-y^2)u_y + ryu = 0 $ where $r$ is some constant and $u(x,y,T) = V(x,y)$ is given. Write an SDE and express $u(x,y,0)$ as the expectation of some function of the path $X_t, Y_t$.
Attempt: I tried to use the multivariate backward equation (2 dimensional) to recover the original SDE's and ended up with $dX_t= \sqrt{x^2 + y^2} dW_t$ and $dY_t = (x-y^2)dt + \sqrt{x^2 + y^2} dW_t$.
The problem I have is recovering the expectation. I'm not too familiar with multidimensional Feynman-Kac, but judging by the $ryu$ term and extrapolating from the one-dimensional case, the desired expectation should have the form E[exp(riemann integral of Y_t)]. Can anyone shed some light on this? Thank you.
EDIT: Oops, wrote the forward equation incorrectly and made a typo, the SDE's have changed
| What do you think about the system of SDEs :
$$dX_t=\sqrt{X_t^2+Y_t^2}dW_t$$
$$dY_t=(X_t-Y_t^2)dt$$
And finally :
$$u(X_t,Y_t,t)=\mathbb{E}[V(X_T,Y_T).e^{-\int_t^TrY_s.ds}|X_t,Y_t]$$
You can check that $u$ is satisfying your PDE, but as always check my calculations as I am used to making errors.
The way I found this is the following :
I set $r=0$, then looking for $u$ as an expectation of $V(X_T,Y_T)$ and deriving its SDE via Itô's lemma and looking for a null drift and then identifying terms with the original PDE with those coming from the drift of $dV$ with $dX_t=a_1(X,Y,t)dt+b_1(X,Y,t)dW_t$ and $dX_t=a_2(X,Y,t)dt+b_2(X,Y,t)dB_t$ gives the solution for $a_1,a_2,b_1,b_2$ when $r=0$ ($B$ and $W$ are independent Brownian motions, which is coming from the intuitive fact that there is no $u_{xy}$ terms in the PDE).
Then two minutes of reflection gives that $F(X_T,Y_T,T)=V(X_T,Y_T).e^{-\int_T^Tr.Y_sds}$ respects the final condition and acts on the drift part of $dF$ by only multiplying the PDE's with a $e^{-\int_t^Tr.Y_s.ds}$ and adds the $rYV$ term which was missing in the solution with $r=0$.
Best regards
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given a matrix, is there always another matrix which commutes with it? Given a matrix $A$ over a field $F$, does there always exist a matrix $B$ such that $AB = BA$? (except the trivial case and the polynomial ring?)
| Another example is the adjoint of $A$:
$$
A \operatorname{adj}(A)= \operatorname{adj}(A) A = \det(A)I
$$
(but for invertible matrices it is equal to the scalar $\det(A)$ multipliying the inverse of $A$, so is trivial that commutes. with $A$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A solvable Lie-algebra of derived length 2 and nilpotency class $n$ Given a natural $n>2$, I want to show that there exists a lie algebra $g$ which is solvable of derived length 2, but nilpotent of degree $n$.
I have seen a parallel idea in groups, but i can't see how i can implement it for Lie-algebras.
Thanks!
| The so called standard graded filiform nilpotent Lie algebra $\mathfrak{f}_{n+1}$ of dimension $n+1$ has nilpotency class $n$, and derived length $2$.
The non-trivial brackets are $[e_1,e_i]= e_{i+1}$ for $i=2,\ldots ,n$. We have
$[\mathfrak{f}_{n+1}, \mathfrak{f}_{n+1}]=\langle e_3,\ldots ,e_{n+1}\rangle$ and
$[[\mathfrak{f}_{n+1}, \mathfrak{f}_{n+1}], [\mathfrak{f}_{n+1}, \mathfrak{f}_{n+1}]]=0$.
| {
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"source": "stackexchange",
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For which value(s) of parameter m is there a solution for this system Imagine a system with one parameter $m$:
\begin{cases}
mx + y = m\\
mx + 2y = 1\\
2x + my = m + 1
\end{cases}
Now the question is: when does this system of equations have a solution?
I know how to do it with the Gaussian method, but how can I do this without the Gaussian method, let's say with Cramer's rule?
| Compute the values of $x$ and $y$ dependent on $m$ for the following system, then solve $2x + my = m + 1$ (the last equation) to find the values of parameter $m$ for $x$ and $y$:
\begin{cases}
mx + y = m\\
mx + 2y = 1\\
\end{cases}
So,
\begin{cases}
2mx + 2y =2 m\\
mx + 2y = 1\\
\end{cases}
Subtracting two equations, will have:
$$mx=2m-1$$
*
*If $m \neq 0$, we may divide by $m$ and get $x = (2m-1)/m$ and $y =
1-m$.
*If $m = 0$, the system has no solution.
Putting $x$ and $y$ in the last equation ($m\neq 0$), we'll have:
$$m^3-3m+2=0 $$
$$(m^3-1)-3m+3=0$$
$$(m-1)(m^2+m+1)-3(m-1)=0$$
$$(m-1)(m^2+m-2)=0$$
Thus the values of parameter $m$ are $m=1$ or $m=-2$.
| {
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A problem about stochastic convergence (I think) I am trying to prove the convergence of the function $f_n = I_{[n,n+1]}$ to $f=0$, but first of all I don't in which way it converges, either in $\mathcal{L}_p$-measure or stochastically, or maybe some other form of convergence often used in measure-theory.
For now I'm assuming it's stochastic convergence, as in the following:
$$ \text{lim}_{n \rightarrow \infty} \, \mu(\{x \in \Re: |f_n(x)-f(x)| \geq \alpha\}\cap A )=0$$
must hold for all $\alpha \in \Re_{>0}$ and all $A \in \mathcal{B}(\Re)$ of finite measure.
I know it must be true since there is no finite $A$ for which this holds. Could someone give me a hint how to start off this proof?
| The sequence $\{f_n\}$ doesn't converge in $\mathcal L^p$ norm, since for all $n$ $$\lVert f_{n+1}-f_n\rVert_{L^p}^p=\int_{\mathbb R}|\mathbf 1_{[n+1,n+2]}-\mathbf 1_{[n,n+1]}|^p =\int_{[n,n+2]}1d\mu =2.$$
This sequence cannot converge in measure since $\mu(\{|f_{n+1}-f_n|\geq \frac 12\})\geq \mu([n,n+1))=1$, but converges pointwise to $0$.
It also converges stochastically to $0$, since if $\alpha> 1$, we have $\{x\in\mathbb R\mid |f_n|\geq \alpha\}=\emptyset$. For $\alpha\leq 1$, and $A\in\mathcal B(\mathbb R)$ with $\mu(A)<\infty$, use the fact that
$$\mu(A)\geq \mu(A\cap \mathbb R_+)=\sum_{n=0}^{+\infty}\mu(A\cap[n,n+1]).$$
| {
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Representing the $q$-binomial coefficient as a polynomial with coefficients in $\mathbb{Q}(q)$? Trying a bit of combinatorics this winter break, and I don't understand a certain claim.
The claim is that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are in $\mathbb{Q}(q)$, the field of rational functions, such that $P_k(q^n)=\binom{n}{k}_q$ for all $n$.
Here $\binom{n}{k}_q$ is the $q$-binomial coefficient. I guess what is mostly troubling me is that $P_k(q^n)$ is a polynomial in $q^n$. I'm sure it's obvious, but why is the claim true? Thanks.
| The $q$-binomial coefficient satisfies the recurrence
$$
\binom{n}{k}_q = q^k \binom{n-1}{k}_q + \binom{n-1}{k-1}_q,
$$
which follows easily from the definition. We can assume inductively that each term on the right is a polynomial and therefore the LHS is a polynomial.
Edit: Unfortunately this does not seem to yield the uniqueness required in the question.
| {
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Showing $\tau(n)/\phi(n)\to 0$ as $n\to \infty$ I was wondering how to show that $\tau(n)/\phi(n)\to 0$, as $n\to \infty$. Here $\tau(n)$ denotes the number of positive divisors of n, and $\phi(n)$ is Euler's phi function.
| Here's a hint: let $Q(n)$ denote the largest prime power that divides $n$. Then prove:
*
*$\displaystyle \frac{\tau(n)}{\phi(n)} \le 2 \frac{\tau(Q(n))}{\phi(Q(n))} \le \frac4{\log2} \frac{\log Q(n)}{Q(n)}$;
*$Q(n) \to \infty$ as $n\to \infty$.
For #1, you'll want to use the fact that $\tau(n)/\phi(n)$ is multiplicative, as well as the explicit evaluations of them on prime powers.
| {
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Find limit of polynomials Suppose we want to find limit of the following polynomial
$$\lim_{x\to-\infty}(x^4+x^5).$$
If we directly put here $-\infty$, we get "$-\infty +\infty$" which is definitely undefined form, but otherwise if factor out $x^5$, our polynomial will be of the form $x^5(1/x+1)$.
$\lim_{x\to-\infty}\frac 1x=0$, so our result will be $-\infty*(0+1)$,which equal to $-\infty$. I have exam in a 3 days and interested if my last procedure is correct? Directly putting $x$ values gives me undefined form, but factorization on the other hand, negative infinity, which one is correct?
| Your factoring method is fine.
In general
given a polynomial, $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1 x+a_0,\quad a_n\ne0,$$
you can factor out the leading term when $x\ne0$:
$$
P(x)= x^n\Bigl(\,a_n+{ a_{n-1}\over x}+ \cdots +{a_1\over x^{n-1}} +{a_0\over x^n} \,\Bigr),\quad x\ne0.
$$
When taking the limit as $x$ tends to an infinity, the parenthetical term above will tend towards $a_n$.
From this,
$$
\lim_{x\rightarrow-\infty} P(x) = \cases{\phantom{-}{\rm sign}(a_n) \infty, & n\text{ even}\cr -{\rm sign}(a_n) \infty, & n\text{ odd}}
$$
and
$$
\lim_{x\rightarrow \infty} P(x) = {\rm sign}(a_n) \infty.
$$
Informally, for large $x$, a polynomial behaves as its leading term. So, to compute a limit "at infinity", you could just drop all but the leading term of the polynomial and take the limit of just the leading term.
| {
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Software to display 3D surfaces What are some examples of software or online services that can display surfaces that are defined implicitly (for example, the sphere $x^2 + y^2 + z^2 = 1$)? Please add an example of usage (if not obvious).
Also, I'm looking for the following (if any):
*
*a possibility to draw many surfaces on the same sheet
*to show cross-sections
| Try these for algebraic surfaces:
*
*surf generates excellent images.
*surfer
*surfex
from http://www.algebraicsurface.net/.
| {
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Why is $\lim\limits_{x \space \to \infty}\space{\arctan(x)} = \frac{\pi}{2}$? As part of this problem, after substitution I need to calculate the new limits.
However, I do not understand why this is so:
$$\lim_{x \to \infty}\space{\arctan(x)} = \frac{\pi}{2}$$
I tried drawing the unit circle to see what happens with $\arctan$ when $x \to \infty$ but I don't know how to draw $\arctan$. It is the inverse of $\tan$ but do you even draw $\tan$?
I would appreciate any help.
| Here's a slightly different way of seeing that $\lim\limits_{\theta\rightarrow {\infty}}\arctan\theta={\pi\over2}$.
Thinking of the unit circle, $\tan \theta ={y\over x}$, where $(x,y)$ are the coordinates of the point on the unit circle with reference angle $\theta$, what happens as $\theta\rightarrow\pi/2$? In particular, what happens to $\tan\theta$ as $\theta\nearrow{\pi\over2}$?
Well, the $x$ coordinate heads towards 0 and the $y$ coordinate heads towards 1.
So in the quotient
$$
y\over x,
$$
the numerator heads to 1 and the denominator becomes arbitrarily small; so the quotient heads to infinity.
Thus, $\lim\limits_{\theta\rightarrow {\pi\over2}}\tan\theta=\infty$ and consequently
$\lim\limits_{\theta\rightarrow {\infty}}\arctan\theta={\pi\over2}$.
| {
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How to prove that geometric distributions converge to an exponential distribution?
How to prove that geometric distributions converge to an exponential distribution?
To solve this, I am trying to define an indexing $n$/$m$ and to send $m$ to infinity, but I get zero, not some relevant distribution. What is the technique or approach one must use here?
| The waiting time $T$ until the first success in a sequence of independent Bernoulli trials with probability $p$ of success in each one has a geometric distribution with parameter $p$: its probability mass function is $P(x) = p (1-p)^{x-1}$
and cumulative distribution function $F(x) = 1 - (1-p)^x$ for positive integers $x$ (note that some authors use a different convention where the random variable is $T-1$ rather than $T$, but that won't make a difference in the limit). The scaled version $p T$ converges in distribution as $p \to 0+$ to an exponential random variable with rate $1$, as for $x \ge 0$
$$ P(p T \le x) = F(x/p) = 1 - (1-p)^{\lfloor x/p\rfloor} \to 1 - e^{-x} $$
| {
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Evaluating a definite integral by changing variables. How can I evalute this integral?
$$\psi(z)=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty [(x-a)^2+(y-b)^2+z^2]^{-3\over 2}f(x,y)\,\,\,dxdy\;.$$
I think we can treat $z$ as a constant and take it out of the integral or something. Maybe changing variables like taking $u={1\over z}[(x-a)^2+(y-b)^2]$? But then how do I change $f(x,y)$ which is some arbitrary function, etc?
Thanks.
| There's no general way to evaluate this integral, for if there were, you could integrate any function $g(x,y)$ by calculating this integral for $f(x,y)=g(x,y)[(x-a)^2+(y-b)^2+z^2]^{\frac32}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is $22/7$ equal to the $\pi$ constant?
Possible Duplicate:
Simple numerical methods for calculating the digits of Pi
How the letter 'pi' came in mathematics?
When I calculate the value of $22/7$ on a calculator, I get a number that is different from the constant $\pi$.
Question: How is the $\pi$ constant calculated?
(The simple answer, not the Wikipedia calculus answer.)
| In answer to your second question, NOVA has an interactive exhibit that uses something like Archimedes method for approximating $\pi$. Archimedes method predates calculus, but uses many of its concepts.
Note that "simple" and "calculus" are not disjoint concepts.
| {
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A question on transcendental numbers Transcendental numbers are numbers that are not the solution to any algebraic equation.
But what about $x-\pi=0$? I am guessing that it's not algebraic but I don't know why not. Polynomials are over a field, so I am guessing that $\mathbb{R}$ is implied when not specified. And since $\pi \in \mathbb{R}$, what is the problem?
| To quote Wikipedia "In mathematics, a transcendental number is a number (possibly a complex number) that is not algebraic—that is, it is not a root of a non-constant polynomial equation with rational coefficients." so the field is $\mathbb{Q}$ and $\pi$ is not included.
| {
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Evaluating Integral $\int e^{x}(1-e^x)(1+e^x)^{10} dx$ I have this integral to evaluate: $$\int e^{x}(1-e^x)(1+e^x)^{10} dx$$
I figured to use u substitution for the part that is raised to the tenth power. After doing this the $e^x$ is canceled out.
I am not sure where to go from here however due to the $(1-e^x)$.
Is it possible to move it to the outside like this and continue from here with evaluating the integral?
$$(1-e^x)\int u^{10} du$$
| let $x=\ln(u)$
$dx=du/u$
$I=\int e^{x}(1-e^x)(1+e^x)^{10} dx$ = $\int ((u(1-u)(1+u)^{10})/u)du$=$\int (1-u)(1+u)^{10}du$
You may want to take it from here...
| {
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A question on Taylor Series and polynomial Suppose $ f(x)$ that is infinitely differentiable in $[a,b]$.
For every $c\in[a,b] $ the series $\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n $ is a polynomial.
Is true that $f(x)$ is a polynomial?
I can show it is true if for every $c\in [a,b]$, there exists a neighborhood $U_c$ of $c$, such that
$$f(x)=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n\quad\text{for every }x\in U_c,$$
but, this equality is not always true.
What can I do when $f(x)\not=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n$?
| As I confirmed here, if for every $c\in[a,b] $, the series $\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n $ is a polynomial, then for every $c\in[a,b]$ there exists a $k_c$ such that $f^{(n)}(c)=0$ for $n>k_c$.
If $\max(k_c)$ is finite, we're done: $f(x)$ is a polynomial of degree $\le\max(k_c)$.
If $\max(k_c)=\infty$ it means there is an infinite number of unbounded $k_c$'s, but $f$ is infinitely differentiable, so (hand waving) the $c$'s can't have a limit point, i.e. although $\max(k_c)=\infty$ it can't be $\lim_{c\to c_\infty}k_c=\infty$ for some $c_\infty\in[a,b]$ because that would mean $k_{c_\infty}=\infty$, i.e. not a polynomial.
So the infinite number of unbounded $k_c$'s need to be spread apart, e.g. like a Cantor set.
Does this suggest a counterexample or can a Cantor-like distribution of $k_c$'s never be infinitely differentiable?
| {
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For which $n\in\mathbf{N}$ do we have $\mathbf{Q}(z_{5},z_{7}) = \mathbf{Q}(z_{n})$? Put $z_{n} = e^{2\pi i /n}$. I am searching for $n \in \mathbf{N}$ so that $\mathbf{Q}(z_{5},z_{7}) = \mathbf{Q}(z_{n})$.
I know that : $z_{5} = \cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5}) $ and $z_{7} =\cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7})$.
Can you give me a hint how to continue my search? Thank you.
| That's one too many hints in the comments, but the OP still seems in doubt, so here is a proof that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_{35})$, where $\zeta_n=e^{2\pi i/n}$ is a primitive $n$th root of unity.
First, let us show that $\mathbb{Q}(\zeta_5,\zeta_7)\subseteq\mathbb{Q}(\zeta_{35})$. Notice that
$$\zeta_{35}^7=(e^{2\pi i/35})^7 = e^{2\pi i/5}=\zeta_5.$$
Thus, $\zeta_5\in \mathbb{Q}(\zeta_{35})$. Similarly, $\zeta_7 = \zeta_{35}^5 \in \mathbb{Q}(\zeta_{35})$. Hence, $\mathbb{Q}(\zeta_5,\zeta_7)\subseteq\mathbb{Q}(\zeta_{35})$.
Next, let us show that $\mathbb{Q}(\zeta_{35})\subseteq \mathbb{Q}(\zeta_5,\zeta_7)$. Indeed, consider
$$(\zeta_5\cdot\zeta_7)^3 = (e^{2\pi i/5}\cdot e^{2\pi i/7})^3 = (e^{2\pi i\cdot 12/35})^3 = e^{2\pi i\cdot 36/35} = e^{2\pi i}\cdot e^{2\pi i/35} = 1 \cdot e^{2\pi i/35} = \zeta_{35}.$$
Thus, $\zeta_{35}=(\zeta_5\cdot\zeta_7)^3\in \mathbb{Q}(\zeta_5,\zeta_7)$, and this shows the inclusion $\mathbb{Q}(\zeta_{35})\subseteq \mathbb{Q}(\zeta_5,\zeta_7)$. Therefore, we must have an equality of fields.
Now, suppose that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_n)$ for some $n\geq 1$. We have just shown that $n=35$ works. Are there any other possible values of $n$ that work? Well, if $n$ is odd, then $\mathbb{Q}(\zeta_n) = \mathbb{Q}(\zeta_{2n})$, so $n=70$ also works.
Finally, one can show that if $\mathbb{Q}(\zeta_m)\subseteq \mathbb{Q}(\zeta_n)$, then $m$ divides $n$ (here neither $m$ or $n$ should be twice an odd number). In particular, since we know that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_{35})=\mathbb{Q}(\zeta_n)$, then $n$ is divisible by $35$, and therefore $\varphi(n)$ is divisible by $24$. If $n>70$ and divisible by $35$, then $\varphi(n)$ would be strictly larger than $24$, and that would be a contradiction, because $\varphi(n)$ is the degree of the extension $\mathbb{Q}(\zeta_n)/\mathbb{Q}$. Hence, $n=35$ or $70$.
| {
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Fourier transform (logarithm) question Can we think, at least in the sense of distribution, about the Fourier transform of $\log(s+x^{2})$? Here '$s$' is a real and positive parameter
However $\int_{-\infty}^{\infty}dx\log(s+x^{2})\exp(iux)$ is not well defined.
Can the Fourier transform of logarithm be evaluated ??
| Throughout, it is assumed that $s>0$ and $u \in \mathbb{R}$.
Define:
$$
\mathcal{I}_\nu(u) = \int_{-\infty}^\infty \left(s+x^2\right)^{-\nu} \mathrm{e}^{i u x} \,\,\mathrm{d} x = \int_{-\infty}^\infty \left(s+x^2\right)^{-\nu} \cos\left(u x\right) \,\,\mathrm{d} x
$$
The integral above converges for $\nu > 0$. We are interested in computing the (distributional) value of $\lim_{\nu \uparrow 0} \left( -\partial_\nu \mathcal{I}_\nu(u)\right)$. Let $\mathcal{J}_\nu(u) = -\partial_\nu \mathcal{I}_\nu(u)$.
Notice that
$$ \begin{eqnarray}
\mathcal{I}_{\nu-1}(u) &=& s \cdot \mathcal{I}_\nu(u) - \partial_u^2 \mathcal{I}_\nu(u) \\
\mathcal{J}_{\nu-1}(u) &=& s \cdot \mathcal{J}_\nu(u) - \partial_u^2 \mathcal{J}_\nu(u)
\end{eqnarray}
$$
whenever integrals are defined.
It's not hard to compute $\mathcal{I}_\nu(u)$ explicitly:
$$
\mathcal{I}_\nu(u) = \sqrt{\pi} \cdot \frac{ 2^{\frac{3}{2}-\nu } s^{\frac{1}{4}-\frac{\nu }{2}} }{\Gamma (\nu )} \cdot |u|^{\nu -\frac{1}{2}} K_{\frac{1}{2}-\nu }\left(\sqrt{s} |u|\right)
$$
One can also compute $J_1\left(u\right)$ by using known expressions for index derivatives of Bessel functions at half-integer order:
$$
\mathcal{J}_1\left(u\right) = \pi \, \frac{\mathrm{e}^{\sqrt{s} |u|} }{\sqrt{s}} \cdot \operatorname{Ei}\left(-2 \sqrt{s} |u|\right) -\pi\, \frac{ \mathrm{e}^{-\sqrt{s} |u|} }{\sqrt{s}} \left(\frac{1}{2} \log \left(\frac{u^2}{4 s}\right)+\gamma \right)
$$
Hence $J_1(u)$ is a continuous function of real argument $u$, and has the following series expansions:
$$
\begin{eqnarray}
\mathcal{J}_1(u) &=& \frac{\pi \log \left(16 s^2\right)}{2 \sqrt{s}}+\pi |u| (\log (u^2)+2 \gamma -2)+\mathcal{o}\left(u\right) \\
\mathcal{J}_1(u) &=& -\frac{\pi}{2} \mathrm{e}^{-\sqrt{s} |u| } \left(\frac{ \left( \log \left(\frac{u^2}{4s}\right) + 2 \gamma \right)}{\sqrt{s}} + \frac{1}{s
|u|}+\mathcal{o}\left(|u|^{-1}\right)\right)
\end{eqnarray}
$$
They show that $\mathcal{J}_1^\prime(u)$ is discontinuous.
In order to express $\mathcal{J}_0(u)$ in terms of distributions we use
$$
\begin{eqnarray}
\int \mathcal{J}_0(u) f(u) \, \mathrm{d} u &=& \int \left( s \mathcal{J}_1(u) - \mathcal{J}_1^{\prime\prime}(u) \right) f(u) \, \mathrm{d} u \\ &=& \int \left( s f(u) - f^{\prime\prime}(u) \right) \mathcal{J}_1(u) \, \mathrm{d} u
\end{eqnarray}
$$
| {
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"url": "https://math.stackexchange.com/questions/93555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determinant of a special kind of block matrix I have a $2\times2$ block matrix $M$ defined as follows:
$$\begin{pmatrix}X+|X| & X-|X| \\ Y-|Y| & Y+|Y|\end{pmatrix}$$
where $X$ and $Y$ are $n\times n$ matrices and $|X|$ denotes the modulus of the entire matrix $X$ that essentially comprises modulus of individual elements of $X$.
How may I find the determinant of the matrix $M$ in terms of $X$ and $Y$? Looking for a simplified solution?
| I shall assume that $X+|X|$ is invertible, although a similar solution exists under the assumption that $Y+|Y|$ is. I shall use $A,B,C,D$ to denote the respective block matrices in your problem to avoid giant equations. The decomposition $$M = \begin{pmatrix}A & B \\ C & D\end{pmatrix} = \begin{pmatrix}A & 0\\ C & I\end{pmatrix} \begin{pmatrix}I & A^{-1}B\\ 0 & D-CA^{-1}B\end{pmatrix} = ST$$ can be verified by simple matrix multiplication (noting that matrix multiplication for block matrices works like multiplying matrices over any other noncommutative ring). The key fact is that $\det(S) = \det(A)$ and $\det(T) = \det(D-CA^{-1}B)$. I shall only prove the first equality (or rather a stronger statement where $I$ is not necessarily the same size as $A$ but the block matrix is still square), as the second can be proved similarly. If $A$ is $1\times 1$, the equation follows from the fact that $S$ is triangular and the product along its diagonal is $\det(A)$. If we assume that it holds for any $n\times n$ matrices $A,C$ then we can apply the Laplace formula to get $$\det(S) = \sum\limits_{j=1}^{n+1} (-1)^{j+1}a_{1j}\det(N_{1j})$$ where $N_{1j}$ is the matrix that results from deleting the first row and $j^{th}$ column of $S$. These matrices satisfy the inductive hypothesis (the identity matrix has not been touched), and so $\det(N_{1j}) = \det(M_{1,j})$ where $M_{1j}$ is the matrix that results from deleting the first row and $j^{th}$ column of $A$. Using the Laplace formula again gives $$\sum\limits_{j=1}^{n+1} (-1)^{j+1}a_{1j}\det(N_{1j}) = \sum\limits_{j=1}^{n+1} (-1)^{j+1}a_{1j}\det(M_{1j}) = \det(A)$$ completing the proof. Since $\det(M) = \det(S)\det(T)$, this gives us $$\det(M) = \det(A)\det(D-CA^{-1}B)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/93621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Software for Galois Theory Background: While studying Group Theory ( Open University M208 ) I had a lot of benefit from the Mathematica Add-on package AbstractAlgebra and later from the GAP software. I am currently self-studying Galois Theory ( using Ian Stewart's Galois Theory ).
Question: Is there a program that calculates the Field Extensions / Galois Group for a ( simple ) polynomial ?
| Canonical answers are Sage, Pari, Magma. The first two are open source, the last one costs money but has an online calculator. Type for example
P<x>:=PolynomialRing(Rationals());
GaloisGroup(x^6+3);
in the online calculator and hit submit. See the online manual on how to interpret the result.
| {
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Non-algebraically closed field in which every polynomial of degree $My problem is to build, for every prime $p$, a field of characteristic $p$ in which every polynomial of degree $\leq n$ ($n$ a fixed natural number) has a root, but such that the field is not algebraically closed.
If I'm not wrong (please correct me if I am) such a field cannot be finite, by counting arguments. But on the other hand, the union of all finite fields (or of any ascending chain of finite fields) of characteristic $p$, which is what I get if I start with $F_p$ and add a root to each polynomial of degree $\leq n$ in each step, is the algebraic closure of $F_p$, hence algebraically closed. I don't see how I can control this process so that in the end I get a field that is not algebraically closed.
Any hint will be welcome. Thanks in advance.
| Let $k$ be a field, $\bar k$ an algebraic closure of $k$. Fix $n>1$ natural. Consider the family $\mathcal{K}_n$ of fields $K$, $k\subset K\subset \bar k$ with the property: there exists a family of intermediate fields
$$k = K_0 \subset K_1 \subset \ldots K_s= K$$
so that $[K_{i+1}\colon K_i]< n$ for all $1\le i \le s$. It is easy to check the following:
*
*$K \in \mathcal{K}_n$, $K\subset L \subset \bar k$, $[L\colon K]< n$ implies $L \in \mathcal{K}_n$
*$K$, $K'\in \mathcal{K}_n$ implies $K K'\in \mathcal{K}_n$.
*$K \in \mathcal{K}_n$, $k \subset K'\subset K $ implies $K'\in \mathcal{K}_n$.
It is easy to see now that the union of the subfields in $\mathcal{K}_n$ is a subfield $k^{(n)}$ and every polynomial of degree $<n$ with coefficients in $k^{(n)}$ splits completely in $k^{(n)}$.
Note that the degree over $k$ of every element in $k^{(n)}$ has all its prime factors $<n$. Therefore, if $k$ is such that there exist elements in $\bar k$ whose degree over $k$ is a prime factor $>n$ (many examples here) then $k^{(n)}\ne \bar k$, that is $k^{(n)}$ is not algebraically closed.
Note: for $n=3$ we get $k^{(n)}$ are the constructible elements of $\bar k/k$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the interesting applications of hyperbolic geometry? I am aware that, historically, hyperbolic geometry was useful in showing that there can be consistent geometries that satisfy the first 4 axioms of Euclid's elements but not the fifth, the infamous parallel lines postulate, putting an end to centuries of unsuccesfull attempts to deduce the last axiom from the first ones.
It seems to be, apart from this fact, of genuine interest since it was part of the usual curriculum of all mathematicians at the begining of the century and also because there are so many books on the subject.
However, I have not found mention of applications of hyperbolic geometry to other branches of mathematics in the few books I have sampled. Do you know any or where I could find them?
| It is my understanding that the principle application of hyperbolic geometry is in physics. Specifically in special relativity. The Lorentz group of Lorentz transformations is a non-compact hyperbolic manifold. But it also shows up in general relativity and astrophysics as the space surrounding black holes is hyperbolic(negative curvature).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/93765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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Understanding bounds on factorions I am trying to understand the upper bound on factorions (in base $10$). The Wikipedia page says:
"If $n$ is a natural number of $d$ digits that is a factorion, then
$10^{d − 1} \le n \le 9!d$. This fails to hold for $d \ge 8$ thus $n$
has at most $7$ digits, and the first upper bound is $9,999,999$. But
the maximum sum of factorials of digits for a $7$ digit number is $9!7
= 2,540,160$ establishing the second upper bound."
Please explain this to me in simple terms, precisely the first part $10^{d − 1} \le n \le 9!d$.
| By definition, $n$ is the sum of the factorials of its digits. Since each digit of $n$ is at most 9, this can be at most $9!\cdot d$, where $d$ is the number of digits of $n$:
If $n$ is a factorion:
$$
n=d_1d_2\cdots d_d\quad\Rightarrow\quad n= d_1!+\,d_2!\,+\cdots+ \,d_d!\le \underbrace{9!+\,9!+\,\cdots+\, 9!}_{d -\text{terms}}\le9!\cdot d
$$
The lower bound is trivial, since $n$ has $d$ digits, it must be at least $10^{d-1}$.
| {
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Trying to figure out how an approximation of a logarithmic equation works The physics books I'm reading gives $$\triangle\tau=\frac{2}{c}\left(1-\frac{2m}{r_{1}}\right)^{1/2}\left(r_{1}-r_{2}+2m\ln\frac{r_{1}-2m}{r_{2}-2m}\right).$$
We are then told $2m/r$
is small for $r_{2}<r<r_{1}$
which gives the approximation$$\triangle\tau\approx\frac{2}{c}\left(r_{1}-r_{2}-\frac{m\left(r_{1}-r_{2}\right)}{r_{1}}+2m\ln\left(\frac{r_{1}}{r_{2}}\right)\right).$$
I can see how $$\frac{2}{c}\left(1-\frac{2m}{r_{1}}\right)^{1/2}\approx\frac{2}{c}$$
but can't see how the rest of it appears. It seems to be saying that$$2\ln\frac{r_{1}-2m}{r_{2}-2m}\approx\left(-\frac{\left(r_{1}-r_{2}\right)}{r_{1}}+2\ln\left(\frac{r_{1}}{r_{2}}\right)\right)$$
I've tried getting all the lns on one side, and also expanding $\ln\frac{r_{1}-2m}{r_{2}-2m}$
to $\ln\left(r_{1}-2m\right)-\ln\left(r_{2}-2m\right)$
and generally juggling it all about but with no luck. Any suggestions or hints from anyone?
It's to do with the gravitational time delay effect. It seems a bit more maths than physics which is why I'm asking it here.
Many thanks
| It actually seems to me they use
$$\frac{2}{c}\left(1-\frac{2m}{r_{1}}\right)^{1/2}\approx\frac{2}{c}\left(1-\frac{m}{r_{1}}\right)$$
and
$$2m\ln\frac{r_{1}-2m}{r_{2}-2m}\approx 2m\ln\left(\frac{r_{1}}{r_{2}}\right) \; .$$
EDIT: Just realized the following:
$$2m\ln\frac{r_{1}-2m}{r_{2}-2m}\approx 2m\ln\left(\frac{r_{1}}{r_{2}}\right) + 2m\left(\frac{2m}{r_2}-\frac{2m}{r_1}\right) \; .$$
Now the last term can be rewritten as
$$2m\left(\frac{2m}{r_2}-\frac{2m}{r_1}\right) = \frac{(2m)^2}{r_1 r_2}(r_1-r_2) = \left(\frac{2m}{r_1}\right)\left(\frac{2m}{r_2}\right)(r_1-r_2) $$
which is negligible.
| {
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Why is $0$ excluded in the definition of the projective space for a vector space?
For a vector space $V$, $P(V)$ is defined to be $(V \setminus \{0 \}) / \sim$, where two non-zero vectors $v_1, v_2$ in $V$ are equivalent if they differ
by a non-zero scalar $λ$, i.e., $v_1 = \lambda v_2$.
I wonder why vector $0$ is excluded when considering the equivalent classes, since $\{0\}$ can be an equivalent class too? Thanks!
| Projective space is supposed to parametrize lines through the origin. A line is determined by two points, so a line through the origin is determined by any nonzero vector.
As Nate's explains, you can certainly include 0, but you will get a different space. Is there a reason to care about it?
One reason we care about the space of lines through the origin is that it is a rich arena for discovering interesting theorems and examples.
In general, projective space is a more natural setting for algebraic geometry than affine space. For instance, theorems have fewer special cases - the most natural one being that two lines always intersect in the projective plane. Others include: Bezout's theorem, the classification of plane conics, 27 lines on a cubic, etc.
There are other reasons it is nice, which have to do with it being compact. Along those lines, we can think of projective space as a natural compactification of affine space, which is designed to catch points that wander off to infinity by assigning to their limit the direction they wandered off in. This is related to how we can use projective space to resolve singularities via blow-ups, by remembering the tangent line along which a point enters the singularity. All of these are natural situations where we care about the space of lines as a geometric object.
Maybe there are natural situations where it also makes sense to include a separate $0$ point, which is the limit of the other points. That doesn't sound too far fetched to me, especially thinking about the blow-up example.
| {
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The Dimension of the Symmetric $k$-tensors I want to compute the dimension of the symmetric $k$-tensors. I know that a covariant $k$-tensor $T$ is called symmetric if it is unchanged under permutation of arguments. Also, I know that the dimension of covariant $k$-tensors is $n^k$ but how can I eliminate non-symmetric the cases? I found this but I could not get the intution. Also, this blog post answers my question but I don't see why we put | between different indices. Any concrete example would also help such as the symmetric covariant 2-tensors in $\mathbb{R^3}$, as I asked in this thread.
| A basis for symmetric tensors, say $\text{Sym}_r(V)$ with $\{v_1,...,v_n\}$ a basis for $V$, is given by the symmetrizations of $\{v_{i_1}\otimes ... \otimes v_{i_r} \ | \ 1\leq i_1\leq...\leq i_r\leq n\}$. You must count the number of non-decreasing sequences (repetitions allowed) of length $r$ with entries in $[1,n]$. I've always heard the method of counting these referred to as stars and bars, i.e. counting the number of multisets of size $r$ with entries from $[1,n]$, and the answer you get is ${n+r-1\choose r}$.
You line up $r$ stars and insert $n-1$ bars, the first bar separating indicies 1 and 2, the second bar separating indicies 2 and 3, ..., the $(n-1)$st bar separating indicies $n-1$ and $n$.
For example, say $r=5$ and $n=3$. Here are some example of non-decreasing sequences of length $r=5$ with entries from $\{1,2,3\}$:
\begin{align*}
11223 \ &: \ **|**|*\\
22333 \ &: \ |**|***\\
11122 \ &: \ ***|**|\\
22222 \ &: \ |*****|\\
\end{align*}
So there are $r+n-1$ ``things'' (stars and bars) and you're choosing $r$ of them to be stars (or $n-1$ of them to be bars).
As for why this determines a basis for symmetric tensors: any pure tensor on the chosen basis determines a symmetric tensor via
$$
S(v_{i_1}\otimes ... \otimes v_{i_r})=\sum_{\pi\in S_n}v_{\pi(i_1)}\otimes ... \otimes v_{\pi(i_r)}
$$
and two pure tensors have the same symmetrization if their indices determine the same multiset (i.e. non-decreasing sequence as described above). I'll leave it to the reader to show that these are independent and that they span the space of symmetric tensors. (On a technical note, the symmetrization needs to be modified in non-zero characteristic and some sources might divide by $n!$.)
| {
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Altitudes of a triangle are concurrent (using co-ordinate geometry)
I need to prove that the altitudes of a triangle intersect at a given point using co-ordinate geometry.
I am thinking of assuming that point to be $(x,y)$ and then using slope equations to prove that the point exists and I can think of another way too by taking two equations (altitudes) forming a family of line which must be equal to the third equation for satisfying the condition given in the question, But thinking is all I am able to do, I am unable to put it on the paper. A hint towards solution would be great.
| I'll prove this proposition by Vector algebra, (not to solve OP's problem, but essentially for other users who might find it useful):
Let $\Delta ABC$ be a triangle whose altitudes $AD$, $BE$ intersect at $O$. In order to prove that the altitudes are concurrent, we'll have to prove that $CO$ is perpendicular to $AB$.
Taking $O$ as the origin, let the position vectors of $A$, $B$, $C$ be $\vec{a}$, $\vec{b}$, $\vec{c}$ respectively. Then $\vec{OA}=\vec{a}$, $\vec{OB}=\vec{b}$ and $\vec{OC}=\vec{c}$.
Now, as $AD \perp BC$, we have $\vec{OA} \perp \vec{BC}$. This means $\vec{OA} \cdot \vec{BC}=0$. This means, $$\vec{a} \cdot (\vec{c}-\vec{b})=0$$
Similarly $\vec{OB} \perp \vec{CA}$ and that gives you, $$ \vec{b} \cdot (\vec{a} -\vec{c})=0$$
Adding these you'll have, $$ (\vec{a}-\vec{b}) \cdot \vec{c} =0$$ This reads off immediately that $\vec{OC} \perp \vec{BA}$. This proves the proposition.
| {
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Proving that $G/N$ is an abelian group
Let $G$ be the group of all $2 \times 2$ matrices of the form $\begin{pmatrix} a & b \\ 0 & d\end{pmatrix}$ where $ad \neq 0$ under matrix multiplication. Let $N=\left\{A \in G \; \colon \; A = \begin{pmatrix}1 & b \\ 0 & 1\end{pmatrix} \right\}$ be a subset of the group $G$. Prove that $N$ is a normal subgroup of $G$ and prove that $G/N$ is abelian group.
Here is my attempt!
To prove $N$ is normal I consider the group homomorphism $f \colon G \to \mathbb R^*$ given by $f(B) = \det(B)$ for all $B$ in $G$. And I see that $f(N)$ is all the singleton $\{1\}$ since $\{1\}$ as a subgroup of $\mathbb R^*$ is normal, it follows that $N$ is also normal. Is this proof helpful here? Then how to prove that $G/N$ is Abelian? I know $G/N$ is a collection of left cosets.
Thank you.
| One way is using first isomorphism theorem.
To do this you should find a group homomorphism such that $\operatorname{Ker} \varphi=N$.
Let us try $\varphi: G\to \mathbb R^*\times \mathbb R^*$ given by
$$\begin{pmatrix} a & b \\ 0 & d\end{pmatrix} \mapsto (a,d).$$
(By $\mathbb R^*$ I denote the group $\mathbb R^*=\mathbb R\setminus\{0\}$ with multiplication. By $G\times H$ I denote the direct product of two groups, maybe your book uses notation $G\oplus H$ for this.)
It is relatively easy to verify that $\varphi$ is a surjective homomorphism. It is clear that $\operatorname{Ker} \varphi=N$. Hence, by the first isomorphism theorem,
$$G/N \cong \mathbb R^*\times\mathbb R^*$$
This is a commutative group.
If you prefer, for any reason, not using the first isomorphism theorem, you could also try to verify one of equivalent definitions of normal subgroup and then describe the cosets and their multiplication.
In this case you have
$$\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}
\begin{pmatrix} 1 & b' \\ 0 & 1 \end{pmatrix}
\frac1{ad}
\begin{pmatrix} d & -b \\ 0 & a \end{pmatrix}=
\begin{pmatrix} 1 & \frac{ab'}d \\ 0 & 1 \end{pmatrix}$$
(I have omitted the computations), which shows that $xNx^{-1}\subseteq N$ for any $x\in G$.
You can find out easily that cosets are the sets of the form
$$\{\begin{pmatrix} x & y \\ 0 & z \end{pmatrix}; y\in\mathbb R\}$$
for $x,z\in\mathbb R\setminus\{0\}$ and that the multiplication of cosets representatives $\begin{pmatrix} x & 0 \\ 0 & z \end{pmatrix}$ is coordinate-wise.
| {
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Generalize the equality $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n\cdot(n+1)}=\frac{n}{n+1}$ I'm reading a book The Art and Craft of Problem Solving. I've tried to conjecture a more general formula for sums where denominators have products of three terms. I've "got my hands dirty", but don't see any regularity in numerators.
Please, write down your ideas.
| I have the following recipe in mind, see how far it helps (leave pointers in this regards as comments):
Let $a_1, a_2, a_3, \cdots, a_n, \cdots$ be the terms of an $A.P$. Let $d$ be the common difference of the given $A.P$. We are interested to find the sum for some $r \in \mathbb{N}$. $$\sum_{k=1}^n \dfrac{1}{a_k a_{k+1} \cdots a_{k+r-1}}$$
Let us denote the sum of the series as $S_n$ and the n-th term of the series, $T_n$. Consider the following definition of the new entity that I'll call $V_n$. (This is not at all a mystery: $V_n$ is obtained by dropping the first of the $r$ entries in the denominator of $T_n$)
$$V_n:=\dfrac{1}{a_{n+1} \cdots a_{n+r-2} a_{n+r-1}}$$
Therefore, $$V_{n-1}:=\dfrac{1}{a_n \cdots a_{n+r-3} a_{n+r-2}}$$
Now (I'll leave the computation that goes here!), you'll have $$V_n-V_{n-1}=T_n(a_n-a_{n+r-1})$$ $$T_n=\dfrac{1}{d(r-1)} \cdot (V_{n-1}-V_n)$$Substituting various values for $n$, you have equations for $T_1, T_2, \cdots, T_n$. Adding these, and noting that common difference is the difference between two consecutive terms taken in an (appropriate!) order, you have, $$S_n=\dfrac{1}{(r-1)(a_2-a_1)}(\dfrac{1}{a_1\cdots a_{r-1}}-\dfrac{1}{a_{n+1} \cdots a_{n+r-1}})$$
For the problem at hand, set $a_1=1,~d=1,~r=3 $. You'll get $$\sum_{k=1}^n{\dfrac{1}{k(k+1)(k+2)}}=\dfrac{1}{4}-\dfrac{1}{(n+1)(n+2)}$$
Have fun doing for $r=4, \cdots$. Hope this helps.
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Subsets and Splits