text stringlengths 83 79.5k |
|---|
H: Showing the bijection between countable dense subset of a metric space $E×\mathbb{Q}→$neighborhoods of $E$ with rational radius
Let $X$ be a separable metric space.
Let $E$ be a countable dense subset of $X$.
Let $p_i$ enumerate $E$ and $q_j$ enumerate $\mathbb{Q}$.
Let $G=\{N(p_i,q_j) \subset X | i,j\in \omega\}$
Then $G$ is a base for $X$.
Can one show the bijection between $E×\mathbb{Q}$ and $G$ in ZF?
I want to show that $G$ is countable, and since $E×\mathbb{Q}$ is countable, if i can show the bijection, proof will be done. Help
I don't know whether $f:(p,q)→N(p,q)$ is a bijection and even if it is a bijection, i have no idea how to show that $f$ is injective..
AI: The map $(p,q)\mapsto N(p,q)$ is not necessarily bijection. Take $X=E=\mathbb Z$ as an example. (For any $z\in\mathbb Z$ the ordered pairs $(z,\frac1n)$ are mapped to the same set $N(z,\frac1n)=\{z\}$.)
However, you can take some well-ordering on $E\times\mathbb Q$ and define
$$N(p,q)\mapsto \min\{(a,b)\in E\times\mathbb Q; N(a,b)=N(p,q)\}.$$
(If you prefer, you can get a well-ordering of order type $\omega$ on the set $E\times\mathbb Q$. This can be obtained using any bijection between $E\times\mathbb Q$ and $\omega$. Since you are given enumeration of $E$ and $\mathbb Q$, exhibiting such a bijection does not need axiom of choice.)
The image of this map is a subset of countable well-ordered set $E\times\mathbb Q$. Therefore it is again a countable well-ordered set. So it is in bijection with some countable ordinal.
Additional question in your comment:
How do I show that $G$ is not finite.
If $E$ is infinite, then $G$ cannot be finite.
Just show by induction on $i$ that for each $p_i$ you can find $r_i\in\mathbb Q$ such that $N(p_i,r_i)$ is different from $N(p_j,r_j)$ for each $j<i$.
This can be done by induction, in the inductive step you choose $r_i < \min \{d(p_i,p_j); j<i\}$. (Note that $ \{d(p_i,p_j); j<i\}$ is a finite set of positive real numbers, so the minimum of this set is a positive real number and there exists a rational number smaller than this minimum.)
Then $i\mapsto N(p_i,r_i)$ is an injective map $\omega\to G$. (The point $p_i$ is contained in $N(p_i,r_i)$, but it does not belong to any of the sets $N(p_j,r_j)$ for $j<i$. Therefore any two sets $N(p_i,r_i)$ and $N(p_j,r_j)$, $i\ne j$, are different.) |
H: What is an example of a vector field that is not left-invariant?
Let $G$ be a Lie group, $L_g$ the left-translation on this group with differential $d L_g$. A vector field $X$ on $G$ is called left-invariant if
$$ X \circ L_g = d L_g \circ X \quad \forall g \in G$$
i.e.
$$ X_{gh} = (d L_g)_h (X_h) \quad \forall g,h \in G. $$
Now, this definition seems so natural to me that I cannot come up with a non-trivial counterexample for a vector field that is $\textit{not}$ left-invariant. In my mind, pushing forward on the tangent space is basically always the same as the group action...
Could you provide me with such a counterexample that helps understand the notion of left-invariance?
AI: You could start taking the Lie group $(\mathbb R^n,+),$ and considering what does it means for a vector field $X$ on $\mathbb R^n$ to be left-invariant. |
H: Find the minimum value of $P=x^2+y^2+\frac{x^2y^2}{(4xy-x-y)^2}$
Given that $\frac{1}{3}<x \le \frac{1}{2}$ and $y\ge1$
Find the minimum value of $P=x^2+y^2+\frac{x^2y^2}{(4xy-x-y)^2}$
AI: As it seems to be a homework question and I don't have Latex-MathType in my job's computer (I'll edit my answer late to make it formal), I will give you just an outline of how to solve it.
You have 3 inequality constraints:
$x > \frac13 \Rightarrow f_1(x,y) = x - \frac13 > 0$
$x \leq \frac12 \Rightarrow f_2(x,y) = x - \frac12 \leq 0$
$f_3(x,y) = y-1 > 0$
Form $F(x,y) = P(x,y) - f_1(x,y) - f_2(x,y) - f_3(x,y)$, and use Karush-Kuhn-Tucker conditions. |
H: Relation between areas of two quadrilaterals
If it is given that:
in a quadrilateral $ABCD$, $X$ is the mid point of the diagonal $BD$, then prove that
area of $AXCB$ = $\frac12$ area of $ABCD$.
I do not know where to start, but I think we can take two cases, something like:
$1$. When $AXCB$ is quadrilateral.
$2$. When $AXCB$ is a triangle, like in case when ABCD is parallelogram.
Now, I want to ask, is this approach correct? Maybe it is long, but is it right?
If it is, then could anyone suggest a shorter and more elegant solution? If it is not then what can I think of, here? like any theorem or any formula or any thing.
Well, rather an elegant solution I would prefer to understand the problem which I think I am not able to do.
AI: Draw a diagram, with $\,A\,$ the upper left vertex of $\,ABCD\,$ and go clockwise, mark point $\,X\,$ on $\,BD\,$ and draw $\,AX\,\,,\,XC\,$.
After a moment of thinking, get convinced that in triangle $\,\Delta AXB\,$ , the height to $\,BX\,$ is exactly the same as the height to $\,DX\,$ in triangle $\,\Delta AXD\,$ , and thus $\,S_{\Delta ADX}=S_{\Delta ABX}\,$ , and exactly the same argument works for the other pair of triangles $\,\Delta BCX\,\,,\,\Delta CXD\,$ |
H: Vector and matrix norm definitions?
I've questions on these four norms whose definitions I'm memorizing like this:
Vector euclidean norm: $(x_1^2+x_2^2+\cdots+x_n^2)^{1/2}$
Vector max norm: $\max\{|x_1|, |x_2|, \ldots, |x_n|\}$
Matrix norm: $\max\limits_{x \neq 0} \frac{\|Ax\|}{\|x\|}$
Matrix max norm: $\max\limits_{1<i<N} \sum\limits_{1<j<N}|a_{ij}|$
How to interpret these? The Vector euclidian norm is a scalar. The vector max norm is also a scalar. The matrix norm is also a scalar but the matrix max norm is a vector? Can you tell me more how to interpret these formulas? Are my formulae correct? Can they be more pedagogically written?
AI: This got too long for a comment:
Using the definition of the vecto $p$ norm
$$
\|\mathbf{x}\|_p := \bigg( \sum_{i=1}^n |x_i|^p \bigg)^{1/p},
$$
you can combine 3.) and 4.) like the following:
Let
$$
\left \| A \right \| _p = \max \limits _{x \ne 0} \frac{\left \| A x\right \| _p}{\left \| x\right \| _p}.
$$
So you get back 3.) with $p=2$. In the case of $p=1$ and $p=\infty$, the norms can be computed as:
$ \left \| A \right \| _1 = \max \limits _{1 \leq j \leq n} \sum _{i=1} ^m | a_{ij} |, $ which is simply the maximum absolute column sum of the matrix.
$ \left \| A \right \| _\infty = \max \limits _{1 \leq i \leq m} \sum _{j=1} ^n | a_{ij} |, $ which is simply the maximum absolute row sum of the matrix
Here I'm not sure, which of both you mean, but none of them is a vector as already pointed out in the comments (taken from Matrix norm/Induced_norm, which also provides some examples, that might help with the interpretation). |
H: Existence of non-trivial solution of Sylvester equation.
I'm trying to solve a special case of Sylvester equation
in my case it looks like
$$A*X=X*B$$
so it can be written in form
$$A*X+X*(-B)=C$$ where C consist of all 0 items.
I tried to solve it in Mathematica with LyapunovSolve
but it give me all =0 trivial solution.
So I want to check if there is non-trivial solution exist. It seems I find out how to test matrices A,B for non-trivial solution(but I'm not sure) if resultant equals to 0, then such solution exist.$$resultant(det(A-λE),det(B-λE),λ)=0$$ I tried real matrices for example A matrix (0 -1 300 1 0 0 0 0 1) B matrix (-0.4009 -1.0787 446.1463 1.6180 0.8875 -159.2272 0.0003 0.0029 1) but there is maybe a problem because B matrix was defined experimentally and it has some small errors, with these matrices I have resultant -6.79 , but due to errors I don't know may be it close enough to 0?
the question is:
can you write condition of existence of non-trivial solution of Sylvester equation not in abstract form but in particular formula, preferably in Mathematica.
Also I don't understand why Mathematica gives me only trivial solution, when I trying to solve this equation with parameters
for example I tried
am = {{a11, a12, a13}, {a21, a22, a23}, {a31, a32, a33}}
bm = {{b11, b12, b13}, {b21, b22, b23}, {b31, b32, b33}}
cm = {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}
LyapunovSolve[am,bm,cm]
gives me all =0
also I tried to multiply matrices and get 9 equations and solve them with function Solve which also give me all =0.
AI: A Sylvester equation like
$$AX + X(-B) = 0$$
can be written as
$$(I_n \otimes A - B \otimes I_n) \operatorname{vec} X = 0$$
so it has a non-trivial solution if and only if $I_n \otimes A - B \otimes I_n$ has a non-trivial null space. An equivalent condition to having a non-trivial null space is having zero as an eigenvalue.
If $\lambda$ is an eigenvalue of $A$ and $\mu$ an eigenvalue of $B$, $\lambda - \mu$ is an eigenvalue of $I_n \otimes A - B \otimes I_n$, and all eigenvalues of $I_n \otimes A - B \otimes I_n$ is on this form. Thus $I_n \otimes A - B \otimes I_n$ has zero as an eigenvalue if and only if $A$ has an eigenvalue $\lambda$ and $B$ has an eigenvalue $\mu$ such that $\lambda - \mu = 0$.
Thus, to see if $AX - XB = 0$ has a solution, you can calculate the spectrums $\sigma(A), \sigma(B)$ and then see if for any $\lambda \in \sigma(A)$ you have $-\lambda \in \sigma(B)$. You can probably write this in some clever way in Mathematica, e.g. calculate the spectrums, multiply one of them with $-1$ and check for overlaps (at the moment I do not have access to Mathematica so I can not test any code.)
The resultant of two polynomials $P$ and $Q$ is zero iff $P$ and $Q$ have a common root, so your thinking seems to be correct. However, you seem to have missed that you should calculate $\det(-B - \lambda E)$, not $\det(B - \lambda E)$. |
H: Showing that $\{0w:w \in A\} \cup \{1w: w \notin A \}$ computes $A$
I'm trying to construct a reduction from $A \in RE \setminus R$ under $\sum=\{0,1\}$ to $B$ which defined: $B=\{0w:w \in A\} \cup \{1w: w \notin A \}$.
I need to show that $B\notin RE \cup co-RE$ with the the defining of $A$ and the use of the theorem that says:
if $L \notin R$ and $\bar{L} \leq_ML$ so $L \notin RE \cup co-RE$
so I though of showing a reduction form A to B which will tell me that $B\notin R$ and to show a reduction from $\bar{B}$ to $B$.
First I have a problem with the following attempt:
Given $M$ Turing for A, Let's construct a Turing machine $M'$ for B:
given input $w$ for $M'$:
simulate $w$ on $A$, if it accepts, $M'$ accepts $0w$, otherwise $M'$ accepts $1w$.
It's not correct, isn't it? since what happens if $M$ stuck in a loop?
Suggestions for correctness? Is it a good plan for this question?
Thanks!
AI: You should (or at least can) work by contradiction. First assume that $B$ is r.e. and show that this implies that $A$ is computable. Then assume $B$ is co-r.e. and show that this also implies that $A$ is computable. Thus you will be reducing $A$ to $B$, and reducing $A$ to the complement of $B$, rather than reducing $B$ to $A$ as suggested in the sketch at the end of the question.
Note that to prove $B$ is not r.e. or co.r.e., it is only necessary to assume that $A$ is not computable, it is not necessary to assume $A$ is r.e. However, one reason that this result is interesting is that you can go farther and prove that $A$ and $B$ are Turing equivalent. So if we do take $A$ to be r.e. but not computable, we get an example of two sets $A$ and $B$ that are Turing equivalent, while $A$ is r.e. and $B$ is not. |
H: Finding $\lim_{E \to U} \frac{1}{4}\left[\frac {U^2}{E(E-U)}\right]\sin^2 k'L$ where $k' = \left[2m(E-U)/\hbar\right]^{1/2}$ and $L$ is a constant
I am trying to find the limit of this equation:
$$\lim_{E \to U} \frac{1}{4}\left[\frac {U^2}{E(E-U)}\right]\sin^2 k'L$$
Isn't there no limit, because $(E-U)$ would be 0, and the denominator would become 0, which means fraction is undefined?
Edit:
$k' = \left[2m(E-U)/\hbar\right]^{1/2}$ and $L$ is some constant
AI: If you are familiar with $\lim_{x\to0}{\sin x\over x}=1$ then you should be able to manipulate your problem to where you can make use of this limit. |
H: Does $\int_{0}^{\infty} \cos (x^2) dx$ diverge absolutely?
I believe it does, but i would like some help formulating a proof.
AI: It's equivalent to the convergence of $\int_\pi^{\infty}\frac{|\cos t|}{\sqrt t}dt$, after having used the substitution $x^2=t$.
We have
$$
\int_{\pi}^{N\pi}\frac{|\cos t|}{\sqrt t}dt=\sum_{n=1}^{N-1}\int_{n\pi}^{(n+1)\pi}\frac{|\cos t|}{\sqrt t}dt$$
Use $\pi$ periodicity of $|\cos|$ and a substitution $s=t-n\pi$ to get bound which doesn't depend on $n$.
Find a good below bound will help to show the divergence.
This argument can be applied for the divergence of $\int_0^{+\infty}|\cos(x^p)|dx$, $p>0$. |
H: Correct order of books for a beginner
what should be the order of the books in which a beginner should do the following books in algebra:
-1.E.J. Barbeau POLYNOMIALS
-2. Polynomials and Polynomial Inequalities (Graduate Texts in Mathematics) - (Springer) - Peter Borwein -Tamas Erdely.
3.Geometry of Polynomials - (American Mathematical Society) - Morris Marden
And if you guys know of any other book then share it.
And in functional equations the following:
1.Functional Equations and Inequalities in Several Variables - (World Scientific Publication) - Stefan Czerwik.
2.Lectures on Functional Equations - (Academic Press) - J. Aczel.
3.Functional Equations: A Problem Solving Approach - (Prism Books) - B.J. Venkatchala.
AI: In the sense that there might be someone who will go through all three polynomial books at some time, that order would likely be Barbeau's Polynomials, which can be read by a high school student, Polynomials and Polynomial Inequalities, Geometry of Polynomials. Although I should mention that the last two don't have a dependence chain or anything, so it doesn't really matter which of the two you take first. But there are many, many, many things to be learned between Barbeau and the others, like calculus, algebra, algebraic geometry, and some analysis. It would be inappropriate for me to suggest that Barbeau would adequately prepare you for the other two.
I'm not familiar with the functional analysis books. |
H: How to `bound' $L^\infty$ by the constant function $1$
Let $(S,\mathcal A, \mu)$ be a measure space and consider the Riesz space $L^\infty=L^\infty(S,\mathcal A, \mu)$ (under point-wise ordering). Let $1_X$ denote the indicator function on $S$ (which is contained $L^\infty$).
Given an arbitrary $f\in L^\infty$, is it possible to find a $\alpha\in\mathbf{R}$ such that $$-\alpha\cdot1_S\le f\le \alpha\cdot 1_S$$ The $\cdot$-symbol denotes the point-wise multiplication, that is $\alpha\cdot 1_S=\alpha\cdot 1_S(x)$ for all $x\in S$.
AI: As $f \in L^\infty$, we have $|f| \le \|f\|_\infty$ almost everywhere. If we let $\alpha = \|f\|_\infty$, we are done. |
H: Evaluating the line integral of $F=\frac{-y}{x^2+y^2}i+\frac{x}{x^2+y^2}j$ along $0 \le t \le 2\pi $
Recently, I had an exam and in that I was asked to evaluate the line integral of the function $$F=\frac{-y}{x^2+y^2}i+\frac{x}{x^2+y^2}j$$ alongside the unit circle, $0 \le t \le 2\pi $ . Moreover, it was asked if this integral could be carried out with using Green's Theorem or not and why?
For the first, I did the following: $$\oint_C F\cdot dr=\int_0^{2\pi}F\big (\cos(t),\sin(t)\big)\cdot\big(-\sin(t),\cos(t)\big)dt$$ which is $2\pi$. But always I have problem with above theorem and I do know I didn't pass this part of question correctly. May I ask to help me? Thank you.
AI: Your calculation of the line integral is correct. No, Green's Theorem cannot be applied directly because the vector field has a singularity at the origin.
Some years ago I taught a multivariable calculus course, and this very vector field was one of the stars of the whole show. You can see for instance $\S 5$ of these notes for an extended discussion of the fact that this vector field is irrotational -- i.e., has zero curl -- but is not conservative and in particular does not have a scalar potential function. Then, if you are feeling especially curious, you can go on to read the next section, which segues from this example to a discussion of De Rham cohomology and the problem that the main character in the film A Beautiful Mind writes on the blackboard in the multivariable calculus class attended by his future wife. |
H: preorders induced by continuous functions to the reals
Any function to a (total) order induces a (total) preorder on its domain. What can be said about the total preorders induced by a continuous function from a "nice" topological space (for instance, a Euclidean space) to the reals?
AI: It is possible to characterize all preorders induced by a real-valued function. The following result is essentially due to Debreu and builds on previous work by Cantor.
Theorem: Let $(T,\preceq)$ be a totally preordered set. The the following are equivalent:
There exists a function $u:T\to\mathbb{R}$ such that $u(x)\leq u(y)$ if and only if $x\preceq y$.
There exists a countable set $C\subseteq T$ such that for all $x\prec y$, there is $c\in C$ with $x\preceq c\preceq y$.
Proof:
(1$\implies$2)$~~~$ Let $u$ be such a function. Call an interval $(r,s)$ a gap if there exists $x,y\in T$ with $u(x)=r$, $u(y)=s$ and $u^{-1}\big((r,s)\big)=\emptyset$. Any two gaps are disjoint and contain a rational number. So there are only countably many gaps. Pic for each gap $(r,s)$ elements $x,y\in T$ with $u(x)=r$, $u(y)=s$ and collect these elements in the countable set $C_1$. For each open interval with rational endpoints $(p,q)$ such that $u^{-1}\big((p,q)\big)\neq\emptyset$, pick an element $x\in u^{-1}\big((p,q)\big)$ and collect them in the countable set $C_2$. Then $C=C_11\cup C_2$ has the desired properties.
(2$\implies$1)$~~~$ Let $C=\{c_1,c_2,\ldots\}$ be such a countable set. For $x\in T$, let $$L_x=\big\{n\in\mathbb{N}:c_n\prec x\big\}$$ and $$U_x=\big\{n\in\mathbb{N}:c_n\succ x\big\}.$$ By letting $$u(x)=\sum_{n\in L_x}\frac{1}{2^n}-\sum_{n\in U_x}\frac{1}{2^n}$$ for all $x\in T$, we get the desired function. $~~\blacksquare$
If $u$ is continuous, we get that the sets $\{y\in T:y\succ x\}$ and $\{y\in T:y\prec x\}$ are open for all $x\in T$. If $T$ is a second countable topological space with a complete preorder on it, this condition in sufficient for the preorder being induced by continuous real valued function. The proof, due to Debreu, is quite complicated, see here (JSTOR). This settles of course the case in which $T$ is a Euclidean space. A book that contains a lot of material around these themes is "Representations of preferences orderings" by Bridges and Mehta. |
H: A mathematical notation for the mean value theorem?
I'm looking for a stringent definition of the mean value theorem i.e. stated with mathematical symbols. I think it should be something like "there exists..." and then the mean value if there is an integral between the surrounding values which to my knowledge is the mean value theorem, stating that the mean has to be somewhere between two points. So could we define it in mathematical notation?
" there exists a point c in $(a, b)$ such that $f'(c)= \frac{f(b)-f(a)}{b-a}$ "
Can we rephrase the above into a "pure" mathematical notation i.e. using ∃ instead of plain words? Is it trivial what I want to do and just replace the "there exists" with ∃ so that it is more like a logical statement?
AI: You could write that as
$$\exists c \in (a,b)\, :\, f'(c) = \frac{f(b)-f(a)}{b-a}$$
Some people might prefer to use other conventions, such as
$$\exists c.\ c \in (a,b) \wedge f'(c) = \frac{f(b)-f(a)}{b-a}$$
Frankly, though, there's nothing wrong with words, especially in something like analysis. It doesn't make it any less rigorous as long as the language used isn't ambiguous.
Edit:
Amused by one of the comments, I thought I'd write out in full the statement of the mean value theorem using only symbols.
$(\forall a,b \in \mathbb{R})(\forall f \in \mathbb{R}^{[a,b]})(((\forall x \in [a,b])(\forall \epsilon > 0)(\exists \delta > 0)$
$(\left|x-y\right|<\delta \Rightarrow \left|f(x)-f(y)\right|<\varepsilon)) \wedge ((\forall x \in (a,b))(\exists f'(x) \in \mathbb{R})$
$(\forall \varepsilon > 0)(\exists \delta > 0)(\forall h>0)((h<\delta) \Rightarrow \left| \frac{f(x+h)-f(x)}{h} - f'(x)\right| < \varepsilon))$
$\Rightarrow (\exists c \in (a,b))(f'(c) = \frac{f(b)-f(a)}{b-a}))$ |
H: Finite $G$ with involutory automorphism $\alpha$ with no nontrivial fixed points. Proving properties of $\alpha$.
The question at hand is:
Let G be a finite group and $\alpha$ an involutory automorphism of G, which doesn't fixate any element aside from the trivial one.
1) Prove that $ g \mapsto g^{-1}g^{\alpha} $ is an injection
2) Prove that $\alpha$ maps every element to its inverse
3) Prove that G is abelian
I think I've found 1).
I assume $ g_1^{-1}g_1^{\alpha} = g_2^{-1}g_2^{\alpha} $ and from this I get $ (g_2g_1^{-1})^\alpha = g_2g_1^{-1} $, so $g_2 = g_1$ (is this correct?)
For 2) I'm sort of stumped though, not sure how to start proving that, so any help please?
AI: 1) is correct as you've written. 3) is an easy consqeuence of 2), so I'll leave that part to you.
For 2) we need to use the fact that $\alpha$ is involutory... Note that $\varphi:g\mapsto g^{-1}g^\alpha$ is injective, so it is bijective (because $G$ is finite). Consider an arbitrary $g$, and $h:=\varphi^{-1}(g)$. What is $\varphi^2(h)$ (in terms of $h$ first, then in terms of $g$)? |
H: What are some examples of vector spaces that aren't graded?
From wikipedia: a vector space $V$ is graded if it decomposes into direct sum $ \oplus_{n \geq 0} V_n$ of vector spaces $V_n$.
So as far as I understand things, any vector space with a countable basis is graded: Let $V$ be a vector space over a field $k$ with basis $\{v_n\}_{n\in\mathbb{N}}$, then $V = \oplus_{n\geq 0} k\cdot v_n$. Then the only vector spaces that I can think of that aren't obviously graded are things like $C(X)$, the space of continuous functions on some manifold $X$
Is this correct? are there any more? or do I not understand something?
Thanks
AI: Grading isn't a property of a vector space: it's extra structure attached to a vector space, in the same way that a multiplication is an extra structure you attach to a set to make it a group. So this is a little like asking "what are some examples of sets that aren't groups?" (As it turns out, every set can be equipped with a group structure, and this is equivalent to the axiom of choice. But this is missing the point.)
Every vector space admits a trivial grading in which $V_0 = V$ and $V_n = 0$ for all $n \ge 1$. But we often encounter vector spaces (such as the space of polynomials) with natural gradings, they are usually nontrivial, and taking advantage of this extra structure is useful in various ways. |
H: Proof of a Proposition on Partitions and Equivalence Classes
I stumbled upon a seemingly rudimentary proposition that I am having trouble writing out a proof for. The proposition goes something like,
Proposition: If $\{A_i|i\in I\}$ is a partition of $\mathcal A$, then there is an equivalence relation on $\mathcal A$ whose
equivalence clases are precisely the sets $A_i, i \in I$.
Where $I$ is some indexing set.
How do I prove the statement ? I can't even decide on a good place to start.
AI: When $I$ is an arbitrary "index set" then a set-vauled function $$f:\quad I\to {\cal P}(A)\ ,\quad i\mapsto A_i\ ,$$ where ${\cal P}(A)$ denotes the power set of $A$, is called a family of subsets of $A$ and is denoted by $(A_i)_{i\in I}\ $. Such a family is called a partition of $A$, if (i) all $A_i$ are nonempty, (ii) the $A_i$ are pairwise disjoint, i.e., $A_i\cap A_j=\emptyset$ when $i\ne j$, and (iii) $\bigcup_{i\in I} A_i=A$.
Given a partition $(A_i)_{i\in I}$ of $A$, for each $x\in A$ there is a unique $i\in I$ such that $x\in A_i$. This defines a function $\iota: A\to I$ which returns for each $x\in A$ the unique $i$ with $x\in A_i$.
Now it is easy to check that
$$x\sim y\quad:\Leftrightarrow\quad \iota(x)=\iota(y)$$
defines an equivalence relation on $ A$ whose equivalence classes are exactly the $A_i$. |
H: Four-parameter Beta distribution and Wikipedia
Sorry if it is not an appropriate place for such questions, but anyway can anybody please confirm that the formula for the density function of the four-parameter Beta distribution is correct in Wikipedia. It seems $(c - a)$ is missing in the denominator. Thank you.
Best regards,
Ivan
AI: Yes, the factor is indeed missing.
Let $X$ be standard 2-parameter Beta random variable. The four-parameter one $Y$ is obtained by affine transformation $Y = (c-a) X + a$ for $c>a$. Then
$$
f_Y(y) = \frac{1}{c-a} f_X\left(\frac{y-a}{c-a}\right) = \frac{1}{c-a} \left(\frac{y-a}{c-a} \right)^{\alpha-1} \left(\frac{c-y}{c-a} \right)^{\beta-1} \frac{\mathbf{1}(a < y <c)}{B(\alpha,\beta)} =
\frac{(y-a)^{\alpha-1} (c-y)^{\beta-1}}{(c-a)^{\alpha+\beta-1}}\frac{\mathbf{1}(a < y <c)}{B(\alpha,\beta)}
$$ |
H: Prerequisites for studying smooth manifold theory?
I am attending first year graduate school in about three weeks and one of the courses I am taking is an introduction to smooth manifolds. Unfortunately, my topology knowledge is minimal, limited to self study. Besides some basic topological definitions, are there specific areas where I should become acquainted with? I have heard multivariable calculus is a 'prerequisite' for the study of smooth manifolds -- is this from an intuition perspective, and if so, what parts and to what degree of rigor are they referring?
Thank you. (My apologies if this is a repeat question, by the way -- I could not find it)
AI: You will need
Calculus/real analysis of functions of one and several variables up to and including the implicit and inverse function theorem. This (the implicit function theorem) is the basic starting point for (smooth) manifold theory, you will not be able to get anywhere without it.
sound knowledge of linear algebra (at least in/for real vector spaces)
1-d integration (Riemann integral will suffice in the beginning), preferably basic knowledge of an n-dimensional integration theory.
basic theory of ODE will sooner or later become important, in order to be able to deal with, e.g, the flow of vector fields (actually one needs existence (Picard-Lindelöf) and rather soon smooth dependence of the solution on the initial values, but the latter may be stated and believed. It'll become difficult without existence of solutions).
Depending on your book of choice everything else will probably be developed in the course of said book, see the other answers for some suggestion. The basics in topology you'll need will likely be known to you, once you really covered the above list. Spivak's calculus on manifolds comes to my mind. |
H: The expected payoff of a dice game
There's a question in my Olympiad questions book which I can't seem to solve:
You have the option to throw a die up to three times. You will earn
the face value of the die. You have the option to stop after each
throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game?
I found a solution here but I don't understand it.
AI: Let's suppose we have only 1 roll. What is the expected payoff? Each roll is equally likely, so it will show $1,2,3,4,5,6$ with equal probability. Thus their average of $3.5$ is the expected payoff.
Now let's suppose we have 2 rolls. If on the first roll, I roll a $6$, I would not continue. The next throw would only maintain my winnings of $6$ (with $1/6$ chance) or make me lose. Similarly, if I threw a $5$ or a $4$ on the first roll, I would not continue, because my expected payoff on the last throw would be a $3.5$. However, if I threw a $1,2$ of $3$, I would take that second round. This is again because I expect to win $3.5$.
So in the 2 roll game, if I roll a $4,5,6$, I keep those rolls, but if I throw a $1,2,3$, I reroll. Thus I have a $1/2$ chance of keeping a $4,5,6$, or a $1/2$ chance of rerolling. Rerolling has an expected return of $3.5$. As the $4,5,6$ are equally likely, rolling a $4,5$ or $6$ has expected return $5$. Thus my expected payout on 2 rolls is $.5(5) + .5(3.5) = 4.25$.
Now we go to the 3 roll game. If I roll a $5$ or $6$, I keep my roll. But now, even a $4$ is undesirable, because by rerolling, I'd be playing the 2 roll game, which has expected payout of $4.25$. So now the expected payout is $\frac{1}{3}(5.5) + \frac{2}{3}(4.25) = 4.\overline{66}$.
Does that make sense? |
H: Properties of homomorphisms of the additive group of rationals
Let $f : (\mathbb{Q},+) \longrightarrow (\mathbb{Q},+)$ be a non-zero homomorphism.
Can we conclude that $f$ is bijective (or, if that fails, that $f$ is injective or surjective)?
Context
The additive group of integers has non-surjective nonzero endomorphisms, such as $n\mapsto 2n$. However, the same formula gives a bijective endomorphism when applied to rationals.
AI: Suppose $\,f\,$ is such a homomorphism and try to work out how knowing $\,f(1)\,$ can help you out, say:
$$\forall\,n\in\Bbb Z\,\,,\,\,f(n)=f(n\cdot 1)=nf(1)$$
$$f(1)=f\left(\frac{n}{n}\right)=nf\left(\frac{1}{n}\right).....etc.$$
So you know the values of $\,f\,$ on the integers and then on the rationals of the form $\,1/n\,$ and thus... |
H: Is it possible to 'approximate' compact, convex sets in $\ell^2$ by the Hilbert cube
Define $H=\{(x_n)_n\in\ell^2:|x_n|\le \frac1n, n\in\mathbf N\}\subset\ell^2$. This set is known as the Hilbert cube and it is well-known that $H$ is compact, convex and non-empty. Let $\overline{\mathrm{conv}}(C)$ denote the closure of the convex hull of a subset $C\subset\ell^2$.
Suppose $S$ is a non-empty, compact, convex subset of $\ell^2$, is it possible to write$$S=\overline{\mathrm{conv}}\left(\bigcup_{n=1}^\infty[ S\cap(n\cdot H)]\right),$$ where (for $n\in\mathbf N$ fixed) $n\cdot H=\{n\cdot x:x\in H\}$.
I think it is possible (since the Hilbert cube keeps getting 'thinner' in each coordinate), but I do not know how to prove it.
AI: Take $S:=\{x\}$, where $x_k=\frac{\ln k}k$. Then $S\cap nH$ is empty for all $n$, since if $S\cap nH\subset S$, the only candidate is $x$, and we can't have $\ln k\leq n$ for all $k$. |
H: The print problem: How to show it is not decidable?
I wonder the following reduction is correct.
I'm trying to show that the following problem "PRINT_BLANK" is not decidable.
Input: (a coding of) Turing machine M.
Question: Does the machine never types "blank" on the stripe when it runs on x?
An attempt for reduction: $AcceptProblem \leq PrintBlank: f(\langle M,x\rangle)= M'.$
Given $M$ and $x$, we'll construct $M'$: For an input $y$ for $M'_x$, $M'_x$ simulates running of M on $w$. if $M$ accepts $w$, $M'_x$ writes "blank" and accepts y, otherwise it writes $x$ itself on the tape and rejects.
Any help?
Thanks!
AI: HINT: to make your reduction work, you should alter $M$ slightly, can you figure out how? If not, let me know.
EDIT: What happens when M already writes a blank during it's execution. |
H: some elementary questions about cardinality
I know little about set theory and while reading some Algebra proof I had difficulty on some details. So my questions are :
If $X$ is an infinite set and $Y$ is the set of all finite subsets of $X$, they have the same cardinality. How can I prove it ?
If $X$ is infinite, then it has the same cardinality of the product $X \times \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers. This seems pretty obvious, specially if you think separately when $X$ is countable and when it's not, but I cant prove this in a rigorous way.
Consider only infinite sets now. I know that if $X$ is countable, then it has the same cardinality as $X \times X$. Is this true in general (I mean, I know there is bijection between $\Bbb R$ and $\Bbb R^2$, but probably there is a counterexample that proves this is wrong for an arbitrary infinite set, or if not, I cannot prove it either) ?
Thank you so much for the help !
AI: The results described below freely use the Axiom of Choice.
Theorem: If $X$ and $Y$ are infinite sets, then the cardinality of $X \times Y$ is the maximum of the cardinality of $X$ and the cardinality of $Y$. For a proof see e.g. Theorem 8 and the following discussion in these notes.
This result answers your third question and also your second question, because every infinite set contains a countably infinite subset.
As for your first question: one way to do it is to write the set $Y$ of all finite subsets of your infinite set $X$ as a countable union of sets $Y_n$ for $n \in \mathbb{N}$, where each $Y_n$ is the set of subsets of $X$ having at most $n$ elements. There is a natural surjection $X^n \rightarrow Y_n$ which shows that $\# Y_n \leq \# X^n = \# X$. Thus $Y$ is a countable union of sets each having cardinality at most $X$, so $Y$ has cardinality at most $X$ (you can think of this in terms of $\# X \times \# \mathbb{N} = \# X$, for instance). On the other hand, the subset $Y_1$ of one element subsets of $X$ is naturally in bijection with $X$ itself, so also $\# X \leq \# Y$. It follows from Cantor-Bernstein that $\# X = \# Y$. |
H: Formation sequence for a logic formula
I will start with some definitions from An Introduction to Mathematical Logic and Type Theory: To Truth through Proof by Peter B. Andrews then give the exercise that I am working along with my attempt at a proof. My actual question will be at the bottom, if you wish to skip ahead.
Definitions:
(1) The set of wffs is the intersection of all sets $\mathscr{S}$ of formulas such that:
(i) $\mathbf{p} \in \mathscr{S}$ for each propositional variable $\mathbf{p}$.
(ii) For each formula $\mathbf{A}$ if $\mathbf{A} \in \mathscr{S}$, then $\mathord{\sim} \mathbf{A} \in \mathscr{S}$.
(iii) For all formulas $\mathbf{A}$ and $\mathscr{B}$, if $\mathbf{A} \in \mathscr{S}$ and $\mathbf{B} \in \mathscr{S}$, then $\left[\mathbf{A} \lor \mathbf{B} \right] \in \mathscr{S}$.
(2) A wff is a member of the set of wffs.
Theorems:
1000 Principle of Induction on the Construction of a Wff. Let $\mathscr{R}$ be a property of formulas, and let $\mathscr{R}(\mathbf{A})$ mean that $\mathbf{A}$ has property $\mathscr{R}$. Suppose
(1) $\mathscr{R}(\mathbf{q})$ for each propositional variable $\mathbf{q}$.
(2) Whenever $\mathscr{R}(\mathbf{A})$, then $\mathscr{R}(\mathord{\sim} \mathbf{A})$.
(3) Whenever $\mathscr{R}(\mathbf{A})$ and $\mathscr{R}(\mathbf{B})$, then $\mathscr{R}([\mathbf{A} \lor \mathbf{B}])$.
Then every wff has property $\mathscr{R}$.
Exercise:
A formation sequence for a formula $\mathbf{Y}$ is a finite sequence $\mathbf{Y}_1, ..., \mathbf{Y}_m$ of formulas such that $\mathbf{Y}_m$ is $\mathbf{Y}$ and for each $i$ ($1 \le i \le m$), one of the following conditions holds:
(a) $\mathbf{Y}_i$ is a propositional variable;
(b) there is $j < i$ such that $\mathbf{Y}_i$ is $\sim \mathbf{Y}_j$;
(c) there are indices $j < i$ and $k < i$ such that $\mathbf{Y}_i$ is $\left[ \mathbf{Y}_j \lor \mathbf{Y}_k \right]$.
Prove that a formula $\mathbf{Y}$ is well-formed iff it has a formation sequence. (Hint: Metatheorem 1000 may be useful.)
Partial Proof:
Suppose $\mathbf{Y}$ is a wff. If $\mathbf{Y} = \mathbf{p}$ for some proprositional variable $\mathbf{p}$ then $\mathbf{Y}_1 = \mathbf{p}$ is a formation sequence for $\mathbf{Y}$. Now assume that formulas $\mathbf{A}$ and $\mathbf{B}$ have formation sequences $\mathbf{A}_1, ..., \mathbf{A}_m$ and $\mathbf{B}_1, ..., \mathbf{B}_n$, respectively. If $\mathbf{Y} = \mathord{\sim}\mathbf{A}$ then $\mathbf{A}_1, ..., \mathbf{A}_m, \mathbf{Y}$ is a formation sequence for $\mathbf{Y}$. If $\mathbf{Y} = \left[ \mathbf{A} \lor \mathbf{B} \right]$ then $\mathbf{A}_1, ..., \mathbf{A}_m, \mathbf{B}_1, ..., \mathbf{B}_n, \mathbf{Y}$ is a formation sequence for $\mathbf{Y}$. By Metatheorem 1000, we can conclude that $\mathbf{Y}$ has a formation sequence.
Question:
Now I am stuck on the proof in the other direction. I know I need to start with "Suppose $\mathbf{Y}$ is a formula with a formation sequence $\mathbf{Y}_1, ..., \mathbf{Y}_m$." I think I need to continue from there with "Let $\mathscr{S}$ be a set of formulas satisfying conditions (i), (ii), and (iii) of the definition of the set of wffs." Now I need to show that $\mathbf{Y} \in \mathscr{S}$. I believe I need to do this by induction. I can see the idea in my head, but I am having trouble formulating it on paper (or on screen, as the case may be). I can start with a base case: "If $\mathbf{Y} = \mathbf{Y}_m = \mathbf{p}$ for some propositional variable $\mathbf{p}$, then $\mathbf{Y} \in \mathscr{S}$." The next case to consider is "If $\mathbf{Y} = \mathbf{Y}_m = \mathord{\sim} \mathbf{Y}_j$ for some $j < m$..." then what? I haven't stated a clear inductive hypothesis, and that's where I'm truly stuck. Should I proceed by induction on $m$, the length of the formation sequence for $\mathbf{Y}$? Or should I use structural induction similar to what I did for the first half of the proof?
Any pointers here would be greatly appreciated. Thanks.
AI: Hint: Prove the result by (strong) induction on the length of the formation sequence. So we suppose the result is true for any formula in a formation sequence of length $\lt n$, and show the result holds for any formula in a formation sequence of length $n$.
More precisely, we want to show (by the aforementioned induction) that the collection of end formulas of our sequence of length $n$ is closed under (i),(ii), and (iii), and is therefore an element of any set $\mathscr{S}$ in the definition of the set of wffs. Once one decides what has to be done, the actual details more or less write themselves. |
H: General conditions for $f \in K[\zeta_p] \Rightarrow f(\zeta_p) \in K$
suppose there is a polynomial $f$ in $\zeta_p$ (root of unity) with coefficients in $K$, $\text{char}(K)>0$. Are there conditions when $f(\zeta_p) \in K$?
AI: Suppose that $f(\zeta_p) = k \in K$. Then $f(\zeta_p) - k = 0,$ and
so $f(X) -k $ is a polynomial which has $\zeta_p$ as a zero, i.e. is divisible by the minimal polynomial of $\zeta_p$ over $K$.
Thus $f(\zeta_p) \in K$ if and only if $f(X) \equiv \text{ a constant } \bmod$ the minimal poly. of $\zeta_p$ over $K$.
This is a fairly tautological answer, but I'm not sure what else one should say, since the minimal polynomial is a factor of $(X^p -1)/(X-1)$ which depends very much on the particular choice of the field $K$. |
H: Relation of compositum of fields
Let $E/k$ be a finite field extension, $\operatorname{char}(k)=p>0$. Suppose that $E^p k = E$. Is it then true that $E^{p^n}k = E$ for any positive integer $n$? If yes, why?
Thanks.
AI: Yes, it is true. I will show that $E=E^{p^2}k$ and leave to you the proof of the general case $E=E^{p^n}k$.
Since $E=E^{p}k$, any element $e\in E$ can be written as $e=\sum q_ie_i^p\;$ (for some $e_i\in k, q_i\in k$) .
[This is due to the fact that the ring formed by the sums on the right is already a field, because that ring is a $k$-subalgebra of the algebraic extension $E/k$]
In the same way, each $e_i$ can be written as $e_i=\sum q_{ij}e_{ij}^p$.
Substituting yields $$e=\sum q_i(\sum q_{ij}e_{ij}^p)^p=\sum q_iq_{ij}^pe_{ij}^{p^2}$$
which shows that indeed $E=E^{p^2}k$ |
H: Point set topology
Some time back, I tried reading Rudin's Principles of Mathematical Analysis and I found no trouble with the introductory chapter. In chapter II I encountered point set topology. The number of theorems packed into some pages kind of overwhelmed me. I got stuck there as I felt I did not fully understand everything geometrically. I thought topology was supposed to be geometric.
So, I request you to suggest something in this regard. I am contemplating learning some basic point set topology from elsewhere (I don't know where from) and I am wondering if this is the solution to my problem. (Yes, as a preparation for analysis. Even some good notes would be nice or perhaps a great book)
Any help is appreciated.
PS: I hate compromising rigor.
AI: You're right, I think Rudin's Chapter 2 is probably not the best place to first learn point-set topology due to how dense and concise his writing is.
James Munkres' Topology is one of the most common introductions to general topology, and it has some nice pictures in Chapter 2 to give some geometric intuition where topological spaces are first introduced. I like this book. The majority of the exercises are not overly challenging, so it helps to get familiarity with the subject.
I also like Stephen Willards General Topology which is similar to Munkres, but I'd say it's slightly more difficult than Munkres' book.
Finally, although a little older, Kelley's General Topology is a good reference on general/point-set topology, but probably better suited for use after going through some of the previously mentioned books. |
H: Find the area of the region inside the limaçon
I'm struggling to figure out the answer to this:
Find the area of the region inside the limaçon, $r=3 + \sin(\theta)$
Could someone please help me out?
AI: It would help to plot it first.
Now use this formula $$A = \frac{1}{2}\int_{\alpha}^{\beta}f(\theta)^2\ d\theta$$
From the picture we see that the curve runs from $0$ to $2\pi$
$$A = \frac{1}{2}\int_{0}^{2\pi}(3 + \sin(\theta))^2\ d\theta = \frac{19\pi}{2}$$ The computation is trivial and it's something you can look up on WolframAlpha if you aren't sure |
H: Combination - How many different ways
I am stomped on the following question
How many different ways are there to draw $6$ cards from a standard deck of cards and obtain $4$ kings and $2$ jacks? (The Answer is $6$)
I believe I am starting the question all wrong since I am doing this for
How many different ways are there to draw $6$ cards from a standard deck of cards
No. of ways = $\frac{52!}{46!6!} $
Any suggestions on how I would solve this problem ?
AI: I believe the total number of hands is completely irrelevant for answering this question. All you need to know is that you need 6 cards. 4 cards need to be kings. 4 cards need to be jacks. You only have 4 kings, and you only have 4 jacks. The number of ways you can draw 4 kings out of a set of 4 kings is 4C4 = 1. The number of ways you can draw 2 jacks from a set of 4 is 4C2 = 6.
This makes your answer 4C4 * 4C2 = 1 * 6 = 6 |
H: Graph Homology and Rank-Nullity Theorem
Let $G$ be a connected, directed graph with $v$ vertices and $e$ edges. According to Massey (Ch. VIII, Section 3), the euler characteristic satisfies
\begin{align}
v - e = \chi(G) = \text{rank} \, H_0(G) - \text{rank} \, H_1(G) = 1 - \text{rank} \, H_1(G).
\end{align}
Based on basic definitions, $H_{1}(G) = \ker \partial_1 / \text{im} \, \partial_2$, where $\partial_*$ is the (chain) boundary map. I'd like to compute $\text{dim} \ker \partial_1$ and $\text{dim} \, \text{im} \, \partial_1$.
From what I understand, if $\rho(\partial_1)$ is a matrix representation of $\partial_1$, then it has size $e \times v$. By Rank-Nullity, $\dim \ker \partial_1 + \dim \text{im} \, \partial_1 = v$. Since $G$ has no $2$-chains, then $H_1 \cong \ker \partial_1$. Thus, $\dim \ker \, \partial_1 = e- v + 1$ (also the number of circles in the homotopy type of $G$) and $\dim \, \text{im} \, \partial_1 = 2v - e - 1$.
I have a feeling that I've misunderstood something fundamental. Is my reasoning correct?
Edit: It appears that I switched rank with size, so $\dim \, \text{im} \, \partial_1 = v - 1$ consistent with the rank of $\rho(\partial_1) = v - 1$, as wckronholm points out. Now the question boils down to the following :Is $\dim \ker \partial_1 = 1$ or $e - v + 1$? It seems now it is $1$. However, $\text{rank} \, H_1(G) = 1 - \chi(G) = e - v + 1$, but $\text{rank} \, H_1(G) = \dim \ker \partial_1 - \dim \, \text{im} \, \partial_2$ and $\dim \, \text{im} \, \partial_2 = 0$, no?
AI: The graph is connected (and finite, evidently) so the rank of $H_0(G)$ is $1$. Hence $\mathrm{dim}\; H_0(G) = \mathrm{dim} \; \mathrm{ker}\; \partial_0 - \mathrm{dim} \; \mathrm{im}\; \partial_1$. But $\mathrm{dim}\;\mathrm{ker} \;\partial_0 = v$ so this gives $\mathrm{dim}\;\mathrm{im}\;\partial_1 = v-1$.
It seems to me that, using your notation, the rank of $\rho(\partial_1)$ is $v-1$.
Edited to address question edits:
As you computed above, $\mathrm{dim}\;\mathrm{ker}\; \partial_1 = e-v+1$. This is consistent with rank-nullity since $\mathrm{dim}\;\mathrm{im}\; \partial_1 + \mathrm{dim}\;\mathrm{ker}\;\partial_1 = (v-1) + (e-v+1) = e = \mathrm{dim}\; C_1(G)$. |
H: When is the geometric multiplicity of an eigenvalue smaller than its algebraic multiplicity?
I was kinda crushed to discover that two different matrices with different properties can actually share the same characteristic polynomial ($-\lambda^3-3\lambda^2+4$):
$A=\begin{pmatrix}
1 & 2& 2\\
-3 &-5 &-3 \\
3& 3 & 1
\end{pmatrix} , B=\begin{pmatrix}
2 & 4& 3\\
-4 &-6 &-3 \\
3& 3 & 1
\end{pmatrix}$
$A$ has an eigenline and an eigenplane (and thus an eigenbasis), whereas $B$ has two eigenlines (so no eigenbasis). The repeated eigenvalue -2 of B corresponds to an eigenspace with basis {(-1,1,0)}.
When is the geometric multiplicity of an eigenvalue smaller than its algebraic multiplicity (as in case B)? Are there general conditions to look for?
Thanks!
AI: The general condition is the presence of nontrivial Jordan blocks. |
H: Proving inequality $\prod _{i=1}^n\frac {1-a_i} {a_i}\geqslant \left( n-1\right) ^n$
Let $a_1,a_2,\ldots ,a_n\in \left( 0,1\right)$ be real numbers such that $\sum\limits_{i=1}^n a_i=1$. Prove that
$$\prod _{i=1}^n\dfrac {1-a_i} {a_i}\geqslant \left( n-1\right)^n.$$
AI: By AM-GM
$$\dfrac {1-a_i} {a_i} = \frac{a_1+a_2+..+a_{i-1}+a_{i+1}+..+a_n}{a_i} \geq (n-1) \frac{\sqrt[n-1]{a_1a_2..a_{i-1}a_{i+1}a_n}}{a_i}$$
Multiply them together and observe that everything in $\prod_{i=1}^n \frac{\sqrt[n-1]{a_1a_2..a_{i-1}a_{i+1}a_n}}{a_i}$ cancels. |
H: Surjective and injective functions
So let's say that we have some function $f: \mathbb{A} \rightarrow \mathbb{B} $
Is it possible to have some function such that not all elements of A map to some value in B?
Like for example, in these pictures for various surjective and injective functions:
Would it be possible to have some function that has elements in A that don't map to any values of B? Like in example 1, just have the 3 in A without mapping to the element in B?
AI: By definition, $f$ is a function from $A$ to $B$ if it assigns to each element $a \in A$ an element $f(a) \in B$. A partial function from $A$ to $B$ is exactly what you're after: it is a function assignment to some elements $a \in A$ values $f(a) \in B$. In a context when partial functions are discussed, if you want to emphasize that a function is not partial, then you call it a total function. |
H: Calculating: $\lim_{n\to \infty}\int_0^\sqrt{n} {(1-\frac{x^2}{n})^n}dx$
Possible Duplicate:
Prove: $\lim\limits_{n \to \infty} \int_{0}^{\sqrt n}(1-\frac{x^2}{n})^ndx=\int_{0}^{\infty} e^{-x^2}dx$
I need some help calculating the above limit.
What i have observed so far is that:
For all x the limit of the sequence inside the integral as n tends to infinity is $e^{-x^2}$
I can use Dini's theorem to show that: $f_n(x) = {(1-\frac{x^2}{n})^n}$ uniformly converges to $e^{-x^2}$
Consequently i can use the theorem regarding "the integral of the limit is the limit of integrals" for the function sequence $f_n(x)$ and it's limit function $f(x) = e^{-x^2}$
All this is well and good, but i would be ignoring the fact that the interval integrated upon is [0,$\sqrt{n}$], and so also affected by the limit.
What should i do to resolve this?
AI: We can apply dominated convergence theorem, since $e^{—x}\geq 1-x$ for each $x\geq 0$ and $e^{-x^2}$ is integrable on $(0,+\infty)$. Define $g(x)=e^{-x}-(1-x)$. Then $g'(x)=-e^{-x}+1\geq 0$ for $x\geq 0$ hence $g(x)\geq g(0)=0$. Now, we have for $0\leq x\leq \sqrt n$,
$$0\leq \left(1-\frac{x^2}n\right)^n\leq \left(e^{-\frac{x^2}n}\right)^n=e^{-x^2}.$$ |
H: Should this be $\int u^2 \sqrt{u+1} du$ after substitution of $u=e^x$?
Should this be $\int u^2 \sqrt{u+1} du$ after substitution of $u=e^x$?
http://www.wolframalpha.com/input/?i=integrate+%28exp%28x%29%29^2*sqrt%281%2Bexp%28x%29%29 -> show steps ->3rd row
AI: No, when you do the substitution $u=e^x$ then $du=e^x dx$ so $dx=du/u$ which is why it's $u$ and not $u^2$. |
H: easy "show this is a subset" question
I am having a hard time showing this simple relation. Here, all the letters below are in $\mathbb{N} \cup \{0\}$.
$$A = \{(a,b) : a + 2b \leq n+2\}$$
$$B = \{(a,c+1) : a + 2c \leq n\}$$
How to show that $B \subset A$?
This is something I got from a sum over the above indices. I wanted a lower bound on the sum (which is where $B$ came in). Obviously I set $b = c+1$ in $A$ to get $B$ but how to show that it's a subset?
Thanks
AI: I take it you mean for $n$ to be some fixed positive integer.
To show $B \subset A$, you want to show that any ordered pair of $B$ can also be found in $A$. To that end, let $(a, c+1) \in B$. That is, $a + 2c \leq n$.
We want to know whether $(a,c+1) \in A$. In other words, is $a + 2(c+1) \leq n + 2$? Just follow your nose:
$$
\begin{align*}
a + 2(c+1) &= [a + 2c] + 2\\
&\leq [n] + 2,
\end{align*}
$$
where the last line comes from our assumption about $(a,c+1) \in B$ (the brackets are just for emphasis). |
H: Proving inequality $\left( \frac {x} {y}\right) ^{x}\left( \frac {1-x} {1-y}\right) ^{1-x}\geqslant 1$ for $x, y\in (0, 1)$
Let $x,y\in \left( 0;1\right).$ I want to prove that
$$\left( \dfrac {x} {y}\right) ^{x}\left( \dfrac {1-x} {1-y}\right) ^{1-x}\geqslant 1$$
AI: You need to show $y^x(1 -y)^{1 - x} \leq x^x(1 - x)^{1-x}$ for all $0 < y < 1$, given a fixed $0 < x < 1$. So maximize $y^x(1 -y)^{1 - x}$ with respect to $y$ using your favorite maximization technique. Logarithmic differentiation works for example. You'll get $y = x$ as the max. |
H: Need to clarify the "At-least Concept" in Combination.
I managed to solve this question but I had some inquiries regarding the solution.
If two cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the two cards if at least one card is a jack, queen or a king?
Here is how I solved it:
Jack/Queen/King = $4 \times 3 = 12$ cards
Other Cards = $52-12 = 40$ cards
Now we can only pick two cards so:
It can be either from the Jack/King/Queen so $\binom{12}{1}=12$
Both cards can be from Jack/King/Queen so $\binom{12}{2} = 66$
One card can be from the remaining stack (Non - jack,king or queen) so
$\binom{40}{1}=40$
Now the only problem I have with this question is when getting the final value initially I was doing
$1$ from Remaining Cards $\times$ [ ($1$ from Jack/King/Queen) + 2(from Jack/King/Queen) ]= $\binom{40}{1} \times ( \binom{12}{1} + \binom{12}{2} ) $
But The actual answer comes if we do the following
[$1$ from Remaining Cards $\times$ $1$ from Jack/King/Queen ] + 2(from Jack/King/Queen) ]= $( \binom{40}{1} \times \binom{12}{1}) + \binom{12}{2} $
I would really appreciate it if someone could clarify why do we do it the second way and not the first way ? Which part is added/multiplied to which part ? Is there an easier way to know how its done. Am I missing some important concept here ?
Edit :
While trying to understand this I also looked up the definition of disjoint events which means "Two events are disjoint if they can't both happen at the same time" so then again
here is what I did (Special here means Jack/king/Queen)
(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special)
$(\binom{40}{1} \times \binom{12}{1}) + (\binom{40}{1} \times \binom{12}{2}) ) $
Which simplifies to $\binom{40}{1} \times ( \binom{12}{1} + \binom{12}{2} ) $
So is the representation of the problem using
(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special) wrong ?
AI: If you expand your first answer, you have
$$
\binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2} \cdot \binom{40}{1}.
$$
Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.
A rule of thumb is to turn "at least" problems into several instances of "exactly".
For the problem at hand, "at least one jack/queen/king" translates into "either exactly one jack/queen/king or exactly two jack/queen/king".
The number of ways to get exactly one jack/queen/king and one other card is
$$
\binom{12}{1} \cdot \binom{40}{1}.
$$
We multiply here because you choose a card from the jack/queen/king pile and a card from the "other" pile.
The number of ways to get exactly two from jack/queen/king is
$$
\binom{12}{2}.
$$
Since these cases are disjoint (this is important), we simply add the results from the two cases. Thus, the number of ways to get at least one Jack, Queen, or King is
$$
\binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2}.
$$ |
H: Is there notation denoting that one sigma-algebra is sub-sigma-algebra of another?
The question is self-describing.
AI: I have never seen such a notation, per se. The closest I've seen is something like:
Let $\mathcal{F}, \mathcal{G}$ be $\sigma$-algebras, with $\mathcal{F} \subset \mathcal{G}$.
That is, using $\subset$ (to indicate containment as sets), where it is made clear elsewhere (or from context) that the sets in question are $\sigma$-algebras. |
H: Projectile Motion
Hello Stack Exchangers!
I'm developing a video game. One feature requires an archer be able to target an enemy and shoot an arrow at it. I've looked around and found plenty of guides on how to do this with the quadratic formula with the appropriate variables and am receiving what I believe to be the correct output, but the arrow trajectory isn't what it should be (it doesn't hit the enemy). Here is what I have (with a sample of values taken from the program).
Origin of the shot (608,-352);
Target of the shot (1280, -705);
x = 672;
y = -353;
g = -275 (this is what I'm using for gravity);
v = 1000; (The velocity the arrow is shot at);
setting up the variables:
a = $.5g(x/v)^2 = -62.0928$
b = $x = 672$
c = $.5g(x/v)^2+y = -415.0928$
$ angle 1 = atan(\cfrac{-b + \sqrt {b^2 - 4ac}}{2a}) = 0.5817$
$ angle 2 = atan(\cfrac{-b - \sqrt {b^2 - 4ac}}{2a}) = 1.4727$
Then in order to get the individual X/Y velocities from that angle (which I believe is in radians):
For Angle 1
$v_x = v cos \theta = 835.50$
$v_y = v sin \theta = 549.48$
For Angle 2
$v_x = v cos \theta = 97.90$
$v_y = v sin \theta = 995.19$
Unfortunately, the arrow doesn't arc through the target location. The second angle (from the - part of the quadratic equation) doesn't fare any better.
Any help out there?
AI: The basic equations for a projectile without air resistance are (assuming initial time is set to $0$):
$$y = y_0 + v_0 \sin \theta t + \frac12gt^2$$
$$x = x_0 + v_0 \cos \theta t$$
From the second one we get $t = \frac{x-x_0}{v_0 \cos \theta}$. Plug that in in the first one, do some manipulation and we get the following:
$$
\Delta y = \tan \theta \Delta x + \frac{g (\Delta x)^2}{2 v_0^2}+\frac{g (\Delta x)^2}{2 v_0^2}\tan^2 \theta
$$
What I call $\Delta x$ and $\Delta y$ are what you call $x$ and $y$, that is, the difference between target and origin of the shot. From here you can get $\theta$. It seems to me that you simply have a sign error in $c$: It should be $c = \frac{g (\Delta x)^2}{2 v_0^2} - \Delta y$, as seen from the above equation. Let me know if this works. |
H: What does "the orthogonal basis vectors spanning the subspace perpendicular to vector $\vec{e}_1$" mean?
I am reading a paper titled "A Robust Real-Coded Genetic Algorithm using Unimodal Normal
Distribution Crossover Augmented by Uniform Crossover : Effects
of Self-Adaptation of Crossover Probabilities" by Ono, Kita, and Kobayashi. You can download the paper from CiteSeerX.
In the discussion about the UNDX operator on page 2, Ono et. al use the phrase "vectors $\vec{e}_2$, ..., $\vec{e}_{n_{param}}$ are the orthogonal basis vectors spanning the subspace perpendicular to vector $\vec{e}_1$".
I do not have the mathematical background to interpret this phrase. I need to understand it so I can code UNDX and similar genetic crossover operators. What does it mean, and how do I do it?
My best interpretation after consulting Wikipedia pages for concepts like basis and spanning set is that I'm supposed to build a rotation and translation of an identity matrix so that one of my points is the origin and the vector to one of the other points is one of my axes. I do not know whether this interpretation is correct. Trying to follow this idea lead to a bunch of 3D specific answers for building rotation matrices from two vectors, but I need a general approach that can handle any number of dimensions.
Am I even on the right track?
EDIT
I have read the Wikipedia pages on the Gram-Schmidt process. It seems easy enough to apply Gram-Schmidt to the UNDX process in the paper, but there is also a UNDX-m process proposed in this paper by Kita, Ono, and Kobayashi. Another paper by Deb, Joshi, and Anand also describes UNDX-m and proposes a modification to it called PCX. Both of these processes allow for an arbitrary number of points to be used in the recombination. Both of them use the same sort of language to describe this a set of vectors $\left\{\vec{e}^{\left(i\right)} \forall 1 \le i \le n_{param} \right\}$, and I am confused as to how I can apply the Gram-Schmidt process to a set of arbitrary points.
I will use the example of PCX since access to that paper is free. Suppose that my number of points $\mu=10$, and I choose the ten blue parent points marked as blue diamonds in the figure below. The red square represents the mean vector $\vec{g}$ of the ten parents. The green arrow represents the direction vector $\vec{d}^{\left(p\right)}$. PCX requires knowing the $\vec{e}^{\left(i\right)}$ vectors that are the "orthonormal bases that span the subspace perpendicular to $\vec{d}^{\left(p\right)}$."
How do I determine the $\vec{e}^{\left(i\right)}$ vectors described above from the vector $\vec{d}^{\left(p\right)}$ and the remaining nine parent points? Applying Gram-Schmidt using all possible $\vec{d}^{\left(i\right)} \equiv \vec{p}^{\left(i\right)} - \vec{g}$ as the set of vectors does not seem to work...
EDIT
I think I figured out the answer to my edit. I'm not sure why it took so long to click >_<
I had been assuming that there had to be something special about the basis calculated so that points outside of those described could not be generated. But that is not the case; the only necessity is that this basis be able to generate vectors in the same directions as the points used. This means that if all the points chosen happen to lie on a hyperplane, there should only be enough orthogonal vectors needed to describe the plane.
If this is true, then it does not matter which $n_{param} - 1$ of the remaining $\mu - 1$ points are chosen for the Gram-Schmidt process. So long as the vectors do not lie along a line, they will generate a new basis vector, and in conjunction they will describe a spanning set possible of describing any and all of the points generated.
I can see why people might have found my bewilderment confusing now : )
AI: This is some very basic linear algebra. The vectors $\vec{v}_i$ are orthogonal: this means that $\vec{e}_j^T \vec{e}_k = 0$ for all $j \ne k$. They are a basis of the subspace orthogonal to $\vec{v_1}$: this means that every vector orthogonal to $\vec{v}_1$ can be
written in exactly one way as a linear combination $\sum_{i=2}^{n_{param}} c_i \vec{v}_i$ for some scalars $c_i$. Thus together with $\vec{v_1}$ they are a basis of the whole vector space: every vector in the vector space can be written in exactly one way as a linear
combination $\sum_{i=1}^{n_{param}} c_i \vec{v}_i$.
It's not clear that you are "supposed to" do anything, but if you want to find such a basis the Gram-Schmidt process is often a good way to do it. |
H: "L'Hôpital's rule" vs. "L'Hospital's rule"?
I know this is not strictly a mathematical question, and I considered putting it on Linguistics SE, but I decided that seeing as this is most probably a mathematical history question, it would be better placed here on math SE.
My question is:
Why is "L'Hôpital's rule" often referred to as "L'Hospital's Rule" in english mathematical literature?
I am aware that the translation from French to English of "L'Hôpital" is "The Hospital", but I haven't seen any cases of other french names which correspond to proper nouns being translated into english, so why the special case here?
Again, sorry if this is completely the wrong place to put this question, and moderators feel free to migrate this question to a more appropriate SE board if one exists, but as I said, I believe this to be the most appropriate board.
AI: There was a change in French orthography in the mid 18th century, where some mute s's were dropped and replaced by the circumflex accent. In the Marquis's own time (1661-1704), his name was spelled "l'Hospital".
Edit: Apparently in at least one letter the Marquis spelled his name "Lhospital".
The 1716 edition of "Analyse des infiniment petits" has "l'Hospital".
The 1768 edition of the book has "l'Hôpital". |
H: Set notation: subtracting elements with given cardinality from the powerset
I have a set $S = \{1,2,\ldots,n\}$ of $n$ elements and I denote with $P(S)$ the powerset of $S$.
Which is a correct and accepted notation to say that the set $Z$ is composed by all the elements in $P(S)$ with the exception of all the subsets of cardinality $h$?
Example: if $S=\{1,2,3\}$ then $P(S) = \{\emptyset, \{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$.
If $h = 1$, then it should be $Z = \{\emptyset, \{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}=P(S)\setminus\{\{1\},\{2\},\{3\}\}$.
How can it be expressed in a formal and concise way, for any value of $n$ and $h$?
Thanks
AI: $$ Z = \left \{ x | x \in P(S) , |x| \neq h \right\} $$
An alternate, more compact version:
$$Z = \{x \in P(S) : |x| \neq h\}$$ |
H: Local rings and flatness
Let $A \rightarrow B$ be a flat and local homomorphism of commutative local rings.
Let $M,N$ be two $B$-modules which are free of finite rank as $A$-modules.
Consider the product $M \otimes_B N$ as an $A$-module. Is this $A$-module flat?
AI: In general the tensor product may not be torsion-free even. Let $B=k[[x,y]]/(xy)$ and $A=k[[x-y]]\subset B$. Let $M=B/xB$ and $N=B/yB$. |
H: A basic group question
Let $G=\{0, \cdot\}$.
I'm arguing with someone over if $G$ is a group with the regular multiplication since I don't see why it isn't.
Addition:
Now, $G=\{\mathbb{Z},\triangle \}$ with $x \triangle y=x+y+xy$. Is it true that $G$ is not a group and the only subset of $\mathbb{Z}$ to form a group with $\triangle$ is $\{0\}$?
AI: Yes.
Closure $0\cdot0=0 \in G$
2.Associativity $(0\cdot0)\cdot 0 = 0(\cdot0\cdot0)$
3.Identity element is $0$.
4.Inverse element holds, because if not, eist $x\neq0 \in G$ that don't have Inverse element. Absurd. |
H: Using the pumping lemma to show that a language is not regular (Computer Science)
Show that $L=\{a^{n^2} | n \ge 0\}$ is not regular
Hey guys. I'm taking a CS class and this stuff is really new to me so bear with me.
I tried to look if I get some contradiction by using the pumping lemma for regular languages and I worked it out like this:
Suppose $L$ is regular. Then there must be a natural number $m$ for all words $z$ in $L$ with length $|z| \ge m$ and there exists a decomposition $z = uvw$, $|uv| \le m$, $|v| > 0$, so that $uv^iw$ is in the language for any $i \ge 0$.
Consider the string $a^{m^2}$.
Then $uv = a^{k^2} = a^{x+y}$, for some $k \le m$ and $x = (k-1)^2$.
Then $v = a^y = a^{2k-1}$.
Let $i = 2$. Then $uv^2w$ = $a^{x+2y}$. But $\sqrt{x+2y}$ is not necessarily a natural number $\Rightarrow$ Contradiction! Hence, $L$ can not be regular.
Well, I know that this way is unnecessarily complicated and you can prove it differently (I already know the most simple solution). But my question here is: Is my proof valid as well or does it contain any flaw? Is it formally correct?
I appreciate any feedback! Thanks!
AI: Here are my comments. I leave some of the details to the reader.
Let's look at your proof starting from the beginning:
Suppose $L$ is regular. Then there must be a natural number $m$ for all words $z$ in $L$ with length $|z| \ge m$...
So far so good. You are following the logical structure of the Pumping Lemma. In order to continue this, I would change the rest of this paragraph as follows (changes italicized):
...such that $z$ can be decomposed as $z = uvw$ where $|uv| \le m$, $|v| > 0$, and $uv^iw$ is in the language for any $i \ge 0$.
Next:
Consider the string $a^{m^2}$.
This is perfect. You could state the obvious fact that this string is in $L$. Now you need decompose the string in such a way that there is a contradiction to the Pumping Lemma.
Then $uv = a^{k^2} = a^{x+y}$, for some $k \le m$ and $x = (k-1)^2$.
You are on the right track here. You are trying to decompose the chosen string. However, you are only decomposing it into $u$ and $v$, but haven't mentioned the remaining bits $w$. Even though this $w$ isn't used as part of the contradiction, you need to show it as part of the decompositions.
The other problem I see is that the word "Then" implies that you are deducing something. However, the exact decomposition is left to you as a choice rather than a logical deduction. With that in mind, I would continue with something like
Let $u = ...$, $v = ...$, and $w = ...$. Note that $a=uvw$.
I'll let you fill in the blanks in the above sentence. Next you consider some string of the form $uv^iw$ for some $i \ge 0$ and show that this string is no longer in $L$. I believe your ideas will fit into this framework. |
H: Find the domain of $f(x)=\frac{3x+1}{\sqrt{x^2+x-2}}$
Find the domain of $f(x)=\dfrac{3x+1}{\sqrt{x^2+x-2}}$
This is my work so far:
$$\dfrac{3x+1}{\sqrt{x^2+x-2}}\cdot \sqrt{\dfrac{x^2+x-2}{x^2+x-2}}$$
$$\dfrac{(3x+1)(\sqrt{x^2+x-2})}{x^2+x-2}$$
$(3x+1)(\sqrt{x^2+x-2})$ = $\alpha$ (Just because it's too much to type)
$$\dfrac{\alpha}{\left[\dfrac{-1\pm \sqrt{1-4(1)(-2)}}{2}\right]}$$
$$\dfrac{\alpha}{\left[\dfrac{-1\pm \sqrt{9}}{2}\right]}$$
$$\dfrac{\alpha}{\left[\left(\dfrac{-1+3}{2}\right)\left(\dfrac{-1-3}{2}\right)\right]}$$
$$\dfrac{\alpha}{(1)(-2)}$$
Now, I checked on WolframAlpha and the domain is $x\in \mathbb R: x\lt -2$ or $x\gt 1$
But my question is, what do I do with the top of the problem? Or does it just not matter at all.
AI: Note that $x^2+x-2=(x-1)(x+2)$. There is a problem only if $(x-1)(x+2)$ is $0$ or negative. (If it is $0$, we have a division by $0$ issue, and if it is negative we have a square root of a negative issue.)
Can you find where $(x-1)(x+2)$ is $0$? Can you find where it is negative? Together, these are the numbers which are not in the domain of $f(x)$.
Or, to view things more positively, the function $f(x)$ is defined precisely for all $x$ such that $(x-1)(x+2)$ is positive.
Remark: Let $g(x)=x^2+x-2$. We want to know where $g(x)$ is positive. By factoring, or by using the Quadratic Formula, we can see that $g(x)=0$ at $x=-2$ and at $x=1$.
It is a useful fact that a nice continuous function can only change sign by going throgh $0$. This means that in the interval $(-\infty, -2)$, $g(x)$ has constant sign. It also has constant sign in $(-2,1)$, and also in $(1,\infty)$.
We still don't know which signs. But this can be determined by finding $g(x)$ at a test point in each interval. For example, let $x=-100$. Clearly $g(-100)$ is positive, so $g(x)$ is positive for all $x$ in the interval $(-\infty,0)$.
For the interval $(-2,1)$, $x=0$ is a convenient test point. Note that $g(0) \lt 0$, so $g(x)$ is negative in the whole interval $(-2,1)$. A similar calculation will settle things for the remaining interval $(1,\infty)$.
There are many other ways to handle the problem. For example, you know that the parabola $y=(x+2)(x-1)$ is upward facing, and crosses the $x$-axis at $x=-2$ and $x=1$. So it is below the $x$-axis (negative) only when $-2 \lt x \lt 1$.
Or we can work with pure inequalities. The product $(x+2)(x-1)$ is positive when $x+2$ and $x-1$ are both positive. This happens when $x \gt 1$. The product is also positive when $x+2$ and $x-1$ are both negative. This happens when $x \lt -2$. |
H: Generalization of metric spaces
The Wikipedia article for metrics mentions several generalizations of metric spaces, but all of them seem to have the property that the metric must be non-negative for all x and y. To me it seems like a space where distances don't have to be non-negative would be an obvious generalization (removing the symmetry property would also be necessary). For example, consider the real line with the "distance function"
$$y-x$$
under which "distances" to numbers on one side of y would be negative while "distances" to numbers on the other side would be positive.
Does this generalization have a name, or is it useless enough not to have been studied?
AI: Will Jagy's comment deserves to be an answer:
Look up Minkowski space and Lorentz manifolds in general.
Minkowski space in $n+1$ dimensions is $\mathbb R^{n+1}$ with the "distance function"
$$ d(\langle x_1,\ldots,x_n, t\rangle, \langle y_1,\ldots,y_n,u\rangle) =
\sqrt{(t-u)^2-(x_1-y_1)^2-(x_2-y_2)^2\cdots-(x_n-y_n)^2}$$
Here $d(x,y)=d(y,x)$ and distances cannot be negative, but they can be null or purely imaginary! (For mathematical sanity, one usually considers the square of this distance function, such as not to be troubled with the multi-valuedness of the square root, though).
Minkowski space is the basic fabric of relativity. Indeed the fundamental postulate of the Special Theory of Relativity could be phrased as:
The stage on which physics plays out can be given the structure of $3+1$-dimensional Minkowski space, such that all fundamental laws of nature are preserved by every Minkowski isometry. (Or at least by every Minkowski isometry that can be "smoothly" turned into the identity).
(And, by the way, light rays connect points whose mutual Minkowski distance is 0).
Beware, however, that the Minkowski distance is not used to define the topology of Minkowski space. One uses the ordinary Euclidean topology on $\mathbb R^{n+1}$.
Minkowski isometries are better known as Lorentz transformations, though often that name is only used for isometries that fix the point $\langle 0,\ldots,0,0\rangle$.
Lorentzian manifolds generalize Minkowski space to curved spacetimes for General Relativity. |
H: Combination - Inverse Way of solving this problem
Regarding my previous post , I'll repeat the question
A five member committee is to be selected from among four Math teachers and five English teachers. In how many different ways can the committee be formed if the committee must contain at least three Math teachers.
I know it could be solved like this
$\binom{4}{3}\binom{5}{2} + \binom{4}{4} \binom{5}{1} = 45 $ Ans
I wanted to know How I would solve this the other way round. Like for example if it was for at least 1 math teacher it would be
$\binom{9}{5} - \binom{5}{5} $ how would I use the same method but instead calculating for at least 1 I would be calculating for at least 3 ?
AI: If I understand you correctly, you want to use that sort of subtraction technique, where you overcount and then take out the complement of the desired probability, to solve the "at least 3" math teacher case.
Then you might do ${9 \choose 5} - \left( {\color{#10C}{5 \choose 5}} + \color{#070}{{5 \choose 4}{4 \choose 1}} + \color{#C01}{{5 \choose 3}{4 \choose 2}}\right)$, where $\color{#10C}{blue}$ counts the number of ways of choosing 5 English people, $\color{#070}{green}$ counts the number of ways of choosing 4 English and 1 mathie, and $\color{#C01}{red}$ counts the number of ways of choosing 3 English and only 2 mathies.
Of course, this is nothing more than saying that ${9 \choose 5} = {5 \choose 5} + {5 \choose 4}{4 \choose 1} + { 5 \choose 3}{4 \choose 2} + {5 \choose 2}{4 \choose 3} + { 5 \choose 1}{4 \choose 4}$. |
H: A qualifying exam question concerning compactness
Here is a qualifying exam question which I hope someone can help me with. I have done all of it except having problem to "visualize" $\partial S$ and hence have no idea, is the intersection of $K \cap \partial S$ empty or not?
Here is the question
Assume $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is a continuous function such that
(a) there exist points $x_0$ and $x_1\in \mathbb{R}^n$ with $f(x_0)=0$ and $f(x_1)=3$,
(b) there exist positive constants $C_1$ and $C_2$ such that $f(x)\geq C_1|x|-C_2$ for all $x\in \mathbb{R}^n$.
Let $S:=\{ x\in\mathbb{R}^n : f(x)<2 \}$ and let $K:=\{ x\in\mathbb{R}^n : f(x)\leq 1 \}$. Define the distance from $K$ to $\partial S$ (the boundary of $S$) by the formula
$$
\text{dist} (K, \partial S) := \inf_{p\in K, q\in \partial S} |p-q|.
$$
Prove that the dist$(K,\partial S) >0$. Then give an example of a continuous function $f$ satisfying (a) but dist$(K,\partial S) =0$.
End of question.
As you can see $K=f^{-1}((-\infty, 1])$ and $K\subset \{ x\in \mathbb{R}^n : ||x||\leq \frac{1+C_2}{C_1} \}$ is closed and bounded, surely $\partial S$ is closed, hence I just needed the empty intersection to draw the final conclusion.
Also is there any easy counterexample that needed to satisfied (a)?
Thanks
AI: First, observe that $S$ is open. Next, use the continuity of $f$ to show that $f \le 2$ on the closure of $S$. Using the definition of $\partial S$, conclude that $f = 2$ on $\partial S$. In particular, $\partial S$ is disjoint from $K$. |
H: Basic help with sigma algebras and borel sets
In non-rigorous, intuitive terms, can someone briefly define:
(i) a measurable set
(ii) a borel set
(iii) a sigma algebra
(iv) a borel sigma algebra
Im studying these concepts independently in preparation for a course in the fall and want to make sure I have a functional intuitive idea before learning them rigorously. Im not looking for references to textbooks, or textbook definitions, just a quick intuitive description from someone who is familiar.
AI: A $\sigma$-algebra is, like a topology, a set of subsets of some space $X$. It's both bigger and smaller than a topology, though: smaller, because it's only required to be closed under countable unions, instead of all unions, but bigger, because it's also closed under complementation, and thus by de Morgan's law, countable intersections. So it's both the open and closed sets you'd get starting from the base of a topology if you only took countable unions but also allowed countable intersections.
A measurable set is just an element of some $\sigma$-algebra on $X$. The content comes in when you define measures, which are functions from the $\sigma$-algebra to $[0,\infty]$ that satisfy a few obvious properties of a generalization of length.
The Borel algebra on some topological space $X$ is the $\sigma$-algebra generated by its topology: take all the closed and open sets, countable unions and intersections of those, complements of those, countable unions and intersections of those, and so on. A Borel set is just an element of the Borel algebra.
Note in this case what I said about a $\sigma$-algebra being smaller than a topology does not hold at all! The Borel algebra is important specifically because it's the smallest $\sigma$-algebra containing the topology: we obviously would like to have open and closed sets of reals measurable, and to accomplish that we've got to let at least all the Borel sets be measurable as well. You can think of the Borel sets as every reasonable set; in particular the rationals are Borel in $\mathbb{R}$ with the standard topology, though they're neither closed nor open. |
H: Prime numbers $(x,c,p)$ such that $x^3-p x^2-cx-5c=0$
How should I proceed to find all prime numbers $x,c,p$ such that
$$x^3-px^2-cx-5c=0$$
AI: If $x^3-px^2-cx-5c=0$, then $x$ divides $5c$. Since $x$ and $c$ are prime, we have the two possibilities $x=c$ and $x=5$.
Suppose $x=c$. Substitute. We get $c^3-(p+1)c^2-5c=0$, so $c^2-(p+1)c-5=0$, so $c$ divides $5$, so $c=5$. Therefore definitely $x=5$.
Substitute. We get $125-25p-10c=0$. Since $25$ divides $10c$, it follows that $c=5$. Thus $p=3$. |
H: limit of exponential function and applying l'Hospital's rule
The eqaution goes the following:
$$\lim_{x\to \infty} \left(\frac {x}{e^x-1}\right)^2 e^x = \ ? $$
The first question is I tried to use l'Hopistal's rule, but unsure whether this is the right approach, as the limit goes to the infinity. (and this way did not produced a right answer, which seems to be zero.)
And what is the right way for solving this? (I know what this is when the limit $x$ goes to $0$, which is $1$, but unsure of this case.)
AI: $\displaystyle \lim_{x\to \infty} \left(\frac {x}{e^x-1}\right)^2 e^x = \lim_{x\to \infty} \left(\frac {x^2}{e^x-1}\right)\left(\frac{ e^x}{e^x-1}\right) $ and both the terms in the product are now easy to evaluate, $\displaystyle\lim_{x\to \infty} \left(\frac {x^2}{e^x-1}\right)=0$ and $\displaystyle\lim_{x\to \infty} \left(\frac{ e^x}{e^x-1}\right)=1$, by applying L'hopital's rule on both these limits.
Can you now see why the limit is zero? |
H: Nonexistence of Pythagorean triple with largest side prime.
The Problem: In the Pythagorean triplets (a,b,c) when a < b then b can't be a prime number.
The Background: While searching the properties of Pythagorean triplets in web I saw quite a few listed, but didn't see the above one which I thought was true, because I had developed a proof.
The Request: As discussed many a times in this site I would request some alternate proofs (or counterexamples) before I share mine for a review.
AI: Hint $\rm\,\ a^2\! + p^2 = c^2\:\Rightarrow\: p^2 = (c\!-\!a)(c\!+\!a).\:$ Unique factorization $\:\Rightarrow \begin{eqnarray}\rm\:c\!-\!a &=&1\\ \rm c\!+\!a &=&\rm p^2\end{eqnarray}\:$ contra $\rm\,a<p$ |
H: Pick out the correct statements
Pick out the correct statements from the following list:
a. A homomorphic image of a UFD (unique factorization domain) is again a
UFD.
b. The element $2 ∈ \Bbb{Z}[\sqrt{−5}]$ is irreducible in $\Bbb{Z}[\sqrt{−5}].$
c. Units of the ring $\Bbb{Z}[\sqrt{−5}]$ are the units of $\Bbb{Z}.$
d. The element $2$ is a prime element in $\Bbb{Z}[\sqrt{−5}].$
I think (d) is not correct. But I'm not sure about the others.
AI: Hints: For (a) note that $\mathbb{Z}_4$ is a homomorphic image of $\mathbb{Z}$.
For (b) and (c), use properties of norm. You are right about (d), since $2$ divides $(1+\sqrt{-5})(1-\sqrt{-5})$. |
H: Nilpotent Element And Jacobson Radical
I am looking for a ring with nilpotent elements such that $J(R)=0$ where $J(R)$ is Jacobson radical.
Any suggestion?
AI: The best examples are the matrix rings over a field. These are simple, so they've got trivial Jacobson radical, and yet already the $2\times 2$ matrices have nilpotent elements $e_{12},e_{21}$.
There arguably is one commutative example: in the trivial ring, $1=0$ is nilpotent but the Jacobson radical is, naturally, $0.$ |
H: $im(I)=im(R)$ implies what?
I'm studying an ideal $I \trianglelefteq R$ and noticed that for a certain non-injective, non-zero homomorphism $\varphi: R \rightarrow S$ I can show that $\varphi(I)=\varphi(R)$. I'm wondering if this implies that $I=R$.
It holds for the one little example I could think of. Let $R=\mathbb{Z}$, $S=\mathbb{Z}$, $I=k\mathbb{Z}$, and $\varphi_{m}(n)=mn$, $m > 1$. Suppose $\varphi_{m}(I)=\varphi_{m}(R)$. Then $\varphi_{m}(I)=\varphi_{m}(k\mathbb{Z})=mk\mathbb{Z}=\varphi_{m}(R)=\varphi_{m}(\mathbb{Z})=m\mathbb{Z}$. Thus, $mk\mathbb{Z}=m\mathbb{Z} \Rightarrow k=1 \Rightarrow I=\mathbb{Z}=R$.
(Rings not necessarily commutative and/or may not contain 1.)
AI: Nah. Consider the quotient map of $\mathbb{Z}$ onto, say, $\mathbb{Z}_2$. The image of $3\mathbb{Z}$ is the whole ring. Since you went to the trouble of mentioning this shouldn't be injective, I'll note we could get non-surjectivity without breaking a sweat by making the codomain $\mathbb{Z}_2[X]$. |
H: Proving an identity involving the derivative of the Laguerre polynomials with respect to $n$
I've recently come across the following equality in a paper: suppose one defines an analytic function $L(n,x)$ which is equal to the $n$th Laguerre polynomial for $n\in\{0,1,\ldots\}$, and let* $L^{(1,0)}(n,x) = \frac{\partial}{\partial n}L(n,x)$. Then apparently
$$L^{(1,0)}(-1,-x) = -[\gamma_E + \Gamma(0,-x) + \log(-x)]e^{-x}$$
where $\gamma_E$ is the Euler-Mascheroni constant and $\Gamma$ is the incomplete gamma function.
Numerically, this checks out, but I would be interested in a symbolic proof that the equality holds. I've been playing around with the generating function for the Laguerre polynomials (and some other representations), but I haven't figured out a way to make the connection. Can anyone help me out?
*$L^{(1,0)}$ is called the "multivariate Laguerre polynomial" in the paper; I couldn't find a source to tell me whether this is a widely recognized function or not.
AI: Laguerre function is defined in terms of the hypergeometric ${}_1F_1$ as follows:
$$L_\nu(x) = {}_1F_1\left(-\nu; 1; x\right)$$
Thus:
$$
\left.\frac{\partial}{\partial \nu} L_\nu(z)\right|_{\nu = -1} = \sum_{n=0}^\infty \frac{(1)_n}{n!} \frac{z^n}{n!} \left( \psi(n+1) - \psi(1) \right) = \sum_{n=1}^\infty \frac{z^n}{n!} H_n = \sum_{n=1}^\infty \frac{z^n}{n!} \sum_{m=1}^n \frac{1}{m} \\= \sum_{m=1}^\infty \frac{1}{m} \sum_{n=m}^\infty \frac{z^n}{n!} = \sum_{m=1}^\infty \frac{1}{m} \mathrm{e}^z \frac{\gamma(m,z)}{\Gamma(m)} = \mathrm{e}^z \int_0^z \sum_{m=1}^\infty \frac{t^{m-1}}{m!} \mathrm{e}^{-t} \mathrm{d} t = \mathrm{e}^z \int_0^z \frac{1- \exp(-t)}{t} \mathrm{d} t = \mathrm{e}^z \left( \gamma + \log(z) + \Gamma(0,z) \right)
$$ |
H: Let A be an $n\times n$ matrix with complex entries which is not a diagonal matrix. Pick out the cases when A is diagonalizable
Let A be an $n\times n$ matrix with complex entries which is not a diagonal
matrix. Pick out the cases when A is diagonalizable.
(a) A is idempotent.
(b) A is nilpotent.
(c) A is unitary.
I think (c) is true. and (b) is false .not sure about (a), though order $2$ idempotent matrices are diagonalizable since it has two distinct eigenvalues $0,1$.but what is the general case .
AI: (a) : $A$ is idempotent hence $A=A^{2}$ i.e. $A^{2}-A=0$ hence $x^{2}-x=x(x-1)$
is a polynomial $P$ s.t $P(A)=0$. Since the minimal polynomial of
$A$ divides the polynomial we conclude that it is one of the following
$x,x-1,x(x-1)$ .
In all cases $A$ is diagonalize since the minimal polynomial splits
into a product of coprime linear factors.
(b) : $A$ is nilpotent does not guarantee $A$ is diagonalize, as a counter
example take $A=\begin{pmatrix}0 & 1\\
0 & 0
\end{pmatrix}$ ($A$ is a $2\times2$ jordan block corresponding to the value $0$
, i.e. $A=J_{2}(0)$ ).
(c) : $A$ is unitary and in particular it is normal hence diagonalize |
H: Is this a legitimate way to show that $[-ze^{-z^2/2}]_{-\infty}^{\infty}=0$? (Proving a statement about a function of a normal random variable)
The problem is, let Z be a standard normal variable and $n\geq1$ be an integer. Show that $E[Z^{n+1}]=nE[Z^{n-1}]$. Here's what I've got so far, miraculously:
$E[Z^{n+1}]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^{n+1}e^{-z^{2}/2}dz
$
Doing integration by parts with $u=z^n$ and $dv=ze^{-z^{2}/2}$ gives
$$[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty+n\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-z^{2}/2}z^{n-1}dz=[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty +nE[Z^{n-1}]$$
Now my Calc II teacher actually told me never to do this $[f(x)]_{-\infty}^\infty$ but my probability textbook does it so I'm just gonna throw caution to the wind. I'm thinking I turn $[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty$ into $$\frac{1}{2\pi}(\lim_{z\rightarrow\infty}-z^{n}e^{-z^{2}/2}-\lim_{z\rightarrow -\infty}-z^{n}e^{-z^{2}/2})$$Wolfram Alpha tells me that each of those limits is 0 but I don't quite see how—as far as I can see both of those limits are in indeterminate forms. What I was thinking is maybe since $n$ is an integer greater than 1 that I can argue that if I write $lim_{z\rightarrow\infty}-z^{n}e^{-z^{2}/2}$ as $lim_{z\rightarrow\infty}-z^n/e^{z^2 /2}$ then repeated applications of l'Hopital's rule will eventually put a 0 in the numerator while you'll still have some exponential function in the denominator, and so that'll be zero and hence the limit is zero. But that feels cumbersome and I'm not sure what I can do beyond just asserting that, unless I want, like, prove that by induction on n or something. Am I missing some obvious better way to evaluate this limit?
AI: The calculation using integration by parts is the important part. Now to the main question.
Your idea of using L'Hospital's Rule is good. Think of the limit of $|z|^n e^{-z^2/2}$ as $|z|\to \infty$. (This is just to collapse the two limit calculations into one.)
In order to make the repeated differentiation easier, one might as well let $u=z^2/2$. So $|z^n|=2^{n/2}u^{n/2}$. We are trying to find
$$\lim_{u\to \infty}\frac{2^{n/2}u^{n/2}}{e^u}.$$
It is enough to show that for any positive integer $k$,
$$\lim_{u\to\infty}\frac{u^k}{e^u}=0.$$
This is a routine application of L'Hospital's Rule.
Much more informally (but enough for probability), $e^{z^2/2}$ grows enormously faster in the long run than any power of $z$. |
H: Suppose $G$ is $2$-connected. Show that there exists a path from $x$ to $y$ containing $z$.
I'm studying for a graph theory exam and am stumped on one of the practice questions:
Suppose $G$ is $2$-vertex-connected. Show that for any distinct vertices $x$,
$y$, $z$ of $G$ there exists a path from $x$ to $y$ containing $z$.
I've tried working with ear decompositions and Menger's theorem (since those are the two major focuses we had in class about 2-connected graphs) but I keep ending up with a walk from $x$ to $y$ containing $z$, which isn't good enough because there's no guarantee that $z$ will exist in the resulting path.
AI: Since $G$ is 2-connected, there must exist a cycle $C$ containing $y$ and $z$.
Now, we construct the path $P_{xzy}$ as follows:
Pick any path $P_{xz}$ from $x$ to $z$. If the only vertex it shares with $C$ is $z$ we're done because $P_{xzy}$ is constructed by first following $P_{xz}$ and then following along the cycle from $z$ to $y$.
Otherwise, follow $P_{xz}$ until we find a vertex shared by $P_{xz}$ and $C$, and then follow along $C$ to get $z$ and then $y$.
That is, we pick the shortest prefix $P_{xz}'$ = $x u_1 u_2 \cdots u_k$ of the path $P_{xz}$ such that the first shared vertex between $P_{xz}'$ and $C$ is $u_k$. Then the cycle $C$ is $u_k v_1 \cdots v_m z v_{m+1} \cdots v_j y v_{j+1} \cdots v_n u_k$ for some vertices $v_i \neq y$.
Hence we set $P_{xzy} = x u_1 u_2 \cdots u_k v_1 \cdots v_m z v_{m+1} \cdots v_j y$. |
H: Compute the series : $\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!}$
How would you compute the following series? I'm interested in some easy approaches that would allow me to work it out.
$$\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!}$$
AI: It suffices to calculate the sum with the summation index running from 0, not from 1. Then
$$\begin{align*}
\sum_{n=0}^{\infty} \frac{4^n n!}{(2n)!}
&= \sum_{n=0}^{\infty} \frac{4^n}{(2n)!}\int_{0}^{\infty}t^n e^{-t}\;dt\\
&= \int_{0}^{\infty}\sum_{n=0}^{\infty} \frac{4^n t^n}{(2n)!} e^{-t}\;dt\\
&= \int_{0}^{\infty} \cosh(2\sqrt{t}) e^{-t}\;dt\\
&= \int_{0}^{\infty} 2u\cosh(2u) e^{-u^2}\;du\qquad(t=u^2)\\
&= \left[-\cosh(2u)e^{-u^2}\right]_{0}^{\infty} + 2\int_{0}^{\infty} \sinh(2u) e^{-u^2}\;du\\
&= 1 + 2\int_{0}^{\infty} \sinh(2u) e^{-u^2}\;du \\
&= 1 + \int_{0}^{\infty} \left(e^{2u} - e^{-2u}\right) e^{-u^2}\;du \\
&= 1 + e \int_{0}^{\infty} \left(e^{-(u-1)^2} - e^{-(u+1)^2}\right)\;du.
\end{align*}$$
Now it is clear how to express the last integrals in terms of the error function, yielding
$$ \sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!} = e\sqrt{\pi} \mathrm{erf}(1).$$ |
H: The greatest possible geometric multiplicity of an eigenvalue
Wikipedia claims that
"Given an n×n matrix A.... both algebraic and geometric multiplicity are integers between (including) 1 and n."
But how can the geometric multiplicity possibly be n? Since $(A-\lambda I)$ is a square matrix (as opposed to a matrix with more columns than rows), each of A's eigenspaces $Nul (A-\lambda I)$ has at most $(n-1)$ dimensions, isn't it?
I.e. The geometric multiplicity of an eigenvalue must be a number between between $1$ and $(n-1)$, right?
AI: If $A=\lambda I$, then $A-\lambda I=0$, and the null matrix obviously has kernel dimension $n$. |
H: Hatcher problem 1.2.3 - technicality in proof of simply connectedness
I am trying to prove that $\Bbb{R}^n$ minus finitely many points $x_1,\ldots,x_m$ is simply connected, where $n \geq 3$. For days now I have tried many different arguments but I have found flaws in all of them. I have finally come up with one, except that there is some small detail that I need to know how to prove.
I prove that $\Bbb{R}^n$ minus finitely many points is simply connected by inducting on the number of points that I remove. If I remove one point (the case $m=1$), I get that $\Bbb{R}^n - \{x_1\} \cong S^{n-1} \times \Bbb{R}$ which upon applying $\pi_1$ shows me that $\Bbb{R}^n - \{x_1\}$ is simply connected.
Inductive Hypothesis: Now suppose that $\Bbb{R}^n -\{x_1,\ldots,x_k\}$ is simply connected for all $ k <m$.
Suppose now I have $m$ points $x_1,\ldots,x_m$ lying in $\Bbb{R}^n$. Now if I write each of these points out in coordinates, I know that there is at least one $j$ with $1 \leq j \leq n$ such that the $j-th$ coordinate of all my points are not all the same (otherwise my points are all just one point and there is nothing to prove!).
Because I have only finitely many points, suppose without loss of generality that $x_1$ is the point whose $j-th$ coordinate is the greatest (this does not mean that such a choice is unique, I only want to know if it exists). Say that the $j-th$ coordinate of $x_1$ is $c$. Now I consider the plane
$$\mathbf{x}_j = \{\mathbf{x} \in \Bbb{R}^n : \text{$j$ -th coordinate of $\mathbf{x}$ is equal to $c$} \}.$$
My idea now is to apply the Seifert-Van Kampen Theorem together with the induction hypothesis as follows: I set
$$A = \Bigg\{\mathbf{x} \in \Bbb{R}^n- \{x_1,\ldots,x_m\} : \text{$j$ -th coordinate of $\mathbf{x}$ is greater than $c-\varepsilon$} \Bigg\}$$
where $\varepsilon$ is chosen such that $A$ does not enclose all my points and
$$B = \Bigg\{\mathbf{x} \in \Bbb{R}^n- \{x_1,\ldots,x_m\} : \text{$j$ -th coordinate of $\mathbf{x}$ is less than $c $} \Bigg\}.$$
Here's a picture of what I'm trying to do in the case of $\Bbb{R}^3$:
Then $A$ is open and so is $B$, clearly $A$ and $B$ are path connected and their intersection which is just a "cuboid" being a convex set is path connected as well.
My Problem: I want to apply my inductive hypothesis to $A$ and $B$ in order to deduce that $\pi_1(A) = \pi_1(B) = 0$. The problem now is that the inductive hypothesis is for $\Bbb{R}^n$ and not "chopped off bits" of $\Bbb{R}^n$ like $A$ and $B$. How do I get around this? Can I say that $A$ and $B$ are somehow deformation retracts of $\Bbb{R}^n$ minus finitely many points?
Thanks.
AI: The problem is, in fact, a simple induction. The base case with $m = 1$ is easily dealt with, as you did. Now assume that $m > 1$ and divide the points in $S = \{x_1, \ldots, x_m\}$ in two sets of smaller size (no matter how), say $A$ and $B$.
For convenience, let's assume that $A$ and $B$ are separated by the hyperplane $\mathcal{H}$, and that $N_{+}$ and $N_{-}$ are two open neighborhoods of the half-spaces that result. For an arbitrary base-point $x_0 \in \mathcal{H}$, Van Kampen theorem applies, giving a surjection from $\pi_1(N_{+} \backslash A) \ast \pi_1(N_{-} \backslash B)$ to $\pi_1(\mathbb{R}^n - S)$. Now just use the induction hypothesis to conclude that $\pi_1(N_{+} \backslash A) = \pi_1(N_{-} \backslash B) = \pi_1(\mathbb{R}^n - S) = 0$, as you wish.
Edit: Of course that $N_+ \backslash A$ and the other set are homeomorphic (or, if you want, homotopy equivalent) to $\mathbb{R^n} \backslash A$ in an obvious way. Such geometric observations are not a difficulty once you understood how to apply the theorem. |
H: $C(X)$ with the pointwise convergence topology is not metrizable
I need to show that if $X$ is an uncountable Tychonoff space, then $C(X)$ is not metrizable. All I've been able to show so far is that that $F(X)$, the space of all functions with pointwise topology, is homeomorphic to $\mathbb{R}^X$ (the product) which is not metrizable, but I can't seem to get much further. Thanks.
AI: Hint: If the pointwise convergence topology was metrizable, it would be first countable, so $0$ would have a countable neighbourhood base. Thus there would be a sequence of open neighbourhoods $U_i$ of $0$ such that every neighbourhood of $0$ contains some $U_i$.
By definition of the topology of pointwise convergence, for each $i$ there exist some finite set $J_i$ and some $\epsilon_i > 0$ such that $f \in U_i$ if $|f(x)| < \epsilon_i$ for all $x \in J_i$. Now define another neighbourhood $U$ of $0$, and use the fact that $X$ is completely regular to show that for any $i$ there is some $f \in C(X)$ such that $f \in U_i$ but $f \notin U$. |
H: Show that any open subsets of the real line are $F_\sigma$-sets.
This is an exercise from a topological book. It is this:
Show that any open subsets of the real line are $F_\sigma$-sets.
Could anybody help to solve it?
AI: First note that open subsets of the real line are countable unions of (pairwise disjoint) open intervals. Next show that every open interval is an F$_\sigma$-set. As countable unions of F$_\sigma$-sets are also F$_\sigma$, this gives the result. |
H: Gre Question Complex Number (plug and chug)
This seems like it should be easy, but I can't seem to simplify it: If $z=e^{i\frac{2\pi}{5}}$, then what is $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9$. The choices are $0, 4e^{i\frac{3\pi}{5}}, 5e^{i\frac{4\pi}{5}}, -4e^{i\frac{-2\pi}{5}}, -5e^{i\frac{3\pi}{5}},$ with the answer being $-5e^{i\frac{3\pi}{5}}.$ I can plug in the given $z$ into the equation and get $5+10e^{-i\frac{2\pi}{5}}+5e^{i\frac{2\pi}{5}}+5e^{i\frac{-4\pi}{5}}+5e^{i\frac{4\pi}{5}}$, but have been unsuccessful in simplifying it so far.
AI: Note that $z^5=1$, as $z$ is a fifth root of unity, so the expression simplifies to
$$
\begin{align}
1+z+z^2+z^3+5z^4+4+4z+4z^2+4z^3+5z^4 &=5+5z+5z^2+5z^3+10z^4\\
&=5(1+z+z^2+z^3+z^4)+5z^4
\end{align}
$$
However, either by using the formula for the geometric series, or the fact that $1+X+X^2+X^3+X^4$ is the fifth cyclotomic polynomial, it follows that $1+z+z^2+z^3+z^4=0$. So the final answer is
$$
5z^4=5\exp(i8\pi/5)=-5\exp(i3\pi/5).
$$ |
H: Period of $f'(x)$
It is easy to prove that if $f(x)$ is periodic with period $T$,then $f'(x+T)=f'(x)$. However I don't know how to prove $T$ is the shortest period for $f'(x)$.
Can anyone help me? Or if it is not true, can anyone give me a counterexample?
AI: It seems that the hint I have given before was too obscure. So I now give a full solution.
By assumption $f'$ is the derivative of a $T$-periodic function $f$, where $T>0$ is a primitive period of $f$. Therefore $f'$ has period $T$ as well.
If $f'$ were a constant then this constant would be $0$ (or $f$ would not be periodic). As a consequence $f$ would be constant and would not have a primitive period.
Therefore $f'$ has a primitive period $p>0$, which then necessarily is of the form $p={T\over n}$ for some $n\geq1$. Put $\int_0^p f'(t)\ dt=:c$. Then $$\int_x^{x+p}f'(t)\ dt=c$$ for any $x$; furthermore one has $$0=f(T)-f(0)=\int_0^T f'(t)\ dt=n\int_0^p f'(t)\ dt=n\, c\ .$$
Therefore $c=0$, and this in turn implies that $$f(x+p)-f(x)=\int_x^{x+p} f'(t)\ dt=0$$ for all $x$; whence $f$ has period $p$. This implies $p= m T$ for some $m\geq1$, so that we in fact have $p=T$. |
H: group of order 30
What are the steps in showing a group of order 30 is solvable/non-solvable?
I don't know how to proceed. All I know is that the group either has a group of order $5$ or $3$. I don't need all the steps for this problem, just an outline what to do.
AI: Some ideas:
1) Show that such a group always has one unique group of order 3 or one unique group of order 5
2) Using the above show that such a group always has a subgroup of order 15
3) Now use the following: if a group $\,G\,$ has a normal sbgp. $\,N\,$ s.t. both $\,N\,$ and $\,G/N\,$ are solvable, then $\,G\,$ is solvable |
H: Can an eigenvalue (of an $n$ by $n$ matrix A) with algebraic multiplicity $n$ have an eigenspace with fewer than $n$ dimensions?
Is it possible for a matrix with characteristic polynomial $(λ−a)^3$ to have an eigenline (one-dimensional eigenspace)?
I know that geometric multiplicity can generally be smaller than algebraic multiplicity. But I was wondering if algebraic multiplicity $n$ might be a special case. This question is motivated by my earlier one The greatest possible geometric multiplicity of an eigenvalue, where I learnt that $A$ has an $n$-dimensional eigenspace iff. $A=\lambda I$.
AI: Sure. Consider the matrix
$$\begin{pmatrix} 1 & 1 \\
0 & 1\end{pmatrix}$$
which has a single eigenvalue $1$ of multiplicity $2$, yet the only solutions to the equation
$$\begin{pmatrix} 1 & 1 \\
0 & 1\end{pmatrix}\begin{pmatrix} x \\
y\end{pmatrix}=\begin{pmatrix} x +y \\
y\end{pmatrix}=\begin{pmatrix} x \\
y\end{pmatrix}$$
come when $y=0$, and so the eigenspace is $1$-dimensional. |
H: Ring homomorphism with $\phi(1_R) \neq1_S$
Let $R$ and $S$ be rings with unity $1_R$ and $1_S$ respectively. Let $\phi\colon R\to S$ be a ring homomorphism. Give an example of a non-zero $\phi$ such that $\phi(1_R)\neq 1_S$
In trying to find a non-zero $\phi$ I've done the following observation:
Since for $\forall r\in R$
$\phi(r) = \phi(r\times1_R) = \phi(r)\times\phi(1_R)$ we must have that $\phi(1_R)$ is an identity of $\phi(R)$ but not an identity of $S$. We must therefor construct a $\phi$ that is not onto and which have this property. I can't come up with any explicit example though, please help me.
AI: Let $R$ be any ring and $S=R\times R$. Then the inclusion map $r\mapsto (r,0)$ gives you such a homomorphism. (Note that some authors require that $\phi(1_r)=1_S$ for $\phi$ to be a homomorphism). |
H: Show that if $\kappa$ is an uncountable cardinal, then $\kappa$ is an epsilon number
Firstly, I give the definition of the epsilon number:
$\alpha$ is called an epsilon number iff $\omega^\alpha=\alpha$.
Show that if $\kappa$ is an uncountable cardinal, then $\kappa$ is an epsilon number and there are $\kappa$ epsilon numbers below $\kappa$; In particular, the first epsilon number, called $\in_0$, is countable.
I've tried, however I have not any idea for this. Could anybody help me?
AI: Here is a slightly easier way:
Lemma: For every $\alpha,\beta$ the ordinal exponentiation $\alpha^\beta$ has cardinality of at most $\max\{|\alpha|,|\beta|\}$.
Now use the definition of $\omega^\kappa=\sup\{\xi^\omega\mid\xi<\kappa\}$, since $|\xi|<\kappa$ we have that $\omega^\kappa\leq\kappa$, but since $\xi\leq\omega^\xi$ for all $\xi$, $\omega^\kappa=\kappa$.
Here is an alternative way (a variation on the above suggestion):
Lemma: If $\alpha$ is an infinite ordinal then there is some $\varepsilon_\gamma\geq\alpha$ such that $|\alpha|=|\varepsilon_\gamma|$
Hint for the proof: Use the fact that you need to close under countably many operations, and by the above Lemma none changes the cardinality.
Now show that the limit of $\varepsilon$ numbers is itself an $\varepsilon$ number, this is quite simple:
If $\beta=\sup\{\alpha_\gamma\mid\alpha_\gamma=\omega^{\alpha_\gamma}\text{ for }\gamma<\tau\}$ (for some $\tau$ that is) then by definition of ordinal exponentiation $$\omega^\beta=\sup\{\omega^{\alpha_\gamma}\mid\gamma<\tau\}=\sup\{\alpha_\gamma\mid\gamma<\tau\}=\beta$$
Now we have that below $\kappa$ there is a cofinal sequence of $\varepsilon$ numbers, therefore it is an $\varepsilon$ number itself; now by induction we show that there are $\kappa$ many of them:
If $\kappa$ is regular then every cofinal subset has cardinality $\kappa$ and we are done;
if $\kappa$ is singular there is an increasing sequence of regular cardinals $\kappa_i$, such that $\kappa = \sup\{\kappa_i\mid i\in I\}$. Below each one there are $\kappa_i$ many $\varepsilon$ numbers, therefore below $\kappa$ there is $\sup\{\kappa_i\mid i\in I\}=\kappa$ many $\varepsilon$ numbers. |
H: Veryify that the union of a co-dense set and a nowhere dense set is a co-dense set
Veryify that the union of a co-dense set and a nowhere dense set is a co-dense set. Give an example to show that the union of two co-dense sets if not necessarily a co-dense set.
Note: A co-dense set $A$ in the topological $X$ denotes its complementary set is dense in $X$. And a nowhere dense set $A$ in $X$ if $\overline{A}$ is co-dense.
What I've tried: For the second question, I find an example. The real line $R$ is the whole space. $Q$ denotes all the rational numbers and $P=R-Q$ denotes all the irrational numbers. We see they are both co-dense sets. However, their union is $R$, and hence is not a co-dense. The first question is still tough for me.
Could anybody help me? New example is also welcome.
AI: This is a slightly different take, using some facts from a previous answer of mine, and also the characterisation that $B \subseteq X$ is nowhere dense iff for every nonempty open $U \subseteq X$ there is a nonempty open $V \subseteq U$ such that $V \cap B = \emptyset$.
To show that $A \cup B$ is co-dense, we need only show that $\mathrm{Int} ( A \cup B ) = \emptyset$.
If $U = \mathrm{Int} ( A \cup B ) \neq \emptyset$, then as $B$ is nowhere dense there is a nonempty open $V \subseteq U$ such that $V \cap B = \emptyset$. As $A$ is co-dense, then $V \not\subseteq A$ (since $\mathrm{Int} (A) = \emptyset$), and so there is an $x \in V \setminus A$. But then $x \in U \setminus ( A \cup B )$, contradicting that $U \subseteq A \cup B$! |
H: Two sums with Fibonacci numbers
Find closed form formula for sum: $\displaystyle\sum_{n=0}^{+\infty}\sum_{k=0}^{n} \frac{F_{2k}F_{n-k}}{10^n}$
Find closed form formula for sum: $\displaystyle\sum_{k=0}^{n}\frac{F_k}{2^k}$ and its limit with $n\to +\infty$.
First association with both problems: generating functions and convolution. But I have been thinking about solution for over a week and still can't manage. Can you help me?
AI: For (2) you have $F_k = \dfrac{\varphi^k}{\sqrt 5}-\dfrac{\psi^k}{\sqrt 5}$ where $\varphi = \frac{1 + \sqrt{5}}{2} $ and $\psi = \frac{1 - \sqrt{5}}{2}$ so the problem becomes the difference between two geometric series.
For (1) I think you can turn this into something like $\displaystyle \sum_{n=0}^{\infty} \frac{F_{2n+1}-F_{n+2}}{2\times 10^n}$ and again make it into a sum of geometric series.
There are probably other ways. |
H: Singular or non-singular matrices
Which of the following matrices are non-singular?
$I + A$ where $A$ not equal to $0$ is a skew-symmetric real $n\times n$ matrix, $n\geq 2$.
Every skew-symmetric non-zero real $5 \times 5$ matrix.
Every skew-symmetric non-zero real $2 \times 2$ matrix.
AI: We will use the properties $\det A=\det(A^t)$ and $\det(-A)=(-1)^d\det A$, where $d$ is the dimension of the matrix.
We have $\det(A+I)=\det(A^t+I)=\det(I-A)$. Let $x$ be such that $(A-I)x=0$. Then
$$\langle x,x\rangle=\langle x,Ax\rangle=\langle A^tx,x\rangle=-\langle Ax,x\rangle=-\langle x,x\rangle,$$
which proves that $x=0$. We have seen that if $(A-I)x=0$ then $x=0$,
hence $A-I$ is invertible. This implies that $\det(I-A)\neq 0 $. Since
$\det(I-A)=\det(A+I)$ we get that $\det(A+I)\neq 0$, which proves that $A+I$ is invertible.
Using the mentioned properties, for a skew-symmetrix matrix of odd dimension the determinant is $0$.
In dimension 2, a non-zero skew-symmetric matrix is of the form $A:=\pmatrix{0&x\\-x&0}$ where $x\neq 0$. Its determinant is $x^2\neq 0$ hence $A$ is necessarily invertible.
For the other even dimensions (say $d=2N$), we can construct a non-zero skew-symmetric matrix which is not invertible. Indeed, consider the block matrix defined by
$$
\pmatrix{A&0_{2,2N-2}\\ 0_{2N-2,2}& 0_{2N-2,2N-2} }
$$
where $0_{i,j}$ denoted the matrix with $i$ rows and $j$ columns and all its entries are zero as $A= \pmatrix{0&1\\-1&0}$. |
H: Morphisms between cyclic groups
I'm trying to solve a group theory question involving morphisms:
How many different morphisms do there exist from $ C_n $ to $ C_m $?
Am I correct in saying if $f$ is a morphism, then $f(0) = 0$ and $f(a+b) = (a+b)f(1)$?
If yes, where do I go from here? And if no, how do I start?
Thanks!
Also, as a follow up question there's:
How many automorphisms exist from $C_n$ onto itself?
Since automorphisms map generators to generators, I have to map $1$ to a generator, so I have $\phi(n)$ choices, with $\phi$ being Euler's totient function?
AI: You are right in your assumption (the phrase used by mathematicians is "A homomorphism is uniquely determined by its image on the generators"), and from there you just have to make sure that the image of the generators behave like they originally do; in this case the one generator $1 \in C_n$ fulfills $n\cdot 1 = 0$
By your assumtion, the homomorphism $\psi$ is completely determined by $\psi(1)$. So the question reads "Where can I send $1$?", and the only limiting factor here is that in $C_n$, $n\cdot 1 = 0$, so we must have $n\cdot \psi(1) = 0$ in $C_m$. Now the question reads "Which elements in $C_m$ has order a divisor of $n$?". This is when we leave algebra and enter number theory, where the same question sounds "Which integers $k$ fulfills $k\cdot n \equiv 0 \mod{m}$?" And the answer is any multiple of $\frac{m}{\gcd(m, n)}$, of which there are $\gcd(m, n)$, including $0$.
As to your follow-up, you can map $1$ to any element $j$ with $\gcd(n, j) = 1$, of which there are $\phi(n)$. If you map it to any other element, it will not generate a surjection, since $1$ would not be in the image, and thus not an isomorphism. So you are right. |
H: Why is the following language decidable? $L_{one\ right\ and\ never\ stops}$
I can't understand how the following language can ever be decidable:
$L= \{ \langle M \rangle | M \ is \ a \ TM \ and\ there\ exists\ an\ input\ that\ in\ the\ computation\ $ $of\ M(w)\ the\ head\ only\ moves\ right\ and\ M\ never\ stops \}$,
but apparently it is.
We need to approve that there's at least one input as requested, so even if run on all inputs in parallel, we can't say something as "if it never stops, accept", it's paradox, and even if we could do that it was showing that the language is in $RE$- recursive enumerable.
so how come it is recursive?
Thanks a lot
AI: As sdcvvc hints, this can be done due to how restricted the TMs in L are. The problem can be decided by examining the set of rules of the TM, which is finite. Firstly check if for some state $q_n$ the TM moves right on a blank input. This is the only way to go right infinitely as the tape contains only a finite number of symbols, other than a blank one. Secondly check that $q_n$ can be reached from $q_0$. For a normal TM this would not be decidable, but since this one can only go right, all of its input is the initial symbols of the tape.
Consider a graph representing this TM. The vertices are states and edges are symbols on which state changes. The direction is always right and the written symbol is irrelevant, so they do not need to be represented. This problem is equivalent to first finding a cycle that only uses blank symbol edges and then a path from starting vertex to this cycle in the graph. |
H: Best way to denote some trigonometric functions ("tg" vs "tan", "ctg" vs "cot")
What is the best way to denote tangent and other trigonometric functions: tg or tan, ctg or cot. What notation is commonly used and standardized?
AI: In current US textbooks, $\tan$ and $\cot$ are commonly used and standardized. Also: $\sin, \cos, \sec, \csc$. In other countries, and in the 19th century, you will find others. |
H: Duality, Symmetry, Dual Spaces
The following is from the book "Tensor Methods in Statistics", which could be downloaded there
http://www.stat.uchicago.edu/~pmcc/tensorbook/
I have a question regarding section 0.3.1, titled "Duality and dual spaces", there it is said
Let $V$ be a vector space with basis $\{ e_1, \ldots, e_n \}$. In the usual expression for $v \in V$ as a linear combination of the basis vectors, we write $v = v^i e_i$. The notation exhibits a certain formal symmetry between the components $v^i$ and the basis vectors $e_i$, but the interpretation, as usually given, is quite asymmetric: $v^i$ is a 'scalar', whereas $e_i$ is a 'vector' in $V$. However $\{v^i, \ldots, v^n\}$, as linear functionals, form a basis in $V^*$.
Here comes my first question, how could the scalar $v^i$ be regarded as a linear functional, and furthermore, how could $\{ v^i, \ldots, v^n\}$ be interpreted as a basis for $V^*$ when its just a set of scalars?
He continues
Thus, the expression $v^i e_i$ could equally well be interpreted as a point in $V^*$ with components $(e_1, \ldots, e_n)$ relative to the dual basis.
Now he interprets the basis vectors $e_i$ as components in the dual basis, but I thought linear functionals, i.e. element of $V^*$ could be represented equally by $n$-tuples of numbers.
AI: When the author says that $v^i$ can be thought of as a functional, they are not talking about the scalar $v^i$ attached to any particlar vector $v$. Rather, they mean the scalar valued function $v \mapsto v^i$ (sometimes called "projection onto the $i$th component").
This is a scalar-valued function on the set of vectors $\{v \, | \, v = (v^1,\ldots, v^n)\},$ i.e. it is a functional.
Although it is a little confusing to denote this functional by $v^i$, this is similar to the kind of confusion that can arise in calculus, where we might use $x$ to denote both a particular number and a variable, and hopefully won't be hard to get used to.
The resulting basis $v^i$ of $V^*$ is called the dual basis, since the value of $v^i$ on $e_j$ is $\delta_{ij}$. (This is just a fancy way of saying that the $i$th component of $e_j$ is $\delta_{ij}$.)
The expresssion $\sum_i v^i e_i $ then has a symmetry: we can interpret it as originally intended: the $e_i$ are basis vectors for $V$, and the $v^i$ are scalars giving the coordinates of a vector in terms of this basis --- or we can think of the $v^i$ as denoting the dual basis of $V^*$, and the $e_i$ as being coordinates of a vector in $V^*$ in terms of this basis.
How are the $e_i$ coordinates on a vector of $v^*$? Well, we just have to reverse engineer the above process of interpreting the scalar valued function $v^i$ as a vector, namely: given a functional $v^*$, we can evaluate it on the vector $e_i$ to get a scalar, and so we can think of this evaluation process as being a functional on $V^*$. In this way the vector $e_i \in V$ becomes thought of as a functional on $V^*$. (This identification of vectors in $V$ with functionals in $V^*$, via evaluation, is often called double duality.)
It's easy to check that in this way, $e_i$ just becomes the functional on $V^*$ which gives the $i$th coordinate of a functional $v^*$ in terms of the basis $(v^i)$. This explains the author's new interpretation of the expression $\sum_i v^i e_i$. |
H: What is a PL mapping?
In a proof of the Borsuk-Ulam theorem I've encuntered the notion of a PL mapping between two n-spheres. Does anyone know what it means?
AI: PL stands for "piecewise linear". Roughly, it means a map which can be decomposed as a piecewise linear map with respect to suitable triangulations of source andtarget. For a more precise description, see the wikipedia entry, or these lecture notes of Jacob Lurie. (Note that is probably implicit in the discussion you are reading that the $n$-spheres are given their standard PL structures.) |
H: Is this a correct way to write the summation?
I have values such as 2,4, 7, 10 , that is not sequential but are stored in array w. Can I use the summation
$$ {\sum_{i = 1}^{n} w[i]} $$
or there is another way to write it down ?
AI: in programming if we have array $w={1,2,3,4....}$,then we write $w[i]$,in mathematics you can simple write $w_i$ |
H: How to invert this symmetric tridiagonal Toeplitz matrix?
What's the best way to invert a simple symmetric tridiagonal Toeplitz matrix of the following form?
$$
A = \begin{bmatrix} 1 & a & 0 & \ldots & \ldots & 0 \\\
a & 1 & a & \ddots & & \vdots \\\
0 & a & 1 & \ddots & \ddots& \vdots \\\
\vdots & \ddots & \ddots & \ddots & a & 0\\\
\vdots & & \ddots & a & 1 & a\\\
0 & \ldots & \ldots & 0 & a & 1 \end{bmatrix}
$$
AI: Usually, an eigendecomposition is the least efficient way to generate the inverse of a matrix, but in the case of the symmetric tridiagonal Toeplitz matrix, we have the nice eigendecomposition $\mathbf A=\mathbf V\mathbf D\mathbf V^\top$, where $$\mathbf D=\mathrm{diag}\left(1+2a\cos\frac{\pi}{n+1},\dots,1+2a\cos\frac{k\pi}{n+1},\dots,1+2a\cos\frac{n\pi}{n+1}\right)$$ and $\mathbf V$ is the symmetric and orthogonal matrix whose entries are $$v_{j,k}=\sqrt{\frac2{n+1}}\sin\frac{\pi jk}{n+1}$$ Thus, to generate the inverse, use $\mathbf A^{-1}=\mathbf V\mathbf D^{-1}\mathbf V^\top$, and inverting a diagonal matrix is as easy as reciprocating the diagonal entries. |
H: conjecture regarding the height of polynomial's square-free part
About some time I am struggling with the following interesting problem:
There is a well-known theorem of Mignotte which says that for a polynomial $f\in\mathbb{Z}[x]$ of degree $n$ and height (coefficient size) $2^\tau$, the height of its divisors is bounded by:
$2^n\mathcal{M}(f)=\mathcal{O}(2^{n+\tau})$,
see, e.g., http://arxiv.org/abs/0904.3057
In other words, the height of polynomial's divisors can be larger than the height of a polynomial itself. Example could be:
$x^5+3x^4+2x^3-2x^2-3x-1=(x^4+4x^3+6x^2+4x+1)(x-1)$.
My conjecture is that this cannot happen for polynomial's square-free part. To be precise, for a polynomial $f\in\mathbb{Z}[x]$ of height $2^\tau$, its square-free part $f^*=f / \gcd(f,f')$ cannot have height larger than $2^\tau$.
Simple example: $f=(x-1)^3=x^3-3x^2+3x-1$ and $f^*=x-1$.
I appreciate if someone has ideas how to prove that or find a counterexample.
Why it's an interesting problem is because theoretical complexity of many algorithms from algebraic geometry use such bounds. On a similar note, modular approaches use these bounds to estimate the number of primes needed for computation.
AI: The squarefree part of $$(x^4+7x^3+8x^2+7x+1)(x-1)^2=x^6+5x^5-5x^4-2x^3-5x^2+5x+1$$ is $$(x^4+7x^3+8x^2+7x+1)(x-1)=x^5+6x^4+x^3-x^2-6x-1$$ which has greater height. It's clear the coefficients and the degree can be varied greatly to get as many counterexamples as desired. |
H: Global dimension of free algebra.
Is there any easy way to see the global dimension of a free algebra
$$
A=k\langle x_{1},\dots,x_{n} \rangle
$$
is 1?
AI: In "Modules over coproducts of Rings", Bergman proved that the global dimension of a free product is the supremum over the global dimensions of the factors (unless all factors have global dimension 0). |
H: Goldbach's conjecture and number of ways in which an even number can be expressed as a sum of two primes
Is there a functon that counts the number of ways in which an even number can be expressed as a sum of two primes?
AI: See Goldbach's comet at Wikipedia.
EDIT: To expand on this a little, let $g(n)$ be the number of ways of expressing the even number $n$ as a sum of two primes. Wikipedia gives a heuristic argument for $g(n)$ to be approximately $2n/(\log n)^2$ for large $n$. Then it points out a flaw with the heuristic, and explains how Hardy and Littlewood repaired the flaw to come up with a better conjecture. The better conjecture states that, for large $n$, $g(n)$ is approximately $cn/(\log n)^2$, where $c>0$ depends on the primes dividing $n$. In all cases, $c>1.32$.
I stress that this is all conjectural, as no one has been able to prove even that $g(n)>0$ for all even $n\ge4$. |
H: How to show that $\mathbb R^n$ with the $1$-norm is not isometric to $\mathbb R^n$ with the infinity norm for $n>2$?
Could you please give me a hint to prove that
$\mathbb{R}^n$ with the 1-norm $\lvert x\rvert_1=\lvert x_1\rvert+\cdots+\lvert x_n\rvert$ is not isometric to $\mathbb{R}^n$ with the infinity-norm $\lvert x\rvert_\infty = \max_{i=1,\ldots,n}(\lvert x_i\rvert)$ for $n\gt2$.
I do not see how the critical case $n=2$ enters the picture.
Thank you!
AI: To wrap the comments up (assuming $n \geq 2$ since the cases $n=0,1$ are uninteresting):
From the Mazur–Ulam theorem we know that a surjective distance-preserving map between normed spaces is necessarily affine. After composing with a translation we may assume that our isometry preserves $0$, so we may assume that the isometry is linear right from the start.
A surjective linear isometry must map the extreme points of the unit ball of the domain onto the extreme points of the unit ball of the range. A moment's thought shows that the unit ball in the $\ell^1$-norm has $2n$ extreme points, namely the $n$ coordinate vectors and their $n$ negatives; and that the unit ball in the $\ell^\infty$-norm has $2^n$ extreme points: the vectors all of whose entries are $\pm 1$. Therefore isometry is only possible if $n =2$, where there is an isometry, given by $e_1 \mapsto e_1+e_2$ and $e_2 \mapsto -e_1 + e_2$. |
H: Help me find equation of this graph
I need to find equations of this list: $[1,1,1,0,0,0,1,1,1,0,0,0, ...]$ (it's periodic)
The closest equation I've got is $\left\lceil \sin (\frac{\pi}{3}x)\right\rceil $, which looks like this:
_ _ _
_| |_| |_| |_
But I need it to look like this:
_ _ _
_/ \_/ \_/ \_
Can you help me?
AI: You may try the simple function
$$f(x)=|(x-s) \bmod 6 -3|-1$$
(the 'shift' $s$ is $0$ for the first list , $1$ for the second and so on...)
by replacing values over $1$ by $1$ and under $0$ by $0$
Or don't you allow tests? |
H: Puzzle: Guessing the bigger number!
Consider the following interesting puzzle:
"Alice writes two distinct real numbers between 0 and 1 on two sheets of paper. Bob selects one of the sheets randomly to inspect it. He then has to declare whether the number he sees is the bigger or smaller of the two. Is there any way he can expect to be correct more than half the times Alice plays this game with him?"
SPOILER ALERT: In the linked site below, the solution to this puzzle is right below the posed question.
The puzzle is lifted from here.
In the following questions, I have just given my hunch. My calculations could be wrong as I am not very strong at probability.
Questions:
1) Is there an assumption here that Bob can play the game many times with Alice (from the last statement)? Are the real numbers assumed to be fixed?
My hunch: If it is played in time, then fixed real numbers dont make sense. So perhaps we have to think over different realizations. Is this right?
2)[Variant] If Bob knows that Alice samples the two real numbers uniformly and independently from [0,1] every time she plays then can he design a strategy that works more than half of the times?
My opinion: I have a strategy that works two-thirds of the time. However I am not sure if that is the best! Is there a strategy that works better?
3) Is there a distribution from which if Alice samples her numbers independently, then she can ensure that Bob can never do better than getting right half the time?
My opinion: No! Although I am not sure. I guess the independence assumption is what allows Bob to get a good strategy.
4) If Alice can sample two real numbers in a correlated fashion every time she plays, can she ensure that Bob can never do better than getting right half the time?
My opinion: Yes! I think I have an answer.
P.S: I will post my version of the answers after a few days, if nobody comes up with the same idea. Thanks! :)
P.P.S: This is not a homework problem, but just a musing :)
AI: 1) That's not well-specified in the formulation of the problem.
2) You can attain $3/4$ by choosing the first number iff it's above $1/2$.
3) No, she can't but that has nothing to do with independence; see 4).
4) No, she can't; see the links in my comment under the question. |
H: Drawing a monkey saddle surface in matlab?
I'd like to draw a monkey saddle surface using matlab. But how do I plot a function of several variables in matlab? I never did that before. I can define $x$ and $y$ as two vectors and then according to wikipedia the monkey saddle equation is $x^3-3xy^2$ so all I wanna do is plot that function? Are there any more monkey saddle surfaces that might be nicer than this one?
AI: The easiest way is to use the surf command:
x = min_x:step:max_x;
y = min_y:step:max_y;
[X,Y] = meshgrid(x,y);
Z = X.^3-3*X.*Y.^2
surf(X,Y,Z);
should work. I don't have my MATLAB install on this computer, so I cannot verify. |
H: Is there a difference between a model and a representation?
I'm thinking of models in logic here, vs. e.g. group representations.
Is there a difference between a model and a representation?
Could one not explain both at the same time?
A model gives an interpretation, but this might be viewed as a side effect to the work you do. You have some abstract axioms and you model/realize them with certain objects, which are part of another theory (e.g. the ordered pair concept is modeled via sets and $\in$). I don't see a real difference to e.g. a group representation, when you have some abstract multiplication laws and these come to live via a matrix representation of a specific dimension, say.
Also,
What kind of realizations do representable functors deal with?
I mean beyond the realization of groups like above.
Edit, from the comments: Like an example would be to consider integers $\mathbb{Z}$ and build the factor group $\mathbb{Z/2Z}$, this would be representing what is called $\mathbb{Z_2}$ (abstractly defined by the four relations between its two elements). The logic analog would be the sets and the abstract idea of a pair with its characterizing feature.
AI: Well, I mean, they are representations of different kinds of things. But one can think of both as functors out of a given category.
For group representations, a group $G$ may be regarded as a category with one object and morphisms the elements of $G$. A group action of $G$ is then precisely a functor $G \to \text{Set}$; similarly, a linear representation of $G$ is precisely a functor $G \to \text{Vect}$. There are endless variations on this.
For models, a collection of axioms ought to describe category which is roughly speaking the free category containing a model of the axioms. This is easiest to explain for axioms nice enough that they can be described using a Lawvere theory, but I believe this formalism is more general than that. For example, the theory of groups can be described using a Lawvere theory which is described here. Given a Lawvere theory $C$, a model of $C$ is then precisely a product-preserving functor $C \to \text{Set}$. Again there are endless variations on this. See the Wikipedia article on categorical logic.
In the first example, the representable functor gives you the action of $G$ on itself (exercise), which is roughly speaking the free $G$-action on one element. In the second example, the representable functors for the Lawvere theory of groups give you free groups on finite sets (exercise).
Edit: Instead of reducing everything to category theory perhaps I should reduce everything to logic. I claim that representations are a special case of models.
For example, let $G$ be a group. Write down a first-order theory with one unary operation for every element of $G$ and one universally quantified axiom for every entry $g \times h = gh$ in the multiplication table of $G$. Then a model of this theory is precisely a set on which $G$ acts. (A similar but more complicated construction gives linear representations also as a special case of models.) Of course, much of this is redundant: it suffices to specify an operation for every element of a fixed set of generators of $G$ and an axiom for every relation in a fixed presentation relative to the generators. |
H: Show that the sequence $\left\{ x^n \right\}$ of functions converges uniformly on $[0,k],k<1$
Show that the sequence of functions $\left\{ x^n \right\}$ converges uniformly on $[0,k],k<1$, but non-uniformly on $[0,1]$.
For $x\in[0,1)$
$$\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}x^n=0$$
and for $x=1$
$$\lim_{n\to \infty}f_n(x)=1.$$
Therefore, the limit function is
$$f(x)=\cases{
0 &$\quad 0\leq x <1$\\
1 &$\quad x=1$}$$
Let $\epsilon >0$ be arbitrary. Then, for each $x\in (0,1)$
$$|f_n(x)-f(x)|<\epsilon\implies x^n<\epsilon \implies n\ln x<\ln\epsilon \implies n>\frac{\ln\epsilon}{\ln x}$$
For $x=0,1$
$$|f_n(x)-f(x)|=0<\epsilon,\quad\forall n\geq1
$$
Choose a natural number $N(\epsilon,x)=\left[\frac{\ln\epsilon}{\ln x}\right]+1$ Here $[\cdot]$ is the box function.
Note that $N(\epsilon,x)\to +\infty$ as $x\to 1^-$. So, $N(\epsilon,x)$ is not bounded on $[0,1]$.
Hence, the convergence of $\left\{ x^n \right\}$ is not uniform on $[0,1]$.
I faced problems in showing uniform convergence of $\left\{ x^n \right\}$ on $[0,k]$, $k<1$.
AI: Given $\varepsilon>0$ take $n_0$ such that $k^{n_0} <\varepsilon$. As $x<k<1$, we have
\begin{equation}
|x^n-0|=|x^n|<|k^{n}|<|k^{n_0}|<\varepsilon \quad \forall x \in[0,k], \forall n\ge n_0.
\end{equation} |
H: What do the $+,-$ mean in limit notation, like$\lim\limits_{t \to 0^+}$ and $\lim\limits_{t \to 0^-}$?
I'm working on Laplace Transforms and have got to a section where they are talking about zero to the power plus or minus and that they are different. I can't remember what this means though.
It's generally used in limits.
$\lim\limits_{t\to 0^-}$ or $\lim\limits_{t\to 0^+}$
Any help would be much appreciated.
AI: Say we let
$$H(x)=\begin{cases} 0, & x < 0, \\ 1, & x > 0, \end{cases}$$
and let $H(0)$ be not defined.
Say I would like to approach $0$ on this function. However, a problem arises! Looking at the plot of the function, it is clear that if one were to approach from the right hand side, the limit is $1$, whilst if one approaches from the left, the limit is $0$ and thus the two-sided limit does not exist (both sides should be approaching the same number for this limit to exist)! This can also be easily seen by plugging in numbers:
$$H(1)=1$$
$$H(.1)=1$$
$$H(.000000000001)=1$$
etc. But, doing the same thing from the left hand side, we find
$$H(-1)=0$$
$$H(-.1)=0$$
$$H-(.000000000001)=0$$
Thus we need to define a different type of limit for functions with similar discontinuities so we may approach from either side. This limit is the "one-sided limit" and is used generally when a two-sided limit does not exist, like in the above case. $\lim_{x \to x_0^+}f(x)$ represents the right handed limit of $f(x)$ to $x_0$ whilst $\lim_{x \to x_0^-}f(x)$ represents the left hand limit. So we see that $\lim_{x \to 0} H(x)$ does not exist, but
$$\lim_{x \to 0^+}H(x)=1$$
$$\lim_{x \to 0^-}H(x)=0$$ |
H: How to integrate $\sec^3 x \, dx$?
Possible Duplicate:
Indefinite integral of secant cubed
How to integrate $\sec^3 x \, dx$?
Can someone please give a method, I tried separating $\sec^3 x$ as $\sec x(\sec^2 x)$ then applying by-parts method but it didn't yield anything useful
AI: $$\sec^3(x)=\frac{1}{\cos^3(x)}=\frac{\cos(x)}{(1-\sin^2(x))^2} $$
$u=\sin(x)$. |
H: How to convert any non-negative matrix into a doubly stochastic matrix?
Given a non-negative real matrix $A \in \Bbb R_+^{m \times n}$, how do I convert it to a doubly stochastic matrix (each row and column sums to $1$)
$$\sum_{j=1}^n A_{ij}= 1, \qquad \forall i = 1, \dots, m \tag{row sum}$$
$$\sum_{i=1}^m A_{ij}= 1, \qquad \forall j = 1, \dots, n \tag{column sum}$$
Is the conversion possible? If not, can we find a nearest matrix that is doubly stochastic matrix?
AI: There is a paper by Richard Sinkhorn: A relationship between arbitrary positive matrices and doubly stochastic matrices, The Annals of Mathematical statistics, 35 (1964), 876–879.
There he proves the following
Theorem. If $A$ is a square matrix with strictly positive entries then there are a unique doubly stochastic matrix $T_A$ and diagonal matrices $D_1$, $D_2$ such that $T_A=D_1AD_2$. The matrices $D_1$ and $D_2$ are themselves unique up to a scalar factor. |
H: Condensation points and derived sets
Let $(X,d)$ be compact metric space and $C$ be the set of condensation points of $X$. Following the notations here, is the equality $\bigcap\limits_{n \geq 1} X^{(n)}=C$ true ?
I succeeded in showing the inclusion $C \subset \bigcap\limits_{n \geq 1} X^{(n)}$ but the other inclusion seems to be more difficult.
Have you an idea?
AI: Consider the ordinal $X=\omega^\omega+1$ with order topology. It is a successor ordinal, so it is compact. It is also countable, so it is metrizable, and has no condensation points (it can be seen either using abstract fact that a second-countable compact Hausdorff space is Polish, or the fact that any countable ordinal embeds into rationals as a totally ordered set).
On the other hand, notice that $X'$ is the set of limit ordinals within $X$. Similarly, $X''$ is the set of limits of limit ordinals, etc., so for any $n$, we have $\omega^n\in X^{(n)}$, and for $X^{(\omega)}=\bigcap_n X^{(n)}$ we have $\omega^\omega\in X^{(\omega)}$, so $X^{(\omega)}\neq C=\emptyset$.
On the other hand, by Cantor-Bendixson theorem, we can see that for any compact metrizable space (indeed, any Polish space), there exists a countable ordinal $\alpha$ such that $X^{(\alpha)}=X^{(\infty)}=P=C$, which is $\omega+1$ in the above case (the minimal such $\alpha$ is called the Cantor-Bendixson rank, and is of interest in e.g. model theory, where it is one of the interpretations of the Morley rank). |
H: a question about a function $g(x) = \sum f(2^nx)$
Let $f(x)$ be non-negative and decreasing for $ x > 0$. Suppose that $\int_0^\infty f(x)dx < \infty$. Let $g(x) = \sum_{n=1}^\infty f(2^nx)$. How do I prove that $\int_0^\infty f(x) dx = \int_0^\infty g(x) dx$?
AI: The conditions imposed on $f$ mean that we can swap the sum and integral signs to get
$$\int_0^{\infty} g(x)\, dx = \sum_{n=1}^{\infty} \int_0^{\infty} f(2^nx)\, dx$$
Substituting $y=2^nx$ into the integrals on the RHS gives
$$\int_0^{\infty} g(x)\, dx = \sum_{n=1}^{\infty} \int_0^{\infty} 2^{-n}f(y)\, dy$$
We can then use linearity properties of the integral and sum on the RHS to show that this is equal to $\int_0^{\infty} f(x)\, dx$, as required.
[Hint: think geometric series] |
H: Answer of $5 - 0 \times 3 + 9 / 3 =$
According to order of operations the answer should be $\mathbf{2}$
But Google and Wolfram calculates as 8
This is last proccess: $5-0+3$
This is how I think: $5-(0+3)$
This is how Google answers: $(5-0)+3$
So, question is which operation is first $+$ or $-$?
AI: You proceed from left to right down the hierarchy.
$$5- 0\cdot3 + 9/3 = 5 - 0 + 3 = 5 + 3 = 8.$$
You make the mistake of distributing the $-$ to two terms in the absence of parentheses. |
H: Same arrow between distinct objects of a category
I'm a bit bothered by something I've come across, and I'd like to know if the misunderstanding is my own (likely) or the author's (unlikely).
Let $\mathcal{C}$ be a category. An arrow $e : A \to A$ in $\mathcal{C}$ is idempotent if $e \circ e=e$. If $\mathcal{E}$ is a class of (not necessarily all) idempotents in $\mathcal{C}$ then we can form a category $\mathcal{C}[\check{\mathcal{E}}]$ whose objects are the members of $\mathcal{E}$ and whose arrows $f : (e:A \to A) \to (d:B \to B)$ are arrows $f : A \to B$ in $\mathcal{C}$ satisfying $d \circ f \circ e=f$.
Something confuses me here. Say $1_A, 1_B \in \mathcal{E}$ and $(e,d) \ne (1_A,1_B)$. We have $d\circ f \circ e=f=1_B \circ f \circ 1_A$. So in the definition of $\mathcal{C}[\check{\mathcal{E}}]$ given above, $f$ is both an arrow $e \to d$ and an arrow $1_A \to 1_B$.
To me this seems absurd. But is it? Can we declare $f : e \to d$ and $f : 1_A \to 1_B$ to be distinct arrows in $\mathcal{C}[\check{\mathcal{E}}]$ despite having the same underlying arrow in $\mathcal{C}$?
AI: Why shouldn't we?
For a motivating example, consider for example category $Set$, and the objects $\{0\}$, $\{0,1\}$, and the arrows $f,g$ from $\{0\}$ to $\{0\}$ and to $\{0,1\}$, respectively, both of which correspond to the map $x\mapsto 0$. They both have the same "underlying mechanism", interpretation, but are nonetheless distinct arrows. |
H: Are there any secure ciphers you can use without a computer?
I have some kids that like encryption schemes such as the Caesar cipher and the Vigenère cipher. I would like to teach them something that's not easily breakable by todays maths and computers, but I want them to be able to use it just using pen and paper.
RSA probably isn't the way to go, since specially the prime finding part is quite intensive.
A symetric cipher like AES would also be fine, but the current AES standard requires lots of rounds of calculations to be effective.
One-time pads are nice, but not really a proper cipher.
Are there any cute modern schemes that are really cheap, but still have a reasonable amount of security?
AI: You might get a kick out of the
solitaire cypher that appeared in Neal Stephenson's novel, Cryptonomicon. The novel itself might be heavy going for kids, but the cypher is cool, since it's not only pretty robust but, even better, uses an ordinary deck of playing cards. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.