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H: Are there any examples of rings $R$ such that $\mathrm{End}(R,+,0)\not\cong R$?
In a handful of examples, I've noticed that the endomorphism ring $\mathrm{End}(R,+,0)$ is isomorphic to the ring $R$ itself. For instance, $\mathrm{End}(\mathbb{Z},+,0)\cong\mathbb{Z}$ and $\mathrm{End}(\mathbb{Z}/n\mathbb{Z},+,0)\cong\mathbb{Z}/n\mathbb{Z}$.
Is this true in general, or are there examples of rings which are not isomorphic to the endomorphism ring as above? If not, is it at least always true for $R$ a field? Thanks.
AI: The correct statement is that a ring $R$ is precisely the endomorphism ring of $R$ as a right $R$-module ("Cayley's theorem for rings"). When $R$ is any quotient of $\mathbb{Z}$, a morphism of right $R$-modules is just a morphism of abelian groups, which gives the examples you describe.
In general, preserving right $R$-module structure is stronger. For example, if $R = \mathbb{Z} \times \mathbb{Z}$ then the endomorphisms of the additive group of $R$ are given by $2 \times 2$ integer matrices. For a field counterexample, if $R = \mathbb{R}$ then $R$, as an abelian group, is a vector space over $\mathbb{Q}$ of uncountable dimension and therefore it admits an enormous endomorphism ring. (For a simpler field counterexample, if $R = \mathbb{Q}(i)$ then $R$, as an abelian group, is $\mathbb{Q}^2$, so its endomorphism ring is $2 \times 2$ rational matrices.) |
H: Rational numbers and cardinality of some subset set of them.
Let $G$ be the set of rational numbers of the form $m/n$ , where $m,n$ are positive integers and $n \leq g $ for some possitive integer $g$. Suppose it is bounded by $1/k$ , k is a positive integer less than g.
Prove or disprove that, cardinality of set $G$ is finite.
AI: More generally, let $g$ be a positive integer and $M$ a positive real number, and let $G$ be the set of rational numbers $m/n$ such that $m$ and $n$ are positive integers, $m/n\le M$, and $n\le g$; then $G$ is finite.
To see this, note first that there are only $g$ possible denominators for members of $G$, namely, the integers $1,2,\dots,g$. Let $n$ be one of these denominators. Then $m/n\le M$ if and only if $m\le Mn$. In other words, $m/n\in G$ if and only if $m$ is one of the $Mn$ integers $1,2,\dots,Mn$. In particular, since $n\le g$, we must have $Mn\le Mg$. Thus, there are only $g$ possible denominators, and for each of them there are at most $Mg$ possible numerators, so altogether there are at most $g(Mg)=Mg^2$ possible fractions in $G$. |
H: Proof of Non-Ordering of Complex Field
Let $\mathcal F$ be a field. Suppose that there is a set $P \subset \mathcal F$ which satisfies the following properties:
For each $x \in \mathcal F$, exactly one of the following statements holds: $x \in P$, $-x \in P$, $x =0$.
For $x,y \in P$, $xy \in P$ and $x+y \in P$.
If such a $P$ exists, then $\mathcal F$ is an ordered field.
Define $x \le y \Leftrightarrow y -x \in P \vee x = y$.
Exercise: Prove that the field of complex numbers $\mathbb C$ cannot be given the structure of an ordered field.
My Work So Far: (Edit 1 note: This section and the Question is at the beginning, simply leaving this up for reference as to where I started)
Let $i$ be such that $i \in P, i \ne 0 \Rightarrow i > 0$. But $i^2 = -1 \notin P$.
My Question: I am not sure how much I need to redefine, and how I go about rigorously making this patchwork argument airtight. I am aware that I have not addressed how I assumed that $-1 \notin P$, but I'm not sure how to distinguish between $1$ and $i$ in this proof.
Edit #1
1st Step: Showing that $-1 \notin P$, observe that $(-1)(-1) = 1$ therefore if $-1 \in P$, both $x, -x \in P$, a contradiction.
2nd Step: To show $i \notin P$, we have that if $i \in P \Rightarrow i^2 \in P$, but $i^2 = -1 \notin P$, so $i \notin P$.
3rd Step: To show $-i \notin P$, we have $(-i)(-i) = i^2 \notin P$, so $-i$ cannot be in $P$.
Conclusion: Since $i \ne 0$, and $i, -i \notin P$, there is no set $P \subset \mathbb C$ that satisfies the above properties, thus $\mathbb C$ is not ordered.
Thank you André Nicolas and Eric Stucky for your help!
AI: To show that $-1$ is not in $P$, note that if $-1\in P$ then $(-1)(-1)\in P$, which contradicts the fact that if $x \ne 0$ exactly one of $x$ and $-x$ is in $P$.
Next we show that $i\notin P$. Suppose to the contrary that $i\in P$. Then $i^2\in P$, which contradicts the fact that $-1\notin P$.
The same argument shows that $-i\notin P$. This contradicts the fact that if $x\ne 0$, then exactly one of $x$ and $-x$ is in $P$. |
H: Please help me understand conditional probability in relation to dice
If 2 fair dice are rolled, what is the probability that the sum is 6, given both dice show odd numbers?
When the sum is 6, there combinations are: (1,5),(2,4),(3,3),(4,2),(5,1). Probability = 5/36
Probability of 2 odd numbers is 1/2.
What I'm not sure about is when we work out Pr(sum is 6|both numbers odd). I initially thought that the top line was Pr(sum is 6), but then the result would be 5/2. This can't be right.
So, does that mean that we only choose combinations which have odd pairs? So, the answer would be 3/2?
Still makes no sense as the number is greater than 1???
Can somebody shed some light on this for me?
AI: What you're trying to compute is the probability that two dice sum to $6$, given that both dice have odd values. Note that there are three ways for pairs of dice with odd values to sum to six, namely $(1,5),(3,3),(5,1)$. There are three odd values each die can take, so there are nine ways for pairs of dice to both have odd values. Thus the probability you are looking for is $3/9=1/3$. |
H: The Lie algebra of a group of matrices
How does one find the Lie Algebra of a Lie group G which is given by matrices
$$\left(\begin{array}{cccc} \cos \theta & -\sin \theta & x & y \\ \sin \theta & \cos \theta & z & w \\ 0 & 0 & \cos \theta & -\sin \theta \\ 0 & 0 & \sin \theta & \cos \theta \end{array} \right)?$$
From the shape of the matrices, it seems G is isomorphic to the semidirect product of $SO(2)$ with $\mathbb{R}^4$.
I read in previous posts that one might try to identify which matrices exponentiate to matrices of this form, although this seems complicated. Is there another approach to tackle the problem?
AI: You can see which matrices exponentiate to matrices of this form. For example if $A \in \mathfrak o(2)$, i.e. $A$ is a skew-symmetric $2\times 2$ matrix then the block matrices
$$
\left(\begin{array}{c|c}
A & 0 \\\hline
0 & 0
\end{array}\right), ~~ \left(\begin{array}{c|c}
0 & 0 \\\hline
0 & A
\end{array}\right)
$$
exponentiate to
$$
\left(\begin{array}{c|c}
e^A & 0 \\\hline
0 & Id
\end{array}\right), ~~ \left(\begin{array}{c|c}
Id & 0 \\\hline
0 & e^A
\end{array}\right)
$$
which are elements of your group since $e^A \in SO(2)$. Similarly, for any $2\times 2$ matrix $B$,
$$
\left(\begin{array}{c|c}
0 & B \\\hline
0 & 0
\end{array}\right)
$$
exponentiates to
$$
\left(\begin{array}{c|c}
Id & B \\\hline
0 & Id
\end{array}\right) \in G
$$ since it is nilpotent. By dimensionality, the span of these must be the entire Lie algebra. |
H: Convention on comparing cardinality
When showing that two sets $A$ and $B$ have the same (finite or infinite) cardinality, it is usually done by constructing a function and showing that it is a bijection. However, in some cases, constructing such a bijection is not always easy. This can be worked around by constructing two functions, in a number of different combinations, for instance: $\phi:A\rightarrow B$ and $\psi:B \rightarrow A$ both injective or both surjective.
My question is, in the cases where it's easier, why is this not done? An example would be between the Cantor set and the closed unit interval. Constructing a bijection means taking great care, since any number with trailing zeros can also be written with trailing $($base $-\:1)$s.
Constructing an injection each way is easy, though. The Cantor set can by construction be seen as a subset of the unit interval, so there's inclusion. Any number in the unit interval has a binary expansion, so pick one of the expansions for each number and this will give an element in the cantor set by going left and right for 0 and 1.
AI: One note: depending on whether you're taking the Axiom of Choice for granted, $\phi,\psi$ both being surjective isn't enough to guarantee a bijection. Schroeder-Bernstein Theorem tells us that it is enough if $\phi,\psi$ are both injective, though.
I more often find myself constructing two injections than one bijection, personally, and many texts do the same. |
H: Find perimeter of trapezium
Find perimeter of a trapezium with 3 sides given and distance b/w parallel sides given.
Let the given sides be a,b,c and distance b/w parallel lines be h. How to go about this problem?
AI: There are four possible variatoins for the given situation:
$\hspace{2cm}$
Thus, depending on which of these variations holds, we get the perimeter to be one of the four following values:
$$
2a+b+c\pm x_b\pm x_c=2a+b+c\pm\sqrt{b^2-h^2}\pm\sqrt{c^2-h^2}
$$ |
H: What does the notation $[0,1)$ mean?
I am studying the procedure for bucket sort from Introduction To Algorithms by Cormen et al, which assumes that the input is generated by a random process that distributes the elements uniformly and independently over the interval $[0,1).$ What does this mean? Why there is no "]" closing bracket for the interval?
AI: The notation $[0,1)$ refers to the set of all real numbers $x$ such that $0\le x\lt 1$. Another common notation for this set is $[0,1[$; which is more common often depends on the language in which the author was educated. |
H: What is the link (if there is one) between separability and the separation axioms?
I was wondering about this topic. Is there a connection between the $T_n$ separation axioms and separability itself?
AI: No; there is no real connection between the two notions. There are both separable and non-separable spaces with any of the separation axioms. |
H: Count permutations.
Hi I have a compinatorial exercise:
Let $s \in S_n$. Count the permutations such that
$$s(1)=1$$ and
$$|s(i+1)-s(i)|\leq 2 \,\, \mathrm{for} \, \, i\in\{1,2, \ldots , n-1 \}$$
Thank you!
AI: This is OEIS A038718 at the On-Line Encyclopedia of Integer Sequences. The entry gives the generating function
$$g(x)=\frac{x^2-x+1}{x^4-x^3+x^2-2x+1}$$
and the recurrence $a(n) = a(n-1) + a(n-3) + 1$, where clearly we must have initial values $a(0)=0$ and $a(1)=a(2)=1$.
Added: I got this by calculating $a(1)$ through $a(6)$ by hand and then looking at OEIS. For completeness, here’s a brief justification for the recurrence. Start with any of the $a(n-1)$ permutations of $[n-1]$. Add $1$ to each element of the permutation, and prepend a $1$; the result is an acceptable permutation of $[n]$ beginning $12$, and every such permutation of $[n]$ is uniquely obtained in this way. Now take each of the $a(n-3)$ permutations of $[n-3]$, add $3$ to each entry, and prepend $132$; the result is an acceptable permutation of $[n]$ beginning $132$, and each such permutations is uniquely obtained in this way. The only remaining acceptable permutation of $[n]$ is the unique single-peaked permutation: $13542$ and $135642$ are typical examples for odd and even $n$ respectively.
From here we can easily get the generating function. I’ll write $a_n$ for $a(n)$. Assuming that $a_n=0$ for $n<0$, we have $a_n=a_{n-1}+a_{n-3}+1-[n=0]-[n=2]$ for all $n\ge 0$, where the last two terms are Iverson brackets. Multiply by $x^n$ and sum over $n\ge 0$:
$$\begin{align*}
g(x)&=\sum_{n\ge 0}a_nx^n\\
&=\sum_{n\ge 0}a_{n-1}x^n+\sum_{n\ge 0}a_{n-3}x^n+\sum_{n\ge 0}x^n-1-x^2\\
&=xg(x)+x^3g(x)+\frac1{1-x}-1-x^2\;,
\end{align*}$$
so
$$\begin{align*}g(x)&=\frac{1-(1-x)-x^2(1-x)}{(1-x)(1-x-x^3)}\\
&=\frac{x-x^2+x^3}{1-2x+x^2-x^3+x^4}\;.
\end{align*}$$
This is $x$ times the generating function given in the OEIS entry; that one is offset so that $a(0)=a(1)=1$, and in general its $a(n)$ is the number of acceptable permutations of $[n+1]$; my $a_n$ is the number of acceptable permutations of $[n]$. (I didn’t notice this until I actually worked out the generating function myself.) |
H: Intersection of Two Simplices
How to find vertices a the polytope-intersection of two simplices, if I know the vertices of these simplices.
More precisely:
Let $T_1$ and $T_2$ be two regular $n-1$ dimensional simplices with vertices $(t,0,\ldots,0), (0,t,\ldots, 0),\ldots, (0, 0, \ldots, t)$ and $(t-n+1,1,\ldots, 1), (1, t-n+1, \ldots, 1), \ldots, (1,1, \ldots, t-n+1)$ resp.
The intersection of these simplices is a polytope $P$.
How to find vertices of it?
AI: The first simplex is the intersection of the half-spaces $x_j \ge 0$ and the hyperplane $\sum_j x_j = t$.
The second is the intersection of $x_j \le 1$ and $\sum_j x_j = t$. So $P$ is the intersection of the hypercube $0 \le x_j \le 1$ with the hyperplane $\sum_j x_j = t$.
Note that any point in $P$ that has more than one non-integer coordinate can't be an extreme point of $P$. Let's suppose $m < t < m+1$ for some integer $m$, $0 < m < n$. Then the vertices of $P$ will be the $n {{n-1} \choose m}$ points that have $m$ coordinates $1$, $n-m-1$ coordinates $0$ and one $t - m$. |
H: Tensor product algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$
I want to understand the tensor product $\mathbb C$-algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$. Of course it must be isomorphic to $\mathbb{C}\times\mathbb{C}.$ How can one construct an explicit isomorphism?
AI: An explicit isomorphism of $\mathbb C$-algebras is given (on generators) by $ \mathbb C\otimes _\mathbb R \mathbb C\stackrel {\cong }{\to} \mathbb C\times \mathbb C: z\otimes w \mapsto (z\cdot w,z\cdot\bar w)$.
Here $ \mathbb C \otimes _\mathbb R \mathbb C$ is considered as a $\mathbb C$-algebra through its first factor: $z_1\cdot (z\otimes w)=z_1 z\otimes w $ |
H: How does one construct a 'sign chart' when solving inequalities?
I'm working on solving inequalities for an assignment. The instructions also request that I draw a 'sign chart' along with each solution. I've never heard of a 'sign chart' before, and the internet also seems to have a limited amount of information. From what I can gather...
PurpleMath prevents hotlinking...
PurpleMath prevents hotlinking...
(both from PurpleMath at this page and this page, respectively.)
PurpleMath never actually refers to these diagrams as sign charts either, but I assume these 'factor charts' are synonymous.
The first image is a sign chart for (x + 4)(x – 2)(x – 7) > 0. I know that this has roots at -4, 2, and 7. These are indicated in between columns. The factors are written along rows. The -/+ is determined by evaluating a test point in a range dependent on the columns in each factor. (This is just my understanding - I could by all means be very, very incorrect.) For example - cell "B3" - the range is -4 < TEST < 2. 0 is in this range, so I will evaluate (x-2) for 0. The resulting sign is -, so I mark that in cell B3. I evaluate all other cells in that column, except for the top. The top cell of each column is determined by the resulting sign of all signs below it (in column 3, - * - = +).
However, my intuition fails me when the problems become more complex. Take, for example,
4x^2 + 8x < 6. When solved using the quadratic equation, x equals a messy (-2 +/- sqrt(10) ). I don't understand what to put along the columns and rows. What goes where?
AI: When I hear the phrase 'sign chart,' I do not think of the factor charts that you have included above.
For example, I would do the following for the inequality $(x+2)(1-x) \geq 0$.
Find the roots. The roots of this polynomial are $x=-2,1$.
Draw a number line, and on this line label my roots.
Decide whether the polynomial is positive or negative between every pair of roots.
So here, my sign chart would look like (except not drawn in GIMP):
To determine the $+$ and $-$ signs, we just test any point in that region. When $x = 0$, we have $(2)(1) > 0$. Plugging in $-10$, we have $-10(11) < 0$, and plugging in $10$ we have $(12)(-9)<0$.
If this looks familiar, then that's what you should do. If not, though, then you'll have to ask your teacher/consult your book about what a sign chart is. |
H: Proving a certain set in a compact metric space is open
The question is:
Let $(X,d)$ be a compact metric space. Let $f : X \rightarrow X$ be continuous. Fix a point $x_0 \in X$, and assume that $d(f(x), x_0)\geq1$ whenever $x\in X$ is such that $d(x,x_0)=1$. Prove that $U\setminus f(U)$ is an open set in $X$, where $U=\{ x\in X: d(x,x_0) <1\}$.
This is one of the analysis qual question. (I am going to sit for one in two days time)
I am having problem visualize the set in the question, whenever I have this sort of feeling, I will have great difficulties answering the question. Any suggestion? Thanks.
AI: Hint: The thing to notice here is that all elements $x\in \partial U$ satisfy $d(x,x_0) = 1$. So by assumption $f(x)\notin U$ for all $x\in \partial U$.
This then implies that $f(\overline U) \cap U = f(U) \cap U$.
In general, if you read the condition "$X$ compact, Hausdorff", a thing on your mind should be that all continuous maps are automatically closed. |
H: Determine characteristics for the method of characteristics
I'm trying to understand the method of characteristics to solve first-order PDEs. As an example in his course, my professor solve this PDE for $u(x,y)$:
$$x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}=R $$
with $u(s,s)=f(s)$ given along the parametrized curve $\Gamma$ defined by $x(s)=s$ and $y(s)=s$.
To determine the characteristics he first writes:
$$\frac{dx}{x}=-\frac{dy}{y}$$
And then immediately jumps to:
$$\int^x_s \frac{dx'}{x'}=-\int^y_x\frac{dy'}{y'}$$
I guess he simplifies a lot the resolution of the simple ODE but I don't really understand how he does that and how he finds the characteristics that way.
AI: The basic idea of the method of characteristics is to differentiate the unknown $u$ along a path $(x(t),y(t))$: You will find
$$\frac{d}{dt}u(x(t),y(t))=\dot x u_x+\dot y u_y$$
where I use dots for differentiation wrt $t$, and subscripts for partial derivatives.
Now, to get some help from the given PDE, we need the vector $(\dot x,\dot y)$ to be parallel to the vector $(x,-y)$ from the coefficients of the PDE. So long as $x$ and $y$ are nonzero this requirement boils down to
$$\frac{\dot x}{x}=-\frac{\dot y}{y}$$
which, if we multiply by $dt$, becomes
$$\frac{dx}{x}=-\frac{dy}{y}$$
just as your professor noted.
Whether you wish to think of $y$ as a function of $x$, or $x$ as a function of $y$, this is the standard form of a separable ODE, and it is solved by integrating as shown. |
H: Generator of the multiplicative groups of units in $\mathbb{Z_5[x]}/(x^{2}+3x+3)$
I would like to find a generator of the multiplicative groups of units in $\mathbb{Z_5[x]}/(x^{2}+3x+3)$. This is a field since $x^{2}+3x+3$ is irreducible, so every coset with $bx+a\not=0$ as a representative should be a unit...
I do not understand how to go from here though... Since $b\not=0$ we have $4$ diffrent choices for $b$ and $5$ different options for coefficient $a$ hence $24$ elements in the multiplicative group of units.
Should any $bx+a$ with order $24$ be a generator? How do I go from here?
AI: It suffices to rule out the maximal factors of $24$, i.e. $8=24/3$ and $12=24/2$ as possible orders. The (coset of) $x$ fits the bill, because
$$
x^4=(-x^2)^2\equiv (3x+3)^2=4(x^2+2x+1)=-x^2+3x+4\equiv6x+7=x+2,
$$
and therefore neither
$$
x^8=(x^4)^2\equiv(x+2)^2=x^2+4x+4\equiv x+1
$$
not
$$
x^{12}=x^8\cdot x^4\equiv(x+2)(x+1)=x^2+3x+2\equiv-1
$$
are equal to $1$.
The point is that other possible orders ($2,3,4,6$) that are less than $24$ are factors of either $8$or $12$.
Now that we settled that $x$ is a generator, the other generators can be found by reducing the powers $x^j$, $0<j<24, \gcd(j,24)=1$ modulo the generator $x^2+3x+3$. |
H: Reference request: stability theory in infinite dimensional dynamical systems/ partial differential equations
I am looking for some references (text books, elementary review papers, journal articles etc) regarding the phenomenon of breakdown in stability for (nonlinear) partial differential equations, i.e if we start off with a partial differential equation and we have a steady state solution and suppose we perturb it, how do we analyse if this leads to stability or instability, I suppose we use semigroup methods and spectral theory amongst other things. What are the various tools, techniques etc that are available. When does linear stability imply nonlinear stability? When does breakdown in stability occur? Are there ways/ theorems to that establish necessary / sufficient conditions for nonlinear instability?
Thank you.
AI: If you have access, look at the paper
Optimal gap condition for invariant manifolds, Continuous and Discrete Dynamical
Systems, 5 (1999) 233-268
By Latushkin and Layton.
Many of the results are based on methods presented in the book by Daletskii and Krein. This is a book for ODE-s, but in many parts it can be extended nicely to PDEs. And, you can understand the ideas before going into the technicalities. |
H: Solving $f'(x) +f(x)=cf(x-1)$
To show that $f(x) =Ae^{nx}$ for constant $n$ and $A$ starting with this thing:
$$f'(x) +f(x)=cf(x-1)$$
Where $c$ is constant and $c\not= 0$.
If it wasn't for the $f(x-1)$ bit, I would just use the integrating factor where $I=e^x$ and plug it into the equation. But the $f(x)$ throws me off, so I would have to put it into a form like $\frac{dy}{dx}+y=?$, in order to feel ok about it.
EDIT: Oh as somebody rightly pointed out this is only for the condition when $n$ satistfies $$n+1 = ce^{-n}$$
AI: If you plug in $f(x) = A e^{kx}$ you find that this satisfies your equation whenever $k+1 = c e^{-k}$.
So for every (real or complex) $k$ satisfying: $$k+1 = c e^{-k}$$ You have a solution of the form $A e^{kx}$
By linearity, arbitrary linear combinations of solutions are solutions. Even if you're just looking for real solutions, you still have to consider complex $k$ because if $$k = a + i b$$ satisfies:
$$k+1 = c e^{-k}$$
Then it follows, that $\cos(bx) e^{ax}$ and $\sin(bx) e^{ax}$ are solutions.
Since (for $c \ne 0$) $k + 1 = c e^{-k}$ has infinitely many complex solutions, this gives an infinite-dimensional family of solutions of your equation. |
H: How low can the approval rating of a majority candidate be?
“Ostrogorski's paradox” describes a strange situation in which voters decide on candidates based on issues in platforms, but on each issue of the platform, the majority of voters disapprove of the majority winner.
What is the lowest possible approval rating for a majority winner?
Example
For example, if there are 3 issues, then the voters may divide into four camps: one camp of “all issues are important”, and three camps of “only this issue is important, the others detract from it.” A sample poll may reveal:
\begin{array}{r|rrr}\newcommand{\t}[1]{\text{#1}}
& \t{Size} & \t{Issue 1} & \t{Issue 2} & \t{Issue 3} \\ \hline
\t{Camp 0} & 40\% & \t{Yes} & \t{Yes} & \t{Yes} \\
\t{Camp 1} & 20\% & \t{Yes} & \t{No} & \t{No} \\
\t{Camp 2} & 20\% & \t{No} & \t{Yes} & \t{No} \\
\t{Camp 3} & 20\% & \t{No} & \t{No} & \t{Yes} \\ \hline
\t{Favored}& & 60\%\t{ Yes}& 60\%\t{ Yes}& 60\%\t{ Yes}
\end{array}
Candidate A looks at the polls and sees that $60\%$ of people support each issue, and so decides his platform is to support all three issues. Candidate B is contrary and decides his platform is to reject all three issues. The voters in camp 0 vote for A, obviously. The voters in camp 1 see that candidate B agrees with them on 2 out of 3 issues, and so vote for B. Similarly camps 2 and 3 vote for B.
Candidate B wins $60\%$ to $40\%$ a huge margin of victory! However, once in office, polls reveal that only $40\%$ of voters approve of his handling of issue 1. Similarly for issue 2 and 3. Nobody likes the majority candidate!
Formal problem
Let $I$ be a finite set (of issues), and $\mathcal{S}=\wp(I)$ be the set of all subsets of issues, so that each $S \in \mathcal{S}$ represents the issues that are supported. A voter profile is a function $\mu$ from $\mathcal{S}$ to the closed interval $[0,1]$ such that $\sum_{S \in \mathcal{S}} \mu(S) = 1$. An election is a subset $C=\{C_1,C_2\}$ of $\mathcal{S}$ of size 2 (the two candidates) along with a voter profile. The outcome $O(C_1,C_2,\mu)$ of an election is $$O(C_1,C_2,\mu) = \sum \left\{ \mu(S) : S \in \mathcal{S} ~\mid~ \left|S \oplus C_1 \right| \leq \left|S \oplus C_2\right| \right\}$$
where $S \oplus C_j$ is the symmetric difference of $S$ and $C_j$, that is, the set of elements where $S$ and $C_j$ differ. The approval rating of candidate $C_j$ on issue $i$ is $$A(C_j, i, \mu) = \sum \left\{ \mu(S) : i \notin S \oplus C_j \right\}$$
and the maximum approval rating of candidate $C_j$ is:
$$A(C_j, \mu) = \max\left\{ A(C_j,i,\mu) : i \in I \right\}$$
What is the infimum of the maximum approval rating $A(C_1,\mu)$ given that $O(C_1,C_2,\mu)\geq \tfrac12$?
Partial result
Let $|I|=2n+1$ be large and set $\mu(I)=0.5$, $\mu(S) = \frac{1}{2 \binom{2n+1}{n}}$ if $|S|=n$, and $\mu(S) = 0$ otherwise; so that half of the voters support every issue and half support just under half of the issues. Let $C_1 = \{\}$ support nothing, and $C_2 = I$ support everything. Then $O(C_1,C_2,\mu) = 0.5$ but $A(C_1,i,\mu)$ is the sum of about half of the $S$ of size $n$:
$$A(C_1,i,\mu) = \sum\left\{ \mu(S) : i \notin S, S \in \mathcal{S} \right\} = \frac{1}{2 \binom{2n+1}{n}} \cdot \binom{2n}{n} = \frac{1}{4} \cdot \frac{2n+2}{2n+1} \to \frac14$$
This is as bad as I could make it. Is it as bad as possible?
AI: Given any $I$, $\mu$ and $C$ with $O(C_1,C_2,\mu)\ge\frac12$, we can modify them as follows without changing the maximum approval rating. First, for all issues $i$ with $A(C_1,i,\mu)\lt A(C_1,\mu)$, we can let voters change their mind on this issue in favour of $C_1$ until $A(C_1,i,\mu)=A(C_1,\mu)$. This doesn't change $A(C_1,\mu)$, and it cannot decrease the majority for $C_1$. Next we can remove all issues on which the candidates agree from the issue set; these issues don't affect the majority, and since $C_1$ now has the same approval rating for all issues, they also don't affect the maximum approval rating. Further (this is just for convenience), we can flip all issues on which $C_1$ is in favour (by negating the question on the ballot) without any substantial change. So now we have $C_1$ against all issues, $C_2$ in favour of all issues, and $C_1$ with the same approval rating for all issues, and the maximum approval rating hasn't changed and the majority hasn't decreased. Since at least half the voters must agree with $C_1$ on at least half the issues it's clear that the maximum approval rating can't be below $\frac14$. |
H: What is the tensor product $M_n(L)\otimes_K L$, where $L/K$ is a quadratic extension?
What is the tensor product $M_n(L)\otimes_K L$, where $L/K$ is a quadratic extension?
Let $K$ be a field of characteristic $0$, $L/K$ a quadratic extension.
Let $\rho\in \operatorname{Gal}(L/K)$ denote the nontrivial element of the Galois group.
Let $M_n(L)$ denote the algebra of $n\times n$ matrices over $L$.
The "conjugation" $\rho$ acts on $M_n(L)$ by conjugating each matrix element.
I want to understand the tensor product $M_n(L)\otimes_K L$
as an algebra over the right-hand $L$,
with conjugation coming from the action of $\rho$ on $M_n(L)$.
Of course it must be $M_n(L)\oplus M_n(L)$ (and the conjugation must send $(X,Y)\in M_n(L)\oplus M_n(L)$ to $(Y^\rho,X^\rho)$).
Why is it so, and how can I construct an explicit isomorphism of $L$-algebras $M_n(L)\otimes_K L\to M_n(L)\oplus M_n(L)$
(in light of the answers to my previous question Tensor product algebra $\mathbb{C}\otimes_\mathbb{R} \mathbb{C}$)?
AI: You may take the $\operatorname{Gal}(L/K)$-equivariant isomorphism of $L$-algebras given by $$M_n(L)\otimes_K L\xrightarrow {\cong } M_n(L)\times M_n(L):A\otimes l\mapsto (A\cdot l,A^\rho\cdot l) $$ |
H: Proving that $\lim\limits_{n\to\infty}n\left( \int_0^1 f(t)\, dt -\frac1n\sum_{k=0}^{n-1}f\left(\frac k n\right) \right)=\frac{f(1)-f(0)}{2}$
Let $f\in C^2([0,1])$. Prove that $$ \lim_{n\to+\infty}n\left( \int_0^1
f(t)\, dt -\frac1n\sum_{k=0}^{n-1}f\Big(\frac k n\Big)
\right)=\frac{f(1)-f(0)}{2}. $$
The second term is clearly the Riemann sum of the function $f$; since the function $f$ is integrable (it is continuous) $\displaystyle \frac1n\sum_{k=0}^{n-1}f\Big(\frac k n\Big)$ converges to $\displaystyle\int_0^1 f(t)\, dt$ when $n \to + \infty$.
So we have an indeterminate form, "$\infty \cdot 0$". How can we start? I thought we should use Taylor expansion ($f$ is $C^2$) but I cannot see how. Would you please help me?
Thanks in advance.
AI: One can check by integrating by parts that
$$
f\left(\frac{k}{n}\right)-n\int\limits_{(k-1)/n}^{k/n}f(t)dt=
n\int\limits_{(k-1)/n}^{k/n}f'(t)\left(t-\frac{k-1}{n}\right)dt=
\int\limits_{0}^1\frac{t}{n}f'\left(\frac{t+k-1}{n}\right)
$$
So using dominated convergence theorem we get
$$
\lim\limits_{n\to+\infty}n\left(\int_0^1f(t)dt -\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac k n\right)
\right)=
\lim\limits_{n\to+\infty}\int\limits_{0}^1\frac{t}{n}\sum_{k=0}^{n-1}f'\left(\frac{t+k-1}{n}\right)=
$$
$$
\int\limits_{0}^1t\lim\limits_{n\to+\infty}\frac{1}{n}\sum_{k=0}^{n-1}f'\left(\frac{t+k-1}{n}\right)=
\int_{0}^{1}t\left(\int\limits_{0}^1 f'(s)ds\right)dt=\frac{f(1)-f(0)}{2}
$$
Note that for this proof it is enough to require that $f\in C^1([0,1])$ |
H: (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior
Since the specific space $\mathbb{R}^k$ is given, this might be provable in ZF.
Let $\{F_n\}_{n\in \omega}$ be a family of closed subset of $\mathbb{R}^k$, of which the union is $\mathbb{R}^k$.
Suppose $\forall n\in \omega, {F_n}^o=\emptyset$ ($o$ denotes interior)
Fix $x_0 \in \mathbb{R}^k$
Then by assumption, $\forall 0<r\in \mathbb{R}$, there exists $y\in \mathbb{R}^k$ such that $y\in B(r,x_0)\setminus (F_n \cup \{x_0\})$ for every $n\in \omega$.
I tried to show that for every $r$, there exists $y\in \mathbb{Q}^k$ that satisfies the condition.
(Once the existence is guaranteed, since $\mathbb{Q}^k$ can be well-ordered lexicographically, $y$ can be chosen uniquely, hence recursion theorem would be appliable)
But i couldn't. Is there any way to show this?
If it is true, let $y_0 \in B(r,x)\setminus (F_0 \cup \{x_0\})$ and $y_{n+1} \in B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$.
Then, $\forall n\in \omega$, $B(d(y_n,x_0),x_0) \setminus (F_{n+1} \cup \{x_0\})$ is nonempty.
Here, again, i don't know how to show that 'there exists $y\in \mathbb{R}^k$ such that $y\notin \bigcup_{n\in \omega} F_n$.
Please help.
If my argument is wrong, please give me a proper proof.
AI: This is essentially the Baire category theorem. Indeed in ZF it holds for separable complete metric spaces.
The argument is as follows:
Suppose that $(X,d)$ is a separable complete metric space, and $\{a_k\mid k\in\omega\}=D\subseteq X$ is a countable dense subset.
By contradiction assume that we can write $X=\bigcup F_n$ where $F_n$ are closed and with empty interior, we can further assume that $F_n\subseteq F_{n+1}$.
Define by recursion the following sequence:
$x_0 = a_k$ such that $k=\min\{n\mid a_n\notin F_0\}$;
$r_0 = \frac1{2^k}$ such that $d(F_0,x_0)>\frac1k$, since $x_0\notin F_0$ such $k$ exists.
$x_{n+1} = a_k$ such that $k=\min\{n\mid a_n\in B(x_n,r_n)\setminus F_n\}$;
$r_{n+1} = \frac1{2^k}$ such that $d(F_n,x_{n+1})>\frac1k$, the argument holds as before.
Note that $x_n$ is a Cauchy sequence, therefore it converges to a point $x$. If $x\in F_n$ for some $n$, first note that $d(x_k,F_n)\leq d(x_k,x)$, by the definition of a distance from a closed set.
If so, for some $k$ we have that $d(x,x_k)<r_n$, in particular $d(F_n,x_k)<r_n$. First we conclude that $n<k$, otherwise $d(F_n,x_n)>r_n$. Now we note that:
$$d(F_n,x_n)\leq d(x,x_n)\leq d(x,x_k)+d(x_k,x_n)\leq r_n+r_n=2r_n$$
It is not hard to see that $2r_n< d(F_n,x_n)$, which is a contradiction to the choice of $x_n$.
Added: Why is there $a_k$ in every step of the inductive definition?
Note that $D$ is dense therefore $\overline{D}=X$. In particular, if $F_n$ is closed and has a dense subset then $F=X$. Since we assume that the interior of $F_n$ is empty we have that $F_n\cap D$ is not dense in $X$, otherwise $F_n=X$ and has a non-empty interior.
We have that $D\setminus F_n$ is dense, since $D\cap F_n$ is not dense. In particular this means that in every open set there is some $a_k\in D\setminus F_n$, and thus the induction can be carried in full. |
H: Every Ring is Isomorphic to a Subring of an Endomorphism Ring of an Abelian Group
Show that for every ring $(R,+,\cdot)$, there is an abelian group, $(A,+)$, such that $R$ is isomorphic to a subring of $(\operatorname{End}(A),+,\circ)$.
$(\operatorname{End}(A),+,\circ)$ is the set of homomorphisms of $A$ that form a ring under function addition and composition.
I am thinking to let $\operatorname{End}(A)$ be the group such that $A$ is the abelian group $(R,+)$ and create a ring homomorphism from $(R,+,\cdot)$ into $\operatorname{End}((R,+))$.
Thoughts?
AI: You are correct, (madame or) sir.
This is essentially the ring-theoretic analogue of Cayley's Theorem for groups.
Also, this issue (as a question) came up a while back on Math Overflow.
Added: I had missed that the explicit definition of the map was not contained in the OP's question. A natural ring embedding from $R$ to $\operatorname{End}(R,+)$ is
$r \mapsto \bullet r: (x \in R \mapsto xr)$.
[Or possibly $r \mapsto r \bullet: (x \in R \mapsto rx)$, depending upon your conventions on composition.] |
H: Any infinite set of a compact set $K$ has a limit point in $K$?
I'm reading principles of mathematical analysis and have a question about a theorem 2.37.
Theorem 2.37
If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
The proof is
If no point of $K$ were a limit point of $E$, each $q \in K$ would have a neighborhood $V_q$ which contains at most one point of $E$.
It is clear that no finite subcollection of $\{V_q\}$ can cover $E$.
The same is true of $K$, since $E \subset K$.
This contradicts the compactness of $K$.
I understand the first part that states that no finite subcollection of $\{V_q\}$ can cover $E$, in other words, $E$ is not compact. But I don't understand why it means that no finite subcollection can cover $K$. Is the author saying that if a subset of a set K is not compact, then $K$ is not compact? If that's the case, a compact set may have open subsets which are not compact, so I'm confused.
Thanks in advance.
AI: A set with no limit points is necessarily a closed set. Being a closed subset of a compact space, it is compact. On the other hand, you're looking at a proof that $E$ is not compact, so you've got a contradiction.
Alternatively, look at this set of neighborhoods that cover $E$ and add one more open set to this collection: the complement of $E$. That set is open, since as noted above, $E$ is closed. Now you've got an open cover of $K$. It must therefore have a finite subcover. But every finite subset of this cover fails to cover all of $E$, so again you have a contradiction. |
H: Proof that $1$ is the least natural number using the Principle of Well Ordering
I'm doing a question from a PDF on Abstract Algebra.(The pdf can be found here http://abstract.pugetsound.edu/ 2012 edition). I have to show that the Principle of Well-Ordering implies that $1$ is the smallest natural number; this is only part of the whole question.(Question 14 of Chapter 2).
This is what I got so far.
Proof.
Let $S$ be an arbitrary subset of the Natural Number excluding one. By the Principle of Well-Ordering there exist a least element $X \in S.$
Now add $1$ to $S.$ $1$ is now the least element. Since there is no $X < 1$ in any such set $S,$ $1$ is the smallest element of any set $S$ and therefore the smallest natural number.
AI: By the Well-Ordering principle, there is a least element $a$ in $\mathbb{N}$. Suppose, to the contrary, that $a<1$. Multiplying both sides of the inequality by $a$, we have $a^2<a$. Since $\mathbb{N}$ is closed under multiplication, $a^2\in\mathbb{N}$, which contradicts the assumption that $a$ is the least element in $\mathbb{N}$. |
H: A specific example of the GNS construction
In an introduction to the GNS construction, I'm told that the GNS construction is a generalization of the way that $L^{\infty} (X, \mu)$ has a representation on $L^2$ where $\mu$ is a measure on $X$. Can someone demonstrate that, with the appropriate notions of equivalence in place (that I hope you will specify), indeed the GNS construction applied to the unital C* algebra $L^\infty$ with respect to some choice positive linear functional, does furnish the Hilbert space $L^2$ and then the representation as stated? As a very brief reminder, a representation is a $*-homomorphism$ coming out of a unital C* algebra into $B(H)$ preserving the identity, and in the GNS construction, one starts out with a unital C* algebra and a positive linear functional on it, which is then necessarily continuous. (Although I always mean everything is continuous.) You take the induced sesquilinear form and quotient by its null space, and then take the Hilbert space completion of the resulting inner product space. The representation is, roughly speaking, by left multiplication. (Maybe that's reversed if you use the reverse convention for what it means to be a sesquilinear form.)
AI: The positive linear functional is integration. Note that if you want the integral of the identity to be finite then $\mu$ needs to be a finite measure.
In that case, integration induces the usual $L^2$ inner product
$$\langle f, g \rangle = \int_X \overline{f(x)} g(x) \, d \mu$$
on $L^{\infty}(X, \mu)$. Null with respect to either $L^2$ or $L^{\infty}$ means equal to zero a.e. so this inner product is already positive-definite. By finiteness, $L^{\infty}$ contains the indicator function of every measurable subset of $X$, so the completion of $L^{\infty}$ with respect to the $L^2$ inner product is just $L^2(X, \mu)$ and $L^{\infty}$ acts on it by left multiplication as desired. |
H: Is there a "simple" proof of the isoperimetric theorem for squares?
"Simple" means that it doesn't use any integral or multivariable calculus concepts.
A friend of mine who's taking a differential calculus course came up with the problem
Prove that among all the quadrilaterals with a given perimeter, the one with the biggest area is the square.
I solved the problem with Lagrange multipliers, using $2h + 2b$ as the function and $hb = A$ as the constraint. But I'm the one who took the multivariable calculus course, not him.
So I'd like to know if there's a way of proving this theorem using differential calculus concepts, or even geometry and trigonometry.
AI: In the case of rectangles, here's a solution of the dual problem: find the rectangle of smallest perimeter for a given area. The shapes should be the same.
Just complete the square (if you can pardon the expression). The width is $w$; the height is $A/w$ (where $A$ is the area). So the semi-circumference is
$$
w + \frac A w = \left(w - 2\sqrt{A}+ \frac A w\right) + 2\sqrt{A} = \left(\sqrt{w} - \sqrt{\frac{A}{w}}\right)^2 + 2\sqrt{A}.
$$
This is as small as possible when the expression that gets squared is $0$. So that $=0$ when $w=\text{what?}$
Later edit: Now let's try it more directly. The perimeter is $4\ell$. You have a rectangle with two opposite sides of length $k$ and and two of length $2\ell-k$. The area is
$$
\begin{align}
A & = k(2\ell-k) = -k^2 + 2k\ell = -\Big(k^2 - 2k\ell\Big) = -\Big(k^2 -2k\ell + \ell^2\Big) +\ell^2 \\[8pt]
& = -\Big(k-\ell\Big)^2 + \ell^2.
\end{align}
$$
This is as big as possible when $k=\ell$, so you have a square.
We still have the case of non-rectangles to deal with. |
H: Why can't $\int_0^1\sin(x^2) dx$ be equal to $2$?
Why can't $\int_0^1\sin(x^2) dx$ be equal to $2$?
What makes this true? Intuitively, it makes sense. But why?
AI: While Peter's answer is right, I think there's a more intuitive estimate. We know that $|\sin(y)| \leq 1$, and thus $\displaystyle \left|\int_0^1 \sin(x^2)dx \right| \leq \int_0^1 1 dx = 1$, the area of a square of side-length $1$. |
H: How to evaluate the integral $\iint_C \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$
I would like to evaluate
$$\iint\limits_C \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$$
where $C$ is the first quadrant, i.e.
$$\int\limits_0^\infty\int\limits_0^\infty \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$$
where
$$\int\limits_0^\infty \sin x^2\, dx=\int\limits_0^\infty \cos x^2\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$
AI: Note that
$$
\int_0^\infty\int_0^\infty \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy=
\operatorname{Im}\left[\int_0^\infty\int_0^\infty \exp (i(x^2+y^2))\,\mathrm dx\,\mathrm dy\right]
$$
so, recalling that $\int f(x)\, \mathrm dx=\int f(y)\, \mathrm dy$
$$
\int_0^\infty\int_0^\infty \exp (i(x^2+y^2))\,\mathrm dx\,\mathrm dy=
\int_0^\infty \exp(ix^2) \,\mathrm dx\int_0^\infty \exp (iy^2)\,\mathrm dy=
\left(\int_0^\infty \exp(ix^2) \,\mathrm dx \right)^2
$$
Expanding the $\exp (ix^2)$ into $\cos x^2+i\sin x^2$, we find
$$
\left(\int_0^\infty \exp(ix^2) \,\mathrm dx \right)^2=
\left(\int_0^\infty \cos x^2+i\sin x^2 \,\mathrm dx \right)^2=\left((1+i)\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)^2=
2i \cdot \frac{\pi}{8}=
i\frac{\pi}{4}
$$
So, taking the imaginary part, we see the answer is $\frac{\pi}{4}$. As an added bonus, we take the real part of that answer ($0$) to determine that $\int_0^\infty\int_0^\infty \cos (x^2+y^2)\,\mathrm dx\,\mathrm dy=0$
Here is an alternate proof:
If we use polar coordinates, we see ($x=r\cos \theta$, $x=r\sin \theta$, $\mathrm dx\,\mathrm dy = r\mathrm dr\,\mathrm d\theta$)
$$
\int_0^\infty\int_0^\infty \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy=
\int_{0}^{2\pi}\int_0^\infty r\sin (r^2)\,\mathrm dr\,\mathrm d\theta
$$
which diverges. So, we introduce a "dummy function," $\exp(-\delta (x^2+y^2))$ that equals $1$ when $\delta \to 0$. Then,
$$
\int_0^\infty\int_0^\infty \exp(-\delta (x^2+y^2))\sin (x^2+y^2)\,\mathrm dx\,\mathrm dy=
\int_{0}^{\pi/2}\int_0^\infty r\exp(-\delta r^2)\sin (r^2)\,\mathrm dr\,\mathrm d\theta=$$
$$=\frac{\pi}{2} \int_0^\infty r\exp(-\delta r^2)\sin (r^2)\,\mathrm dr
$$
Substituting $r^2=u$, $r \mathrm dr = \frac{1}{2}\mathrm du$, the integral becomes
$$\frac{\pi}{4} \int_0^\infty \exp(-\delta u)\sin (u)\,\mathrm du= \frac{\pi}{4(\delta^2+1)}$$
and we see that when $\delta \to 0$ the integral converges to $\frac{\pi}{4}$. |
H: What's the difference between a set and a language?
I'm very familiar with languages such as $L$={$a^n$ where $n$ is a prime number}. I know this language is decidable. However, I ran across a similar problem that asked if the set $X$ = {$p$ where $p$ is a prime number} is decidable or not. It probably is, but what exactly are they talking about when they say "set". How is it different from a language?
Thanks all.
AI: A "set" is usually understood to be a collection of things that all share some common property, such as the set of all red fruits, the set of all animals that have kidneys, or the set of all numbers that are prime. A language is a set of strings, nothing more or less.
A number can be written as a numeral in any of several natural ways. For example, the number 135 might be written as 135, or in base 2 as 10000111, or in "unary" as 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111.
Such a string can be written onto the input tape of a Turing machine, and then you can run the machine and see if the machine accepts the input.
You can then ask the question of whether a particular set of numbers is decidable by asking the analogous question about the language consisting of the strings that represent those numbers. For example, is there a Turing machine that accepts the strings 0, 2, 4, …, 1788, … and rejects the strings 1, 3, 5, …, 1789, …? Yes, there is, and so we say that the set of even numbers is decidable.
In questions of decidability it does not matter whether we represent numbers with base-10 numerals, or base-2 numerals, or "unary" numerals, since a Turing machine can easily convert between these representations. So we allow ourself to forget about the strings that represent the numbers, and just focus on the numbers themselves. The language consisting of strings that represent even numbers is decidable, for any reasonable way of representing numbers as strings, so we just say that the set of even numbers is decidable, or we say that the property of evenness is decidable.
Your question is asking about the set of prime numbers. Is there a Turing machine which will accept the strings 2, 3, 5, 7, 11, 13, …, 1013,… and reject the others? |
H: Does $x^2 \equiv 211\pmod{ 159}$ have a solution?
Note that 159=3*53. The answer to this question is yes. I managed to find two of the solutions. They are $x=23,136$ but there are two more. The main question that I have is whether there is an easier way to find all solutions of a quadratic congruence when the modulus is not prime. Is it always the case that there will be four solutions when the modulus is composite? By solutions I mean the least residues. Thank you!
AI: If you know how to find the solutions when the modulus is prime, then, when the modulus is composite, solve the congruence modulo the primes, and use the Chinese Remainder Theorem to put them together to get solutions modulo the composites.
The number of solutions is (give or take a few exceptional circumstances) $2^r$, where $r$ is the number of distinct primes dividing the modulus. In your case, there are 2 solutions modulo 3, and 2 solutions modulo 53, yielding 4 solutions modulo 159. |
H: Cardinal exponentiation problem from Halmos' Naive Set Theory
In chapter 24 of Halmos' Naive Set Theory the following problem is posed as an exercise (page 96):
Prove that if $a, b$ and $c$ are cardinal numbers such that ${a}\le{b}$, then $a^c\le{b^c}$. Prove that if $a$ and $b$ are finite, greater than $1,$ and if $c$ is infinite, then $a^c=b^c$.
My problem lies in proving the second proposition. The identity ${a}\cdot{a}=a$ for infinite cardinal numbers is not covered yet at that point of the text so that the direct proof ${2}\le{a}\le{2^c}=>{2^c}\le{a^c}\le{2^{{c}\cdot{c}}=2^c}$ is not available. Doing a bit of research, wikipedia points to the use of the axiom of choice on its cardinal number article. I let A, B, and C be sets such that card(A)=a, card(B)=b, card(C)=c, and $a\le{b}$. I then tried to apply Zorn's lemma to the set of functions $f$ such that: ${domf}\subset{B^C}$, ${ranf}\subset{A^C}$, and $f$ is one-to-one; which is partially ordered by inclusion. This yields a maximal element say $\phi$. That's where I am stuck.
AI: If $a=b$, there’s nothing to prove, so assume that $a<b$. From the first part you know that $a^c\le b^c$. Since $a$ is finite and greater than $1$, and $b$ is finite, there is a positive integer $n$ such that $a^n\ge b$, and it follows from the first part that $b^c\le (a^n)^c$. If you can show that $a^c=(a^n)^c$, the result will follow from the Schröder-Bernstein theorem.
To show this, show that $nc=c$ for finite, non-zero $n$ and infinite $c$, and show that in general $(x^y)^z=x^{yz}$. I don’t own a copy of Halmos, but judging from what I can see at Google books, the latter has been stated but not proved. You should go ahead and prove it; there’s a very natural bijection between $\left(X^Y\right)^Z$ and $X^{Y\times Z}$. The former seems not to have been stated yet, but it can be proved by induction on $n$. The only hard step is showing that $2c=c$; after that the induction step just uses that same result. Just after the problem on which you’re working, Halmos proves that $c+c=c$, and there’s a natural bijection between $c+c$ and $2c$, so everything that you need is there. |
H: $H$ is normal whenever $Ha\not = Hb \implies aH\not =bH$
Topics in Algebra- Hernstein pg-47,Q.9
If $H$ be a subgroup of a group $G$ such that $Ha \not=Hb$ implies that $aH\not=bH$. Then how can I show that $gHg^{-1}\subset H$ $\forall$ $g\in G$?
This is what I have done:
For any $h\in H$,$(ghg^{-1})(gh^{-1}g^{-1})=e$ where $e$ is the identity element of the group $G$.
And for $h,k\in H$, $(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$ which is in $gHg^{-1}$ as $hk\in H$.
Thus $gHg^{-1}$ is a group.
AI: Using @user29743 hint, let $a = bh$. Then $h = b^{-1}a \in H$ means $ba^{-1} = bh^{-1}b^{-1} \in H$, and since $H$ is a subgroup, $(bh^{-1}b^{-1})^{-1} = bhb^{-1} \in H$. Replacing $b$ as $g$ we have: $ ghg^{-1} \in H$.
It's actually the case that $gHg^{-1} = H$ for every $g$, since if $gHg^{-1} \subset H$, by letting $y = g^{-1}$, we get $yHy^{-1} \subset H \subset y^{-1}Hy =gHg^{-1}$. |
H: Proving $(n+1)^n > 2^n n!$
If $n>1$ is a natural number, how to prove that
$$(n+1)^n > 2^n n!$$
AI: By the AM GM inequality, $$\eqalign{
& \root n \of {n!} = \root n \of {1 \cdot 2 \cdot 3 \cdot 4 \cdots n} \leq \frac{{1 + 2 + 3 + \cdots + n}}{n} \cr
& \frac{{\left( {n + 1} \right)n}}{{2n}} = \frac{{n + 1}}{2} \cr} $$
Then $$n! \leq {\left( {\frac{{n + 1}}{2}} \right)^n}$$ as desired. |
H: Recursively enumerable set? Any hints?
I don't mean to be pulling answers out of you, but I'm stuck. Any advice on the right direction would be appreciated.
I have the following set
$X$ ={$n$ where $n$ is a number of a turing machine $M$ that does not halt when given $n$ as input}
My gut instinct is that it's not. And that's because the question asks about the set of all x's that are not partially decidable. Recursively enumerable languages ARE partially decidable, so it can't be REL.
Is this correct? And is this sufficient reasoning?
Thanks.
AI: If $X$ is r.e., there is a Turing machine $T$ such that $T$ halts on input $n$ iff $T_n$ does not halt on input $n$. Say $T=T_m$. Then $T_m$ halts on input $m$ iff $T_m$ does not halt on input $m$. Thus, $X$ cannot be r.e. |
H: Artinian local algebras over a complete local noetherian ring.
So let $\Lambda$ be a complete local noetherian ring.
The author claims that $\Lambda[t]/(t^i)$ is an Artinian local $\Lambda$-algebra with the same residue field as $\Lambda$.
I don't see this. For example, of $\Lambda$ is a complete DVR with uniformizer $\pi$, and $i = 2$, then the descending sequence $(\pi,t)^n$ doesn't seem to stabilize. The first few terms are:
$(\pi,t) \supsetneq (\pi^2,\pi t) \supsetneq (\pi^3,\pi^2t)\supsetneq\cdots$
which obviously doesn't stabilize...or am I missing something?
This is from the very beginning of chapter 2.1 in:
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.132.2930
thanks
AI: Note that $\Lambda[t]/(t^i)$ is a free $\Lambda$ module spanned by $1,\ldots,t^{i-1}$, so it is not Artinian unless $\Lambda$ is. Thus the example in your reference
is incorrect. In the notation of your reference, this ring is in $\widehat{\mathcal C}_{\Lambda}$ rather than $\mathcal C_{\Lambda}$, unless itself
$\Lambda$ is Artinian.
Presumably this doesn't affect anything that follows, and you may do well to bear in mind that the author (now an accomplished and highly respected expert in arithmetic geometry) wrote this as part of his master's thesis, and he explicitly states in the link you give that he hasn't tried to correct it since he first wrote it (and warns that it contains "foolish mistakes").
Incidentally,
there are many additional places that you can read deformation theoretic ideas, and learn the roles of the categories $\mathcal C_{\Lambda}$ and $\widehat{\mathcal C}_{\Lambda}$. |
H: How to show $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$?
From Berkeley Problems in Mathematics, Spring 1999, Problem 17.
Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\ge 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Give that $[L:\mathbb{Q}]=n!$, prove that $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$.
(BELOW IS NONSENSE, SHOULD BE IGNORED)
Following Gerry Myerson's advice, since $L$ is the splitting field of $f$, $L$ must be Galois over $\mathbb{Q}$. Thus the Galois group for $[L:\mathbb{Q}]=S_{n}$. Gerry now asserts that either $\mathbb{Q}(\alpha^{4})$ is of degree 2 over rationals or it must be $\mathbb{Q}$. But assuming this one can rule out the case $\mathbb{Q}$, since this would imply $\alpha$ is the root of a 4th or 2nd degree polynomial over $\mathbb{Q}$. But we know $n\ge 3$. So $\alpha$ must be the root of a 4th degree polynomial. We know $S_{n}$ has normal subgroups $A_{n}$ when $n\not=4$ and $V,A_{4}$ when $n=4$. I do not know how to proceed any further.
Now assuming $\mathbb{Q}(\alpha^{4})$ is a degree 2 abelian extension over $\mathbb{Q}$. We should have a chain of normal extensions $\mathbb{Q}(\alpha)\supset \mathbb{Q}(\alpha^{2})\supset \mathbb{Q}$. This would imply $S_{n}$ has a normal subgroup which has a normal subgroup. But we know $A_{n}$ for $n\ge 5$ are simple. So the only possibility is $\mathbb{Q}(\alpha)$ has a Galois group isomorphic to $V$. In this case $n=4$ as well. I can only proceed to here.
AI: Look at an example. Let $f(x)=x^3-2$. You should be able to work out the splitting field, and see that it has degree 6 over the rationals. If you have any problem doing this, come back and let us know where you get stuck.
EDIT: much of what I wrote a few hours ago was not right. Let me try again.
$L$ is normal over the rationals, over $K={\bf Q}(\alpha^4)$, and over $E={\bf Q}(\alpha)$. The group of $L$ over the rationals is $S_n$, the group of $L$ over $E$ is $S_{n-1}$, so the group of $L$ over $K$ is a subgroup $H$ of $S_n$ containing $S_{n-1}$. We're trying to prove that $H=S_{n-1}$.
We can rule out $H=S_n$, as follows. If $H=S_n$, then $\alpha^4=q$ is rational, and the minimal polynomial for $\alpha$ over the rationals is $x^4-q$ (or some factor of that polynomial), and we're not talking about a polynomial with Galois group $S_n$.
So all we have to show is that there is no proper subgroup of $S_n$ properly containing $S_{n-1}$, and we're done. And, hey, there's a proof on m.se, so we're done.
I think what follows can safely be ignored.
Back to the ${\bf Q}(\alpha^4)$ question: if ${\bf Q}(\alpha^4)$ is a proper subfield of ${\bf Q}(\alpha)$, then it's either the rationals or degree 2 over the rationals. It's easy to rule out the first case (details left to you - I have to teach a class in a few minutes). In the second case, it's normal over the rationals, and ${\bf Q}(\alpha)$ is normal over it, and that says something about normal subgroups of the Galois group of $L$ which don't fit with that group being the symmetric group, $S_n$ (that group doesn't have a lot of normal subgroups).
I know I've left a lot out. I hope it's of some use. But if you haven't done the Galois Theory, my answer won't do much for you. |
H: Prove $\sum_{n=2}^{k} \frac{1}{n}= \sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+k-1}\right)$
The question is how to prove this equality
$$\sum_{n=2}^{k} \frac{1}{n}= \sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+k-1}\right)$$
I wasn't sure how to start proving this.
AI: Write out some terms of the righthand side:
$$\begin{align*}&\left(\frac12-\frac1{k+1}\right)+\left(\frac13-\frac1{k+2}\right)+\ldots+\left(\frac1k-\frac1{2k-1}\right)\\
&+\left(\frac1{k+1}-\frac1{2k}\right)+\left(\frac1{k+2}-\frac1{2k+1}\right)+\ldots
\end{align*}$$
Notice that all of the negative terms are cancelled out by positive terms later in the series: $-\frac1{k+1}$ in the first term by $\frac1{k+1}$ in the $k$-th term, $-\frac1{k+2}$ in the second term by $\frac1{k+2}$ in the $(k+1)$-st term, and so on. The only terms that are left uncancelled are the positive parts of the terms in the top line above:
$$\frac12+\frac13+\ldots+\frac1k\;.$$
This is precisely the sum on the lefthand side.
Added: The sketch above is informal and ignores the issue of convergence of the infinite series on the righthand side of the identity. For $m\ge 2$ let $$s_m=\sum_{n=2}^m\left(\frac1n-\frac1{n+k-1}\right)\;.$$ For $m\ge k$ we can carry out the cancellation above to write
$$\begin{align*}s_m&=\sum_{n=2}^k\frac1n-\sum_{n=m-k+2}^m\frac1{n+k-1}\\
&=\sum_{n=2}^k\frac1n-\sum_{n=m+1}^{m+k-1}\frac1n\;.
\end{align*}$$
Now $$0\le\sum_{n=m+1}^{m+k-1}\frac1n\le\sum_{n=m+1}^{m+k-1}\frac1{m+1}=\frac{k-1}{m+1}\to 0\text{ as }m\to\infty\;,$$ so $$\lim_{m\to\infty}s_m=\sum_{n=2}^k\frac1n\;,$$ exactly as we expected from the informal argument. |
H: What does the $A(m)$ denote in Engelking's book?
Recently, I'm reading Engelking's book. In this book, it always uses as the example the space $A(m)$, see for instance the example 1.1.8, page 15.
For this space, I'm not very clear, for the description is complex to me. Sometimes I view it as a sequence with a limit point when $m=\omega$. But I am not sure I'm completely right. So, could anybody knowing this example help me to describe it more clearly? Thanks ahead.
AI: Let $D$ be the discrete space of cardinality $\kappa$, and let $x_0$ be a point not in $D$. Let $A=D\cup\{x_0\}$ be the one-point compactification of $D$, with $x_0$ as the added point. The compact subsets of $D$ are precisely the finite sets, so open nbhds of $x_0$ in $A$ have the form $\{x_0\}\cup(D\setminus F)$ for finite $F\subseteq D$. Let $\mathcal{O}$ be the topology on $A$; what, exactly, is $\mathcal{O}$?
Let $S\subseteq A$. If $x_0\notin S$, then $S\subseteq D$, so $S\in\mathcal O$ iff $S$ is open in $D$. But $D$ is discrete, so $S$ is open in $D$ $-$ all subsets of $D$ are open in $D$ $-$ and therefore $S\in\mathcal{O}$. If $x_0\in S$, then $S$ must contain an open nbhd of $x_0$, so there must be a finite $F\subseteq D$ such that $\{x_0\}\cup(D\setminus F)\subseteq S$. But then $$A\setminus S\subseteq A\setminus\Big(\{x_0\}\cup(D\setminus F)\Big)=D\setminus(D\setminus F)=F\;,$$ so $A\setminus S$ is finite. In other words, $S\in\mathcal O$ iff either $x_0\notin S$, or $x_0\in S$ and $A\setminus S$ is finite. Finally, this is the same as saying that $S\in\mathcal O$ iff $x_0\notin S$, or $A\setminus S$ is finite, which is exactly the definition of the topology $\mathcal O$ given in Example 1.1.8. Thus, $A(\kappa)$ is, as azarel said, the one-point compactification of the discrete space of cardinality $\kappa$.
$A(\omega)$ is indeed homeomorphic to $\omega+1$, a simple sequence with a limit point. If $\kappa>\omega$, it’s similar, but it’s not a sequence: it’s a set of $\kappa$ isolated points and one limit point, $x_0$, with the property that every open nbhd of $x_0$ contains all but finitely many of the isolated points. This means, among other things, that if $S$ is an infinite set of isolated points, then $x_0\in\operatorname{cl}S$: every open nbhd of $x_0$ must contain all but finitely many points of $S$. It also means that $A(\kappa)$ is compact: once you’ve covered the point $x_0$, there are at most finitely many points left to be covered. (Of course this is also clear from the fact that $A(\kappa)$ is a one-point compactification!) |
H: When collapsing a cardinal, what ordinal does it become?
Work in $V$. Let $P = \text{Col}(\omega, \omega_1)$ and suppose that $G$ is generic for $P$ over $V$. Then $V[G]\models |\omega_1^V|=\aleph_0$ and $\omega_2^V=\aleph_1$. In particular, $V[G]\models\omega_1^V$ is an ordinal between $\omega$ and $\omega_1^{V[G]}$. What ordinal is it? I would guess that the answer depends on $G$, and the best that we can say is that it is a limit ordinal.
AI: I think that you're missing the point of the Levy collapse.
Forcing [over transitive models] does not add ordinals. It does not remove ordinals either. What the collapse does is to add a bijection between $\omega$ and $\omega_1$.
Note that all the "definable" ordinals ($\varepsilon_0$, etc.) and so are very small, and $\omega_1^V$ is far far beyond them.
And to your comment, yes. In fact this is what we would call it: $\omega_1^V$. If $\alpha$ is a countable ordinal and we know that there is an inner model $M$ in which $\alpha=\omega_1^M$ then we immediately know two things:
$\alpha$ is quite a large countable ordinal; and
$\omega_1^M$ is an excellent way to name it.
In the case of forcing we actually start with the inner model.
For the comment:
Definability is a strong word. If the ground model was $L$, certainly $\omega_1^L$ is a definable ordinal. It is the least ordinal that there is no bijection between him and $\omega$ which satisfies the constructibility axiom. Furthermore, we now know that the ground model is definable with parameters. This means that $\omega_1^V$ is definable from parameters in $V[G]$ by the same trick.
Note that ordinal arithmetics are not changed by forcing. This means that $\omega_1^V$ is an $\varepsilon$ number in $V[G]$ since $\omega^{\omega_1}=\omega_1$ in $V$; furthermore it is a fixed point of $\varepsilon$ numbers, for the same reasons. Namely if $\alpha=\omega_1^V$ then $\alpha=\varepsilon_\alpha$, which in $V$ is the $\omega_1$-th and in $V[G]$ is not.
Since $\omega_1^V$ is an $\varepsilon$ number its Cantor normal form is in fact $\omega_1^V$, so there is no simpler way of writing it. |
H: A tricky but silly doubt regarding the solutions of $x^2/(y-1)^2=1$
Motivation :
I have been confused with some degree 2 equation. I suddenly came across a simple equation and couldn't get the quintessence behind that.
I have an equation $$\dfrac{x^2}{(y-1)^2}=1 \tag{1}$$ and I was looking for its solutions. It was asked by some kid ( of $9$nth standard ) to me.
I did some manipulation and got $$x^2=(y-1)^2 \tag{2} $$ finally. And one can see that $(0,1)$ satisfies the Equation $[2]$ well. But I was happy, and within small time,
I realized that the same solution set can't satisfy the equation $[1]$ . If you substitute $(0,1)$ in $[1]$ you get $\dfrac{0}{0}=1$ which is wrong.
The answer that convinced me finally :
We can see the same equation as this $x^2. \dfrac{1}{(y-1)^2}=1$ . We know that the set of integers form a ring. So the product of two numbers is one if one number is the inverse of other number. '$1$' present on the R.H.S is the identity element. So the product of the entity with its inverse always gives us the identity.
So when $x$ is $0$, the $0$ doesn't have an inverse in the integers. So the case is to be emitted.
Still persisting questions :
But the thing that makes me surprise is that the Wolfram Alpha gives me this solution .
In the picture you can clearly see that they both intersect at $(0,1)$ . But what is that confusion ?
We omitted that solution, but in fact $(0,1)$ is the intersection of the two lines.
Questions that are to be answered by learned people :
What is the value of term $\dfrac{0}{0}$ ? Isn't it $1$ ?
Why the solution pair $(0,1)$ satisfies $x^2=(y-1)^2$ but not $\dfrac{x^2}{(y-1)^2}=1$ ? We know that both of them are manifestations of each other in a simple manner.
If we need to omit that solution, why do the lines intersect at $(0,1)$ ?
Thank you everyone for giving your time.
AI: The equations $$x^2=(y-1)^2\tag{1}$$ and $$\frac{x^2}{(y-1)^2}=1\tag{2}$$ do not have the same solution set. Every solution of $(2)$ is a solution of $(1)$, but $\langle 0,1\rangle$ is a solution of $(1)$ that is not a solution of $(2)$, because $\frac00$ is undefined.
The reason is that $(1)$ does not imply $(2)$. Note first that $(2)$ does imply $(1)$, because you can multiply both sides of $(2)$ by $(y-1)^2$ to get $(1)$. In order to derive $(2)$ from $(1)$, however, you must divide both sides of $(1)$ by $(y-1)^2$, and this is permissible if and only if $(y-1)^2\ne 0$. Thus, $(1)$ and $(2)$ are equivalent if and only if $(y-1)^2\ne 0$. As long as $(y-1)^2\ne 0$, $(1)$ and $(2)$ have exactly the same solutions, but a solution of $(1)$ with $(y-1)^2=0$ need not be (and in fact isn’t) a solution of $(2)$.
As far as the graphs go, the solution of $(1)$ is the union of the straight lines $y=x+1$ and $y=-x+1$. The solution of $(2)$ consists of every point on these two straight lines except their point of intersection. |
H: Finding the minimum number of selections .
Lets assume I have three boxes that contains only apples, only oranges and a mix of apples and oranges respectively and all of them are mislabelled . What is the minimum number of selections that I need to make so that I can be certain which box contains what fruits assuming I can pick one fruit at a time from any of the boxes ?
P.S: It is not a homework task . So don't put that tag in here . I was asked this in an interview. I was looking for a interview tag but couldn't find one .
AI: You can do it by picking one fruit from the box labelled Mixed.
Since the box labelled Mixed is mislabelled, it actually contains either all apples or all oranges, and you’ll know which as soon as you pick one fruit from it. Suppose that you pick an apple. (The other case is similar.) Then one of the remaining boxes contains only oranges, and the other is mixed. They bear the labels Apples and Oranges, and both are mislabelled, so the one labelled Oranges must contain mixed fruit, and the one labelled Apples must contain oranges. |
H: Computable function and decidable sets: For a computable $g$ and decidable set $A$ , Does $g(A), g^{-1}(A)$ necessarily decidable?
I'm trying to solve the following exercise from an old exam:
For a computable function $G: \mathbb{N} \to \mathbb{N} $, and a set of numbers $A$, we define $g^{-1}(A)=\{x|g(x) \in A\} $ and $g(A)=\{g(x)| x\in A\}$. If $A$ is a decidable group so..
here there were 4 options and I chose $g(A)$ is not necessarily decidable, and $g^{-1}$ is certainly decidable.
I chose this option since we are not sure that $g$ is a total computable function, and it might not stop on one of $x \in A$, but why does $g^{-1}(A)=\{x|g(x) \in A\} $ have to be decidable? we need to use $g$ to find out that $g(x) \in A$, no?
AI: Your reasoning works out for $g(A)$, but $g^{-1}(A)$ needn't be decidable. Just take $A=\mathbb{N}$: then in particular $g^{-1}(A)$ decidable means the domain of every partial computable function is decidable. But the recursively enumerable sets are exactly the domains of computable functions, so we'd have every r.e. set recursive.
Thanks for sharing the specific choices that were offered you in your comment below. None of them hold for an arbitrary computable function (though see below for a total one,) and we can show this without taking $A$ to be anything fancier than $\mathbb{N}$ itself. First, $g(\mathbb{N})$ may be decidable, for instance, if it's finite. $g^{-1}(\mathbb{N})$ may be decidable. For instance, if $g$ is total, $g^{-1}(\mathbb{N})=\mathbb{N}$.
$g(\mathbb{N})$ may be undecidable: take $g$ to be identity on the diagonal set $D=\{n: f_n(n) \textrm{is defined} \}$, where $f_n$ is some Godel numbering of the computable functions, and $0$ elsewhere. $g^{-1}(\mathbb{N})$ may be undecidable: take $g$ again to be identity on $D$, but now undefined elsewhere. It's immediate that these last two sets are undecidable by the diagonalization argument that you've likely seen at some point-I'll elaborate if not.
As we've agreed below, the questioner must have intended $g$ to be total. Then we see something like the first diagonal function defined above has $g(\mathbb{N})$ undecidable, but $g^{-1}(\mathbb{N})$ is decidable for decidable $A$ by simply computing $g(n)$ and seeing whether it's in $A$. |
H: Is this a Polygon?
According to the definition of Polygon, If a Poly-line's first and last points are connected then it is called Polygon. See the image below. I have P1, .... P5 Polyline. If I draw a line from P5 to P1 then it will be called polygon or not? I am confused here.
AI: I think conventionally you would expect the area of a polygon to be non-zero.
Generally your lecturer/teacher should specify in lectures the exact definition of things, and you should use that to answer his/her questions.
If you find loopholes you should ask them about it.
According to the link for Wikipedia you supplied:
In geometry a polygon is a flat shape consisting of straight lines
that are joined to form a closed chain or circuit.
Digging a little deeper you can find this at http://en.wikipedia.org/wiki/Closed_polygonal_chain:
In some cases it is important to draw a distinction between a
polygonal area and a polygonal chain. |
H: Lower bound of $J=\frac{x^TAx}{x^TBx}$
Consider two symmetric positive semi-definite matrices $A, B \in \mathbb{R}^{n\times n}$.
Suppose that $A$ and $B$ have the same null space $\mathcal{N}\subset \mathbb{R}^n$.
Now consider the objective function
$$J=\frac{x^TAx}{x^TBx}$$
where $x\in\mathbb{R}^n$ is an arbitrary unit-norm vector with $x^Tx=1$ and $x\notin \mathcal{N}$.
Although $x^TAx\ne 0$ since $x\notin \mathcal{N}$, it is possible $x^TAx\rightarrow 0$ when $x$ is very close to $\mathcal{N}$. So neither $x^TAx$ nor $x^TBx$ has lower bound.
But under what condition does the objective function $J$ have a positive lower bound? I mean what special properties (e.g., eigenvalues) should $A$ and $B$ have to make $J$ have a positive lower bound?
AI: I am not exactly sure what you are looking for, as $J(x) \geq 0$ for all $x$ such that $J$ is defined. In fact, both $x^TAx \geq 0$ and $x^TBx \geq 0$, so both of these have lower bounds as well.
However, if $x \notin \mathcal{N}$ you can get a better lower bound:
We have $\mathcal{N} = \ker A = \ker B$. Let $\underline{\lambda}_A$ be the smallest non-zero eigenvalue of $A$. Then if $x = x_1 + x_2$, where $x_1 \in \mathcal{N}$, and $x_2 \in \mathcal{N}^\bot$, then we have
$$x^TAx = x_2^T A x_2 \geq \underline{\lambda}_A \| x_2 \|^2, \ \ \ x^TBx = x_2^TBx_2 \leq \|B\| \|x_2\|^2.$$
So, if $x \notin \mathcal{N}$, then $x_2 \neq 0$, which gives the estimate
$J(x) = \frac{x_2^T A x_2}{x_2^TBx_2} \geq \frac{\underline{\lambda}_A}{\|B\|} > 0.$ |
H: Evaluating $\lim\limits_{n\to\infty} \frac1{n^3}\sum\limits_{\ell=1}^{n-1}\sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)}$
Along the way to proving a solution for this stubborn question of mine, I've come upon this expression which I would like to evaluate:
$$
\lim_{n\to\infty} \frac{1}{n^3}\sum_{\ell=1}^{n-1}\sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)}
$$
Assuming consistency+correctness of the rest of my work, I would love for it to turn out that the limit is $\frac{2}{3}$, but to be honest I'm not certain of how to continue. I see no reason for there to be a nice closed form for the sum (and W|A appear to agree).
AI: For every $1\leqslant\ell\leqslant n-1$,
$$
n^2-\ell^2\leqslant\sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)}\leqslant n^2-(\ell-1)^2,
$$
hence the sums $S_n$ you are interested in are such that $R_n\leqslant S_n\leqslant T_n$ for every $n\geqslant1$, with
$$
R_n=\frac1{n^3}\sum_{\ell=1}^{n-1}(n^2-\ell^2),\qquad T_n=\frac1{n^3}\sum_{\ell=0}^{n-2}(n^2-\ell^2).
$$
The rest should be easy (and the limit is indeed $\frac23$). |
H: Want to learn differential geometry and Riemannian geometry by myself
I want to follow a Ph.D. programme on geometric analysis. The main focus are Monge-Ampere equations and convex level set of $p$-harmonic equations. The theory of PDEs is very familar to me. However, I have little knowledge of the geometry.
Therefore, I want some readable texts on this issue. Since (linear) functional analysis and nonlinear analysis are my stong points, I want the texts to describe the geometry in terms of analysis.
AI: Thierry Aubin, A Course in Differential Geometry would be suitable for your goals. It is self-contained (includes the necessary background from differential topology and reaches the Riemannian geometry), and features a chapter with an introduction to Yamabe problem (one of the central topics in the geometric PDEs). A delicious part of this book is a lot of solved exercises that make your acquaintance with the subject much faster. |
H: Where is the symmetric group hidden in the Yoneda lemma?
In extension to the question
Yoneda-Lemma as generalization of Cayley`s theorem?,
can someone point out to me where, in the categorical notation and analyzation of the Cayley's theorem, the symmetrc group of transformations of the group elements is?
Put another way: I see that the group structure is found under the set of all the transformation of the group elements and so the group is isomorphic to a subgroup of the symmetric group. But the symmetric group is a very big object (e.g. for $\mathbb{Z_3}$ with its 3 elements, $S_3$ has 3!=6 elements) and I don't find it in the Yoneda lemma.
Put another way: Cayley's theorem says something about a subset, basically that $G$ is isomorphic to some transformation on $G$, let's name it $\lambda(G)$, and the insight is $\lambda(G)\le S(G)$. As the Yoneda lemma only has an equal (or isomorphic) sign, I wonder where the $\le$ symbol is in the categorical language.
I can only assume that the lemma $F(A)=\text{nat}(\text{hom}(A,-),F)$ translates to $G=\lambda(G)$ if one plugs in the hom functor for $F$ and the resulution to my question would then be to see what the $S(G)$ is in terms of $\text{nat}(\text{hom}(A,-),F)$ and why. (I don't get much out of it as there seems to be only the one object $A$ I can play around. And I don't conceptualize natural transformations very good, I'm afraid.)
AI: $S(G)$ is the set of all bijections in $\mathcal{Set}$ from $|G|$ to itself, where I denote by $|G|$ the underlying set of $G$. This is exactly the permutation group on $|G|$.
Now, the application of Yoneda that gets us Cayley is, as you say, taking $F=\hom_G(B,-),$ where $\hat{G}$ denotes the one-object category with $G$ as its arrows. Call its only object $X$. So, for every $A,B,$ we get $F(A)=\hom(B,A)=\textrm{nat} (\hom(A,-),\hom(B,-))$. But the only candidate for $A$ or $B$ is $X$, so all we really have is $\hom(X,X)=\textrm{nat} (\hom(X,-),\hom(X,-))$. Now by the construction of $\hat{G}, \hom(X,X)=G,$ so now we just
want to see why the natural transformations from $\hom(X,-)$ to itself are contained in the bijections on $|G|$.
First, the objects of the image of $\hom(X,-)$ are just $\hom(X,X)=|G|.$ Let's interpret the images of morphisms in $\hat{G}$, which are the elements of $g,$ by the right action:
$\hom(X,g): h \in |G| \mapsto hg$. Now a natural transformation $\alpha$ needs a component morphism at each object in the image of the functor, but since our functors have singleton images let's identify $\alpha$ with $\alpha_{|G|}$. The naturality $\alpha$ needs is given by this diagram:
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{ccc}
|G|&\ra{\alpha}&|G|\\
\da{g}&&\da{g}\\
|G|&\ra{\alpha}&|G|\\
\end{array}
$$
That is, we need $\alpha(hg)=(\alpha(h))g$ for each $h,g$. We can obviously accomplish this for a set of $\alpha$ isomorphic to $G$ by letting each $k$ in $G$ act on the left. Note that any natural $\alpha$ will have to be a bijection on $|G|,$ in short because the right action of $G$ on itself is transitive. So we see that the admissible bijections are some subset of $S(|G|)$; by Yoneda, they're exactly $G$, so that $G \leq S(|G|)$. |
H: Relationship between cross product and outer product
Since inner products $(V)$ are generalisations of dot products $(\mathbb{R}^n),$ then are outer products $(V)$ also related to cross products $(\mathbb{R}^3)$ in some way?
A quick search reveals that they are—yet the outer product of two column vectors in $ \mathbb{R}^3$ is a $3\times3$ matrix, not another column vector. What is the connection?
AI: Cross product is much more related to exterior product which is in fact a far going generalization.
Outer product is a matricial description of tensor product of two vectors. |
H: Calculating the norms of a triangle based pyramid
Hi I have the following co-ordinates, which make up my triangle based pyramid. I need to calculate the normals of each face. However Im struggling to find the best simplest way to do this?
-0.5, 0, 0.5,
0, 0, -0.5,
0.5, 0, 0.5,
0, 0, -0.5,
0.5, 0, 0.5,
0, 1, 0,
-0.5, 0, 0.5,
0, 0, -0.5,
0, 1, 0,
0.5, 0, 0.5,
-0.5, 0, 0.5,
0, 1, 0
AI: Given three non aligned point of a face, say $A, B, C$, build the vectors
$$
\mathbf{u}=B-A,\\
\mathbf{v}=C-A.
$$
The vector $\mathbf{n}=\mathbf{u}\times\mathbf{v}$ is normal to the given face, you should only normalize its length.
Take
$$
A=(-1/2, 0, 1/2),\\
B=(0, 0, -1/2),\\
C=(1/2, 0, 1/2),
$$
then build
$$
\mathbf{u}=B-A=( 0, 0, -1/2)-(-1/2, 0, 1/2)=(1/2,0,-1),\\
\mathbf{v}=C-A=(1/2, 0, 1/2)-(-1/2, 0, 1/2)=(1,0,0).
$$
and
$$
\mathbf{n}=\mathbf{u}\times\mathbf{v}=\left|
\begin{matrix}
i &j &k\\
1/2 &0 &-1\\
1 &0 &0
\end{matrix}
\right|=(0,-1,0)
$$
The resulting vector is, in this particular case, already normalized. |
H: Wedge Sum of Axes Not a Manifold
Note: for this question a topological manifold is defined to be a locally Euclidean, second countable, Hausdorff space. Also, I am using the subscript $x$ and $y$ just to keep track of which copy of $\mathbb{R}$ I am calling the $x$ and $y$ axis, respectively.
Intuitively I can see why the wedge sum of the $x$ and $y$ axes are not a manifold, since they do not have a well defined dimension. I am having trouble making this idea more rigorous.
Let $\mathbb{R}_x$ denote the $x$-axis and $\mathbb{R}_y$ denote the $y$-axis. Also, let $0_x$ and $0_y$ be the zero on the $x$ and $y$ axis, respectively. The wedge sum of the $x$ and $y$ axes is defined as
$$\mathbb{R}_x\vee\mathbb{R}_y=\mathbb{R}_x\amalg\mathbb{R}_y/(0_x \sim 0_y).$$
Any open neighborhood of $0$ in $\mathbb{R}_x\vee\mathbb{R}_y$ would have to be open when intersected with both $\mathbb{R}_x$ and $\mathbb{R}_y$. This leads me to believe that the open neighborhoods of $0$ in $\mathbb{R}_x\vee\mathbb{R}_y$ are wedge sums of intervals containing $0_x$ and $0_y$ in the $x$ and $y$ axes. Then is $\mathbb{R}_x\vee\mathbb{R}_y$ not a topological manifold because it is not locally Euclidean around $0$?
AI: Yes, that's correct. To finish the proof, you need to show that a wedge sum of two open intervals is not homeomorphic to an open disc in $\mathbb{R}^n$ for any $n$. This can be done, for example, by considering what happens if you delete the origin from the wedge sum: the resulting space will have four connected components. Deleting a point from an open disc in $\mathbb{R}^n$ will leave nothing (if $n=0$), two components (if $n=1$), or one component (if $n\ge 2$), but never four components. |
H: Checking Graph Isomorphism via Generator of Automorphism Group
Let $G$ and $H$ be two connected graphs and $K$ be the disjoint union of them. Also assume that we can find generating permutation for $Aut(X)$ or any graph $X$. $G$ and $H$ are isomorphic if and only if some generator of $Aut(K)$ interchanges the two connected components. How do we prove this last statement? Please give the intuitive explanation.
Thank You.
AI: Since the automorphisms of a graph are graph isomorphisms, an automorphism on $K$ is an isomorphism from $K$ to $K$. Now, assume that some automorphism $\phi$ of $K$ is given.
If $\phi(G) = H$ and $\phi(H) = G$, then (since $\phi$ is an injective graph homomorphism by construction and surjective by assumption), we have found a graph isomorphism from $G$ to $H$ via $\phi|_G$ (and, conversely, its inverse is $\phi|_H$). On the other hand, we can reverse this construction to turn a graph isomorphism $\psi: G \to H$ into an automorphism of $K$. |
H: How to find matrix form of an operator on a vector space V which is direct sum of its two subspaces?
I am studying a lecture notes where I found this result:
Let $A$ be an operator in vector space $V$ . If $V$ is equal to direct sum of its two subspace $U$ and $W$ ie $V = U\bigoplus W$, and if the basis $\{e_1,. . . . .e_n\}$of $V$ is such that the first $k$ vectors $\{e_1,. . . . .e_k\}$ is basis of $U$ and vectors
$\{e_{k+1},...,e_n\}$ is a basis of $W$, Then the matrix of $A$ has the following form
$\{\left(
\begin{array}{cc}
B & 0 \\
0 & C \\
\end{array}
\right)\}$
It has not mentioned anything about matrix $B$ and $C$.
I am finding it difficult to understand this. Could anybody help me with this.
Thanks a ton
AI: Claim. Assume we are given decomposition $V=U\oplus W$ and a linear operator $A:V\to V$. Let $\mathbf{e}_U=\{e_1,\ldots,e_k\}$ be a basis of $U$ ($U=\mathrm{span}\{e_1,\ldots,e_k\}$), and $\mathbf{e}_W=\{e_{k+1},\ldots,e_n\}$ be a basis of $W$ ($W=\mathrm{span}\{e_{k+1},\ldots,e_n\}$). Then the following conditions are equivalent
1) $A(U)\subset U$, $A(W)\subset W$
2) matrix $[A]$ of operator $A$ in basis $\mathbf{e}_V=\{e_1,\ldots,e_n\}$ is block diagonal.
Proof. $\Longrightarrow$ Let $i\in\{1,\ldots,k\}$ then $e_i\in U$ and $A(e_i)\in A(U)\subset U$. Since $U=\mathrm{span}\{e_1,\ldots,e_k\}$, we have $A(e_i)=\sum_{j=1}^k a_{j,i} e_j$. This means that coordiante reperesentation of vector $A(e_i)$ in basis $\mathbf{e}_V$ is
$$
[A(e_i)]=\begin{pmatrix}a_{1,i}\\ a_{2,i}\\...\\a_{k,i}\\ 0\\ 0\\ \ldots\\ 0\end{pmatrix}\tag{1}
$$
Hence $n-k$ last coordinates of $A(e_i)$ in basis $\mathbf{e}_V$ are zeros.
Now let $i\in\{k+1,\ldots,n\}$ then $e_i\in W$ and $A(e_i)\in A(W)\subset W$. Since $W=\mathrm{span}\{e_{k+1},\ldots,e_n\}$, we have $A(e_i)=\sum_{j=k+1}^n a_{j,i} e_j$. This means that coordiante reperesentation of vector $A(e_i)$ in basis $\mathbf{e}_V$ is
$$
[A(e_i)]=\begin{pmatrix}\\ 0\\ \ldots\\ 0\\ 0\\a_{k+1,i}\\ a_{k+2,i}\\...\\a_{n,i}\end{pmatrix}\tag{2}
$$
Hence $k$ first coordinates of $A(e_i)$ in basis $\mathbf{e}_V$ are zeros.
Recall that matrix of operator in some basis is a collections of columns $[A(e_i)]$, so the matrix $[A]$ of operator $A$ is
$$
[A]=
\left(\begin{array}{cccc|cccc}
a_{1,1} & a_{1,2} & \ldots & a_{1,k} & 0 & 0 & \ldots & 0\\
a_{2,1} & a_{2,2} & \ldots & a_{2,k} & 0 & 0 & \ldots & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\
a_{k,1} & a_{k,2} & \ldots & a_{k,k} & 0 & 0 & \ldots & 0\\ \hline
0 & 0 & \ldots & 0 & a_{k+1,1} & a_{k+1,2} & \ldots & a_{k+1,n}\\
0 & 0 & \ldots & 0 & a_{k+2,1} & a_{k+2,2} & \ldots & a_{k+2,n}\\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots\\
0 & 0 & \ldots & 0 & a_{n,1} & a_{n,2} & \ldots & a_{n,n}
\end{array}\right)\tag{3}
$$
This is exactly means that $[A]$ is block diagonal.
$(\Longleftarrow)$ Assume that matrix $[A]$ is block diagonal, i.e. of the form $(3)$.
Let $i\in\{1,\ldots,k\}$, then coordinates of $A(e_i)$ is of the form $(1)$. This equivalent to say that $A(e_i)=\sum_{j=1}^k a_{j,i} e_j\in\mathrm{span}\{e_1,\ldots,e_k\}=U$. Take arbitrary $x\in U$, then $x=\sum_{j=1}^k x_j e_j$, then $A(x)=\sum\limits_{j=1}^k x_j A(e_i)\in\mathrm{span}\{A(e_1),\ldots,A(e_k)\}\subset\mathrm{span}\{e_1,\ldots,e_k\}=U$. Since for all $x\in U$ we have $A(x)\in U$ we conclude $A(U)\subset U$.
Let $i\in\{k+1,\ldots,n\}$, then coordinates of $A(e_i)$ is of the form $(2)$. This equivalent to say that $A(e_i)=\sum_{j=k+1}^n a_{j,i} e_j\in\mathrm{span}\{e_{k+1},\ldots,e_n\}=W$. Take arbitrary $x\in W$, then $x=\sum_{j=k+1}^n x_j e_j$, then $A(x)=\sum\limits_{j=k+1}^n x_j A(e_i)\in\mathrm{span}\{A(e_{k+1}),\ldots,A(e_n)\}\subset\mathrm{span}\{e_{k+1},\ldots,e_n\}=W$. Since for all $x\in W$ we have $A(x)\in W$ we conclude $A(W)\subset W$. |
H: Relation between Interior Product, Inner Product, Exterior Product, Outer Product
Following my previous question Relationship between cross product and outer product where I learnt that the Exterior Product generalises the Cross Product whereas the Inner Product generalises the Dot Product, I was wondering if the simple map that I have drawn below is at all an accurate representation of the links between these different products?
Vertical lines denote generalisation-specification, horizontal lines denote "in opposition to". I'm just trying to get a quick overview before I dive in. Thanks.
AI: I'm not sure what you're reading, but it certainly contains a lot of terms talked about in Clifford algebra (or "geometric algebra").
Clifford algebras were born of a synthesis of inner product spaces and Grassmann's exterior algebras, both of which have geometric applications.
A Clifford algebra is constructed from an inner product space $(V,Q)$ by generating an associative algebra (whose product is a descendant of the tensor product in the tensor algebra for $V$). These are compatible in a sense made clear in the Wiki.
Suppose we are in a real inner product space, and let's denote the algebra multiplication with $\otimes$ and the inner product with $\cdot$. One identity that holds for parallel vectors $v,w$ is $v\otimes w=v\cdot w$. Part of the reason this is true is that $v\otimes w=w\otimes v$ when $v,w$ are parallel.
When they are not parallel, then there is a skew component to $v\otimes w$. This can be retrieved explicitly by computing $\frac{1}{2}(v\otimes w-w\otimes v):=v\wedge w$. With this notation, $v\otimes w=v\cdot w+v\wedge w$ for all vectors $v,w$. This $\wedge$ is called the exterior or outer product in this algebra.
They key thing to know is that this inner and this outer product don't uniquely extend to the whole algebra (well the outer product has a slightly more natural extension than the inner product). They are not "natural" to the algebra really: they are mostly just a notational convenience with some algebraic properties that make them easy to view as products.
The cross product arises too in the discussion of Clifford algebras, but it seems to be regarded as second class to the outer product. I've seen this attributed to the cross-product somehow not carrying the correct information for physical applications.
I went and read a little and found out that "interior product" looks a lot like an important extension of the inner product in a Clifford algebra which is sometimes called the contraction.
In summary, I think these various products might become less mysterious if you read some of these linked articles and see where they are used. It's going to be especially enlightening when you find out where they are all defined (in the vector space $V$ or in an algebra containing $V$.) |
H: How I Prove The Cardinality of the Symmetric difference of $n$ sets
I try to prove that the cardinality of the the symmetric difference of $n$ sets is
$$
\sum\limits_{i}|A_i|-2\sum\limits_{i<j}|A_i\cap A_j|+4\sum\limits_{i<j<k}|A_i\cap A_j\cap A_k|-\ldots
$$
I try to use the Characteristic Function $\chi(x)$.
I know that for a Symmetric difference $A\Delta B$ the Characteristic Function is $(\chi_A-\chi_B)^2$ and the cardinality of each set is $|X|=\sum_x\chi(x)$.
but I do not know how to prove it for $n$ sets? So in that way I will have the characteristic function of each of one.
Many Thanks.
AI: As Asaf said in the comments, this is bookkeeping and induction. I’m going to write out the induction step in full, but I’m going to do it in a very compact fashion that may not be easy to follow at first. I’m doing this for two reasons. First, you really ought to try to work out the induction step on your own, so I don’t want to make it too easy to read. (If you have trouble doing it in general, I suggest writing out the argument for $n=3$ and perhaps $n=4$ first; at that point you should be able to see fairly clearly what’s going on, and the main problem will likely be expressing it clearly in the general case.) Secondly, at some point you’ll want to get used to reading this sort of compact argument.
First let’s write out the sum in full:
$$\left|\bigoplus_{k=1}^nA_k\right|=\sum_{k=1}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\left|\bigcap_{i\in S}A_i\right|\;,\tag{1}$$
where I’ve used $\oplus$ for symmetric difference, and $[n]=\{1,\dots,n\}$. To avoid having too many subscripts, let $\chi_k$ be the characteristic function of $A_k$, and let $\varphi_n$ be the characteristic function of $\bigoplus_{k=1}^nA_k$. Then $(1)$ will certainly be true if
$$\varphi_n=\sum_{k=1}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\prod_{i\in S}\chi_i\;,\tag{2}$$
since $\prod_{i\in S}\chi_i$ is the characteristic function of $\bigcap_{i\in S}A_i$.
You can prove $(2)$ by induction on $n$. You need to use the fact that for any characteristic function $\chi$, $\chi^2=\chi$. You already know that $\chi_{A\oplus B}=(\chi_A-\chi_B)^2$. Now for the induction step you have
$$\begin{align*}
\varphi_{n+1}&=\left(\varphi_n-\chi_{n+1}\right)^2\\
&=\varphi_n^2+\chi_{n+1}^2-2\varphi_n\chi_{n+1}\\
&=\varphi_n+\chi_{n+1}-2\varphi_n\chi_{n+1}\\
&=\sum_{k=1}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\prod_{i\in S}\chi_i+\chi_{n+1}-2\chi_{n+1}\sum_{k=1}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\prod_{i\in S}\chi_i\tag{3}
\end{align*}$$
for starters. The first term in $(3)$ can be split as
$$\sum_{i\in[n]}\chi_i+\sum_{k=2}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\prod_{i\in S}\chi_i\tag{4}$$
and the first term of $(4)$ can be combined with the $\chi_{n+1}$ term in $(3)$ to yield
$$\begin{align*}
\varphi_{n+1}&=\sum_{i\in[n+1]}\chi_i+\sum_{k=2}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\prod_{i\in S}\chi_i-2\chi_{n+1}\sum_{k=1}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\prod_{i\in S}\chi_i\\
&=\sum_{i\in[n+1]}\chi_i+\sum_{k=2}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\prod_{i\in S}\chi_i-2\sum_{k=1}^n(-1)^{k-1}2^{k-1}\sum_{S\subseteq[n]\atop|S|=k}\prod_{i\in S}\chi_i\chi_{n+1}\tag{5}\;.
\end{align*}$$
Now let $S$ be a subset of $[n+1]$ such that $|S|=k$. If $k=1$, then $\prod_{i\in S}$ is one of the $\chi_i$, and it appears in the first term of $(5)$ with coefficient $1=(-1)^02^0=(-1)^{k-1}2^{k-1}$.
Suppose now that $1<k\le n$. If $n+1\notin S$, $\prod_{i\in S}\chi_i$ is counted in the second term of $(5)$, where it appears with coefficient $(-1)^{k-1}2^{k-1}$. If $n+1\in S$, then $\prod_{i\in S}\chi_i$ appears in the last term of $(5)$ as $\prod_{i\in S'}\chi_i\chi_{n+1}$, where $S'=S\setminus\{n+1\}$, and $|S'|=k-1$. In this case the coefficient of $\prod_{i\in S}\chi_i$ is $-2(-1)^{k-2}2^{k-2}=(-1)^{k-1}2^{k-1}$.
Finally, suppose that $k=n+1$. Then $S=[n+1]$, and $\prod_{i\in S}\chi_i=\prod_{i\in[n]}\chi_i\chi_{n+1}$ appears in the last term of $(5)$ with coefficient $-2(-1)^{n-1}2^{n-1}=(-1)^n2^n=(-1)^{k-1}2^{k-1}$.
It follows that $(5)$ (and hence $\varphi_{n+1}$) is equal to
$$\sum_{k=1}^{n+1}(-1)^{k-1}2^{k-1}\sum_{S\subset[n+1]\atop|S|=k}\prod_{i\in S}\chi_i\;,$$
which completes the induction step.
By the way, a simpler approach is to show that $a\in\bigoplus_{k=1}^nA_k$ iff $|\{k\in[n]:a\in A_k\}|$ is odd, which is an easy induction, and then to show that that the righthand side of $(1)$ counts the elements of $\bigcup_{k=1}^nA_k$ that are in an odd number of the $A_k$. This is a fairly easy consequence of the binomial theorem. If an element $a$ is in exactly $m$ of the $n$ sets, for each $k$ from $1$ through $m$ it’s in $\binom{m}k$ $k$-fold intersections, so it contributes
$$\begin{align*}
\sum_{k=1}^m\binom{m}k(-1)^{k-1}2^{k-1}&=-\frac12\sum_{k=1}^m\binom{m}k(-2)^k\\
&=-\frac12\left(\sum_{k=0}^m\binom{m}k(-2)^k-1\right)\\
&=\frac12\Big(1-(1-2)^m\Big)\\
&=\frac{1-(-1)^m}2\\
&=\begin{cases}
1,&\text{if }m\text{ is odd}\\
0,&\text{if }m\text{ is even}
\end{cases}
\end{align*}$$
to the righthand side of $(1)$. |
H: The emptiness problem for "lunatic" and "crazy" Turing machines
Crazy Turing Machine is the same as Turing machine with one stripe , except of the fact that after each ten steps the head jumps back to the beginning of the stripe.
Lunatic Turing Machine is the same as Turing machine with one stripe , except of the fact that after each 10, 100, 1000, ... steps the head jumps back to the beginning of the stripe.
How come that for a crazy Turing machine we can decide the emptiness problem($L(M)= \emptyset$) and for the lunatic one it stays at $co-RE \setminus R $?
I didn't come with any smart conclusion.
AI: The crazy Turing machine can never get beyond the $10^{th}$ cell of the tape, and has ultimately only finitely many configurations. The complete spectrum of behavior of such a machine is therefore computable by a regular Turing machine, which can build the graph of all configurations and compute which configurations reach others and then find if there is an accepting path through these configurations. In particular, the emptiness problem for the crazy Turing machines will be decidable.
According to my understanding of the lunatic Turing machines, however, they get longer and longer periods of time during which they may perform "normal" computation, and so they can systematically make progress on deciding a semi-decidable question. Whenever the lunatic behaviro sets in, this will be detectable, and the machine can simply return to where it was and continue its calculation further. It was merely interrupted by a small hiccup. So these machines are fully as powerful as regular Turing machines, and so their emptiness problem is co-c.e. (since non-emptiness is verified by an accepting instance). |
H: Is this a norm? (triangle inequality for weighted maximum norm)
I've been trying to prove that the following is a norm, but wasn't successful. I also cannot find a counterexample. So help is greatly appreciated.
Let $x \in \mathbb{R}^N, \ w_i \in \mathbb{R}_+,\ i=1,\ldots,N$.
$$\lVert x \rVert_w := \max \lvert w_i x_i\rvert$$
Basically, this is the maximum norm with positive weights assigned to each dimension.
It must be shown that:
$$\max \lvert w_i (x_i+y_i) \rvert \leq \max \lvert w_j x_j \rvert + \max \lvert w_k y_k \rvert$$
AI: Since $|w_ix_i|\leqslant \lVert x \rVert_w$ and $|w_iy_i|\leqslant \lVert y \rVert_w$ for every $i$,
$$
|w_i(x_i+y_i)|\leqslant|w_ix_i|+|w_iy_i|\leqslant\lVert x\rVert_w+ \lVert y \rVert_w,
$$
for every $i$. This proves that $\max\limits_i|w_i(x_i+y_i)|\leqslant\lVert x\rVert_w+ \lVert y \rVert_w$, QED. |
H: Simple Limit Divergence
I am working with a definition of a divergent limit as follows:
A sequence $\{a_n\}$ diverges to $-\infty$ if, given any number $M$, there is an $N$ so that $n \ge N$ implies that $a_n \le M$.
The sequence I am considering is $a_n = -n^2$, which I thought would be pretty simple, but I keep running into a snag.
My Work:
For a given $M$, we want to show that $n \ge N \Rightarrow -n^2 \le M$. So $n^2 \ge -M$. But here is where I run into trouble, because I can't take square roots from here. What should I do?
AI: If $M>0$ inequality $n^2\geq -M$ always holds. So you can take any $N$ you want.
If $M\leq 0$, then $n\geq\sqrt{-M}$, and you can take $N=\lfloor\sqrt{-M}\rfloor +1$. |
H: What is unitary space
In https://www.encyclopediaofmath.org/index.php/Unitary_space, unitary space seems to be Hilbert space. But in http://www.answers.com/topic/unitary-space, "finite dimensional" is required. My question is, which definition of unitary space is commonly used?
AI: Answers.com is wrong. Unitary space is an archaic name for complex inner product space. |
H: An entire function as the projection in a fibre bundle
Let $f$ be a non-constant entire function. I'd like to view $f$ as the projection mapping in a fiber bundle, where the base space is $X =f(\mathbb{C})$ (which according to Picard is either $\mathbb{C}$ or $\mathbb{C} \setminus \{z_0\}$), and the bundle space is the collection $\bigcup_{x\in X}f^{-1}(x)$. In words, the fiber over each $f(z)=x\in X$ is the (discrete) set of points which $f$ maps to $x$. Is anyone familiar with a reference treating this aspect ?
AI: A pedantic remark is that it won't always be a fiber bundle (which in this case is more commonly called a covering space as the fibers will be discrete) over the image. For instance, if $f(z) = z^2$, the fiber over the origin doesn't look like the other fibers.
Anyway, you are describing the Riemann surface attached to the problem of inverting $f$. For example, if $f(z) = z^2$ you are describing the "Riemann surface of the square root function."
http://en.wikipedia.org/wiki/Riemann_surface |
H: Vector analysis can anyone clarify whether my assumptions make sense?
In this problem there are three particles, with velocities $\vec{v_1}, \vec{v_2}$ and $\vec{u}$.
Relative to the particle moving at $u$, the velocities $v_1$ and $v_2$ are of equal magnitude and are perpendicular. Accordingly, show:
$$\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{v_2})\right|^2$$
I couldn't work out whether to put this into components or not. I only got it to work for $v_1\not=v_2$
I have taken $$\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|$$
To mean $$\left(\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right)$$
Then I expand out the LHS that to get:
$$|\vec{u}|^2-\vec{u}\cdot\vec{v_1}-\vec{u}\cdot\vec{v_2}+\frac{1}{4}(\vec{v_1}+\vec{v_2})^2$$
Then I am left with something different to the RHS unless I make some certain assumptions.
AI: The relative velocities are defined as $\vec{v}_1-\vec{u}$ and $\vec{v}_2-\vec{u}$.
If we start from
$$\left|\vec{u}-\frac{1}{2}(\vec{v_1}+\vec{v_2})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{v_2})\right|^2$$
and separate $\vec{u}=\frac{1}{2}\vec{u}+\frac{1}{2}\vec{u}$ on the LHS and add $\frac{1}{2}\vec{u}-\frac{1}{2}\vec{u}=\vec{0}$ on the RHS, we have, grouping the terms opportunely,
$$\left|-\frac{1}{2}(\vec{v_1}-\vec{u})-\frac{1}{2}(\vec{v_2}-\vec{u})\right|^2=\left|\frac{1}{2}(\vec{v_1}-\vec{u})-\frac{1}{2}(\vec{v_2}-\vec{u})\right|^2$$
if we now evaluate the square of the modulus as $|\vec{a}+\vec{b}|^2=\vec{a}^2+\vec{b}^2+2\vec{a}\cdot\vec{b}$, we have
$$\frac{1}{4}(\vec{v_1}-\vec{u})^2+\frac{1}{4}(\vec{v_2}-\vec{u})^2=\frac{1}{4}(\vec{v_1}-\vec{u})^2+\frac{1}{4}(\vec{v_2}-\vec{u})^2$$
where we have taken into account $(\vec{v}_1-\vec{u})\cdot(\vec{v}_2-\vec{u})=0$, as per hypothesis. |
H: How do you calculate the derivative of an integral
How can you calculate $dy/dx$ here?
$$y=\int_{2^x}^{1}t^{1/3}dt$$
I get that the anti derivative is $3/4t^{4/3}$, but I don't understand what I'm supposed to do next.
The answer is
$$\int_x^1\sqrt{1-t^2}dt + 2$$
I have no idea how to get there
AI: What about using the Fundamental Theorem of Calculus?
$$\begin{align*}\dfrac{dy}{dx} &= \dfrac{d}{dx}\left(\int_{2^x}^1t^{1/3}dt\right)\\
&= - \dfrac{d}{dx}\left(\int^{2^x}_1t^{1/3}dt\right)\\
&= -((2^x)^{1/3})\dfrac{d}{dx}(2^x)\\
&= -2^{4x/3}\ln 2.
\end{align*}$$ |
H: Complete course of self-study
I am about $16$ years old and I have just started studying some college mathematics. I may never manage to get into a proper or good university (I do not trust fate) but I want to really study mathematics.
I request people to tell me what topics an undergraduate may/must study and the books that you highly recommend (please do not ask me to define an undergraduate).
Background:
Single variable calculus from Apostol's book Calculus;
I have started IN Herstein's topics in algebra;
I have a limited knowledge of linear algebra: I only know what a basis is, a dimension is, a bit of transpose, inverse of a matrix, determinants defined in terms of co-factors, etc., but no more;
absolutely elementary point set topology. I think open and closed balls, limit points, compactness, Bolzano-Weirstrass theorem (I may have forgotten this topology bit);
binomial coefficients, recursions, bijections;
very elementary number theory: divisibility, modular arithmetic, Fermat's little theorem, Euler's phi function, etc.
I asked a similar question (covering less ground than this one) some time back which received no answers and which I deleted. Even if I do not manage to get into a good university, I wish to self-study mathematics. I thanks all those who help me and all those who give me their valuable criticism and advice.
P.S.: Thanks all of you. Time for me to start studying.
AI: This is a recapitulation and extension of what we talked about in chat.
Whatever you do, I recommend that you try a variety of areas in order to find out what you like best. Don’t feel obliged to stick to the most common ones, either; for instance, if you find that you’ve a taste for set theory, give it a try.
My own interests are outside the undergraduate mainstream, so in mainstream areas others can probably give better recommendations. I do know that you’re working through Herstein for algebra; although it’s a little old-fashioned, it’s still a fine book, and anyone who can do the harder problems in it is doing well.
You mentioned that you’d prefer books and notes that are freely available. The revised version of Judy Roitman’s Introduction to Modern Set Theory is pretty good and is available here as a PDF. You can also get it from Barnes & Noble for $8.99. Introduction to Set Theory by Hrbacek & Jech is also good, but it’s not freely available (or at least not legitimately so).
I’ve not seen a freely available topology text that I like; in particular, I’m not fond of Morris, Topology Without Tears, though I’ve certainly seen worse. If you’re willing to spend a little and like the idea of a book that proves only the hardest results and leaves the rest to the reader, you could do a lot worse than John Greever’s Theory and Examples of Point-Set Topology. It’s out of print, but Amazon has several very inexpensive used copies. (This book was designed for use in a course taught using the so-called Moore method. It’s excellent for self-study if you have someone available to answer questions if you get stuck, but SE offers exactly that. In the interests of full disclosure I should probably mention that I first learned topology from this book when it was still mimeographed typescript.) If I were to pick a single undergraduate topology book to serve both as a text and a reference, it would probably be Topology, by James Munkres, but I don’t believe that it’s (legitimately) freely available. You might instead consider Stephen Willard, General Topology; it’s at a very slightly higher level than the Munkres, but it’s also well-written, and the Dover edition is very inexpensive.
I can’t speak to its quality, but Robert B. Ash has a first-year graduate algebra text available here; it includes solutions to the exercises, and it introduces some topics not touched by Herstein. He has some other texts available from this page; the algebra ones are more advanced graduate level texts, but the complex analysis text requires only a basic real analysis or advanced calculus course.
This page has links to quite a collection of freely available math books, including several real analysis texts; I’ve not looked at them, so I can’t make any very confident recommendations, but if nothing else there may be some useful ancillary texts there. I will say that this analysis text by Elias Zakon and the companion second volume look pretty decent at first glance. For that matter, the intermediate-level book on number theory by Leo Moser available here looks pretty good, too, apart from having very few exercises. Oh, come to think of it there is one real analysis book that I want to mention: DePree and Swartz, Introduction to Real Analysis, if only for its wonderful introduction to the gauge integral. |
H: Motion profile, the maximum velocity allowed from stop
I´m trying to calculate what the maximum allowed velocity for a mechanical axis traveling towards a stop. The maximum velocity must be a secure speed so the axis have is able to stop in time.
I have:
current position
current distance to stop
current velocity
acceleration/deceleration constant
jerk constant
I did found this formula
$$\frac{v^2}{2a}+\frac{va}{2j}=d$$
where
$a$ = acceleration,
$v$ = velocity
$j$ = jerk
$d$ = stop distance
but I don't know how to calculate the $v$ instead.
Is this the correct formula to use and in that case, how should I use it?
AI: I presume $a$ is the maximum allowed acceleration, $j$ is the maximum allowed rate of change in acceleration, and you want to arrive at the stop with zero velocity and zero acceleration. You can run it backwards to find the velocity attained when you start at the stop. Then you ramp the acceleration up to maximum from zero. This phase lasts until $t_1=\frac aj.$ During this phase, $$v_1(t)=\int_0^t a(t')dt'=\frac{jt^2}2$$
$$d_1(t)=\int_0^t v(t')dt'=\frac {jt^3}6$$ so the allowable velocity is $$v_{1a}=\sqrt[\uproot {4}\scriptstyle 3]{\frac {9d_1^2j}2}$$covering a distance $$D=\int_0^{\frac aj}v(t)dt=\left. \frac {jt^3}6 \right|_0^{\frac aj}=\frac{a^3}{6j^2}$$
at a final velocity of $$V=\frac {a^2}{2j}$$ When you are farther away, you accelerate at $a$ and the $j$ doesn't enter. You then have $$v_2(t)=V+a(t-t_1)$$ and reach $$d_2(t)=D+\int_{t_1}^t v_2(t')dt'=D+\frac 12 a(t-t_1)^2$$ so the allowable velocity is now $$v_{2a}=V+\sqrt{2a(d_2-D)}$$ |
H: An element conjugate to an odd number of elements
I have the following problem I've encountered while studying for quals that I just can't seem to tackle:
(revised, apologies - left out an important detail)
An element g of order greater than 2 such that the conjugacy class of G has an odd number of elements. Prove that g is not conjugate to $g^{-1}$.
AI: If $g$ is conjugate to $g^{-1}$, then whenever $g$ is conjugate to $h$ it is also conjugate to $h^{-1}$. If $g \ne g^{-1}$, the elements conjugate to $g$ (including, of course, $g$ itself) split up into pairs $(h, h^{-1})$. However, if $g = g^{-1}$, $g$ could be conjugate to any number of elements (which, of course, are each equal to their inverses). |
H: Is similarity surjection?
How to prove that similarity $f:\mathbb R^n \to \mathbb R^n$, i.e there is $0<r<1$ such that $d(f(x),f(y))=r \space d(x,y)$ for any $x,y\in\mathbb R^n$, is surjection? Intuitively this is the case, but i can't prove it rigorously or give any counterexample. Thanks.
AI: You can compose a similarity of $\mathbf R^n$ with scaling (a bijection) to obtain an isometry, so you can assume without loss of generality that $f$ is an isometry.
$\mathbf R^n$ as a metric space has the property that for any two points, there is a unique geodesic line connecting them: the straight line. Isometries preserve distances, so they must also preserve geodesics, and hence also straight lines, so they are affine.
Therefore, if you compose an isometry $f$ with translation by $-f(0)$ (again a bijection), you obtain a linear isometry of $\mathbf R^n$, so without loss of generality $f$ is a linear isometry.
To check that a linear isometry is bijective, you just need to check that it transforms the sequence of basis vectors to a linearly independent sequence. That's not hard to do: if you had some $\sum_{i<n}\alpha_if(e_i)=0$, then, because $f$ is a linear isometry, $\sum_{i<n}\alpha_ie_i=0$, so $\alpha_i$ are all zero, and we're done.
This also shows that any similiarity is a composition of a linear isometry, translation and scaling, which, with some more work, can be used to show that it is also a composition of several reflections and scaling. |
H: Conjugate elements of G that are in the center of a Sylow p-subgroup
I've had a lot of trouble with this question, as well as my advisor.
Let x and y be two elements of $Z(P)$, where $P$ is a Sylow $p$-subgroup of $G$. If $x$ and $y$ are conjugate in $G$, prove that $x$ is conjugate to $y$ in $N_G(P)$, the normalizer of $P$ in $G$.
AI: Hint: Use Sylow's theorem in $C_G(y)$, the centralizer of $y$ in $G$.
If $y = x^g$ and $x,y$ are in $Z(P)$, then $C_G(y)$ contains both $P$ (because $y \in Z(P)$) and $P^g$ (since $y=x^g \in Z(P^g)$). By Sylow's theorem in $C_G(y)$, $P^g = P^z$ for some $z \in C_G(y)$, so $gz^{-1} \in N_G(P)$ and $x^{gz^{-1}} = y^{z^{-1}} = y$. |
H: Question about continuous functions and indefinite integrals
My math books, in the introductory chapter of indefinite integrals (they call them primitives, and a primitive of a function is any function who's derivative is the original function) concludes the following thing:
Any continuous function on the reals admits primitives on it's domain (that means that there exist indefinite integrals).
However I can easily come up with a counter-example. The function $f(x) = x^x$ is continuous on $[0,1]$ however there exists no such function $F(x)$ such that $F'(x) = x^x$.
Did I misinterpret the conclusion?
AI: Your book deserves a bit of criticism, in my humble opinion. The best way to go in a safe way is to assume that you are always working with functions defined on an interval. Indeed, it may be troublesome to find a primitive of the continuous function $f \colon \mathbb{Z} \to \mathbb{R}$ defined by $f(n)=n$. You would need to define the primitive on an open set that contains $\mathbb{Z}$, but then you'd lose uniqueness up to a constant.
However, if $f$ is defined on some interval $I$ and you look for a function $F \colon I \to \mathbb{R}$ such that $F'=f$ everywhere in $I$, then the continuity of $F$ in $I$ is a sufficient condition for $F$ to exist. |
H: If $R$ is a ring s.t. $(R,+)$ is finitely generated and $P$ is a maximal ideal then $R/P$ is a finite field
Let $R$ be a commutative unitary ring and suppose that the abelian
group $(R,+)$ is finitely generated. Let's also $P$ be a maximal ideal
of $R$.
Then $R/P$ is a finite field.
Well, the fact that the quotient is a field is obvious. The problem is that I have to show it is a finite field. I do not know how to start: I think that we have to use some tools from the classification of modules over PID (the hypotesis about the additive group is quite strong).
I found similar questions here and here but I think my question is (much) easier, though I don't manage to prove it.
What do you think about? Have you got any suggestions?
Thanks in advance.
AI: As abelian groups, both $\,R\,,\,P\,$ are f.g. and thus the abelian group $\,R/P\,$ is f.g....but this is also a field so if it had an element of additive infinite order then it'd contain an isomorphic copy of $\,\Bbb Z\,$ and thus also of $\,\Bbb Q\,$, which of course is impossible as the last one is not a f.g. abelian group. (of course, if an abelian group is f.g. then so is any subgroup) |
H: Appropriate Notation: $\equiv$ versus $:=$
With respect to assignments/definitions, when is it appropriate to use $\equiv$ as in
$$M \equiv \max\{b_1, b_2, \dots, b_n\}$$
which I encountered in my analysis textbook as opposed to the "colon equals" sign, where this example is taken from Terence Tao's blog :
$$S(x, \alpha):= \sum_{p\le x} e(\alpha p) $$
Is it user-background dependent, or are there certain circumstances in which one is more appropriate than the other?
AI: An "equality by definition" is a directed mental operation, so it is nonsymmetric to begin with. It's only natural to express such an equality by a nonsymmetric symbol such as $:=\, .\ $ Seeing a formula like $e=\lim_{n\to\infty}\left(1+{1\over n}\right)^n$ for the first time a naive reader would look for an $e$ on the foregoing pages in the hope that it would then become immediately clear why such a formula should be true.
On the other hand, symbols like $=$, $\equiv$, $\sim$ and the like stand for symmetric relations between predeclared mathematical objects or variables. The symbol $\equiv$ is used , e.g., in elementary number theory for a "weakened" equality (equality modulo some given $m$), and in analysis for a "universally valid" equality: An "identity" like $\cos^2 x+\sin^2 x\equiv1$ is not meant to define a solution set (like $x^2-5x+6=0$); instead, it is expressing the idea of "equal for all $x$ under discussion". |
H: Euler characteristic of a quotient space
I have a question relating to an answer on MathOverflow.net. The cited answer says:
Let $X$ be a topological space for which [the Euler characteristic] $\chi(X)$ is defined and behaves in the expected way for unions, Cartesian products, and quotients by a finite free action. ... [Then] $$\chi(X^{(2)}) = \frac{\chi(X \times X) - \chi(\operatorname{Diag}(X))}{2} + \chi(X) = \frac{\chi(X)^2 + \chi(X)}{2}$$ [where $X^{(2)}$ denotes the symmetric square of $X$].
Question: Does anyone know a reference for this result, or, failing that, a short proof? For the application that I have in mind I need the result for algebraic varieties over an algebraically closed field (whose characteristic may be positive), but a more general result would be nice to see.
AI: The Cartesian product is the union of the diagonal and its complement, and a finite group $C_2$ acts freely on that complement with quotient the symmetric square minus a copy of $X$, so
$$\chi(X \times X) = \chi(\text{Diag}(X)) + 2 (\chi(X^{(2)}) - \chi(X))$$
and the conclusion follows. (This is essentially combinatorics; run through the argument for $X$ a finite discrete space if this part is unclear.)
In practice, though, it seems to me that all of the work goes into verifying the "behaves in the expected way" hypothesis, and if that's what you were asking about then I have nothing useful to say (except that if the hypothesis is difficult to verify for the definition of Euler characteristic that you're using then consider using a different one). |
H: How to maintain equal proportion of investment after buying out a partner
Let's say I have five investors, each owning 20% of a company that is worth $1,000,000. One investor wants to be bought out. So the other investors would, to maintain their equal proportion after the buyout, each pay $50,000 and hence would now have 25% of the company.
This makes sense to me. However, when the numbers are not equal, I am unsure how to figure out how much each person would pay.
For example, let's say:
Investor 1 owns 50%
Investor 2 owns 20%
Investor 3 owns 15%
Investor 4 owns 12%
Investor 5 owns 3%
Investor 2 wants to sell. If the company is worth $1,000,000, how would one calculate how much each investor would have to pay Investor 2 to keep their proportionate share in the company?
AI: Let $S$ be the sum of the percentages controlled by the other investors. In this case, $S=0.8=80\%$.
For what Investor $1$ pays, calculate
$$(200000)\frac{0.5}{S},$$
and similarly for the others. If you prefer, you can work with percentages. Investor $1$ pays $(200000)\frac{50}{80}$, Investor $3$ pays $(200000)\frac{15}{80}$, and so on.
So all of the quantities you calculated should be divided by $0.8$, or equivalently multiplied by $1.25$. The idea is that although your calculation gives the right proportions, the sum is only $160000$. To get it up to the required $200000$, you need to multiply each payment you calculated by $\frac{200000}{160000}$. |
H: Characteristic function of a sum of Uniform random variables
Suppose I have $S=U_1+U_2+\dots+U_n$ where $U_i$ are distributed Uniform$[-1,1]$.
I am trying to show a couple of things. First, what is the characteristic function. I can show this easily enough for a single uniform RV, but for some reason the sum is not coming to me.
Second, does the inversion formula apply as long as $n\ge2$? and if so, how can I show this?
Finally, how can I use the inversion formula to show a real-valued integral for the density (i.e getting rid of the imaginary part of the characteristic function)?
I have worked through how to derive a single $U$, but these extensions escape me.
Many thanks for the help~
AI: The characteristic function is an expectation:
$$
\varphi_S(t) = \mathbb{E}\left(\exp(i S t)\right) = \mathbb{E}\left(\exp\left(i \left(U_1 + U_2 + \cdots + U_n \right) t\right)\right)
$$
Now, if $U_i$ is independent, the expectation factors into product of expectations, because :
$$
\varphi_S(t) = \mathbb{E}\left(\mathrm{e}^{i t U_1}\cdot \mathrm{e}^{i t U_2} \cdots \mathrm{e}^{i t U_n}\right) = \mathbb{E}\left(\mathrm{e}^{i t U_1} \right)\cdot \mathbb{E}\left(\mathrm{e}^{i t U_2} \right) \cdots \mathbb{E}\left(\mathrm{e}^{i t U_n}\right)
$$
Since $U_i$ have identical distributions, these expectations are the same, thus:
$$
\varphi_S(t) = \left(\mathbb{E}\left(\mathrm{e}^{i t U} \right)\right)^n = \left( \frac{\sin(t)}{t}\right)^n
$$
According to the inversion formula, there is one-to-one correspondence between cumulative distribution function, and the characteristic function. Since $\varphi_S(t)$ is a characteristic function for all $n \in \mathbb{N}$ by construction, the inversion formula applies for all natural $n$.
Moverover, if $\varphi_S(t)$ is integrable, then $F_S(x)$ is absolutely continuous, i.e. the notion of the probability density function is well-defined. Notice that $\varphi_S(t)$ is absolutely integrable for $n \geqslant 2$, by virtue of $|\varphi_S(t)| \leqslant \min\left(1, t^{-n}\right)$, and hence integrable. However, $\varphi_S(t)$ is also conditionally integrable for $n=1$ (the integral is known as Dirichlet integral).
Applying the inversion formula to obtain an explicit expression for the probability density is somewhat involved. This earlier post of mine contains the derivation.
To conclude, let me remark, that the sum of iid uniform continuous random variable follows what is known as the Irwin-Hall distribution. |
H: Image of complex circle under polynomial
The question is, how many times does $f(C_r)$ go around $0$ for $C_r = \{x \in \mathbb{C}: |x|=r\}$ and a polynomial with complex coefficients $f$.
The answer is clear to me when $f(x) = ax^n + b$ but I don't see how exactly middle terms affect the result. How to see this without a graph? Can this be found for any $r$?
Many thanks
AI: Look at the Argument Principle (eg, Conway's, "Functions of one complex variable", V.3.4).
You wish to compute $n(f \circ \gamma, 0)$, where $\gamma(t) = r e^{i t}$, and $n$ is the winding number. The Argument Principle gives:
$$n(f \circ \gamma, 0) = \sum n(\gamma, z_i) - \sum n(\gamma, p_j),$$
where $z_i$ are the zeros of $f$ and $p_j$ are the poles. Since $f$ is a polynomial, there are no poles, so you have $n(f \circ \gamma, 0) = \sum n(\gamma, z_i)$, ie, the number of zeros of $f$ inside $C_r$, counted according to multiplicity.
(The above assumes that $f$ has no zeros on the circle $C_r$.) |
H: Complement of co-dense set.
Asaf's argument : (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior
Let $X$ be a separable complete metric space.
Let $D$ be a countable debse subset.
Let $F$ be a closed subset with empty interior.
Here, how do i show that $D\setminus F$ is dense?
Ive been trying to figure it out for a day, and still stuck here..
AI: Let $U$ be a non-empty open subset of $X$. Since $F$ is closed and has an empty interior we have that $U\setminus F$ is non-empty and open.
By density of $D$ we know that $D\cap(U\setminus F)=(D\setminus F)\cap U = (D\cap U)\setminus F$ is non-empty, therefore $D\setminus F$ is dense. |
H: Trigonometry word problem?
Two ships have radio equipment with a range of 200 km.One is 155km North42 degrees 40 minutes east and the other is 165 km north 45 degrees 10 minutes west of a shore station. Can the two ships communicate directly?
How would I solve this problem I know I have to make a triangle but I am not sure how the triangle would look like.
AI: We will assume that the Earth is flat, and that we are not too far North. Draw a picture. Let the share station be at $C$. For the first ship, go $42^\circ40'$ East from due North, and travel $155$ km. Call the resulting point $A$. So we face North from $C$, turn $42^\circ40'$ clockwise, and sail $155$ km.
For the second, go $45^\circ10'$ West from due North, and travel $165$ km. Call the resulting point $B$.
Then $\triangle ABC$ has $CA=155$, $CB=165$, and $\angle C=42^\circ40'+45^\circ10'=87^\circ 50'$.
By the Cosine Law,
$$(AB)^2=155^2+165^2-2(155)(165)\cos C.$$
Calculate. It turns out that the distance is about $222$ km.
Remark: We can proceed more informally without the Cosine Law, and with a little crossing of the fingers. Note that $\angle C$ is almost a right angle. If we pretend it is a right angle, we can use the Pythagorean Theorem to estimate the distance. That gives $226$ km. Not very different from $222$. |
H: Questions about adjointness of quantifiers in first-order logic
I have a category theory homework problem which asks: "In first-order logic, why does $\forall$ not have a right adjoint?" The typical argument is that:
If for some operator $\cdot$ it could be asserted $\forall \dashv \cdot$, then $\cdot$ would be a kind of coproduct.
If $\forall$ preserves coproducts, then $\forall y(\phi y \vee \psi y) \dashv \vdash \forall y \phi y \vee \forall y \psi y$.
Since 2 does not hold, $\forall$ must not preserve coproducts, and therefore $\forall$ does not have any kind of right adjoint.
I have a few questions about this:
Why exactly does 2 not hold? Is it because the expression is equivalent to $\forall y(\phi y \vee \psi y) \dashv \vdash \forall y_1 \phi y_1 \vee \forall y_2 \psi y_2$?
The meaning of $\dashv$ and $\vdash$ are unclear to me in the context of expressions. I understand what $* \dashv \forall$ means, but I need help understanding adjointness in terms of expressions.
How can the relation of $\forall$ to the particular coproduct $\vee$ be generalized to all coproducts, such that all possible right adjoints to $\forall$ are ruled out?
AI: I'm only answering question 1 because I don't know the answers to the others.
For question 1, it suffices to give a counter example. A counter example is where $\psi$ is the negation of $\phi$. Then the left side says "for all $y$, either $\phi y$ or not $\phi y$" which is true for any $\phi$ by the law of the excluded middle. The right hand side says "Either for all y $\phi y$, or for all $y$ not $\phi y$" which claims that $\phi$ is true either for all $y$ or for none of them. Which is clearly wrong in general, and therefore it would be disastrous if you could derive it from the tautology on the left. |
H: Using the invariance principle: how to solve $n+d(n)+d(d(n))=m$?
Let $d(n)$ be the digital sum of $n$. How to solve $n+d(n)+d(d(n))=m$, where $n$ and $m$ are natural?
AI: Not a solution, simply a large size observation. May be deleted once a full solution is posted.
Notice that the left-hand-side is always a multiple of $3$. Indeed, let
$$
n = \sum_{k=0}^{\ell} d_k \cdot 10^k
$$
Observe, that $n \equiv d(n) \mod 3$, by virue of $10^k \equiv 1 \mod 3$.
Then
$$
\left(n + d(n)\right) \bmod 3 = \sum_{k=0}^{\ell} d_k \cdot \left(10^k +1 \right) \bmod 3 = \left(-\sum_{k=0}^{\ell} d_k \right) \bmod 3 = \left(-d(n) \bmod 3\right) = \left(-d(d(n))\right) \bmod 3
$$
Thus the equation is only solvable if $m$ is a multiple of $3$.
Here is a plot of the left-hand-side for $1 \leqslant n \leqslant 3^4$, which shows that the solution is not unique.
It also seems, empirically, that $\left(n + d(n) + d(d(n))\right) \bmod 9 = 3\left(n \bmod 3\right)$, although I do not know how to prove that. |
H: Multiplying exponents, solving for n
When solving for n in this equation I get stuck.
Question: What is the smallest value of n such that an algorithm with running time of $\ 100n^2 $ runs faster than an algorithm whose running time is $\ 2^n $ on the same machine?
Straight out of CLRS chapter 1. Class starts in 2 months wanted to get a head start.
My approach:
$\ 100n^2 = 2^n
$
$\ \sqrt(100n^2) = \sqrt(2^n)
$
$\ 10n = (2^n)^{1/2}
$
$\ 10n = (2^{n/2})
$
Is this last step correct? I know I add exponents when multiplying but this is raising an exponent to an exponent so I should multiply. I'm still unsure how to bring the n down out of the exponent on the two so I can solve for it.
AI: Sometimes in cases like this, it's useful to simply try a few and see what happens:
Trying $n=20$ gives $$10n=200$$ and $$2^{n/2}=2^{10}=1024.$$
Since the rate of increase of $n\mapsto 10n$ is constant, while $n\mapsto 2^{n/2}$ grows more and more quickly for larger $n,$ it follows that $10n<2^{n/2}$ whenever $n\ge 20,$ so we're looking for some $n<20.$
Trying $n=14$ gives us $$10n=140$$ and $$2^{n/2}=128,$$ so by similar reasoning, we need some $n\ge 14$.
From there, the solution's fairly quickly found to be $n=15$.
Note: $n=15$ is not the solution to the equation $100n^2=2^n$, but it is the least $n$ for which $100n^2<2^n$. |
H: Integration substitution problem
$$\int_{-1}^1 x^2 ~dx = \frac{2}{3}.$$
Now let us substitute $u=x^2$. Then $du=2x ~dx$ then the definite integration becomes from $1$ to $1$, i.e. $0$. I know I need to partition this from $-1$ to $0$ and $0$ to $1$. But my question is when we substitute anything we do not concern ourselves with the fact that what we are substituting is even or odd. Then why do we need to partition this?
AI: Carry out the substitution. As you point out, we get $du=2x\,dx$. But we need to express everything in terms of $u$, in particular $dx$. We cannot say $2x=2\sqrt{u}$, for that is incorrect when $x$ is negative. So the integral will have to be broken up at $x=0$. |
H: Reasonable strategy for simple game
I have a very simple probabilistic process which I have to deal with in the software project I'm involved in, yet I can't figure how to do it. I can describe the situation as a money game (in reality it is not, but this is the most simple way to go):
At each moment $t$ my program gets a probability value $P_t$ on input. I have to output the stake $X_t$ ($X_t$ can be, say, from 0 to 10) and then with probability $P_t$ I will win (get $XA$, where $A > 1$) and with probability ($1 - P_t$) I will lose (get nothing, lose my stake $X$).
What strategy can be considered reasonable here? Make stake proportional to $P$? What should I read to understand the topic?
AI: It rather depends on your utility function for wealth.
To maximise your expected return, bet everything when the odds benefit you, i.e. when $AP_t \gt 1$. This also gives you a good chance of losing all your money.
For a safer strategy which adjusts your bet size to the extent the odds are in your favour and avoids losing everything, read about the Kelly criterion, for example at Wikipedia. This is largely based on logarithmic utility. |
H: Are all infinities equal?
A friend of mine was trying to explain to me how all infinities are equal. For example, they were saying that there are the same amount of numbers between $0$–$1$ as there are between $0$–$2$.
The way they explained it, you could prove that there are the same amount of numbers by getting a match to each number.
For example, for any number in the range from $0$–$2$, you can find a matching number in the range $0$–$1$ by dividing the number by 2. In every case, the ending number will end up between $0$–$1$. In addition, if you multiplied any number in the range of $0$–$1$ by 2, you will always end up with a number between $0$–$2$ Therefore, there should be the same infinite amount of numbers between $0$–$1$ as there are between $0$–$2$.
Is this principle accurate?
The problem that I had with it was, if in-fact there are the same amount of numbers between 0 and 1 as there are between 0 and 2, why is 2 greater than 1?
AI: At first, it is probably a good idea to specify more clearly what is meant when you say "infinite". There are many different concepts of infinity which are quite different. For example, in the context of limits you can say that a quantity or function "goes to infinity", but in that case it just means "it gets (and remains) arbitrary large". That's a completely different type of infinity than the infinity you are speaking about when you say "how many". "How many elements does this set have" is called the cardinality of that set, and that's what the argument of your friend is about. Note that "how large is" may or may not refer to cardinality, i.e. "how many" (I'll come to another notion later). This other notion also has a concept if infinite, but that concept is different from the cardinality concept. Since your friend's argument was about cardinality ("how many"), in the following I'll use "infinity" in the cardinality sense, as "infinitely many".
OK, so what does it mean if you ask "how many"? Well, with finite sets, there's a well known way to answer that question, and that way is known as "counting": You choose one element of that set and say "one". You choose another element of that set and say "two". As soon as you run out of elements, the last number you've said is the number of elements in that set.
Now let's look at what you have done that way: You've chosen one object and assigned it the number $1$ (that is, from now on, it is the "first object"). You chosen another object and assigned it the number $2$, and so on. So finally, assuming there are $n$ objects in the set, every object got a number between $1$ and $n$ inclusive, and to each number there's a unique object which got that number. In other words, by counting the objects, you've established a one-to-one relation between that set and the objects in the set "numbers from 1 to $n$". Of course you could as well start counting with $0$ (as you might do if you are a C programmer used to zero-based arrays, or a mathematician used to natural numbers to include the $0$), and end up with $n-1$ as last number. But that doesn't matter because there are as many natural numbers between $0$ and $n-1$ as there are between $1$ and $n$, as you can easily check by counting them (either count the set $\{0,\dots,n-1\}$ starting with $1$, or count the set $\{1,\dots,n\}$ starting with $0$; in both case what you do is to establish a one-to-one relation between the sets). Indeed, you can even define the natural numbers as sets of all natural numbers below, with $0$ being the empty set, in that case, the number $n$ has exactly $n$ elements, and therefore there's a one-to-one relation between the number $n$ and any $n$-element set. (Side remark: This construction of the natural numbers can also be extended to infinite numbers and gives rise to yet another concept of infinity, which I won't talk about here.)
So the point is, whenever you have a one-to-one relation between two sets, they have the same number of elements. This is a useful definition because it can be used even for infinite sets, where the literal procedure of counting one after the other would never end. So two sets have, by definition, the same number of elements if there exists a one-to-one relation between the two sets.
With this knowledge we can now see that, in the context of cardinality, the question "are all infinities equal" means "does there exist a one-to-one mapping between any two infinite sets?" And in the concrete case of the numbers between $0$ and $1$ (that is, the interval $(0,1)$) and the numbers between $0$ and $2$ (that is, the interval $(0,2)$), the question can be rephrased as: "Is there a one-to-one mapping between the numbers in the interval $(0,1)$ and the interval $(0,2)$"?
It is not true that all infinities are equal, but it is true that the interval $(0,1)$ contains as many numbers as the interval $(0,2)$. However, while there are infinitely many numbers in $(0,1)$ and also infinitely many natural numbers, there are more numbers in the interval $(0,1)$ than there are natural numbers.
To see that there are as many numbers in $(0,1)$ and in $(0,2)$, just consider the function $f:x\mapsto 2x$. That function gives for each number in $(0,1)$ a number in $(0,2)$, and each number from $(0,2)$ can indeed be reached. Thus there's a one-to-one mapping between the numbers in $(0,1)$ and $(0,2)$.
However there's no one-to-one mapping between the integers and the numbers in $(0,1)$. The classic proof for this is Cantor's diagonal argument: Assume you'd have a one-to-one mapping between the natural numbers and the numbers in $(0,1)$, i.e. a mapping $\mathbb N\to (0,1), n\mapsto a_n$. Then write the numbers in ($0,1)$ in decimal. Then you can construct a number $x\in(0,1)$ by the following rule: The number starts with $0.$, and the $n$-th decimal digit of $x$ is $3$, unless the $n$-th decimal digit of $a_n$ is $3$, in which case you choose $5$ instead. Now the number $x$ is clearly in the interval $(0,1)$, therefore it should be one of the $a_n$s. However for each $n$, it differs from $a_n$ in the $n$th decimal, thus it cannot be in the list, and therefore the list cannot be complete.
OK, but given that there are as many numbers between $0$ and $1$ as there are between $0$ and $2$, so how can $2$ be larger than $1$? Well, the simplest answer is that the number is not an indication of how many numbers are below it (this is different to the natural numbers, where there are indeed exactly $n$ natural numbers below $n$). However there's defined an order between real numbers, which coincides with that of the natural numbers embedded with it. Basically, every positive number $x$ is larger than $0$ and larger than any other positive number between $0$ and $x$, and the reverse is true for negative numbers.
However, you may still insist that the interval $(0,2)$ is twice as large as the interval $(0,1)$. And you're right! But how does that fit with the fact that there are as many numbers in $(0,1)$ as there are in $(0,2)$? Well, the point is that when determining the size of the interval, you do not count the numbers in it (indeed, as shown above in the diagonal argument, you cannot count them). Instead you define the size of the interval (and of more general sets of numbers). Such a function which tells you how large a set is is called measure. You just have to make sure that the measure has some obvious properties: The size of the set should of course not change if you just move it around, the empty set (that is, when you have no numbers at all) should have size $0$, and if you have two distinct sets (like the numbers between $1$ and $2$ and in addition the numbers between $4$ and $6$), the total size should be the sum of the sizes. With those basic properties, and the assumption that $(0,1)$ should have a finite size (this implies that a set consisting of a single number, i.e. a single point on the real line, has size $0$), you already get that the interval $(0,2)$ is twice as large as the interval $(0,1)$: You get the numbers between $0$ and $2$ as the numbers between $0$ and $1$, the number $1$ (but that has size $0$), and the numbers between $1$ and $2$ (but those are just the numbers between $0$ and $1$ shifted to the right by $1$). So if the interval $(0,1)$ has size $x$, the interval $(0,2)$ has size $x+0+x=2x$, so it is indeed twice as large. |
H: Baire Category Theorem
This is Asaf's argument;
(ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior
Suppose that $(X,d)$ is a separable complete metric space, and $\{a_k\mid k\in\omega\}=D\subseteq X$ is a countable dense subset.
By contradiction assume that we can write
$X=\bigcup F_n$ where $F_n$ are closed and with empty interior, we can further assume that $F_n\subseteq F_{n+1}$.
Define by recursion the following sequence:
$x_0 = a_k$ such that $k=\min\{n\mid a_n\notin F_0\}$;
$r_0 = \frac1{2^k}$ such that $d(F_0,x_0)>\frac1k$, since $x_0\notin F_0$ such $k$ exists.
$x_{n+1} = a_k$ such that $k=\min\{n\mid a_n\in B(x_n,r_n)\setminus F_n\}$;
$r_{n+1} = \frac1{2^k}$ such that $d(F_n,x_{n+1})>\frac1k$, the argument holds as before.
I understand his proof till here, ($x_n,r_n$ are well-defined recursively)
Here, I don't know how to show that $\{x_n\}$ is a Cauchy Sequence and don't understand the rest of the proof..
$\{r_n\}$ is not monotonic.
This is the first time im facing with somewhat recursion and sequence are mixed.. and i really want to understand this. Please explain the rest of the proof in detail..
I think 'understanding different proofs' is 'learning techiniques'. I don't insist on this proof, but do want to learn this technique.
Latter part of the argument;
Note that $x_n$ is a Cauchy sequence, therefore it converges to a point $x$. If $x\in F_n$ for some $n$, first note that $d(x_k,F_n)\leq d(x_k,x)$, by the definition of a distance from a closed set.
If so, for some $k$ we have that $d(x,x_k)<r_n$, in particular $d(F_n,x_k)<r_n$. First we conclude that $n<k$, otherwise $d(F_n,x_n)>r_n$. Now we note that:
$$d(F_n,x_n)\leq d(x,x_n)\leq d(x,x_k)+d(x_k,x_n)\leq r_n+r_n=2r_n$$
It is not hard to see that $2r_n< d(F_n,x_n)$, which is a contradiction to the choice of $x_n$.
AI: You can change the set-up slightly so that in the recursive choices we instead take
[4.] $r_{n+1} = 2^{-k}$ where $k \in \mathbb{N}$ is least such that (a) $2^{-k} < r_n$; (b) $2^{-k} < r_n - d( x_{n+1}, x_n )$; and (c) $d ( F_{n+1} , x_{n+1} ) > k^{-1}$.
(Note that since the $r_n$ are all of the form $2^{-k}$, then (a) implies that that $r_{n+1} \leq \frac{r_n}{2}$; (b) will imply that $B (x_{n+1},r_{n+1}) \subseteq B ( x_n,r_n )$, and therefore for $m < n$ we have that $d ( x_m , x_n ) \leq r_m$.)
Induction should show that $r_n \leq 2^{-n}$, and as $x_{n+1} \in B ( x_n, r_{n} )$ we have that $d ( x_{n+1} , x_n ) < r_n \leq 2^{-n}$. From here you should be able to show that the sequence $(x_n)_n$ is Cauchy.
The basic idea is that since $(x_k)_k$ is Cauchy, it converges to some $x$. We now want to show that $x \notin \bigcup F_n$. So assume that $x \in F_n$ for some $n$.
As $( x_k )_k$ converges to $x$, there must be a $K$ such that $d ( x_k , x ) < r_n$ for all $k \geq K$. Take $k = \max ( K , n + 1 )$. Note that we also have that $d( x_k , F_n ) \leq d ( x_k , x ) < r_n$.
We'll derive a contradiction by estimating $$d ( F_n , x_n ) \leq d ( x , x_n ) \leq d ( x , x_k ) + d ( x_k , x_n ).$$
From the above we have $d ( x_k , x ) < r_n$.
Since $k > n$, by an observation above we have $d ( x_k , x_n ) \leq r_n$.
Therefore $d(F_n,x_n) \leq 2 r_n$. However it is relatively easy to show that $2 r_n \leq n^{-1} < d ( F_n , x_n )$, as stipulated in the recursive choices! |
H: Sylow 7-subgroups in a group of order 168
Another question from my qual studying that's been stumping me. I'm still a little lost on normalizers. The question is:
Let G be a group of order 168, and let P be a Sylow 7-subgroup of G. Show that either P is a normal subgroup of G or else the normalizer of P is a maximal subgroup of G.
AI: Suppose $P$ is not normal. Then by Sylow's theorems, $G$ has $8$ $7$-Sylows, all of them conjugate to $P$. Thus the normalizer $N$ of $P$ has index $8$ in $G$, and hence order $21$. Suppose $N$ is not maximal, so there's a subgroup $H$ of $G$ strictly between $N$ and $G$. Then $H$ must have order $42$ or $84$. Sylow's theorems show that $H$ has a normal $7$-Sylow, which must be $P$. But then $P$ is normal in $H$, contradicting the fact that $H$ properly contains the normalizer $N$ of $P$. |
H: How to prove this ln inequality?
I have the following inequality, which (supposedly) holds for every $x\in\mathbb{R}$:
$$
1+x\ln\left(x+\sqrt{1+x^{2}}\right)\geq\sqrt{1+x^{2}}
$$
I've been struggling to find some known inequalities involving logarithms that could be applied here (and lack of condition for $x$ to be non-negative doesn't help either). I would be very helpful for hints on how this should be approached.
AI: Just to illustrate the trick for differentiating in my comment, using the same $f(x)$ as Norbert but writing $y=\sqrt{1+x^2}$:$$f(x)=1+x\log(x+y)-y$$ and knowing that $$\frac{dy}{dx}=\frac x y$$ we compute:
$$f'(x) = 0 + \log(x+y) +\frac {x (1+\frac x y)} {x+y} -\frac x y $$
And simplify ... which works neatly essentially for the same reason that Davide Giraudo's substitution works, to give first and second derivatives which are easy to work with:
So we get $$f'(x) = \log(x+y)$$ and $$f''(x)=\frac {1+\frac x y} {x+y}=\frac 1 y$$
We find that there is a minimum at $x=0$ with $f(0)=0\text{ and } f''(x) \text { positive throughout, which gives us what we need ...}$ |
H: Is the identity matrix an elementary matrix?
My reasoning is yes, as you can switch row i with row i in the matrix... But I'm not sure if it's a "legal" elementary operation to switch a row with itself.
AI: This is a question of convention, but I would certainly consider the identity an elementary matrix, as I think most other mathematicians would. It corresponds to the elementary row operation of "doing nothing", which is about as simple as it gets. |
H: If $\lim_n f_n(x_n)=f(x)$ for every $x_n \to x$ then $f_n \to f$ uniformly on $[0,1]$?
This is a self-posed question, so I do not know the answer and I would like to know what you think about.
Let $f,f_n:[0,1]\to \mathbb R$ be continuous functions. Assume that for every sequence $(x_n)_{n\in \mathbb N}$ converging to a point $x\in [0,1]$, we have
$$\lim_n f_n(x_n)=f(x).$$
Can we say that $f_n$ converges uniformly to $f$ in $[0,1]$?
First of all, do you think the question is well-posed? I'm stuck in thinking about it and I don't manage to prove it neither to find a counterexample. Thanks for your help.
AI: Yes, this is true:
Suppose that the convergence $f_n \to f$ is not uniform but that $f(x) = \lim_{n\to\infty}f_n(x_n)$ for every convergent sequence $x_n\to x$ in $[0,1]$.
After passing to a subsequence of the $f_n$ we may assume that there is $\varepsilon \gt 0$ such that for all $n \in \mathbb{N}$ we have $\sup_{x \in [0,1]} \lvert f(x) - f_n(x) \rvert \gt \varepsilon$. This implies that there is a sequence of points $x_n \in [0,1]$ such that $\lvert f(x_n) - f_n(x_n) \rvert \gt \varepsilon$ for each $n$. After passing to further subsequences, we may assume that $x_n \to x \in [0,1]$ by compactness of $[0,1]$. But then we're in trouble: our hypotheses tell us that
$$\tag{$\ast$}
\lim_{n\to \infty} \lvert f(x) - f_n(x_n)\rvert = 0,
$$
which is incompatible with $f(x_n) \to f(x)$ and $\lvert f_n(x_n) - f(x)\rvert \gt \varepsilon$.
More formally: continuity of $f$ ensures together with $x_n \to x$ that there is $N$ such that $\lvert f(x) - f(x_n)\rvert \lt \varepsilon/2$ for all $n \geq N$ so that the triangle inequality gives us for all $n \geq N$ that
$$
\begin{align*}
\lvert f_n(x_n) - f(x) \rvert &= \lvert [f_n(x_n) - f(x_n)] - [f(x) - f(x_n)]\rvert \\
& \geq \underbrace{\lvert f_n(x_n) - f(x_n)\rvert}_{\gt \varepsilon} - \underbrace{\lvert f(x) - f(x_n)\rvert}_{\lt \varepsilon/2} \\& \gt \varepsilon/2 \gt 0,
\end{align*}
$$
which is absurd in view of $(\ast)$.
Comments:
The mode of convergence $f_n(x_n) \to f(x)$ for all $x_n \to x$ is called continuous convergence. Continuity of the $f_n$ is never used in the above argument. In fact, $f_n(x_n) \to f(x)$ for all $x_n \to x$ ensures continuity of $f$ independently of the continuity of the $f_n$ (exercise).
You can find a generalization of the above result and a more detailed discussion in the section Compact and continuous convergence (Chapter 3, §1, section 5) starting on page 98 in R. Remmert's Theory of complex functions. The precise result reads that $f_n$ converges to $f$ continuously if and only if $f$ is continuous and $f_n \to f$ uniformly on compact sets. More details in loc. cit. (don't miss the historical notes preceding and following that section!). |
H: Difference of two Cauchy Sequences
Suppose that $\{a_n\}, \{b_n\}$ are both Cauchy sequences of real numbers, and that $a_n \le b_n \ \forall n$. Prove that $\lim_{n \to \infty} a_n \le \lim_{n \to \infty} b_n$.
The definition of a Cauchy sequence I am using is
A sequence $\{a_n\}$ is Cauchy if, given $\varepsilon > 0, \ \exists N$ so that $$ |a_n - a_m| \le \varepsilon \text{ if } n,m \ge N$$
My Work
Since $\{a_n\}, \{b_n\}$ are both Cauchy, they have finite limits $A, B$, so considering the sequence $\{c_n\}$ where $c_n = b_n - a_n$, we have
$$ a_n \le b_n \ \forall n \Rightarrow 0 \le b_n - a_n$$
$$\lim_{n\to \infty} c_n = \lim_{n \to \infty} b_n - \lim_{n \to \infty} a_n = B - A \ge 0 \Rightarrow A \le B$$
Therefore, $\lim_{n \to \infty} a_n \le \lim_{n \to \infty} b_n$.
Steps Left Out: By a limit theorem, I know the limit of the difference of two convergent sequences is the difference of their limits, and I know that $\lim_{n\to \infty} c_n \ge 0$ because by assumption each term of $\{c_n\} \ge 0$.
Have I done this proof correctly?
AI: Your proof is correct, and here's another proof.
Let $a = \lim_{n \to \infty} a_n$ and $b = \lim_{n \to \infty} b_n$. Suppose by contradiction that $a > b$. Set $\epsilon = \frac{a-b}{2}$, and choose $N \in \mathbb{N}$ such that $n \ge N$ implies $|a_n - a| < \epsilon$ and $|b_n - b| < \epsilon$. Then for this $n$,
$$ a_n > a - \epsilon = a - \frac{a -b}{2} = b + \frac{a - b}{2} = b + \epsilon > b_n$$
which is a contradiction. |
H: How do I find the derivative of a definite integral in which the variable of differentiation is a limit in the integral?
The derivative I would like to find is
$$\frac{\partial}{\partial z} \int_{-z_0}^z (z - z^{\prime}) \; f(z^{\prime}) \; dz^{\prime}$$
where $f(z^{\prime})$ is some arbitrary function of $z^{\prime}$. So $z$ actually appears twice: once in the upper limit of integration, and once in the integrand.
AI: Similar questions are this one and this one.
The integral depends on the variable $z$ only, because $z'$ is a dummy variable. So you want to find the derivative of the integral and not its partial derivative. This is a special case of the following more general rule (Leibniz rule):
If
$$I(z)=J(u(z),v(z),z)=\int_{u(z)}^{v(z)} f(z,z')dz',\tag{1}$$
i.e. $$\ J(a,b,z)=\int_{a}^{b} f(z,z')dz',\qquad \text{ with }a=u(z),b=v(z),$$
then, under suitable conditions, we have
$$I^{\prime }(z)=\displaystyle\int_{u(z)}^{v(z)}\dfrac{\partial f(z,z')}{\partial z}dz'+f(z,v(z))v^{\prime }(z)-f(z,u(z))u^{\prime }(z).\tag{2}$$
It is a consequence of
$$\frac{dI}{dz}=\frac{\partial J}{\partial z}+\frac{\partial J}{\partial u}
\frac{du}{dz}+\frac{\partial J}{\partial v}\frac{dv}{dz}.$$
The first term is the differentiation under the integral sign
$$\frac{\partial J}{\partial z}=\int_{u(z)}^{v(z)}\dfrac{\partial f(z,z')}{\partial z}dz'$$
and the other two are a consequence of the Fundamental Theorem of Calculus:
$$\frac{\partial J}{\partial v}
=f(z,v(z)),\qquad\frac{\partial J}{\partial u}
=-f(z,u(z)).$$
In your case you want do find $$\frac{d }{d z}\int_{-z_{0}}^{z}g(z,z')dz',$$ where $$g(z,z^{\prime })=(z-z^{\prime
})f(z^{\prime }).$$ Since the lower limit of integration is constant, we have two terms only
$$\frac{d }{d z}\int_{-z_{0}}^{z}g(z,z^{\prime
})dz'=\int_{-z_{0}}^{z}\frac{\partial g(z,z^{\prime })}{\partial z}dz^{\prime
}+g(z,z)=\int_{-z_{0}}^{z}f(z')dz^{\prime
},$$
because $g(z,z)=0$ and $\frac{\partial g(z,z^{\prime })}{\partial z}=f(z').$ |
H: Why do the rationals, integers and naturals all have the same cardinality?
So I answered this question: Are all infinities equal? I believe my answer is correct, however one thing I couldn't explain fully, and which is bugging me, is why the rationals $\mathbb Q$, integers $\mathbb Z$ and naturals $\mathbb N$ all have cardinality $| \mathbb Q |=| \mathbb Z | = |\mathbb N| = \aleph_0$, when $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$.
The basic proof of the other question, from my answer, is that for any $r$ in $\mathbb Q$ there is a single function $f(r)$ such that $f(r) \in \mathbb Z$, and the same holds between $\mathbb Z$ and $\mathbb N$. Because there is this 1:1 transformation possible, there must be the same number of numbers in the three sets, because otherwise there would be a number in one set for which the bijection could not produce a number of the other set, and this is not so.
Now, Belgi explained why $1<2$ in the other question by defining the value 0 as the cardinality of the empty set $\emptyset$, 1 as the cardinality of a set of sets containing only the empty set, and 2 as the set of sets containing the empty set and a set containing the empty set, then proceeding as follows:
Now that we have defined the natural numbers we can define when one number
is smaller than another. The definition is $x<y$ is $x\subset y$
and $x\neq y$ .
Clearly $\{\emptyset\}\neq\{\emptyset,\{\emptyset\}\}$ and $\{\emptyset\}\subset\neq\{\emptyset,\{\emptyset\}\}$
So this a proof, by definition, of why $1<2$.
... however, by the same definition, because $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$, then $| \mathbb N | < | \mathbb Z | < |\mathbb Q|$ and thus a maximum of one of these quantities can be the ordinal quantity $\aleph_0$.
AI: The first thing you need to ask yourself, about finite sets, is this: When do two sets have the same cardinality?
The way mathematics works is to take a property that we know very well, and do our best to extract its abstract properties to describe some sort of general construct which applies in as many cases as possible.
So how do we compare the sizes of two finite sets? If we can write a table, in one column the set $A$ and in the other the set $B$, and each element from $A$ appears in a unique cell; and each element of $B$ appears in a unique cell. If this table have no rows in which there is only one element, then the sets $A$ and $B$ are equal. For example:
$$\begin{array}{lc}
\text{Two equal sets:} &
\begin{array}{c|c|c}A & a_1& a_2\\\hline B & b_1& b_2\end{array} \\
\text{Non-equal sets:} &
\begin{array}{c|c|c|c}A & a_1 & a_2 & a_3\\\hline
B & b_1 & b_2\end{array} &
\end{array}$$
It is clear that this methods captures exactly when two sets have the same size. We don't require one set to be the subset of another; nor we require that they share the same elements. We only require that such table can be constructed.
Well, the generalization is simply to say that there exists a function from $A$ to $B$ which is injective and surjective, namely every element of $A$ has a unique element of $B$ attached to it; and every element of $B$ has a unique element of $A$ attached to it.
It turns out, however, that this notion has a quirky little thing about infinite sets: infinite sets can have proper subsets with the same cardinalities.
Why is this happening? Well, infinity is quite the strange beast. It goes on without an end, and it allows us to "move around" and shift things in a very nice way. For example consider the following table:
$$\begin{array}{c|c|c|c|c|c}
\mathbb N&0&1&\cdots&n&\cdots\\\hline
\mathbb N\setminus\{0\}& 1&2&\cdots & n+1&\cdots
\end{array}$$
It is not hard to see that this table has no incomplete rows and that every element of the left set ($\mathbb N$) appears exactly once, and every element of the right set ($\mathbb N\setminus\{0\}$) appears exactly once!
This can get infinitely more complicated, and so on and so on.
One can ask, maybe we are thinking about it the wrong way? Well, the answer is that it is possible. We can define "size" in other ways. Cardinality is just one way. The problem is that there are certain properties we want the notion of "size" to have. We want this notion to be anti-symmetric and transitive, for example.
Namely, if $A$ is smaller or equal than $B$ and $B$ is smaller or equal in size than $A$, then $A$ and $B$ have the same size; if $B$ also has the same size as $C$ as well, then $A$ and $C$ are of the same size too. It turns out that the notion described by functions has these properties. Other notions may lack one or both. Some notions of "size" lack anti-symmetry, others may lack transitivity.
So it turns out that cardinality is quite useful and it works pretty fine. However it has a peculiarity... well, who hasn't got one these days?
To overcome this, we need to change the way we think a bit: proper subset need not have a strictly smaller cardinality. It just should have not a larger cardinality. This is the right generalization from the finite case, rather than the naive "strict subset implies strictly smaller".
To read more:
Is there a way to define the "size" of an infinite set that takes into account "intuitive" differences between sets? |
H: Calculate closest point without calculating hypotenuse or linear length in 2D space.
Given $n$ number of $(x, y)$ points, I want to find out which is the closest (linear distance) to the point of origin $(1, 1)$. Both $x$ and $y$ will always be positive integers.
Due to the magnitude of numbers I am dealing with, multiplication, division, square roots are out of the question. Due to this, Pythagorean theorem is out.
Adding, subtracting or comparing $x$ and $y$ will work but I'm not sure how to achieve this.
I do not want to calculate the actual distance using Pythagorean theorem. I just want to find out which of the points are the closest to the origin. Is there a way to determine this with the restrictions above?
AI: if $$a^2+b^2>c^2+d^2$$
for points $(a,b),(c,d)$, then
$$b^2-d^2>c^2-a^2$$
$$(b-d)(b+d)>(c-a)(c+a)$$
$$\ln(b-d)+\ln(b+d)>\ln(c-a)+\ln(c+a)$$
I don't know if you're okay with taking logarithms, but this should help you compare distances to the origin if you don't want to multiply.
Check $c-a$ and $d-b$. If both are positive, $(a,b)$ is closer. If both are negative, $(c,d)$ is closer. If $c-a$ is negative and $d-b$ is positive, so both $\ln$ terms are negative, multiply the inequality by $-1$:
$$(a-c)(a+c)>(d-b)(b+d)$$
$$\ln(a-c)+\ln(a+c)>\ln(d-b)+\ln(b+d)$$
So use the absolute value signs in your computation (to save time), but note that if both of those are negative that you should interpret your answer as saying the opposite, since the inequality is flipped.
The base makes no difference, so this can be in $\log_2$. I don't know of a simple expression for $\log(a+b)$ in terms of $\log a $ and $\log b$ though.
If you want to know which of $(0,0),(1,1)$ a given point is closer to, just check if $$x+y>1$$
Since $x+y=1$ is the line equidistant from both points. |
H: Visualizing quotient groups: $\mathbb{R/Q}$
I was wondering about this. I know it is possible to visualize the quotient group $\mathbb{R}/\mathbb{Z}$ as a circle, and if you consider these as "topological groups", then this group (not topological) quotient is topologically equivalent to a circle.
But then, what does $\mathbb{R}/\mathbb{Q}$ look like?
AI: So, you say that the group (not topological) quotient of $\mathbb{R}/\mathbb{Z}$ is topologically equivalent (i.e., homeomorphic) to the circle. However, this doesn't make any sense unless you have a topology on $\mathbb{R}/\mathbb{Z}$! More the point is that a topological group like $\mathbb{R}$ has both a topological structure and a group structure. Now, when you form the group quotient $\mathbb{R}/\mathbb{Z}$, it can be given a topological space in a natural way, in particular, via the quotient topology. Notice that when we do this we again get a topological group (i.e., the quotient group operations are continuous with respect to the quotient topology). Furthermore, the quotient $\mathbb{R}/\mathbb{Z}$ (as a topological space) is homeomorphic to the circle.
Now, in the case of your question, the quotient topology on $\mathbb{R}/\mathbb{Q}$ is the trivial topology. This is not hard to prove since preimages of open sets must be open and saturated. Thus if such a preimage is nonempty, it contains an open interval, and since it is saturated, it must contain all real numbers which differ by a rational from a point in this interval. It is then easy to see that this set must be all of $\mathbb{R}$. Thus the only saturated open sets of $\mathbb{R}$ are $\emptyset$ and $\mathbb{R}$ itself. Hence the quotient topology is trivial. Furthermore, it is trivial that any map into a space with the trivial topology is continuous, so the quotient group operations on $\mathbb{R}/\mathbb{Q}$ are again continuous. So we again have a topological group, albeit not a very interesting one because it isn't very interesting as a topological space. As far as what this space "looks" like, it is similar to a one point space for the reason Ricky mentioned in the comments. However, it is not really easy to visualize since it is not homeomorphic to any subspace of $\mathbb{R}^n$ equipped with the subspace topology (because it is not Hausdorff, or any one of a number of other reasons).
Edit: I should have added that whenever you have a topological group and form the quotient in the way we did above the result is always a topological group. However, unless the original normal subgroup is closed, the resulting quotient group will not even be $T_0$ as a topological space. Thus it is only really interesting to form the quotient when the set by which you quotient out is closed. This explains why $\mathbb{R}/\mathbb{Z}$ is interesting as a topological group, but $\mathbb{R}/\mathbb{Q}$ is not. |
H: A qualitative, yet precise statement of Godel's incompleteness theorem?
I read online a statement to the effect that (I'm paraphrasing):
Goedel's incompleteness theorem shows that we cannot even have a complete and consistent theory for the natural numbers.
I am under the (qualitative) impression that this statement is true within the axioms of natural numbers themselves, so that
if one expanded the of axioms one could prove all of the true statements that can be expressed solely in terms of natural numbers.
Note that this larger system itself is not complete and consistent.
Does Godel's incompleteness theorem have the feature that it shows that
these larger systems are somehow representable with the axioms of the natural numbers?
AI: For any recursively axiomatized (consistent) theory $T$ that extends first-order Peano arithmetic, there is a sentence in the language of Peano arithmetic that is neither provable nor refutable in $T$. Note that the theory $T$ can be over a language that extends the usual language of Peano arithmetic.
This states a version of the First Incompleteness Theorem. One can strengthen this, and at the same time make it more informal, by replacing the condition "$T$ is recursively axiomatized" by "there is an algorithm for enumerating the axioms of $T$."
Remark: One can weaken the result, by omitting reference to Peano Arithmetic. Let $L_0$ be a language whose non-logical symbols are $0$, $S$ (for successor), $+$, and $\times$, and let $L$ be an extension of $L_0$. Then there is no (consistent) recursively axiomatized theory $T$ over $L$ such that all sentences of $L_0$ that are (under the usual interpretation of the non-logical symbols) true in $\mathbb{N}$ are theorems of $T$. |
H: Definition of the rank of the isometry group of a manifold.
Let $M$ be a manifold (which you can assume compact if it helps) and consider the natural action of the isometry group of $M$, Iso($M$) on $M$. The we can define the symmetry rank of $M$ as the rank of Iso($M$). Here is were I have trouble: I think I'm misunderstanding the definition because apparently the following should be clear or obvious:
Symrank($M$)= $k$ iff there exists a k-torus $T^{k}$ acting isometrically on M.
Please correct me if I'm wrong: Is the rank of a group the minimum number of generators?
AI: The rank of a compact Lie group $G$ is the dimension of any maximal torus $T \subset G$. This is well-defined, since by standard results any two maximal tori are conjugate.
In your setup, it seems that $\mathrm{Symrank}(M)$ is equal to $k$ if there is a $k$-torus acting effectively and isometrically on $M$, and there does not exist an effective isometric action of a $k'$-torus on $M$ for any $k' > k$. |
H: Showing that a function is not differentiable
I want to show that $f(x,y) = \sqrt{|xy|}$ is not differentiable at $0$.
So my idea is to show that $g(x,y) = |xy|$ is not differentiable, and then argue that if $f$ were differentiable, then so would $g$ which is the composition of differentiable functions $\cdot^2$ and $g$.
But I'm stuck as to how to do this. In the one variable case, to show that $q(x) = |x|$ is not differentiable, I can calculate the limit $\frac{|x + h| - |x|}{h}$ as $h\to 0^+$ and $h\to 0^-$, show that the two one-sided limits are distinct, and conclude that the limit
$$\lim_{h\to 0}\frac{|x + h| - |x|}{h}$$ does not exist.
The reason this is easier is that I do not have to have in mind the derivative of the function $q$ in order to calculate it.
But in the case of $g(x,y) = |xy|$, to show that $g$ is not differentiable at $0$, I would need to show that there does not exist a linear transformation $\lambda:\mathbb{R}^{2}\to\mathbb{R}$ such that
$$\lim_{(h,k)\to (0,0)}\frac{\left||hk| - \lambda(h,k)\right|}{|(h,k)|} = 0$$
I thought of assuming that I had such a $\lambda$, and letting $(h,k)\to (0,0)$ along both $(\sqrt{t},\sqrt{t})$ and $(-\sqrt{t},-\sqrt{t})$ as $t\to 0^{+}$, but this didn't seem to go anywhere constructive.
AI: Note: My previous answer was incorrect, thanks to @Lubin for catching that.
Simplify your life and show that $\phi(x) = f(x,x) = |x|$ is not differentiable at $0$. It will follow from this that $f$ is not differentiable at $0$.
Look at the definition of differentiability for this case, which is that $\lim_{h \to 0, h \neq 0} \frac{\phi(h)-\phi(0)}{h} $ exists. We have $\phi(0) = 0$, and $\phi(h) = |h|$, so we are looking at the limit of $h \mapsto \mathbb{sign}(h)$ as $h \to 0$.
If you choose $h_n = \frac{(-1)^n}{n}$, it is easy to see that $\frac{\phi(h_n)-\phi(0)}{h_n} = (-1)^n$, hence it has no limit. It follows that $f$ is not differentiable at $0$. |
H: Homotopy equivalence involving cell - complexes
I have two small questions related to some point-set topology involved in the following. My question is inspired after reading the proof of Proposition 1.26 of Hatcher.
Suppose I have a space $X$ that is built out of a subspace $A$ that is path-connected by attaching some $n$ - cells $e_\alpha^n$ for $n \geq 3$ via attaching maps $\varphi_\alpha : S^{n-1} \to A$. Suppose for each $\alpha$ I choose a point $y_\alpha \in D_\alpha$. Now suppose I set
$$U = X - \left\{\bigcup_{\alpha} y_\alpha\right\}.$$
I am interested to know why $(1)$ $U$ is open in $X$. It could be that a union of infinitely many point sets is not closed, from which we cannot conclude that $U$ is open. How can I see this fact?
My second question is somewhat related to the proposition as well in that I believe:
$U$ is homotopy equivalent to my subspace $A$ from which $X$ is built out of.
Now I don't know if this is true and I am trying to prove it. I already have an inclusion map $i : A \to U$ to use. However, I don't necessarily have a map going the other way from $U$ to $A$. Is my belief true and if it is, how would I go along proving it? At the end of the day, I would like to be able to fill in the details of this by myself.
Thanks.
Edit: I should say that my $\alpha$'s run over an arbitrary index set, not necessarily countable or anything.
AI: I'm first going to answer the question I think you meant to ask. If I'm mistaken, please say so and I'll remove this.
Looking at the section in Hatcher you indicated, I think you meant your $D_\alpha$ to be open cells, i.e. images of open disks with the characteristic maps (this is what Hatcher denotes $e^n_\alpha$). In this case it is simple to see that $U$ is open. Recall that a subset of a CW complex is open/closed iff it intersects each closed cell in an open/closed set (this is a defining characteristic of cell complexes; Hatcher discusses this in the appendix, I think). The intersection of your set $U$ and a closed cell in $X$ is simply the cell, minus an interior point. Pulling back to the disk, this is clearly open, so $U$ must be open.
Similarly, if your $D_\alpha$ are open cells, the answer to your second question is yes. To prove this, it suffices to show that $U$ deformation retracts onto $A$. Let me give a quick sketch of this in the case there is only one (aka finitely many) cell(s). We have removed a point from the interior of the single cell. Pulling this cell back, its preimage with its characteristic map is a punctured disk (remember that the characteristic map is a homeomorphism when restricted to the interior of the disk). There is an obvious deformation retraction of this onto the boundary $S^1$. Now just take this deformation retraction and combine it with the characteristic map to get a deformation retraction of the (closed) cell onto its boundary in $X$. Since this boundary lies in $A$, we are done.
As I said, the same works for finitely many cells. When dealing with infinitely many cells, you might need to use a (fairly standard) contrivance to be able to contract all of the cells in "finite time". This method is also shown in Hatcher, probably somewhere around the point where he introduces $S^\infty$.
If you insist on having $D_\alpha$ be closed cells, I expect things can go wrong. I think $U$ might not be open in this case, but I'm too rusty at the moment to think of a counterexample.
Certainly what you conjecture in your second question is false in this case. Take a closed interval as $A$ and attach an $n$-cell at one of the boundary points. Let that boundary point be the point $y$ you remove. Then $U$ isn't even connected and can't be homotopy equivalent to $A$. |
H: Help with a problem with matrices
Hi everyone I've been messing with a problem with matrices and I can't get it to work. I think I am not seeing it the right way because else I get an 8 variable quadratic system of equations.
If $A$ and $B$ are $2\times 2$ matrices and $A^2 = B^2 = I$ and $$AB =
\begin{pmatrix}
0 & -1 \\
1 & 2 \end{pmatrix}$$
and $$BA =
\begin{pmatrix}
2 & 1 \\
-1 & 0 \end{pmatrix}$$
calculate
$(A+B)^2$.
Any help will be very appreciated
AI: Please, type in TeX. Consider $(A+B)^2=(A+B)(A+B)=AA+AB+BA+BB$ and do the math. |
H: Congruence inequality
Given $n>2$, by calculation or otherwise deduce that $5^{2^{n-3}} \neq -1 \pmod {2^n}$
Note:The problem arose when I tried to deduce $\langle5\rangle \cap \langle2^n-1\rangle=\{1\}$ in the group $\mathbb{Z}^{*}_{2^n}$, I have showed $\operatorname{ord}(5)=2^{n-2}$
AI: We show by induction that if $n \ge 3$, then $5^{2^{n-3}}\equiv 1+2^{n-1}\pmod{2^n}$. And it is clear that $1+2^{n-1}\not\equiv -1 \pmod{2^n}$ if $n \ge 3$.
The result holds when $n=3$. Now we do the induction step. Suppose that we know that for a certain $k$, we have $5^{2^{k-3}}\equiv 1+2^{k-1}\pmod{2^k}$. We show that
$5^{2^{k-2}}\equiv 1+2^{k}\pmod{2^{k+1}}$.
By assumption, $5^{2^{k-3}}=1+2^{k-1} +t2^k$ for some integer $t$. Square both sides, and simplify modulo $2^{k+1}$. We get
$$5^{2^{k-2}}\equiv (1+2^{k-1})^2=1+2^k+2^{2k-2}\pmod{2^{k+1}}.$$
But $2^{2k-2}$ is divisible by $2^{k+1}$, since $2k-2 \ge k+1$ when $k \ge 3$. The result follows. |
H: Apostol's Calculus - Prove the set is open (inequality)
This is from Apostol's Calculus, Vol. II Section 8.3 #3(a).
Prove that $S=\{(x,y,z) \mid z^2 - x^2 - y^2 -1>0\}$ is open.
The only way to prove that a set is open which has been covered so far is to prove that for an arbitrary point $\mathop a\limits^{\small \to} \in S$ there is an open ball $B(\mathop a\limits^{\small \to} , r) \subset S$ by finding an explicit $r$ and showing that $\mathop x\limits^{\small \to} \in B(\mathop a\limits^{\small \to} , r) \implies \mathop x\limits^{\small \to} \in S$. Nothing topological (compactness, completeness, etc.) has been discussed. For some reason I just am stuck trying to figure out an $r$ which would work.
Hints > Complete answers
AI: Choose arbitrary $(a,b,c)\in S$. Assume without loss of generality that $a,b,c\geq 0$. Put $k=c^2-a^2-b^2-1$.
Then, for any $(a',b',c')\in B((a,b,c),r)$ (with $0<r<c$, $r$ to be specified later -- notice that $c>0$) we have that
$$c'^2-a'^2-b'^2-1\geq (c-r)^2-(a+r)^2-(b+r)^2-1=k-2r(a+b+c)-r^2$$
Thus, we can see that if we choose $r>0$ small enough to have $r^2+2r(a+b+c)<k$ (and $r<c$), $B((a,b,c),r)$ will be contained in $S$. |
H: Why does input size matter in NP theory?
When my prof introduced us to the N/NP topics, the first thing he mentioned is input size, which he defines as the number of bytes needed to describe and write a problem's input into a file. Could someone please explain to my why the input size matters to these topics? Thank you.
EDIT: I'm concerned about why this way of measuring size is important, esp. to this P/NP topic. My prof mentioned the pseudo-polynomial running time (of the knapsack problem) which is somewhat relevant to this way of counting input size. I'm not sure how it is connected to the NP picture, mostly because right after redefining input size, he went into reduction examples and there is no mentioning of input size since. And for NP-hard problems, since there is no known way to solve them efficiently, why should we care about input anyway?
AI: The simplistic is answer is that complexity classes are defined in terms of the size of the input (of course there are other ways, via descriptive complexity and so on, but I'll stick to the basics). So introducing the size of the input is essential to defining the classes.
Naturally the question then becomes why define these things in terms of the size of the input? The whole raison d'être for these classes is to work out how much of a resource we need to expend to solve a problem. Perhaps we're interested in how much time it takes, or how much space, or some other resource, but above all we want to know what's possible and how much it'll cost us. Like the other comments and answer have said, it's pretty reasonable to expect that giving a bigger input will consume more of the resource we're interested in. Moreover, it's kind of uninteresting to see how much time/space/etc. it takes to solve a single instance - what we want is a classification that tells us how much we need to expend to solve any possible instance, without just trying it and seeing (imagine checking how long it took to solve a particularly hard instance of a particularly hard problem and finding out it took 10 years, or worse, the lifetime of the universe).
So then of course we need some way of relating the instance to the amount of resources needed to solve it. The first natural measure is the size of the instance - if you give me a bigger instance, it takes more time/space/etc. |
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