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H: Errata for Mendelson's Introduction to Mathematical Logic (5th ed)?
I'm hoping that there exists a prepared errata list for Elliott Mendelson's Introduction to Mathematical Logic (5th ed). If so, how does one locate it? (It does not seem to be available from the publisher (CRC Press). If no such errata resource exists, here's a question about what I think is an omission in a particular formula:
In the working up to a proof of Gödel's incompleteness theorem, Section 3.4 (p. 194) produces a long list of correlations between wfs and their Gödel numbers. Well into this process there is defined – by means of "course of values recursion" – this special relation (which I will try to parse below):
$$
\textrm{Ff}(u,v,w) := \textrm{Trm}(u) \wedge \textrm{EVbl}(2^v) \wedge \textrm{Fml}(w) \wedge
$$
$$
\{ \textrm{Atfml}(w) \wedge [?]
$$
$$
(\exists y)_{y<w}(w=2^3\ast2^9\ast y \ast 2^5 \wedge \textrm{Ff}(u,v,y)) \vee
$$
$$
(\exists y)_{y<w}(\exists z)_{z<w}(w=2^3\ast y \ast 2^{11} \ast z \ast 2^5 \wedge \textrm{Ff}(u,v,y)\wedge \textrm{Ff}(u,v,z)) \vee
$$
$$
(\exists y)_{y<w}(\exists z)_{z<w}[w=2^3\ast 2^3 \ast 2^{13} \ast 2^z \ast 2^5 \ast y \ast 2^5 \wedge \textrm{EVbl}(2^z) \wedge ( z\ne v \Rightarrow \textrm{Ff}(u,v,y)) \wedge (\textrm{Fr}(u,z) \Rightarrow \neg \textrm{Fr}(y,v))]\;\}
$$
The definition is intended to provide a relation which holds precisely when $u$ is the Gödel number of a term that is free for the variable with Gödel number $v$ in the wf with Gödel number $w$. Thus, in the top line above, $\textrm{Trm}(u)$ ensures that the first input is indeed the Gödel number of some term, $\textrm{EVbl}(2^v)$ ensures that the second input is the Gödel number of a variable symbol, and $\textrm{Fml}(w)$ ensures that $w$ is the Gödel number of a wf. The remaining lines (within the braces) are meant to encode the condition that the term referenced by $u$ (which I will label $U$) is free for the variable referenced by $v$ (which I label $V$) within the wf referenced by $w$ (labeled $W$), but here is where there seems to be a problem with the printed text, noted above by my insertion of the question mark.
It appears that setting up the condition is broken into four cases: the second line above captures the case that $W$ is atomic; the third line deals with the case in which $W$ has the form $\neg(Y)$ for some lower-ordered wf $Y$, with Gödel number $y$; the fourth line the case in which $W$ has the form $(Y \Rightarrow Z)$ (where $Z$ has Gödel number $z$); and the bottom two lines the final case in which $W$ has the form $((\exists Z)Y)$, where here, $Z$ is a variable symbol differing from $V$.
All but the second line in the definition makes sense. Since the presence of the conjunction symbol is contradictory, there appears to be an omission at the position of the question mark where some other wf should appear; this formula would express the statement that $U$ is free for $V$ in the atomic formula $W$. For instance, the missing expression might be something like
$$
(\exists y)_{y<w}(\exists \alpha)_{\alpha<w}(\exists \beta)_{\beta<w}[\textrm{Gd}(\alpha)\wedge \textrm{Trm}(y)\wedge \textrm{Gd}(\beta)\wedge w=\alpha\ast y \ast \beta \wedge \textrm{Fr}(y,v)]
$$
followed by a disjunction that ties this condition to the remaining cases that follow.
Does this sound right?
AI: I think what you quote as "$\land [?]$" should simply have been a $\lor$.
Then the second line simply says that every term is free for every variable in an atomic formula (which is intuitively true because an atomic formula contains no variable binders).
By the way, the same misprint (except for the "$[?]$", which I suspect is something you inserted?) appears in the fourth edition, p. 195.
It's not as if very many readers are going to be checking the details at this point. I believe most readers who has gotten through the preceding just think "yeah, yeah, I've convinced myself that things with intuitive interpretations of this general shape are primitive recursive already" and skip over the actual formulas. |
H: Prove that $(\mathbb{Z},d)$ is a metric space
I got this from Mendelson:
Let $\mathbb {Z}$ be the set of integers.Let $p$ be a positive prime integer. Given distinct integers $m$, $n$ there´s a unique integer $t=t(m,n)$ such that:
$$ m-n=p^tk $$
where $k$ is an integer not divisible by $p$. Define a function $d:\mathbb {Z} \times \mathbb {Z}\rightarrow \mathbb {R}$ by the correspondence
$$d(m,m)=0$$
and
$$d(m,n)=\frac{1}{p^t}$$
from $m \neq n.$
Prove that $(\mathbb {Z,d})$ is a metric space.
I would appreciate a better explanation to this question. I didn´t get the $t(m,n)$. This is also a distance,right?
AI: Let $x, y, z$ be integers.
It suffices to prove that $d(x, z) \leq d(x, y) + d(y, z)$.
If two of $x, y, z$ are equal, this is trivial.
So we can assume they are distinct.
Let $x - y = p^k a$, where $a$ is not divisible by $p$.
Let $y - z = p^s b$, where $b$ is not divisible by $p$.
We can assume $k \leq s$.
Then $x - z = (x - y) + (y - z) = p^k a + p^s b = p^k(a + p^{s-k}b)$.
Suppose $x - z = p^r c$, where $c$ is not divisible by $p$.
Then $r \geq k$.
Hence $d(x, z) = 1/p^r \leq 1/p^k = d(x, y) \leq d(x, y) + d(y, z)$.
And we are done. |
H: Least quadratic non residue algorithm
I am trying to implement Tonelli-Shanks algorithm and at one of the steps I have to find the least quadratic non residue. I've searched the web for a while for some kind of algorithm but so far I've seen only paper with lot's of math behind them.
Are there any algorithms for finding least non negative residues our there? Or could someone advise me a paper with good explanations and a couple of examples so I can formalize the algorithm myself?
AI: I would not worry about it. The least quadratic non-residue modulo a fixed prime $q$ is also a prime, so just check the primes $2,3,5,7,11,13,17, \ldots$ in order until the Legendre symbol says you have a non-residue. The important thing is that the first non-residue is really, really small compared to the prime itself. See OEIS.
Oh, why is the first nonresidue prime? Call the prime $q$ and the first nonresidue $N.$ Since $1$ is always a residue, we have $N > 1.$ Since exactly half the numbers from $1$ to $q-1$ are residues, half nonresidues, we know $N < q.$ If $N$ were also composite, we would have $N = ab$ with $1 < a,b < N.$ Since $N$ is the smallest nonresidue, this means $a,b$ would be residues. But the product of quadratic residues is another quadratic residue, which would mean $N=ab$ would need to be a quadratic residue. This is a contradiction, so $N$ is prime. |
H: Compactness of $\mathcal K$ in the Hausdorff distance
Possible Duplicate:
The Class of Non-empty Compact Subsets of a Compact Metric Space is Compact
Let $(M,d)$ be a metric space and let $\mathcal K(M)$ denote the set of all non-empty compact subsets of $M$. This collection is a metric space when equipped with the Hausdorff distance $h$.
I want to prove$$(M,d)\mbox{ is compact}\implies(\mathcal K,h)\mbox{ is compact}.$$ The statement is true according to the book [V. I. Istratescu, Fixed Point Theory: An Introduction], but the proof is omitted. I have already shown that $M$ is complete implies that $\mathcal K$ is complete.
AI: Since $M$ is totally bounded, for each $\epsilon > 0$ there is a finite set $\{x_1, \ldots, x_n\} \subseteq M$ such that $\min_{i=1}^n d(x,x_i) < \epsilon$ for all $x \in M$.
For any nonempty compact $C \subseteq M$, if $S = \{x_i: d(C, x_i) < \epsilon\}$ we have
$h(S,C) < \epsilon$. Thus every $C \in {\cal K}(M)$ is within Hausdorff distance $\epsilon$ of one of the finitely many nonempty subsets of $\{x_1,\ldots,x_n\}$. This shows that ${\cal K}(M)$ is totally bounded. Since you already know that ${\cal K}(M)$ is complete, it is compact. |
H: Define integral for $\gamma,\zeta(i) i\in\mathbb{N}$ and Stirling numbers of the first kind
Consider the integral
$$\int\limits_0^{\infty}e^{-x}x^k\ln(x)^n\dfrac{dx}x$$
For $n=3$ we have
$$(-\gamma^2-2\zeta(3)-3\zeta(2)\gamma)\genfrac{[}{]}{0pt}{}{k}{1}+3(\gamma^2+\zeta(2))\genfrac{[}{]}{0pt}{}{k}{2}-6\gamma\genfrac{[}{]}{0pt}{}{k}{3}+6\genfrac{[}{]}{0pt}{}{k}{4}$$
where $\genfrac{[}{]}{0pt}{}{n}{k}$ is Stirling number of the first kind. For different n there are simmilar formulas. In the general case
$$\int\limits_0^{\infty}e^{-x}x^k\ln(x)^n\dfrac{dx}x\in\bigoplus_{j=1}^{n+1}\genfrac{[}{]}{0pt}{}{k}{j}\mathbb{Z}[\gamma,\zeta(i)]_{n+1-j}$$
where $$\mathbb{Z}[\gamma,\zeta(i)]=\bigoplus_{j=0}^{\infty}\mathbb{Z}[\gamma,\zeta(i)]_{j}$$ is graded ring.
Is ir possible to generalize this formula to multiple zeta values?
AI: This is not a complete answer, rather two observations:
Let $\mathcal{I}_{k,n}$ denote the integral. Integration by parts, assuming $k \geqslant 2$ and $n \geqslant 1$ gives the recurrence equation:
$$
\mathcal{I}_{k+1,n} = n \mathcal{I}_{k,n-1} + k \mathcal{I}_{k,n} \tag{1}
$$
which we can use to reduce the problem to $k=1$. Let $\mathcal{J}_{k,n} = \frac{1}{n!}\mathcal{I}_{k,n}$. It is readily seen that $\mathcal{J}_{k,n}$ satisfies the recurrence relation of the unsigned Stirling numbers of the first kind:
$$
\mathcal{J}_{k+1,n} = \mathcal{J}_{k,n-1} + k \mathcal{J}_{k,n} \tag{2}
$$
One the other hand:
$$
\frac{\mathrm{d}^n}{\mathrm{d} s^n} x^{s-1} = x^{s-1} \log(x)^n
$$
Thus:
$$
\int_0^\infty \mathrm{e}^{-x} x^{k-1} \log(x)^n \mathrm{d} x = \left. \frac{\mathrm{d}^n}{\mathrm{d} s^n} x^{s-1} \int_0^\infty \mathrm{e}^{-x} x^{s-1} \mathrm{d} x \right|_{s=k} = \left. \frac{\mathrm{d}^n}{\mathrm{d} s^n} \Gamma(s) \right|_{s=k} = \frac{\mathrm{d}^n}{\mathrm{d} k^n} \exp(\log\Gamma(k) )
$$
Now use Faà di Bruno formula:
$$
\frac{\mathrm{d}^n}{\mathrm{d} k^n} \exp(\log\Gamma(k) ) = \Gamma(k) \sum_{m=0}^n B_{n,m}\left( \psi(k), \psi^{(1)}(k), \ldots, \psi^{(n-1)}(k)\right)
$$
where $\psi(x)$ is the digamma function, and $\psi^{(m)}(x)$ is the polygamma function, and where $B_{n,k}$ stand for the Faà di Bruno polynomials, which are homogeneous:
$$
B_{n,k}\left(\lambda x_1, \lambda x_2, \ldots, \lambda x_n\right) = \lambda^k B_{n,k}\left(x_1, x_2, \ldots, x_n\right)
$$
$$
B_{n,k}\left(\lambda x_1, \lambda^2 x_2, \ldots, \lambda^n x_n\right) = \lambda^n B_{n,k}\left(x_1, x_2, \ldots, x_n\right)
$$
Assume that recurrence relation $(1)$ was used to reduce to $k=1$ case. For this $k$, values of polygammas are related to $\zeta$-values:
$$
\psi(1) = -\gamma \qquad \psi^{(m)}(1) = \left. \frac{\mathrm{d}^m}{\mathrm{d} s^m} \psi(s) \right|_{s=1} = (-1)^{m+1} m! \sum_{n=1}^\infty \frac{1}{n^{m+1}} = (-1)^{m+1} m! \zeta(m+1)
$$
Thus
$$
\int_0^\infty \mathrm{e}^{-s} \log(x)^n \mathrm{d} x = (-1)^n \sum_{m=0}^n B_{n,m} \left(\gamma, \zeta(2), 2! \zeta(3), \ldots, n! \zeta(n+1)\right)
$$
and this explains the grading. |
H: Uniform convergence of infinite series
Suppose $f$ is a holomorphic function (not necessarily bounded) on $\mathbb{D}$ such that $f(0) = 0$. Prove the the infinite series $\sum_{n=1}^\infty f(z^n)$ converges uniformly on compact subsets of $\mathbb{D}$.
I met this problem on today's qual. Here is what I have so far, since $f(0) = 0$, we can write $f(z) = z^m h(z)$ for some integer $m$. Then $f(z^n) = z^{nm}h(z^n)$. We might then use Cauchy's criterion for uniform convergence to finish the proof.
AI: What you did seems correct.
An alternative way is the following: let $M_R:=\sup_{|z|\leq R}|f(z)|$. For a fixed $R<1$, define $g(z):=\frac 1{1+M_R}f(Rz)$, from the open unit disk to itself. Then by Schwarz lemma, we have $|g(z)|\leq |z|$ for all $z\in D$, hence $|f(Rz)|\leq (1+M_R)|z|$ for all $z\in D$. We get that $|f(z)|\leq \frac{1+M_R}R|z|$ for all $z$ in the closed ball of center $0$ and radius $R$, which shows that the series $\sum_nf(z^n)$ is normally convergent on this set.
More simply, for a fixed $R\in (0,1)$, we have
$$\sup_{|z|\leq R}|f(z^n)|\leq \sup_{|z|\leq R^n}|f(z)|.$$
Then we use continuity at $0$ of $f$. |
H: There exist infinitely many values $n\in \mathbb{N}$ such that $8n^{2}+5$ is divisible by $7$ and $11$, without using diophantine equations
Show that there exist infinitely many values $n\in \mathbb{N}$ such that $8n^2+5$ is divisible by $7$ and $11$, without using diophantine equations.
I have $8n^2+5=8(n^2-9)+77$ and since $7\mid 77$ and $11\mid77$, I am searching for those natural numbers $n$ such that $7\mid(n^{2}-9)$ and $11\mid(n^2-9)$. From the first one I get $n=4+7k$ and the second one $n=8+11l$. Then I have to solve the following diophantine equation $7k-11l=4$, which I know that got infinitely many solutions.
But I would like to know a way to show this without the diophantine equations. Is that possible?
AI: Hint $\rm\ 7,11\mid (n\!-\!3)(n\!+\!3)$ $\rm\!\iff\! 77\mid \color{#0a0}{n\!-\!3}\ \ or\ \ 77\mid \color{#0a0}{n\!+\!3}\ \ $ $\rm or \ \begin{align}\rm 7\mid \color{#c00}{n\!-\!3}\\ \rm 11\mid \color{#c00}{n\!+\!3}\end{align}\ \ {\rm or}\ \begin{array}{}\ \ \rm 7\mid\color{#c00}{n\!+\!3}\\ \rm 11\mid\color{#c00}{n\!-\!3}\end{array}\ $
$\!\!\begin{align}
\rm So\ \ \ \ \bmod (7,11)\!:\,\ &\rm n\equiv \color{#0a0}{\pm(3,3)},\, \color{#c00}{\pm(-3,3)}\\[.1em]
\iff \rm \bmod\ \ \ \, 77\!:\quad\ &\rm n\equiv\ \ \color{#0a0}{\pm\! 3},\ \ \ \ \ \ \ \color{#c00}{\pm\! 25}
\end{align}\ $ by CRT.
Remark $ $ Above we computed $\sqrt9 \bmod 77,\,$ i.e. all roots of $\,x^2\equiv 9\pmod{\!77}.\,$ More generally see here and here and here for general methods to compute modular roots of polynomials using CRT = Chinese Remainder Theorem. |
H: Let $M$ be a maximal ideal in $R$ such that for all $x\in M$, $x+1$ is a unit. Show that $R$ is a local ring with maximal ideal $M$
So I'm trying to prove that if $M$ is a maximal ideal in $R$ such that for all $x\in M$, $x+1$ is a unit, then $R$ is a local ring with maximal ideal $M$, that is to say $R$ has a unique maximal ideal.
I've been at this one for a day now and I just can't figure it out. I have that $R$ being a local ring is equivalent to there being a proper ideal $I$ of $R$ which contains all non-units of $R$, and also equivalent to the set of non-units of $R$ being an ideal.
The set of $x+1$ for $x\in M$ is itself multiplicative, but I'm not sure where to go with that since inverting that set just gives back $M$ (since they're all units). I haven't been successful at proving anything about elements of $R$ which are not either in $M$ nor of the form $x+1$ for $x\in M$.
I also tried just assuming there was some other maximal ideal $N$ and then trying to draw out a contradiction by looking at the ideal $M+N$, clearly if $M+N$ doesn't contain $1$ then I've got my contradiction, but I don't seem to have enough information to pursue that path.
Can anyone give me some guidance? Thanks.
AI: Actually upon writing this up I believe I've solved it:
Assume there existed another maximal ideal $N$. Then if $1\in M+N$ then there exists $1 = m+n\in M+N$ and thus $n = -m+1$ and since $m\in M$ implies $-m\in M$, this means that $n = -m+1$ is a unit and thus $N=R$. Therefore $M+N$ must not contain $1$ and thus we have found a proper ideal of $R$ which contains $M$, contradicting the fact that $M$ is maximal. |
H: Quadrilateral Top Point
Suppose you are given a quadrilateral - orient so that one point is the "bottom" (ie. like a diamond). Given three points: the bottom point, the left point and right point, I want to solve for the top point. The only other things I know are the lengths of the top left and right sides. I know I can construct quite a few equations to describe the top point, but I'm having difficulty putting it all together.
My attempt at a solution: I can "cut" right triangles into the inside of the diamond, and from those use Pythagoras and trig. identities, but I don't know where to go from there. Eg. let the top left edge of the diamond be of length $p$, then:
$p^2 = a_p^2+b_p^2$ and $tan(\alpha)=b/a$ and $\alpha+\beta = \pi/2$ and similarly for the top right edge.
Here, $b$ is the height of the right triangle with hypotenuse $p$ and $a$ is the width; $\alpha$ the left angle and $\beta$ the top.
Where do I go from here? Thanks!
AI: The bottom point is irrelevant. Given two points ($P_1$ "left" and $P_2$ "right"), and the desired distances $r_1$, $r_2$ respectively of the "top" point $P_3$ from those, the "top" point
must be one of the intersections of the circle with radius $r_1$ centred at $P_1$ and the circle with radius $r_2$ centred at $P_2$. You can use the law of cosines to find the angles in the triangle whose vertices are $P_1$, $P_2$ and $P_3$. |
H: Bounded power series without convergence on the boundary
Does there exist a holomorphic function $f\in H^\infty$ (i.e. holomorphic and bounded on the open unit disc) such that its power series expansion at $0$ doesn't converge anywhere in $\{|z|=1\}$?
AI: To every $f \in H^\infty$ there corresponds $f^\star \in L^\infty({\mathbb T})$ such that
$f^\star(\omega) = \lim_{r \to 1-} f(r \omega)$ almost everywhere. The functions $f^\star$ thus obtained are the members of $L^\infty$ whose Fourier coefficents
$c_n = (2 \pi)^{-1} \int_{0}^{2\pi} f^\star(e^{it}) e^{-int}\ dt$ are $0$ for integers $n < 0$, and the Maclaurin series of $f$ on $\mathbb T$ is the Fourier series of $f^\star$.
But by Carleson's theorem the Fourier series of any function in $L^2$ converges almost everywhere. |
H: Showing $f(x)=\sum_{n=1}^{\infty}{\sin\left(\frac{x}{n^2}\right)}$ is continuous.
Let
$$f(x)=\sum_{n=1}^{\infty}{\sin\left(\frac{x}{n^2}\right)}.$$
a) Show that the series converges for $x\in [0,\pi/2]$.
b) Show that $f$ is monotone and continuous on this interval.
This is what I have for (a). Is it right? I am stuck on continuity for part (b).
For (a), I showed that since
$$\frac{x^n}{n!}>\frac{x^{n+2}}{(n+2)!},$$
we get that for $x\in[0,\pi/2]$,
$$0\leq \sin x = x-\sum_{n\in I}{\frac{x^n}{n!} - \frac{x^{n+2}}{(n+2)!}}$$ where $I=\{3,7,11,\ldots\}$. Since the last term is negative or zero, $\sin x\leq x$ for $x\in [0,\pi/2]$.
Thus,
$$\sum_{n=1}^{\infty}{\left|\sin\left(\frac{x}{n^2}\right)\right|} \leq \sum_{n=1}^{\infty}{\frac{x}{n^2}}$$
which converges. So the series converges.
b) I was able to show $f$ is monotone. Is there a nice way to show $f$ is continuous?
AI: Let $\displaystyle M_n=\frac{\pi}{2n^2}$. Suppose your series is $\sum u_n(x)$. From what you showed, $$|u_n(x)|\leq \frac{x}{n^2}\leq \frac{\pi}{2n^2}$$ on all of $[0,\frac{\pi}{2}]$. Since the seires $\sum M_n$ converges, the Weierstrass M-test shows that the series converges uniformly, and since the partial sums are all continuous the limit is continuous as well. |
H: Given the pairwise distances between $n$ points, how can I find plausible coordinates for the points?
If I have three points $A, B, C$, and I know the distances between $A$ and $B$, $B$ and $C$, and $A$ and $C$,
(1) How can I find (one possible value for) the coordinates of $A$, $B$, and $C$?
(2) If the distance function is a viable distance metric, must a solution always exist?
(3) How can I generalize this to $n$ points?
AI: Because the distances do not change with translations, reflection and rotations, we can (even if we are in 3D) restrict to the $x,y$ plane, and fix $A=(0,0)$ , $B=(b_1,0)$, $C=(c_1,c_2)$
Then $b_1 = d_{AB}$. Also, $c_1^2 + c_2^2 = d_{AC}^2$ and $(c_1-b_1)^2 + c_2^2 = d_{BC}^2$
This gives: $$c_1 = \frac{d_{AC}^2 - d_{BC}^2 + d_{AB}^2}{2 d_{AB}}$$
and
$$ c_2^2 = d_{AC}^2 - c_1^2$$
which must be positive to have a solution (well, two). From this we get the general points by rotations, translations and reflections. I'm not sure if there is a more elegant or generalizable way. |
H: The continuity of multivariable function
$F$ is a function on $\mathbb R^n$ such that for every smooth curve $\gamma:[0,1] \rightarrow \mathbb R^n, \gamma(0)=0 $, we have $\mathop {\lim }\limits_{t \to 0} F(\gamma (t)) = 0$, is it necessary that $\mathop {\lim }\limits_{\left| x \right| \to 0} F(x) = 0$ ?
AI: Good question!
For $n=2$, you may find a positive answer to this question in Remarks on the continuity of functions of two variables, by Michael McAsey and Libin Mou. The proof is relatively difficult.
In the paper, the following theorem is proven:
Let $D$ be an open subset of $\mathbf R^2$ containing $(0,0)$, and $f$
a real-valued function on $D$. If $f$ is discontinuous at $(0,0)$,
then
There exists a continuously differentiable convex function $z \in C^1$ such that either $f(x,z(x))$ or $f(z(x),x)$ is discontinuous at
$x=0$;
For any integer $m\geq 0$, there exist functions $x(t), y(t) \in C^m$ such that $f(x(t),y(t))$ is discontinuous at $t=0$.
The authors state a similar theorem for $n=3$ (which also implies a positive answer to your question for $n=3$) and indicate how these results can be further generalized. |
H: Halting problem on finite set of programs
As I understand the halting problem, it imply the fact that there doesn't exist one program which can answer the halting problem for every computable program and it rely on Cantor diagonalization to build the proof.
However, Cantor diagonalization would not seem to be practicable on finite set of programs.
If we are only concerned in solving the halting problem on a finite number of programs, is it still true that there are set of programs for which there exist no program that solve the halting problem?
AI: If you get to choose which programs are in that finite set (e.g., all of the ones that halt) then it is trivial to create such a halting-detecting program (HDP). Otherwise, if the set of programs is arbitrary, it might as well be infinite. I.e., the set of programs being presented to your HDP is being chosen from the infinite set of all programs. This would include a theoretical program from the diagonalization. You don't get to change the program after the fact.
On a practical level, it is possible to create an HDP for a very large number of programs, but that returns the answer "don't know" for some of them. In theory, the "don't knows" would outnumber the "halts" or "doesn't halt" answers, but in practice (assuming well formed programs) it should be feasible to make that percentage fairly small.
Edit to add:
As a corollary to Steven's point, any single program will have an HDP that correctly predicts whether it will halt. Consider two HDPs: one always says its input program halts, and the other one always says it doesn't halt. For any given input program, one of those two HDPs is correct.
Similarly, if we're considering the "real world" where we don't have true Turing machines but machines with a finite amount of memory (including disk space)—although if we include writing to other machines on the internet that "finite" can get quite large—then if you want an HDP that can reliably solve any program running on those machines all you need is an even bigger machine—one with $2^n$ amount of memory, where $n$ is the finite amount of memory available to the machines whose programs you're analyzing. Of course, you're also going to need a Universe with more matter than our current one has. |
H: How to write a good mathematical paper?
I hesitate to ask this question. However I read many advices from math.stackexchange, and I couldn't find anything similar.
A good time always goes too fast! Two years are fled. In the third year of PHD, my major is general topology and I'm facing with graduation from PHD. I do enjoy research, however the pressure to publish makes me be agitated and not quite, for I haven't publish any paper. I find, sometimes, doing research and to publish are contradictory.
Here is my question: How to write a good mathematical paper? Could anybody give me some suggestions?
Thanks ahead.
AI: As someone who is currently working on my first mathematical paper, I've found this guide from MIT to be very helpful. It covers both writing a clear and precise paper in general as well as the specific challenges presented by a mathematical paper. It's also fun to read! For example, the author likes to illustrate common mistakes within the text. One of my favorites is:
Don’t string adjectives together, especially if
they are really nouns. Many high quality pure mathematics original research journal
article sentences illustrate this problem. |
H: Describing the effect on $ax^2$ by manipulating $a$
Please take, for example, $y = x^2$ and $y = 2x^2$.
Graphs: Wolfram Alpha
What is the most appropriate way to describe the effect of $a$? "$a$ causes the parabola to open at $1/a$ the rate of $y = x^2$"?
AI: First of all, the sign of $a$ impacts if it opens up or down.
And the way I think about how it affects the shape is that $ax^2$ is $x^2$ stretched vertically by a factor of $|a|$. So on $y=x^2$ there is the point (1,1), on $y=2x^2$ the $y$ value is scaled by a factor of 2 so the graph includes the point (1,2). |
H: Computing sign table for $2^k$ factorial experiment design
all!
I need to compute the sign table for a generic $2^k$ factorial design. For $k$ factors we compute $2^k$ experiments and need to compute a $2^k \times 2^k$ matrix, as the following example for $k=3$:
\begin{matrix}
& I & A & B & C & AB & AC & BC & ABC\\
1 & +1 & -1 & -1 & -1 & +1 & +1 & +1 & -1\\
2 & +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\
3 & +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\
4 & +1 & +1 & +1 & -1 & +1 & -1 & -1 & -1\\
5 & +1 & -1 & -1 & +1 & +1 & -1 & -1 & +1\\
6 & +1 & +1 & -1 & +1 & -1 & +1 & -1 & -1\\
7 & +1 & -1 & +1 & +1 & -1 & -1 & +1 & -1\\
8 & +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\
\end{matrix}
It is easy to compute the symbol's ($A$, $B$, $C$) columns seeing the experiment number as a bit array, where a bit zero maps to $-1$ and a bit one to $+1$. The following columns are calculated as the product of the combined symbols.
I want to compute the matrix directly, looping through $i$ and $j$, for any generic $k$. How can I find to which combination (and thus, which symbol's signs to multiply) a given $j$ corresponds to?
I hope it is clear enough; if not, please ask. Thanks for any attention!
AI: It's hard to see the bit structure in the table the way you've ordered it. If you write it like this:
$$
\begin{matrix}
& I & A & B & AB & C & AC & BC & ABC\\
8 & +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\
7 & +1 & -1 & +1 & -1 & +1 & -1 & +1 & -1\\
6 & +1 & +1 & -1 & -1 & +1 & +1 & -1 & -1\\
5 & +1 & -1 & -1 & +1 & +1 & -1 & -1 & +1\\
4 & +1 & +1 & +1 & +1 & -1 & -1 & -1 & -1\\
3 & +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\
2 & +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\
1 & +1 & -1 & -1 & +1 & -1 & +1 & +1 & -1\\
\end{matrix}
$$
then the entries are just the parity of the AND of the row and column indices from $0$ to $2^k-1$, which you can most efficiently get by precomputing a lookup table of size $2^k$ for the parities and indexing it with the AND.
If you want the table in the order you showed it in, the reversal of the rows is trivial (the index is $2^k$ minus your row label), but the rearrangement of the columns isn't – you've got the columns sorted by first subset size and then lexicographical order; you can get that either by sorting the columns in the end, or by making another lookup table of size $2^k$ beforehand that translates between the two column indexing schemes. |
H: Contragredient representation of a finite group
I am reading Serre's Linear Representations of Finite Groups and in an exercise in there he asks to show if $\rho$ is a representation of a finite group on $\textrm{GL}(V)$ with $V$ a finite dimensional complex vector space that there exists a unique linear representation $\rho^\ast:G \to \textrm{GL}(V^\ast)$ such that
$$(\rho^{\ast}_sf)(\rho_sx) = f(x) $$
for all $s \in G, x \in V$, $f \in V^\ast$. Now I believe I have shown existence as follows: Let $\rho^\ast_sf$ be the map from $G$ to $\textrm{GL}(V^\ast)$ defined by saying that given any $f \in V^\ast$, $x \in V$ we have that
$$(\rho^\ast_sf)(x) = f\left(\rho_{s^{-1}}(x)\right).$$
Now we see that $\rho_s^\ast$ is a linear map for each $s$ and we can write down a matrix for $\rho^\ast_s$ in the standard basis $\{\Lambda_i\}$ of $V^\ast$. Here I am taking $\{e_i\}$ as my standard basis for $V$, so as usual we have that
$\Lambda_i(e_j) = \delta_{ij}$.
I have found that the matrix of $\rho_s^\ast$ in the basis I stated above is the matrix $(\rho_{s^{-1}})^T$. Now once I have this, it is easy to see that $\rho^\ast_s$ defined a homomorphism from $G$ to $GL(V^\ast)$ and that
$$\textrm{Tr}(\rho^\ast_s) = \overline{\textrm{Tr}(\rho_s)}$$
where the bar indicates complex conjugation.
The problem I am having now is in showing uniqueness. I can't seem to get uniqueness from the property that such a representation must satisfy. What I have tried is this. Suppose we have another homomorphism $\sigma: G \to GL(V^\ast)$. Then we must have that
$$(\sigma_sf)(\rho_sx) = fx$$
for all $f \in V^\ast$, $s \in G$ and $x \in V$. Then by definition of an element in $\textrm{End}(V^\ast)$ we must have that $ f(\sigma_s\rho_s(x)) = f(x)$. Since this must hold for all $f \in V^\ast$ and $x \in V$, I get that
$$\sigma_s\rho_s = \textrm{Id}.$$
Now don't I want $\sigma = (\rho_{s}^{-1})^T$ from here? I can't seem to get it from the expression above. How can I get uniqueness?
Thanks.
AI: If $u,v$ are maps with the same domain and codomain, and $u(x)=v(x)$ for every $x$ in the domain, then $u=v$ as maps. This applies to the symmetries of a function space just the same:
$$\begin{array}{c l} & \forall f\in V^*, & \forall x\in V: & (\rho_s^*f)(\rho_sx)=f(x)=(\sigma_sf)(\rho_sx) \\ \iff & \forall f\in V^*, & \forall v\in V: & (\rho_s^*f)(v)=(\sigma_sf)(v) \\ \iff & \forall f\in V^*: & & \rho_s^*f=\sigma_sf \\ \iff & & &\rho_s^*=\sigma_s. \end{array}$$
Note the substitution $u=\rho_sx$ ($\rho_sV=V$ because $\rho_s$ is always invertible). Applies for every $s\in G$.
The moral: recognize reparametrizations of universal quantification, and recognize that universally quantified equalities in a space pass up higher to equalities in the relevant function space. |
H: Derivatives of vectors involving the expectation operator - Part I
So, I am trying to take the derivative of the following equation, because it is needed in an optimization problem. I want to make sure I am on the right track. The equation is:
$$
-3 \mathbb E[(w^Tz)^2]^2
$$
So my question is, what is:
$$
\frac{\delta (-3 \mathbb E[(w^Tz)^2]^2)}{\delta w} = ?
$$
Please assume here that $w$ is a 2-dimensional column vector, just like $z$. $z$ is also a zero mean, unit variance (joint) random variable. ($w$ is a deterministic vector).
I would like a break down of the steps for evaluating the derivative here - I half syspect the chain rule is involved, however I am getting thrown off by the presence of the expectation operator.
Thanks!
AI: Notice that $(w^T z)^2 = w^T z z^T w = w^T (z z^T) w$. Thus $f(w) = E_z (w^T z)^2 = w^T E_z (z z^T) w$. The derivative is given by $\frac{\partial f(w)}{\partial w} = 2 w^T E_z (z z^T)$. Just as a reminder, $\frac{\partial (w^TAw)}{\partial w} = (A+A^T)w$. This becomes $2Aw$ when $A$ is symmetric.
You wished to compute the derivative of $\phi(w) = -3 f(w)^2$. This can be computed using the usual calculus rules as $\frac{\partial \phi(w)}{\partial w} = -6 f(w) \frac{\partial f(w)}{\partial w} = -12 \, (w^T E_z (z z^T) w) \, w^T E_z (z z^T)$.
Now $E_z(zz^T) = I$ because $z$ is zero mean with covariance $I$. Hence, the final answer is: $-12 \, (w^T w) \, w^T = -12||w||_2^2 w^T$, where $||w||_2$ is the $L2$-norm of $w$. |
H: Why isn't this a regular language?
I'm stuck as to figuring out why $L_1$={$n^p$ | $p$ = a prime number} is not a regular language but $L_2$={$n^p$ | $p$ = a prime number bounded by some fixed number f} is. I can see that $L_2$ is a finite language so it's a regular language. But I can't seem to find a way to build a finite state machine (that is, draw a transition diagram) that would model this.
Thanks all.
AI: $L_1=\{ n^p\quad |\quad \text{p=prime number} \}$ is not regular language ,can easily be proved by pumping lemma.
Let $k$ be the pumping lemma constant.
and the string $\alpha \beta \gamma$ as $\epsilon$ , $a^p$ and $\epsilon$ respectively such that $\alpha \beta \gamma \in L_1$ and $|\beta|\geq k$.
So ,if we decompose $\beta$ into $\beta _1(|\beta_1|=r),\beta _2(|\beta_2|=t),\beta _3(|\beta_3|=s)$ such that $|\beta_2| \neq \phi $,$|\beta_2| \leq k$.
Then choose $i=p+1$ ,we will get $|\alpha \beta_1 \beta_2^i\beta_3\gamma|=p(1+t)$ which is a composite.
But if we bound the prime number by $f$ then enumerate all $p<f$ and build a NFA that first guesses the prime then take that many steps to reach the final state. |
H: Subgame Perfect Nash Equilibrium
My homework question is summarized below:
There are 7 players (say P1,P2,...,P7) trying to split 100 dollars. The game starts with P1 proposing an allocation of the 100 dollars to each player, that is, P1 assigns a dollar amount [0, 100] to each player, including himself. Then the remaining players P2..P7 all vote "Yes" or "No" to P1's allocation, simultaneously. If half or more vote "Yes", then the allocation is passed and everyone's payoff is the dollar amount they were allocated. If less than half vote "Yes", then the current player allocating (P1 in the first round of this game) receives a payoff of -1 and is removed from the subsequent rounds of the game. The game then continues onto the next round but with P2 attempting the next allocation for players P2...P7. The game continues until either the proposed allocation is passed (half or more vote "Yes"), or until P7 is the only player left in the game, in which he receives all 100 dollars.
Using the solution concept of subgame perfect Nash equilibrium, what happens?
I'm not quite sure how to start this question since I am not given an Extensive form game, and if I tried to make one myself, I wouldn't know the payoffs. Also the size of this Extensive form game (around $2^0 + 2^1 + ... + 2^5$ nodes) is quite impractical to solve.
My attempt is below and admittedly it is lacking. I'm fairly positive I'm taking the wrong approach.
I tried starting from the very last round of the game and working upwards. The way I made the games was to consider a Px allocating and I would draw the extensive form game of players Px+1 .. P7 voting "yes" or "no". To make these games work for any allocation, I assigned a payoff of [0,100] to paths where the allocation is passed.
I first considered the case where P6 is allocating and P7 is voting and made the extensive form game of this. I found subgame perfect nash equilibrium and the proposal to fail. Next I made an extensive form game when P5 allocates and P6 and P7 votes. Some payoffs were found by using the result of the previous game. The proposal passes this time. But now I couldn't find the subgame perfect Nash equilibrium when P4 allocates and P5, P6, P7 vote because I didn't know which payoffs were better for some players (because I was allocating [0,100] to paths where proposals passed. To be more specific, there were paths where one strategy would lead to a payoff of [0,100] that P4 allocated and the other would lead to [0,100] that P5 allocated).
The whole point of me making these games was to hopefully find some pattern to each player's actions, but the case when P4 allocates stopped that plan. The only pattern I figured out was P7 always votes "No" when there are 2 or less voters.
Besides this, I don't know how to proceed. Any help at all is appreciated.
EDIT: I forgot an important part of the question. The question states: focus on subgame perfect Nash equilibria when players vote "no" when they are indifferent between voting for or against it. I apologize for my carelessness.
AI: Below the line is an answer to the original question, which was interesting because of subtle issues of indifference. The question has now been edited to include the prescription that indifferent players vote "no".
In that case there is no equilibrium in the continuous version of the game: At the last stage, player $7$ always votes "no", so the payoff is $(-1,100)$. At the penultimate stage, player $6$ always votes "yes", so the payoff is $(100,0,0)$. At the stage before that, there is no equilibrium, since offering something to player $6$ or $7$ is strictly dominated by offering half as much, and offering nothing to either of them is strictly dominated by offering something to one of them.
In the discrete case, the analysis is the same for the last two stages. Player $4$ must offer $1$ to either player $6$ or player $7$, with payoff $(99,0,1,0)$ or $(99,0,0,1)$. Player $3$ must offer $1$ to the two players who would otherwise get nothing, with payoff $(98,0,1,0,1)$ or $(98,0,1,1,0)$. Player $2$ must offer $1$ to the two players who would otherwise get nothing and $2$ to one of the players who would otherwise get $1$, with payoffs $(96,0,1,2,1,0)$, $(96,0,1,0,1,2)$, $(96,0,1,2,0,1)$ or $(96,0,1,0,2,1)$. Finally, player $1$ must offer $1$ to the players who would otherwise get $0$ and $2$ to one of the players who would otherwise get $1$, with payoffs $(96,0,1,2,0,0,1)$, $(96,0,1,0,0,2,1)$, $(96,0,1,2,1,0,0)$, $(96,0,1,0,1,2,0)$, $(96,0,1,2,0,1,0)$, $(96,0,1,0,0,1,2)$ or $(96,0,1,0,1,0,2)$. In all cases, the payoffs for players $1$ through $3$ are $96$, $0$ and $1$, respectively, and the remaining $3$ are divided in various ways among players $4$ through $7$.
In both versions, continuous and discrete, the question "Using the solution concept of subgame perfect Nash equilibrium, what happens?" doesn't make sense, since there is no unique subgame perfect Nash equilbrium in either.
This is a great question; it's interesting and quite relevant to understanding the underlying concepts, and you put considerable effort into describing what you've tried; I wonder why it hasn't received more upvotes.
crf's solution is the "right" solution in practical terms, since one would always offer otherwise indifferent players a small incentive to make sure they don't take an adverse decision; but I understood the question to be more formally about how to treat indifference in finding a (weak) subgame perfect Nash equilibrium. The answer to that is somewhat different, namely that the only subgame perfect Nash equilibrium outcome of the continuous version of this game is where player $1$ assigns herself all the money, without any infinitesimal bribes subtracted, whereas the discrete version (where the proposals have to be integer dollar amounts) has a plethora of subgame perfect Nash equilibria, in which various players are bribed with various small amounts of money if they threaten to vote "no" otherwise.
The reason lies in how the solution concept of a subgame perfect Nash equilibrium treats credible and non-credible threats and promises (let's call them announcements). An announcement is not credible if the player would be worse off by actually carrying it out. Whereas a plain Nash equilibrium allows strategies involving non-credible announcements, a subgame perfect Nash equilibrium doesn't. However, it still allows announcements about what you'll do when you're indifferent; that is, if in a subgame a player is indifferent between two options, either of the two options can be part of a strategy in a subgame perfect Nash equilibrium.
In the continuous version of the present game, none of the solutions found by crf are equilibria, since for any given $\epsilon\gt0$ player $1$ could always choose a smaller $\epsilon$ and get a higher payoff. So the question is whether there's a subgame perfect Nash equilibrium where player $1$ gets all the money. The crucial point here is that the condition for a (weak) Nash equilibrium (subgame perfect or not) is only that the players can't profit from switching strategies. Whether strategies are in equilibrium is not affected by whether one player switching their strategy would make it profitable for another player to also switch their strategy.
In the present case, if player $1$ allocates all the money to herself and everyone else's strategy is to always vote "yes", then no player can profit from switching strategies, so this is a (weak) subgame perfect Nash equilibrium. The fact that players threatening to vote "no" would make it profitable for player $1$ to offer them something doesn't matter; it's not profitable, given the given strategies, for player $1$ to switch strategies.
Here's the analysis in detail. At the last stage, with two players left, the equilibrium strategy profiles are: player $6$ makes any allocation and player $7$ votes "no" no matter what player $6$ does; or player $6$ allocates all the money to player $7$ and player $7$ votes yes if and only if player $6$ allocates all the money to her. The payoffs are $(-1,100)$ in the first case and $(0,100)$ in the second.
At the penultimate stage, with three players left, subgame perfection eliminates non-credible threats by player $6$ to vote "no" to any proposal allocating her any money (since she'd get at most $0$ if the proposal fails), so there can't be an equilibrium allocating any money to player $6$, since player $5$ would be better off allocating half as much to player $6$. Thus in equilibrium player $5$ must allocate all the money to herself. Of course player $7$ votes "no", but player $6$ can't profit from voting "no", so player $6$ voting "yes" yields an equilibrium. (This is independent of which of the two equilibria for the last stage we select; for $(0,100)$ player $6$ is indifferent and for $(-1,100)$ she even profits from voting "yes".) The strategy of player $6$ in case player $5$ offers her something can be anything since it doesn't affect the outcome, so there's a continuum of equilibrium strategy profiles, but they all have the same payoff, $(100,0,0)$.
The analysis proceeds analogously at all stages. With player $4$ proposing, player $5$ will always vote "no", and there can't be an equilibrium where player $4$ offers anything to player $6$ or $7$ because she could offer half as much instead (non-credible threats to vote "no" in that case being eliminated by subgame perfection); so the only equilibrium is that player $4$ allocates all to herself, player $5$ votes "no" and players $6$ and $7$ vote "yes"; again the strategies of players $6$ and $7$ if player $4$ offers them something are arbitrary and don't affect the outcome, which is always $(100,0,0,0)$.
At the next stage, with player $3$ proposing, player $4$ votes "no", and any of the four subsets of players $5$ through $7$ with at least two players voting "yes" yields an equilibrium (as always, the strategies in case player $3$ offers them something being arbitrary). The payoff is $(100,0,0,0,0)$, and the analysis continues, with more different strategy profiles in equilibrium at each stage since there are more majority subsets at each stage, but the payoff is always the same: $(100,0,0,0,0,0)$ at the second stage and $(100,0,0,0,0,0,0)$ at the first.
In the discrete version, this is still a subgame perfect Nash equilibrium, but now there are also ones similar to crf's result.
Again, at the last stage, there are two equilibrium payoffs, $(0,100)$ and $(-1,100)$. At the penultimate stage, in the $(-1,100)$ case player $6$ will always vote "yes" to any proposal, so the only equilibrium payoff is $(100,0,0)$, whereas in the $(0,100)$ case player $1$ is indifferent between $(100,0,0)$ and $(0,100)$ and can therefore threaten to vote "no" when offered $0$, but can't threaten to vote "no" when offered $1$, so in this case there's an additional equilibrium with payoffs $(99,1,0)$.
With player $4$ proposing, in the $(100,0,0)$ case there are additional equilibria with payoffs $(99,0,1,0)$, $(99,0,0,1)$ and $(98,0,1,1)$, where in each case the players offered $1$ threaten to vote "no" if offered $0$. In the $(99,1,0)$ case, there are additional equilibria with payoffs $(98,0,1,1)$, $(98,0,2,0)$ and $(97,0,2,1)$. In summary, the possible equilibrium payoffs at this stage are $(100-x-y,x,y)$, with $0\le x\le2$ and $0\le y\le1$.
At the next stage, with player $3$ proposing, there is for the first time a choice of majorities, and player $3$ can pick the cheapest majority to bribe. The equilbrium payoffs are:
$$
\begin{array}{c|c}
\text{stage }4&\text{stage }3\\\hline
(100,0,0,0)&(100,0,0,0,0),(99,0,0,0,1),(99,0,1,0),(99,1,0,0),\\&(98,0,1,1),(98,1,0,1),(98,1,1,0)\\\hline
(99,0,0,1)&(99,0,0,0,1),(98,0,0,1,1),(98,0,1,0,1),(97,0,1,1,1)\\\hline
(99,0,1,0)&(99,0,0,1,0),(98,0,0,1,1),(98,0,1,1,0),(97,0,1,1,1)\\\hline
(98,0,1,1)&(98,0,0,1,1),(97,0,0,1,2),(97,0,0,2,1),(97,0,1,1,1),\\&(96,0,1,2,1),(96,0,1,1,2)\\\hline
(98,0,2,0)&(98,0,0,2,0),(97,0,1,2,0),(97,0,0,2,1),(97,0,1,2,1)\\\hline
(97,0,2,1)&(97,0,0,2,1),(96,0,1,2,1),(96,0,0,2,2),(96,0,1,2,2)\\
\end{array}
$$
I'll leave the boring details of the first two stages to you :-) |
H: Is this operator monotone?
Consider a convex optimization problem.
$$\min_{u\in\Re^k} f(u)$$
s.t. $g_i(u)\leq0,\ i=1,\ldots,m$
Let $F(x)=F(u,\lambda)=(f'(u)+\sum_{i=1}^m\lambda_ig_i'(u),-g_1(u),\ldots,-g_m(u)):\Re^n\rightarrow\Re^n$, ($n=k+m$)
$G=\{x=(u,\lambda)\in\Re^n:\lambda\geq0\}$.
$f$ and $g_i$ are all convex. For simplicity, assume they belong to $C^1$.
Is $F(x)$ a monotone operator in $G$,
i.e., $(F(x_1)-F(x_2))^T(x_1-x_2)\geq0$ always satisfied in G?
Thanks a lot.
AI: First of all, notice that we have a sum of operators: $(f', 0)$ and $m$ operators of the form $(\lambda g' , -g)$. I claim that each of them is monotone. For the first one this is clear: it is essentially the gradient of convex function $f$. To deal with the rest, recall the inequality $g(v)\ge g(u)+g'(u)(v-u) $ which expresses the convexity of $g$. It follows that in the expression $$(\lambda g'(u)- \mu g'(v)) (u-v) + (-g(u)+g(v)) (\lambda -\mu)$$
the coefficients of $\lambda$ and of $\mu$ are both nonnegative. $\Box$ |
H: 'Quantum' approach to classical probability
Quantum mechanics defines a state of a system as a superposition of 'classical' states with complex coefficients, thus reducing many problems to linear algebra. Can classical statistics be approached to that way? Is this approach useful?
AI: Yes. You can define a(n algebra dual to a) noncommutative probability space as a complex $^{\ast}$-algebra $A$ together with a positive linear $^{\ast}$-functional $\mathbb{E} : A \to \mathbb{C}$ (the expectation). The quantum examples occur when $A$ is a suitable $^{\ast}$-algebra of linear operators on a Hilbert space and the classical examples occur when $A$ is a suitable $^{\ast}$-algebra of integrable functions on a measure space $X$.
I am still thinking about the merits of this approach for classical probability but I think it is better suited for making certain kinds of more algebraic arguments (e.g. it is perfectly suited for thinking about convergence in the sense of moments). Terence Tao's notes on free probability contain some good motivation for this point of view.
Let me give an example. Suppose I want to randomly choose a conjugacy class in a compact Hausdorff group $G$. Classically I would have to construct Haar measure on $G$ to do this. In the algebraic setting, by Peter-Weyl the span of the characters of finite-dimensional unitary irreducible representations of $G$ is dense in the space of class functions, and the integral of any such character over $G$ is uniquely determined by representation-theoretic concerns, so as long as I only care about integrating class functions on $G$ that I know how to express in terms of characters, I can do representation theory instead of constructing the Haar measure!
In this way the problem of, for example, computing moments of various random variables (e.g. the trace in some irreducible representation $V$) reduces to the problem of decomposing tensor products of representations of $G$. For a more specific example see my answer to this MO question.
This observation addresses an interesting question, which is why the category $\text{Rep}(G)$ of, say, finite-dimensional unitary representations of $G$ behaves like a categorified inner product space, with $\text{Hom}_G(V, W)$ categorifying the inner product and so forth. The answer is that what it actually behaves like is a categorified (algebra dual to a) noncommutative probability space, namely Haar measure on the conjugacy classes of $G$ (it more or less categorifies the algebra of class functions), and any such thing is a semi-inner product space with semi-inner product given by
$$\langle a, b \rangle = \mathbb{E}(a^{\ast} b).$$
Here the expectation $\mathbb{E}$ is categorified by taking invariant subspaces $\text{Inv}$, and we have $\text{Hom}_G(V, W) \cong \text{Inv}(V^{\ast} \otimes W)$ naturally.
Let me give a perhaps more accessible example. Suppose I have a random variable $X$ and I want to talk about an independently identically distributed sequence $X_1, X_2, ...$ of random variables all of which are distributed like $X$. Classically I would have to construct an appropriate measure space on which all of the $X_i$ are functions. Algebraically it is enough to start with an algebra $A$ as above in which $X$ sits and consider the "tensor product"
$$A_1 \otimes A_2 \otimes ...$$
of countably many copies of $A$. It is worth emphasizing that this does not agree with the tensor product of underlying vector spaces; its elements are linear combinations of finite tensor products $a_{i_1} \otimes ... \otimes a_{i_n}$ where $a_{i_j} \in A_{i_j}$ and $i_1 < i_2 < ... < i_j$. The expectation is defined multiplicatively on pure tensors (so that the variables coming from different $A_i$ are independent as desired). In fact this strategy works with no modification for an arbitrary collection of i.i.d. variables, whereas classically my impression is that there are some technical issues with defining an arbitrary product of measure spaces (although I could be mistaken here). |
H: Solving $|a| < |b|$
I apologize if this question in general, but I've been having trouble finding solutions as Google discards absolute value signs and inequality symbols.
I am looking for a way to eliminate absolute value functions in $|a| < |b|$.
I can solve $|a| < b$ and $|a| > b$, but I am unsure what method / combination of methods to use to eliminate absolute value signs from both sides.
Thank you!
An example problem:
$$|x + 2| < |x - 4|$$
AI: We have $|a| \lt |b|\,$ if any of these is true:
(i) $\,a$ and $b$ are $\gt 0$ and $a \lt b$
(ii) $a\lt 0$ and $b\ge 0$ and $-a \lt b$
(iii) $b \lt 0$ and $a \gt 0$ and $a \lt -b\,$
(iv) $\,a\lt 0$ and $b\lt 0$ and $-a\lt -b$. We can rewrite this as $b \lt a$.
Four cases! Not surprising, since eliminating a single absolute value sign often involves breaking up the problem into $2$ cases.
Sometimes, one can exploit the simpler $|a| \lt| b|\,$ iff $\,a^2\lt b^2$. But squaring expressions generally makes them substantially messier.
Added: With your new sample problem, squaring happens to work nicely. We have $|x+2| \lt |x-4|$ iff $(x+2)^2 \lt (x-4)^2$. Expand. We are looking at the inequality
$$x^2+4x+4 \lt x^2-8x+16.$$
The $x^2$ cancel, and after minor algebra we get the equivalent inequality $12x \lt 12$, or equivalently $x\lt 1$. The squaring strategy works well for any inequality of the form $|ax+b| \lt |cx+d|$.
But the best approach for this particular problem is geometric. Draw a number line, with $-2$ and $4$ on it. Our inequality says that we are closer to $-2$ than we are to $4$. The number $1$ is halfway between $-2$ and $4$, so we must be to the left of $1$. |
H: Generating random string
I'm working on a small web service that will require users to upload files. The file type is P12. When the file is uploaded and I need to save it on my server, I need to generate a random name for it. I can have the server generate a long random string of letters and numbers called a UUID. Here's a UUID I generated:
5025e8c423cad (so the file would be named 5025e8c423cad.p12)
I read online that a UUID would have to be generated 1 billion times per second for the next 100 years in order to have a 50% chance of being generated again. I'm assuming that Facebook uses something similar, if not UUIDs, to name photos when people upload them. If it stayed around for a very long time, Facebook could meet the maximum (and photos would be overwritten if they did not do something).
This might be a stupid question, but if I changed the name from uuid.p12 to uuid-uuid.p12, what would be the probability of generating the same one after one hundred years? I was quick to think 25%, but after spending a minute thinking about it, I would assume the percentage would be much, much lower.
5025e8c423cad.5025ea6c54cd1.p12
AI: Suppose that there are $n$ IDs available, and we generate $m$ IDs at random. If $m$ is much smaller than $n$, then the probability of no duplicate is approximately
$$e^{-\frac{m^2}{2n}}.\tag{$1$}$$
For an informal description of this result, and some references, see the Wikipedia article on the Birthday Problem.
In your case, you can compute the $n$ for your current ID scheme. To determine how many times we can randomly generate IDs so that the probability of no duplicate is about $50\%$, we find the $m$ that satisfies the equation
$$1-e^{-\frac{m^2}{2n}}=0.5.$$
Suppose now that we double the length of IDs. The number $N$ of IDs available is now $n^2$. Let $M$ be the new number of times we can generate IDs with the probability of a duplicate equal to $50\%$. Then we want
$$\frac{m^2}{2n}=\frac{M^2}{2N}.$$
Using $N=n^2$, we find that
$$M= m\sqrt{n}.$$
So the number of times we can "safely" sample (if $50\%$ is safe!) is multiplied by the extremely large number $\sqrt{n}$. If you change the definition of "safe" to allow only a probability $0.001$ of a duplicate, the same relative calculations holds, one multiplies by $\sqrt{n}$.
To calculate the probability of no duplicate if we double the length of IDs, leaving $m$ alone, simply replace $n$ by $n^2$ in formula $(1)$. Thus the probability of a duplicate is $1-e^{-\frac{m^2}{2n^2}}$. But when $x$ is close to $0$, $e^x$ is approximately $1+x$, so the probability of a duplicate with the double ID scheme is approximately
$$\frac{m^2}{2n^2}.$$
Since $\frac{m}{2n}$ is approximately $0.5$, we get the estimate $\dfrac{0.5}{n}$ for the new probability of no duplicate.
Remark: The formula we have used is only approximate, but it is a pretty good approximation. You may want to check what probability it predicts for the classical Birthday Problem, where $n=365$ and the "break even" $m$ is about $23$. |
H: What is the structure of invariant matrix polynomials?
Let the field $\mathbb F$ be either $\mathbb R$ or $\mathbb C$ and $M_n(\mathbb F)$ all $n \times n$ matrixes.
We denote by $I_n(\mathbb F)$ the space of all functions:
$$P : M_n(\mathbb F) \rightarrow \mathbb F$$
which are polynomial (in the sense that $P(A)$ is a polynomial in the entries of $A$), and which are invariant under the conjugation, i.e. $P(gAg^{-1}) = P(A)$ for all $g \in GL_n(\mathbb F)$.
For each $p \ge 0$, we define $\Sigma_p \in I_n(\mathbb F)$ as $\Sigma_p(A)=\mathrm{Tr}(A^p)$.
Then do we have an isomorphism of algebras
$I_n(\mathbb F) \cong \mathbb F[\Sigma_0,\Sigma_1,\dots,\Sigma_n]$?
AI: Yes. This is almost true in general, but not quite for two reasons: one involving the size of the field and the other involving characteristic.
The correct statement over an infinite field $F$ is that the algebra of invariant polynomials is generated by the elementary symmetric functions of the eigenvalues, or more concretely by the coefficients of the characteristic polynomial $\det (I - At)$. To see this, pass WLOG to the algebraic closure $\bar{F}$ (which we can do because $P$ is polynomial). The identity $P(gAg^{-1}) = P(A)$ is a polynomial identity in the coefficients of $A$, the coefficients of $g$, and $\det g$, and since $F$ is an infinite field two polynomials which are equal when all possible elements of $F$ are plugged in are equal identically. Consequently, this identity continues to hold over $\bar{F}$. (This part of the argument can be ignored over $\mathbb{C}$.)
Now suppose that $M$ is a diagonalizable matrix. Then $P(M)$, being conjugation-invariant, is necessarily a polynomial, in fact a symmetric polynomial, in the eigenvalues of $M$. Since the diagonalizable matrices over $\bar{F}$ are Zariski-dense (as they contain the matrices with distinct eigenvalues, which is a Zariski-open condition), it follows that $P$ must be a symmetric polynomial in the eigenvalues of $M$ identically. Hence $P$ is a polynomial in the elementary symmetric polynomials with coefficients in $\bar{F}$, but using the fact that $P$ is defined over $F$ and letting $M$ be a suitable collection of companion matrices shows that $P$ is a polynomial in the elementary symmetric polynomials with coefficients in $F$.
The above argument does not work over a finite field because the identity $P(gAg^{-1}) = P(A)$ is no longer guaranteed to continue to hold over the algebraic closure, although I don't know a counterexample in this setting.
The result you want is true in characteristic zero or characteristic greater than $n$ by Newton's identities but false in small positive characteristic because using Newton's identities will require dividing by zero. For example, in characteristic $2$, unlike in higher characteristic, it is not possible to express
$$e_2 = \sum_{i < j} \lambda_i \lambda_j$$
in terms of
$$p_1 = \sum_i \lambda_i$$
and
$$p_2 = \sum_i \lambda_i^2 = (\sum_i \lambda_i)^2 = p_1^2$$
(where $\lambda_i$ are the eigenvalues). |
H: A basic question related with the positive definite matrix
I have a one doubt related with positive definite matrices.
Suppose that we have an arbitrary non zero matrix $A$ . Can we find such matrix $B$ which may depend on $A $such that product $AB$ is always a positive definite matrix irrespective of the nature of matrix $A$?
I need help with this.
Thanks
AI: It's not entirely clear from the question whether $B$ can depend on $A$.
If it can, the answer is "almost"; you can choose $B=A^\top$; then
$$x^\top ABx=x^\top AA^\top x=(A^\top x)^\top(A^\top x)\ge0\;,$$
so $AB$ is positive semi-definite, though not definite if $A$ is singular.
If it can't, the answer is "no", since if $AB$ is positive definite then $(-A)B$ is negative definite. |
H: Determing if two k subsets are disjoint given the product of their elements
Consider the following problem (phrased with the use of a black box).
You choose $n$ numbers $X = \{x_1,\ldots,x_n\}$ and pass it to a black box that returns a list $Y = \{y_1,\ldots,y_m\}$ where each of the elements is a product of the elements of some $k-subset$ of $X$. That is for every $1 \leq i \leq m$ the black box picks an element $ Y_i \in {X \choose k}$ and sets $$y_i = \prod_{p \in Y_i} p.$$
Given that I am able to choose the elements of $X$ I am wondering how should I pick them in order to be able to compute the number of intersecting $(Y_i,Y_j)$ pairs as efficiently as possible.
One thing is take $X$ to contain only pairwise coprime elements and then check the if the pairs $(y_i,y_j)$ are coprime or not.
I am wondering, if there is something even more slicker that can be done that would then allow to compute the number of intersecting pairs in say $O(m)$ or at least faster than $O(m^2)$?
AI: I'm not sure if this answers your question, but you could try making each $x_i=2^{2^{i-1}}$. That is, $X=\{2,4,16,256,\ldots\}$ , and each element is the square of the previous one. Now, each of the $y_i$ will be a power of 2, where the binary representation of the exponent will indicate which of the $x_i$ comprise the product. I'm not sure what order the computation would be to evaluate this, but I imagine that it would be better than prime factorisation. |
H: Approximate rational dependence
After seeing question Why is $10\frac{\exp(\pi)-\log 3}{\log 2}$ almost an integer? I wonder if there is an algorithm that can find approximate rational dependence?! I pick any irrational numbers $\alpha$, $\beta$, $\gamma$, $ \delta$ and the task is to find successively good approximations for rational numbers $q_i$ in
$q_1\alpha+q_2\beta+q_3\gamma+q_4\delta\approx0$
Basically it's like continued fraction approximations, but for more irrational numbers.
What is such an algorithm called?
AI: It's called an integer relation algorithm. In addition to LLL mentioned in the comments, there's also PSLQ and others. The Wikipedia article on integer relation algorithms will get you started on understanding these algorithms. |
H: Any software package with Mordell equations implemented?
I admit I have no idea how to tag this post, but I'm looking for a CAS/number theory software package that would implement a decent algorithm for computing the integral solutions to
$x^2 = y^3 - k$,
$k \in \mathbb{Z}_+$ a parameter taking values up to order of $10^5-10^6$, say? Since these are so frequent in "math contests", it would be a help for people working with such things, not to say that it's also of interest in itself.
This came to my mind some time ago and it'll be nice to see it implemented somewhere, since I couldn't find anything of the kind. Maybe I should try in a cryptography-related area, but where to start?
You can close this if software recommendations are contrary to the site rules. If this is "too localized" and trivial, at least I'd like to know if it stays within the confines of classical elliptic curves theory and algorithms.
AI: Maple has a package called ellpack for handling elliptic curves. Also if you type $$\rm Cremona\quad elliptic$$ into the web, many useful links will appear.
I also recommend the paper, Gebel, Petho and Zimmer, On Mordell's equation. According to the summary, they solve Mordell for all $k$ up to 10,000, so you may be asking for a bit much if you want to go up to 1,000,000. |
H: Order of a set $X$ acted upon transitively by the Symmetric Group
Suppose the symmetric group $S_n$ acts transitively on a set $X$, i.e. for every $x, y \in X$, $\exists g \in S_n$ such that $gx = y$.
Show that either $|X| \le 2$ or $|X| \ge n$.
Small steps towards the solution:
As $S_n$ acts transitively on a set $X$, the whole of $X$ is one single orbit under the action of $S_n$.
By the Orbit-Stabilizer Theorem, then, $|X|$ = $|S_n : \text{Stabilizer of }x|$ for any $x \in X$.
We also know that the Stabilizer of any $x \in X$ is a subgroup of $S_n$.
When $|X| = 2$, the Stabilizer of $x$ is the alternating group $A_n$.
I'm halfway but can't get the final result. Any help would be much appreciated, as always. Thank you.
AI: For $n < 4$ the result is clear. For $n = 4$ the result is false - we have a surjection $S_4 \to S_3$ by killing the unique normal subgroup of order 4, given by the products of transpositions (thanks to cocopuffs for fixing an error in this before).
For $n > 4$, the map $S_n \to \text{Aut}_\text{Set}(X) = S_{\# X}$ induced by your action must either be injective or have kernel $A_n$ or $S_n$. In the first case we must have $\#X \ge n$ and in the others we have $\#X = 2$ or $1$ respectively. |
H: Graph isomorphism as permutation matrix.
The automorphism group of a graph (lets us consider undirected) is the set of all permutation on vertices that
preserve the adjacency. It is claimed: automorphism group of graph may be equivalently
defined as the set of permutation matrices $\pi$ which commute with the adjacency
matrix. How can we justify this claim.
Thank you.
AI: Fun facts: (i) The $(i,j)$th entry of any matrix $M$ is equal to $e_i^TMe_j$, where $e_i$ is the vector with $1$ in the $i$th entry and $0$ elsewhere. (ii) $e_{\pi(i)} = Pe_i$, where $P$ is the matrix corresponding to the permutation $\pi$. (iii) Every permutation matrix $P$ is orthogonal, i.e. $P^T = P^{-1}$.
Let the adjacency matrices of the original and permuted graphs be $A$ and $B$. We want the $(\pi(i),\pi(j))$th entry of $B$ to be the same as the $(i,j)$th entry of $A$ is $1$. Equivalently, we want $(Pe_i)^TB(Pe_j) = e_i^TAe_j$.
For this to hold for all $i$ and $j$, we must have $P^TBP = A$, or $B = PAP^T$.
If the permutation preserves adjacency, then $A = B = PAP^T$, so $AP = PAP^TP = PA$. Therefore $P$ commutes with $A$. |
H: What are the solutions of $x^n+a^n=0$?
For what values $a$, the equation $x^n+a^n=0$ has $n$ different solutions? what are the solutions?
(the question refers to complex solutions).
AI: By the Fundamental theorem of algebra, since for every value of $n>0$ this is a polynomial of degree $n>0$ there are exactly $n$ roots (counting multiplicy), in particular there is always $z\in\mathbb{C}$ that satisfies $z^n+a^n=0$.
In order to get all solutions you can use de Moivre's formula
Edit: For $a=0$ we have $x^n=0$ hence $x=0$ is the only solution, for $a\neq 0$ there are $n$ different solutions. |
H: Are there diagonalisable endomorphisms which are not unitarily diagonalisable?
I know that normal endomorphisms are unitarily diagonalisable. Now I'm wondering, are there any diagonalisable endomorphisms which are not unitarily diagonalisable?
If so, could you provide an example?
AI: $$\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$$ will do it. This matrix isn't normal (so by the Spectral theorem, it can't be unitarily diagonalisable) but it is diagonalisable with eigenvalues $1 \pm \sqrt{2}$. |
H: Are compacta in a complete infinite dimensional normed space nowhere dense?
Let $X$ be an infinite dimensional Banach space. I want to show that any compact subset $\varnothing\neq A\subset X$ is nowhere dense.
I've been able to prove the statement for $X=(C[0,1],\|\cdot\|_\infty)$ by using an Arzelà-Ascoli argument. But this is not easy to generalize to an arbitrary infinite dimensional Banach space.
AI: In an infinite-dimensional normed space the (closed) unit ball is always non-compact: using Riesz's Lemma you can construct a sequence of unit vectors $(e_n)_{n=1}^\infty$ such that $\lVert e_n - e_m \rVert \geq 1/2$ for all $m \neq n$. This sequence can't have a convergent subsequence. It follows that no closed ball is compact. More details on the linked Wikipedia page.
Assuming the set $A \neq \emptyset$ is closed and not nowhere dense, it must contain an open ball $B_r(x_0) \subset A$ by definition of “nowhere dense”. Thus, if $X$ is infinite-dimensional, $A$ cannot be compact: otherwise the closed ball $\bar{B}_{r/2}(x_0)$ would have to be compact as a closed subset of $A$, contradicting 1. |
H: Does a polynomial exist such that $|P(x)| < a$ for some real $a$ and all real $x$?
I'm pretty sure the answer is in the negative.
Can someone show me the proof?
AI: The answer is no.
Write
$$
P(x)=ax^n+\text{terms of lower degree}
$$
with $a\neq0$ and $n\geq1$. Then
$$
\lim_{x\to\infty}P(x)=\pm\infty
$$
according to the sign of $a$. This shows that $|P(x)|$ is unbounded. |
H: Rational Functions are Determined By Locations and Multiplicities of Zeros and Poles (why?)
There is a theorem that says rational functions in the extended complex plane are exactly the meromorphic functions.
After this, my textbook draws the corollary:
"...as a consequence, a rational function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeros and poles."
I don't see how it follows. Plus I think I have an example. Take a rational function with no zeros and two poles of multiplicity one, at 0 and 1.
At least two such functions exist:
$f(x)=\frac{1}{x} + \frac{1}{x-1}$
$g(x)=\frac{1}{x} + \frac{2}{x-1}$
AI: The precise statement is the other way round: a function meromorphic on the whole Riemann sphere $S$ is rational.
This is completely false on $\mathbb C$: the function $e^z$ is meromorphic (even holomorphic) on $\mathbb C$, but certainly not rational on $\mathbb C$, nor anywhere else.
Now if $f,g\in Rat(S)$ are rational on the Riemann sphere $S$ and have the same zeros and poles (counted with multiplicities), then the quotient $\phi=f/g$ having neither zeros nor poles is holomorphic on the whole of $S$ and thus constant (say because $\phi$ is bounded on $S$ by compacity, hence bounded on $\mathbb C$ and Liouville's theorem applies).
"Yes,Georges, but I gave you a counterexample"
Your rational functions don't have the same zeros: $$f(\frac {1}{2})=0\neq g(\frac {1}{2})=-2 $$
A very optional remark
That meromorphic functions on $S$ are rational is the simplest example of a very profound result in algebraic geometry: Serre's GAGA principle. |
H: A question about the regular languages being closed under Boolean operation (how to generalize)
I know that if $L_{1},L_{2}$ are regular languages then so is $L_{1}\cap L_{2},L_{1}\cup L_{2}$
are regular languages, I also know that $L$ is regular $\implies L^{c}$
is regular
.
It is easy to generalize in this manner:
Let $L_{1},...L_{n}$ be regular
languages and $x\in\{0,1\}^{n}$ a vector, then $$L_{1}^{x_{1}}\cap...\cap L_{n}^{x_{n}}, L_{1}^{x_{1}}\cup...\cup L_{n}^{x_{n}}$$
is regular, where $L^{0}$
denotes $L^{c}$ and $L^{1}$ denotes $L$, I can also generalize
by changing some of the $\cap$ to $\cup$ .
My question is: How can this be generalized ?
It seems that both statements have some
kind of Boolean operations in the scene that given $L_{1},...,L_{n}$
I can represent those languages with binary vectors (I also noted
that $A\cup B=(A^{c}\cap B^{c})^{c}$ so the two generalizations I
gave seem to be the equivalent to each other)
AI: As you noted, regular languages are closed under arbitrary finite Boolean operations, which gives the following easy corollary:
Given $n$ regular languages $L_1 , \ldots , L_n$ and any function $f : \mathcal{P} ( \{ 1 , \ldots , n \} ) \to \{ 0 , 1 \}$ the language $L$ given by $w \in L$ iff $f ( \{ i \leq n : w \in L_i \} ) = 1$ is also regular.
(To see this, note -- using the notation in the OP -- that all languages of the form $L_1^{x_1} \cap \cdots \cap L_n^{x_n}$, where the $x_i$ are in $\{0,1\}$, are regular. Given $f$ as above, let $( x_{1,1} , \ldots , x_{1,n} )$, $\ldots$, $( x_{k,1} , \ldots , x_{k,n} )$ enumerate the $n$-tuples in the pre-image of $1$. Then the language in question is $$\bigcup_{j=1}^k ( L_1^{x_{j,1}} \cap \cdots \cap L_n^{x_{j,n}} ),$$ which is regular, being a finite union of regular languages.)
Without going into other operations (concatenation, Kleene-star, etc.), I'm not certain if anything more general can be said. |
H: Solving $\phi (n) < (n-1) \cdot \frac{15499}{94744} $
I am working on challenge 243 from Project Euler (PE 243). The question is:
$$\text{Solve } \phi (n) < (n-1)\cdot \frac{15499}{94744}$$
I can calculate $\phi(n)$ for any $n$, but I think the $n$ which solves the problem is larger than the range I can brute force. I have never before worked with $\phi(n)$ before, but I'd love to learn how to solve this kind of problem.
Research on Google gave me definitions of $\phi(n)$, which I already know, but nothing to help me solve the problem. If you could give me any tips in the right direction, and NOT the answer. Thanks in advance.
Edit: I found a clue: $\phi(n) \ge \sqrt{n}$ This should give me a limit where $n$ will always give me a number larger than the desired result. I'm working on it.
AI: Some information that may be useful: Let $n$ have prime power decomposition
$$n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k},$$
where the $p_i$ are distinct primes. Then
$$\varphi(n)=(p_1-1)p_1^{a_1-1}(p_2-1)p_2^{a_2-1}\cdots (p_k-1)p_k^{a_k-1}.$$
(For details about the $\varphi$-function, see this.)
By using the above formula for $\varphi(n)$, we can see that
$$\frac{\varphi(n)}{n-1}=\frac{n}{n-1}\frac{\varphi(n)}{n}=\frac{n}{n-1}\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_k}\right).$$
So to make $\varphi/(n-1)$ smallish "cheaply" it is good to use small primes, all to the first power. Decorating the primes with powers $a_i$ greater than $1$ has only a minor effect on the ratio. |
H: How do I transform the equation based on the condition?
If $q$ and $w$ are the roots of the equation $$2x^2-px+7=0$$
Then $q/w$ is a root of ?
P.s:- It is an another question of How do I transform the equation based on this condition?
AI: The following result (and its generalizations to higher degree polynomials) is very useful. Consider the equation $ax^2+bx+c=0$, where $a\ne 0$. Then the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$. In our case, we have
$$q+w=\frac{p}{2}\quad\text{and}\quad qw=\frac{7}{2}.$$
We will find a quadratic equation whose roots are $\frac{q}{w}$ and $\frac{w}{q}$ without calculating $\frac{q}{w}$ or $\frac{w}{q}$.
Note that
$$\frac{q}{w}+\frac{w}{q}=\frac{q^2+w^2}{qw}=\frac{(q+w)^2-2qw}{qw}=\frac{(q+w)^2}{qw}-2.$$
Substitute our known values of $q+w$ and $qw$. We get
$$ \frac{q}{w}+\frac{w}{q}=\frac{p^2}{14}-2=\frac{p^2-28}{14}.$$
Note that trivially the product of $\frac{q}{w}$ and $\frac{w}{q}$ is $1$.
So the numbers $\frac{q}{w}$ and $\frac{w}{q}$ are roots of the quadratic
$$x^2 -\left(\frac{p^2-28}{14}\right)x +1=0.$$
If we wish, we can change to the equivalent
$14x^2+(28-p^2)x+14=0$. |
H: Nonzero elements of splitting field
Let $F$ be a splitting field of $x^{p^{n}} - x \in \mathbb{Z}_p[x]$.
How is it that the nonzero elements multiply to $-1$ and sum to $0$? I don't get how we get that result.
AI: We appeal to the structure theorem for finite fields: the field $\mathbb{F}_{p^n}$ is precisely the splitting field of the polynomial $x^{p^n} - x$ over the prime field $\mathbb{F}_p$. We thus have a factorisation in $\mathbb{F}_{p^n}$
$$x^{p^n} - x = x (x-a_1) \cdots (x-a_{p^n - 1})$$
and expanding the right hand side and comparing coefficients, assuming $p^n > 2$, we find
\begin{align}
a_1 + \cdots + a_{p^n - 1} & = 0 \\
a_1 \cdots a_{p^n - 1} & = (-1)^{p^n}
\end{align}
which is what we wanted. (Note that the polynomial $x^{p^n} - x$ is separable because its derivative is the constant $1$, so its zeros are all distinct and constitute all the elements of $\mathbb{F}_{p^n}$.) |
H: Pseudocompactness does not imply compactness
It is well known that compactness implies pseudocompactness; this follows from the Heine–Borel theorem. I know that the converse does not hold, but what is a counterexample?
(A pseudocompact space is a topological space $S = \langle X,{\mathfrak I}\rangle$ such that every continuous function $f:S\to\Bbb R$ has bounded range.)
AI: A favorite example (and counterexample) to many things is the first uncountable ordinal $\omega_1$ in its order topology: $[0,\omega_1)$. It is pseudo-compact but not compact. |
H: Constrained optimization: equality constraint
I have this very general problem (for $n>2$):
$$
\begin{align}
& \max Z = f(x_1,\ldots ,x_n) \\[10pt]
\text{s.t. } & \sum_{i=1}^{n} x_i = B \\[10pt]
& x_i \geq 0
\end{align}
$$
Assume that the function $f(x_1,\ldots,x_n)$ is not a "nice" function, non-linear, non-convex, hardly differentiable. Therefore, in general I cannot use the Lagrangian multipliers method.
For finding the $x_i$'s, I am thinking to numerical optimization approaches.
My questions are:
1) Is there a common, smart and effective way to deal with such an equality constraint in numerical optimization?
2) Am I missing something? Do you have any other direction to suggest?
Thanks
AI: The methods to be used will be highly dependent on the character of $f$. If it is non-convex, there are many such algorithms; see this MO post for instance.
You might also wish to look into evolutionary algorithms, such as genetic algorithms and simulated annealing. These algorithms are often much slower, but have the feature that they can sometimes "bump" you out of local extrema. They are also fairly easy to implement.
You can also hybridize approaches: combine an evolutionary algorithm with a standard convex optimization approach on a locally convex subdomain.
Finally, with an equality constraint, you essentially reduce the dimensionality of your problem by 1. That is, one variable is completely determined by the others.
$$x_k = B - \sum_{i=1,\ i\neq k}^n x_i.$$
And then, depending on the character of your function, you might be able to use any sort of algorithm.
But to answer your questions, 1.) there are many common and smart ways, but they depend on your function $f$. I would start with the most simple, conventional approach, and then see whether it is effective. 2.) I don't think you're missing anything. Numerical optimization is a big topic and there are many ways to go about it. |
H: Cantor Bendixson Theorem
Cantor Bendixson Theorem: Every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable.
This definition differs a bit from that in wikipedia.
I have proved that 'If $X$ is a separable metric space and $E$ is a uncountable subset and $P$ is the set of all condensation points of $E$, then $P$ is perfect and $P^c \cap E$ is at most countable'.
Then, you can see that 'every uncountable set in a separable metric space is the union of a nonempty perfect set and a set which is at most countable, and sets are disjoint' (Since $E= P\cup (P^c \cap E)$)
Here, i didn't use the condition 'closed' at all!
Where did i go wrong?
AI: I'll just put my comment into an answer.
The problem comes in the assertion that $E = P \cup ( P^c \cap E)$. In your construction, you have that $P$ is the set of all condensation points of $E$, but for arbitrary $E$ not all condensation points of $E$ belong to $E$. (As a simple example, note that the set of all condensation points of the open interval $(0,1)$ in the real line is the closed interval $[0,1]$.)
If one were to, instead, consider $(P \cap E ) \cup ( P^c \cap E)$ we quickly run into the problem that $P \cap E$ is not perfect -- as the above example would show.
As a conclusion of sorts, one can easily show that $(0,1)$ cannot be represented as the union of a perfect set and a countable set in the real line: Suppose that $P \subseteq (0,1)$ is any perfect set. As $P$ is bounded(-below) and closed, it must have a minimum element, $a$, and $0 < a < 1$. As we all know, $(0,a)$ is uncountable, as we're done as $(0,a) \subseteq (0,1) \setminus P$! |
H: General solution of constant-coefficient second-order linear ODE
I am trying to look a bit deeper into the mathematics the equation of motion used in physics and engineering. I have some specific questions at the end, but please correct me if I make a mistake in
my statement.
I have an ordinary, second-order, linear, homogeneous differential equation:
$$
M \ddot{y} + C \dot{y} + K y = 0
$$
where $M$, $C$ and $K$ are real valued.
I am familiar with an Ad-hoc method of solving this equation. Namely, I assume a solution
takes form of
$$
y_k(t) = Y_k e^{\omega_k t}
$$
substitute the form into the ODE,
$$
\left( M \omega_k^2 + C \omega_k + K\right) y_k = 0,
$$
and solve for $\omega_k$ such that $y_k \neq 0$.
If $y$ is in dimension $N$,
we generally have $2N$ solutions, where for each $k$, $\omega_k$ and its complex
conjugate $\tilde{\omega}_k$ are solutions.
Since our ODE is linear, the general solution is a combination of the individual solutions
$$
y(t) = \sum_{k = 1}^N Y_k e^{\omega_k t} + \tilde{Y}_k e^{\tilde{\omega}_k t}
$$
The values of $Y_k$ and $\tilde{Y}_k$ are determined from initial values, i.e. $y(0) = y_0$,
and further restrictions common in physics, such as $y(t)$ must be real for all $t \geq 0$.
Now here are my questions:
1 - If we select the form $y_k(t) = Y_k e^{ \mathbf{i}\omega_k t}$, am I correct to say that the roots $\omega_k$ and $\tilde{\omega}_k$ are no longer complex conjugates, but $-1$ times their complex conjugates?
2 - How do we know that the general solution cannot contain terms which are not representable by the exponential form? In other words, is our general solution truly general?
AI: Here is another approach to the second part of your question:
Let $x_1 = \dot{y}, x_2 = y$. Assuming $M \neq 0$, you can write your equation as a pair of first order equations:
$$ \dot{x_1} = -M^{-1} (C x_1 + K x_2), \ \ \ \dot{x_2} = x_1, \ \ \ x_1(0) = \dot{y}(0), \ \ \ x_2(0) = y(0). $$
This is a time-invariant linear system, so if we let $x = (x_1,x_2)^T \in \mathbb{R}^{2 }$, then the equation can be written as:
$$\dot{x} = A x = M^{-1}\begin{bmatrix} -C & -K \\ 1 & 0 \end{bmatrix} x, \ \ \ x(0) = x^0 = (\dot{y}(0), y(0))^T.$$
By the existence and uniqueness theorem for ODEs (this is a classical result, for example see Marsden, "Elementary Classical Analysis", Theorem 7.6), the system has a unique solution passing through the point $x^0 \in \mathbb{R}^{2}$.
Hence the initial condition $x^0$ completely defines the solution.
Since this system is linear, the solutions are linear functions of the initial state, which simplifies matters considerably. We need only consider solutions starting from a basis of $\mathbb{R}^2$. In particular, all solutions are linear combinations of these two solutions (because of linearity).
To see why these two solutions have exponentials, you need to consider two
cases:
(1) If $A$ is diagonalizable, then the eigenvectors of $A$ can form a basis of $\mathbb{R}^2$, and it is straightforward to check that $t \mapsto e^{\lambda_i t} v_i$ are solutions, where $v_i$ are the eigenvectors corresponding to the eigenvalues $\lambda_i$.
(2) If $A$ is not diagonalizable, then it has only one eigenvalue, say $\lambda$, and we can find a basis $u,v$ of $\mathbb{R}^2$ that satisfies $A v = \lambda v$ and $A u = \lambda u +v$ (cf. the Jordan form). Then it is straightforward to check that $t \mapsto e^{\lambda t}v$ and $t \mapsto e^{\lambda t} (u + tv)$ are solutions.
Hence all solutions are of exponential form (including, of course, multiplying by $t$ as well). |
H: Continuity of $f:X\to [0,1]$ where $X$ is homeomorphic to the Cantor set.
This is an exercise from Mendelson's Introduction to Topology, page 101.
THEOREM
For each $n\in \Bbb N$, let $X_n$ be the discrete two-point topological space $\{0,2\}$. Define the product space $X=\prod\limits_{n\in \mathbb N}X_n\;$. Let $f:X\to [0,1]$ be defined as
$$f(x)=\sum_{n=1}^\infty \frac{x(n)}{3^n}$$
Then $f$ is one-one and continuous.
PROOF (Revised)
We first prove $f$ is one-one. To this end, suppose we had $$\sum_{k\geqslant 1}\frac{x_n}{3^n}=0$$ with $x_n\in\{0,-2,2\}$. I claim first there cannot be any $K$ with $x_K=-2$. Indeed, let $K$ be the least index with $x_K=-2$. I claim that $x_k=0$ if $k<K$. Indeed, because $$ - \sum\limits_{k = K}^\infty {\frac{2}{{{3^k}}}} = - \frac{1}{{{3^K}}}\sum\limits_{k = K}^\infty {\frac{2}{{{3^{k - K}}}}} = - \frac{2}{{{3^K}}}\sum\limits_{k = 0}^\infty {\frac{1}{{{3^k}}}} = - \frac{1}{{{3^K}}}$$
there is no way the sum would equal zero if some $x_k$ with $k<K$ was nonzero. But with this out of the way, it is impossible the sum is zero. By the same token, in the extreme case we have $$ - \frac{2}{{{3^K}}} + \sum\limits_{k = K + 1}^\infty {\frac{2}{{{3^k}}}} = - \frac{2}{{{3^K}}} + \frac{1}{{{3^{K + 1}}}} < 0$$
This means there cannot exist $k$ with $x_K=-2$. Thus $x_k\in \{0,2\}$. Since the sum is zero, all nonnegative terms must be zero, so the sequence is identically zero.
We now prove $f$ is continuous. Let $a\in X$ and $\epsilon >0$ be given. The claim is that given the open ball $B(f(a);\epsilon)$, we can choose $k$ so that
$$\tag 1 P_k=\bigcap_{i=1}^k p_i^{-1}(a(i))\subset f^{-1}(B)$$
where $p_i:X\to \{0,2\}$ is the projection to the $i$th coordinate. Since $f$ is one-one, $(1)$ is the same as
$$f(P_k)\subset B$$
Note that for any point $x\in P_k$, the difference $|f(x)-f(a)|$ is at most $3^{-k}$, precisely when $x(n)=2$ and $a(n)=0$ for $n\geq k+1$ (or viceversa):
$$\sum\limits_{n = k + 1}^{ + \infty } {\frac{2}{{{3^n}}}} = \frac{1}{{{3^k}}}$$
Thus we may choose $k$ such that $3^{-k}<\epsilon$. It will follow that $f(P_k)\subset B$, and $f$ will be continuous.
AI: Hint: Pick $N$ so that $$\sum_{n=N}^\infty \frac{1}{3^n}<\epsilon$$ |
H: What does asymptotically optimal mean for an algorithm?
What does it mean to say that
heap sort and merge sort are asymptotically optimal comparison sorts .
I know What the Big O , Big Omega($\omega)$ and Theta($\theta$) notations are and I also know why these two sorts are called comparison sorts ? But I am not able to understand why they are called asymptotically optimal ? Why doesn't this asymptotically optimal property also apply to quick sort ?
P.S: I had a look in this question but still I am not clear about the definition of asymptotically optimal .
AI: Computing a simple function that characterizes the behavior of an algorithm for any input is difficult. So upper and lower bounds are used, in addition to probabilistic methods.
The term 'asymptotic' is used because the bound is typically only valid for large $n$, as an accurate formula for small $n$ is more complicated, and not really interesting from a performance perspective.
To answer your question, you can prove that any working comparison sort takes a lower bound of $\Omega(n \lg n)$ on running time. Consequently, if you can show that a sorting algorithm (such as heapsort or merge sort) has an upper bound of $O(n \lg n)$ on running time, then in fact you have a running time of $\Theta(n \lg n)$ which is, by definition, an asymptotically tight bound for the running time (See Cormen, Leiserson & Rivest's "Introduction to Algorithms", Chapter 2, Theorem 2.1, and the section on sorting).
Quicksort has a worst-case running time of $\Theta(n^2)$, which is larger than the above bound, hence is not optimal.
Note, however, that quicksort has an average running time of $\Theta(n \lg n)$. (This, of course, assumes that all inputs of length $n$ are equally likely.) |
H: Congruence for embeddings of a poset into semilattices
Let $\mathfrak{A}$ be a poset, $\mathfrak{B}$ and $\mathfrak{C}$ be meet-semilattices with least elements. Let $f:\mathfrak{A}\rightarrow\mathfrak{B}$ and $g:\mathfrak{A}\rightarrow\mathfrak{C}$ are order embeddings.
Can we warrant that $f(x)\cap^{\mathfrak{B}}f(a) = f(y)\cap^{\mathfrak{B}}f(a) \Leftrightarrow g(x)\cap^{\mathfrak{C}}g(a) = g(y)\cap^{\mathfrak{C}}g(a)$ for every $x,y,a\in\mathfrak{A}$? How to prove this?
AI: I think this is a counterexample:
The posets are $\mathfrak A, \mathfrak B$ and $\mathfrak C$ respectively, from left to right, with the obvious order embeddings $f: \mathfrak A \hookrightarrow \mathfrak B$ and $g: \mathfrak A \hookrightarrow \mathfrak C$. Let $a \in \mathfrak A$ be the element in the middle and $x,y \in \mathfrak A$ the other two elements to the left and right of $a$.
Then $f(x) \cap^{\mathfrak B} f(a) = f(y) \cap^{\mathfrak B} f(a)$, but $g(x) \cap^{\mathfrak C} g(a) \neq g(y) \cap^{\mathfrak C} g(a)$. |
H: Questions regarding the quadratic form $q: \bigwedge^2 \mathbb{R}^4 \to \bigwedge^4 \mathbb{R}^4, x \mapsto x \wedge x$
An exercise of my last year's linear algebra class asks as follows:
Determine the type of the quadratic form $q: \bigwedge^2 \mathbb{R}^4 \to \bigwedge^4 \mathbb{R}^4, x \mapsto x \wedge x$.
Question 1: We have only introduced quadratic forms of bilinear forms and I can't recall our professor talking about different "types" of quadratic forms. What types of quadratic forms are there?
Question 2: In the standard solution, one is adviced to proceed as follows: We have $\dim(\bigwedge^2 \mathbb{R}^4) = 6$ and thus the quadratic form will be represented by a 6x6-matrix no matter what basis we choose. We now choose the basis
$$\mathcal{B} = (e_1 \wedge e_2, e_1 \wedge e_3, e_1 \wedge e_4, e_2 \wedge e_3, e_2 \wedge e_4, e_3 \wedge e_4)$$
and by plugging $\mathcal{B}$ into the quadratic form, we arrive at the matrix
$$M_\mathcal{B}(q) = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}.$$
I don't understand how we arrive at this matrix. Can anyone explain?
AI: The only way to obtain a nontrivial 4-form in $\mathbb{R}^4$ is to have some scalar multiple of the top-form $e_1 \wedge e_2 \wedge e_3 \wedge e_4$. As you examine the basis you have provided you'll see that this top-form is only obtained via certain products.
Denoting $f_1 = e_1 \wedge e_2$ etc... $f_6 = e_3 \wedge e_4$ we calculate
$f_1 \wedge f_6 = e_1 \wedge e_2 \wedge e_3 \wedge e_4$. On the other hand, $f_1 \wedge f_j=0$ for $j \neq 6$.
Continuing, examine $f_2 = e_1 \wedge e_3$ with $f_5=e_2 \wedge e_4$ gives the nontrivial $f_2 \wedge f_5 = e_1 \wedge e_3 \wedge e_2 \wedge e_4 = -e_1 \wedge e_2 \wedge e_3 \wedge e_4$ and in contrast, $f_2 \wedge f_j = 0$ for $j \neq 5$.
You can easily see that $f_3 \wedge f_4 = e_1 \wedge e_2 \wedge e_3 \wedge e_4$ and $f_3 \wedge f_j = 0$ for $j \neq 4$.
Since the wedge product of two forms commutes it follows that the results above capture all possible products. Let the matrix $M$ be defined implicitly via the equation $M_{ij}e_1 \wedge e_2 \wedge e_3 \wedge e_4 = f_i \wedge f_j$.
Abusing notation considerably, $Q(x) = x^TMx$ where I am replacing the two-form $x$ with it's column vector and identifying the top-form $e_1 \wedge e_2 \wedge e_3 \wedge e_4$ with $1$. As an example, $x = f_1+f_6 = [1,0,0,0,0,1]^T$ and we calculate,
$$ Q(x) = [1,0,0,0,0,1]\left[ \begin{array}{cccccc}
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 1 \\
0 & -1 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 1
\end{array} \right]\left[ \begin{array}{c} 1 \\ 0 \\0 \\0 \\ 0 \\ 1 \end{array} \right] = 2 $$
Which is precisely what we expect from direct computation of $Q$;
$$(f_1+f_6) \wedge (f_1+f_6) = f_1 \wedge f_1 + f_1 \wedge f_6+f_6 \wedge f_1+ f_6 \wedge f_6 = 2e_1 \wedge e_2 \wedge e_3 \wedge e_4.$$ |
H: Second degree Diophantine equations
I found a question whether there are general methods to solve second degree Diophantine equations. I was unable to find an answer so is this known? In particular, the original writer wants to know whether one can find all integers satisfying $x^2 + x = y^2 + y + z^2 + z$.
AI: About algorithm: There is an algorithm that will determine, given any quadratic $Q(x_1,\dots,x_n)$ as input, whether or not the Diophantine equation $Q(x_1,\dots,x_n)=0$ has a solution. This is something that I (and others) observed quite a long time ago. I have no knowledge about a nice algorithm.
Set one machine $M_1$ to search systematically for solutions. Another machine $M_2$ simultaneously checks whether there is a real solution (easy) and then checks systematically for every modulus $m$ whether there is a solution modulo $m$.
By the Hasse Principle (which in this case is a theorem), if our equation has "local" solutions (real and modulo $m$ for every $m$) then it has an integer solution. So either $M_1$ will bump into a solution or $M_2$ will find a local obstruction to a solution. Thus the algorithm terminates.
The corresponding question for cubics is unsolved. The same question for quartics (in arbitrarily many variables) is equivalent to the general problem of testing a Diophantine equation for solvability, so is recursively unsolvable.
Added: I think that the details are written out in the book Logical Number Theory I by Craig Smorynski. Very nice book, by the way. |
H: Quantum Information: Tensor with Outer Product
If $\mid\phi\rangle=a_0\mid 00\rangle + a_1\mid 01\rangle +a_2\mid 10\rangle +a_3\mid 11\rangle$ and $P_0=\mid 0\rangle\langle 0\mid \otimes I$, how do we show that $\langle\phi \mid P_0\mid \phi\rangle =\langle\phi \mid a_0\mid 00\rangle+\langle \phi\mid a_1\mid 01\rangle =|a_0|^2+|a_1|^2$
I get the following if we only consider the first quibit (assuming that's what $\otimes I$ does,.... I'm still not sure on that either) :
$(a_0 \langle 0\mid +a_1\langle 0\mid)(\mid 0\rangle\langle0\mid)(a_0\mid0\rangle + a_1\mid0\rangle)=$
$(a_0 \langle 0\mid +a_1\langle 0\mid)(a_0\mid 0\rangle\langle0\mid0\rangle+a_1\mid 0\rangle\langle0\mid0\rangle$)$ = a_0^2+2a_0a_1+a_1^2$
AI: Here's the complete calculation:
First, use linearity to get the individual terms:
$$\begin{aligned}
\langle\phi\rvert P_0\lvert\phi\rangle
&= \left(a_0^*\langle00\rvert + a_1^*\langle01\rvert + a_2^*\langle10\rvert
+ a_3^*\langle11\rvert\right) P_0 \left(a_0\lvert00\rangle
+ a_1\lvert01\rangle + a_2 \lvert10\rangle+a_3\lvert11\rangle\right)\\
&= a_0^*a_0\langle00\rvert P_0\vert00\rangle
+ a_0^*a_1\langle00\rvert P_0\vert01\rangle
+ \dots + a_3^*a_3\langle11\rvert P_0\vert11\rangle
\end{aligned}$$
Note that $\lvert ij\rangle = \lvert i\rangle\otimes\lvert j\rangle$. Thus we have
$$\begin{aligned}
\langle ij\rvert P_0 \lvert kl\rangle
&= \langle ij\rvert\left(\lvert0\rangle\langle0\rvert\otimes I\right)\lvert kl\rangle\\
&= \langle i\rvert0\rangle\langle0\rvert k\rangle\cdot \langle j\rvert I\lvert l\rangle\\
&= \delta_{i0}\delta_{k0}\delta_{jl}
\end{aligned}$$
That is, if all factors are tensor products, you just evaluate every term separately and get the tensor product of the results (the tensor product of numbers is just the normal product of numbers). This is a general rule, so for example $(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$, or $(A\otimes B)(\lvert\phi\rangle\otimes\lvert\psi\rangle) = (A\lvert\phi)\otimes (B\lvert\psi\rangle)$.
Now it's just a matter of inserting that in the first equation and noticing that the only non-zero terms are those with $i=k=0$ and $j=l$, and the corresponding terms have prefactors $a_0^*a_0=\left|a_0\right|^2$ and $a_1^*a_1=\left|a_1\right|^2$. Thus you finally get
$$ \langle\phi\rvert P_0\lvert\phi\rangle
= \left|a_0\right|^2 + \left|a_1\right|^2$$ |
H: Geometric meaning of a nondegenerate critical point
Let $f\!:M\!\rightarrow\!\mathbb{R}$ be a smooth function on a manifold and $p\!\in\!M$. Is there any way to geometrically/visually characterize the conditions
$p$ is a critical point (i.e. $D(f)_p\!=\!0$) and
$p$ is a nondegenerate critical point (i.e. $\det D^2(f)_p\!\neq\!0$)?
If $f(x)=\langle x,a\rangle$ is a height function on a surface, then $f$ is linear, so $D(f)_p\!=\!f\!:\, T_pM\rightarrow T_p\mathbb{R}\!=\!\mathbb{R}$. Thus $f$ is the zero map at those $p$ for which $T_pM$ is perpendicular to $a$, i.e. the critical points of $f$ are those points at which $T_pM=a^\bot$. But what about higher dimensions and different $f$s?
And what about nondegeneracy?
AI: Consider the case where $M$ is an open subset of $\mathbb{R}^2$ and visualize $f$ by visualizing its graph in $\mathbb{R}^3$. In the neighborhood of a point $p$, the function $f$ has an expansion into Taylor series. Then
$p$ is a critical point if this expansion has no linear terms. This occurs whenever $f$ attains a local minimum or a local maximum but it can also occur e.g. if $p$ is a saddle point.
$p$ is a nondegenerate critical point if the quadratic terms of the Taylor series expansion give a nondegenerate quadratic form. There are three cases: $p$ is a local minimum (so think locally $f(x, y) = x^2 + y^2$) $p$ is a local maximum (so think locally $f(x, y) = - x^2 - y^2$, or $p$ is a saddle point (so think locally $f(x, y) = x^2 - y^2$). This follows from the three possible distributions of signs of eigenvalues of the Hessian.
The reason we care about nondegeneracy is that if $p$ is a degenerate critical point then the behavior of $f$ in a neighborhood of $p$, even in a very qualitative sense and in a very small neighborhood, may depend on higher-order terms in the Taylor series (e.g. consider $f(x, y) = x^2 + y^4$ vs. $f(x, y) = x^2 - y^4$). Nondegeneracy allows us to only consider the linear and quadratic terms. |
H: Figuring out the percentage of A in relation to B
I have two numbers
13.49 and 7.8
I want to know how to figure out how much 7.8 is of 13.49 in percentage
e.g. 7.8 is 80% of 13.49
What is the best way to do this?
AI: The percentage is $\frac {7.8}{13.49}\cdot 100 \%\approx 57.82 \%$ Note that multiplying by $100 \%$ is multiplying by $1$. |
H: Class of manifolds is a set?
Is the class of all 2-countable manifolds a set?
I think so: each such space is a countable union of sets of cardinality $|\mathbb{R}^n|\!=\!|\mathbb{R}|$, i.e. a manifold has cardinality continuum, and there are less than $\mathbb{R}^\mathbb{R}$ ways of gluing it together.
Why then is this set still called a class?
AI: No, the class of all manifolds (2nd countable) that are is not quite a set, as it is a proper class. To see this, let us consider the "isomorphism class" of the connected zero dimensional manifolds. This is just a point. But their are as many singletons as their are sets. To each set $S$, we may associate a singleton, $\{S\}$, which may be regarded as a manifold. The upshot is that all of the UP TO ISOMORPHISM, their is only one such point. A similar thing may be said of 2nd countable manifolds. Their is a class worth of these things, but if you select one representitive from each equivalence class, you do get a set. One might say that the category of 2nd countable manifolds is not small, but ESSENTIALLY SMALL. |
H: Projecting a vector component out of vector $a$
Suppose a vector $a$ is given. What is precisely meant by projecting the component of vector $b$ out of $a$? Does that mean that the resulting vector $a_1$ (obtained by "projecting out the component of $b$ from $a$") is orthogonal to $b$? How could one achieve this?
I read that Gram-Schmidt process is used for this; however, it involves the notion of "normalization", and I'm not sure if that means that the vector $a_1$ would be normalized.
AI: It means if you decompose $b$ into a vector $b_1$ parallel to $a$ and a vector $b_2$ perpendicular to $a$, $b_1$ would be the answer. Gram Schmidt (GS) is used to find $b_2$, which is normal to $a$, and then $b_1$ can be found by subtracting.
The purpose (context) for which GS is introduced in most textbooks is to take a number of linearly independent vector (a basis for some vector space) and transform them into an orthogonal basis for the same space. In the case of two vectors above, you go from the basis $(a, b)$ to the basis $(a, b_2)$. In the case of a three-basis $(a, b, c)$ you first find $b_2$, so that you have $(a, b_2, c)$. Then you apply GS on $c$ to find a vector $c_2$ normal to the space given by $(a, b_2)$, but so that $(a, b_2, c_2)$ is the same vector space as $(a, b, c)$ |
H: Subgroup generated by additive integers: Expressing d as an integer combination of ra + sb.
$$a = 123,\ b = 321.$$
The first part of the question I'm trying to solve wants me to find $d = \gcd(a,b)$. I can quickly solve this by using the Euclidean algorithm, and $d$ is $3$.
However, the second part requires me to express $d$ as an integer combination $ra + sb$, for some integers $r$ and $s$.
Is there any way to do this other than mashing some multiples of $a$ and $b$ on a calculator and praying the difference comes out as $|3|$?
AI: The answer is YES, using the Euclidean algorithm backwards.
For $b=321$ and $a=123$, the Euclidean algorithm yields:
$b=2\cdot a+75$
$a=1\cdot75+48$
$75=1\cdot48+27$
$48=1\cdot27+21$
$27=1\cdot21+6$
$21=3\cdot6+3$ and $3$ divides $6$
Hence the greatest common divisor of $a$ and $b$ is $d=3$. Furthermore:
$d=21-3\cdot6$ by (6)
$6=27-21$ by (5), hence $d=21-3\cdot(27-21)=4\cdot21-3\cdot27$
$21=48-27$ by (4), hence $d=4\cdot(48-27)-3\cdot27=4\cdot48-7\cdot27$
$27=75-48$ by (3), hence $d=4\cdot48-7\cdot(75-48)=11\cdot48-7\cdot75$
$48=a-75$ by (2), hence $d=11\cdot(a-75)-7\cdot75=11\cdot a-18\cdot75$
$75=b-2\cdot a$ by (1), hence $d=11\cdot a-18\cdot(b-2a)=47\cdot a-18\cdot b$
Finally, $d=47a-18b$. |
H: Given a random sequence of input points does it always produce a random output, excluding $f(x)$ = constant?
Assume I have $y = f(x) \ne \mathrm{constant} $ and $(x_1 , \ldots ,x_N)$ a sequence N random input points, is there a set of criteria or a theorem that tells me that the output sequence $(y_1, \ldots, y_N)$ is a random sequence as well.
Can it be extended to multivariable and vector functions $f(\vec{x})$ and $\vec{f}(\vec{x})$?
AI: The answer is yes. It doesnt matter if you have a deteministic or stochastic function $f$. If your function is of deterministic type then you have
$$f_Y(y)=f_X(g^{-1}(y))|\frac{d g(y)^{-1}}{d y}|$$
where $g$ is the function that you applied $f_X$ is the previous density with respect to $X$ and $f_Y$ is the density afther you apply the function. |
H: Existence/Finiteness in $\mathbb{R}^2_+$ using Fubini or Tonelli
I am having some trouble understanding how to apply Fubini and/or Tonelli Theorems to determine whether a Lebesgue integral over $\mathbb{R}^2_+$ exists and if it is finite.
If someone could help me by showing the explicit steps for the examples below I would be grateful. I have a long list of exercises I have found online (this is self-study) and instead of posting a bunch of examples here I thought a few simple ones would help me learn how to do these problems going forward.
The examples in hand are: for each of the functions, use Fubini or Tonelli to show the existence/finiteness of the function's Lebesgue integral over $\mathbb{R}^2_+$.
$f_1(x,y)=\frac{\sin xy}{xy}$
$f_2(x,y)=e^{-(1+x^2)y}$
Many thanks in advance!
AI: Assume $f_1$ is integrable in $\mathbb{R}^2_+$ (which I assume is $\mathbb{R}\times \mathbb{R}_+$), by Fubini's theorem $f_1^x(y)=f_1(x,y)$ would be (Lebesgue!) integrable for almost every $x\in \mathbb{R}$, but
$$
\int_{\mathbb{R}_+} f_1^x(y) dy= \int_0^{\infty} \frac{\sin(xy)}{xy} dy
$$
and this last is not a Lebesgue integral for $x\neq 0$ (the positive and negative parts of the function $\sin(z)/z$, when integrated give $\infty$). So $f_1$ is not integrable in $\mathbb{R}^2_+$.
Since $f_2$ is positive everywhere, Tonelli's theorem guarantees that we can integrate first over $y$ and then over $x$ to get
$$
\int_{\mathbb{R}^2_+} f_2(x,y)d(x,y) = \int_{\mathbb{R}} \int_0^\infty e^{-(1+x^2)y}dydx= \int_{-\infty}^{\infty} \frac{1}{1+x^2} dx <\infty
$$ |
H: Finding an interpolating polynomial and natural cubic spline for a given accuracy
I'm trying to make an exercise but I don't know how to start. Is there somebody that can give me a hint so that I can start with the exercise. The exercise is:
Consider the function $f(x) = \sin(x)$ on the interval $[0,\pi]$. Evaluate $f(x)$ in enough points to find an interpolating polynomial $p(x)$ and a natural cubic spline approximation $s(x)$ such that the error functions $|(f-p)(x)|$ and $|(f-s)(x)|$ satisfy
$|(f-p)(x)| \leq 0.005$ for $0 \leq x \leq \pi$
$|(f-s)(x)| \leq 0.005$ for $0 \leq x \leq \pi$
Give the number and the location of the data points for $p$ and $s$. This can be different for $p$ and $s$. Why is the second derivative of the spline function in begin and endpoint equal to zero a good choice?
AI: Wherever you get the formulas for constructing the interpolating polynomial and spline should give you an error estimate. For the polynomial, Wikipedia gives an error limit $f(x)-p_n(x)=\frac {f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n (x-x_i)$ where the $x_i$ are the interpolating points. For $\sin x$ the derivatives are limited to $\pm 1$, so you can use that for your error limit. To minimize the product, you can use Chebyshev nodes. For the last question, what is the derivative of $\sin x$ at $0$ and $\pi$? |
H: Global irreducible decomposition of an analytic set
Let $M$ be a complex manifold (or a complex analityc space) and $Z$ be an analytic subset of $M$. By Noetherianity of the rings of germs of analytic functions at a point we know that $Z$ has finitely many irreducible germs at any $p \in Z$.
What about the "global" irreducible components?
How does one prove that a decomposition into irreducible analytic sets exists?
(I have looked into Gunning and Rossi but couldn't find the relevant reference off-hand.)
I also have a related question: is it possible that an analytic set is connected but has infinitely many irreducible components?
AI: a) Yes, a connected analytic space may have infinitely many irreducible components:
Take a disjoint family of projective lines $(P_i)_{i\in \mathbb N}$ with distinguished points $0_i $ and $1_i$.
Then identify each $1_{i}$ with $0_{i+1}$ .
b) A variant is to take in $\mathbb C^2$ the union of the line $y=0$ and the lines $x=i\; (i\in \mathbb Z)$ . That subspace is connected and has infinitely many irreducible components.
c) The decomposition of an analytic subspace into irreducibles is proved on page 217 of Łojasiewicz's Introduction to Complex Analytic Geometry |
H: Evaluate $\int_0^\infty (u\log(a^2+u^2))/(e^u-1)\, du$
Does the following definite integral have a known "closed form" value?
$$\int_0^\infty\frac{u\log(a^2+u^2)}{e^u-1}~du,$$
or can anyone see a way to integrate it?
AI: I was only able to find the following formula: for $a = 2\pi\alpha$,
$$\begin{align*}
\int_{0}^{\infty} \frac{u \log(4\pi^2\alpha^2+u^2)}{e^u - 1} \; du
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\
&\quad + \pi^2 \alpha \left(2\alpha \log\alpha - \alpha + 2 - 4 \log\Gamma(\alpha+1)\right) \\
&\quad + 4\pi^2 \int_{0}^{\alpha} \log\Gamma(u+1)\;du.
\end{align*}$$
Indeed, let
$$I(a) = \int_{0}^{\infty} \frac{u \log(a^2+u^2)}{e^u - 1} \; du.$$
Then we have
$$\begin{align*}I(0)
& = 2 \int_{0}^{\infty} \frac{u \log u}{e^u - 1} \; du = 2 \left. \frac{d}{ds}\zeta(s)\Gamma(s) \right|_{s=2} \\
& = 2(\zeta'(2) + \zeta(2)\psi_0(2)) = 2\left(\zeta'(2) + \frac{\pi^2}{6}(1-\gamma)\right).
\end{align*}$$
Also,
$$\begin{align*}
I'(a)
&= \int_{0}^{\infty} \frac{2au}{u^2+a^2} \frac{du}{e^u-1} \\
&= 2a \int_{0}^{\infty} \sum_{n=1}^{\infty} e^{-nu} \left(\int_{0}^{\infty} \sin(ux)e^{-ax}\;dx\right)\;du \\
&= 2a \int_{0}^{\infty} \sum_{n=1}^{\infty} \left(\int_{0}^{\infty} \sin(xu)e^{-nu}\;du\right)e^{-ax}\;dx \\
&= 2a \int_{0}^{\infty} \left(\sum_{n=1}^{\infty} \frac{x}{x^2+n^2}\right)e^{-ax}\;dx \\
&= a \int_{0}^{\infty} \left(\pi\coth(\pi x) - \frac{1}{x}\right)e^{-ax}\;dx
\end{align*}$$
Proceeding,
$$\begin{align*}
I'(a)
&= a \left[ \left(\log\sinh(\pi x)-\log x \right)e^{-ax} \right]_{0}^{\infty} + a^2 \int_{0}^{\infty} \left(\log\sinh(\pi x)-\log x \right)e^{-ax} \; dx \\
&= -a\log\pi + a^2 \int_{0}^{\infty} \left(\pi x + \log\left(1 - e^{-2\pi x}\right)-\log2 + \log a -\log ax \right)e^{-ax} \; dx \\
&= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right)+a^2 \int_{0}^{\infty} e^{-ax}\log\left(1 - e^{-2\pi x}\right) \; dx \\
&= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a^2 \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n} e^{-(a+2n\pi)x} \; dx \\
&= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a \sum_{n=1}^{\infty} \frac{\frac{a}{2\pi}}{n\left(n + \frac{a}{2\pi}\right)} \\
&= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a \left(\gamma + \psi_0 \left( \frac{a}{2\pi} +1 \right)\right) \\
&= \pi + a\left(\log\left(\frac{a}{2\pi}\right) - \psi_0 \left( \frac{a}{2\pi} +1 \right)\right).
\end{align*}$$
Thus integrating,
$$\begin{align*}
I(a)
&= I(0) + \int_{0}^{a} I'(t) \; dt \\
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) + \pi a + \int_{0}^{a} t \left(\log\left(\frac{t}{2\pi}\right) - \psi_0 \left( \frac{t}{2\pi} +1 \right)\right) \; dt \\
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) + 2 \pi^2 \alpha + 4\pi^2 \int_{0}^{\alpha} u \left(\log u - \psi_0 (u+1)\right) \; du \\
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\
&\quad + \pi^2 \alpha \left(2 \alpha \log \alpha - \alpha + 2 \right) - 4\pi^2 \int_{0}^{\alpha} u \psi_0 (u+1) \; du \\
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\
&\quad + \pi^2 \alpha \left(2 \alpha \log \alpha - \alpha + 2 \right) - 4\pi^2 \left[ u \log\Gamma(u+1) \right]_{0}^{\alpha} \\
&\quad + 4\pi^2 \int_{0}^{\alpha} \log\Gamma(u+1) \; du,
\end{align*}$$
which gives the desired result. You may verify this formula numerically with the following Mathematica code
a = 8;
NIntegrate[(u Log[4 Pi^2 a^2 + u^2])/(Exp[u] - 1), {u, 0, Infinity}]
2 Zeta'[2] + Pi^2/3 (1 - EulerGamma) +
Pi^2 a (2 a Log[a] - a + 2 - 4 LogGamma[1 + a]) +
4 Pi^2 NIntegrate[LogGamma[1 + u], {u, 0, a}]
Clear[a];
In fact, we can give a neater form as follows:
$$ \begin{align*}
\int_{0}^{\infty} \frac{u \log(a^2+u^2)}{e^{2\pi u} - 1} \; du
&= \zeta'(-1) + \frac{a}{2}\left(1 - \frac{a}{2}\right) + \frac{a^2}{2}\log a - a \log\Gamma(a+1)\\
&\quad + \int_{0}^{a} \log\Gamma(t+1) \; dt.
\end{align*}$$ |
H: Solving inequality $x^3-4x>0$
I was sort of finding the roots by doing $x^3-4x>0=x(x^2-4)$
$x = 0, x = -2,x = 2$ for $$x(x-2)(x+2)>0$$
Then I stopped and thought; maybe I shouldn't be doing that? I am doubting myself! Can anyone confirm if I am doing the right thing to solve the equality?
AI: The polynomial $x^3-4x$ is $0$ at $x=-2, 0,$ and $2.$ At $-3$ the value of the polynomial is $< 0$; at $3$ the value is $>0.$ At $-1,$ the value is $>0.$ At $1$ the value is $<0.$
From this we see the polynomial is $> 0$ for
$$-2<x<0 \text{ and } x>2.$$ |
H: Let X and Y be random variables with joint pdf $f(x,y)=x+y$ for $0
I am using the fact that $X$ and $Y$ are independent if and only if $f_X(x)f_Y(y)=f(x,y)$. So I have
$$f_{X}
(x)=\int_{0}^{1}f(x,y)dy\\=\int_{0}^{1}x+ydy\\=[xy+\frac{y^{2}}{2}]_{y=0}^{y=1}\\=x+\frac{1}{2}$$
and by basically exactly the same math, $f_Y(y)=y+\frac{1}{2}$. Then $$f_X(x)f_y(y)=(x+\frac{1}{2})(y+\frac{1}{2})=xy+\frac{1}{2}(x+y)+\frac{1}{4}\ne f(x,y)$$
And hence they are not independent. But can that be right? Why would the value of X have anything to do with the value of Y? It's not like one is a function of the other.
Or have I made a simple mistake? I looked through a couple of times and I'm pretty sure my math is right...
AI: Note that the conditional density f(x|y) depends on y. f(x|y)=f(x,y)/f(y).
You showed that f(y)=y+1/2 and f(x,y)=x+y. So f(x|y)=(x+y)/(y+1/2)
for any 0<=x<=1 and 0<=y<=1.
It clearly is not independent of y. So knowing y does addect the probability that X is in an fixed interval about x. |
H: References for elliptic curves
I just finished reading Silverman and Tate's Rational Points on Elliptic Curves and thought it was very interesting.
Could any of you point me to some more references (e.g. books, articles) on elliptic curves? Someone suggested Silverman's The Arithmetic of Elliptic Curves, but I don't have any background in algebraic geometry, so I'm not sure how well I would understand it (going to start studying commutative algebra - does that help?). Thank you!
AI: Starting on Monday I will be teaching a (first) graduate course on the arithmetic of elliptic curves. The two texts that I will be using are Silverman's Arithmetic of Elliptic Curves and Cassels's Lectures on Elliptic Curves.
The course does not have any algebraic geometry as a prerequisite. Some students have seen a little algebraic geometry or will be taking a first course in that subject concurrently; a few have seen a lot of algebraic geometry. But at least a few have never taken and will not concurrently be taking any algebraic geometry whatsoever. One of them asked me about this, and I confirmed that the course should still be appropriate for students like him.
If you want to learn about elliptic curves beyond the undergraduate level, you will need to start engaging with some rudiments of algebraic geometry: for instance, really understanding what is going on behind the group law on an elliptic curve requires (in my opinion, at least!) a discussion of the Riemann-Roch Theorem on an elliptic curve. However, elliptic curve theory is concrete enough and the algebraic geometric input is (at the beginning) limited enough so as to make it an excellent opportunity to learn some algebraic geometry from scratch. (I think you will get a taste of that subject faster by learning some elliptic curve theory than by learning commutative algebra, although of course the latter has an essential place in the long run.)
Further, Silverman's book is especially excellently written with respect to this issue: he puts all the algebraic geometry into the first two chapters. I would -- and will! -- recommend that you begin by reading through Chapter 1 on basic algebraic geometry: it is written with a very nice, light touch and mostly serves to introduce terminology and very basic objects. Then I would skip past Chapter 2 and come back to portions of it as needed in the rest of the text. For instance, if you've never seen differentials before, I wouldn't read about them in Chapter 2 until you get to the material on invariant differentials on elliptic curves in Chapter 3.
If it freaks you out to page past two chapters on algebraic geometry, than I would recommend starting with Cassels's text. He takes a more gradual, lowbrow approach to the geometric side, but he is just as much an arithmetic geometer as Silverman, so the approach he takes is quite compatible with a more explicitly geometric perspective which may come later.
I honestly think that these two texts are so excellent that you need look no farther. If it helps, many people around here can tell you that I am very fond of writing my own lecture notes for the graduate courses I teach. However, I wouldn't dream of doing so in this case: what Cassels and Silverman have already done is essentially optimal. |
H: If the sum of positive integers $a$ and $b$ is a prime, their gcd is $1$. Proof?
I feel this is an intuitive result.
If, for example, I was working with the prime number $11$, I could split it in the following ways: $\{1, 10\}$, $\{2, 9\}$, $\{3, 8\}$, $\{4, 7\}$, $\{5, 6\}$.
Then clearly there is no way that the $2$ numbers can have a $\gcd$ of anything other than $1$. However, I am sort of lost on how to start a formal proof for this. Any pointers would be much appreciated.
AI: Let's show the contrapositive, because why not?
So we want to show that if $a,b>0$ and $\gcd(a,b) \neq 1$, then their sum is not prime.
Suppose that $\gcd(a,b) = d > 1$. Then $a = a'd$ and $b = b'd$ for some $a',b'$ natural numbers. But then $a + b = da' + db' = d(a' + b')$, and as each of $d,a',b' \geq 1$, we have that $a + b \geq 2d$, but is divisible by $d$. Thus it is not prime. $\diamondsuit$
Thus if the sum of positive integers is prime, then their gcd is $1$. |
H: If $V_0$ is the subspace of matrices of the form $C=AB-BA$ for some $A,B$ in a vector space $V$ then $V_0=\{A\in V|\operatorname{Trace} (A)=0\}$
If $V_0$ is the subspace consisting of matrices of the form $C=AB-BA$ for some $A,B$ in a vector space $V$ then $V_0=\{A\in V|\operatorname{Trace}(A)=0\}$.
The problem above is one of the past qualifying exam problems. I can prove that $V_0\subset\{A\in V|\operatorname{Trace}(A)=0\}$ but I do not know what to do with converse.
Thank you in advance.
AI: Here is a geometric approach:
Let $\langle A, B \rangle = \mathbb{tr} (A^* B)$. It is straightforward to check that this is an inner product on the set of $p \times p$ matrices.
Let $V_1 = \{A\in V|\operatorname{Trace}(A)=0\}$. It is clear that $V_0 \subset V_1$; we need to show the reverse inclusion.
Note that $V_1 = \{ A | \langle I, A \rangle = 0 \}$, where $I$ is the identity. Also note that $V_0, V_1$ are both (closed) subspaces. To show inclusion it suffices to show that $V_0^\bot \subset V_1^\bot$. It is easy to see that $V_1^\bot = \mathbb{sp} \{I \}$.
Suppose $X \in V_0^\bot$. Then, for any matrices $A,B$, we have $\langle X, AB \rangle = \langle X, BA \rangle$. Let $E_{ij} = e_i e_j^T$, and choose $A= E_{ij}, B = E_{mn}$. Then we get $\langle X, AB \rangle = \langle X, E_{ij} E_{mn} \rangle = \delta_{jm} [X]_{in}$. Similarly, we get $\langle X, BA \rangle = \delta_{in} [X]_{mj}$. Consequently we have $\delta_{jm} [X]_{in} = \delta_{in} [X]_{mj}$ for all $i,j,m,n$. Now choose $j=m$, this simplifies to $ [X]_{in} = \delta_{in} [X]_{jj}$. If $i \neq n$, this gives $[X]_{in} = 0$, if $i = n$, this shows that all diagonal elements are equal, so we have $X = \lambda I$, for some $\lambda$. Consequently $X \in V_1^\bot$. |
H: Irreducible minimal polynomial implies every invariant subspace has an invariant complement
Full version of the problem is following:
Let T be a linear transformation on a finite dimensional vector space $V$ over a field $\mathbb{F}$. If the minimal polynomial $p_t$ of T is irreducible, then every T invariant subspace $W$ has a T-invariant complement $W'$
I used Cyclic decomposition Theorem which states that
"The finite dimensional vector space V can be expressed a s a decomposition of T-cyclic subspaces $Z(\alpha_1;T)\oplus Z(\alpha_2;T)\oplus...\oplus Z(\alpha_k;T)$ and their annihilators $p_1,...p_k$ have properties;
(1) $p_k|...|p_1$, (2)$p_T=p_1, f_T=p_1\cdot p_2\cdot ...\cdot p_k$ where $f_T$ is the characteristic polynomial of T."
Can I say this? Since $p_T$ is irreducible, there is a cyclic vector $\alpha$ such that $V=Z(\alpha;T)$
and V=W and $W'=\{0\}$.
Therefore $W'$ for each W is T-invariant.
Is my way correct?
Thank you in advance.
AI: Let $\mathbb{F}[X]$ be the polynomial ring with one variable.
$V$ can be regarded as an $\mathbb{F}[X]$-module by defining $Xv = T(v)$ for every $v \in V$. $\mathbb{F}[X]$-submodules of $V$ are none other than $T$-invariant subspaces of $V$.
Let $K = \mathbb{F}[X]/(p_t)$.
Since $p_t$ is irreducible, $K$ is a field.
Since $p_t V = 0$, $V$ can be regarded as a $K$-module.
Let $W$ be a $T$-invariant subspace of $V$.
$W$ can be regarded as a $\mathbb{F}[X]$-submodule of $V$.
Since $p_t W = 0$, $W$ can be regarded as a $K$-submodule of $V$.
Hence there exists a $K$-submodule $W'$ such that $V = W \oplus W'$.
Since $W'$ is a $\mathbb{F}[X]$-submodule, it is $T$-invariant.
This completes the proof. |
H: General topology multiple choice question
Let $X$ be a normal Hausdorff space. Let $A_1$, $A_2$, and $A_3$ be closed subsets of $X$ which are pairwise disjoint. Then there always exists a continuous real valued function $f$ on $X$ such that $f(x) = a_i$ if $x$ belongs to $A_i$, $i=1,2,3$
iff each $a_i$ is either 0 or 1.
iff at least two of the numbers $a_1$, $a_2$, $a_3$ are equal.
for all real values of $a_1$, $a_2$, $a_3$.
only if one among the sets $A_1$, $A_2$, $A_3$ is empty.
AI: The third choice is correct. Recall Urysohn's lemma, which states that a space is normal iff disjoint closed sets can be separated by a function. Since $A_1$ and $A_2$ are disjoint closed sets, we have a continuous function $g:X\to [0,1]$ such that $g(x)=0$ for $x\in A_1$ and $g(x)=1$ for $x\in A_2$. Since $A_2\cup A_3$ and $A_3$ are closed disjoint sets, we have a continuous function $h:X\to [0,1]$ such that $h(x)=0$ if $x\in A_1\cup A_2$ and $g(x)=1$ if $x\in A_3$. Thus we have the continuous function $(g+2h):X\to \mathbb R$ which satisfies
$$(g+2h)(x)=\begin{cases}
0 &\text{if } x\in A_1\\
1 &\text{if } x\in A_2\\
2 &\text{if } x\in A_3\\
\end{cases}$$
and so composing this with your favorite function $p:\mathbb R\to\mathbb R$ which sends $0$ to $a_1$, $1$ to $a_2$, and $2$ to $a_3$ gives the desired function $f$. |
H: A bit of direct sum confusion
So I'm reading through Serre's "Linear Representations of Finite Groups," and I'm a bit confused by what's probably a fairly minor point. However, subsequent proofs are hinging on it, so I figure I'll turn to you guys for clarification.
So he defines the direct sum (I guess in this case the internal one), and then introduces the projection map. So if $V=W+W'$, then $p(x=w+w')=w$. Then, for some converse implicating, he assumes we have some map $p$ from $V$ to $V$ whose image is $W$ and whose restriction to $W$ yields $p(x)=x$. Then we can show that $V$ is the direct sum of $W$ and the kernel of $p$. It's the next line that trips me up. He says "A bijective correspondence is thus established between the projections of $V$ onto $W$ and the complements of $W$ in $V$." My brain can't fill the gap in the logic, nor can I really understand what that statement is all about.
Apologies for the logorrhea. Any and all help is very much appreciated!
AI: Every projection $\,p:V\stackrel{\text{onto}}\rightarrow W\,$ determines a complement of $\,W\,$ , namely $\,\ker p\,$, since $\,V=W\oplus \ker p\,$, and the way around: for every decomposition $\,V=W\oplus U\,$ there exists a projection $\,p_U:V\to W\,$ , defined by $\,p_U(v=w\oplus u):= w\,$ . There you have your bijective correspondence. |
H: Equivalent definition for a collection of simplices to be a simplicial complex
I am reading the following lemma from Munkres' Elements of Algebraic Topology:
Lemma 2.1 A collection $K$ of simplices is a simplicial complex if and only if the follow hold:
Every face of a simplex of $K$ is in $K$.
Every pair of distinct simplices of $K$ have disjoint interiors.
Now I am trying to follow the proof that if 1 and 2 above hold then $K$ is a simplicial complex. Although munkres does not define what he means by "distinct simplices", I am told by Mixedmath that this means two simplices that don't have a common vertex. The proof according to Munkres goes as follows:
Let $\sigma$ and $\tau$ be two distinct simplices of $K$ such that they have disjoint interiors. Let $\sigma'$ be the face of $\sigma$ that is spanned by those vertices $b_0,\ldots,b_m$ of $\sigma$ that lie in $\tau$. The claim now is that $\sigma \cap \tau$ is equal to $\sigma'$. Now one direction I understand the other which I don't is when he shows that
$$\sigma \cap \tau \subseteq \sigma'.$$
The line I don't understand is this:
Let $x \in \sigma \cap \tau$. Then $x \in \textrm{Int}\, s \cap \textrm{Int} \hspace{2mm} t$ for some faces $s$ of $\sigma$ and $t$ of $\tau$.
How does this follow from the assumption of (2) above?
AI: "Distinct simplices" means that they are distinct (that is, are not identical). The line you don't understand doesn't use (2); it uses the fact that if $x \in \sigma$ then $x \in \text{int}(s)$ for a face $s$ of $\sigma$ (which is actually unique). $s$ is precisely the minimal face of $\sigma$ (under inclusion) containing $x$ (if $x$ is not contained in the interior of $s$ then it is in the boundary of $s$ which is contained in a strict face of $s$). |
H: Error in manual derivative calculation
I need to compute the derivate of
$g(x) = \ln\left(\frac{1}{x \sqrt{x+1}}\right)$
Simplifying this I get
$g(x) = \ln(1) - (\ln x + \frac{1}{2}\ln(x+1))$
Evaluating the derivate I get
$g'(x) = 0-(\frac{1}{x} + \frac{1}{2x+2})$
I have compared my result to WolframAlphas answer and while my denominator is the same my numerator is different.
My result is $-3x+2$
WA result is $-x+2$.
I assume my mistake is in the multiplication of the negative sign, but why is this wrong?
Thank you
AI: Wolfram Alpha says the same thing. Take a common denominator and check for yourself, your derivative is fine.
I just checked and got -(3x+2) from WA. |
H: Does this graph have two DISTINCT Euler circuits but no Hamiltonian circuit?
I know it doesn't have a Hamiltonian circuit because vertices c and f will be traversed twice in order to return to a. Just confirming this.
I mainly want to know whether I have the definition of distinct Euler circuits in a graph right, and whether the graph below is an example of this, i.e. {a,b,c} and {f,g,h}, being the 2 distinct Euler circuits.
AI: As you correctly state, the graph has not Hamiltonian Cycle, since it has a cut vertex (actually it has two, namely $c$ and $f$).
Regarding the number of distinct Eulerian Circuits, there are more than two if you take the definition of distinct to be
Two Eulerian Circuits are distinct, iff they are not exact reversals of each other.
In this case, the three circuits
$$(a,c,d,f,g,h,f,e,c,b,a)$$
$$(a,c,d,f,h,g,f,e,c,b,a)$$
$$(a,c,e,f,g,h,f,d,c,b,a)$$
are examples of distinct Eulerian Circuits.
As stated in Brian's answer, there are exactly 4 distinct Eulerian Cycles, namely
$$a,c,d,f,g,h,f,e,c,b$$
$$a,c,d,f,h,g,f,e,c,b$$
$$a,c,e,f,g,h,f,d,c,b$$
$$a,c,e,f,h,g,f,d,c,b$$
To count these, and make sure there are no more than 4, look at the number of times the walk "crosses itself". Since there are only two vertices ($c$ and $f$) where such a crossing can happen, the number of crossings can be 0,1 or 2. There is only a single walk with no crossing, namely
$$a,c,d,f,g,h,f,e,c,b$$
There is also only a single walk with two crossings, namely
$$a,c,e,f,g,h,f,d,c,b$$
and there are two walks with a single crossing, the one with a crossing in $c$
$$a,c,e,f,h,g,f,d,c,b$$
and the one with a crossing in $f$
$$a,c,d,f,h,g,f,e,c,b.$$ |
H: How to prove $x^{n}$ is not uniformly continuous
How to prove $x^{n}$ is not uniformly continuous on the interval $[0, +\infty)$ for $n \in \mathbb{N}$ and $n>1$?
AI: You want to show that there exists some $\epsilon>0$ such that there is no $\delta>0$ for which
$$|x-y|<\delta\implies |x^n-y^n|<\epsilon.$$
Try $\epsilon=1$. Thus you want to show that for any $\delta>0$, there is some pair $x,y\in [0,\infty)$ such that $|x-y|<\delta$ yet $|x^n-y^n|\geq 1$. Let's try letting $x=y+\delta/2$. Then we have $|x-y|=\delta/2<\delta$ and
$$|x^n-y^n|=(y+\delta/2)^n-y^n\geq (y+\delta/2)y^{n-1}-y^n=\delta/2\cdot y^{n-1}.$$
Can you find some $y$ such that $\delta/2\cdot y^{n-1}$ is at least $1$? |
H: Regular expression arithmetics
What are the rules of regular expression arithmetics ?
For example: Let $\Sigma=\{0,1\}$
$1. 1+01=(\epsilon+0)1$.
$2. (\epsilon+00)^*=(00)^*$
AI: Off the top of my head you have at least the following:
$\epsilon\alpha=\alpha=\alpha\epsilon$.
$\alpha+\beta=\beta+\alpha$.
$\alpha+\alpha=\alpha$.
$\alpha\beta+\alpha\gamma=\alpha(\beta+\gamma)$ and $\beta\alpha+\gamma\alpha=(\beta+\gamma)\alpha$.
$(\epsilon+\alpha)^*=\epsilon+\alpha^*=\alpha^*$.
$\epsilon^*=\epsilon$.
If your formalization includes $\varnothing$ (no string) distinct from $\epsilon$ (empty string), you have $\varnothing+\alpha=\alpha$. |
H: Using implicit differentiation on trigonometric fraction and logarithm
I am supposed to solve the following using implicit differentiation.
$$\sin\left(\frac{x}{y}\right)+\ln(y)=xy$$
This is what I have so far:
$$\cos\left(\frac{x}{y}\right)\frac{d}{dx}\left(\frac{x}{y}\right)+\frac{d}{dx}\big(\ln(y)\big)=y+x\frac{dy}{dx}$$
My problem here is differentiating the fraction as well as the $\ln y$, especially since $\ln y$ does not contain $x$ to differentiate towards.
Thanks for your help in understanding this.
AI: Keep doing what you did on the right-hand side to get
$$\cos\left(\frac{x}{y}\right)\frac{d}{dx}\left(\frac{x}{y}\right)+\frac{d}{dx}\big(\ln(y)\big)=y+x\frac{dy}{dx}\;.\tag{1}$$
You have
$$\frac{d}{dx}\left(\frac{x}y\right)=\frac{y\cdot1-x\frac{dy}{dx}}{y^2}=\frac1y-\frac{x}{y^2}\frac{dy}{dx}$$
and
$$\frac{d}{dx}\ln y=\frac1y\frac{dy}{dx}\;;$$
both of these are just applications of the chain rule (together with the quotient rule and the rule for differentiating the natural log). Now for convenience I’ll write $y'$ for $dy/dx$; then $(1)$ becomes
$$\cos\left(\frac{x}y\right)\left(\frac1y-\frac{x}{y^2}y'\right)+\frac{y'}y=y+xy'\;,$$
and all that remains is to solve for $y'$. |
H: Product integral of $x$ between $1/a$ and $a$ "should" equal 1 but doesn't
I was looking at the Product Integral. I'm defining it here with the Gamma symbol like this:
$\Gamma _a^bf(x)dx=e^{\int_a^b \log (f(x)) \, dx}$
It occured to me that the product integral of $f(x) = x$ in the range $[\frac {1} {a}, a]$ "should" be 1, because every element in that range has an inverse in that range. But Mathematica argues otherwise:
E^Integrate[Log[x], {x, 1/2, 2}]
$e^{\frac{1}{2} (\log (32)-3)}$
Which is about 1.2622. The general solution for a > 1 is:
$e^{\frac{1}{a}-a+\left(\frac{1}{a}+a\right) \log(a)}$
This is definitely not 1, so my assumption is probably naive. The first explanation that comes to mind (besides the integral not being $0$) is the size difference. Although every element in the range $[\frac {1}{a}, a]$ does have an inverse in that set, $[\frac {1}{a},1]$ is a smaller range than $[1,a]$. So perhaps in some way this imbalance comes out in the product integral.
I have two questions then:
1) Is the product integral telling us anything here? If $\Gamma _{\frac{1}{a}}^{a}x\,dx \approx 1.2622$, does that say anything about the range $[\frac {1}{a},a]$?
2) Is there an obvious product integral which would yield 1 for the function $f(x)=x$ in the range $[\frac {1}{a},a]$?
AI: The fact that the sizes of the intervals $[1/a,1]$ and $[1,a]$ are different indeed points to the right direction. To see this, let us try to show that the integral is zero because the contributions of $f(x)=\log(x)$ and $f(1/x)=\log(1/x)=-\log(x)$ would compensate each other (they do not but let us see what happens when we try to show they do).
A natural idea is to consider a subdivision $(x_k)_{0\leqslant k\leqslant n}$ of the interval $[1,a]$, since the integral of $\log f$ on $[1,a]$ is roughly the Riemann sum
$$
S=\sum_{k=1}^n(x_k-x_{k-1})\log f(x_k).
$$
The symmetry of $\log f$ suggests to consider the subdivision $(z_k)_{0\leqslant k\leqslant n}$ made with the inverses of the points of the first subdivision, more precisely, $z_k=x_{n-k}^{-1}$ for every $0\leqslant k\leqslant n$. The integral of $\log f$ on $[1/a,1]$ is roughly the Riemann sum
$$
S'=\sum_{k=1}^n(z_k-z_{k-1})\log f(z_{k-1})=\sum_{k=1}^n(x_{n-k}^{-1}-x_{n-k+1}^{-1})\log f(x_{n-k+1}^{-1}),
$$
that is
$$
S'=-\sum_{k=1}^n\frac{x_k-x_{k-1}}{x_{k}x_{k-1}}\log f(x_{k}).
$$
Since the coefficient of $\log f(x_k)$ in $S'$ is not the length of the interval $x_k-x_{k-1}$ as in $S$, but a biased version of this length, the argument does not lead a priori to $S'=-S$.
To sum up, the function $\log f$ is indeed symmetric with respect to the transformation $x\to1/x$ but the lengthes of the subintervals of $[1,a]$ and $[1/a,1]$ are not (and in particular, but this is only the tip of the iceberg, the lengthes of $[1,a]$ and $[1/a,1]$ themselves do not coincide). Still in other words, Area = Width $\times$ Height and Height does not change but Width changes, hence Area changes as well.
To save the day, the transformation $x\to1/x$ should leave Width unchanged. As explained in comments, the only way to do this is that the length is measured locally at $x$ by $\dfrac{\mathrm dx}x$. The biased Riemann sums become
$$
\tilde S=\sum_{k=1}^n\frac{x_k-x_{k-1}}{x_k}\log f(x_k),\qquad \tilde S'=\sum_{k=1}^n\frac{z_k-z_{k-1}}{z_{k}}\log f(z_{k-1})=-\tilde S.
$$ |
H: Generator of multiplicative group of $\mathbb{Z}/p\mathbb{Z}$
Possible Duplicate:
Finding a primitive root of a prime number
Given a field $F_p = \mathbb{Z}/p\mathbb{Z}$ I want to find all $x$ such that $\{x^n|n\in \mathbb{N}\} = F_p \setminus \{0\}$. Note that $x=2$ seems to work for a lot of primes, but only $3$ and $5$ work for $7$. I don't see a pattern, not until one such $x$ is found, that is. Is there a way to find such elements without brute force?
AI: Finding a primitive root is a difficult problem. There is a vast mathematical literature on the problem, which others more expert than me could point you to. Let me focus first on the issue of whether $2$ can be a primitive root to illustrate some points. It is known that $2$ is a quadratic residue (mod $p$) if $p \equiv \pm 1$ (mod 8). This means that for such primes $p,$ $2$ has multiplicative order at most $\frac{p-1}{2}$ in $\mathbb{Z}/p\mathbb{Z}$, and consequently has no chance to be a primitive root (mod $p$). In general, quadratic reciprocity can be used to exclude many possibilities for primitive roots (mod $p$). Given an integer $m$ with $1 \leq m \leq p-1$, it is in principle routine to determine whether $m$ is a quadratic residue (mod $p$). If it is a residue, it can't be a primitive root. However, if $m$ is a quadratic non-residue, that is no guarantee that $m$ is a primitive root (mod $p$). The only case where that is guaranteed is when $p$ is a Fermat prime, for then the order of $m$ must be a power of $2,$ and if $m$ is not the square of something else, it must generate the multiplicative group of $\mathbb{Z}/p\mathbb{Z}.$
This is only the beginning of the story, but it does explain why $2$ is not a primitive root (mod $7$), and for example, you can quickly see that $3$ and $5$ are quadratic non-residues (mod $17$), so they are both primitive roots (mod $17$). In fact, if $p$ is any Fermat prime greater than $5$, we see that $$\left( \begin{array}{cc} \underline{3}\\{p} \end{array} \right) = \left( \begin{array}{cc} \underline{p}\\{3} \end{array} \right) = \left( \begin{array}{cc} \underline{2}\\{3} \end{array} \right)= -1,$$ and $$\left( \begin{array}{cc} \underline{5}\\{p} \end{array} \right) = \left( \begin{array}{cc} \underline{p}\\{5} \end{array} \right) = \left( \begin{array}{cc} \underline{2}\\{5} \end{array} \right)= -1,$$ so that both $3$ and $5$ are primitive roots (mod $p$), However, life is rarely this easy when searching for primitive roots! |
H: Evaluating $\int_0^1 \! C(x) \, \mathrm dx$ through integration by parts
$$
\int_0^1 \! C(x) \, \mathrm{d} x.
$$
where $C(x) = \int_0^x \cos(t^2) \, \mathrm{d} t$.
I am really not quite sure how to go about this one, especially given that it needs to be calculated using integration by parts.
My lecturer has an example (done using integration by parts) for $\int_0^1 \! xC(x) \, \mathrm{d} x$.
In this example, he let $u = C(x)$, so that $\frac{du}{dx}\ = \cos(x^2)$, which worked out nicely.
In the 2nd last line of his solution, he had the term:
$$
\int_0^1 \! \sin(x^2) \, \mathrm{d} x.
$$
and simply finished with leaving this part as $S(1)$.
However, in this question, I don't seem to have terms which I can choose as $u$ and $dv$?
Is anyone able to give me some direction?
Many thanks!
AI: It's easier to do integration by parts here:
$$\int C(x)\mathrm dx=x\,C(x)-\int x\cos(x^2)\mathrm dx$$
Can you take it from here? |
H: Area enclosed by lines and curve .
I am given $y=x$ and $y=\frac{x}{8}$ for the lines. The curve is $y=\frac{1}{x^2}$ . This is more of a question about the question though. I have no idea what it is asking.
My instinct is to integrate over some limits but I am not sure where this circle features and whether I am over complicating things.
Should I just be adding 3 indefinite integrals here?
AI: Make a graph showing all three curves. You’ll see that there is just one finite region bounded by these curves. It has two straight sides along the lines $y=x$ and $y=x/8$ and one curved side along $y=1/x^2$. If that third side ran straight from $\langle 1,1\rangle$ to $\left\langle 2,\frac14\right\rangle$, the region would be the triangle with vertices at the origin, at $\langle 1,1\rangle$, and at $\left\langle 2,\frac14\right\rangle$.
(In case it isn’t clear, $\langle 1,1\rangle$ is the point of intersection of $y=x$ and $y=1/x^2$, and $\left\langle 2,\frac14\right\rangle$ is the point of intersection of $y=x/8$ and $y=1/x^2$.)
You’ll want to set it up as two integrals, one from $x=0$ to $x=1$, the other from $x=1$ to $x=2$. Your vertical strips $dA$ of area for the first integral will run between the two straight lines; for the second integral they’ll run from the line $y=x/8$ to the curve $y=1/x^2$. |
H: $m \times n$ persons stand in $m$ rows and $n$ columns
$m \times n$ persons stand as a rectangle of $m$ rows and $n$ columns.
Each one is taller than the one next to his left.
Now, if commander orders each column to sort by height, prove that after such sorting, each one is still taller than the one next to his left.
I'm reading a Combinatorics book on my own, but some exercises I can't solve... many thanks!
AI: For a person $a$ standing in column $i$, there are $k$ persons in that same column which are taller than him, and since for each person in this column, there is a right neighbor in the next column that's even taller, we now that the $k + 1$-th tallest person in the next column is taller than $a$. After sorting, the right neighbor of $a$ (the $k + 1$-th tallest person of column $i$) is the $k + 1$-th tallest person in column $i + 1$, and this one is thus taller than $a$. |
H: Geometry/Triangles Problem
Let $A,B,C$ be a triangle's vertices (FIG.1), and $D$ is a point on $\overline {BC}$. It is given that $\angle B= 2 \angle C $, $\overline{AB}=\overline{DC}$ and also that $\overline{AD}$ is the bisector of angle $A$.. We have to prove that angle $\angle BAC=72°$.
$\hspace{0.54 cm}$
$$\begin{align}\text{FIGURE 1} &\end{align}$$
AI: Let $BE$ bisect $\angle B$ into a pair of angles congruent to $\angle C$. As $\angle CBE$ is a member of that pair, we have an isosceles triangle with $BE=EC$ (blue edges).
Draw $DE$ and note that, because $\angle ABE = \angle C$, we have $\triangle ABE \cong \triangle DCE$ ("side-angle-side"). In particular, $\angle CDE \cong \angle A$ (red-green angles).
Also, $DE\cong EA$ (green edges), so that $\triangle DEA$ is isosceles with $\angle EAD \cong \angle EDA$. As $\angle A$ was bisected by $AD$, we have $\angle EDA \cong \angle BAD$, so that $DE \parallel BA$ ("alternate interior angles", or what I call the "Z theorem").
Consequently, $\angle CDE \cong \angle CBA$ ("corresponding angles", or the "F theorem"), which implies that all of my angle-marking dots match. With five dots in the angles of $\triangle ABC$, each dot represents measure $180^{\circ}/5=36^{\circ}$, so that the two-dot $\angle A$ has measure $72^{\circ}$. QED |
H: Choose 100 numbers from 1~200 (one less than 16) - prove one is divisible by another!
Prove that if 100 numbers are chosen from the first 200 natural
numbers and include a number less than 16, then one of them is
divisible by another.
How to prove this? many thanks....
AI: Assign pigeonholes by considering that $a, b$ are in the same pigeonhole if $a/b$ is a power of two (including negative powers, in case $a<b$). Now each pigeonhole has an odd number, plus that same odd number multiplied by each power of two. Since there are 100 odd numbers in $\{1,\ldots,200\}$, there are 100 pigeonholes. If two or more of the chosen numbers are in the same pigeonhole, then we're done. If not, then there must be one chosen number in each pigeonhole.
In the latter case, let $n$ be the chosen number that's less than 16. Consider these four cases.
If $n$ is an odd number, then it divides whatever number was chosen in the pigeonhole with $3n$.
If $n$ is twice an odd number, then EITHER it divides whatever number was chosen in the pigeonhole with $3n/2$, OR $3n/2$ itself was chosen, which divides whatever was chosen in the pigeonhole with $9n/2$.
If $n$ is 4 or 12, then consider what number might have been chosen in the pigeonhole with 9. If it's 36 or greater, we're done. If it's 9 or 18, then consider what number was chosen in the pigeonhole with 27. If it's 54 or higher, we're done. If it's 27, then it divides whatever number was chosen in the pigeonhole with 81.
If $n$ is 8, then consider what number might have been chosen in the pigeonhole with 3. If it's 24 or greater, we're done; but if it's 3, 6 or 12, then we've already covered this case in one of the earlier cases above.
So in all cases, there'll be a number in one pigeonhole that divides a number from another pigeonhole. |
H: Boolean Algebra simplification - odd number terms
I'm new to boolean algebra and having problems simplifying expressions with odd number terms,
Expressions such as:
1.
A'B'C'D + A'B'CD + AB'C'D + AB'CD + ABC'D
2.
A'BC + AB'C' + A'B'C' + AB'C + ABC
Here is my logic for both expressions:
A'B'C'D + A'B'CD + AB'C'D + AB'CD + ABC'D
A'B'D'(C+C') + AB'D(C+C') + ABC'D
A'B'D + AB'D + ABC'D
B'D(A'+A) + ABC'D
B'D + ABC'D
I never touch the last term, and I don't know what rule I'm missing. Same thing happens on the second expression:
A'BC + AB'C' + A'B'C' + AB'C + ABC
A'BC + B'C'(A+A') + AC(B'+B)
A'BC + B'C + AC
Again, one term untouched..
For 1. the result should be AC'D + B'D
For 2. the result should be B'C + BC + AC
I could maybe use Karnaugh maps but I also would like to understand the algebra logic.
AI: It’s not too hard to check the results after you know them:
$$\begin{align*}B'D+ABC'D&=(B'+ABC')D\\
&=(B'+AB'C'+ABC')D\\
&=\Big(B'+A(B'+B)C')\Big)D\\
&=(B'+AC')D\\
&=B'D+AC'D
\end{align*}$$
and
$$\begin{align*}A'BC + B'C + AC&=B'C+(A'B+A)C\\
&=B'C+(A'B+A+AB)C\\
&=B'C+\Big(A+(A'+A)B\Big)C\\
&=B'C+(A+B)C\\
&=B'C+AC+BC\;.
\end{align*}$$
That second one can be further simplified to $AC+C=C$, since $B'C+BC=(B'+B)C=C$.
In both calculations I used the absorption law: $B'=B'+AB'C'$ in the first, and $A=A+AB$ in the second.
When you have no more than three or four proposition letters, you may find Venn diagrams helpful. |
H: Inequality. $a^2+b^2+c^2 \geq a+b+c$
Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
AI: Using Cauchy-Schwarz inequality we get
$$
a+b+c=a\cdot 1+b\cdot 1+c\cdot 1+\leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}\tag{1}
$$
From AM-GM we obtain $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}=3$, so
$$
\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\tag{2}
$$
From $(1)$ and $(2)$ it follows
$$
a+b+c\leq\sqrt{a^2+b^2+c^2}\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2
$$ |
H: An eigen value problem
Let $A$ be a $3×3$ matrix with $\operatorname{trace} (A) = 3$ and $\det (A) = 2$. If $1$ is an eigenvalue of $A$, then what are the eigenvalues of the matrix $A^2 - 2I$?
AI: Hint: Can you find the other eigenvalues of $A$ ? what do you know about the sum of all eigenvalues of $A$ ? what about their product ? |
H: Probability density function. Find $k$
Let $f_{X,Y}(x,y) = kxy$, for $0 ≤ x, y ≤ 1,$ otherwise $f_{X,Y}(x,y) = 0$.
(a) Determine $k$ such that $f_{X,Y}(x,y)$ is a PDF.
AI: Selected from PDF at WikiPedia.
The probability density function is nonnegative everywhere, and its integral over the entire space is equal to one.
As implied, $\int_{R^2} f_{X,Y}(x,y)=1$. (Property of PDF)
Thus $\int_0^1\int_0^1 kxy dxdy=1$, with the rest being mere integration. |
H: Coin toss sequences
Suppose $20$ tosses of a coin yield $8$ heads, $12$ tails.
Let $X$ be the # of sequences of a head followed by exactly $2$ tails.
Let $Y$ be the # of sequences of a head followed by at least $2$ tails.
What are the values of $E[X]$ and $E[Y]$ ?
My efforts are meeting dead ends.
AI: The probability that HTTH will appear in a particular position is $\dfrac{16 \choose 6}{20 \choose 8} = \dfrac{308}{4845}$ and there are 17 possible positions so $$E[X] = \frac{308}{285} \approx 1.0807.$$
The probability that HTT will appear in a particular position is $\dfrac{17 \choose 7}{20 \choose 8} = \dfrac{44}{285}$ and there are 18 possible positions so $$E[Y] = \frac{264}{95} \approx 2.7789.$$ |
H: Question about definition of Sobolev spaces
I'm trying to understand the following definition:
which can also be found here on page 136.
Question 1: Closure with respect to what norm? It's not given in the definition.
Question 2: Do I have this right: I can view $C^\infty$ as a dense subspace of $H^k$ via the map (embedding) $f \mapsto (D^\alpha f)_\alpha$ where the tuple $f$ is mapped to consists of all derivatives $D^\alpha f$ such that $|\alpha| \leq k$. Then this is cool because if we have this we can extend any linear operator $T: C^\infty \to C^n$ continuously to all of $H^k$ so that anything we can do to smooth functions we can also do to Sobolev functions. That is, even if the functions don't have a strong $\alpha$-th derivative we can treat them as if they did.
Thanks for your help.
AI: We can endow $V$ with the norm
$$\lVert (v_j)_{1\leq j\leq K(k)}\rVert_V:=\sqrt{\sum_{j=1}^{K(k)}\lVert v_j\rVert_{L^2(\mathbb T^d)}^2}$$
(it's an usual norm in product of $L^2$ spaces).
Yes, by definition $C^{\infty}(\mathbb T^d)$ is dense in its closure. When you say "anything we can do with smooth function, we can do it for Sobolev functions", careful. The property true for smooth functions has to be preserved by taking the limit in $\lVert\cdot\rVert_V$. |
H: What is the formula for this exponentially growing "stairs"?
I'm looking for a formula that, given a linear $x$ input, would yield values of $y$ in a "stairs" shape so to speak, in such a way that as the value of $x$ grows higher, the difference between each step is bigger, and it also takes longer to reach the next the step, as this graph (hopefully) illustrates:
As you have probably guessed by now I'm not a mathematician, but I'm fairly sure this has an easy solution. I've been playing around with modulus and powers but I couldn't quite get the graph above so far.
Edit: Graph updated.
AI: My first intuition is $y=2^{\lfloor\log_2\,x\rfloor}$. It's very pretty though: |
H: Extension of an embedding to a field extension by a transcendental element
Let $K$ be a subfield of a field $K'$ and suppose we have an embedding $\phi:K \rightarrow L$ of $K$ into an algebraically close field $L$. Let $x \in K'$. If $x$ is algebraic over $K$, then we can extend $\phi$ to $K(x)$. What if $x$ is transcendental over $K$? Can we then extend $\phi$ to $K(x)$?
Thanks.
AI: Note: this is an edited version of a previous incomplete (and wrong....) answer. While I was editing. Gerry wrote his answer giving the very same counterexample. Sorry for the mishap ... :)
If there is a $z\in L$ transcendental over $\phi(K)$ you can extend $\phi$ by letting $x\mapsto z$.
But it is possible that such a $z$ doesn't exist: for instance $K=\Bbb Q$ embeds in $L=\overline{\Bbb Q}$ but there's no way to extend this to an embedding of $K^\prime=\Bbb Q(\pi)$ into $L$. |
H: Find the number of different sets C such that $B\subset C\subset A$
Assume that A consists of n elements and $B\subset A$ consists of k elements. Find the number of different sets C such that $B\subset C\subset A$. I am reading the A. Shen and N. K. Vereshchagin book, Basic Set Theory.
I try to use Combinatorics for solve the question but still I can not find a solution for the general case.
Many Thanks.
AI: So one must have $k \leq n$. $A - B$ is a set of $n - k$ elements. If $P \subset A - B$, then $A \subset B \cup P \subset A$. On the other hand if $B \subset C \subset A$, then $P = C - B$ is a subset of $A - B$.
So all such $C$ are in correspondence with subsets of $A - B$. Since $A - B$ is a set of size $n - k$. There are $2^{n -k}$ possibilities for $C$. |
H: How to see $\sin x + \cos x$
$$\sin x + \cos x = \sqrt{2} \sin(x + \pi/4)$$
Is there an easy way to visualize this identity or to convert the left-hand side to the right-hand side?
In general, can $p \sin x + q \cos x$ for some integers $p$ and $q$ be easily expressed as $\sin($something$)$ or $\cos($something$)$? If so, how can that be done?
AI: Given $p \sin x + q \cos x$, divide the expression by $\sqrt{p^2+q^2}$ to get $a \sin x + b \cos x$ for $a^2 + b^2 = 1$. Now name $a = \cos \alpha$ and $b = \sin \alpha$. Notice that what you got is the expansion of $\sin(x+\alpha)$. In the given case, $a = b = \frac{\sqrt{2}}{2}$ so $\alpha = \frac{\pi}{4}$. |
H: Evaluating $\int_{I^n} \left( \min_{1\le i \le n}x_i \right)^{\alpha}\,\, dx$
Let $\alpha \in \mathbb R$ and let's call $I:=[0,1]$. Evaluate
$$
\int_{I^n} \left( \min_{1\le i \le n}x_i \right)^{\alpha}\,\, dx.
$$
Well, the case $n=1$ is easy and the integral equals $\frac{1}{\alpha+1}$, for every $\mathbb R \ni \alpha \ne - 1$.
I've done also the case $n=2$ and, if I'm not wrong, it's $\displaystyle \frac{2}{(\alpha+1)(\alpha+2)}$.
My big problem is that I cannot understand how to deal with the general case. Any ideas?
Thanks in advance.
AI: Denote $a_n^{\alpha}$ the integral. We have, since $\alpha\neq -1$, that
\begin{align}
a_n^{\alpha}&=n!\int_{\{0<x_1<\dots<x_n<1\}}(\min_{1\leq j\leq n}x_j)^{\alpha}dx_1\dots dx_n\\
&=n!\int_{\{0<x_1<\dots<x_n<1\}}x_1^{\alpha}dx_1\dots dx_n\\
&=n!\int_{\{0<x_2<\dots<x_n<1\}}\frac{x_2^{\alpha+1}}{\alpha+1}dx_2\dots dx_n\\
&=\frac 1{\alpha+1}na_{n-1}^{\alpha+1}.
\end{align}
By induction, $$a_n^{\alpha}=n!a_1^{\alpha+n-1}\prod_{j=1}^{n-1}\frac 1{\alpha+j}=n!\prod_{j=1}^n\frac 1{\alpha+j}.$$ |
H: Surface integral
Without getting into the whole question, I was asked to evaluate a surface integral
$\iint\limits_S f(x,y,z) da$
where S is the cylinder $x^2 + y^2 = x$
between $z=a$ and $z=b$
Now normally I would parametize this as a cylinder and it would be easy peasy but I'm worried about the equation of the cylinder, as my normal equation would be more like $x^2 + y^2 = C$ with C being some constant .. Any thoughts ? thanks a lot
AI: Your cylinder is offset from the axis. $x^2+y^2=x$ becomes $(x-\frac 12)^2+y^2=\frac 14$. You can substitute $u=x-\frac 12$ to get back on axis if you want. |
H: Newton algorithm for a function in $\mathbf{R}^n\rightarrow \mathbf{R}$
I am curious on how the Newton algorithm would work to solve an equation of the type: $f(x_1,\dots,x_n)=0$.
As far as I understand, in dimension $1$, one solves $f(x)=0$ by starting with some $x_0$, and building the sequence $(x_n)$ defined by $x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}$.
My concern is that I don't see how to generalize this in multiple dimension. Indeed, if I start the algorithm with $x^0=(x_1^0,\dots,x_n^0)$, then I will approximate my equation $f(x_1,\dots,x_n)=0$ by $f(x_1^0,\dots,x_n^0)+\displaystyle\sum_{i=0}^n \frac{\partial f(x^0)}{\partial x_i}(x^1_i-x^0_i)=0$, and then I don't see how I can find $x^1$ as this equation will have an infinite number of solutions...
Please let me know if I am missing something obvious, or if there is a smart trick to choose $x^1$.
Thanks!
AI: For a smooth function of more than one variable, the solution to the equation
$$f(x_1,\ldots,x_n)=0$$
is not a single point in $\mathbb{R}^n$ but zero or more hypersurfaces (that only in some particular situation could reduce to a single point).
For example, in the simple case of two variable $f(x,y)=0$ represents a curve. |
H: Roots of rational equation with multiple variables?
Let's say we have a rational polynomial in $k$ variables. We are only interested in rational solutions. If $k = 1$, name the variables ${x}$, if $k = 2$, name them ${x,y}$.
For $k = 1$, it can be done very fast. The Rational Root Theorem gives a set of candidates. But what for $k=2$? How can I split a polynomial into factors in this case?
Example: $x^2-y^2$ should be split to $(x-y)(x+y)$.
Post also available on https://stackoverflow.com/questions/11922956/roots-of-rational-equation-with-multiple-variables.
AI: Polynomials in more than one variable do not generally split into factors, even if you allow complex coefficients. For example, $x^2 + y^2 - 1$ doesn't split in this way.
For quadratic polynomials you can appeal to the Hasse-Minkowski theorem, the Chevalley-Warning theorem, and Hensel's lemma to determine when a solution exists; this argument is described in more detail in the beginning of Cassels' Lectures on Elliptic Curves.
Beyond the quadratic case, this problem is open. Already for cubic polynomials in two variables it is not known whether there exists an algorithm which provably solves this problem, although there appear to be algorithms which work reasonably well in practice. Bjorn Poonen's Computing Rational Points on Curves contains a good discussion of the issues involved. See also, for example, this MO question.
Note that Fermat's Last Theorem can be phrased as the problem of finding rational points on the family of Fermat curves $x^n + y^n = 1$, so there's no reason to expect that this is an easy problem if you believe that Fermat's Last Theorem is difficult. |
H: Showing $\lim_{n \to +\infty} \log(n!)/(n\log n) = 1$ without using Stirling approximation
As a passage of a bigger limit I have to show that $$ \lim_{ n \to \infty } \frac{\log(n!)}{n\log(n)} = 1. $$
I think it could be done using Stirling approximation, but I'm wondering if there's a way without that formula.
AI: We compute $$ \lim_{n\to\infty} \frac{ \log( (n+1)! ) - \log (n!) }{ (n+1)\log(n+1) - n\log n} = \lim_{n\to\infty} \frac{ \log (n+1) }{ \log (n+ \theta_n)+1}=1$$
where $\theta_n \in (0,1)$ is obtained from the Mean Value Theorem. It follows from the Stolz-Cesàro theorem that the limit in question is also equal to $1.$ |
H: Proof by induction of Bernoulli's inequality $ (1+x)^n \ge 1+nx$
I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:
Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality:
$$ (1+x)^n \ge 1+nx$$
Proof:
Base Case: For $n=1$, $1+x = 1+x$ so the inequality holds.
Induction Assumption: Assume that for some integer $k\ge1$, $(1+x)^k \ge 1+kx$.
Inductive Step: We must show that $(1+x)^{k+1} \ge 1+(k+1)x$
Proof of Inductive Step:
$$\begin{align*}
(1+x)^k &\ge 1+kx \\
(1+x)(1+x)^k &\ge (1+x)(1+kx)\\
(1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\
1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\
\Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square
\end{align*}$$
AI: What you have is perfectly acceptable. The calculations could be organized a little more neatly:
$$\begin{align*}
(1+x)^{k+1}&=(1+x)(1+x)^k\\
&\ge(1+x)(1+kx)\\
&=1+(k+1)x+kx^2\\
&\ge1+(k+1)x\;,
\end{align*}$$
since $kx^2\ge 0$. This completes the induction step. |
H: Some questions about distribution theorem
Given an equation $P(D)u=0$, where $P$ is a polynomial (not equal to a constant). Here are some basic information about the distributional solution $u$:
If $P$ has at least one real root, then there exist a (non zero) solution $u \in S'$(temperate distribution);
It never has a (non zero) solution in $\epsilon'$(distribution with compact support).
These two properties can be checked easily by Fourier transform.
My question is why it always has a distributional solution(non zero) and a $C^{\infty}$ solution(non zero)?
For the distributional solution,it's only needed to check that $P(D)C_{0}^{\infty }(\mathbb{R}^{n})$ is a strictly smaller than $C_{0}^{\infty}(\mathbb{R}^{n})$,than Hahn-Banach theorem can be used.
For the $C^{\infty}$ solution, I don't know how to deal with it.
AI: Let $P(D)u=0$ with $u\in \mathcal{D}'$. Pick $\phi\in\mathcal{D}$. Then $\phi*u\in C^\infty$ and
$$
P(D)(\phi*u) = \phi*P(D)u = \phi*0=0.
$$ |
H: Express Expectation and Variance in other terms.
Let $X \sim N(\mu,\sigma^2)$ and
$$f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}.$$
where $-\infty < x < \infty$.
Express $\operatorname{E}(aX + b)$ and $\operatorname{Var}(aX +b)$ in terms of $\mu$, $\sigma$, $a$ and $b$, where $a$ and $b$ are real constants.
This is probably an easy question but I'm desperate at Probability! Any help is much appreciated as I'm not even sure where to start.
AI: If $a,b$ are constants, i.e. not random, then
$$
\mathbb{E}(aX+b) = a\mathbb{E}(X)+b,
$$
$$
\operatorname{var}(aX+b) = a^2 \operatorname{var}(X).
$$
Now plug in $\mu$ and $\sigma^2$ in the appropriate places. |
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