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H: A question about the convergence of the integral
Define $u : (-1,1) \to \Bbb R$ by $$ u(x) = \left( \log \frac{1}{1-x} \right)^{\alpha} ( 1/2 \leqslant x < 1 ), \;\;\;\;\;u(x) = (\log 2 )^{\alpha} \;( -1 < x \leqslant 1/2 ) $$
If $0 < \alpha < 1/2$ then I want to prove that $$ \int_{-1}^1 (1-x^2 ) | u'(x) |^2 dx + \int_{-1}^1 |u(x)|^2 dx < \infty . $$
AI: Split your integrals on the two intervals $(-1,1/2)$ and $(1/2,1)$. For the function on the interval $(-1,1/2)$, it is easy to see that they are finite. On the interval $(1/2,1)$ we have
$\int_{1/2}^{1} |u(x)|^2\, dx $, and you can do the same technique with the other one,
$$\int_{1/2}^{1} |u(x)|^2\, dx = \int_{1/2}^{1} \left({\ln\left(\frac{1}{1-x}\right)}\right)^{2\alpha}\, dx = \int _{\ln \left( 2 \right) }^{\infty }\!{t}^{2\,\alpha}{{\rm e}^{-t}
}{dt}
< \int _{ 0 }^{\infty }\!{t}^{2\,\alpha}{{\rm e}^{-t}
}{dt} = \Gamma(2\alpha + 1) < \infty \,,$$
where $\Gamma(2\alpha +1)$ is the gamma function, $0<\alpha<1/2$. We used the change of variables, $ x=1-{\rm e}^{-t} $. You can use the same change of variable and the same technique with the other integral. |
H: Probability density function of a quotient of two normal random variables
I have the following expression:
$$R=\frac{\sigma_1^2\nu_1(t)-\sigma_2^2\nu_2(t)}{\sigma_1^2\nu_1(t)+\sigma_2^2\nu_2(t)}$$
where:
$$[\nu_1(t),\nu_2(t)]$$ are two independent normally distributed random variables.
My question is: how can I find an expression for the probability density function $(pdf)$ of $R$?
Thanks.
AI: Let $\nu_1$ and $\nu_2$ be independent standard normal random variables. Then $U=\frac{\nu_1}{\nu_2}$ is well known to follow Cauchy distribution with pdf:
$$
f_U(u) = \frac{1}{\pi} \frac{1}{1+u^2}
$$
Let $X = \frac{\sigma_1^2 - U \cdot \sigma_2^2}{\sigma_1^2 + U \cdot \sigma_2^2}$. Assuming $\sigma_1>0$ and $\sigma_2 > 0$, it is evident that the mapping $u \mapsto \frac{\sigma_1^2 - u \cdot \sigma_2^2}{\sigma_1^2 + u \cdot \sigma_2^2}$ maps $\mathbb{R}\backslash \{ -\frac{\sigma_1^2}{\sigma_2^2} \}$ to $\mathbb{R}\backslash \{-1\}$. Indeed, for $x \not= -1$,
$$
\frac{\sigma_1^2 - u \cdot \sigma_2^2}{\sigma_1^2 + u \cdot \sigma_2^2} = x \qquad \implies \qquad u(x) = \frac{1}{1+x} \left( \frac{\sigma_1^2}{\sigma_2^2} - x \right)
$$
Thus we readily read off $f_X(x)$ from the measure:
$$ \begin{eqnarray}
\mathrm{d} F_U(u) &=& f_U(u) \mathrm{d} u = \frac{1}{\pi} \frac{|u^\prime(x)|}{1+u^2(x)} \mathrm{d}x = \frac{2}{\pi} \frac{\sigma_1^2 \sigma_2^2}{(1+x)^2 \sigma_2^4 + \sigma_1^4(1-x)^2} \mathrm{d}x \\
&=& \frac{2}{\pi} \frac{\sigma_1^2 \sigma_2^2}{\left(\sigma_1^4 + \sigma_2^4\right)\left(x - \frac{\sigma_1^4-\sigma_2^4}{\sigma_1^4+\sigma_2^4} \right)^2 + \frac{4 \sigma_1^4 \sigma_2^4}{\sigma_1^4+\sigma_2^4}} \mathrm{d}x = \mathrm{d}F_X(x)
\end{eqnarray}
$$
We therefore see that $X$ follows a Cauchy distribution with location parameters $\mu = \frac{\sigma_1^4-\sigma_2^4}{\sigma_1^4+\sigma_2^4}$ and scale parameter $\gamma = \frac{2 \sigma_1^2 \sigma_2^2}{\sigma_1^4 + \sigma_2^4}$. |
H: Sum of n different positive integers is less than 100. What is the greatest possible value for n?
I am stumped on the following question:
The sum of n different positive integers is less than 100. What is the greatest possible value for n?
a) 10, b) 11, c) 12, d) 13, e) 14
The answer is d).
Any idea on how to solve it ?
AI: Sum of different numbers is least when it's consecutive numbers from beginning from $1$. The sum would be
$$ {n(n+1) \over 2} \leq 100 $$
This inequality gives $ n \leq 13 $. |
H: construct $\{a_n\}$, for which exists $\{n_k\}$ and $\sum_{n=1}^{n_k}a_nx^n$ converges uniformly to $f$, where $f\in C[0,1]$, $f(0)=0$
Construct the sequence $\{a_n\}\subset\mathbb R$ such, that for every $f\in C[0,1]$ with $f(0)=0$, there exists a sequence $\{n_k\}$ for which
\begin{equation}
\lim_{k\to\infty}\sup_{x\in[0,1]}\left|\sum_{n=1}^{n_k}a_nx^n-f(x)\right|=0
\end{equation}
AI: Here is an outline: First you need a lemma:
If $f\in C[0,1]$ with $f(0)=0$ and $\epsilon>0$ and a natural number $n$ are given, there exists a polynomial $p$ with $\lvert x^np(x)-f(x)\rvert<\epsilon$ for all $x\in[0,1]$.
Then you pick a dense sequence $\{f_n\}$ in $\{f\in C[0,1]\colon f(0)=0\}$ where,moreover, every function in the sequence occurs infinitely often in the sequence. (Edited: There is no need for all this repetition, since every tail of a dense sequence is itself dense!)
Let $\{\epsilon_k\}$ be a sequence of positive numbers converging to $0$.
Now we can start the work: Pick a polynomial $p_1$ approximating $f_1$ better than $\epsilon_1$.
Let $p_1$ have degree $n_1-1$. Pick a polynomial $q_2$ so that $x^{n_1}q_2(x)$ approximates $f_2-p_1$ better than $\epsilon_2$. Let $p_2(x)=p_1(x)+x^{n_1}q_2(x)$. Note that $p_2$ approximates $f_2$ better than $\epsilon_2$.
Repeat the procedure of the previous paragraph. The coefficients of the polynomials $p_k$ as $k\to\infty$ will become the sequence $\{a_n\}$. |
H: How to find rate of depreciation in this problem?
The value of a machine is estimated to be 27,000 at the end of 1994 and 21,870 at the beginning of 1997. Supposing it depreciates at a constant rate per year of it's value at the beginning of the year, calculate:
1) Rate of depreciation
2) The value of the machine at the end of 1997 and at the beginning of 1994
How do I proceed? Is the time 2 or 3 years?
Where P = principal, CI = compound interest, n = number of years, R = rate
AI: The end of the year is basically the same as the beginning of the next year.
So, if $V_n$ is the value at the beginning of year $n$, then you have the formula $V_{n+1} = (1-r) V_n$ (this formula works 'backwards' as well, that is, you can compute $V_n$ knowing $r$ and $V_{n+1}$), where $r$ is the depreciation rate. You are given $V_{1995}$ and $V_{1997}$, from which you can compute $r$. Then, knowing $r$ you need to calculate $V_{1998}$ and $V_{1994}$. |
H: How many rectangles can fit in a polygon with n-sides?
I am trying to write an algorithm to solve a problem I have. I have a few ideas of what the algorithm might be like but I am posting to see if anyone else has a better more efficient solution or any helpful suggestions. For context, this algorithm is going to be for maximising the number of solar panels that can be placed on a roof.
Given a polygon with n-sides with coordinates $(x_1, y_1), (x_2, y_2) .....(x_n, y_n)$, how many rectangles of dimension (for example) $1\times3$ can fit within this shape?
The orientation of the rectangle can be portrait or landscape but must be consistent throughout. They are also to be arranged in a grid pattern.
At the moment my half solution is to "cut" the polygon into "strips" of, let's say, $1$ unit and see how many can fit on that and continue until all the strips are done. Alternatively, for simple polygons with a low "$n$", finding the largest rectangle that can fit within the polygon and fill it? I am keen to work out the coordinates of the rectangle that can fit in the polygon in the final solution!
Any help or suggestions would be much appreciated,
Kelvin.
EDIT: I am probably looking towards a more simple and crude solution rather than a complex and more accurate method (as I am a geographer and therefore my maths skills are a bit limited!)
AI: As has been mentioned over at SO, this problem is very difficult from a computational perspective. It may help to see what methods have been implemented for the packing problem. There are some algorithms out there, but you may have to fine-tune one to fit your needs.
Even a naive, greedy approach may end up being pretty inefficient. Would you take into account things like rotating the shape? There are all kinds of trade-offs to be made, but your "slices" (first fit) approach may work if you need something quick and dirty. |
H: Is the problem of calculating the induced norm *difficult*?
Is the problem of calculating the induced norm of a linear operator (in a finite or infinite-dimensional space) generally a difficult one ?
And by difficult I mean, that there are no closed formulas or no general procedure that always yields the induced norm.
Of course, for the usual spaces with the usual norms, there are formulas, that makes ones life very, so one can take shortcuts in calculating the induced norm of a operator (instead of trying to use the definition of the induced norm : $$ ||A||=\sup_{||x||=1} ||Ax||).$$
But is there also a procedure how to calculate $||A||$ for some very weird vector norms, or for some unsual infinitedimensional spaces (since in finite dimensions we can at least use the fact, that every vectors space is isomorphic to $\mathbb{K}^n$ for some $n$) ?
EDIT: I think the user tomasz bst described what I meant. Are there vector norms such that for their induced operator norm it is proven that there isn't a closed expression ?
AI: Following on from @tomasz's comment: There are examples of operators whose induced norm is NP-hard to approximate even though the norm in the original vector space can be immediately computed.
Here, we'll work with the real vector space $\ell_p^n$. Consider the problem of computing the norm of a matrix $T: \ell_p^n \to \ell_p^n$,
$$
\Vert T \Vert_{p \to p} = \sup_{\Vert x \Vert_p =1} \Vert Tx \Vert_p.
$$
If $p \in \{1, \infty\}$, then the exact value of this norm is easily computed. If $p=2$, then it can be efficiently approximated (where "efficiently" depends on the size $n$ of the problem and the desired accuracy $\varepsilon$). However, this is not true for other values of $p$ even when we insist that the map satisfies the condition $T_{ij} \in \{-1,0,1\}$.
The following theorem is taken from http://arxiv.org/abs/0908.1397 .
Theorem: For any rational $p ∈ [1,\infty)$ except $p = 1,2$, unless P = NP, there
is no algorithm which computes the p-norm of a matrix with entries in $\{−1, 0, 1\}$ to
relative error $\varepsilon$ with running time polynomial in $n, 1/\varepsilon$ .
In particular, there should be no "nice", "easy" formula in terms of $T_{ij}$ that immediately tells you the value of the induced norm $\Vert T \Vert_{p \to p}$. |
H: How to find the original population if it increases with a constant rate?
The population increases by 5% every year. What was the population in 1982, if in 1985 it was 1,85220?
My working:
population in 1985 = 1,85,220
rate=5%
time = 3yrs
( A=P(1-R/100)^n )
therefore, population in 1982 = 185220(1 - 5/100)^3
= 158802.9975
obviously that's wrong, but where's the problem?
AI: You are making a basic error: If you increase a quantity $P_1$ by $y\%$ of itself to obtain the new value $P_2$, then decreasing $P_2$ by $y\%$ of itself does not get you back to $P_1$ (because you also decrease by $y\%$ of the increase previously gained). For example, increasing $100$ by $10\%$ of itself gives you $110$, but decreasing $110$ by $10\%$ of itself gives you $99$.
You need to start with the 1982 value, call this $x$, and then find an equation in $x$ and solve. William's answer explains how to do this.
(Incidentally, why do you say your answer is "obviously wrong"?) |
H: (ZF) Every nonempty perfect set in $\mathbb{R}^k$ is uncountable.
This is the part of proof in Rudin PMA p.41
Let $P(\subset \mathbb{R})$ be a perfect set. Since $P$ has limit points, $P$must be infinite. Suppose that $P$ is countable. Then, we can denote the points of $P$ by $x_1, x_2,...$.
Let $V_1$ be any neighborhood of $x_1$.(i.e. open ball). Suppose $V_n$ is constructed. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ of some point $x_m \in P$ such that (i) $\overline {V_{n+1}}$ $\subset V_n$ and (ii) $x_n \notin \overline {V_{n+1}}$ and (iii) $V_{n+1} \cap P ≠ \emptyset$. Then form a sequence $\{V_n \subset \mathbb{R}^k | n\in \omega \}$.
Here, Axiom Of dependent choice is used.
I have tried some other ways, but ,informally speaking, proof by 'squeezing' region requires AC. (Forming a decreasing sequence)
I want a proof without AC.
Help..
AI: The trick is simple. Use the fact the $P$ is well-ordered and that the rationals are well-ordered.
In the induction step, instead of taking "some $x_m\in P$" take $x_m$ such that $m$ is the least $k$ for which $x_k$ has an open neighborhood etc. etc.
Also require that the open neighborhoods are open balls of rational radius. Well ordering the rationals we can require the radius to be of the least rational in a fixed enumeration such that a ball around $x_m$ with the wanted properties exists.
Now we have a canonical choice of the open neighborhoods, and we can show that their intersection is non-empty because it is equal to the intersection of the closures - which is compact (by a previous question of yours) and therefore contains a point. This point is not in $P$. |
H: How many int. values of n will the expression be greater than 1
How would I solve this problem:
How many integer values of n will the expression $4n+7$ be an integer greater than 1 and less than 200 a)48 b)49 c)50 d)51 e)52 ? Ans 50
I am trying to solve this by using inequalities and doing something like this
$200 > 4n + 7 > 1$
$193 > 4n > -6 $
$\frac{193}{4} > n > \frac{-6}{4}$
However this just gives me the range. How do i get the number of integers 50 ?
AI: We can work with inequalities as you did. We want
$$1\lt 4n+7 \lt 200,$$
which can be rewritten as
$$-6 \lt 4n \lt 193,$$
and then as
$$\frac{-6}{4} \lt n \lt \frac{193}{4}.$$
Note that $\,\frac{-6}{4}=-1.5\,$ and $\,\frac{193}{4}=48.25\,$
So all integers from $-1$ to $48$ inclusive work, and no others. There are $\,50
\,$ of them: the integers $-1$ and $0$, plus the $48$ integers from $1$ to $48$. |
H: Show the result of the following infinite sum, based on a binomial random variable conditioned on a Poisson random variable
$$\sum_{n=0}^\infty \binom{n}{k}p^k(1-p)^{n-k}\frac{\lambda^ne^{-\lambda}}{n!} = \frac{(\lambda p)^ke^{-\lambda p}}{k!}$$
That is, given a random variable $X$ with Poisson distribution $X \sim \operatorname{Poisson}(\lambda)$, and, given $X = n$, random variable $Y$ is distributed by a binomial such that $Y \sim B(n,p): p \in [0,1]$. Given $n$, $P(Y = k|X = n) = \binom{n}{k}p^k(1-p)^{n-k}$. Given $\lambda$, $P(X=n) = \dfrac{\lambda^ne^{-\lambda}}{n!}$. Therefore, $P(Y = k|\lambda) = \sum_{n=0}^\infty P(Y = k | X = n)P(X = n | \lambda) = \sum_{n=0}^\infty \binom{n}{k}p^k(1-p)^{n-k}\dfrac{\lambda^ne^{-\lambda}}{n!}$. What properties and identities can be used to demonstrate the equality above?
AI: A start: We want $\Pr(Y=k)$, so we should end up with an expression that involves $k$. Let $m=n-k$. Effectively we are summing from $m=0$ to $\infty$.
Take the terms that involve only $k$, $\lambda$, and $p$ to the outside, not forgetting that $\lambda^n=\lambda^{m+k}=\lambda^k\lambda^m$ and $\binom{m+k}{k}\cdot \frac{1}{(m+k)!}=\frac{1}{k!}\cdot\frac{1}{m!}$. The stuff that remains on the inside looks like
$$\sum_{m=0}^\infty \frac{(\lambda(1-p))^m}{m!}.$$
From the familiar series expansion of $e^x$, we see that this sum is $e^{\lambda(1-p)}$.
Remark: The above approach is crudely computational. A much better conceptual justification has been given in the answer by Robert Israel. |
H: Using Simpson's Rule to approximate $\int_0^3{\sqrt{9-x^2}dx}$
I have to use Simpson's rule:
$$\int_a^b f(x) \, dx \approx S_n = \frac{1}{3}h[f(a)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+\cdots+4f(x_{n-1})+f(b)]$$
when $n=6$ to approximate the integral:
$$\int_0^3{\sqrt{9-x^2}dx}$$
to four decimal places.
I've gotten $$f(0) = 3$$
$$ 4f\left(\frac{1}{2}\right)=11.831$$
$$2f(1)=5.6569$$
$$4f\left(\frac{3}{2}\right)=10.3923$$
$$2f(2)=4.4721$$
$$4f\left(\frac{5}{2}\right)=6.6332$$
$$f(3)=0$$
and then..
$$\int_0^3{\sqrt{9-x^2} \, dx}\approx S_6 = \left (\frac{1}{3}\right )\left(\frac{1}{2}\right )\left (41.9857\right ) = 6.9976 $$
Did I do this right and is this the correct answer?
AI: Since $h=\dfrac {b-a}n$ and $a=0, b=3$:
$$\int_a^b{f(x)dx}\approx S_n = \frac{1}{3}h[f(a)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+4f(x_{n-1})+f(b)]\\=\frac 16 ( 3+ 11.8321+ 5.6568+10.3923+4.4721+6.6332+0)\\=6.9977$$ |
H: Are bounded open regions in $\mathbb{R}^n$ determined by their boundary?
Let $U$ and $V$ be two bounded open regions in $\mathbb{R}^n$, and let us further assume that their topological boundaries are nice enough that they are homeomorphic to finite simplicial complexes.
Assume $\partial U$ is homeomorphic to $\partial V$.
Is $U$ be homeomorphic to $V$?
A weaker question: if $U$ is contractible, is $V$ contractible?
For $n=1$, the answer to both is yes, since $\partial U\simeq \partial V$ are sets of $2i$-many points, which both divide $\mathbb{R}^1$ into $i$-many open intervals.
For $n=2$, the answer to both is yes, since $\partial U\simeq \partial V$ are sets of $i$-many topological circles, which both divide $\mathbb{R}^2$ into $i$-many open discs.
AI: No for $n=2$. Consider a disk with two circular holes (red in the picture below) versus a disk inside an annulus (blue) versus three disks (green). In all cases the boundary consists of three disjoint circles. |
H: What is smallest possible integer k such that $1575 \times k$ is perfect square?
I wanted to know how to solve this question:
What is smallest possible integer $k$ such that $1575 \times k$ is a perfect square?
a) 7, b) 9, c) 15, d) 25, e) 63. The answer is 7.
Since this was a multiple choice question, I guess I could just put it in and test the values from the given options, but I wanted to know how to do it without hinting and testing. Any suggestions?
AI: Consider the prime factorization of $1575$: $1575=3^2 \cdot 5^2 \cdot 7$. Then $\sqrt{1575}=\sqrt{3^2 \cdot 5^2 \cdot 7}=\sqrt{3^2}\sqrt{5^2}\sqrt{7}=3\cdot 5 \cdot \sqrt{7}$. Clearly the problem here is that $7$ is not a perfect square. So what's the smallest possible integer $k$ that we can multiply $1575$ by so that $1575k$ is a perfect square? Well, we have to fix the problem of $7$ not being a perfect square, so let's multiply by $7$. Is $7$ the smallest number that makes $1575k$ a perfect square? Well, multiplying by $1$ clearly doesn't help, multiplying by $2$ doesn't help since we'll end up with a $\sqrt{2}$ which isn't an integer, etc. Multiplying by $3$ or $5$ won't help since we'd end up with a $\sqrt{3^3}$ or $\sqrt{5^3}$, which does not yield an integer. So we can safely conclude that $7$ is the least integer $k$ such that $1575k$ is a perfect square.
In short, since we can split the square root function apart across the prime decomposition of a number, each distinct prime power must be a perfect square in order for the whole number to be a perfect square. Hope this helps. |
H: What function has a graph that looks like this?
I delete my file which I used to produce this graph. Does anybody have some idea how to produce it again?
Thanks for a while.
AI: For example :
$$f(x,y)=\sqrt{x^2+y^2}\sin(8 \arctan(y/x))$$ |
H: Is there a Cantor-Schroder-Bernstein statement about surjective maps?
Let $A,B$ be two sets. The Cantor-Schroder-Bernstein states that if there is an injection $f\colon A\to B$ and an injection $g\colon B\to A$, then there exists a bijection $h\colon A\to B$.
I was wondering whether the following statements are true (maybe by using the AC if necessary):
Suppose $f \colon A\to B$ and $g\colon B\to A$ are both surjective, does this imply that there is a bijection between $A$ and $B$.
Suppose either $f\colon A\to B$ or $g\colon A\to B$ is surjective and the other one injective, does this imply that there is a bijection between $A$ and $B$.
AI: For the first one the need of the axiom of choice is essential. There are models of ZF such that $A,B$ are sets for which exists surjections from $A$ onto $B$ and vice versa, however there is no bijection between the sets.
Using the axiom of choice we can simply inverse the two surjections and have injections from $A$ into $B$ and vice versa, then we can use Cantor-Bernstein to ensure a bijection exists.
The second one, I suppose should be $f\colon A\to B$ injective and $g\colon A\to B$ surjective, again we need the axiom of choice to ensure that there is a bijection, indeed there are several models without it where such sets exist but there is no bijection between them. Using the axiom of choice we reverse the surjection and use Cantor-Bernstein again.
It should be noted that without the axiom of choice it is true that if $f\colon A\to B$ is injective then there is $g\colon B\to A$ surjective. Therefore if the first statement is true, so is the second, and if the second is false then so is the first.
Another interesting point on this topic is this: The Partition Principle says that if there is $f\colon A\to B$ surjective then there exists an injective $g\colon B\to A$. Note that we do not require that $f\circ g=\mathrm{id}_B$, but simply that such injection exists.
This principle implies both the statements, and is clearly implied by the axiom of choice. It is open for over a century now whether or not this principle is equivalent to the axiom of choice or not.
Lastly, as stated $f\colon A\to B$ injective and $g\colon B\to A$ surjective cannot guarantee a bijection between $A$ and $B$ with or without the axiom of choice. Indeed the identity map is injective from $\mathbb Z$ into $\mathbb R$, as well the floor function, $x\mapsto\lfloor x\rfloor$ is surjective from $\mathbb R$ to $\mathbb Z$ but there is no bijection between $\mathbb Z$ and $\mathbb R$. |
H: Generating function with binomial coefficients
I want to derive formula for generating function $$\sum_{n=0}^{+\infty}{m+n\choose m}z^n$$ because it is very often very useful for me. Unfortunately I'm stuck:
$$ f(z)=\sum_{n\ge 0}{m+n\choose n}z^n= \\ \sum_{n\ge 1}{m+n-1\choose n-1}z^n+\sum_{n\ge 0}{m+n-1\choose n}z^n= \\ zf(z)+\sum_{n\ge 0}{m+n-1\choose n}z^n$$
here, I'm afraid that there is a need for something more sophisticated than above trivial identity. Can you help me?
AI: The function $\left(1 + z\right)^k$ has series expansion $$\left(1 + z\right)^k = \sum_{n\ge0}\binom{k}{n}z^n$$
We also have the binomial coefficient identity $$\binom{n}{k} = (-1)^k\binom{-n + k - 1}{k}$$ which together gives
$$\sum_{n\ge 0}\binom{m+n}{n}z^n = \sum_{n\ge 0}\binom{-m-1}{n}(-z)^n = \frac{1}{(1-z)^{m+1}}$$
which is your required generating function. |
H: Different ways to wedge spaces
I am not a topologist, so please excuse me if the question is trivial.
Suppose I am given three nice, path-connected spaces. Then I can think of two ways to wedge these spaces: join all three at a common basepoint, or have one space in the "middle" with two basepoints (so I guess there are four ways in total). Do these constructions result in the same space up to homotopy? My guess would be to take a path between the two basepoints in the "middle" space and contract it to a point, but I understand that it is sometimes a subtle matter to tell if this sort of thing is a homotopy equivalence.
AI: Generally these things aren't homotopy-equivalent.
As Qiaochu mentions, when you wedge together only two spaces, the homotopy-type of the wedge depends on your choice of basepoint (even if they're path connected).
What you need is that your input spaces are path connected and moreover, given your two choices of base-point you want a homotopy-equivalence $(X,x_0) \simeq (X,x_1)$.
I don't know if these kinds of spaces have a name or not, it seems like homogeneous would be fairly appropriate. |
H: Distribution function and $L^p$ spaces.
I saw the result below without a proof and I would like to see it.
Result:
Let $g$ a nonnegative and measurable function in $\Omega$ and $\mu_{g} $ its distribuction function, i.e.,
\begin{equation}
\mu_{g}(t)= |\{x\in \Omega : g(x)>t\}|, t>0.
\end{equation}
Let $\eta>0$ and $M>1$ be constants. Then, for $0<p<\infty,$
\begin{equation}
g \in L^{p}(\Omega) \Leftrightarrow \sum_{k\ge 1} M^{pk} \mu_{g}(\eta M^k) = S < \infty.
\end{equation}
AI: A consequence of Fubini's theorem for non-negative functions is that for any $p>0$,
$$\int_{\Omega}g(x)^pdx=\int_0^{+\infty}p\mu_g(t)t^{p-1}dt=\sum_{k=0}^{+\infty}\int_{\eta M^k}^{\eta M^{k+1}}p\mu_g(t)t^{p-1}dt.$$
Let $a_k:=\int_{\eta M^k}^{\eta M^{k+1}}p\mu_g(t)t^{p-1}dt$. Consider the case $p\geq 1$. Since $\mu_g$ is decreasing and $t\mapsto t^{p-1}$ is increasing, we have
$$\mu_g(\eta M^{k+1})\eta M^k(M-1)\eta^p (M^k)^{p-1}\leq a_k\leq \mu_g(\eta M^k)(M-1)M^k\eta^{p-1}(M^{k+1})^{p-1},$$
hence
$$C_1\mu_g(\eta M^{k+1})M^{(k+1)p}\leq a_k\leq C_2\mu_g(\eta M^k)M^{kp}.$$
This gives the wanted equivalence.
When $p<1$, a similar argument applies. |
H: Finitely generated module over $\mathbb{Q}$.
Consider the following problem.
Problem: Suppose $M$ is a module over $\mathbb{Q}[x]$ such that $M$ is finitely generated over $\mathbb{Q}$. Prove that there is a non-zero polynomial $p(x) \in \mathbb{Q}[x]$ and a non-zero $m \in M$ such that $p(x)*m = 0$.
My attempt: Since $M$ is finitely generated over a field, $M \simeq \mathbb{Q}^n$ as a $\mathbb{Q}$-module. If $M$ is torsion-free over $\mathbb{Q}[x]$ too, then $M \simeq \mathbb{Q}[x]^m$ as $\mathbb{Q}[x]$-modules. Now restricting the action to $\mathbb{Q}$ again, we get $\mathbb{Q}^n \simeq \mathbb{Q}[x]^m$ as $\mathbb{Q}$-modules. At this point, I feel this is absurd, but can't quite explain why.
My questions: Is the above reasoning correct? How do I finish the argument? If not, what would be the right way to prove this?
Thanks.
AI: $m, xm, x^2m,...$ cannot all be linearly independent over $\mathbb{Q}$. hence for every $m\in M$ there is a $p_m(x)\in\mathbb{Q}[x]$ with $p_m(x)m=0$ |
H: Blocks in sequences from {1,...,k}
In Lecture Notes on Enormous integers Harvey M. Friedman introduces
"... longest finite sequence $x_1,...,x_n$ from $\{1,...,k\}$ such that for
no i < j <= n/2 is $x_i,...,x_{2i}$ a subsequence of $x_j,...,x_{2j}$.
For k ≥ 1, let n(k) be the length of this longest finite
sequence."
Then, the author evaluates this function
"Paul Sally runs a program for gifted high school students at
the University of Chicago.
He asked them to find n(1), n(2), n(3).
They all got n(1) = 3. One got n(2) = 11. Nobody reported
much on n(3)."
which I fail to confirm. Consider 12 character word
001011111101
neither of its starting subsequences
00
010
1011
is contained in "doubled" subsequences and suggest n(2) = 13. Am I missing anything?
AI: $00$ is a subsequence of $010$. You seem to be thinking of substring/subword. |
H: basic question on integral schemes
If $X$ is a (reduced) scheme and $P$ is a point of $X$ (not necessarily closed) such that the local ring $\mathcal{O}_{X,P}$ is a regular domain, then must there exist an open affine neighborhood $U = \text{Spec }A$ of $P$ such that $A$ is an integral domain?
I'm almost certain this is true, since the local ring being regular means that it doesn't sit in the intersection of irreducible components, and hence it must be "locally" irreducible...but I can't think how to prove it.
If needed, we can assume $X$ is also Noetherian.
AI: The theory of irreducible components can be tricky when $X$ is not Noetherian,
so I will suppose that $X$ is Noetherian, and so is the union of finitely many
irreducible components.
Now the components on which $P$ lies are in bijection with the minimal primes
of $\mathcal O_{X,P}$, and hence $P$ lies on a single component. The complement of all the other components is an integral open subscheme of $X$ containing $P$,
and (like an scheme) it contains an open affine n.h. containing $P$, which will also be integral.
If we don't assume that $X$ is Noetherian, there can be topological complications
in the theory of irreducible compoments. E.g. if $A$ is any Boolean ring (i.e. a ring of characteristic $2$ in which every element is idempontent), then each local ring of Spec $A$ is a domain, in fact a copy of $\mathbb F_2$, but Spec $A$ can contain points that are not contained in any integral open subscheme; one concrete example is given by $A = \prod_{n = 1}^{\infty} \mathbb F_2$ (and there are many others). |
H: Solving $5^n > 4,000,000$ without a calculator
If $n$ is an integer and $5^n > 4,000,000.$ What is the least possible value of $n$? (answer: $10$)
How could I find the value of $n$ without using a calculator ?
AI: \begin{eqnarray}
& 5^n &>& 4.000.000\\
\Leftrightarrow & 5^n &>& 5^6 \cdot 2^8 \\
\Leftrightarrow & 5^{n-6} &>& 256.\\
\end{eqnarray}
Then, $n=10$. |
H: Geometric difference between two actions of $GL_n(\mathbb{C})$ on $G\times \mathfrak{g}^*$
Let $G=GL_n(\mathbb{C})$.
Scenerio 1: Let $G$ act on $T^*(G)=G\times \mathfrak{g}^*$ by
$$
g.(x,y)=(gx,y).
$$
Scenerio 2: Let $G$ act on $T^*(G)=G\times \mathfrak{g}^*$ by
$$
g.(x,y)=(xg^{-1},gyg^{-1}).
$$
Is there a geometric difference between the two scenerios?
AI: The actions are equivalent. Consider the map $\phi:G\times\frak{g}^*$ $\to G\times\frak{g}^*$ defined by $(x,y)\mapsto (x^{-1},xyx^{-1}),$ where we consider the domain with the first action, and the codomain with the second. Then this map is equivariant, since $$\phi(g\cdot(x,y)) = \phi((gx,y))=(x^{-1}g^{-1},(gx)y(x^{-1}g^{-1}))= g\cdot(x^{-1},xyx^{-1})=g\cdot\phi(x,y).$$
To see that $\phi$ is a bijection is straightforward, thus the $G$-actions are equivalent. |
H: I want to prove $ \int_0^\infty \frac{e^{-x}}{x} dx = \infty $
How can I prove this integral diverges?
$$ \int_0^\infty \frac{e^{-x}}{x} dx = \infty $$
AI: $$
\int_{0}^{\infty}\frac{e^{-x}}{x}= \int_{0}^{1}\frac{e^{-x}}{x}+\int_{1}^{\infty}\frac{e^{-x}}{x} \\
> \int_{0}^{1}\frac{e^{-x}}{x} \\
> e^{-1}\int_{0}^{1}\frac{1}{x}
$$
which diverges. |
H: Proof that both $ \int_0^\infty | \ell_0 (x) |^2 e^{-x} dx$ and $\int_0^\infty | \ell_1 (x) |^2 e^{-x} dx $ diverge
Define $ \ell_0 (x)$ by $$ \ell_0 (x) := \int_1^x \frac{e^{\zeta}}{\zeta} d \zeta $$ and define $\ell_1 (x)$ by $\ell_1 (x) := (1-x) \ell_0 (x) + x \ell_0 ' (x).$ Then I want to prove that $$ \int_0^\infty | \ell_0 (x) |^2 e^{-x} dx \quad \text{and} \quad \int_0^\infty | \ell_1 (x) |^2 e^{-x} dx $$ both diverge.
AI: Note that $e^{\zeta/2}>\zeta$ for $\zeta\geq 2$, and so $\ell_0(x)\geq \int_2^x e^{\zeta/2}d\zeta=2e^{x/2}-2e$. Thus
$$\int_0^\infty|\ell_0(x)|^2e^{-x}dx= \int_0^\infty \frac{4e^x-4e^{x/2+1}+4e^2}{e^x}dx$$
and since the limit of the integrand as $x\to\infty$ is clearly $4$ (hence not $0$), this integral cannot converge. I'll leave the second one to you. |
H: Constructing a local nested base at a point in a first-countable space
I am trying to prove the following:
Let $X$ be a first countable space and $x$ a member of $X$. Prove that there is a local nested basis $\{S_n\}_{n=1}^\infty$ at $x$.
Since $X$ is first countable there is a countable local base $\mathcal{B}_x$ at $x$. Constructing a nested sequence of subsets of $\mathcal{B}_x$ is easy. Let $B_1 \in \mathcal{B}_x$. Then $B_1$ is an open set containing $x$, and so contains a member $B_2$ of $\mathcal{B}_x$ by the definition of a local base. Then $B_2$ is an open set containing $x$ and so contains a member $B_3$ of $\mathcal{B}_x$. Continuing in this fashion we obtain a nested sequence $\{B_n\}_{n=1}^\infty$ of members of $\mathcal{B}_x$ containing $x$.
I'm having trouble showing that this is a local base, in that I don't see how to prove that every open set in X contains some member of this sequence. Of course it's possible that this isn't true, and there's a different way to construct the nested sequence so that this can be done, but I'm having trouble with this as well.
Which way do I need to proceed, and how do we complete the proof? Thanks.
AI: Note that although you are correct that there is some $B_2 \in \mathcal{B}_x$ such that $B_2 \subseteq B_1$, since $B_1 \in \mathcal{B}_x$ it might be that $B_2 = B_1$, and this could continue ad infinitum. If this were to happen (at any point along the inductive construction) it would almost certainly not generate a basis at $x$ (unless $x$ has a smallest open neighbourhood and we picked it at some point).
To avoid this problem we begin by first fixing any countable local base $\{ B_i \}_{i \in \mathbb{N}}$ at $x$. Next, for each $i \in \mathbb{N}$ we define $$S_i := B_1 \cap \cdots \cap B_i.$$
Note for each $i \in \mathbb{N}$ that
as the set $S_i$ is a finite intersection of open neighbourhoods of $x$, it is itself an open neighbourhood of $x$; and
$S_{i+1} \subseteq S_{i}.$
Therefore $\{ S_i \}_{i \in \mathbb{N}}$ is a nested family of open neighbourhoods of $x$, so we need only show that it is a local base at $x$. But this follows immediately from the fact that $S_i \subseteq B_i$ for each $i$, and we chose $\{ B_i \}_{i \in \mathbb{N}}$ to be a local base at $x$. |
H: A calculation involving two circles
I have two circles, one of which is completely within the other. They do not touch, but are not necessarily concentric. I am given the sum of their circumferences, and the difference in their areas (ie, the area of the space inside the outer circle and outside the inner circle). I need to find the average distance between the circles or, equivalently, the radii of the circles.
In other words, given $C = C_o + C_i$, and $A = A_o - A_i$, find $R_o$ and $R_i$.
I've gotten as far as expressing $A$ in terms of $C_o$ and $C_i$, but then my long-unused algebra fails me. Any help? Thanks...
AI: Recall the area is $\pi r^2$ and circumference is $2\pi r.$
The sum of circumferences gives you equation:
$$ r_1 + r_2 = \frac{C}{2\pi}. \tag{1}$$
The difference of areas gives a second equation:
$$ r_1^2 - r_2^2 = \frac{A}{\pi}. \tag{2}$$
Factor LHS of $(2)$ into $(r_1 - r_2)(r_1 + r_2),$ substitute the value of $r_1 + r_2$ from $(1),$ and you will then have 2 linear equations in two unknowns, which you can solve. |
H: Properties and identities of $\text{ord}_{p}(n)$
$\mathrm{ord}_{p}(a+b)\ge\mathrm{min}(\mathrm{ord}_{p}a,\mathrm{ord}_{p}b)$ with equality holding if $\mathrm{ord}_{p}a\ne \mathrm{ord}_{p}b$. is a the statement that prompted this question.
It was found in Ireland & Rosen's Elements of Number Theory (precurser to their book A Classical Introduction to Modern Number Theory) is the book that I am working through and it asks for its proof.
After some research I'm learned that this function is completely additive($\mathrm{ord}_{p}(ab)=\mathrm{ord}_{p}(a)+\mathrm{ord}_{p}(b)$) among other things and managed to get the following equation out that if I didn't make a mistake, solves the first part of the question:
$\mathrm{ord}_{p}(a+b)=\mathrm{ord}_{p}(dm+dn)=\mathrm{ord}_{p}(d)+\mathrm{ord}_{p}(m+n)\ge \mathrm{ord}_{p}(d)=\mathrm{ord}_{p}(a,b)=\mathrm{min}(\mathrm{ord}_{p}a,\mathrm{ord}_{p}b)$
Where $(a,b)$=$\mathrm{gcd}(a,b)$ is the ideal/greatest common factor and is equal to $d$, and $a=dm$, $b=dn$ where $m$,$n$ are relatively prime.
So my question is where can I learn more about the properties and identities of the functions $\mathrm{ord}_{p}(n)$ and $\mathrm{v}(n)$ (the latter seems to be called valuation or related to such), especially a resource that would include relations similar to those above. If the resource included identities for the lcm, gcd, and min/max functions in relation to the ord function and by themeslves also, that would be wonderful.
Also what are some identities that could solve the "with
equality holding if $\mathrm{ord}_{p}a\ne \mathrm{ord}_{p}b$." part and how are they derived? Edit(This part has been answered, looking for some good resources that expound on ord,v, and related functions.)
Thank you for your assistance.
AI: Gerry's answer is great, though I can't help but add that this is a deep and beautiful idea that generalises and is extremely useful in geometric situations as well (algebraic geometry and commutative algebra). See here, for example. |
H: The orientation of quotient manifold
If $T$ is a torus and $\mathbb Z_2$ acts on it by $(z_1,z_2)\rightarrow(z^{-1}_1,-z_2)$, then is $T/\mathbb Z_2$ orientable?
AI: This argument is morally the same as Matt E's, but interpreted in de Rham cohomology. We denote the quotient by $X$; observe that $X$ inherits the structure of a smooth manifold from $T$ via the quotient map $h : T \to X$. Let $f : T \to T$ be the above-mentioned involution. It is not hard to check that $f$ is an orientation-reversing isometry of $T$, and that $h \circ f = h$.
Now, a smooth manifold $X$ is orientable if and only if there exists a nowhere-vanishing differential 2-form on $X$, so let us consider an arbitrary 2-form $\omega$ on $X$. The pullback 2-form $h^* \omega$ on $T$ satisfies $f^* h^* \omega = h^* \omega$. Now take de Rham cohomology classes; noting that $f^* \mu = - \mu$ where $\mu$ is the volume form on $T$ and the fact that $\mu$ generates $H^2 (T, \mathbb{R})$, we conclude that $h^* \omega$ must be cohomologous to $0$. In particular, $h^* \omega$ must vanish somewhere. But that means $\omega$ also vanishes somewhere – so $X$ is not orientable. |
H: Definition and example of a partition
Partition of a Set is defined as "A collection of disjoint subsets of a given set. The union of the subsets must equal the entire original set." For example, one possible partition of $(1, 2, 3, 4, 5, 6 )$ is $(1, 3), (2), (4, 5, 6).$ Rudin, while defining integral on page $120$ starts like this,
Definition Let $[a, b]$ be a given interval. By a partition $P$ of
$[a,b]$ we mean a finite set of points $x_0, x_1,..., x_n$, where
$a=x_0\leq x_1\leq...\leq x_n=b$.
if all the points from $a$ to $b$ are in partition $P$ then where is the other partition, is it $\phi$?, not mentioned in the book.
AI: This is a slightly different kind of partition. Here the idea is that the interval $[a, b]$ is being partitioned into sub-intervals $[x_0, x_1], [x_1, x_2], \ldots$.
As with the kind of partition you defined, the sub-intervals here completely cover the original set $[a, b]$. Unlike with the kind of partition you defined, the sub-intervals here are not exactly disjoint. Instead they are almost disjoint, since they overlap only at their endpoints.
Rudin says that the points $x_0, x_1,\ldots$ "are" the partition, but that is just because once you know those points, you know everything there is to know about the way that $[a,b]$ has been divided into sub-intervals. In a more general setting, with the definition you quoted, that is not the case. |
H: Differentiation continuous iff domain is finite dimensional
Let $A\subset C([0,1])$ a closed linear subspace with respect to the usual supremum norm satisfying $A\subset C^1([0,1])$. Is $D\colon A\rightarrow C([0,1]), \ f\rightarrow f'$ continuous iff $A$ is finite dimensional?
If $A$ is finite dimensional $D$ is continuous of course. But is the other implication true at all? Wouldn't something like $A:=\overline{\text{span}\{\sin{\left(t+\frac{1}{n}\right)} \ | \ n\in\mathbb{N}\}}$ be a counterexample?
AI: From the dicussion above it follows that it is enough to show equicontinuiuty of $\mathrm{Ball}_A(0,1)$ provided $D$ is continuous.
Fix $\varepsilon>0$ and take $\delta=(2\Vert D\Vert)^{-1}\varepsilon$. From mean value theorem it follows that for all $x_1,x_2\in [0,1]$ such that $|x_2-x_1|\leq\delta$ and all $f\in\mathrm{Ball}_A(0,1)$ we have
$$
|f(x_2)-f(x_1)|=|f'(\xi)(x_2-x_1)|=|f'(\xi)||x_2-x_1|\leq\Vert f'\Vert_\infty\delta\leq
$$
$$
\Vert D\Vert\Vert f\Vert_\infty\delta\leq\Vert D\Vert\cdot 1\cdot \frac{\varepsilon}{2\Vert D\Vert}<\varepsilon
$$
This means that $\mathrm{Ball}_A(0,1)$ is equicontinuous. |
H: A question on the compact subset
This is an exercise from a topological book.
Let $X$ is Hausdorff and $K$ is a compact subset of $X$. $\{U_i:i=1,2,...,k\}$ is the open sets of $X$ which covers $K$. How to prove that there exist compact subsets of $X$: $\{K_i:i=1,2,...,k\}$ such that $K=\cup^k_{i=1}K_i$ and for any $i\le k$, $K_i \subset U_i$?
What I've tried: I try to let $K_i = K\cap U_i$, then it is obvious $K=\cup^k_{i=1}K_i$, however, I'm not sure such $K_i$ is still compact in $X$. I don't know how to go on.
Could anybody help me? Thanks ahead:)
AI: If $x \in U_i$, since $K \backslash U_i$ is compact we can take disjoint open neighbourhoods $V$ and $W$ of $x$ and $K \backslash U_i$ respectively. Then the closure of $V$ is contained in $U_i$. And so each $x \in K$ has an open neighbourhood $V_x$ whose closure is contained in some $U_i$. These form an open cover of $K$, so we can take a finite subcover $V_1, \ldots, V_m$. Let $K_i$ be the union of the closures of those $V_j$ whose closures are contained in $U_i$. |
H: L'hospital rule for two variable.
How to use L'hospital rule to compute the limit of the given function
$$\lim_{(x,y)\to (0,0)} \frac{x^{2}+y^{2}}{x+y}?$$
AI: There is no L'Hopital's Rule for multiple variable limits. For calculating limits in multiple variables, you need to consider every possible path of approach of limits.
What you can do here:
Put $x=r\cos\theta$ and $y=r\sin\theta$, (polar coordinate system) and $(x,y)\to (0,0)$ gives you the limits $r\to 0$ and no limits on $\theta$.
Now we need to substitute these in your problem which then becomes $$\lim_{r\to 0}\dfrac{r^2}{r(\cos\theta+\sin\theta)}=\lim_{r\to 0}\dfrac{r}{(\cos\theta+\sin\theta)}$$ Now for paths where $\cos\theta\to-\sin\theta$ or $\cos\theta=-\sin\theta$, the denominator $\to0$ or $=0$ while the numerator is just tending to $0$ but not exactly $0$.Since the second case exists $\implies$ the limit doesn't exist.
If extra condition that $(x+y)\neq 0$ is applied, even then the limit does not exist as for path $\theta=0$, limit is $0$ while for path $\sin\theta=-\cos\theta+r\cos^2\theta$, limit is something which can be non-zero. |
H: The metrizable space may be not locally compact
My text book said:
Not every metrizable space is locally compact.
And it lists a counterexample as following: The subspace $Q=\{r: r=\frac pq; p,q \in Z\}$ of $R$ with usual topology, i.e., $Q$ is the set of all rationals. It said: for any open ball of any point $r \in Q$, the closure is not compact. I can't understand this sentence. Why the closures of the open balls are not compact.
Could anybody help me to understand this sentence. Thanks ahead:)
AI: Let $I$ be any nontrivial interval, and $r$ be an irrational number in $I$. We want to show that $I\cap\mathbb{Q}$ is not compact in $\mathbb{Q}$. Let $O_n=(-\infty,r-1/n)\cap\mathbb{Q}$ and $U_n=(r+1/n,\infty)\cap\mathbb{Q}$. Show that $$\{O_n:n\in\mathbb{N}\}\cup\{U_n:n\in\mathbb{N}\}$$ is an open cover without finite subcover. |
H: $\int_{-\infty}^\infty e^{ikx}dx$ equals what?
What would $\int\limits_{-\infty}^\infty e^{ikx}dx$ be equal to where $i$ refers to imaginary unit? What steps should I go over to solve this integral?
I saw this in the Fourier transform, and am unsure how to solve this.
AI: $\int\limits_{-\infty}^\infty e^{ikx}dx$ is a (EDIT: scaled by $\frac1{2\pi}$) representation of Dirac's $\delta(k)$ function. For the antiderivative see the other answers... |
H: the rank of an interesting matrix
Let $A$ be a square matrix whose off-diagonal entries $a_{i,j} \in (0,1)$ when $i \neq j$. The diagonal entries of $A$ are all 1s. I am wondering whether $A$ has a full rank.
AI: If
$$A=\begin{pmatrix}
1 & \frac34 & \frac12 &\frac34 \\
\frac34 & 1 & \frac34 & \frac12 \\
\frac12 &\frac34 & 1 & \frac34 & \\
\frac34 & \frac12 & \frac34 & 1
\end{pmatrix}$$
then $(1,-1,1,-1)A=0$, so the matrix is singular. |
H: Proof of existence of zero point
In the proof of existence of zero point, $f(x)$ is continuous in $[a,b]$, where $f(a)<0$ and $f(b)>0$.
It is shown on the proof process in the textbook that when we define a set $V$ as follows:
$$V=\{x |f(x)<0,x\in[a,b]\},$$
so, there exists the supremum for $V$. Take $\xi=\sup V$.
Then I was confused with the following step:
take $x_{n}\in V (n=1,2,... \ )$, $x_{n}\rightarrow\xi$ (when $n\rightarrow \infty$) then
$$f(\xi)=\lim_{n\rightarrow\infty}{f(x_{n})} \le0$$
I know that $f(\xi)=\lim_{n\rightarrow\infty}{f(x_{n})}$ cause $f(x)$ in continuous in $[a,b]$
but why $f(\xi)=0$?
AI: Well, you know $f(\xi)\leq 0$. Assume it is smaller, then by continuity there is a whole neighbourhood of $\xi$ where $f<0$. Therefore $\xi$ is not the supremum of $V$, which is a contradiction.
Edit: If you also don't understand why $f(\xi)\leq 0$, note that it follows from the more general fact that for a converging series where all (but finitely many) elements are smaller than some given $L$, then the limit is smaller or equal to $L$. |
H: Quaternions and Rotations
Two of the interesting achievements in Mathematics are Classification of platonic solids, and also classification of finite groups acting on the unit sphere in $\mathbb{R}^3$, and they are very nicely connected to each other. These objects also enter in the classification of finite subgroups of $GL(2,\mathbb{R}), GL(3,\mathbb{R})$.
When studying these groups with geometry, I visited their classification by various ways: by solving some Diophantine equations, and also using geometry of complex numbers; in particular multiplication by complex numbers.
Some finte subgroups of $GL(3,\mathbb{R})$ can be obtained from a multiplication in quaternions $\mathbb{H}$ by unit quaternions, and these are connected with rotations in $\mathbb{R}^3$; for a pure quaternion $a$, and unit quaternion $q$, the map $a\mapsto qaq^{*}$ is a rotation of $\mathbb{R}^3$, where $\mathbb{R}^3$ is identified with the space of pure quaternions. Many books/notes show this connection, but have not explained ideas behind considering multiplication by only unit quaternion, pure quaternions and multiplication in this specific way ($a\mapsto qaq^{*}$).
Can anybody explains ideas behind them, and suggest good reference for them (except Conway's book). (Thanks in advance..)
AI: John Stillwell's book Naive Lie Theory motivates the subject in the first chapter using an exploration of quaternions acting by conjugation on the unit sphere. I think his explanation is great for someone at any point in their undergraduate career, however the subsequent chapters may be extremely slow for anyone already acquainted with lie theory, group theory or general topology.
http://www.amazon.com/Naive-Theory-Undergraduate-Texts-Mathematics/dp/0387782141
There is a much more in depth coverage of the details of this in another book which I am trying to recall the name of.
edit: the book Finite Möbius Groups, Minimal Immersions of Spheres, and Moduli by Toth begins with a chapter on platonic solids and finite rotation groups all very explicitly, in terms of Möbius transformations. Once you understand how to interchange between Möbius transforms and quaternions acting on $R^3$ from stillwell, this may be helpful. |
H: Vector Taylor series
From pg. 35 of Classical Electrodynamics 3rd edition, Jackson,
$$\begin{aligned} \nabla^{2} \Phi_{a}(\mathbf{x}) &=-\frac{1}{4 \pi \epsilon_{0}} \int \rho\left(\mathbf{x}^{\prime}\right)\left[\frac{3 a^{2}}{\left(r^{2}+a^{2}\right)^{5 / 2}}\right] d^{3} x^{\prime} \end{aligned}$$
"Choose R such that $\rho(\mathbf{x'})$ changes little over the interior of the sphere... With a Taylor series expansion of the well-behaved $\rho (\mathbf{x'})$ around $\mathbf{x'} = \mathbf{x}$ one finds ..."
\begin{align}
\nabla^{2} \Phi_{a}(\mathbf{x}) &=-\frac{1}{\epsilon_{0}} \int_{0}^{R} \frac{3 a^{2}}{\left(r^{2}+a^{2}\right)^{5 / 2}}\left[\rho(\mathbf{x})+\frac{r^{2}}{6} \nabla^{2} \rho+\cdots\right] r^{2} d r+O\left(a^{2}\right),
\end{align}
where $r = |\mathbf{x'} -\mathbf{x}|$.
Could someone explain how to derive this Taylor series for a function of a vector? I've never seen this before and am at a loss.
AI: enzotib has already provided the expansion of a real-valued function of a vector up to second order. Now we can make use of the fact that the function is being integrated over a spherical volume, multiplied by a spherically symmetric factor. The integral containing the linear term vanishes by symmetry. For the quadratic term, the Hessian can be split into a component proportional to the identity and a traceless part:
$$\def\H{\mathbf H}\H=\def\tr{\operatorname{tr}}\frac{\tr\H}3\mathbf I+\left(\mathbf H-\frac{\tr\H}3\mathbf I\right)\;.$$
The integral containing the traceless part vanishes by symmetry, and the integral containing the identity yields your quadratic term, since the trace of the Hessian is the Laplacian.
Please update the question to reflect the context that I used in the answer. Thanks. |
H: How to evaluate $\int \frac{\mathrm dx}{\sqrt[3]{\tan\,x}}$?
Please show me the steps of the following integration. I got an answer in Wolfram, but I need steps..
$$\int \frac{\mathrm dx}{\sqrt[3]{\tan\,x}}$$
AI: We try the substitution $t^3 = \tan^2 x$. Therefore, $3t^2 dt = 2 \tan x \sec^2 x dx$, giving us $\frac{dx}{\sqrt{t}} = \frac{3 dt}{2(1+t^3)}$.
Thus, we will only evaluate $\int \frac{3 dt}{1+t^3} $, divide by $2$ and substitute back. Note that $3 = (1-t+t^2) + (2-t)(1+t)$, reducing our integral to
$$ \int \frac{dt}{1+t} + \int \frac{(2-t)dt}{1-t + t^2} $$
I won't elaborate further, since our integrals are already in standard forms. |
H: How can I generate the binary representation of any real number?
In p. 30 of Baby Rudin, I find a reference to the fact that the binary representation of a real number implies the uncountablity of the set of real numbers. But I have two questions:
Does every real number have a binary representation? If yes, how do I prove it?
How can I generate the binary representation of a given real number $a$?
I'm aware of the binary representation of integers, but had never thought of a binary representation of real numbers earlier.
AI: Every real number $\alpha>0$ (to make things simpler) can be developed into an infinite binary expansion of the form $$110\ldots10\,.\,1011101\ldots\ ,$$
where to the left of the "decimal" point we only have finitely many digits and to the right an unending sequence of digits. Numbers of the form $n/2^N$ $\ (n, N\in{\mathbb N}_{\geq0})$ have exactly two such representations (this is the $0.999\ldots=1.000\ldots$ phenomenon in decimal), all other numbers have exactly one representation. In fact it is possible to construct the full real number system (with addition, multiplication and order) as such an "uncountable list" of binary fractions. This has been sketched by Gowers here: http://www.dpmms.cam.ac.uk/~wtg10/decimals.html
A real number $\alpha>0$ can be considered as "given" when for each $N$ it is possible to name an integer $a_N$ such that
$${a_N\over 2^N}\leq \alpha<{a_N+1\over 2^N}\ .$$
Using induction one easily proves that
$$2 a_N\leq a_{N+1}\leq 2a_N+1\ ,$$
and this implies that the finite binary representation of $a_{N+1}$ is obtained from the representation of $a_N$ by appending a $0$ or a $1$. Now the quotients $a_N/2^N$ approximate the given number $\alpha$. Writing $a_N$ in binary
and separating the last $N$ digits by a "decimal" point we therefore get a finite binary approximation of $\alpha$, and things work out such that for $N'>N$ the first $N$ places (after the "decimal" point) of $a_N$ and $a_{N'}$ coincide. In this way we get an ever better approximation of $\alpha$ just by adding new "binary places" to the right. |
H: holomorphic function on the complex plane
let $ f(z) \in Hol(\mathbb{C}) $ holomorphic function such that for each $ z_0 \in \mathbb{C} $ there exists $ N(z_0) $ such that $ f^{(N(z_0))}(z_0) = 0 $
Prove: that f is a polynom
AI: Modifying Davide's argument but avoiding Baire's category theorem: Let $\overline{\mathbb E}$ be the closed unit ball and
$$F_n := \{ z \in \overline{\mathbb E} : f^{(n)}(z) = 0 \}.$$
Then, by assumption, $\overline{\mathbb E} = \bigcup_n F_n$. Since $\overline{\mathbb E}$ is uncountable, not all $F_n$ can be finite. Let $n \in \mathbb N$ s.t. $F_n$ is infinite. Since $\overline{\mathbb E}$ is compact, $F_n$ has a limit point in $\overline{\mathbb E}$, i.e. there is a $z_0 \in \overline{\mathbb E}$ and a sequence $(z_k)$ in $F_n \setminus \{z_0\}$ that converges to $z_0$. We have $f^{(n)}(z_k) = 0$ for all $k$, hence $f^{(n)} = 0$ by the identity theorem, so $f$ is a polynomial. |
H: Arc direction in given point
I have an arc with a given center, start angle, end angle, and radius.
I want to draw an arrow showing the arc direction in the arc middle point. What is the easiest way to calculate this direction (it does not need to be precise, it's only a representation issue)?
I've found the arc middle point so far.
My only idea is to find a point very close to the middle point with the direction the arc has. Then, take the direction of the line connecting these two points and draw an arc to this line. Unfortunately, this needs much computation, and as I want to use it in a C++ application I'm looking for better way of determining arrow vector direction.
AI: The middle point of the arc given by
$$(x,y)=(x_0,y_0)+r(\cos t,\sin t),\qquad t\in[\theta_1,\theta_2] $$
is relative to the value of $t$ given by $t=(\theta_1+\theta_2)/2$, and the direction of the tangent is obtained adding $\pi/2$ (modulo $\pi$ if you don't care about the versus):
$$t=\frac{1}{2}(\theta_1+\theta_2+\pi).$$ |
H: Integral:$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $
How to evaluate:
$$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $$
Maybe we can evaluate it using the well-known result:$\int_{0}^{\frac{\pi}{2}} \ln{\sin t} \text{d}t=\int_{0}^{\frac{\pi}{2}} \ln{\cos t} \text{d}t=-\frac{\pi}{2}\ln{2}$
But how do I evaluate it, using that ?
AI: Letting $x=\tan t$ leads to $$-4 \int^{\pi/2}_0 \log(\cos t) dt = 2\pi \log 2.$$ |
H: Tips for an adult to learn math -- from the beginning.
First let me start with I am an adult and I can't do simple maths. I some how got through all of my math courses in University (after several attempts) but I honestly couldn't tell you how...
I cannot do these:
Add/Subtract with decimals
Add/Subtract fractions
Basic multiplication
Basic division
More advanced math that uses these principles
How can I go about learning these things now? Is there a particular book that I can study from (with worksheets). I'd really rather not do exercises that involve connecting dots to create pictures, or coloring things...
As a side note, I believe I suffer from dyscalculia.
AI: Check out the following four playlist in order. These are made for people who don’t know how to add or subtract and it even explains the decimal number system.
http://www.youtube.com/playlist?list=PL301908982CBFE20D&feature=plcp
http://www.youtube.com/playlist?list=PL50D1D09ABE9CE271&feature=plcp
http://www.youtube.com/playlist?list=PL1C68557896CFABA8&feature=plcp
http://www.youtube.com/playlist?list=PLE23E2FDF6E935778&feature=plcp
I hope this helped. It helped me to learn other stuff. Of course you can start from a more advanced playlist if you feel you already know one of them. |
H: Complex polynomial and the unit circle
Given a polynomial $ P(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_0 $, such that
$\max_{|z|=1} |P(z)| = 1 $
Prove: $ P(z) = z^n $
Hint: Use cauchy derivative estimation
$$ |f^{(n)} (z_0)| \leq \frac{n!}{r^n} \max_{|z-z_0|\leq r} |f(z)| $$ and look at the function $ \frac{P(z)}{z^n} $
It seems to be maximum principle related, but I can't see how to use it and I can't understand how to use the hint.
AI: Let $g(z):=z^nP\left(\frac 1z\right)$. It's a polynomial whose leading term is $a_0$ and constant coefficient is $1$. We have that $g(0)=1$ and $\max_{|z|=1}|g(z)|=1$, hence by maximum modulus principle, $g$ is constant equal to $1$. This gives the wanted result. |
H: What's wrong with this conversion?
I need to calculate the following limes:
$$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
$$
My first intuition was that the answer is $x$, but after a bit of fiddling with the root I got thoroughly confused. I know that below conversion goes wrong somwhere, but where?
$$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2*n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2*n^2}}{n}
= \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n}
= 0
$$
AI: $$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n}
\neq \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n}
$$
$$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n}
= \lim_{n\rightarrow\infty} \frac{n\sqrt{\frac{1}{n^2}+x^2}}{n}
$$ |
H: Is the product of two sets well-defined if one is empty
Let $X$ be a set. What is $X\times \emptyset$ supposed to mean? Is it just the empty set?
AI: And more can be said: a cartesian product is empty if and only if one of the two factors is empty. |
H: How to conclude $\Re $ is zero?
I'm in a Hilbert space $H$ and for $z,v, h \in H$ and $t \in \mathbb C$ I have
$$ \|z\|^2 \leq \|h−(tv+y)\|^2 = \|z−tv\|^2 =\|z\|^2 −2\Re(t⟨v,z⟩)+|t|^2\|v\|^2$$
According to my notes it follows from this that $\Re(t⟨v,z⟩) = 0$ for all $t$. How does that follow? I can't seem to show it. Thanks for your help.
AI: We get that for all $t\in\Bbb C$,
$$2\Re(t\langle v,z\rangle)\leq |t|^2\lVert v\rVert^2.$$
For an integer $n$, replacing $t$ by $\frac tn$, we get:
$$\frac 2n\Re(t\langle v,z\rangle)\leq \frac 1{n^2}|t|^2\lVert v\rVert^2,$$
hence
$$2\Re(t\langle v,z\rangle)\leq \frac 1n|t|^2\lVert v\rVert^2.$$
Letting $n\to +\infty$, we show the result. |
H: Show that if X is a discrete random variable such that for nonnegative integers m,n, $P(X>m+n|X>m)=P(X>n)$, then X is geometric
Here's my attempt, and I'm not sure if it is even close to the right direction.
If $P(X>m+n|X>m)=P(X>n)$ then if F is the cumulative distribution function of X, we have
$$\frac{P(X>m+n)}{P(X>m+n)+P(m<X<m+n)} = P(X>n)$$$$\frac{1-F(m+n)}{1-F(m+n)+[F(m+n)-F(m)]} = 1-F(n)$$$$
\frac{1-F(m+n)}{1-F(m)} = 1-F(n)$$$$
F(n) = 1-\frac{1-F(m+n)}{1-F(m)}$$$$
F(n) = \frac{F(m)+F(m+n)}{1-F(m)}$$
Is this even anything?
I was hoping to be able to show that this function must be something like $$F(n)=1-(1-p)^n$$ where $p$ is the probability of a success but I don't know how to do that. Is this a dead end? If so what would be a good way to attack this?
AI: It is better to think in terms of $1-F(n) = \mathbb{P}(X>n)$. Then
$$
\mathbb{P}(X>m+n |X>m) = \frac{\mathbb{P}(X>m+n,X>m)}{\mathbb{P}(X>m)} \stackrel{n\geqslant 0}= \frac{\mathbb{P}(X>m+n)}{\mathbb{P}(X>m)} = \frac{1-F(n+m)}{1-F(m)}
$$
Thus the complementary cdf $1-F_X(n)$ satisfies a functional equation:
$$
\left( 1-F(n+m)\right) = \left( 1-F(m)\right) \left( 1-F(n)\right)
$$
Hence the solution is $ 1-F(n) = q^n$ for some $0 < q \leqslant 1$, i.e. $X$ is geometric with $p=1-q$. |
H: How to measure the volume of rock?
I have a object which is similar to the shape of irregular rock like this
I would like to find the volume of this. How to do it?
If I have to find the volume, what are the things I would need. eg., If it is cylindrical, I would measure length and diameter. But, it is irregularly shaped. Like the above rock.
Where should I start? Couple of google search says something related to integration and contours. Somebody pls give me some handle :) I would say i'm very beginner level in math.
Many Thanks :)
Edit:
60 to 70% accuracy would be helpful.
AI: As your comment indicates, you're not interested in rocks so much as tumors.
One possible approach is to use a tomographic technique. Many medical imaging tools image the body using tomography: that is, examining the body one "slice" at a time.
If you have access to such tools, or can derive such an example, then what you want to do is cut the rock/tumor into many slices along some axis, and then compute the area of the tumor at that slice. This is a bit easier to do that doing it in three dimensions.
Then, you move forward by some $\Delta z$ along the $z$-axis (or whatever axis), and repeat.
Multiply each surface area by $\Delta z$, sum them, and you will get a good estimate. The estimate is better the smaller your $\Delta z$.
To compute the area of each "slice", you can do many things: fit a simpler shape to the data, perform Monte Carlo integration, or decompose the shape into a series of piecewise linear segments. |
H: Is this a correct definition of a line integral?
This comes from the beginning chapter on line integrals in the book Mathematical Methods for Science Students:
Suppose $y=f(x)$ is a real single-valued monotonic continuous function of $x$ in some interval $x_1<x<x_2$. Then if $P(x,y)$ and $Q(x,y)$ are two real single-valued continuous functions of $x$ and $y$ for all points of C, the integrals $$\int_C P(x,y)dx,\quad\int_C Q(x,y)dy$$ and, more frequently, their sum $$\int_C \Bigg\{P(x,y)dx+Q(x,y)dy\Bigg\}$$ are called curvilinear integrals or line integrals, the path of integration C being along the curve $y = f(x)$ from A to B.
Is this a correct definition of a line integral?
It doesn't appear to resemble the wikipedia definition of a line integral over a scalar field, which makes sense to me:
For some scalar field $f : U\subseteq R^n → R$, the line integral along a piecewise smooth curve $C \subset U$ is defined as$$\int_C f ds = \int^b_af(r(t))|r'(t)|dt$$
where r: [a, b] → C is an arbitrary bijective parametrization of the curve C such that r(a) and r(b) give the endpoints of C and a
The function f is called the integrand, the curve C is the domain of integration, and the symbol ds may be intuitively interpreted as an elementary arc length. Line integrals of scalar fields over a curve C do not depend on the chosen parametrization r of C.
AI: The first is a line integral over a vector field (presented quite horribly), defined as
$$\int_a^bF(r(t))\cdot r'(t) dt$$
$F$ represents the formula for the field, and $r$ represents the path. If we write $r$ as $(x(t),y(t))$, then we get that $r'(t)=(\frac{dx}{dt},\frac{dy}{dt})$
If $F(x,y)$ represents the equation defining the vector field, we can write it in component form as $F(x,y)=(P(x,y),Q(x,y))$. so $P$ defines the $x$-component of the vector field at each point $(x,y)$, and $Q$ does the same for the $y$-component. Expanding the dot product:
$$\int_a^b P(x,y)\frac{dx}{dt} dt + Q(x,y)\frac{dy}{dt} dt$$
$$\int_a^b P(x,y)dx + Q(x,y)dy$$
Gives your version.
This is useful for determining the work done by a "field type" force (an electromagnetic field, for example) on a moving object.
The second is a line integral over a scalar field. This is just the normal integral really, just extended so it can be defined over any curve, and not just the $x$-axis in the 2D case which is rather limiting.
We call them both line integrals (since we integrate over a curve), but one is over a vector field and the other is over a scalar field, leading to different definitions. |
H: Residue of $\frac{\tan(z)}{z^3}$
What is the easiest way to calculate the residue of $\dfrac{\tan(z)}{z^3}$ at zero? I could either use the line integral theorem, or expand it out as a series. Is there a right way to do it?
AI: You don't have to calculate the series expansion. The $z^{-1}$-coefficient in the series expansion of $\tan(z)/z^3$ is the $z^2$-coefficient in the series expansion of $\tan(z)$. But this is zero since $\tan$ is an odd function. |
H: Is there a version of the Gershgorin circle theorem that is suitable for nearly triangular matricies?
The Gershgorin circle theorem, http://en.wikipedia.org/wiki/Gershgorin_circle_theorem, gives bounds on the eigenvalues of a square matrix, and works well for nearly diagonal matrices.
For a triangular matrix, however, the bounds are not useful in general, despite the fact that the eigenvalues are known to be the diagonal elements. Is there a version (or can some helpful person develop a version) of the Gershgorin circle theorem that gives more useful bounds in the nearly triangular case?
AI: If $A = (a_{ij})$ is your $n \times n$ matrix, and $\alpha > 0$, let $R_i(\alpha) = \sum_{j \ne i} \alpha^{i-j} |a_{ij}|$. Then every eigenvalue $\lambda$ has $|\lambda - a_{ii}| \le R_i(\alpha)$ for some $i$. Note that if $A$ is upper triangular, $R_i(\alpha) \to 0$ as $\alpha \to \infty$, so we get the eigenvalues exactly in that limit.
This is just the regular Gershgorin theorem applied to $U A U^{-1}$ where $U$ is the diagonal matrix with entries $u_{ii} = \alpha^i$. |
H: what is $c$ in Mandelbrot set?
The Mandelbrot Set is an extremly complex object that shows new structure at all magnifications. It is the set of complex numbers $c$ for which the iteration indicated nearby remains bounded.
$$z_0=c$$
$$z_{n+1}=z_n^2+c$$
what is $c$ in Mandelbrot set?
isn't $c$ complex number
AI: When you look at an image of the Mandlbrot set it shows a region of the complex plane. The starting ones usually run from $-2$ to $\frac 12$ on the real axis or so and from $-2i$ to $2i$ or so on the imaginary axis. For each point in the region (at the desired grid spacing) we take the value $c$ and see if the iteration stays bounded. We then plot at $c$ black if it does stay bounded and a color that represents how quickly it goes to infinity if it does not. So for $c=1$ the iteration gives $1, 2, 5$ and we know if it ever gets bigger than $2$ it diverges, so we quit here and plot the color for 2 or 3 iterations. For $c=i$ we get $i,-1+i,-i, -1+i, $ etc and it stays bounded, so we plot black. |
H: Equivalents norms in Sobolev Spaces
I know that this is classical but I have never do the calculations to show that the norms in the sobolev space $W^{k,p}(\Omega)$
\begin{equation}
\|u\|_{k,p,\Omega}= \Bigl(\int_{\Omega} \sum_{|\alpha| \le k}|D^{\alpha} u |^{p} dx \Bigr)^{1/p}
\end{equation}
and
\begin{equation}
\|u\|_{W^{k.p}(\Omega)} =\sum_{|\alpha|\le k}\|D^{\alpha}u\|_{p}
\end{equation}
are equivalents.
AI: Let $N$ the number of $\alpha$ such that $|\alpha|\leq k$, and enumerate them as $\alpha^{(1)},\dots,\alpha^{(N)}$. Since all the norms are equivalent on $\Bbb R^N$, we can find two constants $C_1,C_2>0$ such that for all $(a_1,\dots,a_N)\in\Bbb R^N$,
$$C_1\sum_{j=1}^N|a_j|\leq \left(\sum_{j=1}^N|a_j|^p\right)^{\frac 1p}\leq C_2\sum_{j=1}^N|a_j|.$$
Then apply this inequality to $a_j:=\left(\int_{\Omega}|D^{\alpha^{(j)}}u(x)|^pdx\right)^{\frac 1p}$. |
H: Two forms of quantified conditional statement: equivalent?
There seems to be two forms of the conditional statement in predicate logic.
$$\forall x\,(P(x)\Rightarrow Q(x))$$
versus
$$(\forall x\in S)\Rightarrow Q(x)$$
$$S=\{x:P(x)\}$$
Are these equivalent? I'm a bit confused in the second form because It looks like $Q(x)$ is an open sentence however I know that this is just the universal statement:
$$(\forall x\in S)Q(x)$$
albeit written more explicitly. Are these two statements the same? Is the scope different? If the scope is restricted, or if they are not the same, what is their difference? To me the first says "for every $x$, if $x$ is $P(x)$ then $x$ is a $Q(x)$" and the second says "for every $x$ that is a $P(x)$, aforementioned $x$ is a $Q(x)$". Is more presupposed in this case?
AI: Your second formula "$(\forall x \in S)\Rightarrow Q(x)$" is not well-formed at all. What comes to the left of $\Rightarrow$ must be a complete formula, and "$\forall x\in S$" is not a complete formula -- it's a dangling quantifier with no body formula.
It is true, however, that $(\forall x \in S) Q(x)$ (without the spurious $\Rightarrow$) is the same as $(\forall x)(x\in S \Rightarrow Q(x))$ -- usually the former is considered a mere abbreviation of the latter. |
H: Correct interpretation of $E \subseteq V^{(2)}$
I'm trying to learn more about graph theory, but I'm getting confused by the initial definition:
"A graph $G = (V, E)$ is an ordered pair of finite sets. Elements of V are called vertices or nodes, and elements of $E \subseteq V^{(2)}$ are called edges or arcs. We refer to V as the vertex set of G, with E being the edge set."
I interpret this as "elements E in the subset of V are called edges". This seems to conflict with the next sentence, which defines a graph as being edges and vertices...i.e.two separate sets of things? It doesn't make intuitive sense for edges to be a kind of vertex. Whats the correct interpretation of this?
AI: If you look at the bottom of page 2 of your text, you will see the following explanation:
Notation If $S$ is a set, let $S^{(n)}$ denote the set of unordered $n$-tuples (with possible repetition). We shall sometimes refer to an unordered $n$-tuple as an $n$-set.
Now when the text later says "$E \subseteq V^{(2)}$." This means that $E$ is a set of unordered 2-tuples (or pairs) of vertices.
With that detail out of the way, I strongly suggest that you try to get an intuitive idea of what graphs are. Generally we represent graphs as "connect-the-dot" drawings. The vertices are drawn as dots and the an edge is drawn as a line (or curve) connecting the two dots. With this kind of picture in mind, we can now make sense of the precise definition of an edge: an edge is defined by the two vertices it connects.
I hope that you take the time to study the diagrams of example graphs in your textbook as they will make the above description clearer. |
H: How to disprove there exists a real number $x$ with $x^2 < x < x^3$
I realize that the only method is to show various cases:
I must test for $x > 1$, $x < -1$, $0 \leq x \leq 1$, and $-1\leq x \leq0$.
But even with this, I don't understand how to inject the properties of these four distinct possible $x$'s into the inequality (from the title) in order to show that none of these work.
Thanks for any help
AI: First let $x^3\gt x^2\implies x(x+1)(x-1)\gt 0$ which has solutions $(-1,0)\cup (1,\infty)$. Now, taking $x\gt x^2\implies x(1-x)\gt0$ which has solution set $(0,1)$. Th intersection of these two solution sets is $\emptyset$. Thus we can't have $x^3\gt x^2$ and $x\gt x^2$ at the same time. |
H: modified $\sum{k{n \choose k}}$ closed form expression
There is probably something stupidly simple I'm missing, but I'm trying to find a closed form for:
$$
2\sum_{k=1}^{(n-1)/2} k \, {n \choose k} \hspace{1cm} (n\textrm{ is odd})
$$
Anyone know how to do this?
I've figured out that since $n$ is odd,
$$
2+2\sum_{k=1}^{(n-1)/2}{n \choose k} = 2^n
$$
Thanks...
AI: If you want a complete solution, using the equation in the above comments i.e. $k \binom{n}{k} = n \binom{n-1}{k-1}$, we get that if $n= 2m+1$, $$2\sum_{k=1}^{m} k \binom{2m+1}{k} = 2(2m+1) \sum_{k=0}^{m-1} \binom{2m}{k} = (2m+1) \left(\sum_{k=0}^{2m} \binom{2m}{k} - \binom{2m}{m} \right) = n \left(2^{n-1} - \binom{n-1}{(n-1)/2} \right)$$ |
H: What digit appears in unit place when $2^{320}$ is multiplied out
Is there a way to answer the following preferably without a calculator
What digit appears in unit place when $2^{320}$ is multiplied out ? a)$0$ b)$2$ c)$4$ d)$6$ e)$8$
---- Ans(d)
AI: Quickly look at the last digit of $2^n$, for $n=1$, $2$, $3$, and so on. So we keep multiplying by $2$. For determining the last digit of $2^{n+1}$, only the last digit of $2^n$ matters.
We get $2$, $4$, $8$, $6$, $2$ and the pattern starts all over again. The pattern is periodic with period $4$. So at $n$ a multiple of $4$, we get a $6$, and $320$ is a multiple of $4$.
Remark: If we wanted the last digit of $2^{999}$, note that $996$ is a multiple of $4$. So at $996$ we get a $6$. Now count forward: $2$, $4$, $8$: the answer is $8$. Or else $1000$ is a multiple of $4$. Go backwards one step from $6$: the last digit is $8$. |
H: How to prove that a series expansion of pi has converged to a certain accuracy?
Wikipedia has a great article on methods for calculating pi with arbitrary precision, using for example Machin's infinite series expansion:
$\frac{\pi}{ 4} = 4$ arccot $5 - $arccot $ 239 $
where
arccot $x = \frac{1}{x} - \frac{1}{3 x^3} + \frac{1}{5 x^5} - \frac{1}{7 x^7} + \dots$
The only problem is that they never mention how to prove how many digits you have accurately computed pi to. How is this done? In general, how do you prove this for any series that converges?
Edit: I used pi with an alternating series as an example, but I'm more interested in general techniques for any convergent series.
AI: The series that you mention are alternating series. That means that the terms alternate in sign, go down in absolute value, and the absolute values of the terms approach $0$.
For alternating series, the error made by truncating at a particular place has absolute value less than the first omitted term. This criterion is easy to use, and ordinarily one cannot do substantially better. The sign of the first neglected term also tells you the direction of the error. If it is negative, truncation produces an overestimate. If it is positive, then truncation produces an underestimate. |
H: How to find this moment generating function
I am trying to find the moment generating function of a random variable $X$, which has probability density function given by
$$f_{X}\left( x\right) =\dfrac {\lambda ^{2}x} {e^{\lambda x}}$$
Where $x>0$ and $λ>0$.
The standard approach to doing this i believe is to follow the following algorithm
The moment generating function (mgf) of a continuous rv X, which is not necessarily
non negative, is defined as
$$M_{x}\left( t\right) =E\left[ e^{tX}\right] = \int _{-\infty }^{\infty }e^{tx}f_{X}\left( x\right) dx$$
Since $x>0$ for the given probability density function this is the same as Laplace transform but with $t > 0$.
If this integral does not converge, which i believe is the case for the given probability density function, then we switch to using Characteristic functions( Fourier transform). Given as
$$\phi_{x}\left( t\right) =E\left[ e^{itX}\right] = \int _{-\infty }^{\infty }e^{itx}f_{X}\left( x\right) dx$$
When this integral converges we substitute $t=0$ and viola we have our mgf. As per my readings this integral is always meant to converge, but i believe this does not converge for the given probability density function. This is what i managed to compute
$$\int _{-\infty }^{\infty }e^{x(it-\lambda)}\lambda ^{2}xdx = \lambda ^{2}\left[ \dfrac {x+1} {e^{x\left( \lambda-ti\right) }\left( it-\lambda \right) }\right] _{-\infty }^{\infty }$$
I was hoping some one could point out what am i doing wrong in calculating this integral. I used Integration by parts treating $i$, $t$ and $\lambda$ as constants.
If it does indeed not converge then if there are any more steps one could take to get the mgf ?
Thanks in advance.
AI: You did not quite write the density function correctly. It is zero for $x\lt 0$, and $\dots$. So you will be integrating from $0$ to $\infty$, and it will be quite doable by integration by parts.
If $f_X(x)$ is our density function, then indeed we want
$$\int_{-\infty}^\infty e^{tx}f_X(x)\,dx.$$
This is equal to $\int_{-\infty}^0 (0)e^{tx}\,dx\,$ (which is $0$) plus
$$\int_0^\infty \lambda^2 x e^{(t-\lambda)x}\,dx.$$ |
H: Can the range of this operator be closed?
Given an operator $T:H\rightarrow L$, where $H$ is a finite-dimensional Hilbert space and $L$ an infinitedimensional one, is the range of $T$, $T(H)$, a closed set in $L$ ?
I know that the image doesn't have to be closed if $H$ were infinite, but I'm not sure in this case.
AI: $T(H)$ is a finite dimensional subspace and any finite dimensional subspace is closed since it is isomorphic to $\mathbb R^n$ for some $n$ and hence complete. |
H: Equality in rng with no zero divisors.
I'm working on this problem, but I'm missing some manipulation.
Suppose $R$ is a rng without zero divisors and has elements $a$ and $b\neq 0$ such that $ab+kb=0$ for some $k\in\mathbb{N}$ (that is,
$ab+\underbrace{b+b+\cdots+b}_{k\ \textrm{times}}=0$.) I'm trying to show that $ca+kc=0=ac+kc$ for all $c\in R$.
I observe that
$$
(ca+kc)b=cab+kcb=c(ab+kb)=c\cdot 0=0
$$
so $ca+kc=0$ since $b\neq 0$ and $R$ has no zero divisors. I can't get the other equation. I note it's equivalent to showing $ac=ca$ for all $c\in R$, so I was trying to multiply $ac-ca$ by something nonzero to get $0$ to conclude but with no luck. Does anyone see what to do to get the other equality? Thanks.
AI: Since you've already shown $ca+kc=0,$ let's ignore the information about $b$ now and work directly with $c \neq 0$ (noting the result is trivial for zero.)
Try this: $c(ac+kc)=cac+kc^2=(ca+kc)c=0$ by the previous result, and since the rng's integral, $ac+kc=0.$ |
H: Finding the order of another element of a group
If in a group $G$, $a^5 = e$ , $e$ is the identity element of $G$. If $a,b \in G$ and $aba^{-1}=b^2$ then find the order of b
I have got no clue how to go forward with it, except that perhaps we can construct a cyclic group of order 5 with a and thus the order of $G$ will be $5k$ , but how to go forward from there, I have got no inkling.
Soham
AI: Here are some hints:
Since $a^5=e$, what integer $n$ will satisfy $(a^{-1})^n=e$
Now combine what you know about $a$ and $a^{-1}$ along with the equation $aba^{-1}=b^2$ to find an equation where the $a$s disappear. |
H: Cocountable fibers
Let $C$ be an uncountable set.
Can we construct a set $A \subseteq C^2$ such that it has a cocountable number of cocountable horizontal fibers, and a cocountable number of countable vertical fibers?
AI: Let $C = \omega_1$ and $<$ be the ordinal ordering of $C$. Since $C = \omega_1$, it is the least uncountable ordinal (cardinal). Hence for all $\eta \in \omega_1$, $\{\alpha < \eta\}$ is countable. Define $A \subset C^2$ as follows:
$A = \{(\alpha,\beta) \in C^2 : \alpha < \beta\}$.
Then for any fixed $\alpha$, $\{\beta : (\alpha, \beta) \in C^2\} = \{\beta : \beta > \alpha\}$ which is uncountable. However for any fixed $\beta$, $\{\alpha : (\alpha, \beta) \in C^2\} = \{\alpha : \alpha < \beta\}$ which is countable.
This $A$ is the desired set. |
H: Extension of an operator defined on (not necessarily closed) subspaces
Is there always an extension of an operator $T:U\rightarrow W$, defined on (not necessarily closed) subspaces of the infinitedimensional Hilbert spaces $H\supseteq U,L\supseteq W$, to the operator $$T':cl(U)\rightarrow cl(W),$$ where $cl$ denotes the closure.
Is this extension unique if we require uniform continuity ?
What about the case where $U$ is finite-dimensional ?
AI: Kevin already mentioned that the case where $U$ is finite dimensional is trivial.
You can show that if $T$ is continuous (which implies that it is uniformly continuous) then there is a unique continuous extension. This uses completeness of $L$.
If $T$ is not continuous (or you do not care if the extension is continuous), and you are not asking for any particular properties of the extension, then this is a linear algebra problem. If $\mathrm{cl}(U)=U + V$ with $V$ a subspace of $\mathrm{cl}(U)$ such that $U\cap V=\{0\}$, write $x\in \mathrm{cl}(U)$ as $x=u+v$ with $u\in U$ and $v\in V$, and define $T'(x)=T(u)$. (The complementary subspace $V$ exists assuming the axiom of choice.) Such an extension is not unique unless $U$ is closed. |
H: Integral Galois Extensions (Lang)
I have trouble understanding an argument in the proof of Proposition 2.5, p. 342, of Lang's Algebra. The setup of my question is the following:
Let $A$ be integrally closed in its quotient field $K$ and $B$ be its integral closure in a finite Galois extension $L$ of $K$, with group $G$. Let $p$ be a maximal ideal of $A$ and $\beta$ a maximal ideal of $B$ lying above $p$. Denote $\bar{B}=B/\beta$ and $\bar{A}=A/p$.
In his proof, Lang takes an element $\bar{x} \in \bar{B}$ that generates a separable extension of $\bar{A}$. He shows that $\bar{B}$ is normal over $\bar{A}$ and that $[\bar{A}(\bar{x}):\bar{A}] \le [K(x):K] \le [L:K]$, where $x$ is a representative of $\bar{x}$ in $B$. So far so good.
Then he says that "...this implies that the maximal separable subextension of $\bar{A}$ in $\bar{B}$ is of finite degree over $\bar{A}$ (using the primitive element theorem of elementary field theory)".
Question 1:
How can we use the P.E.T. to see that?
The statement of the primitive element theorem assumes that the underlying field extension is finite.
Question 2:
Is $\bar{B}$ over $\bar{A}$ separable? Why is Lang referring to the "Galois group" of $\bar{B}/\bar{A}$?
AI: The primitive element theorem implies that every finite separable subextension of $\overline B/\overline A$ is simple, hence its degree over $\overline A$ is at most $[L:K]$ by the previous considerations. Let $M$ be a finite separable subextension of maximal degree. If $x \in \overline B$ is separable then $M(x)/\overline{A}$ is a finite separable extension, hence equal to $M$ by the maximality of the degree of $M$. This shows that $M$ is actually the maximal separable subextension of $\overline B/\overline A$.
Concerning you second question: For every extension of fields $L/K$ (without the assumption of separability or normality) one may consider the group of $K$-automorphisms of $L$, i.e. the automorphisms $L\to L$ that leave every element of $K$ fixed. Usually this group is only called the Galois group of $L/K$ if the extension is Galois. However, in the proof of Proposition 2.5 Lang seems to call this the Galois group even though the extension is not necessarily Galois. In the case where $L/K$ is normal it can be shown (as Lang claims) that any $K$-automorphism of $L$ is uniquely determined by its restriction to the maximal separable subfield. |
H: Solution space to a functional equation
This question comes from my attempts at understanding an example presented by Bill Gasarch on his blog. The example is of a continuous strictly increasing function whose derivative is zero almost everywhere. The example is apparently discussed in the book Probability and Measure by Patrick Billingsley, but I currently do not have access to it.
Gasarch explains the background very well. Given a parameter $p \in (0,1)$, he describes a continuous function $F:[0,1]\to[0,1]$ which is increasing, $F(0) = 0$, $F(1) = 1$, but such that $F' = 0$ a.e. The derivative $f = F'$, which is only defined almost everywhere, must be nonnegative and should satisfy the following functional equation $$f(x) = \begin{cases} 2pf(2x) & \text{when $x \in (0,1/2)$,} \\ 2(1-p)f(2x-1) & \text{when $x \in (1/2,1)$,} \end{cases}$$ almost everywhere. It would seem, from the claimed example, that a nonnegative measurable function that satisfies this functional equation almost everywhere must be $0$ almost everywhere. However, this is not true since every constant function satisfies this functional equation when $p = 1/2$. Thinking about the dynamic properties of the transformation $$T(x) = \begin{cases} 2x & \text{when $x \in [0,1/2)$,} \\ 1/2 & \text{when $x = 1/2$ (say),} \\ 2x-1 & \text{when $x \in (1/2,1]$,} \end{cases}$$ it does seem that the space of solutions to the above equation is heavily constrained.
Is there a nice characterization of the space of solutions to the above functional equation? A general characterization would be best but a characterization for special cases (e.g. $f \geq 0$, $p = 1/2$) would be welcome.
AI: Here is the proof that $F'=0$ almost everywhere, if $p\neq \frac 1 2$ :
We have
$F(x) = pF(2x)$ if $x \in [0,0.5]$ and $F(x) = p + (1-p)F(2x-1)$ if $x \in [0.5,1]$.
Let $x \in (0,1)$ and suppose $x$ is normal in binary. Then $F'(x)=0$ :
Let $k>0$, and suppose the first $k$ digits of $x$ contain $a$ digits $0$, $b$ digits $1$, and that the length of the streak of consecutive $0$s (or consecutive $1$s) at the end of those $k$ digits is $c$. For example suppose the $k$th digit is $0$ (so there are $c$ consecutive $0$s)
If $2^{-(k+1)} \le |x-y| < 2^{-k}$, then $y$ has the same $k-c$ first digits as $x$,
thus $|F(x)-F(y)| = p^{a-c} (1-p)^b |F(?)-F(?)| \le p^{a-c} (1-p)^b$,
so the slope between those points is bounded by $\frac{|F(x)-F(y)|}{|x-y|} \le p^{a-c} (1-p)^b 2^{k+1} = 2 (2p)^a (2(1-p))^b p^{-c} = 2(4p(1-p))^{k/2} (\frac{1-p}p)^{b-a}p^{-c}$.
When $p \neq \frac 1 2$, $4p(1-p) < 1$.
Moreover, since $x$ is normal, $(a-b)/k$ converges to $0$ as $k$ grows, as does $c/k$, so after taking logarithms it becomes clear that this quantity converges to $0$ as $k$ grows.
Now, about the positive nonzero solutions of your functional equation.
If you write $x$ in binary, the orbit of $x$ when acted on by the two transformations of the functional equation is the set of $y$ such that the binary expansions of $x$ and $y$ are the same up to removing/adding/changing a finite number of digits. Basically they are equivalent if you can truncate them (not necessarily at the same place) so that you are left with two identical infinite strings.
One problem is that if a string is ultimately periodic (so if $x \in \mathbb Q$), then it makes a nontrivial cycle.
The functional equation implies that $f(x) = (2p)^a (2(1-p))^b f(x)$ where the repeating portion of the expansion of $x$ has $a$ digits $1$ and $b$ digits $0$.
If this multiplier $(2p)^a (2(1-p))^b$ is not one, then $f(x)$ has to be $0$.
Thus, for most values of $p$, $f=0$ on $\mathbb Q$.
Then from this point, $f$ is determined by its value for any set of representants of the equivalence classes of binary strings.
if (again) $p \neq \frac 1 2$, then $f$ is unbounded on any open interval, and if $\log {\frac p {1-p}}$ is irrational, its graph is even dense in $[0;1] \times \mathbb R^+$ :
Suppose $x \in (0,1)$ and that $F(x) > 0$. The functional equation basically says that if we plug $a$ digits $0$s and $b$ digits $1$ in front of its binary expansion to get a number $y$ (in any order), then $F(y) = (2p)^{-a}(2(1-p))^{-b}F(x)$. By fixing the first digits of $y$ you can force $y$ to be in any open interval you want, and then by putting additional digits, since $\log(2p)$ and $\log(2(1-p))$ are of different sign, you can make the multiplier as high (or as low) as you need to get above (or under) a specific value.
And in case their quotient is irrational you can make it as close to a specific value as you need.
If $p= \frac 1 2$, then the value of $F(x)$ doesn't change if you change/add/remove a finite amount of digits in the binary expansion of $x$, so the solutions are simply functions from the set of equivalence classes of binary strings into $\mathbb R$. They are still not good-looking, except the continuous ones which are constant. |
H: Simple inequality for measures
Let $(X,\mathscr B_X)$ and $(Y,\mathscr B_Y)$ be two measure spaces and $(Z,\mathscr B_Z)$ be their product space. Consider two finite measures (not necessarily product measures) $\mu,\nu$ on $(Z,\mathscr B_Z)$. Suppose that for any $A\in \mathscr B_X$ and for any $B\in \mathscr B_X$ it holds that
$$
\mu(A\times B)\geq \nu(A\times B).
$$
Does it mean that $\mu(C)\geq \nu(C)$ for any $C\in \mathscr B_Z$?
Some thoughts: clearly, the question can be equivalently stated as suppose that a measure $\lambda$ on $(Z,\mathscr B_Z)$ is non-negative on rectangles. It it a non-negative measure?
I was going to apply monotone class-like arguments, but I do not know what to do with complements. Clearly, the inequality is preserved under countable disjoint unions, though.
AI: Edit: This answer is not quite correct as stated; see the comments for corrections.
The answer is yes. We'll consider your formulation: if $\lambda$ is a signed measure which is nonnegative on all rectangles, then $\lambda$ is a nonnegative measure.
First observe the following consequence of the Carathéodory extension theorem:
Fact. Let $(X, \mathcal{F})$ be a measurable space, and $\mu$ a positive finite measure on $\mathcal{F}$. Suppose $\mathcal{A}$ is an algebra that generates $\mathcal{F}$. Then for any $B \in \mathcal{F}$, we have
$$\mu(B) = \inf \{\mu(A) : A \in \mathcal{A}, B \subset A\}.$$
(Consider $\mu$ as a premeasure on $\mathcal{A}$ and let $\mu^*$ be the corresponding outer measure. The right side of the above equation is $\mu^*(B)$, but the uniqueness in the Carathéodory theorem says that $\mu = \mu^*$ on $\mathcal{F}$.)
Now apply the above fact with $\mu = |\lambda|$ the total variation, and $\mathcal{A}$ the collection of all finite unions of rectangles. If $B$ is any measurable subset of the product, then for any $\epsilon > 0$ we may find $A$ a finite union of rectangles with $B \subset A$ and $|\lambda|(A \setminus B) < \epsilon$. But then $|\lambda(A) - \lambda(B)| = |\lambda(A \setminus B)| \le |\lambda|(A \setminus B) < \epsilon$. We have $\lambda(A) \ge 0$ by assumption so $\lambda(B) \ge -\epsilon$. $\epsilon$ was arbitrary so $\lambda(B) \ge 0$. |
H: Spivak Exercise involving operator norm
The exercise as stated:
If $T:\mathbb{R}^{m}\to \mathbb{R}^{n}$ is a linear transformation, show that there is a number $M$ such that $|T(h)|\leq M\cdot |h|$ for all $h\in \mathbb{R}^{n}$.
Hint: Estimate $|T(h)|$ in terms of $|h|$ and the other entries in the matrix of $T$.
By the way, here $|\cdot|$ denotes the standard Euclidean norm.
It is clear that $M$ is just the operator norm of $T$.
Indeed take $M$ to be the supremum of $|T(h)|$ over all $h$ in the unit ball of $\mathbb{R}^{n}$, which exists since $T$ is continuous (it is linear and defined on a finite dimensional space) and the unit ball of $\mathbb{R}^{n}$ is compact. Then using linearity it is easy to show that this $M$ works as desired.
However, I'm dodging the point here, and I'd like to come up with a direct proof, so I can gain the benefit of solving this problem. So I tried using the hint:
In the simple case where $n = 2$, I tried writing the matrix of $T$ as $(a_{ij})$ and $h = (h_1, h_2)$.
Then $T(h) = (a_{11}h_{1} + a_{12}h_2, a_{21}h_{1} + a_{22}h_{2})$.
Even in the relatively simple case of $2$ dimensions, the norm of this is ugly.
After simplification, I get:
$$|T(h)| = \sqrt{(a_{11}^{2} + a_{21}^{2})h_{1}^{2} + 2(a_{11}a_{12} + a_{21}a_{22})h_{1}h_{2} + (a_{12}^{2} + a_{22}^{2})h_{2}^{2}}$$
Any suggestions on how I can estimate this in terms of $|h| = \sqrt{h_{1}^{2} + h_{2}^{2}}$?
Thanks as always for your attention.
AI: By Cauchy Schwarz inequality
$$
\left|\sum\limits_{j=1}^m a_{ij}h_j\right|^2\leq
\left(\sum\limits_{j=1}^m a_{ij}^2\right)\left(\sum\limits_{j=1}^m h_{j}^2\right)
$$
so
$$
|T(h)|=
\left(\sum\limits_{i=1}^n|T(h)_i|^2\right)^{1/2}=
\left(\sum\limits_{i=1}^n\left|\sum\limits_{j=1}^m a_{ij}h_j\right|^2\right)^{1/2}\leq
\left(\sum\limits_{i=1}^n\left(\sum\limits_{j=1}^m a_{ij}^2\right)\left(\sum\limits_{j=1}^m h_{j}^2\right)\right)^{1/2}=
\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^m a_{ij}^2\right)^{1/2}\left(\sum\limits_{j=1}^m h_{j}^2\right)^{1/2}=
\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^m a_{ij}^2\right)^{1/2}|h|
$$
And the desired constant is
$$
M=\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^m a_{ij}^2\right)^{1/2}
$$
Note that it is not the best possible constant in this inequality. |
H: Fast algorithms for calculating the Möbius inversion
Recall the Möbius inversion formula: if we have two functions $f,g \colon \mathbf{N} \to \mathbf{N}$ such that
$$g(n) = \sum_{k=1}^n f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$$
holds for every $n \in \mathbf{N}$, then the values of $f$ can be recovered as
$$f(n) = \sum_{k=1}^n \mu(k) g\left(\left\lfloor \frac{n}{k} \right\rfloor\right),$$
where $\mu(k)$ is the Möbius function.
I am interested in fast algorithms for calculating a single value $f(n)$, assuming that $g$ can be calculated in $O(1)$ time.
The best algorithm I know of is as follows: There are $O(\sqrt{n})$ different values $\left\lfloor \frac{n}{k} \right\rfloor$, $1 \le k \le n$. If we have already calculated $f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$ for $2 \le k \le n$, then
$$f(n) = g(n) - \sum_{k=2}^n f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$$
can be calculated in $O(\sqrt{n})$ time by counting the multiplicities of the terms in the sum. By calculating the $f\left(\left\lfloor \frac{n}{k} \right\rfloor\right)$ from bigger to lower $k$ the total time required will then be $O(n^{3/4})$ while using $O(\sqrt{n})$ space.
I'm also interested in possible improvements to the above algorithm, even if they reduce the required time just by a constant factor.
AI: You can improve the $O(n^{3/4})$ algorithm by a constant factor of roughly three by computing only the odd summands of $k$ by using the identity:
$$\begin{array}{lll}
f \left( n \right) & = & g \left( n \right) - g \left( \frac{n}{2} \right) -
\sum_{3 \leq k \leq n, k \:\mathrm{odd}} f \left( \left\lfloor \frac{n}{k}
\right\rfloor \right)
\end{array}$$
which can be derived by subtracting $g(\frac{n}{2})$ from $g(n)$. The odd-only approach is hinted at in [Lehman 1960] (search for "in practice").
You can further improve the time complexity to $O(n^{2/3+\epsilon})$ if you can compute $f(k)$ for each $k\in \left\{1,..,u \right\}$ in $O(u^{1+\epsilon})$, as is typical for arithmetic functions by using a (possibly segmented) sieve.
Two references for the $O(n^{2/3})$ approach are [Pawlewicz 2008] and [Deleglise 1996]. |
H: Cartesian products of families in Halmos' book.
I'm studying some set theory from Halmos' book. He introduces the generalization of cartesian products by means of families. However, I can't understand what is going on. I get the first introduction "The notation..." to "... one-to-one correspondence". What I'm having trouble is with
If $\{X_i\}$ is a family of sets $(i\in I)$, the Cartesian product of the family is, by definition, the set of all families $\{x_i\}$ with $x_i\in X_i$ for each $i$ in $I$.
Could you explain to me the motivation of this definition? I know families are itself functions $f:I\to X$ such that to each $i$ there corresponds a subset of $X$, $x_i$. Instead of this we write them succintly as $\{x_i\}_{i\in I}$ to put emphasis on the range (indexed sets) of the function and the domain (indexing set) in question.
For example, in my case, the family is $f:I\to X$ with $f(i)=A_i$ with ${\rm dom} f=\{0,1,2,3\}$ and ${\rm ran} f =\left\{ {{A_0},{A_1},{A_2},{A_3}} \right\}$.
I'm thinking that we can talk about the cartesian product of sets as a set of tuples. However, I can't understand the definition for families of sets.
I leave the page in question:
$\hspace{1 cm} $
AI: Your intuition is exactly right. If $I$ is a set and you have a collection of sets $X_i$ for each $i \in I$, then the cartesian product is like a tuple. For example, in the case where you have two sets, $X_0$ and $X_1$, your index set is the finite ordinal $2 = \{0,1\}$. $X_0 \times X_1 = \{(a,b) : a \in X_0 \text{ and } b \in X_1\}$. Another way of thinking about this is $X_0 \times X_1$ is the collection of all functions from $2 = \{0,1\} \rightarrow X_0 \cup X_1$ such that $f(0) \in X_0$ and $f(1) \in X_1$. Instead of tuple, the cartesian product here is a correspondence $f$ between the index set $2 = \{0,1\}$ and an element such that $f(i) \in X_i$ for $i \in \{0,1\}$.
Now to generalize, you want the cartesian product to be the set of correspondence between the index set $I$ and elements in $\bigcup_{i \in I} X_i$ such that $f(i) \in X_i$. So formally, the cartesian product $\prod_{i \in I} X_i = \{f : I \rightarrow \bigcup_{i \in I} X_i : f(i) \in X_i\}$. As you can see, this is a generalization of the tuple concept. |
H: Computing $\frac{d^k}{dx^k}\left(f(x)^k\right)$ where $k$ is a positive integer
Does anyone know a formula for the derivative $$\frac{d^k}{dx^k}\left(f(x)^k\right)$$ where $k$ is some positive integer?
I started trying to work it out but it got messy.
AI: Apply Faà di Bruno's formula to get
$$\frac{d^n}{dx^n}(g(x)^n)=\sum \frac{n!^2}{m_1!\dots m_n!(n-\sum_{j=1}^nm_j)!}g(x)^{n-\sum_{j=1}^nm_j}\prod_{j=1}^n\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j},$$
where the sum is taken over the $n$-uples $(m_1,\dots,m_n)$ of integer satisfying $\sum_{k=1}^nkm_k=n$.
(I changed the notations to be conform with the link) |
H: Fundamental matrix and exponential of matrix using Laplace Transform
I'm trying to work out how to find $$\exp(At)$$ for a system of linear differential equations $$x'=Ax.$$
I know that the solution is a fundamental matrix of the system such that $$\exp(At)=I$$
at time $0$.
What is the method for solving this using the Laplace transform? The only method I can figure out is finding the eigenvalues and diagonalzing the matrix.
AI: The Laplace transform of $exp(At)$ is $(sI-A)^{-1}$. So compute that latter and then take its inverse Laplace transform. |
H: Zariski topology on prime $\mathrm{Spec}$ of a ring $R$
Let $R$ be a commutative unital ring. Let $\mathrm{Spec}(R) = \{ \mathfrak p \subset R \mid \mathfrak p \text{ a prime ideal of } R \}$. We define a set $C$ to be closed in this space if and only if there is an ideal $I$ such that $C(I) = \{\mathfrak p \mid I \subset \mathfrak p, \mathfrak p \text{ a prime ideal of } R \}$.
Now I'd like to show that these sets form a topology on $\mathrm{Spec}(R)$:
(i) For the zero ideal we get $C(0) = \mathrm{Spec}(R)$ and for $R$ we get $C(R) = \varnothing$.
(ii) Arbitrary intersections are closed again: $\bigcap_\alpha C(I_\alpha) = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, I_\alpha \subset \mathfrak p \text{ for all } \alpha \} = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, \sum_\alpha I_\alpha \subset \mathfrak p \} = C(\sum_\alpha I_\alpha)$.
(iii) We want to show that finite unions are closed again. It's enough to show it for two ideals $I,J$: $C(I) \cup C(J) = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, \text{ such that either } I \subset \mathfrak p \text{ or } J \subset \mathfrak p \}$
Now I'm stuck. How do I express the "either or" in terms of operations on ideals? Thanks for your help.
AI: Suppose $\mathfrak p \in C(IJ)$ then $IJ \subset \mathfrak p$. Since $\mathfrak p$ is prime, $I \subset \mathfrak p$ or $J \subset \mathfrak p$. Thus $\mathfrak p \in C(I) \cup C(J)$.
Suppose $\mathfrak{p} \in C(I) \cup C(J)$ then $I \subset \mathfrak p$ or $J \subset \mathfrak p$. Since ideals are closed under products $IJ \subset \mathfrak p$. Hence $\mathfrak p \in C(IJ)$. |
H: Is the function $\theta(a,b) = a-2ab+b$ a bijection from $\{0,1\}\times\mathbb{N}$ to $\mathbb{Z}$?
Consider the function $\theta:\{0,1\}\times\mathbb{N}\rightarrow\mathbb{Z}$ defined as $\theta(a,b) = a-2ab+b$. Is this function bijective?
For injective, I tried doing the contrapositive by supposing $\theta(a,b)=\theta(c,d)$, then $a-2ab+b=c-2cd+d$, but I have no idea where to go from there. I tried solving for a and b separately and plugging it back in, but that just turned into a huge algebraic mess.
I haven't figured what I'm going to do for surjective yet.
AI: Formulas are nice, but let's find out what's really going on: $\theta(0,b)=b\,$ and $\theta(1,b)=1-b$. Much clearer.
Could $x=1-y\,$ where $x$ and $y$ are natural numbers? Depends on what one means by natural number. It can certainly happen if we allow $0$ to be a natural number. I am inclined to allow that, out of loyalty to logic, where it is a standard convention. But I am in a minority. Certainly it cannot happen if $x$ and $y$ are positive.
Finally, for surjectivity (love that word) is every integer of the form $x$ or $1-y$ where $x$ and $y$ are positive integers? Sure. |
H: Please explain how to do this proof (involving functions)
Suppose that $f:A \longrightarrow B$ and $g:B \longrightarrow C$ are functions.
If $g \circ f$ is onto and $g$ is one-to-one, then prove that $f$ is onto.
How do I go about proving this?
From $g \circ f$ is onto, I know that there exists an $a \in A$ such that $g(f(a))=c$ and I also know that if $g(b_1)=g(b_2)$ then $b_1=b_2$ from the definition of one-to-one, but I'm not sure where to go from there. Any help would be great :)
AI: Take $b$ in $B$. Let $c=g(b)$. Since $g\circ f$ is onto, you know.... Since $g$ is one-one, you can conclude from $(g\circ f)(a)=c$ and $g(b)=c$ that.... |
H: Prove by induction: $2^n = C(n,0) + C(n,1) + \cdots + C(n,n)$
This is a question I came across in an old midterm and I'm not sure how to do it. Any help is appreciated.
$$2^n = C(n,0) + C(n,1) + \cdots + C(n,n).$$
Prove this statement is true for all $n \ge 0$ by induction.
AI: Marvis's hint suffices. Anyways:
Base case at $n = 0$ is trivial to show: $2^0 = 1 = \dbinom{0}{0} = 1.$
Inductive hypothesis: assume
$$ 2^k = \binom{k}{0} + \binom{k}{1} + \ldots + \binom{k}{k} \tag{1} $$
Inductive step: we need to show
$$ 2^{k+1} \stackrel{?}{=} \binom{k+1}{0} + \binom{k+1}{1} + \ldots + \binom{k+1}{k+1} \tag{2} $$
From $(1),$ we have
$$ 2^{k+1} = 2^k + 2^k = \\ \binom{k}{0} + \color{blue}{\binom{k}{0} + \binom{k}{1}} + \color{red}{\binom{k}{1} + \binom{k}{2}} + \ldots + \color{blue}{\binom{k}{k-1} + \binom{k}{k}} + \binom{k}{k} $$
Now group the colored terms using the identity
$$ \dbinom{k}{r-1} + \binom{k}{r} = \binom{k+1}{r} $$
We get
$$ 2^{k+1} = \binom{k}{0} + \color{blue}{\binom{k+1}{1}} + \color{red}{\binom{k+1}{2}} + \ldots + \color{blue}{\binom{k+1}{k}} + \binom{k}{k} $$
Finally we use $$\binom{k}{0} = \dbinom{k+1}{0} = 1 \quad \text{and} \quad \binom{k}{k} = \dbinom{k+1}{k+1} = 1.$$ So
$$ 2^{k+1} = \color{red}{\binom{k+1}{0}} + {\binom{k+1}{1}} + {\binom{k+1}{2}} + \ldots + {\binom{k+1}{k}} + \color{red}{\binom{k+1}{k+1}}. \square $$ |
H: Higher dimensional analogue of an arc of a circle
What is the higher dimensional analogue for the arc of a circle?
I'd like to work with the set of all points lying within a certain distance of a given point on an n-sphere, and I'd like to describe these sets by the (generalised) solid angles which they subtend.
AI: By request:
It seems you want either the spherical cap or the spherical wedge, or hyperspherical versions thereof. |
H: Combinatorics question: Prove 2 people at a party know the same amount of people
I recently had an assignment and got this question wrong, was wondering what I left out.
Prove that at a party where there are at least two people, there are two people who know the same number of other people there. Be sure to use the variable "n" when writing your answer.
My answer:
n >= 2
Case1: another person comes to the party, but doesn't know either of the first two. So the original two still only know the same number of people.
Case2: another person comes, and knows one out of the original 2, so therefore the >newcommer, and the one that doesnt know the newcommer both know the same number of people.
Case 3: another person comes and knows both of them, implying that they all know each >other, and therefore they all know the same number of people.
So therefore if n>=2, where n and n-1 know each other, in either case whether n+1 joins, >there will be at least two people who know the same amount of people.
Many thanks in advance. Have a test coming up.
AI: You are trying to prove this by induction, but it is not easy to do it that way. Say you had a party where two people each know four others. Now if a new guest arrives who knows one of them and not the other, you have broken the pair. You have to show that there is a pair in the new party.
I would approach it directly. If there are $n$ people at the party, each one knows somewhere from $0$ to $n-1$ of the others, which is $n$ possibilities. But if one person knows all the others, nobody knows nobody else (we are assuming knowing is reflexive here, otherwise the statement is not true-each person could know precisely those who arrived before him). So the number each person knows is either in the range $[0,n-2]$ or $[1,n-1]$. In either case, the pigeonhole principle says there are two that match. |
H: Increasing Sequence of Rationals
Let $x$ be any real number. Construct a sequence $x_n$ of rational numbers such that
$$x = \sup\{x_n : n \in \mathbb{N} \}.$$
I was trying $x_n = [ 10^n x ]/10^n$, but is it actually monotone increasing?
If so, how to prove it analytically?
Thanks for any help.
AI: Yes, your sequence $$x_n = \dfrac{\lfloor 10^n x \rfloor}{10^n}$$ will do the job. Your sequence is also monotone increasing.
HINT:
To prove monotone increasing, try to prove the following first:
$$a \lfloor x \rfloor \leq \lfloor ax \rfloor$$ where $a \in \mathbb{Z}^+$ and $x \geq 0$. |
H: Approximating an $L^2$ function in the Riemann sense
$\newcommand{\R}{\Bbb R}$
Consider the Lebesgue measure in $\R$ and the following proposition:
P. For each representative of a function class $f\in L^2[0,1]$ there is a sequence of continuous functions $(f_n)_{n\in\Bbb N}$ such that:
$|f_n-f|$ is Riemann integrable on $[0,1]$, for all $n\in\Bbb N$.
$\lim\limits_{n\to\infty} \int\limits_0^1 |f_n(x)-f(x)|^2\ \mathrm d x=0$.
There is no reason why this proposition should be true, but I cannot find a counterexample.
AI: This proposition is not true.
Let $f:[0,1]\to\mathbb R$ be defined by $f(x)=1$ iff $x\in\mathbb Q$ and $f(x)=0$ otherwise. $f$ is Lebesgue integrable but not Riemann integrable.
Suppose $|f_n-f|$ is Riemann integrable and $\int_0^1|f_n-f|dx<1/4$.
Then there is a partition $0=t_0<t_1<\dots<t_n=1$ of the unit interval such that
$$\sum_{i=1}^n(t_i-t_{i-1})\sup\{|f_n(x)-f(x)|:x\in[t_{i-1},t_i]\}<1/4.$$
Now there is $i\in\{1,\dots,n\}$ such that $\sup\{|f_n(x)-f(x)|:x\in[t_{i-1},t_i]\}<1/4$.
The function $f_n$ is not continuous on $[t_{i-1},t_i]$ since on a dense subset of that interval, $f_n<1/4$ and on a dense set, $f_n>3/4$. |
H: Why isn't this square root $+$ or $-$?
I was tasked with proving the identity $\tan(\frac x 2) = \dfrac {\sin(x)}{1+\cos(x)}$
I used the quotient identity for tangent and the half angle identities for sine and cosine to get $ \pm \dfrac {\sqrt{\dfrac {1-\cos(x)}{2}}}{\sqrt{\dfrac {1-\cos(x)}{2}}}$
which I reduced to $\pm \sqrt{\dfrac {1-\cos(x)}{1+\cos(x)}}$
I multiplied the fraction (within the square root) by $ \dfrac {1+ \cos(x)}{1+\cos(x)}$
Resulting in $\pm \sqrt{\dfrac {1-\cos^{2}(x)}{(1+\cos(x))^2}}$
Using the Pythagorean identity, I get $\pm\sqrt{\dfrac {\sin^{2}(x)}{(1+\cos(x))^2}}$
Taking the square root of the numerator and denominator I further reduced to $\pm \dfrac {\sin(x)}{1+\cos(x)}$
I thought I was done but when I checked my work in the answer book, it showed $ \left|\dfrac {\sin(x)}{1+\cos(x)}\right|$
Where do they get the absolute value from?
AI: You are tasked with proving that $\tan(\frac x 2) = \dfrac {\sin(x)}{1+\cos(x)}$. If we suppose that this is not an impossible task (i.e. that the identity is correct) then neither $\tan(\frac x 2) =\pm\dfrac {\sin(x)}{1+\cos(x)}$ nor $\tan(\frac x 2) =\left|\dfrac {\sin(x)}{1+\cos(x)}\right|$ are satisfactory places to end up. The first version, which you arrived at, is not exactly incorrect, but rather incomplete, because it doesn't say which sign applies. The second version, from the answer book, is incorrect when $\tan(\frac x2)<0$.
In general, $a=\pm b$ gives you the same information as $|a|=|b|$. However, in this case we can determine which sign applies. Your work essentially shows that $\left|\tan(\frac x 2)\right| =\left|\dfrac {\sin(x)}{1+\cos(x)}\right|$. But note that $1+\cos(x)\geq 0$ so it does not affect the sign, and $\sin(x)$ always has the same sign as $\tan(x/2)$. To see that the last point is true, it is enough to work in the interval $(-\pi,\pi)$ by periodicity. Both $\tan(x/2)$ and $\sin(x)$ are positive when $x$ is in $(0,\pi)$, and both are negative when $x$ is in $(-\pi,0)$.
That said, I agree with lab bhattacharjee that avoiding methods that require later working out signs is a good idea.
Here is a side issue, which in the original version of my answer was all I posted:
$\sqrt{x^2}=|x|$ is an identiy for real numbers $x$. The reason is that for a nonegative real number $a$, $\sqrt{a}$ is defined to be the unique nonnegative squareroot of $a$. Since $|x|^2=x^2$, it follows that $|x|$ is the nonnegative real number whose square is $x^2$.
Therefore $\sqrt{\left(\dfrac{\sin x}{1-\cos x}\right)^2}=\left|\dfrac{\sin x}{1-\cos x}\right|$. |
H: Some exact sequence of ideals and quotients
I saw an exact sequence of ideals
$$0 \rightarrow I \cap J\rightarrow I \oplus J \rightarrow I + J \rightarrow 0$$In this sequence, maps are ring homomorphisms or module homomorphisms?
And how the above sequence yield the exact seqeunce
$$0 \rightarrow R/I \cap J\rightarrow R/I \oplus R/J \rightarrow R/(I + J) \rightarrow 0$$
AI: The maps are module homomorphisms; more precisely, they are:
$I \cap J \mapsto I \oplus J $ via $x \mapsto (x,x)$,
and
$I\oplus J \mapsto I + J $ via $(x,y) \mapsto x - y$.
If you embed this exact sequence as a subsequence of the obvious short exact sequence
$ 0 \to R \to R\oplus R \to R \to 0$
(with maps defined by the same formulas), then the quotient is the second short
exact sequence that you ask about. |
H: Are there sub sigma algebras of the borel sigma algebra on the real line that are not sigma finite?
Im wondering if all sub sigma algebras of sigma finite measure space are also sigma finite?
AI: One example is $\{\emptyset,\mathbb R\}$.
Another is the $\sigma$-algebra of countable or cocountable sets. |
H: $\lim_{n\to\infty}n^{n\cdot a_{n}}=1$ as a sufficient condition for $\sum a_{n}<\infty$
[Not HW]
Let $\left(a_{n}\right)_{n=1}^{\infty}$
a sequence of real numbers such that, for all $ n\in\mathbb{N}$
, $0<a_{n}\leq\frac{1}{2}$
.
Why the following statements don't imply the convergence of $\sum_{n=1}^{\infty}a_{n}$
:
$$\lim_{n\to\infty}n^{n\cdot a_{n}}=1$$
and the second one:
$$\frac{a_{n+1}}{a_{n}}<1-\frac{1}{n+1}$$
For the second one, since the ratio test calls for the $\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$
to be strictly less than 1, it's obvious that the above statement does not imply convergence, but I didn't find a counterexample neither.
At the exam, I did mark statement 1 as a sufficient condition for convergence, but it's not true. So we need to find $a_{n}$
such that:
$\lim_{n\to\infty}n^{n\cdot a_{n}}=1$
$\iff$
$\lim_{n\to\infty}n\cdot a_{n}\ln n=0$
$0<a_{n}\leq\frac{1}{2}$
$\sum_{n=1}^{\infty}a_{n}$
diverges.
I thought of the following sequence:
$$a_{n}=\frac{1}{2},\frac{1}{2},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{16}\dots$$
$\sum a_{n}$
clearly diverges but I'm not sure about $\lim_{n\to\infty}n^{n\cdot a_{n}}$
.
Thanks guys for your help.
AI: Your reasoning for the second statement is invalid. While the given condition that $$\tag{1}{a_{n+1}\over a_n}<1-{1\over n+1}$$ is allowing the possibility that the limit of $a_{n+1}\over a_n$ is 1, the Ratio Test does not imply anything concerning divergence of the corresponding series in this case.
Note that condition $(1)$ is not allowing $a_{n+1}\over a_n$ to approach 1 "too fast".
In fact, if we demand the stronger condition that $$\tag{2}{a_{n+1}\over a_n}<1-{p\over n+1}$$ for some $p>1$ and sufficiently large $n$, then the corresponding series converges. This is known as Raabe's Test (one version of this is excersise 28 here).
With $p=1$, as in your condition, the series may diverge: for example, take take $a_n={1\over n\ln n}$, $n>1$. Then for $n$ sufficiently large, ${a_{n+1}\over a_n}= {n\ln n\over (n+1)\ln (n+1)}<{n\over n+1}=1-{1\over n+1}$. The integral test will show that $\sum{1\over n\ln n}$ is divergent. |
H: How to determine units in a partial differential equation
How do we determine the units used in a differential equation? Yes, in theory a PDE has nothing to do with units, but I'm interested in this question from a modeling point of view. By units, I mean the following. In an ordinary differential equation, finding the correct units seems relative straightforward. For example, if we have a function $u : [0,T] (seconds) -> \mathbb{R} (meters)$, we know that taking the derivative with respect to time gives velocity, $(meters)/(seconds)$. Taking two derivatives with respect to time gives acceleration, $(meters)/(seconds)^2$. Therefore, when we write an ODE
$$
\frac{\partial^2}{\partial t^2} u = f,
$$
we know that $f$ should have units $(meters)/(seconds)^2$. In a PDE, this same trick doesn't seem to work. For example, say we have a function $u : [0,T] (seconds) \times \Omega (meters^2) \rightarrow \mathbb{R} (celsius)$ where $\Omega \subseteq \mathbb{R}^2$. Then, we write the heat equation
$$
\frac{\partial}{\partial t} u - k \frac{\partial^2}{\partial x^2} u - k\frac{\partial^2}{\partial y^2} u = f
$$
or more simply as
$$
\frac{\partial}{\partial t} u - k\Delta u = f.
$$
Now, using the above trick, the term $\frac{\partial}{\partial t} u$ has units $(celsius)/(seconds)$. However, the term $\Delta u$ has units $(celsius)/(meters^2)$. In this context, it doesn't make sense to add the two terms. It also doesn't give clear insight into what the units of the forcing function $f$ need to be. As such, what are the correct units for the heat equation and what's the general rule for establishing units for an arbitrary PDE?
AI: $k$ has units $[distance^2 / time]$, so $$\frac{\partial}{\partial t}u - k\Delta u = \left[\frac{temp}{time} - \frac{distance^2}{time} \frac{temp}{distance^2}\right] = \left[\frac{temp}{time}\right] $$
You are already performing the dimensional analysis correctly! |
H: complex function with real values on the real interval
let $ B(0,1) = \{ z\in \mathbb{C} | |z|<1\} $
and $ f $ be an holomorphic function on $ B(0,1) $ such that $ f(z)\in\mathbb{R} \iff z\in\mathbb{R} $
Prove: $ f $ has at most 1 root in $ B(0,1) $
i think this exercise requires rouche theorem or the argument principle theorem
but i cant see how to use it
AI: Let $g:(-1,1)\to\mathbb R$ be the restriction of $f$. All zeros of $f$ are zeros of $g$. If $g$ has two distinct zeros, then $g'(c)=0$ for some $c\in(-1,1)$ by Rolle's theorem. This implies that $f'(c)=0$, and that $h(z)=f(z)-f(c)$ has a zero of order at least $2$ at $c$. The argument principle implies that $h(z)$ takes on real values at least $4$ times as you go around a small enough circle centered at $c$, and only $2$ of these have real inputs. |
H: Finding the number of combinations of two dates from n dates where one precedes the other?
Suppose that I am given n consecutive dates for buying and selling shares . So what are the number of ways of choosing a pair of buy date and sell date such that buy date always precede sale date ?
AI: Pick any two distinct dates. There are $\binom{n}{2}=\frac{n(n-1)}{2}$ ways to do this. Label the earlier one buy, the later one sell.
I have interpreted "earlier" as meaning different. It is always possible to buy one day and sell later the same day. With that interpretation, there are $n$ additional possibilities. |
H: Number of line segments intersecting diagonals are divided into in a convex polygon
I am currently self-studying introductory combinatorics, and I don't fully get an example in the book.
The question was as follows:
If no three diagonals of a convex decagon meet at the same point, into how many line segments are the diagonals divided by their intersections?.
So I understand that the total number of intersections would be $\binom {10} {4} = 210$ since for every 4 vertices there will be 1 intersection, and also the number of diagonals would be $10 \choose 2$-10 = 35. I also understand that the number of line segments are $k+1$ when there are $k$ intersections along a line. I however don't get the answer the author gave which was $35+2\times210$. Why were the number of diagonals added?, why did he multiply the number of intersections by 2?, shouldn't it be 4 since for every intersection there will be 4 segments?. If anyone can explain this to me, I'd be very grateful. Also if someone can provide a different way to solve this problem, that would be awesome.
AI: You start with 35 diagonals. Each intersection point adds a segment to both of the intersecting diagonals. Therefore answer is 35 + twice number of intersections. |
H: How to show the subset $Y$ is closed discrete?
The example is following:
Let $Y=\{(0,y):y \in R\}$. Let $E \subset R^2$, i.e., the subset of the real plane, and $E=Y\cup \{ (\frac 1n, \frac k{n^2}): n\in Z^+, k \in Z\}$. The topology on $E$ is this:
The point $(\frac 1n, \frac k{n^2})$ is open;
$\{U_n(y_0): n=1,2,...\}$ are the nbhds of the point $(0,y_0)$ of $Y$ is defined as following:
$$U_n(y_0)=\{(x,y): x \le \frac 1n, |y-y_0|\le x\}$$.
My text book said $Y$ is closed discrete in $E$. I could see $Y$ is closed: for any point $y \in Y^C$, the set $\{y\}$ is open which is disjoint with $Y$. However, I fail to show that $Y$ is a discrete space.
Could anybody help me? Thanks ahead:)
AI: Given any $y_0 \in \mathbb{R}$ and $n \in \mathbb{Z}^+$, if $(0,y) \in U_n(y_0)$ then we must have $| y - y_0 | \leq 0 \leq \frac{1}{n}$, and therefore $y = y_0$. It follows that $U_n (y_0) \cap Y = \{ (0,y_0) \}$ for all $y_0 \in \mathbb{R}$ and $n \in \mathbb{Z}^+$. |
H: What is the $\tau$ symbol in the Bourbaki text?
I'm reading the first book by Bourbaki (set theory) and he introduces this logical symbol $\tau$ to later define the quantifiers with it. It is such that if $A$ is an assembly possibly contianing $x$ (term, variable?), then $\tau_xA$ does not contain it.
Is there a reference to this symbol $\tau$? Is it not useful/necessary? Did it die out of fashion?
How to read it?
At the end of the chapter on logic, they make use of it by... I don't know treating it like $\tau_X(R)$ would represent the solution of an equation $R$, this also confuses me.
AI: Adrian Mathias offers the following explanation here:
Bourbaki use the Hilbert operator but write it as $\tau$ rather than $\varepsilon$, which latter is visually too close to the sign $\in$ for the membership relation. Bourbaki use the word assemblage, or in their English translation, assembly, to mean a finite sequence of signs or leters, the signs being $\tau$, $\square$, $\lor$, $\lnot$, $=$, $\in$ and $\bullet$.
The substitution of the assembly $A$ for each occurrence of the letter $x$ in the assembly $B$ is denoted by $(A|x) B$.
Bourbaki use the word relation to mean what in English-speaking countries is usually called a well-formed formula.
The rules of formation for $\tau$-terms are these:
Let $R$ be an assembly and $x$ a letter; then the assembly $\tau_x(R)$ is obtained in three steps:
form $\tau R$, of length one more than that of $R$;
link that first occurrence of $\tau$ to all occurrences of $x$ in $R$
replace all those occurrences of $x$ by an occurrence of $\square$.
In the result $x$ does not occur. The point of that is that there are no bound variables; as variables become bound (by an occurrence of $\tau$), they are replaced by $\square$, and those occurrences of $\square$ are linked to the occurrence of $\tau$ that binds them.
The intended meaning is that $\tau_x(R)$ is some $x$ of which $R$ is true. |
H: Question about inverse limit
I'm puzzled by the definition of inverse limit in this Wolfram article.
I thought if an object was defined by a universal property it meant that the object is unique up to unique isomorphism. This would mean that in the article I linked, $\alpha$ should be a unique isomorphism not just a homomorphism as written in the article. Is this a typo?
And I have another question: if someone asks me what an inverse limit is, would I be giving a correct answer if I answered as follows:
The inverse limit of an inverse system $(X_i, f_{ij})$ is the unique pair $(X, \pi_i)$ (where $X$ is an object and $\pi_i$ are morphisms such that $f_{ij} \circ \pi_i = \pi_j$) such that for any pair $(Y, \pi_i^\prime)$ (with $f_{ij} \circ \pi_i^\prime = \pi_j^\prime$) there exists a unique isomorphism $u^{-1}: X \to Y$ such that the following diagram commutes:
Thanks for your help.
AI: You are mixing two things here:
1) In general there is only a unique homomorphism not an isomorphism.
2) The inverse limit is still unique up to unique isomorphism.
The reason that you still have 2) is that if you had inverse limits $X$ and $Y$ then you had the universal property for both of them. So you have unique homomorphisms $X\to Y$ and $Y\to X$. You can show that their compositions are the identities (by invoking the universal property for a third time.)
Edit To clarify regarding the comments: You have the universal propery for an object $X$. Then you can test it for any objects $Y$ which comes with the according $\pi_i$. You get a homomorphism $f:Y\to X$. Note that $Y$ can be a fairly arbitrary object. Only if $Y$ also has the universal property you get a map in the other direction which then turns out to be an inverse.
Edit2: Your example is pretty good so I'll work with it. So let $X_n=[-\frac1n,\frac1n]$ with the obvious inclusions as maps between them.
Which candidates for the inverse limit do we have? First note that $X$ comes with maps (i.e. inclusions) to all of the $X_n$, so $X$ is a subset of every $X_n$ and therefore a subset of the intersection. In our example $X$ can only be $\{0\}$ or $\emptyset$, since those are the only subsets of the intersection. Note that the class of all candidates at this point is the class of all "$Y$" at which we have to test in the next step.
Now we turn to the universal property: for any test-set $Y$ which is a subset of the intersection we want to have a morphism (inclusion) $Y\subset X$. Thus $X$ must be contained in the intersection (step 1) and also contain all other sets with this property (step 2), which leaves us with $X=\{0\}$.
This is somewhat the whole idea of what is going on there. In general $X$ has to be "small enough" to come with maps into all the $X_n$, and then it must be the "biggest" among all objects with this property. |
H: $I\cdot J$ principal implies $I$ and $J$ principal?
Let $R$ be a Noetherian domain, and let $I$ and $J$ be two ideals of $R$ such that their product $I\cdot J$ is a non-zero principal ideal. Is it true that $I$ and $J$ are principal ideals ? This seems an easy question to settle, but I can't find an answer.
Any idea is welcome, thanks !
Thanks a lot for your enlightening answers. I admit I'm more interested in a geometric setting (i.e. when $R$ is an algebra, finitely generated, or a localization of that). I fail to adapt examples coming from number theory to this setting. What do you think ?
AI: At Lierre's request , here is a geometric example.
Consider an elliptic curve $\bar E$ (say over $\mathbb C$), a point $P\in \bar E$ of order $2$ in the group $\bar E(\mathbb C)$ (there are 3 such) and the complement $E=\bar E \setminus \lbrace O\rbrace $ of the origin in $\bar E$.
Like all non complete integral curves $E$ is affine, with ring $R=\Gamma (E, \mathcal O_E)$.
The ideal $I=\mathfrak m_P\subset R$ of functions vanishing at $P$ is not principal because $\mathcal O_E(-P)$ is a non-trivial line bundle (use Abel-Jacobi's theorem).
However $I^2=\mathfrak m_P^2$ is principal because the line bundle $\mathcal O_E(-P)$ has as its square $\mathcal O_E(-2P)=\mathcal O_E(0)=\mathcal O_E= $ the trivial line bundle, so that $I$ is an example of non-principal ideal with $I\cdot I$ principal. |
H: Is the set of dyadic rationals a field?
I recently learned that the dyadic rationals is the set of rational numbers of the form
$$\frac{p}{2^q}$$
where $p$ is an integer and $q$ is greater than or equal to zero.
I think the set of dyadic rationals is not a field. Here's why:
One of the requirements for a set to be a field is this:
Every element $a$ in the set has exactly one reciprocal such that $a$ multiplied by the reciprocal equals $1$.
I think that this dyadic rational does not have a reciprocal:
$$\frac34$$
The reciprocal is $4/3$, but that is not a dyadic rational because there is no integer $q$ greater than or equal to $0$ such that $2^q=3$.
Therefore the dyadic rationals fails one of the requirements for being a field.
Therefore the dyadic rationals are not a field.
Ha! How about that logic. Am I thinking correctly? Am I correct?
AI: Correct.
The ring of dyadic rationals is obtained from the integers $\Bbb Z$ inverting $2$. But that is not enough to invert all non-zero integers.
Besides, there's no smaller field of the rationals $\Bbb Q$ containing $\Bbb Z$. |
H: Explanation of matrix elements containing integers modulo a prime
From Cormen et all:
The elements of a matrix or vectors are numbers from a number system, such as the real numbers , the complex numbers , or integers modulo a prime .
What do they mean by integers modulo a prime ? I thought real numbers and complex numbers together make up all the elements of a matrix . Why did they put this additional one ?
AI: You can actually form matrices with entries in any ring, although sometimes you won't have the same nice properties.
The ring of integers modulo a prime, sometimes denoted $\Bbb F_p$, is the ring where you perform modular arithmetic modulo $p$.
The reason to stick to a prime number (modular arithmetic can in fact be done modulo any natural number $N$) is that some nice properties are mantained, such as the possibility to find an inverse matrix whenever the determinant is not zero. |
H: Question on statistics
I have a kind of weird question. But this wont be a harder one.
Actually, i feel it is incomplete. I don't have much experience on statistics. But some advance user will be able to understand this (may be guess the incomplete parts). I'll post all the information i have. Please try to give an answer to this.
This is about a set of calls..
Number of calls taken, $N = 1000$
Mean value, $\mu = 240$
Standard deviation, ${\sigma} = 50$ (Can this be in seconds???)
Given above details, the question is something like "What is the ***" of 1% of all are shorter:
And the answer is 124s
Can someone please help me with this. Please explain what can be the question and show me how to solve it.
I have some some knowledge on statistics and normal distribution..
Any help is greatly appreciated...
Thank you!
AI: Consider the normal distribution with mean 240 and standard deviation 50. Then 123.68 (so approx. 124) is the 1%-quantile of this distribution. So if you assume that your observations are normally distributed with the above mean and standard deviation, approximately 1% of them should be 124 or less. So, loosely speaking, the question your are looking for might be something like "What is the length x, such that 1% of the observations are shorter than x?" |
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