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H: Discrete valuation ring associated with a prime ideal of a Dedekind domain
Let $A$ be a Dedekind domain.
Let $K$ be the field of fractions of $A$.
Let $P$ be a non-zero prime ideal of $A$.
Let $v_P$ be the valuation of $K$ with respect to $P$.
Then the localization $A_P$ of $A$ at $P$ is the valuation ring of $v_P$.
How would you prove this?
AI: The ring $A_P$ is a discrete valuation ring because it is a local Dedekind domain, and one can show that, in general, a Dedekind domain with finitely many primes is a principal ideal domain (the argument uses the CRT). It follows that any element $x\in K^\times$ can be uniquely written as $\pi^nu$ where $\pi\in A_P$ is a uniformizer and $u\in A_P^\times$. By definition, $v_P(x)=n$, so $v_P(x)\geq 0$ if and only if $n\geq 0$, if and only if $x\in A_P$.
Alternatively, the valuation ring of $v_P$ is clearly a local extension of $A_P$, and since $A_P$ is a discrete valuation ring itself, the two must coincide, since a valuation ring is maximal for the relation of domination on local subrings of $K$ with fraction field $K$. |
H: The $100$th derivative of $(x^2 + 1)/(x^3 - x)$
I am reading a collection of problems by the Russian mathematician Vladimir Arnol'd, titled A Mathematical Trivium. I am taking a stab at this one:
Calculate the $100$th derivative of the function $$\frac{x^2 + 1}{x^3 - x}.$$
The derivative is non-trivial (in the sense that I computed it for a few rounds, and it only became more assertive). My first thought was to let
$$f(x) = x^2 + 1, \text{ } g(x) = \frac{1}{x^3 - x}$$
and apply the Leibnitz rule for products,
$$fg^{(n)}(x) = \sum_{k=0}^n {n\choose k} f^{(n-k)}(x)g^{(k)}(x) .$$
Since $f$ is vanishing after the third differentiation, we get
$$fg^{(100)}(x) = {100 \choose 2}f^{(2)}g^{(98)} + {100 \choose 1}f^{(1)}g^{(99)} {100 \choose 0}f^{(0)}g^{(100)} \\= 9900g^{(98)} + 200xg^{(99)} + (x^2 + 1)g^{(100)}$$
This would be great if we could compute the last few derivatives of $g$. Indeed, we can boil this down: notice that
$$g(x) = h(x)i(x)j(x), \hspace{4mm} h(x) = \frac{1}{x-1}, \text{ } i(x) = \frac{1}{x}, \text{ } j(x) = \frac{1}{x+1};$$
further, $h, i,$ and $j$ have friendly behavior under repeated differentation, e.g. $h^{(n)}(x) = \frac{(-1)^n n!}{(x-1)^{n + 1}}$.
So overall, it is possible to use Leibnitz again to beat a lengthy derivative out of this function, (namely,
$$g^{(n)}(x) = \sum_{k=0}^n {n \choose k} h^{(n-k)}(x) \Bigl(\sum_{l=0}^k {k \choose l} i^{(k-l)}(x) j^{(l)}(x)\Bigr)$$
with the details filled in).
However, this is really pretty far from computing the derivative.
So, my question: does anyone know how to either improve the above argument, or generate a new one, which can resolve the problem?
AI: We have a partial fraction decomposition
$$
\frac{x^2+1}{x^3-x}=\frac{-1}{x}+\frac{1}{x+1}+\frac{1}{x-1}
$$
It follows that
$$
\left(\frac{d}{dx}\right)^{100}\frac{x^2+1}{x^3-x}=\frac{-100!}{x^{101}}+\frac{100!}{(x+1)^{101}}+\frac{100!}{(x-1)^{101}}
$$ |
H: Hilbert class field of $\mathbb{Q}(\sqrt{65})$
Let $K = \mathbb{Q}(\sqrt{65})$.
Let $L = \mathbb{Q}(\sqrt{5}, \sqrt{13})$.
Is $L$ the Hilbert class field of $K$?
If yes, how would you prove this?
AI: First, compute the class number of $K$; the answer is $2$.
Now $L$ is a quadratic extension of $K$, which is unramified except possibly at primes above $5$ (write $L = K(\sqrt{5})$ ) and is also unramified except possibly at primes above $13$ (write $L = K(\sqrt{13})$). Thus $L/K$ is quadratic and unramified everywhere (including at infinity, since it is a totally real extension), and so must be the Hilbert Class Field of $K$. |
H: Lifting projective indecomposable modules - Benson's vol. 1
I'm reading Benson's Representations and Cohomology I, Section 1.9.
Could someone please clarify to me the following sentence
at page 18 lines -7,-6,-5:
"So given a simple $\bar\Lambda$-module $S_j$, it has a projective cover $P_j=\bar Q_j$ for some projective indecomposable $\Lambda$-module $Q_j$ unique up to isomorphism.''
$\hskip250pt$ Thanks in advance!
AI: I don't have the book available, and the link didn't work for me, so I can't pin down the precise context, but I think I can answer your question:
First let me suppose for a moment that $\Lambda = \overline{\Lambda}$.
Then the statement just becomes the following:
If $S$ is a simple $\Lambda$-module, then it admits a projective cover which is indecomposable.
Why is this? Well, any module admits a projective cover $P \to S$ which is unique up to isomorphism, characterized by the property that if $M$ is any non-zero submodule of $P$, then the induced map $M \to S$ is non-zero. Suppose
now that $M$ is a non-zero direct summand of $P$: then $M \to S$ is non-zero, and hence surjective (because $S$ is simple). Also, $M$ is projective, being a summand of a projective module. Thus $M$ is also a projective cover of $S$, and slightly more argument (essentially, the same arguments that show uniqueness of projective covers) shows that in fact $M = P.$ This proves the indecomposability of $P$.
Returing to your context: since $S$ is a simple $\overline{\Lambda}$-module, it is also a simple $\Lambda$-module. If we let $Q$ denote its projective cover
as a $\Lambda$-module, then I think that $\overline{Q}:= \overline{\Lambda}\otimes_{\Lambda} Q$ will be its projective cover as an $\overline{\Lambda}$-module. (This just uses the surjectivity of $\Lambda \to \overline{\Lambda}$.) The preceding argument shows that $Q$ is indecomposable,
and $Q$ is unique up to isomorphism, since it is the projective cover of $S$ over $\Lambda$. |
H: The probability of bricks arranged randomly.
If someone have $4$ red bricks and $8$ blue bricks and arranges them randomly in a circle,
what is the probability that two red bricks are not side by side?
Is the sample space $\frac{12!}{4!\,8!} \,=\, 495$ ?
and the answer $P(\text{Reds non-adjacent}) \:=\:\frac{70}{495} \:=\:\frac{14}{99}$ ?
AI: Suppose there $r$ red bricks, and $b$ blue bricks. Counting clock-wise, let
$$
(k_1, k_2, \ldots, k_r)
$$
be the tuples, representing the number of blue bricks between two red bricks. The number of favorable configurations is:
$$
N_F = \sum_{k_1=1}^{b-r+1} \sum_{k_2=1}^{b-r+1} \cdots \sum_{k_r=1}^{b-r+1} \delta_{k_1+k_2+\cdots+k_r, b} = [t^b] \left( \frac{t-t^{b-r+2}}{1-t} \right)^r = [t^{b-r}] \left( \frac{1-t^{b-r+1}}{1-t} \right)^r = [t^{b-r}] \left(1-t\right)^{-r} = \binom{b-1}{r-1}
$$
and the number of total configurations:
$$
N_T = \sum_{k_1=0}^{b} \sum_{k_2=0}^{b} \cdots \sum_{k_r=0}^{b} \delta_{k_1+k_2+\cdots+k_r, b} = [t^b] \left( \frac{1-t^{b+1}}{1-t} \right)^r= [t^b] \left(1-t\right)^{-r} = \binom{b+r-1}{b}
$$
Hence the probability:
$$
p =\frac{N_F}{N_T} = \frac{\binom{b-1}{r-1}}{\binom{b+r-1}{b}} = \frac{b! (b-1)!}{(b-r)! (b+r-1)!}
$$
With $b=8$ and $r=4$, we have $N_F = \binom{7}{3} = 35$ and $N_T = \binom{11}{8} = 165$, thus
$$
p = \frac{N_F}{N_T} = \frac{35}{165} = \frac{7}{33}
$$ |
H: Rating system incorporating experience
I am looking to rank players on something and right now I am using a win-loss percentage to "rank" the players.
The issue is that I have some people who just beat out the 5 game threshold and have a win/loss ratio of say 5 wins, 1 loss for 83% and then I have someone who has 54 wins and 10 losses for ~82%. I want to "rank" the more experienced player higher than the other, but I have no idea how to include "experience" as part of the formula.
I know there are many many ways this can be done, I am just interested in a few of the most simplistic accurate and common approaches and perhaps links to rating theory discussion when only knowing the wins and losses.
AI: One option is to use something like the beta-binomial model.
The general idea is that players' true success rates (or winning frequency) come from an underlying distribution (e.g. a beta distribution). As a player plays more games and you get actual information on wins and losses, the prior information from the beta distribution is combined with the wins/losses information which is expected to follow a binomial distribution (based on the true success rate $p$), and a posterior estimate of the success rate made as
$$\hat{p}=\frac{n_{\text{wins}}+\nu\rho}{n_{\text{games}}+\nu}$$
where the beta distribution essentially has the effect of a prior information equivalent to $\nu$ games with $\rho$ success rate.
The advantage of this method is that for playes with few games, the estimated success rate is shrinked towards the population mean; extreme success rates due to highly uncertain success rate estimates for playes with few games are avoided.
Due to the similarity with Bayesian methods, this type of approach is often referred to as empirical Bayes. However, the parameters $\nu$ and $\rho$ used to specify the beta distribution are estimated using traditional frequentist methods (moment or maximum likelihood estimates).
NB: I'm assuming, although it wasn't stated clearly, that these are solitaire games: i.e. one-player games. If they are two-player games, the strength of the opponents must be taken into account. A similar approach may still be used to provide a prior distribution for the strength of new players, but that would have to be a rather different model than the beta-binomial.
As requested in the comments, here are some more details on how this is done. I'll do it using the beta-binomial model, but the same principle applies to a number of similar models (e.g. see Wikipedia's article on Empirical Bayes).
Let's assume we have $N$ players, where player $k$ has played $n_k$ times and won $x_k$ of them. We assume that player $k$ has an underlying probability $p_k$ of winning each time (the game is still solitaire, so no opponents on which this could depend) and so the number of wins, $x_k$, should come from a binomial distribution $\text{Bin}(n_k,p_k)$.
We could, of course, estimate $p_k$ based on $x_k$ and $n_k$ alone: i.e. $\hat{p}_k=x_k/n_k$. However, if $n_k$ is small, this estimate would be very uncertain.
In order to reduce the uncertainty of the estimate, we assume that the winning probabilities, $p_k$, come from a beta distribution. Why a beta distribution, you might ask? Because it is conjugate to the binomial distribution, which makes the maths easier.
Note that the beta distribution is usually defined as
$$\text{Beta}(\alpha,\beta)
\sim \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}
\mbox{ where }
B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.
$$
Often parameters $\rho=\alpha/(\alpha+\beta)$ and $\nu=\alpha+\beta$ are used instead (which are the ones I used above), however I'll stick to $\alpha$ and $\beta$ here to reduce confusion.
If we assume the underlying (but unknown to us) winning probability, $p_k$, is drawn from a beta distribution with parameters $\alpha, \beta$, and then $x_k$ is drawn from $\text{Bin}(n_k,p_k)$, the posterior probability distribution is $\text{Beta}(\alpha+x_k,\beta+n_k-x_k)$. The expected values of $p_k$ given $x_k$ and $n_k$ is thus
$$\text{E}[p_k|x_k,n_k]=\frac{\alpha+x_k}{\alpha+\beta+n_k}.$$
The tricky part is finding good estimates for $\alpha$ and $\beta$. Remember, these parameters govern how the underlying probabilities, $p_k$, are distributed: e.g. the expected value and the variance.
I believe the preferred method will often be maximum likelihood estimates: i.e. the values of $\alpha$ and $\beta$ that maximise
$$
\prod_{k=1}^N \Pr(x_k|\alpha,\beta,n_k)
=\prod_{k=1}^N \int_0^1 \Pr(x_k|n_k,p_k)\cdot\Pr(p_k|\alpha,\beta)\, dp_k
$$
where the right-hand side contains the probability functions for the binomial distribution (of $x_k$) times that of the beta distribution (for $p_k$).
Another alternative are moment estimates.
Here's a quick walk-through of the computations.
The unknown parameters, $\alpha$ and $\beta$ of the beta distribution, describe the distribution of success probabilities of players, $p_k$ for player $k$:
$$
\Pr(p_k|\alpha,\beta)=\frac{p_k^{\alpha-1}(1-p_k)^{\beta-1}}{B(\alpha,\beta)}
$$
where
$$
B(\alpha,\beta)=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}
=\frac{\Gamma(\alpha)\Gamma(\beta}{\Gamma(\alpha+\beta)}
$$
and $\Gamma(x)$ is the gamma-function ($\Gamma(x)=(x-1)!$ for $x\in\mathbb{N}$).
Given the success probability $p_k$ of player $k$ and the number $n_k$ of games played, the probability of $x_k$ wins is
$$\Pr(x_k|p_k,n_k)={n_k\choose x_k}p_k^{x_k}(1-p_k)^{n_k-x_k}.$$
The likelihood of $x_k$ wins given $\alpha$ and $\beta$, but treating $p_k$ as an unknown, is then given by
$$
\begin{split}
\Pr(x_k|\alpha,\beta,n_k)=&\int_0^1 \Pr(x_k|p_k,n_k)\cdot\Pr(p_k|\alpha,\beta)\,dp_k\\
=&\int_0^1 {n_k\choose x_k}
\frac{p_k^{\alpha+x_k-1}(1-p_k)^{\beta+n_k-x_k-1}}{B(\alpha,\beta)}\,dp_k\\
=&{n_k\choose x_k}\frac{B(\alpha+x_k,\beta+n_k-x_k)}{B(\alpha,\beta)}.
\end{split}
$$
If there are $N$ players, i.e. $x=(x_1,\ldots,x_N)$ and $n=(n_1,\ldots,n_N)$ are the number of wins and games played by each player, the likelihood function, i.e. the probability of $x$ given $\alpha$ and $\beta$ is
$$
L(x|\alpha,\beta)
=\prod_{k=1}^N \Pr(x_k|\alpha,\beta,n_k)
=\prod_{k=1}^N {n_k\choose x_k}\frac{B(\alpha+x_k,\beta+n_k-x_k)}{B(\alpha,\beta)}.
$$
The maximum likelihood method basically consist of finding the parameters, $\alpha$ and $\beta$, that maximises the likelihood function: i.e. the parameters that make the data as likely as possible. In order to do this, it is easier to write out the log-likelihood, i.e. the logarithm of the likelihood function, entering the expression for $B$:
$$
\begin{split}
l(x|\alpha,\beta)=&\ln L(x|\alpha,\beta)\\
=&\sum_{k=1}^N \ln{n_k\choose x_k}+\ln B(\alpha+x_k,\beta+n_k-x_k)-\ln B(\alpha,\beta)\\
=&\sum_{k=1}^N \Big[\ln{n_k\choose x_k}\\
&\qquad+\gamma(\alpha+x_k)+\gamma(\beta+n_k-x_k)-\gamma(\alpha+\beta+n_k)\\
&\qquad-\gamma(\alpha)-\gamma(\beta)+\gamma(\alpha+\beta)\Big]\\
\end{split}
$$
where $\gamma(x)=\ln\Gamma(x)$.
Finding the $\alpha, \beta$ that maximise $l(x|\alpha,\beta)$ is done numerically, e.g. using Newtons method which involves solving
$\frac{\partial}{\partial\alpha}l(x|\alpha,\beta)
=\frac{\partial}{\partial\beta}l(x|\alpha,\beta)=0$. Note that the binomials ${n_k\choose x_k}$ are constant with respect to $\alpha$ and $\beta$ and so can be ignored.
For the numerical solution, it is helpful to have to digamma function $\Psi(x)=\gamma'(x)=\Gamma'(x)/\Gamma(x)$ available since the derivatives of the log-likelihood is a sum of these. For large $x$ (not close to zero), $\Psi(x)\approx\ln(x-1/2)$; for small $x$, one may use $\Psi(x)=\Psi(x+k)-\sum_{j=0}^{k-1}\frac{1}{x+j}$, although I suppose there may be more efficient ways of computing it.
The estimated values of $\alpha$ and $\beta$, i.e. the values that maximise $l(x|\alpha,\beta)$, are usually denoted $\hat\alpha$ and $\hat\beta$ to distinguish them from the actual (but unknown) values. These estimated values are then plugged in to get an estimate for the success rate of player $k$:
$$\hat{p}_k=\frac{x_k+\hat\alpha}{n_k+\hat\alpha+\hat\beta}.$$ |
H: Does inversion reverse order for positive elements in a unital C* algebra?
Let's say that in a unital C* algebra, we have $b \geq a \geq 0$ and $a$ is invertible. Then $b$ is also invertible. Can we conclude that $a^{-1} \geq b^{-1}$? If so, why? Can any related statement be made if we only assume $a \leq b$? (And maybe that $a$, and hence $b$ are self adjoint.)
AI: Yes. If $b \ge a \ge 0$ and $a$ is invertible, then $a^{-1/2} b a^{-1/2} \ge a^{-1/2} a a^{-1/2} = 1$
so $ a^{1/2} b^{-1} a^{1/2} = (a^{-1/2} b a^{-1/2})^{-1} \le 1$ and then $b^{-1} = a^{-1/2} a^{1/2} b^{-1} a^{1/2} a^{-1/2} \le a^{-1}$.
Without the $\ge 0$ it's false, e.g. try $a = -1$ and $b = 1$. |
H: What is $\aleph_0$ powered to $\aleph_0$?
By definition $\aleph_1 = 2 ^{\aleph_0}$. And since $2 < \aleph_0$, then $2^{\aleph_0} = {\aleph_1} \le \aleph_0 ^ {\aleph_0}$. However, I do not know what exactly $\aleph_0 ^ {\aleph_0}$ is or how I could compute it.
AI: No. By definition $\aleph_1$ is the least uncountable $\aleph$ number. $2^{\aleph_0}$ can be quite a large $\aleph$, or it could be $\aleph_1$. For example, many forcing axioms (e.g. the proper forcing axiom) prove that $2^{\aleph_0}=\aleph_2$.
The assertion $2^{\aleph_0}=\aleph_1$ is known as The Continuum Hypothesis and was proven unprovable from the usual axioms of set theory. We can therefore add axioms which decide the continuum hypothesis, e.g. itself or the aforementioned forcing axiom.
On the the other hand:
$$2^{\aleph_0}\leq\aleph_0^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}= 2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$$
To read more:
Here are some links to answers discussing the cardinality of the continuum:
Can the real numbers be forced to have arbitrary cardinality?
bound on the cardinality of the continuum? I hope not
Implications of continuum hypothesis and consistency of ZFC
How do we know an $ \aleph_1 $ exists at all? |
H: Puzzle - reordering people in a line
Eighty players numbered I through $80$ are standing in a row, one behind the other, in the
increasing order of their numbers. The physical director of the players performed eight successive
inspections of the players and in each of the respective inspections he sent the first $10$, $20$, $30$, $40$,
$50$, $60$, $70$ and $80$ players, from the front of the row to the end of the row. Each time the players
being sent back one after another. After these eight rounds of inspections what is the position of
the player numbered $5$?
Every time I solve it I gets confused, but still is there any general solution to this problem
AI: Let there are $n$ players.
And say $j$ players are sent at the back, so the position of the $i^\text{th}$ player will be $$k\equiv(i-j)\pmod n$$
Repeat it $p$ number of times we get
$$k_1\equiv (i_1-j_1) \pmod n$$
$$k_2\equiv (k_1-j_2) \pmod n $$
$$\vdots$$
$$k_p\equiv (k_{p-1}-j_{p}) \pmod n$$
Adding all of them and noticing the all the $k_i's$ will get cancelled except $k_p$.
We will get
$$k_{p} \equiv(i_1-\sum _{q=1}^p j_q ) \pmod n$$
In your case $i_1=5 , \sum j=360,p=8\text{ and }n=80$.So the answer will be $k_8\equiv-355\pmod {80}\equiv45$. |
H: Spivak Calculus 3rd ed. $|a + b| \leq |a| + |b|$
I'm working through the first chapter of Michael Spivak's Calculus 3rd ed.
Towards the end of the chapter he proves $ |a + b| ≤ |a| + |b| $ using the observation that $|a|= \sqrt{ a^2 }$ when $a$ is $ ≥ 0 $ .
$ |a + b| ≤ |a| + |b| $
$$ (|a + b|)^2 = (a + b)^2 $$ $$= a^2 + 2ab + b^2 $$
$$ ≤ a^2 + 2|a| |b| + b^2 $$
$$ = |a|^2 + 2|a| |b| + |b|^2 $$
$$ = (|a| + |b|)^2 $$
I am unsure about what's going on with the equality sign. How does it go from $=$ to $≤$ on line 3 when a and b are changed to their absolute value and back to $=$ again on line 4 when $a^2$ and $b^2$ are changed to their absolute values?
AI: To extend Ted's comment, Spivak is claiming that:
$$a^2 + 2ab + b^2 \leq a^2 + 2|a| |b| + b^2$$
(this follows because $x \leq |x|$ for every $x \in \mathbb{R}$) and that:
$$a^2 + 2 |a| |b| + b^2 = |a|^2 + 2 |a| |b| + |b|^2$$
(this follows because $x^2 = |x^2| = |x|^2$ for all $x \in \mathbb{R}$ since $x^2 \geq 0$ for all $x \in \mathbb{R}$).
You can then "string these statements together" to conclude that
$$a^2 + 2ab + b^2 \leq |a|^2 + 2 |a| |b| + |b|^2.$$
(The general principle being $x \leq y$ and $y = z$ implies that $x \leq z$.)
Hope that clarifies your confusion. |
H: Application of Bezout identity
From Wikipedia: en.wikipedia.org/wiki/Chinese_remainder_theorem
Suppose
$x \equiv a_1 \pmod{n_1} \\
x \equiv a_2 \pmod{n_2}$
where $n_1$ and $n_2$ are coprime to each other.
How does this result in the following:
$n_2 [n_2^{-1}]_{n_1} + n_1 [n_1^{-1}]_{n_2} = 1$
where the notation $ [a^{-1}]_b$ denote the multiplicative inverse of $a \pmod{b}$
AI: By using the Euclidean Algorithm (backward), you can prove that for any $n$ and $m$, there exists $r$ and $s$ such that
$rm + sn = \text{gcd}(m,n)$
This is often called Bezout Identity.
If $m$ and $n$ are coprime, then $gcd(m,n) = 1$. So you get
$rm + sn = 1$
Note that $rm = -sn + 1$ implies that $rm \equiv 1 \text{ mod } n$. Thus $r$ is inverse of $m$ modulo $n$. Similarly $s$ is the inverse of $n$ modulo $m$. |
H: Finding Satisfiability, Unsatisfiability and Valid well formed formula
I have a confusion regarding how to check whether a wff is satisfiable, unsatisfiable and valid.
As far as I understood, valid means the truth table must be a tautology, otherwise it is not a valid wff. For eg, p or (not p) is valid since it is a tautology, whereas p and (not p)
is a contradiction. So it is not valid.
Unsatisfiable means the truth table is a contradiction. ie. A wff which is not valid is unsatisfiable. Eg. p and (not p)
Satisfiable can be a tautology or even if there is one True value in the truth table. Eg. p or (not p) is tautology and not(p implies q) has one True. So, they are satisfiable.
Am I correct about the terms? If not, please explain with simple examples like what I've used. Thanks in advance.
UPDATED:
Please check out [satisfiability][1] In that there is a definition about satisfiability which confuses me.
It says, a formula 'A' is satisfiable if and only if 'not(A)' is not a tautology
AI: The following may seem rather strange, but I hope it'll help you think about these three concepts.
Let $\Bbb B$ denote the boolean set $\{ T, F \}.$ Now, think of the WFF containing $n$ propositions as a function: $\Bbb{B}^n \to \Bbb{B}.$ I will use the following WFF as my running example: $p \land \lnot q.$ That's a "function" $\mathbb{B}^2 \to \Bbb{B}.$ It's a function that takes two boolean inputs (each being true or false) and produces a boolean value $f(p, q) = p \land \lnot q.$ The truth table is just a tabulation of this "function" for all (four) possible values of $p$ and $q.$
This "function" could be true for some values of $p, q$. It could also be false for some other values of $p, q.$ Some functions are always (constant) true or always (constant) false. This is exactly the distinction between satisfiable, valid, and unsatisfiable, respectively.
Now, a WFF is valid if $f(p, q) = T$ for all values of $p, q.$ It's unsatisfiable if $f(p, q) = F$ for all values of $p, q.$ The former two cases resemble constant functions. A WFF is satisfiable if $f(p, q) = T$ for some values of $p, q.$ It could be true/false for all other values; that doesn't matter. What matter is that it's possible to find values of $p, q$ to satisfy the WFF (satisfy a WFF = making the WFF true). |
H: Primitive Element for Field Extension of Rational Functions over Symmetric Rational Functions
A rational function $f$ in $n$ variables is a ratio of $2$ polynomials,
$$f(x_1,...x_n) = \frac{p(x_1,...x_n)}{q(x_1,...x_n)}$$
where $q$ is not identically $0$. The function is called symmetric if
$$f(x_1,...,x_n) = f(x_{\sigma(1)},...,x_{\sigma(n)})$$
for any permutation $\sigma$ of $\{1,\ldots,n\}$.
Let $F$ denote the field of rational functions and $S$ denote the subfield of symmetric rational functions. Suppose the coefficients of polynomials are all real numbers.
Show that $F = S(h)$, where $h = x_1 + 2x_2 + ... + nx_n$. In other words, show that $h$ generates $F$ as a field extension of $S$.
Attempt at Solution:
Can't seem to get very far with this one. I know that $F$ is a finite extension of $S$ of degree $n!$ and the Galois group of the extension is $S_n$.
Using $h$ and the 1st symmetric function $s_1 = x_1 + x_2 + \ldots + x_n$, we see that $h - s_1 = x_2 + 2x_3 + \ldots (n-1)x_n \in S(h)$.
Can't seem to find a good way to use the other symmetric functions $s_2,\ldots, s_n$.
Any help would be greatly appreciated. Thank you.
AI: According to Galois theory, since $S \subset S(h) \subset F$, $S(h)$ is $F^H$, the field of elements of $F$ fixed by some subgroup $H$ of $S_n$. Since $h$ is only fixed by the identity automorphism, $H = \{id \}$, and $S(h) =F$.
In more detail :
Let $P$ be the minimal polynomial of $h$ over $S$ and let $\sigma$ be in $S_n$, so that $\sigma(h) = x_{i_1} + 2 x_{i_2} + \ldots + n x_{i_n}$. Since the coefficients of $P$ are in $S$, $\sigma(P) = P$, so $0 = \sigma(0) = \sigma(P(h)) = \sigma(P)(\sigma(h)) = P(\sigma(h))$, thus $\sigma(h)$ is also a root of $P$.
Since all the $\sigma(h)$ are pairwise distinct, $P$ has degree at least $n!$, thus the extension $S(h)$ over $S$ is at least of degree $n!$
But $S(h) \subset F$, and $F$ is also of degree $n!$ over $S$, thus those two fields are equal. |
H: Evaluate $\frac{d}{dx}ⁿx$.
Evaluate $\frac{d}{dx}ⁿx$ which $ⁿx=x^{x^{x^{...}}}$, total $n$ $x$'s.
I have tried to observe when $n=1,2,3,4,5$, but it's difficult to see the pattern. Can anyone get it? Thank you.
AI: As I said in my comment above, define $y= ^n\!\!x$. Then
$$\log y=\log ^nx=\log x^{^{n-1}x}=\, ^{n-1}x\log x.$$
Differentiating implicitly, we get
$$\frac{d y}{d x}\frac{1}{y}=\frac{d}{dx}(\, ^{n-1}x)\log x+\, ^{n-1}x\frac{1}{x},$$
from which we see
$$\frac{d}{dx}(\, ^{n}x)=\frac{\, ^{n}x}{x}\left(x\log x\frac{d}{dx}(\, ^{n-1}x)+\, ^{n-1}x\right).$$
For the first few values of $n$, we get
$n=1:\quad\frac{d}{dx}(\, ^{1}x)=\frac{d}{dx}(x)=1$
$n=2:\quad\frac{d}{dx}(\, ^{2}x)=\frac{\, ^{2}x}{x}\left(x\log x\frac{d}{dx}(\, ^{1}x)+\, ^{1}x\right)=x^x(\log x+1)$
$n=3:\quad\frac{d}{dx}(\, ^{3}x)=\frac{\, ^{3}x}{x}\left(x\log x\frac{d}{dx}(\, ^{2}x)+\, ^{2}x\right)=x^{x^x}\left(x^x(\log x+1)\log x+x^{x-1}\right)$ |
H: Mathematics base for data mining and artificial intelligence algorithms.
Could you give me some clarification about data mining and artificial intelligence algorithms? What mathematics base they used for? Could you give me starting point, in mathematics, to understand these types of algorithms?
AI: As Paxinum mentioned, AI algorithms are sort of solving optimization problems. You would need to solve maximum and minimum problems.
You may use neural network or classical programming methods or both of them in the same program.
It depends on which problem you struggle. If Image processing or voice recognition includes in your program then neural network would be the best choice because the classical programs would have many steps to solve the problem and it may produce unacceptable delays for some inputs because many comparing steps will be required in it for all possibilities but the neural networks would act at the same speed (approximately) for all kind of inputs.One of the big disadvantage of neural networks is required many self learning steps to know how it must react for an input. Many input and output examples can be needed to get the best AI's network. If we do not teach enough with many examples, some results may be very absurd . That can be big risk for critical AI applications such as airplane control system. Some problems can be needed graph approach because teaching process can be very long and neural network can be enormous.
For example, If you want to do a chess program, you may give values of chess pieces.Thus, The computer can decide which step is more valuable.
while doing my chess program, I used the following values for chess pieces.This will help to give metrics on branch of the graph.
The king : uncountable (Theoretically infinity because you cannot lose it.)
The queen :8
The rook :5
The bishop:4
The knight:3
The pawn :1
After that you would need to play for each pieces for yourself and to see which pieces of opponent can be gotten. You need to add positive value of the piece that you got. And then you need to play for the opponent but you will assign negative value of the piece that you lost . You can apply more and more deepness to enlarge the graph to see benefits and loses. I have added 2 deepness (Play for me-Play for opponent-Play for me-Play for opponent, It lasts about few second to move) You will get a final value for each first step. Now you need data mining to find max benefit and minimum lose.
I found max value after playing each steps and then If there are more than one the same value, I also focused on maximum possible step for my side and minimum possible step for the opponent side. Thus, I would have maximum playing area but the opponent can have minimum area.There can be created many different strategy for better play. I just wanted give an example. There can be found many different strategies. I do not claim it is the best strategy for chess game. The chess game is very complex game and you may need to add the best known starting strategies and some database moves that are already known as the best moves. Mostly during final of the game, you will be alone with your AI software if you cannot add all possible known best moves. I think such database will be enormous if you want add many best known moves and it is not practical games applications because a huge database will be needed.
Finally, AI is very deep subject in math world , we are on the shore of that huge ocean yet. |
H: Compute $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{{1}/{2}}\cdots\left(1+\frac{n}{n}\right)^{{1}/{n}}$
Compute the limit:
$$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots\left(1+\frac{n}{n}\right)^{\frac{1}{n}}$$
AI: Note at the onset that $1+\frac{k}n\leqslant\mathrm e^{k/n}$ for every $k$ hence the $n$th product $P_n$ is such that $P_n\leqslant\mathrm e$, in particular, the sequence $(P_n)_{n\geqslant1}$ is bounded.
To show that $(P_n)_{n\geqslant1}$ actually converges and to identify its limit, note that, for every $n$,
$$
\log(P_n)=\frac1n\sum\limits_{k=1}^nf\left(\frac{k}n\right),
\qquad\text{with}\quad f(x)=\frac{\log(1+x)}x.
$$
The function $f$ is continuous on $[0,1]$ (define $f(0)=1$) hence its Riemann sums converge to its integral and $P_n\to\mathrm e^\ell$ with
$$
\ell=\int_0^1f(x)\mathrm dx=\int_0^1\left(\sum_{n\geqslant1}(-1)^{n+1}\frac{x^{n-1}}n\right)\mathrm dx=\sum_{n\geqslant1}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}.
$$ |
H: Exact sequence of four sheaves in Beauville: associated l.e.s.?
This question is about an exact sequence of four sheaves on a smooth projective surface $S$ over $k=\mathbb{C}$, to be found in Beauville: complex algebraic surfaces, theorem I.4, page 3 (second edition).
It comes down to this: Denote the exact sequence by
$$
0 \rightarrow A \rightarrow B \rightarrow C \rightarrow D \rightarrow 0
$$
then Beauville concludes
$$
\chi(A)-\chi(B)+\chi(C)-\chi(D) = 0
$$
where $\chi(A) = \sum(-1)^{i}h^i(S,A)$ where $S$ is the surface.
How does he come to this conclusion? Of course an exact sequence of three sheaves yields a l.e.s. in cohomology, from which a similar result follows, but how to proceed with four sheaves as above?
It might be easy, but i know just the basics of sheaf cohomology.
(Also see Exact sequence in Beauville's "Complex Algebraic Surfaces" for an earlier question about the same sequence)
If necessary i can provide the exact sheaves used in the sequence, but i hope that there's a general solution. Also i have an exam coming up, so prefer to save time (btw this is not homework). The sequence is also to be found in the link i provided.
Thanks in advance.
AI: As usual, the trick is to break up an exact sequence into short exact sequences. First, consider a general short exact sequence of f.g. abelian groups
$$0 \longrightarrow K \longrightarrow M \longrightarrow Q \longrightarrow 0$$
The rank–nullity theorem from linear algebra implies
$$\operatorname{rank} K - \operatorname{rank} M + \operatorname{rank} Q = 0$$
and so by induction, if we have any long exact sequence of f.g. abelian groups
$$0 \longrightarrow M^0 \longrightarrow M^1 \longrightarrow \cdots \longrightarrow M^r \longrightarrow 0$$
we must have
$$\sum_{i=0}^{r} (-1)^i \operatorname{rank} M^r = 0$$
which is a baby version of the result you want.
Now, let $A^\bullet$ be a bounded-below cochain complex in an abelian category $\mathcal{A}$, and suppose
$$A^{n-1} \longrightarrow A^n \longrightarrow A^{n+1}$$
is exact for every integer $n$; if we define $Z^n = \operatorname{ker}(A^n \to A^{n+1})$, that means we have a short exact sequence
$$0 \longrightarrow Z^n \longrightarrow A^n \longrightarrow Z^{n+1} \longrightarrow 0$$
for each integer $n$. Now, suppose $H^\bullet : \mathcal{A} \to \textbf{Ab}$ is any cohomological $\delta$-functor (this just means we get long exact cohomology sequences); then we get an exact sequence
$$\cdots \longrightarrow H^p (A^q) \longrightarrow H^p (Z^{q+1}) \longrightarrow H^{p+1} (Z^q) \longrightarrow H^{p+1} (A^q) \longrightarrow \cdots$$
for each natural number $p$ and each integer $q$. Let us write $h^p (M)$ for $\operatorname{rank} H^p(M)$. Assuming that all the sequences in question vanish in sufficiently high degree and all the groups are f.g., our previous calculation says
$$\sum_p (-1)^p h^p(A^q) = \sum_p (-1)^p h^p (Z^{q+1}) + \sum_p (-1)^p h^p (Z^q)$$
and so
$$\sum_{p, q} (-1)^{p+q} h^p (A^q) = \sum_{p, q} (-1)^{p+q+1} h^p (Z^q) + \sum_{p,q} (-1)^{p+q} h^p (Z^q) = 0$$
so if we define $\chi (A^q) = \sum_p (-1)^p h^p (A^q)$, it follows that
$$\sum_{q} (-1)^q \chi (A^q) = 0$$
as required. |
H: Calculate width in a grid with lines
This is a web related question but is a mathematical problem.
Look at this:
|O|O|O|O|
|O|O|O|
|O|O|
A vertical line represents a 1 pixel wide horizontal line.
A 'O' represents a space between the lines.
All the three rows should have the same width. The space 'O' is streched out to fill the space equally within the row to fill it up. (The last row will get more space than the first.)
The space width can't contain decimals.
I want to find the perfect width for all three lines between 960 and 1140 pixels wide.
Questions
What is the row width?
How did you calculate it?
What I know so far
First line have 5 pixels of lines, the second 4 pixels and the third 3 pixels.
960 and 1140 are perfect numbers because they can be divided by 3, 4 and 2, but this case includes borders which makes it tricky.
AI: Well let the width of the space be n pixels, and $a$, $b$ and $c$ the widths of "O" in the three cases. You need to solve the three equations:
$$4a+5=n=4(a+1)+1$$
$$3b+4=n=3(b+1)+1$$
$$2c+3=n=2(c+1)+1$$
So you need $n$ to be 1 more than a multiple of 2,3, and 4 - so $n=12r+1$ will work for any $r$ with $a+1=3r; b+1=4r; c+1=6r$ - which gives you plenty of options to work with. |
H: Predicting the number of solutions of a system of equations without solving it explicitely
$$
\left\{
\begin{array}{l}
x + y + az + at = a \\
2x + y - a^2z + 2at = 1 \\
4x + 3y + a^2 z + 4a^2 t = 1
\end{array}
\right.
$$
Is there a way to predict the number of solutions $(x, y, z, t)$, depending on the value of the parameter $a$, without solving the system explicitely?
AI: Yes, by reducing the augmented matrix of the system to row echelon form. For your system, the augmented matrix is
$$ \left[ \begin{array}{cccc|c}
1 & 1 & a & a & a \\
2 & 1 & -a^2 & 2a & 1 \\
4 & 3 & a^2 & 4a^2 & 1
\end{array}\right] $$
and the row echelon form
$$ \left[ \begin{array}{cccc|c}
1 & 1 & a & a & a \\
0 & 1 & a^2+2a & 0 & 2a-1 \\
0 & 0 & 2(a^2-a) & 4(a^2-a) & -2a
\end{array}\right]. $$
First thing to look for is consistency, so that the last row does not correspond to $0=$non-zero. This guarantees existence of at least one solution. After that, the dimension of the solution space = the number of free variables = $n - \rm{rank}(A)$, where $n$ is the number of variables of the system, $A$ is the augmented matrix, whose rank is given by the number of non-zero rows in the row echelon form. (Of course if you've calculated the row echelon form, you're only 30 seconds from the explicit solution. If you need to know the dimension of the solution space, I don't think there's any alternative: you need the rank of $A$.) |
H: Books on AI, programming, optimization
I'm studying math (just started) and I like programming as well (just started this too), is there a career or a branch of research including deep aspect of this two aspects? Is there someone among you who is doing this perhaps?
The sensation I have is that there's much more "application" (improving skills in coding) than "pure research" (and when there is it's often about computability and limit of computability). I'm sure I'm wrong about this but that's quite all i can see now.
Can you suggest me a book that can introduce me to the mathematical aspects of artificial intelligence? I would prefer text with examples and exercises.
AI: Artificial Intelligence is a broad field. To get to the mathematical foundation of the field you will have to dive deeper than most introductory books will go - so for the sake of simplicity I will assume that you are not interested in aspects like general decision making, planing and pathfinding (all of which will be covered in basic AI books - but without much mathematical background).
If you want to combine the both fields Math and AI, you best bet is (in my opinion) with Genetic Algorithms or Neural Networks. I can only name two books that might be to your liking - hopefully someone else will add more.
An Introduction to Genetic Algorithms by Melanie Mitchell is certainly basic enough for you to read right away. As an introductory book it will not go into too much detail with calculations (or even omit them, including the results, alltogether), but the books skims a lot of interesting topics that might give you a feel of what can be done with proper mathematics in the field of Genetic Algorithms. Also it mentions some aspects like evolutionary activity that I have yet to see in another book. It contains a lot of exercises, half of which are thought exercises and half of which are programming exercises though, so not too much math there...
Spiking Neuron Models by Wulfram Gerstner and Werner Kistler on the other hand is a rather advanced book ("for advanced undergraduate or graduate students"). It covers the modeling of networks of neurons which communicate by a series of spikes instead of a scalar intensity as it is done in most practical AI uses. As such it is much closer to the real world counterpart of actual braincells and indeed starts with the Hodgkin-Huxley model of the neuron, that has been derived from the behaviour of the neurons of a squid. There is plenty of math in this book, hence the difficulty to study it though. It is an very interesting introduction to somewhat recent developments in the field - but you need at least basic knowledge in differential equation theory and should be comfortable with more advanced mathematical notations / concepts (partial differentiation, dirac-delta, big-O-notation etc.). |
H: can one derive the $n^{th}$ term for the series, $u_{n+1}=2u_{n}+1$,$u_{0}=0$, $n$ is a non-negative integer
derive the $n^{th}$ term for the series $0,1,3,7,15,31,63,127,255,\ldots$
observation gives, $t_{n}=2^n-1$, where $n$ is a non-negative integer
$t_{0}=0$
AI: You already have an expression for the $n^{\text{th}}$ term for the sequence:
$$t_{n}=2^{n}-1\tag{1}$$
So let's prove this closed form by induction. Let $P(n)$ be our proposition that $t_{n}=u_{n}$, $\forall n\in\mathbb{N}\cup\{0\}$. So let us examine our basis case: $P(0)$:
$$t_{0}=2^{0}-1=1-1=0=u_{0} \implies P(0) \text{ is true}$$
Let us now show that if $P(k)$ is true, it follows immediately that $P(k+1)$ must also be true:
$$t_{k+1}=2^{k+1}-1=2\cdot2^{k}-1=2(2^{k}-1)+1=2u_{k}+1=u_{k+1}\implies \text{ if } P(k) \text{ is true, then } P(k+1) \text{ is true}$$
Therefore, as we have shown that $P(0)$ is true, and that $P(k)\implies P(k+1)$. $P(n)$ is true, $\forall n\in\mathbb{N}\cup\{0\}$.
However, if you are interested in how to actually come up with a closed form in general, a good place to start is to look at summation factors.
We can reduce a general recurrence relationship of the form:
$$a_{n}T_{n}=b_{n}T_{n-1}+c_{n} \tag{2}$$
To a sum, by multiplying both sides by a summation factor, denoted as $s_{n}$, such that:
$$s_{n}b_{n}=s_{n-1}a_{n-1}$$
In general, we can find a suitable $s_{n}$ using any multiple of the following:
$$s_{n}=\frac{a_{n-1}a_{n-2}\cdots a_{1}}{b_{n}b_{n-1}\cdots b_{2}}\tag{3}$$
We then have a summation recurrence, to which the solution can be found to be:
$$T_{n}=\frac{1}{s_{n}a_{n}}\left(s_{1}b_{1}T_{0}+\sum_{k=1}^{n}s_{k}c_{k}\right)\tag{4}$$
In your case, we have a recurrence of the form given in $(2)$:
$$a_{n}=1\qquad b_{n}=2\qquad c_{n}=1$$
Therefore, using $(3)$ we have $s_{n}=\frac{1}{2^{n}}$. And we can therefore plug this into $(4)$ to give:
$$T_{n}=2^{n}\left(0+\sum_{k=1}^{n}{\frac{1}{2^{k}}}\right)=2^{n}\left(1-\frac{1}{2^{n}}\right)=2^{n}-1$$
Which is $(1)$, the closed form you got by observation.
If you want a more complete look at solving recurrence relationships, I'd recommend the first few chapters of Concrete Mathematics by Graham, Knuth and Patashnik.
Hope this helps! |
H: Sequence of indicator functions is weakly convergent.
Let $E\subseteq [0,1]$ be a measurable set.
Then we recall that
$\chi_E(x)=\begin{cases} 1&\text{ if } x\in E,\\ 0 & \text{ if } x\not\in E.\end{cases}$
We are given a sequence of measurable subsets $\{E_n\}_n$ of $[0,1]$ and a measurable subset $E\subseteq [0,1]$ such that $\chi_{E_n}\rightharpoonup \chi_E$ is weakly convergent in $L^2[0,1]$.
Then prove that $\chi_{E_n}\rightarrow \chi_E$ is strongly convergent in $L^2[0,1]$.
I don't need a complete solution, even an hint is welcomed.
Thank you for your help.
AI: Hint:
$$
\begin{eqnarray*}
\|\chi_{E_n}-\chi_{E}\|^2&=&\|\chi_{E_n}\|^2-2\langle\chi_{E_n},\chi_{E}\rangle+\|\chi_{E}\|^2\\[5pt]
&=&\langle \chi_{E_n}, \chi_{[0,1]}\rangle -2\langle\chi_{E_n},\chi_{E}\rangle+\langle \chi_{E}, \chi_{[0,1]}\rangle\\
\end{eqnarray*}
$$ |
H: 2 jars with 50 balls each. Pick any ball with $>50\%$ probability.
You are blindfolded and placed in front a table with two jars. One jar has $50$ red balls and other has $50$ blue balls.
What should be your strategy so that you pick up the red ball with more than $50\%$ probability.
AI: Take one ball from each jar. You're guaranteed to get a red. |
H: Truth Tables - How to identify normals in Knights, Knaves and Normals problems?
To describe my question, I'll illustrate an example of a Knights, Knaves and Normals problem and the way I solve it.
Question
Knights always tell the truth.
Knaves always lie.
Normals sometimes lie and sometimes tell the truth.
Given the following statements, identify who is a knave, a knight or a normal:
A: C is a knave or B is a knight.
B: C is a knight and A is a knight.
C: If B is a knave, then A is a knight.
Solution
Statement A (SA) = ¬C or B
Statement B (SB) = C and A
Statement C (SC) = ¬B -> A
Information A (IA) = (SA = A)
Information B (IB) = (SB = B)
Information C (IC) = (SC = C)
Puzzle Solution (PS) = (IA & IB & IC)
Where 1 is true, and 0 is false.
| A | B | C || SA | SB | SC || IA | IB | IC || PS |
---------------------------------------------------
| 1 | 1 | 1 || 1 | 1 | 1 || 1 | 1 | 1 || 1 |
| 1 | 1 | 0 || 1 | 0 | 1 || 1 | 0 | 0 || 0 |
| 1 | 0 | 0 || 1 | 0 | 1 || 1 | 1 | 0 || 0 |
| 0 | 0 | 0 || 1 | 0 | 1 || 0 | 1 | 0 || 0 |
| 1 | 0 | 1 || 0 | 1 | 1 || 0 | 0 | 1 || 0 |
| 0 | 0 | 1 || 0 | 0 | 1 || 1 | 1 | 1 || 1 |
| 0 | 1 | 1 || 1 | 0 | 0 || 0 | 0 | 0 || 0 |
| 0 | 1 | 0 || 1 | 0 | 0 || 0 | 0 | 1 || 0 |
Using a Wolfram Knights and Knaves problem generator, the correct answer is:
A: Knave
B: Normal
C: Knight
My answer would have been:
A: Knave
B: Knave
C: Knight
What I cannot understand, is how do we know that B is a normal, based on the truth table? I assume that since the program can generate who the normal is, that there is a systematic method of identifying who the normal is, using a truth table. Am I correct in my assumption? Is a systematic technique present?
This question has stumped me for a while, and I would be most grateful for an explanation. Thank you.
AI: Your answer fails because C's statement is false and he is a knight. If somebody is Normal, the only way to ID it from the truth table is 1)he makes two statements, one true and one false or 2) somebody else's statements make him Normal. In this case, A shows he is not a Knight and C shows he is not a Knave. |
H: Clarification of a Proof in Lee's An Introduction to Topological Manifolds
I'm having a bit of difficulty understanding the following proof from Lee's An Introduction to Topological Manifolds:
I don't understand the deduction that since $e_0\cap e \subseteq \overline{e}\setminus e$ and $\overline{e}\setminus e$ is contained in the union of finitely many cells of dimension less than $n$, that $e_0\cap e$ is empty since $e_0$ is an $n$-cell.
Why is this exactly?
AI: By defintion (the definition I learned) two cells of a CW complex are either disjoint or equal. Hence if you have cells of different dimensions they are disjoint. Since the author proved that $e_0\cap\bar{e}$ is contained in the finite union of cells of dimension less than the one of $e_0$, $e_0$ has empty intersection with each of them, hence with the union, hence the conclusion. |
H: Poisson's summation formula
It is said that the Fourier transform $\hat{f}(\omega)$ of a function $f(t)$ and the Fourier transform $\hat{b}(\omega)$ of its samples $b(k)=f(t)|_{t=k}$ are related by Poisson's summation formula and it's given by
$$\hat{b}(\omega)=\displaystyle\sum_{k\in{\mathbb{Z}}}\hat{f}(\omega+2\pi{k})$$
where
$$\hat{b}(\omega)=\displaystyle\sum_{k\in\mathbb{Z}}b(k)e^{-i\omega k}, \omega\in{\mathbb{R}}.$$
I just fail to see why, is this equation even right?
AI: One informal interpretation of the Poisson summation formula is that
$$\hat{b}(\omega) = \sum_{k \in \mathbb Z}\hat{f}(\omega + 2\pi k)$$
is a periodic function of $\omega$ of period $2\pi$, and therefore
representable as a Fourier series. The coefficients of this Fourier
series are the $b(k)$. If you write out the integral formula for the
Fourier coefficient and manipulate it a bit, then for reasonably
well-behaved functions, you get the relationship to the function
$f(t)$. |
H: Equivalent metrics determine the same topology
Suppose that there are given two distance functions $d(x,y)$ and $d_1 (x,y)$ on the same space $S$. They are said to be equivalent if they determine the same open sets.
Show that $d$ and $d_1$ are equivalent if to every $\varepsilon$ there exists a $\delta>0$ such that $d(x,y)<\delta$ implies $d_1 (x,y)<\varepsilon$ and vice versa.
I have no idea how to handle this kind of problem.
AI: Since open sets are the sums of open balls, it suffices to check if those metrices determine the same open balls.
Let us denote by $B_d(x,r)$ an open (wrt $d$) ball with a centre in $x\in S$ and a radius $r\gt0$, i.e. $B_d(x,r)=\{y\in S\mid d(x,y)<r\}$.
We will prove an arbitrary open ball $B_d(x_0,r)$ is $d_1$-open. In order to do that, we will show that for every $x\in B_d(x_0,r)$ there exists $r_x$ such that $B_{d_1}(x,r_x)\subset B_d(x_0,r)$.
What we have is $$(*)\quad\forall_{a\in S} \forall_{\varepsilon>0}\exists_{\delta>0}\forall_{b\in S}\ \ d_1(a,b)\lt\delta\implies d(a,b)\lt\varepsilon.$$
(our $a$ is actually $x$ and $\delta$ will be desired $r_x$.)
As has been said, fix an arbitrary $x\in B_d(x_0,r), x\neq x_0$ and set $\varepsilon=\min{(d(x,x_0),r-d(x,x_0))}>0.$ From $(*)$ we know that there exists $\delta>0$ such that for all $y\in S$ $$(1)\quad d(x,y)<\min{(d(x,x_0),r-d(x,x_0))}\le r-d(x,x_0)$$
as long as $d_1(x,y)\lt\delta$.
Now we could reformulate (1) by adding $d(x,x_0)$ to both sides and applying triangle inequality:
$$d(x_0,y)<d(x,y)+d(x,x_0)<r.$$
Thus, what we proved is:
$$\forall_{x\in B_d(x_0,r)}\exists_{\delta>0} \forall_{y\in S}\ y\in B_{d_1}(x,\delta) \implies y\in B_d(x_0,r)$$
and, in other words, it means that every $x\neq x_0$ from $d$-ball is contained there with some small $d_1$-ball, that is $d$-ball is open in the sense of metric $d_1$. For $x_0$, the existence of such $d_1$-ball is ensured by $(*)$ directly (take $\varepsilon=r)$.
Entirely analogously, one can prove second part of the problem. |
H: What makes a sufficient proof?
This question is related to the question posted here. Would a shorter proof to those in the answers, such as:
Take the subsequence $\{a_m\}$ of $\{a_n\}$ where $m > 0$. By induction on $m$ $$\forall{m}( a_m < a_{m+1})$$ Therefore $$ \forall{\epsilon} > 0 \ \exists{N} \ \forall{m_0,m_1} \ (m_0,m_1 \geq N \ \wedge \ m_0 \geq m_1 \ \ \Rightarrow \ \ m_0 - m_1 < \epsilon)$$ and $\{a_m\}$ is Cauchy. Hence it converges, so its parent sequence converges. Therefore its parent sequence is Cauchy.
be acceptable?
Furthermore, what is the minimum amount of detail required in a proof? I am curious because I am switching into a maths program, however all my maths is self taught. I am not used to having my proofs reviews or having to prove to anyone but myself.
AI: This answer solely regards the latter portion of the question.
Whenever you write a proof, it is vitally important that you know your audience. This is key because a proof is simply an argument intended to convince the reader that the purported statement is true (in general, only automatic theorem provers actually show all the details). So the question then becomes: "how much detail should I leave out?" This is precisely where your audience comes into play. For example, a professor of mathematics can, in general, make greater logical leaps than an undergraduate (for a variety of reasons such as familiarity with the material, etc). Thus the proofs presented to the undergraduate student would need to be more detailed than those presented to the other party.
However, there is one significant exception to the above rule: proofs written for coursework. In this case, the student is writing the proof for the professor to read, and so the professor could handle larger logical leaps. But, the student shouldn't write this way because the professor cannot tell if the student could actually fill in the details omitted. So, the student should write a proof as if he/she is communicating it to his/her classmates. Doing this should correspond to a level of detail where the audience (the professor) knows that the student has sufficient understanding of the material.
In summary, the level of detail required depends on the audience for whom you are writing. |
H: Lipschitz Continuous $\Rightarrow$ Uniformly Continuous
The Question: Prove that if a function $f$ defined on $S \subseteq \mathbb R$ is Lipschitz continuous then $f$ is uniformly continuous on $S$.
Definition. A function $f$ defined on a set $S \subseteq \mathbb R$ is said to be Lipschitz continuous on $S$ if there exists an $M$ so that $$\frac{|f(x) - f(c)|}{|x - c|} \le M$$ for all $x$ and $c$ in $S$ such that $x \ne c$.
My Heuristic Interpretation: if $f$ is Lipschitz continuous then the "absolute slope" of $f$ is never unbounded i.e. no asymptotes.
Definition. A continuous function $f$ defined on $\mathrm{Dom}\, (f)$ is said to be uniformly continuous if for each $\varepsilon > 0 \ \exists \ \delta > 0$ s.t. $\forall \ x, c \in \mathrm{Dom}\, (f)$ $$ |x - c| \le \delta \ \Rightarrow \ |f(x) - f(c)| \le \varepsilon$$
Proof:
$f$ Lipschitz continuous $\Rightarrow$ $|f(x) - f(c)| \le M|x - c|$. Since we suppose $|x - c| \le \delta$ for uniform continuity, we have $x$ within $\delta$ of $c$, so $|x| \le |c| + \delta$. So taking $\delta = \varepsilon/M$
\begin{align*}
|f(x) - f(c)| &\le M|x - c| \\
& \le M\delta \\
& = \varepsilon
\end{align*}
My Question: Is my proof valid with the assumptions taken?
AI: It’s not very well organized, and it has some extraneous clutter, but it also has the core of the argument. You want to show that for each $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-f(c)|<\epsilon$ whenever $x,c\in\operatorname{dom}f$ and $|x-c|<\delta$, so in a polished version of the argument your first step should be:
Suppose that $f$ is Lipschitz continuous on some set $S$ with Lipschitz constant $M$, and fix $\epsilon>0$.
You’ve already worked out that $\epsilon/M$ will work for $\delta$, so you can even start out with:
Fix $\epsilon>0$ and let $\delta=\frac\epsilon{M}$.
Now you want to show that this choice of $\delta$ does the job.
Clearly $\delta>0$. Suppose that $x,c\in S$ and $|x-c|<\delta$. Then by the Lipschitz continuity of $f$ we have $|f(x)-f(c)|\le M|x-c|<M\delta=\epsilon$, so $f$ is uniformly continuous on $S$. $\dashv$
Added: Your heuristic interpretation of Lipschitz continuity is inaccurate enough that it may well lead you astray at some point. Consider the function
$$f(x)=\begin{cases}
x\sin\frac1x,&\text{if }x\ne0\\
0,&\text{if }x=0\;.
\end{cases}$$
This function has no vertical asymptotes, but it’s not Lipschitz continuous: for $n\in\Bbb Z^+\setminus\{0\}$ we have
$$\begin{align*}
\left|\frac{f\left(\frac1{2n\pi}\right)-f\left(\frac1{2n\pi+\frac{\pi}2}\right)}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}\right|&=\left|\frac{\frac1{2n\pi+\frac{\pi}2}}{\frac1{2n\pi}-\frac1{2n\pi+\frac{\pi}2}}\right|\\
&=\left|\frac1{\frac{2n\pi+\frac{\pi}2}{2n\pi}-1}\right|\\
&=\left|\frac{2n\pi}{\pi/2}\right|\\
&=4|n|\,,
\end{align*}$$
which can be made as large as you want. This function has very, very steep bits, but they’re also very, very short. |
H: Prove that square root of 2 is irrational using the principle of Mathematical Induction
How do I prove that the square root of 2 is irrational using the principle of mathematical induction?
AI: The usual proof starts out something like "suppose for a contradiction $\sqrt 2$ were rational, and write it as $p/q$ in lowest terms ...".
The "in lowest terms" hides an instance of induction, which you can unfold to get a proof in the shape of
Theorem. For all positive integers $p$ and $q$, it holds that $(p/q)^2\ne 2$.
Proof. By long induction on $q$. If $p$ and $q$ have a common factor $n>1$ then $(\frac pq)^2 = (\frac{p/n}{q/n})^2$, and since $q/n<q$, the induction hypothesis guarantees that $(\frac pq)^2\ne 2$. Now we consider the case that $p$ and $q$ are relatively prime ... |
H: a question about permutation in the digits in the decimal system
This is an exercise from a textbook in Portuguese.
Choose a symbol (number) $abc$ in the decimal system ($a$ being the
hundreds digit, $b$ the tens digit, and $c$ the units digit), in a way
that the hundreds digit $a$ and the units digit $c$ differ by at
least $2$ units. Consider the numbers $abc$ and $cba$ and subtract
the smaller from the bigger in order to get a number $xyz$. Show that
the sum of $xyz$ with $zyx$ is 1089.
I know that $abc=a10^{2}+b10+c$, $cba=c10^{2}+b10+a$, and $c=a+i$, $i\geq 2$ (or $a=c+i$, $i\geq 2$). So if $c=a+i$, $i\geq 2$, then $$cba=(a+i)10^{2}+b10+(c-i)=a10^{2}+b10+c+(i10^{2}-i)=a10^{2}+b10+c+(99i)\;.$$ So $xyz=cba-abc=99i$, or $9 \mid xyz$ wich give me that $x+y+z=9k$, and $11 \mid xyz$ wich give me a condition, but I still can use this.
I would appreciate your help! Sorry for the lousy title.
AI: You're almost there. $xyz = 99i = 100i - i$ which has digits $x = i-1$, $y = 9$, $z = 10-i$ and so $$xyz + zyx = 100(i-1) + 90 + (10-i) + 100(10-i) + 90 + (i-1)$$$$= 100(9) + 90 + 90 + 9 = 1089.$$ |
H: Possible series of typos in Apostol's Calculus Vol.II involving gradients and directional derivatives
The question as presented, this is from Calculus Vol. II Section 8.14 #4
A differentiable scalar field $f$ has, at the point $(1,2)$, directional derivatives $+2$ in the direction toward $(2,2)$ and $−2$ in the direction toward $(1,1)$. Determine the gradient vector at $(1,2)$ and compute the directional derivative in the direction toward $(4,6)$.
Right off the bat, I think there must be a typo in the question because the directional derivatives in the direction of $(2,2)$ should be equal to the directional derivatives in the direction of $(1,1)$. So instead, I thought I would look at the answer for $(4,6)$ and try to find out what the typo was in the original problem. In the back of the book we have the answers as:
The gradient vector at $(1,2)$ is $(2,2)$, and the directional derivative in the direction toward $(4,6)$ is $\frac {14} 5$.
These don't seem to be consistent either, with each other or with the original question. Is this a series of typos, or am I misinterpreting something?
AI: By "in the direction toward," I believe what is meant is that the displacement vector between the base point and the second point is the direction in the directional derivatives. |
H: Geometrically interpret the set $\{z \in \Bbb C :$ $|z+a|+|z-a| \leq 2b,$ $b\in\Bbb R^+,$ $|a|
Geometrically interpret and determine the following set of complex numbers:
$$\{z \in \Bbb C : |z+a|+|z-a| \leq 2b , b\in\Bbb R^+,|a|<b\}.$$
I understand it means the sum of distance between $z$ and $a,$ and between $z$ and $-a$ is less than or equal to $2b.$ I drew a circle with centre at origin and with radius $b,$ a lies within the circle, $-a$ is the reflection of a across the origin, and I'm stuck.
AI: Hint: you need to know one fact from traditional geometry, which is the fact that if a point moves so that the sum of its distances from two fixed points is constant, then the locus is an ellipse, and the two fixed points are the foci of the ellipse.
Wikipedia has some details here.
For your example, the "less than or equal to" just gives the interior of the ellipse. |
H: How to solve $1/2 \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0$?
How to solve trigonomtry function involving $\sin x \cos x$ and $\sin 2x$:
$$\frac{1}{2} \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0. $$
AI: Hint:
Using the identity $\sin(2x) = 2 \sin x \cos x$ we have
$$ \sin x \cos x + \sin x + 2\cos x + 2 = 0$$
Factor
$$ (1 + \cos x) \sin x + 2(1 + \cos x) = 0 \\
(1 + \cos x)(2 + \sin x) = 0
$$
So either $1 + \cos x = 0$ or $2 + \sin x = 0.$
Solve for $x$ in each case. |
H: Lipschitz Condition on a Set of Measure 0
The question is as follows:
Let $f:[0,1] \rightarrow R$ be a non-decreasing continuous function, and let $A \subset [0,1]$ be a set of Lebesgue measure $0$.
(a)Suppose that $f$ satisfies the following condition:
(*) For some constant $C \geq 0$, $|f(s)-f(t)| \leq C|s-t|$ for all $s,t \in [0,1].$ Prove that $f(A)$ has measure zero.
This part is ok, I had no problem proving that. But part (b) has been giving me some issues:
(b) Suppose that $f$ does not satisfy condition (*). Does the conclusion in part (a) still hold?
Me and my office mate feel that if the lipschitz condition does not hold, then $f(A)$ will not have measure zero. We have tried to come up with counterexamples, but are having problems defining a continuous function on a set of measure 0.
AI: Consider $A$ – the standard embedding of the Cantor set $2^\omega$ into $[0,1]$ and the standard surjection $g:A\to [0,1]$, $g(\sum_n 3^{-n}a_n)=\sum 2^{-n}a_n/2$. Extend it to a continuous nondecreasing function from $[0,1]$ to $[0,1]$ (in the only possible way, segmentwise constant outside $A$). Note that $A$ has zero measure. |
H: Modern formula for calculating Riemann Zeta Function
Possible Duplicate:
How to evaluate Riemann Zeta function
I have an amateur interest in the Zeta Function. I have read Edward's book on the topic, which is perhaps a little dated. I would like to know any modern methods for computing estimates of
$\zeta(s)$, specifically for the critical strip.
My goal is to convert the method into C++ code, and actually for my purposes the accuracy and convergence are less important than the ease of writing the code.
Thanks in advance.
AI: Euler-Maclaurin Summation
This was one of the first techniques used to approximate the zeta function ans was in fact used by Euler to approximate $\zeta(2)$. However, this method is only used on the remainder after a certain number terms of the zeta series has been computed.
$$\zeta(s)=\sum_{n=1}^N \frac{1}{n^s}+\frac{N^{1-s}}{s-1}+\frac{N^{-s}}{2}+\sum_{r=1}^{q-1}\frac{B_{2r}}{(2r)!}s(s+1) \cdots(s+2r-2)N^{-s-2r+1}+\epsilon_{2q}(s)$$
where
$$|\epsilon_{2q}(s)| < \left|\frac{s(s+1) \cdots(s+2r-2)N^{-\operatorname{Re}[s]-2r+1}}{(2q)!(s+2q-1)}\right|$$
Alternating Series
The alternating zeta series is given by
$$\zeta_a (s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}=(1-2^{1-s})\zeta(s)$$
Then, by accelerating this sum, we have
$$e_k = \sum_{j=k}^n {n \choose j}$$
$$\zeta(s)=\frac{1}{(1-2^{1-s})}\left(\sum_{k=1}^n \frac{(-1)^{k=1}}{k^s}+\frac{1}{2^n }\sum_{k=n+1}^{2n} \frac{(-1)^{k=1}e_k}{k^s}\right)+\epsilon_n(s)$$
where
$$\epsilon_n(s) < \frac{(1+|t/\operatorname{Re}[s]|)\exp(|t|\pi/2)}{8^n|1-2^{1-s}|}$$
Another, different acceleration gives a slightly faster, but more complicated result
$$d_k = n \sum_{j=k}^n \frac{(n + j − 1)!4^j}{(n − j)!(2j)!}$$
$$\zeta(s)=\frac{1}{d_0(1-2^{1-s})}\sum_{k=1}^n \frac{(-1)^{k-1}d_k}{k^s}+\epsilon_n(s)$$
where
$$|\epsilon_n(s)| \le \frac{2}{(3+\sqrt{8})^n |\Gamma(n)(1-2^{1-s})|}$$
This paper gives pseudocode (pg. 41) for the Zeta function using the Euler Maclaurin summation method and a Mathematica implementation (Appendix D, pg. 57). |
H: Proof of Uniqueness Theorem
I have the following question on the theorem below, which I've been working on while I study for quals.
Suppose $\Omega$ is an open and connected set, with $f$ analytic in $\Omega$, and $a \in \Omega$. Then:
(1) If $f(a)=f'(a)=f''(a)=\ldots=0$ then $f$ is identically zero in $\Omega$.
(2) If $z_0$ is an accumulation point of $\lbrace z|f(z)=0\rbrace$, then $f$ is identically zero.
I've had no trouble proving (1), but (2) is giving me some issues. I've tried re-writing $f(z)$ as $g(z)(z-a)^m$, where $m$ is the smallest number such that $f^{(m)}(a)\neq 0$, but when I attempt to work out the details I usually get flipped around.
AI: (2) is often called or used to prove analytic continuation. I will give a proof below:
Suppose $\alpha$ is an accumulation point of the set $\{z : f(z) = 0\}$. Since $f$ is a holomorphic on $\Omega$, let
$f(z) = \sum a_n(z - \alpha)^n$
by the power series expansion at $\alpha$. The claim is that $f$ is zero in a neighborhood of $\alpha$. Suppose not, then there must exists a smallest $m$ such that $a_m \neq 0$. Hence, the power series takes the form
$f(z) = \sum a_m (z - \alpha)^m (1 + h(z - \alpha)) \ \ \ (*)$
for some holomorphic function $h$ such that $h(z - \alpha) = 0$ as $z \rightarrow \alpha$. ($h$ is clearly what you get when you factor out $a_m(z - \alpha)^m$.)
Now since $\alpha$ is a limit point of the set of zeros. There exists a sequence of zeros of $f$, $(y_n)$ such that $y_n \rightarrow \alpha$. But then by $(*)$, it is clear that $f(y_n) \neq 0$ for any $n$. But you choose $y_n$ to be a sequence of zeros of the function $f$. Contradiction.
Now let $U = \text{Int}(\{z : f(z) = 0\})$. $U$ is nonempty open set by the argument above. $U$ is also closed since if $(y_n)$ is a sequence in $U$ with limit point $y$, then $f(y) = \lim_{n \rightarrow \infty} f(y_n) = 0$ by continuity. By the argument above, applied to the point $y$ in place of $\alpha$, $f$ vanishes in a neighborhood of $y$. Hence $y \in U = \text{Int}(\{z : f(z) = 0\})$. So $U$ is closed. By assumption $\Omega$ is connected, so the only clopen subset is $\emptyset$ or $\Omega$. Since $\alpha \in U$, one has that $U = \Omega$. So $f$ vanishes on all of $\Omega$. |
H: Understanding how to state the Karush-Kuhn-Tucker Conditions for a given problem
I'm trying to understand an example given by Nocedal & Wright (1999), pg 329, Example 12.4. According to a definition given earlier in this book:
At a feasible point x, the inequality constraint $i\in I$ is said to be active if $c_i(x)=0$ and inactive if the strict inequality $c_i(x)>0$ is satisfied.
So, in Example 12.4 of the same book, a constrained minimization problem is given as:
$argmin$ $(x_1-\frac{3}{2})^2 + (x_2-\frac{1}{2})^4$
such that
$c_1 \equiv 1-x_1-x_2\ge0$
$c_2 \equiv 1-x_1+x_2\ge0$
$c_3 \equiv1+x_1-x_2\ge0$
$c_4 \equiv1+x_1+x_2\ge0$
The authors provide a picture which clearly show that the solution is $x^*=(1,0)$ and state that constraints $c_1$ & $c_2$ are active at this point.
I have two questions regarding this problem:
First Question
Since complementarity implies that the lagrange multipliers associated with constraints $c_3$ and $c_4$ are zero (i.e. inactive), can I state the Karush-Kuhn-Tucker Conditions , as follows:
$\nabla f(x^*)=\lambda_1\nabla c_1(x^*) + \lambda_2\nabla c_2(x^*)$
$\Rightarrow \left(
\begin{array}{c}
-1\\
-\frac{1}{2}\\
\end{array}
\right) =
\left(
\begin{array}{c}
-\lambda_1-\lambda_2\\
-\lambda_1+\lambda_2\\
\end{array}
\right)
$
Second Question
If the optimal solution were not provided, would I consider all constraints as active?
My concern is how to state the KKT conditions, in general. That is, do I need to discern the set of active constraints ahead of time to setup the KKT conditions? If so, how would I without knowing the optimal solution apriori? Obviously, because of complementarity I know that the lagrange multipliers of inactive constraints will inevitably become zero, but is there a way to know which will be inactive ahead of time? If the inactive constraints were known, would it simplify the process of obtaining the active lagrange multipliers in general?
AI: 1) Yes, since $c_3$ and $c_4$ are inactive at this particular $x^*$ the KKT conditions will require $\lambda_3 = \lambda_4 = 0$.
2) If you don't know $x^*$, you have to consider all possibilities for which constraints are active. You would write the KKT conditions as
$$ \eqalign{\nabla f(x^*) &= \lambda_1 \nabla c_1(x^*) + \lambda_2 \nabla c_2(x^*) + \lambda_3 \nabla c_3(x^*) + \lambda_4 \nabla c_4(x^*) \cr
c_i(x^*) &\ge 0,\ i=1\ldots 4\cr
\lambda_i &\ge 0,\ i=1\ldots 4\cr
\lambda_i c_i(x^*) &= 0,\ i=1\ldots 4\cr}$$ |
H: Proof of uniform continuity of $\frac{1}{x}$
Show that the function $f(x) = \frac{1}{x}$ is not uniformly continuous on the interval $(0,\infty)$ but is uniformly continuous on any interval of the form $(\mu, \infty)$ if $\mu > 0$.
My Work
Referring to the definition of uniform continuity, I have that $f$ is unif. cts. if for each $\epsilon > 0$ there is a $\delta > 0$ so that for all $x, c$ in the domain of $f$ $|x - c| \le \delta \ \Rightarrow \ |f(x) - f(c) | \le \epsilon$.
From this definition, it is clear that if $f$ is uniformly continuous, it will be uniformly continuous on its domain, $(-\infty, \infty) \backslash \{0\}$. So for $\mu > 0$, $(\mu, \infty) \subset \mathrm{Dom}\,(f)$. Additionally, $f$ cannot be unif. cts on $(0, \infty)$ because $0 \notin \mathrm{Dom}\, (f)$. (Sorry about the longwindedness)
Now to find the $\delta$:
\begin{align*}
|f(x) - f(c)| = \left|\frac{1}{x} - \frac{1}{c}\right| &= \left|\frac{x - c}{cx}\right| \\
\text{since }x\text{ is within }\delta\text{ of }c \ \Rightarrow \ &\le \frac{\delta}{|cx|}\\
x, c>0 \ \Rightarrow \ &= \frac{\delta}{cx}
\end{align*}
This is where I am stuck. Should I use that $x \le c + \delta$, or should I break this up into two cases, one where $cx < 1$ and one where $cx \ge 1$?
Edit (due to Brian M. Scott)
It was pointed out that $0 \notin (0, \infty)$ so my above argument is senseless.
AI: Since $0\notin(0,\infty)$, $0$ is completely irrelevant to the question of whether $f$ is uniformly continuous on $(0,\infty)$. To show that $f$ is not uniformly continuous on $(0,\infty)$, you should show that there is some $\epsilon>0$ such that no matter what $\delta>0$ you pick, you can find points $x,y\in(0,\infty)$ such that $|x-y|\le\delta$, but $|f(x)-f(y)|>\epsilon$. HINT: You can take $\epsilon=1/2$. Now consider values of $x$ of the form $\frac1n$ for $n\in\Bbb Z^+$.
To prove that $f$ is uniformly continuous on $(\mu,\infty)$ for $\mu>0$, you need what is really the key insight for both questions: for $x>0$, the graph of $y=\frac1x$ gets steeper and steeper as $x$ gets smaller and smaller. Given $x,y\in(\mu,\infty)$ with $|x-y|\le\delta$, where $\delta$ is some as yet unspecified positive real number, can you find an upper bound on $|f(x)-f(y)|$? How does it compare with $\frac{\delta}{\mu^2}$? (Consider $f'(x)$.) |
H: Getting square root of negative in completing the square problem
I try to solve the equation $f(x) = 7x - 11 - 2x^2 = 0$ for $x$, but run into troubles. I've gone through it over and over again as well as similar problems, but can't find what I'm doing wrong.
$$f(x) = 7x - 11 - 2x^2 = 0$$
$$\iff x^2 - \frac{7}{2}x + \frac{11}{2} = 0 $$
$$\iff \left(x + \frac{7}{4}\right)^2 = \left(\frac{7}{4}\right)^2 - \frac{11}{2}$$
$$\iff x + \frac{7}{4} = \pm \sqrt{\left(\frac{7}{4}\right)^2 - \frac{11}{2}}$$
$$\iff x = -\frac{7}{4} \pm \sqrt{\frac{49}{16} - \frac{88}{16}}$$
$$\iff x = -\frac{7}{4} \pm \sqrt{\frac{-39}{16}}$$
I should be able to continue but I'm stuck (seeing as it's a negative number). What am I doing wrong?
AI: You did everything fine but your quadratic equation has no real solutions, which you could have found out way more easily had you first calculated the equation's discriminant:
$$\Delta:=b^2-4ac=7^2-4(-2)(-11)=49-88=-39<0$$ |
H: Question about Cauchy sequence in $C(K)$
In my lecture, I need prove the existence of the anyone Cauchy sequence $(f_n)_{n\in \mathbb{N}}$ belonging $C(K)$ with the norm ${\Vert \cdot \Vert}_{\infty}=\sup_{x \in K}|f(x)|$ where $K$ is a compact space. How I can do it? I'm trying that below
Fixing $\varepsilon>0$, there exist any $x_0$ such that
${\Vert f_n - f_m \Vert}_{\infty} < |(f_n - f_m)x_0| + \varepsilon/2$. Then for $n,m$ such that
$|(f_n - f_m)x_0|<\varepsilon/2$. Then $(f_n)_{n\in \mathbb{N}}$ is a Cauchy sequence.
Is this correct?
AI: To have an answer, I'll expand on the comments.
You claim that $C(K)$ is complete with respect to $\|\cdot\|_\infty$. What this means is that every Cauchy sequence $f_k$ has its limit in $C(K)$.
So let $f_k$ be a Cauchy sequence in $C(K)$, that is, for $\varepsilon > 0$ there is $N$ such that for $n,m > N$ you have $\|f_n - f_m \|_\infty < \varepsilon$. You want to show that there is an $f$ in $C(K)$ such that $\|f-f_k\|_\infty \to 0$. Your guess for $f$ is that $f$ is the pointwise limit of $f_k$. For this observe first that the pointwise limit exists: for every $x \in K$ fixed, $f_k (x)$ is a Cauchy sequence in $\mathbb R$, $\mathbb R $ is complete hence its limit exists.
Now that we have a candidate for the limit let's denote it by $f(x) = \lim_{n \to \infty} f_n (x)$.
To finish the proof you need to show that $f$ is in $C(K)$ that is, that $f$ is continuous, and also that $f_k$ converges to $f$ in norm.
If you show that $f_k$ converge to $f$ in norm you will get continuity for free by the uniform limit theorem. So let's show that $f_k \to f$ in $\|\cdot\|_\infty$:
Let $N$ be such that for $k \geq N$ you have $\|f_k - f_N \|_\infty < \varepsilon / 2$ (by Cauchyness of $f_k$). Then
$$ |f(x) - f_k(x)| \leq |f(x) - f_N (x)| + |f_N (x) - f_k(x)|$$
The second term is $< \varepsilon / 2$. For the first term observe that
$$ |f(x) - f_N (x)| = |\lim_{n \to \infty} f_n (x) - f_N(x)|$$
So $|f(x) - f_N (x)| \leq \varepsilon /2$, so that
$$ |f(x) - f_k(x)| \leq |f(x) - f_N (x)| + |f_N (x) - f_k(x)| < \varepsilon$$ |
H: Question about a Proof of Triangularizing a Matrix
I was questioned from my friend about a non-proof in a linear algebra textbook.
The author mentions that the following proof of the theorem, that every square matrix in $\Bbb{C}$ is triangularizable, has a gap and thus is invalid:
First, we have two preliminary lemmas.
Lemma 1. Let $I_m$ be the identity matrix of size $m$, $A\in \mathfrak{M}_{m\times m}(\Bbb{C})$, and $U, B \in \mathfrak{M}_{n\times n}(\Bbb{C})$ such that $U$ is invertible. Then
$$ \bigg(\begin{matrix} I_m & 0 \\ 0 & U^{-1} \end{matrix}\bigg) \bigg( \begin{matrix} A & * \\ 0 & B \end{matrix} \bigg) \bigg( \begin{matrix} I_m & 0 \\ 0 & U \end{matrix} \bigg) = \bigg( \begin{matrix} A & * \\ 0 & U^{-1}BU \end{matrix} \bigg). $$
Lemma 2. Let $A \in \mathfrak{M}_{n\times n}(\Bbb{C})$. Then there exists an eigenvalue $\lambda \in \Bbb{C}$ and an invertible matrix $U \in \mathfrak{M}_{n\times n}(\Bbb{C})$ such that
$$ U^{-1}AU = \bigg( \begin{matrix} \lambda & * \\ 0 & * \end{matrix} \bigg).$$
Proof. Let $v_1 \in \Bbb{C}^n$ be a nonzero eigenvector corresponding to $\lambda$. Let $\{v_1, \cdots, v_n\}$ be a basis of $\Bbb{C}^n$ containing $v_1$. Let $U$ be a matrix whose $k$-th column is $v_k$. Then
$$[AU]_{*1} = A[U]_{*1} = Av_1 = \lambda v_1 = \lambda [U]_{*1}$$
and hence
$$ [U^{-1}AU]_{i1} = [U^{-1}]_{ij}[AU]_{j1} = \lambda [U^{-1}]_{ij}[U]_{j1} = \lambda \delta_{i1},$$
which proves Lemma 2.
Now we claim that for each $1 \leq k \leq n$ there exist
an upper-triangular matrix $T_k$ of size $k$,
a square matrix $A_{n-k}$ of size $n-k$, and
an invertible matrix $U_k \in \mathfrak{M}_{n\times n}(\Bbb{C})$
such that
$$ U_k^{-1} A U_k = \bigg( \begin{matrix} T_k & * \\ 0 & A_{n-k} \end{matrix} \bigg).$$
In particular, for $k = n$ this implies that $A$ is triangularizable.
We prove this claim by induction on $k$. For $k = 1$, this is a trivial consequence of Lemma 2. Now assume this holds for $k < n$. Then applying Lemma 2 to $A_{n-k}$, we obtain some $\tilde{\lambda} \in \Bbb{C}$ and an invertible matrix $\tilde{U}$ of size $n-k$ such that
$$ \tilde{U}^{-1}A_{n-k}\tilde{U} = \bigg( \begin{matrix} \tilde{\lambda} & * \\ 0 & A_{n-k-1} \end{matrix} \bigg).$$
Now let $U_{k+1}$ as
$$U_{k+1} = U_k \bigg( \begin{matrix} I_k & 0 \\ 0 & \tilde{U} \end{matrix} \bigg). $$
Then
$$\begin{align*} U_{k+1}^{-1} A U_{k+1}
&= \bigg( \begin{matrix} I_k & 0 \\ 0 & \tilde{U}^{-1} \end{matrix} \bigg) U_k^{-1} A U_k \bigg( \begin{matrix} I_k & 0 \\ 0 & \tilde{U} \end{matrix} \bigg)
= \bigg( \begin{matrix} I_k & 0 \\ 0 & \tilde{U}^{-1} \end{matrix} \bigg) \bigg( \begin{matrix} T_k & * \\ 0 & A_{n-k} \end{matrix} \bigg) \bigg( \begin{matrix} I_k & 0 \\ 0 & \tilde{U} \end{matrix} \bigg) \\
&= \begin{pmatrix} T_k & * \\ 0 & \tilde{U}^{-1}A_{n-k}\tilde{U} \end{pmatrix}
= \begin{pmatrix} T_k & * & * \\ 0 & \tilde{\lambda} & * \\ 0 & 0 & A_{n-k-1} \end{pmatrix}
=: \begin{pmatrix} T_{k+1} & * \\ 0 & A_{n-k-1} \end{pmatrix}.
\end{align*}$$
Therefore the claim follows by induction and the proof is complete. Q.E.D.
Actually, I was convinced by this argument and was unable to find any gap in this proof.
The book says that this non-proof overlooked the importance of $L$-invariant subspaces. So I tried this non-proof with some examples of $A$ where the multiplicity of an eigenvalue $\lambda$ is greater than the dimension of the kernel of $A - \lambda I$ in hopes that these trials would reveal the fault in the argument above. But it ended up only finding that the algorithm above works in those cases. I want to know what I am missing.
AI: I can't find a gap in the proof either, but let me just comment that the proof works far too hard. As Emil Artin once said,
It is my experience that proofs involving matrices can be shortened by 50% if one throws the matrices out.
I think this proof gets shortened even more than that. Let $A : V \to V$ be a linear operator on an $n$-dimensional complex vector space $V$. We will define inductively a basis $v_1, ... v_n$ of $V$ such that $A$ preserves $V_k = \text{span}(v_1, ... v_k)$. This is equivalent to $A$ being upper-triangular with respect to this basis. Define $v_1$ to be an eigenvector of $A$ (say with eigenvalue $\lambda_1$). If $v_1, ... v_k$ have been defined, then $A$ preserves $V_k$, hence acts on $V/V_k$, hence has some eigenvector there (this is the crucial step!). Let $v_{k+1}$ be any element of $V$ mapping to such an eigenvector (say with eigenvalue $\lambda_{k+1}$). Then $A v_{k+1} \equiv \lambda_{k+1} v_{k+1} \bmod V_k$, hence $A$ preserves $V_{k+1}$. The conclusion follows.
(The subspaces $V_k$ form a complete flag, and "preserving a complete flag" is the basis-invariant way to say "upper-triangular.") |
H: Inequality for singular values
I stumbled on the following inequality for singular values (stated without proof), and would like to understand it better:
Let $A,B$ be two $n\times n$ real matrices. Denoting by $\mu_i(C)$ the $i$-th singular value of a matrix $C$ and $by$ $\|\cdot\|$ the operator norm we have for $i=1,\dots,n$
$\mu_i(AB) \ \leq \|A\| \ \mu_i(B)$
and
$\mu_i(AB) \ \leq \|B\| \ \mu_i(A)$
Why is this true? Standard references? Is this inequality a specific property of singular values or does it work also for eigenvalues?
AI: The singular values of $A$ are the square roots of the eigenvalues of $A^T A$ (or equivalently of $A A^T$). Thus the singular values of $AB$ are the square roots of the eigenvalues of $B^T A^T A B$ or $A B B^T A^T$. Now for any vector $v$, $v^T A^T A v
= \|A v\|^2 \le \|A\|^2 v^T v$. Apply that to $v = B w$ and you get
$w^T B^T A^T A B w \le \|A\|^2 w^T B^T B w$. By the Min-Max Theorem http://en.wikipedia.org/wiki/Min-max_theorem it follows that $\mu_i(AB) \le \|A\| \mu_i(B)$. Similarly, using $A B B^T A^T$ instead of $B^T A^T A B$ you get $\mu_i(AB) \le \|B\| \mu_i(A)$. |
H: Does this Dirichlet series converge to zero?
Consider the periodic Dirichlet series that has this iterative definition:
$$\text{a1}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{6}}+...$$
$$\text{a2}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1}{\sqrt{6}}+...$$
$$\text{a3}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1+a2}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1+a2}{\sqrt{6}}+...$$
$$\text{a4}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-\frac{2+a1+a2+a3}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\frac{2+a1+a2+a3}{\sqrt{6}}+...$$
$$...$$
continuing this iteration, will it converge to zero?
AI: If I understand correctly, $a_1 = \sum_{j=0}^\infty (-1)^j \left((3j+1)^{-1/2} - (3j+2)^{-1/2} - 2 (3j+3)^{-1/2}\right)$, and
$a_{j+1} = a_j + a_j b = a_j (1+b)$ where $b = \sum_{j=1}^\infty (-1)^j (3j)^{-1/2}$.
Thus $a_n = a_1 (1+b)^{n-1}$. Now $1+b \approx .6507616$, so $|1+b| < 1$ so the sequence does converge to $0$. |
H: clarification on binary and decimal representations of integers
This is from the book: Elementary number theory, David M. Burton, page 69.
Given an integer $b>1$, any positive integer $N$ can be written
uniquely in terms of powers of $b$ as
$$N=a_{m}b^{m}+a_{m-1}b^{m-1}+\cdots +a_{2}b^{2}+a_{1}b+a_{0},$$ where
the coefficients $a_{k}$ can take on the $b$ different values
$0,1,2,\dots,b-1$.
For the Division Algorithm yields integers $q_{1}$ e $a_{0}$
satisfying $$N=q_{1}b+a_{0},\,0\leq a_{0}< b.$$ If $q_{1}\geq b,$ we
can divide once more, obtaining $$q_{1}=q_{2}b+a_{1},\, 0\leq a_{1}< b.$$ Now substitute >for $q_{1}$ in the earlier equation to get
$$N=(q_{2}b+a_{1})b+a_{0}=q_{2}b^{2}+a_{1}b+a_{0}$$ As long as
$q_{2}\geq b,$ we can continue in the same fashion. Going one more
step: $q_{2}=q_{3}b+a_{2}$, where $0\leq a_{2}<b$; hence
$$N=q_{3}b^{3}+a_{2}b^{2}+a_{1}b+a_{0}$$ Because $N>q_{1}>q_{2}>\cdots
\geq 0$ is a strictly decreasing sequence of integers, this process
must eventually terminate, say, at the $(m-1)$th stage, where
$$q_{m-1}=q_{m}b+a_{m-1},\,0\leq a_{m-1}<b$$ and $0\leq q_{m}<b$.
Setting $a_{m}=q_{m}$, we reach the representation
$$N=a_{m}b^{m}+a_{m-1}b^{m-1}+\cdots +a_{2}b^{2}+a_{1}b+a_{0}$$ which
was our aim.
After that the author proves the uniqueness.
My question is, once the integers $a_{i}$ and $q_{i}$ are unique, is this not enough to ensure the uniqueness of the representation?
AI: The uniqueness of the integers $a_i$ and $q_i$ means that there is a unique base-$b$ representation of $N$ that is obtainable by the described algorithm — that is, the algorithm doesn't involve any arbitrary choices. It doesn't rule out the possibility that there could be some other representation that was not so obtainable.
On the other hand, one way of proving uniqueness would be to argue that any representation of $N$ must be obtainable by applying the algorithm. |
H: Standard terminology for the relation between $A$ and $B$ if $B= Q^t A P$?
Let $A,B$ be two rectangular $m\times n$ matrices related by
$$B= Q^t A P$$
with $P$ an $n\times n$ and $Q$ an $m\times m $ matrix.
Is there a standard terminolgy for this relation? If instead of the transposed $Q^t$ one takes the inverse $Q^{-1}$ above, they are just called "equivalent" according to
http://en.wikipedia.org/wiki/Matrix_equivalence
I know that when $m=n$ (Edit 1: and P=Q) one uses "congruent" (transposed case) and "similar" (inverse case).
Edit 2: $P$ and $Q$ are assumed both to be invertible (sorry for forgetting to write it). As Marc van Leeuwen pointed out, there is no point in distinguishing among the cases $Q$, $Q^t$ and $Q^{-1}$ since $Q$ is arbitrary ( and invertible). It only makes sense when interpreting $Q$ as coordinate change matrix and $A$ as a linear operator ( -> $Q^{-1}$) or bilinear form (-> $Q^t$) (see explanations of Paul Garrett).
Thanks to everybody who contributed to clarify my confused question.
AI: If both $P$ and $Q$ are invertible, then this is just that $A$ and $B$ have the same rank. That notion of "equivalence" is dubious, as are several archaic bits of terminology about matrices, dating back to times when coordinate-independent understanding of linear algebra did not exist.
There are reasons for considering relations $B=Q^\top A Q$ and $B=Q^{-1} A Q$. The case with inverse is change-of-coordinates/basis for a linear map $A$, where the $Q$ is the change-of-coordinates/basis map. The case with transpose is change-of-coordinates/basis for a _quadratic_form_ $A$, with $Q$ again the change-of-coordinates/basis map.
The cases $B=Q^\top AP$ and $B=Q^{-1} A P$ are change of coordinates of, respectively, a bilinear map $A:V\times W\rightarrow \hbox{scalars}$, and a linear map $T:V\rightarrow W$ given by $A$, where in both cases one changes coordinates in possibly-unrelated fashions on both vectorspaces.
Edit: that is, (letting the other shoe drop), these are change-of-coordinates/basis expressions for four specific things: linear maps of a space to itself, quadratic forms on a space, bilinear maps on two spaces, and maps from one space to another. That is, doing such things to matrices is not mere conformity to some inexplicable set of rules, but, in fact, is doing an obvious thing. Understanding which is appropriate is much aided by understanding what the point is. |
H: Existence of upper limit and lower limit
Previous Question; Does upper limit and lower limit exist for any sequence in $\mathbb{R}$?
For every sequence $\{s_n\}$ in $\mathbb{R}$, $\{x\in \overline{\mathbb{R}}| s_{n_k} →x\}$ is nonempty.
However, how do i prove the existence of $\sup \{x\in \overline{\mathbb{R}}| s_{n_k} →x\}$ and $\inf \{x\in \overline{\mathbb{R}}| s_{n_k} →x\}$?
Example;
$a_{2n}=-n^2$ and $a_{2n+1}=(1/2)^n$
$b_{3n}=-n^2$ and $b_{3n+1}=(1/2)^n$ and $b_{3n+2}=n^2$.
What kind of mathematical property do they share? I don't know where to start my argument.
AI: Edit: I see I misread your question. My apologies.
Let us take any $E\subseteq\overline{\Bbb R}$, and show that there exists $y\in\overline{\Bbb R}$ such that $y=\sup E$. (The arguments will be similar to show that an infimum exists.)
If $E$ is empty, we say that $\sup E=-\infty$. Suppose not. If $E$ has an upper bound in $\Bbb R$, then by completeness of $\Bbb R$, $E$ has a supremum in $\Bbb R$. Otherwise, we say by definition that $\sup E=+\infty$. |
H: Random Walk Expected Number of Visits
I'm having a bit of trouble with a question involving a random walk with five vertices. The graph is shown below.
The problem I can't figure out reads:
Suppose a walker starts in vertex C. What is the expected number of visits to B before the walker reaches A?
I've done a bit of work on this problem already, but I can't seem to put it all together. First, I made my transition matrix
$$P = \begin{bmatrix} 0&\frac{1}{3}&\frac{1}{3}&\frac{1}{3}&0 \\\\ \frac{1}{3}&0&\frac{1}{3}&0&\frac{1}{3} \\\\ \frac{1}{2}&\frac{1}{2}&0&0&0 \\\\ \frac{1}{2}&0&0&0&\frac{1}{2} \\\\ 0&\frac{1}{2}&0&\frac{1}{2}&0 \end{bmatrix}$$
where the columns and rows go in the order A, B, C, D, then E. My textbook describes a method to find the expected number of steps from i to j that involves designating the destination state as an absorbing state and redefining the matrix as such, in this case resulting in this matrix:
$$P = \begin{bmatrix} 1&0&0&0&0 \\\\ \frac{1}{3}&0&\frac{1}{3}&0&\frac{1}{3} \\\\ \frac{1}{2}&\frac{1}{2}&0&0&0 \\\\ \frac{1}{2}&0&0&0&\frac{1}{2} \\\\ 0&\frac{1}{2}&0&\frac{1}{2}&0 \end{bmatrix}$$
Next, it says to extract a matrix Q that includes the rows and columns of only the transient states. Here, Q would be
$$Q = \begin{bmatrix} 0&\frac{1}{3}&0&\frac{1}{3} \\\\ \frac{1}{2}&0&0&0 \\\\ 0&0&0&\frac{1}{2} \\\\ \frac{1}{2}&0&\frac{1}{2}&0 \end{bmatrix}$$
where the columns and rows go in the order B, C, D, then E. We're trying to attain a matrix M, where
$$M = (I - Q)^{-1}$$
I calculated M to be
$$M = \begin{bmatrix} \frac{18}{11}&\frac{-6}{11}&\frac{4}{11}&\frac{-8}{11} \\\\ \frac{-9}{11}&\frac{14}{11}&\frac{-2}{11}&\frac{4}{11} \\\\ \frac{6}{11}&\frac{-2}{11}&\frac{16}{11}&\frac{-10}{11} \\\\ \frac{-12}{11}&\frac{4}{11}&\frac{-10}{11}&\frac{20}{11} \end{bmatrix}$$
As I understand it, I can ignore the negatives in this matrix and just make those entries positive. Now here is where I get stumped. I can easily find the expected number of steps to get from C to A, but the inclusion of vertex B in the question really throws me off. I calculated the invariant probability vector to be
$$\pi = (\frac{1}{4}, \frac{1}{4}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6})$$
but I'm not sure that has anything to do with this problem. Any chance you all could help me out? This isn't homework; the semester hasn't begun yet and I'm just trying to get a bit of a head start. Thanks!
AI: First of all, the minus signs don't actually appear in $M$. You must have miscalculated somewhere.
The answer to your question is the $(C,B)$th entry of the matrix $M$. The expected number of visits to $B$ before hitting state $A$, starting at $C$, is 9/11. |
H: What is subgroup $\langle g^d \rangle$?
According to the introductory abstract algebra, it says that $g^n =e$ where $g$ is an element of some finite group. Then, it talks about the subgroup $\langle g^d \rangle$. What is it exactly, and what would be the elements of this group? (where $d$ is some integer.)
AI: The notation $\langle g^d \rangle$ denotes the cyclic group generated by the element $g^d$. More explicitly, we begin with the element $g^d$. Then we consider $g^d \cdot g^d = g^{2d}$. We continue generating more elements of the subgroup and have the set $\{g^d, g^{2d}, g^{3d}, \cdots\}$ with the group operation coming from the finite group.
Of course it requires proof that this set generates a group. I will leave this as a fun exercise for you, but please ask if you need more help. |
H: distinct generators of a group $G$
By saying distinct generator of a group $G$, is this saying that elements produced by the generator differ from elements produced by other generators?
The distinct generators of $G$ are the elements $g^r$ where $1 ≤ r < N $ and $gcd(r,N) = 1$. Thus, there are
$ϕ(N)$ of them, where $ϕ$ is Euler’s phi function.
AI: I'm willing to bet that in this question, $G$ is a cyclic group of order $N$. To that end, I'm answering the question I think you meant to ask. If I'm wrong, please let me know. In the case where we have a cyclic group of order $N$, there are $\varphi(N)$ different generators, each of the form of the question.
To answer your question: No. When we say an element (say $g$) generates a group, it means that every element of the group is a power of that element (of the form $g^n$ for some $n$). If we were to say that the two elements $g,h$ generate $G$, often written $G = \langle g,h \rangle$, we mean that any element of $G$ can be written as $g^n h^m$ for some $m,n$.
So any two generators generate the same set of elements. By distinct generators, we mean that the two generators are different from eachother. For example, if we are working with the cyclic group of order $2$, there are two elements. We might call them $e$ (the identity) and $g$, so that $g^2 = e$. But $g^3$ also generates this group. But $g^3$ is the same as $g$, so they are not distinct even though the 'words' describing them are not the same. |
H: $m\mid rq,sq\Rightarrow m\mid \gcd(r,s)q\,$ and its view via additive order
Let $m,q,r,s$ be integers.
Suppose GCD($r$,$s$) = 1.
Suppose $m|qr$ and $m|qs$.
Show that $m$ divides $q$.
This was a result that was assumed in class. I couldn't see the reasoning behind it though.. please provide some guidance/hints.
What I've tried so far:
I note that since $(r,s)=1$, then either $m$ does not divide $r$, or $m$ does not divide $s$.
I suppose that $m$ does not divide $r$.
Then I note that $m|qr$ but m does not divide $r$.
I try to come up with a contradiction by assuming that $m$ does not divide $q$, but since $m$ isn't prime, this doesn't get me far.
Am I going down the wrong path? Is there something I'm failing to consider?
AI: Hint $\ $ By basic gcd laws $\rm\ m\:|\:qr,qs\:\Rightarrow\:m\:|\:(qr,qs)\, =\, q(r,s)\, =\, q\:$ by $\rm\:(r,s) = 1.\ \ $ QED
Or, instead of gcd laws one may use the Bezout identity: $\rm\:jr + ks = 1\:$ for some $\rm\:j,k\in\Bbb Z,\:$ hence
$$\rm\ m\:|\:qr,qs\:\Rightarrow\:m\:|\:j(qr)\!+\!k(qs) = q(jr\!+\!ks) = q\ \ \ {\bf QED}$$
Note how the proof by gcd laws eliminates the variables $\rm\,j,k,\:$ which only serve to obfuscate.
Remark $ $ It can also be interpreted in terms of orders of elements. Namely, the hypothesis viewed modularly says $\rm\! \bmod m\!:\ r\cdot q\equiv 0\equiv s\cdot q,\:$ which implies that the additive order of $\rm\:q\:$ divides the coprime integers $\rm\:r,s\:$ so it must be $1,\,$ i.e. $\rm\:1\cdot q\equiv 0,\:$ i.e. $\rm\:m\:|\:q.\:$ This viewpoint is discussed further in some of my posts on order ideals. |
H: Is the functional $F(u) = \int_{\Omega} \langle (A_1(x)\chi_{\{u>0\}}+A_2(x)\chi_{\{u\le0\}}) \nabla u, \nabla u \rangle$ convex?
Is the functional
\begin{equation}
F(u) = \int_{\Omega} \langle (A_1(x)\chi_{\{u>0\}}+A_2(x)\chi_{\{u\le0\}}) \nabla u, \nabla u \rangle \, dx
\end{equation}
where $A_i,i=1,2$ is a matrix satisfying
\begin{equation}
\lambda |\xi|^2 \le \langle A_i(x) \xi,\xi \rangle \le \Lambda |\xi|^2, i=1,2.
\end{equation}
with $\lambda>0$ convex? You can assume $\Omega$ convenient such that the expression above make sense. For example, $C^1(\Omega), H^1_0(\Omega)$ or other space such that the functional above be convex. I will appreciate any hint. Thank you.
AI: No. Here is a one-dimensional counterexample, but you can adopt the idea to higher dimensions if you want. Let $\Omega=(0,20)$, $A_1=10$ and $A_2=1$. Define functions $u$ and $v$ by $$u'=\chi_{[0,1]}-\chi_{[19,20]}\quad \text{ and }\quad v'=\sum_{k=1}^9 (-\chi_{[2k-1,2k]}+\chi_{[2k,2k+1]})$$
Both vanish on the boundary of $\Omega$. Since $u\ge 0$ and $v\le 0$, we have $F(u)=20$ and $F(v)=18$. The average $w=(u+v)/2$ is nonnegative because $v\ge -1$ everywhere and $u=1$ on the support of $v$. Since $|w'|\equiv 1/2$, it follows that $F(u)=\int_0^{20}10\cdot\frac14=50$, which is greater than either $F(u)$ or $F(v)$. |
H: A question about a stochastic process being Markov
Let $(X_{s},\mathcal{F}_{s})$ be a stochastic process adapted to a given filtration. I was told that, in order to prove that $X$ is Markov, it suffice to prove that for any nonnegative, Borel-measurable function $f$, $$E[f(X_{t}) \vert \mathcal{F}_{s}]=g(X_{s}) \quad a.s.$$ for $\textbf{some}$ Borel function $g$. I get confused here about the arbitrariness of the function g, shouldn't we expect that $g=P_{s,t} \circ f$, where $P_{s,t}$ is a transition function?
So I'm wondering if the above characterization for Markov processes is correct. I know this may be a stupid question. Any help or comment is greatly appreciated.
AI: Suppose that for $t\geq s$, we have some function $g$ so that
$$E[f(X_{t}) \,\vert\, \mathcal{F}_{s}]=g(X_{s}) \quad \mbox{a.s.}\tag1$$
Conditioning on $X_s$ in (1) gives
$$E[f(X_{t}) \,\vert\, X_s]=g(X_{s}) \quad \mbox{a.s.}\tag2$$ From this we deduce
$$E[f(X_{t}) \,\vert\, \mathcal{F}_{s}]=E[f(X_{t}) \,\vert\, X_s] \quad \mbox{a.s.}\tag3
$$
Equation (3) is one of the equivalent ways of expressing the Markov property. Notice that the arbitrary function $g$ has dropped out of the picture. |
H: Spivak's Calculus (Chapter 2, Exercise 17)
I am having trouble completing exercise 17 of chapter 2 of Spivak's Calculus. In this exercise, the reader is asked to prove that for all natural numbers $n$ and $p$, there exist real numbers $\alpha_1,\ldots,\alpha_p$ such that:
$$ \sum_{k = 1}^n k^p ~=~ \frac{n^{p + 1}}{p + 1} + \sum_{i = 1}^p \alpha_in^i$$
The attempted proof of the identity that I devised uses the binomial theorem with $p$ and $n$ fixed, but does not use complete induction. The proof of the identity provided by Spivak uses the binomial theorem with $n$ fixed, but uses complete induction on $p$ to complete the proof. What follows is my attempted proof.
In this proof, I apply $(x + 1)^{y + 1}$ and $(n + 1)^{p + 1}$ to the binomial theorem in order to derive the first and second terms of the sum on the left-hand side of the identity to be proved. First, note that applying $(x + 1)^{y + 1}$ to the binomial theorem, where $x$ is a real number and $y$ a natural number, demonstrates that the predicate $\phi(x,y)$ is true for all real numbers $x$ and natural numbers $y$:
$$ \phi(x,y) ~\equiv~ (x + 1)^{y + 1} - x^{y + 1} ~=~ (y + 1)x^y + \sum_{i = 0}^{y - 1} {y + 1 \choose i} x^i$$
Now, suppose that $n$ and $p$ are natural numbers. Then the sum of the left-hand sides of the identities $\phi(k,p)$ and the sum of the right-hand sides of the identities $\phi(k,p)$, for $k = 1,\ldots,n$, are equal:
$$(n + 1)^{p + 1} - 1 ~=~ \sum_{k = 1}^n (p + 1)k^p + \sum_{k = 1}^n \left ( \sum_{i = 0}^{p - 1} {p + 1 \choose i} k^i \right )$$
Rewriting $(n + 1)^{p + 1}$ using the identity $\phi(n,p)$ and dividing both sides of the identity by $p + 1$ gives
$$ \sum_{i = 1}^n k^p ~=~ \frac{n^{p + 1}}{p + 1} + \sum_{i = 0}^p {p + 1 \choose i} \left ( \frac{1}{p + 1}\right ) n^i - \left [\frac{1}{p + 1} + \sum_{k = 1}^n \left ( \sum_{i = 0}^{p - 1} {p + 1 \choose i} \frac{k^i}{p + 1} \right ) \right ] $$
Assuming that I didn't make any algebraic errors, I have shown that there are $\alpha_1,\ldots,\alpha_p$ such that the sum of the $p$th powers of the first $n$ natural numbers can be written in the required way. On the other hand, Spivak derives an identity containing the sums of the $r$th powers of the first $n$ natural numbers, for $r \leq p$, then applies an induction hypothesis to obtain the case for $p + 1$. I'm not sure if my proof is wrong, and whether I am missing something that would require a proof by induction. Is there an error?
For reference: I wrote functions that compute the sum of the $p$h power of the first $n$ numbers, and a version of the last identity derived before expanding $(n + 1)^{p + 1}$ using $\phi(n,p)$ in Haskell. Both computed the same numbers for all values of $n \leq 100$ and $p \leq 10$, so for a few numbers there seems to be no problem. Of course, there are more numbers greater than the number of cases that I tested.
AI: He tells you to use induction. In my copy, he just says:
Use the methods of problem $6$ to prove that $$\sum_{k=0}^n k^p$$ may be written in the form
$$\frac{n^{p+1}}{p+1}+An^{p}+Bn^{p-1}+Cn^p+\dots$$
The really important thing here is
$$\frac{n^{p+1}}{p+1}$$
and later on you'll find out why.
Basically, he hints on using the "recursive" trick to obtaining
$$S_{p+1}$$ from $S_p,\dots,S_1$
where $$S_p=\sum_{k=0}^n k^p$$
Say we want $$S_2=\sum_{k=0}^n k^2$$
Then we note that $$(k+1)^3=3k^2+3k+1$$
So that
$$(k+1)^3-k^3=3k^2+3k+1$$
Now we sum $k=1,\dots,n$.
$$\sum_{k=1}^n(k+1)^3-k^3=3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+\sum_{k=1}^n1$$
$$(n+1)^3-1^3=3\sum_{k=1}^nk^2+3\frac{n(n+1)}{2}+n$$
$$\frac{{{{(n + 1)}^3}}}{3} - \frac{{n + 1}}{2} - \frac{{n(n + 1)}}{2} = \sum\limits_{k = 1}^n {{k^2}} $$
(Expansion aside).
So the idea is that you assume the result true for $p=1,\dots,l$ and prove it for $l+1$. Note you shouldn't be too explicit about the coefficients, and the induction is to be done on $p$, not on $n$. |
H: Help with the integral for the variance of the sample median of Laplace r.v.
When we draw $n$ samples of Laplace-distributed random variable such that $n=2k+1$ and the location parameter is zero, the median $x$ (or the $k$-th order statistic) has the following p.d.f.:
$$f_m(x)=\frac{n!}{(k!)^2}2^{-n}\frac{1}{b}e^{-(k+1)|x|/b}(2-e^{-|x|/b})^k$$
where the p.d.f. of the underlying Laplace distribution is given as $f(y)=\frac{1}{2b}e^{-|y|/b}$.
The formula for p.d.f. of the median stems from the usual method of characterizing the distributions of order statistics and is found as equation (2.5.10) in Kotz's volume on Laplace distribution. There is another formula for the case when $n$ is even, but we shall not be concerned with it for now.
I am interested in the variance of the sample median. Since $f_m(x)$ is symmetric about $x=0$, I can get rid of the absolute value and write it as follows:
$$\sigma^2_m=\frac{n!}{(k!)^2}2^{-n}\frac{1}{b}\int_{0}^{\infty}x^2e^{-(k+1)x/b}(2-e^{-x/b})^kdx$$
What I need is to reduce the following definite integral to a more manageable form:
$$\int_{0}^{\infty}x^2e^{-(k+1)x/b}(2-e^{-x/b})^kdx$$
For fixed $k$ this is fairly easy integral to solve. However, I am interested in asymptotics of variance $\sigma_m^2$ as $n\rightarrow\infty$ and $b$ is a linear function of $\sqrt{n}$, and thus need a solution for an arbitrary $k$.
Any ideas?
AI: Ok, I think I understand now. You want to find
$$\sigma^2=\frac{n!}{(k!)^2}2^{-n}\frac{1}{b}\int_{0}^{\infty}\!dx\,x^2e^{-(k+1)x/b}(2-e^{-x/b})^k$$
where $n=2k+1$ and $b \sim \sqrt{n}$ for $n,k \to \infty$. Let us first use the change of variable $y=x/b$. We then find
$$\sigma^2= b^2\frac{n!}{(k!)^2}2^{-n}\int_{0}^{\infty}\!dy\,y^2e^{-(k+1)y}(2-e^{-y})^k.$$
We will use the method of steepest decent to figure out the asymtptotic behavior of $\sigma^2$ for $k,n \to \infty.$ Let us first rewrite the integral as
$$ \int_0^\infty\!dy\,y^2e^{-(k+1)y}(2-e^{-y})^k =
\int_0^\infty\!dy\, y^2 e^{-y} e^{- k f(y)}, \qquad f(y) =y-\log(2-e^{-y}). $$
The function $f(y)$ assumes its maximum at $y=0$. Thus, the integral is asymptotically dominated for values of $y$ close to $0$. To this end, we expand $f(y) = y^2 +O(y^3)$ and we have
$$\int_0^\infty\!dy\, y^2 e^{-y} e^{- k f(y)} \sim
\int_0^\infty\!dy\, (y^2 -y^3 +O(y^4)) e^{-k y^2} = \frac{\sqrt\pi}{4 k^{3/2}} - \frac{1}{2k^2} +O(k^{-5/2}). $$
Additionally, we use that (see central binomial coefficient) $$\frac{n!}{(k!)^2} \sim2 \binom{2k}{k} \sim 2 \frac{4^k}{\sqrt{\pi k}} = \frac{2^n}{\sqrt{\pi k}}.$$
Together, we have (with $b \sim \alpha \sqrt{n} \sim 2 \alpha \sqrt{k} $)
$$\sigma^2 \sim \underbrace{4 \alpha^2 k}_{b^2} \frac{2^n}{\sqrt{\pi k}} 2^{-n} \frac{\sqrt\pi}{4 k^{3/2}} = \frac{\alpha^2}{k}. $$ |
H: Continuous version of Cauchy-Schwarz
Proving the discrete version of Cauchy-Schwarz is easy:
$$ \left(\sum_{i} a_i^2\right) \left(\sum_i b_i^2\right) \geq
\left(\sum_i a_ib_i\right)^2 $$
can be done via the determinant of the quadratic formula.
Now, however, I want to prove the continuous version, which states:
$$ \int a^2 \int b^2 \geq \left(\int ab\right)^2$$
How do I prove this?
AI: First note that $ab \leq \frac{a^2}{2} + \frac{b^2}{2}$. Then take $a = \frac{f}{(\int{f^2})^{\frac{1}{2}}}, b = \frac{g}{(\int{g^2})^{\frac{1}{2}}}$. |
H: Find a side of a triangle given other two sides and an angle
I have a really simple-looking question, but I have no clue how I can go about solving it?
The question is
Find the exact value of $x$ in the following diagram:
Sorry for the silly/easy question, but I'm quite stuck! To use the cosine or sine rule, I'd need the angle opposite $x$, but I can't find that, cause I don't have anything else to help it along.
Thank You!
AI: Call the top point $A$, the point on the bottom left $B$, and the point on the bottom right $C$. Draw the altitude from $A$ to $BC$, and call the foot of the altitude $D$.
Now $\triangle ABD$ has angles $30^\circ$, $60^\circ$, and $90^\circ$ respectively. Therefore $BD$ has length $3$ and $AD$ has length $3\sqrt3$. Furthermore, by the Pythagorean theorem, $DC$ has length $\sqrt{7^2 - (3\sqrt3)^2} = \sqrt{22}$.
Therefore $x = BC = BD + DC = 3 + \sqrt{22}$.
(Forgive my shoddy length notation.) |
H: $f(x) = x^r$ equaling to $r \cdot x$ when $f$ is cyclic group $G \rightarrow G$
Suppose that there is a cyclic group $G$ of order $n$ with a generator $g$. Also suppose that $r$ is a fixed integer. Then define a homomorphism function $f: G \rightarrow G$, $f(x) = x^r$.
How do we get that $f(x) = r \cdot x$?
Quote:
Theorem: Let $G$ be a cyclic group of order $n$ with generator $g$. Fix an integer $r$, and define
$f : G \rightarrow G$
by
$f(x) = x^r$.
This map $f$ is a group homomorphism of $G$ to itself. If $gcd(r, n) = 1$, then $f$ is an isomorphism, and in that
case every $y ∈ G$ has a unique $r$th root. More generally,
order of kernel of $f = gcd(r, n)$,
order of image of $f = n/gcd(r, n)$.
If an element $y$ has an $r$th root, then it has exactly $gcd(r, n)$ of them. There are exactly $n/gcd(r, n)$ $r$th
powers in $G$.
Proof: Since $G$ is abelian the map $f$ is a homomorphism. Use the fact that $G$ is isomorphic to $\mathbb{Z}/n$.
Converting to the additive notation for $\mathbb{Z}/n$-with-addition, f is
$f(x) = r · x$
AI: The notation $r\cdot x$ is just the same as $x^r$, but using additive notation for $G$, which is appropriate if (for instance) $G=\mathbf Z/n\mathbf Z$. You should avoid mixing additive and multiplicative notation for the same group, but it happens that some general fact written multiplicatively applies to a particular group written additively, and you'll have to cope with translating notation in that case. |
H: How to prove Lagrange trigonometric identity
I would to prove that
$$1+\cos \theta+\cos 2\theta+\ldots+\cos n\theta =\displaystyle\frac{1}{2}+
\frac{\sin\left[(2n+1)\frac{\theta}{2}\right]}{2\sin\left(\frac{\theta}{2}\right)}$$
given that
$$1+z+z^2+z^3+\ldots+z^n=\displaystyle\frac {1-z^{n+1}}{1-z}$$
where $z\neq 1$.
I put $z=e^{i\theta}$. I already got in left hand side cos exp in real part, but there is a problem in the right hand side, I can't split imaginary part and real part. Please help me. Thanks in advance.
AI: $2\sin\frac{\theta}{2}\cos r\theta=\sin\frac{(2r+1)\theta}{2} - \sin\frac{(2r-1)\theta}{2}$
Putting r=1,2,....,n,
$2\sin\frac{\theta}{2}\cos \theta=\sin\frac{3\theta}{2} - \sin\frac{\theta}{2}$
$2\sin\frac{\theta}{2}\cos 2\theta=\sin\frac{5\theta}{2} - \sin\frac{3\theta}{2}$
...
$2\sin\frac{\theta}{2}\cos n\theta=\sin\frac{(2n+1)\theta}{2} - \sin\frac{(2n-1)\theta}{2}$
Adding we get,
$\sum_{1≤r≤n}2\sin\frac{\theta}{2}\cos r\theta=\sin\frac{(2n+1)\theta}{2} - \sin\frac{\theta}{2}$
Divide both sides by $2\sin\frac{\theta}{2}$, we shall get,
$\sum_{1≤r≤n}\cos r\theta=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}} - \frac{1}{2}$ (Assuming $\sin\frac{\theta}{2}≠0$ or $\theta≠2s\pi$ where s is any integer.)
Or, $1+\sum_{1≤r≤n}\cos r\theta=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}} + \frac{1}{2}$ (adding 1 to both sides )
Just observe that for $\sum_{r}\cos(A+2rB)$ where A,B are constants and r is an integer, we need to multiply with $2\sin B$ as
$2\cos(A+2rB)\sin B=sin(A+(2r+1)B) - sin(A+(2r-1)B)$
Putting different ranges of values of r & adding them, we shall get their sums in the compact form.
In the current problem, $A=0, 2B=\theta, 1≤r≤n$
Also as, $2\sin B\sin(A+2rB) = cos(A+(2r-1)B) - cos(A+(2r+1)B)$
This can be used for $\sum_{r}\sin(A+2rB)$.
Also using DonAntonio's approach, we know
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i} \implies e^{ix}-e^{-ix}=2i\sin x$$
So,
$$\begin{align}
\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1} &=\frac{e^{(n+1)i\theta/2}\left(e^{(n+1)i\theta/2}-e^{-(n+1)i\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)} \\[0.5em]
&=e^{ni\theta/2}\frac{2i\sin\frac{(n+1)\theta}{2}}{2i\sin\frac{\theta}{2}} \\[0.5em]
&=\frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\left(\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2}\right)\end{align}$$
Its real part is
$$\begin{align}
\frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\cos\frac{n\theta}{2}
&=\frac{2\sin\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}}{2\sin\frac{\theta}{2}} \\
&=\frac{\sin\frac{(2n+1)\theta}{2}+\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}}\\
&=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}}+\frac{1}{2}
\end{align}$$ |
H: Mixture of two mirror-image Gaussians
Suppose we are given a set of points $(x_i, y_i)\in\mathbb{R}^2$ and are told that they are drawn from a normal (Gaussian) distribution. It is a simple matter in that case to find the mean $(\mu_x,\mu_y)$ of the distribution: $(\langle x \rangle, \langle y \rangle)$, where $\langle x \rangle=\frac{1}{N}\sum_i x_i$ and similarly for $\langle y \rangle$, is an unbiased estimator.
Now suppose, instead, that each point is randomly presented as either $(x_i,y_i)$ or $(y_i, x_i)$, with equal probability. Is there still a comparably simple way to estimate the mean of the original distribution? (Assume $\mu_x \le \mu_y$.) This is equivalent to the case where the points are drawn from a mixture of two Gaussians, where one is constrained to be the reflection of the other across the line $y=x$. However, one might hope that the symmetry of the problem leads to some simplification from the general case of two Gaussians.
Note, by the way, that taking $\mu_x=\langle \min(x, y)\rangle$ and $\mu_y = \langle \max (x,y) \rangle$ does not work: this is adequate only when the individual Gaussian distributions are well-separated by the line $y=x$. Otherwise, it introduces a systematic bias away from that line.
AI: By moments:
Let $Z=[Z_1,Z_2]$ be the mixed variable, let $m_k$ be the $k$-th centered moment of the first components (both components are marginally equivalent, actually), $m_k = E[(Z_1 - m_1)^k]$
Let $\alpha = \frac{1}{2}(\mu_x+ \mu_y)$, $\delta=\frac{1}{2}(\mu_y -\mu_x)$, $s=\frac{1}{2}(\sigma_x^2 + \sigma^2_y)$, $d=\frac{1}{2}(\sigma_y^2 - \sigma^2_x)$
Then, we get the following nonlinear system of equations:
$$\begin{eqnarray}
m_1 &=& \alpha\\
m_2 &=& s + \delta^2\\
m_3 &=& 3 \delta d\\
m_4 &=& 3 s^2 + 3 d^2 + 6 s \delta^2 + \delta^4
\end{eqnarray}
$$
This can be solved numerically or analytically (Maxima...)
$$
\delta=
\frac{\sqrt{{\left( \sqrt{2\,m_{4x}^3+3\,m_3^4}+\sqrt{3}\,m_3^2\right) }^{\frac{2}{3}}-{2}^{\frac{1}{3}}\,m_{4x}}}{{2}^{\frac{1}{3}}\,{3}^{\frac{1}{4}}\,{\left( \sqrt{2\,m_{4x}^3+3\,m_3^4}+\sqrt{3}\,m_3^2\right) }^{\frac{1}{6}}}
$$
where $m_{4x} = m_4 - 3 m_2^2$
This can be simplified if both variables have same variance ($d=0$).
$$\delta=\left(\frac{|m_{4x}|}{{2}} \right)^{\frac{1}{4}}$$
Of course, $m_k$ is estimated by the standard estimators (though high order moments might require many samples to get good estimations) and then, computing $\alpha$ and $\delta$ we have the estimated $\mu_x$,$\mu_y$.
I've done some basic testing with Octave/Matlab, code attached
mu = [3,5]' % (mean vector)
cov = [ 1 , 1.5 ; 1.5,4] % covariance matrix
rho = cov(1,2)/sqrt(cov(1,1)*cov(2,2)) % compute correlation factor
N=2000; % number of samples
X=mvnrnd(mu,cov,N); % unmixed samples
W=unidrnd(2,N,1)-1; % mixing flag
Y = flipdim(X,2); % reversed
Z=X; %
Z(W>0,:)=Y(W>0,:); % mixed
delta=(mu(2)-mu(1))/2 % "true" values
alpha=(mu(2)+mu(1))/2
s=(cov(1,1)+cov(2,2))/2
d=(cov(2,2)-cov(1,1))/2
m1=mean(mean(Z)); % estimated moments
m2=mean(mean( (Z-m1).^2));
m3=mean(mean( (Z-m1).^3));
m4=mean(mean( (Z-m1).^4));
m4x = m4 - 3*m2^2;
alpha
m1
s+delta^2
m2
3*delta*d
m3
sum([3*(s^2+d^2),6*s*delta^2,delta^4])
m4
alphaest= m1;
deltaest = deltaest = sqrt((sqrt(2*m4x^3+3*m3^4)+sqrt(3)*m3^2)^(2/3)-2^(1/3)*m4x)/(2^(1/3)*3^(1/4)*(sqrt(2*m4x^3+3*m3^4)+sqrt(3)*m3^2)^(1/6))
mu(2) % compare true and estimated means
alphaest+deltaest
mu(1)
alphaest-deltaest |
H: Clarification on the proof for the Euclidean algorithm
Lemma: Let $m$ and $n$ be positive integers with $m \leq n$. If $r$ is the remainder of dividing $n$ by $m$, then $(n,m) = (m,r)$.
The proof is given as follows:
We have by the division algorithm that $n = sm + r$ with $0 \leq r < m$. Suppose that $d = (n,m)$ and $e = (m, r)$. Since $r = n - sm$ and $d \mid n,$ $d \mid m$ we have $d \mid n - sm = r$.
The part I don't understand is how $d \mid n - sm = r$ is equivalent to $r = n - sm.$
It seems as if it is saying $n$ is the same as $d \mid n$.
AI: Since $d|n$ and $d|m$, we have $d|n-sm$. But $n-sm=r$, so $d|r$. "$d|n-sm=r$" is just a compressed way of writing all that. Does that answer your question? |
H: what is the equation to get this type of a curve?
what must be the equation to get this type of a curve??
the curve must increase from -infinity to 0 and then decrease again to +infinity.
Thank you
AI: You can try multiplying a $\sin$ wave with a Gaussian, e.g:
$$\sin(5x)e^{-x^2/2}$$ |
H: The average surface area of a projection of a randomly rotated planar rectangular shape on a two-dimensional surface
I have a planar rectangular shape, of dimensions $N$ by $M$, positioned in 3-space above a two-dimensional surface. Provided a large number of random 3-space rotational orientations of the shape, what is the average surface area $A$ of a projection of the shape on the two-dimensional surface?
AI: It is equivalent to keep the shape fixed and randomly rotate the plane. The orientation of the plane is completely determined by its normal. So you want to pick the normal of the plane, say $\hat n$, uniformly over the unit sphere. Then the area of the projection is equal to the area of the original shape times $\left|\hat n\cdot\hat z\right|$, where $\hat z$ is the fixed normal of the shape. Using spherical coordinates with $\hat z$ as the zenith, the expected value of $\left|\hat n\cdot\hat z\right|$ as $\hat n$ varies over the unit sphere is
$$\mathrm E[\hat n\cdot\hat z] = \frac{\int_{-\pi}^\pi\int_0^\pi\left|\cos\theta\right|\cdot\sin\theta\,\mathrm d\theta\,\mathrm d\phi}{\int_{-\pi}^\pi\int_0^\pi1\cdot\sin\theta\,\mathrm d\theta\,\mathrm d\phi} = \frac{2\pi\int_0^\pi\left|\cos\theta\right|\cdot\sin\theta\,\mathrm d\theta}{4\pi} = \frac12.$$
So the expected area of the projection is half the area of the shape itself. |
H: Norm of quotients
$\newcommand{\Ker}{\operatorname{Ker}}$
Let $X$ be Hilbert space and let $T:X\to X$ be a bounded operator. Define the operator
$S: X/\Ker T \to X/\Ker T$ via $S(x+\Ker T)=Sx+\Ker T$. I can show that $||S||\leq ||T||$, and I am wondering whether the norms are actually equal. Is this true or false in general?
Edit: Considering the answer below, I am adding the conditions that $\Ker(T)$ is finite dimensional and $\operatorname{Im}(T)$ is not (say $T$ is surjective). Could you please show a counterxample in this situation as well.
AI: Suppose $T^2 = 0$. Then $\text{im}(T) \subseteq \text{ker}(T)$, so...
Edit: The extra conditions do not really change anything. Here is an example where $T$ is surjective and $\text{ker}(T)$ is one-dimensional. Let $T$ act on a separable Hilbert space with orthonormal basis $e_1, e_2, ...$ by
$$T e_1 = 0$$
$$T e_2 = 2 e_1$$
$$T e_n = \frac{1}{2} e_{n-1}, n \ge 3.$$
The point here is that $\text{ker}(T)$ may still contain the part of the image of $T$ which determines its norm. |
H: division by integers
let d,k be integers, with k even.
suppose d|2k
suppose d does not divide 2
suppose d does not divide k
show that d equals 2k.
(I'm really just trying to understand the 2nd last line, in this answer to the question: Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$)
AI: It's not true. Take $k=6$ and $d = 4$ for example. |
H: how to solve these questions with congruences
how can I solve these type of questions : using congruences
find the last digit or last two digit of $27^{27^{26}}$ and find remainder when $53^{103}$ is divided by 7.
I can solve 2nd question and a simpler case of first question with binomial theorem but in questions like 1st this does not seem to work for me, and using binomial makes the solution too lengthy so as I see I have only left congruence method but I do not know how to initiate the solution .
I have burton's book of number theory in which there are some examples and exercises for questions like this but still I am not able to catch the pattern these queations should follow to be solved by congruences ,
I would really appreciate any advice or suggestions or anything that can help me to solve these questions
AI: By Euler's totient theorem, we know
$$
27^{\phi(100)}=27^{40}\equiv 1 \rm{mod} 100,
$$
On the other hand,
$$
27^{\phi(40)}=27^{16}\equiv 1 \rm{mod} 40,
$$
so we have $27^{26}\equiv 27^{10}\equiv 3^{30}\equiv 3^{4\times7}\cdot9\rm{mod}40\equiv 9 \rm{mod}$.
Hence we have
$$
27^{27^{26}}\equiv27^{40n+9}\equiv 27^9\rm{mod} 100=3^{27}\rm{mod} 100
$$
Which is not difficult to conclude the answer 87. |
H: Find the constant K in probability distribution?
Find the constant K?
I know how to find K, if there is only one "k" in the question and the rest are given number. I know that all the x value are add upto 1.
But in this question, I don't know how to start.
One thing I know is $\Sigma(x)=1(k)+2(2k)+3(3k)+4(4k)+5(5k)$
Does this help to find the constant k?
Please help me "Find the constant k"
Appreciate your help.
AI: You are confused. It's pretty fine. Please note that $X=i$ means the event that X equals i. So, let me interpret the table for you.
There's a probability of $k$ that event $X=1$ shall happen.
There's a probability of $2k$ that event $X=2$ shall happen.
There's a probability of $3k$ that event $X=3$ shall happen.
There's a probability of $4k$ that event $X=4$ shall happen.
There's a probability of $5k$ that event $X=5$ shall happen.
No other events are possible.
Important: By Unitarity of probability, the probability of $Pr({\text{$X$ = 1,2,3,4 or 5}})=1$.
$X=1, X=2, X=3, X=4, X=5$ are five mutually exclusive events in the sample space, and there are no other events. So, $$Pr(X=1,2,3,4,5)=\sum_{i=1}^5Pr(X=i)=k+2k+3k+4k+5k = 15k$$
So, what's $k$? |
H: A question about applying Arzelà-Ascoli
An example of an application of Arzelà-Ascoli is that we can use it to prove that the following operator is compact:
$$ T: C(X) \to C(Y), f \mapsto \int_X f(x) k(x,y)dx$$
where $f \in C(X), k \in C(X \times Y)$ and $X,Y$ are compact metric spaces.
To prove that $T$ is compact we can show that $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is bounded and equicontinuous so that by Arzelà-Ascoli we get what we want. It's clear to me that if $TB_{\|\cdot\|_\infty} (0,1)$ is bounded then $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is bounded too. What is not clear to me is why $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ is equicontinuous if $TB_{\|\cdot\|_\infty} (0, 1)$ is.
I think about it as follows: $TB_{\|\cdot\|_\infty} (0, 1)$ is dense in $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ with respect to $\|\cdot\|_\infty$ hence all $f$ in $\overline{TB_{\|\cdot\|_\infty} (0, 1)}$ are continuous (since they are uniform limits of continuous sequences). Since $Y$ is compact they are uniformly continuous. Now I don't know how to argue why I get equicontinuity from this. Thanks for your help.
AI: Take any $S\subseteq C(X)$ which is equicontinuous. Let $\epsilon>0$. We have some $\delta>0$ such that
$$f\in S, \|x-x'\|<\delta\implies \|f(x)-f(x')\|<\epsilon.$$
For any $f\in \bar S$, we have a sequence $f_n\to f$ uniformly with each $f_n\in S$. Then
$$\|x-x'\|<\delta\implies \|f(x)-f(x')\|\leq \|f(x)-f_n(x)\|+\|f_n(x)-f_n(x')\|+\|f_n(x')-f(x)\|$$
which becomes less than $\epsilon$ for sufficiently large $n$. Thus $\bar S$ is equicontinuous. |
H: Lifting of maps to a covering space
I am reading Algebraic topology by W. Massey and I have a problem with the proof of property 5.1:
Let $(\tilde{X},p)$ be a covering space of $X$, $Y$ a connected and arcwise connected space, $\tilde{x}_0 \in \tilde{X}$, $y_0 \in Y$ and $x_0=p(\tilde{x}_0) \in X$. Given a map $\varphi : (Y,y_0) \to (X,x_0)$, there exists a lifting $\tilde{\varphi} : (Y,y_0) \to (\tilde{X},\tilde{x}_0)$ iff $\varphi_* \pi_1(Y,y_0) \subset p_* \pi_1 (\tilde{X},\tilde{x}_0)$.
To construct $\tilde{\varphi}$, let $y \in Y$ and $f : I \to Y$ a path from $y_0$ to $y$. So $\varphi p : I \to X$ is a path from $x_0$ to $\varphi(y)$. There exists only one path $g : I \to \tilde{X}$ such that $g(0)=\tilde{x}_0$ and $pg= \varphi p$. Let define $\tilde{\varphi}(y)= g(1)$.
We can show that $\tilde{\varphi}$ is well defined (that is, $\tilde{\varphi}$ doesn't depend on the choice of $f$) and it is obvious that $p \tilde{\varphi} = \varphi$.
To prove the continuity of $\tilde{\varphi}$, let $y \in Y$ and $U \subset \tilde{X}$ a open neighborhood of $\tilde{\varphi}(y)$. We can suppose $U$ arc connected. Take $U'$ an elementary neighborhood of $p \tilde{\varphi}(y)= \varphi(y)$ such that $U' \subset p(U)$ and $V \subset Y$ arc connected such that $\varphi(V) \subset U'$ ($\varphi$ is continuous).
Finally, the arc component of $p^{-1}(U')$ containing $\tilde{\varphi}(y)$ is included in $U$.
But how do you prove that for all $y' \in V$, $\tilde{\varphi}(y')$ is in the same arc component of $p^{-1}(U')$ than $\tilde{\varphi}(y)$?
AI: Let $f$ be a path from $y_0$ to $y$, let $y' \in V$ and let $f'$ be a path in $V$ (who is arc connected) from $y$ to $y'$. Then by concatenating $f$ and $f'$, you obtain a path $f \oplus f'$ from $y_0$ to $y'$, from which you can compute $\tilde \varphi(y')$ : $\varphi \circ (f \oplus f')$ is the concatenation of a path in $X$ from $x_0$ to $\varphi(y)$ and a path in $U'$ from $\varphi(y)$ to $\varphi(y')$. The lifting of the first part is a path in $\tilde X$ from $\tilde x_0$ to $\tilde \varphi(y)$, and the lifting of the second second part is a path in $p^{-1}(U')$ from $\tilde \varphi(y)$ to $\tilde \varphi(y')$.
Therefore, $\tilde \varphi(y)$ and $\tilde \varphi(y')$ are in the same arc component of $p^{-1}(U')$.
Unfortunately you removed a step from the proof so you can't conclude that $\tilde \varphi(y')$ is in $U$ just from that. Let $W$ be the arc component of $\tilde \varphi(y)$ in $p^{-1}(U')$. It is not necessarily contained in $U$ so we have to do the extra work.
Let $U''$ be an elementary neighborhood of $\varphi(y)$ such that $U'' \subset p(W \cap U)$, then choose $V$ such that $V$ is arc connected and $\varphi(V) \subset U''$ instead. Then $\tilde \varphi(y)$ and $\tilde \varphi(y')$ are in the same arc component of $p^{-1}(U'')$. Since $U'' \subset p(W) \subset U'$, they are also in the same arc component of $p^{-1}(U')$, thus $\tilde \varphi(y') \in W$.
Since $W$ is arc connected and $p(W) \subset U'$ is an elementary neighborhood, $p|_W$ is injective (otherwise there would be a nontrivial loop in $U'$). Since $\varphi(y') \in p(W \cap U)$ and $\tilde \varphi(y') \in W$, $p|_W(\tilde \varphi(y')) = \varphi(y') \in p|_W(W \cap U)$, thus $\tilde \varphi(y') \in W \cap U$ . And so we have shown that $\tilde \varphi(y') \in U$ for all $y' \in V$ |
H: Why this limit of integration?
I've been solving problems from the book by DeGroot and Schervish and I can't understand why m is the upper limit of integration in the solution to this problem. Why not the lower one?
Here is the problem:
Suppose that a random variable X has a continuous distribution for which the p.d.f. f is as follows:
$f(x) = 2x$ for $ 0< x <1, 0 $ otherwise
Determine the value of $d$ that minimizes $E(|X − d|)$.
Here is the solution:
$$ \int_0^m 2x \, dx=0.5 $$
Thank you very much in advance.
AI: The expected value is given by (assuming $d$ is in $[0,1]$):
$$E = \int_0^1{2x|x-d|}dx = \int_0^d{2x(d-x)}dx + \int_d^1{2x(x-d)}dx$$
$$ = -\frac{2}{3} + d - \frac{2d^3}{3}$$
Derive by $d$ and compare to zero:
$$E' = 1 - 2d^2 = 0 \rightarrow d = \frac{1}{\sqrt{2}}$$ |
H: Is $\int_{\mathbb{R}^D}\exp(\mathbf{y}^\top\mathbf{S}\mathbf{y}) \mathbf{y}\,d\mathbf{y}$ equal to $\mathbf{0}$?
I want to show that $\int_{\mathbb{R}^D}\exp(\mathbf{y}^\top\mathbf{S}\mathbf{y})\mathbf{y} \, d\mathbf{y}$, where $\mathbf{S}$ is a symmetric real matrix, is equal to $\mathbf{0}$ .
My intuition (although very immature) hints that it is true, but I can't show that — actually, it looks like the opposite is true.
$\int_{\mathbb{R}^D}\exp(\mathbf{y}^\top\mathbf{S}\mathbf{y})\mathbf{y}d\mathbf{y} = \int_{\mathbb{R}^D}\prod_{i,j}\exp(y_i y_j s_{ij})\mathbf{y} \, d\mathbf{y}=\mathbf{v}$;
$v_i=\int_{\mathbb{R}^D}\prod_{i,j}\exp(y_i y_j s_{ij})y_i \, d\mathbf{y} = \int_{-\infty}^\infty dy_{k\neq i,j}(\int_{-\infty}^\infty(\int_{-\infty}^\infty{\exp(y_i^2+2y_iy_j)}dy_i)\exp(y_j^2)dy_j)$.
$v_i=0\Leftarrow\forall{a}\;\int_{-\infty}^\infty\exp(x^2+2ax)x \, dx=0$.
Seems like the only hope would be for the function under the last integral to be antisymmetric with regard to some $x=b$, but it's clear that it is not for any $a\neq 0$.
Where's my mistake? I want to prove to myself that the title integral is zero...
[UPDATE]
I apologize to the comminuty for not clarifying what is meant by "$\mathbf{y}\;d\mathbf{y}$", which caused some misunderstanding between the commenters, the reason being my apparent ignorance. This question comes from me trying to derive multivariate Gaussian as the probability distribution with maximum entropy (with the given mean and variance); this is given as an exercise in the textbook I'm working through, and the mean is given by the integral $\int{p(\mathbf{x})\mathbf{x}d\mathbf{x}} = \mathbf{\mu}$ ($\mathbf{\mu}$ is a vector with $D$ elements.) I have a vague understanding of this notion, I supposed it means that the $i$-th component of this vector is given by the integral $\int{p(\mathbf{x})x_idx_1dx_2...dx_n}$. Seems like it's more a volume unit integral, not a dot product. Am I right here?
Also, there was some confusion on the Riemann integral vs. v.p. I don't think v.p. is what I need, after some consideration; please excuse me for misleading you.
AI: Essentially, the trouble comes from some mistakes in indexations. With your notations, assuming $S$ is definite negative, one should rewrite the definition of $v_i$ for each $i\leqslant D$ (as it appears at its first occurrence in the post) as
$$
v_i=\iint_{\mathbb R^D}\varphi_i(y_1,\ldots,y_D)\cdot\mathrm dy_1\cdots\mathrm dy_D,
$$
where
$$
\varphi_i(y)=\exp\left(y^TSy\right)\cdot y_i,\qquad y=(y_1,\ldots,y_D).
$$
In particular, one cannot use $i$ both as the index of $v_i$ and as a running index in the product defining the function that $v_i$ is the integral of.
A different intepretation of $v_i$ is proposed on this page, which I am not sure to understand nor to agree with (but, as always, this is the OP's task to clarify the matter).
Now, to prove that each $v_i=0$ (with the definition of $v_i$ above), note that, for every $y$ in $\mathbb R^D$, $\varphi_i(-y)=-\varphi_i(y)$ hence, as the integral of an odd function, $v_i=0$. |
H: Calculating roots of $x^3 = 2x^2 - x$, getting wrong result
I am working with a text book problem, trying to find the solutions to $x^3 = 2x^2 - x$, but I keep ending up with a different result than the book gives. I'm thinking the problem lie in getting the wrong square root towards the end, seeing as the text book gives different results for $x_1$ and $x_2$, but I can't find any mistakes.
$$x^3 = 2x^2 - x$$
$$\iff x^2 = 2x - 1$$
$$\iff x^2 - 2x + 1 = 0$$
$$\iff (x - \frac{2}{2})^2 = (\frac{2}{2})^2 - 1$$
$$\iff (x - 1)^2 = 0$$
$$\iff x - 1 = \pm \sqrt{0}$$
$$\iff x = 1 + \pm \sqrt{0}$$
$$\iff x = 1 + \pm 0$$
$$x_1 = 1 + 0 = 1$$
$$x_2 = 1 - 0 = 1$$
AI: Note that by dividing by $x$, you are implicitly assuming that $x$ is not $0$. It turns out that this is a solution. You can check this directly.
You are right that $1$ is a double root of the equation. If you move everything to one side and factor, you find that
$$x^3-2x^2+x=x(x-1)(x-1)=0$$
which confirms our answer. |
H: Question regarding technicalities in the paper Iterated Prisoner’s Dilemma contains strategies that dominate any evolutionary opponent
For people on this board I have a probably pretty modest question, but since I'm not a mathematician (just an economist), I'm having trouble.
The full pdf can be found here: http://www.pnas.org/content/early/2012/05/16/1206569109.full.pdf+html
The question is regarding the following passage and has to do with linear algebra. They write:
"where Adj(M′) is the adjugate matrix (also known as the classical
adjoint or, as in high-school algebra, the “matrix of minors”). Eq.
2 implies that every row of Adj(M′) is proportional to v.
Choosing the fourth row, we see that the components of v are
(up to a sign) the determinants of the 3 × 3 matrices formed from
the first three columns of M′, leaving out each one of the four
rows in turn. These determinants are unchanged if we add the
first column of M′ into the second and third columns.
The result of these manipulations is a formula for the dot
product of an arbitrary four-vector f with the stationary vector v
of the Markov matrix, v · f ≡ D(p; q; f), where D is the 4 × 4
determinant shown explicitly in Fig. 2B. This result follows from
expanding the determinant by minors on its fourth column and
noting that the 3 × 3 determinants multiplying each fi are just the
ones described above."
To understand the full context you will probably have to read the beginning of the passage, which is also very short. Yet my question is specifically regarding the formulated relationship between the stationary vector v of the Markov transition-matrix M and the Adj(M'), which is Adj.(M-I). As they say: Every row of Adj.(M') is proportional to v, which is sort of intuitive looking at Eq. 2, but I simply do not understand how they got that. Also the immediately following conclusion that the elements of v are the 3x3 column determinants of M' if you were to eliminate from the bottom of the fourth column.
Also to point out a petty mistake but v · f ≡ D(p; q; f) can't be correct as the dimensions do not link up correctly. v' · f ≡ D(p; q; f) is correct. But yet again I grasp that this formulation makes sense, but fail to understand how this can be arrived at.
If you can point me in the direction of a book or can flat-out explain this to me, I would be very obliged.
Thanks in advance
o_s
AI: A very interesting paper!
On every row of $\operatorname{Adj}(\mathbf M')$ being proportional to $\mathbf v$: This isn't fully justified in the paper. In speaking of "the stationary vector $\mathbf v$ of the Markov matrix", they're implicitly using the fact that the matrix has a unique stationary vector. That's not trivial; an irreducible Markov chain has a unique stationary distribution if and only if all its states are positive recurrent. (For that statement and an explanation of the terminology it uses see Wikipedia.) If it does have a unique stationary vector (i.e. a unique left eigenvector with eigenvalue $1$), then all vectors with $\mathbf x^\top M=\mathbf x^\top$, or equivalently $\mathbf x^\top(\mathbf M-\mathbf I)=0$, must be proportional to that vector, and equation [2] says precisely that every row of $\operatorname{Adj}(\mathbf M')$ satisfies $\mathbf x^\top(\mathbf M-\mathbf I)=0$.
On the components of $\mathbf v$ being given by these $3\times3$ determinants: This is a bit imprecise; $\mathbf v$ was introduced as the stationary vector of the Markov matrix, which should be normalized to sum to $1$; equation [2] only says that $\mathbf v$ is proportional to the vector formed from these determinants. (This is because the fourth row of $\operatorname{Adj}(\mathbf M')$ is defined by these determinants, and as discussed above each row of $\operatorname{Adj}(\mathbf M')$ is proportional to $\mathbf v$.) This imprecision is corrected on the next page, where the payoffs are noramlized using $\mathbf v\cdot\mathbf 1$.
On replacing $\mathbf v\cdot\mathbf f$ by $\mathbf v'\cdot\mathbf f$: I'm not sure what you mean by $\mathbf v'$, but if you mean $\mathbf v^\top$, the transpose of $\mathbf v$, note that the dot notation $\mathbf v\cdot\mathbf f$ for the dot product of two column vectors is quite usual; to write it in matrix notation as you seem to have in mind, you should simply write $\mathbf v^\top\mathbf f$ without a dot, since it's unusual to use dots for matrix multiplication. You can see that the dot product is given by this determinant by applying the Laplace expansion to the last column (the one containing $\mathbf f$). |
H: Why does a time-homogeneous Markov process possess the Markov property?
Klenke defines (Definition 17.3, p. 346) a time-homogeneous Markov process independently, rather than as a special case of a stochastic process that possesses the Markov property (Definition 17.1, p. 345) and in addition satisfies certain further constraints. He then goes on to claim that "a time-homogeneous Markov process is simply a stochastic process with the Markov property and for which the transition probabilities are time-homogeneous" (Remark 17.4, p. 346).
How is the Markov property implied by the definition of a time-homogeneous Markov process?
Relevant Definitions
Setting Let $I\subseteq \left[0,\infty\right)$ and $\left(X_t\right)_{t\in I}$ be an $\mathbb{R}^n$-valued stochastic process vis-a-vis $\left(\mathcal{F}_t\right)_{t\in I}$, the natural filtration generated by $\left(X_t\right)_{t\in I}$.
Definition 17.1 We say that $X$ has the Markov property (MP) iff for every $A\in\mathcal{B}\left(\mathbb{R}^n\right)$ and all $s,t\in I$ with $s\leq t$,
$$\mathrm{P}\left[\left.X_t\in A\space\right|\mathcal{F}_s\right]=\mathrm{P}\left[\left.X_t\in A\space\right|X_s\right]$$
Definition 17.3 Let $I$ be closed under addition and assume $0\in I$. A stochastic process $X = \left(X_t\right)_{t\in I}$ is called a time-homogeneous Markov process with distributions $\left(\mathrm{P}_x\right)_{x\in \mathbb{R}^n}$ on the space $\left(\Omega,\mathcal{A}\right)$ iff:
For every $x\in\mathbb{R}^n$, $X$ is a stochastic process on the
probability space $\left(\Omega,\mathcal{A},\mathrm{P}_x\right)$
with $\mathrm{P}_x\left[X_0=x\right]=1$.
The map $\kappa:\mathbb{R}^n\times\mathcal{B}\left(\mathbb{R}^n\right)^{\otimes I}\rightarrow \left[0,1\right]$, $\left(x,B\right)\mapsto\mathrm{P}_x\left[X\in B\right]$ is a stochastic kernel.
$X$ has the time-homogeneous Markov property: For every $A\in\mathcal{B}\left(\mathbb{R}^n\right)$, every $x\in\mathbb{R}^n$ and all $s,t\in I$, we have
$$\mathrm{P}_x\left[\left.X_{t+s}\in A\space\right|\mathcal{F}_s\right]=\kappa_t\left(X_s,A\right)\space\space\mathrm{P}_x\mathrm{-a.s.}$$
Here, for every $t\in I$, the transition kernel $\kappa_t:\mathbb{R}^n\times\mathcal{B}\left(\mathbb{R}^n\right)\rightarrow\left[0,1\right]$ is the stochastic kernel defined for $x\in \mathbb{R}^n$ and $A\in\mathcal{B}\left(\mathbb{R}^n\right)$ by
$$\kappa_t\left(x,A\right):=\kappa\left(x,\left\{y\in \mathbb{R}^I:y\left(t\right)\in A\right\}\right)=\mathrm{P}_x\left[X_t\in A\right]$$
The family $\left(\kappa_t\left(x,A\right),\space t\in I,x\in\mathbb{R}^n, A\in\mathcal{B}\left(\mathbb{R}^n\right)\right)$ is also called the family of transition probabilities of $X$.
AI: Since $X_s$ is $\mathcal F_s$-measurable, the tower property shows that
$$
\mathrm{P}_x\left[\left.X_{t+s}\in A\space\right|X_s\right]=\mathrm E_x\left[Y\left.\space\right|X_s\right],\qquad \text{where}\quad Y=\mathrm{P}_x\left[\left.X_{t+s}\in A\space\right|\mathcal{F}_s\right].
$$
Item (3) of Definition 17.3 asserts that $Y$ is $\sigma(X_s)$-measurable, hence
$$
\mathrm E_x\left[Y\left.\space\right|X_s\right]=Y,
$$
and, in particular, the property in Definition 17.1 holds. |
H: How can I figure out the plot of the set $M:=\{(x,y) \in R:\frac{1}{2}\leq x \land 1
I have some trouble to figure out the plot/graph of a set like
$$M:=\{(x,y) \in R:\frac{1}{2}\leq x \land 1<y<\frac{1}{x}\}$$
On wolframalpha I saw the plot.
Is there a trick how to figure out such kind of graph?
Greetings.
AI: I suppose there's no such trick. Simply, you have to make as much effort as you can to get the more comprehensive view in the structure of the set you are asked to plot.
In this case, I suggest you to start with the fixed range of variables, namely $x\ge \frac{1}{2}$ and $y\gt1$. You see that the plane was divided into the quarters. The one you are interested in is the up-right. Then you can move to plotting the more sophisticated part, i.e. $y\lt \frac{1}{x}$. In order to do that, sketch the $y=\frac{1}{x}$ graph and mark the underlying part of the plane. Once you've finished drawing all parts, you have to think what kind of conjugates were used to describe the set. Notice that in your example one has only "and" operators. Therefore, to obtain the result you have to take an intersection of all previously driven components.
Don't forget to exclude points which were probably marked by drawing boundaries, as it happens often when sketching sets defined with strict inequalities. You could do that by drawing all "edges" with dashed line first and then bolding ones which reflect weak inequalities. In this case, $y=1$ and $y=\frac{1}{x}$ should be excluded.
Edit: To obtain the graph of $y=\frac{1}{x}$, you can start with calculating coordinates of the points from the graph lying on the boundaries of the up-right quarter, namely two points of the form $(x,\frac{1}{x})$, where first has $x$ coordinate equal to $\frac{1}{2}$ and second $y$ one equal to $1$. It provides you with the points $(\frac{1}{2},2)$ and $(1,1)$. The remaining question is: What is the curve I am supposed to draw between the points? If you want the curve to be accurate, you have to be sure about such its features like monotonicity, concavity/convexity... All of them can be termined with tools known from calculus. Let me stop here as I'm not sure if it was exactly what you were asking about. |
H: 64 possible values - continuous or discrete?
I have a response variable in a model that is calculated from binary values of an 8x8 grid of pixels, giving a proportion. Technically, there are only 64 possible values the result can take, so would this count as discrete data? There are many values, and there are not integers. Would it be fair to describe the variable as pseudo-continuous?
The values, I suppose, are 1/64, 2/64. 3/64, 4/64 etc.
AI: There is no right answer to this question. Almost all real-world data is "really" discrete, but it is often more convenient to model it as continuous. You should follow whichever option gives you the easiest model. For example, if the distribution of values appears to be a smooth curve, use a continuous distribution. If you have a huge amount of data and can can simply count how often each possibility appears, use the discrete distribution. |
H: is this a subset of the set of all selfadjoint operators?
Is the set $Q$ of all operators that have the property that the image is orthogonal to the kernel, and the kernel isn't the null space, only a subset of the set $T$ of the selfadjoint operators or equal to it ?
(would this hold in an infinite dimensional space (modulo some closures of the image or kernel) as well ?)
I know that $Q\subseteq T$, but does equality also hold ?
AI: Take any non-symmetric invertible real matrix. Its image is the whole space, its kernel is zero and these are orthogonal. Yet the operator is not self-adjoint for the standard inner product.
[Added] Your edit does not make much difference, just othogonally project onto a proper finite-dimensional subspace and then apply a non-self-adjoint invertible operator of that subspace to itself; now the kernel is nonzero and orthogonal to the image (the space projected onto) but the operator is not self-adjoint. In spite of your "knowing" it, the claim $Q\subseteq T$ is just false. |
H: Find an equivalent to $(P \lor Q) \land (P \to R) \land (Q \to S)$
I need some help regarding solving a logic. The question is to find an equivalent to the following logic.
$$(P \lor Q) \land (P \to R) \land (Q \to S)$$
The choices are
(a) $S \land R$
(b) $S \to R$
(c) $S \lor R$
(d) none of these
The answer given in the book is (c) $S \lor R$ . I tried many ways, but unable to bring the solution. So, kindly explain me the steps. Thanks in advance.
AI: The solution is (d)
The first formula is false if $P$ and $Q$ are false, but all other answers (a), (b) or (c) can be true if $P$ and $Q$ are false (by setting $R$ and $S$ true for example).
EDIT :
$$(P\vee Q)\wedge (P\rightarrow R) \wedge (Q\rightarrow S)$$
$$\Leftrightarrow$$
$$(P\vee Q)\wedge (\neg P\vee R) \wedge (\neg Q\vee S)$$
And you can't just eliminate $P$ and $Q$ to obtain an equivalent formula. However, you can eliminate them if you just try to find an implication...
$$((P\vee Q)\wedge (\neg P\vee R)) \wedge (\neg Q\vee S) \Rightarrow (Q\vee R) \wedge (\neg Q\vee S) \Rightarrow (R\vee S)$$ |
H: Is $z/z$ holomorphic in $z=0$?
To be holomorphic require to have derivative at $z\to0$ but $f(z)$ is undefined.
Does it mean $z/z$ is not holomorphic at $z=0$?
AI: Generally in complex analysis texts, after we are exposed to Riemann's Removable singularity theorem, every function written down that has a removable singularity is assumed to already have it removed, to avoid unnecessary extra clauses "and equal to $1$ at $z=0$" and the like.
For example, $\displaystyle f(z) = \frac{\sin z}{z}$ may be written down and it may be written that $f(0)=1.$ Similarly, if we define $g(z)=z/z$ then without anything extra, one would normally go ahead and declare $g$ is holomorphic in $\mathbb{C}.$ |
H: Follow-up on Sobolev embedding theorem
I asked this question in a comment when I realised that the answerer is away until September so I am posting it here in a new thread.
I've been thinking about the Sobolev embedding theorem, given as follows:
If $k > l + d/2$ then we can continuously extend the inclusion $C^\infty (\mathbb T^d) \hookrightarrow C^l (\mathbb T^d)$ to $H^k (\mathbb T^d) \hookrightarrow C^l(\mathbb T^d) $ where $\mathbb T^d$ is the $d$-dimensional torus and $H^k$ is the closure of $C^\infty$ with respect to the norm $\|(D^\alpha f)_\alpha \| = \sqrt{ \sum_\alpha \|D^\alpha f\|^2} $.
Can you tell me if this is correct?
(i) By definition of $H^k$ we can uniquely and continuously extend any continuous linear operator $T$ that has domain $C^\infty (\mathbb T^d)$ to all of $H^k$.
(ii) What the Sobolev embedding theorem gives us is a continuous inclusion $i: H^k \hookrightarrow C^l$ so that given a continuous linear operator $T: C^l \to X$ (to any linear normed space $X$) we can apply $T$ to $H^k$ via $T \circ i$.
I think I used to mix up (i) and (ii) and now I think that these are two different facts, independent of each other. Thanks for your help.
AI: I'm assuming you are thinking of $C^\infty$ as a normed space (with the Sobolev norm) in i). Then i) is correct yes, but has nothing to do with Sobolev embedding.
The Sobolev embedding theorem(s) is/are a statement about the regularity of the functions in Sobolev spaces, they claim that there is a smooth/continuous function representing an element of a space obtained by completing a space of smooth functions using integral norms. This is important when you want to prove smoothness of solutions of differential equations, the existence of which is often rather easy to prove in Sobolove spaces. (The statement of the embedding theorem usually includes a statement about the compactness of the embedding (when applicable), which is also useful for proving existence theorems). |
H: Neither provable nor disprovable theorem
I wonder about a theorem which can be proven that this theorem is $neither\, provable$ nor $disprovable$ using any kind of mathematical knowledge.
Questions:
1- Is there any such theorem? or is it proposable?
2- If there exists such a theorem about a fact, how should we think about it? does it mean that this fact is independent of everything?
AI: This is a bit of a semantical issue.
Commonly the word theorem refers to a statement which is provable from a certain theory, for example "Zorn's lemma is a theorem of ZFC" is to say that we can prove Zorn's lemma is true from ZFC.
However sometimes the word "sticks" to a certain assertion (such as Zorn's lemma) and it then becomes a part of the name "Cantor's theorem"; "Hahn-Banach Theorem"; "Tychonoff's theorem"; and so on. For example, the Hahn-Banach theorem is not a theorem of ZF, but it is a theorem of ZFC.
Given a collection of axioms which is "nice" (in the sense that we can easily determine if an arbitrary statement is an axiom); as well strong enough to describe the natural numbers, then if this collection is consistent then there are statements which are true and it cannot prove nor disprove. This is The Incompleteness Theorem mentioned by others.
The Continuum Hypothesis is one example of this. There are models of ZFC in which the continuum hypothesis is true; other models of ZFC in which it is false. Similarly the axiom of choice cannot be proved from the axioms of ZF.
However, we are free to add to ZFC other axioms which are sufficient to prove that the continuum hypothesis is true; or we can add axioms which will prove the continuum hypothesis is false. Of course we cannot add both axioms, as that would generate an inconsistency in the system... and inconsistencies are bad.
For example from the axioms of ZFC+$\lozenge_{\omega_1}$ we can prove the continuum hypothesis is true. On the other hand from ZFC+PFA we can prove that the continuum hypothesis is false.
All in all when we say that $\varphi$ is independent from the theory $T$ it means that $T+\varphi$ is consistent (if $T$ was consistent to begin with, of course). In particular this means that $\varphi$ is not independent of the theory $T+\varphi$. This should answer your second question: no statement is independent from all theories. |
H: direct sum to get a vector space
Concerning this question I asked (specifically my last comment), is it true that the formula $$L\oplus \ker S =H$$
holds in a finite dimensional vector space, where $S:H\rightarrow H$ is linear operator, $(b_1,\ldots,b_k)$ is a basis of the image of $S$ and $L$ consists of the preimages of the $b_i$ under $S$, i.e. $l_j\in L$ is such that $l_j\in S^{-1} (b_j)$ ?
If we equip $H$ with an inner product, such that $S$ becomes selfadjoint, would then $$\text{im} S\oplus \ker S =H$$ hold ? (Or can you give a counterexample?)
AI: No, the statement does not hold: $L$ isn't a vector space. If you define $L$ to be the span of (some choice of) preimages of the $b_i$ it will be true: by rank-nullity all you need prove is that $L \cap \ker S = \{ 0 \}$, but this is clear because $f|_L$ is one-to-one.
Suppose $S$ is self-adjoint with respect to some inner product $(,)$. Let $x \in \ker S \cap \operatorname{im} S$. Then $(x,x) = (x,Sy)$ for some $y$, and by self-adjointness, $(x,Sy)=(Sx,y)=(0,y)=0$. So $\ker S \cap \operatorname{im} S = \{0\}$. It follows that $H = \ker S \oplus \operatorname{im} S$. |
H: Counting the genus of a model
I was reading a webpage on Euler-Poincaré Formula and it has a question that I couldn't figure out. It should be simple, but I am not getting the right answer.
The question on the webpage goes like this:
Consider the following model which is obtained by taking out a torus and tube from the interior of a sphere. What is the genus (penetrating hole) of this model?
So, the shape should look something like a sphere but with the red coloured model excluded from the sphere. The above yellowed coloured model shows the interior of the sphere.
Because the intention is to understand its topology, I cannot tear or glue the model. I could only stretch and squash the model.
The answer given was $1$. However, I could only get $0$ as the answer.
I'm imagining the protruding pole in the middle of the sphere could be expanded to filled up the empty space surrounding it and eventually fills up to become a full sphere again. In this case, the genus is equals to $0$, which means, the topology of this model has no penetrating holes. However, I am wrong because the correct answer should be genus equals to $1$.
How is the genus of this model $1$ when I could easily get $0$?
AI: In your picture you are only looking at he half of the remains of the ball. The other half is obtained by reflection and glueing the halves togehter. In the middle of the ball is a connection between the two halves (stemming from the hole of the torus you removed, the pole you are referring to) which makes the complement homotopically nontrivial -- let a curve run around it and try to homotop this to a point. Actually, what you have here is just a deformed solid torus. |
H: Proving a Riemann integral theorem
Question: Prove that for $f$ continuous on $[a,b] \in \Bbb R$, for $c \in [a,b]$, then:
$$ \int_a^b f(x)\, dx = \int_a^c f(x)\,dx + \int_c^b f(x)\, dx$$
My Work:
Let $\{P_k\}$ be a sequence of partitions such that the mesh of $P$ goes to $0$ as $k \to \infty$. For each $k$, denote the points of the partition $P_k$ by $x_0^k < x_1^k < \cdots < x_N^k$. In each subinterval $[x_{i-1}^k, x_i^k]$ choose a point $x_i^{k*}$.
W.L.O.G. let $c \in [a,b]$ be an endpoint of one of the subintervals. If $c$ is not an endpoint, we can refine the partition a point $x_j^k$ at $c$. Then we have:
$$ \int_a^c f(x)\,dx = \sum_{i=1}^j f(x_i^{k*})(x_i^k - x_{i-1}^k) \quad\text{and}\quad \int_c^b f(x)\, dx = \sum_{i=j+1}^N f(x_i^{k*})(x_i^k - x_{i-1}^k) $$
Adding the two partial sums together, we have $$\sum_{i=1}^N f(x_i^{k*})(x_i^k - x_{i-1}^k) =\int_a^b f(x)\, dx $$
Needing Work: I am not sure I can make the WLOG argument I made, but I'm not sure how else to introduce $c$. And it just seems a little lacking.
AI: Just note that $f$ is continuous, so that the sum converges with respect to refinement of partion. That is, you can add a point $c$ to your partition, this will be closer to the limit state, and the limit(ie, the integral) will not change. |
H: Why is a random walk a time-homogeneous Markov process?
Why is a random walk on $\mathbb{R}^d$ (see below) a time-homogeneous Markov process? Specifically, why does it satisfy requirement #2 of definition 17.3 that the map $\left(x,B\right)\mapsto\mathrm{P}_x\left[X\in B\right]$ be a stochastic kernel?
Relevant Definitions
Let $Y_1,Y_2,\dots$ be i.i.d.$\mathbb{R}^d$-valued random variables and let
$$S_n^x=x+\sum_{i=1}^n Y_i\space\space \mathrm{for}\space x\in\mathbb{R}^d\space\mathrm{and}\space n\in\mathbb{N}_0$$
Define probability measures $\mathrm{P}_x$ on $\left(\left(\mathbb{R}^d\right)^{\mathbb{N}_0},\left(\mathcal{B}\left(\mathbb{R}^d\right)^{\otimes\mathbb{N}_0}\right)\right)$ by $\mathrm{P}_x=\mathrm{P}\circ\left(S^x\right)^{-1}$. Then the canonical process $X_n:\left(\mathbb{R}^d\right)^{\mathbb{N}_0}\rightarrow\mathbb{R}^d$ is a Markov chain with distributions $\left(\mathrm{P}_x\right)_{x\in\mathbb{R}^d}$. The process $X$ is called a random walk on $\mathbb{R}^d$ with initial value $x$. [Klenke, Example 17.5, p. 347]
AI: You have to show that, for each $A$ fixed in the countable product $\sigma$-algebra, the map $x \to \mathbb{P}_x(A) = \mathbb{P}(S^x \in A)$ is measurable.
Since $S^x = S^0 + \underline{x}$, where $\underline{x} = (x, x, ..., x, x, ...)$, we have $\mathbb{P}_x(A) = \mathbb{P}_0(A - \underline{x})$. Setting $A_x = A - \underline{x}$, we have $\mathbb{P}_x(A) = \mathbb{P}_0(A_x) = \mathbb{E}_{\mathbb{P}_0}(\chi_{A_x})$.
Fact : If $(X, \mathcal{A})$ is a measurable space, $(\Omega,\mathcal{T}, \mathbb{P})$ is a probability space, and $f : X \times \Omega \to [0,+\infty]$ is a measurable map, then the map $x \to \mathbb{E}_{\mathbb{P}}(f(x, .))$ is measurable.
(if you want to prove it, it's an application of a Dynkin-class argument : prove it first for characteristic functions of measurable rectangles, and then extend the result in the usual way)
So, we are reduced to prove that the map $(x, \omega) \to \chi_{A_x}(\omega)$ is measurable. But this function is just the characteristic function of the set of all $(x, \omega)$ such that $\underline{x} + \omega \in A$, and since the map $(x, \omega) \to \underline{x} + \omega$ is measurable (exercise !), we are done. |
H: Solving a system of linear congruences in 2 variables
Given:
$$6x+7y \equiv 17 \pmod{42} \tag1$$
$$21x+5y \equiv 13 \pmod{42} \tag2$$
Here's my initial attempt at solving the above system.
$(2) \times 35$: $$21x+7y \equiv 35 \pmod{42} \tag3$$ $(3)-(1)$: $$15x
\equiv 18 \pmod{42}$$ $$5x \equiv 6 \pmod{14}$$ $$x \equiv 4
\pmod{14}$$ $$x \equiv 4,18,32 \pmod{42} \tag4$$ Substitute $(4)$ into
$(2)$: $$5y \equiv 13 \pmod{42}$$ $$y \equiv 11 \pmod{42}$$ Hence the
solutions in $\mathbb Z_{42}$ are $(4,11), (18,11), (32,11)$. I know this is correctly the solution set because the answers work, and because I've been told the system has 3 solutions.
Then I tried substituting $(4)$ into $(1)$, and also into $(3)$, and each time I got
$$7y \equiv 35 \pmod{42}$$
$$7y \equiv 35 \pmod{42}$$
$$y \equiv 5,11,17,23,29,35,41 \pmod{42}$$
Now, I don't understand why substituting $(4)$ into $(1)$ (or $(3)$) instead of into $(2)$ created excess solutions. I would really appreciate it if someone could take a look and explain it to me..thanks!
AI: Equation 1. and 3. don't give you enough information to identify $y$, since you can only solve for the expression $7y$ and $7$ isn't a unit modulo $42$. |
H: On the solutions of $y''(t)=f(y(t))$.
This topic has been inspired by some time spending on trying to refining my knowledge about PDE's in general. Then everybody knows how to solve $$y''(t)=\pm y(t).$$ Then I tried to slightly modify the question and I focused on $$(\therefore)\;y''(t)=f(y(t)),\; f\in C^1(\mathbb R,\mathbb R).$$ What I was trying to do was to derive some general properties about the solutions to this equation. In particular I ask to you, since I was not able to answer myself:
Must a solution of $(\therefore)$, not identically zero, have necessarily a finite number of zeroes on $[0,1]$? My idea was to derive an estimate like
$$|y(\eta)-y(\xi)|\leq C|\eta-\xi|^p,\; p>1;$$ Moreover, if a solution $y$ were to have an infinite number of zeroes in $[0,1]$, the the set of zeroes should have an accumulation point, and in this point all the derivatives should be equal to $0$ by continuity, then maybe the function should remain to much squeezed to be different from zero.
Hope you can help me because this interests me a lot.
Many thanks for your attention.
(I tried to post this on mathlinks as well but nobody answered me yet)
AI: This is a familiar type of ODE. In the following I shall deal with the "purely formal" aspect of it. Questions of sign under the square root or behavior near special points have to be treated "at runtime".
Let $F$ be a primitive of the given function $y\mapsto f(y)$. Multiplying the given differential equation by $y'(t)$ we get
$$y''(t)y'(t)=f\bigl(y(t)\bigr)\,y'(t)$$
or
$${d\over dt}\left({1\over2} y'^2(t)-F\bigl(y(t)\bigr)\right)=0\ .$$
Therefore there is a constant $C$ such that
$$y'(t)=\sqrt{2F\bigl(y(t)\bigr)+C}\ .$$
Now the variables can be separated:
$${dy\over\sqrt{2F(y)+C}}= dt\ .$$
This shows that ODEs of the considered type can be solved by quadratures. (The trick of multiplying by $y'$ is absolutely standard in mechanics, where it leads to the principle of "conservation of energy".) |
H: Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$
I want to prove the following inequality :
$$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab, $$
$a,b,c,d \in \mathbb{R}_{+} .$
In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.
Thanks :)
AI: Also notice that:
$$(a+b+c+d)^3 - 16(abc+abd+acd+bcd) = (a+b+c+d)(a+b-c-d)^2 + 4(c-d)^2(a+b) + 4(a-b)^2(c+d) \ge 0$$
Or
$$(a+b)[(a+b-c-d)^2 + 4(c-d)^2] + (c+d)[(a+b-c-d)^2 + 4(a-b)^2]\ge0$$
This way is suggested by a friend of mine. |
H: Historical Development of CW Complexes
I recently started learning about CW complexes. Although my understanding of them is somewhat nascent, I see that one can deduce a number of useful properties of a space if one can show it is a CW complex. For example, if $X$ is a CW complex, then we can immediately conclude that $X$ is paracompact. While I see the utility of CW complexes, I still do not quite see where the definition of a CW complex comes from. In other words, I don't yet see what motivated the development of CW complexes.
My question is the following:
Are there any sources out there which discuss the historical development of CW complexes and the motivation behind their development?
Note: I am aware that History of Topology by I.M. James contains a chapter entitled “Development of the Concept of a Complex” but I away from my university for the next month and thus do not have access to the book.
AI: To give more details than Miha, the two papers are
(CHI) Whitehead, J. H.C., Combinatorial homotopy. I. Bull. Amer. Math. Soc. 55 (1949) 213–245.
(CHII) Combinatorial homotopy. II. Bull. Amer. Math. Soc. 55 (1949) 453–496.
It was the first paper that developed the notion of CW-complex, and proved their most used properties.
However the roots of these papers go back to very original earlier papers of his developing what is now called “Simple homotopy type”, particularly
Whitehead, J. H.C. On incidence matrices, nuclei and homotopy types.
Ann. of Math. 2 42 (1941) 1197–1239.
which introduced the notion of a “membrane complex”, which is basically the notion of a space obtained by attaching cells. But the development of the notion of adjunction space needs a separate account! This last paper was rewritten by Whitehead, using (CHI), (CHII), to become
(SHT) Simple homotopy types.
Amer. J. Math. 72 (1950) 1–57.
which became a foundation paper for algebraic K-theory.
The key aspect of a CW-complex $X$ is that you can develop properties by induction on the skeleta $X^n$; for example the topology on $X$ is arranged so that a map $f: X \to Y$ is continuous if and only if the restrictions $f|X^n$ are continuous for all $n$.
It should be useful for students to study adjunction spaces and the case of finite cell complexes from the book Topology and Groupoids.
The paper (CHII) is no less original, but has not been so extensively used; it is I believe important for the future. It contains a theorem determining $\pi_2(X \cup _\lambda\{e^2_\lambda\},X,x)$ as a free crossed $\pi_1(X,x)$-module; this theorem is sometimes stated but rarely proved in texts on algebraic topology. Some main ideas of the book Nonabelian Algebraic Topology (EMS,2011) derive from and considerably generalise this result and put many other results of CHII, e.g. on relations with chain complexes with operators, in a wider context.
Later: Relevant to this paper is the following:
Ellis, G.J. "Homotopy classification the J.H.C. Whitehead way".
Exposition. Math. 6(2) (1988) 97--110,
which shows how the work of (CHII) includes work published later by P. Olum and others. |
H: How many times does $3$ appear in the $\{3,6,9,\ldots,1002\}$?
A set $S$ is defined as $S = \{3,6,9,\ldots,1002\}.$
How many times does digit three appear in the decimal representations of members of $S$?
I have solved this question by seeing pattern like $3, 6, 9$, but is there a general solution?
AI: Hint: how many numbers are in the set? How many 3's appear in the units place? In the tens place? |
H: A modification of the harmonic series that causes it to converge.
Possible Duplicate:
sum of inverse of numbers with certain terms omitted
We are working in base $b$, a positive integer greater than $1$. Let $S$ be the set of all positive integers that contain a given symbol in their base $b$ expansion. Let's choose the symbol to be $1$, (because every base we are considering uses this symbol). Show that the following series converges:
$$\sum_{n\in({\mathbb{Z}^+}-S)}^\infty \frac{1}{n}$$
For example, if we look at base 10, the sum would start like
$$\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{20}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}+\frac{1}{32}+...$$
Full disclosure: I got this question from my math class. I have a solution to it (largely inspired by the methods of others in my class), which I will post after awhile. I'm interested in seeing different solutions, as is discussed here
http://meta.math.stackexchange.com/questions/4232/is-it-okay-to-ask-mathematical-puzzles-and-problems-i-have-solved
AI: There are $(b-2)(b-1)^{k-1}$ elements $n$ which aren't in $S$ and have exactly $k$ digits, and for each of these you have $\frac{1}{n} \le \frac{1}{b^k}$. So you have an absolutely convergent majorant: $$\sum_{n \notin S} \frac{1}{n} \le (b-2)\sum_{k=1}^{\infty} (\frac{b-1}{b})^k < \infty$$ |
H: Advice for study plan.
I'm studying math by myself, and now I'm wondering what is the most appropriate for me to study. I think my interest goes to rather pure mathematics than applied math or math for engineers.
So far I finished studying Thomas' Calculus (12th Edition), and I'm basically done with reading the first half of Baby Rudin. In the fall semester, I'm thinking of studying the second half of that and also starting something else together with that. The followings are some options that I'm considering.
Linear Algebra.
Algebra (These two seem to be very important in mathematics)
Topology (I kinda liked the chapter of the basic topology in Baby Rudin)
Abstract Algebra (I found it very powerful in problem-solving while preparing for Putnam. However, I don't know if I can study it or not.)
I also need some recommendations for textbooks. I kinda liked Rudin's style, the repetition of Definition, Theorem and proof. Though it is really tough, I like to fill in the gaps and read between the lines. Last but not least, I prefer old books like Rudin to new books.
Any advice and recommendation are welcome :)
AI: My favorites are:
for linear algebra: Axler's "Linear Algebra Done Right"
here is a link to an MIT class that uses it with problems, etc.
http://math.mit.edu/~trasched/18.700.f11/index.html
and for algebra the free Harvard video series by Benedict Gross:
http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra
This follows Artin's "Algebra" which I think gives a very intuitive presentation which is yet rigorous. But for material after that I would switch to "Dummit and Foote"
These are all highly regarded and are widely used. They can give you a good basis for further study. |
H: software tool for accurate visualization of algebraic curves
First of all, I apologize since this is not strictly speaking a "mathematical" question but I could not find a better place for it.
For a work presentation I need a tool for accurate visualization of implicit
algebraic curves in 2D. I tried ImplicitPlot function from Maple but for some complicated
curves the results are not satisfiable. Also, it is not very convenient
to zoom at certain locations along the curve.
I appreciate if someone can recommend me a good software tool for curve visualization
thanks
PS. One of the equations I need to visualize is the following:
(1-(1/5000000000000000)*x-(1/5000000000000000)*y-2*x^2+(1/5000000000000000)*x^3-(1/10000000000)*y^6+x^4-(1/10000000000000000)*y^5-(1/10000000000000000)*x^5+y^4-(1/10000000000)*x^6+(1/5000000000000000)*y^3-2*y^2+(1/1666666666666666)*x*y^2+(1/1666666666666666)*x^2*y+10*y^2*x^2-4*x*y+(5999999/1000000)*x*y^3+(5999999/1000000)*x^3*y+6*y^4*x^4+(3999999/1000000)*y^5*x^3+(499999/500000)*y^6*x^2-(3999999/500000)*y^4*x^2-(999999/500000)*y^5*x-(1/3333334444444814)*y^4*x-(1/1000000)*y^7*x+(3999999/1000000)*x^5*y^3-12*x^3*y^3-(1/2500000625000156)*x^2*y^3+(499999/500000)*y^2*x^6-(3999999/500000)*y^2*x^4-(1/2500000625000156)*y^2*x^3-(999999/500000)*x^5*y-(1/3333334444444814)*x^4*y-(1/1000000)*x^7*y)=0
AI: Check out our xAlci curve renderer at: http://exacus.mpi-inf.mpg.de/cgi-bin/xalci.cgi. It has Flash interface and provides geometrically-correct rasterization of algebraic curves (i.e. guarantees a fixed Hausdorff distance).
You can also save computed images in png format.
I took the liberty to visualize the equation you provided:
If you zoom in at one of the self-intersection points, you will see that
each branch is composed of a pair of curve arcs: |
H: Manifold/Topology Notation
I have a basic notation related doubt as follows:
Let $M\subset \mathbb{R}^N$ be a manifold. What does $C^\infty(M)$ denote in $f \in C^\infty(M)$?
AI: Each $m \in M$ lies in some coordinate chart $(U, \phi)$, where $\phi: U \to U' \subset \mathbb{R}^k$ is a homeomorphism ($k$ may be less than $N$)
Then $f \in C^\infty(M)$ means that $f \circ \phi^{-1}: U' \to \mathbb{R}$ is smooth in the usual sense (ie, infinitely differentiable) for all charts containing $m$, for all $m \in M$.
(There must be some compatibility condition on the charts as well, but this is not what you are asking.) |
H: How do you pronounce the inverse of the $\in$ relation? How do you say $G\ni x$?
If I am talking about sets $G$ and $H$ and I want to say in words that $G\subset H$, I, like everyone else, will say that $G$ is contained in $H$, or that $H$ contains $G$.
But if I am talking about a set $G$ and a single point $x$, I get vaguely uneasy if I say that $x$ "is contained in" $G$ or that $G$ "contains" $x$. The uneasiness is connected to the idea that it would only be correct to say that $\{x\}$ is contained in $G$, and that it is an abuse of terminology to say the same of $x$ itself.
An alternative is to say that $x$ "is an element of" $G$, which I think is quite standard. But this only avails if I want to mention $x$ first. Sometimes the prose works better to put $G$ first, and this is where my problem arises.
Since "$G$ contains $x$" makes me uneasy, and "$G$ contains $\{x\}$" seems circumlocutory, I will often say that "$G$ includes $x$".
Sometimes I will even do this when $x$ comes first, and say that "$x$ is included in $G$" as a synonym for "$x$ is an element of $G$".
Is this some crazy thing that I made up myself, or is it common usage that I have unconsciously absorbed from the literature? Does everyone else say "$G$ contains $x$" in this case, or do others feel a similar unease about it?
[I should clarify that I'm not only interested in how to say this orally, but also in writing.]
AI: Paul Halmos in How to write mathematics
suggests to distinguish between "$G$ contains $x$" and "$G$ includes $x$", the former meaning $x\in G$ and the latter $x\subseteq G$. Mark seems to have the opposite intuition about "contains" and "includes". This shows that Halmos's idea apparently has not caught on.
On the other hand, it is rare that it is not absolutely clear from the context whether "$G$ contains $x$" means $x\subseteq G$ or $x\in G$. And mathematics is usually communicated in writing or spoken language together with something written on the blackboard or on paper. So I think it is ok to say "$G$ contains $x$" for $x\in G$. |
H: Proving an Integral Theorem
Theorem: Let $f$ be a continuous function on $[a,b] \in \Bbb R$. Prove that there exists a $c \in [a,b]$ so that $$ f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx $$
I cannot use derivatives or any other integral theorems, I am doing these proofs using Riemann Sums and continuity.
My Thoughts:
I am planning on approaching this problem using the following theorem:
Theorem: Let $f$ be a continuous function on the interval $[a,b]$. Then $$\left| \int_a^bf(x)\,dx \right| \ \le \ (b-a)\sup\limits_{[a,b]}|f(x)| $$
Or I could show that this $f(c)$ exists, but then I am unsure as to how I would prove that $c \in [a,b]$. Hints would be appreciated.
Edit: So I have realized that I have the following inequality for $m := \inf\limits_{[a,b]}f(x)(b-a)$, $M := (b-a)\sup\limits_{[a,b]}|f(x)|$:
$$ m \le \frac{1}{b-a}\left| \int_a^bf(x)\,dx\right| \le M $$
So now how do I relate this to $f(c)$?
Edit 2: I changed the title as upon considering the existing MVT, I don't think this is the same thing.
AI: Let $I=\frac1{b-a}\int\limits_a^bf$. Assume that the conclusion does not hold, then either $f(x)\lt I$ for every $x$ in $[a,b]$, or $f(x)\gt I$ for every $x$ in $[a,b]$ (otherwise, $f(x)\lt I\lt f(y)$ for some $x$ and $y$ in $[a,b]$ and, by the intermediate value theorem, $f(z)=I$ for some $z$ between $x$ and $y$).
Without loss of generality, the first case happens. Then $\int\limits_a^bf(x)\,\mathrm dx\lt\int\limits_a^bI\,\mathrm dx=(b-a)I=\int\limits_a^bf$, this is absurd. |
H: How to show that e.g. $\cos(z)$ is analytic using Cauchy- Riemann differential equations?
How to show that e.g. $\cos(z)$ is analytic using Cauchy-Riemann differential equations [$u_x(x,y)=v_y(x,y)$ and $u_y(x,y)=-v_x(x,y)$]? Do all analytic functions satisfy Cauchy-Riemann differential equations (CRDE)? What is the relationship between analyticity of complex functions and Cauchy-Riemann differential equations? I know that holomorphic (analytic?) functions satisfy CRDE, but are functions that satisfy CRDE always analytic (holomorphic)?
AI: Start by rewriting: if $z$ is complex, then let $z=x+iy$. Then we have the function $\cos(x+iy)$. Now you can expand that with the rule $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$. (Because you'll be left with bits like e.g. $\cos(iy)$, you'll may want to replace these with hyperbolic functions with e.g. $\cos(ip)=\cosh(p)$ and a similar relationship for $\sin$.) You'll be left with some complex function which we'll call $u+vi$ - i.e. let $u$ be the real part and $v$ the imaginary part. It is these $u$ and $v$ you are differentiating in the Cauchy-Riemann equations. |
H: Closed set and Open set
EXERCISE Show that $F$ is closed set $\iff$ if for all ball centered at $x$ contains points in $F$, then $x\in F$.
Thanks by your replies.
DEFINITION Given a metric space $(X,\rho)$ and a point $x$, an open ball about $x$ with radius $\epsilon$ is the set $B(x,\epsilon)=\{y\in X:\rho(x,y)<\epsilon\}$
DEFINITION Given a metric space $(X,\rho)$ and a point $x$ in $X$, a set $N$ is a neighborhood of $x$ if it contains an open ball about $x$.
DEFINITION Given a metric space $(X,\rho)$ and a subset $O$ of $X$, $O$ is open if it is a neighborhood of each of its points.
DEFINITION Given a metric space $(X,\rho)$ and a subset $F$ of $X$, $F$ is closed if it is a $X\setminus F$ is open.
AI: Let $X$ be a metric space.
Suppose $F$ is closed. Suppose $x \notin F$. Then $x \in X - F$. Since $X - F$ is open, $x$ is an interior point. There exists a ball $B$ containing $x$ such that $B \subset X - F$. Hence $B$ does not contain points of $F$.
Suppose $F$ is not closed. Then $X - F$ is not open. Thus there exists a point $x \in X - F$ which is not an interior point. That is, every ball $B$ containing $x$ is not completely contained in $X - F$. Hence there is a point $x$ such that every ball containing $x$ intersects $F$ but $x \notin F$. |
H: In need of tips/suggestions when to add or multiply probabilities
I am having trouble deciding when to add or when to multiply probabilities as in the following example. I know that by constructing Probability tree diagrams we could multiply along branches and add vertically. However I could definitely use more suggestions/tips that
might help me decide when to multiply and when to add probabilities.
A jar contains $4$ black and $3$ White balls. If you reach into the jar and pick two balls simultaneously , what is the probability that one is black and the other is white ?
This is how I am solving the above : Pr(Black from the total 7 balls)=$\frac{4}{7}$
Pr(White from the remaining 6 balls after choosing a Black ball) $= \frac{3}{6}$
So Ans = $\frac{4}{7} \times \frac{3}{6} = \frac{2}{7}$
AI: You add probabilities when the events you are thinking about are alternatives [Reading score 0 goals or 1 goal or 2 goals in their match] - you are looking for "mutually exclusive" events - things which could not happen at the same time (in the same match).
You multiply probabilities when you want two or more different things to happen "at the same time" or "consecutively" [Reading score 1 and Leeds score 1 and Arsenal score 2]. The key thing here is that the events are independent - they do not affect each other, or the second does not affect the first (etc).
In your example, to get a black ball and a white ball you have two "mutually exclusive" possibilities: $1.$ white first, black second; $2.$ black first, white second.
Possibility $1$. You can choose a white first with probability $\frac 3 7$ - and then you have 6 balls left, four of which are black, so $\frac 4 6$ of choosing a black one. These are independent events so multiply to get $\frac 3 7 \times \frac 4 6 = \frac 2 7$.
Possibility $2$. Choosing black first then white gives $\frac 4 7 \times \frac 3 6 = \frac 2 7$.
Adding the two together gives $\frac 4 7$.
It takes some skill and practice to get these right all the time. Time spent working through and understanding key examples is time well spent. |
H: Suppose that $d$ divides $2^{r+1}$ but does not divide $2^r$ or $2$. Show that $d$ must equal $2^{r+1}$.
Let $r$, $d$ be positive integers, with $d>1$.
Suppose that $d$ divides $2^{r+1}$ but does not divide $2^r$ or $2$. Show that $d$ must equal $2^{r+1}$.
AI: By unique prime factorization, $d = 2^k$ for some $1 \leq k \leq r + 1$. If $k \leq r$, then $d$ divides $2^r$. Hence $k = r + 1$ and you have $d = 2^{r + 1}$. |
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