text stringlengths 83 79.5k |
|---|
H: If $S$ consists of units then $S^{-1}R \cong R$
I want to show that if $S$ consists of units then $S^{-1}R \cong R$.
Can you tell me if my proof is correct?
Since $S$ consists of units, $S$ is zero-divisor free and hence $f: R \to S^{-1}R$, $r \mapsto \frac{r}{1}$ is injective. So we have an isomorphism $h: R \to f(R)$.
Now we construct an isomorphism $S^{-1}R \to f(R)$: For this we pick any $s_0$ in $S$ and denote by $u \in R$ its inverse, i.e. $s_0 u = us_0 = 1$. Then the map $g_u : S^{-1}R \to f(R)$, $\frac{r}{s} \mapsto \frac{ru}{1}$ is an isomorphism. It is injective since if $\frac{ru}{1} = \frac{r^\prime u}{1}$ then $ru = r^\prime u$ and since $u$ is a unit, $r = r^\prime$. And it is surjective since for $\frac{r}{1}$ in $f(R)$, $g(\frac{rs}{s}) = \frac{rsu}{1} = \frac{r}{1}$. Hence $R \cong f(A) \cong S^{-1}R$.
AI: The natural injection is the isomorphism, but I think you don't have to work so hard.
Just note that for any $(r,s)$ in $S^{-1}R$, we have $(r,s)=(rs^{-1},1)$.
So, $S^{-1}R=Im(R)\cong R$. |
H: How do I "learn" more difficult algebra?
I do not really understand where I was suppose to learn this kind of stuff. I am always told that my algebra knowledge is the biggest reason I am so bad at math. But I do not understand where I was suppose to learn it. I started math with pre-algebra and I learned basically no algebra in that, just arithmetic. After that was college algebra and I was told constantly that the biggest reason people fail that class is because of a weakness in algebra.
That doesn't make any sense to me, isn't that where I am suppose to learn algebra? Anyways I will have a semester off of math, where is it that I can actually learn algebra? All the math books before pre-algebra are incredibly basic and problems I can mostly do in my head. Where do I learn those incredibly complex trick, formulas to memorize (pascal's triangle stuff, quadratic formula and all that other stuff?
It just seems like people expect you to either be good at algebra or be bad at it.
AI: Key to learning algebra? I think it depends on what you mean by algebra. To me, algebra is just arithmetic where we're allowed to use symbols that aren't exactly numbers (variables), or represent numbers that are difficult to write (e.g. $\sqrt{2}$ and $\pi$.) I say "just arithmetic" because all of the old operations and distribution works exactly the same way. I have found many students struggle because they never even mastered the basic arithmetic rules. I think if they did, doing it with $x$'s would make no difference.
Incredibly basic exercises I obviously have no idea what's going on in your head about these exercises, maybe you're exactly right in all or some of your thinking about algebra problems. The problem with this is that this isn't useful to anyone unless you can write it down.
I'm not saying you are this type of student, but I have seen some students fall to a sort of overconfidence where they think they understand something, but it's clear they don't, because their work and answers are completely wrong. The only antidote for this is practice and validation of your answers. The clearer your written solutions are, the clearer it will be in your mind.
Memorization Memorization is an extremely blunt tool, and yet many students treat it as their main weapon. Memorization is too often used to avoid or delay thinking about things that really ought to be understood. My wife would say I have an awful memory, so I can probably attest that math is not so much about memory.
Certainly there are some things that need to be memorized, and there is no sense in spending time "understanding" them, like adding and multiplying single digit numbers. The quadratic formula on the other hand is a borderline case that is good to have memorized, but it's also good to understand where it comes from (completing the square on a quadratic equation!).
Contrary to popular belief, we use mathematics to simplify problems. I would casually argue that a core tenet of math is: "You just find the right way to look at it (or represent it) so it becomes simple".
Edit Dejan Govc inspired another thought with his comment about a student being uncomfortable with what exactly a variable is. This is true: it makes people uncomfortable when something new/ambiguous is introduced, like this. This is because the student hasn't "made the jump" to that new way of thinking yet. Being able to make these jumps is important, because as they learn to abstract further, they will encounter this feeling of disorientation all the time. The best thing to do is simply admit to yourself that you don't get it totally, but trust that through practice you will finally gain a feel for the idea. |
H: does there exist a continuous function
$A_1=\{ \text {closed unit disk in plane}\}$ $A_2=\{(1,y):y\in \mathbb{R}\}$ $A_3=\{(0,2)\}$
We need to confirm: there exist always a continuous real valued function $f$ on $\mathbb{R}^2$ such that $f(x)=a_j $ for $x\in A_j$ $j=1,2,3$
$1$. Iff atleast two of these number are equal.
$2$. all are equal.
$3$. for all values of these 3 numbers.
$4$.iff $a_1=a_2$
Is some how I need to use Urysohn's lemma here?
AI: As discussed in the comments you need to have $a_1=a_2$ (note that $(1,0)\in A_1\cap A_2$), so the only options are 2) and 4), where 2) is the stronger assumption. We can show however that 4) suffices.
Indeed we may apply Urysohn to the sets $A=A_1\cup A_2$ and $B=A_3$. Both sets are closed and disjoint, moreover $\mathbb R^2$ is normal. If $a_1=a_3$ choose $f\equiv a_1$. So assume $a_1\neq a_3$.
By Urysohn there exists a continuous function $f:\mathbb R^2\to [0,1]$ with $f(A)=0$ and $f(B)=1$. Postcompose this map with the canonical homeomorphism $[0,1]\to [a_1,a_3]$ if $a_1< a_3$ or the strictly decreasing homeomorphism $[0,1]\to [a_3,a_1]$ if $a_3< a_1$. We are done.
Edit: Maybe this is actually a bit of an overkill. Since your sets are given explicitely you can just define $f(x,y)=\begin{cases}a_1& \text{ if $x\leq 1$
}\\ a_3&\text{ if $x\geq 2$
}\\
a_1+(x-1)(a_3-a_1)&\text{ if $1\leq x\leq 2$
}\end{cases}$ |
H: Choosing the sign of the separation constant for a vibrating string
Suppose we have this PDE problem
$$\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}$$
$$\psi(0,t)=\psi(L,t)=0$$
It represents the vibrations of a string tightly stretched between two points.
The standard technique for the solution is separation of variables $\psi(x,t)=T(t)y(x)$, giving the equations $y''=\frac{1}{c^2}\lambda y$ $T''=\lambda T$.
Every text I consulted assumes then that $\lambda<0$ and goes on. They say that the physics of the problem requires the solution to be a combination of sines and cosines. But is there a more rigorous mathematical way to see this? Are we losing possible solutions? What could happen if $\lambda>0$?
AI: It is true that the physics of the problem requires that $\lambda$ be $< 0 $,
but for a more fundamental reason than the technical requirement that the answer involve trigonometeric functions.
The equation $T'' = \lambda T$ says that the acceleration of the string is proportional to the displacement. If the constant $\lambda$ were positive, it would say that the further the string were from the mean position of rest, the faster it would be accelerating, so the string would not be vibrating back and forth, but would rather be zooming away at a faster and faster rate.
The negative sign in $\lambda$ ensures that the string behave like a simple harmonic oscillator: as it is displaced from its mean position, there is a restoring force that pulls it back towards this mean position (rather than accelerating it further and further away from this position), so that it truly vibrates. |
H: completeness under different metric
would any one tell me whether $C[0,1]$ is complete under these metrics
1.sup norm i mean $\|f\|_{\infty}$
2.$\|f\|_{\infty,1/2}=\|f\|_{\infty}+|f(1/2)|$
3.$\|f\|_{2}=\sqrt{\int_0^1|f|^2dx}$
Under supnorm I know it is complete,I am not sure about the other two.
AI: for the second case the norm you asked is equivalent to the sup norm so the normed space it is induced also complete $ |\!|f|\!| _{\infty} \leq |\!| f |\!|_{\infty , \frac{1}{2}} =|\!|f|\!| _{\infty}+ |f(\frac{1}{2}) | \leq 2|\!|f|\!| _{\infty } $ |
H: Solving (or estimating) $x$ in $\tau=\log_x\left(\frac{x+1}{2}\right)$
How would one find a real value for $x$ that satisfies
$$\tau=\log_x\left(\frac{x+1}{2}\right),$$
given $0 < \tau < 1$ and $\tau \neq \frac{1}{2}$
(PS I'm not that good with math, so if this is impossible, please explain it to me like I'm 5). I don't need an exact solution, just a method of estimating one. Thanks!
Context:
I'm trying to create a nonlinear slider for the UI in a program I'm writing. I want to be able to specify the logarithmic scale by specifying the what the output value should be at the midpoint. If I map all the values into a (0,1) range, I need to find an invertible function $f(y)$ such that $0 < \tau < 1$, $f(0) = 0$, $f(1) = 1$, $f(\frac{1}{2}) = \tau$, and $f'(y) > 0$ for all $0 < y < 1$, which also has a nice curve so it doesn't seem jumpy to the user. I realized one such function is:
$$f(y) = \log_\alpha\left(1+y\left(\alpha-1\right)\right)$$
Now I just need to find the base $\alpha$ given the midpoint $\tau$, which is where all this comes in. I'm not 100% sure my reasoning on this is correct, but it seems like a plausible starting point.
AI: There is no algebraic method to find $x$ but we can approximate it well.
Using the change of base rule for logs, we see $$\log_x \left( \frac{x+1}{2} \right) = \frac{ \ln \left( \frac{x+1}{2} \right)}{\ln x}$$ so if we rearrange, we get the equation $$ \tau \ln x - \ln \left( \frac{x+1}{2}\right) =0.$$
The reason I rewrote it in this because now we have a root finding problem, which is well studied. In particular, we want to find a value $x$ such that $f(x)=0$ where $f$ is defined by $f(z) = \tau \ln z - \ln \left( \frac{z+1}{2}\right).$ There are many ways to find the roots of a function, the simplest examples being the Bisection Method and the Newton-Raphson method.
For this particular problem, $f$ is quite well behaved and has a simple derivative, so trying Newton-Raphson isn't a bad idea. We compute $f'(x) =\displaystyle \frac{\tau}{x}-\frac{1}{x+1}.$
So Newtwon-Raphson iteration gives $$x_{n+1}= x_n - \frac{ \tau \ln x_n - \ln \left( \frac{x_n+1}{2}\right)}{\frac{\tau}{x_n}-\frac{1}{x_n+1}} .$$
As for an initial estimate, for $0<\tau<1,$ $x_0=2$ isn't a bad start. So once you specify $\tau$ you can start approximating $x.$ |
H: When weak convergence implies moment convergence?
Given a sequence $(\mu_n)_n$ of probability measures on $\mathbb R$, which converges weakly to a probability measure $\mu$, when do we have
$$
\tag{1} \lim_{n}\int x^kd\mu_n(x)=\int x^k d\mu(x) \qquad \forall k\geq 0\;?
$$
Is "$\mu$ has compact support" a sufficient condition?
Note that $\mu_n$ converges to $\mu$ weakly if
$$ \int \varphi d\mu_n \to \int \varphi d\mu$$
for all $\varphi$ which is continuous and has compact support. Note that $x^k$ are continuous but not of compact support, so (1) is not immediately obvious.
AI: A condition on the limit measure will never be enough.
The sequence $\left(1-{1\over n}\right)\delta_0+{1\over n}\delta_{x(n)}$ converges to $\delta_0$ weakly, but we can make its moments behave horribly by choosing $x(n)$ to be
very large.
A sufficient condition for your moments to converge is if all the $\mu_n$s have the same compact support. |
H: How to compute sample variance from sample moments
Given $X_1, \dots , X_n$ i.i.d. and the two sample moments
$$M_1 = \frac{1}{n} \sum_{i = 1}^{n} X_i = \bar{X}$$
and $$ M_2 = \frac{1}{n} \sum_{i = 1}^{n} X_i^2$$
how can I compute:
$$ S^2 = \frac{1}{n} \sum_{i = 1}^{n} (X_i - \bar{X})^2$$
such as:
$$S^2 = f(M_1, M_2)$$
Thank you.
AI: $S^2 = \frac{1}{n}\sum_{i=1}^n (X_i - M_1)^2 = \frac{1}{n}\sum_{i=1}^n (X_i^2 - 2 X_i M_1 + M_1^2) =$
$= \frac{1}{n}\sum_{i=1}^n X_i^2 - 2 M_1 \frac{1}{n}\sum_{i=1}^n X_i + \frac{1}{n}\sum_{i=1}^n M_1^2 =$
$= M_2 - 2M_1^2 + M_1^2 = M_2 - M_1^2$ |
H: Sum of Gaussian processes
I would like to prove that the sum of Gaussian processes is also Gaussian, to be precise, $M_t=W_t+W_{t^2}$, where $W_t$ is standard Wiener process. That is kind of obvious, but I am looking for some more rigorous, as short as possible proof, other than just saying that it is the sum of two Gaussian processes. I have some thoughts but something seems to be missing there. $W_t$ and $W_{t^2}$ are dependent, but $M_t=2W_t+(W_{t^2}-W_t)$ if $t^2\geq t$, then the very first proof, since they are independent, would be what I am looking for. But as I know there has to be done something different when dealing with processes, not random variables. Random process $X_t$ is Gaussian $\Leftrightarrow \forall n\geq1,t_1,\dots,t_n,\lambda_1,\dots,\lambda_n$ $\sum^n_{k=1}\lambda_k X_{t_k}$ is Gaussian. Then $\sum^n_{k=1}\lambda_k M_{t_k}=$ $\left[\sum^n_{k=1}\lambda_k W_{t_k}\right]+\left[\sum^n_{k=1}\lambda_k W_{t_k^2}\right]$, by the same proposition both terms are Gaussian since $W_t$ and $W_{t^2}$ are Gaussian. $\left[\sum^n_{k=1}\lambda_k W_{t_k}\right]+\left[\sum^n_{k=1}\lambda_k W_{t_k^2}\right]=\left[\sum^n_{k=1}2\lambda_k W_{t_k}\right]+\left[\sum^n_{k=1}\lambda_k \left(W_{t_k^2}-W_{t_k}\right)\right]$ so maybe now I could conclude by using that proof for random variables?
AI: As requested by the OP, my comment has been converted into an answer.
Perhaps you are making this harder than it is. Isn't $(W_{t_1},W_{t_1^2},W_{t_2},W_{t_2^2},\ldots,W_{t_n},W_{t_n^2})$ a Gaussian vector (meaning the $2n$ random variables are jointly Gaussian) and so any linear transformation applied to this vector results in a Gaussian vector? There is no requirement that the $2n$ random variables in question have to be independent for this linear transformation property to hold. It is joint Gaussianity that is required, and joint Gaussianity is guaranteed since the random variables are from a Gaussian process. |
H: Historic literature
I'm wondering if it's advantageous to read the original works of Gauss, Jacobi, Cauchy and others (in particular, Jacobi).
Many people say that it's worth it to read original (not translated) works of literature - you have the author's own diction and get a better feel of his/her cogitation. I wonder if it's the same way for mathematics textbooks - is it worth it to study some Latin and etc., and struggle through strange notation, to read the masters?
Apologies if such a question has been asked already, I didn't find it here.
AI: If you're looking to understand the concepts well, rather than get some historical background, I would actually advise against this. More modern treatments will use up to date language, tie theories in with newer concepts that the original authors didn't know about, and have the exposition guided by understanding the ideas in a wider context.
I even find that for research topics barely a decade old, a survey article written a few years later is often better to learn from than the first paper - by knowing some of the results that follow, the survey article authors can phrase the definitions and exposition in a way that makes them seem natural, in a way that the original authors couldn't.
That's not to say there aren't good reasons to read these older works of course, but I find it's much more helpful to have a reasonable understanding of the wider theory first. |
H: Computing: $L =\lim_{n\rightarrow\infty}\left(\frac{\frac{n}{1}+\frac{n-1}{2}+\cdots+\frac{1}{n}}{\ln(n!)} \right)^{{\frac{\ln(n!)}{n}}} $
Compute the following limit:
$$L =\lim_{n\rightarrow\infty}\left(\frac{\frac{n}{1}+\frac{n-1}{2}+\cdots+\frac{1}{n}}{\ln(n!)} \right)^{{\frac{\ln(n!)}{n}}} $$
I'm looking for an easy, simple solution here, but not sure yet this is possible. Any hint, suggestion along this way is welcome. Thanks.
AI: I'm not sure that I'm right.
First we have
$\sum_{k=1}^n (n+1-k)/k = (n+1)H_n-n$,
so
$$L = \lim_{n\to\infty} \left(\frac{(n+1)H_n-n}{\ln n!}\right)^{\frac{\ln n!}n}$$
Take logarithm, we have $\ln L = \lim_{n\to\infty} A(n)B(n)$, where
$$A(n) = \frac{\ln n!}n = \frac{n\ln n+O(n)}n = \ln n+O(1)$$
and $B(n) = \ln C(n)$ where
\begin{align*}
C(n)
&= \frac{(n+1)H_n-n}{\ln n!} \\
&= \frac{(n+1)(\ln n+\gamma+O(1/n))-n}{n\ln n-n+O(\log n)} \\
&= \frac{n\ln n-(1-\gamma)n+O(\log n)}{n\ln n-n+O(\log n)} \\
&= \frac{1-\dfrac{1-\gamma}{\ln n}+O(1/n)}{1-\dfrac1{\ln n}+O(1/n)} \\
&= \left(1-\frac{1-\gamma}{\ln n}\right)\left(1-\frac1{\ln n}\right)^{-1}\left(1+O(1/n)\right)^2 \\
&= \left(1-\frac{1-\gamma}{\ln n}\right)\left(1+\frac1{\ln n}+O(1/\log n)^2\right)\left(1+O(1/n)\right) \\
&= 1+\frac\gamma{\ln n}+O(1/\log n)^2
\end{align*}
So
$$B(n) = \ln C(n) = \ln\left(1+\frac\gamma{\ln n}\right)+O(1/\log n)^2 = \frac\gamma{\ln n}+O(1/\log n)^2$$
and
$$A(n)B(n)=\gamma+O(1/\log n)$$
Let $n\to\infty$, we have $\lim_{n\to\infty} A(n)B(n)=\gamma$, so $L = e^\gamma$.
The following equations come from Concrete Mathematics, proved by Euler-Maclaurin formula
$H_n = \sum_{k=1}^n 1/k = \ln n+\gamma+O(1/n)$, where $\gamma$ is Euler-Mascheroni constant.
$\ln n! = n\ln n-n+O(\log n)$. (It's really Stirling's approximation) |
H: Equivalent definition of exactness of functor?
I'll use the following definition:
(Def) A functor $F$ is exact if and only if it maps short exact sequences to short exact sequences.
Now I'd like to prove the following (not entirely sure it's true but someone mentioned something like this to me some time ago):
Claim: $F$ is exact if and only if it maps exact sequences $M \to N \to P$ to exact sequences $F(M) \to F(N) \to F(P)$
Proof:
$\Longleftarrow$: Let $0 \to M \to N \to P \to 0$ be exact. Then $0 \to M \to N$, $M \to N \to P$ and $N \to P \to 0$ are exact and hence $0 \to F(M) \to F(N) $, $F(M) \to F(N) \to F(P)$ and $F(N) \to F(P) \to 0$ are exact. Hence $0 \to F(M) \to F(N) \to F(P) \to 0$ is exact.
$\implies$: This is direction I'm stuck with. I am trying to do something like this: Given $M \to N \to P$ exact, we have that $0 \to ker(f) \to M \to im(f) \to 0$ is exact. Hence $0 \to F(ker(f)) \to F(M) \to F(im(f)) \to 0$ is exact. Then I want to do this again for the other side of the sequence and stick it back together after applying $F$ to get the desired short exact sequence.
How does this work? Perhaps I need additional assumptions on $F$? Thanks for your help.
AI: Any exact sequence can be broken down into short exact sequences (the $C_i$ are kernels/images):
So, since your functor $F$ preserves short exact sequences, you can apply $F$ and the diagonal sequences will remain exact. It's now a general fact that in any such diagram, if the diagonals are exact, then the middle terms are exact as well (by diagram chasing).
EDIT: If $f_i\colon A_i\to A_{i+1}$, then $C_i=\ker(f_i)$ which by exactness is isomorphic to $\operatorname{im}(f_{i-1})$. |
H: Is the set of all probability measures weak*-closed?
Let $(\Omega,\Sigma)$ be a measurable space. Denote by $ba(\Sigma)$ the set of all bounded and finitely additive measures on $(\Omega,\Sigma)$ (see http://en.wikipedia.org/wiki/Ba_space for a definition). Is the set of all probability measures $\mathcal{M}_1(\Sigma)\subseteq ba(\Sigma)$ weak*-closed? The weak*-topology on $ba(\Sigma)$ is the weakest topology such that the maps $l_Z:ba(\Sigma)\rightarrow \mathbb{R}$, mapping $\mu\mapsto \int_\Omega Z d\mu$, are continuous for all bounded and measurable maps $Z:\Omega\rightarrow \mathbb{R}$.
AI: No. Take $\Omega=\mathbb N$ and $\Sigma$ the power set. As wikipedia says, then $ba(\Sigma)=ba=(\ell^\infty)^*$. However, the collection of probability measures is just the collection of $(x_n)\in\ell^1$ (as a measures are countably additive) with $x_n\geq 0$ for all $n$, and $\sum_n x_n=1$. This is not weak$^*$-closed in $(\ell^\infty)^*$. For example, any limit point of the set $\{\delta_n:n\in\mathbb N\}$, where $\delta_n\in\ell^1$ is the point mass at $n$, is a member of $ba \setminus \ell^1$. |
H: How many $k+2$ letter groups in a $n$ letter string
Given an $n$ letter string of identical letters, how many $k+2$ letter words can be formed of adjacent letters?
By observing data I came up with n-(1+k), but I'm at a loss for a descent combinatorial explanation.
For example, if I had a 5 letter string and k=1 and I label the letters for clarity: 'abcde' I get 'abc', 'bcd', 'cde'.
AI: If I understand the problem correctly, each word is uniquely determined by the position of its first letter. The rightmost word's first letter will be in the $n-(1+k)$ position, and each letter before that also determines a word, so there are exactly $n-(1+k)$ in total. |
H: Proof that every element of A_5 is an involution or a product of two involutions?
It can be verified with brute force that the alternating group on 5 elements ($A_5$) has the property that every member is either an involution or can be written as the product of two involutions. Is there a simple proof of this fact?
AI: In $A_5$, we have four different cycle structures: the identity $(1)$, the $3$-cycles, the $5$-cycles and products of two disjoint transpositions.
The identity is the product of an any involution with itself. Products of two disjoint transpositions are involutions.
For $3$-cycles: $(123) = (13)(12) = (13)(45)(12)(45)$.
For $5$-cycles: $(12345) = (13)(45)(12)(35)$.
Just in case it isn't obvious, the general case $(abc)$ and $(abcde)$ follows by replacing $1, 2, 3, 4$ and $5$ by $a, b, c, d$ and $e$ respectively. |
H: Convergence of a function series
Check whether function series is convergent (uniformly):
$\displaystyle\sum_{n=1}^{+\infty}\frac{1}{n}\ln \left( \frac{x}{n} \right)$ for $x\in[1;+\infty)$
I don't know how to do that.
AI: The series doesn't converge. Use integral test or Cauchy-condensation-test via monotonicity of the general term |
H: Taking fractions $S^{-1}$ commutes with taking intersection
Let $N,P$ be submodules of an $R$-module $M$ and let $S$ be a multiplicative subset of $R$. I think I proved $S^{-1}(N \cap P) = S^{-1}N \cap S^{-1} P$ but since my proof is not the same as the one given in Atiyah-MacDonald on page 39 I suspect there is something wrong with it. Can you tell me please what's wrong here:
Claim: $S^{-1}(N \cap P) = S^{-1}N \cap S^{-1} P$
Proof:
$\frac{m}{s} \in S^{-1}N \cap S^{-1} P \iff$ $\frac{m}{s} \in S^{-1}N$ and $\frac{m}{s} \in S^{-1}P \iff m \in N$ and $m \in P \iff m \in N \cap P \iff \frac{m}{s} \in S^{-1}(N \cap P)$.
AI: In your notation, it doesn't have to be the case that $m \in N$. There just needs to be an $s' \in S$ such that $s'm \in N$. For example, take $R = M = \mathbf Z$, $N = 6\mathbf Z$, and $S = \{1, 2, 2^2, \ldots\}$. But having made this change, I think you can complete your proof. |
H: Multiplicative Selfinverse in Fields
I assume there are only two multiplicative self inverse in each field with characteristice bigger than $2$ (the field is finite but I think it holds in general). In a field $F$ with $\operatorname{char}(F)>2$ a multiplicative self inverse $a \in F$ is an element such that
$$ a \cdot a = 1.$$
I think in each field it is $1$ and $-1$. Any ideas how to proof that?
AI: The equation $x^2-1$ is degree $2$ and thus can have at most two solutions in any field. So checking that $1$ and $-1$ satisfy this is enough to know that they are the only self-inverse elements. (As Nate points out, in the field of characteristic $2$ they are also equal to each other, so there is only one self-inverse element in this case). |
H: Proof of a test for series
I would like to prove that given three sequences ${a_n}, {b_n}\text{ and }{c_n}$ and knowing that:
They aren't necessarily of positive terms.
$a_n \leq b_n \leq c_n, \forall n \geq 1$
$$\text{If }\sum_{n = 1}^{+ \infty}{a_n}\text{ and }\sum_{n = 1}^{+ \infty}{c_n}\text{ are both convergent then }\sum_{n = 1}^{+ \infty}{b_n}\text{ converges and}$$
$$\sum_{n = 1}^{+ \infty}{a_n} \leq \sum_{n = 1}^{+ \infty}{b_n} \leq \sum_{n = 1}^{+ \infty}{c_n}$$
However I'm having difficulties. This is what I've done so far:
$$\text{We have that } \sum_{n = 1}^{+ \infty}{a_n}\text{ converges so we know that } \lim_{N\to{+ \infty}}{A_N} = L$$
$A_N$ is the sequence of partial sums.
$$\text{The same holds for }\sum_{n = 1}^{+ \infty}{c_n}\text{;} \lim_{N\to{+ \infty}}{C_N} = L'$$
$$\text{So }L \leq \lim_{N\to{+ \infty}}{B_N} \geq L'$$
Because the limit of $B_N$ is between $L$ and $L'$ then I can say that $\sum_{n = 1}^{+ \infty}{b_n}$ converges.
The thing is I'm not sure about the assertion, moreover the proof looks easy this way, which makes me suspect.
$$\text{The other thing is: because }\sum_{n = 1}^{+ \infty}{a_n}\text{ and }\sum_{n = 1}^{+ \infty}{c_n}\text{ are both convergent, then }\lim_{n\to{+ \infty}}{a_n} = \lim_{n\to{+ \infty}}{c_n} = 0\text{, so by the sandwich principle }\lim_{n\to{+ \infty}}{b_n} = 0\text{, but that of course doesn't allow me to assert that }\sum_{n = 1}^{+ \infty}{b_n}\text{ converges.}$$
Hope you could help me. Thanks.
AI: We have $0\le b_n-a_n\le c_n-a_n$. The series $\sum_{n=1}^\infty(c_n-a_n)$ is convergent because both $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty c_n$ are; moreover, it's terms are positive. The comparison test for series of positive terms proves that $\sum_{n=1}^\infty(b_n-a_n)$ is convergent. Since $b_n=a_n+(b_n-a_n)$, it follows that $\sum_{n=1}^\infty b_n$ is convergent. |
H: About the definition of Cech Cohomology
Let $X$ be a topological space with and open cover $\{U_i\}$ and let $\mathcal F$ be a sheaf of abelian groups on $X$. A $n$-cochain is a section $f_{i_0,\ldots,i_n}\in U_{i_0,\ldots,i_n}:= U_{i_0}\cap\ldots\cap U_{i_n}$; we can costruct the following abelian group (written in additive form):
$$
\check C^n(\mathcal U,\mathcal F):=\!\!\prod_{(i_o,\ldots,i_n)}\!\!\mathcal F(U_{i_0,\ldots,i_n})
$$
Now my question is the following:
we consider oredered sequences $(i_o,\ldots,i_n)$ ? Because in this case in the direct product we have each group repetead $(n+1)!$ times, that is the number of permutations of the set $\{i_o,\ldots,i_n\}$.
AI: Have a look at : http://math.stanford.edu/~conrad/papers/cech.pdf
There are three complexes which are homotopic, and so, induce the Čech Cohomology :
1) Without ordering the open sets : the Čech complex of singular cochains : $C^n(\mathcal U,\mathcal F)=\displaystyle\prod_{i_0,\ldots,i_n}\mathcal F(U_{i_0,\ldots,i_n})$.
2) Considering alternating open sets : the Čech complex of alternating cochains : $C^n_{\mathrm{alt}}(\mathcal U,\mathcal F)$ where $\omega_{\varphi(i_0,\ldots,i_n)}=\varepsilon(\varphi)\omega_{i_0,\ldots,i_n}$ where $\varphi$ where $\varphi$ is in the symmetric group $\mathfrak S_n$ and where $\varepsilon(\varphi)$ is the sign of $\varphi$.
3) Taking care of the order : the Čech complex of ordered cochains : $C^n_<(\mathcal U,\mathcal F)=\displaystyle\prod_{i_0<\ldots<i_n}\mathcal F(U_{i_0,\ldots,i_n})$, for a total order $<$ on $I$ where $\{U_i\}_{i\in I}$.
As shown in this link, these complexes induce the same cohomology which is the usual Čech cohomology.
The advantage of the complex of alternating cochains over singular cochains is that we can easily use refinements for $\mathcal U$ (for the inductive limit) because we doesn't need the order.
The advantage of the complex of singular cochains over alternating cochains is that we can use non-injective refinements.
An other inconvenient of the ordored cochains is that we need a total order $<$. |
H: Lower bound on Tail Probabilities
Inequalities such as Markov's and Chebyshev’s provide upper bounds on tail probabilities. Are there similar inequalities that give lower bounds in the form $P(X \geq \alpha)>\theta$?
AI: Markov's inequality is also called the first moment method. What you want is the second moment method using bounds for the two first moments to derive the desired inequality :
http://en.wikipedia.org/wiki/Second_moment_method |
H: Define a logical formula as another formula
I'm reading Dirk van Dalen's Logic and Structure and noticed that in many parts of his book he defines some formula to be an alias for another formula (he doesn't use the name alias, he just says that some formula will be defined as another formula).
Example Let $\phi$ be a formula of the system and $t$ a term that receives two parameters. Define $\phi$ such that $\phi(n) := t(n, x)$.
What definition in predicate logic allows one to do that? What does this 'define as' stands for? Is it the same of affirming $\phi(n) \leftrightarrow t(n, x)$? Or is it some meta-logical construct?
AI: The definition is not something that happens in predicate logic.
The formal system itself does not know anything about the letter $\phi$. Its only symbols are the primitive ones: variable letters, constant letters, function letters, predicate letters, the various logical connectives quantifiers, and whatever punctuation (such as parentheses) one needs to put them together.
When the author defines a meaning for $\phi$, that happens outside the formal system, and merely provides him with a convenient way of speaking about particular strings of primitive symbols in a succinct and systematic way. So each time a wff involving $\phi$ appears on the page, the reader is to understand that what actually is seen by the core definitions of the formal system is the string of primitive symbols that the letter $\phi$ stands for, not the letter $\phi$ itself.
This is not very different from what happens in elementary algebra. When we set forth a fact such as $a+b=b+a$, one might ask what definition in arithmetic allows one to add the letters $a$ and $b$. But there is none; the variables live outside basic arithmetic and are never really seen by the "addition" operation. However, when we substitute whatever concrete numbers the $a$ and $b$ represent, the formula is going to represent an arithmetic calculation that can actually be carried out. |
H: Number of terms in an Arithmetic progression
1 and 20 are first and last terms of the arithmetic progression. If all the terms of this arithmetic progression are integers, then find the different number of terms that this arithmetic progression can have ?
AI: Here's an obvious generalization which may be interesting to you, and which exposes the reasoning for the question above: $1$ and $(1+p)$ are the first and last terms of an arithmetic progression, where $p$ is prime. If all the terms of this arithmetic progression are integers, how many different number of terms can this arithmetic progression have?
See if you can generalize Henning's comment regarding your original question which was that the number of terms $k$ with step size $s$ must satisfy $1+s(k-1)=20$ in order to answer this question as well. |
H: Normal subgroups of $S_4$
Can anyone tell me how to find all normal subgroups of the symmetric group $S_4$?
In particular are $H=\{e,(1 2)(3 4)\}$ and $K=\{e,(1 2)(3 4), (1 3)(2 4),(1 4)(2 3)\}$ normal subgroups?
AI: In any group, a subgroup is normal if and only if it is a union of conjugacy classes.
In $S_n$, the conjugacy classes are very easy: a conjugacy class consists exactly of all permutations of a given cycle structure. These corresponds to all possible partitions of $n$.
So, consider $S_4$. The conjugacy classes in $S_4$ are:
The class of the $4$-cycles, cycle structure $(abcd)$, corresponding to the partition $4$. There are $6$ elements in this class.
The class of the $3$-cycles, cycle structure $(abc)(d)$, corresponding to the partition $3+1$. There are $8$ elements in this class.
The class of the product of two transpositions, cycle structure $(ab)(cd)$, corresponding to the partition $2+2$. There are $3$ elements in this class.
The class the transpositions, cycle structure $(ab)(c)(d)$, corresponding to the partition $2+1+1$. There are $6$ elements in this class.
The class of the identity, cycle structure $(a)(b)(c)(d)$, corresponding to the partition $1+1+1+1$. There is a single element in this class.
Now, any subgroup that contains all transpositions is the whole group.
So we can consider only subgroups that don't contain the transpositions. Their order must be a divisor of $24$, and since it does not have the transpositions, it is at most $12$. So the order must be $1$, $2$, $3$, $4$, $6$, or $12$. Moreover, the order must the be sum of the sizes of some conjugacy classes, so it must be a sum of some of the numbers $1$, $3$, $8$, and $6$, and must include $1$.
One possibility is the trivial group, order $1$. We cannot get a normal subgroup of orders $2$ or $3$ (in particular, you $H$ cannot possibly be normal). The only way to get a subgroup of order $4$ is to take the class of the identity and the class of the product of two transpositions. This is your $K$; if it is a subgroup, then being a union of conjugacy classes shows that it is normal. So just check if it is a subgroup.
We cannot get a normal subgroup of order $6$, because we can't just take the conjugacy class of $4$-cycles (we need the identity). As for a subgroup of order $12$, we would need to take the identity ($1$ element), the class of products of two transpositions ($3$ elements), and the class of $3$-cycles ($8$ elements). This collection has a very familiar name...
And that's it! You cannot have any other normal subgroups. So, in summary: the trivial group, the whole group, and possibly $K$ (if it is a subgroup), and possibly this last collection (if it happens to be a subgroup). At most $4$, at least $2$. |
H: Under what circumstances is the discrete metric space separable?
Under what circumstances is the discrete metric space separable?
Can anyone help me please?
AI: A space $X$ is separable if it contains a dense countable subset $D$.
Now that we know the definition we need to think about what it means for $D$ to be dense in a discrete space. Dense means that if we pick any point $x$ in $X$ and an open set $O$ containing it, then $O$ will intersect with $D$.
In a discrete space, the singleton set $\{x\}$ is open. The only way this set can have non-empty intersection with $D$ is if we have $x \in D$.
But this means that the only dense subspace of a discrete space $X$ is $X$ itself. Hence, the only way to have a countable dense subset of a discrete space is if the space itself is countable. |
H: How to graph this?
I have a non-right triangle. I will call
the bottom or base edge $b$,
the top left edge $a$,
the top right edge $c$.
Let $ a=c+2 $ and $b=10$. How do I graph a curve where the graph $ x $ and $ y$ coordinates represent the vertex where edges $ a $ and $c$ meet?
AI: Taking the origin as the intersection of $a$ and $b$, with the equation in terms of the angle $\theta$ at that vertex:
Through the cosine law we have:
$$(a-2)^2=a^2+100-20a\cos\theta$$
$$-4a+4=100-20a\cos\theta$$
$$a=\frac {96}{20\cos\theta-4}$$
With our particular setup, $a$ is the radius. So this is a polar equation, where
$$r(\theta)=\frac{96}{20\cos\theta-4}=\frac{24}{5\cos\theta-1}$$
we can convert to cartesian coordinates if you want:
$$5r\cos\theta-r=24$$
$$(5r\cos\theta-24)^2=r^2$$
$$(5x-24)^2=x^2+y^2$$
$$y^2=24x^2-240x+576$$
$$y^2=24(x-5)^2-24$$
$$(x-5)^2-\frac{y^2}{24}=1$$
which is a hyperbola. You want the rightmost branch (so $a>0$), which is valid when $\cos\theta<\frac 1 5$.
EDIT: picture time! |
H: Evaluation of $\lim\limits_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$
One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?
$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$
AI: The statement $\dfrac{\tan(x)-x}{x^3} \to c$ as $x \to 0$ is equivalent to
$\tan(x) = x + c x^3 + o(x^3)$ as $x \to 0$, so this is a statement about a
Taylor polynomial of $\tan(x)$, and I'm not sure what would count as doing
that "without Taylor". However, one thing you could do is start from $$\sin(x) = x + o(x)$$ integrate to get $$\cos(x) = 1 - x^2/2 + o(x^2)$$ then $$\sec(x) = \frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$$ $$\sec^2(x) = \left(1 + x^2/2 + o(x^2)\right)^2 = 1 + x^2 + o(x^2)$$ and integrate again to get
$$\tan(x) = x + x^3/3 + o(x^3)$$ |
H: $N$ is a matrix such that $N^3=0$
Given a $3\times 3$ matrix $N$ such that $N^3=0$, then which of the following are/is true?
$N$ has a non zero eigenvector
$N$ is similar to a diagonal matrix
$N$ has $3$ linearly independent eigenvector
$N$ is not similar to a diagonal matrix
Well, eigenvalues of $N$ are all zeroes and characteristic polynomial is $x^3=0$, clearly not diagonalizable. so only $1$ is true.
AI: "Clearly not diagonalizable" is not correct; if we know that $N^2\neq 0$, then you are correct (that would imply that the minimal polynomial of $N$ is also $x^3$, and since it is not square free then $N$ is not diagonalizable). But just from knowing that $N$ has characteristic polynomial $x^3$, we do not know whether $N$ is diagonalizable or not. It could be diagonalizable. Explicitly, $N$ is diagonalizable if and only if it is the zero matrix; prove it!
As noted, you cannot have 2 and 4 both false, since they are negations of each other and the excluded middle applies here. And 2 and 3 are logically equivalent for a $3\times 3$ matrix.
If the question explicitly states that $N^2\neq 0$, then you know that 2 is false. If the question does not explicitly state so, then you don't. |
H: $f(1)=-3$ and $ f'(x)\geq7$ how small is $f(5)$?
Here is a test question in my textbook.
Suppose : $f(1)=-3$ and $f'(x)\geq7$ How small $f(5)$ can be possibly
:
a)$25$
b)$-21$
c)$28$
e)$31$
f) None of others.
I just have this : because $f'(x)>0$ then $f(x)$ increasing. So, for all $x$ greater than $1$, $f(x)$ will greater than $-3$.
After this conclusion, I don't have any idea.
Because of the above conclusion, I post a,b,c. Because (I think) this problem will not have a fixed solution. So, we should choose which solution is the most approximate.
AI: A previous answer uses an intuitive "let's use a linear function with a minimal derivative" solution.
For a rigorous argument, one can use the mean value theorem (in undergraduate mathematics, this is considered a "basic" theorem). This theorem says that, if $f$ is continuously differentiable*, then there must be an $x\in[1,5]$ (that is, an $x$ between 1 and 5) such that:
$$f'(x)=\frac{f(5)-f(1)}{5-1}$$
First note that I do not know what $x$ is. It is definitely in the range $[1,5]$, but the mean value theorem doesn't tell us what it is.
Also notice that I'm using a strict equality here - mathematicians like that, because we can now solve for $f(5)$ (which is good, to say the least):
$$f(5)=4f'(x)+f(1)$$
But now it is easy! If $f'(x)\geq7$ (for /any/ $x$), then $4f'(x)\geq28$, so $4f'(x)+f(1)\geq28+f(1)=25$, thus answer a) would be correct.
*There are many functions which are not continuously differentiable, but we usually do not consider them. Also, the question implicitly stated that the second derivative exists for all $x$ (namely, for all $x$ it has some value greater than or equal to 7), which is sufficient for the mean value theorem. |
H: Matrix Multiplication and Function Composition
Given the vector space $F^n$ and two linear function $T,S:F^n \rightarrow F^n$ is it true that multiplying the representative matrices according to the standard basis of $T$ and $S$ is equivalent to the composition of their explicit formula's? I.E. Given a vector v in $F^n$ is $[T]_E[S]_Ev = [TS]_Ev = T(S(v)) $ ?
AI: I'm reading this question as: "Does matrix multiplication really give rise to the matrix for the composition of maps that the matrices represent?" I'm always using matrices operating on the left of column vectors, in my notation below.
Define $\pi_i:F^n\rightarrow F$ by projecting onto the $i$th coordinate. Define $\mu_j:F\rightarrow F^n$ by inserting the input value into the $j$th coordinate.
How do these maps relate to the matrix representation of a function? Let $f:F^n\rightarrow F^n$ be represented by $A=[A]_{ij}$ in the standard basis. If you compute $\mu_j(1)$, you get the unit vector with a 1 in the $j$th coordinate. To affect $f$ on this, you multiply on the left with its matrix $A$. So, $A\mu_j(1)$ is the $j$th column of $A$. Now by applying $\pi_i$ after this, you will pick out the $i$th entry of the $j$th row. So, $\pi_i(A\mu_j(1))=A_{ij}$. This shows that $\pi_i f\mu_j=A_{ij}$. This last $A_{ij}$ is the $1\times 1$ matrix representation of a linear transformation of $F$ into $F$.
Can you see that $\sum_{k=1}^n\mu_k(\pi_k(v))=v$ for all $v\in F^n$? This means that $\sum_{k=1}^n\mu_k\pi_k=Id_V$, the identity on $V$.
Let a transformation $g:F^n\rightarrow F^n$ be represented by the matrix $B=[B]_{ij}$ in standard basis. We will now show that $[BA]_{ij}$ is the matrix representing the composition $gf$. This will be true if $\pi_i gf \mu_j=[BA]_{ij}$, by our discussion before.
We compute that $\pi_i gf \mu_j=\pi_i g \mathrm{Id}_V f \mu_j=\pi_i g(\sum_{k=1}^n \mu_k\pi_k)f \mu_j=\sum_{k=1}^n\pi_i g \mu_k\pi_kf \mu_j=\sum_{k=1}^n B_{ik}A_{kj}=[BA]_{ij}$. So there you have it: the components of $gf$ are given by matrix multiplication! |
H: Normal, Non-Metrizable Spaces
We know that every metric space is normal. We know also that a normal, second countable space is metrizable.
What is an example of a normal space that is not metrizable?
Thanks for your help.
AI: Edit: In light of the comments, I thought it prudent to give the precise definition of normality used in the reference. For the list I've provided, a normal space $X$ is one satisfying:
T1 property: For all $x,y \in X$ there exist open sets $U_x$ and $U_y$ containing $x$ and $y$, respectively, such that $x \notin U_y$ and $y \notin U_x$.
If $A$ and $B$ are disjoint closed sets in $X$, there exist disjoint open sets $U_A$ and $U_B$ containing $A$ and $B$, respectively.
The following examples of normal, non-metrizable spaces come from $\pi$-Base, which is a searchable database inspired by Steen and Seebach's Counterexamples in Topology. You can learn more about each space by visiting the search result.
Alexandroff Square
Appert Space
Arens-Fort Space
Baire Product Metric on $\mathbb{R}^\omega$
Bing's Discrete Extension Space
Boolean Product Topology on $\mathbb{R}^\omega$
Closed Ordinal Space $[0,\Omega]$
Concentric Circles
Countable Excluded Point Topology
Deleted Integer Topology
Divisor Topology
Either-Or Topology
Finite Excluded Point Topology
Fortissimo Space
Helly Space
Hjalmar Ekdal Topology
$I^I$
Lexicographic Ordering on the Unit Square
Michael's Closed Subspace
Nested Interval Topology
Odd-Even Topology
One Point Compactification Topology
One-point Lindelofication of $\omega_1$
Open Ordinal Space $[0,\Omega)$
Radial Interval Topology
Right Half-Open Interval Topology
Right Order Topology on $\mathbb{R}$
Rudin's Dowker Space
Sierpinski Space
Single Ultrafilter Topology
Stone-Cech Compactification of the Integers
The Extended Long Line
The Integer Broom
The Long Line
Tychonoff Plank
Uncountable Excluded Point Topology
Uncountable Fort Space |
H: $C \otimes A \cong C \otimes B$ does not imply $A \cong B$
Let $R$ be a commutative unital ring and let $M$ be an $R$-module and let $S$ be a multiplicative subset of $R$.
Today I proved both of the following:
$$ S^{-1} R\otimes_R S^{-1}M \cong S^{-1} M$$ and
$$ S^{-1} R \otimes M \cong S^{-1} M$$
Now I'm slightly confused.
Either my proofs are wrong or $C \otimes A \cong C \otimes B$ does not imply $A \cong B$. But I can't come up with an example. Can someone give me an example? (Or tell me that my proofs are wrong.)
AI: You are correct in noticing that tensoring with a fixed module isn't an injective operation. Various things can go wrong; probably the simplest thing to notice is that tensoring can kill torsion. To repeat the example given in the comments, both $\mathbb{Z}_2\otimes \mathbb{Q}$ and $\mathbb{Z}_2\otimes\mathbb{R}$ (everything in sight is a $\mathbb{Z}$-module) are trivial, while $\mathbb{Q}$ and $\mathbb{R}$ aren't isomorphic. There is a multitude of examples in the same vein, e.g. tensoring any two finitely generated $\mathbb{Z}$-modules of the same rank with $\mathbb{Q}$ will produce isomorphic modules. |
H: Theorems/entailment notation
When defining a predicate logic system with natural deduction, we can define the syntatic entailment with the operator $\vdash$. Generally, I see authors using the formula $\vdash \phi$ to say that $\phi$ is a theorem of the logical system. However I was used to say that $\phi$ is a theorem too even if it is the entailment of another formula, say $\psi$ (we obtain $\phi$ from $\psi$ by using the rules of natural deduction). But that is the same as affirming $\psi \vdash \phi$.
1) So, authors generally define theorems as those formulas which are always derivable regardless of the theory that they are in, and regardless of any formulas present in the system?
2) If 1) is correct, then saying that the Gödel incompleteness theorem asserts that there are theorems which are not true under the standard interpretation of Peano Arithmetic is false, because the formulas which are not true have to be derived from the axioms of the Peano Arithmetic.
AI: A theorem of the predicate calculus is a sentence (or, if you prefer, formula) that can be derived without additional axioms. If $T$ is a theory (that is, a set of sentences) then a theorem of $T$ is a sentence $\varphi$ such that $\psi\vdash \varphi$ for some finite conjunction $\psi$ of sentences of $T$. So a theorem of the predicate calculus is a sentence $\varphi$ which is a theorem of the "empty" theory.
Every theorem of the predicate calculus, over the appropriate language, is true in the natural numbers. I hope that no one claims that "Gödel incompleteness theorem asserts that there are theorems which are not true under the standard interpretation of Peano Arithmetic".
The Incompleteness Theorem says, among other things, that there are sentences $\varphi$ that are true in the natural numbers, but that are not provable in the theory $T$ whose axioms are the usual axioms of (first-order) Peano Arithmetic (if that theory is consistent). |
H: analysis limit question
Let f be an integrable function on $\mathbb{R}$. Show that $\lim_{t\rightarrow 0} \int_{\mathbb{R}}|f(x + t) -f(x)|dx = 0$.
I can make it work once it is shown to be true for $f\in C_c(\mathbb{R})$ but I am having trouble proving this case.
AI: If $f\in C_c(\Bbb R)$, then the support of $f$ is contained in $[-R,R]$ for some $R>0$. Fix $\varepsilon>0$, and $\delta<1$ such that if $|t|\leq \delta$ and $x\in \Bbb R$ then $|f(x+t)-f(x)|\leq \varepsilon$ (it can be quite easily shown that $f$ is uniformly continuous on $\Bbb R$, since it is on $[-R-1,R+1]$. We have for $|t|\leq \delta$ that
$$\int_{\Bbb R}\left|f(x+t)-f(x)\right|dx=\int_{[-R-1,R+1]}\left|f(x+t)-f(x)\right|dx\leq 2\cdot \varepsilon (R+1).$$ |
H: Is there a sequence that contains every rational number once, but with the "simplest" fractions first?
The Calkin-Wilf sequence contains every positive rational number exactly once:
1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, ….
I'd consider 5/1 to be a "simpler" ratio than 8/5, but it appears later in the series.
Is there a mathematical term for the "simpleness" of a ratio? It might be something like the numerator times the denominator, or maybe there are other ways to measure.
Is there a sequence that contains all the positive rational numbers, but with the "simpleness" of the ratios monotonically increasing?
(Small integer ratios are found in Just intonation, polyrhythm, orbital resonance, etc.)
If you use the Calkin-Wilf sequence with the num*den measure, for instance, it looks like this:
AI: A common measure of how "complicated" a (reduced) fraction is is the height:
Definition. Let $\frac{r}{s}$ be a rational number, with $\gcd(r,s)=1$. The height of $\frac{r}{s}$ is $\mathrm{ht}\left(\frac{r}{s}\right)=\max\{|r|,|s|\}$.
Among those of the same height, you can order them by comparing the minimum. For those with the same minimum, you can compare values.
So one possibility is:
If $\frac{r}{s}$ and $\frac{x}{y}$ are positive rationals with $\gcd(r,s)=\gcd(x,y)=1$, then we say $\frac{r}{s}\preceq \frac{x}{y}$ if and only if
$\mathrm{ht}(\frac{r}{s})\lt \mathrm{ht}(\frac{x}{y})$; or
$\mathrm{ht}(\frac{r}{s}) = \mathrm{ht}(\frac{x}{y})$ and $\min(r,s)\lt \min(x,y)$; or
$\mathrm{ht}(\frac{r}{s})=\mathrm{ht}(\frac{x}{y})$, and $\min(r,s)=\min(x,y)$; and $\frac{r}{s}\leq \frac{x}{y}$.
You would get:
1/1, 1/2, 2/1, 1/3, 3/1, 2/3, 3/2, 1/4, 4/1, 3/4, 4/3, 1/5, 5/1, 2/5, 5/2, 3/5, 5/3, 4/5, 5/4, 1/6, 6/1, 5/6, 6/5, ...
Don't know about a closed formula, though.
Note/Clarification: The height is very standard, especially in Diophantine Analysis and Arithmetic Geometry.
I don't know about the rest of the order I present, though it seems like a natural extension (or one could prefer listing larger rationals first in point 3. Inserting the negatives would also allow for several small variations. |
H: Rigorous proof that $\frac{1}{3} = 0.333\ldots$
I'm a PreCalculus student trying to find a rigorous proof that $\displaystyle\frac{1}{3} = 0.333\ldots$, but I couldn't find it. I think (just think) that this proof would start by proving that
$\displaystyle\sum_{i=1}^{\infty}3\cdot10^{-i} = \frac{1}{3}$. My guesses (assuming that proving that $\displaystyle\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i$ converges is trivial):
$\displaystyle\sum_{i=1}^{\infty}3\cdot10^{-i} = 3\cdot\sum_{i = 1}^{\infty}10^{-i} = 3\cdot\sum_{i=1}^{\infty}\left(\frac{1}{10}\right)^i = 3\cdot\left(\frac{1}{1 - \frac{1}{10}}-1\right) = 3\cdot\left(\frac{10}{9}-1\right) = \frac{1}{3}$.
Questions: is this completely rigorous? Which flaws could be found in this proof? How can I improve it?
PS. I'm not sure how to tag this. Feel free to edit, if necessary.
AI: Since $\sum_{i=1}^\infty 3\cdot 10^{-i}$ is what the notation "$0.333...$" means, your argument is perfectly good. It's not just the "start of a proof", it is all there is to it.
Okay, perhaps it is not really trivial to prove that the geometric series converges, but straightforward it is. Just plug in the definition of the sum of a series and crank the handle, using standard tricks to rewrite each of the partial sums in turn. |
H: Localization arguments in Dedekind domains
I am reading Serre's Local Fields, and have questions about the text. Specifically, pages 11 and 12.
1) Consider a Dedekind domain. We want to show that all fractional ideals are invertible. Serre claims that because the image of a fractional ideal under every localization by prime ideals is invertible, that the fractional ideal itself must be invertible. I do not see why this is.
2) In the same way, he argues "by localization" that because the image of an ideal $\mathfrak{a}$ under the localization $A_\mathfrak{p}$ has the form $\mathfrak{p}^{{v_p(\mathfrak{a}})}$, where $v_{\mathfrak{p}}$ is the valuation, and because the exponents are zero except for a finite number of ideals, that $\mathfrak{a}$ factors into a finite number of prime ideals. Again, I do not see how to carry out the details of this argument.
Thanks for the help.
AI: For (1), let $I$ be a non-zero ideal of a Dedekind domain $R$ with fraction field $K$. Let $I^{-1}=\{r\in K:rI\subseteq R\}$. Then $I^{-1}$ is a fractional ideal and $I$ is invertible if and only if $II^{-1}=R$ (in general one has from the definitions that $II^{-1}\subseteq R$). You can verify that localization at a prime $\mathfrak{p}\in\mathrm{mSpec}(R)$ commutes with products of fractional ideals and with formation of $I^{-1}$, that is, upon identifying $R_\mathfrak{p}$ with a subring of $K$, $(I^{-1})_\mathfrak{p}=(I_\mathfrak{p})^{-1}$. Now, local invertibility means $(II^{-1})_\mathfrak{p}=I_\mathfrak{p}(I_\mathfrak{p})^{-1}=R_\mathfrak{p}$ for all $\mathfrak{p}\in\mathrm{mSpec}(R)$. If $N$ and $M$ are $R$-modules with $N\subseteq M$ and $N_\mathfrak{p}=M_\mathfrak{p}$ for all $\mathfrak{p}$ maximal, then $N=M$. So, using this, you get $II^{-1}=R$. So $I$ is invertible. Since principal fractional ideals are obviously invertible and every fractional ideal differs (multiplicatively) from an integral ideal by a principal ideal, it follows that all fractional ideals are invertible.
For (2), take an ideal $I$. You have $I_\mathfrak{p}=(\mathfrak{p}R_\mathfrak{p})^{v_\mathfrak{p}(I)}$, where $v_\mathfrak{p}(I)$ is defined by this equation (using that $R_\mathfrak{p}$ is a discrete valuation ring for all maximal $\mathfrak{p}$). Put $J=\prod_\mathfrak{p}\mathfrak{p}^{v_\mathfrak{p}(I)}$, the product over all maximal ideals of $R$ (granting the fact that $v_\mathfrak{p}(I)$ is non-zero for only finitely many maximal ideals). Compare the localizations of these two ideals. You'll see that $I_\mathfrak{p}=J_\mathfrak{p}$ for all $\mathfrak{p}\in\mathrm{mSpec}(R)$. This implies $I=J$.
EDIT: In answer to the questions posed in the comments, let $r\in I$ and consider the ideal $I^\prime$ of $s\in R$ with $sr\in J$. Because $r/1\in I_\mathfrak{p}=J_\mathfrak{p}$, we have $r/1=t/s$ for $t\in J$ and $s\notin\mathfrak{p}$. So $sr=t\in J$, and thus $s\in I^\prime$. It follows that $I^\prime$ is not contained in any maximal ideal of $R$, so it must be all of $R$. Thus $1\in I^\prime$, so $r=1r\in J$, and thus $I\subseteq J$. By symmetry, $J\subseteq I$ as well. The same argument works for general $R$-modules. This is a standard fact about localizations. For the other question, in the argument I gave, I was assuming $I$ to be an integral ideal throughout, so I had to say something about general fractional ideals at the end. However, the argument applies without change for an arbitrary fractional ideal $I$. |
H: Proof of Turan's theorem
I'm following the proof of Turan's theorem on $\text{ex}(n,K^r)$ in Diestel's Graph Theory book (click to see the page) and something bothers me:
Since $G$ is edge-maximal without a $K^r$ subgraph, $G$ has a subgraph
$K=K^{r-1}$. By the induction hypothesis, $G-K$ has at most
$t_{r-1}(n-r+1)$ edges [...]
But why would $G-K$ be $K^{r-1}-$free ? Couldn't there be other $K^{r-1}$ disjoint subgraphs in $G$ ?
AI: The proof does not claim that $G-K$ is $K^{r-1}$-free, only that it is $K^r$-free. The induction is on $n$, not on $r$, so the assumption is that all smaller $K^r$-free graphs have been classified. |
H: How do mathematicians think about high dimensional geometry?
Many ideas and algorithms come from imagining points on 2d and 3d spaces. Be it in function analysis, machine learning, pattern matching and many more.
How do mathematicians think about higher dimensions? Can intuitions about the meaning of dot-product, angles and lengths transfer from 2d geometry to a 100d?
If so, would it be enough to fully understand the higher dimesions, namely, could the same problem in 100d have properties\behaviours that are not seen in 2d\3d?
AI: It vastly depends on the objects you define. Indeed, when talking about vector spaces, we algebraically think about $\mathbb{R}^n$, build up intuition, and then set $n=100$.
However, when you start adding exotic objects, like knots, it becomes less "easy". For example, some knots in $\mathbb{R}^3$ are trivial loops in $\mathbb{R}^4$ (ie. the trefoil knot falls apart in 4D).
Then again, functions and their orthogonality are computed in $\mathbb{R}^{89270}$ just as they are in $\mathbb{R}$ - nothing strange going on there. It's only when you consider infinite-dimensional spaces that this becomes slightly unintuitive again.
So, in short, it completely depends on the objects you talk about. Most finite-dimensional vector spaces over some field $K$ are equal in almost all aspects. Adding more structure can make it much more difficult, and oftentimes all mathematicians have is algebra. |
H: Relations of language/theory/signature
Say that the language of the first order logic is the collection of symbols that can be used in the formulas + the grammar (the rules that specify how they can be combined)?
1) However, the signature of the system can include additional symbols to be used in the formula. So, it is wrong to say that a formula can be constructed only from the language of predicate logic, because it needs additional symbols from the signature.
2) When specifying a theory, we need to include a signature that models the domain of the theory? So that the theory will be defined as $\mathfrak{T} = (\Sigma, \Gamma)$, where $\Sigma$ is it's signature and $\Gamma$ the collection of formulas of the signature.
Example. The language of predicate logic includes all the standard symbols: $\forall$, $\land$, .... Say that our signature is the signature for Peano Arithmetic, composed of $\Sigma = \{ \times, +, 0, \dots \}$. A formula $\forall x(x + 0 = x)$, can't be composed only from the language of predicate logic, it need's the signature $\Sigma$ too. And the theory of Peano Airthmetic too will need this signature.
AI: You can specify the language of a theory completely by saying that it is "the language of predicate calculus over such-and-such signature" -- assuming that your idea of "signature" includes the names of the predicates. On the other hand, the information contained in a "language" must be enough to allow us to reconstruct the signature.
So distinguishing between the language and the signature is just a matter of pedantry. (Of course this doesn't make it a bad thing; judiciously applied pedantry is the building block of mathematics!)
Specifying a language as part of a theory is also mostly a matter of general tidyness -- it's good form to explicitly define your notation before you start using them, even if the only definition you give is "an uninterpreted $n$-ary predicate/function which has no meaning except what the axioms imply". Telling the reader that he's not supposed to use any preexisting knowledge of the symbol is valuable information for him.
(There's also the technical point that many constructions while working with a formal system are simpler if we can assume that we just know what the possible symbols are instead of needing to figure it out from contextual clues at each step. That could probably be worked around if we really wanted to, though).
Added later: Finally, for ideas such as that of a complete theory to make good sense, one needs to distinguish between symbols that are not part of the language at all and symbols that are in the language but just happen never to be mentioned by the axioms. |
H: proving facts about $\alpha$-Hölder-continuous functions
I am studying myself some facts about $\alpha$-Hölder-continuous functions but I don't get any further by proving the following:
$(1)$ $\forall\alpha\in ]0,1]$ is $C^{0,\alpha}$ dense in $C^0(D)$ concerning the uniform norm and $D\subset\mathbb R^n$.
$(2)$ $\forall\alpha\in ]0,1]$ and compact set $K\subset\mathbb R^n$ is $(C^{0,\alpha}(K),||\cdot||_{C^{0,\alpha}(K)})$ a complete space (with
$||u||_{C^{0,\alpha}(K)}:=||u||_{\sup}+\sup\limits_{{x,y\in K\space\&\space x\ne y}}\frac{|u(x)-u(y)|}{|x-y|^\alpha}$ and $C^{0,\alpha}(K)$ the space of all $\alpha$-Hölder-continuous functions)
$(3)$ All bounded closed subsets of $(C^{0,\alpha}(K),||\cdot||_{C^{0,\alpha}(K)})$ are compact.
So how do you prove one $(1),(2),(3)$ ?
AI: (1) Assume $D$ compact then by Weiertrass approximation you get that polynomials are dense in $C(D)$, since $D\subset B_R(0)$ for some $R>0$ and polynomials are $\alpha$-Hölder in $B_R(0)$ we get the result.
If $D$ is not compact the result is not true: Take $D=(0,1)$ and $f(x)=\sin(1/x)$. Take $0<r$ and $g\in C^{0,\alpha}$ such that $\| g-f\|_\infty <r$, then for every $n\in \mathbb{N}$ there exists $x_n,y_n\in (0,1)$ with $|x_n-y_n|<r$ and $f(x_n)=1$, $f(y_n)=-1$. Then $|g(x_n)-g(y_n)| \geq 2(1-r)>r^{\alpha}>|x_n-y_n|^{\alpha}$ if $r$ is small enough, contradicting $g\in C^{0,\alpha}$.
If $D=\mathbb{R}$ a similar argument works if we use something like $f(x)= \sin(2n\pi(x-n))$ if $x\in [n,n+1/n]$, $n\in \mathbb{N}$ and zero otherwise.
(2) We know that $C(K)$ is complete so we only need to check that the limit function of a Cauchy sequence in $C^{0,\alpha}$ is in this space and that the seminorms converge. For this notice
$$
|f(x)-f(y)|\leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|
$$
So, picking $n$ so that $\| f-f_n\|_{\infty} <\epsilon$ we get, and remembering that $\frac{|f_n(z)-f_n(w)|}{|z-w|^\alpha} \leq [f_n]_\alpha \leq C$,
$$
\frac{|f(x)-f(y)|}{|x-y|^\alpha} \leq C+2\epsilon/|x-y|^\alpha
$$
which gives $[f]_\alpha <C<\infty$.
To see that the seminorms converge just notice
$$
\frac{|f(x)-f_n(x)-f(y)+f_n(y)|}{|x-y|^\alpha} \leq\limsup_m [f_m-f_n]_\alpha
$$
which gives
$$
\limsup_n [f-f_n]_\alpha \leq \limsup_n \limsup_m [f_m-f_n]_\alpha \to 0.
$$
(3) I don't know about this statement as is, but Arzela-Ascoli gives that bounded closed sets in $C^{0,\alpha}(K)$ are precompact in $C(K)$ |
H: Commutator map and the derived series
Let be $G$ a solvable group, let
$$ G=G_0\supset G_1\supset\cdots\supset G_k=1$$
be the derived series for $G.$ Is clear that $G_ {k-1}$ is abelian.
Now take $b\in G_{k-1}$ e $a\in G_{k-2}$ my question is how to see that:
$$a^{-1}b^{-1}ab \in G_{k-1}~~~?$$
this is a crucial step in a demonstration of Riemannian
geometry I'm reading, the theorem of Byers
AI: I apologize for the dumb question, it is obvious,
$G_{k-1}$ is a normal subgroup of $G_{k-2}$ so $a^{-1}b^{-1}a \in G_{k-1}$, then ,
$$a^{-1}b^{-1}ab \in G_{k-1}.$$ |
H: The subset of elements of order dividing $k$ in an abelian group is a subgroup
Suppose $G$ is an abelian group and $k$ is a natural number.
Prove $H = \{ g \in G : g^k = 1 \}$ is a subgroup of G.
I know I need to show that $1_G \in H$, existence of inverse element in group, and closure,
but how?
AI: $1 \in H$ since $1^k = 1$.
Suppose $g \in H$. That is, $g^k = 1$. Then $(g^{-1})^k = (g^{k})^{-1} = 1$. So $g^{-1} \in H$.
If $x,y \in H$, this means $x^k = 1$ and $y^k = 1$. Then $(xy)^k = x^ky^k = 1$ since $G$ is an abelian group. So $xy \in H$.
$H$ is a subgroup. |
H: Differential equation of $y = e^{rx}$
I am trying to find what values of r in $y = e^{rx}$ satsify $2y'' + y' - y = 0$
I thought I was being clever and knew how to do this so this is how I proceeded.
$$y' = re^{rx}$$
$$y'' = r^2 e^{rx}$$
$$2(r^2 e^{rx}) +re^{rx} -e^{rx} = 0 $$
I am not sure how to proceed from here, the biggest thing I am confused on is that I am working with a variable x, with no input conditions at all, and a variable r (the constant) so how do I do this?
AI: The final equation you have is $$2r^2 \exp(rx) + r \exp(rx) - \exp(rx) = 0$$
$$(2r^2 + r - 1)\exp(rx) = 0$$
Now $\exp(rx) \neq 0$, for all $r$ and $x$. Hence, you get that $$2r^2 + r - 1 =0$$
Can you proceed from here by solving the quadratic equation for $r$?
Move your cursor over the gray area below for complete solution.
Note that we can write $2r^2 + r - 1$ as shown below. $$2r^2 + r - 1 = 2r^2 + 2r -r -1 = 2r(r+1)-1 ( r+1) = (2r-1)(r+1)$$ Hence, $2r^2 + r - 1 = 0 \implies (2r-1)(r+1) = 0 \implies r = \dfrac12 \text{ or } r= - 1$. Hence, $y$ is either $\exp(x/2)$ or $\exp(-x)$. In general, we find that $$y = c_1 \exp(x/2) + c_2 \exp(-x)$$ where $c_1,c_2$ are constants, satisfies the differential equation. |
H: $f\colon M\to N$ continuous iff $f(\overline{X})\subset\overline{f(X)}$
Possible Duplicate:
Continuity and Closure
$f\colon M\to N$ is continuous iff for all $X\subset M$ we have that $f\left(\overline{X}\right)\subset\overline{f(X)}$.
I only proved $\implies$.
If $f$ is continuous then for any $X\subset M$,
$$X\subset f^{-1}[f(X)]\subset f^{-1}\left[\overline{f(X)}\right]=\overline{f^{-1}\left[\overline{f(X)}\right]}$$
therefore
$$\overline{X}\subset f^{-1}\left[\overline{f(X)}\right]\implies f\left(\overline{X}\right)\subset \overline{f(X)}.$$
The other side must be the same idea but I don't know why I can't prove it.
Added: With exactly same idea when I proved $\implies$ I did proved $\Longleftarrow$,
Let $F\subset N$ any closed set then:
$$f\left[f^{-1}(F)\right]\subset f\left[ \overline{f^{-1}(F)}\right]\subset \overline{f\left[ f^{-1}(F)\right]}\subset \overline{F}=F$$
in particular
$$f\left[ \overline{f^{-1}(F)}\right]\subset F\implies f^{-1}(F)\supset\overline{f^{-1}(F)}$$
then $f^{-1}(F)=\overline{f^{-1}(F)}$ and $f$ is continuous.
AI: Suppose that $f$ is not continuous. Then there is a closed set $C\subseteq N$ such that $f^{-1}[C]$ is not closed in $M$. Let $X=f^{-1}[C]$. Since $X$ is not closed, there is some $p\in\operatorname{cl}X\setminus X$. Then $f(p)\in f[\operatorname{cl}X]$, but $f(p)\notin C\supseteq\operatorname{cl}f[X]$, so $f[\operatorname{cl}X]\nsubseteq\operatorname{cl}f[X]$. |
H: Proving Continuity in Several Complex Variables
So I don't have a whole lot of experience in general proving continuity for multivariable functions, and I want to make sure I'm going about things correctly.
Prove that the function $B(z,w):=\int_0^1 t^{z-1}(1-t)^{w-1}dt$, for $z,w\in \mathbb{C}$, is continuous.
So I let $\varepsilon > 0$ and basically consider an expression of the form:
$$|\int_0^1 t^{z-1}(1-t)^{w-1}dt - \int_0^1 t^{u-1}(1-t)^{v-1}dt| < \varepsilon$$
Which simplifies into the form:
$$|\int_0^1 t^{z-1}(1-t)^{w-1}[1-t^{u-z}(1-t)^{v-w}]dt| < \varepsilon$$
And now to make this expression true I consider $|u-z|<\delta_1$ and $|v-w|<\delta_2$ and choose $\delta_1$ and $\delta_2$ as small as I need to (based on $\varepsilon$), which will allow me to make this integral as small as a wish, since $t^{u-z}(1-t)^{v-w}\rightarrow 1$ as $\delta_1,\delta_2 \rightarrow 0$.
Is this an acceptable way to go about continuity in several complex variables, should only a single delta equal to $Min[\delta_1,\delta_2]$ be used, are there any other subtleties I should be made aware of in the multivariable case?
AI: when you are talking about continuity you should know your topology. your domain is $\mathbb{C} \mathrm{x} \mathbb{C}\, $ your topology is induced by your norm in your case your norm could be $| \! | (u,w | \! | = |u|+|w| $ or $\| \! | (u,w) | \! | = \sqrt{|u|^2+|w|^2} $ so normally by definition for $ \epsilon > 0$ you should find a $\delta >0 $ such that for $| \! | (u,w)- (u_0,w_0) | \! | |< \delta \Rightarrow |\!|B(u,w)-B(u_0,w_0)|\!|< \epsilon $. So I think now, using these norms what you did is equivalent. |
H: Cooley-Tukey Algorithm?
Why does the Cooley-Tukey Fast Algorithm take $O(n \log n)$ time? The book derives this from the fact that evaluation takes time:
$T(n) = 2T(n/2) + O(n)$
and then uses the Master Theorem to arrive at the Big O Notation Time. Could someone explain how the above equation was derived?
How I see it:
After a recursion, you are dealing with twice the number of polynomials ($A_o(x)$ AND $Ae(x)$) at half the number of points, (since the points are plus-minutes pairs) and each polynomial is of half the size. So why isn't the time:
$T(n/2)+O(n)$
The size of the problem reduces by half (since the degree of $Ao(x)$ and $Ae(x)$ are half of $A(x)$) but your dealing with the same number of problems. ($A_o(x)$ and $A_e(x)$ at half the number of points.) What am I misunderstanding?
Example
Lets say you are evaluating polynomial $A(x)$ at points: $a, -a, b, -b.$ Therefore you have 4 polynomials to evaluate: $A(a), A(-a), A(b), A(-b).$
Using the algorithm, you would instead evaluate the odd and even coefficient polynomials. Therefore you would evaluate:
1) $A_e(a^2)$
2) $A_o(a^2)$
3) $A_e(-a^2)$
4) $A_o(-a^2)$
5) $A_e(b^2)$
6) $A_o(b^2)$
7) $A_e(-b^2)$
8) $A_o(-b^2)$
Each polynomial is half the size of the original polynomial. However $A_e(a^2)$ and $A_o(a^2) = A_e(-a^2)$ and $A_e(-a^2)$. The same thing applies for $b$ and $-b$. Therefore you would actually only need to evaluate 4 polynomials:
1) $A_e(a^2)$
2) $A_o(a^2)$
5) $A_e(b^2)$
6) $A_o(b^2)$
Therefore you end up with the same amount of problems, but each problem is half the size, since $A_o$ and $A_e$ have half the degree of $A(x)$. Therefore 4 problems of size $n$ has been transformed into 4 problems of size $n/2$! The number of problems have stayed the same, only the size has changed, leading me to the conclusion that the time is:
$T(n) = T(n/2)+O(n)$
AI: Using this pseudocode, you can see that lines 5-6 recurse on two roughly equal-sized arrays, hence the first term. The loop in lines 7-11 executes $N/2 = \Theta (n) $ times, explaining the second term. Assuming the rest of the code is negligible work, you're done.
Your example is perfectly valid, but the algorithm does not know that it only needs to evaluate four polynomials. Rather, it continues to recurse until it has reached $N$ problem instances of size $1$ and then iteratively pieces them together. |
H: Random variable with mean $\mu$ and variance $\sigma ^2$
I have never taken probability theory, and I wonder whether one can express some random variable $X$ with mean $\mu$ and variance $\sigma ^2$ in terms of $\mu$ and $\sigma$ only. Or at least something close to it.
Thank you for your help.
AI: No - the mean and variance (if they exist) tell you about the location and scale of a distribution but nothing about its shape.
If a random variable $X$ has mean $\mu$ and variance $\sigma^2$ then the random variable $Y$ defined by $Y=aX+b$ for real $a$ and $b$ has mean $a\mu+b$ and variance $a^2 \sigma^2$.
So given a particular shape of distribution (well behaved enough to have a mean and variance), it is possible to find a distribution which has the same shape but any mean and (positive) variance you specify. |
H: $g^{k}S=S \Rightarrow g$ has finite order
Let $G$ be a group (not necessarily finite), $g \in G$, $g \ne 1$. Suppose that $S \subseteq G$, $S$ is finite, $1 \in S$, and $gS=S$. It follows that $g^{k}S=S$ for all $k \in \mathbb{N}$. Does it also follow that the order of $g$ is finite?
My attempt: Since $S$ is finite, let $|S|=n$. Since $1 \in S$, $g^{k} \in g^{k}S = S$ for all $k \in \mathbb{N}$. Consider $A=\{g,g^{2},\ldots , g^{n}, g^{n+1}\}$. Since for all $k \in \mathbb{N}$ we have that $g^{k} \in S$, it follows that $A \subseteq S$. Since $A$ and $S$ are finite, $|A| \le |S|=n$. Thus, by the Pigeonhole principle, $g^{s}=g^{t}$ for some $s \ne t$. WLOG, suppose $s < t$. Then $g^{t-s}=1$, so the order of $g$ is finite.
Two questions: Is the above proof correct and is there an easier way to do this by letting $G$ act on itself and considering permutation representations?
AI: You are correct.
Perm rep reprising of your argument: Do the same thing as before, but let "x" be in S, instead of 1. Conclude $g^s x = g^t x$, so that $g^{t-s}$ stabilizes $x$. However, in a regular permutation representation, only the identity stabilizes any element, so $g^{t-s}=1$. |
H: Finding direction vector
Can someone please explain how the direction vector was found in problem $2$ of this worksheet?
Below is an image of the problem $2$ of the worksheet.
AI: You want to know the tangent line to the ellipse at the given point along the plane $y=2$, so go ahead and plug in $y=2$ to obtain $4x^2 + 8 + z^2 = 16$. Take the total differential to obtain
$$ 8x dx + 2 z dz = 0.
$$
This means
$$
2z dz = -8x dx, \mbox{ or } \dfrac{dz}{dx} = -\dfrac{8x}{2z} = -\dfrac{4x}{z}.
$$
Since
$$
\dfrac{dz}{dx}|_{(x,z)=(1,2)} = -\dfrac{4}{2} = -2,
$$
the line tangent to the intersection of the ellipsoid and the plane at the point $(1,2,2)$ has the direction vector
$$
(1,0,-2). $$
Note that the tangent line is parallel to the $y=2$ plane so the vector should not be changing in the $y$-direction. Moreover, any nonzero scalar multiple of $(1,0,-2)$ will also work.
Thus the tangent line is
$$
l(t) = (1+t, 2, 2-2t),
$$
or you could also write it as
$$
l(t) = (1,2,2)+ t(1,0,-2),
$$
where $t$ varies over the reals. |
H: Cauchy-Schwarz Inequality proof (for semi-inner-product A-module).
I am reading a proof of the following Cauchy-Schwarz Inequality and I don't understand one part of the proof:
Theorem:
Let $A$ be a $C^*$-algebra and let $E$ be a semi-inner-product $A$-module. Then
$$
\langle x,y \rangle ^* \langle x,y \rangle \leq \| \langle x,x \rangle \| \langle y,y \rangle \text{ for all } x,y\in E
$$
The proof starts of by saying, without loss of generality we can assume $ \| \langle x,x \rangle \| = 1$ But I don't understand why we can assume this.
The rest of the proof is just some calculations which I understand: for $a\in A$, $x,y\in E$ we have
$$
0 \leq \langle xa-y,xa-y \rangle = a^* \langle x,x \rangle a - \langle y,x \rangle a - a^* \langle x,y \rangle + \langle y,y \rangle \leq a^*a - \langle y,x \rangle a - a^* \langle x,y \rangle + \langle y,y \rangle
$$
and by letting $a=\langle x,y \rangle $ we get the desired result.
I just need to understand why we can assume $\| \langle x,x \rangle \| = 1$. My guess is that if it does not equal 1 then we can replace it with an equivalent norm that equals 1, but I don't know if this is the correct reason.
AI: For any fixed $x$ and $y$ and $t > 0$, if one replaces $x$ with $tx$, both sides of the inequality you want to prove scale by $t^2$. So for all $t > 0$, the inequality holds for a given $x$ and $y$ if and only if it holds for $tx$ and $y$. Since
$$
\|\langle tx, tx\rangle\| = t^2 \|\langle x,x\rangle\|,
$$
if $\langle x,x\rangle$ is nonzero, we can (by taking $t = \|\langle x,x\rangle\|^{-1/2}$) indeed reduce to the case that $\|\langle x, x\rangle\| = 1$.
It remains to handle the case that $\langle x,x\rangle = 0$, which can happen for nonzero $x$ (I think that's the point of the "semi" in "semi-inner-product space"). In this case, I don't see any obvious reason why $\langle x,y\rangle$ should be zero, but if one echoes the argument you gave for the $\|\langle x,x\rangle\| = 1$ case, one sees that for all $a$ in $A$ one has
$$
0 \leq \langle xa - y, xa - y\rangle = -\langle y,x\rangle a - a^* \langle x,y\rangle + \langle y,y\rangle.
$$
For any $t > 0$, by putting $a = \frac{t}{2} \langle x,y\rangle$ in the above one deduces
$$
t \langle x,y\rangle^* \langle x,y\rangle \leq \langle y,y\rangle.
$$
One gets a contradiction if $\langle x,y\rangle$ is nonzero by taking $t$ large enough (if $0 \leq a \leq b$ in a $C^*$ algebra, then $\|a\| \leq \|b\|$), so it must be that $\langle x, y \rangle = 0$, which is what we want in this case.
There is probably a simpler way to handle the $\langle x,x\rangle = 0$ case that I'm not seeing at the moment, but anyway, I think the above works. |
H: Deriving parameterization for hyperboloid
I know there is a parameterization of a hyperboloid $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ in terms of $\cosh$ and $\sinh$, but I don't see how these equations are derived. I would appreciate it if either someone could explain to me how such a parameterization is derived or recommend a reference.
AI: This derivation has been done by André Nicolas!
The parametrization of the hyperbola is
$$x(t)=\cosh t$$
$$y(t)=\pm \sinh t$$
A circle of radius $r$ is parametrized as:
$$x(t)=\cos t$$
$$y(t)=\sin t$$
Rotating the hyperbola above around a circle of radius $\cosh$ (distance of a regular hyperbola from y axis):
$$x(u, v)=\cosh v \cos u$$
$$y(u, v)=\cosh v \sin u$$
$$z(u, v)=\sinh v$$
It is easy to imaging the hyperboloid from two ways - from the top and from the side. This helped me understand the derivation. |
H: Dimensionality of null space when Trace is Zero
This is the fourth part of a four-part problem in Charles W. Curtis's book entitled Linear Algebra, An Introductory Approach (p. 216). I've succeeded in proving the first three parts, but the most interesting part of the problem eludes me. Part (a) requires the reader to prove that $\operatorname{Tr}{(AB)} = \operatorname{Tr}{(BA)}$, which I was able to show by writing out each side of the equation using sigma notation. Part (b) asks the reader to use part (a) to show that similar matrices have the same trace. If $A$ and $B$ are similar, then
$\operatorname{Tr}{(A)} = \operatorname{Tr}{(S^{-1}BS)}$
$= \operatorname{Tr}(BSS^{-1})$
$= \operatorname{Tr}(B)$,
which completes part (b). Part (c) asks the reader to show that the vector subspace of matrices with trace equal to zero have dimension $n^2 - 1$. Curtis provides the hint that the map from $M_n(F)$ to $F$ is a linear transformation. From this, I used the theorem that $\dim T(V) + \dim n(T) = \dim V$ to obtain the dimension of the null space. Part (d), however, I'm stuck on. It asks the reader to show that subspace described in part (c) is generated by matrices of the form $AB - BA$, where $A$ and $B$ are arbitrary $n \times n$ matrices. I tried to form a basis for the subspace, but wasn't really sure what it would look like since an $n \times n$ matrix has $n^2$ entries in it, but the basis would need $n^2 - 1$ matrixes. I also tried to think of a linear transformation whose image would have the form of $AB - BA$, but this also didn't help me. I'm kind of stuck...
Many thanks in advance!
AI: One way of proving this: Note that for all $A,B$ matrices, $AB−BA$ has trace equal to zero. Denote by $E_{ij}$ the matrix with entry $1$ in row $i$, column $j$ and $H_{ij}:=E_{ii}−E_{jj}$. Then $\{E_{ij}:i\ne j\}∪\{H_{i,i+1}:1\le i\le n-1\}$ form a basis for the space. Also, $H_{ij}=E_{ij}E_{ji}−E_{ji}E_{ij}$ and $2E_{ij}=H_{ij}E_{ij}−E_{ij}H_{ij}$. So, you have a basis formed by elements of the form $AB−BA$. |
H: circles and linear fractional transformations
I'm realizing how little (in some respects) I know about circles. Here's something that emerged out of something I was fiddling with. My question is whether this is
"well known" in the way that $229\times983=225107$ is "well known" (don't publish it unless you're publishing a table); or
well known in the sense that every book includes it (for suitable values of "every"); or
well known in the sense that everybody knows it (for at least moderately reasonable values of "everybody").
I'm looking at the circle $|z|=1$ in $\mathbb{C}$. Let $$f(z)=\dfrac{-3z+1}{z-3}.\tag{This is $f$.}$$ This of course fixes $\pm 1$ and leaves the circle invariant, and maps $\pm i$ to $\dfrac{-3\pm4i}{5}$. If we draw a circle through those two images of $\pm i$ meeting the unit circle at a right angle, it is centered at $-5/3$ and has radius $4/3$. That circle meets the real axis at $-1/3$. So look at the line $\operatorname{Re}=-1/3$. Look at the point on that line where $\operatorname{Im} = y$. Draw the line through that point and the aformentioned center $-5/3$. That line crosses the circle twice. It would seem that those two points are $f(z)$ and $f(-\bar z)$, where $z$ and $-\bar z$ are the two points on the unit circle with imaginary part $y$.
This gives us a simple geometric picture of how $f$ behaves. That allows us to use routine Euclidean geometry to show that $\theta\mapsto f(e^{i\theta})$ satisfies the differential equation
$$
\left|\dfrac{dg}{d\theta}\right| = \text{constant}\cdot\operatorname{Re}\left(g-\left(-\dfrac 5 3\right)\right)
$$
subject to the constraint that the values of $g$ are on the unit circle. (The equation says the rate at which $g$ moves along the circle is proportional to a certain affine function of the real part.)
AI: In my opinion:
I don't believe your construction is "well-known" in any of your three versions of "well-known". I am pretty sure that it is the sort of thing that could appear as an exercise in any of the well-known classical texts on complex analysis, but I am pretty sure I have never done anything quite like this, despite having done many exercises from standard texts.
A very quick search through some texts (Ahlfors, Burckel, Lang, Conway) shows nothing quite like it.
I suspect (without doing any analysis yet) that your results depend on the fact that your transformation is of the form $z\mapsto\frac{z-a}{1-\bar{a}z}$, and would be keen to hear if you have investigated further. |
H: Finding the fixed point of a function
Let $p:A \times B \to \mathbb{R}$ be a nonnegative real-valued function on $A \times B$,
where $A$ and $B$ are arbitrary set.
Assume $f:A \to B$ and $g:B \to A$ are such that
\begin{align*}
f(a) &= \operatorname*{arg\,min}_b~p(a,b) \\
g(b) &= \operatorname*{arg\,max}_a~p(a,b)
\end{align*}
Does the map $f \circ g:B \to B$ have a fixed point?
Which conditions are necessary for this fixed point to exists?
This question arised while searching an equilibrium point for a game.
For my game, I empirically noticed that repeated iteration of the map $f \circ g$ eventually gives me a fixed point.
So I start looking for some theoretical justification of this observation.
I dig hard, but I couldn't really find one.
My game is the fair pricing of an insurance product.
The set $A$ is a convex subset of $\mathbb{R}^n$, and is the set of action of the policyholder.
While the set $B=[0,1]$ is the set of premium of the insurer.
The function $f$ represents the objective function of a fair insurer who wants the price to be zero.
While the function $g$ is the objective function of a rational policyholder who wants to arbitrage the insurer.
The function $p$ is such that $p(a,b)=\operatorname{E}(X \mid \text{Action=a, Premium=b})$
where $X=|\text{claim}-\text{premium}|$ is the random variable of the product cash flow.
I looked into Banach fixed point theorem, Nash equilibrium existence theorem
and Kakutani's fixed point theorem.
Any help or pointer appreciated.
AI: I'm not sure if this works or not, but I think your general problem can basically just be turned into a normal form game.
You have two players with strategy sets $A$ and $B$ where both players have the same utility function $p:A\times B\rightarrow\mathbb{R}$. The players best response functions are $f$ and $g$ respectively as you defined them above.
Right off the bat one could point out that these aren't in general functions but a correspondence (eg. what if it's a constant function?), and take for example if you defined $p:\mathbb{R}-\{0\}\times\mathbb{R}-\{0\}\rightarrow\mathbb{R}$ by $p(a,b) = \frac{1}{a} + \frac{1}{b}$, what's $f(2)$ for example? It may not even be a correspondence.
Now, if $g \circ g$ has a fixed point iff $f \circ f$ has a fixed point then I think $g \circ g$ will have a fixed point iff the game given above has a Nash equilibrium (the game's best response correspondence has a fixed point) using the strategies considered above (ie. if A and B are pure strategy sets then it would require the game has a pure strategy Nash equilibrium which is not in general true, whereas if A and B were probability distributions over some other arbitrary sets (mixed strategies) then by Nash's theorem there would exist at least one Nash equilibria).
So basically you need to consider what game is formed more specifically with your example, at least I think you could argue it this way. |
H: Is there any closed-form expression to calculate each element of the inverse of a matrix?
Considering a generic square matrix $A=(a_{i,j})$ we want to compute its inverse $A^{-1}=\left[a^{(-1)}_{i,j}\right]$.
Is there a way to express each $a^{(-1)}_{i,j}$ using a closed form expression?
AI: The $ij$ entry of $A^{-1}$ is $(-1)^{i+j}$ times the determinant of the matrix $C_{ji}$ obtained by deleting row $j$ and column $i$ from $A$, all divided by the determinant of $A$. I don't know whether you consider that to be a closed form. |
H: Find the imaginary part of this sum
Let $$S = e^{i\alpha} + \frac{e^{i3\alpha}}{3} +
\frac{e^{i5\alpha}}{3^2} + \cdots$$
Find Im$(S)$ and show that it is equal to the sum
$$I = \sin(\alpha) + \frac{\sin(3\alpha)}{3} + \frac{\sin(5\alpha)}{3^2} + \cdots$$
So, I found that $S = \frac{3(3e^{i\alpha} - e^{-i\alpha})}{10 - 6\cos(2\alpha)}$ using the formula for geometric series.
I have a provided answer of $\frac{6\sin(\alpha)}{5 - 3\cos(2\alpha)}$ which I can see that I get if I just take the $\sin(\alpha)$ terms out of the $e$ terms in my numerator; however, why don't I have to change the $\cos(2\alpha)$ term in the denominator? Isn't this part of Re$(S)$? I was confused about his part and was trying to change this term before I looked at the answer.
AI: $$
\begin{align}
S
&=e^{ia}\sum_{k=0}^\infty\left(\frac{e^{i2\alpha}}{3}\right)^k\\
&=\frac{3e^{i\alpha}}{3-e^{i2\alpha}}\cdot\frac{3-e^{-i2\alpha}}{3-e^{-i2\alpha}}\\
&=\frac{9e^{i\alpha}-3e^{-i\alpha}}{10-6\cos(2\alpha)}\\
&=\frac{6\cos(\alpha)+12i\sin(\alpha)}{10-6\cos(2\alpha)}\\
&=\color{red}{\frac{3\cos(\alpha)}{5-3\cos(2\alpha)}}+i\color{green}{\frac{6\sin(\alpha)}{5-3\cos(2\alpha)}}
\end{align}
$$
Therefore,
$$
\mathrm{Im}(S)=\color{green}{\frac{6\sin(\alpha)}{5-3\cos(2\alpha)}}
$$
Since the imaginary part of $\color{blue}{\dfrac{e^{i(2k+1)\alpha}}{3^k}}$ is $\color{orange}{\dfrac{\sin((2k+1)\alpha)}{3^k}}$ we get your second sum
$$
\begin{align}
\mathrm{Im}(S)
&=\mathrm{Im}\left(\sum_{k=0}^\infty\color{blue}{\frac{e^{i(2k+1)\alpha}}{3^k}}\right)\\
&=\sum_{k=0}^\infty\color{orange}{\frac{\sin((2k+1)\alpha)}{3^k}}
\end{align}
$$ |
H: Infinite products - reference needed!
I am looking for a small treatment of basic theorems about infinite products ; surprisingly enough they are nowhere to be found after googling a little. The reason for this is that I am beginning to read Davenport's Multiplicative Number Theory, and the treatment of L-functions in there requires to understand convergence/absolute convergence of infinite products, which I know little about. Most importantly I'd like to know why
$$
\prod (1+|a_n|) \to a < \infty \quad \Longrightarrow \quad \prod (1+ a_n) \to b \neq 0.
$$
I believe I'll need more properties of products later on, so just a proof of this would be appreciated but I'd also need the reference.
Thanks in advance,
AI: I will answer your question
"Most importantly I'd like to know why
$$
\prod (1+|a_n|) \to a < \infty \quad \Longrightarrow \quad \prod (1+ a_n) \to b \neq 0.
"$$
We will first prove that if $\sum \lvert a_n \rvert < \infty$, then the product $\prod_{n=1}^{\infty} (1+a_n)$ converges. Note that the condition you have $\prod (1+|a_n|) \to a < \infty$ is equivalent to the condition that $\sum \lvert a_n \rvert < \infty$, which can be seen from the inequality below.
$$\sum \lvert a_n \rvert \leq \prod (1+|a_n|) \leq \exp \left(\sum \lvert a_n \rvert \right)$$
Further, we will also show that the product converges to $0$ if and only if one of its factors is $0$.
If $\sum \lvert a_n \rvert$ converges, then there exists some $M \in \mathbb{N}$ such that for all $n > M$, we have that $\lvert a_n \rvert < \frac12$. Hence, we can write $$\prod (1+a_n) = \prod_{n \leq M} (1+a_n) \prod_{n > M} (1+a_n)$$ Throwing away the finitely many terms till $M$, we are interested in the infinite product $\prod_{n > M} (1+a_n)$. We can define $b_n = a_{n+M}$ and hence we are interested in the infinite product $\prod_{n=1}^{\infty} (1+b_n)$, where $\lvert b_n \rvert < \dfrac12$. The complex logarithm satisfies $1+z = \exp(\log(1+z))$ whenever $\lvert z \rvert < 1$ and hence $$ \prod_{n=1}^{N} (1+b_n) = \prod_{n=1}^{N} e^{\log(1+b_n)} = \exp \left(\sum_{n=1}^N \log(1+b_n)\right)$$
Let $f(N) = \displaystyle \sum_{n=1}^N \log(1+b_n)$. By the Taylor series expansion, we can see that $$\lvert \log(1+z) \rvert \leq 2 \lvert z \rvert$$ whenever $\lvert z \rvert < \frac12$. Hence, $\lvert \log(1+b_n) \rvert \leq 2 \lvert b_n \rvert$. Now since $\sum \lvert a_n \rvert$ converges, so does $\sum \lvert b_n \rvert$ and hence so does $\sum \lvert \log(1+b_n) \rvert$. Hence, $\lim_{N \rightarrow \infty} f(N)$ exists. Call it $F$. Now since the exponential function is continuous, we have that $$\lim_{N \to \infty} \exp(f(N)) = \exp(F)$$
This also shows that why the limit of the infinite product $\prod_{n=1}^{\infty}(1+a_n)$ cannot be $0$, unless one of its factors is $0$. From the above, we see that $\prod_{n=1}^{\infty}(1+b_n)$ cannot be $0$, since $\lvert F \rvert < \infty$. Hence, if the infinite product $\prod_{n=1}^{\infty}(1+a_n)$ is zero, then we have that $\prod_{n=1}^{M}(1+a_n) = 0$. But this is a finite product and it can be $0$ if and only if one of the factors is zero.
Most often this is all that is needed when you are interested in the convergence of the product expressions for the $L$ functions. |
H: Why a connected subspace of a locally connected space X is locally connected if X is the real line?
Why a connected subspace of a locally connected space X is locally connected if X is the real line?
Is this true if X is an arbitrary locally connected space?
Thanks for your help
AI: This is true if $X=\mathbb{R}$ because connected subsets in this case are precisely the intevals.
For a counterexample for other spaces take $X=\mathbb{R}^2$ and $Y=\{ (x,y): x=0,\ 0\leq y\leq 1\} \cup \{ (x,y) :x\in [0,1],\ y=\sin\left(\frac{1}{x}\right) \}=Y_1\cup Y_2$. Then points in $Y_1$ don't have connected open neighbourhoods (in $Y$). |
H: Relation between defining polynomials and irreducible components of variety
I've been puzzled about some basic facts in (classical) algebraic geometry, but I cannot seem to find the answer immediately:
Let $V=V(f_1,\ldots,f_n)$ be a variety over some field $k$, and let $n > 1$. Suppose that $V$ turned out to be reducible, i.e. it is the union of irreducible varieties $V_1,\ldots,V_m$ for some $m > 1$. Must it be the case that at least one of the $f_i$s are reducible polynomials?
Take one of the irreducible components, $V_1$, say. Do the defining equations for $V_1$ have anything to do with the polynomials $f_1,\ldots,f_n$?
Suppose varieties $V = V(f_1,f_2)$ and $W = V(f_3)$ shared a common component. Does that mean that $f_3$ shares a factor with either one of $f_1$ or $f_2$? If not, what's a counterexample?
Thanks so much!
AI: 1) No. The variety $V=V(y,y-x^2+1)\subset \mathbb A^2_k $ is reducible and consists of the two points $(\pm1,0)$ (if $char.k\neq 2$), but the polynomials $y,y-x^2+1$ are irreducible.
2) The question is not very precise. In a sense the answer is "yes", because you can obtain $V_1$ by adding polynomials $g_1,...,g_r$ to the $f_i$'s and write $V_1=V(f_1,...,f_n;g_1,...,g_r)$
3) Yes if $k$ is algebraically closed :
Decompose $f_3$ into irreducibles: $f_3=g_1^{a_1}\ldots g_s^{a_s}$. The irreducible components of $V(f_3)$ are the $V(g_j)$'s. Say $V(g_1)$ is also an irreducible component of $V_1=V(f_1,f_2)$. Then, since $f_1$ vanishes on $V_1$, the polynomial $g_1$ divides $f_1$ (by the Nullstellensatz) and similarly $g_1$ divides $f_2$. So actually $f_3$ shares the same factor $g_1$ with both $f_1$ and $f_2$, which is more than you asked for.
Caveat If $k$ is not algebraically closed, the answer to 3) may be no: for example over $\mathbb R$ the varieties $V=V(x+y,x^4+y^4)$ and $W=V(x^2+y^2)$ are both equal to the irreducible subvariety of the plane consisting of just one point: $V=W=\lbrace (x,y)\rbrace \subset \mathbb A^2(\mathbb R)=\mathbb R^2$ .
However $x^2+y^2$ is irreducible in $\mathbb R[x,y]$ and shares no factor with $x+y$ nor $x^4+y^4$ since $x^2+y^2$ does not divide those polynomials. |
H: Find all functions with $f(x + y) + f(x - y) = 2 f(x) f(y)$ and $\lim\limits_{x\to\infty}f(x)=0$.
Determine all functions $f \colon\mathbb{R}\to\mathbb{R}$ satisfying the following two conditions:
(a) $f(x +y) + f(x - y) = 2 f(x) f(y)$ for all $x, y\in\mathbb{R}$;
(b) $\lim\limits_{x\to\infty}f(x) = 0$.
I found this problem in IMO 1985 longlist. I have been able to figure out that there is one solution of the form $f(x) = 0$ for all $x$.
Any other function satisfying the above has to have $f(0) = 1$ and $f(x) = f(-x)$.
However I don't know how to proceed.
Any thoughts?
AI: For any $x_0,y\in\mathbb{R}$, let $x=y+x_0$, then we have
$f(2y+x_0)+f(x_0)=2f(y+x_0)f(y)$
Hence,
$f(x_0) = 2f(y+x_0)f(y)-f(2y+x_0),\quad \forall y\in\mathbb{R}$
Consequently,
$\displaystyle f(x_0) = \lim_{y\to\infty}\left[f(x_0)\right] = \lim_{y\to\infty}\left[2f(y+x_0)f(y)-f(2y+x_0)\right] = 0$. |
H: Laplace transform having this unusual property in convolution?
Here is the problem
Solve $y'(t) = 1 - \int_{0}^{t} y(t - v)e^{-2v}dv$
The solution sets $\mathcal{L}(y) = Y(s)$ and does the following
Notice that in step 1, they have $$Y(s)\dfrac{1}{s+2}$$
Are they implying $$\mathcal{L}(y(t) * e^{-2t}) = \mathcal{L}(y(t)) \mathcal{L}(e^{-2t}) = Y(s)\mathcal{L}(e^{-2t}) = Y(s)\dfrac{1}{s+2}$$
AI: Yes they are implying that. Applying the Laplace transform to a convolution gives a product. Write
$$\mathcal{L}\{f*g\}(s):=\int_0^\infty \left(\int_0^x f(u)g(x-u)du\right)e^{-sx}dx=\iint_D f(u)g(x-u)e^{-sx}dudx$$
The region of integration in the $xu$-plane is $D=\{(x,u):0\le u\le x\}$, an infinite triangle. Writing out the substitution $v=x-u$ our region of integration in the $uv$-plane is simply $(0,\infty)\times(0,\infty)$, so
$$=\iint_{(0,\infty)^2} f(u)g(v)e^{-s(u+v)}dudv=\int_0^\infty f(u)e^{-su}du\int_0^\infty g(v)e^{-sv}dv=\mathcal{L}\{f\}(s)\cdot \mathcal{L}\{g\}(s).$$
Note the Jacobian determinant of the transformation $(x,u)\mapsto(u,x-u)$ is simply $1$ (in abs. value). |
H: Name of this angle?
Given a planet and a point $P$, is there an existing name for the angle $\theta$ as seen in the diagram below? If not, what would you call it? ("Angle of elevation"?)
Thanks!
AI: I think you want the "angular radius" of the planet, although people more commonly use the "angular diameter" instead. |
H: Permuting 15 books about 2 shelves, with at least one book on each shelf.
From Discrete and Combinatorial Mathematics: An Applied Introduction:
Pamela has 15 different books. In how many ways can she place her books on two selves so that there is at least one book on each shelf? (Consider the books in each arrangement to be stacked one next to the other, with the first book on each shelf at the left of the shelf.)
Initially, I thought that this would be fairly straight forward. There are two cases where either shelf could be empty and so we would subtract 2 from the total number of permutations which is obviously $15!$. It is obvious now that both of these statements are false.
My previous thoughts didn't take into account the fact that there are two shelves and that the books could be stacked in such a way that 5 are on the top shelf and 10 are on the bottom, or 8 on top with 7 on bottom, or any number of other ways. I believe there are $2^{14}$ ways the books can be distributed amongst the two shelves (keep in mind that each shelf has to have at least one book, $15 - 1 = 14$) if the ordering of the books didn't matter, but it does. Is this correct?
My book also talks about the rule of product:
If a procedure can be broken down into first and second stages, and if there are $m$ possible outcomes for the first stage and if, for each of these outcomes, there are $n$ possible outcomes for the second stage, then the total procedure can be carried out, in designated order, in $mn$ ways.
If we consider that the top shelf represents the first "stage" and the bottom represents the second and that $b_{top}$ represents the number of books on the top shelf and $b_{bottom}$ represents the number of books on the bottom shelf, then, for each of the $2^{14}$ ways the books can be distributed amongst the shelves, there are $b_{top}! * b_{bottom}!$ ways the books can be ordered, where $b_{top} + b_{bottom} = 15$.
However, I can't find a way to turn this into a number. This leads me to believe that there is a formula that I should be using that I'm overlooking. What am I missing? Please keep in mind that I'm in the very early stages of this book (page 12, to be precise), so nothing too advanced. :)
AI: Breaking it into stages is the right idea, but the stages aren’t the two shelves. The first stage is putting all $15$ books onto the first shelf in some particular order. The second stage is deciding where to ‘cut’ that line of books to transfer the tail end to the second shelf. Can you finish it from there?
Added: There is a way to get the result by thinking first of how to split the books between the shelves, but it’s a bit messier. There are $\binom{15}k$ ways to decide which $d$ books to put on the first shelf. Once we’ve put them there, we can sort them in $k!$ ways, and independently we can sort the $n-k$ books on the second shelf in $(n-k)!$ ways. Since we have to have at least one book on each shelf, $k$ can range from $1$ through $14$, and the desired number is
$$\sum_{k=1}^{14}\binom{15}kk!(n-k)!=\sum_{k=1}^{14}\frac{15!}{k!(n-k)!}k!(n-k)!=\sum_{k=1}^{14} 15!=14\cdot15!\;.$$ |
H: Constructing the sequence: $0\rightarrow (x-y)^{S_2} \stackrel{f}{\rightarrow} k[x+y,xy]\stackrel{g}{\rightarrow} k[y]$
Let $S_2$, a group of two elements, act on $k[x,y]$ by permuting $x$ and $y$.
It is clear that
$$
0\rightarrow (x-y) \rightarrow k[x,y]\rightarrow \dfrac{k[x,y]}{(x-y)}\cong k[y] \rightarrow 0
$$
is exact.
Taking its invariant subrings, we obtain
$$
0\rightarrow (x-y)^{S_2} \rightarrow k[x,y]^{S_2}\rightarrow k[y]^{S_2},$$
which simplifies as
$$
0\rightarrow (x-y)^{S_2} \stackrel{f}{\rightarrow} k[x+y,xy]\stackrel{g}{\rightarrow} k[y].
$$
What are $f$ and $g$ concretely?
AI: The elements of the ideal $(x-y)$ are of the form $u=p(x,y)(x-y)$, and $p$ is antisymmetric if and only if the element $u$ is $S_2$-invariant. As such, $u$ being a symmetric polynomial, it can be written as a polynomial in the elementary symmetric polynomials $e_1(x,y)=x+y$ and $e_2(x,y)=xy$, hence we can let $f$ simply be the inclusion map. Just as before, $g$ is the evaluation map $x,y\mapsto y$. |
H: Why should a topological space itself be open?
For convenience, let $X$ be our space.
Specifically, can anyone name a few desirable properties or theorems that would fail if $X$ weren't required to be open? More generally, is there a part of topology that would completely fall apart?
It seems to me that we mainly want closure under arbitrary unions and finite intersections, which appears to be the more natural part of the definition (unlike "forcing" $\varnothing$ and $X$ to be open, which feels rather contrived). Of course, we get the empty set for free if we take an arbitrary union of nothing ($\bigcup \varnothing = \varnothing$), so that part really doesn't need to be in the definition.
Let's define a new word: A tolology on $X$ is a subset of $\mathcal P(X)$ that is closed under arbitrary unions and finite intersections. By the to(p/l)ological closure of a set I'm referring to the smallest to(p/l)ology containing it.
Let $\mathscr T$ be a topology on $X$ and consider $\mathscr T' = \mathscr T \setminus \{X\}$. If $\bigcup \mathscr T' = X$, then closure under arbitrary unions forces us to throw $X$ back in anyway, so the topological closure coincides with the tolological closure, nothing interesting here. Otherwise (this is what bothers me), we have $\bigcup \mathscr T' \subsetneq X$, then $\mathscr T'$ is still closed under arbitrary unions and finite intersections, so it's a tolology but not a topology. But throwing in $X$ adds nothing to the richness of the to(p/l)ology at all.
In fact, let's say $\bigcup \mathscr T' \subsetneq X$. Then the topological closure of $\mathscr T'$ is $\mathscr T$, but we still have a pretty boring space. For, if $\left|X \setminus \bigcup \mathscr T' \right| = 1$, then our space is not $T_1$, and if $\left|X \setminus \bigcup \mathscr T' \right| \geq 2$, then it's not even $T_0$. (Actually, any $T_1$ space must satisfy $\bigcup \mathscr T' = X$ by definition, and that probably covers just about every theorem in topology.)
Since un-requiring $X$ to be open doesn't give us any fewer theorems than we already have, and those spaces whose topology and tolology are different are as uninteresting as it gets, why can't we replace the definition of topology with that of tolology for simplicity's sake?
The only argument I can think of against this is that the first Kuratowski closure axiom says $\overline \varnothing = \varnothing$, so $\varnothing$ is closed, which means $X$ is open. But why do we need that first axiom?
AI: I think that it is reasonable to expect that a constant function from any space to any space would be continuous, and this of course true if and only if the all space is an open set.
Edit: I just realized that more is true: Suppose $X$ is a space in which $X$ is not open. Then there are no continuous maps from $X$ to spaces in which the all space is open, because if there is such a map $f:X\to Y$, then we must have $f^{-1}(Y) = X$ is open, which is false. |
H: Show that this function is not increasing on any interval containing $0$:
$$f(x) = \begin{cases}x + 2x^2\sin\left(\frac1x\right),& x\ne 0\\0,& x = 0\;.\end{cases}$$
I am having a tough time answering this question in a rigorous mathematical way, here is what I have tried:
I have proved in a previous part of this question that $f\,'(0) = 1$.
The derivative when $x\ne 0$ is,
$$f\,'(x) = 1+4x\sin\left(\frac1x\right)−2\cos\left(\frac1x\right)\;.$$
I have $$\lim\limits_{x\to 0} f\,'(x) = 1 - 2\cos\left(\frac1x\right)\;,$$ which oscillates between $3$ & $-1$.
Since $f\,'(x)$ containing $0$, is not continuous, it cannot be increasing on the interval. Am I on the right track here?
Thanks in advance!
AI: You're on the right track, but the equation $$\lim\limits_{x\to 0} f\,'(x) = 1 - 2\cos\left(\frac1x\right)$$
doesn't make sense as it stands. It almost makes sens, though, in that the omitted term goes to zero as $x\to0$. So it is definitely true that the derivative oscillates infinitely much near the origin, with upper and lower bounds that approach $3$ and $-1$. For your purposes, though, you just need a sequence of points approaching $0$ where the derivative is negative. Good candidates for such a sequence are points $x$ where $\cos(1/x)=-1$. I trust you can take it from there on your own. |
H: How to resolve Skolem's Paradox by realizing what can be said of a set is relative to what is in the domain of some model?
I apologize in advanced if I'm hopelessly confused...
Skolem's Paradox, I suppose, can be put like this:
$M$ is a countable model of ZFC and $M$ implies the existence of uncountable sets.
I suppose that people find this initially paradoxical because they assume the statement is said from a single, absolute perspective. However, there are (necessarily) two perspectives involved: the inside perspective on $M$ and the outside perspective on $M$. Once these perspectives are separated, we realize that there is no paradox. Consider:
The former conjunct "$M$ is a countable model of ZFC" is necessarily said from an outside perspective on $M$ -- as discussed here. Actually, $M$ can't express its own cardinality at all.
(Is $M$'s inability to express its own cardinality related to $M$'s being a proper class -- namely, that there is no function in $M$ that takes one of $M$'s members onto the universe of $M$?)
Continuing on…Let $N$ be the outside perspective on $M$ such that there is a bijection $f\in N$ between the domain M of $M$ and $\omega^N\in N$.
The latter conjunct "$M$ implies the existence of uncountable sets" is obviously said from $M$'s inside perspective -- after all, $M$ is a model of ZFC and thus must satisfy Cantor's Theorem.
So, we can separate the perspectives of the paradoxical statement above here:
From $M$'s perspective, $M$ is a proper class and there is some $A\in M$ such that $A$ is uncountable in $M$.
From $N$'s perspective, $M$ is countable.
These two statements are jointly consistent when we realize that what can be said of a set $B$ is relative to what some model has to say about $B$. And so, the paradoxical statement isn't so paradoxical.
Is there anything wrong with what I have written above? (It took me a long time to learn this stuff, esp. with zero background in set-theory, higher-math, higher-logics, model-theory, etc., so specifically telling me where I am going wrong, if I am, will be a great help and a great relief.)
What I'm really interested in is what can we say about $A$ from $N$'s perspective? Are the following possible:
$M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as finite.
$M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as countable.
$M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as uncountable.
Under what conditions might (1) - (3) be individually possible (obviously they can't be jointly possible)?
I suspect this question might be rather simple. For example, (2) could be possible when $N$ recognizes a bijection both between $M$ and $\omega^N$ and between $A$ and $\omega^N$. (3) could be possible when $N$ recognizes a bijection between $M$ and $\omega^N$ but doesn't recognize a bijection between $A$ and $\omega^N$.
My goal here is to understand/stress the fact that truth in model theory is relative to what particular models have to say about their members. So, I'm trying to see that while $M$ may take $A$ to be uncountable, $N$ can take $A$ to be of any cardinality even under the condition that $N$ sees $M$ as countable.
AI: The second and third are easily describable:
Suppose that $N$ is a model of ZFC and $N$ thinks that $M$ is a countable transitive model of ZFC ($M$ may not be such model, but internally to $N$ this assertion holds).
This means that $N$ thinks that $M$ is countable, and that every element of $M$ is countable. $M$, on the other hand, knows some sets which are uncountable to it. So we have $\omega_1^M$ is a countable ordinal in $N$, so the second situation holds.
Suppose now that we have a nice model of ZFC, $N$, which is uncountable and it knows about uncountable sets. If we take $\omega_1^N$ we can consider $M$ a countable elementary submodel of $N$ such that $\omega_1^N\in M$. By elementarity $M$ and $N$ agree on $\omega$ and both agree that there is no bijection between that set and $\omega_1^N$. So we have the third situation in which both models agree that some set is uncountable.
Lastly to address the first situation, what we want is to have an ill-founded model of ZFC which thinks of an uncountable set as its $\omega$, but that it will also know about some model which is nicer. I am not sure how to address this situation since for $N$ to think that $M$ is a model of ZFC, $N$ would have to assert that $M$ have certain properties in $N$. These properties may make it impossible to make the jump from infinite to finite in this manner.
There is a possibility that $N$ will not be aware that $M$ is a model of ZFC, but that defeats the purpose because then we talk from an external point of view about both these models. |
H: Matrix with no eigenvalues
Here is another problem from Golan.
Problem: Let $F$ be a finite field. Show there exists a symmetric $2\times 2$ matrix over $F$ with no eigenvalues in $F$.
AI: The solution is necessarily split into two cases, because the theory of quadratic equations has a different appearance in characteristic two as opposed to odd characteristic.
Let $p=\mathrm{char}\, F$. Assume first that $p>2$. Consider the matrix
$$
M=\pmatrix{a&b\cr b&c\cr}.
$$
Its characteristic equation is
$$
\lambda^2-(a+c)\lambda-(ac-b^2)=0.\tag{1}
$$
The discriminant of this equation is
$$
D=(a+c)^2-4(ac-b^2)=(a-c)^2+(2b)^2.
$$
By choosing $a,c,b$ cleverly we see that we can arrange the quantities $a-c$ and $2b$ to have any value that we wish. It is a well-known fact that in a finite field of odd characteristic, any element can be written as a sum of two squares. Therefore we can arrange $D$ to be a non-square proving the claim in this case.
If $p=2$, then equation $(1)$ has roots in $F$, if and only if $tr((ac-b^2)/(a+c)^2)=0.$
By selecting $a$ and $c$ to be any distinct elements of $F$, we can then select $b$ in such a way that this trace condition is not met, and the claim follows in this case also. |
H: Infimum of a union
I have a set $X$ and a function
\begin{equation}
f: X \rightarrow \mathbb{R}
\end{equation}
and I am interested in the value
\begin{equation}
\inf\limits_{x \in X} f(x) \,.
\end{equation}
I can represent $X$ as
\begin{equation}
X = \bigcup\limits_{i \in I} X_i \,,
\end{equation}
where the index set $I$ is uncountable. Now I wonder whether
\begin{equation}
\inf\limits_{x \in X} f(x) = \inf\limits_{i \in I} \left( \inf\limits_{x \in X_i} f(x) \right) \,.
\end{equation}
Is this true? If so, how can I see this?
AI: Yes, it’s true. Let $\alpha=\inf\limits_{x\in X}f(x)$, and for each $i\in I$ let $\alpha_i=\inf\limits_{x\in X_i}f(x)$. Clearly $\alpha\le\alpha_i$ for each $i\in I$, so $\alpha\le\inf\limits_{i\in I}\alpha_i$.
On the other hand, for each $x\in X$ there is an $i\in I$ such that $x\in X_i$ and therefore $\alpha_i\le f(x)$, so $\inf\limits_{i\in I}\alpha_i\le\inf\limits_{x\in X}f(x)=\alpha$. (Alternatively if $\alpha<\inf\limits_{i\in I}\alpha_i$, then there is some $x\in X$ such that $$\alpha\le f(x)<\inf_{i\in I}\alpha_i\;.$$ But then $f(x)<\alpha_i$ for all $i\in I$, which is clearly impossible.)
Note that the cardinality of $I$ doesn’t matter. |
H: Linear dependence of linear functionals
Problem: Let V be a vector space over a field F and let $\alpha$ and $\beta$ be linear functionals on $V$. If $\ker(\beta)\subset\ker(\alpha)$, show $\alpha = k\beta$, for some $k\in F$.
A proposed solution is in the answers below.
AI: If $\alpha$ is the zero functional, we are done, because we take $k=0$.
Otherwise, consider a basis $\{v_\alpha\}$ of $V$. Let $\{v_p\}$ be the vectors that $\beta$ maps to nonzero scalars, and the $\{v_r\}$ the basis vectors mapped to zero. Then $\alpha$ must also map every vector in $\{v_r\}$ to 0, by hypothesis.
If $\{v_p\}$ contains just one vector we are done, because we can just scale $\beta$ so that $\alpha$ and $\beta$ agree on this basis vector. Otherwise, choose two vectors $v_1$ and $v_2$ in this set. Let $\beta(v_1)=b_1$, $\beta(v_2)=b_2$, $\alpha(v_1)=a_1$, and $\alpha(v_2)=a_2$. We want to show that $b_1/a_1=b_2/a_2$. Assume not. Then consider the vector $b_2v_1-b_1v_2$. We see that $\beta$ maps this to 0, but $\alpha$ does not, a contradiction. |
H: complex polynomial satisfying inequality
Each of the polynomial of the form $p(z)=a_0+\dots+a_{n-1}z^{n-1}+z^n$ satisfies the inequality
$\sup\left\{\,|p(z)|\,\big\vert\,|z|\le 1\,\right\}\ge 1$
Is this statement true or false that we have to find. well MMP says that sup will be attained at $|z|=1$ so when $|z|=1$ we have $|p(z)|=|a_0+a_1\dots+ a_{n-1}+1|\le |a_0|+\dots+|a_{n-1}|+|1|$ I can not conclude more.plz help.
AI: Using Rouche theorem if
$$| z^n+a_0+\cdots +a_{n-1}z^{n-1}| < |z^n|+|a_0+\cdots +a_{n-1}z^{n-1}|$$
for every $z$ with $|z|=1$ then these $z^n$ and $a_0+\cdots +a_{n-1}z^{n-1}$ would have the same amount of roots inside the circe, so this cannot be happening cause of the degrees. so there is a point of equality and you get what you want. |
H: Finding the Laplace transform of $f(x)=|\cos(x)|$
I have function $f(x)=|\cos(x)|, x≥0$ and like to derive its Laplace transform. I am told that $f(x+\pi)=f(x)$. Help me please.
AI: Your function is periodic ($T=\pi$) so you can easily use the formula:
$L(f(x))=\frac{1}{1-e^{-sT}}\int_{0}^{T}e^{-sT}f(x)dx$
Note that your function is a piecewise function broken at $\frac{\pi}{2}$. |
H: Question on Topological vector space 1
I have numbered this question as (1) because I will be posting series of questions where I don't understand. I hope its allowed.
I want to prove the following :
If $X$ is a topological vector space then : If $A\subset X $ then
$\bar A = \cap(A+V)$, where $V$ runs through all nbd of $0$.
I tried like this :
$x\in \bar A\subset A+V$ $\implies x\in A+V $ for all V.
The other way , $x\in A+V$ $\implies$ $x\in A$ which means $x\in \bar A$.
Is my argument correct? If it fails what should I take care of? Thank you very much.
AI: For the other direction it's helpful to remember that $x \in \bar{A}$ if and only if every neighbourhood $U$ of $x$ intersects $A$. So let $U$ be an arbitrary neighbourhood of $0$, whence $x + U$ is an arbitrary neighbourhood of $x$. If $x \in \bigcap (A + V)$, then in particular $x \in A - U$ so that we can write $x = a - u$ for some $a \in A$, $u \in U$. Thus $a \in x + U$ and we see that $x + U$ intersects $A$. |
H: Finite set of matrices closed under multiplication
The following problem is from Golan's linear algebra book. I have been unable to make headway.
Problem: Let $n\in\mathbb{N}$ and $U$ be a non-empty finite subset of the $n\times n$ matrices over $\mathbb{C}$ which is closed under the multiplication of matrices and contains more than just the zero matrix. Show there exists a matrix $A$ in $U$ satisfying $tr(A)\in \{1,...,n\}$
EDIT: As noted in the comments, this problem is incorrect as stated. Perhaps it is correct if we allow $0$ to be in the set of desired values, or if we require the set to contain a non-singular matrix? Any help reformulating the problem would be much appreciated.
AI: Pick any matrix $A$ in your finite set, and look at its powers.
Since they are in a finite set, they repeat at some point :
there exists $p,k \ge 1$ such that $A^p = A^{p+k}$.
In particular, if $m$ is a multiple of $k$ and greater than $p$, then $A^{2m}=A^m$.
Thus $A^m$ is the matrix of a projection. It shouldn't be too hard now to show that its trace is an integer between $0$ and $n$
And the trace is $0$ if and only if $A$ was nilpotent in the first place. Thus you would need to accept $0$ or require that the set contains non nilpotent matrices for the statement of the exercise to be correct. |
H: Simple harmonic motion and trigonometry
Here's the question I would like help with:
A particle is moving in simple harmonic motion according to $x=6\sin
\left (2t+\frac{\pi }{2} \right )$. Find the first two times when the
velocity is maximum, and the position then.
Here is my working. I then let $x=0$ and did the following:
$$0=6\sin \left (2t+\frac{\pi }{2} \right )$$
$$\pi =2t+\frac{\pi }{2} $$
$$\frac{\pi}{2}=2t$$ $$\frac{\pi}{4}=t$$
According to my textbook, the answer I got is incorrect. The provided answer is $t=\frac{3\pi}{4}, \frac{7\pi}{4}$.
Could some one please identify where I went wrong and explain the proper way of solving this question?
AI: The velocity is given by $v=12\cos\left(2t+\pi/2\right)$, which is maximised when the cosine of the part in brackets is 1. This happens when $2t+\pi/2$ is a whole multiple of $2\pi$. The first two such values of $t$ are $3\pi/4$ and $7\pi/4$. |
H: Question about generating functions
I have question about generating functions.
I need to make this equation:
$(\frac{1}{1+x})^n\centerdot(1+x)^{2n} = (1+x)^n$
in this form:
$\sum\limits_{i=0}^{k}(-1)^iD(?,?)\binom{?}{?} = \binom{n}{k}$
How can I do this?
Thanks in advance
AI: The coefficient of $x^k$ in $(1+x)^n$ is $\binom{n}k$, so you want to work out the coefficient of $x^k$ on the lefthand side. You know that $$(1+x)^{2n}=\sum_{i\ge o}\binom{2n}ix^i\;,$$ and you probably know that $$\frac1{(1-x)^n}=\sum_{i\ge 0}\binom{n-1+i}ix^i\;,$$ so that $$\frac1{(1+x)^n}=\sum_{i\ge 0}(-1)^i\binom{n-1+i}ix^i\;.$$
Thus, $$\frac{(1+x)^{2n}}{(1+x)^n}=\left(\sum_{i\ge 0}(-1)^i\binom{n-1+i}ix^i\right)\left(\sum_{i\ge 0}\binom{2n}ix^i\right)\;.\tag{1}$$
Now just expand to find the coefficient of $x^k$. I’ve done the rest below but spoiler-protected it; mouse-over to see it.
The coefficient of $x^k$ in the product $$\left(\sum_{i\ge 0}a_ix^i\right)\left(\sum_{i\ge 0}b_ix^i\right)$$ is $$\sum_{i=0}^ka_ib_{k-i}\;,$$ so the coefficient of $x^k$ in $(1)$ is $$\sum_{i=0}^k(-1)^i\binom{n-1+i}i\binom{2n}{k-i}\;.$$ |
H: Finding Big-Theta
I need to use the Master Theorem to find $\Theta(f(n))$ if $f(n)=f(n/2)+3n$ and $f(1)=3$
I don't know how to use the MT in this case, can anyone provide help?
AI: The Master Theorem concerns relations of the form
$$f(n) = a f(n/b) + g(n)$$
which fits your problem with $a=1$, $b=2$ and $g(n)=3n$.
Begin by computing the quantity $\log_ba = \log_21=0$, and check which (if any) of the following three quantities are satisfied:
$g(n)=O(n^{\log_ba-\epsilon})$ for $\epsilon>0$
$g(n)=\Theta(n^{\log_ba} \log^k n)$ for $k\geq 0$
$g(n)=\Omega(n^{\log_ba+\epsilon})$ for $\epsilon>0$, and $ag(n/b)\leq cg(n)$ for some constant $c$ and $n$ sufficiently large
If one of these conditions holds, then you can apply the relevant section of the Master Theorem. |
H: Bat and ball calculations
How would you work this out in MS-Excel?
A bat and ball cost a dollar and ten cents. The bat costs a dollar more than the ball. How much does the ball cost?
The answer is that the bat costs $1.05 and the ball costs $0.05.
Source: http://gizmodo.com/5918045/why-smart-people-are-actually-dumb
AI: You can do this mathematically by using the following variables: $T$ (the total cost of the items), $B$ (the cost of the bat), and $b$ (the cost of the ball).
We know that $T=B+b=\$1.10$, and we also know that $B=\$1.00+b$, we plugging this into our original equation, we have:
$$T=\$1.00+2b=\$1.10\implies 2b=\$0.10 \\ \therefore b=\frac{\$0.10}{2}=\$0.05$$
So we now know the ball costs $5¢$, so to get the price of the bat, we can put:
$$B=\$1.00+\$0.05=\$1.05$$
So we now know that the bat must cost $\$1.05$, and the ball costs $\$0.05$.
You could do this in excel by doing each of the calculuations in a cell, but that would simply be superfluous, it's quite easy to do this mentally, or on paper, as you can see.
Hope this clears things up for you. |
H: Expression for $n$-th moment
I stumbled upon an expression in an article of statistics for an $n$-th moment with $X$ being a random variable over $[0, \infty)$.
$$\mathbb{E} X^{n} = \int^{\infty}_{0} nz^{n-1}\; \text{Pr}(X > z) \; \text{dz}$$
Could someone enlighten me on why the above is true? It indeed works for the exponential distribution.
AI: First, use Tonelli's theorem to conclude that (write the probability as an integral and interchange the two integrals)
$$
E[X^n]=\int_0^\infty P\left(X^n> z\right)\, \mathrm{d} z.
$$
Now write $P\left(X^n> z\right)=P\big(X> z^{1/n}\big)$ and use change of variables with $t=z^{1/n}$. |
H: Characterization of an element being algebraic over $\mathbb{Q}$.
Let $Aut(\mathbb{C}/\mathbb{Q})$ be the set of field automorphisms of $\mathbb{C}$ over $\mathbb{Q}$ (in short, all field automorphisms of $\mathbb{C}$). Let $x$ be an element of $\mathbb{C}$ such that the set $\{\sigma(x)|\sigma \in Aut(\mathbb{C}/\mathbb{Q})\}$ is finite. Is it true then that $x$ is algebraic over $\mathbb{Q}$ (and if so, why?) ?
It's being used in the following paper about elliptic curves to prove that a elliptic curve with complex multiplication has modular invariant $j$ which is algebraic over $\mathbb{Q}$:
http://www.math.tifr.res.in/~eghate/cm.pdf
Any help would be appreciated.
AI: All the permutations of any transcendence basis can be lifted to an automorphism $\sigma$ of $\mathbb{C}$. The cardinality of any transcendence basis of $\mathbb{C}/\mathbb{Q}$ is uncountable. So if $x$ is transcendental over $\mathbb{Q}$, it can be included to a transcendence basis, and hence the set
$$\{\sigma(x)\mid\sigma\in Aut(\mathbb{C})\}$$
would be uncountable. The claim follows. |
H: Research in algebraic topology
I have started studying algebraic topology with the help of Armstrong(Basic), Massey, and Hatcher. If I plan to do research in algebraic topology in future:
What else should I study after completing homology(basic), cohomology(basic) and homotopy theory(basic)?
After completing Hatcher how far I would be (in terms of time and effort) from tackling a research problem?
I have average background in Algebra and never studied Category theory in detail.However I feel comfortable working with algebra. I would like to work in those areas which require more algebraic machinery than any other area and which are more Geometric in flavour.
Are there other areas to which I should switch over to like Geometric topology or algebraic geometry?
AI: Let me attempt to answer this question. I should mention that I am not a research algebraic topologist. In fact, I am a student of algebraic topology and I hope to one day become a researcher in the area. I am currently on the path toward this goal.
Let me begin by saying that you are definitely on the right track by reading Hatcher's textbook. I think that the most fundamental topics of algebraic topology are covered in Hatcher's textbook and a knowledge of these topics will be very useful to you as a research mathematician no matter in which area of mathematics you specialize. I will assume that you have completed Hatcher's book and you are interested in further topics in algebraic topology.
I think the next step in algebraic topology (assuming that you have studied chapter 4 of Hatcher's book as well on homotopy theory) is to study vector bundles, K-theory, and characteristic classes. I think there are many excellent textbooks on this subject.
My favorite book in K-theory is "K-theory" by Michael Atiyah although some people object because they feel that the proof of Bott periodicity in this book is not very intuitive but rather long and involved (and I agree). However, you may as well assume Bott periodicity on faith if you read this book as the techniques used in proving Bott periodicity are not used or mentioned elsewhere in the book (although minor exceptions may show this statement to be false). I think a very slick proof of Bott periodicity is discussed in the paper "Bott Periodicity via Simplicial Spaces" by Bruno Harris. I would recommend you to read this paper if you are interested in a proof of Bott periodicity.
Alternatively, you may wish to learn from Hatcher's textbook entitled "Vector Bundles and K-theory" (available free online from his webpage) or the textbook by Max Karoubi entitled "K-theory: An Introduction". Hatcher's book discusses the image of the J-homomorphism (in stable homotopy theory) which is an important an interesting application of K-theory. I don't think that this is discussed in Atiyah's textbook. Similarly, Hatcher has a more detailed description of the Hopf-invariant one problem than that of Atiyah's book. Thus a good plan would be to read Atiyah's textbook and supplement it with a reading of the Hopf-invariant one problem and the J-homomorphism in Hatcher's book. Alternatively, you could read Karoubi's book which is much lengthier than the two (combined) but is an excellent textbook as well.
If you learn vector bundles and K-theory very well, then you should also learn the theory of characteristic classes. I believe that this is discussed in some detail in Hatcher's book (the same one entitled "Vector Bundles and K-theory") and the most basic properties of characteristic classes are proved. However, a more detailed discussion of characteristic classes can be found in the book entitled "Characteristic Classes" by Milnor and Stasheff. I would recommend reading the latter book if you have time and wish to learn about characteristic classes fairly thoroughly. Otherwise, the minimal treatment of characteristic classes in Hatcher's book is also sufficient in the short-term.
A good topic to learn about at this stage is spectral sequences. Spectral sequences furnish an extremely useful and efficient computational tool in algebraic topology. I can't really recommend the good book on spectral sequences because there are many but you might wish to look at "A User's Guide to Spectral Sequenes" by John McCleary and Hatcher's book on spectral sequences (available free online on his webpage).
Finally, you should now learn homotopy theory in more depth. An excellent place to do this is "Stable Homotopy and Generalized Homology" by Frank Adams. Unfortunately, this is as far as I can advise you because this is as far as I have progressed in algebraic topology. I think once you finish the book "Stable Homotopy and Generalized Homology" by Frank Adams the next step could be to start reading research papers (which you have to do sooner or later). Of course, advice on reading research mathematics papers is long and involved so I won't go into details in this answer as we are discussing algebraic topology. But, the books I suggested should keep you busy at least in the short term.
I hope this helps! |
H: Is there any difference between the absolute values operators $|z|$ and $\|z\|$?
Is there any difference between the absolute values operators $|z|$ and $\|z\|$ where $z=a+ib$?
AI: Usually, no. But if you see both notation used in the same discussion, it is possible that the author intended to define $\|z\|$ to be a different norm; in that case it will strictly depend on context. |
H: How to solve motion question?
A particle is moving at $x=3\cos\left(2t\right)$. Find the expression for velocity in terms of $x$.
I'm not sure where to start.
AI: All right, here's a slightly cleaner version of previous answers. We have
$$x=3\cos 2t,\frac{dx}{dt}=-6\sin 2t$$
Again, we use the trig identity
$$\sin^2u+\cos^2u=1$$
Substituting $u=2t$ we have
$$\sin^2 2t+\cos^2 2t=1$$
Now we multiply this equation by 9.
$$9\sin^22t+9\cos^22t=9$$
$$9\sin^22t+x^2=9$$
$$9\sin^22t=9-x^2$$
$$3\sin2t=\pm\sqrt{9-x^2}$$
$$\frac{dx}{dt}=-6\sin 2t=\pm2\sqrt{9-x^2}$$
The plus or minus depends on the value of $t$. |
H: Is a matrix multiplied with its transpose something special?
In my math lectures, we talked about the Gram-Determinant where a matrix times its transpose are multiplied together.
Is $A A^\mathrm T$ something special for any matrix $A$?
AI: The main thing is presumably that $AA^T$ is symmetric. Indeed $(AA^T)^T=(A^T)^TA^T=AA^T$. For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real.
Moreover if $A$ is invertible, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$
Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors.
Last but not least if one is interested in how much the linear map represented by $A$ changes the norm of a vector one can compute
$$\sqrt{\left<Ax,Ax\right>}=\sqrt{\left<A^TAx,x\right>}$$
which simplifies for eigenvectors $x$ to the eigenvalue $\lambda$ to
$$\sqrt{\left<Ax,Ax\right>}=\sqrt \lambda\sqrt{\left<x,x\right>},$$
The determinant is just the product of these eigenvalues. |
H: Prove: if $A(x)$ is divisible by $(x-a)^m$, then $A'(x)$ is divisible by $(x - a)^{m-1}$ [Derivative decrements multiplicity]
I have this question in my textbook:
If the polynomial $A(x)$ is divisible by $(x - a)^m$, then it's derivative is divisible by $(x - a)^{m - 1}$. Prove this.
I have really no clue on how to tackle this question. I probably have to use the formula $D(x^n) = nx^{n-1}$, but I have no clue were to start applying it.
Any tips would be greatly appreciated.
AI: Hints: First, $(PQ)'=P'Q+PQ'$. Second, the derivative of $(X-a)^m$ is $m(X-a)^{m-1}$. |
H: What was the book about birds and sets?
Possible Duplicate:
KY Birds…which book is that from.
Does anyone know the title of the book that taught sets (I think) through examples with the birds?
AI: Possibly you’re thinking of Raymond Smullyan’s To Mock a Mockingbird, though its topic is combinatory logic, not set theory. |
H: Composition of Analytic Functions
I have a basic question in my mind and wish to consult your ideas:
Suppose $\Omega_1$ and $\Omega_2$ are regions, $f$ and $g$
are nonconstant functions defined in $\Omega_1$ and $\Omega_2$,
respectively, and $f(\Omega_1) \subset \Omega_2$. Define $h=g \circ f$.
What can we say about the third function if
(a) both $g$ and $f$ are analytic;
(b) both $g$ and $h$ are analytic;
(c) both $h$ and $f$ are analytic.
Here I consider all possible cases.
I think in part (a) $h$ is analytic being
the composition of two differentiable functions.
Actually to my mind, analyticity of $g$
implies analyticity of $h$, am I correct ? Otherwise, I can't find counterexamples on each cases.
What is your suggestion?
Thank you.
AI: (a) Function $g \circ f$ is analytic : standard.
(b) Cannot deduce $f$ analytic: $g=17$, $f$ non-analytic.
(c) Cannot deduce $g$ analytic$: f=17$, $g$ non-analytic. |
H: An inequality involving integrals
Let be $f:[0,1] \longrightarrow R $, $f$ is an integrable function such that:
$$\int_{0}^{1} f(x) \space dx = \int_{0}^{1} xf(x) \space dx=1$$
I need to prove that:
$$\int_{0}^{1} f^2(x) \space dx\geq4$$
AI: Note that if $h(x)=-2+6x$ then $$\int_0^1 h(x)\, dx = \int_0^1 xh(x)\, dx =1.$$ Moreover
$$
\int_0^1 (h-f)^2 dx\ge 0
$$
The rest is simple. Also Cauchy--Schwarz works with a slightly modified proof. |
H: Why is the following % profit answer wrong
The question is:
A merchant buys an old carpet for $25 dollars.He spends 15 dollars to have it returned to to good condition and then sells it for 50 dollars . What is percent profit of his total investment ?
The answer is 20% and i get 40% . What am i doing wrong ?
Here is how i solved it:
Buying price = $25
Selling price = 50
Profit = 25 - 15 = 10
%change = (Diff/Orig) x 100 = (10/25) x 100 = 40%
AI: I would have said that he made a $25$% profit: he spent $\$40$ and got back $\$50$, for a profit of $\$10$, so his profit is $\frac{10}{40}=\frac14$, or $25$%, of his investment. The only way that I can see to get a profit of $20$% is to figure profit on the basis of the selling price: his profit is $\frac{10}{50}=\frac15$, or $20$%, of his selling price.
Your basic error is not realizing that you have to count all of his costs together: the effective cost to him of the saleable rug was $\$25+\$15=\$40$. |
H: Linear algebra: orthogonal projection?
(a) Find the orthogonal projection of $(-1, 0, 8)$ onto the normal vector to the plane $x-2y+z=0$.
Is this question saying to find the orthogonal projection in other words? The way the question is phrased "onto the normal vector to the plane" is confusing me..
(b) Find the distance from the point $(-1, 0, 8)$ to the plane $x-2y+z=0$.
Answer: The distance from the point $(-1, 0, 8)$ to the plane is the length of the projection onto the normal direction of any vector $v$ which connects a point on the plane to the point $(-1, 0 , 8)$. Since the plane contains the origin, we can choose $v = (-1, 0, 8)$ and compute the projection.
I don't get what this is saying.. is the question asking us to find the $u$ if $v=w+u$? Also, how come we can choose $(-1, 0, 8)$ just because the plane contains the origin?
AI: In the first part, they want you to first find the normal vector to the plane provided. Let this vector be $N$, and now find the orthogonal projection of $(-1,0,8)$ on $N$.
For the second part they want you to find the distance from a point to a plane. The distance from a point to a plane can be found by taking any vector $v$ from the plane to the point, and then projecting this vector $v$ onto a vector which is normal to the plane. Since the origin is in the plane $x-2y+z=0$, you can consider $v$ as the vector from the origin to the point. If the plane did not pass through the origin, you would have had to choose a different point on the plane first.
Hint: In the first part, you found the orthogonal projection of $(-1,0,8)$ onto a normal vector to the plane, so you can save yourself some work in the second part. |
H: Length of the side of a discrete equilateral triangle from area
Firstly I haven't practised any mathematics in a long time, I understand that this might be pretty basic for math.stackexhcange, but I cannot seem to find any answers on the internet anywhere!
I've come across this problem at work, where basically if you are given $X$ amount of glasses to form into an equilateral triangle, how would you calculate the length of the sides - Using glasses as the unit of measurement?
I realise that you need to figure out if the number is triangular, and that there is only one answer for this problem as there is a standard scale for the sizes of the triangles - 3 glasses for the smallest Triangle, 6 for the next, 10, 15, 21 ect...
Formula for testing if the number is triangular is:
$$
(n/2) × (n + 1)
$$
Then working backwards from the equation (assuming the number is triangular):
$$
\text{Area}=\text{Side}{^2} \frac{\sqrt3}{4}
$$
And what I've been using to try to figure out the potental value of the side is:
$$
\text{Side}={\sqrt\frac{A}{(\frac{\sqrt3} 4)}}
$$
I'm not sure if this formula works in this instance as what I'm measuring isn't using standard units of measurements, as there cannot be fractions of glasses. - All the results I've got from this are wrong.
AI: If I understand you correctly, someone has made an 'equilateral triangle' out of some number $A$ of glasses, and you want to know how many glasses there are in the base. You realized the number must be triangular, so that $A = \dfrac{n(n+1)}{2}$.
Then I recommend that you solve the quadratic equation $2A = x(x+1) = x^2 + x$, i.e. $x^2 + x - 2A = 0$.There will be two solutions, and the positive one will give you the number of glasses to a side. |
H: A question about similar triangles.
Please, I would like help in solving this problem: The sides of a triangle measure 2,3 and 4 cm respectively. The perimeter of a similar triangle is 36 cm. I want to find the length of each side of the second triangle.
I did this by trial and error and I got the sides to 8, 12 and 16. I would like to know how to arrive at the answer formally.
thanks.
AI: The sides of the second triangle must be $2a$, $3a$ and $4a$, and they must add up to 36, so:
$$2a + 3a + 4a = 36,$$
then $9a = 36$, and so $a=4$, giving sides 8, 12, 16. |
H: Is it possible to divide an equilateral triangle into 12 congruent triangles?
Can you divide an equilateral triangle into exactly 12 congruent triangles?
interesting question i haven't yet been able to work on.
The sides can be of any length.
AI: Here is a hint for one way to do it: |
H: A question on symmetric matrix and application of Spectral theorem.
Today in the class Prof. applied spectral theorem and wrote $A$ a semidefinite positive matrix as
$A=\sum \lambda_i v_i\times v_i$ , where $v_i$ are the eigenvectores and $\lambda_i$ are corresponding eigen values.
I think $"\times"$ should be some kind of tensor product or so ( i don't know ) .
I don't know spectral theorem. Any kind of explanation would be nice .
AI: It's the outer product: writing $v_i$ as a column you have $v_i v_i^T$.
The spectral theorem says that a self-adjoint matrix (over $\mathbb{R}$, a symmetric matrix) admits a basis of orthonormal eigenvectors $v_j$: $Av_j = \lambda_j v_j$.
You can easily show that $(\sum \lambda_j v_j v_j^T) v_i = \lambda_i v_i$.
Since the two matrices $A$ and $\sum \lambda_j v_j v_j^T$ map the basis $v_i$ to the same vectors, they are equal. |
H: Method for sketching $y = (1 – 3t + 2t^{2})e^{3t}$
I am doing some examination practice, and I've faced the following question:
Another particular solution which satisfies $y = 1$ and $\frac {dy}{dx} = 0$ when $t = 0$, has equation $$y = (1 – 3t + 2t^{2})e^{3t}$$
For this particular solution draw a sketch graph of y against t, showing where the graph crosses the t-axis.
However I am having trouble drawing the graph. I know how to draw the two parts individually but the only thing I can think of when drawing $y = (1 – 3t + 2t^{2})e^{3t}$ is that $(1 – 3t + 2t^{2})$ will grow exponentially bigger every time.
What method or thinking strategy can I use to draw that graph?
(answer)
AI: Presumably the only things they care about in the exam are:
a) Where are the values positive, where negative?
b) What is $f(0)$?
c) Where are the zeros, maxima and minima, are there saddle points?
d) What is the behaviour for $x\to\pm\infty$
e) Is the function continuous or even smooth?
Most of these questions can be solved very elementary and some might even be trivial. As soon as you solved those questions you know how to draw it, it doesn't matter if you hit the value for an arbitrary point, say $\frac 78$ correctly.
Did I miss any important properties?
Edit: In your particular example you will find $f(0)=1$, you will find two zeros ($1$ and $\frac 12$), a maximum and a minimum (derive!) the graph is smooth and goes to infinity for $x\to \infty$ quite rapidly and to $0$ for $x\to - \infty$. So just draw these points and connect the dots smoothly. |
H: Is there a word to describe the set of permutations of each member of the powerset of a set?
Just what it says on the tin:
For a set, X, is there a word to describe the union of sets of permutations of each member of the powerset of X?
AI: Your phrasing is a little unclear. If $X$ is our set and $\mathcal{P}(X)$ is its power set, I think you either mean
$$\operatorname{Aut}(\mathcal{P}(X))=\{\text{permutations of the set }\mathcal{P}(X)\}$$
or $$\bigcup_{S\in \mathcal{P}(X)}\operatorname{Aut}(S)=\bigcup_{S\in\mathcal{P}(X)}\{\text{permutations of the set }S\}$$
In the first case, I would just call it "permutations of the set of subsets of $X$", while in the second case, this set appears to be known as the set of partial permutations of $X$. |
H: Visibility of the surface of a sphere
If you are $N$ radii above a sphere, what fraction of the hemisphere below you can you see?
The answer is so nice that it prompted another question: is there an intuition behind it, in the sense that one might have guessed it before going into the details of the computation? I'll be content with the answer, but I'm really after the intuition, if any springs to mind.
AI: This is essentially a comment to Rahul Narain's answer, about why areas of spherical caps are proportional to the height.
Imagine dropping the sphere into a cylinder into which the sphere fits snugly. Project the sphere onto the cylinder, with parallels of latitude projecting onto circles. This mapping is area-preserving. That can be shown by working with elements $ds$ of length. The same thing was done rigorously by Archimedes about $2200$ years ago, without (explicit) calculus, in On the Sphere and Cylinder. |
H: For set $X$ of integers, why is the square of the sum of its elements equal to the sum of pairwise products?
title pretty much says it all:
$\sum_{i \in X} \sum_{j \in X} ij = (\sum_{i \in X}i)^2$
I'm trying to find out why two ways of writing the same formula are identical, and this is what it comes down to. I find this to be true for all cases I look at (and I assume it is, because the equality of the two original formulations is pretty well-established), but I have no idea why, nor how to proceed such a kind of question.
AI: This is simply the distributive law in action. Take each term of the left factor and multiply by each term in the right factor where both factors are the sum of the elements of the set. |
H: $|G|=12$ and it is isomorphic to $A_4$?
During reading a book, I have faced to this problem telling:
$G$ is a group of order $12$ such that $Z(G)$ has no element of order $2$ . Then $G≅A_4$.
Obviously, this group is not abelian and I think some information about $S_4$ is involved here because of the desired deduction. Can we say $|\frac{G}{Z(G)}|≠3$? And if so, is it useful for the problem?
AI: If $H$ is a subgroup of order $3$, then $[G:H] = 4$ and there is a homomorphism from $G$ to $S_4$ with the kernel contained in $H$. Show that the kernel is trivial, implying that $G$ is isomorphic to a subgroup of $S_4$ of order $12$, which implies that $G \cong A_4$. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.