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H: Help calculating $\lim_{x\to 0} \tan(5x) / \tan(11x)$
I'm trying to calculate
$$\lim_{x\to0}\frac{\tan(5x)}{\tan(11x)}.$$
It seems simple but I cannot figure it out. Should $\tan$ be converted to $\sin$ and $\cos$?
AI: $$\lim_{y\to0}\frac{\tan y}y=\lim_{y\to0}\frac{\sin y}y\cdot\frac1{\lim_{y\to0}\cos y}$$
We know, $\displaystyle\lim_{y\to0}\frac{\sin y}y=1$
$$\text{Now,}\frac{\tan5x}{\tan11x}=\frac5{11}\cdot\frac{\frac{\sin5x}{5x}}{\frac{\sin11x}{11x}}\cdot\frac1{\cos5x}\cdot {\cos11x}$$ |
H: Solution to a tricky inequality (math analysis)
Let $p>1$ and put $q=\frac{p}{p-1}$, so $1/p+1/q=1$. Show that for any $x>0$ and $y>0$, we have
$$ xy \le \frac{x^p}{p}+\frac{y^q}{q}$$
And find where the equality holds.
So far, I have simply tried to multiply through the RHS of the above expression and see what would happen, plugged in for $q$ and I got this:
$$ pxy \le x^p+(p-1)y^\frac{p}{p-1} $$
We also know that $q>1$ by its definition and using $p>1$, but I am not quite sure how to proceed. Any suggestions?
Thank you for the help
AI: Hint: Since $x > 0$, write $$x = e^{t/p}$$ for some $t$. Likewise $y = e^{s/q}$. Then
$$xy = e^{t/p + s/q} \le \frac{e^t}{p} + \frac{e^s}{q}$$
since the exponential is convex. |
H: Linear Transformation Brief Question
$T:P_{3}\rightarrow P_{3}$ defined by $T(p(t))=tp'(t)+p(0)$ is a linear transformation.
Determine whether $T$ is invertible.
If yes, find $T^{-1}(q(t))$, where $q(t)$ is a polynomial of degree at most three.
One last thing... I have got that if $p(t)=at^3+bt^2+c^t+d$, then $T(p(t))=3at^3+2bt^2+c^t+d$. I understand that I can write the inverse of T as a matrix $[\frac{1}{3}, \frac{1}{2}, 1, 1]$, but how to put that into a nice form like the one $p(t)$ is in?
AI: So if you let $q = \frac 13 at^3 + \frac 12 b t^2 + ct + d$ (the coefficients are just the inverses), what can you say about $T(q)$? |
H: How come the following equality for the integral
There is a function $h(t)=\int\limits_0^{+\infty}\exp\left\{- \frac{t^2}{x^2}-x^2 \right\}dx, t>0$
It's stated that $h'(t)=-2h(t)$.
But $h'(t)=-2\int\limits_0^{+\infty}\exp\left\{- \frac{t^2}{x^2}-x^2 \right\}\frac{t}{x^2}dx$ and i don't know how to get the result above.
Could you give me a hint please?
AI: Make the substitution $u = \frac{t}{x}$. |
H: Prove Sine integral exists as improper Riemann integral but is not Lebesgue-integrable.
I got to prove that $$\int_0^1 \frac{1}{t}\sin\left(\frac{1}{t}\right)dt,$$ exists as an improper Riemann integral, yet that $$f(t)=\frac{1}{t}\sin\left(\frac{1}{t}\right)\notin \mathcal{L}_1((0,1),\mathbb{B},\lambda),$$
i.e. that $$\int_{[0,1]}\left|\frac{1}{t}\sin\left(\frac{1}{t}\right)\right|d\lambda=\infty.$$
Attempt: We know that $\forall \epsilon, 0<\epsilon \le 1$, $f(t)$ is continuous in $[\epsilon,1]$, so $\int_{\epsilon}^1 f(t)dt$ exists. Now, with the change of variable $z=1/t$ the integral becomes
$$\int_1^{\frac{1}{\epsilon}}
\frac{\sin(z)}{z}dz.$$
I have to prove the limit of the above expression converges as $\epsilon \rightarrow 0$.
For the second part I was told I should approximate $|f(t)|$ by simple functions and show that their integrals go to $\infty$, but I can't quite grasp the procedure.
Any ideas or insight would be greatly appreciated.
AI: For the non integrability in the Lebesgue sense, one can use the intervals $$
I_n=\left(\dfrac1{n\pi+3\pi/4},\dfrac1{n\pi+\pi/4}\right).
$$
On each $I_n$, $|\sin(1/t)|\geqslant1/\sqrt2$ and $t\leqslant1/(n\pi)$ hence $|\sin(1/t)/t|\geqslant n\pi/\sqrt2$. The length of $I_n$ is of order $1/n^2$ hence the integral of $|t\sin(1/t)|$ on $I_n$ is at least of order $1/n$. Summing these shows the Lebesgue integral diverges.
Proofs that $\displaystyle\int_1^x\frac{\sin(t)}t\mathrm dt$ converges when $x\to+\infty$ are all over the site. |
H: Quadratic equations tricky question
Solve for x :
$x^4-6x^3+12x^2-12x+4=0$
Tried to factorize and used substitution but no result.
AI: I’m assuming that the problem is to solve
$$x^4-6x^3+12x^2-12x+4=0\;.$$
Here’s a very elementary approach that will work if there is a nice solution. Since there is no rational root, we have no easy way to find a root $r$ and divide out $x-r$ to reduce the problem to that of solving a cubic equation. We can hope, though, that the lefthand side is a product of the form
$$(x^2+ax+c)(x^2+bx+d)\;,\tag{1}$$
where clearly $cd=4$. If we’re really fortunate, the coefficients will be integers. For no good reason I decided to try first the possible factorization
$$(x^2+ax+2)(x^2+bx+2)\;,$$
setting it equal to the original quartic and equating coefficients:
$$(x^2+ax+2)(x^2+bx+2)=x^4+(a+b)x^3+(ab+4)x^2+2(a+b)x+4\;,$$
so
$$\left\{\begin{align*}
&a+b=-6\\
&ab+4=12\\
&2(a+b)=-12\;.
\end{align*}\right.$$
The third equation is equivalent to the first, so we want to solve the system
$$\left\{\begin{align*}
&a+b=-6\\
&ab=8\;.
\end{align*}\right.$$
It has a nice solution, so I’ll leave the rest to you, since it’s all quite straightforward. Note that I was fortunate: I picked the right factorization right off the bat. Had I tried
$$(x^2+ax+1)(x^2+bx+4)\;,$$
for instance, I’d have found myself with an inconsistent system. (Of course I was extremely lucky that there actually is a factorization of the form $(1)$ with integer $c$ and $d$.) |
H: Number of ways of assigning jobs to a group (simple counting)
We have 4 different jobs to be done, and a group of 12 people from which to choose workers. How many different ways are there to choose 3 workers for the 4 jobs assuming that one worker does two jobs?
My thinking is that there are ${12}\choose{3}$ ways of choosing the 3 people from the group. Then there are $4\choose2$ ways of selecting two jobs to be merged into one job. Finally there are $3!$ ways to distribute jobs to the chosen people, so the product of these numbers should be the answer.
Is this correct? Is there a different approach for this problem?
AI: Yes, that’s fine. There are slightly different ways of thinking about it. For instance, you could start by selecting the two jobs to be assigned to one person; as you say, that can be done in $\binom42$ ways. Call the other jobs $J$ and $K$. Now you can pick one of the $12$ for the double job, one of the remaining $11$ for $J$, and one of the remaining $10$ for $K$, for a total of $$\binom42\cdot12\cdot11\cdot10$$ ways of assigning the jobs. Of course
$$\binom{12}33!=12\cdot11\cdot10\;,$$
so this isn’t really all that much different from your approach.
Or you could really stretch it out by first choosing the person to get two jobs, then choosing his pair of jobs, then choosing a person to get one job, then choosing his job, and finally choosing the third worker, who will of course get the only remaining job:
$$12\cdot\binom42\cdot11\cdot2\cdot10\;.$$
But all of these approaches are pretty similar: it’s just a question of the order in which you make the choices. |
H: differentiability of unit sphere
I want to show that
$$x^2 + y^2 +z^2 = 1$$ is differentiable. I have first made the equation a function of $x,y$ that is $$f(x,y) = \sqrt{(1 - x^2 - y^2)}$$ However, now I am confused, can I show that the partial differentials are differentiable and hence the total derivative is differentiable or need I plug directly into the definition: if so, how do I go about doing this? I get stuck trying to show that the function indeed goes to zero using the general definition of differentiability.
AI: Remember that it is functions that may or not be differentiable. Remember too, that if $f,g$ are both (somewhere) differentiable, then the composition $f\circ g$ is differentiable when $g(x)$ is in the set where $f$ is differentiable. Can you write your function f as a composition of (somewhere)-differentiable functions? |
H: Negation of a maximal element in a partially ordered set
Let $(P,\le )$ be a partially ordered set. We call a $m\in P$ a maximal element, if $$p\in P,m\le p\Rightarrow m=p$$
What is the negation of this property? So we assume $m\in P$ is not a maximal element. I would say the following: There exists $p\in P,m\le p \nRightarrow m=p$, i.e. $m< p$. Is this correct?
AI: Not entirely, no: you simply want to say that there is an element of $P$ that is greater than $m$. That’s simply $\exists p\in P(m<p)$ or, if you haven’t defined $<$ yet, $\exists p\in P(m\le p\land m\ne p)$.
The point is that the negation of $m\le p\to m=p$ is $m\le p\land m\ne p$: more generally, $\varphi\to\psi$ is equivalent to $\neg\varphi\lor\psi$, whose negation is $\neg(\neg\varphi\lor\psi)\equiv\varphi\land\neg\psi$ by one of the De Morgan laws. |
H: showing every ideal of some quotient ring is principal.
Let $\mathbb F$ be a field and $A=\mathbb F[t]/(t^2)$, where $(t^2)$ is the ideal of $\mathbb F[t]$ (This quotient ring is not an integral domain as you know), and I write an element of $A$ by $[at+b]$, where $[at+b]$ denotes $at+b+(t^2)$.
My goal is to show that every ideal $I$ of $A$ is a principal ideal.
I think, we can divide the problem by $4$ cases.
$I=([0])$
$I$ (ideal) has an element of the form $[at]$, $a \neq 0$: Then, $I=([at])$
$I$ has an element of the form $[b]$, $b\neq 0$: Then, for every element $[ct+d] \in A$, $[ct+d]=[b][b^{-1}(ct+d)]$. Hence, $A=I$.
$I$ has an element of the form $[at+b]$, $a, b \neq 0$: Then for every element $[ct+d] \in A$, $[ct+d]=[at+b][mt+n]$ where $n=b^{-1}d$, and $m=b^{-1}(c-an)$. Hence $A=I$.
So, I guess that $A$ has only $3$ ideals, namely, $([0])$, $([at])$, $A$.
Is my answer is correct?
AI: You are basically right; your description just needs a little fine-tuning to remove some clutter.
If you know what is a vector space, then the following can be a more compact answer:
$k[t]/(t^2)$ is a vector space of dimension two over $k$, with $\{1,t\}$ being a basis. Therefore it can be checked that $(0),\ (t),\ (1)$ are the only ideals. |
H: How do I motivate myself to do math again?
I have been thinking of asking for help for a few months now but posting in a public forum like this is intimidating.
Still, I am currently in a university studying mathematics as an undergrad. I took quite a few knocks a few months back when I failed to qualify for the universities of my choice and all the hard work for mathematical olympiads and hours of practice went up in smoke.Out of frustration and disappointment, I left math for a few months,doing nothing but staring at the ceiling. After coming to the university,I then tried doing math again and after a months, I tried to study linear algebra and analysis. I studied 30 pages of Rudin and then stopped; and then I studied those 30 pages, each time those 30 pages seemed to be as difficult as before but less interesting and now I am stuck.
I came to the conclusion that I cannot do math OR I have lost the motivation to. How do I get back to studying math?
Thank you everyone for your answers and comments. I have read them and I will try to get back up.
AI: Disclaimer:
If you didn't consult your friends and/or parents on whether your situation is serious enough (i.e. requires specialist attention), then do it now. Major depressive disorder may have detrimental effects on your life and you should treat it accordingly. On the other hand, mild depression, melancholia, etc. may be temporary and may not require any professional treatment. Often those are normal human reactions and are nothing to be ashamed of.
Thanks to @EricLippert for his comment.
It depends on what motivates you generally. Naturally, first
Get enough sleep (but not too much).
Keep you body healthy (within reason, esp. if you have some disabilities).
As to the motivation, I observe three main factors:
Pleasure – Math is fun and beautiful.
Need – we (the world) need your math.
Community – there are some cool guys among mathematicians.
You need to find what exactly works for you, but here are some tips:
Math is fun and beautiful.
Look for some nice proofs, like those in Proofs from the Book.
Read Gödel, Escher, Bach.
There are also comics about math, e.g. Logicomix (one of the authors is a world-class researcher in theoretical computer science).
Find some other nice mathematical books, you can find some recommendations here.
Find some beautiful mathematical facts, you can find some here.
Try some nice online presentations like How to Fold a Julia Fractal.
There are even television series like Numb3rs; to put it mildly, it's not my favorite, but who knows, maybe it would work for you (it doesn't really matter what the IMDb ranking is, only if it remotivates you again).
Look at some highly voted questions and answers here, at math.SE, some are real pearls.
I don't think this needs much commentary. There are many, many articles on why math is important and why it would be good for you to know it (and I think that would be even more so in the near future). I've never been interested in those, so I'm unable to recommend any, but try searching for it.
Sometimes doing things is fun when in group, and math is no exception.
Imagine how music bands stick together, mathematicians very often collaborate and this is for a reason. One of them is that doing something with peers is just more fun than being alone in the cave.
Mathematicians like Tim Gowers and Terence Tao write blogs. You can learn a lot from them (but it might be complicated).
It's easier to get motivated when you have some examples of other motivated people (but be aware that you might also get demotivated).
Why don't you hang around with "us, the cool guys"? Answering and asking questions might also be a motivating experience (if you ignore those lazy bums that come only to get their homework solved).
Finally, this goes without saying, but I feel that is should be emphasized here: there are other domains, one of these might suit you better, do you really want to study math? If the only reason is that you do not wish to throw away all the hard work you did, then there are some good news: the skill you have acquired will stay with you in some form, or simply put, it made you smarter. Moreover, areas like physics, engineering or computer science use math a lot, and the math you know will directly help you there. Mathematics is everywhere, you don't need to be labeled as a mathematician to do mathematics.
I hope this helps and good luck! $\ddot\smile$ |
H: Exponential map on $SO(3)$
(1) As I read some article in here ( I cannot found ), so we know that $$ {\rm exp} \ (T_eSO(3)) \neq SO(3) $$
( ${\rm diag}(-1,-1,1)$ cannot be covered by ${\rm exp}$ )
But there exists some open ball $B$ in $T_eSO(3)$ so that ${\rm exp}\ B$ is open.
Then can we choose $\{ g_1, ... , g_n \}\subset SO(3)$ such that $\{ g_i B\}$ is open cover ? If so, can we find smallest $n$ ?
(2) My second question is : What is ${\rm exp}\ (T_eSO(3))$ ?
$$\{ {\rm exp}\ tE_{12} | t\in {\bf R} \}=SO(2)$$ where $E_{12}$ is a matrix having only nonzero $(1,2),\ (2,1)$-entries which are $1$ and $-1$. That is $\{ {\rm exp}\ tE_{12} | t\in {\bf R} \}$ is a closed geosic in $SO(3)$. But what is ${\rm exp}\ (T_eSO(3))$ ?
AI: Rethink your assumptions - $\rm diag(-1,-1,1)$ is the exponential of $$\left(\begin{matrix} 0& \pi& 0\\ -\pi& 0& 0\\ 0& 0& 0\end{matrix}\right).$$ In fact the exponential map is surjective - this is true for any compact connected Lie group. |
H: Integration of 3x/... function - Substitution
How do I integrate the function $\frac{3x}{\sqrt(16x^4+8x^2+1)}$ ?
I thought about doing this:
$\int \frac{3x}{\sqrt(16x^4+8x^2+1)} dx = \int \frac{3x}{(2x)^2+(1)^2}$
then: Substitute $2x = u \rightarrow dx = du/2 $
However, I do not come closer to the end, as you can see:
$\int \frac{3x}{\sqrt(16x^4+8x^2+1)} dx = \int \frac{3u}{2*2*(u^2+1)}du = \int \frac{3u}{(2u)^2 + (2)^2} $
AI: Let$$u=x^2$$
then,
$$du=2xdx$$
Hence
$$I=\int \frac{1.5}{\sqrt(16u^2+8u+1)} du $$
or
$$I=\int \frac{1.5}{4u+1} du $$
Use this now:
$$\int \frac{dx}{ax+b} = {1\over a}{{ln|ax+b|}}+C$$ |
H: Which form of Euler-Maclaurin formula to use?
This question may be rather elementary, but I am sort of confused about various forms of the Euler-Maclaurin summation formula and their use.
For instance, let us suppose that we want to approximate the sum $$\sum_{k=0}^n \sin\frac{k\pi}{n}$$ by the corresponding integral $$\int_0^{\pi}\sin x\,\mathrm{d}x.$$
Which form of Euler-Maclaurin formula is the best choice to use now? On the one hand, there is a form of the Euler-Maclaurin formula that allows us to approximate the sum by the (clearly equivalent) integral
$$\int_0^n \sin \frac{x\pi}{n}\,\mathrm{d}x.$$
I believe that this is possible, but is this the most efficient way of doing so? Although the integral itself can be clearly transformed to the simple integral above, the computation of the reminder terms may not be so simple.
On the other hand, there is a form of the Euler-Maclaurin formula that is intended for approximating sums on the fixed interval. However, the step size is considered to be fixed here.
So my question essentially is: which form of the Euler-Maclaurin formula would you use to approximate sums of the above type: that is, essentially, sums where we sum according to some set of points in a fixed interval, where the set becomes denser and denser.
My apologies, if this question does not make much sense.
AI: I would use Euler summation for $\sum_{k=0}^n f(k)$ with $f(k)=\sin(kx)$, and $\int_0^n f(x)dx$. How efficient this is, depends on the error term. There is also a formula for the above sum, i.e.,
$$
\sum_{k=1}^n\sin(kx)=\frac{\cos \frac{x}{2}-\cos \frac{(n+1)x}{2}}{2\sin \frac{x}{2}}.
$$ |
H: $\mathbb{Q}[X]/\left(X^{2}-1\right)$ isomorphic to $\mathbb{Q} \times \mathbb{Q}$?
i need to prove that:
$\mathbb{Q}[X]/\left(X^{2}-1\right)$ isomorphic to $\mathbb{Q} \times \mathbb{Q}$, but i don't know where to start. I first wanted to use the Chinese value theorem, but i can't see how that should fit in. Hints/tricks/other theorems would be much apreciated!
Thanks!
AI: Hint: Recall that $X^2 - 1 = (X+1)(X-1)$, so let us consider the map $T \colon \mathbb Q [X] \ni q \mapsto \bigl(q(1), q(-1)\bigr)\in \mathbb Q^2$. What can you say about $T$? Is it onto? What is its kernel? |
H: Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$
Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ in which every number is made by inserting $48$ in the middle of previous as indicated, each number is the square of an integer.
AI: Without words:
$$\begin{align}
\left(6\frac{10^k-1}{9}+1\right)^2 &= 36 \frac{10^{2k} - 2\cdot 10^k + 1}{81} + 12\frac{10^k-1}{9} + 1\\
&= 4\frac{10^k-1}{9}\cdot 10^k - 4 \frac{10^k-1}{9} + 12 \frac{10^k-1}{9} + 1\\
&= 4\frac{10^k-1}{9}\cdot 10^k + 8 \frac{10^k-1}{9} + 1.
\end{align}$$ |
H: Is there the exclusive list of Pythagorean triples? If there is, what is the general pattern for it?
Is there the exclusive list of Pythagorean triples? If there is, what is the general pattern for it?
What about Euclid's formula? Does it generate all the Pythagorean triples?
More explicitly, if $a,b,c$ is a Pythagorean triple, then does it satisfy
$$
a = m^2+n^2, b=2mn, c= m^2 + n^2
$$
for some numbers $m,n$?
If you don't mind, could you give me a full rigorous statement of what I tried to describe right above (that is, the necessary condition of Pythagorean triples)?
AI: A direct quote from the Wikipedia article you linked to:
Despite generating all primitive triples, Euclid's formula does not produce all triples. This can be remedied by inserting an additional parameter $k$ to the formula. The following will generate all Pythagorean triples uniquely:
$$a = k\cdot(m^2 - n^2) ,\ \, b = k\cdot(2mn) ,\ \, c = k\cdot(m^2 + n^2)$$
where $m, n,$ and $k$ are positive integers with $m > n$, $m − n$ odd, and with $m$ and $n$ coprime. |
H: Expressing how many of $a,b,c$ can be zero
My scenario is that I need to express with mathematical syntax the following condition:
There are three integers: ${a, b}$ and ${c}$.
Case 1: two of the three can be zero.
Case 2: only one can be zero.
Case 3: none can be zero.
The last condition is easy: ${a, b, c \neq 0}$ but I can't figure out the other. I was looking for something easy to read, if the mathematical expression is too messy I prefer to write it by word. Thank you.
AI: For something like this you'd usually be alright writing it in english, but if you must use logical notation then the first condition can be written as:
$$
a \neq 0 \vee b \neq 0 \vee c \neq 0
$$
And the second can be phrased with material implications as:
\begin{align}
a = 0 \implies b,c \neq 0 \\
b = 0 \implies a,c \neq 0 \\
c = 0 \implies a,b \neq 0
\end{align}
You could also do this with set notation and the $\exists!$ quantifier, read as "there exists exactly one", but the english statements are probably the nicest. |
H: Variant of the birthday problem
I would like to ask help regarding an example given in the book of V. Rohatgi and A. Saleh. I think this is a variant of the birthday problem. Here it goes:
Consider a class of $r$ students. The birthdays of these $r$ students form a sample of size $r$ from the 365 days in the year. Next suppose that each of the $r$ students is asked for his or her birth date in order, with the instruction that as soon as a student hears his or her birth date the student is to raise a hand. Let us compute the probability that a hand is first raised when the $k^{th}$ ($k = 1, 2, $...$ , r$) student is asked his or her birth date. Let $p_k$ be the probability that the procedure terminates at the $k_{th}$ student. Then $$p_1 = 1-\left(\frac{364}{365}\right)^{r-1}$$ and $$p_k=\frac{_{365}P_{k-1}}{365^{k-1}}\left(1-\frac{k-1}{365}\right)^{r-k+1}\left[1-\left(\frac{365-k}{365-k+1}\right)^{r-k}\right], k=2,3,...,r$$
What I would like to ask then is how were these answers obtained? I think it would be better that someone help me in understanding the problem "concretely" since I actually cannot get the problem. Thanks.
AI: Let $N=365$. A hand is first raised at time $k$ if:
The birth dates of the $k$ first students are different.
The birth date of each of the last $r-k$ students is not one of the $k-1$ birth dates of the $k-1$ first students.
The birth date of at least one of the last $r-k$ students is also the birth date of student $k$.
Call $A_k$, $B_k$ and $C_k$ these three events. Then,
$$
P[A_k]=\left(1-\frac1N\right)\left(1-\frac2N\right)\cdots\left(1-\frac{k-1}N\right).
$$
Assuming $A_k$, $B_k$ happens if $r-k$ independent draws from a set of $N$ avoid $k-1$ values, thus
$$
P[B_k\mid A_k]=\left(1-\frac{k-1}N\right)^{r-k}.
$$
Assuming $A_k$ and $B_k$, $C_k$ does not happen if $r-k$ independent draws from a set of $N-k+1$ avoid $1$ value, thus
$$
P[C_k^c\mid A_k,B_k]=\left(1-\frac{1}{N-k+1}\right)^{r-k}.
$$
Finally,
$$
p_k=P[A_k,B_k,C_k]=P[A_k]\cdot P[B_k\mid A_k]\cdot (1-P[C_k^c\mid A_k,B_k]).
$$ |
H: Is the ring $\mathbb{Z}_5[x]$ isomorphic to the ring of polynomial functions from $\mathbb{Z}_5$ to $\mathbb{Z}_5$?
Is the ring $\mathbb{Z}_5[x]$ isomorphic to the ring of polynomial functions from $\mathbb{Z}_5$ to $\mathbb{Z}_5$?
If not, what is a good counterexample?
If yes, how can we prove that there's a bijection between the two rings?
Thanks in advance.
AI: No, it is not and the difference is a rather important, formal one.
For example: take the elements $\;f(x)=x\;,\;\;g(x)=x^5\;$ . Clearly, $\;f\neq g\;$ in $\;\Bbb F_5[x]\;$ , yet as functions $\;\Bbb F_5\to\Bbb F_5\;$ they are identical. |
H: Is this $3\times 3$ matrix positive definite?
I have three non-null and non-parallel column vectors $z_i \in \mathbb{R}^3$, and I have the next symmetric matrix:
$$
M =
\begin{bmatrix}
||z_1||^2 & z_1^Tz_2 & z_1^Tz_3 \\
z_2^Tz_1 & ||z_2||^2 & z_2^Tz_3 \\
z_3^Tz_1 & z_3^Tz_2 & ||z_3||^2
\end{bmatrix}
$$
Looking at the two first principal minors, they are positive since
$$
||z_1||^2 > 0, \quad ||z_1||^2\,||z_2||^2 > ||z_1^Tz_2||^2
$$
I suspect that the
$$
det(M) = ||z_1||^2\, ||z_2||^2\, ||z_3||^2 + 2\,z_1^Tz_2 z_2^Tz_3z_3^Tz_1 - ||z_1||^2(z_2^Tz_3)^2 - ||z_2||^2(z_1^Tz_3)^2 - ||z_3||^2(z_2^Tz_1)^2
$$
is also positive, but I can not see how to prove it (or to find a counter-example). Any hints?
I suspect that this article is what I am looking for:
https://secure.ele-math.com/buy/mia-11-52
but I can not check it
AI: Let
$$Z := \begin{bmatrix} z_1 & z_2 & z_3 \end{bmatrix}.$$
What would be $Z^T Z$? |
H: Why is connection a map from $\Gamma(E)$ to $\Omega^1\otimes\Gamma(E)$?
On the site Vector Bundle Connection, it gives two definitions of a connection.
One is view a connection as a linear map from a section of $E\otimes TM$ to a section of $E$:
$$
D:\Gamma(E\otimes TM)\rightarrow\Gamma(E)
$$
I can understand this definition, thinking a connection as a directional directive:
$$
v\otimes w\mapsto D_vw
$$
However, I cannot understand the other definition, and which is seemingly more common:
$$
D:\Gamma(E)\mapsto\Gamma(E\otimes T^*M)=\Gamma(E)\otimes\Omega^1
$$
Can anyone explain to me how such map works? Given a vector in $\Gamma(E)$, what is the image of it?
In addition, in the site above there is an example about the connection in a trivial bundle, saying that
Any connection on the trivial bundle $E=M\times\mathbb{R}^k$is of the form $\nabla s=ds+s\otimes\alpha$ where $\alpha$ is a one-form with value in Hom($E,E$).
However, I don't understand. I think $ds$ is an element in the dual bundle of $E$, but $s\otimes\alpha$ is not, although I cannot even point out to what space $s\otimes\alpha$ belongs, how can they be added?
AI: It's not really correct to think of connections as maps $\Gamma\left(E\otimes TM\right)\to\Gamma\left(E\right)$,
because the interplay between the $E$ and $TM$ inputs is not $C^{\infty}\left(M\right)$-multilinear.
More precisely, $s\otimes\left(fu\right)=\left(fs\right)\otimes u$,
so if the connection factored via the tensor product we should have
$D_{fu}s=D_{u}fs$; but $D_{fu}s=fD_{u}s$ is not equal to $D_{u}fs=\left(uf\right)s+f$$D_{u}s$
in general. Thus if you want to "keep the inputs together", you can't go any further than thinking of the connection as a map $\Gamma\left(E\right)\times\Gamma\left(TM\right)\to\Gamma\left(E\right)$.
Now, the other interpretation is indeed correct and common - given
a section $s\in\Gamma\left(E\right)$, the derivative of $s$ is a
$E$-valued one-form. For a section $s\in\Gamma\left(E\right)$, its
image is the map $Ds:\Gamma\left(TM\right)\to\Gamma\left(E\right):u\mapsto D_{u}s$.
This map is $C^{\infty}\left(M\right)$-linear, so we can in
fact interpret it as an element of $\Gamma\left(E\otimes T^{*}M\right)$. |
H: Check whether the following is a metric
I got this question on an internal today,
Check whether
$e(x,y)$ = $d(f(x),f(y))$ for any function $f:X \rightarrow X$ is a metric on $(X,d)$.
I think that I have messed it up.
My argument was that, because the identity map is always injective therefore it should be a metric.
But apparently some of my classmates thought otherwise, so i have become a little doubtful of my argument. Could someone tell me whether i am right or not ?
AI: It is not always true that for any function $f$ from metric space $(X,d)$ to itself that:
$$ e(x,y) = d(f(x),f(y)) $$
will also be a metric. The simplest counterexample would be if $f$ is not injective, say $f(x)=f(y)$ for some $x\neq y$. Then $e(x,y) = 0$ but this contradicts the property of a metric that having zero distance implies equality of two points.
The question of which maps $f$ will give us a metric $e(x,y)$ is an interesting one. The above shows $f$ must be injective, a necessary condition. If $f$ were an isometry, then by definition $e(x,y)=d(x,y)$ gives back the same metric, a sufficient condition. In between these extremes there are other possibilities to investigate. |
H: Digital root of prime numbers,why the digital root of this two prime are the same?
A is prime greater than 5, B is A*(A-1)+1,if B is prime,
then digital root of A and B must the same.(OEIS A065508)
Sample: 13*(13-1)+1 = 157
13 and 157 are prime and have same digital root 4
AI: hint: consider mod 9. What happens in each of these cases?
Use the fact that B us prime, and in particular not a multiple of 3 |
H: What distribution do the rows of the Stirling numbers of the second kind approach?
In wikipedia about the Pascal triangle:
Relation to binomial distribution
"When divided by 2n, the nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where p = 1/2. By the central limit theorem, this distribution approaches the normal distribution as n increases."
How can I find the distribution approached by the rows of the Stirling numbers of the second kind?
$$\begin{array}{llllllll}
1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 3 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 7 & 6 & 1 & \text{} & \text{} & \text{} & \text{} \\
1 & 15 & 25 & 10 & 1 & \text{} & \text{} & \text{} \\
1 & 31 & 90 & 65 & 15 & 1 & \text{} & \text{} \\
1 & 63 & 301 & 350 & 140 & 21 & 1 & \text{} \\
1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1
\end{array}$$
After all the Stirling numbers of the second kind satisfy the recurrence:
$$\left\{{n+1\atop k}\right\} = k \left\{{ n \atop k }\right\} + \left\{{n\atop k-1}\right\}$$
Where as the Pascal triangle satisfies the recurrence:
$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$
If such a distribution does exist I would like to find the x value for the maximum of the distribution for each row and see if it says anything about x/LambertW(x).
AI: I think this is what you are looking for:
Stirling Behavior is Asymptotically Normal,
L. H. Harper,
The Annals of Mathematical Statistics
Vol. 38, No. 2 (Apr., 1967), pp. 410-414
Link |
H: Confused with notation regarding integers modulo $3$
$G$ denotes the equivalence classes of integers (mod $3$).
So $G = \{[0], [1], [2]\}.$
I have to show that $(G, +)$ forms a group but my understanding of the notation is holding me back. Here are a few questions I have.
Does $[0] = \{3k\mid k\in \mathbb{Z}\}$ ?
If so, wouldn't that mean $G = \bigl\{\{3k\mid k\in \mathbb{Z}\}, \{3k+1\mid k\in \mathbb{Z}\}, \{3k+2\mid k\in \mathbb{Z}\}\bigr\}$, meaning I have $3$ elements which are sets?
If instead, $[0]$ represesented the elements of $\{3k\mid k\in \mathbb{Z}\}$ without the braces then I would have $G = \{\ldots-6, 3, 0, 3, 6\ldots, \ldots -5, -2, 1, 4, 7\ldots, \ldots-4, -1, 2, 5, 8\ldots\}$ or in other words $G$ is a set of all integers.
If this second case is true, where $G$ is a set of all integers, then to prove the condition of closure I could say that the sum of any $2$ integers (e.g. $8+2$) is another integer, which is in $G$. That would be that -- no need for any further application of modulo $3$ to my answer before showing the result is in $G$. I understand that this is probably not the case and in fact I do have to include the use of modulo $3$ into the workings of my answer.
Could I get some clarification as to what $[0]$ actually means and what the set $G$ would look like if it was expanded into long form? Finally could the property of closure please be proven the proper way since my method is probably incorrect?
Thanks for bearing with me.
AI: Your first interpretation is correct. $G$ is, in fact, a set of three particular subsets of $\Bbb Z$.
In general, for any $n\in\Bbb Z,$ we have $[n]_3=\{3k+n: k\in\Bbb Z\},$ so that $[m]_3=[n]_3$ if and only if $3\mid(m-n)$. (You should prove this if you haven't already.) The definition of addition is: $$[m]_3+_3[n]_3=[m+n]_3$$ To show that this is a well-defined operation (that closure holds), we must show that if $[m]_3=[m']_3$ and if $[n]_3=[n']_3,$ then $[m]_3+_3[n]_3=[m']_3+_3[n']_3.$ In other words, we must show that if $3\mid(m-m')$ and $3\mid(n-n')$, then $3\mid\bigl((m+n)-(m'+n')\bigr).$ Can you take it from there? |
H: Letter-Sending probability problem
Here is another question from the book of V. Rohatgi and A. Saleh. I would like to ask help again. Here it goes:
Consider a town with $N$ people. A person sends two letters to two separate people, each of whom is asked to repeat the procedure. Thus for each letter received, two letters are sent out to separate persons chosen at random (irrespective of what happened in the past). What is the probability that in the first $n$ stages the person who started the chain letter game will not receive a letter?
I am thinking of solving this through complementation, i.e. what I have so far is that for $n=1$, $$P\{\text{the person who started the chain letter game will receive a letter}\}=0$$ for $n=2$, $$P\{\text{the person who started the chain letter game will receive a letter from the 1st sender}\}$$ $$=P\{\text{the person who started the chain letter game will receive a letter from the 2nd sender}\}$$ $$=\frac{_{N-2}C_1}{_{N-1}C_2}$$ $$P\{\text{the person who started the chain letter game will receive letters from both 2nd senders}\}$$ $$=\left(\frac{_{N-2}C_1}{_{N-1}C_2}\right)^2$$ Then I would simply calculate the probability of the union of the elementary events, then continue for each $n$. Now, I find it too tedious and I am not so sure if this correct. Can anyone help please? Thanks.
AI: Let stage $k=0$ denote the person who started the chain (say A) sending out the first two letters to two different people. I assume you are asking for the probability that after stage $k=n$, A does not receive a letter.
Further, as per the problem statements, even if a person receives $m>1$ letters in a stage or across stages, all that is taken care of is that each letter results in a pair of letters going out to two separate people, irrespective of whether they were recipients of other letters by the same sender.
In this situation, we worry only about how many such letter pairs are sent across stages $k=1,2, ...n$. This is $L = 2+4+... 2^n = 2(2^n-1)$.
Each time a letter is received, there are $(N-1)(N-2)$ ways to select two separate people to send a letter pair to. If A is not to be one of them, this choice reduces to $(N-2)(N-3)$. Hence the probability for A not receiving a letter at this time is
$p = \dfrac{(N-2)(N-3)}{(N-1)(N-2)} = \dfrac{N-3}{N-1}$.
As this event occurs independently $L$ times, the total probability is $p^L = \left( \dfrac{N-3}{N-1}\right)^{2(2^n-1)}$. |
H: Calculus continuity question.
show that the function
f(x,y)= |x-1| + |y-1| is continuous at (2,2)
using epsilon delta definition.
The way I have done this is as follows.
|f(x,y)-f(2,2)
= ||x-1|+|y-1|-(1+1)|
= ||x-1|+|y-1|-2|
<= ||x-1|| + ||y-1|| + |2|
<= sqrt( (x-1)^2+(y-1)^2) + sqrt((x-1)^2+(y-1)^2)+2
<= 2sqrt( (x-1)^2+(y-1)^2)+2
<2*e/2 +2=e+2
AI: You want to prove that when $(x,y) \, \rightarrow \, (2,2)$, then $f(x,y) \, \rightarrow \, f(2,2) = 2$.
For $(x,y) \in \mathbb{R}^{2}$, you have :
$$
\begin{eqnarray*}
\vert f(x,y) - f(2,2) \vert & = & \Big\vert \vert x-1 \vert + \vert y-1 \vert - 2 \Big\vert \\
& = & \Big\vert \left( \vert x-1 \vert - 1 \right) + \left( \vert y-1 \vert - 1 \right) \Big\vert \\
& \leq & \Big\vert \vert x-1 \vert - 1 \Big\vert + \Big\vert \vert y-1 \vert - 1 \Big\vert \\
\end{eqnarray*}
$$
By triangular inequality, you have $\Big\vert \vert x-1 \vert - 1 \Big\vert \leq \vert x-2 \vert$ and $\Big\vert \vert y-1 \vert - 1 \Big\vert \leq \vert y-2 \vert$. So :
$$ \vert f(x,y) - f(2,2) \vert \leq \vert x-2 \vert + \vert y-2 \vert $$
Let $\varepsilon > 0$ and let $\delta = \frac{\varepsilon}{2} > 0$. Then, the following is true :
$$ \forall (x,y) \in \mathrm{B}\Big( (2,2), \delta \Big), \; \vert f(x,y) - f(2,2) \vert \leq \varepsilon $$
where $\mathrm{B}\Big( (2,2),\delta \Big) = \left\{ (x,y) \in \mathbb{R}^{2}, \; \Vert (x,y) - (2,2) \Vert \leq \delta \right\}$. |
H: An unbounded subset of the reals containing no intervals with the following property.
I'm trying to find an unbounded subset of the reals, $A$, with the property that:
$m(A\cap I)<0.9m(I) $ for all intervals, $I$. And $m(A)\neq0$. Where $m(X)$ denotes the outer measure of $X\subset\mathbb{R}$.
Thus far, I have concluded that if such a set does exist, clearly it must be uncountable, it must contain no intervals otherwise you cannot satisfy "$m(A\cap I)<0.5m(I) $", but I can't think of much else.
AI: Let $0\lt\alpha\lt1$, and suppose $A$ is a subset of $\mathbb R$ such that $m(A\cap I)\lt\alpha\ m(I)$ for every interval I, where $m$ is Lebesgue outer measure on $\mathbb R$.
Consider a fixed interval $I$. Choose an open set $B$ such that $A\cap I\subseteq B$ and $m(B)\lt\alpha\ m(I)$. Write $B=\bigcup_{n=1}^{\infty}I_n$ where the $I_n$'s are pairwise disjoint open intervals, so that $m(B)=\sum_{n=1}^{\infty}m(I_n)$. Then $m(A\cap I)\le\sum_{n=1}^{\infty}m(A\cap I_n)\lt\sum_{n=1}^{\infty}\alpha\ m(I_n)=\alpha\ m(B)\lt\alpha^2\ m(I)$.
Repeating this argument, we have $m(A\cap I)\lt\alpha^km(I)$ for every interval $I$ and every natural $k$, whence $m(A)=0$. |
H: Find the expected number of '01's in a string
This is an interview question:
For strings of length m + n, with m 0's and n 1's. Find the expected number of switches from 0 to 1 (a switch can be thought of as presence of '01' in the given string).
I have tried solving it as follows:
Let Xi denote a random variable such that a '01' pattern occurs in the string for a particular '0'. Then
W = X1 + X2 + ............. + Xm
and we have to find E[W].
Now
E[W] = E[X1] + E[X2] + E[X3] + ............... + E[Xm]
Now E[X1] is same as finding expectation such that 0 comes before any 1. It should be n/n+1. This holds for all Xi's. Therefore E[W] = mn/n+1. And this is our final answer.
Please let me know if this is the right approach.
AI: Your answer fails a basic sanity check: the number of switches must certainly be at most $m$ and also must certainly be at most $n$. Your formula does deliver a result that is less than $m$; but, taking (for instance) $n=1$ and $m=4$ yields a result bigger than $n$. So, back to the drawing board!
The issue here is that your $X_i$ are not identically distributed, as you've claimed. The last zero ($X_m$) is less likely to have a one after it than the others... because it could be in the last position!
So, my hint is this. Instead of considering your variables $X_1,\ldots,X_m$, consider the variables $Y_1,\ldots,Y_{n+m-1}$, where $Y_i$ is the indicator that there is a zero in position $i$ and a one in position $i+1$. These ARE identically distributed, their sum is still $W$, and they can lead you to the right answer. |
H: Units of the ring $\{a+b\epsilon:a,b \in k \}$ where $ϵ^2=0$.
I'm making exercises to prepare for my ring theory exam:
Let $k$ a field and $R=k[\epsilon]$ the set
$$\{a+b\epsilon:a,b \in k \}$$
where $$(a+bϵ)+(c+dϵ)=(a+c)+(b+d)ϵ \quad \quad\quad
(a+bϵ)(c+dϵ)=ac+(bc+ad)ϵ$$
What are the units of this ring ?
After calculation I get that $(a+bϵ)^{-1}=a^{-1}+-ba^{-2}ϵ$. This is only defined if and only if $a≠0$. Is this the answer ?
AI: Yes, indeed! Nicely done.${}{}$ |
H: $ 1^k + 2^k + \cdots + (p-1)^k \equiv \begin{cases} -1 \mod p, \text{ if } p-1 | k, \\ 0 \mod p, \text{ if } p-1 \not | k. \end{cases}$
If $p-1 | k \Rightarrow k = (p-1) n$, for some $n \in \mathbb{N}$. Then we $ 1^{(p-1)^n} + 2 ^{(p-1)^n} + \cdots + (p-1)^{(p-1)^n} = \underbrace{1 + 1 + \cdots + 1}_{p-1}
= p-1 \equiv -1 \mod p.$
The second part of the question did not come out. Induction: $k=1$, which is true because we $1 + 2 + \cdots + (p-1)= \frac{(p-1)(p)}{2} \equiv 0 \mod p$. Suppose true for $k = n$, then $1^n + 2^n + \cdots + (p-1)^n \equiv 0 \mod p$. How to finish? Thank you.
AI: The group $\mathbb{F}_p^*$ is cyclic, say generated by $\zeta$ with $\zeta^{p-1}=1$. Then your sum is $\sum_{i=0}^{p-2} (\zeta^k)^i$, which is just a geometric series. If $p-1 | k$, we get $\sum_{i=0}^{p-2} 1 = p-1 = -1$, and otherwise $\zeta^k \neq 1$ so that we get $\dfrac{{(\zeta^k})^{p-1} - 1}{\zeta^k -1} = 0$. |
H: homomorphism from group G to group H
Let f be a homomorphism from group G to group H, K be the kernel of f, which of the following statement is right?
I.K is a normal subgroup in G
II. f(G) is a normal subgroup in H
III. G/K is isomorphic to a subgroup in H
AI: I would say
I. K is a normal subgroup in G. Let $g\in K$ and $a,a^{-1}\in G$ then
$f(aga^-1)=f(a)f(g)f(a^-1)=f(a)ef(a)^-1=e$.
III. $G/K$ is isomorphic to the image $f(G)$ (by the 1st isomorphism theorem) Of course $f(G)$ is a subgroup of H thus $G/K$ is isomorphic to a subgroup of H.
About II. Suppose $b=f(g)$ for some $g\in G$ and $h,h^-1\in H$.
$hbh^-1$ may not always belong to $f(G)$ so $f(G)$ is not always a normal subgroup of H. |
H: Say I have a function $f(x) = x^2$. Can this be a surjective function?
If I'm getting the unto and 1-1 concepts right, $f(x) = x^2$ is always 1-1 as $x$ always maps distinct objects in codomain ($x^2$).
But it's not a surjective function since you can't get all $x$ in codomain. But if I restrict my domain $x$ to be in set of values in codomain, then this function $f(x) = x^2$ can be surjective, depending on the context.. is that right?
AI: It depends on what the domain and codomain are specified to be, so yes, the status of a function being one-to-one and/or onto depends on "context"; in particular, it depends on the function's definition, and part of that definition includes a specification of the domain and codomain.
For example, suppose we define $f$ as follows: $$f:\mathbb R \to \mathbb R,\quad f(x) = x^2$$
Then $f$ is not one-to-one, nor onto. Why not?
Not One-to-one:
What is $f(-3)$? What is $f(3)$? Note that $f(-3) = f(3),$ but $\;-3\neq 3.\;$ So the function is not one-to-one.
Recall: A function is one-to-one if and only if for each $a, b$ in the domain, $$f(a) = f(b) \implies a = b$$ or equivalently, if $$a \neq b \implies f(a)\neq f(b)$$
Nor is it surjective: Recall that a function $f$ is surjective if and only if for every element $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$. In the case at hand, for example, there is no $x$ such that $y = f(x) = x^2 = -4$.
However, if you restrict the domain and codomain to non-negative reals in this case, which we can denote $\mathbb R_{\geq 0},$ then the function $f$ defined by: $$f: \mathbb R_{\geq 0} \to \mathbb R_{\geq 0}, \quad f(x) = x^2$$ is both one-to-one and onto. |
H: Check the convergence of a sequence
Let $X$ =$[0,1]$ and $d(x,y)=|x-y|/(1+|x-y|)$ be the metric defined on $X$. Then check whether the sequence ${x_n = 1/n^2}$
A) Converges in $(X,d)$
B)Does not converge in $(X,d)$
My attempt :
I was trying to use the fact that if, $dist(z,${$x_n$ |n $\in$ S}$)$ =0, (for $z \in X$) for every infinite subset $S$ of $N$ then the sequence is convergent in $X$.
So when i applied the metric i got,
$d(0,1/n^2)$ = $1/(1+n^2)$ which would always be greater than zero and hence its infimum would also always be greater than zero,and hence using the definition given above i concluded that the given sequence does not converge in $(X,d)$.
Am i correct ?
AI: No, you are not. The infimum of a set $X$ of positive numbers may be zero. Example:
$X=(0,\infty)$
$X=\{1/n:n\in\mathbb{N}\}$
$X=\{1/(1+n^2):n\in\mathbb{N}\}$
The infimum of all those sets is $0$, but $0\not\in X$. The sets have an infimum but not a minimum. |
H: Converting 2 sat formula into an implication graph.
Both wikipedia and my lecturer explained how the 2 satisfiability problem work. However, I am finding it really hard understanding how this formula:
xvy≡ ¬x-->y ≡ ¬y -->x
Then breaks down the following conjectures :
(¬x v y) & (¬y v z) & (¬z v w) & (¬w v ¬x) &
(x v ¬y) & (y v ¬z) & (z v ¬w) & (w v x)
is converted to an implcation graph.
Heres my attempt:
(¬x v y) = (¬y-->x)
= (¬x-->y)
but this cannot be right, as they have diffrent truth tables:
(¬y-->x)
1 0 0 0
1 0 1 1
0 1 1 0
0 1 1 1
(¬ x-->y)
1 0 0 0
0 1 1 1
1 0 0 0
0 1 1 1
I understand once you have the conjectures converted to implication, how to construct the implication graph and find out if its not satisfiable (bad loops).
Could someone please explain clearly how to break down the conjectures to implications?
Thanks in advance!
AI: A disjunction $A \vee B$ is equivalent to $\neg A \rightarrow B$ and also to $\neg B \rightarrow A$. This holds generally, i.e. not only when A, B are variables.
Therefore, by applying the above on $A=\neg x$ and $B=y$, you can get: $\neg x \vee y \sim \neg \neg x \rightarrow y \sim x\rightarrow y$ and also $\neg x \vee y \sim \neg y \rightarrow \neg x$.
($\sim$ denotes equivalence of propositions)
Similarily for any other disjunction of two literals. |
H: Intuition for Geometric Transformations
I've been making a lot of effort over the past few hours to gain some intuition into the art of geometric transformation but to little avail. I would really like to be able to look at a transformation matrix and have a pretty good idea of what it does to a shape (or at least a vector)
I sketched a simple box on a graph, and applied two simple transformations to them. After doing the first, I expected the second to produce a sort of mirror, or opposite of it, but that didn't happen (and I suppose it makes sense now thinking about it).
However, I would really appreciate if someone could give me an approach to thinking of these things intuitively and understanding where inside the matrix does the magic happen.
Below is my simple box (all drawn in Microsoft Publisher :) and the transformation matrix I used on each point (only one is labeled, the others are obvious)
I applied a very similar transformation in the second case but got a very different result.
AI: This might help,
Lets first look at what the matrix does to vectors. A useful way of looking at a matrix is to think of each column as being the result of applying the matrix to one of the basis vectors.
$$
\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]
\left[ \begin{array}{cc} 1 \\ 0 \end{array} \right]
=\left[ \begin{array}{cc} a \\ c \end{array} \right]
\qquad
\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]
\left[ \begin{array}{cc} 0 \\ 1 \end{array} \right]
=\left[ \begin{array}{cc} b \\ d \end{array} \right]
$$
When you have more than one nonzero component for a column vector the matrix is applied to each piece and the results are added. In other words multiplying a matrix by a column vector adds columns of the matrix weighted by the components of the vector.
$$
\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]
\left[ \begin{array}{cc} 3 \\ 4 \end{array} \right]
=\left[ \begin{array}{cc} 3a + 4b\\ 3c + 4d \end{array} \right]
$$
Now lets use this to see what your matrix does to the (x,y) ordered pairs,
$$
\left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right]
\left[ \begin{array}{cc} x \\ y \end{array} \right]
=\left[ \begin{array}{cc} x \\ x+y \end{array} \right]
\qquad
\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]
\left[ \begin{array}{cc} x \\ y \end{array} \right]
=\left[ \begin{array}{cc} x+y \\ y \end{array} \right].
$$
So you can see that when the first transformation is applied to a point it leaves the $x$ coordinate alone and then makes a new $y$ coordinate by adding the old $x$ and $y$ values. You can see this in your figure the shape is deformed vertically but not horizontally.
The second transformation does the same thing by with the roles of $x$ and $y$ reversed. Looking at this you can see that the $y$ coordinates do not change as a result of the transformation.
I hope this helps your intuition a bit. I'm not sure of you background but here are some general guidelines:
Your matrix will always treat $(0,0)$ as a special point. So when you are looking at what the transformation does you should keep in mind that it won't treat all squares equally.
When visualizing geometric transformations like this it is helpful to apply it to every point in the xy-grid. This gives you a new deformed grid which gives you an idea of what it does where. Think of this as stretching/compressing the xy plane.
If your transformation is diagonalizable you should find its eigenvalues and eigenvectors. These provide invaluable information about the transformation. |
H: Show that $\lim\limits_{k\rightarrow \infty} x^k = 0$ for $x \in (0,1)$
I want to show that $\lim\limits_{k\rightarrow \infty} x^k = 0$ for $x \in (0,1)$.
This makes intuitive sense but I can't figure out how I should go about proving it. Using the epsilon-delta definition I'd need to show that for every $\epsilon >0$ there is $M \in N$ such that
$$|x^k| < \epsilon \text { where } k \geq M$$
but I just can't figure out where to start.
AI: Hints :
Show that it is a decreasing sequence bounded below.
Notice that $a_0 = x$, and $a_k = f(a_{k-1})$ where $f(y) = xy$. By (1), we know that $L = \lim a_k$ exists. Now
$$
a_k \to L \Rightarrow f(a_k) \to f(L)
$$
since $f$ is continuous. However, $f(a_k) = a_{k+1}$, and so the the sequences $\{f(a_k)\}$ and $\{a_k\}$ are actually the same sequence. Since a sequence cannot converge to two different points,
$$
L = f(L) = xL
$$ |
H: Defining conditional quantum probability
My knowledge of quantum mechanics is very limited, but I will try to ask a purely mathematical question here. If there is a text or resource that explains this, I would definitely appreciate any pointers! I have been unable to find any explanations online (most quantum mechanical resources seem too focused on the physics to ask such questions...).
I am coming at this from a computer science angle, so I will talk about qubits. To make things very simple and concrete, suppose I have two qubits in some quantum state:
$$ a_{00} \left|00\right\rangle + a_{01} \left|01\right\rangle + a_{10} \left|10\right\rangle + a_{11} \left|11\right\rangle $$
where $a_{ij} \in \mathbb{C}$, $\sum_{ij} |a_{ij}|^2 = 1$.
Suppose I now measure the first bit. Given that I measure $0$, what is the quantum state of my system? Given that I measure $1$, what is the quantum state of my system?
A naive extension of Baye's rule leads me to conjecture that, for instance, if I measure zero for the first bit, then the quantum state is
$$ \sqrt{\frac{|a_{00}|^2}{|a_{00}|^2 + |a_{01}|^2}} \left|00\right\rangle + \sqrt{\frac{|a_{01}|^2}{|a_{00}|^2 + |a_{01}|^2}} \left|01\right\rangle .$$
But is this really correct?
The general question would be that if I have a quantum state $\sum_i a_i \left|\phi_i\right\rangle$, and I "partially observe" $\phi_i$ (so I guess one would say I "partially" collapse the wave function?), how is the new quantum state mathematically determined?
AI: Any observation in QM is tied to a hermitian operator, and the collapse of the wave function is always to the eigenstate of that operator corresponding to the measurement you got (at least as long as the possible outcomes are discrete). In your case we're dealing with a four-dimensional vector space over $\Bbb C$. Let's represent your quantum state as the column vector
$$
\begin{pmatrix}a_{00}\\ a_{01} \\ a_{10}\\ a_{11}\end{pmatrix}
$$
Then your "reading the first qbit"-observation is tied to the operator represented by the matrix
$$
\begin{pmatrix}0 & 0 & 0 & 0\\ 0 &0 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}
$$
When reading the first qbit as $1$, the state collapses to the eigenspace spanned by the eigenvectors with eigenvalue $1$, and while I do not know exactly how to handle degenerate states (i.e. that the eigenspace has more than one dimension), I would believe it collapses to
$$
\begin{pmatrix}0\\ 0 \\ a_{10}\\ a_{11}\end{pmatrix}
$$
which after a renormalization becomes
$$
\begin{pmatrix} 0 \\ 0 \\ a_{10}\bigg/\sqrt{|a_{10}|^2 + |a_{11}|^2} \\ a_{11}\bigg/\sqrt{|a_{10}|^2 + |a_{11}|^2}\end{pmatrix}
$$
Edit Wiki seems to confirm this, by the way. The vector
$$
\begin{pmatrix}0\\ 0 \\ a_{10}\\ a_{11}\end{pmatrix}
$$
is the projection of your original state onto the subspace of $C^4$ spanned by the eigenvectors of eigenvalue $1$ (that is, the last two coordinates). |
H: D=P/gv^3 what units is the D
(Sorry for my bad english)
I got a task to do:
In the research, scientists noticed that the necessary power P is obtained from P=Dgv^3, where D is constant, g is the earth's standard acceleration due to gravity and v is the speed. What is the D and what units it is, if the power is 75.0kW, speed 1.41m/s and acceleration of gravity 9.81m/s^2?
So it should go D=P/gv^3
D=75kW/(9.81m/s^2)*(1.41m/s)^3
where D is about 2.73kWs^5/m^4
But looks like the units are wrong, could someone please explain me how it should go and what units D has?
AI: Hint: a Watt is given by
$$\mathrm{W = \frac{J}{s} = \frac{N\cdot m}{s} = \frac{kg\cdot m^2}{s^3}} $$
So a $\mathrm{kW}$ is given by: $\;\;?$
Now, try to determine a different representation of units for $D$, with appropriate cancellations. The units may not necessarily be wrong, as you listed, but they may simply be represented differently in the solution. |
H: If $m,n$ are coprime positive integers and $m-n$ is odd, then $(m-n),(m+n),m,n,$ are coprime each other?
If $m,n$ are coprime positive integers and $m-n$ is odd, then $(m-n),(m+n),m,n,$ are coprime each other?
How do I prove it?
Especially how do I prove $(m-n), (m+n)$ are coprime?
AI: HINT:
$(m-n,n)=(m-n+n,n)=(m,n)$
$(m+n,n)=(m+n-n,n)=(m,n)$
If $d$ divides $m+n,m-n$
$d$ will divide $m+n+(m-n)=2m$ and $m+n-(m-n)=2n$
$\implies d$ will divide $(2m,2n)=2(m,n)$ |
H: How do I compute the summation where k is greater than or equal to $0$ of $\frac{1}{k+1}{99 \choose k}$ ${200 \choose 120-k}$
How do I compute
$$
\sum_{k=0}^{\infty}\frac{1}{k+1}\binom{99}{k}\binom{200}{120-k}.
$$
I have expanded it to this:
$$
\frac{1}{k+1}\cdot\frac{99!}{k!(99-k)!}\cdot\frac{200!}{(120-k)!(80+k)!}
$$
but am stuck after that. Is there any way to write my answer as an expression involving one or two binomial coefficients?
AI: Consider that $\frac{1}{k+1}\cdot\frac{99!}{k!(99-k)!}=\frac{1}{100}\cdot\frac{100!}{(k+1)!(100-(k+1))!},$ so you can re-write your sum as:
$$\frac{1}{100}\sum_{k=0}^{99}\binom{100}{k+1}\binom{200}{120-k}
=\frac{1}{100}\sum_{k=1}^{100}\binom{100}{k}\binom{200}{121-k}.$$
Now:
$$\sum_{k=0}^{100}\binom{100}{k}\binom{200}{121-k}=\left[x^{121}\right](1+x)^{100}(1+x)^{200}=\binom{300}{121},$$
so your sum is equal to:
$$\frac{1}{100}\left(\binom{300}{121}-\binom{200}{121}\right).$$ |
H: element wise matrix operation problem
I am doing an element wise power calculation, and at a given point, I get a complex value out of real values! I have attached a screen shot from the debugging mode in Matlab
So, one can see that the first operation is giving a complex number, although both A(8,111) and 4/(gamma-1) where gamma=1.4 are real!
Can any one explain that? I guess for more details, the screen shot speaks for itself.
Thank you in advance
AI: $A(8,111)$ is a negative number. $4/(\gamma - 1)$ should be the even integer $10$, but because of roundoff error its computed value is not quite equal to $10$.
> f = 4/(1.4 - 1)
f =
10.0000
> f - 10
ans =
1.7764e-15
A negative number to a non-integer power will get you a complex number. |
H: Convergence of series $ \sum_{n=1}^{\infty} \sin\left( \frac{n\pi}{6}\right)$
Decide if the series $ \sum_{n=1}^{\infty} \sin\left( \frac{n\pi}{6}\right)$ converges or not.
I've tried to use the ratio test but I had no success. I don't see how other convergence tests could work.
Thanks for your help!
AI: Hint: What is $\lim_{n\to\infty}\sin\left(\frac{n\pi}{6}\right)$? |
H: Combinatorial probability of multiple dice rolls
I am trying to come up with a probability for a combinatorial problem that I'm working on and I'm stuck. I've managed to distil it down to a toy dice-roll problem:
What is the probability of any two dice adding up to 7 on a throw of n 6-sided dice?
Or being more general:
What is the probability of any two dice adding up to x+1 on a throw of n x-sided dice?
Note that there could be more than one 2-combination in the n dice that add to 7 - so if n = 3 and x = 6 a roll of 6, 6, 1 would satisfy the criteria.
You can assume that x is an even number, and thus there are n/2 2-combinations that sum to x+1. For example for a 6-sided dice there are 3 (x/2) combinations of 2 dice that will produce a 7: 6 & 1, 5 & 2, 4 & 3. I've tried using binomial and multinomial expressions on the expansion of these combinations but can't get this right. Can someone give me some pointers? Help gratefully received.
Thanks
Chris
AI: The 6-sided dice case can be encoded as a Markov chain on the state space $\{0,1,2,3,\partial\}$ where each state $0\leqslant i\leqslant3$ means that $i$ pairs from the set $\{\{1,6\},\{2,5\},\{3,4\}\}$ have already been visited but none is completed yet, and state $\partial$ means that at least one of these pairs has been completed. For example, after the results 26622 the state of the chain is $2$, after the results 266224246 the state of the chain is $3$, and after the results 2662242465 the state of the chain is $\partial$.
The probability that any two dice add to 7 on a throw of $n$ is $P_0[T\leqslant n]$ where $T$ is the hitting time of $\partial$ and the subscript $0$ refers to the starting point of the chain.
To characterize $P_0[T\leqslant n]$ for every $n$, one usually computes the generating functions $u_i=E_i[s^T]$ for $0\leqslant i\leqslant3$, where $|s|\leqslant1$. Writing down carefully the transition probabilities of the chain, one sees that the Markov property after one step yields the relations
$$
u_0=su_1,\quad u_1=s(\tfrac16u_1+\tfrac46u_2+\tfrac16),\quad u_2=s(\tfrac26u_2+\tfrac26u_3+\tfrac26),\quad u_3=s(\tfrac36u_3+\tfrac36).
$$
Solving this Cramér system, one gets (something similar to)
$$
u_0=\frac{s^2(6+3s+s^2)}{(2-s)(3-s)(6-s)}.
$$
Thus, $u_0$ is a rational fraction with respect to $s$, which can be decomposed as
$$
u_0=\frac{c_2}{2-s}+\frac{c_3}{3-s}+\frac{c_6}{6-s}+as+b,
$$
for some suitable constants $a$, $b$, $c_2$, $c_3$ and $c_6$. Expanding each fraction as a power series in $s$ and collecting all the terms of degree at least $n$ yields, for every $n\geqslant2$,
$$
P[T\geqslant n]=\sum_k\frac{c_k}{k-1}\frac1{k^n},
$$
where the sum runs over $k$ in $\{2,3,6\}$.
If one uses some $2z$-sided dice instead, one finds similarly
$$
P[T\geqslant n]=\sum_k\frac{c_k}{k-1}\frac1{k^n},
$$
for some suitable constants $c_k$, where the sum runs over $k=2z/\ell$ with $1\leqslant\ell\leqslant z$. In particular, when $n\to\infty$,
$$
P[T\geqslant n]\sim\frac{c_2}{2^n}.
$$
Going back to the 6-sided case, $c_2=\lim\limits_{s\to2}(2-s)u_0=16$ hence
$$
P[T\geqslant n]\sim\frac{16}{2^{n}}.
$$ |
H: Let $R$ be an infinite commmutative ring which contains a zero divisor. Show that there exists infinite many zero divisors.
I'm making exercises to prepare for my ring theory exam.
Let $R$ be an infinite commmutative ring which contains a zero
divisor. Show that there exists infinite many zero divisors.
Let $a\in R$ a zero divisor. Then $a⋅b=0$ for some element $b≠0$ in $R$. If $a$ or $b$ has infinite order, then I can find infinite zero divisors. But otherwise, I don't see why this should be true. I think I need to do something with the fact that $R$ is commutative. A hint or a detailed solution are both appreciated.
AI: Let $a$ be a zero divisor. Consider two cases:
$Ra$ is an infinite set.
$Ra$ is a finite set.
(If $R\to Ra$ has finite range, $xa=ra$ has infinitely many solutions $x$ for some $r\in R$, so...) |
H: maximize $\int \limits ^b_a (24 - 2x - x^2)^{1/3}$
I need to find $a$ and $b$ in $\int \limits ^b_a $(24 - 2$x$ - $x^2$)$^{1/3}dx$ such that the value of the integral is maximized.
I know I need to solve the integral, plug in $a$ and $b$, and then treat the result as a function of two variables, but I can't figure out how to solve it.
AI: Hint: You need to find the largest interval on which your function is non-negative. No integration!
Remark: We have tacitly assumed that $a\le b$. If we allow $a\gt b$. there is no maximum. For example, by taking $b=4$ and $a$ large positive, we can make $\int_a^4 f(x)\,dx$ arbitrarily large. |
H: Complete Condition in Banach Fixed Point Theorem
Can someone provide an example to show that for the Banach fixed point theorem, that is if $T : X → X$ is a contraction in a complete metric space $(X, d)$ then $T$ has a unique fixed point that $X$ is complete is an essential condition?
AI: $$T \colon \mathbb{R}\setminus\{0\} \to \mathbb{R}\setminus\{0\};\; T(x) = x/2.$$
Basically, all examples are of that form, since a contraction is uniformly continuous, and can thus be extended to the completion of $X$, and the extension is also a contraction. So the extension $\widehat{T}\colon \widehat{X}\to\widehat{X}$ satisfies the premises of Banach's fixed point theorem, and has a unique fixed point. |
H: How to solve $\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2}$
Solve the following question :
\begin{eqnarray}
\\\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2}\\
\end{eqnarray}
The answer should be $\frac{1}{128}$.
I try that:
\begin{eqnarray}
\\\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2} &=& \lim_{x\to 16} \frac{(4-\sqrt{x})(4+\sqrt{x})}{(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})}\\
\\ &=& \lim_{x\to 16} \frac{16-x}{(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})}\\
\end{eqnarray}
What can I do?
Thank you for your attention.
AI: Putting $\sqrt x=h,$
We have $$\lim_{h\to4}\frac{4-h}{16h^2-h^4}=\lim_{h\to4}\frac{4-h}{h^2(16-h^2)}=\lim_{h\to4}\frac{4-h}{h^2(4-h)(4+h)}=\lim_{h\to4}\frac1{h^2(4+h)}=\cdots$$ Cancelling out $4-h$ as $4-h\ne0$ as $h\to4$ |
H: Is this divisibility test for 4 well-known?
It has just occurred to me that there is a very simple test to check if an integer is divisible by 4: take twice its tens place and add it to its ones place. If that number is divisible by 4, so is the original number.
This result seems like something that anybody with an elementary knowledge of modular arithmetic could realize, but I have noticed that it is conspicuously missing on many lists of divisibility tests (for example, see here, here, here, or here). Is this divisibility test well-known?
AI: Yes, it is well known; do you know modular arithmetic? Assuming you do, we have a number $abc=a\cdot 10^2+b\cdot 10^1+c\cdot 10^0$. Now $$a\cdot 10^2+b\cdot 10^1+c\cdot 10^0\equiv 2\cdot b+c\pmod{4}.$$ Many people know the multiples of $4$ for numbers less than $100$, so it is commonly just said if the last two digits (as a number) is divisible by $4$, then the number is divisible by $4$. |
H: Indefinite integral: $\int \frac{\mathrm dx}{\sqrt x(1+\sqrt[3]x)}$
I need :
$$ \int \frac{\mathrm dx}{\sqrt{x}(1+\sqrt[3]{x})}$$
Can you give me a hint, which way would suit best? I tried substitution, but don't have an idea what to substitute.. :/
AI: HINT:
As the Integrand involves $\displaystyle \sqrt x=x^{\frac12},\sqrt[3]x=x^{\frac13}$
and $\displaystyle\mathrm{gcd}\left(\frac12,\frac13\right)=\frac1{\mathrm{lcm}(2,3)}=\frac16$
put $\displaystyle y=x^{\frac16}\implies \sqrt x=y^3,\sqrt[3]x=y^2$ and $\displaystyle x=y^6$ |
H: Is Apostol's Calculus a Good Book to Use for GRE Math Subject Test Prep
I am going to take the math subject GRE, and after looking at the main topics (calculus, diff eqns, linear algebra, prob, and algebra), I got to wondering if Apostol's text on Calculus might not be a bad book to work from (since it contains calc, prob theory, linear algebra and diff eqns). I am not sure if this text has the depth I would need, but past posts suggest that working problems from Stewart is good practice for the calc part and I can't imagine Stewart would be more advanced than Apostol (although the former might have more problems).
Was wondering if anyone having experience with the tests could confirm my notions.
Thanks,
Matt
AI: I recently took the Math GRE subject test and found it quite challenging. My recommendation is that you do as many problems as you can, as quickly as you can. I recommend placing emphasis on calculus first, as it takes up 50% of the exam, then linear algebra/abstract algebra and then the other topics.
I have never used Apostol's book but here are my first impressions (bare in mind I started looking at it 5 minutes ago). It contains the standard material that is assessed in the GRE, this makes it useful for brushing up on any things you may have forgotten or misunderstood. It also seems like there are quite a lot of exercises but a lot look like standard computational exercises.
You need to be able to do these type of questions, but don't expect them to come up the GRE. The questions in the GRE are more complicated and require a deeper familiarity with the material than basic calculations. I would recommend taking a selection of exercises from a variety of sources: do some basic computations from Apostol and make sure you can do them quickly and accurately.
Then try to find harder exercises, maybe some of these can be found in Apostol or try a different calculus textbook like Stewart or Spivak. Then try to look at some hard problems; aside from previous GRE's, the best sources I can come up with right now are the "Berkley Problems in Mathematics" books but they are definitely above the level of the GRE (but hey, who doesnt like looking at hard math problems!). |
H: Did I do this limit right? $\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1}$
This is how I did this limit:
$$\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1} =\lim_{x \to 0}\frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)} = \lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)(\sqrt{x^2+2}+\sqrt{2})}{(x^2+1-1)(\sqrt{x^2+2}+\sqrt{2})} = \lim_{x \to 0}\frac{({x^2+2}-2)(\sqrt{x^2+1}+1)}{(x^2+1-1)(\sqrt{x^2+2}+\sqrt{2})} =\frac{(\sqrt{0^2+1}+1)}{(\sqrt{0^2+2}+\sqrt{2})}=\frac{1}{2\sqrt{2}}$$
But Wolframalpha gave me another answer!
So did I do it right?
AI: If you correctly simplify the very last line in your solution, you'll see you get the same answer: $\frac 2{2\sqrt 2}$
$$\frac{(\sqrt{0^2+1}+1)}{(\sqrt{0^2+2}+\sqrt{2})}=\frac{2}{2\sqrt{2}} = \frac 1{\sqrt 2}$$ |
H: Help me with this riddle.
We have two variables from 2 to 99. Their sum is A. Their product is B. I know A, the other person knows B. We have to get the numbers without asking each other. But I call the other person and say that I have no idea what these numbers can be. The other person tells that he knew for sure that I wouldn't understand the numbers. Then as I hear that he knew I wouldn't get the right answer , I say that I understood the answer just now. Hearing me the other person says the same. How is this possible?
AI: The only way A can know the numbers at the beginning is if they are $(2,2),(2,3),(98,99),(99,99)$. If the numbers were $(4,6)$ or $(3,8)$, for example, $B$ would know the product was $24$ and would know that $A$ doesn't know the numbers. Therefore $A$ cannot determine the numbers from $B$'s remark.
I suspect the first remark is supposed to come from the person who knows the product. In that case, the link lab bhattacharjee gave has the solution. |
H: bilinear form defined on $ X \times Y \to Z$ where $X$ is banach using uniform boundedness principl
This question has a similar that was asked before, but it's not exactly equal. Please help me with this.
Let $X$ be a banach space, $Y,Z$ a normed spaces, let $B: X \times Y \to Z$ be a bilinear map, such that it's continuous in each variable, show that $B$ is continuous.
I tried to use the Uniform Boundedness Principle. For each $y\in Y$ the function $f_y:X\to Z$ defined by $f_y(x)=B(x,y)$ is continuous by assumption. Also for each $x\in X$ the function $f_x :Y\to Z$ defined by $f_x(y)=B(x,y)$ is continuous.
Since each $f_y$ is continuous $|f_y(x)|\le K_y ||x||$ and similar for every $x$ $|f_x(y)|\le K_x ||y||$
We'll use the boundedness principle on the family $f_y$ (since are defined on a Banach space). For each $x\in X$
$|f_y(x)|\le K_y ||x||$ But this I need a bound that depends only on $x$ and not on $y$
Please help me
AI: Consider the family
$$\mathscr{F} = \{ f_y : y \in B_Y \}.$$
By the continuity of the $f_x$, the family is pointwise bounded,
$$\sup_{y\in B_Y} \lVert f_y(x)\rVert_Z = \lVert f_x\rVert < \infty.$$
By the uniform boundedness principle, the family is uniformly bounded, i.e.
$$\sup_{y\in B_Y} \lVert f_y\rVert = K < \infty.$$
And that says
$$\sup_{x\in B_X,\, y\in B_Y} \lVert f(x,y)\rVert_Z = K < \infty,$$
hence
$$\lVert f(x,y)\rVert_Z \leqslant K\cdot \lVert x\rVert_X\cdot \lVert y\rVert_Y.$$ |
H: Random Dynamical Systems: Intuitive Understanding
I am having trouble understanding this definition, [Arnold: Random Dynamical Systems]:
Definition: A measurable random dynamical system on the measurable space $(X,\mathcal{B})$ over a metric dynamical system $(\Omega,\mathcal{F},\mathbb{P},(\theta(t))_{t\in\mathbb{T}})$ as a map $\phi$ with time $\mathbb{T}$ is a mapping
$$\phi:\mathbb{T}\times\Omega\times X\to X,$$
where $\phi$ is measurable and satisfies the cocycle property. $\blacksquare$
I understand that a metric dynamical system is a family of iterative mappings on the space $(\Omega,\mathcal{F},\mathbb{P})$ that act as endomorphisms, but I don't understand how it all ties together. As far as I can tell, $\phi$ is a deterministic map.
I don't understand how randomness is described. Can anyone enlighten / inspire me?
AI: For every $(t,\omega,x)$ in $\mathbb T\times\Omega\times X$, $\phi(t,\omega,x)$ is some point in $X$ hence, for every $(t,x)$ in $\mathbb T\times X$, $\phi(t,x):\omega\mapsto\phi(t,\omega,x)$ is a random variable with values in $X$.
Furthermore, one asks that $\phi(t+s,x)=\phi(t,\phi(s,x))$ for every $(t,s,x)$ in $\mathbb T\times\mathbb T\times X$, in the sense that $\phi(t+s,\omega,x)=\phi(t,\omega,\phi(s,\omega,x))$ for every $(t,s,\omega,x)$ in $\mathbb T\times\mathbb T\times\Omega\times X$. In other words, each $\phi(\ ,\omega,\ )$ is a flow on $\mathbb T\times X$.
Edit: Here is a dynamical system. Assume that $\mathbb T=\mathbb N_0$, $X=\mathbb R$, that $\Omega=\{0,1\}^\mathbb N$ is endowed with its cylindrical sigma-algebra, and consider the shifts $(\theta_n)_{n\geqslant0}$ defined by $\theta_n(\omega)=(\omega_{k+n})_{k\geqslant1}$ for every $n\geqslant0$ and $\omega=(\omega_k)_{k\geqslant1}$ in $\Omega$, for example, $\theta_6(\omega_1,\omega_2,\omega_3,\ldots)=(\omega_7,\omega_8,\omega_9,\ldots)$. Finally, define $\phi(1,\omega,x)=A_{\omega_1}(x)$ for every $x$ in $X$ and every $\omega=(\omega_k)_{k\geqslant1}$ in $\Omega$, where $A_0$ and $A_1$ are some given functions on $\mathbb R$. In words, $\phi(1,\ ,\ )$ applies $A_0$ or $A_1$ to $x$, the index $0$ or $1$ being chosen uniformly randomly.
Let me suggest that you check that, for every $n\geqslant0$,
$$
\phi(n,\omega,x)=A_{\omega_n}\circ \cdots\circ A_{\omega_2}\circ A_{\omega_1}(x),
$$
and that the cocycle property holds in this setting. |
H: Question on Category
i have a question how to prove that
the category of groups and morphisms of groups is realy a category ?
Thank you.
AI: If $f$ and $g$ are functions then $f\circ g$ is by definition the
function $x\mapsto f\left(g\left(x\right)\right)$. This way we have
$h\circ\left(f\circ g\right)\left(x\right)=h\left(f\circ g\left(x\right)\right)=h\left(f\left(g\left(x\right)\right)\right)=\left(h\circ f\right)\left(g\left(x\right)\right)=\left(h\circ f\right)\circ g\left(x\right)$ showing that $h\circ\left(f\circ g\right)$ and $\left(h\circ f\right)\circ g$ are the same functions. |
H: Question about adjoint map and strong operator topology (SOT)
I am wondering
if there is any condition one can apply (e.g. uniform boundedness?) that ensure the adjoint of a net of SOT-continuous elements is again SOT-continuous?
My major question is
$\{v_t\}$ is a SOT continuous path of bounded operators, P is compact operator, then when is $\{v_tPv_t^*\}$ a norm continuous path of operators?
I got some partial answer in my previous question: Question about SOT and compact operators
which shows me that $\{v_tP\}$ is norm-continuous, but I am not sure how to apply that to solve my major question. I was hoping the first question I raised in the current post may help answer it. Thank you!
AI: Preliminaries. Let $H$ be a Hilbert space. Consider bilinear operator
$$
\bigcirc: H\times H^{cc}\to\mathcal{F}(H):(x,y)\mapsto (z\mapsto \langle z,y\rangle x)
$$
where $H^{cc}$ is a complex conjgate Hilbert space and $\mathcal{F}(H)$ is a normed space of all finite rank operators with operator norm. One can check that $\bigcirc$ is bounded bilinear operator of norm one and its image is all rank one operators. For a given $x,y\in H$ this is straightforward to check that
$$
A(x\bigcirc y)B=A(x)\bigcirc B^*(y)\tag{1}
$$
Lemma Let $H$ be a Hilbert space and $(T_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $T\in\mathcal{B}(H)$. Then, $(T_n:n\in\mathbb{N})$ is norm bounded in $\mathcal{B}(H)$ by some constant $c_T$.
Proof. Since $(T_n:n\in\mathbb{N})$ strongly converges to $T$, then for all $x\in X$ the sequence $(T_n(x):n\in\mathbb{N})$ is norm bounded in $H$. By Banach-Steinhaus theorem the sequence $(T_n:n\in\mathbb{N})$ is norm bounded in $\mathcal{B}(H)$ by some constant $c_T>0$
Lemma. Let $H$ be a Hilbert space $A\in\mathcal{K}(H)$ and $(T_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $T\in\mathcal{B}(H)$, $(S_n:n\in\mathbb{N})\subset\mathcal{B}(H)$ strongly converges to $S\in\mathcal{B}(H)$ then so $(T_nAS_n^*:n\in\mathbb{N})$ converges to $TAS^*$ in the norm topology of $\mathcal{B}(H)$.
Proof. Let $x,y\in H$. Since $(T_n:n\in\mathbb{N})$ and $(S_n:n\in\mathbb{N})$ converges to $T$ and $S$ respectively in the strong operator topology, then $(T_n(x):n\in\mathbb{N})$ and $(S_n(y):n\in\mathbb{N})$ converge to $T(x)$ and $S(y)$ respectively in the norm topology of $H$. By properties of bilinear operator $\bigcirc$ we conclude that $(T_n(x)\bigcirc S_n(y):n\in\mathbb{N})$ converges to $T(x)\bigcirc S(y)$ in the norm topology of $\mathcal{B}(H)$. Therefore from $(1)$ we get that
$$
\lim\limits_{n\to\infty} T_n(x\bigcirc y) S_n^*=T(x\bigcirc y)S^*\tag{2}
$$
in the norm topology of $\mathcal{B}(H)$. Now take arbitrary $F\in\mathcal{F}(H)$. Since every finite rank operator is a sum of rank one operators, then there exists $(x_1,\ldots,x_k)\subset H$ and $(y_1,\ldots,y_k)\subset H$ such that $F=\sum_{i=1}^k x_k\bigcirc y_k$. So using $(2)$ we get
$$
\lim\limits_{n\to\infty} T_nF S_n^*
=\sum_{i=1}^k\lim\limits_{n\to\infty} T_n(x_k\bigcirc y_k) S_n^*
=\sum_{i=1}^k T(x_k\bigcirc y_k)S^*
=TFS^*\tag{3}
$$
in the norm topology of $\mathcal{B}(H)$.
Finally, consider compact operator $A\in\mathcal{K}(H)$. Since $H$ is a Hilbert space, then $A$ is limit of finite rank operators in the norm topology of $\mathcal{B}(H)$. Fix $\varepsilon>0$. From the previous note we have that there exist $F\in\mathcal{F}(H)$ and $D\in\mathcal{K}(H)$ such that $A=F+D$ and $\Vert D\Vert<\varepsilon$. Then using previous lemma we get
$$
\Vert T_n DS_n^*-TDS^*\Vert
\leq\Vert T_n DS_n^*\Vert+\Vert TDS^*\Vert
\leq\Vert T_n\Vert \Vert D\Vert \Vert S_n^*\Vert+\Vert T\Vert\Vert D\Vert \Vert S^*\Vert
\leq c_Tc_S \varepsilon +\Vert T\Vert\Vert S\Vert \varepsilon
$$
$$
0\leq\Vert T_n AS_n^*-T AS^*\Vert
\leq \Vert T_n FS_n^*-T FS^*\Vert+\Vert T_n DS_n^*-T DS^*\Vert\\
\leq \Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon
$$
Now we take $\limsup$ when $n\to\infty$ and use $(3)$ to get
$$
0\leq\limsup\limits_{n\to\infty}\Vert T_n AS_n^*-T AS^*\Vert
\leq \limsup\limits_{n\to\infty}\Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon\\
\leq \lim\limits_{n\to\infty}\Vert T_n FS_n^*-T FS^*\Vert+(c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon
\leq (c_Tc_S + \Vert T\Vert\Vert S\Vert)\varepsilon
$$
Since $\varepsilon>0$ is arbitrary from previous inequalities we conclude that $\lim\limits_{n\to\infty}\Vert T_n AS_n^*-T AS^*\Vert$ exists and equals to $0$. Hence we proved that
$$
\lim\limits_{n\to\infty} T_n AS_n^*=T AS^*
$$
in the norm topology of $\mathcal{B}(H)$.
Proposition Let $H$ be a Hilbert space and $P\in\mathcal{K}(H)$. Assume $\gamma:[0,1]\to\mathcal{B}(H):t\mapsto v_tpv_t^*$ is a path continuous in strong operator topology, then it is continuous in norm topology.
Proof Since $[0,1]$ is a first-countable topological space it is enough to show sequential continuity. For a given sequence $(t_n:n\in\mathbb{N})\subset[0,1]$ convrgent to $t\in[0,1]$ define $S_n=T_n=v_{t_n}$ and apply previous lemma to get that $v_{t_n}Pv_{t_n}^*$ converges to $v_tPv_t^*$ in norm. Hence $\gamma$ is norm continuous path. |
H: Is a closed n-dimensional disk compact necessarily compact?
As the title asks, is a closed n-dimensional disk compact necessarily compact? I'm thinking the answer would be no. If you consider the case in $\mathbb{R}^1$ then can you define the radius to be infinite. As such you would get all of $\mathbb{R}$, which is both closed, but not compact (since it isn't bounded). The same construction can be applied for $\mathbb{R}^n$ by taking an infinite radius. Is this correct or am I missing something?
AI: As AndréNicolas points out, a disk necessarily has finite radius.
It is true by the Heine-Borel theorem that, in the Euclidean space $\mathbb R^n$, a set is compact if and only if it is closed and bounded, and therefore every closed disk is compact.
If you want a good exercise, try proving the Heine-Borel theorem on your own. If you get stuck, you can check wikipedia or an entry-level analysis or topology text for hints/an outline. |
H: Math notation for summing up the rows in a matrix
Lets say I have a 100 row by 200 column matrix $\phi$, is there any standard notation or something which defines the vector $\Phi$ which has the same amount of columns (i.e. 200 columns), but all the rows summed up into just one row?
Something like we use $\sum$ for elements of a set or iterating over a sequence.
AI: Simply put, $\Phi=(1,1,\ldots,1)\phi$. |
H: Verify that the transform of $y(t) = t^2e^{at}$ is $Y(s) = \frac{2}{(s-a)^3}$
I made the distinction to amplify "=" 3 times for easier readability.
I tried:
$$F(s) === \int_0^\infty t^2e^{(a-s)t}dt === \frac{1}{a-s}e^{(a-s)t}t^2\Big|_0^\infty \ - \frac{2}{a-s}\int_0^\infty te^{(a-s)t}dt$$
But we have $$\int_0^\infty te^{(a-s)t}dt = \frac{1}{a-s}e^{(a-s)t}t\Big|_0^\infty - \frac{1}{a-s}\int_0^\infty e^{(a-s)t}dt$$
But the Laplace transform of $\int_0^\infty e^{(a-s)t}dt$ is just $\frac{1}{s-a}$. I tried plugging back in to get
$$\int_0^\infty te^{(a-s)t}dt = \frac{1}{a-s}e^{(a-s)t}t\Big|_0^\infty - \frac{1}{s-a}$$
and we know that integral converges when $(0 < s) \land (a < s)$, and since $t < e^t$ for large $t$ as well. Therefore the integral is just $0 - 0 + \frac{1}{a-s}$.
Then I plug back in again to get
$$\int_0^\infty t^2e^{(a-s)t}dt === \frac{1}{a-s}e^{(a-s)t}t^2\Big|_0^\infty - \frac{2}{a-s}[\frac{1}{(a-s)^2}]]$$
The integral on the RHS is just 0 as well, so the answer should be $-\frac{2}{(a-s)^3}$. But that's not the transform, it is $\frac{2}{(a-s)^3}$. Where did I go wrong?
AI: $\mathbf{I}:$
$$\int t^2e^{(a-s)t}dt=\frac{t^2}{a-s}e^{(a-s)t}-\frac{2t}{(a-s)^2}e^{(a-s)t}+\frac{2}{(a-s)^3}e^{(a-s)t}$$
$\mathbf{II:}$
$$\mathcal{L}[e^{at}f(t)]=\mathcal{L}[f(t)]\Big|_{s\to s-a}$$ and you know that $\mathcal{L}(t^n)=\frac{n!}{s^{n+1}}$. |
H: curvature of curve in $\mathbb R^2$
I'm taking a course in differential geometry this semester and I'm stuck with one of my first exercises.
Let $\alpha(s)=(x(s),y(s))$ be a curve such that $|\alpha'(s)|=1$. Prove that the curvature is given by $k(s)=|x'(s)y''(s)-x''(s)y'(s)|$.
So far we defined the curvature of such curves as $k(s)=|\alpha''(s)|$, but this doesn't really get me to the formula I'm supposed to prove. Since we haven't really done anything in the course yet, I don't think that I have to do anything really complicated here, but still I don't know what to do.
Anyone who can help me with this? Thanks.
AI: Since $|\alpha'|^2=(x')^2+(y')^2$ is constant, $\alpha''$ is orthogonal to $\alpha'$.
But $(-y',x')$ is also orthogonal to $\alpha'=(x',y')$, so it is parallel to $\alpha''=(x'',y'')$. What does that tell you about the inner (or dot) product between the two vectors? |
H: Modular arithmetic of numbers
let us consider two integers a,b that are co prime to a prime number p
Then is there any relation between a%p, b%p and ab%p ? % = modulo operator
AI: In general, if $r_m(a)$ denotes the remainder of $a$ when divided by $m$, you have that $$r_m(ab)=r_m(r_m(a)\cdot r_m(b))$$ |
H: For any open subset $A\subseteq \mathbb{R}$, $\operatorname{int}(\overline{A})=A$?
In the quiz of a class in MIT OCW, there is a T/F problem :
For any open subset $A \subseteq \mathbb{R}$, $\operatorname{int}(\overline{A})=A$?
The hompage of the class also provided a answer, and I saw the answer of the above.
The answer is False, because (the writer said) $\operatorname{int}(\overline{A})$ does not contain all isolated points of $A$.
But I think the reason is incorrect, because A is a open subset of $\mathbb{R}$.
(Every open subset of $\mathbb{R}$ consists of interior points.)
Although the original answer of the above question may be False, but is the reason that writer said incorrect?
AI: False. $A=(0,1)\cup(1,2)$, $A$ is open, but $\overline{A}=[0,2]$ and $\operatorname{int}(\overline{A})=(0,2)$. It may be the case that the real reason is, in fact, that $\operatorname{int}(\overline{A})$ may contain isolated points of the complement of $A$. |
H: Probability of X
There can be only 2 possibilities either X or Y. If X occurs first then prob.of X occurring next is 0.4, and if Y occurs 1st prob. of Y occurs next is 0.3. So if X occurs at 5th instant what is the prob. that x will occur at 7th instant?
AI: i think $0.6\times0.7+0.4\times0.4=0.58$ |
H: Continuous in $I\times J$ - then continuous in $I$ and in $J$?
Maybe it is a silly question, but:
If I have a function $f$ that is continuous on $I\times J\subseteq\mathbb{R}^2$; does this imply that $f$ is continuous in $I$ and in $J$?
My intuitive answer is: Yes, of course, because if one looks at the criterion of continuity that uses series, i.e.
$$
(x_k,y_k)\to (x,y)\Rightarrow f((x_k,y_k))\to f((x,y))
$$
that implies
$$
x_k\to x\Rightarrow f(x_k)\to f(x)
$$
respectively
$$
y_k\to y\Rightarrow f(y_k)\to f(y).
$$
Or am I totally wrong and confused?
Regards
AI: It is easy to check that the functions $i_y: I \to I \times J$ and $j_x: J \to I \times J$ defined by $i_y(x) = (x,y)$ and $j_x(y) = (x,y)$ are continuous. Hence the functions $f \circ i_y$, $f \circ j_x$ will be contiinuous. |
H: Finding the kernel , the image and the rank of $[A\ A]$ for an invertible $A$
Let $A$ be an invertible matrix of order $n$. What are the kernel, the image, and the rank of the matrix $\begin{bmatrix} A & A \end{bmatrix}$ (of order $n \times 2n$)?
AI: Hint:
$$\begin{bmatrix} A & A \end{bmatrix} \begin{bmatrix} x \\ x \end{bmatrix} = Ax + Ax = 2Ax.$$ |
H: Approximating an integral by evaluating the cumulated sum
I am using a cumulated sum to approximate an integral.
My initial thought was that the integral in the interval from a to b by evaluating the cumulated sum at b and a, and subtract.
When I do this, however, I always seem to underestimate the integral leading me to think that the integral is better evaluated as $C(b)-C(a-1)$ instead.
Can this really be true? If so, why?
E.g. in IPython
data = (0, 0, 1, 2, 1, 0)
csum = cumsum(data)
A = csum[4]-csum[2]
A gives 3 and not 4 (which is the sum of 1, 2 and 1).
AI: You need to describe what method you are using. It appears you are suspecting a fencepost error, which is possible. Another source of underestimate is that we tend to draw functions that are increasing. If one element of your sum is $(x_{i+1}-x_i)f(x_i)$ you may have a systematic error. Better is to take as that element $(x_{i+1}-x_i)(f(x_i+\frac 12(x_{i+1}-x_i))$ or $(x_{i+1}-x_i)\frac 12(f(x_i)+f(x_{i+1}))$ Both are exact for linear functions. |
H: How to find the subgroups of S4 generated by these sets.
How do I find the subgroups of $S4$ generated by these sets in each case?
$A = {(1,3),(1,2,3,4)}$
$B = {(1,2,4),(2,3,4)}$
$C = {(1,2),(1,3),(1,4)}$
AI: I am giving you some points about A. I hope you can use them well. Let $a=(1,3),~b=(1,2,3,4)$ , we know that $$S_4=\langle x,y\mid x^2=y^4=(xy)^3=1\rangle$$ and beause of the final relation inside the presentation, we can show every element of $S_4$ as $x^iy^j$ where $i=0,1,~~j=0,1,2,3$. Here, $|a|=2,~~|b|=4$ and $$a^2=id\\ b^2=(1,3)(2,4)\\ b^3=(1,4,3,2)\\ab^2=(2,4)\\ab^3=(1,2)(3,4)\\ ab=(1,4)(2,3)$$ This means that $\langle a,b\rangle=\{id,a,b,ab,ab^2,ab^3,b^2,b^3\}$ |
H: What does it mean to be $0.9-$Dimension?
We can visualize $1\mathrm D$, $2\mathrm D$, $3\mathrm D$ and we can think of a higher $M-\mathrm {Dimension}$ where $M\in \mathbb N^+$as a vector. But I recently learned that there are non integer dimensions such as $\frac{\log3}{\log2}$ for Sierpinski triangle and $\frac{\log2}{\log3}$ for Cantor set.
I have a three part question:
(a) What does it mean?
(b) Do we have a vector or other way to represent that?
(c) Are there negative dimensions too?
Thanks a lot.
AI: There are many generalizations of the usual notion of dimension, and they are there to capture different properties. Having said that, the intuition behind the dimension is that it describes the number of degrees of freedom you have, e.g.
A point on a line has one degree of freedom.
A point on a plane has two degrees of freedom.
A point of two-dimensional manifold (e.g. sphere) has two degrees of freedom (e.g. latitude and longitude), even if the manifold itself is defined as a subset of some more-dimensional space (e.g. $\mathbb{R}^3$).
An infinite-dimensional space has infinite degrees of freedom (e.g. the $L_2$ space).
However, the space defined may be more complex than that, e.g. in some areas might have one degree of freedom and in some other more, for example take a union of a disk and a line. Still, things can get even weirder, e.g. take the Hilbert curve, is it yet a line-like, or already a plane-like? In this way, you can think of the Sierpiński triangle as something that is a bit like a triangle, but not exactly yet.
To see how this works, consider the Cantor set and the following informal argument. Observe that to describe a point of $[0,1]$ with precision of $\frac{1}{3^n}$ you need $3^n$ possibilities, and thus
in base $3$ at least $n$ digits. However, to describe the elements of
the Cantor set you only need $2^n$ possibilities (the Cantor set is the
set of all the points that doesn't use the number $1$ in their base-3 representation). But to describe $2^n$ possibilities you need only $\log_3(2^n) = n\frac{\log 2}{\log 3}$ digits. But we needed $n$ digits to represent an point in $[0,1]$ which is of dimension $1$, and so we arrive at $\frac{\log 2}{\log 3}$.
In the case of Hausdorff dimension there is yet another observation related to scaling, but this is already covered by @PedroTamaroff's answer and this article of Wikipedia.
As to negative dimension, I'm not really sure what "negative number of degrees of freedom" could mean. Right now I'm not aware of any useful related notions.
On the other hand, you might be interested to know dimension is not the only seemingly integral characteristic, e.g. there are differentials of fractional order and those certainly can be negative (you get integrals then). In similar spirit a space of negative dimension would have to "remove the degrees of freedom" when joined with some other space, e.g. we have that $\dim(\mathbb{R} \times \mathbb{R}) = 2\dim(\mathbb{R})$, so possibly $\dim(\mathbb{R} \times X) = 0$? This might happen for example in a case where you require that points of your space have some property, and then all points of $\mathbb{R}$ would have that property, but only a countable subset of $\mathbb{R} \times X$ would work for you. Still, we need a new notion of dimension for that (one that incorporates your property).
I hope this helps $\ddot\smile$ |
H: set of all reg exp vs set of all languages
$\Sigma$ = {a,b,c,d,e}
V = {A,B,C,D,E,F,G,H}
According to the notes:
The set of all regular expressions over $\Sigma$ is infinite and countable.
The set of all languages over $\Sigma$ is infinite and uncountable.
I don't get why reg exp is countable whereas language is uncountable.
I guess i can't seem to be grasping the subtle difference between them.
What's the difference?
AI: A language is just a set of strings. A language $L$ over $\Sigma$ is a subset of $\Sigma^\ast$, which is the set of all strings made from letters in $\Sigma$. $L$ contains some of the strings in $\Sigma^\ast$, but not necessarily all; it might contain, or not contain, any string of letters from $\Sigma$. $\Sigma^\ast$ itself is a countable set, and a countable set has an uncountable number of possible subsets, by Cantor's theorem. So the family of all possible languages over $\Sigma$ is uncountable.
However, a regular expression is an expression involving the operators $\cup,\cdot,\ast$ that represents a certain language. For example, the regular expression $(\mathtt{0}^\ast\cup\mathtt{11^\ast01})\mathtt{01}$ represents the language $\{
\mathtt{01},
\mathtt{001},
\mathtt{0001},
\mathtt{00001},
\mathtt{10101},
\mathtt{000001},
\mathtt{110101},
\mathtt{0000001},
\mathtt{1110101},
\ldots
\}$.
The language that a regular expression represents is always a regular language. And a fairly simple argument shows that the collection of all regular expressions over $\Sigma$ is countable: every regular expression is finite, so has some length $\ell$. Let $R_\ell$ be the set of regular expressions of length $\ell$. Every set $R_\ell$ is finite. (It has no more than $N^\ell$ elements, where $N$ is the size of $\Sigma$ plus the other symbols $\cup,\cdot,\ast, ), (,$ etc.) The set of all regular expressions is just the union of all the $R_\ell$, and is therefore a countable union of finite sets and so countable.
(This provides one possible proof that there are languages that are not represented by regular expressions, and which are therefore not regular—indeed that most languages are not regular. One simple example is the language of all strings of $\mathtt{x}$es that have length $p^2$ for some integer $p$.) |
H: Folland PDE typo's, help follow proof that $||u||_s \leq C(||Lu||_{s-k} + ||u||_{s-1})$
I'm reading Folland's Introduction to Partial Differential Equations, and he makes a few claims that I don't understand and I think may be typos in the book.
Firstly let's fix
$$
L = \sum_{|\alpha|=k}a_\alpha \partial^{\alpha}
$$
and assumme the $a_\alpha$ are constants (Folland does it in more generality, but I just want to look at the constant case)
In the following proof, he computes $$\widehat{Lu}(\xi) = (2\pi i)^k \sum_{|\alpha|=k}a_\alpha\xi^\alpha \widehat{u}(\xi)$$
and then claims that by the assumed ellipticity condition
$$
\left|\sum_{|\alpha|=k}a_\alpha \xi^\alpha \right| \geq A|\xi|^k
$$
we have
$$
(1+|\xi|^2)^s|\widehat{u}(\xi)|^2 \leq 2^k(1+|\xi|^2)^{s-k}(1+|\xi|^2)^k|\widehat{u}(\xi)|^2
$$
$$
\leq 2^k A^{-1}(1+|\xi|^2)^{s-k}|\widehat{Lu}(\xi)|^2+2^k(1+|\xi|^2)^{s-k}|\widehat{u}(\xi)|^2.
$$
I have been fiddling with it for a few hours now and I have no idea how the second inequality follows. I suspect the $2^k$ in the first was a typo (it would be equality without it), so maybe he was trying to claim
$$
(1+|\xi|^2)^{s-k}(1+|\xi|^2)^k|\widehat{u}(\xi)|^2 \leq 2^k A^{-1}(1+|\xi|^2)^{s-k}|\widehat{Lu}(\xi)|^2+2^k(1+|\xi|^2)^{s-k}|\widehat{u}(\xi)|^2
$$
but I have no idea what he did in order to reach the conclusion. Could someone enlighten me as to what Folland meant to write?
AI: I figured it out, he was using $(1+|x|)^k \leq 2^k(1+|x|^k)$. It is easy to check for $|x|<1, |x|=1$ and $|x|>1$. |
H: Prove that the min and max of 2 continuous function are continuous
Prove that if $f$ and $g$ are continuous functions the so are $\min\{f(x),g(x)\}$ and $\max\{f(x),g(x)\}$
I know this is true when $f$ and $g$ are not intersect each other, then I can compare them. However, I don't know how to prove it's true when they are intersect.
AI: Let $h(x) = \min\{f(x),g(x)\}$. Suppose $x_0$ is such that $f(x_0) = g(x_0)$. We want to show $h$ is continuous at $x_0$.
Take $\epsilon > 0$, then there is a $\delta_f$ so that $|f(x) - f(x_0)| < \epsilon$ for $|x-x_0| < \delta_f$, and similarly for $g$ and some $\delta_g$ (with the same $\epsilon$).
Use this, and the fact that $h(x_0) = f(x_0) = g(x_0)$ to show that $|h(x) - h(x_0)| < \epsilon$ whether $h(x) = f(x)$ or $h(x) = g(x)$ as long as $|x-x_0| < \delta$ for some $\delta$. |
H: Calculate an unconditional probability given only the conditional probability
Let X and Y be jointly absolutely continuous random variables. Suppose X~Exponential(2) and P(Y > 5 | X = x) = $$e^{-3x}$$. Compute P(Y > 5).
I've computed P(Y > 5) as follows:
P(Y > 5) = P(Y > 5 | X = $x_1$) + ... + P(Y > 5 | X = $x_n$) = $\int_0^\infty e^{-3x} dx = \frac13$
However, I'm not sure the answer is correct, since I did not have to use the fact that X~Exponential(2). Can someone help?
AI: If $P\{Y > 5\mid X = x\} = \exp(-3x)$, then
$$P\{Y > 5\} = \int_{-\infty}^\infty P\{Y > 5\mid X = x\} f_X(x)\,\mathrm dx
= \int_0^\infty \exp(-3x) \cdot 2\exp(-2x)\,\mathrm dx$$
and so the answer does depend on the distribution of $X$. The above
is an application of
the law of total probability for continuous random variable as @Patrick says in
his answer, but your answer of $\frac{1}{3}$ is incorrect. You are
adding up conditional probabilities conditioned on different events,
and that is not what the law of total probability does. |
H: How do you solve a recurrence with a summation function inside: $t(n) = 1 + \sum\limits_{j=0}^{n-1} t(j)$
Show that
$$t(n) = 1 + \sum_{ j=0}^{n-1} t(j)$$
is the same as
$$t(n) = 2^n$$
Initial condition $t(0) = 1$
AI: Use induction:
Base case (n = 1): $$t_1 = 1 + \sum_{ j=0}^{n-1} t_j = 1 + t_0 = 1 + 1 = 2 = 2^1$$
Inductive step.
Assume that
$$t_k = 2^k$$
Then
\begin{align}
t_{k+1} &= 1 + \sum_{ j=0}^{(k+1)-1=k} t_j \\
& = 1 + \left(\sum_{ j=0}^{k-1} t_j\right) +t_k \\
& = 1 +(t_k - 1) + t_k
\end{align}
(
since $$t_{k+1} = 1 + \sum_{ j=0}^{k} t_j$$
which means that $$\sum_{ j=0}^{k} t_j = t_{k+1} - 1$$
and so $$\sum_{ j=0}^{k-1} t_j = t_{k} - 1.$$
)
Hence $$t_{k+1} = 1 + t_k -1 + t_k = 2t_k = 2^{k+1}$$
(by the inductive hypothesis).
The result follows by induction. |
H: showing algebraic inequality with arithmetic and harmonic means
Let x, y, z be positive real numbers. Prove that $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge \frac{3}{2}$$
This problem appears to be simple, but upon further work and lots of failed attempts, I am stuck. I have tried using arithmetic and harmonic means (which I am sure are the key) to show that there exists some number which is fits between these two, thus proving the inequality. I have also tried multiplying it out and simplifying and obtained:
$$1 + \frac{x^3 + y^3 + z^3 + xyz}{(x+y)(y+z)(z+x)} \ge \frac{3}{2} $$ This didn't really seem to help. Any guidance is greatly appreciated!
AI: $$A=\dfrac a{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}$$
$$B= \dfrac b{b+c}+ \dfrac{c}{c+a}+\dfrac{a}{a+b}$$
$$C=\dfrac c{b+c}+\dfrac{a}{a+c}+\dfrac{b}{b+a}$$
By A-G inequality
$$A+B=\dfrac{a+b}{b+c}+\dfrac{b+c}{a+c}+\dfrac{c+a}{b+a}\ge 3$$
$$A+C=\dfrac{a+c}{b+c}+\dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}\ge3$$
so
+
$$(A+B)+(A+C)\ge6$$
$B+C=3$, we get $2A\ge3$, yeah |
H: Prove by induction that for all natural numbers, n, either 3|n or 3|n+1 or 3|n+2?
That is prove that for all natural numbers, n, either 3 is a factor of n or n+1 or n+2
AI: Base case: n = 1.
Then 3 divides 3, which is n+2.
Inductive hypothesis:
if the proposition is true for n: Then proof by cases
1). If n is divisible by 3, this implies that (n+1) + 2 is also divisible by 3.
2). If n +1 is divisible by 3, then (n+1) is also divisible by 3.
3). If n+2 is divisible by 3, then (n+1) + 1 is also divisible by 3.
Thus, for all natural numbers ,n, the 3 divides either n, n+1, or n+2. |
H: Prove that for all $a\in \mathbb Z$, $a>2 \Rightarrow a \nmid b$ or $a \nmid (b +1)$
Prove that for all $a\in \mathbb Z$, $a>2 \Rightarrow a \nmid b$ or $a \nmid (b +1)$
I understand direct proof techniques and contrapositive proofs but I'm stumped on how to go about this.
AI: Instead of using a bunch of notation, let's say what it means: It means that if $a$ is an integer larger than $2$, then $a$ can't divide both $b$ and $b + 1$.
As a hint, suppose that $a | b$ and $a | b + 1$. Then it's also true that $a$ must divide the difference, $(b + 1) - b = 1$, so what can $a$ possibly be? |
H: Does a continuous function commute with $\limsup$ of bounded sequences?
Let $f$ be a continuous function on $\Bbb R$ and let $(a_n)_n$ be a bounded sequence. Prove or disprove:
$$\limsup_{n \to \infty}f(a_n)=f(\limsup_{n\to\infty}a_n)$$
I know that $\limsup_n(\alpha a_n)= \alpha (\limsup_n a_n)$. How can I relate this to functions?
AI: Let $a_n = (-1)^n$ and $f(x) = 2-x$. Then $\limsup f(a_n) = 3 \ne 1 = f(\limsup a_n)$. |
H: Covering a chess board with $2$ missing places with $31$ dominoes
I am reading a book that is intended to a wide audience (and not just
mathematicians etc'), the book is, of course, about mathematics (Its
still not clear about what exactly, I am only in page $2$).
The author gives the following riddle:
Take an $8\times8$ bored, and remove the bottom right piece and the
upper left one, can you cover it with $31$ dominoes ?
The answer is given - it is no. The author explains that if we were
to paint the board as a chess board then the removed pieces are of
the same color, but since each domino will cover exactly one black
piece and one white piece then we can't cover the $30$ white pieces
and $32$ black pieces.
Then the author poses an exercise:
Prove that this can be done if the removed pieces are of different
color
I don't know how to start with this, I can look at a given board and
try to cover it, but I can't give a methodology or a proof for this.
Can someone please hint me in the right direction ?
AI: Draw a hamiltonian cycle on the chessboard. Any cycle will do, and the following one suffices:
Pick (any) one square and call it $s_0$. Number the following squares in the hamiltonian cycle $s_1, s_2,\ldots s_{63}$:
Note that the even $s_i$ are all one color (say pink) and the odd $s_i$ are the other color (say brown). So far this is all very easy.
Now suppose two squares of the opposite colors have been deleted; for example:
The cycle has been cut into two paths, both of which have even length. Such a path is easily covered with dominoes: say one path is $s_k, s_{k+1}, \ldots, s_{k+2j+1}$, where the subscripts are understood mod 64. Then the dominoes easily go onto $(s_k, s_{k+1}), (s_{k+2}, s_{k+3}), \ldots (s_{k+2j}, s_{k+2j+1})$. Similarly the other path is tiled by $(s_{k+2j+3}, s_{k+2j+4}), \ldots, (s_{k-3}, s_{k-2})$.
This theorem and the proof I have given appear on pages 111–112 of Solomon W. Golomb Polyominoes (revised ed.), Princeton University Press, 1994. The proof is attributed to Ralph Gomory. |
H: Joint distribution of two marginal normal random variables
Question: Suppose we have:
\begin{align*} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} \sim N\left(\begin{bmatrix} 6 \\ 3 \end{bmatrix}, \begin{bmatrix} 12 & 3 \\ 3 & 2 \end{bmatrix} \right) \end{align*}
and let $Y = X_1 - 3X_2$, then what is the joint distribution of $X_1$ and $Y$?
My working: The marginal distribution of $X_1$ is:
\begin{align*} X_1 \sim N(6, 12) \end{align*}
The marginal distribution of Y is:
$$Y \sim N\left(-3, 12 \right)$$
since \begin{gather*}
E\left(Y\right) = 6 - (3)(3) = -3 \\
V\left( Y\right) = 12+ (-3)(2)(-3) + 2(3)(-3) = 12
\end{gather*}
So we have $X_1 \sim N\left(6, 12\right)$ and $Y \sim N\left(-3, 12\right)$, first we need to work out the covariance between $X_1$ and $Y$:
\begin{align*}
cov\left(X_1,Y\right) & = E(X_1Y) - E(X_1)E(Y) \\
& = E\left(X_1\left(X_1-3X_2 \right) \right) - 6(-3) \\
& = E(X_1^2) - 3E(X_1X_2) + 18 \\
& = V(X_1) + E(X_1)^2 -3\left[cov(X_1, X_2) + E(X_1)E(X_2) \right] + 18 \\
& = 12+36-3\left[3+(6)(3) \right] + 18\\
& = 3
\end{align*}
Hence, the joint distribution between $X_1$ and $Y$ is given by:
\begin{align*} \begin{bmatrix} X_1 \\ Y \end{bmatrix} \sim N\left(\begin{bmatrix} 6 \\ -3 \end{bmatrix}, \begin{bmatrix} 12 & 3 \\ 3 & 12 \end{bmatrix} \right) \end{align*}
Query:
I feel my answer is not right because I assumed that if we have two marginal normal random variables, say, $X_1 \sim N(\mu_1, \Sigma_{11})$ and $X_2 \sim N(\mu_2, \Sigma_{22})$, then $X_1$ and $X_2$ is jointly normal. That is,
\begin{align*} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} \sim N\left(\begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}, \begin{bmatrix} \Sigma_{11} & cov(X_1, X_2) \\ cov(X_1, X_2) & \Sigma_{22} \end{bmatrix} \right) \end{align*}
where $cov(X_1, X_2)$ denotes the covariance between $X_1$ and $X_2$.
However, I know in general just because $X_1$ and $X_2$ are both marginally normal, this does not mean they are jointly normal. How else can I find the joint distribution between $X_1$ and $Y$?
Thanks
AI: Recall the following fundamental fact about normal distributions:
If a random vector $X$ is normal $(M,\Sigma)$ then, for every matrix $A$, $AX$ is normal $(AM,A\Sigma A^T)$.
In your case, $A=\begin{pmatrix}1 & 0\\ 1 & -3\end{pmatrix}$ and $X=\begin{pmatrix}X_1\\ X_2\end{pmatrix}$ yield $AX=\begin{pmatrix}X_1\\ Y\end{pmatrix}$ hence you simply must compute the mean vector $AM$ and the covariance matrix $A\Sigma A^T$.
Note that the hypothesis in this exercise is not that $X_1$ and $X_2$ are normal but that the random vector $X=\begin{pmatrix}X_1\\ X_2\end{pmatrix}$ is normal. In the former situation, you are right to stress that chaos could ensue, but not in the latter. |
H: How do manifolds have enough structure to do calculus?
I am referring, of course, to to differentiable manifolds. I've seen a few different definitions. The one I like best is the one which says it's a topological space such that every point has a neighborhood homeomorphic to an open set in $\mathbb{R}^n$ and the overlap functions are diffeomorphisms.
I can see that we want the transition maps are diffeomorpisms, so the calculus we do on each coordinate chart agrees with the others, but how do homeomorphisms to Euclidean space give enough structure to develop a derivative, for instance, in a coordinate patch? I mean, the circle and the sphere are homeomorphic but I'd only call one of those smooth. How does smoothness on overlaps do it for the whole manifold? I've seen it said that once a differentiable structure is defined, the homeomorphisms become diffeomorphisms. Is this correct and could somebody expand on that if it is?
Furthermore, I've seen definitions given for manifolds with bijections rather than homeomorphisms. Can that possibly work? Bijections don't have to be particularly nice at all, so I can't for the life of me see how you'd get enough structure out of that. Please correct me if I've misunderstood some part of this.
AI: You don't do calculus directly on the manifold. Everything is done in the world of coordinates.
Your coordinate maps let you push things down into $\mathbb{R}^n$ where we can take limits, compute partials, find Jacobian matrices etc.
Requiring the coordinate maps to be homeomorphisms is redundant. Why? Take your manifold to be a set equipped with some altas (bijective coordinate charts). Then you just declare that these coordinate charts (bijections) are homeomorphisms. In that way you force a topology on the set and turn in into a topological space (now equipped with homeomophism coordinate charts).
The trick to manifold theory is that you're never really working directly on the manifold itself. You always translate to the world of coordinates and work there.
Edit: To address @studiosus comment...
Fair enough, my last paragraph is a bit overstated. Maybe I should say instead...
To do concrete "calculus" computations, you must (usually) translate to the world of coordinates and work there. This is much like it is in geometry. You can do a lot with synthetic geometry, but when you want to calculate...add coordinates and start computing. |
H: Evaluating limit making it $\frac{\infty}{\infty}$ and using L'Hopital Rule
Let $P(x)=x^n+\displaystyle\sum\limits_{k=0}^{n-1}a_kx^k$. Find $$ \lim_{x \to +\infty} ([P(x)]^{1/n}-x) $$
I know that in order to solve this problem I need to multiply it by something that will make it $\frac{\infty}{\infty}$ and then use L'Hopital Rule. I also know that the answer should be $1/n$ if I am not mistaken. I have tried multiplying the expression by $\frac{e^x}{e^x}$ and then using L'H. Rule but with not much success. Any suggestions on how I should proceed? Thank you for the help.
AI: An idea:
$$P(x)^{1/n}-x=x\left[\left(1+\frac{a_0}{x^n}+\ldots+\frac{a_{n-1}}x\right)^{\frac1n}-1\right]=\frac{\left(1+\frac{a_0}{x^n}+\ldots+\frac{a_{n-1}}x\right)^{\frac1n}-1}{\frac1x}$$
Now you can use l'Hospital (can you see why?) and get that your limit equals
$$\lim_{x\to\infty}\frac{\frac1n\left(-\frac{na_0}{x^{n+1}}-\ldots-\frac{a_{n-1}}{x^2}\right)\left(1+\frac{a_0}{x^n}+\ldots+\frac{a_{n-1}}x\right)^{\frac1n-1}}{-\frac1{x^2}}=$$
$$=\frac1n\;\lim_{x\to\infty}\,\left(\frac{na_0}{x^{n-1}}+\ldots+\frac{2a_{n-2}}x+a_{n-1}\right)\left(1+\frac{a_0}{x^n}+\ldots+\frac{a_{n-1}}x\right)^{\frac1n-1}=\frac{a_{n-1}}n$$ |
H: How many paths are there from A to B?
How many paths are there from A to B?
8 is wrong answer.
AI: At each node (starting from $B$) yo can write down the number of ways from there to $B$.
Then whenever you find a node where the path counts from all its successors are known, you know that the path count from that node is the sum over the successors.
This way you finally arrive at $A$ and should find a value of $5$ (the top successor has $2$, the bottom has $3$). |
H: Bijection on finite fields?
I am trying to solve a problem about bijections on finite fields, but I am stuck on this problem and don't see how can I solve it. So, here it is:
Given is a polynomial $g(y):=y^{3}+3y^{2}+3y+3\in \mathbb{Z}[y]$. Show that it defines a bijection from $\mathbb{F}_{11^{31415}}$ to itself.
To show the bijection $\mathbb{F}_{11^{31415}}\rightarrow \mathbb{F}_{11^{31415}}$, I've been thinking that I have to find an inverse map, but I don't know how to do it. I think this should be better than proving that the map is injective and surjective. I also noticed that this polynomial is irreducible over $\mathbb{Z}$ and that the number $31415=5\times 61\times 103$, but I don't know what that brings to the problem.
AI: Observe that $g(y)=(y+1)^3+2$, so its inverse is something like $g^{-1}(x)=\sqrt[3]{x-2}\,-1$.
The only problematic part is the cubic root, so all you shoud show is that $x\mapsto x^3$ is a bijection in the given field. Since that is finite, its multiplicative group is cyclic, of order $11^{31415}-1$.
Since we have $11\equiv -1 \pmod3$, $\ 11^{31415}\equiv -1\pmod{3}$, so that the group order is coprime to $3$, and therefore $x\mapsto x^3$ is indeed a bijection:
By the Bezout identity, there are integers $a,b$ such that $a\cdot N+b\cdot 3=1$ where $N$ denotes the order of the group now. So that $x\mapsto x^b$ will be an inverse for $x\mapsto x^3$. |
H: Relation between compactness and closure?
How are compactness and closure related? Suppose $K \subset M$. Suppose the intersection of a collection of open covers $\{G_a\}$ in $M$ yields a finite subcover of $K$. Then this is the closure in some sense of the cover.
I'm attempting to find out how closure and compactness are related; is my argument heading in the right direction?
AI: An obvious way in which closure and compactness are related is this: a closed subset of a compact set is compact.
I'm not quite sure what you mean about the closure of a cover though. You fix some subset of a set $M$, and then take a collection of open covers, the intersection of which 'yields' a finite subcover of $K$. In what sense can an intersection of covers yield a finite subcover?
This is the closure in some sense of the cover.
What cover are you talking about now? You just had a collection of open covers. And what is 'this'?
If you clarify your question, and maybe give an example or two, I'll be happy to give you an answer. |
H: $f, g$ are convex and positive $\Rightarrow f(x)g(y)$ is convex?
Prove or provide a counterexample: if $f$ and $g$ are real convex positive functions on some intervals, then $f(x)g(y)$ is convex.
AI: $f(x)=x$, $g(x)=1-x$ is a counterexample on $[0,1]$. |
H: Minimizing the length of wire between two poles?
There are two poles (lets say poles A and B) $50$ feet apart and the poles are $15$ and $30$ feet tall. There is a wire which runs from the top of pole A to the ground, and then to the top of pole B so that two triangles are made. What is the minimum length of wire needed to set up this configuration?
I tried finding this by setting up an equation for the length of the wire and taking the first derivative, but I ended up with an ugly polynomial.
I also tried drawing in one more triangle and used the law of cosines, but that didn't turn out too well either.
The most likely source of this problem (altough it had different numbers) is from "Maxima and Minima Witouth Calculus" by Ivan Niven, but I'm not sure.
AI: Geometric method
If one mirror reflects $A$ to the $A_1$, then obviously $AC+CB = A_1C+BC$. Latter is minimal when $A_1$, $B$ and $C$ are aligned. One can even find mirror image of $B$ as well, so final $l = \sqrt{(a+b)^2 + h^2}$. One can use another triangles, triangle similarities, etc, but this one is quite visual.
Algebraic method
\begin{align}
l &= \sqrt{a^2+x^2} + \sqrt{b^2+(h-x)^2} \\
\frac {dl}{dx} &= \frac x{\sqrt{a^2+x^2}} - \frac {h-x}{\sqrt{b^2+(h-x)^2}} = 0
\end{align}
from latter one can find that
$$
x^2 \left [ b^2+(h-x)^2\right ] = (a^2+x^2)(h-x)^2 \\
x^2 b^2 + x^2 (h-x)^2 = a^2(h-x)^2 + x^2(h-x)^2 \\
xb = a(h-x)
$$
So,
\begin{align}
A_0C &= x = \frac {ah}{a+b} \\
B_0C &= h-x = \frac {bh}{a+b}
\end{align}
and
\begin{align}
l &= \sqrt{a^2+\frac {a^2h^2}{(a+b)^2}} + \sqrt{b^2+\frac {b^2h^2}{(a+b)^2}} = \sqrt{1+\frac {h^2}{(a+b)^2}} (a+b) = \sqrt{(a+b)^2+h^2}
\end{align} |
H: Show that $f(x, y) = x^{1/3}\, y^{1/3}$ is not differentiable at the origin
I have a function $f(x, y) = x^{1/3}\, y^{1/3}$ and I have to show they it is not differentiable at $(0,0)$.
So I thought I would show $\lim_{(x, y) \to (0, 0)} f(x, y)$ does not exist. But I keep getting $0$ as the result. What are some other ways to approach this?
AI: Hint: Since $f'_x(0,0) = f'_y(0,0) = 0$, differentiability is equivalent to $f(x,y) = o(\sqrt{x^2+y^2})$, i.e.
$$\lim_{(x,y) \to (0,0)} \frac{x^{2/3}y^{2/3}}{x^2+y^2} = 0,$$
which is false. |
H: The sequence of prime gaps is never strictly monotonic
I have an assignment question that asks me to show that the sequence of prime gaps is never strictly monotonic. I'm also allowed to assume the Prime Number Theorem.
I've managed to show that it cannot be strictly decreasing by considering the numbers $N!+2, N!+3,...,N!+N$ which gets arbitrarily large as $N\rightarrow\infty$.
However, I seem to not be able to show the strictly increasing bit. I have an idea but I'm not sure if it works and whether it is a suitable usage of the Prime Number Theorem.
Here is my idea: Suppose $\pi(n_0)=k$ and $d(n)$ is strictly increasing $\forall n\geq n_0$. Then we consider the "worst case scenario" (call it $\pi_1$) of finding primes at gaps $2,4,6,...$ after $n_0$, meaning that $n_0+2,n_0+2+4,n_0+2+4+6,...$ are the primes.
Then what I'm saying is $\pi_1(n_0+q(q-1))=k+q-1$ because of the arithmetic progression. I don't know if we can give any comparisons between $\pi(n)$ and $\pi_1(n)$ but I also don't see if this actually leads to anything. If I'm totally off track, I'll be glad if somebody can point me in the right direction.
I can't seem to find any literature regarding this either!
AI: As per your ides, there cannot be essentially more than $q$ primes below $q^2$ (give or take low order stuff like your constants $k$ and $n_0$), i.e.
$\pi(q^2 )<q$. As $\pi(q^2)\approx \frac{q^2}{\ln q^2}=\frac q2\cdot \frac q{\ln q}\approx \frac q2\pi(q)$, this leads to $\pi(q)<2$, which is absurd. (Check that dropping low order terms was indeed allowed!) |
H: About The Selected Identity $ J_\nu(z)=\frac z {2 \nu} (J_{\nu-1}(z)+J_{\nu+1}(z));$
I found this
$$ J_\nu(z)=\frac z {2 \nu} (J_{\nu-1}(z)+J_{\nu+1}(z));$$
at the Selected Identities at Wikipedia's site on the Bessel function. How to prove this? The rest of the site doesn't give me anything to start with...
AI: This indentity can be proven directly from the power series expansion from the Bessel function:
$$J_\nu(z) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\nu+1)} {\left(\frac{z}{2}\right)}^{2m+\nu}$$
Writing the power series for $J_{\nu-1}(z)$ and $J_{\nu+1}(z)$, and summing them, we have:
$$J_{\nu-1}(z)+J_{\nu+1}(z) =
\sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\nu)} {\left(\frac{z}{2}\right)}^{2m+\nu-1}+\sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\nu+2)} {\left(\frac{z}{2}\right)}^{2m+\nu+1} =
\frac{1}{\Gamma(\nu)}\left(\frac{z}{2}\right)^{\nu-1}+\sum_{m=0}^{\infty}\left[\frac{(-1)^{m+1}}{(m+1)! \, \Gamma(m+\nu+1)}+\frac{(-1)^{m}}{m! \, \Gamma(m+\nu+2)}\right]\left(\frac{z}{2}\right)^{2m+\nu+1} =
\frac{1}{\Gamma(\nu)}\left(\frac{z}{2}\right)^{\nu-1}+\sum_{m=0}^{\infty}(-1)^{m+1}\left[\frac{m+\nu+1-(m+1)}{(m+1)! \, \Gamma(m+\nu+2)}\right]\left(\frac{z}{2}\right)^{2m+\nu+1}=
\sum_{m=0}^\infty \frac{(-1)^m\nu}{m! \, \Gamma(m+\nu+1)} {\left(\frac{z}{2}\right)}^{2m+\nu-1}=\frac{2\nu}{z}J_\nu(z)
$$ |
H: Help with algebra sets?
Can someone help me with this, please?
Let $A, B$ be sets. Show the following is true:
$$P(A) \cup P(B) \subseteq P(A \cup B)$$
$P$ is a powerset.
Do I just expand $P(A \cup B)$ to $P(A) \cup P(B)$?
$$P(A) \cup P(B) \subseteq P(A) \cup P(B)$$
So therefore it's true?
Seems too simple.
AI: No, $P(A\cup B)$ is the powerset of $A\cup B$, i.e. the set of all subsets of $A\cup B$. So we don't have $P(A\cup B)=P(A)\cup P(B)$, only the containment
$$P(A)\cup P(B)\ \,\subset\ \,P(A\cup B)\,.$$
For example take $A:=\{a_1,a_2\}$ and $B:=\{b_1,b_2\}$, then $\{a_1,b_1\}\ \in\ P(A\cup B)\,\setminus\,\left(P(A)\cup P(B)\right)$. |
H: Is it logically valid to prove DeMorgan's laws using the duality of boolean algebra?
I'm taking an introductory course in boolean algebra, and have been assigned the task of proving DeMorgan's Laws (so, disclaimer, this is homework). One line of reasoning I came up with is the following:
By applying the duality principle to two valued boolean algebra, $X \cdot Y = 1 \Leftrightarrow X + Y = 0$
$X+Y = 0 \Leftrightarrow \overline{X + Y} = 1$
Therefore, $X \cdot Y = 1 \Leftrightarrow \overline{X + Y} = 0$
By transitivity of equality, $X \cdot Y = \overline{X + Y}$
where $\cdot$ and $+$ are the conjunction and disjunction operators respectively.
EDIT: As pointed out by Lord_Farin, this result is incorrect, since the conclusion conflicts with DeMorgan's law. Where am I going wrong?
Rest of original question:
Now, the part I'm unsure about is the last step. Have I only managed to prove 4 in the specific case where the claim $X \cdot Y = 1$ is true, or is the proof valid regardless of whether $X \cdot Y = 0$? To my understanding, I have only made claims about the implications of the statement, $X \cdot Y = 1$ but haven't actually claimed whether it is true or false.
Is there a flaw in this proof?
AI: Your error results from misquoting the duality theorem. Let me state it for you here:
If $T$ is a theorem about boolean algebras, then so is $T^*$, the statement obtained by carrying out the replacements $+ \leftrightarrow \cdot$ and $1 \leftrightarrow 0$.
But "$X\cdot Y =1$" is certainly not a theorem of BAs; it's contingent (it may be either true or false).
So while you can use duality to conclude the other part of De Morgan's laws:
$\overline X \cdot \overline Y = \overline{X+Y}$
$\overline X + \overline Y = \overline{X \cdot Y}$
when you've proven one of them (and in fact, duality is an elegant and efficient method to do it), it cannot be used to prove both of them. |
H: Exercice 8, p.922 from Stewart's Calculus: Concepts and Contexts
I try to do this exercice: $\int_C \sin x\, dx + \cos y\, dy$, where $C$ consists of the top half circle $x^2+y^2=1$ from $(1,0)$ to $(-1,0)$ and the line segment from $(-1,0)$ to $(2,3)$.
I'm fine with the segment, but over the circle my fist guess was to use polar coodrinates, but I got something ugly... very ugly...
Unfornatly, time is running out, exam tomorow... so if you guys can give me the steps it will help me
thank you
AI: With polar coordinates: $x=\cos t$, $y=\sin t$, so that $dx=-\sin tdt$, $dy=\cos tdt$ and the integral become:
$$\int_0^\pi \sin(\cos t)(-\sin t) dt+\int_0^\pi \cos(\sin t)\cos t dt=-\cos(\cos t)+\sin(\sin t)|_0^\pi=0$$. |
H: Combinatorial proof for $a(n-a) \binom{n}{a} = n(n-1) \binom{n-2}{a-1}$
Prove $a(n-a) \binom{n}{a} = n(n-1) \binom{n-2}{a-1}$ by a combinatorial proof.
This is what I tried:
There is a set $X$ of $n$ elements. There is a subset $Y$ of $a$ elements.
LHS, we choose 1 element from $Y$. Then choose 1 element not from $Y$. And choose $a$ elements from $X$.
RHS, we choose 2 elements from $X$, one of which is from $Y$ and the other is not from $Y$. Then choose $a-1$ elements from the remaining elements of $X$.
But the way I understand it, we end up with $2 + a$ elements on the LHS, while we end up with $2+a-1$ elements on the RHS.
AI: Suppose that you have a group of $n$ players. The lefthand side is the number of ways to pick a team of $a$ of these players, designate one member of the team as captain, and then pick one reserve player from the remaining $n-a$ people. The righthand side is the number of ways to pick the captain, then the reserve player, and then the other $a-1$ members of the team. |
H: Solving an equation with floor function
I need some help regarding this question.
Solve the following equation in natural number $x$, where $m,k$ are fixed naturals:
$m\left\lfloor \sqrt{\dfrac{x}{k}}\right\rfloor = x$.
I think the answer is $x = m\left\lfloor\dfrac{m}{k}\right\rfloor$. But I do not know how to prove this claim.
I would be happy if somebody would help me solve this problem.
Thank you,
Isomorphism
AI: Since $x$ must be a multiple of $m$, write $x=my$.
Then the equation becomes $$y=\left\lfloor\sqrt{\frac {my}k}\right\rfloor$$
and equivalent to
$$y\le \sqrt{\frac {my}k}<y+1,$$
i.e.
$$y^2\le \frac {my}k <y^2+2y+1$$
or (using $y\ne 0$, though in fact $x=0$ is a trivial solution)
$$y\le \frac {m}k <y+2+\frac1y\le y+3.$$
Therefore $y=\left\lfloor \frac mk\right\rfloor$, $y=\left\lfloor \frac mk\right\rfloor-1$, and in rare cases $y=\left\lfloor \frac mk\right\rfloor-2$ are the solutions - the latter only if $m\ge 3k$ and $\frac mk-\left\lfloor \frac mk\right\rfloor<\frac 1{\left\lfloor \frac mk\right\rfloor-2}$. |
H: Summing $dn$ floor functions = $d$ times Summing $n$ floor functions
Fix integer $d>1$, and assume real number $x\in[0,1]$.
I claim the following statement:
$\sum_{k=1}^{dn}\lfloor kx\rfloor=d\sum_{k=1}^n\lfloor kx\rfloor$ is true iff $x\in[0,\frac{1}{dn}]$.
I can check it in various cases, and the if-direction is obvious. But my combinatorics skills are not sharp, so is there a counterexample?
AI: You must have $x \in \left[0,\frac{1}{dn}\right)$, for if $x = \frac{1}{dn}$, then the left hand side is $1$, and the right hand side is $0$.
Generally, if $\frac{1}{dn} \leqslant x < \frac{1}{n}$, then
$$d\sum_{k=1}^n \lfloor kx\rfloor = 0 < \lfloor dnx\rfloor \leqslant \sum_{k=1}^{dn} \lfloor kx\rfloor,$$
so in that case you don't have equality. And if $x \geqslant \frac1n$, then you have
$$\begin{align}
\sum_{k=1}^{dn} \lfloor kx\rfloor - d\sum_{k=1}^n \lfloor k\rfloor &= \sum_{j=0}^{d-1} \left(\sum_{k=1}^n \left(\lfloor (k+jn)x\rfloor - \lfloor kx\rfloor\right)\right)\\
&\geqslant\sum_{j=0}^{d-1} \sum_{k=1}^n\left(\lfloor kx + j\rfloor - \lfloor kx\rfloor\right)\\
&\geqslant \sum_{j=0}^{d-1} nj\\
&= n\frac{d(d-1)}{2}\\
&> 0.
\end{align}$$ |
H: Multiplying last row of matrix with the matrix's inverse
Let $A$ be a square $n\times n$ invertible matrix, $a_n$ denote its last row, and $e$ be the $n\times 1$ (column) vector of $1$'s. This is likely a simple linear algebra property, but I am not so sure why
$$a_n^TA^{-1}e=1.$$
Writing out the terms of $A^{-1}$ as $b_{ij}$, this reduces down to
$$a_{n1}(b_{11}+b_{12}+...)
+
a_{n2}(b_{21}+b_{22}+...)
+
...
+
a_{nn}(b_{n1}+b_{n2}+...)
=1.$$
Unfortunately, this doesn't help me see the relationship between the $b_{ij}$'s and $a_{nj}$'s, let alone how some of the entries would cancel together. Is there a better approach to this problem?
AI: Consider the last row of $AA^{-1}=I$; we conclude that $a_nA^{-1}=e_n$, where $e_n$ is all zeroes except a 1 in the last position. Lastly, $e_ne=1$ is clear. |
H: Find a second order ODE given the solution
Find a second order differential equation so that
$$y=C_1e^{-3x}\cos(4x)+C_2e^{-3x}\sin(4x)+4e^{3x}$$
solves the differential equation for any choice of $C_1$and $C_2$.
The answer should be in the form of $ay''+by'+cy=f$
Here's my work:
$y=C_1e^{-3x}\cos(4x)+C_2e^{-3x}\sin(4x)$ is the solution of the homogeneous equation and
$y=4e^{3x}$ is the particular solution. But how do I proceed from here to figure out the second order ODEs?
AI: You have to "solve" the equation backwards. As you pointed, there are solutions to the homogeneous equation and a particular solution. To get the homogeneous equations, we notice that the roots of the characteristic polynomial equation are $-3+4i$ and $-3-4i$. and the characteristic polynomial equations is therefore
$$(x+3-4i)(x+3+4i)=x^2+6x+25$$
So, the homogeneous equation is
$$y''+6y'+25y=0$$
To get the full equation, we substitute the particualr solution
$$(4e^{3x})''+6(4e^{3x})'+25(4e^{3x})=208e^{3x}$$
We get therefore, the equation
$$y''+6y'+25y=208e^{3x}$$ |
H: Disc. Math Proof help?
I am enorolled in a community college level discrete mathematics course and am having problems with a proof I have for homework.
Suppose a, b, c, and d are integers and a≠c. Suppose also that x is a real number that satisfies the equation:
(ax + b) / (cx + d) = 1
must x be rational? If so express x as a ratio of two integers.
This is what i have so far:
let a, b, c, and d be PBAC integers
let x = (p/q)
(cx + d) ≠ 0
(a(p/q) + b) / (c(p/q) + d) = 1
AI: You’ve got off on the wrong foot, I’m afraid.
HINT: Solve the equation
$$\frac{ax+b}{cx+d}=1$$
for $x$; you’ll need to use the hypothesis that $a\ne c$. Then use the hypothesis that $a,b,c$, and $d$ are all integers to decide whether $x$ must be rational. |
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