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H: Linearizing $\, y''+\sin(2x+\cos(2y'+y))+1-\sin(y+3y')=0$
I'd really love your help with the following problem. I need to make linearization and then to find a general solution of the accepted equation for the differential equation:
$$y''+\sin(2x+\cos(2y'+y))+1-\sin(y+3y')=0$$
I usually get a system of two equations, derive them and go on. What should I do in this case? is there a way to turn it into a system of two equation of order one? How it's been done?
Thanks a lot
AI: The easiest linearization I can think of in this case is to use $\cos(\theta) \approx 1$, and $\sin(\theta) \approx \theta$, where $\theta$ is small enough. In your case, these assumptions boil down to making the assumption that $y,y' \approx 0$.
With these assumptions, your differential equation now becomes $$y'' + \sin(2x+1) + 1 - (y+3y') = 0$$ which in turn becomes $$y'' - 3y' - y = - \left(1 + \sin(2x+1) \right)$$
Now you can use your differential equation tricks to solve the above equation to get
$$y(x) = c_1 \exp \left( \left(\dfrac{3-\sqrt{13}}2 \right) x\right) + c_2 \exp \left( \left(\dfrac{3+\sqrt{13}}2 \right) x\right) + \dfrac{5 \sin(2x+1) - 6 \cos(2x+1) + 61}{61}$$ |
H: Number of well-ordering relations on a well-orderable infinite set $A$?
Given a well-orderable infinite set $A$, can we always say that the set $$\left\{R\subset A\times A:\langle A,R\rangle\, \text{is a well-ordering}\right\}$$ has cardinality $2^{|A|}$? How much Choice is required for the proof of this?
I believe that where $A$ is countably infinite, we can proceed without any use of Choice. Is that correct?
AI: This is a delicate matter. Do you mean all the well orders or just up to isomorphism?
For the former observe, for example, that if a set $A$ has one well-ordering then any permutation induces a different well-ordering, although of the same order type. For the natural numbers there are $2^{\aleph_0}$ many permutations so there are at least continuum many well-orders, and that is just of one isomorphism type! On the other hand, there can only be continuum many relations, so we have exhausted the cardinality.
We can continue by induction on the $\aleph$-cardinals, in fact, this is the only way we can resume. Why? Well, if a set has any well-ordering then it has to be in bijection with an ordinal, and if it is infinite this ordinal can be an $\aleph$-number. The argument for $\omega_1$, $\omega_2$ and so on is the same as above and this would require no choice at all.
If a set cannot be well-ordered, well... it has no well-ordering! However if we assume the axiom of choice then every set can be well-ordered and going through cardinals is enough. So arguing this claim for all sets is to require the full axiom of choice.
On the other hand, if you are interested in order types rather than mere orders than the claim that there are $2^A$ many is in fact to assert the Generalized Continuum Hypothesis, since for a set of cardinality $\kappa$ there are only $\kappa^+$ many ordinals of cardinality $\kappa$, and therefore only $\kappa^+$ many order types.
Whether or not $2^\kappa=\kappa^+$ is undecided in ZFC. |
H: Rate of Change Question
I have this equation for a rate of change problem below
$$s = t^
4
− 4t^
3
− 20t^
2
+ 20t,\qquad t \geq 0$$
The question asks me
At what time does the particle have a velocity of 20 m/s?
How do i solve this? Basically what steps do I take to find at what time the particle has that velocity?
AI: If $s$ is the distance as a function of time $t$, then the velocity is given by $v = \dfrac{ds}{dt}$.
Set this derivative $\dfrac{ds}{dt}$ to $20$ and solve the cubic, with the constraint $t>0$, to get the value of $t$.
Note that the cubic you obtain can be factored easily.
Move your cursor over the gray area for the complete answer.
$$\dfrac{ds}{dt} = 4t^3 - 12 t^2 - 40 t + 20$$ Setting this to $20$ gives us, $$4t^3 - 12 t^2 - 40 t + 20 = 20 \implies 4t^3 - 12 t^2 - 40 t = 0 \implies 4t(t^2 - 3t - 10)=0$$ This gives us $$t(t-5)(t+2) = 0$$ Since $t > 0$, we get that $t=5$. |
H: Which number was removed from the first $n$ naturals?
A number is removed from the set of integers from $1$ to $n$. Now, the average of remaining numbers turns out to be $40.75$. Which integer was removed?
By some brute force, I got $61$. I want to know if there's any analytic approach?
AI: The average of the integers $1$ through $n$ is $\frac12(n+1)$. Removing a number smaller than this will increase the average, and removing a number larger than this will lower it. In particular, removing $1$ will cause the maximum increase in the average, to
$$\frac1{n-1}\left(\frac{n(n+1)}2-1\right)=\frac{n^2+n-2}{2(n-1)}=\frac{(n+2)(n-1)}{2(n-1)}=\frac12(n+2)\;,$$
and removing $n$ will cause the maximum decrease in the average, to
$$\frac1{n-1}\left(\frac{n(n+1)}2-n\right)=\frac{n^2-n}{2(n-1)}=\frac{n}2\;.$$
The new average of $40.75$ therefore must be between $\frac{n}2$ and $\frac12(n+2)=\frac{n}2+1$, inclusive. Life becomes easier if we double everything: $81.5$ must be between $n$ and $n+2$, inclusive. That is, $$n\le 81.5\le n+2\;,$$ and therefore $$79.5\le n\le 81.5\;.$$ Since $n$ must be an integer, the only possibilities are $n=80$ and $n=81$.
The sum of the integers $1$ through $80$ is $3240$, so if $n=80$, you need to find $k$ in the range from $1$ to $80$ inclusive so that $$\frac{3240-k}{79}=40.75\;.$$ However, the solution to this equation is not an integer, so $n$ must be $81$.
The sum of the integers $1$ through $81$ is $3321$, so this time you want $k$ satisfying $$\frac{3321-k}{80}=40.75\;,$$ which is easily solved to find that $k=61$. |
H: Limit of $\prod_{i=1}^n (1-1/2^i)$
I am trying to find the limit of the sequence $$s_n:=\displaystyle \prod_{i=1}^n \left(1-\frac{1}{2^i} \right)$$
The sequence is decreasing and bounded below by $0$. I guess that the limit is $0$, is there any way to show this ? Or, is there any argument which shows that the limit is not zero ?
AI: The limit is definitely non-zero. Your product converges to a non-zero value, which is approximately $\approx 0.289$.
You should look up my answer here, why such infinite products can never be zero unless one of the term is itself zero. The product in your question is $\left(\dfrac12; \dfrac12\right)_{\infty}$ where $(a;q)_n$ is the Pochhammer symbol. |
H: Proof $\lim\limits_{n \rightarrow \infty} {\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}=2$ using Banach's Fixed Point
I'd like to prove $\lim\limits_{n \rightarrow \infty} \underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\textrm{ square roots}}=2$ using Banach's Fixed Point theorem.
I think I should use the function $f(x)=\sqrt{2+x}$. This way, if I start the iterations for example with $x_0=0$, I will have $x_1=\sqrt2$. When I calculate $x_2$ I will get $\sqrt{2+\sqrt{2}}$. And $x_3 = \sqrt{2+\sqrt{2+\sqrt{2}}}$ and so on. I can see that these iterations are monotone increasing, but how can I show that this converges to 2?
Pseudo-related formula I found: http://en.wikipedia.org/wiki/Vieta_formula
Many thanks in advance!
Following clark's advice, here's my proof this is a contraction. I'm using the interval $D=[0, 2]$.
$f'(x)=\frac{1}{2\sqrt{x+2}}$, which is monotone decreasing. This means its highest value in $D$ is $0$. $f'(0)=\frac{1}{2\sqrt{2}} < 1$. The rate $M$ of the contraction is then $\frac{1}{2\sqrt{2}}$.
AI: In order to use Banach's fixed pointed theorem you have to show $ |f(x)-f(y)| < M(|x-y|)$ in some interval, say $[a,b]$. Then your work would to prove that starting with $x_0=c\,\,$ then$x_{n+1}=f(x_n)$ stays in that interval, i.e.: $a \leq x_n \leq b$,(so your function is well defined $f:[a,b] \rightarrow \mathbb{R}$. That $M $ can be found $ f'(y_0) = M$ and bound the derivative. Then you will know the limit is the solution $f(k)=k$
EDIT: Since you took the interval [0,2] you need to prove that for $y \in [0,2] $
$0 \leq f(y) \leq 2$ the first holds trivially. For the second you have $\sqrt{2+ \sqrt {2}} \leq 2 \Leftrightarrow \sqrt {2} \leq 2$ which holds. Now you are done because you have that every $x_n$ stays in the interval you choosed. So Banach's fixed point theorem can be applied. (Note that you defined $f$ on $D$ so the previous step is to make sure that the $f$ you took is well defined, because every $x_n$ is used by $f$ to define $x_{n+1}$). |
H: Opening Doors Puzzle
You are in a corridor which has N doors all of the doors are opened.
Every time someone passes through the corridor, he closes randomly
with equal probability a certain number of doors between 1 and the
number of remaining opened doors. What is the expected number of
people that need to pass before all the doors are closed? What about
when they instead close 0 to N doors each time rather than 1 to N?
So currently I have:
Let $P_N$ = number of people needed to close N doors.
$$E(P_N) = 1 + \frac{1}{N}(\sum\limits_{i=1}^N E(P_{N-i}))$$
Since you add 1 for the one guy coming through, and then since each amount of doors happens of equal probability its just that summation term times the probability of each event happening (which is $1/N$).
I'm not quite sure how to continue to simplify this though...
Thanks for any help.
AI: It is probably easier to write your expression as
$$E_N = 1 + \frac{1}{N}\sum\limits_{i=0}^{N-1} E_i$$ starting at $E_0=0$. This is the same as
$$N E_N = N + \sum\limits_{i=0}^{N-1} E_i.$$ But you also have
$$(N-1) E_{N-1} = N-1 + \sum\limits_{i=0}^{N-2} E_i$$ and taking the difference and tidying up gives a simple expression for $E_N - E_{N-1}$ and thus a known sequence for $E_N$.
For your alternative question you instead start with $$A_N = 1 + \frac{1}{N+1}\sum\limits_{i=0}^{N} A_i$$ and proceed in a similar way to get a similar but slightly different result. It is worth working $A_1$ out explicitly as you cannot calculate $\frac00$. |
H: ${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as which of the following?
${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as ?
A. ${21 \choose 5}$
B. ${20 \choose 5}-{11 \choose 4}$
C. ${21 \choose 5}-{10 \choose 5}$
D. ${20 \choose 4}$
Please give me a hint. I'm unable to group the terms.
By brute force, I'm getting ${21 \choose 5}-{10 \choose 5}$
AI: HINT 1 :
Add $\dbinom{10}5$
HINT 2:
Make use of the identity $$\dbinom{n}{r} + \dbinom{n}{r-1} = \dbinom{n+1}{r}$$ |
H: Differentiable map conserving geodetic lines which is no isometry
I am looking for a differentiable map $f: S^n\rightarrow S^n$, which conserves the geodetic lines of the standard metric on $S^n$, but is no isometry.
The geodetic lines on $S^n$ should be the great circles, I think.
Unfortunately, I was only able to think of examples, which are not well-defined...
I would be grateful for any help.
AI: $S^n$ can be identified with $\mathbb{R}^{n+1}/\mathbb{R}_+$ (where $\mathbb{R}_+$ acts on $\mathbb{R}^{n+1}$ by multiplication). Geodesics come from $2$-dimensional vector subspaces of $\mathbb{R}^{n+1}$. The group $GL(n+1,\mathbb{R})$ acts on $\mathbb{R}^{n+1}/\mathbb{R}_+$ and maps geodesics to geodesics. If you take a matrix $A\in GL(n+1,\mathbb{R})$ which is not a constant multiple of an orthogonal matrix, then its action does not preserve the metric. |
H: Normed linear space question
Suppose $x_1, x_2, \ldots, x_n$ are linearly independent elements of a normed linear space $X$. Show that there is a constant $c>0$ with the property that for every choice of scalars $\alpha_1, \ldots, \alpha_n$ we have $$\|\alpha_1x_1+\cdots+\alpha_nx_n\|\geq c(|\alpha_1|+\cdots+|\alpha_n|)$$
I tried doing this by contradiction but I am stuck.
AI: Here's an outline:
1) Use a "normalization argument" to show that it suffices to prove your result for $|\alpha_1|+\cdots+|\alpha_n|=1$.
2) Define a map from the unit sphere of $\ell_1^n$ to $\Bbb R$ via $(\alpha_1,\ldots,\alpha_n)\mapsto\Vert\alpha_1 x_1+\cdots+\alpha_n x_n \Vert$.
3) Using the axioms of norm, show that this function is continuous.
4) Using the fact that the unit sphere of $\ell_1^n$ is compact, show that this function attains a minimum value.
5) Using the independence of the $x_i$, show that this minimum value is positive. |
H: Find the sum of this series :$ \frac{1}{{1!2009!}} + \frac{1}{{3!2007!}} + \cdots + \frac{1}{{1!2009!}}$
Find the sum of this series :
$$\sum\limits_{\scriptstyle 1 \leqslant x \leqslant 2009 \atop
{\scriptstyle x+y=2010 \atop
\scriptstyle {\text{ }}x,y{\text{ odd}} }} {\frac{1}{{x!y!}}} = \frac{1}{{1!2009!}} + \frac{1}{{3!2007!}} + \cdots + \frac{1}{{1!2009!}}$$
I tried converting it into binomial coefficients and I'm getting sort of $\dfrac{2^{2009}}{2009!}$
Please help me.
AI: You’ve the right idea. First,
$$\sum_{k=0}^{1004}\frac1{(2k+1)!(2010-2k-1)!}=\frac1{2010!}\sum_{k=0}^{1004}\binom{2010}{2k+1}\;.$$
Now that last summation is simply the number of odd-sized subsets of a set of $2010$ elements. Since half the subsets of any non-empty set have odd cardinality, it’s simply $2^{2009}$. Thus, the desired sum is $$\frac{2^{2009}}{2010!}\;.$$ |
H: Exponentiating cubic functions (and so on)
The exponential distribution follows from exponentiating a linear function, while the normal distribution follows from exponentiating a quadratic function. Do the distributions that follow from exponentiating functions of 3rd or higher degree have names or uses?
AI: Such distributions would be part of the exponential family. I don't think they have a special name, though. |
H: first order logic question model
suppose we have a model for a language in first order logic $ M=<D,I> $
such that D is the domain and I is the interpetation such that for every $ a \in D $ we have a closed noun (a noun with no free variables) $ t_0 $ such that $ I[t_0] = a $
prove or disprove the following:
a) if $ \forall x A $ is a sentence than $ M \models \forall x A $ iff for each closed noun $t$ $ M \models A\{t/x\} $ where $ A\{t/x\} $ is substitution of $t$ in place of $x$ in $A$
b) if $ \forall x A $ is a formula than $ M \models \forall x A $ iff for each closed noun $t$ $ M \models A\{t/x\} $
c) if $ \exists x A $ is a sentence than $ M \models \exists x A $ iff there exists a closed noun $t$ such that $ M \models A\{t/x\} $
d) if $ \exists x A $ is a formula than $ M \models \exists x A $ iff there exists a closed noun $t$ such that $ M \models A\{t/x\} $
i think a and c are true because in the case where x is the only free variable in A I can be thought as assignment to $ A\{t/x\} $ (where t is closed noun) but in b and d i have no idea
please help me with this exrecise
AI: a) and c) are true, like you said, and I don't think I need to write down the gory details here.
For b) first denote by $B$ the universal closure of $A$ for all variables except $x$. Then we have $M\models \forall x A\iff M\models \forall x B$ (because they're the same thing up to a permutation of universal quantifiers), and the latter is equivalent (by a) ) to that for each closed noun $t$ we have $M\models B\{ t/x\}$, but $t$ is closed, so $B\{t/x\}$ is the universal closure of $A\{t/x\}$, so we have $M\models B\{t/x\}\iff M\models A\{t/x\}$, so b) is true.
For d) we have a counterexample: consider $M=\mathbf Z$ (or any partial order without maximal elements) with $A=(x>y)$. Then we have $M\models \exists x A$, because for any element there is one larger, but for no $k\in Z$ do we have $M\models k>y$, since, for example, we do not have $M\models k>k$. |
H: Integrability of the Newton potential
For $x\in\mathbb{R}^n$ we define the Newton potential as follows:$$N(x) = \begin{cases} \frac{\log|x|}{2\pi}, & n=2 \\[10pt] \frac{|x|^{2-n}}{(2-n)\omega_n}, & n>2\end{cases}$$
where $\omega_n$ denoted to the volume of the n-ball. Moreover, let $\chi_r$ denote to the characteristic function of the ball $B(0,r)$.
Now my lecture notes say (for $n>2$)
$\chi_r N \in L^1(\mathbb{R}^n)$ (or, more general, $L^p$, where $p<\frac{n}{n-2}$) and
$(1-\chi_r)N \in L^\infty(\mathbb{R}^n)$ (or, more general, $L^p$, where $p>\frac{n}{n-2}$)
Since $N \in L_\mathrm{loc}^1(\mathbb{R}^n)$, it follows immediately that $\chi_r N \in L^1(\mathbb{R}^n)$ for any $r>0$.
How do I see the other claims?
AI: We use the following formula, which is true for a radial function $f$ (that is, which depends only of the norm of the argument):
$$\int_{B(0,r)}f(x)dx=c_n\int_0^r\widetilde f(t)t^{n-1}dt,$$
where $c_n$ is a constant I leave to determine, and $\widetilde f$ is such that $f(x)=\widetilde f(|x|)$, where $|\cdot|$ is the euclidian norm.
For $n=2$, $\chi_r N\in L^p$ if and only if $\int_0^rt^{n-1}|\ln t|^pdt $ is finite that is, if and only if $\int_0^1e^{u(n-1)}e^uu^pdu$ is finite, which is always the case.
For $n>2$, $\chi_rN\in L^p$ if and only if $\int_0^1t^{n-1}t^{(2-n)p}dt<\infty$ which is equivalent to $\int_0^1\frac 1{t^{1-(n+(2-n)p)}}dt<\infty$, which occurs if and only if $n+(2-n)p>0$.
We have that
\begin{align}
(1-\chi_r)N\in L^p&\Leftrightarrow \int_1^{+\infty} t^{n-1}t^{(2-n)p}dt<\infty\\
&\Leftrightarrow \int_1^{+\infty}\frac 1{t^{1-(n+(2-n)p)}}dt<\infty \\
&\Leftrightarrow n+(2-n)p<0.
\end{align} |
H: Finding $3$ distinct prime numbers $a| (bc+b+c)$, $b|(ac+a+c)$, $c|(ab+a+b)$
How to find $3$ prime numbers $a,b$ and $c$ such that:
$$a| (bc+b+c)$$
$$b|(ac+a+c)$$
$$c|(ab+a+b)$$
AI: There are no such primes:
Without loss of generality assume $a<b<c$. Clearly 2 cannot be one of the primes, so $a\geq 3$ , $b\geq 5$ and $c\geq 7$.
Now $abc+ab+bc+ca+a+b+c$ is divisible by $a$, by $b$ and by $c$ so it is also divisible by $abc$,
but this is impossible since
$1<\frac{abc+ab+bc+ca+a+b+c}{abc}<\frac{a+1}{a}\frac{b+1}{b}\frac{c+1}{c}\leq\frac{4}{3}\frac{6}{5}\frac{8}{7}<2$ |
H: Uniform convergence of $\sum\limits_{n=1}^\infty \frac{x}{x^2+n^2}$
How to prove that $\sum\limits_{n=1}^\infty \frac{x}{x^2+n^2}$ is uniformly convergent for every $x$?
I was trying all sort of ways, but it think the answer might be in solving the problem for $|x|<1$ and then for $|x|>1$.
Its easy to show that for $|x|<1$ , $\sum\limits_{n=1}^\infty \frac{x}{x^2+n^2} \leq \sum\limits_{n=1}^\infty \frac{1}{n^2}$ and using Weierstrass M-Test that the sum is uniformly convergent.
But for $|x|>1$ its a different story.
Does any one have a simple solution? I'm stuck...
AI: We assume $x$ is in some bounded subset of $\mathbb{R}$, i.e., $|x|\le R$ for some real $R$.
Then
$$\begin{equation*}
\left|\frac{x}{x^2+n^2}\right| \le \frac{R}{x^2+n^2} \le \frac{R}{n^2}
\end{equation*}$$
But the sum $\sum_{n=1}^\infty 1/n^2$ is convergent.
Thus, the series converges uniformly on any bounded subset of $\mathbb{R}$ by the Weierstrass M-test.
Addendum: The series converges uniformly only on bounded subsets and not on $\mathbb{R}$ as @did shows in his answer. |
H: Convergence of sequence: $X_n = \frac{n}{n^2 + 1} + \frac{n}{n^2 + 2} + \frac{n}{n^2 + 3} + \cdots + \frac{n}{n^2 + n}$
What I am trying is finding $A_n < X_n < B_n$, proving that $A_n$ and $B_n$ converge, and then $X_n$ converges.
I first found that $A_n = \frac{n}{n^2 + 1}$ converges because the limit is zero.
I still need to find $B_n$.
AI: Take the upper bound$B_n=n \frac{n}{n^2+1}$ and the lower bound $A_n= n \frac{n}{n^2+n}$ Hence $ \lim_{ n \rightarrow \infty} X_n =1$ |
H: Does the specification of a general sequence require the Axiom of Choice?
Many results in elementary analysis require some form of the Axiom of Choice (often weaker forms, such as countable or dependent). My question is a bit more specific, regarding sequences.
For example, consider a standard proof of the boundedness theorem which states that a function continuous on a closed interval $I$ is bounded on that interval. In the first step of the proof, one specifies a sequence as follows:
Suppose for contradiction that $f$ is unbounded. Then for every $n\in\mathbb{N}$ there exists $x_n$ such that $f(x_n) > n$. This specifies a sequence $(x_n)$.
I'm not sure if the above example requires choice. To me, it certainly feels like it does. More specifically, I think that we are specifying a sequence of sets $$A(n) = \left\{x\in I\mid f(x) > n\right\}$$ and claiming the existence of a choice function $g$ such that $g(n) \in A(n)$ so that this example specifically requires the axiom of countable choice. Please clarify whether my reasoning is correct. More specifically, does the construction of any general sequence (such as one defined as above, or perhaps recursively) then require some form of choice? Thanks for any help.
AI: Actually, if $f$ is continuous from a closed interval (say $[0,1]$) into $\mathbb R$ then you do not need the axiom of choice to prove it is bounded.
First observe that closed and bounded intervals are still compact even without the axiom of choice. The proof of this is quite nice and simple, let $\mathcal B$ be an arbitrary open cover of $[0,1]$, simply consider $x=\sup\{y\in[0,1]\mid [0,y]\text{ has a finite subcover in }\mathcal B\}$, deduce that $[0,x]$ is finitely covered as well, and then argue that we have to have $x=1$ (by the same reason).
We can deduce from the above that a subset of $\mathbb R$ is compact if and only if it is closed and bounded. If it is compact it cannot be unbounded, and it has to be closed since $\mathbb R$ is a Hausdorff space; on the other hand, if it is closed and bounded it is a subset of a closed interval, therefore closed in a compact space and thus compact.
It is still true that if $f$ is a continuous function from a compact set into a metric space then its image is compact. To see this simply note that every open cover of the image can be translated into an open cover of the compact domain, therefore we can take a finite subcover, and this translate to a finite subcover of the image.
Therefore the image of a continuous image of a closed interval is compact and the image attains minimal and maximal values (since the image is a closed set).
Also note that $A(n)$ as you specified it is simply intersection of $I$ with an open set which is the preimage of $(n,\infty)$. Choosing from open sets is doable without the axiom of choice [3].
In general, when we simply produce one sequence we can sometimes avoid choice if we have a method of calculating the next element in a uniform way (induction is not a uniform way!). If we simply "take another element" then we end up using choice, but we can sometimes avoid these things (for example, instead of arbitrary $\delta$ take $\frac1k$ for the least $k$ fitting). It may even be possible, when needed just one sequence, to use the rational numbers. Those are countable and in particular well-orderable and we can choose from those as much as we want.
However sometimes we want to argue that non-trivial sequences exist, and for this we indeed have to have some choice. For example the proof that $\lim_{x\to a}f(x)=f(a)$ implies $f\colon A\to\mathbb R$ is continuous at $a$ may break, because we need to argue for all sequences and not produce just one.
It may be the case that $A$ itself is Dedekind-finite, and every sequence has only finitely many terms (at least one of those repeating, of course) so in the above case $x_n\to a$ implies that almost always $x_n=a$, but we can make sure that $f$ is not continuous at $a$. Indeed in such $A$ there are only finitely many rational numbers, and pulling the trick of choosing rationals no longer works.
Further reading:
Continuity and the Axiom of Choice
Axiom of choice and calculus
Open Sets of $\mathbb{R}^1$ and axiom of choice |
H: Non-$C^{*}$ Banach algebras?
It suddenly occurred to me almost every Banach algebra I know is actually a $C^{*}$ algebra. Several kinds of function algebras are definitely $C^{*}$ algebras. So is the matrix algebra. Although one gets a non-$C^*$ algebra by focusing on the upper triangular matrices, the norm still satisfies the $C^*$ identity.
The algebra of operators on a general banach space is not $C^*$, but at least for me this is a too abstract class that do not provide much intuition.
Thus I wonder whether someone has some good examples of banach algebras that fail the $C^*$ identity but are on the other hand elementary enough to provide intuition and direct computation, like the function algebras.
Thanks!
AI: Adding the assumption that you're asking about Banach $*$-algebras with isometric involution, your question Why is $\ell^1(\mathbb{Z})$ not a $C^{*}$-algebra? has the example $\ell^1(\mathbb Z)$.
Another is the algebra of bounded analytic functions on the unit disk with sup norm and involution $f^{*}(z)=\overline{f(\overline z)}$.
You mention matrices, and implicitly you seem to be assuming that $M_n$ is given the operator norm from acting as operators on $\mathbb C^n$ with the standard inner product, or equivalently $\|a\|=\sqrt{\text{the spectral radius of }a^*a}$, where $a^*$ is the conjugate transpose of $a$. But if you give $M_n$ another submultiplicative norm that makes the conjugate transpose norm-preserving, then you will not have a $C^*$-algebra. One example is the Frobenius norm, a.k.a. the Hilbert–Schmidt norm, $\|a\|=\sqrt{\mathrm{Trace}(a^*a)}$.
You say that the upper-triangular matrices satisfy the $C^*$-identity, but it is not clear what that means when they don't have an involution. If you are asking about $*$-algebras, then subalgebras of $*$-algebras that are not closed under the involution are out of the picture.
The first example above, $\ell^1(\mathbb Z)$, fits into a bigger picture of considering the Banach $*$-algebra $L^1(G)$ of a locally compact Hausdorff group $G$ with Haar measure, which sometimes arises in the study of group representations.
One thing that makes $\ell^1(\mathbb Z)$ more interesting than $M_n$ with the Frobenius norm as an example is that $\ell^1(\mathbb Z)$ is not even isomorphic as an algebra to any $C^*$-algebra. |
H: Parallel lines passing through $(d, 3)$ & $(-2,1)$ and $(5,d)$ & $(1,0)$.
There are two parallel lines with one passing through $(d, 3)$ & $(-2,1)$ and the second line passes through $(5,d)$ & $(1,0)$.
Find the two values for $d$.
I found one value would be $-4$. But how would I do this to find the second value?
AI: If two lines are parallel, then the slope of both the lines are equal.
If a line passes through the points, $(x_1,y_1)$ and $(x_2,y_2)$, then the slop of the line is given by $$\text{Slope } = \dfrac{y_1 - y_2}{x_1 - x_2}$$
The first line passes through the points $(d,3)$ and $(-2,1)$. Hence, the slope of the first line is $$\dfrac{3-1}{d-(-2)}$$
The second line passes through the points $(5,d)$ and $(1,0)$. Hence, the slope of the second line is $$\dfrac{d-0}{5-1}$$
Since the lines are parallel, the slope are equal. Equate the two to get $$\dfrac2{d+2} = \dfrac{d}4 \implies d^2+2d = 8 \implies d^2 +2d-8 = 0$$
Can you now finish it off?
Move your mouse over the gray area for the final answer.
$$d^2 + 2d - 8 = 0 \implies (d+4)(d-2) = 0 \implies d= 2 \text{ or } -4$$ |
H: Polar equation of $y = 2$
Maybe I do not understand what is going on here but I cannot get the right answer.
$$y = 2 $$
$$y^2 + x^2 = r^2$$
$$4 + 0^2 = r^2$$
$$ r = 2$$
$$y = r \sin \theta$$
$$1 = \sin \theta$$
$$\theta = \pi/2$$
This is wrong but I do not see anything wrong with my logic.
AI: Just because $y=2$, it does not mean that $x =0$. In particular, you want that, for any $x$, $y=2$, but $x$ is not $0$. Since you already know that
$$y=2$$
and that to change to polar coordinates, you can use
$$y= \rho \sin \theta$$,
we plug in $y=2$, which gives
$$2= \rho \sin \theta$$
or
$$2 \csc \theta= \rho $$
Note that we don't consider the variable $x$ and $$x= \rho \cos \theta$$, since it suffices to use only the $y$-equation above, because it fully describes the dependence between $\theta$ and $\rho$.
In particular note the radius can't be always $2$, else you would get a circle! Everytime you state something such as $x=0$ or $r=2$, think the implications it carries, and it might help you to spot the flaw. |
H: Odd-dimensional complex skew-symmetric matrix has eigenvalue $0$
There is the standard proof using $$\det(A)=\det( A^{T} ) = \det(-A)=(-1)^n \det(A)$$
I would like a proof that avoids this. Specifically, there is the proof that for $A$ a $\bf{real} $ matrix, the transpose is the same as the adjoint, which gives (using the complex inner product) $\lambda \|x\|^2 =\langle Ax, x \rangle= \langle x, -Ax \rangle=-\overline{\lambda } \|x\|^{2}$, so any eigenvalue is purely imaginary. Then we conclude that, since any odd-dimensional real matrix has a real eigenvalue, that eigenvalue must be zero. This argument doesn't work for a general complex skew-symmetric matrix. Is there something I'm missing, is there a way to modify this argument to get that zero is an eigenvalue for the complex case? Also, can somebody please give a geometric reason why odd-dimensional skew-symmetric matrices have zero determinant (equiv., a zero eigenvalue)?
Thanks!
AI: Here is a "geometric" argument for the real case. The skew-symmetric condition is equivalent to $\langle Av, v\rangle =0$ for all vectors $v$. Geometrically, $Av$ is orthogonal to $v$. If $Av$ is never zero for $v\ne 0$, then we get a nonvanishing vector field on the unit sphere, contradicting the Hairy Ball theorem. |
H: Polar equation of cartesian $y = 1 + 3x$
I have no idea at all what to do on this
I got $$\cos^{-1} \left(\frac{r\sin \theta+1}{3} \right) = \theta$$
Which can be
$$\cos^{-1} \left(\frac{r\sin \left(\cos^{-1} \left(\frac{r\sin \theta+1}{3} \right) \right)+1}{3} \right) = \theta$$
$$\cos^{-1} \left(\frac{r\sin \left(\cos^{-1} \left(\frac{r\sin \left(\cos^{-1} \left(\frac{r\sin \theta+1}{3} \right) \right)+1}{3} \right) \right)+1}{3} \right) = \theta$$
So that obviously seems wrong, other than that 30 minutes of work has produced nothing.
AI: All you need to do in problems like these is to just substitute $x= r \cos(\theta)$ and $y = r \sin(\theta)$ and write out the resulting equation. In this case, you get $$r \sin(\theta) = 1 + 3 r \cos(\theta)$$which on rearranging gives us $$r (\sin(\theta) - 3 \cos(\theta)) = 1 \implies r = \dfrac1{\sin(\theta) - 3 \cos(\theta)}$$
This is all we need to do to convert it into polar form. |
H: What is the meaning of "unitize a vector"?
The expression "to unitize a vector" is often use in computational geometry. What does it mean?
AI: Yeah, it's a perversion of normalize. If $v\not = 0$, we normalize v as follows
$$w = {v\over \|v\|}.$$
Why this ugly neologism is needed is beyond me. |
H: What does the letter $t$ stands for in "$t$-value of a curve"?
The "$t$-value of a curve" is the parameter of a point on a curve. For example if a curve's domain is form $0$ to $1$, then a t-value of $0.5$ would be a mid point. What does the letter $t$ stand for?
AI: $t$ is the underlying parameter. Any "nice" one-dimensional curve can be parameterized by a single parameter. For instance, if you want a line connecting pointing a point $\vec{r}_1$ and $\vec{r}_2$, then the equation of the line is given as $$\vec{r}(t) = (1-t) \vec{r}_1 + t \vec{r_2} $$
What this means is for any value of $t$, $\vec{r}(t)$ will lie on the straight line connecting the points $\vec{r}_1$ and $\vec{r}_2$.
Taking $t=0$, we get that $\vec{r}(0) = \vec{r}_1$, which essentially says that the point $\vec{r}_1$ lies on the line connecting $\vec{r}_1$ and $\vec{r}_2$, which is not surprising. Similarly, taking $t=1$, we get that $\vec{r}(1) = \vec{r}_2$, which essentially says that the point $\vec{r}_2$ lies on the line connecting $\vec{r}_1$ and $\vec{r}_2$, which again is not surprising. The midpoint is given by taking $t = 1/2$ i.e. we get that $\vec{r}(1/2) = \dfrac{\vec{r}_1 + \vec{r}_2}2$ which is nothing but the mid-point of the line-segment connecting $\vec{r}_1$ and $\vec{r}_2$.
Similarly, if we want to parameterize a unit circle in $2$D, we can do as follows. We know that the equation of the unit circle in $2$D is $x^2 + y^2 = 1$. This suggests that we can take $x = \cos(t)$ and $y = \sin(t)$. Hence, the vector $$\vec{r}(t) = (\cos(t),\sin(t))$$ is a point on the circle and parameterizes the circle since every point on the circle can be written as $(\cos(t), \sin(t))$ and conversely all the points of the form $(\cos(t), \sin(t))$ lie on a unit circle. |
H: A (potential) projection operator
Let $I$ be a 3-by-3 identity matrix and $\hat n$ be a unit vector orthonormal to some surface. What then does $I-\hat n\hat n$ mean geometrically? Also what does it mean to multiply this matrix/operator by a vector $v$? Some sort of projection?
Also, I don't understand what $\hat n\hat n$ means. Dimensionally it should be a 3-by-3 matrix because $I$ is such, right?
AI: If $n$ is a column vector (as usual in most linear algebra textbooks), then $\hat n\hat n$ would probably mean $\hat n \hat n^{T},$ which is the projection in the direction of $\hat n.$
And $(I - \hat n \hat n) v$ is the vector that resembles the projection vector from the tip of $v$ into the plane. You can see it if you expand $(I - \hat n \hat n) v$. You get $v - v_n$ where $v_n$ is the projection of $v$ onto the direction of $n.$
You can see in the following image: $v - v_n = e$ |
H: What are the $\{x,y,z\}$ values of a vector?
A vector is often described as $\{x,y,z\}$ similarly to a $3$D point's cartesian coordiantes in CAD tools which is quite confusing. What are the $x$, $y$ and $z$ values in the case of a vector?
AI: First, $\{x,y,z\}$ denotes a set of things with no order. Coordinates are typically defined as ordered sets. So they're denoted $(x,y,z)$ instead.
Mathematically speaking a vector is defined as an abstract object living in a space. Only when you choose a "basis" for the space, we can then attach a "coordinate" to the vector. In this case, the coordinates could be $(x,y,z),$ or "length and angles from the axes," or "length and direction" etc.
To answer your question, I assume you're working in a 3D space equipped with the Euclidean norm, and the standard basis. Let the your vector be $v = (x, y, z).$ Think of $v$ as the oriented line segment (pointy arrow) from the origin $(0,0,0)$ to the point $(x,y,z).$ The amplitude of the vector (aka length) is $$\| v \| = \sqrt{x^2 + y^2 + z^2}.$$ The direction of $v$ would be the unit vector in the direction of $(x,y,z).$ That is,
$$\widehat v = \frac{v}{\| v \|} = ( \frac{x}{ \sqrt{x^2 + y^2 + z^2} }, \frac{y}{ \sqrt{x^2 + y^2 + z^2} }, \frac{z}{ \sqrt{x^2 + y^2 + z^2} } ).$$
Here is a not pretty picture of the situation in 2D |
H: How does partial fraction decomposition avoid division by zero?
This may be an incredibly stupid question, but why does partial fraction decomposition avoid division by zero? Let me give an example:
$$\frac{3x+2}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$$
Multiplying both sides by $x(x+1)$ we have:
$$3x+2=A(x+1)+Bx$$
when $x \neq -1$ and $x \neq 0$.
What is traditionally done here is $x$ is set to $-1$ and $0$ to reveal:
$$-3+2=-B \implies 1=B$$
and
$$2=A$$
so we find that
$$\frac{3x+2}{x(x+1)}=\frac{2}{x}+\frac{1}{x+1}$$
Why can $x$ be set equal to the roots of the denominator (in this case, $0$ and $-1$) without creating a division by zero problem?
AI: $\begin{align}{\bf Hint}\quad \dfrac{f(x)}{h(x)}=\dfrac{3x\!+\!2}{x(x\!+\!1)} &= \dfrac{a(x\!+\!1)+bx}{x(x\!+\!1)}=\dfrac{g(x)}{h(x)}\\[.3em]
\Rightarrow\ \ \ \ \ 3x\!+\!2\,\ &=\ a(x\!+\!1)+bx\ \ {\rm for\ all\ }\, x\neq 0,-1\\[.3em]
\Rightarrow\ \ \ \ \ 3x\!+\!2\,\ &=\ a(x\!+\!1)+bx\ \ {\rm for\ all\ }\, x \ \ \ (\,\color{#c00}{0,-1 \ \rm included\:\!)}
\end{align}$
since their difference $\, f(x)- g(x)\,$ is a polynomial with infinitely many roots (all $\,x\neq 0,-1)$ so it is the zero polynomial (recall a nonzero polynomial over a field has no more roots than its degree).
Generally $ $ If $\,f,g\,$ and $\,h\!\ne\! 0\,$ are polynomial functions over $\,\mathbb R\,$ (or any $\rm\color{#0a0}{infinite}$ field) then
$$\begin{eqnarray} \smash[b]{\dfrac{f(x)}{h(x)} = \dfrac{g(x)}{h(x)}} \,&\Rightarrow&\ f(x) = g(x)\ \ {\rm for\ all}\,\ x\in\mathbb R\, \ {\rm such\ that}\,\ h(x)\ne 0\\[.2em]
&\Rightarrow&\ f(x) = g(x)\ \ {\rm for\ all}\ \,x\in \mathbb R
\end{eqnarray}\qquad$$
by $\,p(x) = f(x)\!-\!g(x) = 0\,$ has $\rm\color{#0a0}{infinitely}$ many roots [all $\,x\in \mathbb R\,$ except finite #roots of $\,h(x)$], $ $ hence $\,p=f-g\,$ is the zero polynomial, so $\, f = g.$
Thus to solve for coef's $\,a,b\,$ that occur in $\,g\,$ it is valid to evaluate $\,f(x) = g(x)\,$ at any $\,x\in \mathbb R,\,$ since it holds true for all $\,x\in \mathbb R\,$ (including all real roots of $\, h).$
Remark $ $ The method you describe is known as the Heaviside cover-up method. It can be generalized to higher-degree denominators as I explain here.
See here for some analogous methods applied in other contexts. |
H: Translation request: theorem concerning automorphisms of $p$-groups
I want to translate the following theorem from German to English:
5.12 Satz. Sei $\mathfrak{P}$ eine $p$-Gruppe und $\alpha$ ein Automorphismums von $\mathfrak{P}$ von su $p$ teilerfremdder Ordnung.*
Für $p\gt 2$ lasse $\alpha$ alle Elemente der Ordnung $p$ von $\mathfrak{P}$ fest.
Für $p\gt 2$ lasse $\alpha$ alle Elemente der Ordnungen $2$ und $4$ von $\mathfrak{P}$ fest.
Dann ist $\alpha =1$.
I used google translator and get the following:
Let B be a p-group and S be an automorphism of B of order coprime to p.
a) For p> 2, let S fix all the elements of B of order p.
b) for p = 2, let S fix all elements of orders 2 and 4 of B.
Then S = 1
I found the following translation on some research
For a finite $p$ group $P$, we write
$$\Omega(P) = \left\{\begin{array}{ll}
\Omega_1(P) &\text{if }p\gt 2\\
\Omega_2(P) &\text{if }p=2
\end{array}\right.$$
where
$$\Omega_i(P) = \Bigl\langle x\in P \;\Bigm|\; |x| \bigm| p^i\Bigr\rangle.$$
Lemma 2.1 ([6] Kap. IV, Satz 5.12]). Suppose that a finite $p'$-group $A$ acts by automorphisms on a finite $p$-group $P$. If $A$ acts trivially on $\Omega_1(P)$ for $p\neq 2$ or on $\Omega_2(P)$ for $p=2$,then $A$ acts trivially on $P$.
I want to know if the last translation is true because I am not sure that if an automorphism comes from a $p^{'}$-group, then this automorphism is of order coprime to $p$.
AI: A better translation of the Theorem is:
Theorem 5.12. Let $P$ be a $p$-group, and let $\alpha$ be an automorphism of $P$ of order coprime to $p$.
If $p\gt 2$ and $\alpha$ fixes all elements of order $p$; or
If $p=2$ and $\alpha$ fixes all elements of orders $2$ or $4$;
then $\alpha$ is the identity.
The quoted lemma is equivalent to this result.
To see that the lemma follows from "Theorem 5.12", note that the fact that $A$ acts by automorphisms on $P$ is equivalent to saying that we have a group homomorphism $A\to \mathrm{Aut}(P)$. The conditions given on $A$ imply that if $\alpha\in A$, then the image of $\alpha$ in $\mathrm{Aut}(P)$ has order coprime to $p$ and satisfies the hypothesis of Lemma 5.12, hence $\alpha$ is the identity. Thus, the image of $A$ is trivial, so $A$ acts trivially on $P$.
Conversely, if the quoted Lemma is true, then Theorem 5.12 follows. Let $\alpha$ be an automorphism of $P$ of order coprime to $p$ that fixes each of the elements of order $p$ if $p\gt 2$, and the elements of orders $2$ and $4$ if $p=2$. Let $A=\langle \alpha\rangle$, viewed as a subgroup of $\mathrm{Aut}(P)$. Then $A$ acts on $P$ and satisfies the hypothesis of the lemma, hence acts trivially. In particular, $\alpha$ acts trivially; i.e., $\alpha$ is the identity automorphism. This proves "Theorem 5.12". |
H: Comparing the time in the time-distance problem
The question is:
Two trains start at point A and B and travel towards each other at a speed of $50$km/hr and $60$Km/hr respectively. At the time of meeting the second train has traveled $120$ km more than the first train. Now the distance between them is:
Now I did manage to solve it with a little help and its like this:
First Train starting from $A$:
$t = x/50$
Second Train starting from $B$:
$t = (120+x) / 60$
Comparing $A$ and $B$ we get $x$ and then using the value of $x$ we can calculate the total distance between them which is $1320$.
My question is why are we comparing $A$ and $B$. The only reason we would compare them is if they were equal. I don't understand how time could be equal when the two trains meet. I would appreciate it if someone could kindly clarify this concept.
AI: It is important that both the trains start at the same time instant. Hence, when they meet, both trains would have taken the same time. |
H: Odd primes $\neq5$ divide a integer made up of all $1$s. Ditto for integers coprime to $10$.
Okay, so I have worked on this problem and even though I can see it is true I just don't know how to show it.
Here it goes.
Question:
Show that every odd prime except $5$ divides some number of the form $111 \ldots 11$ ($k$ digits, all ones).
Thank you my friends!
AI: Given an odd prime $p$ it is known (Fermat's Little Theorem) that $p$ divides $10^{p-1}-1$, hence, $(10^{p-1}-1)/9$ except you have to note 3 divides 111). |
H: Set of convergence is measurable.
Possible Duplicate:
pointwise convergence in $\sigma$-algebra
Problem: Prove that the set of points at which a sequence of measurable real functions converges is a measurable set. (I believe the problem means functions from the reals to the reals.)
Source: W. Rudin, Real and Complex Analysis, Chapter 1, exercise 5.
I have posted a proposed solution in the answers.
AI: Let the sequence of functions be $\{f_n(x)\}$. The $\lim \inf$ and $\lim\sup$ of this sequence of functions are measurable (extended-valued) functions. Denote them $h(x)$ and $g(x)$. The set $A$ where $g$ and $h$ are both positive infinity or both negative infinity is measurable, as they are each measurable functions.
Consider the function $p(x)=h\chi_{\mathbb{R}-A}-g\chi_{\mathbb{R}-A}$. It is zero precisely where the original sequence of functions has a limit. Then $E=p^{-1}(\{0\})$ is measurable, so and $E\cup A$ is measurable, and it is the set of points where the sequence has a limit, so we are done. |
H: Hensel lifting square roots $\!\bmod p\,$ to $\!\bmod p^2$
I've been working on this problem for a while, but hit a dead end.
Here's the problem:
Suppose $p$ is an odd prime. Also let $b^2 \equiv a \pmod p$ and $p$ does not divide $a$. Prove there exists some $k \in \mathbb{Z}$ such that $(b+kp)^2 \equiv a \pmod {p^2}$.
Here's what I've tried so far:
$$(b+kp)^2 \equiv b^2 + 2bkp + k^2p^2 \equiv a \pmod {p^2}$$
Here, I need to find such $k$ that satisfy this congruence. Equivalently, I need to find such $k$ so that $p^2$ divides $(b^2-a) + 2bkp + p^2k^2$ or equivalently show that $p^2$ divides $(b^2-a) + 2bkp$ for some $k \in \mathbb{Z}$.
So far, since $p$ divides $(b^2-a)$, then by definition, there exists some $x \in \mathbb{Z}$ such that $b^2-a = px$. I got stuck here, and I've tried some examples, but I haven't seen any pattern that pertains to this problem.
Any insight would be helpful.
AI: You seek a $k$ such that $p^2\mid \big((b^2-a)+2kbp\big)$. Now, $b^2-a$ is already a multiple of $p$, say $\ell p$. Note that $uv\mid uw\iff v\mid w$, so we deduce (using $u=v=p$) that
$$p^2\mid \big((b^2-a)+2kbp\big)\iff p\mid (\ell+2bk)\iff \ell+2bk\equiv0 \bmod p.$$
Since $p$ is odd and $p\not\mid b$, you can solve for $k$ using modular arithmetic ($2b$ is invertible modulo $p$). |
H: Possible errors in my professor's notes, Abel summation
In my professor's notes he has written this:
$$\int_1^N \frac{\{t\} - \frac{1}{2}}{t}dt = \int_1^N\frac{1}{t}d \left(\int_1^t B(y)dy \right) = \int_1^t B(y)dy|_1^N + \int_1^N \frac{\int_1^t B(y)dy}{t^2}dt$$
Where $\{x\}$ indicates the fractional part of $x$ and $B(x) = \{x\} - \frac{1}{2}$.
This is Abel summation with the Riemann-Stieltjes integral.
What I don't understand is I feel first that the lower bound of the $\displaystyle \int_1^t$ integrals should all be $\dfrac{1}{2}$ and not $1$, and second since this is integration by parts there should be a $\dfrac{1}{t}$ in front of the $\displaystyle \left. \int_1^t B(y)dy \right \vert_1^N$ term.
Am I missing something? Thanks.
AI: There should be a $\dfrac1t$ infront of the integral $\displaystyle \left( \int_1^{t} B(y) dy \right)$.
$$\int_1^{N} \dfrac{\{t\} - \dfrac12}t dt = \int_1^{N} \dfrac1t d \left( \int_1^t B(y) dy\right)\\ = \left. \left(\dfrac1t \left( \int_1^{t} B(y) dy\right) \right) \right \rvert_{1}^{N} + \int_1^{N} \dfrac1{t^2} \left( \int_1^{t} B(y) dy\right) dt$$
The lower bound for the inner integral $\displaystyle \int_1^t (\cdot) dy$ can in principle be anything since $$d \left( \int_a^t B(y) dy\right) = B(t) dt = \left(\{t\} - \dfrac12 \right) dt$$ where $a$ is any fixed real number. |
H: a) Prove that $f$ has a removable singularity if $f'$ does; b) Evaluate $\int_0^\infty\frac{\log x}{(1+x)^3}\,dx$
a) Let $\,f\,$ be an analytic function in the punctured disk $\,\{z\;\;;\;\;0<|z-a|<r\,\,,\,r\in\mathbb R^+\}\,$ . Prove that if the limit $\displaystyle{\lim_{z\to a}f'(z)}\,$ exists finitely, then $\,a\,$ is a removable singularity of $\,f\,$
My solution and doubt: If we develop $\,f\,$ is a Laurent series around $\,a\,$ we get
$$f(z)=\frac{a_{-k}}{(z-a)^k}+\frac{a_{-k+1}}{(z-a)^{k-1}}+\ldots +\frac{a_{-1}}{z-a}+a_0+a_1(z-a)+\ldots \Longrightarrow$$
$$\Longrightarrow f'(z)=-\frac{ka_{-k}}{(z-a)^{k+1}}-\ldots -\frac{a_{-1}}{(z-a)^2}+a_1+...$$
and since $\,\displaystyle{\lim_{z\to a}f'(z)}\,$ exists finitely then it must be that
$$a_{-k}=a_{-k+1}=...=a_{-1}=0$$
getting that the above series for $\,f\,$ is, in fact, a Taylor one and thus $\,f\,$ has a removable singularity at $\,a\,$ .
My doubt: is there any other "more obvious" or more elementary way to solve the above without having to resource to term-term differentiating that Laurent series?
b) Evaluate, using some complex contour, the integral
$$\int_0^\infty\frac{\log x}{(1+x)^3}\,dx$$
First doubt: it is given in this exercise the hint(?) to use the function
$$\frac{\log^2z}{(1+z)^3}$$Please do note the square in the logarithm! Now, is this some typo or perhaps it really helps to do it this way? After checking with WA, the original real integral equals $\,-1/2\,$ and, in fact, it is doable without need to use complex functions, and though the result is rather ugly it nevertheless is an elementary function (rational with logarithms, no hypergeometric or Li or stuff).
The real integral with the logarithm squared gives the beautiful result of $\,\pi^2/6\,$ but, again, I'm not sure whether "the hint" is a typo.
Second doubt: In either case (logarithm squared or not), what would be the best contour to choose? I though using one quarter of the circle $\,\{z\;\;;\;\;|z|=R>1\}\,$ minus one quarter of the circle $\,\{z\;\;;\;\;|z|=\epsilon\,\,,0<\epsilon<<R\}\,$, in the first quadrant both, because
$(i)\,$ to get the correct limits on the $\,x\,$-axis when passing to the limits $\,R\to\infty\,\,,\,\epsilon\to 0\,$
$(ii)\,$ To avoid the singularity $\,z=0\,$ of the logarithm (not to mention going around it and changing logarithmic branch and horrible things like this!).
Well, I'm pretty stuck here with the evaluations on the different segments of the path, besides being baffled by "the hint", and I definitely need some help here.
As before: these exercises are supposed to be for a first course in complex variable and, thus, I think they should be more or less "elementary", though this integral looks really evil.
For the time you've taken already to read this long post I already thank you, and any help, hint or ideas will be very much appreciated.
AI: For part b) I think for this type of question you are typically supposed to use the residue theorem, so here you should look for a contour that goes around $z=-1$. Since the real integral is from $0\to \infty$ we'll likely still want to integrate along the real axis, around the circle $|z|=R$, then back along below the real axis with a small loop to avoid $z=0$.
$$
\int_{x=\epsilon}^R f(x^+)+\int_{\theta= 0}^{2\pi}f(Re^{i\theta})-\int_{x=\epsilon}^R f(x^-)-\int_{\theta=0}^{2\pi}f(\epsilon e^{i\theta}) = 2\pi i~\mathrm{Res}(f,-1)
$$
Using the hint and taking $R\to\infty,\epsilon\to 0$ the circular integrals vanish, the first integral along the real line has $\log^2x$ in the numerator, and the second has the next branch value $(\log x+2\pi i)^2$. When you take the difference the $\log^2 x$ terms cancel and you end up with your desired integral in terms of another integral that is easy to evaluate and the residue. |
H: How do I prove this equality?
A vector x $\in \mathbb{R}^3$ makes angles $a, b, c$ with the
three axes. Using the dot product with standard basis vectors, show:
$$\cos^2a + \cos^2b + \cos^2c = 1$$
I wasn't sure how to start this one. Conceptually, I can see what was going on, but I would think to create a number of simultaneous equations using $a + b = \frac{\pi}{2}$ etc. I wasn't sure what to do with standard basis vectors.
AI: HINT 1:
Recall that if $\vec{x} = x_a \hat{e}_a + x_b \hat{e}_b + x_c \hat{e}_c$ and $\vec{y} = y_a \hat{e}_a + y_b \hat{e}_b + y_c \hat{e}_c$, then $$\vec{x} \cdot \vec{y} = x_a y_a + x_b y_b + x_c y_c = \Vert x \Vert_2 \Vert y \Vert_2 \cos(\theta_{xy})$$ where $\theta_{xy}$ is the angle between the vectors $\vec{x}$ and $\vec{y}$.
HINT 2:
Look at the norm of the vector, which make angles $a,b,c$ with the three axes.
Move the cursor over the gray area for a complete answer.
Let the vector be $\vec{x} = x_a \hat{e}_a + x_b \hat{e}_b + x_c \hat{e}_c$. Note that $$x_a = \vec{x} \cdot \hat{e}_a = \Vert x \Vert_2 \cos(a)$$ $$x_b = \vec{x} \cdot \hat{e}_b = \Vert x \Vert_2 \cos(b)$$ $$x_c = \vec{x} \cdot \hat{e}_c = \Vert x \Vert_2 \cos(c)$$ Further, $$\Vert x \Vert_2^2 = \vert x_a \vert^2 + \vert x_b \vert^2 + \vert x_c \vert^2.$$ Hence, $$\Vert x \Vert_2^2 = \Vert x \Vert_2^2 \cos^2(a) + \Vert x \Vert_2^2 \cos^2(b) + \Vert x \Vert_2^2 \cos^2(c)$$ Canceling off $\Vert x \Vert_2^2$, we get $$\cos^2(a) + \cos^2(b) + \cos^2(c) = 1$$ |
H: Antiderivative of $x \mapsto \operatorname{tr}\{(\mathbf{A}x+\mathbf{B})^{-1}\mathbf{C}\}$
I am wondering how to integrate the function
$$x \mapsto \operatorname{tr}\bigl\{(\mathbf{A}x+\mathbf{B})^{-1}\mathbf{C}\bigr\}$$
In my case, the matrices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ are (strictly) positive definite. If $\mathbf{C} = \mathbf{A}$ it is easy to see that
$$\ln\det(\mathbf{A}x+\mathbf{B})$$
is an antiderivative. But what if $\mathbf{C} \neq \mathbf{A}$?
Thanks,
jens
AI: By the linearity of integration, trace, and matrix multiplication, we can pull $\int$ inside:
$$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx=\mathrm{tr}\left( \int (Ax+B)^{-1}dx\, C\right).$$
I originally thought $\log(Ax+B)\;A^{-1}$ would be the antiderivative for this, but oenamen helpfully points out this can't be said in full generality ($\log XY=\log X+\log Y$ does not necessarily hold if the matrices $X$ and $Y$ don't commute!); instead, we need a special $B^{-1}$ factor. We have:
$$\frac{d}{dx}\log(I+xU) =\quad U(I+xU)^{-1} =(I+xU)^{-1}U.$$
Hence, applying the above alongside $X^{-1}Y^{-1}=(YX)^{-1}$:
$$\begin{array}{c l} \frac{d}{dx}\log \big(B^{-1}(Ax+B)\big) & =\frac{d}{dx}\log(I+xB^{-1}A) \\ & = (I+xB^{-1}A)^{-1}B^{-1}A \\ & = \big(B\cdot(I+B^{-1}Ax)\big)^{-1}A \\ & = (Ax+B)^{-1}A. \end{array}$$
Therefore we conclude
$$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx =\mathrm{tr}\left(\log\big(I+xB^{-1}A)\; A^{-1}C\right)+const. $$ |
H: a question about conditional expectation
Suppose that X and Y are random variables such that E(Y/X)=aX+b,how determine expressions for a and b in terms of E(X),E(Y),Var(X),and Cov(X,Y).assuming that Cov(X,Y) exists and Var(X)>0.
AI: Using the tower property, recall that $E[E[Y|X]] = E[Y]$. You should try to work that out in order to obtain an equation in terms of $E[X]$ and $E[Y]$,
$$E[Y] = E[E[Y|X]] = E[aX + b] = aE[X] + b$$
Also, recall that $Cov(X,Y) = E[XY] - E[X]E[Y]$. Hence you can try to write $E[XY]$ in terms of $Var[X]$ and $E[X]$ to get another equation. Recall that $E[XY] =E[E[XY|X]] = E[X \cdot E[Y|X]]$,
$$E[XY] = E[E[XY|X]] = E[X \cdot E[Y|X]] = E[aX^{2} + bX] = a(Var[X]+E[X]^2) + bE[X]$$
Using this expression, obtain:
$$Cov(X,Y) = a(Var[X]+E[X]^2) + bE[X] - E[X]E[Y]$$
Finally, you can obtain the desired relation solving for $a$ and $b$ using the two equations. |
H: Conditions for random variables to be negligible in their sums
The followings are from Kai Lai Chung's A Course in Probability Theory, on page 207
$X_{n,j}, j=1,\dots, k_n, n=1,\dots,$ are random variables, where $k_n \to \infty$ as $n\to \infty$.
For every $\epsilon > 0$:
(a) for each $j$, $\lim_n P(|X_{nj}|>\epsilon) = 0$;
(b) $\lim_n \max_j P(|X_{nj}| > \epsilon) = 0$; (holospoudic)
(c) $\lim_n P(\max_j |X_{nj}| > \epsilon) = 0$;
(d) $\lim_n \sum_j P(|X_{nj}| > \epsilon) = 0$;
It is clear that (d) => (c) => (b) => (a).
I can understand (d) => (c), because $$\sum_j P(|X_{nj}| > \epsilon) \geq P(\max_j |X_{nj}| > \epsilon).$$
I can understand (b) => (a), because $$\max_j P(|X_{nj}| > \epsilon) \geq P(|X_{nj}|>\epsilon).$$
Questions:
I wonder why (c) => (b)? Neither do I understand why $$P(\max_j
|X_{nj}| > \epsilon) \geq \max_j P(|X_{nj}| > \epsilon),$$ if it
is true.
Also why when $X_{nj}, j=1,\dots,k_n$ are independent for each $n$,
(d) $\equiv$ (c)? This is in the exercise 1 on page 214.
Thanks and regards!
AI: Fix $\varepsilon$, and consider some random variables $(Y_k)_k$ and $Y=\sup\limits_kY_k$. For every $i$, $Y\geqslant Y_i$ hence $[Y_i\gt\varepsilon]\subseteq[Y\gt\varepsilon]$. This implies that $\mathrm P(Y_i\gt\varepsilon)\leqslant\mathrm P(Y\gt\varepsilon)$. This inequality holds for every $i$ and the RHS does not depend on $i$ hence $\sup\limits_i\mathrm P(Y_i\gt\varepsilon)\leqslant\mathrm P(Y\gt\varepsilon)=\mathrm P(\sup\limits_iY_i\gt\varepsilon)$.
Edit: When, furthermore, the random variables $(Y_k)_k$ are assumed independent, the estimation of the distribution of $Y$ proceeds as usual. First, for every $\varepsilon$, $[Y\leqslant\varepsilon]=\bigcap\limits_k[Y_k\leqslant\varepsilon]$ hence
$$
\mathrm P(Y\gt\varepsilon)=1-\prod\limits_k(1-\mathrm P(Y_k\gt\varepsilon)).
$$
Now, $1-x\leqslant\mathrm e^{-x}$ for every $x$ hence
$$
\mathrm P(Y\gt\varepsilon)\geqslant1-\exp\left(-\sum\limits_k\mathrm P(Y_k\gt\varepsilon)\right),
$$
which shows that (c) implies (d). On the other hand, (d) implies (c) because $[Y\gt\varepsilon]=\bigcup\limits_k[Y_k\gt\varepsilon]$ and the probability of the union is at most the sum of the probabilities. |
H: Simplify $\frac{9}{2}(1 + \sqrt 5)\sqrt{10 - 2\sqrt 5} + 9\sqrt{5 + 2\sqrt 5}$
Simplify $\displaystyle{\frac{9}{2}(1 + \sqrt 5)\sqrt{10 - 2\sqrt 5} + 9\sqrt{5 + 2\sqrt 5}}$.
I get this when I was doing another Q,
but I don't know how to further simplify it.
Can anyone help me, please?
AI: Hint: Let's note $o:=\frac {1+\sqrt{5}}2$, $a:=\sqrt{10-2\sqrt{5}}$ and $b:=\sqrt{5+2\sqrt{5}}$
then $ab=\sqrt{30+10\sqrt{5}}=5+\sqrt{5}$
Compute $(o\cdot a+b)^2$ to conclude. |
H: Need help about $P\Gamma L_2(q)$, $q=4,3$
I am asking kindly,
For which values of $n$ we have $$S_n≅P\Gamma L_2(3),S_n≅P\Gamma L_2(4)$$ This may be correct if we replace $S_n$ by $A_n$.
Any help will be appreciated. :)
Edit (JL): Adding the definition of the group. Let $\Omega$ denote the set $GF(q)\cup\{\infty\}$. Then the group $P\Gamma L_2(q)$ consists of bijections from $\Omega$ to itself the type
$$
P\Gamma L_2(q)=\{f:\Omega\rightarrow\Omega\mid f(z)=\frac{az^\sigma+b}{cz^\sigma+d}; a,b,c,d\in GF(q); ad-bc\neq0; \sigma\in Aut(GF(q))\}.
$$
This group has order $q(q^2-1)\cdot |Aut(GF(q))|$.
AI: By definition $P\Gamma L_2(q)$ is a subgroup of $\mathrm{Sym}(\Omega)\simeq S_{q+1}$.
The claims would follow immediately from the knowledge of the order of the group $P\Gamma L_2(q)$:
When $q=3$, the group $Aut(GF(3))$ is trivial, and we have $|P\Gamma L_2(3)|=24=|S_4|$, so it follows that $P\Gamma L_2(q)$ must be all of $S_4$.
When $q=4$, the group $Aut(GF(4))$ is cyclic of order tow, and we have
$$|P\Gamma L_2(4)|=4\cdot(4^2-1)\cdot2=120=|S_5|.$$ Again we see that $P\Gamma L_2(4)$ must be all of $S_5$.
So the real question is to prove the formula for the order of the group in these two cases.
In the case $q=3$ we observe that the transformations $z\mapsto z+1$ and $z\mapsto 2z$
give the permutations $\alpha=(012)(\infty)$ and $\beta=(12)(0)(\infty)$ of $\Omega$ respectively.
Clearly these two generate the group $S_3$ acting on the affine part $GF(3)$.
The transformation $z\mapsto 1/z$ gives the permutation $\gamma=(0\infty)(1)(2)$. Together with $\alpha$ we get a group that acts transitively on $\Omega$, because $\gamma(\infty)=0$,
and $(\alpha^i\gamma)(\infty)=i$ for $i=1,2$. Therefore the group generated by all of
$\alpha,\beta,\gamma$ must have at least $4\cdot|\langle \alpha,\beta\rangle|=4\cdot6=24$
elements, and hence be all of $S_4$.
For the case $q=4$ I write $GF(4)=\{0,1,a,b=a+1=a^2\}$. Here $F:x\mapsto x^2$ is the
only non-trivial automorphism. It corresponds to the permutation $\phi=(0)(1)(ab)(\infty)$.
In addition to that we have $z\mapsto z+1$ giving us $\alpha=(01)(ab)(\infty)$, and
$z\mapsto az$ giving us the permutation $\beta=(0)(1ab)(\infty)$. The elements $\phi$ and $\beta$
fix both $0$ and $\infty$, and we see that they generate a copy of $S_3$ acting on $\{1,a,b\}$. The group of translations $K$ generated by $\alpha$ and $\beta\alpha\beta^{-1}$, the latter
corresponding to $z\mapsto z+a$, gives us a copy of the Klein four group acting transitively on the finite part $GF(4)$. As in the preceding case we thus see that the group
$H=\langle \alpha,\beta,\phi\rangle$ generated by these three transformations must act as $S_4$ on $GF(4)=\Omega\setminus\{\infty\}$ because it has at least $4\cdot6=24$ elements.
It remains to show that adding the generator $\gamma=(0\infty)(1)(ab)$ gives us a group of size at least $120$. This follows from transitivity as in the case $q=3$. We can map $\infty$ to $0$ with $\gamma$, so $\gamma$ followed by an appropriate element of $K$
maps $\infty$ to any other element of $\Omega$. Therefore the order of the group
$G=P\Gamma L_2(4)$ is at least $5\cdot|H|=120$. As $G$ was known to be a subgroup of $S_5$, the isomorphism $G\simeq S_5$ follows. |
H: solving differential equation: $y''-2y'\tan x=\frac{1}{\cos^3x}$
I'd really love your help with solving we following differential equation $$y''-2y'\tan x=\frac{1}{\cos^3x}.$$
First I tried to do it with $z=y'$ but it's just impossible,$z$ is a big and not nice expression, and to integrate it would be very hard problem. Then I though of Euler equations, but it's not in the correct form for doing it.
What should I do?
Thanks a lot!
AI: I'm not sure it's as hard as you're making it out to be. First, let's find an integrating factor. We want
$$py''-2py'\tan x=(py')'$$
$$-2p\tan x=p'$$
$$\frac p{p'}=-2\frac{\sin x}{\cos x}$$
$$\ln p=2\ln(\cos x)=\ln(\cos^2x)$$
$$p=\cos^2x$$
Multiplying through by our integrating factor, we get
$$y''\cos^2x-2y'\sin x\cos x=(y'\cos^2x)'=\sec x$$
$$y'\cos^2x=\ln(\sec x+\tan x)+C$$
$$y'=\sec^2x\ln(\sec x+\tan x)+C\sec^2x$$
$$y=\int\sec^2x\ln(\sec x+\tan x)dx+C\int\sec^2xdx$$
The second part is simply $C\tan x$. For the first integral, we'll use integration by parts. Obviously, we want that logarithm to go away, so that's the part we'll take the derivative of.
$$u=\ln(\sec x+\tan x),du=\sec x$$
$$dv=\sec^2xdx,v=\tan x$$
$$\int\sec^2x\ln(\sec x+\tan x)dx=\tan x\ln(\sec x +\tan x)-\int\sec x\tan xdx=$$
$$\tan x\ln(\sec x+\tan x)-\sec x$$
Putting everything together, we have
$$y=\tan x\ln(\sec x+\tan x)-\sec x+k_1\tan x+k_2$$ |
H: Example of an infinite sequence of irrational numbers converging to a rational number?
Are there any nice examples of infinite sequences of irrational numbers converging to rational numbers?
One idea I had was the sequence: $ 0.1001000010000001\cdots,0.1101000110000001\cdots,\cdots,0.1111000110000001\cdots,$ etc.
Where the first term in the sequence has ones in the place $i^2$ positions to the right of the decimal point. $(i=1,2,3,\dots)$ For the second term, we keep all the ones from the first term and add one in place of the first zero after the decimal point. We then add ones $i^3$ places after the decimal point. (The logical sum 1+1=1, i.e a number $i^2=j^3$ spaces after the decimal place has the value 1).
It is clear that this will converge to $1/9$, and I don't think the decimal expansion repeats at all.
AI: Pick any irrational number $\alpha$ you like, then consider the sequence $\{x_n=\alpha/n\}_{n=1}^\infty$. Then each term of the sequence $x_n$ is irrational and it converges to zero as $n$ tends to infinity. |
H: $\int_{\partial \Omega}\frac{\partial u}{\partial N}d\sigma $ if $\Delta u = 0$ in $\Omega$
Can you help me please with this problem?
What do you know about $\int_{\partial \Omega}\frac{\partial u}{\partial N}d\sigma $ if $\Delta u = 0$ in $\Omega$?
Is it always possible to solve an equation $u''=0$ in $[0,1]$ if $u'(0)$ and $u'(1)$ are given?(What is connection to first question?) Is the solution - unique?
Thanks!
AI: For the first one
From the divergence theorem for the vector field $ \text{grad} u$ we get that:
$\displaystyle{ \int_{\Omega} \text{div(grad u}) \nu = \int_{\partial \Omega} < \text{grad}u ,N> \sigma} \quad (\bigstar)$
where $\nu$ is the volume element.
Using now the ypothesis that $ 0=\Delta u := \text{div(grad u)}$ and that $\displaystyle{\frac{\partial u}{\partial N} := <\text{grad g ,N}>}$
The conclusion now follows from $(\bigstar)$. |
H: Volume of the rotation of the area between two curves
Suppose I have two function $f(x)$ and $g(x)$ such that for $x \in (\alpha, \beta )$ we have $f(x) \ge g(x) $.
I found an "exercise solution" that state that the volume given by the rotation of the area between $f$ and $g$ is:
$$\pi\int_\alpha^\beta (f(x) - g(x))^2 dx$$
but i think it's wrong and that it should be:
$$\pi\int_\alpha^\beta f(x)^2 - g(x)^2 dx$$
Who is right?
AI: I'll assume that one of the candidates is the correct solution. Now, what is the volume of a cylindrical shell of radii $R>r$ and height $1$? It is the difference
$$
\pi R^2 - \pi r^2 = \pi \left(R^2-r^2 \right).
$$
This does not coincide with $\pi \left( R-r \right)^2$. |
H: linear algebra: inverse of a matrix
The inverse of the matrix
$A=\left( \matrix{1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} }\right)$
is
$A^{-1}=\left( \matrix{ 9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180 } \right)$.
Then, perhaps the matrix
$B=\left( \matrix{1 & \frac{1}{2} & ... & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & ... & \frac{1}{n+1} \\ ... & ... & & ...\\ \frac{1}{n} & \frac{1}{n+1} & ... & \frac{1}{2n-1} }\right)$
is invertible and $B^{-1}$ has integer entries. How can I prove it?
AI: Here you will find your answer and many other things about Hilbert matrices : http://www.jstor.org/stable/2975779 |
H: How could I calculate this sum of series
Could you please help me to calculate this finite sum ?
\begin{align}
S=\sum _{n=1}^{\text{Nmax}} n*P^n
\end{align}
where $P\leq 1$ and $\text{Nmax}>1$.
AI: $$S=P.\sum_{n=1}^N nP^{n-1}=P.\frac{d}{dP}\sum_{n=1}^NP^n=P.\frac{d}{dP}\frac{P(1-P^N)}{1-P}$$ So $$S=\frac{(1-P)(1-(N+1)P^N)+P(1-P^N)}{(1-P)^2}$$
Simplify to get the answer.
Another way to do this ( and this does not need derivatives ) is
$$S=1.P+2.P^2+3.P^3\cdots +(N-1)P^{N-1}+N.P^N$$
Now $$S.P=1.P^2+2.P^3\cdots+(N-1)P^N+(N+1)P^{N+1}$$
So $$S-SP=P+P^2+P^3+\cdots+P^N+(N+1)P^{N+1}$$
Which implies $$S(1-P)=\frac{P(1-P^N)}{1-P}+(N+1)P^{N+1}$$
Now simplify to get the value of $S$ |
H: How to find the product of n terms of an arithmetic progression where common difference is not unity?
How can we find the product of $n$ terms of an arithmetic progression where common difference is not unity?
I just want to know the last $3$ digits of $7 \times 23 \times 39 \times \ldots \times 2071$ where common difference is $16$.
AI: There are two (unrelated) questions here. An answer to the first question is
$$\prod_{k=1}^n(ak+b)=a^n\frac{\Gamma\left(n+1+\frac{b}a\right)}{\Gamma\left(1+\frac{b}a\right)}$$
To answer the second question, consider $n=\prod\limits_{k=0}^{129}(7+16k)$. Note that $7+16k=0\pmod{5}$ for every $k=3\pmod{5}$. Using this for $k=3$, $k=8$ and $k=13$ yields $n=0\pmod{125}$, that is, $n=125m$ for some integer $m$.
Likewise, $7+16k=-1\pmod{8}$ and the number of terms in the product which defines $n$ is $130$, which is even, hence $n=1\pmod{8}$. And $125=5\pmod{8}$ hence $n=5m\pmod{8}$, which implies $m=5n\pmod{8}$, that is, $m=5\pmod{8}$.
Finally, $n=125\cdot5\pmod{125\cdot8}$, that is, $n=\underline{\ \ \ \ \ \ \ \ }\pmod{1000}$. |
H: Deriving the Formula for Average Speed (Same distance).
Let me start of by specifying the question:
A and B are two towns. Kim covers the distance from A to B on a scooter at 17Km/hr and returns to A on a bicycle at 8km/hr.What is his average speed during the whole journey.
I solved this problem by using the formula (since the distances are same):
$$ \text{Average Speed (Same distance)} = \frac{2xy}{x+y} = \frac{2\times17\times8}{17+8} =10.88 \text{Km/hr}$$
Now I actually have two questions:
Q1- I know that $$ Velocity_{Average}= \frac{\Delta S }{\Delta T} $$
Now here does $$\Delta S$$ represent $$ \frac{S_2+S_1 }{2} \,\text{or}\, S_2-S_1 ?$$
Where S2 is the distance covered from point A to point B and S1 is the distance covered from point B to point A
Q2. How did they derive the equation:
$$ Velocity_{Average(SameDistance)} = \frac{2xy}{x+y} $$
Could anyone derive it by using
$$ Velocity_{Average}= \frac{\Delta S }{\Delta T} $$
AI: If one traveled distance $d_k$ at speed $v_k$, this took time $t_k=\dfrac{d_k}{v_k}$. It took time $T=\sum\limits_kt_k$ to travel distance $D=\sum\limits_kd_k$ and the average speed $V$ solves $D=VT$, hence $$V=\frac{\sum\limits_kd_k}{\sum\limits_k\frac{d_k}{v_k}}.
$$
In the particular case when there are $n$ distances which are all equal, one gets $V=\dfrac{n}{\sum\limits_{k=1}^n\frac1{v_k}}$, or
$$\frac1V=\frac1n\sum\limits_{k=1}^n\frac1{v_k}.
$$ |
H: Fixed Block is an orbit?
Reviewing some of my old questions here, I am stuck at a comment in which Prof. Holt gave me an interesting example (A small one) about non-transitive $1/2-$transitive group. Here is the link {http://math.stackexchange.com/q/138937/8581}. Of course, he noted a complete answer by giving two non-trivial of these kinds of groups.
Meanwhile, it came to my mind if we could find any fixed block in this small group, $G=\{1\}$ acting on $\Omega=\{1,2\}$ which is not an orbit? I see that we get $\Omega^G=\Omega$ in this example and so, $\Omega$ itself is a fixed block but not an orbit of the action.
In fact, $1^G=\{1\}$ and $2^G=\{2\}$. That $\Omega^G=\Omega$ is achieved, is obvious so I am trying to find any other proper subset of $\Omega$.
Can someone give me other example in which we have a fixed block in any group actions that is not an orbit? Thanks.
Let a group $G$ acts on a set $\Omega$ and $\Delta⊆\Omega$. $\Delta$ is said to be a Fixed Block of $G$ if for any $g\in G, \Delta^g=\Delta$.
AI: Let $G$ act on a set $\Omega$. Suppose $\Delta\subseteq\Omega$ is a subset invariant under the $G$-action. i.e. $g\Delta=\Delta$ for any element $g\in G$. If $x\in\Delta$ then $gx\in\Delta$ for any $g$, so the orbit $Gx\subseteq \Delta$, plus $x\in Gx$. Therefore
$$\Delta=\bigcup_{x\in\Delta}Gx$$
is a union of orbits. To find a fixed block that is not itself an orbit, one must find two or more disjoint orbits and take their union. If the $G$-set $\Omega$ only has two orbits, then the only non-orbit fixed block is the set $\Omega$ itself, i.e. it has no proper subset fixed blocks. In particular, the trivial group acting on a two element set has precisely two orbits.
(P.S. A trivial example of a proper subset fixed block: $G=1,~\Omega=\{1,2,3\},~\Delta=\{1,2\}$.) |
H: A form of cumulative distribution
Let $f(x)$ and $g(x)$ be two probability density functions. Does the expression:
$$
C = 2\int _{-\infty}^{\infty}\left[f(x)\int _{-\infty}^{x}g(y)\,dy\right]\,dx
$$
have any meaningful graphical representation when $E[Y]>E[X]$, where $X$ has $pdf$ $f(x)$ and $Y$ has $pdf$ $g(x)$? Or, can it be expressed in a simpler fashion?
From some numerical simulations, it seems to approximately describe the "overlap" area of the two $pdf$'s. And if we let $g(x)==f(x)$ then $C$ is close to 1...
The overlap is defined as,
$$\int_{-\infty}^{\infty} \min(f(x),g(x)) dx $$
AI: If $X$ and $Y$ are independent continuous random variables with densities $f(x)$ and $g(y)$ then the probability that $Y$ is less than or equal to $X$ is
$$\Pr (Y \le X) = \int _{x=-\infty}^{\infty}[f(x)\int _{y=-\infty}^{x}g(y)dy]dx$$ and graphically is the probability of being below the line $y=x$.
$C$ is twice this. If $g(x)=f(x)$ for all $x$ then it is the probability of being above or below the line and (ignoring the zero measure of being on the line) is $1$. |
H: solve ODE solution with intial value
Consider the initial value problem $y' + \frac{2}{3}y = 1-\frac{1}{2}t, y(0) = y_0$ Find the value of $y_0$ for which the solution touches, but does not cross, the $t$-axis. Solving this equation with $\mu(t)=e^{\frac{2}{3}t}$ we get $y=\frac{21}{8}- \frac{3}{4}t + C e^{-\frac{2}{3}t}$, using $y(0)=y_0$ we get, $C=y_0-\frac{21}{8}$, so the solution that passes through $(0, y_0)$ is $y=\frac{21}{8}- \frac{3}{4}t + (y_0-\frac{21}{8}) e^{-\frac{2}{3}t}$, this solution has slope zero when it just touches the $t$-axis, so $y'=-\frac{3}{4}+(y_0-\frac{21}{8})e^{-\frac{2}{3}t}.(-\frac{2}{3})=0$, using $y(t)=0$ we have $(y_0-\frac{21}{8}) e^{-\frac{2}{3}t}=\frac{3}{4}t-\frac{21}{8}$ and using $y'=0$ we have $(y_0-\frac{21}{8})e^{-\frac{2}{3}t}=\frac{9}{8}$ therefore, $\frac{3}{4}t-\frac{21}{8}=\frac{9}{8} \Rightarrow t=5$ using $(y_0-\frac{21}{8})e^{-\frac{2}{3}(5)}=\frac{9}{8}$ we get $y_0=34.18$, but the answer is $y_0=-1.6428$
AI: The solution is
$$
y(t)=\frac{21}{8}-\frac{3}{4}t + \left( y_0 - \frac{21}{8} \right)e^{-\frac{2}{3}t}.
$$
The two conditions $y(t)=0$ and $y'(t)=0$ read
$$
\begin{cases}
\frac{21}{8}-\frac{3}{4}t + \left( y_0-\frac{21}{8} \right) ^{-\frac{2}{3}t} =0 \\
-\frac{3}{4}-\frac{2}{3} \left( y_0-\frac{21}{8} \right)e^{-\frac{2}{3}t}=0.
\end{cases}
$$
By substitution,
$$
-\frac{3}{4}-\frac{2}{3} \left( \frac{3}{4}t-\frac{21}{8} \right)=0,
$$
and $t=2$. Then
$$
y_0 = \frac{3}{8} \left(7-3 e^{\frac{4}{3}} \right) \approx -1.64288
$$ |
H: A lemma about extension of function
Definition
Suppose that $f(M)$ is a $\mathcal C^n$-function whose domain is $\mathcal X$. If $f^*(M)$ is a $\mathcal C^n$-function whose domain is $\mathcal X^*$, and $f(M)=f^*(M)$ whenever $M\in\mathcal X\cap\mathcal X^*$, we call that $f^*$ is an ($\mathcal C^n$-)extension of $f$, and $f$ is ($\mathcal C^n$-)extended into $\mathcal X^*$.
Lemma
Given that $f(x,y)$ is $\mathcal C^n$ ($n\ge1$) function on some bounded open set $\mathcal M\subset\Bbb R^2$, whose boundary is $\mathcal L$, and for each point on $\mathcal L$, there's a neighborhood into which $f$ could be $\mathcal C^n$-extended. We conclude that $f$ could be extended into $\Bbb R^2$.
Source
Григорий Михайлович Фихтенгольц
I found that the proof on the book was so complicated for me to understand, so I'm looking for some explanation, as intuitive as possible. Can anyone help me? Thanks a lot!
AI: As $\bar M=M\cup L$ is compact I suggest working with a so called partition of unity:
For each $p\in \bar M$ there is an $\epsilon>0$ (depending on $p$) such that $f$ can be smoothly extended to $U_{3\epsilon}(p)$. The family $\bigl(U_\epsilon(p)\bigr)_{p\in L}$ is an open covering of $\bar M$; therefore there exist points $p_k\in \bar M$ $\ (1\leq k\leq N)$ such that $$\bar M\subset\Omega:=\bigcup_{k=1}^N U_{\epsilon_k}(p_k)\ .$$
For each $k\in[N]$ there is a $C^n$- (even a $C^\infty$-) function $\phi_k: {\mathbb R}^2\to [0,1]$ with $\phi_k(z)\equiv1$ for $z\in U_{\epsilon_k}(p_k)$ and $\phi_k(z)\equiv0$ outside $U_{2\epsilon_k}(p_k)$. (First one has to construct once and for all a $C^n$-function $\chi$ which is $\equiv1$ for $0\leq t\leq1$ and $\equiv0$ for $t\geq2$. Then put $\phi_k(z):=\chi\bigl({|z-p_k|\over\epsilon_k}\bigr)$. )
For technical reasons we need the auxiliary function
$$\phi_*(z):=\prod_{k=1}^N \bigl(1-\phi_k(z)\bigr)\in[0,1]\ .$$
It is $\equiv0$ on $\Omega\supset \bar M$ and $=1$ in all points $z$ where the $\phi_k$ simultaneously vanish. Now put
$$\psi_k(z):={\phi_k(z)\over \phi_*(z)+\sum_{k=1}^N\phi_k(z)}\qquad(1\leq k\leq N)\ .$$
Each $\psi_k:{\mathbb R}^2\to[0,1]$ is smooth and $\equiv0$ outside $U_{2\epsilon_k}(p_k)$, and what is essential: The $\psi_k$ sum to $1$ on $\Omega$.
For each $k$ choose an extension $f_k$ of $f$ to $U_{3\epsilon_k}(p_k)$ and put
$$g_k(z):=\cases{\psi_k(z)f_k(z) &$\bigl (z\in U_{3\epsilon_k}(p_k)\bigr)$\cr 0 & (otherwise)\cr}\ .$$
The $g_k$ are smooth in all of ${\mathbb R}^2$, and one easily checks that
$$f_*(z):= \sum_{k=1}^N g_k(z)$$ is a smooth extension of $f$ to all of ${\mathbb R}^2$. |
H: Calculation of $ \frac{\partial^2 g(f) }{\partial x_1 \partial x_2} $
Let $f : \mathbb R^2 \to \mathbb R$. $ g : \mathbb R \to \mathbb R$. Assume that all partial derivatives exist. Then is this statement right? $$ \frac{\partial^2 g\circ f }{\partial x_1 \partial x_2} = \frac{\partial}{\partial x_1} \left( \frac{\partial g}{\partial f} \frac{\partial f}{\partial x_2} \right) = \frac{\partial g}{\partial f} \frac{\partial^2 f}{\partial x_1 \partial x_2} + \frac{\partial^2 g}{\partial f^2} \frac{\partial f}{\partial x_1} \frac{\partial f}{\partial x_2} $$
AI: While the equation is in principle correct the notation you are using is kind of unusual (I did see it before, though), which may cause irritations, as you may infer from Siminore's comment. I'd prefer (abbreviating $x=(x_1, x_2))$ one (any) of the following:
$$ \frac{\partial^2 (g(f(x)))}{\partial x_1 \partial x_1} = \frac{\partial^2 (g\circ f)}{\partial x_1 \partial x_1} (x) $$
for the left hand side of your identity and assuming $g=g(u)$
$$ \frac{dg}{du}(f(x))\frac{\partial^2 f(x)}{\partial x_1 \partial x_1} + \frac{d^2g}{du^2}(f(x)) \frac{\partial f(x)}{\partial x_1} \frac{\partial f(x)}{\partial x_2} $$
or $$ g^{\prime}(f(x))\frac{\partial f^2}{\partial x_1 \partial x_1} (x)+ g^{\prime \prime}(f(x)) \frac{\partial f}{\partial x_1}(x) \frac{\partial f}{\partial x_2} (x) $$
for the right hand side. However only if $g$ depends only on one variable. In that case $\frac{\partial g}{\partial u}$ is not wrong but not common. There are other notation conventions in use. |
H: Finding area of triangle
if the sides of the triangle are given by 20 cm, 30 cm, and 60 cm find the area of the triangle.
I tried a long time.
Apparently, Heron's formula does not seem to work
$\sqrt{s(s-a)(s-b)(s-c)}$
where $s = (a+b+c)/2$
In the above problem $s=55$ and thus we end up with a negative number inside square root. I am not sure if there is any other formula to be applied to this problem .
AI: Heron's formula works if the triangle exists. This triangle does not exist. By the triangle inequality applied to this triangle, we should have $60\le20+30$, which is false.
If the triangle inequality holds, then each of $s-a=\frac{b+c-a}{2}$, $s-b=\frac{a+c-b}{2}$, and $s-c=\frac{a+b-c}{2}$ is non-negative and so $s(s-a)(s-b)(s-c)\ge0$. |
H: Show that $6m \mid (2m+3)^n + 1$ if and only if $4m \mid 3^n + 1$
Let $m$ and $n$ be positive integers. How to prove that $$6m \mid (2m+3)^n + 1$$ if and only if $$4m \mid 3^n + 1$$
AI: $\newcommand{\jaco}[2]{{\left(\frac{#1}{#2}\right)}}$I have tried to solve this - I hope I did not make too many mistakes there. Maybe there is much easier solution than this one.
Let us start by a few easy observations.
Since $2m+3 \equiv 3 \pmod m$, we see that
$$m\mid (2m+3)^n+1 \Leftrightarrow m\mid 3^n+1. \tag{1}$$
It is obvious that
$$3\mid (2m+3)^n+1 \Rightarrow 3\nmid m. \tag{2}$$
Since $3\nmid x$ implies $x^2\equiv 1 \pmod 3$ (Little Fermat's theorem), we get (using this fact for $x=2m+3$) that
$$3\mid (2m+3)^n+1 \Rightarrow n\text{ is odd.}\tag{3}$$
Now we get
$$3\mid (2m+3)^{n}+1 \Rightarrow m\equiv1\pmod3, \tag{4}$$
since $m\equiv2\pmod3$ $\Rightarrow$ $2m+3\equiv1\pmod3$ $\Rightarrow$ $(2m+3)^n\equiv1\pmod3$ $\Rightarrow$ $(2m+3)^n+1\equiv2\pmod3$.
It is easy to see that
$$4\mid 3^n+1 \Leftrightarrow n\text{ is odd}\tag{5}$$
and
$$m\mid 3^n+1 \Rightarrow 3\nmid m.\tag{6}$$
For odd $n=2k+1$ we have $3^{2k+1}+1=(3+1)(3^{2k}-3^{2k-1}+\dots+3^2-3+1)$, which shows that $$\frac{3^{2k+1}+1}4\equiv 1 \pmod 2.$$ Hence
$$4m\mid 3^n+1 \Rightarrow m\text{ is odd.}\tag{7}$$
Now a few more facts with a little less elementary proof.
We will use Legendre symbol and quadratic reciprocity.
We first show:
$$4m\mid 3^n+1 \Rightarrow m\equiv1 \pmod 3.\tag{8}$$
We know that $n$ is odd, let $n=2k+1$. It suffices to show that each odd prime factor $p$ of $3^{2k+1}+1$ fulfills $p\equiv1\pmod 3$.
Suppose that $p\mid 3^{2k+1}+1$ and, consequently $p\mid 3^{2(k+1)}+3$. Therefore $-3$ is a quadratic residue modulo $p$ and $\jaco{-3}p=1$.
We get:
$$\jaco{-3}p=\jaco{-1}p\jaco3p$$
$$1=(-1)^{\frac{p-1}2}(-1)^{\frac{p-1}2}\jaco{p}3.$$
Hence $\jaco{p}3=1$, which implies $p\equiv1\pmod{3}.$
Now we are ready to show that $6m\mid (2m+3)^n+1$ $\Leftrightarrow$ $4m\mid 3^n+1$.
$\boxed{\Rightarrow}$ If $6m\mid (2m+3)^n+1$ then:
$\gcd(m,4)=1$ by (7)
$n$ is odd by (3)
$4\mid 3^n+1$ by (5)
$m\mid 3^n+1$ by (1)
Together we have $4m\mid 3^n+1$.
$\boxed{\Leftarrow}$ If $4m\mid 3^n+1$ then:
$n$ is odd by (5)
$\gcd(m,6)=1$ by (6), (7)
$m\mid (2m+3)^n+1$ by (1)
$2\mid (2m+3)^n+1$ since $2m+3$ is odd
$3\mid (2m+3)^n+1$ since $m\equiv1 \pmod 3$ by (8), which yields $2m+3\equiv 2\pmod 3$ and $(2m+3)^n\equiv2\pmod3$ (using the fact that $n$ is odd).
Together we have $2\cdot 3 \cdot m \mid (2m+3)^n+1$. |
H: APB is congruent to DPC. Find $x$
I wrote in $x+23$ for DPC but then I'm stuck heres the image since I can't embed:
APB is congruent to DPC.
APB is $x+23$
BPC is $x$
BPD = $4x - 39$
Which can be seen here:
AI: HINT: BPC + CPD = BPD. So we have that $2x + 23 = 4x - 39$. Solve for $x$. |
H: Failure of uniqueness for linear ODE
Let's say we have a linear ODE with polynomial coefficients $p_j(x)$:
$$
p_n(x) y^{(n)}(x)+\dots+p_1(x)y'(x)+p_0(x)y=0
$$
and let's say $x_0$ is a root of $p_n(x)$. What can be said about uniqueness of the IVP for this ODE at $x_0$? In particular, if I succeeded to prove that an analytic solution $y_0(x)$ satisfies $y_0(x_0)=y'(x_0)=\dots=y^{(n-1)}(x_0)=0$, does it necessarily follow that $y_0(x)\equiv 0$?
AI: No. Consider Cauchy problem $xy'-y=0\;$, $y(0)=0\;$. It has solutions $y=Cx\;$, $\ C\in\mathbb R\;$. |
H: A problem about a matrix in $SL_2(\mathbb Q)$
If a matrix $A\in SL_2(\mathbb Q)$ has finite multiplicative order, then $\operatorname{tr}(A)\le 2$. Does anyone know a demonstration of this fact?
AI: Using the triangle inequality, if $|\lambda_1|,|\lambda_2|\le1$ then $|\mathrm{tr}(A)|=|\lambda_1+\lambda_2|\le|\lambda_1|+|\lambda_2|\le2$.
If $\lambda,v$ is an eigenpair* of a matrix $A$, and $A^n=I$, then $A^nv=\lambda^nv=Iv=v$, hence $\lambda^n=1$, which implies $|\lambda|=1$ (in particular it is an $n$th root of unity). This applies to all $\lambda$, so $|\mathrm{tr}(A)|\le2$.
Note this applies to all of $\mathrm{GL}_2(\Bbb C)$, not just $\mathrm{SL}_2(\Bbb Q)$, and bounds the trace in all of $\Bbb C$ instead of one side of the real line. More generally, if $A\in\mathrm{GL}_m(\Bbb C)$ satisfies $A^n=I$, then $|\mathrm{tr}(A)|<m$ except when $A=zI$ with $z$ is an $n$th root of unity (think geometrically!). Also, $(\det A)^n=1$.
*All eigenvalues of a matrix have a corresponding eigenvector. For if $\lambda$ satisfies $\det(\lambda I-A)=0$, then $\lambda I-A$ must be singular, hence its image is a proper subspace aka nontrivial cokernel, and by the rank-nullity theorem $\lambda I-A$ has nontrivial kernel; any element of the kernel is an eigenvector. |
H: What is the difference between the symbols $\cap$ and $\setminus$?
What is the difference between $\cap$ and $\setminus$ symbols for operations on sets?
AI: Their definition is different:
$A\cap B=\{x\mid x\in A\text{ and } x\in B\}$, we take all the elements which appear both in $A$ and in $B$, but not just in one of them.
$A\setminus B=\{x\mid x\in A\text{ and } x\notin B\}$, we take only the part of $A$ which is not a part of $B$.
Amongst the different properties, the intersection ($\cap$) is commutative and associative while difference ($\setminus$) is not. Namely it is generally true that:
$$A\cap B=B\cap A\\ A\setminus B\neq B\setminus A$$
and similarly:
$$A\cap (B\cap C) = (A\cap B)\cap C\\ A\setminus(B\setminus C)\neq (A\setminus B)\setminus C$$ |
H: To prove an elementary statement
I have an elementary doubt, Sorry for disturbing you all. I have a statement of this sort. $$r^2-1=p^a(f(p))=(r+1)(r-1). \tag{1}$$
Where $r$ is an even number, and $p$ is an odd prime. $f(p)$ is a degree $n$ polynomial in $p$ with integer coefficients ( Courtesy : André Nicolas ) , $a \in \mathbb{Z}$ is a variable $>0$.
Now from the statement $(1)$ can we infer that $p^a \mid (r+1)(r-1)$ ?.
If so, $p$ being a prime, can't divide both of them (from Euclid's theorem ) . So $p^a$ divides either
$(r+1)$ or $(r-1)$.
So then can we conclude that $r=kp^a \pm 1 $ for some $k>0 $?.
I told the same thing with some professor, and he is saying that it is false, and we can't write $$[\ r^2-1=p^a(f(p))=(r+1)(r-1) \ ]\implies r=kp^a \pm 1.$$
He gave me a counter example , which is $8\cdot6=48=2^4\cdot3$. He is now saying that my statement is like saying either $8 \text{ or } 6 $ divides $3$ which is utterly false.
So my learned elders in this community, please judge who is correct , me or the professor ?.
In any of the case, please do provide some good explanation.
Thank you.
AI: Your statement is correct. $p^a$ divides either $r+1$ or $r-1$, so either there exists $k>0$ such that $kp^a = r+1 \Rightarrow r = k p^a-1$, or else there exists $k$ such that $kp^a=r-1 \Rightarrow kp^a+1$.
The "counterexample" given is not applied properly. Here $r=7$ and $p^a$ should be a power of an odd prime that divides $48$. In this case $3^1$ is the only choice, and your claim is that $7=k3^1\pm 1$ for some $k>0$, which is clearly satisfied by $k=2$. |
H: What is the usual meaning of the notation $ \| f \|_{C^k (A)}$?
If $f \in C^k ( \mathbb R)$ and $A \subset \mathbb R$, what is the usual meaning of the notation below?
$$ \| f \|_{C^k (A)} $$
AI: Assuming you meant $A\subseteq\mathbb{R}$, the notation usually means
$$\|f\|_{C^k(A)}=\|f\|+\|f'\|+\cdots+\|f^{(k)}\|$$
where $f^{(i)}$ is the $i$th derivative of $f$, and where for a function $g:A\to\mathbb{R}$, the notation $\|g\|$ refers to the sup norm:
$$\|g\|=\sup_{x\in A} \,|g(x)|$$
I've also seen the definition
$$\|f\|_{C^k(A)}=\max\left\{\|f\|,\|f'\|,\ldots,\|f^{(k)}\|\right\}$$
which (if I remember correctly) induces the same topology. |
H: What is the difference between direct product and direct sum of a finite number of group representations.
I have been reading in Fulton & Harris's book on representation theory and it talks about things like the decomposition of a direct product of representations $ V \otimes V $ into a direct sum of representations. It seems to me there is a real difference between a finite direct sum of representations and a direct product even though they agree on vector spaces. Is this difference real and if so , what is it?
AI: $\otimes$ denotes the tensor product, not the direct product. This is different even for vector spaces; the tensor product of vector spaces of dimensions $m, n$ has dimension $mn$ rather than $m+n$ for the direct product or direct sum. |
H: The Determinant of Transition Matrices Between Orthonormal Bases
I am trying to show that if $A$ and $B$ are orthonormal bases of a real finite dimensional vector space $V$ and $P$ is the change of basis matrix between $A$ and $B$ then $\det(P) = \pm 1$. Here is my progress:
When working on this problem, I discovered the following fact: If $G_B$ denotes the Gram matrix of the inner product relative to $B$ and $G_A$ denotes the Gram matrix of the inner product relative to $A$ then the Gram matrices and the transition matrix $P$ are related through
$$
G_B = P^T G_A P.
$$
The original claim is then a corollary of this fact:
$$
G_B = P^T G_A P \implies \det(G_B) = \det(P^T) \det(G_A) \det(P) \implies \det(P) = \pm 1
$$
where I have used basic properties of the determinant and the fact that the determinant of the Gram matrix relative to an orthonormal basis is $1$.
So, I think this proves the original claim but I am now stuck with proving the more general claim that $G_B = P^T G_A P$. What would be a good way to approach this?
AI: $$\begin{eqnarray*}
(P^T G_A P)_{il} &=& (P^T)_{ij} (G_A)_{jk} P_{kl} \\
&=& P_{ji} \langle x_k,x_j\rangle P_{kl} \\
&=& \langle P_{kl} x_k, P_{ji} x_j\rangle \\
&=& \langle (P^T x)_l, (P^T x)_i\rangle \\
&=& (G_B)_{il}
\end{eqnarray*}$$
$\def\det{\mathrm{det}\,}$
Addendum: On second reading I may have misinterpreted the meaning of "this" in the last sentence of your question.
Let $X_{ij} = (x_i)_j$ be the matrix whose rows are the orthonormal basis vectors.
$X$ transforms like $X\to P^T X$.
We have $X X^T = \mathbb{I}$, since the basis is orthonormal.
Then
$$(P^T X)(P^T X)^T = P^T X X^T P = P^T P = \mathbb{I},$$
since the new basis is orthonormal.
Therefore, $\det(P^T P) = (\det P)^2 = 1$, so $\det P = \pm 1$. |
H: linear algebra linear transformation
I would really appreciate if someone could help me with this question, I know that you have to row reduce each matrix and put it in vector parametric form to find $U_1$ and $U_2$. any help would be much appreciated. thank you
Define a real linear transformation $L_1 : \mathbb R^4 \to \mathbb R^2$ by
$L_1(x_1; x_2; x_3; x_4) = (3x_1 + x_2 + 2x_3 - x_4; 2x_1 + 4x_2 + x_3 - x_4)$
and let $U_1$ denote the kernel of $L_1$
Define a real linear transformation $L_2 : \mathbb R^2 \to \mathbb R^4$ by
$L_2(y_1; y_2) = (y_1 - y_2; y_1 - 3y_2; 2y_1 - 8y_2; 3y_1 - 27y_2)$
and let $U_2$ denote the image of $L_2$
Construct bases for $U_1$, $U_2$, ${U_1 \cap U_2}$, and, $U_1 + U_2$
AI: You can find a basis for $U_1$ simply by solving the linear system
$$
\begin{align}
3x_1 + x_2 + 2x_3 - x_4&=0\\
2x_1 + 4x_2 + x_3 - x_4&=0
\end{align}$$
Of course, there are many ways to solve this system, e.g. we can try to use elementary row operations:
$$
\begin{pmatrix}
3 & 1 & 2 & -1 \\
2 & 4 & 1 & -1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & -3 & 1 & 0 \\
2 & 4 & 1 & -1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & -3 & 1 & 0 \\
1 & 7 & 0 & -1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & -3 & 1 & 0 \\
-1 & 7 & 0 & 1 \\
\end{pmatrix}
$$
Now we see that the original system is equivalent to the the system of equations $x_1-3x_2+x_3=0$, $-x_1+7x_2+x_4=0$; i.e.
$x_3=-x_1+3x_2$, and $x_4=x_1-7x_2$. We see that for any choice of $x_1$, $x_2$ we obtain a solution of the form
$$(x_1,x_2,-x_1+3x_2,x_1-7x_2)=x_1(1,0,-1,1)+x_2(0,1,3,7),$$
which means that $U_1=[(1,0,-1,1),(0,1,3,7)]$. I.e., these two vectors are the basis for $U_1$. (Of course, there are many different basis for the same subspace. So might get different vectors. It is also useful to check whether the vectors in basis fulfill the original system - if not, there must be a mistake somewhere.)
The space $U_2$ is generated by vectors $L_2(1,0)$ and $L_2(0,1)$, hence $U_2=[(1,1,2,3),(-1,-3,-8,-27)]$. These vectors are linearly independent, so they form a basis. (To check whether two vectors are independent you only need to check whether one of them is a multiple of the other one.)
Now let us try $U_1+U_2$. We know that $U_1+U_2=[(1,0,-1,1),(0,1,3,7),(1,1,2,3),(-1,-3,-8,-27)]$, but we don't know whether these vectors are independent. So we try to make row echelon form.
$$
\begin{pmatrix}
1 & 0 & -1 & 1\\
0 & 1 & 3 & 7\\
1 & 1 & 2 & 3\\
-1 &-3 &-8 &-27
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 1\\
0 & 1 & 3 & 7\\
0 & 0 & 0 & 1\\
-1 &-3 &-8 &-27
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 0\\
0 & 1 & 3 & 0\\
0 & 0 & 0 & 1\\
-1 &-3 &-8 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 0\\
0 & 1 & 3 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{pmatrix}.$$
Now we know that vectors $(1,0,-1,0)$, $(0,1,3,0)$, $(0,0,0,1)$ form a basis of $U_1+U_2$ and that $d(U_1+U_2)=3$.
Since $d(U_1)+d(U_2)=d(U_1+U_2)+d(U_1\cap U_2)$, we know that $d(U_1\cap U_2)=1$.
If we notice that $(2,4,10,30)=(1,1,2,3)+(1,3,8,27)$ and $2(1,0,-1,1)+4(0,1,3,7)=(2,4,10,30)$, we see that the vector $(1,2,5,10)$ belongs to $U_1\cap U_2$. Since $U_1\cap U_2$ is one-dimensional, this implies that $U_1\cap U_2=[(1,2,5,10)]$.
Now let's try to think how we could find the basis for $U_1\cap U_2$ without guesswork.
One possibility is to have a closer look at what we have done when doing row operations. If we denote the rows of the above matrix by $\vec a_1$, $\vec a_2$, $\vec b_1$, $\vec b_2$, then we have can find out (using row operation similar to the above) that:
$\vec a_1+\vec a_2-\vec b_1=(0,0,0,5)$
$\vec a_1+3\vec a_2+\vec b_2=(0,0,0,-5)$
Obviously by combining the above two vectors we get $2\vec a_1+4\vec a_2+\vec b_2-\vec b_1=\vec 0$, which is the same as
$$2\vec a_1+4\vec a_2 = \vec b_1-\vec b_2,$$
hence this vector belongs to the intersection.
The above was still based on the lucky fact that we only had to find one vector. Can we approach this in a more systematic way?
Of course we can. To find vectors in $U_1\cap U_2$ we can solve the linear system of equations given by $x_1\vec a_1+x_2 \vec a_2=y_1\vec b_1+y_2 \vec b_2$. Once we know all possible pairs $(x_1,x_2)$, we can obtain all vectors belonging to $U_1\cap U_2$.
Based on your comment I've guess you got the following when solving the system:
$$
\begin{pmatrix}
3 & 1 & 2 & -1 \\
2 & 4 & 1 & -1 \\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & 7/10 & -3/10 \\
0 & 1 & -1/10 & -1/10
\end{pmatrix}
$$
From the last matrix you can get that the solutions of this system form the subspace
$U_1=[(-7/10,1/10,1,0),(3/10,1/10,0,1)]=[(-7,1,10,0),(3,1,0,10)]$.
This is another basis for the same subspace $U_1$. (And this is also correct.) |
H: Moment Generating Function for Sum of Independent Random Variables
I'm taking a graduate course in probability and statistics using Larsen and Marx, 4th edition and looking specifically at estimation methods this week. I ran into a homework problem that is related to moment generating functions and I can't quite connect the dots on how they arrived at the solution.
If you have three independent random variables $$Y_{1}, Y_{2}, Y_{3}$$ and you would like to determine the moment-generating function of $$W = Y_{1} + Y_{2} + Y_{3}$$ knowing that each of the three independent random variables have the same pdf $$f_{y} = \lambda y e^{-\lambda y}, y \geq 0$$
The easy part of the this problem is applying the theorem that says for $$W = W_{1} + W_{2} + W_{3}$$ the moment generating function of the sum is: $$M_{W}(t) = M_{W_{1}}(t)* M_{W_{2}}(t)* M_{W_{3}}(t)$$
Where I run into trouble is getting the individual moment generating functions for the Y's. The problem directs you to apply yet another theorem where you would let, for example, another random variable V equal to $$aY_{1}+b$$ and it follows that $$M_{V}(t) = e^{bt}M_{W}(at)$$
The solution states that if you allow $$V = (1/\lambda)*W$$ then the pdf of V then becomes $$f_{V}(y) = ye^{-y}, y \geq 0$$ and subsequently, you can get the moment generating function using a simple integration by parts but I can't quite follow the application of the theorem used to get to the pdf of V.
Any insight? Likely a fundamental property I missed along the way...
AI: You have not made use of the definition of a moment generating function. The moment generating function for any random variable $X$ is usually defined as
$$M_X(t) = \mathbb{E} \left( e^{tX} \right)$$
EDIT
Adding more details.
First, your $f_y$ is incorrect. It should be $f_Y(y) = \lambda e^{- \lambda y}$. We get $$M_Y(t) = \displaystyle \int_0^{\infty} e^{ty} \lambda e^{-\lambda y} dy = \displaystyle \int_0^{\infty} \lambda e^{(t-\lambda) y} dy = \dfrac{\lambda}{\lambda-t}$$
Hence, the moment generating function for $Y = Y_1 + Y_2 + Y_3$ is $M_Y(t) = \dfrac{\lambda_1}{\lambda_1-t} \dfrac{\lambda_2}{\lambda_2-t} \dfrac{\lambda_3}{\lambda_3-t}$ |
H: Have there been efforts to introduce non Greek or Latin alphabets into mathematics?
As a physics student, often I find when doing blackboard problems, the lecturer will struggle to find a good variable name for a variable e.g. "Oh, I cannot use B for this matrix, that's the magnetic field".
Even ignoring the many letters used for common physical concepts, it seems most of the usual Greek and Latin letters already have connotations that would make their usage for other purposes potentially confusing, for instance one would associate $p$ and $q$ with integer arguments, $i,j,k$ with indices or quaternians, $\delta$ and $\varepsilon$ with small values, $w$ with complex numbers and $A$ and $B$ with matrices, and so forth.
It then seems strange to me that there's been no effort to introduce additional alphabets into mathematics, two obvious ones, for their visual clarity, would be Norse runes or Japanese katakana.
The only example I can think of offhand of a non Greek or Latin character that has mainstream acceptance in mathematics would be the Hebrew character aleph ($\aleph$), though perhaps there are more.
My question then, is have there been any strong mainstream efforts, perhaps through using them in books, or from directly advocating them in lectures or articles, to introduce characters from other alphabets into mathematics? If there have been, why have they failed, and if there haven't been, why is it generally seen as unnecessary?
Thank you, and sorry if this isn't an appropriate question for math.stackexchange.com, reading the FAQ made it appear as if questions of this type were right on the borderline of acceptability.
AI: In advanced number theory arithmeticians have introduced the russian letter Ш, pronounced "shah".
But this is very localized.
Apart from the Greek alphabet, the only different alphabet I know of used in a Latin environment is Fraktur, popularly known as Gothic.
It is massively used in algebra for ideals in rings.
Actually, essentially all standard references in commutative algebra and algebraic geometry make use of Fraktur: Atiyah-Macdonald, Dieudonné-Grothendieck's EGA, Görtz-Wedhorn, Hartshorne, Jacobson, Matsumura, Qing Liu, Shafarevitch, Zariski-Samuel,...
Edit The $\LaTeX$ command for Fraktur is $\text {\mathfrak}$. For example:
Let $\mathfrak p$ be a prime ideal, $\mathfrak q$ a primary ideal and $\mathfrak a,\mathfrak b, \mathfrak c \:$ arbitrary ideals of the ring $A$, then... |
H: Analysis operator $T_\Phi$ is injective and has a closed range
Definition of the problem
Let $\mathcal{H}$ be a separable Hilbert space on $J\subset\mathbb{N}$
an index set. Let $\Phi:=\left(\varphi_{j}\right)_{j\in J}\subset\mathcal{H}$
be a frame for $\mathcal{H}$.
I have to prove that the analysis operator $T_{\Phi}:\mathcal{H}\rightarrow\ell_{2}\left(J\right)$
of the frame $\Phi$, defined by
$$
T_{\Phi}x:=\left(\left\langle x,\varphi_{j}\right\rangle \right)_{j\in J},\quad x\in\mathcal{H},
$$
is injective and has a closed range.
Effort to prove closed range
We need to show that
$$
ran\, T_{\Phi}closed\Leftrightarrow\forall x_{n}\in\mathcal{H}:\,\lim_{n\rightarrow\infty}T_{\Phi}x_{n}\in\ell_{2}\left(J\right).
$$
Let $x_{n}\in\mathcal{H}$. We have that
$$
\lim_{n\rightarrow\infty}T_{\Phi}x_{n}=\lim_{n\rightarrow\infty}\left(\left\langle x_{n},\varphi_{j}\right\rangle \right)_{j\in J}\overset{?}{\in}\ell_{2}\left(J\right).
$$
For this statement to hold, we have a look at
$$
\sum_{j\in J}\left|\lim_{n\rightarrow\infty}\left\langle x_{n},\varphi_{j}\right\rangle \right|^{2}.
$$
Using the fact that $\Phi$is a frame for $\mathcal{H}$,
$$
\sum_{j\in J}\left|\lim_{n\rightarrow\infty}\left\langle x_{n},\varphi_{j}\right\rangle \right|^{2}\overset{???}{=}\sum_{j\in J}\lim_{n\rightarrow\infty}\left|\left\langle x_{n},\varphi_{j}\right\rangle \right|^{2}=\lim_{n\rightarrow\infty}\sum_{j\in J}\left|\left\langle x_{n},\varphi_{j}\right\rangle \right|^{2}\leq\lim_{n\rightarrow\infty}B\left\Vert x_{n}\right\Vert ^{2}.
$$
My question 1
How could I use that to show that $T_{\Phi}$has a closed range? Would an upper bound help me at all with this?
Effort to show that $T_{\Phi}$is injective
Denote $\left\{ e_{i}:i\in I\right\} $be an orthonormal basis. Let
$x,y\in\mathcal{H}$. Assume $T_{\Phi}x=T_{\Phi}y$. We have that
$$
\sum_{i\in J}\left\langle x,\varphi_{i}\right\rangle e_{i}=\sum_{i\in J}\left\langle y,\varphi_{i}\right\rangle e_{i}
$$
$$
\Leftrightarrow\forall i\in J:\quad\left\langle x,\varphi_{i}\right\rangle =\left\langle y,\varphi_{i}\right\rangle .
$$
My question 2
Am I allowed to make the assumption on the orthornormal basis? How can I show that such an orthonormal basis exists? How could I go any further showing that it is injective? We do not know if the inner product $\left\langle \cdot,\cdot\right\rangle $
is one-to-one?!
Thank you, Franck!
AI: 2) Since $T_\Phi$ is a linear operator, its injectivity amounts to having the trivial kernel $\{0\}$. Use the lower frame bound.
1) To show the range is closed, it suffices prove that $Tx_n\to 0$ implies $x_n\to 0$. Use the lower frame bound for this too. |
H: Does a closed form formula for the series ${n \choose n-1} + {n+1 \choose n-2} + {n+2 \choose n-3} + \cdots + {2n - 1 \choose 0}$ exist.
$${n \choose n-1} + {n+1 \choose n-2} + {n+2 \choose n-3} + \cdots + {2n - 1 \choose 0}$$
For the above series, does a closed form exist?
AI: Your series is
$$\sum_{k=0}^{n-1}\binom{n+k}{n-k-1}=\sum_{k=0}^{n-1}\binom{2n-1-k}k\;,$$
which is the special case of the series
$$\sum_{k\ge 0}\binom{m-k}k$$
with $m=2n-1$. It’s well-known (and easy to prove by induction) that
$$\sum_{k\ge 0}\binom{m-k}k=f_{m+1}\;,$$
where $f_m$ is the $m$-th Fibonacci number: $f_0=0,f_1=1$, and $f_n=f_{n-1}+f_{n-2}$ for $n>1$. Thus, the sum of your series is $f_{2n}$.
The Binet formula gives a closed form for $f_{2n}$: $$f_{2n}=\frac{\varphi^{2n}-\hat\varphi^{2n}}{\sqrt5}\;,$$ where $\varphi=\frac12\left(1+\sqrt5\right)$ and $\hat\varphi=\frac12\left(1-\sqrt5\right)$. A computationally more convenient expression is $$f_{2n}=\left\lfloor\frac{\varphi^{2n}}{\sqrt5}+\frac12\right\rfloor\;.$$ |
H: Is there any number in $\mathbb Z[\sqrt{5}]$ with norm equals 2?
Is there any number $a+b\sqrt{5}$ with $a,b \in \mathbb{Z}$ with norm (defined by $|a^2−5b^2|$) equal 2?
AI: No.
Since $a^2-5b^2\equiv a^2\pmod{5}$, the values of $a^2-5b^2$ must be congruent to either $0$, $1$, or $4$ modulo $5$, since those are the only squares modulo $5$. But neither $2$ nor $-2$ satisfy this condition, so it is impossible to have $a^2-5b^2=2$ or $a^2-5b^2=-2$ with $a,b\in\mathbb{Z}$. |
H: System of differential equations.
\begin{equation}
\dot{x}(t)=\left(1-\frac{x(t)}{E_{0}}\right) x(t) - a\ y(t)
\end{equation}
\begin{equation}
\dot{y}(t)=\left(1-\frac{a\ y(t)}{x(t)}\right) y(t)
\end{equation}
I am trying to solve the system of differential equations above, but (I think) it is nontrivial to calculate its solution. Any suggestions?
AI: Solve the first equation for $y$, substitute into the second. You get a second-order differential equation for $x(t)$:
$$ - \left( {\frac {d^{2}}{d{t}^{2}}}x \left( t \right) \right) x
\left( t \right) {E_{{0}}}^{2}+ \left( {\frac {d}{dt}}x \left( t
\right) \right) ^{2}{E_{{0}}}^{2}- \left( x \left( t \right)
\right) ^{3}E_{{0}}+ \left( x \left( t \right) \right) ^{4}
=0$$
Maple solves this explicitly:
$$x \left( t \right) =4\,{\frac {{{\rm e}^{ta+b}}{a}^{2}{E_{{0}}}^{2}}{-
4\,{a}^{2}{E_{{0}}}^{2}+{{\rm e}^{2\,ta+2\,b}}+4\,{{\rm e}^{ta+b}}E_{{0
}}+4\,{E_{{0}}}^{2}}}
$$
where $a$ and $b$ are arbitrary constants. |
H: Hellinger distance between Beta distributions
I am interested in calculating the Hellinger distance $H(f,g)$ between two Beta distributions $f$ and $g$ of which I already know the parameters for. I am aware that you can calculate it directly using the 2-norm of discrete distributions. But it would be nicer to have full analytical expression.
The wikipedia page that I link to give nice expression for Gaussian, exponential, Weibull, and Poisson distributions. But how to derive similar expressions for:
the 2-parameter Beta distribution defined on $[0,1] \in \mathbf{R}$?
the 4-parameter Beta distribution having arbitrary support in $\mathbf{R}$?
The last option is also known (atleast for me) as the Pearson Type I distribution (was this the origin of the Beta distribution?). I have used the pearsrnd function in MATLAB and much of my data seems to fit a type I distribution.
I just need it for univariate statistics. Please remember the factor $1/2$ as I need the distance in the range $[0,1]$. Whether or not the expression gives me the squared distance is not so important.
Addendum:
I tried to solve it directly in Mathematica 7 usintg Integrate. I created two functions: (1) hellingerDistanceA that implements the integration directly and (2) hellingerDistanceB that evaluates the expression given by Sasha below. The answer by Sasha seems to be correct:
hellingerDistanceA[a_, b_, c_, d_] :=
1 - Integrate[
Sqrt[
Times[
PDF[BetaDistribution[a, b], x],
PDF[BetaDistribution[c, d], x]
]
],
{x, 0, 1},
Rule[Assumptions, {Element[x, Reals]}]
]
hellingerDistanceB[a_, b_, c_, d_] :=
1 - Divide[
Beta[(a + c) / 2, (b + d) / 2],
Sqrt[
Times[
Beta[a, b],
Beta[c, d]
]
]
]
hellingerDistanceA[1/2, 1/2, 5, 1] // N (* gives 0.251829... *)
hellingerDistanceB[1/2, 1/2, 5, 1] // N (* also gives 0.251829... *)
hellingerDistanceA[2, 2, 2, 5] // N (* gives 0.148165... *)
hellingerDistanceB[2, 2, 2, 5] // N (* also gives 0.148165... *)
AI: Let $X$ and $Y$ be independent beta random variables, such that $X \sim \operatorname{Beta}(a_1, b_1)$ and $Y \sim \operatorname{Beta}(a_2, b_2)$. Then
$$ \begin{eqnarray}
H(X, Y) &=& 1 - \int_0^1 \sqrt{ f_X(t) f_Y(t) } \mathrm{d} t \\ &=& 1 - \frac{1}{\sqrt{B(a_1,b_1) B(a_2,b_2)}} \int_0^1 t^{ (a_1 + a_2)/2 -1} (1-t)^{(b_1 + b_2)/2-1} \mathrm{d} t \\ &=& 1 - \frac{B\left(\frac{a_1+a_2}{2}, \frac{b_1+b_2}{2}\right)}{\sqrt{B(a_1,b_1) B(a_2,b_2)}}
\end{eqnarray}
$$
Calculations with 4-parameter beta random variables will be similar. The integrals might be expressible in terms of incomplete beta functions. |
H: Finding Limits of Trig Functions: $\lim_{\theta \rightarrow 0}\frac {\sin^2\theta}{\theta}$
I am asked find the following limit
$$\lim_{\theta \rightarrow 0}\frac {\sin^2\theta}{\theta}$$
I recognize that $$\lim_{\theta \rightarrow 0}\frac{\sin\theta}{\theta}=1$$
But because I have $sin^2\theta$ in the numerator, I am left with...
$$\lim_{\theta \rightarrow 0}1(\sin\theta)$$
When I think about what this implies, I reason that the ratio of the opposite side over hypotenuse of the angle $\theta$ must approach approach zero, but for this to happen the opposite side would have a value of zero, which means the triangle formed would have no x component.
$$\lim_{\theta \rightarrow 0}1(\sin\theta)=0$$
Is my reasoning correct? Am I thinking about this question in a constructive manner?
AI: $$
\lim_{\theta\to0} \frac{(\sin\theta)^2}{\theta} = \lim_{\theta\to0} \left(\sin\theta\frac{\sin\theta}{\theta}\right).
$$
This is equal to
$$
\left(\lim_{\theta\to0} \sin\theta\right)\left(\lim_{\theta\to0} \frac{\sin\theta}{\theta}\right)
$$
if both of these last two limits exist.
Later note: "Exist" in this case means both are real numbers, not things like $\infty$ or $-\infty$. |
H: Solving $y'' - \frac{1}{x} y' + (1+\frac{\cot x}{x}) y = 0$ by rank reduction
With the substitution $$y(x) = \sin x \int u(x) \, dx\tag{*}$$ I managed to get to$$u'(x) = \left(\frac{1}{x}-2\cot x\right)u(x)$$ Solving which gave me $$u(x) = C_1 \frac{x}{\sin^2 x}$$
Inserting that back into $(*)$ $$y(x) = (\sin x) C_1\int \frac{x}{\sin^2 x} \, dx = (\sin x) C_1 \left(\log(\sin x) - \frac{x}{\cot x} + C_2\right)$$
Which doesn't seem to be the correct solution(s). I don't know where I went wrong though.
AI: We have a theorem:
There exists a fundamental set of solutions for a homogenous linear 2th-differential equation $a_2(x)y''+a_1(x)y'+a_0(x)y=0$ on an interval $I$.
You have a second order homogenous differential equation which is linear as well. This is essential to have an independent set of solutions $\{y_1(x),y_2(x)\}$ for further aim. As you noted, there is a method in which we can construct a second solution from a known solution such that the latter set is a fundamental set (called reducing the order). It can be proved that if $y_1(x)$ is a known solution then the second one satisfying the theorem is $$y_2(x)=y_1(x)\int\frac{e^{-\int\frac{a_1(x)}{a_2(x)}dx}}{y_1^2(x)}dx$$
So in your equation we have $$y_2=\sin(x)\int\frac{e^{\int\frac{1}{x}dx}}{\sin^2(x)}dx=\sin(x)\int\frac{x}{\sin^2(x)}dx$$ $$y_2=\sin(x)(-x\cot(x)+\ln(\sin(x)))=-x\cos(x)+\sin(x)\ln(\sin(x))$$ Now, your general solution is as $y(x)=C_1y_1(x)+C_2y_2(x)$. |
H: Tensor contraction with multiple indices
I think I missed a rule somewhere, because I can contract the following expression in multiple ways.
$\epsilon^{\alpha \beta} \sigma_{\dot{\alpha} \alpha} \epsilon^{\dot{\alpha} \dot{\beta}} = \sigma_{\dot{\alpha} \alpha} \epsilon^{\alpha \beta} \epsilon^{\dot{\alpha} \dot{\beta}}=\sigma_{\dot{\alpha}}^{~\beta}\epsilon^{\dot{\alpha} \dot{\beta}}=\sigma^{ \dot{\beta}\beta}$
I also know that $\epsilon^{\alpha \beta}=-\epsilon^{ \beta \alpha}$, so
$\epsilon^{\alpha \beta} \sigma_{\dot{\alpha} \alpha} \epsilon^{\dot{\alpha} \dot{\beta}} = -\epsilon^{ \dot{\beta} \dot{\alpha}}\sigma_{\dot{\alpha} \alpha} \epsilon^{\alpha \beta} =-\sigma^{\dot{\beta}}_{~\alpha}\epsilon^{\alpha \beta}=-\sigma^{ \dot{\beta}\beta}$
What is going wrong here?
AI: Both calculations are wrong, I'm afraid. Since when does $\epsilon^{\lambda\eta}\sigma_{\eta\nu} = \sigma^\lambda{}_\nu$? This is valid only for metric tensor, but $\epsilon^{\lambda\eta}$ ($\begin{pmatrix}0&1\\-1&0\end{pmatrix}$) is not a metric tensor, assuming $\epsilon^{\lambda\eta}$ is the two-dimensional Levi-Civita symbol.
Suppose $T^{\beta\delta} = \epsilon^{\alpha\beta}\sigma_{\gamma\alpha}\epsilon^{\gamma\delta}$, and the metric is identity. Since there are only 4 distinct elements, we could carry out the computation directly:
$T^{00} = \epsilon^{10}\sigma_{11}\epsilon^{10} = \sigma_{11} = -\sigma_{00}$
$T^{01} = \epsilon^{10}\sigma_{01}\epsilon^{01} = -\sigma_{01}$
$T^{10} = \epsilon^{01}\sigma_{10}\epsilon^{10} = -\sigma_{10}$
$T^{11} = \epsilon^{01}\sigma_{00}\epsilon^{01} = \sigma_{00} = -\sigma_{11}$
or use the identities $\epsilon_{ab}=\epsilon^{ab}$ and $\epsilon_{ab}\epsilon^{cd} = \delta_a^c\delta_b^d - \delta_a^d\delta_b^c$ to arrive at
$$ T^{\beta\delta} = (\delta^{\alpha\gamma}\delta^{\beta\delta} - \delta^{\alpha\delta}\delta^{\beta\gamma})\sigma_{\gamma\alpha} =\delta^{\beta\delta}\operatorname{tr}(\sigma) -\sigma^{\beta\delta} = -\sigma^{\beta\delta}. $$
Edit: If you define $\epsilon^{\mu\nu}$ to be the metric, then your first formula is wrong, because index raising is done by
$$\huge x^{\color{red}\mu} = g^{\color{red}\mu\color{green}\nu} x_{\color{green}\nu} \tag{Correct} $$
and not
$$\huge x^{\color{red}\mu} = g^{\color{#FF4000}\nu\color{#808000}\mu} x_{\color{green}\nu} \tag{Wrong} $$
especially when your metric $g^{\mu\nu}$ is asymmetric. |
H: Name Drawing Puzzle
There is a party with 20 people, and everyone writes their name down
on a piece of paper and puts it into a bag. The bag is mixed up, and
each person draws one piece of paper. If you draw the name of someone
else, you are considered to be in his "group". What is the expected
number of groups after everyone draws?
So basically if we have a loop where each person draws someone else's name, and the last person draws the first person in that list's name, we have a group. Not quite sure how to approach this problem.
Thanks for any help.
AI: What you’re asking for is the average number of cycles in a random permutation of $[n]$ in the case $n=20$. (Here $[n]=\{1,\dots,n\}$.)
Let $h(n)$ be the average number of cycles in a random permutation of $[n]$; I claim that $h$ satisfies the recurrence $$h(n)=\frac{n-1}nh(n-1)+\frac1n\Big(h(n-1)+1\Big)\;.\tag{1}$$
Proof: Let $\pi$ be any permutation of $\{2,3,\dots,n\}$ written in cycle form. Say that the entries in $\pi$ from left to right, ignoring parentheses, are $\pi_1,\dots,\pi_{n-1}$. Now insert $1$ into $\pi$ in one of the following two ways.
To the left of $\pi_1$ as $(1)$, forming a cycle of its own.
Immediately after one of the $\pi_k$, $k=1,\dots,n-1$, in the same cycle as $\pi_k$.
Every permutation of $[n]$ can be uniquely obtained in this way from a unique permutation $\pi$ of $\{2,\dots,n\}$.
The average number of cycles of a random permutation of $\{2,\dots,n\}$ is of course $h(n-1)$. The recurrence $(1)$ is now an immediate consequence of the fact that operation (1) above increases the number of cycles by $1$ and accounts for $\frac1n$ of all cases, while operation (2) leaves the number of cycles unchanged and accounts for the remaining $\frac{n-1}n$ cases. $\dashv$
Now rewrite $(1)$ as $$h(n)=h(n-1)+\frac1n$$ and note that $h(1)=1$. Then $h(2)=1+\frac12$, $h(3)=h(2)+\frac13=1+\frac12+\frac13$, and an easy induction verifies that in general $$h(n)=H_n=\sum_{k=1}^n\frac1k\;,$$ the $n$-th harmonic number. It is known that $$H_n=\ln n+\gamma+\epsilon_n\;,$$ where $\gamma\approx 0.5772$ is the Euler-Mascheroni constant and $\epsilon_n\sim\frac1{2n}$.
Added: $H_{20}=\dfrac{55835135}{15519504}\approx 3.59774$. The numerator is taken from A001008 at OEIS and the denominator from A002805. |
H: Confusion about Completion of Metric Space Proof
I have started studying Functional Analysis following "Introduction to Functional Analysis with Applications". In chapter 1-6 there is the following proof
For any metric space $X$, there is a complete metric space $\hat{X}$ which has a subspace $W$ that is isometric with $X$ and is dense in $\hat{X}$
(Page 1 & 2) https://i.stack.imgur.com/S7rbn.png
(Page 3 & 4) https://i.stack.imgur.com/QGL3C.png
I think I understand parts (a) and (b). At the top of page 3, section (c) where it is proving $\hat{X}$ is complete it states:
Let $(\hat{x_{n}})$ be any Cauchy Sequence in $\hat{X}$. Since $W$ is dense in $\hat{X}$, for every $\hat{x_{n}}$, there is a $\hat{z_{n}}\varepsilon W$ such that $\hat{d}(\hat{x_{n}},\hat{z_{n}}) < \frac{1}{n}$
I do not understand why we choose $ \frac{1}{n}$. Would some ε > 0, for each $\hat{x_{n}}$, not suffice? I assume it must not, but I don't see why, so I must not understand this proof. Maybe i'm not sure on what n is referring to because of the subscripts n on the lefthand side. Is n the index of the Cauchy sequence in $\hat{X}$, $(\hat{x_{n}})$? Is it the index of the Cauchy sequence in X, $\hat{x_{n}}$?
Any help would be greatly appreciated, i am pretty dumb and this has puzzled me for a couple days.
AI: As far as I can tell the author, at that place, is just in need of a (any) sequence $\epsilon_n$ tending to zero as $n\rightarrow \infty$. You could choose any $\epsilon_n$ instead with the property that $\epsilon_n \rightarrow 0$, but $1/n$ is easy to write down and handle. Note though the dependency between the index of the elements of the sequence and the sequence tending to zero. You cannot just take any fixed (independent of $n$) $\epsilon >0$ here. |
H: Largest number that divides $n^2(n^2 - 1)(n^2 - n - 2)$ for all $n$
Obtain the greatest natural that divides $n^2(n^2 - 1)(n^2 - n - 2)$ for all natural numbers $n$.
What should be the approach in these type of questions?
Should I equate with prime factorization $2^a 3^b 5^c \cdots$ etc?
AI: \begin{align}
f(n) & = n^2(n^2-1)(n^2-n-2)\\
& = n^2(n+1)(n-1)(n+1)(n-2)\\
& = n^2(n+1)^2(n-1)(n-2)
\end{align}
$n=3$ gives us $9 \times 16 \times 2 = 288$. So the natural number dividing $n^2(n+1)^2(n-1)(n-2)$ for all $n$ should be a divisor of $288$.
Also, note that $24$ has to divide $n^2(n+1)^2(n-1)(n-2)$ since $24$ divides $(n-2)(n-1)n(n+1)$ since it is a product of $4$ consecutive integers.
Further, $2$ divides $n(n+1)$. Hence, we know that at-least $48$ divides $f(n)$. Hence, the desired number should be a multiple of $48$ and must divide $288$.
Can you finish it from here by looking at $f(n)$ for couple of other values?
Move the mouse over the gray area for the answer.
Since the largest number must be a multiple of $48$ and must divide $288$. The possible options are $$\{48, 96, 144, 288\}$$
$$\text{$f(4) = 2400$. And $144,288$ doesn't divide $2400$. Hence, the largest number cannot be $144$ or $288$.}$$
$$\text{$f(5) = 10,800$. And $96$ doesn't divide $10,800$. Hence, the largest number cannot be $96$. Hence, the largest number is $48$.}$$ |
H: Lexicographical order - posets vs preorders
I found the following definition for lexicographical ordering on Wikipedia (and similar definitions in other places):
Given two partially ordered sets $A$ and $B$, the lexicographical order on the Cartesian product $A \times B$ is defined as $(a,b) \le (a',b')$ if and only if $a < a'$ or ($a = a'$ and $b \le b'$). The result is a partial order. If $A$ and $B$ are totally ordered, then the result is a total order as well.
Does this definition work equally well if $A$ and $B$ are preorders, rather than posets? In other words, does the anti-symmetric property of the ordering relations on A and B make any difference here?
AI: Yes, if $\langle A,\le_A\rangle$ and $\langle B,\le_B\rangle$ are preorders, the same construction yields a preorder $\langle A\times B,\preceq\rangle$. Specifically, for $\langle a,b\rangle,\langle c,d\rangle\in A\times B$ define $\langle a,b\rangle\preceq \langle c,d\rangle$ iff $a<_A c$, or $a\le_A c$ and $b\le_B d$. Clearly $\preceq$ is reflexive, so you have only to check that it’s transitive.
Added: I should have said explicitly what I mean by $a<_A c$ for the preorder $\le_A$. I don’t mean that $a\le_A c$ and $a\ne c$. Rather, I mean that $a\le_A c$ and $c\not\le_A a$. Equivalently, I mean that $a\le_A c$ and $a\not\sim_A c$, where $a\sim_A c$ iff $a\le_A c$ and $c\le_A a$. Here $\sim_A$ is the equivalence relation on $A$ whose equivalence classes are sets of $\le_A$-indistinguishable members of $A$, and the relation induced on $A/\sim_A$ by $\le_A$ is a partial order. The point is that if $a\sim_A a'$, I want to treat $a$ and $a'$ exactly alike.
If we were dealing with partial orders $\le_A$ and $\le_B$, I could define $\langle a,b\rangle\preceq \langle c,d\rangle$ iff $a<_A c$, or $a=c$ and $b\le_B d$, using the usual definition of lexicographic order. Note, though, that for partial orders that definition is equivalent to saying that $\langle a,b\rangle\preceq \langle c,d\rangle$ iff $a<_A c$, or $a\le_A c$ and $b\le_B d$: if $\langle a,b\rangle\preceq \langle c,d\rangle$ by the latter definition, then either $a<_A c$, in which case $\langle a,b\rangle\preceq \langle c,d\rangle$ by the usual definition as well, or $a\le_A c$, $a\not<_A c$, and $b\le_B d$, in which case $a=c$ and $b\le_B d$, and again $\langle a,b\rangle\preceq \langle c,d\rangle$ by the usual definition. For preorders the two are not equivalent, because a preorder need not be antisymmetric. For preorders the equivalent formulation is that $\langle a,b\rangle\preceq \langle c,d\rangle$ iff $a<_A c$, or $a\sim_A c$ and $b\le_B d$.
To that end suppose that $\langle a,b\rangle\preceq\langle c,d\rangle\preceq\langle e,f\rangle$; clearly $a\le_A c\le_A e$. If either $a<_A c$ or $c<_A e$, then $a<_A e$, and $\langle a,b\rangle\preceq\langle e,f\rangle$. Otherwise we have $b\le_B d$ and $d\le_B f$, so $b\le_B f$, and again $\langle a,b\rangle\preceq\langle e,f\rangle$. |
H: Question about cardinals without GCH
Without assuming the Generalized Continuum Hypothesis, how to show that there exists a uncountable cardinal $\kappa$ such that, for every $\lambda < \kappa$, one have $2^\lambda < \kappa$. With assumption of GCH, for $\kappa = \aleph_{\omega}$ the affirmation holds. I would appreciate any hint.
AI: The answer lies in $\beth$ numbers.
The definition is as follows:
$\beth_0=\aleph_0$.
$\beth_{\alpha+1}=2^{\beth_\alpha}$.
If $\delta$ is a limit ordinal and $\beth_\gamma$ was defined for $\gamma<\delta$ then $\beth_\delta=\sup\{\beth_\gamma\mid\gamma<\delta\}$.
Now you can prove that $\beth_\omega$ is a strong limit, and if $\delta$ is a limit ordinal then $\beth_\delta$ is a strong limit cardinal. You can even show that for every $\mu$ there is a strong limit $\kappa$ such that $\mu<\kappa$.
(Note that GCH is the assertion $\beth_\alpha=\aleph_\alpha$ for all $\alpha$.) |
H: Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous, show that $\{x:f(x)\leq g(x)\}$ is closed in $X$
Let $Y$ be an ordered set in the order topology with $f,g:X\rightarrow Y$ continuous. Show that the set $A = \{x:f(x)\leq g(x)\}$ is closed in $X$.
I am completely stumped on this problem. As far as I can tell I've either got the task of proving $A$ gathers its limit points, or showing that there is a closed set $V \subset Y$ such that $f^{-1}(V)$ or $g^{-1}(V)$ is equal to A, which would prove $A$ closed since both $f$ and $g$ are continuous.
The latter strategy seems like the more likely to succeed. Unfortunately I can't find any way of constructing this set $V$. The fact that I don't know if $f$ or $g$ are injective means I keep running into the problem of having $f^{-1}(V)$ or $g^{-1}(V)$ give me extra points not in $A$. And even if I were able to construct $V$ this wouldn't guarantee it was closed.
I suspect I may need to use the continuity of both $f$ and $g$ together in some way but I can't see how. Can anyone give me some guidance on this problem? Thanks.
AI: HINT: Let $h:X\to Y:x\mapsto\max\{f(x),g(x)\}$.
Show that $h$ is continuous.
Note that $A=\{x\in X:h(x)=g(x)\}$. |
H: image of complex continuous function cannot be a curve
Let $ f : U \subset \mathbb C \to \mathbb C $ a continuous function where $ U $ is an open and connected set.
Prove that the image of $ f$ cannot be a curve on compex plane.
AI: Are you sure there are no additional constraints, e.g. the Cauchy-Riemann equations are satisfied (continuity + C.R. -> analytic). One could simply take $f(z) = \Re(z)$, which is continuous but not analytic -- the image of an open set (say the ball of radius 1 around the origin) is a curve (the interval $[-1,1]$).
The reason this cannot hold for an analytic function is because of the open mapping theorem. Analytic functions are open mappings, so an open set $U$ has to map to an open set $f(U)$, which cannot be a curve. |
H: Finding horizontal tangents for trig function
I am asked to find the points on the curve at white the tangent is horizontal, for the function:
$$y=\frac{\cos x}{2+\sin x}$$
To find the points at which the tangent is horizontal, I need to know what values will result in the derivative of the function equaling zero, so I differentiate using quotient rule:
$$y'=\frac{(-2\sin x-\sin^2 x)-\cos^2 x}{(2+\sin x)^2}$$
The algebra in simplifying the derivative from here is what is holding be back. Also, have I differentiated this correctly?
AI: Yes, you computed the derivative correctly (assuming you meant to write $(\sin x)^2$ or $\sin^2 x$, not $\sin x^2$).
You need to find the points where the derivative is zero. First note that since $\sin$ is bounded in absolute value by 1, the denominator in your expression for the derivative is never zero. So, the zeroes of the derivative are precisely the zeroes of the numerator of your expression. You thus have to solve the equation
$$
(-2\sin x -\sin^2 x)-\cos^2 x=0;
$$
a task that is made tractable upon using the Pythagorean identity $\sin^2 x+\cos^2 x=1$. Utilizing this (and a bit of algebra) produces the equivalent equation
$$
-2\sin x-1=0.
$$
Can you take it from here? |
H: If $(G : H) = r$ and there exists $K \triangleleft G$ contained in $H$ such that $(G : K) = r!$
Possible Duplicate:
How to prove that if $G$ is a group with a subgroup $H$ of index $n$, then $G$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$
I'm trying to prove the following: "If $G$ is an infinite group and $H < G$ satisfies $(G:H) = r$, then there exists a normal subgroup $K$ of $G$ contained in $H$ such that $(G:K) \leq r!$". The equality is trivial if we if we find $K \triangleleft G$ such that there exists an injective action of $K$ on the cosets of $H$. However, I can't think of any such action, considering $K < H$.
AI: I think you're very near the following argument. The whole group $G$ acts by left translation on the set of cosets $G/H$, and we can view this [see Theorem 1.6 of Keith Conrad's handout] action as a homomorphism $G \to \operatorname{Perm}(G/H) \approx S_r$. Let $K$ be the kernel of this homomorphism. Elements of $K$ must in particular fix the coset $H$, so what does this tell you about $K$? What do you know about the group $G/K$?
[Note that we don't need $G$ to be infinite, although the corollary, "Every finite index subgroup contains a subgroup which is normal and of finite index in the whole group," is less surprising in the finite case.] |
H: Is the inverse of a one-to-one, total function itself one-to-one and total?
The problem, from Boolos and Jeffrey's Computability and Logic, states the following definitions:
"If a function $f(a)$ (from $A$ to $B$) is defined for every element $a$ of $A$, then it is called total"
"A correspondence between sets $A$ and $B$ is a one-to-one, total function from $A$ onto $B$."
"Two sets are said to be equinumerous if and only if there is a correspondence between $A$ and $B$."
Then, it asks the reader to show that if $A$ is equinumerous with $B$, then $B$ is equinumerous with $A$. So far, the work I've done is to show that, for a function $f : A \rightarrow B$, there exists a function $f^{-1} : B \rightarrow A$ that is onto (which follows from $f$ being total), and one-to-one (which follows from $f$ existing). However, it remains to be shown that $f^{-1}$ is total. This seems to require that $f$ is onto, which doesn't seem to be stated by the problem. I would appreciate any hints on how to show that $f^{-1}$ must be total, given the constraints.
AI: It is common to write that the function is "onto $B$" to mean that it is surjective (onto). Otherwise, we would say that a function $f$ is a [total] function "from $A$ to $B$".
So when it says that it is a total one-to-one, function from $A$ onto $B$ it is telling you (i) the domain is all of $A$; (ii) the function is one-to-one; and (iii) the image is all of $B$. |
H: How many k-digit ternary strings are there in which no digit occurs exactly twice? The occurrences do not need to be consecutive.
How many k-digit ternary strings are there in which no digit occurs exactly twice? The occurrences do not need to be consecutive.
AI: This can be computed using an inclusion-exclusion argument. Let $s(k)$ be the number of such strings of length $k$.
There are $3^k$ strings altogether. In $\binom{k}22^{k-2}$ of them there are exactly two $0$’s. There are the same number with exactly two $1$’s and the same number again with exactly two $2$’s. However, there are $\binom{k}2\binom{k-2}2$ strings with exactly two $0$’s and exactly two $1$’s, the same number with exactly two $0$’s and exactly two $2$’s, and the same number with exactly two $1$’s and exactly two $2$’s. Finally, if $k=6$ there is one string with exactly two of each symbol; otherwise there are no such strings. For $k\ne 6$, therefore, the desired total is
$$\begin{align*}s(k)&=3^k-3\binom{k}22^{k-2}+3\binom{k}2\binom{k-2}2\\
&=3^k-3k(k-1)2^{k-3}+\frac34k(k-1)(k-2)(k-3)\;;\end{align*}$$
for $k=6$ this must be decreased by $1$, yielding $$s(6)=3^6-3\cdot6\cdot5\cdot2^3+\frac34\cdot6\cdot5\cdot4\cdot3-1=278\;.$$
The first few values, in tabular form (assuming that I’ve made no arithmetic errors):
$$\begin{array}{r|cc}
k:&0&1&2&3&4&5&6&7&8\\ \hline
s(k):&1&3&6&9&27&93&278&801&2445
\end{array}$$ |
H: Probability: selecting exactly one from each group
A coaching held separate workshops for training in three languages.
50 students participated in workshops and learnt one or more languages. 13 students learnt a single language, 25 students learnt 2 languages and 12 students learnt 3 languages.
If three students are selected at random, find the probability that exactly one has attended 3 workshops, exactly one has attended two workshops and one has attended a single workshop?
AI: Please check my attempt for correctness:
choosing 3 out of 50 children can be done in 50C3 ways.
And to satisfy the condition, i can make a selection of 1 from 13, 1 from 25 and 1 from 12 students.
so i have 12*13*25/(50C3). |
H: Calculate dimensions of inner box
This is my first post here and also i'm bad with math so don't be mad at me :)
Here is my issue, i have a box and i have another box inside, now i want that box inside to be at exactly same distance from each side, no matter what size it parent is.
Now i know how to calculate the distance from the sides, but i can't figure how to get right box dimensions. To make thing more clear i draw this example:
http://www.part.lt/img/f2204b0558a21a3695c6698de6e0f7f4835.png
I hope i make some sense here :)
AI: If your outer box is $x \times y$ and you want the inner one at distance $d$, the inner dimensions are $x-2d \times y-2d$ and the centers need to match. Is that what you are after? |
H: Chain Rule Confusion
Below i have a function that i need to use the chain rule on. My friend showed me his answer which was correct which was $-8x^7\sin(a^8+x^8)$.
$$y = \cos(a^8 + x^8)$$
I am really confused as how he got that. I know that in the chain rule you bring whats outside to the front. So why is $a^8$ not in this solution?
AI: You’re trying to treat the chain rule as a mechanical manipulation of symbols instead of understanding what it actually says. It says that when you differentiate a composite function, say $g\big(f(x)\big)$, you first take the derivative of $g$ as if $f(x)$ were the independent variable, and then you multiply by $f\,'(x)$.
Here you have $h(x)=\cos(a^8+x^8)$, and you want $h'(x)$. First pretend that what’s inside the cosine is a single variable; call it $u$, if you like so that $u=a^8+x^8$ and $h(x)=\cos u$. Now differentiate with respect to $u$ to get $-\sin u$. But you weren’t really differentiating with respect to $u$: you were differentiating with respect to $x$. The chain rule says that in order to compensate for this distinction, you must now multiply by $\frac{du}{dx}$. Since $a^8$ is a constant, its derivative (with respect to anything!) is $0$, and therefore $\frac{du}{dx}=8x^7$. The chain rule now tells you that $$h'(x)=\Big(-\sin(a^8+x^8)\Big)\Big(8x^7\Big)=-8x^7\sin(a^8+x^8)\;.$$ |
H: Necessary and sufficient condition for being a covering map
Suppose I have a map $f:X\rightarrow Y$ continuous, $X$ is compact, connected and also $f$ is a local homeomorphism, what condition should we include in $X$ so that $f$ becomes a covering map? Am I making any sense?
AI: If $X$ and $Y$ are Hausdorff, then compactness of $X$ and surjectivity of $f: X\to Y$ are already sufficient conditions for $f$ to be a covering map (edit: just to be clear, this is in addition to $f$ being a local homeomorphism). See this question.
Note that $f$ is necessarily surjective if $Y$ is assumed to be connected, because $f$ is closed and open (still under the hypothesis that $Y$ is Hausdorff). |
H: Inequalities involving some common functions
I often see the following inequality is used over and over again
$$
1−x⩽e^{−x}
$$
for $x \in \mathbb{R}$, for proving or deriving various statements.
As a layman, I haven't seen this inequality appearing in any class I have taken in my life. So it seems quite unnatural to me, and seems just a special result for linear function and exponential function. It is not yet part of my instinct to use it for solving problems. So I want to fill up this indescribable gap within my knowledge.
I wonder if there are other similar results for possibly other commonly seen functions (elementary functions?).
Is there some source listing such results?
Thanks and regards!
AI: This inequality occurs for two reasons. The line $y = 1-x$ is tangent to the curve $y=e^{-x}$ at $(0,1)$ and $x\mapsto e^{-x}$ is concave up everywhere. Hence the line lies below the curve. |
H: When will these two trains meet each other
I cant seem to solve this problem.
A train leaves point A at 5 am and reaches point B at 9 am. Another train leaves point B at 7 am and reaches point A at 10:30 am.When will the two trains meet ? Ans 56 min
Here is where i get stuck.
I know that when the two trains meets the sum of their distances travelled will be equal to the total sum , here is what I know so far
Time traveled from A to B by Train 1 = 4 hours
Time traveled from B to A by Train 2 = 7/2 hours
Now if S=Total distance from A To B and t is the time they meet each other then
$$\text{Distance}_{\text{Total}}= S =\frac{St}{4} + \frac{2St}{7} $$
Now is there any way i could get the value of S so that i could use it here. ??
AI: We do not need $S$.
The speed of the train starting from $A$ is $S/4$ while the speed of the train starting from $B$ is $S/(7/2) = 2S/7$.
Let the trains meet at time $t$ where $t$ is measured in measured in hours and is the time taken by the train from $B$ when the two trains meet. Note that when train $B$ is about to start train $A$ would have already covered half its distance i.e. a distance of $S/2$.
Hence, the distance traveled by train $A$ when they meet is $\dfrac{S}2 + \dfrac{S \times t}4$.
The distance traveled by train $B$ when they meet is $\dfrac{2 \times S \times t}7$.
Hence, we get that $$S = \dfrac{S}2 + \dfrac{S \times t}{4} + \dfrac{S \times 2 \times t}{7}$$ We can cancel the $S$ since $S$ is non-zero to get $$\dfrac12 = \dfrac{t}4 + \dfrac{2t}7$$ Can you solve for $t$ now? (Note that $t$ is in hours. You need to multiply by $60$ to get the answer in minutes.) |
H: Poincaré Lemma Contractible Hypothesis
Poincaré's Lemma is often stated as saying that a closed differential form on a star-shaped domain is exact. More generally, it is true that a closed differential form on a contractible domain is exact.
What I am wondering is if there is an easy example of a closed differential form on a simply connected domain which is not exact.
AI: Let $U=\mathbb R^n\setminus \lbrace0 \rbrace\subset \mathbb R^n$, a simply connected domain for $n\geq3$, which we assume from now on.
The $(n-1)$ form $\omega \in \Omega^{n-1}(U)$ defined by
$$\omega (x)=\frac {1}{\mid \mid x\mid \mid^n} \sum _{i=1} ^n (-1)^{i-1}x_idx^1\wedge...\wedge \widehat{dx^i} \wedge dx_n$$ for $x\in \mathbb R^n\setminus \lbrace0 \rbrace)$ is closed but not exact. It is thus an example of what you want.
More precisely, its cohomology class $[\omega ]$ generates the one-dimensional $(n-1)$-th De Rham cohomology vector space of $U$, namely $H^{n-1}_{DR}(U)=\mathbb R\cdot[\omega ]$
NB
a) This form can also be seen as the pull-back $\omega =r^*(vol)$ of the canonical volume form $vol\in \Omega ^{n-1}(S^{n-1})$ of the unit sphere under the map $$r:U\to S^{n-1}:x\mapsto \frac {x}{\mid \mid x\mid \mid}$$
b) To be quite explicit, the value of the alternating form $\omega (x)$ on the $(n-1)$-tuple of vectors $v_1,...,v_{n-1}\in T_x(U)=\mathbb R^n$ is $$\omega (x)(v_1,...,v_{n-1})=\frac {1}{\mid \mid x\mid \mid^n}\cdot det[x\mid v_1\mid ...\mid v_{n-1}]$$
where of course $x, v_1, ...,v_{n-1}$ are seen as column vectors of size $n$. |
H: A space that has a countably locally finite basis but not second countable
[Munkres, Ch40, p252] Find a nondiscrete space that has a countably locally finite basis but does not have a countable basis.
I solved this problem. My solution is $X=\mathbb{R} \cup \{0'\}$ with topology given by basis $\mathfrak{B}$ consisting of all the nonzero points and {0,0'}. This is almost discrete, except for zero points. This basis $\mathfrak{B}$ is locally finite, so countably locally finite. And this space has no countable basis, since all the nonzero points(which are uncountable) should be in the basis.
But I think that this space can be regarded discrete by passing to the quotient space. Is there any more truly nondiscrete example?
AI: Note that Munkres’ terminology is non-standard: what he calls countably locally finite is usually called $\sigma$-locally finite.
Here’s an example in which discreteness is used in a more subtle way.
Let $D$ be an uncountable discrete space, and let $C=\{0,1\}^\omega$ be the product of countably infinitely many copies of the discrete two-point space. $C$ is homeomorphic to the familiar middle-thirds Cantor set and is a compact metric space without isolated points. It also has a $\sigma$-discrete (and hence $\sigma$-locally finite) base of clopen sets. Now let $X=D\times C$; then $X$ is easily seen to have a $\sigma$-discrete base as well, but it clearly has no countable base.
Indeed, any space with a $\sigma$-locally finite base could be substituted for $C$. Let $Y$ be such a space, and let $\mathscr{B}=\bigcup_{n\in\omega}\mathscr{B}_n$ be a base for $Y$ such that each $\mathscr{B}_n$ is locally finite. For $n\in\omega$ let $$\mathscr{B}_n'=\Big\{\{x\}\times B:x\in D\text{ and }B\in\mathscr{B}_n\Big\}\;,$$
and let $\mathscr{B}\,'=\bigcup_{n\in\omega}\mathscr{B}_n'$; then $\mathscr{B}\,'$ is a $\sigma$-locally finite base for $X$.
More generally yet, you know from the Nagata-Smirnov theorem that a $T_3$ space is metrizable iff it has a $\sigma$-locally finite base. Thus, every non-second-countable metric space is an example. For example, let $\kappa$ be any uncountable cardinal number, and let $D$ be a set of cardinality $\kappa$. The metric $$d:D\times D\to\Bbb R:\langle x,y\rangle\mapsto\begin{cases}0,&\text{if }x=y\\1,&\text{if }x\ne y\end{cases}$$ generates the discrete topology on $D$. Now let $X=D^\omega$; the function $$\rho:X\times X\to\Bbb R:\Big\langle\langle x_n:n\in\omega\rangle,\langle y_n:n\in\omega\rangle\Big\rangle\mapsto\sum_{n\in\omega}\frac{d(x_n,y_n)}{2^n}$$ is a metric on $X$, and the weight (= minimum cardinality of a base) of $X$ is $\kappa$. |
H: Is the zero map (between two arbitrary rings) a ring homomorphism?
I was looking at the definition under Wikipedia, which states that for arbitrary rings $R,S$, a ring homomorphism $f:R\to S$ must satisfy $f(1)=1.$ Here, I assume they mean $1$ as the multiplicative identity. Certainly then, this implies the zero map is not a ring homomorphism?
This seems somehow intuitionally false; that is, we would want the zero map to be a ring homomorphism, as it is a group homomorphism between groups, a continuous function between reals, a smooth function between manifolds, etc. Could someone help explain why the zero map in the category of rings seems to be an exception to this pattern?
AI: It's a matter of convention.
For many authors, "rings" are required to be unital rings (have a multiplicative unit); when rings are required to be unital, it makes sense to require the homomorphisms to be unital as well (viewing rings as general algebras with two binary, $+$ and $\times$, one unary, $-$, and two nullary operations, $0$ and $1$, homomorphisms are required to respect all the operations). This is the convention followed, for example, by Lam in his A First Course in Noncommutative Rings (to give a highly regarded, professional ring theorist example of someone who would agree with Wikipedia).
Another way to justify this is to recall that a monoid homomorphism is not merely a semigroup homomorphism between monoids: if $M$ and $N$ are monoids, a monoid homomorphism is a map $f\colon M\to N$ such that $f(ab)=f(a)f(b)$ and $f(e_M) = e_N$ holds. Thus, for instance, the map $(\mathbb{N},\times)\to (\mathbb{N}\times\mathbb{N},\cdot)$, where $\mathbb{N}$ are the nonnegative integers under multiplciation, and $\cdot$ is the coordinatewise multiplication, given by $a\mapsto (a,0)$, is not a monoid homomorphism, even though it is a semigroup homomorphism. (In a sense, it is a "happy accident" that any semigroup homomorphism between groups is also a group homomorphism; but it should really be defined as requiring that it map inverses to inverses and the identity to the identity).
If you view a ring as a set that has a structure of an abelian group under $+$ and a monoid under $\times$, with the two structures connected via the distributive laws, then it makes sense to require the homomorphisms between rings to simultaneously be group homomorphisms of the additive structure, and monoid homomorphisms of the multiplicative structure... and this requires the homomorphisms to map $1$ to $1$.
Under these requirements, the only time that the zero map can be a ring homomorphisms $\zeta\colon R\to S$ is when $S=\{0\}$ is the trivial ring.
For other authors, rings are not required to be unital; when the rings are not required to be unital, you certainly cannot expect homomorphisms to be unital. In that case, the zero map is always a homomorphism between two rings. This is the convention followed, for example, by ring theorists who do radical theory. |
H: For order-reversing maps between posets/lattices, is the inverse always order-reversing?
I am having trouble with the following lemma from J. Rotman's Galois Theory book:
Lemma $83$. If $L$ and $L'$ are lattices and $\gamma: L \rightarrow L'$ is an order reversing bijection $[a \leq b$ implies $\gamma(b) \leq \gamma(a)]$, then $\gamma(a \wedge b) = \gamma(a) \vee \gamma(b)$ and $\gamma(a \wedge b) = \gamma(a) \vee \gamma(b)$.
The proof is fairly obvious if you know that when $\gamma$ is order reversing, then $\gamma^{-1}$ is order reversing also. Rotman says that this is "easily seen", but I just don't see it. I think it would be easy if his definition was that $\gamma$ is order reversing when $a \leq b$ if and only if $\gamma(b) \leq \gamma(a)$ instead. In a set theory book I have, isomorphisms between two posets $A$ and $B$ are defined as $f: A \rightarrow B$ such that $a \leq b$ if and only if $f(a) \leq f(b)$.
Am I missing something really easy or should the definition be different? Is there a counterexample? That is, is there a bijective map $f: A \rightarrow B$ between two lattices/posets $A$ and $B$ such that $a \leq b$ implies $f(b) \leq f(a)$, but $f^{-1}$ does not have this property?
AI: This is not true. If you have the lattices $L=\{\bot,a,b,\top\}$ (with $\bot<a,b<\top$) and $L'=\{0,1,2,3\}$ with the usual order, the order-preserving map
$$f(\top)=3\\
f(a)=1\qquad f(b)=2\\
f(\bot)=0$$
is of course a bijection, but $1<2$ yet $f^{-1}(1)=a$ and $f^{-1}(2)=b$ are incomparable.
In other words, $f$ needs to satisfy [$a\le b$ iff $f(a)\le f(b)$] to be an isomorphism for the order structure (indeed, $L$ and $L'$ are not isomorphic).
And this is also a necessary condition for the conclusion of your lemma, since when the order-preserving analogue of the lemma holds, $f(a)\le f(b)$ implies $f(a)=f(a)\wedge f(b)=f(a\wedge b)$ so that because $f$ is injective $a\le b$.
Edit: As said in the comments, when $L$ and $L'$ are isomorphic and finite, it is however true that any bijective $f$ satisfying [$a\le b$ implies $f(a)\le f(b)$] is an isomorphism between $L$ and $L'$. This is because the image of the relation $<_L$ is a subset of $<_{L'}$; but they both are finite sets of equal cardinal so they must be equal.
In fact we have this beautiful generalization of the well-known statement about finite sets:
An injection between two finite sets of equal cardinal is a bijection.
to finite posets:
An order-preserving injection between two finite posets of equal order types is an isomorphism. |
H: Orthogonal Operators and Inner Products
This question is a follow-up/extension to this question. Suppose $P$ is an orthogonal operator on a finite dimensional inner product space $V$. By definition, this means that
$$
\langle Pu, Pv \rangle = \langle u, v \rangle
$$
for all $u,v \in V$. I want to show that $P^TP = I$.
By definition of the transpose,
$$
\langle u, v \rangle = \langle Pu, Pv \rangle = \langle u, P^T(Pv) \rangle
$$
and by bilinearity of the inner product it follows that
$$
\langle u , (I - P^TP)v \rangle = 0
$$
If $v \neq 0 \neq u$, I want to use the fact that the inner product is nondegenerate to conclude that $I - P^TP = 0$ which would prove the claim. However, I'm not so sure that this applies here for the nondegenerate criterion is a statement about all vectors and not just a particular vector. That is, I don't believe it necessarily true that $\langle x, y \rangle = 0$ and $x \neq 0 \implies y = 0$.
On the other hand, without invoking that the inner product is nondegenerate one can see from inspection that $P^TP = I$ satisfies
$$
\langle u, v \rangle = \langle u , (P^TP)v \rangle
$$
Is this enough?
AI: You're almost there. Notice that in fact, given $v \in V$
$$\langle u, (I - P^TP)v \rangle = 0$$
for all $u \in V$. In particular, setting $u = (I - P^TP)v$ gives
$$\|(I - P^TP)v\|^2 = 0$$
which implies
$$(I - P^TP)v = 0$$ |
H: Count the number of solutions of the inequality $x + y + z \leq N$
Problem
Given $A, B, C $ and $N$. How many integer solutions are there of the following inequality:
$$x + y + z \leq N$$
where $0 \leq x \leq A, 0 \leq y \leq B, 0 \leq z \leq C$?
When $A + B + C \leq N$, the solution is obvious $(A + 1) \cdot (B + 1) \cdot (C + 1)$
The general formula for the number of solutions of the form $x + y + z = N$, for $x, y , z \geq 0$ is $\dbinom{N + r -1}{r - 1}$. Unfortunately, I couldn't find a way to bring this idea into place. Any suggestion?
AI: The main trick is to add a fourth variable to take up the slack between $x+y+z$ and $N$. You know that there are $\binom{N+3}3$ solutions to $w+x+y+z=N$ in non-negative integers, so there are also $\binom{N+3}3$ solutions to $x+y+z\le N$ in non-negative integers. Now we have to throw out the solutions that exceed one of the caps $A,B$, and $C$.
Counting non-negative solutions to $w+x+y+z=N$ in which $x>A$ is the same as counting non-negative solutions to $w+x+y+z=N-A-1$, of which there are $$\binom{N-A-1+3}3=\binom{N-A+2}3\;.$$ Similarly, there are $\binom{N-B+2}3$ non-negative solutions to $w+x+y+z=N$ in which $y>B$ and $\binom{N-C+2}3$ in which $z>C$. Thus, a better approximation to the desired number is
$$\binom{N+3}3-\binom{N-A+2}3-\binom{N-B+2}3-\binom{N-C+2}3\;.\tag{1}$$
However, it’s possible that a non-negative solution to $w+x+y+z=N$ exceeds two caps, and $(1)$ overcorrects for those unwanted solutions by subtracting each of them twice. The same reasoning used to calculate the correction terms above shows that there are $$\binom{N-A-B+1}3$$ solutions exceeding the $A$ and $B$ caps, $$\binom{N-A-C+1}3$$ exceeding the $A$ and $C$ caps, and $$\binom{N-B-C+1}3$$ exceeding the $B$ and $C$ caps. Thus, we must add to the expression in $(1)$ the sum
$$\binom{N-A-B+1}3+\binom{N-A-C+1}3+\binom{N-B-C+1}3\;.\tag{2}$$
Finally, we have to subtract the number of solutions that exceed all three caps, which is $$\binom{N-A-B-C}3\;.\tag{3}$$
The final count is $(1)+(2)-(3)$, where the parenthesized numbers refer to the numbered expression above. |
H: Does $\int_{\mathbb R^n} (1 + |x|^2)^{-1} dx$ converge in the Riemann sense?
Recently I was reading an article and came across the following integral:
$$\int_{\mathbb R^n}\dfrac{1}{1+|x|^2}\ dx$$
Is this integral convergent in the Riemann sense?
AI: The integrand is positive and smooth so Riemann or Lebesgue makes no difference. Change coordinates as follows.
$$\int_{\mathbb{R}^n} {dx\over 1 + |x|^2} = \int_0^\infty\int_{S^{n-1}} {r^{n-1}\,d\sigma\,dr\over 1 + r^2} = c_n \int_0^\infty {r^{n-1}\,dr\over 1 + r^2}$$
This converges if $n = 1$. If $n\ge 2$, it diverges. |
H: Square units of area in a circle
I'm studying for the GRE and came across the practice question quoted below. I'm having a hard time understanding the meaning of the words they're using. Could someone help me parse their language?
"The number of square units in the area of a circle '$X$' is equal to $16$ times the number of units in its circumference. What are the diameters of circles that could fit completely inside circle $X$?"
For reference, the answer is $64$, and the "explanation" is based on $\pi r^2 = 16(2\pi r).$
Thanks!
AI: Let the diameter be $d$. Then the number of square units in the area of the circle is $(\pi/4)d^2$. This is $16\pi d$. That forces $d=64$.
Remark: Silly problem: it is unreasonable to have a numerical equality between area and circumference. Units don't match, the result has no geometric significance.
"The number of square units in the area of" is a fancy way of saying "the area of." |
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