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H: prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Assume: $a,b,c >0$ prove that : $$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ AI: Replace $(a,b,c)$ by $(x_1,x_2,x_3)$ for notational convenience, and start with the inequality $$ \sum_{i,j}(x_i-x_j)\cdot\left(\frac1{x_j^2}-\frac1{x_i^2}\right)\geqslant0, $$ which holds because every term in the sum is nonnegative. Expanding the LHS, one gets $$ \sum_{i,j}\frac{x_i}{x_j^2}\geqslant\sum_{i,j}\frac1{x_i}=3\cdot\sum_i\frac1{x_i}. $$ Separating the terms such that $i=j$ from those such that $i\ne j$ in the LHS yields $$ \sum_{i\ne j}\frac{x_i}{x_j^2}\geqslant\color{red}{2}\cdot\sum_i\frac1{x_i}, $$ which is strictly stronger than the inequality to prove thanks to the factor $\color{red}{2}$ in the RHS. Furthermore, this inequality is strict except when all the $x_i$s are equal. Finally, the same proof works for $n$ terms instead of $3$, yielding a factor $n-1$ instead of the factor $\color{red}{2}$.
H: Finding an accurate rating between 1 and 10 based on numerous data points I'm going to provide a fictitious setup that aligns with my mathematical needs. I am a company that is measuring the bounciness of balls and providing two 1-10 rating to each ball based on how it compares to all other balls. The first rating is based on more bounces being good. The second rating is based on less bounces being good. Is there a way I can take all data points (number of ball bounces) and find an accurate 1-10 rating (for each ball) for both scenarios? Thank you for your time and help! (If I used incorrect tags, I apologize) AI: Here is one way. Let $M$ be the maximum and $m$ be the minimum number of bounces among all the balls. Then the rating is just a matter of rescaling $m \le x \le M$ to $1 \le y \le 10$. The following function does this: $$ f(x) = 1+\left \lfloor 10 \frac{x-m}{M+1-m} \right \rfloor. $$ Then $f$ will map a ball's number of bounces, $x$, to an integer between 1 and 10, inclusive, with the property that if $x_1>x_2$, then $f(x_1) \ge f(x_2)$. More bounces results in a higher rating. To get a lower rating with more bounces, just subtract $f(x)$ from 11: $$ g(x) = 10-\left \lfloor 10 \frac{x-m}{M+1-m} \right \rfloor. $$
H: Axiomatization of $\mathbb{Z}$ via well-ordering of positives. Though I've seen several cool axiomizations of $\mathbb{R}$, I've never seen any at all for $\mathbb{Z}$. My initial guess was that $\mathbb{Z}$ would be a ordered ring which is "weakly" well-ordered in the sense that any subset with a lower bound has a minimal element. However, after seeing this definition of a discrete ordered ring, I'm less sure. I made that guess under the impression that the fundamental characteristic of $\mathbb{Z}$ is that every nonzero element has exactly one representation of the form $\pm (1+1+\dots+1)$, but that seems to be shared by every DOR. Presumably, this definition wouldn't exist if every instance of it was isomorphic to $\mathbb{Z}$, so can someone give me an example of another discrete ordered ring? More to the point, what is a sufficient characterization of $\mathbb{Z}$? (and a proof sketch of uniqueness would be nice) I'm aware that $\mathbb{Z}$ is pretty easily constructible from $\mathbb{N}$, but I want to use this for a seminar and given the audience I am expecting, I would rather not deal with Peano. (And I guess it feels like cheating to say "$\mathbb{N}$ is a well-ordered rig") AI: Second-order quantification allows us to talk about properties of subsets of the ring, much like the completeness axiom of the real numbers (which is too a second-order statement). We can adjoin the usual theory of ordered rings the following axiom: $$\forall A(A\neq\varnothing\land\exists x\forall a(a\in A\rightarrow x<a)\rightarrow\exists y(y\in A\land\forall x(x\in A\rightarrow y\leq x)))$$ Saying that for non-empty every set $A$, if there is a lower bound for $A$ then $A$ has a minimal element. We can also follow Zhen Lin's suggestion in the comments. Notice that $\mathbb Z$ is the unique free additive group which has only one generator. That is: $$\exists x(x\neq 0\land\forall A(x\in A\land\forall a\forall b(a\in A\land b\in A\rightarrow a+b\in A)\land\forall a(a\in A\rightarrow -a\in A)\rightarrow\forall y(y\in A)$$ This is a very complicated way of saying that there exists some $x$ which is non-zero and every $A$ in which $x$ is an element, and $A$ is closed under addition and negation imply that $A$ is everything. In $\mathbb Z$ this is true because $x=1$. However this is not true for any other ordered ring.
H: Does this equation imply the Class Equation? At this time, I am reading the following theorem. Let $G$ be a group acting transitively on a set $\Omega$. Then \begin{equation} \|G\|=\sum_{g\in G}\chi(g) \tag{$\clubsuit$} \end{equation} wherein $\chi(g)=\|\{\alpha \in \Omega :\alpha^g=\alpha\}\|$. It seems to me that the equation $(\clubsuit)$ is very similar to the class equation for a finite group $G$, \[ \|G\|=\|Z(G)\|+\sum_{i=1}^{k}\frac{\|G\|}{\|C_G(x_i)\|}.\] Are these equations connected to each other? Thanks. AI: The class equation is really a special case of the fact that if $G$ acts (not necessarily transitively) on a set $\Omega,$ then $\|\Omega\|$ is the sum of the size of the orbits. In the class equation, $G$ acts on itself by conjugation, the orbits are the conjugacy classes, and the size of the conjugacy class of $x$ is $[G:C_{G}(x)].$ The euation as you have written it is often known as the modified class equation. In that, all the conjugacy classes of size $1$ have been collected together, since the element $x$ is in a conjugacy class of size $1$ if and only if $x \in Z(G).$ There is a connection in as much as when a finite group $G$ acts transitively on a set $\Omega,$ then all point stabilizers have the same size$\|G_{\alpha}\|$ for any $\alpha \in \Omega$ so we have $\|\Omega\| = [G:G_{\alpha}].$ Thus $\sum_{ \alpha \in \Omega }\|G_{\alpha}\| = \|G\|.$ But that sum is the number of order pairs $(\alpha, g) \in \Omega \times G$ such that $\alpha.g = \alpha.$ If we sum this over $g \in G$ first, it's the same as $\sum_{g \in G} \# (\alpha \in \Omega : \alpha.g = \alpha).$
H: Applying a contraction to balls' centers increases the size of the balls' intersection? The following statement seems clearly true, but I'm having a hard time proving it: Fix $\alpha\in[0,1]$. Let $\mu$ be Lebesgue measure. For $r\ge 0$, let $B(c,r)\equiv[c-r,c+r]$. Fix $r_1,\ldots,r_n\in[0,\infty)$, and $c_1,\ldots,c_n\in\mathbb R$. Then, $\mu\left(\bigcap_{i=1}^{n}B(\alpha c_{i},r_{i})\right)\ge\mu\left(\bigcap_{i=1}^{n}B(c_{i},r_{i})\right)$ In words, the statement just says that contracting the 1-dimensional balls' centers towards $0$, by a constant proportion, while leaving the radii unchanged, increases the amount of overlap. This seems straightforward enough: if the centers were contracted all the way to the origin, then the intersection would just be the smallest ball. There can't be more overlap than that. Can anyone suggest a simple proof, or point me to a relevant theorem? AI: Let $B(c,r)=\bigcap_{i=1}^n B(c_i,r_i)$. We have $\|c_i-c\|\le r_i-r$ for all $i$. Hence, $\|\alpha c_i-\alpha c\|\le r_i-r$, which implies $B(\alpha c,r)\subseteq \bigcap_{i=1}^n B(\alpha c_i,r_i)$. Explanation: a ball $B$ is contained in ball $B'$ if and only if the distance between centers is at most radius(B')-radius(B).
H: Checking divisibility by large numbers I am currently familiar with the method of checking if a number is divisible by $2$, $3$, $4$, $5$, $6$, $8$, $9$, $10$, $11$. However is there a way to check if a no is divisible by $23$ ? I read something at this link regarding this matter but couldn't really figure out what the author was trying to say. Any help in this regard would be appreciated. Thanks. AI: The idea behind "all those checkings" is to look at divisibility by computing the number you want to check divisibility modulo the divisor. For instance, the trick for computing divisibility by $11$ all reduces to $$ a \times 10^3 + b \times 10^2 + c \times 10 + d \equiv -a + b - c + d \pmod{11} $$ so that you can look at the alternated sum of the numbers instead of looking at the big one. You can develop a similar trick modulo $23$, for instance since $100 \equiv 8 \pmod {23}$, you have $$ a \times 100^2 + b \times 100 + c \equiv a \times 64 + b \times 8 + c \equiv a \times (-5) + b \times 8 + c \pmod {23}. $$ Perhaps the trick doesn't look quite as good, but that's because the tricks work well when the numbers are small or pretty (such as : it is easy to look at divisibility by $10000000000$). Hope that helps,
H: What does sup mean? I found this formula regarding calories burned: Rate per Pound (Cal/lb-min)=A+BV+CV.sup.2 +KDV.sup.3 where: V=Running Speed (mph)--limited to a minimum of 3 mph and a maximum of 14 mph A=0.0395 B=0.00327 C=0.000455 D=[0.00801(W/154).sup.0.425 ]/W W=Weight (lbs) K=0 or 1 (0=Treadmill; 1=Outdoors) But what do the .sup.2 and .sup.3stand for? What does it mean? AI: It means superscript (usually an exponent). For example, the first equation is supposed to read: $$\text{Rate per pound (Cal/lb-min)} = A + BV + CV^2 + KDV^3$$ In my opinion, nobody really writes like that unless they have no way to insert a superscript, which I suspect is the case.
H: Maximum point of a polar function I have a curve C with polar equation $$r^2 = a^2\cos{2\theta} $$ And I am looking to find the length $x$ when $r=max$ Judging from the equation: $$r = \sqrt{a^2\cos{2\theta}} $$ R will be maximum at $\cos{2\theta}=1$ So the maximum value of $r$ is: $$r = \sqrt{a^2} =a$$ However the derivative disagrees as: $$x^2=r^2\sin^2{\theta}=a^2\cos{2\theta}\,\sin^2{\theta} \\ \frac{d}{d\theta}\left (a^2\cos{2\theta}\,\sin^2{\theta} \right )=a^2(2\sin{\theta}\cos{\theta}-8\sin^3{\theta}\cos{\theta}) \\ \sin{\theta}=\frac{1}{2} \\ \theta= \frac{\pi}{6} \\ r= \frac{a}{\sqrt{2}}$$ What am I doing wrong? AI: Differentiating $r$ wrt to $\theta$ gives $\theta \mapsto \|a\|\frac{\sin 2 \theta}{\cos 2 \theta}$, which is zero when $\cos 2 \theta = \pm 1$. No disagreement there! From this compute $x = r \cos \theta$. Since $\theta = 0$ maximizes $r$, the corresponding $x = \|a\|$.
H: game show problem Possible Duplicate: The Monty Hall problem You are in a game show. You have to choose between three buttons, A, B and C. Pressing one of them will give you £200,000, and pressing either of the other two will give you a free mousemat. You choose a button at random (button A). The gameshow host doesn't tell you if you have won or not, but he does tell you that button B was one of the wrong buttons. He also gives you a chance to change your mind. Should you stick with button A, or switch to button C? Why? Or does it make no difference? Can any one suggest what should be the possible answer for this. Does it really make any difference If I change my mind? So what are the possibilities here? AI: This is an very well-known problem. Your probability of success is maximised by changing. You can read about it all day here: http://en.wikipedia.org/wiki/Monty_Hall_problem , but it is instructive to think about it yourself if you are actually studying probability. Writing out rigorously why switcihing is better can be tricky. The one intuitive explanation that really stuck with me is the thought experiment in which you imagine there are, say 100 buttons. You pick one, say button No. 1. The gameshow host reveals to you that buttons 2 through 56 and buttons 58 through 100 are wrong buttons. Now do you want to switch to door 57 which for some reason the host has not mentioned? Or do you really think you would do just as well sticking with button No. 1? It seems clear to me now that in sticking with door No. 1 you have a 1/100 chance of being correct.
H: Are Different Areas of Number Theory That Different I was wondering if number theorists are "number theorists," or eventually resolve themselves into one of the various branches - i.e., algebraic, analytic, etc. Also out of curiosity, I was wondering if there is a particular branch or area that draws the most attention today. (Full disclosure: this is just of interest. I am by no means in a position to utilize such an answer, but in that I am pretty much in math isolation, this is my only forum to find out anything.) Thanks AI: Yes, different areas of number theory can be very, very different. At my university, I'm surrounded by many different number theorists, and half of us can barely understand the other half. I consider myself a rising analytic number theorist and I'll be working close to 4 other analytic number theorists (Jeff Hoffstein's 4 students and me, a hopeful 5th). And between the five of us, we don't come close to understanding a single thing done by the other major corps of number theorists here (Joe Silverman's 4 students). To skimp on lots of details, we do analytic number theory and they do algebraic number theory. And this is all different from the number theory I looked into during my undergrad at Georgia Tech. I would now call that Additive Combinatorics or Additive Number Theory, but I think of most of the people who research in that area as analytic number theorists (just different than my cup of analytic-number-theory-tea). And somehow, there are many disparate parts of number theory. I would say that the three areas I mentioned are also pretty big in drawing lots of attention: the Langlands Program is pretty big right now and has a lot to do with automorphic forms. Elliptic Curves are also pretty big. And Terry Tao does a lot with arithmetic and additive number theory, and thus a lot tends to get done. But I should mention, to not give you the completely wrong idea, that there is some interplay between the different branches of number theory. I like to think that I'm going to do a very algebraic form of analytic number theory, and questions related to elliptic curves might even come up now and again. But number theory is a large field, all the larger because it tends to be classified by content rather than method (as opposed to group theory, analysis, harmonic analysis, ode and pde, etc.).
H: Simplifying the expression of exponential and logarithms I want to simplify the following expression. $$Y=\text{Bottom} + \frac{\text{Top}-\text{Bottom}}{1+10^{((\log EC50-X))}}$$ $\log$ is base of $10$. Some may know that it's a dose response curve, and I want to solve for $EC50$. I tried simplifying it but I forgot math long time ago and I couldn't find the answer online. My problem was the exponential and logarithms part. Can someone simplify the expression to solve for $EC50$? Thank you very much. Edit : And if someone could explain simplifying exponential and logarithm on this one, it would be great as well. I forgot all the math and I would appreciate it. AI: Starting with $$Y = \text{Bottom} +\frac{\text{Top}-\text{Bottom}}{1+10^{\log_{10} EC50 -X}}$$ then using $10^{\log_{10} EC50 -X} = EC50 /10^X$ (if that is what you meant) you get $$Y -\text{Bottom} =\frac{\text{Top}-\text{Bottom}}{1+EC50 /10^X}$$ or $$1+EC50 /10^X =\frac{\text{Top}-\text{Bottom}}{Y -\text{Bottom}}$$ i.e $$EC50 =10^X\frac{\text{Top}-Y}{Y -\text{Bottom}}.$$
H: "Mixing" the diagonals of a positive semidefinite matrix I have arrived at this result from a very different perspective (quantum operations) but, being a completely algebraic result, I was hoping that there would be a simple algebraic way of looking at it too. Let $P$ be a positive semidefinite matrix. Let $E$ be a diagonal matrix with real entries such that - Tr$(E)=0$ Diag$(P+E) \succeq 0 $ [That is, for all $i$ , $P_{ii}+E_{ii}\geq 0$] Prove that $P+E$ is positive semidefinite. Thanks in advance! AI: Unless I'm missing something, the result is not true. Take $P:=\pmatrix{2&1\\1&1}$, and $E=\pmatrix{1&0\\0&-1}$. Then $P$ is symmetric, positive definite, $E$ is diagonal, $P+E=\pmatrix{3&1\\ 1&0}$. Each diagonal entry of $P+E$ is non-negative, and the trace of $E$ is $0$ but $P+E$ is not positive semi-definite (consider $x=\pmatrix{1\\-3}$).
H: convert values from one coordinate system (x,y) to another coordinate system (x', y') Following is a graph that contains both coordinate systems (x,y) and (x',y'). x, y, x', and y' are all axes y' ^ \| +----------> x \| \| \| \| +--+--+--+ \| \| \| \| +--+--+--+ \| \| \| \| +--+--+--+-> x' \| v y Above is a picture of a graph. For example, the following is graph or grid of the (x',y') coordinate system. y' ^ \| +--------+ \| \| \| \| +--+--+--+ \| \| \| \| +--+--+--+ \| \| \| \| +--+--+--+-> x' What is the transformation T which transforms coordinates (x,y) to coordinates (x',y')? Please include simple, intuitive explanations and steps. Note: the solution, T, is probably a 2d matrix. AI: Note that your $x'$ is the same as your $x$. To get $y'$, first note that $y$ is in the opposite direction of $y'$. Hence, $y = c - y'$ where $c$ is some constant. The constant is obtained by finding the value of $y$ at say $y'=0$, in your case the constant is $3$.
H: Divisibility criteria of 24. Why is this? I am currently familiar with the method of checking if a number is divisible by $2, 3, 4, 5, 6, 8, 9, 10, 11$. While Checking for divisibility for $24$ (online). I found out that the number has to satisfy the divisibility criteria of $3$ and $8$. I agree this gives the answer. But why cant I check the divisibility using the divisibility criteria of $6$ and $4$ ? Is there a rule to this criteria ? AI: The problem here might be something like $12$. You see, we have that $12$ is divisible by both $6$ and $4$, but it's not divisible by $24$. The reason they suggest $3$ and $8$ is because they are relatively prime, meaning that you can't have the sort of overlap in the case of $6$ and $4$. This all has to do with the Fundamental Theorem of Arithmetic, which says that each number can be written uniquely as a product of primes, and primes have the special characteristic (or as Marvis points out, they are defined to be exactly those numbers with the characteristic) that if $p\|ab$, then $p\|a$ or $p\|b$. So if $3$ and $8$ divide a number, then $24$ divides that number. But $6$ and $4$ dividing a number doesn't even guarantee that $8$ divides that number.
H: If $k \equiv 1 \pmod 4$ and $q > 3$, does it follow that ${q^k}\sigma(q^k) \equiv q\sigma(q) \equiv 2 \pmod {q - 1}$? If $k \equiv 1 \pmod 4$ and $q > 3$ (where $q$ is prime), does it follow that ${q^k}\sigma(q^k) \equiv q\sigma(q) \equiv 2 \pmod {q - 1}$? Observe that $q\sigma(q) \mid {q^k}\sigma(q^k)$ when $k \equiv 1 \pmod 4$. I got $q\sigma(q) \equiv 2 \pmod {q - 1}$ (for $q > 3$) from this Wolfram link. Edit: After taking to anon, I would like to add that, for the problem I am considering, I actually have $q \equiv k \equiv 1 \pmod 4$. anon's answer gave $k \equiv 1 \pmod {q - 1}$ as a condition to check for the validity of the conjecture above. AI: Given $q$ is prime, we can use the fact that $$q\equiv (q-1)+1\equiv 1\mod (q-1)$$ in order to obtain the slightly more general result (for any nonnegative integers $r,k\ge0$) $$\begin{array}{c l} q^r\sigma_1(q^k) & =q^r\big(q^0+q^1+q^2+\cdots+q^k\big) \\[3pt] & \equiv (1)^r(\underbrace{1+1+1+\cdots+1}_{k+1}) \\ & \equiv k+1 \mod (q-1).\end{array}$$ Of course plugging in $k=1$ and $r=1$ gives $q\,\sigma_1(q)\equiv 2\mod (q-1)$ as in the Wolfram link. The only reason the link says $q\ne3$ is because then the symbol '$2$' actually means $0$ modulo $3-1=2$. (This means the conjecture is not generally true unless $k\equiv1\bmod (q-1)$ happens to hold.)
H: Vector as argument of a function Given a function $f(x)=y$ is correct to say that $f\left(\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right]\right)=\left[\begin{array}{c} y_1 \\ y_2 \\ y_3 \end{array}\right]$? AI: Yes, assuming that $x \in A^3$ and $y \in B^3$ for some suitable spaces $A,B$ you just give the explicit representation of those elements which is equivalent to actually writing $x$ or $y$. If you want to make it even more formal you can say for example that the $x_i$ represent the coordinate entries of $x$, but that is common notation anyways and most people will understand it.
H: Why are the solutions to second order differential equations summed? I'm an A-Level maths student, and earlier in the year we learnt about second order differential equations in the form $ A\frac{d^2y}{dx^2} + B \frac{dy}{dx} + Cy = 0$ and $ A\frac{d^2y}{dx^2} + B \frac{dy}{dx} + Cy = f(x)$ For the first form, we let $y = e^{mx} $ (1), then $\frac{dy}{dx} = me^{mx}$ and $\frac{d^2y}{dx^2} = m^2e^{mx} $. This means $Am^2 + Bm + C = 0$ (2). Substituting the solutions into (1) gives us $y=e^{m_0 x}$ and $y=e^{m_1 x}$, where $m_0$ and $m_1$ are the solutions to (2). From this, we somehow conclude that $y = c_1e^{m_0 x} + c_2e^{m_1 x}$ But what I don't understand is, where to the constants come from, and why are the two solutions summed? Regarding the second form of the equation, the above method breaks down because we cannot solve the quadratic. Why can we ignore the $f(x)$ at this stage? When $f(x)$ is present, we come up with a different solution (with no constants) called the particular integral. Where does the particular integral come from, and why is that also added to the other solution we obtained by ignoring the $f(x)$? AI: The constants you get are arbitrary integration constants. When you take indefinite integrals, you always have a $+C \,$ term at the end, which can be any arbitrary constant. The method you use to solve differential equations is fundamentally the same, so you're left with a few integration constants. In this case, that number is 2, since the highest order derivative is a second order one. Both multiplication by constants and differentiation are linear operators. Hence, $A(c_1 y_1+c_2 y_2)''+B(c_1 y_1+c_2 y_2)'+C(c_1y_1+c_2y_2)=c_1(Ay_1''+By_1'+Cy_1)+c_2(Ay_2''+By_2'+Cy_2)$. That is to say, if $y_1$ and $y_2$ both satisfy the differential equation $Ay''+By'+Cy=0$, then so does $c_1 y_1+c_2 y_2$. Since we know there should be two independent integration coefficients, if $y_1$ and $y_2$ are independent then this is the general solution. To solve the inhomogenous equation with $f(x)$ on the right hand side, remember again the linearity property. So if $c_1 y_1+c_2 y_2$ is the general solution for $Ay''+By'+Cy=0$, and $y_p$ is any solution to $Ay''+By'+Cy=f(x)$, then $c_1 y_1+c_2 y_2+y_p$ will also solve $Ay''+By'+Cy=f(x)$ since $A(c_1 y_1+c_2 y_2+y_p)''+B(c_1 y_1+c_2 y_2+y_p)'+C(c_1y_1+c_2y_2+y_p)=c_1(Ay_1''+By_1'+Cy_1)+c_2(Ay_2''+By_2'+Cy_2)+Ay_p''+By_p'+Cy_p=f(x)$. This works in reverse also, if $y_p$ and $y_q$ are two solutions to the inhomogenous equation, then their difference must be of the form $c_1 y_1+c_2 y_2$ for some $c_1, c_2$, so this is the most generic solution possible.
H: Group structure on $\mathbb R P^n$ For which positive integers $n$ can $\mathbb R P^n$ be given the structure of a topological group? I believe that $\mathbb R P^n$ cannot be given a Lie group structure for even $n$, since then it is not orientable. But, this doesn't necessarily imply it doesn't have a topological group structure (which is not smooth); moreover, it tells us nothing about odd $n$. And ideas? AI: As Olivier has soon this can be reduced to the question for spheres. One way to prove this is to note that for a topological group $G$ we have that $G$ is homotopy equivalent to $\Omega BG$, the loopsapce of the classifying space. For example $\Omega BS^1 = \Omega \mathbb{C}P^\infty = \Omega K(\mathbb{Z},2) = K(\mathbb{Z},1) = S^1$. Thus the question is which spheres are loop spaces of classifying spaces. Adams' work showed this is true only for $n=0,1,3$.
H: Finding the matrix of multiplication by $\theta^2$, where $\theta^3 - 3\theta + 1 = 0$ This is a problem from a on-line source which yet comes with a solution (self-studier; not h.w.). Let $E = \mathbb Q(\theta)$, where $\theta$ is a root of the irreducible polynomial \[ X^3 -3X + 1. \] I was wondering how one gets the matrix of the endomorphism $E \to E$ induced by multiplication by $\theta^2$. I can see that $\theta^2 \cdot 1 = \theta^2$, $\theta^2 \cdot \theta = \theta^3 = 3\theta - 1$, and $\theta^2 \cdot \theta^2 = \theta \cdot \theta^3 = 3\theta^2 - \theta$. This is where I fall apart. The solution gives the matrix \begin{bmatrix} 0&- 1&0 \\ 0&3&-1 \\ 1&0&3 \end{bmatrix} I can that see if you multiply this by a $3 \times 1$ column vector of $1$'s, $4\theta^2$ + $2\theta$ $- 1$ emerges. But I don't see how to derive the matrix; and, further, if it is unique. That is to say, if the $-1$ in the top row was in a different position in the top row, it seems you would still get $-1$ in the top position in the product. Only a little of this is mine — except for the misstatements and confusion. Thanks. AI: Let's first check that $X^3 - 3X + 1$ is irreducible over $\mathbb Q$. Since it's a cubic, it's enough to show that it does not have a root in $\mathbb Q$. The rational root theorem tells you that the only possible roots in $\mathbb Q$ are $\pm 1$, and neither of these works. Some general facts: if I have a field $F$ and an element $\alpha$ algebraic over $F$ with minimal polynomial $p(X)$ of degree $n$, then $\{1, \alpha, \ldots, \alpha^{n - 1}\}$ is a basis for $F(\alpha)$ over $F$. A proof should be in any reference, but it isn't so hard to come up with on your own: the relation $p(\alpha) = 0$ allows you to ignore higher powers of $\alpha$, and an honest linear dependence would give you a non-zero polynomial of degree $< n$ having $\alpha$ as a root, which is impossible. The given matrix, as Olivier pointed out, is using exactly this sort of basis for both the source and target. And you've actually calculated the entries of this matrix — the second column, for example, gives the coefficients of $\theta^2 \cdot \theta = (-1) \cdot 1 + 3 \cdot\theta + 0 \cdot \theta^2$. Of course, if you use different bases then your matrix will change, but that is now a question of linear algebra.
H: ODE series solution with IVP I am a little rusty with initial conditions. If anyone can confirm that I nailed this then I would be very happy. i need to write about five terms in my series to the solution of $xy'' + y = 0$ $y(1) = 5$ $y'(1) = 0$ So my solution will look something like $y = \sum_{n=0}^{\infty}a_n (x - 1)^n$ $y' = \sum_{n=1}^{\infty}na_n (x - 1)^{n-1}$ $y'' = \sum_{n=2}^{\infty}n(n-1)a_n (x - 1)^{n-2}$ $\begin{align}xy'' + y & = x\sum_{n=2}^{\infty}n(n-1)a_n (x - 1)^{n-2} + \sum_{n=0}^{\infty}a_n (x - 1)^n \\ & = (x - 1 + 1) \sum_{n=2}^{\infty}n(n-1)a_n (x - 1)^{n-2} + \sum_{n=0}^{\infty}a_n (x - 1)^n \\ & = \sum_{n=2}^{\infty}n(n-1)a_n (x - 1)^{n-1} + \sum_{n=2}^{\infty}n(n-1)a_n (x - 1)^{n-2} + \sum_{n=0}^{\infty}a_n (x - 1)^n \\ & = \sum_{n=1}^{\infty}n(n+1)a_{n+1} (x - 1)^n + \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} (x - 1)^n + \sum_{n=0}^{\infty}a_n (x - 1)^n \\ & = \sum_{n=0}^{\infty}n(n+1)a_{n+1} (x - 1)^n + \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} (x - 1)^n + \sum_{n=0}^{\infty}a_n (x - 1)^n \\ & = \sum_{n=0}^{\infty}(x - 1)^n[n(n+1)a_{n+1} +(n+2)(n+1)a_{n+2} +a_n]\\ & = 0 \end{align}$ So it is 0 if and only if $n(n+1)a_{n+1} +(n+2)(n+1)a_{n+2} +a_n = 0 \iff a_{n+2} = -\dfrac{a_n}{(n+1)(n+2)} -\dfrac{na_n}{n+2}$ After solving I got this pattern $a_2 = \dfrac{-a_0}{2}$ $a_3 = \dfrac{-a_1}{2\cdot 3}-\dfrac{a_2}{3}$ $a_4 = \dfrac{-a_2}{3\cdot 4}-\dfrac{2a_3}{4}$ For the initial conditions, is this correct? $y(1) = \sum_{n=0}^{\infty}a_n (1 - 1)^n = a_0 = 5$ because all the other terms go to 0 and the limit of $0^0$ is 1 Same process $y'(1) = y' = \sum_{n=1}^{\infty}na_n (1 - 1)^{n-1}= a_1 = 0$ because all the other terms go to 0 and the limit of $0^0$ is 1 If I am correct with I am doing so far then my solution should be $y \approx 5 - \frac{5}{2}x^2 + \frac{5}{6}x^3 - \frac{5}{24}x^4$ Thank you very much for taking the time to read AI: Your solution is correct. When the number of terms to determine is given in advance, I prefer not to write out the entire series at all. My quick-and-dirty solution would be $$y=5+a(x-1)^2+b(x-1)^3+c(x-1)^4+O((x-1)^5)$$ $$y''=2a+6b(x-1)+12c(x-1)^2+O((x-1)^3)$$ $$xy''=y+(x-1)y''=2a+(2a+6b)(x-1)+(6b+12c)(x-1)^2+O((x-1)^3)$$ Hence $2a+5=0$, $2a+6b=0$, $a+6b+12c=0$, and the solution is $a=-5/2$, $b=5/6$, $c=-5/24$.
H: $U\subset{R}^m$ open $f\colon U\to N$ local homeo and $y\in N$ with $\operatorname{card}\big(f^{-1}(\{y\})\big)$ is infinite then $f$ is not closed. Let $U$ be an open set. If $f\colon U\subset{R}^m\to N$ is a local homeomorphism and exists $y\in N$ such that $\operatorname{card}\big(f^{-1}(\{y\})\big)$ is infinite prove that $f$ is not closed. I proved that $f^{-1}(\{y\})$ is closed and discrete, I proved also that $f$ must be continuous and open. But I don't know how to use the fact that $U$ is open in $\mathbb{R}^m$ and the fact that $\operatorname{card}\big(f^{-1}(\{y\})\big)$ is infinite. AI: Let $\{x_n\}$ be a sequence of preimages of $y$. Each of them has a neighborhood $U_n$ in which $f$ is a homeomorphism. Pick a sequence $y_n$ in $N\setminus\{y\}$ such that $y_n\to y$ and for each $n$ there exists $x_n'\in U_n$ such that $f(x_n')=y_n$. Then: the set $\{x_n'\}$ is closed in $U$ the set $\{f(x_n')\}$ is not closed in $f(U)$.
H: Cardinality of varieties Does a (connected?) projective (or just complete?) variety over a finite field have cardinality congruent to 1 mod the size of the field? Do (connected?) affine varieties have cardinality a power of the size of the field? I think the answers to these questions are supposed to be yes, but I suspect some sort of better formulation of the question (missing hypotheses perhaps) is needed. Maybe there needs to be some mention of rational points over an algebraic closure. Feel free to fill such things in. It would be nice if there was any sort of converse, like "if a variety is connected and has cardinality 1 mod \|K\|, then it is complete", but I'm less sure something like this exists. AI: Nope. The general case is very far from the case of projective and affine space. Already this is false for elliptic curves; in general we only have the Hasse bound $$q + 1 - 2 \sqrt{q} \le \|E(\mathbb{F}_q)\| \le q + 1 + 2 \sqrt{q}$$ and neither lower nor upper bound can be improved conditional on the Sato-Tate conjecture. See also the Weil conjectures. In the affine case a simple counterexample is the variety $xy = 1$, which has cardinality $q - 1$ over $\mathbb{F}_q$. In the projective case, any curve of genus $0$ with at least one point over $\mathbb{F}_q$ is isomorphic to $\mathbb{P}^1$ and hence has $q + 1$ points over $\mathbb{F}_q$, but it may happen that it has no such points, e.g. take the projective closure of $x^4 + y^4 = -1$ over $\mathbb{F}_5$.
H: Eigenvalue For A 2-Fold Tensor This is problem 4 from page 258 of Curtis's Linear Algebra: An Introductory Approach. I seem to be having trouble understanding something needed to solve the problem, which reads Suppose A and B are matrices in triangular form, with zeros above the diagonal. Show that A $\times$ B has the same property, and hence that every eigenvalue of A $\times$ B can be expressed in the form $\alpha\beta$, where $\alpha$ is an eigenvalue on A and $\beta$ is an eigenvalue of B. In the actual problem, there's a little dot above the cross in the "matrix product" A $\times$ B to signify that the resulting matrix represents a transformation on the vector space of tensors $V \otimes W$. Because this matrix has entries given by $$\pmatrix{ \alpha_{11}\mathbf{B} & \alpha_{12}\mathbf{B} & ... & \alpha_{1n}\mathbf{B} \\ ... \\ \alpha_{n1}\mathbf{B} & \alpha_{n2}\mathbf{B} & ... & \alpha_{nn}\mathbf{B} }$$ when represented against the basis $\lbrace v_1 \otimes w_1 , ... , v_1 \otimes w_m , v_2 \otimes w_1 , ... , v_2 \otimes w_m , ... , v_n \otimes w_m \rbrace$, I can understand the first claim in the problem (i.e. that the matrix A $\times$ B is also triangular). I'm having trouble proving that every eigenvector of the above matrix can be represented as a product of eigenvectors of A and B. Because of how linear transformations on tensor products are defined, we know that $$ (\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{a} \otimes \mathbf{b}) = (\mathbf{Aa} \otimes \mathbf{Bb}) $$ Also, if $\gamma$ is an eigenvalue of A $\times$ B, then $$ (\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{a} \otimes \mathbf{b}) = \gamma (\mathbf{a} \otimes \mathbf{b}). $$ But at this point, I'm stuck... AI: The eigenvalues of a triangular matrix are its diagonal values. You can prove this by computing the characteristic polynomial, but you can also without too much effort write down the eigenvectors inductively.
H: Finding the Number of Ordered Triples $(x_1,x_2,x_3)$ such that $x_1 + 2x_2 + 3x_3 = n$ I would like to find the number of ordered triples $(x_1,x_2,x_3)$ such that $x_1 + 2x_2 + 3x_3 = n$. For each $n$ call this number $r(n)$. So after reading some similar questions on this site I'm pretty sure this is just some form of the partition function, and my book states explicitly that it is equivalent to asking for all partitions $x_1 + x_2 + x_3 = n$ where $x_1 \leq x_2 \leq x_3$, unfortunately I don't fully understand the wikipedia article on the partition function, so let's just leave that aside for now.. What I was first asked to do was show that $\sum_0^{\infty}r(n)z^n = \frac{1}{(1-z)(1-z^2)(1-z^3)}$. So I successfully did this using the Cauchy Product Formula, and then I was asked to find the partial fraction decomposition of this rational equation into a particular form, which I did: $$ -\frac{\frac{1}{6}}{(z-1)^3} + \frac{\frac{1}{4}}{(z-1)^2} - \frac{\frac{17}{72}}{z-1} + \frac{\frac{1}{8}}{z+1} - \frac{\frac{1}{9}e^{\frac{2\pi i}{3}}}{z-e^{\frac{2\pi i}{3}}} - \frac{\frac{1}{9}e^{\frac{-2\pi i}{3}}}{z-e^{\frac{-2\pi i}{3}}}$$ So now I'm asked to show that $r(n)$ is the integer nearest $\frac{(n+3)^2}{12}$. My thinking was that I was supposed to use this decomposition and convert each fraction back to series form, combine them all, and then the coefficients of this series would be given by $\frac{(n+3)^2}{12}$. But I can't seem to simplify the series(plural) even close to that, can anyone take a look at this for me? Thanks. AI: Life is more pleasant if you set up the partial fractions this way: $${1\over(1-z)(1-z^2)(1-z^3)}={1/6\over(1-z)^3}+{1/4\over(1-z)^2}+{1/4\over1-z^2}+{1/3\over1-z^3}$$ Can you take it from there?
H: Looking for some simple topology spaces such that $nw(X)\le\omega$ and $\|X\|>2^\omega$ I believe there are some topology spaces which satisfying the network weight is less than $\omega$, and its cardinality is more than $2^\omega$ (not equal to $2^\omega$), even much larger. Network: a family $N$ of subsets of a topological space $X$ is a network for $X$ if for every point $x \in X$ and any nbhd $U$ of $x$ there exists an $M \in N$ such that $x \in M \subset U$. Here I want to look for some simple topology spaces which are familiar with us. However, a little complex topology space is also welcome! Thanks for any help:) AI: If $X$ is $T_0$ and $nw(X)=\omega$ (network weight) then $\|X\|\leq 2^\omega$. Let $\mathcal N$ be a countable network. For each $x\in X$ consider $N_x=\{N\in\mathcal N: x\in N\}$. Since $X$ is $T_0$ it follows that $N_x\ne N_y$ for $x\ne y$. Thus, $\|X\|\leq \|P(\mathcal N)\|=2^\omega$.
H: Isomorphism between quotient modules Is it true for a commutative ring $R$ and its ideals $I$ and $J$ that if the quotient $R$-modules $R/I$ and $R/J$ are isomorphic then $I=J$? AI: No, just take a polynomial ring $k[X]$ for some field $k$ (which is a commutative domain), and consider the ideal generated by $X$ and the ideal generated by $X-1$. The quotient rings are both (isomorphic to) $k$. (added later, by mixedmath) This answer asserts that $R/I \simeq R/J$ (as rings) does not imply that $I = J$. However, the OP asks about $R/I \simeq R/J$ (as $R$-modules), which is covered in the other answer.
H: Given Linear function, find function describing decrease in the rate of change x vs y Given a linear function such as $y = 1.62*x - 0.49$ Scenario 1 If $x = .5$ then $y = .32$ If we then increase $x$ by $10$% ($x=.55$), then $y=.401$, which results in $y$ increasing by $\approx .25$% Scenario 2 If $x = .6$ then $y = .482$ If we then increase $x$ by $10$% ($x=.66$), then $y=.579$, which results in $y$ increasing by $\approx .20$% Question I am looking to explain how an improvement in x results in an improvement in y. Clearly, the relationship is not linear. I do not remember enough calc to model this relationship, if that is possible. I am looking for a mathematical way to say "as you improve x, you will see an improvement in y; however, if x is already high then an improvement in y is less impactful. Therefore, increasing x has the affect of improving y, but the return on investment is greater if x is initially very low." or "if x is bad and you improve it you can expect this return in y, but is diminishes as the starting x is higher" I will also allow that my original premise is faulty. Thanks! AI: Let $f(x) = mx + b$ be any linear function. Suppose we start from two points $x_1$, and $x_2$, with images $f(x_1) = mx_1 + b$ and $f(x_2) = mx_2 + b$. Then we consider a proportional increment $\Delta$ in each case. That is, we transform $x_1 \mapsto (1 + \Delta)x_1$ and $x_2 \mapsto (1 + \Delta)x_2$. The images of these values will be: $f\big((1 + \Delta)x_1\big) = m(1 + \Delta)x_1 + b$ and $f\big((1 + \Delta)x_2\big) = m(1 + \Delta)x_2+b$. If we want to find the proportional increase in the images, we have: $\displaystyle \frac{f\big((1 + \Delta)x_1\big)}{f(x_1)}-1 = \frac{m(1 + \Delta)x_1 + b}{mx_1 + b}-1$ and $\displaystyle \frac{f\big((1 + \Delta)x_1\big)}{f(x_1)} -1= \frac{m(1 + \Delta)x_2 + b}{mx_2 + b}-1$ Now, these numbers don't have to be equal at all, even though $\Delta$, $m$ and $b$ are the same in both equations. The proportional increase in $y$, for a proportional increase in $x$ of $\Delta$ is given by: $\displaystyle P_\Delta(x)= \frac{f\big((1 + \Delta)x\big)}{f(x)} -1= \frac{m(1 + \Delta)x + b}{mx + b}-1 = \frac{m x\Delta}{mx + b}= \frac{1.62x\Delta}{1.62x-0.49}$ What you have observed is that $P_\Delta(x)$ is decreasing somewhere. We can verify this by taking the derivative and finding where it's negative, and also note that it will tend to $0$ with bigger and bigger $x$, by taking the limit as $x \to \infty$. If we want to know when $P_\Delta(x) \geq \Delta$, i.e., the increase in $y$ is bigger or equal than the increase in $x$: $\displaystyle P_\Delta(x) = \frac{m x\Delta}{mx + b} \geq \Delta \Longrightarrow \frac{mx\Delta - (mx+b)\Delta}{mx+b} = -\frac{b\Delta}{mx+b} \geq 0$ This happens when: $b\Delta \leq 0$ (i.e. $b$ and $\Delta$ have different signs) and $mx+b \gt 0 \Longleftrightarrow x \gt -\frac{b}{m}$; and when $b \Delta \geq 0$ (i.e. $b$ and $\Delta$ have the same sign) and $mx+b \lt 0 \Longleftrightarrow x \lt -\frac{b}{m}$.
H: Compact resolvent VS certain boundedness condition The following question is motivated by the definition of spectral triples in noncommutative geometry. This question was split in the following parts: First: Could somebody give diverse examples of operators on Hilbert spaces, having compact resolvent? Now, suppose one has certain algebra $A$ acting as operators on a Hilbert Space $X$. Certain self-adjoint operator $D$ on $X$ is imposed to satisfy, in particular, following axioms in order to be a deemed (generalized) Dirac operator, in an abstract sense: $[D,a]$ is bounded for each $a\in A$ and $(D^2+1)^{-1/2}$ is a compact operator. Second: could somebody explain with an example, why conditions 1. and 2. tend to contradict each other? AI: I'm not sure what you mean by "tend to contradict each other", but here is one relevant point. An easy way to satisfy 1 would be for $D$ to be a bounded operator and $A$ to consist of other bounded operators. However, this would make 2 nearly impossible. For if $D$ is bounded, then so is $T := (D^2 + 1)^{1/2}$. If $T^{-1}$ is compact then it is a homeomorphism, which means the closed unit ball of $H$ is compact. But this happens iff $H$ is finite dimensional. So in this sense, boundedness properties like 1 tend to be in tension with compact resolvent properties like 2.
H: Injective linear map between modules If you have an injective linear map between two free modules of equal dimension, is the determinant of the matrix representing the map necessarily nonzero? If not is there an obvious counterexample? (Everything is over a multivariate polynomial ring over a field.) Thanks! AI: Over an integral domain $D$ this is straightforward as Jack Schmidt says in the comments. Suppose $T : D^n \to D^n$ is a morphism such that $\det(T) = 0$. Then $T \otimes \text{Frac}(D) : \text{Frac}(D)^n \to \text{Frac}(D)^n$ has the same property, where $\text{Frac}(D)$ is the fraction field. Since we are now over a field, $T \otimes \text{Frac}(D)$ is not injective, and so it annihilates some nonzero vector in $\text{Frac}(D)^n$. Scaling this vector suitably puts it in $D^n$, and so $T$ is not injective. Over a general commutative $R$ I have a proof for $n = 2$ (consider the adjugate) but it doesn't seem to generalize. I'll get back to you.
H: Eckmann-Hilton and higher homotopy groups How does the Eckmann-Hilton argument show that higher homotopy groups are commutative? I can easily follow the proof on Wikipedia, but I have no good mental picture of the higher homotopy groups, and I can't see how to apply it. Wikipedia mentions this application in one sentence, with no explanation. (Motivation: I'm thinking of giving a talk on algebraic topology for a general audience of math majors. The purpose would be to try to explain how purely "algebraic" methods can be used to gain serious insight into purely "geometric" problems. This would be a great example, if I could do the proof!) Thanks in advance! AI: Recall that the Eckmann-Hilton argument says that if you have two different unital monoid structures on a set such that both are homomorphisms for the other and both have the same unit, then they're equal and commutative. We want to apply that to see that higher homotopy groups are commutative. The two different monoid structures we want to consider are 'vertical' and 'horizontal' concatenation of maps from $I^n$: if I have two maps $f,g:I^n\to X$ for some topological space $X$ so that $f(\partial I^n)=g(\partial I^n)=x_0$, then I can concatenate them in the following ways as long as $n>1$: $f\cdot g=\begin{cases} f(2x,y,\cdots) &\text{ for } 0\leq x\leq 1/2\\ g(2x-1,y,\cdots) &\text{ for } 1/2\leq x\leq 1 \end{cases}$ $f \star g= \begin{cases} f(x,2y,\cdots) &\text{ for } 0\leq y\leq 1/2\\ g(x,2y-1,\cdots) &\text{ for } 1/2\leq y\leq 1 \end{cases}$ Now, if I can show that these are homomorphisms for each other, then I'll have that $\pi_n$ is commutative for $n>1$. But this is easy: both $(f\cdot g)\star (h\cdot i)$ and $(f\star h)\cdot (g\star i)$ are the following map: $(x,y,...)\mapsto\begin{cases} f(2x,2y,\cdots) &\text{ for }(x,y)\in [0,1/2]\times [0,1/2]\\ g(2x-1,2y,\cdots) &\text{ for } (x,y)\in [1/2,1]\times [0,1/2]\\ h(2x,2y-1,\cdots) &\text{ for } (x,y)\in [0,1/2]\times [1/2,1]\\ i(2x-1,2y-1,\cdots) &\text{ for } (x,y)\in [1/2,1]\times [1/2,1] \end{cases}$ Now, by the Eckmann-Hilton argument, these two sorts of composition are the same and are in fact commutative.
H: Prove $F(x,y,z)=o(\|\|(x,y,z)\|\|)(x,y,z)$ has a vector potential This is an unsolved exercise given in my textbook, which I am having trouble with. The exercise seems simple, but for some reason I can't solve it. Help would be very nice! Let $o(t)$ be a real continuous positive function in $[0,\infty)$, and $F(x,y,z)=o(\\|(x,y,z)\\|)(x,y,z)$ ($\\|\cdot \\|$ means the norm, e.g. $\sqrt{x^2+y^2+z^2}$). We need to prove that $F$ has a vector potential in $\mathbb{R}^3$. Attempt: If $G$ is the potential, we know $G_x = o(\\|(x,y,z)\\|)x$, so I figured $G$ could be of the form $G(x,y,z)=\int_{0}^{x} o(\\|(t,y,z)\\|)t dt + h(y,z)$. So now, to find $h(y,z)$ I need to solve: $$ G_y(x,y,z)=yo(\\|(x,y,z)\\|)=(\int_{0}^{x} o(\\|(t,y,z)\\|)t dt + h(y,z))_y $$ and $$ G_z(x,y,z)=zo(\\|(x,y,z)\\|)=(\int_{0}^{x} o(\\|(t,y,z)\\|)t dt + h(y,z))_z. $$ But I don't really know how to differentiate $\int_{0}^{x} o(\\|(t,y,z)\\|)t$ by $y$ or $z$, which gets me stuck. Thanks for reading. Please help me figure this out! AI: Since $o$ is continue in $[0,+\infty)$, let us put :$\Phi(t)=\int_0^t \tau o(\tau) d \tau $, for all $t \in[0,+\infty)$. Then we know that $\Phi$ is a class $C^1$ function on $[0,+\infty)$ and has derivative : $\Phi'(t)=t o(t)$ forall $t \in [0,+\infty)$. if we consider the function $V$ such that :$$\forall X\in \mathbb R^3 \quad V(X)= \Phi (\\|X\\|)$$ it is clear that $V$ in differentiable on $\mathbb R^3 \backslash \{(0,0,0)\}$ and , putting: $X=(x,y,z)$: $$\frac{\partial V}{\partial x} (X)= \frac{x}{\\|X\\|} \Phi'(\\|X\\|)=\frac{x}{\\|X\\|} \\|X\\| o(\\|X\\|)= x o(\\|X\\|) $$ With the same way we have too:$$\frac{\partial V}{\partial y} (X)= y o(\\|X\\|) \quad \text{and} \quad\frac{\partial V}{\partial z} (X)= z o(\\|X\\|)$$ Then : $\bf{Grad}$$ (V) (x,y,z) = F(x,y,z)$
H: Applying Dominated Convergence and/or Monotone Convergence for $\|Z\|^{1/n}$ Problem: I am self-learning about DCT and MCT and related Lemma's. I understand the theorems as constructed, but I am struggling to apply them. As an example: $X_n=\|Z\|^{1/n}$ where $Z\sim N(0,1)$ I am trying to do the following: a)Identify $X$ such that $X_n\rightarrow X$ a.s b)Determine if $E(X_n)\rightarrow E(X)$ as $n\rightarrow\infty$ I don't think the answer requires exact calculation of expectation, but seeing that done explicitly might help me get my footing. Work/attempt: This is not a homework question, so any level of detail is extremely appreciated and welcomed. This is what I know of MCT/DCT: Monotone Convergence: $f_1,f_2...f_n\uparrow f$ as $n\rightarrow\infty\Rightarrow \lim_{n\rightarrow\infty}\int f_nd\mu=\int fd\mu$ Dominated Convergence: Given measurable functions $f_1,f_2...$ and $\|f_n\|\le g$ for some $g$, and if $f_n\rightarrow f \mu-a.e.$, then $\lim_{n\rightarrow\infty}\int\|f_n-f\|d\mu=0$ and $\lim_{n\rightarrow\infty}\int f_nd\mu=\int fd\mu$ So in my example $X_n$ is a series of nth roots of standard normal RVs. I'm not seeing any monotone convergence, so I look to DCT. Now the trick is to find a function, $g$, that is always greater than or equal to my series. I'm not sure what $X$ this converges to, which makes it hard to find $g$. My first guess is something like $g=\|Z\|^n$, but I have no theoretical grounding for this choice, and it's not clear to me that it even meets my criteria. Help? AI: a) $X=\mathbf 1_{Z\ne0}$ (consider separately the cases $Z(\omega)=0$ and $Z(\omega)\ne0$ and proceed). b) Since $X_n\leqslant1+\|Z\|$ for every $n\geqslant1$ (consider separately the cases $\|Z(\omega)\|\lt1$ and $\|Z(\omega)\|\geqslant1$ and proceed), Lebesgue dominated convergence theorem implies that $\mathrm E(X_n)\to\mathrm E(X)=\mathrm P(Z\ne0)=1$. Note: One can also apply Lebesgue monotone convergence theorem twice, separately to the random variables $U_n=\|Z\|^{1/n}\,\mathbf 1_{\|Z\|\leqslant1}$ and $V_n=\|Z\|^{1/n}\,\mathbf 1_{\|Z\|\gt1}$. To see this, note that $X_n=U_n+V_n$ for every $n\geqslant1$, the sequence $(U_n)$ is nondecreasing, the sequence $(V_n)$ is nonincreasing, $U_n\to U=\mathbf 1_{0\lt\|Z\|\leqslant1}$ and $V_n\to V=\mathbf 1_{\|Z\|\gt1}$. Hence $\mathrm E(X_n)\to\mathrm E(U)+\mathrm E(V)=1$.
H: Group of arbitrary order Can we construct a group of order $n$ for any $n \in \mathbb{Z}^+$ i.e set of positive integers? Are there theorems which characterize the order of any finite group? What is the smallest possible restriction you can have on a group such that there does not exist such a group of particular order. AI: The last part of the question is not entirely clear to me, but if you impose the restriction that the group be nonabelian then there are no qualifying groups of orders 1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 17, 19, 23, 25, 29, .... These numbers are tabulated at the Online Encyclopedia of Integers Sequences.
H: Discriminant for $x^n+bx+c$ The ratio of the unsigned coefficients for the discriminants of $x^n+bx+c$ for $n=2$ to $5$ follow a simple pattern: $$\left (\frac{2^2}{1^1},\frac{3^3}{2^2},\frac{4^4}{3^3},\frac{5^5}{4^4} \right )=\left ( \frac{4}{1},\frac{27}{4},\frac{256}{27},\frac{3125}{256} \right )$$ corresponding to the discriminants $$(b^2-4c, -4b^3-27c^2,-27b^4+256c^3,256b^5+3125c^4).$$ Does the pattern for the ratios extend to higher orders? (An online reference would be appreciated.) AI: Yes. Sketch: $b$ is a symmetric polynomial of degree $n-1$ in the roots and $c$ is a symmetric polynomial of degree $n$, whereas the entire discriminant is a symmetric polynomial of degree $n(n-1)$. It follows that the discriminant is a linear combination of $b^n$ and $c^{n-1}$, and the coefficients can be determined by setting $b = 0, c = -1$ and then $b = -1, c = 0$ and reducing to the computation of the discriminant of $x^n - 1$.
H: Does an injective endomorphism of a finitely-generated free R-module have nonzero determinant? Alternately, let $M$ be an $n \times n$ matrix with entries in a commutative ring $R$. If $M$ has trivial kernel, is it true that $\det(M) \neq 0$? This math.SE question deals with the case that $R$ is a polynomial ring over a field. There it was observed that there is a straightforward proof when $R$ is an integral domain by passing to the fraction field. In the general case I have neither a proof nor a counterexample. Here are three general observations about properties that a counterexample $M$ (trivial kernel but zero determinant) must satisfy. First, recall that the adjugate $\text{adj}(M)$ of a matrix $M$ is a matrix whose entries are integer polynomials in those of $M$ and which satisfies $$M \text{adj}(M) = \det(M).$$ If $\det(M) = 0$ and $\text{adj}(M) \neq 0$, then some column of $\text{adj}(M)$ lies in the kernel of $M$. Thus: If $M$ is a counterexample, then $\text{adj}(M) = 0$. When $n = 2$, we have $\text{adj}(M) = 0 \Rightarrow M = 0$, so this settles the $2 \times 2$ case. Second observation: recall that by Cayley-Hamilton $p(M) = 0$ where $p$ is the characteristic polynomial of $M$. Write this as $$M^k q(M) = 0$$ where $q$ has nonzero constant term. If $q(M) \neq 0$, then there exists some $v \in R^n$ such that $w = q(M) v \neq 0$, hence $M^k w = 0$ and one of the vectors $w, Mw, M^2 w,\dots, M^{k-1} w$ necessarily lies in the kernel of $M$. Thus if $M$ is a counterexample we must have $q(M) = 0$ where $q$ has nonzero constant term. Now for every prime ideal $P$ of $R$, consider the induced action of $M$ on $F^n$, where $F = \overline{ \text{Frac}(R/P) }$. Then $q(\lambda) = 0$ for every eigenvalue $\lambda$ of $M$. Since $\det(M) = 0$, one of these eigenvalues over $F$ is $0$, hence it follows that $q(0) \in P$. Since this is true for all prime ideals, $q(0)$ lies in the intersection of all the prime ideals of $R$, hence If $M$ is a counterexample and $q$ is defined as above, then $q(0)$ is nilpotent. This settles the question for reduced rings. Now, $\text{det}(M) = 0$ implies that the constant term of $p$ is equal to zero, and $\text{adj}(M) = 0$ implies that the linear term of $p$ is equal to zero. It follows that if $M$ is a counterexample, then $M^2 \mid p(M)$. When $n = 3$, this implies that $$q(M) = M - \lambda$$ where $\lambda$ is nilpotent, so $M$ is nilpotent and thus must have nontrivial kernel. So this settles the $3 \times 3$ case. Third observation: if $M$ is a counterexample, then it is a counterexample over the subring of $R$ generated by the entries of $M$, so We may assume WLOG that $R$ is finitely-generated over $\mathbb{Z}$. AI: Yes, such an injective morphism has non-zero determinant. Actually, if $M$ is a finitely generated free module over the commutative ring $R$ and $u:M\to M$ is an endomorphism, one has the precise equivalence: $$u \;\text {is injective}\iff \det(u) \; \text {is not a zero divisor in}\; R.$$ The proof is based on the fact that elements $m_1,m_2, \ldots ,m_n\in M$ form a linearly independent set iff there exists a non-zero $0\neq \lambda\in R$ with $\lambda (m_1\wedge m_2\wedge \ldots\wedge m_n)=0\in \Lambda^nM$. You can find the details in Bourbaki, Algebra, III, §7, Proposition 3, page 524.
H: Why is it legit to evaluate $\lim_{x\rightarrow 1} \frac{(x-1)(x+1)}{x-1}$ by cancelling common factors? I haven't ever taken an analysis course, so maybe that's where I would really learn this, but I've always wondered why it's okay to do this when evaluating a limit. I guess it's the case that there is a theorem which says that the limit of a rational function as $x\rightarrow a$ is equal to the limit of that function in lowest terms as $x\rightarrow a$, regardless of division by zero. Is there a name for that theorem? Or does it follow from some other property of limits or rational functions? AI: The important thing to remember about the limit process in something like this is that whenever we evaluate something like $\lim_{x\rightarrow 1} f(x)$, we don't require the function $f(x)$ to be defined at $x=1$, so we look at the values of $f(x)$ as $x$ approaches 1, but taking values which are not equal to 1. In your example, it means that as $x$ approaches 1 without actually equalling 1, the factor may be cancelled because it is definitely not zero. I don't know of any general result, though, and have always done things like this by examining each case on its merits, by asking if we can be sure that cancellation is allowable in a particular case.
H: Prove that: $\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$ I'm interested in proving the following integral inequality: $$\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$$ According to W\|A the result of this integral isn't pretty nice, and involves the exponential integral. AI: Let $f(x) = \dfrac{\ln(x)}{\ln^2(x) + 1}$. $f(x)$ is concave in $[1,\sqrt{2}]$. Hence, the area of $f(x)$ from $1$ to $\sqrt{2}$ i.e. the area under the blue curve is greater than the area of the triangle between the two points at which the blue curve and the red curve intersect i.e. \begin{align} \int_1^{\sqrt{2}} f(x) dx & \geq \dfrac12 \times (\sqrt{2} - 1) \times (f(\sqrt{2}) - f(1)) = \dfrac{\sqrt{2}-1}2 \times \dfrac{\ln(\sqrt{2})}{\ln^2(\sqrt{2})+1}\\ & = \dfrac{(\sqrt{2}-1)\ln(2)}{\ln^2(2)+4} \approx 0.06408 > \dfrac1{20} \end{align}
H: What is the total number of combinations of 5 items together when there are no duplicates? I have 5 categories - A, B, C, D & E. I want to basically create groups that reflect every single combination of these categories without there being duplicates. So groups would look like this: A B C D E A, B A, C A, D A, E B, C B, D B, E C, D . . . etc. This sounds like something I would use the binomial coefficient $n \choose r$ for, but I am quite fuzzy on calculus and can't remember exactly how to do this. Any help would be appreciated. Thanks. AI: There are $\binom{5}{1}$ combinations with 1 item, $\binom{5}{2}$ combinations with $2$ items,... So, you want : $$\binom{5}{1}+\cdots+\binom{5}{5}=\left(\binom{5}{0}+\cdots+\binom{5}{5}\right)-1=2^5-1=31$$ I used that $$\sum_{k=0}^n\binom{n}{k}=(1+1)^n=2^n$$
H: group homomorphism from $S^1\times S^1$ to itself Could any one give me a hint how to show that if the kernel of a group homomorphism from $S^1\times S^1$ to itself is finite then it must be cyclic subgroup of $S^1\times S^1$? AI: Denote the kernel $K=\ker\varphi$ of a homomorphism of $S^1\times S^1$ to itself. Suppose it is finite. There exist two projection maps $ S^1\times S^1\to S^1$ of the $1$st and $2$nd coordinates respectively, say $\pi$ and $\rho$. Since $K$ is finite, $\pi(K)$ and $\rho(K)$ are finite, and hence cyclic. Notice that $K$ must be a subgroup of $\pi(K)\times \rho(K)$ (since it is a subset), i.e. a subgroup of a direct product of two cyclic groups. Hence $K$ is either itself cyclic or a direct product of just two cyclic groups. Both possibilities are realizable; using the circle group $\Bbb T$ ($z\in \Bbb C$ with $\|z\|=1$ under multiplication) as $S^1$, we have that the map $\varphi:(z,w)\mapsto (z^n,w^m)$ has kernel isomorphic to $C_n\times C_m$ (note $C_k$ denotes the cyclic group of order $k$); choosing one of $n,m$ to be $1$ will result in $K$ isomorphic to a single cyclic group. (Aside: to see that finite subgroups of $S^1$ are cyclic, say one has order $n$ and, using $\Bbb T$ again, notice it must be a subgroup of the $n$th roots of unity; the subgroups of a cyclic group are cyclic.)
H: How to find large prime factors without using computer? What is the largest prime factor of the number 600851475143 ? This is the third problem of Project Euler. How to approach this mathematically (without computer programming) ? AI: Well the point of Project Euler is to program. This is a problem that you could solve by hand but it would take you quite a while. Factorisation is not a simple thing to do by hand for numbers this big (unless you have some special insight into divisibility by certain special primes). Just use the bog standard test, square root and check divisibility of primes upto this. Once you find one, divide out and start again. In the end you will have a number that you cannot do this process to. This will be the largest prime factor.
H: What reference contains the proof of the classification of the wallpaper groups? Background: I am doing a course on Groups and Geometry ( Open University M336 ). One of the topics is the classification of the plane symmetries, a.k.a. The Wallpaper Groups. Question: What reference contains the original proof that there exists only 17 wallpaper groups? AI: The wiki page contains lots of info on this. It is quite interesting. http://en.wikipedia.org/wiki/Wallpaper_group There is also the book by Conway, "The Symmetries of Things". I believe this has a nice discussion of how the proof works but I have never read it properly so cannot vouch for the rigour.
H: Lebesgue Integral, existence, improper integrals, etc. Problem: At the request of another user, I am taking an older question and specifically addressing one problem. I am self-learning about Lebesgue integration, and am just starting to try and apply some examples of the existence of the integral. For the function below, does the Lebesgue integral exist on $(0,\infty)$, and if it does, is it finite? $$f(x)=\sum_{k=0}^\infty e^{-2^kx}$$ Since this is self-learning from scratch, I would be grateful if someone could help me break this down bit by bit: 1) What does it mean for the integral to "exist"? Is this just saying that $\int f(x)$ is finite? 2) How do you calculate the integral explicitly? Any help is always appreciated. Thanks! AI: The integral exists means it is finite! Set $S_n(x)=\sum_{k=0}^n\exp(-2^kx)$. Then $S_n (x)\le f(x)$ and $S_n(x) \to f(x)$ for every $x>0$. Thus \begin{eqnarray} \int_0^\infty f &=&\lim_{n\to \infty}\int_0^\infty S_n=\lim_{n\to \infty}\sum_{k=0}^n\int_0^\infty\exp(-2^kx)dx=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{2^k}\cr &=&\lim_{n\to\infty}\frac{1-(1/2)^{n+1}}{1-1/2}=\frac{1}{1-1/2}=2 \end{eqnarray}
H: Compute: $\lim_{n\to\infty} \{ (\sqrt2+1)^{2n} \}$ Compute the following limit: $$\lim_{n\to\infty} \{ (\sqrt2+1)^{2n} \}$$ where $\{x\}$ is the fractional part of $x$. I need some hints here. Thanks. AI: Consider $$ (\sqrt2+1)^{2n} + (\sqrt2-1)^{2n} $$ Try to show that it is an integer and hence this fractional part you are looking for is $1 - (\sqrt2-1)^{2n}$ Now the limit becomes easy.
H: Find all primes $p$ such that $(2^{p-1}-1)/p$ is a perfect square Find all primes $p$ such that $(2^{p-1}-1)/p$ is a perfect square. I tried brute-force method and tried to find some pattern. I got $p=3,7$ as solutions. Apart from these I have tried for many other primes but couldn't find any other such prime. Are these the only primes that satisfy the condition? If yes, how to prove it theoretically and if not, how to find others? Thanks in advance! AI: Hint: Let $p=2k+1$ where $k \in \mathbb{N},$ then $2^{2k}-1=(2^k-1)(2^k+1)=pm^2.$ We know that $\gcd(2^k-1,2^k+1)=1$ since they are consecutive odd integers, so the equation breaks into $2^k-1=px^2, 2^k+1=y^2$ or $2^k-1=x^2, 2^k+1=py^2.$ Easy investigation shows that the only solutions are $p=3,7.$ I will leave it to you to fill the gaps. You may also interested in this.
H: Definition of S(n) for graded ring S In Hartshorne's Algebraic Geometry, the twisting sheaf of Serre $\mathscr{O}(n)$ is defined to be $S(n)^\sim$, where $S$ is an $\mathbb{N}$-graded ring. But I couldn't find the definition of $S(n)$ anywhere in the book (if any one knows where it is, please let me know). It is presumably the regrading of $S$ by $n$, but I wanted to make sure there is absolutely no room for misunderstanding. For example, regrading in which direction? It is probably $S(n)_i=S_{n+i}$, but did Hartshorne use this convention? And presumably $S(n)_i=0\quad\forall i<n$ ? AI: Indeed, he uses this definition. Have a look at II.7, he describes $\mathcal{O}_X(1)$ there, which implies the rest.
H: Expectation of a minimum Why is this: $E(\tau \wedge T) = \int_0^Ttf_\tau(t)dt+T(1-P(\tau\leq T))$, where $f_\tau(t)$ is the pdf of $\tau$. I am thinking its because $E(\tau \wedge T) = \tau P(\tau\leq T)+T(1-P(\tau\leq T))$ since $\tau \wedge T =\min(\tau,T) = \tau$ if $\tau\leq T)$ and T if $\tau> T)$. But the first part is I cannot figure out. AI: We have as you wrote \[ \tau \wedge T = \chi_{\{\tau \le T\}} \cdot \tau + T \cdot \chi_{\{\tau > T\}} \] where $\chi_A$ denotes the indicator function of $A$. We have $\chi_{\\{\tau \le T\\}} = \chi_{[0,T]} \circ \tau$, therefore \begin{align} E(\tau \wedge T) &= E\bigl((\chi_{[0,T]}\cdot \mathrm{id}_{[0,\infty)}) \circ \tau\bigr) + T \cdot E(\chi_{\\{\tau > T\\}})\\ &= \int_0^\infty \chi_{[0,T]}(t) \cdot t \cdot f_\tau(t) \, dt + T \cdot P(\tau > T)\\ &= \int_0^T t f_\tau(t)\, dt + T \bigl(1 - P(\tau \le T)\bigr) \end{align}
H: Similar Matrices and Equivalence Relations Since similarity of matrices' is an equivalence relation, doesn't that imply that given any polynomial equation involving similar matrices you can substitute in any similar matrices' and the equation will still hold? For example, given $A,B,C\in M^F_{n\times n}$ if $B \cong C$ then: $$A \cong B^2+5B+3I \iff A \cong C^2+5C+3I$$ Correct? AI: What you say is correct (although it doesn't follow just from the fact that similarity is an equivalence relation, but from the fact it is an equivalence relation preserved by polynomials, that is if $A\cong B$, then for any polynomial $p$, $p(A)\cong p(B)$). If $B=PCP^{-1}$ and for some polynomial $p$, $p(B)=DAD^{-1}$, then $p(B)=p(PCP^{-1})=Pp(C)P^{-1}$, so $p(C)=(P^{-1}D)A(D^{-1}P)=(P^{-1}D)A(P^{-1}D)^{-1}$
H: Calculate: $\lim_{n\to\infty} \int_{0}^{\pi/2}\frac{1}{1+x\tan^{n} x }dx$ I'm supposed to work out the following limit: $$\lim_{n\to\infty} \int_{0}^{\pi/2}\frac{1}{1+x \left( \tan x \right)^{n} }dx$$ I'm searching for some resonable solutions. Any hint, suggestion is very welcome. Thanks. AI: Note that the integrand is bounded in $[0,\pi/2]$, so if $$\lim_{n\to \infty} \frac{1}{1+x\tan^nx}$$ exists a.e. then we may apply the Dominated Convergence Theorem to show $$\lim_{n\to \infty} \int_0^{\pi \over 2}\frac{1}{1+x\tan^nx}dx = \int_0^{\pi \over 2}\lim_{n\to \infty} \frac{1}{1+x\tan^nx}dx.$$ If $x<\pi/4$ then the integrand converges to 1, and if $x>\pi/4$ then it converges to 0. Thus we have the integral equals $$ \int_0^{\pi \over 4} 1dx + \int_{\pi \over 4}^{\pi \over 2} 0dx = \frac{\pi}{4}. $$
H: Extension of Yau's theorem to general bundles Calabi-Yau manifolds have the nice property that $c_1(TM) = 0$ implies there is a Ricci flat metric: $\text{Ric}(\omega)$. Is it possible to construct a similar theorem vor a Vector Bundle over a Calabi-Yau manifold? i.e. $c_1(V) = 0$ implies that there exists some flat connection on the bundle? Or something related? AI: There is a theorem by Uhlenbeck and Yau, according to wiki, which says that "they proved the existence and uniqueness of Hermitian–Einstein metrics (or equivalently Hermitian Yang–Mills connections) for stable bundles on any compact Kähler manifold". You can look at their paper entitled "On the Existence of Hermitian-Yang- Mills Connections in Stable Vector Bundles" for more details.
H: An elementary congruence question in number theory If ${q^k}\sigma(q^k) \equiv a \pmod b$ and $\displaystyle\frac{\sigma(q^k)}{n} \neq \displaystyle\frac{\sigma(n)}{q^k}$, does it follow that $n\sigma(n) \not\equiv a \pmod b$? Here, $q$ is prime and $n$ is composite. AI: We can generate counterexamples at will. Take for example $n=4$. Then $n\sigma(n)=28$. Take $q=11$, $k=1$. Then $q^k\sigma(q^k)=132$. We find $a$ and $b$ such that $28\equiv a \pmod{b}$ and $132\equiv a\pmod{b}$. Choose $a=2$. Then $28\equiv 2\pmod{13}$ and $132\equiv 2\pmod{13}$. It is too easy to arrange for two numbers to be each congruent to $a$ modulo $b$ if we are allowed to choose $a$ and $b$. A simpler class of counterexamples would use $b=1$. Added: For an example with $a=2$, $b=4$ as asked for in a comment, let $q=17$, $k=1$, and $n=15$. Then $q^k\sigma(q^k)\equiv 2\pmod{4}$ but $n\sigma(n)\not\equiv 2\pmod{4}$.
H: Subtraction of numbers with arbitrary bases Possible Duplicate: How to do +, -, *, / with number in a base b? I am reading research papers in the category of recreational mathematics on the topic of numbers similar to Kaprekar number. Almost all the time we come across subtration of numbers with arbritrary bases such as 11, 13, 8 etc.. Can someone please explain me as to how this kind of calculation is to be done? Is there any free calculator/software available on net that does this kind of calculation? AI: Let us subtract 313663 from 403155 in base 11. Digits in base 11 are 0,1,2,3,4,5,6,7,8,9,A. In base 11, instead of writing 10 + 1 = 11, we write A + 1 = 10. And instead of writing 10 + 10 = 20, we write A + A = 19. I will write base-11 numerals in this special font with the gray background, so that they are easy to recognize. 4 0 3 1 5 5 - 3 1 3 6 6 3 -------------------- Let's call the columns, in order from right to left, columns $C_0, C_1,\ldots C_5$. The algorithm is the same as in base 10, except that we use 11 instead of 10. We start in column $C_0$. 5-3 is 2: 4 0 3 1 5 5 - 3 1 3 6 6 3 -------------------- 2 In column $C_1$, we have 5-6, which is negative, so we borrow an 11 from the 1 in column $C_2$ and change the 5 to a 15. 4 0 3 0 15 5 - 3 1 3 6 6 3 -------------------- 2 Note that the 15 in column $C_1$ is not the base-ten 15; it is the base-eleven 15, which is base-ten 16. 15-6 = A. 4 0 3 0 15 5 - 3 1 3 6 6 3 -------------------- A 2 Now we can't take 6 from 0, so we borrow an 11 from the 3 in the $C_3$ column: 4 0 2 10 15 5 - 3 1 3 6 6 3 -------------------- A 2 10-6 = 5: 4 0 2 10 15 5 - 3 1 3 6 6 3 -------------------- 5 A 2 Now in column $C_3$ we have 2-3, so we sould like to borrow from $C_4$, but $C_4$ has a 0, so there is nothing to borrow. So instead we borrow 11 from $C_5$ to $C_4$: 3 10 2 10 15 5 - 3 1 3 6 6 3 -------------------- 5 A 2 And then we can borrow from $C_4$ to $C_3$: 3 A 12 10 15 5 - 3 1 3 6 6 3 -------------------- 5 A 2 Now in column $C_3$ we have 12-3 = A, and in column $C_4$ we have A-1=9: 3 A 12 10 15 5 - 3 1 3 6 6 3 -------------------- 9 A 5 A 2 And finally in column $C_5$ we have 3-3 = 0, so we are done, and the answer is 403155 - 313663 = 9A5A2.
H: Solving $ \frac{3x+3}{\sqrt{x}}=4+\frac{x+1}{\sqrt{x^{2}-x+1}} $ Solve in $\mathbb{R}$: $$ \frac{3x+3}{\sqrt{x}}=4+\frac{x+1}{\sqrt{x^{2}-x+1}} $$ AI: Note that $\frac{3x+3}{\sqrt{x}}\ge 6$, with equality only at $x=1$. This comes down to the inequality $(\sqrt{x}-1)^2 \ge 0$. Or else we can simply quote the fact that if $t$ is positive then $t+\frac{1}{t}\ge 2$. Now look at the right-hand side. It is not hard to show, using calculus or, more easily, without using calculus, that for positive $x$ $$4+\frac{x+1}{\sqrt{x^2-x+1}}\le 6,$$ with equality only at $x=1$. This comes down to showing that $\frac{x+1}{\sqrt{x^2-x+1}}\le 2$, or equivalently $(x+1)^2 \le 4(x^2-x+1)$, that is, $3(x^2-2x+1) \ge 0$. So the only real solution is $x=1$. Remark: Or else we could start squaring and rearranging and perhaps squaring again. Not an appealing prospect! It would likely turn something that has a reasonably nice shape into a mess.
H: Density character of a subspace of a topological space. Let $(X,\tau)$ be a topological space. Suppose $dc(X)=\kappa$ and let $D\subset_{dense} X$ be a dense subset of $X$ of cardinality $\kappa$. Is it true that $X\setminus D$ has density character $\kappa$, as a subspace of $X$ with the restricted topology? AI: Not necessarily. Let $X=\beta\omega$: $X$ is separable, with $\omega$ as dense subset, but $X\setminus\omega$ is not. An even easier example is a Mrówka $\Psi$-space. Let $\mathscr{A}$ be a maximal almost disjoint family of subsets of $\omega$, and let $X=\omega\cup\mathscr{A}$ with the following topology: points of $\omega$ are isolated, and basic open nbhds of a point $A\in\mathscr{A}$ are of the form $\{A\}\cup(A\setminus F)$ for finite subsets $F$ of $A$. $X$ is separable, since $\omega$ is dense in $X$, but $X\setminus\omega=\mathscr{A}$ is an uncountable discrete set.
H: Find the limit of: $\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}$ Could be the following limit computed without using Stirling's approximation formula? $$\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}$$ I know that the limit is $e$, but I'm looking for some alternative ways that doesn't require to resort to the use of Stirling's approximation. I really appreciate any support at this limit. Thanks. AI: The Stolz–Cesàro theorem implies that if the limit exists, then it is equal to $\lim\limits_{n\to\infty}\dfrac{n}{\sqrt[n]{n!}}$. Some ways to evaluate the latter limit, including a method that uses the Stolz–Cesàro theorem again, are included in the answers to the question Finding the limit of $\frac {n}{\sqrt[n]{n!}}$. This leaves existence of the original limit to be proved.
H: What is $\overline{\sin({z})} $ equal to? What is $\overline{\sin(z)}$ equal to? AI: I will write $\exp(x)$ instead of $e^{x}$, they are synonyms. (Just a notational warning!) Well, recall Euler's formula $$ \exp(i\theta)=\cos(\theta)+i\cdot\sin(\theta).$$ Then we see $$\exp(i\theta)-\exp(-i\theta)=2i\cdot\sin(\theta)$$ allows us to write $$\frac{\exp(i\theta)-\exp(-i\theta)}{2i}=\sin(\theta).$$ So replacing $\theta$ with $z=x+iy$ in this case would produce $\exp(iz)-\exp(-iz)=2i\cdot\sin(z)$, where $z=x+iy$. Addendum: Now consider the following: $$\overline{\sin(z)} = \overline{\frac{\exp(iz)-\exp(-iz)}{2i}}$$ But look, this is just $$\overline{\sin(z)}=\overline{\sin(x+iy)}$$ and the only place the imaginary part plays any role is the $iy$, we have $$\overline{\sin(z)}=\sin(x-iy)$$ and this is precisely $\sin(\bar{z})$. Looking on the right hand side, how can we say this? Well, we just change all the signs for $i$ and replace $z$ with $\bar{z}$, writing $$\overline{\frac{\exp(iz)-\exp(-iz)}{2i}}=\frac{\exp(-i\bar{z})-\exp(i\bar{z})}{-2i}$$ But look, we may multiply the top and bottom by $-1$ producing $$\overline{\frac{\exp(iz)-\exp(-iz)}{2i}}=\frac{\exp(-i\bar{z})-\exp(i\bar{z})}{-2i}=\frac{-\exp(-i\bar{z})+\exp(i\bar{z})}{2i}$$ which is precisely $\sin(\bar{z})$.
H: How to prove $\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$ How could we prove that $$ \frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$$ I have reduced it the form $$\frac{4^{\ln(4)/\ln(3/4)}}{3^{\ln(3)/\ln(3/4)}}$$ I am not sure what to do next to get snappy solution. Any ideas? AI: This refers to the original question, which had the left hand side equal to $\frac{1}{2}$ instead of $\frac{1}{12}$. They are not equal. Your simplification is correct. Then we can rewrite the left hand side as $$\left(\frac{4^{\ln(4)}}{3^{\ln(3)}}\right)^{1/\ln(3/4)}$$ so raising both sides of the equation to the $\ln(3/4)$ power, we get that the equation would be equivalent to $$\frac{4^{\ln(4)}}{3^{\ln(3)}} \stackrel{?}{=} \frac{1}{2}^{\ln(3/4)}.$$ Rewriting $4^{\ln(4)}$ as $e^{(\ln 4)^2}$, $3^{\ln(3)}$ as $e^{(\ln(3))^2}$, and $\left(\frac{1}{2}\right)^{\ln(3/4)}$ as $e^{-\ln(2)\ln(3/4)}$, the equality would be equivalent to $$\left(\ln 4\right)^2 - \left(\ln 3\right)^2 \stackrel{?}{=} -\ln(2)\ln\frac{3}{4}.$$ Now, $\ln(4) = 2\ln(2)$, and $\ln\frac{3}{4} = \ln 3 - 2\ln 2$. So the left hand side is equal to $$4(\ln 2)^2 - (\ln 3)^2$$ while the right hand side is equal to $$-\ln(2)(\ln 3 - 2\ln 2) = 2(\ln 2)^2 - (\ln 2)(\ln 3).$$ But $$4(\ln 2)^2 - (\ln 3)^2 \approx 0.714863$$ and $$2(\ln 2)^2 - (\ln 2)(\ln 3) \approx 0.199406$$ As corrected, the right hand side now be, after the simplification $$ \left(\frac{1}{12}\right)^{\ln(3/4)} = \exp\left(-\ln(12)\ln(3/4)\right).$$ The exponent can be simplified: $$\begin{align} -\ln(12)\ln(3/4) &= -\left(\ln(3)+2\ln(2)\right)\left(\ln(3)-2\ln(2)\right)\\ &= \left(\ln(3)+2\ln(2)\right)\left(2\ln(2)-\ln(3)\right)\\ &= \left(2\ln(2)\right)^2 - \left(\ln 3\right)^2\\ &= 4(\ln 2)^2 - (\ln 3)^2. \end{align}$$ Since this is the same as the exponent of $e$ on the left hand side, we do indeed have $$\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}.$$ There's nothing special about $3$ and $4$. Replacing them with arbitrary positive numbers $a$ and $b$ will lead to $$\exp\left((\ln(a))^2 - (\ln(b))^2\right) \stackrel{?}{=} \exp\left(-\ln(ab)(\ln(b/a)\right)$$ which of course holds, since $$-\ln(ab)\ln(b/a) = (\ln a + \ln b)(\ln a - \ln b)$$ giving the equality you have in the comment: $$\frac{a^{1/\log_a(b/a)}}{b^{1/\log_b(b/a)}} = \frac{1}{ab}.$$
H: How can I convert between powers for different number bases? I am writing a program to convert between megabits per second and mebibits per second; A user would enter 1 Mebibits p/s and get 1.05 Megabits p/s as the output. These are two units of computer data transfer rate. A megabit (SI unit of measurement in deny) is 1,000,000 bits, or 10^6. A mebibit (IEC binary prefix) is 1,048,576 bits or 2^20. A user will specify if they have given a number in mega-or-mebi bits per second. So I need to know, how can I convert between these two powers? If the user inputs "1" and selects "mebibits" as the unit, how can I convert from this base 2 number system to the base 10 number system for "megabits"? Thank you for reading. AI: If you have $x \text{ Mebibits p/s}$, since a Mebibit is $\displaystyle \frac{2^{20}}{10^6} = 1.048576$ Megabits, you have to multiply by $1.048576$, getting $1.048576x \text{ Megabits p/s}$. Likewise, if you have $y\text{ Megabits p/s}$, since a Megabit is $\displaystyle \frac{10^6}{2^{20}} = 0.95367431640625$ Mebibits, you have to multiply by $0.95367431640625$, getting $0.95367431640625y\text{ Mebibits p/s}$. Round up as necessary. To find these conversion factors, you can see that a Mebibit is $2^{20}$ bits, and a Megabit is $10^6$ bits. Therefore a Mebibit is $\displaystyle 2^{20} \text{ bits} = \frac{2^{20}}{10^6} \text{ Megabits}$, and the other direction is analogous.
H: Borel regular measure I stuck with this question, can you help me please. Is it exist $ \mu$ - Borel regular measure in $[0,1]$ so that to all polynomial $p$ one has: $\int_{[0,1]}p(t)d \mu(t)=p'(0)$? Thanks a lot! AI: For such a measure $\mu$ we have, for $n=0$ and $n\geq2$, $$\int_{[0,1]}t^n d \mu(t)=0$$ and hence $\int_{[0,1]}p(t) d \mu(t)=0$ for all polynomials without linear term. The sequence $p_n(t)=1-\sum_{j=1}^n \left\|{1/2 \choose j}\right\| (1-t^2)^j$ converges to $t$ uniformly on $[0,1]$, which implies $$0=\int _{[0,1]}p_n(t)d \mu(t)\to\int _{[0,1]} t\,d \mu(t)=1; $$ a contradiction.
H: A computation in a Hilbert space Can someone give me an idea, why $\forall x: \left<\sum_j \lambda_j \left< x,e_j\right> e_j,x\right>\geq 0$, where the $\lambda_j$'s are fixed, implies that all $\lambda_j$ are $\geq0$,? (The $x$'s belong to a Hilbert space,the $e_j$'s are an orthonormal basis and the $\lambda_j$'s are real or complex .) AI: I understand that this inequality is to hold for all $x$. (Otherwise, $x=(1,1)$ with $\lambda_1=1,\lambda_2=-1$ witnesses that it is false.) Suppose that there is some $\lambda_i$ not real or not greater than or equal to zero. Then for $x=e_i$ we have $$\langle\sum_j\lambda_j\langle x,e_j\rangle e_j, x\rangle=\langle \lambda_i e_i, e_i\rangle=\lambda_i \not\geq 0$$ so we have a contradiction.
H: Combination of n sets that produces a set of n-tuple Given n sets with $3$ elements: $X_i=\{a_i,b_i,c_i\}$ where $\{i\in\mathbb{N}\ \|\ 1\leq i\leq n\}$. How can I define a n-tuple based on combination of this sets that produces the set $S$ with $3^n$ elements ($n$-tuples) as following: $S=\{(a_1,a_2,\cdots,a_n),(a_1,a_2,\cdots,b_n),(a_1,a_2,\cdots,c_n),\cdots,(c_1,c_2,\cdots,c_n)\}$. AI: If I understand your question correctly, what you want is simply the Cartesian product of the $n$ sets: $$\prod_{i=1}^nX_i=\Big\{\langle x_1,\dots,x_n\rangle:x_1\in X_i,x_2\in X_2,\dots,x_n\in X_n\Big\}\;.$$
H: Computing the derivative from the definition Using the limit definition of the derivative which I know is: $$f'(x)=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$ I am trying to solve this problem $$f(x)= \frac{x}{x+2} $$ How do I go about properly solving this, I seemed to get $$\frac{x}{x+2}\ $$ as my answer again? What are the steps I should follow? I am trying to find the derivative of $f(x)= \frac{x}{x+2}$ using the definition of the derivative. AI: I'm supposing you want to find the derivative using the limit definition. $\begin{align}f'(x) &= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0 } \frac{\frac{x+h}{x+h+2} - \frac{x}{x+2}}{h} = \lim_{h \to 0 }\frac{x^2+2x+hx+2h-x^2-xh-2x}{h(x+h+2)(x+2)} = \\ &= \lim_{h \to 0 }\frac{2h}{h(x+h+2)(x+2)} = \lim_{h \to 0 }\frac{2}{(x+2)(x+h+2)} = \frac{2}{(x+2)^2} \end{align}$ You should probably check your algebra. Subtracting the fractions may be a bit tricky.
H: Help with general integration problem I have a specific problem that Ive generalized here for simplicity. Let $F(x)=\int^{g(x)}_0h(x,y)dy $ Suppose $F(0)=0$ (with $g(0)>0$) Now suppose that $h$ is increasing in $y$. Then, it follows that: $F(x) \leq g(x) h(x,g(x)) $ Does it therefore have to be the case that $h(0,g(0)) = 0$? AI: Not at all. Let $g(x) = 1$ be constant and $h(x,y) = 2y - 1$. Then $h(0,g(0)) = 1$.
H: infinity understanding problem? between 0 meter -> 1 meter there are 100 cm. but each cm has infinite numbers : for example between 0..1 cm there are : 0.000000000001 .. 0.00000000000111 .. 0.000000000001111111 and more numbers and combinations... .. .. .. 1.0 to each number I can add another digit to the right there are infinity of numbers question : how can a person walk 3 cm if he had to go through an infinite series of numbers ? it not seems logic any help ? AI: This is an old problem. Zeno of Elea is credited with some classical pointed formulations of it about 2500 years ago. (Note: not "Xeno" as one commenter above spelled him). There are infinitely many different places to be at between 0 cm and 1 cm, but by the same token there are also infinitely many different instants in, say, one second, so they match up nicely. Now, whether space and time can physically be subdivided infinitely finely is not a mathematical question. It's just that the most common mathematical model of them allows arbitrarily fine divisions, because that is much easier to deal with than the alternative and consistently seems to lead to useful results in practice. It is perfectly conceivable that actual time or space cannot be divided indefinitely; that would just mean that the mathematical model is not an accurate description at small enough scales. (Again, this would not be a mathematical problem. The model might describe something else, or describe no physical situation, and it would be no worse as mathematics for that). As a physical question, the last few hundred year's physics has shown that matter cannot be subdivided indefinitely; a a scale of around 0.00000001 cm you find atoms that cannot be cut up without fundamentally changing what they are. However, the atoms are still thought to move around in a fundamentally continuous space. That might change with the next unpredictable revolution in physics, of course.
H: How do you Compute a Squareroot limit? Possible Duplicate: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt\[n\]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ I am trying to compute the limit here, i am not sure how to do it if i just plug it in or do i do the conjugate and solve? For some reason i got infinity as my answer yet I was told theres an actual solution to it. Perhaps someone can help me out here? $\lim\limits_{x\rightarrow \infty}\sqrt{16x^2-5x+10{}}-4x$ AI: To paraphrase Tom Lehrer: Rationalize, rationalize, rationalize. We use the fact that $(a-b)(a+b) = a^2-b^2$. So if we multiply $$\sqrt{16x^2-5x+10} - 4x$$ by $$\sqrt{16x^2-5x+10}+4x$$ that will eliminate the square root. Unfortunately, we are not allowed to just multiply by whatever we please (that changes the function), so we must be sure that what we do, in total, does not amount to anything. So in addition to multiplying by $\sqrt{16x^2-5x+10}+4x$, we also divide by it so that it all amounts to multiplying by $1$ (which doesn't do anything). Then we can do the limit the usual way, by noting that the highest degree of $x$ that shows up in both numerator and denominator is $x$. (I know it looks like an $x^2$ in the denominator, but it is inside a square root, so it amounts to just $x$): $$\begin{align} \lim_{x\to\infty}\left(\sqrt{16x^2-5x+10}-4x\right) &= \lim_{x\to\infty}\frac{(\sqrt{16x^2-5x+10}-4x)(\sqrt{16x^2-5x+10}+4x)}{\sqrt{16x^2-5x+10}+4x}\\ &= \lim_{x\to\infty}\frac{16x^2-5x+10 - (4x)^2}{\sqrt{16x^2-5x+10}+4x}\\ &= \lim_{x\to\infty}\frac{-5x+10}{\sqrt{16x^2-5x+10}+4x}\\ &=\lim_{x\to\infty}\frac{\frac{1}{x}(-5x+10)}{\frac{1}{x}(\sqrt{16x^2-5x+10}+4x)}\\ &= \lim_{x\to\infty}\frac{-5 + \frac{10}{x}}{\frac{1}{x}\sqrt{16x^2-5x+10}+4}\\ &= \lim_{x\to\infty}\frac{-5+\frac{10}{x}}{\sqrt{\frac{16x^2-5x+10}{x^2}}+4}\\ &= \lim_{x\to\infty}\frac{-5+\frac{10}{x}}{\sqrt{16 - \frac{5}{x}+\frac{10}{x^2}} + 4}\\ &= \frac{-5+0}{\sqrt{16-0+0}+4}\\ &= \frac{-5}{4+4}\\ &= -\frac{5}{8}. \end{align}$$
H: How is this expression simplified? I have this: $$\sqrt{(dx)^2 + (dy)^2}$$ And my book simplified it as: $$\sqrt{1 + \Big(\frac{dy}{dx}\Big)^2} \times dx$$ I don't have even a close idea how he did it. If it helps, is about path lenght whit integration. AI: $$a\sqrt{r} = \sqrt{a^2(r)}\quad\text{if }a\gt 0\text{ and } r\gt 0.$$ So, using changes instead of differentials: $$\begin{align} \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \Delta x &= \sqrt{\left(\Delta x\right)^2\left(1 + \left(\frac{\Delta y}{\Delta x}\right)^2\right)}\\ &= \sqrt{(\Delta x)^2 + (\Delta y)^2}. \end{align}$$ Taking limits as $\Delta x\to 0$ converts $\Delta x$ to $dx$, $\Delta y$ to $dy$, and $\frac{\Delta x}{\Delta y}$ to the derivative $\frac{dy}{dx}$.
H: Differential equations and family of function solutions I can't follow what Stewart is doing in his book. I can easily follow his work but his conclusion doesn't make any sense to me. "Show that ever member of the family of functions $$y = \frac{1+ce^t}{1 - ce^t}$$ is a solution of the differential equation $$y' = \frac{1}{2} (y^2 - 1)$$ He starts by differentiating the right side of the first term and then just setting that equal to the $y'$ from the question. I do not see how these are equal or what this means or what is going on at all really, and the book doens't feel the need to explain this anyways, so maybe it isn't important and maybe I just need to memorize that a solution is just the differential. But I don't see hwy this is important or how this helps anything. For his final answer he gets $$y' = \frac{2ce^t}{(1-ce^t)^2}$$ AI: Plug in the value of $y$ into $\frac{1}{2}(y^2-1)$. Verify that you get $\frac{2ce^t}{(1-ce^t)^2}$. In other words, what you are doing is plugging in the proffered solutions into the equation and verifying that you get an equality. It's exactly as if you had been asked to verify that $x=4$ is a solution to the equation $$x^2 - 5x + 6 = 2x^2 - 8x + 2.$$ All you need to do is plug in $x=4$ into the left hand side and compute, $4^2-5(4) + 6 = 16-20+6 = 2$; then plug it into the right hand side and compute, $2(4)^2 - 8(4) + 2 = 32 - 32 + 2 = 2$. And then say: "yes, it's a solution, because it satisfies the equality." You don't have to solve anything, just plug and verify.
H: Teaching abstract maths concepts to young children. I am interested in opinions and, if possible, references for published research, about the pros and cons of teaching abstract maths concepts to young children. My younger brother (five years old) understands negative numbers and square roots so I was thinking of trying to teach him about complex numbers and maybe some other concepts, but my elder brother (who is doing a maths/stats degree) said it was a crazy idea (without elaborating, but that's what he's like). Update: I quizzed my brother on why his thinks it is crazy and his response was "Don't you think there is a reason why 99% of maths teachers have a degree in maths ?" I'm at high school by the way. AI: I’d not set out to teach him anything; I’d make accessible mathematics available to him and let him choose what interests him. Enzensberger’s The Number Devil introduces a wide variety of interesting mathematical ideas in a very accessible way. If (or when) his reading is up to it, Martin Gardner’s collections of columns from Scientific American are good. The main point is that it should be up to him. There’s all manner of accessible mathematics that might prove to be more interesting or more fun: Fibonacci numbers and their patterns come to mind immediately. Representation in other bases can be fun early on; I especially like binary (as the system that arises naturally when you want the most efficient set of counterweights for an equal-arm balance when the object being weighed and the weights must go in opposite pans) and balanced ternary (as the system that arises naturally when the weights may also be placed in the same pan as the object).
H: Gaussian curvature in $S^3$ I'm trying to read a survey paper on the Willmore conjecture and I'm missing a lot of basic knowledge. In particular, let $u: \mathcal{M} \rightarrow S^3 \rightarrow \mathbb{R}^4$ be a smooth immersion of a compact orientable two dimensional surface into the standard 3-sphere, and let $\mathcal{M}$ take the metric induced by the ambient space. Writing the principal curvatures as $k_1$ and $k_2$, we have the Gaussian curvature given by $1+k_1k_2$. I don't understand where the 1 in $K = 1+k_1k_2$ is coming from. The metric on $\mathbb{R}^4$ is the standard $g_{ij} = \delta_{ij}$, and restricting it to $S^3$ yields the round metric. $S^3$ is our ambient space, and restricting to $u$ gives us our induced metric. I can think of two ways of showing this. The first would simply be to do everything out in local coordinates: for any $p \in \mathcal{M}$, we have a chart $\varphi: U \rightarrow V \subset \mathbb{R}^2$. Writing $u(p) = u(\varphi^{-1}(x^1,x^2)) = (y^1,y^2,y^3,y^4)$ and then projecting stereographically $\sigma: S^3 \rightarrow \mathbb{R}^3$, $\sigma(\vec{y}) = (\frac{y^1}{1-y^4},\dots,\frac{y^3}{1-y^4})$, we could recover $K$ via typical computations. Alternatively, since $2K = \mathcal{R}$, the Ricci scalar, we could compute the curvature tensor and find it that way (unless there is some shortcut? I don't have much experience with these diffgeo objects). I'm wondering if either these approaches would get me what I want, or if there is a naive reason for where the $1$ is coming from. Any help is appreciated. The relevant stuff is on p. 365 of the document, or 5th page from the start, section 2: the S^3 framework. AI: For $3$- manifolds of constant curvature $K_0$ (your $S^3$ for which $K_0$ equals $1$) the Gauss curvature of a hypersurface is $ K=k_1*k_2+K_0 $, with $k_i$ being the principal curvatures - see, e.g., Volume 4 (chapter 7) of Spivak's 'Comprehensive Introduction to Differential Geometry', which contains a comprehensive (! - nomen est omen) discussion of the relevant equations, also including a discussion about higher dimensional manifolds. See in particular the proof of propostion 24. Alternatively Theorem 5.5 in Gallot, Hulin, Lafontaine, 'Riemannian Geometry'. I'd expect to find similar results in many books of differential geometry. Search for Gauss formulas and Gauss curvature. (And it's 'principal curvature', not 'principle curvature').
H: The pointwise liminf of a sequence of upper semicontinuous functions is upper semicontinuous. Find the flaw in my counterexample? Let $f_n: \mathbb{R} \to \mathbb{R}$ be defined as follows: $f_n$ is even. $f_n(0) = \frac{1}{2}$ $f_n(x) = 0$ if $0<x< \frac{1}{n}$ $f_n(x) = 1$ if $x> 2/n$ $f_n$ is linear on $(\frac{1}{n}, \frac{2}{n})$, and continuous on $\mathbb{R} - \{0\}$ I think each $f_n$ is USC, but the pointwise liminf is not. How am I wrong? AI: There’s nothing wrong with your example; it’s the theorem that’s wrong. What is true is that the pointwise infimum of upper semicontinuous functions is upper semicontinuous, as is the pointwise limit if the sequence of functions is non-increasing.
H: Probability of sequence being longer than some length We are given random bit generator which generates 0s and 1s with equal probability 1/2. We have an algorithm which generates random numbers using this random bit generator in this way: we look for subsequences of 1s, and length of each subsequence gives us one random number (lets forget that this is really bad random generator). So for example: 0011011100010 gives us random numbers [2, 3, 1] Task is to show that probability of seeing number greater than $c+\log_2n$ is falling exponentially for $c\in\mathbb{N}$ Intuitively every time we increased $c$ for one, there is $1/2$ probability that next bit is going to be 1, so probability is falling exponentially, but I guess I have to prove that expected longest run of ones in sequence is going to be $\log_2n$ Here is an article about distribution of longest run in coin flip (we can se our random bit generator as fair coin flip). On page 6 is given an approximation for length of longest run, but thing that bothers me is that in an article, n is number of all trials, while in this task, n is number of subsequences of ones (equals number of "random numbers"), while number of trials can be much larger. AI: Your question based on the number of $1$s [each with probability $\frac12$], rather than the number of trials, is equivalent to the paper's "longest run of heads or tails" applied to a fair coin, and in particular where it gives $$B_n (x) = 2 A_{n-1}(x-1)\qquad \qquad (2)$$ and explains this by saying ... the distribution of the longest run of heads or tails is simply the distribution of the longest run of heads alone for a sequence containing one fewer coin toss, shifted to the right by one.
H: Limit to Infinity I have a question to compute a limit, i understand that the limit is infinity but How can i show the solution to get this answer to infinity. I understand that it will be large pos #'s on top and small #'s on bottom making it go to infinity but how can i write it out to prove this? At least how would i make it look on a test. $$\lim_{x\rightarrow \ \frac{1}{2}^+} \frac{10+x}{(2x-1)^3}$$ AI: You can, for example, remark that :$$(1) \quad \forall x \in \left(\frac 12,1 \right) \quad \frac{10+ x}{(2x-1)^3 }\geq \frac{1 }{2x-1}$$ since $(2x-1)^3 \leq 2x-1$ because $0 < 2x-1 < 1$ From $(1)$ , since $$(2) \quad \displaystyle \lim_{x \to \frac{1}{2}^+} \frac1{2x-1}=+\infty$$, we can deduce the rest. Here is a proof to $(2)$ using the definition of limit (test). We must proof : $$(3) \quad \quad \forall A>0 \quad \exists \alpha >0 \quad \frac 12 < x < \frac 12 + \alpha \Rightarrow \frac{1}{2x-1} > A$$ It holds since we remark that : $$\frac 1{2x-1} > A \Leftrightarrow 2x-1 < \frac 1A \Leftrightarrow x < \frac 12 + \frac 1{2A} $$ then $\alpha =\frac 1{2A}$ is a convenant response making $(3)$ true.
H: Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$ Here is another interesting integral inequality : $$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$ According to W\|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference is so small. AI: You can actually just evaluate the integral explicitly. You can divide $x^2 -1$ into $x^4$ and get $$\frac{x^4}{x^2 - 1} = x^2 + 1 + \frac {1}{x^2 - 1}$$ So the integral is the same as $$\int_0^1 (x^2 + 1)\log(x)\,dx + \int_0^1 \frac{\log(x)}{x^2 - 1}\,dx $$ The second integral is related to the famous dilogarithm integral, and as explained in Peter Tamaroff's answer can be evaluated to $\frac{\pi^2}{8}$. For the first term, just integrate by parts; you get $$({x^3 \over 3} + x)\log(x)\big\|_{x = 0}^{x =1} - \int_0^1 ({x^2 \over 3} + 1)\,dx$$ The first term vanishes, while the second term is $-{10 \over 9}$. So the answer is just ${\pi^2 \over 8} - {10 \over 9}$ which is less than ${1 \over 8}$. A way of doing the whole integral in one fell swoop occurs to me. Note that ${\displaystyle {1 \over 1 - x^2} = \sum_{n=0}^{\infty} x^{2n}}$. So the integral is $$-\sum_{n = 0}^{\infty} \int_0^1 x^{2n + 4}\log(x)\,dx$$ $$= -\sum_{m = 2}^{\infty} \int_0^1 x^{2m}\log(x)\,dx$$ Integrating this by parts this becomes $$\sum_{m = 2}^{\infty} \int_0^1 {x^{2m} \over 2m + 1}$$ $$= \sum_{m = 2}^{\infty} {1 \over (2m + 1)^2}$$ This is the sum of the reciprocals of the odd squares starting with $5$. The sum of the reciprocals of all odd squares is ${\pi^2 \over 8}$, so one subtracts off $1 + {1 \over 9} = {10 \over 9}$. Hence the result is $ {\pi^2 \over 8} - {10 \over 9} $.
H: Are Tonelli's and Fubini's theorem equivalent? I can derive Fubini's theorem for interated integrals of complex functions from Tonelli's theorem for iterated integrals for unsigned functions. I was wondering whether there is a way to go backwards. I do not think so, because Fubini's theorem assumes the integrals are finite, whereas Tonelli's theorem allows the value of the integral to be $+\infty$. But maybe we can use a limiting argument? This is where I am not clear. So: is it possible to derive Tonelli's theorem from Fubini's theorem? If so, I would appreciate a proof (or a outline of a proof). AI: If we have Tonelli's theorem, then by considering positive parts and negative parts separately we immediately obtain Fubini's theorem. Conversely, assuming Fubini's theorem, Tonelli's theorem follows by monotone convergence argument applied to cut-off functions $f_k(x) = \min \{k, f(x)\} \chi_{B_k}(x)$. You can also find the detail at the Chapter 6.2 of the celebrated textbook Measure and Integral by Wheeden and Zygmund.
H: Are there any conditions for $G$ until above action has non-trivial kernel? Let $G$ is a group and $H$ be a subgroup of it. Then $G$ can act on the following set $$\Omega= \{Hg\|g\in G\}$$ by $\forall Hg\in\Omega$ and $x\in G$; $(Hg)^x=Hgx$ (I don't know if I can call this action right regular representation of $G$?). It can easily be found that the kernel of this action is: $$N=\{x\in G\|(Hg)^x=Hg\}=\bigcap_{g\in G}g^{-1}Hg$$ Clearly, if $H=\{1\}$ then $N=\{1\}$, so the action is faithful. Now, I am thinking about the condition(s) that we can consider for $G$ until above action has non-trivial kernel. For example, if the group be cyclic, abelian or our subgroup is normal in $G$ then $N=H$. Of course I assume $H\neq\{1\}$. Does this problem make any sense? Thanks. AI: The subgroup $$\cap_{g\in G}g^{-1}Hg$$ is known as the core of $H$ in $G$. It is the largest normal subgroup of $G$ that is contained in $H$. Therefore, the kernel of the action is trivial if and only if $H$ is corefree in $G$: it does not contain any nontrivial normal subgroup of $G$.
H: Proving that $\|d\|$ is not a prime number Assume $a,b,c$ are integers and $a+b+c=0$. If $d=a^{1433}+b^{1433}+c^{1433},$ prove that $\|d\|$ is not a prime number. AI: Note that $1433$ is prime. It follows by Fermat's theorem that $a^{1433}\equiv a\pmod{1433}$, with similar results for $b$ and $c$. Thus $$d=a^{1433}+b^{1433}+c^{1433}\equiv a+b+c\equiv 0\pmod{1433}.$$ We conclude that $\|d\|$ is divisible by $1433$. Finally, we need to show that $\|d\|$ cannot be equal to $1433$. This is essentially obvious, since $1433$-th powers of anything other than $0$, $1$, and $-1$ are very far apart. Added: Using Fermat's Theorem was ludicrous overkill! For note that $a+b+c$ is even, and the parity of $x^n$ is the same as the parity of $x$. So $d$ is even. Since $1433$ is odd, $\|d\|\ne 2$, so $d$ cannot be prime.
H: Tuple definition Is it correct? $S=\{\langle t,h\rangle:t\in\{0,\Delta t,2\Delta t,\cdots,24\},h\in\{0,\Delta h,2\Delta h,\cdots,H\}\}$ I would like to say that $S$ is a 2-tuple. The first tuple can vary from $0$ to $24$, with $\Delta T$ step, and the second one can vary from $0$ to $H$, with $\Delta H$ step. AI: $S$ is not a tuple, it's a set of $2$-tuples. I think what you want to say is that $S$ is a set of ordered pairs ($2$-tuples), each of which has a first entry that can vary from $0$ to $24$ in steps of length $\Delta t$; and second entry that can vary from $0$ to $H$ in steps of length $\Delta h$. As written, it would only make sense if $24$ is evenly divisible by $\Delta t$ and $H$ by $\Delta h$; that is, if and only if there exist integers $k$ and $\ell$ such that $24 = k\Delta t$ and $H=\ell\Delta h$. If the latter is not the case, then I would say that the first entry will take the values $k\Delta t$ with $k=0,1,2,\ldots,\lfloor\frac{24}{\Delta t}\rfloor$, and the second entry will take the values $\ell\Delta h$ with $\ell=0,1,2,\ldots,\lfloor\frac{H}{\Delta h}\rfloor$. Note, however, that $S$ is a set of tuples (not a tuple), and it is the first/second entry of the elements of $S$ that we are describing, not a "first tuple" and "second tuple": the elements of $S$ are themselves not ordered, so it doesn't make sense to talk about a "first tuple" and a "second tuple", when $S$ is a set.
H: What's an intuitive explanation of the max-flow min-cut theorem? I'm about to read the proof of the max-flow min-cut theorem that helps solve the maximum network flow problem. Could someone please suggest an intuitive way to understand the theorem? AI: Imagine a complex pipeline with a common source and common sink. You start to pump the water up, but you can't exceed some maximum flow. Why is that? Because there is some kind of bottleneck, i.e. a subset of pipes that transfer the fluid at their maximum capacity--you can't push more through. This bottleneck will be precisely the minimum cut, i.e. the set of edges that block the flow. Please note, that there may be more that one minimum cut. If you find one, the you know the maximum flow; knowing the maximum flow you know the capacity of the cut. Hope that explains something ;-)
H: Uncountability of $\overline{\mathbb{F}_p}$. In the following MathOverflow question, it has been pointed out that $\overline{\mathbb{F}_p}$ is an uncountable set. Whereas according to http://press.princeton.edu/chapters/s9103.pdf (see page 4 theorem 1.2.1) the closure $\overline{\mathbb{F}_p}$ is $\cup_{n=1}^{\infty}\mathbb{F}_{p^n}$, which I think is a countable union of finite sets and hence countable. Where am I going wrong in this? Also, in the same document before the same theorem its mentioned that if $\mathbb{F}_q$ has characteristic $p$ then its closure is same as that of $\mathbb{F}_p$ but I think that the set $\cup_{n=1}^{\infty}\mathbb{F}_{q^n}$ is a proper subset of $\cup_{n=1}^{\infty}\mathbb{F}_{p^n}$ since $q$ is a power of $p$, thus they are not the same. Again where is the fault in my reasoning? AI: The field discussed on MO is $\mathbb{F}_p((t))$, the field of formal Laurent series over $\mathbb{F}_p$. This has uncountable algebraic closure. The algebraic closure of $\mathbb{F}_p$ is countable, as you have correctly stated in the question.
H: Simplifying $\frac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$ I'm stuck in the follow equation: $$\dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$$ As all the bases are equal, I got $\dfrac{3n + 5}{2n - 3}$ Where I've to go now ? Thanks EDIT: Then, my initial idea was totally wrong, starting again, in the right way I got: $$\dfrac{2^n(2^4 + 2^2 + 2^{-1})}{2^n(2^{-2} + 2^{-1})}$$ but it's still wrong, I didn't get the right idea on the divisions you have shown to me in the answers. AI: You don't have an equation, you are probably asked to simplify. Divide top and bottom by $2^{n-2}$. When you do that, at the bottom you will have $1+2$. On top you will have $2^6+2^4+2$.
H: Doubt on displacement of a parabola(Again) In another exercise is given: Find the parabola which is a displacement of $y = 2x^2 - 3x + 4$ which passes though the point $(2, -1)$ and has $x = 1$ as its symmetry axis. I've reduced the based equation to the form: $y = 2(x - \frac{3}{4})^2 + \frac{23}{8}$, so the vertex is $(\frac{3}{4}, \frac{23}{8})$. As the displaced equation has its symmetry axes on $x = 1$ it's known that its vertex is $(1, x)$ what can I do more to find the $y$ of the displaced equation ? Thanks in advance. EDIT: Corrected the typo AI: Your equation for the original parabola should read $y = 2\left(x-\dfrac{3}{4}\right)^2+\dfrac{23}{8}$. The displaced parabola will have the equation $y = 2(x-x_v)^2 + y_v$, where $(x_v, y_v)$ is the vertex. Since the axis is $x=1$, $x_v$ is $1$. The point $(2,-1)$ is on the parabola, so we have $-1 = 2(2-1)^2 + y_v \Rightarrow y_v = -1 -2 = -3$. Then, the equation of the displaced parabola is $y = 2(x-1)^2 -3$.
H: Convex cone of functions of the form $a\log(1+bx + ?)$ Given the set of all functions \begin{align} f_{a,b}(x): [0, \infty) &\to [0, \infty) \\ x &\mapsto a\log(1+bx) \end{align} Is there a way of making the set of these functions $S =\{f_{a,b}: a,b \ge 0\}$ have the property that $\left[f_{a,b} + f_{a',b'}\right] \in S$? Can this be done by adding a fixed number of extra parameters to the definition? AI: No: the functions $\log(1+bx)$ for different $b > 0$ are linearly independent, so you would need infinitely many parameters. A way to see that they are linearly independent is to look at their derivatives, $\dfrac{b}{1+bx}$, which have poles at different points $x=-1/b$.
H: Integrating with respect to different variables I have started reading a book on differential equations and it says something like: $$\frac{dx}{x} = k \, dt$$ Integrating both sides gives $$\log x = kt + c$$ How is it that I can 'integrate both sides here' when I am integrating one side with respect to $x$ yet I am integrating the other side with respect to $t$? AI: What is going on there is what is called an abuse of notation. What you really have there is an equation in $t$. $x=x(t)$ is a function dependent on $t$. So what we're doing is the following - I presume this is the original equation: $$\frac{dx}{dt}=k x $$ This is the same as $$x'(t)=k x(t) $$ $$\frac{x'(t)}{x(t)}=k $$ Now we integrate wrt to $t$ $$\int\frac{x'(t)}{x(t)}dt=kt+C $$ But we note letting $X=x(t)$ so $dX = x'(t) dt$ gives $$\int\frac{dX}{X}=kt+C $$ $$\log X = kt+C$$ So switching back $$\log x(t) = kt+C$$ $$x(t)=C e^{kt}$$ What we actually do, in some sense, is integrate with respect to "only" $t$ in one side, and "$x(t)$" in the other (which is done implicitly). The notation is very useful and suggestive, so we use it, understanding what we're doing is the above.
H: In which interval there is a solution by Bolzano theorem? Let a function of domain equal to $\mathbb{R}$ be $f(x)=e^x-3$. In which of the follows intervals, by the Bolzano theorem, we can say that $f(x)=-x-\frac{3}{2}$ have at least one solution? $A) \left ]0,\frac{1}{5} \right[$ $B) \left ]\frac{1}{5},\frac{1}{4} \right[$ $C) \left ]\frac{1}{4},\frac{1}{3} \right[$ $D) \left ]\frac{1}{3},1 \right[$ At first I tried to solve $\space e^x-3=-x-\frac{3}{2}\space$ in order of $x$. I riched the equation $\space e^x+x=\frac{3}{2} \space$that I don't know how to solve, with the tools that I have learned. I know that the$\space e^x-3=-x-\frac{3}{2}\space$ solution's is the intersept point beteewn the two functions. But I can't manage the solution using the Bolzano theorem principles. Thanks for the help AI: You don't have to solve the equation, because you only want to know if there is a solution. Remember that Bolzano's theorem says that if $f$ is a continuous function in $[a,b]$, and the signs of $f(a)$ and $f(b)$ are different (one positive and the other negative), then there is at least one root of $f$ in $(a,b)$, i.e. a value of $x \in (a,b)$ with $f(x) = 0$. So what you want to do is first come up with a function that will work: since $e^x - 3 = - x - \frac{3}{2} \Leftrightarrow e^x - 3 + x + \frac{3}{2} = 0$, you can pick $f(x) = e^x - 3 + x + \frac{3}{2}$. Now for each one of the intervals, calculate the images of the endpoints, and if the signs are different then there is a root of $f$ in that interval. Note that there could be a root in those intervals even if the images of the endpoints have different signs, but we can't know by Bolzano's theorem. $f(0) \lt 0$ and $f\left(\frac{1}{5}\right) \lt 0$, so we can't assure by Bolzano's theorem that there's a root in $\left(0,\frac{1}{5}\right)$. $f\left(\frac{1}{5}\right) \lt 0$ and $f\left(\frac{1}{4}\right) \gt 0$, so Bolzano's theorem tells us that there's at least one root in $\left(\frac{1}{5},\frac{1}{4}\right)$. $f\left(\frac{1}{4}\right) \gt 0$ and $f\left(\frac{1}{3}\right) \gt 0$, so again we can't be sure if there's a root in $\left(\frac{1}{4},\frac{1}{3}\right)$. $f\left(\frac{1}{3}\right) \gt 0$ and $f(1) \gt 0$, so we can't be sure if there's a root in $\left(\frac{1}{4},1\right)$.
H: Why Zariski topology? Why in algebraic geometry we usually consider the Zariski topology on $\mathbb A^n_k$? Ultimately it seems a not very interesting topology, infact the open sets are very large and it doesn't satisfy the Hausdorff separation axiom. Ok the basis is very simple, but what are the advantages? AI: To appreciate the Zariski topology it helps to have a fairly broad view about what a topological space is. Topological spaces in full generality are, confusingly, not very topological in the naive sense! As discussed in this math.SE question, I think it is better to think of point-set topology as being about semidecidable properties (which are the open sets). The familiar kind of topology induced by a metric is about the specific property of being close in a metric sense, but other kinds of topologies are about different kinds of properties. The Zariski topology is about the property of non-vanishing of polynomials. The semidecidable properties here are the properties "this set of polynomials does not vanish here." Intuitively speaking the reason this is semidecidable is that you can compute the value of a polynomial at a point to finite precision and once you show that it is sufficiently different from zero it cannot be zero. The fact that the Zariski topology isn't Hausdorff isn't a weird property of the Zariski topology; it tells you something important about how vanishing of polynomials behaves, namely that the behavior of a polynomial on a few points can tell you a lot about its behavior at seemingly far-away points. This is intrinsic to the nature of algebraic geometry and pretending that the Zariski topology doesn't exist won't make it go away. Okay, so what can you actually do with it? Here are a couple of things: If two polynomials agree on a Zariski-dense subset, then they agree identically. This is a surprisingly useful way to prove polynomial identities; for example, it can famously be used to prove the Cayley-Hamilton theorem. Moving to the Zariski topology on schemes allows the use of generic points. I am not familiar with examples of this technique in use though. Serre famously made use of the Zariski topology to introduce sheaf cohomology to algebraic geometry, which was (as I understand it) a crucial innovation. To really appreciate the Zariski topology it helps to generalize it to arbitrary commutative rings. An important motivational example: if $X$ is a compact Hausdorff space and $C(X)$ is the ring of continuous functions $X \to \mathbb{R}$, then the maximal spectrum of $C(X)$ not only can be identified with $X$, but has the same topology! (This is an exercise in Atiyah-MacDonald.) The rings one gets in this way are precisely the real subalgebras of complex commutative C*-algebras by the commutative Gelfand-Naimark theorem, and in fact you get a (contravariant) equivalence of categories. Moreover, by the Serre-Swan theorem, the category of real vector bundles on $X$ is naturally equivalent to the category of finitely-generated projective modules over $C(X)$. It helps to think about this example like a physicist. Think of $X$ as the set of possible states of some physical system and the elements of $C(X)$ as observations one can make about the system; the value of a function at a point is the result of the observation in a fixed state. The Zariski topology here captures all semidecidable properties that you can decide using the observations in $C(X)$. For example, if one of the functions in $C(X)$ is called "temperature," there is a corresponding semidecidable property "the temperature of the system is between $0$ and $100$ degrees inclusive," which you can decide by computing the temperature to finite precision. (What if $X$ is not compact? Then if you work with the ring $C_b(X)$ of bounded continuous functions on $X$, there are consistent sets of possible values of the observables which do not arise from an actual state of your system; they are points in the Stone-Čech compactification $\beta X$ instead.) Here's another example that I like: let $B$ be a Boolean ring, which is a ring satisfying $b^2 = b$ for all $b \in B$. Then every element of $B$ can be identified with a subset of its maximal spectrum. This idea can be used to prove Stone's representation theorem for Boolean algebras, deduce the existence of ultrafilters from the existence of maximal ideals in rings, and prove the compactness theorem in propositional logic (without proving the completeness theorem)! For a discussion, see my blog post Boolean rings, ultrafilters, and Stone's representation theorem.
H: Understanding a Markov Chain I am using a Markov Chain to get the 10 best search results from the union of 3 different search engines. The top 10 results are taken from each engine to form a set of 30 results. The chain starts at State x, a uniform distribution of set S = {1,2,3,...30}. If the current state is page $i$, select page $j$ uniformly from the union of the results from each search engine. If the rank of $j$ < rank of $i$ in 2 of the 3 engines that rank both $i$ and $j$, move to $j$. Else, remain at $i$. I understand the above no problem. The sorting point is where I am stuck however. The paper I am using that explains this says: This is known as a Markov process, where the transition matrix $P$ has $$P(i, j) = \frac{1}{n}$$ if a majority of the input rankings prefer j to i, and $$P(i, i) = 1−\sum_{j≠i} P(i, j)$$ Under certain conditions, this process has a unique (up to scalar multiples) limiting distribution $x$ that satisfies $x = xP$, where $x(i)$ gives the fraction of time the process spends at element $i$. Dwork et al. propose sorting the elements by non-increasing $x(i)$ values. To ensure that the process has a unique limiting distribution $x$, we use a "random jump": with probability $δ > 0$, we will choose a random element and move to this element (regardless of whether this element is preferred to the current element). In our experiments we have used $δ= \frac{1}{7}$ , which is the value of $δ$ that is often chosen in the literature for PageRank implementations. Could someone please explain this to me in plain english because I am completely lost with it at this stage. I don't really understand the whole concept. This paper can be found here with this specific part on page 43. It may help if I add a small example. I have 5 result pages from 3 search engines. These are their rankings: engine1 engine2 engine3 Page 1 1 2 2 Page 2 4 3 1 Page 3 2 4 5 Page 4 5 5 3 Page 5 3 1 4 If page 1 is the start point, to move to page 2 it has to have a lower rank than page 1 in 2 out of 3 of the search engines. The probability of this is $\frac{x}{n}$. I don't know if n should be 3 or 5? That's my first major problem! AI: We have $n=\|S\|=5$: the states are the individual pages, so for 3 engines returning 5 possibly duplicate pages, you could have anywhere from 5 to 15 states. Define the relation between pages $j\prec i$ if and only if $j$ is ranked better than $i$ in a majority of engines. This relation is a partial order, and lower elements are "better". The intuitive goal is to build a Markov chain following this relation to start from a random page and gradually improve the quality of the result. Accordingly, the definition of $P$ is such that any path in the Markov chain must be decreasing in the sense of $\prec$, and it can always be extended until we land on a minimal element. Such an element is not unique in general: for example, $i$ could be ranked (1,2,3) and $j$ could be ranked (1,3,2), so that $i\not\prec j$ and $j\not\prec i$. Once we are at a minimal element, we can no longer move: we say that the distribution assigning probability 1 to $i$ is stationary for the chain (a distribution is simply a probability measure on the set of states, that is a set of weights summing to 1). But because there are several stationary distributions, where we land depends on where we started from: we say that there is no equilibrium distribution. This is annoying because we have to pick a sensible starting state, and this won't give good results anyway since we will only get a single non-zero probability page in the stationary distribution. So to fix that, we "shuffle" things a bit and replace the original transition with probability $\delta>0$ with a transition to a uniform random state: $$P'(i,j)=\delta/n + (1-\delta)P(i,j)$$ As a result, the expected time to visit state $i$ starting from any distribution of states is at most $n/\delta<\infty$: we say that state $i$ is positive recurrent. It's a standard property of Markov chains that when this holds for all states, there is a unique equilibrium distribution, and furthermore it has non-zero probability for each state. This means that no matter which state you start from, if you move along the chain sufficiently many times the distribution of states will get arbitrarily close to the equilibrium distribution. So better ranked pages will tend to be "visited" more frequently by the chain: if you compute the equilibrium distribution, you then have a natural order with which to rank the whole set of pages. The equilibrium distribution $\pi$ satisfies $$P'\pi = \pi\\ \sum_{i=1}^n \pi_i=1$$ which is a linear system of equations (and we know this sytem has a unique solution): most mathematical software libraries can easily solve that to give you $\pi$.
H: Triples of positive real numbers $(a,b,c)$ such that $\lfloor a\rfloor bc=3,\; a\lfloor b\rfloor c=4,\;ab\lfloor c\rfloor=5$ Find the all ordered triplets of positive real numbers $(a,b,c)$ such that: $$\lfloor a\rfloor bc=3,\quad a\lfloor b\rfloor c=4,\quad ab\lfloor c\rfloor=5,$$ where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. AI: Hint: Note that $a$, $b$, and $c$ are $\ge 1$. By the first condition, all of our numbers are in the interval $[1,2)$, or two of them are in $[1,2)$ and the other is in $[2,3)$. So there are only $4$ cases to consider, and each gives us an explicit system of ordinary equations (no floors). And it's not really even $4$.
H: Prove Axiom $10$ (Vector Spaces) independent of the others Possible Duplicate: Is it possible to construct a quasi-vectorial space without an identity element? In Apostol Multivariable Calculus, $1.5$ exercise $30 b$, he asks the reader to prove that Axiom $10$ is independent of the other axioms for vector spaces. Axiom $10$ says that $$1 * x=x$$ To do this, it is equivalent to prove that $(-1)x\not = -x$, since $(-1)x=-x$ implies that $1x=x$ (we can prove it using the other axioms. My approach is to find $(V,+,)$ that satisfies the first $9$ axioms and contradicts the axiom $10$. My attempts so far: First, I tried setting $V=\mathbb{R}$, redefining $ax$ to be variations on multiplication. For example, I tried $ax=\|ax\|$ but ran into problems with $\|a(x+y)\|\not=\|ax\|+\|ay\|$ in general. Then, I tried setting $V=\mathbb{R^2}$ and $ax=ax^T$ if $a$ is positive and $ax=ax$ if $a$ is not positive. However, I ran into problems with $(a+b)x\not=ax+b*x$ for $a=-1$,$b=1$,$x=(1,0)$. AI: Take any additive group $V$ and let $F$ be a field. Define a scalar product by $k \cdot v = 0$ for all $k \in F$ and $v \in V$, I believe this satisfies all the other axioms except $1 \cdot v = v$.
H: If $\sum a_n z^n = f(z) = \sum b_n z^n$, What Can Be Said About the Coefficients $a_n$ and $b_n$ I apologize for the cryptic title, this issue came up while obtaining a solution to my question here. I was given a power series: $$\sum_{n=0}^{\infty}r(n)z^n = \frac{1}{(1-z)(1-z^2)(1-z^3)}$$ I was then asked to perform a partial fraction decomposition on the closed form and obtained: $$ -\frac{\frac{1}{6}}{(z-1)^3} + \frac{\frac{1}{4}}{(z-1)^2} - \frac{\frac{17}{72}}{z-1} + \frac{\frac{1}{8}}{z+1} - \frac{\frac{1}{9}e^{\frac{2\pi i}{3}}}{z-e^{\frac{2\pi i}{3}}} - \frac{\frac{1}{9}e^{\frac{-2\pi i}{3}}}{z-e^{\frac{-2\pi i}{3}}}$$ Now upon individually converting these fractions back into power series, in short, what I obtained is a power series with coefficients $\frac{(n+3)^2}{12} + x_n$ for $\frac{1}{3} \leq x_n \leq \frac{2}{3}$. This was expected, since my book states that $r(n)$ is given by the integer closest to $\frac{(n+3)^2}{12}$. What I don't understand is how can we conclude such a close correspondence between $r(n)$ and $\frac{(n+3)^2}{12} + x_n$? Just because two power series converge to the same closed form function, can't their coefficients not still be very different for any particular $n$? I know Taylor series are unique, but I don't think my latter power series is a Taylor series, and even if the power series expansion of a function is always unique, it still has that extra $x_n$ added to or subtracted from each coefficient. What is going on here? Edit: Ok so $r(n)$ is supposed to be the number of ordered triples $(x_1,x_2,x_3)$ such that $x_1 + 2x_2 + 3x_3 = n$. I formed this by employing the Cauchy product $(\sum z^n)(\sum z^{2n})(\sum z^{3n})$. AI: You have two power series expansions of the same function, one with coefficients $r(n)$, one with coefficients $(1/12)(n+3)^2+x_n$. By the uniqueness of power series expansions, $r(n)=(1/12)(n+3)^2+x_n$. Unless, of course, you have edited your post several more times while I was typing this.
H: At which points tangent to a curve is parallel to given plane? At which points on the curve $\alpha(t):=(3t-t^3,3t^2,3t+t^3)$ the corresponding tangent lines are parallel to the plane $3x+y+z+2=0$? AI: Hints: The tangent line at $t$ has direction $\alpha'(t)$ The normal of a plane given by $ax + by + cz + d = 0$ is $(a,b,c)$. In order for a line to be parallel to a plane, its direction vector must be orthogonal to the normal to the plane. This will give you a polynomial equation in $t$. The tangent line at $t$ has slope $\alpha'(t) = (3-3t^2,6t, 3 + 3t^2)$, and for it to be parallel to the plane given by the equation $3x+y+z +2= 0$, with normal $(3,1,1)$, we must have $(3-3t^2,6t, 3 + 3t^2) \cdot (3,1,1) = 0$. Then $\begin{align}&(3-3t^2,6t, 3 + 3t^2)\cdot (3,1,1)=9-9t^2+6t+3+3t^2=-6t^2+6t+ 12=0\\ &\Leftrightarrow t^2-t-2=0 \Leftrightarrow(t+1)(t-2)=0\end{align}$ And we get $t = -1$ or $t=2$.
H: How many $1$'s could there be in this sequence? Matrix, operator? For each $(i,j)\in \mathbb{N}^2$, $a(i,j)=1$ or $0$, and 1) $a(i,i)=0$ for all $i$; 2)for fixed $i$, there is at most one $j$ such that $a(i,j)=1$. Suppose we know that there is a finite $\kappa$ such that \begin{equation} \sum_{i\in S, j\in S^{C}}a(i,j)\le\kappa\end{equation} for any subset $S$ of $\mathbb{N}$,then one can show $\sum a(i,j)$ is bounded by $3\kappa$. I am asking what is the largest possible value of $\sum a(i,j)$. For people who are interested in the background, I am considering an operator $T\in\mathcal{L}(\ell^2)$. With respect to the natural basis, the entries of $T$ are $a(i,j)$. The existence of $\kappa$ in the first inequality is the same as \begin{equation}rank(TP-PTP)\le\kappa\end{equation} for all orthogonal projections $P$ in the masa (maximal abelian, self-adjoint subalgebra) of diagonal operators. In a recent paper by Popov, Marcoux and Radjavi (Almost Invariant Halfspaces and Approximate Commutation), they have shown $\sum a(i,j)$ is bounded by $3\kappa$ (Thm 4.7), but in almost all examples I can think of, $\sum a(i,j)\le 2\kappa$, so I wonder whether the coefficient can be improved. Again, I think the problem has been reduced to a purely algebraic or number theoretic problem in the first two paragraph of this post. Thanks very much! AI: I believe there is a small counterexample with $\kappa=3$: $$\left( \begin{array}{ccccc} 0&1&1&0&0\\ 0&0&1&1&0\\ 0&0&0&1&1\\ 1&0&0&0&1\\ 1&1&0&0&0 \end{array} \right).$$ More generally, it is easy to prove an upper bound of $4 \kappa$, and I am inclined to believe it is asymptotically best possible (I guess the Alon paper cites a bound of the form $4 \kappa - O(\sqrt{\kappa}))$. For the modified problem, where each row can have at most one 1, the stated upper bound $3\kappa$, assuming it is now accurate, is achieved by taking multiple disjoint copies of $\left( \begin{array}{ccc} 0&1&0\\ 0&0&1\\ 1&0&0\\ \end{array} \right)$. This satisfies the constraint using $\kappa$ copies for a total of $3\kappa$.
H: Upper-tail inequality for t-distribution I am interested in upper tail bounds (or bounds on deviation from the mean) for t-distribution with n degrees of freedom (http://en.wikipedia.org/wiki/Student's_t-distribution) A bound that is of the form similar to the tail bound for standard normal distribution Proof of upper-tail inequality for standard normal distribution would be great. Does anyone know where to find such a result? AI: Since $\frac{\mathrm d}{\mathrm dt}(1+t^2/n)^{-(n-1)/2} = -\frac{(n-1)}{2}(1+t^2/n)^{-(n+1)/2}\frac{2t}{n} = -\frac{(n-1)}{n}\cdot t \cdot (1+t^2/n)^{-(n+1)/2}$, the method used by @cardinal in deriving a bound on the upper tail probabilityof the normal distribution seems to work here as well. We have $$\begin{align} P\{X > x\} &= \int_x^{\infty} \frac{c}{\left(1 + \frac{t^2}{n}\right)^{(n+1)/2}} \mathrm dt\\ &\leq \int_x^{\infty}\frac{t}{x} \frac{c}{\left(1 + \frac{t^2}{n}\right)^{(n+1)/2}} \mathrm dt\\ &= \frac{c}{x}\cdot\frac{n}{n-1} \left[-(1+t^2/n)^{-(n-1)/2} \right\|_x^{\infty}\\ &= \frac{n}{n-1}\cdot \frac{1}{x}\cdot \frac{c}{\left(1 + \frac{x^2}{n}\right)^{(n-1)/2}} \end{align}$$
H: Learning Model Theory What books/notes should one read to learn model theory? As I do not have much background in logic it would be ideal if such a reference does not assume much background in logic. Also, as I am interested in arithmetic geometry, is there a reference with a view towards such a topic? AI: I really like Introduction to Model Theory by David Marker. It starts from scratch and has a lot of algebraic examples.
H: Trying to reverse engineer this pattern... this is my first post on the mathematics node of stack exchange, so please forgive me if I'm not posting an appropriate question, but I'm not sure where else to address this. I'm trying to figure out what equation might generate the following pattern. (keep in mind the numbers might be slightly off, not exact) 0 :: 0, 0 10 :: 19, -16 (0.17453) 20 :: 41, -28 (0.34906) 30 :: 64, -37 (0.52359) 40 :: 89, -41 (0.69813) 50 :: 113,-41 (0.87266) 60 :: 138,-37 (1.04719) 70 :: 162,-28 (1.22) 80 :: 182,-16 (1.392) 90 :: 201, 0 (1.570) 100:: 217,19 (1.745) 110:: 229,41 (1.919) 120:: 238,64 (2.094) 130:: 241,89 (2.268) 140:: 241,113 (2.443) 150:: 238,138 (2.617) 160:: 229,162 (2.792) 170:: 217,182 (2.792) 180:: 201,201 (2.967) 190:: 182,217 (3.316) 200:: 162,229 (3.490) 210:: 138,238 (3.665) 220:: 113,241 (3.839) 230:: 89,241 (4.0142) 240:: 64,238 (4.188) 250:: 41,229 260:: 19,217 270:: 0,201 280:: -16,182 290:: -28,160 300:: -37,137 310:: -41,113 320:: -41,89 330:: -37,64 340:: -28,41 350:: -16,19 360:: 0,0 as you math wizez may have guessed it has to do with adjusting coordinates for rotation around an axis. Any help would be great! AI: Just plot your data and make a guess. Let $f(x) = 142 \sin \frac{\pi (t -45)}{180}+100$. Then your data is reasonably well approximated by $\{t, f(t), f(360-t)\}_{t\in \{0,\cdots,360\}}$. Here is a plot of $f$ and the first two columns of the data above:
H: Proving a commutative ring can be embedded in any quotient ring. Here's the exercise, as quoted from B.L. van der Waerden's Algebra, Show that any commutative ring $\mathfrak{R}$ (with or without a zero divisor) can be embedded in a ''quotient ring" consisting of all quotients $a/b$, with $b$ not a divisor of zero. More generally, $b$ may range over any set $\mathfrak{M}$ of non-divisors of zero which is closed under multiplication (that is, $b_1$, $b_2$ is in $\mathfrak{M}$ when $b_1$ and $b_2$ are). The result is a quotient ring $\mathfrak{R}_{\mathfrak{M}}$. I'm not sure if I am stuck or if I am overthinking. My answer goes like this: Commutative rings without zero divisors are integral domains as defined in Algebra and, by removing all $a/b$ which are not in the the commutative ring $R$ from the field $R \hookrightarrow Q$ where $Q$ is the field of all quotients $a/b$, one shows that any commutative ring without zero divisors can be embedded in a quotient ring. (This is more rigorously outlined in Algebra itself.) Now what has me confused is how this case differs from a case with zero divisors. Doesn't the exact same logic hold for a commutative ring with zero divisors? The only thing that I think zero divisors would interfere with is solving equations of the form $ax=b$ and $ya=b$ since there aren't inverses of zero divisors. However, we do not need to show that they can be embedded in a quotient field. We're only showing they can be embedded in a quotient ring. So, there's really no issue here and we apply the same approach as above. Now, as for the last part, I think that all that needs to be said is that if the set $\mathfrak{M}$ wasn't closed under multiplication, neither would the "ring" be and hence it would not be a ring by definition. Therefore, $\mathfrak{M}$ has to be closed and any commutative ring where $b$ ranges over any set of non-divisors of zero can be embedded in a quotient ring. I guess my real issue here is that I can't tell if I'm overthinking or underthinking. Does anyone care to elucidate this for me? AI: I'd say you're essentially on the right track, but I think you should think more carefully about how to rigorously define "$a/b$" (take a look at the page on localization if you need help). The "quotient ring" the question is asking about is called the total ring of fractions of $\mathfrak{R}$, which is a field if and only if $\mathfrak{R}$ is a domain. Side note: I prefer the terminology "ring of fractions" (or "field of fractions") to "quotient ring", which to me is a term that refers to a quotient of a ring by an ideal.
H: To show the function $\frac{1}{x\log x}$ is continuous on $[2,\infty)$ I want to know that how can i show that the function $\displaystyle f(x)= \frac{1}{x\log x}$ continuous? Thanks in advance! AI: If $f$ and $g$ are continuous on a particular interval and $g(x) \neq 0$ for any $x$ on that interval, then $\frac{f(x)}{g(x)}$ is continuous on that interval. $1$ is continuous on that interval. $x\log x$ is continuous and nonzero on that interval. Hence $\frac{1}{x\log x}$ is continuous.
H: For any $n \geq 1$, prove that there exists a prime $p$ with at least n of its digits equal to $0$ For any $n \geq 1$, prove that there exists a prime $p$ with at least n of its digits equal to $0$. I don't even know how to start?? Any help(even a hint) would do. Thanks in advance!! AI: The series $\sum_{p \in \mathbb P} \frac{1}{p}$ of primes diverges (proof). Show that the series $\sum_{n \in A} \frac{1}{n}$ converges, where $A$ is the set of integers with at most $k$ zeroes (modify this proof). Therefore $\mathbb P \not \subset A$.
H: Is this right? $ \| \nabla f \|^k \leqslant \sum_{i=1}^n \| \partial_i f \|^k $ Is this right if $ k \geqslant 1$ ? Then why? $$ \| \nabla f \|^k \leqslant \sum_{i=1}^n \| \partial_i f \|^k \; $$ AI: This is essentially just the triangle equality for the so called $p$-norm (in your case $p=k$) together with the monotonicity of the $k$-th root. See this wikipedia article. Edit: I was wrong, this now doesn't seem to have to do with $p$-norms. What was I thinking? What you want to show is that for a vector $x=(x_1,\dots,x_n)\in\mathbb R^n$ with all components $\geq 0$ the following holds: $$(x_1^2+\dots+x_n^2)^{k/2}\leq x_1^k+\dots+x_n^k$$ Since everything is $\geq 0$, this inequality holds iff it holds for the squares of both sides. So you want to show $$(x_1^2+\dots+x_n^2)^k\leq(x_1^k+\dots+x_n^k)^2.$$ But this is actually not true: Suppose $x_1=x_2=1$ and $k=3$. Then $$(x_1^2+x_2^2)^k=2^3=8\not\leq (x_1^k+x_2^k)^2=2^2=4.$$ You get a counterexample to your inequality from this by choosing a function $f:\mathbb R^2\to\mathbb R$ with $\partial f/\partial x_1=\partial f/\partial x_2=1$.
H: Question on soluble groups I'd like to ask for a clarification. I came across the following: Lemma 2. Let $G$ be a finite solvable group and let $p$ be a prime number dividing $\|G\|$. Suppose that $M_1$ and $M_2$ are inconjugate maximal subgroups of $G$, both of which have $p$-power index in $G$, and neither of which is normal in $G$. Then $(M_1\cap M_2)\mathbf{O}^p(G)=G$. Furthermore, if $P_0\in \mathrm{Syl}_p(\mathbf{O}^p(G))$, then $(M_1\cap P_0)(M_2\cap P_0) = P_0$. Proof. Note that every maximal subgroup containing $\mathbf{O}^p(G)$ is normal of index $p$. To prove that $(M_1\cap M_2)\mathbf{O}^p(G)$, it therefore suffices to show... I don't understand the first line of the proof. Any help would be appreciated. AI: The quotient $G/\mathbf{O}^p(G)$ is a $p$-group. By the basic properties of $p$-groups its maximal subgroups are of index $p$. Also all maximal subgroups of a $p$-group are normal (a subgroup of index equal to the smallest prime factor of the order of the group is normal). So it seems to me that the first line of the proof follows from this and the correspondence principle: the subgroups of $G/\mathbf{O}^p(G)$ are in 1-1 correspondence with subgroups of $G$ containing $\mathbf{O}^p(G)$, normality is preserved.
H: How to prove this, $ \| \Delta f \| \leqslant n \| \nabla^2 f\| $ I hope to prove this, $$ \| \Delta f \| \leqslant n \| \nabla^2 f\| $$ where $f : \mathbb R^n \to \mathbb R $. AI: Use Cauchy-Schwarz inequality $$ \|\Delta f\|=\left\|\sum\limits_{i=1}^n \partial_i^2 f\right\|= \left\|\sum\limits_{i=1}^n 1\cdot\partial_i^2 f\right\|\leq \left(\sum\limits_{i=1}^n 1^2\right)^{1/2}\left(\sum\limits_{i=1}^n (\partial_i^2 f)^2\right)^{1/2}= \sqrt{n}\|\nabla^2 f\|\leq n\|\nabla^2 f\| $$
H: $\lim\limits_{x\to\infty}x^a\sin(1/x)$ $$\lim_{x \rightarrow \infty} x^a\sin{\frac{1}{x}}$$ for this limit ,it was showed on the textbook that $$\lim_{x \rightarrow \infty} x^a\sin{\frac{1}{x}}=\begin{cases} 0 & a<1 \\ 1 & a=1 \\ \infty & a>1 \end{cases}$$ in my opinion ,it's obviously that $\lim\limits_{x \rightarrow \infty}\sin{\frac{1}{x}}=0$ and that $\lim\limits_{x \rightarrow \infty}x^a=\infty$ I wonder what's wrong with my view AI: There is no need for L'Hospital here if you already know that $\lim_{x\to0}\frac{\sin x}{x}=1$. From the above identity we can get the following: $$\lim_{x\to\infty}x^a\sin\frac1x=\lim_{x\to\infty}x^{a-1+1}\sin\frac1x=\lim_{x\to\infty}x^{a-1}x\sin\frac1x=\lim_{x\to\infty}x^{a-1}\frac{\sin\frac1x}{\frac1x}$$ If $a=1$ then we have the limit above is equal to 1 ($x^0=1$ and set $t=\frac1x$, now you have $\frac{\sin t}t$ as $t\to 0$). If $a<1$ then we have $x^{a-1}=\frac1{x^{1-a}}$ approaches zero, and we have a multiplication of a bounded limit by a zero limit. Thus zero. I $a>1$ then $a-1>0$ and so $x^{a-1}$ approaches $\infty$ and the $\sin$ part is finite non-zero. So the limit is infinity.

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