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H: What are rational integer coefficients? I have a question about the following excerpt from Atiyah-Macdonald (page 30): “A ring $A$ is said to be finitely generated if it is finitely generated as a $\mathbb Z$-algebra. This means that there exist finitely many elements $x_1,\dotsc,x_n$ in $A$ such that every element of $A$ can be written as a polynomial in the $x_i$ with rational integer coefficients.” I suspect one should delete "rational" and then it says that $A$ is called finitely generated if $A = \mathbb Z [a_1, \dots a_n]$ for some $a_i \in A$, that is, every element in $A$ can be written as a polynomial in $a_i$ with integer coefficients. If this is a typo it is not mentioned on MO but perhaps it is not and I misunderstand the definitions. If I do: What are rational integer coefficients? AI: Sometimes the members of $\mathbb{Z}$ are called "rational integers" to distinguish them from $\mathbb{Z}[i]$ or other rings.
H: Darboux's theorem In Dunham's "Calculus Gallery", it introduces Darboux's theorem: If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then there exists a c in (a,b) such that $f'(r)=c$ Proof: [summarized] Introduce $g(x)=f(x)-rx$. There is a point $c$ in $[a,b]$ where g takes a minimum. Because $g'(a)=f'(a)-r<0$ and $g'(b)=f'(b)-r>0$ we see that a minimum cannot occur at a or b, and so c lies in (a,b)... I don't see why the minimum can't occur at a or b. Consider $f(x)=x^2, r=1.5,[a,b]=[1,2]$. The minimum value of $g'(x)=2x-1.5$ in $[1,2]$ does indeed seem to be $x=1$ or $x=a$. Does he mean $min(\|g'(x)\|)$? AI: In your example, $f'(x)=2x$, so $f'(1)=2\gt r$. The minimum can't occur at $a$ or $b$ because $g'(a)\lt0$ and $g'(b)\gt0$, so $g$ decreases from the boundary inwards on both sides.
H: Evaluating $\int_{0}^{\infty}\frac{\arctan \sin^2x}{x}dx$ Seems to be a hard nut: $$I=\int_{0}^{\infty}\frac{\arctan \sin^2x}{x}dx$$ Any hint? AI: It is not hard to show that \begin{equation}\int_{0}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty, \quad \cdots \quad (1)\end{equation} and it is also easy to show that $$\arctan x \geq Cx \quad \text{for} \quad 0 \leq x \leq 1 \quad \cdots \quad (2)$$ for some positive constant $C > 0$. Now it is clear that these together imply \begin{equation}\int_{0}^{\infty} \frac{\arctan \sin^2 x}{x} \; dx \geq C \int_{0}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty\end{equation} Indeed, we first show that $(1)$ diverges. It suffices to show that $$ \int_{2012}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty. $$ By integration by parts, we have $$ \begin{align} \int_{2012}^{R} \frac{\sin^2 x}{x} \; dx &= \left[ \frac{1}{2} - \frac{\sin 2x}{4x}\right]_{2012}^{R} + \int_{2012}^{R} \left( \frac{1}{2x} - \frac{\sin 2x}{4x^2}\right) \; dx\\ &= \frac{1}{2}\log R + O(1), \end{align}$$ which proves $(1)$ by letting $R\to\infty$. Now we prove $(2)$. by examining second derivative of arc-tangent function, we find that it is concave on $[0, 1]$. Thus on this interval we have $$\arctan x \geq (\arctan 1) x,$$ which proves $(2)$ with $C=\arctan 1 =\frac{\pi}{4}$.
H: Infinite series: $1/2 + 1/(1\cdot 2 \cdot 3) + 1/(3\cdot 4 \cdot 5) + \ldots$ How do I calculate this: $$\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\dots $$ I have not been sucessful to do this. AI: Hint: $$ \frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} $$ and $$ 1 - \frac12 + \frac13 - \frac14 + \dotsb = \ln 2$$
H: Minimizing a functional on $L^2$ Let $$ \mathcal{M} := \left\{f \in L^2([0,\pi]): \int_0^\pi f(x)\cos x dx = \int_0^\pi f(x)\sin x dx = 1\right\}. $$ Solve this problem: $$ \tag{P} \min_{\mathcal M} \int_0^\pi [f(x)]^2dx $$ Using Cauchy-Schwarz, I get $$ 1 = \langle f(x), \sin{x} \rangle \le \Vert f \Vert_2 \Vert \sin{x} \Vert_2 = \Vert f \Vert_2 \sqrt{\frac{\pi}{2}} $$ i.e. $$ \Vert f \Vert_2^2 \ge \frac{2}{\pi}, \qquad \forall f \in \mathcal M. $$ Is this correct? Now what can I do to prove that this is the minimum? I would like to find a function $f \in \mathcal{M}$ which realizes that value, but I didn't manage to find it. Can you help me? Thanks. AI: If $g\in L^2([0,\pi])$ is orthogonal to $\sin(x)$ and $\cos(x)$ and $f$ is the minimum then for any $\epsilon\in \mathbb R$ $$ \langle f + \epsilon g, \sin(x) \rangle = \langle f, \sin(x)\rangle = 1\\ \langle f + \epsilon g, \cos(x) \rangle = \langle f, \cos(x)\rangle = 1 $$ So for each $\epsilon\in \mathbb R$, $f + \epsilon g$ belongs to $\mathcal M$ and $\phi(\epsilon):= \lVert f + \epsilon g \rVert ^2$ must have a minimum at $\epsilon = 0$. Since $$ \left . \frac {d\phi} {d\epsilon} \right\vert_{\epsilon = 0} = 2\langle f, g\rangle = 0 $$ $f$ is orthogonal to $g$ and therefore (because of freedom in the choice of $g$) it is a linear combination of $\sin(x)$ and $\cos(x)$. The condition $f\in \mathcal M$ allows us to determine the coefficients $$ f = \frac 2 \pi (\sin(x) + \cos(x)) $$
H: Proof by Induction $n^2+n$ is even I'm not entirely sure if I'm going about proving $n^2+n$ is even for all the natural numbers correctly. $P(n): = n^2+n$ $P(1) = 1^2+1 = 2 = 0$ (mod $2$), true for $P(1)$ Inductive step for $P(n+1)$: $\begin{align}P(n+1) &=& (n+1)^2+(n+1)\\ &=&n^2+2n+1+n+1\\ &=&n^2+n+2(n+1)\end{align}$ Does this prove $n^2+n$ is even as it's divisible by $2$? Thanks! AI: I see other answers provide different (possibly simpler) proofs. To finish off your proof: by the induction hypothesis $n^2+n$ is even. Hence $n^2 + n = 2k$ for some integer $k.$ We have $$n^2+n+2(n+1) = 2k + 2(n+1) = 2(k+n+1) = 2\times\text{an integer} = \text{even}.$$ Does this prove $n^2+n$ is even as it's divisible by $2$? The key here is to remember stating & using the induction hypothesis.
H: Set of limit points of continuous functions Let $x_0$ be an accumulation point of the set $D \subset \mathbb{R}$. We say that $y$ is a limit point of a function $f:D \rightarrow \mathbb{R}$ in $x_0$ iff there exists a sequence $(x_n)$, where $x_n \in D\setminus \{x_0\}$ for $n\in \mathbb{N}$ and $x_n \rightarrow x_0$, such that $y=\lim_{n\rightarrow \infty} f(x_n)$. I don't know how to prove the following lemma: If $f: (0,c) \rightarrow \mathbb{R}$ be continuous then the set of all limits points of $f$ in $0$ is the interval $$[\liminf_{x\rightarrow 0}f(x), \limsup_{x\rightarrow 0}f(x)].$$ AI: Note: I assume you require that $x_n\to x_0$ in your definition, otherwise this lemma is false. Suppose $y$ is a limit point of $f$ in $0$, so we have a sequence $(x_n)$ in $(0,c)$ such that $x_n\to x$ and $f(x_n)\to y$. Then $$\liminf\limits_{n\to\infty} f(x_n)=y=\limsup\limits_{n\to\infty} f(x_n)$$ so we have that $$\liminf\limits_{x\to 0} f(x)\leq \liminf\limits_{n\to\infty} f(x_n)=y=\limsup\limits_{n\to\infty} f(x)\leq \limsup\limits_{x\to 0} f(x)$$ by definition. Thus $y\in [\liminf\limits_{x\to 0} f(x),\limsup\limits_{x\to 0} f(x)]$. On the other hand, suppose $y\in [\liminf\limits_{x\to 0} f(x),\limsup\limits_{x\to 0} f(x)]$. Take sequence $(x_n)$ and $(x_n')$ in $(0,c)$ converging to $0$ such that $\lim\limits_{n\to\infty} f(x_n)=\liminf\limits_{x\to 0} f(x)$ and $\lim\limits_{n\to\infty} f(x_n')=\limsup\limits_{x\to 0} f(x)$. If $y$ is either $\liminf\limits_{x\to 0} f(x)$ or $\limsup\limits_{x\to 0} f(x)$ we are done. Otherwise, we have some $N$ such that $$n\geq N\implies f(x)<y<f(x_n')$$ so by the Intermediate value theorem we have some $y_n$ between $x_n$ and $x_n'$ such that $f(y_n)=y$. Since $x_n\to 0$ and $x_n'\to 0$ it follows that $y_n\to 0$, and clearly $f(y_n)\to y$, thus $y$ is a limit point of $f$ in $0$.
H: Is there any graphical explanation of the derivative of $\sin x$? I'm trying to understand in a practical/graphical view the derivative of $\sin(x)$ (that results in $\cos(x)$). Is there any animation or illustration explaining that? AI: MIT OCW's single variable calculus course has a interactive mathlet explaining the derivates of sines and cosines (and few others) graphically. Please refer to the worked example at here.
H: Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$ $$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$$ I got no idea how to find the solution to this. Can someone put me on the right track? Thank you very much. AI: Divide both terms by two and use the fact $\sin(30) = \frac{1}{2}$ and $\cos(30) = \frac{\sqrt{3}}{2}$. Then you just need to use the formulas for $\sin(a+b)$ and $\sin(a-b)$ to find the solution.
H: How can I determine which series comparison test to use? In my textbook, there is a section of questions that's instructions reads "Test for convergence or divergence, using each one of the following tests once," and the test choices it gives me are nth-Term Test p-Series Test Integral Test Limit Comparison Test Geometric Series Test Telescoping Series Test Direct Comparison Test Now, I was able to get most of the problems in this section right on the first try, but when I compared my answers to those in the solutions manual, I found that often the way I chose was not the same as theirs. Sometimes, my solution was simpler, but most of the time, the method they chose was quicker. For one of the problems, I did the limit comparison test and I ended up with a $b_{n}$ that was one, so I was really just doing some convoluted mix between the Limit Comparison Test and the nth-Term Test. How can I develop skills to help me to decide which method to choose? Is there some "trick" to deciding which method will be the simplest? AI: The p-series and geometric series tests are for specific types of sequences, and it is clear when you can apply them. Use the integral test for positive, decreasing functions or negative, increasing functions only (do not forget this condition). Telescoping series always look like $\sum f(x+1)-f(x)$, so like the other series, they are for a particular type of series but watch out for the series $\sum \frac{1}{n(n+1)}$ and similar series that can be made into a telescoping series using partial fractions. The best thing to do is to simply practice exercises until you have mastered their usage.
H: $x$ algebraic over $K$, $v$ a polynomial in $x$ then $v$ algebraic? In the proof of proposition 5.23 Atiyah-Macdonald on page 66 use that if $x$ is algebraic over $K$ and $v = a_n x^n + \dots + a_1 x + a_0$ then $v$ is algebraic over $K$ (where $K$ is the field of fractions of $A$ and $a_i \in A$). I tried to prove it because it seems that it should be easy to prove but didn't manage. Can someone show me how to prove it please? I posted the statement and the first half of the proof here but my question is about the second part of the proof: And I have a second question about this proof: in the first half they use the word transcendental to mean transcendental over the ring $A$ whereas in part (ii) of the proof they use algebraic to mean algebraic over the field of fractions. Is the first a mistake? Usually transcendental means transcendental over a field. AI: $x$ is algebric if and only if $K[x]$ is finite dimensional over $K$. Now if $v$ is polynomial in $x$ then $K \subset K[v] \subset K[x]$, thus if $K[x]$ is finite extenioon, so is $K[v]$.
H: How to prove the convergence for function of function series How to prove the convergence for function of function series? Say, here're two examples Given $x_1>0, x_{n+1}=\ln(1+x_n)$, Prove $\lim_{n\to\infty}nx_n=2$ Given $0<x_1<1, x_{n+1}=\sin x_{n}$, Prove $\lim_{n\to\infty}\sqrt{n}x_n$ exist, and give this limit. I've written programs to check the above two problems, and it seems the assertions are true, however I found proving this kind of problems extreamly hard, since simply expanding this function of function series usually make me totally lost. Any one can give me some suggestions on such problems? Thanks a lot! AI: Since $$f(x)=\log(1+x)-x\,,\,x>0$$ is a decreasing function, we get that $\,f(x)\geq f(y)=f(0)\,\,,\,\forall 0<x<y\,$ , and it follows from here that the sequence $\,\{x_n\}=\{\log(1+x_{n-1})\}\,$ is decreasing and obviously bounded from below by say $\,\log 1 =0\,$ , so the sequence's limit exist and you can find with arithmetic of limits and using the continuity of $\,\log x\,$: $$\text{if}\,\,\alpha=\lim_{n\to\infty}x_n\,\,,\,\,\alpha=\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty}\log(1+x_n)=\log(1+\alpha)\Longrightarrow$$ $$\Longrightarrow e^\alpha=1+\alpha\Longrightarrow e^\alpha-\alpha-1=0$$ Do you recognize $\,\alpha\,$? :) Try the second one on the same lines as above (decreasing sequence and etc.)
H: Bounding Variance of a Convolution Let $G$ be a group with finite subsets $A,B \subseteq G$. Let $k$ be an integer between $1$ and $\|A\|$ (or \|A\|/2 if it helps), and let $C$ be a random subset of $A$ of size $k$, chosen uniformly out of all such sets. We take $\mu_C = \frac{\|A\|}{k}1_C$ (where $1_X$ is an indicator function for a set $X$). For $f,g : G \rightarrow \mathbb{C}$, we have the convolution $fg(x) = \sum_{y \in G}f(y)g(y^{-1}x)$. I was able to show that $\mathbb{E}\left[\mu_C 1_B\right] = 1_A * 1_B(x)$, but it's also supposed to be "easy to see" that $\operatorname{Var}(\mu_C * 1_B(x)) \leq \frac{\|A\|}{k} 1_A1_B(x)$. The bounds for $\operatorname{Var}(\mu_C 1_B(x))$ that I can come up with all seem to be off by at least an extra factor of $\frac{\|A\|}{k}$. Any help is appreciated, thanks! AI: For each $x$, the convolution $\mu_C1_B(x)$ is a sum over certain values of $\mu_C$. Since one of these values being non-zero doesn't raise the chance of other values being non-zero (in fact lowers it unless $k=\|A\|$), the correlation between different terms of this sum is non-positive, so the variance of the sum is at most the sum of the variances, which we get by pulling $1_B(x)$ out of the variance. Then we just have $\mu_C$, which is $\frac{\|A\|}k$ times an indicator function, with $$\operatorname{Var}\mu_C=\langle \mu_C^2\rangle-\langle \mu_C\rangle^2=\left(\frac{\|A\|}k\right)^2\left(\langle1_C\rangle-\langle 1_C\rangle^2\right)\le\left(\frac{\|A\|}k\right)^2\langle1_C\rangle=\frac{\|A\|}k\mu_C\;,$$ and then the convolution with $1_B$ yields $\frac{\|A\|}k1_A*1_B$.
H: Why do we say the harmonic series is divergent? If we have $\Sigma\frac{1}{n}$, why do we say it is divergent? Yes, it is constantly increasing, but after a certain point, $n$ will be so large that we will be certain of millions of digits. If we continue to let $n$ increase, we will end up with a number so large in the denominator that there will be an infinite amount of digits locked into place. What we would have left would just be an irrational number, correct? $\pi$ is an irrational number, but we still consider the value of that to be know. The common estimation of $\pi$ is 3.141592, and we can calculate it past 1,000,000 decimal places, so why can we just assume that we know the first few million places of the harmonic series, slap an irrational label on it, and call it a day? After all, the series $\Sigma\frac{1}{n^n}$ is convergent, and it basically does the same thing, it just gets there a lot faster. I feel like argument has probably been made before, so I feel like there's probably a proof somewhere proving me wrong, if someone could point me to that. AI: The sum $\frac{1}{3}+\frac{1}{4}$ is $\gt \frac{1}{2}$. The sum $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}$ is $\gt \frac{1}{2}$. This is easy, we have $4$ terms, each $\ge \frac{1}{8}$, and all but one $\gt \frac{1}{8}$. The sum $\frac{1}{9}+\frac{1}{10}+\cdots+\frac{1}{16}$ is $\gt \frac{1}{2}$. We have $8$ terms, each $\ge \frac{1}{16}$, and all but one $\gt \frac{1}{16}$. The sum $\frac{1}{17}+\frac{1}{18}+\cdots +\frac{1}{32}$ is $\gt \frac{1}{2}$. The sum of the terms $\frac{1}{33}$ to $\frac{1}{64}$ is $\gt \frac{1}{2}$. The sum of the terms $\frac{1}{65}$ to $\frac{1}{128}$ is $\gt \frac{1}{2}$. And so on. Counting in the first two terms, if we add until the term $\frac{1}{4}$ our sum is $\gt 2$. If we add up to $\frac{1}{8}$, our sum is $\gt 2.5$. Adding to $\frac{1}{16}$ puts us beyond $3$. At $32$ terms, we are beyond $3.5$. At $64$ terms, we are beyond $4$. At $256$, we are beyond $5$. At $1024$, we are beyond $6$. At $4096$, we are beyond $7$. Painfully slow! But if we are patient enough, after a while (a long long while) we will be beyond $10$, beyond $100$, beyond $1000$. But the universe may end first. Remarks: On a calculator, the sum is finite! Because of roundoff, after a while we are just adding $0$. The answer dealt with the series $\sum \frac{1}{n}$. It turns out that for any positive $\epsilon$, the series $\sum \frac{1}{n^{1+\epsilon}}$ converges. We can take for example $\epsilon=0.0001$. So one can say that $\sum \frac{1}{n}$ diverges extremely reluctantly, and that close neighbours converge.
H: Differential equation $d^n/dx^n f(x)=\pm k^2f(x)$ How to solve this differential equation: $$\frac{d^nf(x)}{dx^n}=\pm k^2f(x)$$ For $n=1,2,3$ and $\forall n\in\mathbb{N}$, and both signs, if this is possible. I encounter these often in physics, with solutions but no derivations. I would like to know how they are solved. AI: This equation is linear in $f(x)$, so if you just find $n$ linearly independent solutions by wild-guessing it should be enough to prove that this is the general solution of this equation. For instance, since you know easily that for $n = 1$ you need to substitute $f(x) = e^{rx}$, a wild guess would be the same guess, which leads to $$ r^n e^{rx} = \pm k^2 e^{rx} \quad \Longrightarrow \quad r^n = \pm k^2 \quad \Longrightarrow \quad r = \sqrt[n]{\pm k^2} e^{\frac{2\pi i j}n}, \quad 0 \le j \le n-1. $$ Since all those values of $r$ are distinct, you have $n$ solutions (for a fixed $n^{\text{th}}$ root of $+k^2$ or $-k^2$). Tadam! Hope that helps,
H: Swatting flies with a sledgehammer Prompted by a recent exchange with Gerry Myerson, I was wondering if anyone has a favorite example of a relatively simple problem with a rather elementary (though perhaps complicated) answer for which there's another answer that relies on an elegant use of a powerful result that's almost certainly beyond the background of the poser of the question. AI: If $2^{1/n}$ were a rational $a/b$, with $n>2$, then $a^n=b^n+b^n$, which would contradict Wiles proof of Fermats Last Theorem.
H: Determine if it is possible to fit 2 circles in a rectangle I have the following problem: Given a Rectangle with $L$ length and $W$ width and $2$ circles with $r_1$ and $r_2$ radius, determine if it's possible to fit these two circles inside the rectangle. I realized that: If $2r_1 > L$ or $2r_1 > W$ or $2r_2 > L$ or $2r_2 > W,$ then it is not possible to fit the circles in the rectangle. Thus if $2 r_1 + 2r_2 \leq L$ and $2r_1 + 2r_2 \leq W,$ they can fit vertically or horizontally. My doubt is to check if they can fit diagonally? How do I get it ? AI: Here is my thought process here: It's true that the best way to proceed will always be to stick one circle 'in one corner' of the rectangle and to try to fit the other circle 'in the opposite corner.' I take this for granted. It is very easy to understand where the center of a circle will go when the circle is 'tucked in a corner,' as both sides of the rectangle will be tangent to the circles. For example, if the rectangle is at coordinates $(0,0), (0,h), (l,0),$ and $(l,h)$, and a circle of radius $r_1$ is in the bottom-left corner, then the coordinates of its center will be $(r_1, r_1)$. If another circle, this one of radius $r_2$, is tucked in the top-right corner, the coordinates of it's center will be at $(l-r_2, h-r_2)$. So one has to check two things: one has to make sure that both circles fit in the rectangle itself (without worrying about overlap), and then one has to make sure that the distance between the centers is greater than the sum of the radii. So $2r_1, 2r_2 \leq l,h$ and $d[(r_1,r_1), (l-r_2,h-r_2)] \geq r_1 + r_2$ should be necessary and sufficient, where I let $d(\cdot, \cdot)$ denote the distance function.
H: Use the Division Algorithm to show the square of any integer is in the form $3k$ or $3k+1$ Use Division Algorithm to show the square of any int is in the form 3k or 3k+1 What confuses me about this is that I think I am able to show that the square of any integer is in the form $Xk$ where $x$ is any integer. For Example: $$x = 3q + 0 \\ x = 3q + 1 \\ x = 3q + 2$$ I show $3k$ first $$(3q)^2 = 3(3q^2)$$ where $k=3q^2$ is this valid use of the division algorithm? If it is then can I also say that int is in the form for example 10k for example $(3q)^2 = 10*(\frac{9}{10}q^2)$ where $k=(\frac{9}{10}q^2)$ Why isn't this valid? Am I using the div algorithm incorrectly to show that any integer is the form 3k and 3k+1, if so how do I use it? Keep in mind I am teaching myself Number Theory and the only help I can get is from you guys in stackexchange. AI: By the division algorithm, $$x = 3q + r,\text{ where } r \in \{0, 1, 2\}.$$ So express $$x^2 = 9q^2 + r^2 + 6qr = 3(3q^2 + 2qr) + r^2.$$ For a given $x$ if $r = 0$ or $1,$ then we're done. If $r = 2$ then $r^2 = 4 = 3 + 1,$ and hence $$x^2 = 3\times\text{integer} + 3 + 1 = 3\times(\text{integer} + 1) + 1.$$ We are done.
H: For which values of $n$ is the polynomial $p(x)=1+x+x^2+\cdots+x^n$ irreducible over $\mathbb{F}_2[x]$? For which values of $n$ is the polynomial $p(x)=1+x+x^2+\cdots+x^n$ irreducible over $\mathbb{F}_2[x]$ ? E.g. $x+1$, $x^2+x+1$ are irreducibles. Subcase of this question Factor by irreducible is field. Does it help ? AI: First of all, $n+1$ must be a prime number (otherwise your polynomial is reducible even in $\mathbb{Z}[x]$, to a product of cyclotomic polynomials). If $n+1$ is prime, say $n+1=p$, then the polynomial is irreducible iff $2$ generates the multiplicative group $\mathbb{F}_p^*$, i.e. if the smallest $k$ s.t. $2^k\equiv 1$ mod $p$ is $k=p-1$. See https://math.stackexchange.com/a/167492/8268 for the reason.
H: Why do these two methods of calculating the probability of winning a best-of-7 series give the same answer? I was having a discussion with a friend about the probability, and we came up with very different methods to solve it that lead to the same answer. The problem is pretty simple: you have two teams A and B playing a best of 7 series where wins are independent and the probability that team B wins is .35. What's the probability of team B winning th series? Friend says, "this is just a binomal random variable $X$ with $n=7$ and $p=.35$ and we are looking for $P(X\geq 4)$, which according to my TI-83 here is .1998". I was rather convinced that this must be wrong. If we think of the series as a sequence of A's and B's, then we are looking for the probability of obtaining a sequence of length $n=4,5,6,7$ with 4 B's and the last element is a B. I was sure that his method will include the probability of obtaining, say, {B,A,B,B,B,A,A}, which we are not interested in and in fact couldn't even ever occur. So I figure that what we really want for a sequence of length $n$ is the probability of obtaining a sequence of length $n-1$ with exactly 3 B's, and then tacking a B onto the end. So for a sequence of length $n$, the probability should be $\binom{n-1}{3}.35^4 .65 ^{n-4}$, and then the answer should be $\sum_{n=4}^{7}\binom{n-1}{3}.35^4 .65 ^{n-4}$. I was confident that I was right and he was wrong, but then I plugged that into Wolfram Alpha and got... .1998. What's going on here? Is it a coincidence? AI: This is not a coincidence, you are looking at the same problem in two different ways. The situation {B,A,B,B,B,A,A} is irrelevant in your friend's argument, because you are not counting the number of different possible sequences of events. Rather, you are attempting to estimate the probability of at least $n$ successes in $k$ trials, which is exactly what the binomial distribution does. In your second approach, you are essentially estimating the number of possible legal sequences terminating in a victory condition, and then computing the total fraction out of all possible sequences. Both approaches are equivalent. One is a little easier to implement ;)
H: The relation between the roots of a polynomial equation and a matrix equation (I call matrix equation for example this $X^2-X+1=0$. Does it mean so?) When a polynomial equation is solved for $x$, we get a complex number. If we solve the "same" equation (same coeficients), such that the unknown is inside a matrix, is there any relation between the matrix and the roots? Maybe I have not expressed it well, so I'll put an example: $$ { x }^{ 2 }-x-1=0 $$ So $$ x=\frac { 1\pm \sqrt { 5 } }{ 2 } $$ Now the unknown $x$ is inside a matrix and $\alpha,\gamma,\zeta$ are known numbers: $$ X=\left[\begin{array}{cc}x & \gamma\\\alpha&\zeta\end{array} \right]$$ $$ X^2-X-I =0$$ If it can be solved, is there any relation between $x$ and $X$ ? AI: First, note that a "polynomial matrix equation" may have more solutions than its degree! For example, the equation $X^2-1=0$ has only two complex solutions, $x=-1$ and $x=1$, but it has many matrix solutions, among them: $$\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right),\quad \left(\begin{array}{rr}-1&0\\0&-1\end{array}\right),\quad \left(\begin{array}{rr}1&0\\0&-1\end{array}\right),\quad \left(\begin{array}{rc}-1 & 0\\0&1 \end{array}\right),$$ and more. In fact, there are infinitely many solutions: for any $a,b,c,d$ such that $ad-bc\neq 0$, we have that $$\left(\begin{array}{rr} \frac{ad}{ad-bc} & \frac{bd}{ad-bc}\\ -\frac{ac}{ad-bc} & -\frac{bc}{ad-bc} \end{array}\right)$$ is a solution. And there are plenty more! That said, if $\alpha$ is a complex solution to $p(x)=0$, then $\alpha I$ is a matrix solution to $p(X)=0$. Moreover, if $M$ is a matrix solution to $p(X)=0$, then the minimal polynomial of $M$ is a divisor of $p(x)$, and this means that every eigenvalue of $M$ must be a root of $p(x)$. Thus, the roots of $p(x)=0$ in $\mathbb{C}$ give you the possible eigenvalues of any solution to the matrix polynomial equation $p(X)=0$. In particular, any diagonal matrix whose diagonal entries are chosen from among the roots of $p(x)$ will satisfy the equation. Also, if $M$ is any solution and $Q$ is any invertible matrix, then $QMQ^{-1}$ is also a solution (this is the source of my "infinite family of examples" above; but note that none of the examples in the first displayed equation line are related in this way). Conversely, a diagonalizable matrix is a solution to $p(X)=0$ if and only if the eigenvalues of $X$ are roots of $p(x)=0$. Added. None of which directly addresses your question, on second reading, since you are asking about the unknown being in the matrix, rather than being the matrix. But you can do a bit with the information above. For example, consider the situation you describe. We know that if $X$ is a solution to $p(X)=0$, then the eigenvalues of $X$ must be either $\frac{1+\sqrt{5}}{2}$ or $\frac{1-\sqrt{5}}{2}$. That means that the characteristic polynomial of $X$, which is $$(t-x)(t-\zeta)-\alpha\gamma$$ must either be equal to $t^2+t+1$, to $(t-\phi)^2$, or to $(t-\psi)^2$, where $\phi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$. This allows us to solve for $x$: we must have either: $$\begin{align} x\zeta - \alpha\gamma &= 1\\ x+\zeta &= -1 \end{align}$$ (if the characteristic polynomial is $t^2+t+1$); or $$\begin{align} x\zeta - \alpha\gamma &= \frac{3+\sqrt{5}}{2}\\ x+\zeta &= -1-\sqrt{5} \end{align}$$ (if the characteristic polynomial is $(t-\phi)^2$); or $$\begin{align} x\zeta - \alpha\gamma &= \frac{3-\sqrt{5}}{2}\\ x+\zeta &= -1+\sqrt{5} \end{align}$$ (if the characteristic polynomial is $(t-\psi)^2$). This gives you three possible values of $x$ (if the system is consistent). If the first system is consistent, then the Cayley-Hamilton theorem guarantees that the resulting matrix is a solution to the original polynomial. The other systems are more difficult: it turns out that the only way they can be solutions to the original $p(X)=0$ is if $\alpha=\gamma=0$ and $\zeta=\phi$ or $\psi$, respectively; this follows from the theory of minimal polynomials: we would necessarily have that the minimal polynomial of the matrix is either $t-\phi$ or $t-\psi$, but this would force $X$ to be a scalar multiple of the identity. I think this kind of question is better approached from the point of view of the minimal polynomial; a bit of theory can simplify the search (as in the previous paragraph) or even determine that no search is necessary (because no solution is possible). I didn't make use the trace, for example, which also gives you information about the eigenvalues. (I guess the very short answer is "yes, there is a relation, through the eigenvalues and minimal polynomial")
H: How to I show that $\lim_{x\to0} \frac{1}{x^2}\left(\frac{\sinh x}{x}-1\right) = \frac{1}{6}$ I can do this limit with a symbolic calculator and get the result. $$\lim_{x\to0} \left[ \frac{1}{x^2}\left(\frac{\sinh x}{x} - 1\right) \right] = \frac{1}{6}$$ But how would I do it by hand, and show why it is so. I know that $\lim_{x=0}\frac{\sinh(x)}{x}=1$ but that does not help here. This is not homework, and it is related to the deflection of axially loaded beams. AI: HINT: $$\sinh (x) = \dfrac{e^x - e^{-x}}2 = x + \dfrac{x^3}6 + \mathcal{O}(x^5)$$
H: Finding more details about a triangle using the given details. In the triangle $ABC$ we have $\tan{\frac{A}{2}}=\frac{1}{3}$ $b+c=3a$ Specify which of the following answers is correct: $a) m(\angle B)=\frac{\pi}{2}$ or $m(\angle C)=\frac{\pi}{2}$ $b) m(\angle A)=m(\angle B)$ $c) m(\angle A)=\frac{\pi}{2}$ $d) m(\angle B)=\frac{\pi}{4}$ or $m(\angle C)=\frac{\pi}{4}$ $e) m(\angle A)=m(\angle C)$ $f) m(\angle A)=\frac{\pi}{3}$ I'm lost here. I don't know how I can use $\tan{\frac{A}{2}}=\frac{1}{3}$ so I get to an answer. Can someone give me a solution? I have many exercises involving the $\tan{\frac{A}{2}}$ function and I can't continue. Thank you very much! AI: I will assume that in this multiple choice question only one of the answers can be correct. If we do not assume that, there is additional work to do. It is handy but not necessary to know the formula $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}.$$ Putting $x=y=\frac{A}{2}$, after a while we get $\tan A =\frac{3}{4}$. This is what we get with the good old $3$-$4$-$5$ right triangle with sides $3k$, $4k$, $5k$. And by a miracle we have in that case that $4k+5k=3(3k)$. So the second condition $b+c=3a$ is met. So a right angle at $B$ or $C$ will do the job. If there is a unique answer, that answer is a). Remark: We could get there with a calculator. Use it to find (approximately) $A/2$, and then $A$. Not very informative. But then we may get the lucky idea of computing $\tan A$, (or $\sin A$, or $\cos A$.) The calculator gives $0.75$, (or $0.6$, or $0.8$) and then we recognize the familiar triangle. If we are not allowed to assume a unique answer to the multiple choice question, we need to rule out the other possibilities. Some are very quick to rule out. The possibilities b), d), and e) take a little longer.
H: Unbounded element in $R^\infty$ Let $R^\infty$ be the vector space of all sequence $\{a_j\}$ of real numbers. Put $\\|\{a_j\}\\|_n:= \sum_{j=0}^n \|a_j\|$. This collection of semi norms make this as Frechet space. A set $B$ is bounded if every continuous seminorm is bounded on $B$. That is bounded set will be as $\{a=\{a_i\}: \sum_{i=0}^k \|a_i\|<M_k\text{ for some } M_k>0 \text{ and } \forall k\in \mathbb N\}.$ Can I have an explicit example of one bounded set and one unbounded set. AI: We can check that a subset $B$ of $\Bbb R^{\infty}$ is bounded if and only if all the subsets of $\Bbb R$ defined by $$B_k:=\{x_k,(x_n)_{n\in\Bbb N}\in B\}$$ are bounded. For example, the set $\{(x_n)_{n\geq 1}\},\|x_n\|\leq n\}$ is bounded.
H: Compute Hilbert function of a monomial ideal I'd like to know whether there exist easy methods that compute the Hilbert function of a graded $k$-algebra, without computer programs. My homework asks to me to compute the Hilbert function of $R/I$, where $R=k[x_0, \dots, x_5]$ and $$ I = (x_0 x_3, x_0 x_4, x_0 x_5, x_1 x_4, x_1 x_5, x_3 x_5) $$ For me, it is quite difficult to decide what monomials are not in $I$, therefore I thought of compute a free graded resolution of $R/I$, but for me it is hard to calculate syzygies, because there are too many indeterminates. Some suggestions? Thanks to all! AI: Easy computations, by using the algorithm from the book of Eisenbud, show that the Hilbert series of $R/I$ is $H_{R/I}(t)=\frac{1+3t}{(1-t)^3}$. In order to compute the Hilbert function one uses that $$\frac{1}{(1-t)^3}=\sum_{n\ge 0}{{n+2}\choose{2}}t^n.$$ Of course, any package dedicated to Commutative Algebra can solve your problem immediately.
H: What kind of book would show where the inspiration for the Laplace transform came from? I'm trying to find out where to learn about integral transforms and inversions like the Laplace transform and the Bromwich integral. I'm looking for a book that describes how you can find (derive) that the inverse of the Laplace transform is the Bromwich integral and where people got the idea that integral transforms could do anything in the first place. The books I have just present the Laplace transform like it was handed down from heaven completely out of the blue. I want to know where it came from, how it was derived, and why it has its properties. I also want to know the same things about Fourier transforms and integral transforms in general. I've looked up some types of books on the internet and I am wondering which would more likely give me what I need. I looked up operator theory books, spectral theory books, and operational calculus books but haven't bought anything yet because I don't know exactly what to study. Can anyone tell me if I'm on the right track, which type of book would tell me what I want to know, or suggest a good book? AI: As far as I know an early reference for a thorough mathematical theory (in terms of today's mathematical language) of the Laplace transform and its inversion are the books by Gustav Doetsch. (There are several: A three volume handbook, some books on applications, a practical guide...). Probably one of this books also provide insight about the history (i.e. early references, Heaviside formal treatment,...) but I don't know which of them is available in English. Moreover, there are two articles called "The development of the Laplace transform" by Deakin here and here that could be helpful. You probably already found the Monthly article "What is the Laplace transform?"?
H: Binomial/Geometric Distribution explanation I've found the following exercise in my Stats coursework. I only have solutions to it, but no explanation. And I would really like to know how to get to the answer. An urn holds 5 white and 3 black marbles If two marbles are drawn at random without replacement and X denotes the number of white marbles find the probability distribution of X. Could someone please explain this to me? Thanks! AI: The possible outcomes are $X=0,X=1$, and $X=2$. In order to get no white marbles ($X=0$), you must draw two black marbles. There are $\binom82$ pairs of marbles altogether, and there are $\binom32$ pairs of black marbles, so the probability of drawing a black pair is $$\frac{\binom32}{\binom82}=\frac3{28}\;.$$ That is, $\Bbb P(X=0)=\frac3{28}$. There are $\binom52$ pairs of white marbles, so the probability of drawing one of those pairs is $$\frac{\binom52}{\binom82}=\frac{10}{28}=\frac5{14}\;.$$ That is, $\Bbb P(X=2)=\frac5{14}$. The total probability of all three possible outcomes must be $1$, so we must have $$\Bbb P(X=1)=1-\frac3{28}-\frac5{14}=1-\frac{13}{28}=\frac{15}{28}\;,$$ and we have our probability distribution: $$\begin{align} \Bbb P(X=0)&=\frac3{28}\\ \Bbb P(X=1)&=\frac{15}{28}\\ \Bbb P(X=2)&=\frac5{14}\;. \end{align}$$ We can double-check this by calculating $\Bbb P(X=1)$ directly. Each of the $5$ white balls can be paired with any of the $3$ black balls to make a pair with one white ball, so there are $15$ such pairs, and as we’ve seen already, there are $\binom82=28$ pairs altogether, so the probability of drawing a pair with one ball of each color is indeed $\frac{15}{28}$.
H: Question on Cauchy Criterion of Series We know that the Cauchy Criterion of a series is as follow (proof is taken as excerpt from an analysis book): Theorem: A series $\sum_{j=1}^{\infty}a_j$ converges iff for all $\epsilon>0$ there is an $N\in \mathbb{N}$ so that for all $n\ge m \ge N$ we have $\|\sum_{j=m}^{n} a_j\|< \epsilon$. Let $s_n$ denotes partial sum $\sum_{j=1}^n a_j$. We know that the proof of $\Rightarrow$ direction makes use of the fact that convergent series implies the sequence of partial sum converges and thus is Cauchy in $\mathbb{R}$. So here is part of the proof, since sequence of partial sum is Cauchy, therefore given $\epsilon>0$, there exists $N \in \mathbb{N}$ such that for all $n \ge m \ge N$, $\|s_n - s_{m-1}\|< \epsilon$. I do not get the $s_{m-1}$ part, i.e. I do not get why you can pick $m-1$, since there is possibility for $m=N$, and thus having $m-1<N$, which in this case you cannot guarantee the sequence of partial sum Cauchy. When I was thinking about this, I was rather confused, and my original thought was following: Starting from the sequence of partial sum being Cauchy, $\|s_n-s_m\|< \epsilon$, for all $n,m\ge N$ (the definition of Cauchy sequence), and without loss of generality we can pick $n>m$ such that $\|s_n- s_m\|=\|\sum_{j=m+1}^{n} a_j\|< \epsilon$, but this differs from the statement of the theorem somehow. I was also thinking of having $m \ne N$ and $m>N$ so that we have $m-1\ge N$ and that the original proof of the theorem that I showed right after the theorem is valid. AI: It makes no difference at all if you use $n\geq m > N$ in the definition of Cauchy sequnces, so I guess your 2 option is a good solution to your problems. It is just a problem of what name you give to the numbers. If the sequence of partial sums is a Cauchy sequences there exists such $N$ as you defined. Now call $N' = N+1$ and conclude what you want.
H: Deriving the formula for the radius of the circle inscribed in an equilateral triangle I am trying to derive the formula for the radius of the circle inscribed in an equilateral triangle from scratch. Given $2*n$ = length of a side $H$ = the altitude of the triangle = $h + a$ $h$ = the long subdivision (from the center of the triangle to a vertex) $a$ = the short subdivision (from the center of the triangle to a side. Also the radius of the inscribed circle) By first deriving the altitude of the triangle $\displaystyle \begin{align} 2 n&=\sqrt{H^2+n^2} \\ H&=\sqrt{(2 n)^2-n^2} \\ &=\sqrt{3}\;n \\ \end{align}$ I have gotten to the reduced equation $n \sqrt(3) - a = \sqrt(a^2+n^2)$ $\displaystyle \begin{align} a+h=\sqrt{3}\;n \\ h=\sqrt{3}\;n-a \\\\ a^2+n^2=h^2 \\ h=\sqrt{a^2+n^2} \\ \end{align}$ Trying to solve for $a$, I know in advance that $a$ is $1/3$ and $h$ is $2/3$ of $H$, with $a = n\sqrt(3)/3$ This is of course the answer I wish to derive. In fact, plugging the equation given above into a system such as Mathematica will provide the correct answer. But I can't find out what the steps are, primarily because I know of now way to extract the $a$ term from within the square root term. Please, no trigonometry. I know there is a fast derivation involving tangents, etc, but this is more properly an algebra problem - how to solve the equation for $a$. AI: Square both sides, and you end up with a quadratic equation in $a$.
H: Geometric explanation of centroid of triangle why is the point where the medians of a triangle meet also the center of mass of the triangle. AI: I think I have it right this time; however the construction is not as simple as I had hoped. The idea is (was) to avoid calculus (explicitly, at least). The basic idea is to use scaling and linearity to determine the center of mass (CM). If an object of mass $m$ with CM at $d$ is partitioned into smaller objects of mass $m_i$ (with $m = \sum m_i$, of course) each with CM at $d_i$, then we must have: $$d m = \sum d_i m_i.$$ Excuse the crudeness of my drawings below, I hope you get the idea. We want to find the CM of the big triangle and show that it is at the intersection of the medians. We want to get the CM of the large triangle. Suppose, for simplicity, that one vertex is at the origin, and that the CM is at position $d \in \mathbb{R}^2$. Now take a similar triangle whose sides are $\frac{1}{3}$ of the big triangle. It should be clear that the CM of the little triangle is at position $\frac{d}{3}$ with respect to the corresponding vertex. Let $m$ be the 'mass' of the big triangle, then the mass of each of the smaller triangles is $\frac{m}{9}$. Trisect the edges of the big triangle and connect the points to get $9$ similar small triangles. Let $a,b \in \mathbb{R}^2$ be the other two vertices of the big triangle. By looking careful at the smaller similar triangles, it should be clear that the red dot is at the intersection of the medians, which is at position $\frac{a+b}{3}$. Now consider the $6$ triangles that touch the red dot. By symmetry, the CM of the $6$ triangles is at the red dot, and the 'mass' of the $6$ triangles taken as a unit is $\frac{6}{9}m$. By linearity, we can compute the CM of the combined $6$ units and the three blue triangles and this must equal $d$, the CM of the big triangle. Doing the computation gives: $$\frac{1}{m}\left[ \frac{6m}{9} \frac{a+b}{3} + \frac{m}{9}(0+\frac{d}{3}) + \frac{m}{9}(\frac{2a}{3}+\frac{d}{3}) + \frac{m}{9}(\frac{2b}{3}+\frac{d}{3}) \right] = d.$$ Simplifying gives: $$\frac{6(a+b)}{3}+\frac{2a+2b}{3}+d = 9d,$$ from which we get the formula $d = \frac{a+b}{3}$, which is the intersection of the medians. Note: There is nothing special about splitting the triangle up into $9$ parts other than with this splitting, the CM and the median intersections fall nicely on the vertex of the smaller triangles.
H: equivalence of continuity Could you give me a hint on this problem? Show that $f:A\subset\mathbb{R}^n\longrightarrow \mathbb{R}^m$ is continuous if and only if for every subset $B$ of $A$ , $f(A\cap\overline{B}) \subset \overline{f(B)}$. Currently I know these definitions of continuity: $(\rm i)$ In terms of pre-image of open sets. $(\rm ii)$ In terms of $\epsilon$-$\delta$ $(\rm iii)$ In terms of convergent sequences. By the statement of the problem I could guess the definition to use in this case is the first one. Maybe you could tell me how to start so I can give it a try, thanks in advance. AI: HINTS: ($\Rightarrow$) Let $x\in A\cap\operatorname{cl}B$. There is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $B$ that converges to $x$. What do you know about $\langle f(x_n):n\in\Bbb N\rangle$? ($\Leftarrow$) Prove the contrapositive: suppose that $f$ is not continuous, and show that there is some $B\subseteq A$ such that $f[A\cap\operatorname{cl}B]\nsubseteq\operatorname{cl}f[B]$. If $f$ is not continuous, there is at least one point $x\in A$ such that $f$ is not continuous at $x$. Can you show that there are an $\epsilon>0$ and a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ that converges to $x$ such that $\\|f(x_n)-f(x)\\|\ge\epsilon$ for every $n\in\Bbb N$? What can you then say about $B=\{x_n:n\in\Bbb N\}$?
H: Books like Grundlagen der Analysis in French I am looking for some recommendations for a mathematics (text)book written in French. I am hoping to learn to read and write mathematics in French since I expect to take some mathematics courses that will be taught in French next year. Basically, I would like a book whose linguistic and mathematical level approaches that of Landau's book Grundlagen der Analysis- the language should be clear and simple (read: not unnecessarily fancy or ornamental) and the mathematics contained in the book should be comprehensible to someone who has taken a course or two in analysis and algebra. It would be a huge plus if you can recommend a book that also contains a French-English vocabulary at the back like in Landau's book but the recommendation should primarily be based on the clarity of the text and the level of mathematical depth. Please note that the book you recommend need not be a textbook- it could be lecture notes that are particularly well-written or monographs exploring some topic or even a popular account of mathematics that is relatively easy to understand. Also, the level of the book need not be at the university level. Depending on the book, it might even be preferable if it was at the high school level. So, any suggestions? AI: Here a collection of French mathematics books choose the ones you want they are among the most famous textbooks and problems solving books. and i hope that will help you. Cours (Text-books) Cours de mathematiques Tome 1 - Algèbre, J.M.Arnaudiès, P.Delezoide, H.Fraysse Cours de Mathematiques Tome 2 - Analyse, J.M.Arnaudiès, P.Delezoide, H.Fraysse Cours de Mathematiques Tome 3 - Complements d'analyse, J.M.Arnaudiès, P.Delezoide, H.Fraysse Cours de mathematiques Tome 4 - Algebre bilineaire et geometrie, J.M.Arnaudiès, P.Delezoide Cours de mathématiques spéciales Tome 1 - Algèbre, E.Ramis, C.Deschamps, J.Odoux Cours de mathématiques spéciales Tome 2 - Algèbre et applications à la géométrie, E.Ramis, C.Deschamps, J.Odoux Cours élémentaire de mathématiques supérieures Tome 1 - Algèbre, J.Quinet Cours élémentaire de mathématiques supérieures Tome 2 - Fonctions usuelles, J.Quinet Cours élémentaire de mathématiques supérieures Tome 3 - Calcul intégral et séries, J.Quinet Cours élémentaire de mathématiques supérieures Tome 4 - Equations différentielles, J.Quinet Cours élémentaire de mathématiques supérieures Tome 5 - Géométrie, J.Quinet Eléments d'analyse Tome 1 - Fondements de l'analyse moderne, Jean Dieudonné Eléments d'analyse Tome 2 - Analyse fonctionnelle linéaire, Jean Dieudonné Eléments d'analyse Tome 3 - Analyse sur les variétés, Jean Dieudonné Eléments d'analyse Tome 4 ,Jean Dieudonné Eléments d'analyse Tome 5 - Groupes de Lie compacts et semi-simples, Jean Dieudonné Eléments d'analyse Tome 6 - Analyse harmonique, Jean Dieudonné Eléments d'analyse Tome 7 - Equations fonctionnelles linéaires, 1ère partie - Opérateues pseudo-différentiels, Jean Dieudonné Eléments d'analyse Tome 8 - Equations fonctionnelles linéaires, 2ème partie - Problèmes aux limites, Jean Dieudonné Eléments d'analyse Tome 9 - Topologie algébrique, topologie différentielle élémentaire, Jean Dieudonné Mathématiques 3 - Analyse cours et Exercices, E.Azoulay, J.Avignant Mathématiques 4 - Algèbre, E.Azoulay, J.Avignant Mathématiques L3 - Algèbre cours complet avec 400 tests et exercices corrigés, A.Szpirglas Mathématiques L3 - Analyse Cours complet avec 600 tests et exercices corrigés, Jean-Pierre Marco Roger Godement - Analyse Mathématique 1, Convergence, fonctions élémentaires Roger Godement - Analyse Mathématique 2, Calcul différentiel et intégral,séries de Fourier,fonctions holomorphes Roger Godement - Analyse Mathématique 3, Fonctions analytiques, différentielles et variétés, surfaces de Riemann Roger Godement - Analyse Mathématique 4, Intégration et théorie spectrale, analyse harmonique, le jardin des délices modulaires Mathematique 2eme annee Cours et Exercices Corriges, C.Deschamps, A.Warusfel Mathématiques Algebre et geometrie 50% et 50% exercices, G.Auliac, J.Delcourt, R.Goblot Mathématiques générales 1 ère année cours et exercices corrigés, J.Vélu Mathématiques méthodes et exercices MP, J.M Monier Mathématiques TOUT-EN-UN 1ère année cours et exercices corigés MPSI, PCSI, Série E.Ramis, C.Deschamps, A.Warusfel TD Analyse, 70% Applications 30% Cours, Jean-Pierre Lecoutre, Philippe Pilibossian Topologie et analyse fonctionnelle cours de licence et 240 exercices et 30 problèmes corrigés, Y.Sonntag Calcul diffrentiel et calcul intégral 3eme année Cours et exercices avec solutions, Marc Chaperon Complement d'analyse, K.Arbenz, A.Wohlhauser Cours d'Algébre avec énoncés 40 Exercices et 300 Problèmes, J.Querré Cours de Mathematiques MP-MP', Jean Voedts Problems solving (Exercices resolus) Algèbre Exercices avec solutions, E. Ramis, C. Deschamps, J. Odoux Analyse Tome 1 - Exercices avec solutions, E. Ramis, C. Deschamps, J. Odoux Analyse Tome 2 - Exercices avec solutions, E. Ramis, C. Deschamps, J. Odoux Exercices resolus Tome 1 - Algebre du cours de mathematiques Tome 1, J.M.Arnaudiès Exercices resolus Tome 2 - Analysee du cours de mathematiques, J.M.Arnaudiès, P.Delezoide Exercices resolus Tome 3 - Complements d'analyse du cours de mathematiques, J.M.Arnaudiès Exercices resolus Tome 4 - Algebre bilineaire et geometrie du cours de mathematiques, J.M.Arnaudiès, P.Delezoide, H.Fraysse Exercice d'algebre 1, B.Calvo, J.Doyen, A.Calvo, F.Boschet Exercices d'analyse 1, B.Calvo, J.Doyen, A.Calvo, F.Boschet Problèmes d'Analyse Tome 1 - Nombres réels suites et séries, Exercices corrigés,Wieslawa J.Kaczor, Maria T.Nowak Problèmes d'Analyse Tome 2 - Continuité et dérivabilité, Exercices corrigés,Wieslawa J.Kaczor, Maria T.Nowak Problèmes d'Analyse Tome 3 - Intégration, Exercices corrigés,Wieslawa J.Kaczor, Maria T.Nowak 275 Exercices et Problèmes Résolus de Mathématiques Supérieures, A.Abouhazim, A.Abkari, S.R.Nsiri, M.El Mountassir
H: Can someone resolve my confusion about uniqueness of diagonalization? I am a bit confused about diagonalization. I have $A$ which I know is diagonalizable. I want to find $P$ such that $A = P \Sigma P^{-1}$ where $\Sigma$ is diagonal. Under what circumstances is $P$ unique, if ever? If it is not unique, is it at least unique up to some operation? AI: For any diagonalizible matrix $n\times n$, for $n\geq 2$, $P$ is not unique. If the eigenvalues of $A$ are not all equal, then $\Sigma$ is not unique as well. If you change the columns of $P$ that will correspond to changing the appropriate columns in $\Sigma$. Even if, suppose, you fix $\Sigma$, even then you can change the columns in $P$ that correspond to eigenvectors of the same eigenvalue, or you can multiply them by scalar or make linear combinations. In short, in this case (when you fix $\Sigma$), $P$ will be unique up to elementary matrix column operations, but only between columns that correspond to the same eigenvalue.
H: My book states that $\sum_{n=1}^{\infty}r^{-n} = \frac{1}{r-1}$ for $r > 1$ On page 9 of Edwards' Riemann's Zeta Function, he uses the equality $\sum_{n=1}^{\infty}r^{-n} = \frac{1}{r-1}$ for $r > 1$ to prove an identity connecting the gamma function and the Riemann zeta function. But how can this equality be right? Taking the inverse of $r$ we obtain: $$\sum_{n=1}^{\infty}(\frac{1}{r})^n = \frac{1}{1-\frac{1}{r}} = \frac{r}{r-1} \neq \frac{1}{r-1}$$ However, I double checked the identity (third down) on wikipedia and in the book it's correct as stated, thus from my, apparently erroneous, perspective Edwards derives the correct formula from the incorrect application of the geometric series summation, thus I must be missing something, where have I gone wrong? AI: If $r > 1$, you have $$\sum_{n=1}^\infty r^{-n} = {1/r\over 1 - 1/r} = {1\over r - 1}.$$ However, $$\sum_{n=0}^\infty r^{-n} = {1\over 1 - 1/r} = {r\over r - 1}.$$ Remember for a geometric series $\sum_n g_n$ that decays to zero, you have $$\sum_{n} g_n = {\hbox{first term}\over 1 - \hbox{common ratio}}$$
H: What is the order of discontinuity of this function? Consider the function f(x) such that f(x) = 0 for all rational x and f(x) = 1 for all irrational x. It would seem that the number of 'jumps' up is uncountably infinite and the number of 'jumps' down is countably infinite; or is the other way around? Shouldn't the number of jumps up have the same order as the number of jumps down? Would the total number of jumps up or down have the same cardinality as the set of all real numbers? Is there a formal 'measure' for the order of discontinuity of a function? Thanks in advance for any guidance you can provide. AI: Normally we say that a function $f$ has a jump at a point $a$ if $\lim\limits_{x\to a^-}f(x)$ and $\lim\limits_{x\to a^+}f(x)$ exist and are not equal. In the case of that function, neither limit exists at any point, so there are no jumps in this sense.
H: What makes elementary row operations "special"? This is probably a stupid question, but what makes the three magical elementary row operations, as taught in elementary linear algebra courses, special? In other words, in what way are they "natural" (as opposed to "arbitrary")? It seems that they're always presented in a somewhat haphazard manner ("these are the three legendary elementary row operations, don't ask why, they just are"). From what I understand, they satisfy some nice properties, such as the inverse of each being an operation of the same type, etc. But is there something that characterizes them, i.e. is there some definition of what constitutes "elementary" that's only satisfied by the three types of elementary matrices, and no other matrix? AI: They're not special. They're just convenient. It's relatively easy to tell what happens to a matrix when you apply an elementary row operation to it, and this isn't quite as true for more complicated types of operations. In the language of group theory, elementary matrices form a set of generators for the group of invertible square matrices. You could choose a different set of generators if you wanted to, but again, the elementary matrices are convenient.
H: For what value of h the set is linearly dependent? For what value of $h$ set $(\vec v_1 \ \vec v_2 \ \vec v_3)$ is linearly dependent? $$\vec v_1=\left[ \begin{array}{c} 1 \\ -3 \\ 2 \end{array} \right];\ \vec v_2=\left[ \begin{array}{c} -3 \\ 9 \\ -6 \end{array} \right] ;\ \vec v_3=\left[ \begin{array}{c} 5 \\ -7 \\ h \end{array} \right]$$ Attempt: After row reducing the augmented matrix of $A\vec x=\vec 0$ where $A=(\vec v_1 \ \vec v_2 \ \vec v_3)$: $$\begin{bmatrix} 1 & -3 & 5 & 0 \\ -3 & 9 & -7 & 0 \\ 2 & -6 & h & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -3 & 5 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & h-10 & 0 \end{bmatrix} $$ I am not sure whether the set is linearly dependent when $h=10$ or for any $h$. Help please. AI: That reduced matrix shows you that the set of vectors is linearly dependent for every value of $h$. If $h\ne 10$, the system has no solution, and if $h=10$, it has infinitely many, so there is no value of $h$ that gives it exactly one solution. Indeed, you can see this directly from the vectors themselves: $v_2=-3v_1$.
H: Sufficient conditions for a function $g: A \subset \mathbb{R}^n \to \mathbb{R}^n$ to be locally bounded Let $g: A \subset \mathbb{R}^n \to \mathbb{R}^n$ be an injective continuously differentiable function such that $\forall x \in A, \det g'(x) \neq 0$. Can I say that $g$ is locally bounded? By "$g$ is locally bounded", I means "for all $x \in A$, there is a neighborhood $U$ of $x$ in $A$ and a $M > 0$ such that for all $u \in U$, $\|g(u)\| < M$". AI: Any continuous function $g:A\to\mathbb R^n$ is locally bounded. This is because for any $x\in A$, we have some open ball $B_r(x)\subset A$ and thus the closed ball $C:=\overline{B_{r/2}(x)}\subset A$ as well. Since any closed ball around a point in $\mathbb R^n$ is compact, we have that $g\|_C$ is uniformly continuous, as for any $\epsilon>0$ we have open balls $B_i$ such that $x,y\in B_i\implies \\|g\|_C(x)-g\|_C(y)\\|<\epsilon$ which form an open cover of $C$ hence there exists a finite subcover, so if we let $\delta$ be the minimum of their radii we have $\\|x-y\\|<\delta\implies \\|g\|_C(x)-g\|_C(y)\\|<\epsilon$. Letting $\epsilon=1$ we see that $g\|_C$ varies by at most $r/\delta$ over $B_{r/2}(x)$ thus $g$ is bounded on a neighborhood of $x$.
H: When antidifferentiating, are we impliclty restricting to an interval? I was asked this question by a student I am tutoring and I was left a little puzzled because his textbook only defines antiderivatives on intervals (which leads me to believe its author would answer the question in the title in the affirmative). To my understanding, finding an antiderivative of $f$ means finding a function $F$ with $F' = f$. It does not matter if the domain of $f%$ is not connected. For example, $\int \frac{dx}{x} $ denotes an antiderivative on all of $\mathbb{R} - 0$, and not just on some arbitrary interval $I \subseteq \mathbb{R} - 0 $. Am I mistaking? AI: You can use whatever conventions you want; the author is free to choose hers and you are free to choose yours. One issue with defining antiderivatives on (nice) subsets of $\mathbb{R}$ which are not connected is that they are no longer just only unique up to constants; they are only unique up to locally constant functions. For example, on $\mathbb{R} \setminus \{ 0 \}$ any function which has one constant value when $x$ is negative and another when $x$ is positive has zero derivative. This is the kind of subtlety that I suspect it would be a good idea to avoid in a calculus course. Another issue is that you'd like to write antiderivatives down using definite integrals, but for example if a function $f$ is not defined in the interval $(-1, 1)$ then it is unclear what an integral such as $\int_{-2}^x f(x) \, dx$ would mean for $x > 1$...
H: Solving the differential equation $\frac{dy}{dx} =\sqrt{7x^3}$ $$\frac{dy}{dx} =\sqrt{7x^3}$$ I need to use substitution on the $7x^3$ but I'm a little stuck. My car broke down on the way to class and I missed the lesson! I get this far.. $$u =7x^3$$ $$du = 21x^2 dx$$ $$dx = \frac{du}{21x^2}$$ $$dy = \sqrt{u} \frac{du}{21x^2}$$ AI: We can use the technique of separation of variables. The method is as follows: $$dy=\sqrt{7x^3}dx$$ $$\int dy = \int\sqrt{7x^3}dx$$ The justification of this method is a little more nuanced than simply "multiplying" $dx$ and then integrating--we are taking the antiderivative of both sides: $\int\frac{dy}{dx} dx = \int f'(x) dx = f(x)+C$ So $$y = \sqrt{7}\int x^{3/2} dx$$ Thus, we have $$y = \frac{2\sqrt{7}}{5}x^2\sqrt{x}+C$$ for some constant $C$. This is the general solution to this differential equation.
H: Calculating a number when its remainder is given I am having difficulty solving the following problem: Marge has n candies , where n is an integer between 20 and 50.If marge divides the candies equally among 5 children she will have 2 candies remaining . If she divided the candies among 6 children she will have 1 candy remaining.How many candies remain if she divides the candies among 7 children ? The equation I got was $$5q-6k+1=0$$ I tried using this method but I dont think it will work here. Any suggestions on how I could solve this ?I would appreciate it if the method involves equations instead of plugging in numbers and testing. AI: This is a classic Chinese Remainder Theorem. We know that $$\begin{align} n &\equiv 2\pmod{5}\\ n &\equiv 1\pmod{6}. \end{align}$$ By the Chinese Remainder Theorem, there is a unique number $n$ modulo $30=5\times 6$ that satisfies both equations. We can compute it directly: $n=5q+2$ since it leaves a remainder of $2$ when divided by $5$. That means that we must have $5q+2\equiv1\pmod{6}$, which means $5q\equiv -1\equiv 5\pmod{6}$, hence $q\equiv 1\pmod{6}$. So $n=5q+2 = 5(6k+1)+2 = 30k+5+2 = 30k+7$. That is, the solution is $n\equiv 7\pmod{30}$. Since the number must be between $20$ and $50$, the number is $37$. When divided among $7$ children, she will have $2$ candles left over. If you must avoid congruences (really, you are only avoiding them explicitly, shoving them "under the carpet"), $n$ must be of the form $6k+1$. We can rewrite this as $n=6k+1 = 5k + (k+1)$. Since, when $n$ is divided by $5$ the remainder is $2$, that means that when we divide $k+1$ by $5$ the remainder is $2$. Therefore, the remainder when dividing $k$ by $5$ must be $1$ (so that the remainder of $k+1$ will be $1+1=2$), so $k=5r+1$. So $n=6k+1 = 6(5r+1)+1 = 30r+7$, and we are back in what we had above.
H: on sequences of Lebesgue measurable subsets of a compact set In preparation for the real analysis qualifying exam at my grad school, I've been working through the recommended textbook Modern Real Analysis, by Zimmer (it's free online here). For the last few days, I've been trying to figure out problem 4.27 (on p124 of the text, which is p132 of the pdf), which asks us to prove the following: Theorem. If $\{E_k\}_{k=1}^\infty$ is a sequence of Lebesgue measurable subsets of a compact set $K\subseteq \mathbb{R}^n$ such that $\inf_{k\geq 1} \lambda(E_k)>0$, then there's some point which belongs to infinitely many $E_k$'s (i.e. there's some point which belongs to $E_k$ for infinitely many $k$). Definitions, Etc. Lebesgue measure is denoted by $\lambda$. The Lebesgue outer measure $\lambda^$ is defined to be $$\lambda^(A):= \inf\left\{\sum_{I\in\mathcal{S}} \mathrm{vol}(I) \right\},$$ where the infimum is taken over countable all covers $\mathcal{S}$ of $A$ by sets of the form $[a_1,b_1]\times \cdots \times [a_n,b_n]$ (which Zimmer calls closed $n$-dimensional intervals, or just closed intervals if the dimension is clear). A set $A\subseteq\mathbb{R}^n$ is defined to be Lebesgue measurable if $$\lambda^(A)=\lambda^(A-S)+\lambda^*(A\cap S)$$ for any set $S\subseteq \mathbb{R}^n$. My Attempt If there are finitely many distinct $E_k$'s, this is trivial. So, assume there are infinitely many distinct $E_k$'s (I don't know if this assumption will be useful, but since we're working with compactness, I figured it might). From each $E_k$, choose a point $x_k$. Then because $K$ is compact, there's a subsequence $\{x_{k_j}\}_{j=1}^\infty$ of $\{x_k\}_{k=1}^\infty$ which converges to some point $x\in K$. At this point, I'm stuck. My gut tells me that this point $x$ (or some similarly constructed point) should be in infinitely many $E_k$'s, but I can't for the life of me figure out how to prove this. Could someone point me in the right direction? AI: Suppose every point is contained in at most finitely many $E_k$, so the indicator functions $\chi_{E_k}:K\to \mathbb R$ converge to $0$ pointwise. Since $K$ has finite measure, we can apply Egorov's theorem, so for any $\epsilon>0$ we have a set $S\subseteq K$ such that $\lambda(K\setminus S)<\epsilon$ and $\chi_{E_k}\|_S$ converges uniformly to $0$. But since $\chi_{E_k}\|_S$ takes only the values $0$ and $1$, we must have that this sequence is eventually identically $0$ hence $E_k\cap S=\emptyset$ for sufficiently large $k$. But then $$\lambda(E_k)\leq \lambda(E_k\cap S)+\lambda(K\setminus S)=\epsilon$$ and since $\epsilon$ is arbitrary this violates the fact that $\inf_{k\geq 1} \lambda(E_k)>0$. Thus some point is contained in infinitely many $E_k$.
H: Why is associativity required for groups? Why is associativity required for groups? I'm doing a linear algebra paper and we're focusing on groups at the moment, specifically proving whether something is or is not a group. There are four axioms: The set is closed under the operation. The operation is associative. The exists and identity in the group. Each element in the group has an inverse which is also in the group. Why does the operation need to be associative? Thanks AI: The formalist's answer is: it is just a definition. You could just as well consider studying algebraic structures that satisfy all the axioms for a group except for associativity, and you would be then studying loops. Now the question might be: why is the study of groups more ubiquitous than the study of loops? There are historical reasons (surely others with greater knowledge can expand upon this), and the fact that most loops that arise naturally when doing math are in fact groups is probably a reason too.
H: Closed set in a Hausdorff topological space Possible Duplicate: $X$ is Hausdorff if and only if the diagonal of $X\times X$ is closed I'm trying to prove: If $X$ is a Hausdorff topological space and $\Delta \subset X\times X$ such that $\Delta=\{(x,y): x=y\}$, prove that $\Delta$ is closed. I can not use sequences because I don't have a metric. I thought in accumulation points, but I am not sure how to use this. Thanks for your help. AI: The result is not true in general. Let $X$ be a countably infinite set with the cofinite topology, i.e., $U\subseteq X$ is open iff $U=\varnothing$ or $X\setminus U$ is finite. Let $p=\langle x,y\rangle$ be any point of $X\times X$, and let $U$ be any open neighborhood of $p$ in $X\times X$. Then there are open sets $V$ and $W$ in $X$ such that $x\in V,y\in W$, and $V\times W\subseteq U$. Since $V$ and $W$ are open in $X$, there are finite sets $F_V$ and $F_W$ such that $V=X\setminus F_V$ and $W=X\setminus F_W$. $F_V\cup F_W$ is still finite, so $V\cap W=X\setminus(F_V\cup F_W)\ne\varnothing$. Pick any point $z\in V\cap W$; then $$\langle z,z\rangle\in(V\times W)\cap\Delta\subseteq U\cap\Delta\;,$$ so $U\cap\Delta\ne\varnothing$. $U$ was an arbitrary open neighborhood of $p$, so $p\in\operatorname{cl}\Delta$. And $p$ was an arbitrary point of $X\times X$, so $\operatorname{cl}\Delta=X\times X\ne\Delta$, and therefore $\Delta$ is not closed. But $X$ is countable, so it is certainly separable: it is a countable dense subset of itself. In order to prove that $\Delta$ is closed, you want to assume that $X$ is Hausdorff, meaning that if $x,y\in X$, and $x\ne y$, then there are open sets $U$ and $V$ such that $x\in U,y\in V$, and $U\cap V=\varnothing$. Now suppose that $p=\langle x,y\rangle\in (X\times X)\setminus\Delta$. Then $x\ne y$, so there are disjoint open sets $U$ and $V$ in $X$ such that $x\in U$ and $y\in V$. I leave it to you to show that $U\times V$ is then an open neighborhood of $p$ disjoint from $\Delta$. This will show that every point of $(X\times X)\setminus\Delta$ has an open neighborhood disjoint from $\Delta$ and hence that $\Delta$ is closed.
H: How to put this problem into equation? I start with a value A. I decrease it by M each month m. Every year (ie, m mod 12 =0), I calculate the average of what A has been throughout the year, multiply that by t, and add it to A. This goes on until A reaches 0 (t and M are set in such a way that A decreases every year). How many months m would that take ? Similarly, given A and M, if I want A to reach 0 after m months, what does the value of t need to be ? In other words, I have a capital A that has been given to me by a bank B1. B1 asks me to return M dollars every month, until it reaches a value N that was predefined by B1 (B1 also calculates this in terms of a interest percentage, but I don't understand the definition of that interest value). I put the money A that B1 gave me into a bank B2. What does the yearly interest t in bank B2 need to be so that it is the threshold between winning and losing money ? These may seem like 3 different questions, but I think that once put into equation, I think it's easy to answer all of these by just expressing one variable as a function of others. concrete example: I ask a bank B1 to lend me A=10,000 dollars. In exchange for that 'donation', B1 wants me to give M=100 dollars for 10 years (m=120). In the end, as you can see, I end up giving more (12000 dollars) than I was given. However, I do not plan to use these 10,000 dollars. Instead, I just put it into a bank B2 and wait for interest rates to inflate my money. If don't know how this is done in your country, but in mine, I think the interest rates are given at the end of the year, but calculated monthly (I think that one other way to say that is to say that it is calculated yearly, but with the average of what you had in your balance throughout the year, hence my explanation in the first paragraph). Interests over interests don't apply monthly, but yearly. This is to say that if you have 100 dollars the first year, with an interest rate of 2%, then the second year you have 100+2, but the third you have 1001.021.02=1.0404 instead of just 100+2+2=104. So if I could have these 10,000 dollars for 10 years, I would have 10000*t^(m/12). But I don't have A=10000 dollars for 10 years, because they keep decreasing as I have to give M=100 dollars away every month. So first of all, I'm trying to find a formula (not a estimate by excel, but an exact mathematical formula) that says what is my balance after m months. Then I'm trying to find what value t (the interest rate from bank B2) needs to be (depending on the value of M) so that I end up winning money by asking B1 to give me money and placing this money into B2. AI: Writing the equation is following what you have written, but there are a few details not perfectly specified. Let $B(m)$ be the balance owing to the first bank in month $m$. The first sentence tells us that $B(0)=A$. Then you say that $B(m)=B(m-1)-M$ unless $m$ is a multiple of $12$. It is not clear what values are averaged for the yearly increment, but it looks like we should average B(0) through B(11). Do you also decrement by M in month 12? It would seem that $B(12)=B(11)-M+\frac t{12}\sum_{i=0}^{11}B(i)$. Then we go back to $B(m)=B(m-1)-M$ until month $24$, when $B(24)=B(23)-M+\frac t{12}\sum_{i=12}^{23}B(i)$ and so on. Presumably you deposit the A in the second bank. You are using $t$ for a constant in the bank B1 equation as well as the interest paid by bank B2. Do you know that they will be the same? Probably not, so I will use $s$ as the annual interest paid by B2 and assume that it is compounded monthly. Then let $C(m)$ be the balance in B2. We have $C(0)=A, C(m)=C(m-1)(1+\frac s{12})-M$. Your net position at any month is $C(m)-B(m)$. This is made for a spreadsheet-make columns for $m, B(m), C(m), C(m)-B(m)$ with $t,s$ in cells you reference by name or by fixed reference (Excel uses dollar signs). Given $A, M, t$, you can look to find the month you pay off the loan. Then Excel will let you goal seek to make the balance zero by changing $s$. If you use a spreadsheet without that capability, you can just do it by hand, looking for the correct $s$.
H: Evaluating $\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$ How do I integrate this expression: $$\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$$.I got this in a book.I do not know how to evaluate integrals of this type. AI: One uses trigonometric substitution: $t = \tan\left(\frac{x}{2}\right)$. Then $$ \sin(x) = \frac{2t}{1+t^2} \quad \cos(x) = \frac{1-t^2}{1+t^2} \quad \mathrm{d} x = \frac{2}{1+t^2} \mathrm{d} t $$ Hence: $$\begin{eqnarray} \int \frac{ \ell \sin(x) + m \cos(x)}{(a \sin(x)+ b \cos(x))^2} \mathrm{d}x &=& \int \frac{ \ell \frac{2 t}{1-t^2} + m \frac{1-t^2}{1+t^2}}{\left(a \frac{2 t}{1+t^2}+ b \frac{1-t^2}{1+t^2}\right)^2} \frac{2}{1+t^2}\mathrm{d}t \\ &=& \int \frac{2 \ell t + m (1-t^2)}{\left(2 a t + b (1-t^2)\right)^2} 2 \mathrm{d}t \end{eqnarray}$$ The resulting rational function can be integrated using partial fraction decomposition of the integrand, for example.
H: Groups such that every finitely generated subgroup is isomorphic to the integers What are examples of groups such that every finitely generated subgroup is isomorphic to $\mathbb{Z}$? AI: Suppose any nontrivial finitely generated subgroup of $G$ is isomorphic to $\Bbb{Z}$, and that $G$ itself is nontrivial. Choose some nonzero $e \in G$. For any $g \in G$, the subgroup $\langle e, g\rangle$ is isomorphic to $\Bbb{Z}$, so in particular $ne=mg$ for some integers $m,n$. Define a map $f:G \to \Bbb{Q}$ by $g \mapsto n/m$. Then $f$ is a well-defined group homomorphism; since $G$ is torsion-free and $e$ nonzero, it is straightforward to check that $f$ is injective. So $G$ is isomorphic to a nontrivial subgroup of $\Bbb{Q}$, which are classified, e.g., in this paper.
H: Why $g(x^{p})=(g(x))^{p}$ in the reduction mod $p$? In one of the proof in the book "Abstract Algebra'' by Dummit and Foote (Theorem 41, pg. 554) we have a monic polynomial $g(x)\in\mathbb{Z}[x]$, and the book claims that $g(x^{p})=(g(x))^{p}\mod p$ Can someone please explain why this is true ? I know that $\forall a\in\mathbb{F}_{p}:a=a^{p}$, but I don't see how this imply the equality as polynomials AI: First, note that the claim is clearly true for the polynomial $x\in\mathbb{Z}[x]$, as well as for any constant polynomial $a\in\mathbb{Z}$ (the latter is just Fermat's Little Theorem). Now note that any element of $\mathbb{Z}[x]$ can be obtained as a combination of sums and products of the polynomial $x$ and constant polynomials. For any $f,g\in\mathbb{Z}[x]$, we have that $$(f+g)^p=f^p+\binom{p}{1}f^{p-1}g+\cdots+\binom{p}{p-1}fg^{p-1}+g^p\equiv f^p+g^p\bmod p$$ and $(fg)^p=f^pg^p$, so we certainly have that $(fg)^p\equiv f^pg^p\bmod p$. Thus, the claim is true for all polynomials in $\mathbb{Z}[x]$ (the condition that $g$ be monic is unnecessary).
H: Integrating definite integrals in terms of area Lately, I've been trying to come up with tricks to solve integrals quickly. So let's say I have $$\int_{0}^{2\pi} \cos^2 \theta d\theta$$ Now if I were to look at this integral in polar coordiantes, I get $$\frac{1}{2}\int_{0}^{2\pi} \cos^2 \theta d\theta$$ The integrand is a circle in polar coordinates $r = 2(1/2)\cos\theta$ with radius $1/2$ So the integral $$\frac{1}{2}\int_{0}^{2\pi} \cos^2 \theta d\theta = \frac{\pi}{2}$$ But this doesn't make sense to me because the area should be $\pi (1/2)^2 = \pi/4$ What's going on? I am trying to extend this idea to $$\int_{0}^{2\pi} \sin^2 \theta d\theta$$ and linear combinations of sine and cosines squared AI: Your integral does not "know" that $r$ cannot be negative. So from $\pi/2$ to $3\pi/2$ it cheerfully keeps "adding up" $\cos^2 \theta$, not realizing there is no curve there. Effectively the integral traverses the circle twice. It really should know that the integration should be done from $0$ to $\pi/2$, and from $3\pi/2$ to $2\pi$. Or equivalently that it should be integrating $\frac{1}{2}f^2(\theta)$, where $f(\theta)=0$ on $(\pi/2,3\pi/2)$, and $f(\theta)=\cos\theta$ elsewhere on $[0,2\pi]$. But nobody told it. Remark: There is not universal agreement that the point that has polar coordinates address $(r,\theta)$ makes no sense if $r \lt 0$. One common interpretation in that case is that you graph $(r,\theta)$ and then reflect the result across the origin. That interpretation sometimes produces nicer pictures. If we take that interpretation, then $r=\cos\theta$ sweeps out our circle as $\theta$ travels from $0$ to $\pi$, so integrating from $0$ to $\pi$ gives the area.
H: Reference for upper and lower bounds on $e^x$ I'm looking for a reference for deriving the following commonly used upper and lower bounds for $e^x$: $$1 - x \le e^{-x}$$ and, assuming $x \le 1/2$, $$1 - x \ge e^{-2x}. $$ AI: First Inequality: Let $f(x)=e^{-x}-(1-x)$. Then $f'(x)=-e^{-x}+1$. This is $0$ at $x=0$, negative if $x \lt 0$, and positive for $x \gt 0$. So $f(x)$ reaches an absolute minimum at $x=0$. The minimum value is $0$, so $f(x)\ge 0$ for all $x$. Second Inequality: Let $g(x)=(1-x)-e^{-2x}$. Then $g'(x)=-1+2e^{-2x}$. This is $0$ for $x=(\ln 2)/2$. We find that $g'(x)$ is positive for $x\lt (\ln 2)/2$, and negative for $x \gt (\ln 2)/2$. The function $g(x)$ is increasing up to $x=(\ln 2)/2$, then decreasing. We have $g(0)=0$. Since $g(x)$ is increasing until $(\ln 2)/2$, it follows that $g(x)$ is negative when $x$ is negative, so our inequality fails if $x \lt 0$. For positive $x$, $g(x)$ increases up to $x=(\ln 2)/2$, and then decreases. After a while it will become negative. Let us check whether $g(x)$ is still positive at $x=\frac{1}{2}$. Calculate. We get $g(1/2)\approx 0.132$. So $g(x) \ge 0$ at least in the interval $[0,1/2]$. In fact $g(x) \ge 0$ from $0$ to a bit beyond $0.79$. Remark: There will be few standard North American calculus books that don't do the first inequality. Usually it is done in the equivalent version $1+x \le e^x$. It is an early application of the connection between increasing functions and positive derivatives. The inequality (among many others) is mentioned here. The second inequality uses the same method, but the details involve more work.
H: Proving that $\Phi_{n}$ is irreducible (a problem with the proof) I am trying to follow the proof in the book Abstract Algebra by Dummit and Foote (Theorem 41, pg. 554) that $\Phi_n$ is an irreducible monic polynomial in $\mathbb{Z}[x]$ of degree $\varphi(n)$. What I understand is that if it is not irreducible, than we can factor $\Phi_{n}=fg$ where $f$ is the minimal polynomial of $\zeta_{n}$ (a primitive $n$-th root of unity). I also agree that if $p$ is a prime s.t. $(p,n)=1$ then $f(\zeta_{n}^{p})=0$. The question is why if $(a,n)=1$ then $f(\zeta_{n}^{a})=0$ (I don't understand why this is true even for $a=p_{1}p_{2}$). Can someone please explain why we can move from primes to their product? AI: $\zeta_n$ is an arbitrary primitive $n$th root of unity which is a root for $f.$ Given that $f(\zeta_n^p ) =0,$ we know that $\zeta_n^p$ is also a primitive $n$th root of unity, so that for any second prime $q$ not dividing $n$ we can again apply the proposition to $\zeta^p$; $f(\zeta^{pq})=f((\zeta^p)^q)=0.$ If $(a,n)=1,$ we can write $a=\displaystyle\prod_{i=1}^n p_i$ (not necessarily distinct primes) and apply an easy inductive argument.
H: Help with set notation? I want to describe the set of all words in the following format: a0w1 where a represents EITHER 0 or 1, and w represents {0,1}* So 00011 is valid as is 1010011, etc. etc. I'm really new to set notation, so I'm not sure what I can do. Is L = {a,0,w,1 \| a = 0 or 1, w $\in$ {0,1}} valid for what I want to describe? Thanks! AI: You can describe that set in many ways. For instance, it corresponds to the regular expression $(0\lor 1)0(0\lor 1)^1$ (in one common formalism for regular expressions). However, if you want to describe it as a set of strings using standard set notation, you want something like $$\Big\{a0w1\in\{0,1\}^:a\in\{0,1\}\text{ and }w\in\{0,1\}^\Big\}\;.$$ In particular, you don’t want the commas: words over an alphabet are normally written without, unless some of the symbols in the alphabet require more than one typographical symbol to represent them.
H: Group of groups The product $\times$ of two groups is associative and commutative and there's a neutral element $\{1\}$. Let's say I create "virtual groups" which are inverses with respect to $\times$ (like getting $\mathbb{Z}$ from $\mathbb{N}$). Then I have a group $G$ whose elements are all groups. This isn't allowed, though, is it? Since a set can't be a member of itself. AI: The Cartesian product is not commutative: if $G\ne H$, $G\times H\ne H\times G$, though these groups are isomorphic. It isn’t associative, either, because $\big\langle\langle x,y\rangle,z\big\rangle\ne\big\langle x,\langle y,z\rangle\big\rangle$, though once again there’s an isomorphism between $(G\times H)\times K$ and $G\times(H\times K)$. That aside, there’s no problem in principle with having a group $G$ whose elements are themselves groups; $G$ simply won’t be one of those elements. What you can’t do is form a group $G$ whose elements are all possible groups, any more than you can form a set of all sets. What you can do in ZF is write down a first-order formula $\varphi(x)$ that ‘says’ that $x$ is a group, meaning an ordered pair $\langle y,z\rangle$ such that $y$ is a set and $z$ is a function from $z\times z$ to $z$ with the properties needed to make it a group operation on $y$. You can then write down formulas $\psi(x,y,z)$ that ‘say’ that $x,y$, and $z$ are groups and that $z$ is related to $x$ and $y$ in some specific fashion. I shouldn’t be at all surprised if it were possible to find such a formula for which one could prove $$\forall x,y\Big(\big(\varphi(x)\land\varphi(y)\big)\to\exists!z\psi(x,y,z)\Big)\;,$$ i.e., $\psi$ defines what would be a binary operation on groups if the set of all groups existed, $$\forall u,v,w,x,y,z\left(\Big(\psi(x,y,u)\land\psi(y,z,v)\Big)\to\Big(\psi(u,z,w)\leftrightarrow\psi(x,v,w)\Big)\right)\;,$$ i.e., that ‘operation’ is associative, and the rest of the group axioms. Then one could talk informally about the proper class $\bf{G}$ of groups and a proper class operation making $\bf G$ into a group.
H: Is there a monotonic function discontinuous over some dense set? Can we construct a monotonic function $f : \mathbb{R} \to \mathbb{R}$ such that there is a dense set in some interval $(a,b)$ for which $f$ is discontinuous at all points in the dense set? What about a strictly monotonic function? My intuition tells me that such a function is impossible. Here is a rough sketch of an attempt at proving that such a function does not exist: we could suppose a function satisfies these conditions. Take an $\epsilon > 0$ and two points $x,y$ in this dense set such that $x<y$. Then, $f(x)<f(y)$ because if they are equal, then the function is constant at all points in between, and there is another element of $X$ between $x$ and $y$, which would be a contradiction. Take $f(y)-f(x)$. By the Archimedean property of the reals, $f(y)-f(x)<n\epsilon$ for some $n$. However, after this point, I am stuck. Could we somehow partition $(x,y)$ into $n$ subintervals and conclude that there must be some point on the dense set that is continuous? AI: Such a function is possible. Let $\Bbb Q=\{q_n:n\in\Bbb N\}$ be an enumeration of the rational numbers, and define $$f:\Bbb R\to\Bbb R:x\mapsto\sum_{q_n\le x}\frac1{2^n}\;.\tag{1}$$ The series $\sum_{n\ge 0}\frac1{2^n}$ is absolutely convergent, so $(1)$ makes sense. If $x<y$, there is some rational $q_n\in(x,y)$, and clearly $f(y)\ge f(x)+\frac1{2^n}$, so $f$ is monotone increasing. However, $f$ is discontinuous at every rational: $$\lim_{x\to {q_n}^-}f(x)=\sum_{q_k<q_n}\frac1{2^k}<\sum_{q_k\le q_n}\frac1{2^k}=f(q_n)\;.$$ Thus, $f$ is discontinuous on a set that is dense in $\Bbb R$ (and in every open interval of $\Bbb R$).
H: When is this quotient by an action on the product of a variety with itself non-singular Let $X$ be a smooth projective geometrically connected variety over a field $k$. Let the cyclic group $G=\{e,a\}$ with two elements act on $X \times X$ via $a\cdot (x_1,x_2) = (x_2,x_1)$. When is the quotient $X\times X/ G$ nonsingular? I wrote down the case of $X=\mathbf{A}^1$. The answer is that $X\times X/G$ is given by nonsingular scheme $\mathrm{Spec} (k[xy,x+y]) \cong \mathbf{A}^2$, where $\mathbf{A}^1_x\times \mathbf{A}^1_y = X\times X = \mathrm{Spec} (k[x]\otimes k[y])$. (The subscript $x$ and $y$ indicate the coordinate used.) I'm very interested in the case $\dim X = 1$. AI: Over an algebraically closed field: The quotient $X^{(2)}=X\times X/C_2$ is called the symmetric square of $X$. If $X$ is an irreducible (quasi)projective nonsingular curve, then $X^{(2)}$ is always smooth and, in fact, this generalizes to $X^{(n)}=X\times\cdots\times X/S_n$, the obvious quotient by the symmetric group on $n$ letters. If $X$ is an irreducible (quasi)projective nonsingular surface, then $X^{(n)}$ is singular for all $n\geq2$. Lothar Göttsche wrote notes for a short course he gave in ICTP a few years ago on the subject of Hilbert schemes of points, with title Hilbert schemes: local properties and Hilbert scheme of points. There you should find proofs of all the above statements. (The Hilbert scheme in all this is a «less singular» replacement for what I wrote $X^{(n)}$ which, in the case of smooth surfaces, turns out to be itself a desingularization of the symmetric power; in general, it is also singular, though)
H: A question about a weak form of Hilbert's Nullstellensatz Corollary 5.24 on page 67 in Atiyah-Macdonald reads as follows: Let $k$ be a field and $B$ a finitely generated $k$-algebra. If $B$ is a field then it is a finite algebraic extension of $k$. We know a field extension $E$ over $F$ is algebraic if it's finite, that is, $E = F[e_1, \dots, e_n]$. By definition, a finitely generated $k$-algebra is of the form $k[b_1, \dots , b_n]$. So the corollary above seems to directly follow from these two facts. I hope I misunderstand something fundamental because I worked through the propositions and proofs this corollary is using and it was rather lengthy and not very enjoyable. What am I missing? AI: Perhaps the statement will become more clear in the following language: A ring homomorphism $R \to S$ is finite if $S$ is finitely generated as a module over $R$. A ring homomorphism $R \to S$ is called of finite type if $S$ is finitely generated as an algebra over $R$. Clearly, finite implies of finite type. The converse is not true, in general. We have that finite <=> integral and of finite type. But for fields, the converse is true: Every field extension which is of finite type, is already finite (and therefore algebraic). This is an easy consequence of Noether's normalization lemma. It is not a consequence of the definitions, because it is not clear a priori that our algebra generators are algebraic.
H: Can two collections of different size have same A.M., G.M. and H.M? After following Can two sets have same AM, GM, HM? and the sublime answer of Micah, I am tempted ask the solution of the same question when size of these two sets are not same. AI: Sure, why not? The general statement that my answer was a special case of is: If $\{r_1,\dots,r_n\}$ are the roots of the polynomial $x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$, then their arithmetic mean is $-a_{n-1}/n$, their geometric mean is $\sqrt[n]{(-1)^na_0}$, and their harmonic mean is $-na_0/a_1$. This can be shown by examining the Vieta formulae in exactly the same way as the degree-$4$ case, and gives a recipe for building sets of any size at least 3 that have whatever AM, GM, and HM you want. The only possible worry is that the roots might not all be real and positive, in which case strictly speaking the GM isn't well-defined; for example, this will happen if you try to build a set that violates the AM-GM-HM inequalities. For a concrete-ish example, the set $\{1,2\}$ has AM $3/2$, HM $4/3$, and GM $\sqrt{2}$. By the above formulae, so do the roots of $x^3-9/2x^2+9\sqrt{8}/4-\sqrt{8}$ (which WolframAlpha says are $\left\{\sqrt{2},\frac{1}{4}\left(9-2\sqrt{2}\pm\sqrt{57-36\sqrt{2}}\right)\right\}$).
H: physics related question i am trying to calculate simple problem from physic,but i am getting somehow wrong answer.problem is that what is a mass of bag which is hold by child with mass $50$KG,if there is force of heaviness on both which is equal $600$N so in shortly,we know that on child and bag,there works $600$N force of heaviness,we have to find mass of bag. as i know force of heaviness on mass on earth is $F=mg$ if we have two mass,then i think formula would be $F=m_1m_2*g$ from which $m_2=1.2$,but in book answer is $m_2=10kg$,please help me to clarify why is so? AI: Your formula is wrong. Take $F=(m_1+m_2)g=(50kg+10kg)9.81\frac{m}{s^2}\approx 600N$.
H: Proof of error propagation formula? In my course we have stated and used the error propagation formula: $$\|y-y_0\|\approx\|f^\prime(x)\|\cdot\|x-x_0\|$$ But it was presented with no proof and I wonder if you can help me understand the formula holds? AI: Possibly the best way to understand it is via the mean value theorem: $$f(x)-f(x_0)=f'(c)\cdot(x-x_0)$$ for some $c$ between $x_0$ and $x$. If $f'$ is continuous, $f'(c)$ can be expected to be close to $f'(x)$ or $f'(x_0)$ when $\lvert x-x_0\rvert$ is small enough.
H: How many such squares can be formed? $S$ is a set of all points $(a, b)$ such that $0 ≤ a$, $b ≤ k$. How many squares are there such that all the $4$ vertices are from set $S$? For diagonal squares, a square must contain odd points on its side. so that we can join the mid points of each side. suppose we take $5$ points on each side, $$55 will make = 1^{2}$$ $$33 will make = 3^{2}$$ etc so when $n$ is odd, its summation of all odd squares till $k$. & for even similar approach. Can you guide me further? Please help. AI: I think it's this sequence. It says, "$a(n) =$ number of squares with corners on an $n\times n$ grid."
H: Special dot-product I have been wondering if the following dot product definition for the $n$-coordinate vectors $a$ and $b$ has a name: $$<a\backslash b> = \sum_{i=1}^{n} a_ib_{n-i+1},$$ rather than the classical dot product: $$<a\backslash b> = \sum_{i=1}^n a_ib_{i}.$$ Did you already seen it use somewhere? Thanks a lot. AI: You can diagonalize this bilinear form so that it becomes the sum of $p$ squares minus the sum of $q$ squares where $p$ and $q$ are $n/2$ rounded up and down, respectively. This is called a quadratic form of signature $(p,q)$, and the space in which this norm is the inner product is denoted $\mathbb R^{p,q}$. For example, see http://en.wikipedia.org/wiki/Indefinite_orthogonal_group
H: Is there an unique "minimal enclosing group" for any two groups? I'm not sure I'm using the correct terms, therefore let me define what I mean: Given a set of groups $G_i$, $i\in I$, I call an enclosing group of those groups any group $G$ so that for all $i\in I$ there exists a subgroup $H_i$ of $G$ which is isomorphic to $G_i$. As an example, the direct product of groups is an enclosing group of those groups. Also each group trivially encloses all its subgroups. I call an enclosing group minimal if there's no subgroup of $G$ which also encloses the given set of groups. For example, unless I'm mistaken, the group $U(2)$ is a minimal enclosing group of $SU(2)$ and $U(1)$. My question now is: Is there, up to isomorphism, an unique minimal enclosing group for any set of groups? There are a few simple cases where this is obviously the case: If the set is empty, obviously the trivial group $\{e\}$ is the only minimal enclosing group. If the set has only one element, that element is the only minimal enclosing group. If all members of the set are subgroups of some specific member of that set, that member is the only minimal enclosing group. AI: As stated in the comments, this is false. Considering the two cyclic groups $\mathbb{Z}_2$ and $\mathbb{Z}_3$, both $\mathbb{Z}_2\times\mathbb{Z}_3$ and $S_3$ are enclosing groups and both are minimal, due to cardinality constraints.
H: Using the substitution $p=x+y$, find the general solution of $dy/dx=(3x+3y+4)/(x+y+1)$. Using the substitution $p=x+y$, find the general solution of $$\frac{dy}{dx}=(3x+3y+4)/(x+y+1)$$ Here are my steps: Since $p=x+y$, $$\frac{3x+3y+4}{x+y+1}=\frac{3p+4}{p+1}=\frac{1}{p+1}+3$$ Therefore, integrate both sides $$y=\ln(p+1)+3p+c$$ $$y=\ln(x+y+1)+3(x+y)+c$$ But the answer in my book is $$x+y-\frac{1}{4}\ln(4x+4y+5)=4x+c$$ Is that correct? AI: Since $p(x)=x+y(x)$ therefore $y(x)=p(x)-x$. Thus $$ dy/dx=y'(x)=p'(x)-1. $$ So the new equation is $$ \frac{dp}{dx}=p'(x)=\frac{1}{p(x)+1}+4=\frac{4p(x)+5}{p(x)+1}=\frac{4p+5}{p+1}. $$ This equation is separable. Using the usual method $$ \frac{p+1}{4p+5}dp=1 dx, $$ integrating $$ \frac{p}{4}-\frac{1}{16}\log(4p+5)+C=x $$ Substituting $p(x)=x+y(x)$ we obtain the solution that is the same as in your book.
H: A question about a proof of a weak form of Hilbert's Nullstellensatz I'm trying to prove the following (corollary 5.24 page 67 in Atiyah-Macdonald): Let $k$ be a field and let $B$ be a field that is a finitely generated $k$-algebra, i.e. there is a ring homomorphism $f: k \to B$ and $B = k[b_1, \dots , b_n]$ for $b_i \in B$. Then $B$ is an algebraic (and hence, in this case, finite) extension of $k$. There is a proof in Atiyah-Macdonald but it's more like a hint and I'm not sure I understand the details. Can you tell me if this detailed version of the proof is correct? Here goes (thanks!): We need to show that $b_i$ are algebraic over $k$. Since $k$ is a field we know that $f$ is injective so we may view $k$ as a subfield of $B$ ($f$ is our embedding). Then $k \subset B$ are integral domains and $B$ is finitely generated so we are in the position to apply proposition 5.23 which tells us the following: If $b$ is a non-zero element in $B$ then we can find a non-zero element $c$ in $k$ such that if $f: k \to \Omega$ is a homomorphism into an algebraically closed field $\Omega$ such that $f(c) \neq 0$ then there exists an extension $g: B \to \Omega$ of $f$ such that $g(b) \neq 0$. We observe that $1$ is a non-zero element of $B$. The inclusion $i: k \hookrightarrow$ of $k$ into its algebraic closure $\overline{k}$ is a ring homomorphism such that $i(1) \neq 0$. By the previously stated proposition we hence can find a ring homomorphism $g: B \to \overline{k}$ such that $g(1) \neq 0$. Although the fact that $g(1) \neq 0$ doesn't interest us. But since $g$ is a ring homomorphism defined on a field we know that it's injective hence we can view $B$ as a subfield of $\overline{k}$. And now we are done since we have $k \subset B \subset \overline{k}$, hence $B$ is contained in the algebraic closure of $k$ and hence every element of $B$ is algebraic over $k$. In particular, $b_i$. I wonder why it's called "Nullstellensatz". It doesn't seem to have anything to do with roots of polynomials. Here is an image of proposition 5.23: AI: Your proof looks okay. Here is a reason why it is called the Nullstellensatz: Assume that your ground field $k$ is algebraically closed. Let $B = k[x_1,...,x_n]$ where the $x_i$ are indeterminates. Let $m \subset B$ be a maximal ideal. Then $B/m$ is a field extension of $k$ which is also clearly a finitely generated $k$ algebra. Then by what you proved above, $k \hookrightarrow B/m$ is a finite ring map. However, as $k$ is algebraically closed, this means that the ring map must be an isomorphism. Let $a_i \in k$ be those elements that are mapped to $x_i + m$ in $B/m$. Then it follows that $x_i - a_i \in m$, which implies that $(x_1 - a_1, ..., x_n - a_n) \subset m$. But, $(x_1 - a_1, ..., x_n - a_n)$ is a maximal ideal, so we must have $(x_1 - a_1, ..., x_n - a_n) = m$. What we have just proved is usually referred to as Hilbert's Weak Nullstellensatz (or so I think). You can use the Weak Nullstellensatz to prove the Strong Nullstellensatz (a reference would be Mumford's Red Book), which says that following: For an algebraically closed field $k$, given an ideal $\alpha \subset k[x_1,...,x_n]$, we have $I(V(\alpha)) = \sqrt{\alpha}$, where \begin{equation} V(\alpha) = \{\vec{a} \in k^n: \forall f \in \alpha, f(\vec{a}) = 0 \} \end{equation} and \begin{equation} I(Z) = \{f \in k[x_1,...,x_n]:\forall \vec{a} \in Z, f(\vec{a}) = 0\} \end{equation} for $Z \subset k^n$. You can now see what this has to do with the zeroes of polynomials. As a side note, the Strong Nullstellensatz also has an equivalent algebraic formulation that amounts to proving that $k[x_1,...,x_n]$ is a Jacobson ring, that is a ring where any radical ideal is the intersection of the maximal ideals containing it. Hope this helps, and let me know if you are confused by anything. $\textbf{Later edit:}$ I should mention that the Weak Nullstellensatz is also a very geometric statement. It clearly implies that the elements of any proper ideal of $k[x_1,...,x_n]$ have a common zero. Also, while I like the proof of the statement that I refer to as the Strong Nullstellensatz using Rabinowitsch's Trick, I find it difficult to remember the trick. Here is an alternate proof to the algebraic version of the Strong Nullstellensatz using techniques that I am sure you are familiar with, since you are reading chapter $5$ of A-M. $\textbf{The Nullstellensatz:}$ Let $k$ be any field, not necessarily algebraically closed, and let $A$ be a nonzero finitely generated $k$ algebra. Then we have the following: (a) $\sqrt{(0)} \subset A$ is the intersection of all maximal ideals containing it. (b) Given an ideal $I \subset A$, $\sqrt{I}$ is the intersection of all maximal ideals containing it. $\textbf{Proof:}$ (a) I will let $M$ denote the set of maximal ideals of $k[x_1,...,x_n]$. Since $\sqrt{(0)}$ is the intersection of all prime ideals containing it, it follows that $\sqrt{(0)} \subset \cap_{m \in M} m$. Let $f$ be an element contained in every maximal ideal, and suppose that $f \notin \sqrt{(0)}$. Then $A_f$ is a nonzero ring, hence contains a maximal ideal $m$. Note that $A_f$ is also a finitely generated $k$ algebra (why?), and as a result, so is $A_f/m$. By Zariski's Lemma, this means that $A_f/m$ is a finite extension of $k$. Let $p: A \rightarrow A_f/m$ be the standard map. Then we have the inclusions $k \subset A/p^{-1}(m) \subset A_f/m$. Since $A_f/m$ is integral over $k$, it follows that $A/p^{-1}(m)$ is also integral over $k$. But $A/p^{-1}(m)$ is a domain, so it must be a field. It follows that $p^{-1}(m)$ is a maximal ideal of $A$, and by construction $f \notin p^{-1}(m)$, which contradicts our choice of $f$. This proves (a). (b) follows from (a) by passing to the quotient $A/I$ (do you see why?). $\blacksquare$ As an exercise try to show that $I(V(\alpha)) = \sqrt{\alpha}$ over an algebraically closed field $k$, using the above theorem.
H: homeomorphism of a subset of $GL_3(\mathbb{R})$ with $GL_2(\mathbb{R})$ and connectedness Suppose I denote $G_3$ be the set of all $3\times 3$ matrices with positive determinant, and consider the map $\pi:G_3\rightarrow \mathbb{R}^3\setminus\{0\}$ define by $\pi(g)=ge_1$ where $e_1=(1,0,0)$ then i want to know about the set $\{g:ge_1=e_1\}$, is it connected? is this set homeomorphic to $G_2\times\mathbb{R}^2$? well, is the map surjective submersion? AI: Your set is homeomorphic to $G_2\times \mathbb{R}^2$ (and hence connected). Notice that $ge_1 = e_1$ emplies that the first column of $g$ is $e_1$, but the other entries are free, modulo the fact that the bottom right $2\times 2$ matrix must have positive determinant. Then a map from your set to $G_2\times \mathbb{R}^2$ is given by mapping the bottom $2\times 2$ matrix to $G_2$ and the top middle and top right entries to $\mathbb{R}^2$. In the "usual" topology on $G_3$ (obtained by viewing it as an open subset of $\mathbb{R}^9$), this map is easily seen to be a homeomorphism. Further, the map is surjective. To see this, pick any $v\in \mathbb{R}^3-\{0\}$ and extend it to a positively oriented basis of $\mathbb{R}^3$. Then there is a unique linear transformation mapping map $e_1$ to $v$, and $e_i$ to the other basis vectors you chose. Then this transformation is in $G_3$ and sends $e_1$ to $v$. (I've run out of time, but I'll think and post on the submersion question later if no one else does)
H: Integral-Summation inequality. The following question was in an entrance exam: Show that, if $n\gt0$, then: $$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{2}{n^2{\rm e}}$$ You are allowed to assume $\lim_{x\to\infty}{\frac{\ln{x}}{x}}=0$. Hence explain why, if $1\lt a\lt b$, then: $$\int_{b}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}\lt\int_{a}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$$ Deduce that: $$\sum_{n=1}^{N}{\frac{1}{n^{2}}}<\frac{\rm e}{2}\int_{{\rm e}^{1/N}}^{\infty}{\left(\frac{1-x^{-N}}{x^{2}-x}\right)\ln{x}\:dx},$$ Where $N\in\mathbb{N}:N\gt1$. The first part I believe I integrate by parts, such that: $$\int{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{1}{n}\int{x^{-n-1}\:dx}-\frac{x^{-n}\ln{x}}{n}$$ Clearly $\int{x^{-n-1}\:dx}=-\frac{x^{-n}}{n}$, so we have: $$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\left.\frac{x^{-n}(n\ln{x}+1)}{n^{2}}\right\|_{{\rm e}^{1/n}}^{\infty}=\frac{{\rm e}^{-\frac{n}{n}}(\frac{n}{n}+1)}{n^{2}}-0=\frac{2}{n^{2}{\rm e}}$$ As required. To prove the next inequality, all that is required is to demonstrate that $\left.\frac{\ln{x}}{x^{n+1}}\right\|_{x=1}\geq0$, $\lim_{x\to\infty}{\frac{\ln{x}}{x^{n+1}}}\geq0$, and $\left(\frac{\ln{x}}{x^{n+1}}\right)'\neq0$, $\forall x\in[1,\infty)$ and $\forall n\gt0$. The first inequality is verified simply by observing that $\ln{1}=0$, therefore $\frac{\ln{1}}{1^{n+1}}=0$, $\forall n$.The second inequality can be written as: $$\lim_{x\to\infty}{\left(\frac{1}{x^{n}}\frac{\ln{x}}{x}\right)},$$ And as we know $\lim_{x\to\infty}{\frac{\ln{x}}{x}}=0$, and $\lim_{x\to\infty}{\frac{1}{x^{n}}}=0$, the second inequality must also be true. To verify the second inequality we simply differentiate using the quotient rule and look for critical points: $$\frac{d}{dx}{\left(\frac{\ln{x}}{x^{n+1}}\right)}=\frac{x^{n}-(n+1)x^{n}\ln{x}}{x^{2n+2}}=\frac{1-(n+1)\ln{x}}{x^{n+2}}$$ As $\ln{x}$ is a monotonically increasing function, and at $x=1$, $\ln{x}=0$, we can show that $\forall n\gt 0$, and $x\gt1$, $\left(\frac{\ln{x}}{x^{n+1}}\right)\leq 0$, therefore, the function is positive for all values of $x\gt 0$, which means by the fundamental theorem of calculus that if $1\lt a\lt b$, then $\forall n\gt0$: $$\int_{b}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}<\int_{a}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$$ As required. However, I am stuck for the final part of the question. My first thought was that as $\frac{1}{n^{2}}$ is monotonically decreasing, $\forall n\gt0$; any integral performed over the region $(0,\infty)$ will have a positive error term, therefore, we can replace the summation with $\int_{1}^{N}{\frac{1}{n^2}\:dn}=\left.-\frac{1}{n}\right\|_{1}^{N}=-\frac{1}{N}+1$. I am not sure how to perform the second integral, however. Thanks in advance! AI: The equation $$\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{2}{n^2{\rm e}}$$ can be rewritten as $$\frac{1}{n^2}=\frac{\rm e}2\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}$$ Now, simply sum this for $n=1,2,\ldots,N$ and get: $$\begin{align}\sum_{n=1}^N\frac{1}{n^2}=&\sum_{n=1}^N\frac{\rm e}2\int_{{\rm e}^{1/n}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}<\sum_{n=1}^N\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}{\frac{\ln{x}}{x^{n+1}}\:dx}=\\=&\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}\sum_{n=1}^N{\frac{\ln{x}}{x^{n+1}}\:dx}=\frac{\rm e}2\int_{{\rm e}^{1/N}}^{\infty}\ln{x}\sum_{n=1}^N{\frac{1}{x^{n+1}}\:dx}\end{align}$$ (To get the inequality, we used that the integrand is positive on $(1,\infty)$, which shows that $\int_{{\rm e}^{1/n}}^{\infty}\leq\int_{{\rm e}^{1/N}}^{\infty}$ for $N\geq n$ and equality holds if and only if $N=n$.) The sum here is just the sum of a finite geometric progression. Thus, we have $$\sum_{n=1}^N{\frac{1}{x^{n+1}}}=\sum_{n=1}^N{(x^{-1})^{n+1}}=x^{-2}\sum_{n=1}^{N}{(x^{-1})^{n-1}}=x^{-2}(\frac{(x^{-1})^N-1}{x^{-1}-1})=\frac{x^{-N}-1}{x-x^2}$$ This gives us $$\sum_{n=1}^{N}{\frac{1}{n^{2}}}<\frac{\rm e}{2}\int_{{\rm e}^{1/N}}^{\infty}{\left(\frac{1-x^{-N}}{x^{2}-x}\right)\ln{x}\:dx},$$ which is exactly what we wanted.
H: Are there an infinite set of sets that only have one element in common with each other? In a card game called Dobble, there are 55 cards, each containing 8 symbols. For each group of two cards, there is only one symbol in common. (The goal of the game being to spot it faster than the other players, which is not the point of my question). If I translate that to mathematical language, I would say that: $S = [S_1, S_2, ..., S_{55}]$. $S_n = [n_1, n_2, ..., n_8]$. For $S_n, S_m \in S$ there is one and only one $n_a = m_b$ My double (dobble) question is: Are there a finite or infinite number of sets and elements that allows such a property? I know there is one more with 30 sets containing 6 elements each (because of Dobble Kids, a lighter version of the game). How can I calculate the number of sets, the number of elements in the sets, how many different elements there are in all the sets and which elements go in which sets? Is there a formula or is it simply a step-by-step try and fail method? EDIT I realise that having sets like {1, 2, 3, 4}, {1, 5, 6, 7}, {1, 8, 9, 10}, ... answers the question (with 1 being the only element in common in each set). There is one more restriction: Each element used in the sets must appear the same number of times (for example, in 7 given sets). In the game, there are 50 symbols altogether. (55 cards, 8 symbols per card, 50 symbols altogether). I have figured out a simple example with 4 sets of 3 elements (6 elements overall): $$S_1 = [1, 2, 3], S_2 = [1, 4, 5], S_3 = [2, 5, 6], S_4 = [3, 4, 6]$$ Each element is present twice. AI: This is not a complete explanation, but a summing up of my observations in the comments above. Consider the projective plane ${\Bbb P}^2({\Bbb F}_q)$ over the field with $q$ elements ($q$ must be of the form $q=p^f$ for some prime $p$). Then the following facts follow more or less trivially from the definitions: ${\Bbb P}^2({\Bbb F}_q)$ consists of $1+q+q^2$ points; ${\Bbb P}^2({\Bbb F}_q)$ contains $1+q+q^2$ lines, each of them containing $q+1$ points; every two lines meet at a single point; every point is contained in exactly $1+q$ lines. Thus, if we call "symbols" the points, and "cards" the lines we have a situation which is exactly thatdescribed in the question. The problem is that the numeric data do not correspond: if we take $q=7$ so to match the $8$ symbols in each card, the number of cards and of symbols should be $1+7+7^2=57$. Then, either you lost 2 cards [ :-) ], or I'm left clueless.
H: Transition Kernel of a Markov Chain Supposing $X_t$ is a Markov Process, can the transition kernel be defined by $$K_t(x,A):= P(X_{t+1} \in A \| X_t = x)?$$ Assume that $X_t : \Omega \to \mathbb{R}^n$. The issue is that under the normal definition of conditional probability, r.h.s is defined as $$P(X_{t+1} \in A \| X_t = x) =\frac{P( (X_{t+1} \in A) \cap (X_t = x))}{P(X_t = x)}$$ and the denominator is zero for most random variables. Even if this is assumed to be $E[I_A(X_{t+1}) \| \sigma(X_t = x)]$, the conditional expectation can take arbitrary values on the set $\{X_t =x\}$ if $P(X_t = x) =0$. Another definition I could gather from the web is that $K_t(x,A)$ is called a transition kernel if $$K_t(X_t(\omega),A) := E[I_A(X_{t+1})\|X_t](\omega)~\forall \omega \in \Omega.$$ Also, $K_t(x,\cdot)$ should be a probability measure so that the conditional expectations should be regular (if I am not wrong). The book (page 18) I am reading uses the first definition given above. Thanks for the help. AI: The transition kernel $K_t$ is defined by some measurability conditions and by the fact that, for every measurable Borel set $A$ and every (bounded) measurable function $u$, $$ \mathrm E(u(X_t):X_{t+1}\in A)=\mathrm E(u(X_t)K_t(X_t,A)). $$ Hence, each $K_t(\cdot,A)$ is defined only up to sets of measure zero for the distribution of $X_t$, in the following sense: if $K_t$ is a transition kernel for $X_t$ and if, for every measurable Borel set $A$, $X_t$ is almost surely in $C_A$, where $$ C_A=\{x\in\mathbb R^n\,\mid\,K_t(x,A)=\tilde K_t(x,A)\}, $$ then $\tilde K_t$ is also a transition kernel for $X_t$.
H: Numerical Methods for Linear Matrix Equation How can I solve (numerically) the linear equation $AB=0$. where $A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times m}$? How much is the computational cost? AI: You may treat $B$ one column at a time: \begin{equation} B = \Bigg[b_1\;\;b_2\;\;\ldots\;\;b_m\Bigg] \end{equation} where the $b_i$ are column vectors of length $n$ (and there are $m$ of them). If $A$ is invertible, it spans $\mathbb{R}^n$ and the only solution is $B=0$. Otherwise you are looking for vectors $b_i$ in the nullspace of $A$. Suppose $A$ has rank $r$ (number of independent columns). Then dim N($A$) = n-r. This means that there are $n-r$ independent vectors that give $Ax=0$; these x's form a basis for N($A$). So every matrix $B$ whose columns are linear combinations of these x's will give $AB=0$. As for computational cost, the effort is in finding a basis for N($A$). Usually this is an operation of order $O(n^2)$. Reducing $A$ to it's echelon form, the columns which do not look like the identity matrix will contain the basis for the nullspace (in MATLAB this is rref(A) and in Mathematica it is RowReduce[A]).
H: Closure of the set of all polynomial with variable $x\in [0,1]$ Let ${P}$ denote the set of all polynomial with variable $x\in [0,1]$, I need to know what is the closure of ${P}$ in $C[0,1]$? Well, Stone-Weierstrass theorem says: If $f\in C[0,1]$ then there exists a sequence of polynomials $p_n(x)$ which converges uniformly to $f$. So can I say that $closure{P}=C[0,1]$? AI: Yes. The Stone-Weierstrass approximation theorem tells you that $P$ is dense in $C[0,1]$ with respect to $\\|\cdot\\|_\infty$ and a set $D$ is dense in a set $S$ if the closure of $D$ equals $S$. (by definition)
H: What is the Least Common Multiple of $(a-b)$ and $(b-a)$? What is the Least Common Multiple of $(a-b)$ and $(b-a)$? The question is simple. What's the answer? I'm an Engineering student, but looks like I forgot my basics. AI: $\|a-b\|$ is the LCM of $(a-b), (b-a).$ Recall LCM is the smallest positive integer that is integer multiple of both $(a-b)$ and $(b-a).$ We have $$ \|a-b\| = +1\times(a-b) \\ \|a-b\| = -1\times(b-a).$$ (Of course, the signs above assuming $a > b.$ Flip the signs if $b < a.$) So $\|a-b\|$ is an integer multiple of both $(a-b), (b-a).$ To prove it's the smallest, assume $\ell < \|a-b\|$ is the LCM. Then $$ \ell = k(a-b) = -k(b-a)$$ for some integer $k.$ (again with the convenient choice $a > b.$) Comparing with the equations above, we can conclude that $k < 1$ and simultaneously $-k < -1 \implies k > 1.$ So $k = 0,$ and hence $\ell$ is not an LCM. Contradiction. There exists no non-zero common multiple $< \|a-b\|.$
H: Sequence of continuous functions with bounded derivative Let $f_n$ be a sequence of continuous functions on $[0,1]$, and continuously differentiable on $(0,1)$. Assume $\|f_n\|\le 1$ and $f_n'\le 1$ $\forall x\in [0,1]$ and $n$. Then $f_n$ is a convergent sequence in $C[0,1]$ $f_n$ has a convergent subsequence in $C[0,1]$ well, by Bolzano-Weirstrass theorem (every bounded sequence has a convergent subsequence) we can say $2$ is correct, I am not able to say true or false against $1$, please help. AI: Consider the case $f_n(x)=(-1)^n$ for all integer $n$ and all $x\in [0,1]$. Take $f_n(x):=-x^n$. Then $\|f_n(x)\|\leq 1$ and $f'_n(x)=-nx^{n-1}\leq 0\leq 1$. We have that $f_n$ converges pointwise to the function which is $0$ in $[0,1)$ and $-1$ at $1$. A uniformly converging subsequence would converge to this map, which is not possible. However, if we replace the condition "$\forall x\in[0,1], f'_n(x)\leq 1$" by "$\forall x\in[0,1], \|f'_n(x)\|\leq 1$", Arzelà-Ascoli theorem applies.
H: An inequality of red and black balls We have a box containing red and black balls. If we draw two at random the probability of getting both of them red is $1/2$. Which basically means: \begin{equation} \frac{r}{r+b} \cdot \frac{r-1}{r+b-1} = \frac{1}{2} \end{equation} Then we have for a positive number of red and black balls, $r$ and $b$ respectively: \begin{equation} \frac{r}{r+b} > \frac{r-1}{r+b-1} \end{equation} and the following inequality follows: \begin{equation} \left(\frac{r}{r+b}\right)^2 > \frac{1}{2}>\left(\frac{r-1}{r+b-1}\right)^2 \end{equation} How can I derive this inequality? AI: If $x$ and $y$ are distinct positive numbers with $xy = z$, then one of $x$ and $y$ must be bigger than $\sqrt z$ and one must be smaller. If both $x$ and $y$ were bigger than $\sqrt z$, their product would be bigger than $z$, and if both were smaller, their product would be smaller. So if you have $x>y$, then you must have $x > \sqrt z > y$, and so $x^2 > z > y^2$. Here you have $x={r\over r+b}$, $y = {r-1 \over r+b-1}$ and $z=\frac12$.
H: Explain why $E(X) = \int_0^\infty (1-F_X (t)) \, dt$ for every nonnegative random variable $X$ Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show, $$E(X) = \int_0^\infty (1-F_X (t)) \, dt$$ when $X$ has : a) a discrete distribution, b) a continuous distribution. I assumed that for the case of a continuous distribution, since $F_X (t) = \mathbb{P}(X\leq t)$, then $1-F_X (t) = 1- \mathbb{P}(X\leq t) = \mathbb{P}(X> t)$. Although how useful integrating that is, I really have no idea. AI: For every nonnegative random variable $X$, whether discrete or continuous or a mix of these, $$ X=\int_0^X\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\gt t}\,\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\geqslant t}\,\mathrm dt, $$ hence, by applying Tonelli's Theorem, $$ \mathrm E(X)=\int_0^{+\infty}\mathrm P(X\gt t)\,\mathrm dt=\int_0^{+\infty}\mathrm P(X\geqslant t)\,\mathrm dt. $$ Likewise, for every $p>0$, $$ X^p=\int_0^Xp\,t^{p-1}\,\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\gt t}\,p\,t^{p-1}\,\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\geqslant t}\,p\,t^{p-1}\,\mathrm dt, $$ hence $$ \mathrm E(X^p)=\int_0^{+\infty}p\,t^{p-1}\,\mathrm P(X\gt t)\,\mathrm dt=\int_0^{+\infty}p\,t^{p-1}\,\mathrm P(X\geqslant t)\,\mathrm dt. $$
H: Equation involving an integral depending on two parameters I have some difficulty to find possible solutions of the following equation: $$\int_0^\tau dx \frac{1}{x^\alpha+1}=\beta$$ where $\tau \gt 0,$ $\alpha\in \mathbb N$ ($\alpha=1,2,3,\dots$) and $\beta$ a given real valued constant. Is it possible to find values of $\alpha$ and $\tau$ satisfiyng the equation? Thanks AI: For every $\alpha$ in $[0,1]$, the function $g_\alpha:\tau\mapsto\int\limits_0^\tau\frac{\mathrm dx}{1+x^\alpha}$ is increasing from $g_\alpha(0)=0$ to $\lim\limits_{\tau\to+\infty}g_\alpha(\tau)=+\infty$. Hence, for each $\beta\geqslant0$, there exists a unique $\tau$ such that $g_\alpha(\tau)=\beta$. If $\alpha\gt1$, the same result holds provided $\beta\lt\ell_\alpha$, where $\ell_\alpha=\lim\limits_{\tau\to+\infty}g_\alpha(\tau)$ is finite.
H: Iterating the transform $(a,b)\mapsto(a+b+\sqrt{a^2+b^2} ,a+b-\sqrt{a^2+b^2})$ Assume that $a_0=-2$, $b_0=1$, and that, for every $n\ge0$, $$a_{n+1}=a_n+b_n+\sqrt{a^2_n+b^2_n} \qquad b_{n+1}=a_n+b_n-\sqrt{a^2_n+b^2_n}$$ How to find $a_{2012}$? AI: Here is a 7-steps plan: Stop asking questions with no indication whatsoever about which similar problems you can solve, what you tried to solve the present one, and why you think your attempts failed. Stop ignoring comments asking you to add the pieces of information mentioned in 1. Define $s_n=a_n+b_n$ and $p_n=a_nb_n$ and, for every $n\geqslant0$, express $s_{n+1}$ and $p_{n+1}$ in terms of $s_n$ and $p_n$. Compute $s_0$ and $p_0$. For every $n\geqslant1$, express $a_n$ and $b_n$ in terms of $s_n$ and $p_n$. Conclude. Memorize 1. and 2. for your next questions on the site.
H: Subgroups between $S_n$ and $S_{n+1}$ Lets look at $S_n$ as subgroup of $S_{n+1}$. How many subgroups $H$, $S_{n} \subseteq H \subseteq S_{n+1}$ there are ? AI: None. Let $S_n<H<S_{n+1}$ and suppose that $H$ contains a cycle $c$ involving $n+1$, say $$ c=(a,\cdots,b,n+1). $$ Then by composing to the left with a suitable permutation $\sigma\in S_n<H$ such that $\sigma(a)=b$ we have $$ \sigma c=\sigma^\prime (b,n+1) $$ where $\sigma^\prime(n+1)=(n+1)$, i.e. $\sigma^\prime\in H$. Thus the transposition $(b,n+1)$ is in $H$. Of course we may assume that $b=1$ and so all transpositions $(1,2)$, $(1,3)$, ..., $(1,n+1)$ are in $H$. These transpositions are known to generate $S_{n+1}$.
H: All valuations equal one : unit? Let $F$ be a global field with integers $o$, and let $x \in F$. Does $\|x\|_v =1$ for all non-archimedean valuations of $F$ imply that $x \in o^\times$. AI: Let $x\cal O$ the (possibly fractional) ideal generated by $x$. If it is non-trivial, i.e. $\cal O\neq x\cal O$, it must be a product of (possibly some inverted) prime ideals. The valuation of $x$ at those primes is not $1$. Thus.....
H: How can i compute the probability that a cloud of points was created by a probability distribution? I have a number of probability distributions that describe a number of points. like this: Now if i have draw a one point out of each distribution, i get a bunch of points randomly set on the 2-dimensional surface. How can i, given the set of probability distributions and a set of points compute the probability that i would draw something like that again? What is the correct mathematical term for what i want to compute? AI: If I understand your question right, then what you want is to develop a statistical test. Essentially, given some sampling of points $P = \left\{ p_1, p_2,\ldots,p_n\right\}$, you need to estimate the probability that this sampling of points resulted from some distribution. In this case, your null hypothesis is that $P$ was drawn from your presumed distribution. You will want to set a threshold for statistical significance. In other words, there is no concrete, deterministic way to distinguish between a set of points happening to randomly align with your chosen distribution by sheer luck, or whether they actually came from your distribution. There are many such tests; a basic example is a chi-squared test. Off-hand, I don't know which test is best for your application, but the Wikipedia page should direct you to some good information.
H: Short matrix algebra question $A$ is a square matrix, and there is a matrix $D$ such that $AD=I$. I need help to prove either (a) or (b) below: (a) Show that for each $\vec b$, $A\vec x=\vec b$ has a unique solution. OR (b) Show that $A$ is invertible. For (a), $AD\vec b=I\vec b = \vec b$, so obviously the given equation has at least one solution $\vec x =D\vec b$. But how to show that the solution is unique? For (b), I'm guessing we should show either that $DA=I$ or that $DA=AD$ (and thus $D=A^{-1}$), but I'm not sure how to do this without assuming that A is invertible, which is what we are needing to show. AI: Here is a proof which does not use group theory or vector spaces: I will establish equivalence of the following. The details are left for you to be filled in. Once you have established (b) by the sketch below, establishing (a) is trivial. The following are equivalent for any square matrix $A$: $A$ has a left inverse. $Ax = 0$ has only the trivial solution. The reduced row echelon form of $A$ is the identity matrix. $A$ is a product of elementary matrices. $A$ is invertible. A has a right inverse. $1\implies 2$: If $A$ has a left inverse $L$ then $Ax=0 \implies LAx=L0\implies Ix=x=0$. $2\implies 3$: The augmented matrix for $Ax=0$ must have been reduced to $Ix=0$ by Gauss Jordan elimination. $3\implies 4$: Since $E_1\cdots E_kA=I$ and each elementary matrix is invertible (and the inverse is also an elementary matrix) so $A=E_k^{-1}\cdots E_1^{-1}$. $4\implies 5$: Each elementary matrix is invertible. $5\implies 6$: Trivial. $6\implies 1$: If $R$ is the right inverse of $A$, then $A$ is the left inverse of $R$ and by $1\implies 5$ $R$ is invertible with inverse $A$, following which $R$ is the left inverse of $A$.
H: Proving an elementary integral inequality (in early Dirichlet space material) I'm reading up on Dirichlet spaces using this document here and, on page two, am stumped by a particular integral inequality. If this is trivial and/or I'm missing something blatant, I apologize. First we define, for any analytic function $f$ defined on the unit disk $\mathbb{D}$, the quantity $$ D(f) = \frac{1}{\pi} \int_{\mathbb{D}} \| f' \|^2 dA = \frac{1}{\pi} \int_{\mathbb{D}} \| f' \|^2 dxdy $$ (The Dirichlet space is the set of all functions $f$ as before such that $D(f)<\infty$. $D(f)$ is only a semi-norm, as $D(c)=0$ for any constant $c$.) Next we define, for any $\zeta \in \partial \mathbb{D}$, the quantity $$ L(f,\zeta) = \int_0^1 \|f'(r \zeta)\|dr $$ The author now says If $D(f)<\infty$ we can apply the Cauchy-Schwarz inequality to show that $$ \int_0^{\pi} L(f,e^{i\theta} ) d\theta \leq c D(f) < \infty $$ and so $L(f,e^{i\theta})<\infty$ for almost all $\theta$. This is what I can not show. Attempt: The C-S inequality reads $$ \left\| \int h(x) \overline{g(x)} dx \right\|^2 \leq \int \|h(x)\|^2dx \int \|g(x)\|^2dx $$ and applies because, by above, all functions in the Dirichlet space are square integrable. We are trying to show $$ \int_0^{\pi} \int_0^1 \|f'(r e^{i\theta})\|dr d\theta \leq c \frac{1}{\pi} \int_{\mathbb{D}} \| f' \|^2 dxdy $$ which, when we transfer to polars, gives $$ \int_0^{2\pi} \int_0^1 \|f'(r e^{i\theta})\|dr d\theta \leq A \int_0^{2\pi} \int_0^1 \|f'(re^{i\theta})\|^2rdrd\theta $$ So I defined $h(r) = \|f'(r e^{i\theta})\| \sqrt{r}$ to match C-S with what we have on the right, and then attempted so pick $g$ so that the left hand sides would match, too. This gives $g(r) = r^{-1/2}$. Subbing all this in we get $$ \left\| \int_0^1 \|f'(r e^{i\theta})\| \sqrt{r} \frac{1}{\sqrt{r}} dx \right\|^2 \leq \int_0^1 \|f'(r e^{i\theta})\|^2 rdr \int_0^1 \|r^{-1/2}\|^2dx $$ $$ \left\| \int_0^1 \|f'(r e^{i\theta})\| dx \right\|^2 \leq \int_0^1 \|f'(r e^{i\theta})\|^2 rdr \int_0^1 r^{-1}dr $$ When we integrate over $\theta$ we kind of get what we're looking for (the square shouldn't matter as everything on the right above is finite and all we're trying to show is that the left is finite). The problem is the term $$ \int_0^1 r^{-1}dr = + \infty $$ Any thoughts? Apologies again if this is silly. AI: Your approach is right, but you should integrate, say, from 1/2 to 1 (instead of from 0 to 1). The part within radius 1/2 is of no concern anyway since the derivative is bounded on compact subsets. And @Davide is right - the inequality is false as stated, due to wrong homogeneity.
H: pseudo numbers and surreal numbers A surreal number $\{x_L\\|x_R\} \in No$ is a number when for all $\xi\in x_L$ and all $\eta \in x_R$ we have $\eta > \xi$. All the things $\{x_L\\|x_R\}$ which are not of that form are called "pseudo-numbers" and usually ignored (except in certain game-theoretic applications). Questions: Is it possible (in an algebraic, analytic or topological sense) to make sense of these pseudo-numbers ? Can one obtain a "filling of the gaps" of $No$ by extending $No$ with pseudo-numbers ? Is there research in which the pseudo numbers are studied ? When you study $\mathbb{A}^{1,an}_{No}$ as the set of (bounded, non-archimedean) seminorms on $No[T]]$, is there a connection between pseudonumbers and the set $\mathbb{A}^{1,an}_{No} - No$, also known as the "hyperbolic space" $\mathbb{H}^{1,an}_{No}$ among those who a acquainted with the Berkovich terminology ? AI: Fleshed out to an answer since it got too big for a comment: 'things which are not of that form are called "pseudo-numbers" and usually ignored.' 'Is there research in which the pseudo-numbers are studied?' These two statements seem rather at odds with each other, but yes — as Ted's answer notes, the subject you're looking for is Combinatorial Game Theory and it devotes quite a bit of study to these objects. While the order on games is only a partial order, there are games that can be compared with all numbers and do fall into the gaps you're speaking about - for instance, the game $\uparrow = \{0\|\{0\|0\}\}$ can be shown to be larger than zero but smaller than any positive number, even infinitesimal ones (this is a nice exercise!). Note that even after plugging in games that can be ordered with respect to all numbers, there are still gaps that can be filled; this is usually done by taking $X_L$ and $X_R$ as proper classes rather than sets, although another tricky appraoch that shows up in Winning Ways is to give up the axiom of foundation (in their notation, go beyond the class of enders) and consider, e.g., games like $\mathbf{On}=\{\mathbf{On}\|\}$ (which can easily be seen to be larger than any 'proper' game in a suitable sense). I believe this is equivalent for all practical purposes to defining $\mathbf{On}=\{V\|\}$, where $V$ denotes 'the universe', but abandoning foundation allows games to be considered that couldn't be expressed otherwise. Unfortunately, I don't know anything about the topological properties of these objects - I wasn't even aware that the Surreals had gotten much study in a topological sense! That's a subject that I'd like to know quite a bit more about myself.
H: Calculating the minimum of $\cos x \sin y$ I am about to start university in October, to study computer science, and have been asked by my university to complete a number of problem sheets. I have become stuck on the following question, and therefore would appreciate any help possible. The numbers $x$ and $y$ are subject to the constraints $x+y=\pi$. Find the values of $x$ and $y$ for which $\cos x\sin y$ takes its minimum value. Using this question as a starting point towards a solution, I have the following steps attempted so far. \begin{align} \Lambda(x, y, \lambda)&=\cos x\sin y+\lambda(x+y-\pi)\\ \frac{\partial\Lambda}{\partial x}&=-\sin x\sin y+\lambda=0\\ \frac{\partial\Lambda}{\partial y}&=\cos x\cos y+\lambda=0\\ \frac{\partial\Lambda}{\partial\lambda}&=x+y-\pi=0\\ x&=\cos^{-1}\left(\frac{\lambda}{\sin y}\right)\\ y&=\cos^{-1}\left(\frac{-\lambda}{\cos x}\right)\\ \cos^{-1}\left(\frac{\lambda}{\sin y}\right)+\cos^{-1}\left(\frac{-\lambda}{\cos x}\right)&=\pi\\ \end{align} Unfortunately, such maths is way beyond my abilities, as I have only studied A-Level Maths and Further Maths, and I am working from the first answer in the referenced question and the Wikipedia articles on Lagrange Multipliers and Partial Derivatives. I'm unsure of the correct tags to apply, so any help there would also be wonderful. Edit: After receiving a number of hints, this is part of my solution, however, I'm not sure on how to properly phrase the last bit of the question with respect to properly solving the inequality or expressing values for $y$. \begin{align} \sin x\cos y&=\sin x\cos(\pi-x)\\ &=-\sin x\cos x\\ &=\sin x\cos x\\ &=\frac12\sin2x\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac12\sin2x\right)&=\cos2x\\ \cos2x&=0\Rightarrow x=\frac12\left(n\pi-\frac\pi2\right),n\in\mathbb{Z}\\ \frac{\mathrm{d}}{\mathrm{d}x}\cos2x&=-2\sin2x\\ -2\sin2x>0&\Rightarrow\sin2x<0\\ \end{align} AI: You've made this unnecessarily complicated. Note that the constraint $x+y=\pi$ means that you can substitute $y=\pi-x$ into your original expression. Thus, you need only minimize the $1$-variable function $\cos(x)\sin(\pi-x)$, which can be achieved through methods of basic calculus. Edit: One can make this even simpler by using the identities $\sin(\pi-x)=\sin(x)$ and $2\cos(x)\sin(x)=\sin(2x)$.
H: proving convergence for a sequence defined recursively The sequence $\left \{ a_{n} \right \}$ is defined by the following recurrence relation: $$a_{0}=1$$ and $$a_{n}=1+\frac{1}{1+a_{n-1}}$$ for all $n\geq 1$ Part 1)- Prove that $a_{n}\geq 1$ for all $n\geq 0$ Part2)- Prove that the sequence $\left \{ a_{n} \right \}$ converges to some real number $x$, and then calculate $x$ For the first part, I could prove it using induction. For the second part: The problem is how to prove that this sequence is convergent. Once the convergence is proved, then from the recurrence relation we can deduce that $x=\sqrt{2}$. In order to prove it is convergent, I tried to see how this sequence converges to $x$. I calculated the terms $a_{0}$, $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$. I can see that the sequence is neither decreasing nor increasing, so the monotone convergence theorem cannot be applied. I can see that the distance between two consecutive terms is getting smaller and smaller, so I tried to prove that this sequence is contractive. $\left \| a_{n+1} -a_{n}\right \|=\frac{1}{\left \| 1-a_{n} \right \|\left \| 1+a_{n} \right \|}\left \| a_{n}-a_{n-1} \right \|$, and obviously, $\frac{1}{\left \| 1+a_{n} \right \|}\leq \frac{1}{2}$. I need to prove that $\frac{1}{\left \| 1-a_{n} \right \|}\leq \alpha $ where $0< \frac{\alpha }{2}< 1$, and hence the sequence is contractive and therefore it is convergent. If you have any idea how to prove $\frac{1}{\left \| 1-a_{n} \right \|}\leq \alpha $ or any other idea please share... AI: Hint: Find $c$ such that $$ \frac{a_{n+1}-\sqrt2}{a_{n+1}+\sqrt2}=c\,\frac{a_{n}-\sqrt2}{a_{n}+\sqrt2}. $$
H: Proof of Hilbert's Nullstellensatz I'm working through my notes and I'm stuck in the middle of the proof of Hilbert's Nullstellensatz. (Hilbert's Nullstellensatz) Let $k$ be an algebraically closed field. Let $J$ be an ideal in $k[x_1, \dots , x_n]$. Let $V(J)$ denote the set of $x$ in $k$ such that all $f$ in $J$ vanish on them. Let $U \subset k^n$ and let $I(U)$ denote the set of $f$ in $k[x_1, \dots , x_n]$ that vanish on $U$. Then $$ r(J) = I(V(J))$$ where $r$ denotes the radical of $J$. Let me go through the proof as far as I understand it: $\subset$: Easy. Let $p \in r(J)$. Then $p^k \in J$ which means $p(x)^k = 0$ for all $x$ in $V(J)$. Hence $p(x) = 0$ for $x$ in $V(J)$ hence $p \in I(V(J))$. $\supset$: Assume $f \notin r(J)$. Then for all $k>0$, $f^k \notin J$. We know that there exists a prime ideal $p$ such that $J \subset p$ and $f^k \notin p$ for all $k>0$. To see this we use the same argument used in the proof of proposition 1.8. on page 5 in Atiyah-MacDonald: Let $\Sigma$ be the set of all ideals that do not contain any power of $f$. We order $\Sigma$ by inclusion and use Zorn's lemma to get a maximal element $p$. We claim $p$ is prime. Assume neither $x \notin p$ nor $y \notin p$ (then we want to show $xy \notin p$). Then $p + (x), p + (y)$ are ideals properly containing $p$ hence neither of them is in $\Sigma$ hence $f^n \in p + (x)$ and $f^m \in p + (y)$. Now $f^{n+m} \in (p + (x)) (p + (y)) = p^2 + (x)\cdot p + (y)\cdot p + (xy) \subset p + (xy)$ so $p + (xy) \notin \Sigma$. Hence $p$ is properly contained in $p + (xy)$ hence $xy$ cannot lie in $p$. So we have $p$ is a prime ideal containing $J$. Now consider the map $$ k[x_1, \dots, x_n] \xrightarrow{\pi_1} k[x_1, \dots, x_n]/p \xrightarrow{i} (k[x_1, \dots, x_n]/p) [\overline{f}^{-1}] =: B([\overline{f}^{-1}]) \xrightarrow{\pi_2} B[\overline{f}^{-1}] /m$$ where $\overline{f}$ denotes $\pi_1 (f)$ and $m$ is some maximal ideal in $B[\overline{f}^{-1}]$. We may assume $f \neq 0$ so that $\overline{f} \neq \overline{0}$. $\overline{f}^{-1}$ is an element of the field of fractions of $B$ so we may adjoin it to $B$ to get a new ring. Since we only adjoined one element and otherwise only took quotients, the thing coming out on the RHS is a finitely generated $k$-algebra (because $ k[x_1, \dots, x_n]$ is). Now by theorem 5.24 in Atiyah-MacDonald we know that $k \cong B[\overline{f}^{-1}] /m$. The proof now finishes as follows: "Let $t_1, \dots, t_n$ denote the images of $x_1 , \dots, x_n$ under this composite ring homomorphism. ()By construction, $g \in J \implies g \in p \implies g(t_1, \dots, t_n) = 0 \implies (t_1 , \dots, t_n ) \in V(J)$. ()On the other hand, $f(t_1 , \dots, t_n )$ is precisely the image of $f$ in $B[\overline{f}^{-1}] /m$, which is a unit. $\implies f(t_1 , \dots, t_n ) \neq 0 \implies f \notin I(V(J))$." Question 1: What is the line () showing? I think we want to show $f \notin I(V(J))$, where does this come in here? Question 2: Why is $f(t_1 , \dots, t_n )$ a unit? Thank you for your help. AI: This is a comment made into an answer. Suppose $f$ is any polynomial not in $r(J)$. Question 2 $f(t_1,\dots,t_n)$ is a unit because $f$ is (by construction) invertible in $B[f^{-1}]$ and remains so in the homomorphic image $B[f^{-1}]/m$. But the image of $f$ is precisely $f(t_1,\dots,t_n)$. Question 1 The point of $()$ is that there is one distinguished point $(t_1,\dots,t_n)$ in $B[f^{-1}]/m×⋯×B[f^{-1}]/m≃k^n$ that kills all the polynomials in $J$. Put another way, $$\mathrm{the~point~ of~}()\mathrm{~ is~ that~}(t_1,\dots,t_n)\in V(J)$$ and thus all the polynomials in $I(V(J))$ must be killed by it. By $()$, $f$ does not take this point to $0$ and so $$\mathrm{the~point~ of~}()\mathrm{~ is~ that~}f\notin I(V(J)).$$ By now the authors have shown that $f\notin r(J)\Rightarrow f\notin I(V(J))$ i.e. $k[x_1,\dots,x_n]\setminus r(J)\subset k[x_1,\dots,x_n]\setminus I(V(J))$ i.e. $$I(V(J))\subset r(J).$$
H: Algebraic Topology pamphlets? I'm looking to self-learn some Algebraic Topology and have found the books I've looked at so far (ie. Hatcher) to be rather tome-like for my tastes. Does anyone know of a good slim lecture notes style book (or, indeed, an actual set of lecture notes) I can have a look at? I need to catch up on the Cambridge part II course (I foolishly dropped it about half way through before realising the following term that I actually quite liked Geometry) for which the syllabus is as follows The fundamental group Homotopy of continuous functions and homotopy equivalence between topological spaces. The fundamental group of a space, homomorphisms induced by maps of spaces, change of base point, invariance under homotopy equivalence. Covering spaces Covering spaces and covering maps. Path-lifting and homotopy-lifting properties, and ther application to the calculation of fundamental groups. The fundamental group of the circle; topological proof of the fundamental theorem of algebra. Construction of the universal covering of a path-connected, locally simply connected space. The correspondence between connected coverings of X and conjugacy classes of subgroups of the fundamental group of X. The Seifert–Van Kampen theorem Free groups, generators and relations for groups, free products with amalgamation. Statement and proof of the Seifert–Van Kampen theorem. Applications to the calculation of fundamental groups. Simplicial complexes Finite simplicial complexes and subdivisions; the simplicial approximation theorem. Homology Simplicial homology, the homology groups of a simplex and its boundary. Functorial properties for simplicial maps. Proof of functoriality for continuous maps, and of homotopy invariance. Homology calculations The homology groups of Sn, applications including Brouwer’s fixed-point theorem. The Mayer-Vietoris theorem. Sketch of the classification of closed combinatorical surfaces; determination of their homology groups. Rational homology groups; the Euler–Poincar´e characteristic and the Lefschetz fixed-point theorem AI: http://tartarus.org/gareth/maths/notes/ has the notes from our year.
H: Why is $B[x]/M$ algebraic over $B/m$? Let $B$ be a subring of some field $K$, $x$ some element in $K$, $m$ a maximal ideal in $B$ and $m[x]$ the extension of $m$ in $B[x]$ and $M$ a maximal ideal in $B[x]$ such that $m[x] \subset M $ and $M \cap B = m$. Why is $B[x]/M$ algebraic over $B/m$? Thank you. AI: The $(B/\mathfrak m)$- algebra $B[x]/M$ is finitely generated (by $\bar x$) and is a field. Hence by Zariski's lemma it is finite-dimensional and a fortiori algebraic over $B/\mathfrak m$.
H: proving existence of a sequence such that the limit exists? Can anyone prove the existence of a sequence $(n_{k})_{k\in \mathbb{N}}$ of distinct positive integers such that the limit: $\lim_{k\rightarrow \infty }\sin(n_{k})$ exists in $\mathbb{R}$ I can definitely construct a sequence $(n_{k})_{k\in \mathbb{N}}$ such that $\frac{1}{2}\leq \sin(n_{k})\leq 1$, but this doesn't imply that this sequence is convergent. Any suggestions? AI: Every bounded sequence has convergent subsequence, see Bolzano-Weierstrass theorem. If you apply this to the sequence $(\sin n)_{n=0}^\infty$, you get the desired result. (Or you can mimic the proof of Bolzano-Weierstrass theorem, if you prefer.)
H: Which of the following are compact sets? Which of the following are compact sets? $\{\operatorname{trace}(A): A \text{ is real orthogonal}\}$ $\{A\in M_n(\mathbb{R}):\text{ eigenvalues $\|\lambda\|\le 2$}\}$ Well, orthogonal matrices are compact, but the trace of them may be any $x\in\mathbb{R}$, so I guess 1 is non compact. Let $x$ be an eigenvector corresponding to the eigenvalue $\lambda$; then $Ax=\lambda x$, then $\\|Ax\\|= \|\lambda\|\cdot\\|x\\|\le \\|A\\|\cdot\\|x\\|$ so $\\|A\\|\ge 2$ so $2$ is also non compact as unbounded? AI: The map $$\operatorname{trace}\colon\mathcal M_n(\Bbb R)\to \Bbb R$$ is linear, and from a finite dimensional vector space, hence continuous. Such mapping map compact sets to compact one, and the orthogonal group is compact, hence the first set is compact. The second set is not bounded. The matrices $A_N:=\pmatrix{0&0&\dots&0&N\\ 0&0&\dots&0&0\\ \vdots&\vdots&&\vdots&\vdots\\ 0&0&\dots&0&0}$ is in the second set, but the norm is $N$ for the norm subordinated to the supremum norm for example, is $N$. The only eigenvalue of $A_N$ is $0$ and $\{A_N,N\geq 1\}$ is not bounded hence cannot be compact. Note that we can take any norm we want, since $\mathcal M_n(\Bbb R)$ is finite-dimensional, and the choice of $2$ in the text of the exercise is not important (we can replace it by $M\geq 0$).
H: Under which conditions does $a^n \equiv 1\mod(b) \Rightarrow\ a^{n^m} \equiv 1\mod(b) $? Can you prove it? Under which conditions does $a^n \equiv 1\mod(b) \Rightarrow\ a^{n^m} \equiv 1\mod(b) $? What about viceversa? What is the strongest result(s) that can be proved regarding this kind of thing? I'm kind of getting stuck with following a passage in an example, and anyway I have a general confusion in my head about this which I haven't been able to dispel with googling, searching and grepping, hence asking as my last resort. Thanks a lot. AI: For any $m\geq 1$, if $a^n\equiv 1\bmod b$, then $$a^{(n^m)}=a^{n\cdot (n^{m-1})}=(a^n)^{n^{m-1}}\equiv (1)^{n^{m-1}}\equiv 1\bmod b.$$
H: Prove that a monotonically increasing continuous function is invertible. Let $f:[a,b]\rightarrow[f(a),f(b)]$ be strictly increasing continuous function (i.e $x>y \implies f(x)>f(y)$). Prove that f is invertible. Proving that the function is one-to-one was simple enough. I need some guidance on proving it's onto. I'm new to $\epsilon-\delta$ proofs since I just started self-studying some metric spaces. AI: Let $y \in (f(a), f(b))$ and let $A = \{ x \in [a, b] : f(x) < y\}$. Since A is nonempty, and $A \subset [a, b]$, $s =\sup A$ exists. To show that $f(s) = y$, choose a sequence $x_n \in A$ that converges to $s$. Since we have $s \in [a, b]$, and by continuity, $$f(s) = \lim_{n \to \infty}f(x_n) \le y$$ By contradiction, suppose $f(s) < y$. Then $y - f(x)$ is continuous on $[a, b]$, and $y - f(s) > 0$. We can choose $\delta > 0$ and $u > s$ such that $y-f(u) > \delta$, but this implies that $u \in A$, a contradiction.
H: Units in number fields with complex embeddings Assume that we have an algebraic number field with integers $o$, and with a complex embedding $\iota$. What can be said about the image $\iota( o^\times)$ under $\iota$? Is it discrete? Is infinite? AI: As has been noted in comments, the important result here is Dirichlet's Unit Theorem. E.g., (taken from Daniel Marcus's Number Fields): Dirichlet's Unit Theorem. Let $U$ be the group of units in a number ring $\mathcal{O}_K = \mathbb{A}\cap K$ (where $\mathbb{A}$ represents the ring of all algebraic integers). Let $r$ and $2s$ denote the number of real and non-real embeddings of $K$ in $\mathbb{C}$. Then $U$ is the direct product $W\times V$, where $W$ is a finite cyclic group consisting of the roots of $1$ in $K$, and $V$ is a free abelian group of rank $r+s-1$. In particular, there is some set of $r+s-1$ units, $u_1,\ldots,u_{r+s-1}$ of $\mathcal{O}_K$, called a fundamental system of units, such that every element of $V$ is a product of the form $$u_1^{k_1}\cdots u_{r+s-1}^{k_{r+s-1}},\qquad k_i\in\mathbb{Z},$$ and the exponents are uniquely determined for a given element of $V$.
H: Is R with $j_d$ topology totally disconnected? Let consider the topological space $R_j=(\mathbb{R},j_r)$ where $j_r$ is generated by right side open intervals, i.e $[a,b)$ for $a,b \in \mathbb{R}$; note that this topology includes the euclidean topology. $R_j$ is not a connected space, because given $x \in \mathbb{R}$ we have that $(-\infty,x)$ and $[x,+\infty)$ provide a disconnection or $R_j$. Given the arbitrariness in choosing $x$, is it possible that $R_j$ is totally disconnected? Let consider a certain $y:x<y<+\infty$; I can progressively reduce the open interval containing $x$ $$[x...)...)...)...)...y)$$ so that the connected component which contains $x$ is (strictly) included in $[x,x+\varepsilon)\;\forall\varepsilon>0.$ if we suppose there is a certain limit in this reduction, then we would have to accept that $[x-\varepsilon /2,x+\varepsilon /2)$ is a connected component, which is a contradiction with the beginning of this discussion. What can we conclude, from this? AI: Your suspicion that $R_j$ is totally disconnected is correct. Note that since $R_j$ is strictly finer than the standard real topology, then it has fewer connected sets (or at least no extras). Thus, our only connected candidates are intervals, rays, singletons, the whole line, and the empty set. Well, you've already shown that the whole line isn't connected, and a similar approach will show that no interval or ray is connected. Thus, the only non-empty connected sets are singletons, as desired.
H: Is the space of continuous function with given order of decrease closed? For example, denote $O^1$ the space of continuous function with property $\lim_{x\to\infty}{\|x\|f(x)}=0$ or $f(x)=o\left(\|x\|^{-1}\right)$ as $x\to\infty$. It's obviously an intermediate vector space between $C_c$ and $C_0$. Is it closed? Assume knowledge of completeness of $C_0$, then for a convergent sequence $f_n\to f$, we know $f\in C_0$. Now we prove $f\in O^1$. $\forall\varepsilon,\exists N_1,\text{s.t. }d_{\sup}{(f,f_{N_1})}<\varepsilon /2,\exists N_2,\text{s.t. } \|xf_{N_1}(x)\|<\varepsilon /2\text{ if }\|x\|>N_2,\text{ then }...$ Ah, I see the argument break down here. $O^1$ can't be closed since if it's closed, it will contradict the fact that $C_0=\overline{C_c}$. My question is: is there any example of a sequence of continuous functions with given order of decrease uniformly convergent to a limit function with less order of decrease? I have an additional question: Is space $O=\cap_{n=1}^{\infty}{O^n}$ closed? Any proof or example? Third question: Is space $\cup_{n=1}^{\infty}{O^{1/n}}=C_0$? PS: $C_c$ means continuous function with compact support, $C_0$ means continuous function vanishing at infinity. $O^n$ is the space of continuous functions with property $\lim_{x\to\infty}{\|x\|^nf(x)}=0$ AI: The closure of $O^n$ under uniform convergence is precisely $C_0$. There are functions in $C_0\setminus O^n$, such as $1/(1+\|x\|^{1/2})$. Given a function $f$, let $f_n$ be the function obtained by multiplying pointwise by the indicator function of $[-n,n]$, then convolving with a standard mollifier. You can show that $f\in C_0$ implies $f_n\to f$ uniformly.
H: Extension of a continuous function in $\mathbb{R}^2$ Well, for (a) I have no idea how to extend, I feel that there will be a continuous extension. For example I can define $f=g$ when $f$ takes values from upper boundary of the disk and upper half plane and same way for lower boundary and lower half plane. For (b) I have no idea. (c) is true by Uryshon's lemma? As the plane is normal, and the given sets are closed and disjoint. AI: For (a) you can use Tietze's extension theorem: $B$ is a closed subset of a normal space. For (b) I think you could take any function with a discontinuity at distance one from the origin (like for example $\frac{1}{1 - \|x^2 + y^2\|}$) For (c) your answer is correct: the first set is a circle around $0$ of radius $1.5$, the second set is the closed unit disk union the complement of an open disk of radius $2$ so that we have two closed disjoint sets and can apply Urysohn.
H: which of the following are homeomorphic? well, I have forgotten how to identify ellipse, hyperbola,circle straightline from the general equation of conic, so is there any other way to identify these homeomorphic or not? a) B is an ellipse, b) B is an hyperbola, c) B is an complement of a closed ellipse. please help. AI: A is a circumference in a. and b. while a disk in c. so... a. Yes, homeomorphism by isometric transformation to overlay the centers, and then by projection of one onto the other. b. No, the circumference is compact, while the hyperbola is not c. No, same reason as b
H: What are the differences between class, set, family, and collection? In school, I have always seen sets. I was watching a video the other day about functors, and they started talking about a set being a collection, but not vice-versa. I also heard people talking about classes. What is their relation? Some background would be nice. It has to do with something called Russell's paradox, but I don't know what that is. I think that the difference between a family and a set is that the former is a function and the latter is a set. Is this right? AI: The idea behind a "collection" is simply a notion of a bunch of mathematical objects which are collected into one big pile. Think of it as a big bin full of trash, diamonds and empty bottles of beer, it doesn't have to make sense what is in this collection, it's just a collection. One of the problem to explain these things to people who are not mathematicians (or trying to "outsmart a set theorist", as I ran into several of those) is that the notion of a collection is not fully formal unless you already know what sets and class are, and even then it's not exactly what we mean. Let me start over now. Doing mathematics we often have an idea of an object that we wish to represent formally, this is a notion. We then write axioms to describe this notion and try to see if these axioms are self-contradictory. If they are not (or if we couldn't prove that they are) we begin working with them and they become a definition. Mathematicians are guided by the notion but they work with the definition. Rarely the notion and the definition coincide, and you have a mathematical object which is exactly what our [the mathematicians] intuition tells us it should be. In this case, a collection is a notion of something that we can talk about, like a mystery bag. We might know that all the things inside this mystery bag are apples, but we don't know which kind; we might know they are all Granny Smith, but we cannot guarantee that none of them is rotten. A collection is just like that. We can either know something about its elements or we don't, but we know that it has some. Mathematician began by describing these collections and calling them sets, they did that in a relatively naive way, and they described the axioms in a rather naive way. To the non-mathematician (and to most of the non-set theorists) everything is still a set, and we can always assume that there is a set theorist that assured that for what we need this is true. Indeed, if we just wanted to discuss the real numbers, there is no worry at all we can assume everything we work with is a set. This naive belief can be expressed as every collection is a set. It turned out that some collections cannot be sets, this was expressed via several paradoxes, Cantor's paradox; Russell's paradox; and other paradoxes. The exact meaning is that if we use that particular axiomatic description of "what is a set" then we can derive from it contradiction, which is to say that these axioms are inconsistent. After this happened several people began working on ways to eliminate this problem. One method in common was to limit the way we can generate collections which are sets. This means that you can no longer derive such contradiction within the theory, namely you cannot prove that such collection even exists - or rather you can prove it doesn't. The common set theory nowadays called ZFC (named after Zermelo and Fraenkel, the C denotes the axiom of choice) is relatively close to the naive way from which set theory emerges, and it still allows us to define collections which are not sets, though, for example "the collection of all sets". These collections are called classes, or rather proper classes. What is definable? This is a whole story altogether, but essentially it means that we can describe it with a single formula (perhaps with parameters) of one free variable. "$x$ is taller than 1.68m" is an example to such formula, and it defines the class of all people taller than said height. So in ZFC we may define a collection which is not a set, like the collection of all the singletons, or the collection of all sets. These are not sets because they are too big, in some sense, to be sets, but they are classes, proper classes. We can talk about collections which are not definable but that requires a lot more background in logic and set theory to get into. To sum up Classes are collections which can be defined, sets are particular classes which are relatively small and there are classes which are not sets. Collections is a notion which is expressed via both these mathematical objects, but need not be well-defined otherwise. Of course when we say defined we mean in the context of a theory, e.g. ZFC. In this sense, sets are things which "really exist" whereas classes are collections we can talk about despite their possible nonexistence. One last thing remains, families. Well, as you remarked families are functions. But functions are sets, so families are sets. We can make a slight adjustment to this, and we can, in fact, talk about class functions, and an index which is not a set but a proper class. We, therefore, can talk about families which are classes. Generally, speaking, if so, a family is a correspondence from one collection into another which uses one collection as indices for elements from another collection. To read more What is the difference between a class and a set? Why is "the set of all sets" a paradox, in layman's terms?
H: a question on normal subgroup of $GL_n(\mathbb{C})$ and $GL_n(\mathbb{R})$ I am really sorry that I am not able to solve this one, thank you for your help. AI: $a$ and $c$ are true by abstract group theory. The first one shoudn't be too hard. Just write it out. The third one basically says that the path component of identity in a topological group is a normal subgroup. To show it, for arbitrary $\varphi:[0,1]\to G$, $a\in G$, consider $\varphi_a(x)=axa^{-1}$. The second one isn't true. To see this notice that conjugation in general linear group is just a change of basis, so \begin{pmatrix}1 &1\\ 0 &1\end{pmatrix} and \begin{pmatrix}1 &0\\ 1 &1\end{pmatrix} are clearly conjugate. Edit: I just noticed that the hint I've dropped for c only shows how to see that $H$ is normal, not that it is actually a subgroup. For that, take $a,b\in H$ corresponding to paths $\varphi,\psi$ and consider $\varphi\cdot \psi$ (pointwise multiplication) and for each $a\in H$ consider $\varphi^{-1}$ (not the inverse function, but the pointwise inverse to $\varphi$). This proof shows us something slightly stronger: if $G$ is an arbitrary group, and $N\lhd G$ arbitrary, then the union of path components of all elements of $N$ is again a normal subgroup. In this case we have $N$ as the trivial subgroup, which is of course always normal.
H: Linear dependence of a set for what h? I asked the same question yesterday, but this one is a bit different in terms of computations. It is from my exam I took an hour ago. For what $h$ the columns of this matrix are linearly dependent? $$\begin{bmatrix} 1 & -3 & 4 \\ -4 & 7 & h\\ 2 & -6 & 8 \end{bmatrix}$$ Attempt: after row reducing, but not completely: $$\begin{bmatrix} 1 & -3 & 4 & 0 \\ -4 & 7 & h & 0 \\ 2 & -6 & 8 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -3 & 4 & 0 \\ 0 & -5 & h+16 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$ My guess was that if $h=-16;-\frac{28}{3}$ the system is linearly dependent. And I just guessed the -16. Hints please. AI: You are asking: for what values of $h$ are the vectors $$\vec{v_1}=\left(\begin{array}{r}1\\-4\\2\end{array}\right),\quad \vec{v_2}=\left(\begin{array}{r}-3\\7\\-6\end{array}\right),\quad \vec{v_3}=\left(\begin{array}{r}4\\h\\8\end{array}\right)$$ linearly dependent? You seem to be trying to do this by looking at the equation $$\alpha\vec{v_1}+\beta\vec{v_2}+\gamma\vec{v_3}=\left(\begin{array}{c}0\\0\\0\end{array}\right)$$ and trying to determine for what values of $h$ there is a nonzero solution. This leads to the matrix you have: $$\left(\begin{array}{rrr\|c} 1 & -3 & 4 & 0\\ -4 & 7 & h & 0\\ 2 & -6 & 8 & 0 \end{array}\right).$$ Now, since the third equation is a multiple of the first, that equation does not matter: it provides no new information. That means that you have a homogeneous system of two equations in three unknowns. Those systems always have infinitely many solutions. In particular, no matter what $h$ is, the system has infinitely many solutions, and so must have a nontrivial solution. Thus, the vectors are always linearly dependent. To understand what is happening, note that all three vectors lie in the plane $z=2x$. Any two vectors on the plane that are not collinear will span the plane. Since $\vec{v_1}$ and $\vec{v_2}$ are not collinear, and both lie on the plane $z=2x$, any vector that lies on the plane $z=2x$ will be a linear combination of $\vec{v_1}$ and $\vec{v_2}$. Or, put another way, three vectors in a $2$-dimensional space (a plane through the origin) are always linearly dependent. Here you have three vectors that satisfies $z=2x$; every other vector that satisfies that is a linear combination of $\vec{v_1}$ and $\vec{v_2}$: if $(a,b,2a)^t$ lies in the plane, then the system $$\alpha\left(\begin{array}{r}1\\-4\\2\end{array}\right) + \beta\left(\begin{array}{r}-3\\7\\-6\end{array}\right) = \left(\begin{array}{c}a\\b\\2a\end{array}\right)$$ has a solution, namely $\alpha = -\frac{7a+3b}{5}$, $\beta=-\frac{4a+b}{5}$ (obtained by Gaussian elimination). In particular, since no matter what $h$ is $\vec{v_3}$ lies in the plane $2z=x$, then we will have $$\vec{v_3} = -\frac{28+3h}{5}\vec{v_1} - \frac{16+h}{5}\vec{v_2}.$$ Note that this makes sense no matter what $h$ is. This can be read off your row-reduced matrix: you got $$\left(\begin{array}{rrr\|c} 1 & -3 & 4 & 0\\ 0 & -5 & h+16 & 0\\ 0 & 0 & 0 & 0 \end{array}\right).$$ Divide the second row by $-5$ to get $$\left(\begin{array}{rrr\|c} 1 & -3 & 4 & 0\\ 0 & 1 & -\frac{h+16}{5} & 0\\ 0 & 0 & 0 & 0 \end{array}\right),$$ and now add three times the second row to the first row to get $$\left(\begin{array}{rrc\|c} 1 & 0 & 4+\frac{-3h-48}{5} & 0\\ 0 & 1 & -\frac{h+16}{5} & 0\\ 0 & 0 & 0 & 0 \end{array}\right) = \left(\begin{array}{rrc\|c} 1 & 0 & -\frac{28+3h}{5} & 0\\ 0 & 1 & -\frac{h+16}{5} & 0\\ 0 & 0 & 0& 0 \end{array}\right).$$ So $\alpha$ and $\beta$ are leading variables, and $\gamma$ is a free variable. This tells you that the solutions to the original system are: $$\begin{align} \alpha &= \frac{28+3h}{5}t\\ \beta &= \frac{h+16}{5}t\\ \gamma &= t \end{align}$$ Any nonzero value of $t$ gives you a nontrivial solution, and $t=-1$ gives you the solution I give above. Of course, this can be done much more simply noting that since your original matrix has linearly dependent rows (third row is a scalar multiple of the first row), then the dimension of the rowspace is at most $2$ (in fact, exactly $2$), and hence the dimension of the columnspace is at most $2$ (in fact, exactly $2$, since $\dim(\text{columnspace})=\dim(\text{rowspace})$, so the columns are always linearly dependent.
H: Prove that $X´$ is a closed set Let $X\subset \mathbb{R}$.I have to prove,for all $X$ that $X´$,i.e, the set of accumulation points, is a closed set. Well,I know , by definition, that every accumulation point is a point of closure. How a set is said to be closed if $X = \overline X$ ,and how the accumulations points are points of closure,then: $X´=\overline X$. So $X´ $ is closed. Is that right? AI: HINT: Suppose that $x\notin X'$. Then $x$ has an open nbhd $N$ such that $N\cap X\subseteq\{x\}$. (Why?) What can you say about $N\cap X'$?

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