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H: Do eigenvectors of a Toeplitz matrix form an orthogonal set? It is true for a $2 \times 2$ Toeplitz matrix (put values $a$ and $b$ in the first row and $b$ and $a$ in the second and work out), but when I tried it for a $3 \times 3$, it turns out to be a bit difficult. Any help or reference to my question is greatly appreciated. Thanks. AI: Let $A=\pmatrix{0&1&0\\ 0&0&1\\ 8&0&0}$. Then $\det(xI-A)=x^3-8$ and $A$ has three distinct, simple eigenvalues. That is, $A$ has an eigenbasis and all its eigenspaces are one-dimensional. However, $A^TA-AA^T=\pmatrix{63&0&0\\ 0&0&0\\ 0&0&-63}\ne0$. Therefore $A$ is not normal, i.e. it has two different eigenspaces (corresponding to two different eigenvalues) that are not orthogonal to each other.
H: Definite Integral $\int_0^{\pi/4}\frac{\sqrt{\tan(x)}}{\sin(x)\cos(x)}\,dx$ $\displaystyle\int_0^{\pi/4}\frac{\sqrt{\tan(x)}}{\sin(x)\cos(x)}\,dx$ Needed a detailed solution with explanation. The answer is $2$. AI: HINT: $$\frac{\sqrt{\tan x}}{\sin x\cos x}=\frac{\sqrt{\tan x}}{\frac{\sin x\cos x}{\cos^2x}}\frac1{\cos^2x}=\frac{\sqrt{\tan x}}{\tan x}\sec^2x$$ Put $\sqrt{\tan x}=u\implies \tan x=u^2 $
H: $S^1\times S^1$ diffeomoprhic to torus of revolution. I am searching for an diffeomorphism between $$S^1\times S^1$$ and the torus of revolution $$\{(x,y,z)\in\mathbb{R}^3\|z^2+(\sqrt{x^2+y^2}-a)^2=r^2\}.$$ I know it's true, I am just looking for an explicit formula. Thanks a lot! I managed to wirte down a diffeomoprhism between $$S^1\times S^1$$ and $$\mathbb{R}^2/\mathbb{Z}^2$$ but I am stuck with this one. AI: Parametrize the torus of revolution by thinking of it literally as a circle rotated in a circle. The angle of the circle being revolved will be $\phi$ and the angle of the revolutionary circle will be $\theta$. So for a major radius of $R$, the revolutionary circle is $(R\cos\theta,R\sin\theta,0)$. For any $\theta$, and if the minor radius is $r$, the circle being revolved is $r(\cos\phi,0,\sin\phi)$, except rotated by an angle of $\theta$ in order to lie in the plane spanned by $(\cos\theta,\sin\theta,0)$ and the $z$-axis. Thus: $$f(\theta,\phi) = R(\cos\theta,\sin\theta,0) + r(\cos\phi\cos\theta,\cos\phi\sin\theta,\sin\phi).$$ This is a local diffeomorphism from $\mathbb{R}^2$ onto the torus of revolution which is $(2\pi)\mathbb{Z}^2$-periodic, so it descends to a diffeomorphism from $\mathbb{S}^1\times\mathbb{S}^1$. (There are still details to check, but I'll leave them to you.)
H: determinant calculation This question is in my assignment. We are not allowed to use any symbol to represent any elementary row and column operations used in the solution. We must solve it step-by-step. Please help me to check my solution word by word including my spelling and grammar. Question: Given that $$\begin{vmatrix}a& b& c\\ d& e& f\\ g& h& i\end{vmatrix}=2$$ find $$\begin{vmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{vmatrix}.$$ Solution: We interchange the second and third rows of the matrix $\begin{pmatrix}a& b& c\\ d& e& f\\ g& h& i\end{pmatrix}$ to get the matrix $\begin{pmatrix}a& b& c\\ g& h& i\\ d& e& f\end{pmatrix}$ and we have $$\begin{vmatrix}a& b& c\\ g& h& i\\ d& e& f\end{vmatrix}=-\begin{vmatrix}a& b& c\\ d& e& f\\ g& h& i\end{vmatrix}=-2.$$ We interchange the first and third columns of the matrix $\begin{pmatrix}a& b& c\\ g& h& i\\ d& e& f\end{pmatrix}$ to get the matrix $\begin{pmatrix}c& b& a\\ i& h& g\\ f& e& d\end{pmatrix}$ and we have $$\begin{vmatrix}c& b& a\\ i& h& g\\ f& e& d\end{vmatrix}=-\begin{vmatrix}a& b& c\\ g& h& i\\ d& e& f\end{vmatrix}=-(-2)=2.$$ We multiply the second column of the matrix $\begin{pmatrix}c& b& a\\ i& h& g\\ f& e& d\end{pmatrix}$ by 3 to get the matrix $\begin{pmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{pmatrix}$ and we have $$\begin{vmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{vmatrix}=3\begin{vmatrix}c& b& a\\ i& h& g\\ f& e& d\end{vmatrix}=(3)(2)=6.$$ We add $(-2)$ times the third column of the matrix $\begin{pmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{pmatrix}$ to its first column to get the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{pmatrix}$ and we have $$\begin{vmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{vmatrix}=\begin{vmatrix}c& 3b& a\\ i& 3h& g\\ f& 3e& d\end{vmatrix}=6.$$ We add $(-3)$ times the first row of the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{pmatrix}$ to its second row to get the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{pmatrix}$ and we have $$\begin{vmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{vmatrix}=\begin{vmatrix}c-2a& 3b& a\\ i-2g& 3h& g\\ f-2d& 3e& d\end{vmatrix}=6.$$ Finally, we multiply the first column of the matrix $\begin{pmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{pmatrix}$ by 3 to get the matrix $\begin{pmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{pmatrix}$ and we have $$\begin{vmatrix}3c-6a& 3b& a\\ 3i-9c+18a-6g& 3h-9b& g-3a\\ 3f-6d& 3e& d\end{vmatrix}=3\begin{vmatrix}c-2a& 3b& a\\ i-3c+6a-2g& 3h-9b& g-3a\\ f-2d& 3e& d\end{vmatrix}=(3)(6)=18.$$ Thank you. AI: Very well done, down to the meticulous details! Your solution is very easy to read and to follow; the rationale and your calculations are correct, and I couldn't find any grammatical or spelling errors. Remark: I wish all posters were as industrious as you in taking the time to carefully format your work, not just to hand in, but also here in your post at MSE!
H: How to find all vectors so that a vector equation can be solved? Unfortunately, my text book doesn't clarify this process at all. It's asking to find all all vectors [a b] so that the vector equation can be solved. The vector equation is: $c1$ $[3,1]$ + $c2 [6,2]$=$[a, b]$ The linear system would look like: $3c1+6c2=a$ $c1+2c2=b$ My text doesn't give any indication how to solve this and I'm stuck. Any help would be appreciated! Also, apologies since I can't get the vector forms inserted properly! AI: Hint: You have two equations, and two variables (we treat a, b as constants). Set up the associated augmented coefficient matrix, row reduce, and solve for $c_1, c_2$, which can each be expressed as functions of $a, b$. From that, you should also be able to express $a, b$ as functions of the constants $c_1, c_2$. Associated augmented coefficient matrix to your system of equations: $$\begin{pmatrix} 3 & 6 &\mid& a\\ 1 & 2 & \mid & b\end{pmatrix}$$ TIP: Be sure to choose a value for $b$ that will ensure the system is consistent.
H: Finding the fixed point and the suitable range I have to find the fixed point of x$^3$-x$^2$-1=0.Then x=(1/x$^2$)+1 where I chose g(x)=(1/x$^2$)+1 .Then I tried to find a fixed point for g(x).Since I don't know the range of x,I chose x$\in$[1.3,1.6] as the suitable range because in this range 1.3 My question is my chosen range a suitable one? I wrote a python program to find the fixed point and it found the fixed point 1.47 in just 13 iterations.Why do I get from the program 13 and from calculation 62? def fixedPoint(f,epsilon): guess=1.3 count=0 p=f(guess) while abs(p- guess)>=epsilon: guess=p p=f(guess) count+=1 print guess return "The fixed point is "+str(round(guess,2))+" and it took "+str(count)+" iterations" def f(x): k=(1/x**2)+1 return k AI: You used an upper bound on the absolute value of the derivative to find an upper bound on the number of iterations required for the desired precision. The Mean Value Theorem says that if $x_n$ is the $n$-th estimate, and $r$ is the root, then $\|x_{n+1}-r\|$ is equal to $\|x_n-r\|$ times the absolute value of $g'(x)$ somewhere between $x_n$ and $r$. It is true that the derivative has absolute value $\lt \frac{2}{(1.3)^3}$. But that only gives us an upper bound on the "next" error in terms of the previous error. Furthermore, as $x_n$ gets close to $r$, say around $1.4$ or closer, the relevant derivative has significantly smaller absolute value than your $k$. A further huge factor in this case is that the derivative is negative. That means that estimates alternate between being too small and being too big. When $x_n\gt r$, the derivative has significantly smaller absolute value than your estimate $k$. Even at the beginning, the convergence rate is faster than the one predicted from the pessimistic estimate of the derivative, particularly since half the time $x_n\gt r$. After a while, the disparity, for $x_n\lt r$, gets greater. Remark: You know the root $r$ to high precision. It might be informative to modify the program so that at each stage it prints out $\frac{x_{n+1}-r}{x_n-r}$. That way, you can make a comparison between the upper bound $\frac{2}{(1.3)^3}$ on the ratio, and the actual ratio. Even not very large differences, under repeated compounding, can result in much quicker convergence than the one predicted from the upper bound.
H: Differential of a $C^\infty$ function from a manifold $M$ to $\mathbb{R}$ I am studying differentiable manifolds from Warner. In the book, the differential is defined as follows. If $\psi:M\longrightarrow N$ is $C^\infty$, and if $m\in M$, then the differential of $\psi$ at $m$ is the linear map $d\psi:M_m\longrightarrow N_{\psi(m)}$, defined as under : If $v\in M_m$, then, we define $d\psi(v)(g) = v(g\circ\psi)$ where $g$ is a $C^\infty$ function on a neighbourhood of $\psi(m)$. My question is this. They have mentioned this special case : If $f:M\longrightarrow\mathbb{R}$ is a $C^\infty$ function, then $df(v)=v(f)\frac{d}{dr}\|_{f(m)}$. I don't know how to show this. By definition, $df(v)(g)=v(g\circ f)$, where $g$ is differentiable in a neighbourhood of $f(m)$. How is $v(g\circ f) = v(f)\frac{dg}{dr}\|_{f(m)}$. [I understand the fact intuitively too, because $df(v)$ is a tangent vector on $\mathbb{R}$, and this is like the chain rule, but how does one show it?] AI: If $U\subset \mathbb R^n$ is open and $f:U\to \mathbb R,\; g:\mathbb R \to \mathbb R$ are differentiable functions we have for $u_0\in U$ : $$\frac {\partial (g\circ f)(u_0)}{\partial x_i}=g'(r_0)\cdot \frac {\partial f(u_0)}{\partial x_i}$$ where $r_0=f(u_0)$ Your result follows from there if you write $$ v=\sum v_i \frac {\partial }{\partial x_i} \|_{m_0} \quad (\star) $$ for a tangent vector $v\in T_{m_0}M$ written out in terms of a chart $(x_1,...,x_n)$ for $M$ at $m_0$ and apply both sides of $(\star)$ to $g\circ f$.
H: What is the null vector for the vector space of continuous functions $f \colon \mathbf{R} \to \mathbf{C}$? The set of all continuous complex-valued functions of real variable $x$ together with addition of two vectors $\boldsymbol{f} = f(x)$ and $\boldsymbol{g} = g(x)$ defined by \begin{equation} (f + g)(x) \equiv f(x) + g(x) \, , \end{equation} and multiplication of vector $\boldsymbol{f}$ by a scalar $a \in \mathbf{C}$ defined by \begin{equation} (a f)(x) \equiv a f(x) \, , \end{equation} form a vector space. What is the null vector $\boldsymbol{0}$ in this space? It it defined by $\forall x \in \mathbf{R} \colon 0(x) \equiv 0 \, ?$ AI: Yes, you're correct: It is the zero function that always returns $0$ for any input.
H: Intersection of Two Permutations If you are generating 2 digits codes $[X, X]$ where you have 5 choices for $X$. The number of choices for each code is: Order matters so the total number of permutations is $n^r = 5^2 = 25$. If for the first code the choices were [1,5], [4,8] (inclusive) and in the second code the choices were [3, 7], [2, 6]. How do you find the number of codes which are common to the first and the second? AI: Hint: For the first digit to be common, you have choices that are both $[1,5]$ and $[3,7]$. How many is that? How many choices for the second digit? Now multiply, just as you did in the example.
H: Find an example of a positive function that its improper integral converge, but its series diverge I need to find an example of a positive function f, and a constant a>0 such that the improper integral of f from 0 to infinity converge, but the series f(na) from 1 to infinity diverge. AI: To expand on Jyrki's comment, let $$f(x)=\begin{cases} 1 &\text{ if there exists an }n\in\mathbb{N}\text{ with } \|x-na\|\leq \frac{1}{n^2}\\ 0 &\text{ otherwise} \end{cases}$$ The graph looks like a sequence of thinner and thinner rectangles above the points $na$. (1) What is $\sum_{n=0}^\infty f(na)$? (2) Show that $\int_0^\infty f(x)\ dx$ is equal to a convergent series.
H: solving $\frac{x}{3}+{[\frac{x}{3}]} = \sin(x) + [\sin(x)]$ for real x , in an efficient manner $$\frac{x}{3}+{\left[\frac{x}{3}\right]}=\sin(x)+[\sin(x)]$$ I know the answer and the solution. $$-1 \le \sin(x) \le 1\Rightarrow-2\le\sin(x)+[\sin(x)]\le 2 \Rightarrow -2 \le \frac{x}{3}+\left[\frac{x}{3}\right] \le 2 $$ now there are different situations : $x=3$ is unacceptable $$0 \le x \lt 3 \Rightarrow0 \le \frac{x}{3} \lt 1\Rightarrow \left[\frac{x}{3}\right]=0 \land[\sin(x)]=0\Rightarrow\frac{x}{3}=\sin(x)$$ which has 2 acceptable answers . $$-3 \le x \lt 0 \Rightarrow-1 \le \frac{x}{3}\lt 0\Rightarrow \left[\frac{x}{3}\right] =-1\land [\sin(x)] = -1\Rightarrow\frac{x}{3} =\sin (x) $$ which has one acceptable answer. so the equation has 3 real answers . I want to know if there is a better way to solve it. ( beautiful equation , isn't it?!) AI: There is a more holistic way to solve this. The function $y\mapsto y+[y]$ is strictly increasing; in particular, it is one-to-one. Therefore $y+[y]=z+[z]$ if and only if $y=z$. So you can deduce right away that you're looking exactly for solutions to $\frac x3=\sin x$ (which is what you eventually found).
H: Percentile of CDF Function I am trying to find the 0.5 (mean) percentile of a CDF function. $$ F(X) = 1 - e^{-(x/3)^2} $$ In my book's example it says $$ m = 3[-ln(1-0.5)]^{1/2} = 3\sqrt{ln2}=2.498$$ I am not sure how to get to the above equation even using the definition of a percentile of $$F(x_p) = p $$ AI: The $50$-th percentile is the median, not the mean. Call it $m$. We want to solve the equation $$1-e^{-(m/3)^2}=\frac{1}{2},$$ or equivalently $$ e^{-(m/3)^2}=\frac{1}{2}.$$ Take the (natural) logarithm of both sides. We get $$-\left(\frac{m}{3}\right)^2=\ln(1/2)=-\ln 2.$$ So we are solving $\frac{m^2}{9}=\ln 2$. That gives (since $m$ is positive) $m=3\sqrt{\ln 2}$.
H: Closed form for a fixed point of the exponential function? Let $$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} $$ denote the exponential function, which is defined on the entire complex plane. There is a fixed point of this function at $w= a+bi$ where $a \approx 0.31813$ and $b \approx 1.33723$. I am guessing there is no closed form expression for $w$. If there is, please correct me. But if there is not a closed form expression, how does one go about proving something like that? Any pointers to the appropriate literature would be appreciated. AI: $e^z=z$, so $ze^{-z}=1$ or $-ze^{-z}=-1$. Let $u=-z$, so the equation becomes $ue^{u}=-1$. This equation can be solved via the Lambert-W function, getting $u=W(-1)=-0.31813\ldots-1.3372\ldots i$, which upon negation gives you your answer. There are some power series expansions of $W$ but no easy closed form for the solution.
H: Solving $y'' + (ax+b)y = 0$ This is a problem in quantum mechanics when one considers a linear potential; in physics-speak the equation would be written as $$\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}(E-ax)\psi = 0,$$ with $V(x) = ax$. I've been looking at it for a while and can't find a solution, and I thought I'd ask here for a hint before I go running to W\|A for help. Disclaimer: I have no idea if a closed form solution exists. The first thing I thought is to try a series expansion, but the recurrence relation between the coefficients is pretty ugly. Then I tried a Fourier transform. I have no reason to think the solution will be integrable in $\mathbb{R}$, but as a physicist I've been trained not to care about such things. Setting $$ \phi(k) = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \psi(x)e^{-ikx}\ dx $$ we get the equation $$\frac{d\phi}{dk} - \frac{i}{a}\left(\frac{\hbar^2 k^2}{2m}-E\right)\phi = 0$$ which is easily solved to get $$\large \phi(k) = Ae^{-\frac{ik}{a}(\frac{\hbar^2 k^2}{6m}-E)}$$ which not only can't be Fourier-transformed since it doesn't go to zero at infinity, I also have no idea how to integrate. So this is as far as I got. Does anyone know how to solve this equation, and if there's no closed form (which I suspect), how to get a nice series or something? AI: The solution is in term of the Airy functions, $\text{Ai}(x)$, $\text{Bi}(x)$ (which are well defined, have asymptotic formulas and series representation, etc. see for example Abramowitz & Stegun or Szegö). These functions are the linearly indepedent solutions of $$\frac{d^2y(x)}{dx^2}-xy(x)=0 $$ With the change of variable $$\tilde{x} =\mu x +\nu$$ you will be able to transform the initial form to a Airy form. The solution is then directly $$y(x)=\alpha \text{Ai}(\mu x + \nu)+\beta \text{Bi}(\mu x + \nu) $$ where $\alpha,\beta$ are the integration constants.
H: If $X$ is a cone, show that $I(X)$ is homogeneous. The exercise is 1.3(3) from HP Kraft, "Appendix A: Basics from Algebraic Geometry." If a closed subset $X\subseteq \mathbb C^n$ is a cone, show that $I(X)$ is generated by homogeneous functions. Closed means $X=\cap g_i^{-1}(0)$ the zero set for some $g_i\in\mathbb C[x_1,\dots,x_n]$; cone means $tv\in X$ for all $v\in X,t\in \mathbb C$; the ideal $I(X)$ means polynomials that vanish on $X$; the exercise defines homogeneous to mean $f(tv)=t^df(v)$ for all $v\in \mathbb C^n, t\in \mathbb C$ for some $d\in\mathbb N.$ The exercise appears before Nullstellensatz and affine variety are introduced, but I can't find an elementary proof anywhere. Wolfram and Wikipedia (citing "Chow's theorem") give the statement without proof. The converse is trivial of course. Solution (paraphrasing Georges below). Every polynomial can be written as the sum of homogeneous polynomials $p(x) = p_0 (x)+\dots +p_d(x).$ These homogeneous components generate "at least" as much, $$\sum_{p\in I(X)} (p_1)+\dots+(p_d) \supset I(X).$$ To show the reverse inclusion, we fix $p\in I(X)$ and show that each component $p_i\in I(X).$ Let $v\in X$ be given. Since $X$ is a cone, $p$ vanishes on the line $\mathbb C v,$ and $$p(tv) = \sum p_i(tv) = \sum p_i(v) \, t^i=0$$ for all $t\in \mathbb C.$ Thus $p(tv)\in \mathbb C[t]$ must be the zero polynomial with every "coefficient" $p_i(v)=0.$ AI: It is enough to show that if a polynomial $p(T_1,... ,T_n)=\sum p_d(T_1,... ,T_n)$ vanishes on the line $l=\mathbb C\cdot a $ where $a=(a_1,...,a_n)\neq 0\in X\subset \mathbb C^n$, each of the homogeneous components $p_d(T_1,... ,T_n)$ of degree $d$ of the polynomial $p(T_1,... ,T_n)$ vanishes on $l$. But since $p(z\cdot a)=\sum p_d(z\cdot a_1,... ,z\cdot a_n)=\sum z^d\cdot p_d( a_1,... , a_n)=0$ is a polynomial in $z$ vanishing for all $z\in \mathbb C$, that polynomial is zero and we necessarily must have $p_d( a_1,... , a_n)=0$ for all $d $, i.e. all homogeneous components $p_d$ of $p$ vanish on $l$. We have thus proved that $I(X)$ is indeed a homogeneous ideal.
H: Regular Expression that accepts a language L Given, $L=\{ x \in\{0,\,1\}^* \| x=0^n1^m \text{ and }n+m\text{ is a multiple of }3\}$ give a regexp that accepts the language. My thoughts are: $(000)^(\epsilon + 001 + 011)(111)^$ Is this right? AI: HINT: Divide the word into blocks of $3$; the critical point is where the zeroes and ones meet. If $n=3k+1$, say, that block will be $011$. If $n$ is a multiple of $3$, on the other hand, there won’t be a mixed block: you’ll just have $(000)^(111)^$. Look at the possibilities for the mixed block, when it exists.
H: How to treat system of linear first order differential equations with trigonometric function coefficients? I'm having trouble solving the following IVP: $$x_1^\prime = -x_1\tan t + 3\cos^2t$$ $$x_2^\prime = x_1 + x_2\tan t + 2\sin t$$ where $x_1(0) = 4$ and $x_2(0) = 0$. I'm not sure what to do when the coefficients are trigonometric functions. I've dealt with constant coefficients and we just learned reduction of order, but I'm confused about how to approach this problem. AI: Hint: The first equation is a first-order linear equation in $t$ and $x_1$ with the integrating factor $\sec t$. Thus $x_1(t)=3\sin t\cos t+c_1\cos t$. Substituting this into the second one you will get another linear equation in $t$ and $x_2$ with integrating factor $\cos t$. I guess you can continue from this point.
H: uniform continuous proof take a look Hi take a look of my proof and fix if there is wrong or unnecessary part $F(x) = x^2 , \alpha \in [0,\infty] $ and prove $F(x)$ is not uniformly continuous anyway proof assume that this$ f(x)$ is uniformly continuous. then there should be a some limit point(??) exist $L$, and $$y = L + \epsilon/2 , x = L-\epsilon/2$$ we can write this limit point as $$\|x-y\| < \epsilon$$ since $f(x)$ is uniformly continuous there should be exit of $$\|f(x) - f(y)\| < \delta $$ If we proof there exist some $\|f(x)- f(y)\| > \delta $ this proof is done $$\|f(x) - f(y)\| < \delta = \|x^2 - y^2\| < \delta = \|x+y\|\|x-y\|$$ we know $\|x-y\| < \epsilon/2 $ then write $x<y+\epsilon/2 $ $$\|x+y\| = \|y+\epsilon/2 + y \|,$$ then $$x = 2(y+\epsilon)$$ Therefore, $$\|x+y\|\|x-Y\|> \epsilon$$ there is contradiction. AI: I'm having trouble understanding your argument - what are you getting at with the 'limit point'? You also seem to have the 'usual' role of $\delta$ and $\epsilon$ switched. Of course, this is just notational (and not mathematical), but sticking with conventions will make your work easier to read. Basically, I'd say you have some idea what needs to be proved and how to prove it, but the biggest problem is you make no indication of what conditions $x$ and $y$ must satisfy. In particular, given an $\delta$ and $\epsilon$, there are going to be some points where $\|f(x) - f(y)\| < \epsilon$, you have to find a pair where this fails. To show $f$ is not uniformly continuous, here's an outline. 1) If $f$ is uniformly continuous, then for every $\epsilon > 0$, there is $\delta > 0$ so that, for any $x,y$ such that $\|x-y\| < \delta$ implies $\|f(x)-f(y)\| < \epsilon$. So to show it is not uniformly continuous, we need to show that for some $\epsilon > 0$, for every $\delta > 0$, there exists $x,y$ such that $\|x-y\| < \delta$, but $\|f(x) - f(y)\| > \epsilon$. 2) So, start with an $\epsilon$ that is convenient for you (since it has to hold for any $\epsilon$). Try $\epsilon = 1$ to begin with (we can always start over and make it smaller if this doesn't work). 3) Now, choose any $\delta > 0$. We want to find $x,y$ so that $\|x-y\| < \delta$, but $\|x^2 - y^2\| > 1$. Intuitively, we want to pick really big (but close together) $x$ and $y$ (close together so that $\|x-y\|$ is small, but the difference of their squares is big). Let's set $y = x + \delta/2$ (so automatically, we have $\|x-y\| < \delta$) and then decide how big $x$ has to be. As you note, $$\|x^2 - y^2\| = \|x + y\|\|x-y\| = \|x + x + \delta/2\|\|x - (x + \delta/2)\| \\ = \frac{\delta}{2}\left\|2x + \frac{\delta}{2}\right\|\\ > \frac{\delta}{2}\left\|2x\right\| = \delta\|x\|$$ So, if we choose $x$ so that $\|x\| > \dfrac{1}{\delta}$, then we will have $\|x^2 - y^2\| > 1$ as needed.
H: Help find the MacLaurin series for $\frac{1}{e^x+1}$ What is the MacLaurin series up to $x^4$ for $\frac{1}{e^x+1}$? My Attempt: $$\begin{align} \frac{1}{e^x+1} &=(1+e^x)^{-1} \\ &\approx 1 -e^x+(e^x)^2-(e^x)^3+(e^x)^4 \\ \end{align} $$ Since $$ \begin{align} e^x &\approx \frac{1}{24} \, x^{4} + \frac{1}{6} \, x^{3} + \frac{1}{2} \, x^{2} + x + 1 \\ (e^x)^2 &\approx \frac{2}{3} \, x^{4} + \frac{4}{3} \, x^{3} + 2 \, x^{2} + 2 \, x + 1 \\ (e^x)^3 &\approx \frac{27}{8} \, x^{4} + \frac{9}{2} \, x^{3} + \frac{9}{2} \, x^{2} + 3 \, x + 1 \\ (e^x)^4 & \approx \frac{32}{3} \, x^{4} + \frac{32}{3} \, x^{3} + 8 \, x^{2} + 4 \, x + 1 \end{align} $$ Therefore $$ 1 -e^x+(e^x)^2-(e^x)^3+(e^x)^4 \approx \frac{95}{12} \, x^{4} + \frac{22}{3} \, x^{3} + 5 \, x^{2} + 2 \, x + 1 $$ But the answer is $\frac{1}{48} \, x^{3} - \frac{1}{4} \, x + \frac{1}{2}$ Please advise what did I do wrong and what is the correct way to solve this problem. AI: Your initial expansion is an asymptotic series, valid for small values of $e^x$. But when $x$ is small, $e^x \approx 1$, which isn't small at all! If you instead write $$ \frac{1}{e^x+1}=\frac{1}{2+(e^x-1)}=\frac{1}{2}\cdot\frac{1}{1+\frac{1}{2}(e^x-1)}=\frac{1}{2}\left(1-\frac{1}{2}(e^x-1) +\frac{1}{4}(e^x-1)^2-\frac{1}{8}(e^x-1)^3+\frac{1}{16}(e^x-1)^4+\ldots\right), $$ and then fill in the expansion for $e^x-1$ where appropriate, you should find the right answer.
H: Recurrence sequence limit I would like to find the limit of $$ \begin{cases} a_1=\dfrac3{4} ,\, & \\ a_{n+1}=a_{n}\dfrac{n^2+2n}{n^2+2n+1}, & n \ge 1 \end{cases} $$ I tried to use this - $\lim \limits_{n\to\infty}a_n=\lim \limits_{n\to\infty}a_{n+1}=L$, so $L=\lim \limits_{n\to\infty}a_n\dfrac{n^2+2n}{n^2+2n+1}=L\cdot1$ What does this result mean? Where did I make a mistake? AI: You didn't make any mistake with the limits, you get $L=L$, though uninformative, it is true. You can prove by induction that $\forall n\in \Bbb N\left(a_n=\dfrac 3 8\dfrac{n+1}{n}\right)$. Inductive step: Let $n\in \Bbb N$ be such that $a_n=\dfrac 3 8\dfrac{n+1}{n}$. Then the following holds: $$\begin{align} a_{n+1}=a_n\dfrac{n^2+2n}{n^2+2n+1}&=\dfrac 3 8\dfrac{n+1}{n}\dfrac{n^2+2n}{n^2+2n+1}\\ &=\dfrac 3 8\dfrac{n+1}{n}\dfrac{n(n+2)}{(n+1)^ 2}\\ &=\dfrac 3 8\dfrac{n+2}{n+1}.\end{align}$$
H: Something like a field but with 3 operations? I know of Groups, and Rings, and Fields but what about tacking on a 3rd operation. Is there any use in considering some structure that consists of a field but with a 3rd operation (possibly less well behaved than the other two)? The link in the comments is helpful, but to make this question more specific I would be interested in something that is a field under the first two operations. AI: Have a look through A Course in Universal Algebra, by S. Burris & H.P. Sankappanavar. You'll find many examples of algebras with more than two operations there. It builds up a rich theory for them in fact, proving, for instance, the isomorphism theorems in full generality.
H: 2-digit combinations I can count the number 2-digit combinations: each digit has 10 possibilities, so that gives 10*10 = 100 combinations. But what if we write the combinations in one long string like this: 112131 This string is 6 long, but it gives us 5 combinations: 11, 12, 13, 21 and 31 How can I calculate the shortes string that gives me all 100 combinations? How long will it need to be for all n-digit combinations? How can i generate the sequence? I remember hearing about this somewhere but i forgot. If I could just get a name of what it is called. AI: You are probably thinking of De Bruijn sequences. These are often generated for a binary "alphabet", but can be formed for any size $k$ alphabet. With wraparound the minimum sequence length of $k^n$ for all $n$ subsequences on an alphabet of size $k$ can be obtained. If wraparound is not allowed, then the sequence has to be "padded" with the first $n-1$ characters repeated at the end, for a total length of $k^n + n - 1$. The Combinatorial Object Server has a De Bruijn sequence generator (among other things), and for $n=2, k=10$ as in the Question, it produced this (lexicographically least) sequence: 0010203040506070809112131415161718192232425262728293343536373839445464748495565758596676869778798899
H: Where M is a matrix calculate a formula for M^n Let $$M = \begin{bmatrix} -7 & 8 \\ -8 & -7 \end{bmatrix}.$$ Find formulas for the entries of $M^n$ where $n$ is a positive integer. (Your formulas should not contain complex numbers.) Your answer should be in the form of a matrix. I diagonalized to the form $M = P D P^{-1}$ and $M^n = P D^n P^{-1}$ where $P$ is my matrix of eigenvectors and $D$ is my matrix of eigenvalues. My final answer after diagonalization was $$M^n = \begin{bmatrix} .5((-7+8i)^n+(-7-8i)^n) & (i/2)(-(-7+8i)^n+(-7-8i)^n) \\ (.5/i)(-(-7+8i)^n+(-7-8i)^n) & .5((-7+8i)^n+(-7-8i)^n) \end{bmatrix}$$ I can't seem to find an answer not in terms of complex numbers. Can someone show me what I'm missing? AI: Try this: Write your eigenvalues in polar form $re^{i\theta}$ instead of $x + iy$. (so the two eigenvalues are $re^{i\theta}, re^{-i\theta}$). Recall polar coordinates are given by $r^2 = x^2 + y^2$ and $\tan(\theta) = y/x$ (so $\theta = \tan^{-1}(y/x)$ if $x>0$ and $\tan^{-1}(y/x) + \pi$ if $x<0$). Then, when you compute $D^n$, use De Moivre's formula: $$ \left(re^{i\theta}\right)^n = r^n(\cos(\theta) + i \sin(\theta))^n = r^n (\cos(n\theta) + i \sin(n\theta)) \\ \left(re^{-i\theta}\right)^n = r^n(\cos(\theta) + i \sin(-\theta))^n = r^n (\cos(n\theta) + i \sin(-n\theta)) = r^n (\cos(n\theta) - i \sin(n\theta)) $$ by using even and odd properties of $\cos$ and $\sin$. In particular, if you have to add these two values, you'll get $$ \left(re^{i\theta}\right)^n + \left(re^{-i\theta}\right)^n = 2r^n\cos(n\theta) $$
H: Hölder's Inequality and step functions Define functions $f(x) = \sum_{k=0}^\infty a_k \chi_{[k,k+1)}(x)$ and $g(x) = \sum_{k=0}^\infty b_k \chi_{[k,k+1)}(x)$ where $\chi_ {[k,k+1)}$ is the indicator function for the given interval. Let $f,g$ be zero for all negative numbers. Also assume that the $a_k, b_k$ are non negative. Does Hölder's inequality give us that $\sum_{k=0}^{\infty} a_kb_k \leq (\sum_{k=0}^{\infty} a_k^p)^{1/p} (\sum_{k=0}^{\infty} b_k^q)^{1/q}$? ($1<p<\infty$ and $p,q$ Hölder conjugates) It would seem intuitively like this should be true. AI: You need some absolute values in there (or need to assume the $a_k$'s and $b_k$'s are non-negative), but then this holds.
H: Contrapositive Inquiry Suppose you want to show that $A \implies B$. This is equivalent to showing $\neg B \implies \neg A$. But now suppose you show that assuming $\neg B$ it is POSSIBLE that $\neg A$, but not necessarily the case that $\neg A$. Have you succeeded in showing that $A \implies B$? AI: You have to prove that, assuming $\neg B$, it is DEFINITE that $\neg A$. If the case is not certain, then there are some cases that $A$ is true and $B$ is false together, which means it is not the case that $A\implies B$.
H: How to choose between poisson and binomial distributions I don't get this thing... I know that binomial distribution is used to know the probability of a X v.a. that sounds like this: X = "the probability of having 4 blue balls doing 10 extraction from a chest containing 7 blue and 40 white", and I know that poisson distribution and binomial distribution are really similar for lim_(pn)->0(F(x)) (when pn are really small...). I'm reading everywhere that the distribution of poisson is used a lot to approximate a big binomial... but what's its real purpose? what's the cases in wich I must use poisson and not binomial? (beyond the approximation case). thanks in advance. AI: The Poisson distribution is also called the "law of rare events" -- it is the distribution that counts the number of occurrences of an event given that the probability of the event is very small. That should sound a lot like the binomial distribution. In fact, the Poisson distribution can be derived as a limit of the binomial distribution. So what is the point of it? First, the fact that the limit exists gives us a lot of analytically useful results. The Poisson distribution is easier to work with than the binomial distribution. It is easier to compute the pdf and especially the cdf. Its generating functions have nice properties. Etc. Second, in applications, the Poisson distribution serves in ways that the binomial distribution just cannot handle. Consider the case of radioactive decay. You're measuring the rate, using a Geiger counter, on a sample of hundreds of trillions of atoms. The binomial distribution is arguably applicable in this case, but are we really sure that atoms are "discrete" in the same way the integers are? (That is an empirical question, up to science to figure out) I would argue that without a priori knowledge, using the Poisson approximation is not an approximation of the binomial distribution. It is an approximation of the "real" behavior of radioactive decay, whatever that is.
H: Functions that satisfy $f(x,z) = f(x,y) f(y,z)$ I am specifically looking for solutions that are NOT of the form: $f(x,y) = g(x)/g(y)$ since that is an obvious solution, as is $f(x,y) = 0$. I have a suspicion that there may be answers to this question that do not fall into these categories. The functions don't necessarily have to be continuous or differentiable. Edit: $f$ is not necessarily defined on $\mathbb{R}^2$, it can be defined on any space you like. Basically the equation deals with functions of two variables/arguments. Edit #2: Domains are irrelevant to this question. AI: First, either the function is zero everywhere or nowhere. If $f(a,b)=0$, then $f(a,b')=0f(b,b')=0$, and then $f(a',b')=f(a',a)f(a,b')=0$. Since this is for any $(a',b')$, the function is everywhere zero. Second, suppose the function is zero nowhere. Now we have $\frac{f(x,z)}{f(x,y)}=f(y,z)$, is independent of $x$. Define $g(t)=f(1,t)$. We have $\frac{g(z)}{g(y)}=f(y,z)$. Hence the other "obvious" solution you mention is the only other possibility.
H: A non-equality and an inequality involving $y$ and $y_0$ from Spivak Calculus 4th ed. It's Problem 22. from Chapter 1. I'm given: $y_0 \neq 0$ $\|y - y_0\| < \frac{\|y_0\|}{2}$ $\|y - y_0\| < \frac{\epsilon\|y_0\|^2}{2}$ and I must use them to prove that: $y \neq 0$ $\|\frac{1}{y} - \frac{1}{y_0}\| < \epsilon$ I haven't really ever done proof based problems before this book, so I'm having a little hard time.I can do the problems once I get the general direction, but I'm not sure how exactly to start.Can someone offer a hint where to start?I'm also confused by this: $\|y - y_0\| < \frac{\epsilon\|y_0\|^2}{2} , \|y - y_0\| \geq 0 => \frac{\epsilon\|y_0\|^2}{2} > 0 => \epsilon > 0, y_0 > 0$ (you can just use the $\frac{\|y_0\|}{2}$ part, but the 2nd part also tells you about $\epsilon$) However $y_0 \neq 0$ is already granted at the start, even tho it's obviously the first thing you notice, so does it mean the author is hinting at something with this? AI: PROOF: Step 1: $\|y\| > \frac{\|y_{0}\|}{2}$, since $\|y\|-\|y_{0}\| < \|y-y_{0}\| < \frac{\|y_{0}\|}{2}$. Step 2: $\|\frac{y - y_{0}}{y\cdot y_{0}}\|<\frac{\|y-y_{0}\|}{(\frac{\|y_0\|^{2}}{2})} < \varepsilon\cdot \frac{(\frac{\|y_{0}\|^{2}}{2})}{(\frac{\|y_{0}\|^{2}}{2})} = \varepsilon$. I hope this helps.
H: On integrating a "gaussian-like" integral Let the following "gaussian-like" integral: $$ I = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}\|\Sigma\|^{1/2}} \exp \left\{ -\frac{1}{2} (\mathbf{x}-\mathbf{\mu})^T\Sigma^{-1}(\mathbf{x}-\mathbf{\mu}) \right\} \mathbf{x} \,\mathbf{d}\mathbf{x}, $$ where $\mathbf{x}=(x_1,\dots,x_n)^T$, $\mathbf{\mu} = (\mu_1,\dots,\mu_n)^T\in\Re^n$, and $\Sigma\in\mathbb{S}_{++}^{n}$. Our main goal is to evaluate the above integral. To this end, let $\mathbf{x}-\mathbf{\mu}=S\mathbf{y}$, where $S$ is an $n \times n$ orthogonal matrix ($S^T=S^{-1}$) with determinant equal to $1$. Using this change of variable, the quadratic form shown in the integral written as: $$ -\frac{1}{2}(\mathbf{x}-\mathbf{\mu})^T\Sigma^{-1}(\mathbf{x}-\mathbf{\mu})= -\frac{1}{2}\mathbf{y}^T(S^T\Sigma^{-1}S)\mathbf{y}= -\frac{1}{2}\mathbf{y}^T(S^{-1}\Sigma^{-1}S)\mathbf{y}= -\frac{1}{2}\mathbf{y}^TD\mathbf{y}, $$ where $D=\operatorname{diag}\{d_1,\dots,d_n\}$. As a result it is rewritten as follows: $$ -\frac{1}{2}(\mathbf{x}-\mathbf{\mu})^T\Sigma^{-1}(\mathbf{x}-\mathbf{\mu})= -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 $$ Moreover, $\mathbf{x}-\mathbf{\mu}=S\mathbf{y} \Rightarrow \mathbf{x}=S\mathbf{y}+\mathbf{\mu}=[\mathbf{s_1}\:\dots\:\mathbf{s_n}]\mathbf{y}+\mathbf{\mu}=(\mathbf{s_1}\cdot\mathbf{y}+\mu_1,\dots, \mathbf{s_n}\cdot\mathbf{y}+\mu_n)^T,$ where $\mathbf{s}_j$ is the $j$-th column of matrix $S$. Using the above results, the original integral can be rewritten as follows: $$ I = (I_1,\dots,I_n)^T, $$ where the $j$-th element of $I$ is given by: $$ I_j = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}\|\Sigma\|^{1/2}} \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} (\mathbf{s}_j\cdot\mathbf{y}+\mu_j) \,\mathbf{d}\mathbf{y}\\ = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}\|\Sigma\|^{1/2}} \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} \mathbf{s}_j\cdot\mathbf{y} \,\mathbf{d}\mathbf{y}\\ + \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}\|\Sigma\|^{1/2}} \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} \mu_j \,\mathbf{d}\mathbf{y} \Rightarrow\\ I_j = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}\|\Sigma\|^{1/2}} \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} \mathbf{s}_j\cdot\mathbf{y} \,\mathbf{d}\mathbf{y} + \mu_j $$ If we write the dot product $\mathbf{s}_j\cdot\mathbf{y}$ as $$ \mathbf{s}_j\cdot\mathbf{y} = \sum_{r=1}^{n} s_{jr}y_r, $$ then the integral $I_j$ is given by: $$ I_j = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}\|\Sigma\|^{1/2}} \left(\sum_{r=1}^{n} s_{jr}y_r\right) \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} \,\mathbf{d}\mathbf{y} + \mu_j $$ I would like to ask, first, whether the whole approach above is correct or not(if so, please correct me), and, second, how could I evaluate the last integral, $I_j$. Does it converge, like the gaussian integral over $\Re^n$? Thanks in advance! Every useful comment will be extremely appretiated! AI: Consider a vector whose $j$th component is \begin{align} & \phantom{={}}\text{constant}\cdot\int_{\mathbb R^n} \frac{1}{\sqrt{d_1\cdots d_n}} \exp\left(\frac{-1}{2} \sum_{k=1}^n d_k y_k^2 \right) y_j\,dy_1\cdots dy_n \\[12pt] & = c\int_{\mathbb R^n} \prod_{k=1}^n\left(\frac{1}{\sqrt{d_k}} \exp\left(\frac{-1}{2} d_ky_k^2\right)\right) y_j\,dy_1 \cdots dy_k \\[12pt] & = c\prod_{k=1}^n \int_{\mathbb R} \frac{1}{\sqrt{d_k}} \exp\left(\frac{-1}{2} d_ky_k^2\right) y_k\, dy_k. \end{align} So it's reducible to integrals over $\mathbb R^1$, and if you're thinking about this particular problem, you probably know how to evaluate these particular integrals. It can often happen that the purpose of diagonalizing a matrix is to reduce a problem involving a vector in $n$-space to $n$ problems involving scalars.
H: An Unexpected Circle... I played around with $$z=\frac{-1+e^{it}}{\phantom{-}2+e^{it}}$$ and found that, when I draw the real against the imaginary of $z$, it pretty much looks like a circle. But neither ${\frak{R}} z $, nor ${\frak I} z$ look like $\cos $ or $\sin$. Is it due to a kind of transformed argument of $\cos $ and $\sin$? Something like $\cos(f(t))$? AI: Let $a,b,c,d$ be complex numbers and suppose c,d are not both 0 and a is not 0 ${}\ \ ad-bc\ne0$. Let $f(z) = \dfrac{az+b}{cz+d}$. Functions of this kind are called "linear fractional transformations". Then if $z$ moves around a circle in the complex plane, then $f(z)$ follows either a circle, although usually it's not the same circle, or a straight line. It will be a straight line if and only if $z$ passes through some point that makes the denominator $0$. Except when $c=0$, you'll find that it doesn't move at the same angular rate, i.e. when $z$ goes through a quarter of a circle, then $f(z)$ might go through a half circle. In your example, $a=1$, $b=-1$, $c=1$, and $d=2$. Google the terms "linear fractional transformation" and "Riemann sphere". PS: It is also the case that if $z$ moves along a straight line then $f(z)$ moves along either a straight line or a circle.
H: Regular $n$-gon in the plane with vertices on integers? For which $n \geq 3$ is it possible to draw a regular $n$-gon in the plane ($\mathbb{R}^2$) such that all vertices have integer coordinates? I figured out that $n=3$ is not possible. Is $n=4$ the only possibility? AI: This is impossible for $n=3$ as you mentioned, and for all $n > 4$, but the result is nontrivial. See here. Citation: Klobučar, D. (1998). On nonexistence of an integer regular polygon. Mathematical Communications, 3(1), 143-146. Edit: In response to the comment, this is possible for polyhedra in $\mathbb{Z}^3$. An example of an octahedron with integer coordinates can be found in the third slide here. If you are curious about general higher dimension characterizations, though, I suggest you post a separate question.
H: Why does $\sum\limits_{i=1}^n \frac{i-1}{n} = \frac{n-1}{2}$? I am trying to understand the accepted answer to this question: Find: The expected number of urns that are empty And am stuck on the part I mentioned above. I understand that: $\sum\limits_{i=1}^n \frac{i-1}{n}=\frac{0}{n}+\frac{1}{n}+...+\frac{n-1}{n}.$ But why does this equal $\frac{n\choose2}{n}$? I have also seen this sum solved as follows: $\sum\limits_{i=1}^n \frac{i-1}{n}=\frac{1}{n}\sum\limits_{i=1}^n i-1=\frac{1}{n}\sum\limits_{j=0}^{n-1}j=\frac{1}{n}\frac{n(n-1)}{2}=\frac{n-1}{2}$ With this method, I am having trouble seeing what exactly $j$ is and why we all of sudden entered it into the equation. I'd appreciate if someone could shed light on either of these ways of computing the sum. Thanks AI: The equation $$\sum_{i=1}^n(i-1)=\sum_{j=0}^{n-1}j\tag{1}$$ is just a change of variable. Let $j=i-1$; then $i=j+1$, so the lefthand side of $(1)$ becomes $$\sum_{j+1=1}^nj\;,\tag{2}$$ When $j+1=1$, clearly $j=0$, and when $j+1=n$, $j=n-1$. Thus, as $j+1$ runs from $1$ up through $n$, $j$ itself runs from $0$ up through $n-1$, and we can rewrite $(2)$ as the righthand side of $(1)$. An alternative approach that does not involve changing the index variable is to decompose the sum: $$\sum_{i=1}^n(i-1)=\sum_{i=1}^ni-\sum_{i=1}^n1=\frac{n(n+1)}2-n=\frac{n^2+n-2n}2=\frac{n^2-n}2=\frac{n(n-1)}2\;,$$ where the first summation uses the familiar formula for the sum of the first $n$ positive integers, and the second is just the sum of $n$ ones, or $n$.
H: How do i find the inverse laplace? $$ F(s) = \frac{2s-1}{s^2(s+1)^3} $$ If I try to use partial fractions, I end up with 8 constants to solve for! Is there some shortcut I'm not seeing? Am I supposed to simplify it first? Am I even doing the partial decomposition right? AI: You should have five. $$\displaystyle F(s) = \frac{2s-1}{s^2(s+1)^3} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} + \frac{D}{(s+1)^2} + \frac{E}{(s+1)^3}$$ This will produce: $A = 5$ $B = -1$ $C = -5$ $D = -4$ $E = -3$ The inverse Laplace transform, using this table is: $$-\dfrac{1}{2} e^{-t} (3~ t^2+2~ t~ e^t+ 8~t-10~ e^t+10)$$
H: Found transition matrix and state matrix I tried to found a solution for this problem but I can't! Any suggestion?Thank you. Given that A is a 2x2 matrix and that dx/dt=Ax(t) suppose that x(0)=[1 ; -3] implies x(t)=[e^-3t ; -3e^-3t] and that x(0)=[1 ; 1] implies x(t)=[e^t ; e^t] find the transition matrix for the system and find A. AI: Let $X(t) = \begin{bmatrix} x_1(t) & x_2(t)\end{bmatrix}$, where $x_1(t) = e^{-3t} (1,-3)^T$, $x_2(t) = e^{t} (1,1)^T$. Since $X$ is a solution, we have $\dot{X}(t) = A X(t)$, and, in particular, $ \dot{X}(0) = A X(0)$. Hence we get $A = \dot{X}(0) X(0)^{-1}$. We have $\dot{X}(0) = \begin{bmatrix} -3 & 1 \\ 9 & 1 \end{bmatrix}$, ${X}(0) = \begin{bmatrix} 1 & 1 \\ -3 & 1 \end{bmatrix}$, and so $A = \begin{bmatrix} 0 & 1 \\ 3 & -2 \end{bmatrix}$. Since $Y(t)=X(t)X(t_0)^{-1}$ also solves the system, and $Y(t_0) = I$, we see that $\Phi(t,t_0) = X(t)X(t_0)^{-1}$ is the state transition matrix. Since $X(t) = X(0) \begin{bmatrix} e^{-3t} & 0 \\ 0 & e^{t} \end{bmatrix}$, we have \begin{eqnarray} \Phi(t,t_0) &=& X(0) \begin{bmatrix} e^{-3t} & 0 \\ 0 & e^{t} \end{bmatrix} \begin{bmatrix} e^{3t_0} & 0 \\ 0 & e^{-t_0} \end{bmatrix} X(0)^{-1} \\ &=& X(0) \begin{bmatrix} e^{-3(t-t_0)} & 0 \\ 0 & e^{t-t_0} \end{bmatrix} X(0)^{-1} \\ &=& \frac{1}{4} \left( e^{-3(t-t_0)} \begin{bmatrix} 1 & -1 \\ -3 & 3 \end{bmatrix} + e^{t-t_0} \begin{bmatrix} 3 & 1 \\ 3 & 1 \end{bmatrix} \right) \end{eqnarray}
H: Finding the unknown cardinality of a Set given other information How would you go about finding the number of elements in $A$ if you know that the number of elements in $A\cup B$ is $20$, the number of elements in $B$ is $7$ and the number of elements in $A\cap B$ is $3$? I did it this way: I let $x$ be the number of elements in $A$. So $x+7-3=20$ implies that $x+4=20$ and so $x=16$. Does anyone agree with me or is there a different way to solve this? AI: Yes, that's the correct answer. More generally, by an inclusion-exclusion argument, $$\|A \cup B\| = \|A\| + \|B\| - \|A \cap B\| \implies 20 = \|A\| + 7 - 3 \implies \|A\| = 16$$
H: measure of Cantor type set Prove that the Lebesgue measure of the set $\{x\in[0,1]: \text{decimal expansion of $x$ contains only finitely many 7s}\}$ is zero. I have thought that that if i can show that measure of $\limsup A_k$ is 1, then it is proven, where $$A_k=\bigcup_i^{9^{k-1}} \left[\frac{10i+7}{10^k},\frac{10i+7}{10^k}\right]$$ ($A_k:={}$set of $x$ such that $k$th decimal is $7$) But by Borel Cantelli I have found that $\limsup A_k$ has measure zero. I think, the set in the questions is the complement of $\limsup A_k$. Any help or improvement or disproof of my statements are welcome. AI: Let $A$ be the set of numbers in $[0, 1]$ such that the decimal expansion doesn't contain any $7$s. By this question, $m(A) = 0$. Let $B$ be the set of numbers in $[0, 1]$ such that the decimal expansion contains finitely many $7$s, and the rest are all $0$s. Since $B$ is a subset of $\mathbb Q$, it's countable. Let $X$ be the set in your question. We have $$ X \subset \bigcup_{b \in B} (A + b). $$ By the translation invariance of the Lebesgue measure, we have $$ m(X) \le \sum_{b \in B} m(A + b) = 0. $$
H: Exterior product of $\Bbb Z[x,y]$ Let $R$ be the polynomial ring $\Bbb Z[x,y]$ in the variables $x,y$. If $M=R$ (so we are considering the $R$-module $R$) then $\wedge^2 (M)=0$. This was an example in Dummit and Foote's Abstract Algebra, pg. 449. Can someone explain to me why $\wedge^2 (M)=0$? Could you explicitly calculate some exterior products as an example? I'm having a tough time understanding this. AI: It's easy to see that $\wedge^n R\cong \bigotimes^nR/I$, where $I=\{r_1\otimes...\otimes r_n\|r_i=r_j $for $ $some $ i\neq j $}. Now $\bigotimes^2 R=(r_1\otimes r_2\|r_i\in R)=(r_1r_21\otimes1\|r_i\in R)\subset I$. This means that $\wedge^2 R\cong \bigotimes^2R/I=0$
H: Density of probability and Distribution function, how to turn one into the other I know that to know the Distribution function I got to integrate the Density function from "-oo" to "x", but how to do the inverse? AI: Differentiate.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
H: Finding general solution when given only particular solution The differential equation is given by $$ty''+ (1-2t)y'+ (t-1)y = 0.$$ The particular solution give is $e^t.$ I haven't seen an equation like this so far in my class and so far haven't found anything of use by searching on google. I've tried finding the complimentary equation but it doesn't make sense where where $tr^2 + (1-2t)r + (t-1) = 0.$ This isn't a constant coefficients or var of parameters is it? Then I've tried building another solution by using the formula: $$y_2= y_1 \int \frac{e^{-\int P}}{y_1^2}$$ But so far haven't had any success. What should I try doing? Any help would be appreciated. AI: Hint: Let $y_2(t)=e^tv(t)$ and take first an second derivatives and substitute into the equation to find $v(t)$(This metthod is called reduction of order). Your equation for $v$ will be $tv''+v'=0$, from which $v=\ln t$ and so $y_2(t)=e^t\ln t$.
H: A question in the proof of connectedness in $\mathbb{R}$ In my textbook there is a theorem saying that "If $S\subset\mathbb{R}$ is an interval, then $S$ is connected." I can follow most of the arguments provided there except the one indicated below. Can anyone help me understand it, please? To be complete, I briefly state the proof in my textbook. $\textit{Proof.}$ Suppose $S$ is an interval in $\mathbb{R}$ and $S$ is not connected. Then there exist nonempty open sets $U$ and $V$ in $S$ such that $$S=U\cup V, U\cap V=\emptyset.$$ Choose $a\in U$ and $b\in V$. Without loss of generality we may assume that $a<b$. Since $S$ is an interval, $[a, b]\subset S$. $\textit{My question is with the following argument.}$ (Arguments thereafter are omitted.) Let $c=\mathrm{sup}([a, b]\cap U)$. Since $c\in [a, b]\subset S$, so either $c\in U$ or $c\in V$. Why is $c$ must be in $[a, b]$, please? AI: The set $[a,b]$ is closed, so it contains the supremum and the infimum of each of its non-empty subsets. For a little more detail, let $u=\sup\big([a,b]\cap U\big)$; then by the definition of supremum for each $\epsilon>0$ there is a $u_\epsilon\in[a,b]\cap U$ such that $u-\epsilon<u_\epsilon\le u$. Thus, every open nbhd of $u$ contains a point of $[a,b]$, and $u$ must therefore be in the closure of $[a,b]$. But $[a,b]$ is closed, so $u\in[a,b]$.
H: How to know if a MacLaurin/Taylor Series expansion is good? This question is motivated by this question. So, given $\frac{1}{e^x + 1}$, the 4th order MacLaurin series $1 -e^x+(e^x)^2-(e^x)^3+(e^x)^4$, although correct in terms of the algebra manipulations, is not a good expansion. In general, how do we know if a given expansion is a good approximation? Should we be extra careful when the given function is a fraction and has a variable in its denominator or are there other cases where we should pause and see how the given expansion behaves with small values of $x$? AI: I'm guessing you want to then expand in the relevant series for $e^{x}$, $e^{2x}$, etc. as in your other question (getting ultimately something like $\frac{95}{12} \, x^{4} + \frac{22}{3} \, x^{3} + 5 \, x^{2} + 2 \, x + 1$). To see what's going wrong, look at the radius of convergence for the first series you want to use, i.e. $\dfrac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 + O(x^5)$. The radius of convergence is $\|x\| < 1$, so if you want to use this series to calculate $\dfrac{1}{1 + f(x)}$, then $1 - f(x) + (f(x))^2 - (f(x))^3 \dots$ is only a valid expansion if $\|f(x)\| < 1$. If you want to then use a further Taylor series expansion for $f(x)$, you have to worry about the radius of convergence for that series as well (though in this case, $e^x$ converges everywhere, so we're off the hook there). Thus, your series is only good if $\|e^x\| = e^x < 1$, so for $x<0$. If you want a series expansion for other values of $x$, you'll have to do something else (e.g. follow the suggestion in your other question, or just 'start from scratch' by taking derivates, etc.).
H: Intuition for a physical real line vs. a physical "hyperreal line" As a mathematical structure, I have no problem with the hyperreals. But I came across the following from Keisler's book "Elementary Calculus: An Infinitesimal Approach". "We have no way of knowing what a line in physical space is really like. It might be like the hyperreal line, the real line, or neither. However, in applications of the calculus, it is helpful to imagine a line in physical space as a hyperreal line." We get real number answers to physical problems in distance, from integers to transcendental numbers. However, we never get a nonzero number $w$ s.t. $w \lt 1/k, \forall k \in \mathbb N$ coming from a physical calculation, theoretical or applied. Isn't this what makes physical distances real and not hyperreal? Is it really possible that our space was never (locally) Euclidean all this time? Where are these infinitesmals and why are they hiding? Secondly, we say the real line, by construction from completing the rationals, has "no holes". Yet, the reals are a proper subfield of the hyperreals. Where do these infinitesmals "fit" on the real line to make a hyperreal line when there is no room for them to fit? In other words, if we begin by assuming a physical line segment is a hyperreal line segment and then (mathematically) remove all the infinitesmals, we get a hyperreal line segment with "holes" in the form of missing hyperreal points, but this just gives a real line segment, which has no holes. There seems to be problems in assuming physical lines can be hyperreal lines. AI: The assertion that $\Bbb R$ ‘has no holes’ is an informal paraphrase of the mathematically precise statement that $\langle\Bbb R,\le\rangle$ is a complete linear order. In other words, every non-empty $A\subseteq\Bbb R$ that is bounded above has a least upper bound in $\Bbb R$. This in no way prevents us from shoving new elements into $\Bbb R$. For example, I can take some object $p\notin\Bbb R$, let $X=\Bbb R\cup\{p\}$, and define a linear order $\preceq$ on $X$ by $x\preceq y$ iff $x,y\in\Bbb R$ and $x\le y$, or $x\in\Bbb R$, $y=p$ and $x\le 0$, or $x=p$, $y\in\Bbb R$, and $y>0$, or $x=y=p$. This in effect inserts $p$ between $0$ and all of the positive reals. The various constructions of the hyperreals do something similar, but on a grand scale, surrounding each real number with a ‘cushion’ of infinitesimally different hyperreals, and moreover adding whole galaxies of infinite hyperreals at both ends of the line. This is very different from filling existing holes, which is what we do when we complete the rationals to form the reals.
H: Strictly increasing continuous function Prove that any onto strictly increasing map $f: (0,1) \to (0,1)$ is continuous. Since its strictly increasing then for $x<y$ it implies that $f(x) < f(y)$. For continuity I must show that for any $y\in (0,1)$ there exists a $\delta>0$ such that for $\epsilon>0$ then $\|x - y\|<\delta$ implies that $\|f(x) - f(y)\|<\epsilon.$ AI: Let $a\in(0,1)$. Let $\varepsilon>0$. Since $f$ is onto, for some point $b$ we have $f(b)=f(a)-\varepsilon$, and for some point $c$ we have $f(c)=f(a)+\varepsilon$. Let $\delta=\min\{c-a,b-c\}$. Then prove that that value of $\delta$ is small enough.
H: inequality for complex numbers In general, from triangle inequality we have, for $\|z_1+\epsilon~ z_2\| < 1$ and $\|\epsilon\|=1$, $\|z_1+\epsilon~ z_2\|\leq\|z_1\|+\|\epsilon~\| \| z_2\|=\|z_1\|+\|z_2\|$. But we can not guarantee that $\|z_1\|+\|z_2\|$ will necessarily be less than 1. Following statement gives some general statement in this regard. Prove that if $\|z_1+\epsilon~ z_2\| < 1$ , for all $\|\epsilon\|=1$, then $\|z_1\|+\|z_2\| < 1$ , where $z_1,z_2$ are complex numbers. I am thinking about its analytic proof. Can any one help me ? AI: Let $z_1=\rho_1e^{it_1}$ and $z_2=\rho_2e^{it_2}$, and put $\epsilon=e^{it}.$ Now choose $t$ so that $t_1 \equiv t_2+t$ mod $2\pi$, and arrive at $$z_1+\epsilon z_2=(\rho_1+\rho_2)e^{it_1}.$$ This number has norm $\rho_1+\rho_2$ and this norm is assumed less than $1$, so the proof is immediate, since you want to conclude $\rho_1+\rho_2<1$ in the notation used here.
H: Are there any methods to exponentiate a real number with a number from an arbitrary field? How can I take the following exponent, for some real-valued number a? $$a^{3+2j-9k+3i}$$ over the field of quaternions, or any field for that matter? On wikipedia we are given the following formula, which seems relatively straight forward, but I do not know how to find a numerical expansion for any real-number a. AI: (I'm assuming that $a>0$). What axioms are you starting from for exponentiation? One natural approach is to notice that in the reals, for $a>0$ $$a^x = e^{x\log a},$$ so if you have a definition for $e^x$ (for instance using the above power series) it is natural to define exponentiation of $a$ in any field using this formula as well. If $x = x_r + v$ is the decomposition of $x$ into its real and pure quaternionic parts, then by the formula you've listed above (and setting aside the question of checking that the power series of $e^x$ does indeed converge for all quaternions $x$, and does so to the formula you've listed): $$a^x = e^{x_r\log a}\left(\cos [\\|v\\|\log a] + \frac{v}{\\|v\\|}\sin[\\|v\\|\log a]\right).$$
H: Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$ Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$. My Attempt: We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have $$ \begin{cases} x^2+y^2\geq 2xy\\ y^2+z^2\geq 2yz\\ z^2+x^2\geq 2zx\\ \end{cases} $$ So $x^2+y^2+z^2\geq xy+yz+zx$. How can I solve for $(x,y,z)$ after this? AI: Suppose that $(x,y,z)$ is a solution. An even number of these must be odd. If two are odd, say $x$ and $y$, then $x^2+y^2$ has shape $4k+2$, and therefore so does $x^2+y^2+z^2$, since $z^2$ is divisible by $4$. But $2xyz$ has shape $4k$. So $x,y,z$ are all even, say $2u,2v,2w$. Substituting we get $u^2+v^2+w^2=4uvw$. Again, $u,v,w$ must be all even. Continue, forever. We conclude that $x$, $y$, and $z$ are divisible by every power of $2$. It follows that $x=y=z=0$. Remark: The same argument can be used for $x^2+y^2+z^2=2axyz$. This is an instance of Fermat's Method of Infinite Descent, aka induction.
H: Wave equation: show eventually $\int_{\mathbb{R}}u_x^2 = \int_{\mathbb{R}}u_t^2$ Suppose $u$ solves the wave equation in $\mathbb{R}$ and has compactly supported initial data $f(x) = u(x,0)$ and $g(x)=u_t(x,0)$. Show that the "kinetic energy" $\int_{\mathbb{R}}u_t^2$ eventually equals the "potential energy" $\int_{\mathbb{R}}u_x^2$. My attempt so far: When I expand $\int_{\mathbb{R}}u_t^2 - u_x^2$ using d'Alembert's formula, I get $$\int_{\mathbb{R}}f'(x+t)f'(x-t)-g(x+t)g(x-t)\\ + f'(x+t)g(x+t) - f'(x-t)g(x-t)dx.$$ The first two terms will eventually be zero, because at least one of the two factors of each will be zero (since the initial data has compact support). I need to find a way to make the second two terms zero. I'm trying to do it by integration by parts, noting that the last two terms can be written as $$\left.f'g\right]_{x-t}^{x+t}$$ or $$\int_{x-t}^{x+t} f''g+g'f',$$ but this isn't getting me anywhere. AI: Don't do integration by parts. See that the third and fourth term are equal by using substitution. Once you see it, you will see it is very easy.
H: Cantor-Lebesgue function and an increasing function are equal almost everywhere Denote by $\varphi$ the cantor-lebesgue function and suppose $f$ is a certain increasing function defined on [0,1] and such that $f(x)=\varphi (x)$ for all $x\in[0,1]-C$ where $C$ is the cantor set. Prove that $f(x)=\varphi(x)$ for all $x\in[0,1]$. Any help will appreciated. AI: For each $x \in C$, we have $$ \sup\{f(y) : y \le x, \ y \in [0,1]-C\} = \sup\{\varphi(y) : y \le x,\ y \in [0,1]-C\} $$ and $$ \inf\{f(y) : y \ge x, \ y \in [0,1]-C\} = \inf\{\varphi(y) : y \ge x, \ y \in [0,1]-C\} $$ Now determine that $$ \sup\{\varphi(y) : y \le x,\ y \in [0,1]-C\} = \inf\{\varphi(y) : y \ge x,\ y \in [0,1]-C\} = \varphi(x)$$ and you will be done.
H: How to graph aggregate portfolio performance? Here is an example of a performance graph of a "motif" on Motif Investing. The motif is a portfolio containing Apple and Microsoft stocks weighted at 50% each. The graph represents the aggregate performance of the 2 stocks over time (the blue line). The green line is the S&P500. The y-axis is measured in percent. Given the historical stock prices how would you generate this graph (or something similar to it)? AI: To aggregate two stocks, you have to imagine a portfolio with them in some ratio. The natural thing is $\frac 12$ each by starting dollar value, so you bought the same dollar amount of each on day $1$ and held it throughout. Or maybe you rebalance at the end of each year-selling the one that has gained the most and buying the other so you start each year with the same dollar amount. Maybe you like Apple better than Microsoft, so your portfolio is $\frac 34$ Apple and $\frac 14$ Microsoft. The S & P 500 is essentially this with (about) 500 stocks in a ratio of market capitalization. To create the data, just make a column with the Apple and Microsoft prices each day, then evaluate the value of the portfolio under the rules you choose. If it sounds like this is more about what you want to plot than how to do it, you are correct. That is the hard part. Whatever you plot, please state clearly what you are plotting.
H: Given $z=f(x,y)$, what's the difference between $\frac{dz}{dx}$ and $ \frac{\partial f}{\partial x}$? Given $z=f(x,y)$, what's the difference between $\displaystyle\frac{dz}{dx}$ and $\displaystyle \frac{\partial f}{\partial x}$? I got confused when I saw $dz/dx= \partial f/\partial x+\partial f/\partial y*dy/dx$, could somebody explain the difference to me? Thanks AI: When you write $\displaystyle \frac{dz}{dx}$ you are assuming that $z$ is a function of $x$ only. This forces $y$ to be a function of $x$, therefore abusing the notation we write $y=y(x)$. Meanwhile, we have that $f$ continues to be a function of two variables, therefore $\displaystyle \frac{\partial f}{\partial x}$ means the usual partial derivative with respect to $x$. When we take total derivatives we usually write it this way: $$dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy.$$ If we assume $z$ is a function of a certain parameter $t$ then we'd have $$\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}.$$ In this case the parameter is $x$ itself, and we'd abuse notation by writing $$\frac{dz}{dx} = \frac{\partial f}{\partial x} \frac{dx}{dx} + \frac{\partial f}{\partial y} \frac{dy}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}.$$ We are saying that "dx/dx = 1", essentially.
H: Prove or disprove: There exists a group $G$ and a normal subgroup $N$ such that $G$ is non-abelian, but both $N$ and $G/N$ are abelian. Prove or disprove: There exists a group $G$ and a normal subgroup $N$ such that $G$ is non-abelian, but both $N$ and $G/N$ are abelian. Can anyone give me some hint on this question, please? What theorem(s) in abstract algebra is(are) related to this question, please? Thank you! AI: Hint If $H$ has prime order $p$ it is abelian. Moreover, if $G:H=2$ then $H$ is normal and $ G/H$ is also abelian. So look for a non-abelian group of order $2p$ where $p$ is prime....
H: name of curve of cluster of points of the form $(x,x^2...x^n)$ in $R^n$ what is the name of the curve made up of the points $(x,x^2...x^n)$ in $\mathbb {R}^n$ for all $x\in \mathbb R$?? For example: in $\mathbb R^2$ it would just be a parabola. AI: This is basically the rational normal curve of degree $n$. ("Basically" because the usual rational normal curve is $n$-dimensional projective space and has an additional point at infinity.)
H: Prove or disprove: There exists an element in the quotient group $\mathbb{R}/\mathbb{Z}$ of order 6. Prove or disprove: There exists an element in the quotient group $\mathbb{R}/\mathbb{Z}$ of order 6. By counting formula, since $\|\mathbb{R}/\mathbb{Z}\|$ is infinite, it seems possible to have an element of order 6. But how to find one? In addition, what is $\mathbb{R}/\mathbb{Z}$, please? AI: Since the groups we are dealing with are abelian, I will use $+$ as the group operation. The elements of $\mathbb R/\mathbb Z$ are cosets of the subgroup $\mathbb Z \subset \mathbb R$. For example, the identity element is the coset $[0]=0+\mathbb Z$, and $[.5]=.5 + \mathbb Z$ is another element. Note that $[a]=[b]$ if and only if $a-b\in \mathbb Z$. So $[.5]+[.5]=[1]=[0]$. Thus, we see that even though both $\mathbb R$ and $\mathbb Z$ are torsion free, the quotient group still has torsion. I assert that you can rephrase your question as "Does there exist a real number $x$ such that $6x$ is an integer, but no smaller multiple of $x$ is?" When phrased like this, the question should be easier to answer. Understanding why the question can be rephrased this way is an important step in understanding $\mathbb R / \mathbb Z$
H: Let $F:\mathbb{R}^n\to\mathbb{R}^n$ a transformation of class $C^1$. Supose that $dF_{x_0}$ is inyective. Let $F:\mathbb{R}^n\to\mathbb{R}^n$ a transformation of class $C^1$. Supose that $dF_{x_0}$ is inyective. Prove that there exist an $\varepsilon>0$ such that if $G$ is another transformation $C^1$ such that $\|\|dF_{x_0}-dG_{x_0}\|\|<\varepsilon$ then $G$ is locally injective in $x_0$. Please help with this problem. I dont know how to approach it. AI: From the conditions of the problem, we have that the map $DF_{xo}$ from $T_{xo}$ to $T_{F(xo)}$ is injective. Since the map $DF$ is an injective linear map from $\mathbb R^n$ to $\mathbb R^n$ (i.e., between vector spaces of the same dimension) , then it is also surjective, i.e., $DF_{xo}$ is a linear isomorphism between tangent spaces, so that the conditions of the inverse function theorem for the existence of a differentiable local inverse in a 'hood $U_{xo}$ of $xo$. Now, this inverse exists in an open 'hood ( Neighborhood = 'hood; the 'hood ). Now, if $\|\|DG_{xo}-DF_{xo}\|\|< \epsilon$ , for $\epsilon$ small-enough , then $DG_{xo}$ lies in the 'hood $U_{xo} $ where $F$ is invertible, then you can show that $G$ and $DG$ both also satisfy the conditions for the existence of a local inverse, and therefore , $DG_{xo}$ is locally-injective.
H: H acting on G by left translation $H$ is a subgroup of $G$ and acts on $G$ by left translation, describe the orbits. Here is my take, at first I thought well, isn't that just left cosets? But, that seems too easy. By translations, does that mean by adding some element of $H$ on the left of $G$? Am I even close to the solution? AI: You've almost got the right idea. The action of left translation means for $h\in H$, the action $$ is $hg := hg$, where $g$ ranges in $G$. The orbits aren't quite the left cosets of $H$. The orbit of $g\in G$ is defined to be $$\{h*g: h\in H\}.$$ What is this set?
H: Fraction exponential limit in complex plane Let $n$ be an integer. I need to compute the limit $$\lim_{z\rightarrow 2n\pi i}\dfrac{e^z-1}{z-2n\pi i}$$ for complex number $z$. I think I can't use L'Hospital here since $z$ is complex. How can I do it? AI: Hint: use the definition of the derivative at a point. $$ f'(z_0) = \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} .$$
H: Can we express these sets as Cartesian products of two subsets of $\mathbf{R}$? Let sets $A$ and $B$ be given as follows: $$A := \{ (x,y) \in \mathbf{R}^2 \| \ \ x < y \ \ \} $$ and $$B := \{ (x,y) \in \mathbf{R}^2 \|\ \ x^2 + y^2 < 1 \ \ \}.$$ Can we express $A$ or $B$ as a Cartesian product of two subsets of $\mathbf{R}$? If we can, then how? If we can't, then why not? What is the most general statement that we can make regarding an arbitrary given set of points in the plane? AI: We have $X \times Y = \{ (x,y) \| x \in X, y \in Y \}$. In particular, if $(x_1,y_1), (x_2,y_2) \in X \times Y$, then $(x_1,y_2), (x_1,y_2) \in X \times Y$. Neither $A$ nor $B$ can be expressed as a Cartesian product. Take $A$ first. We have $(1,2),(-2,-1) \in A$, but $(1,-1) \notin A$. Hence $A$ cannot be expressed as a Cartesian product. Similarly for $B$. Let $x \in (\frac{1}{\sqrt{2}},1)$. Then $(x,0),(0,x) \in B$, but $(x,x) \notin B$.
H: Why does the fiber coproduct in $\mathbf{Set}$ actually satisfy the universal property? Suppose you have two morphisms $f\colon A\to B$ and $g\colon A\to C$. Then the fiber coproduct of $B$ and $C$ over $A$ exists and is the disjoint union $B\coprod C$ where we identify $fx$ and $gx$ for all $x\in A$. I'm a little confused on how this satisfies the universal property. Suppose you had another set $D$ and morphisms $h\colon B\to D$ and $k\colon D$ such that $hf=kg$, so that the diagram commutes. How would you define the unique morphism $u$ from $B\coprod C\to D$? Is it something along the lines of $u(b)=h(b)$ if $b\notin f(A)$ and $u(c)=k(c)$ if $c\notin g(A)$, and $u(f(x))=u(g(x))=k(f(x))=h(g(x))$ if $f(x)=g(x)$ is one of the points we smashed together in $B\coprod C$? And this definition makes sense since $hf=kg$. AI: That is exactly it. Can you see why the map is unique? (Perhaps it is better to call this just the "pushout" of $B \leftarrow A \rightarrow C$, or more informally the "glueing of $B$ and $C$ along $A$". The name "fibre coproduct" seems a bit weird, because the construction has nothing to do with fibres!)
H: Prove or disprove: There exists a ring homomorphism $\phi: \mathbb{C}\rightarrow \mathbb{R}\times\mathbb{R}$. Prove or disprove: There exists a ring homomorphism $\varphi: \mathbb{C}\rightarrow \mathbb{R}\times\mathbb{R}$. I think it is intuitive to try $\varphi: \mathbb{C}\rightarrow \mathbb{R}\times\mathbb{R}$ defined by $\varphi: a+bi\mapsto (a, b)$. Then let us check if this works. It is trivial that $\varphi[(a+bi)+(c+di)]=(a+b, c+d)=\varphi(a+bi)+\varphi(c+di)$. $\varphi(1+0i)=(1, 0)\neq(1,1)$. $\varphi[(a+bi)\times(c+di)]=\varphi[(ac-bd)+(ad+bc)i]=(ac-bd, ad+bc)$. From the above results, it seems that this definition does not satisfy the requirement of a ring homomorphism. Is there a way to modify it, please? Or is it possible at all? AI: No such homomorphism can exist, so long as ring homomorphisms are assumed to preserve unity. Main Result. There is no ring homomorphism $\mathbb{C} \rightarrow \mathbb{R} \times \mathbb{R}$. Proof. Suppose toward a contradiction that $\varphi : \mathbb{C} \rightarrow \mathbb{R} \times \mathbb{R}$ is a ring homomorphism. Then by Lemma 0 below, it follows that $-1$ has a square root in $\mathbb{R} \times \mathbb{R}$. Thus by Lemma 1 below, it follows that $-1$ has a square root in $\mathbb{R}$. But this contradicts Lemma 2 below. Lemma 0. If a ring homomorphism $\mathbb{C} \rightarrow R$ exists, then $-1$ must have a square root in $R$. Proof. I claim that $\varphi(i)^2$ is always a square root of $-1$ in $R$. To see this, compute: $$\varphi(i)^2 = \varphi(i^2) = \varphi(-1) = -\varphi(1) = -1.$$ Lemma 1. For all rings $R$ and $S$, if $-1$ has a square root in $R \times S$, then it must have a square root in both $R$ and $S$. Proof. Suppose $-1$ has a square root in $R \times S$, call it $(r,s)$. Then $(r,s)^2 = -1$. Thus $(r^2,s^2) = (-1,-1)$. Thus $r^2 = -1$. Thus $-1$ has a square root in $R$. A similar argument shows that it must have a square root in $S$. Lemma 2. The element $-1 \in \mathbb{R}$ does not have a square root in $\mathbb{R}$. Suppose toward a contradiction that it did, call this value $i_\mathbb{R}.$ Then $(i_\mathbb{R})^2 = -1$. Thus $(i_\mathbb{R})^2 + 1 = 0$. Now we know that $\forall x \in \mathbb{R} : x^2 \geq 0$. Thus $\forall x \in \mathbb{R} : x^2 + 1 > 0.$ Thus $(i_\mathbb{R})^2 + 1 > 0$. Ergo $0 > 0$, a contradiction.
H: Combinatorial Proof of Multinomial Theorem - Without Induction or Binomial Theorem I've been trying to rout out an exclusively combinatorial proof of the Multinomial Theorem with bounteous details but only lighted upon this one - see P2. Any other helpful ones? $(x_1+\cdots+x_k)^n =$ Top of Page 39 from UNC: $\sum\limits_{\large{(a_1, ..., a_{k - 1}) \; \ni \; 0 \le a_1+...+ a_{k - 1} \le n}}\dbinom{n}{a_1,...,a_{k-1}}\cdot x_1^{a_1} \cdots x_{k-1}^{a_{k-1}}x_k^{\large{n - a_1 - \cdots - a_{k-1}}}$ $= \sum\limits_{\large{a_1+...+a_k = n \; \& \; a_i \ge 0 }}\dbinom{n}{a_1,...,a_k} x_1^{a_1} \cdots x_k^{a_k}$ Bottom of P1 from MSU. I see that both are the Multinomial Theorem but which one is better? I thought to try this myself, following the combinatorial proof of the Binomial Theorem. So esteem each term (in green) in $(x_1+\cdots+x_k)^n = \underbrace{\color{green}{[x_1+\cdots+x_k]...[x_1+\cdots+x_k]}}_{\text{n terms}}$ as one box from which to choose $x_1, ..., x_k$. Since each term/box (in green) contains $k$ terms, the total number of terms $ = k^n$. First, consider $x_1$. ● For $x_1^n$, must have $a_ 1 = n\qquad \& \qquad a_2 = ... = a_k = 0$. ● For $x_1^{n - 1}$, must have $a_1 = n - 1 \qquad \& \qquad a_2 \text{ OR } ... \text{ OR } a_k = 1$ (the latter due to $a_1+...+a_k = n$ in definition from P1 of MSU) ... ● For $x_1^{1},$ must have $a_1 = 1 \qquad \& \qquad a_2 \text{ OR } ... \text{ OR } a_k = n - 1$ ● For $x_1^{0},$ must have $a_1 = 0 \qquad \& \qquad a_2 \text{ OR } ... \text{ OR } a_k = n$ The above extends to and must hold for all $x_1, ..., x_k$. How to translate all this into combinatorial notation? How to forge ahead and complete please? AI: This is one approach: When you expand $(x_1+x_2+....+x_k)^n$ , you will have a collection of terms $c_ix_1^{n_1}x_2^{n_2}......x_k^{n_k}$ for each "box"/factor $i$ in your green, so that $n_1+n_2+..+n_k=n$. For a fixed given $k$-tuple $(n_1,..,n_k)$ , the assignment of the $n_1, ..., n_k$ can be chosen in: $\dbinom {n}{n_1} \times \dbinom {n-n_1}{n_2} \times.....\dbinom{n-n_1-....-n_{k-1}}{n_k}$ ways. The $n_1$ balls can be chosen from the original $n$. After having used up $n_1$ balls, you only have $n-n_1$ balls left, which you can use to choose the $n_2$ balls for $x_2$, etc.. This expansion is precisely the multinomial coefficient: $$\dbinom {n}{n_1,n_2,....,n_k} $$ The above is true only for the given $k$-tuple $(n_1,..,n_k)$. Now, we do the sum over all $k$-ples $(n_1, n_2,....,n_k)$ with $n_1+n_2+...+n_k=n$ The reason why we sum over all $k$-ples is that, when we expand the product : $$ (x_1+x_2+...+ x_k)^n= \underbrace{(x_1 + x_2+...+x_k) (x_1+x_2...+x_k)....(x_1+x_2+..+x_k)}_{n \text{ times }}$$ , you will ultimately multiply one element $x_{j1}$ from the first copy of $(x_1+x_2+..+x_k)$ with one element $x_{j2}$ in the second copy,... with an $x_{jk}$ from the $k$-th copy of $(x_1+...+x_k)$, i.e., the product will have one element of each copy of the sum. Notice some specific cases/examples, for $k=n=3$; I think this will be clearer than a more formal argument: $(\color{green}{x_1}+x_2+x_3)(x_1+x_2+x_3)(x_1+x_2+x_3)$ : We start with $\color{green}{x_1}$, multiply by some element $x_{j2}$ in the second parenthesis to get $x_1x_{j2}$, and then we multiply this $x_1x_{j2}$ by some element $x_{j3}$ in the third parenthesis (notice that we may have $x_{j2}=x_1, x_2$ or $x_3$). How many different monomials can we have? Well, each monomial will contain 3 terms, possibly repeated, so that the sum of the exponents of $x_1x_{j2}x_{j3}$ will add up to 3, e.g., we will have terms like $x_1x_1x_2=x_1^2x_2$, or $x_1^3$, or $x_1x_2x_3$, etc. How many monomials of this sort can we have? Well, we can have as many as the number of all the 3-tuples (ie triples) of the exponents, $(n_1, n_2, n_3)$ chosen with ordering and with replacement, for the respective terms $x_1,x_2,x_3$, such that $n_1+n_2+n_3=3$ (because each monomial contains $k = 3$ terms). Basically, we are summing over all strings $(x_1^{n_1}x_2^{n_2}...x_k^{n_k})$ , where all the exponents add-up to $n$; this amount of terms is equivalent to the number of solutions of the equation $x_1+ x_2+....+x_k =n$
H: Semicircle contour for integrating $t^2/(t^2+a^2)^3$ Let $a\in\mathbb{R}$. Evaluate $$\int_0^{\infty}\dfrac{t^2}{(t^2+a^2)^3}dt$$ The function is even, so the value of the integral is half of $\int_{-\infty}^{\infty}\dfrac{t^2}{(t^2+a^2)^3}dt$ I'm going to use countour integration along the semicircle in upper-half plane with radius $R$. The integral along the real line is $\int_{-R}^{R}\dfrac{t^2}{(t^2+a^2)^3}dt$. Now for the curved part of the semicircle, I parametrize $z=Re^{it}$ for $0\leq t\leq 2\pi$. Then the integral becomes $$\int \frac{z^2}{(z^2+a^2)^3}dz=\int_0^{2\pi}\frac{R^2e^{2it}}{(R^2e^{2it}+a^2)^3}\cdot Rie^{it}dt$$ And the absolute value of this integral is bounded from above by $\left\|\dfrac{2\pi R^3}{(R^2e^{2it}+a^2)^3}\right\|$. Why does this go to zero as $R\rightarrow \infty$? It seems like it's because the denominator has larger power of $R$, but how to make that rigorous? AI: Standard calculus result. If $p(x),q(x)$ are two polynomials with $\deg(q(x))>\deg(p(x))$, then $$\lim_{x\rightarrow\pm \infty} \frac{p(x)}{q(x)} = 0.$$ You can prove it by dividing top and bottom by the highest power of $x$ appearing in $p$. In your particular case, the numerator is degree 3 in $R$, and the denominator is degree 6 in $R$. Also note that $\|R^2 e^{2it}+a^2\| \geq R^2-a^2$.
H: What is the center of fundamental groupoid? $C$, $D$ are two categories. $F$, $G$ are functors between $C$ and $D$: $F, G: C\rightarrow D$. Let $Nat(F,G)$ be all the natural transformations between F and G. Like what we do for groups, define the center of the a category $C$: $Z(C)=Nat(id_C,id_C)$. Then what is the center of the fundamental groupoid $Z(\pi_{1}(X))$? AI: This is not really a question about centers of fundamental groupoids but centers of groupoids in general. If $\newcommand{\G}{\mathcal{G}}\G$ is a group, considered as a 1-object category, then $Z(\G)$ (in the usual sense) is $\newcommand{\Nat}{\mathrm{Nat}}\newcommand{\id}{\mathrm{id}}\Nat(\id_\G,\id_\G)$. I claim that if $\G$ is a connected groupoid then $\Nat(\id_\G,\id_\G)$ is canonically isomorphic to $Z(\mathrm{Hom}_\G(x,x))$ for any choice of an object $x$ of $\G$. Indeed, pick an element $g \in Z(\mathrm{Hom}_\G(x,x))$. For any object $y$ of $\G$, define an element $g_y \in \mathrm{Hom}_\G(y,y)$ by choosing an arbitrary isomorphism $f: x \to y$ and setting $g_y = f \circ g \circ f^{-1}$. If we had chosen a different $f'$, then there would be an automorphism $h$ of $x$ such that $f' = f \circ h$. But $$ f \circ h \circ g \circ h^{-1} \circ f^{-1} = f \circ g \circ f^{-1},$$ by the assumption that $g$ is in the center, and therefore $g_y$ is well defined independent of the choice of $f$. One can now check that the elements $g_y$ define a natural transformation from $\id_\G$ to itself and that all natural transformations arise this way.
H: The forever moving billiard ball Suppose I have a rectangular table, dimensions $x$ by $y$, and a billiard ball is positioned in the very center. For descriptive convenience, let us impose a coordinate system on this table with an origin (0,0) in the center of the table where I strike the ball. Now say I strike the ball at an angle of $\theta$ with respect to the horizontal. If the ball moves forever after being struck, for what values of $\theta$ will the ball form a closed loop and eventually return to initial conditions and retrace its own path all over again, and for what value of $\theta$ will the ball not form a closed loop and never re-trace its own path? AI: If you unfold the billiard trajectory, the question becomes equivalent to this: Suppose I have a rectangular lattice, where the unit cell has dimensions $x$ by $y$, and a billiard ball is positioned on a lattice point, which we will call the origin $(0,0)$. For what values of $\theta$ will the ball strike another lattice point $(mx,ny)$, where $m$ and $n$ are even integers? Does that help?
H: Large Regression dataset For a project I need a large regression (least squares) dataset: If $n$ is the number of samples and $p$ the number of features, then I need $p < n$ and $p,n$ both very large. For example $n=10,000,000$, $p=1,000,000$. Does anybody know such a dataset or at least where I could get one? I was thinking about natural language processing data but I couldn't find any. AI: Why don't you generate one writing a small piece of code ? You could include noise in the dependent and independent variables ? About large data sets, may be the Census bureau could provide you one or consumer association too. I suggest you have a look at http://www.census.gov/main/www/access.html
H: Question on polynomial. Here is a question from Hoffman: If $F$ is a field and $h$ is a polynomial over $F$ of degree $\ge 1$, show that the mapping $f \rightarrow f(h) $ is a one-one linear transformation of $F[x]$ into $F[x]$. Show that this transformation is an isomorphism of $F[x]$ onto $F[x]$ if and only if deg $h = 1$. I believe that the first part is easy. To show that the mapping $f \rightarrow f(h) $ is a one-one linear transformation of $F[x]$ into $F[x]$, we need to show that if $f(h) = 0 $, then $f=0$. Suppose $f(h) = 0 $. Since $h$ is a polynomial of degree $\ge 1$ and since $f(h)=0$, $f$ must be equal to $0$. The second part is the part I don't understand. To show that it is an isomorphism, we need to show that 1) It is one-one (already shown) 2) Show that the linear structure is preserved. (Easy to show) 3) Show that the transformation is onto. I don't see how condition 3 can be fulfilled because when deg $h=1$, $fh$ can never be mapped to a polynomial with degree $<1$ in $F[x]$ (e.g. $f=cx^0$, where $c$ is a scalar in the field $F$) AI: If $f(x)=c$ then $f(h)=c$ also. Addendum: If $h$ has degree $1$, $h(x)=ax+b$ then the mapping is surjective: every $g(x)$ has a preimage $f(x)=g(\frac{x-b}{a})$. Further, let $f(x)=c_nx^n+\ldots +c_0, \ c_n\ne 0$. If $f(ax+b)=0$ then the leading coefficient $c_na^n$ must be equal to $0$. This is impossible, so $f=0$ and the mapping is injective.
H: limit of an absolute sequence: ${b_n} = \|{a_n} - 1\|$ $$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } {a_n} = 3 \cr & {b_n} = \|{a_n} - 1\| \cr} $$ Hence, $$\mathop {\lim }\limits_{n \to \infty } {b_n} = \|3 - 1\| = 2$$ Is it right to say that? If so, is it sufficent for a proof? AI: If you're asking, then it isn't sufficient. Consider this $$\lim \limits _{n\to \infty}a_n=3\implies \lim \limits _{n\to \infty}(a_n)-1=2\implies \left\|\lim \limits _{n\to \infty}(a_n-1)\right\|=2\mathop{\implies}^{\text{Why?}} \lim \limits _{n\to \infty}\left\|(a_n-1)\right\|=2.$$
H: how to integrate convolution $\int fg$ I am stuck on this question: $f$,$g$ are blocked, continuous functions and $\int_{-\infty}^\infty\|f(x)\|dx<\infty,\int_{-\infty}^\infty\|g(x)\|dx<\infty$. show that: $$\int_{-\infty}^\infty (fg)(t)dt=\int_{-\infty}^\infty f(x)dx\int_{-\infty}^\infty g(x)dx$$ looking for some clues to put me on the right direction. AI: Hint: $$(fg)(t)=\int_{\mathbb{R}}f(u)g(t-u)du$$, so $$\int_{\mathbb{R}}(fg)(t) dt=\int_{\mathbb{R}}\int_{\mathbb{R}}f(u)g(t-u)dudt$$, then we can change the order of integration by Fubini's theorem when you are manipulating the variables.
H: Compactness and Convergence of Subsequences Let $(X,\rho)$ be a metric space. Suppose that $(x_n)_{n\in\mathbb Z_+}$ is such a sequence in $X$ that any subsequence has a further subsequence that is convergent. However, the limits of these sub-subsequences are not necessarily the same. Let $A$ be the image set of the sequence $(x_n)_{n\in\mathbb Z_+}$ (that is, the elements of the sequence viewed as a set, after suppressing repetitions). My question is: Without further assumptions (chiefly, completeness of $(X,\rho)$), $\textbf{is it true that the closure of } A \textbf{ is compact?}$ If so, how can I prove it? My intuition suggests it is, since I think the convergence of the sub-subsequences imposes a bound on both $A$ and its accumulation points, but I cannot seem to formalize this intuition. AI: One can show that $\bar A$ is compact by showing that it is sequentially compact. Let $\{y_n\}\in \bar A$ be any sequence. We can assume that $$\{ y_n\in \bar A \ : n\in \mathbb N\}$$ is not finite (If it is finite $\{y_n\}$ has a convergent subsequence). By definition, there is $x_{m_1}$ such that $d(y_1, x_{m_1}) \leq \epsilon/2$. Write $k_1 = 1$. Assume by induction that $$k_1< k_2 <\cdots < k_n\ ,\ m_1 < m_2 < \cdots <m_n$$ has been chosen such that $d(y_{k_l} ,x_{m_l}) \leq 1 / 2^l$ for $l = 1, \cdots, n$. Then consider the set $$\{y_a:\ \ a> k_n\}\ .$$ This set is not finite by assumption, thus there is $k_{n+1} > k_n$ such that $$y_{k_{n+1}}\notin \{ x_i:\ \ \ 1\leq i\leq m_n\}\ .$$ Thus there is $m_{n+1} > m_n$ such that $d(y_{k_{n+1}} , x_{m_{n+1}}) \leq 1/2^{n+1}$. Thus by induction we have found subsequences $\{y_{k_n}\}$ of $\{y_n\}$ and $\{x_{m_n}\}$ of $\{x_n\}$ such that $d(y_{k_n}, x_{m_n}) \leq 1 / 2^n$. By your assumption, by again picking a subsequence if necessary, $\{x_{m_n}\}$ converges to some $x\in \bar A$. Then for all $\epsilon >0$ there is $N\in \mathbb N$ such that $$d(x, x_{m_n}) <\epsilon,\ \ \ \forall n\geq N\ .$$ Pick a large $N$ such that $1/2^N\leq \epsilon$, then $$d(y_{k_n} , x) \leq d(y_{k_n}, x_{m_n}) + d(x_{m_n}, x) < 2\epsilon,\ \ \forall n \geq N\ .$$ Which means that $\{y_{k_n}\}$ converges to $x$. Thus $\bar A$ is sequentially compact.
H: Calculating expected value for joint density function $$f(x,y) = 2$$ $$0<x \le y < 1$$ I want to calculate $$E(XY)$$ In order to do that, I need to know the range for $x$ and $y$. Are they both $0$ to $1$? AI: The range is $$ \begin{align} 0<&x\leq y \\ 0 <&y < 1 \end{align} $$ If you're unsure about these things you can always check it, since $$ \int\int f(x,y)dydx=1 $$ if the integrals are over the correct domains. With 0 to 1 for both variables, we get $$ \int_0^1\int_0^1f(x,y)dydx=\int_0^1\int_0^12dydx=2\int_0^1dx=2 $$ so it can't be 0 to 1 for both $x$ and $y$ -- integrating over the entire space should give 1. If we use 0 to 1 for $y$ and 0 to $y$ for $x$, instead we get: $$ \int_0^1\int_0^y2dxdy=\int_0^12ydy=1. $$ So that's a quite easy check which is useful when you're uncertain.
H: Combinatorics/Task Dependency Here is a competitive programming question: You have a number of chores to do. You can only do one chore at a time and some of them depend on others. Suppose you have four tasks to complete. For convenience, we assume the tasks are numbered from 1 to 4. Suppose that task 4 depends on both task 2 and task 3, and task 2 depends on task 1. One possible sequence in which we can complete the given tasks is [3,1,2,4] - in this sequence, no task is attempted before any of the other tasks that it depends on. The sequence [3,2,1,4] would not be allowed because task 2 depends on task 1 but task 2 is scheduled before task 1. In this example, you can check that there exactly three possible sequences compatible with the dependencies: [3,1,2,4], [1,2,3,4] and [1,3,2,4]. In each of the cases below, you have N tasks numbered 1 to N with some dependencies between the tasks. You have to calculate the number of ways you can reorder all N tasks into a sequence that does not violate any dependencies. [Task, Dependency(s)] : [1, NA], [2,1], [3,2], [4,1], [5,4], [6, 3 and 5], [7,6], [8,7], [9,6], [10, 8 and 9]. I inferred the following: Any sequence will always start with 1, since it has no dependencies. 6 will always be in the 6th position of any sequence. 10 will always be at the last position. Then, by trial and error, and listing all possibilities for the two separate parts of the sequence: 1 _ _ _ _ 6 and 6 _ _ _ 10 I got 6x3 = 18 possibilities. However, for a larger set of data, these deductions would not be easy. What is the way to calculate this logically and find the number of possibilities, and how can this be integrated into a program? (I have tried to represent the question as clearly as possible, but you can visit this link to view the question (Q. No 4): http://www.iarcs.org.in/inoi/2013/zio2013/zio2013-qpaper.pdf) I am a high school student preparing for a programming competition, and I haven't taken many courses on algorithm design, so this might be a trivial question - please excuse me! EDIT: 2nd Part: [Task, Dependency(s)] : [1, NA], [2,1], [3,2], [4, NA], [5,4], [6, 3 and 5], [7,6], [8,7], [9,6], [10, 9], [11, 8 and 10], [12, 11], [13, 11] Diagram: 1 \ 4 2 / \ 5 3 / \/ 6 / \ 7 9 / \ 8 10 \ / \ / \ / 11 /\ 12 13 I understand that 7,8,9,10 will have 4 C 2 = 6 combinations, and that 12 and 13 will have two combinations. However, what about the starting? How can we include all 1,2,3,4 and 5 and find the number of possibilities for that segment? AI: I would draw a diagram 1 / \ 2 4 \| \| 3 5 \ / 6 / \ 7 \ \| 9 8 / \ / 10 You need to work out how many orders for $2,3,4,5$. This is ${4 \choose 2}=6$. Similarly for $7,8,9$ is ${3 \choose 1}=3$. So in total there are $6 \times 3 = 18$ possibilities. Part (b) is similar: you have to order $1,2,3,4,5$ in ${5 \choose 3}=10$ ways, $7,8,9,10$ in ${4 \choose 2}=6$ ways and $12,13$ in ${2 \choose 1}=2$ ways, making $120$ ways overall. Part (c) would be $\left(\frac{3!}{1!1!1!}\right) \left(\frac{2!}{1!1!}\right) \left(\frac{6!}{3!3!}\right)\left(\frac{1!}{1!}\right) .$
H: Number of even and odd subsets -- wrong question? The book on Discrete Mathematics I'm following poses the following problem: Prove that a nonempty set has the same number of odd subsets (i.e., subsets with odd number of elements) as even subsets. A short proof is also given, but I have a counter-example: Consider the all the subsets of the set $ \{1, 2, 3\} = \{1\}, \{2\}, \{3\}, \{1,3\}, \{2,3\}, \{1,2\}$ and $\{1,2,3\}$. Now, number of odd subsets is 4, while that of even subsets is 3. Where's the catch? AI: $\{\}$ is also a subset of the nonempty set.
H: Show an isomorphism between two quotient spaces We are given 3 vector spaces $U \subseteq W \subseteq V$ We are asked to prove that there is an isomorphism between $V/W$ and $(V/U)/(W/V)$ What we tried (and it is false, I would like to know why): We basically said that an element of $V/W$ is $v+W$ and that an element of $(V/U)/(W/V)$ is $v+U+W+U = v+U+W = v+W$ and so we can define the identity transform, it is invertible, and so an isomorphism exists. The teacher said it was incorrect. Would someone please tell me why and show me the right solution? AI: Hint: Use the formula $dim(A/B)=dim(A)-dim(B)$ if $B\subset A$. About your solution you can't add cosets this way they are different objects.
H: finding an 'e' based limit for this sequence i need to find the limit for $$\lim_{n \rightarrow \infty} \left(1 + \frac{q}{n}\right)^n $$ where $q \in \mathbb{Q}$ how to i get this sequence to resemble $$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n $$ so i can find its limit? AI: $q = \frac{s}{t}$, where $s,t\in\mathbb{N}$, so We can write $$\left(1 + \frac{q}{n}\right)^n = \sqrt[t]{\left(1 + \frac{s}{tn}\right)^{nt}}$$, for the term inside the bracket, i.e. $$\left(1 + \frac{s}{tn}\right)=\frac{tn+s}{tn}=\frac{tn+1}{tn}\cdot\frac{tn+2}{tn+1}...\frac{tn+s}{tn+s-1}=\left(1+\frac{1}{tn}\right)\cdot\left(1+\frac{1}{tn+1}\right)...\left(1+\frac{1}{tn+s-1}\right)$$ Now it should be clear what you should do. Let me know if you still have problems.
H: Solving inequality. $5(y-2)-3(y+4)\geq 2y-30$ I solved this inequality, but at the end I got that $8 \geq 0$. Did I did this right. What does this mean. This is how I solved it: $$ 5(y-2)-3(y+4) \geq 2y-30\\ 5y-10-3y-12 \geq 2y-30\\ 2y-22 \geq 2y-30\\ 2y-2y-22+30 \geq 0\\ 8\geq0\\ $$ Did I did it right? Thank you very much for help! AI: "Solving" the inequality in here means finding the domain of the variable $y$ in which the inequality is satisfied. What you have done is true, and it means the inequality satisfies for all real (or complex) numbers $y$. By the way, you can use "\ge" to show "$\ge$" in latex.
H: Existence function $f$ which isn't uniform continuity, but $f^2$ is? Existence function $f$ which isn't uniform continuity, but $f^2$ is? I don't have more asumptions about function $f$, so let $f: X \rightarrow Y$ where $X,Y \subseteq \mathbb{R}$ with euclidean metrice. In my opinion if $f^2$ is uniform continuity then we have $$\forall \varepsilon > 0 \quad \exists \delta>0 \quad \forall x,y \in X \quad \|x-y\| < \delta \Rightarrow \|f^2(x) - f^2(y)\| < \varepsilon $$ From the last inequality we have that $\|f(x) - f(y)\| < \frac{\varepsilon}{\|f(x)+f(y)\|}$. So can I now conclude that $f$ is uniform continuity? What can I say about $\|f(x)+f(y)\|$ ? The problem for me is that, if $\|f(x)+f(y)\| \rightarrow 0$ then $\frac{\varepsilon}{\|f(x)+f(y)\|} \rightarrow \infty$. AI: HINT: since $f$ itself does not have to be continuous, think about a function $f$ which is not continuous, so that $f^2$ is constant.
H: Multiplicative order I tried to learn about the multiplicative order from different sources, but I want to get sure that I understand it well... As I understand, if we define to numbers, $a \in \Bbb Z$, $n \in \Bbb N^+$, where $a$, $n$ are coprime, then $O_n(a)$ is the smallest positive integer $k$ such that $a^k\;\mathrm{mod}\;n = 1$. Is this true? AI: Yes, that's it. As $1,a,a^2,a^3,\dots$ is a periodic sequence mod $m$ if $\gcd(a,m)=1$, the smallest $k$ with $a^k\equiv 1 \pmod m$ coincides with the number of elements of $$\{a^n\,\text{mod}\, m\mid n\in\Bbb N\}\,.$$
H: Linear Algebra - linear functionals Let $S$ be a set of vectors in $V$. Define $S^0$ to be the set of linear functional that vanishes on $S$. Show that $$S^0 = \langle S\rangle^0$$ I am really confused what is meant by $S^0$ AI: $S^0=\{f:V\to\Bbb R \mid \forall s\in S: f(s)=0\}$. Show that, if $f\in S^0$ then $f\in \langle S\rangle^0$ and the other way around, if $g\in \langle S\rangle^0$ then $g\in S^0$. (One of these is trivial, the other follows by linearity.)
H: Fixed points through a general circle. The circle $C: x^2 + y^2 + kx + (1+k)y - (k+1)=0$ passes through two fixed points for every real number $k$. Find $(i)$ co-ordinates of these two points and $(ii)$ the minimum value of the radius. AI: HINT: This is an arbitrary circle which passes through the intersection of $x^2+y^2+y-1=0$ & $x+y-1=0$ For the second question : $$x^2+y^2+kx+(1+k)y-(k+1)=0$$ $$\implies \left(x+\frac k2\right)^2+\left(y+\frac{1+k}2\right)^2=k+1+\frac{k^2}4+\frac{(k+1)^2}4$$ If $r$ is the radius, $$r^2=k+1+\frac{k^2}4+\frac{(k+1)^2}4=\frac{4k+4+k^2+(k+1)^2}4=\frac{(2k+3)^2+1}8\ge \frac18$$
H: Is every $T_2$ compact monotonically normal space separable? I have a question about compact spaces: Is every $T_2$ compact monotonically normal space separable? Thanks ahead:) AI: Hint: Every linearly ordered topological space is $T_2$ and monotonically normal. Can you think of one which is compact and nonseparable?
H: Formal proving of languages accepted by a finite automata. Suppose $L_1 \cup L_2,L_1 \subseteq E^* $ are languages accepted by finite automata and $L_1\cap L_2 =\emptyset $. We need to prove that $L_2 $ is also accepted by a finite automaton. So I've started with building an intuition, but I have a few questions. Say I've found an automaton presumably representing a language. What's the proper way of proving that it actually is a representation of the language? Is it enough to prove that, given all of the above, $L_2$ has a regular expression (since both $L_1 \cup L_2$ and $ L_1$ should have regular expressions)? Or is that a wrong way of looking at this kind of question? AI: If a language $M$ is regular, then $\overline{M}=E^*\setminus M$ is regular too. Therefore $$ L_2=\overline{\overline{L_1\cup L_2}\cup L_1} $$ is regular.
H: A Summation Question Let $n$ be an odd integer. If $\sin (n)\theta=\displaystyle \sum _{r=0}^n b_r\sin (r\theta)$, for every value of $\theta$, then $b_0=1, b_1=3$ $b_0=0, b_1=n$ $b_0=-1, b_1=n$ $b_0=0, b_1=n^2+3n+3$ How do I approach it? I am not able to find the values of $b_0$ or $b_1$. AI: For example: $$n=1\;,\;\;\theta=0\implies 0=\sin0=b_0+b_1\sin0=b_0$$ and you already ruled out two possibilities. Another one: $$n=1\;,\;\;\theta=\frac\pi2\implies 1=\sin\frac\pi2=b_0+b_1\sin\frac\pi2\implies\ldots ?$$
H: Question about finding a limit with limit arithmetics $lim_{n \to \infty}(\frac{4n^2}{(2n+1)(2n-1)})^{1-n^2}$ When I simplfy it I get: $\frac{1-4n^2}{(1-4n^2)^{n^2}}$ Now is it enough to use limit arithmetics on the denominator as if it's $\frac1{x^n}$ to show that it goes to 0 ? Note: we can't use logs/lns/derive/integrate. AI: The simplification is wrong as $$\frac{4n^2}{(2n+1)(2n-1)}=\frac{4n^2}{4n^2-1}=1+\frac1{4n^2-1}$$ $$\implies \left(\frac{4n^2}{(2n+1)(2n-1)}\right)^{1-n^2}=\left(1+\frac1{4n^2-1}\right)^{1-n^2}=\left(\left(1+\frac1{4n^2-1}\right)^{4n^2-1}\right)^{\frac{1-n^2}{4n^2-1}}$$ As $n\to\infty, 4n^2-1\to\infty$ and we know $\displaystyle\lim_{m\to\infty}\left(1+\frac1m\right)^m=e$ and $\displaystyle\lim_{n\to\infty}\frac{1-n^2}{4n^2-1}=\lim_{n\to\infty}\frac{\frac1{n^2}-1}{4-\frac1{n^2}}=-\frac14$
H: Finding the limit of $ \lim_{k \rightarrow \infty} \left(\frac{2^k + 1}{2^{k-1} + 3}\right) $ $$ \lim_{k \rightarrow \infty} \left(\frac{2^k + 1}{2^{k-1} + 3}\right) $$ I'm trying to prove that the limit of the sequence is $2$ using the squeeze theorem, but with no success. Thanks AI: HINT: Multiply the fraction by $1$ in the carefully chosen disguise $$\frac{1/2^{k-1}}{1/2^{k-1}}\;.$$
H: check if 2 vectors in $\Bbb R^n$ = $\Bbb R^2$ I need to make sure that a system of any 2 vectors in $\Bbb R^n$ can make a lineair combination to $\Bbb R^2$. The way I want to investigate is to row-reduce the matrix to enchelon form. This way I can check for consistentcy and see if the set has a single solution. Is this a valid approach? AI: Compare coordinates one by one. When you find a coordinate that is not zero in both vectors, you can divide them to find out what the scalar factor that takes one to the other must be if it exists at all. Check whether this factor works for the remaining coordinates.
H: How to calculate the rest of $2^{p^r-p^{r-1}+1}$ divided by $p^r$ I have the next problem: $p$ is an odd prime number and $r$ is a natural number, $r>1$. How can I calculate the rest of the division of $2^{p^r-p^{r-1}+1}$ by $p^r$ ? AI: Hint: $\varphi(p^r)=p^r-p^{r-1}$ and $2$ is coprime to $p^r$, so, by Euler-Fermat...?
H: Integrate over a curve (complex) Given $${\Gamma} = [{z(t) = t\sqrt{\frac2\pi\}}e^{it^2}}]$$ for $$0< t < \sqrt{\pi/2}$$ evaluate $$\int_{\Gamma} ze^{z^2}dz$$ The usual process involves parametrization and then substituting z(t) into f(z) which in this case is the stuff under the integral and then multiplying by z'(t). Generally a simple enough process. Here however I end up with a ridiculously messy integral that I have no idea what to do with. Surely there's something I'm missing or some way to simplify ${\Gamma}$. What do? AI: Note that you may simply evaluate the integral at the endpoints of $\Gamma$ because $$\int dz \, z \, e^{z^2} = \frac12 e^{z^2} + C$$ The endpoints are at $z=0$ and $z=e^{i \pi/2} = i$.
H: Quantity of an object after proliferation A small creature called "Charza" lives in blocks of an infinite square. An infinite number of them can live in a block. After one hour , one Charza is divided into 4 Charzas and each one moves into one of the adjacent blocks. we have only one Charza at the first . after 6 hours , how many Charzas will be in a block which has only one common vertex with the first block? ( for example , after one hour , there will be one Charza in each adjacent block ) AI: Impose a coordinate system that makes the originally occupied cell $\langle 0,0\rangle$. Let $c(m,n,t)$ be the number of Charzas in cell $\langle m,n\rangle$ at time $t$ hours; clearly $c(0,0,0)=1$, $c(m,n,0)=0$ for $\langle m,n\rangle\ne\langle 0,0\rangle$, and the recurrence is $$c(m,n,t+1)=c(m-1,n,t)+c(m+1,n,t)+c(m,n-1,t)+c(m,n+1,t)\;.$$ It’s not hard to see that $c(m,n,t)=c(\|m\|,\|n\|,t)$ for all $m,n,t\in\Bbb Z$. A little numerical experimentation suggests that if we set $$d(m,n,t)=\frac{t-(\|m\|+\|n\|)}2$$ for $m,n\in\Bbb Z$ and $t\in\Bbb N$, then $$c(m,n,t)=\begin{cases} \binom{t}{d(m,n,t)}\binom{t}{m+d(m,n,t)},&\text{if }\|m\|+\|n\|\equiv t\pmod 2\\\\ 0,&\text{otherwise}\;.\end{cases}\tag{1}$$ Clearly $(1)$ satisfies the initial conditions on $c$, so it only remains to show that it satisfies the recurrence. Suppose that $m,n\ge 0$ and $m+n\equiv t+1\pmod2$. Then $$\begin{align} \binom{t+1}{d(m,n,t+1)}&=\binom{t+1}{\frac12(t+1-m-n)}\\\\ &=\binom{t}{\frac12(t-1-m-n)}+\binom{t}{\frac12(t+1-m-n)} \end{align}$$ and $$\begin{align} \binom{t+1}{m+d(m,n,t+1)}&=\binom{t+1}{m+\frac12(t+1-m-n)}\\\\ &=\binom{t}{m+\frac12(t-1-m-n)}+\binom{t}{m+\frac12(t+1-m-n)}\;, \end{align}$$ so $$\begin{align} &c(m,n,t+1)=\\\\ &\binom{t}{\frac12(t-1-m-n)}\binom{t}{m+\frac12(t-1-m-n)}+\\\\ &\qquad\binom{t}{\frac12(t-1-m-n)}\binom{t}{m+\frac12(t+1-m-n)}+\\\\ &\qquad\binom{t}{\frac12(t+1-m-n)}\binom{t}{m+\frac12(t-1-m-n)}+\\\\ &\qquad\binom{t}{\frac12(t+1-m-n)}\binom{t}{m+\frac12(t+1-m-n)}=\\\\ &\\\\ &\binom{t}{d(m,n+1,t)}\binom{t}{m+d(m,n+1,t)}+\\\\ &\qquad\binom{t}{d(m+1,n,t}\binom{t}{m+1+d(m+1,n,t)}+\\\\ &\qquad\binom{t}{d(m-1,n,t)}\binom{t}{m-1+d(m-1,n,t)}+\\\\ &\qquad\binom{t}{d(m,n-1,t)}\binom{t}{m+d(m,n-1,t)}=\\\\ &\\\\ &c(m,n+1,t)+c(m+1,n,t)+c(m-1,n,t)+c(m,n-1,t)\;, \end{align}$$ as desired. Added: The specific number desired is $$c(1,1,6)=\binom62\binom63=15\cdot20=300\;.$$
H: Kind find the limit of the following question What is the limit of the following? lim(h->0) ((2+h)^0.5 -(2)^0.5)/h Note: I'm struggling to make the denominator positive. AI: First way: $$\lim_{h\to 0}\frac{\sqrt{2+h}-\sqrt2}h=(\sqrt x)'\|_{x=2}=\frac1{2\sqrt2}$$ Second way: $$\lim_{h\to 0}\frac{\sqrt{2+h}-\sqrt2}h=\frac{2+h-2}{h(\sqrt{2+h}+\sqrt2)}=\frac1{\sqrt{2+h}+\sqrt2}\xrightarrow[h\to 0]{}\frac1{2\sqrt2}$$
H: sum of all coprimes of a number. What is the sum of all coprimes to number less than that number? I found a bit about it: For example we have to find the sum of coprimes of 2016. Therefore, the required sum $S$ is: $2016 = 2^5 * 7 * 3^2$ $S = \frac{2016}2 * 2016 * (1-\frac13)(1-\frac17)(1-\frac12) = 580608$ If this is right, then the formula should be $S = \frac{N^2}2 * (1-\frac1{\text{prime factor 1}})* (1-\frac1{\text{prime factor 2}})...$ If this example and the interpreted formula is right, then how do I prove it? If this is wrong, then what is the actual formula and its proof? AI: Yes, the formula is right and if you reached it by yourself it is remarkable. If $\;\phi(n)\;$ is Euler's Totient Function, then the sum you want is $$\frac n2\phi(n)=\frac{n^2}2\prod_{p\mid n\,,\,p\,\text{a prime}}\left(1-\frac1p\right)$$ Hint for the proof: $\;1\le k <n\;$ is coprime to $\;n\;$ iff $\;n-k\;$ is also coprime with $\;n\;$ ...
H: the smallest value of $\alpha$ $f$ is a continuous function on the interval $[0,1] $ which satisfy: 1) $f(x) \leq \sqrt{5}$ for all $x \in [0,1] $ 2) $f(x) \leq \frac{2}{x}$ for all $x \in [\frac{1}{2},1]$ Then find the smallest real $\alpha$ such that the inequality $ \int^1_0 f(x) dx \leq \alpha$ holds for any such $f$. Actually I'm relatively new to calculus, and hence to do not know how to proceed. I can do the most basic ones, but I haven't got any idea about this. AI: What you need to do is find $\alpha = \int ^1 _0 \min \{\sqrt 5, \frac 2 x\} dx$. You do that by finding $x_1$ where $\frac 2 {x_1} = \sqrt 5$ and then finding $\int_0^{x_1} \sqrt 5 dx + \int _{x_1}^1 \frac x 2 dx$.
H: Can't find the limit of the following I can't find the limit of the the following: $$\lim_{p\to1} \frac{ p^{1/3} - 1 }{p - 1}$$ Any ideas? AI: Putting $p^{\frac13}=q\implies p=q^3$ as $p\to1, q\to1$ $$\lim_{p\to1}\frac{p^{\frac13}-1}{p-1}$$ $$=\lim_{q\to1}\frac{q-1}{q^3-1}$$ $$=\lim_{q\to1}\frac{(q-1)}{(q-1)(q^2+q+1)}$$ $$=\lim_{q\to1}\frac1{q^2+q+1}\text{ as }q-1\ne0\iff q\ne1\text{ as } q\to1$$ $$=\cdots$$
H: How does $\exp(0)=1$ follow from the definition $\exp(z):= \sum_{n=0}^\infty \frac{1}{n!} z^n$ We introduced the Exponential function as follows: $$ \exp: \begin{cases} \mathbb{C} & \longrightarrow \mathbb{C} \\z & \longmapsto \displaystyle \sum_{n=0}^{\infty} \frac{1}{n!}z^n \end{cases}$$ It might be a trivial thing to ask, but in my mind I always identify this function with $e^z$ or in the real domain with $e^x$. It is a very powerful definition because many interesting results in mathematics can be deduced from it in a straightforward manner, such as the Euler Identity and many trigonometric laws. However I don't see how this definition is consistent with the fact that I remember from school about $e^0$ being equal to $1$. Maybe my computation is flawed but if I substitute $z=0$ I get. $$ \exp(0)=\sum_{n=0}^{\infty} \frac{1}{n!}0^n=\underbrace{0^0}_!+0^1+\frac{1}{2}0^2 + \dots $$ where $0^0$ is not defined, because $0^0=0^{-0} = \frac{0}{0}$ Is there something I misunderstand about this definition? Or is my substitution invalid? AI: In this case $0^0$ is not defined if you look at it from the exponentiation (on integers) point of view, however $0^0:=1$ here. The reason for this is pragmatic: it facilitates notation. It's not that the real number $0$ powered to itself equals $1$. No, that's not defined. In the context of series, $0^0=1$ just so it is easier to write stuff. You should translate $\sum \limits_{n=0}^{+\infty}\left(a_n0^n\right)$ to $a_0\cdot 1+\sum \limits_{n=1}^{+\infty}\left(a_n0^n\right)$ There are others instances in which $0^0=1$. For instance in cardinal arithmetic, given two finite cardinals $\alpha$ and $\beta$, $\alpha ^\beta$ denotes the number of functions from $\beta$ to $\alpha$. If $\alpha=\varnothing =\beta$, then $\varnothing ^\varnothing$ is the number of functions from the empty set on itself, and that's $1$, (it's the empty function). You don't even have to define $0^0$ separately here.
H: Best way to explain how the Infimum and Supremum of this function are obtained...? I have the function $\;f(x)=\dfrac{x^{(1/2)}}{2+x}\;$ and I know that $\inf(f)$ does not exist and $\sup(f)=2$ but I don't know how to formally show this rigorously? Anyone got a formal way of showing this, it would be much appreciated... AI: $$f(x) = \dfrac{x^{1/2}}{2 + x}$$ Domain of $f(x)$: $[0, +\infty) \subset \mathbb R$. We find critical points (and associated extrema) by finding $f'(x)$ and solving where $f'(x) = 0$. $$f'(x) = \dfrac{\frac 12x^{-1/2}(2+x) - x^{1/2}}{(2 + x)^2} =\dfrac{ (2+x) - 2x}{2x^{1/2}(2 + x)^2} = \dfrac{2 - x}{2\sqrt x(x+ 2)^2}$$ $$f'(x) = 0 \iff x = 2;\quad x = 2 \implies f(x) = \dfrac{\sqrt 2}{4} = \dfrac 1{2 \sqrt 2}$$ Confirm, using the derivative, the point $(2, \frac 1{2\sqrt 2})$ is a global maximum of $f(x)$. Note also that $$\lim_{x\to 0^+} \dfrac{x^{1/2}}{2 + x} = 0/2 = 0$$ and $$\lim_{x\to +\infty} \dfrac{x^{1/2}}{2 + x} = \lim_{x\to +\infty}\dfrac{\frac{1}{x^{1/2}}}{\frac 2x + 1} = 0$$ $$\therefore \underbrace{0}_{\inf} \leq f(x) \leq \underbrace{\dfrac 1{2\sqrt 2}}_{\sup}$$
H: Is series convergent? Does this series converge? $$\sum_{n=1}^{\infty }\frac{{(-1)}^{n}(n+2)}{{n}^{2}+4}$$ How can i step-by-step calculate it? AI: Check that $$\frac{n+2}{n^2+4}\ge\frac{n+3}{(n+1)^2+4}\iff (n+2)(n^2+2n+5)\ge(n+3)(n^2+4)\iff$$ $$\iff n^3+4n^2+9n+10\ge n^3+3n^2+4n+12\iff$$ $$\iff n^2+5n-2\ge 0$$ and the last inequality is clear since the positive quadratic's root is less than $\;1\;$ . Thus, your sequence without the alternating sign is monotone decreasing and that is thus a Leibniz series so convergent.
H: Classification of groups of order 30 How do I find all the groups of order 30? That is I need to find all the groups with cardinality 30. I know Sylow theorems. AI: Let $G$ be a group of order 30 and let $n_p$ be the number of its $p$-Sylow subgroups. By Sylow, $n_3\in\{1,10\}$ and $n_5\in\{1,6\}$. It is not possible that $n_3 = 10$ and in the same time $n_5 = 6$, since otherwise, there would be 20 elements of order 3 and 24 elements of order 5, contradicting the fact that $G$ contains only 30 elements. So either the $3$-Sylow subgroup $P_3$ or the $5$-Sylow subgroup $P_5$ is normal, implying that $N = P_3 P_5$ is a subgroup of $G$. Because of $\gcd(3,5) = 1$, $P_3\cap P_5 = \{1\}$ and therefore $\lvert N\rvert = \lvert P_3\rvert\cdot\lvert P_5\rvert = 15$. By the classification of $pq$-groups and $5\not\equiv 1\bmod 3$, $N$ is cyclic. Since its index in $G$ is $2$, it is a normal subgroup. Any $2$-Sylow subgroup is a complement of $N$ in $G$. So $G \cong \mathbb Z/15\mathbb Z \rtimes_{\varphi} \mathbb Z/2\mathbb Z$, where $\varphi : \mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/15\mathbb Z)$ is a group homomorphism. We have $\operatorname{Aut}(\mathbb Z/15\mathbb Z) \cong (\mathbb Z/15\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z \times \mathbb Z/5\mathbb Z)^\times \cong \mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$. Since $\varphi$ is a homomorphism and the order of $1$ in $\mathbb Z/2\mathbb Z$ is $2$, the order of its image $\varphi(1)$ divides $2$. In $\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$ there are four possible images, namely $\varphi(1)\in\{(0,0), (1,0), (0,2), (1,2)\}$. Since $\varphi$ is determined by $\varphi(1)$, this shows that there are at most $4$ isomorphism types of groups of order $30$. Now look at the following four groups or order $30$: $$ \mathbb Z/30\mathbb Z,\quad D_{15},\quad \mathbb Z/5\mathbb Z\times S_3,\quad \mathbb Z/3\mathbb Z\times D_5 $$ It is not hard to check that they are pairwise non-isomorphic. So there are exactly these $4$ isomorphism types of groups of order $30$.
H: Is there a continuous function $f:S^1 \to \mathbb R$ which is one-one? Is there a continuous function $f:S^1 \to \mathbb R$ which is one-one? AI: Suppose such a function $f$ exists. $f(S^1)$ must be connected and compact. $f$ is one-one so it cannot be constant. It follows that $f(S^1) = [a, b]$ for $a \ne b$. Pick a point $s \in S^1$ such that $f(s) \not\in \{a, b\}$. $S^1 - \{s\}$ is connected, but $[a, b] - \{f(s)\}$ is not. We conclude that such a function $f$ cannot exist. Alternatively, if you're familiar with the concept of the fundamental group: The restriction $f : S^1 \to f(S^1)$ must be a homeomorphism (being a continuous bijection from a compact space onto a Hausdorff space). But we know that $\pi_1(S^1) = \mathbb Z$ whereas any connected subset of $\mathbb R$ is contractible; hence it has a trivial fundamental group.
H: How to write this as a boolean expression? How can I write the following sentence as boolean expression: $$ \text{If two sides of triangle are the same, then two opposite angles are the same} $$ I konw it should be something like this: $$ a = \text{if two sides of triangle are the same}\\ b = \text{two opposite angles are the same}\\ \\ a <\text{some logical operator}> b\\ $$ But I am not sure how to write $then$ as logical operator. Thanks. AI: Let $\triangle ABC$ be a triangle with vertices at $A, B, $ and $C$. Denote the length of the side opposite angle $A$ by $a$, the length of the side opposite angle $B$ by $b$, and the length of the side opposite angle $C$ by $c$. Denote the measure of an angle $\theta$ by $m\angle \theta$. Then $$(a = b) \rightarrow (m\angle A = m\angle B)$$ $$(a = c) \rightarrow (m\angle A = m\angle C)$$ $$(b = c) \rightarrow (m\angle B = m \angle C)$$ There are many variations for how you might express this relationship. I've specified "lengths" of sides and "measures" of angles to ensure we pin down what we mean by "the same:" Here, I'm assuming we are talking about sides that have equal length, and angles that have equal measures. The term that is most commonly used to pin down what we mean by "the same" in geometry is the term "congruent." As you've formulated the propositions, we have simply "If $a$, then $b$: $\;a \rightarrow b$.
H: find all the partial limits of this sequence i need to find all the partial limits (is that the current terminology? i mean the limits of all possible subsequences for a given sequence) $a_n = \sqrt[n]{4^2 + 2^n} $ find all partial sums,also find $\liminf_{n\rightarrow\infty} a_n$ and $\limsup_{n\rightarrow\infty} a_n$ i couldn't really find any similar questions with regards to these types of questions, if somone could also point me to what to google for that would be great! AI: This sequence's limit was asked just yesterday. Here's my answer (hint, in fact): $$\sqrt[n]{2^n}\le\sqrt[n]{16+2^n}\stackrel{\text{For}\;n\ge 4}\le\sqrt[n]{2\cdot 2^n}$$
H: calculating norm of the following matrix? I have the following matrix and its norm $$ \begin{aligned} A(s) &= \begin{bmatrix} 0 & e^{-sT} \\ e^{-sT} & 0 \end{bmatrix} \\ \lvert \lvert {A(s)} \rvert \rvert_{\infty} &= \sup_{w} \sqrt{\lambda_{\max} (A^{}(jw)A(jw)) } = 1 \end{aligned}$$ I want to know how the norm has been acquired? AI: Let $H=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Then $\hat{A}(s) = e^{-sT} H$, hence $\\|\hat{A}(s)\\| = \|e^{-sT}\| \\| H \\|$. $\\| H\\| = \sqrt{\\|H^H\\|}= \sqrt{\\|H^*H\\|}=\sqrt{\\|I\\|} = 1$. Hence $\\|\hat{A}(s)\\| = \|e^{-sT}\|$. Now, the $H_\infty$ norm is given by $\\|\hat{A}\\|_\infty = \sup_\omega \\|A(j \omega)\\|$, and $\\|\hat{A}(j \omega)\\| =1$ for all $\omega$, hence the desired result.
H: What does it mean for a value to be minimised? I am trying to solve this problem, but I do not understand what is meant bythe following expression being minimised $$ D = P_{k} + P_{j} $$ AI: A value is minimised when it becomes as small as it can. I expect that your $P_k$ and $P_j$ are points. Then $\\|P_k-P_j\\|$ is the distance between any two points. To minimise $\\|P_k-P_j\\|$ is to find the two points that are closest to one another out of all the other possible pairs of points.

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