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H: How to calculate the optimal number of slices to use for a Riemann sum I wrote some code that calculates the area under a curve using Riemann sum. The way I have it working right now is that a user enters the upper and lower bounds for the integral and then the number of slices to use for the calculation. I want a way to calculate the number of slices needed for the area to converge (within a tolerance). It's easy enough to just tell the user to enter in a very large number but I want to see if I can automate this. My initial though was to calculate the area using, for example, $10$ and $100$ slices ( $S_1, S_2$). Then try and fit that to an exponential decay. The area of each of those trials giving me $ A_1, A_2$ Using that I can get the time constant for the decay as $ T = \frac{S_2-S_1}{\ln(A_1/A_2)}$ Then finding the optimal number of slices to use could just be a multiple of the time constant. For example $ 10T$ AI: I can give you a rough upper bound on the decay rate of the error in your approximation to the integral. For concreteness, say we have a smooth function $f : [0,1] \to \mathbb{R}$. Let $L = \\|f'\\|_{\infty} = \sup \{\|f'(x)\| \mid x \in [0,1]\}$, which is a fancy way of saying the largest magnitude of a tangential slope of $f$. Let's say we're using a left-point rectangle approximation, so the approximation using $n$ even slices is given by $$ A_n = \frac{1}{n} \sum_{k = 0}^{n-1} f(k/n) $$ Let $A = \int_0^1 f(x) dx$ be the actual value of the integral. Then I claim that $$ \|A - A_n\| \leq L/n $$ Using the midpoint approximation, this can be improved to $L/(2n)$ (try to figure this out!). I'll give you the main idea: if $x \in [k/n, (k+1)/n]$ is a point, then by the mean value theorem, $f(x) - f(k/n) = (x - k/n) f'(c)$ for some point $k/n \leq c \leq x$. Using what we know, we can bound $$ \|f(x) - f(k/n)\| \leq L/n $$ The rest of the proof is straightforward from there. In case you're curious, using trapezoids is actually a considerably better way to approximate the integral. Using the taylor expansion out to the second order derivative, it's possible to show (and a good exercise, too) that the trapezoid approximation with $n$ equal slices is within $M/n^2$ of the original integral; here $M = \\|f''\\|_{\infty} = \sup\{\|f''(x)\| \mid x \in [0,1]\}$ is the highest magnitude of the second derivative.
H: Using $\lim_{x \to \infty}$ to determine whether $f(x) = \Theta(g(x))$? I'm learning it in the context of Running time complexity. to determine whether $f(x) = O(g(x))$, you can check whether the folloing limit:$$\lim_{x \to \infty} {f(x) \over g(x)} < \infty$$ if so, then you know that $f(x) = O(g(x))$. Is there a similar way to determine whether $f(x) = \Theta(g(x))$? Thanks in advance. AI: Actually, the Theorem that I know says that if $0 < \lim_{x \to \infty}f(x)/g(x) < \infty$, then $f(x) \in \Theta(g(x))$, which it turn implies that $f(x) \in O(g(x))$. I am currently TA-ing a course on Data Structures, and these are the slides that we are using: https://www.student.cs.uwaterloo.ca/~cs240/f13/modules/module01.pdf - If you look at slide 32/48, you'll see the theorem that I just mentioned.
H: Can uniform local integrability implies uniform local absolute continuity? Suppose we have $u\in L_{u,}^{loc}$, i.e., $$ \sup_{x\in \mathbb{R}^n}\int_{\|x-y\|<1}\|V(y)\|dy<\infty $$ then can we obtain that $\forall \epsilon>0$, there exists a $r>0$, such that $$ \sup_{x\in \mathbb{R}^n}\int_{\|x-y\|<r}\|V(y)\|dy<\epsilon $$ ? If this is not true, are there any counterexamples ? thanks in advance. AI: Consider $n = 1$ for simplicity. Let $$V = \sum_{k = 1}^\infty k\cdot \chi_{[3k-1/k,3k+1/k]}.$$ Then $$\sup_{x\in\mathbb{R}} \int_{\lvert x-y\rvert < 1} \lvert V(y)\rvert\,dy = 2,$$ and for all $1 \geqslant r > 0$, we also have $$\sup_{x\in\mathbb{R}} \int_{\lvert x-y\rvert < r} \lvert V(y)\rvert\,dy = 2.$$
H: Does $\sum_{n=1}^{\infty} \sin\left(\frac{\pi}{n}\right)$ converge? I'm not really sure how to start here, the basic tests don't work. AI: $\sin(\pi/n) = \pi/n + O(1/n^2)$ for $n$ large. Now use the fact that the harmonic series diverges.
H: Evaluating Sums $\sum_{i=1}^{n} \sum_{j=0}^{n-i}$ I'm unsure how to evaluate sums when the second sum has $n-i$ on the top. $$\sum_{i=1}^{n} \sum_{j=0}^{n-i} (3j^{2} - 2)$$ $$=\sum_{i=1}^{n} (\sum_{j=0}^{n-i} 3j^{2} - \sum_{j=0}^{n-i}2)$$ $$=\sum_{i=1}^{n} (\sum_{j=0}^{n-i} 3(\frac {n(n+1)(n+2)} 6 ) - 2n)$$ From here I'm lost to what I could do. Could some one please explain how to evaluate this sum in detail? I'm confused. AI: First off, that last step isn’t right, so I’m going to start from the beginning. $$\begin{align} \sum_{i=1}^n\sum_{j=0}^{n-i}\left(3j^2-2\right)&=\sum_{i=1}^n\left(\sum_{j=0}^{n-i}3j^2-\sum_{j=0}^{n-i}2\right)\\\\ &=\sum_{i=1}^n\sum_{j=0}^{n-i}3j^2-\sum_{k=1}^n\sum_{j=0}^{n-i}2\;. \end{align}$$ Now let’s deal separately with these summations. There are $n-i+1$ terms in the inner summation, so $$\begin{align} \sum_{i=1}^n\sum_{j=0}^{n-i}2&=\sum_{i=1}^n2(n-i+1)\\\\ &=2\left(\sum_{i=1}^n(n+1)-\sum_{i=1}^ni\right)\\ &=2\left(n(n+1)-\frac{n(n+1)}2\right)\\\\ &=n(n+1)\;. \end{align}$$ The first one is a little messier: $$\begin{align} \sum_{i=1}^n\sum_{j=0}^{n-i}3j^2&=\sum_{i=1}^n\frac{(n-i)(n-i+1)\big(2(n-i)+1\big)}2\\\\ &=\frac12\sum_{i=1}^n\Big((n-i)(n-i+1)\big(2(n-i)+1\big)\Big)\\\\ &\overset{}=\frac12\sum_{k=0}^{n-1}k(k+1)(2k+1)\\\\ &=\frac12\sum_{k=0}^{n-1}\left(2k^3+3k^2+k\right)\\\\ &=\sum_{k=0}^{n-1}k^3+\frac32\sum_{k=0}^{n-1}k^2+\frac12\sum_{k=0}^{n-1}k\\\\ &=\left(\frac{(n-1)n}2\right)^2+\frac{(n-1)n\big(2(n-1)+1\big)}4+\frac{(n-1)n}4\;. \end{align*}$$ The step marked with an asterisk is accomplished by letting $k=n-i$: as $i$ runs from $1$ through $n$, $n-i$ runs from $n-1$ down through $0$. Now just simplify this last result, combine with the first one, and simplify again to get the final result.
H: Reference for integration Does anyone have a good reference for a book that already assumes knowledge of calculus/analysis and whose main focus is computing more difficult integrals? I'm looking for something which will have a lot of worked examples for differentiation under the integral, tricky substitutions, unusual contours, etc. Most of what I've found in terms of references simply compile a long list of integrals, many of which are more tedious than insightful. My goal is to be able to get better at computing integrals that can be done by hand in a reasonable amount of time, but may require some ingenuity (like in math competitions, for example). So any discussion of heuristics and problem solving techniques in integration is a plus. AI: You might be interested in The Handbook of Integration by Zwillinger. It appears to be the standard reference on integration methods for scientists and engineers. The downside is that it probably doesn't contain the "tricky" techniques you are looking for. Hopefully someone can find a more math contest-oriented book for you. In the meantime, you might try looking at some answers this site. We have many talented integration experts around who are frequently posting and answering extremely challenging integrals and sums (like this). Here's an incomplete list (go to the 'answers' tab on their profile): sos440 (He also maintains a blog, which has many challenging integrals worked out in complete detail.) Mhenni Benghorbal (His answers are especially good for learning Mellin transform techniques.) Ron Gordon (He uses mostly complex analysis. Also check his 'greatest hits' page, available in his profile.) You may also want to check the questions of 'Chris's sis', the asker of the example question I linked above, though many of her questions are not integrals. Bennett Gardiner has mentioned this useful site in a comment. On this site, there is a link to a useful guide.
H: Solve second order ODE with undetermined coefficients method Consider the differential equation $$y''+5y'=-sin(x)-1$$ Find the general solution. Here's my work: I found the solution to the homogeneous equation to be: $y_h(x)=C_1e^{-5x}+C_2$ And for the particular solution, I guessed $y_p(x)=Acos(x)+Bsin(x)+C$ $y_p'(x)=-Asin(x)+Bcos(x)$ $y_p''(x)=-Asin(x)-Bcos(x)$ After I plugged everything back into the differential equation, I got $(5B-A)cos(x)-(B+5A)sin(x)=-sin(x)-1$ What should I do with the constant $-1$? AI: Instead of your try, you should do $y_p(x)=A\cos x+B\sin x+Cx$, so that when you take the first derivative, you still get a constant to handle the constant factor $-1$ in the RHS.
H: Let H be a subgroup of Z(G). Then show that H is normal in G. Hi i am kinda stuck here. How do i show that H is normal in G??? Any tips or suggestions would be appreciated. Thank You! AI: Any subgroup $H$ is normal in $G$ iff $gHg^{-1} \subset H$ for all $g \in G$. This is equivalent to $gH = Hg$ for all $g \in G$. The latter form is what you want most likely because you're dealing with $xy = yx$ type elements in $Z(G)$.
H: Find the set of all vectors in R3 that are orthogonal to (-1,0,2) and (3,1,-2). Write the set in the standard form of a plane through the origin Let $v = (-1,0,2)$ and $u = (3,1,-2)$ so we need to find $x = (x_1,x_2,x_3)$ such that $v \cdot x = 0$ and $u \cdot x = 0$. This gives us the set of linear equations $$ -x_1+2x_3 = 0$$ $$ 3x_1 + x_2 - 2x_3 = 0$$ which, through elementary row operations gets us the two linear equations $$ x_1-2x_3 = 0$$ $$x_2 + 4x_3 = 0 $$ so $x_1 -2x_3 = x_2 + 4x_3$ which simplifies to $x_1 = x_2 + 6x_3$. so the solution set would be {$(x_2+6x_3, x_2, x_3)$ \| $x_3,x_2\in \mathbb R$} But the answer is {$r(2,-1,4)$ \| $r \in R$}. How do i get this/did I make an error? AI: Your mistake was to assume that $x_2$ and $x_3$ are indepedent, what is not true, since you have got the equation $x_2+4x_3=0$. To get the right result, just chose one variable to be parametrized and using both equations you got, find the relation of the other two variables to this parameter.
H: Jordan measurable for closure and interior Give an example of a bounded set $S\subseteq\mathbb{R}^n$ such that $\overline{S}$ and $\operatorname{Int}S$ are Jordan-measurable, but $S$ is not. Jordan-measurability of a set $A$ is equivalent to the condition that $\operatorname{Bd}A$ has measure zero, where $\operatorname{Bd}A$ is a set of points that belong to neither the interior nor the exterior of $A$. We have that $\operatorname{Int}S\subseteq S\subseteq\overline{S}$, so the desired condition is tricky to satisfy. I don't know what set $S$ to pick. AI: Let's take $n = 2$, and identify $\mathbb{R}^2$ with $\mathbb{C}$ for simplicity. Then let $$S = \{ z : \lvert z\rvert < 1\} \cup \bigcup_{\varphi \in \mathbb{Q}} \left\{t\cdot e^{2\pi i\varphi} : t \in [1,2]\right\}.$$ Then $S$ is obviously bounded, $\operatorname{int}(S) = \{ z : \lvert z\rvert < 1\}$ and $\overline{S} = \{ z : \lvert z\rvert \leqslant 2\}$ are Jordan measurable. But $\operatorname{Bd}(S) = \{z : 1\leqslant \lvert z\rvert \leqslant 2\}$ is not a null set, so $S$ itself is not Jordan measurable.
H: Equating coefficients $A-2B\sin x=2-\sin x$ I'm trying to find out how to find $A$ and $B$ for the equation $A-2B\sin x=2-\sin x$ I know I'm supposed to get $A=2$ and $B=\frac{1}{2}$, and I've looked on Google for help but didn't understand how any examples I found would help me solve the problem. Any help would be appreciated. Thanks! AI: This equation must be valid for all $x$. Since the constant function $2$ and $\sin x$, are linearly independent, it must be that the coeficients of both sides of the equation must be equal. So, $A=2$ and $-2B=-1$.
H: Augmenting «$\Bbb Z[x]$ f.g. $\Rightarrow x$ integral» for ${\frak p}[x]$ In KCd's blurb on ideal factorization, page 5: $\hskip 0.3in$ The situation is this: $K$ is a number field, ${\cal O}_K$ its ring of integers, ${\frak p}\triangleleft{\cal O}_K$ a prime ideal, $x\in K$. We assume $x \frak p\subseteq p$ and want to derive from this $x\in{\cal O}_K$. Based on my reading of the blurb above, I tried to run through the argument for how $\Bbb Z[x]$ f.g. over $\Bbb Z$ implies $x$ is integral, but in ${\frak p}[x]$: Let ${\frak p}[x]$ be the set of all ${\frak p}$-coefficient polynomials in $x$. Then ${\frak p}[x]=\frak p$ since $x\frak p\subseteq\frak p$. Thus the non-unital polynomial ring ${\frak p}[x]$ is a free $\Bbb Z$-module, and we have a basis $P_1(x),\cdots,P_n(x)$. Let $M\ge\deg P_1,\cdots,\deg P_m$ and write $x^M=c_1P_1(x)+\cdots+c_mP_m(x)$ with $c_1,\cdots,c_m\in\frak p$; this exhibits an explicit integral relation that $x$ satisfies. The problem is, in order to write $x^M$ using the basis we would need $x^M\in{\frak p}[x]$, but $x^M$ has coefficient $1\not\in\frak p$ so it's not automatic that $x^M\in{\frak p}[x]$ and this argument falls flat. Is there something the author has in mind that I am not getting? Either way, is there a way to augment the usual proof of $\Bbb Z[x]$ f.g. $\Rightarrow x$ integral to this situation to show $x\in{\cal O}_K$? In a different direction: as $\frak p$ is f.g. we know that $x$ acts by an integer matrix - is ${\rm End}_{\Bbb Z}(\Bbb Z^n)$ is integrally closed? AI: I found the answer relatively quickly in Stewart's Algebraic Number Theory pg $109$. Suppose $\theta{\frak a}\subseteq{\frak a}$ for some ideal ${\frak a}=(a_1,\cdots,a_n)$. Then $$\begin{cases} a_1\theta & = b_{11}a_1+\cdots+b_{1n}a_n \\ \vdots & \vdots \\ a_n\theta & =b_{n1}a_1+\cdots+b_{nn}a_n\end{cases}\implies \begin{pmatrix}b_{11}-\theta & b_{12} & \cdots & b_{1n} \\ b_{21} & b_{22}-\theta & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nn}-\theta\end{pmatrix}\begin{pmatrix}a_1 \\ a_2 \\ \vdots \\ a_n\end{pmatrix}=0.$$ Thus $\det(B-\theta I_{n\times n})=0$ where $B$ is an integer matrix, which is an integral relation in $\theta$. I prefer the "write $x^M$ in the basis" approach to proving integrality from $\Bbb Z[x]$ f.g., but it's the determinant version of the argument that needs to be used in this case.
H: absolute and uniform convergence of a Fourier-like series I am following stein's real analysis book and he claims that if $a_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$ where $f\in L^1([-\pi,\pi])$ then $\sum_{n=-\infty}^{\infty} a_n r^{\|n\|}e^{inx}$ converges absolutely and uniformly for each $r$, $0\leq r <1$. I know for a fact that $a_n$ is bounded because $f$ is integrable but am I missing out on a convergence test which validates the above claim? It seems like Abel's test and the Weierstrass M test don't work here... AI: You use Weierstrass: if $0\leq r<1$, then $\left\|a_nr^{\|n\|}e^{inx}\right\|\leq Mr^{\|n\|},$ and the sum of $r^{\|n\|}$ converges.
H: The Image of the Function: $f$: I am a bit confused how they arrived at this answer: The image of the function $f: \mathbb N \to \mathbb R , f(n) := \frac{(-1)^n+1}{3} $ The answer they got was $\{0, \frac{2}{3}\}$ Can someone explain how to arrive at this conclusion? Thank you! AI: If $n$ is even, then write $n = 2k$ and conclude that $$f(n) = \frac{(-1)^{2k} + 1}{3} = \frac{1^k + 1}{3} = \frac{2}{3}$$ If $n$ is odd, write $n = 2k + 1$ and get $$f(n) = \frac{(-1)^{2k + 1} + 1}{3} = \frac{-1 + 1}{3} = 0$$ Since every element of $\mathbb{N}$ is even or odd, we've exhausted the possibilities.
H: Show that: $(E^{\circ})^{\circ}=E^{\circ}$ i.e. the interior of the interior of a set equals the interior of the set. Let $(M,d)$ be a metric space. Show that $(E^{\circ})^{\circ}=E^{\circ}$. I know that $A=A^{\circ}\Leftrightarrow A \text{ is an open set}$ So I want to prove that $E^{\circ}$ is an open set and then conclude that $(E^{\circ})^{\circ}=E^{\circ}$. How do I prove that $E^{\circ}$ is an open set? Or is there a simpler approach to this question? Thanks. AI: Your definition of $E^\circ$ will affect your proof that $E^\circ$ is open. One may define $$E^\circ=\bigcup\{O:O\subset E\text{ and } O \text{ is open }\}$$ That is, $E^\circ$ is the largest open set contained in $E$, for the union of open sets is open. If you define $$E^\circ=\{x:x \text{ is an interior point of E}\}$$ The observe that if $x\in E^\circ$ then there is an open set $O$ such that $x\in O\subseteq E$. But if $y\in O$ then $O$ itself is an open set such that $y\in O\subseteq E$, i.e $y\in E^\circ$. Thus for any $x\in E^\circ$ there is an open set $O$ such that $x\in O\subseteq E^\circ$, and $E^\circ$ is open.
H: Probability of at least 1 out of 4 televisions breaking? If a television has an 8% chance of breaking in its lifetime, and I buy 4, what are the chances of at least 1 breaking in its lifetime? If I only had 2 I suppose I would do: P(A) + P(B) - P(AB). But I have 4. Thanks AI: Note that the event "at least $1$ breaking" is the complement of the event "none break," so the probability that some TV breaks is just $1$ minus the probability that none break. Now to find the probability that none break, note that each TV has a $92\%$ chance of not breaking. Thus the chance that all four survive is $$(0.92)^4$$ and the probability we want to find is $$1 - (0.92)^4$$
H: If a linear transformation is onto and one-to-one, is it always invertible? I know that if $T: \mathbb{R}^m \to \mathbb{R}^n$ is invertible then it is also onto and 1-1. But is it equivalent? In other words, are all linear transformations that are bijective considered invertible? AI: By basic set theory, as alluded to in @Pedro's comment, a bijective linear transformation $T : \mathbb{R}^m \to \mathbb{R}^n$ will have an inverse function $S : \mathbb{R}^n \to \mathbb{R}^m$. The interesting question from a linear algebra standpoint is Is the inverse $S$ also a linear transormation? The answer to this question is yes. Given $x,y \in \mathbb{R}^n$ and $a,b \in \mathbb{R}$, we have $$ ax + by = a TSx + b TSy = T( aSx + bSy) $$ using $TS = I_n$ for the first equality and linearity of $T$ for the second. Applying $S$ to both sides of the above equation and using $ST = I_m$ gives $$ S(ax+by) = aSx + bSy$$ which says $S$ is linear.
H: How to approximate this series? How to approximate this series, non-numerically? $ S_n = \sum_{n=1}^{50} \sqrt{n}$ AI: For a very simple approximation, which is the first step to the Euler-McLaurin formula, use this $f'(n) \sim f(n)-f(n-1) $ so $f(n) \sim \int_{n-1}^{n} f(x) dx $ so $\sum_{n=1}^N f(n) \sim \int_0^{N} f(x) dx $. (Actually, $f(n) \sim \int_{n-1/2}^{n+1/2} f(x) dx $ is more accurate, but this is an approximation, and definitely not the best.) Letting $f(n) =\sqrt{n}$, $\sum_{n=1}^N \sqrt{n} \sim \int_0^{N} \sqrt{x} dx = \frac{x^{3/2}}{3/2}\big \|_0^N =\frac23 N^{3/2} $. Then, let $N = 50$.
H: Elementary Algebra Inequality question $$ \frac{1}{x-1} < -\frac{1}{x+2} $$ (see this page in wolframalpha) Ok, so I think the main problem is that I don't really know how to do these questions. What I tried to do was move $-1/(x + 2)$ to the LHS and then tried to get a common denominator. I ended up with $$ \frac{(x + 2) + (x -1)}{(x-1)(x+2)} < 0. $$ So then I went $$ \frac{2x + 1}{(x-1)(x+2)}<0 $$ and got $x < -1/2$ and $x \ne 1$ and $x\ne -2$. Therefore, the answer should be $x$ is $(-\infty, -2)$ or $(-2, -1/2)$. But it's not. AI: When you have the form $\frac{a}{b} < 0$ you know there are only two options: $a < 0$, $b>0$ and $a>0$, $b<0$, because $a$ and $b$ need to differ in sign for the fraction to be negative. First case: $$ 2x+1 < 0 \implies x < -\frac12 \\ (x-1)(x+2) > 0 \implies x < ... $$ and then repeat for the other case
H: Help with exact value of: $\tan (\sin^{-1}(-1/2) - \tan^{-1}(3/4))$ Ok, so I used the tan formula of difference of angles, and so far I've got to: I found that $$\tan \alpha = \frac{-1/2}{\sqrt{3}/2} = \pm \frac{\sqrt3}{3}$$ so $$\tan \left(\sin^{-1}\left(\frac{-1}{2}\right) - \tan^{-1}\left(\frac{3}{4}\right)\right) = \frac{\sqrt3/3 - 3/4}{1+(\sqrt3/3)(3/4)}$$ I went a little past this but keep running into trouble. Can someone show me a step by step way to solve this from beginning to end so I can see how this is done? Also note it's not a proof, and I can't use exact values in form of decimals. It needs to be rationalized square-roots and fractions. Much appreciated AI: Let $a=\sin^{-1}\left(-\frac{1}{2}\right)$ and $b=\tan^{-1}\left(\frac{3}{4}\right)$. We have $$\sin^2a+\cos^2a=1 \Rightarrow \frac{1}{4}+\cos^2a=1 \Rightarrow \cos a=\frac{\sqrt 3}{2} \Rightarrow \tan a = -\frac{\sqrt 3}{3}$$ since $\sin^{-1}x$ has $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ as image. Also, $$\frac{\sin b}{\cos b}=\frac{3}{4} \Rightarrow \frac{9}{16}\cos^2 b+\cos^2b=1 \Rightarrow \cos b=\frac{4}{5} \Rightarrow \sin b=\frac{3}{5}$$ since $\tan^{-1}x$ has $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ as image. We then have $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b} = \frac{-\frac{\sqrt 3}{3}-\frac{3}{4}}{1-\frac{\sqrt 3}{3} \frac{3}{4}}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}\frac{12+3\sqrt 3}{12+3\sqrt 3}=$$ $$-\frac{108 + 48\sqrt 3 + 27\sqrt 3 +36}{144-27}=-\frac{144+75\sqrt 3}{117}=-\frac{48+25\sqrt 3}{39}$$
H: Basic limit to infinity Trying to work out a limit from a past exam in Calculus and Linear Algebra $1$, now I know the answer is $0$, and I have a worked solution from Wolframalpha is, but the answer is definitely not done the way they did it(computationally) as it was some 20 steps long. My thought is, I take it to be a fraction somehow, and I divide by highest power, or perhaps apply the conjugate of the function. $\lim_{{x}\to{\infty}}$ $x-\sqrt{1+x^2}$ I did the conjugate and got it down to $${-}\frac{1}{x+\sqrt{1+x^2}}$$ Which I can now rationalize to be 0, as dividing by the highest power will definitely give me zero, and I am pretty sure taking the highest power on $\sqrt{1+x^2}$ just gives me 1, which gives me 1+1 on the bottom(Which means the limit does exist). Edit: I want to solve this without L'Hop if possible for the record. How does one work this limit out? Am I on the right track? AI: HINT: $$ \frac{1}{x + \sqrt{1 + x^2}} = \frac{\frac{1}{x}}{1 + \sqrt{\frac{1}{x^2} + 1}} $$ for $x > 0$. Now what happens when $x \to \infty$?
H: Image of $C^1$ function does not contain open set Let $f:\mathbb{R}\rightarrow\mathbb{R}^2$ be a $C^1$ function. Prove that the image of $f$ contains no open set of $\mathbb{R}^2$. So say $f(x)=(g(x),h(x))$. Since $f$ is $C^1$, we have that $g'(x),h'(x)$ both exist and are continuous functions in $x$. To show that $f$ contains no open set of $\mathbb{R}^2$, it suffices to show that $f$ does not contain any open ball in $\mathbb{R}^2$. Suppose, for contradiction, that it contains the ball centered at $(a,b)$ with radius $r$. How can I continue? AI: Let $F:\mathbb{R}^2\rightarrow\mathbb{R}^2$ be given by $F(x,y)=f(x)$. Then every point of $\mathbb{R}^2$ is a critical point of $F$ (that is $\det J(F)(x)=0$ for every $x\in\mathbb{R}^2$, where $J(F)(x)$ denotes the Jacobian of $F$ at $x$). By Sard's theorem, $f(\mathbb{R})=F(\mathbb{R}^2)$ has zero measure. Since non-empty open sets have positive measure, $f(\mathbb{R})$ cannot contain any open set.
H: Find the rank of a matrix representing $p$ distinct polynomials of maximum degree $n$ I can't find the Reduced Row Echelon Form to find the number of pivots because I don't have numbers to work with. I know an upper bound for the rank is the smaller amongst $p$ and $n+1$. Any tips on how to approach this problem? Basically, I need to find the dimension of the column space. AI: You need to choose a basis for your polynomial space (I recommend the standard basis $\{1,x,x^2,\ldots\}$), then represent each polynomial in terms of that basis. That is, polynomial space of degree at most $n$ is isomorphic with $\mathbb{R}^{n+1}$. Once you have your polynomials represented as $(n+1)$-tuples, then you can do row reduction as usual. Example: $u=1+x, v=2+3x$. $E=\{1,x\}$ is the standard basis. Then $[u]_E=[1, 1]$ and $[v]_E=[2,3]$. We can now row reduce $\left[\begin{smallmatrix}1&1\\2&3\end{smallmatrix}\right]$ to get $\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]$. This has two pivots, so $\{u,v\}$ are independent.
H: Find the interval on which $f$ is increasing if the derivative of $f$ is $f'(x)=(x+4)^6(x-3)^7(x-4)^6$. Find the interval on which $f$ is increasing if the derivative of $f$ is $f'(x)=(x+4)^6(x-3)^7(x-4)^6$. Can anybody please verified that it increases at $(3,\infty)$, or do i have to do product rule with the above formula. Please Help!! AI: Hint: A (sufficiently nice) function is increasing if and only if its derivative is positive. Note that $$(x + 4)^6 (x - 4)^6 \ge 0$$ for every $x$, and $$(x - 3)^7 > 0$$ if and only if $x > 3$.
H: How many non-zero quadratic residues are there for $p^k$, where $p$ is an odd prime and $k$ a positive integer? How many non-zero quadratic residues are there for $p^k$, where $p$ is an odd prime and $k$ a positive integer? Hi everyone, just need a bit of help for this practice question, I have proved that for $k=1$ then the number of non-zero quadratic residues is $\frac{p-1}{2}$ (I did this using Fermat's Little Theorem and primitive roots mod p). It's also clear that if there exists an $x$ which satisfies: $x^2 \equiv a \pmod {p^k}$ then it must also satisfy: $x^2 \equiv a \pmod {p}$, I'm also ok with showing the reverse by induction, that if $x^2 \equiv a \pmod {p^k}$ then it also satisfies $x^2 \equiv a \pmod {p^{k+1}}$ how can I express how many quadratic residues there are for $p^k$ in terms of the factorization of $p^k$? Any help would be greatly appreciated! AI: We must consider even and odd primes separately. For $p$ odd, the group of units $(\mathbb{Z}/p^k\mathbb{Z})^\times$ is cyclic of order $p^k-p^{k-1}$. This is a standard but not entirely trivial result—I can find a reference if you like, but this is the part where you will need something like Hensel's lemma. It follows that exactly half of the units are squares. But there may be non-unit squares too, so you will also want to look at elements divisible by $p^2$. So you will end up with a sum like this: $\frac{p^k - p^{k-1}}{2} + \frac{p^{k-2} - p^{k-3}}{2} + \ldots$ For $p$ even, the group of units $(\mathbb{Z}/2^k\mathbb{Z})^\times$ is a tiny bit more complicated: it's product of a cyclic group of order $2$, and another cyclic group of order $2^{k-2}$ (for $k\geq 3$ anyway; for $k\leq 2$ it is cyclic). Again, this is not a completely trivial result, the proof is understandable but not immediate. But you now find that one-quarter of the units, or $2^{k-3}$ of them, are squares. As before, considering non-units as well gives you a sum: $2^{k-3} + 2^{k-5} + \ldots$, where you need to be careful about the last term or two. But now that I've written this up, I realize that it is quite possible to get around the technical lemmas; you can determine the number of squares without knowing the exact multiplicative structures. It goes something like this: Say that $a^2 = b^2 (\operatorname{mod} p^k)$. Then $(a-b)(a+b)$ is divisible by $p^k$. Now, you must consider various cases separately. What if $p^{k-4} \| (a-b)$, and $p^4 \| (a+b)$? The point is to figure out how many elements are squares by figuring out how many different elements, when squared, can give you the same result. Every time you get a repeated square, you know that this lowers the number of distinct squares by 1. If you do this analysis carefully, you should be able to get the same answer I mentioned above.
H: abstract algebra problem.. Let $U = \{1,0,c\}$ be a ring with three elements ($1$ is the unity). Which statements are true? \begin{align} &I.\,\,\,\,\,\,\,\,1 + 1 + 1 = 0 \\ &II.\,\,\,\,\, 1 + 1 = c \\ &III.\,\,c^2 = 1 \end{align} The solution is I II and III are all true. For III, the solution says by multiplicative table, $c^2$ has to be $1$. I know $c1=c$ and $cc=c$ is a bit weird. But I don't think it contradicts the definition of Ring, which just requires closure and associative for multiply. AI: Note that $c = 1 + 1$ must hold, so $$c^2 = 1^2 + 1 + 1 + 1^2 = 1 + 1 + 1 + 1 = 0 + 1$$ To see the very first statement, note that if $1 + 1 = 1$, then $1 = 0$; if $1 + 1 = 0$, then this contradicts that the additive group must have order $3$.
H: use combinatorial reasoning to calculate $ \sum{\binom{100}{a}\binom{200}{b}\binom{300}{c}}$ Given $ a + b + c = 100 $. $a,\ b,\ c $ are non-negative integers. Calculate $$ \sum {\binom{100}{a} \binom{200}{b} \binom{300}{c} } $$ Can someone help me with this question? I have no idea how to start it. AI: This sum is just $\binom{600}{100}.$ To see this, split a set of $600$ elements into three sets of sizes $300,200,$ and $100.$ Then to choose $100$ elements from the big set, we need to choose $a,b,c$ elements from those three sets for some nonnegative integers $a,b,c.$ Once we've decided how many elements we are choosing, we have $\binom{300}{a}\binom{200}{b}\binom{100}{c}$ ways to choose them.
H: Why must this function have a critical point inside the sphere? Suppose we have $f: \mathbf{R}^{3} \to \mathbf{R}$ with the following property: $\langle \nabla f(x), x \rangle > 0$ for every $x \in S^{2}$, that is, it's gradient points outwards the unit sphere. It's asserted that there must a point $p$ inside the sphere with the property $\nabla f (p) = 0$. Here's what I've done so far: suppose there's no such point in $B(0;1)$. Since $f$ is real valued and defined in the compact $\bar{B}(0;1)$, it must attain maximum and minimum. Since $f$ has no critical points in the interior, then these points must lie in $S^{2}$, say $x_{0}$ is the maximum and $y_{0}$ is the minimum. Now the problem seems to be that the gradient cannot point outwards in the minimum point, and from that we could derive a contradiction. But I don't know how to write that down -- using the directional derivative along the line joining the extrema points perhaps? AI: There is a topological argument. Consider the map $$\sigma : S^2 \to S^2 : x \mapsto \frac{\nabla f}{\\|\nabla f\\|}.$$ By the condition on $f$ at the boundary, $\sigma$ is homotopic to the identity map of $S^2$ (because there is a unique geodesic going from $x$ to $\sigma(x)$). Now suppose that $\\|\nabla f\\| \neq 0$ for each $x \in \overline{B}(0,1)$. Consider the map $$\overline{B}(0,1) \to S^2 : x \mapsto \frac{\nabla f}{\\|\nabla f\\|}.$$ Since $\overline{B}(0,1)$ is contractible, this map is homotopic to a constant map. But that is absurd, since its restriction to the boundary $S^2$ is $\sigma$, which is not homotopic to a constant map. Cool, huh?!
H: How to find the limit of recursive sequence? Suppose $a_0 = 0$, $a_1 = 1$, and $$ a_{n+1} = a_n+2 a_{n-1}$$ if $n \ge 1$. Find $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$$ Usually, what's the approach we use for recursive sequence like this? I know another type of recursive question with no fraction involved, but don't know how to do the type with fraction.... AI: Let $\lim\limits_{n\to \infty}\dfrac{a_{n+1}}{a_n}=l$. Now, $l=\lim\limits_{n\to \infty}\dfrac{a_{n+1}}{a_n}=1+\dfrac{2}{l}$. Thus $l^2=l+2$. The root of this equation gives the limit.
H: Prove $\sum_{k=0}^{n}(-1)^k(k+1)\binom{n}{k}=0$ Prove that $\sum_{k=0}^{n}(-1)^k(k+1)\binom{n}{k}=0$. I know $\sum_{k=0}^{n}(-1)^k\binom{n}{k}=0$ and even $\sum_{k=0}^{n}(-1)^k\dfrac{1}{k+1}\binom{n}{k}=0$ because I can multiply left side by n+1 and divided by n+1. Then I have $\dfrac{1}{n+1}\sum_{k=0}^{n}(-1)^k\dfrac{n+1}{k+1}\binom{n}{k}=\dfrac{-1}{n+1}\sum_{k=0}^{n}(-1)^{k+1}\binom{n+1}{k+1}=0$. Now, how do I eliminate the term (k+1) in the equation $\sum_{k=0}^{n}(-1)^k(k+1)\binom{n}{k}=0$? AI: We have that $$x(1+x)^n=\sum_{k=0}^{n}{n\choose k}x^{k+1}$$ Differentiate and evaluate at $x=-1$.
H: Can the direction of Contour Integral be affect to the result of integration? Now i doing the home work about Residue Integration and i doubt that "Can the direction of Contour Integral be affect to the result of integration?" I mean with the same shape of contour but direction of contour changes from clockwise to counterclockwise or counterclockwise to clockwise. AI: Reversing the orientation flips the sign of the result. This follows more or less immediately from the formula to convert a contour integral to a normal integral. One can define orientation for regions of other dimension as well, and you get the same effect. Usually we assume all two-dimensional regions of the plane are positively oriented, thus the need for boundaries to go counter-clockwise. But if we specify the region to be negatively oriented, then integrals over it get their sign flipped, and its boundary must go clockwise. So this still fits in nicely with all of the nice theorems of contour integrals. We can do this for points too. The 0-dimensional integral of $f$ over the point $x$ (oriented positively) is $f(x)$. With negative orientation, it is $-f(x)$. The fundamental theorem of calculus fits in nicely with the fact the boundary of a path from $a$ to $b$ consists of two points: the point $b$ oriented positively and the point $a$ oriented negatively.
H: Difference equation: $y_{k+1} = y_{k} + \frac{c}{2k}$ I'd like to solve this difference equation. Unfortunately, the forcing term is not geometric, so I don't know how to find the solution: $$ y_{k+1} = y_k + \frac{c}{2k}. $$ AI: Move $y_k$ over to the left side of the equation and take the partial sums on both sides since we conveniently have a telescoping series. $$\sum \limits_{k=1}^{n-1} (y_{k+1} - y_k) = \sum \limits_{k=1}^{n-1} \frac{c}{2k} ,$$ $$ y_{n} - y_1 = \frac{c}{2} \sum \limits_{k=1}^{n-1} \frac{1}{k} ,$$ where the sum on the right is just the $(n-1)$th harmonic number, $H_{n-1}$. This gives us the solution form, $$ y_{n} = y_1 + \frac{c}{2} H_{n-1} .$$
H: If a derivative of a continuous function has a limit, must it agree with that limit? Suppose we have a continuous function $f : \mathbb{R} \to \mathbb{R}$. Suppose also that for a certain point $c$, $\lim_{x \to c} f'(x)$ exists. Must $f'(c)$ exist as well, and be equal to this limit? This isn't quite the same as asking if derivatives are always continuous. The well-known function $f(x) = x^2 \sin (1/x)$ is continuous and differentiable everywhere, but its derivative has no limit at $x = 0$. I'm wondering if the derivative of a continuous function can have a discontinuity where its limit does exist. AI: For the limit to make sense, we have to assume that $f'$ exists on some interval around $c$. If $\lim_{x\to c}f'(x)=L$, then $f'(c)$ exists and it is equal to $L$. Indeed, using the Mean Value Theorem we have $$ \frac{f(c+h)-f(c)}h=f'(\xi(h)) $$ for $\xi(h)$ between $c$ and $c+h$. As $h\to0$, $c+h\to c$ and so $\xi(h)\to c$. So $$ \lim_{h\to 0}\frac{f(c+h)-f(c)}h=\lim_{h\to0}f'(\xi(h))=L. $$
H: Find the Radius of Convergence of the Series $\sum a_{n}x^{n^{2}}$ Using $\sum a_{n}x^{n}$? I want to show that $$\sum a_{n}x^{n^2}$$ has radius of convergence of 1, using the fact that the power series $\sum a_{n}x^{n}$ has radius of convergence $R>1$, where $R$ is a real number (actually the problem says $R=2$). I found out that this holds if $$\lim \|a_{n}\|^{1/n}= \dfrac{1}{R},$$ but I couldn't draw the same result using only that $\limsup \|a_{n}\|^{1/n}= \dfrac{1}{R}.$ Obviously the series converges at $x=1$, so its radius of convergence is at least 1, but how do I show that it's exactly $1$? I'm not even sure about it now. AI: Check This ,$\lim \sup\|a_nx^{n^2}\|^{1\over n}<1$ for convergence
H: Prove $n^3+3n$ is even in case $n=2a+1$ Prove that $\forall n \in Z$, $n^3+3n$ is even. Attempt: I am solving this problem using proof by cases. Case 1 is when $n$ is even, i.e. $n=2b$. This one is easy. However, in case 2 when $n$ is odd, i.e. ($n=2a+1$) I am having difficulties with showing that $n^3+3n$ is even. Namely, after plugging in and expanding I am stuck: $$n^3+3n=(2a+1)^3+3(2a+1)=8a^3+12a^2+6a+1+3a+3=\space...$$ I tried regrouping the terms, but my effors did not amount to anything. I guess I am making a mistake somewhere. Help appreiciated. Thank you. AI: You may write $n^3+3n=n(n^2+3)$. Now, if $n$ is even, then $2$ divides $n$, so $2$ divides $n(n^2+3)$. If $n$ is odd, then $n^2$ is odd, so $n^2+3$ is even, for $3$ is odd, so $2$ divides $n^2+3$, so it divides $n(n^2+3)$.
H: How do we say the behavior of the limit of cosine square? $\lim_{n\rightarrow \infty } (cosx)^{2n}$ Is it correct to say that this sequence diverges? $\lim_{n\rightarrow \infty } ((cosx)^{2})^{n}$ $ 0\le (cosx)^{2}\le1$, hence the behavior of $(cosx)^{2n}$ would be like oscillate between 0 and 1? But in some sense, I also feel like it's not continuous at $k\pi $ Thanks for helping me look into it! AI: The sequence of functions converges to a discontinuous function. For any $x$ such that $\|\cos{x}\| < 1$, the sequence tends to $0$, and otherwise (because of the $2$ in the power), it tends to $1$. Hence, the limit function is simply $$\lim_{n \to \infty} (\cos{x})^{2n} = \left\{\begin{array}{lr} 1 &\text{ if } x = n\pi \\ 0 & \text{ else} \end{array}\right.$$ where $n \in \mathbb{Z}$. This is continuous except at the points $n\pi$.
H: Locally Lipschitz implies continuity. Does the converse implication hold? Let $A$ be open in $\mathbb{R}^m$; let $g:A\rightarrow\mathbb{R}^n$. If $S\subseteq A$, we say that $S$ satisfies the Lipschitz condition on $S$ if the function $\lambda(x,y)=\|g(x)-g(y)\|/\|x-y\|$ is bounded for $x\neq y\in S$. We say that $g$ is locally Lipschitz if each point of $A$ has a neighborhood on which $g$ satisfies the Lipschitz condition. Show that if $g$ is locally Lipschitz, then $g$ is continuous. Does the converse hold? For the first part, suppose $g$ is locally Lipschitz. So for each point $r\in A$, there exists a neighborhood for which $\|g(x)-g(y)\|/\|x-y\|$ is bounded. Suppose $\|g(x)-g(y)\|/\|x-y\|<M$ in that neighborhood. Then $\|g(x)-g(r)\|<M\|x-r\|$ in that neighborhood of $r$. Therefore $g(x)\rightarrow g(r)$ as $x\rightarrow r$, and so $g$ is continuous at $r$. This means $g$ is continuous everywhere in $A$. What about the converse? I don't think it holds, but can't come up with a counterexample. AI: Intuitively, a counterexample must be a function which is very steep without having a jump or other sort of discontinuity. Consider, for example, $$g(x) = x^{1/3}$$ at $0$. Then $$\frac{\|g(x) - g(0)\|}{\|x - 0\|} = x^{-2/3}$$ This cannot be bounded in a neighborhood of $0$.
H: Derivative of ${x^{x^2}}$ Studying past exam problems for my exam in ~$4$ weeks, and I came across this derivative as one of the questions. I actually have no idea how to solve it. $$\frac{d}{dx} (x^{x^2})$$ Using the chain rule on it letting $x^2 = u$ led to me getting $2x^{x^2-2}$, which isn't right. The function acts like $e^x$ so I am thinking I have to convert it to this form. So I took it to be: $$\frac{d}{dx} (e^{{x^2}log(x)})$$ Not really sure where to go from here, or if I am going in the right direction. AI: Yes, you've done it correctly so far, and could proceed by the chain rule. On the other hand, set $$y = x^{x^2}$$ Then $\ln{y} = x^2 \ln{x}$; taking a derivative on both sides and using the chain rule for the left leads to $$\frac{y'}{y} = 2x \ln{x} + \frac{x^2}{x}$$
H: number of distinct ways of writing each element of the set $HK$ in the form $hk$ Let $H$ and $K$ be subgroups of the group $G$. The number of distinct ways of writing each element of the set $HK$ in the form $hk$, for some $h \in H$ and $k \in K$ is $\|H \cap K\|$. My thoughts:- Let $\|H \cap K\|=n$. Let $h_i=k_i$ where $i=1,2,3,....,n$.then if $\alpha \in HK$ then $\alpha = hk$ and also $\alpha = (hh_i)(h_i^{-1}k)$. but how can I able to show these are the only representation toprove the problem. AI: If $hk=h'k'$ then $kk'^{-1}=h^{-1}h'\in H\cap K$ and...?
H: what is the limit of $l_p$ at p=0? The p-norm is defined as: $$ \ \\|x\\|_p=\left(\|x_1\|^p+\|x_2\|^p+\dotsb+\|x_n\|^p\right)^{\frac{1}{p}} $$ When $p<1$, this is no longer a "norm" because it violates the triangle inequality (- it is super additive and not subadditive). However, it is still valid to ask, what is its limit when p goes to 0? My guess is that: If all coordinates are 0, then $l_p=0$, and it remains like this when p=0. If exactly one coordinate, say $x_i$, is non-zero, then $l_p=x_i$, and it remains like this when p=0. If more than one coordinate (say, $x_i$ and $x_j$) are non-zero, then $l_p>x_i$, and because the exponent goes to $\infty$, the value of $l_p$ also goes to $\infty$. Is this correct? (Note that this is not equal to the $l_0$ "norm" = the number of nonzero elements. This is also not equal to the scaled norm, in which there is an additional $1/n$ factor). AI: Your guesses are all correct. In particular, the limit is $+\infty$ if there are two or more nonzero components. Suppose WLOG that $x_1\neq 0$ and $x_2\neq 0$, and that $\|x_1\|\leq \|x_2\|$. Then $\\|x\\|_p\geq 2^{1/p}\|x_1\|\to +\infty$ as $p\searrow 0$.
H: Analysis True or False Question. (Simple) First one, I think it's true just by definition. Second one, I think it's true as well. Third one, i dont think sinx = cosx in interval [0,pi/2]. Am i missing anything? If there are counterexamples that I simply ignored, please tell me. Thank yOu AI: The third one is also true, by the intermediate value theorem. Consider $f(x)=\sin x - \cos x$ on $[0,\pi/2]$. We easily see $f(0)=-1$ and $f(\pi/2)=1$, so there must exist a point $p\in [0,\pi/2]$ with $f(p)=0$. Then $\cos p = \sin p$.
H: Does the rectangle contain the point? A rectangle is defined by the 4 points ABCD. How can I tell if a given point, (x,y), is in the interior of the rectangle? My current guess is the following: To be inside the rectangle, the point should be between the lines AB and CD, and between the lines AD and BC. The equation of line AB (and, similarly, the other lines) can be written as: $$ \frac{x-x_A}{y-y_A} = \frac{x_B-x_A}{y_B-y_A} $$ In order to be between line AB and line CD, one of the following must hold: Either: $$ \frac{x-x_A}{y-y_A} > \frac{x_B-x_A}{y_B-y_A} and \frac{x-x_D}{y-y_D} < \frac{x_C-x_D}{y_C-y_D} $$ Or vice versa: $$ \frac{x-x_A}{y-y_A} < \frac{x_B-x_A}{y_B-y_A} and \frac{x-x_D}{y-y_D} > \frac{x_C-x_D}{y_C-y_D} $$ (According to the actual values of A, B, C and D, some of these conditions may be impossible or trivial). In order to be between line AD and BC, one of the following must hold: Either: $$ \frac{x-x_A}{y-y_A} > \frac{x_D-x_A}{y_D-y_A} and \frac{x-x_B}{y-y_B} < \frac{x_C-x_B}{y_C-y_B} $$ Or vice versa: $$ \frac{x-x_A}{y-y_A} < \frac{x_D-x_A}{y_D-y_A} and \frac{x-x_B}{y-y_B} > \frac{x_C-x_B}{y_C-y_B} $$ Is this correct? Is there an easier way to calculate if a point is inside a square? AI: Assuming $ABCD$ are in cyclic order around the rectangle, if $P$ is your point, consider the points as vectors. You want $(A-B)\cdot P$ to be between $(A-B) \cdot A$ and $(A-B) \cdot B$ and $(A-D) \cdot P$ to be between $(A-D)\cdot A$ and $(A-D) \cdot D$.
H: Prove for a connected graph $G=(V,E)$, $\kappa(G)=1+\min_{v\in V}\kappa(G-v)$ $\kappa(G)$ here is the connectivity of $G$. I'm just at a loss for where to start. I thought about induction on the number of vertices since I can see that it clearly holds for trivial cases, but I can't make that work. I thought about induction on $\kappa$ but I can't make that work either. I considered starting with $K_n$ since $\kappa(K_n)=n-1$ and for any $v$, $K_n-v=K_{n-1}$, and $\kappa(K_{n-1})+1=n-1=\kappa(K_n)$, so I thought maybe I could do induction backwards by starting with the complete graph and forming whatever connected graph I want by deleting edges. Or something. But they're all dead ends I think. Can anyone think of even just a hint on how to start this? AI: Hint: Induction is not required. Looking what happens if you delete a vertex of a minimal cutset should give you one direction. Find a vertex $v$ with a minimal $\kappa(G-v)$, then look at its cutset should give the other direction.
H: Does this series converge? Can I use a rearrangemt to prove it doesn't? $$1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}+\frac{2}{6}-\frac{1}{6}+ \cdots$$ I was thinking to use a rearrangment to get the harmonic series, but I don't know if that's valid, because the series should converge absolutely in order to say that all rearrangments converge to the same limit. AI: You don't need to rearrange, just group appropriately. Some of the partial sums of this series are ...
H: Gamma function in the sight of Lebesgue and Riemann integration. I am taking a somewhat hard measure theory course and I was asked to prove this: a) Let $\alpha > 0$ be a real number. Prove that $$\Gamma(\alpha):=\int_0^\infty e^{-x}x^{\alpha-1}dx$$ exists. (We are studying the relationship between being Lebesgue-integrable and Riemann-integrable, so I am not quite sure what kind of integral should I prove exists, but I suspect it's the latter). b)Prove that: $$\lim_{n\to\infty}\int_0^n \left( 1-\frac{x}{n}\right)^n x^{\alpha-1}dx=\Gamma(\alpha).$$ Hint: $(1-\frac{x}{n})^n\le (1-\frac{x}{n+1})^{n+1}$ whenever $\frac{x}{n}<1$. This reeks of some limit theorem such as Monotone Convergence, Fatou or Lebesgue's Dominated Convergence, but those apply to Lebesgue-integrable functions. I the inequality stated in the hints makes me think I should use Monotone Convergence. Any insight would be greatly appreciated. We are using Bartle's Measure Theory Book and we are more or less around the $\mathcal{L}_p$ spaces part. Thank you in advance. AI: 1) Hint: For $x\ge0$ and $n\ge1$, $e^x=\sum\limits_{k=0}^\infty\frac{x^k}{k!}\ge1+\frac{x^n}{n!}$, thus $e^{-x}\le\frac{n!}{n!+x^n}$ 2) Hint: For $m\ge n$, $$ \int_0^m\left(1-\frac xm\right)^m x^{\alpha-1}\,\mathrm{d}x \ge\int_0^n\left(1-\frac xm\right)^m x^{\alpha-1}\,\mathrm{d}x \ge\int_0^n\left(1-\frac xn\right)^n x^{\alpha-1}\,\mathrm{d}x $$ and $\left(1-\frac xn\right)^n\le e^{-x}$ for $x\le n$.
H: prove $\sum_{i=0}^{n}\binom{2n+1}{i}=2^{2n}$ Can someone help me to prove $\sum_{i=0}^{n}\binom{2n+1}{i}=2^{2n}$. The right side means the total number of subsets of $[1,2,3,..,2n]$. Then What does the left side mean? Can someone please help me? Thank you. AI: You have that $$2^{2n+1}=\sum_{i=0}^{2n+1}\binom{2n+1}{i}=\sum_{i=0}^{n}\binom{2n+1}{i}+\sum_{i=n+1}^{2n+1}\binom{2n+1}{i}=$$$$\sum_{i=0}^{n}\binom{2n+1}{i}+\sum_{i=n+1}^{2n+1}\binom{2n+1}{2n+1-i}=$$$$\sum_{i=0}^{n}\binom{2n+1}{i}+\sum_{j=0}^{n}\binom{2n+1}{j}=2\sum_{i=0}^{n}\binom{2n+1}{i}.$$ So, $\displaystyle \sum_{i=0}^{n}\binom{2n+1}{i}=2^{2n}$.
H: Limit goes to infinity, show that the f has a finite minimum. So limit goes to infinity, and I have to show that there exists a finite infimum. how do i show this? AI: Think of a chain you hold by both ends very high. It has to ''drop down'' somewhere. That's roughly the idea. So let's hold that rope ''very high''. For every $M > 0$, there exists $K > 0$ such that for every $x > K$, $f(x) > M$ and for every $x < -K$, $f(x) > M$ ; in other words, for every $x$ satisfying $\|x\| > K$, $f(x) > M$. Assuming $M$ is large enough so that $f^{-1}(]-\infty,M])$ is non-empty, then we know the infimum of $f$ is inside $[-K,K]$ because outside this interval, $f(x) > M$. Now $[-K,K]$ is a compact interval. Since $f$ is continuous we can find the minimum on it. :) For the example of the function $g$, you have very simple examples : for instance, take $$ g(x) \overset{def}= \begin{cases} -x-2 & \text{ if } x < -1 \\ \\ \frac 1x & \text{ if } 0 < \|x\| < 1 \\ \\ 0 & \text{ if } x = 0 \\ \\ x & \text{ if } x > 1. \end{cases} $$ I'm basically just taking what happens with $\frac 1x$, satisfying your conditions, and then "holding the function with a rope" (the two linear parts). Hope that helps,
H: Integration: easy, but how? Once again I have to deal with an Integral, which seems to be not that hard: $ \int x^{n-1} * sin(x^n) dx $ and $(n \in \mathbb{N})$ Do you have any hint for me ? :) AI: Hint: If $u = x^n$, then $$x^{n - 1} dx = \frac{1}{n} du$$
H: Why does the limit of $F_X$ from the left equal $P (X < x)$? I am trying the understand the definition of the cumulative distribution function: We use $F_X(x^-)$ to denote the limit of $F_X$ from the left $ lim_{ 0 < \Delta \rightarrow 0} F_X(x - \Delta )$ and $F_X(x^+) $ the limit from the right $ lim_{0<\Delta \rightarrow 0} F_X (x + \Delta )$. Note that $F_X (x^-) = P (X < x)$. Why do we have $F_X (x^-) = P (X < x)$? AI: By properties of probability measures we have: $$ \begin{align} P(X<x)&=P_X((-\infty,x))=P_X\Big(\bigcup_{n=1}^\infty \big(-\infty,x-\tfrac1n\big]\Big)\\ &=\lim_{n\to\infty}P_X\Big(\big(-\infty,x-\tfrac1n\big]\Big)\\ &=\lim_{n\to\infty} F_X(x-\tfrac1n)=F_X(x^-). \end{align} $$ Here $P_X=P\circ X^{-1}$ is the distribution of $X$.
H: Logical proof of the statement $xy = 0 \implies x=0\text{ or } y=0$ Claim: If $xy=0$, then $x=0$ or $y=0$. My proof is as follows: case 1: $x=0$, so $0y=0$ case 2: $y=0$, so $x0=0$ Either way, $xy=0$. I'm very confused by this myself. So if I let $xy=0$ be $P$, and $x=0$ or $y=0$ be $Q$, then the claim "if $xy=0$, then either $x=0$ or $y=0$" is asking me to prove that $P \implies Q$. But I feel as if the proof I gave is $Q \implies P$, which is very different from $P \implies Q$. Can anyone enlighten me on the subject of proving if-then statements? AI: Your feeling is right: What you have done is the opposite direction. Your argument proves $$(x = 0 \text{ or } y = 0) \implies xy = 0.$$ How can you prove a statement of the form $$A \implies B$$ in general? The direct method is to assume that $A$ is true, and then to conclude that also $B$ is true under this assumption. Let's apply this to prove the statement $$xy = 0 \implies (x = 0 \text{ or } y = 0).$$ In this case \begin{align} A & = \text{''}xy = 0\text{''} \\ B & = \text{''}x = 0\text{ or }y = 0\text{''} \end{align} So we assume that $xy = 0$ is true. Now we have to show that $(x = 0 \text{ or } y = 0)$ is true. The nature of an "or"-statement often involves a case by case study: If $x = 0$, then of course $(x = 0 \text{ or } y = 0)$ is true. Otherwise, we have $x \neq 0$. Now we may divide our assumption (the equation $xy = 0$) by $x$ to get $y = 0$, so $(x = 0 \text{ or } y = 0)$ is true also in this case. As an addition: We have just proven $$xy = 0 \implies (x = 0 \text{ or } y = 0),$$ and the argument in your question proves $$xy = 0 \Longleftarrow (x = 0 \text{ or } y = 0).$$ So in fact, we have an equivalence, which we can write down as $$ xy = 0 \iff (x = 0 \text{ or } y = 0). $$
H: proofing the $ \nabla (FG) $ $\\ F $and$ G$ are vector fields my problem is that why the answer is not $ (\nabla. F)G + (\nabla .G)F $ AI: Following on from above, if $F, G$ are indeed vector fields in $\mathbb{R}^3$ then write them out accordingly, viz; $$ F = F_{1}\hat{i} + F_{2}\hat{j} + F_{3}\hat{k}$$ And apply on this vector (respectively $G$) $$\nabla = \sum_{p=1}^{3}e^{p}\frac{\partial}{\partial x_{p}}$$ Where the $e_{p}$'s represent the standard $\mathbb{R}^{3}$ basis written slightly differently above.
H: Calculus Integration, using the substitution method Please help solve this problem using the u-substitution method. Thanks in advance. AI: Make the change : $2 x^2 - 3 = y^2$. The integral becomes very simple.
H: Co-prime power modulo Is there $p$ such that $a^p\,$mod$\,r=0$ and $r$ is a prime number and $1\le a<r$. I am believing that $a^p\,$mod$\,r=0$ will be not equal to $0$ any value of $p$ for all values of $a$ from $1$ to $r-1$. Is it correct? please provide any counter example if not. AI: If $a^p\equiv 0\mod r$, it means that $r$ divides $a^p$ (or that $a^p=0$ which is not the case). As $r$ is prime, then $r$ must divide $a$, but $a<r$ so $r$ cannot divide $a$. This contradiction proves that the premise is false that there is $p$ such as $a^p\equiv 0\mod r$ for some $a$ with $1\le a<r$.
H: Are there any integers $a,b$ s.t. ${ a }^{ 2 }-{ b }^{ 2 }=8$? For $a$ and $b$ are integers greater than $1$, ${ a }^{ 2 }-{ b }^{ 2 }=8$ holds? AI: Hint: We have $8= a^2 - b^2 = (a+b)(a-b)$. As $a,b \ge 1$, we have $a+b > a-b$, that is either $a+b=8$ and $a-b = 1$, or $a+b=4$ and $a-b=2$.
H: Integrate $\int_0^{\pi/2}\frac{\sin^3t}{\sin^3t+\cos^3t}dt$? How to integrate $$ \int_0^{\pi/2}\frac{\sin^3t}{\sin^3t+\cos^3t}dt\,? $$ I tried to use $\sin^3tdt=-(1-\cos^2t)d\cos t$. But the term $\sin^3t$ in the denominator can not be simplified. Can anyone give me a hand? Thanks. AI: General Hint: By setting $x=\pi/2-y$ you can see that $$\int_0^{\pi/2}\frac{\sin^m(x)}{\sin^m(x)+\cos^m(x)}dx=\pi/4,~~m\in\mathbb R$$
H: Etymology of the term "filter" According to this article, filters were introduced in general topology by Henri Cartan in 1937. I wonder why he called them filters. AI: It may not have been Cartan who named them. According to Michèle Audin, ‘Henri Cartan & André Weil du vingtième siècle et de la topologie’, p. $8$, the concept was originated at a meeting of Bourbaki in $1937$. After a morning of discussion most of the group went for a walk, but Cartan remained behind and came up with filters as a way of removing the countability restriction imposed by use of sequences. He came up with the idea, but not with the name (La chose était inventée mais pas son nom.) I don’t know what the original motivation for the name was was, but it is certainly possible to see filters on a topological space as a way of filtering out points of a space. For example, if $\mathscr{N}(x)$ is the neighborhood filter of a point $x$ in a space $X$, each $N\in\mathscr{N}(x)$ can be thought of as filtering out the points of $X\setminus N$, keeping only those points that are ‘$N$-close’ to $x$. It does seem that the original motivation for filters was to get a general notion of convergence in topological spaces, and that’s precisely the setting in which this metaphor works best. Subsequently, of course, the notion was generalized and many other applications were found, but it wouldn’t surprise me if this or some similar metaphor were the reason for the name.
H: How is the circle that fits beneath two adjacent circles related? This is hard to search and probably easy to solve, but I keep finding articles about intersecting circles, and that is not what I'm after. I don't know what to tag this under, so if you know how to classify this better, please do. I'd like to know how the red circle's radius is related to the radii of the larger circles, and why. AI: Your setup is a special case of Descartes' theorem. In general that theorem describes the radii of four circles which all touch one another, but in your special case one of these circles is a line. Since the key ingredient to the relation are inverse radii, this means that the corresponding term will simply be zero. Long story short: if $r_1$ and $r_2$ are the radii of the given circles, then the small circle has radius $$r_3 = \frac{1}{\dfrac1{r_1}+\dfrac1{r_2}+\dfrac2{\sqrt{r_1r_2}}}$$ The only idea for a proof which comes to my mind just now would involve Lie geometry, which might be a bit beyond the scope of this question. But perhaps referring to Descartes' theorem is proof enough here?
H: How do I solve this limit? $\lim_{x \to \frac{\pi}{4}}\frac{\sin x-\cos x}{\ln(\tan x)}$ I can't really see the right way to solve this limit. My attempt is: $$\lim_{x \to \frac{\pi}{4}}\frac{\sin x-\cos x}{\ln(\tan x)}=\left(\lim_{x \to \frac{\pi}{4}}\frac{\sin x-\cos x}{\ln(\tan x)}\right):\cos x = \lim_{x \to \frac{\pi}{4}}\frac{\tan x-1}{\frac{\ln(\tan x)}{\cos x}} = \frac{0}{\frac{0\cdot{2}}{0\cdot\sqrt{2}}}$$ So this answer is wrong. But I would like to understand how to solve this problem without using derivation or L'hôpital's rule. AI: Hint: your expression can be transformed as $\tan x-1\over \log (1+\tan x -1)$ Now take $y=\tan x -1$ so $y\to 0$ as $x\to{\pi\over 4}$ so now in $y$ variable $\lim_{y\to 0}{y\over \log(1+y)}=\lim_{y\to 0}{1\over{\log(1+y)\over y}}=1$
H: Autocorrelation problem, regression analysis Bit stuck on my econometrics course (old exam q), not big on mathematical statistics, anyway this is the problem: Given some model $y_{it}=\beta_0+\beta_1x_{it}+u_{it}$ and suppose that the idiosyncratic errors are serially uncorrelated with constant variance i.e. $var(u)=\sigma^2$, $E(u)=0$ $corr(u_{it},u_{is})=0\: \forall \: t\neq s $. Show that $corr(u_{it}-u_{it-1},u_{it-1}-u_{it-2})=-0.5$. I've been trying to jerk around a bit with expected value formulas of variance/covariance, but I keep hitting a dead end. Also note that since this is taught by the econ department my matrix algebra in this area is poor, so it would be more helpful if it were dissected in "simple terms". Any help appreciated! Thanks /I AI: Separate the terms. Then you get: $$ Cov(u_{it}-u_{it-1}, u_{it-1}-u_{it-2})=Cov(u_{it}, u_{it-1})-Cov(u_{it}, u_{it-2})-Cov(u_{it-1}, u_{it-1})+Cov(u_{it-1}, u_{it-2}) $$ The first term is 0 by assumption. So are the second and fourth ones as well. The third one is $$ -Cov(u_{it-1}, u_{it-1})=-V(u_{it-1})=-\sigma^2. $$ So then $$ Corr(u_{it}-u_{it-1}, u_{it-1}-u_{it-2})=\frac{Cov(u_{it}-u_{it-1}, u_{it-1}-u_{it-2})}{\sqrt{V(u_{it}-u_{it-1})V(u_{it-1}-u_{it-2})}}=-\frac{\sigma^2}{\sqrt{4(\sigma^2)^2}}\\ =-\frac{1}{2}. $$
H: If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but: $$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$ for some integers $m, n$ $$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 = \frac{m^3}{n^3}$$ $$\implies a + b + 3 \cdot \sqrt[3]{ab}\left(\sqrt[3]{a} + \sqrt[3]{b}\right) = \frac{m^3}{n^3}$$ Since $a +b$, $\sqrt[3]{a} + \sqrt[3]{b}$ are rational, $\sqrt[3]{ab}$ must be rational as well. For convenience let us say $\sqrt[3]{a} = p, \sqrt[3]{b} = q \implies pq$ is rational. This means, for all $i$, $p^iq^i$ is rational. $$\implies (p + q)^2 = \frac{m^2}{n^2}$$ $$ = p^2 + q^2 + 2pq = \frac{m^2}{n^2}$$ Since $pq$ is rational, $2pq$ is rational and so is $p^2 + q^2$. Assume, for some $i$ that $p^i + q^i$ and $p^{i-1} + q^{i-1}$ is rational. $$\implies (p^i + q^i)(p + q) - pq(p^{i - 1} + q^{i - 1}) = p^{i+1} + q^{i+1}$$ is rational as well. So for all $i$, $$a^{\frac{i}{3}} + b^{\frac{i}{3}}$$ and $$a^{\frac{i}{3}}b^{\frac{i}{3}}$$ are rational. I know I'm really close to the answer, but it somehow just keeps slipping through my fingers. AI: As you have proved, $p^2+q^2\in\Bbb Q$ and $pq\in \Bbb Q$, so $$p-q=\frac{a-b}{p^2+pq+q^2}\in \Bbb Q.$$ Combining this with $p+q\in \Bbb Q$, we are done.
H: Step in proof: Sum of euler phi function over divisors (Group Theory) Proof: $$\sum_{d\|N}\phi(d)=N$$ where the sum is over $d\in div(N)$ Let $G$ be the cyclic group $\mathbb Z/N \mathbb Z$. Then $$N = \sum_{g \in G} 1 = \sum_{d\|N}\sum_{g \in G, ord(g)=d} 1=\sum_{d\|N}\phi(d)$$ The step from $$\sum_{d\|N}\sum_{g \in G, ord(g)=d} 1=\sum_{d\|N}\phi(d)$$ is easy since the number of elements in $G$ having order $d$, where $d \| N=\|G\|$ is $\phi(d)$, which I don't proof here. However how does one take the step $$\sum_{g \in G} 1 = \sum_{d\|N}\sum_{g \in G, ord(g)=d} 1$$ I having trouble seeing this. Please give a detailed answer. Thanks. AI: Every element in $G$ has some order $d$, which must divide $n$. It follows that $\displaystyle \# G = \sum_{d\|N} \#\{g\in G : \operatorname{ord}g=d\}$. But $\displaystyle \#G = \sum_{g\in G}1$, and $\displaystyle \#\{g\in G : \operatorname{ord}g=d\} = \sum_{\substack{g\in G, \\ \operatorname{ord} g = d}} 1$.
H: Showing that any sequence in $[0,1]$ has a convergent subsequence. One should show that any sequence in $[0,1]$ has a convergent subsequence. Now before even trying to prove it in general, I take one sample sequence, $x_n = \|\sin(n)\|$. I think that for this sequence, the only convergent subsequence could be a constant. So $x_{nk} = x_{nk+1}$, i.e., $\|\sin(m(k))\| = \|\sin(m(k+1))\|$. As $\sin$ is $2 \pi$ periodic, I would need some strictly increasing function $m(k)$ s.t. $m(k) 2 \pi = m(k+1)$. However, as $m : \mathbb{N} \rightarrow \mathbb{N} $, I find myself unable to find such a function. Which of my assumptions here is wrong? Could there be some convergent subsequence without the subsequence being constant? Or does an appropriate $m(k)$ exist? Thanks for all hints! AI: There will be a non-constant convergent subsequence. There is a sequence $\langle n_k:k\in\Bbb N\rangle$ of positive integers such that $$\langle n_k\pi-\lfloor n_k\pi\rfloor:k\in\Bbb N\rangle\to 0\;.$$ For $k\in\Bbb N$ let $m_k=\lfloor n_k\pi\rfloor$; then $\langle m_k:k\in\Bbb N\rangle$ is a subsequence of the positive integers such that $$\langle\sin m_k:k\in\Bbb N\rangle\to 0\;.$$ However, I really think that it’s easier simply to prove the general theorem than to find a convergent subsequence of some arbitrary sequence.
H: Co-prime binomial coefficient A number 'r' is prime if and only if $\binom{r-1}{k} \equiv(-1)^k \pmod r$ Since 'r' is a prime and it gives non-zero remainder by dividing $\binom{r-1}{k}$ . So $\binom{r-1}{k}$ and 'r' are co-primes If a0,a1,a2,a3,..........,ar-1 are coprimes to r . Then Is $\binom{r-1}{0}$a0+$\binom{r-1}{1}$a1+$\binom{r-1}{2}$a2+......+$\binom{r-1}{k}$ak+......+$\binom{r-1}{r-1}$ar-1 coprime to r ? AI: No, let $r=3,a_1=4$ and $a_0=a_2=5$ then $a_k$ is coprime to $r$, but $$ 5\binom20 +4\binom21+5\binom22=18=r^22\;. $$
H: Linear Algebra, reflected linear image If I have a linear image of the room where v1 and v2 is an image of theirselves and v3 is an image of the null vector. If that gives me the matrix A=(a, b, c; d, e, f; g, h, i;) then A^n = A because you've already applied a vector to the plane. However, if v3 is an reflection (image of -v3). If this gives me the matrice B=(a, b, c; d, e, f; g, h, i;), what would B^n result? My first thought was that it'd be the same as the first (A) matrix, but it seemed to be wrong. Why is it not the same as the first matrix and what would the result be? Sorry if my translation of mathematics terms is incorrect. AI: If $A$ is a reflector along the subspace spanned by $v_3$ (or with respect to the subspace spanned by $v_1$ and $v_2$), i.e., $$Av_1 = v_1, \quad Av_2 = v_2, \quad Av_3 = -v_3,$$ for some linearly independent vectors $v_1$, $v_2$, and $v_3$, then $$A^n = \begin{cases} A, & \text{$n$ is odd}, \\ I, & \text{$n$ is even}. \end{cases}$$ If $A$ is a projector on a subspace spanned by $v_1$ and $v_2$, i.e., $$Av_1 = v_1, \quad Av_2 = v_2, \quad Av_3 = 0,$$ for some linearly independent vectors $v_1$, $v_2$, and $v_3$, then $A^n = A$ for all $n \ge 1$. If I misunderstood your question, please edit it with further explanations about what you want to know.
H: Can I simplify logarithm of logarithm. It it possible to simplify Logarithm of logarithm: log log x (for example, make it log (base 4) x. AI: No, it's not possible to simplify (although it might be possible to make it very, very complicated). As a side note, we have $$ \log_{4} x = \frac{\log x}{\log 4} $$ so there's that.
H: which is bigger $\log\log x$, $\sqrt x$ , $(\log x)^2$ I need to compare these 3 and rank them based on which is bigger: $\log \log x$ $\sqrt x $ $(\log x)^2$ As $x\to\infty$ AI: For every $x>1$, you can find a $t$ such that $x:=\mathrm {exp}(\mathrm {exp}(t)))^2$. Plug this reparameterization into you functions and you find a clear hierarchy: The first grows linearly in $t$, the second one doubly exponentially and the third only exponentially. As $t$ grows bigger (resp. $x$, because the functionality is monotonally increasing), at one point the doubly exponential function only even takes values which are bigger than that of the simple exponential function and the same goes for the even slower linear one.
H: Useful relationships that are true for every norm I am looking for useful identities that are true for every normed vector space $(V,\|\|.\|\|)$ on either $\mathbb{R}$ or $\mathbb{C}$(if your identity is restricted to either one of them, please say so). I am looking for things like $\|\|\|f\|\|-\|\|g\|\|\|\le \|\|f-g\|\|$ and so on.Therefore: Things, that are always true for a norm in every vector space. The best answer will be the identity that appears to be most useful. AI: I think this paper is what you are looking for. It contains strengthenings of triangle inequality and generalizations for multiple vectors.
H: Cumulative pdf, integration problems. Given the cumulative pdf $f(x,y)=\begin{cases}xy/96, & \text{if 0<=x<=4 and 1<=y<=5} \\0, & \text{otherwise} \end{cases}$ Find $P(X+Y<3)$ My attempt: $$P(X+Y<3)=\int_{1}^{5}\int_{0}^{3-y}xy/96 dxdy=1/12$$ $$P(X+Y<3)=\int_{0}^{4}\int_{1}^{3-x}xy/96 dydx=0$$ Shouldn't those two be equal? What did I do wrong? I calculated the double integrals with this: http://www.wolframalpha.com/widgets/view.jsp?id=f5f3cbf14f4f5d6d2085bf2d0fb76e8a AI: limits are wrong Draw a graph and see these: $$P(X+Y<3)=\int_{0}^{2}\int_{1}^{3-x}xy/96 dydx=1/48$$ or $$P(X+Y<3)=\int_{1}^{3}\int_{0}^{3-y}xy/96 dxdy=1/48$$
H: Is the classification of finite simple groups not a bit arbitrary? I've never been able to find any details on what exactly decides what the classifications ought to be for finite simple groups. We have: Cyclic groups Alternating groups Groups of Lie type Sporadic groups But why does the classification stop there? If there are no requirements on what the categories should be, I could classify all finite simple groups by saying "they're all finite and simple" or "they're all either cyclic or non-cyclic". Or maybe the above categories could be broken down even further into sub-categories. Why was it decided that these ought to be the categories into which we classify groups? As a subquestion: what determines whether or not a group of "sporadic"? Who's to say that the existence of sporadic groups isn't just a sign that the classification is bad, and that the categories need to be redefined so as to include a place for the sporadic groups? AI: There might not be a rigorous answer, but I believe we can say we have achieved a full classification of a type of object when we can in some way construct every single one of them and put them in front of us. Say I want the finite, simple groups of order $n$. Then I just look at the cyclic $n$-group (easily constructed, ant it's simple when $n$ is prime). Then I go on to the alternating group of order $n$ (exists for $n = m!/2$ for some $m$, and simple for $m = 3$ or $m \geq 5$). And so on for Lie type groups and sporadic groups. I can give you all simple groups of a given order, because they have all been classified. As for the subquestion on the sporadic groups, after it had been found that the cyclic, alternating and Lie type groups (of appropriate order) are simple, the sporadic groups are the 26 simple, finite groups that do not fit in. Well, we know what groups they are, and it has been proven that there are only 26 of them. Without knowing the history of classification of simple groups, I suspect that that was the last result to be proven. The sporadic groups can be further divided into sub-classifications, but many of the "classes" you'd end up with only have one group in them, so it's not very useful. I understand that you might be frustrated by the fact that there's no "clean" or elegant classification, but that's how mathematics is some times.
H: How to solve $\lim_{x\to 0}\frac{\sqrt{x^2+4x+5}-\sqrt{5}}{x}$? How to solve the following question? \begin{eqnarray} \\\lim_{x\to 0}f(x)&=&\lim_{x\to 0}\frac{\sqrt{x^2+4x+5}-\sqrt{5}}{x}\\ \\&=&\lim_{x\to 0}\frac{\sqrt{\frac{x^2+4x+5}{x^2}}-\sqrt{\frac{5}{x^2}}}{1}\\ \\&=&\frac{\sqrt{\lim_{x\to 0}(1+\frac{4}{x}+\frac{5}{x^2})}-\sqrt{\lim_{x\to 0}(\frac{5}{x^2})}}{1} \\&=& \infty-\infty \\&=&? \end{eqnarray} It can be taken limit and $\lim_{x\to 0}f(x)=0.89...$ but when x=0, y do not exist. Rgiht? Thank you for your attention AI: $$\lim_{x\to 0}f(x) = \lim_{x\to 0}\frac{\sqrt{x^2+4x+5}-\sqrt{5}}{x}$$ Evaluating the limit of $f(x)$ as $x\to 0$ is not be be confused with evaluating $f(x)$ at zero, where, as you may have noted in this case, $f(x)$ is not defined. What we are interested in when taking the limit as $x\to 0$ is the behavior of $f(x)$ as $x$ approaches, gets arbitrarily close to, zero. If it converges to some value, then that is the limit. This is not to be mistaken with $f(0)$, even though many times, you'll find functions for which $\lim_{x \to 0} f(x) = f(0).$ In this case, since we have an indeterminate form $\frac 00$, we can use L'hopital, or we can multiply the numerator and denominator by the conjugate of the numerator to obtain a non-indeterminate form: $$\begin{align} \\\lim_{x\to 0}f(x) & =\lim_{x\to 0}\frac{\sqrt{x^2+4x+5}-\sqrt{5}}{x}\\ \\& =\lim_{x\to 0}\frac{\sqrt{x^2+4x+5} - \sqrt 5}{x}\cdot \frac{\sqrt{x^2 + 4x + 5} + \sqrt 5}{\sqrt{x^2 + 4x + 5} + \sqrt 5}\\ \\ &= \lim_{x\to 0} \frac{(x^2 + 4x + 5) - 5}{x(\sqrt{x^2 + 4x + 5} + \sqrt 5)}\\ \\ & = \lim_{x\to 0} \frac{x(x + 4)}{x(\sqrt{x^2 + 4x + 5} + \sqrt 5)}\\ \\ & = \lim_{x\to 0} \frac{x + 4}{\sqrt{x^2 + 4x + 5} + \sqrt 5} \\ \\ & = \frac 4{2\sqrt 5} = \frac 2{\sqrt 5} \end{align}$$
H: Properties of Digit root Why the digit root of any number calculated in any way remains same...e.g Let $f(x)$denote the digit root of $x$ $f(1237)=f(12+37)=f(49)=f(123+7)=f(130)=4$ I checked numerically with many numbers but I found that the digit root remains constant irrespective of the way the numbers are summed up I think it has something to do with modular math.A general proof would be appreciated AI: Originally, this question was posted about $f$ being the "sum of digits" or the "digit sum". And it was answered accordingly. To answer your original question: ... Because addition is associative, that's why. Suppose we have a number with digits a, b, c, d, expressed as the concatenation $abcd$, in that order, and suppose we're defining "digit sum" to be $f(abcd) = a+b+c+d,$ and this will be the same digit sum of $a \cdot 10^3 + b\cdot 10^2 + c \cdot 10 + d$, or the digit sum of $(a \cdot 10 + b) + (c \cdot 10 + d)$, etc. We can then apply $f$ to $f(f(abcd))$, but again, addition is associative. We can apply $f$ repeatedly, and associatively, to arrive at the same "digit root" as you are calling it. $$\begin{align} f(f(1237)) & = f(f(1 + 2 + 3 + 7)) \\ &= f(f(1+2) + f(3 + 7)) \\ &= f(f(f(1 + 2) + f(3)) + f(7)) \\ &= f(13) = 4 \\ & = \quad \cdots\end{align}$$
H: Expected number of sixes tossed before success Given a fair dice, what is the expected number of sixes that get tossed before throwing 2 sixes in a row? Would I start by finding all possible sets that occur without two sixes in a row and multiply them by the corresponding probability? ie $$\sum_{k=0}^\infty (A+B)^k $$ where A is all sequences that don't contain a six and B is all sequences that contain just one six Thanks for your help AI: The set of sequences prior to the final two sixes that end the game is actually given by $$\sum_{k=0}^\infty (A+BA)^k $$ The generating function that counts the number of sixes and non-sixes prior to the final two sixes is $$C(a,b)=\sum_{k=0}^\infty (a+ba)^k={1\over 1-(a+ba)}.$$ You turn this into a joint probability generating function by replacing $a$ with $5a/6$ and $b$ with $b/6$, then multiplying by $1/36$ to account for the final two sixes. This gives $$P(a,b)={1\over 36-30a-5ab}.$$ The marginal probability generating function for the number of sixes is found by plugging in $a=1$, to obtain $$P(b)={1\over 6-5b}. $$ Finally you get the expected number of sixes by differentiating and setting $b=1$. The expected number of sixes prior to the final two sixes is 5.
H: showing that some quotient ring is a field. To show that $$\mathbb{Z}[x] / \langle 5, x^3+x+1\rangle$$ is a field; I tried to show that $\langle 5, x^3+x+1\rangle$ is a maximal ideal of $\mathbb{Z}[x]$, but I failed. Because I have not seen the maxmal ideal generated by 2 generateors, the problem is hard. Please inform me how to show that. AI: HINT: Note that $\mathbb{Z}[x]/\langle5, x^3+x^2+1\rangle\cong \mathbb{Z}_5[x]/\langle x^3+x^2+1\rangle$. Show that $\langle x^3+x^2+1\rangle$ is a maximal ideal in $\mathbb{Z}_5[x]$. Note that it is enough to show that $x^3+x^2+1$ has no root in $\mathbb{Z}_5$, this would imply that the polynomial is irreducible and hence the ideal is maximal.
H: Show that a language is not regular using Myhill-Nerode Theorem I'd like to show that the language below is not regular using Myhill-Nerode Theorem. How can I do that? Let Σ = {0, 1}. Let L = {ww\|w ∈ Σ*} I am not sure where or how to go about this... AI: HINT: I use the notation and terminology of the Wikipedia article on the Myhill-Nerode theorem. Show that for this language $L$, the equivalence relation $R_L$ has infinitely many equivalence classes. Use the fact that if $u,v\in L$, $u\ne v$, and $\|u\|=\|v\|$, then $uu\in L$, but $vu\notin L$, so $u$ is a distinguishing extension for $u$ and $v$.
H: Inverse of orthogonal matrix is orthogonal matrix? Is inverse of an orthogonal matrix an orthogonal matrix? I know its inverse is equal to its transpose, but I don't see where the orthogonality would come from. AI: If $A^t = A^{-1}$, then taking inverses of both sides, we have $(A^{t})^{-1} = A = (A^t)^t$.
H: True/False about ring and integral domain I have some true or false questions and would like to have your help to check on it. A). in a ring R, if $x^2=x$, $\forall x\in R$, then R is commutative For (A), when looking at $(x+y)^2$, it has $x+y=(x+y)^2=x^2+xy+yx+y^2$ and then yx+xy=0, and from 2x=4x, therefore 2x=0. how this play a role here? B) In an integral domain, it $\exists m\in N, s,t, mx=0, \forall x\in R$, then it's a finite integral domain. I think this one is correct by definition of characteristic for a ring/field. C) commutative ring with unity has at least two elements, and cancellation holds, then it's an integral domain. I think it's correct, first it's a commutative ring with unity, second, it has at least two element, and there's a property saying that an integral domain must have at least two elements, and the third, cancellation holds implies it has no zero divisor. So the three above looks like fit the profile of integral domain. D) f is a homomorphism from group G to group H, then f(G) is a normal subgroup in H I'm not quite sure about this one, kind of remember that normal subg roup of G under homomorphism is normal subgroup of H, but don't know if f(G) will be normal in H Are these answer or argument correct? Thanks for your help. AI: For (A), first remark that $2x=0$ for all $x \in R$, because $2x = (2x)^2 = 4x^2 = 4x$. Then look at $(x+y)^2$. For (B), it is not necessarily true. For instance, in the polynomial ring $\mathbf F_p[X]$, every element is killed by $p$, but $\mathbf F_p[X]$ is infinite. (C) is correct. (D) is false. (Also why do you introduce $K$ if you don't use it?). Take a group $G$ with a non-normal subgroup $M$, and consider the injection of $M$ in $G$...
H: Number of values of x $$a\dfrac{(x-b)(x-c)}{(a-b)(a-c)}+b\dfrac{(x-c)(x-a)}{(b-c)(b-a)}+c\dfrac{(x-a)(x-b)}{(c-a)(c-b)}=x$$ How many values of $x$ satisfy this equation? It is clear that x=a, x=b, x=c do satisfy the equation, but are those the only three possible solutions? AI: This has to be true for all $x$. Firstly, there are at least $3$ solutions, assuming $a,b,c$ are distinct. Secondly, this is a polynomial equation of degree $2$ in one variable $x$. In other words, you can rewrite the equation as $A x^2 + B x + C = 0$. But this can only have at most $2$ solutions, unless $A=B=C=0$. Combining the two facts, the above equation holds for all $x$. Thus, if you multiply things out, it should come out as an identity.
H: Showing symmetry of a relation between a commutative identity ring and a multiplicative subset of the ring R is a commutative ring with identity, not necessarily an integral domain, and S is a multiplicative subset of R\0 containing 1 which is closed under multiplication. A relation ~ is defined on R x S so that $(r_1,s_1)$ ~ $(r_2,s_2)$ if and only if there exists an element t of S so that $t(r_1s_2 – r_2s_1) = 0$. I am trying to show that this relation is an equivalence relation, and I have everything but symmetry, which is tripping me up a little bit. I know that if $(r_1, s_1)$ ~ $(r_2, s_2)$, then there exists a t in S so that $t(r_1s_2 – r_2s_1) = 0$, and it follows that $-t(r_2s_1 – r_1s_2) = 0$, but how can I be sure that -t is in S? AI: $-t(r_2s_1 – r_1s_2) = 0$ implies $t(r_2s_1 – r_1s_2) = 0$.
H: How to solve $\lim_{x\to\infty}\frac{\cos x}{x-1}$? How to solve the following question? \begin{eqnarray} \\\lim_{x\to\infty}f(x)&=&\lim_{x\to\infty}\frac{\cos x}{x-1}\\ \\&=&\lim_{x\to\infty} \frac{\sqrt{1-\sin^2x}}{x-1}\\ \\&=&\lim_{x\to\infty} \frac{ \frac{\sqrt{1-\sin^2x}}{x}}{\frac{x-1}{x}}\\ \\&=&\lim_{x\to\infty} \frac{\sqrt{\frac{1}{x^2}-1}}{1-\frac{1}{x}}\\ \\&=&\frac {\sqrt{-1}}{1}\\ \\&=&\text{Error Math}\\ \end{eqnarray} The ans is 0. AI: As $x\rightarrow \infty$ the fraction $\frac{1}{x-1}\rightarrow 0$ Since $\|cos(x)\|\leq 1$ the limit $\frac{cos(x)}{x-1}\rightarrow 0$ as $x\rightarrow \infty$.
H: Subspaces and span? Let $S$ be the subspace spanned by $(\text{u}_1, \text{u}_2, ... , \text{u}_m)$. Then, $S$ is the smallest subspace containing $(\text{u}_1, \text{u}_2, ... , \text{u}_m)$ in the sense that if $S_2$ is any other subspace containing $(\text{u}_1, \text{u}_2, ... , \text{u}_m)$, then $S \subseteq S_2$. Could someone please explain why the above statement is true? I don't have any work to show, because I have no clue how to even approach the problem. Thanks. Definition of "Subspace Spanned By": Let $(\text{u}_1, \text{u}_2, ... , \text{u}_m)$ be a set of vectors. The subspace spanned by $(\text{u}_1, \text{u}_2, ... , \text{u}_m)$ is defined to be the set of vectors $A$ such that if $v \in A$, $$v = \sum_{i = 1}^{m} a_i\text{u}_i$$ for some scalars $(a_1, a_2,..., a_i)$ AI: Suppose $S_2$ is another subspace containing $(u_1, \dotsc, u_m)$. Choose some element $v \in \text{span}(u_1, \dotsc, u_m)$. Then $$ v = a_1 u_1 + \dotsb + a_m u_m $$ for appropriate scalars $a_i$. Since $u_i \in S_2$ for each $i$, we must have $a_i u_i \in S_2$ for each $i$, and hence $v \in S_2$. This implies $\text{span}(u_1, \dotsc, u_m) \subseteq S_2$.
H: Finding a polynomial $g(x)$ such that $ g(x)g(x-1)=g(x^2)$ Find all polynomials $g(x)$ with real coefficients with the property $$g(x)g(x-1)=g(x^2).$$ My try: I found $$g(x)=(x^2+x+1)^n$$ satisfies the condition; maybe there are other solution? If so, how to prove it (and/or find them)? Thank you. AI: Hint: What can you say about the roots of the polynomial? (The following are steps. Fill in the rest of the details yourself. It should be clear why this works. I removed as much text as possible due to OP's request in comments.) If $g(x) = g$ is constant, then $g^2 = g$. Otherwise, $\deg g \geq 1$. Let $g(\alpha) = 0 $. Step 1: $g( \alpha) = 0 \Rightarrow \| \alpha \| = 0 $ or 1. Step 2: If $\alpha = 0 \Rightarrow g(n^2 ) = 0\, \forall n \in \mathbb{N}$. $\Rightarrow \Leftarrow$ Step 3: If $\|\alpha \|= 1 \Rightarrow \exists n \in \mathbb{N}, \alpha^n = 1$. Step 4: If $\alpha^n = 1$, then for $\beta = (\alpha + 1)^2$, we have $g(\beta) = 0 $. Step 5: $\alpha = ??$ Step 6: If $(x-\alpha)^m \mid g(x) \Rightarrow (x-\beta)^m \mid g(x).$ Step 7: Hence $g(x) = (x^2 + x+1)^n$.
H: Condition for subgroup lattice of $\mathbb{Z}_{n}$ to be a straight line I was asked what was the condition(s) on n for the lattice subgroup of $\mathbb{Z}_{n}$ to be a straight line (ie each subgroup is a subgroup of another). Then the order of each subgroup divides the order of the one it is included in. So I first answered the condition was that n is prime($\mathbb{Z}_{n}$ is then a simple group). It works but there are some n in $\mathbb{N}$ having this property anyway : $\mathbb{Z}_{4}$ for instance, ( $<2>$ and the identity) , and $4$ is not prime. Then I tried to draw the subgroups lattice of $\mathbb{Z}_{n}$ for several n and it seems like the condition on n is that is has to be divisible by only one prime ($\mathbb{Z}_{9}$ and 3, $\mathbb{Z}_{16}$ and 4, $\mathbb{Z}_{25}$ and 5) Any hints ? Am I on the right track ? AI: You are right: $n=p^m$. Indeed, if $p,q\|n, \ p\ne q$ then $\mathbb{Z}_p$ and $\mathbb{Z}_q$ cannot be included in a common chain.
H: Why is $\tan(x)$ a function? A function $f:X\rightarrow Y$ maps each $x\in X$ to some $y \in Y$. So consider $\tan{\frac{\pi}{2}}$ for which $\tan(x)$ is undefined, so in this case, $\tan(x)$ does not map to an element of its range. This conflicts with my understanding of what a function is. Why do we still consider $\tan(x)$ a function? AI: The set $X$ in your definition is the domain of the function. The domain of $\tan(x)$ is typically taken to be $$ X=\bigcup_{k\in\Bbb{Z}} \left(-\frac{\pi}{2}+k\pi,\frac{\pi}{2}+k\pi\right) $$ Thus $\pi/2\notin X$, and so don't need to assign a value to $\tan(\pi/2)$ (or for any $\pi/2+k\pi,k\in\Bbb{Z}$ for that matter).
H: Invertible functions and their properties If an n × n matrix A is singular, then the columns of A must be linearly independent. Is this true? Invertible functions must be bijective Invertible functions must have square matrices Invertible functions must span R^n Also, am I missing some other must conditions of invertible or singular functions? What's special about invertible functions anyway? AI: If an $n \times n$ matrix $A$ is singular, then the columns of $A$ must be linearly independent. Is this true? No, it's exactly the opposite. Invertible functions must be bijective Yes. Invertible functions must have square matrices Invertible linear functions can be represented by square (nonsingular) matrices. Invertible functions must span $\mathbb{R}^n$. Assuming that they are linear, and their codomain is $\mathbb{R}^n$, then yes. Also, am I missing some other must conditions of invertible or singular functions? Determinant, eigenvalues,... What's special about invertible functions anyway? They have inverse.
H: Volume of a Special Pyramid Let $P$ be a plane in $\mathbb{R}^3$ parallel to the $xy$-plane. Let $\Omega$ be a closed, bounded set in the $xy$-plane with $2$-volume $B$. Pick a point $Q$ in $P$ and make a pyramid by joining each point in $\Omega$ to $Q$ with a straight line segment. Find the $3$-volume of this pyramid. I know that the volume with be dependent on $B$ and the distance from $\Omega$ to $P$, and the solution probably involves multiple integration, but beyond that I don't know where to start. Any help would be appreciated. N.b.: Although tagged as "homework," this problem is no longer "live," i.e. the homework has already been turned in. AI: Let's answer to a bit more ambitious question: Let $\Omega$ be an open, bounded and connected subset of an $(n-1)$-dimensional hyperplane $P$ of $\mathbb{R}^n$, and let $x_0\in\mathbb{R}^n$ be a point not contained in $P$. What is the volume of the pyramid with basis $\Omega$ and vertex $x_0$? By a rotation, we may assume $P=\{x\in\mathbb{R}^n\|x_n=0\}$. By a reflection, we may also assume $(x_0)_n>0$. Our pyramid is given by $$\Pi = \{tx_0+(1-t)x\|t\in[0,1],\ x\in\Omega\}$$ and its volume is $$V_\Pi=\int_\Pi dV$$ where $dV=dx_1\ldots dx_n$. We make the following change of variables: $$\begin{align} x_n = & t(x_0)_n\\ x_i = & (1-t)y_i + t(x_0)_i \end{align}$$ where $i=1,\ldots,n-1$. With this we have $dV = (x_0)_n(1-t)^{n-1}dtdy_1\ldots dy_{n-1}$ and after the change of variables our integral becomes $$\begin{align} V_\Pi = & \int_\Omega dy_1\ldots dy_{n-1}\int_0^1 dt\ \bigg((x_0)_n(1-t)^{n-1}\bigg)\\ = & \frac{1}{n}(x_0)_nV_\Omega \end{align}$$ where $V_\Omega$ is the $(n-1)$-dimensional volume of $\Omega$. Notice how this reduces to the following known formulas $n=2$: $\mathrm{area\ triangle} = \frac{1}{2}\mathrm{basis}\times\mathrm{height}$ $n=3$: $\mathrm{volume\ pyramid} = \frac{1}{3}\mathrm{basis\ area}\times\mathrm{height}$
H: Dimension of a subspace of finite-dimensional product space $V$ equals $\dim V - 1$ Suppose $w$ is a nonzero vector in a finite dimensional inner product space $V$. Let $P = \{ v \in V \| \langle v,w\rangle = 0\}$. Show that $\dim P = \dim V - 1$ where $P$ is a subspace of $V$. AI: Let $v=(x_1,x_2,\ldots,x_n), w=(a_1,a_2, \ldots, a_n)$ Then the condition $<v,w>=0$ is equivalent to the equation $a_1 x_1+a_2 x_2+\cdots a_n x_n =0.$ The solutions of the equation forms a vector space. It is easy to see that $n-1$ vectors $(a_1,0,0,-\frac{a_i}{a_1},0,\ldots)$ are linearly independed. Thus $\dim P=n-1.$
H: How many surjective function are there in infinite sets? Say we have two sets $\|A\| = a$ and $\|B\| = b$, where $a \geq b$ and $a$ is infinite. How would you go about proving that the number of surjective functions from A onto B is $b^a$? AI: HINT: Partition $A$ into sets $A_0$ and $A_1$, where $\|A_0\|=b$ and $\|A_1\|=a$. Let $f:A_0\to B$ be any bijection. For each function $g:A_1\to B$, the function $f\cup g$ is a surjection from $A$ to $B$. Use this to show that there are at least $b^a$ surjections from $A$ to $B$. Showing that there are no more than $b^a$ surjections from $A$ to $B$ is straightforward.
H: Question about the roots of an irreducible polynomial How can we deduce that each field is isomorphic? $g(x) := x^4 - 10x^2 + 1$ is an irreducible polynomial with $\alpha_{1} := \sqrt{2}+ \sqrt{3}$ as a zero. Show that $\alpha_{2} := \sqrt{2}- \sqrt{3},\alpha_{3} :=- \sqrt{2}+ \sqrt{3}, \alpha_{4} := - \sqrt{2}- \sqrt{3}$ are also zeros of $g(x)$ in $\mathbb{C}$ and deduce that the four fields $\mathbb{Q}(\alpha _i), \quad i = 1,...,4$ are isomorphic to each other.$ Thanks AI: You have $$ x^4 - 10x^2 + 1 = (x^2 - 5)^2 - 24. $$ Note that $\pm \sqrt 2 \pm \sqrt 3 = \pm 5 \pm 2 \sqrt 6$, so that $$ (\pm \sqrt 2 \pm \sqrt 3)^2 = (\sqrt 2 \pm \sqrt 3)^2 = 2 \pm 2 \sqrt 6 + 3 = 5 \pm 2 \sqrt 6. $$ Now if you remove $5$ and square again you get $24$. So essentially the other roots are the same as the first root up to two signs because you square twice in the process and these "up to a sign" things are destroyed when squaring. Now the automorphisms of both $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ lift to automorphisms of $\mathbb Q(\sqrt 2, \sqrt 3)$, and since each of the automorphisms of $\mathbb Q(\sqrt 2,\sqrt 3)$ must restrict to a pair of automorphisms for both $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$, you know that the four automorphisms are just sign changes, hence giving you these four roots. Hope that helps,
H: Discrete mathematics Relations Question if r2 is in the set of N*N ( natural numbers) with (X,y) in the subset of r2, if and only if x+y=0 is it reflexive? is it symmetric? is it anti symmetric? is it Transitive? i said it is reflective because 0+0 =0 but i'm a little stuck beyond that. the only relation in the set would be (0,0) correct? AI: If I understand this correctly, you have a relation $R_2\subseteq\Bbb N\times\Bbb N$ defined by $\langle x,y\rangle\in R_2$ if and only if $x+y=0$. If this is correct, then $\langle 0,0\rangle$ is the only element of $R_2$, but that does not make $R_2$ reflexive: in order for $R_2$ to be reflexive, $\langle n,n\rangle$ would have to be in $R_2$ for every $n\in\Bbb N$, which is clearly not the case. For instance, $\langle 1,1\rangle\notin R_2$, even though $1\in\Bbb N$. Thus, $R_2$ is not reflexive. $R_2$ is both symmetric and transitive, however, Symmetry: Is it true that if $\langle x,y\rangle\in R_2$, then $\langle y,x\rangle\in R_2$ as well? Yes: if $\langle x,y\rangle\in R_2$, then $x=y=0$, so $\langle y,x\rangle=\langle x,y\rangle=\langle 0,0\rangle\in R_2$. Transitivity: Is it true that if $\langle x,y\rangle\in R_2$ and $\langle y,z\rangle\in R_2$, then $\langle x,z\rangle\in R_2$? The answer is again yes, and the proof is similar; can you complete it?
H: Proof with quantifiers $(\forall x)(\exists y)(x+y=0)$ $x$ and $y$ are real numbers The statement reads: for all $x$ there exists some $y$ such that $x+y=0$ is true. My proof is: take $y=-x$ Is this valid? I'm just paranoid that since the proof is so simple that something is wrong. AI: Yes, you've got the right idea: Simply expand a tad to say: Let $x$ be any arbitrary real number. Then put $y = -x,$ and so $\;x + y = x + - x = 0$. Since $x$ was chosen arbitrarily, it holds for all $x\in \mathbb R$.
H: Finite lebesgue Integral Hi guys I've been trying to prove this for a very long time, if someone could help me i would appreciated very much! let $(X,S,\mu)$ be a mesurable space, if $\mu(X)$ is finite and $f$ is a mesureble non negative function then: if $\int f d\mu$ is finite then $\sum_{n=0}^{\infty} 2^n \mu(\{ x \in X \vert f(x) \ge 2^n \})$ is finite AI: Define the family $(a_{n,k},n,k\geqslant 0)$ of non-negative real numbers by $$a_{n,k}:=2^n\mu\{x\in X,2^k\leqslant f(x)\lt 2^{k+1}\}[k\geqslant n].$$ Then $$\sum_{n\geqslant 0} 2^n\mu\{x\in X,\|f(x)\|\geqslant 2^n\}=\sum_{n,k\geqslant 0}a_{n,k}=\sum_{k=0}^\infty\sum_{n=0}^\infty a_{n,k}=\sum_{k=0}^\infty2^{k+1}\mu\{x\in X,2^k\leqslant f(x)\lt 2^{k+1}\}\leqslant 2\int_X\|f\|\mathrm d\mu.$$
H: discrete mathematics relations question 2 I am a little confused by this relation R3 is a subset of Z×Z defined by (x,y) in the set R3 if and only if x>2y is it reflexive? Symmetric? antisymmetric? or transitive? i say its NOT reflexive because (1,2) is not in the set R3 i say it is NOT Symmetric because (1,2) is not in the set(2,1) i say it is NOT Transitive either because if (4,1) in the set R3, (1,4) is not in the set R3 AI: $R$ is not reflexive, but your reasoning is off. For any non-negative integer, we have that $x \not\gt 2x$. Hence, in those cases, $(x, x) \notin R$. $R$ is not symmetric, since because $3 \gt 2\cdot 1 = 2$, $(3, 1) \in R$, but since $1 \not\gt 2\cdot 3 = 6, \;(1, 3) \notin R$. However, $R$ is transitive. If $x \gt 2y$ and $y \gt 2z$, then $x \gt 2y \iff x \gt 2(2z)$, so certainly, $\frac 12 x \gt 2z \implies x \gt 2z$.
H: Convergence of $ x_n = \left( \frac{1}{2} + \frac{3}{n} \right)^n$ I need to show that the sequence $ x_n = \left( \frac{1}{2} + \frac{3}{n} \right)^n$ is convergent. Using calculus in $ \mathbb{R}$, we could see that $ \lim _{n \to \infty} \left( \frac{1}{2} + \frac{3}{n} \right)^n = \lim_{n\to \infty} e^{n \ln \left( \frac{1}{2} + \frac{3}{n} \right)} = 0 $. I do not need to calculate the limit of the sequence, just to check it convergence. The definition of convergence doesn't look like a good idea, so I've been trying to use a trick like we do in calculus or to show that the sequence is bounded and monotonic. My guess is the last idea is better than the first one. If I show that $ x_{n+1} < x_n $, since $x_n $ is clearly bigger than zero we are done. But how to show it? i've tried induction withou success. Is there a better way to show it? Thanks in advance AI: Hint: $\dfrac12+\dfrac3n \le \dfrac78$, when $n\ge8$.
H: Find all permutations in increasing order Given a set of distinct numbers, say, {1, 2, 3, 4, 5, 6}, find all permutations containing 3 numbers. All the permutations have to be in ascending order. For e.g., some correct permutations would be {1, 2, 3}, {2, 4, 6}, etc. {2, 3, 1} would be incorrect because it is not in ascending order. How does one go about solving these kinds of questions? Say, instead of choosing 3 numbers, we had to choose 4 or 5, or maybe the given set would be different, what would be the general appraach? Thanks. P.S.: You don't just have to tell the number of possible permutations, but also list them. AI: Since all the numbers in the set are distinct hence the permutations containing 3 numbers in ascending order are: $$^6C_3$$ The reason is that any selection of 3 numbers from the set can be arranged in ascending order in only 1 way. Hence the permutations containing 4 numbers in ascending order are: $$^6C_4$$ Also the permutations containing 5 numbers in ascending order are: $$^6C_5$$
H: Is $\mathcal O_L$ an $\mathcal O_K$-lattice in $L$? This is a basic question. Let $L/K$ be a finite extension of algebraic number fields and let $\mathcal O_L$ and $\mathcal O_K$ be their respective rings of integers. Is it true that $$K\otimes_{\mathcal O_K}\mathcal O_L \cong L$$ If so how can we prove it? Thanks in advance. AI: Yes, this is true. First of all the map is injective. The key to proving this is the observation that every element of $K\otimes_{\mathscr{O}_K}\mathscr{O}_L$ is a pure tensor. This follows from a common denominator argument, and ultimately holds because $K\otimes_{\mathscr{O}_K}\mathscr{O}_L$ is a localization of $\mathscr{O}_L$. In fact, an arbitrary element of the tensor product can be written as $1/a\otimes b$ where $a\in\mathscr{O}_K\setminus\{0\}$ and $b\in\mathscr{O}_L$. The image of this element under the map in question is $a^{-1}b\in L$, so if this is zero, $b=0$, and $1/a\otimes b=1/a\otimes 0=0$. For surjectivity, note that if $x\in L$, then because $x$ is algebraic over $K$, for some non-zero $a\in \mathscr{O}_K$, $ax$ is integral over $\mathscr{O}_K$, i.e., $ax=b\in\mathscr{O}_L$. Then the element $1/a\otimes b\in K\otimes_{\mathscr{O}_K}\mathscr{O}_L$ maps to $a^{-1}b=a^{-1}(ax)=x$. All we actually used was that $\mathscr{O}_K$ is a domain with field of fractions $K$ and that $L$ is an algebraic extension of $K$ with $\mathscr{O}_L$ the integral closure of $\mathscr{O}_K$ in $L$.
H: Line bundles of the circle Up to isomorphism, I think there exist only two line bundles of the circle: the trivial bundle (diffeomorphic to a cylinder) and a bundle that looks like to a Möbius band. Although it seems obvious geometrically I did not find a good argument to justify it. Do you have an idea? AI: Line bundles are classified by the first Čech cohomology with coefficients in $\text{GL}^1(\mathbf R)$. By normalizing we can actually use coefficients in $\text{O}^1(\mathbf R)=\pm 1$. By using the usual covering of the circle, one sees immediately that this cohomology group is cyclic of order $2$, generated by the class of the Möbius strip.
H: Limit problems.. http://s23.postimg.org/xedyol4kr/limit.jpg I got stuck on Q2 , could someone shed light on how to do it? The proof bit. Finding limits in terms of α and β is easy. As for Q3, I know it is probably not the way I was supposed to do but is the following method not correct? Lim (A+B)= lim(a)+lim(b) and we are given that lim(a)=∞ and lim(b) is bounded below hence it is converging to some numerical value L. So lim(a)+lim(b)= ∞+L=∞ ? Many thanks AI: Take $a,b\in \Bbb R$ and let $x=a-b$ and $y=a+3b$. Can you express $a$ and $b$ in terms of $x$ and $y$? If you can, then you can use the exact same thing for the sequences and then use the fact that a linear combination of sequence converges to the linear combination of the limits. For $3$, no, your method doesn't work because it supposes the convergence of $b_n$ while taking $b_n=\left(-1\right)^n$ is bounded below and doesn't converge. For this question, you should just use the definition of convergence to $+\infty$. Intuitively, since you can get as big as you want, if you get pulled back by something at most finite, then you can still get as big as you want. Now all you have to do it write that with inequalities and quantifiers.
H: Combinations of two letters from a $124$-element alphabet The telephone company wants to add an area code composed of $2$ letters to every phone number. In order to do so, the company chose a special sign language containing $124$ different signs. If the company used $122$ of the signs fully and two remained unused, how many additional area codes can be created if the company uses all $124$ signs? can this problem be solved using permutation or combination OR only application of number theory is needed ???? AI: Hint: presumably order matters here as AB is different from BA. How many two letter combinations can be formed from $124$ how many from $122?$ Now subtract
H: Choosing a Set of r elements from a set having n elements. Define a set $X$={$1$,$2$,$...$,$n$} . Determine the number of ways of selecting a subset of $X$ such that it contains no consecutive integers . AI: Let $a_n$, be the number of subsets satisfying the condition, including the empty set. By adding one more number to the set, the subsets that can be obtained in this new set are: all that could be obtained in the previous one, that is $a_n$; all that could be obtained with the first $n-1$ elements with the number $n+1$ included, that is $a_{n-1}$; With this we have the recursive relation $a_{n+1}=a_n+a_{n-1}$. As $a_1=2$ ($\emptyset$ and $\{1\}$), $a_2=3$ ($\emptyset$, $\{1\}$ and $\{2\}$) we obtain the recursive relation defining the Fibonacci sequence $F_n$ (http://en.wikipedia.org/wiki/Fibonacci_number) with just a displacement of indices. So the desired number of subsets is $a_n=F_{n+1}$. The explicit Binet's formula for $F_n$ can be seen in the link. PS: if it the recurrence realtion was not the Fibonacci sequence, there would be no problem. See this link (http://en.wikipedia.org/wiki/Recurrence_relation#Solving) for solving linear recurrence relations.
H: Can we find an example of non-mesuarable set which their outer measure could be computed? We know there is non-measuarable set and we know every set has outer measure, so can anyone give me an example of a non-measuarable and there outer measure could be computed ? AI: Let us consider the Lebesgue (outer) measure on $[0,1]$ and let $V\subset [0,1]$ be a Vitali set. Then $V$ is not measurable and for any measurable set $E\subset V$, $E$ is a measure-zero set. Note that $[0,1]\setminus V$ is non-measurable and for every $a\in [0,1]$, $[0,a]\setminus V$ has outer-measure $a$. As a result, there exists $a_0\in[0,1)$, such that $$\{a\in [0,1] : [0,a]\setminus V \text{ is measurable }\}=[0,a_0].$$ Consequently, for every $a\in(a_0,1]$, $[0,a]\setminus V$ is non-measurable with outer-measure $a$. Remark: I guess $a_0=0$, but I have no idea how to prove or disprove it.

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