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Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
I read in Stewart "single variable calculus" page 83 that the limit $$\lim_{x\to 0}{1/x^2}$$ does not exist. How precise is this statement knowing that this limit is $\infty$?. I thought saying the limit does not exist is not true where limits are $\infty$. But it is said when a function does not have a limit at all like $$\lim_{x\to \infty}{\cos x}$$. I read in Stewart "single variable calculus" page 83 that the limit $$\lim_{x\to 0}{1/x^2}$$ According to some presentations of limits, it is proper to write "$\lim_{x\to 0}\frac{1}{x^2}=\infty$." This does not commit one to the existence of an object called $\infty$. The sentence is just an abbreviation for "given any real number $M$, there is a real number $\delta$ (which will depend on $M$) such that $\frac{1}{x^2}\gt M$ for all $x$ such that $0\lt \|x\| \lt \delta$." It turns out that we often wish to write sentences of this type, because they have important geometric content. So having an abbreviation is undeniably useful. On the other hand, some presentations of limits forbid writing "$\lim_{x\to 0}\frac{1}{x^2}=\infty$." Matter of taste, pedagogical choice. The main reason for choosing to forbid is that careless manipulation of the symbol $\infty$ all too often leads to wrong answers. A limit $$\lim_{x\to a} f(x)$$ exists if and only if it is equal to a number. Note that $\infty$ is not a number. For example $\lim_{x\to 0} \frac{1}{x^{2}} = \infty$ so it doesn't exist. When a function approaches infinity, the limit technically doesn't exist by the proper definition, that demands it work out to be a number. We merely extend our notation in this particular instance. The point is that the limit may not be a number, but it is somewhat well behaved and asymptotes are usually worth note. The term "infinite limit" is actually an oxymoron, like "jumbo shrimp" or "unbiased opinion". True limits are finite. However, it is okay to write down "lim f(x) = infinity" or "lim g(x) = -infinity", if the given function approaches either plus infinity or minus infinity from BOTH sides of whatever x is approaching, especially to distinguish this from the situation in which it approaches plus infinity on ONE side and minus infinity on the OTHER side, in which case the ONLY correct answer would be "the limit does not exist". Note that working in the affinely extended real numbers with the induced order topology this limit exists and equals infinity, unambiguously. We also don't need a "special" definition for infinite limits with this method which is convenient.
Search Now showing items 11-20 of 27 Pseudorapidity dependence of the anisotropic flow of charged particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (Elsevier, 2016-11) We present measurements of the elliptic ($\mathrm{v}_2$), triangular ($\mathrm{v}_3$) and quadrangular ($\mathrm{v}_4$) anisotropic azimuthal flow over a wide range of pseudorapidities ($-3.5< \eta < 5$). The measurements ... Correlated event-by-event fluctuations of flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (American Physical Society, 2016-10) We report the measurements of correlations between event-by-event fluctuations of amplitudes of anisotropic flow harmonics in nucleus–nucleus collisions, obtained for the first time using a new analysis method based on ... Centrality dependence of $\mathbf{\psi}$(2S) suppression in p-Pb collisions at $\mathbf{\sqrt{{\textit s}_{\rm NN}}}$ = 5.02 TeV (Springer, 2016-06) The inclusive production of the $\psi$(2S) charmonium state was studied as a function of centrality in p-Pb collisions at the nucleon-nucleon center of mass energy $\sqrt{s_{\rm NN}}$ = 5.02 TeV at the CERN LHC. The ... Transverse momentum dependence of D-meson production in Pb–Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Springer, 2016-03) The production of prompt charmed mesons D$^0$, D$^+$ and D$^{*+}$, and their antiparticles, was measured with the ALICE detector in Pb–Pb collisions at the centre-of-mass energy per nucleon pair, $\sqrt{s_{\rm NN}}$ of ... Multiplicity and transverse momentum evolution of charge-dependent correlations in pp, p-Pb, and Pb-Pb collisions at the LHC (Springer, 2016) We report on two-particle charge-dependent correlations in pp, p-Pb, and Pb-Pb collisions as a function of the pseudorapidity and azimuthal angle difference, $\mathrm{\Delta}\eta$ and $\mathrm{\Delta}\varphi$ respectively. ... Charge-dependent flow and the search for the chiral magnetic wave in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (American Physical Society, 2016-04) We report on measurements of a charge-dependent flow using a novel three-particle correlator with ALICE in Pb–Pb collisions at the LHC, and discuss the implications for observation of local parity violation and the Chiral ... Pseudorapidity and transverse-momentum distributions of charged particles in proton-proton collisions at $\mathbf{\sqrt{\textit s}}$ = 13 TeV (Elsevier, 2016-02) The pseudorapidity ($\eta$) and transverse-momentum ($p_{\rm T}$) distributions of charged particles produced in proton-proton collisions are measured at the centre-of-mass energy $\sqrt{s}$ = 13 TeV. The pseudorapidity ... Differential studies of inclusive J/$\psi$ and $\psi$(2S) production at forward rapidity in Pb-Pb collisions at $\mathbf{\sqrt{{\textit s}_{_{NN}}}}$ = 2.76 TeV (Springer, 2016-05) The production of J/$\psi$ and $\psi(2S)$ was measured with the ALICE detector in Pb-Pb collisions at the LHC. The measurement was performed at forward rapidity ($2.5 < y < 4 $) down to zero transverse momentum ($p_{\rm ... Elliptic flow of muons from heavy-flavour hadron decays at forward rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (Elsevier, 2016-02) The elliptic flow, $v_{2}$, of muons from heavy-flavour hadron decays at forward rapidity ($2.5 < y < 4$) is measured in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The scalar ... Anisotropic flow of charged particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (American Physical Society, 2016-04) We report the first results of elliptic ($v_2$), triangular ($v_3$) and quadrangular flow ($v_4$) of charged particles in Pb--Pb collisions at $\sqrt{s_{_{\rm NN}}}=$ 5.02 TeV with the ALICE detector at the CERN Large ...
Problem 8 Let $G$ be an abelian group and let $H=\{ g \in G \mid \ \mid g \mid$ divides $12 \}$. Prove that $H$ is a subgroup of $G$. Would your proof be valid if the number $12$ were replaced by another number? State the general result. Solution Let $G$ be an abelian group and let $H=\{ g \in G \mid \ \mid g \mid$ divides $12 \}$. (i) Clearly, $H \subseteq G$, since $H$ is defined to contain only elements that are in G. (ii) Let $a,b \in H$. then $\mid a \mid = m, \mid b \mid = n$. hence, there exists $k, l \in \mathbb{Z} ^+$ such that $km=12$ and $ln=12$, by definition of "divides." Now, consider $(ab)^{12}.$ Observe that, since $\mid a \mid$ divides $12$ and $\mid b \mid$ divides $12$, $a^{12} = e$ and $b^{12} =e$. Now, consider $a^{12}b^{12} = ee = e$. Thus, $(ab)^{12} = a^{12} * b^{12} =e$ since $G$ is abelian. Since $(ab)^{12} = e$, the order of $ab$ is divisible by $12$. Therefore, $ab \in H$, so $H$ is closed under $$. (iii) Observe that $e \in G$, since $G$ is a group. Note, $\langle e \rangle = \{ e \}$, so $\mid e \mid =1$. Since $1$ divides $12$, $e \in H$. Now, since $e$ is the identity for all $a \in G$ and $H \subseteq G, e$ is the identity for $H$. (iv) Let $a \in H$. Then, by definition of $H, \exists a^{-1} \in H$ such that $aa^{-1}=a^{-1}a=e$. Since $a\in H$, $\|a\|$ divides 12, so $a^{12}=e$. Observe that $(a^{-1})^n=a^{-n}=(a^n)^{-1}=e$, so $\|a^{-1}\|$ must divide 12. Therefore, the inverse of $a$ is in $H$ for all $a \in H$. All conditions hold, so $H$ is a subgroup of $G$. $\hspace{250pt} \clubsuit$ This would hold even if $12$ were replaced by another element of $\mathbb{Z}^+$.
Pasting Lemma Theorem Let $X$ and $Y$ be topological spaces. Let $A$ and $B$ be open in $X$. $\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$ Then the mapping $f \cup g : A \cup B \to Y$ is continuous. Let $A$ and $B$ be closed in $X$. Then the mapping $f \cup g : A \cup B \to Y$ is continuous. Let $f: \bigcup \mathcal A \to Y$ be a mapping such that: $\forall i \in I : f \restriction A_i$ is continuous Then $f$ is continuous on $\bigcup \mathcal A$. Proof First it is shown that $f \cup g$ is well-defined. It is sufficient to check that $f \cup g$ maps any $x \in A \cap B$ to a single value in $Y$. Let $x \in A \cap B$. Because $x \in A$: $\map {f \cup g} x = \map f x \in Y$ Because $x \in B$: $\map {f \cup g} x = \map g x \in Y$ But by definition of agreement of mappings: $\map f x = \map g x$ so $f \cup g$ maps $x$ to a unique value in $Y$, as was to be shown. $\Box$ Next it is shown that $f \cup g$ is continuous: Observe first that $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U)$. For if $x \in (f \cup g)^{-1}(U) \subseteq A \cup B$, then $(f \cup g)(x) \in U$; but if $x \in A$, then $f(x) = (f \cup g)(x) \in U$ i.e. $x \in f^{-1}(U)$ or if $x \in B$, then $g(x) = (f \cup g)(x) \in U$ i.e. $x \in g^{-1}(U)$. In both cases, $x \in f^{-1}(U) \cup g^{-1}(U)$. Conversely if $x \in f^{-1}(U) \cup g^{-1}(U)$, then without loss of generality consider the case of $x \in f^{-1}(U) \subseteq A$; in this case $(f \cup g)(x) = f(x) \in U$ so that $x \in (f \cup g)^{-1}(U)$. Having established that $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U)$, note that by continuity of $f$ and $g$, we have that $f^{-1}(U)$ is open in $A$ and $g^{-1}(U)$ is open in $B$. So there are an open sets $P$ and $Q$ of $X$ such that $f^{-1}(U) = P \cap A$ and $g^{-1}(U) = Q \cap B$. Since $A$ and $B$ are open in $X$ and intersections of open sets are open, this implies that $f^{-1}(U)$ and $g^{-1}(U)$ are open in $X$. Furthermore, note that $P \cap A$ and $Q \cap B$ are both subsets of $A \cup B$. Hence, $f^{-1}(U) = f^{-1}(U) \cap (A \cup B)$ and $g^{-1}(U) = g^{-1}(U) \cap (A \cup B)$. Thus: $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U) = (f^{-1}(U) \cup g^{-1}(U)) \cap (A \cup B)$ demonstrating that $(f \cup g)^{-1}(U)$ is open in the subspace topology of $A \cup B$. $\blacksquare$
Let $a \in \Bbb{R}$. Let $\Bbb{Z}$ act on $S^1$ via $(n,z) \mapsto ze^{2 \pi i \cdot an}$. Claim: The is action is not free if and only if $a \Bbb{Q}$. Here's an attempt at the forward direction: If the action is not free, there is some nonzero $n$ and $z \in S^1$ such that $ze^{2 \pi i \cdot an} = 1$. Note $z = e^{2 \pi i \theta}$ for some $\theta \in [0,1)$. Then the equation becomes $e^{2 \pi i(\theta + an)} = 1$, which holds if and only if $2\pi (\theta + an) = 2 \pi k$ for some $k \in \Bbb{Z}$. Solving for $a$ gives $a = \frac{k-\theta}{n}$... What if $\theta$ is irrational...what did I do wrong? 'cause I understand that second one but I'm having a hard time explaining it in words (Re: the first one: a matrix transpose "looks" like the equation $Ax\cdot y=x\cdot A^\top y$. Which implies several things, like how $A^\top x$ is perpendicular to $A^{-1}x^\top$ where $x^\top$ is the vector space perpendicular to $x$.) DogAteMy: I looked at the link. You're writing garbage with regard to the transpose stuff. Why should a linear map from $\Bbb R^n$ to $\Bbb R^m$ have an inverse in the first place? And for goodness sake don't use $x^\top$ to mean the orthogonal complement when it already means something. he based much of his success on principles like this I cant believe ive forgotten it it's basically saying that it's a waste of time to throw a parade for a scholar or win he or she over with compliments and awards etc but this is the biggest source of sense of purpose in the non scholar yeah there is this thing called the internet and well yes there are better books than others you can study from provided they are not stolen from you by drug dealers you should buy a text book that they base university courses on if you can save for one I was working from "Problems in Analytic Number Theory" Second Edition, by M.Ram Murty prior to the idiots robbing me and taking that with them which was a fantastic book to self learn from one of the best ive had actually Yeah I wasn't happy about it either it was more than $200 usd actually well look if you want my honest opinion self study doesn't exist, you are still being taught something by Euclid if you read his works despite him having died a few thousand years ago but he is as much a teacher as you'll get, and if you don't plan on reading the works of others, to maintain some sort of purity in the word self study, well, no you have failed in life and should give up entirely. but that is a very good book regardless of you attending Princeton university or not yeah me neither you are the only one I remember talking to on it but I have been well and truly banned from this IP address for that forum now, which, which was as you might have guessed for being too polite and sensitive to delicate religious sensibilities but no it's not my forum I just remembered it was one of the first I started talking math on, and it was a long road for someone like me being receptive to constructive criticism, especially from a kid a third my age which according to your profile at the time you were i have a chronological disability that prevents me from accurately recalling exactly when this was, don't worry about it well yeah it said you were 10, so it was a troubling thought to be getting advice from a ten year old at the time i think i was still holding on to some sort of hopes of a career in non stupidity related fields which was at some point abandoned @TedShifrin thanks for that in bookmarking all of these under 3500, is there a 101 i should start with and find my way into four digits? what level of expertise is required for all of these is a more clear way of asking Well, there are various math sources all over the web, including Khan Academy, etc. My particular course was intended for people seriously interested in mathematics (i.e., proofs as well as computations and applications). The students in there were about half first-year students who had taken BC AP calculus in high school and gotten the top score, about half second-year students who'd taken various first-year calculus paths in college. long time ago tho even the credits have expired not the student debt though so i think they are trying to hint i should go back a start from first year and double said debt but im a terrible student it really wasn't worth while the first time round considering my rate of attendance then and how unlikely that would be different going back now @BalarkaSen yeah from the number theory i got into in my most recent years it's bizarre how i almost became allergic to calculus i loved it back then and for some reason not quite so when i began focusing on prime numbers What do you all think of this theorem: The number of ways to write $n$ as a sum of four squares is equal to $8$ times the sum of divisors of $n$ if $n$ is odd and $24$ times sum of odd divisors of $n$ if $n$ is even A proof of this uses (basically) Fourier analysis Even though it looks rather innocuous albeit surprising result in pure number theory @BalarkaSen well because it was what Wikipedia deemed my interests to be categorized as i have simply told myself that is what i am studying, it really starting with me horsing around not even knowing what category of math you call it. actually, ill show you the exact subject you and i discussed on mmf that reminds me you were actually right, i don't know if i would have taken it well at the time tho yeah looks like i deleted the stack exchange question on it anyway i had found a discrete Fourier transform for $\lfloor \frac{n}{m} \rfloor$ and you attempted to explain to me that is what it was that's all i remember lol @BalarkaSen oh and when it comes to transcripts involving me on the internet, don't worry, the younger version of you most definitely will be seen in a positive light, and just contemplating all the possibilities of things said by someone as insane as me, agree that pulling up said past conversations isn't productive absolutely me too but would we have it any other way? i mean i know im like a dog chasing a car as far as any real "purpose" in learning is concerned i think id be terrified if something didnt unfold into a myriad of new things I'm clueless about @Daminark They key thing if I remember correctly was that if you look at the subgroup $\Gamma$ of $\text{PSL}_2(\Bbb Z)$ generated by (1, 2\|0, 1) and (0, -1\|1, 0), then any holomorphic function $f : \Bbb H^2 \to \Bbb C$ invariant under $\Gamma$ (in the sense that $f(z + 2) = f(z)$ and $f(-1/z) = z^{2k} f(z)$, $2k$ is called the weight) such that the Fourier expansion of $f$ at infinity and $-1$ having no constant coefficients is called a cusp form (on $\Bbb H^2/\Gamma$). The $r_4(n)$ thing follows as an immediate corollary of the fact that the only weight $2$ cusp form is identically zero. I can try to recall more if you're interested. It's insightful to look at the picture of $\Bbb H^2/\Gamma$... it's like, take the line $\Re[z] = 1$, the semicircle $\|z\| = 1, z > 0$, and the line $\Re[z] = -1$. This gives a certain region in the upper half plane Paste those two lines, and paste half of the semicircle (from -1 to i, and then from i to 1) to the other half by folding along i Yup, that $E_4$ and $E_6$ generates the space of modular forms, that type of things I think in general if you start thinking about modular forms as eigenfunctions of a Laplacian, the space generated by the Eisenstein series are orthogonal to the space of cusp forms - there's a general story I don't quite know Cusp forms vanish at the cusp (those are the $-1$ and $\infty$ points in the quotient $\Bbb H^2/\Gamma$ picture I described above, where the hyperbolic metric gets coned off), whereas given any values on the cusps you can make a linear combination of Eisenstein series which takes those specific values on the cusps So it sort of makes sense Regarding that particular result, saying it's a weight 2 cusp form is like specifying a strong decay rate of the cusp form towards the cusp. Indeed, one basically argues like the maximum value theorem in complex analysis @BalarkaSen no you didn't come across as pretentious at all, i can only imagine being so young and having the mind you have would have resulted in many accusing you of such, but really, my experience in life is diverse to say the least, and I've met know it all types that are in everyway detestable, you shouldn't be so hard on your character you are very humble considering your calibre You probably don't realise how low the bar drops when it comes to integrity of character is concerned, trust me, you wouldn't have come as far as you clearly have if you were a know it all it was actually the best thing for me to have met a 10 year old at the age of 30 that was well beyond what ill ever realistically become as far as math is concerned someone like you is going to be accused of arrogance simply because you intimidate many ignore the good majority of that mate
The words “Approach” and “Tend” are often used in calculus and it’s a basic concept to start learning the limits. According to English language, the real meaning of “Approach” or “Tend” is, come near or nearer to something. For example, if a point comes nearly to another point, then it’s said that the point approaches to another point. In calculus, the word “Approach” is symbolically represented by right arrow ($\rightarrow$) symbol. Mathematical, it is written as $x \,\to\, a$ and it is read as two ways. $x$ is a variable and $a$ is a constant. The value of $x$ starts from $5$ but less than $6$ and the value of $a$ is $6$. Therefore, $x \in [5, 6)$ and $a = 6$. Take $x = 5$. The values of $x$ and $a$ are not equal because their quantities are not same ($5 \ne 6$). Therefore, $x \ne a$. Take $x = 5.9$. The value of $5.9$ is not equal to $6$ but its value is closely equal to $6$. In other words, $5.9 \approx 6$. It is expressed in two ways in calculus. Mathematically, it is written as $5.9 \,\to\, 6$. Therefore, $x \,\to\, a$. Remember, the right arrow represents that the value of $x$ closes the value of $a$ but they’re not equal. Now, take $x = 5.99$. In this case also, the value of $5.99$ does not equal to $6$ but approximately equals to $6$. Therefore, $5.99 \approx 6$. Therefore, $x \,\to \, a$ in this case. The meaning of $x \,\to\, a$ is, the value of $x$ can be anything, which is close to the value of $a$ but not equal to $a$. So, the value of $x$ can be any value closes to $6$, for example $5.9$, $5.95$, $5.991$, $5.999$ and so on. Let’s study the concept of approach or tend in calculus with some more examples. Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
Let $a \in \Bbb{R}$. Let $\Bbb{Z}$ act on $S^1$ via $(n,z) \mapsto ze^{2 \pi i \cdot an}$. Claim: The is action is not free if and only if $a \Bbb{Q}$. Here's an attempt at the forward direction: If the action is not free, there is some nonzero $n$ and $z \in S^1$ such that $ze^{2 \pi i \cdot an} = 1$. Note $z = e^{2 \pi i \theta}$ for some $\theta \in [0,1)$. Then the equation becomes $e^{2 \pi i(\theta + an)} = 1$, which holds if and only if $2\pi (\theta + an) = 2 \pi k$ for some $k \in \Bbb{Z}$. Solving for $a$ gives $a = \frac{k-\theta}{n}$... What if $\theta$ is irrational...what did I do wrong? 'cause I understand that second one but I'm having a hard time explaining it in words (Re: the first one: a matrix transpose "looks" like the equation $Ax\cdot y=x\cdot A^\top y$. Which implies several things, like how $A^\top x$ is perpendicular to $A^{-1}x^\top$ where $x^\top$ is the vector space perpendicular to $x$.) DogAteMy: I looked at the link. You're writing garbage with regard to the transpose stuff. Why should a linear map from $\Bbb R^n$ to $\Bbb R^m$ have an inverse in the first place? And for goodness sake don't use $x^\top$ to mean the orthogonal complement when it already means something. he based much of his success on principles like this I cant believe ive forgotten it it's basically saying that it's a waste of time to throw a parade for a scholar or win he or she over with compliments and awards etc but this is the biggest source of sense of purpose in the non scholar yeah there is this thing called the internet and well yes there are better books than others you can study from provided they are not stolen from you by drug dealers you should buy a text book that they base university courses on if you can save for one I was working from "Problems in Analytic Number Theory" Second Edition, by M.Ram Murty prior to the idiots robbing me and taking that with them which was a fantastic book to self learn from one of the best ive had actually Yeah I wasn't happy about it either it was more than $200 usd actually well look if you want my honest opinion self study doesn't exist, you are still being taught something by Euclid if you read his works despite him having died a few thousand years ago but he is as much a teacher as you'll get, and if you don't plan on reading the works of others, to maintain some sort of purity in the word self study, well, no you have failed in life and should give up entirely. but that is a very good book regardless of you attending Princeton university or not yeah me neither you are the only one I remember talking to on it but I have been well and truly banned from this IP address for that forum now, which, which was as you might have guessed for being too polite and sensitive to delicate religious sensibilities but no it's not my forum I just remembered it was one of the first I started talking math on, and it was a long road for someone like me being receptive to constructive criticism, especially from a kid a third my age which according to your profile at the time you were i have a chronological disability that prevents me from accurately recalling exactly when this was, don't worry about it well yeah it said you were 10, so it was a troubling thought to be getting advice from a ten year old at the time i think i was still holding on to some sort of hopes of a career in non stupidity related fields which was at some point abandoned @TedShifrin thanks for that in bookmarking all of these under 3500, is there a 101 i should start with and find my way into four digits? what level of expertise is required for all of these is a more clear way of asking Well, there are various math sources all over the web, including Khan Academy, etc. My particular course was intended for people seriously interested in mathematics (i.e., proofs as well as computations and applications). The students in there were about half first-year students who had taken BC AP calculus in high school and gotten the top score, about half second-year students who'd taken various first-year calculus paths in college. long time ago tho even the credits have expired not the student debt though so i think they are trying to hint i should go back a start from first year and double said debt but im a terrible student it really wasn't worth while the first time round considering my rate of attendance then and how unlikely that would be different going back now @BalarkaSen yeah from the number theory i got into in my most recent years it's bizarre how i almost became allergic to calculus i loved it back then and for some reason not quite so when i began focusing on prime numbers What do you all think of this theorem: The number of ways to write $n$ as a sum of four squares is equal to $8$ times the sum of divisors of $n$ if $n$ is odd and $24$ times sum of odd divisors of $n$ if $n$ is even A proof of this uses (basically) Fourier analysis Even though it looks rather innocuous albeit surprising result in pure number theory @BalarkaSen well because it was what Wikipedia deemed my interests to be categorized as i have simply told myself that is what i am studying, it really starting with me horsing around not even knowing what category of math you call it. actually, ill show you the exact subject you and i discussed on mmf that reminds me you were actually right, i don't know if i would have taken it well at the time tho yeah looks like i deleted the stack exchange question on it anyway i had found a discrete Fourier transform for $\lfloor \frac{n}{m} \rfloor$ and you attempted to explain to me that is what it was that's all i remember lol @BalarkaSen oh and when it comes to transcripts involving me on the internet, don't worry, the younger version of you most definitely will be seen in a positive light, and just contemplating all the possibilities of things said by someone as insane as me, agree that pulling up said past conversations isn't productive absolutely me too but would we have it any other way? i mean i know im like a dog chasing a car as far as any real "purpose" in learning is concerned i think id be terrified if something didnt unfold into a myriad of new things I'm clueless about @Daminark They key thing if I remember correctly was that if you look at the subgroup $\Gamma$ of $\text{PSL}_2(\Bbb Z)$ generated by (1, 2\|0, 1) and (0, -1\|1, 0), then any holomorphic function $f : \Bbb H^2 \to \Bbb C$ invariant under $\Gamma$ (in the sense that $f(z + 2) = f(z)$ and $f(-1/z) = z^{2k} f(z)$, $2k$ is called the weight) such that the Fourier expansion of $f$ at infinity and $-1$ having no constant coefficients is called a cusp form (on $\Bbb H^2/\Gamma$). The $r_4(n)$ thing follows as an immediate corollary of the fact that the only weight $2$ cusp form is identically zero. I can try to recall more if you're interested. It's insightful to look at the picture of $\Bbb H^2/\Gamma$... it's like, take the line $\Re[z] = 1$, the semicircle $\|z\| = 1, z > 0$, and the line $\Re[z] = -1$. This gives a certain region in the upper half plane Paste those two lines, and paste half of the semicircle (from -1 to i, and then from i to 1) to the other half by folding along i Yup, that $E_4$ and $E_6$ generates the space of modular forms, that type of things I think in general if you start thinking about modular forms as eigenfunctions of a Laplacian, the space generated by the Eisenstein series are orthogonal to the space of cusp forms - there's a general story I don't quite know Cusp forms vanish at the cusp (those are the $-1$ and $\infty$ points in the quotient $\Bbb H^2/\Gamma$ picture I described above, where the hyperbolic metric gets coned off), whereas given any values on the cusps you can make a linear combination of Eisenstein series which takes those specific values on the cusps So it sort of makes sense Regarding that particular result, saying it's a weight 2 cusp form is like specifying a strong decay rate of the cusp form towards the cusp. Indeed, one basically argues like the maximum value theorem in complex analysis @BalarkaSen no you didn't come across as pretentious at all, i can only imagine being so young and having the mind you have would have resulted in many accusing you of such, but really, my experience in life is diverse to say the least, and I've met know it all types that are in everyway detestable, you shouldn't be so hard on your character you are very humble considering your calibre You probably don't realise how low the bar drops when it comes to integrity of character is concerned, trust me, you wouldn't have come as far as you clearly have if you were a know it all it was actually the best thing for me to have met a 10 year old at the age of 30 that was well beyond what ill ever realistically become as far as math is concerned someone like you is going to be accused of arrogance simply because you intimidate many ignore the good majority of that mate
I'm familiar with extraneous roots. For example $\sqrt{x} = x - 2$ We solve it by squaring both sides \begin{align} & \implies x = x^2 - 4x + 4\\ & \implies x^2 - 5x + 4 = 0\\ & \implies (x-1) (x-4) = 0\\ & \implies x = 1~\text{or}~x = 4 \end{align} But! Only $x = 4$ satisfies the parent equation, $x = 1$ doesn't. Hence $x = 1$ gets rejected. First I was really surprised, couldn't figure out why $x = 1$ was getting rejected, because $x = 1$ lies in the domain of the parent equation ($\sqrt{x} = x - 2$). Still can't understand why it's getting rejected. Because when solving equations (or inequalities), we take the intersection between the domain (which is, $x \geq 0$) of the parent equation (or inequality) and the set of solutions obtained on solving the equation (or the inequality). And $x = 1$ here belongs to both, the domain of the equation and the set of $x$ values obtained on solving it. So it's not supposed to get rejected?? Also, is it not supposed to satisfy the equation because as I said, it belongs to both, the domain and the set of values obtained on solving. But it doesn't satisfy. Please explain me what's going on here, why doesn't it satisfy even though it belongs to both, as I've mentioned earlier. I looked it up, and came to know that such roots are called extraneous roots, and they occur when we raise both sides of a radical equation to an even power. And hence we must always plug the obtained values into the original equation in order to check whether they satisfy the parent equation or not. My second question is, can extraneous roots occur while solving radical inequalities too? If they can, then how do we get the final solution? Because we have infinite solutions in case of radical inequalities. We can't plug in each and every value to check whether they satisfy or not? Thanks, and sorry for the long post. I wanted to explain as well as I could, what I don't understand.
RSA for key exchange is declining rapidly and is not recommended because it does not provide forward secrecy. Without forward secrecy, if someone breaks into the server and obtains the private key, they will be able to fully retroactively decrypt all recorded traffic encrypted under that key. ECDH does not have that problem because the private and public ... Definition of trapdoor function from Wikipedia;A trapdoor function is a function that is easy to compute in one direction, yet difficult to compute in the opposite direction without special information, called the "trapdoor".The reverse trapdoor function is just the reverse usage of it.Normally, for encryption, we want the encryption easy but the ... Q1: The random selection should be $\sqrt[3]{n}<m<n$ due to cube-root attack?Suppose $n$ is 2048 bits long. Then $\sqrt[3] n < 2^{700}$. If $m$ is uniformly distributed in $\{1, 2, \dots, n - 1, n\}$, what is $\Pr[m < \sqrt[3] n]$? Is this probability large enough that you have to worry about it?Now at the end the document it says;... You aren't using the same salt.The randomness is introduced with the salt: it is generated randomly. The input to the MGF is calculated from the hash and the salt. Thus, if you sign the same message twice, you get different signatures, because the salt is different each time. For the details, see PKCS#1 (RFC 8017) §9.1.1.There are only two ways to use ... Given a 613-bit integer, it is easy to check that it is composite (and we assume that's been done). It can be hard to factor, but that's feasible with GNFS, including with ready-made tools (see Luke Valenta, Shaanan Cohney, Alex Liao, Joshua Fried, Satya Bodduluri, Nadia Heninger's Factoring as a Service), and that's one sure way to demonstrate how many ... It's a translation of $ed \equiv 1 \pmod{\phi(N)}$, namely that $e$ and $d$ are each other's inverse modulo $\phi(N)$, which says that $ed-1$ must be an integer multiple of $\phi(N)$ (recall that $a \equiv b \pmod{c}$ iff $c$ divides $b-a$ in the integers iff $b-a=lc$ for some integer $l$) and some some $k \in \Bbb Z$ exists with $k\phi(N) = ed-1$ (as an ... Without factoring it, you can only tell if the number of factors is one, or greater than one.What you ask is not possible unless at least one of the prime factors of a composite integer is small enough that the integer can be partially factored. If at least one factor is small, then after revealing that prime, a fast primality test could be used on the ... With Paillier, it's easy; generate a random encryption of 0 ($r^n \bmod n^2$ for random $r$ r.p to $n$), and then homomorphically add it to the encryption (that is, $C2 = C1 \cdot r^n \bmod n^2$), and you're done (and all you need is the public key).I don't believe RSA allows this as a possibility... You're correct, the proof isn't precisely correct, because we don't necessarily have $c^{\phi(n)} = 1$, specifically in the case $c \equiv 0 \pmod p$.Here is a more correct approach; we have $c^1 \equiv c \pmod p$ (trivially), and $c^{p-1} \equiv c \pmod p$ for any $c$, prime $p$ (Fermat's little theorem [1]). By induction, we get $c^{k (p-1) + 1} \equiv ... In what's described, nothing makes B sure that A sent the message. And that can't be obtained without some secret on A's side.A common solution is to have A sign the (e.g. encrypted) message, and B check the signature.A PKI (perhaps, implemented using digital certificates) can help ensure B uses A's genuine public key, which is required for this proof or ... You can't. If the user can run the software, they can extract the key from it. There is research on how to make it difficult to extract the key (white box cryptography), but it's not very successful.You need to think about what do you want to accomplish. Does the user encrypt the files to themselves, so that they can decrypt it later? Then you could derive ... Yes, such a group is useful. In particular, when $N= p\cdot q$ where $p,q$ are both strong primes (i.e. $p=2p'+1,q=2q'+1$ where $p',q'$ are also prime numbers).Discrete logarithm in the prime order cyclic groups (order-$p'$ and order-$q'$) is believed to be hard. Such a group is called a group with hidden order.A few examples of application: Fujisaki ... I'm not sure how I didn't realize there are only two cases: $gcd(p, c) = 1$, or $c\equiv 0 \bmod (p)$. Given this, here's the proof I came with.For the former, the proof into the question is valid. For the latter, we have:$$c\equiv 0 \bmod (p)$$ $$c^k\equiv 0 \bmod (p)$$ for any $k$. So, using transitivity:$$c^k\equiv c\bmod (p)$$In this case $k=d\bmod (... TLS_DHE_RSA_WITH_AES_256_CBC_SHA is256-bit AES encryptionSHA-1 message authenticationEphemeral Diffie-Hellman key exchangeSigned with an RSA certificateWe can find the answer in rfc5246Key IV BlockCipher Type Material Size Size------------ ------ -------- ---- -----NULL Stream 0 0 ... I'll try to answer your questions. First I shall write Coppersmith's Theorem.Theorem. Let $0<\varepsilon<1/d$ and $F(x)$ be a monic polynomial of degree $d$ with at least one root $x_0$ in ${\mathbb{Z}}_N$ and $\|x_0\|<X =\lceil 0.5N^{1/d-\varepsilon}\rceil.$ Then, we can find $x_0$ in time $poly(d,1/\varepsilon,\ln{N}).$First, notice that in ... The function, in this case, is collision-resistant but despite this, is it still possible to break the EUF-CMA of RSA-FDH?Yes, because the requirement for security is the hash function acting like a random oracle and not it being collision-resistant.To illustrate this, consider the "hash function" $H(x)=x$ which is clearly collision resistant. Yet, RSA-... Yes, nowadays RSA is not considered the de facto algorithm for all public key cryptography needs we have nowadays. BUT, it is still omnipresent and this is not likely to change soon.For instance, TLS 1.3 has dropped the RSA key exchange because of its lack of forward secrecy.So you might think RSA is loosing some traction, however this doesn't mean ...
Let $\{a_n\}$ be a sequence such that $$\lim_{n\to \infty}\left\|a_n+3\left(\frac{n-2}{n}\right)^n\right\|^\frac{1}{n}=\frac{3}{5}$$ Then calculate $\lim_{n\to \infty}a_n$. First I tried to take logarithm and got $\lim_{n\to \infty}\frac{1}{n}\ln\left\|a_n+3\left(\frac{n-2}{n}\right)^n\right\|=\ln\frac{3}{5}$, then I thought about L Hospital but that did not work. I am unable to dig it further. Can somebody give me a hint or push towards the solution? Thanks.
Statement: Let $G$ be a group, $H$ a subgroup of $G$, and $a,b \in G$. Then: $a \in aH$ $aH = H \iff a \in H$ Either $aH = bH$ or $aH \cap bH = \emptyset$ $aH = bH \iff a^{-1} b \in H$ $aH \leq G \iff a \in H$ $\|aH\| = \|bH\| = \|H\|$ Proof: 1. Since $H$ is a subgroup of $G$, $e \in H$, so $a = ae \in aH$. 2. We must prove this statement both ways since it is an if and only if statement. ($\Rightarrow$) Suppose $aH = H$; we need to show that $a \in H$. From Property #1, we know $a \in aH$, but $aH = H$, so $a \in H$. ($\Leftarrow$) Suppose $a \in H$; we need to show that $aH = H$. In order to do this, we need to show $aH \subseteq H$ and $H \subseteq aH$. First, let $x \in aH$ and define $x = ah$ for some $h \in H$. Since $H$ is a subgroup, we know that it is closed under its operation, and since $a,h \in H, x = ah \in H$ Therefore $aH \subseteq H$. Now let $x \in H$. Since $H$ is a subgroup, we know that it contains the $e$ and $a^{-1} \forall a \in H$. So, $x = ex = aa^{-1}x = a(a^{-1}x)$. Since $a^{-1}$ and $x$ are both in $H$, $a^{-1}x \in H$ as well. Therefore, $x = a(a^{-1}x) \in aH$ making $H \subseteq aH$. With $aH$ and $H$ being subsets of each other, it follows that $aH = H$. 3. We will have either $aH \cap bH = \emptyset$ or they will share at least one element. Let that one elements be called $c$, i.e. $c \in aH \cap bH$. If we can show that if $c \in aH \implies cH =aH$, then we can show that $aH = bH$. To do this, we must show that $cH$ and $aH$ are subsets of each other. Suppose $c \in aH$ and let $x \in cH$. Then, $x = ch_1$ for some $h_1 \in H$. However, since $c \in aH$, $c = ah_2$ for some $h_2$ in $H$. So, $x = ah_{2}h_{1}$ Since $h_{2}h_{1} \in H$ (because $H$ is closed), we can see that $x \in aH$, so $cH \subseteq aH$. Now let $x \in aH$. Then, $x = ah_1$ for some $h_1 \in H$. But since $c \in aH$, we know $c = ah_2$ for some $h_2 \in H$ which implies $a = ch_{2}^{-1}$ so then $x = ch_{2}^{-1}h_{1}$ which is an element of $cH$, making $aH \subseteq cH$. Since both $aH$ and $cH$ are subsets of each other, $aH = cH$. Now that we know $c \in aH \implies cH = aH$, we can use it for both $aH$ and $bH$. So, since $c \in aH \cap bH$, $cH = aH$ and $cH = bH$ which implies $aH = bH$. 4. From Property #3, we proved that if $a \in bH$ then $aH=bH$. Consider the converse of this statement: if $aH = bH$, then $a \in bH$. This turns out also to be true. Suppose $aH = bH$. Since $H$ is a subgroup, it contains $e$, so $a = ae \in aH$. But, since $aH = bH$, $a = ae \in aH = bH$, hence $a \in bH$. Now that we have proved the converse as well as the original statements, it can be written as an if and only if statement, i.e. $aH = bH \iff a \in bH$. Therefore, $bH = aH \iff b \in aH$ which can be written in the form $b = ah$ for some $h \in H$. Also, we know that $H$ contains both the identity and inverse, so $b = ah \iff a^{-1}b = a^{-1}ah \iff a^{-1}b = h \iff a^{-1}b \in H$. 5. We need to prove this statement both ways since it is an if and only if statement. ($\Rightarrow$) Suppose $aH \leq H$; we need to show that $a \in H$. Since $aH$ is a subgroup of $G$, we know that $e \in aH$ which can be written in the form $e = ah$ for some $h \in H$. We also know that $aH$ contains the inverse, so we can multiply both sides by $h^{-1}$ on the right and get $eh^{-1} = ahh^{-1} \implies h^{-1} = ae \implies h^{-1} = a$. Hence $a = h^{-1} \in H$. ($\Leftarrow$) Suppose $a \in H$; we need to show that $aH \leq G$. From Property #2, we know that if $a \in H$, then $aH = H$. So, $a \in aH = H \implies a \in H$ and $H$ is a subgroup of $G$. 6. In order to show the order of two sets to be equal, we need to establish a one-to-one function from one set onto another. Let $\phi: aH \rightarrow bH$ be defined as $\phi(ah) = bh$ for some $h \in H$. Elements in $aH$ are in the form $ah$ for some $h \in H$. Similarly, elements in $bH$ are in the form $bh$ for some $h \in H$. Let $x = ah$ and $y = bh$. Let's check to see if $\phi$ is one-to-one. Suppose $\phi(x_1) = \phi(x_2)$. Then we have $bh_1 = y_1 = y_2 = bh_2$, which shows $\phi$ is one-to-one. Now we need to check to see if $\phi$ is onto. This means that $\forall y \in bH \; \exists x \in aH$ such that $\phi(x) = y$. This is easy to see, just let $x = x \in aH$. We then have $\phi(x) = \phi(ah) = bh = y$. We have shown that $\phi$ is onto as well. Since we have established a one-to-one correspondence between $aH$ and $bH$, we can say that they have the same order, i.e. $\|aH\| = \|bH\|$.
Cantor ternary function The Cantor ternary function (also called Devil's staircase) is a continuous monotone function $f$ mapping the interval $[0,1]$ onto itself, with the remarkable property that its derivative vanishes almost everywhere (recall that any monotone function is differentiable almost everywhere, see for instance Function of bounded variation). The function can be defined in the following way (see, for example, Exercise 46 in Chapter 2 of [Ro]). Given $x\in [0,1]$ consider its ternary expansion $\{a_i\}$, i.e. a choice of coefficients $a_i\in \{0,1,2\}$ such that \[ x = \sum_{i=0}^\infty \frac{a_i}{3^i}\, . \] Define $n(x)$ to be $\infty$ if none of the coefficients $a_i$ takes the value $1$ the smallest integer $n$ such that $a_n=1$ otherwise. It follows trivially from the definition that $f$ is locally constant on the complement of the Cantor set $C$: since the Cantor set is a set of Lebesgue measure zero, the derivative of $f$ vanishes almost everywhere. For the same reason, the distributional derivative of $f$ is a singular measure $\mu$ supported on $C$. The Cantor function is a prototype of a singular function in the sense of Lebesgue, cf. Lebesgue decomposition. The observation that $f$ is a function of bounded variation for which \[f (1)-f(0) \neq \int_0^a f'(t)\, dt\](where $f'$ denotes the classical pointwise derivative) was first made by Vitali in [Vi]. For this reason some authors use the terminology Cantor-Vitali function (see [AFP]). References [AFP] L. Ambrosio, N. Fusco, D. Pallara, "Functions of bounded variations and free discontinuity problems". Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000. MR1857292Zbl 0957.49001 [Co] D. L. Cohn, "Measure theory". Birkhäuser, Boston 1993. [Ro] H.L. Royden, "Real analysis" , Macmillan (1969). MR0151555 Zbl 0197.03501 [Vi] A. Vitali,"Sulle funzioni integrali", Atti Accad. Sci. Torino Cl. Sci. Fis. Mat. Natur., 40 1905 pp. 1021-1034. How to Cite This Entry: Cantor ternary function. Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Cantor_ternary_function&oldid=27842
Cantor ternary function The Cantor ternary function (also called Devil's staircase) is a continuous monotone function $f$ mapping the interval $[0,1]$ onto itself, with the remarkable property that its derivative vanishes almost everywhere (recall that any monotone function is differentiable almost everywhere, see for instance Function of bounded variation). The function can be defined in the following way (see, for example, Exercise 46 in Chapter 2 of [Ro]). Given $x\in [0,1]$ consider its ternary expansion $\{a_i\}$, i.e. a choice of coefficients $a_i\in \{0,1,2\}$ such that \[ x = \sum_{i=0}^\infty \frac{a_i}{3^i}\, . \] Define $n(x)$ to be $\infty$ if none of the coefficients $a_i$ takes the value $1$ the smallest integer $n$ such that $a_n=1$ otherwise. It follows trivially from the definition that $f$ is locally constant on the complement of the Cantor set $C$: since the Cantor set is a set of Lebesgue measure zero, the derivative of $f$ vanishes almost everywhere. For the same reason, the distributional derivative of $f$ is a singular measure $\mu$ supported on $C$. The Cantor function is a prototype of a singular function in the sense of Lebesgue, cf. Lebesgue decomposition. The observation that $f$ is a function of bounded variation for which \[f (1)-f(0) \neq \int_0^a f'(t)\, dt\](where $f'$ denotes the classical pointwise derivative) was first made by Vitali in [Vi]. For this reason some authors use the terminology Cantor-Vitali function (see [AFP]). References [AFP] L. Ambrosio, N. Fusco, D. Pallara, "Functions of bounded variations and free discontinuity problems". Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000. MR1857292Zbl 0957.49001 [Co] D. L. Cohn, "Measure theory". Birkhäuser, Boston 1993. [Ro] H.L. Royden, "Real analysis" , Macmillan (1969). MR0151555 Zbl 0197.03501 [Vi] A. Vitali, "Sulle funzioni integrali", Atti Accad. Sci. Torino Cl. Sci. Fis. Mat. Natur., 40 1905 pp. 1021-1034. How to Cite This Entry: Cantor ternary function. Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Cantor_ternary_function&oldid=27843
Consider a reversible reaction $\ce{A -> B}$, at standard state wher $\ce{A}$ and $\ce{B}$ are at $\pu{1 atm}$. If the free energy of products is greater than the free energy of reactants, $\Delta_{\text{R}} G^\circ > 0$, and since $Q = 1$, the overall free energy for this reaction is $\Delta_{\text{R}} G^\circ > 0$ and this tells me the reaction is non-spontaneous. Yet, being at standard state does not mean that the reaction is at equilibrium, in fact $Q = 1$ could be greater or less than $K$, the products are not at equilibrium, and hence amount of $\ce{A}$ and $\ce{B}$ will adjust to a point where $Q = K$. Wouldn't that have just mean that the reaction is not at equilibrium and hence would spontaneously adjust to reach equilibrium. But yet $\Delta_{\text{R}} G^\circ$ tells me that it would not be spontaneous. Wouldn't this two concepts be conflicting against each other? Note: This first part isn't actually that related to the specific scenario you are giving, but I think it's a misconception that you had that led you to ask this question. If you want a direct answer, go to the last paragraph. Just because it is non-spontaneous, it doesn't mean that no reaction will occur. In fact, the reaction will always occur to some extent, even if only 0.001% of the particles react. Consider this formula, which I'm sure you have seen before: $$\Delta G=-RT\ce{ln}K$$ This makes sense because if the reaction is spontaneous ($\Delta G<0$), $K>0$, so the forward reaction, which is spontaneous is favored over the reverse reaction; vice versa. If we, for example, say $\Delta G = +100$ kJ, at 298 K, $K$ would equal $3.4\cdot10^{-17}$. Thus, the forward reaction pretty much completely dominates the reverse reaction, so we say that the reaction is essentially irreversible. However, the reverse reaction is still happening, to a very small extent. If we happened to make $\Delta G = +10$ kJ, $K$ would equal only about $56$, so the reverse reaction becomes visible. The terms spontaneous and non-spontaneous are only generalizations; it doesn't mean that a non-spontaneous reaction doesn't occur at all. There will always be some particles in that substance that will overcome the activation energy and react. In terms of your question, since the reaction $\ce{A->B}$ is non-spontaneous, the reaction $\ce{B->A}$ is actually spontaneous (think about their equilibrium constants). Thus, the reverse reaction will be favored until the system reaches equilibrium, or when $Q=K$. If delta G is positive, that means if you start a reaction with both A and B in a container already and each has a partial pressure of 1 atm, the reaction is not at equilibrium. forward reaction is not spontaneous. Reverse reaction is. So it means the reaction will start with B changes to A to reach equilibrium. At equilibrium, B partial pressure will be greater than 1 and A partial pressure will be less than 1 atm. This reaction has a k less than 1. It means this reaction favors the reactants. Most kids think that a non spontaneous reaction means if I start with just A in the container, A won’t proceed to change to B. As you learn from equilibrium chapter, if you start with just A and no B, the forward reaction will proceed to change to B. The delta G (products free energy - reactants free energy) at this condition is negative since there is no products yet so The delta G at standard state is for when A and B both are 1 atm. Also remember if we say a reaction is nonspontaneous, we mean the forward reaction is non spontaneous, it also means the reverse reaction is automatically spontaneous.
I have been struggling to find an acceptable answer for this question for my purposes. There are many ways to find similarity between two organic compounds - some of which are particularly popular in chemoinformatics. The seemingly most popular way is to use fingerprints of molecules, which then somehow correlates to the structure/function of various parts of the molecule. This approach seems very good when you're looking for general similarity between molecules. It also has the added benefit of making it fast to compare huge amounts of molecules, as each molecule can be encoded separately (so it's generally speaking $O(n)$), rather than comparing every pair of molecules (which would be $O(n^2)$). However, for some purposes, this fingerprinting approach is not very good. I need a function of two molecules with the following properties: If $A$ and $B$ are two molecules, then $f(A, B) \in [0, \infty).$ $f(A, A) = 0$ $f(A,B) + f(B,C) \geq f(A,C)$ These are similar to the basic requirements for a metric space. (The big omission is that I don't necessarily need $f(A,B) = f(B,A),$ although this certainly couldn't hurt!) The reason that I want such a function is because I'm designing a neural network that essentially outputs a molecule, and I want to have some sort of error on the output. I have played around a bit with Python-RDKit, and its similarity module, but haven't really been able to form a good "error" function from the output. I've also experimented a bit with an algorithm I created that looks for the largest identical connected subgraphs of two molecules, and essentially finds matches for each part of the query molecule. The final output is then how many different parts the molecule needs to be split into to find a match. For instance, if the "true" molecule is ethylbenzene, while the query molecule is m-xylene, then the algorithm would find that the m-xylene needs to be broken into 3 parts for each part to find a perfect match: a benzyl group missing a hydrogen at the third carbon of the aromatic ring, the methyl group that was formerly attached at the third carbon, and the hydrogen that remains from the benzyl group. However, this algorithm suffers from several problems: If the query is a subset of the true answer, then the algorithm will always give the answer as 2 (one part is the subset, and the other part is the hydrogen that caps it - try seeing it with something like query - methane, true - acetic acid). This isn't that big of a concern, as you can simply run the algorithm twice - once comparing the query to the true molecule, and once in reverse. That way, if the two molecules are indeed far apart, it's not hard to see (the superset molecule may need to be broken into many pieces to be identical). This algorithm is slow. Don't really see a way to speed it up. It searches for maps from subsets of the query molecule to the true molecule, then gradually builds up from there. It also can't pick maps randomly as being the best, then growing from there, as picking the wrong direction can easily make the "best" map drastically short. So it has to do all possible maps at the same time. Which is slow. In short, this is a somewhat open-ended question that boils down to: How can we put a (loose) metric on the set of organic molecules?
Disclaimer: this terminology might be different from what you're used to and this is why I'm writing down some definitions first. Let $A$ be an abelian group, $m \in \mathbb{N}^* := \mathbb{N} \setminus \{0\}$ and let $p$ be a prime number. We define: The subgroup of multiples of $m$ in $A$as the set $mA := \{\,ma \, \mid \, a \in A\,\}$. The $m$-torsion of $A$as the set $A[m] := \{\,a \in A \, \mid \, ma = 0\,\}$. The $p$-subgroup of $A$as the set $A(p) := \{\,a \in A \, \mid \, p^k a = 0 \hspace{1mm} \text{for some} \hspace{1mm} k \in \mathbb{N}\,\}$. I've proven these are all subgroups of $A$. It is known that $A(p) = \{\,a \in A \, \mid \, o(a) = p^k \hspace{1mm} \text{for some} \hspace{1mm} k \in \mathbb{N}\,\}$. Now we want to prove the following statement considering $A := \mathbb{Z}_n$. Square brackets (or a bar if it's a single letter) denote the equivalence class of an element in $\mathbb{Z}_n$, whereas the angle brackets denote the subgroup genereated by the element within them and finally the $\simeq$ symbol denotes an isomorphism. $m\mathbb{Z}_n = \gcd(n,m)\mathbb{Z}_n = \langle [\gcd(n,m)] \rangle \simeq \mathbb{Z}_\frac{n}{\gcd(n,m)}$ $\mathbb{Z}_n[m] = \mathbb{Z}_n[\gcd(n,m)] = \langle [\frac{n}{\gcd(n,m)}] \rangle \simeq \mathbb{Z}_{\gcd(n,m)}$ If $n = p^r n'$ and $p \nmid n'$ (not divides), then $\mathbb{Z}_n(p) = \langle n' \rangle \simeq \mathbb{Z}_{p^r}$ I've tried to solve it in this way. First of all, we have that $\gcd(n,m) \mid m$, so there exists $k \in \mathbb{Z}$ such that $m = \gcd(n,m)k$. Given an element $\bar{x} \in m\mathbb{Z}_n$, by definition there exists $\bar{a} \in \mathbb{Z}_n$ such that $\bar{x} = m \bar{a}$ and therefore $\bar{x} = \gcd(n,m)k\bar{a} = \gcd(n,m)[ka]$. This proves that $\bar{x} \in \gcd(n,m)\mathbb{Z}_n$, thus $m\mathbb{Z}_n \subseteq \gcd(n,m)\mathbb{Z}_n$. I was not able to prove the inclusion $\supseteq$ though. Then I guess I solved the equality $\gcd(n,m) = \langle [\gcd(n,m)] \rangle$. In fact (tell me if I'm mistaken somewhere): \begin{align} \gcd(n,m)\mathbb{Z}_n & = \{\,\gcd(n,m)\bar{a} \, \mid \, \bar{a} \in \mathbb{Z}_n\,\} \\ & = \{\,[\gcd(n,m)a] \, \mid \, a \in \mathbb{Z}\,\} \\ & = \{\,a[\gcd(n,m)] \, \mid \, a \in \mathbb{Z}\,\} = \langle [\gcd(n,m)] \rangle \end{align} For the isomorphism part, let $f \colon \langle [\gcd(n,m)] \rangle \to \mathbb{Z}_{\frac{n}{\gcd(n,m)}}$ be the map defined by $f(k[\gcd(n,m)]) := \bar{k}$. This is trivially a surjective homomorphism. It is injective, too, since: \begin{align} \ker{f} & = \{\,k[\gcd(n,m)] \in \langle [\gcd(n,m)] \rangle \, \mid \, f(k[\gcd(n,m)]) = \bar{0}\,\} \\ & = \{\,k[\gcd(n,m)] \in \langle [\gcd(n,m)] \rangle \, \mid \, \bar{k} = \bar{0}\,\} \\ & = \{\,nh[\gcd(n,m)] \in \langle [\gcd(n,m)] \rangle \, \mid \, h \in \mathbb{Z}\,\} \\ & = \{\,h\gcd(n,m)\bar{n} \in \langle [\gcd(n,m)] \rangle \, \mid \, h \in \mathbb{Z}\,\} = \{\bar{0}\} \end{align} Am I on the right track or is there a better way to prove this stuff? And if this is the right way, how can I complete this proof including statement 2. and 3.? Edit: I found a better way to prove all isomorphisms. I use this result (the order of an element $a \in G$ is denoted by $o(a)$): Let $G$ be a cyclic group, $G = \langle a \rangle$. If $o(a) = n$, then $G \simeq \mathbb{Z}_n$. Now one easily verifies the following conditions: $o([\gcd(n,m)]) = \frac{n}{\gcd(n,m)}$ $o([\frac{n}{\gcd(n,m)}]) = \gcd(n,m)$ $o([n']) = p^r$ So I only need to prove the equalities now if my reasoning is correct.
I recently came across the following argument regarding the uniqueness of the zeros of a complex polynomial. Please note that the proof that a complex polynomial of degree $m$ has $m$ zeros has been established at this point. The following is also not the complete proof of the uniqueness of the roots. I just want to focus on a particular passage from the proof that I need help clarifying. Consider the following equation of two factorizations of the same complex polynomial of degree $m$, where $z, \lambda_i, \tau_i\in\mathbb{C}$, for $i=1\dots m$: $$\left(z - \lambda_1\right)\left(z - \lambda_2\right)\dots\left(z - \lambda_m\right) = \left(z - \tau_1\right)\left(z - \tau_2\right)\dots\left(z - \tau_m\right)$$ Therefore, all $\lambda_i$ and $\tau_i$ are the zeros of the polynomial. Moreover, substituting $z=\lambda_i$, the resulting equation implies that $\lambda_i = \tau_j$ for some $j \in \left\{1\dots m\right\}$. To make it simple, let's relabel $\tau_j$ so that $\lambda_i = \tau_i$. Now, consider $i = 1$. Dividing both sides by $z - \lambda_1$, we get $$\left(z - \lambda_2\right)\dots\left(z - \lambda_m\right) = \left(z - \tau_2\right)\dots\left(z - \tau_m\right)$$ for all $z\in\mathbb{C}$ except possibly $z = \lambda_1$. So far, with the exception of the word "possibly", it has been straight-forward and obvious to me. However, what comes next in the argument puzzles me: "Actually, the equation above [after dividing by $z - \lambda_1$] holds for all $z\in\mathbb{C}$ because, otherwise, by subtracting the right side from the left side, we would get a non-zero polynomial that has infinitely many zeros." That passage is a part of the proof of theorem 4.14 in "Linear Algebra Done Right", third edition (S. Axler). Why would the alternative case imply a non-zero polynomial with infinitely many roots? Could someone please kindly show me? Thank you.
Derivative of Arc Length Theorem Then the derivative of $s$ with respect to $x$ is given by: $\dfrac {\d s} {\d x} = \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$ Consider a length $\d s$ of $C$, short enough for it to be approximated to a straight line segment: By Pythagoras's Theorem, we have: $\d s^2 = \d x^2 + \d y^2$ Dividing by $\d x^2$ we have: $\displaystyle \paren {\frac {\d s} {\d x} }^2$ $=$ $\displaystyle \paren {\frac {\d x} {\d x} }^2 + \paren {\frac {\d y} {\d x} }^2$ $\displaystyle $ $=$ $\displaystyle 1 + \paren {\frac {\d y} {\d x} }^2$ Hence the result, by taking the principal square root of both sides. $\blacksquare$ From Continuously Differentiable Curve has Finite Arc Length, $s$ exists and is given by: $\displaystyle s$ $=$ $\displaystyle \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} }^2} \rd u$ $\displaystyle \leadsto \ \ $ $\displaystyle \frac {\d s} {\d x}$ $=$ $\displaystyle \frac {\d} {\d x} \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} } ^2} \rd u$ differentiating both sides WRT $x$ $\displaystyle $ $=$ $\displaystyle \sqrt {1 + \paren {\frac {\d y} {\d x} }^2}$ Fundamental Theorem of Calculus/First Part $\blacksquare$
Equivalence of Definitions of Separated Sets Contents Theorem Let $T = \struct{S, \tau}$ be a topological space. Let $A, B \subseteq S$. $A$ and $B$ are separated (in $T$) if and only if: $A^- \cap B = A \cap B^- = \O$ $A$ and $B$ are separated (in $T$) if and only if there exist $U,V\in\tau$ with: $A\subset U$ and $U\cap B = \varnothing$ $B\subset V$ and $V\cap A = \varnothing$ where $\varnothing$ denotes the empty set. Proof Let $A, B \subseteq S$ satisfy: $A^- \cap B = A \cap B^- = \empty$ Let $U = S \setminus B^-$ be the relative complement of $B^-$. From Empty Intersection iff Subset of Relative Complement, $A \subseteq S \setminus B^- = U$ From Relative Complement of Relative Complement, $S \setminus U = B^-$. From Empty Intersection iff Subset of Relative Complement, $U \cap B = \empty$. Similarly, let $V = S \setminus A^-$ then $V \in \tau$ with: $B \subset V$ and $V \cap A = \empty$ $\Box$ Let $A, B \subseteq S$. Let $U,V \in \tau$ satisfy: $A \subset U$ and $U \cap B = \empty$ $B \subset V$ and $V \cap A = \empty$ From Empty Intersection iff Subset of Relative Complement, $B \subseteq S \setminus U$. From Set Closure is Smallest Closed Set, $B^- \subseteq S \setminus U$. From Empty Intersection iff Subset of Relative Complement, $B^- \cap U = \empty$. Then $\displaystyle B^- \cap A$ $=$ $\displaystyle B^- \cap \paren {U \cap A}$ Intersection with Subset is Subset $\displaystyle $ $=$ $\displaystyle \paren {B^- \cap U} \cap A$ Intersection is Associative $\displaystyle $ $=$ $\displaystyle \empty \cap A$ As $B^- \cap U = \empty$ $\displaystyle $ $=$ $\displaystyle \empty$ Intersection with Empty Set Similarly, $A^- \cap B = \empty$. $\blacksquare$
→ → → → Browse Dissertations and Theses - Mathematics by Title Now showing items 288-307 of 1147 application/pdfPDF (1MB) (2014-05-30)In this work, we prove effective decay of certain multiple correlation coefficients for Weyl chamber actions of semidirect products of semisimple groups with $G$-vector spaces. Using these estimates we get decay for ... application/pdfPDF (1MB) (1993)Ramsey's Theorem states that if $P = \{C\sb1,\...,C\sb{n}\}$ is a partition of ($\omega\rbrack\sp{k}$ (the set of all unordered k-tuples of natural numbers) into finitely many classes, then there exists an infinite set A ... application/pdfPDF (4MB) (1995)We establish a quadratic time algorithm for the word problem and a cubic time algorithm for the conjugacy problem for Coxeter groups of large type. We also give conditions on the relators of a large-type Coxeter group that ... application/pdfPDF (3MB) (1989)Representations of meromorphic functions as quotients of analytic functions have been studied for years. Miles showed that any meromorphic function f can be written as f$\sb1$/f$\sb2$ where each f$\sb{\rm j}$ is entire and ... application/pdfPDF (3MB) application/pdfPDF (817kB) (2004)Chapter 4 is devoted to finding new partition identities inspired by the work of O. Kolberg and S. Ramanujan. We use functions studied by N. J. Fine and R. J. Evans to construct analogues of modular equations, and then ... application/pdfPDF (6MB) (2016-07-14)In this dissertation, we will focus on a few problems in extremal graph theory. The first chapter consists of some basic terms and tools. In Chapter 2, we study a conjecture of Mader on embedding subdivisions of cliques. ... application/pdfPDF (702kB) (2003)The third part contains some applications, among them Hilbert-type inequalities and inequalities of the form fx ≤cnsupy∈ Rfn y for bounded functions f with a spectral gap at the origin. application/pdfPDF (4MB) (1995)A ribbon knot is one which bounds a certain type of singular disk in the 3-sphere. In this work we investigate an enumeration procedure for such disks and study a natural topological grouping of related ribbons which differ ... application/pdfPDF (2MB) application/pdfPDF (1MB) (2012-09-18)This thesis consists of three parts. In the first part, we compute the topological Euler characteristics of the moduli spaces of stable sheaves of dimension one on the total space of rank 2 bundle on P1 whose determinant ... application/pdfPDF (705kB) (1987)Let K be a number of field with ring of integers o, and let G be a fixed finite group. If K$\sb\pi$ is a tame Galois G-extension, the integral closure ${\cal O}\sb\pi$ of o in K$\sb\pi$ is a locally free rank one oG-module, ... application/pdfPDF (2MB) (2002)We explicitly characterize the family of networks for which such a protocol exists. This characterization is given in terms of forbidden rooted minors, which leads to a linear time recognition algorithm for this family of ... application/pdfPDF (5MB) application/pdfPDF (2MB) application/pdfPDF (2MB) (2017-11-28)The equivariant 𝔼∞G operad has the property that 𝔼∞G(n) is the total space for the G-equivariant universal principal Σn bundle. There is a forgetful functor from 𝔼∞G-algebras to 𝔼∞-algebras, where 𝔼∞ is the classic ... application/pdfPDF (263kB) (2004)We classify embeddings of algebraic groups as open orbits in affine varieties, generalizing results from toric geometry to connected reductive groups. In particular, we show that an embedding is determined by the set of ... application/pdfPDF (4MB) (2013-05-24)We improve the error term in the van der Corput transform for exponential sums, \sum g(n) exp(2 \pi i f(n)). For many smooth functions g and f, we can show that the largest factor of the error term is given by a simple ... application/pdfPDF (931kB) (2017-07-19)My dissertation consists of two essays which analyze the impact of public transit usage on obesity. Chapter 1 introduces the backgrounds of this field and layout the general framework of this thesis work. Chapter 2 ... application/pdfPDF (272kB)
OpenCV 3.2.0 Open Source Computer Vision void cv::accumulate (InputArray src, InputOutputArray dst, InputArray mask=noArray()) Adds an image to the accumulator. More... void cv::accumulateProduct (InputArray src1, InputArray src2, InputOutputArray dst, InputArray mask=noArray()) Adds the per-element product of two input images to the accumulator. More... void cv::accumulateSquare (InputArray src, InputOutputArray dst, InputArray mask=noArray()) Adds the square of a source image to the accumulator. More... void cv::accumulateWeighted (InputArray src, InputOutputArray dst, double alpha, InputArray mask=noArray()) Updates a running average. More... void cv::createHanningWindow (OutputArray dst, Size winSize, int type) This function computes a Hanning window coefficients in two dimensions. More... Point2d cv::phaseCorrelate (InputArray src1, InputArray src2, InputArray window=noArray(), double response=0) The function is used to detect translational shifts that occur between two images. More... Adds an image to the accumulator. The function adds src or some of its elements to dst : \[\texttt{dst} (x,y) \leftarrow \texttt{dst} (x,y) + \texttt{src} (x,y) \quad \text{if} \quad \texttt{mask} (x,y) \ne 0\] The function supports multi-channel images. Each channel is processed independently. The functions accumulate can be used, for example, to collect statistics of a scene background viewed by a still camera and for the further foreground-background segmentation. src Input image as 1- or 3-channel, 8-bit or 32-bit floating point. dst Accumulator image with the same number of channels as input image, 32-bit or 64-bit floating-point. mask Optional operation mask. void cv::accumulateProduct ( InputArray src1, InputArray src2, InputOutputArray dst, InputArray mask = noArray() ) Adds the per-element product of two input images to the accumulator. The function adds the product of two images or their selected regions to the accumulator dst : \[\texttt{dst} (x,y) \leftarrow \texttt{dst} (x,y) + \texttt{src1} (x,y) \cdot \texttt{src2} (x,y) \quad \text{if} \quad \texttt{mask} (x,y) \ne 0\] The function supports multi-channel images. Each channel is processed independently. src1 First input image, 1- or 3-channel, 8-bit or 32-bit floating point. src2 Second input image of the same type and the same size as src1 . dst Accumulator with the same number of channels as input images, 32-bit or 64-bit floating-point. mask Optional operation mask. Adds the square of a source image to the accumulator. The function adds the input image src or its selected region, raised to a power of 2, to the accumulator dst : \[\texttt{dst} (x,y) \leftarrow \texttt{dst} (x,y) + \texttt{src} (x,y)^2 \quad \text{if} \quad \texttt{mask} (x,y) \ne 0\] The function supports multi-channel images. Each channel is processed independently. src Input image as 1- or 3-channel, 8-bit or 32-bit floating point. dst Accumulator image with the same number of channels as input image, 32-bit or 64-bit floating-point. mask Optional operation mask. void cv::accumulateWeighted ( InputArray src, InputOutputArray dst, double alpha, InputArray mask = noArray() ) Updates a running average. The function calculates the weighted sum of the input image src and the accumulator dst so that dst becomes a running average of a frame sequence: \[\texttt{dst} (x,y) \leftarrow (1- \texttt{alpha} ) \cdot \texttt{dst} (x,y) + \texttt{alpha} \cdot \texttt{src} (x,y) \quad \text{if} \quad \texttt{mask} (x,y) \ne 0\] That is, alpha regulates the update speed (how fast the accumulator "forgets" about earlier images). The function supports multi-channel images. Each channel is processed independently. src Input image as 1- or 3-channel, 8-bit or 32-bit floating point. dst Accumulator image with the same number of channels as input image, 32-bit or 64-bit floating-point. alpha Weight of the input image. mask Optional operation mask. This function computes a Hanning window coefficients in two dimensions. An example is shown below: dst Destination array to place Hann coefficients in winSize The window size specifications type Created array type Point2d cv::phaseCorrelate ( InputArray src1, InputArray src2, InputArray window = noArray(), double * response = 0 ) The function is used to detect translational shifts that occur between two images. The operation takes advantage of the Fourier shift theorem for detecting the translational shift in the frequency domain. It can be used for fast image registration as well as motion estimation. For more information please see http://en.wikipedia.org/wiki/Phase_correlation Calculates the cross-power spectrum of two supplied source arrays. The arrays are padded if needed with getOptimalDFTSize. The function performs the following equations: \[\mathbf{G}_a = \mathcal{F}\{src_1\}, \; \mathbf{G}_b = \mathcal{F}\{src_2\}\]where $\mathcal{F}$ is the forward DFT. \[R = \frac{ \mathbf{G}_a \mathbf{G}_b^}{\|\mathbf{G}_a \mathbf{G}_b^\|}\] \[r = \mathcal{F}^{-1}\{R\}\] \[(\Delta x, \Delta y) = \texttt{weightedCentroid} \{\arg \max_{(x, y)}\{r\}\}\] src1 Source floating point array (CV_32FC1 or CV_64FC1) src2 Source floating point array (CV_32FC1 or CV_64FC1) window Floating point array with windowing coefficients to reduce edge effects (optional). response Signal power within the 5x5 centroid around the peak, between 0 and 1 (optional).
Fubini's Theorem Theorem Let $\struct {X, \Sigma_1, \mu}$ and $\struct {Y, \Sigma_2, \nu}$ be $\sigma$-finite measure spaces. Let $\struct {X \times Y, \Sigma_1 \otimes \Sigma_2, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_1, \mu}$ and $\struct {Y, \Sigma_2, \nu}$. Let $f: X \times Y \to \R$ be a $\Sigma_1 \otimes \Sigma_2$-measurable function. Suppose that: $\displaystyle \int_{X \times Y} \size f \map {\rd} {\mu \times \nu} < \infty$ Then $f$ is $\mu \times \nu$-integrable, and: $\displaystyle \int_{X \times Y} f \map {\rd} {\mu \times \nu} = \int_Y \int_X \map f {x, y} \map {\rd \mu} x \map {\rd \nu} y = \int_X \int_Y \map f {x, y} \map {\rd \nu} y \map {\rd \mu} x$ Proof Also see Source of Name This entry was named for Guido Fubini.
Research articles for the 2019-07-07 arXiv Let $\psi$ be a multi-dimensional random variable. We show that the set of probability measures $\mathbb{Q}$ such that the $\mathbb{Q}$-martingale $S^{\mathbb{Q}}_t=\mathbb{E}^{\mathbb{Q}}\left[\psi\lvert\mathcal{F}_{t}\right]$ has the Martingale Representation Property (MRP) is either empty or dense in $\mathcal{L}_\infty$-norm. The proof is based on a related result involving analytic fields of terminal conditions $(\psi(x))_{x\in U}$ and probability measures $(\mathbb{Q}(x))_{x\in U}$ over an open set $U$. Namely, we show that the set of points $x\in U$ such that $S_t(x) = \mathbb{E}^{\mathbb{Q}(x)}\left[\psi(x)\lvert\mathcal{F}_{t}\right]$ does not have the MRP, either coincides with $U$ or has Lebesgue measure zero. Our study is motivated by the problem of endogenous completeness in financial economics. SSRN The failure to find fundamentals that co-move with exchange rates or forecasting models with even mild predictive power -- facts broadly referred to as "exchange rate disconnect'' -- stands among the most disappointing, but robust, facts in all of international macroeconomics. In this paper, we demonstrate that U.S. purchases of foreign bonds, which did not co-move with exchange rates prior to 2007, have provided significant in-sample, and even some out-of-sample, explanatory power for currencies since then. We show that several proxies for global risk factors also start to co-move strongly with the dollar and with U.S. purchases of foreign bonds around 2007, suggesting that risk plays a key role in this finding. We use security-level data on U.S. portfolios to demonstrate that the reconnect of U.S. foreign bond purchases to exchange rates is largely driven by investment in dollar-denominated assets rather than by foreign currency exposure alone. Our results support the narrative emerging from an active recent literature that the US dollar's role as an international and safe-haven currency has surged since the global financial crisis. SSRN The broad structural discontinuity known as the Oman Line extends NNE from Oman across the Strait of Hormuz and divides the flysch-rich eugeosynclinal sediments of the Makran Ranges in the east from the miogeosynclinal shelf sediments of the Zagros Mountain Ranges to the west. The Zagros Crush Zone, west of the Oman Line, marks the location of a continent/continent-style active margin where the Arabian Platform has collided with the Eurasian Plate to the north. To the east, the active margin is a continent/ocean-style boundary where the oceanic lithosphere of the Indian Ocean is being subducted beneath the Central Iranian Microcontinent and other more easterly microcontinental blocks. Geological investigations in the Arabian Plate indicate the presence of a NE-SW trending lineament. This lineament is also recognized on geophysical maps by aligned highs and lows, steep contours gradients and linear offset of trends. There are some indications suggesting that this lineament could represent a SW extension of the Oman Line from Oman across the Empty Quarter (Rub al Khali) of Saudi Arabia to eventually form a transform fault in the Red Sea. arXiv Low-frequency historical data, high-frequency historical data and option data are three major sources, which can be used to forecast the underlying security's volatility. In this paper, we propose two econometric models, which integrate three information sources. In GARCH-It\^{o}-OI model, we assume that the option-implied volatility can influence the security's future volatility, and the option-implied volatility is treated as an observable exogenous variable. In GARCH-It\^{o}-IV model, we assume that the option-implied volatility can not influence the security's volatility directly, and the relationship between the option-implied volatility and the security's volatility is constructed to extract useful information of the underlying security. After providing the quasi-maximum likelihood estimators for the parameters and establishing their asymptotic properties, we also conduct a series of simulation analysis and empirical analysis to compare the proposed models with other popular models in the literature. We find that when the sampling interval of the high-frequency data is 5 minutes, the GARCH-It\^{o}-OI model and GARCH-It\^{o}-IV model has better forecasting performance than other models. SSRN We analyse the consequences of predicting and exploiting financial bubbles in an agent-based model, with a risky and a risk-free asset and three different trader types: fundamentalists, noise traders and "dragon riders" (DR). The DR exploit their ability to diagnose financial bubbles from the endogenous price history to determine optimal entry and exit trading times. We study the DR market impact as a function of their wealth fraction. With a proportion of up to 10%, DR are found to have a beneficial effect, reducing the volatility, value-at-risk and average bubble peak amplitudes. They thus reduce inefficiencies and stabilise the market by arbitraging the bubbles. At larger proportions, DR tend to destabilise prices, as their diagnostics of bubbles become increasingly self-referencing, leading to volatility amplification by the noise traders, which destroys the bubble characteristics that would have allowed them to predict bubbles at lower fraction of wealth. Concomitantly, bubble-based arbitrage opportunities disappear with large fraction of DR in the population of traders. SSRN We present a novel partial explanation for the sixfold increase in student borrowing since 2003: precautionary borrowing. Using a stylized model, we show that annual federal loan limits can induce students whose families face unemployment risk to borrow more today. We then use a new dataset of individually-identified student financial aid records from a large US public university to test our theory of precautionary borrowing. Individually-identified student records allow us to control for both student fixed effects and exclude students whose families were directly implicated by unemployment shocks. Among students whose parents' employment status did not change, we find that a 1 standard deviation increase in local unemployment rates corresponds to a 10% increase in the amount borrowed. A back-of-the-envelope calculation suggests that precautionary borrowing accounted for $21 billion of new student borrowing per year during the Great Recession. SSRN This paper argues that the post-crisis infrastructural reform mandating central clearing of standardized over-the-counter derivatives impacts the valuation of a derivative contract and leads to unintended value redistribution effects among market participants. In a theoretical model, we show how the exogenously imposed change in the market structure affects counterparty risk and funding costs of different types of market participants. Specifically, we find that netting is beneficial for relatively high-quality counterparties, but counterparties with low creditworthiness are better off from accumulating larger gross positions. Further, even though a CCP interposes itself between a buyer and a seller of a derivative contract, precisely in times of distress the network of expected exposures between CCP members becomes fully connected. Our results highlight that mutualization of risks and resources in a CCP leads to externalities between the members. SSRN Since the first fiscal quarter of 2018, financial institutions have implemented a new expected credit loss (ECL) model under International Financial Reporting Standard (IFRS) 9 that replaces the International Accounting Standard (IAS) 39 incurred-loss model. The major difference in provisioning approaches between the two models is the incorporation of forward-looking information under the IFRS 9 ECL model. This article examines whether IFRS 9 improves the credit-risk relevance of loan loss provisions (LLP) for credit default swap (CDS) market participants. The findings suggests that LLP are marginally more credit-risk relevant under IFRS 9 than IAS 39. Moreover, LLP under IFRS 9 affect to a greater extent the pricing of credit risk for longer CDS maturities compared to the IAS 39 regime. This finding is consistent with the IFRS 9 ECL model, providing a more forward-looking measure of credit risk. Finally, institutional features play a significant role in the relevance of accounting information. Under IAS 39, LLP were more informative for the pricing of credit risk in countries with strong official supervisory power, suggesting fewer opportunities for an improvement in the credit-risk relevance of LLP for those countries. Consistent with this argument, the analysis shows that the credit-risk relevance of LLP is particularly enhanced by the IFRS 9 implementation in countries with weak official supervisory power.
Let $A \in \mathcal{M}_3(\mathbb{C})$ such that $\mathrm{tr}(A^2) = \mathrm{tr}(A^3) \in \mathbb{Q},$ where $\mathrm{tr}(A)$ is the trace of $A.$ It is possible to prove that $\mathrm{tr}(A^4) \in \mathbb{Q}?$ closed as off-topic by Travis, Especially Lime, John B, Mars Plastic, user21820 Sep 15 at 14:23 This question appears to be off-topic. The users who voted to close gave this specific reason: " This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Especially Lime, John B, Mars Plastic, user21820 We fix some $t\in \Bbb C$. Let $a,b,c$ be the complex roots of the equation $$ X^3-6tX^2+18t^2 X -36 t^3\ . $$ Let $e_n$ be the elementary symmetric polynomial of degree $n$ in $a,b,c$. Let $p_n$ be the Newton symmetric polynomial of degree $n$ in $a,b,c$. By Vieta, $$\left\{ \begin{aligned} e_1 &= a+b+c = 6t\ ,\\ e_2 &= ab+bc+ca = 18t^2\ ,\\ e_3 &= abc = 36t^3\ . \end{aligned} \right. $$ Let $A$ be the matrix with diagonal entries $a,b,c$. Then, using the Newton identities, $$ \begin{aligned} \operatorname{Trace}(A^2) &=a^2+b^2+c^2\\ &=p_2\\ &=e_1p_1-2e_2\\ &=6t\cdot 6t-2\cdot 18t^2=0\ , \\ \operatorname{Trace}(A^3) &=a^3+b^3+c^3\\ &=p_3\\ &=e_1p_2-e_2p_1+3e_3\\ &=0-18t^2\cdot 6t+3\cdot 36t^3=0\ , \\ \operatorname{Trace}(A^4) &=a^4+b^4+c^4\\ &=p_4\\ &=e_1p_3-e_2p_2+e_3p_1-4e_4\\ &=0-0+36t^3\cdot 6t-0\\ &=216t^4\ . \end{aligned} $$ It is clear that we can chose $t$, such that the last expression is not rational. This gives and explicit counterexample. Later edit: The general family of cubics with roots $a,b,c$, so that $p_1=p_1(a,b,c)=e_1=e_1(a,b,c)=s$ and $p_2=p_3=p_2(a,b,c)=p_3(a,b,c)=q$, thus parametrized by $s,q$, is obtained by solving$$\begin{aligned}e_1&=p_1=s\ ,\\2e_2 &=e_1p_1-p_2=s^2-q\ ,\\3e_3 &=e_2p_1-e_1p_2+p_3=\frac 12(s^2-q)s-sq+q\ ,\\\end{aligned}$$and $p_4$ is obtained algebraically as $p_4=e_1p_3-e_2p_2+e_3p_1-4e_4$. (Here, $e_4=0$.) The question "It is still possible to find a counterexample so that $s$ (trace of $A$) is a natural (or even rational) number?" has a negative answer. If both $s,q$ are rational, than $p_4$ is rational, as an algebraic expression in $s,q$.
''Diamond Paradox'' by Diamond (1971) This is a "less-known paradox," usually put as a counter to famous Bertrand paradox. It is a starting point in the literature on informational frictions in consumer markets, and the scientists in the field agree on its significance. Its idea is diametrically opposite to that of Bertrand. Consider the following simple example. There are $2$ firms which produce homogeneous goods at zero marginal cost and compete in prices, $p$. This simultaneously set prices. Also there is a single consumer who supplies a demand given by $1-p$. Importantly, the consumer does not observe prices set by firms and, therefore, needs to search for them sequentially, where search is costly. Suppose that cost of visiting a firm is given by $0 < c \leq \frac{1}{2}$. Then, the unique equilibrium of the market is that both firms charge monopoly price $$p^M= \frac{1}{2}.$$ This is a diametrically opposite result to that of Bertrand. The reasoning behind the result is as follows. Suppose both firms charge $p=0$. Then, the consumer randomly visits one of the firms, say firm $i$, and buys. However, firm $i$ could have charged $c$ and made positive profits as the consumer would have bought goods anyway because she would have suffered cost $c$ had she left firm $i$ in order to buy from the rival firm. By the same argument, one can see that $p=c$ cannot be an equilibrium as now firm $i$ can charge $c+c$ and improve its profit. Continuing this way, it is easy to arrive to an equilibrium where both firms charge $p^M$. A firm does not want to charge $p^M+c$ simply because its profit is maximized at $p^M$. Formal Analysis of the Example Timing: First, the firms simultaneously set prices. Second, the consumer without knowing prices engage into sequential search. The first search is free and the consumer visit each firm with equal probability. The consumer can come back to the previously searched firm for free. The consumer has to observe a price of a firm to buy goods from that firm. Beliefs: In equilibrium, the consumer has correct belief about strategies of firms. If, upon visiting a firm, she observes a price different from an equilibrium one, the consumers assumes that the rival firm has deviated to the same price too. Thus, the consumer has symmetric (out-of-equilibrium beliefs). Note: the results of the game does not change if the consumers has passive beliefs. Strategies: Strategies of the firms are prices. As mixing is allowed, let $F(p)$ represent the probability that a firm charges a price no greater than $p$. Strategy of the consumer is whether to search for the second price, upon observing the first one. This strategy is given by a reservation price $r$, such that upon observing a price lower than $r$ she buys outright, upon observing a price greater than $r$ she searches further, and upon observing a price equal to $r$ she is indifferent between buying immediately and searching further. Equilibrium Notion: Concept of Perfect Bayesian Equilibrium (PBE) is employed. A PBE is characterized by price distribution $F(p)$ for each firm and the consumer's reservation price strategy given by $r$ such that $(i)$ each firms chooses $F(p)$ that maximizes its profit, given the equilibrium strategy of the other firm and the consumer's optimal search strategy, and (ii) the consumer searches according to the reservation price rule $r$, given correct beliefs concerning equilibrium strategies of firms. Theorem: For any $c>0$, there exists a PBE characterized by triple $(p^M, p^M, r)$, where $p^M$'s are charged with probability $1$ and $$r=1.$$ Proof: First, I prove that $r=1$, or that the consumer buys outright when she observes any price lower than $1$. Clearly, if she observes a price greater than $1$ she does not buy from that firm as this yields a negative payoff to the consumer. Now, suppose she observes price $p'<r$. Then, she expects the rival firm to charge $p'$ too. Thus, if she buys outright her payoff is $\int_{p'}^{1}(1-p)dp$, and if she searches she expects a payoff equal to $\int_{p'}^{1}(1-p)dp - c$. As the former is greater than the latter, she better-off when she buys immediately. This proves that $r=1$. Next, I prove that both firms charge $p^M$. Clearly, firms never charge above $1$ as they will never sell. Then, the expected profit of a firm is $\frac{1}{2}(1-p)p$ because the consumer visits a firm half of the time. It is easy to see that the profit is maximized at $p^M$. QED.
ä is in the extended latin block and n is in the basic latin block so there is a transition there, but you would have hoped \setTransitionsForLatin would have not inserted any code at that point as both those blocks are listed as part of the latin block, but apparently not.... — David Carlisle12 secs ago @egreg you are credited in the file, so you inherit the blame:-) @UlrikeFischer I was leaving it for @egreg to trace but I suspect the package makes some assumptions about what is safe, it offers the user "enter" and "exit" code for each block but xetex only has a single insert, the interchartoken at a boundary the package isn't clear what happens at a boundary if the exit of the left class and the entry of the right are both specified, nor if anything is inserted at boundaries between blocks that are contained within one of the meta blocks like latin. Why do we downvote to a total vote of -3 or even lower? Weren't we a welcoming and forgiving community with the convention to only downvote to -1 (except for some extreme cases, like e.g., worsening the site design in every possible aspect)? @Skillmon people will downvote if they wish and given that the rest of the network regularly downvotes lots of new users will not know or not agree with a "-1" policy, I don't think it was ever really that regularly enforced just that a few regulars regularly voted for bad questions to top them up if they got a very negative score. I still do that occasionally if I notice one. @DavidCarlisle well, when I was new there was like never a question downvoted to more than (or less?) -1. And I liked it that way. My first question on SO got downvoted to -infty before I deleted it and fixed my issues on my own. @DavidCarlisle I meant the total. Still the general principle applies, when you're new and your question gets donwvoted too much this might cause the wrong impressions. @DavidCarlisle oh, subjectively I'd downvote that answer 10 times, but objectively it is not a good answer and might get a downvote from me, as you don't provide any reasoning for that, and I think that there should be a bit of reasoning with the opinion based answers, some objective arguments why this is good. See for example the other Emacs answer (still subjectively a bad answer), that one is objectively good. @DavidCarlisle and that other one got no downvotes. @Skillmon yes but many people just join for a while and come form other sites where downvoting is more common so I think it is impossible to expect there is no multiple downvoting, the only way to have a -1 policy is to get people to upvote bad answers more. @UlrikeFischer even harder to get than a gold tikz-pgf badge. @cis I'm not in the US but.... "Describe" while it does have a technical meaning close to what you want is almost always used more casually to mean "talk about", I think I would say" Let k be a circle with centre M and radius r" @AlanMunn definitions.net/definition/describe gives a websters definition of to represent by drawing; to draw a plan of; to delineate; to trace or mark out; as, to describe a circle by the compasses; a torch waved about the head in such a way as to describe a circle If you are really looking for alternatives to "draw" in "draw a circle" I strongly suggest you hop over to english.stackexchange.com and confirm to create an account there and ask ... at least the number of native speakers of English will be bigger there and the gamification aspect of the site will ensure someone will rush to help you out. Of course there is also a chance that they will repeat the advice you got here; to use "draw". @0xC0000022L @cis You've got identical responses here from a mathematician and a linguist. And you seem to have an idea that because a word is informal in German, its translation in English is also informal. This is simply wrong. And formality shouldn't be an aim in and of itself in any kind of writing. @0xC0000022L @DavidCarlisle Do you know the book "The Bronstein" in English? I think that's a good example of archaic mathematician language. But it is still possible harder. Probably depends heavily on the translation. @AlanMunn I am very well aware of the differences of word use between languages (and my limitations in regard to my knowledge and use of English as non-native speaker). In fact words in different (related) languages sharing the same origin is kind of a hobby. Needless to say that more than once the contemporary meaning didn't match a 100%. However, your point about formality is well made. A book - in my opinion - is first and foremost a vehicle to transfer knowledge. No need to complicate matters by trying to sound ... well, overly sophisticated (?) ... The following MWE with showidx and imakeidx:\documentclass{book}\usepackage{showidx}\usepackage{imakeidx}\makeindex\begin{document}Test\index{xxxx}\printindex\end{document}generates the error:! Undefined control sequence.<argument> \ifdefequal{\imki@jobname }{\@idxfile }{}{... @EmilioPisanty ok I see. That could be worth writing to the arXiv webmasters as this is indeed strange. However, it's also possible that the publishing of the paper got delayed; AFAIK the timestamp is only added later to the final PDF. @EmilioPisanty I would imagine they have frozen the epoch settings to get reproducible pdfs, not necessarily that helpful here but..., anyway it is better not to use \today in a submission as you want the authoring date not the date it was last run through tex and yeah, it's better not to use \today in a submission, but that's beside the point - a whole lot of arXiv eprints use the syntax and they're starting to get wrong dates @yo' it's not that the publishing got delayed. arXiv caches the pdfs for several years but at some point they get deleted, and when that happens they only get recompiled when somebody asks for them again and, when that happens, they get imprinted with the date at which the pdf was requested, which then gets cached Does any of you on linux have issues running for foo in *.pdf ; do pdfinfo $foo ; done in a folder with suitable pdf files? BMy box says pdfinfo does not exist, but clearly do when I run it on a single pdf file. @EmilioPisanty that's a relatively new feature, but I think they have a new enough tex, but not everyone will be happy if they submit a paper with \today and it comes out with some arbitrary date like 1st Jan 1970 @DavidCarlisle add \def\today{24th May 2019} in INITEX phase and recompile the format daily? I agree, too much overhead. They should simply add "do not use \today" in these guidelines: arxiv.org/help/submit_tex @yo' I think you're vastly over-estimating the effectiveness of that solution (and it would not solve the problem with 20+ years of accumulated files that do use it) @DavidCarlisle sure. I don't know what the environment looks like on their side so I won't speculate. I just want to know whether the solution needs to be on the side of the environment variables, or whether there is a tex-specific solution @yo' that's unlikely to help with prints where the class itself calls from the system time. @EmilioPisanty well th eenvironment vars do more than tex (they affect the internal id in teh generated pdf or dvi and so produce reproducible output, but you could as @yo' showed redefine \today oor teh \year, \month\day primitives on teh command line @EmilioPisanty you can redefine \year \month and \day which catches a few more things, but same basic idea @DavidCarlisle could be difficult with inputted TeX files. It really depends on at which phase they recognize which TeX file is the main one to proceed. And as their workflow is pretty unique, it's hard to tell which way is even compatible with it. "beschreiben", engl. "describe" comes from the math. technical-language of the 16th Century, that means from Middle High German, and means "construct" as much. And that from the original meaning: describe "making a curved movement". In the literary style of the 19th to the 20th century and in the GDR, this language is used. You can have that in englisch too: scribe(verb) score a line on with a pointed instrument, as in metalworking https://www.definitions.net/definition/scribe @cis Yes, as @DavidCarlisle pointed out, there is a very technical mathematical use of 'describe' which is what the German version means too, but we both agreed that people would not know this use, so using 'draw' would be the most appropriate term. This is not about trendiness, just about making language understandable to your audience. Plan figure. The barrel circle over the median $s_b = \|M_b B\|$, which holds the angle $\alpha$, also contains an isosceles triangle $M_b P B$ with the base $\|M_b B\|$ and the angle $\alpha$ at the point $P$. The altitude of the base of the isosceles triangle bisects both $\|M_b B\|$ at $ M_ {s_b}$ and the angle $\alpha$ at the top. \par The centroid $S$ divides the medians in the ratio $2:1$, with the longer part lying on the side of the corner. The point $A$ lies on the barrel circle and on a circle $\bigodot(S,\frac23 s_a)$ described by $S$ of radius…
$\log_{b}{(m^{\displaystyle n})}$ $\,=\,$ $n\log_{b}{m}$ Power Rule of logarithm reveals that log of a quantity in exponential form is equal to the product of exponent and logarithm of base of the exponential term. $q$ is a quantity and it is expressed in exponential form as $m^{\displaystyle n}$. Therefore, $q \,=\, m^{\displaystyle n}$. The logarithm of quantity to a base ($b$) is written as $\log_{b}{q}$. The value of $\log_{b}{q}$ can be calculated by calculating $\log_{b}{(m^{\displaystyle n})}$. $\log_{b}{q}$ $\,=\,$ $\log_{b}{(m^{\displaystyle n})}$ The value of logarithmic terms like $\log_{b}{(m^{\displaystyle n})}$ can be calculated by power law identity of logarithms. The power law property is actually derived by the power rule of exponents and relation between exponent and logarithmic operations. Take, $m$ is a quantity and it is expressed in exponential form on the basis of another quantity $b$. The total number of multiplying factors of $b$ is $x$ for representing the quantity $m$. $m$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle x \, factors}$ $\implies m \,=\, b^{\displaystyle x}$ $\,\,\, \therefore \,\,\,\,\,\, b^{\displaystyle x} \,=\, m$ $n$ is a quantity and take $n$-th power both sides of the exponential form equation. $\implies {(b^{\displaystyle x})}^{\displaystyle n} \,=\, m^{\displaystyle n}$ As per power rule of exponents, the whole power of a quantity in exponential form is equal to base is raised to the power of product of exponents. $\implies b^{\displaystyle nx} \,=\, m^{\displaystyle n}$ Take $y \,=\, nx$ and $z \,=\, m^{\displaystyle n}$. $\implies b^{\displaystyle y} \,=\, z$ Express equation in logarithmic form as per the mathematical relation between exponents and logarithms. $\implies y \,=\, \log_{b}{z}$ Replace the literals $y$ and $z$ by their respective values. $\implies nx \,=\, \log_{b}{m^{\displaystyle n}}$ $\implies \log_{b}{m^{\displaystyle n}} \,=\, nx$ It is taken in the first step that $b^{\displaystyle x} \,=\, m$ and it can be expressed in logarithmic form as $x \,=\, \log_{b}{m}$. So, replace the value of $x$ in logarithmic form in the above equation for deriving the power rule of logarithms in algebraic form. $\,\,\, \therefore \,\,\,\,\,\, \log_{b}{(m^{\displaystyle n})}$ $\,=\,$ $n\log_{b}{m}$ Therefore, it is proved that logarithm of a quantity in exponential form is equal to product of the exponent and log of the base of the exponent. Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
Determine all eigenvalues of the matrix $$A=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$$ and then determine a base for each eigenspace. It's easy to compute $\chi_A(z)=z^3-1$ so my roots (and therefore eigenvalues) are $z_1=1, z_2=\cos(2\pi/3)+i \sin(2\pi/3)$ and $z_3=\cos(4\pi/3)+i \sin(4\pi/3)$. Usually I would determine the eigenspaces by $E_\lambda=\ker(A-I_n\lambda)$, but having the solution to this problem shows that the result should be $$E_1=\left\langle\begin{bmatrix}1\\1\\1\end{bmatrix}\right\rangle,\qquad\qquad E_{z_2}=\left\langle\begin{bmatrix}z_3\\z_2\\1\end{bmatrix}\right\rangle,\qquad\qquad E_{z_3}=\left\langle\begin{bmatrix}1\\z_2\\z_3\end{bmatrix}\right\rangle.$$ The first one is obvious, but I don't see where the trick is to quickly compute the other two eigenspaces / eigenvectors! Any help would be appreciated!
Finding a $\gamma$ to define a number field like $E=\mathbb{Q}(\zeta_5)(X^5-\gamma)$ By working with eliptic curves, I found that the extension E defined by: E.<a> = NumberField(x^20 - 2x^19 - 2x^18 + 18x^17 - 32x^16 + 88x^15 + 58x^14 - 782x^13 + 1538x^12 + 1348x^11 - 466x^10 - 894x^9 + 346x^8 - 114x^7 - 424x^6 - 88x^5 + 214x^4 + 54x^3 + 14x^2 + 4*x + 1) Is a cyclic Kummer extension of degree $5$ over $\mathbb{Q}(\zeta_5)$, thus by the clasification of Kummer extensions, there exists $\gamma \in \mathbb{Q}(\zeta_5)$ such that $E=\mathbb{Q}(\sqrt[5]{\gamma})$. So, about all the polynomials $f$ that define $E$ how do I find $\gamma$ such that the polynomial $X^5-\gamma$ also defines $E$?
Asked by: Question Hi, Please help me how to load latex equation in windows 8.1 application. I found this https://math4winrt.codeplex.com/SourceControl/latest but not working for all math formulas. I wont use webview becoz i have to load more than 20 equations in my project. please help me by giving source code that it will render into canvas or someother UI element. Thanks sarvesh. sarvesh All replies Hi SARVESH.RVN, Could you do some modification with your equations in your code to display it using the Win2D as the sample? Best regards, Breeze MSDN Community Support Please remember to click "Mark as Answer" the responses that resolved your issue, and to click "Unmark as Answer" if not. This can be beneficial to other community members reading this thread. If you have any compliments or complaints to MSDN Support, feel free to contact MSDNFSF@microsoft.com. Hi SARVESH.RVN, The WebView is a control that hosts HTML content in an app, can you code your equation to be a HTML content? If that, it can display it in the WebView. Best regards, Breeze MSDN Community Support Please remember to click "Mark as Answer" the responses that resolved your issue, and to click "Unmark as Answer" if not. This can be beneficial to other community members reading this thread. If you have any compliments or complaints to MSDN Support, feel free to contact MSDNFSF@microsoft.com. Hi SARVESH.RVN, I am not familiar with your latex, but I don't think it is your last choice to use the WebView. The simplest way is to image your latex and display it in the Image control. Best regards, Breeze MSDN Community Support Please remember to click "Mark as Answer" the responses that resolved your issue, and to click "Unmark as Answer" if not. This can be beneficial to other community members reading this thread. If you have any compliments or complaints to MSDN Support, feel free to contact MSDNFSF@microsoft.com. Hi SARVESH.RVN, Could you see your latex in any other condition(the shape you can see)? You can snip it and save it as a .pngthen put it in your solution Assetsfolder. After that you can display it in the Image control by setting its source to this png file. Best regards, Breeze MSDN Community Support Please remember to click "Mark as Answer" the responses that resolved your issue, and to click "Unmark as Answer" if not. This can be beneficial to other community members reading this thread. If you have any compliments or complaints to MSDN Support, feel free to contact MSDNFSF@microsoft.com. Hi I have latex with me 1) "\sqrt{\frac{7}{15}}" 2) "\epsilon^k=\sqrt{\frac{\int_a^b\gamma(t)}{\sum_{i,j=1}^n\frac{i}{15}g_{ij}}}" 3) "\rtl{{\mathrm S}_55\frac1{2\cdot}}" and it will dynamically change(it will come from server) I have to show in my app, please help me how to do it ? sarvesh Hello, >>it will dynamically change(it will come from server) Could you please tell us what do you mean about it will dynamically changed from server? For example you have this equation: " ". It should render the UI in the Win2D CanvasControl as following if you followed this sample code: \sqrt{\frac{7}{15}} Do you mean that the parameter 7and 15will be changed dynamically or the equation type will be changed dynamically? If so you can use the HttpClient API to get the server response to update your parameter data or your equation type. In order to get the latest data, you may need to use the DispatcherTimer or a BackgroundTask(TimerTrigger) to refresh the server request and get the latest data based on your requirement. If I have misunderstood you, welcome to share your idea in here. Thanks, Amy Peng We are trying to better understand customer views on social support experience, so your participation in this interview project would be greatly appreciated if you have time. Thanks for helping make community forums a great place. Click HERE to participate the survey. Edited by Amy PengMicrosoft employee, Owner Thursday, September 28, 2017 7:23 AM Hi , I already got that code, but it not working for all formulas(I mentioned in my first question about this project) it will update means from server I will get different equations at different time(it is not matter here once we convert latex in to equation) what I am saying is I have to convert dynamically. Thanks sarvesh Hello,Currently there isn’t an in-box equation API in our UWP. You will need to either do all of the formatting by yourself or find a 3rd party library. >>I already got that code, but it not working for all formulas This math4winrt library is the 3rd party library, if this library does not work for all your formulas, you will need to either contact the library’s author or modify it by yourself. >>I wont use webview becoz i have to load more than 20 equations in my project. If you need to use the WebView, you still need to find a 3rd party service or library that can convert the equation into HTML or a bitmap. Besides, you can try to submit your idea and suggestion in our UserVoice website for the equation related API.If it is very urgent, please use your developer account to open a support case by visiting this URL(https://support.microsoft.com/en-us/getsupport?wf=0&tenant=ClassicCommercial&oaspworkflow=start_1.0.0.0&locale=en-us&supportregion=en-us&pesid=15944&ccsid=636422725863896132). You will get 1:1 support on that. Besides, please kindly note that your support ticket will be free if it is Microsoft's issue. Thanks for your understanding. Thanks, Amy Peng We are trying to better understand customer views on social support experience, so your participation in this interview project would be greatly appreciated if you have time. Thanks for helping make community forums a great place. Click HERE to participate the survey. Edited by Amy PengMicrosoft employee, Owner Friday, September 29, 2017 9:03 AM
The value of cot function when angle of right triangle equals to $45^°$ is called cot of angle $45$ degrees. As per sexagesimal system, it is written as $\cot{(45^°)}$ in mathematics. $\cot{(45^°)} \,=\, 1$ The exact value of cot of angle $45$ degrees is equal to $1$ and it is also an integer. $\cot{(45^°)}$ can be written in alternative form as $\cot{\Big(\dfrac{\pi}{4}\Big)}$ in circular system and also written as $\cot{(50^g)}$ in centesimal system. $(1) \,\,\,$ $\cot{\Big(\dfrac{\pi}{4}\Big)} \,=\, 1$ $(2) \,\,\,$ $\cot{(50^g)} \,=\, 1$ Now, you learnt the exact value of cot of angle $45$ degrees and you can also learn how to derive $\cot{\Big(\dfrac{\pi}{4}\Big)}$ in trigonometry. Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
It is a trigonometric expression which contains sine and cosine functions with complementary angles. This trigonometric problem can be simplified in two methods by using complementary angle trigonometric identities. In this method, all trigonometric functions which contain complementary angles are simplified firstly by using cofunction identities. $\dfrac{\cos{(90^°-\theta)}}{1+\sin{(90^°-\theta)}}$ $+$ $\dfrac{1+\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}}$ The sin of angle is equal to cos of complementary angle as per cofunction identity of sin function and the cos of angle is equal to sin of complementary angle as per cofunction identity of cos function. $= \,\,\,$ $\dfrac{\sin{\theta}}{1+\cos{\theta}}$ $+$ $\dfrac{1+\cos{\theta}}{\sin{\theta}}$ The two terms can be simplified by using least common multiple method for making them as a single term trigonometric expression. $= \,\,\,$ $\dfrac{\sin{\theta} \times \sin{\theta} + {(1+\cos{\theta})}{(1+\cos{\theta})}}{\sin{\theta}{(1+\cos{\theta})}}$ $= \,\,\,$ $\dfrac{\sin^2{\theta}+{(1+\cos{\theta})}^2}{\sin{\theta}{(1+\cos{\theta})}}$ The second term in the numerator is a square of sum of two terms and it can be expanded by using square of sum of two terms formula like ${(a+b)}^2$. $= \,\,\,$ $\dfrac{\sin^2{\theta}+1^2+\cos^2{\theta}+2 \times 1 \times \cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$ $= \,\,\,$ $\dfrac{\sin^2{\theta}+1+\cos^2{\theta}+2\cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$ $= \,\,\,$ $\dfrac{1+\sin^2{\theta}+\cos^2{\theta}+2\cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$ The sum of squares of sin and cos functions is one at an angle as per Pythagorean identity of sin and cos functions. $= \,\,\,$ $\dfrac{1+1+2\cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$ $= \,\,\,$ $\dfrac{2+2\cos{\theta}}{\sin{\theta}{(1+\cos{\theta})}}$ $= \,\,\,$ $\dfrac{2(1+\cos{\theta})}{\sin{\theta}{(1+\cos{\theta})}}$ $= \,\,\,$ $\require{\cancel} \dfrac{2\cancel{(1+\cos{\theta})}}{\sin{\theta}\cancel{(1+\cos{\theta})}}$ $= \,\,\,$ $\dfrac{2}{\sin{\theta}}$ According to reciprocal identity of sin function, the reciprocal of sin function is equal to cosecant function. $= \,\,\,$ $2\csc{\theta}$ It is a method for those who have basic knowledge on trigonometry and have a confidence about dealing trigonometric functions which contain complementary angles. In this method, no trigonometric ratio with complementary angle is transformed till last step. $\dfrac{\cos{(90^°-\theta)}}{1+\sin{(90^°-\theta)}}$ $+$ $\dfrac{1+\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}}$ Firstly, combine the terms in the form of trigonometric function by least common multiple method. $= \,\,\,$ $\dfrac{\cos{(90^°-\theta)} \times \cos{(90^°-\theta)} + {(1+\sin{(90^°-\theta)})}{(1+\sin{(90^°-\theta)})}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$ $= \,\,\,$ $\dfrac{\cos^2{(90^°-\theta)} + {(1+\sin{(90^°-\theta)})}^2}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$ Now, expand the square of sum of two terms in the numerator. $= \,\,\,$ $\dfrac{\cos^2{(90^°-\theta)} + 1^2+\sin^2{(90^°-\theta)}+2 \times 1 \times \sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$ $= \,\,\,$ $\dfrac{\cos^2{(90^°-\theta)} + 1 +\sin^2{(90^°-\theta)}+2\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$ $= \,\,\,$ $\dfrac{1+\cos^2{(90^°-\theta)}+\sin^2{(90^°-\theta)}+2\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$ As per Pythagorean identity of sin and cos functions, the sum of squares of sin and cos functions at an angle is equal to one. $= \,\,\,$ $\dfrac{1+1+2\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$ $= \,\,\,$ $\dfrac{2+2\sin{(90^°-\theta)}}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$ $= \,\,\,$ $\dfrac{2(1+\sin{(90^°-\theta)})}{\cos{(90^°-\theta)}{(1+\sin{(90^°-\theta)})}}$ $= \,\,\,$ $\require{cancel} \dfrac{2\cancel{(1+\sin{(90^°-\theta)})}}{\cos{(90^°-\theta)}\cancel{(1+\sin{(90^°-\theta)})}}$ $= \,\,\,$ $\dfrac{2}{\cos{(90^°-\theta)}}$ The reciprocal of cos of angle is equal to secant of angle as per reciprocal rule of cos function. $= \,\,\,$ $2\sec{(90^°-\theta)}$ Finally, the secant of angle is equal to cosecant of complementary angle as per complementary angle identity of secant function. $= \,\,\,$ $2\csc{\theta}$ Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
Are there matrices that satisfy these two conditions? That is, a matrix $A$ such that $$A^T=A^{-1}=-A$$ What I know is that a skew-symmetric matrix with $n$ dimensions is singular when $n$ is odd. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community The matrix $$A=\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$$ is skew-symmetric and orthogonal. In even dimensions, we can always construct a skew-symmetric and orthogonal matrix as the direct sum of multiple copies of $A$. i.e. the matrix $$\bigoplus_{i=1}^k A=\underbrace{\begin{bmatrix} A &&&\\ &A&&\\ &&\ddots & \\ &&&A\\ \end{bmatrix}}_{k \text{ copies}}$$ is a $2k \times 2k$ orthogonal and skew-symmetric matrix In odd dimensions however, there is no real matrices which are skew-symmetric and orthogonal. As you already know, real skew-symmetric matrices are singular in odd dimensions so they must have at least one eigenvalue which is zero. Therefore if $B$ is an $n\times n$ matrix with $n$ odd, it must be the case that $\text{det}(B)=0$. If $B$ is orthogonal however, $B^TB=I$ and so $$\text{det}(B^TB)=\text{det}(B^T)\text{det}(B)=[\text{det}(B)]^2=\text{det}(I)=1$$ thus $\text{det}(B)= \pm 1$ and so $B$ can not be both orthogonal and skew-symmetric in odd dimensions. A general classification: Note that any such matrix is normal, hence unitarily diagonalizable. Any orthogonal matrix is unitary, so all of it's eigenvalues are contained in the unit circle $C=\{z:\|z\|=1\}$ of the complex plane. Similarly any skew-symmetric has eigenvalues in $i \Bbb R$, i.e., purely imaginary. Thus the orthogonal, skew-symmetric matrices are precisely those matrices whose eigenvalues lie in $C \cap i \Bbb R = \{-i,i\}$, and which are unitarily diagonalizable. In odd dimensions, the characteristic polynomial has at least one real root, hence no such matrices exist. In even dimensions, such matrices are precisely those whose eigenbasis is orthonormal, and whose characteristic polynomial is $(z-i)^{d_1}(z+i)^{d_2}$ where $d_1+d_2=d$. If we want the matrix to have real coefficients, then necessarily $d_1=d_2=d/2$, and thus the characteristic polynomial is $(z^2+1)^{d/2}$. This implies (by orthonormal diagonalizability), that the matrix mentioned by @Alex is the only real skew symmetric and orthogonal matrix, up to an orthogonal change of basis (i.e. conjugation in $O(d, \Bbb R)$).
Linear Boltzmann equation and fractional diffusion 1. Laboratoire J.-L. Lions, BP 187, 75252 Paris Cedex 05, France 2. CMLS, École polytechnique, 91128 Palaiseau Cedex, France 3. DPMMS, University of Cambridge, Wilberforce Road, CB3 0WA Cambridge, United Kingdom Consider the linear Boltzmann equation of radiative transfer in a half-space, with constant scattering coefficient $\sigma$. Assume that, on the boundary of the half-space, the radiation intensity satisfies the Lambert (i.e. diffuse) reflection law with albedo coefficient $\alpha$. Moreover, assume that there is a temperature gradient on the boundary of the half-space, which radiates energy in the half-space according to the Stefan-Boltzmann law. In the asymptotic regime where $\sigma\to+∞$ and $ 1-\alpha \sim C/\sigma$, we prove that the radiation pressure exerted on the boundary of the half-space is governed by a fractional diffusion equation. This result provides an example of fractional diffusion asymptotic limit of a kinetic model which is based on the harmonic extension definition of $\sqrt{-\Delta}$. This fractional diffusion limit therefore differs from most of other such limits for kinetic models reported in the literature, which are based on specific properties of the equilibrium distributions ("heavy tails") or of the scattering coefficient as in [U. Frisch-H. Frisch: Mon. Not. R. Astr. Not. 181 (1977), 273-280]. Keywords:Linear Boltzmann equation, radiative transfer equation, diffusion approximation, fractional diffusion. Mathematics Subject Classification:Primary: 45K05, 45M05, 35R11; Secondary: 82C70, 85A25. Citation:Claude Bardos, François Golse, Ivan Moyano. Linear Boltzmann equation and fractional diffusion. Kinetic & Related Models, 2018, 11 (4) : 1011-1036. doi: 10.3934/krm.2018039 References: [1] [2] C. Bardos, F. Golse, B. Perthame and R. Sentis, The nonaccretive radiative transfer equations, existence of solutions and Rosseland approximation, [3] C. Bardos, E. Bernard, F. Golse and R. Sentis, The diffusion approximation for the linear Boltzmann equation with vanishing scattering coefficient, [4] [5] N. Ben Abdallah, A. Mellet and M. Puel, Anomalous diffusion limit for kinetic equations with degenerate collision frequency, [6] A. Bensoussan, J.-L. Lions and G.C. Papanicolaou, Boundary layers and homogenization of transport processes, [7] H. Brezis, [8] [9] [10] S. Chandrasekhar, [11] R. Dautray and J.-L. Lions, [12] L. Desvillettes and F. Golse, A remark concerning the Chapman-Enskog asymptotics, in [13] [14] F. Golse, Fluid dynamic limits of the kinetic theory of gases, in [15] [16] [17] [18] [19] [20] [21] [22] G. C. Pomraning, [23] L. Tartar, [24] A. Weinberg and E. Wigner, show all references References: [1] [2] C. Bardos, F. Golse, B. Perthame and R. Sentis, The nonaccretive radiative transfer equations, existence of solutions and Rosseland approximation, [3] C. Bardos, E. Bernard, F. Golse and R. Sentis, The diffusion approximation for the linear Boltzmann equation with vanishing scattering coefficient, [4] [5] N. Ben Abdallah, A. Mellet and M. Puel, Anomalous diffusion limit for kinetic equations with degenerate collision frequency, [6] A. Bensoussan, J.-L. Lions and G.C. Papanicolaou, Boundary layers and homogenization of transport processes, [7] H. Brezis, [8] [9] [10] S. Chandrasekhar, [11] R. Dautray and J.-L. Lions, [12] L. Desvillettes and F. Golse, A remark concerning the Chapman-Enskog asymptotics, in [13] [14] F. Golse, Fluid dynamic limits of the kinetic theory of gases, in [15] [16] [17] [18] [19] [20] [21] [22] G. C. Pomraning, [23] L. Tartar, [24] A. Weinberg and E. Wigner, [1] Arnaud Debussche, Sylvain De Moor, Julien Vovelle. Diffusion limit for the radiative transfer equation perturbed by a Wiener process. [2] [3] [4] Luis Caffarelli, Juan-Luis Vázquez. Asymptotic behaviour of a porous medium equation with fractional diffusion. [5] Eugenio Montefusco, Benedetta Pellacci, Gianmaria Verzini. Fractional diffusion with Neumann boundary conditions: The logistic equation. [6] Stefan Possanner, Claudia Negulescu. Diffusion limit of a generalized matrix Boltzmann equation for spin-polarized transport. [7] [8] [9] Kenichi Sakamoto, Masahiro Yamamoto. Inverse source problem with a final overdetermination for a fractional diffusion equation. [10] [11] [12] N. Ben Abdallah, M. Lazhar Tayeb. Diffusion approximation for the one dimensional Boltzmann-Poisson system. [13] Marc Briant. Instantaneous exponential lower bound for solutions to the Boltzmann equation with Maxwellian diffusion boundary conditions. [14] [15] Marie Henry, Danielle Hilhorst, Masayasu Mimura. A reaction-diffusion approximation to an area preserving mean curvature flow coupled with a bulk equation. [16] Hao Yang, Fuke Wu, Peter E. Kloeden. Existence and approximation of strong solutions of SDEs with fractional diffusion coefficients. [17] [18] [19] [20] 2018 Impact Factor: 1.38 Tools Metrics Other articles by authors [Back to Top]
X Search Filters Format Subjects Library Location Language Publication Date Click on a bar to filter by decade Slide to change publication date range 1. Precision measurement and interpretation of inclusive W+, W−and Z/γ production cross sections with the ATLAS detector The European Physical Journal C: Particles and Fields, ISSN 1434-6052, 2017, Volume 77, Issue 6, pp. 1 - 62 High-precision measurements by the ATLAS Collaboration are presented of inclusive W+ -> l(+) nu, W- -> l(-) (nu) over bar and Z/gamma* -> ll (l = e, mu)... MONTE-CARLO \| DRELL-YAN \| INTEGRATOR \| PARTON DISTRIBUTIONS \| DECAY \| PROTON \| QCD ANALYSIS \| EP SCATTERING \| COMBINATION \| PHYSICS, PARTICLES & FIELDS \| Protons \| Luminosity \| Large Hadron Collider \| Leptons \| Scattering cross sections \| Distribution functions \| Physics - High Energy Physics - Experiment \| Physics \| High Energy Physics - Experiment \| Engineering (miscellaneous); Physics and Astronomy (miscellaneous) \| Subatomic Physics \| Astrophysics \| Settore FIS/01 - Fisica Sperimentale \| Settore ING-INF/07 - Misure Elettriche e Elettroniche \| Experiment \| Nuclear and particle physics. Atomic energy. Radioactivity \| Science & Technology \| Settore FIS/04 - Fisica Nucleare e Subnucleare \| Phenomenology \| High Energy Physics \| hep-ex, hep-ex \| Nuclear Experiment \| Subatomär fysik \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences MONTE-CARLO \| DRELL-YAN \| INTEGRATOR \| PARTON DISTRIBUTIONS \| DECAY \| PROTON \| QCD ANALYSIS \| EP SCATTERING \| COMBINATION \| PHYSICS, PARTICLES & FIELDS \| Protons \| Luminosity \| Large Hadron Collider \| Leptons \| Scattering cross sections \| Distribution functions \| Physics - High Energy Physics - Experiment \| Physics \| High Energy Physics - Experiment \| Engineering (miscellaneous); Physics and Astronomy (miscellaneous) \| Subatomic Physics \| Astrophysics \| Settore FIS/01 - Fisica Sperimentale \| Settore ING-INF/07 - Misure Elettriche e Elettroniche \| Experiment \| Nuclear and particle physics. Atomic energy. Radioactivity \| Science & Technology \| Settore FIS/04 - Fisica Nucleare e Subnucleare \| Phenomenology \| High Energy Physics \| hep-ex, hep-ex \| Nuclear Experiment \| Subatomär fysik \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Journal Article Tissue Engineering Part A, ISSN 1937-3341, 12/2016, Volume 22, Issue S1, pp. S-1 - S-156 To better understand the disease mechanism, we generated human induced pluripotent stem cell-derived cardiomyocytes (hiPSC-CMs) from the family cohort carrying... Abstracts \| Myocardial infarction \| Heart \| Bone grafts \| Prostheses \| Conduction \| Heart attacks \| Cardiomyopathy \| Matrices (mathematics) \| Surface chemistry \| Radiation \| Cardiovascular disease \| Incidence \| Proteins \| Coding \| Surgery \| Cell adhesion \| Biocompatibility \| Growth factors \| Deoxyribonucleic acid--DNA \| Wound healing \| Crosslinking \| Gene expression \| Carriers \| Chemistry \| Stem cells \| Cytoskeleton \| Mice \| Skin \| Mutation \| Kinetics \| Kidney transplantation \| Surgical implants \| Animal models \| Transplants & implants \| Hydrogels \| Calcium \| Chemical engineering \| Mitosis \| Medical services \| Risk \| Action potential \| Stimulation \| Transplantation \| DNA repair \| Contraction \| Cell adhesion & migration \| Engineering \| Pain \| Energy \| Actin \| Breast implants \| Reaction kinetics \| Heart diseases \| Deposition \| Adenosine triphosphate \| CRISPR \| Kidneys \| Research & development--R&D \| Complications \| Health risks \| Mechanical properties \| Cardiomyocytes \| Histology \| Inflammation \| Rhythms \| Coronary artery disease \| Medicine \| Collagen \| Ulcers \| Breast \| Protein expression \| Medical wastes \| Coating \| Differentiation \| ATP \| Potassium \| Pluripotency \| Combinatorial analysis Abstracts \| Myocardial infarction \| Heart \| Bone grafts \| Prostheses \| Conduction \| Heart attacks \| Cardiomyopathy \| Matrices (mathematics) \| Surface chemistry \| Radiation \| Cardiovascular disease \| Incidence \| Proteins \| Coding \| Surgery \| Cell adhesion \| Biocompatibility \| Growth factors \| Deoxyribonucleic acid--DNA \| Wound healing \| Crosslinking \| Gene expression \| Carriers \| Chemistry \| Stem cells \| Cytoskeleton \| Mice \| Skin \| Mutation \| Kinetics \| Kidney transplantation \| Surgical implants \| Animal models \| Transplants & implants \| Hydrogels \| Calcium \| Chemical engineering \| Mitosis \| Medical services \| Risk \| Action potential \| Stimulation \| Transplantation \| DNA repair \| Contraction \| Cell adhesion & migration \| Engineering \| Pain \| Energy \| Actin \| Breast implants \| Reaction kinetics \| Heart diseases \| Deposition \| Adenosine triphosphate \| CRISPR \| Kidneys \| Research & development--R&D \| Complications \| Health risks \| Mechanical properties \| Cardiomyocytes \| Histology \| Inflammation \| Rhythms \| Coronary artery disease \| Medicine \| Collagen \| Ulcers \| Breast \| Protein expression \| Medical wastes \| Coating \| Differentiation \| ATP \| Potassium \| Pluripotency \| Combinatorial analysis Journal Article Book Book 2014, Gene transfer and gene therapy Web Resource 6. Measurement of W±W± vector-boson scattering and limits on anomalous quartic gauge couplings with the ATLAS detector Physical Review D, ISSN 2470-0010, 01/2017, Volume 96, Issue 1 Journal Article 7. Search for triboson $W^{\pm }W^{\pm }W^{\mp }$ production in pp collisions at $\sqrt{s}=8$ $\text {TeV}$ with the ATLAS detector European Physical Journal. C, Particles and Fields, ISSN 1434-6044, 03/2017, Volume 77, Issue 3 This paper reports a search for triboson W±W±W∓ production in two decay channels (W±W±W∓ → ℓ±νℓ±νℓ∓ν and W±W±W∓ → ℓ±νℓ±νjj with ℓ = e,μ) in proton-proton... PHYSICS OF ELEMENTARY PARTICLES AND FIELDS PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Journal Article The European Physical Journal C: Particles and Fields, ISSN 1434-6052, 2017, Volume 77, Issue 5, pp. 1 - 53 During 2015 the ATLAS experiment recorded $$3.8\,{\mathrm{fb}}^{-1}$$ 3.8 fb - 1 of proton–proton collision data at a centre-of-mass energy of... PHYSICS, PARTICLES & FIELDS \| Collisions (Nuclear physics) \| Protons \| Data acquisition systems \| Physics - High Energy Physics - Experiment \| High Energy Physics - Phenomenology \| Physics \| PARTICLE ACCELERATORS \| Regular - Experimental Physics \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences PHYSICS, PARTICLES & FIELDS \| Collisions (Nuclear physics) \| Protons \| Data acquisition systems \| Physics - High Energy Physics - Experiment \| High Energy Physics - Phenomenology \| Physics \| PARTICLE ACCELERATORS \| Regular - Experimental Physics \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Journal Article 2004, ISBN 9780306485534, xx, 589 Book 10. Constraints on new phenomena via Higgs boson couplings and invisible decays with the ATLAS detector Journal of High Energy Physics, ISSN 1126-6708, 2015, Volume 2015, Issue 11, pp. 1 - 52 The ATLAS experiment at the LHC has measured the Higgs boson couplings and mass, and searched for invisible Higgs boson decays, using multiple production and... Supersymmetry \| Higgs physics \| Dark matter \| Hadron-Hadron Scattering \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| MASSES \| HADRON COLLIDERS \| PARTICLE \| SEARCH \| LHC \| COLLISIONS \| STANDARD MODEL \| EXTRA DIMENSIONS \| DARK-MATTER \| PHYSICS, PARTICLES & FIELDS \| Couplings \| Decay rate \| Searching \| Decay \| Higgs bosons \| Texts \| Channels \| Constraining \| Physics - High Energy Physics - Experiment \| High Energy Physics - Experiment \| supersymmetry \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS \| higgs physics \| hadron-hadron scattering \| dark matter \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Supersymmetry \| Higgs physics \| Dark matter \| Hadron-Hadron Scattering \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| MASSES \| HADRON COLLIDERS \| PARTICLE \| SEARCH \| LHC \| COLLISIONS \| STANDARD MODEL \| EXTRA DIMENSIONS \| DARK-MATTER \| PHYSICS, PARTICLES & FIELDS \| Couplings \| Decay rate \| Searching \| Decay \| Higgs bosons \| Texts \| Channels \| Constraining \| Physics - High Energy Physics - Experiment \| High Energy Physics - Experiment \| supersymmetry \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS \| higgs physics \| hadron-hadron scattering \| dark matter \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Journal Article European Physical Journal C, ISSN 1434-6044, 2017, Volume 77, Issue 7, pp. 1 - 73 The reconstruction of the signal from hadrons and jets emerging from the proton–proton collisions at the Large Hadron Collider (LHC) and entering the ATLAS... Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology \| INTRANUCLEAR-CASCADE CALCULATION \| HADRONIC CALIBRATION \| NUCLEUS \| DETECTOR \| PHYSICS, PARTICLES & FIELDS \| Particle accelerators \| Electromagnetism \| Collisions (Nuclear physics) \| Hadrons \| Large Hadron Collider \| Particle collisions \| Accident reconstruction \| Transverse momentum \| Clusters \| Clustering \| Physics - High Energy Physics - Experiment \| High Energy Physics - Experiment \| PARTICLE ACCELERATORS \| Regular - Experimental Physics \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Nuclear Physics, Heavy Ions, Hadrons \| Measurement Science and Instrumentation \| Nuclear Energy \| Quantum Field Theories, String Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Astronomy, Astrophysics and Cosmology \| INTRANUCLEAR-CASCADE CALCULATION \| HADRONIC CALIBRATION \| NUCLEUS \| DETECTOR \| PHYSICS, PARTICLES & FIELDS \| Particle accelerators \| Electromagnetism \| Collisions (Nuclear physics) \| Hadrons \| Large Hadron Collider \| Particle collisions \| Accident reconstruction \| Transverse momentum \| Clusters \| Clustering \| Physics - High Energy Physics - Experiment \| High Energy Physics - Experiment \| PARTICLE ACCELERATORS \| Regular - Experimental Physics \| Fysik \| Physical Sciences \| Naturvetenskap \| Natural Sciences Journal Article 12. Search for heavy Majorana or Dirac neutrinos and right-handed $W$ gauge bosons in final states with two charged leptons and two jets at $\sqrt{s}$ = 13 TeV with the ATLAS detector Journal of High Energy Physics (Online), ISSN 1029-8479, 09/2018, Volume 2019, Issue 1 Journal Article
$x$ is a variable, which is considered as an angle of a right triangle and the sine function is written as $\sin{x}$ in trigonometric mathematics. The indefinite integral of $\sin{x}$ with respect to $x$ is written as follows to find the integration of sine function in calculus. $\displaystyle \int{\sin{x} \,}dx$ Write the derivative of cos function with respect to $x$ formula for expressing the differentiation of cosine function in mathematical form. $\dfrac{d}{dx}{\, \cos{x}} \,=\, -\sin{x}$ $\implies$ $\dfrac{d}{dx}{(-\cos{x})} \,=\, \sin{x}$ According to differential calculus, the derivative of a constant is always zero. So, it doesn’t affect the process of the differentiation if an arbitrary constant $(c)$ is added to the trigonometric function $-\cos{x}$. $\implies$ $\dfrac{d}{dx}{(-\cos{x}+c)} \,=\, \sin{x}$ The collection of all primitives of $\sin{x}$ function is called the indefinite integral of $\sin{x}$ function, which is written in the following mathematical form in integral calculus. $\displaystyle \int{\sin{x} \,}dx$ In this case, the primitive or an antiderivative of $\sin{x}$ is $-\cos{x}$ and the constant of integration $c$. $\dfrac{d}{dx}{(-\cos{x}+c)} = \sin{x}$ $\,\Longleftrightarrow\,$ $\displaystyle \int{\sin{x} \,}dx = -\cos{x}+c$ $\therefore \,\,\,\,\,\,$ $\displaystyle \int{\sin{x} \,}dx = -\cos{x}+c$ Therefore, it is proved that the antiderivative or indefinite integration of sine function is equal to the sum of the negative cos function and the constant of integration. Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
Reciprocal of Strictly Positive Real Number is Strictly Positive Jump to navigation Jump to search Theorem $\forall x \in \R: x > 0 \implies \dfrac 1 x > 0$ Proof Let $x > 0$. Aiming for a contradiction, suppose $\dfrac 1 x < 0$. Then: $\displaystyle x$ $>$ $\displaystyle 0$ $\displaystyle \leadsto \ \ $ $\displaystyle x \times \dfrac 1 x$ $<$ $\displaystyle 0 \times 0$ Order of Real Numbers is Dual of Order Multiplied by Negative Number $\displaystyle \leadsto \ \ $ $\displaystyle 1$ $<$ $\displaystyle 0$ Real Number Axioms: $\R M4$: Inverse But from Real Zero is Less than Real One: $1 > 0$ Therefore by Proof by Contradiction: $\dfrac 1 x > 0$ $\blacksquare$ Also see
Just so there are no misunderstandings let me first ask whether it is true that: $$ \int_{-\infty}^{\infty}x\delta(x)\mathrm{d}x=0. $$ If that is not true, then I don't know anything about the Dirac delta distribution and I will be off to correct this :) Otherwise I have this question: Why can we take a delta functional of $x$? As far as I've read, the Dirac delta is a tempered distribution and it must act on Schwartz functions. I am not convinced that this is true for $f(x)=x$.
$\dfrac{d}{dx}{\, (c)}$ According to definition of the derivative, the differentiation of $f{(x)}$ with respect to $x$ can be written in limit operation form. $\dfrac{d}{dx}{\, f{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{f{(x+\Delta x)}-f{(x)}}{\Delta x}}$ Take $f{(x)} \,=\, c$, then $f{(x+\Delta x)} \,=\, c$. Now, substitute them in this formula. $\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{c-c}{\Delta x}}$ Now, take $\Delta x = h$ and express the equation in terms of $h$ from $\Delta x$. $\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{c-c}{h}}$ There are two terms in the numerator and they both are equal. So, the subtraction of them is equal to zero. $\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\cancel{c}-\cancel{c}}{h}}$ $\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \Big(\dfrac{0}{h}\Big)}$ The quotient of zero by $h$ is zero mathematically. $\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize (0)}$ Now, calculate the limit of zero as $h$ approaches zero. In this case, there is no $h$ term in the function and the value of the function is zero. Therefore, the differentiation of a constant with respect to a variable is equal to zero. $\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $0$ In this way, the derivative of a constant rule is derived by first principle in the differential calculus. Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
Let a triangle be inscribed in a unit circle, and let $A$ and $B$ mark two vertices. Let $\theta$ be half the length of the arc connecting $A$ and $B$, and let $\ell$ be the length of the chord, you have that from elementary trigonometry $$ \ell / 2 = \sin (\theta) $$ Now let $\theta_1, \theta_2, \theta_3$ be half the lengths of the three arcs demarcated by the vertices of the triangle, from the above we have that the perimeter is equal to $$ 2 \left[ \sin (\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] $$ and we know that $$ \theta_1 + \theta_2 + \theta_3 = \pi $$ while $$ \theta_1, \theta_2, \theta_3 \in [0,\pi] $$ Now from the fact that $\sin$ is a concave function on $[0,\pi]$, we have that the perimeter is equal to $$ 6 * \frac13 \left[ \sin(\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] \leq 6 * \sin (\pi / 3) $$ by Jensen's inequality, with the optimum (being the case of the equality) holding when $\theta_1 = \theta_2 = \theta_3$.
Transformation semi-group Any sub-semi-group of a symmetric semi-group $T_\Omega$, where $T_\Omega$ is the set of all transformations of a set $\Omega$. Particular cases of transformation semi-groups are transformation groups (cf. Transformation group). Two transformation semi-groups $P_1 \subset T_{\Omega_1}$, $P_2 \subset T_{\Omega_2}$ are called similar if there are bijections $\phi : \Omega_1 \rightarrow \Omega_2$ and $\psi : P_1 \rightarrow P_2$ such that $u\alpha = \beta$ ($\alpha, \beta \in \Omega_1$, $u \in P_1$) implies $(\psi u) (\phi\alpha) = \phi\beta$. Similar transformation semi-groups are isomorphic, but the converse is, usually, not true. However, within some classes of transformation semi-groups isomorphism implies similarity. E.g., the class of transformation semi-groups that include all transformations $u$ such that $u\Omega$ consists of one element. The specification of a semi-group as a transformation semi-group includes more information than its specification up to isomorphism. Distinguishing properties of transformation semi-groups that are invariant under isomorphism is of prime importance. For a given class $\Gamma$ of transformation semi-groups, conditions under which a semi-group $S$ is isomorphic to some semi-group from $\Gamma$ are called abstract characteristics of the class $\Gamma$. Abstract characteristics for certain important classes of transformation semi-groups have been found. Every semi-group is isomorphic to some transformation semi-group. A semi-group $S$ is isomorphic to some symmetric semi-group $T_\Omega$ if it is a maximal complete ideal extension (cf. Extension of a semi-group) of any semi-group $A$ with the identity $xy=x$. One distinguishes directions in the general theory of transformation semi-groups in which the set $\omega$ to be transformed is endowed with a certain structure (a topology, an action, a relation in $\Omega$, etc.) and considers transformation semi-groups related to this structure (endomorphisms, continuous or linear transformations, translations of semi-groups, etc.). The study of relations between properties of the structure in $\Omega$ and properties of the semi-groups of corresponding transformations is a generalization of Galois theory. In particular, cases are known in which the indicated translation semi-group completely determines the structure (cf. e.g. Endomorphism semi-group). Properties of left and right translations of semi-groups are used in general semi-group theory. A generalization of the notion of a transformation is that of a partial transformation, mapping some subset $\Omega' \subset \Omega$ into $\Omega$. Binary relations on a set $\Omega$ are sometimes treated as multi-valued (in general, partial) transformations of this set. Single- and multi-valued partial transformations also form semi-groups under the operation of composition (regarded as multiplication of binary relations). It is expedient to regard them as semi-groups endowed with additional structures (e.g. the relation of inclusion of binary relations, inclusion or equality of domains of definition, inclusion or equality of ranges, etc.). References [1] E.S. Lyapin, "Semigroups" , Amer. Math. Soc. (1974) (Translated from Russian) [2] A.H. Clifford, G.B. Preston, "Algebraic theory of semi-groups" , 1–2 , Amer. Math. Soc. (1961–1967) [3] L.M. Gluskin, "Ideals of semigroups" Mat. Sb. , 55 : 4 (1961) pp. 421–428 (In Russian) [4] B.M. Schein, "Relation algebras and function semigroups" Semigroup Forum , 1 : 1 (1970) pp. 1–62 How to Cite This Entry: Transformation semi-group. Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Transformation_semi-group&oldid=36985
Convergence and, in fact, uniform convergence for case (1) follows from the Dirichlet test since $\displaystyle\left\|\int_1^c \cos x \, dx \right\| \leqslant2 $ for all $c > 1$ (uniformly bounded) and $x^{-\alpha} < x ^{-\alpha_0}$ which implies that $x^{-\alpha} \downarrow 0$ monotonically and uniformly for all $\alpha > \alpha_0$. For case (2), we have convergence since the argument for case (1) applies to any $\alpha_0 > 0$. However, the convergence is not uniform. Given sequences $\displaystyle c_n = -\frac{\pi}{4}+2\pi n$ and $\displaystyle d_n = \frac{\pi}{4}+ 2 \pi n$, we have $\cos x > 1/\sqrt{2}$ for $c_n \leqslant x \leqslant d_n$ and$$\left\|\int_{c_n}^{d_n} \frac{\cos x}{x^\alpha} \right\| \geqslant \frac{1}{d_n^\alpha}\int_{c_n}^{d_n} \cos x \, dx \geqslant \frac{1}{d_n^\alpha}\frac{\pi}{2 \sqrt{2}}.$$ Taking the sequence $\alpha_n = ( \log d_n)^{-1},$ we have $d_n^{\alpha_n} = \exp(\log d_n (\log d_n)^{-1})= e$ and, consequently, $$\tag{}\left\|\int_{c_n}^{d_n} \frac{\cos x}{x^\alpha_n} \right\| \geqslant \frac{\pi}{2 \sqrt{2}e}.$$ Since $c_n , d_n \to \infty$ and $\alpha_n \in (0,\infty)$ as $n \to \infty$, the Cauchy criterion for uniform convergence is violated. Note that uniform convergence would require that for any $\epsilon > 0$ there exists $K > 1$ such that for all $d> c> K$ and for any $\alpha \in (0,\infty)$ we have $$\left\|\int_{c}^{d} \frac{\cos x}{x^\alpha} \right\| < \epsilon ,$$ which is contradicted by ().
Good evening everyone, I'd like to discuss with you the following exercise : $\sum\limits_{n=1}^{\infty} (-1)^{n} \frac{n^{2} +3n - \sin(n)}{n^{4}-\arctan(n^{2})}$ I can prove that $\lim\limits_{x \to \infty} a_{n} = 0$ , where $a_{n} = \frac{n^{2} +3n - \sin(n)}{n^{4}-\arctan(n^{2})}$ But I can't still proove its convergence, I'd have used Leibnitz alternating series test (due to $(-1)^{n}$), but I was unable to say $a_{n+1} \leq a_{n}$. Maybe I could study the Absolute convergence and then by Comparison test find that converges ? Any help would be appreciated, Thanks anyway.
Type the following command at the terminal to copy the template file to the current directory (note the period at the end): cp ~cs61as/autograder/templates/hw3.rkt . Or you can download the template here. Here is the fast-expt procedure from earlier in this lesson: (define (even? n) (= (remainder n 2) 0))(define (fast-expt b n) (cond ((= n 0) 1) ((even? n) (square (fast-expt b (/ n 2)))) (else (* b (fast-expt b (- n 1)))))) Design a procedure that evolves an iterative exponentiation process that uses successive squaring and uses a logarithmic number of steps, as does fast-expt. (Hint: Using the observation that ( b ) n/2 cont-frac An infinite continued fraction is an expression of the form: As an example, one can show that where [mathjaxinline]\phi=\frac{1+\sqrt{5}}{2}[/mathjaxinline] is the golden ratio.One way to approximate an infinite continued fraction is to truncate the expansion after agiven number of terms. Such a truncation—a so-called [mathjaxinline]k[/mathjaxinline]-term finite continuedfraction—has the form: Suppose that n and d are procedures of one argument(the term index [mathjaxinline]i[/mathjaxinline]) that return the[mathjaxinline]N[/mathjaxinline] and [mathjaxinline]D[/mathjaxinline] ofthe [mathjaxinline]i[/mathjaxinline]-th term of the continued fraction.Define a procedure cont-frac suchthat evaluating (cont-frac n d k) computes the value of the [mathjaxinline]k[/mathjaxinline]-termfinite continued fraction. Check your procedure by approximating [mathjaxinline]\frac{1}{\phi}[/mathjaxinline]using (cont-frac (lambda (i) 1.0) (lambda (i) 1.0) k) for successive values of k. How large must you make k in order to get an approximation that is accurate to 4 decimal places? If your cont-frac procedure generates a recursive process, write one that generates an iterative process.If it generates an iterative process, write one that generates a recursive process. In 1737, Swiss mathematician Leonhard Euler showed that for the parameters where [mathjaxinline]e[/mathjaxinline] is the base of natural logarithms.Write a program that uses your cont-frac procedure to approximate [mathjaxinline]e[/mathjaxinline]using Euler's expansion. next-perf A perfect number is defined as a number equal to the sum of all its factors less than itself. For example, the first perfect number is 6, because its factors are 1, 2, 3, and 6, and 1+2+3=6. The second perfect number is 28, because 1+2+4+7+14=28. What is the third perfect number? Write a procedure (next-perf n) that tests consecutive integers starting with n until a perfect number is found. Then you can evaluate (next-perf 29) to solve the problem. Note that your procedure should be able to handle any non-negative integer input. Hint: you’ll need a sum-of-factors subprocedure. Note: If you run this program when the system is heavily loaded, it may take half an hour to compute the answer! Try tracing helper procedures to make sure your program is on track, or start by computing (next-perf 1) and see if you get 6. Here is the definition of count-change program from earlier in this lesson: (define (count-change amount) (cc amount `(50 25 10 5 1))) (define (cc amount kinds-of-coins) (cond [(= amount 0) 1] [(or (< amount 0) (empty? kinds-of-coins)) 0] [else (+ (cc amount (bf kinds-of-coins)) (cc (- amount (first kinds-of-coins)) kinds-of-coins))] )) Explain the effect of interchanging the order in which the base cases in the cc procedure are checked. That is, describe completely the set of arguments for which the original cc procedure would return a different value or behave differently from a cc procedure coded as given below, and explain how the returned values would differ. (define (cc amount kinds-of-coins) (cond [(or (< amount 0) (empty? kinds-of-coins)) 0] [(= amount 0) 1] [else ... ] ) ) ; as in the original version Here is the iterative exponentiation procedure from earlier in this lesson: (define (expt b n) (expt-iter b n 1))(define (expt-iter b counter product) (if (= counter 0) product (expt-iter b (- counter 1) (* b product)))) Give an algebraic formula relating the values of the parameters b, n, counter, and product of the iterative exponentiation procedure defined above. (The kind of answer we're looking for is "the sum of b, n, and counter times product is always equal to 37.") For instructions, see this guide. It covers basic terminal commands and assignment submission. If you have any trouble submitting, do not hesitate to ask a TA!
Do numbers really represent the physical reality? For example 1 apple simply tells us that there is only one apple. Now it becomes complicated when simple arithmetic formulaes are applied. Example: 1 apple x 1 apple = 1 apple. Now I don't know where the other apple got lost. 161 Joe J 03-21-2019 05:41 AM ET (US) How to calculate future value when compounding happens every 5 years? If I use 0.2 for n in this formula, would that be correct? A = P (1 + r/n) (nt) 160 EmberCult 09-14-2018 08:57 AM ET (US) compute the inverse Laplace transformation of the following equation. What I managed to get is this \begin{align} Y^{-1}(\frac{1}{(s+a)^m})&=\frac{t^{m-1}e^{-at}}{{(m-1)!}} \end{align} How do I find the Laplace inverse of the series though? Someone suggested me using partial fraction decomposition and gave me this solution, \begin{align}f(s)=\sum_{1\le k\le m\atop 1\le j\le L} {A_{jk}\over(s+a_j)^k}\end{align} But failed to tell how did we achieve it.
In my last post, I wrote about within- and between-period intra-cluster correlations in the context of stepped-wedge cluster randomized study designs. These are quite important to understand when figuring out sample size requirements (and models for analysis, which I’ll be writing about soon.) Here, I’m extending the constant ICC assumption I presented last time around by introducing some complexity into the correlation structure. Much of the code I am using can be found in last week’s post, so if anything seems a little unclear, hop over here. Different within- and between-period ICC’s In a scenario with constant within- and between-period ICC’s, the correlated data can be induced using a single cluster-level effect like $b_c$ in this model: \[ Y_{ict} = \mu + \beta_0t + \beta_1X_{ct} + b_{c} + e_{ict} \] More complexity can be added if, instead of a single cluster level effect, we have a vector of correlated cluster/time specific effects $\mathbf{b_c}$. These cluster-specific random effects $(b_{c1}, b_{c2}, \ldots, b_{cT})$ replace $b_c$, and the slightly modified data generating model is \[ Y_{ict} = \mu + \beta_0t + \beta_1X_{ct} + b_{ct} + e_{ict} \] The vector $\mathbf{b_c}$ has a multivariate normal distribution $N_T(0, \sigma^2_b \mathbf{R})$. This model assumes a common covariance structure across all clusters, $\sigma^2_b \mathbf{R}$, where the general version of $\mathbf{R}$ is \[ \mathbf{R} = \left( \begin{matrix} 1 & r_{12} & r_{13} & \cdots & r_{1T} \\ r_{21} & 1 & r_{23} & \cdots & r_{2T} \\ r_{31} & r_{32} & 1 & \cdots & r_{3T} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ r_{T1} & r_{T2} & r_{T3} & \cdots & 1 \end{matrix} \right ) \] Within-period cluster correlation The covariance of any two individuals $i$ and $j$ in the same cluster $c$ and same period $t$ is \[ \begin{aligned} cov(Y_{ict}, Y_{jct}) &= cor(\mu + \beta_0t + \beta_1X_{ct} + b_{ct} + e_{ict}, \ \mu + \beta_0t + \beta_1X_{ct} + b_{ct} + e_{jct}) \\ \\ &= cov(b_{ct}, b_{ct}) + cov(e_{ict}, e_{jct}) \\ \\ &=var(b_{ct}) + 0 \\ \\ &= \sigma^2_b r_{tt} \\ \\ &= \sigma^2_b \qquad \qquad \qquad \text{since } r_{tt} = 1, \ \forall t \in \ ( 1, \ldots, T) \end{aligned} \] And I showed in the previous post that $var(Y_{ict}) = var(Y_{jct}) = \sigma^2_b + \sigma^2_e$, so the within-period intra-cluster correlation is what we saw last time: \[ICC_{tt} = \frac{\sigma^2_b}{\sigma^2_b+\sigma^2_e}\] Between-period cluster correlation The covariance of any two individuals in the same cluster but two different time periods $t$ and $t^{\prime}$ is: \[ \begin{aligned} cov(Y_{ict}, Y_{jct^{\prime}}) &= cor(\mu + \beta_0t + \beta_1X_{ct} + b_{ct} + e_{ict}, \ \mu + \beta_0t + \beta_1X_{ct^{\prime}} + b_{ct^{\prime}} + e_{jct^{\prime}}) \\ \\ &= cov(b_{ct}, b_{ct^{\prime}}) + cov(e_{ict}, e_{jct^{\prime}}) \\ \\ &= \sigma^2_br_{tt^{\prime}} \end{aligned} \] Based on this, the between-period intra-cluster correlation is \[ ICC_{tt^\prime} =\frac{\sigma^2_b}{\sigma^2_b+\sigma^2_e} r_{tt^{\prime}}\] Adding structure to matrix $\mathbf{R}$ This paper by Kasza et al, which describes various stepped-wedge models, suggests a structured variation of $\mathbf{R}$ that is a function of two parameters, $r_0$ and $r$: \[ \mathbf{R} = \mathbf{R}(r_0, r) = \left( \begin{matrix} 1 & r_0r & r_0r^2 & \cdots & r_0r^{T-1} \\ r_0r & 1 & r_0 r & \cdots & r_0 r^{T-2} \\ r_0r^2 & r_0 r & 1 & \cdots & r_0 r^{T-3} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ r_0r^{T-1} & r_0r^{T-2} & r_0 r^{T-3} & \cdots & 1 \end{matrix} \right ) \] How we specify $r_0$ and $r$ reflects different assumptions about the between-period intra-cluster correlations. I describe two particular cases below. Constant correlation over time In this first case, the correlation between individuals in the same cluster but different time periods is less than the correlation between individuals in the same cluster and same time period. In other words, $ICC_{tt} \ne ICC_{tt^\prime}$. However the between-period correlation is constant, or $ICC_{tt^\prime}$ are constant for all $t$ and $t^\prime$. We have these correlations when $r_0 = \rho$ and $r = 1$, giving \[ \mathbf{R} = \mathbf{R}(\rho, 1) = \left( \begin{matrix} 1 & \rho & \rho & \cdots & \rho \\ \rho & 1 & \rho & \cdots & \rho \\ \rho & \rho & 1 & \cdots & \rho \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \rho & \rho & \rho & \cdots & 1 \end{matrix} \right ) \] To simulate under this scenario, I am setting $\sigma_b^2 = 0.15$, $\sigma_e^2 = 2.0$, and $\rho = 0.6$. We would expect the following ICC’s: \[ \begin{aligned} ICC_{tt} &= \frac{0.15}{0.15+2.00} = 0.0698 \\ \\ ICC_{tt^\prime} &= \frac{0.15}{0.15+2.00}\times0.6 = 0.0419 \end{aligned} \] Here is the code to define and generate the data: defc <- defData(varname = "mu", formula = 0, dist = "nonrandom", id = "cluster")defc <- defData(defc, "s2", formula = 0.15, dist = "nonrandom")defa <- defDataAdd(varname = "Y", formula = "0 + 0.10 * period + 1 * rx + cteffect", variance = 2, dist = "normal")dc <- genData(100, defc)dp <- addPeriods(dc, 7, "cluster")dp <- trtStepWedge(dp, "cluster", nWaves = 4, lenWaves = 1, startPer = 2)dp <- addCorGen(dtOld = dp, nvars = 7, idvar = "cluster", rho = 0.6, corstr = "cs", dist = "normal", param1 = "mu", param2 = "s2", cnames = "cteffect") dd <- genCluster(dp, cLevelVar = "timeID", numIndsVar = 100, level1ID = "id")dd <- addColumns(defa, dd) As I did in my previous post, I’ve generated 200 data sets, estimated the within- and between-period ICC’s for each data set, and computed the average for each. The plot below shows the expected values in gray and the estimated values in purple and green. Declining correlation over time In this second case, we make an assumption that the correlation between individuals in the same cluster degrades over time. Here, the correlation between two individuals in adjacent time periods is stronger than the correlation between individuals in periods further apart. That is $ICC_{tt^\prime} > ICC_{tt^{\prime\prime}}$ if $\|t^\prime – t\| < \|t^{\prime\prime} – t\|$. This structure can be created by setting $r_0 = 1$ and $r=\rho$, giving us an auto-regressive correlation matrix $R$: \[ \mathbf{R} = \mathbf{R}(1, \rho) = \left( \begin{matrix} 1 & \rho & \rho^2 & \cdots & \rho^{T-1} \\ \rho & 1 & \rho & \cdots & \rho^{T-2} \\ \rho^2 & \rho & 1 & \cdots & \rho^{T-3} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \rho^{T-1} & \rho^{T-2} & \rho^{T-3} & \cdots & 1 \end{matrix} \right ) \] I’ve generated data using the same variance assumptions as above. The only difference in this case is that the corstr argument in the call to addCorGen is “ar1” rather than “cs” (which was used above). Here are a few of the expected correlations: \[ \begin{aligned} ICC_{t,t} &= \frac{0.15}{0.15+2.00} = 0.0698 \\ \\ ICC_{t,t+1} &= \frac{0.15}{0.15+2.00}\times 0.6^{1} = 0.0419 \\ \\ ICC_{t,t+2} &= \frac{0.15}{0.15+2.00}\times 0.6^{2} = 0.0251 \\ \\ \vdots \\ ICC_{t, t+6} &= \frac{0.15}{0.15+2.00}\times 0.6^{6} = 0.0032 \end{aligned} \] And here is the code: defc <- defData(varname = "mu", formula = 0, dist = "nonrandom", id = "cluster")defc <- defData(defc, "s2", formula = 0.15, dist = "nonrandom")defa <- defDataAdd(varname = "Y", formula = "0 + 0.10 * period + 1 * rx + cteffect", variance = 2, dist = "normal")dc <- genData(100, defc)dp <- addPeriods(dc, 7, "cluster")dp <- trtStepWedge(dp, "cluster", nWaves = 4, lenWaves = 1, startPer = 2)dp <- addCorGen(dtOld = dp, nvars = 7, idvar = "cluster", rho = 0.6, corstr = "ar1", dist = "normal", param1 = "mu", param2 = "s2", cnames = "cteffect") dd <- genCluster(dp, cLevelVar = "timeID", numIndsVar = 10, level1ID = "id")dd <- addColumns(defa, dd) And here are the observed average estimates (based on 200 datasets) alongside the expected values: Random slope In this last case, I am exploring what the ICC’s look like in the context of random effects model that includes a cluster-specific intercept $b_c$ and a cluster-specific slope $s_c$: \[ Y_{ict} = \mu + \beta_0 t + \beta_1 X_{ct} + b_c + s_c t + e_{ict} \] Both $b_c$ and $s_c$ are normally distributed with mean 0, and variances $\sigma_b^2$ and $\sigma_s^2$, respectively. (In this example $\sigma_b^2$ and $\sigma_s^2$ are uncorrelated, but that may not necessarily be the case.) Because of the random slopes, the variance of the $Y$’s increase over time: \[ var(Y_{ict}) = \sigma^2_b + t^2 \sigma^2_s + \sigma^2_e \] The same is true for the within- and between-period covariances: \[ \begin{aligned} cov(Y_{ict}, Y_{jct}) &= \sigma^2_b + t^2 \sigma^2_s \\ \\ cov(Y_{ict}, Y_{jct^\prime}) &= \sigma^2_b + tt^\prime \sigma^2_s \\ \end{aligned} \] The ICC’s that follow from these various variances and covariances are: \[ \begin{aligned} ITT_{tt} &= \frac{\sigma^2_b + t^2 \sigma^2_s}{\sigma^2_b + t^2 \sigma^2_s + \sigma^2_e}\\ \\ ITT_{tt^\prime} & = \frac{\sigma^2_b + tt^\prime \sigma^2_s}{\left[(\sigma^2_b + t^2 \sigma^2_s + \sigma^2_e)(\sigma^2_b + {t^\prime}^2 \sigma^2_s + \sigma^2_e)\right]^\frac{1}{2}} \end{aligned} \] In this example, $\sigma^2_s = 0.01$ (and the other variances remain as before), so \[ ITT_{33} = \frac{0.15 + 3^2 \times 0.01}{0.15 + 3^2 \times 0.01 + 2} =0.1071\] and \[ ITT_{36} = \frac{0.15 + 3 \times 6 \times 0.01}{\left[(0.15 + 3^2 \times 0.01 + 2)(0.15 + 6^2 \times 0.01 + 2)\right ]^\frac{1}{2}} =0.1392\] Here’s the data generation: defc <- defData(varname = "ceffect", formula = 0, variance = 0.15, dist = "normal", id = "cluster")defc <- defData(defc, "cteffect", formula = 0, variance = 0.01, dist = "normal")defa <- defDataAdd(varname = "Y", formula = "0 + ceffect + 0.10 * period + cteffect * period + 1 * rx", variance = 2, dist = "normal")dc <- genData(100, defc)dp <- addPeriods(dc, 7, "cluster")dp <- trtStepWedge(dp, "cluster", nWaves = 4, lenWaves = 1, startPer = 2) dd <- genCluster(dp, cLevelVar = "timeID", numIndsVar = 10, level1ID = "id")dd <- addColumns(defa, dd) And here is the comparison between observed and expected ICC’s. The estimates are quite variable, so there appears to be slight bias. However, if I generated more than 200 data sets, the mean would likely converge closer to the expected values. In the next post (or two), I plan on providing some examples of fitting models to the data I’ve generated here. In some cases, fairly standard linear mixed effects models in R may be adequate, but in others, we may need to look elsewhere. References: Kasza, J., K. Hemming, R. Hooper, J. N. S. Matthews, and A. B. Forbes. “Impact of non-uniform correlation structure on sample size and power in multiple-period cluster randomised trials.” Statistical methods in medical research (2017): 0962280217734981.
I have had a number of friends ask me “Okay, AlphaGo was impressive, but where else can you use Reinforcement Learning (RL)?”. If you google the question, you will find a legitimate list like this paper or this article. Today I would like to talk about a cool application of RL that is not as often mentioned: combinatorial optimization. For the sake of conciseness, I will quickly go through a few background concepts and discuss the components I think are most interesting for the audience who likes RL. Combinatorial optimization problems are important in computer science and operations research and correspond to many real-life problems. For example, suppose you have to take a low-cost flight without free check-in baggage. Now you need to somehow find a way to pack as many valuable items into your small backpack while keeping the total weight from exceeding the limit. In computer science speak, you are solving an instance of Knapsack problem. Combinatorial optimization is about finding a solution that maximimizes/minimizes some given objective function defined over a set of solutions that live on a discrete space. Some of such problems are NP-hard which, at this point, means no polynomial time exact solution can ever be found (yeah, probably. there is still hope! ). This is a problem, to my knowledge, at least as old as the history of computer science, so there are already many approximate and heuristics algorithms. Is there anything new there is to talk about? (Drumroll…) well, this is exactly where RL can kick in. In this part of the post, we describe some classic approahches. If you are already familiar with them, I suggest moving to the part 2. Traveling Salesperson Problem As a running example, let us considere an NP-hard combinatorial optimization problem Traveling Salesperson Problem (TSP). A traveling salesperson must visit $n$ cities $X = { x_1,\dots, x_n }$ where every pair of cities has a road (edge) connecting them with varying distances (complete, undirected, weighted graph). The goal of the salesperson is to find a route such that a) each city is visited exactly once 1, b) the salesperson returns to the city it started from (a.k.a depot), and c) the total traveling distance (cost) is minimal. That is, we want to solve. Let $\Pi$ denote the set of possible routes and $\pi$ a permutation (route) over cities . Then we define a cost fuction simply by summing the distances of all edges for a given $\pi$. Note the cost function is defined over a discrete space of cities (or its indices) and the distance function is given and symmetric; $(\Pi, d)$ may not even be a metric space though many fortunate results can be derived with the triangle inequality. Exact algorithms How do we solve this? Without constraints, there are $n!$ possible solutions 2. Unless $n$ is really small or the problems belong to some special cases we know how to attack with ease, exact algorithms are likely an impossible dream. Recall factorial asymtotically grows faster than the infamous exponential. The best exact solution that we know is called Held-Karp algorithm. The trick is to capitalize on the invariant that given the optimal route $\pi^\star$ of minimum distance, any sub-route of it is also of minimum distance. If there were a sub-route that isn’t, $\pi^\star$ is not optimal; a contradiction. The problem can be formulated to recursively compute the optimal route given a start vertex, an end vertex (technically, second to last vertex) and a set of intermediate vertices to go through such that no computation is repeated at the expense of memory; i.e. dynamic programming. Since there are $n$ vertices to consider for the start and the end, respectively and $2^n$ subsets for the intermediate, it gives the worst time complexity of $O(n^2 2^n)$ and the space $O(2^n n)$. Better than factorial but, still exponential. Approximation algorithms and heuristics What people use in practice are (local) approximation algorithms and/or search algorithms with heuristics. Approximation algorithms guarantee that its performance won’t be too bad (up to a multiplicative factor/ratio) compared to the optimal performance (sometimes, in expectation). Local search algorithms find a local optimum where heuristics helps them get stuck at a good local optimum. Note heuristics are problem-dependent to varying degrees and one that is generally applicable gets a distinguished status of meta-heuristic. While there have been many various attempts at TSP, it appears there are a few popular ones. For brevity, we just name them here. The simplest is Nearest Neighbor algorithm. Given a start vertex, it iteratively constructs a route such that at $i$-th iteration, it looks at the unvisted neighbors of its most recently added vertex and chooses the nearest neighbor: $x^\star = \arg\min_{x \in X \setminus \pi} d(\pi_i, x)$ and $\pi = \pi \cup \{ x^\star \}$. At first, it has $n-1$ unvisted vertices to evaluate, then $n-2$ until there’s one left. Each evaluation is independent so we add them up to have $O(n^2)$. Since it is cheap to run, it is often used to produce an initial solution and we use other techniques to improve it iteratively. There’s not much of satisfactory performance guarantees. The sad thing is that this algorithm’s approximation ratio is not even bounded; it can produce the worst toue. That said, special instances with the triangle inequality give the approximation ratio of $O(\log n)$. Christofides algorithm is the best known approximation algorithm that guarantees the ratio of 3/2. The result assumes the distance function forms a metric. It begins by constructing a minimum spanning tree $T$ which is a nice start: $T$ is connected and minimal both in the number of edges and the cost. Plus, $L(T) < L(\pi^\star)$ since removing an edge in $\pi^\star$ would lead to a spanning tree. At this point, $T$ is not a valid tour but we can attack the odd-degree edges in $T$ and find a minimum matching, $M$. If we add $M$ back to $T$ (call it $T’$), there must exist a subgraph in $T’$ that visits every edge exactly once. This kind of graph is called an Euler tour. It is nice as it already contains a TSP tour $\tilde{\pi}$. The TSP tour can be found by adding shortcuts for the vertices visited twice–this is where the triangle inequality is used. It follows that $L(\tilde{\pi}) \le L(T’) = L(T) + L(M) \le (1 + 1/2)L(\pi^\star)$. Note adding shortcuts is a non-increasing operation. The time complexity is polynomial with $O(n^3)$ due to the matching step. Another local heuristic algorithm often mentioned together is 2-Opt. The algorithm assumes a tour $\pi$ is given and aims to improve upon it with a series of two-edge exchanges. The trick is to replace a pair of edges that cross over each other (X-shaped) with another pair that does not. It iteratively considers each possible pair of edges that are not adjacent in $\pi$. Remove the pair and we get two paths, $T_1, T_2$. By a simple argument, one can show that there is only one way to combine $T_1,T_2$ to a new tour $\pi’$. We set $\pi = \pi’$ if an improvement can be made such that $L(\pi’) < L(\pi)$. Why does this algorithm work? When we pick two non-adjacent edges in a tour, the pair will either cross or not. In both cases, there’s a unique way to recombine them into a different and valid tour. The only case where the cost decreases is when they cross. The running time for the portion described so far is $O(n^2)$ since there are at most $\binom{n}{2}$ edge pairs to consider and the swap evaluation is $O(1)$. The catch is you may have to run this procedure (for the entire non-adjacent-edge pairs) many times over which can lead to an exponential number of exchanges. It will converge to a local optimum as it changes the graph only when strict improvement is possible; it can’t cycle back to a higher cost solution it has seen before. There is a generalization of this idea called Lin-Kernighan. Google open-sourced a nice tool called OR-tools. In their first solution strategy section, you can find the algorithms above or similar and other more sophisticated ones that produce an initial solution. Furthermore, it features meta-heuristics (local search options) that help find a better local optimum. The tool is designed to handle not only the TSP but other combinatorial optimization problems such as Vehicle Routing Problem (VRP) that subsumes TSP as a special case of having a single vehicle, a single depot and no constraints on things like capacity or time windows. Next, we talk about RL in the part two of this post. The complete connectivity of a graph assures the existence of a tour that visits each vertex only once. Whether the shortest tour will be included in such visit-once tours generally depends on the traingle equality where short tours are short and long tour are long. ↩ If we assume a depot city (the first and last visit to visit) is assigned as a part of the problem definition, there will be $(n-1)!$ ways. Also $n!/2$ if we assume two tours of opposite directions are identical (e.g. $A\rightarrow B, B \rightarrow A$). ↩
..have to do the burn in..? As pointed out in the comments, taking out some of the initial samples is not a mathematical necessity. We take out burn in because it is a computational statistics hack/cheat to avoid having to get more samples to correct the bias of the initial samples. In the small samples case, the bias could be somewhat noticeable. In the many samples case, this bias is minimal.. but if you do not want to take a lot of samples and your sampler takes time to converge* then it is a good idea to throw some initial samples out. ..a MAP estimation and try to find some local mode of the distribution to initialise my MH-MCMC This is perfectly OK, and has been used/recommended in cases where the network is complex enough that convergence may take an obnoxious amount of time to reach. For example, consider a finite mixture of $K$ Normals with latent mixture labels:$$y_i \mid z_i = k \sim \mathcal{N}(\mu_k, \sigma^2_k)\\z_i \mid \boldsymbol \pi \sim \text{Cat}_K(\boldsymbol \pi) \\\boldsymbol \pi \sim \text{Dir}_K(\dots) \\\mu_k, \sigma^2_k \sim \mathcal{N-}\Gamma^{-1}(\dots)$$ Let's pretend that a sampler based on this model would take a long amount of time to converge and let's also ignore this model's identifiability issues.. One possible way to initialize the data would be to identify 'humps' in your histogram and place appropriately scaled normal distributions in these locations. That is, your initialization of these parameters would be pseudo-empirical. Then you might take each datum $y_i$ and assign its corresponding $z_i$ to be the label which corresponds to the Normal parameters $y_i$ has the highest likelihood under. Then you might initialize $\boldsymbol \pi$ as the proportion vector of all the $z_i$ belonging to each of the $K$ labels. If you have a good eye, the sampler with these initializations will not budge far from where you initialized the values. If you looked at trace plots there would be no reason to think that the convergence was not immediate, and so there would be almost no benefit from taking out burn in. We can do this because there are no restrictions on initializations in MCMC. Initializations do not affect the stationary distribution, so there is no 'crime' (e.g., empirical Bayes) in using the data to inform our initializations, as compared to, say, the prior. With that said, I have only once been, and I would hope it would be rare to be, in a situation where implementing a MAP estimate algorithm was noticeably faster than taking some extra samples, in either run time or developer time. I suppose it would be context-specific, data-specific, MAP-algorithm-specific, and sampler-specific. *Of course it is always worth asking in these cases: did my sampler actually converge?
I have a conceptual problem to understand the standard error of the ratio of two random variables after error propagation. Let $X$ and $Y$ be two random variables with means $\bar x$ and $\bar y$ and standard errors $se_x = \frac{\sigma_x}{\sqrt{m}}$ and $se_y = \frac{\sigma_y}{\sqrt{n}}$, where $m$ and $n$ are the sample sizes. I assume that $X$ and $Y$ are normally distributed. Their means do not necessarily have to differ significantly (which I analyse by a Welch test). Currently I calculate the ration by $r = \frac{\bar x}{\bar y}$ and the according standard error by: $$ se_r = r \cdot \sqrt{\left(\frac{se_x}{\bar x}\right)^2 + \left(\frac{se_y}{\bar y}\right)^2} $$ Usually my sample sizes are small (between 3 and 10) which may result in large individual errors $se_x$ and $se_y$. The choice if I calculate $r = \frac{\bar x}{\bar y}$ or $r = \frac{\bar y}{\bar x}$ is arbitrary (r is just a fold change). So what I don't understand is, that $se_r$ would be the same for $r$ and $1/r$, no matter if $\bar y$ and $\bar x$ differ by order of magnitudes. And this is why I am wondering if I calculate $se_r$ correctly and if the error interval $r \pm se_r$ shouldn't be asymmetrical. Does it make in such cases only sense to calculate log-folds (i.e. $\log_{10}(r))$)? Would the standard error in this case be calculated as $\frac{se_r}{r\cdot\ln{10}}$? Any clarification is highly appreciated.
Since no one has come up with an answer yet, here is a rather brute-force computation on homogeneous coordinates to verify this fact. Without loss of generality I'll choose my coordinate system in such a way that the incircle becomes the unit circle, with $K=(1:0:1)$ at the $0°$ position. Then $D$ and $E$ can be described by one peal parameter each, using the half-angle formula. This gives us \begin{align}E&=(t^2-1:2t:t^2+1) & D&=(u^2-1:2u:u^2+1)\end{align} Everything else can be expressed in terms of $t$ and $u$. The unit circle has the diagonal matrix $U=\operatorname{diag}(1,1,-1)$, and multiplying that matrix with the vecotr of a point gives the tangent in that point. Computing the cross product between two lines gives their point of intersection. So you get \begin{align}A&=(U\cdot D)\times(U\cdot K)=(u:1:u) \\B&=(U\cdot E)\times(U\cdot K)=(t:1:t) \\C&=(U\cdot D)\times(U\cdot E)=(tu - 1 : t + u : tu + 1)\end{align} Next, we need $J$. I'd construct that by intersecting the line orthogonal to $AI$ through $A$ with the one orthogonal to $BI$ through $B$. To form a line connecting two points, you again compute the cross product. If you set the last coordinate of the resulting vector to $0$ you obtain a vector which describes a point infinitely far away in a direction orthogonal to the line. Connecting that with another point gives an orthogonal line. Dropping the last coordinate can be formulated by multiplication with the matrix $F=\operatorname{diag}(1,1,0)$. \begin{align}J &= (A\times(F\cdot(A\times I)))\times(B\times(F\cdot(B\times I)))= (tu - 1 : t + u : tu)\end{align} Next we need circles. I computed circles as conics through the ideal circle points $Q=(1:i:0)$ and $\bar Q=(1:-i:0)$. (Usually I'd call these points $I$ and $J$, but those letters are already taken in your problem statement.) You can find the matrix for the circle through three points $a,b,c$ by computing \begin{align}M_1 &= \det(a,c,Q)\cdot \det(b,c,\bar Q)\cdot (b\times Q)\cdot(a\times\bar Q)^T \\M_2 &= (M_1 - \bar M_1)+(M_1 - \bar M_1)^T\end{align} Using this approach, your circles are described by \begin{align}\bigcirc_{ADJ}=\begin{pmatrix}2 t u & 0 & 1 - t u - u^{2} \\0 & 2 t u & - t - 2 u \\1 - t u - u^{2} & - t - 2 u & 2 u^{2} + 2\end{pmatrix}\\\bigcirc_{BEJ}=\begin{pmatrix}2 t u & 0 & 1 - t^{2} - t u \\0 & 2 t u & -2 t - u \\1 - t^{2} - t u & -2 t - u & 2 t^{2} + 2\end{pmatrix}\end{align} Subtracting these matrices, you obtain a degenerate conic which factors into the line at infinity and the line connecting the two points of intersection. That line connecting the intersections is $g=(t + u : 1 : -t - u)$. It's point at infinity is $G=(-1:t+u:0)$. This allows you to find the point $T$ as the intersection of that line with one of the two circles: \begin{align}T&=G^T\cdot\bigcirc_{ADJ}\cdot G\cdot J - 2\cdot J^T\cdot\bigcirc_{ADJ}\cdot G\cdot G\\T&=%(t^2 + 2tu + u^2 - 1, 2t + 2u, t^2 + 2tu + u^2 + 1)((t+u)^2-1 : 2(t+u) : (t+u)^2+1)\end{align} One can already recognize the half angle format of this point, so it will lie on the unit circle. You need one last circle, $ABT$: \begin{align}\bigcirc_{ABT}=\begin{pmatrix}4 t u & 0 & 1 - t^{2} - 2 t u - u^{2} \\0 & 4 t u & -2 t - 2 u \\1 - t^{2} - 2 t u - u^{2} & -2 t - 2 u & 2 t^{2} + 2 u^{2} + 2\end{pmatrix}\end{align} Now you can verify that $T^T\cdot U\cdot T=T^T\cdot\bigcirc_{ABT}\cdot T=0$, so the point $T$ lies on both these circles. Furthermore, the tangent to the circle is the same in both cases as well, up to a scalar factor. So we have a single touching point. \begin{align}U\cdot T \sim \bigcirc_{ABT}\cdot T &=(t^2 + 2tu + u^2 - 1 : 2t + 2u : - t^2 - 2tu - u^2 - 1)\end{align} All through this post, I've cleared common denominators from the homogeneous coordinates and circle matrices to keep things simple. And I'm happy that I could write all of this without needing a single square root. I think about my homogeneous coordinates as column vectors, even though I wrote them as rows to save space. This explains which expressions I transposed in some computations and which I left as columns. Both your assumptions are true, by the way. $K$ lies on $J\times T=(t+u:1:-t-u)$ and the center of the circle $ABT$, which is the point $M=((t+u)^2 - 1 : 2(t+u) : 4tu)$, lies on the line $T\times I=(-2(t+u) : (t+u)^2 - 1 : 0)$. $M$ does not lie on the incircle, even though it does look that way in the figure. Seeing that the half angle tangent parameter of $T$ is $t+u$ tells me that the line $DE$ will intersect $AB$ in a point and that point connected to $T$ will intersect the incircle in another point, which also lies on $KI$. Not sure whether this is of any use to anybody.
Here is a schematized binary channel that neatly conveys a decimal number. $ \require{begingroup}\begingroup \def\T {{ \cal T }} \def \Ti {{ \T \raise5mu{ \text- \scriptsize 1 } }} \def\Bx #1{{ ~ \rightarrow ~ \boxed{\, #1 \,\Large\strut} \: \rightarrow ~ }} \def \BTi {\Bx { \kern 1mu \Ti \kern1mu }} \def \BT { \Bx{ \rlap{\kern6mu \T} \phantom\Ti }} $ $$ 123_{(10)} \BT 1111011_{(2)} \BTi 123_{(10)} $$ But how to keep up with an endless stream of decimal digits? $$ 1, \, 2, \, 3, \, \ldots _{\, (10)} \BT \ldots \, ? \ldots _{\, (2)} \BTi 1, \, 2, \, 3, \, \ldots _{\,(10)} $$ Binary coding with 4 bits for each decimal digit works readily: $$ 1, \, 2, \, 3, \, \ldots _{\, (10)} \BT 0001 ~ 0010 ~ 0011 \, \ldots _{\, (2)} \BTi 1, \, 2, \, 3, \, \ldots _{\,(10)} $$ Yet this wastes the binary channel's capacity, with a bitwise efficiency of just $ \, \frac{\log_2 10}{4} = 0.83^+ $. The waste can be reduced with larger groupings, such as coding 3 digits at a time with 10 bits for an efficiency of $ \, \frac{\log_2 1000}{10} = \frac{3}{10} \, \log_2 10 = 0.996^+ $, which remains measurably less than $1$ and increases maximum lag from 4 bits to 10 bits. Is there a straightforward algorithm pair, $\T$ and $\,\Ti$,with bounded lag, that is asymptotically wastelesswhen given an input stream of evenly random decimal digits? $$ \dfrac {\small\text {cumulative number of decimal digits conveyed}} { \small\text{cumulative number of bits used in transmission}} ~ {\log_2 10} {\large~\xrightarrow{\quad}~} 1 $$ Please ground any algorithms in terms of specific bases,not necessarily decimal and binary, perhaps ternary and binary,that are not rational powers of each other.A general solution would be easy enough to gather from that.This is an inquiry into ideal,not just efficient,conversion of information. Addenda Another straightforward suboptimal approach is to Huffman code each digit as in the following table. This has a respectable average efficiency of $ \, \frac{1}{3.4} \, \log_2 10 = 0.977^+ $ and maximum lag of 4 bits. $$ \small \begin{array}{lrrrrrrrrrr} \rm Decimal & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\rm Binary & 000 & 001 & 010 & 011 & 100 & 101 & 1100 & 1101 & 1110 & 1111 \\\rm Bit~count & 3 & 3 & 3 & 3 & 3 & 3 & 4 & 4 & 4 & 4 \\[1ex] \rm \rlap{Average~bit~count~~~~3.4} & & & & & & & & & & \end{array} $$ Any table-lookup encoding, such as every approach mentioned above, will indeed have bounded lag but nonetheless be measurably inefficient. An optimally efficient approach with unbounded lag,as commented byleonbloy,is arithmetic coding,which interprets sequences as fractionsthat readily convert between decimal and binary.The potential for unlimited lag can be seenalready in the first decimal digit, below,as the binary transmissions for$ \, 5,9,8, \ldots _{\, (10)} \, $and$ \, 6,0,0, \ldots _{\, (10)} \, $are indistinguishable until the 9 bit. th $$ 5,9,8, \ldots _{\, (10)} \Bx{~ 0.598 ... _{(10)} = 0.100110010..._{(2)} \raise-6mu\strut~} 1,0,0,1,1,0,0,1,0, \ldots _{\, (2)} \\[3ex] 6,0,0, \ldots _{\, (10)} \Bx{~ 0.600 ... _{(10)} = 0.100110011..._{(2)} \raise-6mu\strut~} 1,0,0,1,1,0,0,1,1, \ldots _{\, (2)} \endgroup $$
Here is the graph pf the functionIt is apparent that the minimum distance from the origin occurs about $x=\pm4$. Here is the graph of the distance of the points on the parabola as a function of $x$ It is even more convincing that the minimum distance occurs very close to $\pm4$. Let's calculate the distance as a function of $x$. At an $x$ the coordinates of a point on the graph are $(x,\frac12x^2-9)$. That is, the distance form the origin is $$\sqrt{x^2+\left(\frac12x^2-9\right)^2}.$$ If we are after the minimum, we do do not have to take the square root. So, we want to minimize $$x^2+\left(\frac12x^2-9\right)^2=\frac14x^4-8x^2+81$$ whose derivative is $$x^3-16x^2=x(x^2-16).$$ The solutions of the equation $x^3-16x^2=x(x^2-16)=0$ are $0,\pm 4$ as we suspected. The minimum distance is then $\sqrt{17}.$
This is a two part post. The second part depends on the first. Part 1. Consolidation of the Denotational Semantics As a matter of expediency, I've been working with two different versions of the intersection type system upon which the denotational semantics is based, one version with subsumption and one without. I had used the one with subsumption to prove completeness with respect to the reduction semantics whereas I had used the one without subsumption to prove soundness (for both whole programs and parts of programs, that is, contextual equivalence). The two versions of the intersection type system are equivalent. However, it would be nice to simplify the story and just have one version. Also, while the correspondence to intersection types has been enormously helpful in working out the theory, it would be nice to have a presentation of the semantics that doesn't talk about them and instead talks about functions as tables. Towards these goals, I went back to the proof of completeness with respect to the reduction semantics and swapped in the "take 3" semantics. While working on that I realized that the subsumption rule was almost admissible in the "take 3" semantics, just the variable and application equations needed more uses of $\sqsubseteq$. With those changes in place, the proof of completeness went through without a hitch. So here's the updated definition of the denotational semantics of the untyped lambda calculus. The definition of values remains the same as last time: \[ \begin{array}{lrcl} \text{function tables} & T & ::= & \{ v_1\mapsto v'_1,\ldots,v_n\mapsto v'_n \} \\ \text{values} & v & ::= & n \mid T \end{array} \] as does the $\sqsubseteq$ operator. \begin{gather} \frac{}{n \sqsubseteq n} \qquad \frac{T_1 \subseteq T_2}{T_1 \sqsubseteq T_2} \end{gather} For the denotation function $E$, we add uses of $\sqsubseteq$ to the equations for variables ($v \sqsubseteq \rho(x)$) and function application ($v_3 \sqsubseteq v_3'$). (I've also added the conditional expression $\mathbf{if}\,e_1\,e_2\,e_3$ and primitive operations on numbers $f(e_1,e_2)$, where $f$ ranges over binary functions on numbers.) \begin{align} E[\!\| n \|\!](\rho) &= \{ n \} \\ E[\!\| x \|\!](\rho) &= \{ v \mid v \sqsubseteq \rho(x) \} \\ E[\!\| \lambda x.\, e \|\!](\rho) &= \left\{ T \middle\| \begin{array}{l} \forall v_1 v_2'. \, v_1\mapsto v_2' \in T \Rightarrow\\ \exists v_2.\, v_2 \in E[\!\| e \|\!](\rho(x{:=}v_1)) \land v_2' \sqsubseteq v_2 \end{array} \right\} \\ E[\!\| e_1\;e_2 \|\!](\rho) &= \left\{ v_3 \middle\| \begin{array}{l} \exists T v_2 v_2' v_3'.\, T {\in} E[\!\| e_1 \|\!](\rho) \land v_2 {\in} E[\!\| e_2 \|\!](\rho) \\ \land\, v'_2\mapsto v_3' \in T \land v'_2 \sqsubseteq v_2 \land v_3 \sqsubseteq v_3' \end{array} \right\} \\ E[\!\| f(e_1, e_2) \|\!](\rho) &= \{ f(n_1,n_2) \mid \exists n_1 n_2.\, n_1 \in E[\!\| e_1 \|\!](\rho) \land n_2 \in E[\!\| e_2 \|\!](\rho) \} \\ E[\!\| \mathbf{if}\,e_1\,e_2\,e_3 \|\!](\rho) &= \left\{ v \, \middle\| \begin{array}{l} v \in E[\!\| e_2 \|\!](\rho) \quad \text{if } n \neq 0 \\ v \in E[\!\| e_3 \|\!](\rho) \quad \text{if } n = 0 \end{array} \right\} \end{align} Here are the highlights of the results for this definition. Proposition (Admissibility of Subsumption) If $v \in E[\!\| e \|\!] $ and $v' \sqsubseteq v$, then $v' \in E[\!\| e \|\!] $. Theorem (Reduction implies Denotational Equality) If $e \longrightarrow e'$, then $E[\!\| e \|\!] = E[\!\| e' \|\!]$. If $e \longrightarrow^{} e'$, then $E[\!\| e \|\!] = E[\!\| e' \|\!]$. Theorem (Whole-program Soundness and Completeness) If $v' \in E[\!\| e \|\!](\emptyset)$, then $e \longrightarrow^{} v$ and $v' \in E[\!\| v \|\!](\emptyset)$. If $e \longrightarrow^{} v$, then $v' \in E[\!\| e \|\!](\emptyset) $ and $v' \in E[\!\| v \|\!](\emptyset) $ for some $v'$. Proposition (Denotational Equality is a Congruence) For any context $C$, if $E[\!\| e \|\!] = E[\!\| e' \|\!]$, then $E[\!\| C[e] \|\!] = E[\!\| C[e'] \|\!]$. Theorem (Soundness wrt. Contextual Equivalence) If $E[\!\| e \|\!] = E[\!\| e' \|\!]$, then $e \simeq e'$. Part 2. An Application to Compiler Correctness Towards finding out how useful this denotational semantics is, I've begun looking at using it to prove compiler correctness. I'm not sure exactly which compiler I want to target yet, but as a first step, I wrote a simple source-to-source optimizer $\mathcal{O}$ for the lambda calculus. It performs inlining and constant folding and simplifies conditionals. The optimizer is parameterized over the inlining depth to ensure termination. We perform optimization on the body of a function after inlining, so this is a polyvariant optimizer. Here's the definition. \begin{align} \mathcal{O}[\!\| x \|\!](k) &= x \\ \mathcal{O}[\!\| n \|\!](k) &= n \\ \mathcal{O}[\!\| \lambda x.\, e \|\!](k) &= \lambda x.\, \mathcal{O}[\!\| e \|\!](k) \\ \mathcal{O}[\!\| e_1\,e_2 \|\!](k) &= \begin{array}{l} \begin{cases} \mathcal{O}[\!\| [x{:=}e_2'] e \|\!] (k{-}1) & \text{if } k \geq 1 \text{ and } e_1' = \lambda x.\, e \\ & \text{and } e_2' \text{ is a value} \\ e_1' \, e_2' & \text{otherwise} \end{cases}\\ \text{where } e_1' = \mathcal{O}[\!\|e_1 \|\!](k) \text{ and } e_2' = \mathcal{O}[\!\|e_2 \|\!](k) \end{array} \\ \mathcal{O}[\!\| f(e_1,e_2) \|\!](k) &= \begin{array}{l} \begin{cases} f(n_1,n_2) & \text{if } e_1' = n_1 \text{ and } e_2' = n_2 \\ f(e_1',e_2') & \text{otherwise} \end{cases}\\ \text{where } e_1' = \mathcal{O}[\!\|e_1 \|\!](k) \text{ and } e_2' = \mathcal{O}[\!\|e_2 \|\!](k) \end{array} \\ \mathcal{O}[\!\| \mathbf{if}\,e_1\,e_2\,e_3 \|\!](k) &= \begin{array}{l} \begin{cases} e_2' & \text{if } e_1' = n \text{ and } n \neq 0 \\ e_3' & \text{if } e_1' = n \text{ and } n = 0 \\ \mathbf{if}\,e_1'\, e_2'\,e_3'\|\!](k) & \text{otherwise} \end{cases}\\ \text{where } e_1' = \mathcal{O}[\!\|e_1 \|\!](k) \text{ and } e_2' = \mathcal{O}[\!\|e_2 \|\!](k)\\ \text{ and } e_3' = \mathcal{O}[\!\|e_3 \|\!](k) \end{array} \end{align*} I've proved that this optimizer is correct. The first step was proving that it preserves denotational equality. Lemma (Optimizer Preserves Denotations) $E(\mathcal{O}[\!\| e\|\!](k)) = E[\!\|e\|\!] $ Proof The proof is by induction on the termination metric for $\mathcal{O}$, which is the lexicographic ordering of $k$ then the size of $e$. All the cases are straightforward to prove because Reduction implies Denotational Equality and because Denotational Equality is a Congruence. QED Theorem (Correctness of the Optimizer) $\mathcal{O}[\!\| e\|\!](k) \simeq e$ Proof The proof is a direct result of the above Lemma and Soundness wrt. Contextual Equivalence. QED Of course, all of this is proved in Isabelle. Here is the tar ball. I was surprised that this proof of correctness for the optimizer was about the same length as the definition of the optimizer!
We should find the Cauchy principal value integral of the form $$ I=\oint \frac{dz}{(z-z_1)(z-z_2)}~, $$ where both roots $z_1$ and $z_2$ lie on the contour path. My answer is: $$ I=a \left(-\oint \frac{dz}{z-z_1}+\oint \frac{dz}{z-z_2}\right)=a(-i\pi+i\pi)=0~, $$ where $a=1/(z_2-z_1)$. However, in a book, they do not find it to be zero, though they do not explain why. Is there any case for which I should find this integral to be "not zero"? In the specific example of the book, the poles are $z_1=e^{-ik}$, $z_2=e^{ik}$, and the contour is the unit circle. First: putting $\;f(z):=\frac1{(z-z_1)(z-z_2)}\;$ , we obtain (since both poles are simple assuming $\;z_1\neq z_2\;$ ) $$\text{Res}_{z=z_i}(f)=\lim_{z\to z_i}\frac{z-z_i}{(z-z_1)(z-z_2)}=\begin{cases}&\;\;\;\;\;\;\;\;\;\frac1{z_1-z_2}&,\;\;i=1\\{}\\&-\frac1{z_1-z_2}=\frac1{z_2-z_1}&,\;\;i=2\end{cases}$$ Now using the lemma, and its corollary, in the most upvoted answer here, we get that the integral indeed equals zero.
Lebesgue outer measure Set context $p\in \mathbb N$ definiendum $\eta^p:\mathcal P(\mathbb R^p)\to \overline{\mathbb R}$ definiendum $\eta^p(A):=\mathrm{inf}\{\ \sum_{k=1}^\infty\lambda^p(I_k)\ \|\ I\in\mathrm{Sequence}(\mathfrak J^p)\ \land\ A\subset\bigcup_{k=1}^\infty I_k\ \}$ Discussion The Lebesgue outer aims at measuring subspaces of $\mathcal P(\mathbb R^p)$ as approximated by cubes which themselves are measured via Elementary volume of ℝⁿ. Reference Wikipedia: Lebesgue measure
I'll give a somewhat intuitive explanation of why we might expect a chi-squared approximation in large samples (including illustrating a connection to sums of squared normals) and give a couple of references along the way. Let us begin by starting with an ordinary F statistic for one-way ANOVA. This F-distribution of the test statistic arises from having independent chi-squared variates divided by their respective df in the numerator and denominator when the null is true. The numerator is a variance among group means, while the denominator is the variance of observations around their group means (which shouldn't be impacted by the truth or falsity of the null). The size of the numerator is related to the underlying $\sigma^2$, so the denominator is needed in there so that we have some way of judging when the numerator is "too big".Imagine changing from measuring in $\textit{mm}$ to measuring in $m$; the size of the numerator in the F would reduce by a factor of a million. It needs to be scaled to something that is not a function of the variance (more specifically, we need a pivotal quantity). If you replaced the data by their ranks, the numerator is not affected by the scale of the original observations (again imagine changing from measuring in $\textit{mm}$ to measuring in $m$ -- the ranks would be unchanged). We can consider the numerator of an F-statistic applied to the ranks without regard to the variance of the original data (but we will eventually have to deal with variation in the ranks). For simplicity consider sampling from continuous populations (no ties). Now given the null and under the assumption of normality, the numerator of an ordinary ANOVA F-statistic will have a chi-squared distribution divided by its df. Let's just consider the sum of squares term instead and look at what happens when we switch to ranks. That is, we're looking at something like this: $\text{SST} = \sum_{i=1}^g n_i[\bar{R_i}-(N+1)/2]^2$ where $g$ is the number of groups, $n_i$ the number of observations in the $i$-th group, $\bar{R}_i$ is the average rank in the $i$-th group and $N$ is the overall number of observations. Each term in that sum is $n_i$ times the square of a deviation between a group mean rank and an overall mean rank. The overall mean rank is of course simply the average of the integer from $1$ to $N$. The group mean ranks are averages of ranks to each group (under the null these will be sampled without replacement from the full set of ranks). In large samples, those average ranks $\bar{R}_i$ will be approximately normally distributed. If we subtract the overall mean we would have a zero-mean, approximately normal random variable with variance related to the variance of the available ranks and inversely related to $n_i$. So let's multiply that by $\sqrt{n_i}$ and divide by $\sqrt{(N^2-1)/12}$ (the standard deviation of a uniform over the available ranks): $Z_i = \frac{\sqrt{n_i}(\bar{R_i}-(N+1)/2)}{\sqrt{(N^2-1)/12}}$ These $Z_i$ are approximately standard normal. Now if we square them and add them up we get a "standardized" version of the above numerator $\text{SST}/V(N)$. However, we haven't accounted for the a couple of things there, including fact that these $Z_i$ values are not independent (if you know all but one of them the last one is determined). Accounting for this, we can write $H=\frac{N-1}{N}\sum_i Z_i^2$, where $H$ should be approximately chi-squared with $g-1$ d.f. Now simple algebra can be used to transform $H = \frac{(N-1)}{N} \sum_{i=1}^g \frac{n_i[\bar{R_i}-(N+1)/2]^2}{(N^2-1)/12}$ into the more usual form $H = \frac{12}{N(N+1)}\sum_{i=1}^g n_i \bar{R}_{i}^2 -\ 3(N+1)$ or you can just proceed from considering $\sum_i n_i[\bar{R_i}-(N+1)/2]^2$, suitably scaling that and then expanding and simplifying. Hopefully this gives both some motivation for the form of the statistic and some indication for why we'd expect to get a chi-squared approximation by starting from the SS term in the numerator of an F applied to the ranks (and then suitably scaling it). The second edition of Conover's Practical Nonparametric Statistics gives a little more detail for the argument (see p235, including explicit discussion of the effect of sampling without replacement), and a couple of pages later shows that $H$ is monotonic in the F-statistic applied to the ranks. That is, if you were able to produce suitable tables for $F$ applied to the ranks (they won't quite be distributed as F) and tables for $H$ (or directly produced the permutation distribution for both) they should reject in the same circumstances. (Some other details can be gleaned from the original Kruskal and Wallis paper, but I think Conover's explanation provides a clear summary.) Kruskal and Wallis offer several other approximations in their 1952 paper.
Fourier Series CT Fourier Transform Introduction Examples Properties FT of periodic signals Recall: Fourier series representation of a periodic signal $\tilde{x(t)}$ with time period $'T'$ is given by:- Suppose, $x(t)$ is not periodic.Is there a representation for $x(t)$ as a linear combination of complex exponentials? The main idea is to think of $x(t)$ as the limit of $\tilde{x}(t)$ when $T \rightarrow \infty$ i.e $$\lim_{T \to \infty}\tilde{x}(t)$$. Summary:- 1. F.S representation applies to periodic signals i.e A signal contains only frequencies which are integer multiples of a fundamental frequency. 2. F.T representation applies to Non-periodic (and periodic) signals i.e The signal may contain a continuum of frequencies $X(j\omega)$ refers to the F.T,where $\omega$ is a continuously changing variable. So,the Analysis and Synthesis Equations respectively are given by:-(3)
I am developing a program which seeks strategies for the players A, B in any of a family of simple 2-player gambling-games. The program iterates, using a genetic algorithm to determine, from the current iteration's results, the strategies which are to play one another in the next iteration. Below I give an overview of my algorithm. Although I've used pseudocode, I am not asking about how to code an algorithm; rather, I would like to learn what the algorithm should contain. main{ Strategy A[], B[]; input nIters, nStrats, maxTweak, gameRules; A := nStrats random strategies; B := nStrats random strategies; for(t=1..nIters) { make each A play each B; for each A and each B its total score := sum of its scores in its individual games; sort the A's by total score; sort the B's by total score; v := the game's bias in favour of the best A; write v; evolveStrategies(A, nStrats, maxTweak); evolveStrategies(B, nStrats, maxTweak); }}evolveStrategies(X, nStrats, maxTweak){ keep the best few Xs where they are; for(each of the others X[i]) { j := index of one of the best Xs; X[i] := X[j]; tweak the value of each parameter of X[i] by no more than maxTweak; }} The nature of the game is that there is (with best play by A and B) a bias (positive or negative) in favour of A. Its value $V$ is what it is when A's and B's strategies are those at a Nash equilibrium. Let $v$ be my program's estimate of $V$. Ideally my iterations' successive values of $v$ would converge to $V$. For some games in the relevant family, I happen to know what $V$ is; for others, I don't. The trouble I find is that even if maxTweak is very small, the succession of $v$-values written shows oscillation with an amplitude much larger than maxTweak. Choosing a suitably large number of iterations shows that the oscillation, after an initial large rise and fall, is hardly damped if at all. For example on a simple game, maxTweak=$4\times10^{-6}$ gives an initial large rise and fall, then oscillation with amplitude, peak to peak, $\approx 0.2$ and period $\approx 500000$ iterations. One solution is to reduce maxTweak. But this entails using many more iterations, and thus more CPU time. A more radical change to my approach is needed. Given that, a priori, the program does not know the oscillation's period or the limit value, how can my program detect the oscillation and deal with it (e.g. damp it)? If $v$ varied as undamped simple harmonic motion, plus a constant, then $\frac{dv}{dt}$ varies as undamped simple harmonic motion about 0, so $\frac{d^3 v}{dt^3} \propto -\frac{dv}{dt}$. Since the tweaking is random, there will be small-scale oscillations due to that, as well as the larger-scale oscillation. So it seems that attempts to find higher-order derivatives of $v$ as a function of $t$ will fail. I tried estimating the first and second derivatives $\frac{dv}{dt}$ and $\frac{d^2 v}{dt^2}$ by maintaining exponentially-weighted averages of $v$ and my estimate of $\frac{dv}{dt}$. Based on the sign of my estimate of $\frac{d^2 v}{dt^2}$ I evolved just A or B. This increased the number of iterations during $v$'s initial large rise and fall, but the following oscillation was as large and undamped as before. One idea is to start maxTweak at a value which is not small, and adaptively reduce it. But in deciding (at each iteration) whether to reduce it or not, what should I be looking for?
In order for a right-hand limit $\lim\limits_{x \to a^{+}} f(x)$ to make sense, there must exist a $\delta > 0$ such that the function $f$ is defined in the open interval $(a, a + \delta)$. Since (with $k$ denoting an integer)$$f(x) = \frac{\sin [x]}{[x]} = \begin{cases} \dfrac{\sin k}{k} & k \leq x < k + 1,\ k \neq 0, \\ \text{undefined} & 0 \leq x < 1, \end{cases}$$the right-hand limit of $f$ at $0$ makes no sense. (The dashed line in the plot is the graph $y = \sin x/x$ for $x \neq 0$.) That said, it's conceivable the question (somewhat perversely) refers to the continuous extension$$g(x) = \begin{cases} \dfrac{\sin x}{x} & x \neq 0, \\ 1 & x = 0, \end{cases}$$and that $f(x) = g([x])$. If this interpretation were correct,$$f(x) = \frac{\sin [x]}{[x]} = \begin{cases} \dfrac{\sin k}{k} & k \leq x < k + 1,\ k \neq 0, \\ 1 & 0 \leq x < 1, \end{cases}$$and $\lim\limits_{x \to 0^{+}} f(x) = 1$.
What does the degrees of freedom parameter $n$ mean intuitively in a Wishart distribution $\mathcal{W}_p(\mathbf{V},n)$? Does it have any relation to the covariance of different dimensions of the resulting covariance matrix? Why is $n==p$ called a non-informative prior? What does the degrees of freedom parameter $n$ mean If we consider the expectations of a covariance matrix $\Sigma^{-1}$ under out prior assumptions that is follows an inverse-Wishart distribution, we see $E(\Sigma^{-1})=nV$ for inverse covariance matrix $V$. Essentially the degrees of freedom parameter arises from statisticians assuming it is a (positive) integer $\Bbb Z^+$, which means we have a multivariate generalisation of the $\chi_\nu^2$ distribution. Note they do not have to be integers. You will find that as the Chi-squared distribution describes the sums of squares of $n$ draws from a univariate normal distribution, the Wishart distribution represents the sums of squares (and cross-products) of $n$ draws from a multivariate normal distribution. Hence intuitively for some dimension $p$, if we want to assume: (1) We are confident about the true covariance matrix being near some covariance matrix $\Sigma_0$, then we can choose a large value for $n$ say $p+10$ and set $V=[n-p-1]\Sigma_0$, so the distribution of our prior $\Sigma^{-1}$ is concentrated around $\Sigma_0$. (2) We are not too sure about the true covariance matrix being near some covariance matrix $\Sigma_0$, then we can choose a small value for $n$ say $p+2$ and set $V=\Sigma_0$, so the distribution of our prior $\Sigma^{-1}$ is only loosely centered around $\Sigma_0$. As $n$ decreases to $p$, the distribution becomes increasingly loosely centered around $\Sigma_0$. Hence it will be non-informative when the degrees of freedom is equal to the dimension/number of parameters. This is also clear since the first moment of a Wishart distribution does not exist unless $n>p+1$, with an improper distribution at $n=p$. $n$ simply then deals with how much we assume about the distribution of the covaraince. But it also happens that if $\Sigma^{-1}$ follows a Wishart distribution, then so does any sub-matrix. In fact I think it can be shown that the sub-matrix degrees of freedom is independent of $p$, the dimension. So technically the degrees of freedom for the sub-matrix parallels that of the original matrix.
I am interested in knowing wether the following statement is true of false. Let $ (\Omega, \Sigma , \mathbb{P})$ be a probability space and $\mathcal{A}, \mathcal{B} \subseteq \Sigma$ independant subsets. Then $\sigma A(\mathcal{A})$ and $\sigma A(\mathcal{B})$ are independent sigma algebras. I think the statement is true but haven't been able to finish the proof. What I've done thus far is the following. First define $\{M \in \Sigma : \forall A \in \mathcal{A} \text{ it holds } A, M \text{ independent}\}$ and I claim that it's a sigma algebra. Assuming that's true and since it contains $\mathcal{B}$. This implies that it must contain $\Sigma_2 = \sigma A(\mathcal{B})$. We then define the set $\{M \in \Sigma : \forall B \in \Sigma_2 \text{ it holds } B, M \text{ independent}\}$ Then clearly by the last part this set contains $\mathcal{A}$. I think here (were we able to probe that the first set is a sigma algebra) we should be able to recycle the argument to show that this new set is also a sigma algebra. This would imply $$\Sigma_1 \subseteq \{M \in \Sigma : \forall B \in \Sigma_2 \text{ it holds } B, M \text{ independent}\}$$ where $\Sigma_1 = \sigma A(\mathcal{A}) $ so we would be done. Now I just have to show that $\mathcal{O}=\{M \in \Sigma : \forall A \in \mathcal{A} \text{ it holds } A, M \text{ independent}\}$ is a sigma algebra. But I have been unable to do so. The part that I'm having trouble with is the countable union part. UPDATE: So as someone in the comments suggested if we restrict the case to only showing that the countable union of pairwise disjoint elements $M_n$ is still in $\mathcal{O}$ we get the result directly since. \begin{align} \mathbb{P}(A \cap (\cup_{i=1}^\infty M_n)) &= \mathbb{P}( \cup_{i=1}^\infty (A \cap M_n))\\ &= \sum_{i=1}^\infty \mathbb{P}(A \cap M_n)\\ &=\sum_{i=1}^\infty \mathbb{P}(A)\mathbb{P}(M_n)\\ &=\mathbb{P}(A)\mathbb{P}(\cup_{i=1}^\infty M_n) \end{align} However I was still unable to show that it's possible to restrict the general case to that case. I tried using induction on the sequence $M_n' = M_n \setminus (\cup_{i=1}^{n-1} M_i')$ but failed.
In here: Proving that well ordering principle implies Zorn's Lemma. I asked how to finish a proof of this statement. After a few helpful remarks, I think I have managed to finish it. What do you think? Given that on every set, a well ordering can be defined, we should prove that Given a partially ordered set $A$, if every increasing chain in $A$ has an upper bound in $A$, Then $A$ has a maximal element. Proof: Take $A$ partially ordered by $R$. We know that there exists a well ordering $S$ on $A$. Let $k$ be the smallest ordinal s.t. $k=\|A\|$ and let, $k^+=k+1$. Define by transfinite induction, a function, $g:k^+ \rightarrow A$ as follows: $g(0)$ is the first element in $A$ by $S$. For any , $\alpha < k^+$: If $\alpha$ is a successor ordinal, s.t. $\alpha = \beta + 1$, then, define $g(\alpha)$, the first element (by $S$), $a \in A$ such that $g(\beta) <_{R} a$ if $\alpha$ is a limit ordinal, then, The set $\{g(\beta);\beta<\alpha\}$, is linearly ordered by $R$. Therefor it has an upper bound. From all the uppers bounds, we will take the first (By $S$) to be $g(a)$. $g(k^{+})$ is linearly ordered. So, by the lemma assumption, it has an upper bound in $M \in A$. We claim that $M$ is a maximal element of $A$. Because, if there would be $x >_{R} M$ in $A$, by the construction of $g$, $g(k^{+})$ would contain an element which ia greater or equall (by $R$) to $x$, contradicting the fact that $M$ is an upper bound of $g(k^{+})$. The step which I'm not sure of is step 5. I am not sure weather the fact that $\|k\|=\|A\|$ and that $g(k^{+})$ is isomorphic to $k$ are enough. What do you think? Thank you! Shir
Introduction Energy and Power Basic Operations Practice Problems Transformation of signals defined piecewise Even and Odd Signals Commonly encountered signals Definition A signal $x(t)$ is said to be, (1) Even if, (2) Odd if, The following figures illustrate clearly, Any signal $x(t)$ can be written as the sum of an even signal and odd signal.(4) where, $x_e(t)$ is the even part and $x_o(t)$ is the odd part.(5) So,(6) Therefore,(7) And,(8) Therefore,(9) Examples Properties of Even and Odd Signals Addition/Subtraction: Even Signal $\pm$ Even Signal =Even signal_ Odd Signal $\pm$ Odd Signal =Odd signal Even Signal $\pm$ Odd Signal = We can't say anything Multiplication: Even * Even = Even Odd * Odd = Even Even * Odd= Odd Integrals: If $x(t)$ is odd then $\int_{-A}^{A}x(t)dt=0$ Example: $\int_{-1}^{1}sin^3(t)dt=0$ If $x(t)$ is even then $\int_{-A}^{A}x(t)dt=2\int_{0}^{A}x(t)dt$ Conjugate Symmetry: Suppose $x(t)$ is a complex signal $\Rightarrow$ $x(t)=a(t)+jb(t) = r(t)e^{j\theta(t)}$ Even Signal is Conjugate Symmetric Signal if- $x(t)=x^(-t)$ and Conjugate Anti-Symmetric if- $x(t)=-x^(-t) \Rightarrow x^(t)=-x(-t) \Rightarrow -x(t)=-x^(-t)$ If $x(t)$ is real $\rightarrow X(j\omega)$ is conjugate symmetric. If $x(t)$ is real and even $\rightarrow X(j\omega)$ is real. If $x(t)$ is complex, conjugate symmetric $\rightarrow X(j\omega)$ is real.
The 2n law of thermodynamics can be stated in terms of entropy as follows $dS \geq \frac{dQ}{T},$ which holds for all quasistatic processes (reversible and irreversible ones). Is there a generalization of this statement to a general process between two equilibrium states $e_1$ and $e_2$ (a non-quasistatic process)? I.e. can one write down a similar inequality for $\Delta S = S(e_2) - S(e_1)$ (linking it to $\Delta Q$ and so on)? Or at the very least, is it possible to derive the well-known $\Delta S \geq 0$ for an isolated system? I'm aware of the fact that one can always write $\Delta S = \int_{\gamma} \frac{dQ}{T}$ for any reversible process $\gamma$ driving the system from to $e_1$ to $e_2$. However, it's not obvious how to exploit this, if at all.
Search Now showing items 1-2 of 2 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... The ALICE TPC, a large 3-dimensional tracking device with fast readout for ultra-high multiplicity events (Elsevier, 2010-10-01) The design, construction, and commissioning of the ALICE Time-Projection Chamber (TPC) is described. It is the main device for pattern recognition, tracking, and identification of charged particles in the ALICE experiment ...
Our reader Eswar Chellappa has sent his work on the solution of ‘3X+1’ problem, also called Collatz Conjecture. He had been working on the proof of Collatz Conjecture off and on for almost ten years. The Collatz Conjecture can be quoted as follow: Let $\phi : \mathbb{N} \to \mathbb{N}^+$ be a function defined such that: $$\phi(x):= \begin{cases} \frac{x}{2}, & \text{if }… Infinitely many answers questions are possible to the answer, “No”. So, our real task should be to find one of THOSE many, which seems to be a perfect one. A simple and the first ever logical approach of giving answers to a question is to derive answers from the question, that is, replace some words of the question with reasonable ones and… This mathematical fallacy is due to a simple assumption, that $ -1=\dfrac{-1}{1}=\dfrac{1}{-1}$ . Proceeding with $ \dfrac{-1}{1}=\dfrac{1}{-1}$ and taking square-roots of both sides, we get: $ \dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}$ Now, as the Euler’s constant $ i= \sqrt{-1}$ and $ \sqrt{1}=1$ , we can have $ \dfrac{i}{1}=\dfrac{1}{i} \ldots \{1 \}$ $ \Rightarrow i^2=1 \ldots \{2 \}$ . This is complete contradiction to the… Here is an interesting mathematical puzzle alike problem involving the use of Egyptian fractions, whose solution sufficiently uses the basic algebra. Problem Let a, b, c, d and e be five non-zero complex numbers, and; $ a + b + c + d + e = -1$ … (i) $ a^2+b^2+c^2+d^2+e^2=15$ …(ii) $ \dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} +\dfrac{1}{d} +\dfrac{1}{e}= -1$… Ramanujan (1887-1920) discovered some formulas on algebraic nested radicals. This article is based on one of those formulas. The main aim of this article is to discuss and derive them intuitively. Nested radicals have many applications in Number Theory as well as in Numerical Methods . The simple binomial theorem of degree 2 can be written as: $ {(x+a)}^2=x^2+2xa+a^2 \… This is a puzzle which I told to my classmates during a talk, a few days before. I did not represent it as a puzzle, but a talk suggesting the importance of Math in general life. This is partially solved for me and I hope you will run your brain-horse to help me solve it completely. If you didn’t notice,… This is a famous problem of intermediate analysis, also known as ‘Archimedes’ Cattle Problem Puzzle’, sent by Archimedes to Eratosthenes as a challenge to Alexandrian scholars. In it one is required to find the number of bulls and cows of each of four colors, the eight unknown quantities being connected by nine conditions. These conditions ultimately form a Pell equation… This is not just math, but a very good test for linguistic reasoning. If you are serious about this test and think that you’ve a sharp [at least average] brain then read the statement (only) below –summarize it –find the conclusion and then answer that whether summary of the statement is Yes or No. [And if you’re not serious about… Two close friends, Robert and Thomas, met again after a gap of several years. Robert Said: I am now married and have three children. Thomas Said: That’s great! How old they are? Robert: Thomas! Guess it yourself with some clues provided by me. The product of the ages of my children is 36. Thomas: Hmm… Not so helpful clue. Can… Before my college days I used to multiply this way. But as time passed, I learned new things. In a Hindi magazine named “Bhaskar Lakshya”, I read an article in which a columnist ( I can’t remember his name) suggested how to multiply in single line (row). That was a magic to me. I found doing multiplications this way, very faster –… Multiplication is probably the most important elementary operation in mathematics; even more important than usual addition. Every math-guy has its own style of multiplying numbers. But have you ever tried multiplicating by this way? Exercise: $ 88 \times 45$ =? Ans: as usual :- 3960 but I got this using a particular way: 88 45… Let have a Test: You need to make a calculation. Please do neither use a calculator nor a paper. Calculate everything “in your brain”. Take 1000 and add 40. Now, add another 1000. Now add 30. Now, add 1000 again. Add 20. And add 1000 again. And an additional 10. So, You Got The RESULT! Quicker you see the… A triangle $ T $ is divided into smaller triangles such that any two of the smaller triangles either have no point in common, or have a vertex in common, or actually have an edge in common. Thus no two smaller triangles touch along part of an edge of them. For an illustration let me denote the three vertices of… Problem1: Smallest Autobiographical Number: A number with ten digits or less is called autobiographical if its first digit (from the left) indicates the number of zeros it contains,the second digit the number of ones, third digit number of twos and so on. For example: 42101000 is autobiographical. Find, with explanation, the smallest autobiographical number. Solution of Problem 1 Problem 2:… In how many ways can two queens, two rooks, one white bishop, one black bishop, and a knight be placed on a standard $ 8 \times 8$ chessboard so that every position on the board is under attack by at least one piece? Note: The color of a bishop refers to the color of the square on which it sits,… Four friends Matt, James, Ian and Barry, who all knew each other from being members of the Automattic, called Automatticians, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew… Part I: A fox chases a rabbit. Both run at the same speed $ v$ . At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $ \alpha $ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is… Statement A series $ \sum {u_n}$ of positive terms is convergent if from and after some fixed term $ \dfrac {u_{n+1}} {u_n} < r < {1} $ , where r is a fixed number. The series is divergent if $ \dfrac{u_{n+1}} {u_n} > 1$ from and after some fixed term. D’ Alembert’s Test is also known as the ratio test…
I don't think it's on topic, but it's no skin off my back to write a short answer. You need to use the chain rule $(fg)' = fg' + gf'$: $$\begin{align}\frac{\mathrm dP}{\mathrm dv} &= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi \left[v^2\frac{\mathrm d}{\mathrm dv}(\mathrm e^{-mv^2/2kT}) + \mathrm e^{-mv^2/2kT}\frac{\mathrm d}{\mathrm dv}(v^2)\right] \\&= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi \left[v^2\left(\frac{-2mv}{2kT}\mathrm e^{-mv^2/2kT}\right) + \mathrm e^{-mv^2/2kT}(2v)\right] \\&= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi\mathrm e^{-mv^2/2kT}\left(\frac{-mv^3}{kT} + 2v\right) = 0\end{align}$$ Since an exponential cannot be zero, the polynomial factor must be equal to zero: $$\begin{align}\frac{-mv^3}{kT} + 2v &= 0 \\v\left(2 - \frac{mv^2}{kT}\right) &= 0 \\v &= 0, \pm \sqrt{\frac{2kT}{m}}\end{align}$$ You're not interested in the $v = 0$ stationary point, and the negative square root is unphysical (we're talking about a distribution of speeds, which must be non-negative). So the maximum occurs at $v = \sqrt{2kT/m}$. From a chemist's point of view this is the physically important quantity, as it represents the most probable speed of a gas molecule. You could substitute it back into $P(v)$ and find out the value of $P$ for this value of $v$, but I don't know what circumstances this would be of interest in.
Back at the start of the year (which really doesn’t seem like that long a time ago) I was looking at using Dirichlet Processes to cluster binary data using PyMC3. I was unable to get the PyMC3 mixture model API working using the general purpose Gibbs Sampler, but after some tweaking of a custom likelihood function I got something reasonable-looking working using Variational Inference (VI). While this was still useful for exploratory analysis purposes, I’d prefer to use MCMC sampling so that I have more confidence in the groupings (since VI only approximates the posterior) in case I wanted to use these groups to generate further research questions. After working on something else for a few months I came back to this problem with a bit more determination (and more importantly, time) and decided to try implementing my own Gibbs Sampler. I found a really useful R package called BayesBinMix that should have been ideal, since it implements several sampling methods and is specifically designed for binary data. Unfortunately, it is written in pure R and is therefore a bit too slow on the datasets I was using ($N \approx 1000$, $p \approx 50$), although others may find it suits their need. The best remaining option then was to write my own sampler, not least so that I would have full control of how it worked and be able to tailor it to my exact specifications, but more importantly it would be a fantastic learning opportunity as I hadn’t yet coded one up, instead using general purpose libraries like PyMC3 and JAGS. R package Since one of my requirements was speed, I chose to write the sampler inC++ and then build an R package around it as the Rcpp integration isvery user-friendly. All the code used in this post is available onGithub, and can be installedusing devtools (not tested on Windows): devtools::install_github('stulacy/bmm-mcmc') Dataset I’ve provided several simulated datasets with this package. in this post I’ll use one with 1,000 observations of 5 binary variables that are formed from 3 classes. All variables have a different rate in each class - a rather nice simplification that will never occur in real world data. library(bmmmcmc)library(tidyverse)data(K3_N1000_P5) The code used to simulate this data is shown below (also available in R/simulate_data.R). set.seed(17)N <- 1000P <- 5theta_actual <- matrix(c(0.7, 0.8, 0.2, 0.1, 0.1, 0.3, 0.5, 0.9, 0.8, 0.6, 0.1, 0.2, 0.5, 0.4, 0.9), nrow=3, ncol=5, byrow = T)cluster_ratios <- c(0.6, 0.2, 0.2)mat_list <- lapply(seq_along(cluster_ratios), function(i) { n <- round(N * cluster_ratios[i]) sapply(theta_actual[i,], function(p) rbinom(n, 1, p))})mat <- do.call('rbind', mat_list)saveRDS(mat, "data/K3_N1000_P5_clean.rds") Full Gibbs sampler To get started I used a simple finite mixture model as opposed to the infinite Dirichlet Process version I used before. Where $\pi$ are cluster/component weights, $\alpha$ is a vector of concentration parameters (here $\alpha_1, \dots, \alpha_k = \frac{\alpha_0}{k}$), $z$ are the cluster labels, $\theta$ are the Bernoulli parameter which is a $k \times d$ matrix of $k$ clusters and $d$ variables, and $x$ is the data comprising an $n \times d$ binary matrix. I use the same $a$, $b$ priors on all $\theta$, here Jeffrey’s prior where $a = b = 0.5$. The concentration parameter is fixed at $\alpha_0 = 1$. $$\pi \sim \text{Dirichlet}(\alpha_1, \dots, \alpha_k)$$ $$z \sim \text{Cat}(\pi_1, \dots, \pi_k)$$ $$θ_{kd} \sim \text{Beta}(a_{kd}, b_{kd})$$ $$x_{id} \sim \text{Bern}(\theta_{z_{i}d})$$ The first sampler I built was a full Gibbs sampler that samples from all of these random variables: $\pi$: The cluster weights $\theta$: The Bernoulli parameters (one for each cluster/variable pair) $z$: Cluster assignments (hard labels) This is relatively straight forward owing to the use of conjugate priors (see Grantham’s guide here). Since $k$ is fixed I had to assign it a value. I used $k = 3$: the known number of groups - if I couldn’t get a sampler working with a known $k$ then I had no chance when it is unknown! full <- gibbs_full(K3_N1000_P5, nsamples = 10000, K=3, burnin = 1000) The traces below show a) the proportion of the dataset assigned to each of the 3 components along with b) the sampled $\theta$ values, where the line colour denotes a different component. Notice how the proportion in each cluster matches up with the 60%/20%/20% ratio I used to simulate the data, and even better, how $\theta$ match up with the known values. There is more variation in $\theta$ for the smaller two clusters as expected, and while I didn’t run any serious diagnostics, the full Gibbs sampler is able to identify the known components that generated the data for this toy problem as a proof of concept. plot_gibbs(full) Collapsed Gibbs The full sampler can run into problems due to having separate values of $\theta$ for each $k$, when in reality $\theta$ may not differ too much between groups. In this instance there can be multiple observations from different groups having high probabilities of similar $\theta$ values, which then can prevent $\theta$ from updating quickly as it only updates a group at a time. The Collapsed Gibbs sampler aims to reduce this by integrating out $\pi$ so that only the cluster labels $z$ are sampled, see Eq 3.5 in (Neal 2000) for the sampling probabilities. $\theta$ can be estimated afterwards from the observed rates in the different components. collapsed <- gibbs_collapsed(K3_N1000_P5, nsamples = 10000, K=3, burnin = 1000) The traces below show that this sampler has also been able to identify these 3 groups with a much simpler sampling scheme. plot_gibbs(collapsed) Dirichlet Process Now, while implementing these sampling methods was a valuable experience, my overall goal is to use a fully Bayesian approach where $k$ is random or even infinite. In my tinkerings with PyMC3 I was using the stick-breaking formulation of the Dirichlet Process. However, while reading up on sampling DPs I came across a lot of literature on the Chinese Restaurant Process (CRP) and it conceptually made a lot of sense for how it extends the collapsed Gibbs sampler. In it, a new observation gets assigned either to an existing table/cluster with a probability proportional to how strongly it matches those groups (very similar to finite model), or gets assigned to a new table/cluster with a constant probability, where $\alpha$ acts as a tuning parameter for how likely new clusters are. I implemented the CRP using Algorithm 3 from (Neal2000).Aside from the obvious difference in that $k$ is now infinite, $\alpha$ isnow a Gamma random variable with parameters $\alpha \sim \text{Gamma}(1, 1)$. Fortechnical performance I have an argument limiting the number of clustersfound, although in practice as many as the number of observations in thedataset can be found (hence this model being termed an infinitemixture model). dp_1 <- gibbs_dp(K3_N1000_P5, nsamples = 10000, maxK = 20, burnin = 1000, relabel = FALSE) The resulting traces can be seen below, and while they are largely similar to before, there are two areas of concern: The traces change colour every few thousand samples! I.e. the topmost trace in the upper panel changes from red to orange to light blue There are a lot of components with a small number of members. Note the $\theta$ traces that drastically vary: these are from small poorly defined clusters. This is despite the plot being limited to clusters that had 10% membership at each timepoint. Problem 1) is what is termed the “label-switching problem”, and is actually indicative of a healthy sampler, since it means the sampler is exploring the full parameter space. Section 3.1 of (Jasra, Holmes, and Stephens 2005) details this, where Figure 5 (a) (reproduced below) is a Gibbs sampler that has only visited one of the $k!$ modes. This is the behaviour we observed in the finite sampler above, whereby it seemingly picked out the 3 groups perfectly, but in fact this demonstrated the lack of exploration. Indeed, (Jasra, Holmes, and Stephens 2005) demonstrate a modified MCMC procedure to solve this issue and as shown in Figure 5 (b) this ‘tempered Metropolis-Hastings’ method is able to move between modes, the desired behaviour. So it’s not a bad thing that this behaviour is observed, but we’d ideally be able to obtain the original components that we know exist. Fortunately, there are various “relabelling” techniques available with this aim. The second issue is a bit more worrying as it could indicate that alpha isn’t sampling well since lots of small poorly supported components are formed. plot_gibbs(dp_1, cluster_threshold = 0.1) Relabelling I decided to focus on solving the relabelling problem first as it is commonly understood. The most well cited relabelling technique (Stephens 2000). Again, the author of BayesBinMix has made a fantastic package called label.switchingthat performs post-hoc relabelling using a variety of methods, includingStephen’s. However, yet again it is written in R and so is rather slow.Furthermore, since it is post-hoc it requires storage of all clusterassignment probabilities for every observation at every sample. For thisdataset - even if I limit the number of clusters to 20 - this is still30,000 floats to be saved per sample, or 3e8 floats over a 10,000 samplerun. This alone uses 2GB of memory and so is unfeasible for largedatasets or for situations where the maximum number of clusters cannotbe assumed to be quite so small. Furthermore, it’s extremely slow torun. Fortunately, (Stephens 2000) also provides details of an online version in Section 6.2. I therefore decided to write a C++ implementation so that relabelling could be done on the fly. Implementing the relabelling algorithm was actually the mosttime-consuming part of this work, so if anyone finds it useful pleaselet me know. The source code is in src/stephens.cppand makes use of the lpsolve Linear Programming library. With the online relabelling method working the traces (below) look much better, although they are still not perfect. The relabeling succeeded on the most populous cluster but struggled with the 2 smaller ones - note how one group changes from orange to purple (denoting a label switch) around sample 3,5000. This isn’t a major issue as a manual relabelling would work here if needed. This method also still suffers from the issue of small clusters appearing and disappearing rapidly, not even those with just one member but those with at least 100. This is likely due to the manner in which the Dirichlet Process works, as it is known to generate a large number of parasitic components, larger than would be found with a finite mixture model. dp_2 <- gibbs_dp(K3_N1000_P5, nsamples = 10000, maxK = 20, burnin = 1000, relabel = TRUE, burnrelabel = 500)dp_2$z <- dp_2$z_relabelleddp_2$theta <- dp_2$theta_relabelledplot_gibbs(dp_2, cluster_threshold = 0.1) Further Work On the whole, implementing a custom sampler has produced far better results than using the generic PyMC3 sampler that I tried a few months ago. I’ve learnt a whole lot about Gibbs sampling and have also obtained a strong grasp of the existing literature on Bayesian Mixture Models. There is scope for further work to take advantage of more modern sampling methods for mixture models. Firstly though I’d like to implement my own sampler for the stick-breaking formulation, just for completeness’ sake. I’d then like to look at other ways of modelling an unknown $k$ without using the infinite $k$ approach of the Dirichlet Process. The Reversible Jump MCMC (RJMCMC) (Richardson and Green 1997) is a modified Metropolis-Hastings sampler that is designed for situations such as mixture models where the number of dimensions is unknown. On a similar note (Nobile and Fearnside 2007)’s Allocation Sampler is a more recent development of a similar idea, although it only samples from the cluster labels $z$ rather than the full joint-posterior and is therefore more efficient. Finally, the current model I’ve been using assumes that all variables have different occurrence rates in each cluster, which is a big assumption. It’s easy to imagine a scenario where some variables aren’t differently distributed in each subgroup, or some variables that are different in one or two groups but otherwise have a general background frequency. (Elguebaly and Bouguila 2013) describes an attempt to incorporate this information into the model by using feature weighting, so each $\theta_{kd}$ has an associated Bernoulli marker which denotes whether that rate is different for that cluster to the background rate, if it isn’t then $\theta_{kd} = \theta_{d}$, thereby reducing the number of variables that impact on the clustering itself. I’m not sure how much time I will have to work on this over the coming months but hopefully I’ll be able to post an update before too long. References Elguebaly, Tarek, and Nizar Bouguila. 2013. “Simultaneous BayesianClustering and Feature Selection Using RJMCMC-Based Learning of FiniteGeneralised Dirichlet Mixture Models.” Signal Processing.https://doi.org/10.1016/j.sigpro.2012.07.037. Jasra, A, C Holmes, and D Stephens. 2005. “Markov Chain Monte CarloMethods and the Label Switching Problem in Bayesian Mixture Modeling.” Statistical Science. https://doi.org/10.1214/088342305000000016. Neal, Radford. 2000. “Markov Chain Sampling Methods for DirichletProcess Mixture Models.” Journal of Computational and GraphicalStatistics. https://doi.org/10.2307/1390653. Nobile, Agostino, and Alastair Fearnside. 2007. “Bayesian Finite MixtureModels with an Unknown Number of Components the Allocation Sampler.” Statistics and Computing. https://doi.org/10.1007/s11222-006-9014-7. Richardson, Sylvia, and Peter Green. 1997. “On Bayesian Analysis ofMixtures with an Unknown Number of Components.” Journal of the RoyalStatistical Society Statistical Methodology Series B.https://doi.org/10.1111/1467-9868.00095. Stephens, Matthew. 2000. “Dealing with Label Switching in MixtureModels.” Journal of the Royal Statistical Society StatisticalMethodology Series B. https://doi.org/10.1111/1467-9868.00265.
I'm reading Newey & McFadden - Large sample estimation and hypothesis testing (in the Handbook of Econometrics, Volume 4, 1994, page 2176). In the model I'm interestend in has some former estimation done before the estimation of the primary model will take place. Hence the primary model (2nd-step) includes some estimated regressors from the former step (the 1st-step). In order to calculate the asymptotic variance I follow a approach, provided by Newey & McFadden, where the joint GMM-conditions are defined as $\widetilde{g}\left(z,\beta,\alpha\right) = \left[g\left(z,\beta,\alpha\right),m\left(z,\alpha\right)\right]$ where $g\left(z,\beta,\alpha\right)$ are the 2nd-step conditions and $m\left(z,\alpha\right)$ are the 1st-step ones. The asymptotic variance of the 2nd-step estimator $\widehat{\beta}$ has, under the assumption of uncorrelated 1st- and 2nd-step moments and uncorrelated 1st step moments (if there are more then one 1st-step estimator which will be included in the primary model), the following form: $Var(\widehat{\beta}) = G_\beta^{-1}\mathbb{E}\left(g(z,\beta,\alpha)g(z,\beta,\alpha)^T\right)(G_\beta^{-1})^T + G_\beta^{-1}G_\alpha\mathbb{E}\left(m(z,\alpha)m(z,\alpha)^T\right)G_\alpha^T (G_\beta^{-1})^T$ where $G_\beta = \frac{\partial\mathbb{E}\left(\widetilde{g}(z,\beta_0, \alpha_0)\right)}{\partial \beta^T}$, $G_\alpha = \frac{\partial\mathbb{E}\left(\widetilde{g}(z,\beta_0, \alpha_0)\right)}{\partial \alpha^T}$ In order to estimate the population moments we replace them by the corresponding sample moments. Assume a OLS-case where the 1st-step looks like $Z = X\alpha + v$ and the 2nd step like $y = X\beta_1 + F(v)\beta_2 + e$ with $X$ a $ \ n\times k \ $-matrix, $\beta_1$ a $\ k\times 1 \ $-vector of coefficients and $F(v)$ is the cdf of the 1st-step residuals. $\beta_2$ is the corresponding coefficient for this function. If I set $\widetilde{X} = \left[X ; F(v)\right]$ as the design-matrix of the 2nd-step and $\widetilde{\beta} = \left[\beta_1 ; \beta_2\right]$ as the corresponding vector of coefficients then, for the quantities which determine the asymptotic variance of $\widehat{\beta}$, we will get for a given sample size n $\widehat{G}_\beta = \frac{\partial\left(\frac{1}{n}\widetilde{X}^T\left( y - \widetilde{X}\widehat{\beta}\right)\right)}{\partial\widehat{\beta}} = - \frac{1}{n}\widetilde{X}^T\widetilde{X}$ The estimator for $\mathbb{E}\left(g(z,\beta,\alpha)g(z,\beta,\alpha)^T\right)$ will be $\frac{1}{n^2}\widetilde{X}^T\left( y - \widetilde{X}\widehat{\beta}\right)\left( y - \widetilde{X}\widehat{\beta}\right)^T\widetilde{X}$ Analogously for $\mathbb{E}\left(m(z,\alpha)m(z,\alpha)^T\right)$ we get $\frac{1}{n^2}X^T\left( y - X\widehat{\alpha}\right)\left( y - X\widehat{\alpha}\right)^TX$ Question: I'm not sure how to derive the estimator for $G_\alpha$. Since $\widehat{G}_\alpha = \frac{\partial\left(\frac{1}{n}\widetilde{X}^T\left( y - \widetilde{X}\widehat{\beta}\right)\right)}{\partial\widehat{\alpha}}$ and $F(v)$ is a column of $\widetilde{X}$ this should be equal to $\widehat{G}_\alpha = \frac{\partial\left( F(v)\widehat{\beta_1}\right)}{\partial\widehat{\alpha}}$ Since $F(v)$ is the cdf of $v$ this should be somthing like a function of the pdf of $v$ but im not quite sure how start. Because $\widehat{\beta} = \left(\widetilde{X}^T\widetilde{X}\right)^{-1}\widetilde{X}^Ty$ the entry $\widehat{\beta}_1$ should depend upon $\widehat{\alpha}$ too but im not quite sure about this. A hint would be very helpful.
Why is $\infty^\infty=\tilde\infty$? WA's ComplexInfinity is the same as Mathematica's: it represents a complex "number" which has infinite magnitude but unknown or nonexistent phase. One can use DirectedInfinity to specify the phase of an infinite quantity, if it approaches infinity in a certain direction. The standard Infinity is the special case of phase 0. Note that Infinity is different from Indeterminate (which would be the output of e.g. 0/0). Some elucidating examples: 0/0returns Indeterminate, since (for instance) the limit may be approached as $\frac{1/n}{1/n}$ or $\frac{2/n}{1/n}$, resulting in two different real numbers. 1/0returns ComplexInfinity, since (for instance) the limit may be approached as $\frac{1}{-1/n}$ or as $\frac{1}{1/n}$, but every possible way of approaching the limit gives an infinite answer. Abs[1/0]returns Infinity, since the limit is guaranteed to be infinite and approached along the real line in the positive direction. In your particular example, you get ComplexInfinity because the infinite limit may be approached as (e.g.) $n^n$ or as $n^{n+i}$. TLDR: $\infty$ is not a number, and thus $\infty^\infty$ is meaningless, and Wolfram Alpha is using $\tilde\infty$ to represent something I like to to think of in the sense of "one-point compactification", a topological concept A.) The symbol $\infty$ is not a number in its own right. It can represent a lot of things, and many different objects can be "infinitely large". Just think of something infinite as something not finite and you are generally off to a good start. B.) If infinity is not a number, we can't do arithmetic on it that makes sense in all context, and so we most definitely can't exponetiate it meaningfully without some fundamentals first. For example, there are some infinite things where arithmetic is defined (in some sense - check out ordinal numbers), There are some things where basic arithmetic does nothing (i.e. $\infty+1=\infty$) There are some infinite thing where you most definitely cannot do arithmetic in the way you are used to (take, for example, the limit concept of infinity from a basic Calculus class) C) Wolfram Alpha appears to represent a lot of things as $\tilde\infty$ that are ill-defined according to the real-number system you are used to - for example, according to Wolfram Alpha, $\frac{1}{0}=\tilde\infty$, whereas I would say that $\frac{1}{0}$ is undefined. You could stretch this to say that $\frac{1}{0} = \lim_{x\to 0} \frac 1x = \pm\infty$ in the extended-real number system, but this is starting to push things. To really understand what Wolfram Alpha is doing you must first understand the one-point compactification of $\Bbb C$. See my note at the bottom for links and more details. Notes: It has been pointed out in the comments that I ought to mention that Mathematica/WA use $\tilde\infty$ to represent an infinite magnitude number with no defined phase. This seemed more software dependent, though I get where the the commentator is coming from. I chose to focus on the real number system the OP is likely accustomed to, and focus on the concept of $\infty^\infty$ itself, not the software's interpretation. $\tilde\infty$ doeshave meaning in some sense - see for example, this Wikipedia page on "one-point compactification". Just imagine the complex plane as a big sheet where we take the edges and pull them all into one point and call it $\infty$. For more information look at the page on the Riemann Sphere. Similarly, $\infty$ can have a defined meaning in certain contexts outside of the scope of this question - see, for example, the Wikipedia page on the Real Projective Line. In these contexts, one couldperhaps consider $\infty$ a number. For more on this, see the Wikipedia page "Projectively Extended Real Line". An interpretation in $\overline{\mathbb{C}}=\mathbb{C}\cup \{\infty\}$ of $\infty^\infty$ is via limits. For example $$\lim_{z\to 0}\left(\frac{1}{\|z\|}\right)^{\frac{1}{\|z\|}}\underbrace{=}_{\text{symbolic equality}}\infty^\infty=\infty.$$ Note. $-\infty$ and $+\infty$ don't belong to $\overline{\mathbb{C}}$. As far as I can tell: When $x$ and $y$ approach positive infinity, Wolfram Alpha assumes they may do so through the complex plane, as long as the arguments (angles) of $x$ and $y$ approach zero. The argument of $x^{x+i}$ does not converge to any value as $x\to\infty$, even though the arguments of $x$ and $x+i$ go to zero. (This is because the argument is $\ln x$.) Thus, Wolfram Alpha responds with complex infinity (unknown argument) rather than positive infinity.
Topology AtlasDocument # ppae-04 Concerning the dual group of a dense subgroup W. W. Comfort, S. U. Raczkowski and F. Javier Trigos-ArrietaProceedings of the Ninth Prague Topological Symposium(2001)pp. 23-35 Throughout this Abstract, G is a topological Abelian group and[^G] is the space of continuous homomorphisms from G into T in the compact-open topology. A dense subgroup D of G determines G if the (necessarily continuous)surjective isomorphism [^G]\twoheadrightarrow[^D] given byh --> h\|D is a homeomorphism, and G is determined ifeach dense subgroup of G determines G. The principal result in thisarea, obtained independently by L. Auß enhofer and M. J. Chasco, is the following: Every metrizable group is determined.The authors offer several related results, including these. Question.Is there in ZFC a cardinal \kappa such that a compact group G isdetermined if and only if w(G) < \kappa?Is \kappa = \non( There are (many) nonmetrizable, noncompact, determined groups. If the dense subgroup D i determines G i with G i compact, then \oplus i D i determines \Pi i G i. In particular, if eachG i is compact then \oplus i G i determines \Pi i G i. Let G be a locally bounded group and let G + denote G with its Bohrtopology. Then G is determined if and only if G + is determined. Let \non( N) be the least cardinal \kappa such that someX subset or equal \TT of cardinality \kappa has positive outer measure. No compact G with w(G) >= \non( N) is determined; thus if\non( N)=\aleph 1 (in particular if CH holds), an infinite compact group G is determined if and only if w(G)=\omega. N)?\kappa = \aleph 1? Mathematics Subject Classification. 22A10 22B99 22C05 43A40 54H11 (03E35 03E50 54D30 54E35). Keywords. Bohr compactification, Bohr topology, character, charactergroup, Au{\ss}enhofer-Chasco Theorem, compact-open topology, densesubgroup, determined group, duality, metrizable group, reflexive group,reflective group. Document formats AtlasImage (for online previewing) LaTeX 40.2 Kb DVI 57.4 Kb PostScript 227.7 Kb gzipped PostScript 91.1 Kb PDF 269.1 Kb arXiv math.GN/0204147 Metadata Citation Reference list in BibTeX Comments. A full version of this article, with complete proofs, will be submitted for publication elsewhere. Copyright © 2002Charles University andTopology Atlas. Published April 2002.
Can it be generalized for other powers ? Wolfram seems to say it is true for k below 20000. I stumbled upon it randomly when trying to approximate $\sum_{n=1}^{n=+\infty} \frac{1}{n^4}$. My reasoning was : $$\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2=\sum_{n=k}^{n=+\infty} \frac{1}{n^4} + (\text{double products}) \geq\sum_{n=k}^{n=+\infty} \frac{1}{n^4}$$ So $$\sum_{n=1}^{n=+\infty} \frac{1}{n^4} \leq \sum_{n=1}^{n=k-1} \frac{1}{n^4}+\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2 \leq \left(\sum_{n=1}^{n=k-1} \frac{1}{n^4}\right)+\left(\frac{1}{k-\frac{1}{2}}\right)^2$$ where the last inequality comes from An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$. Then I noticed that, perhaps, I could raise the last term to the power of 3 instead of just 2, making the inequality stronger.
Difference between revisions of "Inverse function theorem" (→Statement with symbols) (→Statement with symbols) Line 12: Line 12: ! Version type !! Statement ! Version type !! Statement \|- \|- − \| specific point, named functions \|\| Suppose <math>f</math> is a [[function]] of one variable that is a [[one-one function]] and <math>a</math> is in the [[domain]] of <math>f</math>. Suppose <math>f</math> is continuous in an open interval containing <math>a</math> + \| specific point, named functions \|\| Suppose <math>f</math> is a [[function]] of one variable that is a [[one-one function]] and <math>a</math> is in the [[domain]] of <math>f</math>. Suppose <math>f</math> is continuous in an open interval containing <math>a</math> [[differentiable function\|differentiable]] at <math>a</math>and <math>b = f(a)</math>. Suppose further that the [[fact about::derivative]] <math>f'(a)</math> is nonzero, i.e., <math>f'(a) \ne 0</math>. Then the [[fact about::inverse function]] <math>f^{-1}</math> is [[differentiable function\|differentiable]] at <math>b</math>, and further:<br><math>(f^{-1})'(b) = \frac{1}{f'(a)}</math> \|- \|- − \| generic point, named functions, point notation \|\| Suppose <math>f</math> is a [[function]] of one variable that is a [[one-one function]]. Then, the formula for the [[derivative]] of the [[inverse function]] is as follows: <br><math>\! (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}</math><br> with the formula applicable at all points in the [[range]] of <math>f</math> for which <math>f</math> is continuous ''around'' the point + \| generic point, named functions, point notation \|\| Suppose <math>f</math> is a [[function]] of one variable that is a [[one-one function]]. Then, the formula for the [[derivative]] of the [[inverse function]] is as follows: <br><math>\! (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}</math><br> with the formula applicable at all points in the [[range]] of <math>f</math> for which <math>f</math> is continuous ''around'' the point and <math>f'(f^{-1}(x))</math> exists and is nonzero. \|- \|- \| generic point, named functions, point-free notation \|\| Suppose <math>f</math> is a [[function]] of one variable that is a [[one-one function]]. Then, the formula for the [[derivative]] of the [[inverse function]] is as follows: <br><math>\! (f^{-1})'= \frac{1}{f' \circ f^{-1}}</math><br> with the formula applicable at all points in the [[range]] of <math>f</math> for which <math>f</math> is continuous around the point and <math>f'(f^{-1}(x))</math> exists and is nonzero. \| generic point, named functions, point-free notation \|\| Suppose <math>f</math> is a [[function]] of one variable that is a [[one-one function]]. Then, the formula for the [[derivative]] of the [[inverse function]] is as follows: <br><math>\! (f^{-1})'= \frac{1}{f' \circ f^{-1}}</math><br> with the formula applicable at all points in the [[range]] of <math>f</math> for which <math>f</math> is continuous around the point and <math>f'(f^{-1}(x))</math> exists and is nonzero. Latest revision as of 18:39, 20 January 2013 This article is about a differentiation rule, i.e., a rule for differentiating a function expressed in terms of other functions whose derivatives are known. View other differentiation rules Contents Statement Verbal statement The derivative of the inverse function at a point equals the reciprocal of the derivative of the function at its inverse image point. Statement with symbols Version type Statement specific point, named functions Suppose is a function of one variable that is a one-one function and is in the domain of . Suppose is continuous in an open interval containing as well as differentiable at , and suppose . Suppose further that the derivative is nonzero, i.e., . Then the inverse function is differentiable at , and further: generic point, named functions, point notation Suppose is a function of one variable that is a one-one function. Then, the formula for the derivative of the inverse function is as follows: with the formula applicable at all points in the range of for which is continuous around the point and exists and is nonzero. generic point, named functions, point-free notation Suppose is a function of one variable that is a one-one function. Then, the formula for the derivative of the inverse function is as follows: with the formula applicable at all points in the range of for which is continuous around the point and exists and is nonzero. Pure Leibniz notation using dependent and independent variables Suppose is a variable functionally dependent on . Then, (with domain caveats as above). MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this? The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varyingthe point we are considering. This is the purpose of the second, generic pointversion. One-sided version One-sided versions exist, but we need to be careful about issues of left and right. We state the two cases: Case for behavior of original function at Short version Long version (using specific point, named functions) increasing function from left left hand derivative of is related to left hand derivative of Suppose is an increasing function from the left at a point . Suppose . Suppose further that the left hand derivative is nonzero, i.e., . Then the inverse function is left differentiable at , and further: increasing function from right right hand derivative of is related to right hand derivative of Suppose is an increasing function from the right at a point (in other words, increases on the immediate right of ). Suppose . Suppose further that the right hand derivative is nonzero, i.e., . Then the inverse function is right differentiable at , and further: decreasing function from left right hand derivative of is related to left hand derivative of Suppose is a decreasing function on the left at a point . Suppose . Suppose further that the left hand derivative is nonzero, i.e., . Then the inverse function is right differentiable at , and further: decreasing function from right left hand derivative of is related to right hand derivative of Suppose is a decreasing function from the right at a point . Suppose . Suppose further that the right hand derivative is nonzero, i.e., . Then the inverse function is left differentiable at , and further: Some additional notes: For a point in the interior of the domain at which the function is continuous, being increasing on the immediate left forces the function to be increasing on the immediate right, and vice versa. Similarly for decreasing. More generally, a continuous one-one function on an interval must be either increasing through the interval or decreasing throughout the interval. We have been more specific in our statements in the table above to allow for the possibility of piecewise defined functions with discontinuities as well as to tackle the issue of interval endpoints where only one-sided notions make sense. Infinity-sensitive versions The following version accounts for the infinity cases. We provide only the specific point, named functions version. Assume that is a one-one function that is continuous at a point in its domain, with . There are six cases of interest: Case for Case for Relation between them Increase/decrease? Example undefined, but approaching zero -- Both increasing. is increasing through (with the rate of increase peaking to ) and is increasing through (though the rate of increase dips to zero because it's a point of inflection with horizontal tangent). , , positive positive reciprocals of each other. Both increasing. is increasing through and is increasing through . , , zero undefined, but approaching , i.e., vertical tangent -- Both increasing. is increasing through (though the rate of increase dips to zero because it's a point of inflection with horizontal tangent) and is increasing through (with the rate of increase peaking to ). , , zero undefined, but approaching , i.e., vertical tangent -- Both decreasing. is decreasing through (though the rate of decrease dips to zero because it's a point of inflection with horizontal tangent) and is decreasing through (with the rate of decrease peaking to ). , , negative negative reciprocals of each other Both decreasing. is decreasing through and is decreasing through . , , undefined, but approaching , i.e., vertical tangent zero -- Both decreasing. is decreasing through (with the rate of decrease peaking to ) and is decreasing through (though the rate of decrease dips to zero because it's a point of inflection with horizontal tangent) , , Significance Version type Significance specific point, named functions (two-sided, finite) This tells us that if a one-one function is differentiable at a point with nonzero derivative, then is differentiable at the image of that point under . specific point, named functions (two-sided, infinity-sensitive) This tells us that if a one-one function is either differentiable at a point or has a vertical tangent, then is either differentiable at the image of that point or has a vertical tangent. Moreover, we can pair the possibilities for with the possibilities for using the theorem. specific point, named functions (one-sided version) This tells us that if a one-one function is one-sided differentiable at a point, then the inverse function is one-sided differentiable at the image point, where the side remains the same for an increasing function and gets switched for a decreasing function. generic point, named functions (two-sided, finite) This tells us that the inverse of a differentiable one-one function with nowhere zero derivative is also a differentiable one-one function. generic point, named functions (two-sided, infinity-sensitive) This tells us that the inverse of a one-one function that is differentiable or has a vertical tangent at each point is also a one-one function that is either differentiable or has a vertical tangent at each point. generic point, named functions (one-sided, infinity-sensitive) This tells us that the inverse of a one-one function that is one-sided differentiable or has a (one or two-sided) vertical tangent at each point is also a one-one function that is one-sided differentiable or has a (one or two-sided) vertical tangent at each point. Note two important caveats: The differentiable of at gives us information about the differentiability of , not at , but at . The reciprocation means we have to be careful about zero and infinity. Thus, the inverse of a differentiable one-one function need not be differentiable everywhereon its domain. Computational feasibility significance Version type Significance specific point, named functions Consider a one-one function . It is possible to compute if we know the value of where . specific point, named functions (second version) Consider a one-one function . It is possible to compute if we know the generic expression for and the specific value . generic point, named functions Consider a one-one function . It is possible to find a generic expression for in terms of and . Note: [SHOW MORE] Computational results significance See the section #Infinity-sensitive versions for some of the basic computational results in this direction. Examples Generic point examples Below we list some examples of functions and their inverse functions to which the inverse function theorem can be fruitfully applied. Original function Domain on which it restricts to a one-one function Inverse function for the restriction to that domain Domain of inverse function (equals range of original function) Derivative of original function Derivative of inverse function Explanation using inverse function theorem sine function arc sine function cosine function By the inverse function theorem, the derivative at is . Use that on the range of and to get that tangent function arc tangent function all real numbers secant-squared function By the inverse function theorem, the derivative at is . Use that and get . natural logarithm exponential function all real numbers reciprocal function exponential function By the inverse function theorem, the derivative at is
The bounty period lasts 7 days. Bounties must have a minimum duration of at least 1 day. After the bounty ends, there is a grace period of 24 hours to manually award the bounty. Simply click the bounty award icon next to each answer to permanently award your bounty to the answerer. (You cannot award a bounty to your own answer.) @Mathphile I found no prime of the form $$n^{n+1}+(n+1)^{n+2}$$ for $n>392$ yet and neither a reason why the expression cannot be prime for odd n, although there are far more even cases without a known factor than odd cases. @TheSimpliFire That´s what I´m thinking about, I had some "vague feeling" that there must be some elementary proof, so I decided to find it, and then I found it, it is really "too elementary", but I like surprises, if they´re good. It is in fact difficult, I did not understand all the details either. But the ECM-method is analogue to the p-1-method which works well, then there is a factor p such that p-1 is smooth (has only small prime factors) Brocard's problem is a problem in mathematics that asks to find integer values of n and m for whichn!+1=m2,{\displaystyle n!+1=m^{2},}where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan.== Brown numbers ==Pairs of the numbers (n, m) that solve Brocard's problem are called Brown numbers. There are only three known pairs of Brown numbers:(4,5), (5,11... $\textbf{Corollary.}$ No solutions to Brocard's problem (with $n>10$) occur when $n$ that satisfies either \begin{equation}n!=[2\cdot 5^{2^k}-1\pmod{10^k}]^2-1\end{equation} or \begin{equation}n!=[2\cdot 16^{5^k}-1\pmod{10^k}]^2-1\end{equation} for a positive integer $k$. These are the OEIS sequences A224473 and A224474. Proof: First, note that since $(10^k\pm1)^2-1\equiv((-1)^k\pm1)^2-1\equiv1\pm2(-1)^k\not\equiv0\pmod{11}$, $m\ne 10^k\pm1$ for $n>10$. If $k$ denotes the number of trailing zeros of $n!$, Legendre's formula implies that \begin{equation}k=\min\left\{\sum_{i=1}^\infty\left\lfloor\frac n{2^i}\right\rfloor,\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\right\}=\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\end{equation} where $\lfloor\cdot\rfloor$ denotes the floor function. The upper limit can be replaced by $\lfloor\log_5n\rfloor$ since for $i>\lfloor\log_5n\rfloor$, $\left\lfloor\frac n{5^i}\right\rfloor=0$. An upper bound can be found using geometric series and the fact that $\lfloor x\rfloor\le x$: \begin{equation}k=\sum_{i=1}^{\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor\le\sum_{i=1}^{\lfloor\log_5n\rfloor}\frac n{5^i}=\frac n4\left(1-\frac1{5^{\lfloor\log_5n\rfloor}}\right)<\frac n4.\end{equation} Thus $n!$ has $k$ zeroes for some $n\in(4k,\infty)$. Since $m=2\cdot5^{2^k}-1\pmod{10^k}$ and $2\cdot16^{5^k}-1\pmod{10^k}$ has at most $k$ digits, $m^2-1$ has only at most $2k$ digits by the conditions in the Corollary. The Corollary if $n!$ has more than $2k$ digits for $n>10$. From equation $(4)$, $n!$ has at least the same number of digits as $(4k)!$. Stirling's formula implies that \begin{equation}(4k)!>\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\end{equation} Since the number of digits of an integer $t$ is $1+\lfloor\log t\rfloor$ where $\log$ denotes the logarithm in base $10$, the number of digits of $n!$ is at least \begin{equation}1+\left\lfloor\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)\right\rfloor\ge\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right).\end{equation} Therefore it suffices to show that for $k\ge2$ (since $n>10$ and $k<n/4$), \begin{equation}\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)>2k\iff8\pi k\left(\frac{4k}e\right)^{8k}>10^{4k}\end{equation} which holds if and only if \begin{equation}\left(\frac{10}{\left(\frac{4k}e\right)}\right)^{4k}<8\pi k\iff k^2(8\pi k)^{\frac1{4k}}>\frac58e^2.\end{equation} Now consider the function $f(x)=x^2(8\pi x)^{\frac1{4x}}$ over the domain $\Bbb R^+$, which is clearly positive there. Then after considerable algebra it is found that \begin{align}f'(x)&=2x(8\pi x)^{\frac1{4x}}+\frac14(8\pi x)^{\frac1{4x}}(1-\ln(8\pi x))\\\implies f'(x)&=\frac{2f(x)}{x^2}\left(x-\frac18\ln(8\pi x)\right)>0\end{align} for $x>0$ as $\min\{x-\frac18\ln(8\pi x)\}>0$ in the domain. Thus $f$ is monotonically increasing in $(0,\infty)$, and since $2^2(8\pi\cdot2)^{\frac18}>\frac58e^2$, the inequality in equation $(8)$ holds. This means that the number of digits of $n!$ exceeds $2k$, proving the Corollary. $\square$ We get $n^n+3\equiv 0\pmod 4$ for odd $n$, so we can see from here that it is even (or, we could have used @TheSimpliFire's one-or-two-step method to derive this without any contradiction - which is better) @TheSimpliFire Hey! with $4\pmod {10}$ and $0\pmod 4$ then this is the same as $10m_1+4$ and $4m_2$. If we set them equal to each other, we have that $5m_1=2(m_2-m_1)$ which means $m_1$ is even. We get $4\pmod {20}$ now :P Yet again a conjecture!Motivated by Catalan's conjecture and a recent question of mine, I conjecture thatFor distinct, positive integers $a,b$, the only solution to this equation $$a^b-b^a=a+b\tag1$$ is $(a,b)=(2,5).$It is of anticipation that there will be much fewer solutions for incr...
The Ising model is defined with the Hamiltonian: $$ H = -\sum_{<i,j>}S_i^z\cdot S_j^z $$ What is the difference between quantum version and classical version? My intuition is that the classical version is equal to quantum version in any dimension and any lattice. If we add a transverse field, I think there will be difference, the Hamiltonian is: $$ H = -\sum_{<i,j>}S_i^z\cdot S_j^z - h\sum_i S_i^x $$ My intuition is that the quantum and classical will be different, because in quantum mechanics $[S^x,S^z]\neq0$. However I still don't know the difference in detail.
2018-09-11 04:29 Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-25 06:58 Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-23 11:31 Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-23 11:31 Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-23 11:31 Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-23 11:31 Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-23 11:31 Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-22 06:27 Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-22 06:27 Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 Αναλυτική εγγραφή - Παρόμοιες εγγραφές 2018-08-22 06:27 Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 Αναλυτική εγγραφή - Παρόμοιες εγγραφές
In its simplest form, a SIR model is typically written in continuous time as: \[ \frac{dS}{dt} = - \beta \frac{S_t I_t}{N_t} \] \[ \frac{dI}{dt} = \beta \frac{S_t I_t}{N_t} - \gamma I_t \] \[ \frac{dR}{dt} = \gamma I_t \] Where $\beta$ is an infection rate and $\gamma$ a removal rate, assuming ‘R’ stands for ‘recovered’, which can mean recovery or death. For discrete time equivalent, we take a small time step $t$ (typically a day), and write the changes of individuals in each compartments as: \[ S_{t+1} = S_t - \beta \frac{S_t I_t}{N_t} \] \[ I_{t+1} = I_t + \beta \frac{S_t I_t}{N_t} - \gamma I_t \] \[ R_{t+1} = R_t + \gamma I_t \] The discrete model above remains deterministic: for given values of the rates $\beta$ and $\gamma$, dynamics will be fixed. It is fairly straightforward to convert this discrete model into a stochastic one: one merely needs to uses appropriate probability distributions to model the transfer of individuals across compartments. There are at least 3 types of such distributions which will be useful to consider. This distribution will be used to determine numbers of individuals leaving a given compartment. While we may be tempted to use a Poisson distribution with the rates specified in the equations above, this could lead to over-shooting, i.e. more individuals leaving a compartment than there actually are. To avoid infecting more people than there are susceptibles, we use a binomial distribution, with one draw for each individual in the compartment of interest. The workflow will be: determine a per-capita probability of leaving the compartment, based on the original rates specified in the equations; if the rate at which each individual leaves a compartment is $\lambda$, then the corresponding probability of this individual leaving the compartment in one time unit is $p = 1 - e^{- \lambda}$. determine the number of individuals leaving the compartment using a Binomial distribution, with one draw per individual and a probability $p$ As a example, let us consider transition $S \rightarrow I$ in the SIR model. The overall rate at which this change happens is $\beta \frac{S_t I_t}{N_t}$. The corresponding per susceptible rate is $\beta \frac{I_t}{N_t}$. Therefore, the probability for an individual to move from S to I at time $t$ is $p_{(S \rightarrow I), t} = 1 - e^{- \beta \frac{I_t}{N_t}}$. Poisson distributions will be useful when individuals enter a compartment at a given rate, from ‘the outside’. This could be birth or migration (for $S$), or introduction of infections from an external reservoir (for $I$), etc. The essential distinction with the previous process is that individuals are not leaving an existing compartment. This case is simple to handle: one just needs to draw new individuals entering the compartment from a Poisson distribution with the rate directly coming form the equations. For instance, let us now consider a variant of the SIR model where new infectious cases are imported at a constant rate $\epsilon$. The only change to the equation is for the infected compartment: \[ I_{t+1} = I_t + \beta \frac{S_t I_t}{N_t} + \epsilon - \gamma I_t \] where: individuals move from $S$ to $I$ according to a Binomial distribution $\mathcal{B}(S_t, 1 - e^{- \beta \frac{I_t}{N_t}})$ new infected individuals are imported according to a Poisson distribution $\mathcal{P}(\epsilon)$ individual move from $I$ to $R$ according to a Binomial distribution $\mathcal{B}(I_t, 1 - e^{- \gamma})$ This distribution will be useful when individuals leaving a compartment are distributed over several compartments. The Multinomial distribution will be used to determine how many individuals end up in each compartment. Let us assume that individuals move from a compartment $X$ to compartments $A$, $B$, and $C$, at rates $\lambda_A$, $\lambda_B$ and $\lambda_C$. The workflow to handle these transitions will be: determine the total number of individuals leaving $X$; this is done by summing the rates ($\lambda = \lambda_A + \lambda_B + \lambda_C$) to compute the per capita probability of leaving $X$ $(p_(X \rightarrow ...) = 1 - e^{- \lambda})$, and drawing the number of individuals leaving $X$ ($n_{_(X \rightarrow ...)}$) from a binomial distribution $n_{(X \rightarrow ...)} \sim B(X, p_(X \rightarrow ...))$ compute relative probabilities of moving to the different compartments (using $i$ as a placeholder for $A$, $B$, $C$): $p_i = \frac{\lambda_i}{\sum_i \lambda_i}$ determine the numbers of individuals moving to $A$, $B$ and $C$ using a Multinomial distribution: $n_{(X \rightarrow A, B, C)} \sim \mathcal{M}(n_{(X \rightarrow ...)}, p_A, p_B, p_C)$ odin We start by loading the odin code for a discrete, stochastic SIR model: ## Core equations for transitions between compartments:update(S) <- S - beta * S * I / Nupdate(I) <- I + beta * S * I / N - gamma * Iupdate(R) <- R + gamma * I## Total population size (odin will recompute this at each timestep:## automatically)N <- S + I + R## Initial states:initial(S) <- S_ini # will be user-definedinitial(I) <- I_ini # will be user-definedinitial(R) <- 0## User defined parameters - default in parentheses:S_ini <- user(1000)I_ini <- user(1)beta <- user(0.2)gamma <- user(0.1) As said in the previous vignette, remember this looks and parses like R code, but is not actually R code. Copy-pasting this in a R session will trigger errors. We then use odin to compile this model: ## function (I_ini = NULL, S_ini = NULL, beta = NULL, gamma = NULL, ## user = list(I_ini = I_ini, S_ini = S_ini, beta = beta, gamma = gamma), ## unused_user_action = NULL) ## <an 'odin_generator' function>## use coef() to get information on user parameters Note: this is the slow part (generation and then compilation of C code)! Which means for computer-intensive work, the number of times this is done should be minimized. The returned object sir_generator is a function that will generate an instance of the model: ## <odin_model>## Public:## contents: function () ## initial: function (step) ## initialize: function (user = NULL, unused_user_action = NULL) ## ir: {"version":"1.0.1","config":{"base":"discrete_determinis ...## run: function (step, y = NULL, ..., use_names = TRUE, replicate = NULL) ## set_user: function (..., user = list(...), unused_user_action = NULL) ## transform_variables: function (y) ## update: function (step, y) ## Private:## core: list## discrete: TRUE## dll: discrete_deterministic_sir_9fb76f84## init: 1000 1 0## interpolate_t: NULL## n_out: 0## name: discrete_deterministic_sir## output_order: NULL## ptr: externalptr## update_metadata: function () ## use_dde: NULL## user: I_ini S_ini beta gamma## variable_order: list## ynames: step S I R x is an ode_model object which can be used to generate dynamics of a discrete-time, deterministic SIR model. This is achieved using the function x$run(), providing time steps as single argument, e.g.: ## step S I R## [1,] 0 1000.0000 1.000000 0.0000000## [2,] 1 999.8002 1.099800 0.1000000## [3,] 2 999.5805 1.209517 0.2099800## [4,] 3 999.3389 1.330125 0.3309317## [5,] 4 999.0734 1.462696 0.4639442## [6,] 5 998.7814 1.608403 0.6102138## [7,] 6 998.4604 1.768530 0.7710541## [8,] 7 998.1076 1.944486 0.9479071## [9,] 8 997.7198 2.137811 1.1423557## [10,] 9 997.2937 2.350191 1.3561368## [11,] 10 996.8254 2.583469 1.5911558 The stochastic equivalent of the previous model can be formulated in odin as follows: ## Core equations for transitions between compartments:update(S) <- S - n_SIupdate(I) <- I + n_SI - n_IRupdate(R) <- R + n_IR## Individual probabilities of transition:p_SI <- 1 - exp(-beta * I / N) # S to Ip_IR <- 1 - exp(-gamma) # I to R## Draws from binomial distributions for numbers changing between## compartments:n_SI <- rbinom(S, p_SI)n_IR <- rbinom(I, p_IR)## Total population sizeN <- S + I + R## Initial states:initial(S) <- S_iniinitial(I) <- I_iniinitial(R) <- 0## User defined parameters - default in parentheses:S_ini <- user(1000)I_ini <- user(1)beta <- user(0.2)gamma <- user(0.1) We can use the same workflow as before to run the model, using 10 initial infected individuals ( I_ini = 10): ## function (I_ini = NULL, S_ini = NULL, beta = NULL, gamma = NULL, ## user = list(I_ini = I_ini, S_ini = S_ini, beta = beta, gamma = gamma), ## unused_user_action = NULL) ## <an 'odin_generator' function>## use coef() to get information on user parameters This gives us a single stochastic realisation of the model, which is of limited interest. As an alternative, we can generate a large number of replicates using arrays for each compartments: ## Core equations for transitions between compartments:update(S[]) <- S[i] - n_SI[i]update(I[]) <- I[i] + n_SI[i] - n_IR[i]update(R[]) <- R[i] + n_IR[i]## Individual probabilities of transition:p_SI[] <- 1 - exp(-beta * I[i] / N[i])p_IR <- 1 - exp(-gamma)## Draws from binomial distributions for numbers changing between## compartments:n_SI[] <- rbinom(S[i], p_SI[i])n_IR[] <- rbinom(I[i], p_IR)## Total population sizeN[] <- S[i] + I[i] + R[i]## Initial states:initial(S[]) <- S_iniinitial(I[]) <- I_iniinitial(R[]) <- 0## User defined parameters - default in parentheses:S_ini <- user(1000)I_ini <- user(1)beta <- user(0.2)gamma <- user(0.1)## Number of replicatesnsim <- user(100)dim(N) <- nsimdim(S) <- nsimdim(I) <- nsimdim(R) <- nsimdim(p_SI) <- nsimdim(n_SI) <- nsimdim(n_IR) <- nsim ## function (I_ini = NULL, S_ini = NULL, beta = NULL, gamma = NULL, ## nsim = NULL, user = list(I_ini = I_ini, S_ini = S_ini, beta = beta, ## gamma = gamma, nsim = nsim), unused_user_action = NULL) ## <an 'odin_generator' function>## use coef() to get information on user parameters set.seed(1)sir_col_transp <- paste0(sir_col, "66")x_res <- x$run(0:100)par(mar = c(4.1, 5.1, 0.5, 0.5), las = 1)matplot(x_res[, 1], x_res[, -1], xlab = "Time", ylab = "Number of individuals", type = "l", col = rep(sir_col_transp, each = 100), lty = 1)legend("left", lwd = 1, col = sir_col, legend = c("S", "I", "R"), bty = "n") This model is a more complex version of the previous one, which we will use to illustrate the use of all distributions mentioned in the first part: Binomial, Poisson and Multinomial. The model is contains the following compartments: There are no birth of natural death processes in this model. Parameters are: The model will be written as: \[ S_{t+1} = S_t - \beta \frac{S_t (I_{R,t} + I_{D,t})}{N_t} + \omega R_t \] \[ E_{t+1} = E_t + \beta \frac{S_t (I_{R,t} + I_{D,t})}{N_t} - \delta E_t + \epsilon \] \[ I_{R,t+1} = I_{R,t} + \delta (1 - \mu) E_t - \gamma_R I_{R,t} + \epsilon \] \[ I_{D,t+1} = I_{D,t} + \delta \mu E_t - \gamma_D I_{D,t} + \epsilon \] \[ R_{t+1} = R_t + \gamma_R I_{R,t} - \omega R_t \] \[ D_{t+1} = D_t + \gamma_D I_{D,t} \] The formulation of the model in odin is: ## Core equations for transitions between compartments:update(S) <- S - n_SE + n_RSupdate(E) <- E + n_SE - n_EI + n_import_Eupdate(Ir) <- Ir + n_EIr - n_IrRupdate(Id) <- Id + n_EId - n_IdDupdate(R) <- R + n_IrR - n_RSupdate(D) <- D + n_IdD## Individual probabilities of transition:p_SE <- 1 - exp(-beta * I / N)p_EI <- 1 - exp(-delta)p_IrR <- 1 - exp(-gamma_R) # Ir to Rp_IdD <- 1 - exp(-gamma_D) # Id to dp_RS <- 1 - exp(-omega) # R to S## Draws from binomial distributions for numbers changing between## compartments:n_SE <- rbinom(S, p_SE)n_EI <- rbinom(E, p_EI)n_EIrId[] <- rmultinom(n_EI, p)p[1] <- 1 - mup[2] <- mudim(p) <- 2dim(n_EIrId) <- 2n_EIr <- n_EIrId[1]n_EId <- n_EIrId[2]n_IrR <- rbinom(Ir, p_IrR)n_IdD <- rbinom(Id, p_IdD)n_RS <- rbinom(R, p_RS)n_import_E <- rpois(epsilon)## Total population size, and number of infectedsI <- Ir + IdN <- S + E + I + R + D## Initial statesinitial(S) <- S_iniinitial(E) <- E_iniinitial(Id) <- 0initial(Ir) <- 0initial(R) <- 0initial(D) <- 0## User defined parameters - default in parentheses:S_ini <- user(1000) # susceptiblesE_ini <- user(1) # infectedbeta <- user(0.3) # infection ratedelta <- user(0.3) # inverse incubationgamma_R <- user(0.08) # recovery rategamma_D <- user(0.12) # death ratemu <- user(0.7) # CFRomega <- user(0.01) # rate of waning immunityepsilon <- user(0.1) # import case rate ## function (E_ini = NULL, S_ini = NULL, beta = NULL, delta = NULL, ## epsilon = NULL, gamma_D = NULL, gamma_R = NULL, mu = NULL, ## omega = NULL, user = list(E_ini = E_ini, S_ini = S_ini, beta = beta, ## delta = delta, epsilon = epsilon, gamma_D = gamma_D, ## gamma_R = gamma_R, mu = mu, omega = omega), unused_user_action = NULL) ## <an 'odin_generator' function>## use coef() to get information on user parameters x <- seirds_generator()seirds_col <- c("#8c8cd9", "#e67300", "#d279a6", "#ff4d4d", "#999966", "#660000")set.seed(1)x_res <- x$run(0:365)par(mar = c(4.1, 5.1, 0.5, 0.5), las = 1)matplot(x_res[, 1], x_res[, -1], xlab = "Time", ylab = "Number of individuals", type = "l", col = seirds_col, lty = 1)legend("left", lwd = 1, col = seirds_col, legend = c("S", "E", "Ir", "Id", "R", "D"), bty = "n") Several runs can be obtained without rewriting the model, for instance, to get 100 replicates: ## [1] 366 600 ## S.1 E.1 Id.1 Ir.1 R.1 D.1 S.2 E.2 Id.2 Ir.2## 1 1000 1 0 0 0 0 1000 1 0 0## 2 1000 1 0 0 0 0 1000 1 0 0## 3 1000 0 0 1 0 0 1000 2 0 0## 4 1000 0 0 1 0 0 1000 1 1 0## 5 1000 0 0 1 0 0 1000 1 1 0## 6 1000 0 0 1 0 0 1000 0 2 0 seirds_col_transp <- paste0(seirds_col, "1A")par(mar = c(4.1, 5.1, 0.5, 0.5), las = 1)matplot(0:365, x_res, xlab = "Time", ylab = "Number of individuals", type = "l", col = rep(seirds_col_transp, 100), lty = 1)legend("right", lwd = 1, col = seirds_col, legend = c("S", "E", "Ir", "Id", "R", "D"), bty = "n") It is then possible to explore the behaviour of the model using a simple function: check_model <- function(n = 50, t = 0:365, alpha = 0.2, ..., legend_pos = "topright") { model <- seirds_generator(...) col <- paste0(seirds_col, "33") res <- as.data.frame(replicate(n, model$run(t)[, -1])) opar <- par(no.readonly = TRUE) on.exit(par(opar)) par(mar = c(4.1, 5.1, 0.5, 0.5), las = 1) matplot(t, res, xlab = "Time", ylab = "", type = "l", col = rep(col, n), lty = 1) mtext("Number of individuals", side = 2, line = 3.5, las = 3, cex = 1.2) legend(legend_pos, lwd = 1, col = seirds_col, legend = c("S", "E", "Ir", "Id", "R", "D"), bty = "n")} This is a sanity check with a null infection rate and no imported case: Another easy case: no importation, no waning immunity: A more nuanced case: persistence of the disease with limited import, waning immunity, low severity, larger population:
I have the following equation: $$-\frac{{{\hbar }^{2}}}{2I{}_{r}}\frac{{{d}^{2}}\psi }{d{{\theta }^{2}}}+V\left( \theta \right)\psi =E\psi, $$ with $$V\left( \theta \right)=a+\sum\limits_{n=1}^{N}{{{b}_{n}}\cos \left( n\theta \right)}+\sum\limits_{m=1}^{M}{{{c}_{m}}\sin \left( m\theta \right)}$$ where the notations are straightforward ($\theta \in \left[ 0,2\pi \right)$). I would like to get $K$ eigenvalues of the spectrum. How to integrate it (with some derivations)? Is there any non-commercial program which is able to provide them? I used MSTOR by Truhlar's group, but the potential does not contain the sine-terms.
This is a cute problem! I toyed with it and didn't really get anywhere - I got the strong impression that it requires fields of mathematics that I am not expert in. Indeed, given that the problem seems related to that of counting integer solutions to the equation $f(x,y) = c$, one may need to use arithmetic geometry tools (e.g. Faltings' theorem). In particular if we could reduce to the case when the genus is just 0 or 1 then presumably one could kill off the problem. (One appealing feature of this approach is that arithmetic geometry quantities such as the genus are automatically invariant (I think) with respect to invertible polynomial changes of variable such as $(x,y) \mapsto (x,y+P(x))$ or $(x,y) \mapsto (x+Q(y),y)$ and so seem to be well adapted to the problem at hand, whereas arguments based on the raw degree of the polynomial might not be.) Of course, Faltings' theorem is ineffective, and so might not be directly usable, but perhaps some variant of it (particularly concerning the dependence on c) could be helpful. [Also, it is overkill - it controls rational solutions, and we only care here about integer ones.] This is far outside of my own area of expertise, though... The other thing that occurred to me is that for fixed c and large x, y, one can invert the equation $f(x,y) = c$ to obtain a Puiseux series expansion for y in terms of x or vice versa (this seems related to resolution of singularities at infinity, though again I am not an expert on that topic; certainly Newton polytopes seem to be involved). In some cases (if the exponents in this series expansion are favourable) one could then use Archimedean counting arguments to show that f cannot cover all the natural numbers (this is a generalisation of the easy counting argument that shows that a 1D polynomial of degree 2 or more cannot cover a positive density set of integers), but this does not seem to work in all cases, and one may also have to use some p-adic machinery to handle the other cases. One argument against this approach though is that it does not seem to behave well with respect to invertible polynomial changes of variable, unless one works a lot with geometrical invariants. Anyway, to summarise, it seems to me that one has to break out the arithmetic geometry and algebraic geometry tools. (Real algebraic geometry may also be needed, in order to fully exploit the positivity, though it is also possible that positivity is largely a red herring, needed to finish off the low genus case, but not necessary for high genus, except perhaps to ensure that certain key exponents are even.) EDIT: It occurred to me that the polynomial $f(x,y)-c$ might not be irreducible, so there may be multiple components to the associated algebraic curve, each with a different genus, but presumably this is something one can deal with. Also, the geometry of this curve may degenerate for special c, but is presumably stable for "generic" c (or maybe even all but finitely many c). It also occurs to me that one use of real algebraic geometry here is to try to express f as something like a sum of squares. If there are at least two nontrivial squares in such a representation, then f is only small when both of the square factors are small, which is a 0-dimensional set and so one may then be able to use counting arguments to conclude that one does not have enough space to cover all the natural numbers (provided that the factors are sufficiently "nonlinear"; if for instance $f(x,y)=x^2+y^2$ then the counting arguments barely fail to provide an obstruction, one has to use mod p arguments or something to finish it off...) EDIT, FOUR YEARS LATER: OK, now I know a bit more arithmetic geometry and can add to some of my previous statements. Firstly, it's not Faltings' theorem that is the most relevant, but rather Siegel's theorem on integer points on curves - the enemy appears to be those points $(x,y)$ where $x,y$ are far larger than $f(x,y)$, and Siegel's theorem is one of the few tools available to exclude this case. The known proofs of this theorem are based on two families of results in Diophantine geometry: one is the Thue-Siegel-Roth theorem and its variants (particularly the subspace theorem), and the other is the Mordell-Weil theorem and its variants (particularly the Chevalley-Weil theorem). A big problem here is that all of these theorems have a lot of ineffectivity in them. Even for the very concrete case of Hall's conjecture on lower bounding $\|x^2-y^3\|$ for integers $x,y$ with $x^2 \neq y^3$, Siegel's theorem implies that this bound goes to infinity as $x,y \to \infty$, but provides no rate; as I understand it, the only known lower bounds are logarithmic and come from variants of Baker's method. As such, a polynomial such as $f(x,y) = (x^2 - y^3 - y)^4 - y + C$ for some large constant C already looks very tough to analyse. (I've shifted $y^3$ here by $y$ to avoid the degenerate solutions to $x^2=y^3$, and to avoid some cheap way to deal with this polynomial from the abc conjecture or something.) The analogue of Hall's conjecture for $\|x^2-y^3-y\|$ suggests that $f(x,y)$ goes to $+\infty$ as $x,y \to \infty$ (restricting $x,y$ to be integers), but we have no known growth rate here due to all the ineffectivity. As such, we can't unconditionally rule out the possibility of an infinite number of very large pairs $(x,y)$ for which $x^2-y^3-y$ happens to be so close to $y^{1/4}$ that we manage to hit every positive integer value in $f(x,y)$ without hitting any negative ones. However, one may be able to get a conditional result assuming some sufficiently strong variant of the abc conjecture. One should also be able to exclude large classes of polynomials $f$ from working; for instance, if the curve $f(x,y)=0$ meets the line at infinity at a lot of points in a transverse manner, then it seems that the subspace theorem may be able to get polynomial bounds on solutions $(x,y)$ to $f(x,y)=c$ in terms of $c$, at which point a lot of other tools (e.g. equidistribution theory) become available. Another minor addendum to my previous remarks: the generic irreducibility of $f(x,y)-c$ follows from Bertini's second theorem, as one may easily reduce to the case when $f$ is non-composite (not the composition of two polynomials of lower degree).
This might be a trivial question. Consider a function $f:\mathbb{R^2}\rightarrow \mathbb{R}$ and consider some point $(a,b)\in \mathbb{R^2}$. Suppose we know that all the directional derivatives $D_{\overline{u}}f(a,b)$ for an arbitrary unit vector $\overline{u}=\langle u_1,u_2\rangle$ in $\mathbb{R^2}$ exist $...(1)$ Furthermore, we also know that $\lim_{\overline{u}\to\overline{v}}D_{\overline{u}}f(a,b) = D_{\overline{v}}f(a,b)$ for some other arbitrary arbitrary unit vector $\overline{v}$$...(2)$ So all the directional derivatives can smoothly transition into one another as we change the unit vector $\overline{u}$ in $\mathbb{R^2}$$...()$ This is however $\textbf{not}$ sufficient for the function $f$ to be differentiable at $(a,b)$ since all the directional derivatives must be parallel to one another to trace out the tangent plane at the point $(a,b)$ as we change $\overline{u}$$...()$ $(*)$ As far as my understanding goes. $\textbf{My Question:}$ From $(1)$ & $(2)$ can we conclude however that $f$ is continuous at $(a,b)$ ? The only other ways of establishing continuity at $(a,b)$ in calculus of $2$ variables - that I am aware of - is by applying the$\ $ $\epsilon -\delta$ definition of continuity to $f$ at $(a,b)$ or establishing the differentiability of $f$ at $(a,b)$ which implies continuity.
In looking at the two functions defined: $$\psi_{{0}}(x)=\ln( \operatorname{lcm}(1,2,3,...,\lfloor x \rfloor))$$ $$\psi_{{1}}(x)=\sum _{j=1}^{ \lfloor x \rfloor } \sum _{i=0}^{ \Bigl\lfloor {\frac {\ln \left( x \right) }{\ln \left( p_{{j}} \right) }} \Bigr\rfloor +1}\ln \left( {p_{{j}}}^{i} \right) $$ (where $p_n$ is the $n^{th}$ prime and $lcm$ denontes the lowest common multiple of the arguments enclosed) I am interested in finding the minimum value (if it exists) of $n \in \mathbb N$ that satisfies: $$ {\Biggl\{\frac{\left(\lfloor \ln \left( \psi_{{0}}(n)+\psi_{{1}}(n)\right) \rfloor +1 \right) !}{\lfloor \sqrt n \rfloor!}}\Biggr\} \ne 0$$ where ${\{x}\}$ denotes the fractional part of $x$. some values evaluated: $$\frac{\left(\lfloor \ln \left( \psi_{{0}}(8)+\psi_{{1}}(8)\right) \rfloor +1 \right) !}{\lfloor \sqrt 8 \rfloor!}=12$$ $$\frac{\left(\lfloor \ln \left( \psi_{{0}}(12)+\psi_{{1}}(12)\right) \rfloor +1 \right) !}{\lfloor \sqrt 12 \rfloor!}=20$$ $$\frac{\left(\lfloor \ln \left( \psi_{{0}}(20)+\psi_{{1}}(20)\right) \rfloor +1 \right) !}{\lfloor \sqrt 20 \rfloor!}=5$$ The value is less than 1 at $n=36$ indicating this to be the immeadiate border of the region of $\mathbb N$ for which the inequality Carl mentioned begins to be true (inductively reasoning). Beyond $0<n<32$ I am not as yet able to produce a result, float approximations continue to imply that the value is 0 up to $n=40$, however really what is needed here is someone with more experience in number theory to assess the situation and give an opinion as to whether it is worth pursuing or not. I will try my best to follow along with Carl's answer, he has skipped a few steps that are probably what he may consider too obvious to show, but so far: Because: $$\psi_{{0}}(x)=\ln(\operatorname{lcm}(1,2,3,...,x-1,\lfloor x\rfloor))=\alpha\,\ln \left( 2 \right) +\beta\,\ln \left( 3 \right)... +\sum _{j= 1}^{ \lfloor x \rfloor }\ln \left( p_{{j}} \right) $$ for some $$\alpha, \beta,... \in \mathbb N$$ And similarly: $$\psi_{{1}}(x)=\sum _{j=1}^{ \lfloor x \rfloor } \sum _{i=0}^{ \Bigl\lfloor {\frac {\ln \left( x \right) }{\ln \left( p_{{j}} \right) }} \Bigr\rfloor +1}\ln \left( {p_{{j}}}^{i} \right) =\frac{1}{2}\sum _{j=1}^{ \lfloor x \rfloor }\ln \left( p_{{j}} \right) \left( \Bigl\lfloor {\frac {\ln \left( x \right) }{ \ln \left( p_{{j}} \right) }} \Bigr\rfloor +2 \right) \left( \Bigl\lfloor {\frac {\ln \left( x \right) }{\ln \left( p_{{j}} \right) }} \Bigr\rfloor +1 \right)$$ $$=\alpha'\,\ln \left( 2 \right) +\beta'\,\ln \left( 3 \right)... +\sum _{j= 1}^{ \lfloor x \rfloor }\ln \left( p_{{j}} \right) $$ for some $$\alpha', \beta',... \in \mathbb N$$ The asymptotic relation I think I originally started from, which think is actually false as I originally stated, but again, just curious about the division relation on the naturals I really am new to asymptotics:
Search Now showing items 1-10 of 165 J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV (Springer, 2014-02) Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ... Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV (Elsevier, 2013-04-10) The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (\|y\| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ... Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV (Springer, 2014-12) The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ... Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC (Springer, 2014-10) Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ... Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV (Elsevier, 2017-12-21) We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ... Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV (American Physical Society, 2017-09-08) The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ... Online data compression in the ALICE O$^2$ facility (IOP, 2017) The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ... Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (American Physical Society, 2017-09-08) In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ... J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (American Physical Society, 2017-12-15) We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($\|y\| < 0.9$) ... Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions (Nature Publishing Group, 2017) At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ...
I am a student who finished Part 1 and am interested in applying label smoothing to a problem. I saw that it was taught in Part 2 so figured this would be a good place to ask my question. I was wondering though if label smoothing can be applied to multi-label problems. In addition, typically, as I read about label smoothing online, it seems that they usually are replacing the labels with the smoothed labels, but that is done in the loss function, correct? Coming back to this, I realized I didn’t simplify it like was done for regular multi-label. Here is the simplified version: \begin{aligned} (1-\epsilon)\sum_i (-\frac{\log p_i }{n} ) + \frac{\epsilon}{N} \sum (-\log p_i) \end{aligned} where the last term is the full cross entropy over the entire dataset. I am unsure how to implement this. I see in the notebook there is a loss = reduce_loss(-log_preds.sum(dim=-1), self.reduction) and also nll = F.nll_loss(log_preds, target, reduction=self.reduction). The output seems to be lin_comb(loss/c, nll, self.ε) so that would be self.ε * loss/c + (1- self.ε)*nll Is nll the cross-entropy of the entire dataset, because then shouldn’t it be multiplied by self.ε instead of (1- self.ε)?
To calculate the volume of a pyramid (not tetrahedron!) you've to use the formula $\frac{1}{3}B\cdot H,$ where $B$ is the area of the base and $H$ is the height. My question is: why 1/3? Is a pyramid one-third of a cuboid? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community While using calculus to derive this is a bit heavy-handed, consider this: $V= \int A(h) \,\mathrm{d}h$, where $A$ is the area of a cross-section parallel to the base, but at distance $h$ from the apex. Since the length of the sides of the cross-section grows $\propto h$, $A(h)\propto h^2$. But $B=A(H)$, so $A(h)=\frac{h^2}{H^2}B$. Evaluate the integral: $$ V= \int_0^H A(h)\,\mathrm{d}h = \int_0^H B \frac{h^2}{H^2}\,\mathrm{d}h = \frac{1}{3}BH$$ Consider a triangular prism. It can be divided into three equivalent pyramids, and so the volume of a triangular pyramid is $1/3$ of the volume of a triangular prism with the same base and height (see figure on the left). A pyramid whose base has $n$ sides may be divided into $n-2$ tetrahedrons. It follows that its volume is $1/3$ of the volume of a prism with the same base and height (see figure on the right). Explanation concerning the sketch on the left in response to Byron Schmuland's comment. Removing the front pyramid we are left with two equivalent pyramids (one on the left and the other on the right) whose height is the distance from the upper front corner to the prism back face and whose bases are the two halves of the back face. On the other hand the front pyramid is equivalent to the right pyramid (equal bases and equal heights). Reference: F. G. - M., §496 and §497 of Cours de Géométrie Élémentaire, 1917. The difficult part is to show that the volume is given by such a simple formula involving only the area of the base and the height. All proofs of this formula have to use a limiting argument of some sort. As for the factor ${1\over3}$, this is easy: A cube is the union of 6 congruent pyramids with a face as base and the midpoint of the cube as peak. The volume of a cuboid is $B\cdot H$. So, yes, a pyramid is a third of a cuboid, since $3(\frac{1}{3}B\cdot H)=B\cdot H$. A way to see this is: http://www.korthalsaltes.com/model.php?name_en=three%20pyramids%20that%20form%20a%20cube
Evaluate $ \int \frac {1} {z^2+1} dz $ along the contour $\Gamma$. (Gamma is some closed circle centered around i, no specified radius, and is oriented counter clockwise.) So far, I've used factored the expression into $ \int \frac 1 {(z+i)(z-i)}$, then used partial fraction decomposition, and got $ \frac 1 {2i}[ \frac{1} {z+i}-\frac{1} {z-i}] $. Now I can take the integral of both of these separate terms from 0 to 2pi, but I'm having a lot of trouble with parameterization of this curve before I integrate. I assumed since it's a closed contour centered at i, I would parameterize it by $z(t)=i+e^{it}$? However, when I substitute this into the integral and multiply by it's derivative $z'(t) = ie^{it}$, I don't get the right answer. The book says I should be getting zero for the first integral, but I don't even know how to do this integral. $$ \int_{\Gamma}\frac{1}{z+i}dz =\int_0^{2\pi}(\frac{1}{(i+e^{it})+i})(ie^{it})dt=\int_0^{2\pi}\frac {ie^{it}}{e^{it}+2i}dt$$ Now I have absolutely no clue how to do this integral, but supposedly it is supposed to come out to 0, and the second integral of $$\int_0^{2\pi}\frac{1}{z-i}$$ should be equal to $ 2\pi i$. They then total together using the 3rd equation from the top, and 2i cancels, and the total integral becomes $ \pi $. That should be the total answer. My question mainly is, how do I do that first integral of $\frac{1}{}
I am wondering about the order of terms from equation 34 this pdf by Hitoshi Murayama on second quantization. Specifically, why they are able to pull the H past the $\Psi(\vec x)$. i.e. $$ H\|\psi(t)\rangle= \int dx\,H \Psi(\vec x,t)\phi^\dagger(\vec x)\|0\rangle $$ makes sense, but how are they allowed to pull H in front of $\Psi(\vec x,t)$: $$ \int dx\,H \Psi(\vec x,t)\phi^\dagger(\vec x)\|0\rangle= \int dx\,\Psi(\vec x,t)H\phi^\dagger(\vec x)\|0\rangle $$ However, here we have $\|\Psi(t)\rangle$ already in the x basis, so I would posit it is because H is an operator and we can move $\Psi(\vec x,t)$ around. Yet, I find issue in this, because H frequently involves derivatives, since for a free particle it is $\frac{P^2}{2m}$, which involves a derivative in the x basis. As a side note to help understand my question: For example, when inserting a complete set of states into an inner product, the order of operators, bras and kets are never changed, until we project the kets into a basis (i.e. the x basis): $$\langle\psi\|\hat A\|\phi\rangle= \langle\psi\|\mathbb I \, \hat A \|\phi\rangle= \langle\psi\|\Bigg(\int \|x\rangle\langle x\| \, dx\Bigg)\hat A \|\phi\rangle= \int\langle\psi\|x\rangle\langle x\|\hat A \|\phi\rangle \, dx= \int\psi(x)\phi_A(x)\, dx= \int\phi_A(x)\psi(x)\, dx $$
It is often said that he discovered non-Euclidean geometry. But in which sense? I am reading the book 'geometry' by Brannan et al. They use the disk model as an example of hyperbolic geometry. Did Lobachevsky have a similar model? History of Science and Mathematics Stack Exchange is a question and answer site for people interested in the history and origins of science and mathematics. It only takes a minute to sign up.Sign up to join this community Yes, Lobachevsky was a discoverer of non-Euclidean (more precisely, hyperbolic) geometry. (It was independently discovered by few other people, but Lobachevsky's treatment was by far the most complete). More precisely, he just accepted as an axiom the negation of the "5-th postulate", and developed a comprehensive geometry based on these new axioms. But he did not have any model. It was developed axiomatically, in the same way as Euclid. The role of the later models (Beltrami, Poincare, Klein) was to show rigorously that IF there is no contradiction in Euclidean geometry, THEN these is no contradiction in the Lobachevsky's one. So they are equally "true". (The question whether there is a contradition in Euclidean geometry, or in the rest of mathematics, does not belong to mathematics itself: this is unprovable by the usual mathematical methods). When developing his geometry, Lobachevsky worked exclusively in the (Lobachevskian) plane. It was the Italian mathematician Beltrami who first showed that the geometry of (part of) the Lobachevskian plane coincided with the geometry of a certain surface - namely the pseudosphere. Beltrami's work came some forty-two years after Lobachevsky first formulated and completed his ideas: 1868 vs 1826. Beltrami's interpretation shows that to every statement of Lobachevskian geometry referring to part of the plane there corresponds an immediate fact about the intrinsic geometry of the psuedosphere. However, not all of the Lobachevskian plane is realised on the pseudosphere, but only part of it. It was Klein, in 1870, who first gave an actual interpretation of Lobachevskian geometry on the whole plane and, more generally, his geometry in space. (See, for example, Klein's interpretation in the circle and the sphere.) Source: MATHEMATICS Its Contents, Methods, and Meaning by Aleksandrov, Kolmogorov, and Lavrent'ev. Another important contributions of Lobachevsky was the development of hyperbolic trigonometry (from the axioms: unlike the modern treatments, he did not use any models since he did not have any). An interesting, and often overlooked, application of this development is the uniqueness of the Lobachevskian geometry: Theorem. Suppose that $X_1, X_2$ are spaces satisfying Lobachevsky axioms with the same "curvature normalization": Each ideal triangle has area $c$ in both geometries. (You can take $c=\pi$ if you like, this corresponds to the curvature $-1$.) Then $X_1$ is isometric to $X_2$. I can explain how it follows if you are interested. Edit: Proof. Pick a base point $o_1\in X_1$ and a reference ray $\rho_1$ emanating from $o$. This defines the "polar coordinates" on $X_1$, namely $P(r,\theta)$, where $r$ is the distance from $P$ to $o_1$ and $\theta\in [0,2\pi)$, the (oriented) angle between $o_1P$ and $\rho_1$. Now, do the same in the second space $X_2$: Pick a base-point $o_2$, a ray $\rho_2$, etc. Then, define a map $f: X_1\to X_2$ by sending $P(r,\theta)\in X_1$ to $Q(r,\theta)\in X_2$. It is clearly a bijection, let us prove that it is an isometry. Take two points $A, B\in X_1$. Then their distance in $X_1$ is given by the hyperbolic cosine law (due to Lobachevsky: I did not check, but it is possible that Bolyai also proved it): $$ cosh(\|AB\|)= cosh(\|o_1A\|)cosh(\|o_1B\|) - sinh(\|o_1A\|)sinh(\|o_1B\|) cos(\angle(Ao_1B)). $$ By the construction, the map $f$ preserves all the quantities on the right-hand side of this equation (the distance to the origin and the angle). Since the hyperbolic cosine formula also holds in $X_2$, it follows that $f$ is an isometry. qed This proof uses the notion of an oriented angle which can be easily avoided by working in one half-plane in $X_i$ (defined by the unique line through $\rho_i$ $i=1,2$) at a time.
I remember trying this problem a while ago and was unable to prove it. I think my idea was to create a surjective homomorphism from $\mathbb{Z}_{5}[x]$ to $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$ and use the first isomorphism theorem but it wasn't working because none of my maps were well-defined. What is the correct way to approach this problem? Note that $x^2+1=(x+2)(x+3)$ over $\mathbb{Z}/5\mathbb{Z}$, and so by the Chinese Remainder Theorem $$ (\mathbb{Z}/5\mathbb{Z})[x]/(x^2+1)=(\mathbb{Z}/5\mathbb{Z})[x]/(x+2)(x+3)\simeq (\mathbb{Z}/5\mathbb{Z})[x]/(x+2)\times (\mathbb{Z}/5\mathbb{Z})[x]/(x+3)$$ $$\simeq \mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}$$ For a constructive approach, your idea for a strategy works. Here is how to think through building a well-defined map in this case. First, instead of starting with the quotient $\mathbb Z_5[x]/(x^2+1)$, it is easier to try starting with $\mathbb Z_5[x]$. In order to get a homomorphism $\mathbb Z_5[x]\rightarrow \mathbb Z_5\times \mathbb Z_5$, we simply map $1\mapsto (1,1)$, and choose any value for $x$; say, $x\mapsto (a,b)$. In order for the map to factor through $\mathbb Z_5/(x^2+1)$, we need the image of $x$ to satisfy the equation $x^2+1=0$, hence $(a,b)^2+(1,1)=0$. Solving this, we need $a^2=b^2=4$. The equation $z^2=4$ has solutions $z=2$ or $3$ in $\mathbb Z_5$, so $(a,b)=(2,2), (3,3), (2,3),$ or $(3,2)$ If $a=b$, then $f$ is not surjective (since its image only contains $(c,c)$ for $c\in \mathbb Z_5$), so we can assume $a=2$ and $b=3$ or vice versa. Finally, note that the map $x\mapsto (2,3)$ is surjective: for instance, it is a $\mathbb Z_5$-linear map whose image contains the basis $(1,1),(2,3)$ of $\mathbb Z_5\times \mathbb Z_5$. Thus an isomorphism $\mathbb Z_5[x]/(x^2+1)\rightarrow \mathbb Z_5\times \mathbb Z_5$ can explicitly be given by $a+bx\mapsto (a+2b,a+3b)$ for $a,b\in \mathbb Z_5$.
Consider cellular automaton rules for a two-dimensional universe with two states, where a cell’s new state can depend on its previous state and the states of the cells in its Moore neighborhood. Such a rule can be modeled as a function that takes as input a 3 by 3 matrix of bits, and outputs a bit. Such a rule can be represented easily using a string of 512 bits representing the output for each of the $ 2^{3 \times 3}$ input states. Since this representation is bijective, a rule can also be randomly generated by sampling 512 bits. However, it is sometimes preferable for reasons of aesthetics and comprehensibility that the output of the rule be the same when the input matrix is rotated or flipped horizontally or vertically. The rule for Conway’s Game of Life is an example of a function satisfying this restriction. Given a function $ r : 2^9 \to 2$ which does not necessarily obey this restriction, we can obtain a function that does by mapping the input to the lexicographically smallest matrix obtainable by reflecting or rotating it before passing to $ r$ . However, this provides little insight into the questions I am interested in: How many functions obey this restriction? How can they be compactly representing using a bit string? How can they be efficiently randomly generated? Hello I was wondering if there’s anyway to Under volt CPU on ubuntu for better temperature? I know there are apps on win 10 like Throttlestop that do this, but i was not sure if the saves are written into bios so lptop stays under volt even after reboot and switchign OS to Ubuntu too. So if there’s anyone know a solution me and rest of people looking for this would appreciate this. Because it is a hot topic. Btw my specs : Inter Core i7-8750H CPU @ 2,20GHz A desktop version of Ubuntu shows the desktop environment (gnome) running. What about ubuntu-server. Does the display remain completely blank or does it show some status etc… I want to create an order form with 14 columns and as many rows as possible that will kick off a workflow. my challenge is creating a single List at one go to contain an X amount of rows with a single row of column headings(14) to be saved as a single list and can be editted later. i am very new on infopath and still finding my way around sharepoint. please asisst. Thanks, It’s common to ask whether a particular class of languages $ \mathcal{C} \subseteq \mathcal{P}(\Sigma^)$ , for some alphabet $ \Sigma$ , is closed under complement, or union, or intersection, or concatenation, or the Kleene star. And those questions seem natural to me, because they’re essentially questions about the power of a model of computation that can decide precisely the languages in $ C$ . However, it seems just as natural to me – perhaps even more natural – to ask whether such a class is also closed under composition. Here’s what I mean by that. If I fix some subset $ \mathcal{C} \subseteq \mathcal{P}(\Sigma^)$ of “decidable” languages relative to some model of computation, for any (finite) $ \Sigma$ , then I have also fixed some subset $ \mathcal{F} \subseteq B^A$ of computable functions, for any languages $ A, B$ – because each function $ f : A \rightarrow B$ can be viewed as a function $ \bar{f} : A \times B \rightarrow \{0, 1\}$ . Given that, I would expect that for any reasonable model of computation, if $ f : A \rightarrow B$ and $ g : B \rightarrow C$ are computable, then $ g \circ f : A \rightarrow C$ should be as well. Put in terms of the corresponding languages, I’d expect that if $ L_f = \{(a, b) \in A \times B : f(a) = b\}$ and $ L_g = \{(b, c) \in B \times C : g(b) = c\}$ are decidable, then $ L_{g \circ f} = \{(a, c) : \exists b \in B s.t. f(a) = b \wedge g(b) = c\}$ should be decidable as well. This seems to me like a natural question to ask of any class of languages. Is it in fact a reasonable question? If not, why? If it is, is there a better way to frame it, one that would make it more clear what we’re demanding? I’ve got both Xbox One Controllers the Original (shiny plastic around guide button) and the Second Edition (smooth matt plastic to Guide button) I am connecting these controllers via a USB cable. They are being detected by dmesg and lsusb gives me details about the attached device. Doesn’t get any further than that. I’ve tried installing the suggested profiles but nothings working. Neither of which are being detected in Game or in Steam menus or showing up in the /dev/input list $ > /dev/input/by-id usb-Logitech_Logitech_G930_Headset-event-if03 usb-Logitech_Trackball-event-mouse usb-Logitech_Trackball-mouse usb-Mad_Catz_Saitek_Pro_Flight_X-56_Rhino_Stick-event-joystick usb-Mad_Catz_Saitek_Pro_Flight_X-56_Rhino_Stick-joystick usb-Microsoft_Natural®_Ergonomic_Keyboard_4000-event-kbd usb-Microsoft_Natural®_Ergonomic_Keyboard_4000-if01-event-kbd usb-Oculus_VR_Rift_Sensor_WMTD303A3011WY-event-if00 $ > dmesg [ 389.625051] usb 3-13: new full-speed USB device number 15 using xhci_hcd [ 389.774252] usb 3-13: New USB device found, idVendor=045e, idProduct=02ea, bcdDevice= 3.01 [ 389.774255] usb 3-13: New USB device strings: Mfr=1, Product=2, SerialNumber=3 [ 389.774258] usb 3-13: Product: Controller [ 389.774260] usb 3-13: Manufacturer: Microsoft [ 389.774261] usb 3-13: SerialNumber: 3032363030303235393132373233 [ 429.862555] usb 3-13: USB disconnect, device number 15 drevilish@Linsair:~$ lsusb -d 045e:02ea -v Bus 003 Device 018: ID 045e:02ea Microsoft Corp. Xbox One S Controller Couldn't open device, some information will be missing Device Descriptor: bLength 18 bDescriptorType 1 bcdUSB 2.00 bDeviceClass 255 Vendor Specific Class bDeviceSubClass 71 bDeviceProtocol 208 bMaxPacketSize0 64 idVendor 0x045e Microsoft Corp. idProduct 0x02ea Xbox One S Controller bcdDevice 3.01 iManufacturer 1 iProduct 2 iSerial 3 bNumConfigurations 1 Configuration Descriptor: Linux Hostname 5.3.0-12-generic #13-Ubuntu SMP Tue Sep 17 12:35:50 UTC 2019 x86_64 x86_64 x86_64 GNU/Linux If a Lich, Vampire, or some other undead casts Undead Anatomy on itself, does the spells effect hold true? In this form, you detect as an undead creature (such as with detect undead, but not with magic that reveals your true form, such as true seeing) and are treated as undead for the purposes of channeled energy, cure spells, and inflict spells, but not for other effects that specifically target or react differently to undead (such as searing light). How does the Doc Set permissions work.I have created a some document sets which i have assigned unique users to and then added some files under the document set. I taught with this approach users would only see there folder in the document set as i have locked the document set/folder with unique permissions but it seems users cant see the files under the document sets and no access to files. Any ideas Hello, I have Ubuntu 16 instance under AWS and on login to its console I see message: New release '18.04.2 LTS' available. Run 'do-release-upgrade' to upgrade to it. Code (markup): I tried to run it following messages in commands : Last login: Fri Sep 6 06:39:22 2019 from 213.109.234.130 ubuntu@ip-NN-NNN-NN-NNN:~$ clear ubuntu@ip-NN-NNN-NN-NNN:~$ sudo apt-get update Hit:1 http://us-east-2.ec2.archive.ubuntu.com/ubuntu xenial InRelease Hit:2... Code (markup): How Ubuntu 16 instance under AWS upgrade to 18.04 ? I am having problems in Ubuntu-MATE 18.04 to use the Windows– E shortcut alias Mod4+ E to start caja (the standard “File Explorer” of MATE). this worked flawlessly on multiple Ubuntu-MATE 16.x machines. also: SHIFT+ Mod4+ E does work(!) Who or what captures my Win– E under 18.x? I don’t see any occupancy in the normal keyboard shortcuts list… (PS: If you wonder, why I have to use parameters with caja, that’s explained here)
An article in the July 30, 2016 New York Times argues that high stock prices are the result of low bond yields. To paraphrase the article: The S&P 500 index has a dividend yield of 2.1%, 40% higher than the 1.5% yield on 10-year Treasury notes. Investors have moved to stocks, chasing yield. This post explains why dividend yield and bond yield are not substitutes. Index Dividends and 10-Year Treasury yields I agree with the first half of the argument - Over the past 15 years, the yield on 10-year Treasury notes has declined, while the dividend yield on the S&P 500 index has (on average) increased: This doesn’t mean you should sell your bonds and buy stocks. More important is the total return - the sum of yield and capital gains (i.e. the price going up). Dividend Yield Dividend yield, , can increase for two reasons - either dividends go up, or prices go down: At the index level, dividends don’t move that much and changes in dividends are weakly correlated with dividend yield. There is, however, a strong inverse relationship between dividend yield and index price level: The standard deviation of dividend growth is about one eighth the standard deviation of stock returns, so the magnitude of price changes drowns out changes in dividends: The point is - dividends have been stable, but the price of the underlying index has not. 10-Year Treasury Yield Yields on existing bonds can increase for one reason - the price went down. This is because (assuming the U.S. government always pays its debts), you are guaranteed to get the face value for a Treasury note at maturity. There is no such guarantee with stocks. This doesn’t mean bonds are easier to understand than stocks, as prices can change for many reasons. For example: if interest rates rise, existing bonds go down in price (suppose not, then arbitrageurs would short old bonds, and buy new bonds to earn the difference in yields). Compare the price of 10-year bonds to S&P 500 index. Both series are normalized to 100 in January, 1986: This doesn’t mean you would have earned zero buying the 10-year notes. Below, I plot the total return (assuming you reinvest dividends and reinvest coupon payments) for each asset: The bond investment goes up a little every year, while the stock investment loses nearly 50% of its value twice in the last 30 years. Despite this, the stocks would be worth 4x as much today as the bonds! Timing In a two period asset pricing model:\begin{equation}P_t= E_t \left(\beta \frac{u’(c_{t+1})}{u’(c_t)} x_{t+1} \right)\end{equation}With risk averse investors, an asset whose payoff, , covariates negatively with consumption commands a higher price (and a lower return) than an asset that covaries positively with consumption. This is why people buy insurance: It’s good to get a check when your house burns down. In the financial crisis, bond prices went up (yields went down), while stock prices went down: This is another reason to treat dividend yields and bond yields differently - if you want to hedge against losing your job in a recession, bonds are a better bet (whether job loss is at the extensive or intensive margin during recessions will be the topic of a future blog post). Individual Stocks Let’s look at two dividend paying stocks mentioned in the article, Verizon and AT&T (formerly SBC). We see the same pattern as we did for the index - dividends are relatively stable, while prices are not: Investors might be attracted to these stocks because their dividend yields are now more than double the yield on 10-Year Treasury Notes: It’s important to remember, however, that both these stocks lost over 30% of their value during the financial crisis. If you were forced to sell, (for example, owing to liquidity constraints) you would have taken a big loss, lowering the effective yield below that of the 10-year note. Conclusion There’s something missing from this analysis - the relationship between inflation, bond yields and stock prices. Mild inflation will affect real bond returns more than real stock returns. This omission, however, doesn’t change the main result - even though dividend yields are relatively stable over time, there is no guarantee on the value of the underlying asset. Investors may be able to increase yields 40% by switching from bonds to stocks, but not without substantial additional risk.
The question is: $$\int^{\pi/4}_{0}\left(\cos 2\theta \right) ^{3/2}\cos \theta\, d\theta$$ I tried to write it in terms of $\sin\theta$ and the substitute $\sin\theta=t$ but then got stuck at the subsequent integral. I tried integrating by parts, which was of no use. Any help would be appreciated. Thanks.
CNO Cycle: 4H $\rightarrow$ He 4 protons (i.e. Hydrogen nuclei) combine to form a Helium molecule while releasing 26.7 MeV energy. This is the net result of any of the various fusion pathways for hydrogen to helium. Three Helium molecules combine to form a Carbon molecule, while releasing 7.4 MeV of energy. Since it takes three CNO reactors (and 12 protons) to make a Carbon molecule, we now have a net gain of 87.5 MeV. See the link, this is a long chain of reactions, each successively adding a Helium to get to a bigger element, until we get to Fe-52. The total energy released by this chain is 80.6 MeV. Taking into account the 87.5 MeV to form the initial carbon and 11 $\times$ 26.7 MeV to form all of the Helium, the net energy gain from fusion is 462 MeV, divided by 52 initial protons. Now, Fe-52 is not the most stable element, and you could theoretically get more energy by reacting up to Fe-56 or Ni-62 or something. But, I wasn't able to find a clear path for fusion up to that point. In the real world, creation of these elements is a result of an equilibrium between various fusion reactions and photodisintegration and such. I think this energy estimate is the best for your purposes. Energy released by accretion This is much more difficult to estimate, because there are a lot of factors here, and it depends strongly on the size of your black hole and shape of the accretion disk. However, reworking an estimate of luminosity based on mass transfer rate into a black hole gives:$$E = \frac{\mu m}{R},$$ where $\mu$ is the standard gravitational parameter for the black hole, $m$ is the mass of the object falling into it, and $R$ is the radius of the accretion disk. Lets take the black hole at the center of our galaxy as an example. I calculate $\mu$ to be about $5.7\times10^{26}$ and $r$ about $7.5\times10^{12}$ meters (~ 50 AU). Therefore, each AMU generates about 0.8 MeV as it falls into the accretion disk. Consider this a pretty rough estimate. The problem here is that much of this kinetic energy is either a. carried into the event horizon by the falling particle or b. radiated into the event horizon by the accretion disk. Either way, much of the released energy is unusable. Conclusion You get about 9 MeV from fusion per AMU of protons that you throw into this process, and less than 0.8 MeV from accretion per AMU of protons. Converting to J and kg, we get 870 TJ per kg from fusion, and less than 77 TJ per kg from accretion. So, you are looking at something in the range of 900 TJ per kg of hydrogen.
Colloquia/Fall18 Contents 1 Mathematics Colloquium 1.1 Spring 2018 1.2 Spring Abstracts 1.2.1 January 29 Li Chao (Columbia) 1.2.2 February 2 Thomas Fai (Harvard) 1.2.3 February 5 Alex Lubotzky (Hebrew University) 1.2.4 February 6 Alex Lubotzky (Hebrew University) 1.2.5 February 9 Wes Pegden (CMU) 1.2.6 March 2 Aaron Bertram (Utah) 1.2.7 March 16 Anne Gelb (Dartmouth) 1.2.8 April 6 Edray Goins (Purdue) 1.3 Past Colloquia Mathematics Colloquium All colloquia are on Fridays at 4:00 pm in Van Vleck B239, unless otherwise indicated. Spring 2018 date speaker title host(s) January 29 (Monday) Li Chao (Columbia) Elliptic curves and Goldfeld's conjecture Jordan Ellenberg February 2 (Room: 911) Thomas Fai (Harvard) The Lubricated Immersed Boundary Method Spagnolie, Smith February 5 (Monday, Room: 911) Alex Lubotzky (Hebrew University) High dimensional expanders: From Ramanujan graphs to Ramanujan complexes Ellenberg, Gurevitch February 6 (Tuesday 2 pm, Room 911) Alex Lubotzky (Hebrew University) Groups' approximation, stability and high dimensional expanders Ellenberg, Gurevitch February 9 Wes Pegden (CMU) The fractal nature of the Abelian Sandpile Roch March 2 Aaron Bertram (University of Utah) Stability in Algebraic Geometry Caldararu March 16 Anne Gelb (Dartmouth) Reducing the effects of bad data measurements using variance based weighted joint sparsity WIMAW April 4 (Wednesday) John Baez (UC Riverside) TBA Craciun April 6 Edray Goins (Purdue) Toroidal Belyĭ Pairs, Toroidal Graphs, and their Monodromy Groups Melanie April 13 Jill Pipher (Brown) TBA WIMAW April 16 (Monday) Christine Berkesch Zamaere (University of Minnesota) TBA Erman, Sam April 20 Xiuxiong Chen (Stony Brook University) TBA Bing Wang April 25 (Wednesday) Hitoshi Ishii (Waseda University) Wasow lecture TBA Tran date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty Spring Abstracts January 29 Li Chao (Columbia) Title: Elliptic curves and Goldfeld's conjecture Abstract: An elliptic curve is a plane curve defined by a cubic equation. Determining whether such an equation has infinitely many rational solutions has been a central problem in number theory for centuries, which lead to the celebrated conjecture of Birch and Swinnerton-Dyer. Within a family of elliptic curves (such as the Mordell curve family y^2=x^3-d), a conjecture of Goldfeld further predicts that there should be infinitely many rational solutions exactly half of the time. We will start with a history of this problem, discuss our recent work (with D. Kriz) towards Goldfeld's conjecture and illustrate the key ideas and ingredients behind these new progresses. February 2 Thomas Fai (Harvard) Title: The Lubricated Immersed Boundary Method Abstract: Many real-world examples of fluid-structure interaction, including the transit of red blood cells through the narrow slits in the spleen, involve the near-contact of elastic structures separated by thin layers of fluid. The separation of length scales between these fine lubrication layers and the larger elastic objects poses significant computational challenges. Motivated by the challenge of resolving such multiscale problems, we introduce an immersed boundary method that uses elements of lubrication theory to resolve thin fluid layers between immersed boundaries. We apply this method to two-dimensional flows of increasing complexity, including eccentric rotating cylinders and elastic vesicles near walls in shear flow, to show its increased accuracy compared to the classical immersed boundary method. We present preliminary simulation results of cell suspensions, a problem in which near-contact occurs at multiple levels, such as cell-wall, cell-cell, and intracellular interactions, to highlight the importance of resolving thin fluid layers in order to obtain the correct overall dynamics. February 5 Alex Lubotzky (Hebrew University) Title: High dimensional expanders: From Ramanujan graphs to Ramanujan complexes Abstract: Expander graphs in general, and Ramanujan graphs , in particular, have played a major role in computer science in the last 5 decades and more recently also in pure math. The first explicit construction of bounded degree expanding graphs was given by Margulis in the early 70's. In mid 80' Margulis and Lubotzky-Phillips-Sarnak provided Ramanujan graphs which are optimal such expanders. In recent years a high dimensional theory of expanders is emerging. A notion of topological expanders was defined by Gromov in 2010 who proved that the complete d-dimensional simplical complexes are such. He raised the basic question of existence of such bounded degree complexes of dimension d>1. This question was answered recently affirmatively (by T. Kaufman, D. Kazdhan and A. Lubotzky for d=2 and by S. Evra and T. Kaufman for general d) by showing that the d-skeleton of (d+1)-dimensional Ramanujan complexes provide such topological expanders. We will describe these developments and the general area of high dimensional expanders. February 6 Alex Lubotzky (Hebrew University) Title: Groups' approximation, stability and high dimensional expanders Abstract: Several well-known open questions, such as: are all groups sofic or hyperlinear?, have a common form: can all groups be approximated by asymptotic homomorphisms into the symmetric groups Sym(n) (in the sofic case) or the unitary groups U(n) (in the hyperlinear case)? In the case of U(n), the question can be asked with respect to different metrics and norms. We answer, for the first time, one of these versions, showing that there exist fintely presented groups which are not approximated by U(n) with respect to the Frobenius (=L_2) norm. The strategy is via the notion of "stability": some higher dimensional cohomology vanishing phenomena is proven to imply stability and using high dimensional expanders, it is shown that some non-residually finite groups (central extensions of some lattices in p-adic Lie groups) are Frobenious stable and hence cannot be Frobenius approximated. All notions will be explained. Joint work with M, De Chiffre, L. Glebsky and A. Thom. February 9 Wes Pegden (CMU) Title: The fractal nature of the Abelian Sandpile Abstract: The Abelian Sandpile is a simple diffusion process on the integer lattice, in which configurations of chips disperse according to a simple rule: when a vertex has at least 4 chips, it can distribute one chip to each neighbor. Introduced in the statistical physics community in the 1980s, the Abelian sandpile exhibits striking fractal behavior which long resisted rigorous mathematical analysis (or even a plausible explanation). We now have a relatively robust mathematical understanding of this fractal nature of the sandpile, which involves surprising connections between integer superharmonic functions on the lattice, discrete tilings of the plane, and Apollonian circle packings. In this talk, we will survey our work in this area, and discuss avenues of current and future research. March 2 Aaron Bertram (Utah) Title: Stability in Algebraic Geometry Abstract: Stability was originally introduced in algebraic geometry in the context of finding a projective quotient space for the action of an algebraic group on a projective manifold. This, in turn, led in the 1960s to a notion of slope-stability for vector bundles on a Riemann surface, which was an important tool in the classification of vector bundles. In the 1990s, mirror symmetry considerations led Michael Douglas to notions of stability for "D-branes" (on a higher-dimensional manifold) that corresponded to no previously known mathematical definition. We now understand each of these notions of stability as a distinct point of a complex "stability manifold" that is an important invariant of the (derived) category of complexes of vector bundles of a projective manifold. In this talk I want to give some examples to illustrate the various stabilities, and also to describe some current work in the area. March 16 Anne Gelb (Dartmouth) Title: Reducing the effects of bad data measurements using variance based weighted joint sparsity Abstract: We introduce the variance based joint sparsity (VBJS) method for sparse signal recovery and image reconstruction from multiple measurement vectors. Joint sparsity techniques employing $\ell_{2,1}$ minimization are typically used, but the algorithm is computationally intensive and requires fine tuning of parameters. The VBJS method uses a weighted $\ell_1$ joint sparsity algorithm, where the weights depend on the pixel-wise variance. The VBJS method is accurate, robust, cost efficient and also reduces the effects of false data. April 6 Edray Goins (Purdue) Title: Toroidal Belyĭ Pairs, Toroidal Graphs, and their Monodromy Groups Abstract: A Belyĭ map [math] \beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C) [/math] is a rational function with at most three critical values; we may assume these values are [math] \{ 0, \, 1, \, \infty \}. [/math] A Dessin d'Enfant is a planar bipartite graph obtained by considering the preimage of a path between two of these critical values, usually taken to be the line segment from 0 to 1. Such graphs can be drawn on the sphere by composing with stereographic projection: [math] \beta^{-1} \bigl( [0,1] \bigr) \subseteq \mathbb P^1(\mathbb C) \simeq S^2(\mathbb R). [/math] Replacing [math] \mathbb P^1 [/math] with an elliptic curve [math]E [/math], there is a similar definition of a Belyĭ map [math] \beta: E(\mathbb C) \to \mathbb P^1(\mathbb C). [/math] Since [math] E(\mathbb C) \simeq \mathbb T^2(\mathbb R) [/math] is a torus, we call [math] (E, \beta) [/math] a toroidal Belyĭ pair. The corresponding Dessin d'Enfant can be drawn on the torus by composing with an elliptic logarithm: [math] \beta^{-1} \bigl( [0,1] \bigr) \subseteq E(\mathbb C) \simeq \mathbb T^2(\mathbb R). [/math] This project seeks to create a database of such Belyĭ pairs, their corresponding Dessins d'Enfant, and their monodromy groups. For each positive integer [math] N [/math], there are only finitely many toroidal Belyĭ pairs [math] (E, \beta) [/math] with [math] \deg \, \beta = N. [/math] Using the Hurwitz Genus formula, we can begin this database by considering all possible degree sequences [math] \mathcal D [/math] on the ramification indices as multisets on three partitions of N. For each degree sequence, we compute all possible monodromy groups [math] G = \text{im} \, \bigl[ \pi_1 \bigl( \mathbb P^1(\mathbb C) - \{ 0, \, 1, \, \infty \} \bigr) \to S_N \bigr]; [/math] they are the ``Galois closure of the group of automorphisms of the graph. Finally, for each possible monodromy group, we compute explicit formulas for Belyĭ maps [math] \beta: E(\mathbb C) \to \mathbb P^1(\mathbb C) [/math] associated to some elliptic curve [math] E: \ y^2 = x^3 + A \, x + B. [/math] We will discuss some of the challenges of determining the structure of these groups, and present visualizations of group actions on the torus. This work is part of PRiME (Purdue Research in Mathematics Experience) with Chineze Christopher, Robert Dicks, Gina Ferolito, Joseph Sauder, and Danika Van Niel with assistance by Edray Goins and Abhishek Parab.
Alexandrou, C. and Forcrand, Ph. de and D'Elia, M. and Panagopoulos, H. (2000) Efficiency of the UV-filtered Multiboson algorithm. Physical review. D, Particles, fields, gravitation, and cosmology, 61 . 074503. ISSN 1550-7998 Abstract We study the efficiency of an improved Multiboson algorithm with two flavours of Wilson fermions in a realistic physical situation ($\beta = 5.60$, $\kappa = 0.156$ on a $16^3 \times 24$ lattice). The performance of this exact algorithm is compared with that of a state-of-the-art HMC algorithm: a considerable improvement is obtained for the plaquette auto-correlation time, while the two algorithms appear similarly efficient at decorrelating the topological charge. Item Type: Article Additional Information: Imported from arXiv Subjects: Area02 - Scienze fisiche > FIS/02 - Fisica teorica, modelli e metodi matematici Divisions: UNSPECIFIED Depositing User: dott.ssa Sandra Faita Date Deposited: 08 Aug 2013 15:00 Last Modified: 08 Aug 2013 15:00 URI: http://eprints.adm.unipi.it/id/eprint/1589 Repository staff only actions View Item
Historically, Special Relativity was motivated by apparent inconsistencies between Maxwell's Electrodynamics and Newtonian Mechanics. In Einstein's well known paper "On the electrodynamics of moving bodies" he explains quite well his motivations. Central objects of the theory are the Lorentz transformations. If one forgets motivations, history and intuition the Lorentz transformations are formally defined as the linear transformations $\Lambda:\mathbb{R}^4\to \mathbb{R}^4$ such that $$\eta(\Lambda v,\Lambda w)=\eta(v,w),$$ where $\eta = \operatorname{diag}(-1,1,1,1)$. Furthermost, it seems that before this definition they were defined as the transformations which keep the speed of light the same in all frames. My question is: how did Lorentz transformations get this modern definition? How were they first defined, how did they relate to Einstein's paper, and how did they get the modern definition as "transformations which preserve the spacetime inner product"? Specifically I'm interested in knowing how from the motivations for relativity physicists got to the definition of Lorentz transformations as the transformations $\Lambda$ such that $\eta(\Lambda x,\Lambda y) = \eta(x,y)$
Use proof by induction to prove that that $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$, .\Base case: $$\frac{1}{4}=\frac{1}{24}\leq \frac{1}{2^4-1}$$ Inductive hypothesis: Assume there exists $k\in \mathbb{N}$ s.t. $$ \frac{1}{k!}\leq\frac{1}{2^k-1} $$ Inductive step: Show that:$$ \frac{1}{(k+1)!}\leq\frac{1}{2^{k+1}-1} $$ Now, $$\frac{1}{(k+1)!}=\frac{1}{k!}\cdot\frac{1}{k+1}$$ Using the hypothesis $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{k+1}$$ Because $n\geq4, \frac{1}{k+1}<\frac{1}{2}$ $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{2}=\frac{1}{2^{k+1}-2}\leq\frac{1}{2^{k+1}-1}$$ Hence by mathematical induction we have proved that $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$ Firstly I need to know if the proof is correct, secondly it has to be as concise as possible hence I would like to know if there are any lines I can change/delete And lastly can anyone explain to me why every sentence starts with "\" It looks perfectly fine in www.sharelatex.com ;(
Introduction Engineering and math modeling applications are an important part of modern software. These applications use extremely complicated math and require presenting results in the form of charts, schemes, 3D models, formulae. The result presentation must be simple to perceive and understand. Therefore, developing math and engineering applications with appropriate GUI is an essential task. WPF technology is one of the suitable GUI developing frameworks for solving the task. There are many free and commercial components for presenting scientific results in beautiful and proper form. Working with math formula is one of the possible requirements for scientific applications. There are several common scenarios of dealing with formulae. The most frequently used are: evaluating formula value and draw formulae in natural math form. The article describes a simple way of realizing math formula evaluation and drawing in WPF applications using two available C# libraries. The first library is used for drawing formulae in natural math form. The second evaluates formula value. Using the libraries together allows creating simple ‘formula calculator’ which evaluates string expressions and shows the result as math formula. Drawing Math Formulae We will use the WPF Math project for drawing math formula. This .NET library realizes rendering math expressions, written in TeX format. The TeX format is widely used for creating scientific articles and documents. For drawing math formula, there is one simple WPF component ‘ FormulaControl’ with string property ‘ Formula’. Assigning value to the property the formula is rendered on the component. For example, assigning this value ‘\frac{(2n-1)!}{2^{n+1}} \sqrt{\frac{\pi}{a^{2n+1}}}’ to the property draws math formula as follows: $\frac{(2n-1)!}{2^{n+1}} \sqrt{\frac{\pi}{a^{2n+1}}}$ Another example is the next TeX expression ‘{A}\cdot{sin\left({n\cdot x}\right)}+\frac{B}{2} \cdot {e^{m \cdot y}}’ which produces the following formula in math form: ${A}\cdot{sin\left({n\cdot x}\right)}+\frac{B}{2} \cdot {e^{m \cdot y}}$ As one can see from the examples above, all formulae drawn in natural math form that is easy to read and understand. One small drawback of such way of writing formula for drawing is that it requires knowing of expression format. The format is like scripting language and uses special commands and lexemes to define math symbols like fraction, square root, sum, product, integral, derivative and so on. The advantage of the format is that all formula can be written as plain text without using any ‘non-keyboard’ symbols and special editors. TeX Evaluating Math Expressions We will use Analytics project (http://sergey-l-gladkiy.narod.ru/index/analytics/0-13) for evaluating math expressions. The ANALYTICS framework is a general purpose symbolic library. The main functionality of the library is evaluating symbolic (analytical) math expressions. For example, there is a simple code for dealing with math expressions: Translator t = new Translator(); t.Add("A", 0.5); t.Add("x", 12.4); string f = "Aln(Pix)"; double y1 = (double)t.Calculate(f); t.Variables["x"].Value = 27.3; double y2 = (double)t.Calculate(f); Here, the is one of the main classes of the Translator Analytics framework that realizes most of the symbolic capabilities. The code adds two variables to the instance: ‘ Translator A’ and ‘ x’. Then it evaluates the symbolic expression using current variable values, changes the value of the ‘ x’ variable and evaluates the same expression again. As can be seen from the example above, the analytics library provides simple interface to evaluate math expressions. The expressions written as simple C# strings and can contain literals ( constants), variables, math operators, functions. The framework supports all algebraic operators (addition, multiplication, power and so on), comparison operators (equality, inequality, less than, greater than and so on) and logical operators (not, and, or). Some special math operators also supported, such as sum ( ∑), product ( ∏), square root ( √) and other. These special operators written in expressions as Unicode symbols and their notation is close to the equivalent math formula. One drawback of using these special symbols is that they must be provided by special editor. The format of expressions in the Analytics framework is closer to math expressions than the TeX expressions are. Nevertheless, this is not natural format math formulae should be written with. For example, power operation for the Analytics expression must be written as ‘ x^y’, which is not what we expect for presenting equivalent math formula x . y Realizing Simple Formula Calculator Therefore, we have one .NET library for evaluating math expressions and one WPF component for drawing math expressions as natural math formula. And we would like to build simple ‘formula calculator’ that would evaluate math expression and show the result as math formula. The only problem left is that the evaluation library Analytics works with expressions written with its own format and the drawing library WPF Math requires formula in TeX format. Fortunately, the Analytics library contains special classes to convert internal formulae into the TeX format. Here is an example code for such conversion: string f = "Asin(nx)+B/2e^(my)"; AnalyticsConverter converter = new AnalyticsTeXConverter(); string texf = ""; try { texf = converter.Convert(f); } catch (Exception ex) { } In the code above, an instance of the AnalyticsTeXConverter class created. This class has the Convert method for converting symbolic expressions to the TeX format. For the expression, result string in TeX format is the following: {{{A}\cdot{{sin}\left({{n}\cdot{x}}\right)}}+{{\frac{B}{2}}\cdot{{e}^{{m}\cdot{y}}}}} Now we have all ingredients for realizing our simple ‘formula calculator’. The main functionality of formulae evaluation and drawing can be realized with one method as follows: string f = inputTextBox.Text; try { if (translator.CheckSyntax(f)) { object v = translator.Calculate(f); string texf = converter.Convert(f); string vs = Utilities.SafeToString(v); formulaControl.Formula = texf + " = {" + vs + "}"; } } catch (Exception ex) { formulaControl.Formula = ex.Message; } The simple WPF application for formulae evaluation and drawing is presented in the picture below: This application allows input math expression written in Analytics format, evaluate its value and draw the result as math formula. There are some other examples of formulae that can be evaluated and drawn with the application. Multiplication of fractions ‘ (x-2)/2(x+1)/A/B(x-A)(x+B/2)’: Parametric (logarithmic) functions ‘ Alog{3}(x+m)-B/log{2}(x+n)’: Complicated formula ‘ Ax^2/(m-1)!+Be^sin(y+1)/n!+C*log{n+m}(z^2)’ As one can see from the examples above, all formulae drawn in natural math format. The presented approach can be used in various engineering and math applications where results of evaluated expressions must be shown as math formulae. It should be noted here that the same approach was used in early versions of Computer Algebra Systems (CAS). For example, Maple system used ‘console’ input, where the expressions must be written as plain text, but the result of evaluation was presented as math formula. Now CAS systems use interactive editors to write all math formulae in natural form. Realization of such interactive editor could be the next developments of the libraries used here. Conclusions This article introduced one approach for realizing formulae evaluation and drawing in WPF applications. Two libraries were used: one for math expression evaluation and one for formulae drawing. The approach demonstrated with simple WPF application which allows input math expression as plain text and then evaluate value of the expression and draw the result as math formulae. The same approach can be used in other math end engineering applications where results of math formulae evaluation should be presented in proper and readable form. History 10 th March, 2019: Initial version
Let $\displaystyle f: [0,1] \rightarrow \mathbb{R}$ given by $$f(x) = \begin{cases} 0 & x \notin \mathbb{Q} \\ \\ 0 & x = 0 \\ \\ \frac{1}{q_x} & x = \frac{p_x}{q_x} \in \mathbb{Q} \backslash \{0\}, \ p_x \in \mathbb{Z}, \ q_x \in \mathbb{N}, \ \text{gcd}(\|p_x\|, q_x) = 1 \end{cases}$$ Is $\displaystyle f$ Riemann integrable? I am trying to use the equivalent statements $\displaystyle g:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable and $\displaystyle \forall \epsilon >0 \ \exists$ step functions $\displaystyle \rho, \psi$ with $\displaystyle \rho \leq g \leq \psi$ such that $\displaystyle \int_a^b (\psi - \rho) \leq \epsilon$. I guess that means I would have to somehow show that given $\displaystyle \epsilon$, there exists only a finite amount of $\displaystyle x \in [0,1]$ with $\displaystyle f(x) \geq \epsilon$? Is it recommended that I consider something else instead? If not, how should I do this?
I am confused, as not clear except by multiplying both terms on the r.h.s, and showing that all cancel out except the two on the l.h.s., as below: $(x)(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)= x^k -1$ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community I am confused, as not clear except by multiplying both terms on the r.h.s, and showing that all cancel out except the two on the l.h.s., as below: $(x)(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)= x^k -1$ Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. Here's another approach using roots of unity and the fundamental theorem of algebra. The zeros of $x^k - 1$ are $\{e^{\frac{n2\pi i}{k}}\}$ for $0 \leq n < k$. Therefore, we can rewrite $x^k - 1$ as: $$x^k - 1 = M\prod_{n=0}^{k-1} (x-e^{\frac{n2\pi i}{k}})$$ for some $M$ Now, let's find the zeros of $P(x) = (x-1)(x^{k-1} + ... + 1)$ Clearly, $x = 1 = e^{\frac{0*2\pi i}{k}}$ is a zero. What about $Q(x) = (x^{k-1} + ... + 1)$? Let's substitute our $k-1$ roots of unity, $r_n$ where $n > 1$ $$Q(r_n) = \sum_{j=0}^{k-1} (r_n)^j = \sum_{j=0}^{k-1} (e^{\frac{n2\pi i}{k}})^j = \frac{1 - (e^{\frac{n2\pi i}{k}})^k}{1 - e^{\frac{n2\pi i}{k}}} = \frac{0}{1 - e^{\frac{n2\pi i}{k}}} = 0$$ Using the sum of a finite geometric series and the fact that: $(e^{\frac{n2\pi i}{k}})^k = e^{n2\pi i} = 1^n$ Therefore, we can rewrite $Q(x)$ as: $$Q(x) = D' \prod_{n=1}^{k-1} (x-e^{\frac{n2\pi i}{k}})$$ and $P(x)$ as: $$P(x) = D\prod_{n=0}^{k-1} (x-e^{\frac{n2\pi i}{k}})$$ for some constant $D$. Now, we have that $P(x)$ agrees with $x^k - 1$ at $k$ points, which are the $k$ roots of unity. However, we also have that $P(0) = -1$ and $0^k - 1 = -1$ We can sub this back into the factored expressions to see that $M = D$, or we can also conclude that because $(x-1)(x^{k-1} + ... + 1)$ and $x^k - 1$ agree at $k + 1$ distinct points, $(x-1)(x^{k-1} + ... + 1) = x^k - 1$ for all $x$. For fun: Let $x$ be real, and consider the geometric series: 1)$\sum_{i=0}^{k-1} x^i = 1+x +x^2 +..+x^{k-1};$ 2) $ x \sum_{i=0}^{k-1}x^i = \ \ x+x^2+...+x^{k-1} +x^k;$ Subtract : 2)-1): $(x-1)\sum_{i=0}^{k-1} x^{i}= x^k-1$. This one cries out for a simple proof by induction: If $k = 1$ we evidently have $x^1 - 1 = (x^1 - 1)(1), \tag 1$ and if $k = 2$: $x^2 - 1 = (x -1)(x + 1) = (x - 1)\left ( \displaystyle \sum_0^1 x^i \right ); \tag 2$ if we now suppose that the formula holds for some positive $m \in \Bbb Z$, $x^m - 1 = (x - 1) \left ( \displaystyle \sum_0^{m - 1} x^i \right ), \tag 3$ then $x^{m + 1} - 1 = x^{m + 1} - x^m + x^m - 1$ $= x^m(x - 1) + (x - 1) \left ( \displaystyle \sum_0^{m - 1} x^i \right ) = (x - 1) \left ( x^m + \displaystyle \sum_0^{m - 1} x^i \right ) = (x - 1) \left ( \displaystyle \sum_0^m x^i \right ), \tag 4$ which shows that the formula $x^k - 1 = (x - 1) \left ( \displaystyle \sum_0^{k -1} x^i \right ) \tag 5$ is valid for all positive integers $k$. It's like this: $(x-1)(x^{k-1}+x^{k-2}+\ldots+1)$ $=x(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)$ $=x^{k-1+1}+x^{k-2+1}+x^{k-3+1}+x^{k-4+1}...+x^3+x^2+x-x^{k-1}-x^{k-2}-x^{k-3}-...-x-1$ $=x^{k}+x^{k-1}+x^{k-2}+x^{k-3}...+x^3+x^2+x-x^{k-1}-x^{k-2}-x^{k-3}-...-x^2-x-1$ $=x^{k}+(x^{k-1}+x^{k-2}+x^{k-3}...+x^3+x^2+x)-(x^{k-1}+x^{k-2}+x^{k-3}+...+x^2+x)-1$ $=x^k-1$ You can also prove it by reversing the solution above.

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