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I am trying to prove A property of a cyclic order on a ring. In order to do it, I need the last two properties (lemmas 1.11 and 1.12) in this question. I separated them from the original theorem because they are not related to rings, and applicable to any cyclically ordered group. Please help me to find if the statements are correct. Some of the intermediate lemmas are also used in the Quadrants of a cyclically ordered group. If lemmas 1.11 and 1.12 are correct, then there is a question about the "magic" number $4$: it looks like any circle with a positive element can always be split onto not more than $4$ positive segments (The rule of three steps for a cyclically ordered group); at the same time, the logic of lemmas 1.11 and 1.12 works only up to $4$ positive elements. Is this just a coincidence? I am using the Cycle notation for cyclic orders wherever it is convenient. Definition 1.1. An element $x$ of a cyclically ordered group is positive iff $[0, x, -x]$. Definition 1.2. An element $x$ of a cyclically ordered group is negative iff $[0, -x, x]$. Lemma 1.1. If an element $a$ of a cyclically ordered group is positive ( negative), then element $-a$ is negative ( positive). Proof: follows from the definitions of positive and negative elements. Lemma 1.2. If $a$ and $b$ are positive elements of a cyclically ordered group, then $a + b \ne 0$ and $b + a \ne 0$. Proof: Assuming $a + b = 0$ or $b + a = 0$: $a = -b$; $b = -a$; Lemma 1.1: $a$ and $b$ are negative, contradiction. Corollary 1.2. If element $a$ of a cyclically ordered group is positive, then $2a \ne 0$. Lemma 1.3. If $a$ and $b$ are positive elements of a cyclically ordered group, and $a + a = b + b$, then $a = b$. Proof: Let $a + a = b + b = c$, $[0, a, b]$; Compatibility with $a$: $[0, a, b] \implies [a, c, a + b]$; Compatibility with $b$: $[0, a, b] \implies [b, a + b, c]$; 4-Cycle: $[a, c, a + b] \land [b, a + b, c] \iff [a, c, b, a + b] \implies [a, c, b]$; 4-Cycle: $[0, a, b] \land [a, c, b] \iff [0, a, c, b] \implies [b, 0, c]$; Compatibility with $−b$: $[b, 0, c] \implies [0, -b, b]$; $b$ is negative, contradiction. Lemma 1.4. If $a$ is a positive element of a cyclically ordered group, and $[0, a', a]$ for an element $a'$, then $a'$ is positive. Proof: Negation: $[0, a', a] \iff [0, -a ,-a']$; Transitivity: $[0, a, -a] \land [0, -a, -a'] \implies [0, a, -a']$; Transitivity: $[0, a', a] \land [0, a, -a'] \implies [0, a', -a']$. Lemma 1.5. If $a$ and $b$ are positive elements of a cyclically ordered group such that $[0, a, b]$, and $[a, c, b]$ for an element $c$, then $c$ is positive. Proof: 4-Cycle: $[0, a, b] \land [a, c, b] \iff [0, a, c, b] \implies [0, c, b]$; Lemma 1.4: $[0, c, b] \implies$ $c$ is positive. Lemma 1.6. If $a$ and $b$ are positive elements of a cyclically ordered group, then $[0, a, a + b]$, $[0, a, b + a]$, $[0, b, a + b]$, $[0, b, b + a]$. Proof: Lemma 1.2: $a + b \ne 0$, $b + a \ne 0$; Assuming $[0, a + b, a]$ or $[0, b + a, a]$: Compatibility with $–a$: $[-a, b, 0]$; Cyclicity: $[-a, b, 0] \iff [0, -a, b]$; Lemma 1.4: $[0, -a, b] \implies -a$ is positive; Lemma 1.1: $a$ is negative, contradiction. Assuming $[0, a + b, b]$ or $[0, b + a, b]$: Compatibility with $–b$: $[-b, a, 0]$; Cyclicity: $[-b, a, 0] \iff [0, -b, a]$; Lemma 1.4: $[0, -b, a] \implies -b$ is positive; Lemma 1.1: $b$ is negative, contradiction. Corollary 1.6. If $a$ is a positive element of a cyclically ordered group, then $[0, a, 2a]$. Lemma 1.7. If $a$ is a positive element of a cyclically ordered group, and $[0,b,a]$ for an element $b$, then $(a - b)$ is positive, and $[0, a - b, a]$, where $a - b \equiv a + (-b)$. Proof: Compatibility with $-b$: $[0, b, a] \implies [-b, 0, a - b]$; Lemma 1.4: $[0, b, a] \implies b$ is positive; Lemma 1.6: $[0, a, a + b]$; Compatibility with $-b$: $[0, a, a + b] \implies [-b, a - b, a]$; 4-Cycle: $[−b, 0, a − b] \land [−b, a − b, a] \iff [−b , 0, a − b, a] \implies [0, a − b, a]$; Lemma 1.4: $[0, a - b, a] \implies (a - b)$ is positive. Lemma 1.8. If $a$ and $b$ are positive elements of a cyclically ordered group, and $[0, a', a]$ for an element $a'$, then $[0, a' + b, a + b]$ and $[0, b + a', b + a]$. Proof: Compatibility with $b$: $[0, a', a] \implies [b, a' + b, a + b]$ (or $[b, b + a', b + a]$); Lemma 1.6: $[0, b, a + b]$ (or $[0, b, b + a]$); 4-Cycle: $[b, a′ + b, a + b] \land [0, b, a + b] \iff [0, b, a′ + b, a + b] \implies [0, a′ + b, a + b]$; (or $[b, b + a′, b + a] \land [0, b, b + a] \iff [0, b, b + a′, b + a] \implies [0, b + a′, b + a]$). Corollary 1.8. If $b$ is a positive element of a cyclically ordered group, and $[0, a, b]$ for an element $a$, then $[0, 2a, a + b]$, $[0, 2a, b + a]$, $[0, a + b, 2b]$, and $[0, b + a, 2b]$. Lemma 1.9. If $a$ and $b$ are positive elements of a cyclically ordered group, and $[0, a', a]$ and $[0, b', b]$ for some elements $a'$ and $b'$, then $[0, a' + b', a + b]$ and $[0, b' + a', b + a]$. Proof: Lemma 1.4: $[0, b', b] \implies b'$ is positive; Lemma 1.8: $[0, a', a] \implies [0, a' + b', a + b']$ (or $[0, b' + a', b' + a]$); Lemma 1.8: $[0, b', b] \implies [0, a + b', a + b]$ (or $[0, b' + a, b + a]$); Transitivity: $[0, a' + b', a + b'] \land [0, a + b', a + b] \implies [0, a' + b', a + b]$ (or $[0, b' + a', b' + a] \land [0, b' + a, b + a] \implies [0, b' + a', b + a]$). Corollary 1.9. If $b$ is a positive element of a cyclically ordered group, and $[0, a, b]$ for an element $a$, then $[0, 2a, 2b]$. Lemma 1.10. If $[0, a', a]$, $[0, b', b]$, and $a + b = 0$ for some elements $a$, $a'$, $b$, $b'$ of a cyclically ordered group, then $a' + b' \ne 0$. Proof: $a + b = 0 \implies b = -a$; Assuming $a' + b' = 0$; $b' = -a'$; Applying 1 and 2 to $[0, b', b]$: $[0, -a', -a]$; Negation: $[0, a', a] \iff [0, -a, -a']$, contradiction. Lemma 1.11. If $a$, $b$, and $c$ are positive elements of a cyclically ordered group such that $a + b + c = 0$, and $[0, a', a]$, $[0, b', b]$, $[0, c', c]$ for some elements $a'$, $b'$, $c'$, then $a' + b' + c' \ne 0$. Proof: Lemma 1.9: $[0, a', a] \land [0, b', b] \implies [0, a' + b', a + b]$; Lemma 1.10: $[0, a' + b', a + b] \land [0, c', c] \land a + b + c = 0 \implies a' + b' + c' \ne 0$. Lemma 1.12. If $a$, $b$, $c$, $d$ are positive elements of a cyclically ordered group such that $a + b + c + d = 0$, and $[0, a', a]$, $[0, b', b]$, $[0, c', c]$, $[0, d', d]$ for some elements $a'$, $b'$, $c'$, $d'$, then $a' + b' + c' + d' \ne 0$. Proof: Lemma 1.9: $[0, a', a] \land [0, b', b] \implies [0, a' + b', a + b]$; Lemma 1.9: $[0, c', c] \land [0, d', d] \implies [0, c' + d', c + d]$; Lemma 1.10: $[0, a' + b', a + b] \land [0, c' + d', c + d] \land a + b + c + d = 0 \implies a' + b' + c' + d' \ne 0$. Note: The logic cannot be extended onto five or more positive elements: in $\mathbb Z_{10}$ element $4$ is positive, $4 + 4 + 4 + 4 + 4 = 0$, $[0,2,4]$, but $2 + 2 + 2 + 2 + 2 = 0$.
For the full paper with proofs, click here. Abstract Editor’s note: This paper represents the second installment of a masters thesis by Jonathan Johnson. This paper continues the development of the theory of summation chains of sequences. The concept of sum-related is defined: two sequences are sum-related if one sequence appears in the summation chain of the other. The main result is a theorem to determine if two given sequences are sum-related. Introduction Editor’s note: Previously, the concept of a summation chain of sequences was defined. We reproduce the example and introduction here: Given a complex-valued sequence (a_n)^{\infty}_{n=1}, the sequence of partial sums of (a_n) is given by the sequence The sequence of differences of (a_n) is given by the sequence The processes of finding the sequence of partial sums and finding the sequence of differences of a sequence are inverses of each other so every sequence is the sequence of differences of its sequence of partial sums and the sequence of partial sums of its sequence of differences. Every sequence has a unique sequence of partial sums and a unique sequence of differences so it is always possible to find the sequence of partial sums of the sequence of partial sums and repeat the process ad infinitum. Similarly, we can find the sequence of differences of the sequence of differences and repeat ad infinitum. The result is a doubly infinite sequence or “chain” of sequences where each sequence is the sequence of partial sums of the previous sequence and the sequence of differences of the following sequence. \begin{array}{r\|ccccccccc}\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\a^{(2)} & 1 & 1 & 2 & 2 & 3 & 3 & 4 & 4 & \cdots\\a^{(1)} & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & \cdots\\a^{(0)} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & \cdots\\a^{(-1)} & 1 & -2 & 2 & -2 & 2 & -2 & 2 & -2 & \cdots\\a^{(-2)} & 1 & -3 & 4 & -4 & 4 & -4 & 4 & -4 & \cdots\\\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\\end{array} Example Let a^{(0)} be the sequence defined by a^{(0)}_n=(-1)^{n+1} for all n\in\mathbb{N}. For all integers m>0, let a^{(m)} be the sequence of partial sums of a^{(m-1)}, and for all integers m<0, let a^{(m)} be the sequence of differences of a^{(m+1)}. Every sequence can be used to create a unique chain of sequences. Editor’s note: This installment examines the relationship between sequences and their summation chains, and provides a way to determine whether two sequences are sum-related.
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Let $X$ be a variety over $\mathbb{C}$, say separated. According to Deligne's results, there is a "mixed Hodge structure" on the total cohomology $H^\bullet(X(\mathbb{C}), \mathbb{Z})$. One component of this is a "weight filtration" on $H^\bullet(X(\mathbb{C}), \mathbb{Q})$. I haven't read Deligne's "Theorie de Hodge" and don't really understand all this, but I believe that in the case of a smooth projective variety, this reduces to usual Hodge theory and the weight filtration is the filtration by grading, and the extension to singular varieties comes by some sort of simplicial resolution by smooth objects. Let $Y_0$ be a variety over a finite field $\kappa$. Given a mixed perverse sheaf $K_0$ on $Y_0$, there is a canonical (and functorial) weight filtration on $K_0$, such that the sucessive subquotients are pure complexes of increasing weight (in the sense of Weil II). What do these to have to do with each other? In section 6 of BBD (asterisque 100), it seems that the authors are using the functoriality of the weight filtration over finite fields to deduce results about the weight filtration over $\mathbb{C}$. Namely, I'd be interested if, given a perverse sheaf $K$ (say, of geometric origin) on a smooth, proper scheme $X$ over $\mathbb{C}$ which can be "spread out" to perverse sheaves of "reduction of $X$ mod a prime" there is some way in which the weight filtration on the cohomology of $K$ (actually, I'm not sure that this exists, it seems to in the constant case at least) can be viewed as a completion of the weight filtrations in finite characteristic. Here is the specific result in BBD: Let $f: X \to Y$ be a separated morphism of schemes of finite type over $\mathbb{C}$. Suppose that the stalks of $R^n f_ \mathbb{Q}$ are $H^n(X_y, \mathbb{Q})$, and that these form a local system. Then the weight filtration on these stalks form a locally constant filtration of the local system $R^n f_* \mathbb{Q}$. This appears to be proved by reducing mod a prime, where one has a Frobenius and the perverse weight filtration makes sense. (One reason to think these might be related is that if $X_0$ is a proper smooth scheme over $\mathbb{\kappa}$, then the cohomologies $H^i(X, \mathbb{Q}_l)$ have weight $i$ by the Weil conjectures, and this has some correspondence with how the weight filtration was defined for projective, smooth schemes over $\mathbb{C}$.) *Which is done by reducing the field $\mathbb{C}$ of definition to some finitely generated ring over $\mathbb{Z}$, and then working from there.
Let's say that a formula $\Phi(x, y)$ is a satisfaction predicate for $\mathcal L_\in$ if is satisfies the Tarski clauses; that is, if: ($\in$) $\Phi(``x_i\in x_j", a) \leftrightarrow a(i)\in a(j)$ ($=$) $\Phi(``x_i = x_j", a) \leftrightarrow a(i) = a(j)$ ($\wedge$) $\Phi(``\phi \wedge \psi", a) \leftrightarrow \Phi(``\phi", a) \wedge \Phi(``\psi", a)$ ($\neg$) $\Phi(``\neg\phi", a)\leftrightarrow \neg\Phi(``\phi", a)$ ($\exists$) $\Phi(``\exists x_i\phi", a)\leftrightarrow \exists a'=_{i} a\Phi(``\phi", a')$ where $\phi,\psi\in\mathcal L_\in$, $a$ is a function with domain $\omega$, and $a=_{i} a'$ means that $a$ and $a'$ agree except possibly at $i$. We assume that $\Phi(x, y)$ only applies to formula/assignment pairs. Then we want to show that there is such a predicate in second-order set theory. I like to do this in two steps. First, I define $X$ to be a satisfaction class for $\phi\in\mathcal L_\in$ if $X$ satisfies the Tarski clauses for subformulas of $\phi$ and I prove by induction that any two satisfaction classes for $\phi$ are co-extensive. Second, I prove by induction that every formula in $\mathcal L_\in$ has a satisfaction class. One can then conclude that $\Phi(x, y) =$ "there is a satisfaction class $X$ for $x$ such that $\langle x, y\rangle\in X$" is a satisfaction predicate for $\mathcal L_\in$. Here's how the second induction goes for the $\exists$ case. Suppose that $X$ is a satisfaction class for $\phi$. Now by predicative comprehension, we define $Y = X \cup \{\langle ``\exists x_i\phi", a\rangle: \exists a'=_{i} a(\langle ``\phi",a'\rangle\in X)\}$. Since $X$ is a satisfaction class for $\phi$, it is easy to see that $Y$ is a satisfaction class for $\exists x_i\phi$, as required. (It might be interesting to note that this argument doesn't require the full power of MK. In particular, it only uses predicative instances of comprehension. What it requires beyond NBG is separation for $\Sigma^1_1$ formulas (in order to run the inductions in the two steps). But this is an assumption about the sets, not about the classes. In other words, the argument will go through even if you think there are relatively few classes, so long as you think that sets satisfy separation for $\Sigma^1_1$ formulas.)
Let $X$ be a smooth projective variety over $\mathbf{Q}$, and $\mathcal{X}$ a smooth projective model over $\mathbf{Z}[1/N]$ for $N$ large enough. Call $\eta$ the generic point $\text{Spec}(\mathbf{Q}) \to\text{Spec}(\mathbf{Z}[1/N]) =: S$, and $s\in S$ any closed point. There is a specialization map $$\text{NS}(X_{\overline{\eta}})\to \text{NS}(X_{\overline{s}})$$ Does there exist $s$ such that it is an isomorphism? By the proper and smooth base change theorem in $\ell$-adic cohomology, it is injective, so the issue is surjectivity.
Our second anniversary, and it's a bit of a love-in. Sorry to disappoint. Colin and his partner have a new baby, Fred! Happy anniversary to us! Number of the podcast: $2+\sqrt{3}$, because it comes up in the Lucas-Lehmer test for Mersenne primes. A brief digression on the test, Mersenne primesRead More → I stumbled, recently, on a blog post about "Egoless programming", a feature of a book called The Psychology of Computer Programming. The points struck me as good general advice -- but also good advice for mathematicians at any level, so I thought I'd rework them here. I've made minor adjustmentsRead More → How would I find the limit of $\frac{1}{x^2} - \cosec^2(x)$ as $x$ goes to 0? There are several valid approaches to this; two that spring to mind are L'Hôpital's rule, which I like because it's got an accent and two apostrophes, and series expansions, which I like because I likeRead More → A reader (not, in fact, a Core 4 student) wrote in to ask: I have an ellipse in my spreadsheet program, using the formula $y = \frac ba \sqrt{a^2 - x^2}$, and I want to know the angle the normal to the ellipse makes with the horizontal at any valueRead More → Someone on the internet asks: I don't get why $\ln(e^x) = e^{\ln(x)}$. Can you explain? Of course I can! Or at least, I can try; the easy answer is to say 'by definition', but that doesn't help you much. $\log_n(x)$ answers the question "what power would I raise $n$ to,Read More → There's a lot of dead time to fill up when you're covering the Boat Race. Once you've interviewed one gaggle of inebriated revellers from each of the two universities, you've interviewed them all. Once you've explained that the best strategy is to go faster than the other boat, there's notRead More → A student asks: What's the best way to use flashcards for revision? Flashcards are very popular in the USA, but have never really taken off over here. Any student with an exam to study for has a large stack of things they're trying to learn: a question on one side,Read More → At the East Dorset MathsJam Christmas party, @jussumchick (Jo Sibley in real life) posed the following question: There are two ways to draw a 16-gon with rotational symmetry of order 8 inside a unit circle, as shown. What's the ratio of their areas? Typically, I look at this sort ofRead More → This is an excerpt from my upcoming book, The Maths Bible, (which will eventually be available from all good bookstores.) Udo of Aachen (c. 1200-1270) was a Benedictine monk, scholar, poet and mathematician. His best-known poetical work is Fortuna Imperatrix Mundi, which is usually know by its choral title, ORead More →
As you know, the definition of random vector is the following: Let $(\Omega, \mathcal{F})$ be a measurable space. Then function $\mathbf{X}: \Omega \to \mathbb{R}^n$ is called random vector on $(\Omega, \mathcal{F})$ if it is measurable-$\mathcal{F}$, i.e. $~\{\omega: \mathbf{X}(\omega) \in B\} \in \mathcal{F}, ~~ \forall B \in \mathcal{B}(\mathbb{R}^n)$. Here $\mathcal{B}(\mathbb{R}^n)$ is Borel sigma-algebra on $\mathbb{R}^n$. And it is well-known fact that every component $X_i: \Omega \to \mathbb{R}$ of random vector $\mathbf{X} = (X_1, \ldots, X_n)$ is also measurable-$\mathcal{F}$ function on $(\Omega, \mathcal{F})$. But there are no words about measure in the aforementioned definition. Let's define measure $P$ on $(\Omega, \mathcal{F})$ and get a probability space $(\Omega, \mathcal{F}, P)$. After that let's define random vector $\mathbf{X}$ on this "domain" probability space. Is it correct to assume that components $X_1, \ldots, X_n$ must be defined on the same probability space $(\Omega, \mathcal{F}, P)$ as vector $\mathbf{X} = (X_1, \ldots, X_n)$ ? I said above that $\mathbf{X}$ and its components $X_1, \ldots, X_n$ must be defined on the same measurable space $(\Omega, \mathcal{F})$. But if we consider probability space instead of measurable space, then should this space be the same for $\mathbf{X}$ and every component $X_i$? P.s. Keep in mind that I don't talk about distributions of $\mathbf{X}$ and $X_i$ (i.e. about induced measures). I know that they can be different. I am talking about the "domain" measure $P$ on $(\Omega, \mathcal{F})$. EDIT Indeed, in few probability textbooks and wikipedia there is a requirement that components $X_i$ must be defined on the same probability space as each other, $(\Omega, \mathcal{F}, P)$. Also it is interesting to look at the well-known formula for the random vector $\mathbf{X} = (X_1, \ldots, X_n)$ with independent components. This formula has the following form: $$P(\mathbf{X} \in B) = \prod_{i=1}^n P(X_i \in B_i).$$ Note that the same measure $P$ is used in both left and right sides of the expression. I think this means that random vector $\mathbf{X}$ and every its component $X_i$ are defined on the same "domain" probability space $(\Omega, \mathcal{F}, P)$. What do you think about it? Should we always assume that components $X_1, \ldots, X_n$ must be defined on the same probability space $(\Omega, \mathcal{F}, P)$ as vector $\mathbf{X} = (X_1, \ldots, X_n)$ ?
1 Review of concepts related to hypothesis tests 1.1 Type I and Type II errors In hypothesis testing, there are two types of errors Type I error:reject null hypothesis when it is true Type I error rate P(reject $H_0$ \| $H_0$ true) When testing $H_0$ at a pre-specified level of significance$\alpha$, the Type I error rate is controlled to be no larger than $\alpha$. Type II error:accept the null hypothesis when it is wrong. Type II error rate P(accept $H_0$ \| $H_0$ wrong). Power :probability of rejecting $H_0$ when it is wrong Power = P(reject $H_0$ \| $H_0$ wrong) = 1 - Type II error rate. 1.2 What determines the power? The power of a testing procedure depends on Significance level $\alpha$ - the maximum allowable Type I error - the larger $\alpha$ is , the higher is the power. Deviation from $H_0$ - the strength of signal - the larger the deviation is, the higher is the power. Sample size: the larger the sample size is, the higher is the power. 2 Power of an F-test 2.1 Power calculation for F-test Test $H_0$ : $\mu_1$ = $\cdots$ = $\mu_r$ under a single factor ANOVA model: given the significance level $\alpha$ : Decision rule $$\left\{\begin{array}{ccc}{\rm reject} H_0 & if & F^{\ast}> F(1-\alpha;r-1,n_T-r)\\{\rm accept} H_0 & if & F^{\ast} \leq F(1-\alpha;r-1,n_T-r)\end{array}\right.$$ The Type I error rate is at most $\alpha$. Power depends on the noncentrality parameter $$ \phi=\frac{1}{\sigma}\sqrt{\frac{\sum_{i=1}^r n_i(\mu_i-\mu_{\cdot})^2}{r}}.$$ Note $\phi$ depends on sample size (determined by the $n_i$'s) and signal size (determined by the $(\mu_i - \mu.)^2$'s). 2.2 Distribution of F-ratio under the alternative hypothesis The distribution of F* under an alternative hypothesis. When the noncentrality parameter is $\phi$, then $$ F^{\ast} \sim F_{r-1,n_T-r}(\phi), $$ i.e., a noncentral F-distribution with noncentrality parameter $\phi$. Power = P($\sim F_{r-1,n_T-r}(\phi)$ > F(1 - $\alpha$;r - 1, $n_T - r$)). Example: if $\alpha$ = 0.01, r = 4, $n_T$ = 20 and $\phi$ = 2, then Power = 0.61. (Use Table B.11 of the textbook.) 2.3 How to calculate power of the F test using R The textbook defines the noncentrality parameter for a single factor ANOVA model as $$ \phi = \frac{1}{\sigma} \sqrt{\frac{\sum_{i=1}^r n_i (\mu_i - \mu_{\cdot})^2}{r}} $$ where r is number of treatment group (factor levels), $\mu_i$'s are the factor level means, $n_i$ is the sample size (number of replicates) corresponding to the i-th treatment group, and $\sigma^2$ is the variance of the measurements. For a balanced design, i.e., when $n_1$ = $\cdots$ = $n_r$ = n, the formula for $\phi$ reduces to $$ \phi = \frac{1}{\sigma} \sqrt{(n/r) \sum_{i=1}^r (\mu_i - \mu_{\cdot})^2}~. $$ Table B.11 gives the power of the F test given the values of the numerator degree of freedom $v_1$ = r - 1, denominator degree of freedom $v_2$ = $n_T - r$, level of significance $\alpha$ and noncentrality parameter $\phi$. Example: For r = 3, n = 5, (so that $v_1$ = 2 and $v_2$ = 12), $\alpha$ = 0.05 and $\phi$ = 2, the value of power from Table B.11 is 0.78. However, if you want to use R to compute the power of the F-test, you need to be aware that the noncentrality parameter for F distribution in R is defined differently. Indeed, compared to the above setting, the noncentrality parameter to used in the function in R will be r x $\phi^2$ instead of $\phi$. Here is the R code to be used for computing the power in the example described above: r = 3, n = 5, $\alpha$ = 0.05 and $\phi$ = 2: Critical value for the F-test when $\alpha$ = 0.05, $v_i$ = r - 1 = 2 and $v_2$ = $n_T$ - r = 12 is F.crit = qf(0.95,2,12) Then the power of the test, when will be computed as F.power = 1 - pf(F.crit, 2, 12, 3$*$2^2) Note that the function qf is used to compute the quantile of the central F distribution. Its second and third arguments are the numerator and denominator degrees of freedom of the F distribution. The function pf is used to calculate the probability under the noncentral F- density curve to the left of a given value (in this case F.crit). Its second and third arguments are the numerator and denominator degrees of freedom of the F distribution, while the fourth argument is the noncentrality parameter r x $\phi^2$ (we specify this explicitly in the above example). The values of F.crit and F.power are 3.885294 and 0.7827158, respectively. 3 Calculating sample size God: find the smallest sample size needed to achieve a pre-specified power$\gamma$; with a pre-specified Type I error rate$\alpha$; for at least a pre-specifiec signal levals. The idea behind the sample size calculation is as follows: On one hand, we want the sample size to be large enough to detect practically important deviations ( with a signal size to be at least s) from $H_0$ with high probability (with a power at least $\gamma$), and we only allow for a pre-specified low level of Type I error rate (at most $\alpha$) when there is no signal. On the other hand, the sample size should not be unnecessarily large such that the cost of the study is too high. 3.1 An example of sample size calculation For a single factor study with 4 levels and assuming a balanced design, i.e., the $n_1 = n_2 = n_3 = n_4$ (=n, say), the goal is to test $H_0$: all the factor level means $\mu_i$ are the same. Question: What should be the sample size for each treatment group under a balanced design, such that the F-test can achieve $\gamma$ = 0.85 power with at most $\alpha$ = 0.05 Type I error rate when the deviation from $H_0$ has at least $s=\sum_{i=1}^{r}(\mu_i-\mu_{\cdot})^2=40$ ? One additional piece of information needed in order to answer this question is the residual variance $\sigma^2$. Suppose from a pilot study, we know the residual variance is about $\sigma^2$ = 10. Use a trial-and-error strategyto search Table B.11. This means, for a given n (starting with n = 1), (i) calculate $\phi = (1/\sigma) \sqrt{(n/r)\sum_{i=1}^r(\mu_i - \mu_{\cdot})^2} = (1/\sigma) \sqrt{(n/r) s}$; (ii) fix the numerator degree of freedom $v_1$ = r - 1 = 3; (iii) check the power of the test when the denominator degree of freedom $v_2 = n_T - r$ (where $n_T$ = nr), with the given $\phi$ and $\alpha$ ; (iv) keep increasing n until the power of the test is closest to (equal or just above) the given value of $\gamma$. 3.2 An alternative approach to sample size calculation Suppose that we want to determine the minimum sample size required to attain a certain power of the test subject to a specified value of the maximum discrepancy among the factor level means. In other words, we want the test to attain power $\gamma$ (= 1 - $\beta$, where $\beta$ is the probability of Type II error) when the minimum range of the treatment group means \[ \Delta = \max_{1\leq i \leq r} \mu_i - \min_{1\leq i \leq r}\mu_i ~. \] Suppose we have a balanced design, i.e., $n_1 = \cdots = n_r$ = n, say. We want to determine the minimum value of n such that the power of the F test for testing $H_0$ : $\mu_1 = \cdots = \mu_r$ is at least a prespecified value $\gamma = 1 - \beta$. We need to also specify the level of significance $\alpha$ and the standard deviation of the measurements $\sigma$. Table B.12 gives the minimum value of n needed to attain a given power 1 - $\beta$ for a given value of $\alpha$, for a given number of treatments r and a given "effect size" $\Delta/\sigma$. Example : For r = 4, $\alpha$ = 0.05, in order that the F-test achieves the power 1 - $\beta$ = 0.9 when the effect size is $\Delta/\sigma$ = 1.5, we need n to be at least 14. That is, we need a balanced design with at least 14 experimental units in each treatment group. Contributors Yingwen Li (UCD) Debashis Paul (UCD)
Event detail Analysis and PDE Seminar: The effect of threshold energy obstructions on the $L^1 \to L^\infty$ dispersive estimates for some Schrödinger type equations Seminar \| October 15 \| 4:10-5 p.m. \| 740 Evans Hall Ebru Toprak, UIUC and MSRI In this talk, I will discuss the differential equation $iu_t = Hu, H := H_0 + V$ , where $V$ is a decaying potential and $H_0$ is a Laplacian related operator. In particular, I will focus on when $H_0$ is Laplacian, Bilaplacian and Dirac operators. I will discuss how the threshold energy obstructions, eigenvalues and resonances, effect the $L^1 \to L^\infty $ behavior of $e^{itH} P_{ac} (H)$. The threshold obstructions are known as the distributional solutions of $H\psi = 0$ in certain dimension dependent spaces. Due to its unwanted effects on the dispersive estimates, its absence have been assumed in many work. I will mention our previous results on Dirac operator and recent results on Bilaplacian operator under different assumptions on threshold energy obstructions.
I always organize my courses with the totality of the course set from the outset whenever I can. I see this as being closely tied to your question concerning homeworks assigned.The benefits I see:Keeps me and my students on track. The semester invariably gets busy, it's nice to have a go-to place where everything is set from the outset. I can always ... If $\gcd(a,b)=1$, there exists a multiplicative inverse for $a$ modulo $b$. (Otherwise, look at the $b-1$ multiples of $a$, namely $a,2a,3a,\dots,(b-1)a$. They must fall into congruence classes that aren't 0 or 1, but there are only $b-2$ of those.)$R(3,3)\leq 6$, and other Ramsey-style argumentsGive any domino tiling of a $6\times 6$ checkerboard, there ... I would not recommend putting false proofs onto the board unless you immediately (within the same class period) point out their falsity, and make an assignment to find it.For smaller mistakes that I make unintentionally, I give the students 'donut points'. 30 donut points (over the semester) and I bring in donuts for the class. I tell them that my main ... It seems that the key term here may be the somewhat non-specific-sounding special functions.By googling for a few examples (Erf, Si, Li) I came across a Table of Special Functions and, on the Lists of integrals wikipage, there is a sub-section on Special Functions.As a related remark, one reason that functions may be presented and/or defined in terms of ... The gamma function is very useful in counting problems (among others) and is seen as an extension of the factorial function into the reals. It is defined as:$$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\,.$$(Incidentally, this is the example of how to use MathJax in the help section.) Since your question is very broad, here is a somewhat broad answer: Read about problem posing.Three key pieces are:Silver, E. A. (1994). On Mathematical Problem Posing. For the learning of mathematics, 14(1), 19-28.and the bookBrown, S. I., & Walter, M. I. (2005). The art of problem posing. Psychology Press.The latter is a re-print of a ... Edit (Dec 2016): Encouraged by a few comments on SE, and a few direct emails about this post, I wrote up the ideas below for a journal of math education. The citation, and linked pre-print, are:Dickman, B. (2017). Enriching Divisibility: Multiple Proofs and Generalizations. Mathematics Teacher, 110(6), 416-423. Pre-Print (no pay-wall).(Adapted from a ... A place that Lagrange Multipliers comes up is in the proof of the real spectral theorem.Namely, let $A$ be a symmetric matrix.Define $f: \mathbb{R}^n \to \mathbb{R}$ by $f(v) = v^\top A v$.If you maximize $f$ on the unit sphere in $\mathbb{R}^n$, the Lagrange Multiplier condition will show that the maximum is achieved at an eigenvector of $A$. This is ... Yes! I have used these a lot in an "intro to proofs" course. Typically, each weekly homework assignment has at least one problem of this variety, and I've written many like this for assignments and a text. Some thoughts about your posed questions and other ideas:Does this help? I think yes in several ways, but it's hard to tell. I don't put questions like ... Application 1: Every rational number has a repeating decimal expansion.Application 2: Each infinite decimal expansion has the property that there exists a $10^{100}$-length sequence of digits that is repeated infinitely often in the expansion.Application 3: If $x$ is irrational, then at least two digits appear infinitely often in the decimal expansion of ... My first take is$$\ln(x) = \int_1^x\frac1t dt.$$Granted, some texts introduce the natural log of the inverse of $\exp$ but other texts define $\ln$ as above and the $\exp$ as the inverse. If I remember correctly, the definition of the logarithm by the integral was historically first. ProblemEach point in the plane is colored one of $n$ colors. Prove that thereexists a rectangle whose four vertices are the same color.Both the problem and the solution are very simple, yet those unfamilar with the Pigeonhole Principle would likely be at a complete loss to solve it.SolutionConsider a grid of points with $n+1$ rows and $n^{n(n+1)/... The heuristic described here is one manifestation of what Polya (1945) and others thereafter refer to as trying a special case. I do not know of a more specific term for the context that you have put forth, but this is often how one approaches a problem if you are initially unsure as to how to solve it in generality.It is a good way to get an initial foot-... Here are some suggestions for problem sources in English. Some of them are appropriate for very bright students studying geometry or Algebra II, but might nonetheless prove too difficult for students accelerated to this extent.-Mathematical Circles, Fomin et al.-Mathematical Problems: An Anthology, Dynkin et al.-Problems in Elementary Mathematics, ... The function $\displaystyle \text{Li}(x) = \int_2^x \frac{1}{\log t} \,\, dt$ comes up in the study of the distribution of primes. Specifically, the number of prime numbers less than $x$ is asymptotic to $\text{Li}(x)$ and a major consequence of the Riemann Hypothesis would be the sharpest possible bound for the difference between these two functions as $x \... In the example you mention of computing truth tables rather than use algebraic manipulations to answer questions about boolean expressions/sets, it's actually a wonderful situation where the students chooses the long and tedious way, not noticing a much more convenient method works much more nicely. In such cases, I let the student waste time getting the ... There are a number of resources I like out there. I tend to use a variety of different sites for different needs -- I don't think there is one that "does it all" but here are some of my favorites. Most of them are pretty random, though the first few below that have lots of customization options tend to make very usable worksheets, and for some applications ... My method for creating a concept-testing question is as follows. Start with the most open-ended question possible on the topic, then slowly refine it until it has a "correct answer." I've outlined an example for a conceptual derivative question below.Start with the generic question: "Tell me about the derivative," and recognize that the generic question is ... I think there are several good problems that can be explored (e.g., using the Moore method) by beginning with a word or term and trying to axiomatize it. I happen to think that assembling several of these words/terms and axiomatizing them would make for a nice textbook, but I digress...Back to the question: Some examples.What should "bigger than" mean?(... I did this for an introductory calculus course at a US state university. My reason was that I wanted to assign more homework than my colleagues typically do. On the first day of class I warned students that there was a lot of homework, explained why I think it's important, and told the students that they could see exactly what was expected of them.The same ... Nice question! Let me add one reference to your list:Silver, Edward A. "On mathematical problem posing." For the learning of mathematics (1994): 14(1) 19-28.(PDF download link.)Silver cites Hadamard's famous book, The Psychology of Invention in the Mathematical Field, as recognizing that isolating key research questionsis a sign of exceptional ... Wondering Why are induction proofs so challenging for students?,I thought of trying this as a possible assignment after introducing induction:Find the flaw in this induction proof.Claim: $3n = 0$ for all $n \ge 0$.Base: for $n=0$, $3n = 3(0) = 0$.AssumeInduction Hypothesis (IH): $3k = 0$ for all $0 \le k \le n$.Write $n+1 = a + b$ where $a > 0$ ... No one has still mentioned Fresnel functions:$S(x)=\int_0^x \sin(t^2)dt$ and $C(x)= \int_0^x \cos(t^2)dt$They are (of course) very relevant in signal analysis and in studying diffraction. What is less often mentioned is that the parametric function $x\mapsto (S(x),C(x))$ gives you a beautiful curve, the Cornu spiral, which is used by engineers in roads ... A possibility, requiring one definition: What is a tiling of the plane with an infinite supply of congruent copies of a single tile (technically,a monohedral tiling). This can go as deep as you'd like, perhaps in stringing together several mini-sessions.Can every triangle tile the plane? (Yes.)Form parallelograms, then argue that a parallelogram tiles ... For me there are perhaps three main types of problems which I assign:Routine skill building: either are modeled on a computation which I have shown similar problems solved, or, are a proof problem which is just a natural consequence of definition with little extra technique required. For a proof course, many problems are little more than an invitation to ... This only works in some cases:Ask not only for the final result but also some intermediate results which are only produced by the new method. This way the new method becomes more feasible, since the students are required to use it anyway to produce the intemediate results. In your example on finding extrema, you could, e.g., ask your students to also ... Here are a couple problems to shatter misconceptions about homomorphisms, while introducing the student to constructive thinking in group theory.Are the following statements true? Prove them or provide a counterexample.If $K_1$ and $K_2$ are isomorphic subgroups of $G$, then $G/K_1$ is isomorphic to $G/K_2$.If $\varphi:G\rightarrow H$ is a ... I disagree with the "diagonalize a matrix" type exam questions. Such questions call for accurate, longish computation, not understanding. That I leave for homework. In an exam I'd ask why you'd diagonalize a matrix (e.g. explain how to do something using this, step by step, not do it). Set questions up so that computation mistakes don't invalidate the work....
These are five important constructions and I would like to know how they are related. The $n$th unordered configuration space of a space $X$ is $$ \operatorname{UConf}_n(X):=\{\text{embeddings of $\{1,...,n\}$ into $X$}\}/(\text{$n$th symmetric group}), $$ topologized as a subquotient of $X^n$. The Ran space of $X$ is the set $\operatorname{Ran}(X)$ of finite subsets of $X$ with the topology generated by sets $$ \nabla_{U_1,...,U_n}:=\{S\in\operatorname{Ran}(U_1\cup\cdots\cup U_n)\mid S\cap U_i\ne\varnothing, i=1,...,n\} $$ where $U_i$ are disjoint open subsets of $X$. The free topological semilattice $\operatorname{Sl}(X)$ on $X$ is the value on $X$ of the left adjoint to the forgetful functor from topological semilattices to topological spaces. The Vietoris space $\mathscr VX$ of $X$ is the set of some (depending on the context) subsets of $X$ topologized by the same kind of $\nabla_{U_1,...,U_n}$ except that they are not required to be disjoint. Finally, one may choose some nice embedding $I$ of some subcategory of spaces that contains $X$ into a topos in various ways, and consider there the power object $\Omega^{IX}$. Usually it is not in the image of $I$. There are versions like $\operatorname{Sub}_{\mathrm{fin}}(IX)\rightarrowtail\Omega^{IX}$ of objects of finite (say, Kuratowski finite) subobjects of $IX$ which might be. (Note that $\operatorname{Sub}_{\mathrm{fin}}$, with Kuratowski finiteness, is the free internal semilattice functor on any topos whatsoever.) As a variation on the latter - say, $X$ is a simplicial set; since simplicial sets readily form a topos we have simplicial sets $\operatorname{Sub}_{\mathrm{fin}}(X)\rightarrowtail\Omega^X$. Questions: Is $\operatorname{UConf}_n(X)$ (homeomorphic to) a subspace of $\operatorname{Ran}(X)$? There is a topology on $\bigcup_n\operatorname{UConf}_n(X)$ with $\{x_1,...,x_n,x_{n+1}\}$ close to $\{x_1,...,x_n\}$ when $x_{n+1}$ is close to $x_n$ in $X$. Is this homeomorphic to $\operatorname{Ran}(X)$? The same two questions with $\operatorname{Sl}$ in place of $\operatorname{Ran}$. Is $\operatorname{Ran}(X)$ homeomorphic to $\operatorname{Sl}(X)$? Are $\operatorname{Ran}(X)$, $\operatorname{UConf}_n(X)$ or $\operatorname{Sl}(X)$ subspaces in $\mathscr VX$ for some nice spaces $X$? Are there known embeddings of some categories of spaces into toposes such that the image of the embedding is closed under taking $\operatorname{Sub}_{\mathrm{fin}}$? In particular, can $\operatorname{Sub}_{\mathrm{fin}}(IX)$ be isomorphic to $I(\operatorname{Sl}(X))$ for some such $I$? How does the geometric realization of $\operatorname{Sub}_{\mathrm{fin}}(X)$ relate to $\operatorname{Ran}$, $\operatorname{UConf}_n$, $\operatorname{Sl}$ and $\mathscr V$ of the geometric realization of $X$ for a simplicial set $X$? $\ \ \,$0.$\ $Are these considered together and compared somewhere in the literature?
I'm trying to solve a problem related to waves on a string. Say I have an infinite string, with tension $T$ and mass density $\mu$. To the string, at $x=0$ (seeing as it's infinite, the specific point doesn't actually matter, so long as it's constant), I attach a spring $k$ (the string is horizontal whilst the spring is vertical). I am attempting to calculate the reflected and transmitted waves, resulting from an incident wave: $$y_{inc}(x,t)=A_{inc}\,e^{i(x-\omega\,t)}$$ My attempt at a solution was to write $$\mu\,\frac{\partial^2y}{\partial t^2}\|_{x=0}=T\,\frac{\partial^2y}{\partial x^2}\|_{x=0}-k\,y(0,t)$$ which (I think) is, generally speaking, true. I am not sure how (if at all) I can use this to find an expression for the reflected wave. So, my question is how does the spring affect the reflection?
Matrix Mechanics In this lesson, we'll cover some of the fundamental principles and postulates of quantum mechanics. These principles are the foundation of quantum mechanics. The eigenvalues are the values that you measure in an experiment: for example, the position or momentum of a particle. Because the eigenvalues are what you measure, it wouldn't make physical sense if the eigenvalue of an observable had an imaginary part. In this lesson, we'll prove that the eigenvalue of any observable is a real number. The three operators—$\hat{σ}_x$, \hat{σ}_y\), and \hat{σ}_z\)—are associated with the measurements of the $x$, $y$, and $z$ components of spin of a quantum particle, respectively. In this lesson, we'll represent each of these three operators as matrices and solve for the entries in each matrix. These three matrices are called the Pauli matrices. In this lesson, we'll derive an equation which will allow us to calculate the wavefunction (which is to say, the collection of probability amplitudes) associated with any ket vector $\|\psi⟩$. Knowing the wavefunction is very important since we use probability amplitudes to calculate the probability of measuring eigenvalues (i.e. the position or momentum of a quantum system). In this lesson, we'll mathematically prove that for any Hermitian operator (and, hence, any observable), one can always find a complete basis of orthonormal eigenvectors. Schrodinger's Equation The wavefunction $\psi(L,t)$ is confined to a circle whenever the eigenvalues L of a particle are only nonzero on the points along a circle. When the wavefunction $\psi(L,t)$ associated with a particle has non-zero values only on points along a circle of radius $r$, the eigenvalues $p$ (of the momentum operator $\hat{P}$) are quantized—they come in discrete multiples of $n\frac{ℏ}{r}$ where $n=1,2,…$ Since the eigenvalues for angular momentum are $L=pr=nℏ$, it follows that angular momentum is also quantized. Newton's second law describes how the classical state {$\vec{p_i}, \vec{R_i}$} of a classical system changes with time based on the initial position and configuration $\vec{R_i}$, and also the initial momentum $\vec{p_i}$. We'll see that Schrodinger's equation is the quantum analogue of Newton's second law and describes the time-evolution of a quantum state $\|\psi(t)⟩$ based on the following two initial conditions: the energy and initial state of the system. In this section, we'll begin by seeing how Schrodinger's time-independent equation can be used to determine the wave function of a free particle. After that, we'll use Schrodinger's time-independent equation to solve for the allowed, quantized wave functions and allowed, energy eigenvalues of a "particle in a box"; this will be useful later on as a qualitative understanding of the quantized wave functions and energy eigenvalues of atoms. In general, if a quantum system starts out in any arbitrary state, it will evolve with time according to Schrödinger's equation such that the probability $P(L)$ changes with time. In this lesson, we'll prove that if a quantum system starts out in an energy eigenstate, then the probability $P(L)$ of measuring any physical quantity will not change with time.
How can you select best cruise altitude (altitude for best range velocity) and maximum speed altitude? The specific case is a turboprop aircraft of gross weight 8000 lb. Aviation Stack Exchange is a question and answer site for aircraft pilots, mechanics, and enthusiasts. It only takes a minute to sign up.Sign up to join this community Generally, climbing as high as you can would be close to the optimum strategy. In the end, the altitude is given by the optimum lift coefficient in cruise, which can be approximated as $$c_L = \sqrt{\frac{1-n_v}{n_v+3}\cdot \pi \cdot AR \cdot \epsilon \cdot c_{D0}}$$ Now it will be important to get the thrust over speed dependency right, which is expressed with $n_v$. A turboprop should have $n_v \approx -0.7$. The rest is determined by the aerodynamics of your plane. Nomenclature: $c_L \:\:\:$ lift coefficient $n_v \:\:\:$ thrust exponent, as in $T \sim v^{n_v} $ $\pi \:\:\:\:\:$ 3.14159$\dots$ $AR \:\:$ aspect ratio of the wing $\epsilon \:\:\:\:\:$ the wing's Oswald factor $c_{D0} \:$ zero-lift drag coefficient If you have the optimum $c_L$, you need to find the altitude where level flight at cruise power will require this lift coefficient. Depending on the range of your small aircraft, you might spend most of the flight in climb and descent. Maximum speed altitude: Draw the envelope at max power, and look for the rightmost point which can be trimmed. The equation above uses the quadratic polar approximation which works well at high lift coefficients, but for maximum speed the details of friction and parasitic drag are important, and they cannot be captured with a simple formula.
Is it possible an algorithm complexity decreases by input size? Simply $O(1/n)$ possible? Consider an algorithm with some running time bounded by $f(n)$ and suppose that $f(n) \in O(1/n)$. That means that there is some constant $c$ such that for sufficiently large values of $n$, it holds that $$f(n) \leq c\frac{1}{n}.$$ Clearly, for any fixed $c$ and sufficiently large $n$, the right side will be strictly less than $1$, which requires $f(n)=0$, since $f$ maps to $\mathbb{N}$. In my understanding, even an algorithm that immediately terminates, takes at least $1$ step (namely to terminate), i.e., $\forall n\colon f(n)\ge 1$. So no such algorithm can exist. Bucket sort's insertion sort step is O(1/n) on average. Reference - CLRS Section 8.4 BUCKET-SORT(A)1 n ← length[A]2 for i ← 1 to n3 do insert A[i ] into list B[nA[i ]]4 for i ← 0 to n − 15 do sort list B[i ] with insertion sort6 concatenate the lists B[0], B[1], . . . , B[n − 1] together in order $T(n) = \theta(n) + \sum_{i=0}^{n−1} O(n_{i}^{2})$ Taking expectations of both sides and using linearity of expectation, we have $E[T(n)] = \theta(n) + \sum_{i=0}^{n−1} O(E[n_{i}^{2}])$ It is proved that $E[n_{i}^{2}] = 2 − 1/n$ Hence step 5 takes $2 - 1/n$ time. Overall complexity of bucket sort is $\theta(n) + n\cdot O(2 − 1/n) = \theta(n)$. By definition, all functions (even the trivial) must perform at least 1 operation. It is impossible to perform half of a step; computers are discrete machines, and work incrementally. As such, the complexity $\Omega(1)$ is the lower time bound of all computational operations. In addition, for any operation accepting any N inputs, it must deal with at least a constant number of them at once (even if that number is zero). It's impossible to start at using one unit of space and halve that space infinitely; at some point your algorithm will reach the atomic unit of space for the computation (one bit, one electron, one string, whatever) and from that point you will use either zero or one of those atomic units, thus arriving at a constant base case. Thus, $\Omega(1)$ is also the lower space bound of all algorithms. Now, it is possible, trivial even, for an algorithm to produce an output on that order of magnitude or cardinality. Given N elements, you can return 1/N of them ( array[x] produces an output 1/N the size of the input array). But, this takes constant time (calculate the location of the element by offsetting the address of the array by the number and size of its elements, and return the bits beginning at that offset and continuing for sizeof(elementType)).
Faulhaber polynomial of order $p \in \Bbb{N}$ is defined as the unique polynomial of degree $p+1$ satisfying $$ S_{p}(n) = \sum_{k=1}^{n} k^p $$ for $n = 1, 2, 3, \cdots$. For example, \begin{align} S_0(x) &= x, \\ S_1(x) &= \frac{x(x+1)}{2}, \\ S_2(x) &= \frac{x(x+1)(2x+1)}{6}, \\ S_3(x) &= \frac{x^2 (x+1)^2}{4}. \end{align} In order to grasp some intuition on the partial decomposition of $1/S_p (x)$, I tried plotting the complex zeros of $S_p (x)$. The following graphics shows the distribution of the zeros of $S_{800}(x)$. (The precision of the calculated zeros $z_j$ of $S_{800}(z)$ above satisfy $\|f(z_j)\| \leq 10^{-300}$.) It turns out that they exhibits a very neat, yet still a strange pattern as seen above. So far I have never heard of the topic related to this pattern, and I want to know (out of curiosity) if there are some results concerning the pattern of zeros of $S_p(x)$. Addendum. The previous fuzzy graphics turned out to be a result of numerical error. Now I removed those errors.
(Disclaimer: I peronally really don't care if one uses $\tau$ or $\pi$, both are just numbers for me.)But I would strongly recommend to use $\pi$. Why?In every technical literature, in many popular literature, the people always use $\pi$ (even worse: $\tau$ is used for different things than $\tau=2\pi$ which would confuse when reading those literature). ... I suppose my preference would be4. The differential equations road.That is, define $x(t) = \cos t$ and $y(t) = \sin t$ to be the solutions to the following initial value problem:$$\frac{dx}{dt} = -y,\qquad \frac{dy}{dt} = x,\qquad x(0) = 1,\qquad y(0)=0.$$This assumes that you've proven the existence and uniqueness theorem ... I feel that it is perhaps a little irresponsible to teach $\tau$ instead of $\pi$. As a first introduction, it is the norm which should be taught: teaching a rare alternative to $\pi$ only serves to confuse students, especially when almost all available resources use $\pi$ instead of $\tau$. Imagine a student's confusion when they see $\tau$ in class and $\... The specific identity\begin{equation}\tag{A}\tfrac{1}{1 - \sin{x}} + \tfrac{1}{1 + \sin{x}} = 2\sec^{2}{x}\end{equation}as such is probably not often encountered, but simplifications akin to \begin{equation}\tag{B}\tfrac{1}{1-t} + \tfrac{1}{1 + t} = \tfrac{2}{1 - t^{2}}\end{equation}occur frequently. For example, integration via partial fractions ... An identity always holds for some values of the free variables. Sometimes the allowed values are all real numbers, sometimes something else. In this case the identity holds whenever $\sec x$ and $\tan x$ are defined (they are defined in the same set). In a certain limiting sense the identity is also true at $x=\pi/2$, but you probably don't want to go into ... Regarding the second part of your question:Would I be doing my students a terrible disservice if I introduced them to trig functions treating them as going clockwise from 12 o'clock?I think it's important to stress that the convention is just a convention, and there is no intrinsic reason why one convention is better than another. But at the same time,... One should teach $\pi$. One might discuss that there is a choice that is made that is somewhat arbitrary, and there are also reasons for a different choice but I do not see this as that relevant to make much ado about this.It should perhaps also be noticed that there are two conflicting proposals for $\tau$. Eagle (1958) proposed $\pi/2$ and Palais (2001) ... Absolutely!In fact, in my opinion, the most important "math skill" that should be taught in conjunction with, and using, word problems is checking whether the answers make sense. This is an absolutely invaluable part of making any practical use of mathematics, as opposed to just blindly applying formulas for the sake of passing an exam.There are several ... More of a comment than an answer but: They are all composites of more basic functions. In fact, all of the trig functions could can expressed in terms of sine, linear changes in coordinates, and rational functions. For instance:$$\tan(\theta) = \frac{\sin(\theta)}{\sin( \frac{\pi}{2}-\theta)}$$We certainly don't want children to have to memorize 20 ... $\tau$ should be taught in schoolsThere's plenty of material arguing why $\tau$ is a much more intuitive and easier to teach concept (some of my favorites: 1,2,3) and I don't want to rehash their arguments, but if you think that $\tau$ is just to make equations look nicer, please check out those resources.The question at hand is whether math educators ... The answers to a word problem should in my opinion make sense (within reasonable limits).The goal you mentioned that students should check their answers is one shared by many, as also witnessed by our recent question How to award points for sense-making at the end of a problem?Now, if it is not a given any more that the solution does indeed make sense, ... My strongly held opinion is that some exact solutions are conceptually fundamental. Knowing that $\sin 45^\circ = \frac{1}{\sqrt{2}}$ (rather than approximately 0.7071) may not be important for applications in engineering — although it is certainly necessary in higher mathematics, and I wouldn't be surprised to see it in, say, theoretical physics &... I cannot imagine a context where it would benefit a student to know about $\tau$ over and above knowing about $\pi$. Every text book a student will ever encounter will exclusively talk about $\pi$. Every calculator a student will ever use will have the constant $\pi$ pre-programmed in.Although there are perfectly lovely arguments to show that $\tau$ is a ... Many high school geometry textbooks define an angle as simplythe union of two rays with a common endpointThe advantage of this definition is its simplicity. Among its disadvantages:It does not serve well for capturing the idea of a "direction": That is, there is no way to distinguish between a clockwise and a counterclockwise rotation.It more or ... As others said, degrees are taught, since they are still used. So, the question becomes why are they still used.To purely work with fractions would not be very convenient for various somewhat everyday things, since many people are more used to/better at operating with integers. So to really use $\tau$ and fractions thereof seems incovenient, and one ... I feel as though a lot of this misses the point. Teaching the unit circle and trig functions in terms of "one turn" makes a lot more intuitive sense to young students than do arbitrary variables tau or pi. It just so happens that Tau is equivalent to one turn. Once students grasp the concept of the unit circle, then the pi conventions can be explored. I agree this memorization is not necessary.If students understand how the trigonometric functions are defined (unit circle) and know several basic identities, everything else can be derived. I think the pythagorean identity and the sine and cosine of sum of angles are sufficient.$\sin^2 \alpha + \cos^2 \alpha = 1$$\sin(\alpha + \beta) = \sin(\... It is possible to treat degrees and radians as units, just as we treat inches and centimeters as units. Just as we can convert inches to centimeters (1 in = 2.54 cm), we can convert degrees to radians (1 degree = $\pi/180$ radians).But this point of view overlooks an important fact about radians: The radian measure of an angle is the ratio of two lengths (... I disagree with your statement on the usefulness of reference angles, they have more use in the plane of the unit circle. So to answer the question specifically, yes we should still teach them and one idea on relevance is below in a Physics I problem.Trigonometric functions in a complete unit circle are not one to one so the concept of a reference angle is ... As several respondents have indicated, one could choose many different trigonometric functions to serve as the basic elements in terms of which other trigonometric functions are expressed and in (ancient) history other choices were made than those standard now. What we know now that was not known to the ancients that leads us to use cosine and sine is the ... This aspect is discussed in great detail in §7.2 of A Course in Calculus and Mathematical Analysis by Ghorpade and Limaye. The discussion must be accessible to students undergoing their first rigorous course in analysis.The discussion is well-motivated: the authors point out in §7.1 that one can integrate all functions of the form $x^r$ for all rational $... My guess is that it comes from drawing complex numbers. Putting real numbers on the $x$-axis from left to right and imaginary numbers on the $y$-axis from bottom to top matches with the way we tend to think (in cultures based on Latin at least). Once you establish that, then the function $x\mapsto e^{2\pi ix}$ takes you anti-clockwise from the positive $x$-... Due to low enrollment, my AP Calc class was filled with the students who otherwise would have taken Pre-Calc this year. So you can imagine that "How much do you really need to know to see the bigger picture in calculus?" has been on my mind lately.Here's where my thoughts have fleshed out in regards to trig so far.Periodic behavior is widespread ... 360 has 24 divisors, more than any smaller number (this is called a highly composite number).None of its prime factors is larger than 5 ("5-smooth").It can be divided by every natural number between 1 and 10, with the exception of 7. No smaller number can.Plus a couple more characteristics that make it very much suited for subdivision. Thus, many ... The notation exists since a long time. It was used already by Irving Stringham in 'Uniplanar Algebra (1893).' This is claimed to be the earliest use on http://jeff560.tripod.com/trigonometry.html giving Cajori vol. 2, page 133 as reference.In this book, Uniplanar Algebra, the notation is used first, as far as I can see, in chapter III (The algebra of ... By their definition (using a right triangle), $\sin 90^\circ$ is undefined, since a triangle cannot have two right angles. You need the definition based on a circle ($\sin \theta = y/r$) to have a value for $\sin 90^\circ$. There is no such thing as a natural sign/direction convention until long after the fact. Consider another answerer's comment that anyone using the left hand rule to construct the cross product of two vectors would be mistreated. Of course, the left hand rule is exactly the correct rule to use for the path of electrons in a magnetic field? Why? Because ...
Fundamental Concepts I am trying something I have seen on many blogs around the internet: Iwill work through all the exercises in Munkres’ textbook, Topology. Ihave never tried anything like this before and it will definitely testmy ability to stay motivated as I tend to get bored easily. So, withoutfurther ado, I present the exercises of Topology, Chapter 1, “SetTheory and Logic”, Section 1, “Fundamental Concepts”. Check the distributive laws for $\cup$ and $\cap$ and DeMorgan’s laws. This amounts to showing the following are true: $A \cap ( B \cup C ) = (A \cap B) \cup (A \cap C)$ $A \cup ( B \cap C ) = (A \cup B) \cap (A \cup C)$ $A - (B \cup C) = (A - B) \cap (A - C)$ $A - (B \cap C) = (A - B) \cup (A - C)$ $A \cap (B \cup C) = \{ x \mid x \in A \mbox{ and } x \in (B \cup C) \}$
14 0 Homework Statement A yo-yo is placed on a conveyor belt accelerating ##a_C = 1 m/s^2## to the left. The end of the rope of the yo-yo is fixed to a wall on the right. The moment of inertia is ##I = 200 kg \cdot m^2##. Its mass is ##m = 100kg##. The radius of the outer circle is ##R = 2m## and the radius of the inner circle is ##r = 1m##. The coefficient of static friction is ##0.4## and the coefficient of kinetic friction is ##0.3##. Find the initial tension in the rope and the angular acceleration of the yo-yo. Homework Equations ##T - f = ma## ##\tau_P = -fr## ##\tau_G = Tr## ##I_P = I + mr^2## ##I_G = I + mR^2## ##a = \alpha R## First off, I was wondering if the acceleration of the conveyor belt can be considered a force. And I'm not exactly sure how to use Newton's second law if the object of the forces is itself on an accelerating surface. Also, I don't know whether it rolls with or without slipping. I thought I could use ##a_C = \alpha R## for the angular acceleration, but the acceleration of the conveyor belt is not the only source of acceleration, since the friction and the tension also play a role. I can't find a way to combine these equations to get the Also, I don't know whether it rolls with or without slipping. I thought I could use ##a_C = \alpha R## for the angular acceleration, but the acceleration of the conveyor belt is not the only source of acceleration, since the friction and the tension also play a role. I can't find a way to combine these equations to get the
Once we have identified two variables that are correlated, we would like to model this relationship. We want to use one variable as a predictor or explanatory variable to explain the other variable, the response or dependent variable. In order to do this, we need a good relationship between our two variables. The model can then be used to predict changes in our response variable. A strong relationship between the predictor variable and the response variable leads to a good model. Figure 9. Scatterplot with regression model. Definition: simple linear regression A simple linear regression model is a mathematical equation that allows us to predict a response for a given predictor value. Our model will take the form of $\hat y = b_0+b_1x$ where b0 is the y-intercept, b1 is the slope, x is the predictor variable, and ŷ an estimate of the mean value of the response variable for any value of the predictor variable. The y-intercept is the predicted value for the response ( y) when x = 0. The slope describes the change in y for each one unit change in x. Let’s look at this example to clarify the interpretation of the slope and intercept. Example $\PageIndex{1}$: A hydrologist creates a model to predict the volume flow for a stream at a bridge crossing with a predictor variable of daily rainfall in inches. Answer \[\hat y = 1.6 +29 x \nonumber\] The y-intercept of 1.6 can be interpreted this way: On a day with no rainfall, there will be 1.6 gal. of water/min. flowing in the stream at that bridge crossing. The slope tells us that if it rained one inch that day the flow in the stream would increase by an additional 29 gal./min. If it rained 2 inches that day, the flow would increase by an additional 58 gal./min. Example $\PageIndex{2}$: What would be the average stream flow if it rained 0.45 inches that day? Answer \[\hat y= 1.6 + 29x = 1.6 + 29(0.45) = 14.65 gal./min \nonumber\] The Least-Squares Regression Line (shortcut equations) The equation is given by $$\hat y = b_0+b_1x$$ where $b_1 = r\left ( \dfrac {s_y}{s_x} \right )$ is the slope and $b_0=\hat y -b_1\bar x$ is the y-intercept of the regression line. An alternate computational equation for slope is: $$b_1 = \dfrac {\sum xy - \dfrac {(\sum x)(\sum y)}{n}} {\sum x^2 - \dfrac {(\sum x)^2}{n}} = \dfrac {S_{xy}}{S_{xx}}$$ This simple model is the line of best fit for our sample data. The regression line does not go through every point; instead it balances the difference between all data points and the straight-line model. The difference between the observed data value and the predicted value (the value on the straight line) is the error or residual. The criterion to determine the line that best describes the relation between two variables is based on the residuals. $$Residual = Observed – Predicted$$ For example, if you wanted to predict the chest girth of a black bear given its weight, you could use the following model. Chest girth = 13.2 +0.43 weight The predicted chest girth of a bear that weighed 120 lb. is 64.8 in. Chest girth = 13.2 + 0.43(120) = 64.8 in. But a measured bear chest girth (observed value) for a bear that weighed 120 lb. was actually 62.1 in. The residual would be 62.1 – 64.8 = -2.7 in. A negative residual indicates that the model is over-predicting. A positive residual indicates that the model is under-predicting. In this instance, the model over-predicted the chest girth of a bear that actually weighed 120 lb. Figure 10. Scatterplot with regression model illustrating a residual value. This random error (residual) takes into account all unpredictable and unknown factors that are not included in the model. An ordinary least squares regression line minimizes the sum of the squared errors between the observed and predicted values to create a best fitting line. The differences between the observed and predicted values are squared to deal with the positive and negative differences. Coefficient of Determination After we fit our regression line (compute b0 and b1), we usually wish to know how well the model fits our data. To determine this, we need to think back to the idea of analysis of variance. In ANOVA, we partitioned the variation using sums of squares so we could identify a treatment effect opposed to random variation that occurred in our data. The idea is the same for regression. We want to partition the total variability into two parts: the variation due to the regression and the variation due to random error. And we are again going to compute sums of squares to help us do this. Suppose the total variability in the sample measurements about the sample mean is denoted by $\sum (y_i - \bar y)^2$, called the sums of squares of total variability about the mean (SST). The squared difference between the predicted value $\hat y$ and the sample mean is denoted by $\sum (\hat {y_i} - \bar y)^2$, called the sums of squares due to regression (SSR). The SSR represents the variability explained by the regression line. Finally, the variability which cannot be explained by the regression line is called the sums of squares due to error (SSE) and is denoted by $\sum (y_i - \hat y)^2$. SSE is actually the squared residual. SST = SSR + SSE $\sum (y_i - \bar y)^2$ = $\sum (\hat {y_i} - \bar y)^2$ +$\sum (\hat {y_i} - \bar y)^2$ Figure 11. An illustration of the relationship between the mean of the y’s and the predicted and observed value of a specific y. The sums of squares and mean sums of squares (just like ANOVA) are typically presented in the regression analysis of variance table. The ratio of the mean sums of squares for the regression (MSR) and mean sums of squares for error (MSE) form an F-test statistic used to test the regression model. The relationship between these sums of square is defined as $$Total \ Variation = Explained \ Variation + Unexplained \ Variation$$ The larger the explained variation, the better the model is at prediction. The larger the unexplained variation, the worse the model is at prediction. A quantitative measure of the explanatory power of a model is $R^2$, the Coefficient of Determination: $$R^2 = \dfrac {Explained \ Variation}{Total \ Variation}$$ The Coefficient of Determination measures the percent variation in the response variable ( y) that is explained by the model. Values range from 0 to 1. An $R^2$ close to zero indicates a model with very little explanatory power. An $R^2$ close to one indicates a model with more explanatory power. The Coefficient of Determination and the linear correlation coefficient are related mathematically. $$R^2 = r^2$$ However, they have two very different meanings: r is a measure of the strength and direction of a linear relationship between two variables; R2 describes the percent variation in “ y” that is explained by the model. Residual and Normal Probability Plots Even though you have determined, using a scatterplot, correlation coefficient and R2, that x is useful in predicting the value of y, the results of a regression analysis are valid only when the data satisfy the necessary regression assumptions. The response variable (y) is a random variable while the predictor variable (x) is assumed non-random or fixed and measured without error. The relationship between yand xmust be linear, given by the model $\hat y = b_0 + b_1x$. The error of random term the values ε are independent, have a mean of 0 and a common variance $\sigma^2$, independent of x, and are normally distributed. We can use residual plots to check for a constant variance, as well as to make sure that the linear model is in fact adequate. A residual plot is a scatterplot of the residual (= observed – predicted values) versus the predicted or fitted (as used in the residual plot) value. The center horizontal axis is set at zero. One property of the residuals is that they sum to zero and have a mean of zero. A residual plot should be free of any patterns and the residuals should appear as a random scatter of points about zero. A residual plot with no appearance of any patterns indicates that the model assumptions are satisfied for these data. Figure 12. A residual plot. A residual plot that has a “fan shape” indicates a heterogeneous variance (non-constant variance). The residuals tend to fan out or fan in as error variance increases or decreases. Figure 13. A residual plot that indicates a non-constant variance. A residual plot that tends to “swoop” indicates that a linear model may not be appropriate. The model may need higher-order terms of x, or a non-linear model may be needed to better describe the relationship between y and x. Transformations on x or y may also be considered. Figure 14. A residual plot that indicates the need for a higher order model. A normal probability plot allows us to check that the errors are normally distributed. It plots the residuals against the expected value of the residual as if it had come from a normal distribution. Recall that when the residuals are normally distributed, they will follow a straight-line pattern, sloping upward. This plot is not unusual and does not indicate any non-normality with the residuals. Figure 15. A normal probability plot. This next plot clearly illustrates a non-normal distribution of the residuals. Figure 16. A normal probability plot, which illustrates non-normal distribution. The most serious violations of normality usually appear in the tails of the distribution because this is where the normal distribution differs most from other types of distributions with a similar mean and spread. Curvature in either or both ends of a normal probability plot is indicative of nonnormality.
${{\boldsymbol X}{(4050)}^{\pm}}$ $I^G(J^{PC})$ = $1^-(?^{? +})$ I, G, C need confirmation. Properties incompatible with a ${{\mathit q}}{{\overline{\mathit q}}}$ structure (exotic state). See the review on non- ${{\mathit q}}{{\overline{\mathit q}}}$ states. Observed by MIZUK 2008 in the ${{\mathit \pi}^{+}}{{\mathit \chi}_{{c1}}{(1P)}}$ invariant mass distribution in ${{\overline{\mathit B}}^{0}}$ $\rightarrow$ ${{\mathit K}^{-}}{{\mathit \pi}^{+}}{{\mathit \chi}_{{c1}}{(1P)}}$ decays. Not seen by LEES 2012B in this same mode after accounting for ${{\mathit K}}{{\mathit \pi}}$ resonant mass and angular structure.
This question regards colourings on edges and vertices on countable directed multigraphs. We start with an example. Let $G=\mathbb Z^2$. We define two functions $a_h$ and $a_v$ from $\mathbb Z^2$ to $\mathbb C$ with the property that for all $\epsilon>0$, $$ \{(n,m)\in\mathbb Z^2 : \|a_h(n,m)a_v(n+1,m)-a_v(n,m)a_h(n,m+1)\|>\epsilon\} $$ is finite. We want to assume as well that both $a_h$ and $a_g$ take only values of absolute value 1 (or that they're bounded, both questions are interesting I think). The question is: can we find a function $d : \mathbb Z^2\to\mathbb C$ such that $d$ agrees asymptotically with $a_h$ and $a_v$, meaning that for all $\epsilon>0$ $$ \{(n,m) :\|d(n+1,m)-d(n,m)a_h(n,m)\|>\epsilon\} $$ and $$\{(n,m) : \|d(n,m+1)-d(n,m)a_v(n,m)\|>\epsilon\} $$ are finite? How is this related to graphs? Draw a picture of $\mathbb Z^2$, and colour in blue the directed vertical edge $(n,m)\to (n,m+1)$ and in red the horizontal ones going right. In this setting, we can think as the functions $a_h$ and $a_v$ as $\mathbb C$-valued weights on the edges, with the property that, asymptotically, one can go first right and then up, or first up and then right, and get the same result when multiplying the weights. The question then transfers to: Can we weigh the vertices so that, asymptotically, the pattern of the weights of the vertices is determined by the one of the edges? Let's formalize this connection. Let $V$ be a set, and suppose that we have partial functions (either finitely or countably many) $s_i: V \to V$, with domain $D(s_i)$. If $\bar t=t_n...t_1$ is a finite word in $\{s_i\}$, we can think as $\bar t$ as a partial function $\bar t: V \to V$. Its domain is $$ D(\bar t)=\{x \in D(t_1): t_i(t_{i-1}(...(t_1(x)..) \in D(t_{i+1}), \; \forall i <n\}. $$ If $\bar t$ and $\bar t'$ are finite words in $\{s_i\}$, let $$ D(\bar t,\bar t')=\{(x,y): x \in D(\bar t) \cap D(\bar t') \text{ and }\bar t(x)=y=\bar t'(x)\} $$ Suppose that for every $i$, a weight function $$ a_i: D(s_i) \to\mathbb C $$ is given. We also assume that $\|a_i(x)\|=1$ for all $a\in D(s_i)$. Define $a_{\bar t}: D(\bar t)\to \mathbb C$ by taking the product (i.e., if $\bar t=t_n...t_1$ then $a_{\bar t}(x)=\prod_{1\leq i\leq n} a_i(t_{i-1}(\cdots(t_1(x))\cdots))$), and assume asymptotic coherence, that is, for all $\bar t, \bar t'$ words in $\{s_i\}$, and $\epsilon>0$, the set $$ \{(x,y) \in D(\bar t,\bar t') : \|a_{\bar t}(x)- a_{\bar t'}(x)\|>\epsilon\} $$ is finite. Is there an assignment $d: V \to\mathbb C$ such that for all $i$ and $\epsilon>0$, the set $$ \{x : \|d(s_i(x))-a_{s_i}(x)d(x)\|>\epsilon\}$$ is finite? (This can all be turned in a graph setting, by drawing a directed edge on colour $i$ between $x$ and $y$ whenever $s_i(x)=y$.). This has a positive answer in case the graph is a tree. In particular, given a countable group $G$ generated by a subset $S$, we can ask this particular question about the Cayley graph of $(G,S)$. We obtain a positive answer to this question if $G=\mathbb F_n$, the free group on $n$ generators and $s_i: V\to V$ is the multiplication by the $i$-th generator. In general, there is a positive answer to this question (i.e., we can find $d$), whenever the corresponding graph is a tree. In this case, to define $d$ is enough to give value $1$ to a specified vertex, and then to "follow the tree" using $a_{s_i}$. A more complicated setting in which this can fit is the one of discrete (semi)groups actions. If $X$ is any set, and $G$ is a (semi)group action, we can construct maps $s_i$ for every element of $G$, by using the action of $G$ on $X$. In this setting, our situation is really a lot reminiscent of "asymptotic untwisting of cocycle", and we have the strong suspect that a positive, or negative, answer might be dependent the structure of the action, or of the semigroup itself. Thanks in advance,
We have shown that a holomorphic map $f: G\to \mathbb{C}$ to be expressed as a power series, which bears a certain similarity to polynomials, and a feature of polynomials are that if $a$ is a root, or zero, for a polynomial $p$, we can factor $p$ such that $p(z)=(z-a)^n q(z)$ where $q$ is another polynomial with the property that $q(a)\neq 0$. Now, does this similarity with polynomials extend to factorization? In fact it does as we shall see. Let $f: G\to \mathbb{C}$ be a holomorphic map that is not identically zero, with $G\subseteq \mathbb{C}$ a domain and $f(a)=0$. It is our claim that there exists a smallest natural number $n$ such that $f^{(n)}(a)\neq 0$. So suppose that there are no such $n$, such that $f^{(k)}(a)=0$ for all $k\in\mathbb{N}$. Let $B_\rho(a)$ be the largest open ball with center $a$ contained in $G$, since we have that \[f(z)=\sum^\infty_{k=0}\frac{f^{(k)}(a)}{k!}(z-a)^k\] we then have that $f$ is identically zero on $B_\rho(a)$. Fix a point $z_0\in G$ and let $\gamma : [0,1]\to G$ be a continuous curve from $a$ to $z_0$. By the paving lemma there is a finite partition $0=t_1 < t_2 <\cdots <t_m=1$ and an $r>0$ such that $B_r(\gamma(t_k))\subseteq G$ for all $k$ and $\gamma([t_{k-1},t_k])\subseteq B_r(\gamma(t_k))$. Note that $B_r(\gamma(t_1))=B_r(a)\subseteq B_\rho(a)$ so $f$ is identically zero on $B_r(\gamma(t_1))$, but since $\gamma([t_1,t_2])\subseteq B_r(\gamma(t_1))$ we must have that $f$ is identically zero on $B_r(\gamma(t_2))$, and so on finitely many times untill we reach $\gamma(t_m)$ and conclude that $f$ is identically zero on $B_r(\gamma(t_m))=B_r(z_0)$ and since $z_0$ was chosen to be arbitrary we must conclude that $f$ is identically zero on all of $G$. A contradiction. Now, let $n$ be the smallest natural number such that $f^{(n)}(a)\neq 0$, then we must have that $f^{(k)}(a)=0$ for $k < n$. We then get, for $z\in B_\rho(a)$: \[\begin{split} f(z) &=\sum^\infty_{k=0}\frac{f^{(k)}(a)}{k!}(a-z)^k \\ &= \sum^\infty_{k=n}\frac{f^{(k)}(a)}{k!}(a-z)^k \\ &= \sum^\infty_{k=0}\frac{f^{(n+k)}(a)}{(n+k)!}(a-z)^{n+k} \\&=(z-a)^n \sum^\infty_{k=0}\frac{f^{(n+k)}(a)}{(n+k)!}(a-z)^{k}, \end{split}\] now, let $\tilde{f}(z)=\sum^\infty_{k=0}\frac{f^{(n+k)}(a)}{(n+k)!}(a-z)^{k}$ and note that $\tilde{f}$ is non-zero and holomorphic on $B_\rho(a)$. We then define a map $g$ given by \[g(z)=\begin{cases} \tilde{f}(z), & z\in B_\rho(a) \\ \frac{f(z)}{(z-a)^n}, & z\in G\setminus \{a\}\end{cases}\] and note that \[f(z)=(z-a)^n g(z),\] showing the existance of a factorization with our desired properties. Showing that this representation is unique is left as an exercise 😉 References 648 {648:SS6NUFWV} items 1 apa default asc Berg, C. (2016). Complex analysis. Copenhagen: Department of Mathematical Sciences, University of Copenhagen. Suppose you have an open map $p$ between topological spaces, and if you have a subet $A$ of $p$’s domain such that $p(A)$ is open. Can you then conclude that $A$ is open? Nope! Consider the following spaces $X=\{x_1,x_2\}$ and $Y=\{y_1,y_2\}$ with topologies $\tau_X=\{\varnothing, X, \{x_1\}\}$ and $\tau_Y=\{\varnothing,Y,\{y_1\}\}$, respectively and let $p: X\times Y\to X$ be the projection onto its first fator. This is an open map. If we consider $A=X\times\{y_2\}$ we see that $A$ is not open in $X\times Y$, but we have that $p(A)=p(X\times\{y_2\})= X$ which is trivially open in $X$. I came across this little problem recently: If $X$ is a topological space with exactly two components, and given an equivalence relation $\sim$ what can we say about its quotient space $X/{\sim}$? It turns out that $X/{\sim}$ is connected if and only if there exists $x,y\in X$ where $x$ and $y$ are in separate components, such that $x\sim y$. Suppose first that there exists $x,y\in X$ such that $x\sim y$. Let $C_1$ and $C_2$ be the two components of $X$ and let $p: X \to X/{\sim}$ be the natural projection. Since $p$ is a quotient map it is surely continuous and since the image of a connected space under a continuous function is connected we have have that, say $p(C_1)$ is connected and so is $p(C_2)$, but since $x\sim y$ we have that $p(C_1)\cap p(C_2)\neq \varnothing$ so $X/{\sim}$ consists of a single component, becuase \[p(C_1)\cup p(C_2) = p(C_1\cup C_2)=p(X)=X/{\sim},\] as wanted. To show the reverse implication, we use the contrapositive of the statement and show: if we for no $x\in C_1$ or $y\in C_2$ have that $x\sim y$, then $X/{\sim}$ is not connected. Assume the hypothesis and note that then $p(C_1)$ and $p(C_2)$ are then disjoint connected subspaces whose union equal all of $X/{\sim}$ (since $p$ is surjective). But then the images of $C_1$ and $C_2$ under $p$ are two components of $X/{\sim}$, showing that $X/{\sim}$ is not connected. As wanted. It’s soon exam time, so I’m practicing proofs in complex analysis. Right now that means Cauchy’s integral formula for $n$’th derivatives. Let $G$ be a domain of the complex numbers and $f: G\to \mathbb{C}$ a holomorphic function. We first want to show that $f$ can be expressed as a power series, such that $$f(z)=\sum^\infty_{n=0} a_n(z-a).$$ for some $a\in\mathbb{C}$, let $B_\rho(a)$ the largest open ball at $a$ contained in $G$. We claim that $$a_n = \frac{1}{2\pi i} \oint\frac{f(z)}{(z-a)^{n+1}}dz.$$ By the Cauchy integral formula we have that, for a fixed $z_0\in B_\rho(a)$ we have $$f(z_0)=\frac{1}{2\pi i} \oint \frac{f(z)}{z-z_0}$$ and by elementary calculations we can, for $z\in \partial B_r(a)$, write $$\frac{1}{z-z_0} = \frac{1}{z-a} \frac{1}{1-\frac{z_0 -a}{z-a}}=\frac{1}{z-a}\sum^\infty_{n=0} \left(\frac{z_0-a}{z-a}\right)^n,$$ and from above we then have $$\begin{split} f(z_0)& = \frac{1}{2\pi i} \oint \frac{f(z)}{z-z_0} \\ &=\frac{1}{2\pi i}\oint\sum^\infty_{n=0} \frac{f(z)(z_0-a)^n}{(z-a)^{n+1}}dz\\ &=\frac{1}{2\pi i}\sum^\infty_{n=0}\oint \frac{f(z)(z_0-a)^n}{(z-a)^{n+1}}dz\\&=\frac{1}{2\pi i}\sum^\infty_{n=0}\oint \frac{f(z)}{(z-a)^{n+1}}dz(z_0-a)^n\\ &=\sum^\infty_{n=0}a_n (z_0-a)^n,\end{split}$$ as wanted. We see that $f$ is a power series and thus infinitely complex differentiable, and the derivatives are $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint\frac{f(z)}{(z-a)^{n+1}}dz,$$ as desired. The observant reader will have noticed that I didn’t check that the sums were uniformly convergent, which is needed in other to switch the sum and integral signs, but this is an easy application of the Weierstraß $M$-test. References 648 {648:SS6NUFWV} items 1 apa default asc Berg, C. (2016). Complex analysis. Copenhagen: Department of Mathematical Sciences, University of Copenhagen.
This is a heuristic explanation of Witten's statement, without going into the subtleties of axiomatic quantum field theory issues, such as vacuum polarization or renormalization. A particle is characterized by a definite momentum plus possible other quantum numbers. Thus, one particle states are by definition states with a definite eigenvalues of the momentum operator, they can have further quantum numbers. These states should exist even in an interactiong field theory, describing a single particle away from any interaction.In a local quantum field theory, these states are associated with local field operators: $$\| p, \sigma \rangle = \int e^{ipx} \psi_{\sigma}^{\dagger}(x) \|0\rangle d^4x$$Where $\psi $ is the field corresponding to the particle and $\sigma$ describes the set of other quantum numbers additional to the momentum.A symmetry generator $Q$ being the integral of a charge density according to the Noether's theorem$$Q = \int j_0(x') d^3x'$$should generate a local field when it acts on a local field:$[Q, \psi_1(x)] = \psi_2(x)$(In the case of internal symmetries $\psi_2$ depends linearly on the components of $\psi_1(x)$, in the case of space time symmetries it depends on the derivatives of the components of $\psi_1(x)$) Thus in general: $$[Q, \psi_{\sigma}(x)] = \sum_{\sigma'} C_{\sigma\sigma'}(i\nabla)\psi_{\sigma'}(x)])$$ Where the dependence of the coefficients $ C_{\sigma\sigma'}$ on the momentum operator $\nabla$ is due to the possibility that $Q$ contains a space-time symmetry.Thus for an operator $Q$ satisfying $Q\|0\rangle = 0$, we have$$ Q \| p, \sigma \rangle = \int e^{ipx} Q \psi_{\sigma}^{\dagger}(x) \|0\rangle d^4x = \int e^{ipx} [Q , \psi_{\sigma}^{\dagger}(x)] \|0\rangle d^4x = \int e^{ipx} \sum_{\sigma'} C_{\sigma\sigma'}(i\nabla)\psi_{\sigma'}(x) \|0\rangle d^4x = \sum_{\sigma'} C_{\sigma\sigma'}(p) \int e^{ipx} \psi_{\sigma'}^{\dagger}(x) \|0\rangle d^4x = \sum_{\sigma'} C_{\sigma\sigma'}(p) \| p, \sigma' \rangle $$Thus the action of the operator $Q$ is a representation in the one particle states. The fact that $Q$ commutes with the Hamiltonian is responsible for the energy degeneracy of its action, i.e., the states $\| p, \sigma \rangle$ and $Q\| p, \sigma \rangle$ have the same energy.This post imported from StackExchange Physics at 2015-06-16 14:50 (UTC), posted by SE-user David Bar Moshe
It’s soon exam time, so I’m practicing proofs in complex analysis. Right now that means Cauchy’s integral formula for $n$’th derivatives. Let $G$ be a domain of the complex numbers and $f: G\to \mathbb{C}$ a holomorphic function. We first want to show that $f$ can be expressed as a power series, such that $$f(z)=\sum^\infty_{n=0} a_n(z-a).$$ for some $a\in\mathbb{C}$, let $B_\rho(a)$ the largest open ball at $a$ contained in $G$. We claim that $$a_n = \frac{1}{2\pi i} \oint\frac{f(z)}{(z-a)^{n+1}}dz.$$ By the Cauchy integral formula we have that, for a fixed $z_0\in B_\rho(a)$ we have $$f(z_0)=\frac{1}{2\pi i} \oint \frac{f(z)}{z-z_0}$$ and by elementary calculations we can, for $z\in \partial B_r(a)$, write $$\frac{1}{z-z_0} = \frac{1}{z-a} \frac{1}{1-\frac{z_0 -a}{z-a}}=\frac{1}{z-a}\sum^\infty_{n=0} \left(\frac{z_0-a}{z-a}\right)^n,$$ and from above we then have $$\begin{split} f(z_0)& = \frac{1}{2\pi i} \oint \frac{f(z)}{z-z_0} \\ &=\frac{1}{2\pi i}\oint\sum^\infty_{n=0} \frac{f(z)(z_0-a)^n}{(z-a)^{n+1}}dz\\ &=\frac{1}{2\pi i}\sum^\infty_{n=0}\oint \frac{f(z)(z_0-a)^n}{(z-a)^{n+1}}dz\\&=\frac{1}{2\pi i}\sum^\infty_{n=0}\oint \frac{f(z)}{(z-a)^{n+1}}dz(z_0-a)^n\\ &=\sum^\infty_{n=0}a_n (z_0-a)^n,\end{split}$$ as wanted. We see that $f$ is a power series and thus infinitely complex differentiable, and the derivatives are $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint\frac{f(z)}{(z-a)^{n+1}}dz,$$ as desired. The observant reader will have noticed that I didn’t check that the sums were uniformly convergent, which is needed in other to switch the sum and integral signs, but this is an easy application of the Weierstraß $M$-test. References Complex analysis. Copenhagen: Department of Mathematical Sciences, University of Copenhagen.
Hello one and all! Is anyone here familiar with planar prolate spheroidal coordinates? I am reading a book on dynamics and the author states If we introduce planar prolate spheroidal coordinates $(R, \sigma)$ based on the distance parameter $b$, then, in terms of the Cartesian coordinates $(x, z)$ and also of the plane polars $(r , \theta)$, we have the defining relations $$r\sin \theta=x=\pm R^2−b^2 \sin\sigma, r\cos\theta=z=R\cos\sigma$$ I am having a tough time visualising what this is? Consider the function $f(z) = Sin\left(\frac{1}{cos(1/z)}\right)$, the point $z = 0$a removale singularitya polean essesntial singularitya non isolated singularitySince $Cos(\frac{1}{z})$ = $1- \frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$$$ = (1-y), where\ \ y=\frac{1}{2z^2}+\frac{1}{4!... I am having trouble understanding non-isolated singularity points. An isolated singularity point I do kind of understand, it is when: a point $z_0$ is said to be isolated if $z_0$ is a singular point and has a neighborhood throughout which $f$ is analytic except at $z_0$. For example, why would $... No worries. There's currently some kind of technical problem affecting the Stack Exchange chat network. It's been pretty flaky for several hours. Hopefully, it will be back to normal in the next hour or two, when business hours commence on the east coast of the USA... The absolute value of a complex number $z=x+iy$ is defined as $\sqrt{x^2+y^2}$. Hence, when evaluating the absolute value of $x+i$ I get the number $\sqrt{x^2 +1}$; but the answer to the problem says it's actually just $x^2 +1$. Why? mmh, I probably should ask this on the forum. The full problem asks me to show that we can choose $log(x+i)$ to be $$log(x+i)=log(1+x^2)+i(\frac{pi}{2} - arctanx)$$ So I'm trying to find the polar coordinates (absolute value and an argument $\theta$) of $x+i$ to then apply the $log$ function on it Let $X$ be any nonempty set and $\sim$ be any equivalence relation on $X$. Then are the following true: (1) If $x=y$ then $x\sim y$. (2) If $x=y$ then $y\sim x$. (3) If $x=y$ and $y=z$ then $x\sim z$. Basically, I think that all the three properties follows if we can prove (1) because if $x=y$ then since $y=x$, by (1) we would have $y\sim x$ proving (2). (3) will follow similarly. This question arised from an attempt to characterize equality on a set $X$ as the intersection of all equivalence relations on $X$. I don't know whether this question is too much trivial. But I have yet not seen any formal proof of the following statement : "Let $X$ be any nonempty set and $∼$ be any equivalence relation on $X$. If $x=y$ then $x\sim y$." That is definitely a new person, not going to classify as RHV yet as other users have already put the situation under control it seems... (comment on many many posts above) In other news: > C -2.5353672500000002 -1.9143250000000003 -0.5807385400000000 C -3.4331741299999998 -1.3244286800000000 -1.4594762299999999 C -3.6485676800000002 0.0734728100000000 -1.4738058999999999 C -2.9689624299999999 0.9078326800000001 -0.5942069900000000 C -2.0858929200000000 0.3286240400000000 0.3378783500000000 C -1.8445799400000003 -1.0963522200000000 0.3417561400000000 C -0.8438543100000000 -1.3752198200000001 1.3561451400000000 C -0.5670178500000000 -0.1418068400000000 2.0628359299999999 probably the weirdness bunch of data I ever seen with so many 000000 and 999999s But I think that to prove the implication for transitivity the inference rule an use of MP seems to be necessary. But that would mean that for logics for which MP fails we wouldn't be able to prove the result. Also in set theories without Axiom of Extensionality the desired result will not hold. Am I right @AlessandroCodenotti? @AlessandroCodenotti A precise formulation would help in this case because I am trying to understand whether a proof of the statement which I mentioned at the outset depends really on the equality axioms or the FOL axioms (without equality axioms). This would allow in some cases to define an "equality like" relation for set theories for which we don't have the Axiom of Extensionality. Can someone give an intuitive explanation why $\mathcal{O}(x^2)-\mathcal{O}(x^2)=\mathcal{O}(x^2)$. The context is Taylor polynomials, so when $x\to 0$. I've seen a proof of this, but intuitively I don't understand it. @schn: The minus is irrelevant (for example, the thing you are subtracting could be negative). When you add two things that are of the order of $x^2$, of course the sum is the same (or possibly smaller). For example, $3x^2-x^2=2x^2$. You could have $x^2+(x^3-x^2)=x^3$, which is still $\mathscr O(x^2)$. @GFauxPas: You only know $\|f(x)\|\le K_1 x^2$ and $\|g(x)\|\le K_2 x^2$, so that won't be a valid proof, of course. Let $f(z)=z^{n}+a_{n-1}z^{n-1}+\cdot\cdot\cdot+a_{0}$ be a complex polynomial such that $\|f(z)\|\leq 1$ for $\|z\|\leq 1.$ I have to prove that $f(z)=z^{n}.$I tried it asAs $\|f(z)\|\leq 1$ for $\|z\|\leq 1$ we must have coefficient $a_{0},a_{1}\cdot\cdot\cdot a_{n}$ to be zero because by triangul... @GFauxPas @TedShifrin Thanks for the replies. Now, why is it we're only interested when $x\to 0$? When we do a taylor approximation cantered at x=0, aren't we interested in all the values of our approximation, even those not near 0? Indeed, one thing a lot of texts don't emphasize is this: if $P$ is a polynomial of degree $\le n$ and $f(x)-P(x)=\mathscr O(x^{n+1})$, then $P$ is the (unique) Taylor polynomial of degree $n$ of $f$ at $0$.
An orbital ring connected to the Earth by space elevators would reduce the cost of going to space to an amount comparable to an airplane ticket. This would cause a boom in the space tourism industry and eventually millions and even billions of people and tons of cargo will be moving from the Earth’s surface to space annually, and vise versa. This would necessitate an expansion in our space-based infrastructure to include space-based solar panels, a lunar mass driver, the routine mining of asteroids, and especially enormous space habitats (for all those billions of people to live in) such as the Standford Torus, the Bernal Sphere, or the O’Neil Cylinder. Orbital rings also allow you to build artificial planets and Dyson spheres, which would allow us to completely colonize the solar system. They would also allow us to build a Birch planet, a single planet with a surface area which exceeds the total surface area of all the planets in the Milky Way galaxy. In this lesson, we’ll give a brief catalog of the various different classes of planets in the universe. We'll discuss Pulsar planets, hot Jupiters, Super Earths, ice and water worlds, archipelago worlds, diamond worlds, and rogue planets. Most of the planets we’ll be discussing were discovered using the Kepler Space Telescope and the transit method. We once believed that the formation of planets was rare and that there probably weren’t many planets beyond our solar system. We couldn’t have been more wrong. In this lesson, we’ll discuss the prospect of life in the Milky Way galaxy beyond the Earth. We'll begin by discussing the speculations made in a paper written by Carl Sagan about the possibility of life in Jupiter's atmosphere. From there, we shall derive a formula which describes the habitable zone of a star. Using this formula and data obtained by the Kepler Space Telescope, we can estimate the total number of "Earth-like" planets in the Milky Way. From there, we discuss the fraction of those planets on which simple and intelligent life evolve; then we'll discuss the fraction of those planets on which advanced communicating civilizations evolve and what fraction of those civilizations are communicating right now. In this lesson, we’ll prove that $\lim_{ϴ→0}\frac{sinϴ}{ϴ}=1$. We'll prove this result by using the squeeze theorem and basic geometry, algebra, and trigonometry. In a future lesson, we'll learn why this result is important: the reason being because knowledge that $\lim_{ϴ→0}\frac{sinϴ}{ϴ}=1$ is required to find the derivatives of the sin and cosine functions. But we'll save that for a future lesson. For a vector field $\vec{F}(x,y)$ defined at each point $(x,y)$ within the region $R$ and along the continuous, smooth, closed, piece-wise curve $c$ such that $R$ is the region enclosed by $c$, we shall derive a formula (known as Green’s Theorem) which will allow us to calculate the line integral of $\vec{F}(x,y)$ over the curve $c$. To find the gravitational force exerted by a sphere of mass $M$ on a particle of mass $m$ outside of that sphere, we must first subdivide that sphere into many very skinny shells and find the gravitational force exerted by anyone of those shells on $m$. We'll see, however, that finding the gravitational force exerted by such a shell is in of itself a somewhat tedious exercise. In the end, we'll see that the gravitational force exerted by a sphere of mass $M$ on a particle of mass $m$ outside of the sphere (where $D$ is the center-to-center separation distance between the sphere and the particle) is completely identical to the gravitational force exerted by a particle of mass $M$ on the other particle of mass $m$ such that $D$ is there separation distance. In previous lessons, we learned that by taking the integral of some function $f(x)$ we can find the area underneath that curve by summing the areas of infinitely many, infinitesimally skinny rectangles. In this lesson, we'll use the concept of a double integral to find the volume underneath any smooth and continuous surface $f(x,y)$ by summing the volumes of infinitely many, infinitesimally skinny columns. The first serious proposal in scientific literature on terraforming other worlds in the universe was about terraforming Venus. The planetary scientist Carl Sagan imagined seeding the Venusian skies with photosynthetic microbes capable of converting Venus's $C0_2$-rich atmosphere into oxygen. Other proposals involve assembling a vast system of orbital mirrors capable of blocking the Sun's light and cooling Venus until this hot and hellish world became very frigid and rained $C0_2$ from its atmosphere. The solleta would also be capable of simulating an Earth day/night cycle. To create oceans and an active hydrosphere on Venus, we could hurl scores of icy asteroids from the Kuiper belt to Venus and, upon impacting the Venusian atmosphere, would rapidly disintegrate releasing enormous quantities of water vapor into the atmosphere which subsequently condense to form the first seas on Venus. Or perhaps Saturn's moon Enceladus—containing a colossal subsurface ocean dwarfing that of the Earth's—could be sacrificed towards the end of creating the first seas on Venus. But even if humans never terraform this hellish world, they could still live their—partially at least—by deploying thousands of blimps into the Venusian skies capable of supporting a long-term, human presence of perhaps over a million people. Venusian sky cities. But eventually, after many millennia of terraforming Venus, a rich ecosystem of life—including us—could live on Venus's surface.
Duke Mathematical Journal Duke Math. J. Volume 116, Number 2 (2003), 189-217. Low-lying zeros of dihedral L-functions Abstract Assuming the grand Riemann hypothesis, we investigate the distribution of the lowlying zeros of the $L$-functions $L(s,\psi)$, where $\psi$ is a character of the ideal class group of the imaginary quadratic field $\mathbb {Q}(\sqrt{-D}) (D\text {squarefree},D>3,D\equiv 3(\mod 4))$. We prove that, in the vicinity of the central point $s = 1/2$, the average distribution of these zeros (for $D\longrightarrow \infty$) is governed by the symplectic distribution. By averaging over $D$, we go beyond the natural bound of the support of the Fourier transform of the test function. This problem is naturally linked with the question of counting primes $p$ of the form $4p = m\sp 2+Dn\sp 2$, and sieve techniques are applied. Article information Source Duke Math. J., Volume 116, Number 2 (2003), 189-217. Dates First available in Project Euclid: 26 May 2004 Permanent link to this document https://projecteuclid.org/euclid.dmj/1085598267 Digital Object Identifier doi:10.1215/S0012-7094-03-11621-X Mathematical Reviews number (MathSciNet) MR1953291 Zentralblatt MATH identifier 1028.11055 Subjects Primary: 11M41: Other Dirichlet series and zeta functions {For local and global ground fields, see 11R42, 11R52, 11S40, 11S45; for algebro-geometric methods, see 14G10; see also 11E45, 11F66, 11F70, 11F72} Secondary: 11F66: Langlands $L$-functions; one variable Dirichlet series and functional equations 11M26: Nonreal zeros of $\zeta (s)$ and $L(s, \chi)$; Riemann and other hypotheses 11N36: Applications of sieve methods 11R42: Zeta functions and $L$-functions of number fields [See also 11M41, 19F27] Citation Fouvry, E.; Iwaniec, H. Low-lying zeros of dihedral L -functions. Duke Math. J. 116 (2003), no. 2, 189--217. doi:10.1215/S0012-7094-03-11621-X. https://projecteuclid.org/euclid.dmj/1085598267
%BEGINLATEXPREAMBLE% \usepackage{amsfonts} \definecolor{Lightmaroon}{rgb}{0.6667,0,0} \definecolor{Cornflowerblue}{rgb}{0,0.4,0.8}%ENDLATEXPREAMBLE% Introduction to %BEGINLATEX{inline="1" color="Lightmaroon"}%\Large\LaTeX%ENDLATEX% %BEGINLATEX{inline="1"}%\LaTeX%ENDLATEX% is a complete typesetting language developed by Leslie Lamport on top of Donald Knuth's %BEGINLATEX{inline="1"}%\TeX%ENDLATEX%. Most of it is implemented using LatexModePlugin , but the main use is producing mathematics, since most other formatting is more conveniently accomplished using Wiki constructs. To produce mathematics in-line , the %BEGINLATEX{inline="1"}%\LaTeX%ENDLATEX% text must be surrounded by %$ and $% . Thus, %$\Gamma \vdash Rx_1\cdots x_n$% looks like this%$\Gamma \vdash Rx_1\cdots x_n$%in the middle of some text. To display mathematics, it must be surrounded by %\[ and \]% . Thus, %\[\mathcal{N}=\langle \mathbb{N},0,1+,\cdot \rangle \subseteq \langle \mathbb{R},0,1+,\cdot \rangle\]% produces this centered display %\[\mathcal{N}=\langle \mathbb{N},0,1+,\cdot \rangle \subseteq \langle \mathbb{R},0,1+,\cdot \rangle\]% in the middle of a paragraph, which is preferable for long formulas or ones requiring emphasis. %BEGINLATEX{inline="1"}%\LaTeX%ENDLATEX% is used by most mathematicians, physicists, and engineers for mathematics, and articles and books are typically submitted for publication in %BEGINLATEX{inline="1"}%\LaTeX%ENDLATEX%. There are therefore many helpful web sites, newsgroups, and books about %BEGINLATEX{inline="1"}%\LaTeX%ENDLATEX%. The standard complete reference is Lamport's LaTeX: A Document Preparation System (2nd Edition) . The place to start on the web is the %BEGINLATEX{color="Cornflowerblue" inline="1"}%\TeX%ENDLATEX% Users Group . Brief Tutorial on %BEGINLATEX{inline="1" color="Lightmaroon"}%\bf\LaTeX%ENDLATEX% All %BEGINLATEX{inline="1"}%\TeX%ENDLATEX% codes begin with a backslash (\). Curly braces ({ }) are used to delimit groupings, except in a few cases where optional arguments must be enclosed in square brackets ([ ]). Many symbols have mnemonic names, like \Alpha , \alpha , \forall , \exists , \lor , \land . A list of some common symbols is given in the LatexSymbols topic. Other off-site lists include: Some Symbols and Some Little-Known Symbols , All the Symbols . Not all the symbols work in the Wiki, but most of them do. A subscript is indicated with an underline (_). If the subscript consists of more than one symbol, it must be enclosed in braces. Similarly, a superscript is indicated with a caret (^). Thus, %$ \alpha _{ \beta _1^2}^{ \gamma ^{ \delta ^ \epsilon }}$% yields %$ \alpha _{ \beta _1^2}^{ \gamma ^{ \delta^ \epsilon }}$%, Document Structure In standard %BEGINLATEX{inline="1"}%\LaTeX%ENDLATEX%2e documents, the main file has the followingstructure: \documentclass[ options]{ class} preamble \begin{document} text \end{document} Any text that normally goes in the preamble or text sections of thetemplate above can be rendered in TWiki. The preamble is typically usedto define new commands, or use LaTeX style-files to extend LaTeX'sfunctionality. The text area contains the actual text to be rendered.One does not need to use the \documentclass or document environment touse LaTeX in TWiki. The ability to render a complete LaTeX document within TWiki, say from anexisting file, is currently under consideration. SeeTWiki:Codev.IncludeExistingLatexDocsInTWiki for details and current status. -- courtesy of TWiki:Main.ShaughanLavine (see also )
Research Open Access Published: On the oscillation of higher order nonlinear neutral difference equations Advances in Difference Equations volume 2019, Article number: 275 (2019) Article metrics 344 Accesses Abstract In this paper, we shall investigate some oscillation criteria for the solutions of mth-order nonlinear neutral difference equation where $m\geq 1$. The results presented here complement some of the known results reported in the literature. Examples are included to illustrate the importance of the main results. Introduction In this paper, we are concerned with the following higher order neutral difference equation: where $m\geq 1$ and Δ is the forward difference operator defined by Throughout this paper, we assume the following conditions to hold: (H1) $\{q (n ) \}$ is a real-valued sequence with $q (n )\geq 0,n\in N$ and $\{q (n ) \}$ is not identically zero. (H2) $\{p (n ) \}$ is a real-valued sequence with $0 \leq p (n )<1,n\in N$. (H3) $\{\tau (n ) \}$ and $\{\sigma (n ) \}$ are nondecreasing sequences such that $\tau (n )< n$ with $\lim_{n\rightarrow +\infty }\tau (n )=+\infty $ and $\sigma (n )< n$ with $\lim_{n\rightarrow +\infty }\sigma (n )=+\infty $. (H4) $f:R\rightarrow R$ is a nondecreasing continuous function such that $xf (x )>0$ for $x\neq 0$ and$$\begin{aligned} -f (-xy )\geq f (xy )\geq f (x )f (y ). \end{aligned}$$(1.2) The factorial expression is defined as $(r )^{ (s )}= \prod_{i=0}^{s-1} (r-i )$ with $(r ) ^{ (0 )}=1$ for all $r\in R= (-\infty , \infty )$ and s, a nonnegative integer. Let $N_{0}$ be a fixed nonnegative integer. By a solution of equation (1.1), we mean a nontrivial real sequence $\{x (n ) \}$ which is defined for all $n\geq \min_{i\geq 0} \{\tau (i ),\sigma (i ) \}$ and satisfies equation (1.1) for $n\geq N_{0}$. A solution $\{x (n ) \}$ is said to be oscillatory if it is neither eventually positive nor eventually negative. Otherwise it is called non-oscillatory. A difference equation is said to be oscillatory if all of its solutions are oscillatory. Otherwise, it is non-oscillatory. In recent years, the oscillation behavior of neutral difference equations has been studied vigorously, for example, see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26] and the references cited therein. This is because of the fact that neutral difference equations find various applications in some variational problems, in natural science and technology. Agarwal et al. [5] considered the $m{th}$ order neutral difference equation In [4], Agarwal and Grace considered the higher order difference equation and obtained some sufficient conditions for the oscillation of all solutions of (1.4). Yasar Bolat et al. [9] have taken even order nonlinear neutral difference equation and established some criteria for oscillation of bounded solutions only. Therefore, it is to be noted that, to the best of our knowledge, there is no paper for higher order nonlinear neutral difference equations which ensures that all the solutions are oscillatory when m is odd. Following this notion, our aim in this paper is to provide sufficient conditions which ensure that all solutions of (1.1) are oscillatory. To obtain our results, we shall need the following lemma. Lemma 1.1 (see [1]) Let $x (n )$ be defined for $n\geq n_{0}\in N$ and $x (n )>0$ with $\Delta ^{n} x (n )$ of constant sign for $n\geq n_{0}$ and not identically zero. Then there exists an integer l, $0\leq l\leq m$, with $(m+l )$ odd for $\Delta ^{m} x (n )\leq 0$ and $(m+l )$ even for $\Delta ^{m} x (n )\geq 0$ eventually such that (i) $l \leq m-1$ implies$(-1 )^{l+k}\Delta ^{k} x (n )>0$ for all$n\geq n_{0}$, $l \leq k\leq m-1$. (ii) $l \geq 1$ implies$\Delta ^{k} x (n )>0$ for all large$n\geq n_{0}$, $1 \leq k\leq l-1$. Main results To obtain the main results, we shall use the following notations. For all large $n\geq n_{0}>0$, let for some nondecreasing function $\eta (n )$ with $\sigma (n )<\eta (n )\leq n, n\geq n_{0}$. Then we shall discuss the following theorems. Theorem 2.1 Assume that conditions (H1)–(H4) hold and Let m be odd. If all the second order equations for $n\geq n_{0}$ are oscillatory and if there exists a nondecreasing sequence $\{\eta (n ) \}$ with $\sigma (n )< \eta (n )\leq n, n\geq n_{0}$ such that the first order difference equation is oscillatory, then every solution of equation (1.1) oscillates. Theorem 2.2 for $n\geq n_{0}$ are oscillatory, then every solution of equation (1.1) oscillates. Let $\{x (n ) \}$ be a non-oscillatory solution of (1.1). Without loss of generality, assume that $x (n )>0, x (\tau (n ) )>0, x (\sigma (n ) )>0$ for all $n\geq n_{0}\geq 0$. Let Then (1.1) becomes From Lemma 1.1, it is easy to check Also, from (2.5), we have $\Delta ^{m} z (n )\leq 0$ eventually. That is, Now the following three cases are considered: $l\in \{1,2,\ldots,m-3 \}, l=m-1, l=0$. Case (i): $l\in \{1,2,\ldots,m-3 \}$. From discrete Taylor’s formula, we have Summing up equation (1.1) from r to $u-1$ and letting $u\rightarrow \infty $, we have From (2.10), we can see that Hence from (2.1), we get Consider the equality with $l\in \{1,2,\ldots,m-1 \}$ and $(l+m )$ is odd. Now from the above, there exists an integer $n\geq n _{3}\geq n_{2}$ such that and Then we can find an integer $N\geq n_{3}$ such that That is, Let $y (n )=\Delta ^{l-1}z (n )$. Then $y (n )>0$ for $n\geq N$ and the above inequality becomes Thus the last inequality has an eventually positive solution. By a well-known result in [14, p. 186, Corollary 7.6.1], we can see that the equation also has an eventually positive solution, which contradicts our assumption. Case (ii): $l=m-1$. Substituting $l=m-1$ in inequality (2.16), we get Set $u (n )=\Delta ^{m-2}z (n )>0$ for $n\geq n_{1}$. Then the above inequality becomes That is, which has an eventually positive solution. Thus we get a contradiction as in Case (i). Case (iii): $l=0$. In this case, m is odd. From discrete Taylor’s formula, we have Considering Lemma 1.1 with $l=0$ and using this in the above equation, we get Then we can find an integer $n_{2}\geq n_{1}$ and a nondecreasing function $\eta (n )$ with $\sigma (n )<\eta (n )\leq n$ such that Let $v (n )= \Delta ^{m-1} z (\eta (n ) )$. Then $v (n )>0$ for $n\geq n_{2}$, and the above inequality becomes for which an eventually positive solution exists. By a well-known result in [14, p. 186, Corollary 7.6.1], we have equation (2.3) also has an eventually positive solution, which contradicts our assumption. This completes the proof. □ Example 2.3 Consider the third order difference equation Here, $0\leq p (n )=\frac{1}{2}<1$, $q (n )=4n$, $\tau (n )=n-1< n$, $\sigma (n )=n-2< n$, and $f (u )=\frac{u}{n}$. Also, One of such solutions is $x (n )= (-1 )^{n}$. Example 2.4 Consider the second order difference equation Here, $0\leq p (n )=\frac{1}{4}<1$, $q (n )= \frac{5n+3}{n-1}$, $\tau (n )=n-2< n$, $\sigma (n )=n-1< n$, and $f (u )=u$. Also, Next we shall present the following theorems. Theorem 2.5 Assume that conditions (H1)–(H4) and (2.1) hold. Let m be odd. If for $j\in \{2,4,\ldots,m-1 \}$ and if there exists a nondecreasing sequence $\{\eta (n ) \}$ with $\sigma (n )<\eta (n )\leq n, n\geq n_{0}$ such that equation (2.3) is oscillatory, then every solution of equation (1.1) oscillates. Theorem 2.6 then all the solutions of (1.1) are oscillatory. Assume that $\{x (n ) \}$ is a non-oscillatory solution of (1.1). Without loss of generality, assume that $x (n )>0, x (\tau (n ) )>0, x (\sigma (n ) )>0$ for all $n\geq n_{0}\geq 0$. Let Case (i): $l\in \{1,2,\ldots,m-3 \}$. From discrete Taylor’s formula, we have From (2.15), there exist $n_{2}\geq n_{1}$ and a positive constant $c>0$ such that and which contradicts (2.18). Case (ii): $l=m-1$. Summing up equation (1.1) from r to $u-1$ and letting $u\rightarrow \infty $, we have Example 2.7 Consider the first order difference equation Here, $0\leq p (n )=\frac{3}{4}<1$, $q (n )= \frac{1}{2}$, $\tau (n )=n-1< n$, $\sigma (n )=n-2< n$, and $f (u )=u$. Example 2.8 Consider the fourth order difference equation Here, $0\leq p (n )=\frac{1}{2}<1$, $q (n )=8 (n+2 )$, $\tau (n )=n-1< n$, $\sigma (n )=n-1< n$, and $f (u )=\frac{u}{n+2}$. Conclusion In this paper, by using discrete Taylor’s formula, the summing averaging technique, and the comparison method, the oscillatory behavior of every solution of equation (1.1) is discussed in Theorems 2.1 and 2.2, Theorems 2.5 and 2.6. Here, some sufficient conditions are proved. These sufficient conditions, which are new, extend and complement some of the known results in the literature. Also, the examples reveal the illustration of the proved results. References 1. Agarwal, R.P.: Difference Equations and Inequalities: Theory, Methods and Applications, 2nd edn. New York (2000) 2. Agarwal, R.P., Bohner, M., Grace, S.R., O’Regan, D.: Discrete Oscillation Theory. CMIA Book Series, vol. 1 3. Agarwal, R.P., Grace, S.R., O’Regan, D.: Oscillation Theory for Difference and Functional Differential Equations. Kluwer Academic Publishers, Dordrecht (2000) 4. Agarwal, R.P., Grace, S.R., O’Regan, D.: On the oscillation of higher order difference equations. Soochow J. Math. 31(2), 245–259 (2005) 5. Agarwal, R.P., Thandapani, E., Wong, P.J.Y.: Oscillations of higher order neutral difference equations. Appl. Math. Lett. 10(1), 71–78 (1997) 6. Agarwal, R.P., Wong, P.J.Y.: Advanced Topics in Difference Equations. Kluwer Academic Publishers, Dordrecht (1997) 7. Bohner, M., Grace, S.R., Sager, I., Tunc, E.: Oscillation of third-order nonlinear damped delay differential equations. Appl. Math. Comput. 278, 21–32 (2016) 8. Bolat, Y., Akin, O.: Oscillatory behaviour of a higher order nonlinear neutral type functional difference equation with oscillating coefficients. Appl. Math. Lett. 17, 1073–1078 (2004) 9. Bolat, Y., Akin, O., Yildirim, H.: Oscillation criteria for a certain even order neutral difference equation with an oscillating coefficient. Appl. Math. Lett. 22, 590–594 (2009) 10. Grace, S.R., Graef, J.R., Panigrahi, S., Tunc, E.: On the oscillatory behavior of even order neutral delay dynamic equations on time-scales. Electron. J. Qual. Theory Differ. Equ. 2012, 96 (2012) 11. Grace, S.R., Graef, J.R., Tunc, E.: On the oscillation of certain third order nonlinear dynamic equations with a nonlinear damping term. Math. Slovaca 67(2), 501–508 (2017) 12. Graef, J.R., Grace, S.R., Tunc, E.: Oscillation of even-order advanced functional differential equations. Publ. Math. (Debr.) 93(3–4), 445–455 (2018) 13. Graef, J.R., Grace, S.R., Tunc, E.: Oscillatory behavior of even-order nonlinear differential equations with a sublinear neutral term. Opusc. Math. 39(1), 39–47 (2019) 14. Gyori, I., Ladas, G.: Oscillation Theory of Delay Differential Equations with Applications. Clarendon Press, Oxford (1991) 15. Kaleeswari, S., Selvaraj, B., Thiyagarajan, M.: A new creation of mask from difference operator to image analysis. J. Theor. Appl. Inf. Technol. 69(1), 211–218 (2014) 16. Kaleeswari, S., Selvaraj, B., Thiyagarajan, M.: Removing noise through a nonlinear difference operator. Int. J. Appl. Eng. Res. 9(21), 5100–5105 (2014) 17. Kelley, W.G., Peterson, A.C.: Difference Equations an Introduction with Applications. Academic Press, Boston (1991) 18. Ladas, G., Philos, C.G., Sficas, Y.G.: Sharp conditions for the oscillation of delay difference equations. J. Appl. Math. Simul. 2, 101–111 (1989) 19. Luo, Z., Shen, J.: New results for oscillation of delay difference equations. Comput. Math. Appl. 41, 553–561 (2001) 20. Selvaraj, B., Gomathi Jawahar, G.: New oscillation criteria for first order neutral delay difference equations. Bull. Pure Appl. Sci. 30E, 103–108 (2011) 21. Selvaraj, B., Kaleeswari, S.: Oscillation of solutions of certain fifth order difference equations. J. Comput. Math. Sci. 3(6), 653–663 (2012) 22. Selvaraj, B., Kaleeswari, S.: Oscillation of solutions of certain nonlinear difference equations. Prog. Nonlinear Dyn. Chaos 1, 34–38 (2013) 23. Selvaraj, B., Kaleeswari, S.: Oscillation theorems for certain fourth order non-linear difference equations. Int. J. Math. Res. 5(3), 299–312 (2013) 24. Selvaraj, B., Kaleeswari, S.: Oscillation of solutions of second order nonlinear difference equations. Bull. Pure Appl. Sci. 32E, 107–117 (2013) 25. Selvaraj, B., Mohankumar, P., Ananthan, V.: Oscillatory and nonoscillatory behavior of neutral delay difference equations. Int. J. Nonlinear Sci. 13(4), 472–474 (2012) 26. Thandapani, E., Selvaraj, B.: Existence and asymptotic behavior of non oscillatory solutions of certain nonlinear difference equations. Far East J. Math. Sci.: FJMS 14(1), 9–25 (2004) Acknowledgements The author would like to thank the referees for the helpful suggestions to improve the presentation of the paper. Funding This research work is not supported by any funding agencies. Ethics declarations Competing interests The author has declared that no competing interests exist. Additional information Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
I'm currently working on exercises of the book "Modal Logic" by A.Chagrov and M.Zakharyaschev (for pleasure, not homework). One exercise asks to prove this version of Gabbay rule (exercise $3.10$): A frame $F$ validates the rule $(\Box p \rightarrow p) \vee \psi$ $/$ $\psi$, with $p$ not appearing in $\psi$, if and only if $F$ is irreflexive. ( Note that: "$p$ not appearing in $\psi$"). I have the $\leftarrow$ part, but I'm having a hard time proving the other implication. I would really like to understand this excercise, and I know that the excercises in this book are sometimes hard. But I think this has to be fairly easy. Until now I tried proving it by reductio ad absurdum: I have one reflexive node $x$, and I try to find a formula $\phi$ that $x$ doesn't validate, but that every other node does validate. If I can prove such formula exists then I'm done, because I plug it in the rule. I hope I was clear, thanks in advance!
Why is the area under the ROC curve the probability that a classifier will rank a randomly chosen "positive" instance (from the retrieved predictions) higher than a randomly chosen "positive" one (from the original positive class)? How does one prove this statement mathematically using integral, giving the CDFs and PDFs of the true positive and negative class distributions? First thing, let's try to define the area under the ROC curve formally. Some assumptions and definitions: We have a probabilistic classifier that outputs a "score" s(x), where x are the features, and s is a generic increasing monotonic function of the estimated probability p(class = 1\|x). $f_{k}(s)$, with $k = \{0, 1\}$ := pdf of the scores for class k, with CDF $F_{k}(s)$ The classification of a new observation is obtained compraing the score sto a threshold t Furthermore, for mathematical convenience, let's consider the positive class (event detected) k = 0, and negative k = 1. In this setting we can define: Recall (aka Sensitivity, aka TPR): $F_{0}(t)$ (proportion of positive cases classified as positive) Specificity (aka TNR): $1 - F_{1}(t)$ (proportion of negative cases classified as negative) FPR (aka Fall-out): 1 - TNR= $F_{1}(t)$ The ROC curve is then a plot of $F_{0}(t)$ against $F_{1}(t)$. Setting $v = F_1(s)$, we can formally define the area under the ROC curve as: $$AUC =\int_{0}^{1} F_{0}(F_{1}^{-1}(v)) dv$$ Changing variable ($dv = f_{1}(s)ds$): $$AUC =\int_{ - \infty}^{\infty} F_{0}(s) f_{1}(s)ds$$ This formula can easiliy be seen to be the probability that a randomly drawn member of class 0 will produce a score lower than the score of a randomly drawn member of class 1. This proof is taken from: https://pdfs.semanticscholar.org/1fcb/f15898db36990f651c1e5cdc0b405855de2c.pdf @alebu's answer is great. But its notation is nonstandard and uses 0 for the positive class and 1 for the negative class. Below are the results for the standard notation (0 for the negative class and 1 for the positive class): Pdf and cdf of the score for negative class: $f_0(s)$ and $F_0(s)$ Pdf and cdf of the score for positive class: $f_1(s)$ and $F_1(s)$ FPR = $x(s) = 1-F_0(s)$ TPR = $y(s) = 1-F_1(s)$ $$\begin{align} \text{AUC} &= \int_0^1 y(x) dx\\ &= \int_0^1 y(x(\tau)) dx(\tau) \\ &= \int_{+\infty}^{-\infty} y(\tau) x'(\tau) d\tau \\ &= \int_{+\infty}^{-\infty} \big( 1-F_1(\tau) \big) \big( -f_0(\tau) \big) d\tau \\ &= \int_{-\infty}^{+\infty} \big( 1-F_1(\tau) \big) f_0(\tau) d\tau \end{align}$$ where $\tau$ stands for threshold. One can apply the interpretation in @alebu's answer to the last expression. The way to calculate AUC-ROC is to plot out the TPR and FPR as the threshold, $\tau$ is changed and calculate the area under that curve. But, why is this area under the curve the same as this probability? Let's assume the following: $A$ is the distribution of scores the model produces for data points that are actually in the positive class. $B$ is the distribution of scores the model produces for data points that are actually in the negative class (we want this to be to the left of $A$). $\tau$ is the cutoff threshold. If a data point get's a score greater than this, it's predicted as belonging to the positive class. Otherwise, it's predicted to be in the negative class. Note that the TPR (recall) is given by: $P(A>\tau)$ and the FPR (fallout) is given be: $P(B>\tau)$. Now, we plot the TPR on the y-axis and FPR on the x-axis, draw the curve for various $\tau$ and calculate the area under this curve ($AUC$). We get: $$AUC = \int_0^1 TPR(x)dx = \int_0^1 P(A>\tau(x))dx$$ where $x$ is the FPR. Now, one way to calculate this integral is to consider $x$ as belonging to a uniform distribution. In that case, it simply becomes the expectation of the $TPR$. $$AUC = E_x[P(A>\tau(x))] \tag{1}$$ if we consider $x \sim U[0,1)$ . Now, $x$ here was just the $FPR$ $$x=FPR = P(B>\tau(x))$$ Since we considered $x$ to be from a uniform distribution, $$P(B>\tau(x)) \sim U$$ $$=> P(B<\tau(x)) \sim (1-U) \sim U$$ \begin{equation}=> F_B(\tau(x)) \sim U \tag{2}\end{equation} But we know from the inverse transform law that for any random variable $X$, if $F_X(Y) \sim U$ then $Y \sim X$. This follows since taking any random variable and applying its own CDF to it leads to the uniform. $$F_X(X) = P(F_X(x)<X) =P(X<F_X^{-1}(X))=F_XF_X^{-1}(X)=X$$ and this only holds for uniform. Using this fact in equation (2) gives us: $$\tau(x) \sim B$$ Substituting this into equation (1) we get: $$AUC=E_x(P(A>B))=P(A>B)$$ In other words, the area under the curve is the probability that a random positive sample will have a higher score than a random negative sample.
In this lesson, we'll cover some of the fundamental principles and postulates of quantum mechanics. These principles are the foundation of quantum mechanics. The eigenvalues are the values that you measure in an experiment: for example, the position or momentum of a particle. Because the eigenvalues are what you measure, it wouldn't make physical sense if the eigenvalue of an observable had an imaginary part. In this lesson, we'll prove that the eigenvalue of any observable is a real number. The three operators—$\hat{σ}_x$, \hat{σ}_y\), and \hat{σ}_z\)—are associated with the measurements of the $x$, $y$, and $z$ components of spin of a quantum particle, respectively. In this lesson, we'll represent each of these three operators as matrices and solve for the entries in each matrix. These three matrices are called the Pauli matrices. In this lesson, we'll derive an equation which will allow us to calculate the wavefunction (which is to say, the collection of probability amplitudes) associated with any ket vector $\|\psi⟩$. Knowing the wavefunction is very important since we use probability amplitudes to calculate the probability of measuring eigenvalues (i.e. the position or momentum of a quantum system). Any ket vector $\|\psi⟩$ can be multiplied by a number $z$ (which, in general can be real or complex) to produce a new vector $\|\phi⟩$: $$z\|\psi⟩=\|\phi⟩.$$ In general, $z=x+iy$. Sometimes the number $z$ will just be a real number with no imaginary part, but in general it can have both a real and imaginary part. The complex conjugate of the number $z$ is represented by $z^$ and is obtained by changing the sign of the imaginary part of $z$; so $z^=x-iy$. Let’s look at some examples of complex numbers and their complex conjugates. Suppose that \z\) is any real number with no imaginary part: $z=x+(i·0)=x$. The complex conjugate of any real number is $z^=x-(i·0)=x$. In other words taking the complex conjugate $z^$ of any real number $z=x$ just gives the real number back. Suppose, however, that $z=x+iy$ is any complex number and we take the complex conjugate twice. Let’s see what we get. $z^$ is just $z^=x-iy$ (a new complex number). If we take the “complex conjugate of the complex conjugate,” we have $(z^)^=x+iy=z$. For any complex number $z$,$z^=z$ . If we multiply any complex number $z$ by its complex conjugate $z^$, we’ll get $$zz^=(x=iy)(x-iy)=x^2-ixy+ixy-i^2y^2=x^2+y^2.$$ For any complex number $z$ the product $z^z$ is always a real number that is greater than or equal to zero. This product is called the modulus squared and, in a very rough sense, represents the length squared of a vector in a complex space. We can write the modulus squared as $\|z\|^2=zz^$. From fig. # we can also see that any complex number can be represented by $z=x+iy=rcosθ+irsinθ=re^{iθ}$. The complex conjugate of this is $z^=x-iy=rcosθ+irsin(-θ)=re^{-iθ}$. The modulus squared of any vector in the complex plain is given by $\|z\|^2=zz^=(re^{iθ})(re^{iθ}=r^2$. If $\|z\|^2=r^2=1$ and hence $\|z\|=r=1$, then the magnitude of the complex vector is 1 and the vector is called normalized. Any vector $\|A⟩$ can be expressed as a column vector: $\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}$. To multiply $\|A⟩$ by any number $z$ we simply multiply each of the components of the column vector by $z$ to get $z\|A⟩=z\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}=\begin{bmatrix}zA_1 \\⋮ \\zA_N\end{bmatrix}$. We can add two complex vectors $\|A⟩$ and $\|B⟩$ to get a new complex vector $\|C⟩$. Each of the new components of $\|C⟩$ is obtained by adding the components of $\|A⟩$ and $\|B⟩$ to get $\|A⟩+\|B⟩=\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}+\begin{bmatrix}B_1 \\⋮ \\B_N\end{bmatrix}=\begin{bmatrix}A_1+B_1 \\⋮ \\A_N+B_N\end{bmatrix}=\|C⟩$. For every ket vector $\|A⟩=\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}$ there is an associated bra vector which is the complex conjugate of $\|A⟩$ and is given by $⟨A\|=\begin{bmatrix}A_1^ &... &A_N^\end{bmatrix}$. The inner product between any two vectors $\|A⟩$ and $\|B⟩$ is written as $⟨B\|A⟩$. The outer product between any two vectors $\|A⟩$ and $\|B⟩$ is written as $\|A⟩⟨B\|$. The rule for taking the inner product between any two such vectors is $$⟨B\|A⟩=\begin{bmatrix}B_1^ &... &B_N^\end{bmatrix}\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}=B_1^A_1\text{+ ... +}B_N^A_N.$$ Whenever you take the inner product of a vector $\|A⟩$ with itself you get $$⟨A\|A⟩=\begin{bmatrix}A_1^ &... &A_N^\end{bmatrix}\begin{bmatrix}A_1 \\⋮ \\A_N\end{bmatrix}=A_1^A_1\text{+ ... +}A_N^A_N.$$ We learned earlier that the product between any number $z$ (which can be a real number but is in general a complex number) and its complex conjugate $z^$ (written as ) is always a real number that is greater than or equal to zero. This means that each of the terms $A_i^A_i$ is greater than or equal to zero and, therefore, will always equal a real number greater than or equal to zero. Suppose we have some arbitrary matrix $\textbf{M}$ whose elements are given by $$\textbf{M}=\begin{bmatrix}m_{11} &... &m_{N1} \\⋮ &\ddots &⋮ \\m_{1N} &... &m_{NN}\end{bmatrix}.$$ To find the transpose of this matrix (written as $\textbf{M}^T$) we interchange the order of the two lower indices of each element. (Another way of thinking about it is that we “reflect” each element about the diagonal.) When we do this we get $$\textbf{M}^T=\begin{bmatrix}m_{11} &... &m_{1N} \\⋮ &\ddots &⋮ \\m_{N1} &... &m_{NN}\end{bmatrix}.$$ The Hermitian conjugate of a matrix (represented by $\textbf{M}^†$) is obtained by first taking the transpose of the matrix and then taking the complex conjugate of each element to get $$\textbf{M}^†=\begin{bmatrix}m_{11}^ &... &m_{1N^} \\⋮ &\ddots &⋮ \\m_{N1}^ &... &m_{NN}^*\end{bmatrix}.$$ We represent observables/measurable as linear Hermitian operators. In our electron spin example, the observable/measurable is given by the linear Hermitian operator $\hat{σ}_r$.
The Mathematical Ninja took one look at the list of the 10 coolest numbers and scoffed. "Those," he said, "are not the ten coolest numbers." "What are, then?" said the student. "Remember, nothing for cultural significance, so you can't have 42. And no physical constants, so $g$ is out." "I wouldn't dream of it," growled the Mathematical Ninja. "Here are the ten coolest numbers." "Six is the smallest semi-prime," said the Mathematical Ninja, "and the smallest perfect number." "I've heard of those," said the student. "The factors - 1, 2, and 3 - add up to the number itself!" "Correct," said the Mathematical Ninja. "And a semi-prime is a number with only two prime factors - factorising them is one of the big unsolved problems in cryptography. Plus, it's the size of the smallest non-Abelian group - where $ab$ doesn't necessarily equal $ba$." "That Abel - smart guy, huh?" "Abel? No good," said the Mathematical Ninja. "You're obsessed with triangles!" The Mathematical Ninja frowned. "This is not obsession. This is focus. And this particular number is the height of an equilateral triangle with unit sides. The simplest non-trivial shape, I would contend." The student, hearing the tone in his voice, decided not to argue. "You and your imaginary friends!" said the student. The Mathematical Ninja's eyebrows lowered like Roger Moore acting. "I see," he growled, "what you did there. I've gone for $i$ because it epitomises the triumph of notation over instinct: when they were first solving cubics, this $\sqrt{-1}$ showed up in some of the equations. Some said 'bah! can't be done;' others said 'let's roll with it,' and found the answers worked out." "Keep going, faith will come?" "Precisely." "That's clearly a cultural reference," said the student, wishing he'd brought his QI klaxen. "You haul 16 tonnes and what do you get?" "Depends on the coefficient of friction," said the Mathematical Ninja. "But no: I've picked 16 because it's $2^4$ and $4^2$, the only integer that's both $a^b$ and $b^a$ for $a \ne b$." "I prefer another day older and deeper in debt." "Oo, I know this one!" said the student. "It's in Pascal's Triangle, isn't it?" "It is," said the Mathematical Ninja. "Eight times, no less. As far as we know, it's the only number that appears more than six times. It's $^{14}C_{6},~^{14}C_{8},~^{15}C_{5},~^{15}C_{10},~^{78}C_{2},~^{78}C_{76},~^{3003}C_{1}$ and $^{3003}C_{3002}$." "Phew! That Pascal, pretty smart guy, huh?" "Pascal? No good," said the Mathematical Ninja. "Singmaster, no good either.1" "The deadliest surd," said the Mathematical Ninja, "and one of the few, possibly the only, that anyone's been murdered over." "There was that lottery scandal where you used to teach." "We do not talk about that. Instead, we talk about poor Hippasus, who proved - legend has it - to Pythagoras that $\sqrt{2}$ is not a rational number. Pythagoras - who believed order in the universe came from whole numbers and fractions - took the huff and threw him into the sea." "You can't write it as a fraction because they both need to be even?" The student knew, if he didn't mention this, he'd suffer a similar fate to Hippasus. "Pythagoras - pretty smart guy, huh?" "Pythagoras? No good," said the Mathematical Ninja. The Mathematical Ninja shrugged. "This one just comes up a lot. And it's easy to do sums with." "It's 0.7, less 1%," said the student. "Very close to $\frac{1}{2}\sqrt{2}$." "Straying into the cultural there, my friend," said the Mathematical Ninja, perhaps projecting a little. "But correct." "Graham's number holds the world record for the largest finite number ever used in a proof," said the Mathematical Ninja. "It's too large to be written using normal mathematical notation, and it's an upper bound on a constant in Ramsay Theory that's believed to be about 13." "It must be pretty close to infinity," said the student, unthinkingly. "Infinity looks at Graham's number and says 'molest me not with this pocket calculator stuff'," said the Mathematical Ninja. The student nodded, chastised. "It's turtles all the way up. So that Graham, pretty smart guy, huh?" "Graham? No good," said the Mathematical Ninja. "Pythagoras? No good. Pascal? No good. Abel? Singmaster? I spit me of them. No mathematicians anywhere any good except me. Paul Erdos and Colin Beveridge not bad. Not good, but not bad.2" "Beveridge's first law of MathsJam probability states '37%' is normally a good guess." "So that Beveridge… oh, you said he wasn't bad." The Mathematical Ninja nodded. "37% is usually a good guess because MathsJam probabilities often work out to $\frac{1}{e}$. That's why it's the second-coolest number of them all." "It's the angle in the middle of a circle," said the Mathematical Ninja. "You may know it as $2\pi$, but $\tau$ is better yet." "I thought it was 360!" said the student. "Get out of my shop," said the Mathematical Ninja.
My question is as follows: What methods can be used to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a certain functional equation. An example of a case where this applies is the following: Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following equation: $$f(x^{3})+f(y^{3})=(x+y)(f(x^{2})+f(y^{2})-f(xy)):\forall x, y\in\mathbb{R}$$ I'm curious as to whether there are general methods (or strategies) for solving this type of question, or whether questions like these should just be handled on a case-by-case basis. Thanks in advance.
Hello guys! I was wondering if you knew some books/articles that have a good introduction to convexity in the context of variational calculus (functional analysis). I was reading Young's "calculus of variations and optimal control theory" but I'm not that far into the book and I don't know if skipping chapters is a good idea. I don't know of a good reference, but I'm pretty sure that just means that second derivatives have consistent signs over the region of interest. (That is certainly a sufficient condition for Legendre transforms.) @dm__ yes have studied bells thm at length ~2 decades now. it might seem airtight and has stood the test of time over ½ century, but yet there is some fineprint/ loopholes that even phd physicists/ experts/ specialists are not all aware of. those who fervently believe like Bohm that no new physics will ever supercede QM are likely to be disappointed/ dashed, now or later... oops lol typo bohm bohr btw what is not widely appreciated either is that nonlocality can be an emergent property of a fairly simple classical system, it seems almost nobody has expanded this at length/ pushed it to its deepest extent. hint: harmonic oscillators + wave medium + coupling etc But I have seen that the convexity is associated to minimizers/maximizers of the functional, whereas the sign second variation is not a sufficient condition for that. That kind of makes me think that those concepts are not equivalent in the case of functionals... @dm__ generally think sampling "bias" is not completely ruled out by existing experiments. some of this goes back to CHSH 1969. there is unquestioned reliance on this papers formulation by most subsequent experiments. am not saying its wrong, think only that theres very subtle loophole(s) in it that havent yet been widely discovered. there are many other refs to look into for someone extremely motivated/ ambitious (such individuals are rare). en.wikipedia.org/wiki/CHSH_inequality @dm__ it stands as a math proof ("based on certain assumptions"), have no objections. but its a thm aimed at physical reality. the translation into experiment requires extraordinary finesse, and the complex analysis starts with CHSH 1969. etc While it's not something usual, I've noticed that sometimes people edit my question or answer with a more complex notation or incorrect information/formulas. While I don't think this is done with malicious intent, it has sometimes confused people when I'm either asking or explaining something, as... @vzn what do you make of the most recent (2015) experiments? "In 2015 the first three significant-loophole-free Bell-tests were published within three months by independent groups in Delft, Vienna and Boulder. All three tests simultaneously addressed the detection loophole, the locality loophole, and the memory loophole. This makes them “loophole-free” in the sense that all remaining conceivable loopholes like superdeterminism require truly exotic hypotheses that might never get closed experimentally." @dm__ yes blogged on those. they are more airtight than previous experiments. but still seem based on CHSH. urge you to think deeply about CHSH in a way that physicists are not paying attention. ah, voila even wikipedia spells it out! amazing > The CHSH paper lists many preconditions (or "reasonable and/or presumable assumptions") to derive the simplified theorem and formula. For example, for the method to be valid, it has to be assumed that the detected pairs are a fair sample of those emitted. In actual experiments, detectors are never 100% efficient, so that only a sample of the emitted pairs are detected. > A subtle, related requirement is that the hidden variables do not influence or determine detection probability in a way that would lead to different samples at each arm of the experiment. ↑ suspect entire general LHV theory of QM lurks in these loophole(s)! there has been very little attn focused in this area... :o how about this for a radical idea? the hidden variables determine the probability of detection...! :o o_O @vzn honest question, would there ever be an experiment that would fundamentally rule out nonlocality to you? and if so, what would that be? what would fundamentally show, in your opinion, that the universe is inherently local? @dm__ my feeling is that something more can be milked out of bell experiments that has not been revealed so far. suppose that one could experimentally control the degree of violation, wouldnt that be extraordinary? and theoretically problematic? my feeling/ suspicion is that must be the case. it seems to relate to detector efficiency maybe. but anyway, do believe that nonlocality can be found in classical systems as an emergent property as stated... if we go into detector efficiency, there is no end to that hole. and my beliefs have no weight. my suspicion is screaming absolutely not, as the classical is emergent from the quantum, not the other way around @vzn have remained civil, but you are being quite immature and condescending. I'd urge you to put aside the human perspective and not insist that physical reality align with what you expect it to be. all the best @dm__ ?!? no condescension intended...? am striving to be accurate with my words... you say your "beliefs have no weight," but your beliefs are essentially perfectly aligned with the establishment view... Last night dream, introduced a strange reference frame based disease called Forced motion blindness. It is a strange eye disease where the lens is such that to the patient, anything stationary wrt the floor is moving forward in a certain direction, causing them have to keep walking to catch up with them. At the same time, the normal person think they are stationary wrt to floor. The result of this discrepancy is the patient kept bumping to the normal person. In order to not bump, the person has to walk at the apparent velocity as seen by the patient. The only known way to cure it is to remo… And to make things even more confusing: Such disease is never possible in real life, for it involves two incompatible realities to coexist and coinfluence in a pluralistic fashion. In particular, as seen by those not having the disease, the patient kept ran into the back of the normal person, but to the patient, he never ran into him and is walking normally It seems my mind has gone f88888 up enough to envision two realities that with fundamentally incompatible observations, influencing each other in a consistent fashion It seems my mind is getting more and more comfortable with dialetheia now @vzn There's blatant nonlocality in Newtonian mechanics: gravity acts instantaneously. Eg, the force vector attracting the Earth to the Sun points to where the Sun is now, not where it was 500 seconds ago. @Blue ASCII is a 7 bit encoding, so it can encode a maximum of 128 characters, but 32 of those codes are control codes, like line feed, carriage return, tab, etc. OTOH, there are various 8 bit encodings known as "extended ASCII", that have more characters. There are quite a few 8 bit encodings that are supersets of ASCII, so I'm wary of any encoding touted to be "the" extended ASCII. If we have a system and we know all the degrees of freedom, we can find the Lagrangian of the dynamical system. What happens if we apply some non-conservative forces in the system? I mean how to deal with the Lagrangian, if we get any external non-conservative forces perturbs the system?Exampl... @Blue I think now I probably know what you mean. Encoding is the way to store information in digital form; I think I have heard the professor talking about that in my undergraduate computer course, but I thought that is not very important in actually using a computer, so I didn't study that much. What I meant by use above is what you need to know to be able to use a computer, like you need to know LaTeX commands to type them. @AvnishKabaj I have never had any of these symptoms after studying too much. When I have intensive studies, like preparing for an exam, after the exam, I feel a great wish to relax and don't want to study at all and just want to go somehwere to play crazily. @bolbteppa the (quanta) article summary is nearly popsci writing by a nonexpert. specialists will understand the link to LHV theory re quoted section. havent read the scientific articles yet but think its likely they have further ref. @PM2Ring yes so called "instantaneous action/ force at a distance" pondered as highly questionable bordering on suspicious by deep thinkers at the time. newtonian mechanics was/ is not entirely wrong. btw re gravity there are a lot of new ideas circulating wrt emergent theories that also seem to tie into GR + QM unification. @Slereah No idea. I've never done Lagrangian mechanics for a living. When I've seen it used to describe nonconservative dynamics I have indeed generally thought that it looked pretty silly, but I can see how it could be useful. I don't know enough about the possible alternatives to tell whether there are "good" ways to do it. And I'm not sure there's a reasonable definition of "non-stupid way" out there. ← lol went to metaphysical fair sat, spent $20 for palm reading, enthusiastic response on my leadership + teaching + public speaking abilities, brought small tear to my eye... or maybe was just fighting infection o_O :P How can I move a chat back to comments?In complying to the automated admonition to move comments to chat, I discovered that MathJax is was no longer rendered. This is unacceptable in this particular discussion. I therefore need to undo my action and move the chat back to comments. hmmm... actually the reduced mass comes out of using the transformation to the center of mass and relative coordinates, which have nothing to do with Lagrangian... but I'll try to find a Newtonian reference. One example is a spring of initial length $r_0$ with two masses $m_1$ and $m_2$ on the ends such that $r = r_2 - r_1$ is it's length at a given time $t$ - the force laws for the two ends are $m_1 \ddot{r}_1 = k (r - r_0)$ and $m_2 \ddot{r}_2 = - k (r - r_0)$ but since $r = r_2 - r_1$ it's more natural to subtract one from the other to get $\ddot{r} = - k (\frac{1}{m_1} + \frac{1}{m_2})(r - r_0)$ which makes it natural to define $\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$ as a mass since $\mu$ has the dimensions of mass and since then $\mu \ddot{r} = - k (r - r_0)$ is just like $F = ma$ for a single variable $r$ i.e. an spring with just one mass @vzn It will be interesting if a de-scarring followed by a re scarring can be done in some way in a small region. Imagine being able to shift the wavefunction of a lab setup from one state to another thus undo the measurement, it could potentially give interesting results. Perhaps, more radically, the shifting between quantum universes may then become possible You can still use Fermi to compute transition probabilities for the perturbation (if you can actually solve for the eigenstates of the interacting system, which I don't know if you can), but there's no simple human-readable interpretation of these states anymore @Secret when you say that, it reminds me of the no cloning thm, which have always been somewhat dubious/ suspicious of. it seems like theyve already experimentally disproved the no cloning thm in some sense.
When you draw a sinusoid in polar coordinates, the results can be beautiful. Depending on the parameters of the sinusoid, we can obtain many different curves. The ones I want to focus on in this post are the rose curves which, as you could've guessed it, resemble a flower. What we'll consider a rose curve will be an equation of the form $$r = a + b \cos (\frac{m}{n}\theta)$$ For simplicity, we'll assume that $a = 0$ and $b = 1$. Also, $\frac{m}{n}$ is a positive rational number, where $\gcd(m,n)=1$. Irrational numbers are not considered because the equation would result in something resembling a disc instead of a curve. (it doesn't have period, and it's not even a full disc as it doesn't go through all points of the disc). The $\cos$ function can be replaced with $\sin$ function, the only difference is that the curve will be rotated by $\frac{\pi n}{2m}$. Feel free to play around with the below interactive GeoGebra applet to get the feel how the curves look like. Simple Case Consider first the simpler case when $n = 1$. By experimentation we can see that the number of petals is $2m$ when $m$ is even, $m$ when $m$ is odd. This may be against intuition at first but starts to make sense once you recall what happens when a point in polar coordinates $(r,\theta)$ has negative distance $r$. We're still drawing $2m$ petals over the range $[0,2\pi)$, however, exactly half of them overlaps when $m$ is odd. Why is that so? Before we start talking about the number of petals we should define what a petal is. Working with ranges will make things unnecessarily harder so we'll define petal as a point $(r,\theta)$ where $r=1$ or $r=-1$. In other words, a petal is a point the furthest from the origin. This is a nice definition because there's only one value of $1$ and one value of $-1$ in one cycle of the $\cos$ (or $\sin$) function. What is more, these values are the maximum and minimum of the function so it's easy to see that they are the vertices of the petals. The task now changes to finding the number of maximums and minimums of function $\cos(m\theta)$ over the range $\theta \in [0,2\pi)$. The period of $\cos(m \theta)$ is $\frac{2 \pi}{m}$. The graph starts redrawing at $\theta = 2 \pi$, hence there are $m$ cycles before the graphs starts redrawing. Note that it is not exactly the period of the graph as the redrawing can happen earlier. For example, the graph of equation $r=\cos \theta$ is the same for $\theta \in [0,\pi)$ and for $\theta \in [0,2\pi)$. However, in this post, I'm going to simplify the situation and call $2\pi$ the period regardless if it starts redrawing at $2\pi$ or $\pi$. We established the number of cycles is $m$. In each cycle there's one minimum and one maximum, so the total number of petals drawn is $2m$. However, some of them overlap. To see that, consider a maximum point $(1,\theta)$. On the graph, this point is equivalent to minimum point $(-1,\theta+\pi)$. Therefore, the petal drawn around $\theta$ will be redrawn at $\theta+\pi$ if point $(-1,\theta+\pi)$ belongs to the graph. This only happens when $m$ is odd as can be seen on the rectangular graph. For more formal proof, see the next section where we'll show how it works for the more general case. Note that when $a \neq 0$ is an integer, we no longer reach any negative values, hence the number of petals will be $m$ -- the number of maximums. More General Case We'll now take a close look at function $$r = \cos (\frac{m}{n}\theta)$$ In rectangular graph, the function has period $\frac{2 \pi n}{m}$. However, the range of $\theta$ is no longer neccesarily $[0,2\pi)$. We need to find the period of the function in polar coordinates. Let's define the period as number of full rotations before the graph starts redrawing itself: the least integer $k > 0$ such that for all $\theta$ $r(\theta)=r(\theta+2k\pi)$. Note that $k$ does not neccesarily has to exist, as would be the case if we were considering irrational numbers. The logic for the definition is that since we start from some point $(r(0),0)$, we want to end up at the same point $(r(0),2k\pi)$. Anyway, in our case: $$\cos(\frac{m}{n}(\theta+2k\pi))=\cos(\frac{m}{n}\theta)$$ Since $\cos$ is periodic and value for $1$ is appearing once per cycle, we can substitue $\theta = 0$ to simplify calculations. $$\cos(\frac{m}{n}(2k\pi))=1$$ This equation is equivalent to the following for any $\ell \in \mathbb{Z}$: $$\frac{m}{n}(2k\pi)=2\ell \pi$$ After some algebraic manipulations we obtain $$k = \frac{n}{m}\ell$$ Which will be the smallest positive integer for $\ell=m$, hence $k=n$ and the period of our polar function is $2\pi n$. Since each cycle of the rectangular function has length $\frac{2\pi n}{m}$, there will be $m$ cycles. We can draw it on paper by repeating the cosine graph $m$ times and marking the last point as $2\pi n$. With the interval established, we can count the number of petals as the number of solutions to two equations $$\cos(\frac{m}{n}\theta)=\pm 1$$ on interval $\theta \in [0,2\pi n)$. After some basic algebraic manipulations we have $\theta = \frac{2kn}{m}\pi$ or $\theta=\frac{n}{m}(2k+1)\pi$. They both lie in the interval when $k \in \{0,1,\ldots,m-1\}$, so there are $2m$ solutions in total. We're not done yet, however, since we're also counting the overlapping petals. There are two ways the two petals can overlap. The first case is when point $(1,\theta)$ is equivalent to $(1,\theta+2k\pi)$ for some $2k\pi<2n\pi$. However, that's impossible since we already determined the period to be $2n\pi$. The other case is when point $(1,\theta)$ is equivalent to $(-1,\theta+(2k+1)\pi)$. Note that since the period is now greater than $2\pi$ we have to consider shifting by odd multiples of $\pi$ which we can interpret as $2k+1$ $180$ degree rotations ending at the same point (remember what $-r$ means in polar coordinates!). Notice that $\theta+(2k+1)\pi$ is one of the solutions to the second equation (for minimum). Therefore, if the equivalent point to a maximum exists, it is a corresponding minimum, and vice versa. For this reason, we'll only consider the points $(1,\theta)$ when finding equivalence, as the steps are very similar for $(-1,\theta)$. Anyway, two petals are equivalent if the following condition is met: $$\cos(\frac{m}{n}(\theta+(2k+1)\pi))=-1$$ In this case $\theta = \frac{2\pi n}{m}a$ for $a \in \{0,1,\ldots,m-1\}$, as we're moving from point $(1,\theta)$ and we already solved for $\theta$ earlier. However, we can't shift too much to not exceed the period $2 \pi n$, but we'll worry about that later. $$\cos(\frac{m}{n}(\frac{2\pi n}{m}a+(2k+1)\pi))=-1$$ $$\frac{m}{n}(\frac{2\pi n}{m}a+(2k+1)\pi)=\pi+2 \ell \pi$$ for some $\ell \in \mathbb{Z}$. After some simplification steps we end up with the following equation: $$2a+(2k+1)\frac{m}{n}=2\ell +1$$ Since $\gcd(m,n)=1$, the term $(2k+1)\frac{m}{n}$ is the only one that has risk of not being an integer. Therefore, $2k+1$ must be a multiple of $n$. However, it can't be any other multiple than $1\cdot n$ otherwise we would end up at point beyond $2\pi n$. So $$2k+1=n$$ It immediately follows that $n$ must be odd. Notice that this must mean there is only one candidate which can be our equivalent petal: $\theta+\pi n$. I find it interesting that we're shifting by exactly half the period. This means the equivalent points can happen only in the second half of the period, hence the graph can only start redrawing itself after either $n\pi$ or $2\pi n$. However, we're not done yet. We still don't know which of our $m$ maximums have equivalent petals. After substituting $n=2k+1$ we are left with $$2a+m=2\ell +1$$ So, the equivalent points can exist only if $2a+m$ is odd. If $m$ is even, the whole expression becomes even, so $m$ can't be even. Therefore $m$ must be odd and indeed it is easy to check that $2a+m$ is odd for odd $m$. Last thing we need to do is to check when we exceed the period. $a \in \{0,1,\ldots,m-1\}$ but we can't shift the angle too much or we'll exceed the period $2\pi n$ and the equivalent petal won't matter to us. So $$\frac{2a\pi n}{m}+n\pi < 2n\pi$$ $$a < \frac{m}{2}$$ Remember that $m$ is odd, so $a \le \lfloor \frac{m}{2} \rfloor$. Summing up, we have a bijection from $\lfloor \frac{m}{2} \rfloor + 1$ maximums in the first half to minimums in the second half of the period. By following similar steps, we also have a bijection from $\lfloor \frac{m-1}{2} \rfloor$ minimums from the first half to the maximums in the second half. These describe two disjoint sets of equivalent petals which together sum up to $m$. $\blacksquare$ Conclusion Rose curve $r = \cos(\frac{m}{n}\theta)$ where $\gcd(m,n)=1$, $m,n>0$: Has period $n\pi$ when both $m$ and $n$ are odd, and it has $m$ petals. Has period $2n\pi$ when either $m$ or $n$ is even, and it has $2m$ petals. Think about what happens to the equation $$r = a + b \cos (\frac{m}{n}\theta)$$ when $a \neq 0$ or $b \neq 1$. Now, don't cite me on that but here's my intuition: $b$ only stretches the values (including maximums and minimums), so the number of petals should remain the same, at least according to the definition of the furthest points from origin. Note, however, there start to appear "smaller petals" as we play with values of $a$ and $b$. When $a$ is large enough to remove $x$-intercepts in the rectangular graph, the number of petals stops to double regardless of value of $m$ and $n$. This might be worth further investigation in the future but for now that's all.
( This may be related to NFAs with more than one initial state ) Consider a dfa $M$ with alphabet $\Sigma$ and states $Q$, typically also characterized by a specific initial state $q_0\in Q$. As the above-cited question points out, nfa's that nondeterministically select among several possible initial states can be converted to equivalent (wrt to their accepted language) nfa's with a unique initial state. And I'm asking if the same is true of dfa's, as follows. Instead of $M$'s input being only a tape/string $s\in\Sigma^\ast$, let its input be $(q_0,s)\in Q\times\Sigma^\ast$, i.e., you give it the initial state $q_0\in Q$ that you want it to start in, as well as a string. I'd think each possible $q_0\in Q$ defines a different language, call it $L_{q_0}\subseteq\Sigma^\ast$, accepted by $M$. But $Q$ can possibly be partitioned into equivalence classes, where all the states in a class accept the same $L$. But if there's more than one such class, then no single dfa with a unique $q_0$ starting state could represent this $M$. So, assuming that's more-or-less correct, my additional question is how to characterize the equivalence classes. That is, given $M$'s transition function, $\delta:\Sigma\times Q\to Q$, how would you determine whether $q\equiv q^\prime$ belong to the same class (besides, that is, individually working out the language each accepts)? Is there a more elegant way to do that?
I'll be updating this as I find answers to your other questions. First question Geiger and Marsden actually did several experiments all very related (and all explained at this website). The one related to your question was done in 1913, to prove relationships that Rutherford calculated and published in a 1911 paper (The Scattering of α and β Particles by Matter and the Structure of the Atom). Geiger and Marsden didn't know what the positive charge of the nucleus of their metals was (they had only just discovered the nucleus after all) but they assumed it was proportional to the atomic weight. They specifically tested whether it was proportional to the atomic weight squared. So what they did was use the apparatus pictured below. They covered the holes in the disc with foils of gold, tin, copper, and aluminum and measured each foil's stopping power by equating it with an equivalent thickness of air. They counted the number of scintillations per minute that the foil produced on the screen. They then divided the scintillations per minute by the foil's air equivalent, and then divided again by the square root of the atomic weight. Therefore, Geiger and Marsden obtained the fixed number of scintillations a fixed number of atoms produces for each metal. Then, for each metal, they divided this number by the square of the atomic weight and found that the ratios were more or less the same. They therefore proved that $s ∝ Q_n^2$ (where $Q_n$ is the positive charge of the atomic nucleus). Second question For your second question, about Moseley's experiment, yes, they were characteristic x-rays. This website gives more information about Moseley's law and this website gives more information about Moseley himself and his other achievements (fun fact: he predicted the existence of element 61, which ended up being named promethium). In answer to the second part of your second question, the spectra of light emitted by atoms is proportional to the square of $Z$, the charge on their nucleus (in the Bohr model of the atom). Moseley was able to confirm that the spectra of light emitted was indeed proportional to $Z$, and he formulated Moseley's law: $${\sqrt f}=k_{1}\cdot \left(Z-k_{2}\right)$$ Where $f$ is the of the main, or $K$ x-ray emission line, and $k_1$ and $k_2$ are constants depending on the type of line. Third question For your third question, he used the beryllium to create radiation. He then aimed the radiation at paraffin wax, and since paraffin wax has a high hydrogen content and therefore offers a target dense with protons, and neutrons have almost equal mass, the protons easily scattered when the radiation hit them. Chadwick looked at the distance the protons scattered and how the radiation impacted atoms of various gases and concluded that the radiation was made up of uncharged particles with around the same mass as the proton - aka the neutron. Now, for why the beryllium was emitting neutrons. (See this website for more information; a summary is given here.) $^9Be$ releases more neutrons than it absorbs - this particular isotope under goes an (n, 2n) reaction, which can be described below as $$\frac {9}{4}Be + n → 2(\frac {4}{2}He) + 2n$$ Neutrons can also be liberated when beryllium nuclei are struck by energetic alpha particles and when beryllium is under bombardment by gamma rays. Fourth question I'm not quite sure I understand this question. Yes, mass spectrometers calculate the charge to mass ratio, but they are also detected by a mechanism such as an electron multiplier and the atoms/ions in the sample can be correlated with known fragmentation patterns. This website and this website might be helpful to you; especially the second link. Hope this helps!
A complex scalar field that describes a quantum mechanical system. The square of the modulus of the wave function gives the probability of the system to be found in a particular state. DO NOT USE THIS TAG for classical waves. The wavefunction is a complex scalar field that describes a quantum mechanical system. The square of the modulus of the wave function gives the probability of the system to be found in a particular state. In the Schrödinger Wave formulation of Quantum Mechanics, the wavefunction can be determined by the schrodinger-equation, which, in it's most general form, can be stated as: $$\hat H\|\Psi\rangle=i\hbar\frac{\mathrm d}{\mathrm d t}\|\Psi\rangle$$ In the case of a "Euclidean Hamiltonian" given by the operator $\hat H=\frac{\hat P^2}{2m}+\hat U$, this becomes: $$\left(\frac{\hat P^2}{2m}+\hat U\right) \|\Psi\rangle=i\hbar\frac{\mathrm d}{\mathrm d t}\|\Psi\rangle$$ Since the momentum operator $\hat P$, in the position bases, is $-i\hbar\nabla$, the Schrödinger equation becomes $$\left(-\frac{\hbar^2}{2m}\nabla^2+U\right)\Psi= i\hbar\frac{\partial\Psi}{\partial t} $$ with $\Psi=\langle x\|\Psi\rangle$. This is known as the time-independent Schrödinger equation. Note, that as the Hamiltonian used was Euclidean, this equation is in fact non-relativistic. The relativistic version of this equation in Relativistic Quantum Mechanics (and also in quantum-field-theory, but there it describes spin-1/2 fields) is the dirac-equation. The wavefunction also appears in Feynman's path-integral formulation of Quantum Mechanics. In the Path Integral formulation, a functional, called the phase is associated with each path: $$\phi = A e^\frac{iS}{\hbar} $$ The Kernel or the Matrix Element, is the path integral of this phase. $$K(x ) =\int\phi\mbox{ } \mathcal{D}x $$ The wavefunction, finally, is given by: $$\Psi(x)=\int_{-\infty}^\infty \left(K(x,x_0)\Psi(x_0) \right) \mbox{d} x_0 $$ It is often surprising to many that the absolute value of the phase squared, $\|\phi\|^2$, is constant for all paths, at $A^2$. However, this actually makes sense, as the position of the particle is initially completely well-defined, so Heisenberg's Uncertainty Principle tells us that we would have no idea about the momentum, and thus no idea about it's future position. However, the next moment, you know absolutely nothing about it's momentum, and so on. This process coarse-grains a particular path, the classical path, which means it is much more probable than the other paths. The mathematical description of this can be obtained by standard procedures (c.f. Feynman, Hibbs, Styer "Quantum Mechanics and Path Integrals", pg 77 - 79) and the final result is the Schrödinger's Equation. DO NOT USE THIS TAG for classical waves. Use the waves tag instead.
ISSN: 1078-0947 eISSN: 1553-5231 All Issues Discrete & Continuous Dynamical Systems - A October 2004 , Volume 11 , Issue 4 Special issue on 'Hamiltonian Systems and Applications' Guest editors: Amadeu Delshams and Antonio Giorgilli Select all articles Export/Reference: Abstract: We describe a method for studying the existence and the linear stability of branches of periodic solutions for a dynamical system with a parameter. We apply the method to the planar restricted 3-body problem extending the results of [A]. More precisely, we prove the existence of some continuous branches of periodic orbits with the energy or the masses of the primaries as parameters, and provide an approximation of the orbits with rigorous bounds. We prove the linear stability or instability of the orbits. Abstract: We consider an example of singular or weakly hyperbolic Hamiltonian, with $3$ degrees of freedom, as a model for the behaviour of a nearly-integrable Hamiltonian near a simple resonance. The model consists of an integrable Hamiltonian possessing a $2$-dimensional hyperbolic invariant torus with fast frequencies $\omega/\sqrt\varepsilon$ and coincident whiskers, plus a perturbation of order $\mu=\varepsilon^p$. We choose $\omega$ as the golden vector. Our aim is to obtain asymptotic estimates for the splitting, proving the existence of transverse intersections between the perturbed whiskers for $\varepsilon$ small enough, by applying the Poincaré-Melnikov method together with a accurate control of the size of the error term. The good arithmetic properties of the golden vector allow us to prove that the splitting function has 4 simple zeros (corresponding to nondegenerate critical points of the splitting potential), giving rise to 4 transverse homoclinic orbits. More precisely, we show that a shift of these orbits occurs when $\varepsilon$ goes across some critical values, but we establish the continuation (without bifurcations) of the 4 transverse homoclinic orbits for all values of $\varepsilon\to0$. Abstract: We consider a singular or weakly hyperbolic Hamiltonian, with $n+1$ degrees of freedom, as a model for the behaviour of a nearly-integrable Hamiltonian near a simple resonance. The model consists of an integrable Hamiltonian possessing an $n$-dimensional hyperbolic invariant torus with fast frequencies $\omega/\sqrt\varepsilon$ and coincident whiskers, plus a perturbation of order $\mu=\varepsilon^p$. The vector $\omega$ is assumed to satisfy a Diophantine condition. We provide a tool to study, in this singular case, the splitting of the perturbed whiskers for $\varepsilon$ small enough, as well as their homoclinic intersections, using the Poincaré--Melnikov method. Due to the exponential smallness of the Melnikov function, the size of the error term has to be carefully controlled. So we introduce flow-box coordinates in order to take advantage of the quasiperiodicity properties of the splitting. As a direct application of this approach, we obtain quite general upper bounds for the splitting. Abstract: This paper describes the global flow of homogeneous polynomial potentials of degree 3 for negative and positive energy. For the negative energy case a blow up of McGehee type is enough to get the complete picture of the flow. In the positive energy case, McGehee blow up fails to give global information about the flow, but comparing with a separable case we are able to obtain all the possible asymptotic behavior of solutions, whenever the coefficients of the normal form of the potential are positive. Abstract: In this note we compare the frequencies of the motion of the Trojan asteroids in the Restricted Three-Body Problem (RTBP), the Elliptic Restricted Three-Body Problem (ERTBP) and the Outer Solar System (OSS) model. The RTBP and ERTBP are well-known academic models for the motion of these asteroids, and the OSS is the standard model used for realistic simulations. Our results are based on a systematic frequency analysis of the motion of these asteroids. The main conclusion is that both the RTBP and ERTBP are not very accurate models for the long-term dynamics, although the level of accuracy strongly depends on the selected asteroid. Abstract: We revisit the celebrated model of Fermi, Pasta and Ulam with the aim of investigating, by numerical computations, the trend towards equipartition in the thermodynamic limit. We concentrate our attention on a particular class of initial conditions, namely, with all the energy on the first mode or the first few modes. We observe that the approach to equipartition occurs on two different time scales: in a short time the energy spreads up by forming a packet involving all low--frequency modes up to a cutoff frequency $\omega_c$, while a much longer time is required in order to reach equipartition, if any. In this sense one has an energy localization with respect to frequency. The crucial point is that our numerical computations suggest that this phenomenon of a fast formation of a natural packet survives in the thermodynamic limit. More precisely we conjecture that the cutoff frequency $\omega_c$ is a function of the specific energy $\epsilon = E/N$, where $E$ and $N$ are the total energy and the number of particles, respectively. Equivalently, there should exist a function $\epsilon_c(\omega)$, representing the minimal specific energy at which the natural packet extends up to frequency $\omega$. The time required for the fast formation of the natural packet is also investigated. Abstract: Fermi-Pasta-Ulam lattice is a classical mechanical system of an infinite number of discrete particles on a line. Each particle is assumed to interact with the nearest left and right neighbors only. We construct travelling waves in the system assuming that the potential has a singularity at zero. The waves appear near the hard ball limit. Abstract: We give a computer-assisted proof for the existence of a renormalization group fixed point (Hamiltonian) with non-trivial scaling, associated with the breakup of invariant tori with rotation number equal to the golden mean. Readers Authors Editors Referees Librarians More Email Alert Add your name and e-mail address to receive news of forthcoming issues of this journal: [Back to Top]
To which extent the adjunction $F\dashv N_\omega$ generated by the $\omega$-nerve described at $n$Lab - oriental (obtained as a particular instance of the nerve-realization paradigm) is linked to the adjunction generated by the functor $O_{[\Theta]}\colon \Theta\to \textbf{Str}\text{-}\omega\text{-}\mathbf{Cat}$ (Joyal's $\Theta$-category), $$\text{Lan}_y(O_{[\Theta]})\dashv N_{[\Theta]},$$ where the functor $N_{[\Theta]}\colon \textbf{Str}\text{-}\omega\text{-}\mathbf{Cat} \to [\Theta^{op}, \mathbf{Sets}]$ sends $C\in \textbf{Str}\text{-}\omega\text{-}\mathbf{Cat}$ to the presheaf $\textbf{Str}\text{-}\omega\text{-}\mathbf{Cat}(O(-), C)$? Is there any reference to learn about, and quote properly, affinities and differences between the two? As you remark, the cellular and simplicial nerves arise from the functors $J \colon \Theta \longrightarrow \omega\text{-Cat}$ and $O \colon \Delta \longrightarrow \omega\text{-Cat}$ respectively, where $J$ is the full inclusion of Joyal's cell category and $O$ is Street's orientals functor. A significant difference between these two is that $J$ is dense, and hence the nerve functor $\omega\text{-Cat} \longrightarrow [\Theta^{op},\text{Set}]$ is fully faithful, whereas this is not true of $O$ and the simplicial nerve induced by the orientals. (For a reference, this comment is made in the introduction to Section 1 of Berger's ' A cellular nerve for higher categories'). Note however that the modification of the simplicial nerve which lands in stratified simplicial sets is fully faithful (see Verity's ' Complicial sets...', whose introduction contains an account of the history of this nerve). The cellular nerve has these nice properties because it arises from general theory; I mean the nerve theorem (see for instance the Introduction and Theorem 1.10 of Berger, Melliès & Weber's ' Monads with arities...'), which applies upon recognising the objects of $\Theta$ as the free $\omega$-categories on the canonical arities for the free $\omega$-category monad on globular sets. Sadly, there is no such nice theoretical description of the orientals functor; the best we can give is an explicit construction using parity complexes or similar. Yet there is still another description of the simplicial nerve of a 2-category (see section 10 on nerves in Lack's ' A 2-categories companion'). The composite functor $\Delta \longrightarrow \text{Cat} \longrightarrow 2\text{-Cat}_{\text{nlax}}$, where $2\text{-Cat}_{\text{nlax}}$ is the category of 2-categories and normal lax 2-functors, is dense and so gives a fully faithful nerve functor $2\text{-Cat}_{\text{nlax}} \longrightarrow [\Delta^{op},\text{Set}]$ which agrees (up to duality) on $2\text{-Cat}$ with the nerve induced by the orientals. So we could hope that, if we had a category $\omega\text{-Cat}_{\text{nlax}}$ whose morphisms are "normal lax $\omega$-functors", the composite functor $\Delta \longrightarrow \text{Cat} \longrightarrow \omega\text{-Cat}_{\text{nlax}}$ would be dense and give a fully faithful nerve functor $\omega\text{-Cat}_{\text{nlax}} \longrightarrow [\Delta^{op},\text{Set}]$ extending the simplicial nerve given by the orientals. This suggests the tentative theoretical description of the $n$-th oriental as the "normal lax $\omega$-functor classifier" of $[n]$. However, I do not know of any definition of normal lax $\omega$-functor, which would likely make use of the orientals anyway. With regard to their affinities, I am not aware of any study in the literature. At the heart of the comparison is the module $\omega\text{-Cat}(J,O) \colon \Theta \nrightarrow \Delta$, corresponding to the functor $\Delta \longrightarrow [\Theta^{op},\text{Set}]$ that gives the cellular nerves of the orientals. One could hope that this module has a nice combinatorial description. Perhaps someone out there has considered this, maybe in the comparison of the models for higher categories based on weak complicial sets and cellular sets. Important edit: A few years ago, Dimitri Ara mentioned to me in private correspondence that he had written a bruteforce program to find objecta in $\Theta$ that do not admit an orientation that can be coherently pasted together. He told me way back then that they found an example in low dimensions very quickly. I do not know if this ended up in any of his published articles yet, but definitively, there is no such extension. He and Maltsiniotis ended up defining a non-full subcategory of $\Theta$ that was coherently orientable, but I don't know the definition. It turns out, due to a series of papers by Richard Steiner and then a later paper by Dimitri Ara and Georges Maltsiniotis, that there doesn't appear to be a meaningful definition for an 'oriental' for any given object of the category $\Theta$. What Ara and Maltsiniotis did figure out was the notion of a lax join of strict $\omega$-categories using Steiner's representation of pasting diagrams as special chain complexes with some additional structure. Essentially the construction is given on "loop-free pasting diagrams", that is to say, directed complexes in the sense of Steiner (which happen to include the pasting diagrams that define $\Theta$ and also the simplicial orientals), then extend it to the category of strict $\omega$-categories by merit of a specialized version of the Day Convolution theorem. On the chain complexes underlying the directed complexes, this is given by the desuspension of the tensor product of the suspensions of two directed complexes, that is, $$A\star B = \Sigma^{-1}(\Sigma A \otimes \Sigma B)$$. In particular, the orientals are obtained by the formula $$\mathcal{O}[n+1] = \mathcal{O}[n] \star [0]$$. This is all contained in the paper of Ara and Maltsiniotis (link) Their definition of the join can also be extended easily to the category of cellular sets using the much simpler (weighted) Day Convolution construction for presheaves. That is, for cellular sets $S,T \in \operatorname{Psh}(\Theta)$, the join $$S\star T (-) = \int^{[a],[b]\in \Theta^+} S^+_a\times T^+_b \times \operatorname{Hom}_{\operatorname{DirComp}^+}( - , [a]\star[b]) $$ where the $+$ means that we're looking at augmented Directed Complexes, augmented cellular sets, and the augmented Disk category $\Theta$. This essentially means attaching an initial object to the categories. There are several other nerves that one can take with respect to different cocellular objects $\Theta\to \operatorname{Str-\omega-Cat}$, but it is important to note that the naive version induced by the fully faithful embedding of $\Theta$ in $\operatorname{Str-\omega-Cat}$ does not induce a Quillen adjunction between strict $\omega$-categories and the various models for weak $\omega$-categories. It turns out that the 'correct' nerve to induce the Quillen adjunction is something a bit more complicated and involves first applying Metayer's resolution by polygraphs to the objects of $\Theta$. Edit: Andrea Gagna informed me by personal correspondence that it appears to be nontrivial and thusfar unproven that the lax join as described above for cellular sets will be associative (and similarly the lax tensor product's extension to cellular sets). The extension of the join is locally biclosed (similarly the extension of the tensor product is truly biclosed), but without further work, the associativity and coherence axioms for monoidal catagories remain to be demonstrated. For example, Andrea gave an interesting example that Dendroidal sets equipped with the Day Convolution-induced tensor product from the tensor product of operads fails to be associative (see the erratum to this paper of Cisinski and Moerdijk), so it's not just free. I have geometric/combinatorial reasons to expect that both will be associative for cellular sets, but it is, as far as I know, an open, if peripheral, question.
Four thousands of years since Aristotle's time, we thought that things that are stationary tend to stay stationary and that things that are moving tend to slow down. This is reasonable since it agrees with out everyday experiences. For example if I push a rock or throw something and it starts moving, it will tend to slow down and eventually come to rest. Isaac Newton stated his first law of motion in his Principia. This law states that if an object is moving at a constant velocity $\vec{v}$, it will tend to keep moving in a straight line at the constant velocity $\vec{v}$ unless it is acted upon by an external force. Contrary to what Aristotle and his successors believed, objects moving tend to keep on moving! For example, if I threw a rock in a cosmic void (vast regions of empty space which can extend for millions of light years), that rock would move in a straight line at a constant velocity $\vec{v}$ practically forever—at least until it reached the nearest galaxy. Also, if you instead simply just let go of the rock without budging it, its velocity would stay as $\vec{v}=0$ practically forever. But why is it that if I threw a rock on Earth's surface, it would eventually slow down and hit the ground? To answer this question, we need to use Newton's second law of motion which he published in his Principia which states $$\sum{\vec{F}}=m\vec{a},$$ where $\sum{\vec{F}}$ is the total force acting on the object, $m$ is the object's mass, and $\vec{a}$ is the object's acelleration. All objects which we'll be encountering in the next several lessons have some positive amount of mass. Thus, according to the equation $\sum{\vec{F}}=m\vec{a}$ where $\vec{a}=\frac{d\vec{v}}{dt}$, the only way that an object can acellerate is if some net force is acting upon it. And when an object acellerates by an amount $\vec{a}$, that object's velocity $\vec{v}$ will change with time according to the derivitive $\frac{d}{dt}\vec{v}$. The reason why when object's are thrown on Earth they slow down and eventually come to rest is because forces are acting upon them. These forces include the Earth's gravity and also the force due to air friction. Newton's second law, at first glance, might appear to lead to a contradiction. When you push on a brick wall, you are clearly applying a force to that wall. Why doesn't that wall accellerate then? This question can be answer using Newton's third law which states that: Whenever an object of mass $m_1$ applies a force $\vec{F}_{1,2}$ to a mass $m_2$, the object of mass $m_2$ applies an equal-and-opposite force $\vec{F}_{2,1}=-\vec{F}_{1,2}$ on the object of mass $m_1$. Therefore, according to Newton's third law, if I push on a wall with a force of $+5N$ to the right, the wall will exert an equal-and-opposite force of $-5N$ on my hand to the left. Since the net force on the hand-wall system is zero, the net acceleration $\vec{a}$ of the system is zero. Thus, both your hand and the wall remain stationary. This article is licensed under a CC BY-NC-SA 4.0 license.
In fractional reserve banking commercial banks create money when they make loans. When these loans are paid back the account is zeroed, the created money disappears, but the bank is still entitled interest. Where does the money to pay the interest come from? Does the central bank necessarily have to inflate the currency to pay it? Does money in circulation generally cover it? Repayment of interest does require an expansion of the money supply, but not in a way that is inflationary. Consider first the way that commercial banks make loans. The whole purpose of loans is to borrow against future output— so if you imagine a firm that wants to purchase a machine that will allow it to produce more stuff— and thus have greater future income— the firm may go to a commercial bank and ask to take out a loan. The commercial bank will price that loan at a premium to the risk-free interest rate that reflects the risk of the loan, so that in expectation (taking into account that some loans will be repaid only in part or not at all), banks making riskier loans will be repaid at a slightly higher rate than ones making safer loans. So firms will be offered an interest rate, and if that rate is less than their expected return on their investment, they'll make the investment. Notably, in this case, they're only borrowing money and investing because their future output will be higher. So when all of this works out (that is, when there isn't a credit bubble in which people are borrowing against future output that won't materialize), future output $Y$ (i.e., GDP) will be higher by an amount that is the amount of additional money $M$ required to repay the interest on the loans. $$ \% \Delta Y \geq \% \Delta M $$ Now if you recall the relationship between the money stock $M$, output $Y$, the price level $P$, and money velocity $V$: $$ PY=MV \rightarrow P=\frac{MV}{Y} $$ It's easy to see that if $V$ is constant, then: $$ \% \Delta P = \frac{\% \Delta M}{\% \Delta Y}$$ Which, when combined with the observation that $ \% \Delta Y \geq \% \Delta M $, implies that an expansion of the money supply that is merely sufficient to allow for repayment of interest would actually be deflationary. This should make perfect sense: in expectation, output must increase by more than the amount captured by banks in the form of interest, otherwise firms in the real economy would never bother investing in fixed capital. So the money supply will in all likelihood increase by an amount strictly greater than the level required to repay interest, yet without any increase in inflation. Loans to consumers are similar, so I won't give them a full treatment. Consumers are borrowing against their own future output, and foreclosures are generally priced in to the interest rate on a loan. You'll note that I left out the case where in the aggregate, future output is less than that required to repay outstanding loans. This is intentional, as it's worth treating as a separate question. The key thing to note is that the interest paid does not simply accumulate at the bank. Nobody runs a business in order to just make a big pile of money to keep in a safe. The banks spend the money back into the economy through staff wages, running costs and dividend payments to shareholders. All these outflows of money will ultimately be spent on real goods and services produced by the rest of the economy. This flow of money is a source of money for the interest payments. You can get an idea of what is going on by imagining an economy with a fixed money supply. The figure below shows a hypothetical flow of the money in the economy. The flow marked "trade" and is simply the money circulating back an forth between households and industry as people earn money and spend it on what has been produced in the factories. In a steady state, the rate of flow of money being created as new loans (shown as "loans" in the diagram) will be equal to the rate of flow of money being paid back in principal repayments (i.e. before interest payments). At the same time the loan interest payments are equal to the spending by banks (those staff wages, running costs and dividend payments to shareholders). As you can see, the flows can balance. Nothing is broken, no new money needs to be added to pay the interest. The system can continue indefinitely. The issue regarding the source of macroeconomic interest (and profits) appears to be unsettled among economists. A free paper on this issue, entitled “What is the Source of Profit and Interest? A Classical Conundrum Reconsidered,” by Gunnar Tomasson and Dirk J. Bezemer, dated January 29, 2010, and posted March 11, 2010, can be found online at https://mpra.ub.uni-muenchen.de/21292/. Personally, although I have not exhaustively researched this issue or economists’ attempts to address it, of the explanations I have studied, I believe that the monetary-circuit approach of Professor Louis-Philippe Rochon most plausibly resolves the conundrum by considering that, in firms’ investment cycles, a cash outflow required for the purchase of capital goods and financed by long-term bank loans occurs in the first period of production in the investment cycle while long-term bank loans may be paid back over multiple periods of production until the end of the investment cycle. Accordingly, based on my understanding of Professor Rochon's view, the central bank does not necessarily have to inflate the currency to pay macroeconomic interest, and money in circulation can generally cover it. Interest is , the reward for a service offered: the renting of (here, financial) capital. From this aspect, it is no different from say, a salary earned for a month's work. This is income too. So, income How salaries are paid?Does the central bank necessarily have to inflate the currency in order for them to be paid? Does money in circulation generally cover it? and in general HowDoes the central bank necessarily have to inflate the currency in order for it to be paid? Does money in circulation generally cover it? any kind of incomeis paid? The answer, simply and perhaps boringly, is, "it depends". On whether the existing money supply covers adequately the transactions needed for the existing level of total income. If interest income increases while salaries fall, it may be the case that no additional money is needed, for example. Or it may be the case that both kinds of income increase, but velocity of money also increases - so again, no need for extra money. Perhaps the most compelling evidence that fractional banking must be supported by a constant expansion of the monetary base by the central bank, is that the Fed while maintaining a somewhat target Fed Funds Rate almost always experiences growth in MB. The demand to pay interest spikes up the interbank lending rate which incites the Fed to put more money into the Fed Funds market. If there were enough money to pay interest, then such pressure should not exist. The ethical implications of this are quite alarming. If the money supply increases and a disproportionate amount goes to support the banking system, then this is an indirect transfer from the poor to the rich. As the system continues, the divide will increase unless alleviated by government actions (like welfare) or by bank failures.
I've been looking around out of curiosity if there is a standard method equivalent to OLS that finds the coefficients satisfying the minimum absolute error criteria as opposed to squared error, but haven't found anything for $$ \min \sum_{i=1}^n \|\|F(x_i)-y_i\|\|. $$ I'm not talking about a LASSO (which is a penalty on the $L_1$ norm of the coefficients). Obviously a simple custom convex optimization could do this, but just curious if I was overlooking something already commonly available by not using the right buzz words/lingo. We know from intuition of the necessity of the Bessel correction that $$\arg\min_x \sum_{j=1}^n (x_j - x)^2 = \bar{x},$$ the sample mean. It similarly turns out that $$\arg\min_x \sum_{j=1}^n \|x_j - x\| = \mathrm{med}(x_1, \dots, x_j),$$ the sample median. Commonly in regression, we minimize the squared error, giving us estimates for the mean, but, if we were to instead minimize the absolute error, we'd get estimates for the median. Of course, when the the regression model is $y \sim \mathcal{N}(X \beta^*, \sigma^2I)$, the median is the mean so the differences between these two methods aren't too pronounced. Indeed, I've commonly seen quantile regression motivated as being useful in the presence of outliers. An interesting history of the use of absolute error in regression is included here (in section 2): Portnoy, S., & Koenker, R. (1997). The Gaussian Hare and the Laplacian Tortoise: Computability of Squared-Error versus Absolute-Error Estimators. Statistical Science, 12(4), 279–300. It's discussed that the idea to use this loss was had long ago, but, due to computational intractability, it didn't gain widespread attention. More generally, we could this is an example of quantile regression, which just installs weights into the absolute error that works for the median.
problems are a type of Second order cone programming (SOCP) convex optimizationproblems. The general form of the problem is \[\begin{array}{ll} \mbox{minimize} & \ f^T x \\ \mbox{subject to} & \lVert A_i x + b_i \rVert_2 \leq c_i^T x + d_i,\quad i = 1,\dots,m \\ & Fx = g \end{array} \] where $f \in \mathbb{R}^n$, $A_i \in \mathbb{R}^{{n_i}\times n}$, $b_i \in \mathbb{R}^{n_i}$, $c_i \in \mathbb{R}^n$, $d_i \in \mathbb{R}$, $F \in \mathbb{R}^{p\times n}$, and $g \in \mathbb{R}^p$. The inequalities, $\lVert A_i x + b_i \rVert_2 \leq c_i^T x + d_i$ are the second order cone constraints. Special cases When $A_i = 0$ for $i = 1,\dots,m$, the SOCP reduces to a linear programming problem. When $c_i = 0$ for $i = 1,\dots,m$, the SOCP is equivalent to a convex quadratically constrained quadratic programming problem. Note that semidefinite programming subsumes second order cone programming since the SOCP constraints can be written as linear matrix inequalities. Online and Software Resources Second Order Cone Programming Solvers on the NEOS Server SOCP example GAMS models
There is indeed a geometric interpretation of $\bf{u}\times\bf{v}$ in terms of the areas of the projections of the parallelogram $\bf{P}$ spanned by $\bf{v}$ and $\bf{w}$ onto the coordinate planes. I'll start from scratch. Motivating problem: We wish to create a vector perpendicular to u,v, i.e. construct w s.t. $\bf{w}\cdot \bf{u} = \bf{w}\cdot \bf{v} = 0$. Two equations in 3 unknowns: we can derive w as $\lambda(u_2 v_3 - u_3 v_2,$ [3&1], [1&2]$)$. So let's $\bf{u}\times \bf{v}$ as ($u_2 v_3 - u_3 v_2$, [3&1], [1&2]). define Observe $u_2 v_3 - u_3 v_2$ is just the signed area of the $(u_2, u_3)$, $(v_2,v_3)$ parallelogram. (2D determinant -- you can work it out with triangles). So we can rewrite our definition: $\bf{u}\times \bf{v}$ as $(A_{yz}, A_{zx}, A_{xy})$ where $A_{\text{plane}}$ = Define signed area of u,v parallelogram projected onto that plane Notice by symmetry, switching u and v just changes the sign of $u_p v_q - u_q v_p$, so we have: $\bf{u}\times \bf{v} = -\bf{v}\times \bf{u}$ Notice also: $\bf{i} \times \bf{j} = \bf{k}$ So we get our 'Right Hand Rule'. Now it makes sense to ask: "Could we have skipped the algebra?" i.e. arrived at this definition purely from geometric insight. And the answer is ! yes Let n be the unit vector perpendicular to u,v. WAIT! There is a problem here -- there are 2 choices: -n also is a valid candidate. So if you are given i,j your choices are k,-k. So let's choose k. Looking at our right-hand, i for thumb, j for index finger and k corresponds to the direction the next finger is pointing. So we have an orientation. Let's call this the Right Hand Rule. So let's choose the n that satisfies the same orientation. We can show that the ratio of the area of the uv-parallelogram to it's yz projection gives $\bf{n}_x$, similarly for xz$\to \bf{n}_y$ and xy$\to \bf{n}_z$: Here is a simplification that will illustrate this... Let's say we have y as depth, z going upwards. We have some near-flat plane (so the normal is pointing nearly straight up). Suppose we draw a unit grid on it. Now we are going to drop each point onto the xy-plane so $(x,y,z)\to(x,y,0)$ and calculate the change in area. Furthermore let's say we have rotated things so that our plane's normal vector has depth component = 0. So alternatively we could set this up by imagining two xy-planes. And rotate one of them slightly around the y-axis, give it a unit grid, and "drop" this grid onto the other (axis-aligned) plane. We want the area of a projected grid square. It should be obvious that the depth-component (y) of any point on our plane is unchanged by this projection. And a simple calculation reveals that the x-component is just $cos(\theta)$, where $\theta$ is the angle between the normals, otherwise known as $\bf{k}\cdot \bf{n}=\bf{n}_z$, where $\bf{k}=(0,0,1)$. So, the xy-area multiplier is $\bf{n}_z$ as required: $A\bf{n}_z = A_{xy}$ Let's write these parallelogram areas using $A$. So we have: $A_{xy} = A \bf{n}_z$, sim. for y & z. So $A\bf{n} = (A_{yz}, A_{zx}, A_{xy})$ (EDIT: Bold rendering incorrectly for $A_{xy}$ in the above line, anyone?) So if we $\bf{u}\times \bf{v}$ as $(A_{yz}, A_{zx}, A_{xy})$ then we have $\bf{u}\times \bf{v} = A \bf{n}$, i.e. $\bf{u}\times \bf{v}$ is perpendicular to define u,v and of length $A$. QED!
I'm a mathematics student with not much background in physics. I'm interested in learning about the path integral formulation of quantum mechanics. Can anyone suggest me some books on this topic with minimum prerequisite in physics? Before answering, please see our policy on resource recommendation questions. Please write substantial answers that detail the style, content, and prerequisites of the book, paper or other resource. Explain the nature of the resource so that readers can decide which one is best suited for them rather than relying on the opinions of others. Answers containing only a reference to a book or paper will be removed! Sources for the path integral You can read any standard source, so long as you supplement it with the text below. Here are a few which are good: Feynman and Hibbs Kleinert (although this is a bit long winded) An appendix to Polchinski's string theory vol I Mandelstam and Yourgrau There are major flaws with other presentations, these are pretty much the only good ones. I explain the major omission below. Completing standard presentations In order for the discussion of the path integral to be complete, one must explain how non-commutativity arises. This is not trivial, because the integration variables in the path integral for bosonic fields or particle paths is over ordinary real valued variables, and these quantities cannot be non-commutative themselves. Non-commutative quantities The resolution of this non-paradox is that the path integral integrand is on matrix elements of operators, and the integral itself is reproducing the matrix multiplication. So it is only when you integrate over all values at intermediate times that you get a noncommutative order-dependent answer. Importantly, when noncommuting operators appear in the action or in insertions, the order of these operators is dependent on exactly how you discretize them--- whether you put the derivative parts as forward differences or backward differences or centered differences. These ambiguities are all important, and they are discussed only in a handful of places (Negele/Orland Yourgrau/Mandelstam Feynman/Hibbs Polchinski and Wikipedia) and nowhere else. I will give the classical examples of this, which are sufficient to resolve the general case, assuming you are familiar with simple path integrals like the free particle. Consider the free particle Euclidean action $$ S= -\int {1\over 2} \dot{x}^2 $$ and consider the evaluation of the noncommuting product $x\dot{x}$. This can be discretized as $$ x(t) {x(t+\epsilon) - x(t)\over \epsilon} $$ or as $$ x(t+\epsilon) {x(t+\epsilon) - x(t)\over \epsilon}$$ The first represents $x(t)p(t)$ in this operator order, the second represents $p(t)x(t)$ in the other operator order, since the operator order is the time order. The difference of the second minus the first is $$ {(x(t+\epsilon) - x(t))^2\over \epsilon} $$ Which, for the fluctuating random walk path integral paths has a fluctuating limit which averages to 1 over any finite length interval, when $\epsilon$ goes to zero. This is the Euclidean canonical commutation relation, the difference in the two operator orders gives 1. For Brownian motion, this relation is called "Ito's lemma", not dX, but the square of dX is proportional to dt. While dX is fluctuating over positive and negative values with no correlation and with a magnitude at any time of approximately $\sqrt{dt}$, dX^2 is fluctuating over positive values only, with an average size of dt and no correlations. This means that the typical Brownian path is continuous but not differentiable (to prove continuity requires knowing that large dX fluctuations are exponentially suppressed--- continuity fails for Levy flights, although dX does scale to 0 with dt). Although discretization defines the order, not all properties of the discretization matter--- only which way the time derivative goes. You can understand the dependence intuitively as follows: the value of the future position of a random walk is (ever so slightly) correlated with the current (infinite) instantaneous velocity, because if the instantaneous velocity is up, the future value is going to be bigger, if down, smaller. Because the velocity is infinite however, this teensy correlation between the future value and the current velocity gives a finite correlator which turns out to be constant in the continuum limit. Unlike the future value, the past value is completely uncorrelated with the current (forward) velocity, if you generate the random walk in the natural way going forward in time step by step, by a Markov chain. The time order of the operators is equal to their operator order in the path integral, from the way you slice the time to make the path integral. Forward differences are derivatives displaced infinitesimally toward the future, past differences are displaced slightly toward the past. This is is important in the Lagrangian, when the Lagrangian involves non-commuting quantities. For example, consider a particle in a magnetic field (in the correct Euclidean continuation): $$ S = - \int {1\over 2} \dot{x}^2 + i e A(x) \cdot \dot{x} $$ The vector potential is a function of x, and it does not commute with the velocity $\dot{x}$. For this reason, Feynman and Hibbs and Negele and Orland carefully discretize this, $$ S = - \int \dot{x}^2 + i e A(x) \cdot \dot{x}_c $$ Where the subscript c indicates infinitesimal centered difference (the average of the forward and backward difference). In this case, the two orders differ by the commutator, [A,p], which is $\nabla\cdot A$, so that there is an order difference outside of certain gauges. The correct order is given by requiring gauge invariance, so that adding a gradiant $\nabla \alpha$ to A does nothing but a local phase rotation by $\alpha(x)$. $$ ie \int \nabla\alpha \dot{x}_c = ie \int {d\over dt} \alpha(x(t))$$ Where the centered differnece is picked out because only the centered difference obeys the chain rule. That this is true is familiar from the Heisenberg equation of motion: $$ {d\over dt} F(x) = i[H,F] = {i\over 2} [p^2,F] = {i/2}(p[p,F] + [p,F]p) = {1\over 2}\dot{x} F'(x) + {1\over2} F'(x) \dot{x}$$ Where the derivative is a sum of both orders. This holds for quadratic Hamiltonians, the ones for which the path integral is most straightforward. The centered difference is the sum of both orders. The fact that the chain rule only works for the centered difference means that people who do not understand the ordering ambiguities 100% (almost everybody) have a center fetishism, which leads them to use centered differences all the time. THe centered difference is not appropriate for certain things, like for the Dirac equation discretization, where it leads to "Fermion doubling". The "Wilson Fermions" are a modification of the discretized Dirac action which basically amounts to saying "Don't use centered derivatives, dummy!" Anyway, the order is important.Any presentation of the path integral which gives the Lagrangian for a particle in a magnetic field without specifying whether the time derivative is a forward difference or a past difference, is no good at all. That's most discussions. A good formalism for path integrals always thinks of things on a fine lattice, and takes the limit of small lattice spacing at the end. Feynman always secretly thought this way (and often not at all secretly, as in the case above of a particle in a magnetic field), as does everyone else who works with this stuff comfortably. Mathematicians don't like to think this way, because they don't like the idea that the continuum still has got new surprises in the limit. Mathematicians are snobby and wrong. The other thing that is hardly ever explained properly (except for Negele/Orland, David John Candlin's Neuvo Cimento original article of 1956, and Berezin) is the Fermionic field path integral. This is a separate discussion, so I will refer to these sources for the time being. "Quantum Mechanics and Path Integrals" by Feynman and Hibbs To get you started here are some lecture notes I like on the path integral: http://bohr.physics.berkeley.edu/classes/221/1011/notes/pathint.pdf (from the page http://bohr.physics.berkeley.edu/classes/221/1011/221.html ) The Book 'Principles of Quantum Mechanics' by R. Shankar has a really good introduction to the path integral formalism (and quantum mechanics in general) with two dedicated chapters about it. Also the book begins with a nice presentation of linear algebra in Bra-ket notation.
Consider the atomic wavefunction: $$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left\|#1\right>} \ket{\psi}=\ket{L,M_L}\ket{S,M_S}$$ let us assume that $\ket{L,0}$ is has a certain symmetry on swapping electrons $i$ and $j$ such that the overall wavefunction $\ket{\psi}=\ket{L,0}\ket{S,M_S}$ is anti-symmetric then will the wavefunction $\ket{L,M_L}$ (for the same $L$) have the same symmetry on swapping electrons $i$ and $j$? i.e. if $\ket{L,0}\ket{S,M_S}$ is anti-symmetric is $\ket{L,M_L}\ket{S,M_S}$ always anti-symmetric for a given $L$, $S$ and $M_S$? The symmetry does not depend on $M_L$. The simplest way to see this is as follows. Start with $\vert L,M_L\rangle$ as a product state. If it is a product of only two states it would be written as $$ \vert L M_L\rangle = \sum_{m_1m_2}C_{\ell m_1;\ell m_2}^{LM_L} \vert \ell m_1\rangle \vert \ell m_2\rangle \tag{1} $$ where $C_{\ell m_1,\ell m_2}^{LM_L}$ is a Clebsch-Gordan coefficient. Permute particles $1$ and $2$, which is same as permuting $m_1$ and $m_2$ and you get $$ C_{\ell m_2;\ell m_1}^{L M_L}=(-1)^{2\ell-L} C_{\ell m_1;\ell m_2}^{L M_L} $$ showing that the phase $(-1)^{2\ell -L}$ and thus the symmetry character does not depend on $M_L$. If you have more than two particles the job is a little more complicated. Start with $\vert L,L\rangle$ as a product state, generalizing (1) to more than one constituent. The coefficients in the linear combinations are no longer CGs but this doesn't matter for now. From the state $\vert L,M_L\rangle$ you reach the state $\vert L,M_L-1\rangle$ by application of the lowering operator$$\hat L_-= \sum_i \hat L_{i,-}= \hat L_{1,-}+\hat L_{2,-}+\hat L_{3,-}\ldots$$Note that this sum is symmetric under permutation of particle numbers, so that, for instance:\begin{align}P_{12}\vert L,M_L-1\rangle &= P_{12}{\cal N}_{M_L}\left(\hat L_{1,-}+\hat L_{2,-}+\ldots \right)\vert L,M_L\rangle\, ,\\&={\cal N}_{M_L}\left(\hat L_{1-}+\hat L_{2-}+\ldots\right)P_{12}\vert L,M_L\rangle\end{align}where ${\cal N}_{M_L}$ is a normalization constant.This shows that the symmetry under permutation of $\vert L,M_L-1\rangle$ is that of $\vert L,M_L\rangle$, and independent of this $M_L$.
Tverberg plus minus Connections for Women Workshop: Geometric and Topological Combinatorics August 31, 2017 - September 01, 2017 Speaker(s):Imre Barany (Alfréd Rényi Institute of Mathematics) Location:MSRI: Simons Auditorium Tags/Keywords Tverberg's theorem sign conditions Primary Mathematics Subject Classification Secondary Mathematics Subject ClassificationNo Secondary AMS MSC 6-Barany We prove a Tverberg type theorem: Given a set $A \subset \R^d$ in general position with $\|A\|=(r-1)(d+1)+1$ and $k\in \{0,1,\ldots,r-1\}$, there is a partition of $A$ into $r$ sets $A_1,\ldots,A_r$ (where $\|A_p\|\le d+1$ for each $p$) with the following property. The unique $z \in \bigcap_{p=1}^r \aff A_p$ can be written as an affine combination of the elements in $A_p$: $z=\sum_{x\in A_p}\al(x)x$ for every $p$ and exactly $k$ of the coefficients $\al(x)$ are negative. The case $k=0$ is Tverberg's classical theorem. This is joint works with Pablo Soberon. 6-Barany H.264 Video 6-Barany.mp4 Download If none of the options work for you, you can always buy the DVD of this lecture. The videos are sold at cost for $20USD (shipping included). Please Click Here to send an email to MSRI to purchase the DVD. See more of our Streaming videos on our main VMath Videos page.
Suppose I have a Marshallian demand function $x_M(p_x^0,p_y,m^0)$. As I understand it, Slutsky compensation is defined as $$T_S = \Delta p_x \cdot x_M(p_x^0,p_y,m^0)$$ Can someone explain why this compensation overcompensates the consumer? Economics Stack Exchange is a question and answer site for those who study, teach, research and apply economics and econometrics. It only takes a minute to sign up.Sign up to join this community Suppose I have a Marshallian demand function $x_M(p_x^0,p_y,m^0)$. As I understand it, Slutsky compensation is defined as $$T_S = \Delta p_x \cdot x_M(p_x^0,p_y,m^0)$$ Can someone explain why this compensation overcompensates the consumer? Here's a figure to explain: Starting from the old price line, where the optimal consumption bundle is point $A$, we increase the price of $y$ to get the new price line. The Slutsky compensation says that we have to give the consumer enough extra income so that he can afford to old bundle ($A$) at the new price. Thus, we shift the new budget constraint out to the dashed line. The reason this "overcompensates" is that, taking the dashed line as the budget constraint, the consumer could afford bundle $B$, which gives him higher utility than he started with. Slutsky compensation makes the original consumption bundle again exactly affordable after the price change. This implies that the original utility level is reachable. But higher utility may also be reachable by changing the consumption bundle. So there may be overcompensation. Hicksian compensation makes the original utility level again exactly reachable after the price change. The consumption bundle provided may differ from the original one. The overcompensation is explained in many textbooks and online sources, e.g. here. For both Slutsky transfer and Hicksian transfer, the only parameter changing in the problem is income $m$. You are adding money in the case of a price increase and removing money in the case of a price decrease. For Hicksian, this means $$T_H = e(p_x^f,p_y,v^o) - m$$ For Slutsky $$T_S = \Delta p x_M$$ To understand why Slutsky overcompensates but Hicksian does not, we need to think more carefully about them. Consider Hicksian compensation. The Hicksian transfer provides the consumer with just enough money to return to his original indifference curve. But he doesn't have enough to buy the original bundle. He only has enough to buy a bundle with the same utility level. This is the same point we would get if we solved $$\min p_x^f x + p_y y$$ subject to $$v^o = U(x,y)$$ In other words, the Hicksian transfer moves us to the bundle that solves that problem. So for this reason Hicks doesn't overcompensate. It is exactly the amount required given the final price and desired utility level. Hence, it is tangent. Now consider the Slutsky transfer. Algebraically, we can see it allows us to purchase the original bundle \begin{align} y &= -\frac{p_x^f}{p_y} x + \frac{m + T_s}{p_y}\\ y &= -\frac{p_x^f}{p_y} x + \frac{m + p_x^f x - p_x^o x}{p_y}\\ p_y y &= -p_x^fx + m + p_x^f x - p_x^o x\\ p_y y &= m + - p_x^o x\\ p_y y + p_x^o x &= m \end{align} So it restores the ability to purchase the original bundle. But this line doesn't have the same slope as our original budget constraint before the price change. This one is parallel to the Hicksian transfer. Yet we already established that the Hicksian transfer was the solution to the expenditure minimization problem and hence tangent. Given that we usually assume strictly convex curves, it must form a chord and thus have bundles that offer greater utility than the Hicksian transfer. Thus, the consumer can move to a higher indifference curve. And so Slutsky transfer overcompensates the consumer because we can now buy bundles on higher indifference curves than prior to the price increase.
Case Study Contents In this example, we estimate the unknown parameters of Constant Elasticity of Substitution (CES) production functions with either additive or multiplicative error terms using Mizon (1977). The CES production function is a neoclassical production function that displays constant elasticity of substitution. In other words, the production technology has a constant percentage change in input proportions (for example, labor and capital) due to a percentage change in marginal rate of technical substitution. This example uses the same optimization model as that in Example 1, except that the optimization is restricted by some CES functions rather than the Cobb-Douglas production function. We introduce two different forms of CES functions. The first one, which we call the MD model, has Multiplicative Disturbance (MD). The standard form with two inputs is $$q_t = \phi[\beta k^{\rho}_t + (1 - \beta)l^{\rho}_t]^{\frac{1}{\rho}}\exp(\mu_t);$$ The second one, which we call the AD model, has Additive Disturbance (AD). The standard form is $$q_t = \phi[\beta k^{\rho}_t + (1 - \beta)l^{\rho}_t]^{\frac{1}{\rho}} + \mu_t,$$ where: $\phi \equiv$ scale parameter, $\phi$ > 0 $\beta \equiv$ value share of capital, $\beta \in (0,1)$ $\rho \equiv$ the substitution parameter, $\rho \in (-\infty,1)$, and $\rho \neq 0$ $k_t$, $l_t$ and $\mu_t$ stand for capital, labor and error term at observation $t$. And we define $\sigma = \frac{1}{1-\rho}$ as the elasticity of substitution in these production functions. As its name suggests, the CES production function exhibits constant elasticity of substitution between capital and labor. Leontief, linear and Cobb–Douglas production functions are special cases of the CES production function. That is, if the substitution parameter $\rho$ equals one, we have a linear or perfect substitutes production function; if $\rho$ approaches zero in the limit, we get the Cobb–Douglas production function; as $\rho$ approaches negative infinity, we get the Leontief or perfect complements production function. In more general cases with more than two inputs, we denote the inputs as $V_{1}, V_{2}, \dots, V_{m}$, $m > 2$. Then the fitting constraints in the MD model are: \[q_t = \phi[\beta_1 V_{1,t}^{\rho} + \beta_2 V_{2,t}^{\rho} + \dots + (1 - \sum_n\beta_n)V_{m,t}^{\rho}]^{\frac{1}{\rho}}\exp(\mu_t),\] and, in the AD model, they are: \[q_t = \phi[\beta_1 V_{1,t}^{\rho} + \beta_2 V_{2,t}^{\rho} + \dots + (1 - \sum_n\beta_n)V_{m,t}^{\rho}]^{\frac{1}{\rho}} + \mu_t.\] We then need to estimate an unknown vector of $\theta = (\phi, \beta_{1}, \beta_{2},\dots, \beta_n, \dots, \beta_{m-1}, \rho)'$ with $m + 1$ elements, where $n = 1, 2, \dots, m-1$ and $m$ is the total number of inputs. Both the standard and general form of the CES production function in the MD model are included in the demo of example 2. Similar to Example 1, we estimate the unknown vector of parameters $\theta$ by minimizing the Sum of Squared Errors (SSE) subject to the CES production function having either additive (AD) and multiplicative (MD) error terms. We report estimator of unknown vector $\hat\theta$, estimated variances based on the diagonal elements of the covariance matrix $\hat{V}_{\hat{\theta}}$, and t-statistics as well as p-values as we did in Example 1. We have implemented a nonlinear least squares model with a CES production function and multiplicative error terms. Click here to experiment with the demo of Example 2. The implementation was inspired by Erwin Kalvelagen's work (2007) on econometric modeling in GAMS. Multiplicative Disturbance Additive Disturbance Mizon, Grayham E. 1977. Inferential Procedures in Nonlinear Models: An Application in a UK Industrial Cross Section Study of Factor Substitution and Returns to Scale. Econometrica 45(5), 1221-1242. Kalvelagen, Erwin. 2007. Least Squares Calculations with GAMS. Available for download at http://www.amsterdamoptimization.com/pdf/ols.pdf. Greene, William. 2011. Econometrics Analysis, 7th ed.Prentice Hall, Upper Saddle River, NJ.
Say, in Hamiltonian mechanics, we know two constants of motion, $A$ and $B$. It could be proven that the quantity $[A,B]$ is also a constant of motion, where $[A,B]$ denotes the Poisson brackets of $A$ and $B$. As an example, consider the Hamiltonian: $$H = \frac 1 2 p_R^2 + \frac{p_\phi^2}{2R^2} + \frac A R.$$ H itself obviously is a constant of motion. Further, we can show that the quantity $p_\phi$ and the quantity $C$, given by: $$C = p_Rp_\phi \sin \phi + \frac{p_\phi^2}{R} \cos \phi + A \cos \phi.$$ are both constants of motion. However, how do we know if $[p_{φ},C]$ is another independent constant of motion? Or more generally, how do we know, amongst the 4 constants of motions (H, $p_\phi$, $C$, $[p_{φ},C]$), how many of them are independent?
${{\boldsymbol Z}_{{c}}{(3900)}}$ $I^G(J^{PC})$ = $1^+(1^{+ -})$ was ${{\mathit X}{(3900)}}$ Properties incompatible with a ${{\mathit q}}{{\overline{\mathit q}}}$ structure (exotic state). See the review on non- ${{\mathit q}}{{\overline{\mathit q}}}$ states. Charged ${{\mathit Z}_{{c}}{(3900)}}$ seen as a peak in the invariant mass distribution of the ${{\mathit J / \psi}}{{\mathit \pi}^{\pm}}$ system by BES III (ABLIKIM 2013T ) in ${{\mathit e}^{+}}$ ${{\mathit e}^{-}}$ $\rightarrow$ ${{\mathit \pi}^{+}}{{\mathit \pi}^{-}}{{\mathit J / \psi}}$ at c.m. energy of 4.26 GeV and by radiative return from ${{\mathit e}^{+}}{{\mathit e}^{-}}$ collisions at $\sqrt {s }$ from 9.46 to 10.86 GeV at Belle (LIU 2013B ). Partial wave analysis of ABLIKIM 2017J determines $\mathit J{}^{P} = 1{}^{+}$ with more than 7$~\sigma $ significance. Neutral ${{\mathit Z}_{{c}}{(3900)}}$ seen in the ${{\mathit J / \psi}}{{\mathit \pi}^{0}}$ invariant mass distribution in ${{\mathit e}^{+}}$ ${{\mathit e}^{-}}$ $\rightarrow$ ${{\mathit \pi}^{0}}{{\mathit \pi}^{0}}{{\mathit J / \psi}}$ at c.m. energies of 4.23, 4.26, and 4.36 GeV by BES III (ABLIKIM 2015U ) and at 4.17 GeV by XIAO 2013A . Peaks in ( ${{\mathit D}}{{\overline{\mathit D}}^{*}}$ )${}^{0,\pm{}}$ reported by BES III (ABLIKIM 2014A , ABLIKIM 2015AB ) are assumed to be related.
I'm trying to recreate some work that a professor explained to me in his office, specifically deriving the free particle propagator going from $(y,0)$ to $(x,T)$ using the Feynman Path Integral. I'm trying to reproduce $$K(x,T;y,0) = \sqrt{\frac{m}{2\pi i\hbar T}}\mathrm{exp}[\frac{im(x-y)^2}{2\hbar T}]$$Here's what I've done so far: $K(x,T;y,0) = \int_y^x\mathscr{D}[x(t)]e^{iS[x(t)]/\hbar}.$ So first I compute the action: $S[x(t)] = \int_0^T \frac{1}{2}m\dot{x}^2 \mathrm{d}t$ We can always split the path $x(t)$ in the following way: $x(t) = x_{cl} (t) +q(t)$, where $x_{cl}(t)$ given by $$x_{cl}(t) = \frac{(x-y)t}{T} + y$$ is the classical path and $q(t)$ is a "quantum fluctuation". Because the endpoints of $x(t)$ and $x_{cl}(t)$ are the same, we get that $q(0)=q(T)=0$, and because any path should be piecewise differentiable, we can represent $q(t)$ in a Fourier Series: $$q(t) = \sum_{n=1}^{\infty} a_n sin(\frac{n\pi t}{T})$$. The action is then $$S[x(t)] = \frac{1}{2}m\int_0^T (\frac{x-y}{T})^2 + 2\frac{x-y}{T} \dot{q} + \dot{q}^2 \mathrm{d}t$$ The first term is trivial, the second term vanishes due to the fundamental theorem of calculus and the fact that $q(t)$ vanishes at the endpoints. Now for the last term we get $$\int_0^T \dot{q}^2 \mathrm{d}t = \int_0^T \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_n a_m (\frac{n\pi}{T}) (\frac{m\pi}{T})cos(\frac{n\pi t}{T})cos(\frac{m\pi t}{T})\mathrm{d}t$$ but due to orthogonality only the $n=m$ terms survive so we get $$ = \sum_{n=1}^{\infty}(\frac{n\pi }{T})^2\int_0^T a_{n}^2cos^2(\frac{n\pi t}{T})\mathrm{d}t = \sum_{n=1}^{\infty}\frac{(n\pi)^2}{2T}a_{n}^2$$. Now to do the actual path integral, "all possible paths" would correspond to "all possible $q(t)$'s" which would mean all possible $a_n$'s. Thus our path integral becomes: $$K(x,T;y) = \lim_{N\to\infty}\int_{-\infty}^{\infty}\mathrm{d}a_1\dotsi\int_{-\infty}^{\infty}\mathrm{d}a_N \mathrm{exp}\{\frac{im}{2\hbar}[\frac{(x-y)^2}{T} + \sum_{n=1}^{\infty}\frac{(n\pi)^2}{2T}a_{n}^2]\}$$ Now the first term in the exponential is clearly the same as the one in the original propagator however for the other integrals, i get an infinite amount of integrals which are infinite! Where does my reasoning or algebra go wrong? PS I know there's probably a simpler way to do it, but since we started out this way I wanna know how it can be done with this method.
Summary: Mechanics problem related with Calculus (differential equations)Hi everyone, I would like some help in that task, if anyone would be willing to help :) Namely I have a problem from particle dynamics. "D:" means given info... so, D: m,g,h,b, miu. We're looking for v0 and S as given... .Above is the figure of the problem.I am trying to solve x(t) and differentiate it to obtain v(t); however, I have difficulty solving the differential equation shown below.$$ v(t)=\int a(t)dt=\int \frac{B(\varepsilon-Blv)d}{Rm}dt \Rightarrow \frac{dx}{dt}=\frac{B\varepsilon... 1. Homework Statement2. Homework Equationseuler##e^{ix} = cos(x) + isin(x)####e^{-ix} = cos(x) - isin(x)##3. The Attempt at a SolutionI'm starting with differential equations and I'm trying to understand this solution including complex numbers:First we determine the zeros. I... 1. Homework Statement2. Homework Equations$$F = \nabla \phi$$3. The Attempt at a SolutionLet's focus on determining why this vector field is conservative. The answer is the following:I get everything till it starts playing with the constant of integration once the... 1. Homework Statementx(dy/dx) = 3y +x4cos(x), y(2pi)=02. Homework EquationsN/A3. The Attempt at a SolutionI've tried a couple different ways to make this separable, but you always carry over a 1/dx or 1/dy term and I can never fully separate this. I've also tried to do a Bernoulli... I am trying to determine an outer boundary condition for the following PDE at ##r=r_m##: $$ \frac{\sigma_I}{r} \frac{\partial}{\partial r} \left(r \frac{\partial z(r,t)}{\partial r} \right)=\rho_D gz(r,t)-p(r,t)-4 \mu_T \frac{\partial^2z(r,t)}{\partial r^2} \frac{\partial z(r,t)}{\partial t} $$... Hey all,I hope this is the correct forum section to post this in.I heard about this problem from a youtube video but I've not been able to simulate it because the video was meant only for an introduction into PID control.Here's the problem:A remote control helicopter is hovering just... 1. Homework StatementFind Orthogonal Trajectories of ##\frac{x^2}{a}-\frac{y^2}{a-1}=1##HintSubstitute a new independent variable w##x^2=w##and an new dependent variable z##y^2=z##2. Homework Equations3. The Attempt at a Solutionsubstituting ##x## and ##y## I get... Hi!When we want to look at different singular points for e.g Bessel's eq. $$u´´(x) + \frac{u'(x)}{x} + (1- \frac{n^2}{x^2})u(x)$$.We usually evaluate the equation letting x= 1/z. But I don't algebraically see how such a substitution ends up with $$w´´(z) +( \frac{2}{z}-... Hey all,I don't understand what makes a differential equation (DE) linear.I found this: "x y' = 1 is non-linear because y' is not multiplied by a constant"but then also this: "x' + (t^2)x = 0 is linear in x".t^2 also isn't a constant.So why is this equation linear? COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081 COPYRIGHT strictly reserved to A. K. Nandakumaran, P. S. Datti & Raju K. George, Department of Mathematics, IISc Bangalore. Duplication PROHIBITED.Lectures: http://www.nptel.ac.in/courses/111108081/Syllabus: http://www.nptel.ac.in/syllabus/syllabus.php?subjectId=111108081
An integral is useful for finding the area underneath a function. Let $f(x)$ be any arbitrary function such that it is smooth and continuous at every point. To find the area underneath $f(x)$, we must go through several steps. First, we'll start off by drawing an $n$ (where $n$ is any positive integer) number of rectangles of equal width underneath $f(x)$ as illustrated in Figure 1. What is the total area of all the rectangles? The area of the first rectangle is $A_1=f(x_1)(x_2-x_1)$; the area of the second rectangle is $A_2=f(x_2)(x_3-x_2)$; and the area of the $n$th rectangle is $A_n=f(x_n)(x_{n+1}-x_n)$. Since every rectangle has the same width, it follows that $x_2-x_2=x_3-x_2=x_{n+1}x_n= Δx$. To find the total area of all the rectangles, let's add up the area of each rectangle: Figure 1 $$A=A_1+A_2+...+A_n=f(x_1)Δx+ f(x_2)Δx+...+ f(x_n)Δx\sum_{i=1}^nf(x_n)Δx.\tag{1}$$ As you can see visually in the animation in Figure 2, as the number of rectangles $n$ increases, the area $A$ becomes closer and closer to equaling the exact area underneath the curve. Using Equations (1), let's take the limit as $n→∞$ to get $$\lim_{n→∞}\sum_{i=1}^nf(x_n)Δx.\tag{2}$$ Let's review the notion of a limit that we covered in an earlier lesson. The value that the limit, $\lim_{z→c}g(x)$, is equal to is the value that $g(x)$ gets closer and closer to while $ z→c$. Take for example the limit, $\lim_{x→2}x^2$, that we looked at in a previous lesson. The value that this limit is equal to is the value that $x^2$ gets closer and closer to as $x→2$. We showed that this value is $4$. Similarly the value of the limit, $\lim_{n→∞}\sum_{i=1}^nf(x_n)Δx$, is the value that $\sum_{i=1}^nf(x_n)Δx$ gets closer and closer to equaling the exact area underneath the curve $f(x)$. Thus, the limit must equal the area underneath $f(x)$ and $$\text{Area underneath f(x)}=\lim_{n→∞}\sum_{i=1}^nf(x_n)Δx.\tag{3}$$ Let's see if there is a simpler way of rewriting the right-hand side of Equation (3). Not to sound too annoyingly repetitive but, again, the limit $\lim_{n→∞}\sum_{i=1}^nf(x_n)Δx$ is equal to the thing that $\sum_{i=1}^nf(x_n)Δx$ gets closer and closer to equaling as $n→∞$. But if you think about it for a moment, the following must be true: if $n→∞$, then the number $n$ of the terms $f(x_i) Δx$ is getting closer and closer to infinity; thus, the finite sum $\sum_{i=1}^n$ (of an $n$ number of terms) is getting closer and closeer to becoming an infinite sum. Let's represent an infinite sum (that's to say, a sum of infinitely many terms) by the symbol "$∫$." As $n→∞$, it is also true that the width $Δx$ is getting closer and closer to becoming infinitely small. Let's represent an infinitely small $Δx$ by the symbol " $dx$ ." With all that said, I'd like to just make a few remarks about the variable $x$, then about the expression $\int{f(x)dx}$, and then we'll see how that ties in with our discussion of the limit $\lim_{n→∞}\sum_{i=1}^nf(x_n)Δx$. Since $x$ is a continuous variable, it can take on an infinite number of values: such as the numbers $2$, $π$, $3.001$, $3.00001$, $3.00000001$, etc. Thus, there are an infinite number of y-values along the curve $f(x)$: including $f(2)$, $f(π)$, $f(3.001$, etc. The term $f(x)dx$ is the area of an infinitely skinny rectangle; and the expression $\int{f(x)dx}$ is the sum of an infinite number of the terms, $f(x)dx$. $\int{f(x)dx}$ gives the infinite sum of all the areas $f(x)dx$ of (infinitely) skinny rectangle. What is $\sum_{i=1}^nf(x_i)Δx$ getting closer and closer to equaling as $ n→∞$? Well, clearly its getting closer and closer to being an infinite sum, and $x_i$ and $ Δx$ are approaching $x$ and $dx$. Thus, the limit $\lim_{n→∞}\sum_{i=1}^nf(x_n)Δx$ must also equal that thing and $$\int{f(x)dx}=\lim_{n→∞}\sum_{i=1}^nf(x_i) Δx.\tag{4}$$ The expression $\int{f(x)dx}$ is called "the integral of $f(x)$ with respect to the variable $x$" and it is equal to two things: first, the area underneath $f(x)$; second, it is also the sum of the infinite number of the terms $f(x)dx$ and is the infinite sum of the area of infinitely many, infinitely skinny rectangles. My apologies, the latter is quite a mouthful. But hopefully this lesson helped give you a better idea of what an integral actually is. In the next several lessons, we'll investigate techniques for solving integrals - that is, finding the area underneath various different functions $f(x)$.
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building) It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore) In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of @TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies \|\| x ( t ) - 0 \|\| < \varepsilon.$$The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $... "If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}\|f(z)\| \leq C_K \\|f\\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed? Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2 Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$ Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight. hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$ for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$ I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything. I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set).theorem ... The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious. (but seriously, the best tactic is over powered...) Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field? Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement? "Infinity exists" comes to mind as a potential candidate statement. Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system @Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity but so far failed Put it in another way, an equivalent formulation of that (possibly open) problem is: > Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object? If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite. My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science... O great, given a transcendental $s$, computing $\min_P(\|P(s)\|)$ is a knapsack problem hmm... By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as: $$P(x) = \prod_{k=0}^n (x - \lambda_k)$$ If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic Thus given $s$ transcendental, to minimise $\|P(s)\|$ will be given as follows: The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases. In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that0<\|x−pq\|<1qn.{\displaystyle 0<\left\|x-{\frac {p}... Do these still exist if the axiom of infinity is blown up? Hmmm... Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum: $$\sum_{k=1}^M \frac{1}{b^{k!}}$$ The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework There's this theorem in Spivak's book of Calculus:Theorem 7Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and$$f'... and neither Rolle nor mean value theorem need the axiom of choice Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set > are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion
You missed a tiny detail when defining $C$: all those languages are RE by assumption; no other language can be an $L(M)$! Let us again look at the language you want to say something about: $\qquad L = \{ \langle M \rangle \mid M \text{ is a TM}, L(M) \subseteq L\bigl(\, 0(0\cup1)^* \,\bigr) \}$. We see that $L \subseteq \Sigma^$ for some alphabet $\Sigma$ that depends on the TM encoding $\langle \_ \rangle$. Now recall the statement of Rice's theorem: the index set $\qquad \Phi(X) = \{ \langle M \rangle \mid M \text{ is a TM}, L(M) \in X \}$ of a language class $X$ is undecidable if $\emptyset \subsetneq X \subsetneq \mathrm{RE}$. Now, it is obvious (read: very easy to prove -- your exercise) that $L = \Phi(C)$ with $\qquad C = \{ Y \in \mathrm{RE} \mid Y \subseteq L\bigl( 0(0+1)^ \bigr) \}$; the requirement $\emptyset \subsetneq C \subsetneq \mathrm{RE}$ is likewise trivial. We don't get to "just say" anything; we have to prove all three conditions -- $L = \Phi(C)$, $\emptyset \subsetneq C$, and $C \subsetneq \mathrm{RE}$ -- but it so happens that all three proofs are very easy here since $L$ has been defined in a very suitable way. For other languages, the proofs can be more intricate (or impossible; then Rice's theorem does not apply). Side note: If we consider $\qquad C' = \{ Y \in \Sigma^* \mid Y \subseteq L\bigl( 0(0+1)^* \bigr) \}$ then we also get $\Phi(C') = L$! While there are non-RE $X$ in $C'$ (a simple if non-constructive argument is that there are uncountably many such $X$), $\Phi$ "ignores" them. However, $C'$ does not fulfill the conditions of Rice's theorem so we can not use it.
Let K be a nontrivial knot in the 3-sphere with the exterior E(K), and u inG(K), the fundamental group of E(K), a slope element represented by anessential simple closed curve on the boundary of E(K). Since the normal closureof u in G(K) coincides with that of the inverse of u, and u and its inverse ucorrespond to a slope r, a rational number or 1/0, we write << r >> = << u >>.The normal closure << u >> describes elements which are trivialized by r-Dehnfilling of E(K). In this article, we prove that << r_1 >> =<< r_2 >> if andonly if r_1 = r_2, and for a given finite family of slopes S = {r_1, ..., r_n},the intersection of << r_1 >> , << r_2>>, ..., and << r_n >> containsinfinitely many elements except when K is a (p, q)-torus knot and pq belongs toS. We also investigate inclusion relation among normal closures of slopeelements. We study chirally cosmetic surgeries, that is, a pair of Dehn surgeries on aknot producing homeomorphic 3-manifolds with opposite orientations. Severalconstraints on knots and surgery slopes to admit such surgeries are given. Ourmain ingredients are the original and the $SL(2,\mathbb{C})$ version of Cassoninvariants. As applications, we give a complete classification of chirallycosmetic surgeries on alternating knots of genus one. We introduce a notion of "quasi-right-veering" for closed braids, which playsan analogous role to "right-veering" for open books. We show that a transverselink $K$ in a contact 3-manifold $(M,\xi)$ is non-loose if and only if everybraid representative of $K$ with respect to every open book decomposition thatsupports $(M,\xi)$ is quasi-right-veering. We also show that severaldefinitions of "right-veering" closed braids are equivalent. For a null-homologous transverse link $\mathcal T$ in a general contactmanifold with an open book, we explore strongly quasipositive braids andBennequin surfaces. We define the defect $\delta(\mathcal T)$ of theBennequin-Eliashberg inequality. We study relations between $\delta(\mathcal T)$ and minimal genus Bennequinsurfaces of $\mathcal T$. In particular, in the disk open book case, under somelarge fractional Dehn twist coefficient assumption, we show that$\delta(\mathcal T)=N$ if and only if $\mathcal T$ is the boundary of aBennequin surface with exactly $N$ negatively twisted bands. That is, theBennequin inequality is sharp if and only if it is the closure of a stronglyquasipositive braid. We study Question 7.9 in the paper "Monoids in the mapping class group" byEtnyre and Van Horn-Morris; whether a symmetric mapping class admitting apositive factorization is a lift of a quasi-positive braid. We answeraffirmatively for mapping classes satisfying certain cyclic conditions. We use the LMO invariant to find constraints for a knot to admit a purely orreflectively cosmetic surgery. We also get a constraint for knots to admit aLens space surgery, and some information for characterizing slopes. Generalizing Howie and Greene's characterization of alternating knots, wegive a topological characterization of almost alternating knots. We show that if the fundamental group of the complement of a rationallyhomologically fibered knot in a rational homology 3-sphere is bi-orderable,then its Alexander polynomial has at least one positive real root. Our argumentcan be applied for a finitely generated group which is an HNN extension withcertain properties. We answer Question 6.12 in the paper "Monoids in the mapping class group"written by Etnyre and Van Horn-Morris. We give a formula of 3-dimensional invariant for a cyclic contact branchedcovering of the standard contact S^{3}. We prove a generalization of the Jones-Kawamuro conjecture that relates theself-linking number and the braid index of closed braids, for planar open bookswith certain additional conditions and modifications. We show that our resultis optimal in some sense by giving several counter examples for naivegeneralizations of the Jones-Kawamuro conjecture. We study a coverings of open books and virtually overtwisted contactmanifolds using open book foliations. We show that open book coverings producesinteresting examples such as transverse knots with depth grater than 1. We alsodemonstrate explicit examples of virtually overtwisted open books. We introduce an essential open book foliation, a refinement of the open bookfoliation, and develop technical estimates of the fractional Dehn twistcoefficient (FDTC) of monodromies and the FDTC for closed braids, which weintroduce as well. As applications, we quantitatively study the `gap' of overtwisted contactstructures and a non-right-veering monodromies. We give sufficient conditionsfor a 3-manifold to be irreducible and atoroidal. We also show that thegeometries of a 3-manifold and the complement of a closed braid are determinedby the Nielsen-Thurston types of the monodromies of their open bookdecompositions. A result of Malyutin shows that a random walk on the mapping class groupgives rise to an element whose fractional Dehn twist coefficient is large orsmall enough. We show that this leads to several properties of random3-manifolds and links. For example, random closed braids and open books arehyperbolic. We give a formula of the colored Alexander invariant in terms of thehomological representation of the braid groups which we call truncatedLawrence's representation. This formula generalizes the famous Buraurepresentation formula of the Alexander polynomial. Using open book foliations we show that an overtwisted disc in a planar openbook can be put in a topologically nice position. As a corollary, we prove thata planar open book whose fractional Dehn twist coefficients grater than one forall the boundary components supports a tight contact structure. We study surface links whose link groups are free abelian, and constructvarious stimulating and highly non-trivial examples of such surface links. We give a topological formula of the loop expansion of the colored Jonespolynomials by using identification of generic quantum sl2 representation withhomological representations. This gives a direct topological proof of theMelvin-Morton-Rozansky conjecture, and a connection between entropy of braidsand quantum representations. We give a lower estimate of the framing function of knots, and prove astrengthened version of Dehn's lemma conjectured by Greene-Wiest. We review a braid theoretic self-linking number formula and study itsapplications. We classify an action of the $n$-strand braid group on the free group of rank$n$ which is similar to the Artin representation in the sense that the $i$-thgenerator $\sigma_{i}$ of $B_{n}$ acts so that it fixes all free generators$x_{j}$ except $j = i,i+1$. We determine all such representations and discussknot invariants coming from such representations. We develop a theory of curve diagrams for Artin groups of type B. We definethe winding number labeling and the wall crossing labeling of curve diagrams,and show that these labelings detect the classical and the dual Garside length,respectively. A remarkable point is that our argument does not require Garsidetheory machinery like normal forms, and is more geometric in nature. We show that an amalgamated free product $G*_{A}H$ admits a discrete isolatedordering, under some assumptions of $G,H$ and $A$. This generalizes theauthor's previous construction of isolated orderings, and unlike knownconstructions of isolated orderings, can produce an isolated ordering with manynon-trivial proper convex subgroups. We show that a non-trivial, non-central normal subgroup of the braid groupscontains a braid whose closure is a hyperbolic knot with arbitrary large genus.This shows that non-faithfulness of a quantum representation implies that thecorresponding quantum invariant fails to detect the unknot. The proof utilizesthe Dehornoy ordering of the braid groups. We establish relations between both the classical and the dual Garsidestructures of the braid group and the Burau representation. Using the classicalstructure, we formulate a non-vanishing criterion for the Burau representationof the 4-strand braid group. In the dual context, it is shown that the Buraurepresentation for arbitrary braid index is injective when restricted to theset of \emph{simply-nested braids}.
Note that $\binom{74}{37}=\frac{74!}{37!37!}$ Note also that $n!$ contains $\sum\limits_{k=1}^\infty \left\lfloor\frac{n}{p^k}\right\rfloor$ factors of a prime $p$ Counting the number of factors of $19$ in the numerator, we get $\lfloor\frac{74}{19}\rfloor+\lfloor\frac{74}{19^2}\rfloor+\dots=3+0+0+\dots=3$ The number of factors of $19$ in the denominator however, we get each $37!$ has $\lfloor\frac{37}{19}\rfloor+\lfloor\frac{37}{19^2}\rfloor+\dots=1+0+0+\dots=1$ so the denominator has $2$ factors of $19$. Thus, one of the factors from the numerator remains uncancelled and so $\binom{74}{37}$ is divisible by $19$. This implies $\binom{74}{37}-2$ is not divisible by $19$ and therefore cannot be divisible by $38$. Similarly, running the same argument for counting the number of factors of $2$, we get a total of $71$ factors of two on the numerator and $68$ factors of two on the denominator, implying that $\binom{74}{37}$ is divisible by four. Thus $\binom{74}{37}-2$ is not divisible by $4$ and therefore cannot be divisible by $36$. For showing $\binom{74}{37}-2$ is divisible by $37^2$, we note first that $37$ is prime. We conjecture that $\binom{2p}{p}-2$ is divisible by $p^2$ for each prime $p$. We notice first that $\binom{2p}{p}=\sum\limits_{j=0}^p\binom{p}{j}^2$ and also that $\binom{p}{j}\equiv 0\pmod{p}$ for $0<j<p$ and $\binom{p}{0}=\binom{p}{p}=1$ otherwise. ( To see this, use a similar technique as above, noting that there is exactly one factor of $p$ in the numerator and no factors of $p$ in the denominator of $\frac{p!}{j!(p-j)!}$ for each $0<j<p$) As the terms in the summation are being squared, we notice further that $\binom{p}{j}^2\equiv 0\pmod{p^2}$ for each $0<j<p$ and $\binom{p}{0}^2=\binom{p}{p}^2=1$ Thus $\binom{2p}{p}\equiv \sum\limits_{j=0}^p\binom{p}{j}^2\equiv 1+(0+0+\dots+0)+1\equiv 2\pmod{p^2}$ Finally, $\binom{2p}{p}-2\equiv 0\pmod{p^2}$ so the claim is true. This proves in particular that $\binom{74}{37}-2$ is divisible by $37^2$
Let $M$ be a riemannian manifold (let $\left\langle \cdot,\cdot \right\rangle_{p}$ be the scalar product on $T_{p}M$). Let $p \in M$ and $\xi \in T_{p}M$. We consider the geodesic $\gamma \, : \, t \, \mapsto \, \exp_{\gamma(0)}(t\xi)$ where $\gamma(0)=p$. Let $\eta \in T_{p}M$. We can introduce a "small perturbation" of the geodesic $\gamma$ : $\gamma_{\varepsilon}(t) = \exp_{\gamma(0)}(t(\xi+\varepsilon \eta))$. Let $f \, : \, (t,\varepsilon) \, \mapsto \, \exp_{\gamma(0)}(t(\xi + \varepsilon \eta))$. We definie $J(t) = \frac{\partial f}{\partial \varepsilon}(t,0)$. Then, $\forall t, \, J(t) \in T_{\gamma(t)}M$. $J$ is a Jacobi vector field. Let $P_{\gamma,0,t} \, : \, T_{p}M \, \rightarrow \, T_{\gamma(t)}M$ be the parallel transport along $\gamma$. I have read that we have the following approximation : for a "small" $\delta t$, $$ \frac{J(\delta t)}{\delta t} - P_{\gamma, 0, \delta t}(\eta) = o(\delta t) $$ The Jacobi vector field $J$ gives a first order approximation of the parallel transport of $\eta$ along the geodesic $\gamma$. I would like to prove this approximation but I don't really know how to start... Considering that $J(t) = \frac{\partial f}{\partial \varepsilon}(t,0)$, we can easily show that $$ J(t) = \mathrm{D}_{t \xi} \left( \exp_{\gamma(0)} \right) \cdot (t \eta) $$ So that, $J(\delta t) = \mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot (\delta t \eta) = \delta t \, \mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot \eta$. Then, $$ \frac{J(\delta t)}{\delta t} = \mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot \eta$$ We want to prove that $\mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot \eta - P_{\gamma,0,\delta t}(\eta)$ goes to $0$ as $\delta t \, \rightarrow \, 0$. One idea I had was to consider $$ \Vert \mathrm{D}_{\delta t \xi} \left( \exp_{\gamma(0)} \right) \cdot \eta - P_{\gamma,0,\delta t}(\eta) \Vert_{\gamma(\delta t)}$$ since the parallel transport is a linear isometry, $\Vert P_{\gamma,0,\delta t}(\eta) , P_{\gamma,0,\delta t}(\eta) \Vert_{\gamma(\delta t)} = \Vert \eta \Vert_{p}$. But it doesn't seem to be very helpful. Can anyone give me a hint ? Thank you for your help.
A classics teacher covering detention asked the question: One angle of a triangle is ${96}^\circ$, find the other two angles $\psi, \theta$ given that: $2\sin4\theta = 2\sin2\psi + 2\cos2\psi + 3\sec3\theta +2\alpha +2\sec2\psi$ Please find the other two angles so you can go home. Addendum: The teacher knows basic arithmetic and geometry, but trigonometry is Greek to him. His proudest mathematical moment occurred during an interview when asked to find the value of, $3\sin3\psi \cdot 3 \tan3\theta$ if $\psi={30}^\circ$ and $\theta={15}^\circ$ answering even before the interviewer had finished speaking. He got the job. He said confidently 9 Hint $4\cosh3\psi + 3\sinh4\theta$
Isosceles triangle (a two-dimensional figure) has two sides that are the same length or at least two congruent sides. Angle bisector of a right isosceles triangle is a line that splits an angle into two equal angles. Circumcircle is a circle that passes through all the vertices of a two-dimensional figure. Congruent is all sides having the same lengths and angles measure the same. Inscribed circle is the largest circle possible that can fit on the inside of a two-dimensional figure. Semiperimeter is one half of the perimeter. a = c x = y x + y + z = 180° Height: $h_a$, $h_b$, $h_c$ Median: $m_a$, $m_b$, $m_c$ - A line segment from a vertex (corner point) to the midpoint of the opposite side Angle bisectors: $t_a$, $t_b$, $t_c$ - A line that splits an angle into two equal angles 3 edges 3 vertexs Area of an Isosceles Triangle formula $\large{ A_{area} = \frac {h\;b} {2} }$ Where: $\large{ A_{area} }$ = area $\large{ b }$ = side $\large{ h }$ = height Circumcircle of an Isosceles Triangle formula $\large{ R = \frac { a^2 } { \sqrt { 4\; a^2 \;-\; b^2 } } }$ Where: $\large{ R }$ = outcircle $\large{ a, b }$ = side Height of an Isosceles Triangle formula $\large{ h = 2 \frac {A_{area}}{b} }$ $\large{ h = \sqrt { a^2 - \frac {b^2}{4} } }$ Where: $\large{ h }$ = height $\large{ a, b }$ = side $\large{ A_{area} }$ = area Inscribed Circle of an Isosceles Triangle formula The radius of a inscribed circle (inner) of an Isosceles triangle if given side $( r )$. $\large{ r = \frac { b } { 2 } \; \sqrt { \frac { 2\;a \;-\; b } { 2\;a \;+\; b } } }$ The radius of a inscribed circle (inner) of an Isosceles triangle if given side and angle $( r )$. $\large{ r = a \; \frac { sine \; \alpha \;x\; cos \; \alpha } { 1 \;+\; cos \; \alpha } = \alpha \; cos \; \alpha \;\;x\;\; tan \frac { \alpha } { 2 } }$ $\large{ r = \frac {b}{2} \;x\; \frac { sine \; \alpha } { 1 \;+\; cos \; \alpha } = \frac {b}{2} \;x\; tan \frac { \alpha } { 2 } }$ The radius of a inscribed circle (inner) of an Isosceles triangle if given side and height $( r ) $. $\large{ r = \frac { b\;h } { b \;+\; \sqrt { 4\;h^2 \;+\; b^2 } } }$ $\large{ r = \frac { h\; \sqrt { a^2 \;-\; h^2 } } { a \;+\; \sqrt { a^2 \;-\; h^2 } } }$ Where: $\large{ r }$ = incircle $\large{ a, b }$ = side $\large{ \alpha }$ (Greek symbol alpha) = angle $\large{ h }$ = height Perimeter of an Isosceles Triangle formula $\large{ P = 2\;a + b }$ Where: $\large{ P }$ = perimeter $\large{ a, b }$ = side Semiperimeter of an Isosceles Triangle formula $\large{ s = \frac { a \;+\; b \;+\; c } { 2 } }$ Where: $\large{ s }$ = semiperimeter $\large{ a, b, c }$ = side Side of an Isosceles Triangle formula $\large{ a = \frac {P} {2} - \frac {b} {2} }$ $\large{ b = P - 2\;a }$ $\large{ b = 2\; \frac {A_{area}} {h} }$ Where: $\large{ a, b }$ = side $\large{ h }$ = height $\large{ P }$ = perimeter $\large{ A_{area} }$ = area Tags: Equations for Triangle
Standing waves for a class of Kirchhoff type problems in ${\mathbb {R}^3}$ involving critical Sobolev exponents Article First Online: Abstract We are concerned with the following Kirchhoff type equation with critical nonlinearity: where $\varepsilon $ is a small positive parameter, $a,b>0$, $\lambda > 0$, $2 < p \le 4$. Under certain assumptions on the potential $$\begin{aligned} \left\{ \begin{array}{ll} - \Bigl ( {{\varepsilon ^2}a + \varepsilon b\int _{{\mathbb {R}^3}} {{{\| {\nabla u} \|}^2}} } \Bigr )\Delta u + V(x)u = \lambda {\| u \|^{p - 2}}u + {\| u \|^4}u{\text { in }}{\mathbb {R}^3}, \\ u > 0,u \in {H^1}({\mathbb {R}^3}), \\ \end{array} \right. \end{aligned}$$ V, we construct a family of positive solutions ${u_\varepsilon } \in {H^1}({\mathbb {R}^3})$ which concentrates around a local minimum of Vas $\varepsilon \rightarrow 0$. Although, critical growth Kirchhoff type problem has been studied in e.g. He et al. [18], where the assumption for $$\begin{aligned} \left\{ \begin{array}{ll} - \Bigl ( {{\varepsilon ^2}a + \varepsilon b\int _{{\mathbb {R}^3}} {{{\| {\nabla u} \|}^2}} } \Bigr )\Delta u + V(x)u = f(u)+{u^5}{\text { in }}{\mathbb {R}^3}, \\ u > 0,u \in {H^1}({\mathbb {R}^3}) \\ \end{array} \right. \end{aligned}$$ f( u) is that $f(u) \sim \|u{\|^{p - 2}}u$ with $4 < p < 6$ and satisfies the Ambrosetti-Rabinowitz condition which forces the boundedness of any Palais-Smale sequence of the corresponding energy functional of the equation. As $g(u): = \lambda {\| u \|^{p - 2}}u + {\| u \|^4u}$ with $2<p\le 4$ does not satisfy the Ambrosetti-Rabinowitz condition ($\exists \mu > 4, 0 < \mu \int _0^u {g(s)ds \le g(u)u}$), the boundedness of Palais–Smale sequence becomes a major difficulty in proving the existence of a positive solution. Also, the fact that the function $g(s)/{s^3}$ is not increasing for $s > 0$ prevents us from using the Nehari manifold directly as usual. Our result extends the main result in He et al. [18] concerning the existence and concentration of positive solutions to the case where $f(u) \sim \|u{\|^{p - 2}}u$ with $4 < p < 6$. Mathematics Subject ClassificationPrimary 35J20 35J60 35J92 Notes Acknowledgments The authors would like to express their sincere gratitude to the referee for all insightful comments and valuable suggestions, based on which the paper was revised. References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.Gilbarg, D., Trudinger, N.S.: Elliptic Partial Differential Equations of Second Order, 2nd edn, vol. 224. Grundlehren Math. Wiss., Springer, Berlin (1983)Google Scholar 16. 17. 18. 19. 20. 21. 22.Kirchhoff, G.: Mechanik. Teubner, Leipzig (1883)Google Scholar 23. 24.Lions, P.L.: The concentration–compactness principle in the calculus of variations, the locally compact case, part 2. Ann. Inst. H. Poincaré Anal. Non. Linéaire 2, 223–283 (1984)Google Scholar 25. 26.Lions, J.L.: On some questions in boundary value problems of mathematical physics. In: Contemporary Developments in Continuum Mechanics and Partial Differential Equations, Proceedings of International Symposium, Inst. Mat., Univ. Fed. Rio de Janeiro, Rio de Janeiro, 1977, North-Holland Math. Stud. vol. 30, North-Holland, Amsterdam, 1978, pp. 284–346Google Scholar 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38.Ramos, M., Wang, Z.Q., Willem, M.: Positive Solutions for Elliptic Equations with Critical Growth in Unbounded Domains, Calculus of Variations and Differential Equations. Chapman & Hall/CRC Press, Boca Raton (2000)Google Scholar 39. 40. 41.Willem, M.: Minimax theorems, Progress in Nonlinear Differential Equations and their Applications, 24. Birkhäuser Boston Inc, Boston (1996)Google Scholar 42. 43.Zhang, J., Chen, Z., Zou, W.: Standing waves for nonlinear Schrödinger equations involving critical growth, arXiv:1209.3074v1 (2012) 44. Copyright information © Springer-Verlag Berlin Heidelberg 2015
Using R, I am trying to simulate how the power of a Monte Carlo two-sample test for central tendency changes with sample size. However, my simulation results does not show power increasing with sample size which is clearly wrong. Could someone please advise where I am going wrong? Below is the basic set-up. The null hypothesis is: $H_0=\mu_X-\mu_Y=0$ The alternative hypothesis is: $H_1=\mu_X>\mu_Y=0$ The test statistic is: $E(x)-E(y)$ For my simulation, I use two samples $X\sim N(\mu_X, \sigma_y)$, $Y\sim N(\mu_Y, \sigma_Y)$ of size $n_X$ and $n_Y$ respectively. I run 100 simulations for every increase in the sample size of X. Below is my coded simulation with comments: mc.pvalue<-function(nx, ny, mu.x, mu.y, sig.x, sig.y){ x<-rnorm(nx, mu.y, sig.x) # Generate samples under H1: mu_x > mu_y y<-rnorm(ny, mu.x, sig.y) obs.diff<-mean(x)-mean(y) # Test-stat is observed diff. in means se.x<-sd(x)/sqrt((length(x))) # Calculate standard errors se.y<-sd(y)/sqrt((length(y))) count=0 # Set counter equal to zero for(i in 1:1000){ # Generate 1000 pairs of samples and calculate x1<-rnorm(nx, mu.y, se.x) # the difference in means between each pair. y1<-rnorm(ny, mu.y, se.y) # The samples have been generated under sim.diff<-mean(x1)-mean(y1) # H0: mu_x = mu_y if(sim.diff<= obs.diff){count=count+1} } count/1000 # Estimated p-value for test statistic}# Calculate 100 estimated p-values for the test statistic. Then find the proportion of times# reject hypothesis at level of significance 0.05calc.pvalues<-function(nx, ny=10, mu.x=21, mu.y=20, sig.x=0.5, sig.y=0.25, n.sim=100){ pvalues<-replicate(n.sim, mc.pvalue(nx, ny, sig.x, sig.y, mu.x, mu.y)) sum(pvalues<0.05)/n.sim }chge.sample.size<-seq(2, 100, by=0.05) # Sample size x increasing from 1 to 100result<-sapply(chge.sample.size, calc.pvalues) # Apply the calc.pvalues function to # estimate p-value for each sample size # of x plot(chge.sample.size, result) # Plot increasing sample size of X # against test power When I an edit the code to calculate the power of a t-test, it works perfectly: calc.pvalue.t<-function(nx, ny, sig.x, sig.y, mu.x, mu.y){ x<-rnorm(nx, mu.x, sig.x) y<-rnorm(ny, mu.y, sig.y) t.test(x, y, alternative="greater", paired=FALSE, conf.level=0.95)$p.value}calc.t<-function(nx, ny=10, mu.x=21, mu.y=20, sig.x=0.5, sig.y=0.25, n.sim=100, alpha=0.05){ p.values<-replicate(n.sim, calc.pvalue.t(nx, ny, sig.x, sig.y, mu.x, mu.y)) sum(p.values<alpha)/n.sim }chge.sample.size<-seq(2, 10, by=1)tpower1<-sapply(chge.sample.size, calc.t)plot(chge.sample.size, tpower1) Something seems to be going wrong with the first part of my code when I try to do the Monte Carlo test
Much of pure mathematics exists to simplify our world, even if it means entering an abstract realm (or creating one) to do it. The isomorphism is one of the most powerful tools for discovering structural similarities (or that two groups are identical structurally) between two groups that on the surface look completely unrelated. In this post, we’ll look at what an isomorphism is. What’s an isomorphism? Pinter’s Abstract Algebra gives a fantastic motivation for studying isomorphic things: we want to know if two objects we’re looking at are really just the same object with, say, different outfits on. Take the two palindromes MADAM and ROTOR. They’re two different words, but if we replace M with R, A with O, and D with T, we can transform MADAM into ROTOR. This tells us that these two palindromes have the same structure. In this case, that they are essentially a “word reflection” about the middle letter. MADAM and ROTOR are just wearing two different outfits. We can replace the letters “M”, “A”, “D” with any three other letters, then reflect the first two about the third to create another word palindrome with the same structure. (Now, picking any three letters at random might not guarantee the palindrome you make is an English word.) Let’s take this into a mathematical realm now. We could create a function f that maps the three letters “M”, “A”, “D” with the structure of the palindrome MADAM onto three new letters, maintaining that structure. So the functionf = \begin{pmatrix}M&A&D\\\downarrow&\downarrow&\downarrow\\R&O&T\end{pmatrix} transforms MADAM into ROTOR. The correspondence is injective (a one-to-one mapping, taking one element of \{M,A,D\} to only one element of \{R,O,T\}, and surjective (every letter “R”, “O”, “T” has something that maps to it), and therefore is called bijective. This means we can actually move back and forth from MADAM to ROTOR by taking the inverse of the function f that mapped us from “M”,”A”,”D”, to “R”,”O”,”T”. Such a function between two groups that preserves the structure of the groups is called an and the two groups are said to be isomorphism isomorphic. Formal Definition of Isomorphism . Suppose G_{1} and G_{2} are groups, and let the operation for G_{1} be denoted \cdot, and the operation for G_{2} be denoted \star. A bijective function f:G_{1}\to G_{2} such that for any two elements a and b in G_{1}f(a\cdot b) = f(a)\star f(b) Definition (isomorphism): is called an isomorphism isomorphic . How do we find out if two groups are isomorphic? Short answer: we find the isomorphism between the two groups. Now, note that since isomorphisms are bijective, we can decide whether we’ll try to find a function from G_{1} to G_{2} or from G_{2} to G_{1}. It’s a simple task to say, but in practice this can actually be extremely difficult, requiring some guesswork and refinement once we find what we think the function might look like. Let’s take an example and look at \mathbb{Z}_{4} and the group \langle G, \cdot\rangle = \langle\{1,-1,i,-i\},\cdot\rangle. (Remember that \mathbb{Z}_{4} is the group of integers modulo 4 under addition.) We’re going to show these two groups are isomorphic by finding the mapping from \mathbb{Z}_{4} to G that preserves the structure of their operation tables. The operation table of \mathbb{Z}_{4} under addition modulo 4 is given by\begin{array}{r\|rrrr}\textcolor{#893B2F}{\langle\mathbb{Z}_{4},+\rangle}&\textcolor{#893B2F}{0}&\textcolor{#893B2F}{1}&\textcolor{#893B2F}{2}&\textcolor{#893B2F}{3}\\\textcolor{#893B2F}{0}&0&1&2&3\\\textcolor{#893B2F}{1}&1&2&3&0\\\textcolor{#893B2F}{2}&2&3&0&1\\\textcolor{#893B2F}{3}&3&0&1&2\end{array} The operation table of G under multiplication is\begin{array}{r\|rrrr}\textcolor{#893B2F}{\langle G,\cdot\rangle}&\textcolor{#893B2F}{1}&\textcolor{#893B2F}{i}&\textcolor{#893B2F}{-1}&\textcolor{#893B2F}{-i}\\\textcolor{#893B2F}{1}&1&i&-1&-i\\\textcolor{#893B2F}{i}&i&-1&-i&1\\\textcolor{#893B2F}{-1}&-1&-i&1&i\\\textcolor{#893B2F}{-i}&-i&1&i&-1\end{array} I wrote the multiplication table for G using that particular order for a reason, mostly to make the isomorphism easier to see. We needed to play with mapping 0, 1, 2, and 3 to the set G such that the table structures are preserved. There are 4!=24 possible mappings. We can try each one, which is a pain. Each person can develop his own strategy, but mine was first to note that the identity element in one group must map to the identity element in the other group. Thus, we know that 0 \in \mathbb{Z}_{4} must map to 1 \in G. From there, I need to figure out what pairs of elements generate the respective identity elements under the respective operations. So whatever element in G that 1 \in \mathbb{Z}_{4} maps to must multiply (remember we have to use the group element of G when we’re over there) with whatever element in G that 3 \in \mathbb{Z}_{4} maps to to give 1 \in G. We can look at the operation table for G and see that i\cdot -i = 1, so we note that perhaps we’ll map 1\in Z_{4} \to i \in G, and 3 \in \mathbb{Z}_{4} \to -i \in G. That leaves us mapping 2\in \mathbb{Z}_{4} \to -1 \in G. We then see if that mapping allows us to interchange the elements of \mathbb{Z}_{4} under mod 4 addition with the elements of G under multiplication 1 to see if we get the same table. As it turns out, we do, so is our isomorphism. (You can also check for yourself that for any elements a and b in \mathbb{Z}_{4}, f((a+b)\bmod 4) = f(a)\cdot f(b).) How do we find out if two groups are not isomorphic? Proving no such function that satisfies the definition of an isomorphism exists for two given groups is far harder in many cases than finding one. 2 Isomorphisms require that the structure and properties of a group be preserved when being mapped to another group. Some possible things to look for that can easily show two groups are not isomorphic: One group may be commutative, and the other isn’t. Commutativity is a structural property, so if the two groups don’t share that property, they can’t have the same structure. One group may have an element (or a few) that are their own inverses, and the other either has none with this property, or a different number of elements with this property. Obvious one: if the two groups can’t be put into 1-1 correspondence (because they have differing numbers of elements), then they cannot be isomorphic. These aren’t the only things to look for by any means. Determining whether or not two groups are isomorphic generally requires quite a bit of lateral thinking. However, a mathematician named Arthur Cayley gave us a classic theorem of modern algebra that allows us to simplify the problem by showing that every group is isomorphic to some group of permutations. In the next article, we’ll explore the theorem and the proof. This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.
Hello guys! I was wondering if you knew some books/articles that have a good introduction to convexity in the context of variational calculus (functional analysis). I was reading Young's "calculus of variations and optimal control theory" but I'm not that far into the book and I don't know if skipping chapters is a good idea. I don't know of a good reference, but I'm pretty sure that just means that second derivatives have consistent signs over the region of interest. (That is certainly a sufficient condition for Legendre transforms.) @dm__ yes have studied bells thm at length ~2 decades now. it might seem airtight and has stood the test of time over ½ century, but yet there is some fineprint/ loopholes that even phd physicists/ experts/ specialists are not all aware of. those who fervently believe like Bohm that no new physics will ever supercede QM are likely to be disappointed/ dashed, now or later... oops lol typo bohm bohr btw what is not widely appreciated either is that nonlocality can be an emergent property of a fairly simple classical system, it seems almost nobody has expanded this at length/ pushed it to its deepest extent. hint: harmonic oscillators + wave medium + coupling etc But I have seen that the convexity is associated to minimizers/maximizers of the functional, whereas the sign second variation is not a sufficient condition for that. That kind of makes me think that those concepts are not equivalent in the case of functionals... @dm__ generally think sampling "bias" is not completely ruled out by existing experiments. some of this goes back to CHSH 1969. there is unquestioned reliance on this papers formulation by most subsequent experiments. am not saying its wrong, think only that theres very subtle loophole(s) in it that havent yet been widely discovered. there are many other refs to look into for someone extremely motivated/ ambitious (such individuals are rare). en.wikipedia.org/wiki/CHSH_inequality @dm__ it stands as a math proof ("based on certain assumptions"), have no objections. but its a thm aimed at physical reality. the translation into experiment requires extraordinary finesse, and the complex analysis starts with CHSH 1969. etc While it's not something usual, I've noticed that sometimes people edit my question or answer with a more complex notation or incorrect information/formulas. While I don't think this is done with malicious intent, it has sometimes confused people when I'm either asking or explaining something, as... @vzn what do you make of the most recent (2015) experiments? "In 2015 the first three significant-loophole-free Bell-tests were published within three months by independent groups in Delft, Vienna and Boulder. All three tests simultaneously addressed the detection loophole, the locality loophole, and the memory loophole. This makes them “loophole-free” in the sense that all remaining conceivable loopholes like superdeterminism require truly exotic hypotheses that might never get closed experimentally." @dm__ yes blogged on those. they are more airtight than previous experiments. but still seem based on CHSH. urge you to think deeply about CHSH in a way that physicists are not paying attention. ah, voila even wikipedia spells it out! amazing > The CHSH paper lists many preconditions (or "reasonable and/or presumable assumptions") to derive the simplified theorem and formula. For example, for the method to be valid, it has to be assumed that the detected pairs are a fair sample of those emitted. In actual experiments, detectors are never 100% efficient, so that only a sample of the emitted pairs are detected. > A subtle, related requirement is that the hidden variables do not influence or determine detection probability in a way that would lead to different samples at each arm of the experiment. ↑ suspect entire general LHV theory of QM lurks in these loophole(s)! there has been very little attn focused in this area... :o how about this for a radical idea? the hidden variables determine the probability of detection...! :o o_O @vzn honest question, would there ever be an experiment that would fundamentally rule out nonlocality to you? and if so, what would that be? what would fundamentally show, in your opinion, that the universe is inherently local? @dm__ my feeling is that something more can be milked out of bell experiments that has not been revealed so far. suppose that one could experimentally control the degree of violation, wouldnt that be extraordinary? and theoretically problematic? my feeling/ suspicion is that must be the case. it seems to relate to detector efficiency maybe. but anyway, do believe that nonlocality can be found in classical systems as an emergent property as stated... if we go into detector efficiency, there is no end to that hole. and my beliefs have no weight. my suspicion is screaming absolutely not, as the classical is emergent from the quantum, not the other way around @vzn have remained civil, but you are being quite immature and condescending. I'd urge you to put aside the human perspective and not insist that physical reality align with what you expect it to be. all the best @dm__ ?!? no condescension intended...? am striving to be accurate with my words... you say your "beliefs have no weight," but your beliefs are essentially perfectly aligned with the establishment view... Last night dream, introduced a strange reference frame based disease called Forced motion blindness. It is a strange eye disease where the lens is such that to the patient, anything stationary wrt the floor is moving forward in a certain direction, causing them have to keep walking to catch up with them. At the same time, the normal person think they are stationary wrt to floor. The result of this discrepancy is the patient kept bumping to the normal person. In order to not bump, the person has to walk at the apparent velocity as seen by the patient. The only known way to cure it is to remo… And to make things even more confusing: Such disease is never possible in real life, for it involves two incompatible realities to coexist and coinfluence in a pluralistic fashion. In particular, as seen by those not having the disease, the patient kept ran into the back of the normal person, but to the patient, he never ran into him and is walking normally It seems my mind has gone f88888 up enough to envision two realities that with fundamentally incompatible observations, influencing each other in a consistent fashion It seems my mind is getting more and more comfortable with dialetheia now @vzn There's blatant nonlocality in Newtonian mechanics: gravity acts instantaneously. Eg, the force vector attracting the Earth to the Sun points to where the Sun is now, not where it was 500 seconds ago. @Blue ASCII is a 7 bit encoding, so it can encode a maximum of 128 characters, but 32 of those codes are control codes, like line feed, carriage return, tab, etc. OTOH, there are various 8 bit encodings known as "extended ASCII", that have more characters. There are quite a few 8 bit encodings that are supersets of ASCII, so I'm wary of any encoding touted to be "the" extended ASCII. If we have a system and we know all the degrees of freedom, we can find the Lagrangian of the dynamical system. What happens if we apply some non-conservative forces in the system? I mean how to deal with the Lagrangian, if we get any external non-conservative forces perturbs the system?Exampl... @Blue I think now I probably know what you mean. Encoding is the way to store information in digital form; I think I have heard the professor talking about that in my undergraduate computer course, but I thought that is not very important in actually using a computer, so I didn't study that much. What I meant by use above is what you need to know to be able to use a computer, like you need to know LaTeX commands to type them. @AvnishKabaj I have never had any of these symptoms after studying too much. When I have intensive studies, like preparing for an exam, after the exam, I feel a great wish to relax and don't want to study at all and just want to go somehwere to play crazily. @bolbteppa the (quanta) article summary is nearly popsci writing by a nonexpert. specialists will understand the link to LHV theory re quoted section. havent read the scientific articles yet but think its likely they have further ref. @PM2Ring yes so called "instantaneous action/ force at a distance" pondered as highly questionable bordering on suspicious by deep thinkers at the time. newtonian mechanics was/ is not entirely wrong. btw re gravity there are a lot of new ideas circulating wrt emergent theories that also seem to tie into GR + QM unification. @Slereah No idea. I've never done Lagrangian mechanics for a living. When I've seen it used to describe nonconservative dynamics I have indeed generally thought that it looked pretty silly, but I can see how it could be useful. I don't know enough about the possible alternatives to tell whether there are "good" ways to do it. And I'm not sure there's a reasonable definition of "non-stupid way" out there. ← lol went to metaphysical fair sat, spent $20 for palm reading, enthusiastic response on my leadership + teaching + public speaking abilities, brought small tear to my eye... or maybe was just fighting infection o_O :P How can I move a chat back to comments?In complying to the automated admonition to move comments to chat, I discovered that MathJax is was no longer rendered. This is unacceptable in this particular discussion. I therefore need to undo my action and move the chat back to comments. hmmm... actually the reduced mass comes out of using the transformation to the center of mass and relative coordinates, which have nothing to do with Lagrangian... but I'll try to find a Newtonian reference. One example is a spring of initial length $r_0$ with two masses $m_1$ and $m_2$ on the ends such that $r = r_2 - r_1$ is it's length at a given time $t$ - the force laws for the two ends are $m_1 \ddot{r}_1 = k (r - r_0)$ and $m_2 \ddot{r}_2 = - k (r - r_0)$ but since $r = r_2 - r_1$ it's more natural to subtract one from the other to get $\ddot{r} = - k (\frac{1}{m_1} + \frac{1}{m_2})(r - r_0)$ which makes it natural to define $\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$ as a mass since $\mu$ has the dimensions of mass and since then $\mu \ddot{r} = - k (r - r_0)$ is just like $F = ma$ for a single variable $r$ i.e. an spring with just one mass @vzn It will be interesting if a de-scarring followed by a re scarring can be done in some way in a small region. Imagine being able to shift the wavefunction of a lab setup from one state to another thus undo the measurement, it could potentially give interesting results. Perhaps, more radically, the shifting between quantum universes may then become possible You can still use Fermi to compute transition probabilities for the perturbation (if you can actually solve for the eigenstates of the interacting system, which I don't know if you can), but there's no simple human-readable interpretation of these states anymore @Secret when you say that, it reminds me of the no cloning thm, which have always been somewhat dubious/ suspicious of. it seems like theyve already experimentally disproved the no cloning thm in some sense.
Third, since $\sf{L} \subseteq \sf{NC}^2$, is there an algorithm to convert any logspace algorithm into a parallel version? It can be shown (Arora and Barak textbook) given a $t(n)$-time TM $M$, that an oblivious TM $M'$ (i.e. a TM whose head movement is independent of its input $x$) can construct a circuit $C_n$ to compute $M(x)$ where $\|x\| = n$. The proof sketch is along the lines of having $M'$ simulate $M$ and defining "snapshots" of its state (i.e. head positions, symbols at heads) at each time-step $t_i$ (think of a computational log). Each step $t_i$ can be computed from $x$ and the state $t_{i-1}$. Because each snapshot involves only a constant-sized string, and there exist only a constant amount of strings of that size, the snapshot at $t_i$ can be computed by a constant-sized circuit. If you compose the constant-sized circuits for each $t_i$ we have a circuit that computes $M(x)$. Using this fact, along with the restriction that the language of $M$ is in $\sf{L}$ we see that our circuit $C_n$ is by definition logspace-uniform, where uniformity just means that our circuits in our circuit family $\{C_n\}$ computing $M(x)$ all have the same algorithm. Not a custom-made algorithm for each circuit operating on input size $n$. Again, from the definition of uniformity we see that circuits deciding any language in $\sf{L}$ must have a function $\text{size}(n)$ computable in $O(\log n).$ The circuit family $\sf{AC}^1$ has at most $O(\log n)$ depth. Finally it can be shown that $\sf{AC}^1 \subseteq \sf{NC}^2$ giving the relation in question. Fourth, it sounds like most people assume that $\sf{NC} \neq \sf{P}$ in the same way that $\sf{P} \neq \sf{NP}$. What is the intuition behind this? Before we go further, let us define what $\sf{P}$-completeness means. A language $L$ is $\sf{P}$-complete if $L \in \sf{P}$ and every language in $\sf{P}$ is logspace reducible to it. Additionally, if $L$ is $\sf{P}$-complete then the following are true $L \in \sf{NC} \iff \sf{P} = \sf{NC}$ $L \in \sf{L} \iff \sf{P} = \sf{L}$ Now we consider $\sf{NC}$ to be the class of languages efficiently decided by a parallel computer (our circuit). There are some problems in $\sf{P}$ that seem to resist any attempt at parallelization (i.e. Linear Programming, and Circuit Value Problem). That is to say, certain problems require computation to be done in a step-wise fashion. For example, the Circuit Value Problem is defined as: Given a circuit $C$, input $x$, and a gate $g \in C$, what is the output of $g$ on $C(x)$? We do not know how to compute this any better than computing all the gates $g'$ that come before $g$. Given some of them may be computed in parallel, for example if they all occur at some time-step $t_i$, but we dont know how compute the output of gates at timestep $t_i$ and time-step $t_{i+1}$ for the obvious difficulty that gates at $t_{i+1}$ require the output of gates at $t_i$! This is the intuition behind $\sf{NC} \neq \sf{P}$. Limits to Parallel Computation is a book about $\sf{P}$-Completeness in similar vein of Garey & Johnson's $\sf{NP}$-Completeness book.
Suppose that we have two probability distributions, $f$ and $g$ on the subsets of a finite set $X$, i.e. $f, g: P(X) \to [0,1]$, with $$ \sum_{A \subseteq X} f(A) = \sum_{A \subseteq X} g(A) = 1. $$ An upper subset of $P(X)$ is just one closed under taking supersets. Definition: $g$ dominates $f$ if, for all upper subsets $U \subseteq P(X)$,$$\sum_{A \in U} f(A) \leq \sum_{A \in U} g(A).$$ Question: Is there an efficient algorithm for determining whether or not $g$ dominates $f$? Obviously, one can't hope for anything much better than $O(2^n)$ when $X$ has $n$ elements. The problem can be translated to a problem concerning the maximum flow in a graph which is (basically) two copies of $P(X)$, with edges of capacity 1 directed from any $A$ in the first copy to all of $A$'s supersets (including $A$ itself) in the second copy (we add a source vertex connected to each $A$ in the first copy with capacity $f(A)$ and a sink vertex from each $B$ in the second copy with capacity $g(B)$). However, the standard max-flow algorithms don't give $O(2^n)$ for that translation, and it feels like there might be a 'trick'.
Back to Nonlinear Least Squares The Levenberg-Marquardt algorithm can be thought of as a trust-region modification of the Gauss-Newton algorithm. Levenberg-Marquardt steps are obtained by solving subproblems of the form for some $\Delta_k > 0$ and scaling matrix $D_k$. The trust-region radius is adjusted between iterations according to the agreement between predicted and actual reduction in the objective function. For a step to be accepted, the ratio $\rho_k = \frac{r(x_k) - r(x_k + s_k)} {r_k - \frac{1}{2} \\|f^\prime (x_k)s_k + f(x_k) \\|_2^2 }$ must exceed a small positive number (typically 0.0001). If this test is failed, the trust region is decreased and the step is recalculated. When the ratio is close to one, the trust region for the next iteration is expanded. Levenberg-Marquardt codes usually determine the step by noting that the solution of (1.2) also satisfies the equation for some $\lambda_k \geq 0$. The Lagrange multiplier $\lambda_k$ is zero if the minimum-norm Gauss-Newton step is smaller than $\Delta_k$; otherwise $\lambda_k$ is chosen so that $\\|D_k s_k \\|_2 = \Delta_k$ Equations (1.3) are simply the normal equations for the least squares problem Efficient factorization of the coefficient matrix in (1.4) can be performed by a combination of Householder and Givens transformations. The Levenberg-Marquardt algorithm has proved to be an effective and popular way to solve nonlinear least squares problems. MINPACK-1 contains Levenberg-Marquardt codes in which the Jacobian matrix may be either supplied by the user or calculated by using finite differences. IMSL, MATLAB, ODRPACK, and PROC NLP also contain Levenberg-Marquardt routines. The algorithms in ODRPACK solve unconstrained nonlinear least squares problems and orthogonal distance regression problems, including those with implicit models and multiresponse data. The simplest case of an orthogonal distance regression problem occurs when we are given observations $y_i$ at times $t_i$ and a model $\phi(t_i, \cdot)$ for each observation. In this case, we aim to find sets of parameters $x \in \mathbb{R}^n$ and corrections $\delta_i$ that solve the minimization problem The ODRPACK algorithms use a trust-region Levenberg-Marquardt method that exploits the structure of this problem, so that there is little difference between the cost per iteration for this problem and the standard least squares problem in which the $\delta_i$ are fixed at zero.
This question already has an answer here: I need to find the time complexity of the following simple algorithm. Calculate the time complexity of the following algorithm: void f(int n) { int i, x; for (i = 1; i <= n; i++) if (i % 2 == 1) x++ } It obvious that the time complexity of this algorithm is $\Theta(N)$, but how I prove it formally? I know that the loop will do $N$ iterations, so the total number of operations will be $N \times \text{(number of operations inside the loop)}$. Here the tricky part, I know that for each iteration will be one operation for the comparison and one operation only if $i$ is odd. So it's something like this: $\sum_{i=1}^{n}(1+\frac{1}{2})$? I'm not sure, it's seems too strange and I'm not sure how to continue. What is the best way to write it formally? Also, I would like to ask if there are books with many examples of analysis of simple algorithms like this and proofs of their complexity. The most recommended book seems to be CLRS, but I didn't found many examples of this kind there and from what I understand it can be very complex topic (depending on the series you need to calculate in the end). Thank you.
@JosephWright Well, we still need table notes etc. But just being able to selectably switch off parts of the parsing one does not need... For example, if a user specifies format 2.4, does the parser even need to look for e syntax, or ()'s? @daleif What I am doing to speed things up is to store the data in a dedicated format rather than a property list. The latter makes sense for units (open ended) but not so much for numbers (rigid format). @JosephWright I want to know about either the bibliography environment or \DeclareFieldFormat. From the documentation I see no reason not to treat these commands as usual, though they seem to behave in a slightly different way than I anticipated it. I have an example here which globally sets a box, which is typeset outside of the bibliography environment afterwards. This doesn't seem to typeset anything. :-( So I'm confused about the inner workings of biblatex (even though the source seems.... well, the source seems to reinforce my thought that biblatex simply doesn't do anything fancy). Judging from the source the package just has a lot of options, and that's about the only reason for the large amount of lines in biblatex1.sty... Consider the following MWE to be previewed in the build in PDF previewer in Firefox\documentclass[handout]{beamer}\usepackage{pgfpages}\pgfpagesuselayout{8 on 1}[a4paper,border shrink=4mm]\begin{document}\begin{frame}\[\bigcup_n \sum_n\]\[\underbrace{aaaaaa}_{bbb}\]\end{frame}\end{d... @Paulo Finally there's a good synth/keyboard that knows what organ stops are! youtube.com/watch?v=jv9JLTMsOCE Now I only need to see if I stay here or move elsewhere. If I move, I'll buy this there almost for sure. @JosephWright most likely that I'm for a full str module ... but I need a little more reading and backlog clearing first ... and have my last day at HP tomorrow so need to clean out a lot of stuff today .. and that does have a deadline now @yo' that's not the issue. with the laptop I lose access to the company network and anythign I need from there during the next two months, such as email address of payroll etc etc needs to be 100% collected first @yo' I'm sorry I explain too bad in english :) I mean, if the rule was use \tl_use:N to retrieve the content's of a token list (so it's not optional, which is actually seen in many places). And then we wouldn't have to \noexpand them in such contexts. @JosephWright \foo:V \l_some_tl or \exp_args:NV \foo \l_some_tl isn't that confusing. @Manuel As I say, you'd still have a difference between say \exp_after:wN \foo \dim_use:N \l_my_dim and \exp_after:wN \foo \tl_use:N \l_my_tl: only the first case would work @Manuel I've wondered if one would use registers at all if you were starting today: with \numexpr, etc., you could do everything with macros and avoid any need for \<thing>_new:N (i.e. soft typing). There are then performance questions, termination issues and primitive cases to worry about, but I suspect in principle it's doable. @Manuel Like I say, one can speculate for a long time on these things. @FrankMittelbach and @DavidCarlisle can I am sure tell you lots of other good/interesting ideas that have been explored/mentioned/imagined over time. @Manuel The big issue for me is delivery: we have to make some decisions and go forward even if we therefore cut off interesting other things @Manuel Perhaps I should knock up a set of data structures using just macros, for a bit of fun [and a set that are all protected :-)] @JosephWright I'm just exploring things myself “for fun”. I don't mean as serious suggestions, and as you say you already thought of everything. It's just that I'm getting at those points myself so I ask for opinions :) @Manuel I guess I'd favour (slightly) the current set up even if starting today as it's normally \exp_not:V that applies in an expansion context when using tl data. That would be true whether they are protected or not. Certainly there is no big technical reason either way in my mind: it's primarily historical (expl3 pre-dates LaTeX2e and so e-TeX!) @JosephWright tex being a macro language means macros expand without being prefixed by \tl_use. \protected would affect expansion contexts but not use "in the wild" I don't see any way of having a macro that by default doesn't expand. @JosephWright it has series of footnotes for different types of footnotey thing, quick eye over the code I think by default it has 10 of them but duplicates for minipages as latex footnotes do the mpfoot... ones don't need to be real inserts but it probably simplifies the code if they are. So that's 20 inserts and more if the user declares a new footnote series @JosephWright I was thinking while writing the mail so not tried it yet that given that the new \newinsert takes from the float list I could define \reserveinserts to add that number of "classic" insert registers to the float list where later \newinsert will find them, would need a few checks but should only be a line or two of code. @PauloCereda But what about the for loop from the command line? I guess that's more what I was asking about. Say that I wanted to call arara from inside of a for loop on the command line and pass the index of the for loop to arara as the jobname. Is there a way of doing that?
What values for theta (0<=theta<=2pi) satify the equation? 2 sin theta cos theta + cos theta = 0 $2\sin\theta\cos\theta+\cos\theta\ =\ 0$ Factor cos θ out of both terms. $(\cos\theta)(2\sin\theta+1)\ =\ 0$ Set each factor equal to 0 and solve for θ . And remember that we only want solutions for θ in the interval [0, 2π] $\begin{array}{} \cos\theta\ =\ 0&\qquad\text{or}\qquad&2\sin\theta+1\ =\ 0\\~\\ \theta\ =\ \frac{\pi}{2}&&2\sin\theta\ =\ -1\\~\\ \theta\ =\ \frac{3\pi}{2}&&\sin\theta\ =\ -\frac{1}{2}\\~\\ &&\theta\ =\ \frac{7\pi}{6}\\~\\ &&\theta\ =\ \frac{11\pi}{6} \end{array}$ So the values for θ in the interval [0, 2π] that satisfy the equation are: $\frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}$_ $2\sin\theta\cos\theta+\cos\theta\ =\ 0$ Factor cos θ out of both terms. $(\cos\theta)(2\sin\theta+1)\ =\ 0$ Set each factor equal to 0 and solve for θ . And remember that we only want solutions for θ in the interval [0, 2π] $\begin{array}{} \cos\theta\ =\ 0&\qquad\text{or}\qquad&2\sin\theta+1\ =\ 0\\~\\ \theta\ =\ \frac{\pi}{2}&&2\sin\theta\ =\ -1\\~\\ \theta\ =\ \frac{3\pi}{2}&&\sin\theta\ =\ -\frac{1}{2}\\~\\ &&\theta\ =\ \frac{7\pi}{6}\\~\\ &&\theta\ =\ \frac{11\pi}{6} \end{array}$ So the values for θ in the interval [0, 2π] that satisfy the equation are: $\frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}$_
Conveners Top Physics There are no conveners in this block Top Physics There are no conveners in this block Top Physics Andrew Ivanov (Kansas State University (US)) Christopher Neu (University of Virginia (US)) Top Physics There are no conveners in this block Nikolaos Kidonakis (Kennesaw State University) 04/08/2015, 14:00 Top Physics I present a calculation of higher-order corrections from NNLL threshold resummation for cross sections and differential distributions in top-antitop pair production and in single-top production. I show that soft-gluon threshold corrections are the dominant contribution to top-quark production and closely approximate exact results through NNLO. I show aN$^3$LO results for the total $t{\bar t}$... Thomas Andrew Schwarz (University of Michigan (US)) 04/08/2015, 14:30 Top Physics The most up-to-date top quark pair production cross section measurements will be presented for p-p collisions with the ATLAS detector at the LHC. Both differential and inclusive measurements will be discussed, as well as analyses focusing on rare decays such as top quark pairs associated with an additional boson or jets. The results will be discussed in the context of new physics, Monte... 331. Top pair production inclusive and differential production cross section measurements in pp collisions Dr Gabriele Benelli (University of Kansas and Fermilab LPC Distinguished Researcher) 04/08/2015, 15:00 Top Physics Top pair production cross-section measurements and theoretical predictions have achieved unprecedented precision. A review is presented of recent top-quark pair inclusive and differential production cross sections in proton-proton collisions at the LHC at centre-of-mass energies of 7 and 8 TeV, using data collected by the CMS experiment in the years 2011 and 2012. The differential... Rebeca Gonzalez Suarez (University of Nebraska (US)) 04/08/2015, 16:00 Top Physics The Run-1 of the LHC was very successful for single top physics in CMS. The main single top production mode, t-channel, is well established. The t-channel cross-section was measured with unprecedented precision and t-channel events have been used for the first time to perform measurement of SM properties, such as: \|Vtb\|, top quark polarization, or W-helicity fractions. The associated... Kenneth Bloom (University of Nebraska (US)) 04/08/2015, 16:30 Top Physics We present the most recent measurements on top-quark physics obtained with Tevatron $p \bar p$ collisions recorded by the D0 experiment at $\sqrt s=1.96$ TeV. The full Run II data set of 9.7 fb$^{-1}$ is analyzed. Both lepton+jets and dilepton channels of top-quark pair production are used to measure the differential and inclusive cross-sections, the forward-backward asymmetries, the... Philip Ilten (Massachusetts Inst. of Technology (US)) 04/08/2015, 17:00 Top Physics Excellent vertexing, tracking, and jet energy resolution enable a variety of forward analyses to be performed using data from the LHCb detector. Recent results on the performance of inclusive b and c-tagging algorithms used by LHCb are presented. These taggers are applied to W+jet data to measure W+b-jet and W+c-jet asymmetries as well as ratios to inclusive W+jet production.... Graham Wilson (University of Kansas (US)) 04/08/2015, 17:30 Top Physics The International Linear Collider and Compact Linear Collider projects aim to build a linear electron-positron collider with a center-of-mass energy well above the top quark pair production threshold. In this contribution an overview is presented of the potential of their top quark precision physics programme. One of the highlights is a precise determination of the top quark mass through a... Zhi Liu (Indiana University) 05/08/2015, 14:00 Top Physics Signals for Lorentz and CPT violation can appear in a wide range of experiments including hadron colliders like the LHC. We present a calculation of the Lorentz-violating cross section for top-quark pair production via gluon fusion. This process dominates at the LHC, and analysis of LHC data should permit sharpening the constraints on top-quark Lorentz violation obtained recently by the D0... Ziqing Hong 05/08/2015, 14:30 Top Physics The forward-backward asymmetry of the top quark pair production at the Tevatron experiments is one of the hottest topics in particle physics in the recent years. It provides unique precision tests of the standard model and of physics beyond the standard model. We present the latest measurement of the top forward-backward asymmetry in the dilepton final state and summarize the legacy results of... 70. Measurement of the charge asymmetry in top quark pair production in 8 TeV $pp$ collision data collected by the ATLAS experiment Daniel Marley (University of Michigan (US)) 05/08/2015, 14:55 Top Physics A summary of results for the top quark charge asymmetry, $\text{A}_{\text{C}}$, measured using 20.3 fb$^{-1}$ of data recorded with the ATLAS detector at a center-of-mass energy $\sqrt{s}=8$ TeV is presented. Events where either both top quarks decay leptonically (dileptonic), or one top quark decays leptonically and the other hadronically (semi-leptonic) are considered. The semi-leptonic... Andrew Brinkerhoff (University of Notre Dame (US)) 05/08/2015, 16:00 Top Physics New measurements of top quark pair production in association with a W or Z boson are presented, using 19.5 fb−1 of 8 TeV pp collision data collected by the CMS experiment at the CERN LHC. Final states with opposite-sign, same-sign, three, and four charged leptons plus b-tagged jets are examined. Signal ttW and ttZ events are identified by reconstructing the top quark pair, yielding the most... 153. Measurement of the production cross sections of top quark pairs in association with a W or Z boson using proton-proton collisions at $\sqrt{s}$=8 TeV with the ATLAS detector Chen Zhou (Duke University (US)) 05/08/2015, 16:25 Top Physics We present a measurement of the production cross sections of top quark pairs in association with a W or Z boson. The measurement uses 20.3 $fb^{-1}$ of data from proton-proton collisions at $\sqrt{s}$=8 TeV recorded by the ATLAS detector at the LHC at CERN. Four different final states are considered: two opposite-sign leptons, two same-sign leptons, three leptons, and four leptons. We report... Andrew Ivanov (Kansas State University (US)) 05/08/2015, 16:50 Top Physics We present a measurement of the production cross-section of top-quark pairs in association with a photon in pp collisions at center-of-mass energy of 8 TeV. The data was recorded at the CMS experiment, corresponding to 19.7 fb-1 integrated luminosity. The measurement is performed in dilepton and lepton+jets final states. Data-driven methods are used to estimate the photon identification... Jesse Alan Heilman (University of California Riverside (US)) 05/08/2015, 17:15 Top Physics A search is presented for standard model (SM) production of four top quarks in pp collisions at the LHC. Using Run-1 data corresponding to an integrated luminosity of 19$fb^{-1}$, a combination of kinematic reconstruction and multivariate techniques is used to distinguish between the small signal ($\sigma^{SM}_{t \bar{t} t \bar{t}} \sim 1 fb$) and backgrounds in the lepton + jets channel. ...
Proof of Lemma 3 Given $i\in N$ , let $\langle v,L\rangle =\mathcal {G}^0_N(i)\bigcap \mathcal {G}^1_N(i)$ , then for each coalition $S\in C^L_{N \backslash \{i\}}$ , we get $v^L(S\cup \{i\})=v^(S)$ and $\sum _{k\in S\cup \{i\}}[v^{L}(S\cup \{i\})-v^L(S\cup \{i\}\backslash \{k\})]=0$ . Thus, the following equation holds, $$\begin{aligned} \sum _{k\in S}[v^L(S\cup \{i\})-v^L(S\cup \{i\}\backslash \{k\})]=0. \end{aligned}$$ In this equation, let $S=\{j\}$ , where $j\ne i$ , we get $v^L(\{i,j\})=v^L(\{i\})$ . By $\langle v,L\rangle \in \mathcal {G}^0_N(i)$ , we deduce $v^L(\{i\})=0$ and $v^L(\{i,j\})=v^L(\{j\})$ . Therefore, $v^L(\{j\})=v^L(\{i,j\})=v^L(\{i\})=0$ . Using mathematical induction on the cardinality of S, we can verify that $v^L(S)=v^L(S\cup \{i\})=0$ exists for each $S\in C^L_{N\backslash \{i\}}$, which means $\langle v,L\rangle $ is the null game, hence $\mathcal {G}^0_N(i)\bigcap \mathcal {G}^1_N(i)=\emptyset $. Note that the dimension of the space $\mathcal {G}_N(i)$ equals $2^n-2$ . According to $\mathcal {G}^0_N(i)\subset \mathcal {G}_N(i)$ , $\mathcal {G}^1_N(i)\subset \mathcal {G}_N(i)$ , it remains to prove that the dimensions of the linear space $\mathcal {G}^0_N(i)$ and $\mathcal {G}^1_N(i)$ are at least $2^{n-1}-1$ , respectively. Define two classes of games as $$\begin{aligned} \mathcal {B}^0_N(i)=\{\langle u_T,L\rangle \|i\notin T, T\ne \emptyset \}, \ \ \ \mathcal {B}^1_N(i)=\{\langle w_T,L\rangle \|i\notin T, T\ne \emptyset \}, \end{aligned}$$ where $\langle w_T,L\rangle $ is defined as in Sect. 3 and $$\begin{aligned} u^L_T(S)=\left\{ \begin{array}{ll} 1, &{} S=T \text { or } S=T\cup \{i\}; \\ 0, &{} \text {otherwise.} \\ \end{array}\right. \end{aligned}$$ Clearly, both $\mathcal {B}^0_N(i)$ and $\mathcal {B}^1_N(i)$ are composed of a set of linearly independent games. Besides, since $\mathcal {B}^0_N(i)\subset \mathcal {G}^0_N(i)$, $\mathcal {B}^1_N(i)\subset \mathcal {G}^1_N(i)$ and the cardinalities of $\mathcal {B}^0_N(i)$ and $\mathcal {B}^1_N(i)$ are respectively $2^{n-1}-1$, we can deduce the cardinalities of $\mathcal {G}^0_N(i)$ and $\mathcal {G}^1_N(i)$ are respectively $2^{n-1}-1$. $\square $ Proof of Theorem 2 It is easy to check that $\varPhi ^\lambda (v,L)$ satisfies component efficiency, additivity, symmetry, rationality, proportionality of Shapley value and proportionality of Solidarity value, the uniqueness will be proved as follow: Step 1. We will prove for each $i\in N$ , there exists $\lambda _i\in \mathbb {R}$ , so that the following equality establish, $\langle v,L\rangle \in \mathcal {G}^L_N$ , $$\begin{aligned} \varPhi _i(v,L)=\lambda _i Sh_i(v,L)+(1-\lambda _i)Sol_i(v,L). \end{aligned}$$ (3) Choose any game $\langle w,L\rangle \in \mathcal {B}^1_N(i)\subset \mathcal {G}^1_N(i)$ such that $Sh_i(w,L)>0$ , let $\lambda _i=\frac{\varPhi _i(w,L)}{Sh_i(w,L)}$ , by proportionality of Shapley value, for all $\langle v,L\rangle \in \mathcal {G}^1_N(i)$ , we have $$\begin{aligned} \varPhi _i(v,L)=\lambda _i Sh_i(v,L). \end{aligned}$$ Similarly, choose any game $\langle u,L\rangle \in \mathcal {B}^0_N(i)\subset \mathcal {G}^0_N(i)$ such that $Sol_i(u,L)>0$ , let $\gamma _i=\frac{\varPhi _i(u,L)}{Sol_i(u,L)}$ , combining proportionality of Solidarity value, for each $\langle v,L\rangle \in \mathcal {G}^0_N(i)$ , the following equality holds: $$\begin{aligned} \varPhi _i(v,L)=\gamma _i Sol_i(v,L). \end{aligned}$$ According to the definitions of $\mathcal {G}^0_N(i)$ and $\mathcal {G}^1_N(i)$ , given $\langle v,L\rangle \in \mathcal {G}^0_N(i)$ , we have $$\begin{aligned} \varPhi _i(v,L)=\lambda _i Sh_i(v,L)+\gamma _i Sol_i(v,L). \end{aligned}$$ (4) Considering game $\langle w,L\rangle \in \mathcal {G}^L_N$ , where $$\begin{aligned} w^L(S)=\left\{ \begin{array}{ll} n, &{} S=N; \\ 0, &{} \text {otherwise,} \\ \end{array}\right. \end{aligned}$$ by component efficiency and symmetry, it follows $\varPhi _i(w,L)=1$ , $i\in N$ . Similarly, it follows that $$\begin{aligned} Sh_i(w,L)=Sol_i(w,L)=1, \ i\in N. \end{aligned}$$ In the Eq. (4 ), let $\langle w,L\rangle =\langle v,L\rangle $ , we get $\gamma _i=1-\lambda _i$ . Consequently, the Eq. (3 ) holds for any $\langle v,L\rangle \in \mathcal {G}_N(i)$ . Now, defining a game $\langle u,L\rangle \in \mathcal {G}^L_N$ as follow: $$\begin{aligned} u^L(S)=\left\{ \begin{array}{ll} 0, &{} S=\emptyset ; \\ 1, &{} S\ne \emptyset . \\ \end{array} \right. \end{aligned}$$ Obviously $\langle u,L\rangle \notin \mathcal {G}_N(i)$ . Note that as linear space, the dimensions of $\mathcal {G}^L_N$ and $\mathcal {G}_N(i)$ are $2^n-1$ and $2^n-2$ respectively, hence, any $\langle v,L\rangle \in \mathcal {G}^L_N$ can be represented as follow $$\begin{aligned} v=v^0+cu, \end{aligned}$$ where $\langle v^0,L\rangle \in \mathcal {G}_N(i)$ and c is a constant coefficient. According to component efficiency, additivity, symmetry, it follows that $$\begin{aligned} \varPhi _i(v,L)=\varPhi _i(v^0,L)+\varPhi _i(cu,L)=\lambda _i Sh_i(v^0,L)+(1-\lambda _i)Sol_i(v^0,L)+\frac{c}{n}. \end{aligned}$$ In addition, we have $$\begin{aligned} \lambda _i Sh_i(v,L)+(1-\lambda _i)Sol_i(v,L)=\lambda _i Sh_i(v^0,L)+(1-\lambda _i)Sol_i(v^0,L)+\frac{c}{n}. \end{aligned}$$ Consequently, Eq. (3 ) holds for any $\langle v,L\rangle \in \mathcal {G}^L_N$ . Step 2. We will prove that coefficient $\lambda _i$ in Eq. (3 ) is independent of i . Assuming that $n\ge 2$ , for $\langle v,L\rangle \in \mathcal {G}^L_N$ , component efficiency implies: $$\begin{aligned} \sum _{i\in N}[\lambda _i Sh_i(v,L)+(1-\lambda _i)Sol_i(v,L)]=v^L(N). \end{aligned}$$ By component efficiency of Solidarity value, this equation is equivalent to $$\begin{aligned} \sum _{i\in N}\lambda _i[Sh_i(v,L)-Sol_i(v,L)]=0. \end{aligned}$$ By component efficiency of the Shapley and Solidarity value, we have $$\begin{aligned}&\sum _{i=2}^n(\lambda _i-\lambda _1)[Sh_i(v,L)-Sol_i(v,L)] \\= & {} \sum _{i\in N}\lambda _i[Sh_i(v,L)-Sol_i(v,L)]-\sum _{i\in N}\lambda _1[Sh_i(v,L)-Sol_i(v,L)] \\= & {} 0. \end{aligned}$$ This means for $\langle v,L\rangle \in \mathcal {G}^L_N$ , $$\begin{aligned} \sum _{i=2}^n\gamma _i[Sh_i(v,L)-Sol_i(v,L)]=0, \end{aligned}$$ (5) where $\gamma _i=\lambda _i-\lambda _1$ , $i=2,3,\ldots ,n$ . Defining games $\langle v_k,L\rangle $ , $k=2,3,\ldots ,n$ , as follow $$\begin{aligned} v^L_k(S)=\left\{ \begin{array}{ll} 1, &{} S=N\backslash \{k\}; \\ 0, &{} \text { otherwise.} \\ \end{array} \right. \end{aligned}$$ According to Eq. (5 ), $$\begin{aligned} \sum _{i=2}^n\gamma _i[Sh_i(v_k,L)-Sol_i(v_k,L)]=0, \ k=2,\ldots ,n \end{aligned}$$ and $$\begin{aligned} Sh_i(v_k,L)=\left\{ \begin{array}{ll} -\frac{1}{n}, &{} \ i=k; \\ \frac{1}{n(n-1)}, &{} \ i\ne k, \\ \end{array} \right. , \ \ Sol_i(v_k,L)=\left\{ \begin{array}{ll} -\frac{1}{n^2}, &{} \ i=k; \\ \frac{1}{n^2(n-1)}, &{} \ i\ne k, \\ \end{array} \right. \end{aligned}$$ we get the linear system $$\begin{aligned} (n-1)\gamma _k-\sum _{i\ne k}\gamma _i-0, \ k=2,3, \ldots , n. \end{aligned}$$ The only solution of the above linear system is $\gamma _2=\gamma _3=\cdots ,\gamma _n=0$ . Then for every $i\in N$, it implies $\lambda _i-\lambda _1=0$, $i=2,3,\ldots , n$. Now we have proved that coefficient $\lambda _i$ is independent of i and $\lambda _i=\lambda _1=\lambda $, $i=2,3,\ldots , n$. Step 3. The coefficient $\lambda $ will be proved belonging to interval [0, 1] in this step. Assuming that $n\ge 2$ , for any player $i\in N$ , considering the following unanimity games: $$\begin{aligned} u^L_{N\backslash \{i\}}(S)=\left\{ \begin{array}{ll} 1, &{} \ S=N \text { or } S=N\backslash \{i\}; \\ 0, &{} \ \text {otherwise.} \\ \end{array} \right. \end{aligned}$$ By applying the property of rationality to $\langle u_{N\backslash \{i\}},L\rangle $ , we obtain $\varPhi _i(u_{N\backslash \{i\}},L)\ge 0$ . In addition, combining the following equations: $$\begin{aligned} Sh_i(u_{N\backslash \{i\}},L)=0, \ \ Sol_i(u_{N\backslash \{i\}},L)=\frac{n-1}{n^2}, \ \lambda _i=\lambda , \end{aligned}$$ we have $$\begin{aligned} 0\le \varPhi _i(u_{N\backslash \{i\}},L)=\lambda Sh_i(u_{N\backslash \{i\}},L)+(1-\lambda )Sol_i(u_{N\backslash \{i\}},L)=\frac{(1-\lambda )(n-1)}{n^2}. \end{aligned}$$ Evidently, we deduce that $\lambda \le 1$ . Considering the game $\langle \bar{w},L\rangle $ , which is defined as follow, $$\begin{aligned} \bar{w}^L(S)=\left\{ \begin{array}{ll} -1, &{} \ S=N; \\ -n, &{} \ S=N\backslash \{i\}; \\ 0, &{} \ \text {otherwise.} \\ \end{array} \right. \end{aligned}$$ Applying the property of rationality to $\langle \bar{w},L\rangle $ , we get $\varPhi _i(\bar{w},L)\ge 0$ . By the conditions: $$\begin{aligned} Sh_i(\bar{w},L)=\frac{n-1}{n}, \ \ Sol_i(\bar{w},L)=0, \end{aligned}$$ we have $$\begin{aligned} 0\le \varPhi _i(\bar{w},L)=\lambda Sh_i(\bar{w},L)+(1-\lambda )Sol_i(\bar{w},L)=\frac{\lambda (n-1)}{n}, \end{aligned}$$ which implies $\lambda \ge 0$ . Then for $n\ge 2$ , we have proven the conclusion. When $n=1$ , the theorem is obviously established. $\square $
A non-Abelian Chern-Simons(C-S) has the action$$S=\int L dt=\int \frac{k}{4\pi}Tr[\big( A \wedge d A + (2/3) A \wedge A \wedge A \big)]$$ We know that the common cases, $A=A^a T^a$ is the connection as a Lie algebra valued one form. $T^a$ is the generator of the Lie group. The well-known case is the well-defined SU(2) C-S theory and SO(3) C-S theory. SU(2) is a compact, simple, simply-connected Lie group. SO(3) is a compact, simple, connected but not simply connected Lie group. Question: what is the minimum requirement on the group structure of $A$ in Chern-Simons theory? Can we have the group of $A$ of C-S theory: (1) to be NOT a Lie group? (2) to be NOT compact? (3) to be NOT connected? (4) to be a Lie group but NOT a simple-Lie group? Please could you also explain why is it so, and better with some examples of (1),(2),(3),(4). ps. Of course, I know C-S theory is required to be invariant under a gauge transformation$$A \to U^\dagger(A-id)U$$with a boundary derives a Wess-Zumino-Witten term. Here I am questioning the constraint on the group. Many thanks!This post imported from StackExchange Physics at 2014-03-06 16:49 (UCT), posted by SE-user cycles
Hello one and all! Is anyone here familiar with planar prolate spheroidal coordinates? I am reading a book on dynamics and the author states If we introduce planar prolate spheroidal coordinates $(R, \sigma)$ based on the distance parameter $b$, then, in terms of the Cartesian coordinates $(x, z)$ and also of the plane polars $(r , \theta)$, we have the defining relations $$r\sin \theta=x=\pm R^2−b^2 \sin\sigma, r\cos\theta=z=R\cos\sigma$$ I am having a tough time visualising what this is? Consider the function $f(z) = Sin\left(\frac{1}{cos(1/z)}\right)$, the point $z = 0$a removale singularitya polean essesntial singularitya non isolated singularitySince $Cos(\frac{1}{z})$ = $1- \frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$$$ = (1-y), where\ \ y=\frac{1}{2z^2}+\frac{1}{4!... I am having trouble understanding non-isolated singularity points. An isolated singularity point I do kind of understand, it is when: a point $z_0$ is said to be isolated if $z_0$ is a singular point and has a neighborhood throughout which $f$ is analytic except at $z_0$. For example, why would $... No worries. There's currently some kind of technical problem affecting the Stack Exchange chat network. It's been pretty flaky for several hours. Hopefully, it will be back to normal in the next hour or two, when business hours commence on the east coast of the USA... The absolute value of a complex number $z=x+iy$ is defined as $\sqrt{x^2+y^2}$. Hence, when evaluating the absolute value of $x+i$ I get the number $\sqrt{x^2 +1}$; but the answer to the problem says it's actually just $x^2 +1$. Why? mmh, I probably should ask this on the forum. The full problem asks me to show that we can choose $log(x+i)$ to be $$log(x+i)=log(1+x^2)+i(\frac{pi}{2} - arctanx)$$ So I'm trying to find the polar coordinates (absolute value and an argument $\theta$) of $x+i$ to then apply the $log$ function on it Let $X$ be any nonempty set and $\sim$ be any equivalence relation on $X$. Then are the following true: (1) If $x=y$ then $x\sim y$. (2) If $x=y$ then $y\sim x$. (3) If $x=y$ and $y=z$ then $x\sim z$. Basically, I think that all the three properties follows if we can prove (1) because if $x=y$ then since $y=x$, by (1) we would have $y\sim x$ proving (2). (3) will follow similarly. This question arised from an attempt to characterize equality on a set $X$ as the intersection of all equivalence relations on $X$. I don't know whether this question is too much trivial. But I have yet not seen any formal proof of the following statement : "Let $X$ be any nonempty set and $∼$ be any equivalence relation on $X$. If $x=y$ then $x\sim y$." That is definitely a new person, not going to classify as RHV yet as other users have already put the situation under control it seems... (comment on many many posts above) In other news: > C -2.5353672500000002 -1.9143250000000003 -0.5807385400000000 C -3.4331741299999998 -1.3244286800000000 -1.4594762299999999 C -3.6485676800000002 0.0734728100000000 -1.4738058999999999 C -2.9689624299999999 0.9078326800000001 -0.5942069900000000 C -2.0858929200000000 0.3286240400000000 0.3378783500000000 C -1.8445799400000003 -1.0963522200000000 0.3417561400000000 C -0.8438543100000000 -1.3752198200000001 1.3561451400000000 C -0.5670178500000000 -0.1418068400000000 2.0628359299999999 probably the weirdness bunch of data I ever seen with so many 000000 and 999999s But I think that to prove the implication for transitivity the inference rule an use of MP seems to be necessary. But that would mean that for logics for which MP fails we wouldn't be able to prove the result. Also in set theories without Axiom of Extensionality the desired result will not hold. Am I right @AlessandroCodenotti? @AlessandroCodenotti A precise formulation would help in this case because I am trying to understand whether a proof of the statement which I mentioned at the outset depends really on the equality axioms or the FOL axioms (without equality axioms). This would allow in some cases to define an "equality like" relation for set theories for which we don't have the Axiom of Extensionality. Can someone give an intuitive explanation why $\mathcal{O}(x^2)-\mathcal{O}(x^2)=\mathcal{O}(x^2)$. The context is Taylor polynomials, so when $x\to 0$. I've seen a proof of this, but intuitively I don't understand it. @schn: The minus is irrelevant (for example, the thing you are subtracting could be negative). When you add two things that are of the order of $x^2$, of course the sum is the same (or possibly smaller). For example, $3x^2-x^2=2x^2$. You could have $x^2+(x^3-x^2)=x^3$, which is still $\mathscr O(x^2)$. @GFauxPas: You only know $\|f(x)\|\le K_1 x^2$ and $\|g(x)\|\le K_2 x^2$, so that won't be a valid proof, of course. Let $f(z)=z^{n}+a_{n-1}z^{n-1}+\cdot\cdot\cdot+a_{0}$ be a complex polynomial such that $\|f(z)\|\leq 1$ for $\|z\|\leq 1.$ I have to prove that $f(z)=z^{n}.$I tried it asAs $\|f(z)\|\leq 1$ for $\|z\|\leq 1$ we must have coefficient $a_{0},a_{1}\cdot\cdot\cdot a_{n}$ to be zero because by triangul... @GFauxPas @TedShifrin Thanks for the replies. Now, why is it we're only interested when $x\to 0$? When we do a taylor approximation cantered at x=0, aren't we interested in all the values of our approximation, even those not near 0? Indeed, one thing a lot of texts don't emphasize is this: if $P$ is a polynomial of degree $\le n$ and $f(x)-P(x)=\mathscr O(x^{n+1})$, then $P$ is the (unique) Taylor polynomial of degree $n$ of $f$ at $0$.
This question already has an answer here: For the recurrence relation $$T(n) = 16T(n/4) + n!\,,$$ I have found that $T(n)\in Θ(n!)$. Can this be deduced using the Master Theorem? Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It only takes a minute to sign up.Sign up to join this community This question already has an answer here: For the recurrence relation $$T(n) = 16T(n/4) + n!\,,$$ I have found that $T(n)\in Θ(n!)$. Can this be deduced using the Master Theorem? Yes. Recall that the Master Theorem deals with recurrences of the form $$ T(n) = a\cdot T\left(\frac{n}{b}\right)+f(n) $$ In your particular case you have $a = 16$, $b=4$ and $f(n) = n!$. Then we decide the complexity by comparing $f(n)$ to three functions of $n$, $a$ and $b$ (really we compare it basically the same function, with a $\log$ wiggle room, but in different ways). In particular the third case holds where $f(n) \in \Omega(n^{c})$ where $c > \log_{b}a$. You have that $\log_{b}a = \log_{4}16 = 2$, so you have to prove (or at least convince yourself, depending on how rigorous you're being) that $n! \in \Omega(n^{c})$ for some $c > 2$. It should be reasonably clear that you can pick any constant $c$ in this instance, and that this case holds. As it does, we get immediately (which is of course the point of the Master Theorem) that $T(n) \in \Theta(f(n))$, which in this case gives $T(n) \in \Theta(n!)$.
-1. Ask what $D^{50} \left\{ \ x^{100} \ \right\}$ is. Stop them from actually multiplying the coefficients; just leave them unmultiplied the way you would for a prime decomposition. Define factorial and show how $\displaystyle \prod_{50}^{100} \bullet = {100! \over 50!}$ shows up naturally in this context of repeated derivatives. This proves that arbitrary number of $D$'s of a polynomial is easy. By fundamental theorem of algebra, $\mathrm{Polynomials}$ cover all functions (so, school algebra can work in the real world) By Taylor's approximation, $\mathrm{Finite\ Polynomials}$ can get you close (so, it actually could work, in the real-real world) Each polynomial is "just a" sequence. (The covector of coefficients to the one basis-vector $(\ldots, x^{-1}, x^0, x^1, x^2, \ldots)$ (with $x^0$ moving the root to wherever you want We can understand sequences. (There is an analogy to digits, possibly negative, in base $x$, but that could be ignored.) We have a shift / lag operator ($D$) on these sequences. (−1) So symbolic differentiation & integration of these letters / strings / symbols, gives us an "Easy way to do calculus" (No hard integrals! Only polynomials). If most functions can be written this way (1), and we can even know beforehand (in the planning phase) how good of a job this approach will do (2), Now play around with them — do ℂ→ℂ Wegert-plots and Cartesian ℝ→ℝ plots, and input/output tables to see how varying the "sliders" (entries in the covector of coefficients) changes the shape of the output in the various views (Cartesian, Wegert, tabular). A computer/interactive setup could be the best here. Just mess around and try to get the output to do cool stuff. exp(ζ)/(ζζζ−ζζ−ζ+i) { ζζζ−ζζ−ζ+i } / { log(ζ)+i } (For the ℂ→arg ℂ plots you want to tell them that arguments swirl around the roots ($\arg \zeta$ takes on every argument at 0) For applications, you can point to some interesting things to do with ℂ maps (dynamical systems, root finders, some of http://arminstraub.com/math/what-is-column). Further applications you can discuss net present value, the geometric series (and the dumb geometric-series trick), the second dumb trick of subtracting two shifted geometric series to get a finite one (since no-one lives forever), and how although rates-of-return might vary in reality, we've now carved out some "easy cases" that we can do back-of-the-envelope, as quickly as $1 \over 1-r$ $- \mathrm{ shift} \left\{ 1 \over 1-r \right\}$ to put some easy bounds on real-world cases. (If your students don't already know how to compute a NPV in Excel, then here is your chance to teach them the only mathematics they will actually need to know in 90% of office jobs.) You can also tell them they now understand (finally) what $e$ is. All the little kids are just made to use it, they can now explain it. (Identify it with the covector $(1,1,1,1,1,\ldots)$ once you adjust by $/ \sim k!$.) And show how $D(e^{\bullet}) = e^{\bullet}$ via $(1,1,1,1,\ldots) \overset{D}{\mapsto} (1,1,1,\ldots)$. Reinforce the shift-arithmetic on symbols p.o.v., mention why this makes $\exp$'s a good basis for ODE's, and mention the "slick" version of instantiating $\sqrt{-1}$ (simply assume it) --- versus seeing it in series. (So which is better? Chance to make a value judgement / form an opinion on mathematical beauty.) It's not a bad time either to talk about the history of $\sqrt{-1}$: it was only possible to discover this "algebraic" / "axiomatic" / "slick" thing (with so many implications) by using stuff from another part of mathematics (analysis, but let's call it series / string-ops). By staring at and doodling around (mention Ramanujan) with the $\exp$ series, only if you've also tried to approximate the circle height function ($\sin$) and its cousin by infinite-approximation (mention Aristotle) To me $\sqrt{-1}$ comes from the intersection of different areas of mathematics, i.e. from different ways of thinking being tied together by writing down on paper. This is a thesis about the way mathematics works as well as about creativity, exploration, proof, dealing with your own ideas, how paper can help you organise your thoughts. (in other domains perhaps, as well) Now the discussion has gone from meaningless symbol-scratching to high-level. The usual approach (Stewart) is, in my opinion, insane. I would show the Stewart chapters as enrichment. This is heavy symbol scratching with no motivation why it matters at all. Yet another exercise in making people do mechanical calculations, instead of teaching them anything about mathematics. Series are here to make our lives easier. They either make the impossible possible (formal power series, Ramanujan stuff) , make hard integrals turn into shift-operators, or they give you some peace-of-mind that (pretty much) any ℝ→ℝ mapping the real world might throw at you, can be reduced to some stuff you could do on paper, in a symbolic differentiator, etc. Reduced to essentially a string / list.
The three traceless Pauli matrices $\sigma_x,\sigma_y,\sigma_z$ are arbitrary in the sense that any three operators with the appropriate commutation relations can be represented with those matrices. This is expressed by the action of the group SU(2) on $\mathbb{C}^2$ and hence also on the space of linear operators $\mathcal{L}(\mathbb{C}^2)$. I want to prove the following: Let $A,B$ two anti-commuting Hermitian unitary traceless operators on a spin. Then there exists unitary U such that $U^\dagger AU=\sigma_z$ and $U^\dagger B U=\sigma_x $ This can be proven quite easily by a solving a system of linear equations on the entries of U. In fact there is even a simpler way. But I am more interested in a simple abstract argument which is more insightful. Perhaps some argument that will have to do with the action of the group $SU(3)\times SU(2) $ where $SU(3)$ expresses the freedom in choosing $A$ to be represented by $Z$ while $SU(2)$ expressed the freedom of representing $B$ as $X$ while keeping the representation of $A$ fixed. As the argument above expressed an intuition regarding the symmetry of Pauli group, it is informal as a proof. Is there a way to formalize this idea in a way which proves the claim above?
In Schwartz's QFT book (eqn 17.31), to find the anomalous magnetic moment of the electron from the form factors, near the end of the calculation the following integral needs to be evaluated: $$ F_{2}(0) = \frac{\alpha}{\pi} \int_{0}^{1} dz \int_{0}^{1} dy \int_{0}^{1} dx \frac{\delta(x+y+z-1) z}{1-z}$$ which gives the result $$ F_{2}(0) = \frac{\alpha}{\pi} \int_{0}^{1} dz \int_{0}^{1-z} dy \frac{z}{1-z}$$ My question is, how do you fill in the gap of this calculation? I am not sure how to use the delta function on a definite integral, especially when there is more than 1 variable. My attempt has been to integrate over $x$ using the delta function to set $x = 1-y-z$, and then evaluate this at the limits $0 \: \& \: 1$, which gives $y = 1-z$ for the upper limit and $y=z=0$ for the lower one. But I'm not sure why this factor appears in the limit of the integral over $y$ and not elsewhere. Any tips on how to approach these kinds of integrals would be appreciated, as they seem to be important when using Feynman parameter.
I want to refer you to Weinberg QFT1, if you have not read it yet. Below is my attempt to answer your question in the formalism addressed there. You are classifying particles by representations of the Poincaré group. A one particle state has to be transformed under an element of this group. Part of this transformation just changes the momentum, the other part makes a particle-specific 'intrinsic' transformation of the state (i.e. rotates the spinor indices). Spin and helicity characterize the latter part and in this sense are rather similar. It happens so that this intrinsic transformation can be understood in terms of what is called the small group. For any momentum $p_\mu$ you can go to a standard reference frame where we will have a standard momentum $k_\mu$ which lies in the same orbit of the Lorenz group as $p$. For example (I am usign signature $+---$), for massive particles, where you have $p^2=M^2$, you can choose the rest frame, and the standard $k_\mu=(M,0,0,0)$. For $p^2=0$ you can go to a reference frame where the momentum becomes $k^\mu=(\kappa,0,0,\kappa)$. Now the small group is the subgroup of the Lorenz group which leaves your standard $k$ invariant. It happens (and for proof of this it is better to read Weinberg, Wigner, etc) that for every Lorenz transformation there is a corresponding element of the small group that determines the 'intrinsic' transformation. Lets talk about the massive case first. We have $k^\mu=(M,0,0,0)$, and the small group is clearly just the rotations $SO(3)$. Irreducible representations of this group are well-known and they are parametrized by spin $s$ which can be integer of half-integer. (Formally, the latter are projective representations, but in QM a phase does not matter.) We can also pick an axis and look at the rotation generator $J_3$, and its eigenvalues, this is the 'spin projection'. As far as we have other rotations at our disposal, we can rotate the particle to change this projection. This is the intuitive reason why we have all spin projections between $-s$ and $s$ (with integer steps). Now, what is different in the case of massless particles. The standard momentum is $k^\mu=(\kappa,0,0,\kappa)$ that is a, say, photon travelling along $z$ direction. Now we still have our $J_3$ rotation, but there is some problem with $J_1$ and $J_2$ -- they will change our vector. However, it is possible to save the situation by adding some boosts to these generators. This fact changes the group in this case, and it becomes isomorphic to $ISO(2)$ -- the group of isometries of 2D euclidian plane. In practical terms, we how have three generators in the small group -- $A,B,J_3$, where $A,B$ is what is left of $J_1$ and $J_2$ and correspond to translations of this plane, while $J_3$ is the rotation along the momentum direction and corresponds to rotations in the plane. Of course, this plane picture is just a mathematical abstraction. It is more or less obvious that $A,B$ correspond to some 2D momentum, and so if they have a non-zero eigenvalue, they have continious spectrum (just rotate it with $J_3$). We do not observe any intrinsic continious degrees of freedom for massless patricles, so we conclude that $A,B$ have zero eigenvalues in the representations of interest. We now only have to think about $J_3$ eigenvalue, which is the projection of spin to the momentum direction. Here I say 'spin projection' because it determines the transformation of wavefunction under rotations around some axis. However, this eigenvalue is actually called helicity. For now, no restrictions on its value. You can see that in the masseless case, the relevant irreducible representation of the Lorenz algebra is in fact one-dimensional (just one state with $A\psi=B\psi=0$, $J_3\psi=\lambda\psi$) and is parametrized by some parameter $\lambda$. Now, it is the topology of the Lorenz group that requires $\lambda$ to be integer or half-integer (in massive case, this requirement grows already from the algebra). Still, no requirements like "there should be also at least $-\lambda$" etc. The physical reason for this is that in massive case, if we knew 'helicity' - the projection of spin to the momentum direction, we could go to the rest frame, rotate the spin as we want, and go back, thus changing the projection. In massless case, there is no rest frame -- when we try to rotate the 'spin', the momentum rotates as well. The helicity is Lorenz-invariant. Why we have, as you say, '$\pm s_{max}$' in massless case, it is because spatial inversion changes the sign of helicity. So, for particles that 'respect $P$-inversion' we have also particles with opposite sign of helicity, and call them the same names. For neutrino (assume them massless), however, we dont have such a nice thing, so there are neutrinos and antinetrinos with different helicities. So, if a mircale happened and I explained it in a correct as well as understandable way, I think this should clarify points 1,2,3. 4. It is the same as should we call spin-up electron and spin-down electron the same particle? If you believe that rotations are symmetries of our world -- yes. Now, helicity? If you believe that spatial inversion ($P$) is a symmetry -- yes. Well, we know it is violated, but, if we take QED alone, it is a symmetry, so it is reasonable to call it so. Also, the mathematical formalism ($A_\mu$-field) highly suggests this.. 5. Go to the rest frame. (The square is a casimir, so it commutes with boosts) Then this is, up to a coefficient, $m^2\vec{J}^2$. Where $J_i$ is the generator of rotations. 1. I would say that you are expecting some continuity that is not here..
Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order that equals 109. You may use the operations $x + y$, $x - y$, $x \times y$, $x \div y$, $x!$, $\sqrt{x}$, $\sqrt[\leftroot{-2}\uproot{2}x]{y}$ and $x^y$, as long as all operands are either $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(\sqrt{8*2})!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $109$ will get plus one from me. Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 \times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get 109, but will not mark them as correct. Here are some examples to this problem: Use 2 0 1 and 8 to make 67 Make numbers 93 using the digits 2, 0, 1, 8 Make numbers 1 - 30 using the digits 2, 0, 1, 8 many thanks to the authors of these questions for inspiring this question.
The wavefunction $\psi(L,t)$ is confined to a circle whenever the eigenvalues L of a particle are only nonzero on the points along a circle. When the wavefunction $\psi(L,t)$ associated with a particle has non-zero values only on points along a circle of radius $r$, the eigenvalues $p$ (of the momentum operator $\hat{P}$) are quantized—they come in discrete multiples of $n\frac{ℏ}{r}$ where $n=1,2,…$ Since the eigenvalues for angular momentum are $L=pr=nℏ$, it follows that angular momentum is also quantized. Newton's second law describes how the classical state {$\vec{p_i}, \vec{R_i}$} of a classical system changes with time based on the initial position and configuration $\vec{R_i}$, and also the initial momentum $\vec{p_i}$. We'll see that Schrodinger's equation is the quantum analogue of Newton's second law and describes the time-evolution of a quantum state $\|\psi(t)⟩$ based on the following two initial conditions: the energy and initial state of the system. In this section, we'll begin by seeing how Schrodinger's time-independent equation can be used to determine the wave function of a free particle. After that, we'll use Schrodinger's time-independent equation to solve for the allowed, quantized wave functions and allowed, energy eigenvalues of a "particle in a box"; this will be useful later on as a qualitative understanding of the quantized wave functions and energy eigenvalues of atoms. In general, if a quantum system starts out in any arbitrary state, it will evolve with time according to Schrödinger's equation such that the probability $P(L)$ changes with time. In this lesson, we'll prove that if a quantum system starts out in an energy eigenstate, then the probability $P(L)$ of measuring any physical quantity will not change with time. In this lesson, we'll discuss quantum dynamics.
When a hash function is used to derive a key from a shared secret (either by simply hashing the shared secret or using a more robust construct like HKDF) what's the strength of the derived material? If, for example, the shared secret is 256-bit, is the security of the derived result also 256-bit or is it $2^{n/2}$ (that is 128-bit in this case) since as per the birthday problem, it "only" takes $2^{n/2}$ guesses to generate a collision. Thus in this case a collision would mean getting the same output and so the same material of the KDF. You—the adversary—have a way to test whether a candidate key $k_i$ might be the true secret key $k^$. Since $k^$ is uniformly distributed among all 256-bit keys, each candidate $k_i$ has probability $\Pr[k_i = k^] = 1/2^{256}$ of being correct. No matter what order you try things in, the expected number of guesses is $$\sum_{i=1}^{2^{256}} i \cdot \Pr[k_i = k^] = \sum_{i=1}^{2^{256}} i \cdot \frac{1}{2^{256}} = \frac{2^{256} (2^{256} - 1)/2}{2^{256}} = 2^{255} - {\textstyle\frac12}.$$ Why don't collisions and the birthday paradox appear in this analysis? Collisions are relevant when you're looking for any $k_i \ne k_j$ such that $H(k_i) = H(k_j)$, but you don't care what either $k_i$ or $k_j$ are. As you try $k_1, k_2, \dots$, searching for a collision, each new key could potentially collide with every previous key, so the probability of a collision among some pair of $n$ keys grows quadratically—specifically, it is $$1 - \biggl(1 - \frac{1}{2^{256}}\biggr) \biggl(1 - \frac{2}{2^{256}}\biggr) \cdots \biggl(1 - \frac{n}{2^{256}}\biggr) \geq \frac{n^2/4}{2^{256}}.$$ (proof; more on birthday paradox) That said, if the way you can test a candidate key $k_i$ is by testing whether $H(k_i) = h$ where you know $h = H(k^)$, and you actually have many target keys $k^_j$ and hashes $h_j = H(k^_j)$, you can save cost in a batch attack like computing Oeschlin's rainbow tables in parallel, for a total expected cost of about $2^{256}/t$ trials to find the first of $t$ targets, and in the total expected time of as little as about $2^{256}/t^3$ sequential evaluations of $H$ if you parallelize it at least about $t^2$ ways. However, if each user had used a different function, that is if you have $h_j = H_{s_j}(k^_j)$ with a unique salt $s_i$ per user, then the multi-target advantage vanishes, and you're back to the expected cost of about $2^{256}$.
Suppose you have vectors $u$ and $v$. Imagine a table $M$ of the products of each of their entries. $$M = \|u\rangle\langle v\| = \begin{bmatrix}u_0 v_0 & u_1 v_0 & u_2 v_0 & \dots & u_{n-1} v_0 \\u_0 v_1 & u_1 v_1 & u_2 v_1 & \dots & u_{n-1} v_1 \\u_0 v_2 & u_1 v_2 & u_2 v_2 & \dots & u_{n-1} v_2 \\\vdots &\vdots &\vdots & \ddots & \vdots \\u_0 v_{n-1} & u_1 v_{n-1} & u_2 v_{n-1} & \dots & u_{n-1} v_{n-1} \\\end{bmatrix}$$ The convolution $u \ast v$ of the two vectors is a vector of the values you get when summing along each anti-diagonal of this grid. For example: $$(u \ast v)_2 = \sum \begin{bmatrix} & & & & & & u_{n-3} v_0 & 0 & 0\\ & & & & & u_{n-4} v_1 \\ & & & & \ddots \\ & & & u_2 v_{n-5} & \\ & & u_1 v_{n-4} & & \\u_0 v_{n-3} & & & & \\0 &\\0&\text{ }\end{bmatrix}$$ We can also think of the FFT of a vector as summing entries in a grid defined by multiplying two vectors' entries. The second vector is the sequence of roots of unity. $$M = \| u \rangle \langle e^{\tau i k/n} \|\begin{bmatrix}u_0 & u_1 & u_2 & \dots & u_{n-1} \\u_0 e^{\tau/n} & u_1 e^{\tau/n} & u_2 e^{\tau/n} & \dots & u_{n-1} e^{\tau/n} \\u_0 e^{2\tau/n} & u_1 e^{2\tau/n} & u_2 e^{2\tau/n} & \dots & u_{n-1} e^{2\tau/n} \\\vdots &\vdots &\vdots & \ddots & \vdots \\u_0 e^{-\tau/n} & u_1 e^{-\tau/n} & u_2 e^{-\tau/n} & \dots & u_{n-1} e^{-\tau/n}\end{bmatrix}$$ The Fourier transform doesn't track along the diagonals, though. Its row-hop is different from its column-hop. For the $k$'th entry of the output we go through each column $c$ and sum up the cell from that column and the $k \cdot c$'th row (wrapping around). $$FFT(v)_2 = \sum \begin{bmatrix}u_0 \\ &&&&u_4e^{\tau/7} \\ & u_1 e^{2\tau/7} \\ &&&&&u_5e^{3\tau/7} \\ & & u_2 e^{4\tau/7} \\ &&&&&&u_6e^{5\tau/7} \\ &&& u_3 e^{6\tau/7} \\\end{bmatrix}$$ So our question becomes: can we somehow re-arrange our vectors so that the each-stride sums $\sum u_i v_{k \cdot i}$ correspond to the each-diagonal sums $\sum u_i v_{k-i}$? Can we turn index-multiplication into index-adding? We can, as long as we're operating in a context that allows for a logarithm. Suppose our vector size is $n$ and $g$ is a multiplicative generator for the integers modulo $n$. Then we can use the powers of $g$, i.e. the base-$g$ discrete logarithm, to define a permutation of the non-zero elements that does most of what we need. Our transformation goes roughly as follows: Define permuted vectors $a$ and $b$ so that $a_k = u_{g^k}$ and $b_k = v_{g^{k}}$. Compute the convolution $c = a \ast b$. So $c_k = \sum_{d=0}^{n-1} a_d b_{k-d} = \sum_{d=0}^{n-1} u_{g^d} v_{g^{-k} \cdot g^d}$. Now un-permute $c$, and also un-permute the sum over $d$. $x_{g^{-k}} = c_{k}$ and $e=g^d$. So $x_k = c_{\log_g k} = \sum_{e=0}^{n-1} u_{e} v_{k \cdot e}$. In other words: define a permutation $P$ based on when indices are reached when iteratively multiplying by $g$. Permute the input vector and the vector of $e^{\tau k /n}$ powers by $P$. Convolve. Unpermute by $P$. The result is the Fourier transform. ... Except that I've ignored one important issue. Our permutation doesn't work for $k=0$, because $g^k \mod n$ is never 0. Also, $g^0 = g^{n-1} = 1$. Fixing this hole requires special-casing the edges of the grid. But hopefully I've communicated the core idea of performing an FFT by delegating the heavy-lifting to a convolution and a permutation. Note that, if you care about bit complexity instead of the number of arithmetic operations, computing the permutation may also be an obstacle. If multiplying by $g \pmod{n}$ costs $\log n$ time, and you don't have the permutation ahead of time, then overall the algorithm will still cost $O(n \log n)$.
Answer The solution set is $$\theta=\{22.5^\circ+n90^\circ,n\in Z\}$$ Work Step by Step $$4\cos2\theta=8\sin\theta\cos\theta$$ - Recall the identity: $2\sin\theta\cos\theta=\sin2\theta$ Therefore, $$8\sin\theta\cos\theta=4\sin2\theta$$ Apply back to the equation: $$4\cos2\theta=4\sin2\theta$$ $$\cos2\theta=\sin2\theta$$ Here we can divide both sides by $\cos2\theta$ to get $\tan2\theta$. However, we need to prove first that $\cos2\theta\ne0$ in this situation. 1) Prove that $\cos2\theta\ne0$ If $\cos2\theta=0$, then $\sin2\theta=\cos2\theta=0$ However, we know that there are no such values of $\theta$ that have both $\sin2\theta=\cos2\theta=0$ Therefore, $\cos2\theta\ne0$ 2) Divide both sides by $\cos2\theta$ $$1=\frac{\sin2\theta}{\cos2\theta}$$ According to the identity $\tan\theta=\frac{\sin\theta}{\cos\theta}$, we have $$\tan2\theta=1$$ 3) Solve the equation over the interval $[0^\circ,360^\circ)$ Over the interval $[0,2\theta)$, there are 2 values of $2\theta$ where $\tan2\theta=1$, which are $\{45^\circ,225^\circ\}$ Therefore, $$2\theta=\{45^\circ,225^\circ\}$$ 2) Solve the equation for all solutions Tangent function has period $180^\circ$, so we would add $180^\circ$ to all solutions found in part 1) for $2\theta$. $$2\theta=\{45^\circ+n180^\circ,225^\circ+n180^\circ, n\in Z\}$$ However, as we notice, $45^\circ+n180^\circ$ and $225^\circ+n180^\circ$ represents the same values, so we would only pick one of them. Here I would go with $45^\circ+n180^\circ$ $$2\theta=\{45^\circ+n180^\circ, n\in Z\}$$ Thus, $$\theta=\{22.5^\circ+n90^\circ,n\in Z\}$$
Is there any interesting ways to find pi? Thanks. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community Is there any interesting ways to find pi? Thanks. Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. Try the Gregory Series, It uses a sequence to find the value of Pi $$ \pi = \sqrt{6(1 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \frac 1{5^2}...)}$$
By Theorem 13.31 in L. Washington Introduction to Cyclotomic Fields, Second Edition, GTM 83 we know that $\mathcal{X}\sim \Lambda^{r_2}\oplus(\Lambda-$torsion), where $r_2$ is the number of complex embeddings of $\mathbb{Q}(\zeta_p)$, so $r_2=(p-1)/2$, and $\sim$ means "pseudo-isomorphism". In particular, since we know that the plus part is torsion, one deduces that the minus part of $\mathcal{X}$ is huge (finitely generated but not torsion over $\Lambda$). But observe that $\mathcal{X}^+$ "is dual" to $X^-$ where I denote by $X=\mathrm{Gal}(L/K_{\infty})$ the Galois group of the maximal abelian $p$-extension of $K_{\infty}$ which is unramified everywhere. As for my "duality" statement, we have (always quoting Washington) Proposition 13.32 $e_j\mathcal{X}(-1)\cong \mathrm{Hom}_{\mathbb{Z}_p}(e_iA_\infty,\mathbb{Q}_p/\mathbb{Z}_p)$ where $i$ is odd and $i+j\equiv 1\pmod{p-1}$ (you can always assume both $0\leq i,j\leq p-1$ so $j=p-i$ is even and the above is a statement about a piece of $\mathcal{X}^+$ and a piece of $X^-)$; and $A_\infty$ is the inductive limit of $p$-Sylow of class groups along the cyclotomic extension. Proposition 15.34 $\tilde{X}\sim (A_\infty,\mathbb{Q}_p/\mathbb{Z}_p)$ where $\tilde{X}$ is the old $X$ with twisted $\Gamma$-action (see p. 359). Combining the two results, you see the duality I was speaking about. Therefore, rather than asking what is the structure of $\mathcal{X}^-$ which, as seen above, is huge and morally free, one can wonder about the structure of $X^+$. There is an old-standing conjecture by Ralph Greenberg (who never, to my knowledge, baptized this question as "conjecture", but wrote the beautiful paper On the Iwasawa invariants of totally real number fields, American J. of Math, (1976), 263-284, where he studied it) predicting that $X^+\sim 0$, i. e. it is finite. Much work has been done on this (google-it to verify, being aware that Greenberg posed several conjectures in Iwasawa theory, so you should double-check the results) but it remains open and is probably one of the last big open question about the Iwasawa theory of totally real fields, together with Leopoldt. Two last words: first of all, much of the above generalizes almost immediately to an abelian real base field $F$ insted of $\mathbb{Q}$ (just to be safe, suppose may be that $p\nmid [F:\mathbb{Q})$. Secondly, a nice table describing these dualities (and some more involving cyclotomic units) and their relations with different Main Conjectures is depicted on page 349 of Washington's book.
Dear Uncle Colin, Why can't I work out $\int \left( \ln(x) \right)^2 \dx$ using the reverse chain rule? -- Previously Acceptable, Reasonable Technique Stumbles Hello, PARTS, There are two answers to this: the first is, you can't use the reverse chain rule -- which I learned as 'function-derivative' when IRead More → This post is based on work by Mark Ritchings; I know of no finer1 maths tutor in Bury. A few weeks ago, I pointed in the vague direction of a few decimal curiosities -- fractions that spit out lovely patterns in their decimal expansions. Having found one that generated theRead More → Dear Uncle Colin, I recently came across a problem in which I had to integrate $\cos^3(x)$. Somewhere in my mind, I recall that the thing to do is to make it into something involving $\cos(3x)$, but I couldn't put the details together. Could you help? -- Not A Very InspiredRead More → Is This Prime? is probably the most infuriating, addictive, revolting and unbearably simple games that has ever disgraced my computer screen. I love it, hate it, am glad of its existence and wish it had never been written. It's pretty tough to think of a simpler premise: you're given anRead More → Dear Uncle Colin, I was doing a STEP paper and it asked me to calculate $\int_0^1 x^3 \arctan\left(\frac{1-x}{1+x}\right) \dx$, given that $\int_0^1 \frac{x^4}{1+x^2} \dx = \frac{\pi}{4} - \frac{2}{3}$. Nut-uh. College Asked Me Back: Rocked Interview. Daren't Get Excited Hello, CAMBRIDGE! Is this thing on? Even with the given hint, thisRead More → This article has also been published as part of the Relatively Prime zine. The student yelped, and found his wrists and ankles strapped to the wheel before the lesson had even started. "Good morning," said the Mathematical Ninja. "I think you and I need to have a little... talk." TheRead More → Dear Uncle Colin, I'm supposed to use the change of variable $z = \sin(x)$ to turn $\cos(x) \diffn{2}{y}{x} + \sin(x) \dydx - 2y \cos^3(x)= 2\cos^5(x)$ into $\diffn{2}{y}{z} - 2y = 2\left(1-z^2\right)$. Yeah but no but. No idea. Lacking, Obviously, Something Trivial Hello, LOST, Right, yes. Nasty one, this. The mainRead More → Once in a while, a student puts me on the spot; it's not always deliberate. In this case, changing a variable in a second-order differential equation, he blithely said "Well, $\diffn 2yx = \frac{1}{\diffn 2xy}$..." "Whoa whoa whoa..." "... isn't it?" Now, had I not had a series of discussionsRead More → In Episode $n$ of Wrong, But Useful (where $n$ is a semiprime followed by two semiprimes and a square), @reflectivemaths and I discuss: Colin's trip to Center Parcs and whether it is sufficiently central The number of the podcast, which is the tetrahedral angle, $\arccos\left(\frac{1}{3}\right)$ The Other Half podcast OurRead More → Dear Uncle Colin, In Basic Maths For Dummies, you mention a method for multiplying numbers from 6 × 6 to 10 × 10 on your fingers. It's almost magical! Why does it work? -- Does It Guarantee Interesting Times Sums? Hello, DIGITS, I'm glad you're finding the book helpful! TheRead More →
A formula is a string of symbols, arranged according to mathematical grammar. A function is a mathematical object that plays a role in arithmetic operations like "evaluation" or "composition". A key point is that if $f$ and $g$ are expressions that denote two functions with the same domain and codomain, and we have $f(x) = g(x)$ for every $x$ in the domain, then $f$ and $g$ denote the same function. So, in your example of functions $f$ and $g$ on the set $\{ -1, 0, 1 \}$, it is indeed true that $f=g$. Some examples of mathematical grammar Let $f$ be a variable that denotes a function on the reals. Let $x$ be a real-valued variable. Then: $f$ is a function. $f(x)$ is a real number. $x^2 + 3$ is a real number. In particular, it is not a function. Unfortunately, people are frequently grammatically incorrect on this point. :( $f(x) = x^2 + 3$ is an equation that relates two real numbers. and all of the bullet points above are examples of formulas. Recall that I mentioned functions were 'defined' pointwise: if two functions solve "$f(x) = x^2 + 3$ for all $x$" for $f$, then they must be the same function. Because of this, we can use equations like this as a way to specify functions. But as you note, if two different formulas for the right hand side actually give the same values when you substitute values from the domain, then the functions so defined will be the same. Incidentally, it is possible to define functions directly rather than pointwise, although it often isn't pleasant. e.g. the function $f$ defined above is given by $$ f = p \circ ((\mu \circ \Delta), c_3) $$ where I'm using the notation $\Delta$ is the diagonal function $\Delta(x) = (x, x)$ $\mu$ is the multiplication function $\mu(x,y) = xy$ $p$ is the addition function $p(x,y) = x+y$ $c_3$ is the constant function $c_3(x) = 3$ $(,)$ is a binary operation on functions; it's defining property is $(g,h)(x) = (g(x), h(x))$. I don't actually know of standard notation for this; sometimes I see $\times$ in place of $,$. $\circ$ is composition of functions: $(g \circ h)(x) = g(h(x))$.
I have 2 questions - the first is what the title refers to, and the second is something I want a reference on (I thought I'd include them in one post since they are very strongly related). Sorry this post is a bit long, I tried to put as much as detail as I could .. $1$-st question: I'm interested only in the group $GL_n(F_q)$. In Carter's book "Finite Groups of Lie Type: Conjugacy Classes and Complex Characters", in Chapter 7 "The generalized characters of Deligne-Lusztig", the construction of the virtual representations $R_{T, \theta}$ as alternating sums of $l$-adic cohomology of Deligne-Lusztig varieties is given in some details, and a series of formulae are proved in the chapter about these ($T$ a torus, and $\theta$ a character of $T^{F}$). It says that if $\theta \in \widehat{T^{F}}$ is in general position, then $\pm R_{T, \theta}$ is irreducible. The following formula is given (also in http://en.wikipedia.org/wiki/Deligne%E2%80%93Lusztig_theory), where $g=su=us$, $s,u$ being the semisimple and unipotent parts, and $Q_{T}(u) = R_{T, 1}(u)$, $C^{0}(s)$ being the identity connected component of the centralizer of $s$, and $F$ the Frobenius endomorphism. $ R_{T, \theta}(g) = \frac{1}{ \| C^{0}(s)^{F} \|} \sum_{ x \in G^{F}, x^{-1}sx \in T^{F} } \theta ( x^{-1} s x) Q_{x T x^{-1}}^{C^{0}(s)} (u) $ The book then says that $Q_{T}(u)$ is a Green function, depends only on the torus (I understand it will not change if we conjugate the torus in $G^F$ either so essentially corresponds to an element of $S_n$ for the group general linear group of size $n$, which is what I'm most curious about; unless I'm mistaken). The book does not give an explicit formulae for these $Q_{T}(u)$, but it does give orthogonality relations and such - explicit formulae is what I"m looking for: Question: What's an explicit formulae for these $Q_{T}(u)$? How does this relate to the Green function that I've been studying from in Macdonald's book "Symmetric Functions and Hall polynomials", in the chapter "Characters of $GL_n$ over a finite field" -i.e., how do I express the character $ \pm R_{T, \theta}$ as a sum of the irreducible characters described by Green functions in Macdonald's book (or a single irreducible character in the case where $\theta$ is in general position)? In that book, I've learnt that the polynomials correspond to symmetric functions $S_{\lambda}$, via a correspondence that maps $A$, the sums of the representation ring of for all $n$, to $B$, an algebra generated by elementary symmetric functions in independent variables $X_{i,f}$ ($f$ ranges over all irreducible polynomials in $\mathbb{F}_{q}[t]$). I'm sorry I'm being a bit vague right here - it would take pages to define precisely all the notation that Macdonald uses in his book; feel free to work with any alternative explicit definitions of these Green functions (but please include a reference so I know where to look it up). $2$-nd question: I have looked through Carter's book and Digne&Michel's book on the same topic, but I have been unable to find a reference which gives the representing matrices for these virtual representations $\pm R_{T, \theta}$ of these finite Lie type groups (the fact that they are defined with alternating sum complicates matters somewhat). I'm not so interested in the entries of the representing matrices as such, just a construction for the module which enables you to find the representing matrices. Can anyone suggest a good reference for this? The closest I can find is Lusztig's original book "Characters of reductive groups over finite fields", where it mentions that $l$-adic intersection homology can be used as a substitute (this was from what I can see in googlebooks preview); but I hear this book is horrible to learn from, and I'm not entirely certain if what's given there is what I'm looking for (I don't have a copy of the book at present).
ISSN: 1078-0947 eISSN: 1553-5231 All Issues Discrete & Continuous Dynamical Systems - A October 2009 , Volume 23 , Issue 4 A special issue on Asymptotic Description of Natural Phenomena Select all articles Export/Reference: Abstract: This special issue of Discrete and Continuous Dynamical Systems grows out of a focused session at the AIMS 6 meeting held in Poitiers, France, in 2006. The session struck its organizers, who were perhaps not unbiased, as very successful. As a consequence, when Shouchuan Hu approached us about editing a special issue based on the papers delivered in the session, we were enthusiastic at the prospect. The focus of the session was asymptotic models of physical phenomena. This is a large subject, and one special session cannot hope to do it justice. Consequently, some focal point was required, and in the event, most of the lectures in the session were centered upon nonlinear wave equations arising in plasma physics and fluid mechanics. This is also the subject around which most of this special issue turns. For more information please click the "Full Text" above. Abstract: Laser sources nowadays deliver optical pulses reaching few cycles in duration and peak powers exceeding several terawatt (TW). When such pulses propagate in transparent media, they first self-focus in space, until they generate a tenuous plasma by photo-ionization. These pulses evolve as self-guided objects, resulting from successive equilibria between the Kerr focusing process, the defocusing action of the electron plasma and the chromatic dispersion of the medium. Discovered ten years ago, this self-channeling mechanism reveals a new physics, having direct applications in the long-distance propagation of TW beams in air, supercontinuum emission as well as pulse self-compression. This review presents the major progress in this field. Particular emphasis is laid to the derivation of the propagation equations, for single as well as coupled wave components. Physics is discussed from numerical simulations and explained by analytical arguments. Attention is also paid to the multifilamentation instability, which breaks up broad beams into small-scale cells. Several experimental data validate theoretical descriptions. Abstract: Studied here is the large-time behavior and eventual periodicity of solutions of initial-boundary-value problems for the BBM equation and the KdV equation, with and without a Burgers-type dissipation appended. It is shown that the total energy of a solution of these problems grows at an algebraic rate which is in fact sharp for solutions of the associated linear equations. We also establish that solutions of the linear problems are eventually periodic if the boundary data are periodic. Abstract: We present here a highly efficient and accurate numerical scheme for initial and boundary value problems of a two-dimensional Boussinesq system which describes three-dimensional water waves over a moving and uneven bottom with surface pressure variation. The scheme is then used to study in details the waves generated from rectangular sources and the two-dimensional wave patterns. Abstract: We consider a Boussinesq system of BBM-BBM type in two space dimensions. This system approximates the three-dimensional Euler equations and consists of three coupled nonlinear dispersive wave equations that describe propagation of long surface waves of small amplitude in ideal fluids over a horizontal bottom. We show that the initial-boundary value problem for this system, posed on a bounded smooth plane domain with homogeneous Dirichlet or Neumann or reflective (mixed) boundary conditions, is locally well-posed in $H^1$. After making some remarks on the temporal interval of validity of these models, we discretize the system by a standard Galerkin-finite element method and present the results of some numerical experiments aimed at simulating two-dimensional surface wave flows in complex plane domains with a variety of initial and boundary conditions. Abstract: The Benjamin-Ono equation is known as a model of internal long waves in stratified fluids or two-fluid systems. In this paper, we consider the validity of this type of modeling of a two-phase problem for capillary-gravity waves, which is a free boundary problem for the incompressible Euler equation with the irrotational condition. We show that the solutions of the free boundary problem split up into two waves and the shape of each wave is governed by the Benjamin-Ono equation in a slow time scale. Abstract: The regularized long-wave or BBM equation $ u_{t}+u_{x}+u u_{x}-u_{x x t} = 0 $ was derived as a model for the unidirectional propagation of long-crested, surface water waves. It arises in other contexts as well, and is generally understood as an alternative to the Korteweg-de Vries equation. Considered here is the initial-value problem wherein $u$ is specified everywhere at a given time $t = 0$, say, and inquiry is then made into its further development for $t>0$. It is proven that this initial-value problem is globally well posed in the $L^2$-based Sobolev class $H^s$ if $s \geq 0$. Moreover, the map that associates the relevant solution to given initial data is shown to be smooth. On the other hand, if $s < 0$, it is demonstrated that the correspondence between initial data and putative solutions cannot be even of class $C^2$. Hence, it is concluded that the BBM equation cannot be solved by iteration of a bounded mapping leading to a fixed point in $H^s$-based spaces for $s < 0$. One is thus led to surmise that the initial-value problem for the BBM equation is not even locally well posed in $H^s$ for negative values of $s$. Abstract: In this essay, we study the initial-value problem $ u_t+u_x+g(u)_x+Lu_t=0, \qquad x\in\mathbb R,\quad t>0,$ $u(x,0)=u_0(x), \qquad x\in\mathbb R, (0.1)$ where $u=u(x,t)$ is a real-valued function, $L$ is a Fourier multiplier operator with real symbol $\alpha(\xi),$ say, and $g$ is a smooth, real-valued function of a real variable. Equations of this form arise as models of wave propagation in a variety of physical contexts. Here, fundamental issues of local and global well-posedness are established for $L_p$, $H^s$ and bore-like or kink-like initial data. In the special case where $\alpha(\xi)=\|\xi\|^{r}$ wherein $r>1$ and $g(u)=1/2u^2,$ (0.1) is globally well-posed in time if $s$ and $r$ satisfy a simple algebraic relation. Abstract: We study the global well-posedness (GWP) and small data scattering of radial solutions of the semirelativistic Hartree type equations with nonlocal nonlinearity $F(u) = \lambda (\|\cdot\|^{-\gamma}$ * $\|u\|^2)u$, $\lambda \in \mathbb{R} \setminus \{0\}$, $0 < \gamma < n$, $n \ge 3$. We establish a weighted $L^2$ Strichartz estimate applicable to non-radial functions and some fractional integral estimates for radial functions. Abstract: We prove the local-well posedness of the generalized Benjamin-Ono equations in $ H^1(\T) $. Abstract: We consider the Cauchy problem for the damped nonlinear Schrödinger equations, and prove some blowup and global existence results which depend on the size of the damping coefficient. We also discuss the $L^2$ concentration phenomenon of blowup solutions in the critical case. Readers Authors Editors Referees Librarians More Email Alert Add your name and e-mail address to receive news of forthcoming issues of this journal: [Back to Top]
Wikipedia informs me that $$S = \vartheta(0;i)=\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$$ I tried considering $f(x,n) = e^{-x n^2}$ so that its Mellin transform becomes $\mathcal{M}_x(f)=n^{-2z} \Gamma(z)$ so inverting and summing $$\frac{1}{2}(S-1)=\sum_{n=1}^\infty f(\pi,n)=\sum_{n=1}^\infty \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}n^{-2z} \Gamma(z)\pi^{-z}\,dz = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(2z) \Gamma(z) \pi^{-z}\,dz$$ However, this last integral (whose integrand has poles at $z=0,\frac{1}{2}$ with respective residues of $-\frac 1 2$ and $\frac 1 2$) is hard to evaluate due to the behavior of the function as $\Re(z)\to \pm\infty$ which makes a classic infinite contour over the entire left/right plane impossible. How does one go about evaluating this sum?
Quantum mechanical postulates So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms? The mathematical expression for each individual operator is sort of a postulate, but it should not be listed. The postulates define a (more or less) complete theory in that I can derive mathematical facts from it without really knowing anything about the physics. One of the main problems is that the site you link to (following McQuarrie) is in my opinion a very bad site for understanding the postulates of quantum mechanics as they nowadays understodd and which let you penetrate deeper into the theory. Your postulates are in between the first postulates of quantum mechanics, when people were still figuring out the basics and current versions of the postulates that allow for better separation of mathematics and physics. So let me first state them in a more modern form and explain the differences (I'm copying some of this from Wikipedia): Postulate 1: Each physical system is associated with a (topologically) separable complex Hilbert space $H$ with inner product . Rays (one-dimensional subspaces) in $H$ are associated with states of the system. The difference in this with your postulate 1 is that we do not refer to "space". The wave function is an object in an abstract Hilbert space, not necessarily some "object" $\psi(x,t)$ with position and time. Your first postulate already incorporates the notion of "space" and thereby needs to have a few operators "fixed" (see further down). Let's go on: Postulate 2: Physical observables are represented by self-adjoint linear operators on $H$. The expectation value (in the sense of probability theory) of the observable $A$ for the system in state represented by the unit vector $\|\psi\rangle\in H$ is $\langle\psi\| A \|\psi\rangle$. There we go. This is your postulate 2 and 4 combined. Your postulate 2 is the first sentence, the Born rule is essentially the second system. Postulate 3: The Hilbert space of a composite system is the Hilbert space tensor product of the state spaces associated with the component systems. Now, your system doesn't have such a postulate, if I see correctly. If you already operate in space - and since each particle lives in the same space - you do not need this postulate. However, it's easier to consider each particle with a separate wave-function in a separate Hilbert space and combine them by introducing the tensor products (for interactions, see below). Note that this can no longer be uphold in quantum field theory, as particle number is not conserved, but this is unimportant here. This way, you can easily accomodate spin into the theory, which is not possible with your wave-function $\psi(x,t)$. It turns out that to properly describe spin, you have to introduce another parameter that your wave-function depends on. That's not very nice and means that your postulates will not work proberly. The abstract picture has no such problem. Postulate 4: The time evolution of the state is given by a (weakly-)differentiable function from the real numbers, representing instants of time, to the Hilbert space of system states. This map is characterized by the Schrödinger equation. This is your postulate 5. Postulate 6 is not present - it cannot be derived from this set of postulates, but if we enhance quantum mechanics to quantum field theory (note that axiomatic QFT is problematic), then it can be, so I don't really want to focus on this and leave it out. Operators in quantum mechanics Note that in this description, we do not even define "space". The above theory describes a very abstract theory. To what does the position operator correspond? Well, we don't know, because we don't even know what "space" means. However, this describes a full theory and I can derive mathematically meaningful statements from it. For example, I can say that by virtue of postulate 2 (in principle) the Hamilton operator in the system must itself be a self-adjoint operator, because time evolution must preserve $\langle \psi \|\psi \rangle$, as this constitutes a probability measure. However, what about physics? The idea of these postulates is that for any experiment, you now have to specify: the Hilbert space the Hamilton operator the possible observables only if you do that, you can start calculating anything at all. This is somewhat reminiscent of classical theory, where you would normally first define the phase space (maybe, your system is constraint to a one-dimensional motion such as a pendulum. In this case, your phase space would have only one $x$ and one $p$-coordinate). Then, you go on and define the Lagrangian or Hamiltonian according to your problem and then you can calculate everything. You have to do the same in quantum mechanics. This means also that a priori, there is no need whatsoever to have the square of the momentum operator be represented by the negative Laplacian (for example). And - guess what - it's also not true in any system (see e.g. discrete systems, where a "momenumt operator" can still make sense). Let's make an example. Suppose you want to consider a particle in a three dimensional box. It is somehow intuitive to take the Hilbert space $L^2([0,1]^3)$ of functions in the box $[0,1]^3$. The position of the particle should then be just the "position" $x$ in the box, i.e. the position operator as you know it. However, there is a mathematical theorem that tells us that all infinite dimensional (separable) Hilbert spaces are the same, so I could also go on describe the particle in a the box $[0,1]^3$ by a function in the Hilbert spae $L^2(\mathbb{R})$. This would of course be highly counterintuitive. What would the position operator look like? Probably different than the "intuitive" one. This is why the operators are not posutlated: They are part of what constitutes the actual physical system you want to consider! The specific mathematical expression of an operator is therefore not independent of how you choose to describe the theory and therefore, they do not form postulates. However, given your postulates here again, this is no longer true. The first postulate identifies a specific Hilbert space, which basically fixes (for most systems) how the position and momentum operators should look like. This might explain your confusion and hopefully helps you in your quest of understanding quantum mechanics better! How to find and attribute operators? However, it doesn't really answer the real underlying question: How do you actually find these observables? And what about the history? As I said, for any system, you have to find the triple Hilbert space, Hamiltonian, Observables. All of this must somehow come from physics. And here is, where your historical picture is somewhat accurate (and this relates also to Mark Mitchison's comment): Finding Hilbert space, Hamiltonian and observables is usually/often done by analogy. Your historical picture gives the right idea in that it tells a story of how people coming from classical mechanics arrive at the observables by analogy. An example: Given a classical particle in a box, you can write down a Hamilton function in classical mechanics. For a quantum particle, knowing that it has something like "momentum" and "position", you want to take a space that looks rather similar to the classical configuration space, hence $L^2([0,1]^3)$ seems a very good choice. Choosing the position operator as you know it is then a "natural" choice, knowing what "position" means in the classical picture. You then choose the momentum operator such that the basic commutation relations are fulfilled. These commutation relations are "postulated" from the Poisson brackets in Hamiltonian mechanics. In short, it turned out that in classical mechanics, we have $\{p,q\}=1$ with the Poisson brackte, and in quantum mechanics, we have $[P,Q]=i\hbar$ with the commutator. This leads to postulate the rule "take any Poisson bracket to a commutator modulo $i\hbar$", which is not well-defined, but works pretty well in defining operators for many systems (this is called canonical quantization). This is "identifying operators by analogy", but of course, it can only work if also the Hilbert space is identified by analogy. What to do with observables like spin that do not have a classical counterpart? Well, in this case the abstract postulates are very powerful: They allow you to just make something up. People knew that the spin works somewhat like angular momentum so it makes sense to postulate that there is a "spin-part" in the Hilbert space that is just $\mathbb{C}^2$ and the spin-measurements are just $\sigma_i$, the Pauli-matrices. Since the experiments confirm your choice, all is well.
Skills to Develop To study the use of G–test of goodness-of-fit (also known as the likelihood ratio test, the log-likelihood ratio test, or the G 2test) when you have one nominal variable To see whether the number of observations in each category fits a theoretical expectation, and the sample size is large When to use it Use the G–test of goodness-of-fit when you have one nominal variable with two or more values (such as male and female, or red, pink and white flowers). You compare the observed counts of numbers of observations in each category with the expected counts, which you calculate using some kind of theoretical expectation (such as a $1:1$ sex ratio or a $1:2:1$ ratio in a genetic cross). If the expected number of observations in any category is too small, the G–test may give inaccurate results, and you should use an exact test instead. See the web page on small sample sizes for discussion of what "small" means. The G–test of goodness-of-fit is an alternative to the chi-square test of goodness-of-fit; each of these tests has some advantages and some disadvantages, and the results of the two tests are usually very similar. You should read the section on "Chi-square vs. G–test" near the bottom of this page, pick either chi-square or G–test, then stick with that choice for the rest of your life. Much of the information and examples on this page are the same as on the chi-square test page, so once you've decided which test is better for you, you only need to read one. Null hypothesis The statistical null hypothesis is that the number of observations in each category is equal to that predicted by a biological theory, and the alternative hypothesis is that the observed numbers are different from the expected. The null hypothesis is usually an extrinsic hypothesis, where you know the expected proportions before doing the experiment. Examples include a $1:1$ sex ratio or a $1:2:1$ ratio in a genetic cross. Another example would be looking at an area of shore that had $59\%$ of the area covered in sand, $28\%$ mud and $13\%$ rocks; if you were investigating where seagulls like to stand, your null hypothesis would be that $59\%$ of the seagulls were standing on sand, $28\%$ on mud and $13\%$ on rocks. In some situations, you have an intrinsic hypothesis. This is a null hypothesis where you calculate the expected proportions after the experiment is done, using some of the information from the data. The best-known example of an intrinsic hypothesis is the Hardy-Weinberg proportions of population genetics: if the frequency of one allele in a population is $p$ and the other allele is $q$, the null hypothesis is that expected frequencies of the three genotypes are $p^2$, $2pq$, and $q^2$. This is an intrinsic hypothesis, because you estimate $p$ and $q$ from the data after you collect the data, you can't predict $p$ and $q$ before the experiment. How the test works Unlike the exact test of goodness-of-fit, the G–test does not directly calculate the probability of obtaining the observed results or something more extreme. Instead, like almost all statistical tests, the G–test has an intermediate step; it uses the data to calculate a test statistic that measures how far the observed data are from the null expectation. You then use a mathematical relationship, in this case the chi-square distribution, to estimate the probability of obtaining that value of the test statistic. The G–test uses the log of the ratio of two likelihoods as the test statistic, which is why it is also called a likelihood ratio test or log-likelihood ratio test. (Likelihood is another word for probability.) To give an example, let's say your null hypothesis is a $3:1$ ratio of smooth wings to wrinkled wings in offspring from a bunch of Drosophila crosses. You observe $770$ flies with smooth wings and $230$ flies with wrinkled wings. Using the binomial equation, you can calculate the likelihood of obtaining exactly $770$ smooth-winged flies, if the null hypothesis is true that $75\%$ of the flies should have smooth wings ($L_{null}$); it is $0.01011$. You can also calculate the likelihood of obtaining exactly $770$ smooth-winged flies if the alternative hypothesis that $77\%$ of the flies should have smooth wings ($L_{alt}$); it is $0.02997$. This alternative hypothesis is that the true proportion of smooth-winged flies is exactly equal to what you observed in the experiment, so the likelihood under the alternative hypothesis will be higher than for the null hypothesis. To get the test statistic, you start with $L_{null}/L_{alt}$; this ratio will get smaller as $L_{null}$ gets smaller, which will happen as the observed results get further from the null expectation. Taking the natural log of this likelihood ratio, and multiplying it by $-2$, gives the log-likelihood ratio, or $G$-statistic. It gets bigger as the observed data get further from the null expectation. For the fly example, the test statistic is $G=2.17$. If you had observed $760$ smooth-winged flies and $240$ wrinkled-wing flies, which is closer to the null hypothesis, your $G$-value would have been smaller, at $0.54$; if you'd observed $800$ smooth-winged and $200$ wrinkled-wing flies, which is further from the null hypothesis, your $G$-value would have been $14.00$. You multiply the log-likelihood ratio by $-2$ because that makes it approximately fit the chi-square distribution. This means that once you know the G-statistic and the number of degrees of freedom, you can calculate the probability of getting that value of $G$ using the chi-square distribution. The number of degrees of freedom is the number of categories minus one, so for our example (with two categories, smooth and wrinkled) there is one degree of freedom. Using the CHIDIST function in a spreadsheet, you enter =CHIDIST(2.17, 1) and calculate that the probability of getting a $G$-value of $2.17$ with one degree of freedom is $P=0.140$. Directly calculating each likelihood can be computationally difficult if the sample size is very large. Fortunately, when you take the ratio of two likelihoods, a bunch of stuff divides out and the function becomes much simpler: you calculate the $G$-statistic by taking an observed number ($O$), dividing it by the expected number ($E$), then taking the natural log of this ratio. You do this for the observed number in each category. Multiply each log by the observed number, sum these products and multiply by $2$. The equation is: \[G=2\sum \left [ O\times \ln \left ( \frac{O}{E}\right ) \right ]\] The shape of the chi-square distribution depends on the number of degrees of freedom. For an extrinsic null hypothesis (the much more common situation, where you know the proportions predicted by the null hypothesis before collecting the data), the number of degrees of freedom is simply the number of values of the variable, minus one. Thus if you are testing a null hypothesis of a $1:1$ sex ratio, there are two possible values (male and female), and therefore one degree of freedom. This is because once you know how many of the total are females (a number which is "free" to vary from 0 to the sample size), the number of males is determined. If there are three values of the variable (such as red, pink, and white), there are two degrees of freedom, and so on. An intrinsic null hypothesis is one where you estimate one or more parameters from the data in order to get the numbers for your null hypothesis. As described above, one example is Hardy-Weinberg proportions. For an intrinsic null hypothesis, the number of degrees of freedom is calculated by taking the number of values of the variable, subtracting $1$ for each parameter estimated from the data, then subtracting $1$ more. Thus for Hardy-Weinberg proportions with two alleles and three genotypes, there are three values of the variable (the three genotypes); you subtract one for the parameter estimated from the data (the allele frequency, $p$); and then you subtract one more, yielding one degree of freedom. There are other statistical issues involved in testing fit to Hardy-Weinberg expectations, so if you need to do this, see Engels (2009) and the older references he cites. Contributor John H. McDonald (University of Delaware)
The expected value and variance of a random variable are actually special cases of a more general class of numerical characteristics for random variables given by moments. Definition$\PageIndex{1}$ The r th moment of a random variable $X$ is given by $$E[X^r].\notag$$ The rof a random variable $X$ is given by thcentral moment $$E[(X-\mu)^r],\notag$$ where $\mu = E[X]$. Note that the expected value of a random variable is given by the first moment, i.e., when $r=1$. Also, the variance of a random variable is given the second central moment. As with expected value and variance, the moments of a random variable are used to characterize the distribution of the random variable and to compare the distribution to that of other random variables. Moments can be calculated directly from the definition, but, even for moderate values of $r$, this approach becomes cumbersome. The next definition and theorem provides an easier way to generate moments. Definition$\PageIndex{2}$ The moment-generating function (mgf) of a random variable $X$ is given by $$M_X(t) = E[e^{tX}], \quad\text{for}\ t\in\mathbb{R}.\notag$$ Theorem $\PageIndex{1}$ If random variable $X$ has mgf $M_X(t)$, then $$M^{(r)}_X(0) = \frac{d^r}{dt^r}\left[M_X(t)\right]_{t=0} = E[X^r].\notag$$ In other words, the $r^{\text{th}}$ derivative of the mgf evaluated at $t=0$ gives the value of the $r^{\text{th}}$ moment. Theorem 3.9.1 tells us how to derive the mgf of a random variable, since the mgf is given by taking the expected value of a function applied to the random variable: $$M_X(t) = E[e^{tX}] = \left\{\begin{array}{c l} \text{discrete:} & \displaystyle{\sum_i e^{tx_i}\cdot p(x_i)} \\ & \\ \text{continuous:} & \displaystyle{\int\limits^{\infty}_{-\infty}\! e^{tx}\cdot f(x)\, dx} \end{array}\right.\notag$$ We can now derive the first moment of the Poisson distribution, i.e., derive the fact we mentioned in Section 4.1 that the expected value is given by the parameter. We also find the variance. Example $\PageIndex{1}$ Let $X\sim\text{Poisson}(\lambda)$. Then, the frequency function of $X$ is given by $$p(x) = \frac{e^{-\lambda}\lambda^x}{x!}, \quad\text{for}\ x=0,1,2,\ldots.\notag$$ Before we derive the mgf for $X$, we recall from calculus the Taylor series expansion of the exponential function $e^y$: $$e^y = \sum_{x=0}^{\infty} \frac{y^x}{x!}.\notag$$ Using this fact, we find $$M_X(t) = E[e^{tX}] = \sum^{\infty}_{x=0} e^{tx}\cdot\frac{e^{-\lambda}\lambda^x}{x!} = e^{-\lambda}\sum^{\infty}_{x=0} \frac{(e^t\lambda)^x}{x!} = e^{-\lambda}e^{e^t\lambda} = e^{\lambda(e^t - 1)}.\notag$$ Now we take the first and second derivative of $M_X(t)$. Remember we are differentiating with respect to $t$: \begin{align} M'_X(t) &= \frac{d}{dt}\left[e^{\lambda(e^t - 1)}\right] = \lambda e^te^{\lambda(e^t - 1)} \\ M''_X(t) &= \frac{d}{dt}\left[\lambda e^te^{\lambda(e^t - 1)}\right] = \lambda e^te^{\lambda(e^t - 1)} + \lambda^2 e^{2t}e^{\lambda(e^t - 1)} \end{align} Next we evaluate the derivatives at $t=0$ to find the first and second moments of $X$: \begin{align} E[X] = M'_X(0) &= \lambda e^0e^{\lambda(e^0 - 1)} = \lambda \\ E[X^2] = M''_X(0) &= \lambda e^0e^{\lambda(e^0 - 1)} + \lambda^2 e^{0}e^{\lambda(e^0 - 1)} = \lambda + \lambda^2 \end{align} Finally, in order to find the variance, we use the alternate formula: $$Var(X) = E[X^2] - \left(E[X]\right)^2 = \lambda + \lambda^2 - \lambda^2 = \lambda.\notag$$ Thus, we have shown that both the mean and variance for the Poisson$(\lambda)$ distribution is given by the parameter $\lambda$. Note that the mgf of a random variable is a function of $t$. The main application of mgf's is to find the moments of a random variable, as the previous example demonstrated. There are more properties of mgf's that allow us to find moments for functions of random variables. Theorem $\PageIndex{2}$ Let $X$ be a random variable with mgf $M_X(t)$, and let $a,b$ be constants. If random variable $Y= aX + b$, then the mgf of $Y$ is given by $$M_Y(t) = e^{bt}M_X(at).\notag$$ Theorem $\PageIndex{3}$ If $X_1, \ldots, X_n$ are independent random variables with mgf's $M_{X_1}(t), \ldots, M_{X_n}(t)$, respectively, then the mgf of random variable $Y = X_1 + \cdots + X_n$ is given by $$M_Y(t) = M_{X_1}(t) \cdots M_{X_n}(t).\notag$$ Recall that a binomially distributed random variable can be written as a sum of independent Bernoulli random variables. We use this and Theorem 7 to derive the mean and variance for a binomial distribution. First, we find the mean and variance of a Bernoulli distribution. Example $\PageIndex{2}$ Recall that $X$ has a Bernoulli$(p)$ distribution if it is assigned the value of 1 with probability $p$ and the value of 0 with probability $1-p$. Thus, the frequency function of $X$ is given by $$p(x) = \left\{\begin{array}{l l} 1-p, & \text{if}\ x=0 \\ p, & \text{if}\ x=1 \end{array}\right.\notag$$ In order to find the mean and variance of $X$, we first derive the mgf: $$M_X(t) = E[e^{tX}] = e^{t(0)}(1-p) + e^{t(1)}p = 1 - p + e^tp.\notag$$ Now we differentiate $M_X(t)$ with respect to $t$: \begin{align} M'_X(t) &= \frac{d}{dt}\left[1 - p + e^tp\right] = e^tp \\ M''_X(t) &= \frac{d}{dt}\left[e^tp\right] = e^tp \end{align} Next we evaluate the derivatives at $t=0$ to find the first and second moments: $$M'_X(0) = M''_X(0) = e^0p = p.\notag$$ Thus, the expected value of $X$ is $E[X] = p$. Finally, we use the alternate formula for calculating variance: $$Var(X) = E[X^2] - \left(E[X]\right)^2 = p - p^2 = p(1-p).\notag$$ Example $\PageIndex{3}$ Let $X\sim\text{binomial}(n,p)$. If $X_1, \ldots, X_n$ denote $n$ independent Bernoulli$(p)$ random variables, then we can write $$X = X_1 + \cdots + X_n.\notag$$ In Example 30, we found the mgf for a Bernoulli$(p)$ random variable. Thus, we have $$M_{X_i}(t) = 1 - p + e^tp, \quad\text{for}\ i=1, \ldots, n.\notag$$ Using Theorem 7, we derive the mgf for $X$: $$M_X(t) = M_{X_1}(t) \cdots M_{X_n}(t) = (1-p+e^tp) \cdots (1-p+e^tp) = (1-p+e^tp)^n.\notag$$ Now we can use the mgf of $X$ to find the moments: \begin{align} M'_X(t) &= \frac{d}{dt}\left[(1-p+e^tp)^n\right] = n(1-p+e^tp)^{n-1}e^tp \\ &\Rightarrow M'_X(0) = np \\ M''_X(t) &= \frac{d}{dt}\left[n(1-p+e^tp)^{n-1}e^tp\right] = n(n-1)(1-p+e^tp)^{n-2}(e^tp)^2 + n(1-p+e^tp)^{n-1}e^tp \\ &\Rightarrow M''_X(0) = n(n-1)p^2 + np \end{align} Thus, the expected value of $X$ is $E[X] = np$, and the variance is $$Var(X) = E[X^2] - (E[X])^2 = n(n-1)p^2 + np - (np)^2 = np(1-p).\notag$$ We end with a final property of mgf's that relates to the comparison of the distribution of random variables. Theorem $\PageIndex{4}$ The mgf $M_X(t)$ of random variable $X$ uniquely determines the probability distribution of $X$. In other words, if random variables $X$ and $Y$ have the same mgf, $M_X(t) = M_Y(t)$, then $X$ and $Y$ have the same probability distribution. The main application of Theorem 8 is to derive the probability distribution of functions of random variables. One of the most important applications like this is the following, which states that the sum of independent normally distributed random variables is also normally distributed. Sum of Normal Random Variables: If $X_1, \ldots, X_n$ are independent random variables with $X_i\sim\text{normal}(\mu_i,\sigma_i^2)$, for $i=1,\ldots,n$, then the random variable given by their sum is also normally distributed. More specifically, if $Y = X_1 + \cdots + X_n$, then $$Y \sim\text{normal}(\mu,\sigma^2), \quad\text{where}\ \mu = \mu_1 + \cdots + \mu_n\ \text{and}\ \sigma^2 = \sigma_1^2 + \cdots + \sigma_n^2.\notag$$ The above result is extremely useful in the study of statistics.
How would I work out a distribution free 95% confidence interval of the median if given a large sample? To get the 95% CI I use \mbox{estimate} \pm 1.96 \times \mbox{standard error}. I can work out the estimate of the median, which is just the sample median. However, to find the standard error do I use var(m) \approx \frac{1}{4nf(m)^2} and then square root that or would that not be distribution free?? Would I use the normal approximation to the binomial? Bi(n,0.5) \approx^d N(0.5n,0.25n)? To get the 95% CI I use \mbox{estimate} \pm 1.96 \times \mbox{standard error}. I can work out the estimate of the median, which is just the sample median. However, to find the standard error do I use var(m) \approx \frac{1}{4nf(m)^2} and then square root that or would that not be distribution free?? Would I use the normal approximation to the binomial? Bi(n,0.5) \approx^d N(0.5n,0.25n)? Last edited:

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