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I just started self-studying Mathematical Logic by Ebbinghaus. I already knew something about formal languages, but nothing about model theory. There is something I don't understand: Exercise 3.3, page 33, states: Let $P$ be a unary relation symbol and $f$ be a binary function symbol. For each of the formulas: $$\forall v_1 fv_0v_1 \equiv v_0, \hspace{.5cm} \exists v_0 \forall v_1 fv_0v_1 \equiv v_1,\hspace{.5cm} \exists v_0 (Pv_0 \wedge \forall v_1 Pfv_0v_1)$$ find an interpretation which satisfies the formula and one which does not satisfy it. I've done them all but I'm not sure of the real significance of what I just did. Let me clarify with an example. Let's take the first one. It doesn't use the symbol $P$, so I might as well take $S=\{f\}$ to be the set of symbols. As an $S$-structure I'll take $(\mathbb{N}, \cdot)$, and as an assignment for the variables I'll take $\beta(v_i)=0$ for all $i=0,1,2,\dots$. Denote $\mathcal{I}$ the corresponding interpretation. So: $\mathcal{I} \models \forall v_1 fv_0v_1 \equiv v_0$ iff for every $n\in \mathbb{N}$ $\mathcal{I} \frac{n}{v_1} \models fv_0v_1 \equiv v_0$, iff for every $n\in \mathbb{N}$ $0\cdot n=0$. Now I'd like to say that since the last sentence is true, then $\mathcal{I} \models \forall v_1 fv_0v_1 \equiv v_0$, i.e. $\mathcal{I}$ is a model for $\hskip0in$$\forall v_1 fv_0v_1 \equiv v_0$. But why is it true that for every $n\in \mathbb{N}$ $0\cdot n=0$? I mean, of course I know this should hold, but I am lost in the sea of formalism. Is there some set theory underlying here?
$$3\times 4=8$$ $$4\times 5=50$$ $$5\times 6=30$$ $$6\times 7=49$$ $$7\times 8=?$$ We have not managed to solve it so far, all we know is the solution (which was given after we had given up) : $224$ How do we find this solution ? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community $$3\times 4=8$$ $$4\times 5=50$$ $$5\times 6=30$$ $$6\times 7=49$$ $$7\times 8=?$$ We have not managed to solve it so far, all we know is the solution (which was given after we had given up) : $224$ How do we find this solution ? These interview problems are sometimes weird, where notations are bad, rules are arbitrary, and they expect only one answer where several could fit. Here is one, which could be the expected one, but probably not: To compute $a \times b$, take the numerator of $\dfrac{ab^2}{6}$ after simplification of the fraction. I don't see how they could argue it is wrong. Easy, just define $$\begin{array}{rcl}a \times b &=& \hspace{10.5pt}(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/72 + \\&& 25(a-3)(b-4)(a-5)(b-6)(a-6)(b-7)(a-7)(b-8)/18 + \\&& 15(a-3)(b-4)(a-4)(b-5)(a-6)(b-7)(a-7)(b-8)/8 \hspace{5.25pt}+ \\&& 49(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-7)(b-8)/36 + \\&&\hspace{5.5pt}7(a-3)(b-4)(a-4)(b-5)(a-5)(b-6)(a-6)(b-7)/18\end{array}$$ This might be a possible solution. For a positive integer $n$, let $\nu_2(n)$ be the largest $k$ such that $2^k\|n$, and similarly, let $\nu_3(n)$ be the largest $k$ such that $3^k\|n$. Finally let $$h(n)=\frac{n}{3^{\nu_3(n)}2^{1+4\lfloor \nu_2(n)/4\rfloor}}$$ If we consider $$ a\times ~ b {\buildrel \rm def\over =}~b h(ab) $$ then $(k-1)\times k$ coincides with the proposed results for $k=4,5,6,7$ and yields $224$ for $k=8$. The left-hand-side input and the right-hand-side output can be imagined as binary numbers in a kind of truth table: All eight output bits can be calculated from the seven input bits evaluating simple Boolean expressions. Spoiler Alert: (I use the answer given above in the response below. If you don't want to see it, you may want to skip this answer...) I'm replacing $\times$ by $\circ$, as the latter is more commonly used with unknown operations. I hate it when people redefine a common symbol, then "$=$" to describe a relationship. Note that $$\begin{align}3\circ4 &= 4\cdot 2\\ 4\circ 5 &= 5\cdot 10\\ 5\circ 6 &= 6\cdot 5\\ 6\circ 7 &= 7\cdot 7 \\ 7\circ 8 &= 8\cdot 28 \\ \end{align}$$ Thus, we can define: $$a\circ b\quad{\buildrel \rm def\over =}\quad b\cdot x_a$$ Where $x_n$ is some sequence. OEIS yields three possible sequences: $$x_n = \frac{\binom{n+2}{2}\gcd(n,3)}{3},\quad n \ge 0$$ (A234041) $$x_n = \text{denominatorOf}\left(\frac{(n-2)(n+3)}{(n)(n+1)}\right)\quad n \ge 3$$ (A027626: GCD of $n$-th and $(n+1)$st tetrahedral numbers, offset by me for this problem) The last sequence from OEIS is A145911 which is not promising at all. (It's a combination of, what appears to be, $3$ other sequences.) The answer is $42$. $69$ is also the answer. "Purple feelings" is also an answer. The truth of each of these is, of course, vacuous. :) If the question is posed as something other than multiplication, then it is the fault of the questioner for miscommunicating. Although, one could arguably blame the person trying to solve this problem for not doing enough to extract enough requirements from the 'customer' to be able to provide a solution. In some settings, this is an extremely important skill. 56 Did the question explicitly say there was a pattern to be found or is it just like you've presented it here? The symbols for multiplication(x) and equality(=) have well defined mathematical meaning and therefore 7 x 8 = 56 regardless of what misleading noise was written before. It may just be a test of the ability to avoid presumption. $$p(x)=$$ $$-\frac{1486263915627335609976345925580307452480}{198824918770116952269605821139049374259}-\frac{23535858736574459335924875719051524464677 x}{1789424268931052570426452390251444368331}+\frac{1532186339457747628597246965489647712097745599 x^2}{742539494635629574624160683858739355082631760}-\frac{5300973178829466500668773673899060773511329723 x^3}{62373317549392884268429497444134105826941067840}+\frac{425139989729581169917246837619141657974952401 x^4}{374239905296357305610576984664804634961646407040}-\frac{15160892592292573821061148160317799661783 x^5}{7128379148502043916391942565043897808793264896}+\frac{2379833487879115598578638026951579913181 x^6}{1496959621185429222442307938659218539846585628160}-\frac{133849478325585275186149006837381343 x^7}{249493270197571537073717989776536423307764271360}+\frac{9291465647310545015926219743101 x^8}{136087238289584474767482539878110776349689602560}$$ Then $$\begin{array}{\|c\|c\|c\|c\|c\|c\|c\|c\|c\|} \hline x & 12\color{grey}{(3\times 4)} & 20\color{grey}{(4\times 5)} & 30\color{grey}{(5\times 6)} &42\color{grey}{(6\times 7)}&56\color{grey}{(7\times 8)}& \color{grey}{1729}&\color{grey}{2014}&\color{grey}{2015}&\color{grey}{2016}\\ \hline p(x)& 8 & 50 & 30 &49&\color{red}{224}& \color{grey}{1729}&\color{grey}{2014}&\color{grey}{2015}&\color{grey}{2016} \\ \hline \end{array}$$ To learn to play this 'game', read me. Here is something I did which lead me to an incorrect result, but it is still pretty close. Since all the values we are given are of the form $a\times (a+1)$, I decided to make the function $f(a)=a\times (a+1)$. Assuming $f$ is a polynomial of grade $4$ or less we obtain $f$ is equal to $\frac{101 x^3}{6}-233 x^2+\frac{6301 x}{6}-1500$ using interpolation. This function gives us $f(7)=208$, which comes close, but is still not correct. This is what I have so far, it seems a bit more intuitive than Omran's solution. Based on the flip-flopping numbers, I figured the answer has to rely on the prime factorization of the numbers in question. So in particular, we see: $$3 \times 2^2 \Rightarrow 2$$ $$2^2 \times 5 \Rightarrow 25$$ $$5 \times 2 3 \Rightarrow 5$$ $$2 * 3 \times 7 \Rightarrow 7$$ $$7 \times 2^3 \Rightarrow 2^27$$ So my initial hypothesis, which is that you took the highest prime and any primes with power greater than $1$ fails for the first equation. But it does look like a promising lead. One solution is to define the operation $\times$ between two integers as $m \times n = n \cdot \left\{ \begin{array}{ll} \frac{1}{3}\sum_{k=1}^m k &\mbox{if } 3 \mid \sum_{k=1}^m k \\ \sum_{k=1}^m k &\mbox{otherwise.} \end{array} \right.$ The point is, that what remains of the RHS after dividing by $n$ can be recognized as the sum of the first $m$ integers, divided by $3$ should that be possible. numbers sequence 3 . 4 = 4 X 2 = 8 4 . 5 = 5 X 10 = 50 5 . 6 = 6 X 5 = 30 6 . 7 = 7 X 7 = 49 7 . 8 = 8 X 28 = 224 8 . 9 = 9 X 4 = 36 9 . 10 = 10 X 5 = 50 10 . 11 = 11 X 55 = 605 11 . 12 = 12 X 11 = 132 12 . 13 = 13 X 13 = 169 13 . 14 = 14 X 91 = 1274 14 . 15 = 15 X 7 = 105 15 . 16 = 16 X 8 = 128 16 . 17 = 17 X 136 = 2312 17 . 18 = 18 X 17 = 306 18 . 19 = 19 X 19 = 361 19 . 20 = 20 X 190 = 3800 20 . 21 = 21 X 10 = 210 a.(a+1) = (a+1)x((a+1)/2) - even number/2 b.( b+1) = (b+1)x(((a+1)/2 )xc)) - middle of the sandwich c.(c+1) = (c+1)xc d.(d+1) = (d+1)x(d+1) - odd number - copy e.(e+1) = (e+1)x((d+1)x(f/2)) f.(f+1) = (f+1)x(f/2) a+1=b, b+1=c,… The problem is more about finding the patterns and relations between numbers and given equations. 3×4=8 4×5=50 5×6=30 6×7=49 7×8=? There are some assumptions that have to be made: 1) look at the given equations as a sequences of numbers (sequences is plural - not just one sequence) 2) results on the right side can always by divided by the second number on the left side 8:4=2, 50:5=10, 30:6=5, 49:7=7 => the result of the last equations is therefore multiple of 8 => 7x8=8x?=??? (Note: Why did they use "x" when multiplication is clearly not what is done with those numbers? Why not better use symbol ∘ for unknown operations? My guess is - it's also a hint.... multiplication is necessary in the answer. ....so don't try to come up with solutions that are more complex than that ;-) But that's just my guess) 3) we can write down what we assume so far: 3 ∘4 = 4 X 2 = 8 4 ∘ 5 = 5 X 10 = 50 5 ∘ 6 = 6 X 5 = 30 6 ∘ 7 = 7 X 7 = 49 7 ∘8 = 8 X ?=??? 5) we can also say that after 7∘8=8x??? some other equations should follow and the patter we know so far is is: 8 ∘9 = 9 X ?=??? 9 ∘10 = 10 X ?=??? 5) now look at the numbers sequence (fourth number in each equation): 2, 10, 5, 7, ... there are of course many things we can do (2+8=10, 10-5=5, 5+2=7,etc.)... but we also have a possible relation to 3.4, 4.5, 5.6,6.7 6) the easy patter would be "sandwich"- second number = firstthird 7) how to define first and third number? - check the relation with 3.4 and 5.6 and first number also has a relation to 7.7 ..the rest I already explained in the comment section below ;-) The first multiplicant is given. So the open question is "what is the second multiplicant"? The list can be grouped in sixes. So lines 1-6 is one group, the rules count for each group. 4. Increment fm by one on each row result = fm * cm. You only change cm from row to row These are the rules for the six rows The sequence of cm would be 2, 10, 5, 7, 28, 4, 5, 55, 11, 13, 91, 7, 8, 136, 17, 19, 190, 10 I think you can continue like that The answer stares you right in the face. 7 x 8 is a question mark. Now I should add that one moderator apparently believed this was a joke answer. It should be obvious that it isn't. If I wanted to make a joke, I would have added a comment. Mathematics is about the manipulation of symbols, and this is an example of symbol manipulation creating a riddle with the answer hidden in plain sight. The riddle equates various symbols resembling products with other symbols in a rather pointless way. The question of the riddle is what the last symbol "7 x 8" equates. It obviously is meant to equate whatever symbol is to the right of the "=" sign. There is one answer here by maddog2k that I would consider better (that 7 x 8 = 56, since we shouldn't care about all the wrong answers given in the riddle but just give the correct answer). $A\cdot B = C$ $\dfrac{AB²}{\gcd(AB²,6)} = C$ or $\dfrac{\text{lcm}(AB²,6)}6 = C$ Surely even this interviewer meant for $\times$ to be commutative, so I propose $$a \times b = \left( \binom{a+1}{3},\binom{b+1}{3} \right) \cdot \max(a,b)$$ where the outer brackets indicate the gcd. Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). Would you like to answer one of these unanswered questions instead?
Search Now showing items 1-2 of 2 Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE (Elsevier, 2017-11) Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ... Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions (Elsevier, 2017-11) Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ...
Let $\mathcal{A}$ be a commutative unital Banach algebra, $\mathcal{B} \subset \mathcal{A}$ a closed unital subalgebra, $\mathcal{I} \subset \mathcal{B}$ a closed ideal. Is there in general a way to "identify" (soft question, perhaps) the maximal ideal spaces $\Sigma(\mathcal{B})$ and $\Sigma(\mathcal{A}/\mathcal{I})$, in terms of $\Sigma(\mathcal{A})$ and possibly some sort of other data? A couple of examples of the sort of thing I have in mind: If $\mathfrak{X}$ is a Banach space and $M \subset \mathfrak{X}$ a closed subspace, then $M^* \simeq \mathfrak{X}^/M^\perp$ and $(\mathfrak{X}/M)^ \simeq M^\perp$, where $$ M^\perp = \{f \in \mathfrak{X}^* \mid \forall m \in M: \, f(m) = 0\}. $$ In the special case where $\mathcal{A}$ is a $C^$-algebra, the contravariant equivalence of categories with (compact Hausdorff spaces, continuous maps) implies that $C^$-subalgebras of $\mathcal{A}$ correspond to quotients of $\Sigma(\mathcal{A})$, and quotients of $\mathcal{A}$ correspond to closed subspaces of $\Sigma(\mathcal{A})$. Are there some sort of analogous relationships with commutative Banach algebras? An example: Viewing the disc algebra $A(\mathbb{D})$ as a closed subalgebra of $C(\mathbb{T})$, the above analogies might lead us to expect that $\Sigma(A(\mathbb{D}))$ is a quotient of $\Sigma(C(\mathbb{T}))$. But $\Sigma(A(\mathbb{D})) \simeq \overline{\mathbb{D}}$ while $\Sigma(C(\mathbb{T})) \simeq \mathbb{T}$, so it looks like in this case the relationship is a subspace rather than a quotient (and going the other direction).
Basically, I want to know how one can see the $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ symmetry of AdS$_3$ explicitly. AdS$_3$ can be defined as hyperboloid in $\mathbb{R}^{2,2}$ as$$X_{-1}^2+X_0^2-X_1^2-X_2^2=L^2$$where $L$ is the AdS radius. Since the metric of $\mathbb{R}^{2,2}$,$$ds^2=-dX_{-1}^2-dX_0^2+dX_1^2+dX_2^2,$$is invariant under $SO(2,2)$ transformations and also the hyperboloid defined above is invariant we can conclude that AdS$_3$ has an $SO(2,2)$ symmetry. One can probably show with pure group theoretical arguments that the $SO(2,2)$ symmetry is isomorphic to an $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ symmetry. I would like to know however, if one can see this symmetry more explicitly in some representation of AdS$_3$? I suppose a starting point might be, that one can write the hyperboloid constraint equation as$$\frac{1}{L^2}\text{det}\;\begin{pmatrix} X_{-1}-X_1 & -X_0+X_2 \\ X_0+X_2 & X_{-1}+X_1\end{pmatrix}=1$$i.e. there is some identification of the hyperboloid with the group manifold of $SL(2,\mathbb{R})$ itself. However, that does not tell us anything about the symmetries. The only explanation that I have found (on page 12 of this pdf Master thesis) was that the group manifold of $SL(2,\mathbb{R})$ carries the Killing-Cartan metric$$g=\frac{1}{2}\text{tr}\,\left(g^{-1}dg\right)^2$$which is invariant under the actions$$g\rightarrow k_L\, g \qquad\text{and}\qquad g\rightarrow g\, k_R$$with $k_L,k_R\in SL(2,\mathbb{R})$. But how does one get from the metric on $\mathbb{R}^{2,2}$ to this Killing-Cartan metric? Also, I don't find this very explicit and was wondering if there is a more direct way.This post imported from StackExchange Physics at 2014-10-27 19:55 (UTC), posted by SE-user physicus
I really don't know how to tackle this proof because it has mod in it. There's 3 parts to the question. You don't have to answer all three parts (would be cool to check answer with though), I just need a starting point so get this proof rolling. Question: Relatively Prime: Means if two numbers' greatest common divisor is $1$. Let $a$ and $n$ be two natural numbers. Prove that if $a$ and $n$ are relatively prime, there exists a unique natural number $b < n$ such that $ab \equiv_n 1$ by doing the following: a) Prove that $$\exists b \in \mathbb N, b < n \land ab\equiv_n 1. \tag1$$ b) Add to equation ($1$) in part (a) to express that $b$ is unique. c) Now prove $b$ is unique. (If you need a picture of the question Question picture
The $\delta$ function is not continuous, so it's a priori not differentiable. In fact, it's not even well-defined as an ordinary real-valued function, but can be made so in terms of distributions - linear maps on a space of test functions given by $f\mapsto\int\delta f=f(a)$. It's possible to sensibly define derivatives of distributions by looking at representations as limits of functions: If $\delta_i$ is a family of functions so that $\lim_{i\rightarrow\infty}\int\delta_i(x) f(x)\mathrm dx=f(a)$ for any test function $f$, then it can be considered a representation of the Dirac delta. Now, if we take the family of derivatives $\frac{\mathrm d}{\mathrm dx}\delta_i$ we arrive at$$\int\left[\frac{\mathrm d}{\mathrm dx}\delta_i(x)\right]f(x)\mathrm dx=-\int\delta_i(x)\left[\frac{\mathrm d}{\mathrm dx}f(x)\right]\mathrm dx$$through integration by parts and using the fact that $f$ has by definition compact support (which makes the boundary term vanish). As the derivative is linear as well, this defines another linear map $f\mapsto-\int\delta f'$ on the space of test functions, which we call the derivative of our distribution. Symbolically,$$\left[\frac{\mathrm d}{\mathrm dx}\delta(x-a)\right]f(x)=-\delta(x-a)f'(x)$$which you can just plug in into your formula above without any need for actual computation as it holds true by definition.
Addendum. After re-considering the form of your question ( e.g. the M †M in the denominator --- as opposed for instance to a single operator M, which suffices for projectors) and reconsulting my copy of Nielsen and Chaung, here are some supplementary details not covered by my previous answer. (I'm posting this as a separate answer due to length, and because I feel that this is even less of an 'explanation' than my previous answer.) Suppose that our only means of measuring a qubit X is indirect: by a 'weak' interaction with an ancilla A, followed by a measurement on A.We would like to be able to talk about these as being in a sense a way of measuring X. How might we describe such a measurement in terms of X alone? Well: suppose we can easily prepare A in the initial state $\|+\rangle \propto \|0\rangle + \|1\rangle$, and perform a controlled unitary of the following sort, with X as the control and A as the target: $$ U \;=\; \left[\begin{matrix} \quad1\quad & 0 & 0 & 0 \\ 0 & \quad1\quad & 0 & 0 \\ 0 & 0 & \cos(\tfrac{\pi}{12}) & \sin(\tfrac{\pi}{12}) \\ 0 & 0 & -\sin(\tfrac{\pi}{12}) & \cos(\tfrac{\pi}{12}) \end{matrix}\right] $$ We then measure A in the standard basis (so that A now stores the measurement result). This transforms the state of X as follows: $\begin{align} \|\psi_0\rangle_X \;=&\; \alpha\|0\rangle_X + \beta\|1\rangle_X\\\\\mapsto&\; \alpha \|0\rangle_X \otimes \bigl(\tfrac1{\sqrt 2} \|0\rangle_A + \tfrac1{\sqrt 2}\|1\rangle_A\bigr) \quad+\quad \beta \|1\rangle_X \otimes \bigl(\tfrac1{\sqrt 2}\|0\rangle_A + \tfrac1{\sqrt 2}\|1\rangle_A\bigr)\\\\\mapsto&\; \alpha \|0\rangle_X \otimes \bigl(\tfrac1{\sqrt 2} \|0\rangle_A + \tfrac1{\sqrt 2}\|1\rangle_A\bigr) \quad+\quad \beta \|1\rangle_X \otimes \bigl(\tfrac{\sqrt 3}2\|0\rangle_A + \tfrac12\|1\rangle_A\bigr)\\\\=&\; \bigl( \tfrac{\alpha}{\sqrt 2} \|0\rangle_X + \tfrac{\sqrt3\beta}{2} \|1\rangle_X \bigr) \otimes \|0\rangle_A \quad+\quad \bigl( \tfrac{\alpha}{\sqrt 2} \|0\rangle_X + \tfrac{\beta}{2} \|1\rangle_X \bigr) \otimes \|1\rangle_A\\\\\mapsto&\;\begin{cases}\|\psi_1\rangle_X \otimes \|0\rangle_A \;\;\propto\;\; \bigl(\tfrac{\alpha}{\sqrt 2}\|0\rangle_X + \tfrac{\sqrt 3\beta}{2}\|1\rangle_X\bigr) \otimes \|0\rangle_A & \;\text{for the result 0; or } \\\|\psi_1\rangle_X \otimes \|1\rangle_A \;\;\propto\;\; \bigl(\tfrac{\alpha}{\sqrt 2}\|0\rangle_X + \tfrac{\beta}{2}\|1\rangle_X\bigr) \otimes \|1\rangle_A & \;\text{for the result 1.}\end{cases}\end{align}$ In the equations above, note that if the result of the measurement is c, the final state $\|\psi_1\rangle$ of X is proportional to $\|\psi'_1\rangle = M_c \|\psi_0\rangle$, where we define $$ M_0 \;=\; \tfrac{1}{\sqrt 2} \|0\rangle\langle 0\| + \tfrac{\sqrt 3}{2} \|1\rangle\langle 1\|\;,\qquad M_1 \;=\; \tfrac{1}{\sqrt 2} \|0\rangle\langle 0\| + \tfrac{1}{2} \|1\rangle\langle 1\|\;;$$ and we may verify that the probabilities with which we obtain the measurement results are in each case $\langle \psi'_1 \| \psi'_1 \rangle \;=\; \langle \psi_0 \| M_c^\dagger M_c \| \psi_0 \rangle$. This is very close to describing the transformation of X in the same way that we describe projective measurements. But is this any sort of measurement, meaningfully speaking? Well: if we can do statistics on the results of multiple iterations of this procedure, and if X is initially in the standard basis, we would notice that there is a bias in when we obtain the '0' result: we obtain it more often when X is initially in the state $\|1\rangle$.If we can sample enough times to distinguish whether the measurement results are distributed more like $(\frac12,\frac12)$ or $(\frac34,\frac14)$, we can determine with high probability whether the qubit is initially in the state $\|0\rangle$ or the state $\|1\rangle$. The similarity of the probabilities-and-update formulae to those of projective measurement, and the fact that we can use measurement statistics to get information about the state measured, motivates a generalization of the notion of 'measurement' to include procedures such as the one above: we may describe possible measurement outcomes by one, two, or more operators $M_c$ (which are in fact 'Kraus operators', objects associated to CPTP maps), with outcomes described by a slightly generalized Born rule $$ \Pr\limits_{\|\psi_0\rangle}(\text{result}=c) \;=\; \langle \psi_0 \| M_c^\dagger M_c \| \psi_0 \rangle \;, $$ where $M_c$ is a Kraus operator associated with your measurement, and with an update rule given by $$ \|\psi_1\rangle \;=\; \frac{M_c \|\psi_0\rangle}{\sqrt{\langle \psi_0 \| M_c^\dagger M_c \| \psi_0 \rangle}} \;.$$ In order for the probabilities to be conserved (so that with certainty at least one of the measurement results occurs), we require $\sum_c M_c^\dagger M_c = I$. This is the more general form in your question, described by Nielsen and Chaung. (Again, this looks slightly better when describing states by density operators.) General remarks. In general, any time that we introduce an ancilla (or collection of ancillas) A, interact a qubit (or register of several qubits) X unitarily with A, and then perform a projective measurement on A, this gives rise to a sort of measurement of X; the measurement operators can then be described by some collection of positive-semidefinite operators $M_c$ such that $\sum_c M_c^\dagger M_c \;=\; I$ (again so that probability is conserved). The more general, weaker measurements described here are more closely related to POVMs, which allow you to easily describe measurement probabilities 'abstractly', without an explicit choice of transformations $M_c$, by providing operators $E_c = M_c^\dagger M_c$ and allowing you to use these in the Born rule to compute probabilities. As I alluded to both above and in my previous response, POVMs can be regarded as describing statistically-available information about a system. Thinking of measurements in terms of Kraus operators (and in terms of a 'measurement result register' A as above) in this way allows you to subsume the notion of measurement into that of a CPTP map, which is an idea that I enjoy. (However, this doesn't really change things from an analytical standpoint, and isn't something you should worry about if you're not yet comfortable with CPTP maps).
8.1.2.1.3 - Example: Ice Cream Research Question: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%? In a sample of 50 customers 60% preferred chocolate over vanilla. $np_0 = 50(0.80) = 40$ $n(1-p_0)=50(1-0.80) = 10$ Both $np_0$ and $n(1-p_0)$ are at least 10. We can use the normal approximation method. This is a left-tailed test because we want to know if the proportion is less than 0.80. $H_{0}:p=0.80$ $H_{a}:p<0.80$ Test statistic: One Group Proportion $z=\frac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}$ $\widehat{p}$ = sample proportion $p_{0}$ = hypothesize population proportion $n$ = sample size $\widehat{p}=0.60$, $p_{0}=0.80$, $n=50$ $z= \frac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}= \frac{0.60-0.80}{\sqrt{\frac{0.80 (1-0.80)}{50}}}=-3.536$ Our $z$ test statistic is -3.536. This is a left-tailed test so we need to find the area to the right of our test statistic, $z=-3.536$. From the Minitab Express output above, the p-value is 0.0002031 $p \leq.05$, therefore our decision is to reject the null hypothesis. Yes, there is evidence that the percentage of all Creamery customers who prefer chocolate ice cream over vanilla is less than 80%.
Let be the unitary evolution operator of a quantum system be $U(t)=\exp(itH)$ for $t >0$. Then what is the meaning of the equation $$\det\bigl(I-U(t)e^{itE}\bigr)=0$$ where $E$ is a real variable? Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.Sign up to join this community Multiply both sides of your equation \begin{equation} \det(I-U(t)e^{itE})=0 \end{equation} by $e^{-intE}$ where $n$ is the number of dimensions of the state vector space. We obtain \begin{equation} \det(e^{-itE}I-U(t))=0 \end{equation} (See below for how this works.) This is a special case of the following equation \begin{equation} \det(\lambda I - A) = 0 \end{equation} whose solutions $λ_k$ are precisely the eigenvalues of operator $A$. Hence, the meaning of your equation is: Each $e^{-itE}$ satisfying your equation is an eigenvalue of the unitary operator $U(t)$. Note that all eigenvalues of unitary operators are complex numbers with absolute value 1. Above we used the following property of determinant: for any scalar $b$ \begin{equation} b^n\det(A) = \det(bA) \end{equation} Determinant of operator $A$ is defined by Leibniz formula as \begin{equation} \det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{k=1}^{n}a_{\sigma(k),k} \end{equation} which implies that for scalar $b$ \begin{equation} b^n\det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{k=1}^{n}ba_{\sigma(k),k}=\det(bA) \end{equation} I would like to elaborate Adam Zalcman's answer, from a physical angle. What Adam actually showed is that all the eigenvalues of the time-evolution operator are of the form $e^{i\phi}$ where $\phi$ is real. The mathematical implication is that $U$ does not change the norm of states. Let's look at the systems eigen-states, $\{\|n\rangle\}$, which are defined by $$H\|n\rangle=\epsilon_n\|n\rangle$$ These states span the whole Hilbert space, so knowing how $U$ acts on them tells you everything you need to know about time evolution of an arbitrary state. Note that these states are also eigenvectors of $U$, because $$\begin{align} U(t)\|n\rangle&=e^{\frac{i}{\hbar} H t}\|n\rangle = \sum_k \frac{\left(\frac{i}{\hbar}Ht\right)^k}{k!}\|n\rangle\\ & = \sum_k \frac{\left(\frac{i}{\hbar}\epsilon_n t\right) ^k }{k!}\|n\rangle = e^{\frac{i}{\hbar} \epsilon_n t}\|n\rangle \end{align}$$ and, indeed, each eigenvalue is of the form $e^{i\phi}$. This means physically that each eigenstate evolve in a very simple way - simply by changing its phase. An arbitrary state is of the form $$\|\psi\rangle=\sum_n c_n \|n\rangle$$ and its norm is $$\sqrt{\langle\psi\|\psi\rangle}=\sqrt{\sum_{n,m}c_m^c_n \langle m\|n\rangle}=\sqrt{\sum_{nm}c_m^c_n\delta_{mn}}=\sqrt{\sum_n \|c_n\|^2}$$ Since application of $U$ changes each $c_n$ only by its phase, it does not change the norm of $\|\psi\rangle$. BTW, since the absolute phase is not measurable, this implies that if the system is in a pure an eigenstate, it does not evolve in time. However, if the system is in a superposition of eigenstates, each eigenstate evolves with a different phase, according to their different energies, their relative phase changes and this is what causes stuff to change. Here's a nice applet that allows you to play with that. ADDED REMARK: This is an answer to Antillar Maximus' question: "A more general question would be, why is a unitary transformation useful?" My answer.The importance of unitary operators in QM relies upon a pair of fundamental theorems, known as Wigner's and Kadison's theorem respectively. Consider a quantum system described in a Hilbert space ${\cal H}$. Pure states are represented by vectors $\psi \in \cal H$ with $\|\|\psi\|\|=1$. Actually a state is the whole set $[\psi ] := \{e^{ia} \psi \:\|\: a \in R\}$ when $\|\|\psi\|\|=1$, because there is no physical way to fix the phase $e^{ia}$ and all physical quantities are independent form it. Notice that, with the introduced notation, one hence has $[\psi]= [\phi]$ if and only if $\psi = e^{ia} \phi$ for some $a \in R$. Henceforth $S(\cal H)$ will denote the space of pure states in $\cal H$. (It coincides to the complex projective space of $\cal H$ but it does not matter now.) If you have two states $[\psi] \neq [\phi]$, represented by normalized vectors $\psi$ and $\phi$, as is well known $$P([\psi],[\phi])\| := \|\langle \phi\|\psi \rangle\|^2$$ is a physical quantity representing the probability transition from the former to the latter state (in both directions). Notice that, as it must be, that quantity is not affected by changes of the arbitrary phases embodied in the states. There are very important transformations of state called (quantum) symmetries. A symmetry, in the sense of Wigner, is a map from the space of the states to the space of the states, say $S(\cal H) \ni [\psi] \to [\psi'] \in S(\cal H)$ verifying two requirements: (1W) the map is bijective ("into" and "onto"). In other words, with a given symmetry one can moves form a given state to any other state, and the action of the symmetry is reversible; (2W) the map preserves the transition probabilities: $\|\langle \phi\|\psi \rangle\|^2 = \|\langle \phi'\|\psi' \rangle\|^2$. There are lots of physical transformations of states verifying (1W) and (2W). For instance: time evolution, charge conjugation, parity reflection, time reversal, the action of Poincaré's group or Galileo's group, and many many others. Sometimes it turns out that a supposed symmetry actually does not exist (like parity reflection for systems subjected to weak interaction). The natural question is therefore: What is the most general form of a symmetry, i.e. a map from the set of the states to the same set satisfying (1W) and (2W)? The answer was given by Wigner with a celebrated theorem commonly known as Wigner's theorem (even if Wigner gave so many contributions to the mathematical foundation of QM that it sounds very reductive to call that theorem Wigner's theorem simply). THEOREM (Wigner). Let $\cal H$ be a complex separable Hilbert space and $f : [\psi] \to [\psi']$ a map transforming states to states and satisfying (1W) and (2W).Then, there exists either unitary or anti-unitary (depending on $f$) operator:$$U_f : \cal H \to \cal H\:,$$ determined by $f$ up to a constant phase (i.e. $U_f$ can only be modified to $e^{ia}U_f$ with $a\in R$, preserving its properties), such that:$$f ([\psi]) = [U_f \psi]\quad\quad \mbox{for every normalized vector $\psi \in \cal H$}$$ In particular, taking Wigner's theorem into account, we may forget of states as classes of vectors and we can safely use vectors as usual to represent states. (Life is not so easy, since an annoying and difficult problem arises as soon as one tries to represent a group of symmetries this way) Let us come to the other, analogous, important theorem due to Kadison.Everybody knows that there exists another, more general, notion of state in QM. I mean a mixed state. A mixed state is represented by a positive, trace class operator $\rho : \cal H \to \cal H$ with unit trace $tr(\rho)=1$. The set $M(\cal H)$ of mixed states is a convex body in the real linear space of self-adjoint bounded operators on $\cal H$. This means that if $p_1,\ldots, p_n \in [0,1]$ and $\rho_1,\ldots, \rho_n \in M(\cal H)$ verify $\sum_i p_i =1$, then: $$\sum_{i=1}^n p_i \rho_i \in M({\cal H})\:. \qquad (1)$$ The expectation value of an observable $A$ respect to the state $\rho$ is $$\langle A \rangle_\rho := tr(\rho A)\:,$$ (where some hypotheses are necessary on the domain of the self-adjoint operator $A$). Pure states $[\psi]$ are a subcase of mixed ones, those of the form $\|\psi\rangle \langle \psi\|$ (notice that the arbitrary phase affecting $\psi$ does not matter here). Actually it is possible to prove that pure states are all of the states in $M(\cal H)$ such that cannot be decomposed as in (1) (for some non-trivial decomposition): They are the extremal elements of $M(\cal H)$. Every mixed state $\rho$, in general, admits many decompositions like that in (1). One is the most natural, that obtained by the spectral decomposition: $$\rho = \sum_{k=0}^{+\infty} p_k \|\psi_k \rangle \langle \psi_k\|\:,\qquad (2)$$where now the $p_k$s are the eigenvalues of $\rho$, that (by hypotheses on $\rho$)belong to $[0,1]$ and their sum is $1$. (2) has a natural interpretation as a classical mixing, with classical probabilities $p_k$, of pure quantum states $\|\psi_k \rangle \langle \psi_k\|$. Actually in general there are many ways to decompose $\rho$ into such a form, it is untenable separating classical probabilities (carried by the $p_n$s) and quantum probabilities (embodied in the $\|\psi_k \rangle \langle \psi_k\|$s). Within this context, there is another notion of quantum symmetry due to Kadison (actually it was formulated using a dual, equivalent, approach referring to the lattice of orthogonal projectors). A symmetry (in the sense of Kadison) is a map $f: M({\cal H})\ni \rho \mapsto \rho' \in M(\cal H)$ that verifies: (1K) it is bijective; (2K) it is convex linear, i.e.,$$f\left( \sum_{i=1}^n p_i \rho_i \right) =\sum_{i=1}^n p_i f(\rho_i)\quad \mbox{for $p_i \in [0,1]$ with $\sum_i p_i=1$.}$$ The second conditions physically means that if we construct a mixed state $\rho$ by classically mixing some other states $\rho_i$ with respective classical probabilities $p_i$, the action of the symmetry preserves this composition (the classical probabilities) with respect to the transformed states. Such a notion of quantum symmetry seems to be quite different from that proposed by Wigner. Nonetheless the two notions coincide in view of the celebrated (however less known than Wigner's one) Kadison's theorem. THEOREM (Kadison). Let $\cal H$ be a complex separable Hilbert space and $f : M({\cal H}) \ni \rho \to \rho' \in M(\cal H)$ a map transforming states to states and satisfying (1K) and (2K).Then there exists either unitary or anti-unitary (depending on $f$) operator:$$U_f : \cal H \to \cal H\:,$$ determined by $f$ up to a constant phase (i.e. $U_f$ can only be modified to $e^{ia}U_f$ with $a\in R$, preserving its properties), such that:$$f (\rho) = U_f \rho U_f^\dagger \quad\quad \mbox{for every $\rho \in M(\cal H)$.}$$ It is obvious that, looking at the associated (anti) unitary operators, a Kadison symmetry defines a Wigner symmetry and vice-versa. Just a mathematical note in response to the previous answer: $e^{i \hat{H} t /\hbar}$ is not defined as the exp-series, although it is common to define it so in physics textbooks. But it is not possible to do this as the series is generally not converging (in the operator norm). One has to use the spectral calculus, in which the "calculation" $$ e^{i \hat{H} t /\hbar} \|n\rangle = e^{i E_n t /\hbar} \|n\rangle $$ becomes (some kind of) a definition. If the dimension of the state space is finite, say $n$, then your question makes sense since the determinant makes sense. Now suppose that $E$ is a real number such that (for all $t$) $$\det (I-U(t)e^{itE} ) =0.$$ Your equation implies that $I-U(t)e^{itE}$ is not invertible (if it were invertible, its determinant would be non-zero). This implies that there exists a non-zero vector $v$, a state, for which $$v=U(t)e^{itE}v$$ and hence $e^{-itH}v=e^{itE}v$ and hence $$ Hv = -Ev$$ so $v$ is an eigenvector with eignevalue $-E$, i.e., $-E$ is an energy level of the system and when it is in the state $v$ it will, if measured, produce the result of energy$ = -E$ with probability one. But in infinite dimensions you can't be so direct. A more general question would be, why is a unitary transformation useful? A unitary transformation preserves the norm, i.e the norm is invariant under basis transformations (as stated by others above). But why is this useful? This is quite useful because in many cases, we cannot make measurements in the space of interest, but we can always transform to an accessible isomorphic space. From the time evolution expression you have, it is important to identify that the Hamiltonian is the infinitesimal generator of time evolution. The exponential notation is just that... a notation for a particular combination of operators. I disagree with the claim about convergence etc being applicable to an operator. As far as I know, no such formalism exists for operators. If you want to talk about real analysis, you can only do so with respect to a representation of the operator, not the operator itself. This critical distinction is often ignored in introductory QM (and fatally so!). +1 for a very good question that anybody studying intro QM ought to have!
For part $(i)$ you're on the right track, but note that we are concerned with $x_n, y_n$ converging to any arbitrary $c$. Note that your $x_n, y_n$ both converge to $0$ (also that $y_n$ is necessarily even a sequence of irrationals - consider $n = 4$). I'm guessing you are to come up with the sequences yourself, and I leave but a hint for you here: If $c$ is rational consider adding to $c$ some rational sequence that converges to $0$ i.e. find an $a_n$ such that $\{a_n\} \subset \mathbb{Q}$ and $a_n \to 0$. If $c$ is irrational consider first a sequence of decimal approximations for $x_n$ (that is the $n^{th}$ term of $x_n$ has $n$ decimal places expanded out) and for $y_n$ simply consider adding on the same sequence $a_n$ as above, will $c + a_n$ be irrational always? Once you have done this then note that you will have $\lim_{n \to \infty} f(x_n) = (c - 2)^3$ and $\lim_{n \to \infty} = (2 - c)$. Now note $f$ is continuous at $c$ iff the limit is the same no matter how you approach it (meaning, no matter what sequence you use to approach $c$). If $c \neq 2$ what do you get above? As for $(ii)$, we can use the sequential characterization of continuity. That is, $f : \mathbb{R} \to \mathbb{R}$ is continuous at $c$ iff for any $c_n \to c$ we have$$\lim_{n \to \infty} f(c_n) = f(c)$$Now theres three possible types of sequences when talking about rational and irrational sequences: Completely rational sequence (all elements of the sequence are themselves rational) Completely irrational sequence (all elements of the sequence are themselves irrational) Mixed: Some elements are rational, some are irrational Now when considering the case of $c = 2$ what is the limit of a completely rational sequence? What about a completely irrational sequence? As for this third category of sequences, the proof really depends on the level of rigor your professor/teacher wants. If you clue me in on how precise you want this to be I can help you out.
This question already has an answer here: I recently encountered this following proposition: For every polynomial, there is some positive integer for which it is composite. What is the most elementary proof of this? 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 This question already has an answer here: I recently encountered this following proposition: For every polynomial, there is some positive integer for which it is composite. What is the most elementary proof of this? Suppose $P$ is a polynomial, then it is periodic mod $m$: I mean $P(a) = P(a+m) \pmod m$. Suppose it takes on different prime values like $p$ and $q$. Then $P(x) \equiv 0 \pmod p$ for some $x$, and $P(y) \equiv 0 \pmod q$ for some $y$. By periodicity we can find a $z$ such that $pq\|P(z)$ using periodicity. Not quite true, look at the polynomial $17$. And the usual theorem specifies that the polynomial has integer coefficients. Let $P(x)$ be a non-constant polynomial with integer coefficients. Without loss of generality we can assume that its lead coefficient is positive. It is not hard to show that there is a positive integer $N$ such that for all $n\ge N$, we have $P(n)\gt 1$, and such that $P(x)$ is increasing for $x\ge N$. (For large enough $x$, the derivative $P'(x)$ is positive.) Let $P(N)=q$. Then $P(N+q)$ is divisible by $q$. But since $P(x)$ is increasing in $[N,\infty)$, we have $P(N+q)\gt q$. Thus $P(N+q)$ is divisible by $q$ and greater than $q$, so must be composite. Remark: One can remove the "size" part of the argument. For any $b$, the polynomial equation $P(x)=b$ has at most $d$ solutions, where $d$ is the degree of $P(x)$. So for almost all integers $n$, $P(n)$ is not equal to $0$, $1$, or $-1$. Let $N$ be a positive integer such that $P(n)$ is different from $0$, $1$, or $-1$ for all $n\ge N$. Let $P(N)=q$. Consider the numbers $P(N+kq)$, where $k$ ranges over the non-negative integers. All the $P(N+kq)$ are divisible by $q$. But since the equations $P(n)=\pm q$ have only finitely many solutions, there is a $k$ (indeed there are infinitely many $k$) such that $P(N+kq)$ is not equal to $\pm q$, but divisible by $q$. Such a $P(N+kq)$ cannot be prime. I prefer using considerations of size.
Any integer greater than 1 is called a prime number if and only if its positive factors are 1 and the number p itself. The basic ideology involved in this post is flawed and the post has now been moved to Archives. – The Editor Prime Generating Formulas We all know how hard it is to predict a formula for prime numbers! They have… Our reader Eswar Chellappa has sent his work on the solution of ‘3X+1’ problem, also called Collatz Conjecture. He had been working on the proof of Collatz Conjecture off and on for almost ten years. The Collatz Conjecture can be quoted as follow: Let $\phi : \mathbb{N} \to \mathbb{N}^+$ be a function defined such that: $$\phi(x):= \begin{cases} \frac{x}{2}, & \text{if }… My personal mobile number 9565804301 is a prime number. What is a prime number?$ Any integer p greater than 1 is called a prime number if and only if its positive factors are 1 and the number p itself. In other words, the natural numbers which are completely divisible by 1 and themselves only and have no other factors, are called prime numbers. 2$ ,$ 3$… Consider two natural numbers $n_1$ and $n_2$, out of which one is twice as large as the other. We are not told whether $n_1$ is larger or $n_2$, we can state following two propositions: PROPOSITION 1: The difference $n_1-n_2$, if $n_1 >n_2$, is different from the difference $n_2-n_1$, if $n_2 >n_1$. PROPOSITION 2: The difference $n_1-n_2$, if $n_1 >n_2$, is the same… This is a continuation of the series of summer projects sponsored by department of science and technology, government of India. In this project work, I have worked to collect and expand what Ramanujan did with Nested Radicals and summarized all important facts into the one article. In the article, there are formulas, formulas and only formulas — I think this… This mathematical fallacy is due to a simple assumption, that $ -1=\dfrac{-1}{1}=\dfrac{1}{-1}$ . Proceeding with $ \dfrac{-1}{1}=\dfrac{1}{-1}$ and taking square-roots of both sides, we get: $ \dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}$ Now, as the Euler’s constant $ i= \sqrt{-1}$ and $ \sqrt{1}=1$ , we can have $ \dfrac{i}{1}=\dfrac{1}{i} \ldots \{1 \}$ $ \Rightarrow i^2=1 \ldots \{2 \}$ . This is complete contradiction to the… People really like to twist the numbers and digits bringing fun into life. For example, someone asks, “how much is two and two?” : the answer should be four according to basic (decimal based) arithmetic. But the same with base three (in ternary number system) equals to 11. Two and Two also equals to Twenty Two. Similarly there are many ways you can… Here is an interesting mathematical puzzle alike problem involving the use of Egyptian fractions, whose solution sufficiently uses the basic algebra. Problem Let a, b, c, d and e be five non-zero complex numbers, and; $ a + b + c + d + e = -1$ … (i) $ a^2+b^2+c^2+d^2+e^2=15$ …(ii) $ \dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} +\dfrac{1}{d} +\dfrac{1}{e}= -1$… The greatest number theorist in mathematical universe, Leonhard Euler had discovered some formulas and relations in number theory, which were based on practices and were correct to limited extent but still stun the mathematicians. The prime generating equation by Euler is a very specific binomial equation on prime numbers and yields more primes than any other relations out there in… “Irrational numbers are those real numbers which are not rational numbers!” Def.1: Rational Number A rational number is a real number which can be expressed in the form of where $ a$ and $ b$ are both integers relatively prime to each other and $ b$ being non-zero. Following two statements are equivalent to the definition 1. 1. $ x=\frac{a}{b}$… Ramanujan (1887-1920) discovered some formulas on algebraic nested radicals. This article is based on one of those formulas. The main aim of this article is to discuss and derive them intuitively. Nested radicals have many applications in Number Theory as well as in Numerical Methods . The simple binomial theorem of degree 2 can be written as: $ {(x+a)}^2=x^2+2xa+a^2 \… This is the last month of the glorious prime year 2011. We are all set to welcome upcoming 2012, which is not a prime but a leap year. Calendars have very decent stories and since this blog is based on mathematical approach, let we talk about the mathematical aspects of calendars. The international calendar we use is called Gregorian Calendar,… If mathematics was a language, logic was the grammar, numbers should have been the alphabet. There are many types of numbers we use in mathematics, but at a broader aspect we may categorize them in two categories: 1. Countable Numbers 2. Uncountable Numbers The numbers which can be counted in nature are called Countable Numbers and the numbers which can… This is a puzzle which I told to my classmates during a talk, a few days before. I did not represent it as a puzzle, but a talk suggesting the importance of Math in general life. This is partially solved for me and I hope you will run your brain-horse to help me solve it completely. If you didn’t notice,… You might be thinking why am I writing about an individual number? Actually, in previous year annual exams, my registration number was 381654729. Which is just an ‘ordinary’ 9-digit long number. I never cared about it- and forgot it after exam results were announced. But today morning, when I opened “Mathematics Today” magazine’s October 2010, page 8; I was brilliantly… Before my college days I used to multiply this way. But as time passed, I learned new things. In a Hindi magazine named “Bhaskar Lakshya”, I read an article in which a columnist ( I can’t remember his name) suggested how to multiply in single line (row). That was a magic to me. I found doing multiplications this way, very faster –… Multiplication is probably the most important elementary operation in mathematics; even more important than usual addition. Every math-guy has its own style of multiplying numbers. But have you ever tried multiplicating by this way? Exercise: $ 88 \times 45$ =? Ans: as usual :- 3960 but I got this using a particular way: 88 45… We all know that the derivative of $x^2$ is 2x. But what if someone proves it to be just x?
I read that the elliptical wing has the lowest induced drag. Why is that so? I welcome a mathematical and intuitive explanation. Induced drag is caused by the backward inclination of the lift vector. Lift is defined as the aerodynamic force perpendicular to the flow direction, and since lift is created by deflecting this flow downwards, the resulting force is slightly inclined backwards. The formula for the deflection angle (called downwash angle) $\alpha_w$: $$\alpha_w = arctan \left(\frac{2\cdot c_L}{\pi\cdot AR}\right)$$ To arrive at the drag, you now need to multiply the local lift with this angle. For small $\alpha_w$s the arcus tangens can be neglected, and we get this familiar looking equation for the backwards-pointing component of the aerodynamic force: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR} = \frac{\alpha_w^2\cdot\pi\cdot AR}{4}$$ Now consider a wing which will do this unevenly over the wingspan. Some parts create a lot of lift and, therefore, a lot of downward deflection. Others produce less lift and less deflection. Due to the quadratic dependency of induced drag with downwash angle, the sum of all parts will always be higher when there are variations in the local downwash angle over span. Only the wing with a constant downwash angle will have the minimum induced drag for a given lift. This is not necessarily an elliptic wing. By selecting the right twist (incidence variation over wingspan), any planform can be made to produce a constant downwash angle, albeit at only one angle of attack. Only the elliptic wing will combine the constant downwash angle over span with a constant lift coefficient over span such that variations in angle of attack still result in a constant downwash angle. This is strictly true only in inviscid flow and when elastic deformation and the weight of the wing (which contributes to the need for lift and hence induced drag) are not considered. Also, an elliptic wing will produce nasty stall characteristics. Adding handling, friction and structural weight will shift the optimum to wings which produce more lift near the center. The exact optimum also will depend on the absolute size of the wing because the structural mass fraction goes up with size due to scaling laws. Nomenclature: $AR \:\:$ aspect ratio of the wing $c_{Di} \:\:$ Induced drag coefficient $c_L \:\:\:$ lift coefficient $\pi \:\:\:\;$ 3.14159$\dots$
Is there any expression for the direct change in conductivity of metals with temperature ? I am aware of change in resistivity with temperature. Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up.Sign up to join this community Since $$\sigma = \frac1\rho $$ and, for a good approximation on typical temperatures where the metal is solid and and warmer than 180 K (==-90 °C): $$\rho = \rho_0 + \alpha T$$ (with $\alpha$ being the temperature coefficient) it follows that $$\sigma = \frac1{\rho_0 + \alpha T} $$
I am wondering about correct answer to this task from a yesterday’s test: A function Pow which calculates $ y = a^k$ is given, where $ k$ is an integer of length $ n$ bits: function Pow(a, k) { k >= 0 } z := a; y := 1; m := k; while m != 0 do if m mod 2 = 1 then y := y * z; end if m := m div 2; z := z * z; end while return y; end function Calculate the worst case time complexity and the average time complexity of this function. The dominant operation is a compare operation performed in line 6. Describe shortly the value of $ k$ when the worst case occurs. So, I believe the number of comparisons is dependant on length of $ k$ in terms of its bits. Let $ k = 0$ : (binary $ 0$ too, which is $ 1$ bit): $ \Rightarrow 0$ comparisons Let $ k = 1$ : (binary $ 1$ too, which is $ 1$ bit): $ \Rightarrow 1$ comparison Let $ k = 8$ : (binary $ 1000$ which is $ 4$ bits) $ \Rightarrow 4$ comparisons Let $ k = 15$ : (binary $ 1111$ which is $ 4$ bits) $ \Rightarrow 4$ comparisons Let $ k = 16$ : (binary $ 10000$ which is $ 5$ bits) $ \Rightarrow 5$ comparisons I think a pattern can be seen. Any number from the set $ \{2^h, 2^h + 1, \cdots, 2^{h+1} – 2, 2^{h+1} – 1 \} \quad \land \quad h > 0 \quad$ , is $ h + 1$ bits long and hence $ h + 1$ comparison. So I’d believe $ T_{avg}(n) = T_{worst}(n) = n \in O(n)$ But $ n$ is number of bits of $ k$ number. Function takes $ k$ as a parameter, not $ n$ . So my solution is not the one that’s desired, I think. In terms of $ k$ I think it would look like that: $ T_{worst}(k) = \lfloor log_{2}(2k) \rfloor \in O(\log k)$ $ T_{avg}(k) = \lfloor log_{2}(2k)\rfloor \in O(\log k)$ Questions: Is the solution in terms of $ k$ correct? The solution in terms of $ n$ : how would you grade that, personally? Knowing the task’s description from the above. I have posted similiar thread on math stackexchange, but would like to get more opinions on this from CS experts themselves.
In the chapter on Artinian rings in "Introduction to Commutative Algebra" by Atiyah and MacDonald, we have: Proposition 8.6. Let $(A,\mathfrak{m})$ be a local Noetherian ring. Then exactly one of the following holds: $\frak{m}^n\neq\frak{m}^{n+1}$ for all $n \in \mathbb{N}$; $\frak{m}$ is a nilpotent ideal, in which case, $A$ is Artinian. Proposition 8.8. Let $(A,\mathfrak{m})$ be a local Artinian ring. Then the following are equivalent: $A$ is a principal ideal ring; $\frak{m}$ is principal; $\dim_K(\frak{m}/\frak{m}^2)\leq 1$ (where $K=A/\frak{m}$ is the residue field). In the proof of 8.8, A&M quickly boil down to the case: $$\dim_K(\mathfrak{m}/\mathfrak{m}^2)= 1 \Rightarrow A \text{ is a principal ideal ring.}$$ A&M begin by explaining that $\mathfrak{m}$ is nilpotent, which is because $\mathfrak{m}$ equals the Jacobson radical of $A$ (as $A$ is a local ring) and the Jacobson radical of an Artinian ring is nilpotent (as A&M prove earlier). However, doesn't this also follow from 8.6 as Artinian rings are Noetherian (which A&M prove in Proposition 8.5)?
a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached). b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct. However, your calculation of amplitude is not correct. Peak-to-trough is twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly. CALCULATION OF AMPLITUDE OF OSCILLATIONS 1. Equation of Motion Method First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision. Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$ 2. Energy Method Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$. At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$. At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$. By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1. (c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it. Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal. Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result. The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate. Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$ We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off. The total energy at P is the same as at U, so from above ( Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
If the time complexity $T(n)$ of two functions $x(n)$ and $y(n)$ are mutually recursive, as in Time complexity of mutually recursive functions, this gives $T(n)=O(2^n)$. But how do you measure an algorithm with two or more (4 in the example below) recursive functions that are not necessarily mutually recursive? Example: Recursive Step Function $$ \begin{align} w(n)&= \begin{cases} w(n-1)+1 & \text{if } x(n-1)=y(n-1), \\ w(n-1) & \text{otherwise.} \end{cases} \\ x(n)&=8w(n). \\ y(n)&= \begin{cases} 1 & \text{if } w(n) > w(n-1), \\ y(n-1)+1 & \text{otherwise.} \end{cases} \\ z(n) &= \begin{cases} z(n-1)+1 & \text{if } z(n-1) < y(n-1) \text{ and } z(n-1)=1, \\ z(n-1)-1 & \text{if } z(n-1) < y(n-1) \text{ and } z(n-1)\neq 1, \\ z(n-1) & \text{otherwise.} \end{cases} \end{align} $$ These functions loop until $w(n)=I$ and then output the value of $z(n)$, where $I$ is the user input value. This input value could be rather large. I get the worse case scenario is dependent on $w(n)$, as it increases slowly. I have calculated that the function specifically $$w(n)=O\left(\frac{\sqrt{16n+16}}{8}+\frac12\right),$$ but I am new to time complexity. Could anyone help me to understand? Here is the example in pseudocode: SET w to 0SET x to 1SET y to 1SET z to 0GET iREPEAT IF z<y and z=1 THEN z=z+1 ELSEIF z<y and z != 1 THEN z=z-1 ELSE do nothing IF x=y THEN w=w+1 and x=8w and y=1 ELSE x=8w and y=y+1UNTIL w=iPRINT z
$\beta$ in statistical mechanics is equal to $\frac{1}{k_BT}$ in in thermodynamics, but I do not understand why $\beta\propto T^{-1}$ instead of, say, $\beta\propto T$? Because by convention, we want to write $\beta$ as the coefficient in front of energy $E$ in the exponent $\exp(-\beta E)$. Exponents have to be dimensionless so $\beta$ has to have units of inverse energy. That's why it has to be objects such as $\beta=1/kT$ because $kT$ has units of energy. The latter statement holds because the energy per degree of freedom increases with the temperature. At the end, that's the fundamental answer to your question. The exponential is $\exp(-E/kT)$ because $E\sim kT$ and "hot" means "highly energetic". It's linked to the fact that the temperature has an absolute lower bound, the absolute zero, much like energy is bounded from below. Sometimes a definition is just a definition. The underlying math of statistical mechanics demands that we work with $E/kT$ quite a bit, and historically, we've chosen to work with $\beta \equiv 1/kT$. This can make some mathematical manipulations easier: for instance, $\frac{\partial}{\partial \beta} e^{\beta E} = E e^{\beta E}$ is easier to work with than if we explicitly left in a division (but there is "conservation of misery" here, as my advisor likes to say, where the nice things we get for doing this cost in converting from $\partial/\partial \beta$ to $\partial/\partial T$ down the road). You're entirely free to rewrite the whole of statistical mechanics in terms of some different parameter that is proportional to $T$. Call it $\tau$ if you like. $\tau = kT$. This can be an illustrative exercise for the inquisitive mind, seeing how it makes some expressions simpler and others more laborious. $\beta$ is the fundamental quantity that appears in the laws of statistical mechanics, and comparison with the laws of thermodynamics then implies that $\beta$ is inversely proportional to the temperature. This can be seen from the first law, which says $dU=TdS-PdV$, while statistical mechanics provides at constant volume the formula $dS/k_B=\beta dU$. As a consequence, $\beta=1/k_BT$. Therefore a ''temperature'' defined to be proportional to $\beta$ would measure coldness rather than hotness. It is a historical accident that temperature was designed to measure hotness rather than coldness. The correspondence between statistical mechnaics and thermodynamics would be simpler if it were otherwise, but one cannot change history and the resulting tradition deeply rooted in our culture. In statistical thermodynamics, when using the method of Lagrange multipliers, we obtain an expression as $$-\ln \rho = \alpha + \beta H$$ where $\alpha$ and $\beta$ are the multipliers to be determined. Multiplying by the Boltzmann constant and averaging we obtain the entropy $$\langle S \rangle = k_\mathrm{B}\alpha + k_\mathrm{B} \beta \langle H \rangle$$ Comparing with the thermodynamic entropy for a closed system at constant composition (Euler expression) $$S = S_0 + \frac{U}{T}$$ you obtain the value $\beta = 1/k_\mathrm{B}T$. You could try an alternative method by defining a Lagrange multiplier $\beta' = 1/\beta$, $$-\ln \rho = \alpha + H / \beta'$$ Repeating the above procedure you would obtain the value $\beta' = k_\mathrm{B}T$, but I do not find any advantage in this.
Difference between revisions of "User:Nikita2" m (→TeXing) m (7 intermediate revisions by the same user not shown) Line 3: Line 3: − I am Nikita Evseev + I am Nikita Evseev Novosibirsk, Russia. My research interests are in [[Mathematical_analysis \| Analysis]] and [[Sobolev space\|Sobolev spaces]]. My research interests are in [[Mathematical_analysis \| Analysis]] and [[Sobolev space\|Sobolev spaces]]. Line 54: Line 54: I'm keen on turning up articles of EoM into better appearance by rewriting formulas and math symbols in TeX. I'm keen on turning up articles of EoM into better appearance by rewriting formulas and math symbols in TeX. − Now there are ''' + Now there are '''''' (out of 15,890) articles with [[:Category:TeX done]] tag. − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − <asy> <asy> size(0,150); size(0,150); − real tex= + real tex=; real all=15890; real all=15890; pair z0=0; pair z0=0; Line 102: Line 71: arrow("still need TeX",dir(-1(0.5theta+d)),2E); arrow("still need TeX",dir(-1(0.5theta+d)),2E); </asy> </asy> − Latest revision as of 21:58, 17 June 2018 Pages of which I am contributing and watching Analytic function \| Cauchy criterion \| Cauchy integral \| Condition number \| Continuous function \| D'Alembert criterion (convergence of series) \| Dedekind criterion (convergence of series) \| Derivative \| Dini theorem \| Dirichlet-function \| Ermakov convergence criterion \| Extension of an operator \| Fourier transform \| Friedrichs inequality \| Fubini theorem \| Function \| Functional \| Generalized derivative \| Generalized function \| Geometric progression \| Hahn-Banach theorem \| Harmonic series \| Hilbert transform \| Hölder inequality \| Lebesgue integral \| Lebesgue measure \| Leibniz criterion \| Leibniz series \| Lipschitz Function \| Lipschitz condition \| Luzin-N-property \| Newton-Leibniz formula \| Newton potential \| Operator \| Poincaré inequality \| Pseudo-metric \| Raabe criterion \| Riemann integral \| Series \| Sobolev space \| Vitali theorem \| TeXing I'm keen on turning up articles of EoM into better appearance by rewriting formulas and math symbols in TeX. Now there are 3040 (out of 15,890) articles with Category:TeX done tag. $\quad \rightarrow \quad$ $\sum_{n=1}^{\infty}n!z^n$ Just type $\sum_{n=1}^{\infty}n!z^n$. Today You may look at Category:TeX wanted. How to Cite This Entry: Nikita2. Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Nikita2&oldid=29993
a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached). b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct. However, your calculation of amplitude is not correct. Peak-to-trough is twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly. CALCULATION OF AMPLITUDE OF OSCILLATIONS 1. Equation of Motion Method First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision. Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$ 2. Energy Method Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$. At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$. At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$. By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1. (c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it. Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal. Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result. The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate. Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$ We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off. The total energy at P is the same as at U, so from above ( Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
I would like your opinion about the following question: Let $\{f_n\}$ be a sequence of integrable functions in $[a,b]$which uniformly converges to integrable function $f$, then let $F(x)=\int_a^xf(t)dt $, and $F_n(x)=\int_a^xf_n(t)dt$, I want to prove that $F_n$ uniformly converges to $F$ on $[a,b]$. Can I just use the continuous of the integral and say that $$\lim_{n \to \infty} F_n(x)=\lim_{n \to \infty}\int_{a}^{x}f_n(t)=\int_{a}^{x}\lim _{n \to \infty}f_n(t)dt=\int_{a}^{x}f(t)dt=F(x)?$$ Thanks a lot!
Optical depth Let's start by looking at the idea of optical depth. Optical depth is a quantity that describes how light is attenuated as it travels through a medium. There are two commonly used equations for optical depth, $\tau$. In a homogeneous 1 medium, they are$$\tau=n\sigma x,\quad\tau=\kappa\rho x$$Here, $n$ and $\rho$ are the number and mass densities of the medium, $\sigma$ is the cross-sectional area for absorption 2, $\kappa$ is a quantity called the opacity of the medium, and $x$ is the distance traveled by light. As an example, if light travels through a uniform slab of optical depth $\tau(x)$, the intensity of the light after traveling a distance $x$ will be$$I(x)=I_0e^{-\tau(x)}=I_0e^{-n\sigma x}=I_0e^{-\kappa\rho x}$$In stellar astrophysics, the surface of the Sun - the edge of its photosphere - is defined as the radius $r$ at which $\tau=2/3$. In general, we can say that an object is opaque if $\tau\geq 1$. 1 In reality, most mediums are not of uniform density, and so we would rewrite the first equation as$\tau(x)=\int_0^xN(x')\sigma dx'$, where $N(x)$ is the column density. However, on your planet, I'd guess that over short distances, $N$ is approximately uniform. 2 $\sigma$ is usually wavelength-dependent, and should really be written as $\sigma_\lambda$. The same goes for the opacity. Scattering cross-sections When talking about scattering in gases, there are two main regimes. When the scattering of light of wavelength $\lambda$ is due to particles of diameter $d$, the theory of Rayleigh scattering holds when $d\ll\lambda$ and the more general Mie scattering holds when $d\simeq\lambda$. Now, for visible light, $\lambda\sim10^{-7}$ meters. In air, the Rayleigh approximation holds well, but for fog, $d\sim10^{-6}$ meters, and Mie theory would be more appropriate. Regrettably, all solutions for the cross-sections in Mie scattering are numerical, not analytical. I'm going to therefore try to use Rayleigh scattering as an example. Let's calculate some cross-sections. Assuming that air has an index of refraction $n'$ 3 and mean particle diameter $d$, we get$$\sigma_{\text{air}}=\frac{2\pi^5}{3}\frac{d_{\text{air}}^6}{\lambda^4}\left(\frac{n_{\text{air}}'^2-1}{n_{\text{air}}'^2+2}\right)^2$$For air at standard temperature and pressure, $n_{\text{air}}'\approx1.000293$, and $d_{\text{air}}\sim2\times10^{-9}$ meters, generously. If we pick light midway through the visible spectrum - say, green light with $\lambda=550$ nm - then we find that $\sigma_{\text{air}}\sim5\times10^{-33}$ square meters. For fog, let's bite the bullet and use the same approximation. We'll say $d_{\text{fog}}\sim5\times10^{-6}$ meters. I wasn't able to find great figures for the index of refraction, but this group indicates $n_{\text{fog}}'\approx1.5$ for some fogs. Using the same formula gives me $\sigma_{\text{fog}}\sim3\times10^{-6}$ square meters - much larger than $\sigma_{\text{air}}$, as one would expect. 3 I'm using $n'$ and not the normal $n$ here so as to not confuse it with number density, $n$. Putting it together It should now be clear that the dominant source of extinction is due to fog. The number densities of air and fog should be relatively similar, and so $n_{\text{fog}}\sigma_{\text{fog}}\gg n_{\text{air}}\sigma_{\text{air}}$. We can therefore say that the optical depth is largely thanks to fog. Now, say we define "opaque" as meaning that $\tau=1$ at a distance of $x=1$ meters. We then would need to have a number density of$$n_{\text{fog}}=\frac{\tau}{\sigma_{\text{fog}}x}\approx3.32\times10^5\text{ particles/m}^3$$which comes out to $1.7\times10^{-7}$ kg/m$^3$ . . . which is much less dense than water vapor, by a factor of about $10^6$. Clearly, the Rayleigh approximation fails. Applying Mie theory So, fortunately, smart people out there have built tools that let the rest of us calculate some important factors. I used this Mie scattering calculator. Plugging in a radius of $d=5$ microns, a light wavelength of $\lambda=550$ nm, and an index of refraction of $n_{\text{fog}}'=1.5$ (and an imaginary index of refraction of $-0.3$, as given in the linked paper), the calculator gave me $\sigma_{\text{fog}}\sim3\times10^{-8}$ square meters, lower than our Rayleigh result by a factor of 100. This, then, means a number density of $n_{\text{fog}}\approx3.32\times10^7$ particles per cubic meter, and a mass density off of water vapor by a factor of $10^4$, rather than $10^6$. Now, you wouldn't expect the atmosphere to have the density of water vapor. On Earth, fog constitutes a much smaller fraction of air, which is why we don't suffocate on a foggy day. The factor of $10^4$, then, seems somewhat reasonable, though possibly off by an order of magnitude or so. At any rate, Mie theory does, as expected, give a much better result, and it seems like the sort of atmosphere you require would not be unreasonable. What could cause this? The Grand Banks are arguably the foggiest place on Earth, where the warm Gulf Stream mixes with the cold Labrador current. However, that sort of mixing won't occur planet-wide - this is the only place on Earth it occurs on such a large scale - as currents of different temperatures will only meet like this in certain regions. The Swiss Plateau might be a better example. Fully foggy days occur from November to January, and slightly less foggier days occur between October and February. This happens because of a specific kind of temperature inversion, thanks to a wind current called a bise, which interacts with the mountains. Turbulence underneath the inversion layer leads to low-level stratus clouds and, eventually, fog. Again, though, we have the problem that this sort of current won't occur everywhere in the world. San Francisco provides another interesting case. The Bay Area has great conditions for fog: moisture from the Pacific, a large temperature gradient between ocean currents and the land, and mountains to further trap clouds and fog. This sort of coastal region isn't foggy year-round, but when fog develops, it becomes extremely thick. Essentially, the ingredients you want for really thick fog are Some sort of temperature gradient, whether that be colliding air/water currents or temperature differences between land and sea. A source of moisture, such as an ocean. Mountains or valleys to trap the fog and low-level clouds and prevent them from dissipating. Combine elements from these three regions, hand-wave the currents a bit, and you have the potential for some very foggy regions. I'm thinking plenty of coastlines, plenty of mountains and valleys, and lots and lots of water.
How does one prepare for the exams? Read up the textbooks, write down the important formulae? Redo the previous year question papers? Well, each and every student may have their own ways of preparing for the exams. Meanwhile, has anyone ever tried solving the important questions from a subject? How can a student access these important questions? For future reference, the Karnataka 10th board students can have a look at these important questions given below: Class 10 Maths Important Questions 1.If A={1,2,3} and B = { 2, 3, 4, 5 } are the subsets of U = { 1, 2, 3, 4, 5, 6, 7, 8 }, verify that $(A\cap B)’= A’ \cup B'$ 2.. Prove that 2+ $\sqrt{3}$ is an irrational number 3. FInd the sum of all 2 digit natural numbers that are divisible by 5 4. If 2 ($^{n}P_{2}) +$ + 50= $^{2n}P_{2}$, then find the value of n 5. Rationalise the denominator and simplify $\frac{3\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ 6. A box has 4 red and 3 black marbles. Four marbles are picked up randomly. Find the probability that two marbles are red. 7. Calculate standard deviation for the following scores : 5, 6, 7, 8, 9. 8. Find the radius of a circle whose centre is ( – 5, 4 ) and which passes through the point ( – 7, 1 ). 9. In Δ ABC, $DE \left \| \right \| BC$, if AD = 2 cm, DB = 5 cm and AE = 4 cm, find AC. 10. If $Cos \theta =\frac{5}{13}$, then find the value of $\frac{Sin\theta + Cos \theta }{Sin \theta – Cos \theta }$ 11. From the top of a building 20 m high, the angle of elevation of the top of a vertical pole is 30° and the angle of depression of the foot of the same pole is 60°. Find the height of the pole. 12. A solid metallic cylinder of diameter 12 cm and height 15 cm is melted and recast into toys in the shape of a right circular cone mounted on a hemisphere as shown in the figure. If radii of the cone and hemisphere are each equal to 3 cm and the height of the toy is 7 cm, calculate the number of such toys that can be formed. 13. Find three consecutive positive integers such that the sum of the square of the first integer and the product of the other two is 92. 14. At constant pressure certain quantity of water at 24°C is heated. It was observed that the rise of temperature was found to be 4°C per minute. Calculate the time required to raise the temperature of water to 100°C at sea level by using formula. 15. Draw a chord of length 6 cm in a circle of radius 5 cm. Measure and write the distance of the chord from the centre of the circle. 16. A student while conducting an experiment on Ohm’s law, plotted the graph according to the given data. Find the slope of the line obtained. 17. Draw the plan for the information given below : 2 ( Scale 20 m = 1 cm ) 18. A dealer sells an article for Rs. 16 and loses as much per cent as the cost price of the article. Find the cost price of the article 19. A solid is in the form of a cone mounted on a right circular cylinder, both having the same radii as shown in the figure. The radius of the base and height of the cone are 7 cm and 9 cm respectively. If the total height of the solid is 30 cm, find the volume of the solid. 20. State and prove Basic Proportionality ( Thale’s ) theorem. Now, we will take you through the significance of solving Karnataka Class 10 Maths Important Questions. Why to Solve Karnataka Board Class 10 Important Questions of Maths? Class 10 Maths can be a difficult subject to crack and it requires a whole lot of practice. This is where the important questions can help students. Covering all the main topics that falls under the Class 10 Karnataka Board Maths Syllabus, these questions are devised after proper analysis of the previous year papers to see which questions have a possibility to repeat. It provides a lot of practice to students They can self-evaluate their performance Can check the knowledge gap of a subject and prepare accordingly Gain more confidence to write the exams More acquainted with the oft repeated questions Supplements one’s studies and textbook material For much exhaustible list of free study material and to get an idea about the Karnataka Board Exam pattern the student can reach out to BYJU’S.
Let us consider testing the null hypothesis that there is zero correlation between two variables $X_{j}$ and $X_{k}$. Mathematically we write this as shown below: $H_0\colon \rho_{jk}=0$ against $H_a\colon \rho_{jk} \ne 0 $ Recall that the correlation is estimated by sample correlation $r_{jk}$ given in the expression below: $r_{jk} = \dfrac{s_{jk}}{\sqrt{s^2_js^2_k}}$ Here we have the sample covariance between the two variables divided by the square root of the product of the individual variances. We shall assume that the pair of variables $X_{j}$and $X_{k}$ are independently sampled from a bivariate normal distribution throughout this discussion; that is: $\left(\begin{array}{c}X_{1j}\\X_{1k} \end{array}\right)$, $\left(\begin{array}{c}X_{2j}\\X_{2k} \end{array}\right)$, $\dots$, $\left(\begin{array}{c}X_{nj}\\X_{nk} \end{array}\right)$ are independently sampled from a bivariate normal distribution. To test the null hypothesis, we form the test statistic, t as below $t = r_{jk}\sqrt{\frac{n-2}{1-r^2_{jk}}}$ $\dot{\sim}$ $ t_{n-2}$ Under the null hypothesis, $H_{o}$, this test statistic will be approximately distributed as t with n - 2 degrees of freedom. Note! This approximation holds for larger samples. We will reject the null hypothesis, $H_{o}$, at level $α$ if the absolute value of the test statistic, t, is greater than the critical value from the t-table with n - 2 degrees of freedom; that is if: $\|t\| > t_{n-2, \alpha/2}$ To illustrate these concepts let's return to our example dataset, the Wechsler Adult Intelligence Scale. Example 5-5: Wechsler Adult Intelligence Scale Section Using SAS This data was analyzed using the SAS program in our last lesson, (Multivariate Normal Distribution), which yielded the computer output below. Download the: Using Minitab Click on the video below to see how you can find the total variance of the Wechsler Adult Intelligence Scale data. Recall that these are data on n = 37 subjects taking the Wechsler Adult Intelligence Test. This test was broken up into four components: Information Similarities Arithmetic Picture Completion Looking at the computer output we have summarized the correlations among variables in the table below: Information Similarities Arithmetic Picture Information 1.00000 0.77153 0.56583 0.31816 Similarities 0.77153 1.00000 0.51295 0.08135 Arithmetic 0.56583 0.51295 1.00000 0.27988 Picture 0.31816 0.08135 0.27988 1.00000 For example, the correlation between Similarities and Information is 0.77153. Let's consider testing the null hypothesis that there is no correlation between Information and Similarities. This would be written mathematically as shown below: $H_0\colon \rho_{12}=0$ We can then substitute values into the formula to compute the test-statistic using the values from this example: \begin{align} t &= r_{jk}\sqrt{\frac{n-2}{1-r^2_{jk}}}\\[10pt] &= 0.77153 \sqrt{\frac{37-2}{1-0.77153^2}}\\[10pt] &= 7.175 \end{align} Looking at our t-table for 35 degrees of freedom and an $\alpha$ level of .005, we get a critical value of $t _ { ( d f , 1 - \alpha / 2 ) } = t _ { 35,0.9975 } = 3.030$. Therefore, we are going to look at the critical value under 0.0025 in the table (since 35 does not appear to use the closest df that does not exceed 35 which is 30) and in this case it is 3.030, meaning that $t _ { ( d f , 1 - \alpha / 2 ) } = t _ { 35,0.9975 } = 3.030$ is close to 3.030. Note! Some text tables provide the right tail probability (the graph at the top will have the area in the right tail shaded in) while other texts will provide a table with the cumulative probability - the graph will be shaded into the left. The concept is the same. For example, if alpha was 0.01 then using the first text you would look under 0.005 and in the second text look under 0.995. Because $7.175 > 3.030 = t_{35, 0.9975}$, we can reject the null hypothesis that Information and Similarities scores are uncorrelated at the $\alpha$ < 0.005 level. Our conclusion here is that Similarities scores increase with increasing Information scores ( t = 7.175; d.f. = 35; p < 0.0001). You will note here that we are not simply concluding that the results are significant. When drawing conclusions it is never adequate to simply state that the results are significant. In all cases, you should seek to describe what the results tell you about this data. In this case, because we rejected the null hypothesis we can conclude that the correlation is not equal to zero. Furthermore, because the actual sample correlation is greater than zero and our p-value is so small, we can conclude that there is a positive association between the two variables, and hence our conclusion that Similarities scores tend to increase with increasing values of Information scores. You will also note that the conclusion includes information from the test. You should always back up your conclusions with the appropriate evidence: the test statistic, degrees of freedom (if appropriate), and p-value. Here the appropriate evidence is given by the test statistic t = 7.175; the degrees of freedom for the test, 35, and the p-value, less than 0.0001 as indicated by the computer print out. The p-value appears below each correlation coefficient in the SAS output. Confidence Interval for $p_{jk}$ Section Once we conclude that there is a positive or negative correlation between two variables the next thing we might want to do is compute a confidence interval for the correlation. This confidence interval will give us a range of reasonable values for the correlation itself. The sample correlation, because it is bounded between -1 and 1 is typically not normally distributed or even approximately so. If the population correlation is near zero, the distribution of sample correlations may be approximately bell-shaped in distribution around zero. However, if the population correlation is near +1 or -1, the distribution of sample correlations will be skewed. For example, if $p_{jk}= .9$, the distribution of sample correlations will be more concentrated near .9. Because they cannot exceed 1, they have more room to spread out to the left of .9, which causes a left-skewed shape. To adjust for this asymmetry or the skewness of distribution, we apply a transformation of the correlation coefficients. In particular, we are going to apply Fisher's transformation which is given in the expression below in Step 1 of our procedure for computing confidence intervals for the correlation coefficient. Steps Compute Fisher’s transformation $z_{jk}=\frac{1}{2}\log\dfrac{1+r_{jk}}{1-r_{jk}}$ Here we have one half of the naturallog of 1 plus the correlation, divided by one minus the correlation. Note!In this course, whenever log is mentioned, unless specified otherwise, log stands for the natural log. For large samples, this transform correlation coefficient zis going to be approximately normally distributed with the mean equal to same transformation of the population correlation, as shown below, and a variance of 1 over the sample size minus 3. $z_{jk}$ $\dot{\sim}$ $N\left(\dfrac{1}{2}\log\dfrac{1+\rho_{jk}}{1-\rho_{jk}}, \dfrac{1}{n-3}\right)$ Compute a (1 - $\alpha$) x 100% confidence interval for the Fisher transform of the population correlation. $\dfrac{1}{2}\log \dfrac{1+\rho_{jk}}{1-\rho_{jk}}$ That is, one half log of 1 plus the correlation divided by 1 minus the correlation. In other words, this confidence interval is given by the expression below: $\left(\underset{Z_l}{\underbrace{Z_{jk}-\frac{Z_{\alpha/2}}{\sqrt{n-3}}}}, \underset{Z_U}{\underbrace{Z_{jk}+\frac{Z_{\alpha/2}}{\sqrt{n-3}}}}\right)$ Here we take the value of Fisher's transform Z, plus and minus the critical value from the ztable, divided by the square root of n- 3. The lower bound we will call the $Z_{1}$ and the upper bound we will call the $Z_{u}$. Back transform the confidence values to obtain the desired confidence interval for $\rho_{jk}$ This is given in the expression below: $\left(\dfrac{e^{2Z_l}-1}{e^{2Z_l}+1},\dfrac{e^{2Z_U}-1}{e^{2Z_U}+1}\right)$ The first term we see is a function of the lower bound, the $Z_{1}$. The second term is a function of the upper bound or $Z_{u}$. Let's return to the Wechsler Adult Intelligence Data to see how these procedures are carried out. Example 5-6: Wechsler Adult Intelligence Data Section Recall that the sample correlation between Similarities and Information was $r_{12} = 0.77153$. Step 1: Compute the Fisher transform: \begin{align} Z_{12} &= \frac{1}{2}\log \frac{1+r_{12}}{1-r_{12}}\\[5pt] &= \frac{1}{2}\log\frac{1+0.77153}{1-0.77153}\\[5pt] &= 1.024 \end{align} You should confirm this value on your own. Step 2: Next, compute the 95% confidence interval for the Fisher transform, $\frac{1}{2}\log \frac{1+\rho_{12}}{1-\rho_{12}}$ : \begin{align} Z_l &= Z_{12}-Z_{0.025}/\sqrt{n-3} \\ &= 1.024 - \frac{1.96}{\sqrt{37-3}} \\ &= 0.6880 \end{align} \begin{align} Z_U &= Z_{12}+Z_{0.025}/\sqrt{n-3} \\&= 1.024 + \frac{1.96}{\sqrt{37-3}} \\&= 1.3602 \end{align} In other words, the value 1.024 plus or minus the critical value from the normal table, at $α/2 = 0.025$, which in this case is 1.96. Divide by the square root of n minus 3. Subtracting the result from 1.024 yields the lower bound of 0.6880. Adding the result to 1.024 yields the upper bound of 1.3602. Step 3: Carry out the back-transform to obtain the 95% confidence interval for ρ 12. This is shown in the expression below: $\left(\dfrac{\exp\{2Z_l\}-1}{\exp\{2Z_l\}+1},\dfrac{\exp\{2Z_U\}-1}{\exp\{2Z_U\}+1}\right)$ $\left(\dfrac{\exp\{2 \times 0.6880\}-1}{\exp\{2 \times 0.6880\}+1},\dfrac{\exp\{2\times 1.3602\}-1}{\exp\{2\times 1.3602\}+1}\right)$ $(0.5967,0.8764)$ This yields the interval from 0.5967 to 0.8764. Conclusion: In this case, we can conclude that we are 95% confident that the interval (0.5967, 0.8764) contains the correlation between Information and Similarities scores. Note!The interpretation of this interval. We did not say that with 95% probabilitythe correlation between Information and Similarities lies between the interval. This statement would imply that the population correlation is random. In fact, the population correlation is a fixed quantity. The only quantities that are random are the bounds of the confidence interval, which are a function of the random data and so are also random---before the data is observed. After the data is observed and the interval is computed, there is no longer any random quantity. Therefore, we use the word "confidence", rather than "probability", to describe the parameter falling inside the interval.
An image burbled up in my social-media feed the other day, purporting to be a list of “17 Equations that Changed the World.” It’s actually been circulating for a while (since early 2014), and purports to summarize the book by that name written by Ian Stewart. This list is typo-ridden, historically inaccurate and generally indicative of a lousy knowledge-distribution process that lets us down at every stage, from background research to fact-checking to copy-editing. The following comments are meant to be representative, not exhaustive. It’s not known whether Pythagoras proved the theorem we named for him—or if any of the stories about him are more than legends, really. When you go back that far, the history of mathematics and science becomes semi-legendary. The best one can typically do for “evidence” is a fragment of a lost book quoted in another book that happened to survive, and all of it dating to decades or centuries after the events ostensibly being chronicled. Did Pythagoras actually prove the theorem we named after him, or did he merely observe that it held true in a few special cases, like the 3-4-5 right triangle? Tough to say, but the latter would have been easier, and it would seem to appeal to a number mystic, for whom it’s all about the Benjamins successive whole numbers. Pythagoras himself probably wrote nothing, and nothing in his own words survives. It’s not clear whether his contemporaries viewed him as a mathematician or primarily as a propounder of an ethical code. (Even only 150 years after the time he purportedly lived, the ancient authorities disagreed about whether Pythagoras was a vegetarian, with Aristoxenus saying no and Eudoxus yes.) If Pythagoras had never lived, and a cult had attributed their work to that name in ritual self-denial; if the stories of his visiting Egypt and being the son of a Tyrian corn merchant began as parables and were later taken as biography—it would be hard to tell the result from what we have today. (And, in fact, groups of mathematicians do sometimes publish under a collective pseudonym: witness the Bourbaki collective.) Typical, really: Indian and Chinese people do the actual work, and the white guy who likely didn’t gets all the credit. I’ll outsource the criticism of the “logarithms” part: Once again [the] simple attribution to John Napier is exactly that, simplistic and historically misleading. We can find the principle on which logarithms are based in the work of several earlier mathematicians. We can find forms of proto-logarithms in both Babylonian and Indian mathematics and also in the system that Archimedes invented to describe very large numbers. In the fifteenth century Triparty, of the French mathematician Nicolas Chuquet we find the comparison between the arithmetical and geometrical progressions that underlay the concept of logarithms but if Chuquet ever took the next step is not clear. In the sixteenth century the German mathematician Michael Stifel studied the same comparison of progressions in his Arithmetica integraand did take the next step outlining the principle of logarithms but doesn’t seem to have developed the idea further. It was in fact John Napier who took the final step and published the first set of logarithmic tables in his book Mirifici Logarithmorum Canonis Descriptioin 1614. However the Swiss clockmaker and mathematician, Jost Bürgi developed logarithms independently of Napier during the same period although his book of tables, Arithmetische und Geometrische Progress Tabulen, was first published in 1620. The “calculus” line is a mess. For starters, in at least one version circulating online, it’s got an extra “=” thrown in, which makes the whole thing gibberish. The $df$ over $dt$ notation is due to Leibniz, but the list attributes it to Newton, his bitter enemy (and a pretty bitter guy overall, by many accounts). Pierre de Fermat understood quite a bit of the subject before Newton worked on it, getting as far as computing the maxima and minima of curves by finding where their tangent lines are horizontal. And the philosophy of setting up the subject of calculus using limits is really a nineteenth-century approach to its foundations. Credit for the normal distribution should also go to de Moivre (earlier than Gauss) and Laplace (contemporaneous). Maxwell never wrote his Equations in that manner; that came later, with Heaviside, Gibbs, Hertz and vector calculus. The simplification provided by the vector calculus is really nothing short of astonishing. The idea of entropy came via Clausius, who found inspiration in the work of Carnot. The statement that entropy either increases or stays the same, which we could write as $dS \geq 0$, predates Boltzmann. What Boltzmann provided was an understanding of how entropy arises in statistical physics, the study of systems with zillions of pieces whose behavior we can’t study individually, but only in the aggregate. If you want to attribute an equation to Boltzmann in recognition of his accomplishments, it’d be better to use the one that is actually carved on his tombstone, $$S = k \log W.$$ I am not sure that $E = mc^2$ is the proper way to encapsulate the essence of relativity theory. It is a consequence, not a postulate or a premise. The Lorentz transformation equations would do a better job at cutting to the heart of the subject. Note that these formulae are named after Lorentz, not Einstein; to put the history very, very briefly, Lorentz wrote the equations down first, but Einstein understood what they meant. (And the prehistory of $E = mc^2$ is pretty fascinating, too.) Plucking out the Schrödinger equation (the list omits the umlaut because careless) does a disservice to the history of quantum mechanics. There are ways of doing quantum physics without invoking the Schrödinger equation: Heisenberg’s matrix mechanics, the Dirac–Feynman path integral, and the one it’s my day job to work on. In fact, not only did Heisenberg’s formulation come first, but we didn’t know what Schrödinger’s work meant until Max Born clarified that the square of the size of Schrödinger’s complex number $\Psi$ is a probability. The number of names in that last paragraph—and I wasn’t even trying—is a clue that factoids and bullet points are not a good way of learning physics. Yes, Robert May did write about the logistic map, $$x_{t+1} = k x_t(1-x_t),$$ but he was hardly the first to poke at it. In his influential paper “Simple mathematical models with very complicated dynamics,” there’s a moment which expresses pretty well how science happens sometimes: How are these various cycles arranged along the interval of relevant parameter values? This question has to my knowledge been answered independently by at least 6 groups of people, who have seen the problem in the context of combinatorial theory, numerical analysis, population biology, and dynamical systems theory (broadly defined). Also, d’Alembert was not named “d’Almbert.”
that the electron and the proton are a particle pair This may appear inconsistent with current theory, as it is widely assumed that particle pairs created in a single event, must have opposite charge and equal mass (matter/antimatter). This assumption turns out to be wrong. We shall see that the difference in mass between the electron and the proton is due to mass defect, the same kind of mass defect as seen in heavy nuclei which is well understood. The electron is trapped in it's own negative electric potential and dispite having the same mass as the proton, appears to the observer as a smaller particle. It was obvious from these assumptions that the observers potential (Ground Potential) had to lie somewhere between the electrical potential of the proton and the potential of the electron, and the reason is simply that we know of no charge more positive than the proton and likewise no charge more negative than the electron, ergo everything else must lie in-between. This inspired me to go after the equation describing the relationship between electron-potential, ground-potential and proton-potential, it was tricky, but the solution presented itself to me within a couple of days of working on it. \[ a (\gamma) = \frac{1}{2} (c-b) \] Where 'a' is electon-potential, 'b' is ground-potential and 'c' is proton-potential. The problem here is $\gamma$ being the same kind of $\gamma$ as in Einstein's relativity, being velocity dependent, so how does one define gamma? After some further thinking it became apparent that four-velocity was a function of potential and since the proton represented the absolute maximum potential, it could be concidered as a physical constant in the same way as the speed of light, so the full equation now became; \[ a = \frac{(c-b)}{2} \sqrt{1-\frac{b^2}{c^2}} \] So here we have the first law of ground potential, a very simple and elegant statement for the first time defining ground potential. Solving the above equation with known values for electron and proton potential gives a ground potential of 930 million volts. I have chosen volts as the ideal unit to work in, because it is the SI unit which moves one elementary charge 1 meter in 1 second, so it is the perfect unit to define the potentials of electrons and protons. A proton which has a mass of $\frac{938 MeV}{c^2}$ contains 938 MeV of energy which divided by one elementary charge gives a potential of 938 million volts. Likewise the electron can be said to have 0.511 million volts potential. This first law of GP shall be the law upon which all other laws shall rest. Steven
Low Risk 1. INTRODUCTION Low-risk investing is a portfolio strategy, where buying low-risk and selling high-risk stocks have historically delivered superior risk-adjusted returns. More precisely, the reported alphas in several publications vs. the FF-3 factor model are impressive . Some even argue that the low-risk anomaly is the key challenge of the efficient market hypothesis or to cite it directly from Ang (2014) [1]: The risk anomaly is that risk - measured by market beta or volatility - is negatively related to returns. Robin Greenwood ... said in 2010, ‘We keep regurgitating the data to find yet one more variation of the size, value, or momentum anomaly, when the Mother of all inefficiencies may be standing right in front of us - the risk anomaly’ . One of the first studies showing a negative relation between risk and return is Haugen and Heins (1975) [2] and Jensen, Black and Scholes (1972) [3]. In more recent years, Ang et al. (2006) [4] show for US data and Blitz and van Vliet (2007) [5] for global data that the anomaly still exists. 2.OVERVIEW OF DIFFERENT VERSIONS Generally, several versions of low-risk investing are circulating in the finance literature and are used in investment management practice: low volatility, low beta, low idiosyncratic volatility and minimum variance. Low Volatility: The volatility is simply obtained by using the standard deviation of the return series of each stock over a defined time window. Low Idiosyncratic Volatility: In order to obtain the idiosyncratic risk of a stock, we need to correct the return for systematic risks. This can be achieved for stock $i$ by running a time series regression of the following form: $r_{i,t}= \beta_{i,Mkt}r_{mkt,t}+ \beta_{i,smb}r_{smb,t}+ \beta_{i,hml}r_{hml,t}+ \beta_{i,umd}r_{umd,t}+ \epsilon_{i,t}\tag{1}$ and then calculating the standard deviation of the residual from regression above, where both the time window of the regression and the residual needs to be defined. Alternatively, one could use simply a univariate regression with only the market as independent variable. Low beta: Calculate each stock’s beta at each rebalance date. You could simply use the same regression as above and use the market beta $\beta_{Mkt}$ as a proxy or you follow simply the method used in Frazzini and Pedersen (2014) [6], where the authors make use of the beta decomposition: $\beta_{i,Mkt}=\rho_{i,t}\frac{\sigma_{i,t}}{\sigma_{mkt,t}}$ and measure the correlation($\rho_{i,t}$) over a longer period than the $\sigma$’s as they argue volatilities are in general not as persistent as correlations and hence, obtain a more forward looking measure to reduce errors in the measurement of the $\beta$s. Relation of Low vola, low beta and low idiosyncratic volatility: The three concepts are related in the following way: $\underbrace{\sigma_{i,t}}_{Low Vola} = \sqrt{\underbrace{\beta_i^2}_{(\text{Low Beta})^2}\sigma_{mkt,t}^2 + \underbrace{\sigma_{\epsilon}^2}_{(\text{IVol})^2}}\tag{2}$ Minimum variance portfolio: Instead of investing in the lowest decile portfolio simply by the previously described criterion, one might directly construct a minimum variance portfolio, by using the full covariance matrix of all available stocks. The covariance could be estimated by a simple $\beta$ - factorization of returns. However, Blitz and van Vliet (2007) [5] find that by simply using the diagonal elements of the covariance (e.g. simple low volatility) one can achieve greater improvements than by using the full covariance structure. 3. PRACTICAL IMPLEMENTATION The practical implementation of the strategy strongly depends on what type of investor you are. The challenge for a pure equity, long only investor, who is benchmarked against the market is to reduce tracking-error or almost equivalently to increase beta, as typically the low-decile portfolio has a $\beta$ around 0.65 and 0.75. If you are a pension fund or multi-strategy manager with a sufficiently low equity portion in your benchmark life is a little easier. Let’s assume your benchmark is 60% cash and 40% equity, and furthermore, the $\beta$ of your low volatility portfolio is 0.7, you could simply hold a fraction of 0.4 *1/0.7 of the low volatility portfolio in your portfolio (if your constraints allow you to) to obtain a fairly beta neutral position vs. your benchmark. Naturally, the question arises what time window to use to obtain the criteria described above. In general, we recommend to use rather longer time windows to estimate the FF $\beta$’s - a good starting point is three years of weekly (or even daily) data. The time window for the residuals can then still be chosen rather short, say 1-month (of daily) data, as suggested in Ang et al. (2006) [4], or probably more appropriate, from a practical perspective, is a three year window (daily or weekly data) as used in Blitz and van Vliet (2007) [5] to reduce turnover. Additionally, the significant alphas of the long-short portfolio reported in Ang et al. (2006) [4] are driven more by the short position in high volatility stocks than in the long of the low volatility stocks. Blitz and van Vliet (2007) [5] on the other hand observe a more symmetric effects with using three year estimates. Combing this information, we have another argument for choosing a longer time window measuring the volatility. We note however, that the two studies use different set-ups - Blitz and van Vliet (2007) [5] use simple return volatilities, Ang et al. (2006) [4] – the FF-3 factor corrected idiosyncratic volatilities. Some critique has been articulated that low-risk investing is partly a result of sector concentration and that the strategy profits from an implied value bias. Asness, Frazzini and Pedersen (2013a) [7] demonstrate that low-risk investing even does the work best when its implemented in a sectorneutral fashion. The authors conclude that neither the sector, nor value is the driving force of the strong performance of low-risk strategies. If your constraints do not allow to utilize the anomaly, an alternative implementation of the low-risk phenomena could be as proposed in Asness, Frazzini and Pedersen (2013b) [8], where low volatility and low beta contribute to the safety dimension of quality (see the quality section for more details). 4. (POSSIBLE) THEORETICAL EXPLANATIONS What are possible explanations of the anomaly? Blitz and van Vliet (2007) [5] mention that leverage constraints and/or tracking error aversion of investors are the potential reasons. Alternatively, the authors provide a behavioural argument, where high risk stocks are lottery tickets for retail investors and hence, high in demand and consequently inflate prices (deflate returns) of these stocks. Frazzini and Pedersen (2014) [6] present a theoretical model, where investors are constrained in their investment opportunities, and how this can lead to high $\alpha$s for low $\beta$-stocks. 5. SUMMARY Low-risk investing can add value to your portfolio, however, it crucially depends on what type of investor you are and to what extent you can actually exploit the strategy. Exposure to the factor can be obtained by building your own low-risk portfolio (as described above) or alternatively, by simply investing into low-risk ETFs (typically called “Index XYZ low volatility ETF” or “ Index XYZ minimum variance ETF”). References Asset Management: A Systematic Approach to Factor Investing, Ang, Andrew , (2014) Risk and the rate of return on financial assets: Some old wine in new bottles, Haugen, Robert A., and A Heins James , Journal of Financial and Quantitative Analysis, Volume 10, Number 05, p.775–784, (1975) The capital asset pricing model: Some empirical tests, Jensen, Michael C., Black Fischer, and Scholes Myron S. , (1972) The cross-section of volatility and expected returns, Ang, Andrew, Hodrick Robert J., Xing Yuhang, and Zhang Xiaoyan , The Journal of Finance, Volume 61, Number 1, p.259–299, (2006) The Volatility Effect, Blitz, David C., and van Vliet Pim , The Journal of Portfolio Management, Volume 34(1), p.102-113, (2007) Betting against beta, Frazzini, Andrea, and Pedersen Lasse Heje , Journal of Financial Economics, Volume 111, Number 1, p.1–25, (2014) Low-Risk Investing Without Industry Bets, Asness, Cliff, Frazzini Andrea, and Pedersen Lasse H. , Working Paper, (2013) Quality Minus Junk, Asness, Clifford S., Frazzini Andrea, and Pedersen Lasse H. , Available at SSRN 2312432, (2013)
Recently, I read the definition of oxidation state on Wikipedia. It read that a 100% ionic bond is impossible. So what does a 75% ionic and 25% covalent bond mean at all? A "100% ionic" bond would be a bond whose bonding electron(s) were never in the vicinity of the cation, but rather always in unperturbed valence orbitals of the anion. That's not far from the truth in a paradigmatic ionic compound like $\ce{NaCl}$, but no matter how electronegative the anion is, the bonding electrons will still experience some attraction toward the positive charge on the cation, and so a detailed model of their wave functions will show a small but nonzero amplitude near the cation. That's just basic electrostatics; you have a positive charge and a negative charge not that far away from each other, they will attract. Despite that, we can still talk about 100% ionic bonds as a theoretical limit and an acceptable approximation for modeling some compounds. It's kind of like how we can use absolute zero as the theoretical zero point of a temperature scale even though nothing in the universe can ever be that cold. In a "100 %" covalent bond, like the $\sigma_{ss}$ bond in the dihydrogen molecule, the electron probability density is perfectly symmetrically divided between the two bonded nuclei because both have the same electronegativity. Below: the schematised electron density $\psi^2$, for a 100 % covalent bond: But when the two atoms have different electronegativities, the electron probability density will be higher towards the more electronegative atom, see e.g. $\ce{HCl}$. We say the bond is polarised. See below, schematised, the permanent dipole of $\ce{XY}$, with partial charges $\delta +$ and $\delta -$: Below: the schematised electron density $\psi^2$, for a bond between atoms with differing electronegativity (highest electronegativity to the right): If the electronegativity difference is really high, see e.g. $\ce{NaCl}$, the bond starts being so strongly polarised (electron probability density strongly skewed to towards the $\ce{Cl}$ atom) that the bond starts taking on an ionic character. Ionic and covalent must be considered relative to each other: in a "75% ionic" bond the electron distribution is strongly skewed toward the electronegative element. There is however no meaningful way to measure this "percentage". Actually bonds that are $100\%$ ionic, or close to it, are possible. But they require a little ingenuity. Instead of combining atoms of very different electronegativity, we can build multiatomic ions in which the charge is (formally) buried or highly de-localized making covalent bonding unfavorable. We use molecular orbital structures rather than electronegativity to drive the charge separation. One example of this is the cyclopropenyl salt $\ce{C_3H_3^+SbCl_6^-}$ produced by Breslow and Groves in 1970 (http://pubs.acs.org/doi/abs/10.1021/ja00707a040). To form a covalent bond between the ions would require either occupying a highly destabilized antibonding orbital in the cyclopropenyl ring, or an interaction between the ring and poorly overlapping orbitals in the bulky anion. So we have very little covalent bonding; the molecular orbital structure does not favor it. We could get covalent compounds by transferring a chloride ion, but this reaction is energetically unfavorable. The salt, with its ionic bonding between the aromatic cation and complex anion, is surprisingly stable for a species with a three-membered carbon ring. Further stabilization may be achieved and ionicity reinforced by placing substituent on the cyclopropenyl ring. Another example is provided by compounds in which alkali metals are disproportionated into cations and anion (https://en.m.wikipedia.org/wiki/Alkalide). Because the ionization energy of alkali metals (at least from sodium on down) are so small, the cation may be stabilized by forming a complex where the (formal) positive charge is deeply buried and again, no obvious means exists for covalent bonding. The alkalide anion must be bonded by essentially ionic attraction to the cationic complex.
Linear Classifiers are one of the most commonly used classifiers and Logistic Regression is one of the most commonly used linear classifiers. The concepts we are going to learn here will actually extend a lot of other classification methods beyond linear classifiers. We’re going to learn the fundamental concepts, but also the underlying algorithms that let us optimize the parameters of this model to fit your training data. Motivation Suppose, we want to make a spam filter which will filter spams from our email list automatically. So, for every email, we’re going to feed it to a classifier and the classifier will say if this is a spam or maybe we want to visit a restaurant for a specific food e.g. sushi but before visiting the restaurant we want to make sure that the sushi there is good. So, what we want to do is to check the review of others visiting this restaurant about sushi and see if it is positive. It is actually very difficult to manually check all the reviews. Instead, what we can do is to feed these reviews to a classifier and it is going to say, is this a positive sentiment or negative sentiment and then watching the number of positive sentiment and the negative sentiment we will decide whether the sushi there is good to go or not. Linear Classifiers Intuition A linear classifier will take some quantity x as input which in our case will be emails or reviews and is going to make a prediction $\hat{y}$ that says is this a positive statement which means is this a no spam or a positive review in which case $\hat{y} = +1$ or is this a negative statement which implies it is a spam or a negative review in which case $\hat{y} = -1$ . We will use restaurant review classifier for the rest fo the tutorial as our use case. A linear classifier does a little bit more associating every word for weight or coefficient which says how positively or negatively influential this word is for a review. For example, Word Coefficient good 1.0 delicious 1.5 wonderful 2.7 bad -1.0 terrific -2.1 awful -3.3 but, food, the, we, to, have, … 0.0 … … Here good and delicious have a coefficient of $1.0$ and $1.5$ respectively. Wonderful is very positive and has a coefficient of $2.7$. On the negative side, bad and terrific might have a coefficient of $-1$ and $-2.1$ respectively. But awful is just awful, so $-3.3$. There are also some words that are not that relevant to the sentiment of the review, might have $0$ coefficient. Now how will we use these coefficient’s to make a prediction of whether a sentence is positive or negative? Let’s take a review, for example, $$\begin{aligned} &\text{Input } \textbf{x}_i: \\ &\text{ Sushi was} \textbf{ good} \text{, the food was } \textbf{delicious}\text{, but the service was } \textbf{terribfic.}\end{aligned} \\ \text{Score}(\textbf{x}_i) = 1.0 + 1.7 – 2.1 = 0. 6$$ here good, delicious and great have coefficients greater than $0$ and rest of the terms having coefficients greater than $0$ as they are not relevant. After calculating we get a positive score which implies the sentiment in the sentence is positive and $\hat{y} = +1$. This is called a linear classifier because the output is the weighted sum of the inputs. So more generally for a simple linear classifier, we are going to take a review and the coefficient associated with each word and feed that in our classifier and calculate the score for that input. If the score is greater than $0$ the prediction is positive and $\hat{y} = +1$ and if the score is less than $0$ we say the prediction is $\hat{y} = -1$. Now, what we need to do is to train the coefficients or weights of these linear classifiers from data. So given some input training data that includes sentences of reviews labeled with either $+1$ or $-1$. We’re going to split those into some training set and some validation set. Then we’re going to feed that training set to some learning algorithm which is going to learn the weights associated with each word. And then after we learn this classifier, we’re going back and evaluate its accuracy on that validation set. Now, how do we learn this classifier from data? Decision Boundary The decision boundary is a boundary between positive and negative predictions. Let’s say that we have taken our data and trained our linear classifier and every word has zero weight except for two of them. Awesome has weight 1.0 and awful has weight -1.5. That means that the score of any sentence is 1.0 times the number of the word awesome minus 1.5 times the number of times the word awful shows up. We can plot that into a graph which depends on every sentence the number of awesome and the number of awful. So for example, for a sentence, $$\begin{aligned} \text{ Sushi was} \textbf{ awesome} \text{, the food was } \textbf{awesome}\text{, but the service was } \textbf{awful.}\end{aligned} $$ We’re going to plot that into a space where we’re going to have two awesome and one awful. So it gets plotted in the $(2,1)$ point. And then for every sentence that we might have in our training data set say, three awful and one awesome, three awesome and no awful and so on we will have different points and if we plot the dataset we find The classifier that we’ve trained with the coefficients 1.0 and -1.5 will have a decision boundary that corresponds to a line plotted above, where 1.0 times awesome minus 1.5 times the number of awful is equal to zero. Everything below that line has a score greater than zero and any points above that line are going to have scored less than zero. So there’s that line, everything below the line is positive, everything above the line’s negative a linear decision boundary. That is what makes it a linear classifier. Linear Classifiers Model in our previous example that we had with just two features with no zero coefficients, awesome and awful and we have calculated the score as $$\text{Score(x)} = w_0 + w_1 \text{ #awesome} + w_2 \text{ #awful}$$ where $w_0$ was $0$ in our example. Now suppose that we had a third feature with no zero coefficient. Let’s say that the word great. So the score function then will be $$ \text{Score(x)} = w_0 + w_1 \text{ #awesome} + w_2 \text{ #awful} + w_3 \text{ #great}$$ So, the general form of our linear classifier model will be : $$\begin{aligned} &\text{Model: } \hat{y}_i = \text{sign(Score(}\textbf{x}_i)) \\ &\text{Score(}\textbf{x}_i) = w_0 + w_1 \textbf{x}_i[1]+ … + w_d \textbf{ x}_i[d] \end{aligned}$$ where $$\begin{aligned} feature \; 1 &= 1 \\ feature \; 2 &= \textbf{x}[1] \dots e.g. , \#awesome \\ feature \; 3 &= \textbf{x}[2] \dots e.g., \#awful \\ \vdots \\ feature \; d+1 &= \textbf{x}[d]\ \dots e.g., \#ramen \end{aligned}$$ Theses notations that we’ve used so far are without features associated with it. So we’re going to have these functions $h_1$ through $h_D$. They’ve defined some features we might extract from the data and we are going to encode the constant function that is $h_0$. So the more generically our model will be looked like: $$\begin{aligned} \text{Model: } \hat{y}_i &= \text{sign(Score(}\textbf{x}_i)) \\ \text{Score(}\textbf{x}_i) &= w_0 h_0\textbf{x}_i + w_1 h_1\textbf{x}_i+ … + w_D h_D\textbf{x}_i \\ &= \textbf{w}^{T}h(\textbf{x}_i) \end{aligned}$$ where $$\begin{aligned} feature \; 1 &= h_0(\textbf{x}) \dots e.g., 1 \\ feature \; 2 &= h_1(\textbf{x}) \dots e.g. , \textbf{x}[1] = \#awesome \\ feature \; 3 &= h_2(\textbf{x}) \dots \textbf{x}[2] = \#awful \; or, \log(\textbf{x}[7]) \textbf{x}[2] = \log(\#bad) * \#awful \\ \vdots \\ feature \; D+1 &= h_D( \textbf{x}) \dots \text{some other function of} \; \textbf{x}[1], \dots, \textbf{x}[d]\end{aligned}$$ So this is our generic linear classifier model with multiple features. Linear classifiers are a pretty abstract concept. Logistic regression is a specific case of that, where we use what’s called the logistic function to squeeze minus infinity to plus infinity into the interval $(0, 1)$ so we can predict probabilities for every class. Next, we will discuss Logistic Regression. Also published on Medium.
8.2.3.1 - One Sample Mean t Test, Formulas Five Step Hypothesis Testing Procedure Data must be quantitative. In order to use the t distribution to approximate the sampling distribution either the sample size must be large ($\ge\ 30$) or the population must be known to be normally distributed. The possible combinations of null and alternative hypotheses are: Research Question Is the mean different from $ \mu_{0} $? Is the mean greater than $\mu_{0}$? Is the mean less than $\mu_{0}$? Null Hypothesis, $H_{0}$ $\mu=\mu_{0} $ $\mu=\mu_{0} $ $\mu=\mu_{0} $ Alternative Hypothesis, $H_{a}$ $\mu\neq \mu_{0} $ $\mu> \mu_{0} $ $\mu<\mu_{0} $ Type of Hypothesis Test Two-tailed, non-directional Right-tailed, directional Left-tailed, directional where $ \mu_{0} $ is the hypothesized population mean. For the test of one group mean we will be using a $t$ test statistic: Test Statistic: One Group Mean $t=\frac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}$ $\overline{x}$ = sample mean $\mu_{0}$ = hypothesized population mean $s$ = sample standard deviation $n$ = sample size Note that structure of this formula is similar to the general formula for a test statistic: $\frac{sample\;statistic-null\;value}{standard\;error}$ When testing hypotheses about a mean or mean difference, a $t$ distribution is used to find the $p$-value. These $t$ distributions are indexed by a quantity called degrees of freedom, calculated as $df = n – 1$ for the situation involving a test of one mean or test of mean difference. The $p$-value can be found using Minitab Express. If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis. Based on your decision in Step 4, write a conclusion in terms of the original research question. The new few pages will walk you through examples before giving you the opportunity to do two on your own.
In musing on how a family tree is kind of like Pachinko, I wondered if this could be animated somewhat like a Pachinko machine in the family tree visualization I have been playing with (available at https://learnforeverlearn.com/ancestors). I took some first steps towards this by pulling out the Bezier curve parameters d3 is generating for the links between nodes, coupled with Particle in Cell's notes on computing the intersection between Bezier curves and a line , to make some first steps towards a Pachinko-like animation effect. Some details on this implementation are at the bottom of this note. A few youtube videos demonstrating this are below. You can also go directly to the interactive visualization with the tree for Charles II of Spain automatically loaded via this link, and clicking the "Do Pachinko" button. (direct link to this tree is at https://learnforeverlearn.com/ancestors/?file=royal92.txt While I still need to play with the timing and smoothness, it is nevertheless interesting to watch the animations play over and over, reflecting on the huge number of things that had to happen (or not happen) through the centuries to result in me being here today. Shown below is an example of a curve displayed on the screen and the svg path generated by d3: $${\bf{r}}(t) = {(1-t)}^3 {\bf P_0} + 3{(1-t)}^2 t {\bf P_1} + 3{(1-t)} t^2 {\bf P_2} + t^3 {\bf P_3}, \ t \in [0,1]$$ Placing a circle on this curve at a given year $\hat{y}$ (i.e., height) requires finding the $\hat{t}$ so that $$\begin{eqnarray} \hat{y} & = & \text{y-coordinate of } {\bf{r}}(\hat{t}) \\ & = & \text{y-coordinate of } {(1-\hat{t})}^3 {\bf P_0} + 3{(1- \hat{t})}^2 \hat{t} {\bf P_1} + 3{(1-\hat{t})} \hat{t}^2 {\bf P_2} + \hat{t}^3 {\bf P_3} \end{eqnarray} $$ There are optimizations I need to address with respect to quickly finding the paths intersecting a given point in time, but it seems to work ok on small to moderately-sized trees.
In a quasistatic process (reversible), the difference between the external pressure and the internal pressure is infinitesimal for each infinitesimal change in the volume of the system. $P_{ext}=P_{int} \pm dP $, As you know that the term for work in terms of pressure is given by, $dW=PdV$. This means that work obtained will be maximum if the pressure is maximum for each infinitesimal change in volume. In a reversible process, the work done in each step obtained is maximum since the external pressure in only infinitesimally greater (or smaller) than the internal pressure. This enables us to connect the internal pressure and external pressure (since they are almost equal) using the ideal gas law, which in turn enables us to derive work in terms of volume change, without knowing the pressure. $P_{int}V=nRT$ $W_{int}=\int_{v_1}^{v_2} P_{int}dV$ $W_{int}=nRT\int_{v_1}^{v_2} \frac{dV}{V}$ For non-reversible process, such thing is not possible. This process is instantaneous. Internal pressure won't have enough time to become almost equal to external pressure. Only, external pressure is a way to find the work. The work done by internal pressure and external pressure are equal in magnitude. For some processes, external pressure is given. We can either simply calculate the work done by using external pressure and volume change or use to ideal gas equation to remove the pressure term. But the latter is won't be possible, since you do not know how the internal pressure is changing. In case of reversible processes, you had the ideal gas equation to give you a relation between pressure and volume. Hence, we use the constant external pressure (when you put a weight on the piston you get constant external pressure) to calculate the work. Therefore, for non-reversible processes, $W=P_{ext}(V_f-V_i)$ Moreover, work done by a particular pressure is not $\Delta P dV$. For instance, when there are two forces acting on a body the work done by a particular force is not given by the the work done by the net of the two forces but instead the work done by a particular force is given by that force's magnitude and the displacement of the body in that force's direction.
I have a confession: despite knowing a reasonably large amount about formal language theory and automata, I’ve never bothered to learn how to prove how to turn a finite automaton back into a regular expression. My attitude has always been that I have a vague sense of how a proof would go and couldn’t be bothered to sort out the details. I’ve probably skimmed a proof in the past but it looked complicated and uninformative so I didn’t bother. Due to reasons (a certain theorem about how you could represent regular expressions in a particular way required the details of a proof) I found I was forced to care recently, so I decided to actually sit down and figure it out. The standard proofs did indeed seem complicated and uninformative, but fortunately it turns out that the general method of understanding regular language theory fundamentals applies in this case. That method, of course, being to ask “What Would Janusz Brzozowski Do?”. Starting from a Nondeterministic Finite Automaton (without $\epsilon$-transitions) there turns out to be a natural algebraic process which is fairly reminiscent of Gaussian Elimination that results in a fairly small regular expression matching the language. The idea is this: Suppose we have a non-deterministic automaton without $\epsilon$-transitions. Label the states $0, \ldots, n$. We consider the languages $R_i$ of strings matched when starting from state $i$. By looking at the automaton we get a set of algebraic equations for each of these, which we can progressively rewrite and simplify to remove loops in them (e.g. where $R_i$ is defined self-referentially, or the definitions of $R_i$ and $R_j$ depend on each other). Once we’ve successfully done that, we can just substitute in terms until we get an expression for $R_0$, which will be our final result. Note: In what follows we’ll adopt the usual convention of equating regular expressions with their regular language. Variables which correspond to a language will be upper case, ones corresponding to an expression will be lower-case. Let $s_{ij}$ be a regular expression matching the language of strings that cause a tradition from state $i$ to $j$. This is a language consisting of strings of length $1$ – if the character $c$ causes a transition from $i$ to $j$ then $c \in s_{ij}$. If there are no transitions from $i$ to $j$ then $s_{ij}$ is $\phi$, the regular expression matching no strings, otherwise it is a union of finitely many basic regular expressions. Note that in the general case there may be strings matched by both $s_{ij}$ and $s_{ik}$ with $j \neq k$. This is because the automaton is non-deterministic. In a deterministic automaton the expressions $s_{ij}, s_{ik}$ with $i \neq k$ will have no common matches. Let $t_i$ be $\phi$ if state $i$ is not accepting and $\epsilon$ if it is. i.e. $t_i$ is a regular expression matching the set of empty strings matched from state $i$. Now, we have that $R_i = t_i \cup \bigcup\limits_j s_{ij} R_j$ – a string in $R_i$ is either the empty string (if $i$ is accepting), or the result of transitioning to another state and matching from there. We’ll now begin the process of substituting in terms to remove cycles. The key idea is to find $t_i’, s_{ij}’$ such that whenever $j \leq i$, $s_{ij}’ = \phi$. This means there can be no cycles because there are no backwards or self-references between the expressions. It will no longer be the case that $s_{ij}’$ only matches strings of length $1$, but it will be the case that they don’t match the empty string (this is important). How do we construct these modified terms? By induction! Suppose we’ve constructed $t_i$ and $s_{ij}$ for $i < k$. If we just substitute in and rearrange all of those terms into $R_k = t_k \cup \bigcup\limits_j s_{kj} R_j$ then we almost get what we want: The only terms that can be non-empty are the ones where $j \geq k$, because we’ve eliminated all the terms that were $< k$ and replaced them with terms that are $\geq k$. So all we need to do is figure out how to remove the $R_k$ term. Fortunately, there’s a great tool for doing that. it’s called Arden’s theorem: If we have languages $X, A, B$ with $\epsilon \not\in A$ and$X = AX \cup B$, then $X = A* B$ . Conceptually this is fairly intuitive because you can just rerwite it as $X = A(AX \cup B) \cup B$ and continue this infinitely to get $X = B \cup AB \cup AAB \cup \ldots$, which is precisely $X = AB$. However this intuition breaks down somewhat – the condition that $A$ does not contain the empty string is essential, because if we picked $A = \{\epsilon\}$ and $B = \emptyset\} then this reduces to \(X = X$, which every language satisfies. The problem is essentially that the term in the $\ldots$ “does not converge” in this case, so we need to do some more work to actually prove this rigorously. Proof: First note that $AB$ is a possible solution for $X$ because $A* = (A A) \cup \epsilon$, so $A B = (A A* B) \cup \epsilon B = A (A* B) \cup B$. Now note that any language satisfying the equation has the following property: $x \in X$ if and only if either $x \in B$ or $x = uv$ with $u \in A$ and $v \in X$. This is simply by the definition: If $x \not\in B$ then $x \in AX$, so $x$ starts with a string of $A$, which is non-empty by assumption. Conversely, if $x \in B$ then $x \in X$ by definition, and if $x = uv$ with $u \in A$ and $v \in L$. Now, suppose we have two non-identical solutions to the equation. Say $L$ and $M$. Suppose there exists $x \in L \setminus M$. Pick $x$ so that it has minimal length among such words.. Then certainly $x \not\in B$ or it would be in both, so by assumption we can write $x = uv$ with $u \in A$ and $v \in L$. But then we must have $v \in M$, by the minimality of $x$. But if $v \in M$ then necessarily $uv = x \in M$. This contradicts the assumption. Therefore no such $x$ exists, and the two languages are equal. QED So it’s important that all the coefficients on the right hand side don’t match the empty string, but this is again true inductively: The initial $s_{ij}$ do not match the empty string, and every coefficient is either a concatenation or a union of two languages that don’t match the empty string, so in turn does not match the empty string. This means that Arden’s theorem gives us the tool we need to remove the coefficient for $R_k$ on the right hand: All we need to do is to prepend the star of that coefficient to the other terms and remove $R_k$ from the equation. This completes our inductive step, which in turn completes our algorithm: We run the induction over the whole set of equations, we now no longer have loops, and we can just substitute back in to get $R_0$ as desired. Let’s work through an example now to clear up the details. Suppose we have the states 0, 1, 2. Only state 2 is accepting. Our characters are a, b and c. The allowed transitions are: From 0: $a, b \to 1$. From 1: $a \to 1, b \to 2$ From 2: $c \to 0, 1$. This gives us the following three initial equations: $R_0 = (a \| b) R_1$ $R_1 =a R_1 \cup b R_2$ $R_2 = \epsilon \cup c R_0 \cup c R_1$ We now perform substitutions as follows: Our equation for $R_0$ is in the desired form already so nothing needs to be done. Our equation for $R_1$ has $R_1$ on the right hand side, so we use Arden’s theorem to rewrite it as $R_1 = a* b R_2$. It is now in the desired form. We substitute in our previous equation for $R_0$ and get $R_2 = \epsilon \cup c (a\|b) R_1 \cup c R_1 = \epsilon \cup c(a\|b\|\epsilon) R_1$. We substitute in our previous equation for $R_1$ and get $R_2 = \epsilon \cup c(a\|b\|\epsilon) a* b R_2 $. We apply Arden’s theorem one final time and get $R_2 = (c(a\|b\|\epsilon) a* b)* \| \epsilon = (c(a\|b\|\epsilon) a* b)$. We now substitute this into our equation for $R_1$ and get $R_1 = a b (c(a\|b\|\epsilon) a* b)$ We now substitute this into our equation for $R_0$ and get $R_0 = (a \| b) a b (c(a\|b\|\epsilon) a* b)*$ We now have a (not all that pleasant) regular expression matching our original deterministic finite automaton, as desired. In general I’m unclear on whether this is the “best” method for doing this. The above explanation is a lightly modified version of the one presented here, which compares it with several other methods, where it seems to come out ahead of the others. It certainly isn’t optimal, in the sense that it doesn’t always produce a minimal regular expression. However, neither do any of the other standard methods. This seems to be a hard problem, and there aren’t as far as I know any good algorithms for doing it. However, regardless of whether it’s the best, it is certainly the one that seems clearest to me: Once you have the key mechanism of Arden’s theorem, you just perform simple algebraic manipulation and the result pops out at the end. If you want to see this working in practice, I’ve rather arbitrarily added in some code for doing this (albeit only with deterministic finite automata. The principle is the same though) to my FALBS project, which has basically become a dumping ground for all my formal languages code, along with a test that this does the right thing. (Want more posts like this? Why not support my Patreon! It will cause me to write more, give you some influence over what I write, and give you access to early drafts of upcoming posts).
I need to evaluate the following integral \begin{equation} \int_{-1}^1 \frac{d^4P_l(x)}{dx^4}P_n(x)dx\end{equation}. Of course the answer I need is in terms of $l$ and $n$. Does anyone have any idea how to proceed? The facts below should allow you to compute the integral in question in terms of $l$ and $n$: The Legendre polynomials $P_0, \dots, P_n$ form a basis for the space of polynomials of degree at most $n$. The Legendre polynomials are orthogonal: $$ \int_{-1}^{1} P_m(x) P_n(x)\,dx = {2 \over {2n + 1}} \delta_{mn} $$ $\dfrac{d^4P_l(x)}{dx^4}$ is a polynomial of degree $l-4$ if $l\ge 4$ or the zero polynomial otherwise. Let us denote $$\mathcal{I}_{l,n}:=\int_{-1}^1P_l^{(4)}(x)P_n(x)\,dx.$$ If $l\leq n+3$, then $P_l^{(4)}(x)$ is a polynomial of degree at most $n-1$ and $\mathcal{I}_{l,n}$ vanishes because of orthogonality of Legendre polynomials. It also vanishes when $n$ and $l$ have different parity. Hence in the following it will be assumed that $l\geq n+4$ and $l=n\;\operatorname{mod}\;2$. Integrating by parts, we can rewrite the integral as $$\int_{-1}^1P_l(x)P_n^{(4)}(x)\,dx+\text{boundary terms}.$$ The same argument as above shows that the integral in this expression vanishes for $l\ge n+5$, and therefore the answer is completely determined by boundary terms: \begin{align} \nonumber\mathcal{I_{l,n}}&=\left[\color{red}{P_l^{(3)}(x)P_n(x)}-\color{blue}{P_l^{(2)}(x)P_n^{(1)}(x)}+\color{green}{P_l^{(1)}(x)P_n^{(2)}(x)}-\color{magenta}{P_l(x)P_n^{(3)}(x)}\right]\biggl\|_{-1}^{\;1}=\\ \nonumber&=\color{red}{\frac{l(l+3)\left(l^2-1\right)\left(l^2-4\right)}{24}}- \color{blue}{\frac{l(l+2)\left(l^2-1\right)n(n+1)}{8}}+\\ \nonumber&\;+\color{green}{\frac{l(l+1)n(n+2)\left(n^2-1\right)}{8}}- \color{magenta}{\frac{n(n+3)\left(n^2-1\right)\left(n^2-4\right)}{24}}=\\ &=\frac{\left(l-n\right)\left(\left(l+n\right)^2-1\right) \left(\left(l-n\right)^2-4\right)(l+n+3)}{24}.\tag{$\spadesuit$} \end{align} Therefore the only remaining case to consider is $l=n+4$. Here the answer may be computed using Rodrigues formula. Replacing expressions for Legendre polynomials into the initial integral and integrating by parts, we get $$\mathcal{I}_{n+4,n}=2(2n+3)(2n+5)(2n+7).$$ But this is compatible with the previous result ($\spadesuit$), which may therefore be used for all $l\ge n+4$. Let me see if I can give a solution. The orthogonality relation for the Gegenbauer polynomials is $$ \int_{-1}^{1} P_m(x) P_n(x) dx = \frac{2}{2n+1} \delta_{m,n}. $$ The forward shift operator for the Jacobi polynomials, e.g. http://homepage.tudelft.nl/11r49/askey/ch1/par8/par8.html, is given by $$ \frac{d}{dx} P_m^{(\alpha,\beta)}(x) = \frac{n+\alpha+\beta+1}{2} P_{m-1}^{(\alpha+1,\beta+1)}(x). $$ Note that the Gegenbauer polynomials are Jacobi polynomial where $\alpha = \beta = 0$. The connection coefficients for the Jacobi polynomials are $$ P_n^{(\gamma,\delta)}(x) = \sum_{k=0}^n c_{n,k}(\gamma,\delta;\alpha,\beta) P_k^{(\alpha,\beta)}(x), $$ where \begin{equation} \begin{split} c_{n,k}(\gamma,\delta;\alpha,\beta) &= \frac{ (\gamma+k+1)_{n-k} (n+\gamma+\delta+1)_k }{ (n-k)! \Gamma(\alpha+\beta+2k+1) } \Gamma(\alpha+\beta+k+1) \\ &\times {}_3 F_2(-n+k, n+k+\gamma+\delta+1, \alpha+k+1; \gamma+k+1, \alpha+\beta+2k+2; 1). \end{split} \end{equation} Hence \begin{equation} \begin{split} \frac{d^4}{dx^4} P_{\ell}^{(0,0)} &= \frac{(\ell-2)_4}{2^4} P_{\ell-4}^{(4,4)}(x) = \frac{(\ell-2)_4}{2^4} \sum_{k=0}^{\ell-4} c_{\ell-4,k}(4,4;0,0) P_k^{(0,0)}(x). \end{split} \end{equation} By the orthogonality relation we have \begin{equation} \int_{-1}^{1} \frac{d^4}{dx^4} P_{\ell}(x) P_n(x) dx = \frac{(\ell-2)_4}{2^3 (2n+1)} c_{\ell-4,n}(4,4;0,0). \end{equation} Note that the connection coefficient can be found in Ismail, Classical and Quantum Orthogonal polynomials and that the connection coefficients probably can be simplified a lot. I hope I don't divide by zero somewhere so that you have to intrepid it formally. So the last part is homework. ;)
I have to solve a linear system of three equations and 3 unknowns and can't find a way to solve it. Applying Cramer's rule I obtain $\Delta = 0$, $\Delta_x \neq 0$, $\Delta_y \neq 0$, $\Delta_z \neq 0$ so that may exist a solution. How to deal with the system in such a situation? Just solve it directly by applying Gauss algorithm. Assume your equation system is of the form $\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{bmatrix} \begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}$ then you can solve it as follows: First make sure that $a_{11}\not=0$, otherwise you can change 2 rows / columns s.t. the new matrix has the first entry not equal to $0$. Then you subtract the first row multiplicated with $-\frac{a_{21}}{a_{11}}$ from the second. This makes, that the resulting system looks something like this $\begin{bmatrix}a_{11}&a_{12}&a_{13}\\0 & \tilde{a}_{22} & \tilde{a}_{23}\\ a_{31} & a_{32} & a_{33}\end{bmatrix} \begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix}b_1\\\tilde{b}_2\\b_3\end{bmatrix}$ NOTE that for example $\tilde{a}_{22} = a_{22} - \frac{a_{21}}{a_{11}}$ Now repeat: Add the first row multiplicated with $\frac{-a_{31}}{a_{11}}$ to the third row. This makes your system $\begin{bmatrix}a_{11}&a_{12}&a_{13}\\0 & \tilde{a}_{22} & \tilde{a}_{23}\\ 0 & \tilde{a}_{32} & \tilde{a}_{33}\end{bmatrix} \begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix}b_1\\\tilde{b}_2\\\tilde{b}_3\end{bmatrix}$ Note again, that the entries in the third row of the system have changed, so they have a tilde above them now. Now you need one more step. Make sure, that $\tilde{a}_{22}\not=0$ (if not, change the 2nd and the 3rd row or column in your system to make this the case). Now subtract the second row (with $\tilde{a}_{22}\not=0$) multiplicated with $\frac{-\tilde{a}_{33}}{\tilde{a}_{22}}$ from the third row. This makes your system look like $\begin{bmatrix}a_{11}&a_{12}&a_{13}\\ 0 & \tilde{a}_{22} & \tilde{a}_{23}\\ 0 & 0 & \bar{a}_{33}\end{bmatrix} \begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix}b_1\\ \tilde{b}_2\\\bar{b}_3\end{bmatrix}$ Note, that in the third row again something has changed. The modified elements have a bar above them. Now you can see that $ \bar{a}_{33} \cdot z = b_3$, so $z = \frac{\bar{b}_3}{\bar{a}_{33}}$. Inserting this value for $z$ in the second equation, you get the value of $y$. Afterwards insert the values of $z$ and $y$ into the first equation, to get a value for $x$. The vector $\begin{bmatrix}x\\y\\z\end{bmatrix}$ is then a solution to the system. REMARK: If at one point during calculation you get to the equation $0 = 0$, this means that you can chose a variable (e.g. z) as you wish. As you're normally interested in ALL solutions of the system, not just a particular one, you should call it $t$. This parameter $t$ is no variable anymore, so you can bring it to the right side of your system and calculate the values for the remaining variables with respect to this parameter $t$. If at one point during calculation you get a wrong equation, e.g. $1=0$, you either did some mistake or the system has no solution! Good luck :-) Here are hints: This kind of systems may have infinitely many solutions or no solution. Change system into row echelon form. Separate pivot variables and free variables. express pivot variables in terms of free variables.
I dont understand completely a proof of the dominated convergence theorem stated in page 104 of Analysis III of Amann and Escher. I will transcribe here the proof and comment about my thoughts. In the next: $(X,\mathcal A,\mu)$ is a $\sigma$-finite measure space, $E$ is a Banach space and $\mathcal L_1(X,\mu, E)$ is the space of Bochner $\mu$-integrable functions from $X$ to $E$, what is complete under the seminorm defined by $\\|f\\|:=\int_X\|f\|\,d\mu$. We denote by $\|{\cdot}\|$ the norm on $E$, and $\mathcal L_0(X,\mu,E)$ is the space of $\mu$-measurable functions in $E^X$. Let $(f_j)$ a sequence in $\mathcal L_1(X,\mu, E)$ and suppose that there exists $g\in\mathcal L_1(X,\mu,\Bbb R)$ such that a) $\|f_j\|\le g$ $\mu$-almost everywhere for all $j\in\Bbb N$ Suppose also that for some $f\in E^X$ b) $f_j\to f$ $\mu$-almost everywhere Then $f$ is $\mu$-integrable, $f_j\to f$ in $\mathcal L_1(X,\mu,E)$ and $\int_Xf_j\,d\mu\to\int_X f\,d\mu$ in $E$. Proof:define $g_j:=\sup_{k,\ell\ge j}\|f_k-f_\ell\|$, then $(g_j)\to 0$ $\mu$-a.e. in $\mathcal L_0(X,\mu,\overline{\Bbb R}^+)$, and $\|f_k-f_\ell\|\le 2g$ $\mu$-a.e. for all $k,\ell\in\Bbb N$. Hence $\|g_j\|\le 2g$ $\mu$-a.e. From a corollary of Fatou's lemma it follow that $$0\le\varlimsup_j\int_X g_j\,d\mu\le\int_X\varlimsup_j g_j\,d\mu=0$$ Therefore $(\int_Xg_j\,d\mu)$ is a decreasing null sequence. Then for every $\epsilon>0$ there is a $N\in\Bbb N$ such that $$\int_X\|f_k-f_\ell\|\,d\mu\le\int_X g_j\,d\mu<\epsilon$$ for $k,\ell\ge j\ge N$. Hence $(f_j)$ is a Cauchy sequence in $\mathcal L_1(X,\mu,E)$. Up to here all is right to me, but now is the punching line what I dont follow clearly: and the claim follow from the completeness of $\mathcal L_1(X,\mu,E)$ and Theorem 2.18. Theorem 2.18 what say is that if $(f_j)\to f$ in the seminorm then a) There is a subsequence $(f_{j_k})\to f$ $\mu$-a.e., and for each $\epsilon>0$ there is some $A\subset X$ with $\mu(A)<\epsilon$ such that $(f_{j_k})\to f$ uniformly in $A^\complement$. b) The integral $\int_X f_j\,d\mu$ converges to $\int_X f\,d\mu$. What I dont follow is how it follow from the completeness of the space. I mean: we knows that $(f_j)$ is Cauchy in the seminorm, so it converges to some value but, how we knows that it converges to $f$ in the seminorm? I guess that it is because $\|f-f_j\|\le\sup_{k,\ell\ge j}\|f_k-f_\ell\|$, but Im not sure. However this is not related to the completeness of the space of integrable functions. Can someone explain more clear the punching line of the proof? Thank you.
But if you don't want to have a Google account: Chrome is really good. Much faster than FF (I can't run FF on either of the laptops here) and more reliable (it restores your previous session if it crashes with 100% certainty). And Chrome has a Personal Blocklist extension which does what you want. : ) Of course you already have a Google account but Chrome is cool : ) Guys, I feel a little defeated in trying to understand infinitesimals. I'm sure you all think this is hilarious. But if I can't understand this, then I'm yet again stalled. How did you guys come to terms with them, later in your studies? do you know the history? Calculus was invented based on the notion of infinitesimals. There were serious logical difficulties found in it, and a new theory developed based on limits. In modern times using some quite deep ideas from logic a new rigorous theory of infinitesimals was created. @QED No. This is my question as best as I can put it: I understand that lim_{x->a} f(x) = f(a), but then to say that the gradient of the tangent curve is some value, is like saying that when x=a, then f(x) = f(a). The whole point of the limit, I thought, was to say, instead, that we don't know what f(a) is, but we can say that it approaches some value. I have problem with showing that the limit of the following function$$\frac{\sqrt{\frac{3 \pi}{2n}} -\int_0^{\sqrt 6}(1-\frac{x^2}{6}+\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$equal to $1$, with $n \to \infty$. @QED When I said, "So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0." I guess what I'm saying is that (f(x+h)-f(x))/h is not continuous since it's not defined at h=0. @KorganRivera There are lots of things wrong with that: dx=0 is wrong. dy/dx - what/s y? "dy/dx is by definition not continuous" it's not a function how can you ask whether or not it's continous, ... etc. In general this stuff with 'dy/dx' is supposed to help as some kind of memory aid, but since there's no rigorous mathematics behind it - all it's going to do is confuse people in fact there was a big controversy about it since using it in obvious ways suggested by the notation leads to wrong results @QED I'll work on trying to understand that the gradient of the tangent is the limit, rather than the gradient of the tangent approaches the limit. I'll read your proof. Thanks for your help. I think I just need some sleep. O_O @NikhilBellarykar Either way, don't highlight everyone and ask them to check out some link. If you have a specific user which you think can say something in particular feel free to highlight them; you may also address "to all", but don't highlight several people like that. @NikhilBellarykar No. I know what the link is. I have no idea why I am looking at it, what should I do about it, and frankly I have enough as it is. I use this chat to vent, not to exercise my better judgment. @QED So now it makes sense to me that the derivative is the limit. What I think I was doing in my head was saying to myself that g(x) isn't continuous at x=h so how can I evaluate g(h)? But that's not what's happening. The derivative is the limit, not g(h). @KorganRivera, in that case you'll need to be proving $\forall \varepsilon > 0,\,\,\,\, \exists \delta,\,\,\,\, \forall x,\,\,\,\, 0 < \|x - a\| < \delta \implies \|f(x) - L\| < \varepsilon.$ by picking some correct L (somehow) Hey guys, I have a short question a friend of mine asked me which I cannot answer because I have not learnt about measure theory (or whatever is needed to answer the question) yet. He asks what is wrong with \int_0^{2 \pi} \frac{d}{dn} e^{inx} dx when he applies Lesbegue's dominated convergence theorem, because apparently, if he first integrates and then derives, the result is 0 but if he first derives and then integrates it's not 0. Does anyone know?
For building a recommendation system, I also use the Pearson correlation coefficient. This is the definition: $r(x, y)=\frac{\sum_{i=1}^n (x_i-\bar{x})(y_i-\bar{y})}{\sqrt{\sum_{i=1}^n (x_i-\bar{x})^2 \cdot \sum_{i=1}^n (y_i-\bar{y})^2}}$ $x$ and $y$ are part of $\mathbb{R}$. Now for coding, it is important to take care of all potential outcomes. For example, if the denominator is zero, you will have to filter that or throw an exception. I came up with some arguments, one of them being that if all values of $x_i$ and/or $y_i$ were equal to the average of $x$ and/or $y$, then the denominator would be zero. But how can I prove that the coefficient is either undefined (zero denominator) or in between -1 and 1? What is the best approach?
Difference between revisions of "Linear representation theory of symmetric group:S5" (→Character table) (→Family contexts) (11 intermediate revisions by the same user not shown) Line 8: Line 8: ==Summary== ==Summary== + {\| class="sortable" border="1" {\| class="sortable" border="1" ! Item !! Value ! Item !! Value \|- \|- − \| [[Degrees of irreducible representations]] over a [[splitting field]] \|\| 1,1,4,4,5,5,6<br>[[maximum degree of irreducible representation\|maximum]]: 6, [[lcm of degrees of irreducible representations\|lcm]]: 60, [[number of irreducible representations equals number of conjugacy classes\|number]]: 7, [[sum of squares of degrees of irreducible representations equals order of group\|sum of squares]]: 120 + \| [[Degrees of irreducible representations]] over a [[splitting field]] \|\| 1,1,4,4,5,5,6<br>[[maximum degree of irreducible representation\|maximum]]: 6, [[lcm of degrees of irreducible representations\|lcm]]: 60, [[number of irreducible representations equals number of conjugacy classes\|number]]: 7, [[sum of squares of degrees of irreducible representations equals order of group\|sum of squares]]: 120 \|- \|- \| [[Schur index]] values of irreducible representations \|\| 1,1,1,1,1,1,1<br>[[maximum Schur index of irreducible representation\|maximum]]: 1, [[lcm of Schur indices of irreducible representations\|lcm]]: 1 \| [[Schur index]] values of irreducible representations \|\| 1,1,1,1,1,1,1<br>[[maximum Schur index of irreducible representation\|maximum]]: 1, [[lcm of Schur indices of irreducible representations\|lcm]]: 1 Line 23: Line 24: \| Smallest size [[splitting field]] \|\| [[field:F7]], i.e., the field of 7 elements. \| Smallest size [[splitting field]] \|\| [[field:F7]], i.e., the field of 7 elements. \|} \|} − + ==Family contexts== ==Family contexts== Line 29: Line 30: ! Family name !! Parameter values !! General discussion of linear representation theory of family ! Family name !! Parameter values !! General discussion of linear representation theory of family \|- \|- − \| [[symmetric group]] \|\| 5 \|\| [[linear representation theory of symmetric groups]] + \| [[symmetric group]] \|\| 5\|\| [[linear representation theory of symmetric groups]] \|- \|- − \| [[projective general linear group of degree two]] \|\| [[field:F5]] \|\| [[linear representation theory of projective general linear group of degree two]] + \| [[projective general linear group of degree two]] \|\| [[field:F5]]\|\| [[linear representation theory of projective general linear group of degree two ]] \|} \|} Line 81: Line 82: \| Unclear \|\| a nontrivial homomorphism <math>\varphi:\mathbb{F}_{q^2}^\ast \to \mathbb{C}^\ast</math>, with the property that <math>\varphi(x)^{q+1} = 1</math> for all <math>x</math>, and <math>\varphi</math> takes values other than <math>\pm 1</math>. Identify <math>\varphi</math> and <math>\varphi^q</math>. \|\| unclear \|\| <math>q - 1</math> \|\| 4 \|\| <math>(q-1)/2</math> \|\| 2 \|\| <math>(q-1)^3/2</math> \|\| 32 \|\| standard representation, product of standard and sign \| Unclear \|\| a nontrivial homomorphism <math>\varphi:\mathbb{F}_{q^2}^\ast \to \mathbb{C}^\ast</math>, with the property that <math>\varphi(x)^{q+1} = 1</math> for all <math>x</math>, and <math>\varphi</math> takes values other than <math>\pm 1</math>. Identify <math>\varphi</math> and <math>\varphi^q</math>. \|\| unclear \|\| <math>q - 1</math> \|\| 4 \|\| <math>(q-1)/2</math> \|\| 2 \|\| <math>(q-1)^3/2</math> \|\| 32 \|\| standard representation, product of standard and sign \|- \|- − + Total \|\| NA \|\| NA \|\| NA \|\| NA \|\| <math>q + 2</math> \|\| 7 \|\| <math>q^3 - q</math> \|\| 120 \|\| NA \|} \|} Line 87: Line 88: {{character table facts to check against}} {{character table facts to check against}} + + + + + + + + + + + + + + + + + + + + {\| class="sortable" border="1" {\| class="sortable" border="1" − ! Representation/conjugacy class representative and size !! <math>()</math> (size 1) !! <math>(1,2)</math> (size 10) !! <math>(1,2, + ! Representation/conjugacy class representative and size !! <math>()</math> (size 1) !! <math>(1,2)</math> (size 10) !! <math>(1,2,)</math> (size ) !! <math>(1,23)</math> (size ) !! <math>(1,2,3,)</math> (size ) !! <math>(1,2,34,5)</math> (size ) !! <math>(1,2,3,4)</math> (size ) \|- \|- − \| trivial representation \|\| 1 \|\| + \| trivial representation \|\| 1 \|\| \|\| \|\| \|\| \|\| \|\| \|- \|- − \| sign representation \|\| 1 \|\| - + \| sign representation \|\| 1 \|\| -\|\| \|\| \|\| -\|\| \|\| \|- \|- − \| standard representation \|\| + \| standard representation \|\| \|\| \|\| \|\| \|\| \|\| -\|\| \|- \|- − \| product of standard and sign representation \|\| + \| product of standard and sign representation \|\| \|\| -\|\| \|\| \|\| \|\| \|\| \|- \|- − \| irreducible five-dimensional representation \|\| + \| irreducible five-dimensional representation \|\| \|\| \|\| \|\| \|\| \|\| \|\| \|- \|- − \| irreducible five-dimensional representation \|\| + \| irreducible five-dimensional representation \|\| \|\| \|\| \|\| \|\| \|\| \|\| \|- \|- − \| exterior square of standard representation \|\| + \| exterior square of standard representation \|\| \|\| 0 \|\| \|\| \|\| 0 \|\| \|\| \|} \|} Latest revision as of 05:41, 16 January 2013 This article gives specific information, namely, linear representation theory, about a particular group, namely: symmetric group:S5. View linear representation theory of particular groups \| View other specific information about symmetric group:S5 This article describes the linear representation theory of symmetric group:S5, a group of order . We take this to be the group of permutations on the set . Summary Item Value Degrees of irreducible representations over a splitting field (such as or ) 1,1,4,4,5,5,6 maximum: 6, lcm: 60, number: 7, sum of squares: 120 Schur index values of irreducible representations 1,1,1,1,1,1,1 maximum: 1, lcm: 1 Smallest ring of realization for all irreducible representations (characteristic zero) -- ring of integers Smallest field of realization for all irreducible representations, i.e., smallest splitting field (characteristic zero) -- hence it is a rational representation group Criterion for a field to be a splitting field Any field of characteristic not equal to 2,3, or 5. Smallest size splitting field field:F7, i.e., the field of 7 elements. Family contexts Family name Parameter values General discussion of linear representation theory of family symmetric group of degree linear representation theory of symmetric groups projective general linear group of degree two over a finite field of size , i.e., field:F5, so the group is linear representation theory of projective general linear group of degree two over a finite field Degrees of irreducible representations FACTS TO CHECK AGAINST FOR DEGREES OF IRREDUCIBLE REPRESENTATIONS OVER SPLITTING FIELD: Divisibility facts: degree of irreducible representation divides group order \| degree of irreducible representation divides index of abelian normal subgroup Size bounds: order of inner automorphism group bounds square of degree of irreducible representation\| degree of irreducible representation is bounded by index of abelian subgroup\| maximum degree of irreducible representation of group is less than or equal to product of maximum degree of irreducible representation of subgroup and index of subgroup Cumulative facts: sum of squares of degrees of irreducible representations equals order of group \| number of irreducible representations equals number of conjugacy classes \| number of one-dimensional representations equals order of abelianization Note that the linear representation theory of the symmetric group of degree four works over any field of characteristic not equal to two or three, and the list of degrees is . Interpretation as symmetric group Common name of representation Degree Corresponding partition Young diagram Hook-length formula for degree Conjugate partition Representation for conjugate partition trivial representation 1 5 1 + 1 + 1 + 1 + 1 sign representation sign representation 1 1 + 1 + 1 + 1 + 1 5 trivial representation standard representation 4 4 + 1 2 + 1 + 1 + 1 product of standard and sign representation product of standard and sign representation 4 2 + 1 + 1 + 1 4 + 1 standard representation irreducible five-dimensional representation 5 3 + 2 2 + 2 + 1 other irreducible five-dimensional representation irreducible five-dimensional representation 5 2 + 2 + 1 3 + 2 other irreducible five-dimensional representation exterior square of standard representation 6 3 + 1 + 1 3 + 1 + 1 the same representation, because the partition is self-conjugate. Interpretation as projective general linear group of degree two Compare and contrast with linear representation theory of projective general linear group of degree two over a finite field Description of collection of representations Parameter for describing each representation How the representation is described Degree of each representation (general odd ) Degree of each representation () Number of representations (general odd ) Number of representations () Sum of squares of degrees (general odd ) Sum of squares of degrees () Symmetric group name Trivial -- 1 1 1 1 1 1 trivial Sign representation -- Kernel is projective special linear group of degree two (in this case, alternating group:A5), image is 1 1 1 1 1 1 sign Nontrivial component of permutation representation of on the projective line over -- -- 5 1 1 25 irreducible 5D Tensor product of sign representation and nontrivial component of permutation representation on projective line -- -- 5 1 1 25 other irreducible 5D Induced from one-dimensional representation of Borel subgroup ? ? 6 1 36 exterior square of standard representation Unclear a nontrivial homomorphism , with the property that for all , and takes values other than . Identify and . unclear 4 2 32 standard representation, product of standard and sign Total NA NA NA NA 7 120 NA Character table FACTS TO CHECK AGAINST (for characters of irreducible linear representations over a splitting field): Orthogonality relations: Character orthogonality theorem \| Column orthogonality theorem Separation results(basically says rows independent, columns independent): Splitting implies characters form a basis for space of class functions\|Character determines representation in characteristic zero Numerical facts: Characters are cyclotomic integers \| Size-degree-weighted characters are algebraic integers Character value facts: Irreducible character of degree greater than one takes value zero on some conjugacy class\| Conjugacy class of more than average size has character value zero for some irreducible character \| Zero-or-scalar lemma Representation/conjugacy class representative and size (size 1) (size 10) (size 15) (size 20) (size 20) (size 24) (size 30) trivial representation 1 1 1 1 1 1 1 sign representation 1 -1 1 1 -1 1 -1 standard representation 4 2 0 1 -1 -1 0 product of standard and sign representation 4 -2 0 1 1 -1 0 irreducible five-dimensional representation 5 1 1 -1 1 0 -1 irreducible five-dimensional representation 5 -1 1 -1 -1 0 1 exterior square of standard representation 6 0 -2 0 0 1 0 Below are the size-degree-weighted characters, i.e., these are obtained by multiplying the character value by the size of the conjugacy class and then dividing by the degree of the representation. Note that size-degree-weighted characters are algebraic integers. Representation/conjugacy class representative and size (size 1) (size 10) (size 15) (size 20) (size 20) (size 24) (size 30) trivial representation 1 10 15 20 20 24 30 sign representation 1 -10 15 20 -20 24 -30 standard representation 1 5 0 5 -5 -6 0 product of standard and sign representation 1 -5 0 5 5 -6 0 irreducible five-dimensional representation 1 2 3 -4 4 0 -6 irreducible five-dimensional representation 1 -2 3 -4 -4 0 6 exterior square of standard representation 1 0 -5 0 0 4 0 GAP implementation The degrees of irreducible representations can be computed using GAP's CharacterDegrees function: gap> CharacterDegrees(SymmetricGroup(5)); [ [ 1, 2 ], [ 4, 2 ], [ 5, 2 ], [ 6, 1 ] ] This means that there are 2 degree 1 irreducible representations, 2 degree 4 irreducible representations, 2 degree 5 irreducible representations, and 1 degree 6 irreducible representation. The characters of all irreducible representations can be computed in full using GAP's CharacterTable function: gap> Irr(CharacterTable(SymmetricGroup(5))); [ Character( CharacterTable( Sym( [ 1 .. 5 ] ) ), [ 1, -1, 1, 1, -1, -1, 1 ] ), Character( CharacterTable( Sym( [ 1 .. 5 ] ) ), [ 4, -2, 0, 1, 1, 0, -1 ] ), Character( CharacterTable( Sym( [ 1 .. 5 ] ) ), [ 5, -1, 1, -1, -1, 1, 0 ] ), Character( CharacterTable( Sym( [ 1 .. 5 ] ) ), [ 6, 0, -2, 0, 0, 0, 1 ] ), Character( CharacterTable( Sym( [ 1 .. 5 ] ) ), [ 5, 1, 1, -1, 1, -1, 0 ] ), Character( CharacterTable( Sym( [ 1 .. 5 ] ) ), [ 4, 2, 0, 1, -1, 0, -1 ] ), Character( CharacterTable( Sym( [ 1 .. 5 ] ) ), [ 1, 1, 1, 1, 1, 1, 1 ] ) ]
Consider the Segre embedding $(\mathbb P^1)^n \rightarrow \mathbb P^{2^n-1}$. What is the ideal corresponding to the image of this embedding? It is known that it is generated by quadratic relations. Is there a proper reference where the relations are written explicitly for this particular map in terms of $x_i$'s and $y_i$'s ? Let $(x_i:y_i)$ be homogeneous coordinates for the $i$-th copy of $\mathbb{P}^1$. Let $(z_I)$ be homogeneous coordinates for $\mathbb{P}^{2^n - 1}$, where $I$ ranges over all subsets of $[n] := \{1,2,\dots,n\}$. (The first psychological hurdle to overcome is possibly a bias toward wanting a linear ordering for coordinates. This gave me trouble when I was first learning algebraic geometry. But the point is that the most useful combinatorial gadget to index the coordinates on $\mathbb{P}^{2^n-1}$ is subsets of $[n]$, not the elements of $[2^n]$.) The Segre map can be described as $$z_I = \prod_{i \in I} x_i \prod_{j \notin I} y_j.$$ Then we must determine the condition on subsets $I,J,K,L$ of $[n]$ such that $z_I z_L = z_J z_K$ for all points in the image of the Segre map. Then if we think of $I \cup L$, resp. $J \cup K$, as a multiset (counting an element $a \in I \cap L$, resp. $b \in J \cap K$, as having multiplicity 2 in $I \cup L$, resp. $J \cup K$), the condition for $z_I z_L - z_J z_K$ to belong to the homogeneous ideal defining the image of the Segre map is that $I \cup L = J \cup K$ as multisets. As a concrete example, consider $n = 5$, $I = \{1,2,3\}$, $L = \{1,5\}$, $J = \{1,2,5\}$, $K = \{1,3\}$. Then $z_I z_L = z_J z_K$ on the image of the Segre map because $I \cup L = J \cup K = \{1,1,2,3,5\}$ as multisets. It should also be pointed out that there is a natural bijection between subsets of $[n]$ and sequences of $0$'s and $1$'s of length $n$. Given $I \subseteq \ [n]$, associate to it the sequence $(a_1,a_2,\dots,a_n)$ where $a_i = 1$ if $i \in I$ and $a_i = 0$ if $i \notin I$. Thus, some people index homogeneous coordinates on $\mathbb{P}^{2^n - 1}$ by using sequences of $0$'s and $1$'s of length $n$. In that case, the multiset formalism is equivalent to adding the sequences. So in the concrete example given above, you would have $I \leftrightarrow (1,1,1,0,0)$, $L \leftrightarrow (1,0,0,0,1)$, $J \leftrightarrow (1,1,0,0,1)$, $K \leftrightarrow (1,0,1,0,0)$ and $I \cup L = J \cup K \leftrightarrow (2,1,1,0,1)$.
I have to build the stability diagram of mercury and I have a problem with this couple: $\ce{Hg^2+}/\ce{Hg2^2+}$ $E^\circ=0.91\ \mathrm{V}$ The exercise says that a the border the concentration is $C=0.10\ \mathrm{mol \cdot L^{-1}}$ for all ions. So I have : $\ce{2Hg^2+ +2e^- <=> Hg2^2+}$ Then by Nernst relation I have : $E=E^\circ+0.03\ \mathrm{V} \times \log\left(\frac{\left[\ce{Hg^2+}\right]^2}{\left[\ce{Hg2^2+}\right]}\right)$ And in the solution of the exercise they write at the border $\left[\ce{Hg^2+}\right]=\left[\ce{Hg2^2+}\right]=\frac{C}{2}$ I don’t understand why it is $C/2$. I think that’s a “stupid” question but I really don’t understand.
My task is this: Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function and assume that the only stationary point $f$ has is a local max in the point $A = (a,f(a))$. Show that $A$ must be a global max for $f$. My work so far: Since $f$ is differentiable we have that $\ f'(a) = 0$. Since $A$ is the only stationary point and local max for $f$, we get: $$\forall x < a \implies f'(x) > 0 \ \land \forall y > a \implies f'(y) < 0.$$ It then follows that: $$f(x),\ f(y) < f(a).$$ Which is what we wanted to show. Is this good enough? Need someone to verify since there are no answers to this in the book I'm using (Multivariate calculus w/linalg by Tom Lindstrøm).
So I was wondering if anybody could help me tackle this problem that I've been assigned in my real analysis course (book used is Foundations of Analysis by Steven G. Krantz). From my understanding, this text was just recently written. First, I could use some help understanding exactly what this problem is asking. The way it's written (perhaps the notation) is confusing to me. We are in the chapter involving series, and this particular exercise is found in the section on elementary convergence tests. I found a similar question but none of the answers there made sense to me, so I figured I'd see if somebody could do it more justice and explain what is being asked and help me start it. Second if one could maybe give me a hint on how to proceed, as I am somewhat at a loss as to how to prove this particular statement, I would be most grateful. The problem is as follows. "Let $a_j$ be a sequence of real numbers. Define $m_j=\frac{a_1+a_2+\cdots a_j}{j}$. Prove that, if $\lim_{j\rightarrow\infty}a_j=l$ then $\lim_{j\rightarrow\infty}m_j=l$. Give an example to show that the converse is not true." I tried playing around with a few of the different elementary convergence tests (such as the root test and comparison test) and seeing how they could apply here, but in all honesty I don't understand how to apply them to this question. I just feel really off base with everything I've tried, so I figured I'd see if someone here could set me on the right path. Thank you all for your time!
Q1 Find the equation to the straight line passing through the point of intersection of the lines 5x+6y +1 = 0 and 3x + 2y +5 =0 and perpendicular to the line 3x-5y+11 = 0 Q2 Given that vectors P=3i+4j+5k and Q=3i+4j+5k , \[P\dot Q\] is __________ Q3 The value point A(1, -1) inside the circle is \[x^{2}+y^{2}-3x+4y=12\] ____________ Q4 Given that ellipse has equation of , the x and y intercepts of the of the equation are ______ and ______ Q5 The focus of the parabola whose equation is \[y^{2}+32\] is ________ Q6 Given that ellipse has equation of\[9x^{2}+4y^{2}=36\] , the length of the minor axis is __________ Q7 Given the equation \[9x^{2}-16y^{2}=144\], the intersection on x-axis is ___________ Q8 If a force of 30N acting in the east direction and another force of 40N acting in the north direction. Find the sum of the two vectors is _____________ Q9 The directrix of the parabola $$ y= -\frac{1}{2} x^{2}$$ is _________ Q10 The focus of the parabola $$ y= -\frac{1}{2} x^{2}$$ is _________ Q11 If A=i+4j+7k and B=5i-2j+k, find \[A\times B\] Q12 Find the equation of the circle with centre (3, 5) and radius 3 Q13 Find the equation of the tangent at the point (0, 2) to the circle $$x^{2}+y^{2}-4x+2y-8=0$$ Q14 Find the standard form of the equation for the parabola with vertex (3, 4) with focus (5, 4). Q15 Given the equation \[9x^2-16y^2=144\]. Find the coordinate of the foci Q16 Given that ellipse has equation of\ 9x^2+4y^2=36\], Find the length of the major axis Q17 Find the equation of the parabola having vertex (0, 0), axis along the x-axis and passing through (2, -10) Q18 What is the focus in the equation \[y^2=5x\] Q19 What is the equation of the directrix of the parabola whose equation is \[x^2-30y=0\] Q20 Find the equation of the tangent to the circle\[ x^2+y^2-2x+4y+15=0\] at the point (-1, 2) Q21 Find the coordinates of point of contact between the two circles \[x^2+y^2+2x+6y=39\] and \[x^2+y^2-4x-2y+1=0\] Q22 Find the parametric equations of a circle with centre (2, -1) and radius 3 Q23 Find the points of intersection of the circle x^2+y^2-x-3y=0 with the line y=x-1. Q24 Find the equation of the circle with circle with centre (-3, 4) which passes through the point (2, 5) Q25 Find the equation of the circle with centre (�??2, 3) and radius 6 Q26 Find the equation of the line through the point (-1, 2) which is perpendicular to y=2x-1 Q27 Find the angle between the two lines -3x+4y=8 and -2x-8y-14=0 Q28 A straight line has a gradient of 5/3 and it passes through the point (1,3). Find the intercept of the straight line on the y-axis Q29 Find the sum of vectors \[\bar{AK}, \bar{KL}, \bar{LP}~~and ~~\bar{PQ}\] Q30 A straight line has a gradient of 5/3 and it passes through the point (1,3). Find the equation of the straight line Q31 Find the angle of slope of the line joining A(4,3) and B(9,7). Q32 If A(3, 6) and B(4, 8) are two points on a line segment. Evaluate the coordinate of the midpoint of AB Q33 Find the distance between the points A(4, 3) and B(6, 5) Q34 If a=2i+4j+3k and b=i+5j-2k. Find the vector product a and b Q35 Find the magnitude of a components of vector\[\bar{AB}=5i+2j+4k\] expressed in terms of the unit vectors Q36 Given that\[ Z_1=2i-4j,~~Z_2=2i+6j ~~and~~ Z_3=3i-j\]. Evaluate\[ Z_1-Z_2-Z_3\] Q37 Find the sum of vectors \[\bar{AC}+ \bar{CL}-\bar{ML}\] Q38 Evaluate the sum of vectors \[\bar{BC}- \bar{DC}+\bar{DE}+\bar{FE}\] Q39 Find the sum of vectors \[\bar{AK}, \bar{KL}, \bar{LP}~~and ~~\bar{PQ}\] Q40 Find the standard equation for parabola whose directrix is the line x=2 and whose focus is the point (-2, 0) Q41 Find the asymptotes of the hyperbola whose equation is given as \[\frac{x^{2}}{4}-\frac{y^{2}}{9}=1\] Q42 Find equation of an ellipse whose major axis is vertical, with the center located (-1,3) at the distance between the center and one of the covertices equal to 4, and the distance between the center and one of the vertices equal to 6. Q43 Find the equation of the parabola with focus (-1,4) and directrix y=3 Q44 Given the equation \[9x^{2}-16y^{2}=14\], Find the coordinate of the foci Q45 Given the equation \[9x^{2}-16y^{2}=14\], Find the interception at x Q46 Given that ellipse has an equation of \[9x^{2}+4y^{2}=36\] Q47 Find the equation of the parabola having vertex (0,0) axis along the x-axis and pass through (2,-1) Q48 Find the focus in the equation \[y^{2}=5x\] Q49 Find the focus of the parabola whose equation is \[y^{2}+32x\] Q50 Find the directrix of the parabola whose equation is \[x^{2}-30y\] Q51 Find the equation of the circle with its center at the origin with points(-3,4) on the circle Q52 Find the radius of a circle given by \[x=4+2\cos\theta, y=-3+2\sin\theta\] Q53 Find the vector product \[a\times b\]. If a = i + 2j – k and b = 2i + 3j + k Q54 Find the center of a center of a circle given by \[x=4+2\cos\theta, y=-3+2\sin\theta\] Q55 Find the parametric equations of a circle with centre (2,-1) and radius 3 Q56 Find the point of the intersection of the center \[x^{2}+y^{2}-3y=0\] with line y=x-1 Q57 Given the equation of a circle is \[x^{2}+y^{2}+2x-6y-15\]. Find the radius of the circle Q58 Given the equation of a circle is \[x^{2}+y^{2}+2x-6y-15\]. Find the center of the circle Q59 Find the center of the circle \[x^{2}+y^{2}+8x + 6y = 0\]. Q60 Find the radius of the circle \[x^{2}+y^{2}+8x + 6y = 0\]. Q61 Find the equation to the straight line passing through the point of intersection of the lines 5x+6y +1 = 0 and 3x + 2y +5 =0 and perpendicular to the line 3x-5y+11 = 0 Q62 A dot product is said to be distributive,if �?â�?�¦�?â�?�¦.. Q63 If the slope of a line passing through the point A(3, 2) is \[\frac{3}{4}\]. then find points on the line which are 5 units away from the point A. Q64 Find the equation of the line through the point (�??1, 2) which is perpendicular to y = 2x – 1 Q65 Find the equation of the line through the point (â�?�??1, 2) which is parallel to y = 2x + 1 Q66 Find the angle between the two lines 3x+4y = 8 and 2x 8y = 14 0 Q67 A straight line has a gradient of \[\frac{5}{3}\] and it passes through the point (1,3). Find its equation Q68 If A is (3,6) and B is (4,8). Find the coordinate of the midpoint of AB Q69 The gradient of the line A(4,3) and B(8,6) Q70 Find the distance between the point A(5,4) and B(7,6) Q71 A vector having direction opposite to that of vector A,buth with the same magnitude is denoted by ——— Q72 If a=2i+4j+3k and b=i+5j-2k. Find the vector product of a and b Q73 If a=5i+4j+2k, b=4i-5j+3k and c=2i-j-2k. Determine value \[a\dot b\] Q74 If a= 2i+2j-k and b=3i-6j+2k. Find the scalar product a and b Q75 Find the direction cosine [l,m,n] of the r=2i+4j-3k Q76 If \[Z_{1}=2i-4j\],\[Z_{2}=2i+6j\] and \[Z_{3}=3i-j\] find the \[Z_{1}-Z_{2}-Z_{3}\] Q77 If \[Z_{1}=3i+5j\] and \[Z_{2}=7i+3j\], find the \[Z_{1}-Z_{2}\] Q78 Find the sum \[\overline{BC}-\overline{DC}+\overline{DE }+\overline{DE}+\overline{EF}\] Q79 Find the sum \[\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}+\overline{EF}\] Q80 If a = a force of 30N, acting in the east direction. b =a force of 40N, acting in the north direction. find the magnitude of the vector sum r of these forces
My professor in class went a little over chaos theory, and basically said that Newtonian determinism no longer applies, since as time goes to infinity, no matter how close together two initial points are, the distance between them will increase greatly. But why isn't this merely a matter of the imprecision of our measuring instruments? If we can somehow know our initial conditions exactly, wouldn't we still be able to calculate what the system will be like at some time t in the future? My professor in class went a little over chaos theory, and basically said that Newtonian determinism no longer applies, since as time goes to infinity, no matter how close together two initial points are, the distance between them will increase greatly. But why isn't this merely a matter of the imprecision of our measuring instruments? If we can somehow know our initial conditions Well, yes. In a purely mathematical world where you can specify initial conditions exactly, chaotic systems are fully deterministic. It's not like a quantum system with wavefunction collapse, whose evolution can never be specified exactly by the initial conditions. But in practice, we can never specify (or know) the initial conditions exactly. So there will always be some uncertainty in the initial conditions, and it makes sense to characterize the behavior of a system in terms of its response to this uncertainty. Basically, a chaotic system is one in which any uncertainty in the state at time $t=0$ leads to exponentially larger uncertainties in the state as time goes on, and a non-chaotic system is one in which any initial uncertainty in the state decays away or at least stays steady with time. In the former (chaotic) case, given that we can't know the initial conditions to infinite precision, there will always be some time after which predictions of the behavior of the system become essentially meaningless - the uncertainty becomes so large that it fills up most of the state space. This is effectively similar to the behavior of a truly non-deterministic (e.g. quantum) system, in that our ability to make predictions about it is limited, so some people call chaotic systems non-deterministic. Classical mechanics is perfectly integrable for two bodies as a closed or isolated system. However, early on it was found that problems existed, where Newton found he could not find a solution for the motion of the planets in a complete form. He made his famous statement that God had to readjust the solar system now and them. Poincare solved the Sweden prize for a solution to the stability of the solar system by demonstrating no such solution existed. This is what opened the door for chaos theory, where Poincare developed methods of separatrices and perturbations on them. For general systems it turns out that Newtonian mechanics is not integrable, Classical mechanics for systems with three or more bodies can\rq t be solved in closed form. For any $N$ body problem there are 3 equations for the center of mass, $3$ for the momentum, $3$ for the angular momentum and one for the energy. These are $10$ constraints on the problem. An N-body problem has $6N$ degrees of freedom. For $N~=~2$ this means the solution is given by a first integral with degree $2.$ For a three body problem this first integral has degree $8$. This runs into the problem that Galois illustrated which is that any root system with degree $5$ or greater generally have no algebraic roots. First integrals for differential equations are functions which remain constant along a solution to that differential equation. So for $8$ solutions there is some eight order polynomial $p_8(x)~=~\prod_{n=1}^8(x~-~\lambda_n)$, with $8$ distinct roots$\lambda_n$ that are constant along the $8$ solutions. Since $p_8(x)~=~p_5(x)p_3(x)$, a branch of algebra called Galois theory tells us that fifth order polynomials have no general algebraic system for finding its roots, or a set of solutions that are algebraic. This means that any system of degree higher than four are not in general algebraic. At the root of the $N$-body problem Galois theory tells us there is no algebraic solution for $N~\ge~3$. A problem in classical mechanics is the vanishing denominator problem for three bodies. This has a Hamiltonian $H( J,~\theta)~=~H_0(J,~\theta)~+~\epsilon H_1(J,~\theta)$ for $J~=~(J_1,~J_2)$ and $\theta~=~(\theta_1,~\theta_2)$. Here a generating function written according to the variable $J^\prime$ $$ S(J^\prime,~\theta)~=~\theta\cdot J^\prime~+~i\epsilon\sum_{n_1,n_2}{{H_{1,n_1,n_2}}\over{n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)}}e^{ n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)} $$ will be divergent for the resonant condition $n_1\omega_1(J^\prime)~+~n_2\omega_2(J^\prime)~=~0$. This resonance condition has lead many to presume that the solar system can not be stable with resonance conditions. Yet the solar system is replete with near resonance conditions. The number of resonance conditions that exist on the real line are dense. Within any $\epsilon$ neighborhood there will exist a countably infinite number of possible resonance conditions that correspond to rational numbers. As the orbit of a planet drifts it will pass through these resonance conditions and be chaotically perturbed. It is to be expected that for simple rational numbers, such as $1/12$ for Earth and Jupiter, rather than $1003/12000$ strong resonances occur. For more complex rational numbers it might be expected that the instability will be weaker. In other words if the ratio of frequencies are \lq\lq sufficiently irrational\rq\rq$~$so that $$ \Big\|{{\omega_1}\over{\omega_2}}~-~{m\over s}\Big\|~>~{{k(\epsilon)}\over{s^{2.5}}},~\lim_{\epsilon~\rightarrow~0}k(\epsilon)~\rightarrow~0 $$ the orbit is more stable. So an orbit that is removed from a “strong resonance”” condition near a simple rational number will be more stable than an orbit that is near an orbit with a simple rational ratio of frequencies. This is the basis for Greenberg’s Hamiltonian approach to chaos theory. This is called deterministic for the differential equations are time reverse invariant so the motion of a particle is absolutely determined. However. if you have a slight variation in the initial conditions of that particle is may in general end up arbitrarily far away from its starting point. The tiny variation $\delta z~=~(\delta q,~\delta p)$ becomes amplified by an exponential map $\delta z~\rightarrow~exp(\lambda t)\delta z$, for $\lambda$ the Lyapunov exponent. Any error in the specification of the initial conditions of a body results in the amplification of this error. From an algorithmic perspective a truncation results in numerical overflow errors which grow. So the dynamics of a particle can’t be integrated by computer to arbitrary accuracy into the future, even though nature actually does determine its dynamics. W. Zurek took this a bit further and considered how quantum fluctuations, where $\delta z$ is set by the Heisenberg uncertainty principle. I think the following from the wikipedia entry clears up well the terminology: Chaos theory is a field of study in applied mathematics, with applications in several disciplines including physics, economics, biology, and philosophy. Chaos theory studies the behavior of dynamical systems that are highly sensitive to initial conditions; an effect which is popularly referred to as the butterfly effect. Small differences in initial conditions (such as those due to rounding errors in numerical computation) yield widely diverging outcomes for chaotic systems, rendering long-term prediction impossible in general.1This happens even though these systems are deterministic, meaning that their future behavior is fully determined by their initial conditions, with no random elements involved.[2] In other words, the deterministic nature of these systems does not make them predictable.[3] This behavior is known as deterministic chaos, or simply chaos. Bold mine. Note that chaos results in completely deterministic systems when small errors in initial conditions yield highly divergent solutions. It is the "exactly" in your question that is unattainable, that will be a tiny bit off in chaotic situations,(highly nonlinear response to input parameters) even in computer solutions, because one cannot be more accurate than the computer bits. Note also that this does not mean that there are no mathematical methods to study the behavior of such systems. There are, and can be predictive in bulk. I would give as an example the study of Tsonis et al who have studied climate with a neural net chaotic model using as inputs bulk behavior of atmospheric and ocean currents. I think there is some debate about this, at least according to Steven Strogatz. I paraphrase a part of his lecture that I'm watching: Even systems with perfectly known, perfectly deterministic laws, where all the positions of all the particles, and all the forces, are known, can still be unpredictable. It is such systems that are called chaotic. This is his view of it. I wish I could provide a link to the video, but it's on my hard drive. I think this is an important debate to have, if it is still up in the air, because it has a lot of implications for the notion of free will. It's a philosophically deep question. From my point of view, which is that of a lay person, Strogatz's view seems a bit hard to imagine, but then again, he's the expert. I am afraid that your professor is confusing determinismus and computability. Chaotic behaviour is a property of some solutions of a system of non linear ODE. However any non linear ODE is strictly deterministic - e.g if you know the value of X at time t0, the ODE allows you to compute deterministically X at time t0+dt with an arbitrary accuracy. That's why it is often talked about deterministic chaos. The difficulties begin when you want to know X at a time T which is no more infinitesimaly close to t0 or in other words when you want to integrate the ODE system for any t. This is only possible numerically for chaotic systems and it is here that the property of exponential divergence (sensibility to initial conditions) kicks in. Indeed the error made in the value of the initial conditions will increase exponentially with time and prevent you to compute accurately the value of the chaotic variable for all times. But this is merely a computability problem, not a determinismus problem. From the physical point of view, the computability problem will never be solved because of the Heisenberg's uncertainty relation which imposes a non zero value for the uncertainty of the measure of initial conditions. If the uncertainty is equal to the minimum value allowed by QM, it is easy to compute for any chaotic system the time beyond which the uncertainty of the value of the dynamical variable will be of the same order of magnitude as the dynamical variable itself. Beyond this time no chaotic variable can be computed. For instance the orbital parameters of the Earth which are chaotic cannot be known beyond some 10 millions years. Of course, but this is another question, the unknowability of the orbital parameters is not synonymous with the unknowability of the orbit itself which may be but must not be stable. In our case we were lucky because despite the chaos in the orbital parameters of the Earth, the orbit itself has been stable or quasi stable for at least 4 billions years. As a caveat one should add that the exponential divergence of orbits is a necessary but non sufficient condition for a system to be chaotic. For instance a variable Y = exp(t) has the property of exponential divergence of nearby trajectories but is computable with an arbitrary accuracy and is not chaotic. our initial conditions exactly this is possible in Classical Mechanics, but it is not possible in Quantum Mechanics. In Quantum Mechanics when the Halmiltonian Operator has a discret spectrum you can have sensitive dependence on the initial condition in the Heisenberg Picture and not in the Schroedinger Picture. This creates problems, but it is a problem of the formalism, because the quantum counterpart of classical chaotic systems is chaotic according to the experimental facts. The most important thing to learn in Chaos Theory is perhaps the idea that deterministic equations lead to unpredictable dynamics. It is not always a matter of having been able to make the measurements practically accurate. Infinitesimal difference in Initial Conditions make the difference between the paths in the phase space diverge exponentially (jargon: Lyapunov Exponent being the order in which they get separated). And that's what makes it a matter of being unpredictable even in theory.
TL;DR Interkosmos and the first Asian in Space I solved global warming with orbital mechanics! This week in space history I'm always down for a little representation out and about the cosmos! On July 23, 1980, Pham Tuan became the first Asian in space, launching on Soyuz 37 to the Salyut 6 space station. Commander Viktor Gorbatko and Flight Engineer Pham Tuan of Soyuz 37 Tuan was chosen as part of the Soviet Interkosmos Program, a fascinating initiative started in the late 1970s to send cosmonauts from across the Eastern bloc and other pro-Soviet nations into space. While the missions were politically motivated to show off the technological superiority of Communism, Interkosmos nonetheless achieved many firsts in space diversity - the first Latin American / black person in space, the first Indian, even the first Japanese and British astronauts in the 90s at the twilight of the Cold War! "I was always in control" \| Photo credit: Justin Mott, 2011 Pham Tuan's life story is really worth telling - for the Vietnamese Communist party, he was the perfect man to be the first Asian in space, a war hero who rose from a poor village to repel the American invaders during the Vietnam War. As a MiG-21 fighter pilot, he's the only person to have ever downed an American B-52 Stratofortress in aerial combat (though the USAF still disputes this claim), and even today as a retired Lieutenant General, he's still a national hero to his people. But at 72 years old, he's enjoying the quiet life in Hanoi, playing tennis and taking care of his numerous birds and plants. Learn more about him - great article here! Me at Reunification Palace in Ho Chi Minh City (May 2018) \| The poster says, "The friendship of the Soviet Union and Vietnam is forever" Today I learned As an investment banker who dreams of going to space, I often poke fun at my lack of formal science and math background on Astronomical Returns. But today, I get to show off a little! My brother-in-law was complaining about the scorching summer heat, and he asked me, "Hans, how much force would it take to slightly raise Earth's orbit and cool us all down?" Turns out, I actually know how to calculate it! 2 billion billion of these bad boys solves global warming! Here's my answer - if you fire the propulsive equivalent of 2 billion BILLION Saturn V first stages for 60 seconds, you can raise Earth's orbit by 5%!! Let me know if you agree or disagree with my math below!! __________________________________________ Isaac Newton defined the vis-viva equation (learn more here), which provides us the instantaneous velocity of an object in an elliptical orbit around a central body $V = \sqrt{GM(\frac{2}{r} - \frac{1}{a})}$ $V =$ instantaneous velocity $G =$ Newton's gravitational constant $M =$ mass of the central body $r = $ radial distance between the central body and the satellite $a =$ semi-major axis We'll assume Earth's orbit around the Sun is circular (which is basically true), that way we can simplify this equation to just $V = \sqrt{\frac{GM}{r}}$ because $a = r$ at all times Here are some constants we're going to need: Gravitational constant: $G = 6.674 \cdot 10^{-11} \thinspace m^3\cdot kg^{-1}\cdot s^{-2}$ Mass of the Sun: $M_{Sun} = 1.989 \cdot 10^{30} \thinspace kg$ Earth's orbital radius: $r = 1.496 \cdot 10^{11} \thinspace m$ *Going forward, I'll just use G and M to keep the formulas concise So let's calculate Earth's current orbital velocity: $V = \sqrt{\frac{GM}{1.496 \cdot 10^{11}}}= 29,788 \thinspace m/s$ Our goal is to raise Earth's orbit 5%, so the new orbital radius is given by: $r_{new} = (1.496 \cdot 10^{11}) 1.05 = 1.571 \cdot 10^{11}\thinspace m$ Using the vis-viva equation again, we can determine what the new orbital velocity would be a the higher orbit $V_{new} = \sqrt{\frac{GM}{1.571 \cdot 10^{11}}}= 29,070 \thinspace m/s$ To get Earth from the current lower orbit to the new higher orbit, we're going to use what's called a Hohmann Transfer Orbit (learn more here), which simply uses two engine burns at the opposite ends of an intermediate elliptical orbit to reach a higher circular orbit Starting at the blue orbit, properly sized burns at V1 and V2 will get you to the orange orbit Since we already know $r$ and $r_{new}$, we can calculate $a$, the semi-major axis of the Hohmann Transfer Orbit (gray) $a = \frac{(1.496 \thinspace \cdot \thinspace 10^{11}) + (1.571 \thinspace \cdot \thinspace 10^{11})}{2} = 1.533 \cdot 10^{11} \thinspace m$ Now that we have the semi-major axis, we can use the vis-viva equation to calculate the necessary velocities at perihelion (closest approach) and aphelion (furthest approach) $v_{perihelion} = \sqrt{GM(\frac{2}{1.496 \thinspace \cdot \thinspace 10^{11}} - \frac{1}{1.533 \thinspace \cdot \thinspace 10^{11}})} = 30,149 \thinspace m/s$ $v_{aphelion} = \sqrt{GM(\frac{2}{1.571 \thinspace \cdot \thinspace 10^{11}} - \frac{1}{1.533 \thinspace \cdot \thinspace 10^{11}})} = 28,714 \thinspace m/s$ Now we have everything we need to calculate the change in Earth's velocity we must produce to raise its orbit $\Delta v_1 = 30,149 - 29,788 = 361 \thinspace m/s$ $\Delta v_2 = 29,070 - 28,714 = 357 \thinspace m/s$ $\Delta v_{total} = 361 + 357 = 718 \thinspace m/s$ So now we know how much delta-v we need to produce in Earth's orbit. Given the Earth's mass, we can finally calculate the thrust needed to move the Earth. Mass of the Earth: $M_{Earth} = 5.972 \cdot 10^{24} \thinspace kg$ Mass of the Earth: $M_{Earth} = 5.972 \cdot 10^{24} \thinspace kg$ $Force = mass \cdot acceleration = M_{Earth} \cdot \frac{\Delta v}{\Delta T} = 5.972 \cdot 10^{24} \cdot \frac{718}{60} = 7.144 \cdot 10^{25} N$ For context, the first stage of the Saturn V produced $3.510 \cdot 10^{7}N$ Do the math: $7.144 \cdot 10^{25} / 3.510 \cdot 10^{7} = 2.035 \cdot 10^{18}$ Saturn V's needed!! And there you have it - who says investment bankers can't do science?! Space is for everyone! "Give me a lever long enough and a fulcrum on which to place it, and I shall move the world" - Archimedes
Geometric mean formula, as the name suggests, is used to calculate the geometric mean of a set of numbers. To recall, the geometric mean (or GM) is a type of mean that indicates the central tendency of a set of numbers by using the product of their values. It is defined as the nth root of the product of n numbers. It should be noted that you cannot calculate the geometric mean from the arithmetic mean. In statistics, the geometric mean is well defined only for a positive set of real numbers. Example of using the formula for the geometric mean is to calculate the central frequency f 0 of a bandwidth BW= f 2–f 1. Formula for Geometric Mean For GM formula, multiply all the “n” numbers together and take the “nth root of them. The formula for evaluating geometric mean is as follows if we have “n” number of observations. x̄ geom = $\sqrt[n]{\prod_{i=1}^{n}x_{i}}=\sqrt[n]{x_{1}\cdot x_{2}\cdot\cdot\cdot x_{n}}$ Notation in the GM Formula x̄ geomis the geometric mean “n” is the total number of observations $\small \sqrt[n]{\prod_{i=1}^{n}x_{i}}$ is the n thsquare of the product of the given numbers. Example Question Using Geometric Mean Formula Question 1: Find the geometric mean of 4 and 3? Solution: Using the formula for G.M., the geometric mean of 4 and 3 will be: Geometric Mean will be √(4×3) = 2√3 So, GM = 3.46 Question 2: What is the geometric mean of 4, 8, 3, 9 and 17? Solution: Step 1: n = 5 is the total number of values. Now, find 1/n. 1/5 = 0.2. Step 2: Find geometric mean using the formula: (4 × 8 × 3 × 9 × 17) 0.2 So, geometric mean = 6.81 Articles Related to Geometric Mean Keep visiting BYJU’S and get various other maths formulas which are explained in an easy way along with solved examples. Also, register now to get maths video lessons on different topics and several practice questions which will help to learn the maths concepts thoroughly. Frequently Asked Questions What is the Geometric Mean of 4 and 25? Using the formula of geometric mean, GM of 4, 25 = √(4×25) = √100 So, geometric mean of 4 and 25 is 10. How to Define Geometric Mean Formula? Geometric mean formula is obtained by multiplying all the numbers together and taking the nth root of the product. Visit BYJU’S to learn more about the formula of geometric mean along with solved example questions. Why Geometric Mean is Better than Arithmetic Mean? Both the geometric mean and arithmetic mean are used to determine average. For any two positive unequal numbers, the geometric mean is always less than arithmetic mean. Now, geometric mean is better since it takes indicates the central tendency. In certain cases, arithmetic mean works better like in representing average temperatures, etc.
9.3 - Lesson 9 Summary Objectives Upon successful completion of this lesson, you should be able to: Identify situations in which the z or t distribution may be used to approximate a sampling distribution Construct a confidence interval to estimate the difference in two population proportions and two population means using Minitab Express given summary or raw data Conduct a hypothesis test for two proportions and two means using Minitab Express given summary or raw data Confidence Interval Test Statistic Two Independent Proportions At least 10 successes and 10 failures in both samples. $(\widehat{p}_1-\widehat{p}_2) \pm z^\ast {\sqrt{\frac{\widehat{p}_1 (1-\widehat{p}_1)}{n_1}+\frac{\widehat{p}_2 (1-\widehat{p}_2)}{n_2}}}$ $z=\frac{\widehat{p}_1-\widehat{p}_2}{\sqrt{\widehat{p}(1-\widehat{p})\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}}$ Two Independent Means Both sample size are at least 30 OR populations are normally distributed. $(\bar{x}_1-\bar{x}_2) \pm t^\ast{ \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$ $Estimated \;df = smallest\; n - 1$ $t=\frac{\bar{x}_1-\bar{x}_2}{ \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$ $Estimated \;df = smallest\; n - 1$
The idea is through conformal transformations satisfying the conditions requested of the problem make this an easier problem to deal,but i don't know which be this transformation. I would appreciate any hint how to solve this. thank you very much Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community The idea is through conformal transformations satisfying the conditions requested of the problem make this an easier problem to deal,but i don't know which be this transformation. I would appreciate any hint how to solve this. thank you very much Looking at my copy of Churchill, Complex Variables with Applications, the appendix on conformal transformations suggests that applying $$ w = f(z) = \log \left(\frac{z-1}{z+1} \right) $$ will transform the region shown (for the case $a = 1$) into a strip in the $w$-plane, where the $V = 1$ rays are mapped to $\Im(w) = \pi$ and the $V = 0$ segment is mapped to $\Im(w) = 0$. If you take $g(z) = \frac{1}{\pi} f(z)$, you map this region to one in which the $V = 1$ segment maps to $\Im(w) = 1$, and the $V = 0$ segment maps to $\Im(w) = 0$. I seem to recall that this is what's needed in this kind of problem, but it's as far as my shoddy memory takes me. (The inverse conformal map for $f$, by the way, is $z = -\coth \frac{w}{2}$.) OK. I went and read the relevant chapter of Churchill ... and found this problem in the exercises! I think that having come up with the transformation, and the potential in the $w$-plane, I'm going to leave the conversion back to a potential in the $z$-plane up to you, although I confess I don't see offhand how to get from the $\log$-expression back to something involving $\arctan$. In comments, I've been asked to add some pics to show off what the log-transform is doing. Let's look at this diagram: Now the "log" that I've used above is really $$ w = \log \|\frac{z-1}{z+1}\| + i \arg \frac{z-1}{z+1}. $$ Let's look at how that gets computed. First, the first ratio is just the ratio of distances to the two point $+1$ and $-1$; I'll come back to that. The second term -- the imaginary part, is the difference of the args of $z-1$ and $z - (-1)$, i.e., the angle labeled $\theta_1 - \theta_2$ in the diagram. What's that look like on (or very near) the real axis? Well, for $z > 1$, $\theta_1$ and $\theta_2$ will both be nearly zero; for $z < -1$, they'll both be nearly $\pi$. For $-1 < z < 1$, we have $\theta_2 \sim 0$ and $\theta_1 \sim \pi$, so the difference is approximately $\pi$. In other words, the red segments of the real axis will be sent to $v = 0$ in the $w$-plane, but the blue segment will be sent to $v = \pi$. Note that the remote ends of the red segments (near $\pm \infty$) will be sent to $u = 0$, while the near ends (near $\pm 1$) are sent off to infinity. What about the ratio of lengths? Well, along the $x = 0$ line, we'll have $r_1 = r_2$, so that will map to $u = 0$. The other $u = constant$ lines will become hyperbolic-like curves in the $xy$-plane. Finally, the transform we really want is $$ w = \frac{1}{\pi}\log \|\frac{z-1}{z+1}\| + \frac{i}{\pi} \arg \frac{z-1}{z+1}, $$ which sends the blue segment to the $v = 1$ line in the $w$-plane. Now since the imaginary part of $w$ is $v$, and we can substitute $x+iy$ for $z$, we get \begin{align} v &= \frac{1}{\pi} \arg \frac{z-1}{z+1} \\ &= \frac{1}{\pi} \arg \frac{(x-1) + iy}{(x+1) + iy} \\ &= \frac{1}{\pi} \arg \frac{x^2 + y^2 - 1 + i 2y}{(x+1)^2 + y^2} \\ &= \frac{1}{\pi} \arg {x^2 + y^2 - 1 + i 2y}\\ &= \frac{1}{\pi} \arctan \frac{2y}{x^2 + y^2 - 1}, \end{align} which is an expression for the potential at point $(x, y)$. Finally, I'm going to show off the effect of the conformal transformation by showing an image in the $z$-plane, and the transformed image in the $w$-plane, courtesy of Matlab. The problem is that the points near $z = \pm 1$ get sent to infinity, so parts of the transformed image will be missing. The starting image (which extends from $-2.5$ to $2.5$ in $x$, and from $0.05$ to $2.0$ in $y$ (with $y = 0$ near the top, because of Matlab's idiosyncracies!) shows a bunch of peppers: The transformed image, extending from -10 to 10 in $u$ and $0$ to $1$ in $v$, is this: If you can make sense of those, then bless you. I just followed the description given here: http://www.mathworks.com/help/images/examples/exploring-a-conformal-mapping.html to make the images. Here's my code: function confImageclose allA = imread('peppers.png');A = A(31:330, 1:500, :);figure, imshow(A)title ('Original Image', 'FontSize', 14);conformal = maketform('custom', 2, 2, [], @conformalInverse, []);uData = [ -10 10]; % Bounds for REAL(w)vData = [ 0 1.0]; % Bounds for IMAG(w)xData = [ -2.5 2.5 ]; % Bounds for REAL(z)yData = [ 0.05 2.0 ]; % Bounds for IMAG(z)B = imtransform( A, conformal, 'cubic', ... 'UData', uData,'VData', vData,... 'XData', xData,'YData', yData,... 'Size', [300 360], 'FillValues', 255 );figure, imshow(B)title('Transformed Image','FontSize',14)function U = conformalInverse(X, t)Z = complex(X(:,1),X(:,2));W = (Z + 1./Z)/2;U(:,2) = (1/pi) * angle((Z - 1) ./ (Z + 1))U(:,1) = (1/pi)* log( abs((Z - 1) ./ (Z + 1))); %real(W); --------------------------------------------------------------- In general we have: $$\frac{dw}{dz}=K(z-a)^{\alpha/\pi-1}(z-b)^{\beta/\pi-1}$$ With $a=-1$ , $b=+1$ and $\alpha=\beta=0$ , giving: $$dw\;=\;K(z-1)^{-1}(z+1)^{-1}dz \;=\;\frac{K}{2}\left(\frac{dz}{z-1}-\frac{dz}{z+1}\right)$$ So indeed, integration gives apart from constants that can be determined (later): $$w=\log\left(\frac{z-1}{z+1}\right)$$
Simulation Tools for Solving Wave Electromagnetics Problems When solving wave electromagnetics problems with either the RF or Wave Optics modules, we use the finite element method to solve the governing Maxwell’s equations. In this blog post, we will look at the various modeling, meshing, solving, and postprocessing options available to you and when you should use them. The Governing Equation for Modeling Frequency Domain Wave Electromagnetics Problems COMSOL Multiphysics uses the finite element method to solve for the electromagnetic fields within the modeling domains. Under the assumption that the fields vary sinusoidally in time at a known angular frequency \omega = 2 \pi f and that all material properties are linear with respect to field strength, the governing Maxwell’s equations in three dimensions reduce to: With the speed of light in vacuum, c_0, the above equation is solved for the electric field, \mathbf{E}=\mathbf{E}(x,y,z), throughout the modeling domain, where \mathbf{E} is a vector with components E_x, E_y, and E_z. All other quantities (such as magnetic fields, currents, and power flow) can be derived from the electric field. It is also possible to reformulate the above equation as an eigenvalue problem, where a model is solved for the resonant frequencies of the system, rather than the response of the system at a particular frequency. The above equation is solved via the finite element method. For a conceptual introduction to this method, please see our blog series on the weak form, and for a more in-depth reference, which will explain the nuances related to electromagnetic wave problems, please see The Finite Element Method in Electromagnetics by Jian-Ming Jin. From the point of view of this blog post, however, we can break down the finite element method into these four steps: Model Set-Up:Defining the equations to solve, creating the model geometry, defining the material properties, setting up metallic and radiating boundaries, and connecting the model to other devices. Meshing:Discretizing the model space using finite elements. Solving:Solving a set of linear equations that describe the electric fields. Postprocessing:Extracting useful information from the computed electric fields. Let’s now look at each one of these steps in more detail and describe the options available at each step. Options for Modifying the Governing Equations The governing equation shown above is the frequency domain form of Maxwell’s equations for wave-type problems in its most general form. However, this equation can be reformulated for several special cases. Let us first consider the case of a modeling domain in which there is a known background electric field and we wish to place some object into this background field. The background field can be a linearly polarized plane wave, a Gaussian beam, or any general user-defined beam that satisfies Maxwell’s equations in free space. Placing an object into this field will perturb the field and lead to scattering of the background field. In such a situation, you can use the Scattered Field formulation, which solves the above equation, but makes the following substitution for the electric field: where the background electric field is known and the relative field is the field that, once added to the background field, gives the total field that satisfies the governing Maxwell’s equations. Rather than solving for the total field, it is the relative field that is being solved. Note that the relative field is not the scattered field. For an example of the usage of this Scattered Field formulation, which considers the radar scattering off of a perfectly electrically conductive sphere in a background plane wave and compares it to the analytic solution, please see our Computing the Radar Cross Section of a Perfectly Conducting Sphere tutorial model. Next, let’s consider modeling in a 2D plane, where we solve for \mathbf{E}=\mathbf{E}(x,y) and can additionally simplify the modeling by considering an electric field that is polarized either In-Plane or Out-of-Plane. The In-Plane case will assume that E_z=0, while the Out-of-Plane case assumes that E_x=E_y=0. These simplifications reduce the size of the problem being solved, compared to solving for all three components of the electric field vector. For modeling in the 2D axisymmetric plane, we solve for \mathbf{E}=\mathbf{E}(r,z), where the vector \mathbf{E} has the components E_r, E_\phi, and E_z. We can again simplify our modeling by considering the In-Plane and Out-of-Plane cases, which assume E_\phi=0 and E_r=E_z=0, respectively. When using either the 2D or the 2D axisymmetric In-Plane formulations, it is also possible to specify an Out-of-Plane Wave Number. This is appropriate to use when there is a known out-of-plane propagation constant, or known number of azimuthal modes. For 2D problems, the electric field can be rewritten as: and for 2D axisymmetric problems, the electric field can be rewritten as: where k_z or m, the out-of-plane wave number, must be specified. This modeling approach can greatly simplify the computational complexity for some types of models. For example, a structurally axisymmetric horn antenna will have a solution that varies in 3D but is composed of a sum of known azimuthal modes. It is possible to recover the 3D solution from a set of 2D axisymmetric analyses by solving for these out-of-plane modes at a much lower computational cost, as demonstrated in our Corrugated Circular Horn Antenna tutorial model. Meshing Requirements and Capabilities Whenever solving a wave electromagnetics problem, you must keep in mind the mesh resolution. Any wave-type problem must have a mesh that is fine enough to resolve the wavelengths in all media being modeled. This idea is fundamentally similar to the concept of the Nyquist frequency in signal processing: The sampling size (the finite element mesh size) must be at least less than one-half of the wavelength being resolved. By default, COMSOL Multiphysics uses second-order elements to discretize the governing equations. A minimum of two elements per wavelength are necessary to solve the problem, but such a coarse mesh would give quite poor accuracy. At least five second-order elements per wavelength are typically used to resolve a wave propagating through a dielectric medium. First-order and third-order discretization is also available, but these are generally of more academic interest, since the second-order elements tend to be the best compromise between accuracy and memory requirements. The meshing of domains to fulfill the minimum criterion of five elements per wavelength in each medium is now automated within the software, as shown in this video, which shows not only the meshing of different dielectric domains, but also the automated meshing of Perfectly Matched Layer domains. The new automated meshing capability will also set up an appropriate periodic mesh for problems with periodic boundary conditions, as demonstrated in this Frequency Selective Surface, Periodic Complementary Split Ring Resonator tutorial model. With respect to the type of elements used, tetrahedral (in 3D) or triangular (in 2D) elements are preferred over hexahedral and prismatic (in 3D) or rectangular (in 2D) elements due to their lower dispersion error. This is a consequence of the fact that the maximum distance within an element is approximately the same in all directions for a tetrahedral element, but for a hexahedral element, the ratio of the shortest to the longest line that fits within a perfect cubic element is \sqrt3. This leads to greater error when resolving the phase of a wave traveling diagonally through a hexahedral element. It is only necessary to use hexahedral, prismatic, or rectangular elements when you are meshing a perfectly matched layer or have some foreknowledge that the solution is strongly anisotropic in one or two directions. When resolving a wave that is decaying due to absorption in a material, such as a wave impinging upon a lossy medium, it is additionally necessary to manually resolve the skin depth with the finite element mesh, typically using a boundary layer mesh, as described here. Manual meshing is still recommended, and usually needed, for cases when the material properties will vary during the simulation. For example, during an electromagnetic heating simulation, the material properties can be made functions of temperature. This possible variation in material properties should be considered before the solution, during the meshing step, as it is often more computationally expensive to remesh during the solution than to start with a mesh that is fine enough to resolve the eventual variations in the fields. This can require a manual and iterative approach to meshing and solving. When solving over a wide frequency band, you can consider one of three options: Solve over the entire frequency range using a mesh that will resolve the shortest wavelength (highest frequency) case. This avoids any computational cost associated with remeshing, but you will use an overly fine mesh for the lower frequencies. Remesh at each frequency, using the parametric solver. This is an attractive option if your increments in frequency space are quite widely spaced, and if the meshing cost is relatively low. Use different meshes in different frequency bands. This will reduce the meshing cost, and keep the solution cost relatively low. It is essentially a combination of the above two approaches, but requires the most user effort. It is difficult to determine ahead of time which of the above three options will be the most efficient for a particular model. Regardless of the initial mesh that you use, you will also always want to perform a mesh refinement study. That is, re-run the simulation with progressively finer meshes and observe how the solution changes. As you make the mesh finer, the solution will become more accurate, but at a greater computational cost. It is also possible to use adaptive mesh refinement if your mesh is composed entirely of tetrahedral or triangular elements. Solver Options Once you have properly defined the problem and meshed your domains, COMSOL Multiphysics will take this information and form a system of linear equations, which are solved using either a direct or iterative solver. These solvers differ only in their memory requirements and solution time, but there are several options that can make your modeling more efficient, since 3D electromagnetics models will often require a lot of RAM to solve. The direct solvers will require more memory than the iterative solvers. They are used for problems with periodic boundary conditions, eigenvalue problems, and for all 2D models. Problems with periodic boundary conditions do require the use of a direct solver, and the software will automatically do so in such cases. Eigenvalue problems will solve faster when using a direct solver as compared to using an iterative solver, but will use more memory. For this reason, it can often be attractive to reformulate an eigenvalue problem as a frequency domain problem excited over a range of frequencies near the approximate resonances. By solving in the frequency domain, it is possible to use the more memory-efficient iterative solvers. However, for systems with high Q-factors it becomes necessary to solve at many points in frequency space. For an example of reformulating an eigenvalue problem as a frequency domain problem, please see these examples of computing the Q-factor of an RF coil and the Q-factor of a Fabry-Perot cavity. The iterative solvers used for frequency-domain simulations come with three different options defined by the Analysis Methodology settings of Robust (the default), Intermediate, or Fast, and can be changed within the physics interface settings. These different settings alter the type of iterative solver being used and the convergence tolerance. Most models will solve with any of these settings, and it can be worth comparing them to observe the differences in solution time and accuracy and choose the option most appropriate for your needs. Models that contain materials that have very large contrasts in the dielectric constants (~100:1) will need the Robust setting and may even require the use of the direct solver, if the iterative solver convergence is very slow. Postprocessing Capabilities Once you’ve solved your model, you will want to extract data from the computed electromagnetic fields. COMSOL Multiphysics will automatically produce a slice plot of the magnitude of the electric field, but there are many other postprocessing visualizations you can set up. Please see the Postprocessing & Visualization Handbook and our blog series on Postprocessing for guidance and to learn how to create images such as those shown below. Attractive visualizations can be created by plotting combinations of the solution fields, meshes, and geometry. Of course, good-looking images are not enough — we also want to extract numerical information from our models. COMSOL Multiphysics will automatically make available the S-parameters whenever using Ports or Lumped Ports, as well as the Lumped Port current, voltage, power, and impedance. For a model with multiple Ports or Lumped Ports, it is also possible to automatically set up a Port Sweep, as demonstrated in this tutorial model of a Ferrite Circulator, and write out a Touchstone file of the results. For eigenvalue problems, the resonant frequencies and Q-factors are automatically computed. For models of antennas or for scattered field models, it is additionally possible to compute and plot the far-field radiated pattern, the gain, and the axial ratio. Far-field radiation pattern of a Vivaldi antenna. You can also integrate any derived quantity over domains, boundaries, and edges to compute, for example, the heat dissipated inside of lossy materials or the total electromagnetic energy within a cavity. Of course, there is a great deal more that you can do, and here we have just looked at the most commonly used postprocessing features. Summary of Wave Electromagnetics Simulation Tools We’ve looked at the various different formulations of the governing frequency domain form of Maxwell’s equations as applied to solving wave electromagnetics problems and when they should be used. The meshing requirements and capabilities have been discussed as well as the options for solving your models. You should also have a broad overview of the postprocessing functionality and where to go for more information about visualizing your data in COMSOL Multiphysics. This information, along with the previous blog posts on defining the material properties, setting up metallic and radiating boundaries, and connecting the model to other devices should now give you a reasonably complete picture of what can be done with frequency domain electromagnetic wave modeling in the RF and Wave Optics modules. The software documentation, of course, goes into greater depth about all of the features and capabilities within the software. If you are interested in using the RF or Wave Optics modules for your modeling needs, please contact us. Comments (0) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science TAGS CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
The ratio wherein the addition of the values to the number of the value is a population mean – if the possibilities are equal. A population mean include each element from the set of observations that can be made. The population mean can be given in the following formula: \[\large \mu=\frac{\sum x_i}{N}\] Where, $x_{i}$ = Sum of the values N=number of the value Solved Examples Question: Find the population mean of the following numbers 1, 2, 3, 4, 5? Solution: Given, $x_{i}$ = 1 + 2 + 3 + 4 + 5 = 15 N = 5 Population Mean = $\frac{\sum x_{i}}{N}$ $\mu$ = $\frac{15}{5}$ $\mu=3$ Population Mean = 3
Lasted edited by Andrew Munsey, updated on June 15, 2016 at 2:08 am. {\| cellspacing="0" cellpadding="0" style="clear:right margin-bottom: .5em float: right padding: .5em 0 .8em 1.4em background: none width: There was an error working with the wiki: Code[1]}" \| __TOC__ \|} Inductance circuit (eg., L circuit) is an electrical network inwhich the electrical elements is primarily composed of inductors. They can both be basically simplified into power source(s) and wire(s). An electrical inductance circuit is a network that has a closed loop and, via an amount of magnetic flux surrounding the circuit, produces a given electric current. The amount of magnetic flux produced by a current depends upon the permeability of the medium surrounded by the current, the area inside the coil, and the number of turns. The greater the permeability, the greater the magnetic flux generated by a given current. Certain (ferromagnetic) materials have much higher permeability than air. If a conductor (wire) is wound around such a material, the magnetic flux becomes much greater and the inductance becomes much greater than the inductance of an identical coil wound in air. When the magnetic flux produced by an inductor links another inductor, these inductors are said to be coupled. Coupling is often undesired but in many cases, this coupling is intentional and is the basis of the Transformer. When inductors are coupled, there exists a mutual inductance that relates the current in one inductor to the flux linkage in the other inductor. Thus, there are three inductances defined for coupled inductors: :L_{11} - the self inductance of inductor 1 :L_{22} - the self inductance of inductor 2 :L_{12} = L_{21} - the mutual inductance associated with both inductors When either side of the transformer is a There was an error working with the wiki: Code[14], the amount of mutual inductance between the two windings determines the shape of the frequency response curve. Although no boundaries are defined, this is often referred to as loose-, critical-, and over-coupling. When two tuned circuits are loosely coupled through mutual inductance, the bandwidth will be narrow. As the amount of mutual inductance increases, the bandwidth continues to grow. When the mutual inductance is increased beyond a critical point, the peak in the response curve begins to drop, and the center frequency will be attenuated more strongly than its direct sidebands. This is known as overcoupling. Mutual inductance is the concept that the current through one inductor can induce a voltage in another nearby inductor. It is important as the mechanism by which Transformers work, but it can also cause unwanted coupling between conductors in a circuit. The mutual inductance, M, is also a measure of the coupling between two inductors. The mutual inductance by circuit i on circuit j is given by the double integral There was an error working with the wiki: Code[15] formula : M_{ij} = \frac{\mu_0}{4\pi} \oint_{C_i}\oint_{C_j} \frac{{ds}_i\cdot{ds}_j}{\|{R}_{ij}\|} See a There was an error working with the wiki: Code[1]. The mutual inductance also has the relationship: :M_{21} = N_1 N_2 P_{21} \! where :M_{21} is the mutual inductance, and the subscript specifies the relationship of the voltage induced in coil 2 to the current in coil 1. :N_1 is the number of turns in coil 1, :N_2 is the number of turns in coil 2, :P_{21} is the There was an error working with the wiki: Code[16] of the space occupied by the flux. The mutual inductance also has a relationship with the coefficient of coupling. The coefficient of coupling is always between 1 and 0, and is a convenient way to specify the relationship between a certain orientation of inductor with arbitrary inductance: :M = k \sqrt{L_1 L_2} \! where :k is the coefficient of coupling and 0 ? k ? 1, :L_1 is the inductance of the first coil, and :L_2 is the inductance of the second coil. Once this mutual inductance factor M is determined, it can be used to predict the behavior of a circuit: : V = L_1 \frac{dI_1}{dt} + M \frac{dI_2}{dt} where :V is the voltage across the inductor of interest, :L_1 is the inductance of the inductor of interest, :dI_1 / dt is the derivative, with respect to time, of the current through the inductor of interest, :M is the mutual inductance and :dI_2 / dt is the derivative, with respect to time, of the current through the inductor that is coupled to the first inductor.}} When one inductor is closely coupled to another inductor through mutual inductance, such as in a Transformer, the voltages, currents, and number of turns can be related in the following way: :V_s = V_p \frac{N_s}{N_p} where :V_s is the voltage across the secondary inductor, :V_p is the voltage across the primary inductor (the one connected to a power source), :N_s is the number of turns in the secondary inductor, and :N_p is the number of turns in the primary inductor. Conversely the current: :I_s = I_p \frac{N_p}{N_s} where :I_s is the current through the secondary inductor, :I_p is the current through the primary inductor (the one connected to a power source), :N_s is the number of turns in the secondary inductor, and :N_p is the number of turns in the primary inductor. Note that the power through one inductor is the same as the power through the other. Also note that these equations don't work if both transformers are forced (with power sources). Self-inductance, denoted L, is the usual inductance one talks about with an There was an error working with the wiki: Code[17]. It is a special case of mutual inductance where, in the above equation, i =j. Thus, : M_{ij} = M_{jj} = L_{jj} = L_j = L = \frac{\mu_0}{4\pi} \oint_{C}\oint_{C'} \frac{{ds}\cdot{ds}'}{\|{R}\|} Physically, the self-inductance of a circuit represents the back-emf described by There was an error working with the wiki: Code[18]. Consider a current loop \delta S with current i(t). According to There was an error working with the wiki: Code[19], current i(t) sets up a There was an error working with the wiki: Code[20] at 'r': :{B}({r},t)= \frac{\mu_{0}\mu_{r} i(t)}{4\pi} \int_{\delta S}{\frac{d{l} \times {\hat r}}{r^2}} Now There was an error working with the wiki: Code[21] through the surface S the loop encircles is: :\Phi(t) = \int_S{B}({r},t) \cdot d{A} = \frac{\mu_{0}\mu_{r} i(t)}{4\pi} \int_S \int_{\delta S}{\frac{d{l} \times {\hat r}}{r^2}} \cdot d{A} = Li(t) From where we get the expression for inductance of the current loop: :L = \frac{\mu_{0}\mu_{r} }{4\pi} \int_S \int_{\delta S}{\frac{d{l} \times {\hat r}}{r^2}} \cdot d{A} where :?0 and ?''r are the same as above :d{l} is the differential length vector of the current loop element :{\hat r} is the unit displacement vector from the current element to the field point 'r' :r is the distance from the current element to the field point 'r' :d{A} There was an error working with the wiki: Code[2] vector element of surface area A, with There was an error working with the wiki: Code[3] small magnitude and direction There was an error working with the wiki: Code[4] to surface S As we see here, the geometry and material properties (if material properties are same in surface S and the material is linear) of the current loop can be expressed with single scalar quantity L. Using There was an error working with the wiki: Code[5], the equivalent There was an error working with the wiki: Code[6] of an inductance is given by: :Z_L = V / I = j L \omega \, where : X_L = L \omega \, is the inductive There was an error working with the wiki: Code[22], : \omega = 2 \pi f \, is the angular frequency, : L is the inductance, : f is the frequency, and : j is the There was an error working with the wiki: Code[23]. The inductance of a circular conductive loop made of a circular conductor can be determined using : L = {r \mu_0 \mu_r \left( \ln{ \frac {8 r}{a}} - 2 \right) } where :?0 and ?''r are the same as above :r is the radius of the loop :a is the radius of the conductor The self-inductance L of such a solenoid can be calculated from: : L = {\mu_0 \mu_r N^2 A \over l} = \frac{N \Phi}{i} where :?0 is the There was an error working with the wiki: Code[7] of free space (4? × 10-7 There was an error working with the wiki: Code[8] per Metre) :?r is the relative permeability of the core (dimensionless) :N is the number of turns. :A is the cross sectional area of the coil in There was an error working with the wiki: Code[24]s. :l is the length of the coil in Metres. :\Phi = BA is the flux in webers (B is the flux density, A is the area). :i is the current in amperes This, and the inductance of more complicated shapes, can be derived from Maxwell's equations. For rigid air-core coils, inductance is a function of coil geometry and number of turns, and is independent of current. However, since the permeability of ferromagnetic materials changes with applied magnetic flux, the inductance of a coil with a ferromagnetic core will generally vary with current. Main: There was an error working with the wiki: Code[25], There was an error working with the wiki: Code[26], There was an error working with the wiki: Code[27], There was an error working with the wiki: Code[28], Transformer Circuits: There was an error working with the wiki: Code[9], There was an error working with the wiki: Code[10], There was an error working with the wiki: Code[11] There was an error working with the wiki: Code[12], There was an error working with the wiki: Code[13] Griffiths, David J. (1998). Introduction to Electrodynamics (3rd ed.). Prentice Hall. ISBN 0-13-805326-X. Wangsness, Roald K. (1986). Electromagnetic Fields, 2nd ed., Wiley. ISBN 0-471-81186-6. Hughes, Edward. (2002). Electrical & Electronic Technology (8th ed.). Prentice Hall. ISBN 0-582-40519-X. Küpfmüller K., Einführung in die theoretische Elektrotechnik, Springer-Verlag, 1959. Heaviside O., Electrical Papers. Vol.1. – L. N.Y.: Macmillan, 1892, p. 429-560. There was an error working with the wiki: Code[1], Wikipedia: The Free Encyclopedia. Wikimedia Foundation.
I doubt that your reasoning goes through without caveats. It's true that when you analytically continue the Minkowski spacetime two-point function to imaginary time, you get a function (the two-point Schwinger function $S(x,y)$) which is symmetric in the two Euclidean spacetime points and is covariant with respect to Euclidean isometries. So if you define$$[\phi (x),\phi(y)]:=S(x,y)-S(y,x)$$for $x$, $y$ being Euclidean spacetime points, then you have $[\phi (x),\phi(y)]=0$. However, it is not clear to me that the left-hand side of the definition above can be interpreted as the commutator of two well-defined operators (or operator-valued distributions). For example, let $\phi$ be a free scalar field, defined initially in Minkowski spacetime by its mode expansion$$\phi(t,\mathbf{x})=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{2\omega(\mathbf{k})}}\left(a(\mathbf{k})e^{-i\omega t+i\mathbf{k}\cdot\mathbf{x}}+a^{\dagger}(\mathbf{k})e^{i\omega t-i\mathbf{k}\cdot\mathbf{x}}\right).$$ If we formally continue this to $t=-i\tau$, we see that the modes of positive (negative) frequency diverge badly for $\tau<0$ ($\tau>0$). (Note: This divergent behavior of the modes for imaginary times also puts your manipulations of the integral representation of the commutator into question.) One can still make sense of expressions like $\langle 0\|\phi (-i\tau_2,\mathbf{x}_2)\phi(-i\tau_1,\mathbf{y}_1) \|0\rangle$ as long as $\tau_2>\tau_1$ and the momentum integrals are understood to be taken only at the end. Using this formalism, one can equate this expression to a path integral in Euclidean spacetime with two "operator insertions":$$\langle 0\|\phi (-i\tau_2,\mathbf{x}_2)\phi(-i\tau_1,\mathbf{x}_1) \|0\rangle = \frac{\int \mathcal D \phi \exp{\left(-S_E[\phi]\right)\phi(x_2)\phi(x_1)}}{\int \mathcal D \phi \exp{\left(-S_E[\phi]\right)}}$$ The path integral automatically takes care of the ordering in imaginary time and we can therefore use it to extend the meaning of the left-hand side to arbitrary $\tau$-ordering, which gives us the Schwinger two-point function for this field expressed as a path integral. Still, this does not mean that we have made sense of the field operator by itself for non-zero imaginary times. The automatic Euclidean time-ordering done by the path integral prevents us from using it to calculate the general matrix elements of the field commutator for unequal Euclidean times. Maybe this is the reason why one usually does not find a calculation of this commutator for Euclidean spacetime in textbooks. (At least I found none in the texts I have available.) For 2-dimensional conformal field theories one finds calculations of commutators $[T,\phi(y)]$ between the field $\phi$ and a conserved charge $T$ making use of the operator product expansion (OPE) in Euclidean spacetime ([1], [2]). Such commutators can be non-zero even if the operator $T$ is built from the field and its derivatives. Your conclusion that the commutator $[\phi(x), \phi(y)]$ is identically zero on Euclidean spacetime would also imply that all commutators of quantities built out of the fields and their derivatives vanish on Euclidean spacetime, wouldn't it? Schwinger mentioned the formal possibility to construct fully (anti)commuting field operators for Euclidean spacetime at the end of [3]. I'm not sure what to make of that. A related question about field operators and two-point functions in Euclidean spacetime has been discussed on mathoverflow: https://mathoverflow.net/q/237647 [1] K. Becker, M. Becker, J. H. Schwarz. String theory and M-theory: A modern introduction. Cambridge University Press, 2007. p. 64f. [2] J. Polchinski. String Theory. Vol. 1 An Introduction to the Bosonic String. Cambridge University Press, 1998. p. 55. [3] J. Schwinger, 1958. Four-dimensional euclidean formulation of quantum field theory. Proceedings, Conference: C58-06-30, p.134-140.
(Original on SE https://physics.stackexchange.com/questions/459086/continuous-transition-of-degrees-of-freedom-in-thermodynamics-with-simple-exampl, though I usually get better answers here.) In thermodynamics books I have read, I have often come across statements about how certain degrees of freedom are relevant only at certain temperatures (such as the vibration degrees of freedom of some molecules only being relevant in certain ranges), but I can't recall a convincing quantitative analysis of this. I tried to set up a simple example to explore this issue, but I'm unsure what goes wrong (though I haven't thought TOO much about it). We know that for a free particle in one dimension at finite temperature, the partition function is given by: $$Z(\beta)=\frac{L}{h}\int \text{d}p e^{-\beta p^2/2m}=\frac{L}{h}\sqrt{\frac{2 \pi m}{\beta}}$$ And then our expected energy is just: $$\langle E \rangle =\frac{1}{2}k_B T$$ Which we'd expect for a particle with one degree of freedom. On the other hand, the partition function for a particle in a 1-D harmonic potential is: $$Z(\beta)=\frac{1}{h}\int \text{d}x \int \text{d}p e^{-\beta(kx^2+p^2/m)/2}=\frac{2\pi}{\beta h\omega}$$ Which gives the expected energy: $$\langle E \rangle = k_BT$$ Here's my problem. If we take a limit of the spring constant to zero ($k \rightarrow 0$), doesn't that just correspond to a free particle? The average energies depend ONLY on temperature, so where exactly does this limit come in? Even though this is a relatively simple example, I encounter similar problems when I try to work out when and how degree of freedom fails to be "relevant" in the examples alluded to in textbooks.
Let $X_n(\Bbb{Z})$ be the simplicial complex whose vertex set is $\Bbb{Z}$ and such that the vertices $v_0,...,v_k$ span a $k$-simplex if and only if $\|v_i-v_j\| \le n$ for every $i,j$. Prove that $X_n(\Bbb{Z})$ is $n$-dimensional... no kidding, my maths is foundations (basic logic but not pedantic), calc 1 which I'm pretty used to work with, analytic geometry and basic linear algebra (by basic I mean matrices and systems of equations only Anyway, I would assume if removing $ fixes it, then you probably have an open math expression somewhere before it, meaning you didn't close it with $ earlier. What's the full expression you're trying to get? If it's just the frac, then your code should be fine This is my first time chatting here in Math Stack Exchange. So I am not sure if this is frowned upon but just a quick question: I am trying to prove that a proper subgroup of $\mathbb{Z}^n$ is isomorphic to $\mathbb{Z}^k$, where $k \le n$. So we must have $rank(A) = rank(\mathbb{Z}^k)$ , right? For four proper fractions $a, b, c, d$ X writes $a+ b + c >3(abc)^{1/3}$. Y also added that $a + b + c> 3(abcd)^{1/3}$. Z says that the above inequalities hold only if a, b,c are positive. (a) Both X and Y are right but not Z. (b) Only Z is right (c) Only X is right (d) Neither of them is absolutely right. Yes, @TedShifrin the order of $GL(2,p)$ is $p(p+1)(p-1)^2$. But I found this on a classification of groups of order $p^2qr$. There order of $H$ should be $qr$ and it is present as $G = C_{p}^2 \rtimes H$. I want to know that whether we can know the structure of $H$ that can be present? Like can we think $H=C_q \times C_r$ or something like that from the given data? When we say it embeds into $GL(2,p)$ does that mean we can say $H=C_q \times C_r$? or $H=C_q \rtimes C_r$? or should we consider all possibilities? When considering finite groups $G$ of order, $\|G\|=p^2qr$, where $p,q,r$ are distinct primes, let $F$ be a Fitting subgroup of $G$. Then $F$ and $G/F$ are both non-trivial and $G/F$ acts faithfully on $\bar{F}:=F/ \phi(F)$ so that no non-trivial normal subgroup of $G/F$ stabilizes a series through $\bar{F}$. And when $\|F\|=pr$. In this case $\phi(F)=1$ and $Aut(F)=C_{p-1} \times C_{r-1}$. Thus $G/F$ is abelian and $G/F \cong C_{p} \times C_{q}$. In this case how can I write G using notations/symbols? Is it like $G \cong (C_{p} \times C_{r}) \rtimes (C_{p} \times C_{q})$? First question: Then it is, $G= F \rtimes (C_p \times C_q)$. But how do we write $F$ ? Do we have to think of all the possibilities of $F$ of order $pr$ and write as $G= (C_p \times C_r) \rtimes (C_p \times C_q)$ or $G= (C_p \rtimes C_r) \rtimes (C_p \times C_q)$ etc.? As a second case we can consider the case where $C_q$ acts trivially on $C_p$. So then how to write $G$ using notations? There it is also mentioned that we can distinguish among 2 cases. First, suppose that the sylow $q$-subgroup of $G/F$ acts non trivially on the sylow $p$-subgroup of $F$. Then $q\|(p-1) and $G$ splits over $F$. Thus the group has the form $F \rtimes G/F$. A presentation $\langle S\mid R\rangle$ is a Dehn presentation if for some $n\in\Bbb N$ there are words $u_1,\cdots,u_n$ and $v_1,\cdots, v_n$ such that $R=\{u_iv_i^{-1}\}$, $\|u_i\|>\|v_i\|$ and for all words $w$ in $(S\cup S^{-1})^\ast$ representing the trivial element of the group one of the $u_i$ is a subword of $w$ If you have such a presentation there's a trivial algorithm to solve the word problem: Take a word $w$, check if it has $u_i$ as a subword, in that case replace it by $v_i$, keep doing so until you hit the trivial word or find no $u_i$ as a subword There is good motivation for such a definition here So I don't know how to do it precisely for hyperbolic groups, but if $S$ is a surface of genus $g \geq 2$, to get a geodesic representative for a class $[\alpha] \in \pi_1(S)$ where $\alpha$ is an embedded loop, one lifts it to $\widetilde{\alpha}$ in $\Bbb H^2$ by the locally isometric universal covering, and then the deck transformation corresponding to $[\alpha]$ is an isometry of $\Bbb H^2$ which preserves the embedded arc $\widetilde{\alpha}$ It has to be an isometry fixing a geodesic $\gamma$ with endpoints at the boundary being the same as the endpoints of $\widetilde{\alpha}$. Consider the homotopy of $\widetilde{\alpha}$ to $\gamma$ by straightline homotopy, but straightlines being the hyperbolic geodesics. This is $\pi_1(S)$-equivariant, so projects to a homotopy of $\alpha$ and the image of $\gamma$ (which is a geodesic in $S$) downstairs, and you have your desired representative I don't know how to interpret this coarsely in $\pi_1(S)$ @anakhro Well, they print in bulk, and on really cheap paper, almost transparent and very thin, and offset machine is really cheaper per page than a printer, you know, but you should be printing in bulk, its all economy of scale. @ParasKhosla Yes, I am Indian, and trying to get in some good masters progam in math. Algebraic graph theory is a branch of mathematics in which algebraic methods are applied to problems about graphs. This is in contrast to geometric, combinatoric, or algorithmic approaches. There are three main branches of algebraic graph theory, involving the use of linear algebra, the use of group theory, and the study of graph invariants.== Branches of algebraic graph theory ===== Using linear algebra ===The first branch of algebraic graph theory involves the study of graphs in connection with linear algebra. Especially, it studies the spectrum of the adjacency matrix, or the Lap... I can probably guess that they are using symmetries and permutation groups on graphs in this course. For example, orbits and studying the automorphism groups of graphs. @anakhro I have heard really good thing about Palka. Also, if you do not worry about little sacrifice of rigor (e.g. counterclockwise orientation based on your intuition, rather than, on winding numbers, etc.), Howie's Complex analysis is good. It is teeming with typos here and there, but you will be fine, i think. Also, thisbook contains all the solutions in appendix! Got a simple question: I gotta find kernel of linear transformation $F(P)=xP^{''}(x) + (x+1)P^{'''}(x)$ where $F: \mathbb{R}_3[x] \to \mathbb{R}_3[x]$, so I think it would be just $\ker (F) = \{ ax+b : a,b \in \mathbb{R} \}$ since only polynomials of degree at most 1 would give zero polynomial in this case @chandx you're looking for all the $G = P''$ such that $xG + (x+1)G' = 0$; if $G \neq 0$ you can solve the DE to get $G'/G = -x/(x+1) = -1 + 1/(x+1) \implies \ln G = -x + \ln(1+x) \implies G = (1+x)e^(-x) + C$ which is obviously not a polyonomial, so $G = 0$ and thus $P = ax + b$ could you suppose that $\operatorname{deg} P \geq 2$ and show that you wouldn't have nonzero polynomials? Sure.
I was assigned this problem as homework, and got it wrong. I have not gotten a chance to ask the teacher about the solution. Can someone tell me why I am wrong, and how to do this correctly? Let $\zeta_n = e^{2\pi i/n}.$ Prove that $\zeta_5 \notin \mathbb{Q}(\zeta_7).$ Since $\zeta_5 = e^{2\pi i/5}$ and $\mathbb{Q}(\zeta_7)$ is a ring, we have $\zeta_5 \cdot \zeta_7 = e^{2\pi i/5} \cdot e^{2\pi i/7} = e^{24\pi i/35} \in \mathbb{Q}(\zeta_7).$ This means $\left(e^{24\pi i/35}\right)^7 = 1.$ But $\left(e^{24\pi i/35}\right)^7 = e^{24\pi i/5} \neq 1.$ So $e^{24\pi i/35} \notin \mathbb{Q}(\zeta_7),$ a contradiction. So it must be that $e^{2\pi i/5} = \zeta_5 \notin \mathbb{Q}(\zeta_7).$
Can someone explain me this proof please: We have $\prod_{i\in I} \overline{A_i}=\cap_{i\in I} F_i$ where $F_i=\prod_{j\in I} B_j$ with $B_i=\overline{A_i}$ and $B_j=E_j$ if $j\neq i$. We have that $E\setminus F_i=\prod_{j\in I} U_j$ with $U_i=E_i\setminus \overline{A_i}$ and $U_j=E_j$ if $j\neq i$. Then $E\setminus F_i$ is open in $E$. So, $F_i$ is closed in $E$, from where $\prod_{i\in I} \overline{A_i}$ is closed in $E$ which contains $A$. Then we have $\overline{A}\subset \prod_{i\in I} \overline{A_i}$. In the other hand, let $x=(x_i)_{i\in I}\in \prod_{i\in I} \overline{A_i}$ and $U=\prod_{i\in I} U_i$ an elementary open of $E$ containing $x$. For each $i\in I,$ $U_i$ in an open of $E_i$ and there exists a finite subset $J$ of $I$ such that for all $i\in I\setminus J$, we have $U_i=E_i$. For all $i\in I$, we have $x_i\in \overline{A_i}\cap U_i,$ then there exist $a_i\in A_i\cap U_i$, where $a=(a_i)_{i\in I}\in A\cap U$. then we have $A\cap U\neq\emptyset.$ And then we have,$x\in \overline{A}$. Finally $\overline{A}=\prod_{i\in I}\overline{A_i}.$ Thank you.
Search Now showing items 21-27 of 27 Measurement of transverse energy at midrapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV (American Physical Society, 2016-09) We report the transverse energy ($E_{\mathrm T}$) measured with ALICE at midrapidity in Pb-Pb collisions at ${\sqrt{s_{\mathrm {NN}}}}$ = 2.76 TeV as a function of centrality. The transverse energy was measured using ... Elliptic flow of electrons from heavy-flavour hadron decays at mid-rapidity in Pb–Pb collisions at $\sqrt{s_{\rm NN}}= 2.76$ TeV (Springer, 2016-09) The elliptic flow of electrons from heavy-flavour hadron decays at mid-rapidity ($\|y\| < 0.7$) is measured in Pb–Pb collisions at $\sqrt{s_{\rm NN}}= 2.76$ TeV with ALICE at the LHC. The particle azimuthal distribution with ... Higher harmonic flow coefficients of identified hadrons in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Springer, 2016-09) The elliptic, triangular, quadrangular and pentagonal anisotropic flow coefficients for $\pi^{\pm}$, $\mathrm{K}^{\pm}$ and p+$\overline{\mathrm{p}}$ in Pb-Pb collisions at $\sqrt{s_\mathrm{{NN}}} = 2.76$ TeV were measured ... D-meson production in $p$–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV and in $pp$ collisions at $\sqrt{s}=7$ TeV (American Physical Society, 2016-11) The production cross sections of the prompt charmed mesons D$^0$, D$^+$, D$^{*+}$ and D$_{\rm s}^+$ were measured at mid-rapidity in p-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm NN}}=5.02$ TeV ... Azimuthal anisotropy of charged jet production in $\sqrt{s_{\rm NN}}=2.76$ TeV Pb–Pb collisions (Elsevier, 2016-02) This paper presents measurements of the azimuthal dependence of charged jet production in central and semi-central $\sqrt{s_{\rm NN}}=2.76$ TeV Pb–Pb collisions with respect to the second harmonic event plane, quantified ... Jet shapes in pp and Pb–Pb collisions at ALICE (Elsevier, 2016) The aim of this work is to explore possible medium modifications to the substructure of inclusive charged jets in Pb-Pb relative to proton-proton collisions by measuring a set of jet shapes. The set of shapes includes the ... Particle identification in ALICE: a Bayesian approach (Springer Berlin Heidelberg, 2016-05-25) We present a Bayesian approach to particle identification (PID) within the ALICE experiment. The aim is to more effectively combine the particle identification capabilities of its various detectors. After a brief explanation ...
Search Now showing items 31-40 of 165 Transverse sphericity of primary charged particles in minimum bias proton-proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV (Springer, 2012-09) Measurements of the sphericity of primary charged particles in minimum bias proton--proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV with the ALICE detector at the LHC are presented. The observable is linearized to be ... Measurement of inelastic, single- and double-diffraction cross sections in proton-proton collisions at the LHC with ALICE (Springer, 2013-06) Measurements of cross sections of inelastic and diffractive processes in proton--proton collisions at LHC energies were carried out with the ALICE detector. The fractions of diffractive processes in inelastic collisions ... Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Elsevier, 2016-03) The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ... Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Springer, 2017-06) The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ... Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV (Elsevier, 2016-03) Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ... Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV (Elsevier, 2016-07) The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ... Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC (Springer, 2017-01) The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ... Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC (Springer, 2017-06) We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ... Transverse Momentum Distribution and Nuclear Modification Factor of Charged Particles in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV (American Physical Society, 2013-02) The transverse momentum ($p_T$) distribution of primary charged particles is measured in non single-diffractive p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV with the ALICE detector at the LHC. The $p_T$ spectra measured ... Measurement of azimuthal correlations of D mesons and charged particles in pp collisions at $\sqrt{s}=7$ TeV and p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Springer, 2017-04) The azimuthal correlations of D mesons and charged particles were measured with the ALICE detector in pp collisions at $\sqrt{s}=7$ TeV and p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV at the Large Hadron Collider. ...
Let $N \unlhd G$. In the factor group $G/N$, the subgroup $N$ acts as identity element. Regard N as being collapsed to a single element, to the identity element. This collapsing of N together with the algebraic structure of G require that other subsets of G, namely the cosets of N, also collapse into a single element in the factor group. This collapsing can be seen in Fig. 15.1 underneath. Recall from Theorem 14.9: $\phi: G \to G/N$ defined by $\phi(g) =gN$ for $g \in G$ is a onto homomorphism. Figure 15.1 resembles Fig. 13.14. But in Fig. 15.1, the image group under the homomorphism is actually formed from G. We can view the "line" $G/N$ at the bottom of the figure as obtained by collapsing to a point each coset of N in another copy of G. Each point of $G/N$ thus corresponds to a whole vertical line segment in the shaded portion, representing a coset of N in G. Remember that multiplication of cosets in $G/N$ can be computed by multiplying in G, using any representative elements of the cosets as shown. (1.) Could someone please flesh out the intuition of the text? What's this picture trying to unfold? What are the nubs? I know what is group G, factor group G/N, identity e, homomorphism $\phi$. (2.) I'm perplexed by all the collapses and collapsing. Why are you authorized to do this? Why are you authorized to change your group like this? (3.) How is a subgroup N collapsed to a single element, to the identity element'? Why want this? Factor groups usually require $N \neq \{id\}$, because $G/\{id\} \cong G$. But this doesn't unfold anything about $G$? Did I muff something? (4.) What do the dotted lines mean? (5.) I can see this looks like Fig 13.14. But can someone please flesh out the similarities and differences? What are the nubs?
Gaussian transformation¶ This tutorial demonstrates the basic working principles of PennyLane for continuous-variable (CV) photonic devices. For more details about photonic quantum computing, the Strawberry Fields documentation is a great starting point. The quantum circuit¶ For this basic tutorial, we will consider a special subset of CV operations:the Gaussian transformations. We work with the following simple Gaussian circuit: What is this circuit doing? We begin with one wire (qumode) in the vacuum state. Note that we use the same notation $\|0\rangle$ for the initial state as the previous qubit tutorial. In a photonic CV system, this state is the vacuum state, i.e., the state with no photons in the wire. We displace the qumode. The displacement gate linearly shifts the state of the qumode in phase space. The vacuum state is centered at the origin in phase space, while the displaced state will be centered at the point $\alpha$. We rotate the qumode. This is another linear transformation in phase space, albeit a rotation (by angle $\phi$) instead of a displacement. Finally, we measure the mean photon number$\langle\hat{n}\rangle = \langle\hat{a}^\dagger \hat{a}\rangle$. This quantity, which tells us the average amount of photons in the final state, is proportional to the energy of the photonic system. The aim of this tutorial is to optimize the circuit parameters $(\alpha, \phi)$such that the mean photon number is equal to one. The rotation gate is actually a passive transformation, meaning that it does not change the energy of the system.The displacement gate is an active transformation, which modifies the energy of thephotonic system. Constructing the QNode¶ As before, we import PennyLane, as well as the wrapped version of NumPy provided by PennyLane: import pennylane as qmlfrom pennylane import numpy as np Next, we instantiate a device which will be used to evaluate the circuit.Because our circuit contains only Gaussian operations, we can make use of thebuilt-in default.gaussian device. dev_gaussian = qml.device("default.gaussian", wires=1) After initializing the device, we can construct our quantum node. As before, we use the qnode decorator to convert our quantum function(encoded by the circuit above) into a quantum node running on the default.gaussiandevice. @qml.qnode(dev_gaussian)def mean_photon_gaussian(mag_alpha, phase_alpha, phi): qml.Displacement(mag_alpha, phase_alpha, wires=0) qml.Rotation(phi, wires=0) return qml.expval(qml.NumberOperator(0)) Notice that we have broken up the complex number $\alpha$ into two realnumbers mag_alpha and phase_alpha, which form a polar representation of$\alpha$. This is so that the notion of a gradient is clear and well-defined. Optimization¶ As in the qubit rotation tutorial, let’s now use one of the built-in PennyLane optimizers in order to optimize the quantum circuit towards the desired output. We want the mean photon number to be exactly one, so we will use a squared-difference cost function: def cost(params): return (mean_photon_gaussian(params[0], params[1], params[2]) - 1.0) ** 2 At the beginning of the optimization, we choose arbitrary small initial parameters: init_params = [0.015, 0.02, 0.005]print(cost(init_params)) Out: 0.9995500506249999 When the gate parameters are near to zero, the gates are close to the identity transformation, which leaves the initial state largely unchanged. Since the initial state contains no photons, the mean photon number of the circuit output is approximately zero, and the cost is close to one. Note We avoided initial parameters which are exactly zero because that corresponds to a critical point with zero gradient. Now, let’s use the GradientDescentOptimizer, and update the circuitparameters over 100 optimization steps. # initialise the optimizeropt = qml.GradientDescentOptimizer(stepsize=0.1)# set the number of stepssteps = 20# set the initial parameter valuesparams = init_paramsfor i in range(steps): # update the circuit parameters params = opt.step(cost, params) print("Cost after step {:5d}: {:8f}".format(i + 1, cost(params)))print("Optimized mag_alpha:{:8f}".format(params[0]))print("Optimized phase_alpha:{:8f}".format(params[1]))print("Optimized phi:{:8f}".format(params[2])) Out: Cost after step 1: 0.999118Cost after step 2: 0.998273Cost after step 3: 0.996618Cost after step 4: 0.993382Cost after step 5: 0.987074Cost after step 6: 0.974837Cost after step 7: 0.951332Cost after step 8: 0.907043Cost after step 9: 0.826649Cost after step 10: 0.690812Cost after step 11: 0.490303Cost after step 12: 0.258845Cost after step 13: 0.083224Cost after step 14: 0.013179Cost after step 15: 0.001001Cost after step 16: 0.000049Cost after step 17: 0.000002Cost after step 18: 0.000000Cost after step 19: 0.000000Cost after step 20: 0.000000Optimized mag_alpha:0.999994Optimized phase_alpha:0.020000Optimized phi:0.005000 The optimization converges after about 20 steps to a cost function value of zero. We observe that the two angular parameters phase_alpha and phido not change during the optimization. Only the magnitude of the complexdisplacement $\|\alpha\|$ affects the mean photon number of the circuit. Continue on to the next tutorial, Plugins and Hybrid computation, to learn how to utilize the extensive plugin ecosystem of PennyLane, build continuous-variable (CV) quantum nodes, and to see an example of a hybrid qubit-CV-classical computation using PennyLane. Total running time of the script: ( 0 minutes 0.193 seconds)
I have a question regarding the proof of the time-independent Schrödinger equation. So if we have a time-Independent Hamiltonian, we can solve the Schrödinger equation by adopting separation of variables method: we write our general solution as $ \psi(r,t) = \psi(r)*f(t)$ and we get to the two equation : one for the f(t) and one for the $\psi(r)$. $$ f(t)=e^{-\frac{i}{\hbar}Ht} $$ and $$ H\psi(r) = E\psi(r) $$ Books in general refer to the second equation as the TISE and it is seen as an eigenvalue problem for the hamiltonian, in order to find the stationary states. Now what i don't understand is why we see that equation as an eigenvalue problem for the hamiltonian, since we have a wave-function $\psi(x)$ which is supposed to be eigenstate for H. So if $\psi(x)$ is eigenstate for H, it means that H is diagonal in coordinate basis, but i know it's not true since H has the term $\frac{P^2}{2m}$ in it, which is not diagonal in coordinate basis. Where am i wrong?? Thank you very much Start with the time-dependent Schrödinger equation $$ \hat H \Psi(r, t) = i \hbar \partial_t \Psi (r, t). $$ Using the ansatz $\Psi(r,t) = \psi(r) f(t)$ yields $$ f(t) \hat H \psi(r) = i \hbar \psi(r) \partial_t f(t)$$ and, via the standard separation of variables trick, $$ i\hbar\frac{\dot f(t)}{f(t)} = \text{const} = \frac{\hat H\psi(r)}{\psi(r)}; $$ the two sides are equal, but the LHS depends only on $t$ while the RHS depends only on $r$, so they each have to be constant. Let us call this constant $E$. Then, for the time-dependent part, we get $$ \dot f(t) = -i \frac{E}{\hbar} f(t) $$ Which is manifestly solved by $$ f(t) = \exp\left\{-i \frac{E}{\hbar} t\right\}. $$ For the spatial part, we find the time-independent Schrödinger equation $$ \hat H \psi(r) = E \psi(r), $$ which as you observe can be viewed as an eigenvalue equation for $\hat H$. This motivates the choice of name for $E$: Physically, the eigenvalue of the Hamiltonian is the energy.
The problem goes as follows: Let $r_{n}$ be an enumeration of the rationals. Using Fubini-Tonelli, show that: $$ F(x) = \sum_{n \geq 1} \frac{1}{n^{2}} \frac{1}{\|r_{n} - x \|^{1/2}}$$ is finite almost everywhere. Attempt at solution: My idea is that I want to show the $F(x)$ is integrable. Then using Fubini-Tonelli, taking the sum outside of the integral: $$ \int_{\mathbb{R}}F(x) dx = \sum_{n \geq 1} \frac{1}{n^{2}} \int_{\mathbb{R}}\frac{1}{\|r_{n} - x \|^{1/2}} dx$$ $$ = \sum_{n \geq 1} \frac{1}{n^{2}} \int_{\mathbb{R}}\frac{1}{\|x \|^{1/2}}dx $$by traslation invariance. Now I know that $\frac{1}{\|x \|^{1/2}}$ locally integrable, but not integrable on all of $\mathbb{R}$. I was also thinking of defining $f_{n}(x) = \frac{1}{\|x \|^{1/2}}$ when $ x$ in $(-n,n)$ and $0$ elsewhere. Then let $$F_{N}(x) = \sum_{n=1}^{N} \frac{1}{n^{2}} f(x-r_{n}) $$ My reasoning starts to fall apart here and I can't seem to make it work. Any hints or solutions are greatly appreciated!
a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached). b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct. However, your calculation of amplitude is not correct. Peak-to-trough is twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly. CALCULATION OF AMPLITUDE OF OSCILLATIONS 1. Equation of Motion Method First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision. Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$ 2. Energy Method Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$. At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$. At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$. By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1. (c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it. Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal. Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result. The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate. Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$ We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off. The total energy at P is the same as at U, so from above ( Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
From my perspective, this is easiest to tackle from a straight forward energy balance. The energy balance determines how energy, heat and work are distributed when you go from State 1 to State 2. Thermodynamic States Just to emphasize this, a state is a fixed point where all properties are set in stone. So state 1 can be an unreacted system, state 2 a reacted system... State 1 can be the inlet of a compressor at a certain T and P and state 2 can be the outlet at a different T and P. State 1 can be a bomb calorimeter before combustion (at a T and V) and state 2 can be the calorimeter after combustion (same V, different T and consequently, different P). You can do an energy balance between any two states, and it will tell you directly what is happening to the energy, heat and work of your system. It is a beautiful thing actually. Energy Balance For an unreacting and open system at steady state $$ \left[ U + \frac{P}{\rho} + KE + PE + W + Q \right]^{\rm State_1} = \left[ U + \frac{P}{\rho} + KE + PE + W + Q \right]^{\rm State_2} $$ Where $U$ is internal energy, $\frac{P}{\rho}$ is flow energy, $KE$ is kinetic energy, $PE$ is potential energy, $Q$ is heat and $W$ is work. I should say each variable is really the summation of all sources contributing. There could be several sources producing heat, same with work. There could be several streams entering and several exiting. Traditionally $State_1$ and $State_2$ represent IN and OUT respectively. If there is flow, then $\frac{P}{\rho}$ is present, and we typically denote it as $Pv$. Thus on both sides we have $U + Pv$ which we write as $H$. It is typical to say changes in $KE$ and $PE$ are negligible. I will also put IN and OUT in place of States Our energy balance is now $$ \left[ H + W + Q \right]^{\rm IN} = \left[ H + W + Q \right]^{\rm OUT} $$ At this point the actual problem naturally falls out. If you want to know what the change in enthalpy is rearrange the sides $$ \Delta H = W_{IN} - W_{OUT} + Q_{IN} - Q_{OUT} $$ It is entirely possible for $\Delta H$ to be positive or negative for either positive $\Delta Q$ or $-\Delta Q$ so I don't see how you can draw any conclusions about $\Delta H$ with respect to what heat is doing. If there are no work interactions, this simplifies the equation $$ \Delta H = Q_{IN} - Q_{OUT} $$With no work interactions the system would need to be exothermic i.e., lose heat for $\Delta H$ to be negative. Let me know if I made a mistake in my derivations, It would not be the first time. But remember all of the assumptions that led us to this point!
Expected Value The expected value is a weighted average of all possible values in a data set. Learning Objectives Compute the expected value and explain its applications and relationship to the law of large numbers Key Takeaways Key Points The expected value refers, intuitively, to the value of a random variable one would “expect” to find if one could repeat the random variable process an infinite number of times and take the average of the values obtained. The intuitive explanation of the expected value above is a consequence of the law of large numbers: the expected value, when it exists, is almost surely the limit of the sample mean as the sample size grows to infinity. From a rigorous theoretical standpoint, the expected value of a continuous variable is the integral of the random variable with respect to its probability measure. Key Terms random variable: a quantity whose value is random and to which a probability distribution is assigned, such as the possible outcome of a roll of a die integral: the limit of the sums computed in a process in which the domain of a function is divided into small subsets and a possibly nominal value of the function on each subset is multiplied by the measure of that subset, all these products then being summed weighted average: an arithmetic mean of values biased according to agreed weightings In probability theory, the expected value refers, intuitively, to the value of a random variable one would “expect” to find if one could repeat the random variable process an infinite number of times and take the average of the values obtained. More formally, the expected value is a weighted average of all possible values. In other words, each possible value the random variable can assume is multiplied by its assigned weight, and the resulting products are then added together to find the expected value. The weights used in computing this average are the probabilities in the case of a discrete random variable (that is, a random variable that can only take on a finite number of values, such as a roll of a pair of dice), or the values of a probability density function in the case of a continuous random variable (that is, a random variable that can assume a theoretically infinite number of values, such as the height of a person). From a rigorous theoretical standpoint, the expected value of a continuous variable is the integral of the random variable with respect to its probability measure. Since probability can never be negative (although it can be zero), one can intuitively understand this as the area under the curve of the graph of the values of a random variable multiplied by the probability of that value. Thus, for a continuous random variable the expected value is the limit of the weighted sum, i.e. the integral. Simple Example Suppose we have a random variable [latex]\text{X}[/latex], which represents the number of girls in a family of three children. Without too much effort, you can compute the following probabilities: [latex]\text{P}[\text{X}=0] = 0.125 \\ \text{P}[\text{X}=1] = 0.375 \\ \text{P}[\text{X}=2] = 0.375 \\ \text{P}[\text{X}=3] = 0.125[/latex] The expected value of [latex]\text{X}, \text{E}[\text{X}][/latex], is computed as: [latex]\displaystyle {\begin{align} \text{E}[\text{X}] &= \sum_{\text{x}=0}^3\text{xP}[\text{X}=\text{x}] \\ &=0\cdot 0.125 + 1\cdot0.375 + 2\cdot 0.375 + 3\cdot 0.125 \\ &= 1.5\end{align}}[/latex] This calculation can be easily generalized to more complicated situations. Suppose that a rich uncle plans to give you $2,000 for each child in your family, with a bonus of $500 for each girl. The formula for the bonus is: [latex]\text{Y} = 1000 + 500\text{X}[/latex] What is your expected bonus? [latex]\displaystyle {\begin{align} \text{E}[1000 + 500\text{X}] &= \sum_{\text{x}=0}^3 (1000 + 500\text{x})\text{P}[\text{X}=\text{x}] \\ &=1000\cdot0.125 +1500\cdot0.375 + 2000\cdot0.375 + 2500 \cdot 0.125 \\ &= 1750 \end{align}}[/latex] We could have calculated the same value by taking the expected number of children and plugging it into the equation: [latex]\text{E}[1000+500\text{X}] = 1000 + 500\text{E}[\text{X}][/latex] Expected Value and the Law of Large Numbers The intuitive explanation of the expected value above is a consequence of the law of large numbers: the expected value, when it exists, is almost surely the limit of the sample mean as the sample size grows to infinity. More informally, it can be interpreted as the long-run average of the results of many independent repetitions of an experiment (e.g. a dice roll). The value may not be expected in the ordinary sense—the “expected value” itself may be unlikely or even impossible (such as having 2.5 children), as is also the case with the sample mean. Uses and Applications To empirically estimate the expected value of a random variable, one repeatedly measures observations of the variable and computes the arithmetic mean of the results. If the expected value exists, this procedure estimates the true expected value in an unbiased manner and has the property of minimizing the sum of the squares of the residuals (the sum of the squared differences between the observations and the estimate). The law of large numbers demonstrates (under fairly mild conditions) that, as the size of the sample gets larger, the variance of this estimate gets smaller. This property is often exploited in a wide variety of applications, including general problems of statistical estimation and machine learning, to estimate (probabilistic) quantities of interest via Monte Carlo methods. The expected value plays important roles in a variety of contexts. In regression analysis, one desires a formula in terms of observed data that will give a “good” estimate of the parameter giving the effect of some explanatory variable upon a dependent variable. The formula will give different estimates using different samples of data, so the estimate it gives is itself a random variable. A formula is typically considered good in this context if it is an unbiased estimator—that is, if the expected value of the estimate (the average value it would give over an arbitrarily large number of separate samples) can be shown to equal the true value of the desired parameter. In decision theory, and in particular in choice under uncertainty, an agent is described as making an optimal choice in the context of incomplete information. For risk neutral agents, the choice involves using the expected values of uncertain quantities, while for risk averse agents it involves maximizing the expected value of some objective function such as a von Neumann-Morgenstern utility function. Standard Error The standard error is the standard deviation of the sampling distribution of a statistic. Learning Objectives Paraphrase standard error, standard error of the mean, standard error correction and relative standard error. Key Takeaways Key Points The standard error of the mean (SEM) is the standard deviation of the sample -mean’s estimate of a population mean. SEM is usually estimated by the sample estimate of the population standard deviation (sample standard deviation) divided by the square root of the sample size. The standard error and the standard deviation of small samples tend to systematically underestimate the population standard error and deviations. When the sampling fraction is large (approximately at 5% or more), the estimate of the error must be corrected by multiplying by a ” finite population correction” to account for the added precision gained by sampling close to a larger percentage of the population. If values of the measured quantity [latex]\text{A}[/latex] are not statistically independent, an unbiased estimate of the true standard error of the mean may be obtained by multiplying the calculated standard error of the sample by the factor [latex]\text{f}[/latex]. The relative standard error (RSE) is simply the standard error divided by the mean and expressed as a percentage. Key Terms regression: An analytic method to measure the association of one or more independent variables with a dependent variable. correlation: One of the several measures of the linear statistical relationship between two random variables, indicating both the strength and direction of the relationship. Quite simply, the standard error is the standard deviation of the sampling distribution of a statistic. The term may also be used to refer to an estimate of that standard deviation, derived from a particular sample used to compute the estimate. For example, the sample mean is the usual estimator of a population mean. However, different samples drawn from that same population would in general have different values of the sample mean. The standard error of the mean (i.e., of using the sample mean as a method of estimating the population mean) is the standard deviation of those sample means over all possible samples (of a given size) drawn from the population. Secondly, the standard error of the mean can refer to an estimate of that standard deviation, computed from the sample of data being analyzed at the time. In regression analysis, the term “standard error” is also used in the phrase standard error of the regression to mean the ordinary least squares estimate of the standard deviation of the underlying errors. Standard Error of the Mean As mentioned, the standard error of the mean (SEM) is the standard deviation of the sample-mean’s estimate of a population mean. It can also be viewed as the standard deviation of the error in the sample mean relative to the true mean, since the sample mean is an unbiased estimator. SEM is usually estimated by the sample estimate of the population standard deviation (sample standard deviation) divided by the square root of the sample size (assuming statistical independence of the values in the sample): [latex]\displaystyle \text{SE}_\bar{\text{x}} = \frac{\text{s}}{\sqrt{\text{n}}}[/latex] where: [latex]\text{s}[/latex] is the sample standard deviation (i.e., the sample-based estimate of the standard deviation of the population), and [latex]\text{n}[/latex] is the size (number of observations) of the sample. This estimate may be compared with the formula for the true standard deviation of the sample mean: [latex]\displaystyle \text{SD}_\bar{\text{x}} = \frac{\sigma}{\sqrt{\text{n}}}[/latex] The standard error and the standard deviation of small samples tend to systematically underestimate the population standard error and deviations. This is due to the fact that the standard error of the mean is a biased estimator of the population standard error. Decreasing the uncertainty in a mean value estimate by a factor of two requires acquiring four times as many observations in the sample. Or decreasing standard error by a factor of ten requires a hundred times as many observations. Standard Error Versus Standard Deviation The standard error and standard deviation are often considered interchangeable. However, while the mean and standard deviation are descriptive statistics, the mean and standard error describe bounds on a random sampling process. Despite the small difference in equations for the standard deviation and the standard error, this small difference changes the meaning of what is being reported from a description of the variation in measurements to a probabilistic statement about how the number of samples will provide a better bound on estimates of the population mean. Put simply, standard error is an estimate of how close to the population mean your sample mean is likely to be, whereas standard deviation is the degree to which individuals within the sample differ from the sample mean. Correction for Finite Population The formula given above for the standard error assumes that the sample size is much smaller than the population size, so that the population can be considered to be effectively infinite in size. When the sampling fraction is large (approximately at 5% or more), the estimate of the error must be corrected by multiplying by a “finite population correction” to account for the added precision gained by sampling close to a larger percentage of the population. The formula for the FPC is as follows: [latex]\displaystyle \text{FPC} = \sqrt{\frac{\text{N}-\text{n}}{\text{N}-1}}[/latex] The effect of the FPC is that the error becomes zero when the sample size [latex]\text{n}[/latex] is equal to the population size [latex]\text{N}[/latex]. Correction for Correlation In the Sample If values of the measured quantity [latex]\text{A}[/latex] are not statistically independent but have been obtained from known locations in parameter space [latex]\text{x}[/latex], an unbiased estimate of the true standard error of the mean may be obtained by multiplying the calculated standard error of the sample by the factor [latex]\text{f}[/latex]: [latex]\displaystyle \text{f}=\sqrt{\frac{1+\rho}{1-\rho}}[/latex] where the sample bias coefficient [latex]\rho[/latex] is the widely used Prais-Winsten estimate of the autocorrelation-coefficient (a quantity between [latex]-1[/latex] and [latex]1[/latex]) for all sample point pairs. This approximate formula is for moderate to large sample sizes and works for positive and negative [latex]\rho[/latex] alike. Relative Standard Error The relative standard error (RSE) is simply the standard error divided by the mean and expressed as a percentage. For example, consider two surveys of household income that both result in a sample mean of $50,000. If one survey has a standard error of $10,000 and the other has a standard error of $5,000, then the relative standard errors are 20% and 10% respectively. The survey with the lower relative standard error has a more precise measurement since there is less variance around the mean. In fact, data organizations often set reliability standards that their data must reach before publication. For example, the U.S. National Center for Health Statistics typically does not report an estimate if the relative standard error exceeds 30%.
Stokes theorem for smooth differential forms is well-known. If $\alpha$ is a smooth differential $n$-form defined on an $(n+1)$-dimensional compact oriented manifold with boundary, then we have $$\int_Md\alpha=\int_{\partial M}\alpha.$$ In physics, one considers a generalized form of this equation, where the manifold need not be compact, and the differential form need not be smooth everywhere; it may have poles. For example, in electrostatics one sees a 2-form field generated by a point charge $Q$ $$E=\dfrac{Q}{4\pi\epsilon_0r^3}\left(x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy\right)=\dfrac{Q}{4\pi\epsilon_0r^2}d\Omega_{S^2},$$ where $d\Omega_{S^2}$ is the volume form of the sphere. I suppose this must be understood as a distribution-valued form, because the physicist computes $$dE=\frac{Q}{4\pi\epsilon_0}\delta^3(r)\,d\text{vol}_{\mathbb{R}^3},$$ The physics discussions I have seen justify this computation by saying it is necessary to make the Gauss's law hold. But where I come from, we don't conspire to ensure theorems hold because we like them, but rather prove from the definitions that theorems follow logically. Could you say how this theory of distribution valued differential forms looks, and state a version of Stokes' theorem for it? I would also accept a reference.
a) Correct. The new equilibrium position occurs at distance $x=mg/k$ below the initial position (where the collision occurred). Alternatively it is $X=(M+m)g/k$ below the top of the unloaded spring (with no cymbal attached). b) The new angular frequency of oscillations can be written down without any calculation : $\omega=\sqrt{\frac{k}{m+M}}$. Correct. However, your calculation of amplitude is not correct. Peak-to-trough is twice the amplitude - ie $2A$. Amplitude is the distance of a peak or trough from the new equilibrium position. Peaks and troughs occur when kinetic energy is zero - ie "when it is not oscillating" as you put it. See Energy Method below. You had the right idea, but your calculation neglects the fact that the cymbal is not released from rest - it has some initial kinetic energy due to the collision. Also your solution does not deal with elastic energy correctly. CALCULATION OF AMPLITUDE OF OSCILLATIONS 1. Equation of Motion Method First work out the speed of the cymbal immediately after the inelastic collision. From conservation of momentum this is $v=\frac{m}{m+M}u$ where $u=\sqrt{2gh}$ is the speed of $m$ immediately before the collision. Next, the oscillation about the new equilibrium position can be described by the equation of motion $$\xi=A\sin(\omega t+\phi), \dot \xi=\omega A\cos(\omega t+\phi)$$ where $\phi$ is some unknown phase angle. Suppose the collision occurs at $t=0, \xi=x$ with $\dot \xi=v$. (It does not matter what time it occurs, we will only get a different phase angle $\phi$.) Then $$x=A\sin\phi, v=\omega A\cos\phi$$ $$A^2=x^2+(\frac{v}{\omega})^2=(\frac{mg}{k})^2+\frac{2ghm^2}{k(M+m)}$$ 2. Energy Method Let point U be the top of the spring when it has no load. I shall measure all potential and elastic energies from this point. Let P be the starting equilibrium position where the load is $M$, let O be the new equilibrium position when the load is $M+m$, and Q be the lowest point of the subsequent oscillations. The amplitude of oscillations is $OQ=A$. Distance $OP=x$ as already calculated. Other distances are $UP=UO-PO=X-x$ and $UQ=UO+OQ=X+A$. At P there is kinetic energy of $\frac12 (m+M)v^2$. The spring is compressed by distance $UP=X-x$ so the elastic energy stored in the spring is $\frac12 k(X-x)^2$ where $X=(M+m)g/k$ as above. Gravitational PE at P relative to U is $-(m+M)g(X-x)=-kX(X-x)$. At Q there is no KE, and gravitational PE is $-(M+m)g(X+A)=-kX(X+A)$. The elastic energy stored here is $\frac12 k(X+A)^2$. By the conservation of energy, the total energy at P is the same as that at Q. Therefore $$\frac12 (m+M)v^2-kX(X-x)+\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2+2AX+A^2)-kX(X+A)$$ $$(m+M)v^2=k(A^2-x^2)$$ $$A^2=x^2+\frac{(m+M)v^2}{k}=x^2+(\frac{v}{\omega})^2$$ as found using Method 1. (c) This part of the question is not clear. I assume that the block moves down with the cymbal as in (a) and (b). However, because it is not fixed to the cymbal it can separate from the cymbal when it rises from Q back above O. The block and cymbal do not 'stick together' because of the collision, they only 'move together' after it. Separation occurs when the downward acceleration of the SHM becomes greater than $g$. Gravity is the only force holding the block in contact with the cymbal, so when gravity is no longer able to supply the required restoring force on the block, it leaves contact with the cymbal. Suppose separation occurs at point R which has displacement $\xi$ above equilibrium position O. The downward acceleration at point R is $\omega^2 \xi=g$. So $$\xi=\frac{g}{\omega^2}=\frac{(m+M)g}{k}=X$$ This means that the lift-off point is always at the top of the spring when it has no load (R=U), whatever the values of $m, M, h, k$. This is a surprising result. The explanation is that at this instant the only force acting on the cymbal and block is gravity, because the spring is no longer compressed or stretched so it exerts no force. Both cymbal and block are in 'free fall' so the force between them is zero. Just before this instant both cymbal and block are moving upwards but accelerating downwards at just less than $g$ because of a small upward push from the spring. After this instant the cymbal is being accelerated downwards at slightly more than $g$ because of a small pull from the spring. But the spring does not pull down on the block so the block is still accelerating downwards at $g$. There is relative acceleration, so the block and cymbal separate. Note that if $A\lt X$ then the cymbal does not rebound to the relaxed position of the spring at U, then there is no lift-off. The condition for lift-off is that $A \gt X$. Substituting from the equations for $A^2$ and $X$ given above we get $$A^2 \gt X^2$$ $$\frac{m^2g^2}{k^2}+\frac{2ghm^2}{k(M+m)} \gt \frac{(m+M)^2g^2}{k^2}$$ $$1+\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}$$ $$\frac{2kh}{g(M+m)} \gt \frac{(m+M)^2}{m^2}-1=\frac{m^2+2mM+M^2-m^2}{m^2}=\frac{M(2m+M)}{m^2}$$ $$h \gt \frac{M(M+m)(M+2m)g}{2km^2}$$ We can also find the maximum height $H$ above R=U reached by the mass $m$ after lift-off. The total energy at P is the same as at U, so from above ( Energy Method) we have $$\frac12 (m+M)v^2+\frac12 k(X-x)^2-kX(X-x)=\frac12 (m+M)V^2$$ $$\frac12 (m+M)(v^2-V^2)=k(X^2-xX)-\frac12 k(X^2-2xX+x^2)=\frac12 k(X^2-x^2)$$ $$v^2-V^2=\frac{k}{(m+M)}(X-x)(X+x)$$ From earlier results we have $$v=\frac{m}{m+M}u=\mu u=\mu \sqrt{2gh}$$ $$v^2=2gh \mu^2$$ $$X-x=\frac{(m+M)g}{k}-\frac{mg}{k}=\frac{Mg}{k}$$ $$X+x=\frac{(m+M)g}{k}+\frac{mg}{k}=\frac{(m+M+m)g}{k}$$ The block rises to a height $H$ above U given by $2gH=V^2$. Making the above substitutions into the equation for $v^2-V^2$ we get $$2g(h\mu^2-H)=\frac{k}{m+M}\frac{Mg}{k} \frac{(m+M+m)g}{k}=\frac{Mg^2}{k}(1+\mu)$$ $$H=h\mu^2-\frac{Mg}{2k}(1+\mu)$$
Consider the problem of finding the maximum number of knights that can be placed on a chessboard without two of them attacking each other. The answer is 32: it's not too difficult to find a perfect matching (the graph induced by knight moves is bipartite, and there's a perfect matching for a 4 × 4 board), which is obviously a minimum edge cover. It's also not difficult to prove that the answer is $\left\lceil \frac{mn}{2} \right\rceil$ for an $m \times n$ chessboard whenever $m,n \geq 3$: it suffices to show matchings for $3 \leq m,n \leq 6$ and do a bit of induction footwork. On the other hand, if the chessboard were toroidal and $m, n$ were even, the proof wouldn't even require showing a matching for small boards: the map $(x, y) \rightarrow (x+1, y+2)$ has only even-length cycles so there must be a perfect matching. Is there any equivalent for rectangular chessboards, i.e. is there any simpler way to show that for sufficiently large $m, n$ there is always a perfect matching of the chessboard? For large boards, the rectangular board and the toroidal board are almost equivalent in the sense that the fraction of missing edges goes to zero, but I'm not aware of any theoretical results that would guarantee a perfect matching in that case. What if, instead of jumping $(1, 2)$ in either direction, a knight jumped $(2, 3)$ squares in either direction? Or, for that matter, $(p, q)$ squares, with $p+q$ odd and $p, q$ coprime? If there is a simple way of proving that the answer is $\left\lceil \frac{mn}{2} \right\rceil$ for sufficiently large $m, n$ (say, $m, n \geq C(p, q)$), what does $C(p, q)$ look like?
1. Not all states produced by $\text{CNOT} \; (H \otimes I)$ are entangled. For the first part of your question: no, not all states which arise as the output of $\text{CNOT} \; (H \otimes I)$ are entangled. Specifically, you can consider$$\begin{align}\|\psi\rangle \;&=\; (H \otimes I) \; \text{CNOT} \;\Bigl[ \tfrac{1}{\sqrt 2}\bigl( \|00\rangle + \|10\rangle \bigr) \Bigr] \\\;&=\; (H \otimes I) \Bigl[ \tfrac{1}{\sqrt 2}\bigl( \|00\rangle + \|11\rangle \bigr) \Bigr] \\\;&=\; \tfrac{1}{2} \bigl( \|00\rangle + \|01\rangle + \|10\rangle - \|11\rangle\bigr),\end{align}$$which is a maximally entangled state (notice from the second line above that it differs from a Bell state by a local unitary). However, by construction, if you apply $\text{CNOT} \; (H \otimes I)$ to $\|\psi\rangle$, you will get back out the state$$ \text{CNOT} \; (H \otimes I) \;\|\psi\rangle \;=\; \tfrac{1}{\sqrt 2}\bigl( \|00\rangle + \|10\rangle \bigr) \;=\; \|+\rangle\otimes\|0\rangle\;,$$which has no entanglement at all. Even if you're interested only in inputs which are product states, we can see that the circuit maps $\|+\rangle \otimes \|0\rangle$ to another product state — specifically, $\|0\rangle\|0\rangle$. 2. All maximally entangled two-qubit states can be easily described by a similar circuit to $\text{CNOT} \; (H \otimes I)$ acting on the standard basis. For the second part of your question: if you allow yourself arbitrary single-qubit unitaries, then for any maximally entangled two-qubit state $\|\psi\rangle$, you can certainly construct a circuit which constructs $\|\psi\rangle$ from standard basis states, and which allows you easily to see that the state is maximally entangled. Every two-qubit state has a Schmidt decomposition,$$ \|\psi\rangle \;=\; u_0\|\alpha_0\rangle\|\beta_0\rangle \;+\; u_1\|\alpha_1\rangle\|\beta_1\rangle \;,$$where $u_0 \geqslant u_1 \geqslant 0$ — in particular, $u_1 = 0$ for $\|\psi\rangle$ a product state — and where $\{ \|\alpha_0\rangle, \|\alpha_1\rangle \}$ and $\{ \|\beta_0\rangle, \|\beta_1\rangle \}$ are each orthonormal bases for $\mathbb C^2$. In order for $\|\psi\rangle$ to be maximally entangled, we need $u_0 = u_1 = \tfrac{1}{\sqrt 2}$. Consider single-qubit unitary matrices $A$ and $B$, such that$$\begin{alignat}{4} A \|x\rangle \;&=\; \|\alpha_x\rangle &\quad& \text{for $x \in \{0,1\}$}, \\[1ex] B \|y\rangle \;&=\; \|\beta_y\rangle && \text{for $y \in \{0,1\}$};\end{alignat}$$then we can describe $\|\psi\rangle = (A \otimes B)\;\text{CNOT}\;(H \otimes I)\;\|0\rangle\|0\rangle$, that is, the effect of applying $(A \otimes B)$ to the Bell state $\|\Phi^+\rangle = \tfrac{1}{\sqrt 2}\bigl( \|00\rangle + \|11\rangle )$.Furthermore, it's not hard to show that any standard basis state is mapped by that circuit to some maximally entangled state similar to $\|\psi\rangle$, and that any circuit of this form (whatever $A$ and $B$ might be) maps any standard basis state to a maximally entangled two-qubit state.
Consider a system of $N$ spin $1/2$ particles. Assume the spin is the only degree of freedom and hence there is no spatial component. Then the dimension of the Hilbert space in this case is $2^N$. This follows since in this case we have $j_1 = j_2 = \ldots = j_N = 1/2. $ And the dimension of the product space is $$ (2j_1 + 1) (2j_2 + 1) \ldots (2j_N + 1). $$ Now suppose these particles obey Fermi-Dirac statistics because they are identical fermions. I'm now being asked to determine the dimension of the vectorspace. I don't really understand the answer. The answer my professor gave is: According to Pauli exclusion principle, no two particles can occupy the same quantum state. In this case we have only two states (spin up, spin down). So there are only two options: $$ N=1 : \qquad \mid + \rangle, \quad \mid - \rangle \qquad \dim = 2 $$ $$ N=2 : \qquad \frac{1}{\sqrt{2}} \bigg( \mid + \rangle \mid - \rangle \ - \ \mid - \rangle \mid + \rangle \bigg) \qquad \dim = 1. $$ I don't understand how one derives these dimensions. I don't see how Pauli principle can be important to determine the dimension. The particles will just fill the lowest energy levels (with no more than two occupying the same state). That doesn't mean there are only two states? They would also be characterized by the quantum number $n$ for example?
In basic chemistry, we are taught that an atom has roughly the same number of neutrons and number of protons, this doesn't seems to hold for larger atoms, but it is always roughly proportional (i.e. you seldom find an atom with 100 protons but 1 neutron, that just does not happen). Why is that? You probably know that the electrons in atoms occupy a series of energy levels, the $1s$, $2s$, $2p$, etc orbitals. Although the structure of nuclei is complicated, basically the same idea applies to nuclei as well as atoms. This happens because you can't put more than one fermion into the same quantum state. The electrons in atoms are fermions, and so are the protons and neutrons in nuclei. However because the proton and neutron are different particles you can put a proton and neutron into the same energy state. Well, that's not really true but the point is that the energy states of the protons and neutrons overlap. Let me try to show this with a diagram: Suppose we have eight nucleons and we arrange them, one per energy level, as shown in the diagram (a). This has some total energy $E_a$. Suppose now that we can pair up the nucleons as shown in (b). This obviously has a lower energy $E_b$. But remember that protons and neutrons can share energy levels because they are different particles. So if we have a nucleus with all protons or all neutrons it would look like (a), while an equal number of protons and neutrons would look like (b) and therefore have a lower energy. But protons and neutrons can interconvert by beta or positron emission or electron capture. So if we started with eight protons as in (a) four of those protons can convert to neutrons to give the energy levels in (b) and this will overall reduce the energy. The nucleus with equal numbers of protons and neutrons will be the one with the lowest energy. Now this is very oversimplified because protons and neutrons don't share exactly the same energy levels and the energy levels aren't equally spaced. But it gives you a feel for why we have approximately equal numbers of protons and neutrons in a nucleus. A phenomenological model can be described in terms of the liquid drop model and semi-empirical mass formula. This ascribes a nucleus a binding energy of $$B(A,Z) = a_v A - a_s A^{2/3} - a_c Z^2A^{-1/3} - a_a \frac{(A-2Z)^2}{A},$$ together with a smaller "pairing term", which I am going to ignore. Here, $A$ is the total number of nucleons and $Z$ is the proton number. The four terms are the volume binding energy associated with the short-range strong nuclear force; a "surface term" which corrects the volume binding energy for the fact that a proportion of the nucleons are at the surface of the nucleus and not surrounded by other nucleons; a coulomb term that accounts for the (long-range) repulsion experienced by protons and an asymmetry term, which is the key to your entire question. If consider a nucleus with a fixed $A$, and I was to ignore the asymmetry term, you can see that the binding energy is maximised (most stable nucleus) if $Z=0$. i.e. No protons and therefore no negative Coulomb energy. But this would mean the nucleus is made up entirely of neutrons and this isn't the case. The explanation of this does not lie in the instability of the neutron, otherwise we would be able to build such nuclei that were stable on the decay time of the neutron. It lies with the asymmetry term. If I now include the asymmetry term and differentiate with respect to $Z$ at fixed $A$. $$ \frac{\partial B}{\partial Z} = -2a_c Z A^{-1/3} +4a_a (A-2Z)$$ At a maximum I would set this to zero and hence find that neutron to proton ratio $$ \frac{(A-Z)}{Z} = 1 + \frac{a_c A^{2/3}}{2a_a}.$$ The empirical values of the constants are $a_c \simeq 0.7$ MeV and $a_a\simeq 23$ MeV, so $$ \frac{(A-Z)}{Z} \simeq 1 + 0.015 A^{2/3}.$$ This means that unless $A$ is much greater than 1, then the neutron to proton ratio is roughly unity and $A-2Z \simeq 0$. In physical terms it is because the asymmetry term would become large if $A \neq 2Z$ for small $A$, but when $A$ (and $Z$) are large, the Coulomb repulsion between protons shifts the balance towards neutrons a bit. This is a sort-of-explanation because it evades explaining where the asymmetry term comes from! John Rennie's answer handles this. You can think of the protons and neutrons in a nucleus as consisting of two fermion gases (neutrons and protons) made of indistinguishable particles. Such particles cannot occupy the same quantum states (because of the Pauli exclusion principle), and they must therefore stack up, filling increasingly higher energy levels (2 per momentum/energy state, with opposite spins). The energy levels are roughly similar for the neutrons and protons because the strong nuclear force is blind to whether neutrons are protons are involved and the differences are smaller than the gap between adjacent energy levels. The total energy of such an arrangement will be the sum of the energies of all the particles. This sum will be approximately minimised when there are the same number of neutrons and protons. If there are more neutrons, then it is possible to achieve a lower energy by beta decay to a proton and an electron and electron anti-neutrino (which escape); if there are more protons, then a lower energy state can be achieved by turning a proton into a neutron and emitting a positron. This is my first answer, be nice please. A different approach (previous answers are already very good) that might give you a better intuition: Back in the day, Weizsäcker proposed a parametrization of the binding energy of nuclei by analogy with a liquid drop such as $BE(A,Z) = a_VA - a_SA^{2/3} - a_CZ^2A^{-1/3} - a_A(A-2Z)^2A^{-1}$ (not considering the pairing term here) where the coefficients $a_V,...$ are fitted on nuclear masses tables (e.g. AME2012), Z is the number of protons and A is the total number of nucleons (protons + neutrons). The different terms in this relation respectively correspond to the bulk energy, the surface energy, the coulomb energy, and the symmetry energy. Considering $BE(A,Z)>0$, the most stable nuclei for a given baryonic number A corresponds to $(A,Z^{eq})$ with $Z^{eq}$ associated to the higher value of the binding energy. You could "reverse" the problem to find the value of A that maximize the binding energy for a given Z. For heavy nuclei, such as uranium, there is a significant difference between the number of protons and the number of neutrons (which is higher). Assuming that our previous relation is true, this is because the balance between the repulsive force between protons and the asymmetry tilts in favor of Coulomb. This is the basics of nuclear phenomenology. There are a lot of microscopic approaches (e.g. DFT, ab initio) that consider the nucleon-nucleon interaction in a different way. protected by Qmechanic♦ Apr 4 '17 at 14:02 Thank you for your interest in this question. 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A portfolio can be viewed as a combination of assets held by an investor. For each asset held, such as company stocks, the logarithmic or continuously compounded rate of return $r$ at time $t$ is given by $$ r_t = \log \left(\frac {P_t} {P_{t-1}} \right) $$ where $P_t$ is the stock price at time $t$, and $P_{t-1}$ is the price in the prior period. The volatility of stock returns, over time $N$ is often estimated by the sample variance $\sigma^2$ $$ \sigma^2 = \sum \limits_{t=1}^N {\frac {(r_t – \bar r)^2} {n – 1}} $$ where $r_t$ is the return realized in period $t$, $\bar r$ is the sample mean, and $N$ is the number of periods. As the variance of returns is in units of percent squared, we take the square root of the variance to determine the standard deviation $$ \sigma = \sqrt { \sum \limits_{t=1}^N {\frac {(r_t – \bar r)^2} {n – 1}}}$$ Suppose an investor has a four stock portfolio comprising of shares on the Australian stock market (AGK, CSL, SPN, and SKT) as listed in figure 1. The companies with stock codes AGK, CSL, SPN, and SKT are: AGK:AGL Energy Ltd – operates Australia’s largest retail energy and dual fuel customer base CSL:CSL Limited – pharmaceutical and diagnostic products SPN:SP Ausnet – electricity transmission and distribution, and gas distribution SKT:SKY Network Television Limited – New Zealand’s preeminent pay television operator These stock were chosen because they have positive returns during the sample period. 1. Descriptive statistics Price data for the four stock is obtained from Yahoo Finance and is filtered for monthly price observations. In figure 1, Column A of the Summary worksheet shows the date for the first trading day of the month. Closing prices for the four stock are in the range B9:J21. The corresponding continuously compounded return series, using the Excel LN function, is calculated in the range G9:J20. Summary statistics from Excel functions: MIN, MAX, AVERAGE, STDEV.P, STDEV.S, VAR.P, and VAR.S are in rows 24 to 29. Descriptive statistics can also be estimated by using the Descriptive Statistics item, from the Data Analysis dialog box (figure 2, item 1), as shown in figure 3. The output for the Descriptive Statistics, with setting from figure 3, is shown in the New Worksheet ply, ATP_DS in figure 4. Rows 7 and 8 of the worksheet in figure 4 have values for the sample standard deviation (row 7) and sample variance (row 8). We will see later, that the Data Analysis > Covariance item returns population values, not sample values. Limitations of the ATP Descriptive Statistics The table of values is static. It is not linked to the source data, and the Descriptive Statistics dialog box must be reactivated to update the values Each row vector in the table has alternate labels and values. This means it is not conducive to using the values as a vector in other calculations such as weighted average portfolio return When assets are held as part of a portfolio, another important consideration is the amount of co-movement between the returns of portfolio components. 2. Covariance The degree of co-movement can be measured by the covariance statistic, and is calculated on a pair-wise basis. The formula for the sample covariance $\sigma_{i,j}$ for the return vectors of stock $i$ and stock $j$ is $$ \sigma_{i,j} = \sum \limits_{t=1}^N {\frac {(r_{i,t} – \bar {r_i})(r_{j,t} – \bar {r_j})} {n – 1}} $$ There are a number of ways the estimation can be operationalised and some techniques are described in this section. Methods include the Analysis Toolpak – Covariance item (figure 2, item 2), and Excel function listed in the following table. Excel covariance functions Description COVARIANCE.P(array1, array2) Returns the population covariance, the average of the product of paired deviations $\sigma_{i,j} = \sum \limits_{t=1}^N {\frac {(r_{i,t} – \bar {r_i})(r_{j,t} – \bar {r_j})} {n}}$ COVARAINCE.S(array1, array2) Returns the sample covariance, the average of the product deviations for each point pair in two data sets $\sigma_{i,j} = \sum \limits_{t=1}^N {\frac {(r_{i,t} – \bar {r_i})(r_{j,t} – \bar {r_j})} {n – 1}}$ The worksheet in figure 6 shows output for the Analysis Toolpak (ATP) covariance item in rows 2 to 5. The covariance table from the ATP is lower triangular, meaning it only returns the main diagonal elements, and the lower left elements. By definition, the covariance of a vector with itself, is the variance of the vector. Thus, the value in cell B2 in figure 6, $\sigma_{AGK,AGK} = 0.001764$ is the same value as the population variance returned by the Excel VARIANCE.P function shown in figure 1 cell G30. To convert the lower triangular table to a full matrix see – lower triangular table to full matrix In figure 7, rows 41 to 44, use the COVARIANCE.P function with Excel range names for each of the return vectors. This is repeated with the COVARIANCE.S function in rows 48 to 51. Construction of the individual cell formulas can be simplified by using range names with the INDIRECT function. To do this: Copy and paste the stock code vector to the range as column headings Using Paste Special > Transpose, paste the transposed stock codes vector to form row headings Enter the formula =COVARIANCE.P(INDIRECT(F$55),INDIRECT($G54))at G55 The returned values, and cell formulas are shown in figure 8. Download the file for this module from the figure 8 Excel Web App #1 link.
Let $M$ come from an ensemble of $N\times N$ matrices. The Wigner surmise is density function $p^W_0(s)=\frac{\pi}{2}se^{-\pi s^2/4}$. From a random matrix point of view, we can write $\rho^W_0(s)=\frac{d^2}{ds^2}E((0,s))$, where $E(I)$ is the eigenvalue gap probability: $M$ has no eigenvalues in interval $I$. $E(I)$ and it's derivatives are intimately related to the correlations between nearest neighbors. Question 1: What are known random matrix ensembles which have their eigenvalue gap probability (in limit) exactly equal to the Wigner surmise? To be specific: either all $N\times N$ matrices have Wigner surmise OR the limiting eigenvalue distribution is exactly Wigner surmise. Question 2: What are known interacting-particle systems which have their particle gap probability (in limit) exactly equal to the Wigner surmise?To be specific: either all $N\times N$ particle systems exhibit Wigner surmise OR the limiting particle distribution is exactly Wigner surmise. One example that I've seen is the real Ginibre ensemble which takes a random Gaussian matrix and focuses only on real eigenvalues. Then the probability of there being an even number of eigenvalues in $[0,s]$ matches the Wigner surmise. This is equivalent to certain statistics of creation/annihilation processes on the line. In addition to this, there are some statistical physics spin systems which seem to give an exact surmise as well. Unfortunately I'm not an expert in the latter area. Some more background: It's a well known fact that many random matrix ensembles exhibit a (limiting) density function of the form: $$p_0(s)=\frac{2u(\pi^2 s^2/4)}{s}\exp\left(-\int_0^{\pi^2 s^2/4}\frac{u(t)}{t}dt\right),$$ where $u$ satisfies a Painleve equation and of which $\rho_0^W(s)$ is a special case. So in short, $\rho^W_0(s)$ is usually an This post imported from StackExchange MathOverflow at 2015-03-12 12:24 (UTC), posted by SE-user Alex R. approximation, not an exact answer. One can certainly derive some conditions on $u$ and the resulting Painleve equation to get an exact Wigner surmise but this doesn't answer from which random matrix ensembles it comes from.
$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\al}{\alpha}$ Let $\Phi:\R^2 \to \R$ be a function which has partial derivatives at the point $(x_0,y_0)$. Let $\al:I \to \R^2$ be a path satisfying: $\al(0)=(x_0,y_0),\dot \al(0)=v=(v_1,v_2)\neq (0,0)$. Assume $\Phi \circ \al$ is differentiable at $t=0$ and the directional derivative of $\Phi$ at direction $v$ ($d_v \Phi(x_0,y_0)$) exists. Is it true that $(\Phi \circ \al)'(0)=d_v \Phi(x_0,y_0)$? Update: As noted by user251257, if $\Phi$ is Lipschitz then both derivatives agree. So, in order to find a counter-example (where they differ) we need to search for non-Lipschitz functions, which still has partial derivatives at the point $(x_0,y_0)$. This is not trivial, since a function with bounded partial derivatives is Lipschitz (as proved here). So, our search is narrowed down to functions which do not have partial derivatives in a neighbourhood of $(x_0,y_0)$, or that there partial derivatives are unbounded in some neighbourhood of it. Note: In general, it can happen that either one of $(\Phi \circ \al)'(0),d_v \Phi(x_0,y_0)$ exists without the other. Example 1: $d_v \Phi(x_0,y_0)$ exists ,$(\Phi \circ \al)'(0)$ does not.(essentially taken from here) $(x_0,y_0)=(0,0), \Phi(x,y) = \begin{cases} \frac{x^2y}{x^4 +y^2}, & \text{if $(x,y) \neq 0$}\\ 0, & \text{if $(x,y) = (0,0)$} \\ \end{cases}, y(x)=x^2 $ $\Phi$ has directional derivatives at every direction at $(0,0)$, but for $\al(x)=(x,x^2)=(x,y(x))$ we get: $F(x)=\Phi \circ \al(x)=\begin{cases} \frac{1}{2}, & \text{if $(x,y) \neq 0$}\\ 0, & \text{if $(x,y) = (0,0)$} \\ \end{cases} $ is not differentiable at $x_0=0$. Example 2: $(\Phi \circ \al)'(0)$ exists ,$d_v \Phi(x_0,y_0)$ does not. Update: this exmaple is wrong! I am not sure yet if a suitable example exists Take $(x_0,y_0)=(0,0),\al(t)=(t,t^3+t),\Phi(x,y)=\sqrt{\|xy\|}$, $\Phi$ has partial derivatives ($\Phi_x(0,0)=\Phi_y(0,0)=0)$. $\Phi(\al(t))=\sqrt{t^4+t^2}$,so $(\Phi \circ \al)'(0)=1$. However, the directional derivative at direction $v=(1,1)$ does not exist.
Attention conservation notice: Unless you fit into a very similar academic niche to me this post likely uses techniques you don’t know to solve problems you don’t care about. Even if you are in that niche this is more of a theoretical curiousity than a useful technique at present, but it was useful for me to think through this to see how some stuff connects up. The basic trick they use is to realise the object we want (an equivalence relationship or a pseudometric in this case) and consider it as an element of some complete lattice, define a monotonic function on that lattice such that a fixed point of it will have the desired properties, and then use the Knaster-Tarski fixed point theorem to show that the set of such fixed points is no-empty. The desired object is then either the smallest or largest fixed point (both of which the theorem guarantees to exist, and indeed shows us how to construct). This turns out to work with deterministic finite automata too! Suppose we have some deterministic finite automaton $M$ with states $S = \{s_1, \ldots, s_n\}$, alphabet $A$, transition function $\tau : A \times S \to S$ and acceptance function $\alpha: S \to \{0, 1\}$. We want to construct the minimal deterministic finite automaton which matches the same language by merging equivalent states. This requires us to define an equivalence relationship $\sim$ on $M_S$ such that $s_i \sim s_j$ if and only if the languages starting from $s_i$ and $s_j$ are the same. This is called the Myhill-Nerode relationship. We can realise this equivalence relationship as follows: We use the observation that the set of equivalence relationships over any set forms a complete lattice (to get the upper bound of a set of equivalence relationships, just take their unions as sets and then take the transitive closure to get back to an equivalence relationship). What we are looking for is an equivalence relationship with the following two properties: If $s_i \sim s_j$ then $\alpha(s_i) = \alpha(s_j)$. If $s_i \sim s_j$ and $a \in A$ then $\tau(s_i, a) \sim \tau(s_j, j)$. i.e. no accepting state is equivalent to a non-accepting state, and the equivalence relationship works with the transition function so that if two states are equivalent they transition to equivalent states given the same character. Any such equivalence relationship is called a bisimilarity. We can find such an equivalence relationship as follows: Let $E$ be the set of equivalence relations over $S$ and define $F: E \to E$ as $F(R) = \{(s, t) \in S^2: \alpha(s) = \alpha(t) \wedge \forall a \in A, (\tau(a, s), \tau(a, t)) \in R\}$. (The fact that if $R$ is an equivalence relationship then so is $F(R)$ is not immediately obvious, but follows from some fairly straightforward mechanical checking that I will leave as an exercise for the interested reader) This definition may seem somewhat non-obvious, so lets unpack it: We require that the resulting equivalence relationship always respects $\alpha$. i.e. we throw away anything we would otherwise get where it disagrees on whether the state should be accepting. We take the new relationship to be the set of pairs that the old relationship cannot distinguish any transitions from. So for example if we had states $s_i, s_j$ such that every transition from them went to an $R$-equivalent state, $F(R)$ would treat them as equivalent. Equally, if you had two R-equivalent states such that $(\tau(a, s_i), \tau(a, s_j)) \not\in R$ (i.e. R distinguishes some transition from them) So you can think of $F$ as taking an equivalence relationship and returning something that is in some sense “a bit closer” to being a bisimilarity. If that intuition is correct, then it should be the case that if $R$ is a fixed point of $F$ then it is a bisimilarity. Suppose $R = F(R)$. Then by construction of $F$ we have the first property satisfied (that if $s \sim t$ then $\alpha(s) = \alpha(t)$. For the second, suppose that we have $s, t, a$ with $(\tau(a, s), \tau(a, t)) \not\in R$. Then by definition of $F$ we musts have that $(s, t) \not\in F(R) = R$. i.e. if $(s, t) \in R$ then $(\tau(a, s), \tau(a, t)) \in R$. So in order to find a bisimilarity we just need to find fixed points of $F$, but this is exactly what the Knaster-Tarski fixed point theorem lets us do. $F$ is montonic, i.e. if $R \subseteq R’$ then $F(R) \subseteq F(R’)$ (another “not totally obvious but a mechanical exercise for the interested reader to check” claim), the set of equivalence relationships form a complete lattice, so the set of fixed points of $F$ also form a complete lattice. In particular there is a greatest fixed point of $F$, called the bisimulation relationship. This is exactly the $\sim$ relationship we were looking for (because if a word causes two states to reach a different language then they could not have been bisimilar, so that relationship is necessarily maximal). Additionally, the proof of the Knaster-Tarski fixed point theorem shows us how to construct this relationship: We just start with the largest equivalence relationship on $S$ and repeatedly apply $F$ to it until we get a fixed point (in general “repeatedly” might require transfinite induction, but because $S$ is finite so is the set of equivalence relationships over $F$ and as a result we only have to iterate finitely many times). But if you squint at it right, this is actually just Hopcroft’s algorithm for DFA minimization! We start with an equivalence relationship which conflates everything, and then we split states apart as we find that they have the wrong label or lead to states that are now known to be equivalent, iterating this until we reach a point where there are no more states to split (i.e. a fixed point of $F$). The steps are slightly different in that we actually just split the states by labels once at the beginning and then don’t have to do so again, but because of the way $F$ behaves in these circumstances the difference is pretty inessential (in particular we always have $F(R) \subseteq R$ for $R$ obtained by iterating $F$ from the largest equivalence relationship, even though this does not hold for arbitrary $R$). Is this observation useful? Well, not really. It’s useful in the sense that it helped me understand why fixed points on lattices are such a useful technique, and it was nice to relate this back to some things I already understood. So it’s a useful exercise in pedagogy. Are there any applications of this? Well, maybe. In the context of the probabilistic work the next step here is to look at other interesting lattices which can use similar structure, and that’s pretty interesting, but I’m not sure how useful it is in the language case. Also most of the ways in which it’s useful are likely better served by transforming it into the probabilistic case.
AliFMDMultCuts () AliFMDMultCuts (EMethod method, Double_t fmd1i, Double_t fmd2i=-1, Double_t fmd2o=-1, Double_t fmd3i=-1, Double_t fmd3o=-1) AliFMDMultCuts (const AliFMDMultCuts &o) AliFMDMultCuts & operator= (const AliFMDMultCuts &o) void Reset () Double_t GetMultCut (UShort_t d, Char_t r, Double_t eta, Bool_t errors) const Double_t GetMultCut (UShort_t d, Char_t r, Int_t etabin, Bool_t errors) const void SetMultCuts (Double_t fmd1i, Double_t fmd2i=-1, Double_t fmd2o=-1, Double_t fmd3i=-1, Double_t fmd3o=-1) void SetMPVFraction (Double_t frac=0) void SetNXi (Double_t nXi) void SetIncludeSigma (Bool_t in) void SetProbability (Double_t cut=1e-5) void Set (EMethod method, Double_t fmd1i, Double_t fmd2i=-1, Double_t fmd2o=-1, Double_t fmd3i=-1, Double_t fmd3o=-1) void Print (Option_t option="") const void FillHistogram (TH2 h) const void Output (TList l, const char name=0) const Bool_t Input (TList l, const char name) EMethod GetMethod () const const char * GetMethodString (Bool_t latex=false) const Cuts used when calculating the multiplicity. We can define our cuts in four ways (in order of priorty) Using a fixed value $ v$- AliFMDMultCuts:: SetMultCuts Using a fraction $ f$ of the most probably value ( $ \Delta_p$) from the energy loss fits Using some number $ n$ of widths ( $ \xi$) below the most probable value ( $ \Delta_p$) from the energy loss fits Using some number $ n$ of widths ( $ \xi+\sigma$) below the most probable value ( $ \Delta_p$) from the energy loss fits Using the $ x$ value for which $ P(x>p)$ given some cut value $ p$ Using the lower fit range of the energy loss fits The member function AliFMDMultCuts::Reset resets all cut values, meaning the lower bound on the fits will be used by default. This is useful to ensure a fresh start: The member function AliFMDMultCuts::GetMethod will return the method identifier for the current method employed (AliFMDMultCuts::EMethod). Like wise will the method AliFMDMultCuts::GetMethodString give a human readable string of the current method employed. Definition at line 38 of file AliFMDMultCuts.h. Set the cut for specified method. Note, that if method is kFixed, and only fmd1i is specified, then the outer rings cut value is increased by 20% relative to fmd1i. Also note, that if method is kLandauWidth, and cut2 is larger than zero, then $\sigma$ of the fits are included in the cut value. Parameters method Method to use fmd1i Value for FMD1i fmd2i Value for FMD2i (if < 0, use fmd1i) fmd2o Value for FMD2o (if < 0, use fmd1i) fmd3i Value for FMD3i (if < 0, use fmd1i) fmd3o Value for FMD3o (if < 0, use fmd1i) Definition at line 63 of file AliFMDMultCuts.cxx. Referenced by AliFMDMultCuts(), AliFMDSharingFilter::AliFMDSharingFilter(), DepSet(), and SetProbability().
A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$.I believe I have numerically discovered that$$\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty$$but cannot ... So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$. Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$. Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow. Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$. Well, we do know what the eigenvalues are... The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$. Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker "a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers. I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd... Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work. @TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now) Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $\|\|v\|\| = \|\|(v)_S\|\|$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism @AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$. Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1) For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism
Because the slab extends infinitely in the xy plane, the electric field lies only along the z direction. The differential form of Gauss' Law for such a one-dimensional electric field is $$\frac{dE}{dz}=\frac{\rho(z)}{\epsilon_0}$$ This can be integrated, using the boundary condition that the electric field must be zero at large distances from the slab, because it is electrically neutral. A more elementary way of solving the problem is to divide the slab into infinitesimally thin layers of thickness $dz$. The electric field of each layer is uniform, independent of distance from the layer, and is $dE=\frac{d\sigma}{2\epsilon}$ pointing away from the layer on each side for +ve surface charge density $d\sigma=\rho(z)dz$. The total electric field at any point inside or outside of the slab is found by superposition of fields from every such layer in the slab. Note that the total electric field at any point due to all layers which are closer to the centre plane $z=0$, is zero. This is because the charge density is anti-symmetric : for every layer of +ve area charge density on one side of $z=0$ there is a layer of -ve charge with the same magnitude of area density on the other side of $z=0$. The electric fields of these two layers cancel out for points which outside of the two layers, in the same way that the total electric field is zero outside of a parallel plate capacitor (if the plate dimensions are very much bigger than the distance from them). From this observation you can see that the electric field outside of the slab is zero, because all layers in the slab are closer to the centre plane. The simplest way of getting the field inside the slab is to apply Gauss' Law using a "pill box" Gaussian surface which has one face A of area S at the surface of the slab $z=a$ (where $E(a)=0$) and the other face B at distance $\|z\|<a$ from the centre plane. The other face(s) of the pill box are parallel to the z direction so the electric flux through them is zero. There is no electric flux through face A; the flux through face B is $E(z)S=\frac{q}{\epsilon_0}$ where by integration $$q=\int_a^z S\rho(z)dz$$ is the total charge inside the pill box. See Electric field in a non-uniformly charged sheet. An answer identical to mine is given in the duplicate question Finding the electric field of a NON uniform slab?
Differentiation \[4x^3\]differentiated is \[4 \times 3 x^{3-1}=12x^2\] We can differentiate a sum usinf the same rule for each term. \[2x^5-4x^7\]when differentiated is \[2 \times 5x^{5-1}-4 \times 7x^{7-1}=10x^4-8x^6\]. This rule 'times by the power and take one off the power' works for \[x\]'s and constants too. To differentiate \[3x\]write as \[3x^1\]then apply the rule to give \[3 \times 1x^{1-1}=3x^0=3 \times 1=1\]since \[x^0=1\]. To differentiate \[4\]write as \[4x^0\]then differentiatie using the above gives \[4 \times 0 x^{0-1}=0\]since anything times 0 equals 0. Finally remember that when you differentiate \[y\]you get \[\frac{dy}{dx}\]. Differentiate \[y=4x^2-6x-4\]. Write \[y=4x^2-6x^1-4x^0\]. We have \[\begin{equation} \begin{aligned} \frac{dy}{dx} &=4 \times 2x^{2-1}-6 \times x^{1-1}-4 \times 0 x^{0-1} \\ & =8x^1-6 \times x^0 \\ &= 8x-6 \times 1 \\ &= 8x-6\end{aligned} \end{equation}\]
The Graphene Revolution: Part 5 In a paper titled “Choosing a Gate Dielectric for Graphene Based Transistors“, the applications of a semiconducting form of graphene are examined. As we have seen before, single-layer graphene is not a semiconductor, it is a zero bandgap conductor (a semimetal). Efforts are well underway to introduce bandgaps to graphene, which would make it semiconducting with a room temperature mobility an order of magnitude higher than silicon. The race is already underway to find applications for such a material once the remaining technical challenges have been overcome. An application of semiconducting graphene is the design of next-generation, fast switching semiconductor metal oxide field effect transistors (MOSFETs). What Is a MOSFET? The basic idea of a MOSFET is to apply a gate voltage to control the drain-to-source resistance and thus the drain current (see image below). At a certain gate-to-source voltage (V GS), and at low drain-to-source voltages (V DS), the drain current is almost linearly dependent on V DS. When V DS increases, the drain current saturates. The level of saturation depends on the gate-to-source voltage and the switching time depends on the mobility of the semiconductor. The higher the mobility of the semiconducting material, the faster the current can be switched on and off. Semiconductor Physics Semiconductor physics is extremely complicated. Strictly speaking, the Boltzmann equation should be solved with Maxwell’s equations in order to describe the device physics completely. Since this is computationally intractable, the most common approach for modeling semiconductors is to solve a set of drift diffusion equations coupled to Poisson’s equation: \frac{\partial n}{\partial t}&=\frac{1}{q}\nabla \cdot \mathbf{J_{n}}-R_n \\ \frac{\partial p}{\partial t}&=-\frac{1}{q}\nabla \cdot \mathbf{J_{p}}-R_p \\ \nabla \cdot{} \left(\epsilon \nabla V\right) &= -q(p-n+N_D^+-N_A^-) \end{aligned} Here n is the number density of electrons, p is the number density of holes, V is the electrostatic potential, R n is the electron recombination rate, Ris the hole recombination rate, p is the electron current, J n is the hole current. Solving this set of equations allows you to construct the voltage-current characteristics of semiconducting devices. The semiconductor equations are highly non-linear and require special numerical methods to solve. J p Our Brand New Semiconductor Module As we recently announced, a dedicated product for modeling semiconductor devices is now available within the COMSOL platform. The Semiconductor Module, as it’s called, allows for detailed analysis of semiconductor device operation at the fundamental physics level. The module is based on the drift-diffusion equations with isothermal or non-isothermal transport models. Two numerical methods are provided: the finite volume method with Scharfetter-Gummel upwinding and a Galerkin least-squares stabilized finite element method. The module provides an easy-to-use interface for analyzing and designing semiconductor devices, greatly simplifying the task of device simulation on the COMSOL platform. Models for semiconducting and insulating materials in addition to boundary conditions for ohmic contacts, Schottky contacts, and gates are provided as dedicated features within the Semiconductor Module. The module includes enhanced functionality for modeling electrostatics. System level and mixed device simulations are enabled through an interface for electrical circuits with SPICE import capability. The Semiconductor Module is useful for simulating a range of practical devices. The built-in Model Library contains a suite of models designed to provide straightforward instruction and demonstrate how to use the interface to simulate your own devices. The Semiconductor Module is particularly relevant for simulating transistors including bipolar, metal semiconductor field-effect transistors (MESFETs), metal-oxide-semiconductor field-effect transistors (MOSFETs), Schottky diodes, and P-N junctions. The Semiconductor Module could hence be used to investigate the device characteristics of graphene-based semiconductors similar to the ones explained in the paper referenced at the beginning of this blog post. This Concludes the Graphene Series… Over the past few months, in the previous four posts of our blog series, we have read lots about the history, applications, and manufacture of graphene. I’ve enjoyed talking about this topic very much — but I’ve only scratched the surface. There are still many sub-topics that haven’t been explored, and I encourage you to keep up with the latest graphene-related developments in technical magazines and publications. Whether you’re interested in the applications or the manufacture of graphene, COMSOL offers a wide range of products that can provide insight and a better understanding of these processes. Further Reading Other Posts in This Series Comments (2) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science TAGS CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
Search Now showing items 1-10 of 33 The ALICE Transition Radiation Detector: Construction, operation, and performance (Elsevier, 2018-02) The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ... Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Elsevier, 2018-02) In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ... First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC (Elsevier, 2018-01) This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ... First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV (Elsevier, 2018-06) The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ... D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV (American Physical Society, 2018-03) The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ... Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV (Elsevier, 2018-05) We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ... Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (American Physical Society, 2018-02) The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ... $\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV (Springer, 2018-03) An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ... J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Elsevier, 2018-01) We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ... Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV (Springer Berlin Heidelberg, 2018-07-16) Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range \|η\| < 0.8 ...
We consider the finite-amplitude instability of incompressible spherical Couette flow between two concentric spheres of radii $R_1$ and $R_2$ $({>}R_1)$ in the narrow-gap limit, $\varepsilon\,{\equiv}\,(R_2-R_1)/R_1\,{\ll}\,1$, caused by rotating them both about a common axis with distinct angular velocities $\Omega_1$ and $\Omega_2$ respectively. In this limit it is well-known that the onset of (global) linear instability is manifested by Taylor vortices of roughly square cross-section close to the equator. According to linear theory this occurs at a critical Taylor number $T_{\rm crit}$ which, remarkably, exceeds the local value $T_c$ obtained by approximating the spheres as cylinders in the vicinity of the equator even as $\varepsilon\,{\downarrow}\,0$. Previous theoretical work on this problem has concentrated on the case of almost co-rotation with $\delta\,{\approx}\,(\Omega_1\,{-}\,\Omega_2)/\Omega_1\,{=}\,\OR(\varepsilon^{1/2})$ for which $T_{\rm crit}\,{=}\,T_c\,{+}\,\OR(\delta^2)\,{+}\,\OR(\varepsilon)$. In this limit the amplitude equation that governs the spatio-temporal modulation of the vortices on the latitudinal extent $\OR(\varepsilon^{1/2}R_1)$ gives rise to an interesting bifurcation sequence. In particular, the appearance of global bifurcations heralds the onset of complicated subcritical time-dependent finite-amplitude solutions. Here we switch attention to the case when $\varepsilon^{1/2}\,{\ll}\,\delta\,{\le}\, 1$. We show that for Taylor numbers $T\,{=}\,T_c+\OR((\delta\varepsilon)^{2/3})$ there exists a locally unstable region of width $\OR((\delta\varepsilon)^{1/3}R_1)$ within which the amplitude equation admits solutions in the form of pulse-trains. Each pulse oscillates at a frequency proportional to its distance from the equatorial plane and consists of a wave propagating towards the equator under an envelope. The pulse drifts at a slow speed (relative to the wave velocity) proportional to its distance (and away) from the equator. Both the wavelength and the envelope width possess the same relatively short length scale $\OR((\varepsilon^{2}/\delta)^{1/3}R_1)$. The appropriate theory of spatially periodic pulse-trains is developed and numerical solutions found. Significantly, these solutions are strongly subcritical and have the property that $T\to T_c$ as $\varepsilon\,{\downarrow}\,0$. Two particular limits of our theory are examined. In the first, $\varepsilon^{1/2}\,{\ll}\,\delta\,{\ll}\,1$, the spheres almost co-rotate and the pulse drift velocity is negligible. A comparison is made of the pulse-train predictions with previously obtained numerical results pertaining to large (but finite) values of $\delta/\varepsilon^{1/2}$. The agreement is excellent, despite the complicated long-time behaviour caused by inhomogeneity across the relatively wide unstable region. Our second special case $\delta\,{=}\,1$ relates to the situation when the outer sphere is at rest. Now the poleward drift of the pulses leads to a slow but exponential increase of their separation with time. This systematic pulse movement, over and above the spatial inhomogeneity just mentioned, necessarily leads to complicated and presumably chaotic spatio-temporal behaviour across the wide unstable region of width $\OR(\varepsilon^{1/3}R_1)$ on its associated time scale, which is $\OR(\varepsilon^{-1/3})$ longer than the wave period. In view of the several length and time scales involved only qualitative comparison with experimental results is feasible. Nevertheless, the pulse-train structure is robust and likely to provide the building block of the ensuing complex dynamics.
Hello, I've a problem to calculate the Position of a pendulum as a function of theta. For example: $\theta (t)$ is a function of time which returns the angle made by the pendulum at a particular instant wrt it's equilibrium Position. So, $$ T = \dfrac 12 m l^2 \dot \theta^2 $$$$ U = - mgl \cos \theta $$ $$ L(\theta, \dot \theta) = \dfrac 12 m l^2 \dot \theta^2 + mgl \cos \theta $$ Using, the Euler - Lagrangian Formula, $$ \dfrac d{dt} \left ( \dfrac{\partial L}{\partial \dot \theta}\right) - \dfrac{\partial L}{\partial \theta} = 0 $$ We get, $$ \boxed{\ddot \theta =- \dfrac gl \sin \theta} $$ which is the equation of motion. But, most of the derivations, I've seen/read go this way: $$ \ddot \theta = \dfrac gl \theta \quad \dots \quad (\text{As, } \sin \theta \approx \theta, \theta \rightarrow 0) \tag{} $$ $$ \theta (t) = \cos \left ( \sqrt{\dfrac gl} t \right) $$ Because it satisfies $()$ So, I've 2 questions here. Other possible solutions of the Second Order Differential Equations exist like $\theta (t) = e^{\left( \sqrt{\dfrac gl}t \right)}$. So, why we choose that only one? One would argue that the sine function oscillates similar to the pendulum, so this makes sense to accept the sine one. But, in general case, when we solve the Lagrangian and get the equation of motion in differential form, then there are tons of complex situation possible, How can you determine which kind is needed? How can we solve the Second Order Differential Equation $\ddot \theta = - \dfrac gl \sin \theta$ and get an exact formula for that? Thanks :)
Plugins and Hybrid computation¶ This tutorial introduces the notion of hybrid computation by combining several PennyLane plugins. We first introduce PennyLane’s Strawberry Fields plugin and use it to explore a non-Gaussian photonic circuit. We then combine this photonic circuit with a qubit circuit — along with some classical processing — to create and optimize a fully hybrid computation. Be sure to read through the introductory qubit rotation and Gaussian transformation tutorials before attempting this tutorial. Note pip install pennylane-sf A non-Gaussian circuit¶ We first consider a photonic circuit which is similar in spirit to the qubit rotation circuit: Breaking this down, step-by-step: We start the computation with two qumode subsystems. In PennyLane, we use the shorthand ‘wires’ to refer to quantum subsystems, whether they are qumodes, qubits, or any other kind of quantum register. Prepare the state$\left\|1,0\right\rangle$. That is, the first wire (wire 0) is prepared in a single-photon state, while the second wire (wire 1) is prepared in the vacuum state. The former state is non-Gaussian, necessitating the use of the 'strawberryfields.fock'backend device. Both wires are then incident on a beamsplitter, with free parameters $\theta$phi`. Here, we have the convention that the beamsplitter transmission amplitude is $t=\cos\theta$, and the reflection amplitude is $r=e^{i\phi}\sin\theta$. See Quantum operations for a full list of operation conventions. Finally, we measure the mean photon number$\left\langle \hat{n}\right\rangle$ of the second wire, where\[\hat{n} = \ad\a\] is the number operator, acting on the Fock basis number states, such that $\hat{n}\left\|n\right\rangle = n\left\|n\right\rangle$. The aim of this tutorial is to optimize the beamsplitter parameters $(\theta, \phi)$ suchthat the expected photon number of the second wire is maximized. Since the beamsplitteris a passive optical element that preserves the total photon number, this to the outputstate $\left\|0,1\right\rangle$ — i.e., when the incident photon from the first wire has been‘redirected’ to the second wire. Exact calculation¶ To compare with later numerical results, we can first consider what happens analytically. The initial state of the circuit is $\left\|\psi_0\right\rangle=\left\|1,0\right\rangle$, and the output state of the system is of the form $\left\|\psi\right\rangle = a\left\|1, 0\right\rangle + b\left\|0,1\right\rangle$, where $\|a\|^2+\|b\|^2=1$. We may thus write the output state as a vector in this computational basis, $\left\|\psi\right\rangle = \begin{bmatrix}a & b\end{bmatrix}^T$. The beamsplitter acts on this two-dimensional subspace as follows: Furthermore, the mean photon number of the second wire is Therefore, we can see that: $0\leq \left\langle \hat{n}_1\right\rangle\leq 1$: the output of the quantum circuit is bound between 0 and 1; $\frac{\partial}{\partial \phi} \left\langle \hat{n}_1\right\rangle=0$: the output of the quantum circuit is independent of the beamsplitter phase $\phi$; The output of the quantum circuit above is maximised when $\theta=(2m+1)\pi/2$ for $m\in\mathbb{Z}_0$. Loading the plugin device¶ While PennyLane provides a basic qubit simulator ( 'default.qubit') and a basic CVGaussian simulator ( 'default.gaussian'), the true power of PennyLane comes from itsplugin ecosystem, allowing quantum computations to be run on a varietyof quantum simulator and hardware devices. For this circuit, we will be using the 'strawberryfields.fock' device to constructa QNode. This allows the underlying quantum computation to be performed using theStrawberry Fields Fock backend. As usual, we begin by importing PennyLane and the wrapped version of NumPy provided by PennyLane: import pennylane as qmlfrom pennylane import numpy as np Next, we create a device to run the quantum node. This is easy in PennyLane; as soon asthe PennyLane-SF plugin is installed, the 'strawberryfields.fock' device can be loaded— no additional commands or library imports required. dev_fock = qml.device("strawberryfields.fock", wires=2, cutoff_dim=2) Compared to the default devices provided with PennyLane, the 'strawberryfields.fock'device requires the additional keyword argument: cutoff_dim: the Fock space truncation used to perform the quantum simulation Note Devices provided by external plugins may require additional arguments and keyword arguments — consult the plugin documentation for more details. Constructing the QNode¶ Now that we have initialized the device, we can construct our quantum node. Likethe other tutorials, we use the qnode decoratorto convert our quantum function (encoded by the circuit above) into a quantum noderunning on Strawberry Fields. @qml.qnode(dev_fock)def photon_redirection(params): qml.FockState(1, wires=0) qml.Beamsplitter(params[0], params[1], wires=[0, 1]) return qml.expval(qml.NumberOperator(1)) The 'strawberryfields.fock' device supports all CV objects provided by PennyLane;see CV operations. Optimization¶ Let’s now use one of the built-in PennyLane optimizers in order tocarry out photon redirection. Since we wish to maximize the mean photon number ofthe second wire, we can define our cost function to minimize the negative of the circuit output. def cost(params): return -photon_redirection(params) To begin our optimization, let’s choose the following small initial values of $\theta$ and $\phi$: init_params = np.array([0.01, 0.01])print(cost(init_params)) Out: -9.999666671111085e-05 Here, we choose the values of $\theta$ and $\phi$ to be very close to zero; this results in $B(\theta,\phi)\approx I$, and the output of the quantum circuit will be very close to $\left\|1, 0\right\rangle$ — i.e., the circuit leaves the photon in the first mode. Why don’t we choose $\theta=0$ and $\phi=0$? At this point in the parameter space, $\left\langle \hat{n}_1\right\rangle = 0$, and $\frac{d}{d\theta}\left\langle{\hat{n}_1}\right\rangle\|_{\theta=0}=2\sin\theta\cos\theta\|_{\theta=0}=0$. Since the gradient is zero at those initial parameter values, the optimization algorithm would never descend from the maximum. This can also be verified directly using PennyLane: dphoton_redirection = qml.grad(photon_redirection, argnum=0)print(dphoton_redirection([0.0, 0.0])) Out: [0.0, 0.0] Now, let’s use the GradientDescentOptimizer, and update the circuitparameters over 100 optimization steps. # initialise the optimizeropt = qml.GradientDescentOptimizer(stepsize=0.4)# set the number of stepssteps = 100# set the initial parameter valuesparams = init_paramsfor i in range(steps): # update the circuit parameters params = opt.step(cost, params) if (i + 1) % 5 == 0: print("Cost after step {:5d}: {: .7f}".format(i + 1, cost(params)))print("Optimized rotation angles: {}".format(params)) Out: Cost after step 5: -0.0349558Cost after step 10: -0.9969017Cost after step 15: -1.0000000Cost after step 20: -1.0000000Cost after step 25: -1.0000000Cost after step 30: -1.0000000Cost after step 35: -1.0000000Cost after step 40: -1.0000000Cost after step 45: -1.0000000Cost after step 50: -1.0000000Cost after step 55: -1.0000000Cost after step 60: -1.0000000Cost after step 65: -1.0000000Cost after step 70: -1.0000000Cost after step 75: -1.0000000Cost after step 80: -1.0000000Cost after step 85: -1.0000000Cost after step 90: -1.0000000Cost after step 95: -1.0000000Cost after step 100: -1.0000000Optimized rotation angles: [1.57079633 0.01 ] Comparing this to the exact calculation above, this is close to the optimum value of $\theta=\pi/2$, while the value of $\phi$ has not changed — consistent with the fact that $\left\langle \hat{n}_1\right\rangle$ is independent of $\phi$. Hybrid computation¶ To really highlight the capabilities of PennyLane, let’s now combine the qubit-rotation QNode from the qubit rotation tutorial with the CV photon-redirection QNode from above, as well as some classical processing, to produce a truly hybrid computational model. First, we define a computation consisting of three steps: two quantum nodes (the qubit rotationand photon redirection circuits, running on the 'default.qubit' and 'strawberryfields.fock' devices, respectively), along with a classical function, that simplyreturns the squared difference of its two inputs using NumPy: # create the devicesdev_qubit = qml.device("default.qubit", wires=1)dev_fock = qml.device("strawberryfields.fock", wires=2, cutoff_dim=10)@qml.qnode(dev_qubit)def qubit_rotation(phi1, phi2): """Qubit rotation QNode""" qml.RX(phi1, wires=0) qml.RY(phi2, wires=0) return qml.expval(qml.PauliZ(0))@qml.qnode(dev_fock)def photon_redirection(params): """The photon redirection QNode""" qml.FockState(1, wires=0) qml.Beamsplitter(params[0], params[1], wires=[0, 1]) return qml.expval(qml.NumberOperator(1))def squared_difference(x, y): """Classical node to compute the squared difference between two inputs""" return np.abs(x - y) ** 2 Now, we can define an objective function associated with the optimization, linking together our three subcomponents. Here, we wish to perform the following hybrid quantum-classical optimization: The qubit-rotation circuit will contain fixed rotation angles $\phi_1$ and $\phi_2$. The photon-redirection circuit will contain two free parameters, the beamsplitter angles $\theta$ and $\phi$, which are to be optimized. The outputs of both QNodes will then be fed into the classical node, returning the squared difference of the two quantum functions. Finally, the optimizer will calculate the gradient of the entire computation with respect to the free parameters $\theta$ and $\phi$, and update their values. In essence, we are optimizing the photon-redirection circuit to return the same expectation valueas the qubit-rotation circuit, even though they are two completely independent quantum systems. We can translate this computational graph to the following function, which combines the three nodes into a single hybrid computation. Below, we choose default values $\phi_1=0.5$, $\phi_2=0.1$: def cost(params, phi1=0.5, phi2=0.1): """Returns the squared difference between the photon-redirection and qubit-rotation QNodes, for fixed values of the qubit rotation angles phi1 and phi2""" qubit_result = qubit_rotation(phi1, phi2) photon_result = photon_redirection(params) return squared_difference(qubit_result, photon_result) Now, we use the built-in GradientDescentOptimizer to perform the optimizationfor 100 steps. As before, we choose initial beamsplitter parameters of$\theta=0.01$, $\phi=0.01$. # initialise the optimizeropt = qml.GradientDescentOptimizer(stepsize=0.4)# set the number of stepssteps = 100# set the initial parameter valuesparams = np.array([0.01, 0.01])for i in range(steps): # update the circuit parameters params = opt.step(cost, params) if (i + 1) % 5 == 0: print("Cost after step {:5d}: {: .7f}".format(i + 1, cost(params)))print("Optimized rotation angles: {}".format(params)) Out: Cost after step 5: 0.2154539Cost after step 10: 0.0000982Cost after step 15: 0.0000011Cost after step 20: 0.0000000Cost after step 25: 0.0000000Cost after step 30: 0.0000000Cost after step 35: 0.0000000Cost after step 40: 0.0000000Cost after step 45: 0.0000000Cost after step 50: 0.0000000Cost after step 55: 0.0000000Cost after step 60: 0.0000000Cost after step 65: 0.0000000Cost after step 70: 0.0000000Cost after step 75: 0.0000000Cost after step 80: 0.0000000Cost after step 85: 0.0000000Cost after step 90: 0.0000000Cost after step 95: 0.0000000Cost after step 100: 0.0000000Optimized rotation angles: [1.20671364 0.01 ] Substituting this into the photon redirection QNode shows that it now produces the same output as the qubit rotation QNode: result = [1.20671364, 0.01]print(photon_redirection(result))print(qubit_rotation(0.5, 0.1)) Out: 0.87319830211464490.8731983044562817 This is just a simple example of the kind of hybrid computation that can be carried out in PennyLane. Quantum nodes (bound to different devices) and classical functions can be combined in many different and interesting ways. Total running time of the script: ( 0 minutes 14.972 seconds)
On non-linear discrete boundary-value problems and semi-groups The following discrete boundary - value problem for non-linear system $x_k=\varphi_k(x_{k-1},y_k),\,y_{k-1}=\psi_k(x_{k-1},y_k)$, $k=\overline{1,\,N},\,\,N<\infty,\,\,x_0=a,\,\,y_N=b, $ is considered. Here the functions $\varphi_k(x,y),\,\psi_k(x,y)\geq0$ are monotone with respect to arguments $ x,\,y\geq0$, satisfing the condition of dissipativity or conservativity: $ \varphi_k(x,y)+\psi_k(x,y)\leq x+y+\gamma_k$, as well as two simple additional conditions. A relation of this problem with multistep processes is demonstrated. Existence and uniqueness of minimal solution of problem is proved. A semi-group approach to solving of problem is developed. The approach is adjoined with V.~Ambartsumian Principle of Invariance and R.Bellman method of On the Central Limit Theorem for Toeplitz Quadratic Forms of Stationary Sequences Let $X(t),$ $t = 0,\pm1,\ldots,$ be a real-valued stationary Gaussian sequence with spectral density function $f(\Lambda)$. The paper considers a question of applicability of central limit theorem (CLT) for Toeplitz type quadratic form $Q_n$ in variables $X(t)$, generated by an integrable even function $g(\Lambda)$. Assuming that $f(\Lambda)$ and $g(\Lambda)$ are regularly varying at $\Lambda=0$ of orders $\alpha$ and $\beta$ respectively, we prove CLT for standard normalized quadratic form $Q_n$ in the critical case $\alpha+\beta=1/2$. We also show that CLT is not valid under the single condition that the asymptotic variance of $Q_n$ is separated from zero and infinity. Generalization of the theorem of de Montessus de Bollore We prove a generalization of the theorem of de Montessus de Bollore for a large class of functional series investigated in [14]. In particular multivalued approximants for power series, as well as for series of Faber and Gegenbauer polynomials are considered. Numerical results are presented. Minimization of Errors of the Polynomial-Trigonometric Interpolation with Shifted Nodes. The polynomial-trigonometric interpolation based on the Krylov approach for a smooth function given on $[-1, 1]$ is defined on the union of $m$ shifted each other uniform grids with the equal number of points. The asymptotic errors of the interpolation in both uniform and $L_2$ metrics are investigated. It turned out that the corresponding errors can be minimized due to an optimal choice of the shift parameters. The study of asymptotic errors is based on the concept of the ''limit function" proposed by Vallee-Poussin. In particular cases of unions of two and three uniform grids the limit functions are found explicitly and the optimal shift parameters are calculated using MATHEMATICA 4.1 computer system. In various applications the problem on separation the original signal and the noise arises. In this paper we consider two cases, which naturally arise in applied problems. In the first case, the original signal permits linear prediction by its past behavior. In the second case the original signal is the values of some analytic function at a points from unit disk. In the both cases the noise is assumed to be a stationary process with zero mean value. Let us note that the first case arises in Physical phenomena consideration. The second case arises in Identification problems for linear systems. RECONSTRUCTION OF CONVEX BODIES FROM PROJECTION CURVATURE RADIUS FUNCTION In this article we pose the problem of existence and uniqueness of convex body for which the projection curvature radius function coincides with given function. We and a necessary and sufficient condition that ensures a positive answer to both questions and suggest an algorithm of construction of the body. Also we find a representation of the support function of a convex body by projection curvature radii. Weighted Classes of Regular Functions Area Integrable Over the Unit Disc This preprint contains some generalizations of the main theorems of M.M.Djrbashian of 1945--1948 which laid ground for the theory of $A^p_\alpha$ $($or initially $H^p(\alpha))$ spaces and his factorization theory of classes $N\{\omega\}$ exhausting all functions meromorphic in the unit disc. Also some later results on $A^p_\alpha$ spaces and Nevanlinna's weighted class are improved. The preprint contains the main analytic apparatus for generalizing almost all known results on $A^p_\alpha$ spaces within a new theory, where instead $(1-r^2)^\alpha dr$ $(-1<\alpha<+\infty$, $0<r<1)$ some weights of the form $d\omega(r^2)$ are used. The obtained results make evident that the theory of $A^p_\omega$ spaces and the factorization theory of M.M.Djrbashian are inseparable parts of a general theory of classes of regular functions associated with M.M.Djrbashian general integrodifferentiation. The author hopes that the publication of this preprint can lead to clarification of some priority misunderstandings in the field.
Hey I want to convert Cartesian to spherical coordinate system. I referred many site and for calculating elevation angle $\theta$ from positive z axis they all used formula $\arctan \frac { \sqrt{x^2+y^2}}{z}$ but my problem is $\theta$ can be anything from 0 to 180 degrees but the range of inverse tan is from -90 to 90 so how is that online calculators gives angle greater than 90. For example take x y z with x and y positive and z negative so angle should be between 90 and 180 but inverse tan gives some negative angle. I tried with x=93.3 y=25 z=-65.8819 $\sqrt{x^2+y^2}$ gives 96.5926 $\arctan(96.5926)/(-65.8819)$ gives -55.7028 but correct answer is 124.30 degrees i guess Thanks Just choose the range of inverse of tangent that fits your need. That is, if $z = 0$, then $\theta = 90^\circ$. If $z < 0$, then $90^\circ<\theta\le180^\circ$ and $\theta$ satisfies $\tan \theta = \frac{\sqrt{x^2+y^2}}{z}$. If $z > 0$, then $0^\circ\le\theta<90^\circ$ and $\tan \theta = \frac{\sqrt{x^2+y^2}}{z}$. These cases come from your application. For easier computation and to let your computer determine the cases for you, you can look up the function atan2 which is very common among programming languages. Then your $\theta$ is $$\theta = \text{atan2}\left(\sqrt{x^2+y^2},z\right)$$
Example 7-15: Swiss Bank Notes Section An example of the calculation of simultaneous confidence intervals using the Swiss Bank Notes data is given in the expression below: $\bar{x}_{1k}-\bar{x}_{2k} \pm \sqrt{\frac{p(n_1+n_2-2)}{n_1+n_2-p-1}F_{p,n_1+n_2-p-1,\alpha}}\sqrt{\left(\frac{1}{n_1}+\frac{1}{n_2}\right) s^2_k}$ Here we note that the sample sizes are both equal to 100, $n = n_{1} = n_{2} =100$, so there is going to be simplification of our formula inside the radicals as shown above. Carrying out the math for the variable Length, we end up with an interval that runs from -0.044 to 0.336 as shown below. Using SAS The SAS program, below, can be used to compute the simultaneous confidence intervals for the 6 variables. Download the SAS program here: swiss11.sasView the video explanation of the SAS code. The downloadable results as listed here: swiss11.lst. Using Minitab At this time Minitab does not support this procedure. Analysis Obs Variable type _TYPE _FREQ_ n1 xbar1 s21 n2 xbar1 s22 1 bottom fake 0 100 100 8.305 0.41321 100 10.530 1.28131 2 diagon fake 0 100 100 141.517 0.19981 100 139.450 0.31121 3 left fake 0 100 100 129.943 0.13258 100 130.300 0.06505 4 length fake 0 100 100 214.969 0.15024 100 214.823 0.12401 5 right fake 0 100 100 129.720 0.12626 100 130.193 0.08894 6 top fake 0 100 100 10.168 0.42119 100 11.133 0.40446 Obs f t sp losim upsim lobon upbon 1 2.14580 2.66503 0.84726 -2.69809 1.75191 -2.57192 -1.97808 2 2.14580 2.66503 0.2551 1.80720 2.32680 1.87649 2.25751 3 2.14580 2.66503 0.09881 -0.51857 -0.19543 -0.47547 -0.23853 4 2.14580 2.66503 0.13713 -0.04433 0.33633 0.00643 0.28557 5 2.14580 2.66503 0.10760 -0.64160 -0.30440 -0.59663 -0.34937 6 2.14580 2.66503 0.41282 -1.29523 -0.63477 -1.20716 -0.72284 Thebounds of the simultaneous confidence intervals are given in columns for losim and upsim. Those entries are copied into the table below: Variable 95% Confidence Interval Length -0.044, 0.336 Left Width -0.519, -0.195 Right Width -0.642, -0.304 Bottom Margin -2.698, -1.752 Top Margin -1.295, -0.635 Diagonal 1.807, 2.327 You need to be careful where they appear in the table in the output. Note!The variables are now sorted in For example, length would be the fourth line of the output data. In any case you should be able to find the numbers for the lower and upper bound of the simultaneous confidence intervals from the SAS output and see where they appear in the table above. The interval for length, for example, can then be seen to be -0.044 to 0.336 as was obtained from the hand calculations previously. alphabetic order! When interpreting these intervals we need to see which intervals include 0, which ones fall entirely below 0, and which ones fall entirely above 0. The first thing that we notice is that interval for length includes 0. This suggests that we can not distinguish between the lengths of the counterfeit and genuine bank notes. The intervals for both width measurements fall below 0. Since these intervals are being calculated by taking the genuine notes minus the counterfeit notes this would suggest that the counterfeit notes are larger on these variables and we can conclude that the left and right margins of the counterfeit notes are wider than the genuine notes. Similarly we can conclude that the top and bottom margins of the counterfeit are also too large. Note, however, that the interval for the diagonal measurements fall entirely above 0. This suggests that the diagonal measurements of the counterfeit notes are smaller than that of the genuine notes. Conclusions Counterfeit notes are too wide on both the left and right margins. The top and bottom margins of the counterfeit notes are too large. The diagonal measurement of the counterfeit notes is smaller than that of the genuine notes. Cannot distinguish between the lengths of the counterfeit and genuine bank notes.
If you need a pdf version of these notes you can get it here We can begin our analysis of this circuit by applying Kirchoff's loop rule to the find potential at any given time. $V=V_{R}+V_{L}+V_{C}$ However an important consequence of the voltages not being at the same phase in the different components is that the peak voltage is not experienced at the same time in each component, and so the peak source voltage $V_{0}$ $V_{0}\neq V_{R0}+V_{L0}+V_{C0}$ The condition of continuity of current demands that the current throughout the circuit should always be in phase, so at any point in this circuit the current will $I=I_{0}\cos\omega t$ To understand the Phasor approach to AC circuits we can follow a very nice set of animations from Physclips. A phasor is a way of representing the voltage across a component taking into account the phase difference between the voltage and the current. As we saw in our last lecture a resistor has a voltage in phase with the current flowing through it. So if we now represent the current as vector moving in a plane we can also represent the voltage across the resistor as a vector of magnitude $V_{R}=I{R}$ which points in the same direction as the current flowing through it. We saw in our last lecture that in a capacitor the current leads the voltage by 90 o. Also we saw that the reactance $X_{C}=\frac{1}{\omega C}=\frac{1}{2\pi f C}$ depends on frequency and so the size of the voltage phasor $V_{0}=I_{0}X_{C}$ also should. We saw in our last lecture that in an inductor the current lags the voltage by 90 o. Also we saw that the reactance $X_{L}=\omega L=2\pi f L$ depends on frequency and so the size of the voltage phasor $V_{0}=I_{0}X_{L}$ also should. To find the total voltage in the circuit at time $t$ we add the phasors for the different components together as we would vectors. We can see that for a current $I=I_{0}\cos\omega t$ flowing through through the circuit the voltage will be offset by a phase $\phi$ $V=V_{0}\cos(\omega t + \phi)$ The peak voltage $V_{0}$ is linked to the peak current $I_{0}$ through the impedance $Z$ $V_{0}=I_{0}Z$ and we can also say $V_{rms}=I_{rms}Z$ The value of $Z$ is found by considering the vector sum of the voltages $V_{0}=\sqrt{V_{R0}^{2}+(VL_{0}-V_{C0})^{2}}=I_{0}\sqrt{R^{2}+(X_{L}-X_{C})^2}$ so $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}$ The phase difference between the current and voltage $\phi$ is obtained from $\tan \phi=\frac{V_{L0}-V_{C0}}{V_{R0}}=\frac{I_{0}(X_{L}-X_{C})}{I_{0}R}=\frac{X_{L}-X_{C}}{R}$ Power in an LRC circuit is only dissipated in the resistor, and so the average power dissipated is given by $\bar{P}=I_{RMS}^{2}R$ but we may want to express this in terms of the impedance of the circuit or the $V_{RMS}$ which is applied. To do this we write $\cos \phi =\frac{V_{R0}}{V_{0}}=\frac{I_{0}R}{I_{0}Z}=\frac{R}{Z}$ which means that $R=Z\cos\phi$ and $\bar{P}=I_{RMS}^{2}Z\cos\phi$ or, as $V_{RMS}=I_{RMS}Z$ $\bar{P}=I_{RMS}V_{RMS}\cos\phi$ The RMS current in the circuit we are considering is given by $I_{RMS}=\frac{V_{RMS}}{Z}=\frac{V_{RMS}}{\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}}$ We can see that the current should be frequency dependent and have a maximum when $(\omega L - \frac{1}{\omega C})=0$ which gives the resonant frequency $\omega_{0}=\sqrt{\frac{1}{LC}}$ While the phasor approach gave us a nice visual picture of how a simple combination of circuit elements should respond to an AC field, a graphical addition approach would rapidly become unmanageable when more elements both in series and in parallel are added.We deal with this by representing our current, voltage and the the impedance of circuit elements in terms of complex numbers. Note, that we will use the math and physics symbol for $i=\sqrt{-1}$ electrical engineers frequently use $j$ instead, for example on this page from hyperphysics. The starting point is Euler's formula. $e^{i\omega t}=\cos\omega t+i\sin\omega t$ Now suppose we want to represent a voltage $V=V_{0}\cos\omega t$ and a current which is offset by a phase $\phi$ $I=I_{0}\cos(\omega t - \phi)$ This can be done by instead writing $V=V_{0}e^{i\omega t}$ and $I=I_{0}e^{i(\omega t - \phi)}$ The advantage of doing this is that we can now express the impedance of a circuit element as $Z=\frac{V_0}{I_{0}}e^{i\phi}=\|Z\|e^{i\phi}$ and we can see that $V=IZ$ In this scheme the impedance of a resistor, capacitor and and inductor become $Z_{R}=R$,$Z_{C}=-\frac{i}{\omega C}$ and $Z_{L}=i\omega L$ When connected in series complex impedances simply add together, the complex math is taking care of the vector addition we had do to previously $Z_{S}=Z_{1}+Z_{2}+Z_{3}+..$ When connected in parallel, impedances, like resistors, add as reciprocals $\frac{1}{Z_{P}}=\frac{1}{Z_{1}} +\frac{1}{Z_{2}}+\frac{1}{Z_{3}}+..$
The paper "Subquadratic Algorithms for 3SUM", by Ilya Baran, Erik D. Demaine, Mihai Patrascu has the following complexity for the 3SUM problem: given a list $L$ of $n$ integers if there are $x,y,z \in L$ such that $x+y=z.$ They state, "On a standard word RAM with $w-$bit words, we obtain a running time of $O(n^2/ \max\{w \log w, \log n (\log \log n)^2 \})$. In the circuit RAM with one nonstandard $AC0$ operation, we obtain $O(n^2/ w^2 \log w)$. In external memory, we achieve $O(n^2/(MB))$, even under the standard assumption of data indivisibility. Cache-obliviously, we obtain a running time of $O(n^2/ MB \log M )$. In all cases, our speedup is almost quadratic in the “parallelism” the model can aﬀord, which may be the best possible. See the Baran, Demaine, Patrascu paper here. Recently, a paper "Threesomes, Degenerates, and Love Triangles" by Grondlund and Pettie has proved that "the decision tree complexity of 3SUM is $O(n^{3/2}\sqrt{\log n})$, and that there is a randomized 3SUM algorithm running in $O(n^2(\log \log n)^2/\log n)$ time, and a deterministic algorithm running in $O(n^2(\log \log n)^{5/3}/(\log n)^{2/3})$ time. These results refute the strongest version of the 3SUM conjecture, namely that its decision tree (and algorithmic) complexity is $Ω(n^2)$." See this second paper here. Clearly, both are important papers. Not being an expert to this area, my question is about how to compare the impact and significance of either, given the different complexity models. Any other insightful comments about this problem is also welcome. For example had the first paper already ruled out the $\Omega(n^2)$ bound?
This was described by EIP 658 which was implemented in the Byzantium fork. The text of the EIP is here, though strangely it doesn't seem to have been formally finalised before the fork. In any case, the relevant text is this: For blocks where block.number >= METROPOLIS_FORK_BLKNUM, the intermediate state root is replaced by a status code, 0 indicating failure (due to any operation that can cause the transaction or top-level call to revert) and 1 indicating success. In terms of your question, then, 1 always equals success. I'm pretty certain that "revert" here doesn't mean "resulting from the revert opcode", but means, essentially, any conditition that causes the state to be reverted - including all the conditions that were formerly called "throws". Now, note that EIP 658 is also called EIP 98, and is also described here EIP98 is authored by Vitalik: Option 3 (update 2017.07.28: we are going with this one): For blocks where block.number >= METROPOLIS_FORK_BLKNUM, the intermediate state root parameter in the receipt should be set to a \x01 byte if the outermost code execution succeeded, or a zero byte if the outermost code execution failed. This confirms that failed transactions (whatever the failure mode) result in 0, only successful should result in 1. It applies only to the "outermost code execution". Finally, for the ultimate authority, see the Yellow Paper update by Yoichi (not yet merged). It defines a status code, s'. It's a bit difficult to read in that form, but I think the relevant definition of the status code is this: Line 775: The account's associated code (identified as the fragment whose Keccak hash is $\boldsymbol{\sigma}[c]_c$) is executed according to the execution model (see section \ref{ch:model}). Just as with contract creation, if the execution halts in an exceptional fashion (i.e. due to an exhausted gas supply, stack underflow, invalid jump destination or invalid instruction), then no gas is refunded to the caller and the state is reverted to the point immediately prior to balance transfer (i.e. $\boldsymbol{\sigma}$). Line 782 I think says, if the state is thus reverted, then s' is zero: +s' & \equiv & \begin{cases} +0 & \text{if} \quad \boldsymbol{\sigma}^{**} = \varnothing \\ +1 & \text{otherwise} Other references to s' are in there, and may shed more light.
As in the one-sample case, the simultaneous confidence intervals should be computed only when we are interested in linear combinations of the variables. If the only intervals of interest, however, are the confidence intervals for the individual variables with no linear combinations, then a better approach is to calculate the Bonferroni corrected confidence intervals as given in the expression below: $\bar{x}_{1k} - \bar{x}_{2k} \pm t_{n_1+n_2-2, \frac{\alpha}{2p}}\sqrt{s^2_k\left(\dfrac{1}{n_1}+\frac{1}{n_2}\right)}$ Carrying out the math we end up with an interval that runs from 0.006 to 0.286 as shown below: $\bar{x}_{1k} - \bar{x}_{2k} \pm t_{2n-2, \frac{\alpha}{2p}}\sqrt{\frac{2s^2_k}{n}}$ $214.959 - 214.813 \pm \underset{2.665}{\underbrace{t_{2\times 100-2, \frac{0.05}{2 \times 6}}}}\sqrt{\dfrac{2 \times 0.137}{100}}$ $(0.006, 0.286)$ Using SAS These calculations can also be obtained from the SAS program as shown below: Download the SAS Program here: swiss11.sas Looking at the data step combine and moving down, we can see that the fourth line sets t=tinv. This calculates the critical value from the t-table as desired. Then the lower and upper bounds for the Bonferroni intervals are calculated under lobon and upbon at the bottom of this data step. The downloadable output as given here: swiss11.lst, places the results in the columns for lobon and upbon. Again, make sure you note that the variables are given in alphabetical order rather than in the original order of the data. In any case, you should be able to see where the numbers in the SAS output appear in the table below: Using Minitab At this time Minitab does not support this procedure. Analysis In summary, we have: Variable 95% Simultaneous Confidence Intervals (Bonferroni corrected) Length 0.006, 0.286 Left Width -0.475, -0.239 Right Width -0.597, -0.349 Bottom Margin -2.572, -1.878 Top Margin -1.207, -0.723 Diagonal 1.876, 2.258 The intervals are interpreted in a way similar as before. Here we can see that: Conclusions Length: Genuine notes are longer than counterfeit notes. Left-width and Right-width: Counterfeit notes are too wide on both the left and right margins. Top and Bottom margins: Counterfeit notes are too large. Diagonal measurement: The counterfeit notes are smaller than the genuine notes.
Because the slab extends infinitely in the xy plane, the electric field lies only along the z direction. The differential form of Gauss' Law for such a one-dimensional electric field is $$\frac{dE}{dz}=\frac{\rho(z)}{\epsilon_0}$$ This can be integrated, using the boundary condition that the electric field must be zero at large distances from the slab, because it is electrically neutral. A more elementary way of solving the problem is to divide the slab into infinitesimally thin layers of thickness $dz$. The electric field of each layer is uniform, independent of distance from the layer, and is $dE=\frac{d\sigma}{2\epsilon}$ pointing away from the layer on each side for +ve surface charge density $d\sigma=\rho(z)dz$. The total electric field at any point inside or outside of the slab is found by superposition of fields from every such layer in the slab. Note that the total electric field at any point due to all layers which are closer to the centre plane $z=0$, is zero. This is because the charge density is anti-symmetric : for every layer of +ve area charge density on one side of $z=0$ there is a layer of -ve charge with the same magnitude of area density on the other side of $z=0$. The electric fields of these two layers cancel out for points which outside of the two layers, in the same way that the total electric field is zero outside of a parallel plate capacitor (if the plate dimensions are very much bigger than the distance from them). From this observation you can see that the electric field outside of the slab is zero, because all layers in the slab are closer to the centre plane. The simplest way of getting the field inside the slab is to apply Gauss' Law using a "pill box" Gaussian surface which has one face A of area S at the surface of the slab $z=a$ (where $E(a)=0$) and the other face B at distance $\|z\|<a$ from the centre plane. The other face(s) of the pill box are parallel to the z direction so the electric flux through them is zero. There is no electric flux through face A; the flux through face B is $E(z)S=\frac{q}{\epsilon_0}$ where by integration $$q=\int_a^z S\rho(z)dz$$ is the total charge inside the pill box. See Electric field in a non-uniformly charged sheet. An answer identical to mine is given in the duplicate question Finding the electric field of a NON uniform slab?
For solving systems of autonomous differential equations we use Jordan normal forms as typically they give us an easy way to calculate the solution to our system. If we have the IVP:$$\left\{\begin{matrix}\frac{dx}{dt} = Ax\\ x(0) = x_0\end{matrix}\right.$$ where $A$ is a constant $n \times n$ matrix and $x_0 \in \mathbb{R}^n$, then the solution is $x(t) = \text{exp}(tA)x_0.$ For a general matrix $A$, this may be a pain to calculate, but using Jordan normal forms we can simplify this process. A property of the matrix exponential is that: If $A$, $B$ and $T$ are $n \times n$ matrices and $T$ is invertible, then if $B = T^{-1}AT$, we have $\text{exp}(B) = T^{-1}\text{exp}(A)T$. This is where Jordan normal forms become useful, because there is a specific "formula" for taking the exponential matrix of Jordan normal forms. (See here for a good explanation of the formula and here for an example of someone calculating the exponential.) For $\dot{x} = Ax$, we can make a linear change of coordinates, $x = Ty$ where $T$ is invertible which transforms our differential equation into $$\dot{y} = T^{-1}ATy.$$ From linear algebra, we can choose $T$ such that $T^{-1}AT$ is in Jordan normal form $B$. Then using the "formula" for the matrix exponential for Jordan normal forms we can easily calculate $\text{exp}(tB).$ Finally, using the property above, the solution to our IVP will be $$x(t) = \text{exp}(tA)x_0 = T\text{exp}(tB)T^{-1}x_0.$$ So in short, Jordan normal forms are used for calculating the solutions of systems of differential equations because they make the calculations easier. Alternative methods would be calculating the exponential term-by-term, not a good idea unless you have a very simple matrix. Or it's possible to get the solution just by looking at the eigenvalues and eigenvectors, but as far as I'm aware that is just an equivalent version of the Jordan normal form method.
I wanted to look into the case where you have an experiment in which your units of analysis are naturally clustered (e.g., households in villages), but you randomize within clusters. The goal is to estimate a treatment effect in terms of difference in means, using design-based principles and frequentist methods for inference. Randomization ensures that the difference in means is unbiased for the sample average treatment effect. Using only randomization as the basis for inference, I know the variance of this estimator is not identified for the sample, as it requires knowledge of the covariance of potential outcomes. But the usual sample estimators for the variance are conservative. If, however, the experiment is run on a random sample from an infinitely large population, then the standard methods are unbiased for the mean and for the variance of the difference in means estimator applied to this population (refer to Neyman, 1990; Rubin, 1990; Imai, 2008; for finite populations, things are more complicated, and the infinite population assumption is often a reasonable approximation). I understand that these are the principles that justify the usual frequentist estimation techniques for inference on population level treatment effects in randomized experiments. The question I had was, how should we account for dependencies in potential outcomes within clusters? The cluster-randomized-experiments literature has made it clear that when the treatment is applied at the cluster level, we certainly need to account for the clustering. This is made explicit by the Moulton factor, which measures the ratio of the true standard deviation of the difference in means with clustered data to a standard error estimate that ignores the clustering. The Moulton factor is given by the square root of the following expression (from Angrist and Pischke, 2008:311), [latex] \frac{V(\hat{\beta})}{V_c(\hat{\beta})} = 1 + \left[\frac{V(n_g)}{\bar{n}}+\bar{n} -1\right]\rho_x\rho_e, [/latex] where $latex V(\hat{\beta})$ is the true variance of the differences in means estimator, $latex \hat{\beta}$, $latex V_c(\hat{\beta})$ is the limit of the variance estimator that ignores the clustering, $latex n_g$ refers to cluster size and $latex \bar{n}$ is the average cluster size, and $latex \rho_x$ and $latex \rho_e$ are the (Kish-type) intra-class correlations for the treatment variable and deviations from predicted means (i.e., the residuals if the difference in means is computed with OLS). From this expression, it would seem that the randomization within clusters would render the need for cluster adjustment unnecessary, given that randomization within clusters ensures that the intra-class correlation of the treatment variable ($latex x$) is zero. But this runs contrary to another intuition: The standard error for your mean estimate for the treated probably needs to take into account the dependence among treated groups within each cluster, and the same is true for your mean estimate for the controls. Heuristically, a more useful summary might be the intra-class correlation among treated units, grouped by cluster, and control units grouped by cluster. Treatment status is constant within those subgroups in each cluster, and so the $latex \rho_x$ equals 1. In that case, substantial residual dependence, measured by the $latex \rho_e$ term, should have some bite. In lieu of working this out analytically, I came up with some simulations to study the performance of cluster robust standard errors (CSEs) versus heteroskedasticity robust standard errors (HSEs) in this situation. In principle, if there is meaningful within-cluster dependence, the CSEs should be bigger than the HSEs, although the opposite may occur in any given sample. So I also looked into the performance of $latex \max(\text{CSE, HSE})$, which I will call “conservative” CSEs (CCSEs). To generate the dependence structure, I constructed potential outcomes as draws from a multivariate normal distribution: $latex \left[ \begin{array}{c} Y^1_{1g}\\ \vdots \\ Y^1_{Mg} \\ Y^0_{1g}\\ \vdots \\ Y^0_{Mg} \end{array} \right] \sim \text{MVN}(\mathbf{0}, \left[ \begin{array}{cc} \Sigma_{11} & \Sigma_{10}\\ \Sigma_{10} & \Sigma_{00} \end{array} \right] )$, where $latex g$ indexes clusters of size $latex M$, $latex Y^1_{gm}$ is the potential outcome under treatment for unit $latex m$ in group $latex g$ and $latex Y^0_{gm}$ is the potential outcome for that unit under control, $latex \Sigma_{11}, \Sigma_{10}, \Sigma_{00}$ specify potential outcomes covariances among treated units, across treated and control units, and among control units, respectively. I kept these matrices simple, with variances for all potential outcomes fixed to 1. I looked into different values for the correlation across potential outcomes for a given unit. I fixed this to $latex \rho_{10,i}$ for all units (the diagonal of $latex \Sigma_{10}$ is an array with this term). I also looked at different values for within-cluster dependence, using common values for the off-diagonals in all three matrices. Call the correlations in the off diagonals, $latex \rho_{1,ij}, \rho_{10, ij}$, and $latex \rho_{0,ij}$, standing for within-cluster correlation between (i) different treated units, (ii) across treated and control outcomes for different units, and (iii) between different control units, respectively. All clusters were given the same covariance structure. A “placebo” treatment was assigned to units via a design that randomly assigned 1/2 of units to treatment within each cluster (i.e., it was a unit-level randomized design blocked by cluster; it was not a cluster-randomized design). The placebo treatment has no causal effect, and so the “true” average treatment effect in this simulated experiment is 0. For the simulations, I fixed the group size ($latex M$) to 10 and the number of groups ($latex G$) to 50 for a total sample of 500. I estimated the average mean difference using OLS. Standard errors were computed as Stata-style finite-sample-adjusted robust standard errors. Specifically, let $latex X$ stand for a matrix that contains the treatment assignment vector in the first column and a vector of ones in the second, and let $latex x_{gm}$ and $latex e_{gm}$ stand for the covariate values and OLS residual for unit $latex m$ in group $latex g$, respectively. Then, HSEs were computed as the square roots of the diagonals of, $latex \frac{GM}{GM-2}(X’X)^{-1} \left( \sum_{g=1}^G \sum_{m=1}^M x_{gm}x_{gm}’ e_{gm}^2 \right)(X’X)^{-1}$ and CSEs were computed as the square roots of the diagonals of, $latex \frac{G}{G-1}\frac{GM-1}{GM-2}(X’X)^{-1} \left( \sum_{g=1}^G X_g’ e_g e_g’X_g \right)(X’X)^{-1}$ CCSEs were computed as the maximum of the two. I evaluated the standard error corresponding to the treatment variable. The table below shows the results from 5,000 runs on different covariance parameter settings. (In all cases, the OLS estimator was unbiased, as expected, so I don’t report that.) The column labeled “S.D. beta” gives the empirical standard deviation of the difference in means estimates. That is our benchmark “truth” against which we compare the different standard error estimates. Whenever there is any cluster dependence, the HSEs perform poorly. For the most part, the CSEs are reliable, although they are not appreciably more reliable than the CCSEs. Indeed, only when we have quite high levels of correlation between unit potential outcomes do they tend to be over-conservative, and even in these cases, the degree of conservatism is not so high on average. Furthermore, in simulations 1, 5, and 9, where HSEs and CSEs should be equal asymptotically, we see that in the majority of cases, CSEs are smaller than HSEs. In such cases, CSEs are anti-conservative (with respect to Type I error). By construction, the CCSEs do not suffer this problem. Thus, if one’s primary concern is to avoid Type I error, CCSEs are a good choice. Thus, in answering our question above, it is clear that yes, we do need to account for the clustering, and to deal with finite sample issues, CCSEs are an appealing option. This is actually in line with what Angrist and Pischke recommend when discussing clustered data. As a second conclusion, it seems the intuition derived from inspecting the Moulton factor here is incorrect. Why might this be the case? Is it because the original derivation is based on a random unit effects model for which, indeed, if we have balance over treatment assignment, then these unit effects will cancel? If so, under more general types of cluster dependence, this cancellation will not occur. Any thoughts on that would be appreciated! UPDATE: Responding to WL’s excellent comments below, I tried another set of simulations. They are similar to what WL suggested in “scenarios two and three,” although rather than using additive village “shocks” I induce dependence via the covariance matrix directly, as discussed above. The difference with the previous simulations is that, rather than having potential outcomes for all groups have a mean of 0, I allowed the group means to vary. To do so, I drew $latex G$ group means, $latex \alpha_1,\hdots,\alpha_G$, independently from $latex \text{N}(0,1)$ and then fixed these group means over all simulation runs. Thus, all potential outcomes for group $latex g$ have mean $latex \alpha_g$ over all simulation runs. This in itself induces an intra-cluster correlation of about 0.5 (recall variances were 1 for all potential outcomes). Then, potential outcomes were drawn using the within-cluster dependence structure described above. Here are the results:In this simulation, using the CCSEs comes at a heavier price than before. When there is either no substantial dependence beyond the group fixed effects (simulations 1, 5, and 9), or if there is substantial dependence between unit-level potential outcomes (simulations 10, 11, and 12), CCSEs strike me as too conservative now. The CSEs continue to perform quite well on average. Simulations 1, 5, and 9 are similar to WL’s scenario two, but note here that the CSEs are much preferable to the HSEs, which wasn’t expected. (Of course, this assumes no coding errors! If you want to check it out for yourself, here is the simulation code (for R): link .)
A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$.I believe I have numerically discovered that$$\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty$$but cannot ... So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$. Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$. Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow. Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$. Well, we do know what the eigenvalues are... The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$. Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker "a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers. I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd... Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work. @TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now) Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $\|\|v\|\| = \|\|(v)_S\|\|$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism @AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$. Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1) For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism
As before, the next step is to determine how these notes differ. This may be carried out using the simultaneous $(1 - α) × 100\%$ confidence intervals. For Large Samples: simultaneous $(1 - α) × 100\%$ confidence intervals may be calculated using the expression below: $\bar{x}_{1k} - \bar{x}_{2k} \pm \sqrt{\chi^2_{p,\alpha}\left(\dfrac{s^2_{1k}}{n_1} + \dfrac{s^2_{2k}}{n_2}\right)}$ This involves the differences in the sample means for the kth variable, plus or minus the square root of the critical value from the chi-square table times the sum of the sample variances divided by their respective sample sizes. For Small Samples: it is better use to the expression below: $\bar{x}_{1k} - \bar{x}_{2k} \pm \sqrt{\dfrac{p(n_1+n_2-2)}{n_1+n_2-p-1}F_{p,\nu,\alpha}}\sqrt{\left(\dfrac{s^2_{1k}}{n_1} + \dfrac{s^2_{2k}}{n_2}\right)}$ Basically the chi-square value and the square root is replaced by the critical value from the F-table, times a function of the number of variables, p, and the sample sizes n 1 and n 2. Example 7-16: Swiss Bank Notes Section An example of the large approximation for length is given by the hand calculation in the expression below: $214.969 - 214.823 \pm \sqrt{12.59 \times (\dfrac{0.15024}{100}+\dfrac{0.12401}{100})}$ $(-0.040, 0.332)$ Here the sample mean for the length of a genuine note was 214.969. We will subtract the sample mean for the length of a counterfeit note, 214.823. The critical value for a chi-square distribution with 6 degrees of freedom evaluated at 0.05 is 12.59. The sample variance for the first population of genuine notes is 0.15024 which we will divide by a sample size of 100. The sample variance for the second population of counterfeit notes is 0.12401 which will also divide by its sample size of 100. This yields the confidence interval that runs from -0.04 to 0.332. The results of these calculations for each of the variables are summarized in the table below. Basically, they give us results that are comparable to the results we obtained earlier under the assumption of homogeneity for variance-covariance matrices. Variable 95% Confidence Interval Length -0.040, 0.332 Left Width -0.515, -0.199 Right Width -0.638, -0.308 Bottom Margin -2.687, -1.763 Top Margin -1.287, -0.643 Diagonal 1.813, 2.321
In the context of superstring compactification on a 6-manifold which admits a covariantly conserved spinor $\eta$, which we normalize so that $\eta^{\dagger} \eta = 1$, I am trying to show that the almost complex structure $$J_m^{\,\,n} = -i \eta^{\dagger} \gamma_{m}^{\,\,n}\gamma \eta$$ satisfies $$J_m^{\,\,n} J_n^{\,\,p} = -\delta_m^p. $$ Here $\gamma_{mn} = \gamma_{[mn]}$ and $\gamma$ is the chirality operator. I'm aware that the Fierz rearrangement identity should be used, $$\chi_\alpha \zeta \xi + \zeta_\alpha \xi \chi + \xi_{\alpha} \chi \zeta= 0$$ but I cannot show $J^2 = -1$, and would appreciate any help. Attempt. Take $\eta$ to be real and to have chirality $+1$ so $\gamma \eta = \eta$. Now, with some spinor indices explicit, $$J_m^{\,\,n} J_n^{\,\,p} = - \eta_{\alpha} (\gamma_m^{\,\,n}\eta)^{\alpha} \eta(\gamma_{n}^{\,\,p}\eta)$$ $$ = \eta_{\alpha}\eta^\alpha (\gamma_{n}^{\,\,p}\eta)(\gamma_m^{\,\,n}\eta)+ \eta_\alpha (\gamma_{n}^{\,\,p}\eta)^\alpha(\gamma_m^{\,\,n}\eta)\eta $$ $$ = 1 \cdot (\gamma_{n}^{\,\,p}\eta)(\gamma_m^{\,\,n}\eta)+ \eta_\alpha (\gamma_{n}^{\,\,p}\eta)^\alpha(\gamma_m^{\,\,n}\eta)\eta.$$ Perhaps there is some gamma matrix algebra which would help me simplify this to $-\delta_m^{\,\,p}$. (Ref: "Vacuum Configurations for Superstrings", Candelas et al. 1985)

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