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In his celebrated paper "Conjugate Coding" (written around 1970), Stephen Wiesner proposed a scheme for quantum money that is unconditionally impossible to counterfeit, assuming that the issuing bank has access to a giant table of random numbers, and that banknotes can be brought back to the bank for verification. In Wiesner's scheme, each banknote consists of a classical "serial number" $s$, together with a quantum money state $\|\psi_s\rangle$ consisting of $n$ unentangled qubits, each one either $$\|0\rangle,\ \|1\rangle,\ \|+\rangle=(\|0\rangle+\|1\rangle)/\sqrt{2},\ \text{or}\ \|-\rangle=(\|0\rangle-\|1\rangle)/\sqrt{2}.$$ The bank remembers a classical description of $\|\psi_s\rangle$ for every $s$. And therefore, when $\|\psi_s\rangle$ is brought back to the bank for verification, the bank can measure each qubit of $\|\psi_s\rangle$ in the correct basis (either $\{\|0\rangle,\|1\rangle\}$ or ${\|+\rangle,\|-\rangle}$), and check that it gets the correct outcomes. On the other hand, because of the uncertainty relation (or alternatively, the No-Cloning Theorem), it's "intuitively obvious" that, if a counterfeiter who doesn't know the correct bases tries to copy $\|\psi_s\rangle$, then the probability that both of the counterfeiter's output states pass the bank's verification test can be at most $c^n$, for some constant $c<1$. Furthermore, this should be true regardless of what strategy the counterfeiter uses, consistent with quantum mechanics (e.g., even if the counterfeiter uses fancy entangled measurements on $\|\psi_s\rangle$). However, while writing a paper about other quantum money schemes, my coauthor and I realized that we'd never seen a rigorous proof of the above claim anywhere, or an explicit upper bound on $c$: neither in Wiesner's original paper nor in any later one. So, has such a proof (with an upper bound on $c$) been published? If not, then can one derive such a proof in a more-or-less straightforward way from (say) approximate versions of the No-Cloning Theorem, or results about the security of the BB84 quantum key distribution scheme? Update: In light of the discussion with Joe Fitzsimons below, I should clarify that I'm looking for more than just a reduction from the security of BB84. Rather, I'm looking for an explicit upper bound on the probability of successful counterfeiting (i.e., on $c$)---and ideally, also some understanding of what the optimal counterfeiting strategy looks like. I.e., does the optimal strategy simply measure each qubit of $\|\psi_s\rangle$ independently, say in the basis $$\{ \cos(\pi/8)\|0\rangle+\sin(\pi/8)\|1\rangle, \sin(\pi/8)\|0\rangle-\cos(\pi/8)\|1\rangle \}?$$ Or is there an entangled counterfeiting strategy that does better? Update 2: Right now, the best counterfeiting strategies that I know are (a) the strategy above, and (b) the strategy that simply measures each qubit in the $\{\|0\rangle,\|1\rangle\}$ basis and "hopes for the best." Interestingly, both of these strategies turn out to achieve a success probability of (5/8) n. So, my conjecture of the moment is that (5/8) n might be the right answer. In any case, the fact that 5/8 is a lower bound on c rules out any security argument for Wiesner's scheme that's "too" simple (for example, any argument to the effect that there's nothing nontrivial that a counterfeiter can do, and therefore the right answer is c=1/2). This post has been migrated from (A51.SE) Update 3: Nope, the right answer is (3/4) n! See the discussion thread below Abel Molina's answer.
Using residues, try the contour below with $R \rightarrow \infty$ and $$\lim_{R \rightarrow \infty } \int_0^R \frac{1}{1+r^n} dr \rightarrow \int_0^\infty \frac{1}{1+x^n} dx$$ I've attempted the residue summation, but my sum did not converge. 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 Using residues, try the contour below with $R \rightarrow \infty$ and $$\lim_{R \rightarrow \infty } \int_0^R \frac{1}{1+r^n} dr \rightarrow \int_0^\infty \frac{1}{1+x^n} dx$$ I've attempted the residue summation, but my sum did not converge. The integral of $$ \int_\gamma\frac1{1+z^n}\mathrm{d}z\tag{1} $$ on the outgoing ray on the real axis tends to $$ \int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{2} $$ On the incoming ray parallel to $e^{2\pi i/n}$, the integral tends to $$ -e^{2\pi i/n}\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{3} $$ For $n\ge2$, the integral on the circular arc vanishes. Therefore, $$ \int_\gamma\frac1{1+z^n}\mathrm{d}z =\left(1-e^{2\pi i/n}\right)\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{4} $$ There is one singularity contained in $\gamma$ at $z_0=e^{\pi i/n}$. The residue of $\frac1{1+x^n}$ at $z_0$ is $\frac1{nz_0^{n-1}}=-\frac{z_0}{n}$. Thus, $$ 2\pi i\left(-\frac{e^{\pi i/n}}{n}\right) =\left(1-e^{2\pi i/n}\right)\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{5} $$ which resolves by division to $$ \int_0^\infty\frac1{1+x^n}\mathrm{d}x=\frac{\pi/n}{\sin(\pi/n)}\tag{6} $$ For $n=1$, the integral diverges and $\frac{\pi}{\sin(\pi)}=\frac\pi0$.
Suppose you have a fair die with 10 sides with numbers from 1 to 10. You roll the die and take the sum until the sum is greater than 100. What is the expected value of this sum? Well, you can make numbers from 1 to 110, that's 110 possible states you can be in. That's a large transition matrix, but not intractable. Define transition matrix $M$ as: $$\begin{align} M_{i,j} = \begin{cases} \frac{1}{10} & \text{ if } 1 \le j - i \le 10 \text{ and } 1 \le i \le 100 \\ 1 & \text{ if } i = j\text{ and } 100 < i \le 110 \\ 0 & \text{ otherwise} \end{cases} \end{align}$$ Define the initial state vector: $$\begin{align} V_{j} = \begin{cases} \frac{1}{10} & \text{ if } 1 \le j \le 10 \\ 0 & \text{ otherwise} \end{cases} \end{align}$$ This matrix compute the probability that you end up in state $j$. I won't wrote out the whole thing here. The steady state $S$ is given by: $$S = VM^{\infty}$$ I don't recommend doing this by hand. Anyway, you get the following probabilities: $$\begin{array} {c\|c} \text{End State} & \text{Probability} \\ \hline 101 & \frac{{ 14545454540916238222253308031039403263876427137099 \atop 728738149747953197899302063661139633020606426446001 }}{10^{101}} \\ \hline 102 & \frac{{ 18181818143103207886299794518678653455112915813572 \atop 471568730433172695421560362344285718841548526446001 }}{10^{101}} \\ \hline 103 & \frac{{ 16363636337149401877562367260240328253764897737948 \atop 745622241485421850822156839220501392260147526446001 }}{10^{101}} \\ \hline 104 & \frac{{ 12727272741576429962125352735631301525861758140731 \atop 984148880861015973102098820410322797857111216446001 }}{10^{101}} \\ \hline 105 & \frac{{ 10909090931874858870242133363925460156752233410373 \atop 601637391699278967219484863685353489177266485446001 }}{10^{101}} \\ \hline 106 & \frac{{ 90909091116377120481295620461022602720063194921283 \atop 52571281564919296780386873223909380629437281346001 }}{10^{101}} \\ \hline 107 & \frac{{ 72727272852713068490158133017227152668140086810036 \atop 91839439446721479480174650845945205326825156836001 }}{10^{101}} \\ \hline 108 & \frac{{ 54545454582842347527531906998645390126303113111522 \atop 24370063982379262253640845972771391003951819875001 }}{10^{101}} \\ \hline 109 & \frac{{ 36363636346255163502142715869656509893927302281916 \atop 05910755643757924951410231819125651609791149217901 }}{10^{101}} \\ \hline 110 & \frac{{ 18181818155610931814042064558296878037883980477975 \atop 93593065135379352069784448816645340273314411495091 }}{10^{101}} \\ \hline \end{array}$$ Expected state is calculated as always, and the final answer is : $$\sum_{k=100}^{109} k\cdot S_{j,k} = \frac{{ 2080000000053214123545556126144601328836935442611255 \atop 847240374822309959920561393884518831211143790906011 }}{2\cdot10^{100}}$$ which is almost exactly $104$ (to eight decimal places). EDIT-- Corrected post, originally answered the question "at least 100" rather than "more than 100". An argument like the one in this answer, shows that the distribution of the final position is very closely approximated by $[10/55,9/55,8/55,\dots ,1/55]$ on the states $[101,102,103,\dots, 110]$. Therefore the average position at the end of the game is very close to $$\sum_{i=1}^{10} (100+i)(11-i)/55=104. $$ Added: Here is some further information on the approximate hitting distribution. Let's express the hitting distribution of $100,101,102,\dots$ as $\sum_{i=0}^9 \pi_i \,\delta_{100+i}$, and the hitting distribution of $101,102, 103,\dots$ as $\sum_{i=0}^9 \pi^\prime_i\, \delta_{101+i}$, where $\pi=(\pi_0,\dots,\pi_9)$ and $\pi^\prime=(\pi^\prime_0,\dots,\pi^\prime_9)$ are distributions on $\{0,1,2,\dots,9\}$. By the strong Markov property, we have $$\pi^\prime=\pi_0 U+\sum_{i=1}^9 \pi_i\delta_{i-1} $$ where $U$ is the uniform distribution on $\{0,1,2,\dots,9\}$. On the other hand, since the process has been running for a long time, we have $\pi\approx \pi^\prime$. If you take this as equality, write $\pi=\pi_0 U+\sum_{i=1}^9 \pi_i\delta_{i-1}$, and solve for $\pi$ you get the required pattern. Introductory remark. As pointed out by @DanielV what follows does not answer the exact question, which asks for a value more than 100 rather than at least 100. Adapting the solution below is left as an exercise to the reader. We can actually say a bit more about this problem using generating functions. Suppose we have an $n$-sided fair die and we ask about the expected value of the sum until it is at least $n^2$. We classify according to the number $k$ of rolls until a value $n^2-q$ is obtained (this is the value before we exceed/hit $n^2$), where $1\le q\le n.$ This scenario is represented by the following generating function: $$g(z) = \sum_{q=1}^n \frac{1}{n} \left(\sum_{p=1}^{n-q+1} z^{n-p+1}\right) z^{n^2-q} [z^{n^2-q}] \left(\frac{1}{n}\right)^k (z+z^2+\cdots+z^n)^k.$$ Summing over $k$ we obtain for the inner term $$\sum_{k\ge 0} \left(\frac{1}{n}\right)^k (z+z^2+\cdots+z^n)^k = \sum_{k\ge 0} \left(\frac{1}{n}\right)^k z^k \left(\frac{1-z^n}{1-z}\right)^k \\ = \frac{1}{1- z/n \times (1-z^n)/(1-z)} = \frac{1-z}{1-z - z/n \times (1-z^n)}.$$ Call this $f(z).$ This yields for the entire generating function $g(z)$ that $$\sum_{q=1}^n \frac{1}{n} \left(\sum_{p=1}^{n-q+1} z^{n-p+1}\right) z^{n^2-q} [z^{n^2-q}] \frac{1-z}{1-z - z/n \times (1-z^n)}.$$ As this is a PGF we may differentiate and set $z=1$ to obtain the expectation. This operation produces $$\sum_{p=1}^{n-q+1} (n-p+1) + (n-q+1)(n^2-q) = \frac{1}{2} (n-q+1) (2n^2+n-q).$$ We obtain the formula $$\frac{1}{2n} \sum_{q=1}^n (n-q+1) (2n^2+n-q) [z^{n^2-q}] \frac{1-z}{1-z - z/n \times (1-z^n)}.$$ This gives for an ordinary die with six faces the value $$37.666667491012523359$$ and for the ten-sided die the value $$103.00000000410210493.$$ A surprising conjecture. The sequence of values of the expected terminal sum with an $n$-sided die rolled until a sum $\ge n^2$ is reached, times three, for $n=5$ to $n=15$ is extremely close to$$79, 113, 153, 199, 251, 309, 373, 443, 519, 601, 689$$which is OEIS A144391, i.e. $3n^2+n-1,$ giving the closed form: $$\frac{1}{3} (3n^2+n-1).$$There are probably upvotes to be had for a proof of this conjecture. This is the proof. The dominant pole of the generating function is at $z=1$ with residue (apply L'Hopital several times)$$-\lim_{z\to 1} \frac{(1-z)^2}{1-z-z/n\times (1-z^n)}= -\lim_{z\to 1} \frac{-2+2z}{-1-1/n+(n+1)/n\times z^n}\\ = -\lim_{z\to 1} \frac{-2+2z}{-(n+1)/n+(n+1)/n\times z^n}= -\lim_{z\to 1} \frac{2}{(n+1)z^{n-1}} = -\frac{2}{n+1}.$$ Therefore $$[z^{n^2-q}] f(z) \sim \frac{2}{n+1} 1^{n^2-q}$$and the dominant contribution to the sum is$$\frac{1}{2n} \sum_{q=1}^n (n-q+1) (2n^2+n-q) \times \frac{2}{n+1}\\ = {n}^{2}+1/3\,n-1/3.$$ Since computer simulations are apparently relevant to this problem I am posting some code that can be used to verify the generating function formula. gf := proc(n) (1-z)/(1-z-z/n(1-z^n)); end; v := proc(n) option remember; 1/2/nadd((n-q+1)(2n^2+n-q)coeftayl(gf(n), z=0, n^2-q),q=1..n); end; ex := proc(n) option remember; local pb, w, res, dist, dist2, term, pot, delta; pb := 1/n; res := 0; dist := 1; do dist := expand(distadd(z^k, k=1..n)); dist2 := 0; delta := 0; for term in dist do pot := degree(term); if pot < n^2 then dist2 := dist2 + term fi; if pot >= n^2 then delta := delta + potcoeff(term, z, pot)pb; fi; od; res := res+delta; if delta > 0 and delta < 10^(-Digits) then break fi; if dist2 = 0 then break fi; pb := pb*1/n; dist := dist2; od; res; end; Some bugs fixed. Needs higher precision (value of Digits) for values like $n=15.$ The version in v that extracts coefficients is not usable for $n>11.$ Using generating functions, there is one way to make each of the numbers $1$ through $10$ on each roll: $$G(x)=1x^1+1x^2+\cdots 1x^{10}=x\frac{x^{10}-1}{x-1}$$ To find the possible ways to get to any number $a$ after $n$ rolls, we need to find the coefficient of $x^a$ in the expansion of $G(x)^n$. So to answer your question, we must find the number of ways to get $101-110$ total, and the number of ways to get each of the outcomes specifically. Thus, we examine the coefficients of $x^{101}$ through $x^{110}$ in the sum of all powers of $G$ (since we don't care about how many rolls it takes). $$G+G^2+\cdots=\frac{G}{1-G}=\frac{x(x^{10}-1)}{(x-1)-x(x^{10}-1)}$$ From here you would have to use a CAS to find the exact result. I got the numbers on Wolfram Alpha, but I have no way to copy them (and they're quite large) so I can't provide the exact answer, but as I mentioned in the comments it will be around $104$. Alternatively, you could try looking at special (or general) cases to find a pattern or formula which would possibly apply to this case. Depending on where you got this question, there is a good chance that it has a much more elegant solution.
Is there somewhere on the internet I can find cosmological redshift data. In particular, I would like to know the redshift around the time when the acceleration of the Universe began to accelerate. One of the main places where data about galaxies gets aggregated is the NASA Extragalactic Database (NED). For example, here's the information page for M101 with the default cosmology in their search form. In particularly you want to look at the redshift-independent distances, and the redshift data points. Using the 'Metric Distance' you can calculate the cosmological redshift it would have if it weren't moving (for a given cosmology) by numerically inverting equation 15 from Hogg's cosmology calculations summary paper (probably have to numerically integrate, too). Note that the peculiar velocity (velocity relative to Hubble flow) is usually around hundreds of kilometers per second. So, for any redshift greater than about $0.01$ (equivalent to a radial velocity of about $3,000\operatorname{km}\operatorname{s}^{-1}$) is almost certainly entirely dominated by the cosmological redshift of the object. There are a lot of databases replete with redshifts of galaxies that stretch back to round $z=1$ for ordinary galaxies, and much further carefully selected galaxies and active galactic nuclei/quasars. For example: Sloan Digital Sky Survey (SDSS), DEEP2, the AGN and Galaxy Evolution Survey (AGES), and Galaxy and Mass Assembly (GAMA). This list is nowhere near complete, of course. Redshift of the time when the universe started to accelerate: From Friedmann's equations: $$\dot{a}=aH=H_0\sqrt{\Omega_{m0}/a+a^2\Omega_{\Lambda 0}}$$ required is: $\ddot{a}>0$. Calculation gets you $$a=\left(\frac{\Omega_{m0}}{2\Omega_{\Lambda 0}}\right)^{1/3} \approx 0.6$$ $\rightarrow z=1/a-1 = 0.67$
Generic classifications by p-group class This page will contain results for generic classes of p-groups. It is very much under construction so the list below is not complete. Contents Cyclic p-groups Morita equivalence classes are labelled by Brauer trees, but it is at present an open problem as to which Brauer trees are realised by blocks of finite groups. Each k-Morita equivalence class corresponds to an unique [math]\mathcal{O}[/math]-Morita equivalence class. For [math]p=2,3[/math] every appropriate Brauer tree is realised by a block and we can give generic descriptions. Tame blocks Erdmann classified algebras which are candidates for basic algebras of tame blocks, i.e., those with dihedral, semidihedral or generalised quaternion defect groups (see [Er90] ) and in the cases of dihedral and semihedral defect groups determined which are realised by blocks of finite groups. In the case of generalised quaternion groups, the case of blocks with two simple modules is still open. These classifications only hold with respect to the field k at present. Abelian 2-groups with 2-rank at most three Let [math]l,m,n \geq 1[/math] be distinct with [math]l,m \neq 1[/math] [math]rk_2(D) \leq 3[/math] [math]D[/math] # [math](\mathcal{O}[/math]-)Morita classes # [math]\mathcal{O}[/math]-Derived equiv classes Always endopermutation source Morita equivalent? References Notes [math]C_{2^n}[/math] 1 1 Yes [Li96] Blocks with cyclic defect groups [math]C_2 \times C_2[/math] 3 2 Yes [Li94] [math]C_{2^m} \times C_{2^m}[/math] 2 2 Yes [EKKS14] [math]C_{2^m} \times C_{2^n}[/math] 1 1 Yes [Pu88] Nilpotent blocks [math]C_2 \times C_2 \times C_2[/math] 8 4 [Ea16] [math]C_{2^m} \times C_{2^m} \times C_{2^m}[/math] 4 4 Yes [WZZ18] [math]C_{2^m} \times C_2 \times C_2[/math] 3 2 [EL18a] [math]C_{2^m} \times C_{2^m} \times C_{2^n}[/math] 2 2 [WZZ18] [math]C_{2^l} \times C_{2^m} \times C_{2^n}[/math] 1 1 Yes [Pu88] Nilpotent blocks Abelian 2-groups Donovan's conjecture holds for 2-blocks with abelian defect groups. Some generic classification results are known for certain inertial quotients. These will be detailed here. Minimal nonabelian [math]2[/math]-groups Blocks with defect groups which are minimal nonabelian [math]2[/math]-groups of the form [math]P=\langle x,y:x^{2^r}=y^{2^r}=[x,y]^2=[x,[x,y]]=[y,[x,y]]=1 \rangle[/math] are classified in [EKS12]. There are two [math]\mathcal{O}[/math]-Morita equivalence classes, with representatives [math]\mathcal{O}P[/math] and [math]\mathcal{O}(P:C_3)[/math]. For arbitrary minimal nonabelian [math]2[/math]-groups, by [Sa16] blocks with such defect groups and the same fusion system are isotypic.
Answer 1.47 kg Work Step by Step We know that $v=\frac{l}{t}$ and that v is equal to $\sqrt{\frac{F}{\frac{m}{l}}}$. Thus, we combine these equations to find: $m = \frac{Ft^2}{l} = \frac{78.6\times.585^2}{18.3} = 1.47 \ kg$ You can help us out by revising, improving and updating this answer.Update this answer After you claim an answer you’ll have 24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback.
In this chapter, we will discuss regarding the Density and Hubble parameters. The Hubble parameter is defined as follows − $$H(t) \equiv \frac{da/dt}{a}$$ which measures how rapidly the scale factor changes. More generally, the evolution of the scale factor is determined by the Friedmann Equation. $$H^2(t) \equiv \left ( \frac{\dot{a}}{a} \right )^2 = \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2} + \frac{\wedge}{3}$$ where, ∧ is a cosmological constant. For a flat universe, k = 0, hence the Friedmann Equation becomes − $$\left ( \frac{\dot{a}}{a} \right )^2 = \frac{8\pi G}{3}\rho + \frac{\wedge}{3}$$ For a matter dominated universe, the density varies as − $$\frac{\rho_m}{\rho_{m,0}} = \left ( \frac{a_0}{a} \right )^3 \Rightarrow \rho_m = \rho_{m,0}a^{-3}$$ and, for a radiation dominated universe the density varies as − $$\frac{\rho_{rad}}{\rho_{rad,0}} = \left ( \frac{a_0}{a} \right )^4 \Rightarrow \rho_{rad} = \rho_{rad,0}a^{-4}$$ Presently, we are living in a matter dominated universe. Hence, considering $\rho ≡ \rho_m$, we get − $$\left ( \frac{\dot{a}}{a} \right )^2 = \frac{8\pi G}{3}\rho_{m,0}a^{-3} + \frac{\wedge}{3}$$ The cosmological constant and dark energy density are related as follows − $$\rho_\wedge = \frac{\wedge}{8 \pi G} \Rightarrow \wedge = 8\pi G\rho_\wedge$$ From this, we get − $$\left ( \frac{\dot{a}}{a} \right )^2 = \frac{8\pi G}{3}\rho_{m,0}a^{-3} + \frac{8 \pi G}{3} \rho_\wedge$$ Also, the critical density and Hubble’s constant are related as follows − $$\rho_{c,0} = \frac{3H_0^2}{8 \pi G} \Rightarrow \frac{8\pi G}{3} = \frac{H_0^2}{\rho_{c,0}}$$ From this, we get − $$\left ( \frac{\dot{a}}{a} \right )^2 = \frac{H_0^2}{\rho_{c,0}}\rho_{m,0}a^{-3} + \frac{H_0^2}{\rho_{c,0}}\rho_\wedge$$ $$\left ( \frac{\dot{a}}{a} \right )^2 = H_0^2\Omega_{m,0}a^{-3} + H_0^2\Omega_{\wedge,0}$$ $$(\dot{a})^2 = H_0^2\Omega_{m,0}a^{-1} + H_0^2\Omega_{\wedge,0}a^2$$ $$\left ( \frac{\dot{a}}{H_0} \right )^2 = \Omega_{m,0}\frac{1}{a} + \Omega_{\wedge,0}a^2$$ $$\left ( \frac{\dot{a}}{H_0} \right )^2 = \Omega_{m,0}(1+z) + \Omega_{\wedge,0}\frac{1}{(1+z)^2}$$ $$\left ( \frac{\dot{a}}{H_0} \right)^2 (1+z)^2 = \Omega_{m,0}(1+z)^3 + \Omega_{\wedge,0}$$ $$\left ( \frac{\dot{a}}{H_0} \right)^2 \frac{1}{a^2} = \Omega_{m,0}(1 + z)^3 + \Omega_{\wedge,0}$$ $$\left ( \frac{H(z)}{H_0} \right )^2 = \Omega_{m,0}(1+z)^3 + \Omega_{\wedge,0}$$ Here, $H(z)$ is the red shift dependent Hubble parameter. This can be modified to include the radiation density parameter $\Omega_{rad}$ and the curvature density parameter $\Omega_k$. The modified equation is − $$\left ( \frac{H(z)}{H_0} \right )^2 = \Omega_{m,0}(1+z)^3 + \Omega_{rad,0}(1+z)^4+\Omega_{k,0}(1+z)^2+\Omega_{\wedge,0}$$ $$Or, \: \left ( \frac{H(z)}{H_0} \right)^2 = E(z)$$ $$Or, \: H(z) = H_0E(z)^{\frac{1}{2}}$$ where, $$E(z) \equiv \Omega_{m,0}(1 + z)^3 + \Omega_{rad,0}(1+z)^4 + \Omega_{k,0}(1+z)^2+\Omega_{\wedge,0}$$ This shows that the Hubble parameter varies with time. For the Einstein-de Sitter Universe, $\Omega_m = 1, \Omega_\wedge = 0, k = 0$. Putting these values in, we get − $$H(z) = H_0(1+z)^{\frac{3}{2}}$$ which shows the time evolution of the Hubble parameter for the Einstein-de Sitter universe. The density parameter, $\Omega$, is defined as the ratio of the actual (or observed) density ρ to the critical density $\rho_c$. For any quantity $x$ the corresponding density parameter, $\Omega_x$ can be expressed mathematically as − $$\Omega_x = \frac{\rho_x}{\rho_c}$$ For different quantities under consideration, we can define the following density parameters. S.No. Quantity Density Parameter 1 Baryons $\Omega_b = \frac{\rho_b}{\rho_c}$ 2 Matter(Baryonic + Dark) $\Omega_m = \frac{\rho_m}{\rho_c}$ 3 Dark Energy $\Omega_\wedge = \frac{\rho_\wedge}{\rho_c}$ 4 Radiation $\Omega_{rad} = \frac{\rho_{rad}}{\rho_c}$ Where the symbols have their usual meanings. The evolution of the scale factor is determined by the Friedmann Equation. H(z) is the red shift dependent Hubble parameter. The Hubble Parameter varies with time. The Density Parameter is defined as the ratio of the actual (or observed) density to the critical density.
Tagged: abelian group Abelian Group Problems and Solutions. The other popular topics in Group Theory are: Problem 616 Suppose that $p$ is a prime number greater than $3$. Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$. Add to solve later (c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575 Let $G$ be a finite group of order $2n$. Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$. Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later Problem 497 Let $G$ be an abelian group. Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$. Also determine whether the statement is true if $G$ is a non-abelian group.Add to solve later Problem 434 Let $R$ be a ring with $1$. A nonzero $R$-module $M$ is called irreducible if $0$ and $M$ are the only submodules of $M$. (It is also called a simple module.) (a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator. Add to solve later (b) Determine all the irreducible $\Z$-modules. Problem 420 In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem. Add to solve later Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$. Problem 343 Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$. Let $\Aut(N)$ be the group of automorphisms of $G$. Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime. Then prove that $N$ is contained in the center of $G$.
As I said in the question I'm trying to find a injection from B to A. I managed to do it in the following way: Fix some value for $N$ and distribute the values of the set $B = \Bigl\{1, 2, \ldots, \frac{N(N+1)}{2} \Bigr\}$ on the upper-triangular part of a $N \times N$ matrix. For instance, to $N = 5$, $B$ is $\{1, 2, 3, ..., 15\}$, so $$\begin{bmatrix}1 & 2 & 4 & 7 & 11\\* & 3 & 5 & 8 & 12\\* & * & 6 & 9 & 13\\* & * & * & 10 & 14\\* & * & * & * & 15\\\end{bmatrix}$$ Now, note that each column $j$ has exactly $j$ elements. So, the first element of a given column $j$ is equal to the number of elements in all the previous column plus one: $(1 + 2 + 3 + ... + j - 1) + 1$. Using the following formula to the sum of elements in a arithmetic progression: $$1 + 2 + 3 + ... + j - 1 = \frac{j(j - 1)}{2}$$ we find that the first element of a given column $j$ is the value $\frac{j(j - 1)}{2} + 1$. With the same analysis, we can see that the second element is $\frac{j(j - 1)}{2} + 2$ and so on. Then, a value $k \in B$ is in column $j$ if and only if $k = \frac{j(j - 1)}{2} + l$ for some $1 \le l \le j$. Using Baskara to invert this equality, we get $j(k) = \frac{1 + \sqrt{1 + 8(k - l)}}{2}$. Since this function $j$ grows assuming non integer values for $l \not = 1$, we can use $l = 1$ and take the floor: $$j(k) = \Big\lfloor \frac{1 + \sqrt{1 + 8(k - 1)}}{2} \Big\rfloor$$ Finally, any value $k \in B$ that lies in a column $j$ is of the form $\frac{j(j - 1)}{2} + i$, then, the index $i$ (the line) is given by $$i(k) = k - \frac{j(k)(j(k) - 1)}{2}$$ and the injection from $B$ to $A$ is $$f(k) = (i(k), j(k))$$
Let $\{ P_{\gamma} \}$ be a parametric family of probability measures on $(\Omega, \mathcal{F})$, such that $P_{\gamma} \ll \mu$ for all $\gamma$, for some $\sigma$-finite $\mu$. Consider the Radon-Nikodym densities $\{ \frac{d P_{\gamma}}{d \mu} \}$. By the Halmos-Savage factorization criterion, a $\sigma$-subalgebra $\mathcal G$ is sufficient for $\{ \frac{d P_{\gamma}}{d \mu} \}$if and only if there exists a measurable $h$, and a $\mathcal G$-measurable family $\{ g_{\gamma} \}$ such that $$ \frac{d P_{\gamma}}{d \mu} = g_{\gamma} \cdot h $$ for each $\gamma$. QuestionGiven $\{ P_{\gamma} \}$, does the family of sufficient $\sigma$-subalgebras depend on the reference measure $\mu$? In other words, suppose $\mu \ll \mu'$ and $\mu' \ll \mu$, is it true that $\mathcal G$ is sufficient for $\{ \frac{d P_{\gamma}}{d \mu} \}$ if and only if it is sufficient for $\{ \frac{d P_{\gamma}}{d \mu'} \}$? Informally, at first glance, the answer must be yes. Sufficiency should be invariant with respect to the measure used to compute the density. Since $$ \frac{d P_{\gamma}}{d \mu'} = \frac{d P_{\gamma}}{d \mu} \frac{d \mu }{d \mu'} = g_{\gamma} \cdot h \frac{d \mu }{d \mu'} = g_{\gamma} \cdot h' $$ where $h' = h \frac{d \mu }{d \mu'}$, the factorization criterion indeed says answer is yes. On the other hand, there is an example below that, apparently, says otherwise. Why the apparent inconsistency here? ExampleConsider the model is given by$$y_t = \rho y_{t-1}, \, t = 1, \cdots, T, \; \epsilon_i \stackrel{i.i.d.}{\sim} \mathcal{N}(0, \sigma^2),$$parametrized by $\gamma = \rho$. Take $\sigma^2$ and $y_0$ as fixed.Let $y = (y_1, \cdots, y_T)'$. Take $\mu$ to be the Lebesgue on $\mathbb{R}^T$. Then the RN density is (see this question) \begin{align} \frac{d P_{\rho}}{d \mu}(y) &= \frac{1}{ ( \sqrt{2 \pi \sigma^2} )^n } e^{-\frac{1}{2\sigma^2} ( \sum_{t = 1}^T (y_t - \rho y_{t-1})^2)} \\\\ &= \frac{1}{ ( \sqrt{2 \pi \sigma^2} )^n } e^{-\frac{1}{2\sigma^2} ((1 + \rho^2) \sum_{t = 1}^T y_{t-1}^2 + y_T^2 - 2 \rho \sum_{t = 1}^T y_{t-1} y_t)^2)}, \end{align} which implies $$ \Phi(y) = (\sum_{t = 1}^T y_{t-1}^2,\, y_T^2,\, \sum_{t = 1}^T y_{t-1} y_t) $$ is a minimal sufficient statistic. Now take $\mu'$ to be $P_1$, where \begin{align} \frac{d P_{1}}{d \mu}(y) &= \frac{1}{ ( \sqrt{2 \pi \sigma^2} )^n } e^{-\frac{1}{2\sigma^2} ( \sum_{t = 1}^T (y_t - y_{t-1})^2)}. \end{align} So \begin{align} \frac{d P_{\rho}}{d \mu'}(y) &= \frac{d P_{\rho}}{d P_1}(y) \\\\ &= \frac{ \frac{d P_{\rho}}{d \mu} (y)}{ \frac{d P_{1}}{d \mu} (y)} \\\\ &= e^{-\frac{1}{2\sigma^2} ( \sum_{t = 1}^T (y_t - \rho y_{t-1})^2) - \sum_{t = 1}^T (y_t - y_{t-1})^2)} \\\\ &= e^{-\frac{1}{2\sigma^2} ( -2 (\rho -1) \sum_{t = 1}^T y_t y_{t-1} + (\rho^2 - 1) \sum_{t = 1}^T y_{t-1})^2}. \end{align} This suggests $$ \Phi'(y) = ( \sum_{t = 1}^T y_t y_{t-1}, \sum_{t = 1}^T y_{t-1}^2 ) $$ is minimal sufficient, apparently contradicting the minimality of $\Phi$.
Is it possible to take the log of an independent variable in a Poisson regression? What to I have to be aware of, when doing so? (The results are getting better, when assuming that the independent variable is with log link.) Thanks for the clarification. I agree with @Greg Snow that any transformation should make sense in the context of the problem. Why are you considering a log transform? Have you tried standardizing your predictors? You want to keep in mind how the transformation changes the assumptions in your model. I'll use $\beta = (\beta_2, \dots, \beta_p)'$ and $X = (X_2, \dots, X_p)$. Your two models are Log transform model: $E(Y\|X_1,X) = \exp(\tilde\beta_1\log(X_1) + X\beta) = X_1^{\tilde\beta_1}\exp(X\beta)$ Original model: $E(Y\|X_1, X) = \exp(\beta_1X_1 + X\beta)$. For convenience I've overloaded $\beta$ slightly, in that their estimates would obviously be different under each model (in general). A simple way to compare the two models is through their relative risk. Suppose we have two observations $y_i, y_j$ with the same covariate values except that $x_{i1} - 1 = x_{j1}$ ($x_{i1}$ is one unit greater than $x_{j1}$). The relative risk $RR=E(Y\|X_1=x_{i1},X)/E(Y\|X_1=x_{i1}-1,X)$ is then the multiplicative change in the rate caused by increasing $x_1$ by one unit. The $RR$ is given by Log transform: $RR = \left(\frac{x_{i1}}{x_{i1} - 1}\right)^{\tilde\beta_1}$ Original model: $RR = \exp(\beta_1)$ RR under the log transform varies over the range of $x_{i1}$ (unless of course $\tilde \beta_1 = 0$). Does that make sense in your problem? In the original model the effect of a unit change in $x_1$ doesn't vary with its magnitude (i.e. increasing $X_1$ one unit has the same effect on the rate whether we move from 4 to 5, or 0 to 1, or 100 to 101, etc). Does that make sense in your problem? The coefficient in the log transformed model is harder to interpret, so unless there is a good reason for the transformation I would pass. You didn't say by what criterion the results are getting better, so it's hard to know for sure than any improvement in fit is "real". But even if it is, it might just be an indication that a Poisson regression is inappropriate. In particular the log transform removes the implicit proportional hazards assumption in the original model. Unfortunately it does so in a very rigid way, so while the overall fit might improve that doesn't necessarily mean you have a good model. Edit: A couple of points re: your comments. Your reference gives another way to interpret the coefficients via partial derivatives. Here, to compare the two models above we would look at $\frac{dE(Y\|X)}{dx_1}$. So let's do that: Log transform: $\frac{dE(Y\|X_1,X)}{dx_1} = \frac{\tilde\beta_1}{x_1}\exp(\tilde\beta_1\log(x_1)+X\beta) = \tilde\beta_1x_1^{\tilde\beta-1}\exp(X\beta)$ Original: $\frac{dE(Y\|X_1,X)}{dx_1} = \beta_1\exp(\beta x_1+X\beta) = \beta_1\exp(\beta_1 x_1)\exp(X\beta)$ Again, these are different models: compare the terms $\tilde\beta_1x_1^{\tilde\beta-1}$ and $\beta_1\exp(\beta_1 x_1)$. You can't interpret the log transformed model in the same way as the original model. However, you could apply that interpretation to $\log(X_1)$; the question is whether or not that's meaningful/reasonable/etc (a percent/unit change on the log scale is very different, obviously). (basically @Greg Snow's original point). If the only reason for the transformation is to reduce the excess variance or improve the residuals then I would look at other aspects of the model first. In terms of decreasing the Pearson residuals: This isn't always a plus. You may be overfitting the data for one, and for another my original point applies - the log transformed predictor might be compensating for a misspecified model , perhaps in a less-than-obvious way. What are the sample mean and variance of $Y$ - are the data over/underdispersed? Have you considered another model, a negative binomial regression for example? There is no problem with taking the log or other transform of predictor/independent variables in a poisson regression so long as the transformation is possible (no 0's or negative numbers) and makes sense given the science.
So I used variation of parameter and concluded that the particular solution follows: $$y_p = 2\cos (2t) \int{\tan (t) \sin (2t) dt} + 2 \sin (2t) \int{\tan(t) \cos(2t) dt}$$ I am not sure how to solve the two integrals $\int{\tan (t) \sin (2t) dt}$ and $\int{\tan(t) \cos(2t) dt}$ since it involves the $2t$ term inside the sine and cosine function. Can anyone help me solve this? First, recall the double-angle trig identity, $\sin(2t)=2\sin(t)\cos(t)$. Thus, the integrand $\tan(t)\sin(2t)=\tan(t)(2\sin(t)\cos(t))=2\sin^2(t)$. The half-angle trig identity allows us to rewrite $2\sin^2(t)$ as $1-\cos(2t)$. Now, we have $$\int{1}dt-\int{\cos(2t)}dt$$ Let $u=2t \Rightarrow du=2dt$ $$\int{1}dt-\int{\cos(2t)}dt=t-\frac{1}{2}\int{\cos(u)}du=t-\frac{1}{2}\sin(u)=t-\frac{1}{2}\sin(2t)=t-\sin(t)\cos(t)$$ The process is similar for $\int{\tan(t)\cos(2t)}dt$. Hope that helps!
I'm writing a program to generate solar systems but I'm having trouble calculating the expected temperature of a planet. I have found a formula to calculate this, but I haven't been able to get a remotely correct answer out of it as it doesn't clearly state what units your supposed to use. This formula I found: $$4 \pi R ^ 2 ơ T ^ 4 = \frac{\pi R ^ 2 L_{\odot}(1 - a)}{(4 \pi d ^ 2)}$$ where $R$ is the planet's radius (not sure what units), $d$ is the distance from the Sun (it mentions AU), $a$ is the albedo, $L_{\odot}$ is the luminosity of the Sun (which I assume can be interchanged with the luminosity of any star), $T$ is the temperature of the planet (kelvin, this is what I'm trying to get), and $ơ$ is the Stefan-Boltzmann constant. The site I found it on is notes for an astronomy college course. Here is the link: Any help would be very much appreciated.
Why do we sandwich operators in quantum mechanics in such a way that the operator acts on the wavefunction and not on its complex conjugate? $$ \int\psi^* \hat{F} \varphi = \int (\hat{F}^+ \psi)^* \varphi $$ also, if F is an Hermitian operator $$ \int \Psi \hat{F} \Psi^* = \int \Psi^* \hat{F}\Psi $$ Take the kinetic energy operator and ground state wave function for a particle in a box with width of $L$ as an example (in 1D) $$\hat{T}=-\frac{\hbar}{2m}\frac{d}{dx^2}$$ and $$\psi=\sqrt{\frac{2}{L}}\sin\frac{\pi}{L}x$$ Obviously, $\hat{T}\psi$ means something and $\psi^{*}\hat{T}$ means something else. Please look at my previous answer for a more complete explanation of the notation.
In free space, $\rho=0$ and $J=0$, so there are no electromagnetic sources/sinks. Maxwell's equations thus reduce to: $\nabla\cdot E = 0$ $\nabla\cdot B = 0$ $\nabla\times E = -\frac{\partial B}{\partial t}$ $\nabla\times B = \mu_0\epsilon_0 \frac{\partial E}{\partial t} $ Suppose I was writing a simple simulation to visualize the electromagnetic field in free space. I have seen people talk about waves propagating in free space, and I know that there is no such thing as electromagnetic waves being created from nothing -- it is usually assumed that such electromagnetic waves propagating in free space are plane waves that originated in a charged source extremely far away. However, when I actually implement the "oscillation" in the EM field, where does that oscillation come from -- speaking practically from a coding point of view? Do you just hardwire, e.g., a sinusoidal source at the location you are interested in probing the EM field? And if there were no such "magical" waves propagating in free space, would the EM field just remain smooth, without any oscillations, vibrations, sinusoids, etc.? In other words, can you have a completely stationary electromagnetic field, or would the last two of Maxwell's equations above prevent such stationary EM fields? But then what, in free space, would cause the initial change in the electric or magnetic field to get the oscillations going? To put it one last way: suppose I wrote a simulation involving the 4 Maxwell's equations above (free space). Would the EM field be stationary for all time, and the only way a propagating wave would appear is if I perturbed, say, the electric field which activated a never-ending loop of the curl equations? So if the initial values of E and B were both 0 in my simulation, then they would stay 0 for all time. But if one or both initial values of E and B were non-zero, then the curl equations would be "activated" and result in a never-ending loop of oscillations?
WDM Filters OptiLayer provides a unique fully automated approach aimed for automatic design of WDM, DWDM, HDWDM and all other types of filters used in modern telecommunication applications. The approach is based on the combination of ideas from classical design approaches with an integer optimization. This approach turns out to be extremely efﬁcient from a computational point of view and makes it possible to construct a set of signiﬁcantly different ﬁlter designs with nearly equivalent spectral properties. It is an important feature of this option that design layers (excluding one or two top anti-reflection layers) are multiple of a quarter-wave optical thickness at the central wavelength of the filter. Due to this fact the error self-compensation mechanism can be explored for the manufacture of designed filters. WDM specifications require high steepness of transmittance slopes in the transition zones between passband and stopband. Standard Fabry-Perot filters cannot meet these demands. For more detail see WDM Filters article. WDM design option is available through Example: Main parameters of WDM Filters: $\lambda_0$ - central wavelength; $\Delta\lambda_p$ - width of passband; $\Delta\lambda_r$ - band width at the rejection level; $SF=\frac{\Delta\lambda_r}{\Delta\lambda_p} $ - shape factor Spectral performance of a WDM filter is entirely specified by Specification of target WDM parameters are organized in a six-step dialog. WDM filters consists of several mirrors and spacer layers which allow one to control pass band width and transmittance slopes simultaneously. $\lambda_0=1550$ nm; $\Delta\lambda_p=0.4$ nm at transmittance level of 89.125%; $\Delta\lambda_r=1.2$ nm at transmittance level of 0.1%; $SF=3$ - shape factor; Layer materials with refractive indices 1.4 and 2.1 For more detail see WDM Filters article. After setting target specifications and layer materials OptiLayer gives you the recommendation on the number of cavities (filter spacer layers). In the example: 4 and more. At the next step OptiLayer lists all possible prototypes. Each prototype can give a design solution. You can also specify spacer material. In the example: Any material is specified for Spacer material. The filter prototype with m=17, k=1 has 143 layers and its optical thickness is 147 QWOT. In order to optimize the prototype with respect to the entered specifications a comprehensive mathematical procedure is used. There are additional settings of the Global Integer Search procedure: Search for designs with symmetrical mirrors only; Search for designs with symmetrical cavities only; The filter can be optimized with respect to the actual refractive index of incident medium or with respect to the incidence medium refractive index equal to that of the substrate; It is possible to use 3H/3L layers for filter mirrors; It is possible to use cavities of a complicated structure; It is possible to consider some mirror layers/cavities consisting of an additional material. At the last step of design process you can adjust the obtained filter solution to the actual incident medium. For this purpose one- or two-layer AR coating can be placed at the top of the design. Result: 126-layer WDM filter; Other solutions: 116-layer WDM filter; 120-layer WDM solution; 112-layer WDM design Learn more: A. V. Tikhonravov and M. K. Trubetskov, "Automated design and sensitivity analysis of wavelengh-division multiplexing filters ," Appl. Opt. 41, 3176-3182 (2002) Narrow bandpass filter The OptiLayer provides additional opportunities for achieving excellent filter spectral properties by using mirrors with 3-quarter wave layers. This gives more freedom in controlling steepness of phase shift upon reflection in filter cavities. Example: narrow bandpass filter for the spectral range from 500 nm to 570 nm. High transmission range is from 525 nm to 535 nm, low transmission range is from 500 nm to 520 nm and from 540 nm to 570 nm. In this case $\lambda_0$=530 nm, $\Delta\lambda_p=10$ nm, $\Delta\lambda_r=15$. OptiLayer allows indicating transmittance values (Y-axis) in linear scale, logarithmic scale and diabatic scale. The number of cavities (spacer layers) recommended by OptiLayer is more than 6. High refractive index material is specified as spacer layer material. 35-layer prototype can be considered. Final result: 36 layer narrow bandpass filter. Two last layers is anti-reflection coating. OptiLayer allows additional settings including the possibility to specify 50% half-width: Learn more in: A. Tikhonravov and M. Trubetskov, "Modern design tools and a new paradigm in optical coating design," Appl. Opt. 51, 7319-7332 (2012).
Logistic regression is widely used in social and behavioral research in analyzing the binary (dichotomous) outcome data. In logistic regression, the outcome can only take two values 0 and 1. Some examples that can utilize the logistic regression are given in the following. We use an example to illustrate how to conduct logistic regression in R. In this example, the aim is to predict whether a woman is in compliance with mammography screening recommendations from four predictors, one reflecting medical input and three reflecting a woman's psychological status with regarding to screening. > usedata('mamm') > head(mamm) y x1 x2 x3 x4 1 1 1 0.22 4 3 2 0 0 0.56 1 1 3 1 0 0.44 4 3 4 0 0 0.33 3 0 5 1 1 0.44 5 0 6 0 1 0.56 5 0 > With a binary outcome, the linear regression does not work any more. Simply speaking, the predictors can take any value but the outcome cannot. Therefore, using a linear regression cannot predict the outcome well. In order to deal with the problem, we model the probability to observe an outcome 1 instead, that is $p = \Pr(y=1)$. Using the mammography example, that'll be the probability for a woman to be in compliance with the screening recommendation. Even directly modeling the probability would work better than predicting the 1/0 outcome, intuitively. A potential problem is that the probability is bound between 0 and 1 but the predicted values are generally not. To further deal with the problem, we conduct a transformation using \[ \eta = \log\frac{p}{1-p}.\] After transformation, $\eta$ can take any value from $-\infty$ when $p=0$ to $\infty$ when $p=1$. Such a transformation is called logit transformation, denoted by $\text{logit}(p)$. Note that $p_{i}/(1-p_{i})$ is called odds, which is simply the ratio of the probability for the two possible outcomes. For example, if for one woman, the probability that she is in compliance is 0.8, then the odds is 0.8/(1-0.2)=4. Clearly, for equal probability of the outcome, the odds=1. If odds>1, there is a probability higher than 0.5 to observe the outcome 1. With the transformation, the $\eta$ can be directly modeled. Therefore, the logistic regression is \[ \mbox{logit}(p_{i})=\log(\frac{p_{i}}{1-p_{i}})=\eta_i=\beta_{0}+\beta_{1}x_{1i}+\ldots+\beta_{k}x_{ki} \] where $p_i = \Pr(y_i = 1)$. Different from the regular linear regression, no residual is used in the model. For a variable $y$ with two and only two outcome values, it is often assumed it follows a Bernoulli or binomial distribution with the probability $p$ for the outcome 1 and probability $1-p$ for 0. The density function is \[ p^y (1-p)^{1-y}. \] Note that when $y=1$, $p^y (1-p)^{1-y} = p$ exactly. Furthermore, we assume there is a continuous variable $y^$ underlying the observed binary variable. If the continuous variable takes a value larger than certain threshold, we would observe 1, otherwise 0. For logistic regression, we assume the continuous variable has a logistic distribution with the density function: \[ \frac{e^{-y^}}{1+e^{-y^}} .\] The probability for observing 1 is therefore can be directly calculated using the logistic distribution as: \[ p = \frac{1}{1 + e^{-y^}},\] which transforms to \[ \log\frac{p}{1-p} = y^.\] For $y^$, since it is a continuous variable, it can be predicted as in a regular regression model. In R, the model can be estimated using the glm() function. Logistic regression is one example of the generalized linear model (glm). Below gives the analysis of the mammography data. glm uses the model formula same as the linear regression model. family = tells the distribution of the outcome variable. For binary data, the binomial distribution is used. link = tell the transformation method. Here, the logit transformation is used. > usedata('mamm') > m1<-glm(y~x1+x2+x3+x4, family=binomial(link='logit'), data=mamm) > summary(m1) Call: glm(formula = y ~ x1 + x2 + x3 + x4, family = binomial(link = "logit"), data = mamm) Deviance Residuals: Min 1Q Median 3Q Max -2.3941 -0.7660 0.3756 0.7881 1.6004 Coefficients: Estimate Std. Error z value Pr(>\|z\|) (Intercept) -1.4466 1.1257 -1.285 0.198754 x1 1.7731 0.4841 3.663 0.000249 * x2 -0.8694 1.1201 -0.776 0.437628 x3 0.5935 0.2058 2.883 0.003933 x4 -0.1527 0.1542 -0.990 0.321946 --- Signif. codes: 0 '*' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 203.32 on 163 degrees of freedom Residual deviance: 155.48 on 159 degrees of freedom AIC: 165.48 Number of Fisher Scoring iterations: 5 > We first focus on how to interpret the parameter estimates from the analysis. For the intercept, when all the predictors take the value 0, we have \[ \beta_0 = \log(\frac{p}{1-p}), \] which is the log odds that the observed outcome is 1. We now look at the coefficient for each predictor. For the mammography example, let's assume $x_2$, $x_3$, and $x_4$ are the same and look at $x_1$ only. If a woman has received a recommendation ($x_1=1$), then the odds is \[ \log(\frac{p}{1-p})\|(x_1=1)=\beta_{0}+\beta_{1}+\beta_{2}x_{2}+\beta_{3}x_{3}+\beta_{4}x_{4}.\] If a woman has not received a recommendation ($x_1=0$), then the odds is \[\log(\frac{p}{1-p})\|(x_1=0)=\beta_{0}+\beta_{2}x_{2}+\beta_{3}x_{3}+\beta_{4}x_{4}.\] The difference is \[\log(\frac{p}{1-p})\|(x_1=1)-\log(\frac{p}{1-p})\|(x_1=0)=\beta_{1}.\] Therefore, the logistic regression coefficient for a predictor is the difference in the log odds when the predictor changes 1 unit given other predictors unchanged. This above equation is equivalent to \[\log\left(\frac{\frac{p(x_1=1)}{1-p(x_1=1)}}{\frac{p(x_1=0)}{1-p(x_1=0)}}\right)=\beta_{1}.\] More descriptively, we have \[\log\left(\frac{\mbox{ODDS(received recommendation)}}{\mbox{ODDS(not received recommendation)}}\right)=\beta_{1}.\] Therefore, the regression coefficients is the log odds ratio. By a simple transformation, we have \[\frac{\mbox{ODDS(received recommendation)}}{\mbox{ODDS(not received recommendation)}}=\exp(\beta_{1})\] or \[\mbox{ODDS(received recommendation)} = \exp(\beta_{1})\mbox{ODDS(not received recommendation)}.\] Therefore, the exponential of a regression coefficient is the odds ratio. For the example, $exp(\beta_{1})$=exp(1.7731)=5.9. Thus, the odds in compliance to screening for those who received recommendation is about 5.9 times of those who did not receive recommendation. For continuous predictors, the regression coefficients can also be interpreted the same way. For example, we may say that if high school GPA increase one unit, the odds a student to be admitted can be increased to 6 times given other variables the same. Although the output does not directly show odds ratio, they can be calculated easily in R as shown below. > usedata('mamm') > m1<-glm(y~x1+x2+x3+x4, family=binomial(link='logit'), data=mamm) > summary(m1) Call: glm(formula = y ~ x1 + x2 + x3 + x4, family = binomial(link = "logit"), data = mamm) Deviance Residuals: Min 1Q Median 3Q Max -2.3941 -0.7660 0.3756 0.7881 1.6004 Coefficients: Estimate Std. Error z value Pr(>\|z\|) (Intercept) -1.4466 1.1257 -1.285 0.198754 x1 1.7731 0.4841 3.663 0.000249 * x2 -0.8694 1.1201 -0.776 0.437628 x3 0.5935 0.2058 2.883 0.003933 x4 -0.1527 0.1542 -0.990 0.321946 --- Signif. codes: 0 '*' 0.001 '' 0.01 '' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 203.32 on 163 degrees of freedom Residual deviance: 155.48 on 159 degrees of freedom AIC: 165.48 Number of Fisher Scoring iterations: 5 > > exp(coef(m1)) (Intercept) x1 x2 x3 x4 0.2353638 5.8892237 0.4192066 1.8103735 0.8583652 > By using odds ratios, we can intercept the parameters in the following. Statistical inference for logistic regression is very similar to statistical inference for simple linear regression. We can (1) conduct significance testing for each parameter, (2) test the overall model, and (3) test the overall model. For each regression coefficient of the predictors, we can use a z-test (note not the t-test). In the output, we have z-values and corresponding p-values. For x1 and x3, their coefficients are significant at the alpha level 0.05. But for x2 and x4, they are not. Note that some software outputs Wald statistic for testing significance. Wald statistic is the square of the z-statistic and thus Wald test gives the same conclusion as the z-test. We can also conduct the hypothesis testing by constructing confidence intervals. With the model, the function confint() can be used to obtain the confidence interval. Since one is often interested in odds ratio, its confidence interval can also be obtained. Note that if the CI for odds ratio includes 1, it means nonsignificance. If it does not include 1, the coefficient is significant. This is because for the original coefficient, we compare the CI with 0. For odds ratio, exp(0)=1. If we were reporting the results in terms of the odds and its CI, we could say, “The odds of in compliance to screening increases by a factor of 5.9 if receiving screening recommendation (z=3.66, P = 0.0002; 95% CI = 2.38 to 16.23) given everything else the same.” > usedata('mamm') > m1<-glm(y~x1+x2+x3+x4, family=binomial(link='logit'), data=mamm) > confint(m1) Waiting for profiling to be done... 2.5 % 97.5 % (Intercept) -3.7161449 0.7286948 x1 0.8665475 2.7871626 x2 -3.1466137 1.2831677 x3 0.2023731 1.0134006 x4 -0.4577415 0.1506678 > exp(confint(m1)) Waiting for profiling to be done... 2.5 % 97.5 % (Intercept) 0.02432757 2.072374 x1 2.37868419 16.234889 x2 0.04299748 3.608051 x3 1.22430472 2.754954 x4 0.63271101 1.162610 > For the linear regression, we evaluate the overall model fit by looking at the variance explained by all the predictors. For the logistic regression, we cannot calculate a variance. However, we can define and evaluate the deviance instead. For a model without any predictor, we can calculate a null deviance, which is similar to variance for the normal outcome variable. After including the predictors, we have the residual deviance. The difference between the null deviance and the residual deviance tells how much the predictors help predict the outcome. If the difference is significant, then overall, the predictors are significant statistically. The difference or the decease in deviance after including the predictors follows a chi-square ($\chi^{2}$) distribution. The chi-square ($\chi^{2}$) distribution is a widely used distribution in statistical inference. It has a close relationship to F distribution. For example, the ratio of two independent chi-square distributions is a F distribution. In addition, a chi-square distribution is the limiting distribution of an F distribution as the denominator degrees of freedom goes to infinity. There are two ways to conduct the test. From the output, we can find the Null and Residual deviances and the corresponding degrees of freedom. Then we calculate the difference. For the mammography example, we first get the difference between the Null deviance and the Residual deviance, 203.32-155.48= 47.84. Then, we find the difference in the degrees of freedom 163-159=4. Then, the p-value can be calculated based on a chi-square distribution with the degree of freedom 4. Because the p-value is smaller than 0.05, the overall model is significant. The test can be conducted simply in another way. We first fit a model without any predictor and another model with all the predictors. Then, we can use anova() to get the difference in deviance and the chi-square test result. > usedata('mamm') > m1<-glm(y~x1+x2+x3+x4, family=binomial(link='logit'), data=mamm) > summary(m1) Call: glm(formula = y ~ x1 + x2 + x3 + x4, family = binomial(link = "logit"), data = mamm) Deviance Residuals: Min 1Q Median 3Q Max -2.3941 -0.7660 0.3756 0.7881 1.6004 Coefficients: Estimate Std. Error z value Pr(>\|z\|) (Intercept) -1.4466 1.1257 -1.285 0.198754 x1 1.7731 0.4841 3.663 0.000249 x2 -0.8694 1.1201 -0.776 0.437628 x3 0.5935 0.2058 2.883 0.003933 x4 -0.1527 0.1542 -0.990 0.321946 --- Signif. codes: 0 '*' 0.001 '' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 203.32 on 163 degrees of freedom Residual deviance: 155.48 on 159 degrees of freedom AIC: 165.48 Number of Fisher Scoring iterations: 5 > > ## method 1 > d_chi <- 203.32 - 155.48 > d_df <- 163 - 159 > > 1 - pchisq(d_chi, d_df) [1] 1.019117e-09 > > ## method 2 > m0 <- glm(y~1, family=binomial(link='logit'), data=mamm) > anova(m0,m1) Analysis of Deviance Table Model 1: y ~ 1 Model 2: y ~ x1 + x2 + x3 + x4 Resid. Df Resid. Dev Df Deviance 1 163 203.32 2 159 155.48 4 47.837 > We can also test the significance of a subset of predictors. For example, whether x3 and x4 are significant above and beyond x1 and x2. This can also be done using the chi-square test based on the difference. In this case, we can compare a model with all predictors and a model without x3 and x4 to see if the change in the deviance is significant. In this example, the p-value is 0.002, indicating the change is signficant. Therefore, x3 and x4 are statistically significant above and beyond x1 and x2 > usedata('mamm') > m1<-glm(y~x1+x2+x3+x4, family=binomial(link='logit'), data=mamm) > m2 <- glm(y~x1+x2, family=binomial(link='logit'), data=mamm) > > anova(m2, m1) Analysis of Deviance Table Model 1: y ~ x1 + x2 Model 2: y ~ x1 + x2 + x3 + x4 Resid. Df Resid. Dev Df Deviance 1 161 168.23 2 159 155.48 2 12.749 > > 1 - pchisq(12.749, 2) [1] 0.001704472 >
In his celebrated paper "Conjugate Coding" (written around 1970), Stephen Wiesner proposed a scheme for quantum money that is unconditionally impossible to counterfeit, assuming that the issuing bank has access to a giant table of random numbers, and that banknotes can be brought back to the bank for verification. In Wiesner's scheme, each banknote consists of a classical "serial number" $s$, together with a quantum money state $\|\psi_s\rangle$ consisting of $n$ unentangled qubits, each one either $$\|0\rangle,\ \|1\rangle,\ \|+\rangle=(\|0\rangle+\|1\rangle)/\sqrt{2},\ \text{or}\ \|-\rangle=(\|0\rangle-\|1\rangle)/\sqrt{2}.$$ The bank remembers a classical description of $\|\psi_s\rangle$ for every $s$. And therefore, when $\|\psi_s\rangle$ is brought back to the bank for verification, the bank can measure each qubit of $\|\psi_s\rangle$ in the correct basis (either $\{\|0\rangle,\|1\rangle\}$ or ${\|+\rangle,\|-\rangle}$), and check that it gets the correct outcomes. On the other hand, because of the uncertainty relation (or alternatively, the No-Cloning Theorem), it's "intuitively obvious" that, if a counterfeiter who doesn't know the correct bases tries to copy $\|\psi_s\rangle$, then the probability that both of the counterfeiter's output states pass the bank's verification test can be at most $c^n$, for some constant $c<1$. Furthermore, this should be true regardless of what strategy the counterfeiter uses, consistent with quantum mechanics (e.g., even if the counterfeiter uses fancy entangled measurements on $\|\psi_s\rangle$). However, while writing a paper about other quantum money schemes, my coauthor and I realized that we'd never seen a rigorous proof of the above claim anywhere, or an explicit upper bound on $c$: neither in Wiesner's original paper nor in any later one. So, has such a proof (with an upper bound on $c$) been published? If not, then can one derive such a proof in a more-or-less straightforward way from (say) approximate versions of the No-Cloning Theorem, or results about the security of the BB84 quantum key distribution scheme? Update: In light of the discussion with Joe Fitzsimons below, I should clarify that I'm looking for more than just a reduction from the security of BB84. Rather, I'm looking for an explicit upper bound on the probability of successful counterfeiting (i.e., on $c$)---and ideally, also some understanding of what the optimal counterfeiting strategy looks like. I.e., does the optimal strategy simply measure each qubit of $\|\psi_s\rangle$ independently, say in the basis $$\{ \cos(\pi/8)\|0\rangle+\sin(\pi/8)\|1\rangle, \sin(\pi/8)\|0\rangle-\cos(\pi/8)\|1\rangle \}?$$ Or is there an entangled counterfeiting strategy that does better? Update 2: Right now, the best counterfeiting strategies that I know are (a) the strategy above, and (b) the strategy that simply measures each qubit in the $\{\|0\rangle,\|1\rangle\}$ basis and "hopes for the best." Interestingly, both of these strategies turn out to achieve a success probability of (5/8) n. So, my conjecture of the moment is that (5/8) n might be the right answer. In any case, the fact that 5/8 is a lower bound on c rules out any security argument for Wiesner's scheme that's "too" simple (for example, any argument to the effect that there's nothing nontrivial that a counterfeiter can do, and therefore the right answer is c=1/2). This post has been migrated from (A51.SE) Update 3: Nope, the right answer is (3/4) n! See the discussion thread below Abel Molina's answer.
What is a ramjet? Was it used on the SR-71 Blackbird? Aviation Stack Exchange is a question and answer site for aircraft pilots, mechanics, and enthusiasts. It only takes a minute to sign up.Sign up to join this community A jet engine compresses air, heats it by mixing it with fuel and burns it, and lets the heated air escape at the end, where it accelerates to more than its initial speed in a convergent-divergent nozzle because the density of the heated gas is lower, thus needing a higher volume at the same pressure. By converting the kinetic energy of the flow into pressure (potential energy), the intake creates high-pressure air to feed the engine. This is called pressure recovery and increases with the square of flow speed. Please see below for a plot: This puts a pressure of 1 at Mach 0.5, which is on the high side for flow speed near the compressor face in a jet engine intake. Note that in static conditions the air needs to be accelerated, so the intake pressure is only 84% of ambient pressure, and at Mach 0.85, the maximum speed of airliners, the intake pressure is 1.37 times higher than ambient pressure. But at supersonic speed things take really off: Pressure recovery for the Concorde was already 6 at Mach 2.0, and for the SR-71 it was 40 at Mach 3.2. If you want a more mathematical approach, the equation for isentropic compression gives: $$p_0 = p_{\infty}\cdot\frac{(1.2\cdot Ma^2)^{3.5}}{\left(1+\frac{5}{6}\cdot(Ma^2-1)\right)^{2.5}}$$ The odd exponents have to do with the ratio of specific heats $\kappa$ of air. 3.5 is actually $\frac{\kappa}{\kappa-1}$ and 2.5 is $\frac{1}{\kappa-1}$. Real compression ratios are slightly below those of the ideal isentropic compression due to friction, but not by much. The exact equation used for the plot above is produced by calculating the ratio to the intake Mach number directly, this time with $\kappa$ = 1.405:$$\frac{p_{intake}}{p_{\infty}} = \left(0.2025\cdot Ma^2 \cdot\left(1-\left(\frac{Ma_{intake}}{Ma_{\infty}}\right)^2\right) + 1\right)^{3.469}$$ Thus, you get already the compression ratio of a J-47, an early turbojet engine, at Mach 2 and that of a GE90, a modern turbofan engine, at Mach 3.2. Beyond that, it does not make much sense to complicate the engine with turbo machinery - just let the ram pressure give you the compression for thrust generation. You need, however, speed up the vehicle by other means first, because the possible thrust is proportional to the pressure recovery, or the square of airspeed. No speed, no thrust! You might have read claims that the J-58 of the SR-71 was a ramjet. This is only half true. Below Mach 2 it worked as a regular turbojet, but it had bypass tubes which ducted some air from the fourth stage of the compressor around the later compressor stages, the combustion chambers and the turbine directly into the afterburner. Now some of the air was compressed in the intake and fed directly to a combustion area and through a convergent-divergent nozzle, so this part worked like a ramjet. Some of the air, however, was still passing through the core engine to keep it running, though. A better example for a ramjet-powered plane is the Lockheed D-21 reconnaissance drone, which used a Marquardt RJ-43 ram jet for propulsion. Its cruise speed was Mach 3.7, back 50 years ago! See below for a picture (source). Note that the same trick which makes a ramjet possible can be used to reduce cooling drag for high-speed piston aircraft. A well-designed cooling duct is slowing and compressing incoming air and heats it by letting it flow through a radiator. The heated air has a higher exit speed, resulting in jet thrust which can compensate cooling drag at higher speeds. The Republic XF-12, a much underappreciated design, made exemplary use of this technique. The ramjet is conceptually the simplest jet engine. It is a duct where air is burned creating a hot jet that provides thrust, it is known aero thermodynamic duct as it is no more than a duct where a thermodynamic cycle is performed. They are no more used for airplanes, as it can not provide thrust at zero airspeed -since he does not have any compressor within the diffuser- and modern turbofan is much more efficient. It is composed by a diffuser , a burner and a nozzle , it uses dynamic compression of ram air within its inlet and then the hot jet expand in a convergent-divergent nozzle as it is supersonic. Airplanes like the Blackbird used a ramjet to reach higher mach speed by starting the ramjet when the where already subsonic. but the short answer A ramjet, sometimes referred to as a flying stovepipe or an athodyd (an abbreviation of aero thermodynamic duct), is a form of airbreathing jet engine that uses the engine's forward motion to compress incoming air without an axial compressor. In other words it uses the thrust it generates to compress the on coming air by "ramming" it into the engine. Since they have no way of drawing air in they do not work in a static (not forward moving) situation. They often work best at supersonic speeds. Aside form the device used to pump fuel into the engine they essentially have no moving parts. A ramjet is a jet engine in which ram air pressure, generated by the forward motion of the air vehicle is employed to compress air before fuel is mixed with it and burned to produce thrust through an increase in temperature and pressure of the expanding gases. This means that a ramjet cannot generate static thrust or operate efficiently below a certain airspeed. Ramjet engines are usually designed to operate at supersonic speeds. A scramjet, or supersonic combustion ramjet is a ramjet in which the airflow through the engine combustion section is supersonic.
Published March 2002,February 2011. When in 1821 the English mathematician Charles Babbage invented what we now recognise as the grandfather of the modern computer he called it the 'Difference Engine' since it was intended to take over the work of making mathematical tables by the techniques described in this article. Photo courtesy of the Charles Babbage Institute, University of Minnesota, Minneapolis. IBM Computer Museum. n 0 1 2 3 4 5 6 7 8 9 n 2 0 1 4 9 16 25 36 49 64 81 If we want to know 5 2 we simply look at the entry under 5. Exercise 1 Produce a similar table of cubes. Mathematicians found a large number of tricks to make the construction and use of tables easier. Here is one of them. Consider our table of squares. Let us take the difference between successive entries as shown below. 0 1 4 9 16 25 36 49 64 81 1 3 5 7 9 11 13 15 17 Now take the difference of successive differences to obtain 0 1 4 9 16 25 36 49 64 81 1 3 5 7 9 11 13 15 17 2 2 2 2 2 2 2 2 2 Finally we take the difference of the differences of the differences to obtain 0 1 4 9 16 25 36 49 64 81 1 3 5 7 9 11 13 15 17 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 Let us try the same thing on a table of 4th powers. n 0 1 2 3 4 5 6 7 n 4 0 1 16 81 256 625 1296 2401 This time successive differencing produces the following table. 0 1 16 81 256 625 1296 2401 1 15 65 175 369 671 1105 14 50 110 194 302 434 36 60 84 108 132 24 24 24 24 0 0 0 Exercise 2 (i) Extend the table of 4th powers n 4 to cover n=8 and n=9. Verify that the pattern suggested above continues to hold. (ii) Find the effect of repeated differencing on your table of cubes. Exercise 3 (i) Try the effect of repeated differencing on 3n 3 -5n 2 . (ii) Try the effect of repeated differencing on a polynomial of your choice. It very much looks as though the effect of repeated differencing a polynomial of degree m is to produce a row of zeros after at most m+1 differences. Let us try our conjecture on the most general quadratic f(n)=An 2 +Bn+C. Here is part of the initial table n k-2 k-1 k k+1 k+2 f(n) A(k 2 -4k+4)+B(k-2)+C A(k 2 -2k+1)+B(k-1)+C Ak 2 +Bk+C A(k 2 +2k+1)+B(k+1)+C A(k 2 +4k+4)+B(k+2)+C and here are the repeated differences Since we could pick any k to investigate, we have verified the conjecture for the general quadratic. Exercise 4 Verify the conjecture for the general cubic. The next exercise requires more algebra. Exercise 5 (i) If f(n)=n k show that f(n+1)-f(n) is a polynomial in n of degree k-1. (ii) If g(n) is a polynomial of degree k show that g(n+1)-g(n) is a polynomial in n of degree k-1. (iii) Prove our conjecture that the effect of repeated differencing a polynomial of degree m is to produce a row of zeros after at most m+1 differences. We now observe that we can reconstruct the whole of a table of differences for a polynomial from a small part of it. For example, suppose we are told that f is a quadratic polynomial and its table of differences looks like ? ? ? 7 ? ? ? ? ? ? ? ? ? 6 ? ? ? ? ? ? ? ? 2 ? ? ? ? ? 0 0 0 0 0 0 0 0 We can immediately fill in the last but one row to obtain ? ? ? 7 ? ? ? ? ? ? ? ? ? 6 ? ? ? ? ? 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 and then the next to obtain ? ? ? 7 ? ? ? ? ? ? 0 2 4 6 8 10 12 14 16 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 Exercise 6 (i) Fill in the top row and check that we have the table of n 2 -n+1 with the leftmost entry corresponding to n=0. (ii) Fill in in the following table of differences for a quadratic. ? ? ? 9 ? ? ? ? ? ? ? ? ? 16 ? ? ? ? ? ? ? ? 6 ? ? ? ? ? 0 0 0 0 0 0 0 0 Check that the table you get is for f(n)=3n 2 -5n-3 with the leftmost entry corresponding to n=0. Now suppose that we want to tabulate f(n)=n 3 -3n 2 +5n+1. Here is a partial tabulation which the reader should check. n -4 -3 -2 -1 0 1 2 3 4 5 f(n) ? ? -29 -8 1 4 7 ? ? ? We can now draw up a partial table of differences which the reader should check. ? ? -29 -8 1 4 7 ? ? ? ? ? 21 9 3 3 ? ? ? ? ? -12 -6 0 ? ? ? ? ? 6 6 ? ? ? ? ? 0 ? ? ? But we can now reconstruct the full table of differences. Here is a partial reconstruction ? -54 -29 -8 1 4 7 16 ? ? ? 39 21 9 3 3 9 ? ? ? -18 -12 -6 0 6 ? ? ? 6 6 6 6 ? ? ? 0 0 0 ? ? Notice that we have extended our original table to read n -4 -3 -2 -1 0 1 2 3 4 5 f(n) ? -54 -29 -8 1 4 7 16 ? ? Exercise 7 Continue the reconstruction of the table of differences to the point where you have the value of f(n) for n=-4, n=4 and n=5. Check your answers by computing f(-4), f(4) and f(5) directly from the definition. Exercise 8 Let $f(n)=3n^2+2n+1$. Compute f(n) for n=-2, -1, 0, 1, 2. Form the associated partial table of differences and then reconstruct the table of differences sufficiently far that you have f(n) for all n with $-5\leq n\leq 4$. Check your answer by computing f(n) directly. The fact that (at least for polynomials) we can use differences to reconstruct a full table from a partial table gives a strong hint as to why table makers were so interested in these ideas. Here is another reason. Suppose we have a table like the following. 0 0 0 0 0 1 0 0 0 0 0 in which all the entries are zero except one. If we compute successive differences we get 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 +1 -1 0 0 0 0 0 0 0 1 -2 1 0 0 0 0 0 1 -3 3 1 0 0 0 1 -4 6 -4 1 0 Exercise 9 (i) Calculate one further row of differences. (ii) Conjecture the form of the nth row of differences. (iii) [Harder and optional] Prove your conjecture. Exercise 10 (i) Find the successive differences for the table 0 0 0 0 0 A 0 0 0 0 0 (ii) Find the successive differences for 13 7 3 1 1 3 7 13 21 (iii) Find the successive differences for 13 7 3 1 1 3 (7+A) 13 21 (iv) Suppose f is a cubic polynomial. What is the fifth line in the table of successive difference for f? Find the fifth line of the table of successive differences whose first line is f(n) f(n+1) f(n+2) f(n+3) f(n+4) f(n+5)+A f(n+6) f(n+7) f(n+8) f(n+9) f(n+10) Explain how you can find A just by looking at the 5th line. (v) The following table is supposed to represent the values of a quadratic but I have made a mistake in one entry. n -1 0 1 2 3 4 5 6 7 8 9 f(n) 8 3 0 -1 0 5 8 15 24 35 49 By successive differencing find the error and correct it. (vi) Explain how you can locate and correct a single error in a table of a polynomial of degree k. (vii) Explain how you would try to locate and correct a single error in a table of a polynomial whose degree you did not know. Successive differencing thus gives an excellent way of finding and correcting errors in tables. So far we have only looked at tables of a function f as a method of finding f(n) where n is an integer. However we really want to know f(x) at all points x. In other words, knowing f(0), f(1),..., f(m) we want to find f(x). We shall see that this can be done but we will need to raise the mathematical level a little bit. The key is a set of observations which go back at least as far as Newton. Exercise 11 (i) Consider the function f 2 (x)=x(x-1)/2!. We tabulate it as follows. n 0 1 2 3 4 5 6 7 8 9 f(n) 0 0 1 3 6 10 15 21 28 36 Find the table of successive differences. (ii) Do the same for f 3 (x)=x(x-1)(x-2)/3!, f 4 (x)=x(x-1)(x-2)(x-3)/4!. Do the same for f 1 (x)=x/1!=x and f 0 (x)=1. (iii) Conjecture the general pattern. (iv) [Harder and optional] Prove your conjecture. Exercise 12 (i) Let f(x)=Af 3 (x)+Bf 2 (x)+Cf 1 (x)+D. Find the table of successive differences as in the previous exercise. If we take it in the form f(0) f(1) f(2) f(3) f(4) f(5) f(6) f(7) f(8) f(9) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? identify ?, ?, ? and ?. (ii) If g is any cubic, show that we can find a, b, c and d such that Hence show that g can be found from its table of successive differences. (iii) Conjecture the general pattern. (iv) [Harder and optional] Prove your conjecture. We can extend the results of the last exercises a bit. Suppose that f is a polynomial of degree m. If we form the the following table of differences (here h> 0) f(a) f(a+h) f(a+2h) f(a+3h) f(a+4h) f(a+5h) f(a+6h) f(a+7h) f(a+8h) ... 0 ? ? ? ? ? ? ? ... ? 1 ? ? ? ? ? ? ... ? 2 ? ? ? ? ? ... ? 3 ? ? ? ? ... ? 4 ? ? ? ... and so on, then $$f(a+k)=\alpha_{0}+\alpha_{1}\frac{k}{h}+ \alpha_{2}\frac{k(k-h)}{2!h^{2}}+ \alpha_{3}\frac{k(k-h)(k-2h)}{3!h^{3}}+\dots $$ * Exercise 13 [Optional] Prove this. If we use formula * for $0\leq k\leq mh$ then we say that we are interpolating . If we use formula * for k outside this region we say that we are extrapolating . The sceptical reader may have noticed that, although we began our discussion by talking about tables of the sine function everything that followed dealt with polynomials. However, as the mathematicians of the 17th century discovered, the kind of `nice' functions that we wish to tabulate look very much like polynomials over small ranges of values. Since they look like polynomials, formula * which applies to polynomials will apply to them (to a very close approximation). We need only calculate our desired function at a small number of values and then we can use equation * to find its values at other points. For example, here are the values of ln x at a certain number of values. x 5.00 5.1 5.2 5.3 5.4 f(x) 1.60944 1.62924 1.64866 1.66771 1.68640 We obtain the following table of differences. 1.60944 1.62924 1.64866 1.66771 1.68640 0.01980 0.01942 0.01905 0.01869 -0.00038 -0.00037 -0.0036 0.00001 0.00001 Thus, in equation * , ? 0 =1.60944, ? 1 =0.01980, ? 2 =-0.00038 and ? 4 =0.0001. We thus hope that \[\ln(5+k)\approx1.60944+0.01980\frac{k}{h}+ -0.00019\frac{k(k-h)}{h^{2}}+ +0.00001\frac{k(k-h)(k-2h)}{6h^{3}}.\] Trying this with $k=0.11$ gives \[\ln 5.11\approx 1.63120\] and this turns out to be correct to the number of digits given. Exercise 14 Compute ln 5.16 using our approximation. Compare this with the correct answer. Repeat the exercise for ln y where you choose y with $5 < y < 5.4$. Although functions like ln behave like polynomials over small ranges of values, they need not behave like polynomials over large ranges. It is thus not surprising that an idea which works well for interpolating (that is, trying to find the value of a function f at a point using values of f at points close by) works less well (or fails entirely) when we try to use it to guess the valueof a function at points far away from those where we know its value. Exercise 15 Compute ln 10 using our approximation. Compare this with the correct answer. There is a much more serious problem associated with differencing which I have avoided mentioning up to now. So far, I have assumed that all the initial tabular values are given exactly. In reality, we usually work to a certain number of decimal places. Suppose that we round to the nearest integer. Then a table which looks like 0 0 0 0 0 0 0 0 with a table of differences 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 could actually represent ${ {1 \over 2} -{1 \over 2}{1 \over 2} -{1 \over 2}{1 \over 2} -{1 \over 2}{1 \over 2} -{1 \over 2} }$ with a table of differences 1/2 -1/2 1/2 -1/2 1/2 -1/2 1/2 -1/2 1 -1 1 -1 1 -1 1 -2 2 -2 2 -2 2 4 -4 4 -4 4 -8 8 -8 8 Thus if the first line of a table of differences is only known to an accuracy of epsilon; the second line is only known to an accuracy of 2 epsilon; and the nth line to an accuracy of 2 n epsilon;. In practice this means that, initially, the entries in successive lines of a table of differences will tend to decrease (just as we saw when we used exact arithmetic) but because the errors are increasing there will come a point when the errors swamp the calculation and the entries in successive lines will tend to increase. As an example consider the entries in a table of sines (x represents degrees). x 10 11 12 13 14 15 16 17 sin x 0.1736 0.1908 0.2079 0.2250 0.2419 0.2588 0.2756 0.2924 Here is the table of differences 0.1736 0.1908 0.2079 0.2250 0.2419 0.2588 0.2756 0.2924 0.0172 0.0171 0.171 0.169 0.0169 0.168 0.0168 -0.0001 0.0000 -0.0002 0.0000 -0.0001 0.0000 0.0001 -0.0001 -0.0002 0.0002 -0.0001 0.0001 -0.0002 -0.0001 0.0004 -0.0003 0.0002 -0.0001 0.0005 -0.0007 0.0005 0.0006 -0.0012 0.0012 It is clear that only the first two lines of the table of differences (and perhaps, if we think hard, the third) carry any information. In the remaining lines the `noise' of the errors drowns out everything else. This means that when we interpolate we must confine ourselves to using the first two lines and our version of * will read \[\sin (10+k)\approx0.1736+0.0172k.\] (However, as the reader can verify, this remains a pretty accurate formula for $k$ with $0\leq k\leq 1$.) Exercise16 Choose a table (or make your own) of some function, form the table of differences, note the line at which error noise becomes dominant. Use the part of the table above that line to interpolate at some point. Check the accuracy of your interpolation. It is very instructive to investigate how the number of significant figures and the spacing of the points of your table affectthe result. We can now see one way in which we can compile tables of a function f like ln or sin with a given level of accuracy. We compute f at a relatively small number of points to much higher accuracy than the table demands. (Of course, these calculations may be quite lengthy but we only need do a few of them.) We obtain the values at f at the remaining points by easy interpolation. Notice that once we have obtained a few points using * we can the build up the rest of the table using the techniques of Exercise 16. For four hundred years science and technology depended on tables. Books of tables made it possible to find the orbits of the planets, to navigate the globe, and to make an industrial revolution. The book of logarithms was as much a symbol of science as the microscope. Yet, even then, few people bothered to celebrate the labours of the maker of tables. Nowadays their work is completely forgotten. I hope that this essay may help the reader appreciate the work of these unsung heros. Dr Tom Körner is a Reader in the Department of Pure Mathematics and Mathematical Statistics at the University of Cambridge and a Fellow of Trinity Hall.
Search Now showing items 1-2 of 2 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... Beauty production in pp collisions at √s=2.76 TeV measured via semi-electronic decays (Elsevier, 2014-11) The ALICE Collaboration at the LHC reports measurement of the inclusive production cross section of electrons from semi-leptonic decays of beauty hadrons with rapidity \|y\|<0.8 and transverse momentum 1<pT<10 GeV/c, in pp ...
Definitions The line integral of a vector function $\mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}$ is said to be path independent, if and only if $P,$ $Q$ and $R$ are continuous in a domain $D,$ and if there exists some scalar function $u = u\left( {x,y,z} \right)$ in $D$ such that \[ {\mathbf{F} = \text{grad}\,u\;\;\;}\kern-0.3pt{\text{ or }\;\;\frac{{\partial u}}{{\partial x}} = P,\;\;\;}\kern-0.3pt {\frac{{\partial u}}{{\partial y}} = Q,\;\;\;}\kern-0.3pt {\frac{{\partial u}}{{\partial z}} = R.} \] If this is the case, then the line integral of $\mathbf{F}$ along the curve $C$ from $A$ to $B$ is given by the formula \[ {\int\limits_C {\mathbf{F}\left( \mathbf{r} \right) \cdot d\mathbf{r}} } = {\int\limits_C {Pdx + Qdy + Rdz} } = {u\left( B \right) – u\left( A \right).} \] (This result for line integrals is analogous to the Fundamental Theorem of Calculus for functions of one variable). Hence, if the line integral is path independent, then for any closed contour $C$ \[\oint\limits_C {\mathbf{F}\left( \mathbf{r} \right) \cdot d\mathbf{r} = 0}.\] A vector field of the form $\mathbf{F} = \text{grad}\,u$ is called a conservative field, and the function $u = u\left( {x,y,z} \right)$ is called a scalar potential. A Test for a Conservative Field The line integral of a vector function $\mathbf{F} = P\mathbf{i} + Q\mathbf{j} $ $+ R\mathbf{k}$ is path independent if and only if \[ {\text{rot}\,\mathbf{F} }={ \left\| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ P&Q&R \end{array}} \right\| }={ \mathbf{0}.} \] It’s implied that each component of $\mathbf{F}$ has continuous partial derivatives of variables $x, y$ and $z.$ If the line integral is taken in the $xy$-plane, then the following formula is valid: \[{\int\limits_C {Pdx + Qdy} }={ u\left( B \right) – u\left( A \right).}\] In this case, the test for determining if a vector field is conservative can be written in the form \[\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}.\] The considered test is the necessary requirement, but generally speaking, it is not sufficient condition for a vector field to be conservative. However, this test is sufficient, if the region of integration $D$ is simply connected. Solved Problems Click a problem to see the solution. Example 1Evaluate the line integral $\int\limits_{AB} {\left( {x + y} \right)dx + xdy} $ for two paths of integration: $AB$ is the line segment from $A\left( {0,0} \right)$ to $B\left( {1,1} \right)$; $AB$ is the parabola $y = {x^2}$ from $A\left( {0,0} \right)$ to $B\left( {1,1} \right)$. Example 2Show that the line integral $\int\limits_{AB} {\left( {3{x^2}y + y} \right)dx }$ $+{ \left( {{x^3} + x} \right)dy}$ is path independent and calculate this integral. The coordinates of the points $A, B$ are $A\left( {1,2} \right),$ $B\left( {4,5} \right).$ Example 3Determine if the vector field $\mathbf{F} = \left( {yz,xz,xy} \right)$ is conservative? Example 4Determine if the vector field $\mathbf{F}\left( {x,y} \right) =$ $\left( {x + y,x – y} \right)$ is conservative? If it is, find its potential. Example 5Determine if the vector field $\mathbf{F}\left( {x,y,z} \right) =$ $\left( {yz,xz + 2y,xy + 1} \right)$ is conservative. If it is, find its potential. Example 1.Evaluate the line integral $\int\limits_{AB} {\left( {x + y} \right)dx + xdy} $ for two paths of integration: $AB$ is the line segment from $A\left( {0,0} \right)$ to $B\left( {1,1} \right)$; $AB$ is the parabola $y = {x^2}$ from $A\left( {0,0} \right)$ to $B\left( {1,1} \right)$. Solution. Consider the first case. Obviously, the equation of the line is $y = x.$ Then using the formula \[ {\int\limits_{AB} {P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy} \text{ = }}\kern0pt {\int\limits_a^b {\left[ {P\left( {x,y} \right) + Q\left( {x,y} \right)\frac{{dy}}{{dx}}} \right]dx},} \] we obtain \[ {{I_1} }={ \int\limits_{AB} {\left( {x + y} \right)dx + xdy} } = {\int\limits_0^1 {\left( {x + x + x \cdot 1} \right)dx} } = {\int\limits_0^1 {3xdx} } = {3\left[ {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right\|_0^1} \right] } = {\frac{3}{2}.} \] If the curve $AB$ is parabola $y = {x^2},$ we have \[ {{I_2} }={ \int\limits_{AB} {\left( {x + y} \right)dx + xdy} } = {\int\limits_0^1 {\left( {x + {x^2} + x \cdot 2x} \right)dx} } = {\int\limits_0^1 {\left( {x + 3{x^2}} \right)dx} } = {\left. {\left( {\frac{{{x^2}}}{2} + \frac{{3{x^3}}}{3}} \right)} \right\|_0^1 } = {\frac{1}{2} + 1 }={ \frac{3}{2},} \] that is we have obtained the same answer. Apply the test ${\large\frac{{\partial P}}{{\partial y}}\normalsize} = {\large\frac{{\partial Q}}{{\partial x}}\normalsize}$ to determine if the vector field is conservative. \[ {\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}},\;\;}\Rightarrow {\frac{{\partial \left( {x + y} \right)}}{{\partial y}} = \frac{{\partial x}}{{\partial x}},\;\;}\Rightarrow {1 \equiv 1.} \] As it can be seen, the vector field $\mathbf{F} = \left( {x + y,x} \right)$ is conservative. This explains the result that the line integral is path independent.
Interior-Point Method In this post the interior point method described in [1] will be discussed. This algorithm solve the nonlinearly constrained optimization problem: \begin{eqnarray} \min_x && f(x), \\ \text{subject to } && c_E(x) = 0,\\ && c_I(x) \le 0, \end{eqnarray} where $x\in \mathbb{R}^n$ is a vector of unknowns, $f$ is called the objective function and $c_E$ and $c_I$ are vectorial functions used to delimit the feasible region. A first version of this trust-region interior-point method was implemented during this week and the guiding principles behind the method will be explained in this post. Applications and test results will be left for future blog posts. Barrier Problem Introducing slack variables $s$ the general nonlinear programming problem can be rewritten as: \begin{eqnarray} \min_x && f(x), \\ \text{subject to } && c_E(x) = 0,\\ && c_I(x) + s = 0, \\ && s \ge 0. \end{eqnarray} The basic idea of the implemented interior-point method is tosolve, instead of the above problem, the equality-constrained barrier problem:\begin{eqnarray} \min_{x, s} && f(x) - \mu \sum_{i} \ln (s_i), \\ \text{subject to } && c_E(x) = 0,\\ && c_I(x) + s = 0, \end{eqnarray} for progressively smaller values of the barrier parameter $\mu$. The previously implemented Sequential Quadratic Programming (SQP) solver (described here) is used to successively solve those problems for $\mu \rightarrow 0$. The way of doing this efficiently described in [1] is not trying to solve the equality problem exactly, but rather to solve it inexactly with increasing accuracy as the problem gets closer to the solution. The name “barrier problem” is due to the the fact the function $-\ln(s_i)$ increases to infinity as $s_i$ approach the value $0$, enforcing the condition $s>0$. By letting $\mu$ converge to zero, the sequence of solutions of the barrier problems converge to a solution point of the original nonlinear programming problem. Outline of the Algorithm The first order optimality conditions [2], p.321,for the barrier problem, guarantee that at a solution point $(x^,~s^)$the following equations are satisfied:\begin{eqnarray} \nabla_x \mathcal{L}(x^, s^, \lambda_E, \lambda_I) &=& 0, \\ \nabla_s \mathcal{L}(x^, s^, \lambda_E, \lambda_I) &=& 0, \\ c_E(x^) &=& 0,\\ c_I(x^) + s^* &=& 0, \end{eqnarray} where $\nabla_x$ and $\nabla_s$ represents the first derivatives regarding, respectively, $x$ and $s$; and, for which $\mathcal{L}(x, s, \lambda_E, \lambda_I)$ represent the Lagrangian: \begin{equation} \mathcal{L}(x, s, \lambda_E, \lambda_I) = f(x) - \mu \sum_{i} \ln (s_i) + \lambda_E^T c_E(x) + \lambda_I^T (c_I(x) + s), \end{equation} and $\lambda_E$ and $\lambda_I$, the Lagrange multipliers. Rather than a exact solution to the barrier problem,we will be contend with an approximation solution satisfying:\begin{eqnarray} \\|\nabla_x \mathcal{L}(x, s, \lambda_E, \lambda_I)\\| &<& \epsilon, \\ \\|\nabla_s \mathcal{L}(x, s, \lambda_E, \lambda_I)\\| &<& \epsilon, \\ \\|c_E(x)\\| &<& \epsilon,\\ \\|c_I(x) + s\\| &<& \epsilon \end{eqnarray} Starting from a point $(x, s)$, for an initial barrier parameter $\mu$ and tolerance $\epsilon$ the algorithm repeats the following steps until a stop criteria: Apply SQP trust-region starting from $(x, s)$ to find approximation solution $(x^{+}, s^{+})$ of the barrier problem (for an barrier parameter $\mu$), satisfying the tolerance $\epsilon$ Update solution $(x, s) \leftarrow (x^+, s^+)$, update the barrier solution $\mu \leftarrow 0.2\mu$ and the tolerance $\epsilon \leftarrow 0.2\epsilon$ Different strategies for decreasing $\mu$ and $\epsilon$ along the iterations are discussed in [3]. Final Comments In this and in the two previous blog posts I tried to describe the main aspects of the algorithm I am implementing.In the first post the projected conjugategradient method was described. This method solve the equality-constrained Quadratic Programming (QP) problem:\begin{eqnarray} \min_x && \frac{1}{2} x^T H x + c^T x + f, \\ \text{subject to } && A x = b. \end{eqnarray} The sequential solution of equality-constrained QP problems is the guiding principle for the SQP solver I described in the second post. This SQP solver is for equality-constrained problems of the form: \begin{eqnarray} \min_x && f(x), \\ \text{subject to } && c(x) = 0. \\ \end{eqnarray} And, in turn, is used as a substep of the interior-point method described in this post.
I suppose you actually suggest independent hugh haphazard aspects; should they be not really separate, then an answer will have to possibly be conveyed with regards to the articulation submission. Solve for $k: e Means \dfrac 04.1$ years With Example, the length of a definite computer aspect gets the rapid circulation having a signify regarding several years ($By \sim Exp(3.Just one)$). The time has an hugh submitting using the normal length of time corresponding to four a matter of minutes. As talked about through @Drew75 inside responses, individuals must remember the fact that the particular necessarily mean essay writer website of any dramatic random varying together with parameter \lambda is equivalent to 1/\lambda. The phone number $at the Equates to 2.71828182846$. As talked about by means of @Drew75 inside comments, you need to understand that the actual signify of any rapid haphazard varying with parameter \lambda is equivalent to 1/\lambda. It is just a quantity employed usually with mathematics. Conversely, when the range of events every model time frame practices some sort of Poisson submitting, then a period involving events comes after the particular hugh syndication. It is a simple and easy wonderful consequence. memoryless residence For any great haphazard variable $X$, your memoryless property owner this declaration that will knowledge of what has happened in days gone by doesn’t have any affect on long term odds. \min(Back button,Y simply)\le x Equals (X\le by) \cup (Ymca \le y), Guess a buyer has got used some units which has a postal worker. On the regular, just how long might all 5 laptop or computer parts continue should they be utilised a person after another? pdf: $f ree p(a) = me^ exactly where \(back button \geq 0$ plus $l 0$ At an interest rate of 5 cars and trucks for each minute, we predict $\dfrac 12$ a few moments to move concerning consecutive cars on average. Half of most industry is completed inside of how long? (Obtain the 40 th percentile) After a person comes, look for the chances that it requires in excess of 5 minutes for buyer being released. Uncover $S(On the lookout for 15) Means 2.4346$ There is a constant \lambda to ensure that P(X \geq testosterone)=e^ t for every single t If you need assistance of any type, essaywriter.org/case-study-help make sure you discover self-paced help on our assistance site. Believe the time of which elapses in one call an additional gets the rapid circulation. Your current coach will report your portions around money. References In order to determine $G(By \leq k$), key in A pair of nd , VARS (DISTR), Chemical:poissoncdf($\lambda, k$). Final observe: As much as I’m sure, there is not any popular good reputation that supply. If there is a legitimate Bing plus code, follow these steps if you wish to get rid of the articles you write, opinions, ballots, and/or account from the Aol product comments message board. Therefore a particularly long delay among a pair of phone calls is not to mean that there might be a short patiently waiting phase for contact. make simpler in addition to remember that Y is usually drastically dispersed and look for it has the parameter. The amount of a short time carry out 50 % of virtually all people wait? All a couple of chances are provided directly by means of F (responding to the leading concern): As brought up through @Drew75 inside the remarks, you ought to do not forget that the particular indicate of an dramatic haphazard diverse along with parameter \lambda comes to 1/\lambda. Media your (-) to the unfavorable. Answer Wiki This generalises effortlessly to your circumstance with well over a couple unbiased dramatic specifics. Lure the chart. and also password to take part. The memoryless property claims that $G(By Six \| X Some) Is equal to Delaware(Times Several)$, therefore we must obtain the probability that the client gets to spend in excess of three moments using a mailing sales person. $f ree p(y) Equates to 2.25e^ (1.30)(A person) Equals 1.30 Equals m$. Find a odds in which exactly 5 calls occur in just a instant. Since a person customer happens every single a couple a few minutes an average of, it will require 6 minutes generally for several visitors to occur. On the average, how much time could several laptop or computer sections last when they are utilised just one after another? At a rate of 5 cars and trucks every minute, we predict $\dfrac 12$ mere seconds to pass through involving effective vehicles generally. The second concept is well known because research the prospect of which impartial variables X along with Y are usually not less than or perhaps comparable to x, distributed by Bucks(1-F_X(by))(1-F_Y(a))Dollar: the actual subtraction through 1 and then gives the subsidiary opportunity which more then one of such aspects is actually a lot less than or perhaps similar to x, that is just what \min(A,Y simply)\le x means. \[P(x Several) Implies A single – W(by x) = 1 -(Just one -e^ = e^ Mass media your (-) with the negative. How long laptop piece will last is exponentially allocated. Allow $Utes =$ the space everyone is ready to travel around mile after mile. X_n are generally unbiased identically-distributed unique issues, exactly what do often be said about the syndication with \min(X_1, . Please note that we are worried just with the velocity of which message or calls come in, and we are overlooked any time used on the cell phone. They may be interchanged out of all former data when F is continuous, yet if not they’ve created a positive change. The dramatic supply offers the memoryless property, which claims that foreseeable future probability never rely upon virtually any recent data. $A single ( blank ) e^(-0.A few) \approx 4.3935$ Relationship between the Poisson as well as the Dramatic Distribution The actual chances in which one or more X_i is smaller when compared with y comes to just one without worrying about probability that X_i are generally more than y. The important (general) t in case and only in case each X_i t. Methodical calculators contain the key “$e^$.Inches When you get into one particular to get $x$, the online car loan calculator will display the worthiness $e$. We discover P(By unces) Equates to One ( space ) F_X(unces) Is equal to One particular ( space ) (One ( space ) e^ z Implies e^ z together with P(B unces) = e^ z Notice the chart is really a declining contour. Cover from the sun the spot to display a probability any particular one scholar has got under Bucks.Forty five in his or her pants pocket as well as handbag. Answer Wiki Then you certainly get that \mathbb (Y x)=\mathbb (X_1 X_2 in which the very last phase comes after from flexibility on the \. Publish the distribution, talk about this possibility density operate, in addition to graph this submission. The number of days in advance vacation goers invest in the airline tickets is usually produced by a good rapid submitting using the ordinary amount of time adequate to 16 times. The following community forum just isn’t administered for any support-related troubles. This means that this possibility of which $X$ meets $by + k$, due to the fact it’s overtaken $x$, matches the actual chance that will $X$ could go beyond $k$ whenever we did not have any information about the idea. Notice now that \mathbb (X_i i and you may almost certainly fill the last details yourself, i actually.at the. Formula Review (The supply operate could possibly be handy). $Delaware(Big t 19 \| Testosterone Equates to 14) = P(To 8) Means 1 — G(T memoryless property or home For an dramatical hit-or-miss diverse \(X$, your memoryless residence is this statement of which expertise in what has happened in the last has no affect on future likelihood. Suppose $X$ has got the Poisson submission by using mean (\lambda\). Answer Wiki X_n) generally speaking? Values on an great hit-or-miss changing occur in these manner. Make Y= the smallest or lowest importance of these 3 random issues. Touch this (:) for your damaging. On the regular, how much time could five computer system sections very last when utilized one particular after another? Using the details in Physical exercise, discover the likelihood that a worker usually spends four to five a few minutes by using a arbitrarily decided on purchaser. $P(any by + p \| Back button back button) Means R(Times p)$ What is the likelihood that your personal computer element lasts more than 7 years? Data through World Earthquakes, The year 2013. Available on the web from http://www.world-earthquakes.com/ (seen 06 13, 2013). After a car passes by, the time typically could it require another seven automobiles to give through? Using the solution through medicare part a, we have seen that it requires $(A dozen)(7) = 84$ seconds for the 6 motor vehicles to give by. Contributed simply by Ann Illowsky & Barbara Dean The particular mailing sales person uses five minutes using the consumers. However this would mean make fish an ancient piece isn’t any more prone to digest with just about any selected time frame over a new element. This memoryless property states that $R(By Six \| Y Some) Equals G(Times Several)$, therefore we only need to chose the likelihood that the shopper usually spends over several a few minutes having a mailing sales person. It could be proven, also, how the importance of the advance that you’ve got in the bank or even wallet around follows a great rapid circulation.
Exercise: Suppose a random sample of size n is drawn from a normal pdf where the mean $\mu$ is known ut the variance $\sigma^{2}$ is unknown. Use the method of maximun likelihood to find a formula for $\hat \sigma^2$ Compare your answer to the maximun likelihood estimator found in Example $5.2.4$ I already this this problem, and I got $\hat\sigma^2_ = \frac{1}{n}\Sigma_{i= 1}^{n} (y_i - \mu)^2$. And that is the book's answer, so it is correct. But it does not mention the comparison.I need to compare my answer to the answer given in Example 5.2.4 This is what example 5.2.4 says Example $5.2.4$ Suppose a random sample of size $n$ is drawn from the two-parameter normal pdf $f_Y(y; \mu, \sigma^2) = \frac{1}{\sqrt(2\pi) \sigma}e^{\frac{(y_{i} - \mu)}{2\sigma^2}}$ Use the method of maximun likelihood to find formulas for $\mu_e$ and $\sigma^2_e$. Answers for Example $5.2.4$ are 1) $\mu_e = \frac{1}{n}\Sigma_{i= 1}^{n}y_i = \bar y$ and 2) $\sigma^2_e = \frac{1}{n}\Sigma_{i= 1}^{n} (y_i - \bar y)^2$. The answers look the same, so I would compare them as being identical. However, I don't know if it's ok to assume that. Can anyone please compare the answers? Thank you
Various geometrical figures in three-dimensional space can be described relative to a set of mutually orthogonal axes O$x$, O$y$, O$z$, and a point can be represented by a set of rectangular coordinates $(x, y, z)$. The point can also be represented by cylindrical coordinates $( ρ , \phi , z)$ or spherical coordinates $(r , θ , \phi )$, which were described in Chapter 3. In this chapter, we are concerned mostly with $(x, y, z)$. The rectangular axes are usually chosen so that when you look down the $z$-axis towards the $xy$-plane, the $y$-axis is $90^\circ$ counterclockwise from the $x$-axis. Such a set is called a right-handed set. A left-handed set is possible, and may be useful under some circumstances, but, unless stated otherwise, it is assumed that the axes chosen in this chapter are right-handed. An equation connecting $x$, $y$ and $z$, such as \[f(x,y,z) = 0 \label{4.1.1} \tag{4.1.1}\] or \[z = z(x,y) \label{4.1.2} \tag{4.1.2}\] describes a two-dimensional surface in three-dimensional space. A line (which need be neither straight nor two-dimensional) can be described as the intersection of two surfaces, and hence a line or curve in three-dimensional coordinate geometry is described by two equations, such as \[f(x,y,z) = 0 \label{4.1.3} \tag{4.1.3}\] and \[g (x,y,z) = 0 . \label{4.1.4} \tag{4.1.4}\] In two-dimensional geometry, a single equation describes some sort of a plane curve. For example, \[y^2 = 4qx \label{4.1.5} \tag{4.1.5}\] describes a parabola. But a plane curve can also be described in parametric form by two equations. Thus, a parabola can also be described by \[x = qt^2 \label{4.1.6} \tag{4.1.6}\] and \[y = 2qt \label{4.1.7} \tag{4.1.7}\] Similarly, in three-dimensional geometry, a line or curve can be described by three equations in parametric form. For example, the three equations \[x = a \cos t \label{4.1.8} \tag{4.1.8}\] \[y = a \sin t \label{4.1.9} \tag{4.1.9}\] \[z = ct \label{4.1.10} \tag{4.1.10}\] describe a curve in three-space. Think of the parameter $t$ as time, and see if you can imagine what sort of a curve this is. We shall be concerned in this chapter mainly with six types of surface: the plane, the ellipsoid, the paraboloid, the hyperboloid, the cylinder and the cone.
In his celebrated paper "Conjugate Coding" (written around 1970), Stephen Wiesner proposed a scheme for quantum money that is unconditionally impossible to counterfeit, assuming that the issuing bank has access to a giant table of random numbers, and that banknotes can be brought back to the bank for verification. In Wiesner's scheme, each banknote consists of a classical "serial number" $s$, together with a quantum money state $\|\psi_s\rangle$ consisting of $n$ unentangled qubits, each one either $$\|0\rangle,\ \|1\rangle,\ \|+\rangle=(\|0\rangle+\|1\rangle)/\sqrt{2},\ \text{or}\ \|-\rangle=(\|0\rangle-\|1\rangle)/\sqrt{2}.$$ The bank remembers a classical description of $\|\psi_s\rangle$ for every $s$. And therefore, when $\|\psi_s\rangle$ is brought back to the bank for verification, the bank can measure each qubit of $\|\psi_s\rangle$ in the correct basis (either $\{\|0\rangle,\|1\rangle\}$ or ${\|+\rangle,\|-\rangle}$), and check that it gets the correct outcomes. On the other hand, because of the uncertainty relation (or alternatively, the No-Cloning Theorem), it's "intuitively obvious" that, if a counterfeiter who doesn't know the correct bases tries to copy $\|\psi_s\rangle$, then the probability that both of the counterfeiter's output states pass the bank's verification test can be at most $c^n$, for some constant $c<1$. Furthermore, this should be true regardless of what strategy the counterfeiter uses, consistent with quantum mechanics (e.g., even if the counterfeiter uses fancy entangled measurements on $\|\psi_s\rangle$). However, while writing a paper about other quantum money schemes, my coauthor and I realized that we'd never seen a rigorous proof of the above claim anywhere, or an explicit upper bound on $c$: neither in Wiesner's original paper nor in any later one. So, has such a proof (with an upper bound on $c$) been published? If not, then can one derive such a proof in a more-or-less straightforward way from (say) approximate versions of the No-Cloning Theorem, or results about the security of the BB84 quantum key distribution scheme? Update: In light of the discussion with Joe Fitzsimons below, I should clarify that I'm looking for more than just a reduction from the security of BB84. Rather, I'm looking for an explicit upper bound on the probability of successful counterfeiting (i.e., on $c$)---and ideally, also some understanding of what the optimal counterfeiting strategy looks like. I.e., does the optimal strategy simply measure each qubit of $\|\psi_s\rangle$ independently, say in the basis $$\{ \cos(\pi/8)\|0\rangle+\sin(\pi/8)\|1\rangle, \sin(\pi/8)\|0\rangle-\cos(\pi/8)\|1\rangle \}?$$ Or is there an entangled counterfeiting strategy that does better? Update 2: Right now, the best counterfeiting strategies that I know are (a) the strategy above, and (b) the strategy that simply measures each qubit in the $\{\|0\rangle,\|1\rangle\}$ basis and "hopes for the best." Interestingly, both of these strategies turn out to achieve a success probability of (5/8) n. So, my conjecture of the moment is that (5/8) n might be the right answer. In any case, the fact that 5/8 is a lower bound on c rules out any security argument for Wiesner's scheme that's "too" simple (for example, any argument to the effect that there's nothing nontrivial that a counterfeiter can do, and therefore the right answer is c=1/2). This post has been migrated from (A51.SE) Update 3: Nope, the right answer is (3/4) n! See the discussion thread below Abel Molina's answer.
My first post of this question is now not on the first page, so I thought I would repost it. Find all possible integer values of $z$ such that the following system of equations has a solution for $z$: $\begin{align} z^n &= 1, \\ \left(z + \frac{1}{z}\right)^n &= 1. \end{align}$ Edit: $z$ can be a complex number, too! Link to first question: https://web2.0calc.com/questions/help-plles $z = r e^{i \theta}\\ z^n = r^n e^{i n \theta}\\ z^n = 1 \Rightarrow r = 1\\ z = e^{i \theta}$ $z + \dfrac 1 z = e^{i\theta}+ e^{-i\theta} = 2\cos(\theta)\\~\\ \left(z + \dfrac 1 z \right)^n = 2^n \cos^n(\theta) = 1\\ \cos^n(\theta) = \dfrac{1}{2^n}\\ \cos(\theta) = \dfrac 1 2\\ \theta = \pm \dfrac{\pi}{3}$ $z = e^{\pm i \pi/3}$.
The two operators are not identical, however they approach to each other in the semiclassical limit. To show that, we can compute matrix elements of the two expressions in a complete set of states and show that the matrix elements approach each other. (Strictly speaking this is weak convergence). I'll perform the computation in one dimension, the generalization to higher dimensions is straightforward. A possible basis is the coherent state basis. The coherent states are eigenstates of the field operator $\Psi(x)$. The coherent states basis is overcomplete and especially convenient for semiclassical approximations. In this basis each Schrodinger particle is labeled by a position and momentum and for $n$ Schroedinger particles, it has the form $\| x_1, ..., x_n; p_1, ..., p_n \rangle $, such that the action of the field operator on the coherent state: $$\Psi(x) \| x_1, ..., x_n; p_1, ..., p_n \rangle = \prod_{i=1}^n \sqrt[\LARGE 4]{\frac {m \omega}{ \pi \hbar }} e^{ \large -\frac{m \omega}{2\hbar}(x-x_i)^2 + \frac{ixp_i}{\hbar}} \| x_1, ..., x_n; p_1, ..., p_n \rangle $$ The factors in the right hand side consist of the coherent state wave functions that Schrodinger wrote in 1926. The constant parameter $\omega$ is a constant having units of angular frequency. The matrix element of the electric current ( I took the liberty of adding the particle electric charge to the expression) can be readily computed: $$\langle x_1, ..., x_n; p_1, ..., p_n \|j(x) \| x_1, ..., x_n; p_1, ..., p_n \rangle = e \sum_{i=1}^n \frac{p_i}{m} \sqrt{ \frac { m \omega}{ \pi \hbar}} e^{-\frac{m \omega}{\hbar}(x-x_i)^2 }$$ The first factors are just the velocities: $$v_i = \frac{p_i}{m}$$ The second term approaches a delta function in the semiclassical limit: $$\lim_{\hbar \to 0} \sqrt{ \frac {m \omega }{ \pi \hbar}} e^{-\frac{m \omega}{\hbar}(x-x_i)^2 } = \delta(x-x_i)$$
I've seen various topics here adressing how to turn math symbols bold, but none of them gives a suitable option for \mathbb characters, as in \mathbb{ABC}. I have tested some options to see which commands turn which symbols in bold. Here is my code: \documentclass{article}\usepackage{amsmath} % some math-related packages, not sure which of them are necessary\usepackage{amssymb}\usepackage{amsfonts}\usepackage{bm} % for \bm\usepackage{fixmath} % for \mathbold\begin{document}\section{Some text $\mathbb{ABC}abc\cosh\div+\alpha$ some text} %1\section{Some text $\boldsymbol{\mathbb{ABC}abc\cosh\div+\alpha}$ some text} %2\section{Some text $\mathbf{\mathbb{ABC}abc\cosh\div+\alpha}$ some text} %3\section{Some text $\pmb{\mathbb{ABC}abc\cosh\div+\alpha}$ some text} %4\section{Some text $\boldmath{\mathbb{ABC}abc\cosh\div+\alpha}$ some text} %5\section{Some text $\bm{\mathbb{ABC}abc\cosh\div+\alpha}$ some text} %6\section{Some text $\mathbold{\mathbb{ABC}abc\cosh\div+\alpha}$ some text} %7\end{document} (The absence of \div in numbers 2 and 6 are not typos, neither is the 'ff' in number 3.) Apparently \pmb is the only command which turns \mathbb{ABC} bold. But it makes the characters quite ugly, take a close look: (on the right is the normal version) If no alternative shows up, I think I'll just use \pmb for the \mathbb characters and \boldsymbol or something for the others. Is there any alternative to turn \mathbb{ABC}bold?
Continuity and Differentiability Second Order Derivative If f'( x) is differentiable, w.r.t. x. Then, the left hand side becomes \frac{d}{dx}\left(\frac{dy}{dx}\right) which is called the second order derivative of y w.r.t. x and is denoted by \frac{d^2y}{dx^2}. The second order derivative of f( x) is denoted by f''( x). It is also denoted by D 2y or y'' or y 2 if y =f(x). View the Topic in this video From 06:00 To 10:26 Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers. If y=\frac{f(x)}{1+\frac{f'(x)}{1+y}}, then \frac{dy}{dx}=\frac{(1+y)f'(x)-y f^{"}(x)}{(1+2y)-f(x)+f^{"}(x)} If y = sin (ax + b) \Rightarrow \frac{d^{n}y}{dx^{n}}=\sin \left(\frac{n\pi}{2}+(ax+b)\right) y = cos (ax + b) \Rightarrow \frac{d^{n}y}{dx^{n}}=\cos \left(\frac{n\pi}{2}+(ax+b)\right) y = e ax + b\Rightarrow \frac{d^{n}y}{dx^{n}}=a^{n} \cdot e^{(ax+b)}
1. Right triangle has legs with lengths $\sqrt7$ on the bottom and 1 on the right side. Use this information to find: $\begin{align} &\sqrt{8}\cdot \cos\left(\arctan\left(\sqrt{7}\right)\right), \\ &\sqrt{8} \cdot \cos\left(\arccos\left(\frac{1}{\sqrt{8}}\right)\right), \\ &\sqrt{8} \cdot \sin\left(\arccos\left(\frac{1}{\sqrt{8}}\right)\right), \end{align}$in the same order. 2. Calculate $\arcsin(1), \arccos\left(\frac{1}{2} \right), \arctan(1), \arcsin\left(-\frac{1}{2}\right)$ in radians and enter them below. 3. A 3-4-5 right triangle has length 3 on the bottom side and 4 on the right side. Use this information to find: $\sin\left(\arctan\left(\frac{3}{4}\right)\right), \cos\left(\arcsin\left(\frac{3}{5}\right)\right), \cot\left(\arctan\left(\frac{4}{3}\right)\right), \sin\left(\arcsin\left(\frac{4}{5}\right)\right)$in the same order. Thank you for your help! I'm having a bit of trouble with calculus. 1. √8 * cos [ arctan (√7 ] This says that we first need to find the cosine of an angle whose tangent = y / x = √7/1 So we have x / √[y^2 + x^2 ] = 1 / √[ (√7)^2 + 1^2] = 1 / √ [7 + 1] = 1 / √8 So... √8 * 1 / √8 = 1 √8 * cos [ arccos (1/ √8) ] Note that cos [ arccos ( 1/√8) ] = 1/√8 So, again.... √8 * 1 / √8 = 1 √8 * sin [ arcos (1/√8)].....this says that we first need to find the sin of an angle whose cosine = 1 / √8 y = √[ r^2 - x^2] = √[ 8 - 1 ] = √7 So sin [ arccos (1 / √8 ) ] = y/r = √7/√8 So √8 * √7/√8 = √7 2. arcsin 1 = we are asking.....where is the sine = 1 ???......this is at pi/2 rads arccos 1/2 = where is the cosine = 1/2 ???.....this is at pi/3 rads arctan 1 = where is the tangent = 1 ???......this is at pi/4 rads arcsin -1/2 = where is the sine = -1/2 ???.....this is at - pi/6 rads P.S. - you should memorize the values of the sine, cosine and tangent at 0, pi/6, pi/4, pi/3 and pi/2 rads Here is a sample of how to solve #3 see the image....think you can now do the others? 3. Note that r = √[3^2 + 4^2] = √25 = 5 sin [ arctan (3/4) ] = we are looking for the sine of an angle whose tangent = 3/4 = y r = 3 / 5 cos [arcsin (3/5)] = we are looking for the cosine whose sine = 3/5 = [ 5^2 - 3^2] / 5 = x/r = √16/5 = 4/5 cot [ arctan ( 4/3) ] = we are looking for the tangent of an angle whose cot = 4/3 = 3/4 sin [arcsin (4/5) ] = 4/5
Some Definitions A function $f\left( x \right)$ defined on an interval $\left[ {a,b} \right]$ is said to be piecewise continuous if it is continuous on the interval except for a finite number of jump discontinuities (Figure $1$). A function $f\left( x \right)$ defined on an interval $\left[ {a,b} \right]$ is said to be piecewise smooth if $f\left( x \right)$ and its derivative are piecewise continuous. Partial Sums of Fourier Series We introduce the Fourier partial sum ${f_N}\left( x \right)$ of the function $f\left( x \right)$ defined on the interval $\left[ {-\pi, \pi} \right]$ as \[{{f_N}\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^N {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)} .}\] In complex form, the $n$th partial sum ${f_N}\left( x \right)$ of a function $f\left( x \right)$ defined on the interval $\left[ {-\pi, \pi} \right]$ is given by \[ {{f_N}\left( x \right) = \sum\limits_{n = – N}^N {{c_n}{e^{inx}}} } = {\int\limits_{ – \pi }^\pi {\left( {\frac{1}{{2\pi }}\sum\limits_{n = – N}^N {{e^{in\left( {x – y} \right)}}} } \right)f\left( y \right)dy}} \] Dirichlet Kernel The function \[{{D_N}\left( x \right) = \sum\limits_{n = – N}^N {{e^{inx}}} }={ \frac{{\sin \left( {N + \frac{1}{2}} \right)x}}{{\sin \frac{x}{2}}} }\] is called the Dirichlet kernel. In Figure $2$ we have graphed Dirichlet kernel for $n = 10.$ The Fourier partial sum of $f\left( x \right)$ can be expressed through the Dirichlet kernel: \[ {{f_N}\left( x \right) }={ \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {{D_N}\left( {x – y} \right)f\left( y \right)dy} } = {\frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {{D_N}\left( y \right)f\left( {x – y} \right)dy} .} \] In this section, we consider three types of convergence: pointwise, uniform and ${L_2}$-convergence. Pointwise Convergence of Fourier Series Let $f\left( x \right)$ be a piecewise smooth function on the interval $\left[ {-\pi, \pi} \right].$ Then for any ${x_0} \in \left[ { – \pi ,\pi } \right]$ \[ {\lim\limits_{N \to \infty } {f_N}\left( {{x_0}} \right) \text{ = }}\kern0pt {\begin{cases} f\left( {{x_0}} \right), \text{if}\,f\left( x \right)\,\;\text{is continuous on}\, \left[ { – \pi ,\pi } \right] \\ \frac{{f\left( {{x_0} – 0} \right) + f\left( {{x_0} + 0} \right)}}{2}, \;\text{if}\,f\left( x \right)\,\text{has a jump discontinuity at}\, {{x_0}} \end{cases}} \] where ${f\left( {{x_0} – 0} \right)}$ and ${f\left( {{x_0} + 0} \right)}$ represent the left limit and the right limit at the point ${x_0}.$ Uniform Convergence of Fourier Series A sequence of the partial sums $\left\{ {{f_N}\left( x \right)} \right\}$ is said to be uniformly convergent to the function $f\left( x \right),$ if the speed of convergence of the partial sums ${{f_N}\left( x \right)}$ does not depend on $x$ (Figure $3$). We say that the Fourier series of a function $f\left( x \right)$ converges uniformly to this function if \[{\lim\limits_{N \to \infty } \left[ {\max\limits_{x \in \left[ { – \pi ,\pi } \right]} \left\| {f\left( x \right) – {f_N}\left( x \right)} \right\|} \right] }={ 0.}\] Theorem. The Fourier series of a $2\pi$-periodic continuous and piecewise smooth function converges uniformly. Convergence of Fourier Series in ${L_2}$-Norm The space ${L_2}\left( { – \pi ,\pi } \right)$ is formed by those functions for which \[\int\limits_{ – \pi }^\pi {{{\left\| {f\left( x \right)} \right\|}^2}dx} < \infty .\] We will say that a function $f\left( x \right)$ is square-integrable if it belongs to the space ${L_2}.$ If a function $f\left( x \right)$ is square-integrable, then \[{\lim\limits_{N \to \infty } \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {{{\left\| {f\left( x \right) – {f_N}\left( x \right)} \right\|}^2}dx} } ={ 0,}\] that is the partial sums ${f_N}\left( x \right)$ converge to $f\left( x \right)$ in the norm ${L_2}.$ The uniform convergence implies both pointwise and ${L_2}$-convergence. But the opposite is not true: the ${L_2}$-convergence implies neither pointwise nor uniform convergence, and the pointwise convergence implies neither uniform nor ${L_2}$-convergence. Gibbs Phenomenon If there is a jump discontinuity, the partial sum of the Fourier series has oscillations near the jump, which might increase the maximum of the partial sum above the function itself. This phenomenon is called Gibbs phenomenon. The amplitude of the “overshoot” at any jump point of a piecewise smooth function is about $18\%$ larger (as $n \to \infty$) than the jump in the original function (Figure $4$). Solved Problems Click a problem to see the solution. Example 1Calculate the integral $\int\limits_{ – \pi }^\pi {{D_N}\left( z \right)dz}.$ Example 2Let the function $f\left( x \right) $ $= {\large\frac{{\pi – x}}{2}\normalsize}$ be defined on the interval $\left[ {0,2\pi } \right].$ Find the Fourier series expansion of the function on the given interval and calculate the approximate value of $\pi.$ Example 3Prove that the Fourier series of the function $f\left( x \right) = {x^2}$ converges uniformly to $f\left( x \right)$ on the interval $\left[ {-\pi, \pi} \right].$ Example 4Prove that the Fourier series of the function $f\left( x \right) = x$ converges to $f\left( x \right)$ in the norm ${L_2}$ on the interval $\left[ {-\pi, \pi} \right].$ Example 5The Fourier series of the function $f\left( x \right) =$ ${\large\frac{{\pi – x}}{2}\normalsize}$ defined on the interval $\left[ {0,2\pi } \right]$ is given by the formula $f\left( x \right) =$ ${\large\frac{{\pi – x}}{2}\normalsize} $ $= \sum\limits_{n = 1}^\infty {\large\frac{{\sin nx}}{n}\normalsize} $ (see Example $2$). Investigate behavior of the partial sums ${f_N}\left( x \right)$ of the Fourier series. Example 1.Calculate the integral $\int\limits_{ – \pi }^\pi {{D_N}\left( z \right)dz}.$ Solution. It is known that \[{{f_N}\left( x \right) \text{ = }}\kern0pt{ \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {{D_N}\left( {x – y} \right)f\left( y \right)dy} .}\] The Dirichlet kernel ${D_N}\left( x \right)$ is an even and $2\pi$-periodic function, so that we may write: \[{{f_N}\left( x \right) \text{ = }}\kern0pt{ \frac{1}{\pi }\int\limits_0^\pi {{D_N}\left( {x – y} \right)f\left( y \right)dy} .}\] Suppose that ${f_N}\left( x \right) = f\left( x \right) = 1$ and plug in this function into the formula above. We obtain \[1 = \frac{1}{\pi }\int\limits_0^\pi {{D_N}\left( {x – y} \right)dy} .\] Make the substitution: $z = x – y.$ Then $y = x – z,$ $dy = dz.$ Find the new limits of integration: when $y = 0,$ we get $z = x,$ and when $y = \pi,$ we have $z = x – \pi.$ As a result, we have \[ {1 = \frac{1}{\pi }\int\limits_x^{x – \pi } {{D_N}\left( z \right)\left( { – dz} \right)} \;\;\text{or}\;\;}\kern-0.3pt {1 = \frac{1}{\pi }\int\limits_{x – \pi}^x {{D_N}\left( z \right)dz} .} \] Due to periodicity of ${{D_N}\left( x \right)}$ we may write: \[1 = \frac{1}{\pi }\int\limits_{ – \pi }^0 {{D_N}\left( z \right)dz} .\] Hence, \[{\int\limits_{ – \pi }^\pi {{D_N}\left( z \right)dz} }={ 2\int\limits_{ – \pi }^0 {{D_N}\left( z \right)dz} }={ 2\pi .}\] There is another way to calculate this integral. Rewrite it in the form \[{I = \int\limits_{ – \pi }^\pi {{D_N}\left( z \right)dz} }={ 2\int\limits_0^\pi {{D_N}\left( z \right)dz} .}\] Since \[ {{D_N}\left( z \right) = \frac{{\sin \left( {N + \frac{1}{2}} \right)z}}{{\sin \frac{z}{2}}} } = {2\left( {\frac{1}{2} + \sum\limits_{n = 1}^N {\cos nz} } \right),} \] we can integrate this series term by term. Then \[ {I = 2\int\limits_0^\pi {{D_N}\left( z \right)dz} } = {4\int\limits_0^\pi {\left( {\frac{1}{2} + \sum\limits_{n = 1}^N {\cos nz} } \right)dz} } = {4\left[ {\left. {\left( {\frac{z}{2} + \sum\limits_{n = 1}^N {\frac{{\sin nz}}{n}} } \right)} \right\|_0^\pi } \right].} \] Here $\sin {nz} = 0$ at $z = 0, \pi.$ Consequently, \[I = 4 \cdot \frac{\pi }{2} = 2\pi .\]
Let $G$ be a real, connected, non-compact, semisimple Lie group with finite center and real rank $1$. Let $G=KAN$ be an Iwasawa decomposition, then $K\subset G$ is a maximal compact subgroup, and $\dim A=1$ (which is the definition of the real rank of $G$ being $1$). Let $M:=Z_K(A)$ be the centralizer of $A$ in $K$. Note that $M$ is a compact subgroup of $G$, being a closed subgroup of $K$. Let $\tau:M\to \mathrm{End}(V)$ be a finite-dimensional complex representation of $M$. Then I would like to know: Under which circumstances does there exist a finite-dimensional complex representation of $K$ that extends $\tau$? I searched for quite a long time now, and all literature about extension of representations of compact Lie groups only deals with normal subgroups. However, $M$ need not be normal in $K$ in general.
Answer The work done by the crate is approximately $1115\text{ foot-pounds}$. Work Step by Step The amount of work done by a force $\mathbf{F}$ on an object from a point A to the point B is denoted by $W=\mathbf{F}\cdot \overrightarrow{AB}$. That is $W=\left\\| \mathbf{F} \right\\|\overrightarrow{\left\\| AB \right\\|}\cos \theta $ ; Where $\theta $ is the angle between the force and the direction of the motion. In the given case, the distance between the crate and the level of floor is 50 feet. $\begin{align} & \left\\| \overrightarrow{AB} \right\\|=50 \\ & \theta =42{}^\circ \end{align}$ And, the magnitude of force $\left\\| \mathbf{F} \right\\|=30$. Therefore, the amount of work done will be calculated as below: $\begin{align} & W=\left\\| \mathbf{F} \right\\|\overrightarrow{\left\\| AB \right\\|}\cos \theta \\ & =\left( 30 \right)\left( 50 \right)\cos 42{}^\circ \\ & =1500\left( 0.743144825 \right) \\ & =1114.71 \end{align}$ This implies $W\approx 1115$. So, work done by the crate is approximately $1115\text{ foot-pounds}$. Hence, the amount of work done by the crate is approximately $1115\text{ foot-pounds}$.
257 0 Greetings, I stumbled across two question that I have no idea on how to answer them. 1) The interaction term in a scalar field theory is [tex]-\frac{\lambda}{4!} \phi^4[/tex] Why should lambda be positive? (they say look at the energy of the ground state...) 2) Write down the Feynman rules for phi^4 I have no clue as to how you get the two intersecting lines that give is -i lambda vertex. I see where the propagator comes from though. Any thougths? I stumbled across two question that I have no idea on how to answer them. 1) The interaction term in a scalar field theory is [tex]-\frac{\lambda}{4!} \phi^4[/tex] Why should lambda be positive? (they say look at the energy of the ground state...) 2) Write down the Feynman rules for phi^4 I have no clue as to how you get the two intersecting lines that give is -i lambda vertex. I see where the propagator comes from though. Any thougths?
I'm currently reading "SAVING RATES AND POVERTY: THE ROLE OF CONSPICUOUS CONSUMPTION AND HUMAN CAPITAL" by Omer Moav and Zvika Neeman, and I encounter some problem deriving one of the equations. Utility function, where $y$ is income, $\widetilde y$ is the belief of $y$ based on $x$ and human capital $h$, $x$ is the product of interest: $$u(y,x) = (y-x)^{1-\lambda} \widetilde y(h,x)^\lambda$$ Differentiate with respect to $x$: $$\frac{\lambda}{1-\lambda} \frac{y-x}{\widetilde y(h,x)} = \frac{1}{d \widetilde y (h,x) / d x}$$ This is the confusing part, how do I get this from above knowing that $y=\widetilde y(h,x)$? Given this $\underline y$, the lowest possible income an individual of $h$ can have ($\pi$ is an uncertain component of income that is not observed and has mean zero): $$\underline y (h) \equiv h + \underline \pi (h)$$ $$\widetilde y (h,x)^{1/(1-\lambda)} - \frac{x}{\lambda} \widetilde y (h,x)^{\lambda / (1-\lambda)} = \underline y (h)^{1/(1-\lambda)}$$ I think it's something to do with utility being indifferent for the marginal person who consume $x>0$ and $x=0$, but I don't know how to work it out to get the equation shown above. Thanks!!
Alpha decay: $(Z,A) \rightarrow (Z-2,A-4)+ ^4_2He $ According to the book: "Nuclear and particle physics" by Williams, $Q_\alpha$ is the measure of available energy to permit an alpha decay. It is defined as: $$ Q_\alpha = M(Z,A)-M(Z-2,A-4)-M(2,4)$$ (in natural units) where M is the nuclear mass, defined as: $M(Z,A) = Zm_p+Am_n-BE(Z,A) $ where BE is the binding energy. Since the number of protons and neutrons doesn't change (in this case), $Q_\alpha$ can be written as: $$ Q_\alpha = BE_{f}-BE_{i}=\Delta BE$$ If $Q_\alpha >0$ then the reaction is possible. I am having difficulties to interpret the $Q_\alpha$-value in alpha decays. I will explain my reasoning hoping that someone can point out my mistake. My reasoning: If $Q_\alpha = \Delta BE > 0$ then it means that the energy of the system increased, and therefore energy from somewhere would be needed to make the reaction possible. Then $Q_\alpha $ wouldn't be the energy available for the reaction to happen, but the extra energy needed for the reaction to happen.
Determinants Area of a Triangle The area of a triangle whose vertices are (x 1, y 1), (x 2, y 2) and (x 3, y 3), is given by \Delta=\frac{1}{2}\begin{vmatrix}x_{1} & y_{1} & 1 \\x_{2} & y_{2} & 1 \\x_{3} & y_{3} & 1 \end{vmatrix} Since area is a positive quantity, we always take the absolute value of the determinant. If area is given, use both positive and negative values of the determinant for calculation. The area of the triangle formed by three collinear points is zero. View the Topic in this video From 00:15 To 06:01 Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers. 1. The area of a triangle whose vertices are (x 1, y 1), (x 2, y 2) and (x 3, y 3), is given by \Delta=\frac{1}{2}\begin{vmatrix}x_{1} & y_{1} & 1 \\x_{2} & y_{2} & 1 \\x_{3} & y_{3} & 1 \end{vmatrix}
I'm currently a third year maths undergrad, writing a dissertation on the application of minimal surfaces in space. I have recently come across the Penrose Conjecture that the mass of a spacetime is: $$ m \geq \sqrt{\dfrac{A}{16 \pi}}.$$ What significance does this have in terms of the geometry of a black hole? Also how does this directly relate to that of minimal surfaces? In that, is it purely to do with the apparent horizons corresponding to a minimal surface?
Let’s recall Morera’s theorem. Suppose that $f$ is a continuous complex-valued function in a connected open set $\Omega \subset \mathbb C$ such thatMorera’s theorem \[\int_{\partial \Delta} f(z) \ dz = 0\] for every closed triangle $\Delta \subset \Omega \setminus \{p\}$ where $p \in \Omega$. Then $f$ is holomorphic in $\Omega$. Does the conclusion of Morera’s theorem still hold if $f$ is supposed to be continuous only in $\Omega \setminus \{p\}$? The answer is negative and we provide a counterexample. Let $\Omega$ be the entire complex plane, $f$ defined as follows \[f(z)=\begin{cases} \frac{1}{z^2} & \text{if } z \neq 0\\ 0 & \text{otherwise} \end{cases}\] and $p$ the origin. For $a,b \in \Omega \setminus \{0\}$ we have \[\begin{aligned} \int_{[a,b]} f(z) \ dz &= \int_{[a,b]} \frac{dz}{z^2}\\ &= \int_0^1 \frac{b-a}{[a+t(b-a)]^2} \ dt\\ &=\left[ -\frac{1}{a+t(b-a)} \right]_0^1 = \frac{1}{a} – \frac{1}{b} \end{aligned}\] Hence for a triangle $\Delta$ with vertices at $a,b ,c \in \Omega \setminus \{0\}$: \[\int_{\partial \Delta} f(z) \ dz = \left( \frac{1}{a} – \frac{1}{b} \right) + \left( \frac{1}{b} – \frac{1}{c} \right) + \left( \frac{1}{c} – \frac{1}{a} \right)=0\] However, $f$ is not holomorphic in $\Omega$ as it is even not continuous at $0$.
Background: In machine learning, we often work with graphical models to represent high dimensional probability density functions. If we discard the constraint that a density integrates (sums) to 1, we get an unnormalized graph-structured energy function. Suppose we have such an energy function, $E$, defined on a graph $G = (\mathcal{V}, \mathcal{E})$. There is one variable $x$ for each vertex of the graph, and there are real-valued unary and pairwise functions, $\theta_i(x_i) : i \in \mathcal{V}$ and $\theta_{ij}(x_i, x_j) : ij \in \mathcal{E}$, respectively. The full energy is then $$E(\mathbf{x}) = \sum_{i \in \mathcal{V}} \theta_i(x_i) + \sum_{ij \in \mathcal{E}} \theta_{ij}(x_i, x_j)$$ If all $x \in \mathbf{x}$ are binary, we can think of an $x$ as indicating set membership and with just a small abuse of terminology talk about submodularity. In this case, an energy function is submodular iff $\theta_{ij}(0, 0) + \theta_{ij}(1, 1) \le \theta_{ij}(0, 1) + \theta_{ij}(1, 0)$. We are typically interested in finding the configuration that minimizes the energy, $\mathbf{x}^* = \arg \min_{\mathbf{x}} E(\mathbf{x})$. There seems to be a connection between minimizing a submodular energy function and monotone boolean functions: if we lower the energy of some $\theta_i(x_i=1)$ for any $x_i$ (i.e., increase its preference to be "true"), then the optimal assignment of any variable $x_i^* \in \mathbf{x}^$ can only change from 0 to 1 ("false" to "true"). If all $\theta_i$ are restricted to be either 0 or 1, then we have $\|\mathcal{V}\|$ monotone boolean functions: $$f_i(\mathbf{\theta}) = x_i^$$ where as above, $\mathbf{x^*} = \arg \min_{\mathbf{x}} E(\mathbf{x})$. Question: Can we represent all monotone boolean functions using this setup by varying the pairwise terms, $\theta_{ij}$? What if we allow $E$ to be an arbitrary submodular energy function? Conversely, can we represent all submodular minimization problems as a set of $\|\mathcal{V}\|$ monotone boolean functions? Can you suggest references that will help me towards better understanding these connections? I'm not a theoretical computer scientist, but I'm trying to understand if there are insights about monotone boolean functions that are not captured by thinking in the submodular minimization terms.
Search Now showing items 1-10 of 27 Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider (American Physical Society, 2016-02) The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ... Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (Elsevier, 2016-02) Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < \|\eta\| < 4.0$) and associated particles in the central range ($\|\eta\| < 1.0$) are measured with the ALICE detector in ... Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV (Springer, 2016-08) The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ... 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 ... 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 ... $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Elsevier, 2016-03) The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ... Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV (Elsevier, 2016-09) The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ... Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV (Elsevier, 2016-12) We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ... Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV (Springer, 2016-05) Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
I found what I believe is a valid solution to the following assignment problem but the answer is bothering me. Let $$H_0(z) = 1+2z^{-1}+3z^{-2}+2z^{-3}+z^{-4}\quad\textrm{and}\quad H_1(z)=H_0(-z).$$ Find causal FIR filters $F_0(z)$ and $F_1(z)$ such that $\hat{x}(n)$ agrees with $x(n)$ except for a possible delay and (nonzero) scale factor. So I came up with the following causal FIR filters for $F_0(z)$ and $F_1(z)$ \begin{align} F_0(z) &= \frac{1}{4}\left(2-5z^{-1}+4z^{-2}-2z^{-3}\right) \\ F_1(z) &= \frac{1}{4}\left(2+5z^{-1}+4z^{-2}+2z^{-3}\right) \end{align} Lets see what happens when I plug them in... First let's determine the transfer function for the system \begin{equation} \hat{X}(z) = X(z)H_0(z)F_0(z) + X(z)H_1(z)F_1(z) \end{equation} then \begin{equation} \frac{\hat{X}(z)}{X(z)} = H_0(z)F_0(z) + H_1(z)F_1(z) \end{equation} Plugging in my answer for $F_0(z)$ and $F_1(z)$ I get \begin{equation} \frac{\hat{X}(z)}{X(z)} = 1 \end{equation} But does this make sense? Shouldn't there always be some delay when I filter? I always thought the best I would be able to do would be linear phase distortion when all of my filters are causal. $F_0$ and $F_1$ are causal right? They are a linear combination of delayed inputs so I don't see how they wouldn't be causal. Any help pointing out where I may have made a mistake would be appreciated. Or is it reasonable to have causal FIR filters composed such that the output has zero phase distortion? Edit: Thanks to Teague's answer I think I understand now. Also I verified using simulink that the output does indeed match the input. Here's the simulation. Edit 2: From Mr. Lyons' request I'm posting the method I used to solve the problem. Disclaimer: The problem gave a hint that I should consider polyphase representations but I couldn't figure out how to get a solution using a polyphase approach so I did it a different way. If anyone has an answer for how to solve this using polyphase I will post a new question so you can answer it. The question asks to find FIR filters $F_0(z)$ and $F_1(z)$ so the first thing I did was assume that FIR filters existed and I thought it was reasonable to think they were no higher order than $H_0(z)$ and $H_1(z)$. I let \begin{equation} F_k(z) = \sum_{m=0}^{4}{b_{km}z^{-m}}, \hspace{10pt} k=0,1 \end{equation} and solve for the 10 coefficients $b_{00}, b_{01}, ... b_{04}, b_{10}, b_{11}, ..., b_{14}$ Then plugging $F_0(z)$ and $F_1(z)$ into the transfer function \begin{align} \frac{\hat{X}(z)}{X(z)} & = H_0(z)F_0(z) + H_1(z)F_1(z) \\ & = (b_{00} + b_{10}) + (2b_{00} + b_{01} - 2b_{10} + b_{11})z^{-1} + ... + (b_{04} + b_{14})z^{-8} \end{align} I want to find the coefficients that result in only a delay and scale. I figured there was no reason mathematically that I couldn't solve for $\hat{X}(z)/X(z)=1$. So I set up the following matrix to zero all the delays except $z^{0}$. \begin{equation} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 & -2 & 1 & 0 & 0 & 0 \\ 3 & 2 & 1 & 0 & 0 & 3 & -2 & 1 & 0 & 0 \\ 2 & 3 & 2 & 1 & 0 & -2 & 3 & -2 & 1 & 0 \\ 1 & 2 & 3 & 2 & 1 & 1 & -2 & 3 & -2 & 1 \\ 0 & 1 & 2 & 3 & 2 & 0 & 1 & -2 & 3 & -2 \\ 0 & 0 & 1 & 2 & 3 & 0 & 0 & 1 & -2 & 3 \\ 0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} b_{00} \\ b_{01} \\ b_{02} \\ b_{03} \\ b_{04} \\ b_{10} \\ b_{11} \\ b_{12} \\ b_{13} \\ b_{14} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} \end{equation} The matrix is a 9x10 rank 9, meaning it is under constrained. Because of this I can arbitrarily add a 10th equation. Because I thought it would be nice if $F_0(z) = F_1(-z)$ I added a constraint that $b_{02} = b_{12} \rightarrow b_{02} - b_{12} = 0$. Adding this to the bottom of the matrix I get \begin{equation} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 & -2 & 1 & 0 & 0 & 0 \\ 3 & 2 & 1 & 0 & 0 & 3 & -2 & 1 & 0 & 0 \\ 2 & 3 & 2 & 1 & 0 & -2 & 3 & -2 & 1 & 0 \\ 1 & 2 & 3 & 2 & 1 & 1 & -2 & 3 & -2 & 1 \\ 0 & 1 & 2 & 3 & 2 & 0 & 1 & -2 & 3 & -2 \\ 0 & 0 & 1 & 2 & 3 & 0 & 0 & 1 & -2 & 3 \\ 0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} b_{00} \\ b_{01} \\ b_{02} \\ b_{03} \\ b_{04} \\ b_{10} \\ b_{11} \\ b_{12} \\ b_{13} \\ b_{14} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} \end{equation} Which makes the matrix a 10x10 rank 10. Solving the linear system yield the solutions for $F_0(z)$ and $F_1(z)$ given above. Interestingly the final constraint is pretty arbitrary as long is it isn't linearly dependent on the other 9 rows. I can make it pretty much anything. For example, if I want the coefficients to sum to 0 then I could make the last row of the matrix all $1$s which yields a different solution than the one I showed above, that is... \begin{align} F_0(z) & = \frac{1}{8} (7 - 16z^{-1} + 17z^{-2} - 10z^{-3} + 3z^{-4}) \\ F_1(z) & = \frac{1}{8} (1 + 4z^{-1} - 1z^{-2} - 2z^{-3} - 3z^{-4}) \end{align} Again if anyone knows the polyphase way to do this problem I am curious. Let me know in the comments and I'll post the question.
We have an equilibrium mixture for the reaction $$\ce{CO (g) + Cl2 (g) <=> COCl2 (g)}$$ We should recognize that $$K_\mathrm{eq} = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$ Note that the reactants and products all have stoichiometric coefficients of 1 so we can write our $K_\mathrm{eq}$ equation out in this way. Because we are given concentrations of reactants in the equilibrium mixture, we can go ahead and plug in the values that we know and solve for $[\ce{COCl2}]$. $$\begin{align}2.9 \times 10^{10} &= \frac{[\ce{COCl2}]}{(\pu{1.8 \times 10^-6 M})(\pu{6.3 \times 10^-7 M})} \\[\ce{COCl2}] &= \pu{0.033 M}\end{align}$$ We are asked how many grams of $[\ce{COCl2}]$ is present in the reaction mixture. Concentration and mass are related via the following equation: $$[\ce{COCl2}] = \frac{n(\ce{COCl2})}{V} = \frac{m(\ce{COCl2})}{M(\ce{COCl2}) \cdot V}$$ or $$m(\ce{COCl2}) = [\ce{COCl2}] \cdot M(\ce{COCl2}) \cdot V$$ where $n$ is the amount of $\ce{COCl2}$ (in g), $M$ is the molar mass (in g/mol), and $V$ is the volume (in L). Let us plug in all of our known values and solve. $$\begin{align}m(\ce{COCl2}) &= (\pu{0.033 mol L-1})(\pu{98.906 g mol-1})(\pu{5.69 L}) \\&= \pu{18.6 g}\end{align}$$
I know how to derive Navier-Stokes equations from Boltzmann equation in case where bulk and viscosity coefficients are set to zero. I need only multiply it on momentum and to integrate it over velocities. But when I've tried to derive NS equations with viscosity and bulk coefficients, I've failed. Most textbooks contains following words: "for taking into the account interchange of particles between fluid layers we need to modify momentum flux density tensor". So they state that NS equations with viscosity cannot be derived from Boltzmann equation, can they? The target equation is $$ \partial_{t}\left( \frac{\rho v^{2}}{2} + \rho \epsilon \right) = -\partial_{x_{i}}\left(\rho v_{i}\left(\frac{v^{2}}{2} + w\right) - \sigma_{ij}v_{j} - \kappa \partial_{x_{i}}T \right), $$ where $$ \sigma_{ij} = \eta \left( \partial_{x_{[i}}v_{j]} - \frac{2}{3}\delta_{ij}\partial_{x_{i}}v_{i}\right) + \varepsilon \delta_{ij}\partial_{x_{i}}v_{i}, $$ $w = \mu - Ts$ corresponds to heat function, $\epsilon$ refers to internal energy. Edit. It seems that I've got this equation. After multiplying Boltzmann equation on $\frac{m(\mathbf v - \mathbf u)^{2}}{2}$ and integrating it over $v$ I've got transport equation which contains objects $$ \Pi_{ij} = \rho\langle (v - u)_{i}(v - u)_{j} \rangle, \quad q_{i} = \rho \langle (\mathbf v - \mathbf u)^{2}(v - u)_{i}\rangle $$ To calculate it I need to know an expression for distribution function. For simplicity I've used tau approximation; in the end I've got expression $f = f_{0} + g$. An expressions for $\Pi_{ij}, q_{i}$ then are represented by $$ \Pi_{ij} = \delta_{ij}P - \mu \left(\partial_{[i}u_{j]} - \frac{2}{3}\delta_{ij}\partial_{i}u_{i}\right) - \epsilon \delta_{ij}\partial_{i}u_{i}, $$ $$ q_{i} = -\kappa \partial_{i} T, $$ so I've got the wanted result.
If we assign probabilities to every elementary event such that they sum up to 1 and they're all positive, can we say that this function is a probability function with its domain equal to the set of all subsets of the union of those elementary events? Or should we also specify that P(A or B) = P(A) + P(B) for all A, B where A and B=Null set? Edit: $\Omega = \{\omega_1,\omega_2,\omega_3\}$ Let $f:\Omega \rightarrow[0,1]$ such that: $f(\omega_1)=0.2$, $f(\omega_2)=0.3$, $f(\omega_3)=0.5$. Let $F:2^\Omega\rightarrow[0,1]$ such that: $F(\{\omega_i\})=f(\omega_i), i\in \{1,2,3\}$, $F(\{\omega_1,\omega_2\})=f(\omega_1)+f(\omega_2)$ $F(\{\omega_1,\omega_3\})=f(\omega_1)+f(\omega_3)$ $F(\{\omega_2,\omega_3\})=f(\omega_2)+f(\omega_3)$ $F(\{\omega_1,\omega_2\,\omega_3\})=f(\omega_1)+f(\omega_2)+f(\omega_3)$ $F(\emptyset)=0$ Then we can say that F is a probability function, right? Can we say that this is the unique probability function generated by f (not speaking formally)?
Perhaps it would be helpful to remind ourself of some of the basic properties of orthogonal bases. Let's for this purpose assume the $\{\|i\rangle\}$ represents an orthogonal basis of a vector space $V$. In addition to the orthogonality condition$$ \langle j\|i\rangle=\delta_{ij} , $$we also have the completeness condition$$ \sum_i \|i\rangle\langle i\| = {\cal I} , $$which can serve as a way to resolve the identity operator ${\cal I}$. The important thing to realize is that this is not the only orthogonal basis that one can define for this vector space. In fact any unitary operation would convert this basis into a new basis, $$ U\|i\rangle = \|m\rangle $$and it is a simple exercise to show the this new basis will also obey similar orthogonality and completeness conditions. Let's now consider this case of an operator that is diagonal in our initial basis. This means we can write this operator as$$ A = \sum_i \|i\rangle\lambda_i\langle i\| . $$What would happen if we convert this expression to the new basis $\{\|m\rangle\}$? To do this we operate on both sides of $A$ with the identity resolved in terms of the new basis (which we'll alternatively denote by either $\|m\rangle$ or $\|n\rangle$). See what happens$$\begin{align} {\cal I}A{\cal I} &= \sum_{mni} \|m\rangle\langle m\|i\rangle\lambda_i\langle i\|n\rangle\langle n\| \\ &= \sum_{mni} \|m\rangle U_{mi} \lambda_i {U^{\dagger}}_{in}\langle n\| \\ &= \sum_{mn} \|m\rangle B_{mn} \langle n\|\,.\end{align}$$It is hopefully clear to see that the matrix$$ B_{mn} = \sum_i U_{mi} \lambda_i {U^{\dagger}}_{in} $$would in general not be a diagonal matrix. In factor, the right-hand side $UDU^{\dagger}$ (where $D$ represent a diagonal matrix) is the spectral decomposition for some matrix. This also implies that if one were to perform this process in reverse, one would be starting with a non-diagonal matrix, and then converting it to a diagonal matrix by an appropriate choice of basis. Let's see how that works. Let's assume I'm given a normal matrix $M$, expressed in some arbitrary basis $\{\|a\rangle\}$. (I'm deliberately using different symbols here to avoid confusion with what we had before.) According to the spectral theorem, one can now express this as$$ M = U D U^{\dagger} , $$where $U$ is a unitary matrix and $D$ is a diagonal matrix. Note that $M$ is still defined in terms of the basis $\{\|a\rangle\}$ in which it is not diagonal. However we can remove the unitary matrices by operating on both sides as follows$$ U^{\dagger} M U = U^{\dagger} U D U^{\dagger} U = D . $$Thus we end up with only the diagonal matrix. In the process we have redefined the basis in which the matrix is expressed. This redefinition comes about through the unitary matrix: $\|a\rangle U = \|i\rangle $ and $ U^{\dagger} \langle a\| = \langle i\|$. Therefore, the unitary matrix that is needed to diagonalize the matrix, also convert the basis to the special one in which the matrix becomes diagonal. Let's look at the explicit questions: "What does it mean for an operator to be diagonal with respect to a basis?" It means that in this particular basis the operator (expressed as a matrix), one has non-zero elements on the diagonal only and these elements then represent the eigenvalues of the matrix. All other elements of the matrix are zero. The phrase "with respect to a basis" means that the rows (and columns) of the matrix are associated with particular element in that basis. "Do they mean that $M$ has a diagonal representation, as above, and, that using the specified basis, the matrix representation of $M$ is a diagonal matrix?" Yes indeed, provided that $M$ is a normal matrix, it always has a diagonal representation. (This is what the Spectral Theorem states.) "So, is the matrix representation of $A$ wrt the basis $\{\|0\rangle,...,\|n\rangle\}$ simply diag$\{\lambda_0,...,\lambda_n\}$?" Well, provided that this basis is the basis in which $A$ is diagonal, then yes, the diagonal matrix contains the eigenvalues on the diagonal.
Normal Nilpotent Matrix is Zero Matrix Problem 336 A complex square ($n\times n$) matrix $A$ is called normal if \[A^* A=A A^,\] where $A^$ denotes the conjugate transpose of $A$, that is $A^=\bar{A}^{\trans}$. A matrix $A$ is said to be nilpotent if there exists a positive integer $k$ such that $A^k$ is the zero matrix. (a) Prove that if $A$ is both normal and nilpotent, then $A$ is the zero matrix. You may use the fact that every normal matrix is diagonalizable. (b) Give a proof of (a) without referring to eigenvalues and diagonalization. (c) Let $A, B$ be $n\times n$ complex matrices. Prove that if $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$, where $I$ is the $n\times n$ identity matrix. Contents Proof. (a) If $A$ is normal and nilpotent, then $A=O$ Since $A$ is normal, it is diagonalizable. Thus there exists an invertible matrix $P$ such that $P^{-1}AP=D$, where $D$ is a diagonal matrix whose diagonal entries are eigenvalues of $A$. Since $A$ is nilpotent, all the eigenvalues of $A$ are $0$. (See the post “Nilpotent matrix and eigenvalues of the matrix” for the proof.) Hence the diagonal entries of $D$ are zero, and we have $D=O$, the zero matrix. It follows that we have \begin{align} A=PDP^{-1}=POP^{-1}=O. \end{align} Therefore, every normal nilpotent matrix must be a zero matrix. (b) Give a proof of (a) without referring to eigenvalues and diagonalization. Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$. We prove by induction on $k$ that $A=O$. The base case $k=1$ is trivial. Suppose $k>1$ and the case $k-1$ holds. Let $B=A^{k-1}$. Note that since $A$ is normal, the matrix $B$ is also normal. For any vector $x \in \C^n$, we compute the length of the vector $B^Bx$ as follows. \begin{align} &\\|B^Bx\\|=(B^Bx)^(B^Bx) && \text{by definition of the length}\\ &=x^B^(BB^)Bx\\ &=x^B^(B^B)Bx && \text{since $B$ is normal}\\ &=x^ (B^)^2B^2x, \end{align} and the last expression is $O$ since $B^2=A^{2k-2}=O$ as $k \geq 2$ implies $2k-2 \geq k$. Hence we have $B^Bx=\mathbf{0}$ for every $x\in \C^n$. This yields that \begin{align} \\|Bx\\|&=(Bx)^(Bx)\\ &=x^B^Bx=0, \end{align} for every $x\in \C^n$, and hence $B=O$. By the induction hypothesis, $A^{k-1}=O$ implies $A=O$, and the induction is completed. So the matrix $A$ must be the zero matrix. (c) If $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$ We claim that the matrix $B$ is normal as well. If this claim is proved, then part (a) yields that $B=O$ since $B$ is a nilpotent normal matrix, which implies the result $A=I$. To prove the claim, we compute \begin{align} B^ B&=(I-A)^* (I-A)\\ &=(I-A^)(I-A)\\ &=I-A-A^+A^A, \end{align} and \begin{align} B B^&=(I-A) (I-A)^\\ &=(I-A)(I-A^)\\ &=I-A^-A+AA^\\ &=I-A^-A+A^A \qquad \text{ since $A$ is normal.} \end{align} It follows that we have $B^ B=BB^*$, and thus $B$ is normal. Hence, the claim is proved. Add to solve later
On separation of gradient Young measures 43 Downloads Citations Abstract. Let $E\subset M^{N\times n}$ be a linear subspace of real $N\times n$ matrices without rank-one matrices and let $K = \{A_i\}_{i = 1}^m\subset E$ be a finite set. Suppose $\Omega \subset \mathbb R^n$ is a bounded arcwise connected Lipschitz domain and $u_j:\Omega \to \mathbb R^N$ is a sequence of bounded vector-valued mappings in $W^{1,1}(\Omega, \mathbb R^N)$ such that ${\rm dist}(Du_j,K_\epsilon)\to 0$ in $L^1(\Omega)$ as $j\to \infty$, where $K_\epsilon = \cup_{i = 1}^m\bar B_\epsilon(A_i)$ is the closed $\epsilon$-neighbourhood and ${\rm dist}(\cdot,K_\epsilon)$ the distance function to $K_\epsilon$. We give estimates for $\epsilon \gt 0$ such that up to a subsequence, ${\rm dist}(Du_j,B_\epsilon(A_{i_0})) \to 0$ in $L^1(\Omega)$ for some fixed $A_{i_0}\in K$. In other words, we give estimates on $\epsilon \gt 0$ such that $K_\epsilon$ separates gradient Young measure. The two point set $K = \{ A_1, A_2\}\subset M^{N\times n}$ with ${\rm rank}(A_2-A_1) \gt 1$ is a special case of such sets up to a translation. KeywordsDistance Function Linear Subspace Lipschitz Domain Real Matrice Young Measure Preview Unable to display preview. Download preview PDF.
$PdV$ is boundary work. $VdP$ is isentropic shaft work in pumps (as you have identified above), gas turbines, etc. Now you must realize that even in a pump or turbine the mechanism of work is still $Pdv$, i.e., the gas pushing on the blade out of its way. But, then there is work required to maintain the flow in and out of the device/control volume, which requires flow work $PV$ so the net reversible work from a steady-flow device turns out to be shaft $vdP$. Why flow work $PV$? To push a packet of fluid with volume $V$ forward into a device you have to do work against the pressure of the fluid already in the device, i.e., overcome the back force of that fluid. This implies the work you do in pushing your new packet of length $L$ and cross-section area $A$ into the device is:\begin{align}\int Fdx = \int_{0}^{L}PAdx = PV\end{align}It must be noted that in a steady-flow device (unlike in a piston) the back pressure $P$ is constant. Now consider the device (e.g., turbine to be a control volume). The energy of the fluid going in is its internal energy and the work invested into the fluid to enter the device:$U_{entry}+P_{entry}V_{entry}=H_{entry}$. Similarly for exit from the device. The net change across the device is $\Delta H$. For a differential device (or across a small change) this is $dH$. The work output from the shaft of then device is the $\delta W= dH$. Now if the device is isentropic, i.e., adiabatic-reversible. The Gibbs equation provides:\begin{align}&dH=TdS+VdP=VdP\\&\delta W =dH=VdP \qquad (\text{isentropic}\; dS=0)\end{align} Therefore $VdP$ is isentropic shaft work from a flowing device. Important points: 1) Both internal energy and enthalpy are state variables, therefore can be measured for a system static or flowing. This is why sometimes there is a tendency to use $U$ and $H$ incorrectly. The true purpose of $H$ is to capture the work required to push/maintain a flow against a back pressure, i.e., it incorporates the $PV$ part. Therefore when you write an energy balance with flows coming in and out, the energy crossing boundary is not just $U$ but $H$ and this distinction must be kept in mind. 2) $VdP$ is isentropic steady-flow shaft work. The isentropic is key here.
The Strauss Point Process Model Creates an instance of the Strauss point process model which can then be fitted to point pattern data. Keywords spatial Usage Strauss(r) Arguments r The interaction radius of the Strauss process Details The (stationary) Strauss process with interaction radius $r$ and parameters $\beta$ and $\gamma$ is the pairwise interaction point process in which each point contributes a factor $\beta$ to the probability density of the point pattern, and each pair of points closer than $r$ units apart contributes a factor $\gamma$ to the density. Thus the probability density is $$f(x_1,\ldots,x_n) = \alpha \beta^{n(x)} \gamma^{s(x)}$$ where $x_1,\ldots,x_n$ represent the points of the pattern, $n(x)$ is the number of points in the pattern, $s(x)$ is the number of distinct unordered pairs of points that are closer than $r$ units apart, and $\alpha$ is the normalising constant. The interaction parameter $\gamma$ must be less than or equal to $1$ so that this model describes an ``ordered'' or ``inhibitive'' pattern. The nonstationary Strauss process is similar except that the contribution of each individual point $x_i$ is a function $\beta(x_i)$ of location, rather than a constant beta. The function ppm(), which fits point process models to point pattern data, requires an argument of class "interact" describing the interpoint interaction structure of the model to be fitted. The appropriate description of the Strauss process pairwise interaction is yielded by the function Strauss(). See the examples below. Note the only argument is the interaction radius r. When r is fixed, the model becomes an exponential family. The canonical parameters $\log(\beta)$ and $\log(\gamma)$ are estimated by ppm(), not fixed in Strauss(). Value An object of class "interact"describing the interpoint interaction structure of the Strauss process with interaction radius $r$. See Also Aliases Strauss Examples Strauss(r=0.1) # prints a sensible description of itself data(cells) ppm(cells, ~1, Strauss(r=0.07), rbord=0.07) # fit the stationary Strauss process to `cells' ppm(cells, ~polynom(x,y,3), Strauss(r=0.07), rbord=0.1) # fit a nonstationary Strauss process with log-cubic polynomial trend Documentation reproduced from package spatstat, version 1.6-2, License: GPL version 2 or newer
Algebraic Geometry Seminar Fall 2016 The seminar meets on Fridays at 2:25 pm in Van Vleck B305. Here is the schedule for the previous semester. Contents Algebraic Geometry Mailing List Please join the AGS Mailing List to hear about upcoming seminars, lunches, and other algebraic geometry events in the department (it is possible you must be on a math department computer to use this link). Fall 2016 Schedule date speaker title host(s) September 16 Alexander Pavlov (Wisconsin) Betti Tables of MCM Modules over the Cones of Plane Cubics local September 23 PhilSang Yoo (Northwestern) Classical Field Theories for Quantum Geometric Langlands Dima October 7 Botong Wang (Wisconsin) Enumeration of points, lines, planes, etc. local October 14 Luke Oeding (Auburn) Border ranks of monomials Steven October 28 Adam Boocher (Utah) Bounds for Betti Numbers of Graded Algebras Daniel November 4 Lukas Katthaen Finding binomials in polynomial ideals Daniel November 11 Daniel Litt (Columbia) Arithmetic restrictions on geometric monodromy Jordan November 18 David Stapleton (Stony Brook) Hilbert schemes of points and their tautological bundles Daniel December 2 Rohini Ramadas (Michigan) Dynamics on the moduli space of pointed rational curves Daniel and Jordan December 9 Robert Walker (Michigan) Uniform Asymptotic Growth on Symbolic Powers of Ideals Daniel Abstracts Alexander Pavlov Betti Tables of MCM Modules over the Cones of Plane Cubics Graded Betti numbers are classical invariants of finitely generated modules over graded rings describing the shape of a minimal free resolution. We show that for maximal Cohen-Macaulay (MCM) modules over a homogeneous coordinate rings of smooth Calabi-Yau varieties X computation of Betti numbers can be reduced to computations of dimensions of certain Hom groups in the bounded derived category D(X). In the simplest case of a smooth elliptic curve embedded into projective plane as a cubic we use our formula to get explicit answers for Betti numbers. In this case we show that there are only four possible shapes of the Betti tables up to a shifts in internal degree, and two possible shapes up to a shift in internal degree and taking syzygies. PhilSang Yoo Classical Field Theories for Quantum Geometric Langlands One can study a class of classical field theories in a purely algebraic manner, thanks to the recent development of derived symplectic geometry. After reviewing the basics of derived symplectic geometry, I will discuss some interesting examples of classical field theories, including B-model, Chern-Simons theory, and Kapustin-Witten theory. Time permitting, I will make a proposal to understand quantum geometric Langlands and other related Langlands dualities in a unified way from the perspective of field theory. Botong Wang Enumeration of points, lines, planes, etc. It is a theorem of de Brujin and Erdős that n points in the plane determines at least n lines, unless all the points lie on a line. This is one of the earliest results in enumerative combinatorial geometry. We will present a higher dimensional generalization to this theorem. Let E be a generating subset of a d-dimensional vector space. Let [math]W_k[/math] be the number of k-dimensional subspaces that is generated by a subset of E. We show that [math]W_k\leq W_{d-k}[/math], when [math]k\leq d/2[/math]. This confirms a "top-heavy" conjecture of Dowling and Wilson in 1974 for all matroids realizable over some field. The main ingredients of the proof are the hard Lefschetz theorem and the decomposition theorem. I will also talk about a proof of Welsh and Mason's log-concave conjecture on the number of k-element independent sets. These are joint works with June Huh. Luke Oeding Border ranks of monomials What is the minimal number of terms needed to write a monomial as a sum of powers? What if you allow limits? Here are some minimal examples: [math]4xy = (x+y)^2 - (x-y)^2[/math] [math]24xyz = (x+y+z)^3 + (x-y-z)^3 + (-x-y+z)^3 + (-x+y-z)^3[/math] [math]192xyzw = (x+y+z+w)^4 - (-x+y+z+w)^4 - (x-y+z+w)^4 - (x+y-z+w)^4 - (x+y+z-w)^4 + (-x-y+z+w)^4 + (-x+y-z+w)^4 + (-x+y+z-w)^4[/math] The monomial [math]x^2y[/math] has a minimal expression as a sum of 3 cubes: [math]6x^2y = (x+y)^3 + (-x+y)^3 -2y^3[/math] But you can use only 2 cubes if you allow a limit: [math]6x^2y = \lim_{\epsilon \to 0} \frac{(x^3 - (x-\epsilon y)^3)}{\epsilon}[/math] Can you do something similar with xyzw? Previously it wasn't known whether the minimal number of powers in a limiting expression for xyzw was 7 or 8. I will answer this and the analogous question for all monomials. The polynomial Waring problem is to write a polynomial as linear combination of powers of linear forms in the minimal possible way. The minimal number of summands is called the rank of the polynomial. The solution in the case of monomials was given in 2012 by Carlini--Catalisano--Geramita, and independently shortly thereafter by Buczynska--Buczynski--Teitler. In this talk I will address the problem of finding the border rank of each monomial. Upper bounds on border rank were known since Landsberg-Teitler, 2010 and earlier. We use symmetry-enhanced linear algebra to provide polynomial certificates of lower bounds (which agree with the upper bounds). This work builds on the idea of Young flattenings, which were introduced by Landsberg and Ottaviani, and give determinantal equations for secant varieties and provide lower bounds for border ranks of tensors. We find special monomial-optimal Young flattenings that provide the best possible lower bound for all monomials up to degree 6. For degree 7 and higher these flattenings no longer suffice for all monomials. To overcome this problem, we introduce partial Young flattenings and use them to give a lower bound on the border rank of monomials which agrees with Landsberg and Teitler's upper bound. I will also show how to implement Young flattenings and partial Young flattenings in Macaulay2 using Steven Sam's PieriMaps package. Adam Boocher Let R be a standard graded algebra over a field. The set of graded Betti numbers of R provide some measure of the complexity of the defining equations for R and their syzygies. Recent breakthroughs (e.g. Boij-Soederberg theory, structure of asymptotic syzygies, Stillman's Conjecture) have provided new insights about these numbers and we have made good progress toward understanding many homological properties of R. However, many basic questions remain. In this talk I'll talk about some conjectured upper and lower bounds for the total Betti numbers for different classes of rings. Surprisingly, little is known in even the simplest cases. Lukas Katthaen (Frankfurt) In this talk, I will present an algorithm which, for a given ideal J in the polynomial ring, decides whether J contains a binomial, i.e., a polynomial having only two terms. For this, we use ideas from tropical geometry to reduce the problem to the Artinian case, and then use an algorithm from number theory. This is joint work with Anders Jensen and Thomas Kahle. David Stapleton Fogarty showed in the 1970s that the Hilbert scheme of n points on a smooth surface is smooth. Interest in these Hilbert schemes has grown since it has been shown they arise in hyperkahler geometry, geometric representation theory, and algebraic combinatorics. In this talk we will explore the geometry of certain tautological bundles on the Hilbert scheme of points. In particular we will show that these tautological bundles are (almost always) stable vector bundles. We will also show that each sufficiently positive vector bundles on a curve C is the pull back of a tautological bundle from an embedding of C into the Hilbert scheme of the projective plane. Rohini Ramadas The moduli space M_{0,n} parametrizes all ways of labeling n distinct points on P^1, up to projective equivalence. Let H be a Hurwitz space parametrizing holomorphic maps, with prescribed branching, from one n-marked P^1 to another. H admits two different maps to M_{0,n}: a ``target curve'’ map pi_t and a ``source curve map pi_s. Since pi_t is a covering map,pi_s(pi_t^(-1)) is a multi-valued map — a Hurwitz correspondence — from M_{0,n} to itself. Hurwitz correspondences arise in topology and Teichmuller theory through Thurston's topological characterization of rational functions on P^1. I will discuss their dynamics via numerical invariants called dynamical degrees. Robert Walker Symbolic powers ($I^{(N)}$) in Noetherian commutative rings are mysterious objects from the perspective of an algebraist, while regular powers of ideals ($I^s$) are essentially intuitive. However, many geometers tend to like symbolic powers in the case of a radical ideal in an affine polynomial ring over an algebraically closed field in characteristic zero: the N-th symbolic power consists of polynomial functions "vanishing to order at least N" on the affine zero locus of that ideal. In this polynomial setting, and much more generally, a challenging problem is determining when, given a family of ideals (e.g., all prime ideals), you have a containment of type $I^{(N)} \subseteq I^s$ for all ideals in the family simultaneously. Following breakthrough results of Ein-Lazarsfeld-Smith (2001) and Hochster-Huneke (2002) for, e.g., coordinate rings of smooth affine varieties, there is a slowly growing body of "uniform linear equivalence" criteria for when, given a suitable family of ideals, these $I^{(N)} \subseteq I^s$ containments hold as long as N is bounded below by a linear function in s, whose slope is a positive integer that only depends on the structure of the variety or the ring you fancy. My thesis (arxiv.org/1510.02993, arxiv.org/1608.02320) contributes new entries to this body of criteria, using Weil divisor theory and toric algebraic geometry. After giving a "Symbolic powers for Geometers" survey, I'll shift to stating key results of my dissertation in a user-ready form, and give a "comical" example or two of how to use them. At the risk of sounding like Paul Rudd from "Ant-Man," I hope this talk will be awesome.
In fact any flat coherent sheaf is locally free; see Tag 00NX implication $(1) \Rightarrow (6)$ for the corresponding algebra statement. Thus, if $\mathscr F$ is flat over $T$, then $\mathscr F$ is locally free over $T$. But because $X_T \to T$ is an isomorphism, we conclude that $\mathscr F$ is locally free on $X_T$ (rather than on $T$). This is the step for which we crucially need that $X = S$. By assumption, the Hilbert polynomial of each fibre $\mathscr F\|_{X_t} = \mathscr F \otimes_{\mathcal O_T} \kappa(t)$ is $\ell$ . Hence, $\mathscr F \otimes_{\mathcal O_T} \kappa(t)$ is an $\ell$-dimensional vector space over $\kappa(t)$. It is then easy to see that the rank of $\mathscr F$ at $t$ is $\ell$. (Note that the fibres of $X_T \stackrel \sim\to T$ are just points, so for any coherent sheaf $\mathscr F$ on $T$, the Hilbert polynomial of the fibre $\mathscr F\|_{X_t} = \mathscr F \otimes_{\mathcal O_T} \kappa(t)$ has degree $\leq 0$, i.e. is constant. This corresponds to the fact that coherent sheaves on the spectrum of a field are just finite-dimensional vector spaces, so the only numerical invariant is their dimension.) (Note that the rank is a priori only locally constant. But you assume that the Hilbert polynomial is constant, which forces the rank to be constant.) Remark. Of course the situation is different once we start considering Quot schemes $\mathfrak Quot_{X/S}$ where $X$ is no longer equal to $S$: then $\mathscr F$ is a sheaf on $X_T$ which is only flat over $T$, so the above method does not give us much.
Given vectors $a_1, \dots, a_n\in \mathbb R^d$ where $n$ is even, I want to find partitions $I$ and $J$ of $[n]$ with $\|I\|=\|J\|=\frac n2$ to minimize $$\left\\| \sum_{i\in I} a_i - \sum_{j\in J} a_j \right\\|.$$ This problem can be written as a binary optimization problem. Given matrix $A = [a_1 \dots a_n]$, I want to minimize $\\|Ax\\|$ over $x\in\{-1,1\}^n$ and $\sum_{i=1}^n x_i=0$. Finding exact global minimum looks NP-hard (in $d$ or $n$). Is it possible to find a nice approximate solution (like $(1+\epsilon)$-approximation for $K$-means)? Convex relaxation does not seems to work because the convex hull of the feasible region contains a trivial global minimizer $x=0$. Any help will greatly appreciated.
Let $C$ be a cocomplete category. Suppose that $X : A \to B$ is a functor, where $A$ is small. Then every functor $F : A \to C$ admits a left Kan extension $\mathrm{Lan}_X(F) : B \to C$, defined by mapping $b \in B$ to $\mathrm{colim}_{X(a) \to b} F(a)$. The action on morphisms is as follows: If $\iota_{\sigma} : F(a) \to \mathrm{Lan}_X(F)(b)$ denotes the colimit inclusion induced by $\sigma : X(a) \to b$, then for a morphism $f : b \to b'$ we have $\mathrm{Lan}_X(F)(f) \circ \iota_{\sigma} = \iota_{f \sigma}$. This is well-known and easy to check. Now I would like to work with $k$-linear (i.e. $\mathsf{Mod(k)}$-enriched) categories instead. So assume that $A,B,C$ are $k$-linear categories and likewise $X,F$ are $k$-linear functors. Then it doesn't seem to be the case that $\mathrm{Lan}_X(F)$, as defined above, is a $k$-linear functor again, right? So how can we explicitly construct a $k$-linear functor $\mathrm{Lan}_X(F)$ which serves as a linear left Kan extension (which I would define by a natural isomorphism of $k$-modules $\hom(\mathrm{Lan}_X(F),T) \cong \hom(F,T \circ X)$, where $T : B \to C$ is a $k$-linear functor)? I'm aware that Kelly treats enriched Kan extensions in his book on enriched categories. But it doesn't seem to help me. I would like to know a down-to-earth construction without delving into general enriched category theory. After all, linear categories should be a very basic example.
Suppose $k$ is a number field and $\sigma:k \hookrightarrow \mathbb{C}$ is an embedding. If $\sim$ is an adequate equivalence relation, we can construct the category of pure motives with rational coefficients, $\textbf{M}_{\sim}(k,\mathbb{Q})$. Let the Hodge realisation functor be $R$, \begin{equation} R:\textbf{M}_{\sim}(k,\mathbb{Q}) \rightarrow \text{MHS}_{\mathbb{Q}} \end{equation} Question 1: Suppose $M$ is a pure motive of $\textbf{M}_{\sim}(k,\mathbb{Q})$ such that its Hodge realisation is the pure Tate object $\mathbb{Q}(0)$ of $\text{MHS}_{\mathbb{Q}}$, what could $M$ be? For example, suppose $\chi: (\mathbb{Z}/N\mathbb{Z})^\times \rightarrow \mathbb{Q}$ is a real Dirichlet character, then the Artin motive $h^0(\text{Spec}\,\mathbb{Q}(\zeta_N))$ has a direct summand $\chi$ whose etale realisation is the Galois representation associated with the Dirichlet character $\chi$. The Hodge realisation of $\chi$ is $\mathbb{Q}(0)$, and when $\chi$ is the trivial character, the associated motive is just the Tate motive $\mathbb{Q}(0)$. But are there any other possibilities? If none, how to prove it? Question 2: Suppose $M_1$ and $M_2$ are two pure motives of $\textbf{M}_{\sim}(k,\mathbb{Q})$ that have isomorphic Hodge realisations, is it true that $M_1$ is isomorphic to $M_2 \otimes \epsilon$, where $\epsilon$ is a pure motive whose Hodge realisation is $\mathbb{Q}(0)$? Remark: If $k$ is algebraically closed and $\sim$ is numerical equivalence relation, then these two questions are immediate consequence of Hodge conjecture. I am also wondering whether they could be answered without citing any conjecture!
When we talked about thin film interference, we said that when light encounters a smooth interface between two transparent media, some of the light gets through, and some bounces off. There we limited the discussion to the case of normal incidence. (Recall that normal means perpendicular to and normal incidence is the case where the direction in which the light is traveling is perpendicular to the interface.) Now we consider the case in which light shining on the smooth interface between two transparent media, is not normally incident upon the interface. Here’s a “clean” depiction of what I’m talking about: and here’s one that’s all cluttered up with labels providing terminology that you need to know: As in the case of normal incidence, some of the light is reflected and some of it is transmitted through the interface. Here we focus our attention on the light that gets through. Experimentally we find that the light that gets through travels along a different straight line path than the one along which the incoming ray travels. As such, the transmitted ray makes an angle $\theta_2$ with the normal that is different from the angle $\theta_1$ that the incident ray makes with the normal. The adoption of a new path by the transmitted ray, at the interface between two transparent media is referred to as refraction. The transmitted ray is typically referred to as the refracted ray, and the angle $\theta_2$ that the refracted ray makes with the normal is called the angle of refraction. Experimentally, we find that the angle of refraction $\theta_2$ is related to the angle of incidence $\theta_1$ by Snell’s Law: \[n_1 \sin\theta_1=n_2 \sin\theta_2 \label{27-1}\] where: $n_1$ is the index of refraction of the first medium, the medium in which the light is traveling before it gets to the interface, $\theta_1$ is the angle that the incident ray (the ray in the first medium) makes with the normal, $n_2$ is the index of refraction of the second medium, the medium in which the light is traveling after it goes through the interface, and, $\theta_2$ is the angle that the refracted ray (the ray in the second medium) makes with the normal. Dispersion On each side of the equation form of Snell’s law we have an index of refraction. The index of refraction has the same meaning as it did when we talked about it in the context of thin film interference. It applies to a given medium. It is the ratio of the speed of light in that medium to the speed of light in vacuum. At that time, I mentioned that different materials have different indices of refraction, and in fact, provided you with the following table: Medium Index of Refraction Vacuum 1 Air 1.00 Water 1.33 Glass(Depends on the kind of glass. Here is one typical value.) 1.5 What I didn’t mention back then is that there is a slight dependence of the index of refraction on the wavelength of the visible light, such that, the shorter the wavelength of the light, the greater the index of refraction. For instance, a particular kind of glass might have an index of refraction of 1.49 for light of wavelength 695 nm (red light), but an index of refraction that is greater than that for shorter wavelengths, including an index of refraction of 1.51 for light of wavelength 405 nm (blue light). The effect in the case of a ray of white light traveling in air and encountering an interface between air and glass is to cause the different wavelengths of the light making up the white light to refract at different angles. This phenomena of white light being separated into its constituent wavelengths because of the dependence of the index of refraction on wavelength, is called dispersion. Total Internal Reflection In the case where the index of refraction of the first medium is greater than the index of refraction of the second medium, the angle of refraction is greater than the angle of incidence. For such a case, look what happens when we increase the angle of incidence, $\theta_1$: The angle of refraction gets bigger… until eventually it (the angle of refraction) gets to be $90^{\circ}$. It can’t get any bigger than that, because, beyond that, the light is not going through the interface. An angle of refraction greater than $90^{\circ}$ has no meaning. But, note that we still have room to increase the angle of incidence. What happens if we continue to increase the angle of incidence? Well, indeed, no light gets through the interface. But, remember at the beginning of this chapter where we talked about how, when light is incident on the interface between two transparent media, some of the light gets through and some of it is reflected? Well, I haven’t been including the reflected ray on our diagrams because we have been focusing our attention on the transmitted ray, but, it is always there. The thing is, at angles of incidence bigger then the angle that makes the angle of refraction $90^{\circ}$, the reflected ray is all there is. The phenomenon, in which there is no transmitted light at all, just reflected light, is known as total internal reflection. The angle of incidence that makes the angle of refraction $90^{\circ}$ is known as the critical angle. At any angle of incidence greater than the critical angle, the light experiences total internal reflection. Note that the phenomenon of total internal reflection only occurs when the light is initially in the medium with the bigger index of refraction. Let’s investigate this phenomenon mathematically. Starting with Snell’s Law: \[n_1 \sin\theta_1=n_2 \sin \theta_2\] solved for the sine of the angle of refraction: \[\sin \theta_2 =\frac{n_1}{n_2} \sin\theta_1\] we note that, since it was stipulated that $n_1>n_2$, the ratio $n_1/n_2$ is greater than 1. The $\sin \theta_1$ is always less than $1$, but, if $\theta_1$ is big enough, $\sin\theta_1$ can be so close to $1$ that the right-hand side of the equation $\sin \theta_2=\frac{n_1}{n_2} \sin\theta_1$ is greater than $1$. In that case, there is no $\theta_2$ that will solve the equation because there is no angle whose sine is greater than 1. This is consistent with the experimental fact that, at angles of incidence greater than the critical angle, no light gets through the interface. Let’s solve for the critical angle. At the critical angle, the angle of refraction $\theta_2$ is $90^{\circ}$. Let’s plug that into the equation we have been working with and solve for $\theta_1$: \[\sin \theta_2=\frac{n_1}{n_2} \sin\theta_1\] evaluated at $\theta_2 = 90^{\circ}$ yields: \[\sin 90^{\circ}=\frac{n_1}{n_2} \sin\theta_1\] \[1=\frac{n_1}{n_2} \sin\theta_1\] \[\sin \theta_1=\frac{n_2}{n_1}\] \[\theta_1= \sin^{-1} \frac{n_2}{n_1}\] This is such a special angle of incidence that we not only give it a name (as mentioned, it is called the critical angle), but, we give it its own symbol. The critical angle, that angle of incidence beyond which there is no transmitted light, is designated $\theta_C$ , and, as we just found, can be expressed as: \[\theta_C=\sin ^{-1} \frac{n_2}{n_1} \label{27-2}\]
The electron-volt is a unit of energy or work. An electron-volt (eV) is the work required to move an electron through a potential difference of one volt. Alternatively, an electronvolt is equal to the kinetic energy acquired by an electron when it is accelerated through a potential difference of one volt. Since the magnitude of the charge of an electron is about $1.602 × 10^{−19}$ C, it follows that an electron-volt is about $1.602 × 10^{−19}$ J. Note also that, because the charge on an electron is negative, it requires work to move an electron from a point of high potential to a point of low potential. Exercise. If an electron is accelerated through a potential difference of a million volts, its kinetic energy is, of course, 1 MeV. At what speed is it then moving? First attempt. \[\frac{1}{2}mv^2=eV\] (Here $eV$, written in italics, is not intended to mean the unit electron-volt, but e is the magnitude of the electron charge, and $V$ is the potential difference ($10^6$ volts) through which it is accelerated.) Thus $v = \sqrt{2eV / m}$ . With $m = 9.109 × 10^{−31}$ kg, this comes to $v = 5.9 × 10^8 \text{m s}^{ −1}$ . Oops! That looks awfully fast! We’d better do it properly this time. Second attempt. \[(\gamma -1)mc^2=eV.\] Some readers will know exactly what we are doing here, without explanation. Others may be completely mystified. For the latter, the difficulty is that the speed that we had calculated was even greater than the speed of light. To do this properly we have to use the formulas of special relativity. See, for example, Chapter 15 of the Classical Mechanics section of these notes. At any rate, this results in $\gamma = 2.958$, whence $β = 0.9411 \text{ and }v = 2.82 × 10^8 \text{m s}^{ −1}$ .
Let $a>0$. Show that $$\int_0^\pi {dx\over a+\sin^2(x)}={\pi\over \sqrt{a(a+1)}}.$$ I'm not sure how to show this. Any help is greatly appreciated. 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 Let $a>0$. Show that $$\int_0^\pi {dx\over a+\sin^2(x)}={\pi\over \sqrt{a(a+1)}}.$$ I'm not sure how to show this. Any help is greatly appreciated. This question appears to be off-topic. The users who voted to close gave this specific reason: METHODOLOGY 1: Complex Analysis and Contour Integration We can simplify the problem by using the double-angle formula for the cosine to write $$\begin{align} \int_0^\pi \frac{1}{a+\sin^2(x)}\,dx&=\int_0^\pi \frac{2}{(2a+1)-\cos(2x)}\,dx \\\\ &=\int_0^{2\pi}\frac{1}{(2a+1)-\cos(x)}\,dx \tag 1\\\\ \end{align}$$ Next, we move to the complex plane by letting $z=e^{ix}$. Then, we have $$\begin{align} \int_0^{2\pi}\frac{1}{(2a+1)-\cos(x)}\,dx&=\oint_{\|z\|=1}\frac{1}{(2a+1)+\frac12(z+z^{-1})}\,\frac{1}{iz}\,dz\\\\ &=\frac{2}i\oint_{\|z\|=1} \frac{1}{z^2+2(2a+1)z+1}\,dz\\\\ &=\frac{2}{i}\,(2\pi i)\text{Res}\left(\frac{1}{z^2+2(2a+1)z+1},z=-(2a+1)+2\sqrt{a(a+1)}\right)\\\\ & =\frac{4\pi}{4\sqrt{a(a+1)}}\\\\ &=\frac{\pi}{\sqrt{a(a+1)}} \end{align}$$ METHODOLOGY 2: Real Analysis and the Weierstrass Substitution As an alternative to using contour integration, write the integral in $(1)$ as $$\int_0^{2\pi}\frac{1}{(2a+1)-\cos(x)}\,dx=2\int_0^\pi \frac{1}{(2a+1)-\cos(x)}\,dx$$ Then, use the classical Tangent Half-Angle Substitution to write $$\begin{align} 2\int_0^\pi \frac{1}{(2a+1)-\cos(x)}\,dx&=2\int_0^\infty \frac{1}{(2a+1)-\frac{1-t^2}{1+t^2}}\,\frac{2}{1+t^2}\,dt\\\\ &=\frac{2}{a+1}\int_0^\infty \frac{1}{t^2+\frac{a}{a+1}}\,dt\\\\ &=\frac{\pi}{\sqrt{a(a+1)}} \end{align}$$ as expected! Hint: use symmetry to make it an integral from $0$ to $2\pi$ (or $-\pi$ to $\pi$ if you prefer), and transform to an integral around the unit circle with $z = e^{ix}$. Use residues. My hint would be to divide the numerator and denominator by $\cos^2 x$. Then in the numerator you have $\tan'x$ and in the denominator you have $a(\tan^2 x + 1) + \tan^2 x$. Use the substitution $y=\tan x$. (To avoid infinities in the middle, switch the limits of integration to $[-\pi/2,\pi/2]$ or divide instead by $\sin^2 x$ to deal with cotangents. Then you'll end up with $\int^\infty_{-\infty} \frac{dx}{a+(1+a)x^2}$ which you'll be able to evaluate. I wanted to just show you this potentially interesting property about your integral: Let$$I(a):=\int_0^\pi \frac{\text{d}x}{a+\sin^2 x}$$ Then $$(-1)^{n+1}I^{(n)}(a)=\int_0^\pi \frac{\text{d}x}{(a+\sin^2 x)^{n+1}}$$ This implies that $$I(a)-I'(a)+I''(a)-I'''(a)\cdots=\int_0^\pi \frac{\text{d}x}{a+\sin^2 x+1}=I(a+1)$$ This is a bit too long for comment, but might help with some solution-finding =) Rewritten as $$\int_0^{\pi}\frac{dx}{a\cos^2x+(a+1)\sin^2x}$$ this is an integral that gets posted about once a week. There are $6$ methods I can think of to attack it, and they all work with the result $\frac{\pi}{\sqrt{a(a+1)}}$. The first time I saw this was in a differential geometry textbook. EDIT: The above might seem a bit dismissive, but that is far from the truth; in fact I love this integral $\heartsuit$. So much, in fact, that I will provide a seventh method! Cue Gary Larson cartoon with free-body diagram of baseball approaching window$\dots$ Now, consider an object subject to a $2$-dimensional harmonic oscillator potential, where the force is proportional to the distance from the force center: $\vec F=-k\vec r$. Now, wait a minute, you are saying: shouldn't such a force in $2$-d be different from that in $1$-d or $3$-d? For example the Coulomb force is spatially invariant in $1$-d (electric field near a large flat sheet of unifort charge or flat-earth gravitation) proportional to $1/r$ in $2$-d (electric field near a long uniformly charged wire) or $1/r^2$ in $3$-d (electric field of a small charged object or round-earth gravitation.) But the reason for the harmonic oscillator potential is that if an object is at its equilibrium position due to the conservative forces acting on it, then the zero-order term of the Taylor series for its potential doesn't contribute to forces ($\vec F=-\vec\nabla V$) and the first order terms are all zero (that's the condition for equilibrium) so the first terms that can act for small displacements from equilibrium are the second-order terms, so we get quadratic potential, linear force in any number of dimensions. Examples are objects near the $L_4$ or $L_5$ points in orbit, or the old experiment with NdFeB magnets and pyrolytic graphite. OK, given that such a restoring force is plausible and assuming $2$-d motion in an isotropic field, we have the potential $V=\frac12kr^2$. The kinetic energy $T=\frac12mv^2=\frac12m(\dot x^2+\dot y^2)$, where $k$ is the force or 'spring' constant, $m$ is the object's mass, $x$ and $y$ its spatial coordinates and $\dot x$ and $\dot y$ their time derivatives. So now we have the Lagrangian of the system $$\mathscr L=T-V=\frac12mv^2-\frac12kr^2=\frac12m(\dot x^2+\dot y^2)-\frac12k(x^2+y^2)$$ And we are ready to derive Lagrange's equations of motion: $$p_x=\frac{\partial\mathscr L}{\partial\dot x}=m\dot x$$ $$\dot p_x=m\ddot x=\frac{\partial\mathscr L}{\partial x}=-kx$$ $$p_y=\frac{\partial\mathscr L}{\partial\dot y}=m\dot y$$ $$\dot p_y=m\ddot y=\frac{\partial\mathscr L}{\partial y}=-ky$$ So there it is, $\vec F=-k\vec r$. For an isotropic harmonic osciallator, the $x$ and $y$ motions are decoupled like this and we know how to solve this system: $$x=c_1\cos\omega t+c_2\sin\omega t$$ $$y=c_3\cos\omega t+c_4\sin\omega t$$ Where $t$ is the time and $\omega=\sqrt{\frac km}$ is the angular frequency of oscillation. This kind of motion would produce Lissajous figures of ellipses on your oscilloscope, or a straight line in exceptional cases. But let's choose an ellipse of semimajor axis $a$ oriented in the $x$ direction and semiminor axis $b$ in the $y$ direction: $x=a\cos\omega t$, $y=b\sin\omega t$ for our model case. Now we are going to attempt solution of the system in polar coordinates where $x=r\cos\phi$ and $y=r\sin\phi$. Then $\dot x=\dot r\cos\phi-r\dot\phi\sin\phi$, $\dot y=\dot r\sin\phi+r\dot\phi\cos\phi$, so $v^2=\dot x^2+\dot y^2=\dot r^2+r^2\dot\phi^2$ and now we have the Lagrangian $$\mathscr L=\frac12mv^2-\frac12kr^2=\frac12m\left(\dot r^2+r^2\dot\phi^2\right)-\frac12kr^2$$ We can find the momentum conjugate to $\phi$ $$p_{\phi}=\frac{\partial\mathscr L}{\partial\dot\phi}=mr^2\dot\phi$$ and its time derivative, $$\dot p_{\phi}=\frac{\partial\mathscr L}{\partial\phi}=0$$ So we find that angular momentum is constant as it is for all central forces. This is Kepler's law of areas. $$mr^2\dot\phi=\ell=\text{constant}$$ Now we are ready to do the integral! Going back to our model solution, $$r^2=x^2+y^2=a^2\cos^2\omega t+b^2\sin^2\omega t$$ At $t=0$, $r=a$ and the tangential velocity $r\dot\phi=v_y=b\omega\cos\omega t=b\omega$. So $\ell=mr^2\dot\phi=mab\omega$ and so $$mr^2\frac{d\phi}{dt}=mab\omega$$ Rewrite as $$\begin{align}\frac{d\phi}{ab}&=\frac{\omega dt}{r^2}=\frac{\omega dt}{a^2\cos^2\omega t+b^2\sin^2\omega t}\\ &=\frac{d\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\end{align}$$ Where we have substituted $\theta=\omega t$. Now we integrate over a full cycle of the motion: $$\int_0^{2\pi}\frac{d\phi}{ab}=\frac{2\pi}{ab}=\int_0^{2\pi}\frac{d\theta}{a^2\cos^2\theta+b^2\sin^2\theta}$$ The seventh method!
I don't think it is difficult to derive analytically the shape of the Earth. Simply look for the shape of the surfaces of equal potential. The geometrical symmetry reduces the calculation to a 2-dimensional problem. Assume the rotation axis is vertical. The potential is the sum of the gravitational plus centrifugal: $\Phi=\Phi_{g}+\Phi_{c}=-\frac{GM_{(x,y)}}{\sqrt{x^2+y^2}}+\frac{\omega^{2}}{2}x^{2}=-\frac{GM_{r}}{r}+\frac{\omega^{2}}{2}r^{2} \cos^{2} l$ The angle $l$ is the same as the latitude, and $M_{r}$ is the mass enclosed by a spherical surface ( but please see footnote) at the point, i.e $M_{r}=\frac{4}{3}\pi \rho r^{3}$ by assuming a constant density model. Therefore, $\Phi= -\frac{4}{3} G \pi \rho r^{2} +\frac{\omega^{2}}{2}r^{2} \cos^{2} l = r^{2}(\frac{\omega^{2}}{2}\cos^{2} l -\frac{4}{3} G \pi \rho)$ Thus, the family of curves of constant (negative) potential $\Phi=-C^{2}$ is: $ -C^{2} = r^{2}(\frac{\omega^{2}}{2}\cos^{2} l -\frac{4}{3} G \pi \rho) = r^{2}(A^{2} \cos^{2} l -B^{2}) $ Let's go back to rectangular coordinates, to see that this is indeed an ellipse: $ C^{2} = r^{2}(B^{2} - A^{2} \cos^{2} l) = (x^{2}+y^{2})(B^{2} - A^{2} \frac{x^{2}}{x^{2}+y^{2}}) = (x^{2}+y^{2})B^{2} - A^{2} x^{2}$ $ C^{2} = (B^{2} - A^{2}) x^{2} + B^{2} y^{2} $ For that equation to be an ellipse, $B^{2} - A^{2}$ must be positive. This is natural, otherwise (see how we defined $A$ and $B$) the angular speed $\omega$ would make the centrifugal force stronger than the gravitational force. The semiaxis are then $1/B$ for the vertical direction, and $1/\sqrt{B^{2} - A^{2}}$, i.e. bigger, in the horizontal direction. Note too, that $A=0$ for $\omega = 0$, that is, you recover the spherical shape if there is no rotation. Thus, an Earth with constant density that rotates as a rigid solid can be approximated by an ellipsoid shape, whose dimension along the rotation axis is smaller. Additionally, we probably don't need the interior of the Earth to be molten, for the hydrostatic equilibrium assumption to be valid. It could be completely cold and solid and the model still would hold, because at that size scales, relative small deviations of matter distribution from the constant potential surfaces give rise to enormous shear stress that rocks, no matter how hard and solid, cannot resist. That is why the liquid model is a valid approximation (but I have not done any numbers on this). NOTE: We have assumed that any point belongs to a spherical surface that is completely full of matter, therefore the potential gravitational energy is the same as if all matter inside that sphere were located at the Earth centre. If the Earth were much more flattened, this approximation would not be valid.
Definition and Methods of Solution An equation of type \[F\left( {x,y,y’} \right) = 0,\] where $F$ is a continuous function, is called the first order implicit differential equation. If this equation can be solved for $y’,$ we get one or several explicit differential equations of type \[y’ = f\left( {x,y} \right),\] that can be solved by methods covered in other sections. Further we suppose that the differential equation can not be solved in the explicit form, so we should use different methods. The main techniques for solving an implicit differential equation is the method of introducing a parameter. Below we show how this method works to find the general solution for some most important particular cases of implicit differential equations. Here we note that the general solution may not cover all possible solutions of a differential equation. Besides the general solution, the differential equation may also have so-called singular solutions. We consider this in more detail on the page Singular Solutions of Differential Equations. Case $1.$ Implicit Differential Equation of Type $x = f\left( {y,y’} \right).$ In this case, the variable $x$ is expressed explicitly in terms of $y$ and the derivative $y’.$ We introduce the parameter $p = y’ $ $= {\large\frac{{dy}}{{dx}}\normalsize}.$ Differentiate the equation $x = f\left( {y,y’} \right)$ with respect to $y.$ This produces: \[{\frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {f\left( {y,p} \right)} \right] }={ \frac{{\partial f}}{{\partial y}} + \frac{{\partial f}}{{\partial p}}\frac{{dp}}{{dy}}.}\] As ${\large\frac{{dx}}{{dy}}\normalsize} = {\large\frac{1}{p}\normalsize},$ the last expression can be written as follows: \[\frac{1}{p} = \frac{{\partial f}}{{\partial y}} + \frac{{\partial f}}{{\partial p}}\frac{{dp}}{{dy}}.\] We obtain an explicit differential equation such that its general solution is given by the function \[g\left( {y,p,C} \right) = 0,\] where $C$ is a constant. Thus, the general solution of the original implicit differential equation is defined in the parametric form by the system of two algebraic equations: \[\left\{ \begin{array}{l} g\left( {y,p,C} \right) = 0\\ x = f\left( {y,p} \right) \end{array} \right..\] If the parameter $p$ can be eliminated from the system, the general solution is given in the explicit form $x = f\left( {y,C} \right).$ Case $2.$ Implicit Differential Equation of Type $y = f\left( {x,y’} \right).$ Here we consider a similar case, when the variable $y$ is an explicit function of $x$ and $y’.$ Introduce the parameter $p = y’ $ $= {\large\frac{{dy}}{{dx}}\normalsize}$ and differentiate the equation $y = f\left( {x,y’} \right)$ with respect to $x.$ As a result, we have: \[ {{\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {f\left( {x,p} \right)} \right] }={ \frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial p}}\frac{{dp}}{{dx}}\;\;}}\kern-0.3pt {\text{or}\;\;p = \frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial p}}\frac{{dp}}{{dx}}.} \] Solving the last differential equation, we get the algebraic equation $g\left( {x,p,C} \right)$ $ = 0.$ Together with the original equation, they form the following system of equations: \[\left\{ \begin{array}{l} g\left( {x,p,C} \right) = 0\\ y = f\left( {x,p} \right) \end{array} \right.,\] which is the general solution of the given differential equation in the parametric form. In some cases, when the parameter $p$ can be eliminated from the system, the general solution can be written in the explicit form $y = f\left( {x,C} \right).$ Case $3.$ Implicit Differential Equation of Type $x = f\left( {y’} \right).$ Here the differential equation does not contain the variable $y.$ Using the parameter $p = y’ $ $= {\large\frac{{dy}}{{dx}}\normalsize},$ it’s easy to construct the general solution of the equation. As $dy = pdx$ and \[{dx = d\left[ {f\left( p \right)} \right] }={ \frac{{df}}{{dp}}dp,}\] then the following relationship holds: \[dy = p\frac{{df}}{{dp}}dp.\] Integrating the last equation gives the general solution in the parametric form: \[\left\{ \begin{array}{l} y = \int {p\frac{{df}}{{dp}}dp} + C\\ x = f\left( p \right) \end{array} \right..\] Case $4.$ Implicit Differential Equation of Type $y = f\left( {y’} \right).$ The equation of this kind does not contain the variable $x$ and can be solved the similar way. Using the parameter $p = y’ $ $= {\large\frac{{dy}}{{dx}}\normalsize},$ we can write $dx = \large\frac{{dy}}{p}\normalsize.$ Then it follows from the equation that \[{dx = \frac{{dy}}{p} }={ \frac{1}{p}\frac{{df}}{{dp}}dp.}\] Integrating the last expression gives the general solution of the original implicit equation in parametric form: \[\left\{ \begin{array}{l} x = \int {\frac{1}{p}\frac{{df}}{{dp}}dp} \\ y = f\left( p \right) \end{array} \right..\] Solved Problems Click a problem to see the solution.
Gated Recurrent Units (GRUs) Gated Recurrent Units (GRUs) are a form of RNN which can capture long range dependencies in a sequential data. A typical RNN unit is represented as the following equation: $a^{\langle t \rangle}$ = tanh($W_a [a^{\langle t-1 \rangle},x^{\langle t \rangle}] + b_a)$ Sentences: (1) The cat, which already ate……………., was full (2) The cats, which already ate……………, were full GRU (simplfied): New parameter: c which can be taken as a memory cell to store long range dependencies. Initially, $c^{\langle t \rangle}$ = $a^{\langle t \rangle}$ At every time step, we are going to consider overwriting $c^{\langle t \rangle}$. The candidate for replacing $c^{\langle t \rangle}$ is $\tilde c^{\langle t \rangle}$. $\tilde c^{\langle t \rangle}$ = tanh($W_c[c{\langle t-1 \rangle},x{\langle t \rangle}] + b_c)$ Gate: A Gate is represented as $\Gamma_u$ which is the Update Gate. $\Gamma_u$ = $\sigma (W_u[c^{\langle t-1 \rangle},x^{\langle t \rangle}]+b_u$…(i) The Gate $\Gamma_u$ whether or not to update $c^{\langle t \rangle}$ with $\tilde c^{\langle t \rangle}$. For singular “cat”, $c^{\langle t \rangle}$ = 1which will be carried forward in the network. The job of the gate $\Gamma_u$ is to decide when to update $c^{\langle t \rangle}$ with $\tilde c^{\langle t \rangle}$. Equation: $c^{\langle t \rangle}$ = $\Gamma_u * \tilde c^{\langle t \rangle} + (1-\Gamma_u)c^{\langle t-1 \rangle}$…(ii) The multiplication in the above equation is the elementwise multiplication. if $\Gamma_u$ = 1, $c^{\langle t \rangle}$ = $\tilde c^{\langle t \rangle}$. if $\Gamma_u$ = 0, $c^{\langle t \rangle}$ = $c^{\langle t-1 \rangle}$. Representation of Simplified GRU Unit «IMAGE» Advantages of GRU: Doesn’t suffer from the vanishing gradient problem Can learn long range dependencies $c^{\langle t \rangle}$ can be a vector. $\Gamma_u$ and $\tilde c^{\langle t \rangle}$ would also be a vector of same dimension. In the multiplication in the equation (ii) it will also be of the same dimension telling you which unit to remember and which to forget (elementwise multiplication) Simplified GRU Equations (1) $\tilde c^{\langle t \rangle}$ = tanh($W_c[c^{\langle t-1 \rangle},x^{\langle t \rangle}] + b_c)$ (2) $\Gamma_u$ = $\sigma (W_u[c^{\langle t-1 \rangle},x^{\langle t \rangle}]+b_u$ (3) $c^{\langle t \rangle}$ = $\Gamma_u \tilde c^{\langle t \rangle} + (1-\Gamma_u)c^{\langle t-1 \rangle}$ Full GRU The Full GRU incorporates additional gates: An Output Gate $\Gamma_o$, Forget gate $\Gamma_f$, Update Gate $\Gamma_u$ and a Relevance Gate $\Gamma_r$. Equations: boldface notations are the standard notations used in literatures (1) $\tilde h$: $\tilde c^{\langle t \rangle}$ = tanh($W_c[\Gamma_rc^{\langle t-1 \rangle},x^{\langle t \rangle}] + b_c)$ (2) $r$: $\Gamma_r$ = $\sigma(W_r[c^{\langle t-1 \rangle},x^{\langle t \rangle}] + b_c)$ (Relevance Gate) (3) $U$: $\Gamma_u$ = $\sigma(W_u[c^{\langle t-1 \rangle},x^{\langle t \rangle}] + b_u)$ (Update Gate) (4) $h$: $c^{\langle t \rangle}$ = $\Gamma_u * \tilde c^{\langle t \rangle} + (1-\Gamma_u)*c^{\langle t-1 \rangle}$ The relevance gate in (1) $\Gamma_r$ tells how relevant is $c^{\langle t-1 \rangle}$ for updating $\tilde c^{\langle t \rangle}$. Source material from Andrew NG’s awesome course on Coursera. The material in the video has been written in a text form so that anyone who wishes to revise a certain topic can go through this without going through the entire video lectures.
10:25 PM Suppose $f,g : [-\infty, \infty] \to [- \infty, \infty]$ are measurable. Prove that the sets $\{x \mid f(x) < g(x) \}$ and $\{x \mid f(x) = g(x)\}$ are measurable. The first is measurable because $\{x \mid f(x) < g(x) \} = \bigcup_{q \in \Bbb{Q}} \bigg[ \{x \mid f(x) < q \} \cap \{x \mid q < g(x)\} \bigg]$; i.e., the set is the countable union of measurable sets. The second set is measurable because $\{x \mid f(x) = g(x)\} = \bigg[ f^{-1}(-\infty) \cap g^{-1}(-\infty) \bigg] \cup \bigg[f^{-1}(\infty) \cap g^{-1}(\infty) \bigg] \cup (f-g)^{-1}(0)$, which means the set is the union of measurable sets and $f-g$ is a measurable function. « first day (1624 days earlier) ← previous day next day → last day (142 days later) »
My code solves the incompressible Navier-Stokes equation in a conducting fluid, together with the induction equation: $ \partial_t u + u \nabla u + 2\Omega \times u = -\nabla p + \nu \Delta u + (\nabla \times b) \times b \\ \partial_t b = \nabla \times (u \times b) + \eta \Delta b$ where $u$ and $b$ are the velocity and magnetic fields, $\eta$ and $\nu$ are diffusion constants, $p$ is the pressure, $\Omega$ the rotation rate of the reference frame. I solve this PDE system in spherical geometry using finite differences in the radial direction and spherical harmonic expansion. Let's focus on the Navier-Stokes equation which contains the Lorentz force $(\nabla \times b) \times b$. I use a Crank-Nicholson semi-implicit scheme for the diffusive terms, while the non-linear terms $u\nabla u$ and $(\nabla \times b)\times b$ are treated explicitely: $Au_{n+1} = Bu_{n} + NL(u_n,b_n)$ where $u_n$ and $b_n$ are respectively my velocity and magnetic field at time-step $n$, and $A$ and $B$ are matrices containing the time-step. (Note that $NL(u_n,b_n)$ actually also includes $2\Omega\times u$ for convenience, and that $\nabla p$ is eliminated by taking the curl) I would like to adjust dynamically my time-step. For Navier-Stokes, the usual CFL criterion can be used, but what about the Lorentz force ? I can proably come up with some specific criterion, but here is my question: Is there a way to adjust the time-step $\Delta t$ knowing only the non-linear term $NL(u_n,b_n)$ and ignoring its analytical form or physical meaning ?Some sort of generalised CFL criterion ? I've thought about something like $NL \Delta t < c\, U$ (where $c$ is a constant). When $NL = u\nabla u \sim \frac{U^2}{\Delta x}$, it reduces to the CFL $\frac{U\Delta t}{\Delta x} < c$. Doest it makes sense ? Is there a better way to think about this ? Are there references on the subject ?
Boundary bubbling solutions for a planar elliptic problem with exponential Neumann data 1. School of Mathematics, Zhejiang University, Hangzhou 310027, China 2. College of Sciences, Nanjing Agricultural University, Nanjing 210095, China $\mathbb{R}^2 $ $\left\{ \begin{gathered} \begin{gathered} - \Delta \upsilon + \upsilon = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\text{in}}\;\;\Omega {\text{,}} \hfill \\ \frac{{\partial \upsilon }}{{\partial \nu }} = {e^\upsilon } - s{\phi _1} - h\left( x \right)\;\;\;{\text{on}}\;\partial \Omega \hfill \\ \end{gathered} \end{gathered} \right.$ $ν$ $\partial \Omega$ $h∈ C^{0,α}(\partial \Omega)$ $s>0$ $\phi_1$ $\phi_1$ $s\to+∞$ Keywords:Nonlinear Neumann boundary condition, boundary bubbling solutions, Lyapunov-Schmidt reduction procedure. Mathematics Subject Classification:Primary: 35B25, 35J65; Secondary: 35B40. Citation:Haitao Yang, Yibin Zhang. Boundary bubbling solutions for a planar elliptic problem with exponential Neumann data. Discrete & Continuous Dynamical Systems - A, 2017, 37 (10) : 5467-5502. doi: 10.3934/dcds.2017238 References: [1] A. Ambrosetti and G. Prodi, On the inversion of some differentiable mappings with singularities between Banach spaces, [2] [3] S. Baraket and F. Parcard, Construction of singular limits for a semilinear elliptic equation in dimension 2, [4] B. Breuer, P. J. McKenna and M. Plum, Multiple solutions for a semilinear boundary value problem: A computational multiplicity proof, [5] H. Brezis and F. Merle, Uniform estimates and blow-up behavior for solutions of $-Δ u=V(x)e^u $ in two dimensions, [6] [7] [8] [9] [10] [11] [12] J. Dávila, M. del Pino and M. Musso, Concentrating solutions in a two-dimensional elliptic problem with exponential Neumann data, [13] D. G. de Figueiredo, P. N. Srikanth and S. Santra, Non-radially symmetric solutions for a superlinear Ambrosetti-Prodi type problem in a ball, [14] [15] [16] [17] P. Esposito, M. Grossi and A. Pistoia, On the existence of blowing-up solutions for a mean field equation, [18] O. Kavian and M. Vogelius, On the existence and "blow-up" of solutions to a two-dimensional nonlinear boundary-value problem arising in corrosion modelling, [19] [20] [21] [22] G. Li, S. Yan and J. Yang, The Lazer-McKenna conjecture for an elliptic problem with critical growth, [23] G. Li, S. Yan and J. Yang, The Lazer-McKenna conjecture for an elliptic problem with critical growth, part Ⅱ, [24] [25] K. Medville and M. Vogelius, Blow up behavior of planar harmonic functions satisfying a certain exponential Neumann boundary condition, [26] R. Molle and D. Passaseo, Elliptic equations with jumping nonlinearities involving high eigenvalues, [27] R. Molle and D. Passaseo, Existence and multiplicity of solutions for elliptic equations with jumping nonlinearities, [28] R. Molle and D. Passaseo, Multiple solutions for a class of elliptic equations with jumping nonlinearities, [29] [30] [31] [32] [33] [34] [35] [36] L. Zhang, Classification of conformal metrics on $\mathbb{R}^2_+ $ with constant Gauss curvature and geodesic curvature on the boundary under various integral finiteness assumptions, show all references References: [1] A. Ambrosetti and G. Prodi, On the inversion of some differentiable mappings with singularities between Banach spaces, [2] [3] S. Baraket and F. Parcard, Construction of singular limits for a semilinear elliptic equation in dimension 2, [4] B. Breuer, P. J. McKenna and M. Plum, Multiple solutions for a semilinear boundary value problem: A computational multiplicity proof, [5] H. Brezis and F. Merle, Uniform estimates and blow-up behavior for solutions of $-Δ u=V(x)e^u $ in two dimensions, [6] [7] [8] [9] [10] [11] [12] J. Dávila, M. del Pino and M. Musso, Concentrating solutions in a two-dimensional elliptic problem with exponential Neumann data, [13] D. G. de Figueiredo, P. N. Srikanth and S. Santra, Non-radially symmetric solutions for a superlinear Ambrosetti-Prodi type problem in a ball, [14] [15] [16] [17] P. Esposito, M. Grossi and A. Pistoia, On the existence of blowing-up solutions for a mean field equation, [18] O. Kavian and M. Vogelius, On the existence and "blow-up" of solutions to a two-dimensional nonlinear boundary-value problem arising in corrosion modelling, [19] [20] [21] [22] G. Li, S. Yan and J. Yang, The Lazer-McKenna conjecture for an elliptic problem with critical growth, [23] G. Li, S. Yan and J. Yang, The Lazer-McKenna conjecture for an elliptic problem with critical growth, part Ⅱ, [24] [25] K. Medville and M. Vogelius, Blow up behavior of planar harmonic functions satisfying a certain exponential Neumann boundary condition, [26] R. Molle and D. Passaseo, Elliptic equations with jumping nonlinearities involving high eigenvalues, [27] R. Molle and D. Passaseo, Existence and multiplicity of solutions for elliptic equations with jumping nonlinearities, [28] R. Molle and D. Passaseo, Multiple solutions for a class of elliptic equations with jumping nonlinearities, [29] [30] [31] [32] [33] [34] [35] [36] L. Zhang, Classification of conformal metrics on $\mathbb{R}^2_+ $ with constant Gauss curvature and geodesic curvature on the boundary under various integral finiteness assumptions, [1] [2] [3] Shengbing Deng, Fethi Mahmoudi, Monica Musso. Bubbling on boundary submanifolds for a semilinear Neumann problem near high critical exponents. [4] Martin Gugat, Günter Leugering, Ke Wang. Neumann boundary feedback stabilization for a nonlinear wave equation: A strict $H^2$-lyapunov function. [5] Alexander Gladkov. Blow-up problem for semilinear heat equation with nonlinear nonlocal Neumann boundary condition. [6] Zuodong Yang, Jing Mo, Subei Li. Positive solutions of $p$-Laplacian equations with nonlinear boundary condition. [7] Jaeyoung Byeon, Sangdon Jin. The Hénon equation with a critical exponent under the Neumann boundary condition. [8] R.G. Duran, J.I. Etcheverry, J.D. Rossi. Numerical approximation of a parabolic problem with a nonlinear boundary condition. [9] Jesús Ildefonso Díaz, L. Tello. On a climate model with a dynamic nonlinear diffusive boundary condition. [10] Mariane Bourgoing. Viscosity solutions of fully nonlinear second order parabolic equations with $L^1$ dependence in time and Neumann boundary conditions. [11] Umberto De Maio, Akamabadath K. Nandakumaran, Carmen Perugia. Exact internal controllability for the wave equation in a domain with oscillating boundary with Neumann boundary condition. [12] Tsung-Fang Wu. Multiplicity of positive solutions for a semilinear elliptic equation in $R_+^N$ with nonlinear boundary condition. [13] M. Eller. On boundary regularity of solutions to Maxwell's equations with a homogeneous conservative boundary condition. [14] Christina A. Hollon, Jeffrey T. Neugebauer. Positive solutions of a fractional boundary value problem with a fractional derivative boundary condition. [15] Antonio Cañada, Salvador Villegas. Optimal Lyapunov inequalities for disfocality and Neumann boundary conditions using $L^p$ norms. [16] Shu Luan. On the existence of optimal control for semilinear elliptic equations with nonlinear neumann boundary conditions. [17] Shouming Zhou, Chunlai Mu, Yongsheng Mi, Fuchen Zhang. Blow-up for a non-local diffusion equation with exponential reaction term and Neumann boundary condition. [18] Zhenhua Zhang. Asymptotic behavior of solutions to the phase-field equations with neumann boundary conditions. [19] Zhongwei Tang. Segregated peak solutions of coupled Schrödinger systems with Neumann boundary conditions. [20] Jason Metcalfe, Jacob Perry. Global solutions to quasilinear wave equations in homogeneous waveguides with Neumann boundary conditions. 2018 Impact Factor: 1.143 Tools Metrics Other articles by authors [Back to Top]
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
$L$ isn't recognizable. We'll first establish a couple of preliminary results I. $\overline{L}$ is recognizable The complement of $L$,$$\overline{L}=\{\langle M\rangle\mid M \text{ halts on at least one input}\} $$is recognizable. Define a recognizer TM as follows: R(<M>) = for n = 1, 2, 3, ... for each x in {x_1, x_2, ... , x_n} // in some canonical order run M on x for one move if M halts return accept It should be clear that $R$ accepts all and only those $\langle M \rangle$ for which $\langle M \rangle\in \overline{L}$ and so $\overline{L}$ is recognizable. Now if $L$ were also recognizable, then we could use the two recognizers to make decider for $L$, which brings us to our second result. II. L is undecidable If $L$ were decidable, then $\overline{L}$ would also be, and conversely. If that were the case, we could define a reduction from the known undecidable language$$HALT = \{(\langle M\rangle \mid M \text{ halts on input }w\}$$to $\overline{L}$ by the mapping$$(\langle M\rangle, w)\rightarrow M_w$$where, as babou has already noted, M_w(y) = erase the input y write w on the input tape simulate M on w Now observe that $M$ halts on $w$ $\Longleftrightarrow$ $M_w$ halts (on every input $y$, in fact) $\Longleftrightarrow$ $M_w\in \overline{L}$. In summary, if $L$ were decidable, then we could $L$'s decider (reversing the roles of accept and reject) to decide the halting problem, a contradiction. Now we can show that $L$ is not recognizable. If it were, then, using (I) we could conclude that $L$ was decidable, a contradiction to result (II).
Tagged: determinant Problem 548 An $n\times n$ matrix $A$ is said to be invertible if there exists an $n\times n$ matrix $B$ such that $AB=I$, and $BA=I$, where $I$ is the $n\times n$ identity matrix. If such a matrix $B$ exists, then it is known to be unique and called the inverse matrix of $A$, denoted by $A^{-1}$. In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition. So if we know $AB=I$, then we can conclude that $B=A^{-1}$. Let $A$ and $B$ be $n\times n$ matrices. Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix. Prove that $BA=I$, and hence $A^{-1}=B$.Add to solve later Problem 452 Let $A$ be an $n\times n$ complex matrix. Let $S$ be an invertible matrix. (a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix. (b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$. Add to solve later (c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$. Problem 438 Determine whether each of the following statements is True or False. (a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$. (b) If the characteristic polynomial of an $n \times n$ matrix $A$ is \[p(\lambda)=(\lambda-1)^n+2,\] then $A$ is invertible. (c) If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible. (d) If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$. (e) If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$. ( Stanford University, Linear Algebra Exam Problem) Read solution Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$. Add to solve later (b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. Problem 391 (a) Is the matrix $A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$? (b) Is the matrix $A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ 4& 3 \end{bmatrix}$? (c) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 0& 2 \end{bmatrix}$? Add to solve later (d) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ -1& 4 \end{bmatrix}$? Problem 389 (a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$. (b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$. (c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$? ( Harvard University, Linear Algebra Exam Problem) Problem 374 Let \[A=\begin{bmatrix} a_0 & a_1 & \dots & a_{n-2} &a_{n-1} \\ a_{n-1} & a_0 & \dots & a_{n-3} & a_{n-2} \\ a_{n-2} & a_{n-1} & \dots & a_{n-4} & a_{n-3} \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{2} & a_3 & \dots & a_{0} & a_{1}\\ a_{1} & a_2 & \dots & a_{n-1} & a_{0} \end{bmatrix}\] be a complex $n \times n$ matrix. Such a matrix is called circulant matrix. Then prove that the determinant of the circulant matrix $A$ is given by \[\det(A)=\prod_{k=0}^{n-1}(a_0+a_1\zeta^k+a_2 \zeta^{2k}+\cdots+a_{n-1}\zeta^{k(n-1)}),\] where $\zeta=e^{2 \pi i/n}$ is a primitive $n$-th root of unity. Problem 363 (a) Find all the eigenvalues and eigenvectors of the matrix \[A=\begin{bmatrix} 3 & -2\\ 6& -4 \end{bmatrix}.\] Add to solve later (b) Let \[A=\begin{bmatrix} 1 & 0 & 3 \\ 4 &5 &6 \\ 7 & 0 & 9 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}.\] Then find the value of \[\det(A^2B^{-1}A^{-2}B^2).\] (For part (b) without computation, you may assume that $A$ and $B$ are invertible matrices.) Problem 338 Each of the following sets are not a subspace of the specified vector space. For each set, give a reason why it is not a subspace. (1) \[S_1=\left \{\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \R^3 \quad \middle \| \quad x_1\geq 0 \,\right \}\] in the vector space $\R^3$. (2)\[S_2=\left \{\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \R^3 \quad \middle \| \quad x_1-4x_2+5x_3=2 \,\right \}\] in the vector space $\R^3$. (3)\[S_3=\left \{\, \begin{bmatrix} x \\ y \end{bmatrix}\in \R^2 \quad \middle \| \quad y=x^2 \quad \,\right \}\] in the vector space $\R^2$. (4)Let $P_4$ be the vector space of all polynomials of degree $4$ or less with real coefficients. \[S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}\] in the vector space $P_4$. (5)\[S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}\] in the vector space $P_4$. (6)Let $M_{2 \times 2}$ be the vector space of all $2\times 2$ real matrices. \[S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\} \] in the vector space $M_{2\times 2}$. (7)\[S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\} \] in the vector space $M_{2\times 2}$. ( Linear Algebra Exam Problem, the Ohio State University) (8)Let $C[-1, 1]$ be the vector space of all real continuous functions defined on the interval $[a, b]$. \[S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\} \] in the vector space $C[-2, 2]$. (9)\[S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$. (10)Let $C^2[a, b]$ be the vector space of all real-valued functions $f(x)$ defined on $[a, b]$, where $f(x), f'(x)$, and $f^{\prime\prime}(x)$ are continuous on $[a, b]$. Here $f'(x), f^{\prime\prime}(x)$ are the first and second derivative of $f(x)$. \[S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$. (11)Let $S_{11}$ be the set of real polynomials of degree exactly $k$, where $k \geq 1$ is an integer, in the vector space $P_k$. (12)Let $V$ be a vector space and $W \subset V$ a vector subspace. Define the subset $S_{12}$ to be the complementof $W$, \[ V \setminus W = \{ \mathbf{v} \in V \mid \mathbf{v} \not\in W \}.\] Add to solve later
Notice: If you happen to see a question you know the answer to, please do chime in and help your fellow community members. We encourage our fourm members to be more involved, jump in and help out your fellow researchers with their questions. GATK forum is a community forum and helping each other with using GATK tools and research is the cornerstone of our success as a genomics research community.We appreciate your help! Test-drive the GATK tools and Best Practices pipelines on Terra Check out this blog post to learn how you can get started with GATK and try out the pipelines in preconfigured workspaces (with a user-friendly interface!) without having to install anything. HC step 4: Assigning per-sample genotypes This document describes the procedure used by HaplotypeCaller to assign genotypes to individual samples based on the allele likelihoods calculated in the previous step. For more context information on how this fits into the overall HaplotypeCaller method, please see the more general HaplotypeCaller documentation. See also the documentation on the QUAL score as well as PL and GQ. Note that this describes the regular mode of HaplotypeCaller, which does not emit an estimate of reference confidence. For details on how the reference confidence model works and is applied in -ERC modes ( GVCF and BP_RESOLUTION) please see the reference confidence model documentation. Overview The previous step produced a table of per-read allele likelihoods for each candidate variant site under consideration. Now, all that remains to do is to evaluate those likelihoods in aggregate to determine what is the most likely genotype of the sample at each site. This is done by applying Bayes' theorem to calculate the likelihoods of each possible genotype, and selecting the most likely. This produces a genotype call as well as the calculation of various metrics that will be annotated in the output VCF if a variant call is emitted. 1. Preliminary assumptions / limitations Quality Keep in mind that we are trying to infer the genotype of each sample given the observed sequence data, so the degree of confidence we can have in a genotype depends on both the quality and the quantity of the available data. By definition, low coverage and low quality will both lead to lower confidence calls. The GATK only uses reads that satisfy certain mapping quality thresholds, and only uses “good” bases that satisfy certain base quality thresholds (see documentation for default values). Ploidy Both the HaplotypeCaller and GenotypeGVCFs (but not UnifiedGenotyper) assume that the organism of study is diploid by default, but desired ploidy can be set using the -ploidy argument. The ploidy is taken into account in the mathematical development of the Bayesian calculation. The generalized form of the genotyping algorithm that can handle ploidies other than 2 is available as of version 3.3-0. Note that using ploidy for pooled experiments is subject to some practical limitations due to the number of possible combinations resulting from the interaction between ploidy and the number of alternate alleles that are considered (currently, the maximum "workable" ploidy is ~20 for a max number of alt alleles = 6). Future developments will aim to mitigate those limitations. Paired end reads Reads that are mates in the same pair are not handled together in the reassembly, but if they overlap, there is some special handling to ensure they are not counted as independent observations. Single-sample vs multi-sample We apply different genotyping models when genotyping a single sample as opposed to multiple samples together (as done by HaplotypeCaller on multiple inputs or GenotypeGVCFs on multiple GVCFs). The multi-sample case is not currently documented for the public but is an extension of previous work by Heng Li and others. 2. Calculating genotype likelihoods using Bayes' Theorem We use the approach described in Li 2011 to calculate the posterior probabilities of non-reference alleles (Methods 2.3.5 and 2.3.6) extended to handle multi-allelic variation. The basic formula we use for all types of variation under consideration (SNPs, insertions and deletions) is: $$ P(G\|D) = \frac{ P(G) P(D\|G) }{ \sum_{i} P(G_i) P(D\|G_i) } $$ If that is meaningless to you, please don't freak out -- we're going to break it down and go through all the components one by one. First of all, the term on the left: $$ P(G\|D) $$ is the quantity we are trying to calculate for each possible genotype: the conditional probability of the genotype G given the observed data D. Now let's break down the term on the right: $$ \frac{ P(G) P(D\|G) }{ \sum_{i} P(G_i) P(D\|G_i) } $$ We can ignore the denominator (bottom of the fraction) because it ends up being the same for all the genotypes, and the point of calculating this likelihood is to determine the most likely genotype. The important part is the numerator (top of the fraction): $$ P(G) P(D\|G) $$ which is composed of two things: the prior probability of the genotype and the conditional probability of the data given the genotype. The first one is the easiest to understand. The prior probability of the genotype G: $$ P(G) $$ represents how probably we expect to see this genotype based on previous observations, studies of the population, and so on. By default, the GATK tools use a flat prior (always the same value) but you can input your own set of priors if you have information about the frequency of certain genotypes in the population you're studying. The second one is a little trickier to understand if you're not familiar with Bayesian statistics. It is called the conditional probability of the data given the genotype, but what does that mean? Assuming that the genotype G is the true genotype, $$ P(D\|G) $$ is the probability of observing the sequence data that we have in hand. That is, how likely would we be to pull out a read with a particular sequence from an individual that has this particular genotype? We don't have that number yet, so this requires a little more calculation, using the following formula: $$ P(D\|G) = \prod{j} \left( \frac{P(D_j \| H_1)}{2} + \frac{P(D_j \| H_2)}{2} \right) $$ You'll notice that this is where the diploid assumption comes into play, since here we decomposed the genotype G into: $$ G = H_1H_2 $$ which allows for exactly two possible haplotypes. In future versions we'll have a generalized form of this that will allow for any number of haplotypes. Now, back to our calculation, what's left to figure out is this: $$ P(D_j\|H_n) $$ which as it turns out is the conditional probability of the data given a particular haplotype (or specifically, a particular allele), aggregated over all supporting reads. Conveniently, that is exactly what we calculated in Step 3 of the HaplotypeCaller process, when we used the PairHMM to produce the likelihoods of each read against each haplotype, and then marginalized them to find the likelihoods of each read for each allele under consideration. So all we have to do at this point is plug the values from that table into the equation above, and we can work our way back up to obtain: $$ P(G\|D) $$ for the genotype G. 3. Selecting a genotype and emitting the call record We go through the process of calculating a likelihood for each possible genotype based on the alleles that were observed at the site, considering every possible combination of alleles. For example, if we see an A and a T at a site, the possible genotypes are AA, AT and TT, and we end up with 3 corresponding probabilities. We pick the largest one, which corresponds to the most likely genotype, and assign that to the sample. Note that depending on the variant calling options specified in the command-line, we may only emit records for actual variant sites (where at least one sample has a genotype other than homozygous-reference) or we may also emit records for reference sites. The latter is discussed in the reference confidence model documentation. Assuming that we have a non-ref genotype, all that remains is to calculate the various site-level and genotype-level metrics that will be emitted as annotations in the variant record, including QUAL as well as PL and GQ -- see the linked docs for details. For more information on how the other variant context metrics are calculated, please see the corresponding variant annotations documentation.
We can of course also define a four-vector version of the acceleration, by taking the derivative of the four-velocity with respect to the proper time. As with the forces, we’ll see that we’re in for some nasty surprises, this time because the proper time derivative acts on the $\gamma(u)$ factor in the velocity as well as on the components: \[\overline{\boldsymbol{a}} \equiv \frac{\mathrm{d} \overline{\boldsymbol{v}}}{\mathrm{d} \tau}=\gamma(v) \frac{\mathrm{d}}{\mathrm{d} t} \gamma(v)\left(c, v_{x}, v_{y}, v_{z}\right)=\left(\gamma^{4}(v) \frac{\boldsymbol{a} \cdot \boldsymbol{v}}{c}, \gamma^{2}(v) \boldsymbol{a}+\gamma^{4}(v) \frac{\boldsymbol{a} \cdot \boldsymbol{v}}{c^{2}} \boldsymbol{v}\right) \label{15.2.1}\] where we used the time derivative of the $\gamma(u)$ function \[\frac{\mathrm{d} \gamma}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t} \frac{1}{\sqrt{1-(v / c)^{2}}}=-\frac{1}{2} \frac{1}{\left(1-(v / c)^{2}\right)^{3 / 2}} \cdot\left(-2 \frac{\boldsymbol{v}}{c^{2}} \cdot \frac{\mathrm{d} \boldsymbol{v}}{\mathrm{d} t}\right)=\gamma^{3}(v) \frac{\boldsymbol{v} \cdot \boldsymbol{a}}{c^{2}} \label{15.2.2}\] and we’ve introduced the (classical) acceleration three-vector as the coordinate time derivative of the velocity three-vector: $\boldsymbol{a}=\mathrm{d} \boldsymbol{v} / \mathrm{d} t$ . As you can see in equation (\ref{15.2.1}), the four-acceleration has terms that scale with $\gamma^{2}$ and terms that scale with $\gamma^{4}$, making it an inconvenient object to work with. Geometrically though, it has a clean interpretation, which comes into view once you consider the inner product between the acceleration and velocity four-vectors: \[\begin{aligned} \overline{\boldsymbol{a}} \cdot \overline{\boldsymbol{v}} &=\gamma^{5}(v)(\boldsymbol{a} \cdot \boldsymbol{v})-\gamma^{3}(v)(\boldsymbol{a} \cdot \boldsymbol{v})-\gamma^{5}(v) \frac{\boldsymbol{a} \cdot \boldsymbol{v}}{c^{2}}(\boldsymbol{v} \cdot \boldsymbol{v}) \\ &=\gamma^{3}(v)(\boldsymbol{a} \cdot \boldsymbol{v})\left[\frac{1-v^{2} / c^{2}}{1-v^{2} / c^{2}}-1\right] \\ &=0 \end{aligned}\label{15.2.3}\] These four-vectors are therefore (in the four-vector sense) always perpendicular! That seems odd from a classical point of view: if you move in the $x$-direction, and speed up, both velocity and acceleration point in the same direction and are thus certainly not perpendicular. We do have a perpendicular case of course: circular motion (with a velocity along the circle, and acceleration inwards). Something similar happens here, if you consider the world line of a particle in a spacetime diagram (see Figure 15.2.1). You can think of this line as a curve that’s parametrized by the proper time $\tau$; points on the curve are then given by the position four-vector at time $\tau$. The velocity four-vector is the normalized tangent to this line (and indeed, by construction, has a fixed length $c$). When you’re moving at constant velocity, the line is straight, but if you change your velocity (i.e., you accelerate), the line curves. The acceleration four-vector both quantifies that curvature, and points in the direction that the curve is bending$^{1}$. Because by definition $\overline{\boldsymbol{p}}=m \overline{\boldsymbol{v}}, \text { and } \overline{\boldsymbol{F}}=\mathrm{d} \overline{\boldsymbol{p}} / \mathrm{d} \tau$, as long as the mass is conserved ($\mathrm{d} m / \mathrm{d} \tau=0$), we do have F¯ Æma¯, so Newton’s second law does hold for four-vectors. This result is not nearly as useful as in classical mechanics though since as we’ve seen, forces transform in unwieldy manners between inertial frames, and the acceleration can only curve the trajectory in spacetime. To see how forces and accelerations can be used for a case where you have no choice but to use them$^{2}$, consider a particle that is under constant acceleration, due to a constant three-force acting on it in the (noninertial!) co-moving frame of the particle (e.g. due to a rocket engine attached to the particle). From the point of view of an inertial rest frame, we have \[\frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} t}=\boldsymbol{F} \quad \text { and } \quad \frac{\mathrm{d} \boldsymbol{F}}{\mathrm{d} t}=0 \label{15.2.4}\] Choose the $x$-axis to be along the direction of $\boldsymbol{F}$, and define a = a_{x} = F_{x}/m\). Then \[a=\frac{\mathrm{d}\left(p_{x} / m\right)}{\mathrm{d} t}=\frac{\mathrm{d} w_{x}}{\mathrm{d} t} \label {15.2.5}\] where $\boldsymbol{w} \equiv \boldsymbol{p} / m=\gamma(v) \boldsymbol{v}$, and, as we have only motion in the positive $x$-direction here, we have $w_{x}=w$, (v_{x}=v\). Solving equation (\ref{15.2.5}) for $w$, we get the velocity of a uniformly accelerated particle: $w(t)=w(0)+a t$ . Now solving for the actually measured velocity in the inertial frame (taking $w(0) = 0$), we find \[\gamma(v(t)) v(t)=w(t)=a t \quad \Rightarrow \quad v^{2}=a^{2} t^{2}\left(1-\frac{v^{2}}{c^{2}}\right) \quad \Rightarrow \quad v=\frac{a t}{\sqrt{1+a^{2} t^{2} / c^{2}}} \label{15.2.6}\] Figure 15.2.2 compares the relativistic velocity with the classical result. Unsurprisingly, they are initially identical, but once the speed starts picking up, we see that the classical results starts to deviate significantly. In particular, the relativistic result confirms that no matter how long we accelerate, our particle will never exceed the speed of light. On a side note, we can also solve for the actual trajectory of our particle: simply integrate $\mathrm{d} x / \mathrm{d} t=v(t)$, which gives \[x(t)=\frac{c^{2}}{a}\left(\sqrt{1+\frac{a^{2} t^{2}}{c^{2}}}-1\right)\label{15.2.7}\] For small values of $t$, we (again) recover the classical result, $x=\frac{1}{2} a t^{2}$. $^{1}$There is a one-on-one correspondence between these ‘world curves’ and affinely parametrized curves in real space of two or more dimensions. There too, you can define a tangent vector as the derivative of the position vector, which for an affinely parametrized curve is always of unit length. The derivative of the tangent vector, known as the normal, is always perpendicular to the tangent, and points in the direction in which the curve is bending; its magnitude quantifies the curvature. $^{2}$I'm sure you’ve noted the obvious pun here.
I am following along well several textbooks (Geophysics) that helps me understand the in-depth physics behind the magnetic field of a dipole magnet. I understand that the basic magnetic potential (when observing a dipole magnet where the one of the poles is far enough to have negligible effect on its partner is: $$W= -\int_r^{\infty}B \space dr = \frac{u_op}{4 \pi r}$$ Then to find the magnetic potential at point P due to both poles are: $$W(\theta, r)=\frac{u_om\space \cos \theta}{4 \pi r^2}$$ Now to find the magnetic field strength at point P, I know its the vector addition of both $B_r$ and $B_\theta$ (radial and tangential respectively), and to find both I need to differentiate the potential with respect to r and $\theta$ Now here is finally where I get to ask my question. To find $B_r$ is easy enough by: $$B_r=\frac{\partial W}{\partial r} = -\frac{2 u_o m\space \cos\theta}{4 \pi r^3}$$ But when I take the potential and differentiate with respect to $\theta$ I should get: $$B_\theta = \frac{\partial W}{\partial \theta} = -\frac{u_o m\space \sin\theta}{4 \pi r^2}$$ but in ALL the textbooks they multiply $\frac{1}{r}$ to get: $$B_\theta = \frac{1}{r}\frac{\partial W}{\partial \theta} = -\frac{u_o m\space \sin\theta}{4 \pi r^3}$$ Which is needed to find the total magnetic strength field ($B_r + B_\theta$) Can anyone please explain where they got that extra $\frac{1}{r}?$ I have a feeling it has something to do with a unit vector, but I can't seem to connect the dots. .
Let $T: \R^n \to \R^m$ be a linear transformation.Suppose that the nullity of $T$ is zero. If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$. Let $V$ denote the vector space of all real $2\times 2$ matrices.Suppose that the linear transformation from $V$ to $V$ is given as below.\[T(A)=\begin{bmatrix}2 & 3\\5 & 7\end{bmatrix}A-A\begin{bmatrix}2 & 3\\5 & 7\end{bmatrix}.\]Prove or disprove that the linear transformation $T:V\to V$ is an isomorphism. Let $G, H, K$ be groups. Let $f:G\to K$ be a group homomorphism and let $\pi:G\to H$ be a surjective group homomorphism such that the kernel of $\pi$ is included in the kernel of $f$: $\ker(\pi) \subset \ker(f)$. Define a map $\bar{f}:H\to K$ as follows.For each $h\in H$, there exists $g\in G$ such that $\pi(g)=h$ since $\pi:G\to H$ is surjective.Define $\bar{f}:H\to K$ by $\bar{f}(h)=f(g)$. (a) Prove that the map $\bar{f}:H\to K$ is well-defined. (b) Prove that $\bar{f}:H\to K$ is a group homomorphism. Let $\calF[0, 2\pi]$ be the vector space of all real valued functions defined on the interval $[0, 2\pi]$.Define the map $f:\R^2 \to \calF[0, 2\pi]$ by\[\left(\, f\left(\, \begin{bmatrix}\alpha \\\beta\end{bmatrix} \,\right) \,\right)(x):=\alpha \cos x + \beta \sin x.\]We put\[V:=\im f=\{\alpha \cos x + \beta \sin x \in \calF[0, 2\pi] \mid \alpha, \beta \in \R\}.\] (a) Prove that the map $f$ is a linear transformation. (b) Prove that the set $\{\cos x, \sin x\}$ is a basis of the vector space $V$. (c) Prove that the kernel is trivial, that is, $\ker f=\{\mathbf{0}\}$.(This yields an isomorphism of $\R^2$ and $V$.) (d) Define a map $g:V \to V$ by\[g(\alpha \cos x + \beta \sin x):=\frac{d}{dx}(\alpha \cos x+ \beta \sin x)=\beta \cos x -\alpha \sin x.\]Prove that the map $g$ is a linear transformation. (e) Find the matrix representation of the linear transformation $g$ with respect to the basis $\{\cos x, \sin x\}$. Suppose that the vectors\[\mathbf{v}_1=\begin{bmatrix}-2 \\1 \\0 \\0 \\0\end{bmatrix}, \qquad \mathbf{v}_2=\begin{bmatrix}-4 \\0 \\-3 \\-2 \\1\end{bmatrix}\]are a basis vectors for the null space of a $4\times 5$ matrix $A$. Find a vector $\mathbf{x}$ such that\[\mathbf{x}\neq0, \quad \mathbf{x}\neq \mathbf{v}_1, \quad \mathbf{x}\neq \mathbf{v}_2,\]and\[A\mathbf{x}=\mathbf{0}.\] (Stanford University, Linear Algebra Exam Problem) Let $V$ be the subspace of $\R^4$ defined by the equation\[x_1-x_2+2x_3+6x_4=0.\]Find a linear transformation $T$ from $\R^3$ to $\R^4$ such that the null space $\calN(T)=\{\mathbf{0}\}$ and the range $\calR(T)=V$. Describe $T$ by its matrix $A$. A hyperplane in $n$-dimensional vector space $\R^n$ is defined to be the set of vectors\[\begin{bmatrix}x_1 \\x_2 \\\vdots \\x_n\end{bmatrix}\in \R^n\]satisfying the linear equation of the form\[a_1x_1+a_2x_2+\cdots+a_nx_n=b,\]where $a_1, a_2, \dots, a_n$ (at least one of $a_1, a_2, \dots, a_n$ is nonzero) and $b$ are real numbers.Here at least one of $a_1, a_2, \dots, a_n$ is nonzero. Consider the hyperplane $P$ in $\R^n$ described by the linear equation\[a_1x_1+a_2x_2+\cdots+a_nx_n=0,\]where $a_1, a_2, \dots, a_n$ are some fixed real numbers and not all of these are zero.(The constant term $b$ is zero.) Then prove that the hyperplane $P$ is a subspace of $R^{n}$ of dimension $n-1$. Let $n$ be a positive integer. Let $T:\R^n \to \R$ be a non-zero linear transformation.Prove the followings. (a) The nullity of $T$ is $n-1$. That is, the dimension of the nullspace of $T$ is $n-1$. (b) Let $B=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}\}$ be a basis of the nullspace $\calN(T)$ of $T$.Let $\mathbf{w}$ be the $n$-dimensional vector that is not in $\calN(T)$. Then\[B’=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}, \mathbf{w}\}\]is a basis of $\R^n$. (c) Each vector $\mathbf{u}\in \R^n$ can be expressed as\[\mathbf{u}=\mathbf{v}+\frac{T(\mathbf{u})}{T(\mathbf{w})}\mathbf{w}\]for some vector $\mathbf{v}\in \calN(T)$. Let $A$ be the matrix for a linear transformation $T:\R^n \to \R^n$ with respect to the standard basis of $\R^n$.We assume that $A$ is idempotent, that is, $A^2=A$.Then prove that\[\R^n=\im(T) \oplus \ker(T).\] (a) Let $A=\begin{bmatrix}1 & 2 & 1 \\3 &6 &4\end{bmatrix}$ and let\[\mathbf{a}=\begin{bmatrix}-3 \\1 \\1\end{bmatrix}, \qquad \mathbf{b}=\begin{bmatrix}-2 \\1 \\0\end{bmatrix}, \qquad \mathbf{c}=\begin{bmatrix}1 \\1\end{bmatrix}.\]For each of the vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$, determine whether the vector is in the null space $\calN(A)$. Do the same for the range $\calR(A)$. (b) Find a basis of the null space of the matrix $B=\begin{bmatrix}1 & 1 & 2 \\-2 &-2 &-4\end{bmatrix}$. Let $A$ be a real $7\times 3$ matrix such that its null space is spanned by the vectors\[\begin{bmatrix}1 \\2 \\0\end{bmatrix}, \begin{bmatrix}2 \\1 \\0\end{bmatrix}, \text{ and } \begin{bmatrix}1 \\-1 \\0\end{bmatrix}.\]Then find the rank of the matrix $A$. (Purdue University, Linear Algebra Final Exam Problem) Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\]where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring homomorphism, called the augmentation map and the kernel of $\epsilon$ is called the augmentation ideal. (a) Prove that the augmentation ideal in the group ring $RG$ is generated by $\{g-e \mid g\in G\}$. (b) Prove that if $G=\langle g\rangle$ is a finite cyclic group generated by $g$, then the augmentation ideal is generated by $g-e$.
Latent Distribution Two-Graph Testing¶ [1]: import numpy as npimport matplotlib.pyplot as pltnp.random.seed(8888)from graspy.inference import LatentDistributionTestfrom graspy.embed import AdjacencySpectralEmbedfrom graspy.simulations import sbm, rdpgfrom graspy.utils import symmetrizefrom graspy.plot import heatmap, pairplot%matplotlib inline Generate a stochastic block model graph¶ We generate a stochastic block model graph (SBM), which is shown below. [2]: n_components = 4 # the number of embedding dimensions for ASEP = np.array([[0.9, 0.11, 0.13, 0.2], [0, 0.7, 0.1, 0.1], [0, 0, 0.8, 0.1], [0, 0, 0, 0.85]])P = symmetrize(P)csize = [50] * 4A = sbm(csize, P)X = AdjacencySpectralEmbed(n_components=n_components).fit_transform(A)heatmap(A, title='4-block SBM adjacency matrix')pairplot(X, title='4-block adjacency spectral embedding') [2]: <matplotlib.axes._subplots.AxesSubplot at 0x128644518> [2]: <seaborn.axisgrid.PairGrid at 0x12881a048> Latent distribution test where null is true¶ Now, we want to know whether the above two graphs were generated from the same latent position. We know that they were, so the test should predict that the differences between SBM 1 and 2 (up to a rotation) are no greater than those differences observed by chance. In other words, we are testing \begin{align} H_0:&X_1 = X_2\\ H_\alpha:& X_1 \neq X_2 \end{align} and want to see that the p-value for the unmatched test is high (fail to reject the null) We generate a second SBM in the same way, and run an unmatched test on it, generating a distance between the two graphs as well as a null distribution of distances between permutations of the graph. We can see this below. [3]: A1 = sbm(csize, P)heatmap(A1, title='4-block SBM adjacency matrix A1')X1 = AdjacencySpectralEmbed(n_components=n_components).fit_transform(A1)pairplot(X1, title='4-block adjacency spectral embedding A1') [3]: <matplotlib.axes._subplots.AxesSubplot at 0x12d1420f0> [3]: <seaborn.axisgrid.PairGrid at 0x12d52da90> Plot of Null Distribution¶ We plot the null distribution shown in blue and the test statistic shown red vertical line. We see that the test static is small, resulting in p-value of 0.94. Thus, we cannot reject the null hypothesis that the two graphs come from the same generating distributions. [4]: ldt = LatentDistributionTest()p = ldt.fit(A, A1)fig, ax = plt.subplots(figsize=(10, 6))ax.hist(ldt.null_distribution_, 50)ax.axvline(ldt.sample_T_statistic_, color='r')ax.set_title("P-value = {}".format(p), fontsize=20)plt.show(); Latent distribution test where null is false¶ We generate a seconds SBM with different block probabilities, and run a latent distribution test comaring the previous graph with the new one. [5]: P2 = np.array([[0.8, 0.2, 0.2, 0.5], [0, 0.9, 0.3, 0.2], [0, 0, 0.5, 0.2], [0, 0, 0, 0.5]])P2 = symmetrize(P2)A2 = sbm(csize, P2)heatmap(A2, title='4-block SBM adjacency matrix A2')X2 = AdjacencySpectralEmbed(n_components=n_components).fit_transform(A2)pairplot(X2, title='4-block adjacency spectral embedding A2') [5]: <matplotlib.axes._subplots.AxesSubplot at 0x1286af128> [5]: <seaborn.axisgrid.PairGrid at 0x12e1eaf98> Plot of Null Distribution¶ We plot the null distribution shown in blue and the test statistic shown red vertical line. We see that the test static is small, resulting in p-value of 0. Thus, we reject the null hypothesis that the two graphs come from the same generating distributions. [6]: ldt = LatentDistributionTest()p = ldt.fit(A, A2)fig, ax = plt.subplots(figsize=(10, 6))ax.hist(ldt.null_distribution_, 50)ax.axvline(ldt.sample_T_statistic_, color='r')ax.set_title("P-value = {}".format(p), fontsize=20)plt.show();
How to solve these equations using an algebraic method? I need to show my working, don't you do something in reverse, like 7 multiplies by something. I haven't done it in class. $$\dfrac{5(3y-4)}{2y}=7$$ 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 How to solve these equations using an algebraic method? I need to show my working, don't you do something in reverse, like 7 multiplies by something. I haven't done it in class. $$\dfrac{5(3y-4)}{2y}=7$$ Multiply each side of the equation by $2y$ to get: $$\dfrac{5(3y-4)}{2y}=7 \iff 5(3y - 4) = 7\cdot 2y = 14 y$$ Now, distribute, and then gather "like terms", and simplify: $$ \begin{align} 5(3y - 4) = 14 y & \iff 15y - 20 = 14y \\ \\ & \iff 15y - 14y = 20 \\ \\ & \iff y = 20. \end{align} $$ $$\begin{equation} \frac{5(3y-4)}{2y}=7.\tag{0} \end{equation} $$ like 7 multiplies by something The given equation is only defined when the denominator of the left-hand side is different from $0$. So assume that $y\ne 0$. Then you can multiply both sides of equation $(0)$ by $2y$ to obtain an equivalent one, i.e. the new equation has the same solution as the original. $$\begin{eqnarray}\frac{5(3y-4)}{2y}\times 2y &=&7\times 2y \tag{1$\mathrm{a}$} \\\Leftrightarrow5(3y-4) &=&14y \tag{1$\mathrm{b}$} \\\Leftrightarrow15y-20 &=&14y,\qquad\text{after expanding the LHS}.\tag{1$\mathrm{c}$}\end{eqnarray}$$ You can subtract $14y$ from both sides. The new equation is equivalent to the previous one:$$\begin{eqnarray}15y-20-14y &=&14y-14y \tag{2$\mathrm{a}$}\\\Leftrightarrow y-20 &=&0.\tag{2$\mathrm{b}$}\end{eqnarray}$$ You can add $20$ to both sides. The new equation is equivalent to the previous one:$$\begin{eqnarray}y-20+20 &=&0+20 \tag{3$\mathrm{a}$}\\\Leftrightarrow y &=&20.\tag{3$\mathrm{b}$}\end{eqnarray}$$ Since the solution $y=20\neq 0$, the multiplication in 1 is valid. Comment. In general to get an equivalent equation one can: add or subtract the same value to and from both sides. Simplify either side according to the algebraic rules as in $(1\mathrm{b})$ to $(1\mathrm{c})$. We are given: $\dfrac{5(3y-4)}{2y}=7$ We must next eliminate the $2y$ from the denominator. We can do this by multiplying $2y$ by the $7$ on the other side of the equation. What we have now is a straightforward solve for y question. $\implies$ $5(3y-4)=7(2y)$ $\implies$ $15y-20=14y$ We must now isolate y. $\implies$ $15y-20=14y\implies15y-14y=20\implies y=20$ We can now check our solution to see if we are correct by plugging the value of y (which is 20) back into the initial equation. $\dfrac{5(3(20)-4)}{2(20)}=7$ $\dfrac{5(60-4)}{40}=7$ $\dfrac{280}{40}=7$ We find that 7 is indeed equal to 7 which proves our answer is correct. $7=7$
A body in free motion does not necessarily rotate about the center of mass. The center of mass might have straight linear motion in addition any rotation. The general motion is a screw motion with a rotation about some instantaneous axis and parallel translation at the same time. Consider an arbitrary body rotating by $\vec{\omega}$ and at some instant the center of mass (point C) has linear velocity $\vec{v}_C$. I can prove that there is always a point A where the linear velocity of the extended rigid body is only parallel to the rotation axis defined by $\vec{\omega}$. The combined motion is a rotation about A with a parallel translation of $\vec{v}_A = h \vec{\omega}$. The scalar $h$ is called pitch. If the body is purely rotating without translation then $h=0$ and if the body is purely translates then $h=\infty$ and $\\|\vec{\omega}\\|=0$. The motion of an arbitrary rigid motion is decomposed as such: Speed of rotation $$\omega = \\| \vec{\omega} \\|$$ Direction of rotation $$\hat{e} = \frac{\vec{\omega}}{\omega}$$ Location of rotation axis $$\vec{r}_A = \vec{r}_C + \frac{\vec{\omega}\times \vec{v}_C}{\omega^2}$$ Screw pitch $$h = \frac{\vec{\omega}\cdot\vec{v}_C}{\omega^2}$$ NOTES: $\cdot$ is the vector inner product, and $\times$ is the vector cross product. Proof The linear velocity at A is found by the frame transformation law $$\vec{v}_A = \vec{v}_C + \vec{\omega} \times (\vec{r}_A-\vec{r}_C)$$Using the location expression from above is $$ \vec{v}_A = \vec{v}_C + \frac{\vec{\omega} \times(\vec{\omega}\times \vec{v}_C)}{\omega^2}$$ Using the Vector Triple Product $$ \vec{v}_A = \vec{v}_C + \frac{\vec{\omega}(\vec{\omega}\cdot\vec{v}_C)-\vec{v}_C (\vec{\omega}\cdot\vec{\omega})}{\omega^2}$$ With the simplification that $(\vec{\omega}\cdot\vec{\omega}) = \omega^2$ and the definition for screw pitch $\vec{\omega}\cdot\vec{v}_C = h \omega^2$ the above is $$ \vec{v}_A = \vec{v}_C + \frac{\vec{\omega}(h \omega^2)-\vec{v}_C (\omega^2)}{\omega^2} = h \vec{\omega}$$ So the velocity at A is parallel to the rotation $\vec{\omega}$ Reverse Proof You can start from a general screw motion at a known point A, with direction $\hat{e}$, speed $\omega$ and pitch $h$ (two parameters for the point as the location along the line does not count, two parameters for the direction as the magnitude does not mater, one for the speed and one for the pitch equals six independent parameters that describe the motion. These will be transformed to the more familiar six motion parameters $\vec{\omega}$ and $\vec{v}_C$ below: Rotational Vector $$\vec{\omega} = \omega \hat{e}$$ Linear Vector $$\begin{align}\vec{v}_C &= \vec{v}_A - \vec{\omega} \times (\vec{r}_A-\vec{r}_C) \\& = h \vec{\omega} + (\vec{r}_A-\vec{r}_C) \times \vec{\omega}\end{align}$$ All six motion parameters are defined now at the center of mass. Sometimes the above is combined into one expression $$ \begin{bmatrix} \vec{v}_C \\ \vec{\omega} \end{bmatrix} = \omega \begin{bmatrix} h \hat{e} + \vec{r}\times \hat{e} \\ \hat{e} \end{bmatrix} $$ which clearly decomposes the motion into the magnitude (speed) in the first part, and the screw axis (geometry) in the second part. The two vectors defining the screw axis are called plücker line coordinates. Example A cylindrical body is rotating about its axis, and translating perpendicular to the axis (red vectors). This motion is described by a pure rotation about the screw axis (purple vectors).
The Annals of Applied Probability Ann. Appl. Probab. Volume 5, Number 1 (1995), 128-139. Minimal Positions in a Branching Random Walk Abstract We consider a branching random walk on the real line, with mean family size greater than 1. Let $B_n$ denote the minimal position of a member of the $n$th generation. It is known that (under a weak condition) there is a finite constant $\gamma$, defined in terms of the distributions specifying the process, such that as $n \rightarrow \infty$, we have $B_n = \gamma n + o(n)$ a.s. on the event $S$ of ultimate survival. Our results here show that (under appropriate conditions), on $S$ the random variable $B_n$ is strongly concentrated and the $o(n)$ error term may be replaced by $O(\log n)$. Article information Source Ann. Appl. Probab., Volume 5, Number 1 (1995), 128-139. Dates First available in Project Euclid: 19 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aoap/1177004832 Digital Object Identifier doi:10.1214/aoap/1177004832 Mathematical Reviews number (MathSciNet) MR1325045 Zentralblatt MATH identifier 0836.60089 JSTOR links.jstor.org Subjects Primary: 60J80: Branching processes (Galton-Watson, birth-and-death, etc.) Citation McDiarmid, Colin. Minimal Positions in a Branching Random Walk. Ann. Appl. Probab. 5 (1995), no. 1, 128--139. doi:10.1214/aoap/1177004832. https://projecteuclid.org/euclid.aoap/1177004832
Current browse context: math.AC Change to browse by: References & Citations Bookmark(what is this?) Mathematics > Commutative Algebra Title: Totalseparierte Moduln (Submitted on 1 Apr 2015) Abstract: Let $(R, \mathfrak{m})$ be a noetherian local ring, $M$ a separated $R$-module (i.e. $\bigcap\limits_{n\geq 1}\mathfrak{m}^n M = 0$) and $\widehat{M} = \lim\limits_{\leftarrow} M/\mathfrak{m}^n M$ its completion. Generally, $M$ is not pure in $\widehat{M}$ and $\widehat{M}$ is not pure-injective. But if $M$ is totally separated, i.e. $X\underset{R}{\otimes} M$ is separated for all finitely generated $R$-modules $X$, the situation improves: In this case, $M$ is pure in $\widehat{M}$ and, under additional conditions, $\widehat{M}$ is even pure-injective, e.g. if $M\cong X^{(I)}$ holds with $X$ finitely generated or $M \cong\coprod_{i=1}^{\infty} R/\mathfrak{m}^i$. In section 2, we investigate the question under which conditions both $M$ and $\widehat{M}$ are totally separated and establish a close connection to the class of strictly pure-essential extensions. In section 3, we replace the completion $\widehat{M}$ in the case $M = \coprod_{i\in I}M_i$ with the $\mathfrak{m}$-adic closure $A$ of $M$ in $P = \prod_{i\in I} M_i$, i.e. with $A = \bigcap_{n \geq 1}(M + \mathfrak{m}^n P)$. We give criteria so that $A/M$ is radical and show that this always holds in the countable case $M = \coprod_{i=1}^{\infty} M_i$. Finally, we deal with the case that $A$ is even totally separated and additionally determine the coassociated prime ideals of $A/M$. Submission historyFrom: Helmut Zoeschinger [view email] [v1]Wed, 1 Apr 2015 10:05:25 GMT (11kb)
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Definition of the Lyapunov Function A Lyapunov function is a scalar function defined on the phase space, which can be used to prove the stability of an equilibrium point. The Lyapunov function method is applied to study the stability of various differential equations and systems. Below, we restrict ourselves to the autonomous systems \[ {\mathbf{X’} = \mathbf{f}\left( \mathbf{X} \right)\;\;\text{or}\;\;}\kern-0.3pt {\frac{{d{x_i}}}{{dt}} = {f_i}\left( {{x_1},{x_2}, \ldots ,{x_n}} \right),\;\;}\kern-0.3pt {i = 1,2, \ldots ,n,} \] with the zero equilibrium $\mathbf{X} \equiv \mathbf{0}.$ We suppose that we are given a continuously differentiable function \[V\left( \mathbf{X} \right) = V\left( {{x_1},{x_2}, \ldots ,{x_n}} \right)\] in a neighborhood $U$ of the origin. Let $V\left( \mathbf{X} \right) \gt 0$ for all $\mathbf{X} \in U\backslash \left\{ \mathbf{0} \right\},$ and $V\left( \mathbf{0} \right) = 0$ in the origin. For example, these are functions of the form \[ {V\left( {{x_1},{x_2}} \right) = ax_1^2 + bx_2^2,\;\;}\kern-0.3pt {V\left( {{x_1},{x_2}} \right) = ax_1^2 + bx_2^4,\;\;}\kern-0.3pt {a,b \gt 0.} \] We find the total derivative of the function $V\left( \mathbf{X} \right)$ with respect to time $t:$ \[ {\frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial {x_1}}}\frac{{d{x_1}}}{{dt}} }+{ \frac{{\partial V}}{{\partial {x_2}}}\frac{{d{x_2}}}{{dt}} + \cdots } + {\frac{{\partial V}}{{\partial {x_n}}}\frac{{d{x_n}}}{{dt}}.} \] This expression can be written as a scalar (dot) product of two vectors: \[ {\frac{{dV}}{{dt}} = \left( {\text{grad}\,V,\frac{{d\mathbf{X}}}{{dt}}} \right),\;\;}\kern-0.3pt{\text{where}\;\;}\kern-0.3pt {V = \left( {\frac{{\partial V}}{{\partial {x_1}}},\frac{{\partial V}}{{\partial {x_2}}}, \ldots ,\frac{{\partial V}}{{\partial {x_n}}}} \right),}\;\; {\frac{{d\mathbf{X}}}{{dt}} = \left( {\frac{{d{x_1}}}{{dt}},\frac{{d{x_2}}}{{dt}}, \ldots ,\frac{{d{x_n}}}{{dt}}} \right).} \] Here, the first vector is the gradient of $V\left( \mathbf{X} \right),$ i.e. it’s always directed toward the greatest increase in $V\left( \mathbf{X} \right).$ Typically, the function $V\left( \mathbf{X} \right)$ increases with the distance from the origin, i.e. provided $\left\| \mathbf{X} \right\| \to \infty .$ The second vector in the scalar product is the velocity vector. At any point, it is tangent to the phase trajectory. Consider the case when the derivative of $V\left( \mathbf{X} \right)$ in a neighborhood $U$ of the origin is negative: \[{\frac{{dV}}{{dt}} \text{ = }}\kern0pt{\left( {\text{grad}\,V,\frac{{d\mathbf{X}}}{{dt}}} \right) }\lt{ 0.}\] This means that the angle $\varphi$ between the gradient vector and the velocity vector is greater than $90^\circ.$ For a function of two variables, it is shown schematically in Figures $1-2.$ Obviously, if the derivative $\large\frac{{dV}}{{dt}}\normalsize$ along a phase trajectory is everywhere negative, then the trajectory tends to the origin, i.e. the system is stable. Otherwise, when the derivative $\large\frac{{dV}}{{dt}}\normalsize$ is positive, the trajectory moves away from the origin, i.e. the system is unstable. We now turn to the strict formulation. Let a function $V\left( \mathbf{X} \right)$ be continuously differentiable in a neighborhood $U$ of the origin. The function $V\left( \mathbf{X} \right)$ is called the Lyapunov function for an autonomous system \[\mathbf{X’} = \mathbf{f}\left( \mathbf{X} \right),\] if the following conditions are met: $V\left( \mathbf{X} \right) \gt 0$ for all $\mathbf{X} \in U\backslash \left\{ \mathbf{0} \right\}$; $V\left( \mathbf{0} \right) = 0$; ${\large\frac{{dV}}{{dt}}\normalsize} \le 0$ for all $\mathbf{X} \in U$. Stability Theorems Theorem on stability in the sense of Lyapunov. If in a neighborhood $U$ of the zero solution $\mathbf{X} = \mathbf{0}$ of an autonomous system there is a Lyapunov function $V\left( \mathbf{X} \right),$ then the equilibrium point $\mathbf{X} = \mathbf{0}$ of the system is Lyapunov stable. Theorem on asymptotic stability. If in a neighborhood $U$ of the zero solution $\mathbf{X} = \mathbf{0}$ of an autonomous system there is a Lyapunov function $V\left( \mathbf{X} \right)$ with a negative definite derivative ${\large\frac{{dV}}{{dt}}\normalsize} \lt 0$ for all $\mathbf{X} \in U\backslash \left\{ \mathbf{0} \right\},$ then the equilibrium point $\mathbf{X} = \mathbf{0}$ of the system is asymptotically stable. As it can be seen, the total derivative ${\large\frac{{dV}}{{dt}}\normalsize}$ must be strictly negative (negative definite) in a neighborhood of the origin for the asymptotic stability of the zero solution. Instability Theorems Lyapunov instability theorem. Suppose that in a neighborhood $U$ of the zero solution $\mathbf{X} = \mathbf{0}$ there is a continuously differentiable function $V\left( \mathbf{X} \right)$ such that $V\left( \mathbf{0} \right) = 0$; ${\large\frac{{dV}}{{dt}}\normalsize} \gt 0$. If in the neighborhood $U$ there are points at which $V\left( \mathbf{X} \right) \gt 0,$ then the zero solution $\mathbf{X} = \mathbf{0}$ is unstable. Chetaev instability theorem. Suppose that in a neighborhood $U$ of the zero solution $\mathbf{X} = \mathbf{0}$ of an autonomous system there exists a continuously differentiable function $V\left( \mathbf{X} \right).$ Let the neighborhood $U$ contain a subdomain ${U_1},$ including the origin (Figure $3$) such that $V\left( \mathbf{X} \right) \gt 0$ for all $\mathbf{X} \in {U_1}\backslash \left\{ \mathbf{0} \right\}$; ${\large\frac{{dV}}{{dt}}\normalsize} \gt 0$ for all $\mathbf{X} \in {U_1}\backslash \left\{ \mathbf{0} \right\}$; $V\left( \mathbf{X} \right) = 0$ for all $\mathbf{X} \in \delta {U_1},$ where $\delta {U_1}$ denotes the boundary of the subdomain ${U_1}$. Then the zero solution $\mathbf{X} = \mathbf{0}$ of the system is unstable. In this case, the phase trajectories in the subdomain ${U_1}$ will move away from the origin. Thus, Lyapunov functions allow to determine the stability or instability of a system. The advantage of this method is that we do not need to know the actual solution $\mathbf{X}\left( t \right).$ In addition, this method allows to study the stability of equilibrium points of non-rough systems, for example, in the case when the equilibrium point is a center. The disadvantage is that there is no general method of constructing Lyapunov functions. In the particular case of homogeneous autonomous systems with constant coefficients, the Lyapunov function can be found as a quadratic form. Solved Problems Click a problem to see the solution.
Layer Sensitivity The You can specify relative errors in layer thicknesses on For each disturbed design, the variations $\Delta MF_i$ with respect to design target \[ \Delta MF_i=\|MF(d_1,...,d_i(1+\delta_{H,L}),...,d_m)-MF(d_1,...,d_m)\|\] or to theoretical spectral characteristic $S$ (Eq.(2)): \[ \Delta MF_i=\|S(d_1,...,d_i(1+\delta_{H,L}),...,d_m)-S(d_1,...,d_m)\|.\] Then, layer sensitivities $LS(i)$ are calculated and ranged from 0 to 100% (Fig. 2, column \[ LS(i)=\frac{\Delta MF_i}{\max_{i=1,...,m}\Delta MF_i}\cdot 100\%\] In the The design is obtained using the WDM option. Stack is a set of thick media separated by gaps. Each surface can be coated by a multilayer. In Figs. 6-8, an example of stack is presented. The stack consists of two substrates, BK7 and B270, separated by an air gap, the surfaces are coated with AR_1, AR2, AR3, and AR4 12-layer coatings. Layer materials are TiO Sensitivities of layers in different coatings are shown by different colors (Fig. 9). The sensitivities were calculated assuming 0.5%, 1%, and 2%, in SiO
18 0 Density?! Thinking of it as mass was only an analogy so that you could know to apply the moment of annulus formula, or parallel axis theorem, or whatever. In post #14 you had ##\sigma\omega\tau r^2B^2\int dA##. But the r should have been inside the integral: ##\sigma\omega\tau B^2\int r^2dA##. That integral is the second moment of area (like moment of inertia, but without the density factor), and in post #23 I gave you the solution for this when it is an annulus: had ##\sigma\omega\tau B^2\frac 12\pi(r_2^4-r_1^4)##. But your shape is only half an annulus, so ##\sigma\omega\tau B^2\frac 14\pi(r_2^4-r_1^4)##. Hihi, So sorry for the delay, I tried again with this new value. It's still giving me way too high values for the torque. I'm really stuck now. Everything in the equation looks fine but it's just the ## \sigma ## which drives up the torque numbers a lot.
berylium? really? okay then...toroidalet wrote:I Undertale hate it when people Emoji movie insert keywords so people will see their berylium page. A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content. 83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X When xq is in the middle of a different object's apgcode. "That's no ship!" Airy Clave White It Nay When you post something and someone else posts something unrelated and it goes to the next page. Also when people say that things that haven't happened to them trigger them. Also when people say that things that haven't happened to them trigger them. "Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life." -Terry Pratchett -Terry Pratchett Huh. I've never seen a c/posts spaceship before.drc wrote:"The speed is actually" posts Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace! It could be solved with a simple PM rather than an entire post.Gamedziner wrote:What's wrong with them?drc wrote:"The speed is actually" posts An exception is if it's contained within a significantly large post. I hate it when people post rule tables for non-totalistic rules. (Yes, I know some people are on mobile, but they can just generate them themselves. [citation needed]) "Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life." -Terry Pratchett -Terry Pratchett OK this is a very niche one that I hadn't remembered until a few hours ago. You know in some arcades they give you this string of cardboard tickets you can redeem for stuff, usually meant for kids. The tickets fold beautifully perfectly packed if you order them one right, one left - zigzagging. When people fold them randomly in any direction giving a clearly low density packing with loads of strain, I just think You know in some arcades they give you this string of cardboard tickets you can redeem for stuff, usually meant for kids. The tickets fold beautifully perfectly packed if you order them one right, one left - zigzagging. When people fold them randomly in any direction giving a clearly low density packing with loads of strain, I just think omg why on Earth would you do that?!Surely they'd have realised by now? It's not that crazy to realise? Surely there is a clear preference for having them well packed; nobody would prefer an unwieldy mess?! Also when I'm typing anything and I finish writing it and it just goes to the next line or just goes to the next page. Especially when the punctuation mark at the end brings the last word down one line. This also applies to writing in a notebook: I finish writing something but the very last thing goes to a new page. "Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life." -Terry Pratchett -Terry Pratchett 83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: ... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature. ON A DIFFERENT NOTE. When i want to rotate a hexagonal file but golly refuses because for some reason it calculates hexagonal patterns on a square grid and that really bugs me because if you want to show that something has six sides you don't show it with four and it makes more sense to have the grid be changed to hexagonal but I understand Von Neumann because no shape exists (that I know of) that has 4 corners and no edges but COME ON WHY?! WHY DO YOU REPRESENT HEXAGONS WITH SQUARES?! In all seriousness this bothers me and must be fixed or I will SINGLEHANDEDLY eat a universe. EDIT: possibly this one. EDIT 2: IT HAS BEGUN. HAS BEGUN. Last edited by 83bismuth38 on September 19th, 2017, 8:25 pm, edited 1 time in total. Actually, I don't remember who I was referencing, but I don't think it was you, and if it was, it wasn't personal.83bismuth38 wrote:... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature. x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$ http://conwaylife.com/wiki/A_for_all Aidan F. Pierce 83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: oh okay yeah of course sureA for awesome wrote:Actually, I don't remember who I was referencing, but I don't think it was you, and if it was, it wasn't personal.83bismuth38 wrote:... you were referencing me before i changed it, weren't you? because I had fit both of those.A for awesome wrote: When people put non- spectacularly-interesting patterns, questions, etc. in their signature. but really though, i wouldn't have cared. When someone gives a presentation to a bunch of people and you knowthat they're getting the facts wrong. Especially if this is during the Q&A section. "Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life." -Terry Pratchett -Terry Pratchett Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X When you watch a boring video in class but you understand it perfectly and then at the end your classmates dont get it so the teacher plays the borinh video again Airy Clave White It Nay 83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: when scientists decide to send a random guy into a black hole hovering directly above Earth for no reason at all. hit; that random guy was me. hit; that random guy was me. 83bismuth38 Posts:453 Joined:March 2nd, 2017, 4:23 pm Location:Still sitting around in Sagittarius A... Contact: When I see a "one-step" organic reaction that occurs in an exercise book for senior high school and simply takes place under "certain circumstance" like the one marked "?" here but fail to figure out how it works even if I have prepared for our provincial chemistry olympiadEDIT: In fact it's not that hard.Just do a Darzens reaction then hydrolysis and decarboxylate. Current status: outside the continent of cellular automata. Specifically, not on the plain of life. An awesome gun firing cool spaceships: An awesome gun firing cool spaceships: Code: Select all x = 3, y = 5, rule = B2kn3-ekq4i/S23ijkqr4eikry2bo$2o$o$obo$b2o! When there's a rule with a decently common puffer but it can't interact with itself "Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life." -Terry Pratchett -Terry Pratchett Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X When that oscillator is just When you're sooooooo close to a thing you consider amazing but miss... not sparky enough. When you're sooooooo close to a thing you consider amazing but miss... Airy Clave White It Nay People posting tons of "new" discoveries that have been known for decades, showing that they've not observed standard netiquette by reading the forums a while before posting, nor done the most minimal research about whether things have been already known, despit repeated posts about where to find such resources (e.g. jslife, wiki, Life lexicon, etc.). People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades. Posts where the quoted text is substantially longer than added text. Especially "me too" posts. People whose signatures are longer than the actual text of their posts. People whose signatures include graphics or pattern files, especially ones that are just human-readable text. Improper grammar, spelling, and punctuation (although I've gotten used to that; long-term use of the internet has made me rather fluent in typo, both reading and writing). Imperfect English is not unreasonable from people for whom English is not a primary language, but from English speakers, it is a symptom of sloppiness that can also manifest in other areas. People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades. Posts where the quoted text is substantially longer than added text. Especially "me too" posts. People whose signatures are longer than the actual text of their posts. People whose signatures include graphics or pattern files, especially ones that are just human-readable text. Improper grammar, spelling, and punctuation (although I've gotten used to that; long-term use of the internet has made me rather fluent in typo, both reading and writing). Imperfect English is not unreasonable from people for whom English is not a primary language, but from English speakers, it is a symptom of sloppiness that can also manifest in other areas. Posts:3138 Joined:June 19th, 2015, 8:50 pm Location:In the kingdom of Sultan Hamengkubuwono X That's G U S T A V O right theremniemiec wrote:People posting tons of "new" discoveries that have been known for decades, showing that they've not observed standard netiquette by reading the forums a while before posting, nor done the most minimal research about whether things have been already known, despit repeated posts about where to find such resources (e.g. jslife, wiki, Life lexicon, etc.). People posting tons of useless "new" discoveries that take longer to post than to find (e.g. "look what happens when I put this blinker next to this beehive"). Newbies with attitudes, who think they know more than people who have been part of the community for years or even decades. Also, when you walk into a wall slowly and carefully but you hit your teeth on the wall and it hurts so bad. Airy Clave White It Nay
Let $k\geq2$ be an integer, a graph $G=(V,E)$ is called $k$-partite if $V$ admits a partition into $k$ classes such that every edge of $G$ has its ends in different classes: vertices in the same class must not be adjacent. Now let $G=(V,E)$ be $k$-partite and $V_1,V_2,\cdots,V_k$ be the $k$ classes into which $V$ Is partitioned such that: $(1)\Delta (G)=2;$ $(2)\|V_i\|=3,i=1,2,\cdots,k;$ $(3)$For any $1\leq i\neq j\leq k$, there does not exist $u\in V_i$ and $v,w\in V_j$ such that $u$ are adjacent with both $u$ and $w$. I want to ask if I can choose $v_i$ from $V_i, i=1,2,\cdots,k$ such that $\{v_1,v_2,\cdots,v_k\}$ is an independent set in $G$.
DA recognizes some frequently used values of a multiply-and-accumulate operation, pre-computes these values, and stores them in a lookup table (LUT). Reading these stored values from ROM rather than calculating them leads to an efficient implementation. It should be noted that the DA method is applicable only to cases where the multiply-and-accumulate operation involves fixed coefficients. Distributed Arithmetic Consider calculating the following expression: $$y = \sum_{i=1}^{N} c_i x_i$$ where the $$c_i$$ coefficients are fixed-valued and $$x_i$$ represents the inputs of the multiply-and-accumulate operation. Assume that these inputs are in two’s complement format and are represented by b+1 bits. Moreover, assume that they are scaled and are less than 1 in magnitude. To keep things simple, we’ll consider the above equation for N=3. Hence, $$y = c_1 x_1 + c_2 x_2 + c_3 x_3$$ Equation 1 Equation 1 Since the inputs are in two’s complement format, we can write $$x_1 = -x_{1,0} + \sum_{j=1}^{b} x_{1,j} 2^{-j}$$ $$x_2 = -x_{2,0} + \sum_{j=1}^{b} x_{2,j} 2^{-j}$$ $$x_3 = -x_{3,0} + \sum_{j=1}^{b} x_{3,j} 2^{-j}$$ where $$x_{1,0}$$ is the sign bit of $$x_1$$ and $$x_{1,j}$$ is the jth bit to the right of the sign bit. The same notation is used for $$x_2$$ and $$x_3$$. If you need help with deriving these equations, read the section entitled “An Important Feature of the Two’s Complement Representation” in my article, "Multiplication Examples Using the Fixed-Point Representation" and note that we have assumed $$ \|x_i\| <1$$. Substituting our last three equations into Equation 1 gives $$\begin{align} y = &- x_{1,0} c_1 + x_{1,1} c_1 \times 2^{-1} + \dots + x_{1,b} c_1 \times 2^{-b} \\ &- x_{2,0} c_2 + x_{2,1} c_2 \times 2^{-1} + \dots + x_{2,b} c_2 \times 2^{-b} \\ &- x_{3,0} c_3 + x_{3,1} c_3 \times 2^{-1} + \dots + x_{3,b} c_3 \times 2^{-b} \end{align}$$ Equation 2 Equation 2 How can we use a LUT to efficiently implement these calculations? For now, let’s ignore the $$2^{-j}$$ terms of Equation 2 and look at this equation as a summation of some columns rather than a summation of some rows. For example, the second column of Equation 2 is $$y_1 = x_{1,1} c_1 + x_{2,1} c_2 + x_{3,1} c_3$$ How many distinct values are there for this expression? Note that $$x_{1,1}$$, $$x_{2,1}$$, and $$x_{3,1}$$ are one-bit values. Hence, $$y_1$$ can have only eight distinct values as given in Table 1 below: Table 1 Table 1 Ignoring the $$2^{-b}$$ term of the last column, we have $$y_b = x_{1,b} c_1 + x_{2,b} c_2 + x_{3,b} c_3$$ Again, we can only have the eight distinct values of Table 1. As you can see, the columns of Equation 2 involve calculating the function given by Table 1 (provided that we ignore the minus sign of the first column and the $$2^{-j}$$ terms). Instead of repeatedly calculating this function, we can pre-calculate the values of $$y_1$$ and store them in a LUT, as shown in the following block diagram: Figure 1 Figure 1 As shown in the figure, the jth bit of all the input signals, $$x_1$$, $$x_2$$, $$x_3$$, will be applied to the LUT, and the output will be $$y_j$$. The output of the ROM is represented by l bits. l must be large enough to store the values of Table 1 without overflow. Now that the LUT is responsible for producing the $$y_j$$ terms, we can rewrite Equation 2 as $$y = - y_0 + 2^{-1} y_1 + 2^{-2} y_2 + \dots + 2^{-b} y_b$$ Therefore, we need to take the $$2^{-j}$$ terms into account and note that the first term must be subtracted from the other terms. Let’s assume that we are using only five bits to represent the $$x_i$$ signals, i.e., $$b=4$$. Hence, $$y = - y_0 + 2^{-1} y_1 + 2^{-2} y_2 + 2^{-3} y_3 + 2^{-4} y_4$$ By repeatedly factoring $$2^{-1}$$, we can rewrite the above equation as $$y = - y_0 + 2^{-1} \Bigg ( y_1 + 2^{-1} \bigg ( y_2 + 2^{-1} \Big ( y_3 + 2^{-1} ( y_4 + 0 \big ) \Big ) \bigg ) \Bigg )$$ Note that a zero is added to the innermost parentheses to further clarify the pattern that exists. The multiply-and-add operation is now written as a repeated pattern consisting of a summation and a multiplication by $$2^{-1}$$. We know that multiplication by $$2^{-1}$$ can be implemented by a one-bit shift to the right. Therefore, we can use the ROM shown in Figure 1 along with a shift register and an adder/subtractor to implement the above equation. The simplified block diagram is shown in Figure 2. Figure 2 Figure 2 At the beginning of the calculations, the shift register SR is reset to zero and the other shift registers are loaded with the appropriate inputs. Then, the registers $$x_1$$, $$x_2$$, and $$x_3$$ apply $$x_{1,4}$$, $$x_{2,4}$$, and $$x_{3,4}$$ to the ROM. Hence, the adder will produce $$acc=a+b=y_4+0=y_4$$. This value will be stored in the SR, and a one-bit shift will be applied to take the $$2^{-1}$$ term into account. (As we’ll see in a minute, the output of the adder/subtractor will generate the final result of the algorithm by gradually accumulating the partial results. That’s why we’ve used “acc”, which stands for accumulator, to represent the output of the adder/subtractor.) So far, $$2^{-1}(y_4+0)$$ has been generated at the output of the SR register. Next, the input registers will apply $$x_{1,3}$$, $$x_{2,3}$$, and $$x_{3,3}$$ to the ROM. Hence, the adder will produce $$acc=a+b= y_3+2^{-1}(y_4+0)$$. Again, this value will be stored in the SR and a one-bit shift will be applied to take the $$2^{-1}$$ term into account, which gives $$2^{-1}(y_3+2^{-1}(y_4+0))$$. In a similar manner, the sum and shift operations will be repeated for the next terms, except that for the last term, the adder/subtractor will be in the subtract mode. Note that the number of shift-and-add operations in Figure 2 does not depend on the number of input signals N. The number of inputs affects only the size of the ROM’s address input. This is a great advantage of the DA technique over a conventional implementation of a multiply-and-add operation, i.e., an implementation in which partial products are generated and added together. However, a large N can lead to a slow ROM and reduce the efficiency of the technique. In the DA architecture, the number of shift-and-add operations depends on the number of bits used to represent the input signals, which in turn depends on the precision that the system requires. Conclusion DA recognizes some frequently used values of a multiply-and-accumulate operation, pre-computes these values and stores them in a lookup-table (LUT). Reading these stored values from ROM rather than calculating them leads to an efficient implementation. It should be noted that the DA method is applicable only to cases where the multiply-and-accumulate operation involves fixed coefficients.
CryptoDB Paper: Approx-SVP in Ideal Lattices with Pre-processing Authors: Alice Pellet-Mary Guillaume Hanrot Damien Stehlé Download: DOI: 10.1007/978-3-030-17656-3_24 Search ePrint Search Google Abstract: We describe an algorithm to solve the approximate Shortest Vector Problem for lattices corresponding to ideals of the ring of integers of an arbitrary number field K. This algorithm has a pre-processing phase, whose run-time is exponential in $$\log \|\varDelta \|$$log\|Δ\| with $$\varDelta $$Δ the discriminant of K. Importantly, this pre-processing phase depends only on K. The pre-processing phase outputs an “advice”, whose bit-size is no more than the run-time of the query phase. Given this advice, the query phase of the algorithm takes as input any ideal I of the ring of integers, and outputs an element of I which is at most $$\exp (\widetilde{O}((\log \|\varDelta \|)^{\alpha +1}/n))$$exp(O~((log\|Δ\|)α+1/n)) times longer than a shortest non-zero element of I (with respect to the Euclidean norm of its canonical embedding). This query phase runs in time and space $$\exp (\widetilde{O}( (\log \|\varDelta \|)^{\max (2/3, 1-2\alpha )}))$$exp(O~((log\|Δ\|)max(2/3,1-2α))) in the classical setting, and$$\exp (\widetilde{O}((\log \|\varDelta \|)^{1-2\alpha }))$$exp(O~((log\|Δ\|)1-2α)) in the quantum setting. The parameter $$\alpha $$α can be chosen arbitrarily in [0, 1 / 2]. Both correctness and cost analyses rely on heuristic assumptions, whose validity is consistent with experiments.The algorithm builds upon the algorithms from Cramer et al. [EUROCRYPT 2016] and Cramer et al. [EUROCRYPT 2017]. It relies on the framework from Buchmann [Séminaire de théorie des nombres 1990], which allows to merge them and to extend their applicability from prime-power cyclotomic fields to all number fields. The cost improvements are obtained by allowing precomputations that depend on the field only. Video from EUROCRYPT 2019 BibTeX @article{eurocrypt-2019-29376, title={Approx-SVP in Ideal Lattices with Pre-processing}, booktitle={Advances in Cryptology – EUROCRYPT 2019}, series={Advances in Cryptology – EUROCRYPT 2019}, publisher={Springer}, volume={11477}, pages={685-716}, doi={10.1007/978-3-030-17656-3_24}, author={Alice Pellet-Mary and Guillaume Hanrot and Damien Stehlé}, year=2019 }
Solutions Solubility and Henry's Law, Ideal and Non-Ideal Solutions, Vapour Pressure and Raoult's Law Solubility & Henry's law: Solution of gas in liquid:- (1) In these solutions gas is solute and liquid is solvent (2) The solubility of a gas in a liquid is influenced by the following factors. (a) Nature of the gas: Easily liquifiable gases (Ex; CO 2, SO 2, HCl, NH 3 etc.,) are more soluble in a solvent when compared with permanent gases, (He, N 2, H 2, O 2) (b) Nature of liquid : A gas more soluble in a liquid in which it undergoes either dissociation (or) association. Ex:- HCl is more soluble in H 2O than in C 6H 6 (C)Temperature:- Here forward reaction is (dissolution of a gas in liquid) exothermic, According to Le Chatelier principle low 'T' are favourable (d)Pressure:- In Forward reaction number of moles of gas decreases (2) According to Le chatelier principle higher "P" are favourable (3) Effect of pressure on solubility of a gas in a liquid can be explained by Henry's law (4) Henry's law: (a)Definition 1:- At a particular "T" for a given volume of a liquid weight of gas dissolved is directly proportional to it's pressure w ∝ P ⇒ w = k HP K H = Henry's constant w = wt of gas dissolved P = pressure of the gas (b)Definition (2):- At a particular "T" for a given volume of a liquid the partial pressure of a gas over a liquid is directly proportional to it's mole fraction in the liquid P ∝ X (gas) P i = K H X gas P i = Partial pressure of the gas over the liquid K H = Henry's constant X gas mole fraction of the gas in liquid Trick:- (a) According to above Definition "2" higher the K H for a particular gas lesser it's solubility (b) (C) Definition (3):- At a particular T for a given volume of liquid solubility (S) of a gas is directly proportional to partial pressure of gas over the liquid S ∝ P Solubility can be considered as any concentration term like molarity, molality, mole fraction. Trick:- Selection of formula (1) or (2) depends up on units of "P" (i) If units of K H = units of "P" we need to consider formula (2) (ii) if units of K H = (units of P) −1 we need to consider formula (1) (5) Limitations of Henry's Law:- (1) Temperature should not be too low (2) Pressure should not be too high (3) Gas should not undergo either dissociation (or) association in the liquid. (4) Gas should not react with liquid. Vapour pressure and Raoult's law:- (1) Vapour pressure:- The pressure exerted by a vapour of a liquid over a liquid when it is in equilibrium with the liquid at a particular "T" is called vapour pressure. (2) Vaporization depends up on (a) Nature of liquid (b) Surface area of liquid (It is directly proportional) (c) Air current over the liquid (d) Temperature. (3) But vapour pressure depends upon only temperature \tt P = K e^{\frac{-\Delta H_{vap}}{RJ}} (4) (5) \log \frac{P_{2}}{P_{1}} = \frac{\Delta H_{vap}}{2.303 R}\left[\frac{1}{T_{1}} - \frac{1}{T_{2}}\right] ΔHvap = Enthalpy of vapourisation R = Universal gas constant. (6) Raoult's law for liquid- liquid Binary solution:- (1) Definition:- For a solution of two/more volatile liquids partial vapour pressure of any liquid in the solution is equal to product of mole fraction of the liquid in the solution and vapour pressure of that pure liquid. (2) Let us assume a solution is formed by the combination of two volatile liquids A & B \tt P_{A} = X_{A}P^{o}_{A} ; P_{B} = X_{B}P^{o}_{B} ∴P S = P A + P B \tt \Rightarrow P_{S} = X_{A}P^{o}_{A} + X_{B}P^{o}_{B} Where P A & P B are partial vapour pressures of liquids A & B X A & X B are mole fractions of liquids A & B \tt P^{o}_{A} \ \& \ P^{o}_{B} are vapour pressure of pure liquids A & B P S is total vapour pressure of solution. \tt \therefore P_{S} = (P^{o}_{A} - P^{o}_{B}) X_{A} + P^{o}_{B} (or) \tt P_{S} = (P^{o}_{B} - P^{o}_{A}) X_{B} + P^{o}_{A} (3) Mole fraction of A & B in liquid & vapour state:- According to Dalton's law of partial pressure \tt P^{o}_{i} = X_{i}P P A = Y A P S Y A = Mole fraction of A in vapour phase \tt \Rightarrow Y_{A} = \frac{P_{A}}{P_{S}} \tt \Rightarrow Y_{A} = \frac{X_{A}P^{o}_{A}}{X_{A}P^{o}_{A} + X_{B}P^{o}_{B}} similarly \tt Y_{B} = \frac{X_{B}P^{o}_{B}}{X_{A}P^{o}_{A} + X_{B}P^{o}_{B}} Ideal and non ideal solutions:- (1) Based on Raoult's law solutions are classified into 2 types (a) Ideal solution :- The solution which obeys Raoult's law at all temperature and concentration is called an ideal solution (b) Non-ideal solution :- The solution which does not obey Raoult's law at all temperatures and concentrations is called a non-ideal solution. (2) Non Ideal solutions are again classified into 2 types. (a) Non-Ideal solution which shows positive deviation from Raoult's law (Type - I) (b) Non-Ideal solution which shows negative deviation from Raoult's law (Type-II) (3) Comparison between ideal & non ideal solution can be shown as follows Ideal Solution Type- I solution Type-II solution (1) \tt P_{A} = X_{A}P^{o}_{A} \tt P_{A} > X_{A}P^{o}_{A} \tt P_{A} < X_{A}P^{o}_{A} (2) \tt P_{B} = X_{B}P^{o}_{B} \tt P_{B} > X_{B}P^{o}_{B} \tt P_{B} < X_{B}P^{o}_{B} (3) \tt P_{S} = X_{A}P^{o}_{A} + X_{B}P^{o}_{B} \tt P_{S} > X_{A}P^{o}_{A} + X_{B}P^{o}_{B} \tt P_{S} < X_{A}P^{o}_{A} + X_{B}P^{o}_{B} (4) \tt \Delta H_{mix} \simeq 0 Δ H mix > 0 (positive) Endothermic ΔHmix < 0 (negative) exothermic (5) \tt \Delta V_{mix} \simeq 0 ΔV mix > 0 ΔV mix < 0 (6) A-B attractions are average of A-A & B-B attractions A-B attractions are weaker when compared with A-A & B-B attractions A-B attractions are stronger when compared with A-A & B-B attractions (7) Fractional distillation:- (1) The process of separating component liquids from a solution by using the concept of distillation is called fractional distillation. (2) During fractional distillation compositions are changed (3) Fractional distillation is possible at a temperature at which compositions of liquids & vapour phases are different. (4) Azeotropic mixture:- (1) An azeotropic mixture behaves as a pure liquid. (2) A solution of two or more liquid which boils at a particular temperature and distills over completely at that particular temperature without under going any change in composition is called azeotropic mixture. (3) Azeotrope with maximum vapour pressure and minimum boiling point is formed by the solution which shows positive deviation from Raoult's law Ex:- 95.57% by mass of Et-OH (4) Azeotrope with minimum vapour pressure and maximum boiling point is formed by the solution which shows negative deviation from Raoult's law Ex:- 6.8% by mass of HNO 3. Trick for completely immiscible liquids: For completely immiscible liquids (1) \tt P_{A}^{o} = Y_{A}P_{mix} \tt P_{B}^{o} = Y_{B}P_{mix} 2) \tt \frac{P_{A}^{o}}{P_{B}^{o}} = \frac{Y_{A}}{Y_{B}} 3) \tt \frac{P_{A}^{o}}{P_{B}^{o}} = \left(\frac{n_{A}}{n_{B}}\right)_{vapour phase} 4) \tt \left(\frac{w_{A}}{w_{B}}\right)_{vapour} = \frac{P_{A}^{0}}{P_{B}^{0}} \times \frac{M_{A}}{M_{B}} Part1: View the Topic in this Video from 0:27 to 59:58 Part2: View the Topic in this Video from 0:07 to 19:05 Part3: View the Topic in this Video from 0:09 to 11:07 Part4: View the Topic in this Video from 0:09 to 8:17 Part5: View the Topic in this Video from 0:07 to 15:38 Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers. 1. Henry's Law \tt p=\left(\frac{1}{k'}\right)x_{2}=k_{H}x_{2} where k H is known as Henry's law constant. 2. Raoult's Law \tt p_{total}=p_{A}+p_{B}=x_{A}\ p_{A}^{\star}+x_{B}\ p_{B}^{\star}
Let me answer the second query. We know $$\mathbf{ B}=\text{curl}\;\mathbf{ A} $$ where $\bf B$ is the magnetic field & $\bf A$ is vector-potential. Now, using Maxwell's equation, we get $$\text{curl}\;(\text{curl}\; \mathbf A)= \mu_0 \mathbf J . $$ By cracking a bit-algebra, we get, $$-\frac{\partial^2 A_x}{\partial x^2}-\frac{\partial^2 A_x}{\partial y^2}- \frac{\partial^2 A_x}{\partial z^2} +\frac{\partial}{\partial x}\left(\frac{\partial A_x}{\partial x}+ \frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)= \mu_0 J_x.$$ Now, for convenience, we take $\text{div}\; \mathbf A= 0\; .$ This makes the above relation looks like $$-\frac{\partial^2 A_x}{\partial x^2}-\frac{\partial^2 A_x}{\partial y^2}- \frac{\partial^2 A_x}{\partial z^2} = \mu_0 J_x\;.$$ The vector potential at $(x_1,y_1,z_1)$ is then given by $$\mathbf A(x_1,y_1,z_1)= \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(x_2,y_2,z_2)\;\mathrm dv_2}{r_{12}}\;.$$ Now, consider a loop of wire carrying current $I$. Now, $\mathrm{d} v_2= a\; \mathrm{d} l$ where $\mathrm dl$ is an infinitesimal section of the wire; $\mathbf{J} \; \mathrm{d} v_2= I\; \mathrm{d}\mathbf{l} .$ Therefore our vector-potential for the thin-wire carrying steady current is given by $$\mathbf A= \frac{\mu_0 I}{4\pi}\int \frac{\mathrm d\mathbf l}{r_{12}}.$$ Let us focus on that section of wire which happens to let the current in the $\hat{\mathbf x}$ direction & is located at the origin of our frame. Then at a certain point $(x,y)$ in $xy$ plane, contribution to the vector-potential from the infinitesimal wire-section at the origin is given by $$\mathrm{d}\mathbf A= \hat{\mathbf x} \;\frac{\mu_0 I}{4\pi} \frac{\mathrm{d}\mathbf l}{\sqrt{x^2 + y^2}} \;.$$ Since, $\mathbf A$ is in $xy$ plane, its curl must point in the $\hat{\mathbf z}$ direction. therefore, \begin{align}\mathrm{d}\mathbf B &= \text{curl}\;\mathrm{d}\mathbf A\\& =\hat{\mathbf z}\left(-\frac{\partial A_x}{\partial y}\right)\\ &= \hat{\mathbf z}\;\frac{\mu_0 I}{4\pi} \frac{ \mathrm{d} l\; y}{(x^2 + y^2)^{3/2}}\\ &= \hat{\mathbf z}\;\frac{\mu_0 I}{4\pi} \frac{\mathrm{d}l\; \sin\varphi}{r^2}\;. \end{align} Now, that being deduced, we can say that this would be valid for any general coordinate system; all that matters is the relative orientation of the element $d\mathbf l$, radius vector $\bf r$ from the element to the concerned point. The contribution to the magnetic field from any element of wire can be taken to be a vector perpendicular to the plane containing $\mathrm{d} \mathbf l$ & $\bf r$ & the angle between them namely $\varphi\; .$ This can be compactly written as $${\mathrm{d}\mathbf B= \frac{\mu_0}{4\pi} \frac{I\; \mathrm{d}\mathbf l \times \hat{ \mathbf r}}{r^2}}\; .$$ And this is Biot-Savart Law, which you wanted.
Period is independent of amplitude. (Vias.org) But given that, Simple harmonic motion can be defined by $$x = A * \sin(\omega t) \tag{1}$$ where $A$ is the amplitude of oscillation, $\omega$ the angular velocity, $t$ the time, and $x$ the displacement from the mean position And, $$T = 2 \pi/\omega = 1/f \tag{2}$$ where, $T$ is the period of motion and $f$ is the frequency of oscillation. Equation (1) can be rearranged to give \begin{align} x &= A \sin (\omega t)\\ \frac{x}{A} &= \sin(\omega ωt) \\ \arcsin\left(\frac{x}{A}\right) &= \omega t\\ \omega &=\frac{\arcsin\left(\frac{x}{A}\right)}{t} \end{align*} Subbing this into (2) gives the following relationship between $T$ and $A$ $$T = \frac{2 \pi t}{\arcsin(x/A)} = \frac{1}{f}$$ Doesn't the fact that both $T$ and $A$ appear in the above equation show that $T$(period) is dependent on $A$(amplitude) ? FYI: Although a physical explanation may be useful, I am particularity interested in why deriving a relationship between $T$ and $A$ doesn't mean that they are dependent. Note there is a similar question here but that is concerned with the physics of the phenomena, and not why the maths can't be used to solve it. This is because I have this exam where we are given a stimulus and based on that stimulus alone are meant to answer questions (i.e. the exam is expected to contain material/principles that we haven't been exposed to before, but should be able to answer given the stimulus). And as I didn't know much about simple harmonic motion, my initial reaction was to see if the formulas link $T$ and $A$.
1IntroductionIn the theory of ordinary differential equations and in particular in the theory of Hamiltonian systems the existence of first integrals is important, because they allow to lower the dimension where the Hamiltonian system is defined. Furthermore, if we know a sufficient number of first integrals, these allow to solve the Hamiltonian system explicitly, and we say that the system is integrable. Almost until the end of the 19th century the major part of mathematicians and physicians believe that the equations of classical mechanics were graph for every independent set I of G , a graph is a fractional independent-set-deletable ( a,b,m )-deleted graph (shortly, fractional ID-( a,b,m )-deleted graph). If a = b = k , then a fractional ID-( a,b,m )-deleted graph is a fractional ID-( k,m )-deleted graph. If m = 0, then a fractional ID-( a,b,m )-deleted graph is just a fractional ID-( a,b )-factor-critical graph.If G has a fractional ( g, f )-factor containing a Hamiltonian cycle, it is said that G includes a Hamiltonian fractional ( g, f )-factor. A graph G is called an ID-Hamiltonian 1IntroductionThe use of computer algebra systems for normal forms computations is considered at present a routine operation. As a general reference see e.g. Sanders et al . [ 36 ] and Meyer et al . [ 32 ]. Nevertheless when we deal with special classes of differential equations, like Poisson or Hamiltonians systems which is our case, it is advisable to employ specific transformations as well as tailored variables for those problems [ 32 ], mostly connected with the symmetries that those systems might possess. More precisely we are interested in the longest path or cycle is required the problem is closely related to well-known hamiltonian problems in graph theory. In the rest of this paper, we will use standard terminology in graphs(see ref.[ 2 ]). It is very difficult to determine that a graph is hamiltonian or not. Readers may refer to [ 4 , 5 , 6 ].2 Definitions and NotationWe follow [ 2 ] for graph-theoretical terminology and notation not defined here. A graph G = ( V,E ) always means a simple graph(without loops and multiple edges), where V = V ( G ) is the vertex set and E = E ( G 1IntroductionThe rotation of a triaxial rigid body in the absence of external torques is known to be integrable [ 1 , 2 ]. In particular, the canonical transformation to Andoyer variables [ 3 ] reduces the free rigid body rotation to an integrable, one degree of freedom Hamiltonian, which immediately shows the preservation of the total angular momentum and allows for the representation of the possible solutions by contour plots of the reduced Hamiltonian [ 4 ]. However, because the solution to the torque-free motion depends on elliptical integrals and coupling with the Korteweg-de Vries equation, which is associated with non-semisimple matrix Lie algebras. In the references [ 10 ] and [ 11 ], its Lax pair and bi-Hamiltonian formulation were presented respectively. It should be noted that its bi-Hamiltonian structure is the first example of local bi-Hamiltonian structures, which lead to hereditary recursion operators in (2+1)-dimensions.Several methods have been developed to find exact solutions of the NLPDEs. Some of these are the homogeneous balance method [ 12 ], the ansatz method [ 13 ], the inverse scattering Poincaré and Arnold, we split the Hamiltonian into two terms:H = H 0 + H 1 ,$$\begin{array}{}\displaystyle{\cal H} = {\cal H}_0 + {\cal H}_1,\end{array}$$where the intermediary 𝓗 0 defines a non-degenerate and simplified model of the problem at hand, which includes the Kepler and free rigid-body as particular cases and 𝓗 1 is usually dubbed as the perturbation. A special realization of an intermediary occurs for the case in which it is an integrable 1-DOF system. The work of Hill on the Moon motion [ 32 ] is, perhaps, the best known example. The 1IntroductionFor motivation and background to this work see [ 1 ]. In this paper, we consider only finite and simple graphs. Let G = ( V ( G ) , E ( G )) be a graph, where V ( G ) denotes its vertex set and E ( G ) denotes its edge set. A graph is Hamiltonian if it admits a Hamiltonian cycle. For each x ∊ V ( G ), the neighborhood N G ( x ) of x is the set of vertices of G adjacent to x , and the degree d G ( x ) of x is \| N G ( x )\|. For S ⊆ V ( G ), we write N G ( S ) = ∪ x∊S N G ( x ). G [ S ] denotes the subgraph of G , and we denote it by div( x , y ), as the functiondiv ( x , y ) = ∂ X ∂ x ( x , y ) + ∂ Y ∂ y ( x , y ) . $$\begin{array}{}\displaystyle{\rm div} (x,y) \, = \, \frac{\partial X}{\partial x}(x,y) \, + \,\frac{\partial Y}{\partial y}(x,y).\end{array}$$System 1 is said to be Hamiltonian if div( x , y ) ≡ 0. In such a case there exists a neighborhood of the origin U and an analytic function H : U ⊆ ℝ 2 → ℝ, called the Hamiltonian, such thatX ( x , y ) = − ∂ H ∂ y and Y ( x , y ) = ∂ H ∂ y . $$\begin{array}{}\displaystyleX(x,y) = - \frac –body problem Physica D 238 2009 563 571 10.1016/j.physd.2008.12.014[26] Llibre, J., Moeckel, R. and Simó, C., Central configurations, periodic orbits and Hamiltonian systems , Advances Courses in Math., CRM Barcelona, Birhauser, 2015. Llibre J. Moeckel R. Simó C. Central configurations, periodic orbits and Hamiltonian systems Advances Courses in Math., CRM Barcelona Birhauser 2015[27] Long, Y. and Sun, S., Four–Body Central Configurations with some Equal Masses , Arch. Rational Mech. Anal. 162 (2002), 24–44. doi 10.1007/s002050100183 Long Y. Sun S. Four–Body Central
In this section we derive how the gas pressure $P$ depends on the height over sea level $h$ in the gravitational field of Earth. If we take an arbitrary gas column with intersection area $S$ and height $h,$ then the weight of this column is given by \[{F = mg = \rho gV }={ \rho ghS,}\] where $\rho$ is the gas density. Then the gas pressure is expressed by the following formula: \[\require{cancel} {P = \frac{F}{S} = \frac{{\rho gh\cancel{S}}}{\cancel{S}} }={ \rho gh.} \] Now imagine such a column in the atmosphere and separate a thin layer of air with the height $dh$ (Figure $1$). It’s clear that such a layer causes the pressure change by the value of \[dP = – \rho gdh.\] We have put the minus sign because the pressure must decrease as the altitude increases. Considering atmospheric air as an ideal gas, we can use the ideal gas law to express the density $\rho$ through pressure $P:$ \[ {PV = \frac{m}{M}RT,\;\;}\Rightarrow {P = \frac{m}{{VM}}RT = \frac{\rho }{M}RT.} \] Here $T$ is the absolute temperature, $R$ is the universal gas constant equal to $8.314\,{\large\frac{\text{J}}{K \cdot \text{mol}}\normalsize},$ $M$ is the molar mass, which is for air equal to $0.029\,{\large\frac{\text{kg}}{\text{mol}}\normalsize}.$ It follows from here that the density is given by the formula \[\rho = \frac{{MP}}{{RT}}.\] Putting this into the differential relation for $dP$ gives: \[ {dP = – \rho gdh }={ – \frac{{MP}}{{RT}}gdh,\;\;}\Rightarrow {\frac{{dP}}{P} = – \frac{{Mg}}{{RT}}dh.} \] We obtain a differential equation describing the gas pressure $P$ as a function of the altitude $h.$ Integrating gives the equation: \[ {\int {\frac{{dP}}{P}} = – \int {\frac{{Mg}}{{RT}}dh} ,\;\;}\Rightarrow {\ln P = – \frac{{Mg}}{{RT}}h }+{ \ln C.} \] Getting rid of the logarithms, we obtain the so-called barometric formula \[P = C\exp \left( { – \frac{{Mg}}{{RT}}h} \right).\] The constant of integration $C$ can be determined from the initial condition $P\left( {h = 0} \right) $ $= {P_0},$ where ${P_0}$ is the average sea level atmospheric pressure. Thus, dependency of the barometric pressure on the altitude is given by the formula \[P = {P_0}\exp \left( { – \frac{{Mg}}{{RT}}h} \right).\] Substituting the known constant values (see Figure $2$), we find that the dependence $P\left( h \right)$ (in kilopascals) is expressed by the formula: \[ {P\left( h \right) }={ 101.325 \cdot}\kern0pt{ \exp \left( { – \frac{{0.02896 \cdot 9.807}}{{8.3143 \cdot 288.15}}h} \right) } = {101.325\exp \left( { – 0.00012\,h} \right)\;}\kern-0.3pt{\left[\text{kPa} \right],} \] where the height $h$ above sea level is expressed in meters. If the pressure is given in millimeters of mercury $\left( \text{mmHg} \right),$ the barometric formula is written in the form: \[{P\left( h \right) }={ 760\exp \left( { – 0.00012\,h} \right)\;}\kern-0.3pt{\left[ \text{mmHg} \right].}\] In case when the height $h$ is given in feet, and pressure in inches of mercury $\left( \text{inHg} \right),$ this formula is written as \[{P\left( h \right) }={ 29.92\exp \left( { – 0.00039\,h} \right)\;}\kern-0.3pt{\left[ \text{inHg} \right].}\] The barometric formula is often used for estimating the air pressure under different conditions, although it gives slightly higher values compared with the real ones. Solved Problems Click a problem to see the solution. Example 1Determine at what altitude the air pressure is twice less than on the sea level? Example 2Find the air pressure in a mine at a depth of $1\,\text{km}$ at the temperature of $40$ degrees Celsius. Example 1.Determine at what altitude the air pressure is twice less than on the sea level? Solution. To estimate the altitude, we use the barometric formula \[{P\left( h \right) }={ {P_0}\exp \left( { – 0.00012\,h} \right).}\] When $h = 0$, the pressure $P\left( h \right)$ is equal to the average atmospheric sea level pressure ${P_0}.$ At a certain altitude $H$, the pressure is twice less: \[{P\left( H \right) = \frac{{{P_0}}}{2} }={ {P_0}\exp \left( { – 0.00012\,H} \right).}\] It follows from here that \[\exp \left( { – 0.00012\,H} \right) = \frac{1}{2}.\] Taking logarithms of both sides, we find the altitude $H:$ \[ {\ln \frac{1}{2} = – 0.00012\,H,\;\;}\Rightarrow {\ln 2 = 0.00012\,H,\;\;}\Rightarrow {H = \frac{{\ln 2}}{{0.00012}} \approx 5780\,\text{m}.} \] Example 2.Find the air pressure in a mine at a depth of $1\,\text{km}$ at the temperature of $40$ degrees Celsius. Solution. The air pressure in the mine can be estimated using the general barometric formula: \[P = {P_0}\exp \left( { – \frac{{Mg}}{{RT}}h} \right).\] We substitute the following values into the formula: $h = – 1000\,\text{m}$ (the sign is minus because the mine is under sea level); $T = 40 + 273.15 $ $= 313.15\,\text{K}.$ The remaining parameters are standard: $M = 0.02896\,\large\frac{\text{kg}}{\text{mol}}\normalsize,$ $R = 8.3143\,\large\frac{\text{N}\cdot\text{m}}{\text{mol}\cdot\text{K}}\normalsize,$ $g = 9.807\,\large\frac{\text{m}}{\text{s}^2}\normalsize.$ After simple calculations we find: \[ {P = {P_0}\exp \left( { – \frac{{Mg}}{{RT}}h} \right) } = {{P_0}\exp \left[ { – \frac{{0.02896 \cdot 9.807}}{{8.3143 \cdot 313.15}}\left( { – 1000} \right)} \right] } \approx {{P_0}\exp \left( {0.109} \right) } \approx {1.115{P_0}.} \] Since the atmospheric sea level pressure is ${P_0} = 760\;\text{mmHg},$ the air pressure in the mine will be $848\;\text{mmHg},$ that is about $12$ percent greater than on the sea level.
A railway train trundles towards the east at speed $ \nu_{1}$, and a passenger strolls towards the front at speed $ \nu_{2}$. What is the speed of the passenger relative to the railway station? We might at first be tempted to reply: “Why, $ \nu_{1}+\nu_{2}$ of course.” In this section we shall show that the answer as predicted from the Lorentz transformations is a little less than this, and we shall develop a formula for calculating it. We have already discussed (in Section 15.6) our answer to the objection that this defies common sense. We pointed out there that the answer (to the perfectly reasonable objection) that “at the speeds we are accustomed to we would hardly notice the difference” is not a satisfactory response. The reason that the resultant speed is a little less than $ \nu_{1}+\nu_{2}$ results from the way in which we have defined the Lorentz transformations between references frames and the way in which distances and time intervals are defined with reference to reference frames in uniform relative motion Figure XV.17 shows two references frames, $ \Sigma$ and $ \Sigma'$, the latter moving at speed $ \nu$ with respect to the former. A particle is moving with velocity $ \bf{u'}$ in $ \Sigma'$, with components $ u'_{x'}$ and $ u'_{y'}$. (“ in $ \Sigma'$ ” = “referred to the reference frame $ \Sigma'$”.) What is the velocity of the particle in $ \Sigma$? Let us start with the $ x$-component. We have: \[ u=\dfrac{dx}{dt}=\dfrac{\left(\dfrac{\partial x}{\partial x'}\right)_{t'}dx'+\left(\dfrac{\partial x}{\partial t'}\right)_{x'}dt'}{\left(\dfrac{\partial t}{\partial x'}\right)_{t'}dx'+\left(\dfrac{\partial t}{\partial t'}\right)_{x'}dt'}=\dfrac{\left(\dfrac{\partial x}{\partial x'}\right)_{t'}u'+\left(\dfrac{\partial x}{\partial t'}\right)_{x'}}{\left(\dfrac{\partial t}{\partial x'}\right)_{t'}u'+\left(\dfrac{\partial t}{\partial t'}\right)_{x'}} \label{15.16.1}\] We take the derivatives from Equations 15.15.3a,b,c,d, and, writing $ \dfrac{\nu}{c}$ for $ \beta$, we obtain \[ u_{x}=\dfrac{u'_{x}+\nu}{1+u'_{x'}\dfrac{\nu}{c^{2}}}. \label{15.16.2}\] The inverse is obtained by interchanging the primed and unprimed symbols and reversing the sign of $ \nu$. The $ y$-component is found in an exactly similar manner, and I leave its derivation to the reader. The result is \[ u_{y}=\dfrac{u'+\nu}{1+u'\dfrac{\nu}{c^{2}}} \label{15.16.3}\] Special cases: If $ u'_{x'}=u'$ and $ u'_{y'}=0$, then \[ u_{x}=\dfrac{u'+\nu}{1+u'\dfrac{\nu}{c^{2}}} \label{15.16.4a}\tag{15.16.4a}\] \[ u_{y}=0 \label{15.16.4b}\tag{15.16.4b}\] If $ u'_{x'}=0$ and $ u'_{y'}=u'$ then \[ u_{x}=\nu \label{15.16.5a}\tag{15.16.5a}\] \[ u_{y}=\dfrac{u'}{\gamma} \label{15.16.5b}\tag{15.16.5b}\] Equation $ \ref{15.16.4a}$ as written is not easy to commit to memory, though it is rather easier if we write $ \beta_{1}=\dfrac{\nu}{c},\ \beta_{2}=\dfrac{u'}{c}$ and $ \beta=\dfrac{u_{x}}{c}$. Then the equation becomes \[ \beta=\dfrac{\beta_{1}+\beta_{2}}{1+\beta_{1}\beta_{2}} \label{15.16.6}\] In Figure XV.18, a train $ \Sigma'$ is trundling with speed $ \beta_{1}$ (times the speed of light) towards the right, and a passenger is strolling towards the front at speed $ \beta_{2}$. The speed $ \beta$ of the passenger relative to the station $ \Sigma$ is then given by Equation $ \ref{15.16.6}$. In Figure XV.19, two trains, one moving at speed $ \beta_{1}$ and the other moving at speed $ \beta_{2}$, are moving towards each other. (If you prefer to think of protons rather than trains, that is fine.) Again, the relative speed b of one train relative to the other is given by Equation $ \ref{15.16.6}$. Example $\PageIndex{1}$ A train trundles to the right at 90% of the speed of light relative to $ \Sigma$, and a passenger strolls to the right at 15% of the speed of light relative to $ \Sigma '$. The speed of the passenger relative to $ \Sigma$ is 92.5% of the speed of light. The relation between $ \beta_{1}$ , $ \beta_{2}$ and $ \beta$ is shown graphically in Figure XV.20. If I use the notation $ \dfrac{\beta_{1}}{\beta_{2}}$ to mean “combining $ \beta_{1}$ \[ \beta_{1}\oplus\beta_{2}=\dfrac{\beta_{1}+\beta_{2}}{1+\beta_{1}\beta_{2}} \label{15.16.7}\] You may notice the similarity of Equation $ \ref{15.16.6}$ $ \beta=\dfrac{\beta_{1}+\beta_{2}}{1+\beta_{1}\beta_{1}}$ to the hyperbolic function identity \[ \tanh(\phi_{1}+\phi_{2})=\dfrac{\tanh\phi_{1}+\tanh\phi_{2}}{1+\tanh\phi_{1}\tanh\phi_{2}} \label{15.16.8}\] Thus I can represent the speed of an object by giving the value of $ \phi$, where \[ \beta = \tanh\phi \label{15.16.9}\] or \[ \phi=\tanh^{-1}\beta=\dfrac{1}{2}\ln\left(\dfrac{1+\beta}{1-\beta}\right) \label{15.16.10}\] The factor $ \phi$ combines simply as \[ \dfrac{\phi_{2}}{\phi_{2}}=\phi_{1}+\phi_{2} \label{15.16.11}\] If you did what I suggested in Section 15.3 and programmed your calculator or computer to convert instantly from one relativity factor to another, you now have a quick way of adding speeds. Example $\PageIndex{2}$ A train trundles to the right at 90% of the speed of light ($ \phi_{1}$ = 1.47222) relative to S, and a passenger strolls to the right at 15% of the speed of light ($ \phi_{2}$ = 0.15114) relative to $ \Sigma'$. The speed of the passenger relative to $ \Sigma$ is $ \phi$ = 1.62336, or 92.5% of the speed of light. Example $\PageIndex{3}$ (Sorry – there is no Figure XV.21.) An ocean liner $ \Sigma'$ sails serenely eastwards at a speed $ \beta_{1}$ = 0.9$ c$ ($ g_{1}$ = 2.29416) relative to the ocean $ \Sigma$. A passenger ambles athwartships at a speed $ \beta_{2}$ = 0.5$ c$ relative to the ship. What is the velocity of the passenger relative to the ocean? The northerly component of her velocity is given by Equation $ \ref{15.16.5b}$, and is 0.21794$ c$. Her easterly component is just 0.9$ c$. Her velocity relative to the ocean is therefore 0.92601$ c$ in a direction 13 o 37' north of east. Exercise $\PageIndex{1}$ Show that, if the speed of the ocean liner is $ \beta_{1}$ and the athwartships speed of the passenger is $ \beta_{2}$, the resultant speed $ \beta$ of the passenger relative to the ocean is given by \[ \beta^{2}=\beta_{1}^{2}+\beta_{2}^{2}-\beta_{1}^{2}\beta_{2}^{2} \label{15.16.12}\] and that her velocity makes and angle $ \alpha$ with the velocity of the ship given by \[ \tan\alpha=\beta_{2}\sqrt{1-\dfrac{\beta_{1}^{2}}{\beta_{1}}}. \label{15.16.13}\] Example $\PageIndex{4}$ A railway train $ \Sigma'$ of proper length $ L_{0}$ = 100 yards thunders past a railway station $ \Sigma$ at such a speed that the stationmaster thinks its length is only 40 yards. (Correction: It is not a matter of what he “thinks”. What I should have said is that the length of the train, referred to a reference frame $ \Sigma$ in which the stationmaster is at rest, is 40 yards.) A dachshund waddles along the corridor towards the front of the train. (A dachshund, or badger hound, is a cylindrical dog whose proper length is normally several times its diameter.) The proper length $ l_{0}$ of the dachshund is 24 inches, but to a seated passenger, it appears to be... no, sorry, I mean that its length, referred to the reference frame $ \Sigma'$, is 15 inches. What is the length of the dachshund referred to the reference frame $ \Sigma$ in which the stationmaster is at rest? We are told, in effect, that the speed of the train relative to the station is given by $ \gamma_{1}$ = 2.5, and that the speed of the dachshund relative to the train is given by $ \gamma_{2}$ = 1.6. So how do these two gammas combine to make the factor $ \gamma$ for the dachshund relative to the station? There are several ways in which you could do this problem. One is to develop a general algebraic method of combining two gamma factors. Thus: Exercise $\PageIndex{2}$ Show that two gamma factors combine according to \[ \gamma_{1}\oplus\gamma_{2}=\gamma_{1}\gamma_{2}+\sqrt{(\gamma_{1}^{2}-1)(\gamma_{2}^{2}-1)}. \label{15.16.14}\] I’ll leave you to try that. The other way is to take advantage of the programme you wrote when you read Section 15.3, by which you can instantaneously convert one relativity factor to another. Thus you instantly convert the gammas to phis. Thus $ \gamma_{1}=2.5\ \Rightarrow\ \phi_{1}=1.56680$ and $ \gamma_{1}=1.6\ \Rightarrow\ \phi_{1}=1.04697$ $ \therefore \qquad \qquad \qquad \phi=2.61377\ \Rightarrow\ \gamma=6.86182$ Is this what Equation $ \ref{15.16.14}$ gets? Therefore, referred to the railway station, the length of the dachshund is $ \dfrac{24}{\gamma}$ = 3.5 inches.
There are many circumstances in which we need to determine the frequency content of a time-domain signal. For example, we may have to analyze the spectrum of the output of an LC oscillator to see how much noise is present in the produced sine wave. This can be achieved by the discrete Fourier transform (DFT). The DFT is usually considered as one of the two most powerful tools in digital signal processing (the other one being digital filtering), and though we arrived at this topic introducing the problem of spectrum estimation, the DFT has several other applications in DSP. Please note that this article tries to give a basic understanding of the DFT in an intuitive way; examining a list of its properties, as is usual in textbooks, is not the goal of this article. Why the DFT? Assume that $$x(t)$$, shown in Figure 1, is the continuous-time signal that we need to analyze. Figure 1. A continuous-time signal for which we need to determine the frequency content. Figure 1.A continuous-time signal for which we need to determine the frequency content. Obviously, a digital computer cannot work with a continuous-time signal and we need to take some samples of $$x(t)$$ and analyze these samples instead of the original signal. Moreover, while Figure 1 shows only the first $$5$$ millisecond of the signal, $$x(t)$$ may continue for hours, years, or more. Since our digital computer can process only a finite number of samples, we have to make an approximation and use a limited number of samples. Therefore, generally, a finite-duration sequence is utilized to represent this analog continuous-time signal which may extend to positive infinity on the time axis. The reader may wonder how many samples, $$L$$, we need in order to estimate the frequency content of a given signal. We will discuss this question in a future article of this series. For the time being, assume that we sample $$x(t)$$ in Figure 1 with a sampling rate of $$8000$$ samples/second and take only $$L=8$$ samples of this signal. The result is shown in red in Figure 2. Figure 2. Sampling allows us to analyze continuous-time signals in a digital computer. Figure 2.Sampling allows us to analyze continuous-time signals in a digital computer. If we normalize the time axis to the sampling period $$T_s=\frac{1}{f_s}$$, we will obtain the discrete sequence $$x(n)$$, representing $$x(t)$$, as follows: $$n$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$x(n)$$ $$0.2165$$ $$0.8321$$ $$0.7835$$ $$0.5821$$ $$0.2165$$ $$-0.5821$$ $$-1.2165$$ $$-0.8321$$ Fourier analysis provides several mathematical tools for determining the frequency content of a time-domain signal, but which tool is a good choice for analyzing $$x(n)$$? There are only two techniques from the Fourier analysis family which target discrete-time signals (see page 144 of this book): the discrete-time Fourier transform (DTFT) and the discrete Fourier transform (DFT). The DTFT of an input sequence, $$x(n)$$, is given by $$X(e^{j \omega})=\sum_{n=- \infty}^{+\infty}x(n)e^{-jn \omega}$$ Equation 1 Equation 1 The inverse of the DTFT is given by $$x(n)=\frac{1}{2 \pi} \int_{- \pi}^{\pi} X(e^{j \omega})e^{jn \omega}d \omega$$ Equation 2 Equation 2 We can use Equation 1 to find the spectrum of a finite-duration signal $$x(n)$$; however, $$X(e^{j \omega})$$ given by the above equation is a continuous function of $$\omega$$. Hence, a digital computer cannot directly use Equation 1 to analyze $$x(n)$$. However, we can use samples of $$X(e^{j \omega})$$ to find an approximation of the spectrum of $$x(n)$$. The idea of sampling $$X(e^{j \omega})$$ at equally spaced frequency points is in fact the basis of the second Fourier technique mentioned above, i.e., the DFT. Note that this sampling is taking place in the frequency domain ($$X(e^{j \omega})$$ is a function of frequency ). Performing the DFT Using Basic Math Let’s proceed with the example started in the previous section. We have an L-sample-long sequence, $$x(n)$$, representing the analog continuous-time signal $$x(t)$$. The goal is to find a set of sinusoids which can be added together to produce $$x(n)$$. As discussed above, the DFT is based on sampling the DTFT, given by Equation 1, at equally spaced frequency points. $$X(e^{j \omega})$$ is a periodic function in $$\omega$$ with a period of $$2 \pi$$. If we take $$N$$ samples in each period of $$X(e^{j \omega})$$, the spacing between frequency points will be $$\frac{ 2\pi }{N}$$. Hence, the frequency of the set of sinusoids that we are looking for will be of the form $$\frac{2 \pi}{N} \times k$$, where we can select $$k=0, 1, ..., N-1$$. Using complex exponentials similar to Equation 1 and 2, the basis functions will be $$e^{j \frac{2 \pi}{N}kn}$$. We are looking for a weighted sum of these functions which will hopefully give us the original signal $$x(n)$$. This means that $$x(n)=\sum_{k=0}^{N-1}X'(k)e^{j\frac{2 \pi}{N}kn}$$ $$n=0, 1, \dots, L-1$$ Equation 3 Equation 3 where $$X'(k)$$ denotes the weight used for the complex exponential $$e^{j \frac{2 \pi}{N}kn}$$. Equation 3 is equivalent to the following set of equations $$x(0)=X'(0)+X'(1)+X'(2)+ \dots +X'(N-1)$$ $$x(1)=X'(0)+X'(1)e^{j\frac{2 \pi}{N}}+X'(2)e^{j\frac{2\pi}{N}2}+ \dots +X'(N-1)e^{j\frac{2\pi}{N}(N-1)}$$ $$x(2)=X'(0)+X'(1)e^{j\frac{2\pi}{N}2}+X'(2)e^{j\frac{2\pi}{N}2\times 2}+ \dots +X'(N-1)e^{j\frac{2\pi}{N}(N-1)\times 2}$$ . . . $$x(L-1)=X'(0)+X'(1)e^{j\frac{2\pi}{N}(L-1)}+X'(2)e^{j\frac{2\pi}{N}2\times(L-1)}+ \dots +X'(N-1)e^{j\frac{2\pi}{N}(N-1)\times(L-1)}$$ Equation 4 Equation 4 For a given $$L$$ and $$N$$, the values of complex exponentials are known and, having the value of the time-domain signal, we can calculate the coefficients $$X'(k)$$. The number of time-domain samples, $$L$$, gives the number of equations in the above set of equations, and the number of frequency-domain samples of $$X(e^{j\omega})$$ determines the number of unknowns in Equation 4. With $$L=N$$, there are $$N$$ linearly independent equations to find the $$N$$ unknowns, $$X'(k)$$. Assuming $$L=N=8$$, we can find the matrix form of Equation 4 and use MATLAB to find the unknowns: $$k$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$X'(k)$$ $$0$$ $$-0.5j$$ $$0.1083-0.0625j$$ $$0$$ $$0$$ $$0$$ $$0.1083+0.0625j$$ $$0.5j$$ What is the interpretation of this analysis? We decomposed the sequence $$x(n)$$ into a sum of complex exponentials given by Equation 3. Let’s substitute the results in the mentioned equation. With $$L=N=8$$ and omitting the exponentials with zero coefficients according to the above table, we obtain $$x(n)=X'(1)e^{j\frac{2 \pi}{8}n}+X'(2)e^{j\frac{2\pi}{8}2n}+X'(6)e^{j\frac{2\pi}{8}6n}+X'(7)e^{j\frac{2\pi}{8}7n}$$ Since these exponential functions are periodic with $$N=8$$, we can further simplify this equation. For example, $$e^{j\frac{2\pi}{8}7n}$$ is equal to $$e^{j\frac{2\pi}{8}(8-1)n}=e^{j\frac{2\pi}{8}8n}e^{-j\frac{2\pi}{8}n}=e^{-j\frac{2\pi}{8}n}$$. After simplifying and rearranging the terms of the above equation, we have $$x(n)=X'(1)e^{j\frac{2\pi}{8}n}+X'(7)e^{-j\frac{2\pi}{8}n}+X'(2)e^{j\frac{2\pi}{8}2n}+X'(6)e^{-j\frac{2\pi}{8}2n}$$ Finally, we obtain $$x(n)=sin(\frac{2\pi}{8}n)+0.125sin(\frac{4\pi}{8}n)+0.2166cos(\frac{4\pi}{8}n)=sin(\frac{2\pi}{8}n)+0.25sin(\frac{4\pi}{8}n+3\frac{\pi}{3})$$ Equation 5 Equation 5 This shows that $$x(n)$$ can be represented by two components, one at the normalized frequency of $$\frac{2\pi}{8}$$ with an amplitude of $$1$$ and the other at $$\frac{4\pi}{8}$$ with an amplitude of $$0.25$$. These frequencies are normalized; if the sampling frequency is $$f_s=8000$$ samples/second, we obtain the frequency of these two components as $$f_1=\frac{2\pi}{8}\frac{f_s}{2\pi}=1 kHz$$ and $$f_2=\frac{4\pi}{8}\frac{f_s}{2\pi}=2 kHz$$. Let’s summarize what we have discussed so far. We started with a continuous-time signal and used a finite number of samples to analyze the frequency content of this continuous-time signal. We saw that the problem of decomposing this discrete sequence boils down to solving a set of linear equations. In the case of this article’s simple example, we even simplified the result (Equation 5) and obtained an explicit equation for $$x(n)$$. There is one point which needs further attention. The analysis started with a sequence of length eight which is shown in Figure 3. Clearly, we know the value of $$x(n)$$ for $$n=0, \dots, 7$$, but we don’t know $$x(n) $$ outside of this range. The above analysis is actually looking for a sum of complex exponentials which can reproduce the values of $$x(n)$$ for $$n=0, \dots, 7$$. Let’s examine Equation 5 outside of this range. To avoid ambiguity, we refer to the sequence given by Equation 5 as $$p(n)$$. $$p(n)$$, shown in Figure 4, is a periodic function with $$N=8$$. As shown in this figure, the values of the original $$x(n)$$ will be repeated every $$8$$ samples. In other words, while we could represent $$x(n)$$ using some complex exponentials, the obtained representation is periodic, and $$x(n)$$ is equal to $$p(n)$$ only in one period. When developing, discussing, and applying the DFT, always remember that we are achieving a representation of the finite-duration sequence using a periodic sequence, where the values in one period of this periodic sequence are equal to those in the finite-duration sequence. Note that the periodic behaviour of $$p(n)$$ can also be understood by recalling that we are sampling $$X(e^{j\omega})$$, the DTFT of $$x(n)$$, in the frequency domain. In a previous article, we discussed how sampling a signal in the time domain leads to replicas of the origin signal in the frequency domain. We observe a similar phenomenon here: sampling $$X(e^{j\omega})$$ in the frequency domain leads to replicas of $$x(n)$$ in the time domain. The periodic form of $$x(n)$$ in Figure 3 is shown as $$p(n)$$ in Figure 4. Figure 3. The analysis started using only these eight samples. Figure 3.The analysis started using only these eight samples. Figure 4. The analysis leads to p(n)which is the periodic form of x(n). Figure 4.The analysis leads to p(n)which is the periodic form of x(n). Deriving the DFT Equations The discussed method for calculating the spectrum of a finite-duration sequence is simple and intuitive. It clarifies the inherent periodic behavior of DFT representation. However, it is possible to use the above discussion and derive closed-form DFT equations without the need to calculate the inverse of a large matrix. To this end, we only need to make a period signal out of the $$N$$ samples of the finite-duration sequence. Then, applying the discrete-time Fourier series expansion, we can find the frequency domain representation of the periodic signal. The obtained Fourier series coefficients are the same as the DFT coefficients except for a scaling factor. Assume that the finite-duration sequence that we need to analyze is as shown in Figure 5 (a). To calculate the N-point DFT, we need to make a periodic signal, $$p(n)$$, from $$x(n)$$ with period $$N$$, as shown in Figure 5(b). Considering the fact that $$p(n)=x(n)$$ for $$n=0, 1, \dots, N-1$$, we obtain the discrete-time Fourier series of this periodic signal $$a_k=\frac{1}{N}\sum_{n=0}^{N-1}x(n)e^{-j\frac{2\pi}{N}kn}$$ Equation 6 Equation 6 where $$N$$ denotes the period of the signal. The time-domain signal can be obtained as follows: $$x(n)=\sum_{k=0}^{N-1}a_{k}e^{j\frac{2\pi}{N}kn}$$ Equation 7 Equation 7 Figure 5. (a)The finite-duration sequence, x(n), to be analyzed. (b) The periodic signal obtained from x(n). Image courtesy of Digital Signal Processing, Fundamentals, and Applications. Figure 5.(a)The finite-duration sequence, x(n), to be analyzed. (b) The periodic signal obtained from x(n). Image courtesy of Digital Signal Processing, Fundamentals, and Applications. Multiplying the coefficients given by Equation 6 by $$N$$, we obtain the DFT coefficients, $$X(k)$$, as follows: $$X(k)=\sum_{n=0}^{N-1}x(n)e^{-j\frac{2\pi}{N}kn}$$ Equation 8 Equation 8 The inverse DFT will be $$x(n)=\frac{1}{N}\sum_{k=0}^{N-1}X(k)e^{j\frac{2\pi}{N}kn}$$ Equation 9 Equation 9 Please note that while the discrete-time Fourier series of a signal is periodic, the DFT coefficients, $$X(k)$$, are a finite-duration sequence defined for $$0 \leq k \leq N-1$$. Summary The DFT is one of the most powerful tools in digital signal processing; it enables us to find the spectrum of a finite-duration signal x(n). Basically, computing the DFT is equivalent to solving a set of linear equations. The DFT provides a representation of the finite-duration sequence using a periodic sequence, where one period of this periodic sequence is the same as the finite-duration sequence. As a result, we can use the discrete-time Fourier series to derive the DFT equations.
Properties of the algorithm: Sequential complexity: [math]O(n^3)[/math] Height of the parallel form: [math]O(n)[/math] Width of the parallel form: [math]O(n^2)[/math] Amount of input data: [math]\frac{n (n + 1)}{2}[/math] Amount of output data: [math]\frac{n (n + 1)}{2}[/math] 1 Properties and structure of the algorithm 1.1 General description The Cholesky decomposition algorithm was first proposed by Andre-Louis Cholesky (October 15, 1875 - August 31, 1918) at the end of the First World War shortly before he was killed in battle. He was a French military officer and mathematician. The idea of this algorithm was published in 1924 by his fellow officer and, later, was used by Banachiewicz in 1938 [7]. In the Russian mathematical literature, the Cholesky decomposition is also known as the square-root method [1-3] due to the square root operations used in this decomposition and not used in Gaussian elimination. Originally, the Cholesky decomposition was used only for dense real symmetric positive definite matrices. At present, the application of this decomposition is much wider. For example, it can also be employed for the case of Hermitian matrices. In order to increase the computing performance, its block versions are often applied. In the case of sparse matrices, the Cholesky decomposition is also widely used as the main stage of a direct method for solving linear systems. In order to reduce the memory requirements and the profile of the matrix, special reordering strategies are applied to minimize the number of arithmetic operations. A number of reordering strategies are used to identify the independent matrix blocks for parallel computing systems. 1.2 Mathematical description Input data: a symmetric positive definite matrix [math]A[/math] whose elements are denoted by [math]a_{ij}[/math]). Output data: the lower triangular matrix [math]L[/math] whose elements are denoted by [math]l_{ij}[/math]). The Cholesky algorithm can be represented in the form [math]\begin{align}l_{11} & = \sqrt{a_{11}}, \\l_{j1} & = \frac{a_{j1}}{l_{11}}, \quad j \in [2, n], \\l_{ii} & = \sqrt{a_{ii} - \sum_{p = 1}^{i - 1} l_{ip}^2}, \quad i \in [2, n], \\l_{ji} & = \left (a_{ji} - \sum_{p = 1}^{i - 1} l_{ip} l_{jp} \right ) / l_{ii}, \quad i \in [2, n - 1], j \in [i + 1, n].\end{align}[/math] There exist block versions of this algorithm; however, here we consider only its “dot” version. In a number of implementations, the division by the diagonal element [math]l_{ii}[/math] is made in the following two steps: the computation of [math]\frac{1}{l_{ii}}[/math] and, then, the multiplication of the result by the modified values of [math]a_{ji}[/math] . Here we do not consider this computational scheme, since this scheme has worse parallel characteristics than that given above. Read more…
I have a question about the moments generated by torsional springs. Consider an inertial reference frame A and a body fixed reference frame B. The frames are connected by 3 rotational springs acting around all three axes. For a simple rotation $\psi$ around the $Z_B$-axis the moment will be in the negative $Z_B$ direction: $$ M_B = \begin{bmatrix} 0 \\ 0 \\ -k_z \psi \end{bmatrix} $$ My question is how this can be extended to an arbitrary rotation. For example, using a ZYX Euler rotation, denoted by angles $\psi$, $\theta$, and $\phi$ (commonly called yaw, pitch, roll), the moment could be described as: $$ M_B = \begin{bmatrix} -k_x \phi \\ -k_y \theta \\ -k_z \psi \end{bmatrix} $$ However, this gives the impression that the moment is dependent on the order of rotation. Any rotation can be represented by 12 different sets of angles. For example, if an XYX rotation was considered (with angles corresponding to the same orientation of the B-frame) then the moment would be different than before (no moment in the Z-axis), while the rotation is the same. Does anybody have an idea how to model such a torsional spring? Am I on the right track? EDIT Maybe I should edit my question. What I want to do is model the rotational behaviour of two bodies (1 and 2). They are connected by three virtual rotational springs (representing a link between them). For a normal (translational) spring the forces on body 1 in X, Y, and Z would be easily calculated as: $$ \mathbf{F}_1 = -\mathbf{K}(\mathbf{q}_1 -\mathbf{q}_2) $$ with $\mathbf{q}_1 = \begin{bmatrix} q_{1x} & q_{1y} & q_{1z} \end{bmatrix}^T$ and $\mathbf{q}_2 = \begin{bmatrix} q_{2x} & q_{2y} & q_{2z} \end{bmatrix}^T$ the displacements of body 1 and 2, respectively. And $\mathbf{K}$ the 3x3 diagonal matrix containing the spring constants. The translational motion of the bodies is then described as: $$ \mathbf{m}_1 \ddot{\mathbf{q}}_1 = -\mathbf{K}(\mathbf{q}_1 -\mathbf{q}_2)\\ \mathbf{m}_2 \ddot{\mathbf{q}}_2 = -\mathbf{K}(\mathbf{q}_2 -\mathbf{q}_1) $$ with $\mathbf{m}_1$ and $\mathbf{m}_2$ the diagnonal mass matrices of body 1 and 2. Now, I want to do something similar for the rotational motion as a result of the torsional springs. So, the motion of by 1 and 2 is described by: $$ \dot{\boldsymbol{\omega}}_1 \mathbf{I}_1 + \boldsymbol{\omega}_1 \times \mathbf{I}_1 \boldsymbol{\omega}_1 = \mathbf{M}_1 \\ \dot{\boldsymbol{\omega}}_2 \mathbf{I}_2 + \boldsymbol{\omega}_2 \times \mathbf{I}_2 \boldsymbol{\omega}_2 = \mathbf{M}_2 $$ with $\boldsymbol{\omega}$ the angular velocities of the bodies (expressed in a body frame), $\mathbf{I}$ the moments of inertia (expressed in a body frame) and $\mathbf{M}_1$ and $\mathbf{M}_2$ are the moments resulting from the torsional springs. I'm looking for a method to calculate these moments.
There is a huge strand of literature on graph theory in finance which analyzes networks summarized here:Allen, F., and A. Babus (2009): “Networks in Finance,” in Network-based Strategies and Competencies, ed. by P. Kleindorfer, and J. Wind, pp. 367–382.First applications of graph theory in networks focused on credit-risk in interconnected banks and how ... (these are just my random opinions. Sorry, I expect many people to disagree with some of them.)Bloomberg makes no effort to make its terminals available to students - quite the opposite. Hence lots of recent graduates don't know the terminal, but do know how to get things done without the terminal, and don't think they need the terminal. It's a problem for ... We assume that the stock price process $\{S_t,\,t>0\}$ satisfies, under the real-world probability measure $P$, an SDE of the form\begin{align}dS_t=S_t\big((\mu-q)dt+\sigma dW_t\big),\end{align}where $\{W_t, \, t >0\}$ is a standard Brownian motion.Here, we need to consider the total return asset $e^{qt}S_t$, that is, the asset with the dividend ... The only difference in the derivation when you have a dividend-yield paying stock lies in the value of the Riskless Portfolio $\Pi_t$.The financial meaning here is the key: to delta-hedge your option you buy a quantity $\Delta$ of the stock $S$, and only the stock is paying you the dividend, so you have to add this contribution in time to your hedge.The ... This is pretty standard fare for a Stats 101 course, so as to rationale, etc. you might benefit from picking up a textbook or otherwise do some reading on this.In brief though, hypothesis testing allows us to assess the likelihood sample estimates are different than theorized values in the absence of actual population values.In the cases above, with a ... At $t_1$, this payoff can be priced using the Margrabe formula as used for pricing an exchange option.See Margrabe Formula hereUsing the notations in the question and those used the hyperlinked document above -$Price_{t_1} = P_{t_1}e^{(\mu_P-r)\tau}\Phi(d_+) -HR \times G_{t_1}e^{(\mu_G-r)\tau}\Phi(d_-) \tag{1}$$Price_0$ is the discounted value of $... Kelly DOES reflect the odds!The simple binary bet form of Kelly is:Kelly Fraction = (p(win) * (odds + 1) - 1) / oddsSo for a 60% chance of a 50% risk, ie 1:1 equals odds 1, that’s 20% of your capital at risk.More formally, Kelly seeks to maximise log-wealth (LW)LW = sum ( Pi * ln(1 + Stake * Payoffi)Maximise LW, then dLW/dStake = 0For each ... The allocation of regulatory capital to a set of trades that makes up the portfolios and sub portfolios of a bank. One theory is Shapley value which is combinatorial index, but complexity runs far deeper.Encryption one might consider inherent to a lot of finance.I suspect one might consider graph theory in terms of money flows / capital flows from ... The formula is multiplicative because interests are compounded (re-invested).Let's say you have 10% interest on a 1 USD deposit, compounded (calculated and paid) twice a year. That would be 5 cents of interest for the first semester. Then you have a deposit of 1.05 USD during the next semester. After one year, your interests are1 * 0.05 + 1.05 * 0.05 = 0.... Does optimizing a solution for a given set of parameters sound like something quants would need to know how to do?Yep!Indeed many things quants do revolve around optimization, and linear programming (and integer programming, multi-integer programming, quadratic programming, etc.) is all about finding the optimal solution to something given some set of ...
I don't think I am putting out something truly novel here either; the answer by @MAFIA36790 and the discussion in the link you yourself post it seem quite satisfactory to me, yet I shall try to put a different spin on things so to speak. Let me define the chemical potential, as the partial molar Gibbs energy of species "$j$": $$\mu_j = \left ( \frac{\partial G}{\partial n_j} \right)_{p,T,n'}$$ and for a system of components A, B, etc. the fundamental equation of chemical thermodynamics gives us: $$\mathrm dG= V~\mathrm dp-S ~\mathrm dT+\Sigma \mu_j~\mathrm dn_j$$ Now, I consider a simple equilibria $$R \rightleftharpoons P $$ and additionally, I introduce a parameter " $\xi$ " called the extent of the reaction (it has the dimensions of amount of substance) for macroscopic changes, the amount of a component $j$ changes by $\nu_j\Delta\xi$ ($\nu_j$ are the "signed" stoichiometric coefficients; positive for products, and negative for reactants) Now, if we defined $$\Delta _rG = \left ( \frac{\partial G}{\partial \xi} \right)_{p,T} $$ This $\Delta _r G$ still holds the meaning you might already be familiar with, namely: $$\Delta _r G = \Delta _f G\textrm{ (products)} - \Delta _f G\textrm{ (reactants)}$$ and I have nearly extended it to mean the derivative defined above so that i can relate it to the extent of a reaction (measured in amounts of substance consumed/produced) and at constant temperature and pressure, from the second equation in this post: $$ \mathrm dG = (\mu_P - \mu_R)~ \mathrm d\xi$$ so, $$ \Delta _rG = \left ( \frac{\partial G}{\partial \xi} \right)_{p,T} = (\mu_P - \mu_R)$$ Now, as already mentioned, if we were to plot the reaction Gibbs energy as a function of the extent of the reaction, a minimum on the plot would correspond to the first derivative defined above being equal to zero. (or, equivalently in the chemical potential of reactants and products being exactly equal). This is the condition for equilibrium. Perhaps, stooping down to layman terminology (caution), one can say $\Delta _r G $ is like a "force" that drives a reaction forward, and the system is in equilibrium when this driving force is zero (and moves in the reverse direction when it changes sign). An appropriate mechanical analogy could perhaps be two weights on either end of a string wrapped across a pulley. If the heavier weight is higher, it would descend raising the lower weight. The system would come to rest when they are in balance.
Diff. Geom, Math. Phys., PDE Seminar: Ali Hyder Date: 11/14/2017 Time: 15:30 University of British Columbia Conformal metrics on \mathbb{R}^n with arbitrary total Q-curvature I will talk about the existence of solution to the Q-curvature problem \begin{align}\label{1} (-\Delta)^\frac n2 u=Qe^{nu}\quad\text{in }\mathbb{R}^n,\quad \kappa:=\int_{\mathbb{R}^n}Qe^{nu}dx<\infty, \end{align} where Q is a non-negative function and n>2. Geometrically, if u is a solution to \eqref{1} then Q is the Q-curvature of the conformal metric g_u = e^{2u}\|dx\|^2 (\|dx\|^2 is the Euclidean metric on \mathbb{R}^n), and \kappa is the total Q-curvature of g_u. Under certain assumptions on Q around origin and at infinity, we prove the existence of solution to \eqref{1} for every \kappa > 0. ESB 2012 Tue 14 Nov 2017, 3:30pm-4:30pm
I'm trying to typeset the following $Q_å=l_å \cdot m$ But LaTeX won't typeset these special or accented letters in my document! Is there a simple way to achieve this? TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. It only takes a minute to sign up.Sign up to join this community Names for accents change in math mode, so you should use \mathring a as follows: \documentclass{article}\begin{document}$Q_{\mathring a}=l_{\mathring a} \cdot m$\end{document} See the comprehensive list of symbols Table 164 for the other accent commands. Changing input encoding usually does not help for this situation. ADDITION If your text italics and math italics agree, which is the case by default but may well change if a font package is loaded, then you can implement @daelif's suggestion as follows: \documentclass{article}\usepackage[utf8]{inputenc}\usepackage[T1]{fontenc}\usepackage{mathtools}\begin{document}$Q_{\textnormal{\itshape å}}=l_{\textnormal{\itshape å}} \cdot m$\end{document} I have used \textnormal to negate any formatting from the surrounding text, inputenc to enable input of å directly, fontenc to ensure a font with a good glyph, and mathtools to ensure the size of the subscript is appropriate. If you're planning on using these accents in math-mode you should use the \accents package. For this specific example, you would use: $Q_{\accentset{\circ}{a}}=l_{\accentset{\circ}{a}} \cdot m$ which would output: The accents package is very useful and is well documented (see: CTAN - Accents Package) You can use this: \documentclass[varwidth]{standalone}\usepackage{amsmath}\begin{document} \begin{equation} Q_å=l_å \cdot m \end{equation} \begin{equation} Q_{\mathring{a}}=l_{\mathring{a}} \cdot m \end{equation}\end{document} Result:
Calculations In this question, I solved for the stresses on an spacecraft passing close to a black hole's event horizon. There isn't some magical barrier that you go through around an event horizon, you just can't get out; you probably wouldn't even notice. I can use the same process to calculate the stresses that a prison slightly inside the event horizon would face. Event horizon of a black hole The event horizon is the distance from a black hole where the escape velocity is equal to $c$. Escape velocity is $$v_e = \sqrt{\frac{2GM}{r}}.$$ A black hole with 40 billion solar masses will have mass $8\times10^{40}$ kg. Solving for r when $v_e=c$ gives$$r = \frac{2GM}{c^2} = \frac{2\cdot6.7\times10^{-11}\cdot2\times10^{38}}{\left(3\times10^{8}\right)^2} = 1\times10^{14} \text{ meters}.$$ Gravity as a function of distance from the black hole A person is 2 m tall, and 'orbiting' just inside the event horizon at $1\times10^{14}$ m from a black hole of mass $8\times10^{40}$ kg. The tidal acceleration between the head and feet of a 2 meter tall person due to the gravity of the black hole is $$\begin{align}a &= \frac{m_{hole}G}{(r+2)^2}-\frac{m_{hole}G}{r^2} \\&= 6.7\times10^{-11}\cdot8\times10^{40}\frac{1}{\left(100000000000002\right)^2}-\frac{1}{\left(1\times10^{14}\right)^2}\\&=-2\times10^{-11} \frac{\text{m}}{\text{s}^2}\end{align}$$ What would that do to a 1 km long cylinder? Conveniently, in my other question, I calculated the tension on a 1km long cylindrical object. Conveniently, this could be a pretty reasonable prison space station. Near the ergosphere of a black hole, the tension forces would destroy any known object, but what about at the even horizon? Following the same math in the other question, and with the same structural assumptions, I get the differential stress on any slice of the prison/station: $$\frac{dF_{slice}}{dl} = \frac{2\times10^{35}}{(1\times10^{14}+l)^2}.$$ Total net force on the rod is $2\times10^{10}$ N, and maximum stress of about $1\times10^{12}$ N. Working backwards using the equation for gravity, we see that the assumption is that the station has a mass of 40000 tons. Depending on the cross-sectional area of the load bearing parts of your station, we can calculate the stresses. If your station is a cylinder 100m in radius, and 1/10 of the available area is taken up by load bearing structures, then the maximum stress on the 3000 m$^2$ of load bearing structure is about 6 MPa. A common structural steel has a yield strength in tension of about 250 MPA, so this isn't too much. If you have the technology to build space stations inside a black hole, then it is reasonable that you could construct it out of materials that won't fall apart. The second question is how long you can maintain your orbit. Using simple Newtonian mechanics (Warning! Not valid near a singularity!) the gravitational pull of $2\times10^{10}$ N will have to be counter-acted by thrust. Now, that is a lot of thrust, about three orders of magnitude greater than a Saturn V. I suppose it really depends what sort of propulsion system you have. Thrust as a function of mass flow rate is given as $T = v\frac{dm}{dt}$. Assuming exhaust at the speed of light from some magical propulsion system, you still need 70 kg tons of propellant passed every second to keep from falling into the black hole. Conclusions Given the small tidal acceleration, even a large object (1 km long, 25000 tons) could reasonably be kept together with known materials at the event horizon of such a large black hole. As for keeping such an object in orbit, for any propulsion system with reaction mass, the propellant usage would be very large (about 250 tons per hour, as calculated above). Given that the mass of the whole station is 40,000 tons, you would burn through the entire station's mass in a week. Just like the tyranny of the rocket equations, the tyranny of a black hole's gravity is oppressive: the more propellant you keep on board, the harder you are pulled in and the more propellant you need. I suppose you could be refueled with propellant, but that is a lot of money to be literally throwing into a black hole. For some sort of reaction-less system, well, I don't know how to measure that. You can't gain momentum out of black hole, so I don't know how you could calculate the thrust given off by a photonic engine. In any case, the 'not falling into the hole' part seems to be the catch, with any propellant based system. Doesn't seem very reasonable, given relatively hard science constraints.

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