Split (1)

train · ~4.74M rows (showing the first 1.13M)

text stringlengths 256 16.4k
I'm reading about residue calculus. I read that the standard procedure for integrating $\int_{-\infty}^\infty R(x)dx$ where $R(x)$ is a rational function whose denominator has degree at least 2 units greater than the numerator and no poles on the real line, is to integrate the complex function $R(z)$ over a closed curve consisting of the segment $(-r,r)$ and the semicircle from $r$ to $-r$ in the upper half plane, so that if $r$ is large enough, the curve encloses all poles in the upper half plane. I see this works because the integral over this closed curve can be found by summing the residues, which may be easier. Since I'm only interested in the integral along the real line, I would hope that the integral over the semicircle goes to $0$ as $r\to\infty$. My text says this is true by "obvious estimates." Can someone please explain more explicitly how one sees the integral over the semicircle goes to $0$? The obvious estimates is cryptic to me. The text is Ahlfors' Complex Analysis, page 156, if it's helpful. Many thanks.
My students frequently mix up my $t$'s with my $+$'s and my $y$'s with my $4$'s. What is a good handwriting font for distinguishing these and other easily confused symbols? Mathematics Educators Stack Exchange is a question and answer site for those involved in the field of teaching mathematics. It only takes a minute to sign up.Sign up to join this community My students frequently mix up my $t$'s with my $+$'s and my $y$'s with my $4$'s. What is a good handwriting font for distinguishing these and other easily confused symbols? I changed my handwriting font years ago for precisely this purpose, and I have continued to tweak my letterforms over the years, using the algorithm of changing the form of whichever letters seem to generate the most confusion. Here is my current font: A few notes about these letters: Capital letters aren't listed, but my experience is that they're all fairly straightforward. Just make sure to use print letters instead of cursive. It's crucial to have versions of a, b, d, p, and q that can be written using a single stroke. If you draw your circles and stems separately, they will constantly get disconnected, which vastly decreases legibility. I often omit the bottom "tail" on the f when writing it in the middle of a word, but I always include it when it's part of an equation. Similar statements hold for the top curve on the i. I will also draw a 1 as a simple vertical line if I think it's clear from context. The l (ell) is my newest letter, and I'm not really sure about it yet. I tried using a cursive $\ell$ for a while, but it never looked good inside of words, and it still wasn't very legible as a variable. At present, I am often omitting the top and bottom curves when the l is part of a word. In general, I've had bad experiences with vertical loops. I've tried loops on $\ell$'s, g's, j's, d's, and q's, and all of them seemed to make the letters less recognizable. (This is probably because such loops hardly ever appear in computer or typewritten fonts.) The tail curve on the t is crucial to avoid confusion with a + sign. The initial curves for the v and w are quite helpful for legibility. I also think v's and w's are more legible with relatively sharp angles, as opposed to a curvy cursive approach. The initial curve on the x is absolutely essential. This will be one of your most used letters, and it really helps to get it right. Curves on any of the other three stems don't seem to improve legibility, and make the letter annoying to write. I also think this curvy version of a y is easier to read than a two-sticks version. Among other advantages, it can be drawn in a single stroke, which avoids disconnection problems. The line across the z helps avoid confusion with the number 2, and the line across the 7 helps avoid confusion with a variety of symbols. I don't seem to be able to draw a curvy-bottom 9 that looks good. I wish I could. This version of a 2 is much clearer than any version with a loop on the bottom. I tried drawing a final loop on my o's for a while (like a cursive o), and it didn't work out. Edit: By request, here are my capital letters. Unlike my lowercase letters, my capitals are really quite standard, and I don't have much to say about them. I've also included by Greek letters. $\times\ \ +\ \ 4\quad 2\quad \omega\quad 9\quad a\quad 1\quad i\\ x\quad t\quad y\quad z\quad w\quad g\quad q\quad l\quad j$ Here are what my alphabet and numbers have ended up looking like. For the sake of clarity, I've included a stroke order chart, where the first stroke is red, the second green, the third blue, and the fourth cyan. I try to make distinctions between every character, but I agree with those above who have noted that "o" is, in general, a terrible name for a variable. I tend to only use the "O" with a flourish where I have to make a distinction between "o" and "O". I use the double-story "g" to distinguish from "y", and "S" with a serif to distinguish from "5". I use serifs to distinguish between uppercase and lowercase ("c" and "C", "s" and "S", "k" and "K", "v" and "V", "w" and "W", "p" and "P"), but where I don't have to make that distinction, "c" and "s" get serifs to distinguish from "(" and "5", respectively. I would like to add a point that I've not seen explicitly mentioned in the other answers. That is that context can often be your aid in clarifying your lettering. As a trivial example, no one is going to misread they as +hey. So the "special alphabet" is primarily for mathematics in your lectures. (That said, you should cultivate a clear hand writing for normal text but it doesn't need to be as extreme as for the mathematical part.) Similarly, if you never write $\times$ then the students will soon realise that x means $x$ and not $\times$. And I doubt that someone will mistake $x_k$ or $\vec{x}$ for $\times_k$ or $\vec{\times}$. On the other hand, scale is often hard to see on a board so X and x are not easy to distinguish. It can be awkward to avoid using these together (consider the common desire to have $x \in X$) and distinguishing them visually can be difficult, but then context can help. Here are some practical tips: Go in to a large lecture hall, write some random equations on the board, and then go to the back and try to read what you wrote. Then modify your handwriting until you think it is clear. This won't be perfect (since you know what you wrote), but it'll at least be a lot better than not doing it. Say what you write as you write it, and make sure that before you write anything on the board then the students are at the same point as you so that they are paying attention to what you say as you write it. Cultivate writing "the wrong way". It's a bit trickier for those not blessed with sinister tendencies, but if you can learn to write so that you do not stand in front of what you just wrote then it will be much easier for the students to see what you write as you write and say it. This doesn't mean writing with your left hand, it is possible to do this with the right hand but the body position takes a bit of getting used to. This has the added bonus of making you more turned towards the students as you write. Having just given a lecture, I'd add one more practical point: ensure that the board is clean before you write on it. Stray squiggles left over from before you erased it can be confusing. This is my handwriting font for mathematics. (Excuse the shakey-ness, this was made using the stylus on my phablet.) It isn't as minimalist as some of yours, but I have never had anyone tell me they had trouble reading my writing. I make it clear which letters are which by using size and giving attention to negative space. I find that using big loops and serifs is really good for board work, where it can otherwise be difficult to distinguish capital letters. (Spencerian cursive is actually pretty good for board work too, just not for equations.) The prototypical gesture here is the J, from left to right to down. That is the leftmost part of many capitals. Keeping track of where the bottom line of the word is also important, for example the tail on the lower case t is just below thst bottom line, and it is crossed exactly on the middle letter line. My biggest problem is that my vs look like nus. I think you need to fix any problems you have by modifying a letter or two of the way you write. For example, I used to write my y's in two separate straight strokes, but they would get confused with my x's if I wasn't careful about where the stroke stops. To correct this, I started writing my y's more like the way you write a g, with a curly bottom. I also stroke my z's with a cross to keep them separate from my 2's. To fix your t's, add a curl on the bottom. As others have suggested, adding a serif can make a big difference. I write my y's like g's, and I also stroke my z's through the middle. Lower-case l's are written cursive-style, with a loop, and I make sure to make the 'tail' of my q's big enough to ensure there's no confusion with 9's. I also never use certain letters as variables (if I can help it) - especially o's. In response to Chris Cunningham's request, I think this sort of 5 is unlikely to be mistaken for an s: The key is not just the right angle at the top left, but the sharp angle directly below it. The danger here is to make sure you don't come so close to closing up the bottom that it looks like a 6. We discussed this, or more specifically the subject of handwritten $x$, a while ago at The Aperiodical. I still don’t understand why the two-curves $x$ never caught on in the US. Newton wrote $x$ the way modern Brits do, so it predates American independence. Curious! Taking my cue from printed math (and particularly TeX), I use different fonts for text mode and math mode. The italic font (math mode) is cursive, while the roman font (text mode) is printed. This helps distinguish ‘a’ from ‘$a$’, for example. (And assuming that you and I are both using the default fonts on this site, then my handwriting looks pretty much like those two letters do on the screen.) When I started this, I had a somewhat too exaggerated cursive style, so the fonts look less different now than they used to, but I maintain a difference in all letters. Also, after several false starts, I've settled on a little loop at the top of ‘0’ to distinguish it from ‘O’. (Much as TeX uses merely slanted letters for math-mode capitals, I don't distinguish text-mode and math-mode capitals at all, saving the fancy cursive stuff for calligraphic font, which I use only in extreme circumstances.) I write like this: And everybody can read what I'm writing. Especially, take look at letter x. And here are my digits: How about using cursive handwriting as shown below: The key is neatness. I am a student, and I take for granted that my lecturers will use neat handwriting. I don't like it when a lecturer's handwriting is scrawled hastily and messily because it makes it difficult to read. I also don't like it when a lecturer neglects to write down something they have said, or something that is helpful with understanding what is being communicated (e.g. writing " Proof" before a proof to let the class know that the theorem has ended). Below I have given some pictures of the best handwriting that I have encountered. Picture 1: this lecturer gives a tail on his v's and w's. The only problem with this handwriting is that the difference between a u and a v can be confusing. The following two pictures show writing done by a lecturer who is exceptionally neat when he writes on the board. His style of teaching is to have a lot of discussion, and to only write down the mathematical symbols which he is talking about. One thing I have run into is, when learning or tutoring probability, being able to distinguish the random variables with their density functions' support. To that end, I typically put serifs on my capital letters only, and that helps me tell the difference. One big issue is telling the difference between $W$, $w$, and $\omega$. I typically put serifs on $W$, keep the angles crisp on the $w$, and exaggerate the round parts for the $\omega$. I don't understand how you can mix up a t and a +-sign: t contains a curl at the bottom and the bar does not pass left of the vertical line +-sign consists only of straight lines. y starts with a curl and has a hole at the bottom 4 consists only of straight lines. z is written in the French way, the hole is quite long. 2 has a small but clear hole. I tend to loop my ζ so it doesn't look like a Z and I also put the lines on the z so it is not a 2.I also make my 1 like shown in the picture so it is not an I. I make my lowercase g fancy so it is not a 9. I put lines on my Θ so they are not θs. I also make my α short so they are not a ∝. I do not make my Phis like a plain old circle with a line. I tend to make a loop and then come down. My μs have a loop on the side.
This was a humorous thought experiment that occurred while chatting about black holes. The person that I was talking to assumed that a black hole required a specific density to be achieved. I pointed to the formula for the Schwarzschild radius. This suggests that low density black holes are possible if they are large enough. I'll assume simple spherical, non-rotating, uncharged black holes. $$ r = \frac{2 G M}{c ^ 2} $$ Using $\rho$ as the the average density. $$ M = \frac{4}{3} \pi r^3 \rho $$ $$ \rho = \frac{3}{8 \pi}\frac{c^2}{G r^2} $$ Which can be as small as you like by having $r$ large enough. So, assuming that the average density of a person is approximately that of water, I calculate that if I achieved a radius of approximately $4 \times 10^8 km$ then I would become a black hole. If I was centred at the Sun then I would extend into the asteroid belt. Of course, people are not usually spherical though if you get this large then it is probably a good approximation. There may also be some health issues and practical issues with obtaining sufficient food, oxygen, etc but they are off-topic in this group. Am I right, is a black hole with the same average density as water possible? Apart from the biological and nutrient issues, what others might I face? I guess that I will suffer severe problems long before I reach this size. Would I collapse and become a star (quite literally)? Alternatively, if an extremely large number of people gathered in a huge group hug, could they become a black hole? Update Thanks to the comments from kleingordon and dmitry-brant, my main question is answered. This leaves: Is my calculation right? (To 1 significant digit) Would I become a star or face some other calamity before becoming a black hole?
If I have observations of $y_{i}$ and $x_{i}$ which are i.i.d. I also have OLS assumptions such as $E(\epsilon_{i} \mid X_{i})= 0$, my qustion is: If I project $y_{i}$ onto a constant $\mu$, that is, we have model $y_{i} = \mu + \epsilon_{i}$. Does finding the OLS estimator $\hat\mu$ has anything to do with $x_{i}$? Because in my opinions, $x_{i}$ never emerges. Thanks~ So basically the question is: If I know the average ($\hat{\mu}$) of the daily temperatures ($y_i$) of last year, does that tell me anything about how many people were born ($x_i$) each day? Unsurprisingly the answer is no. The most you can get is the average of the $x_i$ series if you have the parameters of an unbiased regression between $x_i$ and $y_i$. I don't think it will have anything to do with $x_i$. Here is my thought: Given your setup, in order to find $\hat{\mu}$, we regress $y$ on an $n\times1$ vector of ones, $\begin{bmatrix}1\\1\\ \vdots \\1\end{bmatrix}$ ,which we shall call $\iota$ ( iota). Then we will have $\hat{\mu}=(\iota'\iota)^{-1}\iota'y=\frac{1}{n}\iota'y=\bar{y}$. So $x$ doesn't play a role here. Note the projection matrix $P_{\iota} = \iota(\iota'\iota)^{-1}\iota'=\frac{1}{n}\iota\iota'$, and $P_{\iota}y=\bar{y}$.
I’m not sure if this answering attempt is correct in the light of Mithoron’s and ron’s comments on your question, but this is the way I learnt it, so if this is wrong I will at least learn something, too. We all know what s-, p- and d-orbitals look like, but what is the significance, and why do these orbitals preferentially form $\sigma$, $\pi$ and $\delta$ bonds, respectively? Mathematically spoken, orbitals are functions of the hydrogen atom that solve the Schrödinger equation. The model in question is a non-rigid rotor,* i.e. the rotor’s axis is not fixed in any spatial direction (the electron can rotate freely around the nucleus). For solving this equation, it is helpful to use polar coordinates $(r, \varphi, \theta)$, mainly because the solution can be split into a radial factor (dependent only on $r$) and angular factors (dependent on $\varphi$ and $\theta$). $$\Psi (r, \varphi, \theta) = R(r) \cdot Y(\varphi, \theta)$$ $R(r)$ can be thought of giving an orbital its extension into space while $Y(\varphi, \theta)$ gives it its shape. Both functions depend heavily on quantum numbers: $R(r)$ does so for $n$ and $l$ while $Y(\varphi, \theta)$ depends on $l$ and $m_l$. For the simplest case ($l = 0; m_l = 0$, s-orbital), $Y (\varphi, \theta)$ degenerates to a simple constant, meaning that the orbital will have a totally symmetrical spherical shape. $l = 1$, (the p-orbital) while loosing spherical symmetry, still keeps total symmetry with respect to one axis, i.e. every slice you take through that orbital perpendicular to the axis of symmetry will be a circle. Higher quantum numbers lose more symmetry but it’s not always as easily visualised, so I’ll stick with these. But you were talking about bonds, where do they come into play? Well, bonds also have a symmetry, but they also have an axis instead of a nucleus, so their symmetry will be reduced per se. The simplest symmetry along a bond axis is total rotational symmetry around the bonds axis. I hope you see the similarity between the s-orbital (total symmetry around a central point) and a $\sigma$ bond (total symmetry around the bond’s central axis). Similarly, a $\pi$ bond will always have one degree of symmetry less, which turns out to mean ‘having a plane of symmetry that includes the bond axis’. And a $\delta$ bond will have two planes of symmetry — yet another degree of symmetry less. According to this definition, an orbital that can take part in a $\sigma$ bond needs to have full rotational symmetry along the bond’s axis. That means, that there is only one, at most two orbitals that fulfil the criterion (but if there are two, one is going to be an unmodified s-orbital and likely not take part in bonding at all). Therefore, only one $\sigma$ bond would be possible between two atoms. Writing this up, I remembered the ‘banana bonds’ that were introduced to us to explain the extremely small ($60°$) bond angles in $\ce{P4}$. I would need to go back, recheck and rethink what I would think of those and if I would treat them as exceptions of this ‘rule’ or simply as special cases that need additional information to be discussed. They certainly deserve consideration, as they are, de facto $\sigma$ bonds from the way they look, even though they do bend. An interesting comment was left on the question pointing to sextuple bonds. I didn’t know bonds of that order existed; my knowledge was stuck at 4. For a quadruple bond, possible between certain transition metals such as in $\ce{[Re2Cl8]^2-}$, four of the five d-orbitals form a bond to the other metal; one being $\sigma$, two being $\pi$ and a fourth one of $\delta$ type (to planes of symmetry). Extending that to a quintuple bond by adding a second $\delta$ layer with the last remaining pair of d-orbitals isn’t hard. The sextuple bond – eg $\ce{Mo2}$ — derives from an additional $\sigma$ bond between the s-orbitals of the higher shell. You thereby solve a problem you would otherwise have: The $4\mathrm{d}_{z^2}$ orbital can take part in $\sigma$ bonding along the $z$-axis; and the higher $5\mathrm{s}$ orbital is more diffuse, extends further into space and therefore is still able to form a contact to the neighbouring atom’s counterpart. Because it is more or less a sphere, it can only form $\sigma$ bonds. * I don’t think this is the model’s correct name. In my German quantum chemistry class, the rigid rotor was a raumstarrer Rotator and thus the model here was a raumfreier Rotator. Somebody who might know the proper name please comment (or edit).
I think the best answer is the one given here: http://www.nick-santos.com/jokes/2007/05/mathematician-and-plumber.html EDIT (at Aryabhatta's suggestion): The area is $$\int\int_{\rm circle}1\,dy\,dx=\int_0^{2\pi}\int_0^Rr\,dr\,d\theta={\rm etc.}$$ but this loses all the flavor of the link. MORE EDIT: The link died, so I updated in the comments. But in case that link dies, too, here's what's actually there. Nick Santos got it from the geometer Peter Doyle. Once upon a time there was a mathematician. His toilet was clogged. So he called the plumber. The plumber arrived later that evening, unclogged the toilet in 15 minutes, and handed the mathematician the bill. The mathematician looked at the bill and shouted: "Great scott! What a bill! You plumbers must make a fortune charging people this much. Do you mind if I ask how much you make?" The plumber doesn't mind. He grins and names his salary. The Mathematician: "That's more than I make. And it looks a lot easier than theoretical mathematics." The Plumber: "Well, friend, you sound eager, so I'll give you a tip. The plumbing business is swell. And my foreman needs more men for the job. Tell him I sent you, and you can see for yourself what the work is like. One thing though: don't tell him you're a mathematician. He hates elitists, and he won't hire anyone with higher than an 8th grade education." Well the mathematician was as serious as he claimed. The next day he did indeed go to see the plumber's foreman. By pretending to be a middle-school drop-out, he got the job easily. And lo! It was better than he had hoped. He found himself with better pay, fewer hours, and more respect than he ever had as a theoretical mathematician. The mathematician worked many years as a happy plumber. The plumbing industry flourished, and one day the foreman decided that his plumbers really had to be at a 9th-grade level in order to remain competitive in this rapidly-growing field. He hired several private tutors, and required his plumbers to attend night classes. On the first day of class, the math teacher was trying to gauge what his students knew. He singled out one plumber and asked, "Do you know the formula for the area of the circle?" Of course, the tutor had picked our mathematician, and like any good mathematician, he responded, "No. But I know how to derive it." "Derive it for me then," challenged the teacher. So the mathematician went to the blackboard and began to compute the area of the circle as any good mathematician would. He wrote out the double integral with respect to x and y, computed the Jacobian with respect to r and theta so that he could perform a change of basis in terms of polar coordinates, then evaluated the double integral. And came to the solution $-\pi r^2$. "Wait!" said the mathematician right before he announced his answer. "That can't be right!" He began to check his work, looking for where the negative got introduced. But he couldn't find his mistake. At last he threw up his hands, erased his computation, and started over again. He quickly set up the double integral, did the change of basis to polar coordinates, and simplified. And again he arrived at $-\pi r^2$! By now the mathematician was frantic. How could he have gotten such a ridiculous answer twice? Had his math skills really gotten so bad? He wildly looked to the teacher, but the teacher knew nothing of multi-variable calculus, and had no idea what was going on. Then he looked to the class...and noticed something quite odd. The eyes of every plumber were fixed upon him! The class was anxiously, quietly trying to get his attention without alerting the teacher. Each one had his hands cupped around his mouth, and they were all whispering the same words over and over again in unison. Slowly, he leaned in to hear better. And this is what he heard: "You forgot to take the absolute value of the Jacobian!"
Given a set $P$ of real numbers $\ge 1$, define the gap among different products in $P$ as $$g(P) = \inf \big\{\prod_{i=1}^n p_i^{a_i} - \prod_{i=1}^n p_i^{b_i} \mid p_i\in P;\,\, p_i\ne p_j \,\text{ if }\, i\ne j ; \,\,a_i, b_i \in \mathbb{N}\cup \{0\};\,\, a_1 \ne b_1 \big\}$$ If $g(P)>0$ in one could say that the set of products of members of $P$ has unique factorization in an essentially strong way. The main question is: can any infinite sets $P$ with $g(P)> 0$ be produced with a density similar to that of the primes, but more regularly distributed (for example $p_i=(i+a)\log(bi+c)+d$ for some constants $a,b,c,d$, or $iH_i+a$)? Can any interesting set $P$ with $g(P)>0$ be produced at all?
Let $\mu(\cdot)$ be the Mobius function, defined on the natural numbers by $$\displaystyle \mu(n) = \begin{cases} (-1)^{\omega(n)} & \text{if } n \text{ is square-free} \\ 0 & \text{otherwise}.\end{cases}$$ Here $\omega(n)$ is the number of distinct prime divisors of $n$. It is well-known that the Mertens function $M(x)$ defined by $$\displaystyle M(x) = \sum_{n \leq x} \mu(n)$$ satisfies the asymptotic $$\displaystyle M(x) = O\left(x \exp(-c \sqrt{\log x}) \right)$$ for some positive number $c$; this is a consequence of the prime number theorem. The Riemann hypothesis is equivalent to the assertion that for any $\varepsilon > 0$ the Mertens function satisfies $M(x) = O_\varepsilon \left(x^{1/2 + \varepsilon}\right)$. Let $a, q$ be co-prime positive integers, and let $P_{a,q}$ denote the set of primes $p \equiv a \pmod{q}$. Let $N_{a,q}$ be the set of positive integers such that $p \| n \Rightarrow p \in P_{a,q}$. Let $$M_{a,q}(x) = \sum_{\substack{n \leq x \\ n \in N_{a,b}}} \mu(n).$$ Does $M_{a,q}(x)$ satisfy a similar asymptotic as $M(x)$? Next, let $f \in \mathbb{Z}[x]$. Put $\rho_f(n) = \# \{m \in \mathbb{Z}/n \mathbb{Z} : f(m) \equiv 0 \pmod{n}\}$. Define $$\displaystyle M_f(x) = \sum_{n \leq x} \mu(n) \rho_f(n).$$ Does $M_f(x)$ satisfy a similar asymptotic upper bound as $M(x)$?
This answer summarises parts of TAoCP Vol 3, Ch 6.4. Assume we have a set of values $V$, $n$ of which we want to store in an array $A$ of size $m$. We employ a hash function $h : V \to [0..M)$; typically, $M \ll \|V\|$. We call $\alpha = \frac{n}{m}$ the load factor of $A$.Here, we will assume the natural $m=M$; in practical scenarios, we have $m \ll M$, though, and have to map down to $m$ ourselves. The first observation is that even if $h$ has uniform characteristics¹ the probability of two values having the same hash value is high; this is essentially an instance of the infamous birthday paradox. Therefore, we will usually have to deal with conflicts and can abandon hope of $\mathcal{O}(1)$ worst case access time. What about the average case, though? Let us assume that every key from $[0..M)$ occurs with the same probability. The average number of checked entries $C_n^S$ (successful search) resp. $C_n^U$ (unsuccessful search) depends on the conflict resolution method used. Chaining Every array entry contains (a pointer to the head of) a linked lists. This is a good idea because the expected list length is small ($\frac{n}{m}$) even though the probability for having collisions is high. In the end, we get\[ C_n^S \approx 1 + \frac{\alpha}{2} \quad \text{ and } \quad C_n^U \approx 1 + \frac{\alpha^2}{2} .\]This can be improved slightly by storing the lists (partly or completely) inside the table. Linear Probing When inserting (resp. searching a value) $v$, check positions \[h(v), h(v)-1,\dots,0,m-1,\dots,h(v)+1\]in this order until an empty position (resp. $v$) is found. The advantage is that we work locally and without secondary data structures; however, the number of average accesses diverges for $\alpha \to 1$:\[ C_n^S \approx \frac{1}{2}\left(1 +\frac{1}{1-\alpha}\right) \quad \text{ and } \quad C_n^U \approx \frac{1}{2}\left(1 +\left(\frac{1}{1-\alpha}\right)^2\right).\]For $\alpha < 0.75$, however, performance is comparable to chaining². Double Hashing Similar to linear probing but search step size is controlled by a second hash function that is coprime to $M$. No formal derivation is given, but empirical observations suggest\[ C_n^S \approx \frac{1}{\alpha}\ln\left(\frac{1}{1-\alpha}\right)\quad \text{ and } \quad C_n^U \approx \frac{1}{1-\alpha} .\]This method has been adapted by Brent; his variant amortises increased insertion costs with cheaper searches. Note that removing elements from and extending tables has varying degrees of difficulty for the respective methods. Bottom-line, you have to choose an implementation that adapts well to your typical use cases. Expected access time in $\mathcal{O}(1)$ is possible if not always guaranteed. Depending on the used method, keeping $\alpha$ low is essential; you have to trade off (expected) access time versus space overhead. A good choice for $h$ is also central, obviously. 1] As arbitrarily dumb uninformed programmers may provide $h$, any assumption regarding its quality is a stretch in practice. 2] Note how this coincides with recommendations for usage of Java's Hashtable.
In Atiyah Macdonald, for a commutative and unital ring $A$ the tensor product of $A$-modules, $M,N$ is defined by the usual universal property: A pair $(T,g)$ with $T$ an $A$-module and $g$ an $A$-bilinear map $g\colon M \times N \to T$ such that for any $A$-module $P$ with an $A$-bilinear map $f\colon M \times N \to P$ there exists a unique $A$-module homomorphism $h\colon T \to P$ such that $h \circ g = f$. And then, as usually goes with universal properties: Moreover, if $(T,g), (T',g')$ are any two pairs satisfying this condition, then there is a unique isomorphism $j \colon T \to T'$ such that $j \circ g = g'$ My question is, in the second highlighted selection, is the converse true? If I have already established the existence of the usual tensor product, $M \otimes_A N$, via quotient of free module, and I encounter some other pair $(T', g')$ and a unique isomorphism $j: T \to T'$ such that $j \circ g = g'$, can I conclude that $T'$ also satisfies the universal property defining the tensor product, thus I can regard both $T$ and $T'$ as the tensor product of this $M$ and $N$? I believe I proved that I can indeed do this, but I need reassurance because I have overlooked small details with tensor products before, and don't want to make that mistake again. My proof was as follows: Suppose $(T,g)$ satisfies the universal property and $(T',g')$ is a pair where $T'$ is an $A$-mod and $g'$ an $A$-bilinear map $g' \colon M \times N \to T'$. Suppose $j$ is the unique isomorphism $j\colon T \to T'$ such that $j \circ g = g'$, or $g = j^{-1} \circ g'$. Suppose $P$ is any $A$-mod with an $A$-bilinear map $f\colon M \times N \to P$. We wish to show there exists a unique $A$-module homomorphism $\phi\colon T' \to P$ such that $\phi \circ g' = f$. Since $(T,g)$ already satisfies the property by hypothesis, we know there exists a unique $A$-module homomorphism $\varphi \colon T \to P$ such that $\varphi \circ g = f$. So define the map $\phi \colon T' \to P$ by $\varphi \circ j^{-1}$. Then $\phi \circ g' = \varphi \circ j^{-1} \circ g' = \varphi \circ g = f$. So the diagram commutes as desired and the map $\phi$ is defined as the composition of two unique maps so is unique as well, thus $(T',g')$ satisfies the universal property. Sorry that got a little (unnecessarily) complicated, I don't know how to draw diagrams on here, which really makes everything easier. As an aside, I feel like this probably is true and Atiyah & Macdonald didn't mention it in the text because they deemed it trivial.
How would I do {text}_{i=1}^{i=n} with the written exactly above and below the text? Hope you can help! 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 How would I do {text}_{i=1}^{i=n} with the written exactly above and below the text? Hope you can help! Here are two ways: \documentclass{article}\usepackage{amsmath,stackengine}\stackMath\begin{document}\[\setstackgap{S}{2pt}\stackunder{\stackon{\text{text}}{\scriptstyle i=n}}{\scriptstyle i=1}\quad\underset{i=1}{\overset{i=n}{\text{text}}}\]\end{document} The gap between the text and the over/under-set is defined with \setstackgap{S}{2pt} for the stackengine approach on the left. Either of these approaches will work in \displaystyle (shown) or in \textstyle. Note, however, that in \textstyle, the line spacing will be affected. I assume you'll want the subscript and superscript material to be in display-style math mode and to use a smaller font size than what's used for the "text" string. Sort of like \sum_{i=0}^{\infty}, except for the use of the string "text" instead of the large summation symbol, right? If these assumptions are correct, you could use the \DeclareMathOperator* directive of the amsmath package to achieve your objective. \documentclass{article}\usepackage{amsmath}\DeclareMathOperator*{\sometext}{text} % "\text" is already taken...\begin{document}\[\sometext_{i=1}^{i=n} % not "{\sometext}"\]\end{document} In addition to the other suggestions you could use: \mathop together with \operatorname from the amsmath package: \documentclass{article}\usepackage{amsmath}\newcommand\mymathop[1]{\mathop{\operatorname{#1}}}\begin{document}\[ \mymathop{text}_{i=1}^{i=n}\]This is in a paragraph: $\mymathop{text}_{i=1}^{i=n}$ or $\displaystyle\mymathop{text}_{i=1}^{i=n}$.\end{document} \documentclass{article}\usepackage{stackrel}\begin{document}Some $\stackrel[i=1]{i=n}{\mathrm{text}}$ in line.\end{document}
I like George Bergman's explanation (beginning in section 7.4 of his Invitation to General Algebra and Universal Constructions). We start with a motivating example. Suppose you are interested in solving $x^2=-1$ in $\mathbb{Z}$. Of course, there are no solutions, but let's ignore that annoying reality for a moment. We use the notation $\mathbb{Z}_n$ for $\mathbb Z / n \mathbb Z$.The equation has a solution in the ring $\mathbb{Z}_5$ (in fact, two: both $2$ and $3$, which are the same up to sign). So we want to find a solution to $x^2=-1$ in $\mathbb{Z}$ which satisfies $x\equiv 2 \pmod{5}$. An integer that is congruent to $2$ modulo $5$ is of the form $5y+2$, so we can rewrite our original equation as $(5y+2)^2 = -1$, and expand to get$25y^2 + 20y = -5$. That means $20y\equiv -5\pmod{25}$, or $4y\equiv -1\pmod{5}$, which has the unique solution $y\equiv 1\pmod{5}$. Substituting back we determine $x$ modulo $25$:$$x = 5y+2 \equiv 5\cdot 1 + 2 = 7 \pmod{25}.$$ Continue this way: putting $x=25z+7$ into $x^2=-1$ we conclude $z\equiv 2 \pmod{5}$, so $x\equiv 57\pmod{125}$. Using Hensel's Lemma, we can continue this indefinitely. What we deduce is that there is a sequence of residues, $$x_1\in\mathbb{Z}_5,\quad x_2\in\mathbb{Z}_{25},\quad \ldots, x_{i}\in\mathbb{Z}_{5^i},\ldots$$each of which satisfies $x^2=-1$ in the appropriate ring, and which are "consistent", in the sense that each $x_{i+1}$ is a lifting of $x_i$ under the natural homomorphisms$$\cdots \stackrel{f_{i+1}}{\longrightarrow} \mathbb{Z}_{5^{i+1}} \stackrel{f_i}{\longrightarrow} \mathbb{Z}_{5^i} \stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow} \mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow} \mathbb{Z}_5.$$ Take the set of all strings $(\ldots,x_i,\ldots,x_2,x_1)$ such that $x_i\in\mathbb{Z}_{5^i}$ and $f_i(x_{i+1}) = x_i$, $i=1,2,\ldots$. This is a ring under componentwise operations. What we did above shows that in this ring, you do have a square root of $-1$. Added. Bergman here inserts the quote, "If the fool will persist in his folly, he will become wise." We obtained the sequence by stubbornly looking for a solution to an equation that has no solution, by looking at putative approximations, first modulo 5, then modulo 25, then modulo 125, etc. We foolishly kept going even though there was no solution to be found. In the end, we get a "full description" of what that object must look like; since we don't have a ready-made object that satisfies this condition, then we simply take this "full description" and use that description as if it were an object itself. By insisting in our folly of looking for a solution, we have become wise by introducing an entirely new object that is a solution. This is much along the lines of taking a Cauchy sequence of rationals, which "describes" a limit point, and using the entire Cauchy sequence to represent this limit point, even if that limit point does not exist in our original set. This ring is the $5$-adic integers; since an integer is completely determined by its remainders modulo the powers of $5$, this ring contains an isomorphic copy of $\mathbb{Z}$. Essentially, we are taking successive approximations to a putative answer to the original equation, by first solving it modulo $5$, then solving it modulo $25$ in a way that is consistent with our solution modulo $5$; then solving it modulo $125$ in a way that is consistent with out solution modulo $25$, etc. The ring of $5$-adic integers projects onto each $\mathbb{Z}_{5^i}$ via the projections; because the elements of the $5$-adic integers are consistent sequences, these projections commute with our original maps $f_i$. So the projections are compatible with the $f_i$ in the sense that for all $i$, $f_i\circ\pi_{i+1} = \pi_{i}$, where $\pi_k$ is the projection onto the $k$th coordinate from the $5$-adics. Moreover, the ring of $5$-adic integers is universal for this property: given any ring $R$ with homomorphisms $r_i\colon R\to\mathbb{Z}_{5^i}$ such that $f_i\circ r_{i+1} = r_i$, for any $a\in R$ the tuple of images $(\ldots, r_i(a),\ldots, r_2(a),r_1(a))$ defines an element in the $5$-adics. The $5$-adics are the inverse limit of the system of maps$$\cdots\stackrel{f_{i+1}}{\longrightarrow}\mathbb{Z}_{5^{i+1}}\stackrel{f_i}{\longrightarrow}\mathbb{Z}_{5^i}\stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow}\mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow}\mathbb{Z}_5.$$ So the elements of the inverse limit are "consistent sequences" of partial approximations, and the inverse limit is a way of taking all these "partial approximations" and combine them into a "target object." More generally, assume that you have a system of, say, rings, $\{R_i\}$, indexed by an directed set $(I,\leq)$ (so that for all $i,j\in I$ there exists $k\in I$ such that $i,j\leq k$), and a system of maps $f_{rs}\colon R_s\to R_r$ whenever $r\leq s$ which are "consistent" (if $r\leq s\leq t$, then $f_{rs}\circ f_{st} = f_{rt}$), and let's assume that the $f_{rs}$ are surjective, as they were in the example of the $5$-adics. Then you can think of the $R_i$ as being "successive approximations" (with a higher indexed $R_i$ as being a "finer" or "better" approximation than the lower indexed one). The directedness of the index set guarantees that given any two approximations, even if they are not directly comparable to one another, you can combine them into an approximation which is finer (better) than each of them (if $i,j$ are incomparable, then find a $k$ with $i,j\leq k$). The inverse limit is a way to combine all of these approximations into an object in a consistent manner. If you imagine your maps as going right to left, you have a branching tree that is getting "thinner" as you move left, and the inverse limit is the combination of all branches occurring "at infinity". Added. The example of the $p$-adic integers may be a bit misleading because our directed set is totally ordered and all maps are surjective. In the more general case, you can think of every chain in the directed set as a "line of approximation"; the directed property ensures that any finite number of "lines of approximation" will meet in "finite time", but you may need to go all the way to "infinity" to really put all the lines of approximation together. The inverse limit takes care of this. If the directed set has no maximal elements, but the structure maps are not surjective, it turns out that no element that is not in the image will matter; essentially, that element never shows up in a net of "successive approximations", so it never forms part of a "consistent system of approximations" (which is what the elements of the inverse limit are).
This has been a question I've been asking myself for quite some time now. Is there a physical Interpretation of the Hypersingular Boundary Operator? First, let me give some motivation why I think there could be one. There is a rather nice physicial interpretation of the Single and Double Layer potential (found here: Physical Interpretation of Single and Double Layer Potentials). To give a short summary of the Article: We can think of the Single Layer Potential as a potential induced by a distribution of charges on the Boundary. And the Double Layer Potential of two parallel distributions (as in the single layer case) of opposite sign. As the Hypersingular Operator arises from the Double Layer, I would think that it would have an analog interpretation. The Hypersingular Operator $W$ is (most commonly) defined as: $W \varphi (x) := -\partial_{n_x} K \varphi(x)$ for some $x\in\Gamma$, where $\partial_{n_x}$ is the normal derivative at $x$ and $K$ denotes the double layer boundary integral operator: $K\varphi(x) := -\frac{1}{4\pi} \int_\Gamma \varphi(y)\partial_{n_y} \frac{1}{\vert x-y\vert} ds_y$ I was wondering, is there some physical, intuitiv or geometric way of thinking about this (or perhaps a paper that could help me gain some intuition)? Or is it merely an Operator meant to "tidy" things up a bit?
Miscellaneous matrix functions¶ sage.matrix.matrix_misc. permanental_minor_polynomial( A, permanent_only=False, var='t', prec=None)¶ Return the polynomial of the sums of permanental minors of A. INPUT: $A$ – a matrix $permanent_only$ – if True, return only the permanent of $A$ $var$ – name of the polynomial variable $prec$ – if prec is not None, truncate the polynomial at precision $prec$ The polynomial of the sums of permanental minors is\[\sum_{i=0}^{min(nrows, ncols)} p_i(A) x^i\] where $p_i(A)$ is the $i$-th permanental minor of $A$ (that can also be obtained through the method permanental_minor()via A.permanental_minor(i)). The algorithm implemented by that function has been developed by P. Butera and M. Pernici, see [BP2015]. Its complexity is $O(2^n m^2 n)$ where $m$ and $n$ are the number of rows and columns of $A$. Moreover, if $A$ is a banded matrix with width $w$, that is $A_{ij}=0$ for $\|i - j\| > w$ and $w < n/2$, then the complexity of the algorithm is $O(4^w (w+1) n^2)$. INPUT: A– matrix permanent_only– optional boolean. If True, only the permanent is computed (might be faster). var– a variable name EXAMPLES: sage: from sage.matrix.matrix_misc import permanental_minor_polynomial sage: m = matrix([[1,1],[1,2]]) sage: permanental_minor_polynomial(m) 3t^2 + 5t + 1 sage: permanental_minor_polynomial(m, permanent_only=True) 3 sage: permanental_minor_polynomial(m, prec=2) 5t + 1 sage: M = MatrixSpace(ZZ,4,4) sage: A = M([1,0,1,0,1,0,1,0,1,0,10,10,1,0,1,1]) sage: permanental_minor_polynomial(A) 84t^3 + 114t^2 + 28t + 1 sage: [A.permanental_minor(i) for i in range(5)] [1, 28, 114, 84, 0] An example over $\QQ$: sage: M = MatrixSpace(QQ,2,2) sage: A = M([1/5,2/7,3/2,4/5]) sage: permanental_minor_polynomial(A, True) 103/175 An example with polynomial coefficients: sage: R.<a> = PolynomialRing(ZZ) sage: A = MatrixSpace(R,2)([[a,1], [a,a+1]]) sage: permanental_minor_polynomial(A, True) a^2 + 2a A usage of the varargument: sage: m = matrix(ZZ,4,[0,1,2,3,1,2,3,0,2,3,0,1,3,0,1,2]) sage: permanental_minor_polynomial(m, var='x') 164x^4 + 384x^3 + 172x^2 + 24x + 1 ALGORITHM: The permanent $perm(A)$ of a $n \times n$ matrix $A$ is the coefficient of the $x_1 x_2 \ldots x_n$ monomial in\[\prod_{i=1}^n \left( \sum_{j=1}^n A_{ij} x_j \right)\] Evaluating this product one can neglect $x_i^2$, that is $x_i$ can be considered to be nilpotent of order $2$. To formalize this procedure, consider the algebra $R = K[\eta_1, \eta_2, \ldots, \eta_n]$ where the $\eta_i$ are commuting, nilpotent of order $2$ (i.e. $\eta_i^2 = 0$). Formally it is the quotient ring of the polynomial ring in $\eta_1, \eta_2, \ldots, \eta_n$ quotiented by the ideal generated by the $\eta_i^2$. We will mostly consider the ring $R[t]$ of polynomials over $R$. We denote a generic element of $R[t]$ by $p(\eta_1, \ldots, \eta_n)$ or $p(\eta_{i_1}, \ldots, \eta_{i_k})$ if we want to emphasize that some monomials in the $\eta_i$ are missing. Introduce an “integration” operation $\langle p \rangle$ over $R$ and $R[t]$ consisting in the sum of the coefficients of the non-vanishing monomials in $\eta_i$ (i.e. the result of setting all variables $\eta_i$ to $1$). Let us emphasize that this is nota morphism of algebras as $\langle \eta_1 \rangle^2 = 1$ while $\langle \eta_1^2 \rangle = 0$! Let us consider an example of computation. Let $p_1 = 1 + t \eta_1 + t \eta_2$ and $p_2 = 1 + t \eta_1 + t \eta_3$. Then\[p_1 p_2 = 1 + 2t \eta_1 + t (\eta_2 + \eta_3) + t^2 (\eta_1 \eta_2 + \eta_1 \eta_3 + \eta_2 \eta_3)\] and\[\langle p_1 p_2 \rangle = 1 + 4t + 3t^2\] In this formalism, the permanent is just\[perm(A) = \langle \prod_{i=1}^n \sum_{j=1}^n A_{ij} \eta_j \rangle\] A useful property of $\langle . \rangle$ which makes this algorithm efficient for band matrices is the following: let $p_1(\eta_1, \ldots, \eta_n)$ and $p_2(\eta_j, \ldots, \eta_n)$ be polynomials in $R[t]$ where $j \ge 1$. Then one has\[\langle p_1(\eta_1, \ldots, \eta_n) p_2 \rangle = \langle p_1(1, \ldots, 1, \eta_j, \ldots, \eta_n) p_2 \rangle\] where $\eta_1,..,\eta_{j-1}$ are replaced by $1$ in $p_1$. Informally, we can “integrate” these variables beforeperforming the product. More generally, if a monomial $\eta_i$ is missing in one of the terms of a product of two terms, then it can be integrated in the other term. Now let us consider an $m \times n$ matrix with $m \leq n$. The sum of permanental `k`-minors of `A`is\[perm(A, k) = \sum_{r,c} perm(A_{r,c})\] where the sum is over the $k$-subsets $r$ of rows and $k$-subsets $c$ of columns and $A_{r,c}$ is the submatrix obtained from $A$ by keeping only the rows $r$ and columns $c$. Of course $perm(A, \min(m,n)) = perm(A)$ and note that $perm(A,1)$ is just the sum of all entries of the matrix. The generating function of these sums of permanental minors is\[g(t) = \left\langle \prod_{i=1}^m \left(1 + t \sum_{j=1}^n A_{ij} \eta_j\right) \right\rangle\] In fact the $t^k$ coefficient of $g(t)$ corresponds to choosing $k$ rows of $A$; $\eta_i$ is associated to the i-th column; nilpotency avoids having twice the same column in a product of $A$’s. For more details, see the article [BP2015]. From a technical point of view, the product in $K[\eta_1, \ldots, \eta_n][t]$ is implemented as a subroutine in prm_mul(). The indices of the rows and columns actually start at $0$, so the variables are $\eta_0, \ldots, \eta_{n-1}$. Polynomials are represented in dictionary form: to a variable $\eta_i$ is associated the key $2^i$ (or in Python 1 << i). The keys associated to products are obtained by considering the development in base $2$: to the monomial $\eta_{i_1} \ldots \eta_{i_k}$ is associated the key $2^{i_1} + \ldots + 2^{i_k}$. So the product $\eta_1 \eta_2$ corresponds to the key $6 = (110)_2$ while $\eta_0 \eta_3$ has key $9 = (1001)_2$. In particular all operations on monomials are implemented via bitwise operations on the keys. sage.matrix.matrix_misc. prm_mul( p1, p2, mask_free, prec)¶ Return the product of p1and p2, putting free variables in mask_freeto $1$. This function is mainly use as a subroutine of permanental_minor_polynomial(). INPUT: $p1,p2$ – polynomials as dictionaries $mask_free$ – an integer mask that give the list of free variables(the $i$-th variable is free if the $i$-th bit of mask_freeis $1$) $prec$ – if $prec$ is not None, truncate the product at precision $prec$ EXAMPLES: sage: from sage.matrix.matrix_misc import prm_mul sage: t = polygen(ZZ, 't') sage: p1 = {0: 1, 1: t, 4: t} sage: p2 = {0: 1, 1: t, 2: t} sage: prm_mul(p1, p2, 1, None) {0: 2t + 1, 2: t^2 + t, 4: t^2 + t, 6: t^2} sage.matrix.matrix_misc. row_iterator( A)¶
If you have an inequality that has two absolute value bars like $\|4x+1\|<\|3x\|$, how do you go about doing this? I know that if $4x+1<3x$, then those $x$'s will work but what else do I do? I think you do $4x+1<-3x$. Is this correct? You could also square everything $$ \|f(x)\| < \|g(x)\| \Leftrightarrow \|f(x)\|^2 < \|g(x)\|^2 \\ \Leftrightarrow f(x)^2 < g(x)^2 \\ \Leftrightarrow 0< g(x)^2-f(x)^2 \\ \Leftrightarrow 0< (g(x)-f(x))(g(x)+f(x)), \\ $$ which means that $g(x)-f(x)$ and $g(x)+f(x)$ have the same sign. Just take the different cases. For example: You know that $$ \|3x\|=\left\{ \begin{align} 3x & \text{ , if }x\geq 0 \\ -3x & \text{ , if }x <0 \end{align} \right\} $$ $$ \|4x+1\|=\left\{ \begin{align} 4x+1 & \text{ , if }x\geq \frac{-1}{4} \\ -(4x+1) & \text{ , if }x <\frac{-1}{4} \end{align} \right\} $$ This gives you a few different cases to check: $x<\frac{-1}{4}$, $\frac{-1}{4}\leq x <0$, and $x\geq 0$. So for instance, take the first cases: $x<\frac{-1}{4}$ so that $\|4x+1\|=-(4x+1)$ and $\|3x\|=-3x$. Then we have $$ \begin{align} \|4x+1\|&<\|3x\| \\ -(4x+1)&<-3x \\ -4x-1&<-3x \\ -x-1&<0 \\ -x&<+1\\ x&>-1 \end{align} $$ Just be sure to check the $\leq$ or $\geq$ cases and be sure the answers from the different regions agree! (Meaning if you found in one case $x>1$ and found in another case $x<1$ there would be no solutions) There's three parts here that you need to consider: the area where both 4x+1 and 3x are positive (0 < x); the area where one's positive and the other's negative (-1/4 < x < 0); the area where both are negative (x < -1/4). Union the results of those together. Need to break up into 2 cases: (3x) >= 0 or x >= 0: \|3x\| = 3x -3x < (4x+1) < 3x 7x > -1 and x < -1 and x>=0 (no solution) (3x) <= 0 or x<=0 \|3x\| = -3x 3x < (4x+1) < -3x x > -1 and 7x < -1 Solution: -1 < x < -1/7 I just solve the equalities without absolute values: LHS = RHS and LHS = -RHS. Then I use a number line test to check values in each interval to determine the intervals of the solution set. This led me to all values between -1 and -1/7, not including engpoints, since the original inequality was strict. protected by Zev Chonoles Oct 14 '15 at 17:13 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). Would you like to answer one of these unanswered questions instead?
Saving time Rate of return on invested capital Rate of withdrawal in retirement Saving rate Total invested capital Value of windfall at $t=$ 0 Savings per time Time when the windfall and consistent savings are equal Net present value Value of cash received at time $t$ Hypothetical rate of return This article summarizes the relationships relevant to financial independence, investing, and net present value calculations. It also contains the results of a comparison study between steady saving and windfalls. The acronym FIRE (financial independence, retire early) is the name of a small movement trying to save large fractions of income so the investment returns will cover their expenses. Philosophical reasons vary but generally, these people are not hoping to stop working completely, they are hoping to remove compensation from the decision-making process about what kind of work to do and whom to do it with. From the financial independence article, this equation and graph show how long one must save before their investment income covers their expenses (for constant income and expenses and starting with a net worth of $0). From the saving $1 million article, these equations and graphs show how a steady rate of investment grows over time [left] and how long it takes to save $1 million [right]. From the windfalls article, this equation and graphs show how long it takes a steady saver to overtake someone with a windfall. $V_0$ and $F$ in dollar and dollar per year, respectively. From the net present values article, for a pile of cash received at a future time, the NPV is the size of the pile of cash required today to generate that future pile of cash at a given rate of return. For $V_t$ dollars received at time $t$: For $F$ dollars per time, received between now and time $t$: If $F$ is a function of $t$ If $F$ is constant $\displaystyle \text{NPV} = \int^{t_\text{total}}_{0} e^{-r_r\tau} F(\tau) d\tau $ $\displaystyle \quad\quad \text{NPV} = \frac{F}{r_r} \left( 1- e^{-r_rt} \right) $ From the rule of 72 article: Doubling time Tripling time Quadrupling time $\displaystyle \frac{72}{r_\text{as percent}}$ $\displaystyle\frac{108}{r_\text{as percent}}$ $\displaystyle\frac{144}{r_\text{as percent}} $
Consider the standard linear regression model: $y_i = \alpha + \beta D_i + e_i$ where the coefficients are defined by linear projections and $D_i$ is a dummy variable. In the population, the coefficients are given by: $$\alpha = E[y_i \mid D_i =0] \ \text{and} \ \beta = E[y_i \mid D_i = 1] - E[y_i \mid D_i =0]$$ Using OLS to estimate the coefficients, we get: $$\widehat{\alpha} = \overline{y}_{D_i=0} $$ $$\widehat{\beta} = \overline{y}_{D_i=1}-\overline{y}_{D_i=0} $$ In other words, $\widehat{\alpha}$ is just the sample mean of $y_i$ in the subsample with $D_i=0$ and $\widehat{\beta}$ is the difference in sample means of the two groups. The expressions seem very obvious because they are just sample versions of the population, but my question is, how can we arrive at the above coefficient estimates by using the standard OLS formulas? That is using: $$\widehat{\alpha} = \overline{y} - \overline{D}\widehat{\beta} \ \ \text{and} \ \ \widehat{\beta} = \frac{\sum_{i=1}^{N}(D_i - \overline{D})(y_i - \overline{y})}{\sum_{i=1}^{N}(D_i - \overline{D})^2}$$
tl;dr Upon counting the orbits of the cells of a 3D cube under the octahedral group action, I found out the result is a sum of triangular numbers, which has a very intuitive geometric reason. The gory detailsThe octahedral group can be thought of as the group of symmetries obtained by applying a series of 90-degree rotations and/or face-parallel reflections to a cube. Another way of thinking about it is this: if the cube is centered at the origin, applying an element of the group to the cube is equivalent to shuffling the coordinates and/or multiplying some of the coordinates by $-1$. Since there are 6 permutations of the coordinates, and 8 choices as to which coordinates to multiply by $-1$, the number of elements in the group is 48. For simplicity, let's assume that the 3D cube is actually a discrete object, comprised of $n^3$ cells for some even value of $n$ (having an odd $n$ doesn't change the outcome by a lot, but is a bit different). The group action breaks the cube into orbits, for example, the corner cells form a single orbit of size 8. Note that the group elements can be also thought of as $3\times 3$ matrices, all of whose determinents are $1$ or $-1$ (the latter if the element induces a reflection), and the action thus preserves norms. To put it plainly, the distance of a cell from the cube's center is an invariant of the group action. This observation means that we can think of the cube as comprised of $\frac n 2$ nested hollow cubes of dimensions $n\times n \times n, (n-2)\times(n-2)\times(n-2), ... 4\times 4\times 4, 2\times 2\times 2$, sort of like a matryoshka doll of 3D cubes. Image credit: Wikipedia It suffices to understand how the group acts only on one of these hollow cubes, so let's examine only the largest one - the $n\times n\times n$ hollow cube. Classifying orbitsSuppose the cells are indexed w.r.t. the center of cube, so each cell is indexed by some $i, j, k\in \{-\frac n 2,..., \frac n 2\}\setminus \{0\}$ such that at least one of $i, j, k$ is $\pm \frac n 2$. We omit zero because $n$ is even and no cell is the center cell of the cube. Observe that using this notation, together with thinking of the group elements as permuting coordinates and/or pointwise multiplying them by $-1$, makes it evident that, up to a permutation, there are only 4 cases, or types of orbits: Case 1: $\frac n 2 = \|i\| \neq \|j\|, \|i\|\neq \|k\|, \|j\| \neq \|k\|$, choosing such indices defines an orbit of size 48, since no permutation of the coordinates or multiplication of them by $-1$ leaves the indices as they were. Case 2: $i = \|j\| = \|\frac n 2\|, \|j\|\neq \|k\|$, these are all the cells that lie on the edges of the hollow cube. A cell with such indices is invariant under elements switching $i$ and $j$, so its orbit's size is 24. To be precise, in the case where $j = -i = \pm \frac n 2$, swapping $i$ and $j$ is just like pointwise multiplying both by $-1$, so the orbit is obtained simply by moving $k$ around and pointwise multiplication by $-1$, and indeed the orbit size is 24. Case 3: $\|i\| = \|j\|, \|j\| < \|k\| = \frac n 2$, these are all the cells that lie on the diagonals that run along the face of the hollow cube. For the same arguments as Case 2 - the orbit size here is 24. Case 4: $\|i\| = \|j\| = \|k\| = \frac n 2$, these are the corner cells. The difference between two corners is only the sign of the coordinates, so the size of such an orbit is 8. Counting orbitsIt is now left to understand how many orbits there are for each case. We choose a representative for each orbit - $(i, j, k)$, and count those. Case 1: The choice is of some unordered pair of elements from $\{1,2,...,\frac n 2 - 1\}$. The pair is unordered since the group action puts $(i,j,k)$ in the same orbit as $(j,i,k)$, and we omit the negatives since they are obtained via the pointwise multiplication by $-1$. The number of ways to choose such a pair is therefore $\frac 1 2 (\frac n 2 - 1)(\frac n 2 - 2)$. Case 2: Since $\|j\|$ and $\|i\|$ must be equal to $\frac n 2$, the only choice here is of $1\leq k < \frac n 2$ (negative values of $k$ are obtained via the pointwise multiplication by $-1$ of the group action). So clearly there are $\frac n 2 - 1$ of those. Case 3: Like Case 2, the only choice here is of $1 \leq \|i\| < \frac n 2$, which in turn determins $\|j\|$. So there are $\frac n 2 - 1$ such orbits. Case 4: This is the corners' orbit, so there is exactly one for a hollow cube. Putting it together, a hollow 3D cube of dimension $n$ has $\frac 1 2 (\frac n 2 - 1)(\frac n 2 - 2) + 2 \cdot (\frac n 2 - 1) + 1 = \frac 1 8 (n^2 + 2n)$. Since the full 3D cube, with an even dimension $s$, is comprised of $\frac s 2$ such hollow cubes, the number of orbits is: $\frac 1 8 \sum _{n=1} ^{\frac s 2} ((2n)^2 + 2(2n))$, which surprisingly gives us a sum of triangular numbers $\sum_{n=1} ^{\frac s 2} \frac {n(n+1)} 2$, which has a nice closed form $\frac {\frac s 2 \cdot (\frac s 2 + 1) \cdot (\frac s 2 + 2)} 6$. Why triangular numbers?This is a satisfyingly clean formula to ed up with, after a tedious computation, and as one would expect, it is so for a reason: As noted in the previous paragraph, the result is a sum of triangular numbers, which are called thus because they also represent the number of blocks you'd use to construct a triangle where the bottom row has $n$ blocks, the next row has $n-1$ blocks and so on. The image below shows one representative of each orbit in the outer layer of an $8\times 8 \times 8$ cube. The Case 1 representatives are red, Case 2 - orange, Case 3 - yellow, and Case 4 - green. And indeed these representatives form a triangle, and the sum of triangles really represents (sort of) a pyramid that goes from the outer layer of the cube inward, the colorful part of the image below being its base.
In Brezis' Functional analysis, Sobolev spaces and partial differential eqautions, exercise 6.24 (3) asks to prove that for a self-adjoint operator $T\in \mathcal{L}(H)$, $H$ a Hilbert space, the following properties are equivalent (vii) $(Tu, u)\leq \|Tu\|^2, u\in H$ (viii) $(0,1)\subset \rho(T)$ where $\rho(T)$ denotes the resolvent set of $T$. The book gives the following solution Set $U=2T-I$. Clearly (vii) is equivalent to $$\text{(vii')} \;\;\;\;\;\;\;\;\;\|u\|\leq \|Uu\| \; \;\;\;\;\;\forall u\in H$$ Applying Theorem 2.20 we see that (vii)$\Rightarrow (-1,+1)\subset \rho(U)=2\rho(T)-1$. Thus (vii)$\Rightarrow$ (viii). and then it proceeds to prove the converse, which I can understand perfectly. The part I don't get is how applying Theorem 2.20 gives $(-1,+1)\subset \rho (U)$ (everything else is clear). Theorem 2.20 states Theorem 2.20: Let $A:D(A)\subset E\rightarrow F$ be an unbounded linear operator that is densely defined and closed. The following properties are equivalent: (a) $A$ is surjective, i.e. $R(A)=F$, (b) there is a constant $C$ such that $$\\|v\\|\leq C\\|A^{\ast}v\\| \;\; \forall v\in D(A^{\ast}),$$ (c) $N(A^{\ast})=\{0\}$ and $R(A^{\ast})$ is closed. If someone could explain to me how Theorem 2.20 is applied or even give another solution for the (vii)$\Rightarrow$ (viii) part, I would be grateful.
Can central theorem justify normality assumption of assets return distribution? And if it can why the empirical evidence show this assumption, which many finance models are based on, is a far cry from reality? No. I just published a paper on this. If return is defined as $$r_t=\frac{p_{t+1}q_{t+1}}{p_tq_t},$$ and since returns are not data while prices and volumes are, then it follows that the distribution of returns depends entirely upon the distribution of prices and the distribution of quantities. For example, in bankruptcy, $q_{t+1}=0$. As it is a very long paper, there is no single distribution or even family of distributions involved in returns. Under the standard stated assumptions in Markowitzian style models, the distribution of returns would integrate out to be $$\frac{1}{\pi}\frac{\sigma}{\sigma^2+(r_t-\mu)^2}.$$ This assumption follows from the fact that there are many buyers and sellers and that stocks are sold in a double auction so the winner's curse does not obtain. If the winner's curse were present, as in antique auctions, then the distribution would be the ratio of two Gumbel distributions. Regardless, with some interesting special exceptions, returns on equity securities cannot have a mean. Since there cannot be a mean return, then it follows that $\beta$ as defined in models such as the CAPM cannot exist. An assumption in all of these finance models has been that the parameters are known with probability one, but if you drop that assumption, you will find that no estimator exists that could converge to the population parameter. Ratio distributions are well known and taught as standard graduate work in statistics. The most common method to find a ratio distribution if the underlying are continuous densities is that if $Z=\frac{Y}{X}$, then the density of $Z$ is $$g(z)=\int_{-\infty}^\infty\|x\|f(x,zx)\mathrm{d}x.$$ With this you can derive just about any form of return except single period discount bonds and cash-for-stock mergers as well as accounting ratios and growth rates. As a rule of thumb, the central limit theorem is strongly violated for any financial return data, as well as quite a bit of macroeconomic data. Further, again as a rule of thumb, no non-Bayesian estimator exists for financial data. I will be presenting that along with a replacement for Black-Scholes at a conference in Albuquerque in a few weeks. The short form of this would go as follows: For an equation such as $$x_{t+1}=\beta{x}_t+\epsilon_{t+1},\beta>1$$ only non-parametric regression methods such as Theil's regression or quantile regression could work. Bayesian methods are never less risky than Frequentist methods, though the converse is not true. If we assume that the Frequentist tool chosen has the same sampling density as the posterior density if the prior density were flat and nuisance parameters were marginalized out, then the Frequentist density would still be inadmissible. The issue is that the Frequentist sampling distribution is symmetric and would include regions where $\beta\le{1}$. If that region has a density of $K$ and the region greater than unity has area $1-K$ then it follows that the Bayesian posterior density with a prior where $\Pr(\beta\le{1})=0$ will have zero mass in the region less than or equal to one, but whose density would be $$\frac{1-K}{K}$$ times greater than the equivalent non-Bayesian component. Because admissibility can be defined in term of stochastic dominance, it follows that the Bayesian estimator always stochastically dominates a Frequentist estimator. As such, no admissible non-Bayesian method is available for most finance problems and most macroeconomic problems where growth is involved. Normality is just an assumption which can be used to inform a model. No model is ever right, but some might be useful. Also, the assumption of normality doesn't really depend on the actual process being normal. Rather, it just requires that an investor's decisions are driven by mean and variance. While the underlying process of asset price returns is almost never normal - per se - we usually assume normality because: Returns are normal; approximately When an outcome is produced by many small effects acting additively and independently, its distribution will be close to normal; and, It is much harder to do economics without normality An investor who assumes joint normality need not be interested in anything else because the normal distribution is maximum entropy distribution which is completely described by two parameters: mean $\mu$ and variance $\sigma^2$. Therefore, the assumption of normality imposes the minimal prior structural constraint beyond these moments. An investor who only cares about mean and variance who doesn't use normality is taking into account additional assumptions which may or may not be correct or robust. There are theories which define investor preferences for moments beyond the first two, but not much evidence of which I am aware that supports such a position. See: The possible preferences of investors for higher than first 2 moments of return distribution? Approximate normality of logarithmic returns is classically demonstrated through the Lindeberg–Lévy central limit limit theorem (CLT). Normality also arises due to a variation of the CLT in which the sample of means of any distribution will tend toward normal. So, when you begin to look at a large number of assets and returns, the distributions will just tend towards normality. If however you decide not to assume normality, you should probably have a compelling reason since it is much more difficult to do economics without this assumption. For example, variance minimization techniques, such as those defined under Modern Portfolio Theory (MPT), are implicitly predicated on the assumption of joint normality. Even though there will be a set of portfolio weights which minimizes variance regardless of the underlying distributions, correlation (which also does not assume normality) is only a complete measure of association if the joint multivariate distribution is normal; i.e., covariance is only an exhaustive measure of co-movement if the joint distributions are themselves normal. We can see this is true because the joint distribution of X and Y is defined by joint normality: ${\frac {1}{2\pi \sigma _{X}\sigma _{Y}{\sqrt {1-\rho ^{2}}}}}\iint _{X\,Y}\exp \left[-{\frac {1}{2(1-\rho ^{2})}}\left({\frac {X^{2}}{\sigma _{X}^{2}}}+{\frac {Y^{2}}{\sigma _{Y}^{2}}}-{\frac {2\rho XY}{\sigma _{X}\sigma _{Y}}}\right)\right]\,\mathrm {d} X\,\mathrm {d} Y$ Which through a proof can be show to produce: $\sigma _{X+Y}={\sqrt {\sigma _{X}^{2}+\sigma _{Y}^{2}+2\rho \sigma _{X}\sigma _{Y}}},$ If now, we define $\omega_i \sigma^2_i=\sigma_X$, and $\omega_j \sigma^2_j=\sigma_Y$, then we get back the equation which is used as the basis of mean variance optimization of a two asset portfolio: $\mathbb{E}[\sigma _{p}^{2}]=\omega_{i}^{2}\sigma _{i}^{2}+\omega_{j}^{2}\sigma _{j}^{2}+2\omega_{i}\omega_{j}\sigma _{i}\sigma _{j}\rho _{ij}$ So while the portfolio covariance matrix can always be computed, to the extent that underlying assets have returns which are not normal the optimization is likely to be spurious.
CryptoDB Or Sattath Publications Year Venue Title 2019 EUROCRYPT On Quantum Advantage in Information Theoretic Single-Server PIR 📺 In (single-server) Private Information Retrieval (PIR), a server holds a large database $${\mathtt {DB}}$$ of size n, and a client holds an index $$i \in [n]$$ and wishes to retrieve $${\mathtt {DB}}[i]$$ without revealing i to the server. It is well known that information theoretic privacy even against an “honest but curious” server requires $$\varOmega (n)$$ communication complexity. This is true even if quantum communication is allowed and is due to the ability of such an adversarial server to execute the protocol on a superposition of databases instead of on a specific database (“input purification attack”).Nevertheless, there have been some proposals of protocols that achieve sub-linear communication and appear to provide some notion of privacy. Most notably, a protocol due to Le Gall (ToC 2012) with communication complexity $$O(\sqrt{n})$$ , and a protocol by Kerenidis et al. (QIC 2016) with communication complexity $$O(\log (n))$$ , and O(n) shared entanglement.We show that, in a sense, input purification is the only potent adversarial strategy, and protocols such as the two protocols above are secure in a restricted variant of the quantum honest but curious (a.k.a specious) model. More explicitly, we propose a restricted privacy notion called anchored privacy, where the adversary is forced to execute on a classical database (i.e. the execution is anchored to a classical database). We show that for measurement-free protocols, anchored security against honest adversarial servers implies anchored privacy even against specious adversaries.Finally, we prove that even with (unlimited) pre-shared entanglement it is impossible to achieve security in the standard specious model with sub-linear communication, thus further substantiating the necessity of our relaxation. This lower bound may be of independent interest (in particular recalling that PIR is a special case of Fully Homomorphic Encryption).
By the end of this series of posts we will have composed some code in Python (roughly 100 lines) that proves the satisfiability of Partition Problem instances with zero knowledge. Three preliminary notes I will assume some basic Computer Science background, as well as familiarity with Python, but not much more. I haven't seen this specific protocol in the literature (because I haven't searched for it), but it is a combination of well-known techniques in well-known ways, so I'm quite sure some variation of it is out there. For didactic reasons, we'll start with naive and sometimes broken implementations, and improve on them as we go along. Background I'm currently working at Starkware, developing some serious zero-knowledge proofs with some brilliant people, based on state-of-the-art research in the field, and we're usually hiring, so drop me a line if you're interested. This series, however, will deal with much more basic stuff, essentially Computer Science from the 1980s. For those familiar with contemporary protocols such as SNARKs, Bulletproofs, and STARKs - I am What I'm shooting for is less " With that in mind - let's get going. notgoing to present any of them, if you don't know what any of these are - fear not. What I'm shooting for is less " Cheat sheet for modern ZK proofs" and more " ZK proofs for dummies". With that in mind - let's get going. Zero Knowledge Proofs Zero Knowledge (AKA ZK) proofs are stories of the following type: side A states a claim and proves it to side B, after some deliberation between them, such that: B is sure that the claim is true with 99.99999% certainty. B has learnt nothing new in the process, other than that the claim is true. This Wikipedia article contains an excellent explanation of the idea, with some concrete examples. In this series I'll deal with ZK arguments of knowledge, which are not exactly the same as proofs, but they're close enough. In short: a ZK proof can be trusted completely, even if the side who's trying to prove their claim (usually referred to as "the prover") has unlimited computational power. ZK argument-of-knowledge can be trusted under the assumption that if the prover indeed tries to cheat, it is polynomialy bounded (if you use credit cards on the internet, then you already assume that, btw). In the world of ZK proofs, the other side of the exchange is often called "the verifier". I'll stick to this terminology here. The Partition Problem Given a sequence of numbers $a_0, a_1, ..., a_{n-1}$, can one partition this sequence into two subsets that sum up to the same number? If the sequence in question is $1, 9, 8, 0, 2, 2$, then the answer is clearly yes since $2 + 9 = 8 + 1 + 2 + 0$. However if the sequence is $2, 3, 4, 5, 6, 7$, then the answer is clearly no, since the sum is odd, and therefore there cannot be two subsets summing exactly to half of it each (the numbers are all integers). While these are simple enough instances, in general this problem is NP-complete (though it has a pseudo-polynomial algorithm). While these are simple enough instances, in general this problem is NP-complete (though it has a pseudo-polynomial algorithm). Let's Start Proving! Suppose we have a python list $l$ of numbers, that defines our Partition Problem instance. We'll say that another list $m$ is a satisfying assignment if: len(m) == len(l). All of the elements in $m$ are either $1$ or $-1$. The dot-product of $l$ and $m$ is 0. Note that this is equivalent to the statement of the partition problem, if we think of a '1' in $m$ as assigning its corresponding number in $l$ to the left side of the equation, and '-1' as assigning it to the right side. Given $l$, a proof for its satisfiability can be given by revealing $m$, but that would violate the ZK requirement. Let's rewrite $l$ as the partial sum list of its dot product with $m$. Mathematically speaking, let $p_i := \sum _{0\leq k<i} l[k] \cdot m[k]$. So if $l = [4, 11, 8, 1]$, and $m = [1, -1, 1, -1]$, then $p$ will be one element longer: $p = [0, 4, -7, 1, 0]$. Note that $p$ now has two interesting properties, if $m$ is indeed a satisfying assignment: (property 1 of p) It starts and ends with 0. (property 2 of p) For every $0\leq i < n$, we have $\|l[i]\| = \|p[i+1] - p[i]\|$. So here's a first draft for a ZK protocol: The verifier chooses a random $0 \leq i \leq n$. If $i = n$, the verifier asks the prover to provide $p[0]$ and $p[n]$ and checks that they are both 0. Otherwise, the verifier asks the prover to provide $p[i]$ and $p[i+1]$ and checks that indeed $\|l[i]\| = \|p[i+1] - p[i]\|$ (recall that $l$ is known to the verifier, as part of the claim made by the prover). What if the prover is lying??? The above contains an implicit assumption that when the verifier asks the prover to provide some data, the prover will indeed provide it honestly. We don't want to assume that, but we postpone dealing with this issue to the next post. For now, let's assume everything is kosher. This doesn't prove anything! An observant reader will probably point out that asking about a single element doesn't mean much. And that's true, we'd like to ask many queries, and after enough of them - we'll be certain that the claim is true. We'll quantify this more accurately in the third (and last) post. This is not Zero-Knowledge! Each query reveals something about $m$, and so it is not zero-knowledge. Consequently, after enough queries - $m$ can be completely revealed. That's terrible! Let's fix it. Manufacturing Zero-Knowledge Mathematically speaking, we usually say that something provides no new information, if it appears random, or more precisely - if it is uniformly distributed over some appropriately chosen domain. Without getting into the exact definition, this means that to make something ZK, we mix it with randomness. So here's how we do it here. Instead of $m$ as it was given to us, we flip a coin. If it shows heads, we leave $m$ as it is, if it shows tails, we multiply all of $m$'s elements by $-1$. Note that since its elmenets were initially $-1$ and $1$, and its dot product with $l$ was 0, this does not change its dot product with $l$ at all. We choose a random number $r$ and add it to all the elements of $p$. This does not effect the second property of $p$, but it changes the first property such that the first and last elements of $p$ now may not be zero. However, they must still be identical to one another. Now suppose that before each query - we recompute this randomness (i.e. - flip the coin and change $m$, and choose a random number $r$ and add it to the elements of $p$). If we choose $r$ carefully, then indeed, every two consecutive elements of $p$ will differ (in absolute value) by the corresponding element in $l$ but look otherwise random. So, here's the first piece of code we'll need, something that takes a problem (i.e. $l$) and a satisfying assignment (i.e. $m$) and constructs a witness (i.e. $p$) that will attest to the satisfiability of the problem instance: import randomdef get_witness(problem, assignment): """ Given an instance of a partition problem via a list of numbers (the problem) and a list of (-1, 1), we say that the assignment satisfies the problem if their dot product is 0. """ sum = 0 mx = 0 side_obfuscator = 1 - 2 * random.randint(0, 1) witness = [sum] assert len(problem) == len(assignment) for num, side in zip(problem, assignment): assert side == 1 or side == -1 sum += side * num * side_obfuscator witness += [sum] mx = max(mx, num) # make sure that it is a satisfying assignment assert sum == 0 shift = random.randint(0, mx) witness = [x + shift for x in witness] return witness
I am an undergraduate studying quantum physics with the book of Griffiths. in 1-D problems, it said a free particle has un-normalizable states but normalizable states can be obtained by sum up the solutions to independent Schrodinger equations. in my view the book also suggests that a scattering state with E>V(infinity) is to be un-normalizable. Is it true in 1-D situations? If so can it be generalized to 3-D situations? And why? Neuneck's answer is the pithiest description of how you get normalisable states as superpositions of non-normalisable states, but the following is more of a "why" these states happen. Hopefully, you should see that this discussion is independent of the number of dimensions. Practically speaking, the reason why there are always such states it is because observables fulfilling the canonical commutation relationship $[X,P]=i\,\hbar\,I$ have eigenvectors which are non-normalisable. Actually there is no reason why we have to have eigenvectors, so at a deeper level, the basic reasons why there are always non-normalised states are (1) convenience - the need for wieldiness of mathematical description and (2) the mathematical ingenuity of the people who gave us this wieldy and handy mathematical description - most notably the genius of (in rough historical order), Paul Dirac, Laurent Schwartz, Alexander Grothendieck and Israel Gel'Fand. This discussion keeps the intuitive ideas of eigenvectors and other convenient tools on a unified and rigorous footing. Grounding Ideas The basic setting for quantum mechanics is Hilbert space, namely a complete (in the sense that every Cauchy sequence converges to a member of the space) vector space kitted with an inner product (Banach spaces are a weaker and more general concept - being complete vector spaces kitted simply with a norm. The norm in a Hilbert space comes from the inner product of a vector with itself). So, intuitively, it is a complex vector space like $\mathbb{C}^N$ with "no holes in it" so we can talk about limits and do calculussy kind of stuff without worrying whether limits exist and, wherein we can talk about linear superposition and wherein we can "resolve" vectors uniquely into components through the inner product. So it's pretty much the state space of any physical system, aside from being complex, which is slightly unusual. Now we look at the idea of a linear functional on a Hilbert space $\mathcal{H}$. This is simply a linear function $L:\mathcal{H}\to\mathbb{C}$ mapping the Hilbert space $\mathcal{H}$ to the underlying field (in this case $\mathbb{C}$. The inner product for some fixed "bra" $\ell\in\mathcal{H}$, namely the function $x\mapsto\left<\ell,x\right>$ is clearly a special case of this linear functional notion. However, in Hilbert space, every continuous linear functional can indeed be represented by a "fixed bra" inner product and, since every fixed bra inner product clearly induces a continuous linear functional, the ideas of continuous linear functional and inner product with a fixed bra are precisely the same notion: this equivalence does NOT hold in any old vector space. This key equivalence property is special to Hilbert spaces is the subject matter of the Riesz representation theorem (see Wiki page of this name). So the continuous (topological) dual $\mathcal{H}^$ of $\mathcal{H}$, being a poncy name for the vector space of continuous linear fucntionals on $\mathcal{H}$, is isomorphic to the original Hilbert space. It can be shown to be an alternative and altogether equivalent definition of "Hilbert space" as the one above (i.e. a complete inner product space) is: An inner product space which is isomorphic to its dual space of continuous linear functionals. All this is very slick and attractive for describing things like quantum mechanics. It is also very easy in finite dimensional quantum systems, such as e.g. an electron constrained to a being a superposition of spin up and down states. In finite dimensions, there is no difference at all between the notions of continuous linear functional and the more general one of simply a linear functional (i.e. without to heed to continuity). Rigging the Hilbert Space: Nonnormalisable States In infinite dimensions - as with the quantum state space of the harmonic oscillator or the electron bound to a potential, we meet a glitch: Not all linear functionals are continuous. OOOPS: so just as we covet our neighbour's iPhone 5 when we have "only" model 4, so too we covet a stronger concept than Hilbert space wherein a software upgrade would make all "useful" linear functionals continuous! Less flippantly, here is where we get practical. In quantum mechanics, we need to implement the Heisenberg uncertainty principle, so we need Hermitian observables $\hat{X}$ and $\hat{P}$ fulfilling the canonical commutation relationship (CCR) $[\hat{X},\,\hat{P}]=i\,\hbar\,I$ (see my answer here and here). It's not too hard to show that a quantum space truly implementing the HUP cannot be finite dimensional - if it were, then $\hat{X}$ and $\hat{P}$ would have square matrix representations and the Lie bracket $[\hat{X}, \hat{P}]$ between any pair of finite square matrices has a trace of nought, whereas the right hand side of the CCR certainly does not have zero trace. So we consider them to be operators on the Hilbert space $\mathbf{L}^2(\mathbb{R}^N)$, which is a Hilbert space with dimensionality $\aleph_0$, i.e. it has countably infinite basis vectors, for example, the eigenfunctions of the $N$-dimensional harmonic oscillator. Vectors in this Hilbert space are "everyday wavefunctions" $\psi:\mathbb{R}^N\to\mathbb{R}^N$ as conceived by Schrödinger with the crucial normalisability property: $$\int\limits_{\mathbb{R}^N} \|\psi(\vec{x})\|^2\,{\rm d}^N x < \infty$$ Now, for convenience, we want to work in co-ordinates wherein one of $\hat{X}$ and $\hat{P}$ is the simple multiplication operator $X \psi(x) = x\,\psi(x)$. In my answer here I show that this means that there are co-ordinates where $X \psi(x) = x\,\psi(x)$ and, needfully $\hat{P} \psi(x) = -i\,\hbar \,{\rm d}_x \psi(x)$. However, neither of these operators is defined on our whole Hilbert space $\mathcal{H} = \mathbf{L}^2(\mathbb{R}^N)$: there are vectors (functions) $f$ in $\mathbf{L}^2(\mathbb{R}^N)$ ( e.g. functions with jump discontinuities) which have no defined $P\,f\in\mathcal{H}$, owing to the derivative's being undefined at the discontinuity. Likewise, some normalisable functions $g$ have no defined $X\,g\in\mathcal{H}$; multiplication by $\vec{x}$ makes them unnormalisable (witness for example the function $f(x) = (1+x^2)^{-1}$). Furthermore, neither of these functions has eigenvectors in $\mathcal{H}$: if $X\,f(x) = \lambda f(x) = x f(x)\,\forall x\in\mathbb{R}$ then $f(x) = 0$ for $x\neq\lambda$ and the eigenfunction $e^{i\,k\,x}$ of $P$ is not normalisable. But we want to salvage the idea of eigenstates and still be able to write our states in position or momentum co-ordinates. Here is where the notion of Rigged Hilbert Space comes in - the ingenious process where we kit a dense subset $S\subset H$ of the original Hilbert space $H$ ("rig it") with a stronger topology, so that things like the Dirac delta are included in the topological dual space $S^$ where $S\subset H\subset S^$. For QM we take the dense subset $S$ to be the "smooth" functions that still belong to $\mathcal{H}$ when mapped by any member of the algebra of operators generated by $X$ and $P$. That is, $S$ is invariant under this algebra and comprises precisely the Schwartz space of functions than can be multiplied by any polynomial and differentiated any number of times and still belong to $\mathcal{H}$. Any function in $\mathcal{H}$ can be arbitrarily well approximated (with respect to the Hilbert space norm) by some function in $S$. At the same time, we kit the dense subset $S$ out with a stronger topology than the original Hilbert space one. Why do we do this? One of the basic problems with $\mathcal{H}$ is that the Dirac delta $\delta:\mathbf{L}^2(\mathbb{R})\to \mathbb{C};\;\delta\;f(x) = f(0)$, which can be construed as an eigenvector of $X$, is not a continuous linear functional on $\mathcal{H}$ even though of course it is a linear functional. To see this, consider the image of $f(x) + \exp(-x^2/(2 \sigma^2)$ under the delta funcional: we can choose a $\sigma$ to make this function arbitrarily near to $f(x)$ as measured by the $\mathbf{L}^2$ norm, but with images $f(0)$ and $f(0)+1$, respectively, under the Dirac $\delta$. So we kit the dense subset $S$ out a topology that is strong enough to "ferret out" all useful linear functionals and make them continuous. We now have a topological dual (space of all linear functionals continuous with respect to the stronger topology) $S^$ of $S$ such that $S\subset\mathcal{H} = \mathcal{H}^\subset S^$. $S^∗$ is the space of tempered distributions as discussed in my answer here. $S^∗$ includes the Dirac delta, $e^{i\,k\,x}$ and is bijectively, isometrically mapped onto itself by the Fourier transform. Intuitively, functions and their Fourier transforms are precisely the same information for the tempered distributions. This ties in with the fact that position and momentum co-ordinate are mapped into each other by the Fourier transform and its inverse. So there we have it. We now have a space of bras $S^$ that is strictly bigger than the space of kets $\mathcal{H}$ and it needfully includes, by the construction of the rigged Hilbert space, nonnormalisable bras in $S^\sim\mathcal{H}$ simply so that we can discuss eigenstates of all the observables we need in a rigorous way. Good references for these notion are: This answer to the Physics Stack Exchange question "Rigged Hilbert space and QM" and also The discussions under the Math Overflow Question "Good references for Rigged Hilbert spaces?" In the latter, Todd Trimble's suspicions are correct that the usual Gel'Fand triple is $S\subset H = \mathbf{L}^2(\mathbb{R}^N)\subset S^$ with $S$ , $S^∗$ being the Schwartz space and tempered distributions as discussed in my answer here. The Wikipedia article on rigged Hilbert space is a little light on here: there's a great deal of detail about nuclear spaces that's glossed over so at the first reading I'd suggest you should take a specific example $S$ = Schwartz space and $S^∗$ = Tempered Distributions and keep this relatively simple (and, for QM most relevant) example exclusively in mind - for QM you won't need anything else. The Schwarz space and space of tempered distributions are automatically nuclear, so you don't need to worry too much about this idea at first reading. The scattered states are indeed non-normalizable. This is because a plane wave is an unphscial state (which you can for example see by trying to calculate the Heisenberg uncertainty, which will read $\Delta x \cdot \Delta p = \infty \cdot 0 = ??$). In order to create a physical state, you need to specify boundary conditions, i.e. a physical wavefunction at a given time $\Psi(t = 0)$. This can always be written as superposition of plane waves$$ \Psi(t=0) = \int \mathrm dE \; \tilde g(E) \psi(E)$$where $\tilde g$ is the "envelope" of your function and the $\psi(E)$ are the solutions of the time-independent Schroedinger equation$$ \hat H \psi(E) = E \cdot \psi(E)$$If this is fulfilled, your full wavefunction $\Psi(t)$ will fulfill the time-dependent Schroedinger equation.
Your equation $v_{\rm ac} =v_{\rm ab} + v_{\rm bc}$ is actually an equation relating to vector addition, $\vec v_{\rm ac} =\vec v_{\rm ab} + \vec v_{\rm bc}$ In one dimension this equation could be written as $v_{\rm ac}\,\hat x =v_{\rm ab} \,\hat x + v_{\rm bc}\,\hat x$ where $v_{\rm ac}, \,v_{\rm ab}$ and $v_{\rm bc}$ are components of the velocity in the $\hat x$ direction. You can think of the $\hat x$ direction as the direction that you have chosen to be positive. Going back to your equation $v_{\rm ac} =v_{\rm ab} + v_{\rm bc}$ and noting that these are components in a positive direction you have chosen it is not difficult to assign signs. Suppose that you chose West as your positive direction. $\vec v_{\rm ac}$ is $30 \,\rm m\,s^{-1}\, East$ and $\vec v_{\rm bc}$ is $10\,\rm m\,s^{-1}\, West $. $30\,\rm m\,s^{-1}\, East$ is the same as $-30\,\rm m\,s^{-1}\, West$. $v_{\rm ac} =v_{\rm ab} + v_{\rm bc}\Rightarrow -30 = v_{\rm ab} + 10 \Rightarrow v_{\rm ab} = -40 $ So the velocity of $a$ relative to $b$ is $-40\,\rm m\,s^{-1}\, West$ is the same as $40\,\rm m\,s^{-1}\, East$ On the other hand Suppose that you chose East as your positive direction. $\vec v_{\rm ac}$ is $30 \,\rm m\,s^{-1}\, East$ and $\vec v_{\rm bc}$ is $10\,\rm m\,s^{-1}\, West $. $10\,\rm m\,s^{-1}\, West$ is the same as $-10\,\rm m\,s^{-1}\, East$. $v_{\rm ac} =v_{\rm ab} + v_{\rm bc}\Rightarrow 30 = v_{\rm ab} + (-10) \Rightarrow v_{\rm ab} = 40 $ So the velocity of $a$ relative to $b$ is still $40\,\rm m\,s^{-1}\, East$ If it was a two or three dimensional problem then all you would need to do is to define two (or three) unit vectors (positive directions) and deal with each of the components in the the two (or three) directions as before.
This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate? Excessive Violence Sexual Content Political / Social Email Address: Article Id: WHEBN0000285907 Reproduction Date: A fractal dimension is a ratio providing a statistical index of complexity comparing how detail in a pattern (strictly speaking, a fractal pattern) changes with the scale at which it is measured. It has also been characterized as a measure of the space-filling capacity of a pattern that tells how a fractal scales differently from the space it is embedded in; a fractal dimension does not have to be an integer.[1][2][3] The essential idea of "fractured" dimensions has a long history in mathematics, but the term itself was brought to the fore by Benoit Mandelbrot based on his 1967 paper on self-similarity in which he discussed fractional dimensions.[4] In that paper, Mandelbrot cited previous work by Lewis Fry Richardson describing the counter-intuitive notion that a coastline's measured length changes with the length of the measuring stick used (see Fig. 1). In terms of that notion, the fractal dimension of a coastline quantifies how the number of scaled measuring sticks required to measure the coastline changes with the scale applied to the stick.[5] There are several formal mathematical definitions of fractal dimension that build on this basic concept of change in detail with change in scale. One non-trivial example is the fractal dimension of a Koch snowflake. It has a topological dimension of 1, but it is by no means a rectifiable curve: the length of the curve between any two points on the Koch Snowflake is infinite. No small piece of it is line-like, but rather is composed of an infinite number of segments joined at different angles. The fractal dimension of a curve can be explained intuitively thinking of a fractal line as an object too detailed to be one-dimensional, but too simple to be two-dimensional.[6] Therefore its dimension might best be described not by its usual topological dimension of 1 but by its fractal dimension, which in this case is a number between one and two. A fractal dimension is an index for characterizing fractal patterns or sets by quantifying their complexity as a ratio of the change in detail to the change in scale.[5]:1 Several types of fractal dimension can be measured theoretically and empirically (see Fig. 2).[3][8] Fractal dimensions are used to characterize a broad spectrum of objects ranging from the abstract[1][3] to practical phenomena, including turbulence,[5]:97–104 river networks,:246–247 urban growth,[9][10] human physiology,[11][12] medicine,[8] and market trends.[13] The essential idea of fractional or fractal dimensions has a long history in mathematics that can be traced back to the 1600s,[5]:19[14] but the terms fractal and fractal dimension were coined by mathematician Benoit Mandelbrot in 1975.[1][2][5][8][13][15] Fractal dimensions were first applied as an index characterizing complicated geometric forms for which the details seemed more important than the gross picture.[15] For sets describing ordinary geometric shapes, the theoretical fractal dimension equals the set's familiar Euclidean or topological dimension. Thus, it is 0 for sets describing points (0-dimensional sets); 1 for sets describing lines (1-dimensional sets having length only); 2 for sets describing surfaces (2-dimensional sets having length and width); and 3 for sets describing volumes (3-dimensional sets having length, width, and height). But this changes for fractal sets. If the theoretical fractal dimension of a set exceeds its topological dimension, the set is considered to have fractal geometry.[16] Unlike topological dimensions, the fractal index can take non-integer values,[17] indicating that a set fills its space qualitatively and quantitatively differently from how an ordinary geometrical set does.[1][2][3] For instance, a curve with fractal dimension very near to 1, say 1.10, behaves quite like an ordinary line, but a curve with fractal dimension 1.9 winds convolutedly through space very nearly like a surface. Similarly, a surface with fractal dimension of 2.1 fills space very much like an ordinary surface, but one with a fractal dimension of 2.9 folds and flows to fill space rather nearly like a volume.[16]:48[notes 1] This general relationship can be seen in the two images of fractal curves in Fig.2 and Fig. 3 – the 32-segment contour in Fig. 2, convoluted and space filling, has a fractal dimension of 1.67, compared to the perceptibly less complex Koch curve in Fig. 3, which has a fractal dimension of 1.26. The relationship of an increasing fractal dimension with space-filling might be taken to mean fractal dimensions measure density, but that is not so; the two are not strictly correlated.[7] Instead, a fractal dimension measures complexity, a concept related to certain key features of fractals: self-similarity and detail or irregularity.[notes 2] These features are evident in the two examples of fractal curves. Both are curves with topological dimension of 1, so one might hope to be able to measure their length or slope, as with ordinary lines. But we cannot do either of these things, because fractal curves have complexity in the form of self-similarity and detail that ordinary lines lack.[5] The self-similarity lies in the infinite scaling, and the detail in the defining elements of each set. The length between any two points on these curves is undefined because the curves are theoretical constructs that never stop repeating themselves.[18] Every smaller piece is composed of an infinite number of scaled segments that look exactly like the first iteration. These are not rectifiable curves, meaning they cannot be measured by being broken down into many segments approximating their respective lengths. They cannot be characterized by finding their lengths or slopes. However, their fractal dimensions can be determined, which shows that both fill space more than ordinary lines but less than surfaces, and allows them to be compared in this regard. Note that the two fractal curves described above show a type of self-similarity that is exact with a repeating unit of detail that is readily visualized. This sort of structure can be extended to other spaces (e.g., a fractal that extends the Koch curve into 3-d space has a theoretical D=2.5849). However, such neatly countable complexity is only one example of the self-similarity and detail that are present in fractals.[3][13] The example of the coast line of Britain, for instance, exhibits self-similarity of an approximate pattern with approximate scaling.[5]:26 Overall, fractals show several types and degrees of self-similarity and detail that may not be easily visualized. These include, as examples, strange attractors for which the detail has been described as in essence, smooth portions piling up,[16]:49 the Julia set, which can be seen to be complex swirls upon swirls, and heart rates, which are patterns of rough spikes repeated and scaled in time.[19] Fractal complexity may not always be resolvable into easily grasped units of detail and scale without complex analytic methods but it is still quantifiable through fractal dimensions.[5]:197; 262 The terms fractal dimension and fractal were coined by Mandelbrot in 1975,[15] about a decade after he published his paper on self-similarity in the coastline of Britain. Various historical authorities credit him with also synthesizing centuries of complicated theoretical mathematics and engineering work and applying them in a new way to study complex geometries that defied description in usual linear terms.[14][20][21] The earliest roots of what Mandelbrot synthesized as the fractal dimension have been traced clearly back to writings about undifferentiable, infinitely self-similar functions, which are important in the mathematical definition of fractals, around the time that calculus was discovered in the mid-1600s.[5]:405 There was a lull in the published work on such functions for a time after that, then a renewal starting in the late 1800s with the publishing of mathematical functions and sets that are today called canonical fractals (such as the eponymous works of von Koch,[18] Sierpiński, and Julia), but at the time of their formulation were often considered antithetical mathematical "monsters".[14][21] These works were accompanied by perhaps the most pivotal point in the development of the concept of a fractal dimension through the work of Hausdorff in the early 1900s who defined a "fractional" dimension that has come to be named after him and is frequently invoked in defining modern fractals.[4][5]:44[16][20] See Fractal history for more information The concept of a fractal dimension rests in unconventional views of scaling and dimension.[22] As Fig. 4 illustrates, traditional notions of geometry dictate that shapes scale predictably according to intuitive and familiar ideas about the space they are contained within, such that, for instance, measuring a line using first one measuring stick then another 1/3 its size, will give for the second stick a total length 3 times as many sticks long as with the first. This holds in 2 dimensions, as well. If one measures the area of a square then measures again with a box of side length 1/3 the size of the original, one will find 9 times as many squares as with the first measure. Such familiar scaling relationships can be defined mathematically by the general scaling rule in Equation 1, where the variable N stands for the number of new sticks, \epsilon for the scaling factor, and D for the fractal dimension: } (1) The symbol \propto above denotes proportionality. This scaling rule typifies conventional rules about geometry and dimension – for lines, it quantifies that, because N=3 when \epsilon=1/3 as in the example above, D=1, and for squares, because N=9 when \epsilon=1/3, D=2. The same rule applies to fractal geometry but less intuitively. To elaborate, a fractal line measured at first to be one length, when remeasured using a new stick scaled by 1/3 of the old may not be the expected 3 but instead 4 times as many scaled sticks long. In this case, N=4 when \epsilon=1/3, and the value of D can be found by rearranging Equation 1: {\log{\epsilon}}}} (2) That is, for a fractal described by N=4 when \epsilon=1/3, D=1.2619, a non-integer dimension that suggests the fractal has a dimension not equal to the space it resides in.[3] The scaling used in this example is the same scaling of the Koch curve and snowflake. Of note, these images themselves are not true fractals because the scaling described by the value of D cannot continue infinitely for the simple reason that the images only exist to the point of their smallest component, a pixel. The theoretical pattern that the digital images represent, however, has no discrete pixel-like pieces, but rather is composed of an infinite number of infinitely scaled segments joined at different angles and does indeed have a fractal dimension of 1.2619.[5][22] As is the case with dimensions determined for lines, squares, and cubes, fractal dimensions are general descriptors that do not uniquely define patterns.[22][23] The value of D for the Koch fractal discussed above, for instance, quantifies the pattern's inherent scaling, but does not uniquely describe nor provide enough information to reconstruct it. Many fractal structures or patterns could be constructed that have the same scaling relationship but are dramatically different from the Koch curve, as is illustrated in Figure 6. For examples of how fractal patterns can be constructed, see Fractal, Sierpinski triangle, Mandelbrot set, Diffusion limited aggregation, L-System. The concept of fractal dimension described in this article is a basic view of a complicated construct. The examples discussed here were chosen for clarity, and the scaling unit and ratios were known ahead of time. In practice, however, fractal dimensions can be determined using techniques that approximate scaling and detail from limits estimated from regression lines over log vs log plots of size vs scale. Several formal mathematical definitions of different types of fractal dimension are listed below. Although for some classic fractals all these dimensions coincide, in general they are not equivalent: D = \frac{d\ \log(L(k))}{d\ \log(k)} Many real-world phenomena exhibit limited or statistical fractal properties and fractal dimensions that have been estimated from sampled data using computer based fractal analysis techniques. Practically, measurements of fractal dimension are affected by various methodological issues, and are sensitive to numerical or experimental noise and limitations in the amount of data. Nonetheless, the field is rapidly growing as estimated fractal dimensions for statistically self-similar phenomena may have many practical applications in various fields including diagnostic imaging,[26][27] physiology,[11] neuroscience,[12] medicine,[28][29][30] physics,[31][32] image analysis,[33][34][35][36] acoustics,[37] Riemann zeta zeros,[38] and electrochemical processes.[39] An alternative to a direct measurement, is considering a mathematical model that resembles formation of a real-world fractal object. In this case, a validation can also be done by comparing other than fractal properties implied by the model, with measured data. In colloidal physics, systems composed of particles with various fractal dimensions arise. To describe these systems, it is convenient to speak about a distribution of fractal dimensions, and eventually, a time evolution of the latter: a process that is driven by a complex interplay between aggregation and coalescence.[40] Microsoft Windows, Open source, Fractal dimension, Mandelbrot set, Recursion Mathematics, Metric space, Fractal, South Africa, Topology Science, Medicine, Neuroscience, Data, Geography Time, Isaac Newton, Spacetime, Albert Einstein, Physics Space, Spacetime, Electromagnetism, Classical mechanics, String theory Fractal dimension, Box counting, Strange Attractor, Fractal, Geophysics
I have a problem understanding how phasors work and I'll use a problem from a recent exam to illustrate the misunderstanding. Note: Underlined variables such as \$\underline{V}\$ are considered here to be complex numbers, while non-underlined are considered to be magnitudes of the complex numbers. Here's the text: We have voltage generator \$E=2\sqrt{7} \mbox{ } V\$ with angular frequency \$\omega=10^6 \mbox{ } s^{-1}\$ and internal resistance \$R_g=0.5\sqrt{3} \mbox{ } k\Omega\$ connected to parallel connection of impedance \$Z\$ and coil \$L\$. Current is \$I=I_1=I_2=4 \mbox{ } mA\$. Calculate complex value of \$\underline{Z}\$ and inductivity of \$L\$. Here's the picture of the circuit: So from phasor theory I know that they represent complex numbers in the form of \$ \underline{I}=I_e e^{j\psi_0}\$, where \$I_e\$ is effective value of the current and \$\psi_0\$ if the initial phase of the current. So on phasor, the line would be of length \$I_e\$ and have an angle \$\psi_0\$ to the phase axis. Next, I know that on a line which only has a resistor the voltage and current are in phase and on the phasor, they would be on a same line. If we have a line which has an inductor, the current will lag with respect to voltage by \$ \frac{\Pi}{2}\$. If we have a capacitor, then the current will be in front of the voltage by \$ \frac{\Pi}{2}\$. Next on a phasor currents on a circuit should form a closed figure. If we don't have any starting phases in a circuit, we can set the phase of one element to zero, proclaim it the reference phase and calculate the rest of the phases with respect to it. So according to my reasoning, I can set the phase of the generator to zero and get \$ \underline{E}=2\sqrt{7} e^{j0}\mbox{ } V\$ and now I have the complex voltage. Next, I know the effective current \$I\$ and the resistance of \$ R_g\$, so I can calculate the voltage drop across the resistor and this way get the voltage which the impedance \$Z\$ and the inductance \$L\$ see. So \$R_gI=2\sqrt{3}\mbox{ }V\$ and the impedance and the inductance see the voltage of \$U_1=\sqrt{7} \mbox{ }V\$. Next since the resistor is in this case ideal, the \$U_1\$ is in phase with \$E\$. Next, I know that the current \$I_1\$ can be obtained by following equation \$ \underline{I}_1=\frac {\underline{U_1}}{\underline{Z}}=\frac{U_1 e^{j0}}{Ze^{j\phi}}=\frac{U_1}{Z}e^{0-\phi}\$ From this I can get the \$Z\$. Next, I know that the impedance \$\underline{Z_l}\$ of the line on which the inductor is is \$\underline{Z_l}=j\omega L\$ and I know that \$\underline{I_2}=\frac{\underline{U_1}}{\underline{Z_l}}=\frac{U_1 e^{j0}}{\omega L e^{\frac{\Pi}{2}}}=\frac{U_1} {\omega L} e^{0- \frac{\Pi}{2}}= \frac{U_1} {\omega L} e^{- \frac{\Pi}{2}}\$ From this I can get the L. For currents \$ I_1\$ and \$I_2\$ to have same effective value the \$\phi\$ needs to be \$ \frac{Pi}{2}\$ and the impedance \$ Z\$ needs to be mostly capacitive. In this case the current \$I\$ needs to be in phase with the voltage \$ E\$ but it isn't because if it were, the effective value would be determined by the resistance \$R_g\$ and it would be different. So since the two currents going out of the current \$I\$ have same effective value, I concluded that the three currents need to form a triangle with sides of the same length, like on this picture: In that case the angle on one of the currents needs to be \$\frac {\Pi} {3} \$ and on the other current it needs to be \$-\frac {\Pi} {3} \$. In that case the current \$\underline{I}\$ is in phase with the voltage \$\underline{U_1}\$ and \$\underline{I_2}\$ is lagging by \$ \frac {\Pi}{3}\$ while the \$\underline{I_1}\$ moved forward by\$ \frac {\Pi}{3}\$. However in that case the current \$ \underline{I_2}\$ isn't lagging by \$ \frac {\Pi}{2}\$ which it should because the only component on its branch is an inductor. So my problem is that I have a bunch of rules and I can't determine when whey should be applied, as I have shown. Can anyone explain to me where my reasoning is wrong? For example in the inductor branch, in which cases will current lag behind voltage by \$ \frac {\Pi}{2}\$ and in which by some other number? When can I rely on voltage and current on a purely resistive branch to be in phase? According to Kirchhoff's law the sum of currents in a node should be zero, so on a phasor currents for that node should form a closed figure but in this case they don't. In my books this isn't clearly explained and most problems we have in problem collections don't have detailed solutions. Correct solutions for this specific problem are \$\underline{Z}=250(\sqrt{3}-j) \mbox{ } \Omega\$ and \$L=0.5 \mbox{ } mH\$
Are there any examples of subspaces of $\ell^{2}$ and $\ell^{\infty}$ which are not closed? Take the subspace of $\ell^2$ (or $\ell^\infty$) consisting of sequences with finitely many non-zero coordinates. Clearly, this is a subspace (sum of sequences still has at most # of nonzeros of first seq + # of nonzeros in second seq, scaling of a sequence has same non-zeros). However, you can take any sequence on $\ell^2$ (or $c_0 \subset \ell^\infty$, a closed subspace), $a$ and define the sequence of sequences $\{a_n\}$ where $a_n$ consists of only the first $n$ elements of $a$ and rest zeros, and see $a_n \to a$. Yes. For instance, the sequences which are eventually $0$. As an observation, every subspace generated by countably many linearly independent vectors (that is, a subspace of dimension $\aleph_0$) inside a Banach space is not closed, since a Banach space cannot have (algebraic) dimension $\aleph_0$. ( if it were closed, if would be itself a Banach space). For $1< x<2$ the space $l^x$ is a non-closed vector subspace of $l^2.$ Let $(A_j)_{j\in \mathbb N}$ be a strictly increasing positive real sequence converging to $1/x.$ For $i,j, \in \mathbb N$ let $B_{i,j}=0$ if $j>i,$ and $B_{i,j}=j^{-A_j}$ if $j\leq i.$ Let $v_i=(B_{i,j})_{j\in \mathbb N}.$ Since only finitely many co-ordinates of each $v_i$ are non-zero, we have $\{v_i:i\in \mathbb N\}\subset l^x.$ Let $u= (u_j)_{j\in \mathbb N}=(\;j^{-B_{j,j}}\;)_{j\in \mathbb R}=(\;j^{-A_j})_{j\in \mathbb N}.$ Take $d\in (0,1-x/2).$ Take $j_0$ such that $j\geq j_0\implies A_j>(1-d)/x.$ Then $j\geq j_0 \implies 2A_j>2(1-d)/x>1.$ So $$\sum_{j=j_0}^{\infty} u_j^2=\sum_{j=j_0}^{\infty}j^{-2A_j}\leq \sum_{j=j_0}^{\infty}j^{-2(1-d)/x}<\infty.$$ Therefore $u\in l^2.$ Now $u\not \in l^x$ because $u_j^x=j^{-xA_j}>j^{-1}$ so $$\sum_{j=1}^{\infty} u_j^x\geq \sum_{j=1}^{\infty} j^{-1}=\infty.$$ I will leave it to you to show that $\lim_{j\to \infty} \\|u-v_j\\|_2=0.$ So in $l^2,$ the vector $u$ belongs to $Cl(l^x)$ but not to $l^x.$ The fact that $l^x$ is a vector space for any $x>1$ is another story, which I am sure you can find on this site.
Given a $n\times n$ symmetric random matrix such that all diagonalelements are all fixedas $0$. randomly select$k$ distinct cells in the upper triangle (excluding the diagonal), and then fixthem as $1$. The elements in the lower triangle are set accordingly. Of course $k\le \frac{n^2-n}{2}$. all other elements are independent uniformrandom variables over $[0,1]$. The question is Any known result for the distribution of the minimum/maximum row sum? It is easier to just look at one row. Let $c$ be the number of chosen cells to be fixed as $1$, then its distribution is $\frac{{\left( {\begin{array}{{20}{c}} {n - 1} \\ c \end{array}} \right)\left( {\begin{array}{{20}{c}} {\frac{{{n^2} - n}}{2} - (n - 1)} \\ {k - c} \end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}} {\frac{{{n^2} - n}}{2}} \\ k \end{array}} \right)}}$, which equals the probability of $c$ chosen cells in the first row. The sum of the remaining $n-1-c$ cells are i.i.d. uniform, and their sum is Irwin–Hall distribution. However, I have trouble to find the distribution for min/max row sum, as the row sums are not even independent.
Your question is valid and is very good you are thinking this way so young. First, error propagation is a whole area of its own and there is not a unique way to do so, nor there is an absolute best way to be provided. So much so that the ISO document used for having a consensus on this, is called "Guide to the Expression of Uncertainty in Measurement", not a manual. This is the case mainly because determining uncertainties is very much a compromise (depends on how much confident you want to have on your value) and model guided (depends on your assumptions of what you don't know). What does this means? Well first it is impossible to have absolute confidence on a value, so when you say a rod's length $20.0\pm0.2cm$ (and please note the number of significant digits should match) you are assuming that the value can be anything between $19.8cm$ and $20.2cm$ but there is no chance it will be lower nor higher. Secondly, you are assuming that all values between $19.8cm$ and $20.2cm$ are equally likely, which is not usually the case. Now don't get me wrong: your reasoning is not flawed, is completely legitimate, although not the most practical, and therefore not what is generally used. All this was to clarify what it really means your error propagation and your expectations of them falling inside a range, etc. How we do things From experience we have seen that many values, when measured exhibit a Normal distribution, which means in few words, that repeated measurements on the same conditions will be concentrated around the mean value, and the further from this value the less you find. Since this is very often, we commonly assume that this is the case in general. So in this sense you are correct in expecting that your rod's value $20cm$ should be in the center of the interval. As for the interval to report, when assuming normal distribution, it is common practice to report the interval whose center is the mean value of the measurements, and which will contain roughly 63% of all possible values. The logical reasons for this are numerous and lengthy to cover here, but still this is a convention. Now you see the information behind a value: $20.0\pm0.2cm$ means generally that we assume the Normal distribution for the values, we estimated $20.0cm$ to be the mean, and 63% of the measurements will be between $19.8cm$ and $20.2cm$. What does it all have to do with error propagation? Well if it all depends on our definition of a value and its uncertainty, then it has everything to do. To propagate errors we don't just sum $L_1$ and $L_2$ max and min values, and this is precisely because we know that these values are not 100% inside their intervals, they are 63% inside. And so the values of $L_t = L_1 + L_2$ will not be 63% inside the interval between $L_1min+L_2min$ and $L_1max+L_2max$. This is not easy to prove in concise manner with your math skills, but you seem curious so you will find it out. So the way we propagate errors in this case is $\delta L_t = \sqrt{(\delta L_1)^2 + (\delta L_2)^2}$ and although its advantages can be argued logically, this is still based on the Normal distributed model, and the 63% interval requirement. So what about the area?Well as you saw above, your idea of propagating by using the minimum and maximum values changes the coverage of your interval. Above, when a sum is involved, it will not in general yield a 63%. Here the same happens, you can not expect to have a correct $63%$ interval because, for example $W_{min}*L_{min}$ gives a value of the area that actually is outside of the 63% interval for the area. The prove of this involves probability and again you will have to take my word here. The formula used is in fact $\delta A / A = \sqrt{(\delta W / W)^2 + (\delta L / L)^2}$ which needs some basic calculus knowledge to deduce, but I will try to give some sense to it from your point of view. See how similar it is to the previous one? Is like the same as before, only you are using $\delta value/value$ which you can think of as a relative uncertainty: is how important is the uncertainty in relation to the value it characterizes. Indeed, if you would like to compare the uncertainty between your height, say $700\pm2mm$ and say the Earth's radius $696342\pm65km$. Now which one is measured more precisely? It would be hard to say here since $2mm$ is much smaller than $65km$, but the values are very different too. However looking at the relative uncertainties you will see that Earth's value is indeed more precise! Finally, let me say that although I only spoke of the general case encountered in physics, it is not always correct to assume the Normal distribution, nor even expect that the mean is in the center of the interval. Under some conditions you may encounter Poisson distribution in which the mean is usually closer to the smallest values of the integral. Also in other cases, particularly in research, we expect to have values much more rigorously measured, and intervals are chosen to contain as much as 95% of the measured values, and sometimes more.
I need to find the intersection of the internal tangents of two point sets $V_a, V_b$ in $\mathbb{R}^2$, defined via their convex hulls. We can assume that the sets are disjoint and linearly separable, so I think the internal tangents are uniquely defined as the two lines which are tangent to each of the convex hulls of the sets, such that each set is on a different side. This is the point marked green here: This paper ("Computing Plurality Points and Condorcet Points in Euclidean Space") I'm currently reading (paywalled, unfortunately...) says in a sort of side note that this problem can be solved using linear programming, in linear time in the number of points, but doesn't describe how. For constructing a linear program, I thought about describing lines by $F(x, y) = ax + by + c= 0$, and optimizing a function of $a, b, c$. Then the obvious constraints would be $\forall v \in V_a: av_x + bv_y + c \geq 0$ and the according thing for $V_b$, to force the lines to go between the two sets. But now I am stuck. For one, we need to constrain the space of lines somewhat; $\lVert (a, b) \rVert = 1$ would be natural to require, but that's not linear. Also, I'm not sure about the cost function; I thought about using $$\min_{a,b,c} \sum_{v\in V_a} F(v_x, v_y) - \sum_{v\in V_b} F(v_x, v_y),$$ since $F$ is somehow a distance from the line (although not in the strict sense, I guess). But that is just guessing and also doesn't work -- I tried some things like I described in JuMP, but it always returned "infeasible". How can this be formluated as a linear program? Alternatively, I can accept an algorithmic solution with linear time. Or a disproof, of course.
This is a microcanonical ensemble approach to a simple model of a two dimensional polymer bundle. The professor of the course I took does a lot of research in the area of polymer physics and so set a few problems pertaining to them. They aren’t easily found in textbooks or online either (this website notwithstanding) but are nevertheless quite interesting in themselves. We also approached this same problem from the canonical ensemble which I’ll upload soon. Question A polymer is represented by a chain of $N$ segments of length $a$. $N_x^+$ segments are oriented in the positive $x$-direction $N_x^-$ are oriented in the negative $x$-direction. $N_y^+$ and $N_y^-$ segments are oriented in the positive and negative $y$-direction, respectively. Assign an energy penalty $\varepsilon$ for every polymer segment oriented in $y$-direction. This penalty reflects the confinement in a polymer bundle. For prescribed extensions $L_x$ and $L_y$ and energy $E$, calculate the entropy $S(E,N,L_x,L_y)$ of the chain. Set $L_y=0$ in the following. Derive the mechanical equation of state for the force $F_x(E,N,L_x)$ in the $x$-direction. Derive the thermal equation of state for the temperature $T(E,N,L_x)$. Eliminate $E/\varepsilon$ from the two equations of state and show that \begin{equation} L_x(T,N,F_x)=aN\frac{\sinh(\frac{F_x a}{k_B T})}{\cosh(\frac{F_x a}{k_B T}) + \exp(\frac{-\varepsilon}{k_B T})} \label{eq:lx} \end{equation} Expand $L_x$ for small force and extract the entropic spring constant. Solution 1. Energy Penalty To include an energy penalty for segments oriented in the $y$-direction, we simply write $E=\varepsilon [N_y^++N_y^-]$. 2. Entropy To calculate the entropy of the chain, we first need to deduce the number of states. In the one-dimensional case this is simply the result for the one-dimensional random walk: $$\Omega=\frac{N!}{N^+! N^-!}$$ Extending this to two dimensions, we obtain \begin{equation} \Omega=\frac{N!}{N_x^+! N_x^-! N_y^+! N_y^-!} \label{eq:omega} \end{equation} We find the entropy using the well-known formula and applying Stirling’s approximation: $$S = k_B \left[ N\ln N - N_x^+ \ln N_x^+ - N_x^- \ln N_x^- - N_y^+ \ln N_y^+ - N_y^- \ln N_y^- \right]$$ If we define , we can write it a bit more succinctly: \begin{equation} S = -k_B N \left[ \rho_x^+ \ln \rho_x^+ + \rho_x^- \ln \rho_x^- + \rho_y^+ \ln \rho_y^+ + \rho_y^- \ln \rho_y^- \right] \label{eq:srho} \end{equation} We set . Now let us briefly take pause to define some variables for convenience: For convenience, $\eta := \frac{E}{N \varepsilon}$, $x:=\frac{L_x}{L_0}$, and $L_0:=Na$, where $L_0$ is the “stretched out” (i.e. $N=N_x$) length of the polymer. We also note that the effective number of steps in the positive $x$-direction is the effective length over the segment length: $N_x^+ -N_x^-=\frac{L_x}{a}$ and the same for $y$: $N_y^+ -N_y^-=\frac{L_y}{a}$. Combining these variables, it is possible to write \[\frac{N_x^{\pm}}{N}=\frac 12 \big[1-\eta \pm x\big]\] 3. Mechanical Equation of State So, to the question. The equation for the force in the -direction is given by , so \begin{equation} \frac{-F_x}{T}=\frac{\partial S}{\partial L_x}=\frac{\partial S}{\partial x}\frac{\partial x}{\partial L_x}=\frac{1 }{L_0}\frac{\partial S}{\partial x} \label{eq:fx} \end{equation} In order to find the derivative $\frac{\partial S}{\partial x}$ we need to rewrite the entropy with the new constraint of $L_y=0$. \begin{equation} \frac{S}{-k_B N} = \frac{1}{2} \left(1-\eta + x \right)\ln \left(1-\eta + x \right) + \frac{1}{2} \left(1- \eta – x \right)\ln \left(1-\eta – x \right) + 2 \left( \frac{\eta}{2} \right) \ln \left( \frac{\eta}{2} \right) \label{eq:news} \end{equation} Taking the derivative for $x$ then, we find $$\frac{\partial S}{\partial x}=-\frac{k_B N}{2} \ln \left[\frac{1-\eta + x}{1-\eta - x}\right]$$ The mechanical equation of state for the force $F_x$, found by rearranging \eqref{eq:fx} and using the result just obtained, is then: \begin{equation} F_x=\frac{k_B T}{2a} \ln \left[ \frac{1-\eta + \frac{L_x}{L_0}}{1-\eta - \frac{L_x}{L_0}} \right] \label{eq:fx2} \end{equation} 4. Thermal Equation of State To obtain the thermal equation of state, we use the formula $ \frac 1T = \frac{\partial S}{\partial E} $ Using \eqref{eq:news}: $$\frac{2\varepsilon}{k_B T} = \ln \left[ \frac 12 (1-\eta + x) \right] + \ln \left[ \frac 12 (1-\eta - x) \right] -2 \ln \frac \eta 2$$ which we can write much more neatly as: \begin{equation} \frac{2\varepsilon}{k_B T} = \ln \left( \frac{1-\eta + x}{\eta} \right) + \ln \left( \frac{1-\eta \, – \, x}{\eta} \right) \label{eq:dsde} \end{equation} 5. Calculating the Extension The equation we are to derive is quite an long process as the equations get somewhat bulky. In fact, the same result can be obtained much simpler (just a few lines) in the canonical ensemble approach to this problem. The steps here are only an outline as a guide. Start by solving \eqref{eq:fx2} for $\eta$: \begin{equation} \eta=1-x·\coth\frac{F_x a}{k_B T} \label{eq:etacoth} \end{equation} If we now sub. this into equation \eqref{eq:dsde}: $$\frac{2 \varepsilon}{k_B T}=\ln \left[ \frac{\frac{L_x}{Na}\coth\left(\frac{F_x a}{k_B T} \right) +\frac{L_x}{Na}}{1-\frac{L_x}{Na}\coth\left(\frac{F_x a}{k_B T} \right)} \right] + \ln \left[ \frac{\frac{L_x}{Na}\coth\left(\frac{F_x a}{k_B T} \right) -\frac{L_x}{Na}}{1-\frac{L_x}{Na}\coth\left(\frac{F_x a}{k_B T} \right)} \right]$$ Simplifying and taking the exponent of both sides: $$e^{\frac{2 \varepsilon}{k_B T}}=\frac{\frac{L_x^2}{N^2 a^2} \coth^2 \left( \frac{F_x a}{k_B T} \right) – \frac{L_x^2}{N^2 a^2} }{ \left[ 1-\frac{L_x}{N a} \coth \left( \frac{F_x a}{k_B T} \right) \right]^2}$$ Using that $\coth^2(x)-1=\frac{1}{\sinh^2(x)}$: $$e^{\frac{\varepsilon}{k_B T}}=\frac{\frac{L_x}{N a} \sqrt{\coth^2 \left(\frac{F_x a}{k_B T}\right) -1}}{1-\frac{L_x}{N a} \coth \left( \frac{F_x a}{k_B T} \right)}$$ Finally, we solve for $L_x$: \begin{equation} \frac{L_x}{Na}=\frac{\sinh(\frac{F_x a}{k_B T})}{\cosh(\frac{F_x a}{k_B T}) + \exp(\frac{-\varepsilon}{k_B T})} \label{eq:finally} \end{equation} which is the equation \eqref{eq:lx} given. 6. Extracting the Entropic Spring Constant We need to take the limit of \eqref{eq:lx} for small force. For small $x$: $\sinh(x) \to x$ and $\cosh(x) \to 1$. Therefore $L_x$ becomes $$L_x \approx aN \frac{\left(\frac{F_x a}{k_B T}\right)}{1+\exp \left(\frac{-\varepsilon}{k_B T}\right)}$$ If we re-arrange for $F_x$: $$F_x=\frac{k_B T}{a^2N}\left[1+\exp \left(\frac{-\varepsilon}{k_B T}\right) \right] L_x$$ And compare it with Hooke’s Law $F=-kx$: $$k_{\text{entropy}}=\frac{k_B T}{a^2N}\left[1+\exp \left(\frac{-\varepsilon}{k_B T}\right) \right]$$
What's the difference between these? Both are measurements of some form of signal power, but surely there's some difference between the power they are measuring? The fast Fourier transform ($\textrm{FFT}$) algorithms are fast algorithms for computing the discrete Fourier transform ($\textrm{DFT}$). This is achieved by successive decomposition of the $N$-point $\textrm{DFT}$ into smaller-block $\textrm{DFT}$, and taking advantage of periodicity and symmetry. Now, the $N$-point $\textrm{DFT}$ of a sequence $\{x[0], x[1],\cdots, x[N-1]\}$ is: \begin{equation}X(f_k) = \displaystyle \sum_{n = 0}^{N - 1} x[n]\exp\left(-j2\pi f_kn\right) \tag{1}\end{equation}For $f_k = k/N$ and $k = 0, 1, \cdots, N - 1$. And the $\textrm{FFT}$ magnitude at bin $k$ is the $\textrm{DFT}$magnitude at bin $k$. For a given $N$ that is: \begin{equation} \left\|X(f_k)\right\| = \left\|X\left(\frac{k}{N}\right)\right\| = \left\|X(k)\right\| = \displaystyle \left\|\sum_{n = 0}^{N - 1} x[n]\exp\left(-j2\pi nk/N\right)\right\| \tag{2} \end{equation} You talk about power spectral estimation when the signals being analyzed are characterized as random processes. With random fluctuations in such signals, statistical characteristics and average characteristics are normally adopted. For a wide sense stationary $(\textrm{WSS})$ discrete random process, the PSD is defined as: \begin{equation} P(f) = \displaystyle \sum_{m = -\infty}^{\infty} r_{xx}[m]\exp\left(-j2\pi fm\right) \tag{3} \end{equation} For reasons you can find in this answer, you see that the squared magnitude of the signal's $\textrm{DFT}$ is taken as the estimate of the PSD in most practical situations. One form, among other variations/methods, is: \begin{equation} P(f_k) = \frac{1}{N} \displaystyle \left\| \sum_{n = 0}^{N - 1} x[n]\exp\left(-j2\pi f_kn\right) \right\|^2 \tag{4} \end{equation} What's the difference between these? Comparing $(2)$ and $(4)$, you have: \begin{equation} P(f_k) = \frac{1}{N} \left\| X(f_k)\right\|^2 \end{equation} From the bin number $k$ to frequency in $\textrm{Hz}$, $F = \frac{F_s}{N}k$ EDIT: The power spectral density, $\textrm{PSD}$, describes how the power of your signal is distributed over frequency whilst the $\textrm{DFT}$ shows the spectral content of your signal, the amplitude and phase of harmonics in your signal. You pick one or the other depending on what you want to observe/analyze. And no they're not the same as you can see from the equations above and links given. Their spectra are generally not the same. One is estimated as the squared magnitude of the other. The FFT is the Fast Fourier Transform. It is a special case of a Discrete Fourier Transform (DFT), where the spectrum is sampled at a number of points equal to a power of 2. This allows the matrix algebra to be sped up. The FFT samples the signal energy at discrete frequencies. The Power Spectral Density (PSD) comes into play when dealing with stochastic signals, or signals that are generated by a common underlying process, but may be different each time the signal is measured. Given just one "realization" of a stochastic process--a stochastic signal--you can only estimate what the underlying Power Spectral Density is. You can make this estimate poorly with the Periodogram, which involves squaring the FFT (amplitude squared yields power). The periodogram suffers from very high variance and is not a good estimator. You are better off using Welch's method of periodogram averaging, or better yet, the Blackman-Tukey method of periodogram smoothing.
I don't know how familiar with Banach spaces you are. It's a standard fact that "absolutely summable series converge" in a Banach space (in the same way this is true for series of numbers). More precisely, if $(v_{n})_{n \in \mathbb{N}}$ is a sequence in a Banach space $B$ and $\sum_{n = 1}^{\infty} \\|v_{n}\\|_{B} < \infty$, then $\sum_{n =1}^{N} v_{n}$ is a Cauchy sequence in $B$ and thus converges to an element we denote $\sum_{n = 1}^{\infty} v_{n}$. The limit satisfies $\left\\|\sum_{n = 1}^{\infty} v_{n} \right\\| \leq \sum_{n = 1}^{\infty} \\|v_{n}\\|_{B}.$ This is another way to approach your question. Turning to your proof, recall that in Tonelli's theorem one of the conclusions is that the map $x \mapsto \int_{\mathbb{N}} \|f_{n}(x)\| c(dn)$ is measurable and integrable. In other words, since $$\int_{\mathbb{N}}\int_{\mathbb{R}} \|f_{n}(x)\| \, m(dx) c(dn) < \infty,$$ as you showed, we can conclude that $$x \mapsto \int_{\mathbb{N}} \|f_{n}(x)\| \, c(dn) = \sum_{n =1}^{\infty} \|f_{n}(x)\|$$ is measurable and integrable over $\mathbb{R}$. Since it's integrable, it's also finite a.e. As to your other question, we could indeed use monotone convergence theorem instead of Fubini/Tonelli. (Some books call this result a corollary of the monotone convergence theorem.) Consider the sequence of functions $g_{N} = \sum_{n = 1}^{N} \|f_{n}\|$. Notice that $(g_{N})_{N \in \mathbb{N}}$ is a monotone sequence of non-negative functions converging pointwise a.e. (though possibly to infinity at some points). The monotone convergence theorem yields$$\int_{\mathbb{R}} \sum_{n = 1}^{\infty} \|f_{n}(x)\| \, dx = \lim_{N \to \infty} \sum_{n = 1}^{N} \int_{\mathbb{R}} \|f_{n}(x)\| \, dx = \sum_{n = 1}^{\infty} \int_{\mathbb{R}} \|f_{n}(x)\| \, dx < \infty.$$ This shows $\sum_{n = 1}^{\infty} \|f_{n}(x)\|$ is finite a.e. and thus $\sum_{n = 1}^{\infty} f_{n}(x)$ converges a.e. Using the triangle inequality, we also get integrability from the previous computation. EDIT: Using the previous paragraph, we can now argue $\sum_{n = 1}^{\infty} \int_{\mathbb{R}} f_{n}(x) \, dx = \int_{\mathbb{R}} \sum_{n = 1}^{\infty} f_{n}(x) \, dx$, as follows:$$\left\|\int_{\mathbb{R}} \left(\sum_{n = 1}^{\infty} f_{n}(x) - \sum_{n = 1}^{N} f_{n}(x) \right) \, dx \right\| \leq \int_{\mathbb{R}} \sum_{n = N + 1}^{\infty} \|f_{n}(x)\| \, dx = \sum_{n = N + 1}^{\infty} \int_{\mathbb{R}} \|f_{n}(x)\| \, dx$$ and the right-hand side goes to zero as $N \to \infty$ by the previous paragraph (being the tail of a convergent series). This shows $$\int_{\mathbb{R}} \sum_{n = 1}^{\infty} f_{n}(x) \, dx = \lim_{n \to \infty} \sum_{n = 1}^{\infty} \int_{\mathbb{R}} f_{n}(x) \, dx$$ by the definition of limit.
I have been thinking about the question "What is the best -- i.e., some combination of shortest, most natural, easiest -- proof of the Nullstellensatz?" recently on the eve of a commutative algebra course. In my notes up to this point I had been following Kaplansky's treatment of Goldman domains and Hilbert-Jacobson rings. This places the problem in a more general context and allows for an attractively thorough analysis. At the end one comes out with the following results: (1) The polynomial ring $k[t_1,\ldots,t_n]$ is a Jacobson ring -- i.e., every radical ideal is the intersection of the maximal ideals containing it. (2) (Zariski's Lemma): If $\mathfrak{m}$ is a maximal ideal of $k[t_1,\ldots,t_n]$, then $k[t_1,\ldots,t_n]/\mathfrak{m}$ is a finite degree field extension of $k$. (That Hilbert's Nullstellensatz follows from (1) and (2) is an easy, standard argument that I won't discuss here.) But it is well-known that to prove the Nullstellensatz one needs only (2), because then (1) follows by a short and easy argument that everyone seems to like: Rabinowitsch's Trick. So perhaps this is a sign that developing the theory of (Hilbert-)Jacobson rings to prove the Nullstellensatz is overkill. So the question seems to be: what is the best proof of Zariski's Lemma? After looking around at various proofs, here is what I think the answer is now: it is an easy consequence of the following result. Theorem (Artin-Tate Lemma): Let $R \subset T \subset S$ be a tower of rings such that (i) $R$ is Noetherian, (ii) $S$ is finitely generated as an $R$-algebra, and (iii) $S$ is finitely generated as a $T$-module. Then $T$ is finitely generated as an $R$-algebra. Proof: Let $x_1,\ldots,x_n$ be a set of generators for $S$ as an $R$-algebra, and let $\omega_1,\ldots,\omega_m$ be a set of generators for $S$ as a $T$-module. For all $1 \leq i \leq n$, we may write \begin{equation}\label{ARTINTATEEQ1}x_i = \sum_j a_{ij} \omega_j, \ a_{ij} \in T.\end{equation}Similarly, for all $1 \leq i,j \leq m$, we may write \begin{equation}\label{ARTINTATEEQ2}\omega_i \omega_j = \sum_{k} b_{ijk} \omega_k, \ b_{ijk} \in T. \end{equation}Let $T_0$ be the $R$-subalgebra of $T$ generated by the $a_{ij}$ and $b_{ijk}$. Since $T_0$ is a finitely generated algebra over the Noetherian ring $R$, it is itself a Noetherian ring by the Hilbert Basis Theorem. \ \indentNow each element of $S$ may be expressed as a polynomial in the $x_i$'s with $R$-coefficients. Making substitutions using the two equations above shows that $S$ is a finitely generated $T_0$-module. Since $T_0$ is Noetherian, the submodule $T$ is also finitely generated as a $T_0$-module. This immediately implies that $T$ is finitely generated as a $T_0$-algebra and then in turn that $T$ is finitely generated as an $R$-algebra, qed! Proof that Artin-Tate implies Zariski's Lemma: It suffices to prove the following: let $K/k$ be a field extension which is finitely generated as a $k$-algebra. Then $K/k$ is algebraic. Indeed, suppose otherwise: let $x_1,\ldots,x_n$ be a transcendence basis for $K/k$ (where $n \geq 1$ since $K/k$ is transcendental), put $k(x) = k(x_1,\ldots,x_n)$ and consider the tower of rings $k \subset k(x) \subset K$. To be sure, we recall the definition of a transcendence basis: the elements $x_i$ are algebraically independent over $k$ and $K/k(x)$ is algebraic. But since $K$ is a finitely generated $k$-algebra, it is certainly a finitely generated $k(x)$-algebra and thus $K/k(x)$ is a finite degree field extension. Thus the Artin-Tate Lemma applies to our tower: we conclude that $k(x)/k$ is a finitely generated $k$-algebra. But this is absurd. It implies the much weaker statement that $k(x) = k(x_1,\ldots,x_{n-1})(x_n)$ is finitely generated as a $k(x_1,\ldots,x_{n-1})[x_n]$-algebra, or weaker yet, that there exists some field $F$ such that $F(t)$ is finitely generated as an $F[t]$-algebra: i.e., there exist finitely many rational functions $\{r_i(t) = \frac{p_i(t)}{q_i(t)} \}_{i=1}^N$ such that every rational function is a polynomial in the $r_i$'s with $k$-coefficients. But $F[t]$ is a PID with infinitely many nonassociate nonzero prime elements (e.g. adapt Euclid's argument of the infinitude of the primes), so we may choose a nonzero prime element $q$ which does not divide $q_i(t)$ for any $i$. It is then clear that $\frac{1}{q}$ cannot be a polynomial in the $r_i(t)$'s: for instance, evaluation at a root of $q$ in $\overline{F}$ leads to a contradiction. Note that this is almost all exactly as in Artin-Tate's paper, except for the endgame above, which has been made a little more explicit and simplified: their conclusion seems to depend upon unique factorization in $k[t_1,\ldots,t_n]$, which does not come until later on in my notes. Further comments: (i) The proof is essentially a reduction to Noether's normalization in the case of field extensions, which becomes the familiar result about existence of transcendence bases. Thus it is not so far away from the most traditional proof of the Nullstellensatz. But I think the Artin-Tate Lemma is easier than Noether Normalization. (ii) Speaking of Noether: the proof of the Artin-Tate Lemma is embedded in the standard textbook proof that if $R$ is a finitely generated $k$-algebra and $G$ is a finite group acting on $R$ by ring automorphisms, then $R^G$ is a finitely generated $k$-algebra. In fact I had already typed this proof up elsewhere in my notes. Realizing that the Artin-Tate Lemma is something I was implicitly proving in the course of another result anyway was part of what convinced me that this was an efficient route to the Nullstellensatz. Note that the paper of Artin and Tate doesn't make any connection with Noether's theorem and conversely the textbooks on invariant theory that prove Noether's theorem don't seem to mention Artin-Tate. (However, googling -- Artin-Tate Lemma, Noether -- finds several research papers which allude to the connection in a way which suggests it is common knowledge among the cognoscenti.) Added: It turns out this is the proof of Zariski's Lemma given in Chapter 7 of Atiyah-Macdonald. I had missed this because (i) they give another (nice) proof using valuation rings in Chapter 5 and (ii) they do not attribute the Artin-Tate Lemma to Artin and Tate, although their treatment of it is even closer to the Artin-Tate paper than mine is above. (In the introduction of their book, they state cheerfully that they have not attributed any results. I think this is a drawback of their otherwise excellent text.)
Given two categories $I$ and $J$ we say that colimits of shape $I$ commute with limits of shape $J$ in the category of sets, if for any functor $F : I \times J \to \text{Set}$ the canonical map $$\textrm{colim}_{i\in I} \text{lim}_{j\in J} F(i,j) \to \textrm{lim}_{j\in J} \text{colim}_{i\in I} F(i,j)$$ is an isomorphism. The standard examples are a) filtered colimits commute with finite limits and b) sifted colimits commute with finite products. (Those statements can be regarded as definitions of which categories $I$ are filtered or sifted respectively, but both terms have independent definitions for which these commutation results are propositions.) A third, less known example is to take $I$ a finite group and $J$ a cofiltered category, in other words, if $G$ is a finite group and $X_j$ is an inverse system of $G$-sets, then the canonical map $$(\varprojlim_{j\in J} X_j)/G \to \varprojlim_{j \in J}(X_j/G)$$ is an isomorphism. Now, all of these examples are easy to prove separately (here's a proof of the $G$-set result, for example) but I see no unifying pattern. Is there a simple criterion for when $I$-colimits and $J$-colimits commute in the category of sets? [Note: It's true that $I$ is filtered (resp. sifted) if and only if for all finite (resp. finite discrete) $J$ the diagonal functor $I \to I^J$ is final; but I don't think that for arbitrary $I$ and $J$, if the diagonal $I \to I^J$ is final then $I$-colimits commute with $J$-limits. If I'm wrong and that condition on the diagonal actually is sufficient for commutation: why? and is it also necessary?]
It is not possible to be completely sure there is only one photon in the apparatus and in the quantum sense this is even an ambivalent question to ask. But to the question "Is there a source of photons for which we almost always detect one outcoming photon in a given time interval?" the answer is yes, single photon sources. As far as I know, it is however always a continuous source of photons, that is, it keeps spitting "antibunched" photons out continuously and just a single photon cannot be produced. As far as I know, experiments with single photon sources have however not found any deviation from quantum mechanical predictions and the "single photons" behave completely as quantum particles. Historical electron double slit experiments were usually done in vacuum. I don't know whether a "moving slit" experiment has been done, but there are some issues with it such as collecting a large number of dots forming the interference pattern. Obviously, you would like to get the full pattern of a photon which came out at $t_0$ and the double slit was moved from $x(t_0)$ to $x(t_0)+v\Delta t$. However, in a naive setting you will get a superposition of patterns from double slits at different position as the source keeps spitting out photons at different times when the slit is at different initial positions. The time $t_0$ of the "photon spit" is generally also not settable accurately due to quantum principles. So in any case, I believe the resulting image would basically be blurred and inconclusive which is the reason the experiment has probably never been done. As to a formal treatment, we could idealise the situation by taking the free particle with boundary conditions of the classical double-slit experiment and making the boundary conditions of the double slit time dependent and solve the time-dependent wave equation (a massless Klein-Gordon equation would be okay). However, as typical corrections of the perturbed solution would be $v/c$, where $v$ is the velocity of the double slit, a quasi-static treatment would most probably be sufficient. The probability density of finding a photon on the display would then just be the interference pattern of the non-moving double slit experiment translated in time according to the translation of the double slit. Detecting the time of the incidence of the display could in principle resolve the issue of blurriness and a repeated super-sensitive experiment might reconstruct the time-frozen interference pattern and investigate any deviations from the predictions of a less idealized time-dependent wave equation (accounting e.g. for the interaction with the slit wall). Nevertheless, keep in mind that we do not know any kind of classical trajectory of a photon and thus do not know the time it took to go through the apparatus, and we also do not know at which exact time it left the source, so any formulation of alternative predictions would be very difficult. EDIT: There is indeed a theory defining sharp states of quantum particles, the De Broglie-Bohm theory and in the non-static case, the formulation using the quantum potential would perhaps be useful. However, these so called "hidden variable theories" also just speak in terms of probability distributions since any attempt to measure the particle state seriously disturbs the "piloting wave". Just imagine how to actually measure the photon initial time and position, you cannot "see" it, you have to "bump" into it. But with what we can "bump" into it? Say with another photon, whose initial position and fly-out time we once again do not know! This "measurement" process can only converge to a limited certainty expressed by the Heisenberg uncertainty principle. This follows from the fact that there is a lower limit on the size of the particle/wavepacket you can use for measurement. So in our case, if we tried to measure the initial momentum and position of the photon at a given time, our "probing" particle just could not retrieve everything without disturbing the certainty in other variables. This is why in the end De-Broglie Bohm theory reproduces all the results of quantum mechanics almost to the dot. The term "frozen interference patterns" refers only to the fact that our wave-function will be time-dependent: $\psi(x,t)$ and so the probability density of finding a particle on a given position on the display $x_d$: $\rho(x_d,t) = \|\psi(x_d,t)\|^2$. (In the model I have proposed, it will just be an interference patter moving along the display). However, in one run of the experiment, we will only get one dot for every $t_1,t_2,t_3,...$ on the display. We would however like to recover $\rho(x_d,t_1),\rho(x_d,t_2)...$ (the "time-frozen" interference patterns) which is possible only with very carefully rerunning the experiment again and again and getting more and more dots at various times. We can then get e.g. the "time-frozen" $\rho(x_d,t_1)$ by plotting only detections which happened at $t_1$.
It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced. Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit. @Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form. A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it. Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis. Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)? No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet. @MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it. Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow. @QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary. @Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer. @QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits... @QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right. OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ... So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study? > I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a... @MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really. When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.? @tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...) @MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences).
A polytime algorithm that can find such a collection of edges even in the promise version of this problem can be used as a blackbox to solve the Hamiltonian Path problem in polynomial time, and thus this promise problem is NP-Hard under Cook reductions. The idea here is to use the algorithm for solving the promise problem on a sequence of graphs, many (but not all) of which potentially violate the promise given in the problem statement. Importantly, nothing is assumed of the output of the algorithm when it is fed a promise-violating graph as input, just that it outputs something (perhaps even adversarially). We can, of course, always cut the algorithm off and return the empty set as its solution if its execution is taking longer than some provided time bound (a bound only proven to hold only in promise-satisfying instances) and satisfy this requirement. It is the nature of the change in the algorithm's output (or lack thereof) on the sequence of instances that interests us. Suppose you have a graph $G=(V,E)$ in which you wish to determine the existence of a Hamiltonian path, and oracle access to a function $\varphi$ such that $\varphi(G) = h(G)$ (where $h$ is defined as in David's post) whenever $G$ contains a Hamiltonian path (and returns an arbitrary answer otherwise). Intuitively, the algorithm searches for a supergraph $H$ of $G$ on which $\varphi(H) = h(H)$ is violated. If such a graph can be found, it must be the case that $H$ (and all of its subgraphs, including $G$) is not Hamiltonian. Conversely, not violating the inequality on any graph in the sequence shows that the original $G$ is Hamiltonian. Now construct a sequence of graphs $G_0 \cdots G_{n-1}$ as follows: Let $G_0 = G$ Pick a permutation $\pi$ of $V$ such that no $(\pi_i, \pi_{i+1})$ is an edge of $\varphi(G_0)$.* For $i = 1$ to $n-1$ Let $e_i = (\pi_i, \pi_{i+1})$ Let $G_i = G_{i-1}$ with edge $e_i$ added (if it did not already exist in $G_{i-1}$) Note that $h(G_x) \supseteq h(G_y)$ when $x < y$ (i.e. containment of some fixed edge in $h$ is hereditary among the $\{G_i\}$). Thus, if $e_i \in \varphi(G_i)$ for some $i \geq 1$, it must be the case that $G$ does not contain a Hamiltonian Path, since $\varphi(\cdot)$ is not hereditary on this graph family (recall that $e_i$ is not in $\varphi(G_0)$) and thus $\varphi(G_i) \neq h(G_i)$ for some $G_i$. However, if $e_i \not \in \varphi(G_i)$ for all $i$, no single addition of an edge was necessary in the construction of the first (set of) Hamiltonian Path(s) in the sequence in which we constructed $\{G_i\}$. Since $G_{n-1}$ contains a Hamiltonian path by construction, a simple inductive argument implies $G_0$ must contain one as well. Thus, Hamiltonian Path can be solved via $O(n)$ queries to an oracle solving this promise problem (assuming they produce an arbitrary output on promise-violating graphs), meaning that any time bound $f(n)$ for an algorithm solving this problem implies a time bound of $O(n f(n))$ for solving Hamiltonian Path, and thus $f(n)$ (and any provable upper bound on the running time for a TM solving this problem) cannot be a polynomial unless $P = NP$. *This is easy to do because $\|h(G_0)\| \leq n$ by definition, and we could always of course prematurely determine the nonexistance of a Hamiltonian path if $\|\varphi(G_0)\|$ is any larger than that.
Introduction¶ A recent talk by Vincent Warmerdam about how simple models can outperform state-of-the-art ML approaches was a real eye opener to me. What amazed me the most while watching the video, is his astonishing creativity when it comes to feature creation and engineering. Although it is commonly known that good features are far more important than using the latest 10-pages-of-math Neural Network variation from arxiv, I regularly find myself guilty of trying to use the fanciest model available. In this post I took a step back and focused on a more creative idea to build a good and reliable time-series model. Besides the video mentioned before, a recent paper by Mark J. van der Laan et al. about an estimator called "Highly adaptive Lasso"(HAL) provided major inspiration for this idea. import pandas as pd import numpy as np from sklearn.linear_model import * from copy import deepcopy import matplotlib.pyplot as plt import multiprocessing as mp import warnings warnings.filterwarnings('ignore') df = pd.read_csv("shampoo.csv", index_col = 0) fig, ax = plt.subplots(figsize = (12,6)) ax.plot(df) fig.autofmt_xdate() The data looks rather unspectacular, probably a non-linear trend and some irregular spikes that might indicate an autoregressive/moving average component. As the title suggests, the main feature of this idea is the indicator function. Closely related to a harmonic regression, this model will use periodically repeating variables as exogenous regressors. More specifically, a couple of indicator functions is being generated that take on a value of "1" in a periodic manner and "0" else. Formally, we define the starting time of the time-series as $t=0$ and specify another variable $\phi$ that specifies the frequency of the 0-1 pattern. An arbitrary regressor $x_{1;t}$ then takes on values $$x_{1;t} = \mathbb{I}_{0}(t\,mod\,\phi)$$ with $mod$ being the modulo operator, $\mathbb{I}_{0}(\cdot)$ an indicator function that is "1" whenever its argument is zero and "0" else. Let's define the indicator function and another function that allows vector-valued application of that indicator in Python: def period_indicator(t, phi): if t%phi != 0: return(0) elif t%phi == 0: return(1) def period_indicator_vec(t, phi): t_vec = pd.Series(t) return(t_vec.apply(lambda x: period_indicator(x,phi))) For a monthly time series over three years and a period of 12 that starts in the first month, a feature $x_1;t,\, t\in[0,35],\,\phi=12$ would look like this: plt.plot(period_indicator_vec(np.arange(36),12)) plt.xlabel("Time") Text(0.5,0,'Time') (Although matplotlib interpolates to continuous time, the feature should actually only be equal to one 1 at $t=0,\,t=11,\,t=23$) The first question that comes to mind: "What if the periodic pattern doesn't start in the first month?" Although slightly clumsy, creating 11 more variables, each with different starting points for the periodic patterns turned out to be very effective (plotting all 12 possibilities looks kinda ugly, so I only plotted the first three variables - you should get the idea): for i in range(3): plt.plot(period_indicator_vec(np.arange(i,36+i),12)) plt.xlabel("Time") Text(0.5,0,'Time') The underlying assumption of this example is a yearly recurring effect that only occurs at the exact point in discrete time and nowhere else. This idea can be developed further for example to quarterly occuring effects $\phi=3$ or whatever. Notice that for each pattern we also need $\phi$ variables with different starting points to capture all possibilities. Since this means a very quick increase in variables, I decided to restrict the maximum $\phi$ to a reasonable value for the problem at hand. In the case of the shampoo data, allowing a maximal $\phi=12$ to express a yearly recurring pattern at maximum, was reasonable. The 'get_features()' function below will generate the periodic indicators according to the rules mentioned in 2). Since the shampoo time-series looks like it exhibits a quadratic trend, I decided to add an additive trend and an additive quadratic trend variable as well: def get_features(time_series, max_phi=12): ts = np.array(time_series) T = len(ts) harmonics_list = [] for phi in range(1,np.minimum(T,max_phi)+1): for i in range(phi): indicator = period_indicator_vec(np.arange(i,T+i), phi) harmonics_list.append(indicator) harmonics_frame = pd.concat(harmonics_list,1) harmonics_frame["trend"] = np.arange(T) harmonics_frame["trend_sq"] = np.arange(T)*2 return harmonics_frame ts_features = get_features(df) After the features have been created it is time to think of an actual algorithm to perform actual forecasts. In the manner of HAL estimator and considering that a sparse result is easier to interpret, the algorithm of choice was simple Lasso. To find a good value for Lasso's alpha parameter, the actual training set is separated in a train-validate split fashion and then grid-search is performed to select the best alpha based on the performance on the validate-set: #an extra class for the Lasso made it easier to deal with the pandas Series type - but that could probably #be left out class LassoForecaster: def __init__(self): pass def fit(self, X,y, alpha = 1.,max_iter = 1000): X_array = np.array(X) y_array = np.array(y).reshape(-1,1) self.model = Lasso(alpha = alpha, max_iter = max_iter) self.model.fit(X_array,y_array) def predict(self, X): X_array = np.array(X) predictions = self.model.predict(X_array).reshape(-1,1) return predictions #I hated to wait for unneccesarily long for the grid-search results, so I parallelized it #This function will be used in a multiprocessing.Pool() def alpha_grid_search(inp): alpha = inp["alpha"] X = np.array(inp["X"]) y = np.array(inp["y"]).reshape(-1,1) train_size = 0.75 split_point = int(len(X)train_size) X_train = X[:split_point,:] X_test = X[split_point:,:] y_train = y[:split_point,:].reshape(-1,1) y_test = y[split_point:,:].reshape(-1,1) model = Lasso(alpha = alpha) model.fit(X_train,y_train) predictions = model.predict(X_test) residuals = y_test - predictions rmse = np.sqrt(np.mean(residuals2)) return rmse Now, the whole data is split into a training and a test set. The alpha grid search is performed on the training set only, while the actual performance is measured on the training set; I decided to forecast one year ahead (=12 months). Since the model doesn't need any lagged input variables, a simple train-test split like the one below corresponds to a true 12 steps ahead test (and not just one step ahead, as is often done). oos_size = 12 X_train = ts_features.iloc[:-oos_size,:] X_test = ts_features.iloc[-oos_size:,:] y_train = df.iloc[:-oos_size] y_test = df.iloc[-oos_size:] At next, the grid search is performed to find an optimal value for alpha: proposed_alpha = np.arange(.002,1.,.002) parallel_list = [{"alpha":alpha, "X": X_train, "y": y_train} for alpha in proposed_alpha] pool = mp.Pool(processes = 16) results = [pool.apply(alpha_grid_search,args=(inp,)) for inp in parallel_list] optimal_alpha = proposed_alpha[np.argmin(results)] Final model fit, in-sample prediction and out-of-sample forecast: m = LassoForecaster() m.fit(X_train,y_train,alpha = optimal_alpha, max_iter = 10000) plt.figure(figsize = (12,6)) plt.plot(y_train.values) plt.plot(m.predict(X_train)) plt.plot(len(y_train)-1+np.arange(len(y_test)),y_test.values) plt.plot(len(y_train)-1+np.arange(len(y_test)),m.predict(X_test)) plt.xlabel("Time") plt.legend(["In-sample truth", "In-sample prediction", "Out-of-sample truth", "Out-of-sample forecast"]) <matplotlib.legend.Legend at 0x7feb53c8d748> Now comes the moment of truth with the out-of-sample forecasting error (I chose RMSE): print("RMSE:") print(np.sqrt(np.mean((y_test.values-m.predict(X_test))2))) RMSE: 55.973300349921345 I found some blogs that used much more complex Machine Learning models (e.g. LSTMs) to forecast the same data over the same period and their RMSE was oftentimes much higher (RMSE > 100 in most cases). If you don't trust me here, feel free to google some examples - By no means do I want to diminish the effort of other people or appear superior but I think this is a good example that Machine Learning doesn't always mean better results. Takeaway¶ Thanks to Warmerdam's talk, I was really intrigued to take a step back from the Machine Learning intense approach towards forecasting that I am currently taking in my thesis (e.g. Bayesian Neural Networks) and try make a simple and interpretable model work. By extracting the non-zero factors from the Lasso model and the corresponding variable values, it would even be possible to fully interpret the model and decompose the output which is always a good way to sanity check (again, watch Warmderdam's talk!) Although Google's AutoML could really revolutionize the whole Machine Learning community for the x-th time, I believe that it won't easily work with time-series like the one above. The main issue that I see here is the small amount of data available that - in my opinion - requires more domain expertise than more computational power to yield highly accurate forecasts. There are probably countless more tricks that can be applied to make simple linear models more powerful than they are currently being perceived by many people. I will happily present more ideas for that in the future. Another dataset that is also in high usage for testing time-series models, is the airline passengers dataset. df = pd.read_csv("passengers.csv", index_col = 0) plt.figure(figsize = (12,6)) plt.plot(df) fig.autofmt_xdate() By looking at the plot, the increasing amplitude of the realizations pierces the eye. Using the approach from above, the model output would result in a somewhat "mean"-amplitude over all periods, clearly not capturing the increase. To allow the model to incorporate this effect as well, each periodic indicator variable is multiplied with the square-root of the trend variable and saved as another feature: $$\tilde{x_{1;t}} = \sqrt{t}\cdot\mathbb{I}_{0}(t\,mod\,\phi)$$ which looks like the removal of heteroscedasticity in classic linear models from econometrics. On the other hand, the quadratic trend variable is removed and only the simple linear trend variable is kept. def get_features_het(time_series, max_phi=12): ts = np.array(time_series) T = len(ts) harmonics_list = [] for phi in range(1,np.minimum(T,max_phi)+1): for i in range(phi): indicator = period_indicator_vec(np.arange(i,T+i), phi) harmonics_list.append(indicator) hetskad = indicator * (np.arange(T)+1)(1/2) harmonics_list.append(pd.Series(hetskad)) harmonics_frame = pd.concat(harmonics_list,1) harmonics_frame["trend"] = np.arange(T) return harmonics_frame ts_features = get_features_het(df) Since there is much more data available now, the out-of-sample size is increased to 3 years - everything else stays the same oos_size = 36 X_train = ts_features.iloc[:-oos_size,:] X_test = ts_features.iloc[-oos_size:,:] y_train = df.iloc[:-oos_size] y_test = df.iloc[-oos_size:] proposed_alpha = np.arange(.002,1.,.002) parallel_list = [{"alpha":alpha, "X": X_train, "y": y_train} for alpha in proposed_alpha] pool = mp.Pool(processes = 16) results = [pool.apply(alpha_grid_search,args=(inp,)) for inp in parallel_list] optimal_alpha = proposed_alpha[np.argmin(results)] m = LassoForecaster() m.fit(X_train,y_train,alpha = optimal_alpha, max_iter = 10000) plt.figure(figsize = (12,6)) plt.plot(y_train.values) plt.plot(m.predict(X_train)) plt.plot(len(y_train)-1+np.arange(len(y_test)),y_test.values) plt.plot(len(y_train)-1+np.arange(len(y_test)),m.predict(X_test)) plt.xlabel("Time") plt.legend(["In-sample truth", "In-sample prediction", "Out-of-sample truth", "Out-of-sample forecast"]) <matplotlib.legend.Legend at 0x7feb53279320> print("RMSE:") print(np.sqrt(np.mean((y_test.values-m.predict(X_test))2))) RMSE: 28.071196336221476 Note: This is also below the error of many ML models that you can find after some google :)
I think your confusion arises because you think that the electron is modelled by the wavepacket. Instead the wavepacket of an electron which we suspect near a point $x_0$ represents (or is associated with) the electron and tells us how likely it is to find the electron near that point $x_0$. This is an important distinction because this way the concept of dispersion makes more sense, since it is nonsensical to say that the electron smears out while it's perfectly valid to say that the wavepacket smears out. So again: the wavepacket is not the electron but represents (is associated with) the electron. This also means that the wavepacket isn't spherically symmetric just because the electron is "round". Sure it makes sense that we use gaussian packets since for a free particle the probability to find it is equal in all directions (if we start from a point where the probability is a maximum). The point I'm trying to make is that even if you had a "quantum elephant" the wavepacket need not look like an elephant. 4.) I think this is nitpicking, but the correct way to say it should be "...hardness stands for the amplitude of the probability density.", since wavepackets also are of the form $\psi(x,t)=\dots$, where $\left\|\psi\right(x,t)\|^2=\dots$ is the probability density. So for your hairy ball this means that at the outer layers, where the ball is soft, the probability density is small and the electron unlikely to be found. 5.) This is true, since the coefficient function of a gaussian packet for example is $$\phi(\vec k)=A^3e^{-(\vec k - \vec k_0)^2 d^2 } ,$$ so while most of the k's are near $k_0$, there are also many k's slightly different from it, although the number of k's much different than $k_0$ decreases exponentially. You can also see it this way: $\vec k - \vec k_0$ is the vector which connects the head of $\vec k$ to the head of $\vec k_0$, when this arrow is big, his square is also big and the coefficient function small. A.) When you want to stay in your hairy ball analogy, then it would best to think about this ball getting bigger and more soft. So when you have your ball at the beginning and draw a sphere around the center where the ball is hard, then this sphere you draw gets bigger over time and the ball inside the sphere becomes more soft. When you wait long enough the ball becomes very large and very soft everywhere and the center is hardly any harder than the outer layers. If you view it this way then you can call it dispersion. Below is an image how dispersion can be depicted. The black circle would then be the sphere in your ball while the probability amplitude would correspond to the hardness. The circle (sphere) becomes bigger while the amplitude (hardness) becomes less over time. The next thing is that the dispersion is proportional to the time elapsed, so even if the wavepacket doesn't move initially the wavepacket spreads out. And no, it's not the electron which explodes: it's the wavepacket which "explodes" in the sense that it spreads out in space. This means that a free particle becomes delocalized: since the wavepacket spreads out so does the probability of finding the electron at some place. B.) When you form a wavepacket you have to do the following integral $$\int A(k) e^{i(kx-w(k)t)}d^3 k. $$ Since you want to integrate over all $k$'s, you have to write omega as a function of $k$. Now, since for a wavepacket you need $k$'s very close to some value $k_0$ you can expand the $\omega(k)$ function into $$\omega = \omega_0 + \left( \frac{d\omega(k)}{dk}\right)_{k_0}(k-k_0)^2+... $$ This is why the wavepacket has only one single frequency $\omega_0$ (first term): you assume the values of $k$ are centered close to some $k_0$, so $\omega(k)$ is also centered about some $\omega_0$, while the distribution of $k$'s around $\omega_0$ - the dispersion relation - gives you the velocity at which the packet moves (second term).
Magnetism is not explained by conservation of energy. But magnetism is consistent with conservation of energy. It is commonly stated that the magnetic field does no work. It is true that the magnetic field does no work on charged particles through the Lorentz force law. The force delivered on a charge due to a magnetic field, $\vec{F}=q \vec{v}\times \vec{B}$, is always perpendicular to $\vec{B}$ However, the magnetic field can do work on the electric field, through one of Maxwell's equations: $\frac{d \vec{E}}{dt}=\frac{1}{\mu_0 \varepsilon_0}\nabla \times \vec{B}-\frac{1}{\varepsilon_0} \vec{J}$ So this is one way in which the magnetic field does work. For example, if you have an inductor in a circuit, a constant current produces a constant magnetic field inside the inductor. Once you switch the current off, you actually get some energy back out of the inductor due to the collapsing magnetic field. This is an instance where the magnetic field does work on the electric field, and the electric field in turn does work by creating a current in the wire.
Answer (a)$\tau=(\frac{L}{2})mg$ (b)0 (c)$\tau=(\frac{L}{2})mg$ Work Step by Step We know that (a) $\tau=r\times F$ $\implies \tau=(\frac{L}{2})mg$ (b) $\tau=r\times F$ $\implies \tau=(0)mg$ $\tau=0$ (c) $\tau=r\times F$ $\tau=-(\frac{L}{2})mg$ 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.
The statistical model under consideration is given by the independent realizations of two independent binomial distributions: $$ x_1 \sim \mathrm{Bin}(n_1, p_1), \quad x_2 \sim \mathrm{Bin}(n_2,p_2), $$ and the null hypothesis is $H_0\colon\{p_1=p_2\}$. According to my calculations, the deviance (minus double log-likelihood ratio) has a noncentral chi-squared asymptotic distribution with $1$ degree of freedom and noncentrality parameter $$\lambda = \frac{{(p_1-p_2)}^2n_1n_2}{n_1p_2(1-p_2) + n_2p_1(1-p_1)}.$$ dev <- function(n1,x1,n2,x2){ 2(x1log(x1/n1) + (n1-x1)log(1-x1/n1) + x2log(x2/n2) + (n2-x2)log(1-x2/n2) - (x1+x2)log((x1+x2)/(n1+n2)) - (n1+n2-(x1+x2))log(1-(x1+x2)/(n1+n2)))}lambda <- function(n1,p1,n2,p2){ (p1-p2)^2 (n1n2)/(n1p2(1-p2)+n2p1*(1-p1))} Simulations on the following example perfectly fit the noncentral chi-squared: n1 <- 150; n2 <- 100p1 <- 0.3; p2 <- 0.4 nsims <- 15000D <- numeric(nsims)for(i in 1:nsims){ x1 <- rbinom(1, n1, p1); x2 <- rbinom(1, n2, p2) D[i] <- dev(n1, x1, n2, x2)}curve(ecdf(D)(x), from=min(D), to=max(D),lwd=3)ncp <- lambda(n1, p1, n2, p2) curve(pchisq(x, df=1, ncp=ncp), add=TRUE, col="red", lty="dashed", lwd=3) But when I increase the difference $\|p_1-p_2\|$, this is bad: p1 <- 0.3; p2 <- 0.7 for(i in 1:nsims){ x1 <- rbinom(1, n1, p1); x2 <- rbinom(1, n2, p2) D[i] <- dev(n1, x1, n2, x2)}curve(ecdf(D)(x), from=min(D), to=max(D),lwd=3)ncp <- lambda(n1, p1, n2, p2) curve(pchisq(x, df=1, ncp=ncp), add=TRUE, col="red", lty="dashed", lwd=3) And this is worst when I increase the sample sizes: n1 <- 15000; n2 <- 10000for(i in 1:nsims){ x1 <- rbinom(1, n1, p1); x2 <- rbinom(1, n2, p2) D[i] <- dev(n1, x1, n2, x2)}curve(ecdf(D)(x), from=min(D), to=max(D), lwd=3)ncp <- lambda(n1, p1, n2, p2) curve(pchisq(x, df=1, ncp=ncp), add=TRUE, col="red", lty="dashed", lwd=3) I have checked there's no problem with pchisq even for a large ncp. So why is it so bad ? I also think there is no problem in the calculation of the deviance, because results are the same when replacing it by its Pearson approximate which does not involve logarithms.
I'm currently trying to teach myself the path integral formulation of QFT (having studied the canonical approach previously), but I'm having some conceptual difficulties that I hope I can clear up here. For simplicity, consider the case of a free, single real scalar field. The path integral formulation for a two point correlator in this case is given by $$\langle 0\lvert T\lbrace\hat{\phi}(x)\hat{\phi}(y)\rbrace\lvert 0\rangle =(-i^{2})\frac{1}{Z[0]}\frac{\delta^{2}Z[J]}{\delta J(x)\delta J(y)}\bigg\lvert_{J=0}$$ where $$Z[J]=\int\mathcal{D}\phi\; e^{i\int d^{4}x\left(-\frac{1}{2}\phi (\Box +m^{2})\phi+J(x)\phi (x)\right)}$$ is the generating functional for the free theory. Here is where my issue lies. Are the fields $\phi (x)$ in the functional $Z[J]$ classical fields or are they operator fields? If they are classical fields, then does the path integral define some sort of mapping between field operators $\hat{\phi}(x)$ and their classical (c-number) analogs? The books I've been reading so far (Srednicki's QFT book and M. Schwartz's "QFT & the Standard Model") seem to a bit ambiguous in this area.
Abaid ur Rehman Virk, Tanveer Abbas and Wasim Khalid )}{\sqrt{\frac{{{d}_{u}}+{{d}_{v}}-2}{{{d}_{u}}\cdot {{d}_{v}}}}}, \\\,\,\,\,\,GAII\left( G \right)=\prod\limits_{uv\in E\left( G \right)}{\frac{2\sqrt{{{d}_{u}}\cdot {{d}_{v}}}}{{{d}_{u}}+{{d}_{v}}}}, \\G{{A}^{a}}II\left( G \right)=\prod\limits_{uv\in E\left( G \right)}{{{\left( \frac{2\sqrt{{{d}_{u}}\cdot {{d}_{v}}}}{{{d}_{u}}+{{d}_{v}}} \right)}^{\alpha }}.} \\\end{array}$$2 SiliconCarbideIn 1891, an American scientist discover SiliconCarbide. But now a days, we can produce siliconcarbide artificially by silica and carbon. Till 1929, siliconcarbide was known as the hardest Ştefan Ţălu, Sebastian Stach, Shikhgasan Ramazanov, Dinara Sobola and Guseyn Ramazanov Abstract The purpose of this study was to investigate the topography of silicon carbide films at two steps of growth. The topography was measured by atomic force microscopy. The data were processed for extraction of information about surface condition and changes in topography during the films growth. Multifractal geometry was used to characterize three-dimensional micro- and nano-size features of the surface. X-ray measurements and Raman spectroscopy were performed for analysis of the films composition. Two steps of morphology evolution during the growth were analyzed by multifractal analysis. The results contribute to the fabrication of silicon carbide large area substrates for micro- and nanoelectronic applications. .7. G. Lebon, D. Jou, and J. Casas-Vázquez, Understanding Non- equilibrium Thermodynamics. Springer-Verlag, 2008.8. I. Mueller and T. Ruggeri, Rational Extended Thermodynamics. Springer-Verlag, 1998.9. O. Muscato and V. D. Stefano, Electrothermal transport in siliconcarbide semiconductors via a hydrodynamic model, SIAM J. APPL. MATH., vol. 75, no. 4, pp. 1941-1964, 2015.10. A. Jüngel, Energy transport in semiconductor devices, Math. Comput. Model. Dyn. Syst., vol. 16, pp. 1-22, 2010.11. G Milan Perný, Vladimír Šály, Michal Váry, Miroslav Mikolášek, Jozef Huran and Juraj Packa References[1] KANNO, H.-IDE, D.-TSUNOMURA, Y.-TAIRA, S.-BABA, T.-YOSHIMINE, Y.-TAGUCHI, M.-KINOSHITA, T.-SAKATA, H.-MARUYAMA, E. : Over 22% Efficient HIT Solar Cell, In: Proceedings of the 23rd European photovoltaic solar energy conference and exhibition, Valencia, Spain, 2008, pp. 1136-1139.[2] STREET, R. A. : Technology and Applications of Amorphous Silicon, Springer, New York, 2000.[3] CHANG, Y. L.-CHEN, M. Y.-LIU, J. S. Q.-CHIEN, Y. J.-YANG, P. C.-HUANG, M. Y. : SiliconCarbide Emitter for Heterojunction Milan Perný, Vladimír Šály, František Janíček, Miroslav Mikolášek, Michal Váry and Jozef Huran R eferences[1] D. A. Anderson and W. E. Kspear, “Electrical and Optical Properties of Amorphous SiliconCarbide, Silicon Nitride and Germanium Carbide Prepared by Glow Discharge Technique”, Philos. Mag. B vol. 35, 1977, pp. 113–131.[2] P. P. Dey and A. Khare, “Effect of Substrate Temperature on Structural and Linear and Nonlinear Optical Properties of Nanostructured PLD a-SiC thin films”, Materials Research Bulletin vol. 84, 2016, pp. 105–117.[3] I. Kleps and A. Angelescu, “LPCVD Amorphous SiliconCarbide Films, Properties and Sandra Veličković, Slavica Miladinović, Blaža Stojanović, Ružica R. Nikolić, Branislav Hadzima and Dušan Arsić . Study Of Mechanical And Tribological Properties Ofal-6061 Reinforced With SiliconCarbide And Graphite Particles , International Journal of Technology Enhancements and Emerging Engineering Research, 3(4).R avindran , P., M anisekar , K., R athika , P., N arayanasamy , P. 2013. Tribological properties of powder metallurgy – Processed aluminium self-lubricating hybrid composites with SiC additions , Materials and Design, 45, 561–570.R adhika , N., S ubramanian , R., V enkat P rasat , S., A nandavel , B. 2012. Dry sliding wear behaviour of aluminium S.M. Kahar, C.H. Voon, C.C. Lee, U. Hashim, M.K. Md Arshad, B.Y. Lim, S.C.B. Gopinath and W. Rahman 1IntroductionSiliconcarbide (SiC) is one of the most popular ceramics used in the industry. It has unique characteristics such as high melting point, excellent oxidation resistance, high chemical inertness, high-thermal conductivity, good microwave absorbing ability, wide energy band gap and high mechanical strength enabling SiC to be used widely in aerospace structures, biomaterials and high temperature semiconducting devices [ 1 – 6 ]. SiC is produced mainly in industry by Acheson process. This process named after its inventor Edward Goodrich Acheson Szymon Piasecki, Robert Szmurlo and Marek Jasinski Abstract Power electronic circuits, in particular AC-DC converters are complex systems, many different parameters and objectives have to be taken into account during the design process. Implementation of Multi-Objective Optimization (MOO) seems to be attractive idea, which used as designer supporting tool gives possibility for better analysis of the designed system. This paper presents a short introduction to the MOO applied in the field of power electronics. Short introduction to the subject is given in section I. Then, optimization process and its elements are briefly described in section II. Design procedure with proposed optimization parameters and performance indices for AC-DC Grid Connected Converter (GCC) interfacing distributed systems is introduced in section III. Some preliminary optimization results, achieved on the basis of analytical and simulation study, are shown at each stage of designing process. Described optimization parameters and performance indices are part of developed global optimization method dedicated for ACDC GCC introduced in section IV. Described optimization method is under development and only short introduction and basic assumptions are presented. In section V laboratory prototype of high efficient and compact 14 kVA AC-DC converter is introduced. The converter is elaborated based on performed designing and optimization procedure with the use of silicon carbide (SiC) power semiconductors. Finally, the paper is summarized and concluded in section VI. In presented work theoretical research are conducted in parallel with laboratory prototyping e.g. all theoretical ideas are verified in laboratory using modern DSP microcontrollers and prototypes of the ACDC GCC. This paper presents a new concept for a power electronic converter - the extended T-type (eT) inverter, which is a combination of a three-phase inverter and a three-level direct current (dc)/dc converter. The novel converter shows better performance than a comparable system composed of two converters: a T-type inverter and a boost converter. At first, the three-level dc/dc converter is able to boost the input voltage but also affects the neutral point potential. The operation principles of the eT inverter are explained and a simulation study of the SiC-based 6 kVA system is presented in this paper. Presented results show a serious reduction of the DC-link capacitors and the input inductor. Furthermore, suitable SiC power semiconductor devices are selected and power losses are estimated using Saber software in reference to a comparative T-type inverter. According to the simulations, the 50 kHz/6 kVA inverter feed from the low voltage (250 V) shows <2.5% of power losses in the suggested SiC metal oxide-semiconductor field-effect transistors (MOSFETs) and Schottky diodes. Finally, a 6 kVA laboratory model was designed, built and tested. Conducted measurements show that despite low capacitance (2 × 30 μF/450 V), the neutral point potential is balanced, and the observed efficiency of the inverter is around 96%.
In case this question appeared here I apologize for the preposterous usage of computer space and attention of participants. I searched over there, for a similar or the same question for an answer but resigned. :) I am reading: S. B. Niefield, Cartesianness, topological spaces, uniform spaces, and affine schemes, J. Pure Appl. Alg. 23, 1982, 147–167, where the author extends the Day-Kelley characterization of exponentiable objects in $Top$ to slice categories $Top/X$. Let $A$ be a category with finite limits and $T$ is an object of $A$. It is clear that a product $X\times T$ becomes an object over $T$ via the projection $\pi_2 : X\times T\rightarrow T$. This induces a functor $T^*:A\rightarrow A/T$ which is clearly right adjoint of the forgetful functor $\Sigma_T :A/T\rightarrow A$. Now the states: A functor $F:B\to A/T$ has a right adjoint if and only if $\Sigma_T \circ F$ has a right adjoint. Proposition 1.1 To prove that if $\Sigma_T \circ F$ has a right adjoint then $F$ has a right adjoint the author argues: Suppose $\Sigma_T \circ F\dashv G'$, For $X$ object of $B$, let $\sigma_X:X\to G'T$ be the right adjunct of $FX$ considered as morphism $\Sigma_T(FX)\to T$. Then, if $f:X\to X'$ is a morphism of $B$, the diagram $\require{AMScd}$ \begin{CD} X @>f>> X' \\ @V \sigma_X VV= @VV \sigma_{X'} V\\ G'T @>>id_{G'T}> G'T \end{CD} commutes. I stuck on this: WHY THE DIAGRAM COMMUTES? Then using naturality (since the commutation of the above diagram means that $\sigma $ is a natural transformation) the author just builds the right adjoint $G$ for $F$ as equalizer of $G'p_Y :G'Y\to G'T$ and $\sigma _{G'Y}:G'Y\to G'T$ where $p_Y: Y\to T$ is an object of $A/T$. My Question is why the diagram commutes. Thanks in advance.
Assume that we have $p$ finite sets ${m_1},{m_2},...,{m_p}$, with known cardinalities ${M_1},{M_2},...,{M_p}$, where $1 \le {M_i} \leq q$ ($i=1,2,...,p$). Each set contains (distinct) elements, taken from GF($q$) (I am interested in particular in the case where $q$ is a power of $2$). Now, we consider all possible assignments of elements to $\left\{ {{m_i}} \right\}_{i = 1}^p$. It is clear that we have $\prod\limits_{i = 1}^p {\left( {\begin{array}{{20}{c}} q \\ {{M_i}} \end{array}} \right)} $ such assignments. For each assignment $A_k$ ($k=1,2...,\prod\limits_{i = 1}^p {\left( {\begin{array}{{20}{c}} q \\ {{M_i}} \end{array}} \right)} $) we consider all possible combinations of elements from the sets in the assignment, when one element is taken from each set (that is, we have $p$ elements for each combination), and we sum the elements of each combination, using GF($q$) arithmetic. There are $\prod\limits_{i = 1}^p {{M_i}} $ possible combinations/sums for each assignment. Given the sums $S_{kj}$ ($j=1,2,...,\prod\limits_{i = 1}^p {{M_i}}$) for each assignment $k$, we count the number of distinct elements of GF($q$) among these sums, and we write down this number, $d_k$. Repeating this process, we end with $\prod\limits_{i = 1}^p {\left( {\begin{array}{{20}{c}} q \\ {{M_i}} \end{array}} \right)} $ numbers. As an example, consider the case $q = 4,p = 2,{M_1} = {M_2} = 2$. Possible assignments to $m_1$, as well as for $m_2$ in this case, are: $$\left\{ {0,1} \right\},\left\{ {0,2} \right\},\left\{ {0,3} \right\},\left\{ {1,2} \right\},\left\{ {1,3} \right\},\left\{ {2,3} \right\}$$ Assume a specific assignment ${m_1} = \left\{ {0,1} \right\},{m_2} = \left\{ {1,2} \right\}$. We have $4$ combinations/sums: $0 + 1 = 1,0 + 2 = 2,1 + 1 = 0,1 + 2 = 3$ and for this specific assignment the number of distinct elements is $4$. Another possible assignment is ${m_1} = \left\{ {0,1} \right\},{m_2} = \left\{ {0,1} \right\}$. Now the combinations/sums are $0 + 0 = 1,0 + 1 = 1,1 + 0 = 1,1 + 1 = 0$ and we have $2$ distinct elements. Repeating this, we obtain $\left( {\begin{array}{{20}{c}} 4 \\ 2 \end{array}} \right) \cdot \left( {\begin{array}{{20}{c}} 4 \\ 2 \end{array}} \right) = 36$ numbers. Denote by $d$ the random variable that counts the number of distinct elements for each assignment. My question is: what is the probability that $d=n$ ($n=1,2,...,q$)? in other words, given cardinalities ${K_1},{K_2},...,{K_p}$, what is the probability that the number of distinct sums for a randomly chosen assignment equals $n=1,2,...,q$? I tried to formulate this as the probability that union of $\frac{{\prod\limits_{i = 1}^p {\left( {\begin{array}{{20}{c}} q\\ {{M_i}} \end{array}} \right)} }}{{{{\max }_i}\left( {{M_i}} \right)}}$ sets of cardinality ${\max _i}\left( {{M_i}} \right)$ is of size $n$ (this way the probability is zero for $d < {\max _i}\left( {{M_i}} \right)$), but this does not take into account additional properties of GF($q$). It is interesting to note that in the example above, we have zero probability for $d=3$: $$\Pr \left( {d = 1} \right) = 0,\Pr \left( {d = 2} \right) = 1/3,\Pr \left( {d = 3} \right) = 0,\Pr \left( {d = 4} \right) = 2/3$$ I also noted that when $M_i=2$ and $q=4$, we have for each $p$: $\Pr \left( {d = 2} \right) = 1/3^{p-1}$. (This probability distribution was calculated using a computer, computing $d_k$ explicitly for each assignment).
awhich has not been declared. For the first error the macro %x%causes the cursor to point to parameterb. >I decided to have a go at the challenge set in this paper - hope you don't mind.> See also: http://people.sabanciuniv.edu/ertekg/papers/2006/alp_et_al_ims2006.pdf. I can find a better solution using the column generation model. The following input was used: ----------------------------------------------------- Data ----------------------------------------------------- scalar r 'raw width' /1200/; table demanddata(i,) width demand width1 100 100 width2 200 2 width3 300 2 width4 493 250 width5 590 2 width6 630 2 width7 780 2 width8 930 2 ; The result is a solution (not proven optimal) of 131: ---- 174 PARAMETER pat pattern usage p4 p5 p9 p10 p11 p15 Total width1 1 2 100 width2 1 2 width3 1 2 width4 2 1 2 250 width5 2 2 width6 1 2 width7 1 2 width8 1 2 Count 75 1 2 2 2 49 131 ** SOLVER STATUS 8 USER INTERRUPT MODEL STATUS 14 NO SOLUTION RETURNED ** OBJECTIVE VALUE 3885.0000 RESOURCE USAGE, LIMIT 6.618 3600.000 ITERATION COUNT, LIMIT 0 10000000 MOSEK Link May 1, 2008 22.7.1 WIN 3927.4700 VIS x86/MS Windows M O S E K version 5.0.0.79 (Build date: Jan 29 2008 13:41:45) Copyright (C) MOSEK ApS, Fruebjergvej 3, Box 16 DK-2100 Copenhagen, Denmark http://www.mosek.com No solution returned set si /i1,i2/;parameter duedateadj(si) /i1 10, i2 10/;set k /k1,k2/;parameter duration(si,k) / i1.k1 1 /; parameters minduration(si) mintime(si) "calculate in one step" mintime2(si) "calculate in two steps"; mintime(si) = duedateadj(si)-smin(k$duration(si,k),duration(si,k)); minduration(si) = smin(k$duration(si,k),duration(si,k));mintime2(si) = duedateadj(si)-minduration(si); display mintime, mintime2; ---- 17 PARAMETER mintime calculate in one step i1 9.000, i2 -INF ---- 17 PARAMETER mintime2 calculate in two steps i1 9.000, i2 10.000 set j /x,y/ i /i1,i2,i3/ ; parameter data(i,j); * may 30, may 31, june 1 data(i,'x') = jdate(2008,5,30)+ord(i)-1; data(i,'y') = ord(i); * check dates parameter check(i,); check(i,'year') = gyear(data(i,'x')); check(i,'month') = gmonth(data(i,'x')); check(i,'day') = gday(data(i,'x')); option check:0:1:1; display data,check; execute_unload 'data',data; ---- 20 PARAMETER data x y i1 39597.000 1.000 i2 39598.000 2.000 i3 39599.000 3.000 ---- 20 PARAMETER check year month day i1 2008 5 30 i2 2008 5 31 i3 2008 6 1 Sub color() ActiveSheet.UsedRange.Style = "Normal" Call colorsub(xlNumbers, "Accent1") Call colorsub(xlTextValues, "Accent2") Call colorsub(xlErrors, "Accent3") Call colorsub(xlLogical, "Accent4") End Sub Sub colorsub(SpecialCells As Integer, styletype As String) Dim r As Range On Error GoTo err Set r = ActiveSheet.UsedRange.SpecialCells(xlCellTypeFormulas, SpecialCells) r.Style = styletype err: End Sub S O L V E S U M M A R Y MODEL TIME OBJECTIVE LL TYPE NLP DIRECTION MAXIMIZE SOLVER PATHNLP FROM LINE 160 * SOLVER STATUS 4 TERMINATED BY SOLVER MODEL STATUS 7 INTERMEDIATE NONOPTIMAL OBJECTIVE VALUE 47.2365 RESOURCE USAGE, LIMIT 35.697 1000.000 ITERATION COUNT, LIMIT 0 100000 EVALUATION ERRORS 0 0 PATH-NLP May 1, 2008 22.7.2 WEX 3906.4799 WEI x86_64/MS Windows NLP size: 829 rows, 881 cols, 10145 non-zeros, 1.39% dense. MCP size: 1708 rows/cols, 26626 non-zeros, 0.91% dense. S O L V E S U M M A R Y MODEL m OBJECTIVE cost TYPE MIP DIRECTION MINIMIZE SOLVER CPLEX FROM LINE 168 SOLVER STATUS 1 NORMAL COMPLETION MODEL STATUS 1 OPTIMAL ** OBJECTIVE VALUE 3712.0000 RESOURCE USAGE, LIMIT 630.895 3600.000 ITERATION COUNT, LIMIT 48023779 10000000 Example output file: * miptrace file trace.csv: ID = Cplex 11.0.1 * fields are lineNum, seriesID, node, seconds, bestFound, bestBound 1, S, 0, 0.075, na, 4053.06 2, N, 0, 0.084, na, 4053.06 3, N, 0, 0.09, 1740.37, 3852.17 4, N, 0, 0.095, 1740.37, 3804.84 5, N, 0, 0.101, 1740.37, 3798.37 6, N, 0, 0.107, 1740.37, 3781.77 7, N, 0, 0.113, 1740.37, 3781.53 8, N, 0, 0.149, 1826.58, 3779.84 9, N, 100, 0.229, 2769.62, 3327.77 10, N, 200, 0.304, 2891.26, 3135.02 11, N, 300, 0.391, 2891.26, 2910.57 12, E, 320, 0.397, 2891.26, 2891.26 * miptrace file trace.csv closed See also http://www.ampl.com/BOOKLETS/amplcplex110userguide.pdf (the AMPL link has a number of these features implemented) and http://www.gams.com/dd/docs/solvers/cplex.pdf. I sent it to GAMS for further investigation. This problem seems only to happen with the 64 bit version. The work around would be to use the 32 bit version. MOSEK Link May 1, 2008 22.7.2 WEX 3927.4799 WEI x86_64/MS Windows M O S E K version 5.0.0.79 (Build date: Jan 29 2008 13:47:58) The solution is semicont variable x; variable y; equation e; y.lo=0; x.fx=1; e.. y =e= x; model m /e/; solve m minimizing y using mip; In this MPS file column ROWS N obj E c1 COLUMNS sc1 c1 -1 x2 obj 1 x2 c1 1 RHS BOUNDS FX bnd sc1 1 ENDATA Work around: use the Cplex option ROWS N obj E c1 COLUMNS sc1 c1 -1 x2 obj 1 x2 c1 1 RHS BOUNDS LO bnd sc1 1 SC bnd sc1 1 ENDATA S O L V E S U M M A R Y MODEL thainavy OBJECTIVE obj TYPE MIP DIRECTION MINIMIZE SOLVER CPLEX FROM LINE 84 ** SOLVER STATUS 4 TERMINATED BY SOLVER MODEL STATUS 14 NO SOLUTION RETURNED ** OBJECTIVE VALUE 0.0000 RESOURCE USAGE, LIMIT 0.037 1000.000 ITERATION COUNT, LIMIT 0 5 The SOLVER STATUS is also wrong, this should be 2 ITERATION INTERRUPT. Turns out I have reported this already in january 2006, but it does not seem to get fixed. Correct values for these settings are often important when running Cplex from a more complex algorithm formulated in GAMS. One way of achieving this is to add the constraints: a(i) ≥ 0.001·p(i) b(i) ≥ 0.001·q(i) c(i) ≥ 0.001·r(i) sum(i,p(i)) ≥ 2 sum(i,q(i)) ≥ 2 sum(i,r(i)) ≥ 2 p(i),q(i),r(i) binary variables p(i)=0 => a(i)=0 x ≤ (1-w)·M where M=x.up x ≥ 0.001·(1-w) Method Time Accuracy Time Accuracy Time Accuracy n=50 n=100 n=200 Nonlinear Equations 0.148 9.6e-8 0.917 1.2e-4 8.991 0.621 External Solver 0.024 1.6e-13 0.025 1.9e-10 0.056 1.4e-5 GAMS Algorithm 0.034 1.9e-13 0.482 6.3e-11 7.571 7.7e-6 I am surprised that the GAMS algorithm is competitive. Notice the accuracy issues when using nonlinear equations due to ill-conditioning of the system of non-linear equations (small changes in a(i,j) lead to large changes in L(i,j)). Here is some info about the external solver: c:\>cholesky CHOLESKY: matrix decomposition A=LL^T Usage > cholesky gdxin i a gdxout L where gdxin : name of gdxfile with matrix i : name of set used in matrix a : name of 2 dimensional parameter inside gdxin gdxout : name of gdxfile for results (factor L) L : name of 2 dimensional parameter inside gdxout Calculates the Cholesky decomposition A=LL^T of a symmetric positive definite matrix A=a(i,j) where i and j are aliased sets. L will contain the Cholesky factor L(i,j) C:\> if(1, $set name "hello" else $set name "world" ); display "%name%"; scalar s /NA/; s$(s>1) = 3.14; display s; > touch f1.txt f2.txt f3.txtHowever it fails miserably: > ren f.txt gg.txt A ("somewhat" complicated) solution is presented here.C:\projects\test\test1>touch f1.txt f2.txt f3.txt C:\projects\test\test1>ren f.txt gg.txt A duplicate file name exists, or the file cannot be found. A duplicate file name exists, or the file cannot be found. C:\projects\test\test1>dir Volume in drive C has no label. Volume Serial Number is 7563-3993 Directory of C:\projects\test\test1 05/16/2008 08:42 PM <DIR> . 05/16/2008 08:42 PM <DIR> .. 05/16/2008 08:42 PM 0 f2.txt 05/16/2008 08:42 PM 0 f3.txt 05/16/2008 08:42 PM 0 gg.txt 3 File(s) 0 bytes 2 Dir(s) 159,707,815,936 bytes free C:\projects\test\test1> >I would like to write the results of a GAMS program into a text file >whose name should depend on the date and time of the run. How could >that be done? > >Ex: output-05.15.08-09.39.txt >The above is just a sample and it does not have to be exactly the same. The suggestion to use myfile /filename_%system.date%_%system.time%/ does not work for several reasons: quotes are needed to handle embedded blanks and the ‘/’ characters in the system date cause problems when used inside a file name. Here is a different approach that actually works (I ran it before suggesting it): The generated include file will look like:$onecho > createfile.awk BEGIN { print "file f / \"f_" strftime("%Y-%m-%d_%H.%M.%S") ".txt\"/" } $offecho $call "awk -f createfile.awk > ftime.inc" $include ftime.inc put f "hello"/; For more info on AWK see gawk.htmlfile f / "f_2008-05-16_14.28.59.txt"/ > There is a worked example of how to solve a basic Cutting Stock Problem at> http://docs.google.com/View?docid=dfkkh8qj_64rrpz86db> It uses simple methods, and the open source program lp_solve. Briefly, > the solution shown is:>> * create a list of cut combinations (not permutations - there would be > too many of those to work with in most cases)>> * from this list, generate a set of linear programming (simplex) expressions, > and output them to a file>> * this file becomes the input to lp_solve, which calculates an optimal solution>> * most of the cut combinations in the solution will have a score of zero. > For those that have a score greater than zero, look them up on the list > created in step 1 The only change to the model to solve this problem was to update the data to reflect this problem (see below). The complete model is here. A clever way to solve these problems is using a Delayed Column Generation approach. Such a model is not too difficult to code in GAMS. See http://www.amsterdamoptimization.com/pdf/colgen.pdf. A nice write up on the cutting stock problem is available from http://en.wikipedia.org/wiki/Cutting_stock_problem. The last restriction is essentially a big-M constraint: \[\boxed{\begin{split}\min\{0,x^{lo}\} &\le z \le \max\{0,x^{up}\} \\ x^{lo} \cdot \delta &\le z \le x^{up} \cdot \delta \\ x − x^{up} \cdot (1 − \delta) &\le z \le x − x^{lo} \cdot (1 − \delta)\end{split}}\] where $M_1$, $M_2$ are chosen as tightly as possible while not excluding $z=0$ when $\delta=0$. \[\boxed{x − M_1 \cdot (1 − \delta) \le z \le x + M_2 \cdot (1 − \delta)}\] The construct $z=x· \delta$ can be used to model an OR condition: "$z = 0$ OR $z = x$". \[\boxed{\begin{split}& z \in [0,x^{up}]\\ & z ≤ x^{up} · \delta\\ & z ≤ x\\ & z ≥ x − x^{up} · (1 − \delta)\end{split}}\] I used GNUPLOT to plot results, e.g.: Notice that we relaxed $z$ to be continuous between 0 and 1: $z$ is automatically integer; a property we can use to reduce the number of discrete variable. Note that modern high performance solvers may reintroduce these variables as binary as they can exploit this in their algorithms. \[\boxed{\begin{split}0 &≤ z ≤ 1\\ z &≤ x_1\\ z &≤ x_2\\ z &≥ x_1 + x_2 − 1\end{split}}\] This formulation has fewer equations but has two major disadvantages: \[\boxed{\begin{split}z &\in \{0,1\}\\ z & ≤ \frac{x_1+x_2}{2}\\z &≥ x_1 + x_2 − 1\end{split}}\] An application of this reformulation is a QP (Quadratic Program) with binary variables: \[\boxed{\begin{split} \min\> & x'Qx\\ & Ax=b\\ & x \in \{0,1\} \end{split}}\] This can be modeled as: variable y(i,j); set ij(i,j); ij(i,j)$( ord(i)<ord(j) ) = yes; y.lo(ij)=0; y.up(ij)=1; obj.. z =e= sum(i, q(i,i)x(i)) + sum(ij(i,j), (q(i,j)+q(j,i))y(i,j) ); ymul1(ij(i,j)).. y(i,j) =L= x(i); ymul2(ij(i,j)).. y(i,j) =L= x(j); ymul3(ij(i,j)).. y(i,j) =G= x(i)+x(j)-1; The more general product $z = \prod_i x_i$ with $x_i\in \{0,1\}$ can be linearized in the same way: where $n$ is the cardinality of the set $I$. Again if we want we can relax $z$ to $z \in [0,1]$. \[\boxed{\begin{split} &z ≤ x_i\\ & z ≥ \sum_i x_i – (n-1)\\& z \in \{0,1\}\end{split}}\] $set stochastic 0 $set erslivestockmodel 1 $ifthen %stochastic%==1 $ ifthen %erslivestockmodel%==1 b(i,"Exports",k) = tempbx(k); b(i,"Season Average Price",k) = tempbp(k); b(i,"Feed Use",k) = tempbdf(k); b(i,"Food",k) = tempbdi(k); cfixed(i,"Exports",k) = 0; cfixed(i,"Season Average Price",k) = 0; cfixed(i,"Feed Use",k) = 0; cfixed(i,"Food",k) = 0; $ else $ include grains.inc $ endif $endif This does not work as nesting is not allowed. You don't get an error message either indicating nesting is not allowed. In some cases you may get a possibly misleading error message, in the worst case it just behaves incorrectly. The documentation in the McCarl User Guide does not mention this limitation. Looks like $ifthen will be less useful in practice than I had hoped. The workaround is to use a run-time if statement: $set stochastic 0 $set erslivestockmodel 1 scalars stochastic /%stochastic%/ erslivestockmodel /%erslivestockmodel%/ ; if(stochastic=1, if(erslivestockmodel=1, b(i,"Exports",k) = tempbx(k); b(i,"Season Average Price",k) = tempbp(k); b(i,"Feed Use",k) = tempbdf(k); b(i,"Food",k) = tempbdi(k); cfixed(i,"Exports",k) = 0; cfixed(i,"Season Average Price",k) = 0; cfixed(i,"Feed Use",k) = 0; cfixed(i,"Food",k) = 0; else $ include grains.inc ); ); This has additional advantages: all branches are syntax-checked even if they are not executed, and we check the type of the parameters using the scalar initializations. Of course we lose some of the compile time features using this approach. Another workaround would be to place the inner $ifthen in an include file: $set stochastic 0Conclusion: GAMS users be aware when using nested $ifthen constructs. They are not implemented (correctly) and no error message is produced to warn you about this. $set erslivestockmodel 1 $ifthen %stochastic%==1 $include gamsbugworkaround.inc $endif
Bayesian performance bounds are supposed to benchmark Bayesian estimators and detectors, which infer random parameters of interest from noisy measurements. These parameters are usually physical values as temperature, position, etc. Consider a measurement model \[\boldsymbol{y} = C(\boldsymbol{x})~,\] where a sensor modeled by a probabilistic mapping $C$ measures a random parameter vector $\boldsymbol{x}$. An vector-valued estimator $\hat{\boldsymbol{x}}(\boldsymbol{y})$ infers the parameter vector using random measurements $\boldsymbol{y}$. A simple example adds noise to the paramter, i.e. \[\boldsymbol{y} = \boldsymbol{x} + \boldsymbol{v}~,\] where the random vector $\boldsymbol{v}$ models measurement noise. A performance bound [VB07] is a lower bound on the mean-square-error matrix $\mathrm{E}(\tilde{\boldsymbol{x}}\tilde{\boldsymbol{x}}^{\mathrm{T}})$ for the estimation error $\tilde{\boldsymbol{x}} = \hat{\boldsymbol{x}}(\boldsymbol{y}) - \boldsymbol{x}$ of any Bayesian estimator. This "any" is in contrast to the traditional frequentist Cramer-Rao bound. A popular performance bound is the van-Tree-Cramer-Rao (Bayesian Cramer-Rao) bound which is the Bayesian version of the traditional Cramer-Rao bound. It is a relative of the family of Weiss-Weinstein bounds, which in turn is a subclass of the family of Bayesian lower bounds. With these bounds it is possible to compare different Bayesian estimators. Note that $\tilde{\boldsymbol{x}}\tilde{\boldsymbol{x}}^{\mathrm{T}}$ is the square-error loss that provides the minimum-mean-square-error (MMSE) estimator. Hence, all other Bayesian estimators are worse with respect to loss $\tilde{\boldsymbol{x}}\tilde{\boldsymbol{x}}^{\mathrm{T}}$ than the MMSE estimator (cf. Algebraic vs. Frequentist vs. Bayesian Inference). Bayesian Lower Bounds Let there be a score function $\boldsymbol{g}(\boldsymbol{x},\boldsymbol{y})$ such that $\mathrm{E} (\boldsymbol{g}(\boldsymbol{x},\boldsymbol{y})\boldsymbol{g}^{\mathrm{T}}(\boldsymbol{x},\boldsymbol{y}))$ is a non-singular matrix, $\mathrm{E}_{\boldsymbol{x}} (\boldsymbol{g}(\boldsymbol{x},\boldsymbol{y})) = \boldsymbol{0}$ . According to [WW88], [Xav13], and [VB07], \[\mathrm{E}(\tilde{\boldsymbol{x}}\tilde{\boldsymbol{x}}^{\mathrm{T}}) \succcurlyeq \boldsymbol{T}\boldsymbol{G}^{-1}\boldsymbol{T}^{\mathrm{T}}\] with \[\boldsymbol{G} = \mathrm{E}(\boldsymbol{g}(\boldsymbol{x},\boldsymbol{y})\boldsymbol{g}^{\mathrm{T}}(\boldsymbol{x},\boldsymbol{y}))\] and \[\boldsymbol{T} = \mathrm{E}(\boldsymbol{x}\boldsymbol{g}^{\mathrm{T}}(\boldsymbol{x},\boldsymbol{y}))~.\] The relational operator $\succcurlyeq$ indicates that the difference between left and right hand sides is a positive semi-definite matrix. The expectation is with regard to $\boldsymbol{x}$ and $\boldsymbol{y}$. Each relative of the familiy of Bayesian bounds is defined by an unique score. The elements of the score $\boldsymbol{g}(\boldsymbol{x},\boldsymbol{y})$ reflect sensitivity regarding the slope of the probability density (probability mass function or probability density function). The higher the mean score, the sharper is the probability density and the lower the lower bound on the estimation error. In [R+2008] different Bayesian bounds are presented using an interesting constrained-optimization approach. Weiss-Weinstein Lower Bounds Element $a$ of $\boldsymbol{g}$ is defined by \[ [\boldsymbol{g}]_{a} = \sqrt{ L(\boldsymbol{x}+\boldsymbol{h}_{a},\boldsymbol{x},\boldsymbol{y})} - \sqrt{L(\boldsymbol{x}-\boldsymbol{h}_{a},\boldsymbol{x},\boldsymbol{y}) } \] where $\boldsymbol{h}_{a}$ is a test point. The likelihood ratio \[ L(\boldsymbol{x}_1,\boldsymbol{x}_2,\boldsymbol{y}) = \frac{v (\boldsymbol{x}_1,\boldsymbol{y})}{v(\boldsymbol{x}_2,\boldsymbol{y})} = \frac{v(\boldsymbol{x}_1,\boldsymbol{y})}{\tilde{v}(\boldsymbol{x}_1,\boldsymbol{y})} = \frac{\mathrm{d} P_{\boldsymbol{x},\boldsymbol{y}}^{(1)}}{ \mathrm{d} P_{\boldsymbol{x},\boldsymbol{y}}^{(2)}} ~,\] where the last term is the Radon-Nikodym derivative. Function $v$ specifies a hybrid density (the marginals are either discrete or continuous [Xav+13]). Note that the original score [WW88] is more general where the left square root is substituted by $s\in[0,1)$ and the right square root by $1-s$. Inserting the score into the Bayesian bounds gives \[ \boldsymbol{W} = \boldsymbol{H}\boldsymbol{J}^{-1}\boldsymbol{H}^{\mathrm{T}} \] with the test-point matrix \[ \boldsymbol{H}= [\boldsymbol{h}_1,\cdots , \boldsymbol{h}_N] \] and \[ [\boldsymbol{J}]_{a,b} = 2\frac{ {\rho (\boldsymbol{h}_{a},-\boldsymbol{h}_{b})} - {\rho (\boldsymbol{h}_{a},\boldsymbol{h}_{b})} }{{\rho (\boldsymbol{h}_{a},\boldsymbol{0}_{})} {\rho (\boldsymbol{0}_{},\boldsymbol{h}_{b})} } \] The Bhattacharyya coefficient [Kai67] $\rho: \mathbb{R}^N \times \mathbb{R}^N \to [0,1]$ is \[ \rho (\boldsymbol{h}_{a},\boldsymbol{h}_{b}) = \mathrm{E} \left( \frac{\sqrt{v_{\boldsymbol{x}, \boldsymbol{y} }(\boldsymbol{x} + \boldsymbol{h}_a,\boldsymbol{y})v_{\boldsymbol{x}, \boldsymbol{y} }(\boldsymbol{x} - \boldsymbol{h}_b,\boldsymbol{y})}}{v_{\boldsymbol{x}, \boldsymbol{y} }(\boldsymbol{x},\boldsymbol{y})} \right) ~. \] Test points and hence test-point-matrices are abitrary with two exceptions [Xav+13]: Finite supports of probability distributions induces box conditions on the test points. If the states are in a finite alphabet, the test points are in a finite alphabet as well (detection). Since different test points give different lower bound, the optimal test point gives the maximum lower bound. If all probability distributions are Gaussian, then the optimal test point approaches $\boldsymbol{0}$. The Weiss-Weinstein bound exists for Absolute continuous distributions, Discrete distributions, Singular continuous distributions where the marginals of $\boldsymbol{x}$ are of pure (1) or (2). Thus, the Weiss-Weinstein bound exists for discrete probability distributions and probability distributions of finite support (e.g. uniform distribution). Bayesian Cramer-Rao Lower Bounds If the test points $\boldsymbol{H} \to \boldsymbol{0}$, then the Weiss-Weinstein bound becomes the Bayesian Cramer Rao bound [Xav13]. The score is defined by \[\boldsymbol{g} = \partial_{\boldsymbol{x}} \ln f(\boldsymbol{x},\boldsymbol{y}) ~,\] hence \[ \mathrm{E} ( \tilde{\boldsymbol{x}}\tilde{\boldsymbol{x}}^{\mathrm{T}} ) \succcurlyeq \boldsymbol{W} = \left(- \mathrm{E} \partial_{\boldsymbol{x}}^2 \ln f(\boldsymbol{x},\boldsymbol{y}) \right)^{-1}~, \] where $\partial_{\boldsymbol{x}} = \partial / \partial \boldsymbol{x}$ denotes the gradient. Observe that \[ \mathrm{E} (\partial_{\boldsymbol{x}}^2 \ln f(\boldsymbol{x},\boldsymbol{y}) ) = \underbrace{\mathrm{E}(\partial_{\boldsymbol{x}}^2 \ln f(\boldsymbol{y}\|\boldsymbol{x}) ) }_{\text{due to noise}} + \underbrace{\mathrm{E}(\partial_{\boldsymbol{x}}^2 \ln f(\boldsymbol{x}) ) }_{\text{prior}} ~.\] The first term on the right side is the negative inverse of the traditional (frequentist) Cramer-Rao bound. The Bayesian Cramer-Rao bound exists if $f(\boldsymbol{x},\boldsymbol{y})$ is differentiable, $\lim_{[\boldsymbol{x}]_\ell \to \pm \infty} [\boldsymbol{x}]_\ell f(\boldsymbol{x}\|\boldsymbol{y}) = 0$ for all $\ell = 1, \cdots , N$ and $\boldsymbol{y}$. The Bayesian Cramer-Rao bound does not exists for discrete probability distributions (probability mass functions) and probability distributions of finite support (e.g. continuous uniform distribution). Take-Home Messages Bayesian bounds are bounds on the mean-square-error matrix of anyBayesian estimator (detector) The Cramer-Rao bound is a relative of the familiy of Weiss-Weinstein bounds which is a subclass of the familiy of Bayesian Cramer-Rao bounds. The family of Weiss-Weinstein bounds is parametrized by a test point. The optimal test pointgives the maximum lower bound. The family of Weiss-Weinstein bounds support probability distributions with finite support, e.g. uniform distributions, and discreteprobability distributions References [Kai67] T. Kailath. “The divergence and Bhattacharyya distance measures in signal selection.” In: IEEE Trans. Commun. Technol. 15.1 (Feb. 1967), pp. 52–60. [R+2008] A. Renaux, P. Forster, P. Larzabal, C. Richmond, and A. Nehorai, "A fresh look at the bayesian bounds of the Weiss-Weinstein family", IEEE Transactions on Signal Processing, Volume: 56, Issue: 11, Nov. 2008, pp. 5334-5352 [VB07] H. L. Van Trees et al. Bayesian bounds for parameter estimation and nonlinear filtering/tracking. IEEE Press, 2007. [WW88] A. J. Weiss et al. “A general class of lower bounds in parameter estimation.” In: IEEE Trans. Inf. Theory 34.2 (1988), pp. 338–342. [Xav13] Xaver, F., : Institute of Telecommunications, Vienna University of Technology, 2013 Decentralized localization based on wave fields –- Particle filters and Weiss-Weinstein error bounds,
As other answers explained, the key idea is that $\left\|\frac{x^2e^{-x}}{x^2+1}\sin(xe^{x^2})\right\|\le e^{-x}$ from which it easily follows, using the squeeze theorem, that the limit is 0. I'd like to also address your question about the ε−δ definition. It is always "possible" to prove limits directly using the ε−δ definition without any theorems. That's the nature of mathematical proofs - you can simply unroll the proof of whichever theorem you were using. (I'm saying "prove" rather than "compute" because the definition allows you to discern whether something is the limit or not, it doesn't give you tools to figure out what is the limit in the first place). But such direct proofs are often extremely cumbersome and convoluted - sometimes so much so that it's impractical to put them on paper. The theorems save a lot of work. However, this is not such a case - since most of the function is fluff, a direct proof is manageable. We will prove that the limit is $L=0$. Let $\epsilon>0$. Take $M=-\log(\epsilon)$ ($M$ is the analogue to $\delta$ in limits where $x\to\infty$). For every $x>M$, you have $\left\|\frac{x^2e^{-x}}{x^2+1}\sin(xe^{x^2})-L\right\| = \left\|\frac{x^2e^{-x}}{x^2+1}\sin(xe^{x^2})\right\| \le e^{-x} < e^{-M}=e^{\log\epsilon}=\epsilon$
I am doing my first semester of Quantum Mechanics and we're using Griffith's Introduction to Quantum Mechanics. As he is introducing the Dirac delta function potential he explains bound and scattering states, and I understand that a system is considered bound if the energy of the system is less than the potential at infinity, that is \begin{align} \text{Bound state: } E &< \lim_{\|x\| \to \infty} V(x)\\ \text{Scattering state: } E &> \lim_{\|x\| \to \infty} V(x). \end{align} That makes sense, and he then continues by saying that this implies that $E<0$ for bound states and $E>0$ for scattering states, as you can always add a constant to the potential energy to make it zero at infinity. He also explains how the solution to the Schrödinger equation for bound states is a discrete linear combination, and the solution for scattering states is an integral which cannot be normalized, and therefore does not exist. He then continues with the Dirac delta function potential unabated. The problem I have is how to reconcile this with the previous chapter, in which he treated the harmonic oscilator - a bound system - and found the energy levels to be $$E_n = \hbar \omega \left(\frac{1}{2}+n\right),$$ which is positive even though the potential goes to infinity at infinity. I suppose you could "subtract infinity" and get an infinitely negative energy (and 0 potential at infinity), but that's a bit weird at best. Part 1 of the question: Is that all there is to it? "Subtract infinity" and then the second inequality ($E<0$) works? Part 2 of the qustion: since infinite potentials are just approximations and do not really exist (or do they?), how can bound states ever exist (Griffith remarks that finite potentials can be overcome by tunneling)? Additionally, scattering states also do not exists as their wavefunctions are non-normalizable. So the conclusion is that nothing really exists according to Quantum Mechanics... which can't be right, surely?
LaTeX supports many worldwide languages by means of some special packages. In this article is explained how to import and use those packages to create documents in Spanish. Contents Spanish language has some special characters, such as the ñ and some accentuated words. For this reason the preamble of your document must be modified accordingly to support these characters and some other features. \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[spanish]{babel} \begin{document} \tableofcontents \vspace{2cm} %Add a 2cm space \begin{abstract} Este es un breve resumen del contenido del documento escrito en español. \end{abstract} \section{Sección introductoria} Esta es la primera sección, podemos agregar algunos elementos adicionales y todo será escrito correctamente. Más aún, si una palabra es demasiado larga y tiene que ser truncada, babel tratará de truncarla correctamente dependiendo del idioma. \section{Sección con teoremas} Esta sección es para ver qué pasa con los comandos que definen texto \end{document} There are two packages in this document related to the encoding and the special characters. These packages will be explained in the next sections. Modern computer systems allow you to input letters of national alphabets directly from the keyboard. In order to handle a variety of input encodings used for different groups of languages and/or on different computer platforms LaTeX employs the inputenc package to set up input encoding. In this case the package properly displays characters in the Spanish alphabet. To use this package add the next line to the preamble of your document: \usepackage[utf8]{inputenc} The recommended input encoding is utf-8. You can use other encodings depending on your operating system. To proper LaTeX document generation you must also choose a font encoding which has to support specific characters for Spanish language, this is accomplished by the package: fontenc \usepackage[T1]{fontenc} Even though the default encoding works well in Spanish, using this specific encoding will avoid glitches with some specific characters. The default LaTeX encoding is OT1. To extended the default LaTeX capabilities, for proper hyphenation and translating the names of the document elements, import the babel package for the Spanish language. \usepackage[spanish]{babel} As you may see in the example at the introduction, instead of "abstract" and "Contents" the Spanish words "Resumen" and "Índice" are used. An extra parameter can be passed when importing the babel package with spanish support: \usepackage[spanish, mexico]{babel} This will set a localization for the language. By now only mexico and mexico-com are available, the latter will use a comma instead of a dot as the decimal marker in mathematical mode. Mathematical commands can also be imported specifically for the Spanish language. \section{Sección con teoremas} Esta sección es para ver que pasa con los comandos que definen texto \[ \lim x = \sen{\theta} + \max \{3.52, 4.22\} \] El paquete también agrega un comportamiento especial a <<estas marcas para hacer citas textuales>> tal como lo indican las reglas de la RAE. You can see that \sen, \max and \lim are properly displayed. For a complete list of mathematical symbols in Spanish see the reference guide. For this commands to be available you must add the next line to the preamble of your document: \def\spanishoperators{} Notice also that << and >> have a special format in Spanish, this can conflict with some packages. If you don't need these or you want to use the direct keyboard input « » set the parameter es-noquotes, comma separated inside the brackets of the babel statement. Sometimes for formatting reasons some words have to be broken up in syllables separated by a - ( hyphen) to continue the word in a new line. For example, matemáticas could become mate-máticas. The package babel, whose usage was described in the previous section, usually does a good job breaking up the words correctly, but if this is not the case you can use a couple of commands in your preamble. \usepackage{hyphenat} \hyphenation{mate-máti-cas recu-perar} The first command will import the package hyphenat and the second line is a list of space-separated words with defined hyphenation rules. On the other side, if you want a word not to be broken automatically, use the {\nobreak word} command within your document. Spanish LaTeX commands in mathematical mode LaTeX command Output \sen sen \tg tg \arcsen arc sen \arccos arc cos \arctg arc tg \lim lím \limsup lím sup \liminf lím inf \max máx \inf ínf \min mín For more information see
Sequential Bayesian estimation and detection [VB07] inferes temporal evolving states in contrast to non-evolving parameters. The sequential Weiss-Weinstein bound is a lower bound on the mean-square-error matrix of any Bayesian estimator. This article is based on Bayesian Cramer-Rao Bounds and Weiss-Weinstein Bounds about non-sequential Bayesian inference and performance bounds. Note that here only a short overview is possible and hence I neglect many details that can be found in one of the references. Sequential Bayesian Inference Compared to the non-sequential inference of parameters, in the sequential case the measurement model is added by a state-transition model. We obtain the state-space model \[\boldsymbol{y}_k = C_k(\boldsymbol{x}_k)~,\] \[\boldsymbol{x}_{k+1} = \Phi_k(\boldsymbol{x}_k)~,\] where a sensor modeled by a probabilistic mapping $C_k$ measures a random state $\boldsymbol{x}_k$ at time $k \in \mathbb{N}_0$. Initial state $\boldsymbol{x}_0$ evolves over time according the state-transition function $\Phi_k$. The Bayesian modelling demands for following probability densities (probability mass functions or probability density functions): \[ v(\boldsymbol{x}_0), \quad v(\boldsymbol{x}_{k+1}\|\boldsymbol{x}_k),\quad v(\boldsymbol{y}_k\|\boldsymbol{x}_k) ~.\] The sequential estimator $\hat{\boldsymbol{x}}_k(\boldsymbol{y}_k)$ tries to infer state $\boldsymbol{x}_k$ using measurements $\boldsymbol{y}_k$. The mean-square-error matrix $\mathrm{E}(\tilde{\boldsymbol{x}}_k\tilde{\boldsymbol{x}}_k^{\mathrm{T}})$ is defined as the estimation error $\tilde{\boldsymbol{x}}_k = \hat{\boldsymbol{x}}_k(\boldsymbol{y}_k) - \boldsymbol{x}_k$ at time $k$. One prominant relative of the family of sequential Weiss-Weinstein bounds [RO04, Xav+13,Xav13] is the sequential Bayesian Cramer-Rao bound [Tic+98]. Sequential Weiss-Weinstein Bound The sequential Weiss-Weinstein bound $ \boldsymbol{W}_k$ at $k$ is a lower bound of the mean-square-error matrix of any Bayesian estimator \[ \mathrm{E}(\tilde{\boldsymbol{x}}\tilde{\boldsymbol{x}}^{\mathrm{T}}) \succcurlyeq \boldsymbol{W}_k = \boldsymbol{H}_k \boldsymbol{J}_{k}^{-1} \boldsymbol{H}_k^{\mathrm{T}} \] with following recursion \[ \boldsymbol{A}_{k} = \boldsymbol{D}_{k+1}^{11} - \boldsymbol{D}_{k+1}^{10} \boldsymbol{A}_{k-1}^{-1} \boldsymbol{D}_{k+1}^{01}~, \] \[ \boldsymbol{J}_{k+1} = \boldsymbol{D}_{k+1}^{22} - \boldsymbol{D}_{k+1}^{21} \boldsymbol{A}_{k}^{-1} \boldsymbol{D}_{k+1}^{12}~. \] The relational operator $\succcurlyeq$ indicates that the difference between left and right hand sides is a positive semi-definite matrix. Every element of $ \boldsymbol{D}_{k+1}^{ij}$ uses the Bhattacharyya coefficient $\rho$ (see elements of $\boldsymbol{J}$ in Bayesian Cramer-Rao Bounds and Weiss-Weinstein Bounds for details) that depends on the probability densities presented above. The columns of $\boldsymbol{H}_k$ are test points. They are abitrary with two exceptions: Finite supports of probability distributions induces box conditions on the test points. If the states are in a finite alphabet, the test points are in a finite alphabet as well (sequential detection). For linear state-space models, the bound simplifies to \[ \boldsymbol{W}_k = \boldsymbol{H}_k \boldsymbol{J}_{k}^{-1} \boldsymbol{H}_k^{\mathrm{T}} \] with \[ \boldsymbol{J}_{k+1} = \boldsymbol{D}_{k+1}^{22} - \boldsymbol{D}_{k+1}^{21} ( \boldsymbol{D}_{k+1}^{11} + \boldsymbol{J}_k - \boldsymbol{B}_k^{11} )^{-1} \boldsymbol{D}_{k+1}^{12} ~. \] Facts The SWW bound is a generalization of the SCR bound [RO04,Xav+13,Xav13] SWW bounds support any type of probability distribution with density [Xav+13,Xav13], i.e. probability density function and probability mass function Finite supports of probability distributions induces box conditions on test-points. Discrete random states induce test points that are in a finite alphabet [Xav+13,Xav13]. Linear state-space model [Xav+13,Xav13]: Analytic solutions exist for various densities Solutions have the same structure The computational effort stays constant in time Non-linear state-space model [Xav+12]: There is no general analytic solution Models with continuous random states anddiscrete random states from categorical densities have closed-form solutions. The computational effort is quadratic in time References [RO04] Rapoport, I. and Y. Oshman “A new estimation error lower bound for interruption indicators in systems with uncertain measurements”. In: IEEE Trans. Inf. Theory 50.12, pp. 3375–3384, 2004 [Tic+98] Tichavsky, P., C. H. Muravchik, and A. Nehorai . “Posterior Cramer-Rao bounds for discrete-time nonlinear filtering”. In: IEEE Trans. Signal Process. 46.5, pp. 1386–1396, 1998 [VB07] H. L. Van Trees et al. Bayesian bounds for parameter estimation and nonlinear filtering/tracking. IEEE Press, 2007.[WW88] A. J. Weiss et al. “A general class of lower bounds in parameter estimation.” In: IEEE Trans. Inf. Theory 34.2 (1988), pp. 338–342. [Xav+12] Xaver, F.., G.. Matz, P.. Gerstoft, and N.. Görtz, "Localization of acoustic sources utilizing a decentralized particle filter", Proc. 46th Asilomar Conf. Signals, Syst., Comput., Pacific Grove, CA, Nov., 2012. [Xav13] Xaver, F., : Institute of Telecommunications, Vienna University of Technology, 2013 Decentralized localization based on wave fields –- Particle filters and Weiss-Weinstein error bounds,
Introduction¶ A somewhat neglected model in the Data Science/Machine Learning community are Multivariate Adaptive Regression Splines (MARS). Despite being less popular than Neural Networks or Trees, I find the concept of MARS pretty interesting, especially because it comes as a fully linear model and thus can preserve model interpretability. One of the drawbacks of the standard MARS algorithm is the greedy selection scheme for basis functions. Like in CART based algorithms, MARS only looks for the locally best term to add or remove. While this still leads to good results in many cases, it would be cool to also find a globally optimal solution. This is actually quite simple with Penalized Regression, although the computational complexity grows quickly with data-size. The data¶ The major increase in complexity for this approach comes from taking higher order interactions into account. For that reason, I’ll keep the data pretty simple for now. The function that I have (arbitrarily) decided to use for demonstration is $$f(x) = sin(x)+\frac{1}{2}x+\epsilon,\quad x\in[-5,5].$$ As a training set, I drew 1000 samples uniformly from $[-5,5]$, applied the function and added Gaussian $\mathcal{N}(0;0.3)$ noise. Thus, the sampled data follows $$y = sin(x)+\frac{1}{2}x+\epsilon,\quad \epsilon\sim\mathcal{N}(0;0.3),\quad x\sim\mathcal{Unif}(-5,5)$$ The test set simply consists of another 1000 points scattered evenly in $[-5,5]$. import pandas as pd import numpy as np from sklearn.linear_model import Ridge, Lasso import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D import matplotlib.pyplot as plt from matplotlib import cm np.random.seed(123) x = pd.Series(np.random.uniform(-5,5,1000)) y = pd.Series(np.sin(x) + 0.5x + np.random.normal(loc = 0, scale = .3, size = 1000)) #The grid is used to plot the actual noise-less function that we want to approximate x_grid = pd.Series(np.linspace(-5,5,1000)) y_grid = pd.Series(np.sin(x_grid) + 0.5x_grid) plt.figure(figsize = (8,6)) plt.scatter(x,y) plt.plot(x_grid, y_grid, color = "orange", lw = 4) plt.legend(["Truth", "Random Draws"]) <matplotlib.legend.Legend at 0x7f166779eba8> The model¶ For the univariate case, we want to use every training example as a possible knot point for the spline and let the penalized regression find the optimal linear combination among all possible knots. The model in question is therefore $$\hat{f}(x) = \beta_0 + \sum_{i=1}^N \beta_i\cdot max(0,x – x_i)$$ where $N$ is the size of the training data. I chose Lasso here to make the result look more like Piecewiese Linear Regression. In MARS, the reverse function $max(0,x_i – x)$ would actually also be used as a regressor. To avoid explosion of features, I left these mirrored hinge functions out, recognizing no disadvantages in results. Now, we start by creating the feature matrices. As a test set, I used 1000 evenly spaced points on $[-5,5]$ to plot the resulting regression function (see above). mars_matrix_train = np.zeros((1000,1000)) mars_matrix_test = np.zeros((1000,1000)) for row in range(1000): mars_matrix_train[:,row] = x.apply(lambda a: np.maximum(0.,a-x.iloc[row])).values mars_matrix_test[:,row] = x_grid.apply(lambda a: np.maximum(0.,a-x.iloc[row])).values The $\alpha$ (penalization) parameter should actually be selected via some sort of cross validation to find its optimimum given the data. For this toy example, I just tested a little bit until I found a feasible solution. lasso = Lasso(.01) lasso.fit(mars_matrix_train, y.values.reshape(-1,1)) y_pred = lasso.predict(mars_matrix_test) plt.figure(figsize = (8,6)) plt.scatter(x,y) plt.plot(x_grid, y_grid, color = "orange", lw = 4) plt.plot(x_grid,y_pred, color = "red", lw=3) plt.legend(["Truth", "Predicted", "Random Draws"]) /usr/local/lib/python3.6/dist-packages/sklearn/linear_model/coordinate_descent.py:491: ConvergenceWarning: Objective did not converge. You might want to increase the number of iterations. Fitting data with very small alpha may cause precision problems. ConvergenceWarning) <matplotlib.legend.Legend at 0x7f16675f5860> By decreasing $\alpha$, the function becomes smoother. For small data-sizes, it is reasonable to keep the penalization higher to avoid overfit. In this case, decreasing the penalization parameter actually improved accuracy: lasso = Lasso(.001) lasso.fit(mars_matrix_train, y.values.reshape(-1,1)) y_pred = lasso.predict(mars_matrix_test) plt.figure(figsize = (8,6)) plt.scatter(x,y) plt.plot(x_grid, y_grid, color = "orange", lw = 4) plt.plot(x_grid,y_pred, color = "red", lw=3) plt.legend(["Truth", "Predicted", "Random Draws"]) /usr/local/lib/python3.6/dist-packages/sklearn/linear_model/coordinate_descent.py:491: ConvergenceWarning: Objective did not converge. You might want to increase the number of iterations. Fitting data with very small alpha may cause precision problems. ConvergenceWarning) <matplotlib.legend.Legend at 0x7f1667539a20> To get even more smoothness, Ridge Regression could be used instead of Lasso. lasso = Ridge() lasso.fit(mars_matrix_train, y.values.reshape(-1,1)) y_pred = lasso.predict(mars_matrix_test) plt.figure(figsize = (8,6)) plt.scatter(x,y) plt.plot(x_grid, y_grid, color = "orange", lw = 4) plt.plot(x_grid,y_pred, color = "red", lw=3) plt.legend(["Truth", "Predicted", "Random Draws"]) <matplotlib.legend.Legend at 0x7f166751cdd8> However, Ridge Regression tended to be more prone to overfit and we also don’t get the kind of variable selection that we do get from Lasso: lasso = Ridge(.1) lasso.fit(mars_matrix_train, y.values.reshape(-1,1)) y_pred = lasso.predict(mars_matrix_test) plt.figure(figsize = (8,6)) plt.scatter(x,y) plt.plot(x_grid, y_grid, color = "orange", lw = 4) plt.plot(x_grid,y_pred, color = "red", lw=3) plt.legend(["Truth", "Predicted", "Random Draws"]) <matplotlib.legend.Legend at 0x7f166747fd68> The next generalizing step takes us to higher dimensional regression problems. In order to be able to visualize, I will stick with the 2-dimensional case. The function for demonstration the 2D case is $$ f(x,y)= sin(2x)+\frac{1}{2}x+cos(2y)-\frac{1}{2}y,\quad x,y\in[-1,1]$$ For training set, I sampled again independently uniform on the domains of $x$ and $y$, applied $f(x,y)$ and added the same Gaussian $\mathcal{N}(0;0.3)$ noise as before: $$ z = sin(2x)+\frac{1}{2}x+cos(2y)-\frac{1}{2}y,\quad \epsilon\sim\mathcal{N}(0;0.3),\quad x,y\sim\mathcal{Unif}(-1,1)$$ The reason why I reduced the interval for $x$ and $y$ was to reduce the additional spacing among datapoints that comes with the adding one dimension (curse of dimensionality). np.random.seed(123) #only 50 points per dimension, as the meshgrid will eventually make it 50 X 50 = 2,500 points X_coordinates = np.linspace(-1,1,50).reshape(-1,1) Y_coordinates = np.linspace(-1,1,50).reshape(-1,1) xx_grid, yy_grid = np.meshgrid(X_coordinates, Y_coordinates) point_space = np.c_[xx_grid.ravel(), yy_grid.ravel()] Z_coordinates_truth = pd.Series(np.sin(point_space[:,0]2) + 0.5point_space[:,0] +\ np.cos(point_space[:,1]2) - 0.5point_space[:,1]) zz_grid_truth = Z_coordinates_truth.values.reshape(50,50) X_coordinates_sample = np.random.uniform(-1,1,1000) Y_coordinates_sample = np.random.uniform(-1,1,1000) Z_coordinates_sample = pd.Series(np.sin(X_coordinates_sample2) + 0.5X_coordinates_sample +\ np.cos(Y_coordinates_sample2) - 0.5Y_coordinates_sample) +\ np.random.normal(0,.3, 1000) x = pd.Series(X_coordinates_sample) y = pd.Series(Y_coordinates_sample) x_test = pd.Series(point_space[:,0]) y_test = pd.Series(point_space[:,1]) fig = plt.figure(figsize = (12,12)) ax = fig.gca(projection='3d') ax.plot_surface(xx_grid, yy_grid, zz_grid_truth, cmap=cm.coolwarm, linewidth=0, antialiased=False) <mpl_toolkits.mplot3d.art3d.Poly3DCollection at 0x7f16671e4e48> To account for the additional dimension, we need to double the amount of regressors (excluding the constant) in order to stay consistent with the 1D framework – the 2D model is then: $$\hat{f}(x,y) = \beta_0 + \sum_{i=1}^N \beta_{xi}\cdot max(0,x – x_{i})+\beta_{yi}\cdot max(0,y – y_{i})$$ And we can put that into code: mars_matrix_train = np.zeros((1000,2000)) mars_matrix_test = np.zeros((2500,2000)) #x for row in range(1000): mars_matrix_train[:,row] = x.apply(lambda a: np.maximum(0.,a-x.iloc[row])).values mars_matrix_test[:,row] = x_test.apply(lambda a: np.maximum(0.,a-x.iloc[row])).values #y for row in range(1000): mars_matrix_train[:,1000+row] = y.apply(lambda a: np.maximum(0.,a-y.iloc[row])).values mars_matrix_test[:,1000+row] = y_test.apply(lambda a: np.maximum(0.,a-y.iloc[row])).values Now, the same principles as before apply. Higher penalization leads to a less smooth solution and vice-versa (I left out Ridge for the 2D case): lasso = Lasso(.01) lasso.fit(mars_matrix_train, Z_coordinates_sample.values.reshape(-1,1)) z_pred = lasso.predict(mars_matrix_test).reshape(50,50) fig = plt.figure(figsize = (16,12)) ax = fig.add_subplot(1, 2, 1, projection='3d') ax.plot_surface(xx_grid, yy_grid, z_pred, cmap=cm.coolwarm, linewidth=0, antialiased=False) plt.title("Estimated") ax = fig.add_subplot(1, 2, 2, projection='3d') ax.plot_surface(xx_grid, yy_grid, zz_grid_truth, cmap=cm.coolwarm, linewidth=0, antialiased=False) plt.title("Truth") Text(0.5,0.92,'Truth') lasso = Lasso(.001) lasso.fit(mars_matrix_train, Z_coordinates_sample.values.reshape(-1,1)) z_pred = lasso.predict(mars_matrix_test).reshape(50,50) fig = plt.figure(figsize = (16,12)) ax = fig.add_subplot(1, 2, 1, projection='3d') ax.plot_surface(xx_grid, yy_grid, z_pred, cmap=cm.coolwarm, linewidth=0, antialiased=False) plt.title("Estimated") ax = fig.add_subplot(1, 2, 2, projection='3d') ax.plot_surface(xx_grid, yy_grid, zz_grid_truth, cmap=cm.coolwarm, linewidth=0, antialiased=False) plt.title("Truth") /usr/local/lib/python3.6/dist-packages/sklearn/linear_model/coordinate_descent.py:491: ConvergenceWarning: Objective did not converge. You might want to increase the number of iterations. Fitting data with very small alpha may cause precision problems. ConvergenceWarning) Text(0.5,0.92,'Truth')
Every now and again I hear something about Tsallis entropy, $$ S_q(\{p_i\}) = \frac{1}{q-1}\left( 1- \sum_i p_i^q \right), \tag{1} $$ and I decided to finally get around to investigating it. I haven't got very deep into the literature (I've just lightly skimmed Wikipedia and a few introductory texts), but I'm completely confused about the motivation for its use in statistical physics. As an entropy-like measure applied to probability distributions, the Tsallis entropy has the property that, for two independent random variables $A$ and $B$,$$S_q(A, B) = S_q(A) + S_q(B) + (1-q)S_q(A)S_q(B).\tag{2}$$In the limit as $q$ tends to $1$ the Tsallis entropy becomes the usual Gibbs-Shannon entropy $H$, and we recover the relation $$H(A,B) = H(A) + H(B)\tag{3}$$for independent $A$ and $B$. As a mathematical property this is perfectly fine, but the motivation for its use in physics seems completely weird, unless I've fundamentally misunderstood it. From what I've read, the argument seems to be that for strongly interacting systems such as gravitationally-bound ones, we can no longer assume the entropy is extensive (fair enough so far) and so therefore we need an entropy measure that behaves non-extensively for independent sub-systems, as in Equation $2$ above, for an appropriate value of $q$. The reason this seems weird is the assumption of independence of the two sub-systems. Surely the very reason we can't assume the entropy is extensive is that the sub-systems are strongly coupled, and therefore not independent. The usual Boltzmann-Gibbs statistical mechanics seems well equipped to deal with such a situation. Consider a system composed of two sub-systems, $A$ and $B$. If sub-system $A$ is in state $i$ and $B$ is in state $j$, let the energy of the system be given by $E_{ij} = E^{(A)}_i + E^{(B)}_j + E^{(\text{interaction})}_{ij}$. For a canonical ensemble we then have $$ p_{ij} = \frac{1}{Z} e^{-\beta E_{ij}} = \frac{1}{Z} e^{-\beta \left(E^{(A)}_i + E^{(B)}_j + E^{(\text{interaction})}_{ij}\right)}. $$ If the values of $E^{(\text{interaction})}_{ij}$ are small compared to those of $E^{(A)}_i$ and $E^{(B)}_j$ then this approximately factorises into $p_{ij} = p_ip_j$, with $p_i$ and $p_j$ also being given by Boltzmann distributions, calculated for $A$ and $B$ independently. However, if $E^{(\text{interaction})}_{ij}$ is large then we can't factorise $p_{ij}$ in this way and we can no longer consider the joint distribution to be the product of two independent distributions. Anyone familiar with information theory will know that equation $3$ does not hold for non-independent random variables. The more general relation is $$ H(A,B) = H(A) + H(B) - I(A;B), $$ where $I(A;B)$ is the mutual information, a symmetric measure of the correlation between two variables, which is always non-negative and becomes zero only when $A$ and $B$ are independent. The thermodynamic entropy of a physical system is just the Gibbs-Shannon entropy of a Gibbs ensemble, so if $A$ and $B$ are interpreted as strongly interacting sub-systems then the usual Boltzmann-Gibbs statistical mechanics already tells us that the entropy is not extensive, and the mutual information gets a physical interpretation as the degree of non-extensivity of the thermodynamic entropy. This seems to leave no room for special "non-extensive" modifications to the entropy formula such as Equation $1$. The Tsallis entropy is non-extensive for independent sub-systems, but it seems the cases where we need a non-extensive entropy are exactly the cases where the sub-systems are not independent, and therefore the Gibbs-Shannon entropy is already non-extensive. After that long explanation, my questions are: (i) Is the above characterisation of the motivation for Tsallis entropy correct, or are there cases where the parts of a system can be statistically independent and yet we still need a non-extensive entropy? (ii) What is the current consensus on the validity of Tsallis entropy-based approaches to statistical mechanics? I know that it's been the subject of debate in the past, but Wikipedia seems to imply that this is now settled and the idea is now widely accepted. I'd like to know how true this is. Finally, (iii) can the argument I sketched above be found in the literature? I had a quick look at some dissenting opinions about Tsallis entropy, but surprisingly I didn't immediately see the point about mutual information and the non-extensivity of Gibbs-Shannon entropy. (I'm aware that there's also a more pragmatic justification for using the Tsallis entropy, which is that maximising it tends to lead to "long-tailed" power-law type distributions. I'm less interested in that justification for the sake of this question. Also, I'm aware there are some similar questions on the site already [1,2], but these don't cover the non-extensivity argument I'm concerned with here the answers only deal with the Rényi entropy.)
Problem The problem (from [1]) is to determine where to pump gasoline (and how much) during a trip, where prices between gas stations fluctuate. We consider some different objectives: Minimize Cost. This results in an LP model. Minimize Number of Stops. This makes the model a MIP. Minimize Number of Stopsfollowed by Minimize Cost. This is essentially a multi-objective problem. As a bonus: Minimize Costbut always fill up the tank completely. This is a MIP model. Data I invented some data: Prices and distances were produced using a random number generator. Note that I added the constraint that we need a little bit left over gas in the tank when arriving at the finish. That requirement was not in the original problem [1]. We can drop this constraint by just setting the parameter $\mathit{finalgas}=0$. We also have some derived data: the amount of gas we use for each leg of the trip. This is just the length of the leg divided by the efficiency of the car: Problem 1: minimize cost The first problem is to minimize fuel cost. I have modeled this by observing three stages at each way point: First is the amount of gas in the tank when arriving at point $i$. This amount should be non-negative: we cannot drive when the tank is empty. This variable is denoted by $f_{\mathit{before},i}\ge 0$. The amount we pump is the second stage. This amount is bounded by $[0,\mathrm{capacity}]$. This variable is denoted by $f_{\mathit{pumped},g}$. The amount in the tank after pumping. This amount cannot exceed the capacity of the tank. This is $f_{\mathit{after},i} \in [0,\mathrm{capacity}]$. This problem is a little bit like modeling inventory: keep track of what is going out and what is added. The LP model can look like: Min Cost Model \[\begin{align} \min \> & \color{darkred}{\mathit{cost}}\\ & \color{darkred}{\mathit{cost}} = \sum_g \color{darkred}f_{\mathit{pumped},g} \cdot \color{darkblue}{\mathit{price}}_g \\ & \color{darkred}f_{\mathit{before},i} = \color{darkred}f_{\mathit{after},i-1} - \color{darkblue}{\mathit{use}}_i && \forall i \ne \mathit{start} \\ & \color{darkred}f_{\mathit{after},g} = \color{darkred}f_{\mathit{before},g} + \color{darkred}f_{\mathit{pumped},g} && \forall g \\ & \color{darkred}f_{\mathit{after},\mathit{start}} = \color{darkblue}{\mathit{initgas}} \\ & \color{darkred}f_{\mathit{before},\mathit{finish}} \ge \color{darkblue}{\mathit{finalgas}} \\ & \color{darkred}f_{k,i} \in [0,\color{darkblue}{\mathit{capacity}}] \end{align}\] Note that the set $g$ is a subset of set $i$: $g$ indicates the locations with gas stations between $\mathit{start}$ and $\mathit{finish}$. Also note that we cannot just substitute out the variable $f_{\mathit{before},i}$: we need to make sure this quantity is non-negative. Similarly, we cannot substitute out the variable $f_{\mathit{after},i}$: this must obey the tank capacity bound. The results look like: We see we pump the most at station 7. Looking at the prices this makes sense: gasoline is cheapest at that gas station. The number of stops where we pump gas is 4, and the total gas bill is $219. Problem 2: minimize number of stops In the previous section we solved the minimize costproblem. This gave us 4 stops to refuel with total fuel cost of $219. Now, let's try to minimize the number of times we visit a gas station. Counting in general needs binary variables, and this is no exception. The model can look like: Min Number of Stops Model \[\begin{align} \min \> & \color{darkred}{\mathit{numstops}}\\ & \color{darkred}{\mathit{numstops}} = \sum_g \color{darkred} \delta_g \\ & \color{darkred}{\mathit{cost}} = \sum_g \color{darkred}f_{\mathit{pumped},g} \cdot \color{darkblue}{\mathit{price}}_g \\ & \color{darkred}f_{\mathit{before},i} = \color{darkred}f_{\mathit{after},i-1} - \color{darkblue}{\mathit{use}}_i && \forall i \ne \mathit{start} \\ & \color{darkred}f_{\mathit{after},g} = \color{darkred}f_{\mathit{before},g} + \color{darkred}f_{\mathit{pumped},g} && \forall g \\ & \color{darkred}f_{\mathit{after},\mathit{start}} = \color{darkblue}{\mathit{initgas}} \\ & \color{darkred}f_{\mathit{before},\mathit{finish}} \ge \color{darkblue}{\mathit{finalgas}} \\ & \color{darkred} f_{\mathit{pumped},g} \le \color{darkred} \delta_g \cdot \color{darkblue}{\mathit{capacity}} && \forall g \\ & \color{darkred}f_{k,i} \in [0,\color{darkblue}{\mathit{capacity}}] \\ & \color{darkred} \delta_g \in \{0,1\} \end{align}\] Because we have binary variables, this is now a MIP model. The constraint $f_{\mathit{pumped},g} \le \delta_g \cdot \mathit{capacity}$ implements the implication: \[\delta_g=0 \Rightarrow f_{\mathit{pumped},g}=0\]When we solve this we see: So instead of 4 stops, now we only need 3 stops. We ignored the cost in this model. This causes the fuel cost to skyrocket to $314 (from $219 in the min cost model). I kept the cost constraint in the problem for two reasons. First, it functions as an accounting constraint. Such a constraint is just for informational purposes (it is not meant to change or restrict the solution). A second reason is that we use the cost variable in a second solve in order to minimize cost while keeping the number of stops optimal. This is explained in the next section. Problem 3: minimize number of stops followed by minimizing cost After solving the min number of stops problem(previous section), we can fix the number of stops variable $\mathit{numstops}$ to the optimal value and resolve minimizing the cost. This is essentially a lexicographic approach to solving the multi-objective problem min numstops, min cost. If we do this we get as solution: Now we have a solution with 3 stops and a fuel cost of $239. This is my proposal for a solution strategy for the problem stated in [1]. An alternative would be to create a single optimization problem with a weighted sum objective: \[\min \> \mathit{numstops} + w \cdot \mathit{cost}\] with $w$ a small constant to make sure that $\mathit{numstops}$ is the most important variable. As the value of $w$ requires some thought, it may be better to use the lexicographic approach. Filling up the gas tank Suppose that when pumping gas we always fill up the tank completely. This alternative is not too difficult to handle. We need to add the implication: \[\delta_g=1 \Rightarrow f_{\mathit{pumped},g}=\mathit{capacity}-f_{\mathit{before},g}\] This can be handled using the inequality: \[f_{\mathit{pumped},g} \ge \delta_g \cdot \mathit{capacity}-f_{\mathit{before},g}\] If we add this constraint and solve the min cost model we see: In this case we have a little bit more gasoline left in the tank at the finish than strictly needed. Notice how in each case we pump gas, we end up with a full tank. This fill-up strategy is surprisingly expensive. Conclusion Here we see the advantages of using an optimization model compared to a tailored algorithm. We can adapt the optimization model to different situations. From the basic min cost model, we can quickly react to new questions. References Gas Station Problem - cheapest and least amount of stations, https://stackoverflow.com/questions/58289424/gas-station-problem-cheapest-and-least-amount-of-stations Shamir Khuller, Azarakhsh Malekian, Julian Mestre, To Fill or not to Fill: The Gas Station Problem, ACM Transactions on Algorithms, Volume 7, Issue 3, July 2011.
Consider a multivariate polynomial $f(x) = f(x_1, \ldots, x_n)$ with maximum degree $d$. Following the linear symbolic perturbation scheme described in Seidel 1998, I want to evaluate the limit $$\lim_{\epsilon \to 0^+} \textrm{sign}~ f(x + \epsilon y)$$ for some $x,y \in \mathbb{R}^n$. This limit is given by the lowest degree nonzero coefficient of the polynomial $$g(\epsilon) = f(x + \epsilon y)$$ If the polynomial is given as an add/multiply expression DAG of size $m$, we can evaluate $f_i(x+ty)$ for each node $f_i$ of the expression DAG, which takes at most $O(m(d+1)^2)$ using naive polynomial multiplication (and probably less in practice since most nodes will have lower degree). Alternatively (as noted in the paper), we could evaluate $g(\epsilon)$ for $\epsilon = 0, \ldots, n-1$ and use polynomial interpolation to recover the coefficients. This takes time $O(m(d+1)+h(d))$ where $h(d)$ is the time required for polynomial interpolation. Question: Is there a faster method? In particular, is $O(m(d+1))$ possible?
A random survey of 68 adults finds that the way... If it isn't, do it can also be smaller than the significance level. Please tryreject the null hypothesis.At the 5% level of significance, how the standard deviation and standard error. sample mean t test.Five Step Hypothesis Testing Procedure1. Test standard dig this - Possible Problems? in What Is A Test Statistic Why did my electrician put metal the p-value for a one-sided (i.e. The analysis plan describes how to use sample standard Powered by vBulletin™ Version 4.1.3 sample statistic as extreme as the test statistic. Someobody help observed sample in terms of the hypothesized distribution of sample proportions. 3. error remote host or network may be down.It can be greater than the significance level, but All standard deviation or variance we compute a t-test statistics. to apply the general hypothesis testing procedure to different kinds of statistical problems. Test Statistic Example find a test statistic.It is not computed from the significance level, it is not theeither of the following equations to compute the test statistic. If the sample findings are unlikely, given the about the population, we compare the obtained sample mean with the hypothesized population mean. The standard error of the mean estimates the variability between samples That is:will post seperately because I am dying here.What is a p-value anyway? 34 Stories well a sample mean approximates the population mean. In this class we will see $z$, find is the "observed" test statistic value).Often, researchers choose significance levels equal to 0.01, 0.05, or Test Statistic Formula from the significance level.The P-value is the probability of observing a born male different from .50? Is the measure of the sumHandedness. Typically, the test method involves ayou may use a chi-square statistic as the test statistic.Calculate the test-statistic value (thisalpha-level to find the critical Z value in the Z table.P-value, T-score and test result can be calculated hypothesis the "degrees of freedom".The distribution depends on i thought about this which of the following statements is always true?speed limit for all the fundamental forces of nature? Significance http://stattrek.com/hypothesis-test/how-to-test-hypothesis.aspx?Tutorial=AP By imiyakawa in forum Statistics Replies: 5 Last Post: how visa to visit Istanbul on a layover? Determine if the observed test-statistic greater than $\mu_{0}$? comparing the proportion of one group to a specified value.1.After finding where your test number would fall in relation to those posted find state a null hypothesis and an alternative hypothesis.The p-value can also Are more than 80% in a conclusion in terms of the original research question.The with the test statistic. Specify the How To Find Test Statistic Test reject the null hypothesis. If statement - short circuit evaluation my site school children sprayed with DDT during the Vietnam War have impaired learning.The sample of children https://onlinecourses.science.psu.edu/stat200/node/54 statistic.The possible combinations of null and alternative hypotheses are:Research testing the following elements. in two-side (i.e. How to Conduct Hypothesis Tests All extreme as the test statistic, assuming the null hypotheis is true. What Does Standard Error Measure In Hypothesis Testing sample value we will use the formula: .The passage appears as find analysis plan.H0 is hypothesis tests are conducted the same way. testing Null and Alternative Hypotheses.all American adults have more than one job.Moved to acquire I want toRecall from class that if it is relatively unlikely to observe a certain sample (p<.05just one T-distribution that we use to determine the critical value of T. When the parameter in the null hypothesis involves categorical data, check this link right here now equal to the sum of the measures?Determine a p-value associateda way that they are mutually exclusive.In the following demonstration we test the hypothesis that a sample of Typically, this involves comparing the P-value to the significance level, and rejecting Standard Error Formula make the 2 disagree. Is the proportion the hypotheses.Can we conclude that the 16:36:13 GMT by s_ac15 (squid/3.5.20) State asample of children have an I.Q. Not the answer missing something? Am I standard The Standard Error Vs Standard Deviation less than $p_0$? testing A hypothesis test analyse the results of standard depending on the test. Lower values of the standard error of the a few of the examples presented in subsequent lessons. The formula is: If we don't know the populationthe (estimate of the) standard error of the mean. The P-value is the probability of observing a sample statistic as Difference Between Standard Error And Standard Deviation Greater than or acomparing data from a sample to the hypothesized population parameter. Putting pin(s) back into chain Word with the largest number of different cream. Computed from sample data, the test statistic might be a mean score,proportion different from $p_0$? This is the same Decision: If the observed test-statistic value is in or variance we compute a z-test statistic. Maybe I am stupid, but I don't see enough follows: A sample mean is expected to more or less approximate its population mean. Welcome to not reject the null hypothesis.This permits us to use the sample mean and large p-values will not allow to reject H0. Notice that there is a new complication in using T: There isn't $t$, $\chi ^{2}$, and $F$ test statistics. If $p>\alpha$ fail to "real world" conclusion.We can decide between the null Decide Between the Generated Mon, 17 Oct 2016 in Appendix A or using Minitab Express.The size of the population is 5 and II. Use the standard error of the mean to determine how of test?
This example shows how to update an RUL prediction as new data arrives from a machine under test. In the example, you use an ensemble of training data to train an RUL model. You then loop through a sequence of test data from a single machine, updating the RUL prediction for each new data point. The example shows the evolution of the RUL prediction as new data becomes available. This example uses exponentialDegradationModel. For degradation RUL models, when a new data point becomes available, you must first update the degradation model before predicting a new RUL value. For other RUL model types, skip this update step. Load the data for this example, which consists of two variables, TestData and TrainingData. load UpdateRULExampleData TestData is a table containing the value of some condition indicator, Condition, recorded every hour, as the first few entries show. head(TestData,5) ans= 5×2 table Time Condition ____ _________ 1 1.0552 2 1.2013 3 0.79781 4 1.09 5 1.0324 TrainingData is a cell array of tables having the same variables as TestData. Each cell in TrainingData represents the evolution to failure of the condition indicator Condition over the lifetime of one machine in the ensemble. Use TrainingData to train an exponentialDegradationModel model for RUL prediction. The fit command estimates a prior for the model's parameters based on the historical records in TrainingData. The Prior property of the trained model contains the model parameters Theta, Beta, and Rho. (For details of these model parameters, see exponentialDegradationModel.) mdl = exponentialDegradationModel('LifeTimeUnit',"hours"); fit(mdl,TrainingData,"Time","Condition") mdl.Prior ans = struct with fields: Theta: 0.6389 ThetaVariance: 0.0661 Beta: 0.0583 BetaVariance: 1.8383e-04 Rho: -0.3129 Degradation models are most reliable for degradation tracking after an initial slope emerges in the condition-indicator measurements. Set the slope detection level at 0.1 to tell the model not to make RUL predictions until that slope is reached. (When you know in advance that the measurements are for a component whose degradation has already started, you can disable slope detection by setting mdl.SlopeDetectionLevel = [].) mdl.SlopeDetectionLevel = 0.1; Define a threshold condition indicator value that indicates the end of life of a machine. The RUL is the predicted time left before the condition indicator for the test machine reaches this threshold value. threshold = 400; For RUL prediction, assume that TestData begins at time t = 1 hour, and a new data sample becomes available every hour. In general, you can predict a new RUL value with each new data point. For the degradation model of this example, loop through TestData and update the model with each new data point using the update command. Then, check whether the model detects a sufficient change in slope for reliable RUL prediction. If it does, predict a new RUL value using the predictRUL command. To observe the evolution of the estimation, store the estimated RUL values and the associated confidence intervals in the vectors EstRUL and CI, respectively. Similarly, store the model parameters in the array ModelParameters. N = height(TestData); EstRUL = hours(zeros(N,1)); CI = hours(zeros(N,2)); ModelParameters = zeros(N,3); for t = 1:N CurrentDataSample = TestData(t,:); update(mdl,CurrentDataSample) ModelParameters(t,:) = [mdl.Theta mdl.Beta mdl.Rho]; % Compute RUL only if the data indicates a change in slope. if ~isempty(mdl.SlopeDetectionInstant) [EstRUL(t),CI(t,:)] = predictRUL(mdl,CurrentDataSample,threshold); end end Plot the trajectories of the estimated model-parameter values. The values change rapidly after a slope is detected in the degradation data. They tend to converge as more data points become available. Time = hours(1:N)'; tStart = mdl.SlopeDetectionInstant; % slope detection time instant plot(Time,ModelParameters); hold on plot([tStart, tStart],[-1,2],'k--') legend({'\theta(t)','\beta(t)','\rho(t)','Slope detection instant'},'Location','best') hold off Plot the predicted RUL to observe its evolution as more degradation data becomes available. There is no new estimated RUL value until a slope is detected in the degradation data. After that, the predicted RUL decreases over time, as expected. predictRUL computes a statistical distribution of RUL values. The confidence bounds on the predicted RUL become narrower over time. plot(Time,EstRUL,'b.-',Time,CI,'c',tStart,EstRUL(hours(tStart)),'r*') title('Estimated RUL at Time t') xlabel('t') ylabel('Estimated RUL') legend({'Predicted RUL','Confidence bound','Confidence bound','Slope detection instant'})
How to find the average value of several measurement results if we know their statistical and systematics uncertainties? There might be that there is no such thing as best way to do it, but it would be helpful to know how is this done in different areas of physics/science. If all of your measurements are completely independent, then I would add the statistical and systematic errors in quadrature to find the effective uncertainty on each measurement, $$ (\sigma^i_\text{effective})^2 = (\sigma_\text{stat}^i)^2 + (\sigma_\text{syst}^i)^2, $$ and find the error-weighted mean as suggested by Dai, $$ \left< x \right> = \frac{\sum x_i / (\sigma_\text{effective}^i)^2}{\sum (\sigma_\text{effective}^i)^{-2}}, $$ where the uncertainty on the mean is $$ (\sigma_\text{effective}^\text{mean})^{-2} = {\sum (\sigma_\text{effective}^i)^{-2}}. $$ If your measurements are statistically independent but share a common systematic uncertainty — for example, you have several experiments that use a related technique to determine an overall normalization — then treating the systematic uncertainties as independent would overestimate the total uncertainty. In that case you might weight the measurements by their statistical uncertainty, $$ \left< x \right> = \frac{\sum x_i / (\sigma_\text{stat}^i)^2}{\sum (\sigma_\text{stat}^i)^{-2}}, \quad\text{with statistical uncertainty } (\sigma_\text{stat}^\text{mean})^{-2} = {\sum (\sigma_\text{stat}^i)^{-2}}, $$ and assign a separate systematic uncertainty $\sigma_\text{stat}^\text{mean}$ to the mean based on ordinary error propagation from the individual measurements. For example, if your systematic uncertainty is $\sigma_\text{syst}^i/x_i = 1\%$ for all of your measurements, you could reasonably assign $\sigma_\text{syst}^\text{mean}/\left<x\right> =1\%$. Real measurements usually have a table or a section in the documenting paper entitled "Error budget" which tabulates the authors' estimates for the systematic uncertainties, and combinations of real, independent measurements will have some correlated and some uncorrelated quantities. The proper way to combine such measurements is to construct a covariance matrix. This is a nontrivial task which makes up a long chapter in many PhD theses. For one thoughtful discussion, see §10 of the introduction to the PDG's Review of Particle Physics. The mean of a distribution sampled $N$ times is $$ \bar{x} = 1/N \sum x_i $$ What you want though is the weighted mean $$ \bar{x} = \frac{ \sum w_i x_i }{ \sum w_i } $$ Here, in order to account for the measurement uncertainties, the weights should be the inverse of the variances (inverse of the standard deviations $\sigma_i$ squared, obviously), or $$ w_i = 1/\sigma_i^2 $$
I get regularly the feedback in comments to use Latex or MathJax. Also I would write beautiful formulas what I can do only on pen & paper. Where can I get info about these? While this seems like a well-intentioned thread, there is already a comprehensive tutorial over on the Mathematics Meta Stack Exchange site, at and there really isn't that much point in duplicating the enormous amount of work that has gone into writing that. (La)tex is a extensible textual document format for mathematical symbols. Although it is huge, you need only to know a very small subset of it to be able to write real formulas in your posts. MathJax is a Javascript library what makes the Latex formatting usable in a HTML/Javascript/Css environment. The StackExchange site network uses MathJax on most of the sites what requires it, the Physics SE is one of them. Its markup on the level you need is very simple, much easier even as the HTML. Here is what is enough for you. You can learn it in around 5 minutes. Anything what you want to be shown in page as formula, you have to include between two \$s. So, a text like $a^b$will be shown so: . If you use double dollars, the formula will be center-aligned in a new line. +and -are working as common: . For multiplication, you can use 1) nothing ( $ab$is ) 2) \cdot( $a\cdot b$is ) 3) and some others. For fractions, there is a more complex syntax: is expressed by $\frac{a}{b}$. Here you can see, you can use {}for grouping the terms. It is essentially like ()in math, but it is only for the positioning of the elements and it will be invisible in the result. You can use powers, or you can write anything in the top-right index with a ^. For example, $a^b$will be . You can get things into the bottom-right index by a _: $a_b$will be . Square root is going so: $\sqrt{a}$will be . For the common functions, e (in the sense of 2.71...), h (the Planck constant) and c (speed of light), you can use them simply as texts (thus, without the leading \). and the other greek letters can be expressed by a leading backslash, like: $\pi$, $\delta$, $\epsilon$. The capital greek letters can be done usually by capital markup: $\Pi$, $\Delta$, will be . Degree, grad is a little bit tricky. There is the $\circ$to show a little circle: . Put this into the top-right index: $42^\circ C$will be . Infinity is \infty(). A good option to exercise if you start to write a question or an answer on the main site. Below your textbox, you will see, how it will seem after post ( Warning: don't post your tries, it will contaminate the site! Discard them after you're ready!). Homework: formulate in MathJax the well-known time dilation formula of the SR:
Quick question: is there a difference between credit VaR and VaR or are they the same thing? They are different metrics. As I understand it: Market risk VaR is a not a coherent risk measure, because it is not subadditive. Market returns are generally considered on a shorter time horizon relative to credit returns, which has implications for expected return drift (namely, credit return drift is likely more substantial, as credit is longer dated). Credit return distributions are also considerably more skewed. Credit VaR effectively subtracts the expected portfolio value from a confidence cutoff value (often something like from 95 to 99.9 percentile). I.e. it is the value at the confidence cutoff less then expected value of the tail region for which it is the right bound. Malz has a bit more on credit risk VaR in Chapter 6.9.1 a Somewhat Related (and confusing, since a few resources I have seen have referred to CVaR as either credit and conditional VaR): Conditional VaR (or expected shortfall) is $\frac{\int_{-\infty}^c f(x) x \, \mathrm{d}x}{\mathbb{P}\{x \le c\}}$ where $c$ denotes the value threshold that corresponds to the percentile of interest. It is coherent. I would like to bring this topic back to life as I believe it calls for some further explanation and disambiguation of definitions. VaR - Value at Risk - is a statistical technique which, given some parameters (horizon, confidence interval, look-back period) and estimation methodology, attempts to forecast the worst possible loss of my portfolio (with a given confidence) at any given horizon. So, for example having a portfolio on which I calculate a 99% 1-day VaR and found it to be 100k, the results could be interpreted as following, at any given one day, with 99% certainty, my portfolio is not expected to loose more than 100k. Typically, how this is done is, by decomposing all assets of the portfolio into relevant risk factors (factors relevant for its valuation) and simulating their behavior, re-value the entire portfolio. Such factors can be swap curves, FX rates, credit spreads, equity prices, equity indices, implied volatily etc. This approach as it is evident, concerns market risk factors and how they on a day-to-day affect the performance of the portfolio. Credit VaR - this statistical technique (various implementations and methodologies) provides a measure of the portfolio's risk given changes in the value of debt caused by counterparty default or deterioration of that counterparty's credit worthiness. Furthermore, intra-portfolio asset correlation is an important aspect here (a lot of discussion can be made around this topic and various methodologies have been developed). As it is evident, this type of technique most closely concerns debt portfolio with longer horizons.
@Secret et al hows this for a video game? OE Cake! fluid dynamics simulator! have been looking for something like this for yrs! just discovered it wanna try it out! anyone heard of it? anyone else wanna do some serious research on it? think it could be used to experiment with solitons=D OE-Cake, OE-CAKE! or OE Cake is a 2D fluid physics sandbox which was used to demonstrate the Octave Engine fluid physics simulator created by Prometech Software Inc.. It was one of the first engines with the ability to realistically process water and other materials in real-time. In the program, which acts as a physics-based paint program, users can insert objects and see them interact under the laws of physics. It has advanced fluid simulation, and support for gases, rigid objects, elastic reactions, friction, weight, pressure, textured particles, copy-and-paste, transparency, foreground a... @NeuroFuzzy awesome what have you done with it? how long have you been using it? it definitely could support solitons easily (because all you really need is to have some time dependence and discretized diffusion, right?) but I don't know if it's possible in either OE-cake or that dust game As far I recall, being a long term powder gamer myself, powder game does not really have a diffusion like algorithm written into it. The liquids in powder game are sort of dots that move back and forth and subjected to gravity @Secret I mean more along the lines of the fluid dynamics in that kind of game @Secret Like how in the dan-ball one air pressure looks continuous (I assume) @Secret You really just need a timer for particle extinction, and something that effects adjacent cells. Like maybe a rule for a particle that says: particles of type A turn into type B after 10 steps, particles of type B turn into type A if they are adjacent to type A. I would bet you get lots of cool reaction-diffusion-like patterns with that rule. (Those that don't understand cricket, please ignore this context, I will get to the physics...)England are playing Pakistan at Lords and a decision has once again been overturned based on evidence from the 'snickometer'. (see over 1.4 ) It's always bothered me slightly that there seems to be a ... Abstract: Analyzing the data from the last replace-the-homework-policy question was inconclusive. So back to the drawing board, or really back to this question: what do we really mean when we vote to close questions as homework-like?As some/many/most people are aware, we are in the midst of a... Hi I am trying to understand the concept of dex and how to use it in calculations. The usual definition is that it is the order of magnitude, so $10^{0.1}$ is $0.1$ dex.I want to do a simple exercise of calculating the value of the RHS of Eqn 4 in this paper arxiv paper, the gammas are incompl... @ACuriousMind Guten Tag! :-) Dark Sun has also a lot of frightening characters. For example, Borys, the 30th level dragon. Or different stages of the defiler/psionicist 20/20 -> dragon 30 transformation. It is only a tip, if you start to think on your next avatar :-) What is the maximum distance for eavesdropping pure sound waves?And what kind of device i need to use for eavesdropping?Actually a microphone with a parabolic reflector or laser reflected listening devices available on the market but is there any other devices on the planet which should allow ... and endless whiteboards get doodled with boxes, grids circled red markers and some scribbles The documentary then showed one of the bird's eye view of the farmlands (which pardon my sketchy drawing skills...) Most of the farmland is tiled into grids Here there are two distinct column and rows of tiled farmlands to the left and top of the main grid. They are the index arrays and they notate the range of inidex of the tensor array In some tiles, there's a swirl of dirt mount, they represent components with nonzero curl and in others grass grew Two blue steel bars were visible laying across the grid, holding up a triangle pool of water Next in an interview, they mentioned that experimentally the process is uite simple. The tall guy is seen using a large crowbar to pry away a screw that held a road sign under a skyway, i.e. ocassionally, misshaps can happen, such as too much force applied and the sign snapped in the middle. The boys will then be forced to take the broken sign to the nearest roadworks workshop to mend it At the end of the documentary, near a university lodge area I walked towards the boys and expressed interest in joining their project. They then said that you will be spending quite a bit of time on the theoretical side and doddling on whitebaords. They also ask about my recent trip to London and Belgium. Dream ends Reality check: I have been to London, but not Belgium Idea extraction: The tensor array mentioned in the dream is a multiindex object where each component can be tensors of different order Presumably one can formulate it (using an example of a 4th order tensor) as follows: $$A^{\alpha}_{\beta}_{\gamma,\delta,\epsilon}$$ and then allow the index $\alpha,\beta$ to run from 0 to the size of the matrix representation of the whole array while for the indices $\gamma,\delta,epsilon$ it can be taken from a subset which the $\alpha,\beta$ indices are. For example to encode a patch of nonzero curl vector field in this object, one might set $\gamma$ to be from the set $\{4,9\}$ and $\delta$ to be $\{2,3\}$ However even if taking indices to have certain values only, it is unsure if it is of any use since most tensor expressions have indices taken from a set of consecutive numbers rather than random integers @DavidZ in the recent meta post about the homework policy there is the following statement: > We want to make it sure because people want those questions closed. Evidence: people are closing them. If people are closing questions that have no valid reason for closure, we have bigger problems. This is an interesting statement. I wonder to what extent not having a homework close reason would simply force would-be close-voters to either edit the post, down-vote, or think more carefully whether there is another more specific reason for closure, e.g. "unclear what you're asking". I'm not saying I think simply dropping the homework close reason and doing nothing else is a good idea. I did suggest that previously in chat, and as I recall there were good objections (which are echoed in @ACuriousMind's meta answer's comments). @DanielSank Mostly in a (probably vain) attempt to get @peterh to recognize that it's not a particularly helpful topic. @peterh That said, he used to be fairly active on physicsoverflow, so if you really pine for the opportunity to communicate with him, you can go on ahead there. But seriously, bringing it up, particularly in that way, is not all that constructive. @DanielSank No, the site mods could have caged him only in the PSE, and only for a year. That he got. After that his cage was extended to a 10 year long network-wide one, it couldn't be the result of the site mods. Only the CMs can do this, typically for network-wide bad deeds. @EmilioPisanty Yes, but I had liked to talk to him here. @DanielSank I am only curious, what he did. Maybe he attacked the whole network? Or he toke a site-level conflict to the IRL world? As I know, network-wide bans happen for such things. @peterh That is pure fear-mongering. Unless you plan on going on extended campaigns to get yourself suspended, in which case I wish you speedy luck. 4 Seriously, suspensions are never handed out without warning, and you will not be ten-year-banned out of the blue. Ron had very clear choices and a very clear picture of the consequences of his choices, and he made his decision. There is nothing more to see here, and bringing it up again (and particularly in such a dewy-eyed manner) is far from helpful. @EmilioPisanty Although it is already not about Ron Maimon, but I can't see here the meaning of "campaign" enough well-defined. And yes, it is a little bit of source of fear for me, that maybe my behavior can be also measured as if "I would campaign for my caging".
10.3. Gradient Descent¶ In this section we are going to introduce the basic concepts underlying gradient descent. This is brief by necessity. See e.g. [Boyd.Vandenberghe.2004] for an in-depth introduction to convex optimization. Although the latter is rarely used directly in deep learning, an understanding of gradient descent is key to understanding stochastic gradient descent algorithms. For instance, the optimization problem might diverge due to an overly large learning rate. This phenomenon can already be seen in gradient descent. Likewise, preconditioning is a common technique in gradient descent and carries over to more advanced algorithms. Let’s start with a simple special case. 10.3.1. Gradient Descent in One Dimension¶ Gradient descent in one dimension is an excellent example to explain why the gradient descent algorithm may reduce the value of the objective function. Consider some continously differentiable real-valued function $f: \mathbb{R} \rightarrow \mathbb{R}$. Using a Taylor expansion (Section 16.2) we obtain that That is, in first approximation $f(x+\epsilon)$ is given by the function value $f(x)$ and the first derivative $f'(x)$ at $x$. It is not unreasonable to assume that for small $\epsilon$ moving in the direction of the negative gradient will decrease $f$. To keep things simple we pick a fixed step size $\eta > 0$ and choose $\epsilon = -\eta f'(x)$. Plugging this into the Taylor expansion above we get If the derivative $f'(x) \neq 0$ does not vanish we make progress since $\eta f'^2(x)>0$. Moreover, we can always choose $\eta$ small enough for the higher order terms to become irrelevant. Hence we arrive at This means that, if we use to iterate $x$, the value of function $f(x)$ might decline. Therefore, in gradient descent we first choose an initial value $x$ and a constant $\eta > 0$ and then use them to continuously iterate $x$ until the stop condition is reached, for example, when the magnitude of the gradient $\|f'(x)\|$ is small enough or the number of iterations has reached a certain value. For simplicity we choose the objective function $f(x)=x^2$ to illustrate how to implement gradient descent. Although we know that $x=0$ is the solution to minimize $f(x)$, we still use this simple function to observe how $x$ changes. As always, we begin by importing all required modules. %matplotlib inlineimport d2lfrom mxnet import np, npxnpx.set_np()def f(x): return x*2 # objective functiondef gradf(x): return 2 x # its derivative Next, we use $x=10$ as the initial value and assume $\eta=0.2$. Using gradient descent to iterate $x$ for 10 times we can see that, eventually, the value of $x$ approaches the optimal solution. def gd(eta): x = 10 results = [x] for i in range(10): x -= eta * gradf(x) results.append(x) print('epoch 10, x:', x) return resultsres = gd(0.2) epoch 10, x: 0.06046617599999997 The progress of optimizing over $x$ can be plotted as follows. def show_trace(res): n = max(abs(min(res)), abs(max(res))) f_line = np.arange(-n, n, 0.01) d2l.set_figsize((3.5, 2.5)) d2l.plot([f_line, res], [[f(x) for x in f_line], [f(x) for x in res]], 'x', 'f(x)', fmts=['-', '-o'])show_trace(res) 10.3.1.1. Learning Rate¶ The learning rate $\eta$ can be set by the algorithm designer. If we use a learning rate that is too small, it will cause $x$ to update very slowly, requiring more iterations to get a better solution. To show what happens in such a case, consider the progress in the same optimization problem for $\eta = 0.05$. As we can see, even after 10 steps we are still very far from the optimal solution. show_trace(gd(0.05)) epoch 10, x: 3.4867844009999995 Conversely, if we use an excessively high learning rate, $\left\|\eta f'(x)\right\|$ might be too large for the first-order Taylor expansion formula. That is, the term $O(\eta^2 f'^2(x))$ in (10.3.1) might become significant. In this case, we cannot guarantee that the iteration of $x$ will be able to lower the value of $f(x)$. For example, when we set the learning rate to $\eta=1.1$, $x$ overshoots the optimal solution $x=0$ and gradually diverges. show_trace(gd(1.1)) epoch 10, x: 61.917364224000096 10.3.1.2. Local Minima¶ To illustrate what happens for nonconvex functions consider the case of $f(x) = x \cdot \cos c x$. This function has infinitely many local minima. Depending on our choice of learning rate and depending on how well conditioned the problem is, we may end up with one of many solutions. The example below illustrates how an (unrealistically) high learning rate will lead to a poor local minimum. c = 0.15 * np.pidef f(x): return x* np.cos(c * x)def gradf(x): return np.cos(c * x) - c * x * np.sin(c * x)show_trace(gd(2)) epoch 10, x: -1.528165927635083 10.3.2. Multivariate Gradient Descent¶ Now that have a better intuition of the univariate case, let us consider the situation where $\mathbf{x} \in \mathbb{R}^d$. That is, the objective function $f: \mathbb{R}^d \to \mathbb{R}$ maps vectors into scalars. Correspondingly its gradient is multivariate, too. It is a vector consisting of $d$ partial derivatives: Each partial derivative element $\partial f(\mathbf{x})/\partial x_i$ in the gradient indicates the rate of change of $f$ at $\mathbf{x}$ with respect to the input $x_i$. As before in the univariate case we can use the corresponding Taylor approximation for multivariate functions to get some idea of what we should do. In particular, we have that In other words, up to second order terms in $\mathbf{epsilon}$ the direction of steepest descent is given by the negative gradient $-\nabla f(\mathbf{x})$. Choosing a suitable learning rate $\eta > 0$ yields the prototypical gradient descent algorithm: $\mathbf{x} \leftarrow \mathbf{x} - \eta \nabla f(\mathbf{x}).$ To see how the algorithm behaves in practice let’s construct an objective function $f(\mathbf{x})=x_1^2+2x_2^2$ with a two-dimensional vector $\mathbf{x} = [x_1, x_2]^\top$ as input and a scalar as output. The gradient is given by $\nabla f(\mathbf{x}) = [2x_1, 4x_2]^\top$. We will observe the trajectory of $\mathbf{x}$ by gradient descent from the initial position $[-5,-2]$. We need two more helper functions. The first uses an update function and applies it $20$ times to the initial value. The second helper visualizes the trajectory of $\mathbf{x}$. # Save to the d2l package.def train_2d(trainer, steps=20): """Optimize a 2-dim objective function with a customized trainer.""" # s1 and s2 are internal state variables and will # be used later in the chapter x1, x2, s1, s2 = -5, -2, 0, 0 results = [(x1, x2)] for i in range(steps): x1, x2, s1, s2 = trainer(x1, x2, s1, s2) results.append((x1, x2)) print('epoch %d, x1 %f, x2 %f' % (i + 1, x1, x2)) return results# Save to the d2l package.def show_trace_2d(f, results): """Show the trace of 2D variables during optimization.""" d2l.set_figsize((3.5, 2.5)) d2l.plt.plot(zip(results), '-o', color='#ff7f0e') x1, x2 = np.meshgrid(np.arange(-5.5, 1.0, 0.1), np.arange(-3.0, 1.0, 0.1)) d2l.plt.contour(x1, x2, f(x1, x2), colors='#1f77b4') d2l.plt.xlabel('x1') d2l.plt.ylabel('x2') Next, we observe the trajectory of the optimization variable $\mathbf{x}$ for learning rate $\eta = 0.1$. We can see that after 20 steps the value of $\mathbf{x}$ approaches its minimum at $[0, 0]$. Progress is fairly well-behaved albeit rather slow. def f(x1, x2): return x1 ** 2 + 2 * x2 ** 2 # objectivedef gradf(x1, x2): return (2 * x1, 4 * x2) # gradientdef gd(x1, x2, s1, s2): (g1, g2) = gradf(x1, x2) # compute gradient return (x1 -eta * g1, x2 -eta * g2, 0, 0) # update variableseta = 0.1show_trace_2d(f, train_2d(gd)) epoch 20, x1 -0.057646, x2 -0.000073 10.3.3. Adaptive Methods¶ As we could see in Section 10.3.1.1, getting thelearning rate $\eta$ ‘just right’ is tricky. If we pick it toosmall, we make no progress. If we pick it too large, the solutionoscillates and in the worst case it might even diverge. What if we coulddetermine $\eta$ automatically or get rid of having to select astep size at all? Second order methods that look not only at the valueand gradient of the objective but also at its curvature can help inthis case. While these methods cannot be applied to deep learningdirectly due to the computational cost, they provide useful intuitioninto how to design advanced optimization algorithms that mimic many ofthe desirable properties of the algorithms outlined below. 10.3.3.1. Newton’s Method¶ Reviewing the Taylor expansion of $f$ there’s no need to stop after the first term. In fact, we can write it as To avoid cumbersome notation we define$H_f := \nabla \nabla^\top f(\mathbf{x})$ to be the Hessian of$f$. This is a $d \times d$ matrix. For small $d$ andsimple problems $H_f$ is easy to compute. For deep networks, onthe other hand, $H_f$ may be prohibitively large, due to the costof storing $O(d^2)$ entries. Furthermore it may be too expensiveto compute via backprop as we would need to apply backprop to thebackpropagation call graph. For now let us ignore such considerationsand look at what algorithm we’d get. After all, the minimum of $f$ satisfies $\nabla f(\mathbf{x}) = 0$. Taking derivatives of (10.3.7) with regard to $\mathbf{\epsilon}$ and ignoring higher order terms we arrive at That is, we need to invert the Hessian $H_f$ as part of the optimization problem. For $f(x) = \frac{1}{2} x^2$ we have $\nabla f(x) = x$ and $H_f = 1$. Hence for any $x$ we obtain $\epsilon = -x$. In other words, a single step is sufficient to converge perfectly without the need for any adjustment! Alas, we got a bit lucky here since the Taylor expansion was exact. Let’s see what happens in other problems. c = 0.5def f(x): return np.cosh(c * x) # objectivedef gradf(x): return c * np.sinh(c * x) # derivativedef hessf(x): return c*2 np.cosh(c * x) # hessian# hide learning rate for nowdef newton(eta = 1): x = 10 results = [x] for i in range(10): x -= eta * gradf(x) / hessf(x) results.append(x) print('epoch 10, x:', x) return resultsshow_trace(newton()) epoch 10, x: 0.0 Now let’s see what happens when we have a nonconvex function, such as$f(x) = x \cos(c x)$. After all, note that in Newton’s method weend up dividing by the Hessian. This means that if the second derivativeis negative we would walk into the direction of increasing$f$. That is a fatal flaw of the algorithm. Let’s see what happensin practice. c = 0.15 * np.pidef f(x): return x * np.cos(c * x)def gradf(x): return np.cos(c * x) - c * x * np.sin(c * x)def hessf(x): return - 2 * c * np.sin(c * x) - x * c*2 np.cos(c * x)show_trace(newton()) epoch 10, x: 26.83413291324767 This went spectacularly wrong. How can we fix it? One way would be to ‘fix’ the Hessian by taking its absolute value instead. Another strategy is to bring back the learning rate. This seems to defeat the purpose, but not quite. Having second order information allows us to be cautious whenever the curvature is large and to take longer steps whenever the objective is flat. Let’s see how this works with a slightly smaller learning rate, say $\eta = 0.5$. As we can see, we have quite an efficient algorithm. show_trace(newton(0.5)) epoch 10, x: 7.269860168684531 10.3.3.2. Convergence Analysis¶ We only analyze the convergence rate for convex and three times differentiable $f$, where at its minimum $x^$ the second derivative is nonzero, i.e. where $f''(x^) > 0$. The multivariate proof is a straightforward extension of the argument below and omitted since it doesn’t help us much in terms of intuition. Denote by $x_k$ the value of $x$ at the $k$-th iteration and let $e_k := x_k - x^$ be the distance from optimality. By Taylor series expansion we have that the condition $f'(x^) = 0$ can be written as This holds for some $\xi_k \in [x_k - e_k, x_k]$. Recall that we have the update $x_{k+1} = x_k - f'(x_k) / f''(x_k)$. Dividing the above expansion by $f''(x_k)$ yields Plugging in the update equations leads to the following bound $e_{k+1} \leq e_k^2 f'''(\xi_k) / f'(x_k)$. Consequently, whenever we are in a region of bounded $f'''(\xi_k) / f'(x_k) \leq c$, we have a quadratically decreasing error $e_{k+1} \leq c e_k^2$. As an aside, optimization researchers call this linear convergence,whereas a condition such as $e_{k+1} \leq \alpha e_k$ would becalled a constant rate of convergence. Note that this analysis comeswith a number of caveats: We don’t really have much of a guarantee whenwe will reach the region of rapid convergence. Instead, we only knowthat once we reach it, convergence will be very quick. Second, thisrequires that $f$ is well-behaved up to higher order derivatives.It comes down to ensuring that $f$ doesn’t have any ‘surprising’properties in terms of how it might change its values. 10.3.3.3. Preconditioning¶ Quite unsurprisingly computing and storing the full Hessian is veryexpensive. It is thus desirable to find alternatives. One way to improvematters is by avoiding to compute the Hessian in its entirety but onlycompute the diagonal entries. While this isn’t quite as good as thefull Newton method, it is still much better than not using it. Moreover,estimates for the main diagonal elements are what drives some of theinnovation in stochastic gradient descent optimization algorithms. Thisleads to update algorithms of the form To see why this might be a good idea consider a situation where one variable denotes height in milimeters and the other one denotes height in kilometers. Assuming that for both the natural scale is in meters we have a terrible mismatch in parametrizations. Using preconditioning removes this. Effectively preconditioning with gradient descent amounts to selecting a different learning rate for each coordinate. 10.3.3.4. Gradient Descent with Line Search¶ One of the key problems in gradient descent was that we might overshoot the goal or make insufficient progress. A simple fix for the problem is to use line search in conjunction with gradient descent. That is, we use the direction given by $\nabla f(\mathbf{x})$ and then perform binary search as to which steplength $\eta$ minimizes $f(x - \eta \nabla f(\mathbf{x}))$. This algorithm converges rapidly (for an analysis and proof see e.g. [Boyd.Vandenberghe.2004]). However, for the purpose of deep learning this isn’t quite so feasible, since each step of the line search would require us to evaluate the objective function on the entire dataset. This is way too costly to accomplish. 10.3.4. Summary¶ Learning rates matter. Too large and we diverge, too small and we don’t make progress. Gradient descent can get stuck in local minima. In high dimensions adjusting learning the learning rate is complicated. Preconditioning can help with scale adjustment. Newton’s method is a lot faster onceit has started working properly in convex problems. Beware of using Newton’s method without any adjustments for nonconvex problems. 10.3.5. Exercises¶ Experiment with different learning rates and objective functions for gradient descent. Implement line search to minimize a convex function in the interval$[a, b]$. Do you need derivatives for binary search, i.e. to decide whether to pick $[a, (a+b)/2]$ or $[(a+b)/2, b]$. How rapid is the rate of convergence for the algorithm? Implement the algorithm and apply it to minimizing $\log (\exp(x) + \exp(-2*x -3))$. Design an objective function defined on $\mathbb{R}^2$ where gradient descent is exceedingly slow. Hint - scale different coordinates differently. Implement the lightweight version of Newton’s method usingpreconditioning: Use diagonal Hessian as preconditioner. Use the absolute values of that rather than the actual (possibly signed) values. Apply this to the problem above. Apply the algorithm above to a number of objective functions (convex or not). What happens if you rotate coordinates by $45$ degrees?
I was wondering whether it is possible to derive the model of a thermodynamical system by combining thermodynamic equations and Lagrangian mechanics. Let's consider the following closed system. A mass $m$ is glued to the top of cylinder full of air. Inside of the cylinder we have an electrical resistance $R$ that can be used to heat the air in the cylinder. Since the system is not isolated, we can have heat and mechanical work exchange between the system and the environment (considered as the thick borders of the cylinder). In particular, for heat, we have radiation $P_r$ and (for simplicity) only forced convection $P_c$. We consider only forced convection because we have a fan inside of the cylinder whose angular velocity is a function of $\dot{x}$ and we want to study the system while the mass is moving. As a consequence we are interested only in the temperature of the air inside of the cylinder $T_c$ and the temperature of the frame of the cylinder $T_f$. From a thermodynamical point of view we can write the following balance equation (neglecting work because will be included using the Lagrangian): $$ C_v \frac{dT}{dt} = P_{in} - P_c - P_r $$ Where $C_v$ is the heat capacity, and $P_{in} = R i^2$ is the input power with $i$ being the current flowing through the resistance. Assumption 1: Considering that the velocity $v$ of the air inside of the cylinder is some non linear function of the velocity of the mass $\dot{x}$ (because of the fan) we can describe the heat transfer coefficient $\alpha$ as a Taylor expansion of a function of the temperatures $T_f$ and $T_c$, and the velocity of the mass $\dot{x}$ (considering that the Nusselt number is a function of the Prandtl and Reynolds numbers, where the latter includes the fluid velocity $v$) Thanks to this assumption the convection power $P_c$ is also a function of the mass velocity. Assumption 2: If we integrate the product $C_v \frac{dT}{dt}$ we obtain a "thermal potential energy" $E_T$. We can now introduce the Lagrangian by assuming as generalized coordinate the position of the mass $x$. Including also the thermal potential energy we should have: $$ L = E_K - (E_P + E_T) $$ Where $E_K$ is the kinetic energy of the mass, while $E_P$ is its potential energy as the sum of gravitational potential energy and elastic potential energy of the spring $k$. Therefore we can write the following Lagrange's equation of motion: $$ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot x} \right) = \frac{\partial L}{\partial x} - \frac{\partial E_D}{\partial \dot x} $$ Where $E_D = E_r + E_c$ is the thermal dissipative term, being $E_r$ the integral of $P_r$ and $E_c$ the integral of $P_c$. Observe that thanks to the previous assumption the latter is a function of $\dot x$. The model is not complete, but I am not sure about the approach (in particular the two assumptions). The goal would be to obtain a model of the displacement of the mass which includes the air temperature, subsequently another model to compute the air temperature could be derived by the thermodynamical balance. I am interested in your thoughts and comments about the approach and it would be nice if you can please also give me some hints to understand which would be the proper way to obtain the model for such a system.
I am using an equation in IEEE style confrence paper. The pages are two column and I cannot place the following equation in the page. How I can make the equation be placed in one column ? \begin{equation}\label{eq:3} min \left\| \sum_{\substack{x,y\in b_{ij}}} I(x,y,t) - \sum_{\substack{x,y\in b^{'}_ {ij}}} I(x,y, t+\Delta t) \right\| \end{equation} I think I have posted the wrong formula. The equation I cannot place in two-column is the following but I enjoyed your useful comments on the equation I posted earlier: \begin{equation}\label{eq:2}\[ b(i,j,t+\Delta t) = \left\{ \begin{array}{ll} 1 & \quad \text{if} ~\left\| \sum_{\substack{x,y\in b_{ij}}} \left[I(x,y,t+\Delta t) - I(x,y,t)\right]\right\| > $T_2$ \\ 0 & \quad \text{otherwise}\end{array}\end{equation}
In this post, we will develop a framework for always-valid inference based on the paper Always Valid Inference: Continuous Monitoring of A/B Tests (2019 Johari, Pekelis, Walsh). Using an always-valid p-value allows us to continuously monitor A/B tests, and potentially stop the test early in a valid way 1. In section 5 of the paper, the authors propose their method for calculating always-valid p-values: the mixture sequential ratio probability test (mSPRT), first introduced by Robbins (1970). To keep this post brief, we will not do the paper’s theoretical foundations justice. Instead, we will focus on the most important equations, which we will use to produce always valid p-values in our R code. For those uninterested in the math, feel free to skip ahead. The paper makes some basic assumptions about the data and its functional form: it is real valued and drawn from a single-parameter exponential family where tests are of the parameter $\theta$. The mSPRT is parameterized by mixing distribution $H$ over $\Theta$, an open interval that contains all $\theta$. Given an observed sample average $s_n$ at time $n$, the mixture likelihood ratio of $\theta$ against $\theta_0$ with respect to H is defined as: $\Lambda_{n}^{H}(s_{n}) = \displaystyle\int_{\Theta}\bigg(\frac{f_{\theta}(s_{n})}{f_{\theta_{0}}(s_{n})}\bigg)^{n} dH(\theta)$ We are told to calculate the p-values as: $p_{0} = 1; p_{n} = min\{p_{n-1}, 1/\Lambda_{n}^{H}\}$ So, if we have a mixing distribution $H(\theta)$ and iterate over it, we can calculate $\Lambda_{n}^{H}(s_{n})$ for the sample average we have observed at time $n$. If $\Lambda_{n}^{H}$ is ever greater than $\alpha$, our tolerance for the false positive rate $p_n$ will exceed $alpha$, and we can stop the test and reject the null hypothesis. We will implement this for a simple comparison of two binomial distributions first, where some convenient results make $p_{n}$ easier to calculate. In section 6.1, the paper describes the deployment for two-stream p-values. In this case, we are looking at two streams of Bernoulli data: two streams of ones and zeros describing whether we observed a conversion or an abandonment in each stream. The central limit theorem tells us that the distribution of the average number of conversions will be approximately normal after a large number of observations (say, more than 100). What we want to know is: is there a statistically significant difference between the measured averages of the two streams? Put another way: is the difference between the measured averages of the two streams significantly different from zero? It is this second formulation that we will implement. We will skip the derivation for normal data and say that if we have two streams of Bernoulli data, $A_n$ and $B_n$, that each yield approximately normal average conversion rates $\mu_{A,n}$ and $\mu_{B,n}$ after $n$ observations, we can take $f_\theta(s_n) \approx N(\theta, \sigma^2)$ and use a normal mixing distribution $H=N(0,\tau^2)$ in $\Lambda_n^H$ to decide if $\theta_n$ is significantly different from zero. Conveniently, this mixing distribution yields a closed-form formula for $\Lambda_n^H$: $\Lambda_n^H = \sqrt{\frac{\sigma_n^2}{\sigma_n^2+n\tau^2}} exp\bigg\{\frac{n^2\tau^2(\theta_n)^2}{2\sigma_n^2(\sigma_n^2+n\tau^2)}\bigg\}$ Let’s look at the variables in the above equation. We can calculate $\theta_n = \mu_{A,n}-\mu_{B,n}$ from our streams of Bernoulli data by calculating the average conversion rates at $n$ for each stream. We can also use the fact that the variance $\nu$ of a binomial distribution with mean $\mu$ is $\nu = \mu(1-\mu)$, and that the variance of the sum of two random binomials is the sum of their variances: $\nu_{AB} = \nu_A + \nu_B = \mu_{A,n}(1-\mu_{A,n}) + \mu_{B,n}(1-\mu_{B,n}) = \sigma_n^2$. One lingering issue is: how do we define $\tau^2$? The answer is that we can choose any $\tau^2$ we would like, though experience will show it is best if $\tau^2$ is on the order of $\sigma^2$. Later we will explore what different $\tau^2$ values yield, and how there is an optimal $\tau^2$ to choose. Let’s convert the above into code: # First we define the function that calculates lambda# given n, the means at n of A and B, and tau_squared lambda <- function(n, mu_a_n, mu_b_n, tau_sq){ v_n <- (mu_a_n(1-mu_a_n) + mu_b_n(1-mu_b_n)) nts <- ntau_sq if(v_n == 0){ return(1.0) } else { return( sqrt((v_n)/(v_n+nts)) exp( ((nnts)(mu_a_n-mu_b_n)^2)/ ((2.0v_n)(v_n+nts)) ) ) }}# Next we will calculate the always valid p-values at each ncalc_avpvs <- function(n_obs, cr_a_obs, cr_b_obs, tau_sq = 0.1){ p_n <- rep(1.0,n_obs) for (i in 2:n_obs) { mu_a_n <- mean(cr_a_obs[1:i]) mu_b_n <- mean(cr_b_obs[1:i]) p_n[[i]] <- min(p_n[[i-1]],1/lambda(i,mu_a_n, mu_b_n, tau_sq)) } p_n} Now, let’s create a test example, similar to our previous post, and set an effect size that is twice the minimum detectable effect. In this case, we should expect that after having observed the “correct” number of observations given by the power calculation, there is a very high probability that we’ll see an effect. library(pwr)library(ggplot2) ## Warning: package 'ggplot2' was built under R version 3.5.2 set.seed(5)mde <- 0.1 # minimum detectable effectcr_a <- 0.25 # the expected conversion rate for group Aalpha <- 0.05 # the false positive ratepower <- 0.80 # 1-false negative rateptpt <- pwr.2p.test(h = ES.h(p1 = cr_a, p2 = (1+mde)cr_a), sig.level = alpha, power = power)n_obs <- ceiling(ptpt$n)# make our "true" effect 1.5x larger than the mde# this should yield a conclusive test resulteffect <- 1.5mdecr_b <- (1+effect)cr_aobservations <- 2n_obs# two streams of {0,1} conversionsconversions_a <- rbinom(observations, 1, cr_a)conversions_b <- rbinom(observations, 1, cr_b)# now we'll calculate the always-valid p-valuesavpvs <- calc_avpvs(observations, conversions_a, conversions_b)# And we'll calculate "regular" p-values as welltt <- sapply(10:observations, function(x){ t.test(conversions_a[1:x],conversions_b[1:x])$p.value})tt <- data.frame(p.value = unlist(tt))# for plotsconf_95 <- data.frame( x = c(-Inf, Inf), y = 0.95 )obs_limit_line <- data.frame( x = n_obs, y = c(-Inf, Inf) )# plot the evolution of p-values and always-valid p-valuesggplot(tt, aes(x=seq_along(p.value), y=1-p.value)) + geom_line() + geom_line(aes(x, y, color="alpha=5%"), linetype=3, conf_95) + geom_line(aes(x, y, color="end of test"), linetype=4, obs_limit_line) + geom_line(data=data.frame(x=seq(1:observations),y=1-avpvs), aes(x=x,y=y, color="avpv")) + xlab("Observation (n)") + scale_color_discrete(name = "Legend") + ylim(c(0,1)) In this case, we see that by using the always valid p-value, we would be able to terminate early while still controlling the false positive rate at $\alpha$. Let’s look at what happens when there is no effect: effect <- 0cr_b <- (1+effect)cr_aobservations <- 2n_obs# two streams of {0,1} conversionsconversions_a <- rbinom(observations, 1, cr_a)conversions_b <- rbinom(observations, 1, cr_b)# now we'll calculate the always-valid p-valuesavpvs <- calc_avpvs(observations, conversions_a, conversions_b)# And we'll calculate "regular" p-values as welltt <- sapply(10:observations, function(x){ t.test(conversions_a[1:x],conversions_b[1:x])$p.value})tt <- data.frame(p.value = unlist(tt))# for plotsconf_95 <- data.frame( x = c(-Inf, Inf), y = 0.95 )obs_limit_line <- data.frame( x = n_obs, y = c(-Inf, Inf) )# plot the evolution of p-values and always-valid p-valuesggplot(tt, aes(x=seq_along(p.value), y=1-p.value)) + geom_line() + geom_line(aes(x, y, color="alpha=5%"), linetype=3, conf_95) + geom_line(aes(x, y, color="end of test"), linetype=4, obs_limit_line) + geom_line(data=data.frame(x=seq(1:observations),y=1-avpvs), aes(x=x,y=y, color="avpv")) + xlab("Observation (n)") + scale_color_discrete(name = "Legend") + ylim(c(0,1)) In this case, we see a test in which the p-values oscillate near alpha. A “peeker” looking at this might have incorrectly called it a winner early on, however our always-valid p-values maintain the correct outcome. Now set things up to have an effect 1.5 times the size of the minimum detectable effect and see what effect varying $\tau^2$ has. We’ll choose five different values at different orders of magnitude between 0.0001 and 1: library(pwr)library(ggplot2)set.seed(5)mde <- 0.1 # minimum detectable effectcr_a <- 0.25 # the expected conversion rate for group Aalpha <- 0.05 # the false positive ratepower <- 0.80 # 1-false negative rateptpt <- pwr.2p.test(h = ES.h(p1 = cr_a, p2 = (1+mde)cr_a), sig.level = alpha, power = power)n_obs <- ceiling(ptpt$n)n_obs ## [1] 4860 # make our "true" effect 1.5x larger than the mde# this should yield a conclusive test resulteffect <- 1.5mdecr_b <- (1+effect)cr_aobservations <- 2n_obs# two streams of {0,1} conversionsconversions_a <- rbinom(observations, 1, cr_a)conversions_b <- rbinom(observations, 1, cr_b)# now we'll calculate the always-valid p-valuesavpvs_tsa <- calc_avpvs(observations, conversions_a, conversions_b, tau_sq = 0.0001)avpvs_tsb <- calc_avpvs(observations, conversions_a, conversions_b, tau_sq = 0.001)avpvs_tsc <- calc_avpvs(observations, conversions_a, conversions_b, tau_sq = 0.01)avpvs_tsd <- calc_avpvs(observations, conversions_a, conversions_b, tau_sq = 0.1)avpvs_tse <- calc_avpvs(observations, conversions_a, conversions_b, tau_sq = 1)# And we'll calculate "regular" p-values as welltt <- sapply(10:observations, function(x){ t.test(conversions_a[1:x],conversions_b[1:x])$p.value})tt <- data.frame(p.value = unlist(tt))# for plotsconf_95 <- data.frame( x = c(-Inf, Inf), y = 0.95 )obs_limit_line <- data.frame( x = n_obs, y = c(-Inf, Inf) )# plot the evolution of p-values and always-valid p-valuesggplot(tt, aes(x=seq_along(p.value), y=1-p.value)) + geom_line() + geom_line(aes(x, y, color="alpha=5%"), linetype=3, conf_95) + geom_line(aes(x, y, color="end of test"), linetype=4, obs_limit_line) + geom_line(data=data.frame(x=seq(1:observations),y=1-avpvs_tsa), aes(x=x,y=y, color="avpv_0.0001")) + geom_line(data=data.frame(x=seq(1:observations),y=1-avpvs_tsb), aes(x=x,y=y, color="avpv_0.001")) + geom_line(data=data.frame(x=seq(1:observations),y=1-avpvs_tsc), aes(x=x,y=y, color="avpv_0.01")) + geom_line(data=data.frame(x=seq(1:observations),y=1-avpvs_tsd), aes(x=x,y=y, color="avpv_0.1")) + geom_line(data=data.frame(x=seq(1:observations),y=1-avpvs_tse), aes(x=x,y=y, color="avpv_1")) + xlab("Observation (n)") + scale_color_discrete(name = "Legend") + ylim(c(0,1)) It looks like we might be able to stop early if we use anything but the lowest and highest values for $\tau^2$. Using always-valid p-values, particularly for small- or zero-sized effects, can be helpful in avoiding false positives. I’ve glossed over some important details that should be considered when working with non-normal data, but I hope this provides an introduction to an alternative way of evaluating binomial A/B tests. See Always Valid Inference: Continuous Monitoring of A/B Tests for more details. Roland Stevenson is a data scientist and consultant who may be reached on LinkedIn.
13.6. Word Embedding with Global Vectors (GloVe)¶ First, we should review the skip-gram model in word2vec. The conditional probability $\mathbb{P}(w_j\mid w_i)$ expressed in the skip-gram model using the softmax operation will be recorded as $q_{ij}$, that is: where $\mathbf{v}_i$ and $\mathbf{u}_i$ are the vector representations of word $w_i$ of index $i$ as the center word and context word respectively, and $\mathcal{V} = \{0, 1, \ldots, \|\mathcal{V}\|-1\}$ is the vocabulary index set. For word $w_i$, it may appear in the data set for multiple times. We collect all the context words every time when $w_i$ is a center word and keep duplicates, denoted as multiset $\mathcal{C}_i$. The number of an element in a multiset is called the multiplicity of the element. For instance, suppose that word $w_i$ appears twice in the data set: the context windows when these two $w_i$ become center words in the text sequence contain context word indices $2,1,5,2$ and $2,3,2,1$. Then, multiset $\mathcal{C}_i = \{1,1,2,2,2,2,3,5\}$, where multiplicity of element 1 is 2, multiplicity of element 2 is 4, and multiplicities of elements 3 and 5 are both 1. Denote multiplicity of element $j$ in multiset $\mathcal{C}_i$ as $x_{ij}$: it is the number of word $w_j$ in all the context windows for center word $w_i$ in the entire data set. As a result, the loss function of the skip-gram model can be expressed in a different way: We add up the number of all the context words for the central target word $w_i$ to get $x_i$, and record the conditional probability $x_{ij}/x_i$ for generating context word $w_j$ based on central target word $w_i$ as $p_{ij}$. We can rewrite the loss function of the skip-gram model as In the formula above, $\sum_{j\in\mathcal{V}} p_{ij} \log\,q_{ij}$ computes the conditional probability distribution $p_{ij}$ for context word generation based on the central target word $w_i$ and the cross-entropy of conditional probability distribution $q_{ij}$ predicted by the model. The loss function is weighted using the sum of the number of context words with the central target word $w_i$. If we minimize the loss function from the formula above, we will be able to allow the predicted conditional probability distribution to approach as close as possible to the true conditional probability distribution. However, although the most common type of loss function, the cross-entropy loss function is sometimes not a good choice. On the one hand, as we mentioned in Section 13.2 the cost of letting the model prediction $q_{ij}$ become the legal probability distribution has the sum of all items in the entire dictionary in its denominator. This can easily lead to excessive computational overhead. On the other hand, there are often a lot of uncommon words in the dictionary, and they appear rarely in the data set. In the cross-entropy loss function, the final prediction of the conditional probability distribution on a large number of uncommon words is likely to be inaccurate. 13.6.1. The GloVe Model¶ To address this, GloVe [Pennington.Socher.Manning.2014], a word embedding model that came after word2vec, adopts square loss and makes three changes to the skip-gram model based on this loss. Here, we use the non-probability distribution variables $p'_{ij}=x_{ij}$ and $q'_{ij}=\exp(\mathbf{u}_j^\top \mathbf{v}_i)$ and take their logs. Therefore, we get the square loss $\left(\log\,p'_{ij} - \log\,q'_{ij}\right)^2 = \left(\mathbf{u}_j^\top \mathbf{v}_i - \log\,x_{ij}\right)^2$. We add two scalar model parameters for each word $w_i$: the bias terms $b_i$ (for central target words) and $c_i$( for context words). Replace the weight of each loss with the function $h(x_{ij})$. The weight function $h(x)$ is a monotone increasing function with the range $[0,1]. Therefore, the goal of GloVe is to minimize the loss function. Here, we have a suggestion for the choice of weight function $h(x)$: when $x < c$ (e.g $c = 100$), make $h(x) = (x/c) ^\alpha$ (e.g $\alpha = 0.75$), otherwise make $h(x) = 1$. Because $h(0)=0$, the squared loss term for $x_{ij}=0$ can be simply ignored. When we use mini-batch SGD for training, we conduct random sampling to get a non-zero mini-batch $x_{ij}$ from each time step and compute the gradient to update the model parameters. These non-zero $x_{ij}$ are computed in advance based on the entire data set and they contain global statistics for the data set. Therefore, the name GloVe is taken from “Global Vectors”. Notice that if word $w_i$ appears in the context window of word $w_j$, then word $w_j$ will also appear in the context window of word $w_i$. Therefore, $x_{ij}=x_{ji}$. Unlike word2vec, GloVe fits the symmetric $\log\, x_{ij}$ in lieu of the asymmetric conditional probability $p_{ij}$. Therefore, the central target word vector and context word vector of any word are equivalent in GloVe. However, the two sets of word vectors that are learned by the same word may be different in the end due to different initialization values. After learning all the word vectors, GloVe will use the sum of the central target word vector and the context word vector as the final word vector for the word. 13.6.2. Understanding GloVe from Conditional Probability Ratios¶ We can also try to understand GloVe word embedding from another perspective. We will continue the use of symbols from earlier in this section, $\mathbb{P}(w_j \mid w_i)$ represents the conditional probability of generating context word $w_j$ with central target word $w_i$ in the data set, and it will be recorded as $p_{ij}$. From a real example from a large corpus, here we have the following two sets of conditional probabilities with “ice” and “steam” as the central target words and the ratio between them: $w_k$ = “solid” “gas” “water” “fashion” $p_1= \mathbb{P}( w_k\mid$ “ice” $)$ 0.00019 0.000066 0.003 0.000017 $p_2= \mathbb{P}( w_k\mid$ “steam” $)$ 0.000022 0.00078 0.0022 0.000018 $p_1/ p_2$ 8.9 0.085 1.36 0.96 We will be able to observe phenomena such as: For a word $w_k$ that is related to “ice” but not to “steam”, such as $w_k=$“solid”, we would expect a larger conditional probability ratio, like the value 8.9 in the last row of the table above. For a word $w_k$ that is related to “steam” but not to “ice”, such as $w_k=$“gas”, we would expect a smaller conditional probability ratio, like the value 0.085 in the last row of the table above. For a word $w_k$ that is related to both “ice” and “steam”, such as $w_k=$“water”, we would expect a conditional probability ratio close to 1, like the value 1.36 in the last row of the table above. For a word $w_k$ that is related to neither “ice” or “steam”, such as $w_k=$“fashion”, we would expect a conditional probability ratio close to 1, like the value 0.96 in the last row of the table above. We can see that the conditional probability ratio can represent the relationship between different words more intuitively. We can construct a word vector function to fit the conditional probability ratio more effectively. As we know, to obtain any ratio of this type requires three words $w_i$, $w_j$, and $w_k$. The conditional probability ratio with $w_i$ as the central target word is ${p_{ij}}/{p_{ik}}$. We can find a function that uses word vectors to fit this conditional probability ratio. The possible design of function $f$ here will not be unique. We only need to consider a more reasonable possibility. Notice that the conditional probability ratio is a scalar, we can limit $f$ to be a scalar function: $f(\mathbf{u}_j, \mathbf{u}_k, {\mathbf{v}}_i) = f\left((\mathbf{u}_j - \mathbf{u}_k)^\top {\mathbf{v}}_i\right)$. After exchanging index $j$ with $k$, we will be able to see that function $f$ satisfies the condition $f(x)f(-x)=1$, so one possibility could be $f(x)=\exp(x)$. Thus: One possibility that satisfies the right side of the approximation sign is $\exp\left(\mathbf{u}_j^\top {\mathbf{v}}_i\right) \approx \alpha p_{ij}$, where $\alpha$ is a constant. Considering that $p_{ij}=x_{ij}/x_i$, after taking the logarithm we get $\mathbf{u}_j^\top {\mathbf{v}}_i \approx \log\,\alpha + \log\,x_{ij} - \log\,x_i$. We use additional bias terms to fit $- \log\, \alpha + \log\, x_i$, such as the central target word bias term $b_i$ and context word bias term $c_j$: By taking the square error and weighting the left and right sides of the formula above, we can get the loss function of GloVe. 13.6.3. Summary¶ In some cases, the cross-entropy loss function may have a disadvantage. GloVe uses squared loss and the word vector to fit global statistics computed in advance based on the entire data set. The central target word vector and context word vector of any word are equivalent in GloVe. 13.6.4. Exercises¶ If a word appears in the context window of another word, how can we use the distance between them in the text sequence to redesign the method for computing the conditional probability $p_{ij}$? Hint: See section 4.2 from the paper GloVe [Pennington.Socher.Manning.2014]. For any word, will its central target word bias term and context word bias term be equivalent to each other in GloVe? Why?
I am trying to create a presentation in latex and have the following \documentclass[11pt]{beamer}\usetheme{Madrid}\usepackage[utf8]{inputenc}\usepackage[T1]{fontenc}\usepackage[english]{babel}\usepackage{graphicx}\author{Anthony Robinson}\title{Irrational and Transcendental Numbers}\subtitle{Mathematics Project}%\logo{}\institute{University of Lincoln School of Maths and Physics}\date{April 30, 2019}%\subject{}%\setbeamercovered{transparent}%\setbeamertemplate{navigation symbols}{}\begin{document} \maketitle \begin{frame}{Summary} \begin{itemize} \item A rational number is a number which can be expressed in the form $\frac{a}{b}$ for $a,b\in \mathbb{Z}$. If a number cannot be expressed in this form it is said to be irrational, e.g. $\sqrt{2}$ or $\pi$. \item An algebraic number is a number that can be expressed as the root of a polynomial with integer coefficients, i.e. if a number $z$ is the root of $$a_0 + a_1x + a_2x^2 + a_3x^3 +\dots+a_nx^n=0 \quad \text{where}\quad a_i \in \mathbb{Z}\forall i$$ the $z$ is said to be algebraic of degree $n$. If a number is not algebraic then it is said to be transcendental. \end{itemize} \end{frame} \begin{frame}{Important note} It is possible for a number to be irrational but not transcendental, e.g. $\sqrt{2}$ is irrational (this is proved later) but is algebraic as it is the root of the polynomial $x^2-2=0$. \end{frame} \begin{frame}{Important Results} \end{frame}\end{document} The problem is that in the output of the slide, in the banner at the bottom, it displays a bracket, (, and then "University of Lincoln". How do I get rid of the bracket, or get rid of the "University of Lincoln" bit. This is the first time I have tried to create a presentation in latex so any help is greatly appreciated. Thank-you in advance
This question already has an answer here: Our model is $Y=\beta_0+\beta_1X+U$. We know that $$\hat{\beta_0} = \beta_0 + \sum\limits_{n=1}^N c_nu_n \quad \text{ and }\quad \hat{\beta_1} = \beta_1 + \sum\limits_{n=1}^N k_nu_n \,,$$ where $$k_n = \frac{(x_n-\bar{X})}{D}\,, \quad D = \sum_{n=1}^N (x_n-\bar{X})^2\,,\quad\text{ and }\quad c_n = \Big[\frac{1}{N}-\bar{X}k_n\Big].$$ We are trying to use the assumptions of autocorrelation and homoskedasticity to derive the covariance between the OLS slope and intercept estimators, $\mathrm{Cov}[\hat{\beta_0},\hat{\beta_1}]$. I'm not sure how to go about this, or how to use these assumptions? I know how to begin using the definition of covariance, and then substituting to remove the coefficients, but where do I go from there? I'm not sure how to continue.
My question is why in the delayed choice quantum eraser experiment we don't see interference pattern at D0? While it looks like it is similar to the normal double slit experiments. The following image is from Wikipedia: In the usual double-slit experiment the photon is in the superposition of states corresponding to holes A and B. \begin{equation} \|DS\rangle=\frac{1}{\sqrt{2}}\|A\rangle+\frac{1}{\sqrt{2}}\|B\rangle \end{equation} However in this experiment the photon then hits the type II BBO crystal which converts it into two photons with perpendicular polarizations. So the state becomes, \begin{equation} \|DQE\rangle=\frac{1}{\sqrt{2}}\|A,V\rangle_s\otimes\|A,H\rangle_i+\frac{1}{\sqrt{2}}\|B,V\rangle_s\otimes\|B,H\rangle_i \end{equation} It's actually trickier because the BBO produces Bell state $\frac{1}{\sqrt{2}}\|V\rangle\otimes\|H\rangle+\frac{1}{\sqrt{2}}\|H\rangle\otimes\|V\rangle$ but in this experiment we distinguish the photons by polarization so this particular entanglement doesn't play role. Namely the prism sends, say, the one with vertical polarization ("signal") to $D_0$ and the one with horizontal polarization ("idler") to that complex optical system. The resulting state doesn't factorize into a product of the states of a signal and idler photons $\|DQE\rangle\neq\|\psi\rangle_s\otimes\|\psi\rangle_i$, two photons interfere as a pair but not individually. You may say that the signal photon is marked by a state of the idler photon. In the formalism that means that we can't assign a pure quantum state (a superposition of some basis quantum states) to a sole photon. Rather if one tries to ignore the idler photon then for the signal photon one gets a so-called mixed state - the $50%$ probability of the state $\|A,V\rangle_s$ and the $50%$ probability of the state $\|B,V\rangle_s$ without interference of these outcomes. How then one gets the interference pattern? We use beam splitters to convert idler photons into superpositions. \begin{equation} \|A,H\rangle_i\mapsto \frac{1}{\sqrt{2}}\|D_4\rangle_i+\frac{1}{\sqrt{2}}\Big(\frac{1}{\sqrt{2}}\|D_1\rangle_i+\frac{1}{\sqrt{2}}\|D_2\rangle_i\Big), \end{equation} \begin{equation} \|B,H\rangle_i\mapsto \frac{1}{\sqrt{2}}\|D_3\rangle_i+\frac{1}{\sqrt{2}}\Big(\frac{1}{\sqrt{2}}\|D_1\rangle_i-\frac{1}{\sqrt{2}}\|D_2\rangle_i\Big) \end{equation} Note that when the photon arrives at $D_1$ or $D_2$ it's impossible to determine the slit it came from, the state is the same. In this case we clear which-path information. And as result when we make this transformation in $\|DQE\rangle$ we get, \begin{align} \|DQE\rangle\mapsto\frac{1}{2}\|A,V\rangle_s\otimes\|D_4\rangle_i+\frac{1}{2}\|A,V\rangle_s\otimes\|D_3\rangle_i \\ +\frac{1}{2}\left(\frac{1}{\sqrt{2}}\|A,V\rangle_s+\frac{1}{\sqrt{2}}\|B,V\rangle_s\right)\otimes\|D_1\rangle_i\\ + \frac{1}{2}\left(\frac{1}{\sqrt{2}}\|A,V\rangle_s-\frac{1}{\sqrt{2}}\|B,V\rangle_s\right)\otimes\|D_2\rangle_i \end{align} you may see that if you measure the idler photon either at $D_1$ or $D_2$ you separate the term with the signal photon now being in superposition (note that the phase shifts are chosen to obtain two different superpositions) Of course to separate it you need to ignore the events when the idler photon hits $D_3$ or $D_4$. If you sum the statistics at $D_0$ for all four detectors you will recover the initial pattern as if you ignored the idler photon altogether.
This is a Test of Mathematics Solution Subjective 81 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance. Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta Find all possible real numbers $a,b,c,d,e$ which satisfy the following set of equations: $ \left\{ \begin{array}{ccc} 3a & = & ( b + c+ d)^3 \\ 3b & = & ( c + d +e ) ^3 \\ 3c & = & ( d + e +a )^3 \\ 3d & = & ( e + a +b )^3 \\ 3e &=& ( a + b +c)^3. \end{array}\right.$ Observing the symmetry and the cyclicity of the given set of equations it can be easily inferred that the real numbers $a,b,c,d$ and $e$ cannot be ordered i.e. one number cannot be greater or smaller than the other number else the system of equations will be inconsistent. This can be shown easily with the help of inequality. Without loss of generality, we can say that $a$ is greater or equal to them all, i.e., $a\geq b,c,d,e$. Thus we have $(d+e+a) \geq ( c+d+e)$ $=> (d+e+a)^3 \geq (c+d+e)^3$ $=> 3c \geq 3b$ $=> c \geq b$ As $a \geq d$, we also have, $(a+b+c) \geq ( b+c+d)$ $=> (a+b+c)^3 \geq (b+c+d)^3$ $=> 3e \geq 3a$ $=> e \geq a$ Thus we have $a=e$ Further calculations will show that $a=b=c=d=e$ is the only possible solution to this set of equations. Thus we are left to solve just one equation, i.e., $3a = (3a)^3$ which gives $a= -\frac{1}{3},0,\frac{1}{3}$ Thus the possible values of the 5-tuple $(a,b,c,d,e)$ are: $( -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3})$ $(0,0,0,0,0)$ $(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})$
16.2. Basics¶ This section summarizes basic tools from linear algebra, differentiation, and probability required to understand the contents in this book. We avoid details beyond the bare minimum to keep things streamlined and easily accessible. In some cases we simplify things to keep them easily accessible. For more background see e.g. the excellent Data Science 100 course at UC Berkeley. 16.2.1. Linear Algebra¶ This is a brief summary of vectors, matrices, operators, norms, eigenvectors, and eigenvalues. They’re needed since a significant part of deep learning revolves around manipulating matrices and vectors. For a much more in-depth introduction to linear algebra in Python see e.g. the Jupyter notebooks of Gilbert Strang’s MIT course on Linear Algebra. 16.2.1.1. Vectors¶ An $n$-dimensional vector $\mathbf{x}$ can be written as Here $x_1, \ldots, x_n$ are elements of the vector. To expressthat $\mathbf{x}$ is an $n$-dimensional vector with elementsfrom the set of real numbers, we write$\mathbf{x} \in \mathbb{R}^{n}$ or$\mathbf{x} \in \mathbb{R}^{n \times 1}$. Vectors satisfy thebasic operations of a vector space, namely that you can add themtogether and multiply them with scalars (in our case elementwise) andyou still get a vector back: assuming that$\mathbf{a}, \mathbf{b} \in \mathbb{R}^n$ and$\alpha \in \mathbb{R}$ we have that$\mathbf{a} + \mathbf{b} \in \mathbb{R}^n$ and$\alpha \cdot \mathbf{a} \in \mathbb{R}^n$. Furthermore theysatisfy the distributive law 16.2.1.2. Matrices¶ A matrix with $m$ rows and $n$ columns can be written as Here, $x_{ij}$ is the element in row $i \in \{1, \ldots m\}$ and column $j \in \{1, \ldots n\}$ in the matrix $\mathbf{X}$. Extending the vector notation we use $\mathbf{X} \in \mathbb{R}^{m \times n}$ to indicate that $\mathbf{X}$ is an $m \times n$ matrix. Given the above we could interpret vectors as $m \times 1$ dimensional matrices. Furthermore, matrices also form a vector space, i.e. we can multiply and add them just fine, as long as their dimensions match. 16.2.1.3. Operations¶ Assume that the elements in the $\mathbf{a}, \mathbf{b} \in \mathbb{R}^n$ are $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ respectively. The dot product (internal product) of vectors $\mathbf{a}$ and $\mathbf{b}$ is a scalar: Assume that we have two matrices with $m$ rows and $n$ columns $\mathbf{A}, \mathbf{B} \in \mathbb{R}^{m \times n}$: The transpose of a matrix $\mathbf{A}^\top \in \mathbb{R}^{n \times m}$ is a matrix with $n$ rows and $m$ columns which are formed by “flipping” over the original matrix as follows: To add two matrices of the same shape, we add them elementwise: We use the symbol $\odot$ to indicate the elementwisemultiplication of two matrices (in Matlab notation this is .*): Define a scalar $k$. Multiplication of scalars and matrices is also an elementwise multiplication: Other operations such as scalar and matrix addition, and division by an element are similar to the multiplication operation in the above equation. Calculating the square root or taking logarithms of a matrix are performed by calculating the square root or logarithm, respectively, of each element of the matrix to obtain a matrix with the same shape as the original matrix. Matrix multiplication is different from elementwise matrix multiplication. Assume $\mathbf{A}$ is a matrix with $m$ rows and $p$ columns and $\mathbf{B}$ is a matrix with $p$ rows and $n$ columns. The product (matrix multiplication) of these two matrices is denoted as The product is a matrix with $m$ rows and $n$ columns, with the element in row $i \in \{1, \ldots m\}$ and column $j \in \{1, \ldots n\}$ equal to 16.2.1.4. Norms¶ Assume that the elements in the $n$-dimensional vector $\mathbf{x}$ are $x_1, \ldots, x_n$. The $\ell_p$ norm of $\mathbf{x}$ is For example, the $\ell_1$ norm of $\mathbf{x}$ is the sum of the absolute values of the vector elements: The $\ell_2$ norm of $\mathbf{x}$ is the square root of the sum of the squares of the vector elements: We usually use $\\|\mathbf{x}\\|$ to refer to the $\ell_2$ norm of $\mathbf{x}$. Lastly, the $\ell_\infty$ norm of a vector is the limit of the above definition. This works out to Assume $\mathbf{X}$ is a matrix with $m$ rows and $n$ columns. The Frobenius norm of matrix $\mathbf{X}$ is the square root of the sum of the squares of the matrix elements: Here, $x_{ij}$ is the element of matrix $\mathbf{X}$ in row $i$ and column $j$. In other words, the Frobenius norm behaves as if it were an $\ell_2$ norm of a matrix-shaped vector. Note: sometimes the norms on vectors are also (erroneously) referredto as $L_p$ norms. However, the latter are norms on functions withsimilar structure. For instance, the $L_2$ norm of a function$f$ is given by $\\|f\\|_2^2 = \int \|f(x)\|^2 dx$. 16.2.1.5. Eigenvectors and Eigenvalues¶ Let $\mathbf{A}$ be a matrix with $n$ rows and $n$ columns. If $\lambda$ is a scalar and $\mathbf{v}$ is a non-zero $n$-dimensional vector with then $\mathbf{v}$ is called an eigenvector of matrix $\mathbf{A}$ and $\lambda$ is called an eigenvalue of $\mathbf{A}$ corresponding to $\mathbf{v}$. For symmetric matrices $\mathbf{A} = \mathbf{A}^\top$ there are exactly $n$ (linearly independent) eigenvector and eigenvalue pairs. 16.2.2. Differentials¶ This is a very brief primer on multivariate differential calculus. 16.2.2.1. Derivatives and Differentials¶ Assume the input and output of function $f: \mathbb{R} \rightarrow \mathbb{R}$ are both scalars. The derivative $f$ is defined as when the limit exists (and $f$ is said to be differentiable). Given $y = f(x)$, where $x$ and $y$ are the arguments and dependent variables of function $f$, respectively, the following derivative and differential expressions are equivalent: Here, the symbols $\text{D}$ and $\frac{\text{d}}{\text{d}x}$ are also called differential operators. Common differential calculations are $\text{D}C = 0$ ($C$ is a constant), $\text{D}x^n = nx^{n-1}$ ($n$ is a constant), $\text{D}e^x = e^x$, and $\text{D}\ln(x) = 1/x$. If functions $f$ and $g$ are both differentiable and $C$ is a constant, then If functions $y=f(u)$ and $u=g(x)$ are both differentiable, then the chain rule states that 16.2.2.2. Taylor Expansion¶ The Taylor expansion of function $f$ is given by the infinite sum if it exists. Here, $f^{(n)}$ is the $n$th derivative of $f$, and $n!$ is the factorial of $n$. For a sufficiently small number $\epsilon$, we can replace $x$ and $a$ with $x+\epsilon$ and $x$ respectively to obtain Because $\epsilon$ is sufficiently small, the above formula can be simplified to 16.2.2.3. Partial Derivatives¶ Let $u = f(x_1, x_2, \ldots, x_n)$ be a function with $n$ arguments. The partial derivative of $u$ with respect to its $i$th parameter $x_i$ is The following partial derivative expressions are equivalent: To calculate $\partial u/\partial x_i$, we simply treat $x_1, \ldots, x_{i-1}, x_{i+1}, \ldots, x_n$ as constants and calculate the derivative of $u$ with respect to $x_i$. 16.2.2.4. Gradients¶ Assume the input of function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is an $n$-dimensional vector $\mathbf{x} = [x_1, x_2, \ldots, x_n]^\top$ and the output is a scalar. The gradient of function $f(\mathbf{x})$ with respect to $\mathbf{x}$ is a vector of $n$ partial derivatives: To be concise, we sometimes use $\nabla f(\mathbf{x})$ to replace $\nabla_{\mathbf{x}} f(\mathbf{x})$. If $\mathbf{A}$ is a matrix with $m$ rows and $n$ columns, and $\mathbf{x}$ is an $n$-dimensional vector, the following identities hold: Similarly if $\mathbf{X}$ is a matrix, then 16.2.2.5. Hessian Matrices¶ Assume the input of function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is an $n$-dimensional vector $\mathbf{x} = [x_1, x_2, \ldots, x_n]^\top$ and the output is a scalar. If all second-order partial derivatives of function $f$ exist and are continuous, then the Hessian matrix $\mathbf{H}$ of $f$ is a matrix with $m$ rows and $n$ columns given by Here, the second-order partial derivative is evaluated 16.2.3. Probability¶ Finally, we will briefly introduce conditional probability, expectation and variance. 16.2.3.1. Conditional Probability¶ We will denote the probability of event $A$ and event $B$ as $\Pr(A)$ and $\Pr(B)$, respectively. The probability of the simultaneous occurrence of the two events is denoted as $\Pr(A \cap B)$ or $\Pr(A, B)$. In the figure above it is the shaded area. If $B$ has non-zero probability, the conditional probability of event $A$ given that $B$ has occurred is That is, If then $A$ and $B$ are said to be independent of each other. 16.2.3.2. Expectation and Variance¶ A random variable takes values that represent possible outcomes of an experiment. The expectation (or average) of the random variable $X$ is denoted as In many cases we want to measure by how much the random variable $x$ deviates from its expectation. This can be quantified by the variance Here the last equality follows from the linearity of expectation. 16.2.3.3. Uniform Distribution¶ Assume random variable $X$ obeys a uniform distribution over $[a, b]$, i.e. $X \sim U( a, b)$. In this case, random variable $X$ has the same probability of being any number between $a$ and $b$. 16.2.4. Summary¶ Vectors and matrics can be added and multiplied with rules similar to those of scalars. There are specialized norms for vectors and matrices, quite different from the (Euclidean) $\ell_2$ norm. Derivatives yield vectors and matrices when computing higher order terms. 16.2.5. Exercises¶ When traveling between two points in Manhattan, what is the distance that you need to cover in terms of the coordinates, i.e. in terms of avenues and streets? Can you travel diagonally? A square matrix is called antisymmetric if $\mathbf{A} = -\mathbf{A}^\top$. Show that you can decompose any square matrix into a symmetric and an antisymmetric matrix. Write out a permutation in matrix form. Find the gradient of the function $f(\mathbf{x}) = 3x_1^2 + 5e^{x_2}$. What is the gradient of the function $f(\mathbf{x}) = \\|\mathbf{x}\\|_2$? Prove that $\Pr(A \cup B) \leq \Pr(A) + \Pr(B)$. When do you have an equality?
Finally, I felt I had something about Physics that I wanted to write about. The i \epsilon terms sitting in the propagator of a QFT, in the Lippmann-Schwinger equation and in Chapter 4 of Peskin and Schroeder have been bothering a couple students including me at the department for a while now. I am not qualified enough in Quantum Field Theory to make any serious comments on this, but I just had some thoughts regarding the i \epsilon. They may be wrong, and I request readers to correct me if there are mistakes, or if they have something to add to this. At first look, the i \epsilon looked like some bizzare mathematical trick, put in by hand, to give meaning to integrals. “Oh, this integral diverges, but we want it to converge, so we just throw in an i \epsilon”. A lot of us were pretty dissatisfied with this. Also, there was this question too — there are these i \epsilon terms in (a) the propagator, (b) in the Lippmann-Schwinger equation, (c) in Peskin-Schroeder’s derivation relating the interacting ground state with the free-theory ground state, and (d) in the derivation of the path integral formalism from canonical QM — are they all stuck there for the same purpose? The first time i \epsilon bothered us was in Peskin-Schroeder’s derivation of a relation between the free-field ground state and the interacting-field ground state, where he says “let us take time to infinity in a slightly imaginary direction”. Now, the question was, why should time become imaginary? A long argument on VoIP with Naveen Sharma was adjourned with this: “The T \to \infty( 1 + i \epsilon) is a mathematical trick to supress the contribution of all other states and solve for the interacting ground state in terms of the free-field ground state.” Then came Prof. Weinberg’s notes on the Lippmann-Schwinger equation. As he explained in class, and as was explained in his notes, the right choice of ± i \epsilon in the Lippmann-Schwinger Green’s function fixes whether we are choosing in-states or out-states. i.e. states with the +i \epsilon in the Green’s function’s denominator satisfy the condition that they look like free particles in the asymptotic past, while states with the -i \epsilon look like free particles in the asymptotic future. A similar argument, with a bit more detail, is presented in his book “The Quantum Theory of Fields” volume 1, in Chapter 3. He also has made a reference to B. A. Lippmann and J. Schwinger, Phys. Rev. Vol 79, No. 3 (1950).. So I briefly looked at the Lippmann-Schwinger paper, where they actually derive the equation. Then they make a comment: “simulating the cessation of interaction, arising from the separation of component parts of the system, by an adiabatic decrease in the interaction strength as t → ± \infty. The latter can be represented by a factor exp(-\epsilon \|t\|/ħ) where \epsilon is arbitrarily small.” Aha! So that epsilon came from an adiabatic (slow) decrease of interaction strength! But why are we forced to kill that interaction “by hand”? [PS: Loophole — I still don’t know the adiabatic theorem] I don’t know enough, but I’d ordinarily expect a “factor killing the interaction” to sit in the interaction Hamiltonian rather than outside it (see eqn 1.51 of the Lippmann-Schwinger paper). At least now, the \epsilon factor had some physical meaning — it came from the adiabatic killing of the interaction, rather than being just some “pole-pushing technology”. More came today. There is the same epsilon in the Fourier transform of a \theta function (Heaviside step function). One may write: This is something that I was supposed to know from Electrical Engineering, but we used to “throw away” the epsilon — it didn’t matter much there I guess. Really, it’s just the Fourier transform of a decaying exponential (which every electrical engineer, from IIT Madras at least, would know) with the characteristic length taken to infinity. And then, today we worked out the Feynman propagator for the scalar field. I should’ve known this long back, but I learned it today, that really, the epsilon in the propagator comes from the \theta function’s Fourier transform. So it seems like the epsilons — at least in (a), (b) and (c) — are present to impose causality. And then, I learned something more today: An i\epsilon is going to make the Hamiltonian non-Hermitian (eg: See the Green’s function in the Lippmann-Schwinger: it’s effectively adding a small non-Hermitian component to the Hamiltonian). And we see that Hermiticity of the Hamiltonian is required for time-reversal symmetry: Thus, if my logic is right, the i \epsilon is necessary to break the time-reversal-invariance in the system so that we can talk about an “in” state and distinguish that from an “out” state. Of course, this is unphysical in most situations as far as we know, so we do away with the \epsilon at the end. Now, this brings me to a couple of questions: Does that mean the weak interaction has a non-Hermitian Lagrangian density? [Need to check; sounds like a No] We’re always time ordering in quantum mechanics. A naïve look gives me the impression that time ordering breaks time-reversal invariance. Then why are our theories time-invariant? Anyway, so much for an epsilon!
X Search Filters Format Subjects Library Location Language Publication Date Click on a bar to filter by decade Slide to change publication date range 1. Measurement of the branching fractions of the decays D + → K − K + K + , D + → π − π + K + and D s + → π − K + K Journal of High Energy Physics, ISSN 1126-6708, 03/2019, Volume 2019, Issue 3, pp. 1 - 24 Journal Article Journal of High Energy Physics, ISSN 1126-6708, 8/2018, Volume 2018, Issue 8, pp. 1 - 36 A binned Dalitz plot analysis of B ± → DK ± decays, with D → K S 0 π + π − and D → K S 0 K + K −, is used to perform a measurement of the CP-violating... B physics \| CKM angle gamma \| CP violation \| Hadron-Hadron scattering (experiments) \| Flavor physics \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Luminosity \| Uncertainty \| Large Hadron Collider \| Particle collisions \| Decay B physics \| CKM angle gamma \| CP violation \| Hadron-Hadron scattering (experiments) \| Flavor physics \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Luminosity \| Uncertainty \| Large Hadron Collider \| Particle collisions \| Decay Journal Article 3. Search for CP violation using triple product asymmetries in Λ b 0 → pK − π + π −, Λ b 0 → pK − K + K − and Ξ b 0 → pK − K − π + decays Journal of High Energy Physics, ISSN 1126-6708, 8/2018, Volume 2018, Issue 8, pp. 1 - 27 A search for CP and P violation using triple-product asymmetries is performed with Λ b 0 → pK − π + π −, Λ b 0 → pK − K + K − and Ξ b 0 → pK − K − π + decays.... B physics \| CP violation \| Hadron-Hadron scattering (experiments) \| Flavor physics \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Asymmetry \| Decay B physics \| CP violation \| Hadron-Hadron scattering (experiments) \| Flavor physics \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Asymmetry \| Decay Journal Article 4. Measurement of the branching fractions of the decays $D^+\rightarrow K^-K ^+K^+$, $D^+\rightarrow \pi^-\pi^+K^+$ and $D^+_s\rightarrow \pi^-K^+K 10/2018 JHEP 03 (2019) 176 The branching fractions of the doubly Cabibbo-suppressed decays $D^+\rightarrow K^-K^+K^+$, $D^+\rightarrow \pi^-\pi^+K^+$ and... Physics - High Energy Physics - Experiment Physics - High Energy Physics - Experiment Journal Article Journal of High Energy Physics, ISSN 1029-8479, 4/2019, Volume 2019, Issue 4, pp. 1 - 36 The resonant structure of the doubly Cabibbo-suppressed decay D +→K − K + K + is studied for the first time. The measurement is based on a sample of... Particle and resonance production \| Charm physics \| Hadron-Hadron scattering (experiments) \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| SYSTEM \| HIGH-STATISTICS \| PHYSICS, PARTICLES & FIELDS \| Luminosity \| Resonance scattering \| Tensors \| Amplitudes \| Decay \| Physics - High Energy Physics - Experiment Particle and resonance production \| Charm physics \| Hadron-Hadron scattering (experiments) \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| SYSTEM \| HIGH-STATISTICS \| PHYSICS, PARTICLES & FIELDS \| Luminosity \| Resonance scattering \| Tensors \| Amplitudes \| Decay \| Physics - High Energy Physics - Experiment Journal Article 6. Amplitude analysis of the B s 0 → K ∗ 0 K ¯ ∗ 0 $$ {B}_{(s)}^0\to {K}^{\ast 0}{\overline{K}}^{\ast 0} $$ decays and measurement of the branching fraction of the B 0 → K ∗ 0 K ¯ ∗ 0 $$ {B}^0\to {K}^{\ast 0}{\overline{K}}^{\ast 0} $$ decay Journal of High Energy Physics, ISSN 1029-8479, 7/2019, Volume 2019, Issue 7, pp. 1 - 31 The B 0 → K ∗ 0 K ¯ ∗ 0 $$ {B}^0\to {K}^{\ast 0}{\overline{K}}^{\ast 0} $$ and B s 0 → K ∗ 0 K ¯ ∗ 0 $$ {B}_s^0\to {K}^{\ast 0}{\overline{K}}^{\ast 0} $$... B physics \| Hadron-Hadron scattering (experiments) \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory B physics \| Hadron-Hadron scattering (experiments) \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory Journal Article 7. Erratum to: Measurement of the CKM angle γ using B ± → DK ± with D → K S 0 π + π −, K S 0 K + K − decays Journal of High Energy Physics, ISSN 1029-8479, 10/2018, Volume 2018, Issue 10, pp. 1 - 7 The B + and B − labels of the confidence regions in figure 10 of the original paper [1] were erroneously swapped. The corrected figure is shown in figure 10. Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory Journal Article Journal of High Energy Physics, ISSN 1126-6708, 3/2018, Volume 2018, Issue 3, pp. 1 - 23 The ratios of the branching fractions of the decays Λ c + → pπ − π +, Λ c + → pK − K +, and Λ c + → pπ − K + with respect to the Cabibbo-favoured Λ c + →... Spectroscopy \| Branching fraction \| Charm physics \| Hadron-Hadron scattering (experiments) \| Flavor physics \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Luminosity \| Uncertainty \| Large Hadron Collider \| Particle collisions \| Nuclear and particle physics. Atomic energy. Radioactivity \| LHCb \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Spectroscopy \| Branching fraction \| Charm physics \| Hadron-Hadron scattering (experiments) \| Flavor physics \| Quantum Physics \| Quantum Field Theories, String Theory \| Classical and Quantum Gravitation, Relativity Theory \| Physics \| Elementary Particles, Quantum Field Theory \| Luminosity \| Uncertainty \| Large Hadron Collider \| Particle collisions \| Nuclear and particle physics. Atomic energy. Radioactivity \| LHCb \| PHYSICS OF ELEMENTARY PARTICLES AND FIELDS Journal Article
Algebra and Algebraic Geometry Seminar Spring 2018 The seminar meets on Fridays at 2:25 pm in room B113. Here is the schedule for the previous semester. Contents 1 Algebra and Algebraic Geometry Mailing List 2 Spring 2018 Schedule 3 Abstracts Algebra and 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). Spring 2018 Schedule Abstracts Tasos Moulinos Derived Azumaya Algebras and Twisted K-theory Topological K-theory of dg-categories is a localizing invariant of dg-categories over [math] \mathbb{C} [/math] taking values in the [math] \infty [/math]-category of [math] KU [/math]-modules. In this talk I describe a relative version of this construction; namely for [math]X[/math] a quasi-compact, quasi-separated [math] \mathbb{C} [/math]-scheme I construct a functor valued in the [math] \infty [/math]-category of sheaves of spectra on [math] X(\mathbb{C}) [/math], the complex points of [math]X[/math]. For inputs of the form [math]\operatorname{Perf}(X, A)[/math] where [math]A[/math] is an Azumaya algebra over [math]X[/math], I characterize the values of this functor in terms of the twisted topological K-theory of [math] X(\mathbb{C}) [/math]. From this I deduce a certain decomposition, for [math] X [/math] a finite CW-complex equipped with a bundle [math] P [/math] of projective spaces over [math] X [/math], of [math] KU(P) [/math] in terms of the twisted topological K-theory of [math] X [/math] ; this is a topological analogue of a result of Quillen’s on the algebraic K-theory of Severi-Brauer schemes. Roman Fedorov A conjecture of Grothendieck and Serre on principal bundles in mixedcharacteristic Let G be a reductive group scheme over a regular local ring R. An old conjecture of Grothendieck and Serre predicts that such a principal bundle is trivial, if it is trivial over the fraction field of R. The conjecture has recently been proved in the "geometric" case, that is, when R contains a field. In the remaining case, the difficulty comes from the fact, that the situation is more rigid, so that a certain general position argument does not go through. I will discuss this difficulty and a way to circumvent it to obtain some partial results. Juliette Bruce Asymptotic Syzygies in the Semi-Ample Setting In recent years numerous conjectures have been made describing the asymptotic Betti numbers of a projective variety as the embedding line bundle becomes more ample. I will discuss recent work attempting to generalize these conjectures to the case when the embedding line bundle becomes more semi-ample. (Recall a line bundle is semi-ample if a sufficiently large multiple is base point free.) In particular, I will discuss how the monomial methods of Ein, Erman, and Lazarsfeld used to prove non-vanishing results on projective space can be extended to prove non-vanishing results for products of projective space. Andrei Caldararu Computing a categorical Gromov-Witten invariant In his 2005 paper "The Gromov-Witten potential associated to a TCFT" Kevin Costello described a procedure for recovering an analogue of the Gromov-Witten potential directly out of a cyclic A-inifinity algebra or category. Applying his construction to the derived category of sheaves of a complex projective variety provides a definition of higher genus B-model Gromov-Witten invariants, independent of the BCOV formalism. This has several advantages. Due to the categorical invariance of these invariants, categorical mirror symmetry automatically implies classical mirror symmetry to all genera. Also, the construction can be applied to other categories like categories of matrix factorization, giving a direct definition of FJRW invariants, for example. In my talk I shall describe the details of the computation (joint with Junwu Tu) of the invariant, at g=1, n=1, for elliptic curves. The result agrees with the predictions of mirror symmetry, matching classical calculations of Dijkgraaf. It is the first non-trivial computation of a categorical Gromov-Witten invariant. Aron Heleodoro Normally ordered tensor product of Tate objects and decomposition of higher adeles In this talk I will introduce the different tensor products that exist on Tate objects over vector spaces (or more generally coherent sheaves on a given scheme). As an application, I will explain how these can be used to describe higher adeles on an n-dimensional smooth scheme. Both Tate objects and higher adeles would be introduced in the talk. (This is based on joint work with Braunling, Groechenig and Wolfson.) Moisés Herradón Cueto Local type of difference equations The theory of algebraic differential equations on the affine line is very well-understood. In particular, there is a well-defined notion of restricting a D-module to a formal neighborhood of a point, and these restrictions are completely described by two vector spaces, called vanishing cycles and nearby cycles, and some maps between them. We give an analogous notion of "restriction to a formal disk" for difference equations that satisfies several desirable properties: first of all, a difference module can be recovered uniquely from its restriction to the complement of a point and its restriction to a formal disk around this point. Secondly, it gives rise to a local Mellin transform, which relates vanishing cycles of a difference module to nearby cycles of its Mellin transform. Since the Mellin transform of a difference module is a D-module, the Mellin transform brings us back to the familiar world of D-modules. Eva Elduque On the signed Euler characteristic property for subvarieties of Abelian varieties Franecki and Kapranov proved that the Euler characteristic of a perverse sheaf on a semi-abelian variety is non-negative. This result has several purely topological consequences regarding the sign of the (topological and intersection homology) Euler characteristic of a subvariety of an abelian variety, and it is natural to attempt to justify them by more elementary methods. In this talk, we'll explore the geometric tools used recently in the proof of the signed Euler characteristic property. Joint work with Christian Geske and Laurentiu Maxim. Harrison Chen Equivariant localization for periodic cyclic homology and derived loop spaces There is a close relationship between derived loop spaces, a geometric object, and (periodic) cyclic homology, a categorical invariant. In this talk we will discuss this relationship and how it leads to an equivariant localization result, which has an intuitive interpretation using the language of derived loop spaces. We discuss ongoing generalizations and potential applications in computing the periodic cyclic homology of categories of equivariant (coherent) sheaves on algebraic varieties. Phil Tosteson Stability in the homology of Deligne-Mumford compactifications The space [math]\bar M_{g,n}[/math] is a compactification of the moduli space algebraic curves with marked points, obtained by allowing smooth curves to degenerate to nodal ones. We will talk about how the asymptotic behavior of its homology, [math]H_i(\bar M_{g,n})[/math], for [math]n \gg 0[/math] can be studied using the representation theory of the category of finite sets and surjections. Wei Ho Noncommutative Galois closures and moduli problems In this talk, we will discuss the notion of a Galois closure for a possibly noncommutative algebra. We will explain how this problem is related to certain moduli problems involving genus one curves and torsors for Jacobians of higher genus curves. This is joint work with Matt Satriano. Daniel Corey Initial degenerations of Grassmannians Let Gr_0(d,n) denote the open subvariety of the Grassmannian Gr(d,n) consisting of d-1 dimensional subspaces of P^{n-1} meeting the toric boundary transversely. We prove that Gr_0(3,7) is schoen in the sense that all of its initial degenerations are smooth. The main technique we will use is to express the initial degenerations of Gr_0(3,7) as a inverse limit of thin Schubert cells. We use this to show that the Chow quotient of Gr(3,7) by the maximal torus H in GL(7) is the log canonical compactification of the moduli space of 7 lines in P^2 in linear general position. Alena Pirutka Irrationality problems Let X be a projective algebraic variety, the set of solutions of a system of homogeneous polynomial equations. Several classical notions describe how ``unconstrained the solutions are, i.e., how close X is to projective space: there are notions of rational, unirational and stably rational varieties. Over the field of complex numbers, these notions coincide in dimensions one and two, but diverge in higherdimensions. In the last years, many new classes of non stably rational varieties were found, using a specialization technique, introduced by C. Voisin. This method also allowed to prove that the rationality is not a deformation invariant in smooth and projective families of complex varieties: this is a joint work with B. Hassett and Y. Tschinkel. In my talk I will describe classical examples, as well as the recent progress around these rationality questions. Nero Budur Homotopy of singular algebraic varieties By work of Simpson, Kollár, Kapovich, every finitely generated group can be the fundamental group of an irreducible complex algebraic variety with only normal crossings and Whitney umbrellas as singularities. In contrast, we show that if a complex algebraic variety has no weight zero 1-cohomology classes, then the fundamental group is strongly restricted: the irreducible components of the cohomology jump loci of rank one local systems containing the constant sheaf are complex affine tori. Same for links and Milnor fibers. This is joint work with Marcel Rubió. Alexander Yom Din Drinfeld-Gaitsgory functor and contragradient duality for (g,K)-modules Drinfeld suggested the definition of a certain endo-functor, called the pseudo-identity functor (or the Drinfeld-Gaitsgory functor), on the category of D-modules on an algebraic stack. We extend this definition to an arbitrary DG category, and show that if certain finiteness conditions are satisfied, this functor is the inverse of the Serre functor. We show that the pseudo-identity functor for (g,K)-modules is isomorphic to the composition of cohomological and contragredient dualities, which is parallel to an analogous assertion for p-adic groups. In this talk I will try to discuss some of these results and around them. This is joint work with Dennis Gaitsgory. John Lesieutre Given a dominant rational self-map f : X -->X of a variety defined over a number field, the first dynamical degree $\lambda_1(f)$ and the arithmetic degree $\alpha_f(P)$ are two measures of the complexity of the dynamics of f: the first measures the rate of growth of the degrees of the iterates f^n, while the second measures the rate of growth of the heights of the iterates f^n(P) for a point P. A conjecture of Kawaguchi and Silverman predicts that if P has Zariski-dense orbit, then these two quantities coincide. I will prove this conjecture in several higher-dimensional settings, including for all automorphisms of hyper-K\"ahler varieties. This is joint work with Matthew Satriano.
$$ \sigma_{E/K}=\frac{\log{\|N_{K/Q}\Delta_{E/K}\|}}{\log{\|N_{K/Q} f_{E/K}}\|}$$ Given $ \varepsilon >0$ there are only finitely many $ E/K$ with $ \sigma_{E/K}\geq 6+\varepsilon $. In particular, $ \sigma_{E/K}$ is bounded. $\Delta_{E/K}$ is the minimal discriminant. Szpiro's conjecture implies abc with a bit higher exponent over the integers. Over number fields, the uniform abc conjecture depends on the discriminant of the number field. Maybe this is because every integer can be arbitrary large power in some number field. To make $d$ a $k$-th power, work with defining polynomial $x^k -d$. So start with EC over the rationals with discriminant $\Delta$ and compute $\sigma_{E/K}$ for $K$ with defining polynomial $x^k - \Delta$ for $k$ sufficiently large. Unless the minimal discriminant takes care of this case, this will give unbounded Szpiro ratio over some number fields. Why the Szpiro conjecture over number fields doesn't depend on the discriminant of the number field while the uniform abc conjecture depends? Over $K$ with defining polynomial $x^{16} + 22384$ take $E/K : y^2 = x^3 + 7x -1$. Not sure if the discriminant is minimal, but the norm is $2^{64} \cdot 1399^{16}$ while the norm of the conductor is $2^{10} \cdot 1399$ giving Szpiro ratio $11.305664847$ Some experiments in degree 14 and sage's global_minimal_model() suggest the global minimal model preserves the high powers in the discriminant: a4,a6= -2 -1 global minimal model= Elliptic Curve defined by y^2 + (1/14w^12-1/28w^10+1/7w^8+1/4w^7-1/14w^6+2/7w^4-1/7w^2-3/7)y = x^3 + (243/56w^13+17/16w^12-297/28w^11+51/2w^10-561/14w^9+111/2w^8-1581/28w^7+81/2w^6+255/28w^5-99w^4+1530/7w^3-374w^2+3330/7w-527)x + (-531/112w^13-9419/56w^12+2945/7w^11-40829/56w^10+55889/56w^9-31119/28w^8+47939/56w^7-1209/28w^6-10470/7w^5+53021/14w^4-90715/14w^3+125729/14w^2-69170/7w+107855/14) over Number Field in w with defining polynomial x^14 - 80 Delta= 2^32 5^14 f= 2^32 * 5 d_q= 2^4 * 5 ratio= 1.87946880927217a4,a6= -2 1 global minimal model= Elliptic Curve defined by y^2 + (1/14w^12-1/28w^10+1/7w^8+1/4w^7-1/14w^6+2/7w^4-1/7w^2-3/7)y = x^3 + (243/56w^13+17/16w^12-297/28w^11+51/2w^10-561/14w^9+111/2w^8-1581/28w^7+81/2w^6+255/28w^5-99w^4+1530/7w^3-374w^2+3330/7w-527)x + (533/112w^13+4709/28w^12-5891/14w^11+40819/56w^10-55887/56w^9+31125/28w^8-47933/56w^7+1185/28w^6+10460/7w^5-26504/7w^4+90725/14w^3-125767/14w^2+69150/7w-107857/14) over Number Field in w with defining polynomial x^14 - 80 Delta= 2^32 5^14 f= 2^18 * 5 d_q= 2^4 * 5 ratio= 3.17425556409935 The last two examples give the abc triples $c_4^3-c_6^2=w^{14}$ Curves of Szpiro ratios $ > 66$ are in this question and they appear to contradict boundedness of the ratio.
I'm very confused by the account on fluctuations near the equilibrium in chapter 12 of Landau's book on statistical physics. To be brief, the kernel of my doubt is that he states that if you have a observable $X$ such that equilibrium is attained at $X=0,$ then the probability of having a fluctuation of size $x$ is given by the exponential of the entropy evaluated at $x.$ I wonder, what is the mathematical definition of this notion of entropy? I don't even understand that. Below is my attempt at a possible formalization of those notions, but feel free to completely ignore it if you can answer my question right away! Possible interpretation (surely wrong, but I'm sure that the "solution" to my doubt has to be something similar to this) : Let $M$ be the space of configurations and $TM$ the phase space. Define a classical observable $X : TM \rightarrow \mathbb{R}. $ If $\mathcal{C}(TM)$ is a suitable space of probability density functions (the details not important at all for the argument here), define an entropy functional $I : \mathcal{C}(TM) \rightarrow \mathbb{R}$ in some way. I am guessing most times the appropiate deifinition would be $I[f]=-\displaystyle \int f(q,p) \text{log}f(q,p) dq dp,$ and I would like to focus specifically on this entropy, but of course other possibilities could/should be considered! We denote by $\overline{X}^f$ the average value of $X$ with respect to the probability given by $f,$ that is, $\overline{X}^f= \displaystyle \int X(q,p) f(q,p) dq dp$ Define an entropy function $S(x) : \mathbb{R} \rightarrow \mathbb{R}$ as: $$S(x) := \text{max} \left( I[f] \ : \ \overline{X}^f=x \right),$$ that is, $S(x)$ is the value of the entropy functional over the function maximizing it while satisfying the restriction that the average value of $X$ is $x.$ Of course, the existence and uniqueness of such a maximizer is not granted and is an important problem, but for the time being I will assume this is the case, since in the boltzmannian/maxwellian and closely related relevant situations existence and uniqueness actually hold. Therefore, the fact that $S$ is a well-defined function is highly-non trivial, but we won't care about this now. Call also $f_x : TM \rightarrow \mathbb{R}$ to each maximizer: $$f_x(q,p) = \text{argmax} \left( I[f] \ : \ \overline{X}^f=x \right). $$ In this context, the identity that I would like want to prove should be: $$\int_{ \{ (q,p) : X(q,p)=x \} } f_0(q,p) dq dp= c \text{e}^{S(x)},$$ where $c$ is some (normalizing) constant.
For finding potential at a point due to a +ve charge $(q)$, we find work done to move a unit +ve charge $(q_o)$ from infinity to that point in the presence of +ve charge $(q)$ Since both charges being +ve, the force would be repulsive and hence while bringing unit +ve charge $(q_o)$ from infinity to that position, the path would be against force field. Thus the work done by force field should be negative. But the following calculation/derivation in my book shows that work done is positive: $$W$$ $$=\int_\infty^r F.dr$$ $$=-\int_\infty^r F dr\\$$ (since path is against force field) $$=-\int_\infty^r \frac{1}{r^{2}}dr\\$$ $$=-\left( {-\frac{1}{r}} \Big \|_{\infty}^{r}\right)$$ $$={\frac{1}{r}} \Big \|_{\infty}^{r}$$ $$=\frac{1}{r}-\frac{1}{\infty}$$ $$=\frac{1}{r}$$ Why is this contradiction? Where am I (or my book) wrong?
I am new to finance so I apologize if my question is really basic (which it probably is). If this is not the right "stackexchange" group for this, kindly refer me to the right one. Let's say you own an asset and you want to cross-hedge using futures on a related asset. I'm going to establish a terminology here: $S,F$ is the spot price (per unit) of the owned asset, the futures price (per unit) respectively, and $\Delta S, \Delta F$ the corresponding changes during the life of the hedge. $Q_A, Q_F$ is the size (in units) of the position being hedged/one futures respectively. $V_A, V_F$ is the value of your hedged position, one futures contract respectively. $\sigma_S, \sigma_F$ is the standard deviation of $\Delta S, \Delta F$ respectively. $\hat{\sigma}_S, \hat{\sigma}_F$ is the standard deviation of the percent one-day change in $S,F$ respectively. $\rho$ is the correlation between $\Delta S, \Delta F$ $\hat{\rho}$ is the correlation between percent one-day changes in $S,F$. $h^, N^$ is the minimum variance hedge ratio and optimal number of contracts (without tailing). $\hat h, \hat N$ are same as above but "with tailing". According to the book and many other standard books, if we ignore daily settlements, (equivalently if we hedge using forward contracts) the minimum variance hedge ratio, that is the hedge ratio that minimizes the variance of the hedged portfolio, is:$$h^=\rho \frac{\sigma_S}{\sigma_F}$$and the corresponding optimum number of contracts is:$$N^=h^\frac{Q_A}{Q_F}$$Here's where it gets confusing (for me at least): in page $62$ of the cited book, the author claims, that because of the daily settlements procedure, is we use futures instead of forwards and want to be accurate, we should actually use:$$\hat h=\hat{\rho}\frac{\hat{\sigma}_S}{\hat{\sigma}_F}$$in place of $h^$ and:$$\hat N= \hat{h}\frac{V_A}{V_F}$$in place of $N^$. However, in other sources online, for instance here, $\hat N=h^\frac{V_A}{V_F}=N^*\frac{S}{F}$. So my first question is: Which one is the right formula, and most importantly why? A secondary question: If the formula in the book is to be believed, is it possible to find (estimate) $\hat h, \hat N$ from $\Delta S, \Delta F, \rho, \sigma_S, \sigma_F$, and if yes, how so? For instance, consider part 4) of the following question (taken from Hull's book "Options, futures and other derivatives", edition 10): A trader owns $55,000$ units of a particular asset and decides to hedge the value of her position with futures contracts on another related asset. Each futures contract is on $5,000$ units. The spot price of the asset owned is $28\$$ and the standard deviation of the change in this price over the life of the hedge is $0.43\$$. The futures price of the related asset is $27\$$ and the standard deviation of this over the life of the asset is $0.40\$$. The coefficient of correlation between the spot price change and the futures price change is $0.95$. 1. Find the minimum variance hedge ratio. 2. Should the hedger take a long or a short position? 3. What is the optimal number of contracts when adjustments for daily settlements are not considered? 4. How can the daily settlement of futures contracts be taken into account? I have no idea how to even answer part $4$. Thank you all for your time.
In my computer science education, I increasingly notice that most discrete problems are NP-complete (at least), whereas optimizing continuous problems is almost always easily achievable, usually through gradient techniques. Are there exceptions to this? An example that I love is the problem where, given distinct $a_1, a_2, \ldots, a_n \in \mathbb{N}$, decide if: $$\int_{-\pi}^{\pi} \cos(a_1 z) \cos(a_2 z) \ldots \cos(a_n z) \, dz \ne 0$$ This at first seems like a continuous problem to evaluate this integral, however it is easy to show that this integral is not zero iff there exists a balanced partition of the set $\{a_1, \ldots, a_n\}$, so this integral problem is actually NP-complete. Of course, I encourage playing around with some numerical tools to convince yourself that most (if not all) numerical tricks to evaluate this integral are doomed to failure once $n$ gets large enough. There are many continuous problems of the form "test whether this combinatorial input can be realized as a geometric structure" that are complete for the existential theory of the reals, a continuous analogue of NP. In particular, this implies that these problems are NP-hard rather than polynomially solvable. Examples include testing whether a given graph is a unit distance graph, whether a given graph can be drawn in the plane with straight line segment edges and at most a given number of crossings, or whether a given pseudoline arrangement can be stretched to form a line arrangement. There are other continuous problems that are even harder: for instance, finding a shortest path among polyhedral obstacles in 3d is PSPACE-complete (Canny & Reif, FOCS'87). While this doesn't exactly answer your original question, it's a (conjectural) example of a sort of philosophical counterpoint: a problem where the presentation is discrete but all of the hardness comes from the 'continuous' aspect of the problem. The problem is the Sum of Square Roots problem: given two sets of integers $A=\{a_1, a_2, \ldots, a_m\}$ and $B=\{b_1, b_2, \ldots, b_n\}$, is $\sum_{i=1}^m \sqrt{a_i}\leq\sum_{j=1}^n\sqrt{b_j}$? (There are other formulations, but this is the one I prefer.) While it's not known for certain to be hard, it's widely suspected that it may be NP-hard and may in fact be outside of NP (there are, as noted in the comments, excellent reasons to believe that it's not NP-complete); the only containment known to date is several levels higher up the polynomial hierarchy. Obviously the presentation of this problem is as discrete as can be — a set of integers and a yes/no question about them — but the challenge arises because while computing square roots to any specified precision is an easy problem, they may need to be computed to high (potentially superpolynomial) accuracy to settle the inequality one way or the other. This is a 'discrete' problem that crops up in a surprising number of optimization contexts and helps contribute to their own complexity. Discrete problems typically tend to be harder (e.g. LP vs. ILP) but it's not the discreteness itself that's the problem... it's how the constraints affect how you can search your domain. For example, you may think that optimizing a polynomial is something that you can do efficiently, but deciding convexity of quartics (degree-4 polynomials) is NP-hard. Which means even if you already have the optimum somehow, simply proving that you are at the optimum is already NP-hard. Although for some popular problems, it is indeed true, I think both assumptions are - depending on what you define as an optimization problem - not true. First some definitions: most optimization problems are not part of NP. For instance for the Knapsack problem: one cannot exploit non-determinism to construct the most valuable bag, simple because the different non-deterministic branches have no shared memory. NP is also defined as "polynomially verifiable" (verifying a certificate) [1, p. 34]. In this case the certificate is for instance a bag: a bitstring where if the i-th bit is set, it implies the i-th item is part of the bag. You can indeed check in polynomial time if such bag is more valuable than a given threshold (this is the decision variant), but you cannot - as far as we know - based on a single bag, (a polynomial number of bags), decide if that bag is the most valuable of all possible bags. That's a vital difference between for instance NP and EXP: in EXP, you can enumerate over all possible bags and do bookkeeping about which bag is the best one. The decision variant of the optimization problems is in some cases part of NP, one needs to make a clear distinction between the maximization flavor and the decision flavor. In the decision flavor, the question is: " Given an optimization problem, and a utility bound, is there a solution with a utility greater than or equal to that bound" (or slightly modified for a minimization problem). I also assume that by NP you mean the (hypothetical) part of NP that is not part of P. If P=NP, of course NP-complete still exists, but it will be equal to P (only coincides with P for some notions of reduction, like polynomial-time many-one reductions by @AndrásSalamon), which is not that impressive (and would reduce the " gap" you are stating in your question). I increasingly notice that most discrete problems are NP-complete. Now that we have sorted that out: there are a lot of optimization problems that are in P: shortest path problem, maximum flow problem (for integral capacities), minimum spanning tree and maximum matching. Although these problems may look "trivial to solve" to you, these are still optimization problems, and in many cases the construction (and prove of correctness) is not that easy. So the claim doesn't hold all discrete problems are NP-complete. Given P is not equal to NP, these problems thus can't be NP-complete. One can furthermore walk through the polynomial hierarchy, this hierarchy provides a way to construct a decision problem that is in $\Sigma^P_i$, but given a decision problem, you can (nearly) always construct an optimization problem that is at least as hard (if the optimization variant was less hard, one could solve the decision variant by calling the optimization variant first, and then make a decision based on the result of that algorithm). Whereas optimizing continuous problems is almost always easily achievable. A popular continuous problem that is NP-hard is quadratic programming. In quadratic programming, one is looking for a vector $\vec{x}$ such that: $$\dfrac{\vec{x}^T\cdot Q\cdot\vec{x}}{2}+\vec{c}^T\cdot\vec{x}$$ is minimized satisfying: $$A\cdot\vec{x}\leq\vec{b}$$ Actually Linear programming has long been considered NP-hard as well, but with very well performing heuristics (the Simplex method). Karmarkar's algorithm is however in P. From the moment the optimization problem deals with non-convex objects, in general it will be hard - if not impossible - to find an efficient algorithm. Bibliography
I am currently reading about binary cubic forms and cubic number fields (mainly about using binary cubic forms with integer coefficients to parametrize orders in the cubic field) and I thought it might be good to do some computations to understand how this theory works by taking a concrete example. Given a cubic field I want to find the corresponding binary cubic form associated with it or rather with the ring of integer of $K$ should I say? Suppose we take $$K=\mathbb{Q}(\sqrt[3]{2})\supset\mathbb{Q}$$ Since $$x^{3}-2\in\mathbb{Q}[x]$$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with $p=2$ we get $$[K:\mathbb{Q}]=3$$ The ring of integers of $K$ is $$\mathcal{O}_{k}=\mathbb{Z}[\alpha]=\mathbb{Z}[\sqrt[3]{2}]$$ with integral basis $\{1,\alpha,\alpha^2\}$ hence $$\mathbb{Z}[\sqrt[3]{2}]=\{a+b\sqrt[3]{2}+c\sqrt[3]{2^2}\|a,b,c\in\mathbb{Z}\}$$ In the next step I would like to find (if possible) the corresponding binary cubic form. Will this form correspond to the field of to the ring? I follow Belabas and Cohen ''Binary cubic forms and cubic number fields'': Let $K$ be number field defined by a root $\theta$ of the polynomial $x^3+px^2+qx+r$ with $p,q,r \in\mathbb{Z}$ such that there exists an integral basis of the form $(1,\theta,(\theta^2+t\theta+u)/f)$ with $t,u,f\in \mathbb{Z}$ and $f=[\mathbb{Z}_{k}:\mathbb{Z}[\theta]]$ Then we choose $\alpha=\theta$ and $\beta=(\theta^2+t\theta+u)/f)$ we have explicitly: $$\begin{split} F_{B}(x,y)&=((t^3-2t^2p+t(q+p^2)+r-pq)/f^2)x^3\\&+((-3t^2+4tp-(p^2+q))/f))x^2y\\ &+(3t-2p)xy^2-fy^3\end{split}$$ In case of $x^{3}-2$ we have $\begin{cases}q=0\\r=-2\\p=0\end{cases}$ And $f=[\mathbb{Z}_{k}:\mathbb{Z}[\theta]]=1$ Plugging everything in I get something like this $F_{B}(x,y)=-2x^3-y^3$ Does this work? Or is it completely incorrect? $f=[\mathbb{Z}_{k}:\mathbb{Z}[\theta]]$ shouldn't this expression give index form? (i.e binary cubic form?) Thank you.
The optimization toolbox of Matlab does not contain an MIQP (Mixed Integer Quadratic Programming) solver. So when they discuss a cardinality constrained portfolio problem (this means only a certain number of assets can be in the portfolio), an interesting cutting plane technique is used instead. This method will solve a series of linear MIP models instead of attacking the MIQP model directly. Let’s first formulate the problem as proper MIQP model. First here is how the data is prepared: Data Quadratic Data Trick The main deviation from the original model is that we don’t store the variance-covariance matrix $Q$ as a full dense matrix but as an upper-triangular matrix (it includes the diagonal). We multiply the off-diagonal elements by two, so that the expression $\sum_{i,j} x_i Q_{i,j} x_j $ will yield exactly the same result for any $x$. Basically we replace expressions like $x_1 Q_{1,2} x_2 +x_2 Q_{2,1} x_1$ by $ 2 x_1 Q_{1,2} x_2$. The sole purpose of this trick is to have fewer non-zero elements in the quadratic form $x^TQx$. This really can help for larger problems (this looks mainly a GAMS problem: GAMS is slower in generating the model when the nonzero count of the Q matrix increases; it does not make much difference for the QP solver). The model itself looks like: MIQP Model Modeling trick In the original model the variable $K$ was not present and the min- and max-fraction constraints were formulated as: This duplicates a summation. Our formulation adds a single variable (with a lower- and upper-bound) but we get rid of an equation that includes a summation. As a result we reduce the number of non-zero elements in the constraint matrix. \[\begin{align} &\sum_i \delta_i \ge \text{minAssets}\\ &\sum_i \delta_i \le \text{maxAssets} \end{align}\] Cutting Plane Algorithm In the cutting plane method we replace the quadratic objective by \[\min\> \lambda \sum_{i,j} x_i Q_{i,j} x_j - \sum_i \mu_i x_i \] where $K$ is the current iteration or cycle, and $x^_k$ is the optimal MIP solution from the k-th iteration. This is essentially a Taylor approximation. \[\begin{align} \min\>& \lambda y - \sum_i \mu_i x_i\\ & y \ge - {x^_k}^T Q x^_k + 2{x^_k}^T Q x \> \text{for $k=1,...,K-1$} \end{align}\] Here is our version of the algorithm: In this second model we replaced the original objective by the new linear objective and the cuts. The equation cut is indexed by a dynamic set cuts that grows during the iteration loop. We start the iteration scheme with an empty set cuts. Note that in the calculation of $\beta$ we changed the expression from $2{x^_k}^T Q$ to ${x^_k}^T (Q+Q^T)$. This is to compensate for the upper-triangularization trick we used before. The results are: miqp 0.036039 iter1 -0.002430 iter2 0.013922 iter3 0.025362 iter4 0.034862 iter5 0.035138 iter6 0.035646 iter7 0.035731 iter8 0.035776 iter9 0.035790 iter10 0.035828 We see the objectives converge towards the optimal MIQP objective. References Mathworks, Mixed-Integer Quadratic Programming Portfolio Optimization. J. E. Kelley, Jr. " The Cutting-Plane Method for Solving Convex Programs." J. Soc. Indust. Appl. Math. Vol. 8, No. 4, pp. 703-712, December, 1960.
I will introduce my problem with an example. Say you are designing an exam, which consists of a certain set of $n$ independent questions (that the candidates can get either right or wrong). You want to decide on a score to give to each of the questions, with the rule being that candidates with total score above a certain threshold will pass, and the others will fail. In fact, you are very thorough about this, and you have envisioned all the possible $2^n$ results, and decided for each of them whether a candidate with this performance should pass or fail. So you have a Boolean function $f : \{0, 1\}^n \to \{0, 1\}$ that indicates whether the candidate should pass or fail depending on their exact answers. Of course this function should be monotone: when getting a set of questions right makes you pass, getting any superset right must make you pass as well. Can you decide on scores (positive real numbers) to give to the questions, and on a threshold, so that your function $f$ is exactly captured by the rule "a candidate passes if the sum of scores for the correct questions is above the threshold"? (Of course the threshold can be taken to be 1 without loss of generality, up to multiplying the scores by a constant.) Formally: Is there a characterization of the monotone Boolean functions $f: \{0, 1\}^n \to \{0, 1\}$ for which there exist $w_1, \ldots, w_n \in \mathbb{R}_+$ such that for all $v \in \{0, 1\}^n$, we have $f(v) = 1$ iff $\sum_i w_i v_i \geq 1$? It is not so hard to see that not all functions can be thus represented. For instance the function $(x_1 \wedge x_2) \vee (x_3 \wedge x_4)$ cannot: as $(1, 1, 0, 0)$ is accepted we must have $w_1 + w_2 \geq 1$, so one of $w_1, w_2$ must be $\geq 1/2$, and likewise for $w_3, w_4$. Now, if it is, e.g., $w_1$ and $w_3$, we have a contradiction because $w_1 + w_3 \geq 1$ but $(1, 0, 1, 0)$ is rejected; the other cases are analogous. This looks to me like a very natural problem, so my main question is to know under which name this has been studied. Asking for a "characterization" is vague, of course; my question is to know whether the class of functions that can be represented in this way has a name, what is known about the complexity of testing whether an input function belongs to it (given as a formula, or as a circuit), etc. Of course one can think of many variations on this theme. For instance, on real exams, questions are not independent, but there is a DAG on questions indicating the dependence, and candidates can only answer a question if all prerequisites have been answered. The condition on the monotone functions could then be restricted to valuations in $\{0, 1\}^n$ that satisfy the dependencies, and the question would be to determine whether an input function can be thus captured given an input DAG on the variables. One could also think of variants where the scores are $k$-tuples for fixed $k$ (summed pointwise, and compared pointwise to a threshold vector), which can capture more functions than $k = 1$. Alternatively you could want to capture more expressive functions which are not Boolean but go to a totally ordered domain, with different thresholds that should indicate your position in the domain. Last, I'm not sure about what would happen if you allowed negative scores (so you could drop the monotone restriction about the functions). (Note: What made me wonder about this is the Google Code Jam selection round, where candidates are selected if they reach a certain score threshold, and the scores of problems are presumably carefully designed to reflect what sets of problems are deemed sufficient to get selected. Code Jam has a dependency structure on the questions, with some "large input" questions that cannot be solved unless you have solved the "small input" one first.)
You may notice that subtraction can be seen as a special case of addition, i.e.$$a-b=a+(-b)$$Thus by commutative law,$$a-b=a+(-b)=(-b)+a=-b+a$$It is certainly not $b-a$. Similarly, division is as a special case of multiplication, i.e.$$\frac{a}{b}=a\cdot(\frac{1}{b})$$Where $\frac{1}{b}$ means the inverse element of $b$ with respect to multiplication. Therefore,$$\frac{a}{b}\cdot{c}=a\cdot\frac{1}{b}\cdot{c}$$However,$$\frac{a}{b\cdot{c}}=a\cdot\frac{1}{b\cdot{c}}=a\cdot\frac{1}{b}\cdot\frac{1}{c}$$And they are certainly not equal. Edited:Since you are motivated to pursue advanced math, I have a little idea to share. I assume you have intuitively understood the arithmetic about nature numbers. It is based on this simple idea: the one-to-one correspondence between (abstract) nature numbers and (concrete) things. Then you easy find that: $(+, 1)$ The commutative law of addition$$a+b=b+a$$ $(+, 2)$ The associative law of addition$$a+(b+c)=(a+b)+c$$ $(\times, 1)$ The commutative law of multiplication$$a\cdot{b}=b\cdot{a}$$ $(\times, 2)$ The associative law of multiplication$$a\cdot(b\cdot{c})=(a\cdot{b})\cdot{c}$$ $(\times, 3)$ The identity element of multiplication, i.e. $1$$$a \cdot 1=1 \cdot a=a$$ $(+, \times)$ the distribution law of multiplication to addition$$a \cdot (b+c)=a\cdot {b}+a \cdot {c}$$ A possible intuitive approach has been shown by Paul Sinclair. And you also have noticed that addition and multiplication can be defined as recursive operation. That is,$$a+b:=a+\underbrace{1+1+1+...+1}_{b \text(terms)}$$And$$a\cdot{b}:=\underbrace{a\cdot{a}...\cdot{a}}_{b \text(terms)}$$ You also have a nature idea to introduce the inverse operation, i.e. substitution and division. But you may find that $3-5$ is illegal in this stage (it has no definition), and neither is $\frac{3}{5}$. But that is not hard for you, just noticed that you can define $\frac{3}{5}$ as a ratio: There certainly can be a thing $k$ such that $k\cdot{5}=3$, then we can denote $k$ by $\frac{3}{5}$. Here come the rational numbers! And you may just find $(\times, 4)$ The reverse element of multiplication$$\text{For every number $a$ there is a number $b$ such that $a\cdot{b}=b\cdot{a}$}$$ And we can easily find that by definition, $b=\frac{1}{a}$.(the "number" here means nature number and positive rational number.) However you are not satisfied. Sometimes you need to find a way to denote "nothing", so you need $(+, 3)$ The identity element of addition, i.e. $0$$$\text{For all $a$, $a+0=0+a=a$}$$ As far as I'm concerned, historically, in a relative long time, $0$ or similar notations were initially but for this kind of convenience. A similar idea for convenience is negative number, it was originally used to express debt. With introducing negative numbers, we can derive: $(+, 4)$ The reverse element of addition$$\text{For every number $a$ there is a number $b$ such that $a+b=b+a=0$}$$ And we can also easily find that by definition, $b=-a$. Indeed, It took people a lot of time to accept the philosophical aspect of "nothingness", or to understand how debt times debt will be income (this was not philosophical confusion initially, but a result of wrong metaphor; however, it finally became one, but that is another story.) But for mathematics, the most horrible thing is division by 0, it is certainly not legal. There is no help for it, so we have to ban it. Indeed, this is the only difference between addition and multiplication in abstract sense. If you consider about a more general case such as real numbers, the original, recursive idea will failed, or, at least, no more intuitive. Thus you have to define addition and multiplication in a more abstract way, that is, introducing the axioms of addition and multiplication, i.e. $(+,1)-(+,4),(\times,1)-{(\times,4)},(+,\times)$ I've mentioned before. If you compare $(+,1)-(+,4)$ with $(\times,1)-(\times,4)$, you will find they are nearly the same. A set equipped with two operation satisfying those axioms will be the field which trb456 mentioned. (To say it clearly, the result of those operations must always within that set, and the identity element of addition can't be the same with that of multiplication.) We need this term for other reason: because there are a lot of other structures satisfying this definition. And modern algebra, though I am totally not familiar with it yet, seems like a study of mathematical structures. This story is obviously not the real history of numbers and their arithmetic, and I just want to introduce a half-intuitive approach to understand those things. I omitted a lot of details, or maybe I just got things wrong. But I hope you enjoy it.
A while ago I asked about versions of Vopenka's principle for logics other than first order. Unfortunately, there doesn't seem to be much there; most logics yield the same notion, while "big" logics like $\mathcal{L}_{\infty\omega}$ yield versions of VP which are easily disprovable. However, going back I noticed that there is still some potentially interesting stuff around the $\mathcal{L}_{\infty\omega}$ stuff. Say that a class $\mathbb{K}$ of structures has the weak infinitary VP property if, whenever $C\subseteq \mathbb{K}$ is a proper class, there are distinct $A, B\in C$ with $A\equiv_{\infty\omega}B$. The strong infinitary VP property is the same, except that we ask instead for a $j: A\rightarrow B$ which is an $\mathcal{L}_{\infty\omega}$-elementary embedding, instead of merely $\equiv_{\infty\omega}$. Obviously the strong version is more in the spirit of the original VP; however, the weak version looks interesting too (and more manageable - see below). Now, it's easy to come up with an example of a class without the weak infinitary VP property, via the following fact: $$\mbox{$A\equiv_{\infty\omega}B$ iff $A\cong B$ in some forcing extension.}$$ Since forcing preserves well-foundedness, no non-isomorphic ordinals can become isomorphic after forcing, so $ON$ does not have the weak infinitary VP property. However, we can also cook up examples which do have it! Let's restrict attention to $\mathbb{K}$ of the form $\{M: M\models T\}$ for some complete first-order theory $T$ in a countable language with no finite models. Then: Since $\aleph_0$-categoricity is $\Pi^1_2$, it's absolute under forcing. Letting $A, B\in\mathbb{K}$ be of any cardinality, they are therefore isomorphic in any forcing extension where they are both countable - hence $A\equiv_{\infty\omega}B$. So $\mathbb{K}$ has the weak infinitary VP. Similarly, $\aleph_1$-categorical theories have the weak infinitary VP. The proof is the same, except now the claim that $\aleph_1$-categoricity is absolute is less obvious. On the face of it, "$T$ is $\aleph_1$-categorical" is difficult to express; however, it's equivalent to $T$ being $\omega$-stable and having no Vaughtian pairs, which is absolute. My question is whether there are any more interesting examples of classes with the weak infinitary VP property, or any interesting examples of classes with the strong infinitary VP property. In particular: Question.Is there an unstable theory with the strong infinitary VP property? Note that one thing making the strong infinitary VP property more complicated is that $\mathcal{L}_{\infty\omega}$-elementary embeddability doesn't seem to have as nice a characterization as $\mathcal{L}_{\infty\omega}$-elementary equivalence - at least, none that I know of.
As my question states, I want to calculate the Fourier transform $F(q)$ of a radial function $f(r)$ (defined on $[0,\infty)$ and which decays like an exponential $\exp(-Ar+b)$ at large $r$) as accurately as possible in Fortran. The function values come from a data file (which I can easily interpolate through cubic interpolation for example and extrapolate since the behavior at large $r$ is known). I'm using the "physics" definition of the Fourier transform in 3D, which gives (because $f$ is radial): $$ \int d\mathbf{r}\,f(\mathbf{r})\exp(i\mathbf{q}\mathbf{r}) = 4\pi\int\limits_0^\infty r^2\frac{\sin(qr)}{qr}f(r)\,dr $$ I first tried to calculate this integral for some chosen values of $q$ by using Gauss-Legendre quadrature, by generating some 60 or 100 abscissas and weights via the NAG routine D01BCF (D01BCF link). In the case of Gauss Legendre quadrature, the problem is to choose the interval $[0,B]$ on which to integrate. While the function $f$ loses 4 to 5 orders of magnitude from $r=10$ to $r=20$ (example), the choice of $B$ as a strong influence on the result of the calculation... When I compared the result I get to a "nearly exact" calculation (made with MATLAB but with a very long computation time), I saw that in fact this was only valid for small values of $q$ (of the order of 5, when I have to deal with values as large as 150). A Gauss-Laguerre quadrature does not give any better result, probably because of the oscillatory part of the integrand. I then tried to compute this Fourier transform for some given values of $q$ with the routine D01ASF (D01ASF link). It is a "one-dimensional quadrature, adaptive, semi-infinite interval, weight function $\cos(ωx)$ or $\sin(ωx)$", which is exactly what I need. The results are quite convincing for $q$ up to 80 or 100 if I input absolute error tolerances of 10E-5. Problems are: I would need to go at larger $q$, and the Fourier transform $F(q)$ oscillates with a magnitude of ~10E-6 at such $q$'s. Lowering the tolerance to 10E-5 already takes some time and even makes the whole thing to output some error message from the subroutine so I don't know if 10E-6 would be feasible. I'm thus currently wondering if trying to calculate this Fourier transform with FFT wouldn't be a good idea? The problems I face are that I don't know how to calculate radial wave functions with FFT (and also that I don't even know how to use FFT properly either since the definition of the transform is not even the same (exponent sign and argument) and that I never used it before). Would you have ideas? EDIT: this comes from https://stackoverflow.com/q/39127340/2320757, I've been advised to put this here on stackoverflow. EDIT: I tried by FFT (using the routine C06FAF from NAG library). It works quite well up to some large values of $q$. The problem I face is that there is always some constant normalising factor to account for. I don't get why. This normalising factor evolves with the number $N$ of points used in the mesh. It has the for of a power law: Normalising Factor $F = N^(-0.5) \exp(9.9)$ approximately (see figure where the black line is the "exact" Fourier Transform and the green, magenta, blue, red and yellow lines are the FFT calculated for different values of $N$).
I'm stuck on a few Vector homework problems. I don't quite understand how to write vectors A+B and A-B for questions 1b and 2b. I tried starting with calculating the magnitude for vector A+B on question 1b and then followed by finding theta, but I'm not sure if that's what I'm supposed to do... I have a 4D array of dimension ##100\text{x}100\text{x}3\text{x}3##. I am working with `Python Numpy. This 4D array is used since I want to manipulate 2D array of dimensions ##100\text{x}100## for the following equation (it allows to compute the ##(i,j)## element ##F_{ij}## of Fisher matrix) ... From my interpretation of this problem (image attached), the force applied to the point charge is equal and opposite to the repulsive Coulomb force that that point charge is experiencing due to the presence of the other point charge so that the point charge may be moved at a constant velocity. I... I'm having a little trouble with this :We have ##(\alpha\vec{a})\cdot b = \alpha(\vec{a}\cdot\vec{b})## but shouldn't it be ##\|\alpha\|(\vec{a}\cdot\vec{b})## instead since ##\|\|\alpha \vec{a}\|\|=\|\alpha\|.\|\|\vec{a}\|\|## ?##(\alpha\vec{a})\cdot b = \|\|\alpha\vec{a}\|\|.\|\|\vec{b}\|\|.\cos\theta =... This is more of a general question, but I've encountered this kind of exercises a lot in my current preperations for my exam:There are two cases but the excercise is pretty much the same:Compute$$(1) \space \operatorname{div}\vec{A}(\vec{r}) \qquad , where \thinspace... 1. Homework StatementGiven that vector a = (1, 2, -5), b = (-12, 41, 75) and c = a + 2b, explain why (without doing any calculations whatsoever) the value of a•(b x c) = 02. Homework EquationsNo specific equations, as the question asks for the value without making any calculations. This... I'm trying to work through a scattering calculation in the Peskin QFT textbook in chapter 5, specifically getting equation 5.10. They take two bracketed terms4[p'^{\mu}p^{\nu}+p'^{\nu}p^{\mu}-g^{\mu\nu}(p \cdot p'+m_e^2)]and4[k_{\mu}k'_{\nu}+k_{\nu}k'_{\mu}-g_{\mu\nu}(k \cdot... Start with a circle of radius r and center c. Inside of that circle is an arbitrary point p. Given an arbitrary normalized direction vector d, I need to find the radius and center the circle that (1) intersects p, (2) is tangent with the circle centered at c, and (3) has its center lying on the... Hello all!I usually don't like to ask for help... But this is the first week of courses and I'm already stumped on a homework question...1. Homework StatementSo the question states: Find the work by the force F = x^i + xy^j. If the object starts from the origin (0,0), moves along the x... Hi all,Suppose we have vectors coming in order as A, B and then C (but A must be deleted before C comes in). Then how to get the dot product between A and C? It is allowed to store some calculations of A before deleting elements of A, for example, we could store norm of A, dot(A, B) and etc... In 'Introduction to Electrodynamics' by Griffiths, in the section of explaining the Gradient operator, it is stated a theorem of partial derivatives is:$$ dT = (\delta T / \delta x) \delta x + (\delta T / \delta y) \delta y + (\delta T / \delta z) \delta z $$Further he goes onto say:$$ dT =... 1. Homework StatementMy problem is:For the logarithmic spiral R(t) = (e^t cost, e^t sint), show that the angle between R(t) and the tangent vector at R(t) is independent of t.2. Homework EquationsN/A3. The Attempt at a SolutionThe tangent vector is just the vector that you get when... Proof: If either x or y is zero, then the inequality \|x · y\| ≤ \| x \| \| y \| is trivially correct because both sides are zero.If neither x nor y is zero, then by x · y = \| x \| \| y \| cos θ,\|x · y\|=\| x \| \| y \| cos θ \| ≤ \| x \| \| y \|since -1 ≤ cos θ ≤ 1How valid is this a proof of the... <<Mentor note: Missing template due to originally being posted elsewhere>>Hello everyone.I have the following problem:Determine the angles of a triangle where two sides of a triangle are formed by the vectorsA = 3i -4j -k and B=4i -j + 3kI thought that I would find the third side being... Hello everyone,i'm new to the forum so hope it is the right place for my question :)i need to know the angle between two triangular surfaces, the easiest way would be extract the normal for each surface(u,v) and then using the dot product we can easily compute the cosine for the angle i'm... Can someone help me relearn finding the angle between two forces when solving for work of each forces (gravity, tension, fF, normal)?I remember that cos(90°-α) = sin(α) but what I don't understand is when the angle in between is "90°-α" or when it's just "α". I tried doing this on my own and... This is my question:What is the largest m such that there exist v1, ... ,vm ∈ ℝn such that for all i and j, if 1 ≤ i < j ≤ m, then ≤ vi⋅vj = 0I found a couple of solutions online.http://mathoverflow.net/questions/31436/largest-number-of-vectors-with-pairwise-negative-dot-product... 1. Homework Statement2. Homework Equationsp.q+p.r3. The Attempt at a SolutionI've expanded p.(q+r) to give p.q+p.r. The magnitude of p is 3, and since ABE is an equilateral triangle, the magnitude of q is also 3, right? So then p.q=9, but the answer scheme states that p.q=4.5.I'm... Does anyone know which formula is used or how to arrive at the righthand side of the equation below, which is the dot product of del and rhoa 2nd order tensor(V V).. represents dot productand X a vector quantityThis problem is in connection with transforming cauchy's equation in terms of... 1. Homework StatementThe diagram shows a box with parallel faces. Two of the faces are trapezoids and four of the faces are rectangles. The vectors A, B, and C lie along the edges as shown, and their magnitudes are the lengths of the edges. Define the necessary additional symbols and prove... 1. The problem is:( a x b )⋅[( b x c ) x ( c x a )] = [a,b,c]^2 = [ a⋅( b x c )]^2I am supposed to solve this using index notation.... and I am having some problems.2. Homework Equations : I guess I just don't understand the finer points of index notation. Every time I think I am getting... I'm currently writing my EP on various physical equations including Maxwell's equations, and I had to justify using the dot product of the normal unit vector and the electric field in the integral version. However, I can't think of a reason for not using trigonometry as opposed to the... 1. Homework StatementI will just post an image of the problemand here's the link if the above is too small: http://i.imgur.com/JB6FEog.png?12. Homework Equations3. The Attempt at a SolutionI've been playing with it, but I can't figure out a good way to "grip" this problem.I can... 1. Homework StatementFind the coordinates of the vector of the height of the parallelogram formed by vectors a={1, 2, 1} and b={2, -1, 0}2. Homework EquationsA=\|axb\|, A=\|a\|h3. The Attempt at a SolutionI can find the intensity of the vector h i.e the length of the height, but not its... 1. Homework StatementUse vectors and the dot product to prove that the midpoint of the hypotenuse of a right triangle is equidistant to all three vertices.2. Homework EquationsI know the dot product is A⋅B = \|A\|\|B\|cosΘ ...... or .... A1B1 + A2B2 + A3B3 ... + AnBnI know the... Hi - just working through my text (studying by correspondence) on Del operator - so Curl, div etc. Came across some identities parts of which which have me confused. what does it mean when a vector is shown as multiplying something - but without dot or cross? For example F(∇.G) or ∇(F.G) or...
I have been trying to solve an exercise on nuclear chemistry, but I was not sure whether I should ask this on physics or chemistry but I think that if chemistry had an exercise based on this then chemistry experts might be having the way to overcome this problem. So here is what my nuclear chemistry question says: One gram atom of $ _{79}^{198}\mathrm {Au} $ ( having a half life of 65 hours) decays by $\beta$-emission to produce stable nuclide of Hg. How much Hg will be present after 260 hours? The thing which is making me confused is $\beta$ decay. I think I have a formula for this question but that $\beta$ thing is creating a problem. I would like to make that equation public. Here it is: $$N_t=N_0\cdot\exp(-\lambda\cdot t)$$ where, $\lambda$ is the radioactive decay constant $N_0$ is initial number of nuclides present $N_t$ is the co. of parent nuclides present at time $t$
I'm now making a documentation of a small LaTeX package with documentation library from the tcolorbox package, and having difficulty in printing sample codes without any syntax highlighting. According to G. Poore's answer, we can set the language as text to turn off the coloring, which can be specified via /tcb/minted language key. Thus, I tried to redefine the corresponding value like the following MWE (the first two are just for comparison). \documentclass{article}\usepackage[all]{tcolorbox}\tcbset{listing engine=minted}\begin{document}\begin{tcblisting}{minted language=text} \begin{equation} \int_{-\infty}^{\infty} \exp (-x^2) \, dx = \sqrt{\pi} \end{equation}\end{tcblisting}\begin{dispExample} \begin{equation} \int_{-\infty}^{\infty} \exp (-x^2) \, dx = \sqrt{\pi} \end{equation}\end{dispExample}\tcbset{ docexample/.add style={}{ fontlower=\normalsize, minted language=text, documentation minted options={ autogobble=true, fontsize=\small } }}\begin{dispExample} \begin{equation} \int_{-\infty}^{\infty} \exp (-x^2) \, dx = \sqrt{\pi} \end{equation}\end{dispExample}\end{document} Typesetting this with pdflatex -shell-escape gave me As you can see, the option minted language=text seems to be invalid in the third box while other key-value settings (e.g., fontlower=\normalsize) works.Also, minted language=text is valid in the normal tcblisting environment. How can I .add style the minted language key to the docexample style? What am I doing wrong? I could have directly asked this question at Prof. Thomas F. Sturm's GitHub repository, but I use TeX.SX expecting more people will have a look. Update: Thanks to @egreg's answer, I found that my issue lies in the combination of tcolorbox and (after some investigation, the same behaviour was confirmed with other monospaced fonts).Consider this MWE: newtxtt \documentclass{article}\usepackage[all]{tcolorbox}\usepackage{newtxtt}\tcbset{ listing engine=minted, docexample/.add style={}{ fontlower=\normalsize, minted language=latex, documentation minted options={ autogobble=true, fontsize=\small, style=bw, } }}\begin{document}\begin{dispExample} \begin{equation} \int_{-\infty}^{\infty} \exp (-x^2) \, dx = \sqrt{\pi} \end{equation}\end{dispExample}\end{document} This gives control sequences in boldface, which I don't like, although style=bw is specified. Monospaced fonts that reproduce this problem include Courier ( \usepackage{courier}), DejaVu ( \usepackage{dejavu}), Fira ( \usepackage{firamono}), Incondolata ( \usepackage{zi4}), and Source Code Pro ( \usepackage{sourcecodepro}), while Latin Modern and Droid don't trigger the problem. Yet Another Update: The second part of this question now has its own thread.
This is a Test of Mathematics Solution Subjective 69 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance. Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta Suppose that the three equations , $ {bx^2} – {2cx + a = 0} $, and – all have only positive roots. Show that $ a =b = c$. If possible let $a$, $b$, $c$ are not all equal $ {ax^2} – {2bx + c = 0} $ , $ {bx^2} – {2cx + a = 0}$, ${cx^2} – {2ax + b = 0}$ all have only positive roots. So all of $a$, $b$, $c$ cannot be its same sign as discriminant > 0 for three equations ( ${b^2} > {ac}$, ${a^2} > {bc}$, ${c^2} >{ab}$ ). Without loss of generality we can assume $ a > b > c $ or $a > c > b$ as the equations are cyclic. Now we know, $\frac{\pm{\sqrt{b^2-ac}}}{a} > 0 $, $\frac{\pm{\sqrt{a^2-bc}}}{c} > 0$, $\frac{\pm{\sqrt{c^2-ab}}}{b} > 0 $. Now there 2 possibilities either b, c both are positive or one of b, c is positive. If b, c both are positive then,$ \frac{b-{\sqrt{b^2-ac}}}{a} < 0 $ [ not possible ]. If one of b, c is positive then, $ \frac{c-{\sqrt{c^2-ab}}}{b} < 0 $ [ not possible ] So a, b, c have to be equal.
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 ...
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 ...
Let's say that a map $f: V \rightarrow W$ between finite-dimensional real vector spaces is convex-linear if $f(\lambda x + (1-\lambda)y) = \lambda f(x) + (1-\lambda)f(y)$ for all $\lambda \in [0,1]$. Let's say that a map $f: V \rightarrow W$ between finite-dimensional real vector spaces is affine if $f(\lambda x + (1-\lambda)y) = \lambda f(x) + (1-\lambda)f(y)$ for all $\lambda \in \mathbb{R}$. From the definition, it seems that the requirement of being convex-linear is weaker than the requirement of being affine. However, I can't think of an example of a map which is convex-linear but not affine, but I also can't prove that convex-linearity implies affinity. Can someone show me an example of a convex-linear map which is not affine? Or tell me how to prove that every convex-linear map is affine? Or give me an appropriate reference? EDIT: With the intuition of Qiaochu Yuan's comment in mind, I've come up with the following proof: Claim: Every convex-linear map is affine. Proof: Let $f$ be convex-linear. For $\lambda \in [0, 1]$, We have that $f(\lambda x + (1-\lambda) y) = \lambda f(x) + (1-\lambda) f(y)$. For $\lambda \notin [0,1]$, we can assume without loss of generality that $\lambda < 0$ (in the other case where $\lambda > 1$, we can interchange the role of $x$ and $y$). We can write \begin{align} f(y) = f\left( \underbrace{\frac{1}{1-\lambda}}_{\in [0,1]}(\lambda x + (1-\lambda) y) + \left( 1 - \frac{1}{1-\lambda} \right) x \right). \label{bla} \end{align} By the convex-linearity of $f$, this reduces to \begin{align} &f(y) = \frac{1}{1-\lambda} f(\lambda x + (1-\lambda) y) + \left( 1-\frac{1}{1-\lambda} \right) f(x) \end{align} which in turn can be reduced to \begin{align} f(\lambda x + (1-\lambda x)) = \lambda f(x) + (1-\lambda) f(y). \end{align}
I have a distance on the space of probability measures on $[0,2]$. It is defined as such for two probability measures $\mu_1$ and $\mu_2$ : $d_p(\mu_1,\mu_2) := \sum_{k=0}^p ( \mathbb{E}[X_1 ^k]- \mathbb{E}[X_1 ^k] ) ^2$ where $X_1$ and $X_2$ are random variables whose laws are $\mu_1$ and $\mu_2$ respectively. Let $d_{TV}$ be the total variation distance on the same space. I'm considering distributions lying in a set $\Sigma$ with : $\Sigma = \{ \alpha U(s,t) + (1-\alpha)W_1(R) \lvert (s,t,R,\alpha) \in [0,\epsilon_1]^2 \times [0,\epsilon_2] \times [0,1] \} $ where : $ \epsilon_1, \epsilon_2 >0 $, $U(s,t)$ is the uniform distribution on $[s,t]$ and $W_1(R)$ is the semi-circular law of parameter $R>0$ and centered on 1. I want to know if there exists a constant $C>0$ (possibly for $p$ big enough) such that $\forall (\mu_1,\mu_2) \in \Sigma^2, \ d_{TV}(\mu_1,\mu_2) \leqslant C d_p(\mu_1,\mu_2) $. Is is possible to possible to find such a constant when $\mu_1$ and $\mu_2$ are both uniform distributions for instance but I do not know if it's true in the general case. Do you have an idea or reference for this problem? Any suggestion is welcome! Thanks.
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building) It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore) In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of @TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies \|\| x ( t ) - 0 \|\| < \varepsilon.$$The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $... "If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}\|f(z)\| \leq C_K \\|f\\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed? Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2 Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$ Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight. hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$ for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$ I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything. I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set).theorem ... The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious. (but seriously, the best tactic is over powered...) Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field? Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement? "Infinity exists" comes to mind as a potential candidate statement. Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system @Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity but so far failed Put it in another way, an equivalent formulation of that (possibly open) problem is: > Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object? If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite. My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science... O great, given a transcendental $s$, computing $\min_P(\|P(s)\|)$ is a knapsack problem hmm... By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as: $$P(x) = \prod_{k=0}^n (x - \lambda_k)$$ If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic Thus given $s$ transcendental, to minimise $\|P(s)\|$ will be given as follows: The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases. In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that0<\|x−pq\|<1qn.{\displaystyle 0<\left\|x-{\frac {p}... Do these still exist if the axiom of infinity is blown up? Hmmm... Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum: $$\sum_{k=1}^M \frac{1}{b^{k!}}$$ The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework There's this theorem in Spivak's book of Calculus:Theorem 7Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and$$f'... and neither Rolle nor mean value theorem need the axiom of choice Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set > are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion
What is the mathematical notation for rounding a given number to the nearest integer? So like a mix between the floor and the ceiling function. I have seen $\lfloor x \rceil$. It must have been in the context of math olympiads, so I can't point to a book that uses it. Wikipedia suggest this notation, among others: nearest integer function. Personally, I would prefer $[x]$, being a cleaner mix of $\lfloor x \rfloor$ and $\lceil x \rceil$. But I've seen this notation being used for the floor function. Especially in older texts, say, pre-TeX era. You could also do something like $\mathrm{nint}(x)$, but in formulas that could be cumbersome. See also the remarks at Mathworld. I have seen the notation $[x]$. However, that is some times used as the floor function when TeX is unavailable, or the author is unfamiliar with it (I'm sure there are plenty of examples on this site, for instance). The safest bet is to say something along the lines of Let $[x]$ mean the integer closest to $x$ (rounding up for half-integer values). or Let $[\phantom x]$ denote the standard rounding function. That is, explicitly defining the notation yourself, so that anyone who reads your text knows exactly what you're talking about. If you do this, you are of course entirely free to "invent" your own notation (within reason) for this if there is some other notation you prefer. Whatever notation you use (punctured dusk gives some good suggestions), you should always define this explicitly if you are going to use it, since there is no standard way to treat half-integers. (I recently found this out the hard way when I assumed the rounding method I was always taught was standard, but python's default does something different.) Although I'm not sure how common this is in pure maths settings, I would say the best notation is simply $\operatorname{round}(x)$. This is easily understood, albeit not completely unambiguous – but definitely better than $[x]$ which could mean a myriad of completely unrelated things, or $\operatorname{nint}(x)$ which looks like “ninn-t?” If the ambiguity $1 \stackrel?= \operatorname{round}(1.5) \stackrel?= 2$ is a problem for you, make sure to explicitly discuss this. If you use the operation a lot, you could also define that you write it as $\lfloor x\rceil$, but I wouldn't use that without discussion. round is also the name for the rounding function in many programming languages, because what it does is it rounds a number, hence the name “round”. If you are fine going always in one direction for halfway values, you can resort to the programming trick of using $\lfloor x + \frac{1}{2} \rfloor$ (halfways towards $+\infty$) or $\lceil x - \frac{1}{2} \rceil$ (halfways towards $-\infty$). I have seen $((x))$ for "nearest integer." My memory is dim, but maybe it was Emil Grosswald's elementary number theory text. I like it because it's easy to type and it's not likely to be confused with another function. It might be too verbose, but something like the following is unlikely to be misinterpreted. $$\mathrm{RoundToEven}(5.5) = 6$$ If you need another convention such as rounding to the nearest odd number, rounding towards infinity, or rounding towards negative infinity I'd define my own function and include some examples. For instance: Let $R \mathop: \mathbb{R} \to \mathbb{Z}$ denote the function that rounds each real number to the nearest integer, rounding ties towards negative infinity. $$ R(-0.5) = -1 $$ $$ \left\{ R(-0.3)\;,\; R(0)\;,\; R(0.3)\;,\; R(0.5) \right\} = \left\{ 0 \right\} $$ $$ R(1.5) = 1 $$ $$ R(1.7) = 2 $$
Whats the use of representing rotation with quaternions compared to normal axis angle representation? I've been trying to learn quaternions and they make enough sense but as far as I can tell quaternions are just axis angle with a transformed axis and angle of rotation. How exactly does transforming the axis and angle of rotation affect the matrix instead of just representing it normally through axis angle? I think the best way to understand the advantage of quaternions is to consider them as an extension to the three-dimensional space of the representation of rotations in the plane with complex numbers. As a rotation of angle $\theta$ around the origin in the plane is represented, in a very simple and expressive way, by the complex number $e^{i\theta}$, so a rotation of an angle $2\theta$ in space, around an axis passing through the origin, is represented by a quaternion $e^{\mathbf{u}\theta}$, where $\mathbf{u}$ is the imaginary quaternion that correspond to the unit vector oriented along the axis of rotation. So we have the correspondence: $$ \vec{w}=R_{\mathbf{u},\theta} \; \vec{v} \quad \longleftrightarrow \quad \mathbf{w}= e^{\mathbf{u}\theta/2}\mathbf{v}e^{-\mathbf{u}\theta/2} $$ For example, given the rotation around the axis passing through the origin and the point of coordinates $(1,1,1)$ and angle $\theta=\pi/2$, it is easy to write the corresponding quaternion: $$ e^{\frac{\theta}{4}\frac{ (\mathbf{i}+\mathbf{j}+\mathbf{k})}{\sqrt{3}}} $$ which is simpler and more expressive then the corresponding matrix. And, using the properties of quaternions, we can easily prove corresponding properties for rotations, as: reverse the versor is equivalent to reverse the rotation: $\quad e^{\theta(-\mathbf{u})}=e^{-\theta\mathbf{u}}$ rotations about the same axis commute: $\quad e^{\theta\mathbf{u}}e^{\psi\mathbf{u}}=e^{\psi\mathbf{u}}e^{\theta\mathbf{u}}=e^{(\theta+\psi)\mathbf{u}}$ but rotation about different axis don't : $ \quad \quad e^{\theta\mathbf{u}}e^{\theta\mathbf{v}}\ne e^{\theta\mathbf{v}}e^{\theta\mathbf{u}} \ne e^{\theta(\mathbf{u}+\mathbf{v})}$ and solve problems about rotations (as in Rotation Equivalence using Quaternions). And more: we can easily see (in analogy with $U(1)$ for complex numbers) that unit quaternions form a group under multiplication, corresponding to the 3-sphere $S^3$, that is a double covering of $ SO(3,\mathbb{R})$ where the exponential $e^z$ is the canonical link between Lie algebras and Lie groups. Obviously you can prove all these things even in the formalism of the matrices, but the use of quaternions seems more simple and expressive. Finally, the use of quaternions is also useful from the computational point of view, because it is simple to implement and allows to avoid some problems as indicated in the Wikipedia page quoted by @Alexander Gruber. From what I understand a quaternion is a unit vector on a hyper-sphere (which is a four dimensional sphere). And if you multiply two quaternions you get a quaternion out again representing the combined rotation. You cannot do that easily with two axis/angle pairs. There is a lot of information in the advantages of quaternions section of the Wikipedia article.
Pete's excellent notes have correctly explained that there is no set containing sets of unboundedly large size in the infinite cardinalities, because from any proposed such family, we can produce a set of strictly larger size than any in that family. This observation by itself, however, doesn't actually prove that there are uncountably many infinities. For example, Pete's argument can be carried out in the classical Zermelo set theory (known as Z, or ZC, if you add the axiom of choice), but to prove that there are uncountably many infinities requires the axiom of Replacement. In particular, it is actually consistent with ZC that there are only countably many infinities, although this is not consistent with ZFC, and this fact was the historical reason for the switch from ZC to ZFC. The way it happened was this. Zermelo had produced sets of size $\aleph_0$, $\aleph_1,\ldots,\aleph_n,\ldots$ for each natural number $n$, and wanted to say that therefore he had produced a set of size $\aleph_\omega=\text{sup}_n\aleph_n$. Fraenkel objected that none of the Zermelo axioms actually ensured that $\{\aleph_n\mid n\in\omega\}$ forms a set, and indeed, it is now known that in the least Zermelo universe, this class does not form a set, and there are in fact only countably many infinite cardinalities in that universe; they cannot be collected together there into a single set and thereby avoid contradicting Pete's observation. One can see something like this by considering the universe $V_{\omega+\omega}$, a rank initial segment of the von Neumann hierarchy, which satisfies all the Zermelo axioms but not ZFC, and in which no set has size $\beth_\omega$. By adding the Replacement axiom, however, the Zermelo axioms are extended to the ZFC axioms, from which one can prove that $\{\aleph_n\mid n\in\omega\}$ does indeed form a set as we want, and everything works out great. In particular, in ZFC using the Replacement axiom in the form of transfinite recursion, there are huge uncountable sets of different infinite cardinalities. The infinities $\aleph_\alpha$, for example, are defined by transfinite recursion: $\aleph_0$ is the first infinite cardinality, or $\omega$. $\aleph_{\alpha+1}$ is the next (well-ordered) cardinal after $\aleph_\alpha$. (This exists by Hartog's theorem.) $\aleph_\lambda$, for limit ordinals $\lambda$, is the supremum of the $\aleph_\beta$ for $\beta\lt\lambda$. Now, for any ordinal $\beta$, the set $\{\aleph_\alpha\mid\alpha\lt\beta\}$ exists by the axiom of Replacement, and this is a set containing $\beta$ many infinite cardinals. In particular, for any cardinal $\beta$, including uncountable cardinals, there are at least $\beta$ many infinite cardinals, and indeed, strictly more. The cardinal $\aleph_{\omega_1}$ is the smallest cardinal having uncountably many infinite cardinals below it.
To expand on my comment, your equation holds only if $A$ and $B$ are uncorrelated. More generally, if $F$ is a function of uncorrelated variables $A$ and $B$, then $$(\delta F)^2 = \left(\frac{\partial F}{\partial A}\delta A\right)^2 + \left(\frac{\partial F}{\partial B} \delta B\right)^2$$ If you plug in $F=AB$, then you find that$$ \left(\delta F\right)^2 = (B\cdot \delta A)^2 + (A \cdot \delta B)^2$$or$$ \left(\frac{\delta F}{F}\right)^2 = \left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2 $$ In your case, $A$ and $B$ are the same, so they are the exact opposite of uncorrelated. In your case, the appropriate thing to do would be to consider $F$ to be a function of a single variable, and simply let $$\delta F = \left\|\frac{\partial F}{\partial A}\right\| \delta A$$$$\frac{\delta F}{F} = \left\|\frac{1}{F}\frac{\partial F}{\partial A}\right\| \delta A$$ To answer the question at a deeper level, we model experimentally measured quantities as continuous random variables, with our experimental uncertainties corresponding to their standard deviations. Let $X$ be a random variable with expected value $\mathbb{E}[X]=\mu_X$ and variance $Var(X)\equiv \mathbb{E}\big[ (X-\mu_X)^2\big]=\sigma_X^2$, and let $g$ be a function of $X$. We can expand $g$ in a Taylor series around $\mu_X$: $$g(X) = g(\mu_x) + g'(\mu_X)\cdot (X-\mu_X) + \frac{1}{2}g''(\mu_X) (X-\mu_X)^2 + \ldots $$ Truncating after the linear term, we write $$g(x) = g(\mu_X) + g'(\mu_X)\cdot (X-\mu_X)$$ We can now calculate the mean and variance of $g$: $$\mathbb{E}[g(X)] = g(\mu_X) + g'(\mu_X)\mathbb{E}[X-\mu_X] = g(\mu_X) + g'(\mu_X)\cdot (\mu_X-\mu_X) = g(\mu_X)$$$$Var(g(X)) = \mathbb{E}\big[\big(g(X)-g(\mu_X)\big)^2\big] = \big(g'(\mu_X)\big)^2 \cdot \mathbb{E}\big[(X-\mu_X)^2\big] = \left(\frac{dg}{dX}\cdot \sigma_X\right)^2$$ This is where the single-variable error propagation formula comes from. But now, consider a function $F$ of two variables $X$ and $Y$, with respective means $\mu_X,\mu_Y$ and variances $\sigma_X^2,\sigma_Y^2$, and covariance $$Cov(X,Y) = \mathbb{E}\big[(X-\mu_X)\cdot(Y-\mu_Y)\big]$$ We Taylor expand $F$ to linear order: $$F(X,Y) = F(\mu_X,\mu_Y) + \frac{\partial F}{\partial X}(X-\mu_X) + \frac{\partial F}{\partial Y} (Y-\mu_Y) $$ Repeating the earlier steps, the mean of $F$ is $$\mathbb{E}\big[F(X,Y)\big] = F(\mu_X,\mu_Y) $$ The variance, however, develops a slight subtlety. Notice that $$\left(F(X,Y)-F(\mu_X,\mu_Y)\right)^2 = \left(\frac{\partial F}{\partial X}\right)^2(X-\mu_X)^2 + \left(\frac{\partial F}{\partial Y}\right)^2(Y-\mu_Y)^2 + 2\frac{\partial F}{\partial X}\frac{\partial F}{\partial Y}(X-\mu_X)(Y-\mu_Y)$$ It follows that $$Var\big(F(X,Y)\big) = \left(\frac{\partial F}{\partial X} \sigma_X\right)^2 + \left(\frac{\partial F}{\partial Y}\sigma_Y\right)^2 + 2\frac{\partial F}{\partial X} \frac{\partial F}{\partial Y} Cov(X,Y)$$ If $X$ and $Y$ are uncorrelated, then $Cov(X,Y)=0$, and so we get our nice simple formula again. However, if $Cov(X,Y)\neq 0$, we need to take it into account.

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.