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Determining the Derivatives of the Inverse Trigonometric Functions
Now let's determine the derivatives of the inverse trigonometric functions, \(y = \arcsin x,\) \(y = \arccos x,\) \(y = \arctan x,\) \( y = \text{arccot}\, x,\) \(y = \text{arcsec}\, x,\) and \(y = \text{arccsc}\, x.\)
One way to do this that is particularly helpful in understanding how these derivatives are obtained is to use a combination of implicit differentiation and right triangles. An added benefit of this approach is that it will prepare you to be more successful in a future topic called trigonometric substitution.
Example \(\PageIndex{1}\): Finding the derivative of \(y = \arcsin x\)
Find the derivative of \(y = \arcsin x\).
Solution:
To find the derivative of \(y = \arcsin x\), we will first rewrite this equation in terms of its inverse form. That is, \[ \sin y = x \label{inverseEqSine}\]
Now this equation shows that \(y\) can be considered an acute angle in a right triangle with a sine ratio of \(\dfrac{x}{1}\). Since the sine ratio gives us the length of the opposite side over the length of the hypotenuse, this means that the opposite side has a length of \(x\) and the hypotenuse has a length of \(1\). See Figure \(\PageIndex{1}\).
Figure \(\PageIndex{1}\)
Now let's differentiate Equation \ref{inverseEqSine} implicitly with respect to \(x\).
\[\cos y \cdot \frac{dy}{dx} = 1\]
Then we solve this for \(\dfrac{dy}{dx}\).
\[\frac{dy}{dx} =\frac{1}{\cos y}\]
Looking at Figure \(\PageIndex{1}\), we see that \(\cos y = a\). Now we use the Pythagorean Theorem to find an expression for \(a\) in terms of \(x\) using the other sides of the right triangle that we know.
Thus we have:
\[\begin{align*} a^2 + x^2 &= 1^2 \\[5pt]
a^2 &= 1 - x^2 \\[5pt] a &= \sqrt{1-x^2} \end{align*}\] Figure \(\PageIndex{2}\)
Now using this result, we see that \[\cos y = \sqrt{1-x^2} \]
so
\[\frac{dy}{dx} =\frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}\]
Thus we have found the derivative of \( y = \arcsin x\),
\[\frac{d}{dx}\big( \arcsin x \big) = \frac{1}{\sqrt{1-x^2}}\]
Exercise \(\PageIndex{1}\)
Use the same approach to determine the derivatives of \(y = \arccos x,\) \(y = \arctan x,\) and \(y = \text{arccot} \,x.\)
Answer
\[\begin{align*} \frac{d}{dx}\big( \arccos x \big) &= \frac{-1}{\sqrt{1-x^2}} \\[5pt]
\frac{d}{dx}\big( \arctan x \big) &= \frac{1}{1+x^2} \\[5pt] \frac{d}{dx}\big( \text{arccot} \,x \big) &= \frac{-1}{1+x^2} \end{align*} \]
Example \(\PageIndex{2}\): Finding the derivative of \(y = \text{arcsec}\, x\)
Find the derivative of \(y = \text{arcsec}\, x\).
Solution:
To find the derivative of \(y = \text{arcsec}\, x\), we will first rewrite this equation in terms of its inverse form. That is, \[ \sec y = x \label{inverseEqSec}\]
As before, let \(y\) be considered an acute angle in a right triangle with a secant ratio of \(\dfrac{x}{1}\). Since the secant ratio is the reciprocal of the cosine ratio, it gives us the length of the hypotenuse over the length of the adjacent side, so this means that the hypotenuse has a length of \(x\) and the adjacent side has a length of \(1\). See Figure \(\PageIndex{3}\).
Figure \(\PageIndex{3}\)
Differentiating Equation \ref{inverseEqSec} implicitly with respect to \(x\), gives us:
\[\sec y\tan y \cdot \frac{dy}{dx} = 1\]
Solving this for \(\dfrac{dy}{dx}\), we get:
\[\frac{dy}{dx} =\frac{1}{\sec y\tan y}\]
In order to find \(\tan y\) in terms of \(x\), we need to find the length of the opposite side, \(a\), in terms of \(x\). By the Pythagorean Theorem, we have:
\[\begin{align*} 1^2 + a^2 &= x^2 \\[5pt]
a^2 &= x^2 - 1 \\[5pt] a &= \sqrt{x^2-1} \end{align*}\]
Figure \(\PageIndex{4}\) shows the resulting right triangle.
Figure \(\PageIndex{4}\)
From the right triangle in Figure \(\PageIndex{4}\), we can see that \[\tan y = \sqrt{x^2 - 1}.\]
Since \(\sec y = x\), it appears that
\[\frac{dy}{dx} =\frac{1}{\sec y\tan y} = \frac{1}{x\sqrt{x^2 - 1}}.\]
But this is not completely correct, at least not for negative values of \(x\). Considering the graph of \(y = \text{arcsec}\, x\) in Figure \(\PageIndex{5}\), we see that its slope is always positive. But for negative values of \(x\), the form of the derivative stated above would be negative (and clearly incorrect).
As we'll prove below, the actual derivative formula for this function is:
\[\frac{d}{dx}\big( \text{arcsec}\, x \big) = \frac{1}{|x|\sqrt{x^2 - 1}}\]
Consider the domain and range of the original function, \(y = \text{arcsec}\, x:\)
\[\text{Domain: } (-\infty, -1] \cup [1, \infty) \quad \text{or} \quad |x| \geq 1\]
\[\text{Range: } \big[0, \frac{\pi}{2}\big) \cup \big(\frac{\pi}{2}, \pi\big] \quad \text{or} \quad 0 \leq y \leq \pi, y \ne \frac{\pi}{2}\]
Note that the domain of the derivative is a subset of the domain of the original function, excluding the endpoints, \(x = -1\) and \(x = 1.\)
Now, let's rewrite \(\dfrac{dy}{dx}\) as:
\[\frac{dy}{dx} =\frac{1}{\sec y\tan y} = \frac{\cos y}{1}\cdot\frac{\cos y}{\sin y} = \frac{\cos^2 y}{\sin y}\]
We see that this function allows \(y\) to have all values between \(0\) and \(\pi\), except for \(y = \frac{\pi}{2}\), where the original expression \(\dfrac{1}{\sec y\tan y}\) is undefined. This is a subset of the range of the original function \(y = \text{arcsec}\, x.\) Note that in the derivative, \(y\) cannot take on the values of the endpoints of this interval, while these were part of the range of \(y = \text{arcsec}\, x.\)
Now consider that for all values of \(y\) in this range between \(0\) and \(\pi\), the expression for \(\dfrac{dy}{dx}\) is positive, as both
\(\sin y \gt 0\) and \(\cos^2 y \gt 0\) for all \(y \in (0, \pi)\).
Therefore, we have that,
\[\frac{d}{dx}\big( \text{arcsec}\, x \big) = \frac{1}{|x|\sqrt{x^2 - 1}}.\]
Exercise \(\PageIndex{2}\)
Use the same approach to determine the derivative of \(y = \text{arccsc} \,x.\)
Answer
\[\frac{d}{dx}\big( \text{arccsc}\, x \big) = \frac{-1}{|x|\sqrt{x^2 - 1}}.\nonumber\]
Using the Chain Rule with Inverse Trigonometric Functions
Now let's see how to use the chain rule to find the derivatives of inverse trigonometric functions with more interesting functional arguments.
Example \(\PageIndex{3}\):
Find the derivatives for each of the following functions:
\(\quad y = \arcsin(x^2) \) \(\quad y = \arctan(x^3+1) \) \(\quad y = \text{arcsec}(\ln|x|) \) \(\quad y = \arcsin(\cos x) \) Solution: Using the chain rule, we see that: \[ \frac{d}{dx}\big(\arcsin(x^2) \big) = \frac{1}{\sqrt{1-(x^2)^2}}\cdot \frac{d}{dx}\big(x^2\big) = \frac{2x}{\sqrt{1-x^4}}\] Here we have: \[ \frac{d}{dx}\big(\arctan(x^3+1) \big) = \frac{1}{1+(x^3+1)^2}\cdot \frac{d}{dx}\big(x^3+1\big) = \frac{3x^2}{1+(x^3+1)^2}\] Although it would likely be fine as it is, we can simplify it to obtain: \[ \frac{d}{dx}\big(\arctan(x^3+1) \big) = \frac{3x^2}{x^6+2x^3+2}\] For \( y = \text{arcsec}(\ln|x|) \), we obtain: \[ \frac{d}{dx}\big(\text{arcsec}(\ln|x|)\big) = \frac{1}{|\ln|x||\sqrt{(\ln|x|)^2 - 1}}\cdot \frac{d}{dx}\big(\ln|x|\big) = \frac{1}{x |\ln|x||\sqrt{(\ln|x|)^2 - 1}}\] For \( y = \arcsin(\cos x)\), we obtain: \[ \frac{d}{dx}\big(\arcsin(\cos x) \big) = \frac{1}{\sqrt{1-(\cos x)^2}}\cdot \frac{d}{dx}\big(\cos x\big) = \frac{-\sin x}{\sqrt{\sin^2 x}}\] Note that it may look like the denominator should simplify to \(\sin x\) and the entire derivative to \(\frac{dy}{dx} = -1\). But this is not the case. Remember that \[ \sqrt{x^2} = |x|.\] This means that what we actually obtain here is something more interesting: \[ \frac{d}{dx}\big(\arcsin(\cos x) \big) = \frac{-\sin x}{|\sin x|}\] This function alternates between values of \(-1\) and \(1\), when \(\sin x\) is positive and negative, respectively. See the graph of the function \( y = \arcsin(\cos x)\) below in Figure \(\PageIndex{6}\).
Derivative Formulas
In the same way that we can encapsulate the chain rule in the derivative of \(\ln u\) as \(\dfrac{d}{dx}\big(\ln u\big) = \dfrac{u'}{u}\), we can write formulas for the derivative of the inverse trigonometric functions that encapsulate the chain rule. Note that \(u\) represents a function of \(x\) in these formulas, and \(u'\) represents the derivative of \(u\) with respect to \(x\).
\[\begin{align*} \frac{d}{dx}\big(\arcsin u\big) \quad&=\quad \frac{u'}{\sqrt{1-u^2}} & &\frac{d}{dx}\big(\arccos u\big) \quad=\quad \frac{-u'}{\sqrt{1-u^2}} \\
\frac{d}{dx}\big(\arctan u\big) \quad&=\quad \frac{u'}{1+u^2} & & \frac{d}{dx}\big(\text{arccot}\, u\big) \quad=\quad \frac{-u'}{1+u^2} \\ \frac{d}{dx}\big(\text{arcsec}\, u\big) \quad&=\quad \frac{u'}{|u|\sqrt{u^2-1}} & & \frac{d}{dx}\big(\text{arccsc}\, u\big) \quad=\quad \frac{-u'}{|u|\sqrt{u^2-1}} \end{align*} \] Contributors Paul Seeburger (Monroe Community College) |
Difference between revisions of "Inaccessible"
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A cardinal $\kappa$ is ''hyperinaccessible'' if it is $\kappa$-inaccessible. One may similarly define that $\kappa$ is $\alpha$-hyperinaccessible if it is hyperinaccessible and for every $\beta\lt\alpha$, it is a limit of $\beta$-hyperinaccessible cardinals. Continuing, $\kappa$ is ''hyperhyperinaccessible'' if $\kappa$ is $\kappa$-hyperinaccessible.
A cardinal $\kappa$ is ''hyperinaccessible'' if it is $\kappa$-inaccessible. One may similarly define that $\kappa$ is $\alpha$-hyperinaccessible if it is hyperinaccessible and for every $\beta\lt\alpha$, it is a limit of $\beta$-hyperinaccessible cardinals. Continuing, $\kappa$ is ''hyperhyperinaccessible'' if $\kappa$ is $\kappa$-hyperinaccessible.
−
More generally, $\kappa$ is ''hyper${}^\alpha$-inaccessible'' if for every $\beta\lt\
+
More generally, $\kappa$ is ''hyper${}^\alpha$-inaccessible'' if for every $\beta\lt\$ it is $\kappa$-hyper${}^\beta$-inaccessible.
Every [[Mahlo]] cardinal $\kappa$ is hyper${}^\kappa$-inaccessible.
Every [[Mahlo]] cardinal $\kappa$ is hyper${}^\kappa$-inaccessible.
Revision as of 02:54, 16 May 2017 Inaccessible cardinals are the traditional entry-point to the large cardinal hierarchy, although weaker notions such as the worldly cardinals can still be viewed as large cardinals. If $\kappa$ is inaccessible, then $V_\kappa$ is a model of ZFC and so inaccessible cardinals are worldly, but this is not an equivalence. Every inaccessible cardinal $\kappa$ is an aleph fixed point and a beth fixed point, and consequently $V_\kappa=H_\kappa$. (Zermelo) The models of second-order ZFC are precisely the models $\langle V_\kappa,\in\rangle$ for an inaccessible cardinal $\kappa$. Solovay proved that if there is an inaccessible cardinal, then there is an inner model of a forcing extension satisfying ZF+DC in which every set of reals is Lebesgue measurable there. (citation) Shelah proved that Solovay's use of the inaccessible cardinal is necessary, in the sense that in any model of ZF+DC in which every set of reals is Lebesgue measurable, there is an inner model of ZFC with an inaccessible cardinal. Consequently, the consistency of the existence of an inaccessible cardinal with ZFC is equivalent to the impossibility of our constructing a non-measurable set of reals using only ZF+DC. The uncountable Grothedieck universes are precisely the sets of the form $V_\kappa$ for an inaccessible cardinal $\kappa$. The universe axiom is equivalent to the assertion that there is a proper class of inaccessible cardinals. Contents Weakly inaccessible cardinal
A cardinal $\kappa$ is
weakly inaccessible if it is an uncountable regular limit cardinal. Under the GCH, this is equivalent to inaccessibility, since under GCH every limit cardinal is a strong limit cardinal. So the difference between weak and strong inaccessibility only arises when GCH fails badly. Every inaccessible cardinal is weakly inaccessible, but forcing arguments show that any inaccessible cardinal can become a non-inaccessible weakly inaccessible cardinal in a forcing extension, such as after adding an enormous number of Cohen reals (this forcing is c.c.c. and hence preserves all cardinals and cofinalities and hence also all regular limit cardinals). Meanwhile, every weakly inaccessible cardinal is fully inaccessible in any inner model of GCH, since it will remain a regular limit cardinal in that model and hence also be a strong limit there. In particular, every weakly inaccessible cardinal is inaccessible in the constructible universe $L$. Consequently, although the two large cardinal notions are not provably equivalent, they are equiconsistent. Levy collapse
The Levy collapse of an inaccessible cardinal $\kappa$ is the $\lt\kappa$-support product of $\text{Coll}(\omega,\gamma)$ for all $\gamma\lt\kappa$. This forcing collapses all cardinals below $\kappa$ to $\omega$, but since it is $\kappa$-c.c., it preserves $\kappa$ itself, and hence ensures $\kappa=\omega_1$ in the forcing extension.
Inaccessible to reals
A cardinal $\kappa$ is
inaccessible to reals if it is inaccessible in $L[x]$ for every real $x$. For example, after the Levy collapse of an inaccessible cardinal $\kappa$, which forces $\kappa=\omega_1$ in the extension, the cardinal $\kappa$ is of course no longer inaccessible, but it remains inaccessible to reals. Universes
When $\kappa$ is inaccessible, then $V_\kappa$ provides a highly natural transitive model of set theory, a universe in which one can view a large part of classical mathematics as taking place. In what appears to be an instance of convergent evolution, the same universe concept arose in category theory out of the desire to provide a hierarchy of notions of smallness, so that one may form such categories as the category of all small groups, or small rings or small categories, without running into the difficulties of Russell's paradox. Namely, a
Grothendieck universe is a transitive set $W$ that is closed under pairing, power set and unions. That is, (transitivity) If $b\in a\in W$, then $b\in W$. (pairing) If $a,b\in W$, then $\{a,b\}\in W$. (power set) If $a\in W$, then $P(a)\in W$. (union) If $a\in W$, then $\cup a\in W$.
The
Grothendieck universe axiom is the assertion that every set is an element of a Grothendieck universe. This is equivalent to the assertion that the inaccessible cardinals form a proper class. Degrees of inaccessibility
A cardinal $\kappa$ is
$1$-inaccessible if it is inaccessible and a limit of inaccessible cardinals. In other words, $\kappa$ is $1$-inaccessible if $\kappa$ is the $\kappa^{\rm th}$ inaccessible cardinal, that is, if $\kappa$ is a fixed point in the enumeration of all inaccessible cardinals. Equivalently, $\kappa$ is $1$-inaccessible if $V_\kappa$ is a universe and satisfies the universe axiom.
More generally, $\kappa$ is $\alpha$-inaccessible if it is inaccessible and for every $\beta\lt\alpha$ it is a limit of $\beta$-inaccessible cardinals.
Hyper-inaccessible
A cardinal $\kappa$ is
hyperinaccessible if it is $\kappa$-inaccessible. One may similarly define that $\kappa$ is $\alpha$-hyperinaccessible if it is hyperinaccessible and for every $\beta\lt\alpha$, it is a limit of $\beta$-hyperinaccessible cardinals. Continuing, $\kappa$ is hyperhyperinaccessible if $\kappa$ is $\kappa$-hyperinaccessible.
More generally, $\kappa$ is
hyper${}^\alpha$-inaccessible if it is hyperinaccessible and for every $\beta\lt\alpha$ it is $\kappa$-hyper${}^\beta$-inaccessible, where $\kappa$ is $\alpha$-hyper${}^\beta$-inaccessible if it is hyper${}^\beta$-inaccessible and for every $\gamma<\alpha$, it is a limit of $\gamma$-hyper${}^\beta$-inaccessible cardinals.
Every Mahlo cardinal $\kappa$ is hyper${}^\kappa$-inaccessible. |
Decide if the series $$\sum_{n=1}^\infty\frac{4^{n+1}}{3^{n}-2}$$ converges or diverges and, if it converges, find its sum.
Is this how you would show divergence attempt:
For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n -2} \geq 0$
For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n-2} \geq \frac{4^{n+1}}{3^n} = b_n$
Since $\sum_{n=1}^{\infty} \frac{4^{n+1}}{3^n}$ is a geometric series with $r = \frac{4}{3} > 1$. Therefore it diverges by the geometric series test and by the comparison test $\sum a_n$ diverges too. |
There's not a
lot to say here, but we can say a couple of things. We'll assume $H\le G\le S_n$ and $H$ and $G$ are both transitive. A block system is defined by a choice of a single block $\Delta\ni 1$, so let $B_H$ be the set of blocks for $H$ containing $1$ and $B_G$ be the set of blocks for $G$ containing $1$.
Immediately if $\Delta^g=\Delta$ or $\Delta^g\cap\Delta=\emptyset$ for all $g\in G$ then certainly this holds for all $g\in H$. So a block for $G$ is a block for $H$ which we can write as $B_G\le B_H$.
The inverse equality clearly doesn't always hold (e.g. $H$ any imprimitive group and $G=S_n$). One can show however that blocks of $G$ containing $1$ are in one-to-one correspondence with subgroups of $G$ containing the point stabiliser $G_1$ (specifically $\Delta$ corresponds to $G_\Delta$ the setwise stabiliser of $\Delta$). I claim that a block $\Delta$ of $H$ containing $1$ is a block for $G$ if and only if $G_\Delta\ge G_1$. This means $B_G=\{\Delta\in B_H|G_1\le G_\Delta\}$ - that is a block of $G$ containing $1$ is any block of $H$ containing $1$ fixed by $G_1$.
Proof of Claim:
"Prove $\Delta$ is a block of $G$ implies $G_1\le G_\Delta$" is a standard exercise, so I will only do the reverse implication.
Assume $G_\Delta\ge G_1$ where $\Delta$ is a block for $H$.
Let $g\in G$ and suppose $x\in\Delta\cap\Delta^g$.
Relabelling if necessary we may assume $x=1$.
As $1\in\Delta^g$ there is some $y\in\Delta$ with $y^g=1$.
As $H$ is transitive there is some $h\in H$ with $1^h=y$ so $1^{hg}=1$.
That is $hg\le G_1\le G_\Delta$.
As $\Delta$ is a block for $H$, $h\le H_\Delta\le G_\Delta$.
Hence $g=h^{-1}(hg)\le G_\Delta$, so $\Delta^g=\Delta$ completing the proof. |
I am trying to solve a mean-variance problem with a non-linear market impact cost term in there. This is the problem I am trying to solve
$$ \max_x \left ( \alpha x - \gamma x' \Sigma x - a\sqrt{|x-x_0|} \right ) \quad s.t. \quad \text{unconstrained}.$$
where, $x_0$ is the current portfolio holding.
I am using MATLAB's quadprog program to solve this problem. How can I setup my optimization problem correctly to incorporate the sqrt term in the objective ? I am sure there is a way to do this, but I am not aware of it. |
I think you are interpreting too much into the matter. The $-\frac12\sigma^2$ is just a correction term that comes from Jensen's inequality.You need this when switching from supposedly symmetric returns (normal distribution) to the skewed price process (log-normal distribution).I think there are no deeper truths to be found here.
MorningstarMorningstar partnered with Quantopian, and the latter published the structure of Morningstar's equity fundamentals database:https://www.quantopian.com/help/fundamentalsQuantopian users can use this data for free.
Quantopian provides both the fundamental data (from Morningstar), as well as the backtest platform to reproduce results from the books you mentioned. Here's the introduction to our fundamentals offering: https://www.quantopian.com/posts/fundamental-data-from-morningstar-now-available-for-backtesting(disclosure: I'm the ceo of quantopian)
This is a very subjective question. One thing you need to understand is that there are many types of quants and it is not always about predicting the future returns.Many quantitative analysts are involved in market-making, this is where you sell products at a slight cost to customers and try to stay more or less neutral to market moves. When you read about ...
Quandl has two premium fundamental datasets that may be of interest to you, Robur Global Select Stock Fundamentals and Mergent Global Fundamentals Data. Quandl also has fundamental datasets for specific countries including the US, China, and India.If you have an Interactive Brokers account and can program against their API, you can sign up for Reuters ...
You can use refined methodologies but if you just need a rough estimation of liquidity, you can simply use an average of daily volume over N days. In practice, for equities, people tend to use N = 20 or 30.Once you have the average daily volume (say 100,000 shares), you compare it to your holding (say 50,000 shares) to determine the the size of your ...
Compustatsupports unlimited data exportkeeps the history of disbanded entitiesprovides restatements since 1950 + point-in-time data since 1986coverage since 1950list of variables (data guide)Compustat is a S&P subsidiary. It goes as a plugin for CapitalIQ (also S&P), WRDS, CRSP, and other platforms. Pricing starts from \$3k. A platform ...
Welcome.You're looking at 1995 data. Back then, Edgar was just coming online. They did not have documents in electronic form. If you want data that old, you may have to pay a vendor, such as S&P.If you look at the same Макдак for 2019 https://www.sec.gov/cgi-bin/browse-edgar?action=getcompany&CIK=0000063908 , everything can be downloaded.
The highly respected CXO Advisory Group has done some research on this topic on basis of a suggestion from me. The result is summarized as follows (cited with permission):In summary, evidence suggests that high-dividend stock ETFs mostlygenerate positive alpha with beta less than one relative to SPY, butother performance comparisons to the market ...
One thing to keep in mind here is that the world of risk-free/arbitrage-free models is not necessarily the real world. Specifically, this equation$$\mu = r - \frac{1}{2}\sigma^2$$occurs not because this is the way stocks behave in reality (they don't! For S&P 500, long-run $\mu$ is closer to 6-9%, if I recall correctly), but because using any ...
Let $\Omega$ be the outcome space at some future date and fix a specific outcome $\omega \in \Omega$. Now consider a portfolio that gives one unit of currency if $\omega$ happens and zero otherwise, i.e., with payoff $\mathbb{I}(\omega)$. Any other payoff function can be given as a linear combination of these portfolios. The price of this portfolio today is ...
Examples for cash-settled futures are:Interest Rate futuresFutures on implied Volatility (e.g. on VIX)Futures on Commodity Indices: Indices such as the Dow Jones UBS consist of futures themselves. Furthermore in asset management you usually don't want physical delivery of the underlying (oil, gas, coal, pig, ... ;)Futures on Equity IndicesThe pricing ...
There are two mainly (good) free sources available online:wolphramalpha.comQuandlThey report the mainly market and fundamental data, so you will not find any particular fundamental accounting ratio. In the case you need particular ratio or data, you should get some better financial data provider, as, for instance, Bloomberg or Thompson Reuters.
It turns out that GBM with constant drift and constant volatility is not really used in real life. It is well known that volatility as well as drift may vary over time. Hence, if you want to use a model with time-varying parameters, you need to come up with a model to define $\mu_t$ and $\sigma_t$. There are classic models that use some mean-reverting ...
I think that it may be very simple but since I just started in thequant finance it would be great to have some feedback andrecommendations about how to improve my backtesting.From my experience, most who begin testing a model straight from academia overlook several things that are quite different in the real world. Factoring them in will help to ...
The forex week starts on Monday 7am Wellington time when the Kiwi value date rolls. It ends on Friday 5pm New York. Within that each currency has its own “cut off” time when the value date rolls. The cut off depends on the time zone and liquidity in the currency managed by the dealers who make its market. The convention for an FX day “cut off” for a given ...
There are at least two ways of doing it:1) Resampling them to their median frequency.2) Build one ML model for each data type, then combine the 4 different forecasts into a single meta-ML model.(Courtesy: MARCOS LO´PEZ DE PRADO)
Let $c_t$ be the price of an European call with maturity $T$ and $D_{t,T}$ the discount factor from $T$ to $t$. We assume deterministic rates. Then note that for $s<t\leq T$:$$\begin{align}E^Q_s\left(c_t\right)&=E^Q_s\left(E^Q_t\left(D_{t,T}(S_T-K)^+\right)\right)\\[3pt]&=E^Q_s\left(D_{t,T}(S_T-K)^+\right)\\[3pt]&=E^Q_s\left(\frac{D_{s,t}...
This has been driving me nuts as well!Thanks for providing the spreadsheet. Looking at MSN Money there is a discrepancy of over 11B in market cap between viewing GOOG and GOOGL shares! GOOGL has a market cap associated with 537B and GOOG has a market cap of 526B. I don't understand how one site can list two different market caps based on the class of ...
This is a question of how to aggregate ratios.I see your two options, and raise you one more.Method 1: The mean (or median) value of Price-to-Book values for individual securities$$f(E[x]) = \frac{\sum_{i}^{n} \frac{P_i}{B_i}}{n} $$Pros: Relatively simple to calculate and gives an idea of what the typical company's Price-to-Book is.Cons: ...
I think one of the main liquidity measures is the one from Pastor and Stambaugh (2003).You can use it for both individual stocks or indexes.Just run the following intra-month regression with daily data:$r^e_{i,d+1,t} = \theta_{i,t}+\phi_{i,t}r_{i,d,t}+\gamma_{i,t}sign(r^e_{i,d,t}) \times v_{i,d,t}+\epsilon_{i,d+1,t}$.Where $r^e_{i,d+1,t}$ is the ...
XBRL became mandatory for US filers on June 15th, 2011. The SEC requires XBRL data for:Quarterly and annual reports and transition reportsForm 8-K revisionsLimited Securities Act registration statementsXBRL instances for quarterly and annual reports would typically contain the usual items found on Income statements, Balance Sheets, and Cash Flows.
You don't need to read/parse financial statements in XBRL format in the first place.There exists the method getFundamentals(ticker) provided by the package eodhistoricaldata-api (https://www.npmjs.com/package/eodhistoricaldata-api).The library returns quarterly (and yearly) financial statements (income statements, balance sheets, and cash flow ... |
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Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Highlights of experimental results from ALICE
(Elsevier, 2017-11)
Highlights of recent results from the ALICE collaboration are presented. The collision systems investigated are Pb–Pb, p–Pb, and pp, and results from studies of bulk particle production, azimuthal correlations, open and ...
Event activity-dependence of jet production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV measured with semi-inclusive hadron+jet correlations by ALICE
(Elsevier, 2017-11)
We report measurement of the semi-inclusive distribution of charged-particle jets recoiling from a high transverse momentum ($p_{\rm T}$) hadron trigger, for p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in p-Pb events ...
System-size dependence of the charged-particle pseudorapidity density at $\sqrt {s_{NN}}$ = 5.02 TeV with ALICE
(Elsevier, 2017-11)
We present the charged-particle pseudorapidity density in pp, p–Pb, and Pb–Pb collisions at sNN=5.02 TeV over a broad pseudorapidity range. The distributions are determined using the same experimental apparatus and ...
Photoproduction of heavy vector mesons in ultra-peripheral Pb–Pb collisions
(Elsevier, 2017-11)
Ultra-peripheral Pb-Pb collisions, in which the two nuclei pass close to each other, but at an impact parameter greater than the sum of their radii, provide information about the initial state of nuclei. In particular, ...
Measurement of $J/\psi$ production as a function of event multiplicity in pp collisions at $\sqrt{s} = 13\,\mathrm{TeV}$ with ALICE
(Elsevier, 2017-11)
The availability at the LHC of the largest collision energy in pp collisions allows a significant advance in the measurement of $J/\psi$ production as function of event multiplicity. The interesting relative increase ... |
Exact Differential Equations
We will now look at another type of first order differential equation that we can solve known as exact differential equations which we define below.
Definition: Let $M(x, y)$ and $N(x, y)$ be functions, and suppose we have a differential equation in the form $M(x, y) + N(x, y) \frac{dy}{dx} = 0$. This differential equation is said to be Exact if there exists a function $\psi (x, y)$ such that $\frac{\partial}{\partial x} \psi (x, y) = \psi_x = M(x, y)$ and $\frac{\partial}{\partial y} \psi (x, y) = \psi_y = N(x, y)$.
Now consider a differential equation of the following form where $M$ and $N$ are functions of $x$ and $y$:(1)
Suppose that there exists a function $\psi (x, y)$ such that:(2)
Then this first order differential equation is an exact differential equation. Plugging these equalities into the differential equation gives us:(3)
Now the chain rule for multivariable functions tells gives us a way to compress the formula for our differential equation above. Note that:(4)
When we plug this into our differential equation from above - we see that:(5)
The above equation is the solution to our differential equation when the variable $y$ is isolated.
Now we want to know how we can immediately tell whether or not a differential equation is exact. We will not prove this result, but provided that the functions $M$, $N$, $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$ are continuous in a region $R$, then a differential equation of the form $M(x, y) + N(x, y) \frac{dy}{dx} = 0$ is exact if and only if:(6)
We will now look at an example of solving an exact differential equation.
Example 1 Solve the differential equation $(e^x \sin y - 2y \sin x) + (e^x \cos y + 2 \cos x) \frac{dy}{dt} = 0$.
We first need to check that the differential equation is indeed exact. Let $M(x, y) = e^x \sin y - 2y \sin x$ and let $N(x, y) = e^x \cos y + 2 \cos x$. The partial derivative of $M$ with respect to $y$ and the partial derivative of $N$ with respect to $x$ are respectively:(7)
So indeed this differential equation is exact. Let $\psi (x, y)$ be the function such that:(9)
From the first equation, we have that $\frac{\partial}{\partial x} \psi (x, y) = e^x \sin y - 2y \sin x$. If we integrate both sides with respect to $x$ then we have that:(10)
Note that we do not have a constant of integration but instead a "function" of integration, $h(y)$. This is because if we were to partial differentiate both sides of the equation above with respect to $x$, then any function of $y$ would disappear.
We now take the equation above and partial differentiate it with respect to $y$ to get:(11)
We have two expressions for $\frac{\partial}{\partial y} \psi (x, y)$ - the one directly above, and $\frac{\partial}{\partial y} \psi (x, y) = N(x, y) = e^x \cos y + 2 \cos x$. In comparing these two, we see that $h'(y) = 0$. Therefore $h(y) = C$, and so therefore:(12)
Therefore the solution to our differential equation is of the form $\psi (x, y) = D$, where $D$ is an arbitrary constant. In absorbing the constant $C$ into $D$, we have that then:(13) |
Towards Understanding the Origin of Cosmic-Ray Electrons
We present the precision measurement of the electron flux with a particular emphasis on the behavior at high energies. The measurement is based on 28.1 million electron events collected by AMS from May 19, 2011 to November 12, 2017. This corresponds to a factor of three increase in statistics and factor of two increase in the energy range compared to our results published in 2014 [PRL
113, 121102 (2014)]. The latest precision results on cosmic-ray electrons up to 1.4 TeV reveal new and unexpected features. Our latest data on cosmic-ray electrons and positrons are crucial for providing insights into origins of high energy cosmic-ray electrons and positrons.
We found that in the entire energy range the electron and positron spectra have distinctly different magnitudes and energy dependences. The electron flux exhibits a significant excess starting from $ 42.1_{-5.2}^{+5.4} $ GeV compared to the lower energy trends, but the nature of this excess is different from the positron flux excess above $25.2 \pm 1.8$ GeV. Contrary to the positron flux, which has an exponential energy cutoff of $810_{-180}^{+310} $
Figure 1 shows the latest AMS results on the precision measurements of the electron spectrum and the positron spectrum from the most recent AMS data. These measurements are based on 28.1 million electron and 1.9 million positron events.
Figure 2 shows the latest AMS results on the precision measurements of the electron spectrum. The other most recent measurements are shown for comparison.
Similar to the analysis of the positron flux, we examine the changing behavior of the electron spectrum using the same power law approximation
$$ \label{eq:1} \Phi_{e^{-}}(E)= \begin{cases} C(E/20.04\mbox{ GeV})^{\gamma}, & E \leq E_{0}; \\ C(E/20.04\mbox{ GeV})^{\gamma}(E/E_{0})^{\Delta\gamma}, & E > E_{0}. \end{cases} \tag{1} $$
A fit to data is performed in the energy range [20.04−1400] GeV. The results are presented in Figure 3. The fit yields $E_{0} = 42.1_{-5.2}^{+5.4}$ GeV for the energy where the spectrum behavior changes. The significance of this change is established at 7σ.
To examine the energy dependence of the electron flux in a model independent way, the flux spectral index $\gamma$ is calculated from $\gamma = d[\log(\Phi)]/d[\log(E)]$ over non-overlapping energy intervals which are chosen to have sufficient sensitivity to the spectral index. The energy interval boundaries are 3.36, 5.00, 7.10, 10.32, 17.98, 27.25, 55.58, 90.19, 148.81, 370 and 1400 GeV. The results are presented in Figure 4 together with the positron results. As seen, both the electron and positron indices decrease (soften) rapidly with energy below ∼10 GeV, and then they both start increasing (harden) at >20GeV. In particular, the electron spectral index increases from $\gamma = −3.295\pm0.026$ in the energy range [17.98 − 27.25] GeV to an average $\gamma = −3.180\pm0.008$ in the range [55.58 − 1400] GeV, where it is nearly energy independent.
As seen in Figure 4, the behavior of the electron and positron spectral indices is distinctly different.
New sources of high energy positrons, such as dark matter, may also produce an equal amount of high energy electrons. We test this hypothesis using the source term from our positron analysis. The electron flux is parametrized as a sum of a power law component and the positron source term with the exponential energy cutoff:
$$ \label{eq:2} \Phi_{e^{-}}(E)= C_{e^{-}}(E/E_{1})^{\gamma_{e^{-}}} + f_{e^{-}}C_{S}^{e^{+}}(E/E_{2})^{\gamma_{S}^{e^{+}}}\exp(-E/E_{S}^{e^{+}}) \tag{2} $$
The power law component is characterized by the normalization factor $C_{e^{-}}$ and the spectral index $\gamma_{e^{-}}$ . The constant $E_1= 41.61$ GeV corresponds to the beginning of the fit range, it does not affect the fitted value of $\gamma_{e^{-}}$ . The values of the source term parameters $C_{S}^{e^{+}}$, $\gamma_{S}^{e^{+}}$, $E_2$ , and $E_{S}^{e^{+}}$ are taken from positron data. A fit to the data with the source term normalization $f_{e^{−}}$ fixed to 1 is performed in the energy range [41.61 − 1400] GeV, where the solar modulation effects are negligible. It yields $C_{e^{-}} = (1.965 \pm 0.010) \times 10^{−3} [ \mathrm{m^{2}\,sr\,s\,GeV}]^{-1}$ and $\gamma_{e^{-}} = −3.248 \pm 0.007$ for the power law component with $\chi^2/\mathrm{d.o.f.} = 15.5/24$. The result of the fit is presented in Figure 5(a).
A similar fit of Eq. (\ref{eq:2}) to data, but with $f_{e^{-}}$ fixed to 0, yields $C_{e^{-}} = (2.124 \pm 0.010) \times 10^{−3} [\mathrm{m^{2}\,sr\,s\,GeV}]^{-1}$ and $\gamma_{e^{-}} = −3.186 \pm 0.006$ with $\chi^2/\mathrm{d.o.f.} = 15.2/24$. The result of this fit is presented in Fig. 5(b). Varying the normalization of the source term $f_{e^{-}}$ as a free fit parameter does not improve the $\chi^2$ and yields $f_{e^{-}} = 0.5_{-0.6}^{+1.2}$ . As seen in Figures 5 (a) and (b) the data are consistent both with the charge symmetric positron source term ($f_{e^{-}} = 1$ in Eq. (\ref{eq:2})) and also with the absence of such a term ($f_{e^{-}} = 0$). Therefore, it is not possible to extract any additional information on the existence and properties of the source term with the electron flux.
To investigate the existence of a finite energy cutoff as seen in the positron flux, the electron flux is fitted with
$$ \label{eq:3} \Phi_{e^{-}}(E)= C_{S}(E/41.61\mbox{ GeV})^{\gamma_{S}}\exp(-E/E_{S}) \tag{3} $$
A fit to data in the energy range [41.61, 1400] GeV yields the inverse cutoff energy $1/E_{s} = 0.00_{-0.00}^{+0.08}$ TeV
−1 with $\chi^2/\mathrm{d.o.f.} = 15.2/23$. A study of the cutoff significance shows that $E_{s} < 1.9$ TeV is excluded at the 5σ level. These results are presented in Figure 6.
In addition to a small contribution of secondary electrons produced in the collisions of ordinary cosmic rays with the interstellar gas, there are several astrophysical sources of primary cosmic-ray electrons. It is assumed that there are only a few astrophysical sources of high energy electrons in the vicinity of the solar system each making a power law-like contribution to the electron flux. In addition, there are several physics effects which may introduce some spectral features in the original fluxes. Therefore, it is important to know the minimal number of distinct power law functions needed to accurately describe the AMS electron flux.
We found that in the entire energy range [0.5 − 1400] GeV the electron flux is well described by the sum of two power law components:
$$ \label{eq:4} \Phi_{e^{-}}(E)= \dfrac{E^{2}}{\hat{E}^{2}}[1+(\hat{E}/E_{t})^{\Delta\gamma_{t}}]^{-1}[C_{a}(\hat{E}/E_{a})^{\gamma_{a}}+C_{b}(\hat{E}/E_{b})^{\gamma_{b}}] \tag{4} $$
The two components, $a$ and $b$, correspond to two power law functions. To account for solar modulation effects, the force-field approximation is used, with the energy of particles in the interstellar space $\hat{E}=E+\varphi_{e^{-}}$ and the effective modulation potential $\varphi_{e^{-}}$. The additional transition term, $[1+(\hat{E}/E_{t})^{\Delta\gamma_{t}}]^{-1}$, has vanishing impact on the flux behavior at energies above $E_{t}$ (e.g. < 0.7% above 40 GeV). A fit to the data in the energy range [0.5−1400] GeV is presented in Figure 7. We conclude that in the energy range [0.5 − 1400] GeV the sum of two power law functions provides an excellent description of the data with $\chi^{2}/\mathrm{d.o.f.} = 36.5/68$.
An analysis of the individual components in the electron flux, namely the power law $a$ and $b$ terms, is presented in Figures 8 and 9 together with the corresponding positron data. As seen in Figure 8, at low energies positrons come from cosmic ray collisions, electrons do not. As seen in Figure 9, the positron source term has a cutoff, whereas electrons have neither the source term nor the cutoff.
In the entire energy range the electron and positron spectra have distinctly different magnitudes and energy dependences. The different behavior of the cosmic-ray electrons and positrons measured by AMS is clear evidence that most high energy electrons originate from different sources than high energy positrons. |
Assuming we do not know the Lorenz curve function,
If $(X_k, Y_k)$ are the known points on the Lorenz curve, with the $X_k$ indexed in increasing order $(X_{k – 1} < X_k)$, so that:
$X_k$ is the cumulated proportion of the population variable, for $k = 0,...,n$, with $X_0 = 0, X_n = 1$. $Y_k$ is the cumulated proportion of the income variable, for $k = 0,...,n$, with $Y_0 = 0, Y_n = 1$. $Y_k$ should be indexed in non-decreasing order $(Y_k \geq Y_{k – 1})$
If the Lorenz curve is approximated on each interval as a line between consecutive points, then the area B can be approximated with trapezoids and:
(taken from wikipedia)
I've also found this other formula to calculate the Gini coefficient:
$G=1-\frac{\sum^{n-1}_{k=1}Y_k}{\sum^{n-1}_{k=1}X_k}$
My question is how can I prove that both are equal? I haven't been able to deduce the latter from the former... |
A new construction of rotation symmetric bent functions with maximal algebraic degree
School of Mathematics and Statistics, Henan University, Kaifeng 475004, China
$ n = 2m\ge4 $
$ n $
$ m $
$ f(x_0,x_1\cdots,x_{n-1}) = \bigoplus\limits_{i = 0}^{m-1}(x_ix_{m+i})\oplus \bigoplus\limits_{i = 0}^{n-1}(x_ix_{i+1}\cdots x_{i+m-2} \overline{x_{i+m}} ), $
$ \widetilde{f}(x_0,x_1\cdots,x_{n-1}) = \bigoplus\limits_{i = 0}^{m-1}(x_ix_{m+i})\oplus \bigoplus\limits_{i = 0}^{n-1}(x_ix_{i+1}\cdots x_{i+m-2} \overline{x_{i+n-2}} ), $
$ \overline{x_{i}} = x_{i}\oplus 1 $
$ x $
$ n $ Keywords:Orbit, rotation symmetric Boolean function, Walsh-Hadamard transform, bent function, algebraic degree. Mathematics Subject Classification:Primary: 58F15, 58F17; Secondary: 53C35. Citation:Sihong Su. A new construction of rotation symmetric bent functions with maximal algebraic degree. Advances in Mathematics of Communications, 2019, 13 (2) : 253-265. doi: 10.3934/amc.2019017
References:
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C. Carlet, Boolean functions for cryptography and error correcting codes, in
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A secondary construction and a transformation on rotation symmetric functions, and their action on bent and semi-bent functions,
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$\acute{E}$. Filiol and C. Fontaine, Highly nonlinear balanced Boolean functions with a good correlation-immunity, in
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G. Gao, X. Zhang, W. Liu and C. Carlet,
Constructions of quadratic and cubic rotation symmetric bent functions,
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Search for Boolean functions with excellent profiles in the rotation symmetric class,
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F. MacWilliams and N. Sloane,
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S. Su and X. Tang,
Systematic constructions of rotation symmetric bent functions, 2-rotation symmetric bent functions, and bent idempotent functions,
[19]
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show all references
References:
[1] [2]
C. Carlet, Boolean functions for cryptography and error correcting codes, in
[3]
C. Carlet, G. Gao and W. Liu,
A secondary construction and a transformation on rotation symmetric functions, and their action on bent and semi-bent functions,
[4]
C. Carlet, G. Gao and W. Liu, Results on constructions of rotation symmetric bent and semi-bent functions, in
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$\acute{E}$. Filiol and C. Fontaine, Highly nonlinear balanced Boolean functions with a good correlation-immunity, in
[7] [8] [9]
G. Gao, X. Zhang, W. Liu and C. Carlet,
Constructions of quadratic and cubic rotation symmetric bent functions,
[10]
S. Kavut, S. Maitra and M. Yücel,
Search for Boolean functions with excellent profiles in the rotation symmetric class,
[11] [12]
F. MacWilliams and N. Sloane,
[13] [14] [15] [16] [17] [18]
S. Su and X. Tang,
Systematic constructions of rotation symmetric bent functions, 2-rotation symmetric bent functions, and bent idempotent functions,
[19]
W. Zhang, Z. Xing and K. Feng, A construction of bent functions with optimal algebraic degree and large symmetric group, preprint, Cryptology ePrint Archive, : Submission 2017/229.Google Scholar
[1]
Wenying Zhang, Zhaohui Xing, Keqin Feng.
A construction of bent functions with optimal algebraic degree and large symmetric group.
[2] [3] [4]
Behrouz Kheirfam.
A full Nesterov-Todd step infeasible interior-point algorithm for symmetric optimization based on a specific kernel function.
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Sihem Mesnager, Fengrong Zhang, Yong Zhou.
On construction of bent functions involving symmetric functions and their duals.
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Ken Ono.
Parity of the partition function.
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Jaume Llibre, Y. Paulina Martínez, Claudio Vidal.
Phase portraits of linear type centers of polynomial Hamiltonian systems with Hamiltonian function of degree 5 of the form $ H = H_1(x)+H_2(y)$.
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SelÇuk Kavut, Seher Tutdere.
Highly nonlinear (vectorial) Boolean functions that are symmetric under some permutations.
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Welington Cordeiro, Manfred Denker, Michiko Yuri.
A note on specification for iterated function systems.
[17] [18] [19] [20]
2018 Impact Factor: 0.879
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You may want to save the results of these exercises, since we will revisit in the next two sections. In Exercises [exer:3.1.1}– [exer:3.1.5} use Euler’s method to find approximate values of the solution of the given initial value problem at the points \(x_i=x_0+ih\), where \(x_0\) is the point where the initial condition is imposed and \(i=1\), \(2\), \(3\). The purpose of these exercises is to familiarize you with the computational procedure of Euler’s method.
[exer:3.1.1} \(y'=2x^2+3y^2-2,\quad y(2)=1;\quad h=0.05\)
[exer:3.1.2} \(y'=y+\sqrt{x^2+y^2},\quad y(0)=1;\quad h=0.1\)
[exer:3.1.3} \(y'+3y=x^2-3xy+y^2,\quad y(0)=2;\quad h=0.05\)
[exer:3.1.4} \(y'= {1+x\over1-y^2},\quad y(2)=3;\quad h=0.1\)
[exer:3.1.5} \(y'+x^2y=\sin xy,\quad y(1)=\pi;\quad h=0.2\)
[exer:3.1.6} Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[y'+3y=7e^{4x},\quad y(0)=2\] at \(x=0\), \(0.1\), \(0.2\), \(0.3\), …, \(1.0\). Compare these approximate values with the values of the exact solution \(y=e^{4x}+e^{-3x}\), which can be obtained by the method of Section 2.1. Present your results in a table like Table [table:3.1.1}.
[exer:3.1.7} Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[y'+{2\over x}y={3\over x^3}+1,\quad y(1)=1\] at \(x=1.0\), \(1.1\), \(1.2\), \(1.3\), …, \(2.0\). Compare these approximate values with the values of the exact solution \[y={1\over3x^2}(9\ln x+x^3+2),\] which can be obtained by the method of Section 2.1. Present your results in a table like Table [table:3.1.1}.
[exer:3.1.8} Use Euler’s method with step sizes \(h=0.05\), \(h=0.025\), and \(h=0.0125\) to find approximate values of the solution of the initial value problem \[y'={y^2+xy-x^2\over x^2},\quad y(1)=2\] at \(x=1.0\), \(1.05\), \(1.10\), \(1.15\), …, \(1.5\). Compare these approximate values with the values of the exact solution \[y={x(1+x^2/3)\over1-x^2/3}\] obtained in Example [example:2.4.3}. Present your results in a table like Table [table:3.1.1}.
[exer:3.1.9} In Example [example:2.2.3} it was shown that \[y^5+y=x^2+x-4\] is an implicit solution of the initial value problem \[y'={2x+1\over5y^4+1},\quad y(2)=1. \eqno{\rm(A)}\] Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of (A) at \(x=2.0\), \(2.1\), \(2.2\), \(2.3\), …, \(3.0\). Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual \[R(x,y)=y^5+y-x^2-x+4\] for each value of \((x,y)\) appearing in the first table.
[exer:3.1.10} You can see from Example [example:2.5.1} that \[x^4y^3+x^2y^5+2xy=4\] is an implicit solution of the initial value problem \[y'=-{4x^3y^3+2xy^5+2y\over3x^4y^2+5x^2y^4+2x},\quad y(1)=1. \eqno{\rm(A)}\] Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of (A) at \(x=1.0\), \(1.1\), \(1.2\), \(1.3\), …, \(2.0\). Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual \[R(x,y)=x^4y^3+x^2y^5+2xy-4\] for each value of \((x,y)\) appearing in the first table.
[exer:3.1.11} Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[(3y^2+4y)y'+2x+\cos x=0, \quad y(0)=1; \mbox{\; (Exercise~2.2.~\hspace*{-3pt}\ref{exer:2.2.13})}\] at \(x=0\), \(0.1\), \(0.2\), \(0.3\), …, \(1.0\).
[exer:3.1.12} Use Euler’s method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[y'+{(y+1)(y-1)(y-2)\over x+1}=0, \quad y(1)=0 \mbox{\; (Exercise~2.2.~\hspace*{-3pt}\ref{exer:2.2.14})}\] at \(x=1.0\), \(1.1\), \(1.2\), \(1.3\), …, \(2.0\).
[exer:3.1.13} Use Euler’s method and the Euler semilinear method with step sizes \(h=0.1\), \(h=0.05\), and \(h=0.025\) to find approximate values of the solution of the initial value problem \[y'+3y=7e^{-3x},\quad y(0)=6\]
at \(x=0\), \(0.1\), \(0.2\), \(0.3\), …, \(1.0\). Compare these approximate values with the values of the exact solution \(y=e^{-3x}(7x+6)\), which can be obtained by the method of Section 2.1. Do you notice anything special about the results? Explain.
[exer:3.1.14} \(y'-2y= {1\over1+x^2},\quad y(2)=2\); \(h=0.1,0.05,0.025\) on \([2,3}\)
[exer:3.1.15} \(y'+2xy=x^2,\quad y(0)=3\) (Exercise 2.1. [exer:2.1.38}); \(h=0.2,0.1,0.05\) on \([0,2}\)
[exer:3.1.16} \( {y'+{1\over x}y={\sin x\over x^2},\quad y(1)=2}\); (Exercise 2.1. [exer:2.1.39}); \(h=0.2,0.1,0.05\) on \([1,3}\)
[exer:3.1.17} \( {y'+y={e^{-x}\tan x\over x},\quad y(1)=0}\); (Exercise 2.1. [exer:2.1.40}); \(h=0.05,0.025,0.0125\) on \([1,1.5}\)
[exer:3.1.18} \( {y'+{2x\over 1+x^2}y={e^x\over (1+x^2)^2}, \quad y(0)=1}\); (Exercise 2.1. [exer:2.1.41}); \(h=0.2,0.1,0.05\) on \([0,2}\)
[exer:3.1.19} \(xy'+(x+1)y=e^{x^2},\quad y(1)=2\); (Exercise 2.1. [exer:2.1.42}); \(h=0.05,0.025,0.0125\) on \([1,1.5}\)
[exer:3.1.20} \(y'+3y=xy^2(y+1),\quad y(0)=1\); \(h=0.1,0.05,0.025\) on \([0,1}\)
[exer:3.1.21} \( {y'-4y={x\over y^2(y+1)},\quad y(0)=1}\); \(h=0.1,0.05,0.025\) on \([0,1}\)
[exer:3.1.22} \( {y'+2y={x^2\over1+y^2},\quad y(2)=1}\); \(h=0.1,0.05,0.025\) on \([2,3}\)
[exer:3.1.23} Numerical Quadrature. The fundamental theorem of calculus says that if \(f\) is continuous on a closed interval \([a,b]\) then it has an antiderivative \(F\) such that \(F'(x)=f(x)\) on \([a,b]\) and \[\int_a^bf(x)\,dx=F(b)-F(a). \eqno{\rm(A)}\] This solves the problem of evaluating a definite integral if the integrand \(f\) has an antiderivative that can be found and evaluated easily. However, if \(f\) doesn’t have this property, (A) doesn’t provide a useful way to evaluate the definite integral. In this case we must resort to approximate methods. There’s a class of such methods called
numerical quadrature, where the approximation takes the form \[\int_a^bf(x)\,dx\approx \sum_{i=0}^n c_if(x_i), \eqno{\rm(B)}\] where \(a=x_0<x_1<\cdots<x_n=b\) are suitably chosen points and \(c_0\), \(c_1\), …, \(c_n\) are suitably chosen constants. We call (B) a quadrature formula. Derive the quadrature formula \[\int_a^bf(x)\,dx\approx h\sum_{i=0}^{n-1}f(a+ih) \tag{C}\] where \(h=(b-a)/n)\) by applying Euler’s method to the initial value problem\[y'=f(x),\quad y(a)=0.\] The quadrature formula (C) is sometimes called the left rectangle rule. Draw a figure that justifies this terminology. For several choices of \(a\), \(b\), and \(A\), apply (C) to \(f(x)=A\) with \(n = 10,20,40,80,160,320\). Compare your results with the exact answers and explain what you find. For several choices of \(a\), \(b\), \(A\), and \(B\), apply (C) to \(f(x)=A+Bx\) with \(n=10\), \(20\), \(40\), \(80\), \(160\), \(320\). Compare your results with the exact answers and explain what you find. |
If you sent all the gifts in the Twelve Days of Christmas, then how many would you send?
The gifts for each day are the triangular numbers :
\(\begin{array}{c | cc}
\text{Day}& \text{Count} & \text {Total}\\ \hline 1 & 1 & 1\\ 2 & 1 + 2 & 3\\ 3 & 1 + 2 + 3 & 6\\ 4 & 1 + 2 + 3 + 4 & 10\\ n & 1 + 2 + 3 + 4 +… +n & \frac {n (n + 1) }{2} \end{array}\)
We have done these before :
\( \displaystyle \sum_{k=1}^n k = \frac {n (n + 1)}{2} \)
In English, the sum from 1 to \(n\) of the counting numbers is equal to the last number multiplied by the last number plus one all divided by two.
On the twelfth day you would send :
\(\displaystyle \frac {12 \cdot 13}{2} = 6 \cdot 13 = 78 \) gifts.
This does not answer the question. How many gifts would be sent in total over the period of twelve days?
We have to sum the triangular numbers; that is, what is the sum of ?
\( \displaystyle \sum_{k=1}^n \frac {k(k+1)}{2} \)
This is equivalent to (take out the factor and expand)
\( \displaystyle \frac 12 \sum_{k=1}^n k^2 + k \)
Now, we know the sum of \(k\), so what is the sum of \(k^2\)? Don’t forget that the sum of the odd numbers \((2k + 1, k = 0,1,2,…)\) is \(n^2\). We have also done this before. Pay attention.
When we sum the square numbers we arrive at the square pyramid numbers (cannonballs) :
\( \begin {array}{c |c cc}
n & \text{Square} & \text{Count} & \text{Total}\\ \hline 1 & 1 & 1 & 1\\ 2 &4 & 1 + 4 & 5\\ 3 & 9 & 1 + 4 + 9 & 14\\ 4 & 16 & 1 + 4 + 9 + 16 & 30\\ n & n^2 & 1 + 4 + 9 + 16 + … +n^2 & \frac { n(n + 1)(2n +1)}{6} \end {array} \)
Now that we have all the information we can say that the sum of the triangular numbers is
\( \displaystyle \begin{align}\sum_{k=1}^n \frac{k(k+1)}{2} & = \frac 12 \Bigl (\frac {n(n+1)(2n+1)}{6} + \frac {n(n+1)}{2} \Bigr ) \\ & =\frac {n(n + 1)(n +2)}{6} \end{align}\)
These are the tetrahedral (triangular pyramid) numbers :
\(\begin{array}{c |c c c}
\text{Day} & \text{Triangle} &\text{Count} & \text {Total}\\ \hline 1 &1 & 1 & 1\\ 2 &3 & 1 + 3 & 4\\ 3 &6 & 1 + 3 + 6 & 10\\ 4 &10 & 1 + 3 + 6 + 10 & 20\\ n &\frac{n(n +1)}{2} & 1 + 3 + 6 + 10 + … +\frac{n(n + 1)}{2} & \frac{n(n +1)(n +2)}{6} \end{array} \)
Hence, if you sent all the gifts in the Twelve Day of Christmas, then you would send :
\(\displaystyle \frac{12\cdot13\cdot14}{6} = 2\cdot13\cdot14 = 364 \) gifts.
© OldTrout \(2018\)
No Audio file – it does not translate well |
The question says to solve this equation: $(z+1)^5 = z^5$
I did. Just want to find out if I did it properly and if my run-around logic makes sense.
First I begin my writing the equations as:
$$ (z+1)^5 = z^5$$ $$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$ So $$ \mathbf{Log}(z+1) = \ln|z+1| + \mathbf{Arg}(z+1)i $$ $$ \mathbf{Log}(z) = \ln|z| + \mathbf{Arg}(z)i $$
Now, because the natural logarithm is one-to-one, I write:
$$ \ln |z| = \ln |z+1| \Rightarrow |z| = |z+1|$$
So assign $ z = a +bi$
So that $|z| = \sqrt{a^2 +b^2} = |z+1| = \sqrt{(a+1)^2 +b^2} \Rightarrow a^2 +b^2 = (a+1)^2 +b^2 \Rightarrow a^2 = (a+1)^2 \Rightarrow a = -\frac12 $
So, $z = -\frac12 + bi$ and $z+1 = \frac12 + bi$ for some $b \in \mathbb R$
Now to find $b$
$$ \mathbf{Arg}(z+1) = \mathbf{Arg}(z)$$ $$ \tan^{-1} \frac{b}{\frac12} = \pi - \tan^{-1}\frac{b}{-\frac12}$$
I have a feeling this last part isn't quite right, so I just want to find out if I'm approaching this question properly?
Ultimately, I get $ z = -\frac12$ which upon inspection...is wrong... |
Well, I'm amazed that this works at all as it is in reality not supported input. The syntax for lists is
\begin{enumerate} \item ...
and nothing in between. So putting
\begin{multicols} in that space is at best adventurous.
However, if I run your document with the very latest
multicols from CTAN 1.8g everything works as far as I can see even if you might think the spacing is a bit strange but see below. I've taken your input and repeated the set of questions modifying the number of columns from 5 down to 2 and if I do this I get the following output:
And if I change the order of the environments to have correct input, i.e.,
\begin{multicols}{5}
\begin{enumerate}
\item $(3^{\pi})^{\pi} = \underline{\qquad}$ ; \\
\item $(e^{\pi})^{1/\pi} = \underline{\qquad}$ ; \\
\item $100^{3} \cdot 10^{5} = \underline{\qquad}$ ; \\
\item $(4^x)^2\cdot 4^{x^2} = \underline{\qquad}$ ; \\
\item $\bigl((5^2)^3\bigr)^4 = \underline{\qquad}$ ; \\
\end{enumerate}
\end{multicols}
the result stays (nearly) the same. The difference is that spacing between item label and text seems to change with the result that you text doesn't any longer fit into a single column if we have 5 columns. I guess the reason is that the label correctly now sits withing the columns whereas before it was actually ending up in the column separation because of that (unsupported interaction between the environments).
Making a quick test file that uses both your original code and the one with correctly ordered environments shows that this is indeed the case (I added
\columnseprule=.4pt to make this visible:
\documentclass[9pt]{amsart}
\usepackage[lmargin=.7in,rmargin=.9in,tmargin=1in,bmargin=1in]{geometry}
\geometry{letterpaper}
\usepackage{amsmath}
\usepackage{multicol}
\columnseprule.4pt
\begin{document}
Incorrectly ordered environments (multicols between enumerate and \verb=\item=)
\begin{enumerate}
\item I added an additional item in front to show what's going on: multicol works in the smaller space as the
enumerate indents. But at the same time the labels protrude outside of the space into the column separation.
\begin{multicols}{5}
\item $(3^{\pi})^{\pi} = \underline{\qquad}$ ; \\
\item $(e^{\pi})^{1/\pi} = \underline{\qquad}$ ; \\
\item $100^{3} \cdot 10^{5} = \underline{\qquad}$ ; \\
\item $(4^x)^2\cdot 4^{x^2} = \underline{\qquad}$ ; \\
\item $\bigl((5^2)^3\bigr)^4 = \underline{\qquad}$ ; \\
\end{multicols}
\end{enumerate}
Correctly ordered environments (multicols outside)
\begin{multicols}{5}
\begin{enumerate}
\item $(3^{\pi})^{\pi} = \underline{\qquad}$ ; \\
\item $(e^{\pi})^{1/\pi} = \underline{\qquad}$ ; \\
\item $100^{3} \cdot 10^{5} = \underline{\qquad}$ ; \\
\item $(4^x)^2\cdot 4^{x^2} = \underline{\qquad}$ ; \\
\item $\bigl((5^2)^3\bigr)^4 = \underline{\qquad}$ ; \\
\end{enumerate} \end{multicols}
\end{document}
If we run this we get:
I haven't checked what goes wrong with an older version of multicol. if it is just the horizontal spacing then what happens there I hope is explained in the example above.
But if also the vertical spacing was somewhat wrong then the reason is that I recently fixed a bunch of alignment issues for very special scenarios, precisely cases like multicol interacting with lists rather than with mainly straight text (for which it was originally written).
Note that version 1.8g may not yet automatically update on TeX distributions like TeX Live so you have to pick it up from CTAN. |
Eigenvalues and Eigenvectors Examples 1
Recall from the Eigenvalues and Eigenvectors page that the number $\lambda \in \mathbb{F}$ is said to be an eigenvalue of the linear operator $T \in \mathcal L (V)$ if $T(u) = \lambda u$ for some nonzero vector $u \in V$. The vectors $u$ such that $T(u) = \lambda u$ are called eigenvectors corresponding to the eigenvalue $\lambda$.
We will now look at some examples regarding eigenvalues of linear operators and eigenvectors corresponding to eigenvalues.
Example 1 Suppose that $V$ is a finite-dimensional vector space over $\mathbb{F}$, $T$ is a linear operator on $V$, and $\mathrm{dim} (\mathrm{null} (T)) = m > 0$. Prove that $T$ has at most $\mathrm{dim} (V) - m + 1$ distinct eigenvalues.
Suppose that $\mathrm{dim} (\mathrm{null} (T)) = m > 0$. Then $\mathrm{null} (T) \neq \{0 \}$ and so $\lambda = 0$ is an eigenvalue of the linear operator $T$. If $\lambda = 0$ is the only eigenvalue of $T$, then we are done. If not, then suppose that $T$ has $k$ non-zero eigenvalues, call them $\lambda_1, \lambda_2, ..., \lambda_k \in \mathbb{F}$. We will show that $k ≤ \mathrm{dim} (V) - m$.
Now let $v_i \in V$ be an eigenvector associated with the eigenvalue $\lambda_i$ for $i = 1, 2, ..., k$. This set of eigenvectors $\{ v_1, v_2, ..., v_k \}$ is linearly independent from an earlier theorem, so if we let $W = \mathrm{span} (v_1, v_2, ..., v_n)$ then $\mathrm{dim} (W) = k$.
No notice that $W \cap \mathrm{null}(T) = \{ 0 \}$. We can see this since if $w = a_1v_1 + a_2v_2 + ... + a_kv_k$ and $w \in W \cap \mathrm{null}(T)$ then:(1)
Since $\{ v_1, v_2, ..., v_k \}$ is a linearly independent set, then we have that $a_1\lambda_1 = 0$, $a_2 \lambda_2 = 0$, …, $a_k \lambda_k = 0$. Since $\lambda_i \neq 0$ for $i = 1, 2, ..., k$ (from the hypothesis made earlier) then we have that $a_i = 0$ for $i = 1, 2, ..., k$ and so:(2)
Hence we have that:(3)
Since both $W$ and $\mathrm{null}(T)$ are subspaces of $V$ we have that $\mathrm{dim} (W + \mathrm{null} (T)) ≤ \mathrm{dim} (V)$ and so:(4)
Thus we have that $k ≤ \mathrm{dim} (V) - m$.
Example 2 Let $T$ be a linear operator on $V$ and let $\mathrm{dim} ( \mathrm{range} (T)) = k$. Prove that $T$ has at most $k + 1$ distinct eigenvalues.
Suppose that $\lambda_1, \lambda_2, ..., \lambda_m$ are the distinct eigenvalues of $T$. Let $v_1, v_2, ..., v_m$ be corresponding eigenvectors to these eigenvalues. We note that of these eigenvalues, that at most one can be zero (since these $\lambda$ are distinct). So for $\lambda_j \neq 0$, $j = 1, 2, ..., m$ we have that:(5)
So we see that since at most one of $\lambda_1, \lambda_2, ..., \lambda_m$ equals zero, then we have that at least $m - 1$ of the vectors $v_1, v_2, ..., v_m$ are contained in $\mathrm{range} (T)$. However, these vectors are linearly independent and so:(6)
Therefore we see that $m$, the number of distinct eigenvalues of $T$ is less than or equal to $k + 1$.
Example 3 Suppose that $T$ is an invertible linear operator on $V$. Prove that the $\lambda \neq 0$ is an eigenvalue of $T$ if and only if $\frac{1}{\lambda}$ is an eigenvalue of $T^{-1}$.
$\Rightarrow$ First suppose that $\lambda \neq 0$ is an eigenvalue of $T$. Then for some vector $u \in V$ we have that $T(u) = \lambda u$ and since $\lambda \neq 0$ then $\frac{1}{\lambda} T(u) = u$ so $T \left ( \frac{1}{\lambda} u \right ) = u$.
Since $T$ is invertible, then $T^{-1}$ exists, and applying $T^{-1}$ to both sides of the equation above and we get that:(7)
So for $u \in V$ we have that $T^{-1}(u) = \frac{1}{\lambda} u$ so $\frac{1}{\lambda}$ is an eigenvalue of $T^{-1}$.
$\Leftarrow$ Now suppose that $\frac{1}{\lambda}$ is an eigenvalue of $T^{-1}$. Then for some vector $u \in V$ we have that $T^{-1} (u) = \frac{1}{\lambda} u$ and $T^{-1} (\lambda u) = u$. Since $T^{-1}$ is also invertible with $T$ as its inverse, then applying $T$ to both sides of this equation and we have that:(8)
So for $u \in V$ we have that $T(u) = \lambda u$ so $\lambda \neq 0$ is an eigenvalue of $T$. |
Calculating the Probability of a Sample Containing Bad Parts
Received a question from a reader this morning that will make a nice tutorial.
A box contains 27 black and 3 red balls. A random sample of 5 balls is drawn without replacement. What is the probability that the sample contains one red ball?
So here’s my thinking and two ways to solve this problem. Instead of red and black balls in an urn type problem, which is pretty abstract, let’s say we know 3 bad parts are in a bin of 30 total parts.
We need five parts to build a system and if one is bad, we have to repair the system, which has a cost.
Or, should we sort and test each part which also has a cost and takes time to accomplish?
The details will be in the cost of inspections and the cost of a system repair, yet we also need to know the probability of building a system with one bad part out of the five needed.
Probability and Combinations
My first thought in solving the problem was the nature of the sampling, without replacement. This is realistic in that we need five parts for the build, so selecting one and putting it back in the bin would serve no purpose.
Without replacement means that the number of parts available for selection changes and depending on if a good or bad (black or red) part is selected the number remaining changes respectively.
The other note to consider is the order of the final parts doesn’t matter, so thought about using combinations. Combinations provide a quick way to count the number of collections possible for a given situation. For example, if drawing 5 parts from the bin, there are 30 draw 5 different unique sets of results.
The idea of combinations led me to consider using the hypergeometric distribution to sort out the probability. There are four bits of information, 30 total parts, 27 good parts, 3 bad parts, drawing 5 parts. More on that later.
Another approach is to map out the array of possibilities using a branching diagram. It’s a brute force approach and not always convenient. Let’s explore that first.
Mapping Out All Possible Outcomes
Keep in mind that for this problem there are a finite number of potential outcomes, and the sum of all those possible outcomes must equal one. We are interested in one specific subset of outcomes, those with one, and only one, bad (red) part included in the set of parts.
Here’s my map (I stopped at three parts as it fit nicely on the paper I was using and illustrates the approach).
The first selected part is either good (black) or bad (red). There is a 3 in 30 chance of selecting a bad part and a 27 in 30 chance of selecting a good part. This stands to reason as there are just a few bad parts and many more good parts in the bin.
Let’s follow the path where we have already selected one bad part, as we need to select two good parts to have a result with only one bad part of the three selected.
The second draw is from a bin with 29 parts, and assuming we are selecting the parts where each part has an equal chance of being selected at random (not always a good assumption, btw), we have a 27 out of 29 chance of selecting a good part. All 27 good parts are still in the bin after the first selection.
The third draw is from a bin with 28 parts. And, there is one less good part, so we have a 26 in 28 chance of selecting a good part.
Adding two more selections just continues the map. When all the bad parts have been selected there are none left in the bin so there is no chance of selecting a bad part at that point. You can see the process here and with five draws the map just gets bigger, not more complex.
The probability of the possible outcome where we select a bad part followed by two good parts is the product of the probability for each selection along the path. 3/30 x 27/29 x 26/28 = 0.08645
There are three potential ways to have just one bad part in a set of three: Select the bad part first, or second, or third, and good parts otherwise. Thus, we calculate the probability of those three paths and add them to find the probability of having just one bad part in a collection of three parts.
If my math is correct there is a 25.935% chance of having one bad part out of three selected from a bin containing 3 bad parts and 27 good parts. In other words, if we build the system there is about a 26% chance we will have to conduct a repair.
One more note, to check the math and mapping, the sum of all the potential paths should equal one.
The cost of the repair times 0.26 is the net cost considering the risk of getting one bad one. To fully expand this problem we may want to know the chance of getting at least (one, two, or three) bad parts in the selection, which would be a bit higher than having just one bad part.
Hypergeometric Approach
Keep in mind that a probability is a ratio of some event or set of outcomes and the tally of all the possible outcomes. For example, with a coin flip, we count two possible outcomes, heads or tails. The chance of one coin toss resulting in a head is 1 divided by 2 or 50% (assuming a fair coin and it doesn’t land on its edge or falls into a crack in the floor never to be seen again…)
The hypergeometric distribution uses combination calculations of the counts of good parts, bad parts, and total combinations.
Continuing with the example of drawing three parts (not five) from the bin we have the following known bits of information:
30 total parts in the bin (N) 3 is the sample selected without replacement from the bin (n) We are interested in the probability of finding 1 bad part in the set (x) There are 3 bad parts (m)
$$ \large\displaystyle f(x,N,n,m)=\frac{\left( \begin{array}{l}m\\x\end{array} \right)\left( \begin{array}{l}N-m\\n-x\end{array} \right)}{\left( \begin{array}{l}N\\n\end{array} \right)}$$
where $$ \large\displaystyle \left( \begin{array}{l}m\\x\end{array} \right)=C_{x}^{m}=\frac{m!}{x!(m-x)!}$$
Let’s run the calculations.
$$ \large\displaystyle {f}{(}{1}{,}{30}{,}{3}{,}{27}{)}{=}\frac{\mbox{ $\left({\begin{array}{l}{3}\\{1}\end{array}}\right)$}\mbox{ $\left({\begin{array}{c}{{30}{-}{3}}\\{{3}{-}{1}}\end{array}}\right)$}}{\mbox{ $\left({\begin{array}{c}{30}\\{3}\end{array}}\right)$}}{=}\frac{\mbox{ $\left({3}\right)$}\mbox{ $\left({351}\right)$}}{4060}{=}{0}{.}{25935} $$
Same result, all good.
BTW: I used the Google spreadsheet function combin
(x,y) to do the combination calculations rather than resorting to my calculator and factorials. Summary
Now just follow the above example for a selection of 5 parts. A 5 step map or change from a sample selected from 3 to 5 (n) in the hypergeometric approach.
With a 26% chance of having one bad part out of three installed into a system the repair of the system has be pretty simple and quick and the inspection process for parts before assembly pretty expensive and time consuming before it makes sense to just build and hope you got all good parts.
Also published on Medium. |
The product converges because $\sum n^{-2}$ does. Can you see why? What is the greatest sum that can possibly appear when we expand the product?
In fact, suppose that $a_n\geq 0$ for each $n$. Set $$p_n=\prod_{k=1}^n a_k$$
Then $\log p_n=\sum_{k=1}^n\log a_k$
If $\sum a_k$ converges, then $a_k\to 0$, then since $$\lim_{x\to 0}\frac{\log (1+x)}x=1$$ so that $$\lim \frac{\log (1+a_n)}{a_n}=1$$
the comparison test tells us $\log p_n$ converges, say to $\ell$. By continuity of the logarithm, $p_n$ must converge to $p$ with $\log p=\ell$ so that $\lim p_n=e^\ell$.
Conversely, suppose $\log p_n=\sum_{k=1}^n\log a_k$ converges. This means that $\log(1+a_k)\to 0$ so that $a_k\to 0$. Comparison yields from $$\lim \frac{\log (1+a_n)}{a_n}=1$$ that $$\sum a_k$$ converges too. Thus, we have shown
PROP If $a_n\geq 0$ then $\prod (1+a_k)$ converges if and only if $\sum a_k$ (if and only if $\sum \log(1+a_k)$ does.) |
This post has been cross-posted on the Quansight LabsBlog.
As of November, 2018, I have been working at Quansight. Quansight is a new startup founded by the same people who started Anaconda, which aims to connect companies and open source communities, and offers consulting, training, support and mentoring services. I work under the heading of Quansight Labs. Quansight Labs is a public-benefit division of Quansight. It provides a home for a "PyData Core Team" which consists of developers, community managers, designers, and documentation writers who build open-source technology and grow open-source communities around all aspects of the AI and Data Science workflow.
My work at Quansight is split between doing open source consulting for various companies, and working on SymPy. SymPy, for those who do not know, is a symbolic mathematics library written in pure Python. I am the lead maintainer of SymPy.
In this post, I will detail some of the open source work that I have done recently, both as part of my open source consulting, and as part of my work on SymPy for Quansight Labs.
Bounds Checking in Numba
As part of work on a client project, I have been working on contributing codeto the numba project. Numba is a just-in-timecompiler for Python. It lets you write native Python code and with the use ofa simple
@jit decorator, the code will be automatically sped up using LLVM.This can result in code that is up to 1000x faster in some cases:
In [1]: import numbaIn [2]: import numpyIn [3]: def test(x): ...: A = 0 ...: for i in range(len(x)): ...: A += i*x[i] ...: return A ...:In [4]: @numba.njit ...: def test_jit(x): ...: A = 0 ...: for i in range(len(x)): ...: A += i*x[i] ...: return A ...:In [5]: x = numpy.arange(1000)In [6]: %timeit test(x)249 µs ± 5.77 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)In [7]: %timeit test_jit(x)336 ns ± 0.638 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)In [8]: 249/.336Out[8]: 741.0714285714286
Numba only works for a subset of Python code, and primarily targets code that uses NumPy arrays.
Numba, with the help of LLVM, achieves this level of performance through manyoptimizations. One thing that it does to improve performance is to remove allbounds checking from array indexing. This means that if an array index is outof bounds, instead of receiving an
IndexError, you will get garbage, orpossibly a segmentation fault.
>>> import numpy as np>>> from numba import njit>>> def outtabounds(x):... A = 0... for i in range(1000):... A += x[i]... return A>>> x = np.arange(100)>>> outtabounds(x) # pure Python/NumPy behaviorTraceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 4, in outtaboundsIndexError: index 100 is out of bounds for axis 0 with size 100>>> njit(outtabounds)(x) # the default numba behavior-8557904790533229732
In numba pull request #4432, I amworking on adding a flag to
@njit that will enable bounds checks for arrayindexing. This will remain disabled by default for performance purposes. Butyou will be able to enable it by passing
boundscheck=True to
@njit, or bysetting the
NUMBA_BOUNDSCHECK=1 environment variable. This will make iteasier to detect out of bounds issues like the one above. It will work like
>>> @njit(boundscheck=True)... def outtabounds(x):... A = 0... for i in range(1000):... A += x[i]... return A>>> x = np.arange(100)>>> outtabounds(x) # numba behavior in my pull request #4432Traceback (most recent call last): File "<stdin>", line 1, in <module>IndexError: index is out of bounds
The pull request is still in progress, and many things such as the quality of the error message reporting will need to be improved. This should make debugging issues easier for people who write numba code once it is merged.
removestar
removestar is a new tool I wrote toautomatically replace
import * in Python modules with explicit imports.
For those who don't know, Python's
import statement supports so-called"wildcard" or "star" imports, like
from sympy import *
This will import every public name from the
sympy module into the currentnamespace. This is often useful because it saves on typing every name that isused in the import line. This is especially useful when working interactively,where you just want to import every name and minimize typing.
However, doing
from module import * is generally frowned upon in Python. It isconsidered acceptable when working interactively at a
python prompt, or in
__init__.py files (removestar skips
__init__.py files by default).
Some reasons why
import * is bad:
It hides which names are actually imported. It is difficult both for human readers and static analyzers such aspyflakes to tell where a given name comes from when
import *is used. For example, pyflakes cannot detect unused names (for instance, from typos) in the presence of
import *.
If there are multiple
import *statements, it may not be clear which names come from which module. In some cases, both modules may have a given name, but only the second import will end up being used. This can break people's intuition that the order of imports in a Python file generally does not matter.
import *often imports more names than you would expect. Unless the module you import defines
__all__or carefully
dels unused names at the module level,
import *will import every public (doesn't start with an underscore) name defined in the module file. This can often include things like standard library imports or loop variables defined at the top-level of the file. For imports from modules (from
__init__.py),
from module import *will include every submodule defined in that module. Using
__all__in modules and
__init__.pyfiles is also good practice, as these things are also often confusing even for interactive use where
import *is acceptable.
In Python 3,
import *is syntactically not allowed inside of a function definition.
Here are some official Python references stating not to use
import * infiles:
In general, don’t use
from modulename import *. Doing so clutters the importer’s namespace, and makes it much harder for linters to detect undefined names.
PEP 8 (the official Python style guide):
Wildcard imports (
from <module> import *) should be avoided, as they make it unclear which names are present in the namespace, confusing both readers and many automated tools.
Unfortunately, if you come across a file in the wild that uses
import *, itcan be hard to fix it, because you need to find every name in the file that isimported from the
* and manually add an import for it. Removestar makes thiseasy by finding which names come from
* imports and replacing the importlines in the file automatically.
As an example, suppose you have a module
mymod like
mymod/ | __init__.py | a.py | b.py
with
# mymod/a.pyfrom .b import *def func(x): return x + y
and
# mymod/b.pyx = 1y = 2
Then
removestar works like:
$ removestar -i mymod/$ cat mymod/a.py# mymod/a.pyfrom .b import ydef func(x): return x + y
The
-i flag causes it to edit
a.py in-place. Without it, it would justprint a diff to the terminal.
For implicit star imports and explicit star imports from the same module,
removestar works statically, making use ofpyflakes. This means none of the code isactually executed. For external imports, it is not possible to work staticallyas external imports may include C extension modules, so in that case, itimports the names dynamically.
removestar can be installed with pip or conda:
pip install removestar
or if you use conda
conda install -c conda-forge removestar
sphinx-math-dollar
In SymPy, we make heavy use of LaTeX math in our documentation. For example, in our special functions documentation, most special functions are defined using a LaTeX formula, like
However, the source for this math in the docstring of the function uses RST syntax:
class besselj(BesselBase): """ Bessel function of the first kind. The Bessel `J` function of order `\nu` is defined to be the function satisfying Bessel's differential equation .. math :: z^2 \frac{\mathrm{d}^2 w}{\mathrm{d}z^2} + z \frac{\mathrm{d}w}{\mathrm{d}z} + (z^2 - \nu^2) w = 0, with Laurent expansion .. math :: J_\nu(z) = z^\nu \left(\frac{1}{\Gamma(\nu + 1) 2^\nu} + O(z^2) \right), if :math:`\nu` is not a negative integer. If :math:`\nu=-n \in \mathbb{Z}_{<0}` *is* a negative integer, then the definition is .. math :: J_{-n}(z) = (-1)^n J_n(z).
Furthermore, in SymPy's documentation we have configured it so that textbetween `single backticks` is rendered as math. This was originally done forconvenience, as the alternative way is to write
:math:`\nu` everytime you want to use inline math. But this has lead to many people beingconfused, as they are used to Markdown where `single backticks` produce
code.
A better way to write this would be if we could delimit math with dollarsigns, like
$\nu$. This is how things are done in LaTeX documents, as wellas in things like the Jupyter notebook.
With the new sphinx-math-dollarSphinx extension, this is now possible. Writing
$\nu$ produces $\nu$, andthe above docstring can now be written as
class besselj(BesselBase): """ Bessel function of the first kind. The Bessel $J$ function of order $\nu$ is defined to be the function satisfying Bessel's differential equation .. math :: z^2 \frac{\mathrm{d}^2 w}{\mathrm{d}z^2} + z \frac{\mathrm{d}w}{\mathrm{d}z} + (z^2 - \nu^2) w = 0, with Laurent expansion .. math :: J_\nu(z) = z^\nu \left(\frac{1}{\Gamma(\nu + 1) 2^\nu} + O(z^2) \right), if $\nu$ is not a negative integer. If $\nu=-n \in \mathbb{Z}_{<0}$ *is* a negative integer, then the definition is .. math :: J_{-n}(z) = (-1)^n J_n(z).
We also plan to add support for
$$double dollars$$ for display math so that
.. math :: is no longer needed either .
For end users, the documentation on docs.sympy.org will continue to render exactly the same, but for developers, it is much easier to read and write.
This extension can be easily used in any Sphinx project. Simply install it with pip or conda:
pip install sphinx-math-dollar
or
conda install -c conda-forge sphinx-math-dollar
Then enable it in your
conf.py:
extensions = ['sphinx_math_dollar', 'sphinx.ext.mathjax']
Google Season of Docs
The above work on sphinx-math-dollar is part of work I have been doing to improve the tooling around SymPy's documentation. This has been to assist our technical writer Lauren Glattly, who is working with SymPy for the next three months as part of the new Google Season of Docs program. Lauren's project is to improve the consistency of our docstrings in SymPy. She has already identified many key ways our docstring documentation can be improved, and is currently working on a style guide for writing docstrings. Some of the issues that Lauren has identified require improved tooling around the way the HTML documentation is built to fix. So some other SymPy developers and I have been working on improving this, so that she can focus on the technical writing aspects of our documentation.
Lauren has created a draft style guide for documentation at https://github.com/sympy/sympy/wiki/SymPy-Documentation-Style-Guide. Please take a moment to look at it and if you have any feedback on it, comment below or write to the SymPy mailing list. |
Eigenvalues and Eigenvectors Examples 2
Recall from the Eigenvalues and Eigenvectors page that the number $\lambda \in \mathbb{F}$ is said to be an eigenvalue of the linear operator $T \in \mathcal L (V)$ if $T(u) = \lambda u$ for some nonzero vector $u \in V$. The nonzero vectors $u$ such that $T(u) = \lambda u$ are called eigenvectors corresponding to the eigenvalue $\lambda$.
We will now look at some examples regarding eigenvalues of linear operators and eigenvectors corresponding to eigenvalues.
Example 1 Find the eigenvalues of the linear operator $T \in \mathcal (\mathbb{\mathbb{C}}^2)$ defined by $T(x, y) = (-y, x)$ and the corresponding set of eigenvectors for each eigenvalue.
Let $u = (x, y) \in \mathbb{C}^2$ be a nonzero vector in $\mathbb{C}^2$. We want to find the numbers $\lambda \in \mathbb{F}$ such that:(1)
The equation above gives us the following system of equations:(2)
Plugging the second equation into the first equation and dividing both sides by $y$ (which is nonzero since if $y = 0$ then this implies $x = 0$ and so $(x, y) = (0, 0)$) and:(3)
Therefore the eigenvalues of $T$ are $\lambda_1 = i$ and $\lambda_2 = -i$.
For the eigenvalue $\lambda_1 = i$ we have that $(x, y) = (x, - \lambda_1 x) = (x , -ix)$ where $x \neq 0$ are the corresponding eigenvectors to $\lambda_1 = i$.
For the eigenvalue $\lambda_2 = -i$ we have that $(x, y) = (x, -\lambda_2x) = (x, ix)$ where $x \neq 0$ are the corresponding eigenvectors to $\lambda_2 = -i$.
Example 2 Find the eigenvalues of the linear operator $T \in \mathcal (\mathbb{\mathbb{F}}^2)$ defined by $T(x, y) = (y, x)$ and the corresponding set of eigenvectors for each eigenvalue.
Let $u = (x, y)$. We want to find numbers $\lambda \in \mathbb{F}$ such that:(4)
From the equation above we get the following system of equations:(5)
Plugging the second equation into the first equation and we have that $y = \lambda ( \lambda y )$ and so $y = \lambda^2 y$. Once again, we note that $y \neq 0$ (otherwise $x = 0$ and so $u = (0, 0)$). Therefore can divide by $y$ to get that:(6)
Thus $\lambda_1 = 1$ and $\lambda_2 = -1$ are both eigenvalues of $T$.
For $\lambda_1 = 1$ we have that the set of vectors $(x, y) = (x, \lambda x) = (x, x)$ where $x \neq 0$ are the corresponding eigenvectors of $\lambda_1$.
For $\lambda_2 = -1$ we have that the set of vectors $(x, y) = (x, \lambda x) = (x, -x)$ where $x \neq 0$ are the corresponding eigenvectors of $\lambda_2$. |
@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". |
So my book proves the convergence of $\Gamma(z) = \int_0^{\infty}t^{z-1}e^{-t}dt$ in the right half plane $Re(z) > 0$, and then goes on to prove the initial recurrence relation $\Gamma(z+1)=z\Gamma(z)$ by applying integration by parts to $\Gamma(z+1)$:
$$\int_0^{\infty}t^{z}e^{-t}dt = -t^ze^{-t}|_0^{\infty} + z\int_0^{\infty}t^{z-1}e^{-t}dt$$
The book explicitly states this equality to be true only in the right half plane, since otherwise $-t^ze^{-t}|_0^{\infty} = \infty$, instead of equaling zero. With this initial recurrence relation we are 'supposably' able to analytically continue the Gamma function to $Re(z) > -1$ (not including the origin) by writing the relation in the form:
$$\Gamma(z) = \frac{\Gamma(z+1)}{z}$$
What I don't understand is this relation is still only true in the right half plane, since otherwise $-t^ze^{-t}|_0^{\infty}\neq 0$. I don't see what reason we have to believe that, for instance, $\Gamma(-\frac{1}{2}) = \frac{\Gamma(\frac{1}{2})}{-\frac{1}{2}}$.
Furthermore $\int_0^{\infty}t^{z-1}e^{-t}dt$ is clearly not convergent in the left half plane, so I can't even imagine why it would be plausible to think that a recurrence relation directly based on it could possibly lead to a genuine analytic continuation of its domain. |
Problem at hand (not homework): Let $N, K$ be subgroups of a group $G$, with $N$ normal in $G$. If $N$ and $K$ are abelian groups and $G= NK$, is $G$ the direct product of $N$ and $K$?
So, here's the criteria that I know:
Given normal subgroups $N_1, \cdots N_k$ of $G$ such that any element $g\in G$ can be written uniquely as $g=n_1\cdots n_k$, where $n_i\in N_i$. Then, $G\cong N_1\times\cdots\times N_k$.
If $M, N\trianglelefteq G$ such that $G=MN$ and $M\cap N=\{e\}$, then $G=M\times N$.
I think that conjecture is false, but I don't know how to go about utilizing the facts that I've listed above. I looked at the proofs of the above theorems (we used (1) to prove (2)), and they use normality. |
Eigenvalues and Eigenvectors Examples 3
Recall from the Eigenvalues and Eigenvectors page that the number $\lambda \in \mathbb{F}$ is said to be an eigenvalue of the linear operator $T \in \mathcal L (V)$ if $T(u) = \lambda u$ for some nonzero vector $u \in V$. The nonzero vectors $u$ such that $T(u) = \lambda u$ are called eigenvectors corresponding to the eigenvalue $\lambda$.
We will now look at some examples regarding eigenvalues of linear operators and eigenvectors corresponding to eigenvalues.
Example 1 Let $T \in \mathcal L(\mathbb{F}^3, \mathbb{F}^3)$ be defined by $T(x, y, z) = (3y, 2x, 0)$. Find all eigenvalues of $T$.
Let $u = (x, y, z)$. Then we want to find $\lambda \in \mathbb{F}$ such that:(1)
From the equation above we get the following system of equations:(2)
The third equation implies that $\lambda = 0$ or $z = 0$.
If $\lambda = 0$ then all three equations are satisfied, so $\lambda_1 = 0$ is an eigenvalue of $T$. We will also then have that $x = y = 0$.
Thus the set of vectors $(0, 0, z)$ such that $z \neq 0$ are the corresponding eigenvectors to $\lambda_1 = 0$.
If $z = 0$, then we must still satisfy the first two equations. From the second equation we have that $x = \frac{\lambda y}{2}$. Substituting this into the first equation and we have that:(3)
We note that $y \neq 0$ otherwise $x = y = 0$ and so $u = (0, 0, 0)$. Therefore we can divide both sides by $y$ to get that $\lambda_2 = \sqrt{6}$ and $\lambda_3 = -\sqrt{6}$ are both eigenvalues of $T$.
The set of vectors $\left (x, \frac{\lambda x}{3} ,0 \right ) = \left ( x, \frac{\sqrt{6}}{3} x, 0 \right )$ where $x \neq 0$ are the corresponding eigenvectors to $\lambda_1 = \sqrt{6}$.
The set of vectors $\left ( x, \frac{\lambda x}{3}, 0 \right ) = \left ( x, - \frac{\sqrt{6}}{3}x, 0 \right )$ where $x \neq 0$ are the corresponding eigenvectors to $\lambda_2 = -\sqrt{6}$.
Example 2 Let $T$ be a linear operator on $V$ be such that every vector $v \in V$ is an eigenvector to some eigenvalue of $T$ Prove that then $T = aI$ where $a \in \mathbb{F}$.
Since every vector $v \in V$ is an eigenvector to some eigenvalue of $T$, then for every $v \in V$ we have that $T(v) = \lambda v$ where $\lambda$ is an eigenvalue of $T$. Note that if $v = 0$ then any choice of $\lambda$ will suffice this equation.
Therefore, suppose that $u, v \in V \setminus \{ 0 \}$. We have that $T(u) = \lambda u$ and $T(v) = \mu v$. We want to show that $\lambda = \mu$.
Suppose that $\{ u, v \}$ is a linearly dependent set of vectors. Then $v$ is a scalar multiple of $u$, so $v = bu$ for some $b \in \mathbb{F}$ and thus:(4)
The equation above implies that $\mu = \lambda$.
Now suppose that $\{ u, v \}$ is a linearly independent set of vectors. For the vector $u + v$ let $\gamma$ be the eigenvalue such that $T(u + v) = \gamma (u + v)$. Then we have that:(5)
Since $\{ u, v \}$ is a linearly independent set of vectors, the equation above implies that $\gamma = \lambda$ and $\gamma = \mu$, so $\lambda = \mu$.
Therefore $T$ must be a scalar multiple of the identity operator, that is, $T = aI$ for $a \in \mathbb{F}$.
Example 3 Let $T$ be an operator on the finite-dimensional vector space $V$. Prove that every vector $v \in V$ is an eigenvector of $T$ if and only if $ST = TS$ for all linear operators $S \in \mathcal L (V)$.
$\Rightarrow$ Suppose that every vector $v \in V$ is an eigenvector of $T$. Then from example 2, we must have that $T = aI$ for some $a \in \mathbb{F}$. Therefore we have that:(6)
Therefore $ST = TS$.
$\Leftarrow$ Now suppose that $ST = TS$. If this is true for every linear operator $S \in \mathcal L (V)$ then this implies that $\mathcal M (S) \mathcal M (T) = \mathcal M (T) \mathcal M (S)$, i.e, matrix commutativity holds for all $S$. Therefore we must have that $T = a I$ for some $a \in \mathbb{F}$. Therefore every vector $v \in V$ is an eigenvector of $T$ (that belongs to the eigenvalue $a$). |
Note to Self: Getting in Touch with My Inner Austrian: Memo to Self: A Start of a Model...: How can you be an Austrian, and yet have big swings in the desired capital stock—and thus bigger swings in the desired rate of investment—without having to have massive shocks to the current level of technology? How can you get a big downward shock to desired investment without committing yourself to massive technological amnesia?
This is how:
You can do it with two shocks to the dividends paid on a unit of capital: a shock $ \epsilon $ to the level of $ d_{t} $ and a shock $ \eta $ to the growth rate $ g_{t} $ of $ d_{t} $, thus:
Why this dividend process? Because if we then have a required expected rate of return r and a value function $ V(d_{t},g_{t}) = d_{t}v(g_{t}) $, then the one-period arbitrage condition:
$ (1+r)d_{t}v(g_{t}) = d_{t} + E_{t}\left(d_{t}{v(g_{t+1})}\right) $
leads us to
$ (1+r)v(g_{t}) = 1 + \left(\frac{1+g_{t}}{2}\right)\left((1+\theta_{t}){v(g_{t} + \eta)} + (1-\theta_{t}){v(g_{t} - \eta)}\right) $
$ (r-g_{t})v(g_{t}) = 1 + \left(\frac{1+g_{t}}{2}\right)\left((1+\theta_{t}){v(g_{t} + \eta)} + (1-\theta_{t}){v(g_{t} - \eta)} - 2v(g_{t})\right) $
Then simply postulate:
$ v(g_{t}) = \frac{1}{r-g_{t}} $
And find:
$ 1 = 1 + \left(\frac{1+g_{t}}{2}\right)\left(\frac{1+\theta_{t}}{r - g_{t} - \eta} + \frac{1-\theta_{t}}{r - g_{t} + \eta} -\frac{2}{r - g_{t}}\right) $
$ 0 = \left(\frac{(1+\theta_{t})(r - g_{t} + \eta)}{(r - g_{t})^2 - (\eta)^2} + \frac{(1-\theta_{t})(r - g_{t} - \eta)}{(r - g_{t})^2 - (\eta)^2} \right) $
$ 0 = \frac{r - g_{t} + \eta + \theta_{t}r - \theta_{t}g_{t} + \theta_{t}\eta + r - g_{t} - \eta - \theta_{t}r + \theta_{t}g_{t} + \theta_{t}\eta}{(r - g_{t})^2 - (\eta)^2} - \frac{2}{r - g_{t}} $
$ 0 = \frac{2(r - g_{t}) + 2\theta_{t}\eta}{(r - g_{t})^2 - (\eta)^2} - \frac{2}{r - g_{t}} $
$ \frac{1}{r - g_{t}} = \frac{(r - g_{t}) + \theta_{t}\eta}{(r - g_{t})^2 - (\eta)^2} $
$ 1 = \frac{(r - g_{t})^2 + \theta_{t}\eta({r - g_{t}})}{(r - g_{t})^2 - (\eta)^2} $
$ - \eta = \theta_{t}({r - g_{t}}) $
$ \theta_{t} = - \frac{\eta}{r - g_{t}} $
Why all the $ \theta_{t} $ terms? Because you cannot have the expected growth rate of dividends $ g_{t} $ random walk. There is some chance it will random walk upwards and valuations will explode, and a chance of valuations exploding in the future means that valuations explode now (if they haven't already exploded). If you just have:
Then if you assume the Gordon equation for your value function, you find that it is wrong, for your arbitrage equation leads to:
$ 1 = 1 + \left(\frac{1+g_{t}}{2}\right)\left(\frac{1}{r - g_{t} - \eta} + \frac{1}{r - g_{t} + \eta} -\frac{2}{r - g_{t}}\right) $
which then generates the unpleasant:
$ 0 = \left(\frac{r - g_{t}}{(r - g_{t})^2 - (\eta)^2} -\frac{1}{r - g_{t}}\right) $
Writing, instead:
with:
$ \theta_{t} = - \frac{\eta}{r - g_{t}} $
imparts just enough downward drift to the random walk of the growth rate $ g_(t) $ to make the Gordon equation the right valuation equation.
You can complain that this process is not stable: with a constantly-applied downward drift to the growth rate, for a large enough valute of $ t $ it is almost surely the case that $ r - g_{t} $ is very large, and thus that valuations are almost surely very low relative to dividends because dividends will then be almost surely shrinking fast.
So sue me...
(No calculations yet) Background Previous posts: This ipynb: https://www.dropbox.com/s/3wsggjajbm897zr/2017-11-26%20Getting%20in%20Touch%20with%20My%20Inner%20Austrian.ipynb?dl=0 |
You only need to consider the case $\mathfrak{h}_{s}^\ast(A) \lt \infty$, but you need to be a bit careful in choosing the outer approximations since swapping $\inf$ and $\sup$ certainly isn't allowed without some thinking. If you knew that you can always take the same set $E$ in the $\inf$ (which I will show in \eqref{eqn:ast} below) then you'd be essentially done, but it is impossible to say if you got so far or not.
I'm going to ignore $\alpha_s$ since it only is a scalar factor that doesn't play a rôle in the argument and the formulas will already get cumbersome enough without it.
The following argument is essentially the one from Fremlin's measure theory, Volume 4II, 471D, page 102. $\DeclareMathOperator{\diam}{diam}$ $\newcommand{\hsd}[1]{\mathfrak{h}_{s,#1}^\ast}$
Set $\delta_n = 2^{-n}$ and choose $(A_{i}^n)_{i \in \mathbb{N}}$ of diameter $\leq 2^{-n}$ such that $A \subset \bigcup_{i=1}^\infty A_{i}^n$ and that
\begin{equation}\tag{$1$}\label{eqn:eq1} \sum_{i=1}^\infty (\diam{A_{i}^n})^s \leq \hsd{2^{-n}}(A) +2^{-n},\end{equation}which is possible because of the definition of $\hsd{2^{-n}}$ as infimum.
Observe that there's no reason for the $A_{i}^n$ to be $\hsd{2^{-n}}$-measurable, let alone Borel, so we would like to “blow them up” slightly, so as to get open sets still approximating the $\hsd{2^{-n}}$-measure of $A$ well. The problem is that by doing so we will lose the diameter condition which appears in the definition of $\hsd{2^{-n}}$, but this isn't a serious problem: we can simply choose a larger $n$ and work with the sets we obtain from there. Here are the gory details:
Choose $0 \lt r_{i}^n \lt 2^{-n}$ so small that$$
(\diam{(A_{i}^n)} + 2r_{i}^n)^s \leq (\diam{A_{i}^n})^s + 2^{-n-i}.
$$Now put $U_{i}^n = \{x \in X\,:\,d(x,A_{i}^n) \lt r_{i}^n\}$ and note that $U_{i}^n \supset A_{i}^n$ is an open set whose diameter satisfies\begin{equation}\tag{$2$}\label{eqn:eq2}(\diam{U_{i}^n})^s \leq (\diam{A_{i}^n})^s + 2^{-n-i}.\end{equation}Let$$
B = \bigcap_{n=1}^\infty \bigcup_{i=1}^\infty U_{i}^n
$$and observe that $B$ is a $G_{\delta}$-set (countable intersection of open sets) containing $A$.
Given any $\delta \gt 0$, we can find $N$ such that $3 \cdot 2^{-N} \leq \delta$ so that for all $n \geq N$ we have $\diam{U_{i}^n} \leq \delta$. As $B \subset \bigcup_{i = 1}^\infty U_{i}^n$ we see from \eqref{eqn:eq1} and \eqref{eqn:eq2} that for $n \geq N$$$
\hsd{\delta}(B) \leq \sum_{i=1}^\infty (\diam{U_{i}^n})^s
\leq
\sum_{i=1}^\infty [(\diam{A_{i}^n})^s + 2^{-n-i}] \leq \hsd{2^{-n}}{(A)} + 2 \cdot 2^{-n}
$$Since $\hsd{2^{-n}}(A) \leq \mathfrak{h}_{s}^\ast(A)$ we get for $n \geq N$$$
\hsd{\delta}(B) \leq
\mathfrak{h}_{s}^\ast(A) + 2^{-n}
$$and letting $n \to \infty$ this gives
\begin{equation}\tag{$\ast$}\label{eqn:ast}\hsd{\delta}(B) \leq \mathfrak{h}_{s}^\ast(A)\end{equation}
for every $\delta \gt 0$. [
Note: this is stronger than your condition involving $\inf$ since we can specify the set $E = B$ and thus avoid the infimum]
Taking the $\sup$ over all $\delta$ in \eqref{eqn:ast} this yields$$
\mathfrak{h}_{s}^\ast(B) \leq \mathfrak{h}_{s}^\ast(A)
$$and since $B \supset A$ and $B$ is $\mathfrak{h}_{s}$-measurable (being Borel) we can finally concludethat$$
\mathfrak{h}_{s}(B) = \mathfrak{h}_{s}^\ast(A),
$$as desired. |
Consider the integral operator $T : C([0,1])\to C([0,1])$ given by
$$Tf(t)=\int_0^1 K(t,\tau)f(\tau)d\tau.$$
I'm solving one exercise which is to show this operator is bounded. The exercise is from a mathematical physics course.
Also, I want to show this
without any results from Lebesgue integration. So I'm disconsidering here theorems like the dominated convergence theorem.
My first idea was to consider the supremum norm. Indeed if we were considering it we would have:
$$\left|\int_0^1 K(t,\tau)f(\tau)d\tau\right|\leq \|K(t,\cdot)f\|_{\infty}$$
Now, $f$ is continuous and $[0,1]$ is compact so that $f$ has a maximum. But we have that $t$ dependence on $K$.
Also, although the exercise doesn't say anything about the norm or $K$, I believe the norm being considered here is the one which comes from the usual inner product $\langle, \rangle$ given by
$$\langle f,g\rangle = \int_0^1 f(t)g(t)dt.$$
I've been thinking about this for some time now and I didn't get any further. How can I show this operator is bounded? |
Given that $(ab)^2=(bc)^4=(ca)^x=abc$ Then what is the value of $x$?
$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$
Then I am lost, any other easier way to solve?
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Given that $(ab)^2=(bc)^4=(ca)^x=abc$ Then what is the value of $x$?
$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$
Then I am lost, any other easier way to solve?
Taking logarithm gives: $$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$$
then taking $2(\log a+\log b)=\log a+\log b+\log c$
and $4(\log b+\log c)=\log a+\log b+\log c$
we would get $\log a+\log b-\log c=0$ & $3\log b+3\log c-\log a=0$ and by solving these two equations we get $\log b=-\log c$
similarly $\log a=-\log b$ then the solution becomes obvious......as $x=\frac12$
Hint: Split these two equalities into
$$(ab)^2 = (ca)^x$$ and $$(bc)^4 = (ca)^x$$
Then use $\log$ on both equations and see what happens ;-)
Let’s try to solve this without logarithms, as simply as possible. This allows to extend the solution to negative and complex values of $a$,$b$ and $c$, and includes a more interesting set of solution in the specific case of $|b|=1$.
If $abc=0$, then at least two elements of $\{a,b,c\}$ are $0$, and $x$ can take any value, except $0$ if $0^0=1$.
From now on $abc\neq 0$ : $$(ab)^2=abc \Rightarrow c=ab$$ replacing $c$ by its value transforms the equalities into : $$(ab)^2=\left(ab^2\right)^4=\left(a^2b\right)^x$$ The first equality then gives $$a^2b^6=1$$ Replacing $a^2$ by $\frac1{b^6}$, we have then $$\frac1{b^4}=\left(\frac1{b^5}\right)^x \Leftrightarrow b^4=b^{5x}.$$
If $b=1$, this is true for all $x$ and $a=c=\pm1$
Otherwise, if $|b|≠1$, one has $\boxed{x=\frac45}$ and $a=\pm\frac1{b^3}$, $c=\pm\frac1{b^2}$. This is the solution you were looking for.
But, for $|b|=1$ and $b≠1$, the solution above is still true, but it is not the only one. Let’s define $β≠0$ by $b=e^{iβ}$. The conditions on $x$ then becomes $$5xβ=4β+2kπ, k∈\mathbb Z$$ giving the set of solutions $$x=\frac45+\frac{2kπ}{5β}, k∈\mathbb Z.$$
This includes, for example the non trivial solution where $a=b=i$, $c=-1$ and $x=4$. |
A sine sweep is a sine function that gradually changes frequency over time.
The sine sweep can also be called "sinusoidal sweep," "frequency sweep", or "chirp".
The functions
$$x(t)=\sin(2\,\pi(f_0\, t+\frac{f_1-f_0}{2\,T}t^2))$$
and
$$x(t)=\sin(2\,\pi\,f_0 \,T \frac{(\frac{f_1}{f_0})^{\frac{t}{T}}-1}{\ln(\frac{f_1}{f_0})})$$
for example, change frequency from f
0 to f 1 over the time T. The first function is called a linear sine sweep, as the derivative of the frequency term inside the sine with respect to the time t is linear. Similarly, the second function is an exponential sine sweep.
Both functions have an initial phase of zero. A different initial phase can be added as with any other sine function, if necessary. In both functions, replacing t with k / f
s, where k = 0, 1, 2, … and f s is the sampling frequency produces the discrete time equivalent of the functions. Example sine sweep
The following is an exponential sine sweep that changes frequency from 1 Hz to 50 Hz over 2 seconds.
Sine sweep sound
The following is a linear sine sweep, generated with the formula above, where the beginning frequency f
0 is 50 Hz, the end frequency is f 1 is 1000 Hz, and the time T is 1 second.
Click Play to hear the linear sine sweep.
The following is an exponential sine sweep with the same parameters – sweeping the frequencies between 50 Hz and 1000 Hz over 1 second.
Click Play to hear the exponential sine sweep.
Using the sine sweep
Sine sweeps are useful when obtaining the impulse response of natural reverberations to design an impulse reverb. In practice, one can play a sine sweep in a room, record the result, and deconvolve the result with the sine sweep inverted in time to obtain the impulse of the reverb.
An example of this is below. Typically, a sine sweep used for this purpose will span the full possible frequency spectrum (e.g., from 20 Hz to 22 kHz for the full spectrum of human hearing and to cover the maximum possible under the 44.1 kHz sampling rate). Recommendations for how long the sine sweep should be vary, but are usually between 3 seconds to 20 seconds. For ease of presentation, the example below uses a shorter sine sweep that does not span the full frequency spectrum.
Take the following impulse response a(t).
This graph imitates the impulse response of some reverb. It was produced by running an impulse through a feedforward comb filter and running the output of the comb filter through a Shroeder all pass filter (see All pass filter). This is a 300-point impulse response over the sampling frequency 500 Hz.
Convolve this impulse response with an exponential sine sweep between 1 Hz and 200 Hz. The convolution h(t), which by itself is not very interesting, is as follows.
In practice, when recording natural reverberations of a sine sweep, we will have this result and we will know the sine sweep employed, but we will not know the impulse response. To get the impulse response, we can convolve this result with the same sine sweep but inverted in time. If we do so in this example, we will get the following impulse response.
This impulse response is quite close to the initial impulse response.
There is, however, a good amount of "pre-ringing". This generated impulse response does not jump quickly from zero to the peak of the first notch but does that gradually.
The technique to reduce this pre-ringing is to deconvolve with the time inverted sine sweep with an adjusted amplitude. We reduce the gain of the frequencies in the sine sweep by 6 dB per octave. The gain will be 0 dB (no change) at the high frequency end of the sine sweep and will gradually decrease to
$$-6 \, \log_2 \frac{f_1}{f_0}$$
at the low frequency end. In the current example, since frequencies in the sine sweep change from 1 Hz to 200 Hz, the amplitude over the sine sweep will gradually, linearly change from 0 dB at 200 Hz to -46 dB at 1 Hz.
The new impulse response, after adjusting the amplitude of the sine sweep used in the deconvolution (and after we add some gain to the result), is below.
Note that the pre-ringing, if not adjusted for, may cause a notable decrease in the amplitude of high frequencies in the signal, to which the impulse response is applied later.
Deconvolution in the frequency domain
The deconvolution (the convolution with the time inverted sine sweep) in the example above was performed in the time domain. That is, each value a(k) of the impulse response at sample k was computed as the convolution of the time inverted sine sweep x(k) and the result of the first convolution h(k) as follows.
$$a(k)=\sum_{n=0}^{N} x(n) \, h(k-(N-n))$$
where N is the length of the sine sweep. Sine sweeps, however, and sine sweep recordings in practice could be relatively long, which would make this convolution very slow. An alternative is to use convolution or deconvolution in the frequency domain, relying on the fact that the Fourier transform of a convolution of two functions is the product of the Fourier transforms of the two functions.
$$F(h)=F(a * x)=F(a)\,F(x)$$
This, for example, the impulse response is the inverse Fourier transform of the division of the Fourier transform of the recorded sine sweep and the Fourier transform of the sine sweep. |
I have an unknown $n$-dimensional vector $x$ whose analytical expression depends on the following sum $x = z + Ba$ where the vector $z$ and the matrix $B\in \mathbb{R}^{n\times s}$ are given. So the $s$-dimensional vector $a$ is to be computed to find $x$.
The only assumption that we have is $x=0$ when we project $x$ onto the space spanned by $s$ different rows (that we don’t know their indices) of the matrix $B$ which has $n$ rows. To do this projection we can use $P_s\in \mathbb{R}^{n\times n}$ which is $1$ on the diagonal entries that correspond to the $s$ selected rows of $B$ and $0$ elsewhere. Hence, $P_s x= P_s z + P_s Ba=0 \implies a=-(P_sB)^{-1}P_sz$.
The main issue is that we don’t know the positions of these $s$ rows, so the problem is combinatorial and we need to go through all possible $n\choose s$ projections to find the exact $x$ which corresponds to the least cost $f(x)=\|y-Ax\|_2$ where $\|v\|_2=\big(\sum_iv_i^2\big)^{1/2}$, $y\in \mathbb{R}^{m\times 1}$ and the matrix $A\in \mathbb{R}^{m\times n}$ are given.
So my question is how I can reformulate my problem as a mixed-integer quadratic programming to go through all possible $n\choose s$ submatrices of $B$ formed by the $s$ selected rows and finally find the set of rows which corresponds to the least $f(x)$. |
I've been working through Gelman's Bayesian Data Analysis 3 text and have been trying to understand one of the hierarchical models revolving around rat tumors (Chapter 5). He uses a binomial model with p assigned a beta distribution. The Beta distribution has parameters $\alpha$ and $\beta$ which need a distribution for the fully Bayesian hierarchical model.
In order to create a noninformative distribution he parametrizes the model in terms of $\frac{\alpha}{\alpha+\beta}$ (prior mean) and an approximation of the standard deviation of the beta distribution $(\alpha+\beta)^{-1/2}$ . (described here too http://andrewgelman.com/2009/10/21/some_practical/) On his blog, he mentions not favoring this approach anymore and preferring weakly informative models, but I'd still like to understand the thinking that supports this model.
I have a few questions about this:
Why use an approximation here for the parametrization rather than the actual standard deviation of the Beta distribution? How did he arrive at this particular approximation? Bonus: What connection, if any, does this have to a Pareto distribution? I tried parametrizing this model with a Pareto(1.5,1) distribution for $\alpha+\beta$ and a uniform distribution on $\alpha/(\alpha+\beta)$ and ended up with $p(\alpha,\beta)\propto (\alpha+\beta)^{-3/2}$ but Gelman's approach seems to yield $p(\alpha,\beta)\propto (\alpha+\beta)^{-5/2}$ which disagrees with the gentleman writing into Gelman's blog in the link above.
The most detailed explanation of this problem I could find was here http://streaming.stat.iastate.edu/~stat444x/Class%20Notes/6-HierarchicalModels.pdf but this doesn't get at the questions I have.I've been stumped by this for a while and lecture notes available online at a number of universities seem to gloss over why some of these things are done and jump to the joint prior distribution. Any help or resources that could be offered would be really helpful and if this isn't the right forum for this question or I've displayed poor etiquette please tell me (haven't posted before on this).Thank you! |
Note: This begins [with DS] January 1, 2017.
Note: This resumes with NS January 30, 2019 Note: This starts with DA February 4, 2019 NA complete, Feb 26, 2019 to Sep 24, 2019. 100-digit eigenvalues - the regular pentagon Unit-edged Regular Pentagon, both Dirichlet and Neumann boundary conditionsBob Jones Winter 2017
This work is a numerical refinement of the initial GSVD sweep results.
The numbers on this page are the values of $\lambda$ to solutions of the Helmholtz problem $$ (\Delta+\lambda)\Psi(\mathbf{r}) = 0 \qquad\mathbf{r}\in\Omega , \qquad \mbox{D: }\Psi(\mathbf{r})=0\quad \mbox{or}\quad\mbox{N: } \frac{\partial\Psi}{\partial n}(\mathbf{r})=0 \qquad \mathbf{r}\in\partial\,\Omega $$ inside $\Omega$, the unit-edged regular pentagon, subject to either Dirichlet D or Neumann N boundary conditions on the pentagon's boundary $\partial\,\Omega$, where $n$ is the outward-pointing unit normal vector.
DA eigenvalues $\#(806,004) = 10892 =\mbox{tally}$ $\mathrm{d}(\lambda)=N_w(\lambda)-\#(\lambda)$ cumsum($\mathrm{d}(\lambda))$ |
GolfScript (23 chars)
{:^((1${\.**2^?%}+*}:f;
The sentinel result for a non-existent inverse is
0.
This is a simple application of Euler's theorem. \$x^{\varphi(2^n)} \equiv 1 \pmod {2^n}\$, so \$x^{-1} \equiv x^{2^{n-1}-1} \pmod {2^n}\$
Unfortunately that's rather too big an exponential to compute directly, so we have to use a loop and do modular reduction inside the loop. The iterative step is \$x^{2^k-1} = \left(x^{2^{k-1}-1}\right)^2 \times x\$ and we have a choice of base case: either
k=1 with
{1\:^(@{\.**2^?%}+*}:f;
or
k=2 with
{:^((1${\.**2^?%}+*}:f;
I'm working on another approach, but the sentinel is more difficult.
The key observation is that we can build the inverse up bit by bit: if \$xy \equiv 1 \pmod{2^{k-1}}\$ then \$xy \in \{ 1, 1 + 2^{k-1} \} \pmod{2^k}\$, and if \$x\$ is odd we have \$x(y + xy-1) \equiv 1 \pmod{2^k}\$. (If you're not convinced, check the two cases separately). So we can start at any suitable base case and apply the transformation \$y' = (x+1)y - 1\$ a suitable number of times.
Since \$0x \equiv 1 \pmod {2^0}\$ we get, by induction
\$x\left(\frac{1 - (x+1)^n}{x}\right) \equiv 1 \pmod {2^n}\$
where the inverse is the sum of a geometric sequence. I've shown the derivation to avoid the rabbit-out-of-a-hat effect: given this expression, it's easy to see that (given that the bracketed value is an integer, which follows from its derivation as a sum of an integer sequence) the product on the left must be in the right equivalence class if \$x+1\$ is even.
That gives the 19-char function
{1$)1$?@/~)2@?%}:f;
which gives correct answers for inputs which have an inverse. However, it's not so simple when \$x\$ is even. One potentially interesting option I've found is to add
x&1 rather than
1.
{1$.1&+1$?@/~)2@?%}:f;
This seems to give sentinel values of either \$0\$ or \$2^{n-1}\$, but I haven't yet proved that.
Taking that one step further, we can ensure a sentinel of \$0\$ for even numbers by changing the expression \$1 - (x+1)^n\$ into \$1 - 1^n\$:
{1$.1&*)1$?@/~)2@?%}:f;
That ties with the direct application of Euler's theorem for code length, but is going to have worse performance for large \$n\$. If we take the arguments the other way round, as
n x f, we can save one character and get to
22 chars:
{..1&*)2$?\/~)2@?%}:f; |
I am trying to solve the following trigonometric equation algebraically, where $0\leq \displaystyle x \leq2 \pi$
$$\sin2x = -\frac{1}{2}.$$
My answer must be an exact solution.
Here is what I have tried: If $\displaystyle \sin(30°) = \frac{1}{2}$, then $\displaystyle \sin (2\times 15°) = \frac{1}{2}$.
Sin is negative in quadrants III and IV, so with a reference angle of $15°$ degree, $x$ can equal $180°+15°=\bf{195°}$, and $360°-30°=\bf{345°}$
Feeling confident about my answer, I checked the solution with my graphing calculator. The graph intersects the x-axis at $345°$ ($6.021$ rad), but it doesn't intersect the graph at $195°$.
Where did I make my mistake? I know I can work backwards, by taking the radian values from the graph and finding their "degree equivalents", but I need to be able to solve this algebraically.
Edit
210 degrees/2 =
105 degrees
330 degrees/2 =
165 degrees
210+360/2 =
285 degrees
330+360/2 =
345 degrees
Answer:
x= 105˚, 165˚, 285˚, 345˚ |
Will be glad for a little hint: let x and n be positive integer such that $1+x+x^2+\dots+x^{n-1}$ is a prime number then show that n is prime
Hint $\ $ The sequence $\rm\:f_n = (x^n-1)/(x-1)\:$ is a divisibility sequence, i.e. $\rm\:m\:|\:n\:$ $\Rightarrow$ $\rm\:f_m\:|\:f_n.\:$
In fact it is a strong divisibility sequence, i.e. $\rm\:(f_m,f_n) = f_{\:\!(m,n)},\:$ which implies an intimate relationship between divisibility properties of the $\rm\:f_n\:$ and integers $\rm\:n\:$ (so, in particular, relation between notions associated with divisibility, such as irreducible = prime).
Let $n = kl$. Consider $$ (1 + x + ... + x^{k-1})(1 + x^k + x^{2k} + ... + x^{(l-1)k}) = (1 + x + ... + x^{n-1}) $$
Assuming that you meant that $1+x+x^2+\ldots+x^{n-1}$ is prime, note that this sum is $\frac{x^n-1}{x-1}$. If $n=ab$, we have $$\frac{x^n-1}{x-1}=\frac{(x^a)^b-1}{x-1}=\frac{(x^a-1)(1+x^a+x^{2a}+\ldots+x^{(b-1)a})}{x-1}\;,$$ and $\dfrac{x^a-1}{x-1}=\;$? |
I'm stuck on something in the derivation of the Principal Components.
We have random vectors $X$ of dimension $p$. We want to find linear combinations of $X$, $a'X,$ where $a \in \mathbb{R}^{p}$ that satisfy the constraints:
(1) $\max_{a_{1}\in\mathbb{R}^{p}} \big(Var(a_1' X) \big) $ subject to $a_1'a_1 =1$
(2) $\max_{a_{2}\in\mathbb{R}^{p}} \big(Var(a_2' X )\big)$ subject to $a_2'a_2 =1$ and $Cov(a_1'X, a_2'X) = 0$
and so on.
Letting the covariance matrix of vector $X$ be denoted $\Sigma$, the proof proceeds to satisfy the first constraint by recognizing
$$\max_{a \ne 0} \big(a'\Sigma a \big) = \lambda_1 $$,
where $\lambda_1$ is the largest eigenvalue of $\Sigma$. Since this maximum is known to be achieved by the eigenvector $e_1$ associated with $\lambda_1$, this gives us $a_1 = e_1$. Fine, I understand that. It allows us to find a (unit-length) vector that satisfies the first equation.
But, then it proceeds like this:
$$ max_{a \perp e_1}\big( a'\Sigma a \big) = \cdots$$
Why are we introducing that perpendicular sign? How do we know that the space of all $\{a : a \perp e_1 \}$ is the largest possible space that can provide a maximum variance while satisfying the "uncorrelated with first principal component" constraint? Is there some relationship between orthogonal coeficcients to a linear combination and correlation? I have spent an hour or two failing to derive any such relationship.
(edit based on answer: I don't think there's a relationship between covariance and orthogonality of the coefficients in general, but here $a_1$ is an eigenvector, which gives such a relationship ) |
Vertex Colouring and Chromatic Numbers
Imagine that we could take the vertices of a graph and colour or label them such that the vertices of any edge are coloured (or labelled) differently. This is what we recognize as vertex colouring
Definition: A Good Vertex $k$-Colouring for $k \in \mathbb{Z^+}$ is a function $f: V(G) \to \{ 1, 2, 3, ... k\}$ for all $x, y \in V(G)$, whenever $\{ x , y \} \in E(G)$, $f(x) \neq f(y)$. A graph is said to be Vertex $k$-Colourable if it contains a good vertex $k$-colouring.
Note that we use the numbers $1, 2, ..., k$ to often represent colours.
Now let's look at the following graph for an example:
Clearly, the graph $G$ can have its vertices coloured with $4$ colours since every edge contains two different coloured vertices. We can thus say that $G$ is $4$-colourable. Once again, we note that we can easily replace the colours with numbers or any other labels. The question then arises if the graph $G$ can have its vertices coloured with less colours. In fact, the vertices in $G$ can be coloured with exactly $3$ colours, so $G$ is $3$-colourable.
Unfortunately, this graph is not $2$-colourable, so we say that minimum colourability of this graph is $3$. Often times determining this number is valuable in optimization problems. We will now define this number formally.
Definition: The Chromatic Number of a graph $G$ denoted $\chi (G)$ is the smallest positive integer $k$ such that $G$ is vertex $k$-colourable (and thus $G$ has a good vertex $k$-colouring).
For our graph in the earlier example, we can say that $\chi (G) = 3$. Let's now look at some chromatic numbers for important graphs:
Graph Chromatic Number $\chi (G)$ Explanation Complete Graphs $K_n$ $\chi (K_n) = n$ The chromatic number for complete graphs is n since by definition, each vertex is connected to one another. Hence each vertex must be coloured differently for a good colouring. Even Cycle Graphs, $C_{2n}$ $\chi (C_{2n}) = 2$ For even cycle graphs, we start with one vertex and alternate using the two colours of our choice. We will only ever need at minimum $2$ colours. Odd Cycle Graphs, $C_{2n+1}$ $\chi (C_{2n +1}) = 3$ For odd cycle graphs, we can also start at one vertex and alternate between two colours of our choice. We will eventually end up with one vertex that we cannot colour with the two colours we started with, hence, we will need a third colour. Complete Bipartite Graphs, $K_{r,s}$ $\chi (K_{r, s}) = 2$ By definition, the complete bipartite graphs can have their vertices partitioned into two sets. We can thus take these two sets and colour them with $2$ distinct colours. Tree graphs, $T$ $\chi (T) = 2$ We note that tree graphs are bipartite (and acyclic) allowing us to colour the vertices at minimum with 2 colours. Path Graphs, $P_n$ $\chi (P_n) = 2$ Once again, path graphs are tree graphs which are bipartite. Null Graphs, $N_n$ $\chi (N_n) = 1$ Null graphs contain no edges, so each vertex is isolated and can be coloured the same. |
Is there any practical difference as to how a unity gain buffer using an op-amp might be configured? For example, the input signal can be connected to the noninverting input and the feedback to the inverting input; but it can also be configured the other way, with the inverting input as the signal and the noninverting as the feedback. Practically speaking (noise, stability, frequency response, offset error and so on), is there a difference between these configurations?
A normal opamp has an infinite gain, practically [factor] x 10^5. The difference between + and - terminal determines its output:
Vout = (V+ - V-) * A_ol
For an opamp you will have 2 rules:
No input current. Input terminals share no voltage difference. This can be explained because A_cl for an ideal opamp is infinite, so (V+ - V-) should be 0V, otherwise Vout would be infinite too.
When you make a real circuit, you reduce the open loop gain to a closed loop gain. However, the 2 rules stated only work for negative feedback. If you use positive feedback, they do not apply.
So, if the rule of no input voltage difference doesn't apply, the opamp basically becomes an comperator. An inverting situation would try to get the difference to 0V because of its feedback. Now it will be become a simple comperator with Vout=H if V+ > V-, Vout=L if V+ < V-. In an wrong unity gain buffer, you'll see Vout=L because V+ is lower then the signal you're feeding it with.
Because I couldn't believe both situations would simulate the same, I did it myself:
Just 2 opamps which are internally fed to +/-15V. They follow a 1kHz 10Vpp source. The results are: (Note: Colors are inverted, so green = purple, cyan = red)
Oh so they do amplify correctly. But the ideal opamp has an infinite gain, no offset voltages, no input bias currents, no bandwith limitations (however, we wont notice much of that at 1kHz) etc. If we look at a real opamp, I picked one randomly (TL031):
An now it suddenly clips, because the opamp doesn't have the correct feedback.
If you use a feedback from the output to the non-inverting input you will find that as the output goes too high it causes the difference to increase and it will rail.
You have to use the feedback to the inverting input.
What is really happening?
An op-amp attempts to take the difference from the non-inverting compared to the inverting and multiply it by infinity. Lets say that the gain is only 10 though(instead of the 10^5 you can reasonable expect) and work through an example. There are two major points to remember here, the feedback from the output to the inverting input is nearly instant, this is not always true, but it is a reasonable approximation for low frequencies.
The circuit:
If, lets say, you instantly changed Vi from being 0V to 10V what would happen?
The instant that the voltage change happens your Vi is 10V, your Vo is 0V. This means that the op-amp has a difference of 0-10 V and will attempt to start swinging towards the -100V mark. As the voltage swings towards 0V the voltage it is targeting will go down. This will happen all the way until it reaches 0V. In a no load circuit with no delay, this will be instant.
In the real world you have delay, this can cause overshoot and ringing. In the real world, you have a load, so a higher gain gate will allow a higher speed transition.
The opamp's transfer function is
\$ \mathrm{V_{OUT} = G \times (V_+ - V_-)} \$
In our circuit with the positive feedback that becomes
\$ \mathrm{V_{OUT} = G \times (V_{OUT} - V_{IN})} \$
\$ \mathrm{(-G + 1) \times V_{OUT} = -G \times V_{IN}} \$
\$ \mathrm{V_{OUT} = \dfrac{G}{G - 1} \times V_{IN}} \$
For an ideal opamp \$\mathrm{G}\$ is infinite, then
\$ \mathrm{V_{OUT} = \dfrac{\infty}{\infty - 1} \times V_{IN} = V_{IN}} \$
And because the ideal opamp is infinitely fast it can follow the input voltage perfectly. That's the reason why it works in your simulator.
How are real opamps different? Well, first they don't have infinite gain, and second they're not infinitely fast. Real opamps have a gain in the order of 100 000. But it's speed which will kill our voltage follower. Opamps tend to oscillate, and the early opamps had to be compensated in the designer's circuit, which was a PITA. Current opamps have internal compensation which makes them a lot more user-friendly. The compensation limits the bandwidth, and introduces a propagation delay from input to output.
Let's start with both inputs at 0 V. If \$\mathrm{V_-}\$ rises 1 \$\mu\$V the output (and hence the non-inverting input) won't follow immediately. We get a small negative voltage difference which gets amplified by 100 000 to become -100 mV at the output. That results in a new input difference of -100.001 mV (we had +1 \$\mu\$V on the inverting input :-)), which again gets amplified by 100 000, and the output goes to the negative rail.
This circuit has two stable states: the output to the positive supply rail, and to the negative supply rail. |
Hints will display for most wrong answers; explanations for most right answers. You can attempt a question multiple times; it will only be scored correct if you get it right the first time. To see a new question, reload the page.
I used the official objectives and sample test to construct these questions, but cannot promise that they accurately reflect what’s on the real test. Some of the sample questions were more convoluted than I could bear to write. See terms of use. See the MTEL Practice Test main page to view questions on a particular topic or to download paper practice tests.
MTEL General Curriculum Mathematics Practice
Question 1
Use the expression below to answer the question that follows: \( \large \dfrac{\left( 7,154 \right)\times \left( 896 \right)}{216}\) Which of the following is the best estimate of the expression above?
2,000
Hint:
The answer is bigger than 7,000.
20,000
Hint:
Estimate 896/216 first.
3,000
Hint:
The answer is bigger than 7,000.
30,000
Hint:
\( \dfrac{896}{216} \approx 4\) and \(7154 \times 4\) is over 28,000, so this answer is closest.
If you found a mistake or have comments on a particular question, please contact me (please copy and paste at least part of the question into the form, as the numbers change depending on how quizzes are displayed). General comments can be left here. |
Recall that the exponential function \(f(x)=2^{x}\) produces this table of values
\(x\) -3 -2 -1 0 1 2 3 \(f(x)\) \(\frac{1}{8}\) \(\frac{1}{4}\) \(\frac{1}{2}\) 1 2 4 8
Since the logarithmic function is an inverse of the exponential, \(g(x)=\log _{2} (x)\) produces the table of values
\(x\) \(\frac{1}{8}\) \(\frac{1}{4}\) \(\frac{1}{2}\) 1 2 4 8 \(g(x)\) -3 -2 -1 0 1 2 3
In this second table, notice that
1. As the input increases, the output increases.
2. As input increases, the output increases more slowly. 3. Since the exponential function only outputs positive values, the logarithm can only accept positive values as inputs, so the domain of the log function is \((0,\infty )\). 4. Since the exponential function can accept all real numbers as inputs, the logarithm can output any real number, so the range is all real numbers or \((-\infty ,\infty )\).
Sketching the graph, notice that as the input approaches zero from the right, the output of the function grows very large in the negative direction, indicating a vertical asymptote at
\(x = 0\).
In symbolic notation we write
as \(x \to 0^{+} ,f(x) \to -\infty\), and as\(x \to \infty ,f(x) \to \infty\)
Graphical Features of the Logarithm
Graphically, in the function \(g(x)=\log _{b} (x)\)
The graph has a horizontal intercept at (1, 0)
The graph has a vertical asymptote at \(x = 0\)
The graph is increasing and concave down
The domain of the function is \(x > 0\), or \((0,\infty )\)
The range of the function is all real numbers, or \((-\infty ,\infty )\)
When sketching a general logarithm with base \(b\), it can be helpful to remember that the graph will pass through the points (1, 0) and (\(b\), 1).
To get a feeling for how the base affects the shape of the graph, examine the graphs below.
Notice that the larger the base, the slower the graph grows. For example, the common log graph, while it grows without bound, it does so very slowly. For example, to reach an output of 8, the input must be 100,000,000.
Another important observation made was the domain of the logarithm. Like the reciprocal and square root functions, the logarithm has a restricted domain which must be considered when finding the domain of a composition involving a log.
Example 1
Find the domain of the function \(f(x)=\log (5-2x)\).
Solution
The logarithm is only defined with the input is positive, so this function will only be defined when \(5 - 2x > 0\). Solving this inequality,
\(\begin{array}{l} {-2x>-5} \\ {x<\frac{5}{2} } \end{array}\)
The domain of this function is \(x < \frac{5}{2}\), or in interval notation, \(\left(-\infty ,\frac{5}{2} \right)\)
Exercise
Find the domain of the function \(f(x)=\log (x-5)+2\); before solving this as an inequality, consider how the function has been transformed.
Answer
Domain: {\(x\) | \(x > 5\)}
Transformations of the Logarithmic Function
Transformations can be applied to a logarithmic function using the basic transformation techniques, but as with exponential functions, several transformations result in interesting relationships.
First recall the change of base property tells us that \(\log _{b} x=\frac{\log _{c} x}{\log _{c} b} =\frac{1}{\log _{c} b} \log _{c} x\)
From this, we can see that \(\log _{b} x\) is a vertical stretch or compression of the graph of the \(\log _{c} x\) graph. This tells us that a vertical stretch or compression is equivalent to a change of base. For this reason, we typically represent all graphs of logarithmic functions in terms of the common or natural log functions.
Next, consider the effect of a horizontal compression on the graph of a logarithmic function. Considering \(f(x)=\log (cx)\), we can use the sum property to see
\(f(x)=\log (cx)=\log (c)+\log (x)\)
Since log(\(c\)) is a constant, the effect of a horizontal compression is the same as the effect of a vertical shift.
Example 2
Sketch \(f(x)=\ln (x)\) and \(g(x)=\ln (x)+2\).
Solution
Graphing these,
Note that this vertical shift could also be written as a horizontal compression, since \(g(x)=\ln (x)+2=\ln (x)+\ln (e^{2} )=\ln (e^{2} x).\)
While a horizontal stretch or compression can be written as a vertical shift, a horizontal reflection is unique and separate from vertical shifting.
Finally, we will consider the effect of a horizontal shift on the graph of a logarithm.
Example 3
Sketch a graph of \(f(x)=\ln (x+2)\).
Solution
This is a horizontal shift to the left by 2 units. Notice that none of our logarithm rules allow us rewrite this in another form, so the effect of this transformation is unique. Shifting the graph,
Notice that due to the horizontal shift, the vertical asymptote shifted to \(x = -2\), and the domain shifted to \((-2,\infty )\).
Combining these transformations,
Example 4
Sketch a graph of \(f(x) = 5log(-x + 2)\).
Solution
Factoring the inside as \(f (x) = 5log(−(x − 2))\) reveals that this graph is that of the
common logarithm, horizontally reflected, vertically stretched by a factor of 5, and shifted to the right by 2 units.
The vertical asymptote will be shifted to \(x\) = 2,
and the graph will have domain \((\infty, 2)\) . A rough
sketch can be created by using the vertical asymptote along with a couple points on the graph, such as
\(f (1) = 5log(−1+ 2) = 5log(1) = 0\)
\(f (−8) = 5log(−(−8) + 2) = 5log(10) = 5\)
Exercise
Sketch a graph of the function \(f(x)=-3\log (x-2)+1\).
Answer
transormations of logs
Any transformed logarithmic function can be written in the form
\(f(x)=a\log (x-b)+k\), or \(f(x)=a\log \left(-\left(x-b\right)\right)+k\) if horizontally reflected,
where \(x = b\) is the vertical asymptote.
Example 5
Find an equation for the logarithmic function graphed.
Solution
This graph has a vertical asymptote at \(x = –2\) and has been vertically reflected. We do not know yet the vertical shift (equivalent to horizontal stretch) or the vertical stretch (equivalent to a change of base). We know so far that the equation will have form
\(f(x)=-a\log (x+2)+k\)
It appears the graph passes through the points (–1, 1) and (2, –1). Substituting in (–1, 1),
\(\begin{array}{l} {1=-a\log (-1+2)+k} \\ {1=-a\log (1)+k} \\ {1=k} \end{array}\)
Next, substituting in (2, –1),
\(\begin{array}{l} {-1=-a\log (2+2)+1} \\ {-2=-a\log (4)} \\ {a=\frac{2}{\log (4)} } \end{array}\)
This gives us the equation \(f(x)=-\frac{2}{\log \eqref(4)} \log (x+2)+1\).
This could also be written as \(f(x)=-2\log _{4} (x+2)+1\).
Exercise
Write an equation for the function graphed here.
Answer
The graph is horizontally reflected and has a vertical asymptote at \(x = 3\), giving form \(f(x)=a\log \left(-\left(x-3\right)\right)+k\). Substituting in the point (2,0) gives \(0=a\log \left(-\left(2-3\right)\right)+k\), simplifying to \(k = 0\). Substituting in (-2,-2), \(-2=a\log \left(-\left(-2-3\right)\right)\), so \(\frac{-2}{\log (5)} =a\).The equation is \(f(x)=\frac{-2}{\log (5)} \log \left(-\left(x-3\right)\right)\) or \(f(x)=-2\log _{5} \left(-\left(x-3\right)\right)\).
flashback
Write the domain and range of the function graphed in Example 5, and describe its long run behavior.
Answer
Domain: {\(x\) | \(x\) > -2}, Range: all real numbers; As \(x\to -2^{+} ,f(x)\to \infty\)and as \(x\to \infty ,f(x)\to -\infty\).
Important Topics of this Section Graph of the logarithmic function (domain and range) Transformation of logarithmic functions Creating graphs from equations Creating equations from graphs |
If I have two normally distributed independent random variables $X$ and $Y$ with means $\mu_X$ and $\mu_Y$ and standard deviations $\sigma_X$ and $\sigma_Y$ and I discover that $X+Y=c$, then (assuming I have not made any errors) the conditional distribution of $X$ and $Y$ given $c$ are also normally distributed with means $$\mu_{X|c} = \mu_X + (c - \mu_X - \mu_Y)\frac{ \sigma_X^2}{\sigma_X^2+\sigma_Y^2}$$ $$\mu_{Y|c} = \mu_Y + (c - \mu_X - \mu_Y)\frac{ \sigma_Y^2}{\sigma_X^2+\sigma_Y^2}$$ and standard deviation $$\sigma_{X|c} = \sigma_{Y|c} = \sqrt{ \frac{\sigma_X^2 \sigma_Y^2}{\sigma_X^2 + \sigma_Y^2}}.$$
It is no surprise that the conditional standard deviations are the same as, given $c$, if one goes up the other must come down by the same amount. It is interesting that the conditional standard deviation does not depend on $c$.
What I cannot get my head round are the conditional means, where they take a share of the excess $(c - \mu_X - \mu_Y)$ proportional to the original variances, not to the original standard deviations.
For example, if they have zero means, $\mu_X=\mu_Y=0$, and standard deviations $\sigma_X =3$ and $\sigma_Y=1$ then conditioned on $c=4$ we would have $E[X|c=4]=3.6$ and $E[Y|c=4]=0.4$, i.e. in the ratio $9:1$ even though I would have intuitively thought that the ratio $3:1$ would be more natural.
Can anyone give an intuitive explanation for this?
This was provoked by a Math.SE question |
In this section, we demonstrate which integers have primitive roots. We start by showing that every power of an odd prime has a primitive root and to do this we start by showing that every square of an odd prime has a primitive root.
If \(p\) is an odd prime with primitive root \(r\), then one can have either \(r\) or \(r+p\) as a primitive root modulo \(p^2\).
Notice that since \(r\) is a primitive root modulo \(p\), then \[ord_pr=\phi(p)=p-1.\] Let \(m=ord_{p^2}r\), then \[r^m\equiv 1(mod \ p^2).\] Thus \[r^m\equiv 1(mod \ p).\] By Theorem 54, we have \[p-1\mid m.\] By Exercise 7 of section 6.1, we also have that \[m\mid \phi(p^2).\] Also, \(\phi(p^2)=p(p-1)\) and thus \(m\) either divides \(p\) or \(p-1\). And since \(p-1\mid m\) then we have \[m=p-1 \ \ \mbox{or} \ \ m=p(p-1).\] If \(m=p(p-1)\) and \(ord_{p^2}r=\phi(p^2)\) then \(r\) is a primitive root modulo \(p^2\). Otherwise, we have \(m=p-1\) and thus \[r^{p-1}\equiv 1(mod \ p^2).\] Let \(s=r+p\). Then \(s\) is also a primitive root modulo \(p\). Hence, \(ord_{p^2}s\) equals either \(p-1\) or \(p(p-1)\). We will show that \(ord_{p^2}s\neq p-1\) so that \(ord_{p^2}s=p(p-1)\). Note that \[\begin{aligned} s^{p-1}=(r+p)^{p-1}&=&r^{p-1}+(p-1)r^{p-2}p+...+p^{p-1}\\&=& r^{p-1}+(p-1)p.r^{p-2}(mod \ p^2).\end{aligned}\] Hence \[p^2\mid s^{p-1}-(1-pr^{p-2}.\] Note also that if \[p^2 \mid (s^{p-1}-1),\] then \[p^2\mid pr^{p-2}.\] Thus we have \[p\mid r^{p-2}\] which is impossible because \(p\nmid r\). Because \(ord_{p^2}s\neq p-1\), we can conclude that \[ord_{p^2}s=p(p-1)=\phi(p^2).\] Thus, \(s=r+p\) is a primitive root of \(p^2\).
Notice that 7 has 3 as a primitive root. Either \(ord_{49}3=6\) or \(ord_{49}3=42\). But since \(3^6\not\equiv 1(mod \ 49)\). Hence \(ord_{49}3=42\). Hence 3 is a primitive root of 49.
We now show that any power of an odd prime has a primitive root.
Let \(p\) be an odd prime. Then any power of \(p\) is a primitive root. Moreover, if \(r\) is a primitive root modulo \(p^2\), then \(r\) is a primitive root modulo \(p^m\) for all positive integers \(m\).
By Theorem 62, we know that any prime \(p\) has a primitive root \(r\) which is also a primitive root modulo \(p^2\), thus \[\label{1} p^2\nmid (r^{p-1}-1).\] We will prove by induction that \[\label{2} p^m\nmid (r^{p^{m-2}(p-1)}-1)\] for all integers \(m\geq 2\). Once we prove the above congruence, we show that \(r\) is also a primitive root modulo \(p^m\). Let \(n=ord_{p^m}r\). By Theorem 54, we know that \(n\mid \phi(p^m)\). Also, we know that \(\phi(p^m)=p^m(p-1)\). Hence \(n\mid p^m(p-1)\). On the other hand, because \[p^m\mid (r^n- 1),\] we also know that \[p\mid (r^n-1).\] Since \(\phi(p)=p-1\), we see that by Theorem 54, we have \(n=l(p-1)\). also \(n\mid p^{m-1}(p-1)\), we have that \(n=p^s(p-1)\), where \(0 \leq s\leq m-1\). If \(n=p^s(p-1)\) with \(s\leq m-2\), then \[p^k\mid r^{p^{m-2}(p-1)}-1,\] which is a contradiction. Hence \[ord_{p^m}r=\phi(p^m).\]
We prove now ([2]) by induction. Assume that our assertion is true for all \(m\geq 2\). Then \[p^m\nmid (r^{p^{m-2}(p-1)}-1).\] Because \((r,p)=1\), we see that \((r,p^{m-1})=1\). We also know from Euler’s theorem that \[p^{m-1}\mid (r^{p^{m-2}(p-1)}-1).\] Thus there exists an integer \(k\) such that \[r^{p^{m-2}(p-1)}=1+kp^{m-1}.\] where \(p\nmid k\) because \(r^{p^{m-2}(p-1)}\not\equiv 1(mod \ p^m)\). Thus we have now \[\begin{aligned} r^{p^{m-1}(p-1)}&=&(1+kp^{m-1})^p\\ &\equiv&1+kp^m(mod \ p^{m+1})\end{aligned}\] Because \(p\nmid k\), we have \[p^{m+1}\nmid (r^{p^{m-1}(p-1)}-1).\]
Since 3 is a primitive root of 7, then 3 is a primitive root for \(7^k\) for all positive integers \(k\).
In the following theorem, we prove that no power of 2, other than 2 or 4, has a primitive root and that is because when \(m\) is an odd integer, \(ord_2^km\neq \phi(2^k)\) and this is because \(2^k\mid (a^{\phi(2^k)/2}-1)\).
If \(m\) is an odd integer, and if \(k\geq 3\) is an integer, then \[m^{2^{k-2}}\equiv 1(mod \ 2^k).\]
We prove the result by induction. If \(m\) is an odd integer, then \(m=2n+1\) for some integer \(n\). Hence, \[m^2=4n^2+4n+1=4n(n+1)+1.\] It follows that \(8\mid (m^2-1)\).
Assume now that \[2^k\mid (m^{2^{k-2}}-1).\] Then there is an integer \(q\) such that \[m^{2^{k-2}}=1+q.2^{k}.\] Thus squaring both sides, we get \[m^{2^{k-1}}=1+q.2^{k+1}+q^22^{2k}.\] Thus \[2^{k+1}\mid (m^{2^{k-1}}-1).\]
Note now that 2 and 4 have primitive roots 1 and 3 respectively.
We now list the set of integers that do not have primitive roots.
If \(m\) is not \(p^a\) or \(2p^a\), then \(m\) does not have a primitive root.
Let \(m=p_1^{s_1}p_2^{s_2}...p_i^{s_i}\). If \(m\) has a primitive root \(r\) then \(r\) and \(m\) are relatively prime and \(ord_mr=\phi(m)\). We also have, we have \((r,p^s)=1\) where \(p^s\) is of the primes in the factorization of \(m\). By Euler’s theorem, we have \[p^s\mid (r^{\phi(p^s)}-1).\] Now let \[L=[\phi(p_1^{s_1}), \phi(p_2^{s_2}),...,\phi(p_i^{s_i})].\] We know that \[r^L\equiv 1(mod \ p_k^{s_k})\] for all \(1\leq k\leq m\). Thus using the Chinese Remainder Theorem, we get \[m\mid (r^L-1),\] which leads to \(ord_mr=\phi(m)\leq L\). Now because \[\phi(m)=\phi(p_1^{s_1})\phi(p_2^{s_2})...\phi(p_n^{s_n})\leq [\phi(p_1^{s_1}),\phi(p_2^{s_2}),...,\phi(p_n^{s_n})].\] Now the inequality above holds only if \[\phi(p_1^{s_1}),\phi(p_2^{s_2}),...,\phi(p_n^{s_n})\] are relatively prime. Notice now that by Theorem 41, \[\phi(p_1^{s_1}),\phi(p_2^{s_2}),...,\phi(p_n^{s_n})\] are not relatively prime unless \(m=p^s\) or \(m=2p^s\) where \(p\) is an odd prime and \(t\) is any positive integer.
We now show that all integers of the form \(m=2p^s\) have primitive roots.
Consider a prime \(p\neq 2\) and let \(s\) is a positive integer, then \(2p^s\) has a primitive root. In fact, if \(r\) is an odd primitive root modulo \(p^s\), then it is also a primitive root modulo \(2p^s\) but if \(r\) is even, \(r+p^s\) is a primitive root modulo \(2p^s\).
If \(r\) is a primitive root modulo \(p^s\), then \[p^s\mid (r^{\phi(p^s)}-1)\] and no positive exponent smaller than \(\phi(p^s)\) has this property. Note also that \[\phi(2p^s)=\phi(p^s),\] so that \[p^s\mid (r^{\phi(2p^s)}-1).\]
If \(r\) is odd, then \[2\mid (r^{\phi(2p^s)}-1).\] Thus by Theorem 56, we get \[2p^s\mid (r^{\phi(2p^s)}-1).\] It is important to note that no smaller power of \(r\) is congruent to 1 modulo \(2p^s\). This power as well would also be congruent to 1 modulo \(p^s\) contradicting that \(r\) is a primitive root of \(p^s\). It follows that \(r\) is a primitive root modulo \(2p^s\).
While, if \(r\) is even, then \(r+p^s\) is odd. Hence \[2\mid ((r+p^s)^{\phi(2p^s)}-1).\]
Because \(p^s\mid (r+p^s-r)\), we see that \[p^s\mid ((r+p^s)^{\phi(2p^s)}-1).\] As a result, we see that \(2p^s\mid ((r+p^s)^{\phi(2p^s)}-1)\) and since for no smaller power of \(r+p^s\) is congruent to 1 modulo \(2p^s\), we see that \(r+p^s\) is a primitive root modulo \(2p^s\).
As a result, by Theorem 63, Theorem 65 and Theorem 66, we see that
The positive integer \(m\) has a primitive root if and only if \(n=2,4, p^s\) or \(2p^s\)
for prime \(p\neq 2\) and \(s\) is a positive integer.
Exercises
Which of the following integers 4, 12, 28, 36, 125 have a primitive root.
Find a primitive root of 4, 25, 18.
Find all primitive roots modulo 22.
Show that there are the same number of primitive roots modulo \(2p ^s\) as there are modulo \(p^s\), where \(p\) is an odd prime and \(s\) is a positive integer.
Find all primitive roots modulo 25.
Show that the integer \(n\) has a primitive root if and only if the only solutions of the congruence \(x^2\equiv 1(mod n)\) are \(x\equiv \pm1 (mod \ n)\). |
Disclaimer. This is not an answer! But I hope the following transformations and numerical experiments can be useful in finding the solution.
First, we will get rid of trigonometric functions. Setting $t=\tan \phi$ we obtain:
$$\int_0^\infty \frac{t^a [1+t^2+2(1-t^2) \ln t]}{(1+t^b)(1+t^2)^2}dt$$
Here $a=\frac{1}{2}\sqrt{2}$, $b=2\sqrt{2}$. It makes sense to pose a more general problem with $a=\frac{1}{2}c$, $b=2c$ for some real parameter $c$.
Thus we have:
$$I(c)=\int_0^\infty \frac{t^{c/2} [1+t^2+2(1-t^2) \ln t]}{(1+t^{2c})(1+t^2)^2}dt$$
Here is a plot of $I(c)$. There appear to be only two roots $c= \pm \sqrt{2}$ - the function appears to be even.
It's easy to prove that $I(c)=I(-c)$. Let's set $t=\frac{1}{u}$:
$$I(c)=-\int_\infty^0 \frac{u^{-c/2} [1+u^{-2}-2(1-u^{-2}) \ln u]}{u^2(1+u^{-2c})(1+u^{-2})^2}du$$
Multiplying both sides by $u^2$ we obtain:
$$I(c)=\int_0^\infty \frac{u^{-c/2} [1+u^2+2(1-u^2) \ln u]}{(1+u^{-2c})(1+u^2)^2}du=I(-c)$$
Let's separate the integral into to parts: $$I=\int_0^\infty=\int_0^1+\int_1^\infty$$
Making the substitution $t=\frac{1}{u}$ we obtain:
$$\int_1^\infty \frac{t^{c/2} [1+t^2+2(1-t^2) \ln t]}{(1+t^{2c})(1+t^2)^2}dt=\int_0^1 \frac{u^{3c/2} [1+u^2+2(1-u^2) \ln u]}{(1+u^{2c})(1+u^2)^2}du$$
Thus, we can introduce another form of the integral with finite limits:
$$I(c)=\int_0^1 \frac{t^{c/2}(1+t^c) [1+t^2+2(1-t^2) \ln t]}{(1+t^{2c})(1+t^2)^2}dt$$
The plot of the integrand is presented below:
Because now $0<t<1$ we can expand a part of the integrand into a series:
$$I(c)=\sum_{k=0}^\infty (-1)^k \int_0^1 \frac{t^{(2k+1/2)c}(1+t^c) [1+t^2+2(1-t^2) \ln t]}{(1+t^2)^2}dt$$
Expanding the numerator we obtain $8$ integrals in the form:
$$\int_0^1 \frac{t^\alpha}{(1+t^2)^2} dt \quad \text{and} \quad \int_0^1 \frac{t^\alpha \ln t}{(1+t^2)^2} dt$$
The first integral can be evaluated using the digamma function while the second integral can be obtained by differentiating the first w.r.t. $\alpha$.
Thus, the integral is equal to the following series:
$$I(c)=\frac{1}{16} \sum_{k=0}^\infty (-1)^k \times$$
$$ \times \left[ 4 \left(\psi\left(\frac{c (4 k+1)+2}{8}\right)-\psi \left(\frac{c (4 k+1)+2}{8}+\frac{1}{2}\right)\right)+ \\ +4 \left(\psi \left(\frac{c (4 k+3)+2}{8}\right)-\psi \left(\frac{c (4 k+3)+2}{8}+\frac{1}{2}\right)\right)+ \\ +c (4 k+1) \left(\psi ^{(1)}\left(\frac{c (4 k+1)+2}{8}\right)-\psi ^{(1)}\left(\frac{c (4 k+1)+2}{8}+\frac{1}{2}\right)\right)+ \\ +c (4 k+3) \left(\psi ^{(1)}\left(\frac{c (4 k+3)+2}{8}\right)-\psi ^{(1)}\left(\frac{c (4 k+3)+2}{8}+\frac{1}{2}\right)\right) \right]$$
This seems too complicated, but I have hope that some properties of the digamma function can help in showing that $I(\sqrt{2})=0$.
A little more generalization. Consider the integral with two parameters:
$$I(a,b)=\int_0^\infty \frac{t^a [1+t^2+2(1-t^2) \ln t]}{(1+t^b)(1+t^2)^2}dt=$$
$$=\int_0^1 \frac{(t^a+t^{b-a}) [1+t^2+2(1-t^2) \ln t]}{(1+t^b)(1+t^2)^2}dt$$
$$I(a,b)=I(b-a,b)$$
The curve $I(a,b)=0$ is represented below:
We see it has two branches. So far I haven't been able to find other algebraic pairs on it except for:
$$I \left(\frac{1}{2} \sqrt{2},2\sqrt{2} \right)=0 \\ I \left(\frac{3}{2} \sqrt{2},2\sqrt{2} \right)=0$$
And the trivial one:
$$I(1,2)=0$$
Update:
The integral can be simplified in the following way:
$$\int_0^\infty \frac{t^a [1+t^2+2(1-t^2) \ln t]}{(1+t^b)(1+t^2)^2}dt=\frac{1}{2} \int_0^\infty \frac{t^{a-1} [1+t^2+2(1-t^2) \ln t]}{(1+t^b)(1+t^2)^2}dt^2$$
Introducing $v=t^2$, $\alpha=\frac{a-1}{2}$ and $\beta=\frac{b}{2}$ we need to investigate:
$$J(\alpha,\beta)=\int_0^\infty \frac{v^{\alpha} [1+v+(1-v) \ln v]}{(1+v^\beta)(1+v)^2}dv=\int_0^1 \frac{(v^{\alpha}+v^{\beta-\alpha-1}) [1+v+(1-v) \ln v]}{(1+v^\beta)(1+v)^2}dv$$
$$J(\alpha,\beta)=J(\beta-\alpha-1,\beta)$$
This integral seems much better. We now need to prove that:
$$J\left(\frac{\sqrt{2}-2}{4},\sqrt{2} \right)=0$$ or $$J\left(\frac{3\sqrt{2}-2}{4},\sqrt{2} \right)=0$$
Update 2:
Some further work on the integral. From the comment of @user90369 we introduce a new form of the integral:
$$Y(A):=\int_0^\infty \frac{\cosh Ax}{\cosh 2Ax} \frac{\cosh x-2x \sinh x}{\cosh^2 x} dx$$
Setting $t=e^x$ we obtain:
$$Y(A)=2 I(2A)$$
Separating the integral into two parts and introducing a new parameter $P$ we have:
$$Y_1(A,P):=\int_0^\infty \frac{\cosh Ax}{\cosh 2Ax} \frac{1}{\cosh P x} dx$$
$$Y_2(A,P):=-2\int_0^\infty \frac{\cosh Ax}{\cosh 2Ax} \frac{x \sinh P x}{\cosh^2 P x} dx=2\frac{\partial}{\partial P} Y_1(A,P)$$
Thus, we can work with $Y_1$ first:
$$Y_1(A,P)=\frac{1}{A} \int_0^\infty \frac{\cosh y}{\cosh 2y} \frac{1}{\cosh p y} dy$$
Here $p=P/A$. Setting $e^y=z$ we obtain:
$$Y_1(p):=\int_0^\infty \frac{\cosh y}{\cosh 2y} \frac{1}{\cosh p y} dy=2 \int_0^1 \frac{1+z^2}{1+z^4} \frac{z^p}{1+z^{2p}} dz=\int_0^\infty \frac{1+z^2}{1+z^4} \frac{z^p}{1+z^{2p}} dz$$
Despite @Lucian's comment, I can't represent $Y_1(p)$ in terms of Beta function in
closed from, only as a series, but it might be possible. I would wager a closed form in terms of hypergeometric functions.
For particular values of $p$ there are closed forms. So far I've found closed forms for $p=(0,1/4,1/3,1/2,2/3,1,2,3,4,5)$. They are all algebraic multiples of $\pi$. But to evaluate the original integral, we need the closed form for general $p$. Seethis question.
Update 3:
If we allow double sums we can express the problem in a compact and almost symmetrical way:
Prove that for $B=\sqrt{2}$: $$\sum_{k,n=0}^\infty (-1)^{k+n} \frac{4k+1-(2n+1)B}{(4k+1+(2n+1)B)^2}=-\sum_{k,n=0}^\infty (-1)^{k+n} \frac{4k+3-(2n+1)B}{(4k+3+(2n+1)B)^2}$$ |
There are two major faults in this argument.
The first is the assumption that over an infinitesimal time with an infinitesimal additional force (hence doing approximately 0 work to the object) that the object will be moving with a non-infinitesimal velocity.
The second is the assumption that because the object is moving without accelerating, the object is not gaining energy.Specifically, the statement that
the net work done on the object is 0, so the object doesn't gain any energy even though its height increases is false.
To elaborate on the first point:The smaller the value of $\epsilon$, the longer the time $\delta$ will have to be in order to reach a velocity $v$. In order for the velocity not to be infinitesimal ($\simeq 0$), the value of $\epsilon$ will have to be equal to $m v/\delta$ (where $m$ is the mass of the object), which means that the work being applied by the force will be $(v / \delta) \cdot \frac{1}{2} m v \delta = \frac{1}{2}{m v^2}$. The work is only zero if the velocity is $0$, and if the velocity is $0$, the height will never change.
To elaborate on the second point: With the application of the external, velocity maintaining force, the object most assuredly does gain energy as it gains height, because the work being done on the object by our external applied force that maintains the object's velocity after its initial acceleration is being counteracted by the work of gravity (so that the object does not accelerate). However, because our applied work is external to the object and is applied in the direction of the object's motion, the work done by the gravitational field is
negative (because the force and the object's velocity are in opposite directions), meaning that energy is being stored in the interaction between gravity and the object; i.e., the object's gravitational potential energy is increasing. (If we remove the applied force, then the gravitational field will perform positive work on the object, decreasing the potential energy and increasing its kinetic energy as the object moves downward, in the same direction as the gravitational force.
The only way the object would not gain energy is if the force being used to initially accelerate the object and subsequently maintain its velocity were to come from a store of energy inside the object itself (chemical, electrical, or internal kinetic energy). The otherwise fallacious statement would be true because no external work would have been applied to the object.
If we consider the system as a whole, the energy that is being used to maintain the constant velocity of the object must come from somewhere, so if the system considered as a whole includes the object, the gravitational field, and the source of the applied force, only then can we say that no work is done. Even then, the object gains energy (its gravitational potential energy increases at the expense of the energy used to maintain its constant velocity), but the system as a whole does not. |
Actually the $\partial_\mu$ is a "total derivative", otherwise it wouldn't be a total divergence and we would not be able to get rid of it during the derivation of Lagrange equations
$$\partial_\mu \mathcal{L} \equiv \frac{\partial \mathcal{L}}{\partial \phi} \partial_\mu \phi + \frac{\partial \mathcal{ L}}{\partial ( \partial _\nu \phi)} \partial_\mu \partial _\nu \phi$$
This is a slightly confusing but in the end somewhat practical convention in classical field theory. For instance, you can see that this very same derivative is used in the Lagrange equations
$$\partial_\mu ( \frac{\partial \mathcal{ L}}{\partial ( \partial _\mu \phi)}) - \frac{\partial \mathcal{L}}{\partial \phi} = 0$$
etc. This notation is possible only thanks to the fact that the "real partial derivative" of $\mathcal{L}$ is postulated to be always zero and that there cannot thus be any ambiguity in what is meant by $\partial_\mu \mathcal{L}$.
Some authors prefer to use notation such as $ d/dx^\mu$ or $D/dx^\mu$ to underline the "totalness" but this always feels like notation abuse to me. My opinion is that if anything should change then it is the discussion of the "omitted pullback" $\mathcal{L}(\phi,...) \to \mathcal{L}(\phi(x^\mu),...)$, what does $\partial/\partial (\partial_\mu \phi)$ really mean in term of the pullback and so on. |
Let $A = \begin{pmatrix} \cos \delta & -\sin \delta & 0 \\ \sin \delta & \cos \delta & 0 \\ 0 & 0 & 1 \end{pmatrix}$, and let $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \delta & -\sin \delta \\ 0 & \sin \delta & \cos \delta \end{pmatrix}$ be rotation by $\delta$ along the $z$ and $x$ axes, respectively. In suitable coordinates, a progression down one step in the tree is either $AB$ or $AB^{-1}$.
The trace will fill (a.s.) a dense subset of the surface if and only if the closure of the group generated by $AB$ and $AB^{-1}$ is not a subgroup of $SO(3)$ of dimension zero or one.
The dimension zero closed subgroups of $SO(3)$ are either cyclic, dihedral, or symmetries of Platonic solids, and the dimension one closed subgroups are conjugates of $SO(2)$ and $O(2)$. Therefore, it suffices to determine which values of $\delta$ yield a pair of elements in either a conjugate of $O(2)$, or a conjugate of one of the three Platonic groups (isomorphic to $A_4$, $S_4$, and $A_5$).
In order for $AB$ and $AB^{-1}$ to both lie in a conjugate of $O(2)$ it is necessary and sufficient that they have a common eigenvector with eigenvalue $\pm 1$ - this eigenvector is the axis of rotation. Writing this requirement explicitly yields a polynomial identity in $\sin \delta$ and $\cos \delta$
(whose solutions I haven't enumerated yet). Edit: Some straightforward case elimination with the $z$ coordinate of a common eigenvector shows that $\delta$ must be an integer multiple of $\pi/2$.
For the Platonic solutions, we can narrow down the solution set using the criterion that the rotation $(AB^{-1})^{-1}(AB) = B^2$ lies in the group, and Platonic solids have rotational symmetries of order at most 5. This means $\delta$ is a multiple of $\pi/3$, $\pi/4$ or $\pi/5$.
Since the traces of $AB$ and $AB^{-1}$ are both $(\cos \delta)(2 + \cos \delta)$, we can compare with character table entries to see if that number is the trace of an element in a Platonic group. It was pretty easy to eliminate candidates by eyeball in SAGE.
Conclusion: The only values of $\delta$ where the image is not dense are $0$, $\pm \pi/2$, and $\pi$. |
Theorem: Let $q:X\rightarrow\mathbb{R}$ be a sublinear functional on a real linear space $X$. Let $M$ be a linear subspace of $X$ and suppose that $f:M\rightarrow\mathbb{R}$ is a linear functional such that $f(x)\leqslant q(x)$ for all $x\in M$. Then there exists a linear functional $F:X\rightarrow\mathbb{R}$ such that $F(x)\leqslant q(x)$ for all $x\in X$ and $F(x)=f(x)$ for all $x\in M$. Proof: let $S$ be the collection of all pairs $(M_1,f_1)$, where $M_1$ is a linear subspace of $X$ such that $M\subseteq M_1$ and $f_1:M_1\rightarrow\mathbb{R}$ is a linear extension of $f$ satisfying $f_1(x)\leqslant q(x)$ for all $x\in M_1$. By the claim above, $S\neq\emptyset$. For any $(M_1,f_1),(M_2,f_2)\in S$, define $(M_1,f_1)\preccurlyeq(M_2,f_2)$ if $M_1\subseteq M_2$ and $f_2$ is a linear extension of $f_1$. Then $\preccurlyeq$ is a partial order on $S$. Let $C=\{(M_i,f_i):i\in I\}$ be a chain in $S$. Let $\tilde{M}=\bigcup\limits_{i\in I}M_i$. Then $\tilde{M}$ is a linear subspace of $X$. Define $\tilde{f}:\tilde{M}\rightarrow\mathbb{R}$ by $\tilde{f}(x)=f_i(x)$ if $x\in M_i$ for some $i\in I$.
My questions are:
In Conways's book, $\tilde{M}$ is a linear subspace of $X$ since $C$ is a chain in $S$. I don't get it. $\tilde{M}$ is a linear subspace of $X$ isn't because each $M_i$ is a linear subspace of $X$? Why do we need $C$ is a chain in $S$?
How to show $\tilde{f}$ is linear? My attempt is :
For all $x,y\in\tilde{M}$, there exists $j\in I$ such that $x,y\in M_j$. For all $\alpha\in\mathbb{R}$, we have $$\tilde{f}(x+\alpha y)=f_j(x+\alpha y)=f_j(x)+\alpha f_j(y)=\tilde{f}(x)+\alpha\tilde{f}(y).$$ Hence, $\tilde{f}:\tilde{M}\rightarrow\mathbb{R}$ is linear. But my professor said it is not clear at all!! Can someone help me point out which part is missing?
Thank you in advance! |
My brother asked me to calculate the following integral before we had dinner and I have been working to calculate it since then ($\pm\, 4$ hours). He said, it has a beautiful closed form but I doubt it and I guess he has tried to trick me again (as usual). So I am curious, what is the closed form (
if any) of the following integral:
\begin{equation} \int_{-1}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\,dx \end{equation}
I have tried by parts method, partial fractions (stupid idea), converting into series (nothing familiar), many substitutions such as: $u=2x-1$, $u=1-x$, $x=\cos^2\theta$, etc, but I failed and got nothing. Wolfram Alpha also doesn't give an answer. Either he is lying to me or telling the truth, I don't know. Could anyone here please help me to obtain the closed form of the integral with any methods (
whatever it takes)? Any help would be greatly appreciated. Thank you. |
We present a new approach to the eigensystem multiscale analysis (EMSA) for the Anderson model that relies on the Wegner estimate. The EMSA treats all energies of the finite volume operator in an energy interval at the same time, simultaneously establishing localization of all eigenfunctions with eigenvalues in the energy interval with high probability. It implies all the usual manifestations of localization (pure point spectrum with exponentially decaying eigenfunctions, dynamical localization) for the Anderson model. The new method removes the restrictive level spacing hypothesis used in the previous versions of the EMSA, allowing for single site probability distributions that are H\"older continuous of order $\alpha \in (0,1]$. (Joint work with Alex Elgart.)
In this talk, we first consider quasi-periodic Schr\"odinger operators with finitely differentiable potentials. If the potential is analytic, there are numerous results. But not every result holds if one replaces the analyticity with a smoothness condition. We will give some positive results in this aspect, generalizing some interesting results in the analytic case to the finitely smooth case. This includes the global reducibility results, generalized Chamber's formula and their applications to the study of continuity of the spectra. Finally we will give a recent result on the continuity of spectral measure of multi frequency quasi-periodic Schr\"odinger operators with small analytic quasi-periodic potentials.
We develop the basic spectral theory of ergodic Schrodinger operators when the underlying dynamics are given by a conservativeergodic transformation of a \sigma-finite measure space. Some fundamental results, such as the Ishii--Pastur theorem carry over to theinfinite-measure setting. We also discuss some examples in which straightforward analogs of results from the probability-measure case do not hold. We will discuss some examples and some interesting open problems.
The talk is based on a joint work with M. Boshernitzan, D. Damanik, and M. Lukic.
Understanding the statistical properties of Laplacian eigenfunctions in general and their nodal sets in particular, have an important role in the field of spectral geometry, and interest both mathematicians and physicists. A quantum graph is a system of a metric graph with a self-adjoint Schrodinger operator. It was proven for quantum graphs that the number of points onwhich each eigenfunction vanish (also known as the nodal count) isbounded away from the spectral position of the eigenvalue by the first Betti number of the graph. A remarkable result by Berkolaiko and Weyand showed that the nodal surplus is equal to a magnetic stability index of the corresponding eigenvalue. A similar result for discrete graphs holds as well proved first by Berkoliako and later by Colin deVerdiere.
Both from the nodal count point of view and the magnetic point of view, it is interesting to consider the distribution of these indices over the spectrum. In our work, we show that such a density exists and defines a nodal count distribution. Moreover, this distribution is symmetric, which allows deducing the topology of a graph from its nodal count. Although for general graphs we can not a priori calculate the nodal count distribution, we proved that a certain family of graphs will have a binomial distribution. As a corollary, given any sequence of graphs from that family with an increasing number of cycles, the sequence of nodal count distributions, properly normalized, will converge to a normal distribution.A numerical study indicates that this property might be universal and led us to state the following conjecture. For every sequence of graphs with an increasing number of cycles, the corresponding sequence of properly normalized nodal count distributions will converge to a normal distribution.In my talk, I will present our latest results extending the numberof families of graphs for which we can prove the conjecture.
This talk is based on joint works with Ram Band (Technion) and Gregory Berkolaiko (Texas A&M)
By judiciously constructing local defects in graph models of multi-layer graphene, bound states can be constructed at energies that lie within the continuous spectrum of the associated Schrödinger operator. The layers may be stacked in AA or AB fashion. A necessary condition for this construction is the reducibility of the Fermi surface for the multi-layer structure. This is achieved due to a special reduction of the complex dispersion relation to a function of a single polynomial "composite" function of the quasimomenta.This is joint work with Wei Li and Stephen Shipman at LSU.
We consider discrete Schr\"odinger operators on $\ell^2(\mathbb{Z})$ with bounded random but not necessarily identically distributed values of the potential. The distribution at a given site is not assumed to be absolutely continuous (or to contain an absolutely continuous component). We prove spectral localization (with exponentially decaying eigenfunctions) as well as dynamical localization for this model.
An important ingredient of the proof is a non-stationary analog of the Furstenberg Theorem on random matrix products,which is also of independent interest.
Abstract: In this talk, I will give a brief introduction to several popular topics in the spectral theory of quasi-periodic Schrodinger operators. I will then talk about several sharp results we get recently on these topics (especially for almost Mathieu operator). Our results are based on quantitative almost reducibility, a method originally proposed by Dinaburg and Sinai. Finally I will explain the key points in developing and refining this method to get optimal results. |
Difference between revisions of "Vopenka"
(→Variants: gVP so?)
(One intermediate revision by the same user not shown) Line 76: Line 76:
* For every $Γ$, $VP(κ, Γ)$ for some $κ$ implies $VP(Γ)$.
* For every $Γ$, $VP(κ, Γ)$ for some $κ$ implies $VP(Γ)$.
* $VP(κ, \mathbf{Σ_1})$ holds for every uncountable cardinal $κ$.
* $VP(κ, \mathbf{Σ_1})$ holds for every uncountable cardinal $κ$.
−
* $VP(Π_1) \iff VP(κ, Σ_2)$ for some $κ \iff$ There
+
* $VP(Π_1) \iff VP(κ, Σ_2)$ for some $κ \iff$ There a [[supercompact]] cardinal.
* $VP(\mathbf{Π_1}) \iff VP(κ, \mathbf{Σ_2})$ for a proper class of cardinals $κ \iff$ There is a proper class of supercompact cardinals.
* $VP(\mathbf{Π_1}) \iff VP(κ, \mathbf{Σ_2})$ for a proper class of cardinals $κ \iff$ There is a proper class of supercompact cardinals.
* For $n ≥ 1$, the following are equivalent:
* For $n ≥ 1$, the following are equivalent:
** $VP(Π_{n+1})$
** $VP(Π_{n+1})$
** $VP(κ, \mathbf{Σ_{n+2}})$ for some $κ$
** $VP(κ, \mathbf{Σ_{n+2}})$ for some $κ$
−
** There
+
** There a $C(n)$-[[extendible]] cardinal.
* The following are equivalent:
* The following are equivalent:
** $VP(Π_n)$ for every n.
** $VP(Π_n)$ for every n.
** $VP(κ, \mathbf{Σ_n})$ for a proper class of cardinals $κ$ and for every $n$.
** $VP(κ, \mathbf{Σ_n})$ for a proper class of cardinals $κ$ and for every $n$.
** $VP$
** $VP$
−
** For every $n$, there
+
** For every $n$, there a $C(n)$-extendible cardinal.
===Generic===
===Generic===
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The '''Generic Vopěnka’s Principle''' states that for every proper class $\mathcal{C}$ of structures of the same type there are $B ≠ A$, both in $\mathcal{C}$, such that $B$ elementarily embeds into $A$ in some set-forcing extension.
The '''Generic Vopěnka’s Principle''' states that for every proper class $\mathcal{C}$ of structures of the same type there are $B ≠ A$, both in $\mathcal{C}$, such that $B$ elementarily embeds into $A$ in some set-forcing extension.
−
(Boldface) $gVP(\mathbf{Σ_n})$
+
(Boldface) $gVP(\mathbf{Σ_n})$ and (lightface) $gVP(Σ_n)$ (with analogous definitions for $Π_n$ and $∆_n$) ().
+ + + +
==External links==
==External links==
Revision as of 22:15, 21 September 2019
Vopěnka's principle is a large cardinal axiom at the upper end of the large cardinal hierarchy that is particularly notable for its applications to category theory. In a set theoretic setting, the most common definition is the following:
For any language $\mathcal{L}$ and any proper class $C$ of $\mathcal{L}$-structures, there are distinct structures $M, N\in C$ and an elementary embedding $j:M\to N$.
For example, taking $\mathcal{L}$ to be the language with one unary and one binary predicate, we can consider for any ordinal $\eta$ the class of structures $\langle V_{\alpha+\eta},\{\alpha\},\in\rangle$, and conclude from Vopěnka's principle that a cardinal that is at least $\eta$-extendible exists. In fact if Vopěnka's principle holds then there is a stationary proper class of extendible cardinals; bounding the strength of the axiom from above, we have that if $\kappa$ is almost huge, or even almost-high-jump, then $V_\kappa$ satisfies Vopěnka's principle.
Contents 1 Formalizations 2 Vopěnka cardinals 3 Equivalent statements 4 Other points to note 5 Variants 6 External links 7 References Formalizations
As stated above and from the point of view of ZFC, this is actually an axiom schema, as we quantify over proper classes, which from a purely ZFC perspective means definable proper classes. A somewhat stronger alternative is to view Vopěnka's principle as an axiom in second-order set theory capable to dealing with proper classes, such as von Neumann-Gödel-Bernays set theory. This is a strictly stronger assertion. [1] Finally, one may relativize the principle to a particular cardinal, leading to the concept of a Vopěnka cardinal.
Vopěnka's principle can be formalized in first-order set theory as a schema, where for each natural number $n$ in the meta-theory there is a formula expressing that Vopěnka’s Principle holds for all $Σ_n$-definable (with parameters) classes.[1]
Vopěnka cardinals
An inaccessible cardinal $\kappa$ is a
Vopěnka cardinal if and only if $V_\kappa$ satisfies Vopěnka's principle, that is, where we interpret the proper classes of $V_\kappa$ as the subsets of $V_\kappa$ of cardinality $\kappa$. Because of a characterization of Vopěnka's principle in terms of graphs, a cardinal $\kappa$ is Vopěnka if and only if $\kappa$ is inaccessible and any set $\kappa$-sized set $G$ of $<\kappa$-sized nonisomorphic graphs has some $g_0$ and $g_1$ with $g_0$ a proper subgraph of $g_1$. (Need to cite sources)
As we mentioned above, every almost huge cardinal is a Vopěnka cardinal.
Equivalent statements $C^{(n)}$-extendible cardinals
The schema form of Vopěnka's principle is equivalent to the existence of a proper class of $C^{(n)}$-extendible cardinals for every $n$; indeed there is a level-by-level stratification of Vopěnka's principle, with Vopěnka's principle for a $\Sigma_{n+2}$-definable class corresponds to the existence of a $C^{(n)}$-extendible cardinal greater than the ranks of the parameters (see section "Variants”). [3]
Strong Compactness of Logics
Vopěnka's principle is equivalent to the following statement about logics as well:
For every logic $\mathcal{L}$, there is a cardinal $\mu_{\mathcal{L}}$ such that for any language $\tau$ and any $\mathcal{L}(\tau)$-theory $T$, $T$ is satisfiable if and only if every $t\subseteq T$ such that $|t|<\mu_{\mathcal{L}}$ is satisfiable. [4]
This $\mu_{\mathcal{L}}$ is called the strong compactness cardinal of $\mathcal{L}$. Vopěnka's principle therefore is equivalent to every logic having a strong compactness cardinal. This is very similar in definition to the Löwenheim–Skolem number of $\mathcal{L}$, although it is not guaranteed to exist.
Here are some examples of strong compactness cardinals of specific logics:
If $\kappa\leq\lambda$ and $\lambda$ is strongly compact or $\aleph_0$, then the strong compactness cardinal of $\mathcal{L}_{\kappa,\kappa}$ is at most $\lambda$. Similarly, if $\kappa\leq\lambda$ and $\lambda$ is extendible, then for any natural number $n$, the strong compactness cardinal of $\mathcal{L}^n_{\kappa,\kappa}$ ($\mathcal{L}_{\kappa,\kappa}$ with $n+1$-th order logic) is at most $\lambda$. Therefore for any natural number $n$, the strong compactness cardinal of $n+1$-th order finitary logic is at most the least extendible cardinal. Locally Presentable Categories
Vopěnka's principle is equivalent to the axiom stating "no large full subcategory $C$ of any locally presentable category is discrete." (Sources needed). Equivalently, no large full subcategory of Graph (the category of all graphs) is discrete; that is, for any proper class of simple directed graphs, there is at least one pair of nonequal graphs $G$ and $H$ in the class such that $G$ is a subgraph of $H$. This is a $\Pi^1_1$ statement, so the least Vopěnka cardinals are not even weakly compact (although the least weakly compact cardinal is much, much, much smaller than the least Vopěnka cardinal, if it exists).
Intuitively, a "category" is just a class of mathematical objects with some notion of "morphism", "homomorphism", "isomorphism", (etc.). For example, in Set, the category of all sets, homomorphisms are just injections, and isomorphisms are bijections. In categories of groups and models, homomorphisms and isomorphisms share their actual names.
A "locally small category" $C$ is one with only set-many morphisms between any two objects of $C$. This is one where the objects of $C$ behave "set-like" in the sense that, usually, the number of morphisms between two set-sized objects is at most the number of functions between their universes (like in groups and in graphs). A "locally presentable category" is a locally small category with a couple more really nice properties; you can "generate" all of the objects from set-many objects in the category.
Vopěnka's principle intuitively states that if you have a locally presentable category $C$, then any proper class of objects of $C$ has some nonisomorphic objects $c$ and $d$ where $c$ has a morphism into $d$.
Woodin cardinals
There is a strange connection between the Woodin cardinals and the Vopěnka cardinals. In particular, Vopěnkaness is equivalent to two strengthening variants of Woodinness, namely the Woodin for Supercompactness cardinals and the $2$-fold Woodin cardinals. As a result, every Vopěnka cardinal is Woodin.
Elementary Embeddings Between Ranks
An equivalent statement to Vopěnka's principle is that for any proper class $C\subseteq ORD$, there are $\alpha\in C$, $\beta\in C$, and a nontrivial elementary embedding $j:\langle V_\alpha;\in,P\rangle\rightarrow\langle V_\beta;\in,P\rangle$. Vopěnka's principle quite obviously implies this. The reason the converse holds is because every elementary embedding can be "encoded" (in a sense) into one of these. For more information, see [5].
Other points to note
Whilst Vopěnka cardinals are very strong in terms of consistency strength, a Vopěnka cardinal need not even be weakly compact. Indeed, the definition of a Vopěnka cardinal is a $\Pi^1_1$ statement over $V_\kappa$ (Vopěnka's principle itself is $\Pi^1_1$), and $\Pi^1_1$-indescribability is one of the equivalent definitions of weak compactness. Thus, the least weakly compact Vopěnka cardinal must have (many) other Vopěnka cardinals less than it.
Variants
(Boldface) $VP(\mathbf{Σ_n})$ denotes the fragment of Vopěnka’s Principle for $Σ_n$-definable classes and (lightface) $VP(Σ_n)$ is the weaker principle, where parameters are not allowed in the definition of the class (with analogous definitions for $Π_n$ and $∆_n$).
Vopěnka-like principles $VP(κ, \mathbf{Σ_n})$ for cardinal $κ$ state that for every proper class $\mathcal{C}$ of structures of the same type that is $Σ_n$-definable with parameters in $H_κ$ (the collection of all sets of hereditary size less than $κ$), $\mathcal{C}$ reflects below $κ$, namely for every $A ∈ C$ there is $B ∈ H_κ ∩ C$ that elementarily embeds into $A$.
Results:
For every $Γ$, $VP(κ, Γ)$ for some $κ$ implies $VP(Γ)$. $VP(κ, \mathbf{Σ_1})$ holds for every uncountable cardinal $κ$. $VP(Π_1) \iff VP(κ, Σ_2)$ for some $κ \iff$ There is a supercompact cardinal. $VP(\mathbf{Π_1}) \iff VP(κ, \mathbf{Σ_2})$ for a proper class of cardinals $κ \iff$ There is a proper class of supercompact cardinals. For $n ≥ 1$, the following are equivalent: $VP(Π_{n+1})$ $VP(κ, \mathbf{Σ_{n+2}})$ for some $κ$ There is a $C(n)$-extendible cardinal. The following are equivalent: $VP(Π_n)$ for every n. $VP(κ, \mathbf{Σ_n})$ for a proper class of cardinals $κ$ and for every $n$. $VP$ For every $n$, there is a $C(n)$-extendible cardinal. Generic
(Information in this section from [6])
The
Generic Vopěnka’s Principle states that for every proper class $\mathcal{C}$ of structures of the same type there are $B ≠ A$, both in $\mathcal{C}$, such that $B$ elementarily embeds into $A$ in some set-forcing extension.
(Boldface) $gVP(\mathbf{Σ_n})$ and (lightface) $gVP(Σ_n)$ (with analogous definitions for $Π_n$ and $∆_n$) as well as $gVP(κ, \mathbf{Σ_n})$ are generic analogues of corresponding weakenings of Vopěnka's principle.
Results:
$gVP(Π_n) \iff gVP(κ, \mathbf{Σ_{n+1}})$ for some $κ \iff$ There is an $n$-remarkable cardinal $gVP(Π n ) \iff gVP(κ, \mathbf{Σ_{n+1}})$ for a proper class of $κ \iff$ There is a proper class of $n$-remarkable cardinals
References Bagaria, Joan. $C^{(n)}$-cardinals.Archive for Mathematical Logic 51(3--4):213--240, 2012. www DOI bibtex Perlmutter, Norman. The large cardinals between supercompact and almost-huge., 2010. arχiv bibtex Bagaria, Joan and Casacuberta, Carles and Mathias, A R D and Rosický, Jiří. Definable orthogonality classes in accessible categories are small.Journal of the European Mathematical Society 17(3):549--589. arχiv bibtex Makowsky, Johann. Vopěnka's Principle and Compact Logics.J Symbol Logic www bibtex Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Bagaria, Joan and Gitman, Victoria and Schindler, Ralf. Generic {V}opěnka's {P}rinciple, remarkable cardinals, and the weak {P}roper {F}orcing {A}xiom.Arch Math Logic 56(1-2):1--20, 2017. www DOI MR bibtex |
The Gluing Lemma
The Gluing Lemma
Lemma 1: Let $X$ and $Y$ be topological spaces and let $A, B \subset X$ be closed subsets of $X$ such that $X = A \cup B$. Furthermore, let $f : A \to Y$ and $g : B \to Y$ be continuous maps such that $f(x) = g(x)$ for all $x \in A \cap B$. Then the function $h : X \to Y$ given by $\left\{\begin{matrix} f(x) & \mathrm{if} \: x \in A \\ g(x) & \mathrm{if} \: x \in B \end{matrix}\right.$ is continuous. We require that $f(x) = g(x)$ for all $x \in A \cap B$ so that $h$ is well defined! Proof:To show that $h$ is continuous we will show that for every closed set $V \subseteq Y$ we have that $h^{-1}(V)$ is closed in $X$. Let $V \subseteq Y$ be a closed set. Since $f$ is a continuous map we have that $f^{-1}(V)$ is closed in $A$. But $A$ has the subspace topology (from $X$), and so there exists a closed set $F \subseteq X = A \cup B$ such that:
\begin{align} \quad f^{-1}(V) = A \cap F \end{align}
We're given that $A$ is closed though, and so the intersection $f^{-1}(V) = A \cap F$ is closed in $X$. Similarly, since $g$ is a continuous map we have that $g^{-1}(V)$ is closed in $B$. But $B$ has the subspace topology (from $X$), and so there exists a closed set $G \subseteq X = A \cup B$ such that:
\begin{align} \quad g^{-1}(V) = B \cap G \end{align}
We're given that $B$ is closed though, and so the intersection $g^{-1}(V) = B \cap G$ is closed in $Y$. We last observe that for any set $V \subseteq Y$ that:
\begin{align} \quad h^{-1}(V) = f^{-1}(V) \cup g^{-1}(V) \end{align}
Since $f^{-1}(V)$ and $g^{-1}(V)$ are both closed in $X$ we conclude that $f^{-1}(V)$ is also closed in $X$. $\blacksquare$ |
Let $g_1, g_2, h_1, h_2 : \mathbb{R} \rightarrow \mathbb{R}$ be non-decreasing and right-continuous. Define $$ \begin{align} f_1 & := g_1 - h_1 \\ f_2 & := g_2 - h_2 \end{align} $$ and suppose $f_1 = f_2$. In other words, for every $a, b \in \mathbb{R}$ with $a < b$, the restriction of $g_1 - h_1$ and $g_2 - h_2$ to $[a, b]$ are two Jordan decompositions of the same bounded-variation function.
Denote with $\mu_1, \mu_2$ the (positive) Lebesgue-Stieltjes measures engendered by $g_1, g_2$, respectively, and with $\nu_1, \nu_2$ the (positive) Lebesgue-Stieltjes measures engendered by $h_1, h_2$, respectively. Suppose that either $\mu_1$ or $\nu_1$ is finite (so we may define the signed measure $\mu_1 - \nu_1$, as we do below).
Is it necessarily the case that either $\mu_2$ or $\nu_2$ is finite? (so we may define the signed measure $\mu_2 - \nu_2$, as we do below.)
Suppose that either $\mu_2$ or $\nu_2$ is finite. Define the signed measures $$ \begin{align} \varphi_1 & := \mu_1 - \nu_1 \\ \varphi_2 & := \mu_2 - \nu_2 \end{align} $$ Is it the case that $\varphi_1 = \varphi_2$? |
Let $X_1,...,X_n$ i.i.d random variables, square integrable, and with $E[X_1]=0$.
Let $Y_n = \frac{|X_1 +...+X_n|}{\sqrt{n}}$
I am trying to show that $(Y_n)$ is uniformly integrable, i.e
$\sup_n\mathbb E[Y_n\mathbb 1\{Y_n\gt K\} ] \to 0$ when $K \to +\infty$
I have tried to use Cauchy-Schwarz inequality:
$\mathbb E[Y_n\mathbb 1\{Y_n\gt K\} ]^2 \leq \mathbb E[Y_n^2] \mathbb E[1\{Y_n\gt K\}]=\text{Var}(X_1) \mathbb P(Y_n>K)$
By the CLT, $\mathbb P(Y_n>K) \to \mathbb P(Y>K)$ where $Y$ follows $|\mathcal{N}(0,1)|$.
Also $\mathbb P(Y>K) \to 0$ when $K \to +\infty$.
Ideally I would like to have $\mathbb E[Y_n\mathbb 1\{Y_n\gt K\} ] \leq \epsilon(K) $ where $\epsilon(K) \to 0$ but I don't see how to get it.
Thanks for your help. |
Exercise
Prove the statement using the $\epsilon$, $\delta$ definition of a limit: $$\lim \limits_{x \to 3}{(x^2+x-4)} = 8$$
The Precise Definition of a Limit In case you're not familiar with the definition of "The Precise Definition of a Limit", here it is.
Let $f$ be a function defined on some open interval that contains the number $a$, except possible $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $$\lim \limits_{x \to a}{f(x)} = L$$ if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that $$\text{if } 0 < |x - a| < \delta \text{ then } |f(x) - L| < \epsilon$$
Attempt
$\lim \limits_{x \to 3}{(x^2+x-4)} = 8 \implies \text{if } 0 < |x - 3| < \delta \text{ then } |x^2+x-4 - 8| < \epsilon$
$|x^2+x-4 - 8| < \epsilon \implies |x^2+x-12| < \epsilon \implies |(x-3)(x+4)| < \epsilon$
From here, I quickly get lost.
I notice that $|(x-3)(x+4)| < \epsilon \implies |x-3||x+4| < \epsilon \implies |x-3| < \frac{\epsilon}{|x+4|}$, which means that $\delta = \frac{\epsilon}{|x+4|}$.
In the other exercises, I often end the proof here, because I've solved for $\delta$ purely in terms of $\epsilon$. However, here I have $x$ on the RHS. |
How does one go about showing $ A \cup B \cup C $ is countable if $ A, B $ are countable and $C$ is finite?
I understand most of the confusion for resolving set theory questions online seem to be the definition. For my course we consider the following definitions:
countable: Finite or $A \sim\mathbb{N}$ uncountable: not countable finite: The empty set or $A \sim J_n$ where $n \in \mathbb{N}$ infinite: not finite
So I'm not sure if I have this right but looking at the above definitions the way I'm looking to approach this is to consider 6 different cases.
Where C is the empty set and A, B are both finite sets Where C is the empty set and A is finite and B is countably infinite Where C is the empty set and both A, B are countably infinite
4 - 6. Repeated above but with C being a non-empty finite set
This seems like quite a round about way but intuitively it seems to me like the only way to cover all bases according to the definitions. I'm hoping I might be absolutely wrong on this. Is there a simpler way to prove this? |
Finally, I would like to introduce
reaction-diffusion systems, a particular class of continuous field models that have been studied extensively. They are continuous field models whose equations are made of only reaction terms and diffusion terms, as shown below:
\[\dfrac{\partial{f_1}}{\partial{t}} =R_1(f_1,f_2,...,f_n) + D_1∇^{2}f_1 \label{(13.53)} \]
\[\dfrac{\partial{f_2}}{\partial{t}} = R_2 (f_1,f_2,...,f_n) + D_2∇^{2}f_2 \label{(13.54)} \]
\[ \vdots\]
\[\dfrac{\partial{f_n}}{\partial{t}} =R_n(f_1,f_2,...,f_n) + D_n∇^{2}f_n \label{(13.55)} \]
Reaction terms (\(R_i(...)\)) describe only local dynamics, without any spatial derivatives involved. Diffusion terms (\(D_i∇^{2}f_i\)) are strictly limited to the Laplacian of the state variable itself. Therefore, any equations that involve non-diffusive spatial movement (e.g., chemotaxis) are not reaction-diffusion systems.
There are several reasons why reaction-diffusion systems have been a popular choice among mathematical modelers of spatio-temporal phenomena. First, their clear separation between non-spatial and spatial dynamics makes the modeling and simulation tasks really easy. Second, limiting the spatial movement to only diffusion makes it quite straightforward to expand any existing non-spatial dynamical models into spatially distributed ones. Third, the particular structure of reaction-diffusion equations provides an easy shortcut in the stability analysis (to be discussed in the next chapter). And finally, despite the simplicity of their mathematical form, reaction-diffusion systems can show strikingly rich, complex spatio-temporal dynamics. Because of these properties, reaction diffusion systems have been used extensively for modeling self-organization of spatial patterns. There are even specialized software applications available exactly to simulate reaction-diffusion systems
3.
Exercise \(\PageIndex{1}\)
Extend the following non-spatial models into spatially distributed ones as reaction-diffusion systems by adding diffusion terms. Then simulate their behaviors in Python.
In what follows, we will review a few well-known reaction-diffusion systems to get a glimpse of the rich, diverse world of their dynamics.
Turing Pattern Formation
As mentioned at the very beginning of this chapter, Alan Turing’s PDE models were among the first reaction-diffusion systems developed in the early 1950s [44]. A simple linear version of Turing’s equations is as follows:
\[\dfrac{\partial{u}}{\partial{t}} =a(u-h) +b(v-k) +D_{u}\Delta^{2}u \label{13.56}\]
\[\dfrac{\partial{u}}{\partial{t}} =c(u-h) +d(v-k) +D_{u}\Delta^{2}u \label{13.57}\]
The state variables \(u\) and \(v\) represent concentrations of two chemical species. \(a, b, c,\) and \(d\) are parameters that determine the behavior of the reaction terms, while \(h\) and \(k\) are constants. Finally, \(D_u\) and \(D_v\) are diffusion constants.
If the diffusion terms are ignored, it is easy to show that this system has only one equilibrium point, (\(u_eq,v_eq) = (h,k)\). This equilibrium point can be stable for many parameter values for \(a, b, c,\) and \(d\). What was most surprising in Turing’s findings is that, even for such stable equilibrium points, introducing spatial dimensions and diffusion terms to the equations may destabilize the equilibrium, and thus the system may spontaneously self-organize into a non-homogeneous pattern. This is called
diffusion-induced instability or Turing instability. A sample simulation result is shown in Fig. 13.6.1.
The idea of diffusion-induced instability is quite counter-intuitive. Diffusion is usually considered a random force that destroys any structure into a homogenized mess, yet in this particular model, diffusion is the key to self-organization! What is going on? The trick is that this system has two different diffusion coefficients, \(D_u\) and \(D_v\), and their difference plays a key role in determining the stability of the system’s state. This will be discussed in more detail in the next chapter.
There is one thing that needs particular attention when you are about to simulate Turing’s reaction-diffusion equations. The Turing pattern formation requires small random perturbations (noise) to be present in the initial configuration of the system; otherwise there would be no way for the dynamics to break spatial symmetry to create non-homogeneous patterns. In the meantime, such initial perturbations should be small enough so that they won’t immediately cause numerical instabilities in the simulation. Here is a sample code for simulating Turing pattern formation with a suggested level of initial perturbations, using the parameter settings shown in Fig. 13.6.1:
Figure \(\PageIndex{1}\): Simulation of the Turing pattern formation model with \((a,b,c,d) = (1,−1,2,−1.5)\) and \((D_u,D_v) = (10^−4,6×10^−4)\). Densities of \(u\) are plotted in grayscale (darker = greater). Time flows from left to right.
This simulation starts from an initial configuration \((u(x,y),v(x,y)) ≈ (1,1) = (h,k)\), which is the system’s homogeneous equilibrium state that would be stable without diffusion terms. Run the simulation to see how patterns spontaneously self-organize!
Exercise \(\PageIndex{2}\)
Conduct simulations of the Turing pattern formation with several different parameter settings, and discuss how the parameter variations (especially for the diffusion constants) affect the resulting dynamics.
Exercise \(\PageIndex{3}\)
Discretize the Keller-Segel slime mold aggregation model (Eqs. (13.4.13) and (13.4.14)) (although this model is not a reaction-diffusion system, this is the perfect time for you to work on this exercise because you can utilize Code 13.8). Implement its simulation code in Python, and conduct simulations with \(µ = 10^{−4}\), \(D = 10^{−4}\), \(f = 1\), and \(k = 1\), while varying \(χ\) as a control parameter ranging from \(0\) to \(10^{−3}\). Use \(a = 1\) and \(c = 0\) as initial conditions everywhere, with small random perturbations added to them.
Belousov-Zhabotinsky Reaction
The
Belousov-Zhabotinsky reaction, or BZ reaction for short, is a family of oscillatory chemical reactions first discovered by Russian chemist Boris Belousov in the 1950s and then later analyzed by Russian-American chemist Anatol Zhabotinsky in the 1960s. One of the common variations of this reaction is essentially an oxidation of malonic acid (\(CH_{2}(COOH)_{2}\)) by an acidified bromate solution, yet this process shows nonlinear oscillatory behaviorf or a substantial length of time before eventually reaching chemical equilibrium. The actual chemical mechanism is quite complex, involving about 30 different chemicals. Moreover, if this chemical solution is put into a shallow petri dish, the chemical oscillation starts in different phases at different locations. Interplay between the reaction and the diffusion of the chemicals over the space will result in the self-organization of dynamic traveling waves (Figure \(\PageIndex{2}\)), just like those seen in the excitable media CA model in Section 11.5.
https://chemlegin.wordpress.com/2016/04/ Figure \(\PageIndex{2}\): Belousov-Zhabotinsky reaction taking place in a petri dish. Image used with permission (Chemlignan).
A simplified mathematical model called the
“Oregonator” was among the first to describe the dynamics of the BZ reaction in a simple form [50]. It was originally proposed as a non-spatial model with three state variables, but the model was later simplified to have just two variables and then extended to spatial domains [51]. Here are the simplified “Oregonator” equations:
\[ \begin{align} \epsilon{\dfrac{\partial{u}}{\partial{t}}} &= u(1-u)-\dfrac{u-q}{u+q}fv + D_{u}\Delta^{2}u \label{(13.58)} \\[4pt] \dfrac{∂v}{∂t} &=u-v + D_{u}\Delta^{2}v \label{(13.59)} \end{align}\]
Here, \(u\) and \(v\) represent the concentrations of two chemical species. If you carefully examine the reaction terms of these equations, you will noticethat the presenceof chemical \(u\) has a positive effect on both \(u\) and \(v\), while the presence of chemical \(v\) has a negative effect on both. Therefore, these chemicals are called the “activator” and the “inhibitor,” respectively. Similar interactions between the activator and the inhibitor were also seen in the Turing pattern formation, but the BZ reaction system shows nonlinear chemical oscillation. This causes the formation of traveling waves. Sometimes those waves can form spirals if spatial symmetry is broken by stochastic factors. A sample simulation result is shown in Fig. 13.6.3.
Figure \(\PageIndex{3}\): Simulation of the “Oregonator” model of the BZ reaction with \((\epsilon,q,f) = (0.2,10^−3,1.0)\) and \(D_u = D_v = 10^−5\), starting with a tongue-like initial configuration. The concentration of chemical u is plotted in grayscale (darker = greater). Time flows from left to right (the first row followed by the second row).
Exercise \(\PageIndex{5}\)
Implement a simulator code of the “Oregonator” model of the BZ reaction in Python. Then conduct simulations with several different parameter settings, and discuss what kind of conditions would be needed to produce traveling waves.
Gray-Scott Pattern Formation
The final example is the
Gray-Scott model, another very well-known reaction-diffusion system studied and popularized by John Pearson in the 1990s [52], based on a chemical reaction model developed by Peter Gray and Steve Scott in the 1980s [53, 54, 55]. The model equations are as follows:
\[ \dfrac{∂u}{∂t} =F(1-u) -uv^{2} +D_{u}\Delta^{2}u \label{(13.60)} \]
\[\dfrac{∂v}{∂t} =-(F+k)v +uv^{2}+ D_{u}\Delta^{2}v \label{(13.61)} \]
The reaction terms of this model assumes the following
autocatalytic reaction (i.e., chemical reaction for which the reactant itself serves as a catalyst):
\[u+2v \rightarrow 3v \label{(13.62)}\]
This reaction takes one molecule of \(u\) and turns it into one molecule of \(v\), with help of two other molecules of \(v\) (hence, autocatalysis). This is represented by the second term in each equation. In the meantime, \(u\) is continuously replenished from the external source up to 1 (the first term of the first equation) at feed rate \(F\), while \(v\) is continuously removed from the system at a rate slightly faster than \(u\) ’s replenishment (\(F +k \)seen in the first term of the second equation). \(F\) and \(k\) are the key parameters of this model.
It is easy to show that, if the diffusion terms are ignored, this system always has an equilibrium point at \((u_{eq},v_{eq}) = (1,0)\) (which is stable for any positive \(F\) and \(k\)). Surprisingly, however, this model may show very exotic, biological-looking dynamics if certain spatial patterns are placed into the above equilibrium. Its behaviors are astonishingly rich, even including growth, division, and death of “cells” if the parameter values and initial conditions are appropriately chosen. See Fig. 13.6.4 to see only a few samples of its wondrous dynamics!
Exercise \(\PageIndex{5}\)
Implement a simulator code of the Gray-Scott model in Python. Then conduct simulations with several different parameter settings and discuss how the parameters affect the resulting patterns.
Figure \(\PageIndex{4}\): (Next page) Samples of patterns generated by the Gray-Scott model with \(D_u = 2 × 10^{−5}\) and \(D_v = 10^{−5}\). The concentration of chemical \(u\) is plotted in grayscale (brighter = greater, only in this figure). Time flows from left to right. The parameter values of \(F\) and \(k\) are shown above each simulation result. The initial conditions are the homogeneous equilibrium \((u,v) = (1,0)\) everywhere in the space, except at the center where the local state is reversed such that \((u,v) = (0,1)\).
3For example, check out Ready (https://code.google.com/p/reaction-diffusion/). |
nth Order Ordinary Differential Equations
Definition: Let $D \subseteq \mathbb{R}^{n+1}$ be a domain (a nonempty, open, connected subset of $\mathbb{R}^{n+1}$) and let $h \in C(D, \mathbb{R})$. An n is of the form $y^{(n)} = h(t, y, y^{(1)}, ..., y^{(n-1)})$. th Order Ordinary Differential Equation The notation $y^{(i)}$ is used to the denote the $i^{\mathrm{th}}$ derivative of $y$ with respect to $t$, that is, $\displaystyle{y^{(i)} = \frac{d^iy}{dt^{i}}}$ for all $i = 0, 1, 2, ..., n$. We let $y^{(0)} = y$.
An example of a fifth order ordinary differential equation is:(1)
Using prime notation, the above fifth order ordinary differential equation can be written as:(2)
Definition: A Solution to the nth order ordinary differential equation $y^{(n)} = h(t, y, y^{(1)}, ..., y^{(n-1)})$ on an open interval $J = (a, b)$ is an n-times continuously differentiable function $\phi \in C^n(J, \mathbb{R})$ such that for all $t \in J$ we have that $(t, \phi(t), \phi^{(1)}(t), ..., \phi^{(n-1)}(t)) \in D$ and $\phi^{(n)} = h(t, \phi, \phi^{(1)}, ..., \phi^{(n-1)})$.
We can also characterize initial value problems for nth order ordinary differential equations.
Definition: An Initial Value Problem for an nth order ordinary differential equation is an nth order ODE $y^{(n)} = h(t, y, y^{(1)}, ..., y^{(n-1)})$ with Initial Conditions $y^{(i-1)}(\tau) = \xi_i$ where $i = 1, 2, ..., n$ and $(\tau, \xi_1, \xi_2, ..., \xi_n) \in D$. A Solution to the initial value problem $y^{(n)} = h(t, y, y^{(1)}, ..., y^{(n-1)})$ with initial conditions $y^{(i-1)}(\tau) = \xi_i$ where $i = 1, 2, ..., n$ on the open interval $J = (a, b)$ is an n-times continuously differentiable function $\phi \in C^n (J, \mathbb{R})$ such that for all $t \in J$ we have that $(t, \phi(t), \phi^{(1)}(t), ..., \phi^{(n-1)}(t)) \in D$, $\phi^{(n)} = h(t, \phi, \phi^{(1)}, ..., \phi^{(n-1)})$, and $\phi^{(i-1)}(\tau) = \xi_i$ for $i = 1, 2, ..., n$.
Theorem 1: Every initial value problem of an $n^{\mathrm{th}}$ order ordinary differential equation $y^{(n)} = h(t, y, y^{(1)}, ..., y^{(n-1)})$ with initial conditions $y^{(i-1)}(\tau) = \xi_i$ where $i = 1, 2, ..., n$ on an open interval $J = (a, b)$ ($\tau \in J$) can be expressed as an initial value problem of a system of $n$ first order ordinary differential equations. Proof:Consider the initial value problem: With initial conditions $y^{(i-1)}(\tau) = \xi_i$ where $i = 1, 2, ..., n$ on the open interval $J = (a, b)$. We define $x_1$, $x_2$, …, $x_n$ as: We differentiate each of the equations above with respect to $t$ to get: We can rewrite the above list of equations as: The system of $n$ first order ordinary differential equations is defined for all $(t, x_1(t), x_2(t), ..., x_n(t)) \in D$. Now if $\phi_1$ is a solution to $(*)$ with initial conditions $\phi_1^{(i-1)} (\tau) \xi_i$ for all $i = 1, 2, ..., n$ then $\phi_1^{(n)} = h(t, \phi_1, \phi_1^{(1)}, ..., \phi_1^{(n-1)})$ and from above we have that that $\phi = (\phi_1, \phi_1^{(1)}, ..., \phi_1^{(n-1)})$ is a solution to $(**)$ satisfying $\phi(\tau) = (\xi_1, \xi_2, ..., \xi_n)$. Now if $\phi = (\phi_1, \phi_2, ..., \phi_n)$ is a solution to $(**)$ with initial conditions $\phi_i(\tau) = \xi_i$ for all $i = 1, 2, ..., n$ then: From the equations above we get that: So $\phi_1$ is a solution to $(*)$ and $\phi_1^{(i-1)}(\tau) = \xi_i$ for all $i = 1, 2, ..., n$. $\blacksquare$ |
How can I calculate the Total Variation Distance of a transition Matrix? is there any built in function? I've searched all documentation and haven;t found anything.
** More information:
Let me try to explain it better. let's say we have a transition matrix ($P$), $4\times4$ that describes the probability of going from a, b, c and d to a, b, c or d in 1 step.
We can calculate the stationary distribution of $P$, and that's called $\pi$ in the following equation, and $P_{yx}$ is the probability of going from state $y$ to state $x$ (a,b,c,d):
$$\frac12\sum_x\left|P_{yx}^t-\pi(x)\right|$$
What I want to do is calculate the Total Variation Distance of $P$ from $\pi$ after $n$ steps and starting on a given state.
*** This is what I have so far:
M = {{0.3, 0, 0.5, 0.2}, {0, 0.4, 0.3, 0.3}, {0.3, 0.2, 0, 0.5}, {0.4, 0.1, 0, 0.5}} B = Transpose[M] N[B] // MatrixForm{eVals, eVecs} = Eigensystem[B]eVals // MatrixForm eVecs // MatrixFormeigenvector = eVecs[[1]]Print["Stationary Distribution"]; eigenvector/Total[eigenvector]Print["M after 1 step"]; M2 = MatrixPower[M, 2] |
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Published on Oct 3, 2014
For germs of holomorphic functions $f : \mathbf{C}^{m+1} \to \mathbf{C}, g : \mathbf{C}^{n+1} \to \mathbf{C}$ having an isolated critical point at 0 with value 0, the classical Thom-Sebastiani theorem describes the vanishing cycles group $\Phi^{m+n+1}$ (𝑓 ⊕ 𝑔) (and its monodromy) as a tensor product $\Phi^m(f) \otimes \Phi^n(g)$, where $(f \oplus g)(x,y) = 𝑓 (𝑥)+𝑔(𝑦), $x = (x_0,\cdots,x_m)$, $y = (y_0,\cdots,y_n)$. I will discuss algebraic variants and generalizations of this result over fields of any characteristic, where the tensor product is replaced by a certain local convolution product, as suggested by Deligne. The main theorem is a Künneth formula for 𝑅𝛹 in the framework of Deligne's theory of nearby cycles over general bases. |
I understand that during phase transition nucleation must occur. I'm wondering, once phase equilibrium is established, does nucleation still occur? For instance, in liquid-vapour equilibrium, does independent water molecules just simply enter and leave the liquid-vapour interface, or does some kind of nucleation (say liquid to vapour) have to happen during equilibrium phase transition?
There are actually two questions here. At phase equilibrium, yes independent water molecules do enter and leave the liquid vapour interface. The rate of crossing of individual molecules from one phase to another is characterized by an Arrhenius type rate equation of the form: $$\alpha \exp \left[ - \frac{\Delta E}{kT} \right]$$ where $\Delta E$ is the activation energy and $\alpha$ is the so called rate constant. There is a separate rate equation for liquid-vapour and for vapour to liquid. The difference between the liquid-to-vapour and vapour-to-liquid activation energies is the latent heat per molecule. The rate constant is dependent on a large number of factors and usually has to be experimentally determined. The two exponentials cross at the equilibrium temperature so that at this point the rates of transfer in each direction are equal. Below this temperature the vapour-to-liquid rate is higher, above the liquid-to-vapour rate is higher.
However this rate behaviour is for transfer of molecules across existing phase interfaces and so has nothing to do with
nucleation which concerns the formation of new interfaces. Nucleation requires the formation of sizable clusters of molecules in one phase or the other. This formation is a random process. Exactly how big a cluster has to be to be regarded as a true region of the phase depends a lot on the specific system and state. However at any temperature there is a certain critical size above which the cluster is stable. Although transient cluster formation still occurs at all temperatures, at equilibrium the rate of growth of any phase is effectively zero and so the formation of new phase regions above the critical size is highly unlikely. |
The point is that the ball gets a tangential hit by the ground. This changes the angular momentum of the ball.
Consider a ball thrown with a horizontal speed v. It should also not rotate.
Right before hitting the ground, the ball has an angular momentum of
$$L=mvr$$
This is a result of $\vec{L}=\vec{v}\times\vec{p}$, which is also valid for linear moving objects with respect to a resting observer. Also, the vertical movement of the ball is neglected here.
Now, the ball hits the ground (inelastic) in such a way, that the part touching the ground comes to rest. The ball will still move forward with a reduced speed v', but it will also rotate such that the point facing the ground does not move. This means the ball rotates with a radial speed of $v'$ or $\omega=\frac{v'}{r}$.
The overall momentum of the ball remains constant, because its mass is much lower than that of the earth. So:
$$L=mvr=mvr'+J\omega=mvr'+J\frac{v'}{r}$$
Now, $J=\beta mr^2$ ($\beta$ is that fraction defined by the shape of the body):
$$mvr=mvr'+\beta mv'r=(1+\beta)mv'r$$
$$\frac{v'}{v}=\frac{1}{1+\beta}$$
For a solid ball with $\beta=2/5$, you get $\frac{v'}{v}=\frac{5}{7}\approx0.71$
For a hollow ball with $\beta=2/3$, you get $\frac{v'}{v}=\frac{5}{7}\approx0.6$
Finally, the ball continues moving forward with 60-70% of its initial speed!
In reality, friction limits the maximum change of momentum, so the ball will slide while touching the ground and rotate slower after that. Then, more momentum is left for the forward movement, and the ball is faster. In an extreme case, there is no friction, the ball does not start to rotate, and does not change its speed.
From the side, the ball always flies a curve shaped like a V. A difference in angle is hard to see, and it also depends on the ratio of vertical and horizontal speed.
If you give the ball a rotation when throwing it, you will notice that it will move slower/faster after touching the ground. It's even possible that it comes back to you. |
Taken straight-out from Prof. L. Braile's collection of earthquake hazard information, that explains it quite nicely:
The magnitude scale is really comparing amplitudes of waves on a seismogram, not the STRENGTH (energy) of the earthquakes.
While one unit of magnitude is 10 times the amplitude on a seismogram, one unit of magnitude represents $10^{1.5}$ times (approximately 32 times) the energy, based on the long-standing empirical formula
$$\log{E} \propto 1.5M$$
thus
$$E \propto 10^{1.5M}$$
where $E$ is energy and $M$ is magnitude.
The example set in the webpage is to compare how much bigger would a 9.7 magnitude earthquake be in comparison to a 6.8 magnitude earthquake:
The magnitude scale is logarithmic, so a magnitude 9.7 earthquake is $\frac{10^{9.7}}{10^{6.8}} = 794.328$ times bigger on the seismogram than a magnitude 6.8 earthquake.
Measuring the change in energy, however, shows that the 9.7 earthquake is $\frac{{10^{1.5}}^{9.7}}{{10^{1.5}}^{6.8}} = 10^{1.5 \times (9.7-6.8)} = 22~387$ times the energy of a 6.8 earthquake.
Comparison of earthquake "size" on seismograms: this image from Nevada Seismological Lab's Living with Earthquakes in Nevada
A graphical comparison of earthquake energy release, a video by the US NWS Pacific Tsunami Warning Center (PTWC).
The USGS has a calculator for magnitude difference.
And there's a calculator for seismic energy and several equivalences. Unfortunately, most of the equivalences are related to the USA.
See Brian Hill's answer that explains that with the original Richter scale, one unit of magnitude is apprimately $31^{1}$ times the energy, and two units of magnitude are $31^{2}$ times the energy, and so on.
In short: you are half-right (one unit of magnitude it is 10 times bigger on the seismogram, not in energy), and the book is right (two units of magnitude are 1000 times stronger). |
Kalid Azad uses color for equations [1] Chris Olah uses tables for equations [2]
Here I’m using both color and tables (using the table tag instead of css grid), with labels for the symbols:
\(\frac{dR}{dt}\) = \(\alpha\) \(\cdot\) \(R\) \(-\) \(\beta\) \(\cdot\) \(R\) \(\cdot\) \(F\) rate of change
of rabbits
is birth rate times rabbits minus kill rate times rabbits times foxes
This is highly distracting. Describing the basic symbols is overkill for an audience that’s reading about differential equations! Let’s drop those:
\(\frac{dR}{dt}\) = \(\alpha\) \(\cdot\) \(R\) \(-\) \(\beta\) \(\cdot\) \(R\) \(\cdot\) \(F\) rate of change
of rabbits
birth rate rabbits kill rate rabbits foxes
There are still too many colors. Let’s use color only for rabbits:
\(\frac{dR}{dt}\) = \(\alpha\) \(\cdot\) \(R\) \(-\) \(\beta\) \(\cdot\) \(R\) \(\cdot\) \(F\) rate of change
of rabbits
birth rate rabbits kill rate rabbits foxes
An alternative to describing the
symbols is to describe the expressions. Chris Olah recommends left alignment and extra whitespace.
\( \frac{dR}{dt}\) \(=\) \( \alpha \cdot R\) \(-\) \( \beta \cdot R \cdot F\) rate of change rabbits born rabbits die
I haven’t quite figured out the TeX to make explanations appear underneath the parts of the formula. These don’t line up:
\[ \underbrace{\frac{dR}{dt}}_\text{rate of change} = \underbrace{\alpha \cdot R}_\text{rabbits born} - \underbrace{\beta \cdot R \cdot F}_\text{rabbits die} \]
Also see Shan Carter
[3]’s beautiful version, which has both labels for variables and labels for expressions, and also uses color! I haven’t tried to reproduce this yet. I think it uses CSS grid with borders on an empty cell to create the bracket. |
ASU Electronic Theses and Dissertations
This collection includes most of the ASU Theses and Dissertations from 2011 to present. ASU Theses and Dissertations are available in downloadable PDF format; however, a small percentage of items are under embargo. Information about the dissertations/theses includes degree information, committee members, an abstract, supporting data or media.
In addition to the electronic theses found in the ASU Digital Repository, ASU Theses and Dissertations can be found in the ASU Library Catalog.
Dissertations and Theses granted by Arizona State University are archived and made available through a joint effort of the ASU Graduate College and the ASU Libraries. For more information or questions about this collection contact or visit the Digital Repository ETD Library Guide or contact the ASU Graduate College at gradformat@asu.edu.
Schmidt, Kevin E 7 Arizona State University 3 Beckstein, Oliver 2 Alarcon, Ricardo 2 Lebed, Richard 2 Shumway, John 1 Alarcon, Ricardo O more 1 Alarcón, Ricardo 1 Blyth, David Cooper 1 Chen, Tingyong 1 Comfort, Joseph R 1 Erten, Onur 1 Liu, Jianheng 1 Lynn, Joel Eric 1 Madeira, Lucas 1 Nelson, Garrett 1 Ritchie, Barry G 1 Ros, Robert 1 Sadjadi, Seyed Mahdi 1 Shovkovy, Igor 1 Shumway, John B 1 Spence, John C 1 Thorpe, Michael F 1 Treacy, Michael MJ 1 Weierstall, Uwe J 1 Yu, Hongbin 1 Zhang, Jie 7 English 7 Public Physics 2 Condensed matter physics 2 Nuclear physics and radiation 1 2-Photon Polymerization 1 2D materials 1 Atomic physics 1 Biophysics more 1 Carlo 1 Cold Fermi gases 1 GDVN 1 Hadronic weak interaction 1 Injector 1 Materials Science 1 Monte 1 Nuclear physics 1 Nuclear structure 1 Parity violation 1 Pions 1 Quantum 1 Quantum Monte Carlo 1 Serial Crystallography 1 Theoretical physics 1 Viscous Jet 1 XFEL 1 chiral 1 computational physics 1 condensed matter 1 glasses 1 graphene 1 path integral Monte Carlo 1 physics 1 quantum Monte Carlo 1 quantum wires 1 rigidity 1 silica
One dimensional (1D) and quasi-one dimensional quantum wires have been a subject of both theoretical and experimental interest since 1990s and before. Phenomena such as the "0.7 structure" in the conductance leave many open questions. In this dissertation, I study the properties and the internal electron states of semiconductor quantum wires with the path integral Monte Carlo (PIMC) method. PIMC is a tool for simulating many-body quantum systems at finite temperature. Its ability to calculate thermodynamic properties and various correlation functions makes it an ideal tool in bridging experiments with theories. A general study of the features interpreted by the …
Contributors Liu, Jianheng, Shumway, John B, Schmidt, Kevin E, et al. Created Date 2012
This work presents analysis and results for the NPDGamma experiment, measuring the spin-correlated photon directional asymmetry in the $\vec{n}p\rightarrow d\gamma$ radiative capture of polarized, cold neutrons on a parahydrogen target. The parity-violating (PV) component of this asymmetry $A_{\gamma,PV}$ is unambiguously related to the $\Delta I = 1$ component of the hadronic weak interaction due to pion exchange. Measurements in the second phase of NPDGamma were taken at the Oak Ridge National Laboratory (ORNL) Spallation Neutron Source (SNS) from late 2012 to early 2014, and then again in the first half of 2016 for an unprecedented level of statistics in order …
Contributors Blyth, David Cooper, Alarcon, Ricardo O, Ritchie, Barry G, et al. Created Date 2017
Monte Carlo methods often used in nuclear physics, such as auxiliary field diffusion Monte Carlo and Green's function Monte Carlo, have typically relied on phenomenological local real-space potentials containing as few derivatives as possible, such as the Argonne-Urbana family of interactions, to make sampling simple and efficient. Basis set methods such as no-core shell model or coupled-cluster techniques typically use softer non-local potentials because of their more rapid convergence with basis set size. These non-local potentials are typically defined in momentum space and are often based on effective field theory. Comparisons of the results of the two types of methods …
Contributors Lynn, Joel Eric, Schmidt, Kevin E, Alarcón, Ricardo, et al. Created Date 2013
In this dissertation two kinds of strongly interacting fermionic systems were studied: cold atomic gases and nucleon systems. In the first part I report T=0 diffusion Monte Carlo results for the ground-state and vortex excitation of unpolarized spin-1/2 fermions in a two-dimensional disk. I investigate how vortex core structure properties behave over the BEC-BCS crossover. The vortex excitation energy, density profiles, and vortex core properties related to the current are calculated. A density suppression at the vortex core on the BCS side of the crossover and a depleted core on the BEC limit is found. Size-effect dependencies in the disk …
Contributors Madeira, Lucas, Schmidt, Kevin E, Alarcon, Ricardo, et al. Created Date 2018
Sample delivery is an essential component in biological imaging using serial diffraction from X-ray Free Electron Lasers (XFEL) and synchrotrons. Recent developments have made possible the near-atomic resolution structure determination of several important proteins, including one G protein-coupled receptor (GPCR) drug target, whose structure could not easily have been determined otherwise (Appendix A). In this thesis I describe new sample delivery developments that are paramount to advancing this field beyond what has been accomplished to date. Soft Lithography was used to implement sample conservation in the Gas Dynamic Virtual Nozzle (GDVN). A PDMS/glass composite microfluidic injector was created and given …
Contributors Nelson, Garrett, Spence, John C, Weierstall, Uwe J, et al. Created Date 2015
Spin-orbit interactions are important in determining nuclear structure. They lead to a shift in the energy levels in the nuclear shell model, which could explain the sequence of magic numbers in nuclei. Also in nucleon-nucleon scattering, the large nucleon polarization observed perpendicular to the plane of scattering needs to be explained by adding the spin-orbit interactions in the potential. Their effects change the equation of state and other properties of nuclear matter. Therefore, the simulation of spin-orbit interactions is necessary in nuclear matter. The auxiliary field diffusion Monte Carlo is an effective and accurate method for calculating the ground state …
Contributors Zhang, Jie, Schmidt, Kevin E, Alarcon, Ricardo, et al. Created Date 2014
The structure of glass has been the subject of many studies, however some details remained to be resolved. With the advancement of microscopic imaging techniques and the successful synthesis of two-dimensional materials, images of two-dimensional glasses (bilayers of silica) are now available, confirming that this glass structure closely follows the continuous random network model. These images provide complete in-plane structural information such as ring correlations, and intermediate range order and with computer refinement contain indirect information such as angular distributions, and tilting. This dissertation reports the first work that integrates the actual atomic coordinates obtained from such images with structural …
Contributors Sadjadi, Seyed Mahdi, Thorpe, Michael F, Beckstein, Oliver, et al. Created Date 2018 |
Dear all experts
I have had a problem since many years ago, but I've never tried to find the solution, but now I need it urgently.
Many of friends and classmates would ask me if they could use "condition on variables" in a summation in GAMS. Obviously for me, it was not possible. But all the time my answer was one thing: you have a problem in your model building procedure, try to modify it, it is not standard to have condition on decision variable which its value is obtained through the interaction of parameters, sets, scalars and etc. In my idea, it was a blind loop when range of a summation is dependent on a decision variable!Now, my question is:
I really appreciate any kind of suggestion. Looking for your answers
Bests
asked
Bob Pay
If I'm understanding you correctly, you are asking if it is possible to write a conditional sum, with conditions on the variables instead of the counters, for instance, sum(xi if xi > 2). This is equivalent to the SUMIF function in Excel.
Is this right?
If so, yes, it is possible if you introduce binary variables (resulting in a Mixed-Integer Program).
For the above example, i.e. \(y = \sum_{i=1}^{n} x_i\) with the condition that we only sum terms that obey the predicate \(x_i > 2\), we can state the problem as follows:
$$ \begin{align} &y = \sum_{i=1}^{n} w_i\\ &w_i = \left\{ \begin{array}{ll} x_i, & x_i > 2\\ 0, & x_i \leq 2 \end{array} \right. \end{align} $$
If you're solving a Mixed Integer
$$ \begin{align} &y = \sum_{i=1}^{n} \delta_i x_i\\ & \delta_i \in \{0,1\}\\ & (1-\delta_i)L + (2+ \epsilon)\delta_i \leq x_i \leq \delta_i U + 2(1 - \delta_i) \end{align} $$
If you want to further reduce that to a Mixed-Integer
$$ \begin{align} &y = \sum_{i=1}^{n} z_i\\ & \delta_i \in \{0,1\}, y_i \in \{0,1\}\\ & (1-\delta_i)L + (2+ \epsilon)\delta_i \leq x_i \leq \delta_i U + 2(1 - \delta_i)\\ & Ly_i \leq z_i \leq Uy_i\\ & L(1-y_i) \leq x_i - z_i \leq U(1-y_i) \end{align} $$
(The linearization was done by applying the trick in Section 2.8 of the FICO MIP guide.)
Marco alerted me to another interpretation: suppose \(y = \sum_{i=1}^{n} x_i\) where \(n \in \{2,3,\ldots, 7\}\) is a variable (!). There is a way to handle this, but it can get expensive if the domain of \(n\) is large. We can write:
$$ \begin{align} &y = \sum_{i=1}^{7} \delta_i x_i\\ & \delta_7 \leq \delta_6 \leq \delta_5 \leq \ldots \leq \delta_2\\ & \delta_1 = 1 \end{align} $$
I remember there being a question some time ago about optimization problems over \(\mathbb{R}^n\) where \(n\) is a variable. The above is one of the ways to handle it (but again, it is awfully expensive when the domain of \(n\) is large, especially if linearizations are desired). I would say this is one of those "last resort" type approaches -- in general it's not a good idea to specify optimization problems with decision variable vectors of variable dimension.
I will dissent a bit from the other answers. As they note, doing it directly in a mathematical programming model is not possible, although in some cases there is a work-around if you are willing to add binary variables and/or constraints. More precisely, it is not possible for a typical MP solver to handle it. (You can certainly write conditional summations in algebra; just not in most MP languages.)
It is both possible and common, however, to have summations that depend on the states of variables in a constraint program. CP languages allow, CP solvers can deal with it.
I don't know about GAMS, but some modeling languages (OPL comes to mind) are intended to support both MP and CP, and so will allow the sort of thing you have in mind (possibly conditional on your designating the model a CP).
answered
Paul Rubin ♦♦
What you are looking for would be highly appreciated, but this is not possible in mathematical programming (given, I understand you right). You may consider variables with a different meaning. As a very rough example, in scheduling you may wish to have variables expressing the start time of a job. Summation may depend on the value of the start time (for whatever reason, maybe for a capacity constraint). An alternative is to model with binary variables, for every possible point in time a different variable to state whether the job starts or not. There, the variable value (of the first model's variables) "somehow" goes into the summation.
So: you are on the right side!
answered
Marco Luebbecke ♦
Generally, I think the answer is "Yes, you're right". But it might depend on your approach to solve the problem. If you intend to solve the model via an artificial intelligence method such as a metaheuristic, I don't think such declarations would be wrong. In fact, a metaheuristic could easily handle all these conditions via its encoding and decoding mechanisms. You must note that in this situation, your mathematical model is merely acting as a tool for better understanding the problem and nothing more.
However if you want to solve your problem via mathematical programming methods, the answer is no. All the mathematical programming solution methods require a complete representation of your problem (e.g. in case of MIP and LP model, solvers require a matrix representation of the problem structure).
Personally, I think it would be for the best not to define such conditions so that you could compare your solutions obtained from metaheuristics with exact solutions from solvers. |
This question already has an answer here:
Series in Real Analysis 2 answers
Prove$\sum_{n=1}^\infty a_n \in \mathbb R \mathbb \leftrightarrow \sum_{n=1}^\infty\frac{a_n}{1+a_n}\in \mathbb R$
The hint says that for one direction argue that $b_n = \frac{a_n}{\frac{a_n}{1+a_n}}$ is bounded after some fixed cutoff point. I dont know where the $b_n$ is coming from or which test to use to prove this. |
The Group Isomorphism problem for finitely presentable groups is undecidable. Restricting to finite groups, the problem is certainly decidable, with the trivial algorithm of enumerating all $n!$ permutations (assuming of course, the two groups have the same order).
If the input groups $G, H$ are given by their Cayley tables (which is a generous assumption), we can improve the bound to $n^{O(\log(n))}$ in the following way. First, observe that for a group $G$ of order $n$, there exists a generating set of size $\log_{2}(G)$. To see this, we consider the following algorithm:
-Set $K := \emptyset$. -While $\langle K \rangle \neq G$, select $g \in G \setminus \langle K \rangle$, and add $g$ to $K$.
Each new element we add to $K$ at least doubles the size of $\langle K \rangle$. So $|K| = \log_{2}(n)$.
As $\langle K \rangle = G$, we seek to map the elements of $K$ into a generating set of $H$. There are $|H|^{|K|} = n^{\log(n)}$ such functions. Accounting for some overhead in testing such functions, we achieve the $n^{O(\log(n))}$ bound. This bound has been relatively unscathed, though several classes of groups have polynomial time algorithms for the isomorphism problem. Some of these classes include abelian groups (https://www.sciencedirect.com/science/article/pii/S0022000007000293), groups without abelian normal subgroups (https://link.springer.com/chapter/10.1007%2F978-3-642-31594-7_5), and groups with abelian Sylow towers (http://drops.dagstuhl.de/opus/volltexte/2012/3400/). |
Orthonormal Vectors Review
Orthonormal Vectors Review
We will now review some of the recent content regarding orthonormal vectors.
Recall from the Orthonormal Bases of Vector Spaces page that if $V$ is a finite-dimensional inner product space then an Orthonormal Basisof $V$ is a basis $\{ e_1, e_2, ..., e_n \}$ such that $<e_i, e_j> = 0$ for all $i, j = 1, 2, ..., n$, $i \neq j$ and $<e_i, e_i> = 1$ for all $i = 1, 2, ..., n$ - the first condition giving us orthogonality of the vectors and the second condition giving us unit (normal) vectors. We then look at a very important proposition regarding orthonormal vectors which said that if $V$ is a finite-dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$ and $\{ e_1, e_2, ..., e_n \}$ is an orthonormal basis of $V$ then the norm squared of any vector $v = a_1e_1 + a_2e_2 + ... + a_ne_n$ in $V$ is given by:
\begin{align} \quad \| a_1e_1 + a_2e_2 + ... + a_ne_n \|^2 = \mid a_1 \mid^2 + \mid a_2 \mid^2 + ... + \mid a_n \mid^2 \end{align}
We also obtained the nice property that if $\{ e_1, e_2, ..., e_n \}$ is an orthonormal set of vectors in $V$ then $\{ e_1, e_2, ..., e_n \}$ is a linearly independent set. Furthermore, we noted that if $V$ is an inner product space and $\{ e_1, e_2, ..., e_n \}$ is an orthonormal basis of $V$ then for every $v \in V$:
\begin{align} \quad v = <v, e_1>e_1 + <v, e_2>e_2 + ... + <v, e_n>e_n \end{align}
We then looked at a very important process known as The Gram-Schmidt Process which allows us take a linearly independent set of vectors $\{ v_1, v_2, ..., v_n \}$ in an inner product space $V$ and produce an orthonormal set of vectors $\{ e_1, e_2, ..., e_n \}$ from it where:
\begin{align} \quad e_1 = \frac{v_1}{\| v_1 \|} \: , \end{align}(4)
\begin{align} \quad e_j = \frac{v_j - <v_j, e_1>e_1 - <v_j, e_2>e_2 - ... - <v_j, e_{j-1}> e_{j-1}}{\| v_j - <v_j, e_1>e_1 - <v_j, e_2>e_2 - ... - <v_j, e_{j-1}>e_{j-1} \| } \end{align}
As an important corollary to the Gram-Schmidt process we noted that if $V$ is a finite-dimensional inner product space then $V$ has an orthonormal basis. As another important corollary we had that if $V$ is an finite-dimensional inner product space then any orthonormal set of vectors $\{ e_1, e_2, ..., e_n \}$ can be extended to a basis of $V$ due the linear independence of this set of vectors. On the Orthogonal Complements page we said that if $V$ is an inner product space and $U$ is a subset of $V$ (not needing $U$ to be a subspace of $V$) then the Orthogonal Complement of $U$denoted $U^{\perp}$ is defined to be the set of vectors $v \in V$ such that $<u, v> = 0$ for all $u \in U$. Some nice properties of an orthogonal complement $U^{\perp}$ is that $U^{\perp}$ is a subspace of $V$, $\{ 0 \}^{\perp} = V$, $V^{\perp} = \{ 0 \}$ and if $U_1$ and $U_2$ are subsets of $V$ such that $U_1 \subseteq U_2$ then $U_1^{\perp} \supseteq U_2^{\perp}$ If $U$ is actually a subset of a finite-dimensional vector space $V$ then we also saw that $V = U \oplus U{\perp}$. We also saw that $(U^{\perp})^{\perp} = U$. On the Orthogonal Projection Operators, if $V = U \oplus U^{\perp}$ such that $v = u + w$ where $u \in U$ and $w \in U^{\perp}$ then we defined the Orthogonal Projection Operatorof $V$ onto $U$ to be the linear map $P_U \in \mathcal L(V)$ defined as $P_U(v) = u$. Some of the nice properties of the orthogonal projection operator of $V$ onto $U$ is that $\mathrm{range} (P_U) = U$, $\mathrm{null} (P_U) = U^{\perp}$, $(v - P_U(v)) \in U^{\perp}$, $P_U^2 = P_U$, and that $\| P_U(v) \| ≤\| v \|$. |
As with the sine, we do not know anything about derivatives that allows us to compute the derivatives of the exponential and logarithmic functions without going back to basics. Let's do a little work with the definition again:
\[\eqalign{ {d\over dx}a^x&=\lim_{\Delta x\to 0} {a^{x+\Delta x}-a^x\over \Delta x}\cr& =\lim_{\Delta x\to 0} {a^xa^{\Delta x}-a^x\over \Delta x}\cr& =\lim_{\Delta x\to 0} a^x{a^{\Delta x}-1\over \Delta x}\cr& =a^x\lim_{\Delta x\to 0} {a^{\Delta x}-1\over \Delta x}.\cr }\]
There are two interesting things to note here: As in the case of the sine function we are left with a limit that involves \(\Delta x\) but not \(x\), which means that whatever \( \lim_{\Delta x\to 0} (a^{\Delta x}-1)/\Delta x\) is, we know that it is a number, that is, a constant. This means that \( a^x\) has a remarkable property: its derivative is a constant times itself.
We earlier remarked that the hardest limit we would compute is \( \lim_{x\to0}\sin x/x=1\); we now have a limit that is just a bit too hard to include here. In fact the hard part is to see that \( \lim_{\Delta x\to 0} (a^{\Delta x}-1)/\Delta x\) even exists---does this fraction really get closer and closer to some fixed value? Yes it does, but we will not prove this fact.
We can look at some examples. Consider \( (2^x-1)/x\) for some small values of \(x\): 1, \(0.828427124\), \(0.756828460\), \(0.724061864\), \(0.70838051\), \(0.70070877\) when \(x\) is 1, \(1/2\), \(1/4\), \(1/8\), \(1/16\), \(1/32\), respectively. It looks like this is settling in around \(0.7\), which turns out to be true (but the limit is not exactly \(0.7\)). Consider next \( (3^x-1)/x\): \(2\), \(1.464101616\), \(1.264296052\), \(1.177621520\), \(1.13720773\), \(1.11768854\), at the same values of \(x\). It turns out to be true that in the limit this is about \(1.1\).
Two examples don't establish a pattern, but if you do more examples you will find that the limit varies directly with the value of \(a\): bigger \(a\), bigger limit; smaller \(a\), smaller limit. As we can already see, some of these limits will be less than 1 and some larger than 1. Somewhere between \(a=2\) and \(a=3\) the limit will be exactly 1; the value at which this happens is called \(e\), so that
\[\lim_{\Delta x\to 0} {e^{\Delta x}-1\over \Delta x}=1.\]
As you might guess from our two examples, \(e\) is closer to 3 than to 2, and in fact \(e\approx 2.718\).
Now we see that the function \( e^x\) has a truly remarkable property:
\[\eqalign{ {d\over dx}e^x&=\lim_{\Delta x\to 0} {e^{x+\Delta x}-e^x\over \Delta x}\cr& =\lim_{\Delta x\to 0} {e^xe^{\Delta x}-e^x\over \Delta x}\cr& =\lim_{\Delta x\to 0} e^x{e^{\Delta x}-1\over \Delta x}\cr& =e^x\lim_{\Delta x\to 0} {e^{\Delta x}-1\over \Delta x}\cr& =e^x.\cr }\]
That is, \( e^x\) is its own derivative, or in other words the slope of \( e^x\) is the same as its height, or the same as its second coordinate: The function \( f(x)=e^x\) goes through the point \( (z,e^z)\) and has slope \( e^z\) there, no matter what \(z\) is. It is sometimes convenient to express the function \( e^x\) without an exponent, since complicated exponents can be hard to read. In such cases we use \(\exp(x)\), e.g., \( \exp(1+x^2)\) instead of \( e^{1+x^2}\).
What about the logarithm function? This too is hard, but as the cosine function was easier to do once the sine was done, so the logarithm is easier to do now that we know the derivative of the exponential function. Let's start with \( \log_e x\), which as you probably know is often abbreviated \(\ln x\) and called the "natural logarithm'' function.
Consider the relationship between the two functions, namely, that they are inverses, that one "undoes'' the other. Graphically this means that they have the same graph except that one is "flipped'' or "reflected'' through the line \(y=x\), as shown in Figure \(\PageIndex{1}\).
This means that the slopes of these two functions are closely related as well: For example, the slope of \( e^x\) is \(e\) at \(x=1\); at the corresponding point on the \(\ln(x)\) curve, the slope must be \(1/e\), because the "rise'' and the "run'' have been interchanged. Since the slope of \( e^x\) is \(e\) at the point \((1,e)\), the slope of \(\ln(x)\) is \(1/e\) at the point \((e,1)\).
More generally, we know that the slope of \( e^x\) is \( e^z\) at the point \( (z,e^z)\), so the slope of \(\ln(x)\) is \( 1/e^z\) at \( (e^z,z)\), as indicated in Figure \(\PageIndex{2}\). In other words, the slope of \(\ln x\) is the reciprocal of the first coordinate at any point; this means that the slope of \(\ln x\) at \((x,\ln x)\) is \(1/x\). The upshot is: \({d\over dx}\ln x = {1\over x}.\) We have discussed this from the point of view of the graphs, which is easy to understand but is not normally considered a rigorous proof---it is too easy to be led astray by pictures that seem reasonable but that miss some hard point. It is possible to do this derivation without resorting to pictures, and indeed we will see an alternate approach soon.
Note that \(\ln x\) is defined only for \(x>0\). It is sometimes useful to consider the function \(\ln |x|\), a function defined for \(x\not=0\). When \(x < 0\), \(\ln |x|=\ln(-x)\) and
\[{d\over dx}\ln |x|={d\over dx}\ln (-x)={1\over -x}(-1)={1\over x}.\]
Thus whether \(x\) is positive or negative, the derivative is the same.
What about the functions \( a^x\) and \( \log_a x\)? We know that the derivative of \( a^x\) is some constant times \( a^x\) itself, but what constant? Remember that "the logarithm is the exponent'' and you will see that \( a=e^{\ln a}\). Then \(a^x = (e^{\ln a})^x = e^{x\ln a},\) and we can compute the derivative using the chain rule:
\[{d\over dx} a^x = {d\over dx}(e^{\ln a})^x = {d\over dx}e^{x\ln a} = (\ln a)e^{x\ln a} =(\ln a)a^x.\]
The constant is simply \(\ln a\). Likewise we can compute the derivative of the logarithm function \( \log_a x\). Since \(x=e^{\ln x}\) we can take the logarithm base \(a\) of both sides to get \( \log_a(x)=\log_a(e^{\ln x})=\ln x \log_a e\). Then
\[{d\over dx}\log_a x = {1\over x}\log_a e.\]
This is a perfectly good answer, but we can improve it slightly. Since
\[\eqalign{ a&=e^{\ln a}\cr \log_a(a) &= \log_a(e^{\ln a}) = \ln a\log_a e\cr 1&=\ln a\log_a e\cr {1\over \ln a}&=\log_a e,\cr }\]
we can replace \( \log_a e\) to get \({d\over dx}\log_a x = {1\over x\ln a}\).
You may if you wish memorize the formulas
\[{d\over dx}a^x = (\ln a)a^x \quad \hbox{and}\quad {d\over dx}\log_a x = {1\over x\ln a}.\]
Because the "trick'' \( a=e^{\ln a}\) is often useful, and sometimes essential, it may be better to remember the trick, not the formula.
Example \(\PageIndex{1}\)
Compute the derivative of \( f(x)=2^x\).
Solution
\[\eqalign{ {d\over dx}2^{x} &= {d\over dx}(e^{\ln 2})^x\cr& = {d\over dx}e^{x\ln 2}\cr& = \left({d\over dx} x\ln 2\right) e^{x\ln 2}\cr& = (\ln 2) e^{x\ln 2}=2^x\ln2\cr }\]
Example \(\PageIndex{2}\)
Compute the derivative of \( f(x)=2^{x^2}=2^{(x^2)}\).
\[\eqalign{ {d\over dx}2^{x^2} &= {d\over dx}e^{x^2\ln 2}\cr& = \left({d\over dx} x^2\ln 2\right) e^{x^2\ln 2}\cr& = (2\ln 2) x e^{x^2\ln 2}\cr& = (2\ln 2) x 2^{x^2}\cr }\]
Example \(\PageIndex{3}\)
Compute the derivative of \( f(x)=x^x\). At first this appears to be a new kind of function: it is not a constant power of \(x\), and it does not seem to be an exponential function, since the base is not constant. But in fact it is no harder than the previous example.
\[\eqalign{ {d\over dx}x^x&={d\over dx}e^{x\ln x}\cr& =\left({d\over dx}x\ln x\right)e^{x\ln x}\cr& =(x{1\over x}+\ln x)x^x\cr& =(1+\ln x)x^x\cr }\]
Example \(\PageIndex{4}\)
Recall that we have not justified the power rule except when the exponent is a positive or negative integer. We can use the exponential function to take care of other exponents.
\[\eqalign{ {d\over dx}x^r&={d\over dx}e^{r\ln x}\cr& =\left({d\over dx}r\ln x\right)e^{r\ln x}\cr& =(r{1\over x})x^r\cr& =rx^{r-1}\cr }\] |
A. Enayat, J. D. Hamkins, and B. Wcisło, “Topological models of arithmetic,” ArXiv e-prints, 2018. (under review)
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Abstract. Ali Enayat had asked whether there is a nonstandard model of Peano arithmetic (PA) that can be represented as $\newcommand\Q{\mathbb{Q}}\langle\Q,\oplus,\otimes\rangle$, where $\oplus$ and $\otimes$ are continuous functions on the rationals $\Q$. We prove, affirmatively, that indeed every countable model of PA has such a continuous presentation on the rationals. More generally, we investigate the topological spaces that arise as such topological models of arithmetic. The reals $\newcommand\R{\mathbb{R}}\R$, the reals in any finite dimension $\R^n$, the long line and the Cantor space do not, and neither does any Suslin line; many other spaces do; the status of the Baire space is open.
The first author had inquired whether a nonstandard model of arithmetic could be continuously presented on the rational numbers.
Main Question. (Enayat, 2009) Are there continuous functions $\oplus$ and $\otimes$ on the rational numbers $\Q$, such that $\langle\Q,\oplus,\otimes\rangle$ is a nonstandard model of arithmetic?
By a model of arithmetic, what we mean here is a model of the first-order Peano axioms PA, although we also consider various weakenings of this theory. The theory PA asserts of a structure $\langle M,+,\cdot\rangle$ that it is the non-negative part of a discretely ordered ring, plus the induction principle for assertions in the language of arithmetic. The natural numbers $\newcommand\N{\mathbb{N}}\langle \N,+,\cdot\rangle$, for example, form what is known as the
standard model of PA, but there are also many nonstandard models, including continuum many non-isomorphic countable models.
We answer the question affirmatively, and indeed, the main theorem shows that every countable model of PA is continuously presented on $\Q$. We define generally that a
topological model of arithmetic is a topological space $X$ equipped with continuous functions $\oplus$ and $\otimes$, for which $\langle X,\oplus,\otimes\rangle$ satisfies the desired arithmetic theory. In such a case, we shall say that the underlying space $X$ continuously supports a model of arithmetic and that the model is continuously presented upon the space $X$.
Question. Which topological spaces support a topological model of arithmetic?
In the paper, we prove that the reals $\R$, the reals in any finite dimension $\R^n$, the long line and Cantor space do not support a topological model of arithmetic, and neither does any Suslin line. Meanwhile, there are many other spaces that do support topological models, including many uncountable subspaces of the plane $\R^2$. It remains an open question whether any uncountable Polish space, including the Baire space, can support a topological model of arithmetic.
Let me state the main theorem and briefly sketch the proof.
Main Theorem. Every countable model of PA has a continuous presentation on the rationals $\Q$.
Proof. We shall prove the theorem first for the standard model of arithmetic $\langle\N,+,\cdot\rangle$. Every school child knows that when computing integer sums and products by the usual algorithms, the final digits of the result $x+y$ or $x\cdot y$ are completely determined by the corresponding final digits of the inputs $x$ and $y$. Presented with only final segments of the input, the child can nevertheless proceed to compute the corresponding final segments of the output.
\begin{equation*}\small\begin{array}{rcr}
\cdots1261\quad & \qquad & \cdots1261\quad\\ \underline{+\quad\cdots 153\quad}&\qquad & \underline{\times\quad\cdots 153\quad}\\ \cdots414\quad & \qquad & \cdots3783\quad\\ & & \cdots6305\phantom{3}\quad\\ & & \cdots1261\phantom{53}\quad\\ & & \underline{\quad\cdots\cdots\phantom{253}\quad}\\ & & \cdots933\quad\\ \end{array}\end{equation*}
This phenomenon amounts exactly to the continuity of addition and multiplication with respect to what we call the
final-digits topology on $\N$, which is the topology having basic open sets $U_s$, the set of numbers whose binary representations ends with the digits $s$, for any finite binary string $s$. (One can do a similar thing with any base.) In the $U_s$ notation, we include the number that would arise by deleting initial $0$s from $s$; for example, $6\in U_{00110}$. Addition and multiplication are continuous in this topology, because if $x+y$ or $x\cdot y$ has final digits $s$, then by the school-child’s observation, this is ensured by corresponding final digits in $x$ and $y$, and so $(x,y)$ has an open neighborhood in the final-digits product space, whose image under the sum or product, respectively, is contained in $U_s$.
Let us make several elementary observations about the topology. The sets $U_s$ do indeed form the basis of a topology, because $U_s\cap U_t$ is empty, if $s$ and $t$ disagree on some digit (comparing from the right), or else it is either $U_s$ or $U_t$, depending on which sequence is longer. The topology is Hausdorff, because different numbers are distinguished by sufficiently long segments of final digits. There are no isolated points, because every basic open set $U_s$ has infinitely many elements. Every basic open set $U_s$ is clopen, since the complement of $U_s$ is the union of $U_t$, where $t$ conflicts on some digit with $s$. The topology is actually the same as the metric topology generated by the $2$-adic valuation, which assigns the distance between two numbers as $2^{-k}$, when $k$ is largest such that $2^k$ divides their difference; the set $U_s$ is an open ball in this metric, centered at the number represented by $s$. (One can also see that it is metric by the Urysohn metrization theorem, since it is a Hausdorff space with a countable clopen basis, and therefore regular.) By a theorem of Sierpinski, every countable metric space without isolated points is homeomorphic to the rational line $\Q$, and so we conclude that the final-digits topology on $\N$ is homeomorphic to $\Q$. We’ve therefore proved that the standard model of arithmetic $\N$ has a continuous presentation on $\Q$, as desired.
But let us belabor the argument somewhat, since we find it interesting to notice that the final-digits topology (or equivalently, the $2$-adic metric topology on $\N$) is precisely the order topology of a certain definable order on $\N$, what we call the
final-digits order, an endless dense linear order, which is therefore order-isomorphic and thus also homeomorphic to the rational line $\Q$, as desired.
Specifically, the final-digits order on the natural numbers, pictured in figure 1, is the order induced from the lexical order on the finite binary representations, but considering the digits from right-to-left, giving higher priority in the lexical comparison to the low-value final digits of the number. To be precise, the final-digits order $n\triangleleft m$ holds, if at the first point of disagreement (from the right) in their binary representation, $n$ has $0$ and $m$ has $1$; or if there is no disagreement, because one of them is longer, then the longer number is lower, if the next digit is $0$, and higher, if it is $1$ (this is not the same as treating missing initial digits as zero). Thus, the even numbers appear as the left half of the order, since their final digit is $0$, and the odd numbers as the right half, since their final digit is $1$, and $0$ is directly in the middle; indeed, the highly even numbers, whose representations end with a lot of zeros, appear further and further to the left, while the highly odd numbers, which end with many ones, appear further and further to the right. If one does not allow initial $0$s in the binary representation of numbers, then note that zero is represented in binary by the empty sequence. It is evident that the final-digits order is an endless dense linear order on $\N$, just as the corresponding lexical order on finite binary strings is an endless dense linear order.
The basic open set $U_s$ of numbers having final digits $s$ is an open set in this order, since any number ending with $s$ is above a number with binary form $100\cdots0s$ and below a number with binary form $11\cdots 1s$ in the final-digits order; so $U_s$ is a union of intervals in the final-digits order. Conversely, every interval in the final-digits order is open in the final-digits topology, because if $n\triangleleft x\triangleleft m$, then this is determined by some final segment of the digits of $x$ (appending initial $0$s if necessary), and so there is some $U_s$ containing $x$ and contained in the interval between $n$ and $m$. Thus, the final-digits topology is the precisely same as the order topology of the final-digits order, which is a definable endless dense linear order on $\N$. Since this order is isomorphic and hence homeomorphic to the rational line $\Q$, we conclude again that $\langle \N,+,\cdot\rangle$ admits a continuous presentation on $\Q$.
We now complete the proof by considering an arbitrary countable model $M$ of PA. Let $\triangleleft^M$ be the final-digits order as defined inside $M$. Since the reasoning of the above paragraphs can be undertaken in PA, it follows that $M$ can see that its addition and multiplication are continuous with respect to the order topology of its final-digits order. Since $M$ is countable, the final-digits order of $M$ makes it a countable endless dense linear order, which by Cantor’s theorem is therefore order-isomorphic and hence homeomorphic to $\Q$. Thus, $M$ has a continuous presentation on the rational line $\Q$, as desired. $\Box$
The executive summary of the proof is: the arithmetic of the standard model $\N$ is continuous with respect to the final-digits topology, which is the same as the $2$-adic metric topology on $\N$, and this is homeomorphic to the rational line, because it is the order topology of the final-digits order, a definable endless dense linear order; applied in a nonstandard model $M$, this observation means the arithmetic of $M$ is continuous with respect to its rational line $\Q^M$, which for countable models is isomorphic to the actual rational line $\Q$, and so such an $M$ is continuously presentable upon $\Q$.
Let me mention the following order, which it seems many people expect to use instead of the final-digits order as we defined it above. With this order, one in effect takes missing initial digits of a number as $0$, which is of course quite reasonable.
The problem with this order, however, is that the order topology is not actually the final-digits topology. For example, the set of all numbers having final digits $110$ in this order has a least element, the number $6$, and so it is not open in the order topology. Worse, I claim that arithmetic is not continuous with respect to this order. For example, $1+1=2$, and $2$ has an open neighborhood consisting entirely of even numbers, but every open neighborhood of $1$ has both odd and even numbers, whose sums therefore will not all be in the selected neighborhood of $2$. Even the successor function $x\mapsto x+1$ is not continuous with respect to this order.
Finally, let me mention that a version of the main theorem also applies to the integers $\newcommand\Z{\mathbb{Z}}\Z$, using the following order.
Go to the article to read more.
A. Enayat, J. D. Hamkins, and B. Wcisło, “Topological models of arithmetic,” ArXiv e-prints, 2018. (under review)
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I’m not sure to whom the image or the idea is due. Please comment if you have information. (See comments below for current information.)
The rules will naturally generalize those in Connect-Four. Namely, starting from an empty board, the players take turns placing their coins into the $\omega\times 4$ grid. When a coin is placed in a column, it falls down to occupy the lowest available cell. Let us assume for now that the game proceeds for $\omega$ many moves, whether or not the board becomes full, and the goal is to make a connected sequence in a row of $\omega$ many coins of your color (you don’t have to fill the whole row, but rather a connected infinite segment of it suffices). A draw occurs when neither or both players achieve their goals.
In the $\omega\times 6$ version of the game that is shown, and indeed in the $\omega\times n$ version for any finite $n$, I claim that neither player can force a win; both players have drawing strategies.
Theorem. In the game Connect-$\omega$ on a board of size $\omega\times n$, where $n$ is finite, neither player has a winning strategy; both players have drawing strategies. Proof. For a concrete way to see this, observe that either player can ensure that there are infinitely many of their coins on the bottom row: they simply place a coin into some far-out empty column. This blocks a win for the opponent on the bottom row. Next, observe that neither player can afford to follow the strategy of always answering those moves on top, since this would lead to a draw, with a mostly empty board. Thus, it must happen that infinitely often we are able to place a coin onto the second row. This blocks a win for the opponent on the second row. And so on. In this way, either players can achieve infinitely many of their coins on each row, thereby blocking any row as a win for their opponent. So both players have drawing strategies. $\Box$
Let me point out that on a board of size $\omega\times n$, where $n$ is odd, we can also make this conclusion by a strategy-stealing argument. Specifically, I argue first that the first player can have no winning strategy. Suppose $\sigma$ is a winning strategy for the first player on the $\omega\times n$ board, with $n$ odd, and let us describe a strategy for the second player. After the first move, the second player mentally ignores a finite left-initial segment of the playing board, which includes that first move and with a odd number of cells altogether in it (and hence an even number of empty cells remaining); the second player will now aim to win on the now-empty right-side of the board, by playing as though playing first in a new game, using strategy $\sigma$. If the first player should ever happen to play on the ignored left side of the board, then the second player can answer somewhere there (it doesn’t matter where). In this way, the second player plays with $\sigma$ as though he is the first player, and so $\sigma$ cannot be winning for the first player, since in this way the second player would win in this stolen manner.
Similarly, let us argue by strategy-stealing that the second player cannot have a winning strategy on the board $\omega\times n$ for odd finite $n$. Suppose that $\tau$ is a winning strategy for the second player on such a board. Let the first player always play at first in the left-most column. Because $n$ is odd, the second player will eventually have to play first in the second or later columns, leaving an even number of empty cells in the first column (perhaps zero). At this point, the first player can play as though he was the second player on the right-side board containing only that fresh move. If the opponent plays again to the left, then our player can also play in that region (since there were an even number of empty cells). Thus, the first player can steal the strategy $\tau$, and so it cannot be winning for the second player.
I am unsure about how to implement the strategy stealing arguments when $n$ is even. I shall give this more thought. In any case, the theorem for this case was already proved directly by the initial concrete argument, and in this sense we do not actually need the strategy stealing argument for this case.
Meanwhile, it is natural also to consider the $n\times\omega$ version of the game, which has only finitely many columns, each infinite. The players aim to get a sequence of $\omega$-many coins in a column. This is clearly impossible, as the opponent can prevent a win by always playing atop the most recent move. Thus:
Theorem. In the game Connect-$\omega$ on a board of size $n\times\omega$, where $n$ is finite, neither player has a winning strategy; both players have drawing strategies.
Perhaps the most natural variation of the game, however, occurs with a board of size $\omega\times\omega$. In this version, like the original Connect-Four, a player can win by either making a connected row of $\omega$ many coins, or a connected column or a connected diagonal of $\omega$ many coins. Note that we orient the $\omega$ size column upwards, so that there is no top cell, but rather, one plays by selecting a not-yet-filled column and then occupying the lowest available cell in that column.
Theorem. In the game Connect-$\omega$ on a board of size $\omega\times\omega$, neither player has a winning strategy. Both players have drawing strategies. Proof. Consider the strategy-stealing arguments. If the first player has a winning strategy $\sigma$, then let us describe a strategy for the second player. After the first move, the second player will ignore finitely many columns at the left, including that first actual move, aiming to play on the empty right-side of the board as though the first player using stolen strategy $\sigma$ (but with colors swapped). This will work fine, as long as the first player also plays on that part of the board. Whenever the first player plays on the ignored left-most part, simply respond by playing atop. This prevents a win in that part of the part, and so the second player will win on the right-side by pretending to be first there. So there can be no such winning strategy $\sigma$ for the first player.
If the second player has a winning strategy $\tau$, then as before let the first player always play in the first column, until $\tau$ directs the second player to play in another column, which must eventually happen if $\tau$ is winning. At that moment, the first player can pretend to be second on the subboard omitting the first column. So $\tau$ cannot have been winning after all for the second player. $\Box$
In the analysis above, I was considering the game that proceeded in time $\omega$, with $\omega$ many moves. But such a play of the game may not actually have filled the board completely. So it is natural to consider a version of the game where the players continue to play transfinitely, if the board is not yet full.
So let us consider now the transfinite-play version of the game, where play proceeds transfinitely through the ordinals, until either the board is filled or one of the players has achieved the winning goal. Let us assume that the first player also plays first at limit stages, at $\omega$ and $\omega\cdot 2$ and $\omega^2$, and so on, if game play should happen to proceed for that long.
The concrete arguments that I gave above continue to work for the transfinite-play game on the boards of size $\omega\times n$ and $n\times\omega$.
Theorem. In the transfinite-play version of Connect-$\omega$ on boards of size $\omega\times n$ or $n\times\omega$, where $n$ is finite, neither player can have a winning strategy. Indeed, both players can force a draw while also filling the board in $\omega$ moves. Proof. It is clear that on the $n\times\omega$ board, either player can force each column to have infinitely many coins of their color, and this fills the board, while also preventing a win for the opponent, as desired.
On the $\omega\times n$ board, consider a variation of the strategy I described above. I shall simply always play in the first available empty column, thereby placing my coin on the bottom row, until the opponent also plays in a fresh column. At that moment, I shall play atop his coin, thereby placing another coin in the second row; immediately after this, I also play subsequently in the left-most available column (so as to force the board to be filled). I then continue playing in the bottom row, until the opponent also does, which she must, and then I can add another coin to the second row and so on. By always playing the first-available second-row slot with all-empty second rows to the right, I can ensure that the opponent will eventually also make a second-row play (since otherwise I will have a winning condition on the second row), and at such a moment, I can also make a third-row play. By continuing in this way, I am able to place infinitely many coins on each row, while also forcing that the board becomes filled. $\Box$
Unfortunately, the transfinite-play game seems to break the strategy-stealing arguments, since the play is not symmetric for the players, as the first player plays first at limit stages.
Nevertheless, following some ideas of Timothy Gowers in the comments below, let me show that the second player has a drawing strategy.
Theorem. In the transfinite-play version of Connect-$\omega$ on a board of size $\omega\times\omega$, the second player has a drawing strategy. Proof. We shall arrange that the second player will block all possible winning configurations for the first player, or to have column wins for each player. To block all row wins, the second player will arrange to occupy infinitely many cells in each row; to block all diagonal wins, the second player will aim to occupy infinitely many cells on each possible diagonal; and to block the column wins, the second player will aim either to have infinitely many cells on each column or to copy a winning column of the opponent on another column.
To achieve these things, we simply play as follows. Take the columns in successive groups of three. On the first column in each block of three, that is on the columns indexed $3m$, the second player will always answer a move by the first player on this column. In this way, the second player occupies every other cell on these columns—all at the same height. This will block all diagonal wins, because every diagonal winning configuration will need to go through such a cell.
On the remaining two columns in each group of three, columns $3m+1$ and $3m+2$, let the second player simply copy moves of the opponent on one of these columns by playing on the other. These moves will therefore be opposite colors, but at the same height. In this way, the second player ensures that he has infinitely many coins on each row, blocking the row wins. And also, this strategy ensures that in these two columns, at any limit stage, either neither player has achieved a winning configuration or both have.
Thus, we have produced a drawing strategy for the second player. $\Box$
Thus, there is no advantage to going first. What remains is to determine if the first player also has a drawing strategy, or whether the second player can actually force a win.
Gowers explains in the comments below also how to achieve such a copying mechanism to work on a diagonal, instead of just on a column.
I find it also fascinating to consider the natural generalizations of the game to larger ordinals. We may consider the game of Connect-$\alpha$ on a board of size $\kappa\times\lambda$, for any ordinals $\alpha,\kappa,\lambda$, with transfinite play, proceeding until the board is filled or the winning conditions are achieved. Clearly, there are some instances of this game where a player has a winning strategy, such as the original Connect-Four configuration, where the first player wins, and presumably many other instances.
Question. In which instances of Connect-$\alpha$ on a board of size $\kappa\times\lambda$ does one of the players have a winning strategy?
It seems to me that the groups-of-three-columns strategy described above generalizes to show that the second player has at least a drawing strategy in Connect-$\alpha$ on board $\kappa\times\lambda$, whenever $\alpha$ is infinite.
Stay tuned… |
Given that $\alpha$ is a root (in the field extension) of the irreducible polynomial $X^4+X^3-X+2\in\mathbb{Q}[X]$, I have to find the minimal polynomial of $\alpha^2$. I am thinking about this for a while, but I can't find it. I need some hints. Thank you.
You have $\alpha^4 + 2 = \alpha - \alpha^3$, and squaring both sides gives a polynomial of degree $4$ satisfied by $\alpha^2$. (All the powers of $\alpha$ appearing will be even.) Then show it is the minimal polynomial for $\alpha^2$.
Maybe someone can come up with some clever way, but here is the hard-work method.
Write down $1, \alpha^2, \alpha^4, \alpha^6, \alpha^8$ in terms of $1, \alpha, \alpha^2, \alpha^3$. These five elements should be linearly dependent, so you'll find $c_0 1 + c_1 \alpha^2 + c_2 \alpha^4 + c_3 \alpha^6 + c_8 \alpha^8 = 0$ for some $c_i \in {\mathbb Q}$. This gives you a 4th degree expression in $\alpha^2$ that equals 0.
(Note, if the degree of the extension ${\mathbb Q}(\alpha^2) : {\mathbb Q}$ were 2, you'd already find $1, \alpha^2, \alpha^4$ to be linearly dependent, but you'll see directly that they're not. The degree can't be 3, because it must be a divisor of $[{\mathbb Q}(\alpha) : {\mathbb Q}] = 4$. So it must be 4.)
Look up the Dandelin-Graeffe iteration. In short, $p(x)\cdot p(-x)$ is a polynomial in $x^2$, say $q(x^2)=p(x)\cdot p(-x)$. So if $p(\alpha)=0$, then $q(\alpha^2)=0$.
This gives the same result as the answer by Zarrax, however $q$ need not be irreducible. |
Apart from images of representations of subgroups of SU(2), what are the Lie subgroups of SU(3)? Where should I look for a reference?
A first approximation to an answer is to determine the Lie subalgebras of $\mathfrak{su}(3)$. Here is their Hasse diagram with edges denoting inclusions. One can work this out iteratively by working out the maximal subalgebras and this follows from Dynkin's work.
The notation $\mathfrak{so}(2)_{[\alpha,\beta]}$ means one element of the pencil of one-dimensional subalgebras of $\mathfrak{so}(2)\oplus \mathfrak{so}(2)$ and $\mathfrak{so}(3)_{\text{irr}}$ is a subalgebra which acts irreducibly on the 3-dimensional irreducible representation.
This is essentially just a slower walk through José's diagram.
I'll assume you're interested in closed, connected subgroups $G\leq SU(3)$. I'll write $V$ for $\mathbb{C}^3$ regarded as a faithful representation of $G$ via the inclusion in $SU(3)$. If $G$ is abelian then we can decompose $V$ as a sum of three one-dimensional representations, say $V=V_1\oplus V_2\oplus V_3$, and then $$ G \leq SU(V)\cap (U(V_1)\times U(V_2)\times U(V_3)) \simeq U(1)\times U(1). $$ All remaining questions about the abelian case are now easy, so I'll assume that $G$ is nonabelian.
Now consider the rank of $G$, ie the dimension of its maximal torus. As $G$ is nonabelian and contained in $SU(3)$ the rank must be one or two.
If the rank is one, then it is standard that $G$ is isomorphic to $SU(2)$ or $SU(2)/\{\pm I\}\simeq SO(3)$. Let $W_1$ denote the standard two-dimensional representation of $SU(2)$, and let $W_2$ denote the symmetric tensor square of $W_1$, which has dimension $3$. The action of $SU(2)$ on $W_2$ factors through $SO(3)$. The only irreducible representations of $SU(2)$ of dimension $\leq 3$ are $\mathbb{C}$, $W_1$ and $W_2$. It follows that one of the following holds:
$G\simeq SU(2)$, and $V$ corresponds to $\mathbb{C}\oplus W_1$. This means that there is a one-dimensional subspace $L\leq V$ such that $G=\{g\in SU(V):g|_L=1_L\}$. $G\simeq SO(3)$, and $V$ corresponds to $W_2$. This means that there is a real subspace $X\leq V$ with $V=X\oplus iX$, and $G$ is the evident copy of $SO(X)$ in $SU(V)$.
Suppose instead that the rank is two. This means that any maximal torus in $G$ is also a maximal torus in $SU(V)$, so $G$ is a parabolic subgroup.
Suppose that $V$ is reducible as a representation of $G$. This implies that there is a one-dimensional subspace $L\leq V$ that is preserved by $G$, and thus that $G$ is contained in the image of the homomorphism $\phi:U(L^\perp)\to SU(V)$ given by $$ \phi(g)=g\oplus(\det(g)^{-1}.1_L) $$ As $G$ is connected and nonabelian of rank two, I think it has to be the whole image of $\phi$.
Suppose instead that $V$ is irreducible. I think it then follows from the standard story about parabolic subgroups that $G$ is all of $SU(V)$.
U(2) is also a subgroup of SU(3). It is missing in the answers above. Probably people mean U(2) when they write SU(2)xU(1). But U(2) is a genuine subgroup of SU(3). |
Commutative Rings
Recall that if $(G, *)$ is a group with the additional property that for all $x, y \in G$ we have that $x * y = y * x$ then $(G, *)$ is said to be an abelian group or a commutative group.
Now suppose that we have a ring, $(R, +, *)$ Then $(R, +)$ by definition is an abelian group, but there's no requirement that the operation $*$ needs to be commutative as well Rings for which both operations are commutative are called commutative rings which we formally define below.
Definition: Let $(R, +, *)$ be a ring. Then $(R, +, *)$ is said to be a Commutative Ring if for all $x, y \in R$ we have that $x * y = y * x$, i.e., the operation $*$ is commutative.
There are many examples of commutative rings. For example, $(\mathbb{R}, +, *)$ is a commutative ring, and of course, $(\mathbb{Z}, +, *)$, and $(\mathbb{Q}, +, *)$. This is because multiplication of the real numbers (and the integers/rationals) is commutative.
Of course, not all rings are commutative though. For example, the ring $(M_{nn}, +, *)$ where $M_{nn}$ denotes the set of all $n \times n$ matrices is not commutative for $n \geq 2$ because matrix multiplicative is not commutative in general. |
Here's a proof of Ben's claim--I don't know where in Matsumura Ben is referring, but it's not too hard if you take two well known theorems for granted.
To se things up, let us first make a definition. If $f:X\to Y$ is a morphism, and $\mathscr{F}$ a quasicoherent sheaf on $X$ then define the
flat locus of $f$ (with respect to $\mathscr{F}$) to be the set of $x\in X$ such that $\mathscr{F}_x$ is a flat $\mathcal{O}_{Y,f(x)}$ module.
Then a version of the theorem is then the following:
Theorem: Let $f:X\to Y$ be a finite type morphism between Noetherian schemes, and let $\mathscr{F}$ be a coherent $\mathcal{O}_X$-module. Then, the flat locus of $f$ is open.
The hard facts one needs to assume (and which are certainly in Matsumura I believe, but is easy to find online) are Grothendieck's Generic Freeness Lemma as you guessed:
Lemma(Generic Freeness): Let $A$ be a Noetherian domain, $B$ a finite type $A$-algebra, and $M$ a finitely generated $B$-module. Then, there exists some non-zero $f\in A$ such that $M_f$ is a free $A_f$ module.
as well as this following very technical, but very useful lemma about deducing flatness from flatness over a quotient:
Lemma(Gross Technical): Let $A\to B$ be a morphism of Noetherian rings. Suppose that $I$ is an ideal of $A$ such that $IB$ is contained in the Jacobson radical of $B$, and that $M$ is a finitely generated $B$-module. Then, the following are equivalent
$$\begin{aligned}& (1)\quad M\textit{ is flat over }A\\& (2)\quad M/(IB)M\textit{ is flat over }A/I\textit{ and }\text{Tor}_1^A(M,A/I)=0\end{aligned}$$
The proof of this, I will admit, is really horrendous. Or, at least the one I know is. That said, it does show up somewhat often in certain circles.
Now, we only need one more result before we actually prove the main theorem, but one which is much easier than the above two results
Theorem: Let $A$ be a Noetherian ring, $B$ a finite type $A$-algebra, $M$ a finitely generated $B$-module, $\mathfrak{q}\in\text{Spec}(B)$, and $\mathfrak{p}$ the image of $\mathfrak{q}$ under $\text{Spec}(B)\to\text{Spec}(A)$. Suppose that $M_\mathfrak{q}$ is flat over $A_\mathfrak{p}$, then there is some $g\in B-\mathfrak{q}$ such that $(M/\mathfrak{p}M)_g$ is flat over $A/\mathfrak{p}$ and $\text{Tor}_1^A(M,A/\mathfrak{p})_g=0$.
Proof: Since $A/\mathfrak{p}$ is a Noetherian domain, and $M/\mathfrak{p}M$ a finitely generated $B$-module, we can use Generic Freeness to produce some element $f\in A-\mathfrak{p}$ such that $(M/\mathfrak{p}M)_f$ is a free $A_f$-module.
Now, since $M_\mathfrak{q}$ is flat over $A$, since by assumption $M_\mathfrak{q}$ is flat over $A_\mathfrak{p}$ and $A\to A_\mathfrak{p}$ is flat, we deduce
$$\text{Tor}_1^A(M,A/\mathfrak{p})_\mathfrak{q}\cong\text{Tor}_1^A(M_\mathfrak{q},A/\mathfrak{p})=0$$
We'd be done if we knew $\text{Tor}_1^A(M,A/\mathfrak{p})$ were a finitely generated $B$-module, since we could clearly then find $g_1,\ldots,g_r\in B-\mathfrak{q}$ annihilating the generators of $\text{Tor}_1^A(M,A/\mathfrak{p})$ and then take $g=fg_1\cdots g_r$. To see that $\text{Tor}_1^A(M,A/\mathfrak{p})$ is a finitely generated we merely use the short exact sequence
$$0\to\mathfrak{p}\to A\to A/\mathfrak{p}\to 0$$
to get the long exact sequence
$$\cdots\to\underbrace{\text{Tor}_1^A(M,A)}_{=0}\to\text{Tor}_1^A(M,A/\mathfrak{p})\to M\otimes_A\mathfrak{p}\to M\otimes_A \to M\otimes_A A/\mathfrak{p}\to 0$$
to see that $\text{Tor}_1^A(M,A/\mathfrak{p})$ is a submodule of the finitely generated $B$-module $M\otimes_A\mathfrak{p}$ which, since $B$ is Noetherian, implies that $\text{Tor}_1^A(M,A/\mathfrak{p})$ is finitely generated as desired. $\blacksquare$
With this, we have the following corollary:
Corollary: Let us have the same hypotheses as in the previous theorem. Then, for every prime $\mathfrak{Q}$ of $B$ which contains $\mathfrak{p}$ but which does not contain $g$ has the property that $M_\mathfrak{Q}$ is flat over $A$
Proof: Consider the map $A\to B_\mathfrak{Q}$. Then apply the Gross Technical lemma to the ideal $I=\mathfrak{p}$ and the $B_\mathfrak{Q}$ module $M_\mathfrak{Q}$, in conjunction with the previous theorem. $\blacksquare$
Now, we can finally prove our desired theorem:
Proof: It's clear that this is a local property, and so we may as well assume that we are in affine land, say $X=\text{Spec}(B)$ and $Y=\text{Spec}(A)$, and that $\mathscr{F}$ is $\widetilde{M}$ for some finitely generated $B$-module $M$. Let $U$ denote the flat locus of $f$.
Note first that $U$ is clearly closed under generalization. Then, since our spaces are Noetherian, it suffices to show that $U$ is constructible. But, one form of constructibility is that if $C$ is an irreducible closed subset of $X$, then if $C\cap U$ is dense (i.e. non-empty), then there exists some non-empty open subset $O\subseteq C$ such that $C\cap U\supseteq O$.
To see this holds, suppose that $U\cap V(\mathfrak{p})$ is non-empty. Then, by definition, we know that $M_\mathfrak{q}$ is a flat $A_\mathfrak{p}$-module. Then, by the corollary to the Gross Technical lemma we know that $U\cap V(\mathfrak{p})$ contains $V(\mathfrak{p})\cap D(g)$.
The conclusion follows. $\blacksquare$
I guess, in retrospect, it is a pretty annoying theorem to prove in general. Mostly technical details. |
In Riemannian geometry, a
Jacobi field is a vector field along a geodesic \gamma in a Riemannian manifold describing the difference between the geodesic and an "infinitesimally close" geodesic. In other words, the Jacobi fields along a geodesic form the tangent space to the geodesic in the space of all geodesics. They are named after Carl Jacobi.
Contents Definitions and properties 1 Motivating example 2 Solving the Jacobi equation 3 Examples 4 See also 5 References 6 Definitions and properties
Jacobi fields can be obtained in the following way: Take a smooth one parameter family of geodesics \gamma_\tau with \gamma_0=\gamma, then
J(t)=\left.\frac{\partial\gamma_\tau(t)}{\partial \tau}\right|_{\tau=0}
is a Jacobi field, and describes the behavior of the geodesics in an infinitesimal neighborhood of a given geodesic \gamma.
A vector field
J along a geodesic \gamma is said to be a Jacobi field if it satisfies the Jacobi equation: \frac{D^2}{dt^2}J(t)+R(J(t),\dot\gamma(t))\dot\gamma(t)=0,
where
D denotes the covariant derivative with respect to the Levi-Civita connection, R the Riemann curvature tensor, \dot\gamma(t)=d\gamma(t)/dt the tangent vector field, and t is the parameter of the geodesic. On a complete Riemannian manifold, for any Jacobi field there is a family of geodesics \gamma_\tau describing the field (as in the preceding paragraph).
The Jacobi equation is a linear, second order ordinary differential equation; in particular, values of J and \frac{D}{dt}J at one point of \gamma uniquely determine the Jacobi field. Furthermore, the set of Jacobi fields along a given geodesic forms a real vector space of dimension twice the dimension of the manifold.
As trivial examples of Jacobi fields one can consider \dot\gamma(t) and t\dot\gamma(t). These correspond respectively to the following families of reparametrisations: \gamma_\tau(t)=\gamma(\tau+t) and \gamma_\tau(t)=\gamma((1+\tau)t).
Any Jacobi field J can be represented in a unique way as a sum T+I, where T=a\dot\gamma(t)+bt\dot\gamma(t) is a linear combination of trivial Jacobi fields and I(t) is orthogonal to \dot\gamma(t), for all t. The field I then corresponds to the same variation of geodesics as J, only with changed parameterizations.
Motivating example
On a sphere, the geodesics through the North pole are great circles. Consider two such geodesics \gamma_0 and \gamma_\tau with natural parameter, t\in [0,\pi], separated by an angle \tau. The geodesic distance
d(\gamma_0(t),\gamma_\tau(t)) \,
is
d(\gamma_0(t),\gamma_\tau(t))=\sin^{-1}\bigg(\sin t\sin\tau\sqrt{1+\cos^2 t\tan^2(\tau/2)}\bigg).
Computing this requires knowing the geodesics. The most interesting information is just that
d(\gamma_0(\pi),\gamma_\tau(\pi))=0 \,, for any \tau.
Instead, we can consider the derivative with respect to \tau at \tau=0:
\frac{\partial}{\partial\tau}\bigg|_{\tau=0}d(\gamma_0(t),\gamma_\tau(t))=|J(t)|=\sin t.
Notice that we still detect the intersection of the geodesics at t=\pi. Notice further that to calculate this derivative we do not actually need to know
d(\gamma_0(t),\gamma_\tau(t)) \,,
rather, all we need do is solve the equation
y''+y=0 \,,
for some given initial data.
Jacobi fields give a natural generalization of this phenomenon to arbitrary Riemannian manifolds.
Solving the Jacobi equation
Let e_1(0)=\dot\gamma(0)/|\dot\gamma(0)| and complete this to get an orthonormal basis \big\{e_i(0)\big\} at T_{\gamma(0)}M. Parallel transport it to get a basis \{e_i(t)\} all along \gamma. This gives an orthonormal basis with e_1(t)=\dot\gamma(t)/|\dot\gamma(t)|. The Jacobi field can be written in co-ordinates in terms of this basis as J(t)=y^k(t)e_k(t) and thus
\frac{D}{dt}J=\sum_k\frac{dy^k}{dt}e_k(t),\quad\frac{D^2}{dt^2}J=\sum_k\frac{d^2y^k}{dt^2}e_k(t),
and the Jacobi equation can be rewritten as a system
\frac{d^2y^k}{dt^2}+|\dot\gamma|^2\sum_j y^j(t)\langle R(e_j(t),e_1(t))e_1(t),e_k(t)\rangle=0
for each k. This way we get a linear ordinary differential equation (ODE). Since this ODE has smooth coefficients we have that solutions exist for all t and are unique, given y^k(0) and {y^k}'(0), for all k.
Examples
Consider a geodesic \gamma(t) with parallel orthonormal frame e_i(t), e_1(t)=\dot\gamma(t)/|\dot\gamma|, constructed as above.
The vector fields along \gamma given by \dot \gamma(t) and t\dot \gamma(t) are Jacobi fields. In Euclidean space (as well as for spaces of constant zero sectional curvature) Jacobi fields are simply those fields linear in t. For Riemannian manifolds of constant negative sectional curvature -k^2, any Jacobi field is a linear combination of \dot\gamma(t), t\dot\gamma(t) and \exp(\pm kt)e_i(t), where i>1. For Riemannian manifolds of constant positive sectional curvature k^2, any Jacobi field is a linear combination of \dot\gamma(t), t\dot\gamma(t), \sin(kt)e_i(t) and \cos(kt)e_i(t), where i>1. The restriction of a Killing vector field to a geodesic is a Jacobi field in any Riemannian manifold. The Jacobi fields correspond to the geodesics on the tangent bundle (with respect to the metric on TM induced by the metric on M). See also References
[do Carmo] M. P. do Carmo, Riemannian Geometry, Universitext, 1992.
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Performing Topology Optimization with the Density Method
Engineers are given significant freedom in their pursuit of lightweight structural components in airplanes and space applications, so it makes sense to use methods that can exploit this freedom, making topology optimization a popular choice in the early design phase. This method often requires regularization and special interpolation functions to get meaningful designs, which can be a nuisance to both new and experienced simulation users. To simplify the solution of topology optimization problems, the COMSOL® software contains a density topology feature.
About the Density Method for Topology Optimization
As the name suggests, topology optimization is a method that has the ability to come up with new and better topologies for an engineering structure given an objective function and set of constraints. The method comes up with these new topologies by introducing a set of design variables that describe the presence, or absence, of material within the design space. These variables are defined either within every element of the mesh or on every node point of the mesh. Changing these design variables thus becomes analogous to changing the topology. This means that holes in the structure can appear, disappear, and merge as well as that boundaries can take on arbitrary shapes. In addition, the control parameters are somewhat automatically defined and tied to the discretization.
As of COMSOL Multiphysics® software version 5.4, the add-on Optimization Module includes a density topology feature to improve the usability of topology optimization. The feature is designed to be used as a density method (Ref. 3), meaning that the control parameters change a material parameter through an interpolation function. Interpolation functions for solid and fluid mechanics are built into the feature and used in example models throughout the Application Library in COMSOL Multiphysics.
A bracket geometry is topology optimized, leaving only 50% of the material, which contributes the most to the stiffness. The printed bracket geometry.
The density method involves the definition of a control variable field, \theta_c, which is bounded between 0 and 1. In solid mechanics, \theta_c=1 corresponds to the material from which the structure is to be built, while \theta_c=0 corresponds to a very soft material. By default, the void Young’s modulus is 0.1% of the solid Young’s modulus. In fluid mechanics, convention dictates that \theta_c=1 corresponds to fluid, while \theta_c=0 is a (slightly) permeable material with an inverse permeability factor, \alpha; i.e., a damping term is added to the Navier-Stokes equation:
The damping term is 0 in fluid domains, while a large value is used in solid domains. These different values give a good approximation of the no-slip boundary condition on the interface between the domains.
An Introduction to the Density Model Feature
The
Density Model feature supports regularization via a Helmholtz equation (Ref. 1). This introduces a minimum length scale using the filter radius, R_\mathrm{min}:
Here, \theta_c is the raw control variable, which is modified by the optimizer, and \theta_f is the filtered variable. The mesh edge size is the default value for the filter radius. While this works well in terms of regularizing the optimization problem, it’s important to set a fixed length (larger than the mesh edge size) to get mesh-independent results.
Top: The equation for the Helmholtz filter can be solved analytically for a 1D Heaviside function. Bottom: This plot is taken from the MBB beam optimization model. It shows the raw control variables to the left and the filtered version to the right.
The Helmholtz filter gives rise to significant grayscale, which does not have a clear physical interpretation. The grayscale can be reduced by applying a smooth step function in what is referred to as
projection in topology optimization. Projection reduces grayscale, but it also makes it more difficult for the optimizer to converge. The density topology feature supports projection based on the hyperbolic tangent function, and the amount of projection can be controlled with the projection steepness, \beta.
Here, \theta_{\beta} is the projection point.
Plot showing the filtered field to the left and the projected field to the right.
Projection makes it possible to avoid grayscale, but grayscale can still appear if the optimization problem favors it. If the same interpolation function is used for the mass and the stiffness, grayscale is optimal in volume-constrained minimum compliance problems. It is thus common to use interpolation functions that cause intermediate values to be associated with little stiffness relative to their cost (compared to the fully solid value). You can think of this as a
penalization of intermediate values for the material volume factor, and the Density Model interface (shown below) supports two such interpolation schemes for solid mechanics: solid isotropic material with penalization (SIMP) and rational approximation of material properties method (RAMP) interpolation. Darcy interpolation is provided for fluid mechanics. The interpolated variable is called the penalized material volume factor, \theta_p, and is used for interpolating the material parameters, e.g., for SIMP interpolation, the p_\textsc{simp} exponent can be increased to reduce the stiffness of intermediate values, so that grayscale becomes less favorable.
\theta_p %26= \theta_\mathrm{min}(1-\theta_\mathrm{min})\theta^{p_\textsc{simp}}\\
E_p %26= E\theta_p
\end{align}
Here, E is the Young’s modulus of the solid material and E_p is the penalized Young’s modulus to be used throughout all optimized domains.
The Density Model feature is available under Topology Optimization in Component > Definitions . The mesh edge length is taken as the default filter radius and it works well, but it has to be replaced with a fixed value in order to produce mesh-independent results.
The penalized Young’s modulus can be defined as a domain variable, or (as in the case of the bracket model) it can be defined directly in the materials.
Topology optimization with the density method involves varying the Young’s modulus spatially. In this case, it is achieved by going to the material properties and multiplying the solid Young’s modulus with the penalized material volume factor, dtopo1.theta_p.
In summary, the density topology feature adds four variables. The filtered material volume factor is defined implicitly using a dependent variable.
Symbol Description Equation \theta_c Control material volume factor 0\leq\theta_c\leq1 \theta_f Filtered material volume factor \theta_f = R_\mathrm{min}^2\mathbf{\nabla}^2\theta_f + \theta_c \theta Material volume factor \theta = \frac{\tanh(\beta(\theta_f-\theta_{\beta}))+\tanh(\beta\theta_{\beta})}{\tanh(\beta(1-\theta_{\beta}))+\tanh(\beta\theta_{\beta})} \theta_p Penalized material volume factor \theta_p = \theta_\mathrm{min}+(1-\theta_\mathrm{min})\theta^{p_\textsc{simp}} or \theta_p = \frac{q_\mathrm{Darcy}(1-\theta)}{q_\mathrm{Darcy}+\theta}
When the filtering is disabled, the filtered variable becomes undefined and the projection instead uses the control material volume factor directly. If the projection is disabled, the material volume factor still exists, but it becomes identical to the projection input.
Applying Continuation to Avoid Local Minima
When the topology is not too complicated, the default values of the density topology feature work well. This is the case for the MBB beam optimization and topology optimized hook models. If the optimal design is more complicated (such as for the bracket example shown at the top of this post), there might be many local minima. To avoid these minima, you can use continuation in the SIMP exponent and the projection slope. This can be achieved by modifying the initial value expression in the density topology feature and adding a
Parametric Sweep feature, as shown below. As a result, the solver ramps over the specified parameters, using the optimum from the previous case as the initial value for the next optimization step. That is, it starts with a small SIMP exponent and projection slope and then continues to higher values. It is possible to apply continuation by combining a parametric sweep with a study reference. See the Bracket — Topology Optimization tutorial model for details. Objectives and Constraints in Topology Optimization
If the geometry is optimized for a single load case (as shown below to the left), the resulting design will be optimal with respect to that load case. This can seem obvious, but often designers make assumptions about symmetries and the design topology. Unless these assumptions are formalized as constraints, they will not be respected. Therefore, the design shown to the right below uses eight load cases (two load groups times four constraint groups).
Left: The bracket geometry is optimized for a single load case, resulting in an asymmetric design with two loosely connected halves. Right: The bracket geometry with eight load cases.
Designers often have several objectives that need to be weighted. To make an informed decision about these objectives, a designer can trace the Pareto optimal front using several optimizations with different weights.
The Pareto optimal front for the bracket geometry can be traced by varying the weight in a parametric sweep. Animation of the topology optimized bracket. (Download the glTF™ file from the Application Gallery in GLB-file format to rotate the geometry yourself.) Exporting and Importing Topology Optimization Results
It is possible to analyze the result of a topology optimized design with respect to stress concentration and buckling without remeshing. However, if you want to be completely sure that the void phase does not play a role, you can eliminate it by exporting and importing the resulting design, as shown below. The details of this procedure are discussed in a previous blog post.
The contour (left) for the topology optimized MBB beam design is exported and imported as an interpolation curve (right). Next Steps
To learn more about the built-in tools and features for solving optimization problems, check out the Optimization Module product page by clicking the button below.
Further Resources Try using the density feature for topology optimization with these example models: Read more about topology optimization on the COMSOL Blog: References B.S. Lazarov and O. Sigmund, “Filters in topology optimization based on Helmholtz‐type differential equations,” International Journal for Numerical Methods in Engineering, vol. 86, no. 6, pp. 765–781, 2011. F. Wang, B.S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Structural and Multidisciplinary Optimization, vol. 43, pp. 767–784, 2011. M.P. Bendsøe, “Optimal shape design as a material distribution problem,” Structural Optimization, vol. 1, pp. 193–202, 1989.
glTF and the glTF logo are trademarks of the Khronos Group Inc. Comments (0) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
Even though the title is quite a mouthful, this post is about two really cool ideas:
A solution to the "chicken-and-egg" problem (known as the Expectation-Maximization method, described by A. Dempster, N. Laird and D. Rubin in 1977), and An application of this solution to automatic image clustering by similarity, using Bernoulli Mixture Models.
For the curious, an implementation of the automatic image clustering is shown in the video below. The source code (C#, Windows x86/x64) is also available for download!
While automatic image clustering nicely illustrates the E-M algorithm, E-M has been successfully applied in a number of other areas: I have seen it being used for word alignment in automated machine translation, valuation of derivatives in financial models, and gene expression clustering/motif finding in bioinformatics.
As a side note, the notation used in this tutorial closely matches the one used in Christopher M. Bishop's "Pattern Recognition and Machine Learning". This should hopefully encourage you to check out his great book for a broader understanding of E-M, mixture models or machine learning in general.
Alright, let's dive in!
1. Expectation-Maximization Algorithm
Imagine the following situation. You observe some data set (e.g. a bunch of images). You hypothesize that these images are of different objects... but you don't know which images represent which objects.
Let be a set of
latent (hidden) variables, which tell precisely that: which images represent which objects.
Clearly, if you knew , you could group images into the clusters (where each cluster represents an object), and vice versa, if you knew the groupings you could deduce . A classical "chicken-and-egg" problem, and a perfect target for an Expectation-Maximization algorithm.
Here's a general idea of how E-M algorithm tackles it. First of all, all images are assigned to clusters arbitrarily. Then we use this assignment to modify the parameters of the clusters (e.g. we change what object is represented by that cluster) to
maximize the clusters' ability to explain the data; after which we re-assign all images to the expected most-likely clusters. Wash, rinse, repeat, until the assignment explains the data well-enough (i.e. images from the same clusters are similar enough). (Notice the words in bold in the previous paragraph: this is where the expectation and maximization stages in the E-M algorithm come from.)
To formalize (and generalize) this a bit further, say that you have a set of model parameters (in the example above, some sort of cluster descriptions).
To solve the problem of cluster assignments we effectively need to find model parameters that maximize the likelihood of the observed data , or, equivalently, the model parameters that maximize the log likelihod
Using some simple algebra we can show that for any latent variable distribution , the log likelihood of the data can be decomposed as
\begin{align} \ln \,\text{Pr}(\mathbf{X} | \theta) = \mathcal{L}(q, \theta) + \text{KL}(q || p), \label{eq:logLikelihoodDecomp} \end{align} where is the Kullback-Leibler divergence between and the posterior distribution , and \begin{align} \mathcal{L}(q, \theta) := \sum_{\mathbf{Z}} q(\mathbf{Z}) \left( \mathcal{L}(\theta) - \ln q(\mathbf{Z}) \right) \end{align} with being the "complete-data" log likelihood (i.e. log likelihood of both observed and latent data).
To understand what the E-M algorithm does in the expectation (E) step, observe that for any and hence is a lower bound on .
Then, in the E step, the gap between the and is minimized by minimizing the Kullback-Leibler divergence with respect to (while keeping the parameters fixed).
Since is minimized at when , at the E step is set to the conditional distribution .
To maximize the model parameters in the M step, the lower bound is maximized with respect to the parameters (while keeping fixed; notice that in this equation corresponds to the old set of parameters, hence to avoid confusion let ).
The function that is being maximized w.r.t. at the M step can be re-written as
\begin{align*} \theta^\text{new} &= \underset{\mathbf{\theta}}{\text{arg max }} \left. \mathcal{L}(q, \theta) \right|_{q(\mathbf{Z}) = \,\text{Pr}(\mathbf{Z} | \mathbf{X}, \theta^\text{old})} \\ &= \underset{\mathbf{\theta}}{\text{arg max }} \left. \sum_{\mathbf{Z}} q(\mathbf{Z}) \left( \mathcal{L}(\theta) - \ln q(\mathbf{Z}) \right) \right|_{q(\mathbf{Z}) = \,\text{Pr}(\mathbf{Z} | \mathbf{X}, \theta^\text{old})} \\ &= \underset{\mathbf{\theta}}{\text{arg max }} \sum_{\mathbf{Z}} \,\text{Pr}(\mathbf{Z} | \mathbf{X}, \theta^\text{old}) \left( \mathcal{L}(\theta) - \ln \,\text{Pr}(\mathbf{Z} | \mathbf{X}, \theta^\text{old}) \right) \\ &= \underset{\mathbf{\theta}}{\text{arg max }} \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \theta^\text{old}} \left[ \mathcal{L}(\theta) \right] - \sum_{\mathbf{Z}} \,\text{Pr}(\mathbf{Z} | \mathbf{X}, \theta^\text{old}) \ln \,\text{Pr}(\mathbf{Z} | \mathbf{X}, \theta^\text{old}) \\ &= \underset{\mathbf{\theta}}{\text{arg max }} \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \theta^\text{old}} \left[ \mathcal{L}(\theta) \right] - (C \in \mathbb{R}) \\ &= \underset{\mathbf{\theta}}{\text{arg max }} \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \theta^\text{old}} \left[ \mathcal{L}(\theta) \right], \end{align*}
i.e. in the M step the expectation of the joint log likelihood of the complete data is maximized with respect to the parameters .
So, just to summarize,
Expectationstep: Maximizationstep: (where superscript indicates the value of parameter at time ).
Phew. Let's go to the image clustering example, and see how all of this actually works.
2. Bernoulli Mixture Models for Image Clustering
First of all, let's represent the image clustering problem in a more formal way.
2.1. Formal description
Say that we are given same-sized training images for , each image containing binary pixels (i.e. ).
Assuming that the pixels are conditionally independent from each other (i.e. that is conditionally independent from for each ), the probability distribution of the pixel over all images belonging to a component can be modelled using Bernoulli distribution with a parameter .
To incorporate some prior knowledge about the image assignment to clusters (e.g. the proportions of images in each cluster), the assignments can be treated as being sampled from the multivariate distribution with the parameters (where , ). Each for is called a
mixing coefficient of cluster .
Let say that the model parameters include the pixel distributions of each cluster and the prior knowledge about the image assignments, i.e. , where and .
Then, the likelihood of a single training image is
\begin{align} \,\text{Pr}(\mathbf{x} | \theta) = \,\text{Pr}(\mathbf{x} | \mathbf{\mu}, \mathbf{\pi}) = \sum_{k = 1}^K \pi_k \,\text{Pr}(\mathbf{x}|\mathbf{\mu_k}) \end{align} where the probability that is generated by cluster can be written as \begin{align} \,\text{Pr}(\mathbf{x}|\mathbf{\mu_k}) = \prod_{i = 1}^D \mu_{k, i}^{x_i} (1 - \mu_{k, i})^{1 - x_i}. \end{align}
To model the assignment of images to clusters, associate a latent -dimensional binary random variable with each of the training examples . Say that has a 1-of- representation, i.e. for it must be the case that for and .
Furthermore, let the marginal distribution over be specified in terms of mixing coefficients s.t. , then
\begin{align} \,\text{Pr}(\mathbf{z_n} | \mathbf{\pi}) = \prod_{i = 1}^K \pi_i^{z_{n, i}}. \end{align}
Similarly, let , then
\begin{align} \,\text{Pr}(\mathbf{x_n} | \mathbf{z_n}, \mathbf{\mu}, \mathbf{\pi}) = \prod_{k = 1}^K \,\text{Pr}(\mathbf{x_n} | \mathbf{\mu_k})^{z_{n, k}}. \end{align}
By combining all latent variables into a set , we can write
\begin{equation} \label{eq:probZ} \begin{split} \,\text{Pr}(\mathbf{Z}|\mathbf{\pi}) &= \prod_{n = 1}^N \,\text{Pr}(\mathbf{z_n}|\mathbf{\pi}) \\ &= \prod_{n = 1}^N \prod_{k = 1}^K \pi_k^{z_{n, k}}, \end{split} \end{equation} and, similarly, combining all training images into a set , we can express the marginal training data distribution as \begin{equation} \label{eq:probXgivZ} \begin{split} \,\text{Pr}(\mathbf{X}|\mathbf{Z}, \mathbf{\mu}, \mathbf{\pi}) &= \prod_{n = 1}^N \,\text{Pr}(\mathbf{x_n}|\mathbf{z_n},\mathbf{\mu},\mathbf{\pi}) \\ &= \prod_{n = 1}^N \prod_{k = 1}^K \,\text{Pr}(\mathbf{x_n} | \mathbf{\mu_k})^{z_{n, k}} \\ &= \prod_{n = 1}^N \prod_{k = 1}^K \left( \prod_{i = 1}^D \mu_{k, i}^{x_{n, i}} (1 - \mu_{k, i})^{1 - x_{n, i}} \right)^{z_{n, k}}. \end{split} \end{equation}
From the last two equations and the probability chain rule, the complete data likelihood can be written as:
\begin{equation} \label{eq:probXandZ} \begin{split} \,\text{Pr}(\mathbf{X}, \mathbf{Z}| \mathbf{\mu}, \mathbf{\pi}) &= \,\text{Pr}(\mathbf{X} | \mathbf{Z}, \mathbf{\mu}, \mathbf{\pi}) \,\text{Pr}(\mathbf{Z}| \mathbf{\mu}, \mathbf{\pi}) \\ &= \prod_{n = 1}^N \prod_{k = 1}^K \left( \pi_k \prod_{i = 1}^D \mu_{k, i}^{x_{n, i}} (1 - \mu_{k, i})^{1 - x_{n, i}} \right)^{z_{n, k}}, \end{split} \end{equation}
and thus the complete data log likelihood can be obtained by taking a log of the equation above:
\begin{equation} \begin{split} \mathcal{L}(\theta) &= \ln \,\text{Pr}(\mathbf{X}, \mathbf{Z}| \mathbf{\mu}, \mathbf{\pi}) \\ &= \sum_{n = 1}^N \sum_{k = 1}^K z_{n, k} \left( \ln \pi_k + \sum_{i = 1}^D x_{n, i} \ln \mu_{k, i} + (1 - x_{n, i}) \ln (1 - \mu_{k, i}) \right). \end{split} \end{equation} (Still following? Great. Take five, and below we will derive the E and M step update equations.) 2.2. E-M update equations for BMMs
In order to update the latent variable distribution (i.e. image assignment to clusters) at the
expectation step, we need to set the probability distribution of to .
However, we cannot calculate this distribution exactly, hence we will have to approximate this assignment. A simple way of doing it is to replace the current values of with the expected ones:
\begin{equation} \label{eq:z}
\begin{split} z_{n, k}^\text{new} \leftarrow \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \mathbf{\mu}, \mathbf{\pi}}[z_{n, k}] &= \sum_{z_{n, k}} \,\text{Pr}(z_{n,k} | \mathbf{x_n}, \mathbf{\mu}, \mathbf{\pi}) \, z_{n,k}\\ &= \frac{\pi_k \,\text{Pr}(\mathbf{x_n} |\mathbf{\mu_k})}{\sum_{m = 1}^K \pi_m \,\text{Pr}(\mathbf{x_n} | \mathbf{\mu_m})} \\ &= \frac{\pi_k \prod_{i = 1}^D \mu_{k, i}^{x_{n, i}} (1 - \mu_{k, i})^{1 - x_{n, i}} }{\sum_{m = 1}^K \pi_m \prod_{i = 1}^D \mu_{m, i}^{x_{n, i}} (1 - \mu_{m, i})^{1 - x_{n, i}}}. \end{split} \end{equation}
(Notice that after this update is no longer represented as 1-of- vector, i.e. the same image can be "partially" assigned to multiple clusters.)
In the
maximization step we need to maximize the model parameters (i.e. the mixing coefficients and the pixel distributions) using the update equation from earlier
Observe that
\begin{align} \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \theta^\text{old}} \left[ \mathcal{L}(\theta) \right] &= \sum_{n = 1}^N \sum_{k = 1}^K \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \mathbf{\mu}^\text{old}, \mathbf{\pi}^\text{old}} \left[ z_{n, k} \right] \left( \ln \pi_k + \sum_{i = 1}^D x_{n, i} \ln \mu_{k, i} + (1 - x_{n, i}) \ln (1 - \mu_{k, i}) \right). \end{align} The equation above can be maximized w.r.t. by simply setting its derivative to zero: \begin{align} \frac{\partial}{\partial \mu_{m, j}} \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \theta^\text{old}} \left[ \mathcal{L}(\theta) \right] &= \sum_{n = 1}^N \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \mathbf{\mu}^\text{old}, \mathbf{\pi}^\text{old}} \left[ z_{n, m} \right] \left( \frac{x_{n, j}}{\mu_{m, j}} - \frac{1 - x_{n, j}}{1 - \mu_{m, j}} \right) \\ &= \sum_{n = 1}^N z_{n, m}^\text{new} \frac{x_{n, j} - \mu_{m, j}}{\mu_{m, j} (1 - \mu_{m, j})} = 0 \Leftrightarrow \\ \mu_{m, j} &= \frac{1}{N_m} \sum_{n = 1}^N x_{n, j} z_{n, m}^\text{new}, \end{align} where is the effective number of images assigned to cluster .
Then the full cluster pixel distribution vector can be written as
where is the weighted mean of the images associated with cluster .
To maximize w.r.t. the mixing coefficients (subject to the constraint ) we can use the Lagrange multipliers, yielding the following optimization problem:
\begin{equation*} \Lambda(\theta, \lambda) := \mathbb{E}_{\mathbf{Z} | \mathbf{X}, \theta^\text{old}} \left[ \mathcal{L}(\theta) \right] + \lambda \left( \sum_{k = 1}^K \pi_k - 1 \right). \end{equation*} The optimizing solution can then be found again with simple partial derivatives: \begin{align} \frac{\partial}{\partial \pi_{m}} \Lambda(\theta, \lambda) &= \frac{1}{\pi_m} \sum_{n = 1}^N z_{n,m}^\text{new} + \lambda = 0 \Leftrightarrow \\ \pi_m &= -\frac{N_m}{\lambda}, \end{align} \begin{align} \frac{\partial}{\partial \lambda} \Lambda(\theta, \lambda) &= \sum_{k = 1}^K \pi_k - 1 = 0 \Leftrightarrow \\ \sum_{k = 1}^K \pi_k &= 1. \end{align} By combining these two results , and thus
Done!
2.3. Summary
In summary, the update equations for Bernoulli Mixture Models using E-M are:
Expectationstep: Maximizationstep: where and . 3. References
[Dempster et al, 1977]
A. P. Dempster, N. M. Laird, D. B. Rubin. "Maximum Likelihood from Incomplete Data via the EM Algorithm". Journal of the Royal Statistical Society. Series B (Methodological) 39 (1): 1–38.
[Bishop, 2006]
C. M. Bishop. "Pattern Recognition and Machine Learning". Springer, 2006. ISBN 9780387310732. |
The equation of motion for a fluid parcel in the atmosphere (in Cartesian space) is
$$\dfrac{D\mathbf u}{Dt} = -\dfrac{1}{\rho}\nabla p-2 \mathbf \Omega \times \mathbf u + \mathbf g + \mathbf F,$$
where $\mathbf u$ is the wind, $\rho$ is density, $p$ is pressure, $\mathbf\Omega$ is the angular velocity of the Earth, $\mathbf g$ is gravity and $\mathbf F$ is friction. The derivative is a material derivative (Lagrangian perspective) where
$$\dfrac{D\varphi}{Dt} = \dfrac{\partial\varphi}{\partial t} + \mathbf u\cdot\nabla\varphi.$$
There is a non-dimensional number called the Rossby number ($Ro$) that determines when a flow behaves geostrophically. This number is given by
$$Ro = {U\over fL},$$
where $U$ is a velocity scale, $L$ is a length scale and $f$ is the Coriolis parameter ($f=2\Omega\sin\phi$, $\Omega = 7.2921 \times 10^{-5}\ \text{s}^{-1}$, and $\phi$ is latitude). When $Ro << 1$, the flow exhibits geostrophic balance. This occurs when $L$ becomes large, which it does as the storm grows. When I say the flow is geostrophic, what I really mean is that the net acceleration of a parcel is small. In a tropical cyclone the isobars are roughly circular and this curvature gives rise to the gradient wind balance.
The gradient wind is the balance of the pressure gradient force, Coriolis force and centripetal acceleration. In this flow, just as in geostrophic flow, the wind will follow the isobars, flowing cyclonically around the center of low pressure, though slower than a geostrphic flow with the same pressure gradient. Close to the ocean surface friction plays a role, and for the near surface winds the friction will cause the wind to be slightly deflected toward low pressure, or inward across the isobars.
Close to the center of the storm, in and near the eyewall, the length scale $L$ is reduced and the Coriolis force places a smaller role. Here the flow attains cyclostrophic balance -- a balance between centrifugal and pressure gradient forces.
The gradient and cyclostrophic balances explain the wind rotating around the storm and surface friction will give a radially inward component at low levels, advecting angular momentum toward the center of the storm. Convection in the eyewall will lift air to the tropopause where it will flow anti-cyclonically (due to the thermal wind) away from the storm before subsiding. The core of the storm is warm and the air there is subsiding, creating the cloud free eye. The flows around the storm are balanced flows and the flow inward/upward/outward/downward is a thermodynamic carnot engine.
This gives us a fairly stable setup and this is why once a tropical storm forms it tends to persist rather than tear itself apart. The key to sustaining this balance is a warm ocean surface and weak vertical windshear. If you take away the warm ocean the storm will start to spin down and if you have strong shear you will disconnect the lower and upper level circulations and the storm will weaken. |
Compactness Review
Compactness Review
We will now review some of the recent material regarding compactness in topological spaces.
Recall from the Covers of Sets in a Topological Spacepage that if $X$ is a topological space then an a Coverof $X$ is a collection of sets $\mathcal F$ such that:
\begin{align} \quad X \subseteq \bigcup_{U \in \mathcal F} U \end{align}
If the sets in $\mathcal F$ are all open (or closed) and satisfy the inclusion above then $\mathcal F$ is called an Open Cover(or Closed Cover) of $X$. On the Compactness of Sets in a Topological Spacepage we said that a set $A \subseteq X$ is said to be Compactin $X$ if every open cover of $A$ has a finite subcover. We saw that the set of natural numbers $\mathbb{N}$ is NOT compact in $\mathbb{R}$ with the usual topology. On the Compactness of Finite Sets in a Topological Spacewe saw a very nice theorem regarding the compactness of finite sets. We saw that if $A = \{ x_1, x_2, ..., x_n \}$ is a finite subset of a topological space $A$ is compact in $X$. This is because if we are given an open cover $\mathcal F$ of $A$ then all we need to do is select a set in the open cover associated to each point $x_i \in A$, and so, we will always be able to construct a finite subcover of size less than or equal to $n$. As a result, every subset of a finite topological space is compact. On the Finite Intersection Property Criterion for Compactness in a Topological Spacepage we looked at a nice result with regards to compactness of a topological space. First, if $\mathcal F$ is any collection of sets then we said that $\mathcal F$ has the Finite Intersection Propertyif every finite collection of sets $\{ F_1, F_2, ..., F_n \} \subseteq \mathcal F$ we have that:
\begin{align} \quad \bigcap_{i=1}^{n} F_i \neq \emptyset \end{align}
In other words, the intersection of all sets in a finite subset of $\mathcal F$ is nonempty. We then proved that a topological space $X$ is compact if and only if for every collection of closed sets $\mathcal F$ of $X$ that have the finite intersection property then the intersection of all sets in $\mathcal F$ is nonempty, i.e. $\displaystyle{\bigcap_{F \in \mathcal F} F \neq \emptyset}$. On the Preservation of Compactness under Continuous Mapswe saw that if $f : X \to Y$ is a continuous map and $X$ is a compact topological space then the range $f(X)$ is also a compact space. On the Closed Sets in Compact Topological Spaceswe saw that if $X$ is a compact topological space, $A \subseteq X$, and $A$ is closed in $X$ then $A$ is also compact in $X$. On the Compact Sets in Hausdorff Topological Spacespage we proved that if $X$ is a Hausdorff topological space (i.e., for every pair of distinct points $x, y \in X$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$) then every compact set $A \subseteq X$ is closed. We then began to see how useful the concept of a compact topological space was. On the Homeomorphisms Between Compact and Hausdorff Spacespage we saw that if $f : X \to Y$, $X$ is compact, $Y$ is Hausdorff, and $f$ is a continuous bijective function, then $f$ is necessarily a homeomorphism between $X$ and $Y$. Furthermore, on the Quotients from Equivalence Relations defined by Functions from Compact to Hausdorff Spaceswe first proved that if $f : X \to Y$, $X$ is compact, $Y$ is Hausdorff, and $f$ is a continuous map then $f$ is also a closed map (i.e., the image of any closed set in $X$ is closed in $Y$). We then proved a nice result. We proved that if $f : X \to Y$, $X$ is compact, $Y$ is Hausdorff, and $f : X \to Y$ is continuous and surjective then for the equivalence relation $\sim_f$ defined for all $x_1, x_2 \in X$ by $x_1 \sim_f x_2$ if and only if $f(x_1) = f(x_2)$, we have that the quotient space $X /\: \sim_f$ is homeomorphic to $Y$. |
Let $G$ be a compact connected $n$-dimensional Lie group, and let $1 \le k < n$. Do there exist a non-zero
bi-invariantdifferential $k$-form on $G$?
I know that the answer is positive for $k=n$ (top forms)-but the proof I know does not adapt to the case $k<n$. (That proof shows that every left-invariant $n$-form is in fact bi-invariant).
Edit:
I guess that we can always take a bi-invariant top-form $\text{Vol} \in \Omega^n(G)$ and given some
right-invariant $k$-form $\omega$ on $G$, define
$$ \tilde \omega=\int_G (L_g)^*\omega \cdot \text{Vol},$$
or more explicitly
$$ \tilde \omega_x=\int_G ((L_g)^*\omega)_x \cdot \text{Vol}\text{ for every } x\in G,$$
where the last integral is an integral of a $\bigwedge^k (T_xG)^*$- valued function, hence makes sense. (as we think of the point $x$ as a constant).
Then if I am not mistaken, $\tilde \omega$ should be bi-invariant; this is similar to how one constructs a bi-invariant Riemannian metric on a compact group, but I guess that in the case of a form, we cannot guarantee that the averaging process won't give zero as a result. (This is in contrast to the metric case, where we have positivity, which is preserved by averaging).
It would be nice to see this in a concrete example; the argument given by Nate Eldredge shows that any bi-invariant $1$-form on a simple Lie group is zero. So the averaging process described above should give zero on any such group, e.g. $\text{SO}(5)$.
I wonder if there is a way to "compute directly" the average, thus verifying that it is zero. |
Theorems Regarding Linear Independence and Dependence
Table of Contents
Theorems Regarding Linear Independence and Dependence
Recall from the Linear Independence and Dependence page the definition of linear independence/dependence:
Definition: A set of vectors $V = \{ \mathbf{x_1}, \mathbf{x_2}, ..., \mathbf{x_n} \}$ and the scalars $k_1, k_2, ..., k_n$ has a solution for the vector equation $k_1\mathbf{x_1} + k_2\mathbf{x_2} + ... + k_n\mathbf{x_n} = 0$, namely when $k_1 = k_2 = ... = k_n = 0$. If this is the only solution to the vector equation, then the set $V$ is said to be Linearly Independent. If there exists other solutions where not all $k_i = 0$, then $V$ is said to be Linearly Dependent.
We will now look at some important theorems regarding linear independence/dependence.
Theorem 1: The vector set containing only the zero vector from the vector space $V$ is linearly dependent. Proof:Consider the set of vector set only containing the zero vector, $\{ 0 \}$. We note that the vector equation $a_1\vec{0} = 0$ is satisfied for all $a_1 \in \mathbb{F}$, and so $\{ 0 \}$ is a linear dependent set. $\blacksquare$
Corollary 1: Any set of vectors $\{ v_1, v_2, ..., v_n \}$ containing the zero vector is linearly dependent. Proof:Without loss of generality suppose that $v_1$ is the zero vector. Then the vector equation $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$ has infinitely many sets of scalars $a_1, a_2, ..., a_n$ that satisfy this equation, for example, $100v_1 + 0v_2 + ... + 0v_n = 0$ since $v_1$ is the zero vector, and so $\{ v_1, v_2, ..., v_n \}$ is a linearly dependent set. $\blacksquare$
Theorem 2: If $v_1 \neq 0$ is a vector from the vector space $V$, then the set containing just the vector $v_1$ is linearly independent. Proof:Suppose that $v_1 \neq 0$ (otherwise we know this set would be linearly dependent from theorem 1). Then if we look at the vector equation, $kv_1 = 0$ if and only if $k = 0$ and so $\{ v_1 \}$ is a linearly independent set. $\blacksquare$
Theorem 3: The set of vectors $\{ v_1, v_2 \}$ from the vector space $V$ is linearly independent if $v_1$ is not a scalar multiple of $v_2$. Proof:We will show this by proof by contradiction. Suppose that $\{ v_1, v_2 \}$ is a linearly independent set of vectors and $v_1$ is a scalar multiple of $v_2$, that is $kv_1 = v_2$ for some $k \in \mathbb{F}$. Then it follows from the vector equation that:
\begin{equation} a_1v_1 + a_2v_2 = 0 a_1v_1 + ka_2v_1 = 0 \end{equation}
Therefore if $a_1 = 1$ and $a_2 = -\frac{1}{k}$, then $a_1v_1 + a_2v_2 = 0$, so then $\{ v_1, v_2 \}$ would not be a linearly independent set of vectors which is a contradiction. So then $\{v_1, v_2 \}$ is linearly independent if $v_1$ is not a scalar multiple of $v_2$. $\blacksquare$
Note: It is important to recognize that theorem 3 cannot necessarily be expanded to a set of three or more vectors. That is, a set of three or more vectors is not necessarily guaranteed to be linearly independent if none of the vectors are scalar multiples of one another. For example, consider the set of vectors $(1,0,0)$, $(0,0,1)$ and $(1,0,1)$, and notice that $(0,0,0) = 1(1,0,0) + 1(0,0,1) - 1(1, 0, 1) = 0$, and so $\{ (1, 0, 0), (0, 0, 1), (1, 0, 1) \}$ is a linearly dependent set, however, none of the vectors in this set are scalar multiples of one another.
Theorem 4: Let $\{ v_1, v_2, ..., v_n \}$ be a set of vectors from the vector space $V$. If the vector $v_i$ is a linear combination of the set of vectors $\{ v_1, v_2, ..., v_{i-1}, v_{i+1}, ..., v_n \}$ then $\{ v_1, v_2, ..., v_n \}$ is a linearly dependent set. Proof:Let $\{ v_1, v_2, ..., v_n \}$ be a set of vectors from $V$. Without loss of generality, assume that $v_i = v_1$ is a vector that is a linear combination of $\{ v_2, v_3, ..., v_m \}$. Then for some set of scalars $v_1 = a_2v_2 + a_3v_3 + ... + a_nv_n$, and rewriting this equation we get:
\begin{equation} 1v_1 -a_2v_2 - a_3v_3 - ... - a_nv_n = 0 \end{equation}
Since $a_1 = 1$, we have that this set of vectors is linearly dependent. $\blacksquare$ |
For a discrete random variable \(X\) (a random variable whose range \(\mathcal{X}\) is countable) with probability mass function \(p(x) = P(X = x)\), we can define the (Shannon or discrete) entropy of the random variable as \[ H[X] = -E[\log_{2} p(X)] = - \sum_{x \in \mathcal{X}} p(x) \log_{2} p(x).\] That is, the entropy of the random variable is the expected value of one over the probability mass function.
For a continuous random variable \(X\) (a random variable who admits a density over its range \(\mathcal{X}\)) with probability density function \(f(x)\), we can define the differential entropy of the random variable as \[ h[X] = -E[\log_{2} f(X)] = - \int_{\mathcal{X}} f(x) \log_{2} f(x) \, dx.\]
In both cases, we should make the disclaimer, "If the sum / integral exists." For what follows, assume that it does.
The Shannon / discrete entropy can be interpreted in many ways that capture what we usually mean when we talk about 'uncertainty.' For example, the Shannon entropy of a random variable is directly related to the minimum expected number of binary questions necessary to identify the random variable. The larger the entropy, the more questions we'll have to ask, the more uncertain we are about the identity of the random variable.
A lot of this intuition does not transfer to differential entropy. As my information theory professor said, "Differential entropy is not nice." We can't bring our intuition from discrete entropy straight to differential entropy without being careful. As a simple example, differential entropy can be
negative.
An obvious question to ask (and another place where our intuition will fail us) is this: for a continuous random variable \(X\), could we discretize its range using bins of width \(\Delta\), look at the discrete entropy of this new discretized random variable \(X^{\Delta}\), and recover the differential entropy as we take the limit of smaller and smaller bins? It seems like this should. Let's see why it fails, and how to fix it.
First, once we've chosen a bin size \(\Delta\), we'll partition the real line \(\mathbb{R}\) using intervals of the form \[ I_{i} = (i \Delta, (i + 1) \Delta ], i \in \mathbb{Z}.\] This gives us our discretization of \(X\): map all values in the interval \(I_{i}\) into a single candidate point in that interval. This discretization is a transformation of \(X\), which we can represent by \[X^{\Delta} = g(X)\] where \(g(x)\) is \[ g(x) = \sum_{i \in \mathbb{Z}} 1_{I_{i}}(x) x_{i}.\] We'll choose the points \(x_{i}\) in a very special way, because that choice will make our result easier to see. Choose the candidate point in the interval \(x_{i}\) such that \(f(x_{i}) \Delta\) captures all of the probability mass in the interval \(I_{i}\). That is, choose the \(x_{i}\) that satisfies \[ f(x_{i}) \Delta = \int_{I_{i}} f(x) \, dx.\] We're guaranteed that such an \(x_{i}\) exists by the Mean Value Theorem
1. So in each interval \(I_{i}\), we assign all of the mass in that interval, namely \(p_{i} = f(x_{i}) \Delta\) to the point \(x_{i}\). By construction, we have that \[ \sum_{i \in \mathbb{Z}} p_{i} = \sum_{i \in \mathbb{Z}} f(x_{i}) \Delta = \sum_{i \in \mathbb{Z}} \int_{I_{i}} f(x) \, dx = \int_{\mathbb{R}} f(x) \, dx = 1,\] so the probability mass function for \(X^{\Delta}\) sums to 1 as it must.
We can now try to compute the discrete entropy of this new random variable. We do this in the usual way, \[ \begin{align*} H[X^{\Delta}] &= - \sum_{i \in \mathbb{Z}} p_{i} \log_{2} p_{i} \\ &= - \sum_{i \in \mathbb{Z}} (f(x_{i}) \Delta) \log_{2} (f(x_{i}) \Delta) \\ &= - \sum_{i \in \mathbb{Z}} \{ f(x_{i}) \Delta \log_{2} f(x_{i}) + f(x_{i}) \Delta \log_{2} \Delta \} \\ &= - \sum_{i \in \mathbb{Z}} f(x_{i}) \Delta \log_{2} f(x_{i}) - \log_{2} \Delta \left(\sum_{i \in \mathbb{Z}} f(x_{i}) \Delta \right) \\ &= - \sum_{i \in \mathbb{Z}} f(x_{i}) \Delta \log_{2} f(x_{i}) - \log_{2} \Delta \cdot 1 \\ \end{align*}\] As long as \(f(x) \log_{2} f(x)\) is Riemann integrable, we know that the first term on the right hand side converges to the differential entropy, that is, \[ - \sum_{i \in \mathbb{Z}} f(x_{i}) \Delta \log_{2} f(x_{i}) \xrightarrow{\Delta \to 0} - \int_{\mathcal{X}} f(x) \log_{2} f(x) \, dx = h[X].\] Thus, rearranging terms, we see that \[H[X^{\Delta}] + \log_{2} \Delta \xrightarrow{\Delta \to 0} h[X].\] Thus, the discrete entropy of the discretization of \(X\) doesn't converge to the differential entropy. In fact, it blows up as \(\Delta\) gets smaller. Fortunately, \(\log_{2} \Delta\) blows up in the opposite direction, so the discrete entropy plus the additional term \(\log_{2} \Delta\) is what converges to the differential entropy. This may not be what we want (ideally, it would be nice if we recovered the differential entropy by taking the discretization sufficiently fine), but it is what we get.
This is all well and good, but what does this discretization look like in practice? The text that I'm getting this from, Cover and Thomas's
Elements of Information Theory, gives the example of \(X\) distributed according to various types of uniform and \(X\) distributed according to a Gaussian. The former is pretty boring (and has the disadvantage that the \(x_{i}\) we choose are not unique), and the latter would make for some annoying computation. Let's compromise and choose a density more interesting than a constant, but less involved than a Gaussian. We'll take \(X\) to have the density given by \[f(x) = \left\{\begin{array}{cc}6 x (1-x) &: 0 \leq x \leq 1 \\0 &: \text{ otherwise}\end{array}\right.\] which is a friendly parabola taking support only on \([0, 1]\) and normalized to integrate to 1.
We'll take our discretization / bin width to be \(\Delta = 2^{-n}\), so we'll first divide up the interval into halves, then fourths, then eighths, etc. Since we know \(f(x)\), for each of these intervals we can apply the Mean Value Theorem to find the appropriate representative \(x_{i}\) in each \(I_{i}\). For example, here we take \(n = 2\) and \(\Delta = 1/2\):
The red lines indicate the demarcations of the partitions, and the blue 'lollipops' indicate the mass assigned each \(x_{i}\). As expected, we choose two candidate points that put half the mass to the left of \(x = 1/2\) and half the mass to the right. All points to the left of \(x = 1/2\) get mapped to \(x_{0} = \frac{1}{2} - \frac{\sqrt{3}}{2} \approx 0.211\), all points to the right of \(x = 1/2\) get mapped to \(x_{1} = \frac{1}{2} + \frac{\sqrt{3}}{2} \approx 0.789\).
We can continue in this way, taking \(n\) larger and larger, and we get the following sequence of discretizations:
As we take our discretization to be finer and finer, the discrete entropy of \(X^{\Delta(n)}\) increases without bound, \[\begin{align} H\left[X^{\Delta(1)}\right] &= 1 \\ H\left[X^{\Delta(2)}\right] &= 1.896 \\ H\left[X^{\Delta(3)}\right] &= 2.846 \\ H\left[X^{\Delta(4)}\right] &= 3.828 \\ \vdots \\ H\left[X^{\Delta(4)}\right] &= 9.189, \end{align}\] We need to control this growth by tacking on the \(\log_{2} \Delta(n)\) term, in which case we have \[\begin{align} H\left[X^{\Delta(1)}\right] + \log_{2}(\Delta(1)) &= -1.110 \times 10^{-16} \\ H\left[X^{\Delta(2)}\right] + \log_{2}(\Delta(2)) &= -0.1039 \\ H\left[X^{\Delta(3)}\right] + \log_{2}(\Delta(3))&= -0.154 \\ H\left[X^{\Delta(4)}\right] + \log_{2}(\Delta(4))&= -0.172 \\ \vdots \\ H\left[X^{\Delta(10)}\right] + \log_{2}(\Delta(10))&= -0.1804, \end{align}\] which of course gets closer and closer to the differential entropy \(h[X] \approx -0.180470.\)
As usual, dealing with the continuum makes our lives a bit more complicated. We gain new machinery, but at the cost of our intuition.
In particular, because we have assumed that \(X\) admits a density, we know that \(f\) must be continuous, because otherwise it would
notadmit a density. For example, if it did have discontinuities, then probability mass would leave at those discrete points, and the density would not exist.↩ |
If $X$ is a discrete random variable, its entropy $H(X)$ is usually defined as something along the lines of $-\sum \def\P{\mathbb{P}}\P(x) \log_2( \P(x))$, where the sum ranges over all the possible values $x$ of $X$.
I have seen a few expositions of the problem of extending this definition to continuous random variables. I take it that the standard counterpart to $H$ in the continuous case is the so-called
differential entropy, denoted by $h(X)$, and defined as
$$ \int f(x)\log_2(f(x)) dx\,, $$
where $f$ is the probability density of $X$, and the integral runs over the (continuous) set of values of $X$.
This definition is always contingent on the existence of the integral, but even putting aside this existence question, I'm a bit confused about the proper way to interpret this integral.
None of the derivations (of the passage to the continuous case) has been terribly concerned over mathematical subtleties, such as well-definedness, convergence, etc. A possible clue to this attitude is that all these derivations are based on the Riemann integral formalism, which strikes me as particularly ill-suited for thinking about this particular integral's convergence.
I'm looking for a (hopefully mathematically rigorous) measure-theoretic approach to this generalization of the entropy to the continuous case. I'd appreciate any pointers. |
Chapter Review Exercises Exercise \({1}\)
Let \(f(x+y)=f(x)+f(y)+x^2y+xy^2\) and \(\lim_{x \rightarrow 0} \dfrac{f(x)}{x}=1\). Then find \( f^'(x) \).
Answer
\(1+x^2\)
Exercise \({2}\)
Use the limit definition of the derivative to find \(f'(x)\), where \(f(x)=3x^2\).
(You may use differentiation rules to check your answer). Find the tangent line to \(f(x)=3x^2\) at \(x=1.\) Answer
Under Construction
Exercise \({3}\)
Using the definition of the derivative (Newton's quotient)
find \(f'(x)\) for $$ \displaystyle{f(x)=\frac{1}{\sqrt{2x+5}}} $$ Answer
Under Construction
Exercise \({4}\)
Use the limit denition of the derivative to find \(f'(x),\) where \(f(x) = \frac{1}{x+1}\)
(You may using differentiation rules to check your answer). Then find the equation of
the tangent line to \(f(x)\) at \(x = 2\)
Answer
Under Construction
Exercise \({5}\)
Find the derivatives of the following functions:
1) \(\displaystyle{f(x)=\sqrt{(x^2+x+1)}+{x^2}}\)
2) \(\displaystyle{g(x)=\frac{\cos x}{x+ \sin x}}\)
3) \(\displaystyle{h(t)=t^2+ \csc t+ +\sec^2 t}\)
4) \(\displaystyle{y(x)=\frac{3x^2+5}{2x^2+x-3}}\)
5) \(f(x) =\displaystyle \frac{(x^2+1) \cot x}{3-\cos x \csc x}\)
6) \(\displaystyle{h(t)=\sqrt{\tan(2t-1)}}\)
7) \(\displaystyle{y(x)=\frac{x}{x^2-1}}\)
8) \(\displaystyle{f(x)=\frac{ln(x^2+1)}{x^2}}\)
9) \(f(x) = \left( \displaystyle \frac{x-5}{2x+1} \right)^3\)
10) \(\displaystyle{h(t)=\sin^2(\sqrt{\tan(2t-1)})}\)
11) \(\displaystyle{h(t)=\frac{\sin(t)}{t}}\)
12) \(y= \csc\left(\cot(\sqrt{4x})\right)\)
13) \(\displaystyle{y(x)=(x+5)^7(x^2-1)}\)
14) \(g(x)=\displaystyle \frac{\tan (2x) + \sqrt{x}}{\cos (x)+x^2}\)
15) \(g(x)=\displaystyle \frac{\cos (x) - \sqrt{x}}{\tan (x)+3x^2}\)
16) \( y=\displaystyle{\frac{3x+4}{x^2+5}}$ \)
17) \( y=\displaystyle{\sec(x)\tan(x)}\)
18) \(\displaystyle{y = \sqrt{5+\sqrt{6x}}}\).
19) \(y= x^3 cos(x).\)
20) \(f(x)-\frac{tan(2x)}{2x +1}\)
Answer
Under Construction
Exercise \({6}\)
Find equations of the tangent and normal to $$y= \frac{2}{3-4 \sqrt{x}}$$ at the point \((1,-2)\).
Answer
Under Construction
Exercise \({7}\)
An object is released from rest (its initial velocity is zero) from the Empire State Building at a height of \(1250\) ft above street level. The height of the object can
be modeled by the position functions \(s = f(t) = 1250 - 16t^2\).
a. Verify the object is still falling at \(t = 5s\):
b. Find the object's instantaneous velocity at the time \( t = 5s\)
Answer
Under Construction
Exercise \({8}\)
True or False? Justify the answer with a proof or a counterexample.
1) Every function has a derivative.
2) A continuous function has a continuous derivative.
3) A continuous function has a derivative.
4) If a function is differentiable, it is continuous.
Answers to odd numbered questions
1. False.
3. False.
Exercise \({9}\)
Use the limit definition of the derivative to exactly evaluate the derivative:
1) \(f(x)=\sqrt{x+4}\)
2) \(f(x)=\frac{3}{x}\)
Answers to odd numbered questions
1. \(\frac{1}{2\sqrt{x+4}}\)
Exercise \({10}\)
Find the derivatives of the following functions:
1) \(f(x)=3x^3−\frac{4}{x^2}\)
2) \(f(x)=(4−x^2)^3\)
3) \(f(x)=e^{sinx}\)
4) \(f(x)=ln(x+2)\)
5) \(f(x)=x^2cosx+xtan(x)\)
6) \(f(x)=\sqrt{3x^2+2}\)
7) \(f(x)=\frac{x}{4}sin^{−1}(x)\)
8) \(x^2y=(y+2)+xysin(x)\)
Answers to odd numbered questions
1. \(9x^2+\frac{8}{x^3}\)
3. \(e^{sinx}cosx\)
5. \(xsec^2(x)+2xcos(x)+tan(x)−x^2sin(x)\)
7. \(\frac{1}{4}(\frac{x}{\sqrt{1−x^2}}+sin^{−1}(x))\)
Exercise \({11}\)
Find the following derivatives of various orders:
1) First derivative of \(y=xln(x)cosx\)
2) Third derivative of \(y=(3x+2)^2\)
3) Second derivative of \(y=4^x+x^2sin(x)\)
Answers to odd numbered questions
1. \(cosx⋅(lnx+1)−xln(x)sinx\)
3. \(4^x(ln4)^2+2sinx+4xcosx−x^2sinx\)
Exercise \({12}\)
Find the equation of the tangent line to the following equations at the specified point:
1) \(y=cos^{−1}(x)+x\) at \(x=0\)
2) \(y=x+e^x−\frac{1}{x}\) at \(x=1\)
Answer to even numbered questions
2. \(T=(2+e)x−2\)
Exercise \({13}\)
Draw the derivative for the following graphs.
21)
Answer Exercise \({14}\)
The following questions concern the water level in Ocean City, New Jersey, in January, which can be approximated by \(w(t)=1.9+2.9cos(\frac{π}{6}t),\) where t is measured in hours after midnight, and the height is measured in feet.
1) Find and graph the derivative. What is the physical meaning?
2) Find \(w′(3).\) What is the physical meaning of this value?
Answers to even numbered questions
2. \(w′(3)=−\frac{2.9π}{6}\). At 3 a.m. the tide is decreasing at a rate of 1.514 ft/hr.
Exercise \({15}\)
The following questions consider the wind speeds of Hurricane Katrina, which affected New Orleans, Louisiana, in August 2005. The data are displayed in a table.
Hours after Midnight, August 26 Wind Speed (mph) 1 45 5 75 11 100 29 115 49 145 58 175 73 155 81 125 85 95 107 35
Wind Speeds of Hurricane KatrinaSource: http://news.nationalgeographic.com/n..._timeline.html.
1) Using the table, estimate the derivative of the wind speed at hour 39. What is the physical meaning?
2) Estimate the derivative of the wind speed at hour 83. What is the physical meaning?
Answers to even numbered questions
2. \(−7.5.\) The wind speed is decreasing at a rate of 7.5 mph/hr |
The solution of the model contains constant: $k = \alpha K$, it relates to: (i) probability of getting a fill ($\alpha$) and (ii) market impact ($K$).
Estimating (i). The author proposes that the market order sizes follow a power law distribution:
$$f(x)^Q \propto x^{-1-\alpha}$$
So no problem estimating this.
Estimating (ii). There are two equations of interest here:
$$\Delta p \propto \ln Q \tag{11}$$
$$\begin{align} \lambda (\delta) &= \Lambda \mathbb{P} [\Delta p > \delta] \\ &= \Lambda \mathbb{P}[\ln Q > K \delta] \\ &= \Lambda \mathbb{P} [Q > \exp \left( K \delta \right)] \\ &= \Lambda \int_{\exp (K \delta)}^{\infty} x^{-1-\alpha} dx \\ &= A \exp \left( -k \delta \right) \end{align} \tag{12}$$
According to the above two:
Line 1 in (12) says $\Delta p > \delta$, but we know that $\Delta p = c \ln Q$, therefore we can re-write:
$$\begin{align} \mathbb{P} [\Delta p > \delta] &= \mathbb{P} [c \ln Q > \delta] \\ &= \mathbb{P} [\ln Q > \frac{1}{c} \delta] \end{align}$$
Therefore, $K = \frac{1}{c}$, the inverse of the proportionality constant that we find in relation $\Delta p \propto \ln Q$.
Is this correct? (The estimation of $K$).
Note: I am aware of Sophie Laruelle's paper. But, unfortunately, I do not know French.
Super Note:My understanding of this answer relating to Sophie's paper: Definitions:
$t_0$ - start time (this is the start time point in the interarrival time slots in a Poisson Process). Simply put, if you get hit, that is a new time: $t_1$, which becomes your new start time.
$\delta P$ - distance of your order from the mid price. If the mid price is \$100, and your bid is at \$85, then this quantity is \$15. Same for asks, obviously.
$P^m (t) $ - is your mid price at time $t$.
Procedure: Record start time: $t_0$ and corresponding mid price: $P^m(t_0)$ at this time. Record the time $t_1$ when the bid/ask is hit and the $\delta P$s. The reason I have added plural "s" ending in there is because you might get hit at a number of levels. An example is asked for here. This is your order book (OB):
|--- qty ---| --- size --- |asks $105 10 $104 5bids $100 6 $99 5
Imagine a market order comes in that eats 8 units on the bid size (so it was a market sell). You record the change in price, $\delta P = \$102 - \$100 = \$2$ at first level and the corresponding time of this trade, $t_1$. You also record $\delta P = \$102 - \$99 = \$3$ at the same time, $t_1$ (in this case). If there was a market order that only took portion of the best ask/bid, or full best ask/bid and did not go deeper, we would only collect a single $\delta P$ with its corresponding $t$. Note that you can define the point at which the order is hit differently, this is up to you. - You now have time lengths and corresponding sizes. Interarrival times in a Poisson Process are exponential:
$$\mathbb{P}[X_1 > t] = \mathbb{P}\left[ \texttt{no arrival in time (0, t]} \right]= e^{-\lambda t} $$
So now for each bucket of price changes like: $[\$1, \$1.5]$ i.e. all of the $\delta P$ that are between 1 and 1.5 bucks, you have a list of interarrival times: $[0.3, 0.2, 0.5, 0.7, 1.1]$ in seconds. Fit the above exponential distribution to each data bucket to obtain some empirical estimate of $\lambda$.
You now have figure 1 from Sophie's paper. Well done.
Problems ahead:
What now? So the question is, if you fit the regression to this, what is your $k$, what is your $A$?
Not so important. What is this equation referring to, what is $P_1$, what is $P_2$, what is $P$:
$$k = \mathbb{E}_{P1,P2}\left(\frac{\log\lambda(\Delta P1) - \log(\lambda(\Delta P2)}{\Delta P1 - \Delta P2}\right),\quad A=\mathbb{E}_{P}(\lambda(\Delta P) \exp k \Delta P)$$ |
In the book
Bayesian Data Analysis by Gelman et al. (3rd edition, 2014), a hierarchical model (or one-way random-effects ANOVA) is presented in section 5.4 as follows,
\begin{equation}\label{eq:lme1} y_{ij} = b_0 + \lambda_i + \varepsilon_{ij}, \end{equation}
where the data $y_{ij}$ come from the $i$th measuring entity (e.g., student performance in a school district) collected under the $j$th condition (e.g., a school within the district), $b_0$ is the population mean, $\lambda_i$ is the deviation of the $i$th measuring entity from the population mean, and $\varepsilon_{ij}$ is the measuring error ($i=1, 2, \ldots, k;\ j=1, 2, \ldots, n$).
A posterior inference is derived in the book for the effect of each measuring entity $\theta_i=b_0 + \lambda_i$ based on a Gaussian assumption with a known variance $\sigma^2$ for the residuals $\varepsilon_{ij}$ and a prior distribution $G(0, \tau^2)$ for $\lambda_i$. Specifically, the mean and variance for $\theta_i$ are estimated as below:
\begin{align} {\rm mean}(\theta_i) &= \frac{\frac{n}{\sigma^2}\bar{y}_{i\cdot}+\frac{1}{\tau^2}b_0}{\frac{n}{\sigma^2}+\frac{1}{\tau^2}} \\[7pt] {\rm Var}(\theta_i) &= \frac{1}{\frac{n}{\sigma^2}+\frac{1}{\tau^2}} \end{align}
where $\bar{y}_{i\cdot}=\frac{1}{n}\sum_{j=1}^n y_{ij}$.
Even though the variance for the $\lambda_i$ is assumed to be known, I could solve the model as a mixed-effects model through, for example, function
lmer() in the R package
lme4, and use the estimated variances $\tau^2$ and $\sigma^2$ to obtain the posterior distribution using the formulation above. Is this a reasonable and solid approach?
I know that I could directly obtain the posterior distribution through R packages such as
brms and
rstanarm. However, the computational cost is too heavy in my case, and that's why I'm trying to see if the above closed form is a a reasonable approach to directly obtaining the posterior distribution by plugging the variance estimate $\hat{\sigma}^2$ from
lmer, rather than going through the typical Bayesian route.. |
PCTeX Talk Discussions on TeX, LaTeX, fonts, and typesetting
Author Message Michael Spivak Joined: 10 Oct 2005 Posts: 52
Posted: Tue Oct 11, 2005 3:46 pm Post subject: new version of fonts We are making a new version of the MTPro fonts, which will have Times-Italic-like characters designed into them, so that there will be no need for virtual fonts. This is a good opportunity to ask for new characters, etc.
PLEASE , IF YOU SUBMIT A REQUEST AS A GUEST, ADD AN EMAIL ADDRESS, SO THAT I CAN CONTACT YOU IF I HAVE QUESTIONS ABOUT IT!!!
My email is mikespivak@aol.com.
The following have been suggsted: :=, = with ^ accent above it,
updownarrows and downuparrows---and I think I'll add
updownharpoons, downupharpoons, upharpoons, downharpoons. Also, slanted \sum, \prod [and
presumably \coprod]. We could have \usum, \slsum, etc. to specify
upright or slanted \sum, etc., while \uoperators would normally make
\sum mean \usum, etc., while \sloperators would normally make
\sum mean \slsum, etc. Last edited by Michael Spivak on Wed Nov 02, 2005 9:59 am; edited 2 times in total AnnaD Guest
Posted: Thu Oct 27, 2005 6:26 pm Post subject: Michael,
Here are several things I came across:
1. \rightarrow with \sim on top of it (looks like \simeq but with an arrow)
2. \ast as a big math operator with limits (variable sizes: for in-text mode and display)
3. variable-length corner ( __| ) which works similar to \framebox (vertical line is variable too)
4. wide dual math accents (like \Hat{\Bar{}}, for example) with smaller gap between them kolchin Joined: 27 Oct 2005 Posts: 15 Location: Moscow, Russia
Posted: Thu Oct 27, 2005 8:56 pm Post subject: Hello Michael,
I would like to see
\& as a big math operator with limits, of variable size for text- and
display modes. Nase Joined: 28 Oct 2005 Posts: 1
Posted: Fri Oct 28, 2005 1:28 am Post subject: Re: new version of fonts
Michael Spivak wrote: We are making a new version of the MTPro fonts, which will have Times-Italic-like characters designed into them, so that there will be no need for virtual fonts. This is a good opportunity to ask for new characters, etc.
Dear Michael,
something that would be desirable in analysis
is a mean value integral "\mint". What I need
is an integral sign which is crossed horizontally
in the middle by a short bar. I have got some
TeXnical solution, but this hack only works in
the "\nolimits" case:
\newcommand{\meanbar}[1]{%
\setbox0 = \hbox{$#1 \int$}
\hbox to 0pt{%
\thinspace
\hskip 0.1\wd0
\raise 0.5\ht0
\hbox{%
\lower 0.5\dp0
\hbox{\rule{0.8\wd0}{2\linethickness}}
}%
\hss
}%
}
\newcommand{\palette}[1]{%
\mathchoice{#1 \displaystyle}%
{#1 \textstyle}%
{#1 \scriptstyle}%
{#1 \scriptscriptstyle}%
}
\newcommand{\mean}{\palette \meanbar}
\newcommand{\mint}{\mean \int}
Thank you and all the other folks from PCTeX
for developing a fairly complete mathematical
Times font family, especially for adding suitable
fonts for smaller design sizes!
Best regards,
Jens _________________ Jens Andre Griepentrog
WIAS Berlin (Germany) Guest
Posted: Fri Oct 28, 2005 12:25 pm Post subject: I would suggest a symbol similar to \hbar, but for d (i.e., \dbar). This is useful in thermodynamics as an inexact differential. Michael Spivak Joined: 10 Oct 2005 Posts: 52
Posted: Fri Oct 28, 2005 1:14 pm Post subject:
AnnaD wrote: Michael,
Here are several things I came across:
1. \rightarrow with \sim on top of it (looks like \simeq but with an arrow)
2. \ast as a big math operator with limits (variable sizes: for in-text mode and display)
3. variable-length corner ( __| ) which works similar to \framebox (vertical line is variable too)
4. wide dual math accents (like \Hat{\Bar{}}, for example) with smaller gap between them
1. should be simple
2. A \bigast should be OK, but do you have a sample to show how large it should be?
3. Don't know about \framebox (presumably from LaTeX? about which I
also don't know anything), but this should be doable completely in TeX, without any need for a font character.
4. Will get back to you about dual math accents later. Michael Spivak Joined: 10 Oct 2005 Posts: 52
Posted: Fri Oct 28, 2005 1:15 pm Post subject:
kolchin wrote: Hello Michael,
I would like to see
\& as a big math operator with limits, of variable size for text- and
display modes.
Should it look just like a Time &? Although \& would be a convenient name for the user, it would be simpler to implement things if it had another name, like \ampersand, or \bigampersand, or perhaps you have another name in mind. Michael Spivak Joined: 10 Oct 2005 Posts: 52
Posted: Fri Oct 28, 2005 1:18 pm Post subject: Re: new version of fonts
Nase wrote: something that would be desirable in analysis
is a mean value integral "\mint".
OK, but perhaps it should be called \barint? Michael Spivak Joined: 10 Oct 2005 Posts: 52
Posted: Fri Oct 28, 2005 1:20 pm Post subject:
Anonymous wrote: I would suggest a symbol similar to \hbar, but for d (i.e., \dbar). This is useful in thermodynamics as an inexact differential.
OK. I presume you want a regular d, not a barred partial sign ("eth"). Guest
Posted: Fri Oct 28, 2005 2:57 pm Post subject:
AnnaD wrote: Michael,
wide dual math accents (like \Hat{\Bar{}}, for example) with smaller gap between them
I will add \widehatdown#1#2, which puts a \widehat on #2, but moves
it down by #1. So, for example,
\widehatdown{2pt}{\widehat{a+b+c+d+e+f+g+h}}
will look better. jp Guest
Posted: Sat Oct 29, 2005 12:06 pm Post subject: Symbols Hi,
what I would like to see is the contraction operator,
for example for differential forms: \omega _| X .
:= is definitely very welcome, as would be =: and :<=>
jürgen pöschel
Uni Stuttgart kolchin Joined: 27 Oct 2005 Posts: 15 Location: Moscow, Russia
Posted: Sat Oct 29, 2005 12:07 pm Post subject:
Michael Spivak wrote:
kolchin wrote: Hello Michael,
I would like to see
\& as a big math operator with limits, of variable size for text- and
display modes.
Should it look just like a Time &? Although \& would be a convenient name for the user, it would be simpler to implement things if it had another name, like \ampersand, or \bigampersand, or perhaps you have another name in mind.
Yes, it should look as a big & in Time font; the name I like
is \bigvarland (big alternative logical "and").
Best regards and many thanks.
Andrei
Steklov Inst. Math. Michael Spivak Joined: 10 Oct 2005 Posts: 52
Posted: Sat Oct 29, 2005 4:44 pm Post subject: Re: Symbols
jp wrote: Hi,
what I would like to see is the contraction operator,
for example for differential forms: \omega _| X .
:= is definitely very welcome, as would be =: and :<=>
jürgen pöschel
Uni Stuttgart
Can add these. Do you mean literally :<=> or is that an abbreviation for one or more symbols? kolchin Joined: 27 Oct 2005 Posts: 15 Location: Moscow, Russia
Posted: Sun Oct 30, 2005 12:46 am Post subject: another lowercase "z" in math Dear Michael,
Is it possible to make lowercase mathematical italic "z"
look as in Adobe Times PS font in my printer, with a swash? Al Freed Guest
Posted: Tue Nov 01, 2005 12:45 pm Post subject: more blackboard fonts Hi Michael,
As long as you're soliciting a wish list, here is mine:
1) blackboard bold Greek fonts, upper and lower case
2) slanted blackboard bold fonts, medium weight, upper and lower case
Thanks,
Al Freed
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Covering Space Examples R2 \ {(0, 0)}
Recall from the Covering Spaces page that if $X$ is a topological space then a covering space of $X$ is a pair $(\tilde{X}, p)$ where $\tilde{X}$ is a path connected and locally path connected topological space and $p : \tilde{X} \to X$ is a continuous map such that for every $x \in X$ there exists a path connected open neighbourhood $U$ of $x$ (called an elementary neighbourhood of $x$) such that $p$ restricted to every path component of $p^{-1}(U)$ is a homeomorphism onto $U$.
We will now look at an example of a covering space of $\mathbb{R}^2 \setminus \{ (0, 0) \}$.
Let $\tilde{X} = \mathbb{R}^2$ and let $p : \mathbb{R}^2 \to \mathbb{R}^2 \setminus \{ (0, 0) \}$ be defined for all $(x, y) \in \mathbb{R}^2$ by:(1)
Clearly $p$ is a continuous map since each component of $p$ is continuous.
Observe that $(0, 0)$ is mapped to the point $(1, 0)$. In general, $(0, 2k \pi)$ is mapped to the point $(1, 0)$ for every $k \in \mathbb{Z}$.
Also observe that for a fixed $x_0$, $(x_0, 0)$ is mapped to the point $(e^{x_0}, 0)$.
As we can see, $p$ maps $\mathbb{R}^2$ onto $\mathbb{R}^2 \setminus \{ (0, 0) \}$ an infinite number of times.
If $(x_0, y_0) \in \mathbb{R}^2 \setminus \{ (0, 0) \}$ then there exists an open neighbourhood of $(x_0, y_0)$ which does not contain the puncture from the removal of the origin. In particular, there exists an open disk $U$ centered at $(x_0, y_0)$ that is contained in $\mathbb{R}^2 \setminus \{ (0, 0) \}$. If we take this open disk small enough, then $p^{-1}(U)$ will contain an infinite number of squashed open disks which are disjoint from each other. Clearly the restriction of $p$ onto any of these open disks will be a homeomorphism onto $U$. |
Wronskian Determinants and Linear Homogenous Differential Equations
Recall from the The Principle of Superposition page that if we have the second order linear homogenous differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = 0$ and $y = y_1(t)$ and $y = y_2(t)$ are solutions to this differential equation, then for constants $C$ and $D$, the linear combination $y = Cy_1(t) + Dy_2(t)$ is also a solution to this differential equation. One question to ask if whether or not
all of the solutions to this differential equation are in this form as we do not want to miss any other potential solutions.
Suppose that we are given an initial value problem to a second order linear homogenous differential equation in this form alongside the initial conditions $y(t_0) = y_0$ and $y'(t_0) = y'_0$. Applying the first initial condition $y(t_0) = y_0$ to our solution $y = Cy_1(t) + Dy_2(t)$ and we have that:(1)
Now the derivative of $y = Cy_1(t) + Dy_2(t)$ is $y' = Cy_1'(t) + Dy_2'(t)$. Applying the second initial condition $y'(t_0) = y'_0$ and we have that:(2)
We need the unknown constants $C$ and $D$ to satisfy both of these equations, that is, we want to solve the system $\left\{\begin{matrix} Cy_1(t_0) + Dy_2(t_0) = y_0 \\ Cy_1'(t_0) + Dy_2'(t_0) = y'_0 \end{matrix}\right.$ for the constants $C$ and $D$. We note that we have a system of two equations with two unknowns and thus, a unique solution exists if the determinant of the augmented matrix for this system is nonzero, that is:(3)
Notice though that this determinant $W$ is simple the Wronskian of the functions $y_1$ and $y_2$ evaluated at $t_0$. If $W \neq 0$, then we can apply Cramer's Rule from linear algebra to find the values of the constants $C$ and $D$. We have that:(4)
If this determinant $W$ equals zero, then either no solutions for $C$ and $D$ exist, or infinitely many solutions for the values of $C$ and $D$ may exist. The following theorem summarizes what we have just found.
Theorem 1: Let $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ be a second order linear homogenous differential equation where $p$ and $q$ are continuous functions on an open interval $I$ such that $t_0 \in I$ and with the initial conditions $y(t_0) = y_0$ and $y'(t_0) = y'_0$. If $y = y_1(t)$ and $y = y_2(t)$ are solutions to this differential equation then there exists constants $C$ and $D$ for which $y = Cy_1(t) + Dy_2(t)$ is a solution to the initial value problem if and only if the Wronskian at $t_0$ is nonzero, that is $W(y_1, y_2) \biggr \rvert_{t_0} = y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0) \neq 0$. |
If a team of researchers perform multiple (hypothesis) tests on a given data set, there is a volume of literature asserting that they should use some form of correction for multiple testing (Bonferroni, etc), even if the tests are independent. My question is this: does this same logic apply to multiple teams testing hypotheses on the same data set? Said another way-- what is the barrier for the family-wise error calculations? Should researchers be limited to reusing data sets for exploration only?
I disagree strongly with @fcoppens leap from recognizing the importance of multiple-hypothesis correction within a single investigation to claiming that "By the same reasoning, the same holds if several teams perform these tests."
There is no question that the more studies are performed and the more hypotheses are tested, the more Type I errors will occur. But I think there's a confusion here over the meaning of "family-wise error" rates and how they apply in actual scientific work.
First, remember that multiple-testing corrections typically arose in
post-hoc comparisons for which there were no pre-formulated hypotheses. It is not at all clear that the same corrections are required when there is a small pre-defined set of hypotheses.
Second, the "scientific truth" of an individual publication does not depend on the truth of each individual statement within the publication. A well-designed study approaches an overall
scientific (as opposed to statistical) hypothesis from many different perspectives, and puts together different types of results to evaluate the scientific hypothesis. Each individual result may be evaluated by a statistical test.
By the argument from @fcoppens however, if even
one of those individual statistical tests makes a Type I error then that leads to a "false belief of 'scientific truth'". This is simply wrong.
The "scientific truth" of the
scientific hypothesis in a publication, as opposed to the validity of an individual statistical test, generally comes from a combination of different types of evidence. Insistence on multiple types of evidence makes the validity of a scientific hypothesis robust to the individual mistakes that inevitably occur. As I look back on my 50 or so scientific publications, I would be hard pressed to find any that remains so flawless in every detail as @fcoppens seems to insist upon. Yet I am similarly hard pressed to find any where the scientific hypothesis was outright wrong. Incomplete, perhaps; made irrelevant by later developments in the field, certainly. But not "wrong" in the context of the state of scientific knowledge at the time.
Third, the argument ignores the costs of making Type II errors. A type II error might close off entire fields of promising scientific inquiry. If the recommendations of @fcoppens were to be followed, Type II error rates would escalate massively, to the detriment of the scientific enterprise.
Finally, the recommendation is impossible to follow in practice. If I analyze a set of publicly available data, I may have no way of knowing whether anyone else has used it, or for what purpose. I have no way of correcting for anyone else's hypothesis tests. And as I argue above, I shouldn't have to.
The ''multiple testing'' correction is necessary whenever you 'inflate the type I error': e.g. if you perform two tests, each at a confidence level $\alpha=5\%$, and for the first we test the null $H_0^{(1)}$ against the alternative $H_1^{(1)}$ and the second hypothesis $H_0^{(2)}$ versus $H_1^{(2)}$.
Then we know that the type I error for e.g. the first hypothesis is the probability of falsely rejecting $H_0^{(1)}$ and is it $\alpha=5\%$.
If you perform the two tests, then the probability that at least one of the two is falsely rejected is equal to the 1 minus the probability that both are accepted so $1 - (1-\alpha)^2$ which, for $\alpha=5\%$ is equal to $9.75\%$, so the type one error of having
at least one false rejection is has almost doubled !
In statistical hypothesis testing one can only find statistical evidence for the alternative hypothesis by rejecting the null, rejecting the null allows us to conclude that there is
evidence in favour of the alternative hypothesis. (see also What follows if we fail to reject the null hypothesis?).
So a false rejection of the null gives us false evidence so a false belief of ''scientific truth''. This is why this type I inflation (the almost doubling of the type I error) has to be avoided; higher type I errors imply more
false beliefs that something is scientifically proven. Therefore people ''control'' the type Ierror at a familywise level.
If there is a team of researchers that performs multiple tests, then each time they reject the null hypothesis they conclude that they have found statitiscal evidence of a scientific truth. However, by the above, many more than $5\%$ of these conclusions are a false believe of ''scientific truth''.
By the same reasoning, the same holds if several teams perform these tests (on the same data).
Obviously, the above findings only hold if we the teams work
on the same data. What is different then when they work on different samples ?
To explain this, let's take a simple and very unrealistic example. Our null hypothesis is that a population has a normal distribution, with known $\sigma$ and the null states that $H_0: \mu = 0$ against $H_1: \mu \ne 0$. Let's take the significance level $\alpha=5\%$.
Our sample ('the data') is only one observation, so we will reject the null when the observation $o$ is either larger than $1.96\sigma$ or smaller than $-1.96\sigma$.
We make a type I error with a probability of $5\%$ because it could be that we reject $H_0$ just by chance, indeed, if $H_0$ is true (so the population is normal and $\mu=0$) then there is (with $H_0$ true) a chance that $o \not \in [-1.96\sigma;1.96\sigma$]. So even if $H_0$ is true then there is a chance that we have bad luck with the data.
So if we use the same data, it could be that the conclusions of the tests are based on a sample that was drawn with ''bad chance''. With another sample the context is different. |
Preprints (rote Reihe) des Fachbereich Mathematik Refine Year of publication 1995 (9) (remove)
292
Symmetry properties of average densities and tangent measure distributions of measures on the line (1995)
Answering a question by Bedford and Fisher we show that for every Radon measure on the line with positive and finite lower and upper densities the one-sided average densities always agree with one half of the circular average densities at almost every point. We infer this result from a more general formula, which involves the notion of a tangent measure distribution introduced by Bandt and Graf. This formula shows that the tangent measure distributions are Palm distributions and define self-similar random measures in the sense of U. Zähle.
267
In this paper we investigate two optimization problems for matroids with multiple objective functions, namely finding the pareto set and the max-ordering problem which conists in finding a basis such that the largest objective value is minimal. We prove that the decision versions of both problems are NP-complete. A solution procedure for the max-ordering problem is presented and a result on the relation of the solution sets of the two problems is given. The main results are a characterization of pareto bases by a basis exchange property and finally a connectivity result for proper pareto solutions.
268
In this paper we will introduce the concept of lexicographic max-ordering solutions for multicriteria combinatorial optimization problems. Section 1 provides the basic notions of multicriteria combinatorial optimization and the definition of lexicographic max-ordering solutions. In Section 2 we will show that lexicographic max-ordering solutions are pareto optimal as well as max-ordering optimal solutions. Furthermore lexicographic max-ordering solutions can be used to characterize the set of pareto solutions. Further properties of lexicographic max-ordering solutions are given. Section 3 will be devoted to algorithms. We give a polynomial time algorithm for the two criteria case where one criterion is a sum and one is a bottleneck objective function, provided that the one criterion sum problem is solvable in polynomial time. For bottleneck functions an algorithm for the general case of Q criteria is presented.
265
In multiple criteria optimization an important research topic is the topological structure of the set \( X_e \) of efficient solutions. Of major interest is the connectedness of \( X_e \), since it would allow the determination of \( X_e \) without considering non-efficient solutions in the process. We review general results on the subject,including the connectedness result for efficient solutions in multiple criteria linear programming. This result can be used to derive a definition of connectedness for discrete optimization problems. We present a counterexample to a previously stated result in this area, namely that the set of efficient solutions of the shortest path problem is connected. We will also show that connectedness does not hold for another important problem in discrete multiple criteria optimization: the spanning tree problem.
262
An improved asymptotic analysis of the expected number of pivot steps required by the simplex algorithm (1995)
Let \(a_1,\dots,a_m\) be i.i .d. vectors uniform on the unit sphere in \(\mathbb{R}^n\), \(m\ge n\ge3\) and let \(X\):= {\(x \in \mathbb{R}^n \mid a ^T_i x\leq 1\)} be the random polyhedron generated by. Furthermore, for linearly independent vectors \(u\), \(\bar u\) in \(\mathbb{R}^n\), let \(S_{u, \bar u}(X)\) be the number of shadow vertices of \(X\) in \(span (u, \bar u\)). The paper provides an asymptotic expansion of the expectation value \(E (S_{u, \bar u})\) for fixed \(n\) and \(m\to\infty\). The first terms of the expansion are given explicitly. Our investigation of \(E (S_{u, \bar u})\) is closely connected to Borgwardt's probabilistic analysis of the shadow vertex algorithm - a parametric variant of the simplex algorithm. We obtain an improved asymptotic upper bound for the number of pivot steps required by the shadow vertex algorithm for uniformly on the sphere distributed data.
266 |
Speaker: Misha Tyomkyn (TAU) Title: Lagrangians of hypergraphs and the Frankl-Furedi conjecture Abstract: Frankl and Furedi conjectured in 1989 that the maximum Lagrangian of all r-uniform hypergraphs of given size m is realised by the initial segment of the colexicographic order. For r=3 this was partially solved by Talbot, but for r\geq 4 the conjecture was widely open. We verify the conjecture for all r\geq 4, whenever $\binom{t-1}{r} \leq m \leq \binom{t}{r}- \gamma_r t^{r-2}$ for a constant $\gamma_r>0$. This range includes the principal case $m=\binom{t}{r}$ for large enough $t$.
Date:
Sunday, 30 April, 2017 - 11:00 to 13:00
Repeats every week every Sunday until Sun Jun 25 2017 except Sun Apr 30 2017
Location:
Rothberg B221 (CS building) |
As Twitter and Github and some other sites have switched from square icons to circle icons, I needed to update my logo to fit. I had cropped the previous one, but I decided I should generate new ones from scratch. The main ingredients:
The red blob logo is \(r = 5 + \sin(5\theta)\). The mouth can be drawn with my face generator. The eyes will have to be drawn separately, as my face generator uses circles and my logo uses tall ellipses. 1 Circular frame#
I need to figure out how to size and position the blob so that it fits inside a circle. Since the blob shape is based on polar coordinates, it fits quite nicely!
I had thought it wouldn't. Why? It's because when I tried using my square logo on the sites with circle frames, it didn't fit. I assumed it was because of the circle. However, it was actually because I had previously adjusted the logo to fit in a square.
2 Square frame#
The logo doesn't fit properly in a square.
There are two problems:
It's not vertically centered. It's not as large as it could be.
The blob is centered at 0,0 but because its feet are out at an angle and the head is not, the extent of y values is not centered at 0. The hands are also at an angle so the extent of x values doesn't fill the box.
To find the extent (min/max) analytically, I could take \(r = 5 + \sin(5\theta)\), calculate the coordinates \(x = r \cos(theta), y = r \sin(\theta)\), and then calculate min/max from the derivative of \(x\) and \(y\) with respect to \(\theta\).
Instead of calculating the dimensions analytically, I calculated them in numerically. For a box with range
-100:+100, the blob's x range is
-96:+96 and y range is
-100:+85. That means I can lower the center by
(100-85)/200, or 7.5%, and increase the blob's radius by
200/max(96+96, 100+85), or 4.1%. I also need to decrease the blob's radius by the line width. Let's try it:
Looks much better positioned than before!
However I think it's better to stretch the logo a little bit:
3 Generators#
Oops, the formula I've been using on my site is actually \(5.88 +\) instead of \(5 +\).
Calculate the unaltered shape. For a circle: multiply by radius to fit For a square: stretch, then adjust y position and size to fit autostretch: (ymax - ymin) / (xmax - xmin) y += (ymax + ymin) / 2 y *= autostretch size *= (side - linewidth) / max(ymax - ymin, xmax - xmin)
flex =
size = line = bri =
I used this to generate new icons for Safari, Github, Twitter, etc. |
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
Portuguese.
Contents
Portuguese language has some accentuated words. For this reason the preamble of your file must be modified accordingly to support these characters and some other features.
\documentclass{article} %encoding %-------------------------------------- \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} %-------------------------------------- %Portuguese-specific commands %-------------------------------------- \usepackage[portuguese]{babel} %-------------------------------------- %Hyphenation rules %-------------------------------------- \usepackage{hyphenat} \hyphenation{mate-mática recu-perar} %-------------------------------------- \begin{document} \tableofcontents \vspace{2cm} %Add a 2cm space \begin{abstract} Este é um breve resumo do conteúdo do documento escrito em Português. \end{abstract} \section{Seção introdutória} Esta é a primeira seção, podemos acrescentar alguns elementos adicionais e tudo será escrito corretamente. Além disso, se uma palavra é um caminho muito longo e tem de ser truncado, babel irá tentar truncar corretamente, dependendo do idioma. \section{Segunda seção} Esta seção é para ver o que acontece com comandos de texto que definem \[ \lim x = \theta + 152383.52 \] \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.
If your are looking for instructions on how to use more than one language in a single document, for instance English and Portuguese, see the International language support article.
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 Portuguese 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 format LaTeX documents properly you should also choose a font encoding which supports specific characters for Portuguese language, this is accomplished by the
package: fontenc \usepackage[T1]{fontenc} Even though the default encoding works well in Portuguese, using this specific encoding will avoid glitches with some specific characters, e.g., some accented characters might not be directly copyable from the generated PDF and instead are constructed using the base character and an overlayed shifted accent symbol, resulting in two separate symbols if you copy it. 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 Portuguese language. \usepackage[portuguese]{babel} As you may see in the example at the introduction, instead of "abstract" and "Contents" the Portuguese words "Resumo" and "Conteúdo" are used.
If you need to use the Brazilian Portuguese localization use
brazilian instead of
portuguese as parameter when importing
babel.
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, matemática could become mate-mática. 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-mática 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 or include it in an
\mbox{word}.
For more information see |
Author Message jautschbach Joined: 17 Mar 2006 Posts: 11
Posted: Fri Apr 21, 2006 7:42 am Post subject: small caps Hello
I noticed some inconsistencies between Y&Y and MikTeX on WinXP when using the small caps font with \textsc{}. With Y&Y I get the correct Times SC font in the oouput but with MikTeX it seems like some fake small caps fonts are used. I'm not sure if this is related to mtpro2 or not. Some help would be greatly appreciated! (I have theTimes SC font installed).
I use
\usepackage[LY1]{fontenc} %% needed for YandY TeX
\usepackage{times}
\usepackage[subscriptcorrection,nofontinfo,mtpbbi]{mtpro2}
\usepackage{bm}
Jochen murray Joined: 07 Feb 2006 Posts: 47 Location: Amherst, MA, USA
Posted: Sun Apr 23, 2006 9:26 am Post subject: Re: small caps
jautschbach wrote: Hello
I noticed some inconsistencies between Y&Y and MikTeX on WinXP when using the small caps font with \textsc{}. With Y&Y I get the correct Times SC font in the oouput but with MikTeX it seems like some fake small caps fonts are used. I'm not sure if this is related to mtpro2 or not. Some help would be greatly appreciated! (I have theTimes SC font installed).
I use
\usepackage[LY1]{fontenc} %% needed for YandY TeX
\usepackage{times}
\usepackage[subscriptcorrection,nofontinfo,mtpbbi]{mtpro2}
\usepackage{bm}
Jochen
I see a difference, too, at least when I print the output. (I don't really see the difference when I view with YAP, on the one hand, from MiKTeX or with DVIwindo, on the other hand, from Y&Y TeX.)
I presume the difference is that mtpro2 under MiKTeX is not using actual Adobe Times SC fonts.
I presume you changed the LY1 option to T1 for MiKTeX.
I also noticed you used \usepackage{times}, whereas the docs for mtpro2 show using instead \renewcommand{rmdefault}{ptm}. But I don't detect any difference in output.
Here's my little test document:
\documentclass[12pt]{article}
%\usepackage[T1]{fontenc} %% for MiKTeX
\usepackage[LY1]{fontenc} %% needed for YandY TeX
%\usepackage{times}
\renewcommand{\rmdefault}{ptm}
\usepackage[subscriptcorrection]{mtpro2}
\begin{document}
Start with normal text, then \textsc{Small Caps font text.} Now math:
\[
\int_0^{\pi} \sin \theta\,d\,\theta .
\]
\end{document} zedler Joined: 03 Mar 2006 Posts: 15 murray Joined: 07 Feb 2006 Posts: 47 Location: Amherst, MA, USA
Posted: Sun Apr 23, 2006 3:00 pm Post subject: Re: small caps
But will this package now break Y&Y? There are issues of font naming and font encoding involved, which of course are handled differently by Y&Y vs. other TeX's. zedler Joined: 03 Mar 2006 Posts: 15
Posted: Sun Apr 23, 2006 11:37 pm Post subject: Re: small caps
Quote: But will this package now break Y&Y? There are issues of font naming and font encoding involved, which of course are handled differently by Y&Y vs. You asked for a solution to use the Adobe Times SC fonts with Miktex/tetex. The ptms package won't run under Y&Y.
Michael murray Joined: 07 Feb 2006 Posts: 47 Location: Amherst, MA, USA
Posted: Mon Apr 24, 2006 5:46 am Post subject: Re: small caps
zedler wrote:
Quote: But will this package now break Y&Y? There are issues of font naming and font encoding involved, which of course are handled differently by Y&Y vs. You asked for a solution to use the Adobe Times SC fonts with Miktex/tetex. The ptms package won't run under Y&Y.
Michael I was concerned about renaming the fonts, as you said would be required. jautschbach Joined: 17 Mar 2006 Posts: 11
Posted: Fri Apr 28, 2006 12:46 pm Post subject: Re: small caps
zedler wrote:
Quote: But will this package now break Y&Y? There are issues of font naming and font encoding involved, which of course are handled differently by Y&Y vs. You asked for a solution to use the Adobe Times SC fonts with Miktex/tetex. The ptms package won't run under Y&Y.
Michael
Without small caps I get identical results with Y&Y and MikTeX from the same document when I use LY1 encoding. Isn't there some package option for times or mtpro2 to let the tex system know that Adobe Times SC fonts are installed?
Jochen |
Soil composition
by
V
olume and
M
ass, by phase:
a
ir,
w
ater,
v
oid (pores filled with water or air),
s
oil, and
t
otal.
Water content or moisture content is the quantity of water contained in a material, such as soil (called soil moisture), rock, ceramics, fruit, or wood. Water content is used in a wide range of scientific and technical areas, and is expressed as a ratio, which can range from 0 (completely dry) to the value of the materials' porosity at saturation. It can be given on a volumetric or mass (gravimetric) basis.
Contents Definitions 1 Measurement 2 Direct methods 2.1 Laboratory methods 2.2 Soil moisture 3 Geophysical methods 3.1 Satellite remote sensing method 3.2 Classification and uses 4 Earth and agricultural sciences 4.1 Agriculture 4.1.1 Groundwater 4.1.2 See also 5 References 6 Further reading 7 Definitions
Volumetric water content, θ, is defined mathematically as: \theta = \frac{V_w}{V_\text{wet}}
where V_w is the volume of water and V_\text{wet} = V_h + V_w + V_a is the volume of wet material, that is, host material (e.g., soil particles, vegetation tissue) volume + water volume + air space.
Gravimetric water content [1] is expressed by mass (weight) as follows: u = \frac{m_w}{m}
where m_w is the mass of water and m is the mass of the substance. Normally the latter is taken before drying:
u' = \frac{m_w}{m_\text{wet}}
except for woodworking, geotechnical and soil science applications where oven-dried material is used instead:
u'' = \frac{m_w}{m_\text{dry}}
To convert gravimetric water content to volumetric water content, multiply the gravimetric water content by the bulk specific gravity of the material.
Derived quantities
In soil mechanics and petroleum engineering, the term
water saturation or degree of saturation, S_w is used, defined as S_w = \frac{V_w}{V_v} = \frac{V_w}{V_T\phi} = \frac{\theta}{\phi}
where \phi = V_v / V_T is the porosity and V_v is the volume of void or pore space. Values of
S can range from 0 (dry) to 1 (saturated). In reality, w S never reaches 0 or 1 - these are idealizations for engineering use. w
The
normalized water content, \Theta, (also called effective saturation or S_e) is a dimensionless value defined by van Genuchten [2] as: \Theta = \frac{\theta - \theta_r}{\theta_s-\theta_r}
where \theta is the volumetric water content; \theta_r is the residual water content, defined as the water content for which the gradient d\theta/dh becomes zero; and, \theta_s is the saturated water content, which is equivalent to porosity, \phi.
Measurement Direct methods
Water content can be directly measured using a known volume of the material, and a drying oven. Volumetric water content, θ, is calculated
[3] via the volume of water V_w and the mass of water m_w: V_w = \frac{m_w}{\rho_w} = \frac{m_{\text{wet}}-m_{\text{dry}}}{\rho_w}
where
m_{\text{wet}} and m_{\text{dry}} are the masses of the sample before and after drying in the oven; \rho_w is the density of water; and
For materials that change in volume with water content, such as coal, the water content,
u, is expressed in terms of the mass of water per unit mass of the moist specimen: u' = \frac{m_{\text{wet}} - m_{\text{dry}}}{m_{\text{wet}}}
However, geotechnics requires the moisture content to be expressed with respect to the sample's dry weight (often as a percentage, i.e. % moisture content =
u×100%) u'' = \frac{m_{\text{wet}} - m_{\text{dry}}}{m_{\text{dry}}}
For wood, the convention is to report moisture content on oven-dry basis (i.e. generally drying sample in an oven set at 105 deg Celsius for 24 hours). In wood drying, this is an important concept.
Laboratory methods
Other methods that determine water content of a sample include chemical titrations (for example the Karl Fischer titration), determining mass loss on heating (perhaps in the presence of an inert gas), or after freeze drying. In the food industry the Dean-Stark method is also commonly used.
From the Annual Book of ASTM (American Society for Testing and Materials) Standards, the total evaporable moisture content in Aggregate (C 566) can be calculated with the formula:
p = \frac{W-D}{D}
where p is the fraction of total evaporable moisture content of sample, W is the mass of the original sample, and D is mass of dried sample.
Soil moisture Geophysical methods
There are several geophysical methods available that can approximate
in situ soil water content. These methods include: time-domain reflectometry (TDR), neutron probe, frequency domain sensor, capacitance probe, amplitude domain reflectometry, electrical resistivity tomography, ground penetrating radar (GPR), and others that are sensitive to the physical properties of water . [4] Geophysical sensors are often used to monitor soil moisture continuously in agricultural and scientific applications. Satellite remote sensing method
Satellite microwave remote sensing is used to estimate soil moisture based on the large contrast between the dielectric properties of wet and dry soil. The microwave radiation is not sensitive to atmospheric variables, and can penetrate through clouds. Also, microwave signal can penetrate, to a certain extent, the vegetation canopy and retrieve information from ground surface.
[5] The data from microwave remote sensing satellite such as: WindSat, AMSR-E, RADARSAT, ERS-1-2, Metop/ASCAT are used to estimate surface soil moisture. [6] Classification and uses
Moisture may be present as adsorbed moisture at internal surfaces and as capillary condensed water in small pores. At low relative humidities, moisture consists mainly of adsorbed water. At higher relative humidities, liquid water becomes more and more important, depending or not depending on the pore size can also be a influence of volume. In wood-based materials, however, almost all water is adsorbed at humidities below 98% RH.
In biological applications there can also be a distinction between physisorbed water and "free" water — the physisorbed water being that closely associated with and relatively difficult to remove from a biological material. The method used to determine water content may affect whether water present in this form is accounted for. For a better indication of "free" and "bound" water, the water activity of a material should be considered.
Water molecules may also be present in materials closely associated with individual molecules, as "water of crystallization", or as water molecules which are static components of protein structure.
Earth and agricultural sciences
In soil science, hydrology and agricultural sciences, water content has an important role for groundwater recharge, agriculture, and soil chemistry. Many recent scientific research efforts have aimed toward a predictive-understanding of water content over space and time. Observations have revealed generally that spatial variance in water content tends to increase as overall wetness increases in semiarid regions, to decrease as overall wetness increases in humid regions, and to peak under intermediate wetness conditions in temperate regions .
[7]
There are four standard water contents that are routinely measured and used, which are described in the following table:
Name Notation Suction pressure (J/kg or kPa) Typical water content (vol/vol) Conditions Saturated water content θ s 0 0.2–0.5 Fully saturated soil, equivalent to effective porosity Field capacity θ fc −33 0.1–0.35 Soil moisture 2–3 days after a rain or irrigation Permanent wilting point θ pwp or θ wp −1500 0.01–0.25 Minimum soil moisture at which a plant wilts Residual water content θ r −∞ 0.001–0.1 Remaining water at high tension
And lastly the available water content, θ
a, which is equivalent to: θ a ≡ θ fc − θ pwp
which can range between 0.1 in gravel and 0.3 in peat.
Agriculture
When a soil becomes too dry, plant transpiration drops because the water is increasingly bound to the soil particles by suction. Below the wilting point plants are no longer able to extract water. At this point they wilt and cease transpiring altogether. Conditions where soil is too dry to maintain reliable plant growth is referred to as agricultural drought, and is a particular focus of irrigation management. Such conditions are common in arid and semi-arid environments.
Some agriculture professionals are beginning to use environmental measurements such as soil moisture to schedule irrigation. This method is referred to as
smart irrigation or soil cultivation. Groundwater
In saturated groundwater aquifers, all available pore spaces are filled with water (volumetric water content = porosity). Above a capillary fringe, pore spaces have air in them too.
Most soils have a water content less than porosity, which is the definition of unsaturated conditions, and they make up the subject of vadose zone hydrogeology. The capillary fringe of the water table is the dividing line between saturated and unsaturated conditions. Water content in the capillary fringe decreases with increasing distance above the phreatic surface.
One of the main complications which arises in studying the vadose zone, is the fact that the unsaturated hydraulic conductivity is a function of the water content of the material. As a material dries out, the connected wet pathways through the media become smaller, the hydraulic conductivity decreasing with lower water content in a very non-linear fashion.
A water retention curve is the relationship between volumetric water content and the water potential of the porous medium. It is characteristic for different types of porous medium. Due to hysteresis, different wetting and drying curves may be distinguished.
See also References ^ T. William Lambe & Robert V. Whitman (1969). "Chapter 3: Description of an Assemblage of Particles". Soil Mechanics (First ed.). John Wiley & Sons, Inc. p. 553. ^ van Genuchten, M.Th. (1980). "A closed-form equation for predicting the hydraulic conductivity of unsaturated soils" (PDF). Soil Science Society of America Journal 44 (5): 892–898. ^ Dingman, S.L. (2002). "Chapter 6, Water in soils: infiltration and redistribution". Physical Hydrology (Second ed.). Upper Saddle River, New Jersey: Prentice-Hall, Inc. p. 646. ^ F. Ozcep, M. Asci, O. Tezel, T. Yas, N. Alpaslan, D. Gundogdu (2005). "Relationships Between Electrical Properties (in Situ) and Water Content (in the Laboratory) of Some Soils in Turkey" (PDF). Geophysical Research Abstracts 7. ^ [3] ^ [4] ^ Lawrence, J. E., and G. M. Hornberger (2007). "Soil moisture variability across climate zones". Geophys. Res. Lett. 34 (L20402): L20402. Further reading
Field Estimation of Soil Water Content: A Practical Guide to Methods, Instrumentation and Sensor Technology (PDF), Vienna, Austria: International Atomic Energy Agency, 2008, p. 131,
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I heard a couple of times that there is no dynamics in 3D (2+1) GR, that it's something like a topological theory. I got the argument in the 2D case (the metric is conformally flat, Einstein equations trivially satisfied and the action is just a topological number) but I don't get how it is still true or partially similar with one more space dimension.
The absence of physical excitations in 3 dimensions has a simple reason: the Riemann tensor may be fully expressed via the Ricci tensor. Because the Ricci tensor vanishes in the vacuum due to Einstein's equations, the Riemann tensor vanishes (whenever the equations of motion are imposed), too: the vacuum has to be flat (no nontrivial Schwarzschild-like curved vacuum solutions). So there can't be any gravitational waves, there are no gravitons (quanta of gravitational waves). In other words, Ricci flatness implies flatness.
Counting components of tensors
The reason why the Riemann tensor is fully determined by the Ricci tensor is not hard to see. The Riemann tensor is $R_{abcd}$ but it is antisymmetric in $ab$ and in $cd$ and symmetric under exchange of the index pairs $ab$, $cd$. In 3 dimensions, one may dualize the antisymmetric index pairs $ab$ and $cd$ to simple indices $e,f$ using the antisymmetric $\epsilon_{abe}$ tensor and the Riemann tensor is symmetric in these new consolidated $e,f$ indices so it has 6 components, just like the Ricci tensor $R_{gh}$.
Because the Riemann tensor may always be written in terms of the Ricci tensor and because they have the same number of components at each point in $D=3$, it must be true that the opposite relationship has to exist, too. It is $$ R_{abcd} = \alpha(R_{ac}g_{bd} - R_{bc}g_{ad} - R_{ad}g_{bc} + R_{bd}g_{ac} )+\beta R(g_{ac}g_{bd}-g_{ad}g_{bc}) $$ I leave it as a homework to calculate the right values of $\alpha,\beta$ from the condition that the $ac$-contraction of the object above produces $R_{bd}$, as expected from the Ricci tensor's definition.
Counting polarizations of gravitons (or linearized gravitational waves)
An alternative way to prove that there are no physical polarizations in $D=3$ is to count them using the usual formula. The physical polarizations in $D$ dimensions are the traceless symmetric tensor in $(D-2)$ dimensions. For $D=3$, you have $D-2=1$ so only the symmetric tensor only has a unique component, e.g. $h_{22}$, and the traceless condition eliminates this last condition, too. So just like you have 2 physical graviton polarizations in $D=4$ and 44 polarizations in $D=11$, to mention two examples, there are 0 of them in $D=3$. The general number is $(D-2)(D-1)/2-1$.
In 2 dimensions, the whole Riemann tensor may be expressed in terms of the Ricci scalar curvature $R$ (best the Ricci tensor itself is $R_{ab}=Rg_{ab}/2$) which is directly imprinted to the component $R_{1212}$ etc.: Einstein's equations become vacuous in 2D. The number of components of the gravitational field is formally $(-1)$ in $D=2$; the local dynamics of the gravitational sector is not only vacuous but it imposes constraints on the remaining matter, too.
Other effects of gravity in 3D
While there are no gravitational waves in 3 dimensions, it doesn't mean that there are absolutely no gravitational effects. One may create point masses. Their gravitational field creates a vacuum that is Riemann-flat almost everywhere but creates a deficit angle.
Approximately equivalent theories
Due to the absence of local excitations, this is formally a topological theory and there are maps to other topological theories in 3D, especially the Chern-Simons theory with a gauge group. However, this equivalence only holds in some perturbative approximations and under extra assumptions, and for most purposes, it is a vacuous relationship, anyway. |
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Filomat, ISSN 0354-5180, 2016, Volume 30, Issue 6, pp. 1519 - 1524
Journal Article
Physics Letters B, ISSN 0370-2693, 03/2019, Volume 790, Issue C
We present the first measurements of femtoscopic correlations between the K$^0_S$ and K± particles in pp collisions at $\sqrt{s}=7$ TeV measured by the ALICE...
Journal Article
3. Global asymptotic stability of the higher order equation $$x_{n+1} = \frac{ ax_{n}+bx_{n-k}}{A+Bx_{n-k}}$$ x n + 1 = a x n + b x n - k A + B x n - k
Journal of Applied Mathematics and Computing, ISSN 1598-5865, 10/2017, Volume 55, Issue 1, pp. 135 - 148
In this paper, we investigate the local and global stability and the period two solutions of all nonnegative solutions of the difference equation,...
Computational Mathematics and Numerical Analysis | Semi-cycles | Mathematics of Computing | Appl.Mathematics/Computational Methods of Engineering | Global asymptotic stability | Mathematics | Theory of Computation | Difference equation | Equilibrium point
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4. Dynamics of nonlinear difference equation $$x_{n+1}=\frac{\beta x_{n}+\gamma x_{n-k}}{A+Bx_{n}+C x_{n-k}}$$ x n + 1 = β x n + γ x n - k A + B x n + C x n - k
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Physics Letters. Section B, ISSN 0370-2693, 09/2017, Volume 774, Issue C
We present the first ever measurements of femtoscopic correlations between the KS0 and K± particles. The analysis was performed on the data from Pb–Pb...
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The Lancet, ISSN 0140-6736, 10/2016, Volume 388, Issue 10053, pp. 1459 - 1544
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Internal Medicine | VERBAL AUTOPSY | UNITED-STATES | MEDICINE, GENERAL & INTERNAL | EPIDEMIOLOGIC TRANSITION | PUBLIC-HEALTH | ROAD TRAFFIC INJURIES | DEVELOPMENT ASSISTANCE | 188 COUNTRIES | CLIMATE-CHANGE | INTEGRATED APPROACH | CIVIL REGISTRATION | Mortality, Premature | Global Health | Communicable Diseases - epidemiology | Humans | Mortality - trends | Life Expectancy - trends | Cause of Death | Medical research | Life expectancy | Analysis | Mortality | Questions and answers | Medicine, Experimental | Health aspects | Studies | Global health | Life span | Epidemiology | Public health | Samfunnsmedisin, sosialmedisin: 801 | Community medicine, Social medicine: 801 | VDP | Helsefag: 800 | Health sciences: 800 | Medisinske Fag: 700 | Medical disciplines: 700 | Clinical Medicine | Hälsa och välfärd | Medical and Health Sciences | Klinisk medicin | Medicin och hälsovetenskap | Health and Welfare | Nutrition and Disease | HNE Nutrition and Disease | Chair Nutrition and Disease | Voeding en Ziekte | VLAG | HNE Voeding en Ziekte
Internal Medicine | VERBAL AUTOPSY | UNITED-STATES | MEDICINE, GENERAL & INTERNAL | EPIDEMIOLOGIC TRANSITION | PUBLIC-HEALTH | ROAD TRAFFIC INJURIES | DEVELOPMENT ASSISTANCE | 188 COUNTRIES | CLIMATE-CHANGE | INTEGRATED APPROACH | CIVIL REGISTRATION | Mortality, Premature | Global Health | Communicable Diseases - epidemiology | Humans | Mortality - trends | Life Expectancy - trends | Cause of Death | Medical research | Life expectancy | Analysis | Mortality | Questions and answers | Medicine, Experimental | Health aspects | Studies | Global health | Life span | Epidemiology | Public health | Samfunnsmedisin, sosialmedisin: 801 | Community medicine, Social medicine: 801 | VDP | Helsefag: 800 | Health sciences: 800 | Medisinske Fag: 700 | Medical disciplines: 700 | Clinical Medicine | Hälsa och välfärd | Medical and Health Sciences | Klinisk medicin | Medicin och hälsovetenskap | Health and Welfare | Nutrition and Disease | HNE Nutrition and Disease | Chair Nutrition and Disease | Voeding en Ziekte | VLAG | HNE Voeding en Ziekte
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8. Global, regional, and national comparative risk assessment of 79 behavioural, environmental and occupational, and metabolic risks or clusters of risks, 1990–2015: a systematic analysis for the Global Burden of Disease Study 2015
The Lancet, ISSN 0140-6736, 10/2016, Volume 388, Issue 10053, pp. 1659 - 1724
Summary Background The Global Burden of Diseases, Injuries, and Risk Factors Study 2015 provides an up-to-date synthesis of the evidence for risk factor...
Internal Medicine | MEDICINE, GENERAL & INTERNAL | COST-EFFECTIVENESS | REDUCED MORTALITY | AIR-POLLUTION | CARDIOVASCULAR-DISEASE | BODY-MASS INDEX | ISCHEMIC-HEART-DISEASE | CERVICAL-CANCER | ALCOHOL-CONSUMPTION | URINARY SODIUM | BLOOD-PRESSURE | Cholesterol - blood | Humans | Cost of Illness | Life Expectancy | Substance-Related Disorders - epidemiology | Malnutrition - epidemiology | Risk-Taking | Disabled Persons - statistics & numerical data | Sodium, Dietary - administration & dosage | Hypertension - epidemiology | Environmental Exposure - statistics & numerical data | Occupational Exposure - statistics & numerical data | Body Mass Index | Quality-Adjusted Life Years | Global Health | Risk Assessment | Risk Factors | Biomarkers - blood | Middle East - epidemiology | Unsafe Sex - statistics & numerical data | Alcohol Drinking - epidemiology | Smoking - epidemiology | Africa, Northern - epidemiology | Blood Glucose - metabolism | Africa South of the Sahara - epidemiology | Air Pollution, Indoor - statistics & numerical data | Medicine, Experimental | Medical research | Comparative analysis | Risk assessment | Environmental assessment | Global health | Disease | Mortality | Regional analysis | Epidemiology | Ecological risk assessment | Comparative studies | Trends | Risk exposure | Health risk assessment | Public health | Index Medicus | Abridged Index Medicus | Community medicine, Social medicine: 801 | Health sciences: 800 | VDP | Medical disciplines: 700 | Clinical Medicine | Hälsa och välfärd | Medical and Health Sciences | Klinisk medicin | Medicin och hälsovetenskap | Health and Welfare | Nutrition and Disease | HNE Nutrition and Disease | Chair Nutrition and Disease | Voeding en Ziekte | VLAG | HNE Voeding en Ziekte
Internal Medicine | MEDICINE, GENERAL & INTERNAL | COST-EFFECTIVENESS | REDUCED MORTALITY | AIR-POLLUTION | CARDIOVASCULAR-DISEASE | BODY-MASS INDEX | ISCHEMIC-HEART-DISEASE | CERVICAL-CANCER | ALCOHOL-CONSUMPTION | URINARY SODIUM | BLOOD-PRESSURE | Cholesterol - blood | Humans | Cost of Illness | Life Expectancy | Substance-Related Disorders - epidemiology | Malnutrition - epidemiology | Risk-Taking | Disabled Persons - statistics & numerical data | Sodium, Dietary - administration & dosage | Hypertension - epidemiology | Environmental Exposure - statistics & numerical data | Occupational Exposure - statistics & numerical data | Body Mass Index | Quality-Adjusted Life Years | Global Health | Risk Assessment | Risk Factors | Biomarkers - blood | Middle East - epidemiology | Unsafe Sex - statistics & numerical data | Alcohol Drinking - epidemiology | Smoking - epidemiology | Africa, Northern - epidemiology | Blood Glucose - metabolism | Africa South of the Sahara - epidemiology | Air Pollution, Indoor - statistics & numerical data | Medicine, Experimental | Medical research | Comparative analysis | Risk assessment | Environmental assessment | Global health | Disease | Mortality | Regional analysis | Epidemiology | Ecological risk assessment | Comparative studies | Trends | Risk exposure | Health risk assessment | Public health | Index Medicus | Abridged Index Medicus | Community medicine, Social medicine: 801 | Health sciences: 800 | VDP | Medical disciplines: 700 | Clinical Medicine | Hälsa och välfärd | Medical and Health Sciences | Klinisk medicin | Medicin och hälsovetenskap | Health and Welfare | Nutrition and Disease | HNE Nutrition and Disease | Chair Nutrition and Disease | Voeding en Ziekte | VLAG | HNE Voeding en Ziekte
Journal Article |
So this is from Charles C. Pinter’s “A Book of Abstract Algebra”- specifically, it’s from the second chapter on permutations. The question is:
[Prove that] Let $ \alpha_1$ and $ \alpha_2$ be cycles of the same length. Let $ \beta_1$ and $ \beta_2$ be cycles of the same length. Let $ \alpha_1$ and $ \beta_1$ be disjoint, and let $ \alpha_2$ and $ \beta_2$ be disjoint. There is a permutation, $ \pi\in S_n$ , such that $ \alpha_1\beta_1=\pi\alpha_2\beta_2\pi^{-1}$ .
That’s the question.
But this is actually the last of a five part question, sooooo here are the previous parts – the parts build on each other and it looks like part 4 (which can be found near the bottom) is of greatest relevance but I can’t quite see how to use it here.
Part 1 was:
Let $ \alpha=(a_1,…,a_s)$ be a cycle and let $ \pi$ be a permutation in $ S_n$ . Then $ \pi\alpha\pi^{-1}$ is the cycle $ (\pi(\alpha_1),…,\pi(\alpha_s))$ .
(a solution can be found here Proof for conjugate cycles)
Part 2 was:
Conclude from part 1: Any two cycles of the same length are conjugates of each other.
The idea here is not too hard to formulate and some formalization is given in a hint which suggests defining a permutation, $ \pi$ , as sending any element not in either cycle to itself, sending all of the elements of one of the cycles, denoted by
A, to the other, denoted by B. Finally, define $ \pi$ over B-A as some bijection between B-A and A-B. Now, it is easy to show that $ \pi$ is a permutation and by part 1, that’s enough.
Parts 3 and 4 are straightforward.
Part 3:
If $ \alpha$ and $ \beta$ are disjoint cycles, then $ \pi\alpha\pi^{-1}$ and $ \pi\beta\pi^{-1}$ are disjoint cycles. (this follows directly from part 1)
And
Part 4: Let $ \sigma$ be a product $ \alpha_1…\alpha_t$ of t disjoint cycles of lengths $ l_1,…,l_t$ , respectively. Then $ \pi\sigma\pi^{-1}$ is also a product of t disjoint cycles of lengths $ l_1,…,l_t$ . (an easy generalization of part 3)
In the fifth part, I see how one can find a $ \pi$ such that $ \alpha_1=\pi\alpha_2\pi^{-1}$ . I also see how one can find a $ \pi$ such that $ \beta_1=\pi\beta_2\pi^{-1}$ (we can get this via an immediate application of part 2). What I can’t see is how we can ensure that these two $ \pi$ ‘s will be the same (which is what the question seems to be asking). I also can’t see how to use Part 4 here as, on its own, it doesn’t seem enough- it looks to me like we would need, in addition, to show that the act of taking a conjugate could be made ‘onto’ in a sense (and I don’t get how to do that).
Any help appreciated. |
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