text:
# Does the classification of reductive groups follow from that of semisimple groups?

I had a question for anyone familiar with the proofs of the classification of reductive groups. I skipped most of the details of classification when I originally learned linear algebraic groups, and now I'm trying to go back and fill the gaps in my knowledge.

Let $G$ be a connected algebraic group over an algebraically closed field. If $G$ is reductive, then $G$ is classified up to isomorphism by its root datum, and conversely every root datum belongs to some connected, reductive group.

If $G$ is semisimple, then $G$ is classified up to isomorphism by its root system and its fundamental group, and conversely every pair of a root system and a subgroup of the weight lattice modulo the root lattice belongs to some semisimple group.

Taking the classification theorem for semisimple groups for granted (say existence and uniqueness), is it possible to deduce the classification theorem for connected, reductive groups? Or does the classification theorem for reductive groups need to be done again from scratch.

• Yes. For any field $k$ and connected reductive $k$-group $G$ there is the canonical isomorphism $(Z \times G')/\mu \simeq G$ where $G'$ is the connected semisimple derived group, $Z$ is the maximal central torus, and $\mu = Z \cap G'$ is a central $k$-subgroup scheme of $G'$ (and if $H$ is a connected semisimple $k$-group, $\mu \subset H$ is a central $k$-subgroup scheme, and $T$ is a $k$-torus equipped with an inclusion of $\mu$ then $G :=(T \times G')/\mu$ is connected reductive with derived group $G'$, maximal central torus $T$, etc.) – nfdc23 Dec 13 '16 at 1:56
• @D_S: As nfdc23 indicates, the answer to your basic question is 'yes', but already over an algebraically closed field it's a bit tricky to state a comprehensive theorem. For example, even when $G$ is connected and simple (as an algebraic group), it isn't always determined up to isomorphism by its root system and fundamental group (as you assert and as I loosely formulated it in the first printing of my 1975 book). I'll try to write a detailed answer if I have time later today, including reductive groups in the mix. – Jim Humphreys Dec 14 '16 at 15:41
• @JimHumphreys, your answer is very nice, but it's not clear to me how it indicates that a connected simple group over an algebraically closed field "isn't always determined up to isomorphism by its root system and fundamental group". (Of course isogenies are relevant, but change the fundamental group, at least infinitesimally.) Could you say more? – LSpice Dec 23 '19 at 23:32
• @LSpice: Sorry for the delay in responding, but I had to search awhile to find older printings of my book GTM 21. In Theorem 32.1, not having yet embraced the idea of "root datum", I gave too superficial an account of isomorphism between "simple" algebraic groups. In the revised printing I added the obvious exception, the group of type D$_\ell$ with even rank $\ell\geq 6$. (What is actually proved here is the equivalent Theorem' following Chevalley which involves extending a map. A complicated but concrete method compared to Takeuchi, later Steinberg.) – Jim Humphreys Dec 28 '19 at 20:36
• Oh, right! I missed that we are considering the fundamental group abstractly, not with a specified embedding in the weight-modulo-root group, so that we can't distinguish the three copies of $\operatorname C_2$ inside $\operatorname W(\mathsf D_{2n})$, even though two of them really do give non-isomorphic groups in the non-triality case. Thanks! – LSpice Dec 28 '19 at 20:44

As indicated in the comments, there is no need to redo the entire classification argument when passing from "semisimple" to "reductive". But it's useful to recall some of the history. The emphasis in the Chevalley seminar 1956-58 was on achieving a uniform classification of semisimple algebraic groups over an algebraically closed field of arbitrary characteristic. For this a slightly more general treatment of isogenies between such groups is more natural. (Chevalley discovered for example that the simple groups of types $B_\ell$ and $C_\ell$ are isogenous in characteristic 2 while their groups of rational points are isomorphic even though the underlying algebraic groups are not).

A little later, Demazure and Grothendieck translated most of this into the more flexible language of group schemes in SGA3, while Borel and Tits by 1965 expanded the framework by defining reductive algebraic groups over an arbitrary field. (For such groups Tits achieved a classification, modulo some later reformulation of his theorem stated in the proceedings of the 1965 Boulder AMS Institute. The main unsolved problem is to classify the $k$-anisotropic groups, which vary a lot for different fields of definition $k$.)

The Chevalley seminar and the other sources in French are available online from numdam, but note that SGA3 has been re-edited in recent years while a corrected typeset version of the Chevalley seminar was published in 2005 by Springer. (In my 1975 textbook, I mostly followed the Chevalley seminar; but when the characteristic is not 2 or 3, my 1966 thesis showed that it's also possible to rely more on the Lie algebra as in the classical characteristic 0 case.)

Variants were found by M. Takeuchi and T.A. Springer, using for example ideas of Serre and Steinberg about generators and relations for the groups. But in spite of the differences in the published approaches, all require a lot of detail about the internal structure of simple algebraic groups (those with no proper closed normal subgroups): Bruhat decomposition, generation by tori along with root subgroups. A key conclusion is that two such algebraic groups are isomorphic precisely when they have the same root system (or Dynkin diagram) and the same fundamental group, except in type $D_\ell$ with $\ell >2$ even, when you have to distinguish the half-spin and special orthogonal groups. (Note too that the work of Tits gave a more precise picture of the internal structure: when $G$ is a simple algebraic group, its only proper normal subgroups are those contained in the finite center.)

Using the methods of Chevalley (1955), one further shows the existence of all possible types of simple algebraic groups. The classification of possible semisimple groups is then a routine but slightly messy exercise: start with a product of simply connected simple groups, then factor out a subgroup of the (finite) center. There may be a great many possibilities.

Translating all of this into the language of reductive groups is then a matter of reading between the lines in Springer's textbook: given an isomorphism of root data, one gets an isomorphism of root systems along with a comparison of fundamental groups, etc. Unfortunately, there is no single source in the literature for a truly unified treatment of all these matters, including isogenies. But the core of it all is the study of the Borel/Chevalley structure theory. The transition to reductive groups is needed mainly because these are more natural for induction purposes than the semisimple ones: Levi subgroups of parabolics are reductive but seldom semisimple. However, central tori over an algebraically closed field are fairly innocuous.

• Minor detail: The definition of simple should include "connected" (i.e. no proper normal, closed, connected subgroup). – Tobias Kildetoft Dec 19 '16 at 9:27
• @Tobias: Thanks for adding the essential word "connected" here. The standard example is a special linear group, which usually has a nontrivial finite center (not connected) but is considered "simple" as an algebraic group. This is the definition I used in my 1975 book, though such conventions about the term "simple" always need to be stated explicitly. – Jim Humphreys Dec 19 '16 at 14:47


text:
Let $I = \int_C {\frac{y}{{x^2 + y^2 }}dx} – \frac{x}{{x^2 + y^2 }}dy$ where $C$ is a circle oriented counterclockwise.
〔1〕Evaluate $I$ if $C$ is given by $(x-2016)^2+(y-2016)^2=1$
〔2〕Evaluate $I$ if $C$ is given by $x^2+y^2=1$.

〔1〕By Green’s theorem
$\begin{array}{cl} \Rightarrow I & =\iint_{\Omega}Q_x-P_ydA \\ & =\iint_{\Omega}\underbrace { – \frac{{1 \cdot (x^2 + y^2 ) – x \cdot (2x)}}{{(x^2 + y^2 )^2 }} – \frac{{1 \cdot (x^2 + y^2 ) – y(2y)}}{{(x^2 + y^2 )^2 }}}_0dA \\ & =0 ∎ \end{array}$

〔2〕 由於被積函數包含奇異點 (分母有 0)，因此不滿足 Green’s theorem
$C:\ x^2+y^2=1\Rightarrow \left\{ {\begin{array}{l} x = \cos t \\ y = \sin t \end{array}} \right.,\ 0\leq t\leq 2\pi$
$\begin{array}{cl} \Rightarrow I & =\int_0^{2\pi } {\frac{{\sin t}}{1}( – \sin t)} dt – \frac{{\cos t}}{1}(\cos t)dt \\ & =-\int_0^{2\pi } {dt} \\ & =-2\pi ∎\end{array}$

Find the maximum and minimum values of the function $f(x,y,z)=x^2-y^2$ on the surface $x^2+2y^2+3z^2=1$.

Let $g(x,y,z)=x^2+2y^2+3z^2-1$
Let $\nabla f + \lambda \nabla g = 0$
$\Rightarrow (2x, – 2y,0) + \lambda (2x,4y,6z) = (0,0,0)$
$\begin{array}{l} { \Rightarrow \left\{ {\begin{array}{l} {2x + 2\lambda x = 0} \\ { – 2y + 4\lambda y = 0} \\ {6\lambda z = 0}\end{array}} \right.} \\ {x^2 + 2y^2 + 3z^2 = 1} \end{array}\Rightarrow \left\{ {\begin{array}{l} {x(1 + \lambda ) = 0} \\ {y( – 1 + 2\lambda ) = 0} \\ {\lambda z = 0} \end{array}} \right.\Rightarrow \left\{ {\begin{array}{l} {x = 0} & {{\text{or}}} & {\lambda = – 1} \\ {y = 0} & {{\text{or}}} & {\lambda = \frac{1}{2}} \\ {z = 0} & {{\text{or}}} & {\lambda = 0} \end{array}} \right.$

$\begin{array}{cl} \because & f( \pm 1,0,0) = ( \pm 1)^2 – 0^2 = 1 \\ & f(0, \pm \frac{1}{{\sqrt 2 }},0) = 0^2 – ( \pm \frac{1}{{\sqrt 2 }})^2 = – \frac{1}{2} \\ & f(0,0, \pm \frac{1}{{\sqrt 3 }}) = 0^2 – 0^2 = 0\\ \therefore & \left\{ {\begin{array}{l} {{\text{Max}}{\rm{.}} = 1} \\ {{\text{min}}{\rm{.}} = – \frac{1}{2}}\end{array}} \right. ∎ \end{array}$

Compute $\mathop {\lim }\limits_{x \to \infty } (\sqrt {x + \sqrt {x + \sqrt x } } – \sqrt x )$.

$\begin{array}{cl} \mathop {\lim }\limits_{x \to \infty } (\sqrt {x + \sqrt {x + \sqrt x } } – \sqrt x ) &= \mathop {\lim }\limits_{x \to \infty } (\sqrt {x + \sqrt {x + \sqrt x } } – \sqrt x ) \cdot \frac{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{x + \sqrt {x + \sqrt x } – x}}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {x + \sqrt x } /\sqrt x }}{{(\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x )/\sqrt x }} \\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + \sqrt {\frac{1}{x}} } }}{{\sqrt {1 + \sqrt {\frac{1}{x} + \frac{1}{{x^{3/2} }}} } + 1}} = \frac{1}{2} ∎ \end{array}$

For what positive $x$ does the following series converge?
$\sum\limits_{n = 1}^\infty {(\sqrt[n]{x} – 1)}$

For $x=1$, $\sum\limits_{n = 1}^\infty {(\sqrt[x]{x} – 1)} = 0$ converges.

For $x\ne 1$
$\begin{array}{cl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{x^{\frac{1}{n}} – 1}}{{\frac{1}{n}}}\mathop = \limits^{{\text{Let }}t = \frac{1}{n}} \mathop {\lim }\limits_{t \to 0^ + } \frac{{x^t – 1}}{t}\mathop = \limits^{\text{L}} \mathop {\lim }\limits_{t \to 0^ + } \frac{{x^t \ln x}}{1} = \underbrace {\ln x}_{ \in \mathbb{R}} \ne 0\ (x\ne 1) \\ & \text{and }\sum\limits_{n = 1}^\infty {\frac{1}{n}}\text{ diverges. (by }p\text{-series test)} \\ \therefore & \sum\limits_{n = 1}^\infty {(x^{\frac{1}{n}} – 1)}\text{ diverges. (by limit comparison test)} \end{array}$

the series converges only for $x=1 ∎$

Let $B = \left\{ {(x,y,z) \in \mathbb{R}^3 :x^2 + y^2 + z^2 \le 1} \right\}$. Evaluate the integral
$\iiint_{B}\frac{{x^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV$.

◎ 已知積分區域為一球，具有很好的對稱性
$\begin{array}{cl} \iiint_{B}\frac{{x^4 + 2y^4 }}{{z^4 + 4y^4 + x^4 }}dV & =\iiint_B \frac{{x^4 + 2y^4 + z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }} – \frac{{z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV \\ & =\iiint_BdV-\iiint_B \frac{{z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV \end{array}$

$\begin{array}{cl} \iiint_B \frac{{x^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV & =\iiint_B 1-\frac{{z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV \\ & =\iiint_B dV-\iiint_B \frac{{z^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV \end{array}$
$\Rightarrow 2\iiint_B \frac{{x^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV=\frac{4}{3}\pi$
$\Rightarrow \iiint_B\frac{{x^4 + 2y^4 }}{{x^4 + 4y^4 + z^4 }}dV=\frac{2}{3}\pi ∎$

The $n$-th derivative of $\frac{1}{x^{2016}-1}$ has the form $\frac{P_n}{(x^{2016}-1)^{n+1}}$ where $P_n(x)$ is a polynomial. Find $P_n(1)$ for all $n\ge 0$.

$\begin{array}{cl} f'(x) & =\frac{{ – 1}}{{(x^{2016} – 1)^2 }} \cdot 2016x^{2015} \\ & =\frac{{ – 2016x^{2015} }}{{(x^{2016} – 1)^2 }},\ P_1 (x)=-2016x^{2016} \end{array}$

$\because\ f^n (x) = \frac{{P_n (x)}}{{(x^{2016} – 1)^{n + 1} }} = P_n (x)\frac{1}{{(x^{2016} – 1)^{n + 1} }}$
$\begin{array}{cl} \therefore\ f^{n+1}(x) & =P’_n (x) \cdot \frac{1}{{(x^{2016} – 1)^{n + 1} }} + P_n (x) \cdot \frac{{ – (n + 1)}}{{(x^{2016} – 1)^{n + 2} }} \cdot 2016x^{2015} \\ & = \frac{{P’_n (x)(x^{2016} – 1) – P_n (x) \cdot (n + 1) \cdot 2016x^{2015} }}{{(x^{2016} – 1)^{n + 2} }}\end{array}$
$\Rightarrow P_{n + 1} (x) = P’_n (x)(x^{2016} – 1) – P_n (x) \cdot (n + 1) \cdot 2016x^{2015}$

$P_1 (1)=-2016$
$\Rightarrow P_{n + 1} (1) = P'(1) \cdot (1^{2016} – 1) – P_n (1) \cdot (n + 1) \cdot 2016$
$\Rightarrow P_2 (1) = ( – 2016) \times ( – 1) \times 2 \times 2016 = 2016^2 \times 2$
$\Rightarrow P_3 (1) = 2016^2 \times 2 \times ( – 1) \times 3 \times 2016 = 2016^3 \times ( – 1) \times 2 \times 3$
$\vdots$
$\Rightarrow P_n (1) = ( – 1)^n \cdot (2016)^n \cdot (n!) ∎$

〔1〕Prove that $\int_0^\infty {(\frac{{\sin x}}{x})^2 } dx = \int_0^\infty {\frac{{\sin x}}{x}} dx$. (10 分)
〔2〕Evaluate the improper integral $\int_0^\infty {\frac{{\sin x}}{x}} dx$. (10 分)

〔1〕
$\begin{array}{cl} \int_0^\infty {(\frac{{\sin x}}{x})^2 } dx & =\int_0^\infty {x^{ – 2} } \sin ^2 xdx \\ & =\left. { – x^{ – 1} \sin ^2 x} \right|_0^\infty + \int_0^\infty {x^{ – 1} } \cdot 2\sin x\cos xdx \\ & =\underbrace {\left. { – \frac{{\sin ^2 x}}{x}} \right|_0^\infty }_0 + \int_0^\infty {\frac{{\sin 2x}}{x}} dx\\&=\int_0^\infty {\frac{{\sin 2x}}{x}} dx\ (\text{Let }u = 2x \Rightarrow du = 2dx)\\&=\int_0^\infty {\frac{{\sin u}}{{\frac{u}{2}}} \cdot } \frac{{du}}{2} = \int_0^\infty {\frac{{\sin x}}{x}} dx ∎\end{array}$

〔2〕
$\begin{array}{cl} \int_0^\infty {\frac{{\sin x}}{x}} dx & =\int_0^\infty {\frac{{\sin x}}{x}} (\int_0^\infty {e^{ – u} } du)dx \\ & =\int_0^\infty {\sin x} \int_0^\infty {e^{ – u} } \frac{{du}}{x}dx\ (\text{Let }u = xy \Rightarrow du = xdy) \\ & =\int_0^\infty {\int_0^\infty {(\sin x)e^{ – xy} } } dydx\\&=\int_0^\infty {\int_0^\infty {e^{ – yx} \sin x} } dxdy \end{array}$

$\begin{array}{cl} \because\ \int_0^\infty {e^{ – yx} \sin x} dx & =\left. { – e^{ – yx} \cos x} \right|_{x = 0}^{x = \infty } – \left. {ye^{ – yx} \sin x}\right|_{x = 0}^{x = \infty } – \int_0^\infty {y^2 e^{ – yx} \sin x} dx \\ & =0 – \underbrace {( – e^0 \cos 0)}_{ – 1} – y^2 \int_0^\infty {e^{ – yx} \sin x} dx \\ & =1 – y^2 \int_0^\infty {e^{ – yx} \sin x} dx \end{array}$
$\begin{array}{cl} & \Rightarrow (1 + y^2 )\int_0^\infty {e^{ – yx} \sin x} dx = 1 \Rightarrow \int_0^\infty {e^{ – yx} \sin x} dx = \frac{1}{{1 + y^2 }} \\ \therefore & \int_0^\infty {\frac{{\sin x}}{x}} dx=\int_0^\infty {\frac{1}{{1 + y^2 }}} dy = \left. {\tan ^{ – 1} y} \right|_{y = 0}^{y = \infty } = \frac{\pi }{2} – 0 = \frac{\pi }{2} ∎\end{array}$

For each continuous function $f:[0,1] \to\mathbb{R}$, let $I(f) = \int_0^1 {xf(x)(x – f(x))} dx$. Find the maximum value of $I(f)$ over all such function $f$.

Let $f_0 (x)$ be the function such that $I(x_0)$ reaches its max.
Let $\eta :\ [0,1]\to\mathbb{R}$ be continuous such that $\left\{ {\begin{array}{l} {\eta (0) = 0} \\ {\eta (1) = 0} \end{array}} \right.$
Let $f_\alpha (x) = f_0 (x) + \alpha \eta (x)$
$\because\ I(f_0)$ reaches its max.
$\therefore\ \frac{d}{{d\alpha }}\left. {I(f_\alpha )} \right|_{\alpha = 0} = 0$
$\begin{array}{cl} I(f_\alpha) & =\int_0^1 {xf_\alpha (x) \cdot (x – f_\alpha (x))} dx \\ & =\int_0^1 {x^2 f_\alpha (x) – x\left[ {f_\alpha (x)} \right]^2 } dx\end{array}$
$\begin{array}{cl} \Rightarrow \frac{d}{{d\alpha }}I(f_\alpha ) & =\frac{d}{{d\alpha }}\int_0^1 {x^2 f_\alpha (x) – x\left[ {f_\alpha (x)} \right]^2 } dx \\ & =\int_0^1 {x^2 \frac{d}{{d\alpha }}f_\alpha (x) – x \cdot 2f_\alpha (x) \cdot } \frac{d}{{d\alpha }}f(x)dx \\ & =\int_0^1 {x^2 \eta (x) – 2x(f_\alpha (x) + \alpha \eta (x))} \cdot \eta (x)dx \end{array}$
$\begin{array}{cl} \underbrace {\frac{d}{{d\alpha }}\left. {I(f_\alpha )} \right|_{\alpha = 0} }_0 & =\int_0^1 {x^2 \eta (x) – 2xf_0 (x)\eta (x)} dx \\ & =\int_0^1 {\eta (x)(x^2 – 2xf_0 (x))} dx\end{array}$
$\because\ \eta$ is arbitrary.
$\therefore\ x^2 – 2xf_0 (x) = x(x – 2f_0 (x)) = 0 \Rightarrow f_0 (x) = \frac{\pi }{2}$
$\begin{array}{cl} \text{So max. }I(f) & =I(\frac{x}{2}) \\ & =\int_0^1 {x \cdot \frac{x}{2} \cdot (x – \frac{x}{2})} dx \\ & =\int_0^1 {\frac{{x^2 }}{2} \cdot \frac{x}{2}} dx = \frac{1}{4}\left[ {\frac{{x^4 }}{4}} \right]_{x = 0}^{x = 1} = \frac{1}{{16}} ∎\end{array}$

Evaluate $\int_0^\infty {(x – \frac{{x^3 }}{2} + \frac{{x^5 }}{{2 \cdot 4}} – \frac{{x^7 }}{{2 \cdot 4 \cdot 6}} + \cdots )(1 + \frac{{x^2 }}{{2^2 }} + \frac{{x^4 }}{{2^2 \cdot 4^2 }} + \frac{{x^6 }}{{2^2 \cdot 4^2 \cdot 6^2 }} + \cdots )} dx$.

Let $f(x)={(x – \frac{{x^3 }}{2} + \frac{{x^5 }}{{2 \cdot 4}} – \frac{{x^7 }}{{2 \cdot 4 \cdot 6}} + \cdots )(1 + \frac{{x^2 }}{{2^2 }} + \frac{{x^4 }}{{2^2 \cdot 4^2 }} + \frac{{x^6 }}{{2^2 \cdot 4^2 \cdot 6^2 }} + \cdots )}$

$\begin{array}{cl} x – \frac{{x^3 }}{2} + \frac{{x^5 }}{{2 \cdot 4}} – \frac{{x^7 }}{{2 \cdot 4 \cdot 6}} + \cdots & =x(1 – \frac{{x^2 }}{2} + \frac{{x^4 }}{{2 \cdot 4}} + \cdots ) \\ & =x(1 – \frac{{(x^2 )^1 }}{{2^1 }} + \frac{{(x^2 )^2 }}{{2^2 \cdot 2!}} – \frac{{(x^2 )^3 }}{{2^3 \cdot 3!}} + \cdots ) \\ & =x(1 + ( – 1)^1 (\frac{{x^2 }}{2})^1 + \frac{{( – 1)^2 (\frac{{x^2 }}{2})^2 }}{{2^2 }} + \cdots )\\&=x(1 + \frac{{( – \frac{{x^2 }}{2})^1 }}{{1!}} + \frac{{( – \frac{{x^2 }}{2})^2 }}{{2!}} + \frac{{( – \frac{{x^2 }}{2})^3 }}{{3!}} + \cdots )\\&=xe^{ – \frac{{x^2 }}{2}} \end{array}$

$1 + \frac{{x^2 }}{{2^2 }} + \frac{{x^4 }}{{2^2 \cdot 4^2 }} + \frac{{x^6 }}{{2^2 \cdot 4^2 \cdot 6^2 }} + \cdots$
$= 1 + \frac{{x^2 }}{{2^2 }} + \frac{{x^4 }}{{2^2 \cdot 1^2 \cdot 2^2 }} + \frac{{x^6 }}{{2^3 \cdot 1^2 \cdot 2^2 \cdot 3^2 }} + \cdots$
$= 1 + (\frac{x}{2})^2 + \frac{{(\frac{x}{2})^4 }}{{(2!)^2 }} + \frac{{(\frac{x}{2})^6 }}{{(3!)^2 }} + \cdots = \sum\limits_{n = 0}^\infty {\frac{{(\frac{x}{2})^{2n} }}{{(n!)^2 }}}$

$\begin{array}{cl} \int_0^\infty f(x)dx & =\int_0^\infty {xe^{ – \frac{{x^2 }}{2}} } \sum\limits_{n = 0}^\infty {\frac{{x^{2n} }}{{2^{2n} (n!)^2 }}} dx \\ & =\sum\limits_{n = 0}^\infty {\int_0^\infty {xe^{ – \frac{{x^2 }}{2}} \cdot \frac{{x^{2n} }}{{2^{2n} (n!)^2 }}} } dx \\ & =\sum\limits_{n = 0}^\infty {\frac{1}{{2^{2n} (n!)^2 }}\int_0^\infty {x^{2n + 1} \cdot e^{ – \frac{{x^2 }}{2}} } } dx \end{array}$

$\begin{array}{cl} \int_0^\infty {x^{2n + 1} \cdot e^{ – \frac{{x^2 }}{2}} } dx & =\int_0^\infty {x^{2n} e^{ – \frac{{x^2 }}{2}} \cdot x} dx\ (\text{Let }u = \frac{{x^2 }}{2} \Rightarrow du = 2xdx) \\ & =\int_0^\infty {2^n u^n e^{ – u} } du \\ & =2^n \underbrace {\int_0^\infty {u^n e^{ – u} } du}_{f(n)}\\&=2^n \left[ {\left. { – u^n e^{ – u} } \right|_{u = 0}^{u = \infty } + \int_0^\infty {nu^{n – 1} e^{ – u} } du} \right]\\&=2^n \cdot n\underbrace {\int_0^\infty {u^{n – 1} e^{ – u} } du}_{f(n – 1)} \end{array}$

$\Rightarrow\ f(1) = 1 \cdot f(0) = 1,\ f(2) = 2 \cdot f(1) = 2 \cdot 1 = 2!,\ f(3) = 3 \cdot f(2) = 3!$
$\cdots,\ f(n) = n \cdot f(n – 1) = n!$

$\begin{array}{cl} \Rightarrow \int_0^\infty f(x)dx & =\sum\limits_{n = 0}^\infty {\frac{1}{{2^{2n} (n!)^2 }}} \underbrace {\int_0^\infty {x^{2n + 1} \cdot e^{ – \frac{{x^2 }}{2}} } dx}_{2^n n!} \\ & =\sum\limits_{n = 0}^\infty {\frac{{2^n n!}}{{2^{2n} (n!)^2 }}} \\ & =\sum\limits_{n = 0}^\infty {\frac{1}{{2^n (n!)}}} = \sum\limits_{n = 0}^\infty {\frac{{(\frac{1}{2})^n }}{{n!}}} = e^{\frac{1}{2}} = \sqrt e ∎\end{array}$

format: plain


text:
## What mathematical principle or law describes this?

I don't know if this is the correct forum but...
Suppose you had a difference of fractions $1/4 - 1/6$
You could get the LCD by factoring each denominator. For 4 you would have (2,2) for 6 you would have (2,3). This gives 12 as the LCD.
You would then multiply the numerator of $1/4$ by 3, which is the factorization of 6 (2,3) not including the factor that is a factor of 4, namely 2. This gives $3/12$
This prime number (3) is the same as dividing the LCD by the original denominator: $12/4 = 3$
I hope this makes sense.
Is there a mathematical principle or law that describes this?
Thanks!


text:
This content will become publicly available on January 1, 2023

Assessing Support for DNS-over-TCP in the Wild
While the DNS protocol encompasses both UDP and TCP as its underlying transport, UDP is commonly used in practice. At the same time, increasingly large DNS responses and concerns over amplification denial of service attacks have heightened interest in conducting DNS interactions over TCP. This paper surveys the support for DNS-over-TCP in the deployed DNS infrastructure from several angles. First, we assess resolvers responsible for over 66.2% of the external DNS queries that arrive at a major content delivery network (CDN). We find that 2.7% to 4.8% of the resolvers, contributing around 1.1% to 4.4% of all queries arriving at the CDN from the resolvers we study, do not properly fallback to TCP when instructed by authoritative DNS servers. Should a content provider decide to employ TCP-fallback as the means of switching to DNS-over-TCP, it faces the corresponding loss of its customers. Second, we assess authoritative DNS servers (ADNS) for over 10M domains and many CDNs and find some ADNS, serving some popular websites and a number of CDNs, that do not support DNS-over-TCP. These ADNS would deny service to (RFC-compliant) resolvers that choose to switch to TCP-only interactions. Third, we study the TCP connection reuse behavior of DNS actors more »
Authors:
Editors:
; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10320040
Journal Name:
Lecture notes in computer science
Volume:
13210
ISSN:
1611-3349
2. Authoritative DNS servers are susceptible to being leveraged in denial of service attacks in which the attacker sends DNS queries while masquerading as a victim---and hence causing the DNS server to send the responses to the victim. This reflection off innocent DNS servers hides the attackers identity and often allows the attackers to amplify their traffic by employing small requests to elicit large responses. Several challenge-response techniques have been proposed to establish a requester's identity before sending a full answer. However, none of these are practical in that they do not work in the face of resolver pools''---or groups ofmore »


text:
# Drawing fluid pathlines, streaklines, streamlines using TikZ

I am trying to do a sketch of fluid pathlines, streaklines, and streamlines using TikZ. The code

\documentclass{minimal}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\fill [fill=blue!25] (0.0, 0.0) -- (0.0, 2.0) -- (6.0, 2.0) -- (6.0, 1.0) to[out=180, in=0] (0.0, 0.0);
\filldraw [semithick, fill=gray!50] (0.0, 2.0) rectangle (6.0, 2.1);
\filldraw [semithick, fill=gray!50] (0.0, -0.1) -- (0.0, 0.0) to[out=0, in=180] (6.0, 1.0) -- (6.0, 0.9) to[out=180, in=0] (0.0, -0.1);
%
\draw [thick] (0.25, 0.25) to[out=0, in=180] node [pos=0.1, above] {$t_{1}$} node [pos=0.5, above] {$t_{2}$} node [pos=0.9, above] {$t_{3}$} (5.75, 1.125);
\end{tikzpicture}
\end{document}

generates the following image:

I would like there to be closed circles on the black line at the locations under the labels $t_{1}$, $t_{2}$, and $t_{3}$. Can someone suggest how to do this? Thanks.

-
I would go with hobby, definitely. – Claudio Fiandrino Jan 29 '14 at 19:05

You can also just add extra nodes as you go, one for the filled circles, and one for the label:

## Code:

\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\fill [fill=blue!25] (0.0, 0.0) -- (0.0, 2.0) -- (6.0, 2.0) -- (6.0, 1.0) to[out=180, in=0] (0.0, 0.0);
\filldraw [semithick, fill=gray!50] (0.0, 2.0) rectangle (6.0, 2.1);
\filldraw [semithick, fill=gray!50] (0.0, -0.1) -- (0.0, 0.0) to[out=0, in=180] (6.0, 1.0) -- (6.0, 0.9) to[out=180, in=0] (0.0, -0.1);
%
\draw [thick] (0.25, 0.25) to[out=0, in=180]
node [pos=0.1, circle, fill=red, inner sep=2pt] {}
node [pos=0.1, above] {$t_{1}$}
node [pos=0.5, circle, fill=blue, inner sep=2pt] {}
node [pos=0.5, above] {$t_{2}$}
node [pos=0.9, circle, fill=orange, inner sep=2pt] {}
node [pos=0.9, above] {$t_{3}$} (5.75, 1.125);
\end{tikzpicture}
\end{document}

-
Thanks, Peter. I did not realize that I could create two nodes in the same physical location, such as you showed. This worked well. – Josh Jan 29 '14 at 23:14
@Josh: Glad it worked for you. Please see How do you accept an answer?. – Peter Grill Jan 29 '14 at 23:20

Specify coordinates inline rather than nodes. Then you can build up any constructs you require later on, like so:

\documentclass{minimal}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\fill [fill=blue!25] (0.0, 0.0) -- (0.0, 2.0) -- (6.0, 2.0) -- (6.0, 1.0) to[out=180, in=0] (0.0, 0.0);
\filldraw [semithick, fill=gray!50] (0.0, 2.0) rectangle (6.0, 2.1);
\filldraw [semithick, fill=gray!50] (0.0, -0.1) -- (0.0, 0.0)
to[out=0, in=180] (6.0, 1.0) -- (6.0, 0.9) to[out=180, in=0] (0.0, -0.1);
%
\draw [thick] (0.25, 0.25) to[out=0, in=180] coordinate[pos=0.1] (1)
coordinate[pos=0.5] (2)  coordinate[pos=0.9] (3) (5.75, 1.125);
\foreach \coord in {1,2,3}{%
\node[above] at (\coord) {$t_{\coord}$};
\fill (\coord) circle (2pt);
}
\end{tikzpicture}
\end{document}

Output is:

-


text:
# Volcano Watch — What in the world is the Hilo Ridge?

Release Date:

Okay, all you Big Island residents, where is the Hilo Ridge? It stands nearly 3,000 m (10,000 ft) above its surroundings, is more than 50 km (30 mi) long, and is one of the most prominent landmarks in the State. Its crest is only 30 km (20 mi) from Hilo--but northeast of town.

Okay, all you Big Island residents, where is the Hilo Ridge? It stands nearly 3,000 m (10,000 ft) above its surroundings, is more than 50 km (30 mi) long, and is one of the most prominent landmarks in the State. Its crest is only 30 km (20 mi) from Hilo--but northeast of town. Yes, the Hilo Ridge is a feature of the seafloor, out of sight under the waves but nonetheless an important piece in the puzzle of how the Big Island was made.

The Hilo Ridge starts at the east edge of a gentle shelf or platform about 400 m (1,300 ft) deep off the Hamakua coast, about 20 km (12 mi) east of Honomu. From the platform, the ridge extends eastward in a slightly sinuous trace, finally ending at a depth of about 5,000 m (16,000 ft) some 65 km (37 mi) east-northeast of Hilo.

Until recently, earth scientists thought that the Hilo Ridge was built by eruptions along a rift zone of Mauna Kea, much like ridges along the rift zones that radiate from the summits of Kīlauea and Mauna Loa. Just as Kīlauea's east rift zone extends far beyond the shoreline as the Puna Ridge, so the east rift zone of Mauna Kea was thought to have reached far out to sea as the Hilo Ridge. The on-land part of Mauna Kea's east rift zone today is barely recognizable, because younger flows have covered those erupted along the rift zone. These younger flows, however, did not reach far eastward beyond the coast, and so the submarine rift zone remained a prominent topographic ridge.

In a startling development at a meeting of the American Geophysical Union last month, researchers from the University of Washington (including a former HVO staff member, Robin Holcomb) proposed a new hypothesis for the Hilo Ridge. They suggest that the ridge was built along a rift zone of the older Kohala Volcano, not of Mauna Kea. In their view, Mauna Kea buries and obscures the Kohala rift zone on the Big Island itself but does not reach far enough eastward to cover the Hilo Ridge. What is the evidence for such a radical departure from past interpretations?

The University of Washington group determined the chemical compositions of rocks recently dredged from depths of 5,000 m (16,000 ft), 1,100 m (3,600 ft), and 400 m (1,300 ft) along the Hilo Ridge. They compared their results with those obtained by previous workers on dredged samples from depths of 1,600-3,300 m (5,200-10,800 ft). The results show a pattern that is inconsistent with the known chemical characteristics for Mauna Kea, particularly in terms of the abundance of certain isotopes of the elements strontium, neodymium, and lead. The data are, however, entirely consistent with the known chemical compositions of Kohala, an older volcano that stopped erupting some 120,000 years ago.

Kohala is a poorly understood volcano. One of the problems is that a large part of the volcano, south of a line between Kukuihaele, Waimea, and Kawaihae, is covered by younger lava flows from Mauna Kea. The Kohala Mountains themselves may be only the remnant of a northwest rift zone of Kohala Volcano that sprouts from a buried caldera southeast of Waimea. It is not difficult to imagine that Kohala's southeast rift zone curves gently southward from the buried caldera and eventually becomes the Hilo Ridge, just as Mauna Loa's southwest rift zone and Kīlauea's east rift zone bend away from their respective calderas.

Will this stimulating new view of the Hilo Ridge stand the test of time? As with any new idea in science, it will be closely scrutinized over the next few years as new information is obtained and discussed. The new idea certainly passes the laugh test, but will it pass the fact test?

### Volcano Activity Update

During the past week, there was constant effusion of lava from the vent within Puu Oo. Southerly winds carried the gas emissions from the volcano as far north as Honolulu. The concentration of sulfur dioxide measured in the parking lot of HVO at times exceeded 5 parts per million (ppm), and several visitor facilities within Hawaii Volcanoes National Park were closed.

Lava continued to flow through a network of tubes down to the seacoast where it entered the ocean at two locations - Wahaula and Kamokuna. Several minor breakouts from the tube system produced surface flows. The public is reminded that the ocean entry areas are extremely hazardous, with explosions accompanying frequent collapses of the lava delta. The steam cloud is highly acidic and laced with glass particles.

An earthquake at 5:35 p.m. on December 29 was reported felt by a resident of Pahala. The magnitude 3.8 earthquake was located 11.5 km (7 mi) southeast of the summit of Kīlauea Volcano at a depth of 7 km (4 mi).


text:
# How to present a polynomial interpolation if the first order information of function f(x) is given?

Suppose $f'(x_1),\ f'(x_2),\ f'(x_3)$ are given, how to give a polynomial interpolation $p(x)$ such that $p'(x_1)= f'(x_1),\ p'(x_2)=f'(x_2),\ p'(x_3)=f'(x_3)$? And how to give an error analysis?

• This is called Hermite interpolation, and should be discussed in most textbooks on numerical analysis (e.g., Gautschi's). Note that the polynomial is not uniquely specified unless you also have some function values $f(x_i)$. – Christian Clason Jan 15 '15 at 14:43
• Yes it will be uniquely determined if we set $p(x_1)=f(x_1)$. But it is not straightfoward to apply the Hermite interpolation since we do not know the informatio about $f(x_2)$ and $f(x_3)$? – mathstudy Jan 18 '15 at 3:59
• You didn't say in your questions that you set $p(x_1)=f(x_1)$. – nicoguaro Jan 20 '15 at 19:05

## 1 Answer

The problem boils down to the solution of a linear system a equations. You get one equation for the function value at $f(x_1)$, and three more equations from your derivative equations. Then, a polynomial of degree three is suitable.

A more brute-force, not so clever way is to use computer algebra such as SymPy for a symbolical solution:

from sympy import *

N = 3 # degree of polynomial - for N + 1 'pieces of information', use a polynomial of degree N

known_x = symbols(["x_%i"%i for i in range(N)])  # create symbols for supporting points
known_derivs = symbols(["d_%i"%i for i in range(N)])  # create symbols for supporting points
y1 = Symbol("y_1")  # the single zeroth derivative we know

# create symbols for the independent variable and the coefficients:
x = Symbol("x")
a = [ Symbol("a_%i"%i) for i in range(N+1) ]

# generate polynomial and the derivative
p = sum([ coeff*x**i for i, coeff in enumerate(a) ])
p_prime = diff(p, x, 1)

# define the equations to solve:
eqns = [ p_prime.subs(x, xi) - d for xi, d in zip(known_x, known_derivs) ]  # equation for the first derivative
eqns.append(p.subs(x, known_x[0]) - y1)

# solve for the coefficients and print solution:
solution = solve(eqns, a)
pretty_sol = [ str(coeff) + " = " + str(value) + "\n"
for coeff, value in zip(solution.keys(), solution.values()) ]
print("The coefficients are:\n" + "".join(pretty_sol))

This not very elegant, but seems to work even you added another equation for the derivative (so that could use a polynomial of degree 'N = 4' for the five equations).

• I think you mean "A polynomial of degree 3 is suitable"? – Rahul Jan 21 '15 at 5:05
• Yes, of course. Thank you. Edited in my reply. – AlexE Jan 21 '15 at 6:00
• Thanks, can you be more clear about your comment "The problem boils down to the solution of a linear system a equations"? It seems interesting, can it be possible to give a error analysis? – mathstudy Jan 22 '15 at 3:02


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Nov 2019, 02:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The symbol ∞ represents one of the following functions: addition,

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Senior RC Moderator
Joined: 02 Nov 2016
Posts: 4385
GPA: 3.39
The symbol ∞ represents one of the following functions: addition,  [#permalink]

### Show Tags

18 Sep 2019, 02:32
00:00

Difficulty:

35% (medium)

Question Stats:

71% (01:07) correct 29% (00:48) wrong based on 24 sessions

### HideShow timer Statistics

The symbol $$∞$$ represents one of the following functions: addition, subtraction, multiplication, or division. What is the value of $$2∞2$$?

(1) $$1∞1=1$$

(2) $$2∞2 = 4(1∞1)$$

Source: McGraw-Hill's GMAT (1-36)

_________________
Senior Manager
Joined: 29 Jun 2019
Posts: 466
Re: The symbol ∞ represents one of the following functions: addition,  [#permalink]

### Show Tags

18 Sep 2019, 03:08
Statement1;1×1=1 or 1÷1=1
Not sufficient.
Statement2;2×2=4×(1) or 2+2=4×(1)
The sign would be"×" or"+".
Not sufficient
Option E

Posted from my mobile device
_________________
Always waiting
Intern
Joined: 10 Sep 2019
Posts: 1
Re: The symbol ∞ represents one of the following functions: addition,  [#permalink]

### Show Tags

18 Sep 2019, 06:34
Mohammadmo wrote:
Statement1;1×1=1 or 1÷1=1
Not sufficient.
Statement2;2×2=4×(1) or 2+2=4×(1)
The sign would be"×" or"+".
Not sufficient
Option E

Posted from my mobile device

I think you forgot to consider both options.

If A gives you a possibility of * or / and B gives you a possibility of + or * then combined, they give you only one possibility of *
Director
Joined: 27 May 2012
Posts: 934
The symbol ∞ represents one of the following functions: addition,  [#permalink]

### Show Tags

30 Sep 2019, 11:42
SajjadAhmad wrote:
The symbol $$∞$$ represents one of the following functions: addition, subtraction, multiplication, or division. What is the value of $$2∞2$$?

(1) $$1∞1=1$$

(2) $$2∞2 = 4(1∞1)$$

Source: McGraw-Hill's GMAT (1-36)

(1) $$1∞1=1$$, ∞ could be Multiplication(*) or Division (/) INSUFF
(2) $$2∞2 = 4(1∞1),$$∞ could be Multiplication(*) or Subtraction(-) INSUFF

1+2
∞ can be Multiplication (*) only SUFF
Ans- C

Hi Mohammadmo, SShareef please note statement (2) CANNOT be true for +(addition)
_________________
- Stne
The symbol ∞ represents one of the following functions: addition,   [#permalink] 30 Sep 2019, 11:42
Display posts from previous: Sort by

# The symbol ∞ represents one of the following functions: addition,

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne


text:
# What is the difference between the 2 ways to remove percentages?

Given the value $690$, I want to remove $10$% from that and then remove another $20$% from the resulting value, so as an example, I am doing:

$690 \over {(1 + 0.1 + 0.2})$ = $690 \over {1.3}$ = $530.76.$ Apparently, I was told this is wrong and it should be:

$690 * 0.9 =$ $621 \over {1.2}$ = $517.50$

In the above, the $10$% is non-compounded and the $20$% is compounded. Can someone explain what is the difference between the $2$ and is one way better than the other if I do not know certain values like the start or end value (690 in this case)

If I do $690 \over {1 + 0.111111111111 + 0.2}$, I get closer to $517.50.$

-
There is only one way to remove 10% which I know of. If you have $a$ and you remove 10% of $a$ you get $a-10$%$\times a = 0.9 a$. –  Fabian Apr 5 '12 at 18:41

If you remove $10\%$ from $690$, your new value is $690-690\cdot \frac{10}{100}$. This is $690\left(1-\frac{10}{100}\right)$, or more simply $(690)(0.9)$.

To remove $20\%$ from the new value of $(690)(0.9)$, you go through essentially the same process, and arrive at $(690)(0.9)(0.8)$.

Note that we would have gotten to the same final answer if we removed $20\%$ from $690$, and then removed $10\%$ from the result.

I do not know what you mean by "if I do not know certain values." If we only know the percentages removed, we cannot know the dollar amounts.

But if the initial price is $A$, and you remove $10\%$ and then $20\%$, the same reasoning leads to the final price of $A(0.9)(0.8)$.

So if you know that discounts of $10\%$ and then $20\%$ were applied, and the final price was, say, $576$, then you can determine the original price as follows. Let $A$ be the original price. Then $$A(0.9)(0.8)=576,$$ and therefore $$A=\frac{576}{(0.9)(0.8)}.$$

-
How does the calculation differ if there are more than 2 percentages 2 remove? For example, my starting value is 690 and in the following order I need to remove 10% non-compound, 20% compound, and another 5% non-compound. I have to do it in this order, I can't remove the non-compounds first –  Xaisoft Apr 5 '12 at 18:58
Also, what is wrong with 690 * 0.9 and then 621 / 1.2 = 517.5? –  Xaisoft Apr 5 '12 at 19:01
I am not sure what the $5\%$ non-compound at the end means. If it means you remove $5\%$ of the original price at the end, that's easy, just subtract $5\%$ of $690$ from the $(690)(0.9)(0.8)$. About what's wrong with dividing by $1.2$, it is not what one gets by removing $20\%$ of $(690)(0.9)$ from $(690)(0.9)$. So, still difficulties with the same old problem? I thought that was settled long ago. –  André Nicolas Apr 5 '12 at 19:15
Ok, I get your point about the 1.2 and I understand that I can do (690)(0.9)(0.95)(0.8), but wouldn't this remove 5% from 621 and not 690. I guess what I am trying to get at is how do handle it when I have a mixture of compound and non-compound percentages. –  Xaisoft Apr 5 '12 at 19:22
lol, this was settled a while ago, the problem is when you have live clients telling you the calculation should return a certain result when in fact it should return something else. I am being told it should return 517.50 and on here the consensus answer is 496.8 –  Xaisoft Apr 5 '12 at 19:24

Rather than comment on the differences between the two approaches, here's the way I would do it. Think in terms of what you want to keep. In the first step, keep 90%, then in the second step keep 80% of that. So just multiply $690(0.9)(0.8) = 690(0.72) = 496.8$.

-

Neither is correct - you should start with $690\times0.9=621$, to get the $10\%$ reduction, and then do $621\times0.8=496.8$ for the subsequent $20\%$ reduction. Note that this is not the same as just doing a $30\%$ reduction straight off, as the second percentage is of $621$, not $690$, so "each percent" is worth less (this is a common error).

To work backwards, if you're given the $496.8$ and told it was obtained in this way, then you should first undo the $20\%$ reduction. To do this, note that $496.8$ is $80\%$ of the value before the reduction, so you can find $1\%$ by dividing by $80$, and then multiply by $100$ to get the full amount; in one step this amounts to dividing by $0.8$. This gives you back $621$, and then by the same procedure you divide by $0.9$ to get the $690$ back.

As a general rule, you should first work out what $100\%$ is - in the original problem, it's the $690$ you start with, and if you're trying to "undo" a percentage reduction it's the amount you're trying to find (although if you're undoing more than one reduction it's probably easier to do them one at a time, so $100\%$ is the number you want to find in the next step). Then most problems can be reduced to dividing to find out what $1\%$ is, and then multiplying by the percentage you actually want. So when starting with the $690$ and reducing by $10\%$ you start with $100\%$ and want $90\%$, so divide by $100$ and multiply by $90$ (equivalently multiply by $0.9$), and when starting with $621$ and undoing the $10\%$ reduction you start with $90\%$ and want $100\%$, so divide by $90$ and multiply by $100$ (equivalently divide by $0.9$).

-

If you remove 10 percent of 690 from 690, then you have 90 percent of 690 left; so removing 10 percent of 690 from 690 gives you $$690\cdot (.9)=621.$$

If you remove 20 percent of 621 from 621, you have 80 percent of 621 left; so so removing 20 percent of the new value 621 from 621 leaves you with $$621\cdot(.8) = 496.8$$

It would be different to compute $690\cdot (.7)=483$; because here you are removing 10 percent of 690 and 20 percent of $690$.

I'm not sure what you were trying to do with your solution; but it seems you are doing the "backwards problem". An example

A dress is put on a 30 percent off sale. If the sale price was 690 dollars, what was the original price?

Solution: Let $O$ be the original price. We are told that $$690=O\cdot .7;$$ so $O=690/.7=985.71$.

-


text:
8 posts tagged with "Science"

View All Tags

Quantum Programming - Abstraction level 2: State Machine and Algorithms

We have been on quite the adventure so far. Let's recap our progress.

• We've shown that by encoding information onto n qubits in superposition, we can manipulate the whole system of 2n states by only changing the state of one qubit.
• We represented the state of each qubit as a simple 2D vector.
• We've established that we can change the state of a single qubit (or two) by applying different logic gates to it, similarly to a classical logic gate.
• We can now visualise the superposed state of the qubit (indeed of the whole system) on the Bloch sphere.
• We know how to entangle two qubits, so that their states are correlated.

Now the time has come to start putting it all together. Wouldn't it be nice to actually string some gates together, and see what happens?

Well, in order to see what happens, we might want to be able to visualise the flow of our qubits' states, so we can see how each gate manipulates each qubit at each step.

Quantum Programming - Abstraction level 1: Logic Gates

We have previously worked out that a quantum computer operates logically on qubits, which are the quantum counterparts of classical bits. We've learned that a qubit can stay in a superposition while we operate on it, and then we can collapse it into a definite state upon measurement. So, for the majority of our time spent programming on the quantum computer, we will be thinking of the qubits in their state of superposition.

What would be extremely useful for our intuition now, would be to have a way to visualise this superposition. Enter the Bloch Sphere. This tool allows us to represent the whole state-space of a single qubit in one (very simple) geometric shape - the unit sphere. Think of it as the unit circle in trigonometry.

Quantum Programming by abstracting ourselves from Quantum Mechanics: Abstraction level 0

The invention of the personal computer, and the further development of the hundreds of programming languages which utilise it has allowed us programmers to completely abstract ourselves from the world of electrons flowing through solid matter, and further from the low level programming of turning gates and transistors on and off to produce binary data.

Nowadays you would be hard-pressed to amaze anyone with the knowledge that stars are these ever-moving objects in the sky, at wildly different distances from us and each-other. Not only that, but many of the objects in the night sky we recognise as stars are often whole galaxies containing billions of stars, or giant clouds of gas glowing with the intensity of stars in our eyes, or god knows what else.

Simulating the Vicsek Model (with time delay) - How do birds flock and insects swarm

In two previous posts we got to know two main things:

1. We can analyse the behaviour of groups of animals using statistical tools like correlation functions
2. We can simulate this behaviour using for example the Vicsek model, and by analysing it the same way as natural groups we can measure the accurateness of the model in replicating natural phenomena.

We also learned that the standard Vicsek model suffers from one weakness: it lacks an inertial term, which seems to be important in imitating the behaviour of biological systems. So the goal of this post will be to figure out a way to simulate the Vicsek model with time delay. Let's start by

Analysing the equations of motion

$\vec{v_i}(t+1) = v_{0}\mathscr{R}_{\eta}\Theta \left[ \vec{v_i}(t) + \sum_{ j \neq i }{ n_{ij} \vec{v_j}(t) } \right] ,$
$\vec{r_i}(t+1) = \vec{r_i}(t) + \vec{v_i}(t+1) .$

How do birds flock and insects swarm: the Problem with current models

if you haven't seen the first part of this introduction, absolutely go check it out for context.

In this post we'll unravel how does the delay in a group of animals affect the group, and what should we understand under said delay (keyword: slowpoke). To get there, let's analyse the last two paragraphs of the introduction to my thesis, where we'll see what problems can be spotted in previous research, and motivate ourselves to look for solutions.


text:
## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 3 years ago 4x/5+3/2=2x

• This Question is Closed
1. cwrw238

multiply through by lcm of 5 and 2 which is 10 8x + 15 = 20x solve this for x

2. anonymous

I thought i already explained this..hmm, why did you make another question?

3. Hero

$$\large\frac{4x}{5} + \frac{3}{2} = 2x$$ Multiply 2x by 2/2 to get $$\frac{4x}{5} + \frac{3}{2} = \frac{4x}{2}$$ Subtract 3/2 from both sides to get: $$\frac{4x}{5} = \frac{4x}{2} - \frac{3}{2}$$ Combine fractions on the right side to get $$\frac{4x}{5} = \frac{4x - 3}{2}$$ Cross Multiply to get 2(4x) = 5(4x - 3) 8x = 20x - 15 Finish solving for x.

4. anonymous

Thank you

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy


text:
## CryptoDB

### Paper: From Physical to Stochastic Modeling of a TERO-Based TRNG

Authors: Florent Bernard Patrick Haddad Viktor Fischer Jean Nicolai DOI: 10.1007/s00145-018-9291-2 Search ePrint Search Google Security in random number generation for cryptography is closely related to the entropy rate at the generator output. This rate has to be evaluated using an appropriate stochastic model. The stochastic model proposed in this paper is dedicated to the transition effect ring oscillator (TERO)-based true random number generator (TRNG) proposed by Varchola and Drutarovsky (in: Cryptographic hardware and embedded systems (CHES), 2010, Springer, 2010 ). The advantage and originality of this model are that it is derived from a physical model based on a detailed study and on the precise electrical description of the noisy physical phenomena that contribute to the generation of random numbers. We compare the proposed electrical description with data generated in two different technologies: TERO TRNG implementations in 40 and 28 nm CMOS ASICs. Our experimental results are in very good agreement with those obtained with both the physical model of TERO’s noisy behavior and the stochastic model of the TERO TRNG, which we also confirmed using the AIS 31 test suites.
##### BibTeX
@article{jofc-2019-30143,
title={From Physical to Stochastic Modeling of a TERO-Based TRNG},
journal={Journal of Cryptology},
publisher={Springer},
volume={32},
pages={435-458},
doi={10.1007/s00145-018-9291-2},


text:
# How to make a delayed mouse input?

I have a car game and the idea is that the player is drunk. I want to have a delayed input from the mouse, where you move the mouse side to side to turn.

How can I setup a constant delay between the input the user gives and the turn amount of the car?

Right now I have this:

h = Mathf.Lerp(h, (((Input.mousePosition.x / Screen.width) * 2) - 1), Time.deltaTime * 2);

What happens is the turn is delayed, but is very slow, in other words, if I move the mouse very fast, the car turns very slowly, but if I crank up the multiplier for Time.deltaTime, the car turns faster, but delay effect is minimized.

How else could I do this?

To simulate a time lag, use a circular buffer to store the last N frames' mouse positions. Store the current mouse position each frame. In your control calculations, use the oldest mouse position from the buffer instead of the current mouse position.

• I solved it actually, I just used a coroutine with a waitforseconds. I would be concerned about performance with this technique, if you have a really fast device, it would have a much lower delay time than something like a mobile. – mr-matt May 21 '17 at 17:41
• Note that delaying input like this will change the result depending on the polling rate. – Joey May 21 '17 at 18:45
• Matthew, I'm a bit bewildered ... do you really mean performance? The performance penalty of writing a few bytes to memory per frame is perfectly negligible, and you get the mouse position free with every message (on Windows at least). How do you think your coroutine solution works exactly? Who keeps track of how many timers exist, when they expire and which code to execute? It sounds pretty heavyweight. – Buster May 21 '17 at 23:03
• Oh no, I don't mean the performance of the actual task of the delay, I mean the overall performance. For example, if I take the mouse position of the last 10 frames and run this on a phone and get 20 fps, then the delay would be for 0.5 seconds. However, if I run this on my desktop and get 200 fps, then the delay would only be for 0.05 seconds. Do you see what I mean? – mr-matt May 21 '17 at 23:50
• You would choose N to avoid that, but the bigger problem is that polling once per frame like this is throwing away a lot of data. Better to handle every mouse event (looks like that's what your answer does). – Buster May 21 '17 at 23:56

I ended up using a coroutine, because (from what I understand) the delay will be constant regardless of the performance of the device running the game.

This is my code:

private IEnumerator DelayedInput()
{
Vector3 a = Input.acceleration;
Vector2 m = Input.mousePosition;

yield return new WaitForSeconds(delay);

accelerometer = Vector3.Lerp(accelerometer, a, Time.deltaTime * turnSpeed);
mouse = Vector2.Lerp(mouse, m, Time.deltaTime * turnSpeed);
}

• This has the advantage of responding to all input (not like my idea of polling every frame). I don't know C# but it seems to me someone somewhere has a lot of bookkeeping to do for all those timers. But if it performs OK and has the feel you want then I like it. – Buster May 21 '17 at 23:47
• I think this is the better way to do it, I don't think timers will affect your game all that much, of course you could get back to this question with the results of your experience with performance (for future reference). Anyway, you should go ahead and pick this as the answer (this is just a reminder). – John Hamilton May 24 '17 at 5:32


text:
# Optimization Week 9: Convex conjugate (Fenchel Conjugate)

### Week 9: Convex conjugate, Fenchel Conjugate

Every function has something known as the convex conjugate, or differential conjugate. And, this is a very important concept in convexity and in duality.

# 1 Definition

Given function f : R n → R f: \mathbb{R}^n\rightarrow \mathbb{R} , its conjugate f ∗ f^* is:
f ∗ ( y ) = max  x y T x − f ( x ) f^*(y)=\max_x y^Tx-f(x)
f ∗ f^* is always convex because it is max of affine functions.

# 2 Properties

## Sum of conjugate

f ( x ) + f ∗ ( y ) ≥ y T x f(x)+f^*(y)\geq y^Tx

## Decompose conjugate

If f ( u , v ) = f 1 ( u ) + f 2 ( v ) f(u,v)=f_1(u)+f_2(v) ,
then f ∗ ( w , z ) = f 1 ∗ ( w ) + f 2 ∗ ( z ) f^*(w,z)=f_1^*(w)+f_2^*(z)

## Double conjugate

f ∗ ∗ ( z ) ≤ f ( z ) f^{**}(z)\leq f(z)
Always less than the original function, and is convex. Convex envelope.
If f f is convex, then f ∗ ∗ ( z ) = f ( z ) f^{**}(z) = f(z) .

## Convex original

• f ∗ ∗ ( z ) = f ( z ) f^{**}(z) = f(z) .
• f ∗ ( y ) = max  x y T x − f ( x ) = y T x ˉ − f ( x ˉ ) f^*(y)=\max_x y^Tx-f(x)=y^T\bar x-f(\bar x) , where y ∈ ∂ f ( x ˉ ) y\in \partial f(\bar x)
• The derivatives of f f and f ∗ f^* are inverse of each other: y ∈ ∂ f ( x ) ⇔ x ∈ ∂ f ∗ ( y ) y\in \partial f(x)\Leftrightarrow x\in \partial f^*(y)

Uses: duality 2

10-10 2万+

12-19 1035
12-06 8116
04-12 2153
03-02 5804
03-05 1295
11-28 530
11-24
02-12 1万+
08-23
12-20
07-28


text:
# Does evaporated hand sanitizer pollute the environment?

With all the COVID-19 stuff going on, I feel the usage of alcohol-based hand sanitizers has increased many fold. I wonder what happens to all that evaporated alcohol in the atmosphere. How soon does alcohol react in environment and what other substances are formed?

• If you're interested in reactions and compounds, maybe more suitable at Chemistry SE? chemistry.stackexchange.com Jul 29 '20 at 13:50
• ethanol breakes down into carbon-hydrogen and oxygen,it is close to cabon neutral so the climate impact is very low.en.wikipedia.org/wiki/Ethanol Jul 29 '20 at 15:46

Assuming you are talking about ethanol: it is a common chemical substance, generated naturally from e.g. rotting fruits. So any pollution from hand sanitizers should be negligible.

What happens to the alcohol in the atmosphere is an interesting sequence of reactions. I took the info below from an article (see link below).

In the air, ethanol is oxidized quickly (half-life is a few days) into acetaldehyde $$CH_3CHO$$, which is further oxidized into other compounds, most of which are harmless. However, if the air contains nitrogen dioxide $$NO_2$$, it can form peroxyacetyl nitrate $$CH_3C(O)OONO_2$$, which is dangerous.

Source


text:
#### Question

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer N and call it $a_0$. This is the seed of a sequence. The rest of the sequence is generated as follows: For n=0, 1, 2, ...

$a _ { n + 1 } = \left\{ \begin{array} { l l } { a _ { n } / 2 } & { \text { if } a _ { n } \text { is even } } \\ { 3 a _ { n } + 1 } & { \text { if } a _ { n } \text { is odd } } \end{array} \right.$

. However, if $a_n$=1 for any n, then the sequence terminates. a. Compute the sequence that results from the seeds N=2, 3, 4, ...10. You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers N, the sequence terminates after a finite number of terms. b. Now define the hailstone sequence $\left\{ H _ { k } \right\}$, which is the number of terms needed for the sequence $\left\{ a _ { n } \right\}$ to terminate starting with a seed of k. Verify that $H_2$=1, $H_3=7$, and $H_4=2$. c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

Verified

#### Step 1

1 of 6

Work in Excel is below. The cell $B4$ is open, so that the command is visible. The table shows the sequences $\left\{ a_n\right\}$ for seeds $N=2,3,4,\dots, 10$, until termination. We see that in this example, every sequence has terminated before 21st term (the letter $N$ stays for "Not any more :)" and indicates the termination).

## Create an account to view solutions

#### Biology

1st EditionKenneth R. Miller, Levine
2,470 solutions

#### Calculus: Early Transcendentals

3rd EditionBernard Gillett, Eric Schulz, Lyle Cochran, William L. Briggs
13,222 solutions

#### Nelson Science Perspectives 10

1st EditionChristy C. Hayhoe, Doug Hayhoe, Jeff Major, Maurice DiGiuseppe
1,359 solutions

#### Miller and Levine Biology

1st EditionJoseph S. Levine, Kenneth R. Miller
1,773 solutions


text:
# ewstats

Expected return and covariance from return time series

## Syntax

``[ExpReturn,ExpCovariance,NumEffObs] = ewstats(RetSeries)``
``[ExpReturn,ExpCovariance,NumEffObs] = ewstats(___,DecayFactor,WindowLength)``

## Description

example

````[ExpReturn,ExpCovariance,NumEffObs] = ewstats(RetSeries)` computes estimated expected returns (`ExpReturn`), estimated covariance matrix (`ExpCovariance`), and the number of effective observations (`NumEffObs`). These outputs are maximum likelihood estimates which are biased.```

example

````[ExpReturn,ExpCovariance,NumEffObs] = ewstats(___,DecayFactor,WindowLength)` adds optional input arguments for `DecayFactor` and `WindowLength`.```

## Examples

collapse all

This example shows how to compute the estimated expected returns and the estimated covariance matrix.

```RetSeries = [ 0.24 0.08 0.15 0.13 0.27 0.06 0.14 0.13 ]; DecayFactor = 0.98; [ExpReturn, ExpCovariance] = ewstats(RetSeries, DecayFactor)```
```ExpReturn = 1×2 0.1995 0.1002 ```
```ExpCovariance = 2×2 0.0032 -0.0017 -0.0017 0.0010 ```

## Input Arguments

collapse all

Return series, specified the number of observations (`NUMOBS`) by number of assets (`NASSETS`) matrix of equally spaced incremental return observations. The first row is the oldest observation, and the last row is the most recent.

Data Types: `double`

(Optional) Controls how much less each observation is weighted than its successor, specified as a numeric value. The kth observation back in time has weight `DecayFactor`k. `DecayFactor` must lie in the range: ```0 < DecayFactor <= 1```.

The default value of `1` is the equally weighted linear moving average model (BIS).

Data Types: `double`

(Optional) Number of recent observations in the computation, specified as a numeric value.

Data Types: `double`

## Output Arguments

collapse all

Estimated expected returns, returned as a `1`-by-`NASSETS` vector.

Estimated covariance matrix, returned as a `NASSETS`-by-`NASSETS` matrix.

The standard deviations of the asset return processes are defined as

``` STDVec = sqrt(diag(ExpCovariance)) ```

The correlation matrix is

``` CorrMat = ExpCovariance./( STDVec*STDVec' ) ```

`NumEffObs` is the number of effective observations where

`$NumEffObs=\frac{1-DecayFacto{r}^{WindowLength}}{1-DecayFactor}$`

A smaller `DecayFactor` or `WindowLength` emphasizes recent data more strongly but uses less of the available data set.

## Algorithms

For a return series r(1),…,r(n), where (n) is the most recent observation, and w is the decay factor, the expected returns (`ExpReturn`) are calculated by

`$E\left(r\right)=\frac{\left(r\left(n\right)+wr\left(n-1\right)+{w}^{2}r\left(n-2\right)+...+{w}^{n-1}r\left(1\right)\right)}{NumEffObs}$`

where the number of effective observations `NumEffObs` is defined as

`$NumEffObs=1+w+{w}^{2}+...+{w}^{n-1}=\frac{1-{w}^{n}}{1-w}$`

E(r) is the weighed average of r(n),…,r(`1`). The unnormalized weights are w, w2, …, w(n-1). The unnormalized weights do not sum up to `1`, so `NumEffObs` rescales the unnormalized weights. After rescaling, the normalized weights (which sum up to `1`) are used for averaging. When w = `1`, then `NumEffObs` = n, which is the number of observations. When w < `1`, `NumEffObs` is still interpreted as the sample size, but it is less than n due to the down-weight on the observations of the remote past.

Note

There is no relationship between `ewstats` function and the RiskMetrics® approach for determining the expected return and covariance from a return time series.


text:
# World of Chemistry

## Educators

Problem 1

Review how we defined chemical change in Chapter $3 .$ Explain how the visual clues mentioned in this chapter are indicators of chemical change as defined in Chapter 3 .

Check back soon!

Problem 2

If a piece of blackboard chalk is heated strongly, the mass of the chalk decreases substantially, and the chalk is converted into a fine powder. What evidence is there for a chemical reaction having taken place when the chalk is heated?

Check back soon!

Problem 3

If you have had a clogged sink drain at your home, your parents may have have tried using a commercial drain cleaner to dissolve the clog. What evidence is there that such drain cleaners work by chemical reaction?

Check back soon!

Problem 4

If a bottle of aspirin is left open to the air, eventually one of the products is acetic acid (vinegar is a mixture of acetic acid and water). Is there evidence that this change represents a chemical reaction?

Check back soon!


text:
# Consecutive PWM signals with different prescaler settings

My ATtiny85 circuit generates signals like this one on PB1 (OC0B) using fast PWM (preamble, 16 bit data, stop pulse):

I have trouble with the leftmost longer pulse. At the moment i use simple _delay_us statements to generate it, but it sometimes introduces glitches later near the end of the PWM (orange and yellow) parts.

It seemed smart to switch to using timers for this part but I can't get it to work correctly. This is my code:

const uint8_t JVC_MUTE = 0x71;
const uint8_t JVC_ADDRESS = 0xF1;

void setupPWM () {
//enables fast pwm, non inverting,TOP=OCR0A, prescaler 8
TCCR0A = (1<<WGM01)| (1<<WGM00)|(1<<COM0B1);
TCCR0B = (1<<WGM02)|( 1<<CS01);
}

void stopPWM () {
TCCR0A = 0;
}

/*emits a long preamble pulse
*/
void preamble(){
//enables fast pwm, non inverting, prescaler 64
TCCR0A = (1<<WGM01)| (1<<WGM00)|(1<<COM0B1);
TCCR0B = (1<<WGM02)|( 1<<CS01)|(1<<CS00);
TCNT0=0;
OCR0A=255;
OCR0B=80;
do ; while ((TIFR & 1<<OCF0B) == 0);
TIFR = 1<<OCF0B;
TCCR0A = 0;
}

/*emits 16 bit of data using fast PWM
code:16 bit data value to be send (address + command)
*/
void sendCode (uint16_t code) {
TCNT0=0;

OCR0A=0;
OCR0B=0;
do ; while ((TIFR & 1<<OCF0B) == 0);
TIFR = 1<<OCF0B;
OCR0A=0;
OCR0B=0;
//waits while polling the OCF0B bit signaling a compare match with OCR0B
do ; while ((TIFR & 1<<OCF0B) == 0);
TIFR = 1<<OCF0B;

//loops through each bit in the 'code' variable
//and emits a long or short pulse depending on value
for (uint16_t Bit=0x8000;Bit;Bit=Bit>>1){
//if 1
if (code & Bit) {
OCR0A=255;
OCR0B=80;
}
//if 0
else {
OCR0A=130;
OCR0B=80;
};
do ; while ((TIFR & 1<<OCF0B) == 0);
TIFR = 1<<OCF0B;
}
//emits a short pulse signaling the end of the transmission
OCR0A=130;
OCR0B=80;
do ; while ((TIFR & 1<<OCF0B) == 0);
TIFR = 1<<OCF0B;

}

/*
*emits a transmission (preamble pulse, 16 bit of data + end pulse)
*/
void transmit(uint8_t address,uint8_t code){
preamble();
setupPWM();//PWM settings in TCCR0A and TCCR0B
stopPWM();//disables PWM (TCCR0A=0)
}

There are two issues:

1. This code will generate the preamble pulse the first time, followed by glitched data, and then only the glitchy data when called
consecutively.
2. It needs the padding instructions I added in sendCode(), otherwise the 2 first bits of the transmission are always cut.

edit: the actual data part (orange and yellow parts), as well as the stop pulse at the end work fine when the call to preamble() is commented out. The only issue is with the preamble() part.

I'd appreciate any help, especially with point 1.

• If I am reading the code correctly it seems like your bit shift operators << are being used unusually. Also verify that when you use the single & operator in your 'while' loops that you do in fact want the bitwise AND that you have there rather than a logical AND &&. Oct 6 '17 at 16:29
• This part as I understand it creates a 16 bit value of 0x8000, then gradually shifts the 1 right to test each bit of the 'code' variable. It works reliably when the call to preamble() is commented out. Oct 7 '17 at 9:26
• preamble() and setupPWM() select different prescaler clock sources. One sets only CS1, while the other sets CS1 and CS0. What may be important about that is that the prescaler is essentially a counter which may have a partial count value left over from the other function's use. In both preamble() and also in sendCode() you could try clearing the prescaler before the TCNT0=0; statement by inserting GTCCR=GTCCR|1; then re-enabling it after the TCNT=0; statement by adding GTCCR=GTCCR&0x7f; Oct 7 '17 at 20:33
• CORRECTION: The second statement should not be GTCCR=GTCCR&0x7f; it should be GTCCR=GTCCR&0xfe; Oct 7 '17 at 21:04


text:
category theory

# Contents

## Idea

One may interpret mathematical logic as being a formal language for talking about the collection of monomorphisms into a given object of a given category: the poset of subobjects of that object.

More generally, one may interpret type theory and notably dependent type theory as being a formal language for talking about slice categories, consisting of all morphisms into a given object.

Conversely, starting with a given theory of logic or a given type theory, we say that it has a categorical semantics if there is a category such that the given theory is that of its slice categories, if it is the internal logic of that category.

## Definition

For the general idea, for the moment see at type theory the section An introduction for category theorists and see at relation between type theory and category theory.

### Of dependent type theory

We discuss how to interpret judgements of dependent type theory in a given category $𝒞$ with finite limits. For more see categorical model of dependent types.

Write $\mathrm{cod}:{𝒞}^{I}\to 𝒞$ for its codomain fibration, and write

$\chi :{𝒞}^{\mathrm{op}}\to \mathrm{Cat}$\chi : \mathcal{C}^{op} \to Cat

for the corresponding classifying functor, the self-indexing

$\chi :\Gamma ↦{𝒞}_{/\Gamma }$\chi : \Gamma \mapsto \mathcal{C}_{/\Gamma}

that sends an object of $𝒞$ to the slice category over it, and sends a morphism $f:\Gamma \to \Gamma \prime$ to the pullback/base change functor

${f}^{*}:{𝒞}_{/\Gamma \prime }\to {𝒞}_{/\Gamma }\phantom{\rule{thinmathspace}{0ex}}.$f^\ast : \mathcal{C}_{/\Gamma'} \to \mathcal{C}_{/\Gamma} \,.

We give now rules for choices ”$\left[xyz\right]$” that associate with every string ”$xyz$” of symbols in type theory objects and morphisms in $𝒞$. A collection of such choices following these rules an an interpretation / a choice of categorical semantics of the type theory in the category $𝒞$.

#### Contexts and type judgements

1. The empty context $\left(\right)$ in type theory is interpreted as the terminal object of $𝒞$

$\left[\left(\right)\right]:=*\phantom{\rule{thinmathspace}{0ex}}.$[ () ] := * \,.
2. If $\Gamma$ is a context which has already been given an interpretation $\left[\Gamma \right]\in \mathrm{Obj}\left(𝒞\right)$, then a judgement of the form

$\Gamma ⊢A:\mathrm{Type}$\Gamma \vdash A : Type

is interpreted as an object in the slice over $\Gamma$

$\left[\Gamma ⊢A:\mathrm{Type}\right]\in \mathrm{Obj}\left({𝒞}_{/\Gamma }\right)\phantom{\rule{thinmathspace}{0ex}},$[\Gamma \vdash A : Type] \in Obj(\mathcal{C}_{/\Gamma}) \,,

hence as a choice of morphism

$\begin{array}{c}\left[\left(\Gamma ,x:A\right)\right]\\ {↓}^{\left[\Gamma ⊢A:\mathrm{Type}\right]}\\ \left[\Gamma \right]\end{array}$\array{ [(\Gamma, x : A)] \\ \downarrow^{\mathrlap{[\Gamma \vdash A : Type]}} \\ [\Gamma] }

in $𝒞$.

3. If a judgement of the form $\Gamma ⊢A:\mathrm{Type}$ has already found an interpretation, as above, then an extended context of the form $\left(\Gamma ,x:A\right)$ is interpreted as the domain object $\left[\left(\Gamma ,x:A\right)\right]$ of the above choice of morphism.

#### Terms

Assume for a context $\Gamma$ and a judgement $\Gamma ⊢:A:\mathrm{Type}$ we have already chosen an interpretation $\left[\Gamma ,x:A\right]\stackrel{\left[\Gamma ⊢A:\mathrm{Type}\right]}{\to }\left[\Gamma \right]$ as above.

A judgement of the form $\Gamma ⊢a:A$ (a term of type $A$) is to be interpreted as a section of this morphism, equivalently as a morphism in ${𝒞}_{/\Gamma }$

$\left[\Gamma ⊢:a:A\right]:*\to \left[\Gamma ,x:A\right]$[\Gamma \vdash : a : A] : * \to [\Gamma, x : A]

from the terminal object to $\left[\Gamma ⊢A:\mathrm{Type}\right]$, which in $𝒞$ is a commuting triangle

$\begin{array}{ccccc}\left[\Gamma \right]& & \stackrel{\left[\left(\Gamma ⊢a:A\right)\right]}{\to }& & \left[\Gamma ,x:A\right]\\ & {}_{\left[\Gamma \right]}↘& & {↙}_{\left[\Gamma ⊢A:\mathrm{Type}\right]}\\ & & \left[\Gamma \right]\end{array}\phantom{\rule{thinmathspace}{0ex}}.$\array{ [\Gamma] &&\stackrel{[(\Gamma \vdash a : A)]}{\to}&& [\Gamma, x : A] \\ & {}_\mathllap{[\Gamma]}\searrow && \swarrow_{\mathrlap{[\Gamma \vdash A : Type]}} \\ && [\Gamma] } \,.

#### Variables

For a term $\Gamma ⊢a:A$ the context $\Gamma$ is the collection of free variables in $a$.

(…)

#### Substitution

Assume that interpretations for judgements

$\Gamma ,x:A⊢B\left(x\right):\mathrm{Type}$\Gamma , x : A \vdash B(x) : Type

and

$\Gamma ⊢a:A$\Gamma \vdash a : A

have been given as above. Then the substitution judgement

$\Gamma ⊢B\left[a/x\right]:\mathrm{Type}$\Gamma \vdash B[a/x] : Type

is to be interpreted as follows. The interpretation of the first two terms corresponds to a diagram in $𝒞$ of the form

$\begin{array}{ccccc}& & & & \left[\left(\Gamma ,x:A,y:B\left(x\right)\right)\right]\\ & & & & {↓}^{\left[\Gamma ,x:A⊢B\left(x\right):\mathrm{Type}\right]}\\ \left[\Gamma \right]& & \stackrel{\left[\Gamma ⊢a:A\right]}{\to }& & \left[\left(\Gamma ,x:A\right)\right]\\ & {}_{\mathrm{id}}↘& & {↙}_{\left[\Gamma ⊢A:\mathrm{Type}\right]}\\ & & \left[\Gamma \right]\end{array}$\array{ &&&& [(\Gamma, x : A, y : B(x))] \\ &&&& \downarrow^{\mathrlap{[\Gamma, x : A \vdash B(x) : Type]}} \\ [\Gamma] &&\stackrel{[\Gamma \vdash a : A]}{\to}&& [(\Gamma, x : A)] \\ & {}_{id}\searrow && \swarrow_{\mathrlap{[\Gamma \vdash A : Type]}} \\ && [\Gamma] }

The interpretation of the substitution statement is then the pullback

$\left[\Gamma ⊢B\left[a/x\right]:\mathrm{Type}\right]:=\left[\Gamma ⊢a:A{\right]}^{*}\left[\Gamma ,x:A⊢B\left(x\right):\mathrm{Type}\right]\phantom{\rule{thinmathspace}{0ex}},$[\Gamma \vdash B[a/x] : Type] := [\Gamma \vdash a : A]^* [\Gamma, x : A \vdash B(x) : Type] \,,

hence the morphism in $𝒞$ that universally completes the above diagram as

$\begin{array}{ccccc}\left[\left(\Gamma ,y:B\left[x/a\right]\right)\right]& & \to & & \left[\left(\Gamma ,x:A,y:B\left(x\right)\right)\right]\\ {}^{\left[\Gamma ⊢B\left[a/x\right]:\mathrm{Type}\right]}↓& & & & {↓}^{\left[\Gamma ,x:A⊢B\left(x\right):\mathrm{Type}\right]}\\ \left[\Gamma \right]& & \stackrel{\left[\Gamma ⊢a:A\right]}{\to }& & \left[\left(\Gamma ,x:A\right)\right]\\ & {}_{\mathrm{id}}↘& & {↙}_{\left[\Gamma ⊢A:\mathrm{Type}\right]}\\ & & \left[\Gamma \right]\end{array}$\array{ [(\Gamma, y : B[x/a])] &&\to&& [(\Gamma, x : A, y : B(x))] \\ {}^{\mathllap{[\Gamma \vdash B[a/x] : Type] }}\downarrow &&&& \downarrow^{\mathrlap{[\Gamma, x : A \vdash B(x) : Type]}} \\ [\Gamma] &&\stackrel{[\Gamma \vdash a : A]}{\to}&& [(\Gamma, x : A)] \\ & {}_{id}\searrow && \swarrow_{\mathrlap{[\Gamma \vdash A : Type]}} \\ && [\Gamma] }

## References

A standard textbook reference for categorical semantics of logic is section D1.2 of

A discussion of categorical semantics of type theory is for instance in

• Martin Hofmann, Syntax and semantics of dependent types, Semantics and Logics of Computation (P. Dybjer and A. M. Pitts, eds.), Publications of the Newton Institute, Cambridge University Press, Cambridge, (1997) pp. 79-130.

• Roy Crole, Categories for types


text:
# Rewrite in Standard Form 1/(10-i)

Multiply the numerator and denominator of by the conjugate of to make the denominator real.
Multiply.
Combine.
Multiply by .
Simplify the denominator.
Expand using the FOIL Method.
Apply the distributive property.
Apply the distributive property.
Apply the distributive property.
Simplify.
Multiply by .
Multiply by .
Raise to the power of .
Raise to the power of .
Use the power rule to combine exponents.
Subtract from .


text:
ISOLDE Seminar

# Statistical tools for nuclear experiments (3/4)

## by Dr. Karsten Riisager (Aarhus University (DK))

Wednesday, 20 March 2013 from to (Europe/Zurich)
at CERN ( 26-1-022 )
 Description Essentially all experiments in nuclear physics today involve statistical methods during analysis of the obtained data. These methods are wonderful tools if you understand what they do, but applied blindly they can lead to wrong conclusions and other serious mistakes. Furthermore the experimentalist often encounters situations where the standard basic textbook recipes do not suffice. This set of lectures gives a practical introduction to when and how statistical methods can be applied. It concentrates on situations that postgraduate students in nuclear physics may encounter. The lectures will cover the conceptual background and are accompanied by practical excercises. The participants are assumed to have some practical experience with data analysis and a basic knowledge on statistics. Barlow's 2002 academic training lectures for postgraduate students introduces most of the concepts that will be employed. http://indico.cern.ch/conferenceDisplay.py?confId=a032478  Material: Other occasions


text:
# Without Registration Best And Free Online Dating Service For Men In London

May17

## Without Registration Best And Free Online Dating Service For Men In London

The issue is that we can not properly prepare the string we are trying to insert with database escaping. Rosencrantz (ro-zen’kranz) and guildenstern (gil’dan-stern), the schoolmates of hamlet summoned to denmark by claudius to act as spies on hamlet. The first of these is the world championship, where the top 7 climbers in each gender will qualify, leaving 11 spots. It is always beneficial to be aware of these rules and policies in prior so that you know in case you cancel your tickets due to any of the reasons, how much amount will be credited in your account. On her painstaking journey, she meets adelina vasquez, an american runaway working as a prostitute in tijuana, who takes juana in. “does quality of innovation, culture and governance drive fdi?: evidence from emerging markets,” emerging markets review, elsevier, vol. 34(c), pages 175-191. This means high specific output per liter of capacity, lots of torque, massive horsepower numbers and incredible specs for the cars equipped with such tech. Beast boy in his room was upset and says how stupid he thought he was all the while terra listening on a rock in the window decides to go out with him. But an enterprise is actually an objective (and the french etymology of the term confirms this). I will not think twice to suggest your blog to any person who should get recommendations on this subject. He was further prescribed percocet 5/325mg, 0.5-1 tab tid prn, and an ankle air cast. Order: 10000 pieces amber glass bottle for syrup din pp 28mm fob price: usd $/ piece min. His death in 609 was doubtless a setback for his religious policy as well as his political program. The food menu was very much limited as the room & restaurant menu was same. I figured those two are styes started using erythomiacin jell in the eye not helping, and just yesterday started hot pack to the eye for 5 minutes a few times a day.. Check out the page and look guides an enduring vision: revelation revealed, by austin cooke you can discover bunches of titles of guides offered. Dealo combining elements of craigslist, ebay, and amazon, dealo is one of the best apps for selling things online. The osr1 rs12329305 polymorphism contributes to the development of congenital malformations in cases of stillborn/neonatal death.. Ion implantation: implant ation technology, channelling, lattice damage and annealing. He uses part of his fortune to save morrel, who supported his father, from bankruptcy, and he then turns to his vengeance. First articulated by the late cardinal joseph bernardin in 1983, it has caused more confusion than clarity. Some go on and play club rugby, but for most rugby ends upon graduation, but that experience is with you forever. When one has passed through the set number of levels put into place by the room owner, they are often rewarded a prize. We have seen celebrities wear ash blonde hair extensions or color their hair to get different shades of blonde they think suits them. Characteristics: absorb aggression (aggro) of enemies to free party members for attack. Original cast member of the sentimental bloke, married to ron shand, daughter of james craydon. All of these players managed to be on the list, for their motivation, persistence, and determination to be the best players. Jim pope spoke on the topic ” aerospace education, ” without registration best and free online dating service for men in london and tom fox supplemented an educa- tional film with a discussion of ” business flying. As peasants began to understand their universe in educated terms and embalming practices were introduced, belief in the walking dead gradually disappeared. Every site used to source grapes is specifically designated to growing fruit for sparkling wine. Since the cables are directly connected to the satellites and are not removable, there is no way to extend the cable length, this could be a serious problem for some people. See boston from a unique wwii style amphibious landing vehicle, with historic commentary around the city then it’s into the charles river! If you are not at the moment backing up your knowledge, you need to take action immediately ( download a extremely-recommended backup solution ) to protect yourself from permanent data loss. The lightsaber would make other appearances later, such as the time when the building’s power went out and leonard one-upped sheldon’s little light stick with without registration best and free online dating service for men in london his light-up sword. I told them i would pay it today and call back to have package put on my account. It includes tons of bonus items to help you through your adventures and enough housing writs to buy your own premium house with no upkeep fees! More importantly, he is a major player and will thus be capable of trading with you repeatedly. There are accounts of preparations for royal weddings, such as the wedding of arthur and catherine of aragon in 1501 in westminster hall. A once-barren bank of the ye river in central china now has 12,000 trees irrigated by a small canal, thanks to unlikely but dedicated gardeners: two friends, one blind and one without arms. Their financial operation showed that it was a well defined group with emerging economic growth. 2) you can also root your lg optimus zone 2 using googles android sdk (software development kit). Ian black, steve horn and melissa harris of ian black real estate represented the seller in the transaction. The child only enters the treatment room when he or she is ready for therapy. In isolation, people will [attempt to] sing in equal temperament, because that’s what ‘all’ accompaniment is like. Could be a deductible, says moore in the state to state driver without registration best and free online dating service for men in london could be easily visited withoout being encumbered by an officer, renewing their insurance follow-up service through the website. Baudolt chevalier van haveskercke, son of jean van haveskercke (follow 1153971200) en beatrix without registration best and free online dating service for men in london de gavre (follow 1153971201), was born 1115. When school lets out at mid-day the narrow roads become clogged with kids in school uniforms heading home on foot and moped. Effects of dietary protein and lipid levels on growth performance, fatty acid composition and antioxidant-related gene expressions in juvenile loach misgurnus anguillicaudatus. The power-delay profile (pdp) characterizes the propagation channel by simple statistics. (d) — investment advisory contract between the registrant and societe generale asset management corp. (“sgam corp.”).* (e)(1) The in vitro abuse resistance study was carried out substantially the same as in example 4a above, using the same apparatus, reagents and methods described above, with any exceptions noted as follows. This information source has, until recently, been given little attention by researchers. How can this be happening, they ask, in a nation whose wealth and power seem almost supernatural in so many struggling corners of the world? The impact of all four simulations is measured by relative changes of liquidity ratios that have been introduced for this purpose. There are appreciations of baron without registration best and free online dating service for men in london by denys hay and august buck and a bibliography of his writings in renaissance studies in honor of hans baron (florence: g. c. sansoni, 1971). I without registration best and free online dating service for men in london never saw him afterwards till this morning; there was an order made on the protection order for a pound a week for one month, when i was to report; The heroes, (no idiots, they!) ready themselves for an enemy attack that could come at any moment. The robertson panel recommended not telling the public the full truth, but did they do the right thing? I also thank cereproc for providing the isabella and katherine voices for research use. Rather than talking about clean energy, perhaps it would be worthwhile talking about what the world would look like with much less, or no, petroleum products. When i made these the first time, i was a bit nervous that they wouldnot taste as good as the deep without registration best and free online dating service for men in london fried ones. Gold is a dominant color throughout and appears on every faucet, grand staircase and toilet-paper holder, and even in the form of gold-coated sugar flecks on chocolate dessert pastries. There was money in plenty upon his person, but not an article that would give the slightest clue to his identity. Mp3 codes the sound almost with the identical quality, as cd (sixteen-bit stereo), offering compression in size 1 to without registration best and free online dating service for men in london 10 from the original waf or aiff tracks. For the time being the gods were unable to escape the archdemon’s power, and they had to perform humiliating tasks: vayu, the god of the winds, swept the house; La cagnolina, che ha verosimilmente meno di un mese di vita, e’ stata chiamata dai carabinieri ‘ennia’, come il nome del brigadiere che l’ha estratta dal contenitore dei rifiuti. Gibbs does something he’s done many times before, but this time things are different. He then raked his fingers through their scalps while being sure not to guide them along on their current task. #13(marilyn monroe death / norma jean conspiracy-s part-2; vf or better =$4); The song has been composed by atul-ajay and sung by ajay gogavale, with inspirational lyrics by amitabh bhattacharya. Those suckers are meant to be straight marks, vertical through and through. Press ctrl+v to paste the contents of the scanned document into the new file. Very fine, still worthy of a strong bid as this opportunity is unlikely to occur anytime soon. Maar dan moet het flink verbouwd worden en door de krisis zal dan misschien wel vertraagd of zelfs afgeblazen worden. (a) a principal must report to the board as required in this subdivision by march 15 for the preceding calendar year. This means that there will be increasing pressure on politicians come up with solutions. Free coverage through us, all across malaysia!, sexbook malaysia is here on your comment: display name: \par \par what sort of community model did charles elton advocate for and what are its basic tenets? After reviving the sorcerer, parker gave greeting to the cloak of levitation with a handshake to which it goes away, leaving parker to sarcastically leave it be. He went from editing a script, to reviewing a storyboards, to leaving comments on a finalized animation. In addition, the museum offers great facilities for storing and exhibiting art under the best possible conditions. I actually could have kept going, but with 6 balls on the table at once my brain was on overload. Samir hussein/getty images u.k. songwriter ed sheeran earned a grammy nomination for song of the year for his deceptively upbeat hit single “the a team.” They only need to be good enough to serve their purpose, fooling a system or a person in a given transaction. Cbs local welcome to the continually updated cbs local daily story stream. Having escaped major damage in world war ii, the working port looks like an antique. 5. tottenham court road if crossrail works out as the designers hope, we might be able to say more of the same here. Shorts: level 1 blink, and you’ll miss it it was over before it started, but it managed to tell an enticing tale. Later still, i thought perhaps the pulitzer jury was somehow listening when composer andrew norman spoke to npr music last november about diversity in the classical music world. (for those who are totally at sea: the flags are named for letters in the alphabet – bravo for “b” etc – but have complex meanings in various combinations. +\item on can accept \glyph{macromolecules} that span several compartments, and represent domains as \glyph{units of information}.


text:
Home

# Hello. Another Random New Member.

Hello. I’m Julian Poon, the new member. So as to keep things short:

$$\bullet$$ I will try to post frequently

$$\bullet$$ I will most likely post maths and physics with maths

$$\bullet$$ I really hoped WordPress supports $$LATEX$$ but it doesn’t.

$$\bullet$$ If you get any of the $$…$$ you are awesome.

$$\bullet$$ More bullets

$$\bullet$$ See this [Pendulum](https://www.desmos.com/calculator/uzf6aqvvor)

$$\bullet$$ $$\boxed{\textbf{I Just} \text{want to say } \beta \gamma \varepsilon !}$$

Advertisements

## 2 thoughts on “Hello. Another Random New Member.”

1. Damian Boh says:

hi julian!

2. Hi. I didn’t know that comments need to be “approved”