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# Math Help - Calculation in Rijndael field GF(2^8) and (mod x^4+1)?

1. ## Calculation in Rijndael field GF(2^8) and (mod x^4+1)?

I have a 6th degree polynomial with unknown coefficients, but I need to take it (mod x^4+1) to find some 3rd degree polynomial. Using wolfram alpha and mathematica 9.0, I could only reduce the polynomial to a different 6th degree polynomial that was shorter.

e.g. $a_1x^6 + a_2x^5 + ... a_6x + a_7$ is what I'm given, where each a_i is something like:
$a_1 = s_1 + s_3 + s_4$
$a_2 = s_1 + s_2 + s_3$
$a_3 = s_3 + s_4$
...

for some unknown $s_i$.

Also, the coefficients are to (mod 2).

How can I find $a_1x^6 + a_2x^5 + ... a_6x + a_7$ (mod x^4+1) as a 3rd degree polynomial?

I heard that under mod (x^4 + 1) and (mod 2), I have congruence x^4 = -1 = 1 but that doesn't make sense to me?

2. ## Re: Calculation in Rijndael field GF(2^8) and (mod x^4+1)?

Since x^4+1 = 0 mod (x^4+1), then x^4 = -1 mod (x^4+1).

Let me rewrite your 6th degree polynomial as:

f(x) = a6 * x^6 + a5 * x^5 + ...a1 * x + a0 = (a6,a5,a4,a3,a2,a1,a0) ...in other words concentrate on the coefficients.

Also note the subscript on the coefficient matches the power of x.

Now a4 * x^4 = a4 * (-1) mod (x^4+1). Likewise a5*x^5 = a5*(-x) and a6*x^6 = a6*(-x^2).

So (a6,a5,a4,a3,a2,a1,a0) = (a3,a2,a1,a0) - (a6,a5,a4) = (a3,a2-a6,a1-a5,a0-a4).

So the cubic is: a3 * x^3 + (a2-a6)*x^2 + (a1-a5)*x + (a0-a4).

If you want to work in GF(2^8), just take the coefficients mod (2^8).

3. ## Re: Calculation in Rijndael field GF(2^8) and (mod x^4+1)?

Thanks a lot qmech! It's pretty obvious once you see how it's done...

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# Archimedes Volume of a Sphere

Topic:
Sphere, Volume
In the diagram above, we have a hemisphere with radius , side by side with a cylinder with radius and height . The cylinder has had an inverted cone with the same radius and height removed from its interior. The diagram at the right shows the cross-sections of both figures. To change the height of the plane, use the slider at the bottom of the diagram. Both cross-sections have the same area. Which can be shown using two triangles in the diagram. Since the cross-sections have the same area at every height of the plane, the two figures have the same volume by Cavalieri's Principle. Thus, The volume of the hemisphere is: Making the volume of a sphere:


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## Faculty Bibliography 2000s

#### Title

Polynomial growth solutions of Sturm-Liouville equations on a half-line and their zero distribution

#### Authors

C. E. Shin; A. Tovbis;A. I. Zayed

Math. Nachr.

#### Keywords

Sturm-Liouville equation; polynomial growth; perturbed Bessel functions; Mathematics

#### Abstract

For alpha is an element of [0, 2pi], consider the Sturm-Liouville equation on the half line y"(x) + (lambda - q(x))y(x) 0, 0 less than or equal to x < infinity, with y(0) = sin alpha, y'(0) -cos alpha. For each lambda > 0, denote by phi(x, lambda) the solution of the above initial-value problem. It is known that the condition xq(x) is an element of L-1(R+) is sufficient for phi(x, lambda) to be uniformly bounded by a linear function in x for all x, lambda greater than or equal to 0; however, this condition is not necessary as the Bessel differential equation demonstrates. In this paper we extend this result to the borderline case in which q(x) = O(1/x(2)) as x --> infinity. We show that if q(x) is continuously differentiable and q(x) = O(1/x(2)) as x --> infinity, that is, xq(x) may not be integrable on R+, then there exists a polynomial p(x) such that \phi(x, lambda)\ less than or equal to p(x) for any x is an element of [0, infinity) and lambda is an element of [0, infinity). As a particular example, we consider the perturbed Bessel equation v"(x) + [1 - (nu(2) - 1/4)/x(2) + h(x)] v(x) = 0, where nu is an element of R and h(x) = o(1/x(2)) as x --> infinity. The technique, developed in the paper, allows us to find upper and lower bounds on the distance between consecutive zeroes x(n), x(n+1) of the solution v(x) of the perturbed Bessel equation, as well as the asymptotics of x(n+1) - x(n) as n --> infinity. (C) 2004 WILEY-VCH Verlag GmbH Co. KGaA

#### Journal Title

Mathematische Nachrichten

263

1-1-2004

#### Document Type

Article

http://dx.doi.org/10.1002/mana.200310132

English

204

217

#### WOS Identifier

WOS:000188845500012

0025-584X

COinS


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# Why is an object not defined as identity morphism?

I've seen that there was a single-sorted definition of a category. In some ways, it seems more understandable than the original definition.

I don't know much about category theory. But I would like to know how each definition is useful.

• The two-sorted definition corresponds to how most people think of categories, and how they talk about categories. For example, we speak of the categories of sets (not of functions), of groups (not of group homomorphisms), of topological spaces (not of continuous maps), etc. (If I remember correctly, Ehresmann did write about the categories of functions, of homomorphisms, of continuous maps, etc., but that never caught on.) – Andreas Blass Jul 19 at 16:39
• One can identify an object with its identity morphism. Then we have a category consisting of morphisms that can be composed. Usually the composition is thought of one function for the whole category. However, the composition function can be broken up into a performing compositions, one such place for each object. In higher order category theory there may be 2-morphisms between the usual morphisms. They may also be places or performing compositions. With the appropriate additional axioms higher order category theory fits into this scheme. – Jay Kangel Jul 19 at 17:52
• My impression is that the single-sorted POV is slightly more widespread in the context of groupoids (considered as small categories where every morphism is iso) but I am not a specialist and would be happy to be corrected here by others. However, one often ends up introducing the "unit space" of the groupoid which is the "set of objects", so even then two sorts seem to emerge – Yemon Choi Jul 19 at 18:20
• Another area where the single sorted definition is more widespread is when working with strict $n$-categories or strict $\infty$-categories. Working with the single sorted definition allows to compose arrows of different dimension without writing iterated identities everywhere. It makes the manipulation of expressions a little more bearable, so lots of paper use it (where the "lots" need to be taken relatively as there is not that many paper on that topics anyway) – Simon Henry Jul 19 at 19:54
• Actually I would argue that the homsets-definition (ncatlab.org/nlab/show/category#AFamilyOfCollectionsOfMorphisms) is the one that corresponds to how more people think of categories in practice. – Mike Shulman Jul 21 at 23:26


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# Tag Info

10

No. We might have $P(X=n)=2^{-n}$. Then $\Bbb E[X]=2$ and $\Bbb E[e^X]=\sum_n 2^{-n}e^n=\infty$

8

It might be easier to think about if we abstract a little. Brownian motion is just a continuous, time dependent random variable determined by some probability space $(X,\mathcal F, P)$. Let us view it as a function of two variables, $f(x,t)$. If $x$ is fixed, we have a continuous function, $f_x(t)=f(x,t)$, and we can compute it's integral $F(x,t)=F_x(t)=\... 6 Let$X$denote any value between$1-5$, then the optional sequences are:$XXX66X6X666XX66$Calculate the probability of each sequence:$P(XXX66)=\frac56\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^3}{6^5}P(X6X66)=\frac56\cdot\frac16\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}{6^5}P(6XX66)=\frac16\cdot\frac56\cdot\frac56\cdot\...

6

The moment generating function of a Poisson random variable $X$ with parameter $\lambda$ is $$\mathbb{E}\left[e^{tX}\right]=\sum_{n=0}^\infty \frac{e^{-\lambda}e^{nt}\lambda^n}{n!}=e^{\lambda(e^t-1)}$$ which converges for all $t$, and in fact is the restriction to $\mathbb{R}$ of an entire function. In particular, the power series for the moment generating ...

5

Of course you don't need to learn statistical physics to do probability theory and there are many areas of probability that are unrelated to physics. However, it certainly wouldn't hurt to at least learn some of the basics. For a very mathematical approach to the subject, I highly recommend the ongoing book project Statistical Mechanics of Lattice Systems: a ...

5

To be clear, you want to find $P_n(t)$ for a birth-death process with constant birth rate $q(i,i+1)=\lambda$ and constant death rate $q(i,i-1)=\mu$? This particular birth and death process is exactly the $M/M/1$ queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state.

5

We have to assume that the underlying probability space is complete; otherwise the assertion might fail. So, suppose that $(\Omega,\mathcal{A},\mathbb{P})$ is a complete probability space and $(X_t)_{t \in [0,1]}$ a process with almost surely continuous sample paths, i.e. there exists a null set $N \in \mathcal{A}$ such that $$[0,1] \ni t \mapsto X_t(\omega)... 5$$E[z_1]=\frac{E\Big[{\sqrt{a_1^2-4a_0}}\Big]}2\\=\frac18\left(\int_{-1}^1\int_{-1}^1\sqrt{x^2-4y}dydx\right)\\=\frac1{48}\left(\int_{-1}^1(x^2+4)^{\frac32}-(x^2-4)^{\frac32}dx\right)\\=\frac1{24}\left(\int_0^1(4+x^2)^\frac32dx+i\int_0^1(4-x^2)^\frac32dx\right)\\=\frac{1}{24}\left(\left(\frac{11\sqrt5}4+6\sinh^{-1}\frac12\right)+i\left(\frac{9\sqrt3}4+\pi\...

5

The other solutions work perfectly for a modest number of rolls. If you were interested in a larger number, you might find a recursion helpful. Let $P(n)$ be the probability that your game ends in exactly $n$ rolls. Thus your problem is asking for $P(5)$. We note that $P(1)=0,\;P(2)=\frac 1{6^2}$. For $n>2$ we remark that the first roll is either a $6$...

4

These imply $E[X]=c$ and $E[Y]=d$. However, a constant conditional expectation does not imply independence. Consider any independent random variable $(X,Z)$, and by defining $Y=ZX$ create $(X,Y)$ for which $E[ZX|X]=XE[Z|X]=XE[Z]=d$.$X$ and $Y=ZX$ are not necessarily independent.

4

We see that $$P(Y|X) \cdot P(X) = \frac{P(Y,X)}{P(X)}\cdot P(X) = P(Y,X) = P(Z),$$ Where $Z$ is defined as the variable over the cartesian product of the supports of $Y$ and $X$ (as in, a realization $z$ of $Z$ is in $S_Y \times S_X$). Hence, 1. is the more appropriate answer.

4

Here is a solution that uses SLLN and the additive property of the Poisson distribution. Let $(N_t)_{t \geq 0}$ be a Poisson process of unit rate. Then by the SLLN, together with the inequality $$\frac{N_{[t]}}{[t]+1} \leq \frac{N_t}{t} \leq \frac{N_{[t]+1}}{[t]},$$ it is easy to check that $N_t / t \to 1$ as $t \to \infty$ a.s. Now let $T_k = \lambda_1 ... 4 Each coupon appears$\log n$times on average. If I treat the number for a single coupon as a binomial, the variance is also asymptotically$\log n$. Now treat each of the counts as a normal variable$\mathcal N(\log n,\log n) $. Expectation of the maximum of gaussian random variables shows the maximum of$n$independent standard normal random ... 4$\sum_iY_i=0$. That can only happen for independent variables if they are all almost surely constant. Thus there are no such variables. 4 Fix a nonnegative integer$n$, an integer$m$with$0\leq m\leq n$, and a nonnegative integer$l$. Let$q(n,m,l)$be the probability that the union of random$m$-subsets$S_r$, for$r\in[l]$with$[l]:=\{1,2,\ldots,l\}$, of the set$[n]:=\{1,2,\ldots,n\}$is precisely$[n]$, where the random sets are chosen uniformly randomly and statistically independently.... 4 Since Brownian Motion has continuous paths with probability 1 thus$\int_{0}^{t} W_s ds$is well define. Now set$f(x)=x^3$. By application of Ito's lemma, we have $$f(W_t)=f(W_0)+\int_{0}^{t}f'(W_s)dW_s+\frac{1}{2}\int_{0}^{t}f''(W_s)ds$$ $$W^3(t)=3\int_{0}^{t}W^2_s\,dW_s+3\int_{0}^{t}W_sds$$ therefore $$\int_{0}^{t}W_sds=\frac{1}{3}W^3(t)-\int_{0}^{t}W^2_s\,... 4 The former parts are all right, except you didn't pay attention to the definition region of t. The joint distribution of t and x is:$$ f(x,t)=f_{0}(t\mid x)f(x)=3x/8~~~1\leq x\leq 2, ~x\leq t \leq 2x $$From the graph you draw, we could see the integration should be:$$P[T>3]=\int_3^4 dt \int_{t/2}^{2}3x/8 \, dx=0.171875$$4 An “in real life” interpretation of the last identity is that the number of different chaired even-sized committees from n people equals the number of chaired odd-sized committees from n people. Note that this is not an identity if n=1, as the left side is 1 and the right side is 0. However, it is an identity for n>1. A combinatorial proof (... 4 So both the fourth and the fifth rolls need to be sixes:$$P(4^{th},5^{th}\mbox{ rolls are sixes}) = \frac{1}{6^2}$$There are following possible combinations of rolls for the game to not end until the fifth roll:$$(XXXOO), (XOXOO),(OXXOO)$$where X,O represent non-six and six, respectively. Thus, the probability is:$$P(\mbox{end in fifth trial}) = \... 4 Here is simplified solution. In order experiment to end at 5th trial, the last two rolls must be 6. So we have, NNN66$Pr=5^3/6^5$6NN66$Pr=5^2/6^5$N6N66$Pr=5^2/6^5(N <6)$Add all probabilities. PS. The last two roll must be (6,6), so we get$\frac {1}{6^2}$The third roll can't be 6, so we get$\frac {5}{6}$And first two rolls can be anything ... 4 Since$\int f \, d\mu_n \to \int_0^1 f(x) \, \lambda(dx)$for all continuous bounded funtions$f$, we know that$\mu_n$converges weakly to$\lambda|_{[0,1]}$. Now it follows directly from the portmanteau theorem that $$\int f \, d\mu_n \to \int_0^1 f(x) \, \lambda(dx)$$ for all bounded measurable functions$f:[0,1] \to \mathbb{R}$with$\lambda(D_f)=0$... 4 No, your process does not make any sense. Even if the mean of$X_k$is$1$, this does not mean that $$\frac {X_1 + \dots + X_N}{N} = 1$$ for some$N$. This is a huge misunderstanding; I'll try to briefly explain why is it so. Random variables are not numbers When talking about a random variable, you're talking about a function. When you write$X_1$, ... 4 Actually, the answer to this lies in quite a bit more advanced topic called rough path theory (beware: PDF). A rough path is a way of "enhancing" a$\alpha$-Hölder continuous path with some extra information. A rough path is an ordered pair,$\textbf{X}=(X, \mathbb{X})$where$X\colon [0,T]\to V$where$V$is some Banach space (typically$\Bbb{R}$) and a ... 4 If$n$is the number of tosses, then$T=n-H$, so$H-T=2H-n$. Since$n$is large,$H$has a close to normal distribution, mean$n/2$, and variance$(n)(1/2)(1/2)$. So$2H-n$has close to normal distribution, mean$0$and variance$n$. Let$W$be a normal with mean$0$and variance$n$. Let us find the mean of$|W|\$. This is \int_{-\infty}^\infty \frac{|w|}... 4 Somewhat more generally, let X be any random variable whose moment generating function M(z) = \mathbb E[e^{tX}] is analytic in a neighbourhood of 0. This says that the series \sum_{j=0}^\infty \mathbb E[X^j] t^j/j!  has positive radius of convergence, i.e. |\mathbb E[X^j]| \le C D^j j! for some constants C, D. Then we have uniqueness in the ... 4 You can make a table. The names of the rows are the outcomes of the first dice (d1). The names of the columns are the outcomes of the second dice. In total we have 36 possible outcomes. And the cells marked with \color{red}x are the outcomes where the second dice has a greater outcome than the first dice.  \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{... 4 Since \mathcal F is any \sigma-algebra on [0,1], this is essentially the general case of: Halmos [1] showed that the range of a non-negative, finite measure is a closed subset of real numbers. [1] Halmos, Paul R. On the set of values of a finite measure. Bull. Amer. Math. Soc. 53 (1947), no. 2, 138--141. http://projecteuclid.org/euclid.bams/... 3 First, rewrite the exponenent as \begin{align*} \exp \left( n\left( a\frac{S_n}{n} - b \right) \right) \end{align*} By the Law of Large Numbers, we have a S_n/n- b \to a \mathbb{E}(\xi_1) - b a.s.. So a S_n -bn \to - \infty a.s. if and ony if b>a \mathbb{E}(\xi_1) =0, and thus \begin{align*} \exp( a S_n - b n ) \to 0 \text{ iff } b > 0. \end{... 3 Let X denote the random variable counting the number of times heads occurs in 2n independent flips of a fair coin. Via the binomial distribution: Pr(X=k) = \dfrac{\binom{2n}{k}}{2^{2n}} More specifically, Pr(X=n) = \dfrac{\binom{2n}{n}}{2^{2n}} Via stirling's approximation, \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}} as n grows large, so the ... 3 This can be solved using the general solution of the Gaussian integral \int\!d^nx\,\exp\Bigl(-\frac12 x^T A x\Bigr) = \sqrt{\frac{(2\pi)^n}{\det A}}.$$In your case, you we have that$$\mathbb{E}\left[e^{x^{T}Ax}\right] = \sqrt{\frac{1}{(2\pi)^n \det \Sigma}} \int\!d^nx \exp\Bigl(-\frac{1}{2} x^T \Sigma^{-1} x\Bigr) e^{x^T A x}=\sqrt{\frac{1}{(2\pi)^n \...

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# Entropy for an irreversible adiabatic process?

1. Mar 8, 2010

### faceoclock

1. The problem statement, all variables and given/known data
Calculate the entropy for an irreversible adiabatic expansion of an ideal gas which goes from (Vi, Pi) = (9.01L, 3.00 bar) to (Vf, Pf) = (15.44L, 1.50 bar)

2. Relevant equations

$$\Delta$$S = $$\int dq_{reversible}/T$$

3. The attempt at a solution
I understand that entropy is a state function, and thus independent of path. So the above equation can be used to calculate the change in entropy for an irreversible adiabatic expansion. However, what I don't understand is how the entropy can possibly be non-zero. I mean, in a reversible adiabatic expansion which has the same (Vi,Pi) and (Vf,Pf) as the irreversible case, q = 0 by definition. So if $$dq_{reversible}$$ = 0 then how can $$\Delta$$S be non-zero?

Any help as to what I'm missing here will be highly appreciated!

Edit: having written this question out, I think I might be on to something. I don't think it's possible for the reversible expansion to be adiabatic because U is a path function, and must be the same for reversible and irreversible case. In irreversible case, delta U = w. In the reversible case, w is different, so the remainder of the energy must be necessarily in form of heat, which is q(reversible). Am I on the right track?

Last edited: Mar 8, 2010
2. Mar 21, 2010

### passingthru

Your problem doesn't state how the volume was increased, whether it was done by the gas, or by opening a divider of some sort. . The first law for non-diffusive systems is $d U = d Q -d W$. In this problem, $d U = 0$ so that $d Q = d W = T d S = P dV$. From the ideal gas law, $PV = nRT$. Using this to substitute for $P$, we have $dS = \frac{nRdV}{V}$. Integrating yields $\Delta S = nR\;ln \left( \frac{V_f}{V_i} \right)$. Thus, $Q$ and $S$ are not zero.

The reason I asked about how your volume was increased is because there is an irreversible process called free expansion where a divider is removed in order to increase the volume. The gas does no work in increasing the volume. The kinetic energy of the molecules stays the same. This is a question in the GRE Physcis practice bulletin, question 47. I have three different thermo texts that do not agree. It seems to me that the molecules hit the inner surface of the container with the same amount of force as before, but now the force is applied to a larger inner surface area, so the pressure, which is force per unit area, would go down. Since the kinetic energy of the molecules stays the same, it would seem that the temperature must stay the same because of the kinetic theory of gases. What stumps me is that the texts say that $dQ$ and $dW$ are zero, so the above calculation would not be correct, but they give this result as the correct answer to the question. Can anyone explain this?

3. Mar 23, 2010

### Andrew Mason

The reversible path between these two states involves doing work (ie expansion occurs against an external pressure infinitessimally lower than the internal gas pressure). Since the temperature, hence internal energy does not change ($\Delta U = 0$), $\Delta Q = W$. So heat must be added to reach the final state. If heat is added, entropy increases: dS = dQ/T. Simply put, the reversible path between an adiabatic irreversible expansion is not adiabatic.

AM

Last edited: Mar 23, 2010
4. Mar 23, 2010

### Count Iblis

http://en.wikipedia.org/wiki/Talk:Work_(thermodynamics [Broken])

Last edited by a moderator: May 4, 2017


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RUS  ENG JOURNALS   PEOPLE   ORGANISATIONS   CONFERENCES   SEMINARS   VIDEO LIBRARY   PACKAGE AMSBIB
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 Funktsional. Anal. i Prilozhen.: Year: Volume: Issue: Page: Find

 Funktsional. Anal. i Prilozhen., 2001, Volume 35, Issue 3, Pages 36–47 (Mi faa257)

On Polytopes that are Simple at the Edges

V. A. Timorinab

a Steklov Mathematical Institute, Russian Academy of Sciences
b University of Toronto

Abstract: We study some combinatorial properties of polytopes that are simple at the edges. We give an elementary geometric proof of an analog of the hard Lefschetz theorem for the polytopes for which the distance between any two nonsimple vertices is sufficiently large. This implies that the $h$-vector of such polytopes satisfies the relations $h_{[d/2]}\ge h_{[d/2]+1}\ge\cdots\ge h_d=1$, where $d$ is the dimension of the polytope, which proves a special case of Stanley's conjecture.

DOI: https://doi.org/10.4213/faa257

Full text: PDF file (188 kB)
References: PDF file   HTML file

English version:
Functional Analysis and Its Applications, 2001, 35:3, 189–198

Bibliographic databases:

UDC: 514.172.45+515.165.4

Citation: V. A. Timorin, “On Polytopes that are Simple at the Edges”, Funktsional. Anal. i Prilozhen., 35:3 (2001), 36–47; Funct. Anal. Appl., 35:3 (2001), 189–198

Citation in format AMSBIB
\Bibitem{Tim01} \by V.~A.~Timorin \paper On Polytopes that are Simple at the Edges \jour Funktsional. Anal. i Prilozhen. \yr 2001 \vol 35 \issue 3 \pages 36--47 \mathnet{http://mi.mathnet.ru/faa257} \crossref{https://doi.org/10.4213/faa257} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=1864987} \zmath{https://zbmath.org/?q=an:1009.52020} \elib{http://elibrary.ru/item.asp?id=14150283} \transl \jour Funct. Anal. Appl. \yr 2001 \vol 35 \issue 3 \pages 189--198 \crossref{https://doi.org/10.1023/A:1012374711617} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000172598500004} \scopus{http://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-0035733985}

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• https://doi.org/10.4213/faa257
• http://mi.mathnet.ru/eng/faa/v35/i3/p36

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This publication is cited in the following articles:
1. Karu, K, “Hard Lefschetz theorem for nonrational polytopes”, Inventiones Mathematicae, 157:2 (2004), 419
2. V. M. Buchstaber, “Ring of Simple Polytopes and Differential Equations”, Proc. Steklov Inst. Math., 263 (2008), 13–37
3. Cattani, E, “Mixed Lefschetz Theorems and Hodge-Riemann Bilinear Relations”, International Mathematics Research Notices, 2008, rnn025
4. Esterov A., “Characteristic Classes of Affine Varieties and Plucker Formulas For Affine Morphisms”, J. Eur. Math. Soc., 20:1 (2018), 15–59
•  Number of views: This page: 311 Full text: 170 References: 43


text:
Asked in Math and Arithmetic, Algebra, Mathematical Constants
Math and Arithmetic
Algebra
Mathematical Constants

Why is zero factorial equal to one?

12/20/2010

It is defined to be so, presumably for the sake of convenience. Some people will tell you that it can be proven, but if it could be proven, it would be categorized as a theorem, not a definition. The common notion is that, since n!/n = (n-1)!, we can substitute 1 in for n and we can see that 0! is 1. The problem with this is that, if 0! is not assumed to be 1 (which is an assumption mathematicians do make), this rule will only hold for values of n that are equal to or greater than 2. To see why, let's look at the proof that n!/n = (n-1)!:

n!/n = n!/n | reflexive property

(n)(n-1)(n-2).../n = n!/n | definition of factorial

(n-1)(n-2)... = n!/n | cancelling the common factor of n

(n-1)! = n!/n | definition of factorial

Notice that, in order for n! to be described as (n)(n-1)(n-2)... and proceed to be rewritten as (n-1)! after the n's cancel, the natural number n had to be greater than some natural number for (n-1) to be a factor in the factorial. This means that n must be at least 2, because if n were 1, (n-1) would not have been a factor of the factorial, and the proof would fail unless we assume that n is at least 2. So now you know that this rule cannot prove that 0! is 1 because 1 cannot be substituted into the rule because, as it stands, the rule is only valid for values of 2 or greater. The rule is valid for values of 1 or greater if it is assumed that 0! is 1, but that is what you are trying to prove.


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# deepchem.rl package¶

## deepchem.rl.a3c module¶

Asynchronous Advantage Actor-Critic (A3C) algorithm for reinforcement learning.

class deepchem.rl.a3c.A3C(env, policy, max_rollout_length=20, discount_factor=0.99, advantage_lambda=0.98, value_weight=1.0, entropy_weight=0.01, optimizer=None, model_dir=None, use_hindsight=False)[source]

Bases: object

Implements the Asynchronous Advantage Actor-Critic (A3C) algorithm for reinforcement learning.

The algorithm is described in Mnih et al, “Asynchronous Methods for Deep Reinforcement Learning” (https://arxiv.org/abs/1602.01783). This class requires the policy to output two quantities: a vector giving the probability of taking each action, and an estimate of the value function for the current state. It optimizes both outputs at once using a loss that is the sum of three terms:

1. The policy loss, which seeks to maximize the discounted reward for each action.
2. The value loss, which tries to make the value estimate match the actual discounted reward that was attained at each step.
3. An entropy term to encourage exploration.

This class only supports environments with discrete action spaces, not continuous ones. The “action” argument passed to the environment is an integer, giving the index of the action to perform.

This class supports Generalized Advantage Estimation as described in Schulman et al., “High-Dimensional Continuous Control Using Generalized Advantage Estimation” (https://arxiv.org/abs/1506.02438). This is a method of trading off bias and variance in the advantage estimate, which can sometimes improve the rate of convergance. Use the advantage_lambda parameter to adjust the tradeoff.

This class supports Hindsight Experience Replay as described in Andrychowicz et al., “Hindsight Experience Replay” (https://arxiv.org/abs/1707.01495). This is a method that can enormously accelerate learning when rewards are very rare. It requires that the environment state contains information about the goal the agent is trying to achieve. Each time it generates a rollout, it processes that rollout twice: once using the actual goal the agent was pursuing while generating it, and again using the final state of that rollout as the goal. This guarantees that half of all rollouts processed will be ones that achieved their goals, and hence received a reward.

To use this feature, specify use_hindsight=True to the constructor. The environment must have a method defined as follows:

def apply_hindsight(self, states, actions, goal):
... return new_states, rewards

The method receives the list of states generated during the rollout, the action taken for each one, and a new goal state. It should generate a new list of states that are identical to the input ones, except specifying the new goal. It should return that list of states, and the rewards that would have been received for taking the specified actions from those states.

fit(total_steps, max_checkpoints_to_keep=5, checkpoint_interval=600, restore=False)[source]

Train the policy.

Parameters: total_steps (int) – the total number of time steps to perform on the environment, across all rollouts on all threads max_checkpoints_to_keep (int) – the maximum number of checkpoint files to keep. When this number is reached, older files are deleted. checkpoint_interval (float) – the time interval at which to save checkpoints, measured in seconds restore (bool) – if True, restore the model from the most recent checkpoint and continue training from there. If False, retrain the model from scratch.
predict(state, use_saved_states=True, save_states=True)[source]

Compute the policy’s output predictions for a state.

If the policy involves recurrent layers, this method can preserve their internal states between calls. Use the use_saved_states and save_states arguments to specify how it should behave.

Parameters: state (array) – the state of the environment for which to generate predictions use_saved_states (bool) – if True, the states most recently saved by a previous call to predict() or select_action() will be used as the initial states. If False, the internal states of all recurrent layers will be set to all zeros before computing the predictions. save_states (bool) – if True, the internal states of all recurrent layers at the end of the calculation will be saved, and any previously saved states will be discarded. If False, the states at the end of the calculation will be discarded, and any previously saved states will be kept. the array of action probabilities, and the estimated value function
restore()[source]

Reload the model parameters from the most recent checkpoint file.

select_action(state, deterministic=False, use_saved_states=True, save_states=True)[source]

Select an action to perform based on the environment’s state.

If the policy involves recurrent layers, this method can preserve their internal states between calls. Use the use_saved_states and save_states arguments to specify how it should behave.

Parameters: state (array) – the state of the environment for which to select an action deterministic (bool) – if True, always return the best action (that is, the one with highest probability). If False, randomly select an action based on the computed probabilities. use_saved_states (bool) – if True, the states most recently saved by a previous call to predict() or select_action() will be used as the initial states. If False, the internal states of all recurrent layers will be set to all zeros before computing the predictions. save_states (bool) – if True, the internal states of all recurrent layers at the end of the calculation will be saved, and any previously saved states will be discarded. If False, the states at the end of the calculation will be discarded, and any previously saved states will be kept. the index of the selected action
class deepchem.rl.a3c.A3CLoss(value_weight, entropy_weight, **kwargs)[source]

This layer computes the loss function for A3C.

add_summary_to_tg()

Can only be called after self.create_layer to gaurentee that name is not none

clone(in_layers)

Create a copy of this layer with different inputs.

copy(replacements={}, variables_graph=None, shared=False)

Duplicate this Layer and all its inputs.

This is similar to clone(), but instead of only cloning one layer, it also recursively calls copy() on all of this layer’s inputs to clone the entire hierarchy of layers. In the process, you can optionally tell it to replace particular layers with specific existing ones. For example, you can clone a stack of layers, while connecting the topmost ones to different inputs.

For example, consider a stack of dense layers that depend on an input:

>>> input = Feature(shape=(None, 100))
>>> dense1 = Dense(100, in_layers=input)
>>> dense2 = Dense(100, in_layers=dense1)
>>> dense3 = Dense(100, in_layers=dense2)

The following will clone all three dense layers, but not the input layer. Instead, the input to the first dense layer will be a different layer specified in the replacements map.

>>> replacements = {input: new_input}
>>> dense3_copy = dense3.copy(replacements)

Parameters: replacements (map) – specifies existing layers, and the layers to replace them with (instead of cloning them). This argument serves two purposes. First, you can pass in a list of replacements to control which layers get cloned. In addition, as each layer is cloned, it is added to this map. On exit, it therefore contains a complete record of all layers that were copied, and a reference to the copy of each one. variables_graph (TensorGraph) – an optional TensorGraph from which to take variables. If this is specified, the current value of each variable in each layer is recorded, and the copy has that value specified as its initial value. This allows a piece of a pre-trained model to be copied to another model. shared (bool) – if True, create new layers by calling shared() on the input layers. This means the newly created layers will share variables with the original ones.
create_tensor(**kwargs)[source]
layer_number_dict = {}
none_tensors()
set_summary(summary_op, summary_description=None, collections=None)

Annotates a tensor with a tf.summary operation Collects data from self.out_tensor by default but can be changed by setting self.tb_input to another tensor in create_tensor

Parameters: summary_op (str) – summary operation to annotate node summary_description (object, optional) – Optional summary_pb2.SummaryDescription() collections (list of graph collections keys, optional) – New summary op is added to these collections. Defaults to [GraphKeys.SUMMARIES]
set_tensors(tensor)
set_variable_initial_values(values)

Set the initial values of all variables.

This takes a list, which contains the initial values to use for all of this layer’s values (in the same order retured by TensorGraph.get_layer_variables()). When this layer is used in a TensorGraph, it will automatically initialize each variable to the value specified in the list. Note that some layers also have separate mechanisms for specifying variable initializers; this method overrides them. The purpose of this method is to let a Layer object represent a pre-trained layer, complete with trained values for its variables.

shape

Get the shape of this Layer’s output.

shared(in_layers)

Create a copy of this layer that shares variables with it.

This is similar to clone(), but where clone() creates two independent layers, this causes the layers to share variables with each other.

Parameters: in_layers (list tensor) – in tensors for the shared layer (List) – Layer

## deepchem.rl.ppo module¶

Proximal Policy Optimization (PPO) algorithm for reinforcement learning.

class deepchem.rl.ppo.PPO(env, policy, max_rollout_length=20, optimization_rollouts=8, optimization_epochs=4, batch_size=64, clipping_width=0.2, discount_factor=0.99, advantage_lambda=0.98, value_weight=1.0, entropy_weight=0.01, optimizer=None, model_dir=None, use_hindsight=False)[source]

Bases: object

Implements the Proximal Policy Optimization (PPO) algorithm for reinforcement learning.

The algorithm is described in Schulman et al, “Proximal Policy Optimization Algorithms” (https://openai-public.s3-us-west-2.amazonaws.com/blog/2017-07/ppo/ppo-arxiv.pdf). This class requires the policy to output two quantities: a vector giving the probability of taking each action, and an estimate of the value function for the current state. It optimizes both outputs at once using a loss that is the sum of three terms:

1. The policy loss, which seeks to maximize the discounted reward for each action.
2. The value loss, which tries to make the value estimate match the actual discounted reward that was attained at each step.
3. An entropy term to encourage exploration.

This class only supports environments with discrete action spaces, not continuous ones. The “action” argument passed to the environment is an integer, giving the index of the action to perform.

This class supports Generalized Advantage Estimation as described in Schulman et al., “High-Dimensional Continuous Control Using Generalized Advantage Estimation” (https://arxiv.org/abs/1506.02438). This is a method of trading off bias and variance in the advantage estimate, which can sometimes improve the rate of convergance. Use the advantage_lambda parameter to adjust the tradeoff.

This class supports Hindsight Experience Replay as described in Andrychowicz et al., “Hindsight Experience Replay” (https://arxiv.org/abs/1707.01495). This is a method that can enormously accelerate learning when rewards are very rare. It requires that the environment state contains information about the goal the agent is trying to achieve. Each time it generates a rollout, it processes that rollout twice: once using the actual goal the agent was pursuing while generating it, and again using the final state of that rollout as the goal. This guarantees that half of all rollouts processed will be ones that achieved their goals, and hence received a reward.

To use this feature, specify use_hindsight=True to the constructor. The environment must have a method defined as follows:

def apply_hindsight(self, states, actions, goal):
... return new_states, rewards

The method receives the list of states generated during the rollout, the action taken for each one, and a new goal state. It should generate a new list of states that are identical to the input ones, except specifying the new goal. It should return that list of states, and the rewards that would have been received for taking the specified actions from those states.

fit(total_steps, max_checkpoints_to_keep=5, checkpoint_interval=600, restore=False)[source]

Train the policy.

Parameters: total_steps (int) – the total number of time steps to perform on the environment, across all rollouts on all threads max_checkpoints_to_keep (int) – the maximum number of checkpoint files to keep. When this number is reached, older files are deleted. checkpoint_interval (float) – the time interval at which to save checkpoints, measured in seconds restore (bool) – if True, restore the model from the most recent checkpoint and continue training from there. If False, retrain the model from scratch.
predict(state, use_saved_states=True, save_states=True)[source]

Compute the policy’s output predictions for a state.

If the policy involves recurrent layers, this method can preserve their internal states between calls. Use the use_saved_states and save_states arguments to specify how it should behave.

Parameters: state (array) – the state of the environment for which to generate predictions use_saved_states (bool) – if True, the states most recently saved by a previous call to predict() or select_action() will be used as the initial states. If False, the internal states of all recurrent layers will be set to all zeros before computing the predictions. save_states (bool) – if True, the internal states of all recurrent layers at the end of the calculation will be saved, and any previously saved states will be discarded. If False, the states at the end of the calculation will be discarded, and any previously saved states will be kept. the array of action probabilities, and the estimated value function
restore()[source]

Reload the model parameters from the most recent checkpoint file.

select_action(state, deterministic=False, use_saved_states=True, save_states=True)[source]

Select an action to perform based on the environment’s state.

If the policy involves recurrent layers, this method can preserve their internal states between calls. Use the use_saved_states and save_states arguments to specify how it should behave.

Parameters: state (array) – the state of the environment for which to select an action deterministic (bool) – if True, always return the best action (that is, the one with highest probability). If False, randomly select an action based on the computed probabilities. use_saved_states (bool) – if True, the states most recently saved by a previous call to predict() or select_action() will be used as the initial states. If False, the internal states of all recurrent layers will be set to all zeros before computing the predictions. save_states (bool) – if True, the internal states of all recurrent layers at the end of the calculation will be saved, and any previously saved states will be discarded. If False, the states at the end of the calculation will be discarded, and any previously saved states will be kept. the index of the selected action
class deepchem.rl.ppo.PPOLoss(value_weight, entropy_weight, clipping_width, **kwargs)[source]

This layer computes the loss function for PPO.

add_summary_to_tg()

Can only be called after self.create_layer to gaurentee that name is not none

clone(in_layers)

Create a copy of this layer with different inputs.

copy(replacements={}, variables_graph=None, shared=False)

Duplicate this Layer and all its inputs.

This is similar to clone(), but instead of only cloning one layer, it also recursively calls copy() on all of this layer’s inputs to clone the entire hierarchy of layers. In the process, you can optionally tell it to replace particular layers with specific existing ones. For example, you can clone a stack of layers, while connecting the topmost ones to different inputs.

For example, consider a stack of dense layers that depend on an input:

>>> input = Feature(shape=(None, 100))
>>> dense1 = Dense(100, in_layers=input)
>>> dense2 = Dense(100, in_layers=dense1)
>>> dense3 = Dense(100, in_layers=dense2)

The following will clone all three dense layers, but not the input layer. Instead, the input to the first dense layer will be a different layer specified in the replacements map.

>>> replacements = {input: new_input}
>>> dense3_copy = dense3.copy(replacements)

Parameters: replacements (map) – specifies existing layers, and the layers to replace them with (instead of cloning them). This argument serves two purposes. First, you can pass in a list of replacements to control which layers get cloned. In addition, as each layer is cloned, it is added to this map. On exit, it therefore contains a complete record of all layers that were copied, and a reference to the copy of each one. variables_graph (TensorGraph) – an optional TensorGraph from which to take variables. If this is specified, the current value of each variable in each layer is recorded, and the copy has that value specified as its initial value. This allows a piece of a pre-trained model to be copied to another model. shared (bool) – if True, create new layers by calling shared() on the input layers. This means the newly created layers will share variables with the original ones.
create_tensor(**kwargs)[source]
layer_number_dict = {}
none_tensors()
set_summary(summary_op, summary_description=None, collections=None)

Annotates a tensor with a tf.summary operation Collects data from self.out_tensor by default but can be changed by setting self.tb_input to another tensor in create_tensor

Parameters: summary_op (str) – summary operation to annotate node summary_description (object, optional) – Optional summary_pb2.SummaryDescription() collections (list of graph collections keys, optional) – New summary op is added to these collections. Defaults to [GraphKeys.SUMMARIES]
set_tensors(tensor)
set_variable_initial_values(values)

Set the initial values of all variables.

This takes a list, which contains the initial values to use for all of this layer’s values (in the same order retured by TensorGraph.get_layer_variables()). When this layer is used in a TensorGraph, it will automatically initialize each variable to the value specified in the list. Note that some layers also have separate mechanisms for specifying variable initializers; this method overrides them. The purpose of this method is to let a Layer object represent a pre-trained layer, complete with trained values for its variables.

shape

Get the shape of this Layer’s output.

shared(in_layers)

Create a copy of this layer that shares variables with it.

This is similar to clone(), but where clone() creates two independent layers, this causes the layers to share variables with each other.

Parameters: in_layers (list tensor) – in tensors for the shared layer (List) – Layer

## Module contents¶

Interface for reinforcement learning.

class deepchem.rl.Environment(state_shape, n_actions, state_dtype=None)[source]

Bases: object

An environment in which an actor performs actions to accomplish a task.

An environment has a current state, which is represented as either a single NumPy array, or optionally a list of NumPy arrays. When an action is taken, that causes the state to be updated. Exactly what is meant by an “action” is defined by each subclass. As far as this interface is concerned, it is simply an arbitrary object. The environment also computes a reward for each action, and reports when the task has been terminated (meaning that no more actions may be taken).

Environment objects should be written to support pickle and deepcopy operations. Many algorithms involve creating multiple copies of the Environment, possibly running in different processes or even on different computers.

n_actions

The number of possible actions that can be performed in this Environment.

reset()[source]

Initialize the environment in preparation for doing calculations with it.

This must be called before calling step() or querying the state. You can call it again later to reset the environment back to its original state.

state

The current state of the environment, represented as either a NumPy array or list of arrays.

If reset() has not yet been called at least once, this is undefined.

state_dtype

The dtypes of the arrays that describe a state.

If the state is a single array, this returns the dtype of that array. If the state is a list of arrays, this returns a list containing the dtypes of the arrays.

state_shape

The shape of the arrays that describe a state.

If the state is a single array, this returns a tuple giving the shape of that array. If the state is a list of arrays, this returns a list of tuples where each tuple is the shape of one array.

step(action)[source]

Take a time step by performing an action.

This causes the “state” and “terminated” properties to be updated.

Parameters: action (object) – an object describing the action to take the reward earned by taking the action, represented as a floating point number (higher values are better)
terminated

Whether the task has reached its end.

If reset() has not yet been called at least once, this is undefined.

class deepchem.rl.GymEnvironment(name)[source]

This is a convenience class for working with environments from OpenAI Gym.

n_actions

The number of possible actions that can be performed in this Environment.

reset()[source]
state

The current state of the environment, represented as either a NumPy array or list of arrays.

If reset() has not yet been called at least once, this is undefined.

state_dtype

The dtypes of the arrays that describe a state.

If the state is a single array, this returns the dtype of that array. If the state is a list of arrays, this returns a list containing the dtypes of the arrays.

state_shape

The shape of the arrays that describe a state.

If the state is a single array, this returns a tuple giving the shape of that array. If the state is a list of arrays, this returns a list of tuples where each tuple is the shape of one array.

step(action)[source]
terminated

Whether the task has reached its end.

If reset() has not yet been called at least once, this is undefined.

class deepchem.rl.Policy[source]

Bases: object

A policy for taking actions within an environment.

A policy is defined by a set of TensorGraph Layer objects that perform the necessary calculations. There are many algorithms for reinforcement learning, and they differ in what values they require a policy to compute. That makes it impossible to define a single interface allowing any policy to be optimized with any algorithm. Instead, this interface just tries to be as flexible and generic as possible. Each algorithm must document what values it expects create_layers() to return.

Policy objects should be written to support pickling. Many algorithms involve creating multiple copies of the Policy, possibly running in different processes or even on different computers.

create_layers(state, **kwargs)[source]

Create the TensorGraph Layers that define the policy.

The arguments always include a list of Feature layers representing the current state of the environment (one layer for each array in the state). Depending on the algorithm being used, other arguments might get passed as well. It is up to each algorithm to document that.

This method should construct and return a dict that maps strings to Layer objects. Each algorithm must document what Layers it expects the policy to create. If this method is called multiple times, it should create a new set of Layers every time.


text:
## Basic College Mathematics (10th Edition)

Published by Pearson

# Chapter 6 - Percent - 6.4 Using Proportions to Solve Percent Problems - 6.4 Exercises - Page 422: 74

#### Answer

300g of carbohydrates is the recommended daily value.

#### Work Step by Step

The part is 27. The percent is 9%. The whole is unknown. $\frac{27}{x} = \frac{9}{100}$ $9x = 27\times100$ $\frac{9x}{9} = \frac{2700}{9}$ $x = 300$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
Compound Registration Installation requirements

You need three things to check, before starting to install Compound Registration: minimum system requirements, existence of a home directory and if the database is properly set up.

1. Environment

It is assumed that the application server is properly set up with a compatible version of the operating system, Java environment, and Apache Tomcat. Please make sure your system meets the minimum requirements to run Compound Registration.

2. Compound Registration home directory

Home directory must be set up using the environment variable REGISTRYCXN_HOME. Since CompReg version 17.06.12-1738 e xterna l resources of the registration system will be loaded from the home directory of the application.

Home directory must be an existing directory with read and write access rights. In case the home directory is not provided - or not meeting the requirements above - the application will start, but the user interface will display an error related to invalid application state. The l og will contain the error entry of the exact problem related to the home directory settings. You can specify any location on the disk, but please be sure to specify an absolute path. If Tomcat was running while you defined the environment variable, a restart might be required.

How to set up the environment variable?

Run the Command Prompt as Administrator and type the following:

On Unix based systems: export REGISTRYCXN_HOME=/Users/johndoe/.chemaxon

On Window based systems: setx -M REGISTRYCXN_HOME "C:\path\to\any\directory"

To make sure that the environment variable properly set up restart Command Prompt and type: echo %REGISTRYCXN_HOME% (Windows) or echo $REGISTRYCXN_HOME (Unix). If everything has been entered correctly the previously typed pathway will be displayed. License files This step is just optional. In the following, you will have the opportunity to upload the suitable license file. License files must be located under the directory$REGISTRYCXN_HOME/licenses . In the case of a missing license file, the logs will contain the error of the problem, and the user interface of the application will display an error upon user login.

Since version 17.10.30-2159:

You can also set a license file via URL with the $CHEMAXON_LICENSE_URL environment variable, but you have to make sure, in this case,$REGISTRYCXN_HOME/licenses is empty:


text:
# Definition of Reciprocal equation. Meaning of Reciprocal equation. Synonyms of Reciprocal equation

Here you will find one or more explanations in English for the word Reciprocal equation. Also in the bottom left of the page several parts of wikipedia pages related to the word Reciprocal equation and, of course, Reciprocal equation synonyms and on the right images related to the word Reciprocal equation.

## Definition of Reciprocal equation

Reciprocal equation
Reciprocal Re*cip"ro*cal, a. [L. reciprocus; of unknown origin.] 1. Recurring in vicissitude; alternate. 2. Done by each to the other; interchanging or interchanged; given and received; due from each to each; mutual; as, reciprocal love; reciprocal duties. Let our reciprocal vows be remembered. --Shak. 3. Mutually interchangeable. These two rules will render a definition reciprocal with the thing defined. --I. Watts. 4. (Gram.) Reflexive; -- applied to pronouns and verbs, but sometimes limited to such pronouns as express mutual action. 5. (Math.) Used to denote different kinds of mutual relation; often with reference to the substitution of reciprocals for given quantities. See the Phrases below. Reciprocal equation (Math.), one which remains unchanged in form when the reciprocal of the unknown quantity is substituted for that quantity. Reciprocal figures (Geom.), two figures of the same kind (as triangles, parallelograms, prisms, etc.), so related that two sides of the one form the extremes of a proportion of which the means are the two corresponding sides of the other; in general, two figures so related that the first corresponds in some special way to the second, and the second corresponds in the same way to the first. Reciprocal proportion (Math.), a proportion such that, of four terms taken in order, the first has to the second the same ratio which the fourth has to the third, or the first has to the second the same ratio which the reciprocal of the third has to the reciprocal of the fourth. Thus, 2:5: :20:8 form a reciprocal proportion, because 2:5: :1/20:1/8. Reciprocal quantities (Math.), any two quantities which produce unity when multiplied together. Reciprocal ratio (Math.), the ratio between the reciprocals of two quantities; as, the reciprocal ratio of 4 to 9 is that of 3/4 to 1/9. Reciprocal terms (Logic), those terms which have the same signification, and, consequently, are convertible, and may be used for each other. Syn: Mutual; alternate. Usage: Reciprocal, Mutual. The distinctive idea of mutual is, that the parties unite by interchange in the same act; as, a mutual covenant; mutual affection, etc. The distinctive idea of reciprocal is, that one party acts by way of return or response to something previously done by the other party; as, a reciprocal kindness; reciprocal reproaches, etc. Love is reciprocal when the previous affection of one party has drawn forth the attachment of the other. To make it mutual in the strictest sense, the two parties should have fallen in love at the same time; but as the result is the same, the two words are here used interchangeably. The ebbing and flowing of the tide is a case where the action is reciprocal, but not mutual.

## Meaning of Reciprocal equation from wikipedia

- the Lineweaver–Burk plot (or double reciprocal plot) is a graphical representation of the Lineweaver–Burk equation of enzyme kinetics, described by Hans...
- In thermodynamics, the Onsager reciprocal relations express the equality of certain ratios between flows and forces in thermodynamic systems out of equilibrium...
- In physics, the reciprocal lattice represents the Fourier transform of another lattice (usually a Bravais lattice). In normal usage, the initial lattice...
- The Schrödinger equation is a linear partial differential equation that describes the wave function or state function of a quantum-mechanical system. It...
- number of numbers times the reciprocal of the sum of their reciprocals. The optic equation requires the sum of the reciprocals of two positive integers a...
- This has the same form as an equation for a straight line: y = m x + b , {\displaystyle y=mx+b,} where x is the reciprocal of T. So, when a reaction has...
- theory, the optic equation is an equation that requires the sum of the reciprocals of two positive integers a and b to equal the reciprocal of a third positive...
- equations Gibbs–Helmholtz equation Gibbs' phase rule Kopp's law Kopp–Neumann law Noro–Frenkel law of corresponding states Onsager reciprocal relations Stefan number...
- In mathematics, a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x−1, is a number which when multiplied by x yields the multiplicative...
- the pressure and the reciprocal temperature. This ****umes a temperature-independent heat of vaporization. The Antoine equation allows an improved, but...


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

Here's the question you clicked on:

## anonymous 4 years ago Find the length of a 40 degree arc in a circle with a radius of 4

• This Question is Closed
1. anonymous

|dw:1327427504346:dw|

2. anonymous

C=(180-40)/2=70

3. amistre64

arcLength = 40*4pi/180

4. anonymous

$CH=\cos (70) * radius(4)$

5. anonymous

BC=2CH because AB=AC and CH=BH

6. amistre64

arc length is just part of a circumference; C = 2pi r arc length = angle * radius ; whenere the angle is in radians

7. amistre64

i think ali might of read it wrong to begin with ...

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text:
# Crazy Calculus 5

Calculus Level pending

$\large f\left( \dfrac xy \right) = f( x) - f( y)$

Let $$f(x)$$ be defined for all real $$x > 1$$ such that it satisfies the above functional equation for all real $$x$$ and $$y$$ and that $$f(e) =1$$. Which is of the following options is correct?

• $$P: f \left( x \right)$$ is bounded
• $$Q: f\left( \dfrac 1x \right) \to 0$$ as $$x\to 0$$
• $$R: xf\left( x \right) \to 1$$ as $$x \to 0$$
• $$S: f\left( x \right) =\ln x$$
×


text:
[NTG-context] Re: A few questions regarding ConTeXt

Mats Broberg mats.broberg at arsimprimispress.com
Sun Jul 25 12:19:20 CEST 2004

> -----Original Message-----
> From: ntg-context-bounces at ntg.nl
> [mailto:ntg-context-bounces at ntg.nl] On Behalf Of Peter Münster
> Sent: den 24 juli 2004 19:45

> Hello Mats,
> I agree with you: ConTeXt is certainly superior to LaTeX, but
> sometimes, I know how to do a special thing in LaTeX, but not
> in ConTeXt. Often I don't know, if a special feature is
> already there but undocumented, or not. Sometimes, I find a
> sort of "workaround" to get some LaTeX behaviour, for example
> "\flushbottom". For beginners there is the document
> http://www.berenddeboer.net/tex/LaTeX2ConTeXt.> pdf by Berend
> de Boer. It would be nice, to have something
> like this for advanced topics, such as varioref or lettrine.
> These special "LaTeX to ConTeXt" examples can then be added
> to the ConTeXt-wiki: http://contextgarden.net/From_LaTeX_to_ConTeXt
> Cheers, Peter

Thanks for the info.

Having looked more closely at this, I guess it now stands between
learning LaTeX and the memoir class, or learning ConTeXt. There are
things that talk in favour for both these roadmaps, but I have not
decided yet.

One of the issues is how to solve a problem in ConTeXt if it suddenly
becomes clear this or that feature is not implemented.

Best regards,
Mats Broberg


text:
# Communication complexity of string matching

let us consider Alice and Bob communicate over channel bit by bit. They have one string over $\{0,1\}^{*}$. Let $x$ (uniformly at random) be the string that Bob have and $y$ be the string that Alice have. They want to check whether there strings are same or not.

On simple algorithm is exchange $n$ bits one by one. We want to solve this problem with few bits exchanged between Alice and Bob.

My Idea : We will randomly select $k = O(\log {n})$ many bits and exchange only these bits. In this probability of error is going to be less than $1/2^{k}$ (because probability of making error for one bit is 1/2) and as each time we are randomly selecting the bits overall probability of error will be less than $\frac{1}{2^{k}}$.

Question : Is my solution optimal one to solve the problem described above?

By simply exchanging random indices you cannot hope to do better than $\Omega(n)$ communication if you want constant error probability. If the strings differ in only a single bit, and you exchange $k$ indices, then the probability of finding the index in which they differ is $\frac{k}{n}$, so if you want this to be constant you must have $k=\Omega(n)$.
The randomized communication complexity can indeed be reduced to $O(\log n)$, but you're going to need stronger arguments. One way to do this is to encode both words with an error correcting code having a large distance, and then use your algorithm (naive sampling) on the codewords.
To elaborate on the above idea, the choice of code doesn't really matter, as the only property you need in order to make naive sampling work is large distance. One example is using Reed-Solomon, given $s\in\{0,1\}^n$, think of it as a string of length $k$ over $\mathbb{F_q}$, with $q=n$ and $k=\frac{n}{\log n}$. In that case you can think of $s=s_0,...,s_{k-1}$ as the polynomial $P_s=s_0+s_1x+...+s_k x^{k-1}$ over $\mathbb{F}_q$. If $s_1\neq s_2$ then their corresponding polynomials agree on at most $k-1$ points in $\mathbb{F}_q$, which means they will differ on at least $n-(k-1)=n\left(1-\frac{1}{\log n}\right)+1$ points in $\mathbb{F}_{q=n}$. Thus, with high probability, $P_{s_1},P_{s_2}$ will disagree on a random point in $\mathbb{F}_q$. This protocol requires you to send three elements in $\mathbb{F}_q$ (a point and both polynomials evaluations on it), which results in $O(\log q)=O(\log n)$ communication.
This is only one option, there are many randomized protocols for equality. One example is using a simple public coins protocol, in which both parties share two random strings $r,r'\sim U_n$, and exchange the values of $\langle xr,r'\rangle, \langle yr,r'\rangle$. This succeeds with probability $\frac{1}{2}$, and you can turn it into a private coin protocol with logarithmic overhead in communication, see these notes for details. Another option is to choose a prime $p\le poly(n)$ at random, and exchange $x,y\bmod p$. If $x\neq y$ then with high probability $p$ will not divide $x-y$.


text:
# Calculate EMF of concentration cell consisting of zinc electrodes,one immersed in a solution of 0.01M $Zn^{+2}$ ions and the other in a solution of 0.1M $Zn^{+2}$ ions at $25^{\large\circ}C$.The two solution are separated by a salt bridge

$\begin{array}{1 1}(a)\;0.03V\\(b)\;0.09V\\(c)\;1.2V\\(d)\;-0.03V\end{array}$

The cell may be represented as $Zn/Zn^{+2}(0.01M)\parallel Zn^{+2}(0.1M)/Zn$
The EMF of cell is given by
$E_{cell}=\large\frac{RT}{nF}$$ln \large\frac{C_2}{C_1} \Rightarrow \large\frac{0.0591}{2}$$\log \large\frac{0.1}{0.01}$
$\Rightarrow 0.03$V
Hence (a) is the correct answer.
answered Feb 18, 2014


text:
# Proactively managing volatile market data with the rule of three

Data volumes have spiked on the back of increased market volatility, and many systems across the front and back offices have struggled to keep up. KX explores how financial institutions can optimise data management and analytics platforms to cope with a constantly changing investment landscape

In 2019 a common complaint among FX traders was: ‘Where has all the market volatility gone?’

As if to answer this grumbling, the Covid-19 pandemic emerged, disrupting supply chains and forcing many central banks to row back on monetary tightening. Volatility returned with a vengeance and, so far, shows no signs of abating.

This was good news for many FX traders, who were able to profit on the back of these wild currency swings. It was less welcome news for other market participants, such as corporates, who were forced to re‑examine their currency hedges to ensure they were adequately protected.

Amid this market chaos, many dealers have been doing well from increased client demand to profit from or protect against violent market movements. For example, according to data from Greenwich Associates, global pool revenue on US dollar options trading jumped more than 50% between 2019 and 2020 on the back of market volatility (rising from $3.4 billion to$5.3 billion). Revenue in 2021 (\$4.6 billion) was still at elevated levels.

However, while the return of volatility to the FX markets may be welcome news for some, it also means sensitivity to risk is much higher.

The consequences of putting an FX trade on at the wrong moment can be more severe in volatile market conditions. Moreover, dealers must be more careful over their hedging strategies. This was highlighted earlier this year when sharp moves in the Japanese yen against the US dollar left some FX options dealers nursing their wounds and scrambling to re-hedge.

The upshot of this is that it is becoming increasingly important for dealers to ensure the information on which they base trading decisions is as accurate and as dynamic as possible.

“Traders are increasingly mindful of the fact that, for them to be competitive, they have to remain responsive to changing market dynamics,” says Harry Darrell‑Brown, product manager of financial solutions at KX, which provides time series database technology to financial institutions worldwide. “If they aren’t tech-savvy and don’t move with the times, they can quickly be replaced.”

KX’s data platform, kdb+, allows firms to store, analyse, process and retrieve large data-sets at speed. It has proved particularly popular with high-frequency traders, but has also been deployed in other time-sensitive data applications, such as the energy markets, telecommunications, monitoring equipment used on manufacturing sites and even Formula One racing.

### The rule of three

Data gathering and data analysis have long been within the purview of technical back-office departments, but this has resulted in an inefficient understanding of the information being captured.

Steve Wilcockson, product marketing manager at KX, says that proactively managing volatile market data is all about what he terms the rule of three.

“It’s about having the right analytics for the right group at the right time,” he says.

It’s also about being able to effectively tailor data needs to meet the requirements of different users.

“You have the quants, who design the system; the data engineers, who build it; and the traders, who use it. Each has a keen interest in data analytics, but all from a slightly different perspective. The platform needs to be able to take care of each of these divergence interests,” says Alex Weinrich, FX and crypto solutions specialist at KX.

Gathering the data on a single platform and providing a user-friendly interface to run analytics allows the whole firm to gain a valuable insight into the daily trades being run.

“Capturing the fluidity between these three different roles is absolutely the right thing to do,” says Wilcockson. “Each participant in the group must be able to work together to build, analyse and understand the lifecycle of a trade.”

This firm-wide buy-in is crucial for the successful deployment of the data analytics engine. “Everyone at the firm – from traders to senior management – needs to realise the competitive edge this technology gives them,” says Darrell‑Brown. “If they don’t embrace this new way of doing things but their competitors do, then they are quickly going to be overtaken and will likely be replaced. No one wants that to happen to them.”

### Data optimisation

Processing and managing large volumes of such data is slow and time-consuming. This can lead to increased costs, reduced agility and, ultimately, uncompetitive pricing. While each data-set may hold important insights – upon which trading decisions can be made – the real power is unlocked once this data is brought under a single roof.

Unfortunately, in many organisations, data still exists in a range of different formats across multiple locations. This is typically a legacy from the way the data was captured in the first place.

Some firms have sought to get around the problem of dispersed data-sets by sampling down the data or producing cached calculations, before performing analytics on the data. But this is far from optimal and undermines firms’ abilities to have real-time data at their fingertips.

Analytics needs to be performed at the point at which the data sits, says Weinrich.

“In many cases, firms move the data to a different location before they perform analytics on it – which brings latency into the picture,” he says.

Seamlessly combining historical data with real-time data can also yield more valuable insights.

“All large banks and financial institutions use predictive machine learning these days. This means incorporating historical client behaviour into data algorithms, and adjusting models based on how a particular set of clients or liquidity providers is likely to behave following a market shock,” says Weinrich.

Combining different data-sets in this way allows firms to use the same code during development as they will use when the system is up and running.

“If the code being executed in production or the data is different from the real world, then how do you know what you’re testing is valid? So it has to be the same,” says Darrell‑Brown.

Pre- and post-trade analytics with real-time responsiveness to market change is absolutely vital to the successful integration of any data platform.

“With the FX markets in such a state of flux, traders must be able to hedge their portfolios and make sure they get those FX hedges right,” says Wilcockson. “To this end, they need the ability to quickly recalibrate models to account for sudden shocks to markets.”

This means being able to generate real-time insights at the end of trades – not just at the end of the day. Transaction cost analysis, liquidity and venue analysis and execution analytics are all key to this.

“It’s not just about having the right tools – it’s about being able to use them effectively,” says Wilcockson.

With no signs of market volatility letting up soon, it’s also about being able to future-proof the data going forward.

“Optimised cloud storage, combined with exceptional compression capabilities, can provide a highly cost-effective alternative to on-premises data storage,” says Richard Kiel, global head of FX solutions at KX.

An important part of using the cloud is connectivity. Not only do firms have to be able to access the data as quickly as possible, they also need to be able to do so in a convenient way.

“People need to be able to easily incorporate the cloud-enabled platform into their workflows, using languages and capabilities they are already familiar with,” says Wilcockson.

Typical tools that data scientists and engineers already use include Python, Jupyter notebooks and structured query language – and any new cloud-enabled platform needs to be able to seamlessly interact with these.

“This democratises who can get access to the platform straight away and continue using it – not just today or tomorrow but next year and well beyond that,” says Wilcockson.

### Beyond finance

While KX’s cloud-enabled platform has proved an invaluable addition to the trading teams of many large banks and fund managers, its ability to efficiently handle very large amounts of data means its application extends well beyond the financial world.

“The deep insight you need to have for electronic trading is the same insight you need to improve efficiency and automate processes across a variety of industry sectors,” says Darrell‑Brown. “Again, this comes down to having complete, accurate and timely data, from which real-time insights can be easily extracted.”

One of the most intriguing examples of the platform being used outside of finance is in the world of Formula One motor racing.

“If you’re in a race situation and your competitor comes in for a pit stop, you need to be able to judge whether you need to adapt your tactics to compensate for this,” says Wilcockson. “Timely data with predictive algorithmics can provide this insight.”

Another successful use case of the technology has been in the supply and delivery of commodities such as natural gas, which combines timely data with predictive analytics to spot current supply-and-demand trends.

“The point is that our real-time analytics platform doesn’t have to be confined to finance,” says Weinrich. “Whether for trading analytics, predictive health, predictive maintenance or predictive motor sports, the ‘power of three’ still holds: the right analytics for the right group at the right time, tailored to that winning combination of modeller, data engineer and system user.”

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text:
# [l2h] '\\' (newline) in math-mode not working

Ross Moore ross at ics.mq.edu.au
Mon Mar 31 23:27:55 CEST 2003

On Mon, 31 Mar 2003, Michael De Nil wrote:

> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
> Hey
>
> I have got a latex-file with pieces of src like the following:
>
>         \item
>     	{$> A = > \frac{v_{out}}{v_{in}} \\= > \frac{v_c}{v_b} \\= > \frac{-i_c.(r_0//R_C//R_L)}{i_b.r_\pi+i_e.r_e} \\= > \frac{-i_c.(r_0//R_C//R_L)}{\frac{i_c}{\beta}.r_\pi+i_c.r_e} \\= > \frac{-i_c.(r_0//R_C//R_L)}{i_c.\left(\frac{r_\pi}{\beta}+r_e\right)} \\= > \frac{-(r_0//R_C//R_L)}{r_e'+r_e} >$}
>         \item
>         {$> A_V= > \frac{v_{out}}{v_s} = > \frac{v_{out}}{v_{in}} = > A = > \frac{-(r_0//R_C//R_L)}{r_e'+r_e} >$}
>
> The formula in the upper item should be displayed with every time a
> newline between, and the second one should be printed on one line. When I
> compile the code with latex -> dvi, it looks nice,

Surely you jest.
Here you have so much mathematics that it belongs in a displayed
environment, yet you are trying to put it 'inline'.

Furthermore, you do not line up the = signs properly, and the lines are
cramped to be rather too close together.
Are you working in a narrow column ? Why else would you not allow your
inline math to use the full typesetting width that is available ?

>                  .... but a latex2html
> fails...

Yes, because LaTeX2HTML does not recognise use of \\ within inline
math as being valid markup for mathematics. It needs to be within an
environment designed for alignment and line-breaking, otherwise it will
not be recognised the way you seem to want it to be.

>
> In the upper example all math is printed on one line (like it should be in
> item 2), and this runes the whole layout of the html-document. You can see
> an example on
> http://flex-it.be:8080/?target=1ge/sem2/elektronica-theorie&sv=1 , the src

You have some nice stuff here; but there's too much on a single page.
Break it up over many smaller pages.
Some of the images have poor resolution so need to be shown much larger.

Marking-up for a web-document is generally harder than marking-up for
the printed page, since the graphics model is not so sophisticated,
and the resolution is considerably lower.

> can be found on
> http://flex-it.be:8080/school/1ge/sem2/elektronica-theorie/elektronica-theorie.tar.bz2
>
> Does anyone know a work-around for this, or if it's not possable yet, you
> can see it as a feature-request ;=)

No changes are warranted, arising from this example.

Hope this helps,

Ross Moore

>
>
>
> Thanks
> 	Michael
>
>
> - --
>
> Michael De Nil --- michael at flex-it.be
> Linux lisa.flex-it.be running 36 days
> -----BEGIN PGP SIGNATURE-----
> Version: GnuPG v1.0.6 (GNU/Linux)
> Comment: For info see http://www.gnupg.org
>
> iD8DBQE+iCOYNnTs0s0Nh3sRAtu5AJ9L5VDj9jXg6x/UbR2hnt20qiIM1ACgwjdO
> Y2pQitvug6XVSao1IVNsBq4=
> =mzOq
> -----END PGP SIGNATURE-----
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> latex2html mailing list
> latex2html at tug.org
> http://tug.org/mailman/listinfo/latex2html
>