text:
American Mathematical Monthly Contents—August/September 2013

While many of you enjoy some vacation time during the late summer, the Monthly cruises into August and September with an outstanding collection of five articles and five notes. Key amongst our articles is a piece by David Aldous on how to illustrate undergraduate probability issues using live market data. Also, Peter Kuchment and Sergey Lvin offer an interesting array of identities involving sin x which resulted from work in medical imaging. In our notes section, Ram Murthy and Jaban Meher rewrite Ramanujan's proof of Bertrand's postulate without the use of Stirling's formula. We close with our Problem Section and Joseph Malkevitch's review of The Mathematical Coloring Book, by Alexander Soifer.

Check back for the October issue when Allan Tucker will offer us a review of the history of undergraduate mathematics in the United States. Scott Chapman

Vol. 120, No. 7, pp.583-674.

ARTICLES

Using Prediction Market Data to Illustrate Undergraduate Probability

David J. Aldous

Prediction markets provide a rare setting, where results of mathematical probability theory can be related to events of real-world interest and where theory can be compared to data. The paper discusses two simple mathematical results—the halftime price principle and the serious candidates principle—and corresponding data from baseball and the 2012 Republican presidential nomination race.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.583

Pi, the Primes, Periodicities, and Probability

Stephen D. Casey and Brian M. Sadler

The theory of numbers has repeatedly shown itself to be both practical and beautiful. This paper gives an example of this duality. We present a very efficient (and practical) algorithm for extracting the fundamental period from a set of sparse and noisy observations of a periodic process. The procedure is computationally straightforward, stable with respect to noise, and converges quickly. Its use is justified by a theorem, which shows that for a set of randomly chosen positive integers, the probability that they do not all share a common prime factor approaches one quickly as the cardinality of the set increases. The proof of this theorem rests on a (beautiful) probabilistic interpretation of the Riemann zeta function.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.594

Identities for $$\sin x$$that Came from Medical Imaging

Peter Kuchment and Sergey Lvin

The article describes interesting nonlinear differential identities satisfied by standard exponential and trigonometric functions. They appeared as byproducts of medical imaging research and look like some type of non-commutative binomial formulas. A brief description of the origin of these identities is provided, as well as their direct algebraic derivation. Relations with separate analyticity theorems in several complex variables and some open problems are also mentioned.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.609

Subdivision Using Angle Bisectors Is Dense in the Space of Triangles

Steve Butler and Ron Graham

Starting with any nondegenerate triangle, we can use an interior point of the triangle to subdivide it into six smaller triangles. We can repeat this process with each new triangle, and continue doing so over and over. We show that starting with any arbitrary triangle, the resulting set of triangles formed by this process contains triangles arbitrarily close (up to similarity) to any given triangle when the point that we use to subdivide is the incenter.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.622

Commuting and Noncommuting Infinitesimals

Mikhail G. Katz and Eric Leichtnam

Infinitesimals are natural products of the human imagination. Their history goes back to Greek antiquity. Their role in the calculus and analysis has seen dramatic ups and downs; they have stimulated strong opinions and even vitriol. Edwin Hewitt developed hyperreal fields in the 1940s. Abraham Robinson’s infinitesimals date from the 1960s. A noncommutative version of infinitesimals, due to Alain Connes, has been in use since the 1990s. We review some of the hyperreal concepts, and compare them with some of the concepts underlying noncommutative geometry.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.631

NOTES

Another Proof of $$\zeta(2)=\frac{\pi^{2}}{6}$$ Using Double Integrals

Daniele Ritelli

Starting from the double integral $$\int_{0}^{\infty}\int_{0}^{\infty}\frac{dxdy}{(1+y)(1+x^{2}y)}$$ we give another solution to the Basel Problem $$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$$.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.642

Simple Approximation Formulas for the Birthday Problem

Matthias Arnold and Werner Glaß

We approximate the probability for at least $$k=2$$ or $$3$$ of $$n$$ persons having the same birthday if we ignore leap years and assume that each day is equally likely. Although there are exact formulas available in the literature, a possible drawback to using them in the classroom is that these formulas are comparatively cumbersome. As an alternative, we suggest an easily computable approximation that still provides a close fit. Furthermore, the approximation allows an easy answer to the question of how many people are needed, so that the probability of at least two or three of them having the same birthday exceeds a given value.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.645

Another Irreducibility Criterion

Anthony J. Bevelacqua

For any prime $$p$$ and integers $$a_{1},\dots,a_{n}$$ such that $$p\geq a_{1}\geq\cdots\geq a_{n}\geq1$$, the polynomial $$f=p+a_{1}x+\cdots+a_{n}x^{n}$$ is irreducible in $$\mathbb{Z}[x]$$ if and only if the list $$(p,a_{1},\dots,a_{n})$$ does not consist of $$(n+1)/d$$ consecutive constant lists of length $$d>1$$.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.648

Ramanujan’s Proof of Bertrand’s Postulate

Jaban Meher and M. Ram Murty

We present Ramanujan’s proof of Bertrand’s postulate and in the process, eliminate his use of Stirling’s formula. The revised proof is elegant and elementary so as to be accessible to a wider audience.

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.650

A Simple Proof of Vitali’s Theorem for Signed Measures

T. Samuel

Several theorems are named after the Italian mathematician Vitali. In this note, we provide a simple proof of an extension of Vitali’s Theorem on the existence of nonmeasurable sets. Specifically we show, without using any decomposition theorems, that there does not exist a nontrivial, atom-less, $$\sigma$$-additive and translation invariant set function $$\mathcal{L}$$ from the power set of the real line to the extended real numbers with $$\mathcal{L}([0,1])$$. (Note that $$\mathcal{L}$$ is not assumed to be nonnegative.)

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.654

PROBLEMS AND SOLUTIONS

Problems 11719-11725
Solutions 11584, 11588, 11592, 11611, 11615, 11616, 11621, 11603 (correction)

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.660

REVIEWS

The Mathematical Coloring Book by Alexander Soifer

Reviewed by Joseph Malkevitch

To purchase the article from JSTOR: http://dx.doi.org/10.4169/amer.math.monthly.120.07.670


text:
999 views

Given the following input $(4322, 1334, 1471, 9679, 1989, 6171, 6173, 4199)$ and the hash function $x$ mod $10$, which of the following statements are true?

1. $9679, 1989, 4199$ hash to the same value

2. $1471, 6171$ hash to the same value

3. All elements hash to the same value

4. Each element hashes to a different value

1. I only
2. II only
3. I and II only
4. III or IV
edited | 999 views

Option C is correct answer because the last digit of every digit given is equal in I and II.

edited

This might Help to understand better

Answer C we can simply get it by dividing the keys

1
2

format: plain


text:
Homework Help: Find the General expression for the adiabatic relationship...

1. Apr 8, 2016

grandpa2390

1. The problem statement, all variables and given/known data

edit: I figured out my mistake

Here's the entire problem. But I don't yet need help with all the steps. I'm getting off on the wrong foot (as usual)

a We have shown in class that ideal gas adiabats have the form: $P_I(V_i)^γ = P_f(V_f)^γ$ with the heat capacity ratio: $γ = \frac{C_p}{C_v}$ This problem is designed to find the general expression for the adiabatic relationship between P and V for any gas.

i. The adiabatic relationship between P and V begins with the completely general form: $(\frac{∂P}{∂V})_s$

ii. Apply the permuter to this expression.

iii.Find two Maxwell relations to replace the resulting partial derivative expressions (you will need the inverter for one of them)

iv. Insert the expression we derived in class for the adiabatic relationship between V and T for one of the partials.

v. Use the permuter on the other partial derivative that came from step iii.

v. You should at this point be able to use the relationship: https://www.physicsforums.com/file://localhost/Users/simon/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image008.png [Broken]

vi. Use another Maxwell relation on the remaining partial derivative term from step v.

vii. Apply the permuter (in reverse) to the two partial derivatives to finally arrive at the general expression of the adiabatic relationship between P and V for any gas: https://www.physicsforums.com/file://localhost/Users/simon/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image010.png [Broken]

b. Show for the ideal gas that this equation integrates to the proper expression. (before integrating, I suggest you manipulate the result of https://www.physicsforums.com/file://localhost/Users/simon/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image012.png [Broken]into a form in terms of P and V only.)

2. Relevant equations
Maxwell Relations

3. The attempt at a solution
ii. so first I applied the permuter to $(\frac{∂P}{∂V})_s$

I got $(\frac{∂P}{∂S})_v(\frac{∂S}{∂V})_P$

iii. The adiabatic relationship derived in class between V and T is $(\frac{∂V}{∂T})_s = \frac{-C_v}{\frac{RT}{V}}$

so I inverted the partial of P with respect to S and used the maxwell relation on it to get $-\frac{∂V}{∂T}$ and plugged in the adiabatic relationship. I used the maxwell relation on the partial of S with respect to V to get the partial of P with respect to T.

this is about where I get lost because after I use the maxwell relation and use the permuter I can't plug in the relationship:
$(\frac{∂S}{∂T})_p = \frac{C_p}{T}$
because I don't get a partial of S with respect to T.

I get the partial of P with respect to V (constant T) times the partial of V with respect to T (constant P)

Last edited by a moderator: May 7, 2017
2. Apr 8, 2016

grandpa2390

I'm going to keep playing with it.
I may have it


text:
## Hamiltonian Equation

$H_{\psi }=E_{\psi }$

1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$

Denise Montes 3B
Posts: 15
Joined: Thu Jul 25, 2019 12:17 am

### Hamiltonian Equation

Will we ever have to calculate the double derivative in the hamiltonian equation portion or will that be given to us?

Chem_Mod
Posts: 19169
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 833 times

### Re: Hamiltonian Equation

No, you will not. You will not ever have to make calculations involving Schrodinger's Equation.

### Who is online

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text:
Generate Procedure - Maple Help

CodeTools[ProgramAnalysis]

 GenerateProcedure
 generate a Maple procedure from a ForLoop or WhileLoop

 Calling Sequence GenerateProcedure(loop)

Parameters

 loop - WhileLoop or ForLoop data structure

Description

 • The WhileLoops or ForLoops generated by CreateLoop can be transformed back into a Maple procedure using GenerateProcedure.
 • The returned procedure may not be the same as the original procedure processed using CreateLoop.  Statements that do not affect the loop variables are removed and the loops are put into a canonical form.

Examples

 > $\mathrm{with}\left(\mathrm{CodeTools}\left[\mathrm{ProgramAnalysis}\right]\right):$

Form a procedure:

 > p := proc (A, n)     local i;     for i from 1 to n do        A[i] := A[+1]^2:     end do: end proc;
 ${p}{≔}{\mathbf{proc}}\left({A}{,}{n}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{local}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{i}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{for}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{i}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{to}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{n}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{do}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{A}{[}{i}{]}{≔}{A}{[}{1}{]}{^}{2}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end do}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end proc}}$ (1)

Create the related ForLoop data structure:

 > $\mathrm{loop}≔\mathrm{CreateLoop}\left(p\right)$
 ${\mathrm{loop}}{≔}{\mathrm{Record}}{}\left({\mathrm{ForLoop}}{,}{\mathrm{body}}{=}\left[\left[\left[\right]{,}\left[{{A}}_{{i}}{=}{{A}}_{{1}}^{{2}}\right]\right]\right]{,}{\mathrm{layers}}{=}\left[\left[{i}{,}{1}{,}{1}{,}{n}\right]\right]{,}{\mathrm{initialization}}{=}\left[\right]{,}{\mathrm{variables}}{=}\left[{i}\right]{,}{\mathrm{parameters}}{=}\left[{n}\right]{,}{\mathrm{arrays}}{=}\left[{A}\right]\right)$ (2)

Translate the ForLoop loop back into a procedure:

 > $\mathrm{GenerateProcedure}\left(\mathrm{loop}\right)$
 ${\mathbf{proc}}\left({A}{,}{n}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{local}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{i}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{for}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{i}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{to}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{n}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{do}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{A}{[}{i}{]}{≔}{A}{[}{1}{]}{^}{2}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end do}}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{return}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end proc}}$ (3)

Compatibility

 • The CodeTools[ProgramAnalysis][GenerateProcedure] command was introduced in Maple 2016.


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# Tag Info

15

Gamma is the second partial derivative of the change in the price of the option wrt to the change in the underlying. Said another way, it is the change in delta. If you write down the Black-Scholes pricing formula, you's see the gamma term: $$...\frac{1}{2}\frac{\partial^2C}{\partial S^2}(\Delta S)^2...$$ Notice that the $\Delta S$ (change in stock price) ...

12

I might be misunderstanding your question. My thoughts: being short gamma is being long volatility your comment re gamma increasing regardless of direction only holds for ATM options. For ITM options, being short gamma is being long the underlying. For OTM options, being short gamma is being short the underlying. Some graphs: Below, except as ...

10

Being short gamma simply means that you are short options regardless of whether they are puts or calls. The most common type of investor that is willing to be short gamma is someone who sells options, also known as a premium collector. These investors commonly use strategies such as short puts, covered calls, iron condors, vertical credit spreads, and a ...

10

Delta is the partial derivative of the value of the option with respect to the value of the underlying asset. An option with a delta of 0.5 (here listed as +50 points) goes up \$0.50 if the underlying asset goes up \$1.00. Or goes down \$0.50 if the underlying asset goes down \$1.00. Keep in mind that delta is an instantaneous derivative, so the value will ...

9

This is in fact a tricky matter. As you say one way is to calculate delta by an analytic formula, i.e. calculate the first derivative of the option pricing formula you are using with respect to the underlying's spot price. The second way is to do it numerically, i.e. change the spot price by a small value $dS$, calculate the value of the option and then ...

8

You need to compute your greeks as finite differences, but the full procedure may be pretty tricky. I will use vega $\aleph$ as the example here. Let's begin by designating your Monte Carlo estimator as a function $V(\sigma,s,M)$ where $\sigma$ is the volatility as usual, $s$ is the seed to your random number generator, and $M$ is the sample count. To ...

7

I dusted off my oldest option theory books and searched the indexes for "vega". The oldest reference I found was in Option Volatility and Pricing Strategies (1st ed.) by Sheldon Natenberg, copyright 1988. When discussing the sensitivity of prices to volatility (p. 132), he says, [T]here is no single commonly accepted term for this number. It is ...

7

Pretty much irrelevant for vanilla markets but really cannot be ignored when pricing exotics such as barriers. Basically, if you do not hedge vega you are likely to sell lots of cheap exotics. Webb discusses the practical relevance of vanna and vomma in "The Sensitivity of Vega" (Derivatives Strategy, November (1999), pp. 16 - 19).

7

You are absolutely right to point out that most proactive participants in options markets prefer to be long gamma, and it is typically reactive market makers who take the opposite side of their trades. While the typical options trader (I find it difficult to call anyone trading options an "investor") does not hedge his position, market makers will attempt ...

6

Short gamma is a bet on volatility (expressed as hedging costs) not getting too large. The key concept here is that you get paid to be short gamma. Consider that any option is sold for a bit more more than its intrinsic value (the extra bit is often called volatility value.). If nothing moves, then the option ultimately expires precisely at intrinsic ...

6

For non-interest rate derivatives with not-so-long maturities worrying about rho is uncommon. Think about it: interest-rates do not change that often relative to options expiring next week, next month or at most next year. LEAPS are obviously another turf. You could think about gamma, but the intimate relation of gamma and vega (at least in BS model) makes ...

6

First, my notation. $K$ is the strike price, $S$ is the stock price, $r$ is the continuously compounded risk-free rate, $T$ is time at expiration, $t$ is time at issue, $\sigma$ is volatility, $\delta$ is continuously compounded dividend rate. The Black-Scholes formula for a European call is $C = Se^{-\delta (T-t)} N(d_1) - Ke^{-r(T-t)} N(d_2)$ $d_1 = ... 5 VIX is calculated from a basket of SPX options, and VIX futures expire into following expiration, e.g. September VIX futures that will expire next Wednesday will use SPX October options chain to calculate settlement value. If$B$is the value of the basket then VIX value at expiration is$\sqrt{ B }$. Then VIX futures price is the expectation of the basket ... 5 I have no reference, but it's largely phonetic. Must variables in econ/finance are Greek versions English letter you'd want to use.$\omega$for weight,$\rho$for rate,$\epsilon$for error, and so. Vega is partial derivative of price with respect to V olatility. But there's no Greek letter for V. Vega sounds kind of Greek. 5 I handle volatility curves where moneyness is quoted in delta by an iterative guess: Use an initial guess for delta of 0.5 (call)/-0.5 (put) Look up the volatility on the curve using the guess for delta Calculate delta for the option using the vol found in 2. Repeat using Newton-Raphson, until the difference in delta is small enough. 5 If you get paid enough theta it absolutely makes sense to be short gamma. And the closer to expiration, the faster the time-value flees. Most of the time, most people would prefer to be gamma long though. It's simply a safer bet because of uncertainty: unexpected events can seriously damage your book if you're short vol. 5 FDMs represent PDEs over a simple grid shape; the different implementations are just different recurrence relations to approximate the solutions to the PDE between boundary values (e.g., for options pricing,$T=[t_\mathrm{now},t_\mathrm{maturity}]$and$S=[\mathrm{deep\_itm},\mathrm{deep\_otm}])$. FEM is a general name for a lot of different ... 5 As far as PDEs (deterministic) are concerned we have the notion of a "strong solution" (directly solving the differential operator in the strong formulation of the problem) and the "weak solution" that deals with a weak formulation of the problem. For the strong formulation, finite differences are the way to go since they are the natural discretization of ... 5 most models in financial maths are linear so prices and Greeks just add. This is in particular true of Black--Scholes so Yes. However, once one starts taking into account value adjustments non-linearities appear and it is a lot more complicated. 5 If you want to know what Greeks the market assigns to an option, i.e. the market implied Greeks, then you would use the implied volatility. And that is what traders like to look at. 4 Short gamma is being of the view that realized volatility would be less than the implied volatility for the period in which an option is valid. So if you think realized volatility in the future would be consistently lesser than implied volatility at present, then you'd be short gamma. The premium one would receive by selling an option (call or put) is a ... 4 The most general answer is to shift your input to approximate the first derivative. Given that you need Monte Carlo to price this, it may get expensive. But that's the way it goes as when you have no analytical solutions as there aint't no free lunch ... 4 Short Answer : Futures don't have Greeks Long Answer : Assuming a non strictly mathematical (i.e. false) point of view. Well, having Greeks on VIX Futures is not relevant, VIX value is itself a "Greek" (and Futures don't have Greeks). Sensitivity to Price of the Underlying : Insensitive (ν = 0) Volatility of the Underlying : Delta Δ = 1 (to Volatility ... 4 If your "European vanilla options" are restricted to piece-wise linear pay-offs, then the following may help: Remark: I assume you are looking for a rule of thumb to get the profiles without the use of a computer. All piece-wise linear pay-offs can be decomposed into a sum of digital options and call options with different notional (possibly negative) and ... 4 Simply put, no. Vega depends on a variety of factors (including the level/price of the underlying asset). However, vomma/volga/vega convexity (whatever you want to call dVega/dIV) is always positive. So as IV increases, the vega of an option increases - I think this might have been what you were getting at. It's important to understand that IV is an input ... 4 Automatic Differentiation (aka AD) is a family of methods that are used to evaluate the derivative of a coded function. These methods are far more accurate than finite differences, since they are theoretically exact in the absence of floating point roundoff error. AD is, however, subtly different than symbolic differentiation. The key difference here is ... 4 Since the volatility is not changing, we can assume that the only change is the underlying asset price$S\$. Then \begin{align*} C(S+\Delta) &\approx C(S) + Delta \times\Delta +\frac{1}{2} Gamma \times \Delta^2 \\ &=11.50 + 0.58 \times 0.5 + \frac{1}{2}\times 2 \times (0.5)^2\\ &=12.04. \end{align*}

3

I somewhat disagree (partially) with the other answers so I offer my own. First, most importantly is that you specify exactly what underlying asset you really talk about. Even for non-interest bearing assets an unhedged rho can sometimes have devastating results on your profitability. Imagine a stock denominated in a highly inflated currency. If you buy ...

3

No, there is not. If you are willing to assume the stock price stays constant between now and then, you could do so using the standard formula.

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# mahout-user mailing list archives

##### Site index · List index
Message view
Top
From Ted Dunning <ted.dunn...@gmail.com>
Subject Re: Question Regarding Entropy calculation in Mahout
Date Fri, 23 May 2014 18:31:20 GMT
Yash,

I am not sure how your suggestion will work.

The problem is clustering algorithms tend to make hard assignments.  Thus,
if you try to compute entropy relative to some reference probability
distribution (aka perplexity [1]) then a reference clustering will provide
1 or 0 as the probability.  Any item that gets classified into a different
cluster will cause the Entropy to include a term - 1 log 0 which is
infinite.

One way to deal with this is to assign probability 1-\epsilon to the
cluster an item is in and \epsilon/(k-1) for all the other clusters.  You
then have issues finding a good value of \epsilon which seem to me to be
out of scope for the original question.

Computing entropy relative to the fraction of documents in each cluster is
easier to compute, but much harder to understand.  Computing mutual
information (not entropy) on the confusion matrix between two clusterings
can also be done, but that also seems beyond the original question.

As such, I think that the burden is on the original questioner to describe
the problem more accurately.

On Fri, May 23, 2014 at 11:21 AM, Yash Sharma <yash360@gmail.com> wrote:

> Hi Darshan,
> What i understand from your problem is that:
> - You have clustered few documents
> - You want to verify the accuracy of ur clustering , and you want to use
> entropy for that
> - You are not sure what should be the input for entropy calculation.
>
> Possible solution:
> The entropy would expect a String[] to calculate the information contained
> in the data/sequence.
> One simplest way is to keep all the documents labelled with categories.
> - Cluster the docs as you usually do.
> - For entropy calculation create a String[] for every cluster. Each array
> containing all the labels of the docs in the cluster.
> cluster1 = {"sports", "tech", "tech", "tech", "book", ..}
> cluster2 = {"sports", "drama", "sports", "sports"...}
> etc
>
> - Calculate the entropy of each cluster.
> Entropy would measure the degree of randomness of a system. High entropy
> means there is high degree of randomness in a system.
> Lower Entropy are desirable for validation of accuracy of your clustering
> technique.
>
> P.S. You can use Entropy.java class for your validation purpose but
> its deprecated now.
>
> Having Said that - Kindly be patient while asking questions and provide
> enable all of us to answer quickly & correctly :)
>
> Hope it was helpful. Other Approaches are welcome..!!
>
> Peace,
> Yash
>
>
> On Fri, May 23, 2014 at 10:55 PM, Ted Dunning <ted.dunning@gmail.com>
> wrote:
>
> > I am sorry, but I don't understand your questions or needs sufficiently
> to
> >
> >
> >
> >
> > On Wed, Apr 23, 2014 at 12:21 PM, Darshan Sonagara <
> > darshan.sonagara@gmail.com> wrote:
> >
> > > thanks in advance.
> > >
> > >
> > > On Tue, Apr 22, 2014 at 11:50 PM, Darshan Sonagara <
> > > darshan.sonagara@gmail.com> wrote:
> > >
> > > > waiting for the replay sir .
> > > >
> > > >
> > > > On Tue, Apr 22, 2014 at 7:13 PM, Darshan Sonagara <
> > > > darshan.sonagara@gmail.com> wrote:
> > > >
> > > >> Thnks for the Replay sir,
> > > >>
> > > >> actually i am doing clustering for gathering similar king of
> document
> > in
> > > >> same cluster as much as possible.
> > > >> i can see from output file by cluster dump by observing top term.
> > > >> i also figure out that by varying Distance Measure Technique. it
> > > differs.
> > > >> but i want some mathematical prof that it is better then other
> > > technique.
> > > >> so for that i need to calculate Entropy and pureness of cluster.
> > > >> but i am not able to find any command in mahout which can give me
> > > entropy
> > > >> as a result.
> > > >> i found Entropy.java under mahout common math statistic package.
> but i
> > > >> don't what should i give it as input so that i can find entropy or
> > other
> > > >> parameter. so i can find how much cluster is good or bed.
> > > >>
> > > >>
> > > >>
> > > >> On Tue, Apr 22, 2014 at 7:01 PM, Ted Dunning <ted.dunning@gmail.com
> > > >wrote:
> > > >>
> > > >>> On Tue, Apr 22, 2014 at 12:11 AM, Darshan Sonagara <
> > > >>> darshan.sonagara@gmail.com> wrote:
> > > >>>
> > > >>> > But the problem is that i want check that whether my clustering
> is
> > > >>> good or
> > > >>> > bad. so for that i need to calculate Entropy Value. I am
not
> having
> > > any
> > > >>> > idea how to calculate entropy in mahout or by other technique.
> > > >>> > by finding entropy i can have good conclusion.
> > > >>> > so please can anyone help me with these.
> > > >>> >
> > > >>>
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> > > >
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> > > *Collaborative Platform lead,** SSN Team | Gujarat Section.*
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> > >  : Darshan Sonagara<


text:
Regression - Finding a Function to Fit Your Data

Here’s a simple dataset - just two variables that we’ll call $$x$$ and $$y$$.

It looks like $$x$$ and $$y$$ are related! Let’s say we want to use $$x$$ to predict $$y$$. We can use regression to fit a function to the data. Here is our function plotted over the data:

A pretty good fit! This was a simple regression problem, because for every unit increase in $$x$$, $$y$$ also increases some proportionate amount. The equation for the above line is $$\hat y=2x$$, so we know that for every unit increase in $$x$$, $$y$$ increases by about 2.

What do we do when fitting the data isn’t so easy, such as when our data looks like this?

The equation for this function is $$\hat y=81.6-.1x$$, and it doesn’t fit the data very well. Rather than $$y$$ varying as a function of $$x$$, it looks like $$y$$ varies as a function of $$x^2$$. This means we want to look for a function of the form $$y=\beta_0+\beta_1x+\beta_2x^2$$. Our regression software will take care of finding the best $$\beta$$ values; we just need to give it an $$x$$ variable as well as an $$x^2$$ variable to work with.

So I square my $$x$$ values, and now I have data that looks like this:

I run regression to predict $$y$$ in terms of these two other variables, and I get:

That worked pretty well!

The principle here is to try and understand how my $$y$$ variable varies as a function of my $$x$$ variables, and then transform my $$x$$’s to match that function.

Sometimes, having an understanding of the subject matter that you’re dealing with can help you do this. Take a look at one more example using some made-up data:

Imagine this data comes from a psychology experiment where participants complete tasks of varying difficulty after being kept awake for varying numbers of hours. The experimenter measures how participant response time (in milliseconds) on the tasks changes as a function of task difficulty and sleep deprivation. (Note that we have to visualize this in 3-D, because we now have three variables.)

If we wanted to make predictions based on this data, we would try to predict response time (our $$y$$ variable) using task difficulty and sleep deprivation (our two $$x$$ variables).

Let’s again use regression to fit a function to this data without doing any kind of transformation beforehand. Here’s what we get:

The equation for this function is $$\hat y=22.6x_1+14x_2+316.1$$, and as you were probably expecting, it isn’t a good fit. The bottom right corner of the plane is too high, whereas the top right corner is too low to fit the data correctly.

An experienced researcher might guess that there is an interaction between your $$x$$ variables. This would mean that, say, a difficult task increases response time, but it increases response time even more when you’re sleep deprived. Mathematically, an interaction would look something like $$y=\beta_0+\beta_1x_1+\beta_2x_2+\beta_3x_1x_2$$. Notice the last term of the equation has $$x_1$$ and $$x_2$$ multiplied by each other.

Let’s add an interaction term to our data, then, and see how it fits. As before, this is as simple as creating a new variable that is the product of $$x_1$$ and $$x_2$$. When we run regression to predict $$y$$ in terms of $$x_1$$, $$x_2$$, and $$x_1x_2$$, we get:

This is a good fit! The function now bends to show the interaction between task difficulty and sleep deprivation. Look at the bottom of the plot - for someone who has had plenty of sleep, as task difficulty increases, response time only increases a little bit. But at the top of the plot we can see that for someone who is sleep deprived, as task difficulty increases, response time increases a lot.

Conclusion

These are just a few common transformations you might need to do to your data to fit a regression. It’s also common to do a $$\log$$ transformation, or create higher-order polynomial variables like $$x^3$$ or $$x^4$$.

I hope this helps build your intuition about using regression to fit a function to your data!

Written on January 31, 2017


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# If 5^(k + 1) = 2,000, what is the value of 5^k + 1?

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If 5^(k + 1) = 2,000, what is the value of 5^k + 1?  [#permalink]

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26 Aug 2018, 05:31
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If $$5^{k + 1} = 2,000$$, what is the value of $$5^k + 1$$?

(A) 399
(B) 401
(C) 1,996
(D) 2,000
(E) 2,001

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Re: If 5^(k + 1) = 2,000, what is the value of 5^k + 1?  [#permalink]

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26 Aug 2018, 05:36
B?
$$5^k$$.5 = $$2^4$$.$$5^3$$

$$5^k$$= $$2^4$$.$$5^2$$ = 16 * 25 = 400

$$5^k$$ + 1 = 400 + 1 = 401
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Re: If 5^(k + 1) = 2,000, what is the value of 5^k + 1?  [#permalink]

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26 Aug 2018, 06:48
Bunuel wrote:
If $$5^{k + 1} = 2,000$$, what is the value of $$5^k + 1$$?

(A) 399
(B) 401
(C) 1,996
(D) 2,000
(E) 2,001

$$5^k*5 = 2^4*5^3$$

So, $$5^k = 2^4*5^2$$

Or, $$5^k = 400$$

So, $$5^k + 1 = 401$$ , Answer must be (B)
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Re: If 5^(k + 1) = 2,000, what is the value of 5^k + 1?  [#permalink]

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26 Aug 2018, 09:43
Bunuel wrote:
If $$5^{k + 1} = 2,000$$, what is the value of $$5^k + 1$$?

(A) 399
(B) 401
(C) 1,996
(D) 2,000
(E) 2,001

$$5^{k+1} = 2000$$
$$5^{k+1} = 2^45^3$$
$$5^k5 = 5^32^4$$
$$5^k = 5^22^4$$

$$5^2 = 25$$
$$2^4 = 16$$

25*16 = 400

$$5^k = 400$$

$$5^k$$ + 1 = 401.

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Re: If 5^(k + 1) = 2,000, what is the value of 5^k + 1?  [#permalink]

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26 Aug 2018, 10:53
Bunuel wrote:
If $$5^{k + 1} = 2,000$$, what is the value of $$5^k + 1$$?

(A) 399
(B) 401
(C) 1,996
(D) 2,000
(E) 2,001
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Re: If 5^(k + 1) = 2,000, what is the value of 5^k + 1?  [#permalink]

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26 Aug 2018, 21:56
Bunuel wrote:
If $$5^{k + 1} = 2,000$$, what is the value of $$5^k + 1$$?

(A) 399
(B) 401
(C) 1,996
(D) 2,000
(E) 2,001

$$5^{k + 1} = 2,000$$,

5^k . 5 = 2000,
5^k = 400

Therefore, 5^k + 1 = 400 + 1 = 401.

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Re: If 5^(k + 1) = 2,000, what is the value of 5^k + 1? &nbs [#permalink] 26 Aug 2018, 21:56
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## Thursday, January 7, 2016

### Averaging can produce misleading standardized effect sizes

Recently, there have been many calls for a focus on effect sizes in psychological research. In this post, I discuss how naively using standardized effect sizes with averaged data can be misleading. This is particularly problematic for meta-analysis, where differences in number of trials across studies could lead to very misleading results.

There are two main types of effect sizes in typical use: raw effect sizes and standardized effect sizes. Raw effect sizes are what you typically see in a plot: for instance, the effect of a priming manipulation might be 30ms. The advantage of raw effect sizes are that they are closer to the process of interest and more interpretable. We all know what it means for something to take 30ms to happen.

Another kind of effect size is the standardized effect size. With a standardized effect size, the raw effect is compared to the some measure of variability in the population. For instance, if the standard deviation of childrens’ heights at age 10 were 3 inches, and a “good” diet had an effect of 1.5 inches on average, we could say that the effect of the diet was .5 standard deviations. This is the logic of Cohen’s $d$, for instance. The disadvantage of this is that it is more difficult to understand what an effect of “half a standard deviation” means (and the variance-accounted-for statistics such as $\eta^2$ and $\omega^2$ are even more difficult to interpret); but standardized effect sizes have many good properties, including a close relationship to the concept of statistical power, comparability across paradigms, and the fact that they can often be computed from reported statistics such as $t$ and $F$.

For these reasons, standardized effect sizes are very common in meta-analysis. However, the common practice of averaging over trials in cognitive psychology makes them difficult to compare or even interpret.

Consider a typical cognitive experiment in with 30 participants, each performing 10 response time trials in two conditions. Typically each participant’s data will be averaged to form a single, average response time in each condition; these average RTs are then submitted to a repeated measures ANOVA (in fact, some R packages, such as afex do this automatically).

Hypothetical data is shown in the plot below. These data represent two hypothetical experiments, one with 10 samples per participant and one with 50.

 Error bars represent Morey (2008) adjusted within-subject error bars, computed using this code.

I have generated the data so that everything is the same across the two experiments except the scale of the “error”: the scale of the error in experiment 2 is $1/\sqrt{5}$ times that of experiment 1, because each averaged “observation” represents five times more data. The raw effect size is precisely the same, but our certainty about the effect size is greater in experiment 2. This is exactly as it should be.

We can now perform the typical repeated measures ANOVAs on these two data sets, using the afex. The package will, if requested, compute the common partial $\eta^2$ standardized effect size statistic.

The results of the ANOVA for the “condition” in experiment 1 are:

Effect df MSE F pes p.value
condition 1, 29 0.08 7.72 .21 .009

And the results of the ANOVA for the “condition” in experiment 2 are:
Effect df MSE F pes p.value
condition 1, 29 0.02 38.58 .57 <.0001

This is would not be unexpected by anyone who routinely uses repeated measures ANOVA. Typically, the whole point of running more trials is to get more power. We perform more trials, and we obtained a higher $F$ value for the comparison of interest.

Notice that the sum of squares for the effect is precisely the same for both experiments. That’s because the raw effect is precisely the same. What is driving the higher $F$ value is the lower residual mean square (MSE) for the comparison, which is about 5 times smaller in Experiment 2. Again, this is what we expect. More trials, less “noise”.

But notice what happens to partial $\eta^2$. Because the MSE is smaller, the proportion of variance accounted for by the condition effect is larger. This drives the partial $\eta^2$ from .21 in experiment 1 to .57 in experiment 2. Researchers have previously warned about using partial $\eta^2$ for comparisons across designs (see for instance, Olejnik & Algina, 2003), but these two experiments appear to have the same design; at least, from the perspective of someone used to only analysing averaged data, they do.

This has the potential to wreak havoc on meta-analyses. Suppose someone combs the literature looking for $F$ values and computing partial $\eta^2$ values from the $F$ values (or, alternatively, Cohen’s $d$ from $t$ values). Assume experiment 1 represents a patient group; due to time constraints, the patients only had time for 10 trials per condition. Suppose experiment 2, on the other hand, represents a group of college students, who had time for more trials. The figure below shows the standardized effects in the two experiments.

Although the raw effect size is precisely the same across the two experiments, the standardized effect size is radically different, possibly leading to erroneous conclusions. Even if there are no systematic differences in number of trials across experiments experiments with different kinds of groups, this introduces a new source of variability into estimates, as well as making it nearly impossible to interpret the effect size. What is the “true” standardized effect size? It seems difficult to say. How can we solve this problem?

### Solution 1: Generalized $\omega^2$

One option is generalized $\omega^2$ (see for instance, Olejnik & Algina, 2003) Instead of using the residual variance to standardize against, generalized $\omega^2$ standardizes against all measured (as opposed to manipulated) factors. For instance, the variability in participants is a measured source of variability. These sources of variability are assumed to be stable properties of populations and not effected by mere design choices. We can compute generalized $\omega^2$ again using the afex, which yields $\omega^2_g=0.011$ for the experiment 1 and $\omega^2_g=0.012$ for experiment 2; notice that these very similar. The effect “looks” smaller, because participants vary quite a bit relative to the size of the effect.

One problem with this approach is that the statistics necessary to compute generalized $\omega^2$ are not typically reported, meaning that this solution is useless for meta-analysis of existing literature. Ideally, we’d like a way to use reported statistics to at least compare across studies, when designs are similar enough.

### Solution 2: Adjust the MSE

If we know the relative numbers of trials across two studies that have the same basic design, we should be able to “adjust” the MSE in the formula for the effect size (whichever effect size it happens to be) for the number of trials. In our example, experiment 2 has five times as many trials as experiment 1; we therefore would expect the MSE of experiment 2 to be one-fifth as large as that for experiment 1. To make the effect size computed from experiment 2 comparable to that from experiment 1, we can multiply its MSE by 5 before applying the formula for the effect size of interest. For partial $\eta^2$, this leads to an easy adjustment:
$\eta^2_2 = \frac{1}{c/\eta^2_1 - c + 1}$ where $c$ is the adjustment factor, $\eta^2_1$ is the original partial $\eta^2$, and $\eta^2_2$ is the adjusted partial $\eta^2$.

As an example, take our experiment 2, which had a partial $\eta^2$ of 0.571 and a sample size 5 times larger than experiment 1. Applying the formula above yields
$\begin{eqnarray*} \eta^2_2 &=& \frac{1}{5/0.571 - 5 + 1}\\ &\approx&0.21 \end{eqnarray*}$ which matches the partial $\eta^2$ from experiment 1 very well.

The problem with this approach is that it uses experiment 1 as a “reference” experiment. It is therefore not clear what the standardized effect size means in this case, except as a way to compare across experiments with similar designs. This may be enough to someone performing a meta-analysis — particularly if they can’t obtain the statistics to compute generalized $\omega^2$ — but as a general reporting solution, it is unsatisfactory.

### Wrap up

Although standardized effect sizes have been advocated as a general tool for science and are increasingly reported, they are difficult to interpret because they are affected by trivial, common design decisions. The issues I raise here should be of interest to anyone working with standardized effect sizes, particularly those performing meta-analysis. They affect repeated measures designs with averaged data most acutely; however, between-subjects designs are also affected if each participant contributes an “average” score to the analysis. In the between-subjects case the adjustment would have to be different, but for large number of trials per participant might be acceptably ignored, if the error each participants’ score is small enough.

After a re-read, I want to ensure that I make clear that I'm not implying that the only problem here is with meta-analyses; that's just what drove me to write this post, and how I decided to frame it. But consider this: if an arbitrary decision (driven merely by the resources at hand, such as time or money, or even whim) such as "how many trials will we perform per cell in this experiment?" can cause the standardized effect size to increase almost 200%, that standardized effect size should not be taken to reveal any psychological "truth" and is useless for drawing substantive conclusions.

1. Great post. You might also be interested in:

http://janhove.github.io/reporting/2015/02/05/standardised-vs-unstandardised-es/

http://janhove.github.io/design/2015/03/16/standardised-es-revisited/

2. Thanks! You make some nice observations there, which are particularly relevant in this comment: here. I didn't expand on it too much above, but I think you're right about the difficulties inherent in standardised effect sizes. Although standardised effect sizes do allow for comparison across paradigms, given the difficulties, I wonder if the meaningfulness of such comparisons is merely illusory. It would definitely be profitable to at least consider what meta-analysis would look like based solely on raw effect sizes.

1. Actually, those are not my blog posts, and so not my interesting observations (though I agree with them). So sadly, I cannot take credit.

2. With a mere 14-month delay: Thanks to jwdink for linking to my blog, and thanks to Richard for his comments!

3. Interesting post!
I had recently come across this issue myself, and observed that including random intercepts and slopes in the simulations attenuates the overestimation of the effect sizes (depending on the ratio between the within- and between subjects variability).

I have presented this at our labmeeting only (and do not have a blog to summarize this), but I thought it might be useful to share these slides here, in case you would be interested: https://www.dropbox.com/s/o2ss3iehjnl808f/Labmeeting_12112015.pptx?dl=0

As you highlight in the post-script, I agree one of the most important implications of this is that standardized effect sizes behave in unanticipated ways when researchers first average across trials for each participant before analyzing the data, but this has implications for power analysis, and meta-analysis indeed.

1. Very nice. It looks like we're thinking along the same lines.

4. This comment has been removed by a blog administrator.

5. Richard, can you provide intuition/explanation for why the standardize effect sizes are so much lower In Expt 1 vs. Expt 2? I could see it if it were Cohen's d because Expt 1 would have higher same mean difference but higher standard deviation, thus Expt 1's d would be lower than Expt 2's. How does this work for eta-squared though, which according to Wikipedia is SS_treatment / SS_total? https://en.wikipedia.org/wiki/Effect_size#Eta-squared.2C_.CE.B72

There, you have SS in both numerator and denominator, so why aren't the ratios in both experiments roughly the same? Is it because the SS_total is similar/same in both experiments, and Expt 1 has higher SS_treatment than does Expt 2?

Also, would be interested in seeing code that underlies this running example.

1. edit: "Expt 1 would have higher same mean difference" should be "... have same mean difference"

2. Still wondering about the specifics behind why the standardized effect size is lower in Expt 1, if anyone has an idea...

6. This comment has been removed by a blog administrator.

7. Can Solution 2 be applied also when for adjusting reliability coefficients?
e.g. If I measure reliability using the split-half method, I am really only measuring the reliability of a measure with half as many trials as the actual measure used in analysis.

8. Quantitative data depicts the quality and can be scrutinized, but measuring it precisely is daunting enough; in contrast quantitative data can be easily measured and is depicted in number or amount. See more qualitative analysis with nvivo

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text:
# Minif and Maxif function in calc

On sheet one, I have a list of URLs, with duplicates in column B. In column J, I have dates visited. On sheet two, I have a list of all the URLs from sheet 1, with no duplicates.

On sheet two, I want to lookup the highest date that a URL from sheet one matching the url from sheet two in one column, and the earliest date in another. Something like:

A|B|C

URL|Most recent update|first checked [url]|=maxif(S1.B2:B25001,a2,S1.J2:J25001)|=minif(S1.B2:B25001,a2,S1.J2:J25001)

I realize that there is no minif or maxif functions (or if there are, I haven't found them), but if someone could show me a way to implement something that would work in the same way as they might work, I would really appreciate it.

Oh yeah: Fully updated English language Windows 7 x64 Ultimate edition with Japanese locale, using LibreOffice v4.4.4.3 Calc with en_US locale.

Had no idea what tags I was supposed to use, so please forgive me if I used the wrong ones.

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This post is a wiki. Anyone with karma >75 is welcome to improve it.

Hi

May be a solution in MinIfMaxIf.ods

[EDIT]

I forgot some explanations:

• Most recent update (e.g. Sheet2.B2) is an array formula: needs key combination Shift+Ctrl+Enter
• To copy down with fill handle in the bottom right corner of the cell, needs to hold down the Ctrl key

Regards

more

Version 4.4 is quite outdated.. it might be interesting to you that some later version introduced the MAXIFS and MINIFS spreadsheet functions.

more

I'm using Ubuntu 16.04 (i. e. the current LTS distribution) which installs Libreoffice 5.1.6.2 which still doesn't have an MAXIFS() :-(

( 2017-11-24 12:39:46 +0200 )edit
( 2017-11-24 12:46:07 +0200 )edit

True, were introduced with LibreOffice 5.2. On Ubuntu you can use a PPA.

( 2017-11-24 12:47:38 +0200 )edit

The other answer given here (alas, I cannot see the author's name—weird) nudged me into the right direction. But by specifying a thing like MAX(ROW($Sheet1.$B$2:$B$13)*($Sheet1.$A$2:$A$13=A2)) it only takes the value furthest down (by computing the maximum if the row number). Since in this case the dates are sorted and later ones are indeed further down, this works as well. This might not be the case in other scenarios (as it was in mine).

To really get the maximum of a list of values for which a selector next to it matches a criterion (which would be a true MAXIF()), one can use a formula like this:

=MAX(IF($Sheet1.$A$2:$A$13=A2,$Sheet1.$B$2:$B$13, 0))

And, again: Don't forget to press CTRL-SHIFT-Enter in order to have this code be interpreted as an array function.

Mind also, that this will make all values for which the criterion doesn't match just 0 (which in many cases will not end up to be the maximum). If you, however, try to find the maximum of a bunch of negative numbers, this will fail because then the maximum will be any of the zeros. You need to use a value lower than any of yours:

=MAX(IF($Sheet1.$A$2:$A$13=A4,$Sheet1.$B$2:$B$13, -693595))

(The -693595 being the number for the date 0001-01-01 which is probably before any date you are dealing with.)

more

Since V5.2 (first released week 31, 2016; see here) there are implemented the functions MAXIFS() and MINIFS(). Unfortunately there are not yet (V5.4.3) the respective offline help texts. You need to resort to the function wizard to be guided. (This actually is clear enough.)
(The mentioned functions are not yet specified in "OpenFormula". They were introduced for better compatibility with recent versions of MS-Excel and are expected to work in the same way as SUMIFS and AVERAGEIFS concerning the conditions given as a pair of a range and a criterion each, while the range to get the data from is given as the first parameter.)

more


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#4 | Work Energy Theorem
(Physics) > Work,Energy and Power
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Related Practice Questions :

A steel wire can withstand a load up to 2940 N. A load of 150 kg is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is (2008 E)

(1) 30$°$

(2) 60$°$

(3) 80$°$

(4) 85$°$

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A ball is suspended by a thread of length l. What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension:

(1) gl

(2) 2 gl

(3) $\sqrt{gl}$

(4) $\sqrt{2gl}$

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As per given figure to complete the circular loop what should be the radius if initial height is 5?

(1) 4 m

(2) 3 m

(3) 2.5 m

(4) 2 m

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A stone of mass 1 kg tied to a light inextensible string of length $L=\frac{10}{3}m$ is whirling in a circular path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m/sec2, the speed of the stone at the highest point of the circle is

(1) 20 m/sec

(2) $10\sqrt{3}m/\mathrm{sec}$

(3) $5\sqrt{2}\text{\hspace{0.17em}}m/\mathrm{sec}$

(4) 10 m/sec

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Which one of the following is not a conservative force

(1) Gravitational force

(2) Electrostatic force between two charges

(3) Magnetic force between two magnetic dipoles

(4) Frictional force

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text:
# Open statistical ensemble

Open statistical ensemble (OSE) corresponds to a physical system, which exchanges energy with the environment, being with it in thermal equilibrium, exchanges particles with the environment for a given chemical potential and correctly takes into account the surface terms.

The open statistical ensemble is similar to the grand canonical ensemble (GCE) with the key difference being that the OSE is not spatially fixed (it may expand and contract naturally).

The expression for the partition function of the ensemble is similar to the partition function of the grand canonical ensemble (GCE), with the replacement of the Boltzmann factor in terms of the series on the correlation functions of special form.

The coefficient of surface tension, included in the partition function corresponds to the interface of fluid and hard solid, due to the strict respect of probability and potential limitations.

Unlike the GCE, for OSE the average number of particles in a given volume strictly coincides with the volume term. Also in contrast to the GCE, the correlation functions of the OSE strictly satisfy the requirement of translational invariance.

The expression for the general term of the distribution has the form

$p^v_m = \frac{z^m}{m! \Upsilon_v} \sum_{t=0}^\infty \frac{z^t}{t!} \int \left [ \prod_{i = 1}^m \prod_{j = m+1}^{m+t}\psi^v_i\chi^v_j \right ] {\mathcal B}^{(m,t)}_{1...m+t} d\boldsymbol{r}_1...d\boldsymbol{r}_{m+t},$

where $p^v_m$ is the probability to find $~m$ particles in volume $~v$, $~m \geq 1$, $~z$ is activity, $~\Upsilon_v$ - OSE partition function. $~\psi^v_i$ and $~\chi^v_j = 1 - \psi^v_j$ - characteristic functions, equal to unity inside and outside the system, respectively, and $~{\mathcal B}^{(m,t)}_{1...m+t}$ - partial localization factors, generalize the notions of the Boltzmann and Ursell factors and contain them as extreme cases.

The first term of the series corresponds to the distribution of the GCE with the accuracy up to normalizing factors - partition functions.

Summing the series we obtain the expression

$p^v_m = \frac{1}{m! \Upsilon_v} \int \left [ \prod_{i = 1}^m \psi^v_i \right ] \varrho^{(m)}_{G,1...m}(\chi^v) d\boldsymbol{r}_1...d\boldsymbol{r}_m,$

where $~\varrho^{(m)}_{G,1...m}$ - correlation function, depending on $~z$, $~\chi^v$ and expressed through a series of $~{\mathcal B}^{(m,t)}_{1...m+t}$. The last expression is equivalent to the GCE distribution with the replacement

$\varrho^{(m)}_{G,1...m} \approx z^m \exp(-\beta U^{m}_{1...m})$

and corresponding renormalization, where $~U^{m}_{1...m}$ - $~m$-particle interaction potential, $~1/\beta = k_B T$, $~k_B$ - Boltzmann constant, $~T$ - temperature. This expression shows that the GCE is a low-density approximation of the OSE.

For the partition function of the OSE, we have the expression

$\Upsilon_v = \exp { \sum_{t=1}^\infty \frac{z^t}{t!} \int \left [1 - \prod_{i = 1}^{t} \chi^v_i \right ] {\mathcal U}^{(t)}_{1...t} d\boldsymbol{r}_1...d\boldsymbol{r}_t },$

unlike the partition function of GCE

$\Xi_v = \exp { \sum_{t=1}^\infty \frac{z^t}{t!} \int \left [ \prod_{i = 1}^{t} \psi^v_i \right ] {\mathcal U}^{(t)}_{1...t} d\boldsymbol{r}_1...d\boldsymbol{r}_t },$

where $~{\mathcal U}^{(t)}_{1...t}$ - Ursell factors. Collapsed series of activities for $~\Upsilon_v$ we obtain the alternative representation

$~\Upsilon_v = \exp{\beta [ vP(z,T) + a\sigma (z,T) ]},$

where $~P(z,T)$ is the pressure, $~\sigma (z,T)$ - coefficient of surface tension on the interface of the fluid and hard solid, $~a$ - the surface bounding the system.

It should be stressed that an open system does not singled out, and the surface tension is created due to the fluctuation component of the partition function.

The last expression is exactly consistent with the probability of the formation of the hole volume $~v$ in the fluid

$~p^v_0 = \exp-{\beta [ vP(z,T) + a\sigma (z,T) ]},$

is determined from thermodynamic considerations

$~p^v_0 \propto \exp{( -\beta R_{min})},$

where $~R_{min}$ - minimum work of formation of such fluctuations.

## Some properties of the OSE

• Scale invariance. In contrast to grand canonical ensemble, an open statistical ensemble satisfies the scale invariance requirement: general term of the included subsystem distribution corresponds to that of the original system.
• Application to small systems. This distribution may be applied to however small volumes including those less than the size of a molecule. In this case it degenerates into a Bernoulli distribution with $~p=\varrho v$.
• Separation of fluctuations. When volume is much greater than the size of the molecule squared deviation of the number of particles is divided into bulk and surface terms.

## References

• Zaskulnikov V. M., Open statistical ensemble and surface phenomena: arXiv:0911.3106
• Zaskulnikov V. M., Open statistical ensemble: new properties (scale invariance, application to small systems, meaning of surface particles, etc.): arXiv:1004.0896v1


text:
# Testbook Testseries

0 votes
98 views

A packet has arrived in which the offset value of second frame is 450, the header length is 20 and the value of total length field is 1000. What is the number of the first byte in the first frame and the last byte in the last frame if the size of IP packet is 3500 B.

1. 2620, 6539
2. 2619, 6540
3. 2620, 6540
4. 2622, 6539

I'm confused here that if offset value of second frame is $450$ then 1st frame consisted of $450 \times 8 = 3600B$. that's more than the data we want to send.

0
i am also, struck.... post their solution.
0

...

0
in question he gives that frame, but in answer he considered fragment.
1
There is no offset field in frame
0
0

Shaik Masthan ill formed question?

0

ill formed question

means not well formed?

0
yes

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how is statement iv always true??


text:
# 5.6 Range-Separated Hybrid Density Functionals

Whereas RSH functionals such as LRC-$\omega$PBE are attempts to add 100% LR Hartree-Fock exchange with minimal perturbation to the original functional (PBE, in this example), other RSH functionals are of a more empirical nature and their range-separation parameters have been carefully parameterized along with all of the other parameters in the functional. These cases are functionals are discussed first, in Section 5.6.1, because their range-separation parameters should be taken as fixed. User-defined values of the range-separation parameter are discussed in Section 5.6.2, and Section 5.6.3 discusses a procedure for which an optimal, system-specific value of this parameter ($\omega$ or $\mu$) can be chosen for functionals such as LRC-$\omega$PBE or LRC-$\mu$PBE.


text:
F03 Chapter Contents
F03 Chapter Introduction
NAG Library Manual

# NAG Library Routine DocumentF03BHF

Note:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.

## 1  Purpose

F03BHF computes the determinant of a $n$ by $n$ symmetric positive definite banded matrix $A$ that has been stored in band-symmetric storage. F07HDF (DPBTRF) must be called first to supply the Cholesky factorized form. The storage (upper or lower triangular) used by F07HDF (DPBTRF) is relevant as this determines which elements of the stored factorized form are referenced.

## 2  Specification

 SUBROUTINE F03BHF ( UPLO, N, KD, AB, LDAB, D, ID, IFAIL)
 INTEGER N, KD, LDAB, ID, IFAIL REAL (KIND=nag_wp) AB(LDAB,*), D CHARACTER(1) UPLO

## 3  Description

The determinant of $A$ is calculated using the Cholesky factorization $A={U}^{\mathrm{T}}U$, where $U$ is an upper triangular band matrix, or $A=L{L}^{\mathrm{T}}$, where $L$ is a lower triangular band matrix. The determinant of $A$ is the product of the squares of the diagonal elements of $U$ or $L$.

## 4  References

Wilkinson J H and Reinsch C (1971) Handbook for Automatic Computation II, Linear Algebra Springer–Verlag

## 5  Parameters

1:     UPLO – CHARACTER(1)Input
On entry: indicates whether the upper or lower triangular part of $A$ was stored and how it was factorized. This should not be altered following a call to F07HDF (DPBTRF).
${\mathbf{UPLO}}=\text{'U'}$
The upper triangular part of $A$ was originally stored and $A$ was factorized as ${U}^{\mathrm{T}}U$ where $U$ is upper triangular.
${\mathbf{UPLO}}=\text{'L'}$
The lower triangular part of $A$ was originally stored and $A$ was factorized as $L{L}^{\mathrm{T}}$ where $L$ is lower triangular.
Constraint: ${\mathbf{UPLO}}=\text{'U'}$ or $\text{'L'}$.
2:     N – INTEGERInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{N}}>0$.
3:     KD – INTEGERInput
On entry: ${k}_{d}$, the number of superdiagonals or subdiagonals of the matrix $A$.
Constraint: ${\mathbf{KD}}\ge 0$.
4:     AB(LDAB,$*$) – REAL (KIND=nag_wp) arrayInput
Note: the second dimension of the array AB must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{N}}\right)$.
On entry: the Cholesky factor of $A$, as returned by F07HDF (DPBTRF).
5:     LDAB – INTEGERInput
On entry: the first dimension of the array AB as declared in the subprogram from which F03BHF is called.
Constraint: ${\mathbf{LDAB}}\ge {\mathbf{KD}}+1$.
6:     D – REAL (KIND=nag_wp)Output
7:     ID – INTEGEROutput
On exit: the determinant of $A$ is given by ${\mathbf{D}}×{2.0}^{{\mathbf{ID}}}$. It is given in this form to avoid overflow or underflow.
8:     IFAIL – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).

## 6  Error Indicators and Warnings

If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
 On entry, ${\mathbf{UPLO}}\ne \text{'L'}$ or $\text{'U'}$.
${\mathbf{IFAIL}}=2$
 On entry, ${\mathbf{N}}<1$.
${\mathbf{IFAIL}}=3$
 On entry, ${\mathbf{KD}}<0$.
${\mathbf{IFAIL}}=5$
 On entry, ${\mathbf{LDAB}}<{\mathbf{KD}}+1$.
${\mathbf{IFAIL}}=6$
The matrix $A$ is not positive definite.

## 7  Accuracy

The accuracy of the determinant depends on the conditioning of the original matrix. For a detailed error analysis see page 54 of Wilkinson and Reinsch (1971).

The time taken by F03BHF is approximately proportional to $n$.
This routine should only be used when $m\ll n$ since as $m$ approaches $n$, it becomes less efficient to take advantage of the band form.

## 9  Example

This example calculates the determinant of the real symmetric positive definite band matrix
 $5 -4 1 -4 6 -4 1 1 -4 6 -4 1 1 -4 6 -4 1 1 -4 6 -4 1 1 -4 6 -4 1 -4 5 .$

### 9.1  Program Text

Program Text (f03bhfe.f90)

### 9.2  Program Data

Program Data (f03bhfe.d)

### 9.3  Program Results

Program Results (f03bhfe.r)


text:
When doing some exercise in Java I notice that 30.7 may be printed as 30.700000000000003. This is because floating-point numbers, such as double, are actually stored in binary form. That is, in zeros and ones using only a limited number of numbers. As the number of floating-point numbers is infinite — (in case you’re wondering how infinite?, think how many floating-point or decimal values fit between the numbers 5 and 6 for instance). All of the floating-point numbers simply cannot be represented by a finite number of zeros and ones. Thus, the computer must place a limit on the accuracy of stored numbers.

Normally, account balances, for instance, are saved as integers such that, say, the value 1 represents one cent.


text:
zbMATH — the first resource for mathematics

Ricci flow with surgery on manifolds with positive isotropic curvature. (English) Zbl 1423.53080
Summary: We study the Ricci flow for initial metrics with positive isotropic curvature (strictly PIC for short).
In the first part of this paper, we prove new curvature pinching estimates that ensure that blow-up limits are uniformly PIC in all dimensions. Moreover, in dimension $$n\geq 12$$, we show that blow-up limits are weakly $$\mathrm{PIC}2$$. This can be viewed as a higher-dimensional version of the fundamental Hamilton-Ivey pinching estimate in dimension $$3$$.
In the second part, we develop a theory of ancient solutions that have bounded curvature, are $$\kappa$$-noncollapsed, are weakly $$\mathrm{PIC}2$$, and are uniformly PIC. This is an extension of Perelman’s work; the additional ingredients needed in the higher dimensional setting are the differential Harnack inequality for solutions to the Ricci flow satisfying the PIC2 condition, and a rigidity result due to Brendle-Huisken-Sinestrari for ancient solutions that are uniformly $$\mathrm{PIC}1$$.
In the third part of this paper, we prove a Canonical Neighborhood Theorem for the Ricci flow with initial data with positive isotropic curvature, which holds in dimension $$n \geq 12$$. This relies on the curvature pinching estimates together with the structure theory for ancient solutions. This allows us to adapt Perelman’s surgery procedure to this situation. As a corollary, we obtain a topological classification of all compact manifolds with positive isotropic curvature of dimension $$n \geq 12$$ that do not contain nontrivial incompressible $$(n-1)$$-dimensional space forms.

MSC:
 53C44 Geometric evolution equations (mean curvature flow, Ricci flow, etc.) (MSC2010)
Full Text:
References:
 [1] B\"ohm, Christoph; Wilking, Burkhard, Manifolds with positive curvature operators are space forms, Ann. of Math. (2). Annals of Mathematics. Second Series, 167, 1079-1097, (2008) · Zbl 1185.53073 [2] Brendle, Simon, A general convergence result for the Ricci flow in higher dimensions, Duke Math. J.. Duke Mathematical Journal, 145, 585-601, (2008) · Zbl 1161.53052 [3] Brendle, Simon, A generalization of Hamilton’s differential Harnack inequality for the Ricci flow, J. Differential Geom.. Journal of Differential Geometry, 82, 207-227, (2009) · Zbl 1169.53050 [4] Brendle, Simon, Einstein manifolds with nonnegative isotropic curvature are locally symmetric, Duke Math. J.. Duke Mathematical Journal, 151, 1-21, (2010) · Zbl 1189.53042 [5] Brendle, Simon, Ricci Flow and the Sphere Theorem, Grad. Stud. Math., 111, viii+176 pp., (2010) · Zbl 1196.53001 [6] Brendle, Simon, Ricci flow with surgery in higher dimensions, Ann. of Math. (2). Annals of Mathematics. Second Series, 187, 263-299, (2018) · Zbl 1393.53055 [7] Brendle, Simon; Huisken, Gerhard; Sinestrari, Carlo, Ancient solutions to the Ricci flow with pinched curvature, Duke Math. J.. Duke Mathematical Journal, 158, 537-551, (2011) · Zbl 1219.53062 [8] Brendle, Simon; Schoen, Richard, Manifolds with $$1/4$$-pinched curvature are space forms, J. Amer. Math. Soc.. Journal of the American Mathematical Society, 22, 287-307, (2009) · Zbl 1251.53021 [9] Cheeger, Jeff; Gromoll, Detlef, The splitting theorem for manifolds of nonnegative Ricci curvature, J. Differential Geometry. Journal of Differential Geometry, 6, 119-128, (1971/72) · Zbl 0223.53033 [10] Cheeger, Jeff; Gromoll, Detlef, On the structure of complete manifolds of nonnegative curvature, Ann. of Math. (2). Annals of Mathematics. Second Series, 96, 413-443, (1972) · Zbl 0246.53049 [11] Chen, Bing-Long; Zhu, Xi-Ping, Ricci flow with surgery on four-manifolds with positive isotropic curvature, J. Differential Geom.. Journal of Differential Geometry, 74, 177-264, (2006) · Zbl 1103.53036 [12] Chen, Bing-Long; Tang, Siu-Hung; Zhu, Xi-Ping, Complete classification of compact four-manifolds with positive isotropic curvature, J. Differential Geom.. Journal of Differential Geometry, 91, 41-80, (2012) · Zbl 1257.53053 [13] Chow, Bennett; Lu, Peng, The maximum principle for systems of parabolic equations subject to an avoidance set, Pacific J. Math.. Pacific Journal of Mathematics, 214, 201-222, (2004) · Zbl 1049.35101 [14] Chow, Bennett; Chu, Sun-Chin; Glickenstein, David; Guenther, Christine; Isenberg, James; Ivey, Tom; Knopf, Dan; Lu, Peng; Luo, Feng; Ni, Lei, The Ricci Flow: Techniques and Applications. Part III. Geometric-Analytic Aspects, Math. Surveys Monogr., 163, xx+517 pp., (2010) · Zbl 1216.53057 [15] Fraser, Ailana M., Fundamental groups of manifolds with positive isotropic curvature, Ann. of Math. (2). Annals of Mathematics. Second Series, 158, 345-354, (2003) · Zbl 1044.53023 [16] Gromov, M., Positive curvature, macroscopic dimension, spectral gaps and higher signatures. Functional Analysis on the eve of the 21st Century, Vol. II, Progr. Math., 132, 1-213, (1996) · Zbl 0945.53022 [17] Hamilton, Richard S., Three-manifolds with positive Ricci curvature, J. Differential Geom.. Journal of Differential Geometry, 17, 255-306, (1982) · Zbl 0504.53034 [18] Hamilton, Richard S., Four-manifolds with positive curvature operator, J. Differential Geom.. Journal of Differential Geometry, 24, 153-179, (1986) · Zbl 0628.53042 [19] Hamilton, Richard S., The Harnack estimate for the Ricci flow, J. Differential Geom.. Journal of Differential Geometry, 37, 225-243, (1993) · Zbl 0804.53023 [20] Hamilton, Richard S., The formation of singularities in the Ricci flow. Surveys in Differential Geometry, Vol. II, 7-136, (1995) · Zbl 0867.53030 [21] Hamilton, Richard S., Four-manifolds with positive isotropic curvature, Comm. 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text:
# Internal hom in $(\infty,2)$-categories

Let $X,Y$ be two $(\infty,2)$-categories, viewed as two fibrant objects in $\mathrm{Fun}(\Delta^{op},\mathrm{Set}_\Delta)$ with the complete Segal model structure (one uses the Joyal model structure on $\mathrm{Set}_\Delta$ to define this). It has been proved that the $(\infty,1)$-category 2-$\mathrm{Cat}$ of $(\infty,2)$-categories is Cartesian closed. The category of bisimplicial sets $\mathrm{Fun}(\Delta^{op}, \mathrm{Set}_\Delta)$ is also Cartesian closed.

My question is:

Does $Y^X$ in 2-$\mathrm{Cat}$ coincide with $Y^X$ in $\mathrm{Fun}(\Delta^{op}, \mathrm{Set}_\Delta)$, i.e. is the latter one fibrant?

If not, is there a reference about the relation between the two (more specific than saying the first $Y^X$ is a fibrant replacement of the latter $Y^X$)?

## 1 Answer

Yes because the model structure on Fun(Δ^op, sSet) is a cartesian model structure. This follows from the fact that the Joyal model structure is cartesian, and §10, §11 in Rezk's paper “A model for the homotopy theory of homotopy theory”.

• Thanks Dmitri! I am not very familiar with model categories. Does Cartesian model structure implies something like $Y^X$ is fibrant, when $Y$ is fibrant and $X$ is cofibrant? – Xin Jin Jul 23 '17 at 5:31
• Yes, internal hom is a right Quillen functor in a cartesian model structure, which implies your claim. – Dmitri Pavlov Jul 23 '17 at 14:29


text:
• Corpus ID: 207780663

# Controlled quantum operations and combs, and their applications to universal controllization of divisible unitary operations

@article{Dong2019ControlledQO,
title={Controlled quantum operations and combs, and their applications to universal controllization of divisible unitary operations},
author={Qingxiuxiong Dong and Shojun Nakayama and Akihito Soeda and Mio Murao},
journal={arXiv: Quantum Physics},
year={2019}
}
• Published 5 November 2019
• Physics
• arXiv: Quantum Physics
Unitary operations are a fundamental component of quantum algorithms, but they seem to be far more useful if given with a "quantum control" as a controlled unitary operation. However, quantum operations are not limited to unitary operations. Nevertheless, it is not a priori clear if a controlled form of these general deterministic quantum operations can be well-defined. This paper proposes a mathematically consistent definition of a controlled form of general deterministic quantum operations…

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