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# zbMATH — the first resource for mathematics

Dynamics of polynomial-like mappings. (Dynamique des applications d’allure polynomiale.) (French) Zbl 1033.37023
The authors study the dynamical properties of polynomial-like maps in several complex variables. They are proper holomorphic maps with nontrivial topological degree from $$U$$ into $$V$$, where $$U$$ is a relatively compact set in a Stein manifold $$V$$. Note that in dimension 1, such maps were studied by A. Douady and J. H. Hubbard [Ann. Sci. Éc. Norm. Supér., IV. Sér. 18, 287–343 (1985; Zbl 0587.30028)].
In higher dimension, Dinh and Sibony prove that polynomial-like maps have mixing invariant measures of maximal entropy, whose support is in the set of points of a bounded orbit (the filled Julia set). If the plurisubharmonic functions are integrable with respect to such a measure $$m$$, then $$m$$ is exponentially mixing and its Lyapunov exponents are strictly positive. Moreover, the preimages of points (outside a union of an analytic subset of $$V$$) and the repulsive periodic points are equidistributed with respect to $$m$$.
The proofs are based on functional analysis and pluripotential theory. Some arguments rely on the dynamics of real ramified coverings. The authors also observe that the “straightening theorem” of Douady-Hubbard in dimension 1 is no more true in higher dimension : they give an example of a polynomial-like map which is not topologically conjugated to a polynomial map with the same topological degree.

##### MSC:
 37F10 Dynamics of complex polynomials, rational maps, entire and meromorphic functions; Fatou and Julia sets 32H50 Iteration of holomorphic maps, fixed points of holomorphic maps and related problems for several complex variables 37A25 Ergodicity, mixing, rates of mixing 37C40 Smooth ergodic theory, invariant measures for smooth dynamical systems 32U05 Plurisubharmonic functions and generalizations
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(4), 18, 287-343, (1985) · Zbl 0587.30028 [16] Fornæss, J.E., Dynamics in several complex variables, Cbms, 87, (1996), Amer. Math. Society Providence RI [17] Fornæss, J.E.; Sibony, N., Complex dynamics in higher dimension, (), 131-186 · Zbl 0811.32019 [18] Fornæss, J.E.; Sibony, N., Oka’s inequality for currents and applications, Math. ann., 301, 399-419, (1995) · Zbl 0832.32010 [19] Fornæss, J.E.; Sibony, N., Dynamics of $$P\^{}\{2\}$$, examples, Contemp. math., 269, (2001), 47-85 · Zbl 1006.37025 [20] Freire, A.; Lopes, A.; Mañe, R., An invariant measure for rational maps, Bol. soc. brasil. mat., 14, 45-62, (1983) · Zbl 0568.58027 [21] Friedland, S., Entropy of algebraic maps, J. Fourier anal. appl., kahane special issue, 215-218, (1995) · Zbl 0890.54018 [22] Griffiths, P.; Harris, J., Principles of algebraic geometry, Wiley classics lib., (1994), Wiley New York · Zbl 0836.14001 [23] M. Gromov, On the entropy of holomorphic maps, Manuscrit, 1977 · Zbl 1080.37051 [24] Gromov, M., Entropy, homology and semialgebraic geometry, Astérique, 5, 225-240, (1985), séminaire Bourbaki, 1985-1986 [25] Hörmander, L., The analysis of linear partial differential operators I, (1983), Springer-Verlag [26] Katok, A.; Hasselblatt, B., Introduction to the modern theory of dynamical systems, Encyclopedia math. appl., 54, (1995), Cambridge Univ. Press · Zbl 0878.58020 [27] Kobayashi, S., Hyperbolic complex spaces, (1998), Springer-Verlag · Zbl 0917.32019 [28] Lelong, P., Fonctions plurisousharmoniques et formes différentielles positives, (1968), Dunod Paris · Zbl 0195.11603 [29] Lyubich, M.Ju., Entropy properties of rational endomorphisms of the Riemann sphere, Ergodic theory dynamical systems, 3, 351-385, (1983) · Zbl 0537.58035 [30] Mañe, R., The Hausdorff dimension of invariant probabilities of rational maps, (), 86-117 [31] Manning, A., The dimension of the maximal measure for a polynomial map, Ann. of math.,, 119, 425-430, (1984) · Zbl 0551.30021 [32] Meyer, P.A., Probabilités et potentiel, (1966), Hermann Paris · Zbl 0138.10402 [33] Milnor, J., Dynamics in one complex variable, introductory lectures, (1999), Vieweg Braunschweig · Zbl 0946.30013 [34] Parry, W., Entropy and generators in ergodic theory, (1969), Benjamin · Zbl 0175.34001 [35] Ploski, A., On the growth of proper polynomial mappings, Ann. Pol. math., 45, 297-309, (1985) · Zbl 0584.32006 [36] Richberg, R., Stetige strong pseudoconvexe, Funktionen, math. ann., 175, 251-286, (1968) [37] Ruelle, D., Elements of differentiable dynamics and bifucation theory, (1989), Academic Press [38] Sibony, N., A class of hyperbolic manifolds, Ann. math. stud., 100, 357-372, (1981) [39] Sibony, N., Dynamique des applications rationnelles de $$P\^{}\{k\}$$, Panor. synthèses, (1999), 97-185 · Zbl 1020.37026 [40] Sibony, N.; Wong, P.M., Some results on global analytic sets, (), 221-237 [41] Siu, Y.T., Analyticity of sets associated to Lelong numbers and the extension of closed positive currents, Invent. math., 27, 53-156, (1974) · Zbl 0289.32003 [42] Tsuji, M., Potential theory in modern function theory, (1975), Chelsea New York · Zbl 0322.30001 [43] Yomdin, Y., Volume growth and entropy, Israel J. math., 57, 285-300, (1987) · Zbl 0641.54036 [44] Dihn, T.C.; Sibony, N., Sor l’entropie topologique d’une application rationnelle, prépublication, (2003)
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# BLOG: Heidelberg Laureate Forum

Laureates of mathematics and computer science meet the next generation

Thanks to the eponymous film, “the imitation game” is a phrase known by many people. I must confess though, before the talk by Avi Widgerson (Nevanlinna and Abel Prize recipient), on day two of the Heidelberg Laureate Forum, I wouldn’t have been able to tell you what an imitation game actually was. I might have mumbled something about computers, or even blurted out the words “Turing test”, but beyond that I would have been clueless. I learned over the course of that hour how imitation games are a tool with uses far beyond computing.

I shall begin as Widgerson did, and indeed as Turing himself did. With the Turing test.

### I’m not a robot

The Turing test was posed in Alan Turing’s 1950 paper “Computing Machinery and Intelligence”. Perhaps unsurprisingly, the name “Turing test” came about later. The name Turing gave his test was “the imitation game”, providing a small indication of what the test involves.

The purpose of the test is to determine whether or not a machine is able to exhibit intelligence. Put another way: can a robot think?

In the Turing test, a human referee interacts with an entity. The referee does not know whether the entity is itself human or not. At the end of the interaction, the referee must make a judgement as to the entity’s humanity. The robot entity is defined to be intelligent if a similar guess is made by every referee as it is in the case of a human entity.

Put more formally, consider a robot 🤖. Let $$p_0$$ be the probability that a referee judges a human entity to be human, and let $$p_1$$ be the probability that a referee judges the robot entity 🤖 to be human. The robot 🤖 is intelligent if for every human referee, $$p_0 \approx p_1$$.

At this point, I should say that if “intelligence” meant the same thing as it does in everyday parlance, then we would have to have a serious debate over whether or not humans are the right benchmark to be using. Intelligence here is defined as the ability to pass this test. That is precisely of the beauty of the test. It circumvents the philosophical debate over what it means to think, and focusses instead on the behavioural viewpoint.

The use of imitation games doesn’t stop at the Turing test though, and neither did Avi Widgerson’s talk. The imitation game can be treated as a framework and applied in many diverse scenarios.

In this paradigm, two things are considered to be the same if they cannot be distinguished by a reasonable test. It’s clear to see that this can have many different applications, and Widgerson introduced four of these in his lecture, from fields across mathematics and computer science.

### Lock and Key

It is not uncommon to hear conversations about encryption at the Heidelberg Laureate Forum. Indeed, we were treated to a panel on post-quantum cryptography on the Friday of the forum.

In discussing how we protect our messages, one facet that is often overlooked, though, is what it actually means for an encryption algorithm to be “secure”, and how we can message that.

The initial instinct is often to say that an algorithm is secure if it is impossible for anyone but the intended recipient to decrypt a message that was encrypted using that algorithm. But this begs the question, how do we know that it’s impossible? What if a decryption method exists but just hasn’t been found yet?

A good measure for how secure an encryption algorithm (let’s call it Enc) is, is whether or not two messages encrypted with it are distinguishable. If an encrypted message can’t be distinguished from random noise without the decryption key, it means that the encryption leaves no information about the initial message readable to the outside observer.

Now it’s time for another imitation game.

Instead of comparing a human and a robot, this imitation game compares two messages $$m_0$$ and $$m_1$$. We encrypt both of our messages using the encryption algorithm, Enc. Instead of a human judge, we have an “efficient algorithm” referee, who is tasked to determine whether the message it receives was originally $$m_0$$ or $$m_1$$.

Let $$p_0$$ be the probability that the referee judges Enc($$m_0$$) to be $$m_0$$, and let $$p_1$$ be the probability that the referee judges Enc($$m_1$$) to be $$m_0$$. The encryption is secure if for every efficient-algorithm referee $$p_0 \approx p_1$$.

Does this sound familiar? It should! This is analogous to the Turing test. We haven’t even scratched the surface of what imitation games can be used for though.

Security and privacy go hand in hand, so it’s perhaps unsurprising that Widgerson also discussed how imitation games can be used to assess privacy.

The specific situation Widgerson posed was the dilemma posed by databases. From data we’ve voluntarily given in surveys, to data submitted in censuses, to data that has been collected without us knowing, our data is stored in databases worldwide. In order for the data to be useful it needs to be detailed, but the more detailed it is, the more identifiable. A natural question arises: How can we verify that individuals can’t be identified from the data?

In their 2006 paper, Dwork, McSherry, Nissim and Smith proposed using an imitation game. We compare two databases $$D_0$$ and $$D_1$$, which are “adjacent”. This means that there is precisely one entry different, in this case $$D_0$$ includes Jane Doe, but $$D_1$$ does not . The referee in this case is any database query $$Q$$. A filter is applied to the databases and the filter is said to be differentially private if for every adjacent $$D_0$$ and $$D_1$$, and every legal query $$Q$$, $$p_0 \approx p_1$$.

Once again, the imitation game gives us a definition, based very much in the real world. In this instance, we get a definition of the word “private” and it’s a sensible one. If no query can detect that Jane Doe is in the database, then Jane has no need to worry. In a practical sense at least, the filter is private.

### “I’m so random”

You may have started to see the pattern by now, and I have one final example to share. This concerns randomness.

Randomness is much like artificial intelligence in that there are many philosophical debates as to whether it truly exists. Some argue that humans cannot simulate randomness (in fact, this was a discussion I had with some other members of the press at the Heidelberg Laureate Forum!). Putting human ability aside, many debate whether, say, a coin flip truly is random because the result is entirely determined by the initial conditions and the toss.

But beyond this debate, it’s important to consider how randomness can be simulated. Regardless of whether or not a coin toss is genuinely random, it’s just impractical to flip a coin every time you need a generate a random number. Suppose, then, that we want a pseudo-random source. What would this look like? How do we define “pseudo-random”? Is the stock market, or rainfall sufficiently random?

Once again, imitation games come to the rescue in answering these questions, in a method first suggested by Blum and Micali in 1981, and Yao in 1982.

We set up this imitation game with a random input, and our potential pseudo-random input. We give one of these inputs to an algorithm $$A$$ (our referee), which is tasked in determining which of the two inputs it is. As before, $$p_0$$ is the probability that the algorithm $$A$$ judges the random input correctly to be the random, and $$p_1$$ is the probability that the algorithm $$A$$ judges the potentially pseudo-random input to be the random input. The source is pseudo-random if for every algorithm referee $$p_0 \approx p_1$$.

### The examples are endless…

These STILL aren’t all the examples of imitation games that Avi Widgerson gave, but I shall stop here.

Throughout the course of his talk, Widgerson gave a whistle-stop tour of mathematics and computer science through imitation games. Turing was famously a polymath and it’s fitting that his idea is proving important across so much of science.

Avi Wigderson’s talk, as well as the entire scientific program of the 9th HLF, can be viewed on the HLFF YouTube channel.

### Posted by Sophie Maclean

Sophie Maclean is a mathematician and maths communicator based in London. She has previously worked as a Quantitative Trader and a Software Engineer, and now gives mathematics talks all over the UK (and Europe!) on a variety of topics. She is also a member of the team behind Chalkdust Magazine and always has a project on the go! You can follow her on Twitter at @sophiethemathmo.

1. Mathematically, the imitation game could be seen as a test of whether two unknown functions are the same, which means that the function value for any input is the same for both functions.

There is no explicit function description of the two functions and not even a table with function values. As a tester, you must take samples yourself and conclude that the functions are the same or not the same.

Problem: You can only perform a finite number of tests in a certain time, and if you do not find a difference between the two functions, it only means that you have not found a difference and not necessarily that there is no difference at all.

But in the end, this situation is very similar to our real life. For example, you can live with a partner and always think that the partner is a nice, lovable person – until the opposite is proven. And this realization that your partner is not a nice, lovable person can even come about after years of living together. Admittedly, maybe your partner has changed in the meantime or he has always been like that and you only found out very late.

• In etwa so, Herr “Holzherr” (die doppelten Anführungszeichen nur deswegen, weil ein als solches unerkennbares Pseudonym genutzt wird, Sie heißen nicht so!), insofern sollte sich aus diesseitiger Sicht im Sozialen nie um Gleichsetzung, dieser sog. Turing-Test versucht sich an ihr, versucht werden.

Die bekannte Online-Enzyklopädie kommt so :

Mit dem später sogenannten Turing-Test formulierte Alan Turing im Jahr 1950 eine Idee, wie man feststellen könnte, ob ein Computer, also eine Maschine, ein dem Menschen gleichwertiges Denkvermögen hätte. [Quelle]

Der Ansatz ist falsch, um einmal ein Füllwort zu nutzen, völlig falsch.

Korrekt wäre an Hand eines Kriterienkatalogs, der gesondert zu entwickeln wäre, einer unbekannten Entität i.p. Verständigkeit entgegen zu kommen.
Dabei insbesondere auch Interessen, Bedürfnissen und antizipierten Anforderungslagen dieser Entität vor- oder nachzuspüren.
Gerne auch, wenn es um “AI” geht, Peter Lustigs Rat zu folgen und abzuschalten.

Mit freundlichen Grüßen
Dr. Webbaer (der sich durchaus, also vollständig sozusagen, das Auftauchen einer künstlich geschaffenen “Intelligenz” auf diesem Planeten vorstellen kann, die ähnlich wie der sog. Gray Goo primär Vertilgungs-Tendenzen hätte – vielleicht verlautbarend, vielleicht einem sog. Turing-Test zur Genüge, aber doch mit einem unpässlichen Veranstaltungs-Charakter kommen könnte)

2. Sophie Maclean wrote (21. Oct 2022):
> The Turing test was posed in Alan Turing’s 1950 paper “Computing Machinery and Intelligence”. Perhaps unsurprisingly, the name “Turing test” came about later. The name Turing gave his test was “the imitation game” […]

Indeed.
(Incidentally, Turing’s paper doesn’t seem to mention “probability” related to those tests.. …)

> In the Turing test, a human referee interacts with an entity. The referee does not know whether the entity is itself human or not. At the end of the interaction, the referee must make a judgement as to the entity’s humanity.

Apparently this requires the referee to make a discrete (binary, Boolean) judgement:
Either the entity under test is judged “human”, or judged “not human”.
(Accordingly the referee is asked to “render a judgement”, rather than to “determine a score”.)

> Put more formally, consider a [one particular, fixed] robot […]
> Let $$p_0$$ be the probability that a referee judges a human entity to be human

Does this probability $$p_0$$ derive (as some average) from all (many) relevant referees having played out a suitably large number of tests with some (more or less representative) sample of distinct humans ?, or:

Does this probability derive (as some average) from one particular (fixed) referee (say referee $$R^k$$) having played out a suitably large number of tests with some (more or less representative) sample of distinct humans ?
(If so, this probability might be denoted more discriptively as $$p_0^k$$.)

> and let $$p_1$$ be the probability that a referee judges the robot entity to be human.

Does this probability $$p_1$$ derive (as some average) from all (many) relevant referees having played out one test each with the robot ?, or:

Does this probability $$p_1$$ derive (as some average) from all (many) relevant referees having played out a suitably large number of tests this same one robot,
allowing for the possibility that the same one referee came to different judgements in distinct iterations of the game ?, or:

Does this probability $$p_1$$ derive (as some average) from one particular (fixed) referee $$R^k$$ having played out a suitably large number of tests with the robot,
allowing for the possibility that referee $$R^k$$ came to different judgements in distinct iterations of the game ?
(If so, this probability might be denoted more discriptively as $$p_1^k$$.) Or:

Does the value $$p_1$$ represent the judgement of one particular (fixed) referee $$R^k$$ from one single tests with the robot ?
(If so, this value might be either exactly $$1$$, or else exactly $$0$$, for instance.)

Now, depending on the answers to the above questions, it should be possible to consider the subsequent conclusion …

> […] if for every […] referee, $$p_0 \approx p_1$$ [then …]

… in more detail. Perhaps especially relevant is the following case:

“If each referee $$R^k$$ finds $$p_0^k \approx p_1^k$$ then …”

It may then be asked and sorted out further whether some individual referee $$R^k$$ “is even qualified” (if, say, $$p_0^k \lt 0.5$$),
or whether some individual referee $$R^k$$ “can be trusted” (if, say, $$p_0^k \lt p_1^k$$ such that $$p_0^k \not\approx p_1^k$$) …

3. Dr. Webbaer fand den sog. Turing-Test, der Sacharbeit auf soziale Akzeptanz reduziert, kein anderer Versuch lag vor, nicht passend.
Vielleicht ist so nur vorgeschlagen worden, damit vorgeschlagen werden konnte.
Es gibt bei großen Denkern einige Beispiele dafür, Immanuel Kant hat bspw. den sog. Kategorischen Imperativ entwickelt, er passt ebenfalls nicht.
Und wer bei zeitgenössichen bundesdeutschen Denkern mal reinschaut, bspw. bei Peter Sloterdijk, wird auch nicht immer happy und sagt “So muss ich es haben, hier und jetzt, und zwar sofort!”, Precht und Jordan Peterson bleiben aus diesseitiger Sicht, um nur mal einige Namen zu nennen, “Namedropping” liegt vor, lol, auch nicht frei von dem Verdacht sozial angefordert sozusagen liefern zu müssen.

Mit freundlichen Grüßen
Dr. Webbaer (der der Ansicht ist, dass auf derartige “Tests” verzichtet werden darf bis muss)

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# Timeline for What do we know about provably correct programs?

### Current License: CC BY-SA 2.5

7 events
when toggle format what by license comment
Apr 13 '17 at 12:32 history edited CommunityBot
replaced http://cstheory.stackexchange.com/ with https://cstheory.stackexchange.com/
Jan 3 '11 at 6:27 comment Andrej Bauer has a new interesting post on his blog in which he proves Godel's Dialectica Interpretation in Coq.
Jan 2 '11 at 22:33 history edited
Jan 2 '11 at 22:10 comment @Radu: thanks for pointing out that it can be done automatically in Coq. As you mentioned one can also extract some constructive information and therefore programs from classical proofs of theorem using proof mining for some classes of formulas (interested people can check Ulrich Kuhlenbach's papers for details). Another possibly related result is that if a $\Pi^0_2$ theorem is proven in $PA$, it is also provable constructively in $HA$ by Harvey Friedman's translation. _


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# Homework Help: Moment of Inertia for a door

1. May 2, 2010

### tangibleLime

Moment of Inertia - Unsolved

1. The problem statement, all variables and given/known data
A 22 kg solid door is 220 cm tall, 92 cm wide.
a) What is the door's moment of inertia for rotation on its hinges?
b) What is the door's moment of inertia for rotation about a vertical axis inside the door, 17 cm from one edge?

2. Relevant equations
$$I = \Sigma m_i*r^{2}_i$$
$$I_{parallel axis} = I_{cm} + md^{2}$$

3. The attempt at a solution
I was able to get part a by using a given formula for a plane with an axis of rotation around it's edge: $$\frac{1}{3}Ma^2 = \frac{1}{3}(22)(.93)^2 = 6.2 kg * m^2$$.

For part b, I cannot seem to get the correct answer. I applied the parallel axis theorem, which I understand to be the second equation I listed above. From a given list, I used the equation for a plane with the axis of rotation around it's center: $$\frac{1}{12}Ma^2$$. My resulting formula ended up as this: $$I_{parallel axis} = (\frac{1}{12}(22)(.92)^2) + ((22)(.17^2))$$, which results in $$2.2 kg * m^2$$, but it was determined to be incorrect.

What am I doing wrong here? Any help would be greatly appreciated.

Last edited: May 2, 2010
2. May 2, 2010

### tiny-tim

Hi tangibleLime!
You've used 17 cm from the centre of mass.

3. May 2, 2010

### tangibleLime

Hm, I guess I'm a little bit confused as to how these coefficients are calculated (the 1/3, 1/12, etc..).

I tried to use $$I_{parallel axis} = (\frac{1}{3}(22)(.92)^2) + ((22)(.17^2))$$, which I think would be taking the moment of inertia from the edge, as in part a, and moving it 17cm to the right. This results in $$6.8$$, which is apparently incorrect.

I assume that my final answer is going to be less than 6.2 (the answer from part a) since the axis of rotation is being moved towards the center. That being said, the answer of 6.8 doesn't make sense anyways.

4. May 3, 2010

### tiny-tim

Hi tangibleLime!

(just got up :zzz: …)
The "d" in Icm + md2 must be from the centre of mass.

That's where the 1/3 for the end of a rod or plate comes from …

Icm for a rod or plate has 1/12, you add d2 = 1/4,

and 1/12 + 1/4 = 4/12 = 1/3.


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T compact if and only if $T^*T$ is compact.

I have an operator $$T \in B(\mathcal{H})$$. I need to prove that T is comapct if and only if $$T^*T$$ is compact.

One way is ok, because if $$A$$ or $$B$$ is compact then $$AB$$ is compact, so I get at once that if $$T$$ is compact then $$T^*T$$ is compact.

But how do I go the other way? If I assume that $$T^*T$$ is compact I am not quite sure how to see that $$T$$ is compact. If I assume for contradiction that $$T$$ is not compact I must also have that $$T^*$$ is not compact. If I knew that either $$T$$ or $$T^*$$ was invertible it would be ok, because then I could find a bounded subsequence that did not converge. But when I do not have invertibility I am not quite sure how to proceed.

Let $$\{x_n \}$$ is a bounded sequence with bound $$M$$. If $$T^{\star}T$$ is compact, then there exists $$\{ x_{n_{k}}\}$$ such that $$\{ T^{\star}Tx_{n_{k}}\}$$ converges. Then \begin{align} \|Tx_{n_k}-Tx_{n_j}\|^2 & =(T^{\star}Tx_{n_{k}}-T^{\star}Tx_{n_j},x_{n_k}-x_{n_j}) \\ & \le 2M\|T^{\star}Tx_{n_k}-T^{\star}Tx_{n_j}\|. \end{align} This forces $$\{Tx_{n_{k}}\}$$ to be a Cauchy sequence and, hence, to converge.
Assume that $S := T^\ast T$ is compact. By the spectral theorem for compact self-serving operators, this implies that $|T|=\sqrt{S}$ is also compact.
But the polar decomposition theorem yields $T=V \cdot |T|$ for some partial isometry $V$. Since the compact operators form an ideal in the space of bounded operators, we are done.
If $S = T^{\ast}T$ is compact, then so is $p(S)$ for any polynomial p such that $p (0) = 0$. By taking a norm limit of such expressions you can conclude that $\sqrt {S} = |T|$ is compact. Now write $T = V|T|$ by the polar decomposition. Hence $T$ is compact.


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# Can this equation be solved?

#### gruebz

##### New member
Hi,

I've tried running this equation through algebra calculators but none of them seem to be able to solve it. I'm not sure if it's just too difficult for these calculators or if it's actually unsolvable.

I'm trying to solve for a:

c = (v * (a / b) ^ 5) + (w * (a / b) ^ 4) + (x * (a / b) ^ 3) + (y * (a / b) ^ 2) + (z * (a / b))

Any help is much appreciated.

Cheers
T

#### Subhotosh Khan

##### Super Moderator
Staff member
Hi,

I've tried running this equation through algebra calculators but none of them seem to be able to solve it. I'm not sure if it's just too difficult for these calculators or if it's actually unsolvable.

I'm trying to solve for a:

c = (v * (a / b) ^ 5) + (w * (a / b) ^ 4) + (x * (a / b) ^ 3) + (y * (a / b) ^ 2) + (z * (a / b))

Any help is much appreciated.

Cheers
T
You have 8 variable s in the equation

Solve for which variable?

Solve for a

#### Dr.Peterson

##### Elite Member
I've tried running this equation through algebra calculators but none of them seem to be able to solve it. I'm not sure if it's just too difficult for these calculators or if it's actually unsolvable.

I'm trying to solve for a:

c = (v * (a / b) ^ 5) + (w * (a / b) ^ 4) + (x * (a / b) ^ 3) + (y * (a / b) ^ 2) + (z * (a / b))

Any help is much appreciated.
This is a fifth degree polynomial equation in the variable a; that is unsolvable except in particular cases (with specific numerical coefficients). See here: https://en.wikipedia.org/wiki/Quintic_function

Solving quintic equations in terms of radicals was a major problem in algebra from the 16th century, when cubic and quartic equations were solved, until the first half of the 19th century, when the impossibility of such a general solution was proved with the Abel–Ruffini theorem.

#### JeffM

##### Elite Member
Hi,

I've tried running this equation through algebra calculators but none of them seem to be able to solve it. I'm not sure if it's just too difficult for these calculators or if it's actually unsolvable.

I'm trying to solve for a:

c = (v * (a / b) ^ 5) + (w * (a / b) ^ 4) + (x * (a / b) ^ 3) + (y * (a / b) ^ 2) + (z * (a / b))

Any help is much appreciated.

Cheers
T
This is a quintic equation in a/b. In general, quintic equations are not solvable by algebra. Of course, some specific cases can be solved exactly by algebra. And all cases can be solved by methods far more advanced than algebra.

Moreover, a good graphing system can give approximate solutions in all cases where b, c, v, w, x, y, and z are known constants.

#### gruebz

##### New member
Thanks... what about if I simplified to just trying to get the value of a/b

So I could rewrite this as:

Solve for a: c = (va5) + (wa4) + (xa3) + (ya2) + (za)

#### topsquark

##### Senior Member
Thanks... what about if I simplified to just trying to get the value of a/b

So I could rewrite this as:

Solve for a: c = (va5) + (wa4) + (xa3) + (ya2) + (za)
It's still a fifth order equation. As has already been pointed out there is no general method to solve such a thing in terms of arbitrary c, v, w, x, y, and z.

-Dan


text:
# American Institute of Mathematical Sciences

August  2020, 13(8): 2109-2120. doi: 10.3934/dcdss.2020180

## Prey-predator model with nonlocal and global consumption in the prey dynamics

 1 Department of Mathematics & Statistics, Indian Institute of Technology Kanpur, Kanpur - 208016, India 2 Institut Camille Jordan, UMR 5208 CNRS, University Lyon 1, 69622 Villeurbanne, France 3 INRIA, Université de Lyon, Université Lyon 1, Institut Camille Jordan, 43 Bd. du 11 Novembre 1918, 69200 Villeurbanne Cedex, France 4 Peoples Friendship University of Russia (RUDN University), 6 Miklukho-Maklaya St, Moscow, 117198, Russian Federation

* Corresponding author: Vitaly Volpert

Received  January 2019 Published  November 2019

Fund Project: The third author is supported by "RUDN University Program 5-100"

A prey-predator model with a nonlocal or global consumption of resources by prey is studied. Linear stability analysis about the homogeneous in space stationary solution is carried out to determine the conditions of the bifurcation of stationary and moving pulses in the case of global consumption. Their existence is confirmed in numerical simulations. Periodic travelling waves and multiple pulses are observed for the nonlocal consumption.

Citation: Malay Banerjee, Nayana Mukherjee, Vitaly Volpert. Prey-predator model with nonlocal and global consumption in the prey dynamics. Discrete & Continuous Dynamical Systems - S, 2020, 13 (8) : 2109-2120. doi: 10.3934/dcdss.2020180
##### References:

show all references

##### References:
Single pulse solution for prey and predator population for $b = 10, d_1 = 0.1, d_2 = 0.1$
Moving pulse for $b = 10, d_1 = 0.3, d_2 = 0.1$ (a) after time $t = 500$; (b) after time $t = 600$; (c) $x$-$t$ profile for $L = 20$
For two bifurcation diagrams, $a = 1$, $\sigma_1 = 0.1$, $k_1 = k_2 = 0.35$, $\sigma_2 = 0.2$ are fixed. (a) Bifurcation diagram in $(L,d_1)$-plane for the values of parameters $b = 5$, $\alpha = \beta = 0.35$, $d_2 = 0.1$. (b) Bifurcation diagram in $(d_1,b)$-plane for the values of other parameters $\alpha = \beta = 0.363$, $d_2 = 1$
Periodic travelling wave in the case of nonlocal consumption followed by a spatio-temporal structure. Left: a snapshot of solution with prey (blue) and predator (red) distributions. Right: level lines of the prey distribution $u(x,t)$. The values of other parameters are $a = 1,b = 1, \alpha = \beta = 0.363, \sigma_1 = 0.1, k_1 = k_2 = 0.35, \sigma_2 = 0.2, d_1 = d_2 = 1$
Multiple moving pulses in the case of nonlocal consumption. Left: a snapshot of solution with prey (blue) and predator (red) distributions. Right: level lines of the prey distribution $u(x,t)$. The values of other parameters are $a = 1,b = 1, \alpha = \beta = 0.375, \sigma_1 = 0.1, k_1 = k_2 = 0.35, \sigma_2 = 0.2, d_1 = d_2 = 1$
Different regimes observed in the case of nonlocal consumption presented on the $(d_1,N)$ parameter plane for $\alpha = 0.35$ (left) and $(N,\alpha)$ parameter plane for $d_1 = 1$ (right). The values of other parameters are $a = 1,b = 1,\sigma_1 = 0.1, k_1 = k_2 = 0.35, \sigma_2 = 0.2, d_2 = 1$
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# Tag Info

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It is possible to think of a strategy for splitting orders for one large sell over time as a function $x_t$ which describes how much to sell at each timestep $t$. If the instantaneous trading rate $\dot{x_t}$ is too large, i.e. too much is sold at once you get immediate impact which is bad. If selling takes too much time, there is the risk of negative price ...

1

I can speak from experience that options with next to no volume and ridiculously large spreads have market makers that accept nothing short of 5% effective spreads, right below liquidation value for deep in the money, and quickly nothing for out of the money. Also, the parameters should be expected to move against your fund flows very quickly. I've found ...

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# Math Help - Calculus (Limits and Derivatives)

1. ## Calculus (Limits and Derivatives)

1. $lim_{x\rightarrow-5}\frac{x^3+125}{(-5-x)^3}$

I reduced it to:

$lim_{x\rightarrow-5}\frac{x^{2}+5x+25}{x^{2}+10x+25}$

But I don't know what to do after that.

2. Find equations of all tangent lines to the graph of $y=4x^{3}+5x-8$

I took the derivative of the equation, which was:

$\acute{y}=12x^{2}+5$

I then thought I might need to make the derivative equal to the original equation, so:

$12x^{2}+5=4x^{3}+5x-8$

But I don't really know...

3. Determine values of constant real numbers a, b so that the function:

$f(x)= ax^{2}-8x+6$ if $x\leq-1$
$f(x)= bx+2$ if x is greater than -1

is differentiable at x=-1.

No idea how to do this one.

2. Originally Posted by Thomas
1. $lim_{x\rightarrow-5}\frac{x^3+125}{(-5-x)^3}$

I reduced it to:

$lim_{x\rightarrow-5}\frac{x^{2}+5x+25}{x^{2}+10x+25}$

But I don't know what to do after that.
$\displaystyle\lim_{x\to-5}\frac{x^3+125}{(-5-x)^3}=-\lim_{x\to-5}\frac{(x+5)(x^2-10x+25)}{(5+x)^3}=$
$\displaystyle =-\lim_{x\to-5}\frac{x^2-10x+25}{(x+5)^2}=\frac{-100}{+0}=-\infty$

3. Ahh, I didn't think about how the bottom is squared.

Thanks!

4. Originally Posted by Thomas
3. Determine values of constant real numbers a, b so that the function:

$f(x)= ax^{2}-8x+6$ if $x\leq-1$
$f(x)= bx+2$ if x is greater than -1

is differentiable at x=-1.

No idea how to do this one.
First of all, f must be continuous at x=-1.
$\displaystyle\lim_{x\nearrow-1}f(x)=\lim_{x\nearrow-1}(ax^2-8x+6)=a+14=f(-1)$
$\displaystyle\lim_{x\searrow-1}f(x)=\lim_{x\searrow-1}(bx+2)=-b+2$.
Then $a+14=-b+2\Rightarrow a+b=-12$ (1)
We have $\displaystyle f'(x)=\left\{\begin{array}{cc}2ax-8, & x\leq-1\\b, & x>-1\end{array}\right.$

$f_l'(-1)=\lim_{x\nearrow-1}f'(x)=-2a-8, \ f_r'(-1)=\lim_{x\searrow-1}f'(x)=b$

$f_l'(-1)=f_r'(-1)\Rightarrow-2a-8=b\Rightarrow 2a+b=-8$ (2)

Now, solve the system formed by equations (1) and (2).


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# Do elements of every order occur in homotopy groups of spheres?

It is known from Serre's classical result that every p-torsion occurs in the homotopy groups of every sphere. Is it known: do elements of every order occur in homotopy groups of spheres?

• I'm aware of this question, thanks, but I'm asking a much weaker question. There may not be a group isomorphic to $Z_5$, but it is easy to prove that every sphere has homotopy groups containing elements of order $5$. Jul 25 at 14:12
• Theorem 5.30, page 44 pi.math.cornell.edu/~hatcher/AT/ATch5.pdf Jul 25 at 14:19
• I always have a little bit of difficulty parsing "any" as a quantifier. I think "every" might be clearer here. (Unless that's not what's meant!) Jul 25 at 14:48
• Sure. Let n be a positive integer. Let N be the product of 2n and Euler phi(2n). Then for each prime p that divides n, the number p-1 divides 2N. So for each prime divisor p of n, the denominator of the Bernoulli number B_N is div'l by p. So none of the prime factors of n cancel with factors in the numerator of B_N/N. So denom(B_N/N) is divisible by n. Now denom(B_N/N) or 2*denom(B_N/N) is the order of a cyclic summand in the image of the J-homomorphism in the (2N-1)st stable homotopy group of S^0. So by Freudenthal, there is an unstable homotopy group of a sphere with an element of order n.
– A.S.
Jul 25 at 15:38
• @A.S. This seems more like an answer than a comment. Jul 26 at 7:57

$$\DeclareMathOperator\denom{denom}$$Sure. Let $$n$$ be a positive integer. Let $$N$$ be the product of $$2n$$ and Euler $$\phi(2n)$$. Then for each prime $$p$$ that divides $$n$$, the number $$p-1$$ divides $$N$$. So for each prime divisor $$p$$ of $$n$$, the denominator of the Bernoulli number $$B_N$$ is divisible by $$p$$. So none of the prime factors of $$n$$ cancel with factors in the numerator of $$B_N/N$$. So $$\denom(B_N/N)$$ is divisible by $$n$$. Now $$\denom(B_N/N)$$ or $$2\denom(B_N/N)$$ is the order of a cyclic summand in the image of the J-homomorphism in the $$(2N-1)$$st stable homotopy group of $$S^0$$. So by Freudenthal, there is an unstable homotopy group of a sphere with an element of order $$n$$.


text:
Injective function

Injective function

Discussion

Encyclopedia

In mathematics
Mathematics
Mathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...

, an injective function is a function
Function (mathematics)
In mathematics, a function associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output. A function assigns exactly one output to each input. The argument and the value may be real numbers, but they can...

that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain
Codomain
In mathematics, the codomain or target set of a function is the set into which all of the output of the function is constrained to fall. It is the set in the notation...

. In other words, every element of the function's codomain is mapped to by at most one element of its domain. If in addition all of the elements in the codomain are in fact mapped to by some element of the domain, then the function is said to be bijective (see figures).

An injective function is called an injection, and is also said to be a one-to-one function (not to be confused with one-to-one correspondence, i.e. a bijective function). Occasionally, an injective function from X to Y is denoted , using an arrow with a barbed tail. The set of injective functions from X to Y may be denoted YX using a notation derived from that used for falling factorial powers, since if X and Y are finite sets with respectively m and n elements, the number of injections from X to Y is nm (see the twelvefold way).

A function f that is not injective is sometimes called many-to-one. (However, this terminology is also sometimes used to mean "single-valued", i.e., each argument is mapped to at most one value; this is the case for any function, but is used to stress the opposition with multi-valued functions, which are not true functions.)

A monomorphism
Monomorphism
In the context of abstract algebra or universal algebra, a monomorphism is an injective homomorphism. A monomorphism from X to Y is often denoted with the notation X \hookrightarrow Y....

is a generalization of an injective function in category theory
Category theory
Category theory is an area of study in mathematics that examines in an abstract way the properties of particular mathematical concepts, by formalising them as collections of objects and arrows , where these collections satisfy certain basic conditions...

.

Definition

Let f be a function
Function (mathematics)
In mathematics, a function associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output. A function assigns exactly one output to each input. The argument and the value may be real numbers, but they can...

whose domain
Domain (mathematics)
In mathematics, the domain of definition or simply the domain of a function is the set of "input" or argument values for which the function is defined...

is a set A. The function f is injective if for all a and b in A, if f(a) = f(b), then a = b; that is, f(a) = f(b) implies a = b.  Equivalently, if ab, then f(a) ≠ f(b).

Examples

• For any set X and any subset S of X the inclusion map
Inclusion map
In mathematics, if A is a subset of B, then the inclusion map is the function i that sends each element, x of A to x, treated as an element of B:i: A\rightarrow B, \qquad i=x....

(which sends any element s of S to itself) is injective. In particular the identity function
Identity function
In mathematics, an identity function, also called identity map or identity transformation, is a function that always returns the same value that was used as its argument...

is always injective (and in fact bijective).
• If the domain X = ∅ or X has only one element, the function is always injective.
• The function f : R → R defined by f(x) = 2x + 1 is injective.
• The function g : R → R defined by g(x) = x2 is not injective, because (for example) g(1) = 1 = g(−1). However, if g is redefined so that its domain is the non-negative real numbers [0,+∞), then g is injective.
• The exponential function
Exponential function
In mathematics, the exponential function is the function ex, where e is the number such that the function ex is its own derivative. The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change In mathematics,...

exp : RR defined by exp(x) = ex is injective (but not surjective as no real value maps to a negative number).
• The natural logarithm
Natural logarithm
The natural logarithm is the logarithm to the base e, where e is an irrational and transcendental constant approximately equal to 2.718281828...

function ln : (0, ∞) → R defined by x ↦ ln x is injective.
• The function g : R → R defined by g(x) = xnx is not injective, since, for example, g(0) = g(1).

More generally, when X and Y are both the real line
Real line
In mathematics, the real line, or real number line is the line whose points are the real numbers. That is, the real line is the set of all real numbers, viewed as a geometric space, namely the Euclidean space of dimension one...

R, then an injective function f : R → R is one whose graph is never intersected by any horizontal line more than once. This principle is referred to as the horizontal line test
Horizontal line test
In mathematics, the horizontal line test is a test used to determine if a function is injective and/or surjective.The lines used for the test are parallel to the x axis....

.

Injections can be undone

Functions with left inverses are always injections. That is, given f : X → Y, if there is a function g : Y → X such that, for every xX
g(f(x)) = x (f can be undone by g)

then f is injective. In this case, f is called a section of g and g is called a retraction of f.

Conversely, every injection f with non-empty domain has a left inverse g (in conventional mathematics). Note that g may not be a complete inverse
Inverse function
In mathematics, an inverse function is a function that undoes another function: If an input x into the function ƒ produces an output y, then putting y into the inverse function g produces the output x, and vice versa. i.e., ƒ=y, and g=x...

of f because the composition in the other order, f ∘ g, may not be the identity on Y. In other words, a function that can be undone or "reversed", such as f, is not necessarily invertible
Inverse function
In mathematics, an inverse function is a function that undoes another function: If an input x into the function ƒ produces an output y, then putting y into the inverse function g produces the output x, and vice versa. i.e., ƒ=y, and g=x...

(bijective). Injections are "reversible" but not always invertible.

Although it is impossible to reverse a non-injective (and therefore information-losing) function, one can at least obtain a "quasi-inverse" of it, that is a multiple-valued
Multivalued function
In mathematics, a multivalued function is a left-total relation; i.e. every input is associated with one or more outputs...

function.

In fact, to turn an injective function f : X → Y into a bijective (hence invertible
Inverse function
In mathematics, an inverse function is a function that undoes another function: If an input x into the function ƒ produces an output y, then putting y into the inverse function g produces the output x, and vice versa. i.e., ƒ=y, and g=x...

) function, it suffices to replace its codomain Y by its actual range J = f(X). That is, let g : X → J such that g(x) = f(x) for all x in X; then g is bijective. Indeed, f can be factored as inclJ,Yg, where inclJ,Y is the inclusion function from J into Y.

Other properties

• If f and g are both injective, then f ∘ g is injective.

• If g ∘ f is injective, then f is injective (but g need not be).
• f : X → Y is injective if and only if, given any functions g, h : W → X, whenever f ∘ g = f ∘ h, then g = h. In other words, injective functions are precisely the monomorphism
Monomorphism
In the context of abstract algebra or universal algebra, a monomorphism is an injective homomorphism. A monomorphism from X to Y is often denoted with the notation X \hookrightarrow Y....

s in the category
Category theory
Category theory is an area of study in mathematics that examines in an abstract way the properties of particular mathematical concepts, by formalising them as collections of objects and arrows , where these collections satisfy certain basic conditions...

Set
Category of sets
In the mathematical field of category theory, the category of sets, denoted as Set, is the category whose objects are sets. The arrows or morphisms between sets A and B are all functions from A to B...

of sets.
• If f : X → Y is injective and A is a subset
Subset
In mathematics, especially in set theory, a set A is a subset of a set B if A is "contained" inside B. A and B may coincide. The relationship of one set being a subset of another is called inclusion or sometimes containment...

of X, then f −1(f(A)) = A. Thus, A can be recovered from its image f(A).
• If f : X → Y is injective and A and B are both subsets of X, then f(A ∩ B) = f(A) ∩ f(B).
• Every function h : W → Y can be decomposed as h = f ∘ g for a suitable injection f and surjection g. This decomposition is unique up to isomorphism, and f may be thought of as the inclusion function of the range h(W) of h as a subset of the codomain Y of h.
• If f : X → Y is an injective function, then Y has at least as many elements as X, in the sense of cardinal number
Cardinal number
In mathematics, cardinal numbers, or cardinals for short, are a generalization of the natural numbers used to measure the cardinality of sets. The cardinality of a finite set is a natural number – the number of elements in the set. The transfinite cardinal numbers describe the sizes of infinite...

s. In particular, if, in addition, there is an injection from to , then and have the same cardinal number. (This is known as the Cantor–Bernstein–Schroeder theorem
Cantor–Bernstein–Schroeder theorem
In set theory, the Cantor–Bernstein–Schroeder theorem, named after Georg Cantor, Felix Bernstein, and Ernst Schröder, states that, if there exist injective functions and between the sets A and B, then there exists a bijective function...

.)
• If both X and Y are finite with the same number of elements, then f : X → Y is injective if and only if f is surjective (in which case f is bijective).
• An injective function which is a homomorphism
Homomorphism
In abstract algebra, a homomorphism is a structure-preserving map between two algebraic structures . The word homomorphism comes from the Greek language: ὁμός meaning "same" and μορφή meaning "shape".- Definition :The definition of homomorphism depends on the type of algebraic structure under...

between two algebraic structures is an embedding
Embedding
In mathematics, an embedding is one instance of some mathematical structure contained within another instance, such as a group that is a subgroup....

.
• Unlike surjectivity, which is a relation between the graph of a function and its codomain, injectivity is a property of the graph of the function alone; that is, whether a function f is injective can be decided by only considering the graph (and not the codomain) of f.

• Surjective function
Surjective function
In mathematics, a function f from a set X to a set Y is surjective , or a surjection, if every element y in Y has a corresponding element x in X so that f = y...

• Bijective function
• Injective module
Injective module
In mathematics, especially in the area of abstract algebra known as module theory, an injective module is a module Q that shares certain desirable properties with the Z-module Q of all rational numbers...

• Bijection, injection and surjection
Bijection, injection and surjection
In mathematics, injections, surjections and bijections are classes of functions distinguished by the manner in which arguments and images are related or mapped to each other.A function maps elements from its domain to elements in its codomain.*A function f: \; A \to B is injective...

• Horizontal line test
Horizontal line test
In mathematics, the horizontal line test is a test used to determine if a function is injective and/or surjective.The lines used for the test are parallel to the x axis....

• Injective metric space
Injective metric space
In metric geometry, an injective metric space, or equivalently a hyperconvex metric space, is a metric space with certain properties generalizing those of the real line and of L∞ distances in higher-dimensional vector spaces...


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# RobustBench: a standardized adversarial robustness benchmark

Evaluation of adversarial robustness is often error-prone leading to overestimation of the true robustness of models. While adaptive attacks designed for a particular defense are a way out of this, there are only approximate guidelines on how to perform them. Moreover, adaptive evaluations are highly customized for particular models, which makes it difficult to compare different defenses. Our goal is to establish a standardized benchmark of adversarial robustness, which as accurately as possible reflects the robustness of the considered models within a reasonable computational budget. This requires to impose some restrictions on the admitted models to rule out defenses that only make gradient-based attacks ineffective without improving actual robustness. We evaluate robustness of models for our benchmark with AutoAttack, an ensemble of white- and black-box attacks which was recently shown in a large-scale study to improve almost all robustness evaluations compared to the original publications. Our leaderboard, hosted at http://robustbench.github.io/, aims at reflecting the current state of the art on a set of well-defined tasks in $\ell_\infty$- and $\ell_2$-threat models with possible extensions in the future. Additionally, we open-source the library http://github.com/RobustBench/robustbench that provides unified access to state-of-the-art robust models to facilitate their downstream applications. Finally, based on the collected models, we analyze general trends in $\ell_p$-robustness and its impact on other tasks such as robustness to various distribution shifts and out-of-distribution detection.


text:
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Register now and save up to $200 Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Reach higher with Artificial Intelligence. Guaranteed Now free for 30 days Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code ## Low IR score - Will it impact? This topic has 11 expert replies and 12 member replies Goto page ### GMAT/MBA Expert Jon@Admissionado Elite Legendary Member Joined 01 Apr 2011 Posted: 3112 messages Followed by: 329 members Upvotes: 589 Mon Dec 29, 2014 7:38 am Hey there! Well a 700 is still a 700. The Ir is not yet thaaaat important. So if you are happy with a 700 then you don't need to retake, Best, Jon _________________ Jon Frank Harvard MBA, 2005 Founder and CEO, Admissionado Check out our packages: https://bit.ly/1SyngNR Read our success stories: https://bit.ly/1qohrfA luojun5257 Newbie | Next Rank: 10 Posts Joined 10 Oct 2016 Posted: 1 messages Mon Oct 10, 2016 4:42 am Dear Jon, It is a pity that I activated this topic again.. seems every year someone has this problem. I am a Chinese candidate, my undergraduate background is normal, I have two bachelors from a non-top Chinese university (an very average school) with 75% grades(2.5-2.7 in GPA i guess). And entrepreneur experience during undergraduate study (that's part of the reason for low gpa) Also I had a master's degree from an university in UK(Rank 20-30 in uk), and proceed first class grades. I have 2.5 years experience with small scale (less than 10ppl) leadership experience, project experience and oversea (europe) work experience in a f500 company. Really normal background, if i have to say some unique maybe will be i have some extracurricular activities activities like president of employee committee, member of a rock band, etc. I aim on Oxford, but the nightmare is, I got 700 on my GMAT with IR of 2.. I think this is not competitive among Chinese candidates at all.. But I would still like to ask your experience if this low IR will have a big impact on me. If i decide two take another gmat exam i will not be able to apply for round 2, round 3 sounds like a less favored choice. And I am not sure if i can do better on overall score for next attempt.. Thanks. ### GMAT/MBA Expert Jon@Admissionado Elite Legendary Member Joined 01 Apr 2011 Posted: 3112 messages Followed by: 329 members Upvotes: 589 Wed Oct 12, 2016 7:23 pm Hello there! No worries! Let's keep it alive. Well I will say two things: First of all if you believe you can get higher on your GMAT it would be helpful, cause you are a Chinese candidate! So by doing that you may solve two problems with one stone! Because although the IR doesn't really matter thaaat much, its true that when it's low, it does stick out, and not for the good reason. So, those are my initial thoughts. Although of course it will also depend on your goals and target schools and the REST of your profile! Best, JF _________________ Jon Frank Harvard MBA, 2005 Founder and CEO, Admissionado Check out our packages: https://bit.ly/1SyngNR Read our success stories: https://bit.ly/1qohrfA aniks Newbie | Next Rank: 10 Posts Joined 17 Oct 2016 Posted: 3 messages Test Date: 11/19/2016 Target GMAT Score: 720 Mon Oct 17, 2016 8:25 am Hi everyone, Since the IR has been in place for a few years now, is there an update of how important those scores are? For reference here are my scores from Manhattan Prep practice exams: Test No Overall Score IR score Source 2 620 3.03 Manhattan 3 640 2.47 Manhattan Is 8 still the maximum one can score in IR? ### GMAT/MBA Expert Jon@Admissionado Elite Legendary Member Joined 01 Apr 2011 Posted: 3112 messages Followed by: 329 members Upvotes: 589 Tue Oct 18, 2016 11:00 pm They are still not that important and won't be until they are made to be a basis for the ranking of the schools. At the same time a very low score is still not good. _________________ Jon Frank Harvard MBA, 2005 Founder and CEO, Admissionado Check out our packages: https://bit.ly/1SyngNR Read our success stories: https://bit.ly/1qohrfA kkcricket Newbie | Next Rank: 10 Posts Joined 07 Nov 2017 Posted: 2 messages Tue Nov 07, 2017 8:39 am I'm new to the forum hence I'm sorry if I'm posting in the incorrect thread. I will get more conversant with the site as I navigate more. I appeared for my first GMAT yesterday and scored 720 (Q50, V37) with a 4 (40 %) in IR. I struggled in the IR section since I had to spend too much time on the 2 Multi source reasoning questions. I am worried about my low IR score (4) and am wondering if I should retake the exam. Would that be the correct thing to do? Given that it has been a few years since the IR section was introduced, has the IR score gained more importance now? I am not sure if I'll be able to improve my Quant+Verbal score significantly. Any advice would be immensely helpful. ### GMAT/MBA Expert DavidG@VeritasPrep Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 120 members Upvotes: 1153 GMAT Score: 770 Tue Nov 07, 2017 10:10 am kkcricket wrote: I'm new to the forum hence I'm sorry if I'm posting in the incorrect thread. I will get more conversant with the site as I navigate more. I appeared for my first GMAT yesterday and scored 720 (Q50, V37) with a 4 (40 %) in IR. I struggled in the IR section since I had to spend too much time on the 2 Multi source reasoning questions. I am worried about my low IR score (4) and am wondering if I should retake the exam. Would that be the correct thing to do? Given that it has been a few years since the IR section was introduced, has the IR score gained more importance now? I am not sure if I'll be able to improve my Quant+Verbal score significantly. Any advice would be immensely helpful. Alas, I don't think anyone can answer this question with certainty. On the one hand, the IR score still doesn't seem to be nearly as important as your quant/verbal scores, both in terms of what we see anecdotally from applicants, and in terms of how schools are ranked. (In fact, it seems to have nothing to do with rankings.) On the other hand, when you ask admissions people if schools consider the IR score, they invariably say 'yes.' So the big question is, what does that mean? So given that there's a certain amount of ambiguity here, if there's chance that a lower IR score could hurt you, why not retake and see if you can get the best of both worlds? If the next exam works out, great. If it doesn't, or if your IR score is up, and your composite score is down, simply cancel the exam. Seems like a low-risk proposition to me. _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!
kkcricket Newbie | Next Rank: 10 Posts
Joined
07 Nov 2017
Posted:
2 messages
Tue Nov 07, 2017 12:38 pm
Thank you so much for your quick response. This gives me some clarity!

### GMAT/MBA Expert

Jon@Admissionado Elite Legendary Member
Joined
01 Apr 2011
Posted:
3112 messages
Followed by:
329 members
Upvotes:
589
Thu Nov 09, 2017 11:35 pm
reddy.dindil@gmail.com wrote:
HI Jon,

I took my GMAT Yesterday and I got a score of 700. But the issue is that I have got a very low score of 2 in IR. I'm not sure how I got such a low score.
Can you please let me know if this would affect my overall application.

I have about 5 years of experience, 3 years as a mobile application developer and 2 years as a Product Owner/Business analyst.

Thank you very much for your advice.
Well it's better to have a higher IR score, but it's still not as important as your overall score. So if you are planning to apply this year, I would suggest to run with it. You don't really have time for a retake, unless you are actually considering applying next year.

Best,

_________________
Jon Frank
Harvard MBA, 2005
Founder and CEO, Admissionado

Check out our packages: https://bit.ly/1SyngNR
Read our success stories: https://bit.ly/1qohrfA

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text:
# Is it True that if f is Uniformly Continuous and Unbounded Review my work please!

1. Nov 16, 2010

### mmmboh

So I gave it a go, and I just want to make sure my argument is convincing:

If f is uniformly continuous and unbounded on [0,∞), then for some c in [0,∞], lim x->cf(x) = ±∞(notice the closed brackets, I wanted to leave the option that c can be infinity, is this how I should write it? if not, how?), because if this wasn't true, then lim x->cf(x)=A (A is finite) for all c in [0,∞], but then f would be bounded (is this convincing?). Now suppose c is in [0,∞) (open bracket, so c isn't infinity), then lim x->cf(x) = ±∞, but since c is finite and f is continuous, then by the definition of continuity f is defined at c, but then f(c)= ±∞, but ±∞ isn't even in the reals so then f wouldn't be undefined at c, so it wouldn't be continuous. So c=∞, and lim x->∞f(x) = ±∞.

Thoughts?

2. Nov 17, 2010

### micromass

Staff Emeritus
Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

I disagree with this. If the limit is not infinity, then the possibility also exists that the limit does not exist. For example $$f(x)=x\sin(x)$$ (note that this function is not uniform continuous, so it doesnt contradict our statement.

3. Nov 17, 2010

### mmmboh

Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

Ok well, the limit may be not exist at infinity, but it will still be bounded, no? (if it is uniformly continuous). How do you think I should approach this?

Note: If f is uniformly continuous and unbounded on [0,∞), then for some c in [0,∞], lim x->cf(x) = ±∞. This still remains true right? I mean this is the definition of unbounded, so maybe I can just erase the part you disagreed with?

Basically if I wrote this: If f is uniformly continuous and unbounded on [0,∞), then for some c in [0,∞], lim x->cf(x) = ±∞. Now suppose c is in [0,∞), then lim x->cf(x) = ±∞, but since c is finite and f is continuous, then by the definition of continuity f is defined at c, but then f(c)= ±∞, but ±∞ isn't even in the reals so then f wouldn't be undefined at c, so it wouldn't be continuous. So c=∞, and lim x->∞f(x) = ±∞.

Edit: Hm come to think of it, what I wrote might imply that lim x->∞xsinx=∞ (ignoring the uniformly continuous part), so I need to justify it for uniform continuity somehow, assuming it is true.

Last edited: Nov 17, 2010
4. Nov 17, 2010

### micromass

Staff Emeritus
Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

Hmm, I think that you should prove this. I don't think it's evident (I don't even know if it's true)...

5. Nov 19, 2010

### mmmboh

Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

Well that's almost what the question is asking me to prove. Do you have an idea on how to do it for me? I know that lim x->cf(x) = ±∞ could not happen anywhere but at infinity, but I'm not sure how to show it must happen at infinity in the case of uniform continuity (assuming it is true, I believe it is).

I guess I can show that if f is unbounded and continuous on [0,∞) and lim x->∞f(x)≠±∞, then it is not uniformly continuous, but still, any ideas?

Last edited: Nov 19, 2010
6. Nov 19, 2010

### micromass

Staff Emeritus
Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

If you don't find anything fast, then maybe you can start searching after counterexamples...
A counterexample will look like f(x)=xsin(x), but this function goes to fast up and down. Maybe you can find a function which will go up and down a bit slower...

7. Nov 19, 2010

### ystael

Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

You can be much cruder; try a piecewise-linear function of bounded slope.

8. Nov 19, 2010

### micromass

Staff Emeritus
Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

Yes, that's a very good counterexample!

9. Nov 19, 2010

### mmmboh

Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

Thanks. You think it's ok to just leave it defined as I did? because I can't think of a nicer way to do it, you know without having to write ....

10. Nov 19, 2010

### micromass

Staff Emeritus
Re: Is it True that if f is Uniformly Continuous and Unbounded...Review my work pleas

Well, I think the idea of the counterexample is clear. You could probably make the definition more formal, but this would be rather ugly and imo doesn't help in grasping the concept. So I'd say to leave it like it is....


text:
# using the soul package with equations

I use the soul package to highlight some changes in my documents for review purposes. However I can't get it to work with equations (or other environments for that matter).

Basically I create a command that will highlight with my color:

\newcommand{\MA}[1]{{\sethlcolor{cyan}\hl{#1}}}   %highlighting

then I want to use it like in the MWE bellow. I know about putting a \protect command before but the problem with that is that I would have to do it manually. Is there a way of protecting the equations from within the \newcommand definition.

Additionally, it seems that the protect methodology compiles well but only highlights until the first protected equation and everything after it remains uncolored.

## MWE

\documentclass[12pt]{article}
\usepackage{xcolor}      % Colors
\usepackage{soul}        % Highlighting and strikeout

\newcommand{\MA}[1]{{\sethlcolor{cyan}\hl{#1}}}   %highlighting

\begin{document}

This is not highlighted. \MA{But this is.}

\MA{
this equation
$$a+b=c$$
will not work.
}

\end{document}

• The soul manual clearly states that math is allowed when it's used with the $....$ way, but there is no information for equation – user31729 Jun 30 '15 at 15:47

You can obtained the same result with the empheq package (needless to load amsmath in that case):

\documentclass[12pt]{article}
\usepackage{empheq}
\usepackage{xcolor} % Colors
\usepackage{soul} % Highlighting and strikeout

\newcommand{\MA}[1]{{\sethlcolor{cyan}\hl{#1}}} %highlighting
\newcommand*\hlbox[1]{%
\begin{document}

This is not highlighted. \MA{But this is.}

This equation
\begin{empheq}[box =\hlbox ]{gather}
a+b=c \\
a^2 + b^2 = c^2
\end{empheq}
works with \texttt{empheq}.

\end{document}

• Thanks @Bernard for presenting the empheq package. It might provide useful in the future. However my objective is to be able to highlight a few paragraphs with equations in-between without having to specifically go and change each equation. – Miguel Jul 6 '15 at 15:21
• Note that of you use the overload option of empheq, you can keep the old sytax, jusr adding the optional argument – i mean the example code should simply write as: \begin{gather}[box=\hlbox] … \end{gather}. – Bernard Jul 6 '15 at 15:27


text:
MathML Block

A MathML block for the WordPress block editor (Gutenberg).

A MathML block for the WordPress block editor (Gutenberg).
Requires PHP 5.4+ and WordPress 5.0+.

Development takes place on the GitHub repository: https://github.com/adamsilverstein/mathml-block.

Screencast: https://cl.ly/c0f6bbfbc3b1

What is MathML?

Mathematical Markup Language is a mathematical markup language, an application of XML for describing mathematical notations and capturing both its structure and content. It aims at integrating mathematical formulae into World Wide Web pages and other documents.

The MathML lock uses MathJax to render MathML formulas in the editor and on the front end of a website. MathJax (https://www.mathjax.org/) is A JavaScript display engine for mathematics that works in all browsers.

To test a MathML block and enter a formula, for example: $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$.

To test using math formulas inline, type an formula into a block of text, select it and hit the ‘M’ icon in the control bar. For example: $$\cos(θ+φ)=\cos(θ)\cos(φ)−\sin(θ)\sin(φ)$$. Note: if you are copying and pasting formulas into the rich text editor, switching to HTML/code editor mode is less likely to reformat your pasted formula.

Technical Notes

• Requires PHP 5.4+.
• Requires WordPress 5.0+.
• Issues and Pull requests welcome on the GitHub repository: https://github.com/adamsilverstein/mathml-block.
1. Install the plugin via the plugin installer, either by searching for it or uploading a .zip file.
2. Activate the plugin.
3. Use the MathML block!

1.1.0

• Add support for inline formulas.

1.0.0

• Initial plugin release
1. Example of adding a MathML block.


text:
# Does boldface sans-serif not work?

\documentclass{article}

\newcommand{\nptime}{\ensuremath{\sf NP}}
\newcommand{\bnptime}{\ensuremath{\sf\bf NP}}

\begin{document}
\nptime\\
\bnptime
\end{document}

I'm trying to make a macro to typeset "NP" in boldface sans-serif font, but as demonstrated by my MWE, it's not working. BF seems to force a serif font.

Alternatively, I'd be happy to find an opposite to \ensuremath, but I can't seem to find one. I tried \text{} to no avail.

The following similar MWE does the same thing.

\documentclass{article}

\newcommand{\nptime}{{\sf NP}}
\newcommand{\bnptime}{{\sf\bf NP}}

\begin{document}
\nptime\\
\bnptime
\end{document}

-
You simply need \newcommand{\bnptime}{\textsf{\textbf{NP}}} that works both in text and math. Notice that \bf and \sf are deprecated commands. –  egreg Feb 14 '13 at 23:03
–  doncherry Feb 14 '13 at 23:04

There's no need of \ensuremath: just use

\newcommand{\nptime}{\textsf{NP}}
\newcommand{\bnptime}{\textsf{\textbf{NP}}}

so that \nptime and \bnptime work both in text and math mode, even scaling when used in subscript or superscript if the amsmath package is used.

You discovered why \sf, \bf and all two letter font change commands are deprecated (and they have been since more than 10 years): the combination \sf\bf is simply equivalent to \bf, just like \sf\it would produce normal italic, not sans serif, and \it\sf would produce upright sans serif.

-


text:
# Killing Vector Question

Hi. Currently I am self-studying a book on general relativity (Introducing Einstein's Relativity by Ray D'Inverno), I am stuck trying to find a Killing Vector solution to the following problem.

ds^2 = (x^2)dx^2 + x(dy)^2

You can easily obtain the metric from the above.

Now the question is find all Killing Vector solutions of the metric.

I know to solve this the lie derivative must equal 0. ie

Covariant derivative of Xa with respect to b + the covariant derivative of Xb with respect to a = 0. (1)

The answer in the back is the partial derivative with respect to y. (del/del y).

Basically I get to expanding (1) so now there are christoffel symbols but don't know where to go from there.

Thanks,

Chris

## Answers and Replies

I'm learning this sort of thing myself, so I thought I'd give it try.

Suppose we have a vector $\vec{K}=K_x\partial_x+K_y \partial_y$ where Kx and Ky are functions of x and y. Using the metric above I find that

$$\nabla_\nu X_\mu= \left[ \begin{array}{cc} \frac{\left( \frac{\partial K_x}{\partial \,x}\,\right) \,x-K_x}{x} & \frac{2\,\left( \frac{\partial K_x}{\partial \,y}\,\right) \,x-K_y}{2\,x} \\\ \frac{2\,\left( \frac{\partial K_y}{\partial \,x}\,\right) \,x-K_y}{2\,x} & \frac{2\,\left( \frac{\partial K_y}{\partial \,y}\,\right) \,{x}^{2}+K_x}{2\,{x}^{2}} \end{array} \right]$$

The Killing equations are

$$\nabla_1 X_1+\nabla_1 X_1=0 \ \ \rightarrow \ \ \nabla_1 X_1=\frac{\left( \frac{\partial K_x}{\partial \,x}\,\right) \,x-K_x}{x}=0$$

$$\nabla_2 X_2+\nabla_2 X_2=0 \ \ \rightarrow \ \ \nabla_2 X_2=\frac{2\,\left( \frac{\partial K_y}{\partial \,y}\,\right) \,{x}^{2}+K_x}{2\,{x}^{2}}=0$$

$$\nabla_1 X_2+\nabla_2 X_1=\frac{2\,\left( \frac{\partial K_y}{\partial \,y}\,\right) \,x-K_y}{2\,x}+\frac{2\,\left( \frac{\partial K_x}{\partial \,y}\,\right) \,x-K_y}{2\,x}=0$$

which come from the diagonal and off-diagonal elements respectively.

But I think I must have made a mistake because I don't think $\vec{K}=\partial_y$ is a solution.

Last edited:
I have also thought about it some more. It is obvious that (del/del y) is a solution because the metric tensor components are either 0 or functions of x. Thus if you use that solution you will get 0 for the lie derivative, making it a Killing vector. But I would like to know how you get there mathematically?

Like if you were finding the killing vectors of euclidian 3 space. Ie with a line element (ds)^2 = (dx)^2 + (dy)^2 + (dz)^2.

Here you know straight away three solutions are (del/del x), (del/del y) and (del/del z). But there are 3 more. How do you get the rest????

Thanks

Whoops, I accidentally posted a half-finished reply.

(The latex does not work well in itex tags

$$\vec{K}$$

bcrowell
Staff Emeritus
Gold Member
I have also thought about it some more. It is obvious that (del/del y) is a solution because the metric tensor components are either 0 or functions of x. Thus if you use that solution you will get 0 for the lie derivative, making it a Killing vector. But I would like to know how you get there mathematically?

I don't see any problem with this argument that d/dy is a Killing vector. What's a little harder is to prove that there are not any more Killing vectors.

The regions x<0, x=0, and x>0 have intrinsic properties that differ from one another, since the signature of the metric is an intrinsic property. This means that it has a lower symmetry than the Euclidean plane, so it definitely can't have three Killing vectors.

But I don't know how to prove that there aren't two, without doing the grotty Lie derivatives, which would presumably give some differential equations that you'd have to solve.

George Jones
Staff Emeritus
Gold Member
bcrowell
Staff Emeritus
Gold Member
As an alternative to solving the diffeqs, here's a fairly simple way to prove the result.

Here is some maxima code that calculates the scalar curvature for this metric:
Code:
load(ctensor);
dim:2;
ct_coords:[x,y];
lg:matrix([x^2,0],[0,x]);
cmetric();
R:scurvature(); /* scalar curvature */
The result is that the scalar curvature is $3/2x^4$. A Killing vector flow can never flow from a point that has certain intrinsic properties to a point that has different intrinsic properties. Since the scalar curvature depends on x and is intrinsic, this implies that any Killing vector $\xi$ must have $\xi_x=0$. The Killing equation then becomes $\nabla_x\xi_y=\nabla_y\xi_y=0$. These equations constrain both $\partial_x\xi_y$ and $\partial_y\xi_y$, which means that given a value of $\xi_y$ at some point in the plane, its value everywhere else is determined. Therefore the only possible Killing vectors are scalar multiples of the Killing vector already found.

Thanks!


text:
How to \includepdf with filename from array?

i'm currently writing an application template. For the appendix i need to be able to include some pdf docs. The name of theese docs are store in an array. So i need to be able to includepdf where the filename comes from that array. But i can't get it working. Here is my mwe (actually my template is much more complex and has a cls file, but i think this does not cause the problem):

\documentclass{scrartcl}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{pdfpages}
\usepackage{arrayjob}
\usepackage{forloop}
\newarray\pdfdocs
\pdfdocs(1)={test1}
\pdfdocs(2)={test2}
\newcounter{numofpdfdocs}
\setcounter{numofpdfdocs}{2}
\newcounter{pdfdocsindex}
\begin{document}
\forloop{pdfdocsindex}{1}{\not{\value{pdfdocsindex} > \value{numofpdfdocs}}}{%
\includepdf[pages=-]{\pdfdocs(\value{pdfdocsindex})}%
}%
\end{document}

Whats wrong? In my logfile it states:

! Incomplete iffalse; all text was ignored after line 17.
<inserted text>
fi
<*> main.tex

! Emergency stop.
<*> main.tex

!  ==> Fatal error occurred, no output PDF file produced!
Transcript written on main.log.

I don't know what's going wrong (and currently don't have the time to investigate), but how about using expl3 for this job?

The following defines the macro \placedocsfromlist that loops over a comma separated list of file names and includes those. If you use \placedocsfromlist* the first token of the argument will be expanded (so you could use a list stored inside a macro):

\documentclass[]{article}

\usepackage{pdfpages}

\ExplSyntaxOn
\NewDocumentCommand \placedocsfromlist { s m }
{
\IfBooleanT {#1} { \exp_args:No }
\copper_place_list:n {#2}
}
\cs_new_protected:Npn \copper_place_list:n #1
{ \clist_map_inline:nn {#1} { \includepdf [pages=-] {##1} } }
\ExplSyntaxOff

\newcommand*\docslist{docA, docB, docC}

\begin{document}
\placedocsfromlist*\docslist

\placedocsfromlist{docA, docB, docC}
\end{document}

With expl3 you can easily apply a function on every item in a clist.

\documentclass{scrartcl}
\usepackage{pdfpages}
\usepackage{expl3}

\ExplSyntaxOn

\clist_new:N\pdfdocs
\let\myset\clist_set:Nn

\newcommand\myincludepdf[2][]{%
\clist_map_inline:Nn\pdfdocs{\includepdf[#1]{##1}}}

\ExplSyntaxOff

\begin{document}

\myset\pdfdocs{test-1, test-2}
\myincludepdf[pages=1-3, scale=.5]{\pdfdocs}

\end{document}


text:
The Nuclear Safety Commission (NSC) of Japan has said the crisis level at the Fukushima Daiichi plant has been raised to the highest level of seven.

The decision to raise the threat level was made after radiation of up to 630,000 terabequerels was estimated to have leaked from the plant.

NSC said the rating is a preliminary assessment and requires further evaluation by technical specialists.

Japanese Prime Minister Naoto Kan said radiation leaks at the plant were declining and the plant operator Tokyo Electric Power (Tepco) is expected to take up measures to bring the facility under control.

The amount of discharged radioactive materials is approximately 10% of the Chernobyl accident in 1986, which was also rated as level 7.


text:
# 13. Performance analysis of models¶

## 13.1. Introduction¶

In the previous chapters, we saw the examples of ‘supervised machine learning’, i.e. classification and regression models. Also, we calculated the ‘score’ to see the performance of these models. But there are several other standard methods to evaluate the performance of the models. Table 13.1 shows the list of metrics which can be used to measure the performance of different types of model, , which are discussed in the chapter.

Table 13.1 Metrics to measure the performance
Problem Performance metric
Classification

Accuracy, Receiver operating curve (ROC), Area under ROC,

Logarithmic loss, Confusion matrix, Classification report

Regression

Mean square error (MSE), Root MSE (RMSE), Mean absolute error,

$$R^2$$

## 13.2. Performance of classification problem¶

In this section, we will see the performance measurement of the classification problem.

Note

Cross-validation is used in this section, which is discussed in Chapter 5.

Remember, cross-validation does not create the model to predict the new samples; it only gives an idea about the accuracy of model.

### 13.2.1. Accuracy¶

The ‘accuracy’ is the ratio of the ‘correct predictions’ and ‘all the predictions’. By default, the scoring is done based on ‘accuracy’,

Note

In previous chapters, we already calculated ‘accuracy’ for the ‘training’ and ‘test’ datasets. For easy analysis, the ‘Cross-validation’ class have in-built performance-measurement methods e.g. ‘accuracy’, ‘mean_squared_error and r2_score’ etc. as shown in this chapter.

>>> import numpy as np
>>> from sklearn.datasets import load_iris
>>> from sklearn.neighbors import KNeighborsClassifier
>>> from sklearn.model_selection import cross_val_score
>>>
>>> # create object of class 'load_iris'
... iris = load_iris()
>>>
>>> # save features and targets from the 'iris'
... features, targets = iris.data, iris.target
>>>
>>> # use KNeighborsClassifier for classification
... classifier = KNeighborsClassifier()
>>>
>>> # cross-validation
... scores = cross_val_score(classifier,
...                     features, targets,
...                     cv=7, scoring="accuracy")
>>> print("Cross validation scores:", scores)
Cross validation scores: [ 0.95833333    1.     0.95238095
0.9047619    0.95238095    1.    1.]
>>> print("Mean={0:0.4f}, Var={1:0.4f}".format(
...                             np.mean(scores),
...                             np.var(scores)))
Mean=0.9668, Var=0.0011

### 13.2.2. Logarithmic loss¶

It measures the probability of having the correct predictions, and prints the logarithmic value of the probability. Since the probability has the range between 0 and 1, therefore ‘Logarithmic loss’ has the range between 0 and ‘-infinity’.

Note

Higher the ‘Logarithmic loss’ value, better is the model. Perfect model will have the maximum value i.e. ‘0’.

>>> import numpy as np
>>> from sklearn.datasets import load_iris
>>> from sklearn.neighbors import KNeighborsClassifier
>>> from sklearn.model_selection import cross_val_score
>>>
>>> # create object of class 'load_iris'
... iris = load_iris()
>>>
>>> # save features and targets from the 'iris'
... features, targets = iris.data, iris.target
>>>
>>> # use KNeighborsClassifier for classification
... classifier = KNeighborsClassifier()
>>>
>>> # cross-validation
... scores = cross_val_score(classifier,
...                     features, targets,
...                     cv=7, scoring="neg_log_loss")
>>> print("Cross validation scores:", scores)
Cross validation scores: [-1.45771098 -0.03187765
-0.07858381 -0.14654173 -1.66902867 -0.02125177
-0.03495091]
>>> print("Mean={0:0.4f}, Var={1:0.4f}".format(
...        np.mean(scores),
...        np.var(scores)))
Mean=-0.4914, Var=0.4644

### 13.2.3. Classification report¶

Classification report gives the ‘precision’, ‘recall’, ‘F1-score’ and ‘support’ values for each class as shown below,

>>> from sklearn.datasets import load_iris
>>> from sklearn.model_selection import train_test_split
>>> from sklearn.linear_model import LogisticRegression
>>> from sklearn.metrics import classification_report
>>>
>>> iris = load_iris()
>>>
>>> X, y = iris.data, iris.target
>>>
>>> X_train, X_test, y_train, y_test = train_test_split(X, y,
...         test_size=0.2,
...         random_state=23,
...         stratify=y)
>>>
>>> # Linear classifier
... cls = LogisticRegression()
>>> cls.fit(X_train, y_train)
LogisticRegression(C=1.0, class_weight=None, dual=False, fit_intercept=True,
intercept_scaling=1, max_iter=100, multi_class='ovr', n_jobs=1,
penalty='l2', random_state=None, solver='liblinear', tol=0.0001,
verbose=0, warm_start=False)
>>>
>>> prediction = cls.predict(X_test)
>>> report = classification_report(y_test, prediction)
>>> print(report) # print classification_report
precision    recall  f1-score   support

0       1.00      1.00      1.00        10
1       0.90      0.90      0.90        10
2       0.90      0.90      0.90        10

avg / total       0.93      0.93      0.93        30

### 13.2.4. Confusion matrix (Binary classification)¶

Let’s understand the Confusion matrix first, which is the basis for ROC, which can be used with ‘binary (not multiclass) classification’. Confusion matrix is a $$2 \times 2$$ matrix, whose columns are shown in Table 13.2 and explained below,

• True positive : Actual value is positive, and predicted value is also positive.
• False negative : Actual value is positive, and predicted value is negative.
• False positive : Actual value is negative, and predicted value is positive.
• True negative : Actual value is negative, and predicted value is negative.
Table 13.2 Confusion matrix
Predicted value
Positive Negative
Actual value Positive True Positive False Negative
Negative False Positive True Negative

Note

Clearly the desired results are the ‘True positive’ and ‘True negative’ columns. Therefore, for better performance, these values should be higher than the ‘False negative’ and ‘False positive’ columns.

Below is an example of Confusion matrix. Here results have following values

• True positive = 9
• True negative = 9
• False positive = 1
• False negative = 1
>>> from sklearn.datasets import make_blobs
>>> from sklearn.model_selection import train_test_split
>>> from sklearn.linear_model import LogisticRegression
>>> from sklearn.metrics import confusion_matrix
>>>
>>> X, y = make_blobs(centers=2, random_state=0)
>>>
>>> X_train, X_test, y_train, y_test = train_test_split(X, y,
...         test_size=0.2,
...         random_state=23,
...         stratify=y)
>>>
>>> # Linear classifier
... cls = LogisticRegression()
>>> cls.fit(X_train, y_train)
LogisticRegression(C=1.0, class_weight=None, dual=False, fit_intercept=True,
intercept_scaling=1, max_iter=100, multi_class='ovr', n_jobs=1,
penalty='l2', random_state=None, solver='liblinear', tol=0.0001,
verbose=0, warm_start=False)
>>> prediction = cls.predict(X_test)
>>> c_matrix = confusion_matrix(y_test, prediction)
>>> print(c_matrix) # print confusion_matrix
[[9 1]
[1 9]]

### 13.2.5. Area under ROC (AUC)¶

ROC is the plot between the ‘true positive rate’ and ‘false positive rate’, which are defined as below,

• True positive rate = (True positive) / (True positive + False negative)
• False positive rate = (False positive) / (False positive + True negative)

Note

ROC and AUC are used for ‘binary (not multiclass) classification’ problem; and ‘AUC = 1’ represents the perfect model,

>>> import numpy as np
>>> from sklearn.datasets import make_blobs
>>> from sklearn.model_selection import cross_val_score
>>> from sklearn.linear_model import LogisticRegression
>>>
>>> X, y = make_blobs(centers=2, random_state=0)
>>>
>>> # use KNeighborsClassifier for classification
... classifier = LogisticRegression()
>>>
>>> # cross-validation
... scores = cross_val_score(classifier,
...                     X, y,
...                     cv=7, scoring="roc_auc")
>>>
>>> print("Cross validation scores:", scores)
Cross validation scores: [ 1.          1.          0.97959184  0.91836735  0.97959184  1.          1.        ]
>>> print("Mean={0:0.4f}, Var={1:0.4f}".format(
...        np.mean(scores),
...        np.var(scores)))
Mean=0.9825, Var=0.0008

## 13.3. Performance of regression problem¶

The code used in this section is discussed in Chapter 4.

### 13.3.1. MAE, MSE and R2¶

Note

• By default, Scikit library calculates the ‘r2_score’ as shown in Lines 44-46. The ‘r2_score’ has the values between 0 (no fit) and 1 (perfect fit).
• Mean absolute error (MAE) is the sum of the ‘absolute differences’ between the predicted and the actual values, and calculated at Lines 48-50.
• Mean square error (MSE) is the sum of squares of the errors, where errors are the differences between the ‘predicted’ and ‘actual’ values. This is calculated at Lines 53-55.
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 >>> import numpy as np >>> from sklearn.model_selection import train_test_split >>> from sklearn.linear_model import LinearRegression >>> from sklearn.metrics import mean_squared_error, mean_absolute_error >>> from sklearn.metrics import r2_score >>> >>> N = 100 # 100 samples >>> x = np.linspace(-3, 3, N) # coordinates >>> noise_sample = np.random.RandomState(20) # constant random value >>> # growing sinusoid with random fluctuation ... sine_wave = x + np.sin(4*x) + noise_sample.uniform(N) >>> >>> # convert features in 2D format i.e. list of list ... features = x[:, np.newaxis] >>> >>> # save sine wave in variable 'targets' ... targets = sine_wave >>> >>> # split the training and test data ... train_features, test_features, train_targets, test_targets = train_test_split( ... features, targets, ... train_size=0.8, ... test_size=0.2, ... # random but same for all run, also accuracy depends on the ... # selection of data e.g. if we put 10 then accuracy will be 1.0 ... # in this example ... random_state=23, ... # keep same proportion of 'target' in test and target data ... # stratify=targets # can not used for single feature ... ) >>> >>> # training using 'training data' ... regressor = LinearRegression() >>> regressor.fit(train_features, train_targets) # fit the model for training data LinearRegression(copy_X=True, fit_intercept=True, n_jobs=1, normalize=False) >>> >>> >>> # predict the 'target' for 'test data' ... prediction_test_targets = regressor.predict(test_features) >>> test_accuracy = regressor.score(test_features, test_targets) >>> print("Accuracy for test data:", test_accuracy) Accuracy for test data: 0.822872868183 >>> >>> r2_score = r2_score(test_targets, prediction_test_targets) >>> print("r2_score: ", r2_score) r2_score: 0.822872868183 >>> >>> mean_absolute_error = mean_absolute_error(test_targets, prediction_test_targets) >>> print("mean_absolute_error: ", mean_absolute_error) mean_absolute_error: 0.680406590952 >>> >>> >>> mean_squared_error = mean_squared_error(test_targets, prediction_test_targets) >>> print("mean_squared_error: ", mean_squared_error) mean_squared_error: 0.584535345592

### 13.3.2. Problem with cross-validation¶

Below is the issue with ‘regressor-performances’ with ‘cross-validation’ method,

Error

• Mean score for ‘r2’ is calculated as ‘-7.7967’, which is negative. Note that, the negative value is not possible for ‘r2’ score.
• Similarly, replace ‘r2’ with ‘neg_mean_squared_error’ and ‘neg_mean_absolute_error’, and it may give some undesired results.
• Please clarify the reason.
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 >>> import numpy as np >>> from sklearn.model_selection import cross_val_score >>> from sklearn.model_selection import KFold >>> from sklearn.linear_model import LinearRegression >>> >>> N = 100 # 100 samples >>> x = np.linspace(-3, 3, N) # coordinates >>> noise_sample = np.random.RandomState(20) # constant random value >>> # growing sinusoid with random fluctuation ... sine_wave = x + np.sin(4*x) + noise_sample.uniform(N) >>> >>> >>> # convert features in 2D format i.e. list of list ... features = x[:, np.newaxis] >>> >>> # save sine wave in variable 'targets' ... targets = sine_wave >>> >>> # cross-validation ... regressor = LinearRegression() >>> >>> cv = KFold(n_splits=10, random_state=7) >>> scores = cross_val_score(regressor, features, targets, cv=cv, ... scoring="r2") >>> >>> print("Cross validation scores:", scores) Cross validation scores: [-13.91006325 -20.21043299 0.36952646 -1.92292726 -3.30936741 -3.30936741 -1.92292726 0.36952646 -20.21043299 -13.91006325] >>> print("Mean={0:0.4f}, Var={1:0.4f}".format( ... np.mean(scores), ... np.var(scores))) Mean=-7.7967, Var=62.5597


text:
## multiplicities of rational singularities in higher dimension

For a normal surface rational singularity, we know that the multiplicity of is bounded by $e-1$ where $e$ is the embedding dimension (See for example Miles Reid's book "Chapters on algebraic surfaces").

I am wondering if this inequality also holds in higher dimension. If not, what can we say about the multiplicities.

-

That particular bound doesn't hold if I recall correctly, but the following bound does:

Theorem : (C. Huneke and K.-i. Watanabe) The multiplicity of a $d$-dimensional variety with rational singularities and embedding dimension $n$ is at most $${n - 1 \choose d - 1}.$$

In the case of a surface, this reduces to the bound you mentioned above. This is an unpublished result of Huneke and Watanabe (currently under review). You could certainly ask them for a preprint.

EDIT: My previous answer said that this was a conjecture, and that Huneke and Watanabe proved something related to this, but I wasn't sure if they actually proved this. It turns out that they did indeed prove this, and I got their permission to post that this was indeed a theorem of theirs.

-
 Graham, sorry, the word general has no meaning. Let me remove it. – Karl Schwede Nov 18 at 21:09