text:
• 71

## Goal

You are at a frog contest. The frog with the longest jump wins.
Your frog participates in the contest.

You want to know in which place your frog is, but you are impatient and you find it is too long to wait for the results.

The ranking of the contest is as follows:
- The further the frog lands, the better the frog's place.
- The best place is the First place.
- If two frogs jump exactly to the same distance, they have the same place and that means that the next place does not exist. For example, if there are two frogs in the fifth position, there is no sixth position, we directly take the seventh position.
- Distances are in meters, rounded to the nearest centimeter. That is, distances are rounded to the nearest multiple of 0.01

############################################################

The frog is in a two dimensional world.
There is only one external force which is the gravity. Gravity only applies on the y axis, x coordinate is always 0.
The ground is flat at y = 0.

We will place ourselves in a two-dimensional orthonormal frame, in Galilean land frame of reference.
Given are the initial position of the frog, (x, y), her mass m (in gram), the angle alpha (in degrees) of the speed vector of the frog at time = 0, the initial speed magnitude (in m/s) and the gravity intensity vector (a, b each in m/s²).
You have to do some mechanics and some math calculus to find the position of the frog after her bound.

I should explain the way to solve this puzzle as it was explained in Physics.
Step 1: Calculate the velocity components horizontal and vertical.
Step 2: Integrate the velocity to calculate the position components.
Step 3: Calculate the time when the frog landed after her bound. That is when the vertical expression of the position is equal to 0 meters.
Step 4: Inject the time in the horizontal expression to find the x position of the frog after her bound.

For people who do not want to make physics calculations, here is the formula to know the distance of the jump of your frog. All units here are the same as in the specifications.

speed_x = cos(alpha) * speed
speed_y = sin(alpha) * speed

delta = speed_y^2 - 4 * (b * 1/2) * y
time = (-speed_y - sqrt(delta)) / (2 * b * 1/2)

x final = (a * 1/2 * time^2) + (speed_x * time) + x
Input
Line 1: The number of frogs f at the contest (without your frog).
Line 2: Frogs jump lengths in meters of each frog separated by spaces.

Line 3: Two integers x y separated by space representing the initial position of your frog.
Line 4: An integer m representing the mass of your frog in grams.
Line 5: An integer alpha which is the angle alpha of the speed vector at t = 0 in degree.
Line 6: A float speed representing the norm of the initial speed vector (at t = 0) in meters / seconds
Line 7: Two floats a b separated by space representing the vector of the gravity intensity on the planet where the frog is in meters / seconds².
Output
One line with the place of your frog in the contest.
Constraints
The x is horizontal and y is vertical.

The contest cannot be organized if there are less than 3 frogs or more than 100 frogs. 3 ≤ f ≤ 100

The ground is flat and is a sea level +0.
The mass of the frog is strictly positive. 0 < m ≤1500
The angle can be between 0° and 90°. 0 ≤ alpha ≤ 90
The speed can only be positive. 0 ≤ speed
The gravity intensity is strictly vertical. a = 0 and b < 0

The frog always comes back on the ground.
The frog jumps along the positive direction of the x axis.

For the precision of the distance, we round to two decimal places.
Example
Input
5
0.06 0.01 0.10 0.07 0.08
0 0
125
30
1.02
0 -9.81
Output
2

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text:
# Tutorial Track

## J. Cummings: Singular cardinal combinatorics

The tutorial will focus on the combinatorics of singular cardinals and their successors, with a special emphasis on compactness phenomena.

## M. Hušek: Large cardinals in general topology

Influence of some large cardinals on results in topological structures will be discussed. Mainly Ulam measurable cardinal, sequential cardinal and their "higher" concepts (measurable and submeasurable cardinals) will be considered. For topological structures we take topological spaces, uniform spaces, topological groups, topological linear spaces and their nice subcategories. Among the properties discussed are commutations of extensions with products, simplicity of some classes of structures, sequential continuity versus continuity and productivity of special classes of structures. One of the main tools is factorization of mappings defined in products. In some factorizations $\lambda$-strongly compact cardinals will be used.

## W. Kubiś: Generic structures

Generic structures are mathematical objects with rich automorphism groups, arising as limits of special sequences. These objects can also be described by the existence of a winning strategy in a natural infinite game where two players alternately choose bigger and bigger structures of the same type. The theory of generic structures originates from the work of Roland Fraı̈ssé in 1954 and for a long time it was considered to be part of model theory. On the other hand, one of the first generic structures was discovered by Urysohn: A universal ultrahomogeneous Polish metric space. As it happens, generic objects give rise to examples of Polish groups with very interesting dynamics.

The purpose of this tutorial is to present the theory of generic structures, with selected applications and examples from several areas of mathematics.

## J. Lopez-Abad: Extremely amenable automorphism groups

The goal of this mini course is to present the main points in the study of the topological dynamics of topological groups when represented as automorphism groups of approximate ultrahomogeneous metric structures. More particularly, we will discuss:

• Fraïssé and metric Fraïssé correspondence for (approximate) ultrahomogeneous structures (AuH).
• Representation of polish groups as automorphism groups $\mathrm{Aut}(\mathcal M)$ of (AuH) structures $\mathcal M$.
• The Kechris–Pestov–Todorcevic (KPT) correspondence between extremely amenable groups of automorphisms of (AuH) structures and the approximate Ramsey property.
• The computation of universal minimal flows.
• Examples, including
1. $\mathrm{Aut}(\mathbb Q,<)$,
2. $\mathrm{Aut}(\mathbb B)$, where $\mathbb B$ is the countable atomless boolean algebra,
3. $\mathrm{Aut}(\mathbb G)$, where $\mathbb G$ is the Gurarij space,
4. $\mathrm{Aut}(L_p[0,1])$.

# Research Track

SpeakerTitleAbstract/Slides
Viera ŠottováThe first sketch: $\mathcal{N}$ with respect to ideals on $\omega$ slides
Szymon ŻeberskiSuperperfect Mycielski slides
Serhii BardylaDense and closed subsets of compact-like topological spacesabstract slides
Adam BartošCrookedness and almost homogeneity in categories of compactaabstract slides
Judyta BąkOn the productivity of the countable chain condition slides
Mariam BeriashviliApplications of Mazurkiewicz type sets in the study of measurability properties of sets and functionsabstract slides
José Antonio CoronaOpen coloringsabstract slides
Cesar CorralUniformization properties of ladder systems under MA(S)abstract slides
Matheus Duzi Ferreira CostaTopological Games of Bounded Selectionsabstract slides
Mirna DzamonjaLogics compact at singular cardinals slides
Monroe EskewElementary embeddings and correctnessabstract slides
Miha HabičSome results on ultrapower capturingabstract slides
K P HartYou Chooseabstract slides
Radek HonzikProjective wellorder and the tree propertyabstract slides
Eliza JabłońskaOn $n$-fold sum of a non-flat continuumabstract slides
Michał KorchGeneralized Egorov's statement for idealsabstract slides
Ziemowit KostanaOn countably saturated linear ordersabstract slides
Andrzej KucharskiOn some properties of a base.abstract slides
Ashutosh KumarOrder dimension of Turing degrees slides
Borisa KuzeljevicA long chain of P-pointsabstract slides
Arkady LeidermanQuotients of free topological groupsabstract slides
Marcin MichalskiSmital and Fubini among treesabstract slides
Kaethe MindenMaximality and Resurrectionabstract slides
Sonia Navarro FloresBorel ideals and Ramsey spacesabstract slides
José de Jesús Pelayo GómezBetween Hausdorff and Selective slides
Veronika PitrováHereditary (bi)coreflective subcategories in certain categories of semitopological groupsabstract slides
Grzegorz PlebanekSmall almost disjoint families and their applicationsabstract slides
Márk PoórSets on the plane meeting every line in a set of measure 1abstract slides
Robert RalowskiContinunous images of Bernstein sets slides
Jonathan SchilhanDefinability aspects of ultrafilters slides
Damian SobotaOn the Josefson-Nissenzweig theorem for C(K)-spacesabstract slides
Šárka StejskalováIndestructibility of the tree propertyabstract slides
Jaroslav SupinaOn a level measure slides
Jarosław SwaczynaHaar-small sets in Polish groups slides
Dorottya SzirákiThe Open Dihypergraph Dichotomy for Definable Subsets of Generalized Baire Spacesabstract slides
Alessandro VignatiHeavy liftingabstract slides
Magdalena WłudeckaProducts of gamma setsabstract slides
Thilo WeinertPolarised Partition Relations for Order Typesabstract slides
Wolfgang WohofskyA Sacks amoeba preserving distributivity of $P(\omega)/\mathrm{fin}$ slides
Rafael ZamoraGenerics in a measurable wreath productabstract slides

format: plain


text:
## Topological limits / Kechris p26, Ex 4.23

See here for definitions.

Kechris, p. 26
(B) Exercise (4.23).  Let $(X, d)$ be metric space with $d \le 1$. Show that for nonempty $K, K_n \in K(X)$,
(i) $\delta(K, K_n) \to 0 \Rightarrow K \subseteq \underline{\text{Tlim}}_n K_n$.
Proof.
Suppose $\delta(K, K_n) \to 0$.

Thus, for all $\epsilon > 0$, there exists $N$, such that $n \ge N$ implies $\delta(K, K_n) < \epsilon$.
That is, $\max_{x \in K} d(x, K_n) < \epsilon$.

Let $x \in K$.  Now, choose $M_1 < M_2 < \ldots < M_m < \ldots$ such that $n \ge M_i$ implies $d(x, K_n) < 1/ i$.
Now, choose a sequence $(x_n)$ in $X$, such that $x_n \in K_n$, in the following way:  For $n \in \mathbb{N}$,
$M_i \le n < M_{i + 1}$, for some $i \in \mathbb{N}$.  Pick any $x_n \in K_n$ such that $d(x, x_n) < 1 /i$.

Then $x_n \to x$, so $x \in \underline{\text{Tlim}}_n K_n$. $\blacksquare$

(ii) $\delta(K_n, K) \to 0 \Rightarrow K \supseteq \overline{\text{Tlim}}_n K_n$.
Proof.
Assume $\delta(K_n,K) \to 0$.  Let $x \in \overline{\text{Tlim}}_n K_n \cap K^c$.  Since $x \notin K$
and since $K$ is compact, hence closed, $d(x,K) = \epsilon > 0$.
$x \in \overline{\text{Tlim}}_n K_n \Rightarrow \exists(x_n)[ x_n \in K_n \text{, for all n, and for some subsequence } (x_{n_i}), x_{n_i} \to x]$.
$\exists$ some $N$ such that $n_i \ge N \Rightarrow d(x_{n_i}, x) < \epsilon / 2$, and $\delta(K_{n_i}, K) < \epsilon / 2$.
Therefore, $d(x_{n_i}, K) < \epsilon / 2$.  Therefore $\exists (y \in K)[d(x_{n_i}, y) < \epsilon / 2]$.
Therefore $d(x,y) \le d(x, x_{n_i}) + d(x_{n_i}, y) < \epsilon$, which is a contradiction.
Thus, $\overline{\text{Tlim}}_n K_n \subseteq K$. $\blacksquare$

(i) and (ii) show that $d_H(K_n, K) \to 0 \Rightarrow K = \text{Tlim} K_n$.
To show that the converse fails, consider the real numbers $\mathbb{R}$ with the trivial metric.
Let $K_n = \{ 1/n, 1 \}$, which is clearly compact.  Then $\text{Tlim} K_n = \{ x : \text{every open neighborhood of x meets } K_n \text{ for infinitely many n} \} = \{ 1 \}$
And $\delta(K_n, K) = \max_{x \in K_n} d(x, K) = 1 - 1/n = \frac{n - 1}{n} \to 1$.


text:
# Answer the Following Question. Describe Briefly the Process of Transferring the Charge Between the Two Plates of a Parallel Plate Capacitor When Connected to a Battery. - Physics

#### Question

Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.

#### Solution

Consider a parallel plate capacitor which is connected across a battery. As soon as the charges from the battery reach one plate, due to insulating gap charge is not able to move further to the other plate. Thus, the positive charge is developed at one plate and a negative charge is developed on the other. As the amount of charge increases on the plates, a voltage is developed across the capacitor that is opposite to the applied voltage. Hence, the current flowing in the circuit decreases and gradually becomes zero. Thus, the charge is developed on the capacitor.

Energy Stored in a Charged Capacitor
The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential.
Let us assume that initially, both the plates are uncharged. Now, we have to repeatedly remove small positive charges from one plate and transfer them to the other plate.
Let
q → Total quantity of charge transferred
V → Potential difference between the two plates
Then
q = CV

Now, when an additional small charge dq is transferred from the negative plate to the positive plate, the small work done is given by,

dW = Vdq = q/C dq

The total work is done in transferring charge Q is given by,

W = ∫_0^Q q/C dq = 1/C∫_0^Q qdq = 1/C[q^2/2]_0^Q

W = Q^2/(2C)

This work is stored as electrostatic potential energy U in the capacitor.

U = Q^2/(2C)

U = (CV)^2/(2C)   ...[∵ Q = CV]

U = 1/2 CV^2

Concept: The Parallel Plate Capacitor
Is there an error in this question or solution?


text:
# zbMATH — the first resource for mathematics

The lattice theoretic part of topological separation properties. (English) Zbl 0396.54014

##### MSC:
 54D10 Lower separation axioms ($$T_0$$–$$T_3$$, etc.) 54D15 Higher separation axioms (completely regular, normal, perfectly or collectionwise normal, etc.) 06D99 Distributive lattices 54F05 Linearly ordered topological spaces, generalized ordered spaces, and partially ordered spaces
Full Text:
##### References:
 [1] DOI: 10.2307/2311686 · Zbl 0106.15504 [2] Dowker, Topics in Topology pp 223– (1972)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
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Posted by8 years ago
Archived

## How Many Words Did Shakespeare Know? 66,534

67% Upvoted
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level 1
[deleted]
4 points · 8 years ago

The task of counting words is a nontrivial task; the results are compiled in a concordance of the works of Shakespeare.

Why is this nontrivial?

level 2
1 point · 8 years ago

Judging by the introductory remarks on this page, I think this article intends to have less focus on shakespeare's word count and more on the phenomenon of statistical inference. I'll admit though, if the article does intend to do that (example in statistics) they lose a lot of focus on the topic and don't give the reader the intended effect.

Edit: I've forgotten to answer your question, which is what I was building up to: I think he (or she) is just using it as an example for other things that would not be reasonable to count in this manner, but does not make an entirely accurate comparison.

level 2
1 point · 8 years ago · edited 8 years ago

A few things come to mind. Firstly, there are many different editions of Shakespeare's works, with several differences (variant readings). Most of Shakespeare's works were not even printed until a long time after they were written and performed — until they had become popular. Additionally, English spelling was less standardized than it is now, and sometimes it is difficult to be certain what word Shakespeare intended. Probably, different spellings of the same word shouldn't be counted as distinct. But what about, say, plurals? What about noun forms, adjectival forms, gerunds and so on? You may want to ignore them all, but what if Shakespeare was creating forms that had never existed until then (e.g. by verbing nouns)? Should phrases that are sometimes hyphenated, sometimes compounded, and sometimes left separate be counted as one word or two? [Also see W:Concordance and online Shakespeare concordance: it chooses to treat each word form as distinct.]

After you define the words that exist in the works of Shakespeare, yes, counting them ought to be a trivial task (today, with the help of computers), but the first step takes some effort and literary scholarship… in fact, I'm not sure there's an entirely satisfactory answer.

level 1
1 point · 8 years ago

I really doubt that one person had this kind of vocabulary. Especially when you consider his poor penmanship and his children. Shakespeare never signed his name the same way. Not something you would expect from a writer. Also you would expect someone this educated would educate his children. Not so.

Sussana Shakespeare was illiterate so she would not have attended school or received an education which customary for most girls in the Elizabethan era.

Judith was also illiterate.

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text:
# Mechandising electronic texts

Fri Feb 28 11:56:19 EST 1997

I have sent a letter to the company which is selling "our" electronic
texts. The text is attached.  Let us see what happens.

All the best,
Dominik

--
Dominik Wujastyk               Wellcome Institute for the History of Medicine
email: d.wujastyk at ucl.ac.uk          183 Euston Road, London NW1 2BE, England
<URL: http://www.ucl.ac.uk/~ucgadkw/>                    FAX: 44 171 611 8545

-------------- next part --------------
\documentclass[12pt,a4paper]{letter}
\usepackage{mylet,garamond}
\usepackage[cpdom]{inputenc}
\makelabels
\date{28 Feb 1997}

\begin{document}

\signature{Dr D. Wujastyk,\\Founder of the INDOLOGY network discussion group\\
and WEB site.}
\begin{letter}{H. Boessow PARK GmbH, \\
Widmannstr. 11a, \\
81829 Muenchen,\\
Germany}

\opening{Dear Sir,}

It has come to my attention that your company is marketing the following
electronic texts:

\begin{verbatim}
Rgveda  40,DM
Satapathabrahmana 50,DM
Mahabharata 100,DM
Ramayana 60,DM
Pali Canon 150,DM
Arthasastra 50,DM
Manusmrti 50,DM
Brhatsamhita 50,DM
Buddhacarita 20,DM

Complete package (ca. 20 HDisks'') 500,DM
\end{verbatim}

I and my immediate colleagues at various universities are responsible for
having made these texts available on the internet.  All these texts are
protected by Copyright from unauthorised duplication.  Authorised
duplication includes duplication for private, scholarly purposes.
\textbf{Unauthorised} duplication includes the duplication and
distribution of these texts for sale.

Please confirm to me in writing by return of post that you have indeed
gained written permission from all the copyright holders of these electronic
texts to duplicate and sell them, or that you are currently in the process
of doing so.

withdraw these electronic texts from the market immediately, until such time
as the right to right the files has been successfully negotiated with the

to distribute these texts, and if you do not do so immediately, then the
copyright holders will have no alternative but to consider legal action

While I cannot speak for my colleagues, it is my impression that they will
be prepared to offer you the right to duplicate and sell copies of these
files provided that you satisfy them that the price asked is approximately
equal to the cost incurred in duplication, media, and delivery.  In other
words, it is our wish to help fellow scholars and interested parties who do
not have an internet connection to acquire these files, but it is not our
wish that substantial profit be generated from such a service.

\closing{Yours faithfully,}
\end{letter}

\end{document}


text:
## Elementary Linear Algebra 7th Edition

$A$ is nonsingular whenever $x\neq 8$.
$A$ is singular if $$ad-bc=0 \Longrightarrow 8-x=0\Longrightarrow x=8.$$ Then $A$ is nonsingular whenever $x\neq 8$.


text:
Lemma 54.5.3. Let $X$ be a scheme which is Noetherian, Nagata, and has dimension $2$. Let $f : Y \to X$ be a proper birational morphism. Then there exists a commutative diagram

$\xymatrix{ X_ n \ar[r] \ar[d] & X_{n - 1} \ar[r] & \ldots \ar[r] & X_1 \ar[r] & X_0 \ar[d] \\ Y \ar[rrrr] & & & & X }$

where $X_0 \to X$ is the normalization and where $X_{i + 1} \to X_ i$ is the normalized blowing up of $X_ i$ at a closed point.

Proof. We will use the results of Morphisms, Sections 29.18, 29.52, and 29.54 without further mention. We may replace $Y$ by its normalization. Let $X_0 \to X$ be the normalization. The morphism $Y \to X$ factors through $X_0$. Thus we may assume that both $X$ and $Y$ are normal.

Assume $X$ and $Y$ are normal. The morphism $f : Y \to X$ is an isomorphism over an open which contains every point of codimension $0$ and $1$ in $Y$ and every point of $Y$ over which the fibre is finite, see Varieties, Lemma 33.17.3. Hence there is a finite set of closed points $T \subset X$ such that $f$ is an isomorphism over $X \setminus T$. For each $x \in T$ the fibre $Y_ x$ is a proper geometrically connected scheme of dimension $1$ over $\kappa (x)$, see More on Morphisms, Lemma 37.52.6. Thus

$BadCurves(f) = \{ C \subset Y\text{ closed} \mid \dim (C) = 1, f(C) = \text{a point}\}$

is a finite set. We will prove the lemma by induction on the number of elements of $BadCurves(f)$. The base case is the case where $BadCurves(f)$ is empty, and in that case $f$ is an isomorphism.

Fix $x \in T$. Let $X' \to X$ be the normalized blowup of $X$ at $x$ and let $Y'$ be the normalization of $Y \times _ X X'$. Picture

$\xymatrix{ Y' \ar[r]_{f'} \ar[d] & X' \ar[d] \\ Y \ar[r]^ f & X }$

Let $x' \in X'$ be a closed point lying over $x$ such that the fibre $Y'_{x'}$ has dimension $\geq 1$. Let $C' \subset Y'$ be an irreducible component of $Y'_{x'}$, i.e., $C' \in BadCurves(f')$. Since $Y' \to Y \times _ X X'$ is finite we see that $C'$ must map to an irreducible component $C \subset Y_ x$. If is clear that $C \in BadCurves(f)$. Since $Y' \to Y$ is birational and hence an isomorphism over points of codimension $1$ in $Y$, we see that we obtain an injective map

$BadCurves(f') \longrightarrow BadCurves(f)$

Thus it suffices to show that after a finite number of these normalized blowups we get rid at of at least one of the bad curves, i.e., the displayed map is not surjective.

We will get rid of a bad curve using an argument due to Zariski. Pick $C \in BadCurves(f)$ lying over our $x$. Denote $\mathcal{O}_{Y, C}$ the local ring of $Y$ at the generic point of $C$. Choose an element $u \in \mathcal{O}_{X, C}$ whose image in the residue field $R(C)$ is transcendental over $\kappa (x)$ (we can do this because $R(C)$ has transcendence degree $1$ over $\kappa (x)$ by Varieties, Lemma 33.20.3). We can write $u = a/b$ with $a, b \in \mathcal{O}_{X, x}$ as $\mathcal{O}_{Y, C}$ and $\mathcal{O}_{X, x}$ have the same fraction fields. By our choice of $u$ it must be the case that $a, b \in \mathfrak m_ x$. Hence

$N_{u, a, b} = \min \{ \text{ord}_{\mathcal{O}_{Y, C}}(a), \text{ord}_{\mathcal{O}_{Y, C}}(b)\} > 0$

Thus we can do descending induction on this integer. Let $X' \to X$ be the normalized blowing up of $x$ and let $Y'$ be the normalization of $X' \times _ X Y$ as above. We will show that if $C$ is the image of some bad curve $C' \subset Y'$ lying over $x' \in X'$, then there exists a choice of $a', b' \mathcal{O}_{X', x'}$ such that $N_{u, a', b'} < N_{u, a, b}$. This will finish the proof. Namely, since $X' \to X$ factors through the blowing up, we see that there exists a nonzero element $d \in \mathfrak m_{x'}$ such that $a = a' d$ and $b = b' d$ (namely, take $d$ to be the local equation for the exceptional divisor of the blowup). Since $Y' \to Y$ is an isomorphism over an open containing the generic point of $C$ (seen above) we see that $\mathcal{O}_{Y', C'} = \mathcal{O}_{Y, C}$. Hence

$\text{ord}_{\mathcal{O}_{Y, C}}(a) = \text{ord}_{\mathcal{O}_{Y', C'}}(a' d) = \text{ord}_{\mathcal{O}_{Y', C'}}(a') + \text{ord}_{\mathcal{O}_{Y', C'}}(d) > \text{ord}_{\mathcal{O}_{Y', C'}}(a')$

Similarly for $b$ and the proof is complete. $\square$

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text:
# Convergence of $\sum_{n=1}^{\infty}(-1)^n\frac{n}{f_n}$ where $f_n$ is the $n$'th Fibonacci number

Can we show convergence of$$B=\sum_{n=1}^{\infty}(-1)^n\frac{n}{f_n}$$where $f_n$ is the $n$'th Fibonacci number?

And then can we determine the exact value of $B$?

-
It may be useful to know, for this or similar problems, that $f_n$ grows exponentially for large $n$, particularly $f_n\sim\phi^n$ where $\phi=(1+\sqrt{5})/2$. – Jonathan Jan 19 '13 at 4:09

You can conclude it based on alternating series test i.e. if $a(n)$ is an eventually monotone decreasing sequence, converging to $0$, then $$\sum_{n=1}^{\infty} (-1)^n a(n)$$ converges.

In your case, $a(n) = \dfrac{n}{f_n}$ is a monotone decreasing sequence for some $n>m$, converging to $0$, since $f_n$ grows exponentially.

Hence, we have that $$\sum_{n=1}^N (-1)^n\dfrac{n}{f_n}$$ converges.

EDIT Mathematica gives the sum of the infinite series as $\approx 0.2692050394$ and the inverse symbolic calculator seems to suggest this number is not related to any of the other "well-known constants".

-
I have a question. Could I apply the same solution replacing $f_n$ with $p_n$, where $p_n$ is the $n$'th prime? – sunny Jan 19 '13 at 3:58
Marvis, $p_n$ is roughly $n\log n$. – Gerry Myerson Jan 19 '13 at 4:02
Also, $n/f_n$ is not monotone decreasing, not if you take $f_1=f_2=1$ as is commonly done. – Gerry Myerson Jan 19 '13 at 4:03
@GerryMyerson Oops what a blunder I made! Thanks for pointing it out! Have removed it. – user17762 Jan 19 '13 at 4:04
Finally, isn't the ordinary everyday alternating series test good enough for the original question (after you clear out the non-decreasing part)? – Gerry Myerson Jan 19 '13 at 4:05


text:
15,939 views

Let $G$ be an undirected complete graph on $n$ vertices, where $n > 2$. Then, the number of different Hamiltonian cycles in $G$ is equal to

1. $n!$
2. $(n-1)!$
3. $1$
4. $\frac{(n-1)!}{2}$

both C and D are correct in a final answer key

The number of different Hamiltonian cycles in

complete undirected graph on n vertices is (n − 1)! / 2

complete directed graph on n vertices is (n − 1)!

The number of different Hamiltonian cycles in a complete undirected labeled graph on $n$ vertices is $(n − 1)! / 2$.
Ref : https://en.wikipedia.org/wiki/Hamiltonian_path

If the graph is unlabeled number of different Hamiltonian cycles become $1.$

Since the question does not mention whether the graph is labeled or not, answer can be either (C) or (D).

Official answer key is (D) or (C)

They won't give marks to all but might include C option too. Let's wait.

@Arjun Sir, in PYQ paper 2019 on GO. When we mark option C it shows wrong . Please update it.

Why it is (n-1)!. If we assume one of the initial vertex as fixed ?. If starting vertex is not fixed means it is n!.? But in question nothing mentioned above intial vertex

From any vertex we have n−1 edges to choose from first vertex, n−2 edges to choose from second vertex, n−3 to choose from the third and so on. These being independent choices, we get (n−1)! possible number of choices.

Each Hamiltonian circuit has been counted twice (in reverse direction of each other like these: A→B→C→A and A→C→B→A).

Number of distinct not edge disjoint Hamiltonian circuits in complete graph Kn is (n−1)!/2

by

1 comment

why  A→B→C→A and A→C→B→A are the same? after all, we consider vertices in a distinctive order
Ques never said labelled vertices, please check that wikipedia link which is posted they proved this via taking labelled vertices,

Ans will be 1 only as there is only one hamilton graph possible as all will be isomorphic
by


text:
# [SOLVED] Stuck on the next step

• February 3rd 2010, 11:42 AM
vinson24
[SOLVED] Stuck on the next step
Solve $e^{-2x}-3e^{-x}=-2$ for x without using a calculator

Now the first step i did was make it into a quadratic equation
$e^{-2x}-3e^{-x}+2=0$
Then i let $u=e^{-x}$
which gives me $u^{x}-3u+2=0$
now im kinda stuck on the next step becuase of the $u^{x}$ in the equation, not sure if i started of wrong or if im missing something thats right in front of me
• February 3rd 2010, 11:50 AM
Plato
Multiply by $e^{2x}$ to get $2e^{2x}-3e^{x}+1=0$.
• February 3rd 2010, 12:01 PM
vinson24
at which point in the problem are you multiplying $e^{2x}$ because i dont see where the $2e^{2x}$is coming from and the 1 at the end either
• February 3rd 2010, 12:03 PM
wonderboy1953
Correction
It's not $u^x$, it's $u^2$ in your last equation.
• February 3rd 2010, 01:04 PM
Plato
Quote:

Originally Posted by Plato
Multiply by $e^{2x}$ to get $2e^{2x}-3e^{x}+1=0$.

You must know basic algebra in order to do calculus.
$e^{-2x}(e^{2x})-3e^{-x}(e^{2x})=-2(e^{2x})$.


text:
# Find a Sum of Particles in A Set of Boxes Given a Number of Rows and Columns

This is a code challenge, from Codewars, the details of the specific challenge in depth can be found here: https://www.codewars.com/kata/magnet-particules-in-boxes/javascript.

The summary of it though is a function which gives a sum of particles in a set of boxes given a number of rows and a number of columns to work with. You are given a specific function that calculates the number of particles in the box based off of the specific box that is being calculated. Here is the equation:

$$v(k,n) = \frac{1}{k(n + 1)^{2k}}$$

$$doubles (k_{max}, n_{max}) = \sum_{k=1}^{k_{max}} \sum_{n=1}^{n_{max}} v(k,n)$$

Here is the function I came up with:

function doubles(maxk, maxn) {

let total = 0

for(let k = 1; k <= maxk; k++){
const twoK = 2*k
for(let n = 1; n <= maxn; n++){
total += 1/(k*Math.pow(n+1, twoK))
}
}

}

My function does the job perfectly, and comparing it to other responses marked as 'best practices' it is very similar. But I would like to know if anyone sees obvious ways to improve my function, or I would really love to see something that can improve the time complexity of the function, if that is possible.

Inside the inner loop, $$\1/k\$$ is a constant term. So you could extract that multiplication from the inner loop, and apply it on the computed subtotal. This won't change the order of complexity though.


text:
Meguin Hydraulikoel HLP 10

Description

Hydraulic oils HLP are mineral oils which are used as pressurized liquid In hydraulic systems. They contain Active substances for increasing aging resistance, Corrosion protection and EP properties. by virtue of Their additives, they will meet all And mainly used in hydraulic High thermal stresses occur By water corrosion can form and their pumps Or hydromotors due to the operating conditions oils With wear protection in the case of mixed friction.

Specifications and release:

Hydraulic oil DIN 51524 part 2 HLP 10

Looking for any other product in MEGUIN Lubricants or other Lubricant ? Just give us a call.


text:
# Elliptic curves over $\mathbb C$ have same endomorphism ring but not isomorphic

I am finding elliptic curves over $\mathbb C$ have same endomorphism ring but not isomorphic.

Elliptic curves over $\mathbb C$ can be identified with $\mathbb C/\land$ for some lattice $\land$. And endomorphism ring is just a set of z s.t. z$\land \subseteq \land$.

But I am stuck here, the lattice in my mind is too simple, I guess I need to see more example. Can somebody help me?

What about lattice generated by {$1$,$\sqrt{2} i$} and {$i$,$\sqrt{2}$}? Is this an example?

• The vast majority of elliptic curve over $\Bbb C$ have $\Bbb Z$ as their endomorphism ring... – mercio Dec 6 '15 at 18:33
• @mercio You are absolutely right. So the point here is to find example with endomorphism ring larger than $\mathbb Z$ – user198206 Dec 6 '15 at 23:15

Your two lattices actually give isomorphic elliptic curves, because one can be rotated into the other via multiplication by $i$. So it doesn't work.
It is actually slightly tricky to find an example. You need two lattices which are both closed under multiplication by some subring $R \subseteq \mathbb{C}$ strictly containing $\mathbb{Z}$, but are not multiples of each other. Note that "multiple" can be any complex number.
One way to find an example is to fix a subring $R$, then look for sublattices within $R$ which are closed under multiplication by $R$. By definition, these sublattices are the ideals of $R$. So we are looking for 2 ideals of $R$ that are not multiples of each other by any complex number.
Now we enter the land of algebraic number theory: Two ideals of $R$ are complex multiples of each other iff they belong to the same ideal class. So to find our example, we need a ring $R \subseteq \mathbb{C}$ with more than one ideal class. This is equivalent to the statement that $R$ does not have unique factorization into elements.
The standard example is $R = \mathbb{Z}[\sqrt{-5}]$. It has 2 ideal classes. There is the trivial ideal class represented by $R$ itself, and the non-trivial ideal class represented by the ideal $J = (2, 1 + \sqrt{-5})$. So $R$ and $J$ represent two elliptic curves which are non-isomorphic, but both have endomorphism ring $R$. For some more details on the ideal classes in $\mathbb{Z}[\sqrt{-5}]$, see here.
• I still think that $\Bbb Q(\sqrt{-6}\,)$ gives a clearer example. Then the lattices $\Bbb Z+\sqrt{-6}\Bbb Z$, generated by $1$ and $\sqrt{-6}$; and $2\Bbb Z+\sqrt{-6}\Bbb Z$, generated by $2$ and $\sqrt{-6}$, work just as well, and are more transparent. – Lubin Dec 7 '15 at 23:51


text:
Calculate the margin of error for a derived proportion

moe_prop(num, denom, moe_num, moe_denom)

Arguments

num

The numerator involved in the proportion calculation (an estimate)

denom

The denominator involved in the proportion calculation (an estimate)

moe_num

The margin of error of the numerator

moe_denom

The margin of error of the denominator

Value

A margin of error for a derived proportion


text:
## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 4 Quadratic Functions and Factoring - 4.10 Write Quadratic Functions and Models - 4.10 Exercises - Skill Practice - Page 313: 27

See below.

#### Work Step by Step

If the vertex of a graph is at (m,n) then the general formula for the quadratic function is $f(x)=a(x-m)^2+n$. Hence he used the wrong formula in the first step.

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


text:
# Let R be the region enclosed by f(x) = x^2 + 2 and g(x) = (x - 2)^2. What is the volume of the solid produced by revolving R around the x-axis and then the y-axis?

The only point of intersection is at $x = \frac{1}{2}$


text:
# RESNET 50 LB 0.566 notebook

Sharing this notebook so that more people can get a headstart on how to use pre-trained models to get better results.

# FOODC Resnet50 submission¶

This notebook basically follows the baseline notebook and instead of training a network from scratch we use Resnet 50 model and train it for 25 epochs.

Author - Pulkit Gera

## To open this notebook on Google Computing platform Colab, click below!¶

These include the train test images as well the csv indexing them

In [0]:
!wget -q https://s3.eu-central-1.wasabisys.com/aicrowd-practice-challenges/public/foodc/v0.1/train_images.zip
!wget -q https://s3.eu-central-1.wasabisys.com/aicrowd-practice-challenges/public/foodc/v0.1/test_images.zip
!wget -q https://s3.eu-central-1.wasabisys.com/aicrowd-practice-challenges/public/foodc/v0.1/train.csv
!wget -q https://s3.eu-central-1.wasabisys.com/aicrowd-practice-challenges/public/foodc/v0.1/test.csv

We create directories and unzip the images

In [0]:
!mkdir data
!mkdir data/test
!mkdir data/train
!unzip train_images -d data/train
!unzip test_images -d data/test

## Import necessary packages¶

In [0]:
import torchvision.transforms as transforms
from torch.utils.data.sampler import SubsetRandomSampler
import torch
import torch.nn as nn
import torch.nn.functional as F
from torch.optim import lr_scheduler
from torch.utils.data import TensorDataset, DataLoader, Dataset
import torchvision
from torchvision import models
import torch.optim as optim
import pandas as pd
import numpy as np
import cv2
import os
from sklearn import preprocessing
import matplotlib.pyplot as plt
%matplotlib inline
import time

In pytorch we can directly load our files into torchvision(the library which creates the object) or create a custom class to load data. The class must have __init__ , __len__ and __getitem__ functions. We create a custom dataloader to suit our needs. More info on custom loaders can be read here

In [0]:
class FoodData(Dataset):
def __init__(self,data_list,data_dir = './',transform=None,train=True):
super().__init__()
self.data_list = data_list
self.data_dir = data_dir
self.transform = transform
self.train = train

def __len__(self):
return self.data_list.shape[0]

def __getitem__(self,item):
if self.train:
img_name,label = self.data_list.iloc[item]
else:
img_name = self.data_list.iloc[item]['ImageId']
img_path = os.path.join(self.data_dir,img_name)
img = cv2.resize(img,(256,256))
if self.transform is not None:
img = self.transform(img)
if self.train:
return {
'gt' : img,
'label' : torch.tensor(label)

}
else:
return {
'gt':img
}

We first convert the data labels into encodings using Label Encoders. This basically converts labels into number encodings. This is an important step as without it we cannot train our network

In [0]:
train = pd.read_csv('train.csv')
le = preprocessing.LabelEncoder()
targets = le.fit_transform(train['ClassName'])
ntrain = train
ntrain['ClassName'] = targets

We load our train data and some necessary augementations like converting to PIL image, converting to tensors and normalizing them across channels. We can add more augementations such as Random Flip, Random Rotation, etc more on which can be found here. Augmentation is an important step and helps increasing the data size. The more data the better the model.

In [0]:
transforms_train = transforms.Compose([
transforms.ToPILImage(),
transforms.RandomRotation(90),
transforms.RandomHorizontalFlip(),
transforms.ColorJitter(),
transforms.ToTensor(),
transforms.Normalize( mean = np.array([0.485, 0.456, 0.406]),
std = np.array([0.229, 0.224, 0.225]))
])
train_path = 'data/train/train_images'
train_data = FoodData(data_list= ntrain,data_dir = train_path,transform = transforms_train)

## EDA¶

Let us do some exploratory data analysis. The idea is to see the class distribution, how the images are and much more.

In [0]:
train = pd.read_csv('train.csv')
num = train['ClassName'].value_counts()
classes = train['ClassName'].unique()
print("Percentage of each class")
for cl in classes:
print(cl,'\t',num[cl]/train.shape[0]*100,"%")

Percentage of each class
water 	 9.25667703528907 %
pizza-margherita-baked 	 1.179877721763381 %
broccoli 	 0.9009975329829454 %
egg 	 2.2417676713504235 %
butter 	 3.71125174300118 %
apple 	 2.0486967714255067 %
dark-chocolate 	 0.9439021774107046 %
white-coffee-with-caffeine 	 1.3085916550466588 %
sweet-pepper 	 0.9009975329829454 %
tomato-sauce 	 1.179877721763381 %
cucumber 	 1.1476992384425615 %
cheese 	 1.4694840716507562 %
pasta-spaghetti 	 1.040437627373163 %
rice 	 2.7458972433765956 %
zucchini 	 0.9653544996245843 %
salmon 	 0.5470342164539311 %
mixed-vegetables 	 2.542100182344739 %
espresso-with-caffeine 	 2.0916014158532663 %
banana 	 1.9414351603561086 %
strawberries 	 0.9331760163037649 %
mayonnaise 	 0.4612249275984125 %
almonds 	 0.740105116378848 %
wine-white 	 1.619650327147914 %
hard-cheese 	 1.2013300439772605 %
ham-raw 	 0.7079266330580285 %
tomato 	 3.8399656762844576 %
french-beans 	 0.8044620830204869 %
mandarine 	 0.740105116378848 %
wine-red 	 2.585004826772498 %
potatoes-steamed 	 1.673281132682613 %
croissant 	 0.8044620830204869 %
carrot 	 3.185669848761128 %
salami 	 0.5255818942400515 %
boisson-au-glucose-50g 	 0.9117236940898853 %
biscuits 	 0.7293789552719082 %
corn 	 0.39686796095677357 %
leaf-spinach 	 0.9331760163037649 %
tea-green 	 0.740105116378848 %
chips-french-fries 	 1.4587579105438164 %
parmesan 	 0.7293789552719082 %
beer 	 0.8580928885551861 %
coffee-with-caffeine 	 4.043762737316314 %
chicken 	 1.1369730773356215 %
soft-cheese 	 0.5148557331331117 %
tea 	 1.8985305159283494 %
gruyere 	 0.7615574385927276 %
sauce-savoury 	 0.6542958275233294 %
honey 	 0.6972004719510887 %
mixed-nuts 	 0.868819049662126 %
jam 	 1.7483642604311918 %
water-mineral 	 0.922449855196825 %
onion 	 0.4397726053845329 %
pickle 	 0.3003325109943151 %

We observe that water is the most popular class although the distribution is not that skewed. Let us plot the images of white flour french bread and french fries and have a look at the kind of images we have

In [0]:
imgs = train.loc[train['ClassName'] == 'bread-french-white-flour']
plt.figure(figsize=(10,10))
for i in range(imgs[:16].shape[0]):
path = imgs.iloc[i]['ImageId']
image = cv2.cvtColor(image,cv2.COLOR_BGR2RGB)
plt.subplot(4,4,i+1)
plt.axis('off')
plt.imshow(image)

In [0]:
imgs = train.loc[train['ClassName'] == 'chips-french-fries']
plt.figure(figsize=(10,10))
for i in range(imgs[:16].shape[0]):
path = imgs.iloc[i]['ImageId']
image = cv2.cvtColor(image,cv2.COLOR_BGR2RGB)
plt.subplot(4,4,i+1)
plt.axis('off')
plt.imshow(image)

## Split Data into Train and Validation¶

Now we want to see how well our model is performing, but we dont have the test data labels with us to check. What do we do ? So we split our dataset into train and validation. The idea is that we test our classifier on validation set in order to get an idea of how well our classifier works. This way we can also ensure that we dont overfit on the train dataset. There are many ways to do validation like k-fold,leave one out, etc

We also make dataloaders which basically create minibatches of dataset which are used in each epoch

Although the data is not that imbalanced, one idea we can add is SMOTE which helps in sampling incase of imbalanced classification where it samples multiple times. I havent tried that but it works well. Read more about it here

In [0]:
batch = 64
valid_size = 0.2
num = train_data.__len__()
# Dividing the indices for train and cross validation
indices = list(range(num))
np.random.shuffle(indices)
split = int(np.floor(valid_size*num))
train_idx,valid_idx = indices[split:], indices[:split]

#Create Samplers
train_sampler = SubsetRandomSampler(train_idx)
valid_sampler = SubsetRandomSampler(valid_idx)

train_loader = DataLoader(train_data, batch_size = batch, sampler = train_sampler)
valid_loader = DataLoader(train_data, batch_size = batch, sampler = valid_sampler)

Here we load test images. Note: This file will not have any labels with it

In [0]:
transforms_test = transforms.Compose([
transforms.ToPILImage(),
transforms.ToTensor(),
transforms.Normalize( mean = np.array([0.485, 0.456, 0.406]),
std = np.array([0.229, 0.224, 0.225]))
])
test_path = 'data/test/test_images'
test_data = FoodData(data_list= test,data_dir = test_path,transform = transforms_test,train=False)

Here we check if we have a GPU or not. If we have we just need to shift our data and model to GPU for faster computations.

In [0]:
device = torch.device("cuda:0" if torch.cuda.is_available() else "cpu")

# Assuming that we are on a CUDA machine, this should print a CUDA device:

print(device)

cuda:0

## Define the Model¶

We are going to use ResNet50. The idea was born from the fact that deeper the models are , after a certain point their accuracy becomes poor. This is due to the fact that gradients passed back become smaller and smaller which leads to negligible change in the early layers.
In order to combat that, we add skip connections to layers forward. This way we are able to preserve the gradients and build deeper models.

Pytorch provides a collection of pretrained resnet architectures. We use resnet 50.
To read more about resnet and its variants, this is a good blog
More on pretrained models with pytorch here and making models here.

In [0]:
train_sampler.__len__()

Out[0]:
7459
In [0]:
dataloaders = {}
dataset_sizes = {}
dataset_sizes['train'] = train_sampler.__len__()
dataset_sizes['val'] = valid_sampler.__len__()

## Train¶

Alright enough talk and time to train. We define the number of epochs and train the model. An epoch is a forward pass and backward pass of all the data points. An epoch consists of iterations which depend on batch size. So basically we take a batch, get its output, do a backward pass and let the optimizer take a step. This is the workflow for any pytorch code.

## Validate¶

Now after an epoch ends, we check with validation and do the same steps except backward pass on loss and optimizer step. If we get a reduction in validation loss, we save the model. This is sort of an early stopping.

In [0]:
def train_model(model, criterion, optimizer, scheduler, num_epochs=25):
since = time.time()

best_model_wts = copy.deepcopy(model.state_dict())
best_acc = 0.0

for epoch in range(num_epochs):
print('Epoch {}/{}'.format(epoch, num_epochs - 1))
print('-' * 10)

# Each epoch has a training and validation phase
for phase in ['train', 'val']:
if phase == 'train':
model.train()  # Set model to training mode
else:
model.eval()   # Set model to evaluate mode

running_loss = 0.0
running_corrects = 0

# Iterate over data.
for data in dataloaders[phase]:
inputs = data['gt'].squeeze(0).to(device)
labels = data['label'].to(device)
# inputs = inputs.to(device)
# labels = labels.to(device)

# zero the parameter gradients

# forward
# track history if only in train
with torch.set_grad_enabled(phase == 'train'):
outputs = model(inputs)
_, preds = torch.max(outputs, 1)
loss = criterion(outputs, labels)

# backward + optimize only if in training phase
if phase == 'train':
loss.backward()
optimizer.step()

# statistics
running_loss += loss.item() * inputs.size(0)
running_corrects += torch.sum(preds == labels.data)
if phase == 'train':
scheduler.step()

epoch_loss = running_loss / dataset_sizes[phase]
epoch_acc = running_corrects.double() / dataset_sizes[phase]

print('{} Loss: {:.4f} Acc: {:.4f}'.format(
phase, epoch_loss, epoch_acc))

# deep copy the model
if phase == 'val' and epoch_acc > best_acc:
best_acc = epoch_acc
best_model_wts = copy.deepcopy(model.state_dict())

print()

time_elapsed = time.time() - since
print('Training complete in {:.0f}m {:.0f}s'.format(
time_elapsed // 60, time_elapsed % 60))
print('Best val Acc: {:4f}'.format(best_acc))

# load best model weights
return model

In [0]:
import copy
model_ft = models.resnet50(pretrained=True)
# To only train the last layer
# for param in model_ft.parameters():
#     param.requires_grad = False
num_ftrs = model_ft.fc.in_features

# Alternatively, it can be generalized to nn.Linear(num_ftrs, len(class_names)).
model_ft.fc = nn.Linear(num_ftrs, 61)

model_ft = model_ft.to(device)

criterion = nn.CrossEntropyLoss()

# Observe that all parameters are being optimized
optimizer_ft = optim.Adam(model_ft.parameters(), lr=0.001)

# Decay LR by a factor of 0.1 every 7 epochs
exp_lr_scheduler = lr_scheduler.StepLR(optimizer_ft, step_size=7, gamma=0.1)

Here we define our model object along with our optimizer and error function. Typically for multi class classification we use Cross Entropy Loss. More about different types of losses are here.
We use the popular Adam optimizer with its default parameters. There are other optimizers like SGD, RMSPROP, Adamax,etc. You can have a detailed look at optimizers here

You must be wondering what is a scheduler. A scheduler provides a policy for the decay of learning rate. For example we can say that let the learning rate value decrease after some fixed number of epochs or say if the validation accuracy doesnt change, we can change the learning rate. This helps in faster and better convergence. Read more here

In [0]:
model_ft = train_model(model_ft, criterion, optimizer_ft, exp_lr_scheduler,
num_epochs=25)

Epoch 0/24
----------
train Loss: 3.0510 Acc: 0.2419
val Loss: 2.6348 Acc: 0.3192

Epoch 1/24
----------

## Predict on Validation¶

Now we predict our trained model on the validation set and evaluate our model

In [0]:
# model.load_state_dict(torch.load('best_model_so_far.pth'))
model_ft.eval()
correct = 0
total = 0
pred_list = []
correct_list = []
for images in valid_loader:
data = images['gt'].squeeze(0).to(device)
target = images['label'].to(device)
outputs = model_ft(data)
_, predicted = torch.max(outputs.data, 1)
total += target.size(0)
pr = predicted.detach().cpu().numpy()
for i in pr:
pred_list.append(i)
tg = target.detach().cpu().numpy()
for i in tg:
correct_list.append(i)
correct += (predicted == target).sum().item()

print('Accuracy of the network on the 10000 test images: %f %%' % (
100 * correct / total))

Accuracy of the network on the 10000 test images: 49.141631 %

## Evaluate the Performance¶

We use the same metrics as that will be used for the test set.
F1 score and Log Loss are the metrics for this challenge

In [0]:
from sklearn.metrics import f1_score,precision_score,log_loss
print("F1 score :",f1_score(correct_list,pred_list,average='micro')*100)

F1 score : 49.141630901287556

## Predict on test set¶

Time for the moment of truth! Predict on test set and time to make the submission.

In [0]:
# model.load_state_dict(torch.load('best_model_so_far.pth'))
model_ft.eval()

preds = []
for images in test_loader:
data = images['gt'].squeeze(0).to(device)
outputs = model_ft(data)
_, predicted = torch.max(outputs.data, 1)
pr = predicted.detach().cpu().numpy()
for i in pr:
preds.append(i)

## Save it in correct format¶

In [0]:
# Create Submission file
df = pd.DataFrame(le.inverse_transform(preds),columns=['ClassName'])
df.to_csv('submission.csv',index=False)

## To download the generated in collab csv run the below command¶

In [0]:
from google.colab import files


text:
# zbMATH — the first resource for mathematics

Asymptotic stability of fractional impulsive neutral stochastic partial integro-differential equations with state-dependent delay. (English) Zbl 1293.34102
Summary: We study the asymptotical stability in $$p$$-th moment of mild solutions to a class of fractional impulsive partial neutral stochastic integro-differential equations with state-dependent delay in Hilbert spaces. We assume that the linear part of this equation generates an alpha-resolvent operator and transform it into an integral equation. Sufficient conditions for the existence and asymptotic stability of solutions are derived by means of the Krasnoselskii-Schaefer type fixed point theorem and properties of the alpha-resolvent operator. An illustrative example is also provided.

##### MSC:
 34K37 Functional-differential equations with fractional derivatives 60H15 Stochastic partial differential equations (aspects of stochastic analysis) 35R60 PDEs with randomness, stochastic partial differential equations 34K30 Functional-differential equations in abstract spaces 34K50 Stochastic functional-differential equations 47N20 Applications of operator theory to differential and integral equations 34K45 Functional-differential equations with impulses 34K20 Stability theory of functional-differential equations 34K40 Neutral functional-differential equations 45K05 Integro-partial differential equations
Full Text:


text:
# Black-Scholes Price for Call and Put Options using Matlab

Matlab → Function → Black-Scholes Option Prices

In the next example, call and put option values are calculated using Black-Scholes formula. Remember the option parameter values are given in the command window when we are executing the program. Here we used, $S_0=50,K=52,r=0.1,\sigma=0.35$, and $T=1$.

% Function calculating call and put function [call, put] = CallPut(s,k,r,v,t) d1=(log(s/k)+(r+v^2/2)*t)/(v*sqrt(t)); d2=d1-v*sqrt(t); call=s*normcdf(d1)-k*exp(-r*t)*normcdf(d2) put=k*exp(-r*t)*normcdf(-d2)-s*normcdf(-d1)

To run this program, we need to write function name (for example, CallPut in this example) and input values in the command window. Like,

Command Window
>> CallPut(50,52,0.1,0.35,1) call = 8.3158 put = 5.3673

Posted


text:
# The amount of water in reservoirs is often measured in acre-ft. One acre-ft is a volume that...

## Question:

The amount of water in reservoirs is often measured in acre-ft. One acre-ft is a volume that covers an area of one acre to a depth of one foot. An acre is 43,560 ft{eq}^2 {/eq}. Find the volume in SI units of a reservoir containing 26.0 acre-ft of water.

## Unit Conversion:

The basic types of physical quantities (such as mass, length, and charge) are called dimensions. In order to measure these dimensions, we need standard units (such as kilograms, feet, or Coulombs). We can convert between different units if they measure the same dimension.

For example, let's consider how to convert between inches and centimeters. One inch is equal to 2.54 cm, so the fraction {eq}\frac{1 \ in}{2.54 \ cm} {/eq} is equal to one (it is a thing divided by itself, which is always equal to one). So, we can multiply this fraction by any quantity without changing the value of that quantity (because multiplying a number by one results in the same number. Let's see what happens when we multiply this conversion factor by 7 cm:

{eq}L = 7 \ cm (\dfrac{1 \ in}{2.54 \ cm}) {/eq}

Notice that there is a {eq}cm {/eq} in the numerator and denominator that we can cancel out.

{eq}L = 7 (\dfrac{1 \ in}{2.54})\\ L = 2.76 \ in {/eq}

This is how we can convert between units: by choosing the appropriate conversion factor and multiplying it by the original quantity.

The answer is {eq}32,100 \ m^3 {/eq}.

We can solve this problem using unit analysis. First, for simplicity, let's convert the volume from acre-ft to...

Become a Study.com member to unlock this answer! Create your account


text:
Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01pk02cd40v
 Title: Electronic Structure Tools for Transition Metal Complexes with Many Open Shells Authors: Sayfutyarova, Elvira R. Advisors: Chan, Garnet K.-L. Contributors: Chemistry Department Keywords: automated active space selectiondensity matrix renormalization group theoryEPR spectroscopyspin-orbit couplingtransition metal chemistry Subjects: ChemistryPhysical chemistry Issue Date: 2018 Publisher: Princeton, NJ : Princeton University Abstract: In this thesis we present three different tools to model open shell transition metal electronic structure, commonly referred to as multi-reference (MR) or multi-configurational electronic structure. The first tool allows us to incorporate spin-orbit coupling into a method that also rigorously treats the multiple configurations arising from electrons in open shells. We call this method the state interaction spin-orbit (SISO) coupling method using density matrix renormalization group (DMRG) wavefunctions. We implement the DMRG-SISO scheme using a spin-adapted DMRG algorithm that computes transition density matrices between arbitrary wavefunctions of the interacting electronic states. To demonstrate the potential of this method, we present accurate benchmark calculations for the zero-field splitting (ZFS) of the copper and gold atoms and also calculate, for the first time, the ZFS of a [2Fe-2S] complex. The second tool builds on the first to allow the calculation of electron paramagnetic resonance (EPR) $g$-values, one of the main spectroscopic methods used to characterize paramagnetic metals in transition metal complexes. We apply this to several benchmark systems such as \ce{TiF3} and \ce{CuCl4^2-}, as well as to determine the $g$-tensor for a [2Fe-2S] complex. This work opens up the possibility to model the $g$-tensor of the active site metals in bioinorganic systems. Finally, we introduce the atomic valence active space (AVAS), which is a simple automated technique to identify the important orbitals to treat in a multi-configurational electronic structure method. We discuss the background, theory, and implementation of the idea, and several of its variations are tested. To demonstrate the performance and accuracy, we calculate the excitation energies for various transition metal complexes in typical application scenarios. The described technique makes MR calculations easier to execute, easier to reproduce by any user, and simplifies the determination of the appropriate size of the active space required for accurate results. URI: http://arks.princeton.edu/ark:/88435/dsp01pk02cd40v Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog: catalog.princeton.edu Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: Chemistry


text:
## When Students Get OnLine Help from Wolfram|Alpha

This term I had my intermediate algebra students turn in a graph of $y=-2x^2+42x-180$.  They had to show all their work.  One of my students used Wolfram|Alpha to check his answer.  He came to me with this image.

Wolfram|Alpha Image - Quadratic Function

He said, “I thought you told us that the y-intercept should be $-180$, the constant term.”  It took me a few seconds to see that the vertical axis in the diagram is not the y-axis.  I didn’t notice  the  notation tucked to  the right “(x from 4 to 18).”  Wolfram|Alpha’s graphs aren’t particularly user-friendly. I would suggest something like this:

Wolfram|Alpha Image - Quadratic Function Improved

Now the vertical axis doesn’t look like the y-axis. (The arrow is just there to note the suggested improvement.)

My precalculus students were required to turn in a graph of this function $f(x)= \frac{x^3+0.8x^2-5.39x+3.63}{x^3-4.3x^2-2.68x+19.84}$ which in factored form becomes $f(x) = \frac{(x+3)(x-1.1)^2}{(x+2)(x-3.1)(x-3.2)}$.  Here is part of the Wolfram|Alpha output.

Note that the function is not completely factored.  Also note that it takes two graphs to show all of the important behavior of the function.  I designed my problem to be this way since as I say in the class “You don’t get to copy a graph from your calculator  if you don’t understand it.”  To get full credit for this problem my students needed to discuss the various behaviors around the asymptotes and zeros as we discussed them in class.”  This way they should have eventually found that bit of the graph in the fourth quadrant.  I think the second graph is incomplete.  Here is what I think it should look like.

Wolfram|Alpha Image - Rational Function Improved

When I assign problems to be done with a calculator, I try to make them such that my students really need to understand what they are doing. Too many  assignments in our texts just use the calculator to verify the answers to problems that could have been done by hand.

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## About jrh794

I am a sixty-five year old math instructor at Southern Oregon University. I taught at the College of the Siskiyous in Weed California for twenty-six years. Prior to that I worked as a computer programmer, carpenter and in various other jobs. I graduated from Rice University in 1967 and have a MS in Operations Research from Stanford. In the past I have hand-built a stone house and taken long solo bicycle tours. Now I ride my mountain bike and play golf for recreation.
This entry was posted in Curriculum, Pedagogy, Rants, Uncategorized. Bookmark the permalink.


text:
# 21.(i)   In a class of  $\small 60$  students, $\small 30$ opted for NCC, $\small 32$ opted for NSS and  $\small 24$ opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) The student opted for NCC or NSS.

Answers (1)
H Harsh Kankaria

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

n(S) = 60, n(A) = 30, n(B) =32, n(A $\cap$ B) = 24

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

P(B) = $\inline \frac{32}{60} = \frac{8}{15}$

P(A $\cap$ B) = $\inline \frac{24}{60} = \frac{2}{5}$

(i) We know,

P(A $\cup$ B) = P(A)+ P(B) - P(A $\cap$ B)

$\frac{1}{2} + \frac{8}{15} - \frac{2}{5} = \frac{15+16-12}{30}$

$= \frac{19}{30}$

Hence,the probability that the student opted for NCC or NSS is $\frac{19}{30}$

Exams
Articles
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text:
Nevil Hopley

## Gesamte Ressourcen(10)|TypActivitiesWebinarsVideosLinksfunction elasticsubmit() { $('#type').val($('#type-clone').val()); document.getElementById("elastic-search").submit(); }

Autor: Nevil Hopley

Fach:  STEM

Schlagwörter: E-Learning, Algebra, Modelling

Constructing a scale model of the Earth and Moon in a classroom.

Autor: Nevil Hopley

Fach:  Mathematics

Schlagwörter: Electricity, Physics, STEM

We will show you how to construct simulations of experiments of chance. We shall use the minimum amount of code, aiming to use just the Calculator and Data & Statistics applications as much…

Autor: Nevil Hopley

Fach:  Mathematics

This code simulates a test of association for any two-way contingency table and compares it to the appropriate Chi-Squared distribution.

Autor: Nevil Hopley

Fach:  Mathematics

Schlagwörter: Circle theorems, Area

Supports the conceptual understanding of where the formula A=π*r² comes from.

Autor: Nevil Hopley

Fach:  Mathematics

Schlagwörter: Patterns, Scattergraphs, Linear regression, Modelling

Use the TI-Nspire Handheld (either Numeric or CAS) to investigate and generalise results from numerical patterns.

Autor: Nevil Hopley

Schlagwörter: Fractions, Decimals, Angles, History of maths, Trigonometry

Measuring angles using degrees and radians (both decimals and fractions of π).

Autor: Nevil Hopley

Schlagwörter: Linear, Quadratic, Powers and roots, Coordinate systems, Discriminant

Explore linear, quadratic and cubic functions, their graphs and their different algebraic representations.

Autor: Nevil Hopley

Use graphs and the nsolve(...) command to explore equations' solutions and discover properties of negative indices.

Autor: Nevil Hopley


text:
## Terminal morphism

### Collection

 context $Z:\mathrm{Ob}_{\bf C}$ context $F$ in ${\bf D}\longrightarrow{\bf C}$ definiendum $\langle B,\phi\rangle$ in $\mathrm{it}$ inclusion $A:\mathrm{Ob}_{\bf D}$ inclusion $\phi:{\bf C}[FB,Z]$ for all $B:\mathrm{Ob}_{\bf D}$ for all $\psi:{\bf C}[FA,Z]$ range $f:{\bf D}[A,B]$ postulate $\exists_!f.\ \psi=\phi\circ F(f)$

### Discussion

#### Terminology

Note that the name “terminal morphisms” for $\langle B,\phi\rangle$ is slightly confusing, because the data is not just the morphism $\phi$ going from $F(B)$ to $Z$, but also the object $B$.

#### Idea

You want to specify a smaller world within ${\bf C}$. You do this by setting up another category ${\bf D}$ and mapping it into ${\bf C}$ with a functor $F$. The image of $F$ is your smaller world.

For a fixed object $Z$, a universal morphism $\phi$ is then defined by demanding that every arrow from this small world which tries to connect $Z$, must pass through it. In other words (and now I just state the definition): For all objects $B$ in ${\bf D}$, if $\psi$ is an arrow in ${\bf C}$ from $FB$ to $Z$, then there must be an arrow $f$ within ${\bf D}$, so that $\phi$ is really just a combination of “small world business” $F(f)$, followed by the kind of gate function $\phi$, finally leading to $Z$.

For the reason that everything going on in the image of $F$ must pass to $Z$ only through $\phi$, one says $\phi$ is the terminal morphism “from the functor $F$ to the object $Z$”.

##### Example: The exponential object and currying

So you have a function $\psi:X\times Y\to Z$ and you want to be able to do partial function application, i.e. make sense of just passing an argument of $X$ to $\psi$ and leave the second component unevaluated. In plain English, we need the function space which contains the half-evaluated function. To make use of it, we also need the evaluation function, which when passed the half-evaluated function and an argument of $Y$ returns the same as what $\psi$ would do.

It's done as follows: We have $Z:\mathrm{Ob}_{\bf C}$. Now fix any “second argument space” $Y$. Define the object map of the auto-functor $F$ as $FA:=A\times Y$ and its function map by $F(f):=f\times\mathrm{id}_Y$. Now we are done by asking for the terminal morphism from $F$ to $Z$: This exactly means the category must contain a function object, let's call it $Z^Y$, so that for any $\psi:X\times Y\to Z$ (which is $\psi:FX\to Z$) there is a function $f$, which chooses a function $F(f)(x)$ in $Z^Y$, so that $\psi$ can be written as first passing $\langle x,y\rangle:X\times Y$ to $\langle F(g)(x),y\rangle$ and then evaluates this via $\mathrm{eval}(F(f)(x),y):=\left(F(f)(x)\right)(y)$.

(The left side is in ${\bf D}$, the right is in ${\bf C}$ and $f$ is called $\lambda g$ here.)

As a side note: As this works for all $Z$, this gives rise to a hom-class adjunction of the above functor $F$ and the one mapping $Y$ to $Z^Y$. The categorical currying is hence $\mathrm{hom}(X\times Y,Z)$ being isomorphic to $\mathrm{hom}(X,Z^Y)$. The nice thing is, of course, that this doesn't just work for categories with function, but the same construction also works when reasoning about logic, where currying then is the tautology $\left((A\land B)\implies C\right)\Leftrightarrow \left(A\implies(B\implies C)\right)$.


text:
# What is difference between template's background and background canvas?

What is difference between:

\setbeamertemplate{background}{
\begin{tikzpicture}[remember picture,overlay]
([shift={(0.0cm,0.0cm)}]current page.north west)
rectangle
([shift={(0.0cm,0.0cm)}]current page.south east);
\end{tikzpicture}
}

and:

\setbeamertemplate{background canvas}{
\begin{tikzpicture}[remember picture,overlay]
([shift={(0.0cm,0.0cm)}]current page.north west)
rectangle
([shift={(0.0cm,0.0cm)}]current page.south east);
\end{tikzpicture}
}

First one behaves corectly, the second one behaves randomly on each rebuild? I build it with pdflatex and with gummi (preview). And the second one doesn't show any background when I don't rebuild, but build from scratch. Quite weird for me.

Edit: even first one behaves randomly with different themes...

From beamer user guide:

So, the background canvas goes behind everything, and on that comes background and then main contents.

Compare these codes:

\documentclass{beamer}
\usepackage{tikz}
\setbeamertemplate{background}{
\begin{tikzpicture}[remember picture,overlay]
\node at (current page.center) {here};
\end{tikzpicture}
}
\setbeamertemplate{background canvas}{
\begin{tikzpicture}[remember picture,overlay]
([shift={(0.0cm,0.0cm)}]current page.north west)
rectangle
([shift={(0.0cm,0.0cm)}]current page.south east);
\end{tikzpicture}
}
\begin{document}
\begin{frame}
Some
\end{frame}
\end{document}

and

\documentclass{beamer}
\usepackage{tikz}
\setbeamertemplate{background canvas}{
\begin{tikzpicture}[remember picture,overlay]
\node at (current page.center) {here};
\end{tikzpicture}
}
\setbeamertemplate{background}{
\begin{tikzpicture}[remember picture,overlay]
([shift={(0.0cm,0.0cm)}]current page.north west)
rectangle
([shift={(0.0cm,0.0cm)}]current page.south east);
\end{tikzpicture}
}
\begin{document}
\begin{frame}
Some
\end{frame}
\end{document}

In first you see here at the center and in second it goes behind (as it is inside background canvas.

• And when you use remember picture, you have to compile 2-3 times to settle down. – user11232 Dec 4 '14 at 13:49
• Thanks. You answered the question, as it seems, bug is caused by something else. – kravemir Dec 5 '14 at 10:07


text:
# Differential Equations

- An equation involving derivatives of a function or functions. Solving ODE and PDE
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Aug5-14 11:22 PM Greg Bernhardt 1 242 Can anybody give me some links for plotting bifurcations online? Aug5-14 11:22 PM Greg Bernhardt 1 352 Suppose we have some partial differential equation for a scalar ##f##$$Df = \rho$$taking values in... Aug5-14 11:22 PM Greg Bernhardt 1 747 does negative divergence of gradient tempearature gives to lalace equation...? -div(∇T) = Aug5-14 07:32 PM SteamKing 1 163 I want to be able to map the position of a planet given initial position, velocity, and acceleration. I know the... Jul30-14 08:21 AM Juan Carlos 2 2,223 Hi!, I'm working on a personal project: Solve the heat equation with the semi discretization method, using my own... Jul30-14 08:08 AM Juan Carlos 2 239 Hey there, I have modeled a propagating wave in a 1D dispersive media, in which square and cubic nonlinear terms... Jul30-14 01:14 AM Strum 2 1,963 How to understand Uniqueness and existence theorem for first order and second order ODE's intuitively? Jul28-14 02:51 PM homeomorphic 3 466 Hi. First off, sorry for the not so descriptive title. If one of you finds a better tilte I will edit it. We have... Jul28-14 03:27 AM Strum 1 785 Hello Everyone. I have a question. Suppose I have a differential equation for which I want to find the values at... Jul26-14 12:39 AM crossfire234 1 785 I'm currently reading Tolstov's "Fourier Series" and in page 58 he talks about a criterion for the convergence of a... Jul25-14 02:49 PM jbunniii 2 702 When deriving the Runge-Kutta Method to solve y'=f(x) we need to use Taylor expansion. Hence we need to differentiate... Jul23-14 01:30 AM matematikawan 4 921 My textbook says that: "A differential equation contains both the function and the derivative of the function" ... Jul22-14 04:31 PM johann1301 3 856 I want to show, in a talk, how to build the Predator-Prey (Lotka Volterra) model. It starts off with the "Population... 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