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Area of triangle - How to find the Area of a Triangle (Examples) - BYJUS

# Area of triangle

Triangle is the most basic polygon, which has three sides, three vertices, and three angles. Triangles can be categorized into three based on the properties of their sides. We can find the area of a triangle if we know the length of its base and height by using a specific formula. Learn how to identify the base and height of a triangle and find its area. ...Read MoreRead Less

## What is a triangle ?

A triangle is a two-dimensional geometrical shape with three sides. It is a polygon of three sides. It also has three vertices, three sides and three angles.

The shape of traffic signals, a clothes hanger, Light wooden clothes hanger etc. are examples of the shape of a triangle.

The triangles that we see around us or what we study in the classroom are classified into two types  The first type is based on the sides, and the second type is based on the  angles of a triangle.

Types of triangles based on sides:

There are three types of triangles based on the sides:

Scalene triangle: A triangle in which the three sides are unequal is called a scalene triangle.

Isosceles triangle: A triangle in which two sides are equal and the third side is of a different length is called an isosceles triangle. The two base angles of an isosceles triangle are also equal.

Equilateral triangle: A triangle with three equal sides is called an equilateral triangle. The three angles of an equilateral triangle are also equal in measure.

## Properties of a triangle

1) A triangle has three vertices, three sides and three angles.

2) The sum of the length of any two sides of a triangle is always greater than the length of the third side of the triangle.

3) The angle sum or the summation of the three angles is always equal to two right angles or $$180^{\circ }$$

## What is the area of a triangle ?

The area of a triangle is the space occupied by the sides of a triangle in a two-dimensional plane. It is measured in square units like square inches, square meters, square yards, etc. For any triangle, the area is calculated as half of the product of the base and the height of the triangle.

The area of a triangle can also be calculated from the area of the parallelogram. We draw a parallelogram and cut the parallelogram into two identical triangles along one of the diagonals .The area of one of these triangles is half the area of the parallelogram.

Area of triangle  = $$\frac{1}{2}\times \text{b}\times \text{h}$$

## Formula for the area of triangle

The area (A) of a triangle is one-half of the product of its base (b) and height (h). The area of the triangle formula in an algebraic form is:

Area of triangle (A) = $$\frac{1}{2}\times \text{b}\times \text{h}$$

Where “b” is the base of a triangle and “h” is the height of a triangle.

## Solved examples

Example 1: Find the area of the triangle whose base is 8 inches long and height is 12 inches.

Solution:

Area of triangle = $$\frac{1}{2}\times \text{b}\times \text{h}$$

= $$\frac{1}{2}$$ (8) (12)                                                [Substitute values of b and h]

= $$\frac{1}{2}$$(96)                                                     [Multiply]

= 48                                                           [Multiply]

The area of the triangle is 48 square inches.

Example 2:

Find the area of the triangle shown in the figure.

Solution:

The height of the triangle is 10 feet and the base is 8 feet. Therefore, we use the area of a triangle formula to calculate the area,

Area of triangle = $$\frac{1}{2}\times \text{b}\times \text{h}$$

= $$\frac{1}{2}$$ (8) (10)                                                      [Substitute values of b and h]

= $$\frac{1}{2}$$(80)                                                           [Multiply]

=  40                                                               [Multiply]

The area of the triangle is 40 square feet.

Example 3:

The area of the triangle is 48 square inches. Find the height of the triangle if the base is 8 inches long.

Solution:

The area of the triangle is 48 square inches and the base is 8 inches. Therefore, we use the area of a triangle formula to form an equation to find height

Area of triangle = $$\frac{1}{2}\times \text{b}\times \text{h}$$

$$\frac{1}{2}$$ (8)  (h) = 48                                          [Substitute values of b and h]

(8) (h) = 96                                           [Multiply by 2 both sides]

h = 12                                             [Divide both sides by 8]

The height of the triangle is 12 inches.

Example 4: Larry has brought a plot in New York. The plot is in the shape of a triangle as shown in the diagram. Find the area of the plot.

Solution:

The plot is in the shape of a triangle. The base of the triangle is 50 yards and the height of the triangle is 100 yards. We use the area of a triangle formula to calculate the area,

Area of triangle = $$\frac{1}{2}\times \text{b}\times \text{h}$$

= $$\frac{1}{2}$$ (50) (100)                                                             [Substitute values of b and h]

= $$\frac{1}{2}$$ (5000)                                                                  [Multiply]

= 2500                                                                      [Multiply]

The area of the triangular plot is about 2500 square yards.

Example 5: There is a triangular wall poster on a street wall. Find the area of the poster.

Solution:

The poster is in the shape of a triangle. The base of the triangle is 20 inches and the height of the triangle is about 50 inches. We use the area of a triangle formula to calculate the area,

Area of triangle = $$\frac{1}{2}\times \text{b}\times \text{h}$$

= $$\frac{1}{2}$$ (20) (50)                                   [Substitute values of b and h]

= $$\frac{1}{2}$$  (1000)                                      [Multiply]

= 500                                               [Multiply]

The area of the poster is about 500 square inches.

Heron’s formula is applied to find the area of a triangle when the lengths of all the three sides of a triangle are known.

If a, b, and c are sides of a triangle,

Area of the triangle (A) = $$\sqrt{\text{s(s-a)(s-b)(s-c)}}$$

Here S = $$\frac{\text{a+b+c}}{2}$$    is the semi-perimeter of the triangle.

The area of a equilateral triangle formula is,

A = $$\sqrt{\frac{3}{4}}\times a^{2}$$

Where ‘a’ is the side of the equilateral triangle.

The area formula for a right angled triangle is,

Area of triangle = $$\frac{1}{2}\times \text{b}\times \text{h}$$

Where ‘h’ is the altitude or height of the triangle and ‘b’ is the base of the triangle.

However the area for any other triangle when the three sides are given is,

Area of triangle  A = $$\sqrt{s(s-a)(s-b)(s-c)}$$

in which “s” is the semi-perimeter and$$s=\frac{a+b+c}{2}$$, and a,b,c are the lengths of the sides of the triangle.


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# CL: Mass/Molar Flux

Recall that in terms of mass transfer we are interested in either Ni (called the molar flux) or ni (mass flux) so that we are interesting in the rate at which mass traveles through a certain area in a given time. While we have written this as a flow dived by an area:

$\displaystyle{\mathbf{n} = \frac{\dot m}{A}}$,

we might instead (for something like a pipe flow) write that the mass flow rate is given by:

$\displaystyle{\dot m = \rho v A}$

where $\displaystyle{\rho}$ is the mass concentration (density), v is the (average) velocity in the pipe, and A is the cross section of the pipe so that we can then write the mass flux as:

$\displaystyle{\mathbf{n} = \frac{\dot m}{A} = \frac{\rho v A}{A} = \rho v}$

These simple arguments, in fact, can be applied to any mass flow to determine the total mass flux, so that, in general, the mass flux of component i is given by:

$\displaystyle{\mathbf{n_i} = \rho_i v_i}$

Similar arguments can give us an expression for the molar flux

$\displaystyle{\mathbf{N_i} = c_i v_i}$

where ci is the molar concentration of component i.

Unfortunately, the velocity in these expressions vi is not trivial to calculate (or measure!) without some further consideration. (Think, for example, of measuring the velocity of oxygen in the cool breeze on a summer afternoon instead of measuring the velocity of the air! Clearly, measuring the air velocity would be simple, but distinguishing the velocity of the oxygen from all the other gases that make up air would be quite troublesome.) So, let's think in more detail about the bulk velocity (like the air velocity).

Let's assume for a minute that we do actually know the velocities of each the individual gases that make up air. How might we calculate the bulk or average velocity?

Well, for our purposes there are essentially two methods: averaging by mass, or averaging by mole. If we wanted to calculate the bulk velocity of the air by mass we would add up all of the velocities of the individual gases, weighted by the mass of that particular gas and then divide the total by the mass of the air. In other words, we would calculate a weighted average (by mass) like this:

$\displaystyle{v = \frac{\sum_i \rho_i v_i}{\sum_i \rho_i}}$

We could do the exact same thing using moles

$\displaystyle{V = \frac{\sum_i c_i v_i}{\sum_i c_i}}$

where ci is the molar concentration of component i.

Now that we know how the actual velocity of each component in a mixture is related to the average (bulk) velocity (which we can measure). We can try to determine how to go in the other direction (in other words, how to calculate the actual component velocity from the bulk velocity.)

If we think of a crowd of people walking down the hallway, it is certainly possible that the individuals within the crowd are walking at different speeds. If we draw a box (centered) around the individuals at one time, and then again a little while later, it is clear that the crowd as a whole is moving at some average velocity, V (which is not necessarily equal to the velocity of any of the individuals). If we take the difference between the individual's velocity and the bulk velocity we get a quantitative measure of the individual's deviation (diffusion!) velocity.

$\displaystyle{v_{diff_i} = v_i - v}$ or $\displaystyle{v_{diff_i} = v_i - V}$

In exactly the same way that we defined a total mass (mole) flux using the component's total (actual) velocity, we can then define a component's diffusive flux ji (mass) or Ji (molar) to be:

$\displaystyle{j_i = \rho_i v_{diff_i} = \rho_i(v_i-v)}$
or
$\displaystyle{J_i = c_iv_{diff_i} = c_i(v_i-C)}$

We discussed earlier in the course that the actual origin of diffusive motion is the random fluctuations of molecules "swapping" their way down a concentration gradient. This explanation clearly has much in common with heat flow (down a temperature gradient) and thus it is not surprising that Fick's Law (proposed for binary mixtures) looks much like Fourier's Law

$\displaystyle{J_i = -D\nabla c_i}$

for the diffusive flux, where D is the binary diffusivity of component i in the system (species j. This expression shows that (just like temperature) concentration "flows downhill" (until there isn't a hill!) A more general equation to account for possible changes in overall concentration, ctotal (in non isobaric, isothermal systems), is

$\displaystyle{J_i = -c_{total}D\nabla y_i}$

where yi is the mole fraction.

##### NOTE:

While Fick's Law is formally true for only binary mixtures, it is often used for multi-component systems with a somewhat fudged value of D.

If we then plug this equation into our definition of the diffusive flux

$\displaystyle{J_i = c_i(v_i-V) = -c_{total}D\nabla y_i}$

we can solve for the component's actual velocity vi

$\displaystyle{v_i = \frac{-c_{total}D\nabla y_i}{c_i} +V}$

We can then plug this equation into our expression for the molar flux

$\displaystyle{N_i = c_iv_i = c_i\left ( \frac{-c_{total}D\nabla y_i}{c_i} +V \right ) = -c_{total}D\nabla y_i +c_iV}$

which (finally!) gives us the total molar flux in terms of things we can measure easily. Again, deriving the expression based on mass is not any more difficult

$\displaystyle{n_i = -\rho_{total}D\nabla \omega_i +\rho_i v}$

where $\displaystyle{\omega_i}$ is the mass fraction of component i.

##### OUTCOME:

Explain the difference between the total flux and the diffusive flux

format: plain


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# How to find the magnitude of electric and magnetic fields in an arbitary inertial frame?

In the context of special relativity, and using proper Lorentz transformations:

If the electromagnetic fields have an uniform value [= constant in space and time] in an inertial frame, how will the magnitude of these fields change in a different (arbitrary) inertial frame?

For example, an inertial frame which makes an angle $\alpha$, or moves with a different velocity $v'$ with respect to the original one.

• There was some useful information in the comments here, but the discussion was getting rather long, so I've deleted them. I'll just leave here the link to the help center, and also mention that Physics Meta or Physics Chat are the places to ask for clarification on how the site works. – David Z Dec 1 '15 at 7:45

The relevant transformations, where ${\bf v}$ is a velocity (vector) of a "primed" frame with respect to the stationary "lab frame" and $\gamma$ is the Lorentz factor at speed $v$, are
$$\mathbf {{E}}' = \mathbf{{E}_{\parallel}} + \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right)$$ $$\mathbf {{B}}' = \mathbf{{B}_{\parallel}} + \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right),$$ where the parallel and perpendicular subscripts refer to components of the E- and B-field that are either parallel or perpendicular to ${\bf v}$. i.e. The components parallel to ${\bf v}$ end up unchanged but the perpendicular components are transformed.


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Poster
Bayesian Counterfactual Risk Minimization
Ben London · Ted Sandler

Tue Jun 11th 06:30 -- 09:00 PM @ Pacific Ballroom #113

We present a Bayesian view of counterfactual risk minimization (CRM) for offline learning from logged bandit feedback. Using PAC-Bayesian analysis, we derive a new generalization bound for the truncated inverse propensity score estimator. We apply the bound to a class of Bayesian policies, which motivates a novel, potentially data-dependent, regularization technique for CRM. Experimental results indicate that this technique outperforms standard $L_2$ regularization, and that it is competitive with variance regularization while being both simpler to implement and more computationally efficient.


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# Reflection principle for intuitionistic Zermelo–Fraenkel?

The well-known reflection principle for classical Zermelo–Fraenkel states:

For any formula $$\varphi(x_1,\ldots,x_n)$$ of the language of ZFC with free variables $$x_1,\ldots,x_n$$, ZFC proves $$\forall M_0. \exists M \supseteq M_0. \forall x_1,\ldots,x_n \in M. (\varphi(x_1,\ldots,x_n) \Leftrightarrow \varphi^M(x_1,\ldots,x_n)),$$ where $$\varphi^M$$ is the $$M$$-relativization of $$\varphi$$.

Question. Does the analogous theorem hold with ZFC replaced by IZF, intuitionistic Zermelo–Fraenkel? All proofs of the reflection principle for ZFC that I know of don't obviously carry over to IZF, firstly because of the failure of Scott's trick and secondly because we cannot, unlike in classical texts, assume without loss of generality that formulas don't contain universal quantifiers.

Surely this has been studied before, but I wasn't able to track down a reference.

Motivation. With such a reflection principle in place, we could mimic the definition of ZFC/S, a conservative extension of ZFC useful for category theory, to create IZF/S. The reflection principle is what powers the automatic transfer from a given theorem formulated, for instance, for all small groups, to one formulated for all groups, and this leap shouldn't require non-intuitionistic techniques.

• I added the top-level arXiv tag lo.logic, and a link to the IZF axioms (are they the ones you are thinking of, Ingo?) – David Roberts Dec 20 '18 at 23:46
• For how large a fragment of logic can you prove the reflection principle for IZF? – Andrej Bauer Dec 21 '18 at 7:08
• @IngoBlechschmidt you can prove that using full separation and collection. Define $X := \{ x \in \{0\} | (\exists y)y \in C \}$, then by collection there exists a set $C'$ such that if $X$ is inhabited, then so is $C' \cap C$. Note that $C'$ is not definable, and in fact by a result due to Friedman and Scedrov it is impossible in general for such a $C'$ to be definable. Even with this trick it is difficult to deal with formulas of the form $(\exists y)\phi(x, y)$, with reflection as you stated it, because it is necessary to repeat transfinitely. – aws Dec 24 '18 at 15:25
• @aws Lubarsky provides an answer of the independence of Reflection over $\mathbf{CZF}$ in his article Independence Results around Constructive ZF. – Hanul Jeon Feb 19 '19 at 4:15
• @HanulJeon Thanks. I discussed this with Ingo a bit offlist and he noticed that reflection implies full separation, so in fact reflection is not even provable in $\mathbf{CZF}$ with powerset and $\mathbf{RDC}$. But Lubarsky's proof is interesting too - it shows reflection is not provable even with full separation (but without powerset). – aws Feb 19 '19 at 9:21


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Optimization description

A trader using alternative data who already forecasted expected returns of different stocks want to find what is the optimal weights of his portfolio. Let’s input the users’ data in the optimization.

Our goal is to maximize the expected return given by the user divided by the square root of the variance and find what are the optimal weights for the portfolio.

 goal = expected_return / variance.sqrt()

The optimization is run on 2 years of data from 01/01/2012 to 01/01/2014.  A portfolio is created with 2 stocks: AAPL and GOOG. Expected return is inputted by user. Portfolio’s initial value is \$1 million.

Results

After running the optimzation, we found that the optimal weights are the following for Apple and Google. The first 3 months correspond to the warming period of the model.

We can also backtest our optimize porfotfolio. Here we decided to calculate the variance of the portfolio.

How to install hft Python package

Use Anaconda2, install JDK. Refer yourself to this manual.

conda install -c portfolioeffect hft

Implementation of the strategy using hft Python package

from hft.optimization import *
from hft.metric import *
import pandas as pd
import time
import datetime

#Create Portfolio
portfolio = Portfolio("2012-01-01 09:30:00", "2014-01-01 16:00:00", 'SPY')
portfolio.settings(portfolioMetricsMode='price', inputSamplingInterval='30s',
resultsSamplingInterval='1d', windowLength='50d')

filename = 'return.csv'

#Optimize portfolio's weights
variance = portfolio.variance()
expected_return = portfolio.expected_return()
goal = expected_return / variance.sqrt()
optimizer = Optimizer(goal, direction="max")

for symbol in list(set(data['symbol'])):
temp = data[data.symbol == symbol]
Time = [(time.mktime(datetime.datetime.strptime(i, "%m/%d/%Y").timetuple())+57600)* 1000
for i in temp['date']]
optimizer = optimizer.forecast("ExpReturn", Metric([Time, list(temp['alpha'])], symbol))

optimalPortfolio = optimizer.run()

#Plot optimal weights
optimalPortfolio.get_position('AAPL').weight().plot(title='AAPL Weight')
optimalPortfolio.get_position('GOOG').weight().plot(title='GOOG Weight')

#Plot portfolio's variance
optimalPortfolio.variance().plot(title='Portfolio Variance')


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# Math Help - A simple maths question-simplifying

1. ## A simple maths question-simplifying

Hello

3 log x - log4y

What is this expression simplified?

2. ## Re: A simple maths question-simplifying

Hello

3 log x - log4y

What is this expression simplified?
1. I assume that you mean

$3\cdot \log(x) - \log(4y)$

2. If you want to use logarithms in the simplest form then you'll get:

$3\cdot \log(x) - \log(4) -\log(y)$

3. If you want to collect these terms to one logarithm then you'll get:

$3\cdot \log(x) - \log(4y) = \log\left(\frac{x^3}{4y}\right)$

3. ## Re: A simple maths question-simplifying

Thankyou buddy...it was the last one by the way.

4. ## Re: A simple maths question-simplifying

There are many ways of solving this.
one among them is,
½ log4x2+1/3 log4y (1/2 log x squared to the base 4 + 1/3 log y to the base 4) *4 is the base.
In other way,
-logx+ 2 log(x-1)+3 logx 2
That's -logx+2 (x-1)+3logx squared.

maths homework
maths assignments help


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Blog Entries: 3

## Logic in Computation

Since I know that this will get lost in the Liar's paradox thread I wanted to separate it out from the thread:

"
 Of course, there are other variations on logic than classical logic. I mentioned computability theory: that the alternatives are {true, false, infinite loop} is rather important to the theory. e.g. it gives a way out to the specific construction of the Liar's paradox I mentioned earlier: to refresh: Q(R) := not R(R) P := Q(Q) The naive implementation of the predicate Q and the sentence P clearly results in P evaluating to "infinite loop".
and in post #47 he writes:

 Quote by Hurkyl Third, I think it would be interesting to point out that in the logic of computation, there is no problem with there being a sentence P satisfying "P = P is not true". Here's an implementation in python: Code: def P(): return not P() however, your computer probably cannot do this computation: an equivalent implementation that will not overflow your stack is: Code: def P(): while True: pass

I would like to know where he is going with this line of thought. A related concept I've seen before is the concept of bottom:

"In type theory, a theory within mathematical logic, the bottom type is the type that has no values. It is also called the zero or empty type, and is sometimes denoted with falsum (⊥).
A function whose return type is bottom cannot return any value. In the Curry–Howard correspondence, the bottom type corresponds to falsity.
"
http://en.wikipedia.org/wiki/Bottom_type

So while bottom acts like a third truth value, it is actually a subtype of both true and false. So both true and false can be of type Bottom. However, normally both True and False should not be of type bottom unless their is an error in the computation.

To summarize the intent of this thread: How do the laws of logic relate to computation and what deviations might their be from the law of logic (for instance: The law of the excluded middle).

P.S. Also in the Liar's paradox thread the question of the Halting problem came up:
http://en.wikipedia.org/wiki/Halting_problem

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 One thing that might be of interest to you is a number theory approach to computation. Gregory Chaitin developed a way to take code and convert it to a number theoretic problem. I see this kind of thinking as a critical way to systematically and analytically assess whether something is computable, because if you have a number theory system expression to solve for a particular collection of variables, then if you can show that no solution exists, then you have indirectly (or directly) shown that it can not be computed. We know that trying to compute 2x = (1 MOD 2) for any x is impossible, and such a statement shows that x is completely un-computable. This kind of formalism provides a nice way for showing uncomutability properties for a general algorithm, if it can be turned into a number theory system expression (could be a collection of constraints like you have in the Chinese Remainder Theorem calculations, or one constraints like a simple one line congruence relationship). If each computation was done in this manner and a standard mechanism was used to find the solutions, then we know that the program will not halt if there is no solution to the diophantine system and this is of course very useful. Not only does the above approach standardize computation, it tells us about halting if we are able to show that the system has no solution (trying to use a computer to solve the diophantine system using a search algorithm that only terminates when it finds the solution is the approach I would use). Most of this post just takes Gregory Chaitin's work and uses the idea of having standardized algorithms be put in Diophantine form where no-solution implies that under a standardized algorithm that only returns when it finds a solution, it will only halt if one exists and not halt if there are no solutions. Some food for thought.

Blog Entries: 3
 Quote by chiro This kind of formalism provides a nice way for showing uncomutability properties for a general algorithm, if it can be turned into a number theory system expression (could be a collection of constraints like you have in the Chinese Remainder Theorem calculations, or one constraints like a simple one line congruence relationship).
I bet that this approach has high complexity both because the Chinese remainder theory is for linear equations and computers can deal with quite large numbers.

## Logic in Computation

 Quote by John Creighto I bet that this approach has high complexity both because the Chinese remainder theory is for linear equations and computers can deal with quite large numbers.
It does, but it's beneficial to have it in this kind of form because it means that when the field is developed enough to the point where one can check if solutions exist, then from computer science of view this is very useful in terms of finding whether algorithms ever halt.

Number theory in its current form is very young, but I imagine later that this kind of specification (or a similar kind) is going to be probably the way people look at computation because not only does it standardize it, it offers the opportunity given further research to find whether the algorithm will terminate.


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# Transformer Based Multi-Source Domain Adaptation

Dustin Wright and Isabelle Augenstein

Published in EMNLP, 2020

In practical machine learning settings, the data on which a model must make predictions often come from a different distribution than the data it was trained on. Here, we investigate the problem of unsupervised multi-source domain adaptation, where a model is trained on labelled data from multiple source domains and must make predictions on a domain for which no labelled data has been seen. Prior work with CNNs and RNNs has demonstrated the benefit of mixture of experts, where the predictions of multiple domain expert classifiers are combined; as well as domain adversarial training, to induce a domain agnostic representation space. Inspired by this, we investigate how such methods can be effectively applied to large pretrained transformer models. We find that domain adversarial training has an effect on the learned representations of these models while having little effect on their performance, suggesting that large transformer-based models are already relatively robust across domains. Additionally, we show that mixture of experts leads to significant performance improvements by comparing several variants of mixing functions, including one novel mixture based on attention. Finally, we demonstrate that the predictions of large pretrained transformer based domain experts are highly homogenous, making it challenging to learn effective functions for mixing their predictions.

View paper here

@inproceedings{wright2020transformer,


text:
Elementary Algebra

$\frac{-30x^{5}+20x^{4} -10x^{3}}{-5x^{2}}$ = 6$x^{3}$ - 4$x^{2}$ + 2x To check if this is true or not, we simplify the left side. RECALL: $\frac{b^{n}}{b^{m}}$ = $b^{n-m}$ So, $\frac{-30x^{5}+20x^{4} -10x^{3}}{-5x^{2}}$ = $\frac{-30x^{5}}{-5x^{2}}$ + $\frac{20x^{4}}{-5x^{2}}$ + $\frac{-10x^{3}}{-5x^{2}}$ = $\frac{-30}{-5}$$x^{5-2} + \frac{20}{-5}$$x^{4-2}$ + $\frac{-10}{-5}$$x^{3-2}$ = 6$x^{3}$ - 4$x^{2}$ + 2x Because 6$x^{3}$ - 4$x^{2}$ + 2x is equal to 6$x^{3}$ - 4$x^{2}$ + 2x, this statement is true.


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# Simplify to real and positive numbers

eng8 = [solve(eng==(k-1)*md*Vm1^2/(2*Patm*Vs*((Pc-(P2*s6))-(Pc-(P2*s6))^(1/k))),Vm1) for s6 in srange(r9,r10,s5,include_endpoint=1)]
v4=[v[0].rhs().n() for v in eng8]
v5=[v for v in v4 if v >= 0]

I can't seem to get v5 to output both real and positive numbers, only an empty list. All the inputs are positive, and I know the square root of the squared gives both positive and negative output. Any help?

edit retag close merge delete

Have you verified that the results in eng8 are indeed positive real numbers? The statement "the square root of the squared gives both positive and negative output" does not make mathematical sense. What you probably mean is that the equation $x^2 = a$ has two solutions, one positive and one negative when $a > 0$. These are the principle square root of a (usually represented by sqrt(a)) and the negative of the principle square root of a.


text:
## Cryptology ePrint Archive: Report 2011/437

Approximate common divisors via lattices

Abstract: We analyze the multivariate generalization of Howgrave-Graham's algorithm for the approximate common divisor problem. In the $m$-variable case with modulus $N$ and approximate common divisor of size $N^\beta$, this improves the size of the error tolerated from $N^{\beta^2}$ to $N^{\beta^{(m+1)/m}}$, under a commonly used heuristic assumption. This gives a more detailed analysis of the hardness assumption underlying the recent fully homomorphic cryptosystem of van Dijk, Gentry, Halevi, and Vaikuntanathan. While these results do not challenge the suggested parameters, a $2^{\sqrt{n}}$ approximation algorithm for lattice basis reduction in $n$ dimensions could be used to break these parameters. We have implemented our algorithm, and it performs better in practice than the theoretical analysis suggests.

Our results fit into a broader context of analogies between cryptanalysis and coding theory. The multivariate approximate common divisor problem is the number-theoretic analogue of noisy multivariate polynomial interpolation, and we develop a corresponding lattice-based algorithm for the latter problem. In particular, it specializes to a lattice-based list decoding algorithm for Parvaresh-Vardy and Guruswami-Rudra codes, which are multivariate extensions of Reed-Solomon codes. This yields a new proof of the list decoding radii for these codes.

Category / Keywords: foundations / Coppersmith's algorithm, lattice basis reduction, approximate common divisors, fully homomorphic encryption, list decoding, Parvaresh-Vardy codes, noisy polynomial interpolation

Date: received 12 Aug 2011, last revised 13 Mar 2012

Contact author: nadiah at cs ucsd edu

Available format(s): PDF | BibTeX Citation

[ Cryptology ePrint archive ]


text:
# What is the formula for the compound dinitrogen pentoxide?

${N}_{2} {O}_{5}$
dinitrogen means ${N}_{2}$ and pentoxide(oxide means oxygen) ${O}_{5}$
so chemical formula = ${N}_{2} {O}_{5}$


text:
### Leakage-Resilient Storage

Francesco Davì, Stefan Dziembowski, and Daniele Venturi

##### Abstract

We study a problem of secure data storage on hardware that may leak information. We introduce a new primitive, that we call {\em leakage-resilient storage} (LRS), which is an (unkeyed) scheme for encoding messages, and can be viewed as a generalization of the {\em All-Or-Nothing Transform} (AONT, Rivest 1997). The standard definition of AONT requires that it should be hard to reconstruct a message $m$ if not all the bits of its encoding $\Encode(m)$ are known. LRS is defined more generally, with respect to a class $\Gamma$ of functions. The security definition of LRS requires that it should be hard to reconstruct $m$ even if some values $g_1(\Encode(m)),\ldots,$ $g_t(\Encode(m))$ are known (where $g_1,\ldots,g_t \in \Gamma$), as long as the total length of $g_1(\Encode(m)),\ldots,g_t(\Encode(m))$ is smaller than some parameter $c$. We construct an LRS scheme that is secure with respect to $\Gamma$ being a set of functions that can depend only on some restricted part of the memory. More precisely: we assume that the memory is divided in $2$ parts, and the functions in $\Gamma$ can be just applied to one of these parts. We also construct a scheme that is secure if the cardinality of $\Gamma$ is restricted (but still it can be exponential in the length of the encoding). This construction implies security in the case when the set $\Gamma$ consists of functions that are computable by Boolean circuits of a small size. We also discuss the connection between the problem of constructing leakage-resilient storage and a theory of the compressibility of NP-instances.

Available format(s)
Category
Foundations
Publication info
Published elsewhere. Unknown where it was published
Contact author(s)
stefan @ dziembowski net
History
2010-04-13: last of 2 revisions
See all versions
Short URL
https://ia.cr/2009/399

CC BY

BibTeX

@misc{cryptoeprint:2009/399,
author = {Francesco Davì and Stefan Dziembowski and Daniele Venturi},
title = {Leakage-Resilient Storage},
howpublished = {Cryptology ePrint Archive, Paper 2009/399},
year = {2009},
note = {\url{https://eprint.iacr.org/2009/399}},
url = {https://eprint.iacr.org/2009/399}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
### Homology of the curve complex and the Steinberg module of the mapping class group

Source: Duke Math. J. Volume 161, Number 10 (2012), 1943-1969.

#### Abstract

By the work of Harer, the reduced homology of the complex of curves is a fundamental cohomological object associated to all torsion-free finite index subgroups of the mapping class group. We call this homology group the Steinberg module of the mapping class group. It was previously proved that the curve complex has the homotopy type of a bouquet of spheres. Here, we give the first explicit homologically nontrivial sphere in the curve complex and show that under the action of the mapping class group, the orbit of this homology class generates the reduced homology of the curve complex.

First Page:
Primary Subjects: 57N05
Secondary Subjects: 32G15
We're sorry, but we are unable to provide you with the full text of this article because we are not able to identify you as a subscriber.

Permanent link to this document: http://projecteuclid.org/euclid.dmj/1340801628
Digital Object Identifier: doi:10.1215/00127094-1645634
Zentralblatt MATH identifier: 06063225
Mathematical Reviews number (MathSciNet): MR2954621

### References

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Digital Object Identifier: doi:10.1023/A:1020973910646


text:
# Lipschitz function and uniform continuity

1. Sep 25, 2008

### atm06001

A function f:D$$\rightarrow$$R is called a Lipschitz function if there is some
nonnegative number C such that

absolute value(f(u)-f(v)) is less than or equal to C*absolute value(u-v) for all points u and v in D.

Prove that if f:D$$\rightarrow$$R is a Lipschitz function, then it is uniformly continuous.

I am having trouble proving this, I am not sure if I should suppose not or go about it by some other method?

Last edited: Sep 25, 2008
2. Sep 25, 2008

### mathwonk

try a direct proof.


text:
## Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall

# Chapter 20 - Sections 20.1-20.14 - Exercises - Problems by Topic - Page 993: 56b

#### Answer

The name of the alkyne is $4,5-dimethyl-2-heptyne$.

#### Work Step by Step

The longest carbon chain has 7 carbon atoms (hept-). There are two methyl substituents, one each at positions 4 and 5. The triple bond is at position 2. Hence, the name of the alkyne is $4,5-dimethyl-2-heptyne.$

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:
# An exercise from my brother: $\int_{-1}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\,dx$

My brother asked me to calculate the following integral before we had dinner and I have been working to calculate it since then ($\pm\, 4$ hours). He said, it has a beautiful closed form but I doubt it and I guess he has tried to trick me again (as usual). So I am curious, what is the closed form (if any) of the following integral:

$$\int_{-1}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\,dx$$

I have tried by parts method, partial fractions (stupid idea), converting into series (nothing familiar), many substitutions such as: $u=2x-1$, $u=1-x$, $x=\cos^2\theta$, etc, but I failed and got nothing. Wolfram Alpha also doesn't give an answer. Either he is lying to me or telling the truth, I don't know. Could anyone here please help me to obtain the closed form of the integral with any methods (whatever it takes)? Any help would be greatly appreciated. Thank you.

• What is the ln of a negative number? – André Nicolas Jul 26 '14 at 15:46
• Didn't your mother ever warn you not to do integrals before dinner? – copper.hat Jul 26 '14 at 16:07
• If you restrict the interval of integral to $[\frac12,1]$ where the integrand is real valued, the substitution $x(1-x)=u$ transforms the integral into $\int_{0}^{\frac14}\frac{\ln{\sqrt{1-4u}}}{u^{1/6}(1-4u)^{5/6}}\,du$, which is just a hop, skip and u-substitution away from a derivative of a beta function. Perhaps this is what your brother noticed??? – David H Jul 26 '14 at 16:15
• The integral over $[\frac12,1]$ I mentioned has a nice closed form: $I=-\frac{\pi}{2^{8/3}}\left(\sqrt{3}\,\pi+\ln{432}\right)$. – David H Jul 26 '14 at 16:37

Too long for a comment and this is not the answer too, but following @David H's comment. Perhaps your brother means

$$\int_{\color{red}{\large\frac12}}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\ dx.\tag1$$

If so, then setting $u=2x-1$ to $(1)$ yields $$\frac1{\sqrt[\large 3]{4}}\int_{0}^1\frac{\ln u}{\sqrt[\large 6]{(1+u)(1-u)u^4}}\ du=\frac1{\sqrt[\large 3]{4}}\int_{0}^1\frac{\ln u}{\sqrt[\large 6]{(1-u^2)u^4}}\ du.\tag2$$ Setting $t=u^2$ yields $$\frac1{\sqrt[\large 3]{128}}\int_{0}^1\frac{\ln t}{(1-t)^{\large\frac16}t^{\large\frac56}}\ dt=\frac1{\sqrt[\large 3]{128}}\int_{0}^1(1-t)^{-\large\frac16}t^{\large-\frac56}\ln t\ dt.\tag3$$

Now, consider beta function $$\text{Beta}(x,y)=\int_{0}^1t^{\large x-1}(1-t)^{\large y-1}\ dt.\tag4$$ Differentiating $(4)$ with respect to $x$ yields \begin{align} \frac{\partial}{\partial x}\text{B}(x,y)&=\int_0^1\frac{\partial}{\partial x}\left(t^{\large x-1}(1-t)^{\large y-1}\right)\ dt\\ (\psi(x)-\psi(x+y))\text{B}(x,y)&=\int_0^1 t^{\large x-1}(1-t)^{\large y-1}\ln t\ dt,\tag5 \end{align} where $\psi(\cdot)$ is the digamma function.

Using $(5)$ then $(3)$ turns out to be $$\frac1{\sqrt[\large 3]{128}}\left(\psi\left(\frac16\right)-\psi\left(1\right)\right)\text{B}\left(\frac16,\frac56\right)=\large\color{blue}{-\frac\pi{\sqrt[\large 3]{256}}\left(\pi\sqrt{3}+4\ln2+3\ln3\right)}.$$ Either your brother is teasing you or he just misplaces the interval of integral.

• I like how you substituted $u=1-2*x$ instead of $x(1-x)$ like I chose to, because it demonstrates the underlying symmetry of the integral in these factors. This result is pleasing enough to me to give the brother the benefit of the doubt. +1 – David H Jul 26 '14 at 17:26
• @DavidH Thanks. Anyway, I like your answer here. I tried using derivative of beta function to obtain the integral but I couldn't. +1 – Tunk-Fey Jul 26 '14 at 17:36
• Now that you mention it, you do appear to have an incorrect final result. You can verify numerically the integral is approximately $\approx -5.6947...$, but your value comes to $-28.69956...$. Somewhere you lost a factor of $2^{-7/3}$. – David H Jul 26 '14 at 17:45
• And thank you for the compliment on that answer other you mentioned. I tried and failed to create an answer to this question modeled off of that one, and had to resort to more involved methods. – David H Jul 26 '14 at 17:49
• @DavidH Thanks for pointing that out. I've edited my answer. Now it numerically equals. – Tunk-Fey Jul 26 '14 at 18:06

I highly doubt that one can evaluate that integral by hand, however the integral does have a closed form found by Mathematica to be: $$\frac{2 \left(4 (-1)^{5/6} \, _4F_3\left(1,1,1,\frac{7}{6};2,2,2;\frac{1}{9}\right)+8 (-1)^{5/6} \log (3) \, _3F_2\left(1,1,\frac{7}{6};2,2;\frac{1}{9}\right)+432 (-1)^{5/6} \log ^2(2)-189 (-1)^{5/6} \log ^2(3)+648 (-1)^{5/6} \log (3) \log (2)-54 i \psi ^{(1)}\left(\frac{5}{6}\right)+54 \sqrt{3} \psi ^{(1)}\left(\frac{5}{6}\right)\right)+\pi ^2 \left(-963 \sqrt{3}+1827 i\right)-54 i \left(\sqrt{3}-13 i\right) \pi \log (432)}{1728\ 2^{2/3}},$$ where $pF_q(a;b;z)$ is the Generalized Hypergeometric function, and $\psi^{(n)}$ is the Polygamma function of order $n$.


text:
# Help over

## Definitions

• Chambers's Twentieth Century Dictionary
• Help over to enable to surmount
• ***

## Quotations

• Craig Mary
“The presence of a long-term, conscious goal has helped me maintain stability through the ubiquitous changes of over half a century.”

## Etymology

Chambers's Twentieth Century Dictionary
A.S. helpan, pa.t. healp, pa.p. holpen; Ice. hjálpa, Ger. helfen, to aid.

## Usage

### In literature:

Nature, Paul thought, had often come to their assistance, watching over them, as it were, and helping them when they were weakest.
"The Free Rangers" by Joseph A. Altsheler
"Red Pepper's Patients" by Grace S. Richmond
I gave him my word I'd help him, and I begged him to wait a bit, to give you time to get over that other affair.
"The Odds" by Ethel M. Dell
I got him to help me over the niceties, you understand; but on my lines, lad.
"Poison Island" by Arthur Thomas Quiller-Couch (Q)
That long adobe barracks over yonder used to house the help.
"The Pride of Palomar" by Peter B. Kyne
"The Shoulders of Atlas" by Mary E. Wilkins Freeman
I told Becky to ride over this morning and ask your wife if she could help Mandy.
"The Trumpeter Swan" by Temple Bailey
Carver swearing viciously as he helped to thrust me outward over the wooden bar.
"The Devil's Own" by Randall Parrish
I've sent over for Paulina Maria and your aunt B'lindy to come in an' help.
"Jerome, A Poor Man" by Mary E. Wilkins Freeman
He did not believe, he often said, in feeling sad over things you could not help.
"Sunny Boy and His Playmates" by Ramy Allison White
***

### In poetry:

There was little seeing, that I recall:
We seemed to grope in a cave profound.
They might have come by a painful fall,
Had we not helped them over the ground.
"A Lost Opportunity" by Robert Fuller Murray
'Twas over -- I gazed on the statue --
"Our Father", "Hail Mary" still came;
And to-night faith and love cannot help it,
I must still pray the same -- still the same.
"The Immaculate Conception" by Abram Joseph Ryan
Ef the way a man lights out of this world
Helps fix his heft for the other sp'ere,
I reckon my old friend Golyer's Ben
Will lay over lots of likelier men
For one thing he done down here.
"Golyer" by John Hay
I have taught your young lips the good words to say over,
Which form the petition we call the Lord's Prayer,
And now let me help my dear child to discover
The meaning of all the good words that are there.
"On The Lord's Prayer" by Charles Lamb
And I mean to lay my bones over yonder by the lake--
By and by when I have nothing else to do--
And I'll give the "chicks" the farm, and I know for Mollie's sake,
That the good and gracious Lord will help 'em through.
"The Pioneer" by Hanford Lennox Gordon
Kindly brothers of the East,--
Thou great tiaraed priest,
Thou sanctified Rienzi of Rome and of the earth,--
Or thou who bear'st control
Over golden Istambol,
Who felt for our misfortunes and helped us in our dearth,--
"Ireland" by Denis Florence MacCarthy

### In news:

Marell Booker scored from five yards out on 3rd-and-goal to help lead the Red Team to a 16-10 overtime victory over the White Team in Western Kentucky University's annual spring football game Saturday at L.T.
A man in Canada helped to save the life of an elderly woman over a thousand miles away.
Jefferson helps Jazz orchestrate victory over Los Angeles Lakers.
Continuous innovations in the implants' technologies have helped the orthopedic industry to maintain a steady growth of 7-10% over the last decade.
Ohio State's electric quarterback rushed for 149 yards and a touchdown to help the Buckeyes score a 52-49 win over Indiana on Saturday OSU moved to 7-0 (3-0 Big Ten) while IU (2-4, 0-3) dropped its fourth straight game.
'Hey, Lucinda, can we get a little help over here ?'.
I Need Some Help Over Here .
Over the past four years, hundreds of volunteers have stepped up to help maintain landscaping in Bend's public rights of way.
David Beckham leads Galaxy to a win Posted: Aug 25, 2012 David Beckham helps his side take the 5-2 victory over Isidro Metapan.
I like the defensive adjustments he's helped implement over the last few seasons.
Pour remaining caramel over pepitas , using an offset spatula to help spread caramel if necessary.
Over the past week, we've heard about dozens of worthwhile ways to help the victims of Japan's earthquake and tsunami, but one recent announcement caught our eyes—and nose.
Health Record tailored to helping people over 50 better control their health information.
The group has collected over 9,000 quilts that were donated to help raise money for Alzheimer's research.
Three or four things will occur over the next few years that'll help farmers in their continuing battle to control resistant pigweed .
***

### In science:

Wootters during the joint workshop of the ESF-QIT programme and the Newton institute in Cambridge, July 1999. S.V. would also like to thank Ben Tregenna for some very helpful discussions, and Daniel Jonathan for demonstrating a great deal of patience over the past year.
Ordering States with Entanglement Measures
Averaging over u with the help of the standard replica trick gives the quartic in ψ term in the action.
Nonlinear $\sigma$ model for disordered superconductors
Over the past decade, balloon measurements have provided a wealth of new information to help address questions about cosmic rays from traditional astrophysical sources as well as about exotic cosmic rays.
Particle Astrophysics and Cosmology: Cosmic Laboratories for New Physics (Summary of the Snowmass 2001 P4 Working Group)
Using similar presentation for diagram (2) we calculate the integral over the photon momentum q in Eq.(2) for these two diagrams with the help of Eqs.(A2).
Z alpha expansion for self-energy radiative corrections to parity nonconservation in atoms
The fact that m is not a probability measure can be helped, at least over N, using compactiﬁcation as above, and extending the notion of randomness tests.
Uniform test of algorithmic randomness over a general space
***


text:
# Search by Topic

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### How Many Solutions?

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Find all the solutions to the this equation.

### Rational Roots

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Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.

### Janine's Conjecture

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Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .

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Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.

### Diverging

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Show that for natural numbers x and y if x/y > 1 then x/y>(x+1)/(y+1}>1. Hence prove that the product for i=1 to n of [(2i)/(2i-1)] tends to infinity as n tends to infinity.

### Three Ways

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If x + y = -1 find the largest value of xy by coordinate geometry, by calculus and by algebra.

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Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.

### Target Six

##### Stage: 5 Challenge Level:

Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.

### Pair Squares

##### Stage: 5 Challenge Level:

The sum of any two of the numbers 2, 34 and 47 is a perfect square. Choose three square numbers and find sets of three integers with this property. Generalise to four integers.

### Mechanical Integration

##### Stage: 5 Challenge Level:

To find the integral of a polynomial, evaluate it at some special points and add multiples of these values.

### Square Mean

##### Stage: 4 Challenge Level:

Is the mean of the squares of two numbers greater than, or less than, the square of their means?

### Polynomial Relations

##### Stage: 5 Challenge Level:

Given any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials. Try this one.

### Archimedes and Numerical Roots

##### Stage: 4 Challenge Level:

The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?

### AMGM

##### Stage: 4 Challenge Level:

Can you use the diagram to prove the AM-GM inequality?

### Proof Sorter - Quadratic Equation

##### Stage: 4 and 5 Challenge Level:

This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.

### Leonardo's Problem

##### Stage: 4 and 5 Challenge Level:

A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?

### Interpolating Polynomials

##### Stage: 5 Challenge Level:

Given a set of points (x,y) with distinct x values, find a polynomial that goes through all of them, then prove some results about the existence and uniqueness of these polynomials.

##### Stage: 5 Challenge Level:

Find all positive integers a and b for which the two equations: x^2-ax+b = 0 and x^2-bx+a = 0 both have positive integer solutions.

### Golden Eggs

##### Stage: 5 Challenge Level:

Find a connection between the shape of a special ellipse and an infinite string of nested square roots.

### Unit Interval

##### Stage: 4 and 5 Challenge Level:

Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?

### Thousand Words

##### Stage: 5 Challenge Level:

Here the diagram says it all. Can you find the diagram?

### The Clue Is in the Question

##### Stage: 5 Challenge Level:

This problem is a sequence of linked mini-challenges leading up to the proof of a difficult final challenge, encouraging you to think mathematically. Starting with one of the mini-challenges, how. . . .

### Pythagoras Proofs

##### Stage: 4 Challenge Level:

Can you make sense of these three proofs of Pythagoras' Theorem?

### Mediant

##### Stage: 4 Challenge Level:

If you take two tests and get a marks out of a maximum b in the first and c marks out of d in the second, does the mediant (a+c)/(b+d)lie between the results for the two tests separately.

### Proofs with Pictures

##### Stage: 5

Some diagrammatic 'proofs' of algebraic identities and inequalities.

### Where Do We Get Our Feet Wet?

##### Stage: 5

Professor Korner has generously supported school mathematics for more than 30 years and has been a good friend to NRICH since it started.

### Telescoping Functions

##### Stage: 5

Take a complicated fraction with the product of five quartics top and bottom and reduce this to a whole number. This is a numerical example involving some clever algebra.

### Look Before You Leap

##### Stage: 5 Challenge Level:

Relate these algebraic expressions to geometrical diagrams.

### Transitivity

##### Stage: 5

Suppose A always beats B and B always beats C, then would you expect A to beat C? Not always! What seems obvious is not always true. Results always need to be proved in mathematics.

### More Sums of Squares

##### Stage: 5

Tom writes about expressing numbers as the sums of three squares.

### Multiplication Square

##### Stage: 4 Challenge Level:

Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?

##### Stage: 5 Challenge Level:

Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.

### Tetra Inequalities

##### Stage: 5 Challenge Level:

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.

### An Introduction to Number Theory

##### Stage: 5

An introduction to some beautiful results of Number Theory

### Without Calculus

##### Stage: 5 Challenge Level:

Given that u>0 and v>0 find the smallest possible value of 1/u + 1/v given that u + v = 5 by different methods.

### The Triangle Game

##### Stage: 3 and 4 Challenge Level:

Can you discover whether this is a fair game?

### Impossible Sandwiches

##### Stage: 3, 4 and 5

In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.

### Euclid's Algorithm II

##### Stage: 5

We continue the discussion given in Euclid's Algorithm I, and here we shall discover when an equation of the form ax+by=c has no solutions, and when it has infinitely many solutions.

### Converse

##### Stage: 4 Challenge Level:

Clearly if a, b and c are the lengths of the sides of an equilateral triangle then a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true?

### Modulus Arithmetic and a Solution to Dirisibly Yours

##### Stage: 5

Peter Zimmerman from Mill Hill County High School in Barnet, London gives a neat proof that: 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.

### Continued Fractions II

##### Stage: 5

In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).

### Diophantine N-tuples

##### Stage: 4 Challenge Level:

Can you explain why a sequence of operations always gives you perfect squares?

### Square Pair Circles

##### Stage: 5 Challenge Level:

Investigate the number of points with integer coordinates on circles with centres at the origin for which the square of the radius is a power of 5.

### Composite Notions

##### Stage: 4 Challenge Level:

A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.

### Number Rules - OK

##### Stage: 4 Challenge Level:

Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...

### Mouhefanggai

##### Stage: 4

Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.

### Euler's Formula and Topology

##### Stage: 5

Here is a proof of Euler's formula in the plane and on a sphere together with projects to explore cases of the formula for a polygon with holes, for the torus and other solids with holes and the. . . .

### A Computer Program to Find Magic Squares

##### Stage: 5

This follows up the 'magic Squares for Special Occasions' article which tells you you to create a 4by4 magicsquare with a special date on the top line using no negative numbers and no repeats.

### Basic Rhythms

##### Stage: 5 Challenge Level:

Explore a number pattern which has the same symmetries in different bases.


text:
statgenGxE is an R package providing functions for Genotype by Environment (GxE) analysis for data of plant breeding experiments.

The following types of analysis can be done using statgenGxE:

• Mixed model analysis of GxE table of means
• Finlay-Wilkinson Analysis
• AMMI Analysis
• GGE Analysis
• Identifying mega environments
• Stability measures
• Modeling of heterogeneity of genetic variances and correlations

statgenGxE has extensive options for summarizing and visualizing the results of the analyses. For a full overview of all options it is best to read the vignette

## Installation

• Install from CRAN:
install.packages("statgenGxE")
• Install latest development version from GitHub (requires remotes package):
remotes::install_github("Biometris/statgenGxE", ref = "develop", dependencies = TRUE)


text:
Sandbox

# Recreate modulo [cancelled]

+0
−0

Recreate the modulo operation via function or program.

Mechanics:

• The modulo operator involves division and returns the remainder.
• For a function, use 2 variables as foo(a,b) such that $x \mod y$.
• For a program, take 2 variables as input where a b is $x \mod y$.
• Use standard division and return the remainder without using a built-in modulo operator.

Rules:

• Standard loopholes are forbidden.
• Shortest code in bytes wins.
Why does this post require moderator attention?


text:
# Transitive subgroup of symmetric group $S_n$ containing an $(n-1)$-cycle and a transposition

Suppose $G$ is a transitive subgroup of $S_n$ such that it there exist $\sigma, \tau \in G$ such that $\sigma$ is an $(n-1)$-cycle and $\tau$ is a transposition. Prove that $G = S_n$.

I just don't understand how to mathematically use the transitive nature of the subgroup.

• $\sigma$ is an $n$-cycle I think. – Yilong Zhang Jun 23 '15 at 9:26
• Sorry, I missed "transitive". – Yilong Zhang Jun 23 '15 at 9:28
• @YilongZhang No, it is $n-1$. Just out of curiosity, can you prove the statement if it is $n$? – MathManiac Jun 23 '15 at 9:28
• @YilongZhang, when $n=4$ the $2$-Sylows of $S_4$ contain a transposition and a $4$-cycle, but they are not the whole group. Think of $\langle(1,2,3,4),(1,3)\rangle$ (However the property is true if $n$ is prime). – Clément Guérin Jun 23 '15 at 9:30
• @ClémentGuérin Thanks a lot for your comment. I was just under the impression that every pair of $n$-cycle and transposition can generate $S_n$. – Yilong Zhang Jun 23 '15 at 9:51

Take your subgroup $G$, up to the study of a conjugate $G$ you can assume that the $n-1$-cycle of $G$ is $c=(2,...,n)$. Now if $\tau$ is a transposition in $G$ then $\tau=(i,j)$ with $i\neq j$. Take $\sigma_i\in G$ such that $\sigma_i(i)=1$ (this is where I use the transitivity).
Then I claim that $\sigma_i\tau\sigma_i^{-1}=(1,k)$ where $k\geq 2$. Now you have $\tau_0=(1,k)$ where $k\geq 2$ and $c=(2,...,n)$ in $G$, I claim that :
$$\{c^s\tau_0c^{-s}|s\in\mathbb{N}\}=\{(1,2),...,(1,n)\}$$
This shows that $G$ will contain $(1,2),...,(1,n)$ (because $c^s\tau_0c^{-s}\in G$, $c$ and $\tau_0$ are in $G$). Now it is easy to see that $\{(1,2),...,(1,n)\}$ is a generating set of $S_n$, hence $G=S_n$.


text:
# Explanation of path integrals

1. Nov 24, 2004

### redshift

What exactly does a path integral measure? Is it area between the ends/bounds of the line? Or is it the length of the line? Just started complex analysis and am comletely confused by this.

2. Nov 26, 2004

### shmoe

It's not the length, you can think of the dz as keeping track of direction and distance you move. The integral $\int_{\gamma}dz$ along a path $$z=\gamma(t),\ t\in$a,b$$$ is simply $\int_{a}^{b}\gamma '(t)dt=\gamma(b)-\gamma(a)$ so you actually get the distance and direction your end point is from your start point, not the total distance you've travelled. This integral doesn't even keep track of how the path made it from start to finish, nor how fast it went.


text:
Free Version
Difficult

Refraction in Unknown Material

APPH12-WRLKBZ

Consider a beam of light traveling through water toward an unknown material. The beam of light strikes the surface, as shown in the image, at an incidence angle of ${ 52.1}^{ \circ }$. Although the material is unknown, measurements show that light travels at a speed five times less the speed of light in water.

Does the beam of light refract (cross the boundary) in the unknown material?

A

Yes, by an angle of ${ 11.5 }^{ \circ }$.

B

No, the critical angle is ${ 11.5 }^{ \circ }$ and the incidence angle is greater than the critical angle. Thus total internal reflection occurs and no light refracts.

C

Yes, by an angle of ${ 90.1 }^{ \circ }$.

D

Yes, by an angle of ${ 9.1 }^{ \circ }$.


text:
Math Help - Derivative

1. Derivative

2. Originally Posted by biermann33
From the chain rule, $F'(x) = 5x^4 f'(x^5)$.


text:
In this article, you will learn to write information from your model in a text file using the MATLAB command fprintf.

Let’s say you have a model that you want to check because there have been modifications to it, and you need to make sure it runs. It would be nice to be able to check if the model is running and write the results in a text document automatically so that every time you want to know, you can just run this script. (If you don’t care about the explanation, the script is at the end of this article.)

Assuming you have the following model:

First, let’s check if the model can run:

modelName = 'myVerySimpleModel';
open(modelName);

try sim([modelName '.slx'])
modelRun = 1;
catch
modelRun = 0;
end

The sim command is explained here.

Here, we are running the Simulink model. If it runs modelRun is 1. If it generates an error, modelRun is 0.

Then, we can close the model and write the results in a .txt document.

modelName = 'myVerySimpleModel';
open(modelName);

try sim([modelName '.slx'])
modelRun = 1;
catch
modelRun = 0;
end
close_system(modelName)

% open the file
textFileName = 'checkReport.txt';
fid = fopen(textFileName, 'at');
if modelRun
% \n is to go to the next line
fprintf(fid, 'The model runs.\n');
else
% char(39) is used as a single quotation mark
fprintf(fid, ['The model doesn' char(39) 't run.\n']);
end
% close the file
fclose(fid);

Let’s examine how we wrote content:

• We use the fopen MATLAB command to open the file (‘a’ is for appending data to the end of the file, and ‘t’ is for opening the file in text mode).
• We write the content using the fprintf MATLAB command.
• We close the file using the fclose MATLAB command.

After running this script, a file named “checkReport.txt” is generated:

Let’s add an undefined variable “x” as the simulation time to generate an error:

Finally, after running the script again, we get:


text:
SOCR EduMaterials ModelerActivities MixtureModel 1

SOCR Modeler Activities - SOCR Mixture Model Fitting Activity

This is a SOCR Activity that demonstrates random sampling and fitting of mixture models to data

SOCR Mixture-Model Distribution

The 1D SOCR mixture-model distribution enables the user to specify the number of mixture Normal distributions and their parameters (means and standard deviations). This applet demonstrates how unimodal-distributions come together as building-blocks to form the backbone of many complex processes. In addition, this applet allows computing probability and critical values for these mixture distributions, and enables inference on such complicated processes. Extensive demonstrations of mixture modeling in 1D, 2D and 3D are available on the SOCR EM Mixture Modeling page. The figure below shows one such example of a tri-modal mixture of 4 Normal distributions.

Data Generation

You typically have investigator-acquired data that you need to fit a model to. In this case we will generate the data by randomly sampling using the SOCR resource. Go to the SOCR Modeler and select the Data Generation tab from the right panel.
• Now, click the Raw Data check-box in the left panel, select Laplace Distribution (or any other distribution you want to sample from), choose the sample-size to be 100 (keep the center, Mu (μ = 0)) and click Sample. Then go to the Data tab, in the right panel. There you should see the 100 random Laplace observations stored as a column vector.
• Next, go back to the Data Generation tab from the right panel and change the center of the Laplace distribution (set Mu=20, say). Click Sample again and you will see the list of randomly generated data in the Data tab expand to 200 (as you just sampled another set of 100 random Laplace observations).

Exploratory Data Analysis (EDA)

Go to the Data tab and select all observations in the data column (use CTR-A, or mouse-copy). Then open another web browser and go to SOCR Charts. Choose HistogramChartDemo2, say, clear the default data (Data tab) and paste (CTR-V or mouse paste-in) in the first column the 200 observations that you sampled in the SOCR Modeler Data Generator (above). Then you need to map the values - go to the Mapping tab, select the first column, where you pasted the data (C1), and click XValue. This will move the C1 column label from the left bin to the bottom-right bin. Finally, click Update Chart, on the top, and go to the Graph tab to see your histogram of the 200 (bimodal) Laplace observations. Notice, that you can change the width of the histogram bin to clearly see the bi-modality of the distribution of these 200 measurements. Of course, this is due to the fact that we sampled from two distinct Laplace distributions, one with mean of zero and the second with mean of 20.0.

Model Fitting

Now go back to the SOCR Modeler browser (where you did the data sampling). Choose Mixed-Model-Fit from the drop-down list in the left panel.
• We will now try to fit a 2-component mixture of Gaussian (Normal) distributions to this Bimodal Laplace distribution (of the generated sample). You may need to click the Re-Initialize button a few times. The Expectation-Maximization algorithm used to estimate the mixture distribution parameters is unstable and will produce somewhat different results for different initial conditions. Hence, you may need to re-initialize the algorithm a few times until a visually satisfactory result is obtained.
• Notice the quantitative results of this mixture model fitting protocol (in the Results panel). Recall that we sampled 100 observations from Laplace distribution with mean of zero (not Normal Gaussian, which we could also have done and the fit would have been much better, of course) and then another 100 observations from Laplace distribution with mean = 20.0. In this case, the reported estimates of the means of the two Gaussian mixtures are 0 and 22 (pretty close to the original/theoretical means). We could have also fit in a mixture of 3 (or more) Gaussian mixture components, if we had a reason to believe that the data distribution is tri- (or higher-)modal, and therefore, requires a multi-modal mixture fit.
• There are statistical tests added to assess:
• how statistically significant are the mean values of any pair of Gaussian models $\left \{N(\mu_1,\sigma_1^2), N(\mu_2,\sigma_2^2)\right\}$ in the Mixture Distributions. Normal Z tests are use for this assessment $\left \{ Z_o=\frac{\mu_1-\mu_2}{\sqrt{\frac{\sigma_1^2}{N_1}+ \frac{\sigma_2^2}{N_2}}} \sim N(0,1^2)\right \}$.
• And how good the overall mixture model fir tot he data is, using the Kolmogorov-Smirnoff Test.

Caution

You may need to properly set the values of the sliders on the top of your Graph tab, in the right panel, so that you can see the entire graph of the histogram and the models fit to the data. Also, the random data you generate and the EM algorithm are stochastic and you can not expect to get exactly the same results and charts as reported in this SOCR activity. Everyone that tries to replicate these steps will obtain different results, however, the principles we demonstrate here are indeed robust.


text:
Algebra 2 (1st Edition)

Published by McDougal Littell

Chapter 10 Counting Methods and Probability - 10.1 Apply the counting Principles and Permutations - 10.1 Exercises - Skill Practice - Page 687: 46

360

Work Step by Step

YELLOW has 6 letters, 2 of which are repeated $$6!/2!=\boxed{360}$$

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:
Physics Help Forum lissajous curves/figures

 General Physics General Physics Help Forum

 Jul 11th 2019, 07:11 AM #1 Junior Member   Join Date: Jun 2019 Posts: 1 lissajous curves/figures I have found myself in a position where i need to learn more about lissajous curves/figures. I have come across them before but only ever setup equipment to show the different types of patterns. I now need to understand more on how to use and interpret a lissajous in a practical sense as in how to make sense of the patterns when hooked up to a circuit. Can anyone explain how a lissajous is used and how to make sense of the images produced. Or maybe point me in a direction that may help me to learn more about them with out going too glass eyed?
 Jul 11th 2019, 09:31 AM #2 Senior Member   Join Date: Aug 2010 Posts: 418 What have you already looked up about them? It has been a very long time since I dealt with Lissajous figures but a Google search gave several websites. This one: https://www.bing.com/search?q=lissaj...&setlang=en-US defines them as "graph of a system of parametric equations x=Asin(at+δ), y=Bsin(bt)" (I hadn't remembered the "$\displaystyle +\delta$", I just used x= A sin(at) and y= B sin(bt) so the curves all passed through (0, 0).) topsquark likes this.
 Jul 11th 2019, 10:26 AM #3 Senior Member     Join Date: Jun 2016 Location: England Posts: 932 Alternatively try: "any of a number of characteristic looped or curved figures traced out by a point undergoing two independent simple harmonic motions at right angles with frequencies in a simple ratio." This gives a hint toward the type of usages I have seen them used for, comparing the characteristics of two (related) signals. If you feed a sinusoidally varying signal into the X channel of an oscilloscope and a different sinusoidally varying signal into the Y channel of the oscilloscope then it will display Lissajous figures. The pattern of the figure has a direct relationship with the frequency differences of the signals topsquark likes this. __________________ ~\o/~

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post anani Waves and Sound 2 Dec 26th 2013 07:08 AM Adesinapaul Physics 0 May 25th 2012 01:24 AM Black Kinematics and Dynamics 1 Jan 13th 2010 06:26 PM Morgan82 General Physics 1 Sep 22nd 2008 03:44 PM


text:
Outlook: Community Healthcare Trust Incorporated Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Sell
Time series to forecast n: 13 Jan 2023 for (n+6 month)
Methodology : Modular Neural Network (Market Direction Analysis)

## Abstract

Community Healthcare Trust Incorporated Common Stock prediction model is evaluated with Modular Neural Network (Market Direction Analysis) and Lasso Regression1,2,3,4 and it is concluded that the CHCT stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Sell

## Key Points

1. Prediction Modeling
2. Operational Risk
3. How can neural networks improve predictions?

## CHCT Target Price Prediction Modeling Methodology

We consider Community Healthcare Trust Incorporated Common Stock Decision Process with Modular Neural Network (Market Direction Analysis) where A is the set of discrete actions of CHCT stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Lasso Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Market Direction Analysis)) X S(n):→ (n+6 month) $\begin{array}{l}\int {r}^{s}\mathrm{rs}\end{array}$

n:Time series to forecast

p:Price signals of CHCT stock

j:Nash equilibria (Neural Network)

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## CHCT Stock Forecast (Buy or Sell) for (n+6 month)

Sample Set: Neural Network
Stock/Index: CHCT Community Healthcare Trust Incorporated Common Stock
Time series to forecast n: 13 Jan 2023 for (n+6 month)

According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Sell

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Grey to Black): *Technical Analysis%

## IFRS Reconciliation Adjustments for Community Healthcare Trust Incorporated Common Stock

2. An entity applies IAS 21 to financial assets and financial liabilities that are monetary items in accordance with IAS 21 and denominated in a foreign currency. IAS 21 requires any foreign exchange gains and losses on monetary assets and monetary liabilities to be recognised in profit or loss. An exception is a monetary item that is designated as a hedging instrument in a cash flow hedge (see paragraph 6.5.11), a hedge of a net investment (see paragraph 6.5.13) or a fair value hedge of an equity instrument for which an entity has elected to present changes in fair value in other comprehensive income in accordance with paragraph 5.7.5 (see paragraph 6.5.8).
3. For purchased or originated credit-impaired financial assets, expected credit losses shall be discounted using the credit-adjusted effective interest rate determined at initial recognition.
4. An example of a fair value hedge is a hedge of exposure to changes in the fair value of a fixed-rate debt instrument arising from changes in interest rates. Such a hedge could be entered into by the issuer or by the holder.

*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.

## Conclusions

Community Healthcare Trust Incorporated Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Community Healthcare Trust Incorporated Common Stock prediction model is evaluated with Modular Neural Network (Market Direction Analysis) and Lasso Regression1,2,3,4 and it is concluded that the CHCT stock is predictable in the short/long term. According to price forecasts for (n+6 month) period, the dominant strategy among neural network is: Sell

### CHCT Community Healthcare Trust Incorporated Common Stock Financial Analysis*

Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBa1Baa2
Balance SheetBa3B2
Leverage RatiosCBa3
Cash FlowBa3Baa2
Rates of Return and ProfitabilityBaa2Baa2

*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?

### Prediction Confidence Score

Trust metric by Neural Network: 79 out of 100 with 866 signals.

## References

1. R. Sutton and A. Barto. Introduction to reinforcement learning. MIT Press, 1998
2. Breiman L. 2001a. Random forests. Mach. Learn. 45:5–32
3. Scholkopf B, Smola AJ. 2001. Learning with Kernels: Support Vector Machines, Regularization, Optimization, and Beyond. Cambridge, MA: MIT Press
4. Belloni A, Chernozhukov V, Hansen C. 2014. High-dimensional methods and inference on structural and treatment effects. J. Econ. Perspect. 28:29–50
5. Mnih A, Kavukcuoglu K. 2013. Learning word embeddings efficiently with noise-contrastive estimation. In Advances in Neural Information Processing Systems, Vol. 26, ed. Z Ghahramani, M Welling, C Cortes, ND Lawrence, KQ Weinberger, pp. 2265–73. San Diego, CA: Neural Inf. Process. Syst. Found.
6. G. Theocharous and A. Hallak. Lifetime value marketing using reinforcement learning. RLDM 2013, page 19, 2013
7. Chipman HA, George EI, McCulloch RE. 2010. Bart: Bayesian additive regression trees. Ann. Appl. Stat. 4:266–98
Frequently Asked QuestionsQ: What is the prediction methodology for CHCT stock?
A: CHCT stock prediction methodology: We evaluate the prediction models Modular Neural Network (Market Direction Analysis) and Lasso Regression
Q: Is CHCT stock a buy or sell?
A: The dominant strategy among neural network is to Sell CHCT Stock.
Q: Is Community Healthcare Trust Incorporated Common Stock stock a good investment?
A: The consensus rating for Community Healthcare Trust Incorporated Common Stock is Sell and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of CHCT stock?
A: The consensus rating for CHCT is Sell.
Q: What is the prediction period for CHCT stock?
A: The prediction period for CHCT is (n+6 month)


text:
## anonymous one year ago Solve 1+3/y/5/y^2

• This Question is Open
1. anonymous

2. anonymous

Y=1/20 i guess

3. jiteshmeghwal9

$\frac{\frac{1+3}{y}}{\frac{5}{y^2}}$is this ur question ?

4. anonymous

Yes i guess but how did you do it ??

5. jiteshmeghwal9

latex

6. anonymous

What's that ??

7. anonymous

$1+3\div y \div 5 \div y^2$ is it that huh ?

8. anonymous

now it can be right as $(1+3)y^2 \div 5(y)$

9. anonymous

now it also can be right as (1+3)y/5=4y/5

10. anonymous

so y=5/4

11. Plasmataco

no. 4y/5. It is an expression so you don't have an equal sign

12. anonymous

y=5/4

13. anonymous

Mathlete yes that is the problem. How is that solved