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# Choosing a draw call when not reusing vertex data

Up until now I've always used glDrawElements, and have my vertex and index buffers bound. From what I've read, glDrawElements is generally the way to go, and it works fine. However, I'm starting a new project and am questioning whether this is still the best fit.

For the current project, I'm recomposing a 2D scene from scratch every frame. The player's character runs around, and I only draw the part of the world that is visible, generating 2D quads from the map tiles and building the vertex/index buffers.

Given that (from my understanding) glDrawElements works best when you upload your VBOs and make multiple draw calls against them, wouldn't it be better to use something else in this case, such as glDrawArrays? I've generally read that glDrawArrays is slower, but given that I'm having to push the vertex data every frame anyway, that might not be true here.

I don't want to switch draw calls and assume I made the right choice when I could in fact be using it improperly to begin with, invalidating any performance improvements I see.

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# Sigma algebra of stochastic process

A stochastic process is a collection $(X_t)_{t\in T}$ of random variables from a prob. space $(\Omega,\mathcal{F},P)$ to some measurable space $(E,\mathcal{E})$. Now, in order to understand the whole process as a 'single' random variable one look at it as a map $X:\Omega\to E^T$, where $E^T=\prod_{t\in T}E$ is the set of all functions from $T$ to $E$.

$E^T$ is usually equipped with the product sigma algebra $\mathcal{E}^T$ which is produced by the one-dimensional cylinder sets, i.e. $\mathcal{E}^T=\sigma(\pi^{-^1}(A): A\in\mathcal{E}, t\in T)$ (is is the same as the sigma algebra produced by all finite-dimensional cylinders?), where $\pi_t$ is the projection map from $E^T$ to $E$. Now, if we are interested just in a subspace $S\subset E^T$, then it is often the case that $S\notin\mathcal{E}^T$.

For example take $T=[0,1]$, $E=\mathbb{R}$, $\mathcal{E}=\mathcal{B}_{\mathbb{R}}$ and $C[0,1]$ as a subspace of $\mathbb{R}^{[0,1]}$, then $C[0,1]\notin\mathcal{E}^T$.

By looking at $\mathcal{S}=S\cap \mathcal{E}^T=\{S\cap A:A\in\mathcal{E}^T\}$ one can avoid this problem. Now my main question: If our subspace $S$ is a metric space it is natural to look at the Borel $\sigma$-algebra $\mathcal{B}_S$. Is it always the case that $\mathcal{S}=\mathcal{B}_S$? I think that we need some additional property like separability, but I cannot prove it.

• Take $S=E^T$, endowed with the discrete metric. – Michael Greinecker Apr 15 '18 at 9:20
• And if we assume that $E$ is separable with Borel sigma algebra $\mathcal{B}_E$? – aaaaaaaa Apr 16 '18 at 13:04
• The question is rather what $S$ should be. It works for $S=C[0,1]$, but there is no reason it should work for every metric space. – Michael Greinecker Apr 16 '18 at 14:57
• OK, thank you. Does it also work for the space of cadlag functions $D[0,1]$ if it is equipped with a "right" metric? – aaaaaaaa Apr 16 '18 at 17:47
• I'm not sure. The reason why it works for $C[0,1]$ is that every function in $C[0,1]$ is determined by countably many coordinates. I don't know if this argument transfers to cadlag functions. – Michael Greinecker Apr 16 '18 at 17:55


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# Showing Lyapunov (Lindeberg) condition holds for sum of independent bernoulli distribution with poisson tail probability

I'm trying to find an asymptotic distribution of the follwing random variable

$$Z_n=\sum_{i=0}^n Y_i$$ where $$Y_i = I[T_i with $$T_i \text{~} Gamma(i, \lambda)$$.

Here $$t$$ is a fixed number.

My initial trial was using Linderberg CLT by using the Lyapunov CLT ( or Linderberg CLT). First I noticed that by using integral by parts,

$$E[Y_i]=P[Y_i=1]=P[T_i

which is the tail probabilty of the poisson random variable with mean $$t/\lambda$$.

Let $$p_i:=E[Y_i], R_i:=Y_i-p_i, S_n:=\sum_{i=1}^nR_i, \sigma_n^2:=Var(S_n)=\sum_{i=1}^np_i(1-p_i)$$.

I'm trying to show that $$R_i$$ satisfies the Lyapunov condition where

$$\sigma_n^{-(2+\delta)}\sum_{i=1}^nE[|R_i|^{(2+\delta)}]\to0\ \text{as n\to \infty }$$.

With $$\delta=1$$,

I get

$$E[|R_i|^3]=(1-p_i)^3p_i + p_i^3(1-p_i)^3\le p_i(1-p_i)$$ therefore

$$\sigma_n^{-3}\sum_{i=1}^nE[|R_i|^{3}] \le \sigma_n^{-1}.$$

As long as I can show that $$\sigma_n \to \infty \ \text{as n\to \infty }$$

I'm done with the proof. But I'm stuck with showing the divergence of $$\sigma_n$$.

In other words, I need to show that

$$\sum_{i=1}^n p_i(1-p_i) \to \infty \ \text{as n\to \infty }.$$

I sincerely ask to help me with the proof of the last statement. Or Am i doing something wrong?

Thank you.

• What parametrization are you using for the Gamma distribution? Shape - scale or shape-rate? It would be good to always clarify that. – Alecos Papadopoulos Apr 22 '20 at 16:34

Your treatment appears sound.

The following is improvisation, so most likely it is inefficient. It transforms your question to a simpler one, that of convergence of a sequence.

I will assume $$\lambda =1$$ to simplify notation. The assumption is inconsequential.

The symbol $$p_i$$ represents the cumulative distribution function of a Gamma distribution with integer shape parameter equal to $$i$$, i.e. of an Erlang distribution. We have

$$p_i = e^{-t}\sum_{m=i}^\infty \frac{t^m}{m!}, \;\;p_{i+1} = e^{-t}\sum_{m=i+1}^\infty \frac{t^m}{m!},$$

$$p_{i+1} = p_i - \pi_i,\;\;\; \pi_i \equiv e^{-t}\frac{t^i}{i!},\;\;\; \pi_i\to 0 \;\text {as}\; i\to \infty$$

By the monotone converge Theorem, it also follows that $$p_i$$ will go to zero at the limit, and one can deduce that $$p_i(1-p_i)$$ will go to zero too.

So

$$p_{i+1}\left(1-p_{i+1}) = ( p_i - \pi_i ) (1- p_i + \pi_i) = p_i(1-p_i) -\pi_i(1-2p_i+\pi_i\right)$$

Using this we can decompose the sum we are interested in in two different ways, and we will use both: \begin{align} p_2(1-p_2) &= p_1(1-p_1) - \pi_1(1-2p_1+\pi_1) \\ p_3(1-p_3) & = p_2(1-p_2) - \pi_2(1-2p_2+\pi_2) \\ &= p_1(1-p_1) - \pi_1(1-2p_1+\pi_1) - \pi_2(1-2p_2+\pi_2)\\ etc \end{align} Using the first equality, i.e. simply going one index value back we get

$$\sum_{i=1}^n p_i(1-p_i) = p_1(1-p_1) + \sum_{i=1}^{n-1} p_i(1-p_i) - \sum_{i=1}^{n-1}\pi_i(1-2p_i+\pi_i)$$

and canceling out and rearranging, we get the expression

$$p_1(1-p_1) = p_n(1-p_n) + \sum_{i=1}^{n-1}\pi_i(1-2p_i+\pi_i) \tag{1}$$

Now if we decompose every $$p_i(1-p_i)$$ element down to an expression involving $$p_1(1-p_1)$$, as we indicatively did for $$p_3(1-p_3)$$ previously we get \begin{align} \sum_{i=1}^n p_i(1-p_i) &= np_1(1-p_1) - (n-1)\pi_1(1-2p_1+\pi_1) \\ &- (n-2)\pi_2(1-2p_2+\pi_2)-...-\pi_{n-1}(1-2p_{n-1}+\pi_{n-1}) \tag{2} \end{align}

Use expression $$(1)$$ to substitute for $$p_1(1-p_1)$$ in the right-hand-side of expression $$(2)$$:

\begin{align} \sum_{i=1}^n p_i(1-p_i) &= n\cdot\Big[p_n(1-p_n) + \sum_{i=1}^{n-1}\pi_i(1-2p_i+\pi_i)\Big] - (n-1)\pi_1(1-2p_1+\pi_1) \\ &- (n-2)\pi_2(1-2p_2+\pi_2)-...-\pi_{n-1}(1-2p_{n-1}+\pi_{n-1}) \\ \\\implies \sum_{i=1}^n p_i(1-p_i) &= np_n(1-p_n) + \sum_{i=1}^{n-1}\Big(i\cdot \pi_i(1-2p_i+\pi_i)\Big) \tag{3} \end{align}

We are interested in

$$\text{lim}_{n\to \infty}\sum_{i=1}^n p_i(1-p_i) = \text{lim}_{n\to \infty}\Big[np_n(1-p_n)\Big] + \sum_{i=1}^{\infty}\Big(i\cdot \pi_i(1-2p_i+\pi_i)\Big) \tag{4}$$

Use selectively the expression for $$\pi_i$$ and apply the Ratio Test for the convergence of the infinite series:

$$\text{lim}_{i\to \infty}\left(\frac{(i+1)\cdot e^{-t}\frac{t^{i+1}}{(i+1)!}(1-2p_{i+1}+\pi_{i+1})}{i\cdot e^{-t}\frac{t^i}{i!}(1-2p_i+\pi_i)}\right)$$

$$t\cdot\text{lim}_{i\to \infty}\left(\frac{(i+1)\cdot (1-2p_{i+1}+\pi_{i+1})}{i\cdot(i+1) (1-2p_i+\pi_i)}\right) \to 0 <1.$$

...because the terms in parentheses go to unity. So the infinite series in expression $$(4)$$ is absolutely convergent. Therefore, whether the sum of interest converges or diverges hints on the limit of the sequence $$\{np_n(1-p_n)\}$$.

Comment: after obtaining that the series $$\{p_i(1-p_i)\}$$ is monotonically declining, it would be trivial to realize that

$$np_n(1-p_n) < \sum_{i=1}^n p_i(1-p_i)$$ and so that if we could determine that $$\{np_i(1-p_i)\}$$ diverges the sum of interest would also diverge. But if we were to find out that $$\{np_i(1-p_i)\}$$ converges, we would not be able to conclude on the divergence or convergence of the sum of interest. All the previous work was done to show that both divergence and convergence of it depend on what happens with $$\{np_n(1-p_n)\}$$:What is now needed is to examine whether $$\{np_n(1-p_n)\}$$ is Cauchy or not.


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# Which is smallest distance measurement which makes sense talking about [duplicate]

This might seem a silly question, but here it goes:

Of all possible values for a measure of distance, which is the smallest that makes sense talking about?

I mean, I could talk about $10^{-100000}$ meters but in real life this value is smaller than anything we know (i think), so it wouldn't make any sense talking about this value, now would it? Initially I thought about Planck length, but I'm not sure.

Please not I'm not referring to mathematics, any number can be of use in mathematics. But in real life, small values like $10^{-100000}$ meters don't make sense.

So, what is smallest distance value that makes sense talking about?

Thanks.

-

## marked as duplicate by Brandon Enright, John Rennie, Qmechanic♦Mar 12 '14 at 10:23

Possible duplicates: physics.stackexchange.com/q/28720/2451 and links therein. – Qmechanic Feb 6 '14 at 22:27

According to the generalized uncertainty principle, the Planck length is, in principle, within a factor of order unity, the shortest measurable length – and no improvement in measurement instruments could change that. (Wikipedia)

$$1 \text{ Planck length} = 1.61619926 \times {10}^{-35} m$$

-
Thanks. I should have checked that. – Dayman75 Feb 6 '14 at 23:16
You should note that this is speculative and only holds true for certain physical theories (which remain untested) beyond the standard model. It's better to say that the Planck length is the smallest distance for which our current theories should hold. Physics at scales less than the Planck length is not currently known. – mcFreid Feb 7 '14 at 13:18

As an engineer I would say $10^{-5}\mathrm{m}$ because for mechanical purposes the smoothest surface finishing does not need to be net as this. As a practical chemist I would say $10^{-10}\mathrm{m}$ is small enough since I will never care about the very real diameter of an electron. I guess physicist would like to get a little bit smaller. Basically speaking, there is, at least, some interesting and energetic Electro Magentic wave with smaller wavelengths ...

But you are not expecting a definitive answer to that kind of question, do you ?

-
You're right. It's hard to get a definitive answer. I was just searching if there was any field of study which might use smaller values than Planck Length. But thanks for your answer. – Dayman75 Feb 6 '14 at 23:20
You're welcome, it is not the perfect answer, just a pragmatic point of view. – jlandercy Feb 6 '14 at 23:26
It has been over a decade, but last I looked into it the experiments to determine the size of the electron had all delivered an answer equal to zero within experimental error (which was really really small compared to 10e-10m. So, I lean more towards the Planck length. I'll try to find the relevant papers tomorrow. – Jon Custer Mar 12 '14 at 0:30


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# Comparing Fractions with Like Numerators – Definition With Examples

## What Is a Numerator?

When numbers are written in the form of a fraction, it can be represented as a/b, where a is the numerator and b is the denominator. A numerator is the part of a fraction, which is above the line, and it represents the number of parts out of the whole. On the other hand, a denominator is the total numbers of the parts.

For example: In the fraction $\frac{3}{5}$, the numerator is 3.

In the above image, the number of shaded parts out of total parts are 5. So, the numerator is 5 and the denominator is 6.

## Definition

Let us have a look at two fractions, i.e., $\frac{5}{7}$ and $\frac{5}{9}$. We see that the numerators are the same, i.e., 5. Such numerators are said to be like numerators. In other words, when the numerators of two or more fractions are the same and the denominators are different, they are called the same or like numerators

Let us look at one more example:

In the above example, the first figure is divided into 3 parts and only 1 is shaded whereas in the second figure, there are 4 equal parts and only 1 is shaded. As a child, you remember you did unit fractions. Unit fractions are the fractions in which the fractions have the numerator 1.

• Comparing fractions with like numerators

Comparing the fractions with like or same numerators is an easy task.

Let us compare $\frac{3}{5}$ and $\frac{3}{7}$.

From the models, we can see that $\frac{3}{5}>\frac{3}{7}$.

In order to compare two or more fractions with the same or like numerators, we just have to compare their denominators. The fraction that has the bigger denominator is smaller.

If we compare $\frac{9}{11}$ and $\frac{9}{17}$, we see that the denominator 17 is larger than the denominator 11. So,  $\frac{9}{11}>\frac{9}{17}$.

• Ordering Fractions with Like Numerators
1. Ascending order

Ascending is also known as increasing. So when numbers have their values from smallest to largest, we say that it is in ascending order. Ascending order for the fractions with like numerators is quite easy. The larger the denominator, the smaller the fraction.

For example: Write the numbers in ascending order: $\frac{1}{33},\frac{1}{45},\frac{1}{27},\frac{1}{19}$.

As we know, the larger the denominator, the smaller the fraction.

45 > 33 > 27 > 19

$\Rightarrow\frac{1}{45}<\frac{1}{33}<\frac{1}{27}<\frac{1}{19}$

1. Descending order

Descending is also known as decreasing. So, when numbers have their values from largest to the smallest, we say that it is in descending order. Descending order is just the opposite of ascending order. The smaller the denominator, the greater the fraction.

For example: Write the numbers in descending order: $\frac{1}{33},\frac{1}{45},\frac{1}{27},\frac{1}{19}$.

As we know that the larger the denominator, the smaller the fraction.

19 < 27 < 33 < 45

$\Rightarrow\frac{1}{19}>\frac{1}{27}>\frac{1}{33}>\frac{1}{45}$

• Addition and Subtraction of fractions with like or same numerators

Adding or subtracting the fractions with same or like numerators and unlike denominators is the same as adding or subtracting the fractions with unlike denominators.

Let us suppose we have to have $\frac{2}{3}$ and $\frac{2}{5}$. To add them, you have to first make the denominators equal.

LCM (3, 5) = 15

$\frac{2\times5}{3\times5}=\frac{10}{15}$and$\frac{2\times3}{5\times3}=\frac{6}{15}$

$\frac{10}{15}+\frac{6}{15}=\frac{16}{15}$

Let’s subtract them.

$\frac{10}{15}-\frac{6}{15}=\frac{4}{15}$

## How to Make Like or Same Numerators?

For making the numerators of two or more fractions same or like, we can find out the LCM of the numerators and then multiply the numerators with relevant numbers. It becomes easy to compare if there are the like or same numerators.

For example: We have to compare $\frac{3}{4}$ and $\frac{9}{11}$.

By finding the LCM, you can find the common numerators.

LCM (3, 9) = 9

$\frac{3\times3}{4\times3}=\frac{9}{12}$ and $\frac{9}{11}$

For comparing $\frac{9}{12}$ and $\frac{9}{11}$, we just have to look at the denominator.

Since 12 > 11

$\frac{9}{12}< \frac{9}{11}\Rightarrow\frac{3}{4}<\frac{9}{11}$

## Solved Examples

Example 1: Write the following fractions in descending order.

$\frac{7}{20},\frac{7}{9},\frac{7}{11},\frac{7}{19}$ and $\frac{7}{25}$

Solution: The smaller the denominator, the greater the fraction when the fractions have like or same numerators.

9 < 11 < 19 < 20  25

$\frac{7}{9}>\frac{7}{11}>\frac{7}{19}>\frac{7}{20}>\frac{7}{25}$

Example 2: Find the fractions with the like or same numerators from the following group of fractions.

$\frac{3}{5},\frac{3}{10},\frac{1}{6},\frac{3}{8},\frac{3}{19},\frac{8}{13}$

Solution:

The fractions with the like or same numerators are:  $\frac{3}{5},\frac{3}{10},\frac{3}{8},\frac{3}{19}$.

Example 3: Add $\frac{1}{3}+\frac{1}{5}+\frac{1}{9}$

Solution: LCM(3, 5 and 9) = 45

$\frac{1\times15}{3\times15}=\frac{15}{45};\frac{1\times9}{5\times9}=\frac{9}{45};\frac{1\times5}{9\times5}=\frac{5}{45}$

$\frac{15}{45}+\frac{9}{45}+\frac{5}{45} = \frac{29}{45}$

Example 4: Write the following fractions in ascending order:

$\frac{135}{178},\frac{135}{199},\frac{135}{101},\frac{135}{119}$ and $\frac{135}{229}$

Solution: The greater the denominator, the smaller the fraction when the fractions have like or same numerators.

229 > 199 > 178 > 119 > 101

$\frac{135}{229}<\frac{135}{199}<\frac{135}{178}<\frac{135}{119}<\frac{135}{101}$

## Practice Problems

1

### Which of the following is true?

$\frac{13}{35}\gt\frac{13}{34}$
$\frac{15}{17}\gt\frac{15}{13}$
$\frac{11}{34}\gt\frac{11}{49}$
$\frac{21}{34}\lt\frac{21}{49}$
CorrectIncorrect
Correct answer is: $\frac{11}{34}\gt\frac{11}{49}$
The greater the denominator, the smaller is the fraction when the fractions have like or same numerators.
2

### Which sign will come in between $\frac{2}{7}$ and $\frac{4}{13}$?

$\gt$
$\lt$
=
None of these
CorrectIncorrect
Correct answer is: $\lt$
Let’s make the numerator the same or like.
LCM(2, 4) = 4
$\frac{2\times2}{7\times2}=\frac{4}{14}$ and $\frac{4}{13}$
Since, $14 \gt 13$
$27 \lt \frac{4}{13}$
3

### What will be the fraction $\frac{1}{3}-\frac{1}{4}$ and $\frac{1}{5}-\frac{1}{6}$ called?

Improper Fraction
Equivalent Fraction
Like Fraction
Fraction with Like Numerators
CorrectIncorrect
Correct answer is: Fraction with Like Numerators
LCM (3, 4) = 12
$\frac{1\times4}{3\times4}=\frac{4}{12};\frac{1\times3}{4\times3}=\frac{3}{12}$
$\frac{4}{12}-\frac{3}{12}=\frac{1}{12}$

LCM (5, 6) = 30
$\frac{1\times6}{5\times6}=\frac{6}{30};\frac{1\times5}{6\times5}=\frac{5}{30}$
$\frac{6}{30}-\frac{5}{30}=\frac{1}{30}$
$\frac{1}{12}$ and $\frac{1}{30}$ are the fractions with like or same numerators.


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# XVII International Forum “Fuel and Energy Complex of Ukraine: The Present and The Future”

XVII International Forum “Fuel and Energy Complex of Ukraine: the Present and the Future” was held on November 5-7, 2019 at the International Exhibition Center in Kyiv. The main aim of the Forum was promoting the implementation of the “Energy Strategy of Ukraine up to 2035” to ensure safe and efficient operation of the whole country’s energy system, state energy independence and security; demonstration of achievements in the energy sector aimed at modernization of the energy infrastructure, its technological, organizational, regulatory and scientific development.

In the course of the Forum Nuclear Power Day “Nuclear Power in the period of transformation. How to preserve the potential for future technologies implementation” was conducted on November 6. The event was opened with the speech of Yurii Nedashkovskyi, the President of NNEGC “Energoatom”: “…On one hand the state fully uses economic and ecological benefits of Nuclear Power. Often it “patches holes” and closes its gaps at the expense of the nuclear industry. On the other – the state does not want to see what you have to pay for it”. He emphasized that state institutions set alternative tasks before Nuclear Power: “The State demands from its own Nuclear Power for miserable means to be at the same time effective (i.e. to produce as cheap electric energy as possible) and to execute large-scale programs on modernization and safety improvement, power units lifetime extension and in addition to bear the burden of social charges and to accumulate means for decommissioning withdrawing them to the state budget where they disappear without a trace for the next holes patching”.

Yurii Nedashkovskyi offered to participants to find the answer for the question: “What should be done for breaking this awful format of relationships that hardly can be called cooperation”.

Participants of the Forum tried to find possible variants of problems solving defined by the President of NNEGC “Energoatom”. The Forum was conducted in the format of the round table at three strategic panels:

• State and the Nuclear Power: vectors of cooperation for increase of economy and high social standards;
• Nuclear Power and business. In pursuit of points of increase;
• Nuclear Power in conditions of new energy market.

Representatives of Verkhovna Rada of Ukraine, directors of enterprises of Ukrainian nuclear power, State Nuclear Regulatory Inspectorate of Ukraine, WANO, World Nuclear Association (WNA), NNEGC “Energoatom”, international companies – partners of “Energoatom” and others participated in the Forum. The total number of representatives is 200.

Anatolii Nosovskyi Director of the ISP NPP, Volodymyr Borysenko Head of the Division for Nuclear Power Engineering, Yevheniya Kudin Director Assistant, participated in the Forum.

Цей запис також доступний: Ukrainian, Russian

## Related posts

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# Tag Info

## New answers tagged correlation

0

Correlation provides a metric of the linear dependence between two signals. The normalized correlation coefficient (which is the result of xcorr and then divided by the standard deviation of each of the two waveforms) will provide a result that is between +1 and -1, where +1 means the two waveforms are virtually identical other than perhaps a gain scaling ...

0

In my opinion, using fast Fourier transform (FFT) to circularly deconvolve a maximum length sequence (MLS) makes sense from a development point of view because FFT is typically readily available and because using FFT makes switching to another test signal easy. If only power-of-two-size FFTs are available, then for a length $2^m-1$ MLS, a length $2^{m+1}$ ...

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The square of a positive number is 21 more than 4 times the number. How do you find the number?

Apr 14, 2018

$x = 7$

Explanation:

First, translate the statement into an equation:

${x}^{2} = 21 + 4 x \text{ }$ Subtract 21 and 4x on both sides to get:

${x}^{2} - 4 x - 21 = 0 \text{ }$ Factor the quadratic to get:

$\left(x - 7\right) \left(x + 3\right) = 0 \text{ }$ Set each factor equal to zero:

$x - 7 = 0$ and $x + 3 = 0 \text{ }$ Solve each equation:

$x = 7$ and $x = - 3$

Since the statement said it must be a "positive" number, we go with only 7.


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# American Institute of Mathematical Sciences

June & December 2018, 5(1&2): 1-32. doi: 10.3934/jcd.2018001

## Optimal transport over nonlinear systems via infinitesimal generators on graphs

 1 School for Engineering of Matter, Transport and Energy, Arizona State University, Tempe, AZ 85281, USA 2 Mitsubishi Electric Research Labs, 201 Broadway, Cambridge, MA 02139, USA

* Corresponding author

Fund Project: This work was funded by Mitsubishi Electric Research Labs

We present a set-oriented graph-based computational framework for continuous-time optimal transport over nonlinear dynamical systems. We recover provably optimal control laws for steering a given initial distribution in phase space to a final distribution in prescribed finite time for the case of non-autonomous nonlinear control-affine systems, while minimizing a quadratic control cost. The resulting control law can be used to obtain approximate feedback laws for individual agents in a swarm control application. Using infinitesimal generators, the optimal control problem is reduced to a modified Monge-Kantorovich optimal transport problem, resulting in a convex Benamou-Brenier type fluid dynamics formulation on a graph. The well-posedness of this problem is shown to be a consequence of the graph being strongly-connected, which in turn is shown to result from controllability of the underlying dynamical system. Using our computational framework, we study optimal transport of distributions where the underlying dynamical systems are chaotic, and non-holonomic. The solutions to the optimal transport problem elucidate the role played by invariant manifolds, lobe-dynamics and almost-invariant sets in efficient transport of distributions in finite time. Our work connects set-oriented operator-theoretic methods in dynamical systems with optimal mass transportation theory, and opens up new directions in design of efficient feedback control strategies for nonlinear multi-agent and swarm systems operating in nonlinear ambient flow fields.

Citation: Karthik Elamvazhuthi, Piyush Grover. Optimal transport over nonlinear systems via infinitesimal generators on graphs. Journal of Computational Dynamics, 2018, 5 (1&2) : 1-32. doi: 10.3934/jcd.2018001
##### References:

show all references

##### References:
Approximation of infinitesimal generator $A$. The entry $A_{ij}$ is proportional to flux across $B_i\cap B_j$ from $B_i$ to $B_j$, due to vector field $f$.
(A) Some minimizing geodesics to the origin in the Grushin plane. (B) Analytically computed optimal transport solution between a uniform measure whose support is the disk $\Omega = \{(x, y)|x^2+(y-.8)^2<.15^2\}$, and a measure concentrated at the origin.
(A)-(E) The optimal transport solution in the Grushin plane using graph based algorithm between a measure whose support is the disk $\Omega = \{(x, y)|x^2+(y-.8)^2<.15^2\}$, and delta measure at the origin. The parameters are $m = 10^4, k = 75$. (F) Convergence of optimal transport cost with number of time discretization steps $k$ and grid size $m$.
Particle trajectories with feedback control computed using Eq. (47) from the optimal transport solution in the Grushin plane. Each box contained in the support of uniform initial measure $\mu_0$ is initially populated with 4 particles.
nvariant manifolds and lobe-dynamics in the double-gyre system (reproduced from Ref. [39]).
Optimal transport in the periodic double gyre system (Eqs.(57a-57b)) between measures at $t = 0$ and $t_f = 1$. The parameters are $m = 1800, \Delta t = \dfrac{1}{40}$.
Optimal transport in the periodic double gyre system between measures at $t = 0$ and $t_f = 10, \Delta t = \dfrac{1}{40}$. The optimal transport solution shows a quantization phenomenon. Ten 'packets' are transported via lobe-dynamics from the left side to the right side of the domain. During $2<t<5$, the third packet is transported to the right side via the sequence $F^{-1}(A)\rightarrow A\rightarrow F(A) \rightarrow F^2(A)$, where set $A$ is defined in Fig. 5. The last packet gets transported to the right side over $9<t<10$. Animation available at: https://www.youtube.com/watch?v=Pu7sCkpm4RY
The cost of optimal transport between two measures supported on two AIS for the double-gyre system, as a function of time-horizon of the problem.
Initial and final measures shown on $(x, y)$ plane for two cases of optimal transport in the unicycle model. The green arrows indicate the third coordinate $\theta$. (A) $\mu_0$ is supported on $(0, 0.5, 0)$, $\mu_1$ is supported on $(1, 0, 0)$ and $(1, 1, 0)$. (B) $\mu_0$ is supported on $(0, 0.5, 0)$, $\mu_1$ is supported on $(1, 0, \frac{3\pi}{2})$ and $(1, 1, \frac{\pi}{2}).$
The optimal transport solution of unicycle model shown in the $x-y$ plane for case 1. The grid size is $m = 25^3$, and $k = 20$.
The optimal transport solution of unicycle model shown in the $x-y$ plane for case 2. The grid size is $m = 25^3$, and $k$ = 20.
Illustration of the proof of Theorem 3.7. The existence of an trajectory $f$ approximating a curve $\gamma$ connecting $B_v$ to $B_w$, obtained by piecewise constant control, is guaranteed by the small time local controllability. By continuity, this leads to non-zero transition rates, and hence strong connectivity of the control graph $\mathcal{G}_c$.
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# Is VSupport LLC legit?

by yungman
Tags: legit, ligit, vsupport
 P: 3,898 My wife called VSsupport LLC to help and took over the computer and did a lot of things......while I was asleep!!!! They claimed I have trojen stuff and all. When I woke up, I quickly disconnect the phone call and everything else. They claimed a lot of the system drivers are disabled. My question is 1)Is VSupport ligit? 2)They claimed the trogen house can infect the DSL MODEM and wireless connection. That the Norton anti virus cannot even protect. Is this true? 3)My wife they them took over the computer, what potential damage can they do? 4) We have electronics bank accounts on the computer, any danger for them to hack into our account? We called the bank, they said it's not easy to get into the account. 5) What should I do. I am using an old computer right now, I don't even dare to turn on all the new computers!! Thanks
 P: 3,898 Thanks for the detail reply. I since contact Norton, they said it is not possible to infect the DSL modem and the wireless connection. Is that true. Should I can the Att provider? They suggested and I am running the Norton Power Eraser on the other computer. So I should go onto Control Panel, then Add/Delete programs to look for what is installed today and uninstall that? Again, thanks for the quick reply.
Mentor
P: 26,541
 Quote by yungman My wife called VSsupport LLC to help and took over the computer and did a lot of things......while I was asleep!!!! They claimed I have trojen stuff and all. When I woke up, I quickly disconnect the phone call and everything else.
May I ask why she contacted these people to begin with?
P: 3,898
 Quote by Evo May I ask why she contacted these people to begin with?
i was sleeping!!! She was interested to learn Win 8.....too interested!!!
P: 3,898
I really appreciate your time to type this.
1) I did run the Power Erase and only 6 programs got picked and deleted. Nothing like the guy in VSupport claimed of 500!!!. The computer is running normal.....in fact it never ran abnormal. My wife just lost an email account and one way lead to another, she contacted VSupport before she woke me up. I almost fell on the floor!!!
2)I definitely going to run the HijackThis you suggested and also the Spybot. I am not on that computer, I'll report back if I see anything comes out. Do you have a reliable site to download these, I am getting paranoid in getting into some unreliable sites.
3) I was surfing and I read about the a malware that called "psyb0t" or "Bluepill" that can stay in the modem and router and read the traffic going in and out. But according to the article, the electronic banking communication are encrypted and the psyb0t still cannot read it even they can read the data. You heard anything about this?

Again I really appreciate your help. Can you tell me which is the best way to learn these kind of knowledge about computers? I am absolutely not interested in gaming and all, but I am very interested in learning how to fix computer from infection, getting into the nuts and bolts of how to deal with computer issues. I was an EE and I have knowledge of the hardware, but not on this end.
 P: 3,898 Another concern is my wife gave the VSupply permission to take over the computer and the guy was actually getting into the configuration and all and start typing all sort of things on the screen. Can he still get into the computer. I do have the Norton Fireware, anti virus and all the goodies running already.
P: 9
 Quote by yungman We have electronics bank accounts on the computer, any danger for them to hack into our account? We called the bank, they said it's not easy to get into the account. Thanks
Your bank accounts are not stored on your computer. You are logging onto the bank's website to view them.

The only way they could have accessed your account is if you somehow set your browser to remember your user name and password. Simply change your online banking password if you're paranoid. That will solve that.
P: 3,898
 Quote by Rick21383 Your bank accounts are not stored on your computer. You are logging onto the bank's website to view them. The only way they could have accessed your account is if you somehow set your browser to remember your user name and password. Simply change your online banking password if you're paranoid. That will solve that.
Thanks, you make me feel a lot better. I'll have her change the password.
 P: 3,898 In my wife's defense, I was too desperate yesterday running around, I should have said it all started out she could not get into her email in the new computer and it prompted her to contact the company as the computer is new. One thing lead to another, she allow them to take hold of the computer. She saw me let McAfee took control one time, she thought it's ok to do that. I told her I let McAfee did that because we had their service at the time, and they are well known and I got the number direct from the software installed.....And is McAfee!!! BTW, McAfee sucked!!! I had infections with emphasis "s" under them. Their service sucks. I have Norton now, never have a single problem in over 2 years......with my grandson download and playing video games all the time!!! Only thing I ever had to do is stop and delete the extra Apps after he finish playing the game. I have been using their life chat the last two days and they are nothing but helpful. They even called me on the phone to make sure everything is ok. I used their register clean and the old computer runs much faster.
P: 9
 Quote by yungman Thanks, you make me feel a lot better. I'll have her change the password.
Yea and, rest assured, they are a valid company so I wouldn't worry about them doing anything malicious.

My biggest fear would be them inadvertantly "breaking" something by whatever changes they made to your computer. Like deleting/modifying a file or something in an attempt to fix your computer, and not realizing that another application uses that same file.

If your machine seems to be running ok, then I wouldn't worry about it.
P: 3,898
 Quote by Rick21383 Yea and, rest assured, they are a valid company so I wouldn't worry about them doing anything malicious. My biggest fear would be them inadvertantly "breaking" something by whatever changes they made to your computer. Like deleting/modifying a file or something in an attempt to fix your computer, and not realizing that another application uses that same file. If your machine seems to be running ok, then I wouldn't worry about it.
They sure did break something. The Photoshop program disappeared. In the control panel, it had a strange look program not normal Photoshop icon. It was empty. I tried to reload Photoshop, the CD would not activate the computer and not even read the CD. Luckily I learned how to turn the CD into ISO image and mount it. I ended up had to do it the long route to reload the Photoshop.

P: 94
Sorry for late response, I was busy with some exams and stuff.

2) HijackThis is in sourceforge.net. Spybot is in safer-networking.org. Hint: Wikipedia usually has the legitimate websites links of some software.

3) There always exists the possibility that the virus is programmed to spoof SSL certificates and if it does so, then it can read (and reading means it can copy too) whatever is sent between the supposed to be encrypted page and your computer. Once you know how networks work, spoofing an SSL certificate is kids game. (Why do I always have to be the party pooper and concentrate only on the bad possibilites).

The good news is that from what I have read, psyb0t hasn't been seen doing such even thought to me it appears to have the capability of doing so (I can't find details on the attack tools it posses). Although it is mentioned that it can steal personal information, to me it appears to be targeted more at websites like forums and databases and to spread itself, not at stealing personal people's info.

I'm happy you want to learn. Since you are an EE you will get things really fast . It's just that I'm a bad teacher like all those nerds who look like they have a lot of knowledge but are uncappable of expressing what is in their brains (rendering them useless when communicating). Lets see......... Let's organize everything........ and....... ok, I think I have it. The best way for that is the internet (most info is free and you have it anywhere as long as you pay for the internet service or live close to places with free internet access). To go deeper into programming and other informatics subjects then a university would be the best way in my opinion (or maybe just get the names of the books they use for each course and buy them for self study and then you won't have to pay for the course ).

But let's give a try to the internet first. Since we are talking about Windows, the first thing I would tell you to understand is the many ways a program can be executed at startup. There are services, registry entries, scheduled tasks, etc. Understanding them all is a must to know what to look for when dealing with malware (but is not necesarry to go too deep into services). The other would be to understand that in windows there exists ways for programs to code inject others so you won't necesarily see a virus asking for administrator privileges to get a hold of your computer. Then getting your hands in computer forensics software and learning how it works will give you a huge insight and techniqes that will allow you to better asess a problem and choose the correct tools to solve it.

http://www.bleepingcomputer.com/tuto...tup-locations/
http://www.bleepingcomputer.com/tuto...es-in-windows/
http://technet.microsoft.com/en-us/l.../cc959354.aspx

Once again, since we are talking about windows here are a few keywords to look for on a search engine (google, yahoo, duckduckgo, etc) to get something to learn and a boost to look into other stuff:
windows startup locations, windows startup entries,
start stop windows services, hidden files,
data recovery , computer forensics,
using live cd to remove malware,
types of network attacks, eavesdropping,
spoofing, man-in-the-middle attack, kinds of malware,
code injection......
and I'm leaving a bunch of other concepts, but its ok, you will find them yourself as those keywords will link themselves to other information you may want to know.

Now if you want to go even deeper like really understanding the system then you need to program for the operating system you are trying to understand. Some books that provide insights into the operating systems workarounds are those that teach about driver programming for the operating system of interest. They teach stuff like the kernel and how it handles nearly everything in the OS.
P: 3,898
 Quote by Psinter Sorry for late response, I was busy with some exams and stuff. 2) HijackThis is in sourceforge.net. Spybot is in safer-networking.org. Hint: Wikipedia usually has the legitimate websites links of some software. 3) There always exists the possibility that the virus is programmed to spoof SSL certificates and if it does so, then it can read (and reading means it can copy too) whatever is sent between the supposed to be encrypted page and your computer. Once you know how networks work, spoofing an SSL certificate is kids game. (Why do I always have to be the party pooper and concentrate only on the bad possibilites). The good news is that from what I have read, psyb0t hasn't been seen doing such even thought to me it appears to have the capability of doing so (I can't find details on the attack tools it posses). Although it is mentioned that it can steal personal information, to me it appears to be targeted more at websites like forums and databases and to spread itself, not at stealing personal people's info. I'm happy you want to learn. Since you are an EE you will get things really fast . It's just that I'm a bad teacher like all those nerds who look like they have a lot of knowledge but are uncappable of expressing what is in their brains (rendering them useless when communicating). Lets see......... Let's organize everything........ and....... ok, I think I have it. The best way for that is the internet (most info is free and you have it anywhere as long as you pay for the internet service or live close to places with free internet access). To go deeper into programming and other informatics subjects then a university would be the best way in my opinion (or maybe just get the names of the books they use for each course and buy them for self study and then you won't have to pay for the course ). But let's give a try to the internet first. Since we are talking about Windows, the first thing I would tell you to understand is the many ways a program can be executed at startup. There are services, registry entries, scheduled tasks, etc. Understanding them all is a must to know what to look for when dealing with malware (but is not necesarry to go too deep into services). The other would be to understand that in windows there exists ways for programs to code inject others so you won't necesarily see a virus asking for administrator privileges to get a hold of your computer. Then getting your hands in computer forensics software and learning how it works will give you a huge insight and techniqes that will allow you to better asess a problem and choose the correct tools to solve it. Here are some nice reads: http://www.bleepingcomputer.com/tuto...tup-locations/ http://www.bleepingcomputer.com/tuto...es-in-windows/ http://www.cisco.com/web/about/secur...orm-diffs.html http://technet.microsoft.com/en-us/l.../cc959354.aspx Once again, since we are talking about windows here are a few keywords to look for on a search engine (google, yahoo, duckduckgo, etc) to get something to learn and a boost to look into other stuff: windows startup locations, windows startup entries, start stop windows services, hidden files, data recovery , computer forensics, using live cd to remove malware, types of network attacks, eavesdropping, spoofing, man-in-the-middle attack, kinds of malware, code injection...... and I'm leaving a bunch of other concepts, but its ok, you will find them yourself as those keywords will link themselves to other information you may want to know. Now if you want to go even deeper like really understanding the system then you need to program for the operating system you are trying to understand. Some books that provide insights into the operating systems workarounds are those that teach about driver programming for the operating system of interest. They teach stuff like the kernel and how it handles nearly everything in the OS.
Thanks for the detail answers. I went on live chat with Norton, they said their anti virus and firewall will take care even the infection of the modem. But I don't know how true that is.

Does it help to use a totally different computer even though it is still on the same DSL modem? How about changing the password of the bank account?

I am not sure the modem has been infected, only the VSupport person said it's a possibility. He showed 300 trojen horses in my computer, which make me very suspicion of what he said, particularly he tried to make us pay \$199 to clean up the system.

Is there any way to check for modem infection and clean it up?

Thanks
 PF Gold P: 347 The best thing to do is, save any personal data that is on the local drive on a USB device. Then, reload Windows from scratch. And, it never hurts to change your bank password if you think someone else may have it. Teach your wife never, ever to allow anyone to take control of the computer. In fact, there is a setting to disable this capability. In the Control Panel, open the System applet and then select Remote Settings. Make sure the box is unchecked next to "Allow Remote Assistance connections for this computer".
P: 3,898
 Quote by harborsparrow The best thing to do is, save any personal data that is on the local drive on a USB device. Then, reload Windows from scratch. And, it never hurts to change your bank password if you think someone else may have it. Teach your wife never, ever to allow anyone to take control of the computer. In fact, there is a setting to disable this capability. In the Control Panel, open the System applet and then select Remote Settings. Make sure the box is unchecked next to "Allow Remote Assistance connections for this computer".
Thanks, I disable the remote access.

 Related Discussions Medical Sciences 9 Engineering, Comp Sci, & Technology Homework 8 Special & General Relativity 4 Math & Science Software 2 General Math 15


text:
• # question_answer The work done for the cycle shown in given figure will be A)  45 J    B)  54 J   C)                22.5 J              D)    32.5 J

The area of P-V diagram given work. The area of $\Delta abc$is $A=\frac{1}{2}\times 30\times 3=45\,J=\text{work}\,\text{done}$ Hence, the correction option is (a).


text:
Solve linear partial variation about selling handbags

Question Sample Titled 'Solve linear partial variation about selling handbags'

When Caroline sells ${n}$ handbags in a month, her income in that month is $S$ . It is given that ${S}$ is a sum of two parts, one part is constant and the other part varies as ${n}$ . When ${n}={13}$ , ${S}={11550}$ ; when ${n}={8}$ , ${S}={9800}$ .

Initiating...

HKDSE 數學試題練習平台


text:
# table

The peoples table (around 1725) is a tabular listing of European peoples

A table (from the Latin tabella literally for "[the] tablet" and also translated "[the] tablet") is an orderly compilation of texts or dates . The content to be displayed is divided into rows ( horizontally ) and columns ( vertically ), which are graphically aligned with one another. The first column of a table is called the front column if it describes or explains the contents of the fields in the other columns in the same row. The first lines of a table, which contain explanations of the table content and / or designate or explain the fields in the following lines of the individual columns, form the table header ; if this consists of only one line, this is called the header . Tables also consist of table fields ( cells ). The reference to a field is the addressing ( reference )

Usually there is a semantic relationship between the content of a cell and the row or column in which it is located.

1. In the case of relational databases , the term table stands for a collection of similarly structured data, which are technically mostly stored together (see under database table ).
2. A spreadsheet is a computer program that is particularly suitable for displaying and calculating hypothetical numerical data on a large scale.
3. In propositional logic, there are truth tables that clearly represent various operations and their possible results.

Well-known tables are, for example, the Düsseldorf table and the periodic table .

## Data tables

### construction

Table 1: Absolute frequencies of eye and hair color combinations from 592 statistics students. Source: parcel datasets in R .
Eye color
Hair color brown blue Gray green total
brown 119 84 54 29 286
Blond 7th 94 10 16 127
black 68 20th 15th 5 108
red 26th 17th 14th 14th 71
total 220 215 93 64 592
Table 2: Conditional frequencies of eye colors given a hair color.
Eye color
Hair color brown blue Gray green total
brown 42% 29% 19% 10% 100%
Blond 5% 74% 8th % 13% 100%
black 63% 18% 14% 5% 100%
red 36% 24% 20% 20% 100%
total 37% 36% 16% 11% 100%

The aim of the tables is to provide a brief and concise overview of detailed information. In order to be able to take the relevant information from tables, an appropriate structure or design is necessary. Therefore, some rules should be observed for better readability:

1. Tables should have explanatory text in a heading or a signature.
Because readers may just read the table, and then the most important information should be in the heading or signature. In addition, with an automatic (dynamic) layout, a table is moved to the next page after the descriptive text for reasons of space, and additional information is also important then.
2. For documents with several tables, table headings or the legend should contain their own identifier, e.g. numbering.
3. Then you can clearly refer to a specific table in the accompanying text.
4. A table should also be described in the accompanying text.
A table is often not easy to read, so the accompanying text should refer to the important statements in the table. Readers who are not familiar with the subject matter dealt with would otherwise not be able to understand the information in the table.
5. A table shouldn't have too many columns and rows.
The psychologist George A. Miller demonstrated in 1956 that one can only remember "information" in short-term memory . Therefore, it gives as a guide to the number of columns and rows in each case .${\ displaystyle 7 \ pm 2}$${\ displaystyle 7 \ pm 2}$
6. Rows and columns should be ordered naturally or by size.
In Table 1 , the rows and columns are arranged according to the row or column sums. Alternatively, an alphabetical sorting according to the name of the hair or eye color would have been possible. The order in Table 2 is based on Table 1, as the reader is already familiar with the order in Table 1.
7. Do not use repetitive characters such as B. use "ditto".
Better to leave the space blank or simply repeat the number.
8. Tables should not contain disjointed information.
It is then better to divide the information into two or more tables. For example, Tables 1 and 2 could have been combined into one table, but the information is better presented in two tables.
9. The layout should "guide" the eye:
• Avoid vertical lines
• Avoid double lines
• To support legibility, it is better to use background colors instead of a line "network"
• A good example can be found in the documentation for the LaTeX package booktabs on page 2.
10. Units of measurement should be written in the head of the respective column.
If the units of measurement are omitted, it is unclear in which unit the data are given. If you write it at the bottom of the table, you have to look for it first.
11. For columns and rows, totals (or averages) should be given if possible / necessary.
This saves a reader having to calculate the sums and also allows him to check the table values.
12. It is better to place values to be compared in different columns and not in different rows.
From each column in Table 2, a statistician can immediately deduce from the different values that there is a connection between hair and eye color.
13. Texts should be left-justified.
Readers whose native language is German are used to the fact that text is written from left to right and can therefore grasp left-justified text more quickly. In languages in which the direction of writing is different, e.g. B. Hebrew or Arabic , one will prefer right-aligned texts.
14. The following applies to numbers:
• Large numbers should be set upwards if possible
• They should be right justified
• They should be rounded to two to three significant digits
• They should contain leading zeros, ie "0.1" instead of ", 1"
• The decimal points for numbers in a column should always be below each other.
• Table 2 gives an example of this. However, rounding the entries to whole percentages with commercial rounding leads to the problem that the line total is no longer 100%, but 99% or 101%. Therefore, the sum-preserving rounding was used here.

The rules are intended to help improve understanding and readability of tables and are not absolute. Table 1 shows the eye and hair color data set with the absolute frequencies of statistical students with the aim of showing a relationship between eye and hair color. The rules have been followed as far as possible, e.g. B. the background color changes from line to line to make it easier to read the values in a line, or the eye and hair colors have been sorted according to the size of the column or line sums. Since the table standard in Wikipedia provides for a line network, this has been retained, in contrast to the rules. From table 2 it can be seen that there is a relationship between hair color and eye color.

### Visualization

#### With a heat map

Table 1 with gray values for the corresponding frequencies: the darker, the more often the combination occurs.

There are various options for displaying tables with numerical values in graphical form. One possibility is to replace the numerical entries with symbols or colors.

In the figure, the frequencies in Table 1 have been replaced by rectangles with corresponding gray levels. The darker a rectangle appears, the more often the corresponding combination of characteristics occurred in the 592 statistics students. The combination of brown eyes and hair was the most common, the combination of green eyes and black hair or brown eyes and blonde hair the least common.

#### With a mosaic plot

Tables 1 and 2 visualized with the help of a mosaic plot

Since Tables 1 and 2 show absolute and conditional frequencies, they can also be shown in a so-called mosaic plot. The height of each rectangle is determined by the relative proportion of the corresponding hair color. For the hair color brown it results , i. This means that the rectangles in the 1st line have a height of approx. 48% of the total height of the graphic. The width of the rectangles results from the conditional frequencies in Table 2. The rectangle for brown hair and brown eyes therefore has a width of 42% of the total width of the graphic. The proportion of the area for a combination of features in the total area corresponds exactly to the proportion of this combination of features. So the area for the brown rectangle corresponds to , i. H. approx. one fifth of the total area. ${\ displaystyle 286/592 \ approx 48 \, \%}$${\ displaystyle 48 \, \% \ cdot 42 \, \% \ approx 20 \, \%}$

• List - one-dimensional table
• Matrix - arrangement of numerical values or other objects in tabular form in linear algebra
• Table digits - special digits with the same width ( thickness ) so that the digits are flush with each other.

Commons : Tables  - collection of images, videos and audio files
Wiktionary: table  - explanations of meanings, word origins, synonyms, translations

## Individual evidence

1. a b DIN 5008 : 2020-03, Sections 14.1 Tables of Numbers - General and 14.4 Tables of Numbers - Table header and front column
2. ^ RD Snee: Graphical display of two-way contingency tables . In: The American Statistician . tape 28 , 1974, p. 9-12 .
3. ^ Rüdiger Ostermann, Adalbert FX Wilhelm, Karin Wolf-Ostermann: Presentation of statistical data in nursing, Part 1 - Tables: More complicated than you think . In: Care magazine . tape 1 , 2004, p. 18-21 .
4. Michael Piefel: Golden Rules. Retrieved December 15, 2012 .
5. ^ Edward R. Tufte: The Visual Display of Quantitative Information . Graphics Press, 1983.
6. ^ Edward R. Tufte: Envisioning Information . Graphics Press, 1990.
7. a b Simon Fear: Publication quality tables in LaTeX. (PDF; 194 kB) In: Comprehensive TeX Archive Network. April 14, 2005, accessed January 4, 2013 .
8. George A. Miller: The magical number seven, plus or minus two: Some limits on our capacity for processing information . In: Psychological Review . tape 63 , no. 2 , 1956, p. 81-97 , doi : 10.1037 / h0043158 .


text:
# A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens.

more_vert
A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is :

(1) 0.55 cm away from the lens

(2) 1.1 cm away from the lens

(3) 0.55 cm towards the lens

(4) 0

more_vert

done_all

Ans: (1) 0.55 cm away from the lens

Sol:

${1\over v}-{1\over u}={1\over f}\implies{1\over10}-{1\over-10}={1\over f}$

$\implies f=5cm$

Shift due to slab $=t\left(1-{1\over\mu}\right)$ in the direction of incident ray

$=1.5\left(1-{2\over3}\right)=0.5$

again, ${1\over v}-{1\over-9.5}={1\over5}$

$\implies{1\over v}={1\over5}-{2\over19}={9\over95}$

$\implies v={95\over9}=10.55cm$


text:
Sums, Direct Sums and Gaussian Elimination

Aug-Nov 2020

Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• Finite-dimensional vector space
• $V$ is span of a finite set of vectors
• Basis
• Linearly independent spanning set
• Any two bases have same number of vectors
• Dimension
• Size of basis

Sum of subspaces

Sum of subsets: For subsets $U,W\subseteq V$, the sum $U+W$ is defined as $U+W = \{u+w:u\in U,w\in W\}$

Usually, $U$ and $W$ are taken to be subspaces. $U+W$ is the smallest subspace containing both $U$ and $W$.

• Examples

1. $\text{span}((1,0))+\text{span}((0,1))$

2. $\text{span}((1,2))+\text{span}((2,3))$

3. $\text{span}((1,2))+\text{span}((2,4))$

4. $U=\text{span}((1,2,3,4),(2,3,4,5),(1,-1,2,-2))$, $W=\text{span}((1,1,1,1),(4,1,8,1),(1,0,0,0))$

$\text{span}(u_1,\ldots,u_m)+\text{span}(w_1,\ldots,w_m)=\text{span}(u_1,\ldots,u_n,w_1,\ldots,w_m)$

Find linear dependence in $u_1,\ldots,u_n,w_1,\ldots,w_m$ to reduce.

Intersection of subspaces

Intersection of subspaces: If $U,W$ are subspaces, $U\cap W$ is a subspace.

• Examples

1. $\text{span}((1,0))\cap \text{span}((0,1))=\{0\}$

2. $\text{span}((1,2))\cap \text{span}((2,3))=\{0\}$

3. $\text{span}((1,2))\cap \text{span}((2,4))=\text{span}((1,2))$

4. $U=\text{span}((1,2,3,4),(2,3,4,5),(1,-1,2,-2))$, $W=\text{span}((1,1,1,1),(4,1,8,1),(1,0,0,0))$

$\text{span}(u_1,\ldots,u_n)\cap\text{span}(w_1,\ldots,w_m)=\{v:v=a_1u_1+\cdots+a_nu_n=b_1w_1+\cdots+b_mw_m\}$

Find linear combinations of $u_1,\ldots,u_n,w_1,\ldots,w_m$ that result in 0.

Direct sum of subspaces

Direct sum of subspaces: If $U,W$ are subspaces and $U\cap W=\{0\}$, $U+W$ is denoted $U\oplus W$ and called as direct sum.

Given a subspace $U$ of a finite-dimensional vector space $V$, there exists a subspace $W$ such that $V=U\oplus W$.

• Proof
• Extend basis of $U$ to basis of $V$.
• Define $W$ as span of new vectors needed in extension.

Examples

1. $U=\text{span}((1,2))$
2. $U=\text{span}((1,2,3))$
3. $U=\text{span}((1,2,3,4),(2,3,4,5))$

Subspace decomposition: any $v\in V$ can be written as $v=u+w$, $u\in U$, $w\in W$, in a unique way.

Dimension of sum of two subspaces

If $U,W$ are subspaces of a finite dimensional vector space, $\text{dim}(U+W)=\text{dim} U+\text{dim} W - \text{dim}(U\cap W).$

• Proof
• Extend basis of $U\cap W$ to basis of $U$ and to basis of $W$.
• Show that vectors above form a basis for $U+W$.

Numerical examples require methods for the following:

• Find linear dependence between vectors.
• Extend basis.

Gaussian Elimination: Implementing Linear Dependence Lemma

• Modify a set of vectors to make them look like standard basis without changing the span.
• Modifications are done one step at a time and are reversible.

$S=((1,2,3,4),(2,3,4,5),(3,4,5,6))$

• Pivot at first vector, first coordinate (pivot should be nonzero)
• Replace $(2,3,4,5)$ by $(2,3,4,5) - 2(1,2,3,4) = (0,-1,-2,-3)$
• Replace $(3,4,5,6)$ by $(3,4,5,6) - 3(1,2,3,4) = (0,-2,-4,-6)$

$((1,2,3,4),(0,-1,-2,-3),(0,-2,-4,-6))$

• Pivot at second vector, second coordinate
• Replace $(1,2,3,4)$ by $(1,2,3,4) + 2(0,-1,-2,-3) = (1,0,-1,-2)$
• Replace $(0,-2,-4,-6)$ by $(0,-2,-4,-6) - 2(0,-1,-2,-3) = (0,0,0,0)$
• This means that third vector is linearly dependent on first two, and can be dropped

$((1,0,-1,-2),(0,-1,-2,-3))$ is a linearly independent set with span equal to $\text{span}(S)$

• The above form is called “reduced echelon” form
• Extending to basis of $V$
• $(1,0,-1,2),(0,1,2,3),(0,0,1,0),(0,0,0,1)$


text:
# Talk:Overview of African American Research

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)
Jump to: navigation, search

In the section about "1870", can examples of sources/resources be listed?

Matrayback We are still bringing in all the references from the original book. I'd suggest watching this page and keeping your eye out for those changes. If you still have questions after that, let me know.


text:
# How can we calculate the sum of sines or cosines where the angles are in geometric progression?

For example:

$$\cos\frac{\pi}{7} + \cos\frac{3\pi}{7} + \cos\frac{9\pi}{7}$$

In this example, there are only a few terms, and we can use things like $\cos(9\pi/7) = -\cos(2\pi/7)$ and complex numbers to solve it. I am trying to find if there is a closed formula to a generic case

$$\cos(x) + \cos(x\cdot q) + \cos (x\cdot q^2) + \cdots + \cos(x\cdot q^n)$$

• $$\dfrac{9\pi}7=2\pi—?$$ Use math.stackexchange.com/questions/117114/… – lab bhattacharjee May 1 '18 at 15:48
• I believe there is nothing like a closed formula for the general case; there's no 'simplifying formula' for sums of the form $t^{xq^n}$, after all, and $\cos(xq^n)$ is essentially just a 'special case' of this where $t=e^i$. – Steven Stadnicki May 1 '18 at 16:00


text:
# Prove that $\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$

show that $$\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$$

and I found $$LHs-RHS=0.017\cdots$$

I have post this interesting problem Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$

can someone suggest any other nice method? Thank you everyone.

• It is true that $7^8>8^7$, since $(1+\frac{1}{7})^7<e<7$. Then we have, $\sqrt{7}^8>\sqrt{8}^7$. Does it imply that $\sqrt{7}^\sqrt{8}>\sqrt{8}^\sqrt{7}$. – mtm Jun 28 '13 at 9:26
• No,How can have $\sqrt{7}^8>\sqrt{8}^7$? – math110 Jun 28 '13 at 9:31
• $\sqrt{7}^{16}>\sqrt{8}^{14}$. So if we take $\frac{1}{2}$ power of both sides, we have $\sqrt{7}^{8}>\sqrt{8}^{7}$. Maybe i am wrong. – mtm Jun 28 '13 at 9:37
• $\sqrt{7}^{8}>\sqrt{8}^{7}\Longrightarrow \sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$? I think is wrong. – math110 Jun 28 '13 at 10:58
• @mtm Monotonicity argument like yours will not work here, no matter you take $(\ln t)/\sqrt{t}$ or $(\ln t)/t$. – Shuhao Cao Jun 28 '13 at 15:21

Note that $\sqrt{7}^{\sqrt{8}}\doteq15.673$ and $\sqrt{8}^{\sqrt{7}}\doteq15.656$, so these two are pretty close. Furthermore $\sqrt{7}<e<\sqrt{8}$, and the function $f(x):={\log x\over x}$ has a local maximum at $x:=e$. This excludes the use of monotonicity arguments. The following proof uses integer arithmetic instead.

To begin with we need a rational approximation to $\sqrt{7\over8}$ that is slightly larger than $\sqrt{7\over8}$. Using Mathematica (or continued fractions) one finds that $$7\cdot 31^2=6727<6728=8\cdot 29^2\ ,$$ which implies $$\sqrt{7}\cdot31<29\cdot\sqrt{8}\ .$$ Furthermore one computes $$8^{29}=15\>47425\>04910\>67253\>43623\>90528<15\>77753\>82034\>84580\>66150\>42743=7^{31}\ .$$ It follows that $$\bigl(8^{\sqrt{7}}\bigr)^{31}<\bigl(8^{29}\bigr)^{\sqrt{8}}<\bigl(7^{31}\bigr)^{\sqrt{8}}\ .$$ Now take the $62^{\rm th}$ root on both sides.

• Very instructive! +1 – Markus Scheuer Nov 13 '14 at 7:58

The challenge here is to demonstrate the inequality in a way that can easily be checked by hand. Taking logs, the inequality to be proved is

$$\sqrt7\ln8 \lt \sqrt8\ln7$$

which can be written as

$$3\sqrt{1-{1\over8}}\ln2 \lt \ln(8-1) = 3\ln2+\ln(1-{1\over8})$$

so the inequality to prove is

$$-\ln\left(1-{1\over8}\right)\lt 3\left(1-\sqrt{1-{1\over8}}\right)\ln2$$

We'll use the Taylor series

$$-\ln(1-x)=x+{1\over2}x^2+{1\over3}x^3+{1\over4}x^4+\cdots$$

to obtain

\begin{align} -\ln\left(1-{1\over8}\right) &\lt {1\over8}+{1\over2}\left({1\over8}\right)^2+{1\over3}\left({1\over8}\right)^3+{1\over3}\left({1\over8}\right)^4\\ &={1\over8}+{1\over2}\left({1\over8}\right)^2+{1\over3}\left({1\over8}\right)^3{1\over1-{1\over8}}\\ &={1\over8}+{1\over128}+{1\over21}{1\over64}\\ &\lt{1\over8}+{1\over128}+{1\over16}{1\over64}\\ &={128+8+1\over2^{10}}\\ &={137\over2^{10}} \end{align}

We'll use the fact that all the coefficients in the Taylor series

$$1-\sqrt{1-x}={1\over2}x+{1\over8}x^2+{1\over16}x^3+\ldots$$

are positive to get

$$1-\sqrt{1-{1\over8}}\gt {1\over2}{1\over8}+{1\over8}\left({1\over8}\right)^2={33\over2^9}$$

Finally, we use the easily verified identity

$$\ln2=3\ln\left(1+{1\over4}\right)+\ln\left(1+{3\over125}\right)$$

and the alternating Taylor series

$$\ln(1+x)=x-{1\over2}x^2+{1\over3}x^3-\cdots$$

to get

\begin{align} \ln2&\gt 3\left({1\over4}-{1\over2}\left({1\over4}\right)^2+{1\over3}\left({1\over4}\right)^3-{1\over4}\left({1\over4}\right)^4\right)+{3\over125}\\ &={21\over32}+{1\over64}-{3\over1024}+{3\over125}\\ &\gt{21\over32}+{1\over64}-{3\over1024}+{3\over128}\\ &={21\cdot32+16-3+24\over1024}={709\over1024} \end{align}

hence

$$3\ln2\gt{2127\over1024}\gt{2126\over1024}={1063\over512}$$

so that

$$3\left(1-\sqrt{1-{1\over8}}\right)\ln2\gt{33\over2^9}{1063\over2^9}={33\cdot1063\over2^{18}}$$

Consequently, the inequality to verify is

$${137\over2^{10}}\lt{33\cdot1063\over2^{18}}$$

which is to say,

$$256\cdot137\lt33\cdot1063$$

Eschewing calculators, note that $1063=1024+32+8-1$, so

$$33\cdot1063=(2^5+1)(2^{10}+2^5+2^3-1)=2^{15}+2^{11}+2^8+2^3-1$$

whereas

$$256\cdot137=2^8(2^7+2^3+1)=2^{15}+2^{11}+2^8$$

so the desired inequality holds.

• Reading along, I kept expecting the computation to end, but it never did! A rather intimidating verification. – Ryan Reich Aug 8 '13 at 5:29
• @RyanReich, but it did end! ;-) – Barry Cipra Aug 8 '13 at 6:00

I just want to give you a bigger picture on this question. The fact you point out corresponds to $f(2)>0.$

$f(\text{x$\_$})\text{:=}\left(t=\lfloor \exp (x)\rfloor ;a=t^{1/x};b=(t+1)^{1/x};a^b-b^a\right);$

$\text{Plot}[f(x),\{x,1,3\}]$


text:
## Experimental Mathematics

### Minimal Permutation Representations of Nilpotent Groups

#### Abstract

A minimal permutation representation of a finite group $G$ is a faithful $G$-set with the smallest possible size. We study the structure of such representations and show that for certain groups they may be obtained by a greedy construction. In these situations (except when central involutions intervene) all minimal permutation representations have the same set of orbit sizes. Using the same ideas, we also show that if the size $d(G)$ of a minimal faithful $G$-set is at least $c|G|$ for some $c>0$, then $d(G) = |G|/m + O(1)$ for an integer $m$, with the implied constant depending on $c$.

#### Article information

Source
Experiment. Math. Volume 19, Issue 1 (2010), 121-128.

Dates
First available in Project Euclid: 12 March 2010


text:
# Tangents to parabola

1. Sep 21, 2008

### Mentallic

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

I am given the parabola $$y=\frac{x^2}{2}$$

I need to find the equations of the 2 tangents to the parabola that pass through the point A(5,-2)

2. Relevant equations

$$y-y_{1}=m(x-x_{1})$$

$$a = \frac{1}{2}$$
therefore: tangents pass through the points $$P(p,\frac{p^{2}}{2})$$ and $$Q(q,\frac{q^{2}}{2})$$

$$\frac{dy}{dx}=x$$

3. The attempt at a solution

I began to check if the given point was outside the parabola
i.e. $$y_{1}<\frac{x^{2}_{1}}{2}$$

$$-2<\frac{5}{2}$$ therefore, the point lies outside the parabola and there are 2 lines that will pass through the point, and are a tangent to the parabola.

$$y+2=m(x-5)$$ where there are 2 values of m, each intersecting the parabola only once. i.e. tangent to parabola.

From here I am totally stumped. I can't use the 1st derivative as I don't know the x value for which the gradient will pass through the point.
Any help would be much appreciated.

Last edited: Sep 21, 2008
2. Sep 21, 2008

### Knissp

This means that your equation of your line must pass through two given points.

(5, -2) is one of them, so your equation $$y+2=m(x-5)$$ is a good place to start.

First, what is m? Remember, m is your slope...

Once you got that, now where does that second point have to be? If the line is tangent to the parabola with equation $$y=\frac{x^2}{2}$$...

3. Sep 21, 2008

### HallsofIvy

Staff Emeritus
You want to find (x, y) such that y= m(x- 5)- 2 and y'(x)= m. You should certainly be able to find the derivative of y in terms of x. Putting that in for y' gives three equations for x, y, and m. this will, of course, reduce to a quadratic equation so you can expect two solutions: two tangents.

4. Sep 21, 2008

### Mentallic

Thank you, I managed to solve it.

Last edited: Sep 22, 2008


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# How does the Mermin-Ho theorem handle singularities arising from the hairy ball theorem?

I am reading about the derivation of the Mermin-Ho relation in chapter III. B of (Kamien, 2002) and trying to understand how it fits with the hairy ball theorem.

Assuming you have a unit vector $$\hat{n}(\vec{r})$$ which covers the entire unit sphere and can be presented in the form $$\hat{n}(\vec{r}) = \hat{e}_1(\vec{r}) \times \hat{e}_2(\vec{r})$$, this gives rise to a spin connection $$\Omega(\vec{r}) = \hat{e}_1(\vec{r}) \cdot \nabla \hat{e}_2(\vec{r})$$. The Mermin-Ho relation now establishes, that it's curl is independent of the choice of $$\hat{e}_1(\vec{r})$$, $$\hat{e}_2(\vec{r})$$ via $$[\nabla \times \Omega(\vec{r})]_i = \frac{1}{2} \varepsilon_{\alpha\beta\gamma} n^\alpha(\vec{r}) \varepsilon_{ijk} \partial_j n^\beta(\vec{r}) \partial_k n^\gamma(\vec{r})$$

However, considering $$\hat{n}(\vec{r})$$ as the position on a unit sphere, then $$\hat{e}_1(\vec{r})$$ (as well as $$\hat{e}_2(\vec{r})$$) forms a tangent vector field to the unit sphere and – following the hairy ball theorem/Poicaré-Hopf theorem – has to have a singularity of index 2. This leads to a singularity in $$\Omega(\vec{r})$$ and taking the curl should give an additional $$\delta$$-function of strength $$4 \pi$$.

In other words, why doesn't the Mermin-Ho relation read something like $$[\nabla \times \Omega(\vec{r})]_i = \frac{1}{2} \varepsilon_{\alpha\beta\gamma} n^\alpha(\vec{r}) \varepsilon_{ijk} \partial_j n^\beta(\vec{r}) \partial_k n^\gamma(\vec{r}) + 4 \pi \delta(\vec{r}-\vec{r}^\prime)$$ where $$\vec{r}^\prime$$ captures where the index-2 singularity of $$\hat{e}_1(\vec{r})$$, and therefore of $$\Omega(\vec{r})$$, is placed?

This is also a conceptual question with respect to the physics:

As a $$\Omega(\vec{r})$$ can be interpreted as a vector potential, its curl constitutes a magnetic field $$\vec{B} = \nabla \times \Omega$$. If $$\hat{n}(\vec{r})$$ winds exactly once around the unit sphere, as it is the case for a magnetic skyrmion, the integrated magnetic flux $$\Phi = \int d^2r B(\vec{r}) = -4 \pi$$ (for $$e = \hbar = 1$$). However, this flux would be cancelled exactly by the $$\delta$$-function of strength $$4 \pi$$. So, is there a total magnetic flux or does it vanish?

## 1 Answer

The whole derivation of the Mermin-Ho relation required a choice of $$\hat{e}_1$$ and $$\hat{e}_2$$ so that $$\hat{n} = \hat{e}_1 \times \hat{e}_2$$ everywhere. The construction of $$\Omega$$ requires them to be differentiable.

However, from the hairy ball theorem we know that it is impossible for $$\hat{e}_1$$ and $$\hat{e}_2$$ to be smooth and differentiable. Thus, there are places, where $$\Omega$$ is singular – but right there, the Mermin-Ho relation simply does not apply because our initial assumption, of $$\hat{e}_1$$ and $$\hat{e}_2$$ being smooth, does not hold.

To define the Mermin-Ho relation, poke a hole everywhere on the unit sphere where singularities occur and define $$\hat{e}_1$$ and $$\hat{e}_2$$ everywhere else apart from these holes. This way, one has excluded the singularities.


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## Saturday, January 16, 2016

### How Hot Is The Lounge?

“When you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind”
― Lord Kelvin

Approaching Christmas it was hot, enough to keep checking the lounge room temperature obsessively. To keep the house cool without air conditioning the windows are opened at night then closed during the heat. Recently it was an extreme 37° outside while inside was a more livable 27°.
 These digital thermometers are great but they only tell you the temperature right now.
A couple of digital thermometers at less than A$2 per piece are used to know when it's hotter out than in and then I have to to keep observing so as to close or open the windows appropriately. Obsession though, requires better than spot observations. How cool does it stay inside compared to out, how often is it uncomfortably hot, how cool does it need to be at night to cool the house sufficiently and so on? The house has a lot of thermal mass so how much less effective is this technique when there is successive hot days? Answering these questions requires more data. No problem, I'll get a sensor that records observations and to make analysis easy it should publish the data online.  This sensor from Wireless Loggers looks good and runs on a coin battery but is a hefty A$ 77 per unit.
 These sensors also need a wireless ethernet base station which adds complexity and another A$164 in expense. The closest consumer product I could find was from Wireless Loggers . This was a more expensive solution than I'd anticipated and the data was online but only accessible through a phone app. Sometimes things don't advance as fast as you'd expect. I was putting sensors inside cow's stomachs back in 2004 and watching temperatures remotely as they walked around eating and drinking using similar bespoke technology. Back then we were expecting this would be point and click easy and cost only a few dollars once sensor networks were commercialised, yet 12 years later it still looks difficult. A pencil and paper was looking attractive but I'd also bought a convenient to use NodeMCU version of the esp8266 12-E microcontroller with integrated WiFi as I was reading about them everywhere as being functional and cheap, this one from Ali Express at just US$4.15 delivered.
 The esp8266 12-E modules are tiny, cheap, lots of I/O and reasonably powerful.
 Development is a lot more convenient when the esp8266 12-E is deployed in this NodeMCU format. It can be programmed directly from a USB port on a PC, powered from a USB charger during use and provides easy access to all I/O pins when plugged into a prototyping breadboard.
Could I use this newfangled esp8266 to satisfy my curiosity?

After Christmas I got to spend a few days with Chris, a smart guy I've known since high school. He is living the dream, with the Hawkesbury River 3 metres from the door. It would be hard to get bored with that, but just in case he has a spare house at Chittaway Point, again with a wharf in the back yard. Between the crabbing, bush walking and watching boats pass from the balcony we had a go at learning how to use the esp8266.

The esp8266 first became available as an AT command driven serial modem but now several alternative development environments are available of which  LUA seems the most popular. We opted for Arduino which needs to be configured to program the esp8266 and the easiest way to get started is to follow the "Installing with Boards Manager" instructions. With Arduino, the standard method of testing if everything is set up correctly  is to compile and upload the code to blink a led on the target device. We found an esp8266 version of blink and in a few minutes were watching a blinking led. Trivial perhaps, but satisfying. A lot later there was simulated data periodically being uploaded to an online service.

We'd made more progress than I expected and I hadn't sourced a temperature sensor, so that had to wait until I returned home. A DS18B20 temperature sensor looks ideal at 5 units for A$3.88 and they are also available with a metre long waterproof cable for A$2.22. The DS18B20 reads temperature resolved to 9 bits by default and is configurable to 12 bits with an absolute accuracy of +/- 0.5 degrees. It needs only power, earth and a connection to a single GPIO pin. Multiple sensors can also be connected to the same pin. The DS18B20 documentation says a 4.7K ohm pull up resistor is required between the power and data pins and other implementations specify this also but the internal pull up in the ESP8266 12-E which is reported to range between 30K - 100K ohms has proved sufficient for me with two sensors connected physically close to the esp8266 12-E module. Later I came across a pedant aiming to Save The Earth One Resistor At A Time who analyses why a resistor is rarely required.
 Connecting the DS18B20 temperature is this simple. Most documentation says a 4.7K pull up resistor is required between the data and power lines but the internal pull up resistor is good enough.
Using the sensor requires the Arduino temperature control library, which in turn requires the One Wire library for communication. It is possible to have installed a One Wire library previously so there might be two. If so the version that was installed with the other esp8266 libraries must be used.

We want to capture rapid temperature changes but minimise data streamed, so the algorithm I settled on is:-

• read the temperature every 5 seconds,
• if it hasn't changed much go back to sleep,
• otherwise log measured temperature to a web service.
If no data has been logged for a while usually the temperature hasn't changed but if I don't see new data I worry the logger has failed, so to provide reassurance, log at least every10 minutes.

After implementing this, when I have the sensor running, we can see live data:

or we can view it in a dashboard. Feel free to try this yourself, using our code. Before compiling, config-sample.h needs to be renamed to config.h and all the "TODO" fields replaced with appropriate values. For example, you must enter your WiFi access point name and password.

The problem was solved for only a few dollars if you put zero value on your time but this was as much a learning experience as a solution and I've seen enough for it to be already apparent why the esp8266 has generated so much interest. A couple of bucks for a tiny integrated microcontroller, with lots of I/O and inbuilt communications is what is required for the oft spoken of sensor networks to emerge on a scale proponents have envisaged. CSIRO spent many millions trying to develop something similar, marketed as a Fleck and now relegated to the sensor networks museum.  The esp8266 from Espressif Systems (Shanghai) is an order of magnitude more capable. That capability has attracted a buzz, will surely attract competitors and Espressif are close to releasing an even more capable esp32 that can additionally communicate over Bluetooth Low Energy.

I now know how hot the lounge has been though this is not yet a quality solution and there are others including a Lua version by Eyal Lebedinsky from our local hacker space. Ideally you want a device to arrive in the mail, fit a battery, look at a web browser and see the temperature. Getting a device that works out of the box and also comes with enough information to make tinkering easy would be nice. If time has value, the Wireless Loggers would have been much cheaper this time around but properly developed, the esp8266 could be at the core of a far superior solution and surely someone will do a commercial esp8266 based temperature logger soon.


text:
# Difference Quotient – Definition, Formula, & Examples

When you hear of the difference quotient, what are the math concepts that you have in mind? What’s an algebra concept that involves differences and quotients? Well, if you guessed slope, you’re actually close to the definition of the difference quotient.

The difference quotient measures the slope of a secant line passing through the curve of $f(x)$.

There are two important keywords here that we need to understand: slope and secant line.

A linear function’s slope represents the measure of its “rise over run”. What does this mean mathematically?

Slopes (normally represented as $m$) are calculated by finding the difference between two given points’ $y$-coordinates and $x$-coordinates then taking the quotient of the two.

$m=\dfrac{y_2 – y_1}{x_2 – x_ 1}$

Now, what are secant lines? They are the lines that pass through a figure or a graph at exactly two points.

The graph shown above is an example of a secant line passing through a curve. As you can see, the secant line passes through two points along the curve.

## What is the difference quotient?

The difference quotient of a function measures the average rate of change of $f(x)$ with respect to x given an interval, $[a, a + h]$.

Given a function, $f(x)$, its difference quotient tells us the slope of the line that passes through two points of the curve: $(a, f(a))$ and $((a+h), f(a+h))$. Let’s visualize this to better understand the concept of the difference quotient.

As can be seen from the graph shown, the difference quotient measures the slope of the line passing through $a$ and $a + h$.

### Understanding the difference quotient formula

So, how do we apply this definition and calculate for the difference quotient of a function given a secant line? We use our knowledge of slope to establish the formula for the difference quotient.

Let’s determine the function’s value at $a + h$ and $a$. Once we have the values, we find the difference between the two to find the “rise” of the function or $\Delta y$ at those two points.

$\Delta y = f(a+h) – f(a)$

The difference between the two intervals represents the slope’s run or $\Delta x$. Hence, we have

$\Delta x = (a+h) – a = h$

With these two, we can now write the formula for the difference quotient as

$\dfrac{\Delta y}{\Delta x} = \dfrac{f(a+h) – f(a)}{h}$

### How to evaluate difference quotient?

$\dfrac{\Delta y}{\Delta x} = \dfrac{f(a+h) – f(a)}{h}$

Now that we know the difference quotient’s definition and formula, it’s time that we learn how to actually apply them. Here are four steps to remember when evaluating difference quotients:

1. Find the expression of $f(a+h)$ by substituting $x$ in $f(x)$ with $a + h$.
2. Find the expression of $f(a)$ by substituting $x$ in $f(x)$ with $a$.
3. Find the difference between the two.
4. Divide the expression by $h$.

Let’s try to evaluate $f(x) = \Delta x$. Start by finding $f(a + h)$:

\begin{aligned}f(a) &= 4a \\ f(a + h) &= 4(a + h)\\ &= 4(a) + 4(h)\\ &= 4a + 4h\end{aligned}

Subtract $f(a + h)$ and $f(a)$ then divide the result by $h$:

\begin{aligned}\dfrac{f(a+h) – f(a)}{h} &= \dfrac{4a + 4h – 4a}{h}\\ &= \dfrac{4h}{h} \\&=4\end{aligned}

This means that the difference quotient of $f(x) = 4x$ is $4$.

## How to simplify difference quotient?

We’ve worked out the difference quotient of $f(x) = 4x$. Now, what if we’re working on more complicated functions?

The same process will be applied, but it’s important for us to remember the tricks and techniques in simplifying algebraic expressions.

• When simplifying the numerator, $f(a+h) – f(a)$, make sure to distribute the negative signs correctly.
• Factor out $h$ from the numerator whenever possible.
• When working with quadratic expressions, remember the algebraic property, $(m \pm n)^2 = m^2 \pm 2mn + n^2$.
• Apply the FOIL method whenever possible too.
• When adding two rational expressions, rewrite the functions so that they share a common denominator.
• Do your best to simplify the numerator first. Remember, $\dfrac{f(a+h) – f(a)}{h} = \dfrac{1}{h} \cdot [f(a+h) – f(a)]$, so you can focus on the denominator later.

Now that we have these helpful tips, let’s try to simplify the difference quotient of the function shown below.

$f(x) = \dfrac{1}{x^2}$

We begin by finding the expression for $f(a+h)$. Substitute $a+h$  into the expression for x and apply the algebraic property, $(m \pm n)^2 = m^2 \pm 2mn + n^2$.

$f(a + h) = \dfrac{1}{ (a + h)^2}$

Find the difference between $f(a + h)$ from the expression of $f(a)$. Rewrite the two fractions so that they share a common denominator.

$f(a + h) – f(a) = \dfrac{1}{ (a + h)^2} – \dfrac{1}{ a^2}$

Rewrite the two fractions so that they share a common denominator.

$= \dfrac{a^2}{a^2(a + h )^2}– \dfrac{(a + h )^2}{a^2 (a + h)^2}$

Combine the numerators and expand the factored expressions using $(m \pm n)^2 = m^2 \pm 2mn + n^2$. Don’t forget to distribute the negative signs!

\begin{aligned} = \dfrac{a^2 – (a + h )^2}{a^2 (a + h )^2} &= \dfrac{a^2 – (a^2 + 2ah + h^2)}{a^2 (a + h )^2}\\&= \dfrac{a^2 – a^2 – 2ah – h^2}{a^2 (a + h )^2}\\ &= \dfrac{– 2ah – h^2}{a^2 (a + h )^2}\\ &= \dfrac{-h(2a + h)}{a^2 (a + h )^2}\end{aligned}

To find the difference quotient, let’s divide the resulting expression by $h$.

\begin{aligned}\dfrac{f(a+h) – f(a)}{h } &= \dfrac{\dfrac{1}{(a + h )^2} – \dfrac{1}{a^2}}{ h}\\ &= \dfrac{– h(2a + h)}{a^2 (a + h )^2}{h}\\ &= \dfrac{1}{h} \cdot \dfrac{– h(2a + h)}{a^2 (a + h )^2}\end{aligned}

Cancel out common factors to further simplify the difference quotient.

$\dfrac{f(a+h) – f(a)}{h} = -\dfrac{ (2a + h)}{a^2 (a + h )^2}$

This means that the difference quotient of $f(x) = \dfrac{1}{x^2}$ is$-\dfrac{2a + h}{a^2 (a + h )^2}$.

Don’t worry, the last example will also be easy once we practice more problems involving difference quotients.

Example 1

If $f (a + h) = 4a + 4h + 3$ and $f(a) = 4a + 3$, which of the following statements is not true?

1. The function, $f(x)$, is equal to $4x + 3$.
2. The difference between $f (a)$ and $f (a + h)$ is $4h$.
3. The difference quotient of $f(x)$ is $4$.
4. The difference quotient of $f(x)$ is $4h$.

Solution

To confirm whether $f(x)$ is actually equal to $4x + 3$, let’s compare $f(a)$ and the factored form of $f(a + h)$:

\begin{aligned} f (a) &= 4a + 3 \\f (a + h) &= 4(a + h) + 3\end{aligned}

From this, we can see that $f(x)$ is equal to $4x + 3$. This confirms that (a) is a true statement.

Let’s go ahead and subtract the two given expressions.

\begin{aligned}f (a + h) – f(a) &= 4a + 4h + 3 – (4a + 3)\\ &= 4a + 4h + 3 – 4a – 3\\ &= 4h\end{aligned}

This means that (b) is correct.

Find the difference quotient by dividing the $f (a + h) – f(a)$  by $h$.

\begin{aligned}\dfrac{f (a + h) – f(a)}{h} &= \dfrac{4h}{h}\\ &=4\end{aligned}

This means that the correct statement between (c) and (d) is (c), so the statement that is not correct is (d).

Example 2

What is the difference quotient of the function represented by the graph shown?

Solution

When working with problems that involve the difference quotient, let’s also go back to its definition:

Difference Quotient $=\dfrac{f (a + h) – f(a)}{h}$

Since both $f(a)$ and $f(a + h)$ are already given, let’s go ahead and subtract the two expressions. Note: it’s important to double-check when distributing the negative sign.

\begin{aligned}f(a+h) – f(a) &= 4(a + h) + 9 – (4a + 9) \\ &= 4a + 4h + 9 – 4a – 9\\ &= 4h \end{aligned}

Divide the resulting difference by $h$ to find the difference quotient.

Difference Quotient $= \dfrac{4h}{h} = 4$

This means that the function shown has a difference quotient of $4$.

Example 3

What is the difference quotient of the function represented by the graph shown below?

Solution

This is a similar problem, but this time, we’re working with a more complex expression. Let’s go ahead and write the difference quotient formula:

Difference Quotient $=\dfrac{f (a + h) – f(a)}{h}$

Since both $f(a)$ and $f(a + h)$ are also given, let’s go ahead and subtract the two expressions. Since the first term is a squared binomial, we can expand it using the algebraic property, $(m \pm n)^2 = m^2 \pm 2mn + n^2$.

\begin{aligned} f(a+h) – f(a) &= 3(a + h)^2 + 5 – (3a^2 + 5)\\ &= 3(a^2 + 2ah + h^2) + 5 – 3a^2 – 5\end{aligned}

Distribute $3$ into the parenthesis and combine like terms.

\begin{aligned}&= 3a^2 + 6ah + 3h^2 + 5 –3a^2 – 5\\ &= 6ah + 3h^2\end{aligned}

Once we have the simplified difference, let’s divide the expression by $h$.

\begin{aligned} \text{Difference Quotient} &= \dfrac{f(a+h) – f(a)}{h}\\ &= \dfrac{6ah + 3h^2}{h }\\&= 6a + 3h \end{aligned}

This means that $3x^2 + 5$ has a difference quotient of $6a + 3h$.

Example 4

What is the difference quotient of the function $f(x) = 3(x + 1) – 5$?

Solution

When finding the difference quotient of a function, let’s begin by finding $f(a)$ and $f(a + h)$.

\begin{aligned} f(a) &= 3(a + 1) – 5\\ &= 3a + 3 – 5\\ &= 3a – 2\\\\f(a + h) &= 3(a + h + 1) – 5\\ &= 3a + 3h + 3 – 5\\ &= 3a + 3h – 2\end{aligned}

Find the difference between the two expressions.

\begin{aligned} f(a+h) – f(a) &= 3a + 3h – 2 – (3a – 2)\\ &= 3a + 3h – 2 – 3a + 2\\ &= 3h\end{aligned}

Now that we have the difference, let’s divide the expression by $h$ to find the difference quotient of $f(x)$.

\begin{aligned} \text{Difference Quotient} &= \dfrac{f(a+h) – f(a)}{h}\\ &= \dfrac{3h}{h} \\&= 3\end{aligned}

This means that $f(x) = 3(x + 1) – 5$ has a difference quotient of $3$.

Example 5

What is the difference quotient of the function $f(x) = x^2 – 4x + 3$?

Solution

When finding the difference quotient of a function, let’s begin by finding $f(a)$ and $f(a + h)$.

\begin{aligned}f(a) &= a^2 – 4a + 3\\f(a + h) &= (a + h)2 – 4(a + h) + 3 \end{aligned}

Expand $f(a + h)$ using the algebraic property, $(m ± n)^2 = m^2 \pm 2mn + n^2$. Distribute $4$ in the second group then combine like terms.

\begin{aligned}f(a + h) &= (a^2 + 2ah + h^2) – 4(a + h) + 3\\ &= a^2 + 2ah + h^2 – 4a – 4h + 3\\ &= a^2 + 2ah + h^2 – 4a – 4h + 3\end{aligned}

Find the difference between the two expressions.

\begin{aligned} f(a+h) – f(a) &= a^2 + 2ah + h^2 – 4a – 4h + 3 – (a^2 – 4a + 3)\\ &= a^2 + 2ah + h^2 – 4a – 4h + 3 – a^2 + 4a – 3\\ &= h^2+ 2ah – 4h\end{aligned}

Now that we have the difference, let’s divide the expression by $h$ to find the difference quotient of $f(x)$.

\begin{aligned} \text{Difference Quotient} &= \dfrac{h^2+ 2ah – 4h}{h}\\ &= h + 2a – 4\end{aligned}

This means that $f(x) = x^2 – 4x + 3$ has a difference quotient of $h + 2a – 4$.

Break time! How are you doing so far? Don’t forget to double-check your calculations (especially for tricky numerators). Make use of properties as much as you can and if you think you need some refresher, take your old resources out (our site has a lot or resources too!) and review some techniques that may help.

Let’s continue and work with more complicated expressions and answer some word problems as well.

Example 6

What is the difference quotient of the function $f(x) = \dfrac{1}{4x}$?

Solution

Working with rational expressions may appear intimidating at first when finding the difference quotient of a function. This example will show how, with the right technique, we can actually manipulate the expressions faster. What are we waiting for? Let’s begin!

Find the expressions of $f(a)$ and $f(a + h)$.

\begin{aligned}f(a) &= \dfrac{1}{4a}\\ f(a + h) &= \dfrac{1}{4(a + h)}\end{aligned}

No need to expand $f(a + h)$ first and let’s proceed to finding the difference of $f(a)$ and $f(a + h)$. Rewrite each expression so that they share a common denominator, $4a(a + h)$, then combine the terms in the numerator.

\begin{aligned} f(a+h) – f(a) &= \dfrac{1}{4(a + h)} – \dfrac{1}{4a}\\ &= \dfrac{a}{4a(a + h)} – \dfrac{a + h}{4a(a + h)}\\ &=\dfrac{a – (a + h)}{4a(a + h)}\\ &= \dfrac{a – a – h}{4a (a + h)}\\&= -\dfrac{h}{4a (a + h)}\end{aligned}

Now that we have the difference, let’s divide the expression by $h$ to find the difference quotient of $f(x)$. Remember that we can instead multiply the difference by $\dfrac{1}{h}$ to find the quotient.

\begin{aligned} \text{Difference Quotient} &= \dfrac{h^2+ 2ah – 4h}{h}\\ = \dfrac{1}{h} \cdot -\dfrac{h}{4a (a + h)}\end{aligned}

We can then cancel out $h$ to simplify the quotient.

$=-\dfrac{1}{4a(a + h)}$

This means that $f(x) = \dfrac{1}{4x}$ has a difference quotient of $-\dfrac{1}{4a(a + h)}$ or $-\dfrac{1}{4a^2 + 4ah}$.

Example 7

What is the difference quotient of the function $f(x) = \dfrac{2x}{3-x}$?

Solution

Here’s another rational expression for us to work on. Let’s go ahead and find the expressions of $f(a)$ and  $f(a + h)$.

\begin{aligned} f(a) &= \dfrac{2a}{3 – a}\\ f(a + h) &= \dfrac{2(a + h)}{3- (a+h)} \\ &= \dfrac{(2a + 2h)}{3 – a -h}\end{aligned}

Now, let’s subtract the $f(a)$ and $f(a + h)$. We can do this by first rewriting each fraction so that they share their common denominator, $(3 – a – h)(3 – a)$.

\begin{aligned} f(a+h) – f(a) &= \dfrac{2a + 2h}{3 – a – h} – \dfrac{2a}{3 – a} \\&= \dfrac{(2a + 2h)(3 – a )}{(3 – a – h)(3 – a)} – \dfrac{\dfrac{2a}{3 – a – h}}{\dfrac{3 – a – h}{(3 – a}}\\&= \dfrac{6a – 2a^2 + 6h – 2ha – (6a – 2a^2 – 2ah)}{(3 – a – h)(3 – a)}\\ &= \dfrac{6h}{(3 – a – h)(3 – a)}\end{aligned}

We can stop simplifying the difference here and proceed to finding the difference quotient of $f(x)$. Remember that we can instead multiply the difference by $\dfrac{1}{h}$ to find the quotient.

\begin{aligned} \text{Difference Quotient} &= \dfrac{h^2+ 2ah – 4h}{h}\\ &= \dfrac{1}{h} \cdot \dfrac{6h}{(3- a -h)(3 – a)}\end{aligned}

We can then cancel out $h$ to simplify the quotient.

$= \dfrac{6}{(3 – a – h)(3 – a)}$

This means that $f(x) = \dfrac{2x}{3 – x}$ has a difference quotient of $\dfrac{6}{ (3 – a – h)(3 – a)}$.

Time for a cool down. Now that we’ve worked our mind muscles in working on these problems, let’s go ahead and slow down by answering the conceptual problems.

Example 7

What is the difference quotient of $f(x) = 8$? How about $g(x) = \pi$? What’s the reason behind their difference quotients?

Solution

Let’s quickly find each of the important components in finding their respective difference quotients.

$f(a) = 8$,               $g(a) = \pi$

$f(a + h) = 8$,        $g(a + h) = \pi$

This means that the difference for each is $0$ and consequently, each of their difference quotients are equal to zero.

The constant functions $f(x)$ and $g(x)$ will always remain constant throughout any given interval. This means that the rate of change in any point of their curves will be zero.

Example 8

The general form of linear functions can be expressed as $f(x) = px + q$, where $p$ and $q$ are nonzero coefficients. Show that the difference quotient of any linear function is equal to the coefficient before $x$.

Solution

Let’s go ahead and express $f(a)$ and $f(a + h)$ in terms of $p$ and $q$.

\begin{aligned} f(a) &= pa + q\\ f(a + h) &= p(a + h) + q\\ &= pa + ph + q\end{aligned}

Subtract the two expressions and divide the result by $h$.

\begin{aligned} \text{Difference Quotient} &= \dfrac{h^2+ 2ah – 4h}{h}\\ &= \dfrac{pa + ph + q – (pa + q)}{h}\\ &= \dfrac{ph}{h}\\ &= p\end{aligned}

We’ve just shown that the difference quotient of $f(x) = px + q$ is $p$ or the coefficient before $x$.

### Practice Questions

1. If $f (a + h) = 6a + 6h + 5$ and $f(a) = 6a + 5$, which of the following statements is not true?
a. The function, $f(x)$, is equal to $6(x + 1)$.
b. The function, $f(x)$, is equal to $6x + 5$.
c. The difference quotient of $f(x)$ is $6h$.
d. The difference quotient of $f(x)$ is $6$.
2. What is the difference quotient of the function represented by the graph shown below?

3. What is the difference quotient of the function represented by the graph shown below?

4. What is the difference quotient of the function of the following linear functions?
a. $f(x) = 2x – 5$
b. $f(x) = 5(2x – 3) – 6$
c. $f(x) = \dfrac{x}{3}- 8$
5. What is the difference quotient of the following quadratic functions?
a. $f(x) = x^2– 2x + 1$
b. $f(x) = 6x^2– 4$
c. $f(x) = 3x^2 – 5x – 2$
6. What is the difference quotient of the following rational functions?
a.$f(x) = \dfrac{1}{4x}$
b. $f(x) = \dfrac{1}{(x – 1)^2}$
c. $f(x) = \dfrac{4}{(x + 1)^2}$