text:
I'm trying to plot a number of points such that each has a colour that depends on a certain parameter. I have two lists, points and heights, and I would like to do something like this:
for i in range(len(points)):

This creates len(points) different graphics, so it's not what I want. What do I need to change?


text:
Photovoltaic and Superconductivity

My question deals with experimentation in combining these two technologies. I think it would be a good duo, since PV's are so inefficient, and superconductivity is so efficient. I would like to know if people agree or disagree (and why), where I could find useful information on this topic (experimental and theoretical), or any other ideas on improving PV efficiency.

Related Quantum Physics News on Phys.org
I would submit that, at the bleeding edge, PV aren't as inefficient as you might think, and superconductors aren't as practical as you might think.

f95toli
Gold Member
How exactly would you "combine" superconductivity and PV?
Superconductors are -from an electrical point of view- just very good conductors (at least at DC) so I can't see anyway of improving PVs by making some parts out of a superconducting material.
You CAN make very sensitve photon detectors using superconducting materials; but none of the types I can think of actually generates energy; they are usually read-out by detecting changes in e.g. the impedance or the current-voltage charachteristics.

I am not sure it is possible, I imagine that someone might have tinkered with this. I wish I had a better understanding, but I will give it a go anyways
Theoretical (-math)
I think electrons may jump to there conductive band easier utilizing the phonons in the formation of the cooper pair. The electrons now look more like a conveyor belt than an ort cloud to move away uniformally and uninhibited.
Experimental (?funding)
I would try to use elements which show superconductive properties (high temp superconductors) to make the photovoltaic cells. Then place the cells in a glass container, apply a strong vacuum, and use liquid nitrogen bringing it to superconductive temperatures. Then I would place light on them. I might try filter hot frequencies.
Of course, I dont really know what Im doing. I thought it sounded nice though:).

Of course, I dont really know what Im doing. I thought it sounded nice though:).
No offense, but that's kind of the problem. Just because you can say something that sounds nice by throwing together a bunch of words, it doesn't mean that it's in anyway realistic or even sensical.

f95toli
You can of course use light with $hf>2\Delta$ to split Cooper-pairs into two electrons which can create an "avalanche" effect which generates more electrons (quasiparticles), this is how some superconducting detectors work. However, this process is not efficient enough to result in any net energy production. There are also quite a few other processes that can be used for detector applications; but none that can be used to make PV cells.

format: plain


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• Today, 10:41
Here is this week's POTW: A charge $-q_1$ is located at the origin of a $y$-axis and a charge $-q_2$ is located at $y=d$. At what point along...
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Congratulations to the following members for their correct submission: kaliprasad My solution is as follows: For the entire system...
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• Yesterday, 15:53
mathmari replied to a thread Open cover in Analysis
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mathmari started a thread Open cover in Analysis
Hey!! :o Let $\mathbb{R}$ provided with the metric $d(x,y)=|x-y|$. I want to check if the collections of sets $$S_1=\left \{\left (\frac{x}{2},... 3 replies | 48 view(s) • Yesterday, 09:31 I like Serena replied to a thread Right triangle. in Geometry Google translates triangle as sankakkei. And searching for sankakkei shows that this word is also used for a three dimensional puzzle of triangles.... 19 replies | 193 view(s) • April 24th, 2018, 06:54 My own best guess was to look at the differences: \frac 13, \frac 16, \frac 1{10}. The next difference might be \frac 1{15}, although it does... 9 replies | 141 view(s) • April 23rd, 2018, 15:52 I like Serena replied to a thread Special Relativity in Physics It's the other way around. 11 replies | 135 view(s) • April 23rd, 2018, 15:26 I like Serena replied to a thread Right triangle. in Geometry For the record, something like that is called a sangaku. A drawing that exhibits a proof, without any words, expressions, or symbols. Admittedly my... 19 replies | 193 view(s) • April 23rd, 2018, 15:04 I like Serena replied to a thread Right triangle. in Geometry You mean just drawings? (Wondering) Something like this: \begin{tikzpicture} %preamble \usetikzlibrary{arrows} \def\a{sqrt(17)} ... 19 replies | 193 view(s) • April 23rd, 2018, 13:37 I like Serena replied to a thread Right triangle. in Geometry To be honest, I only used trigonometry because it means I have to use fewer words to explain those ratios! (Rofl) 19 replies | 193 view(s) • April 23rd, 2018, 13:20 I like Serena replied to a thread Right triangle. in Geometry Trigonometry is only used in the first step. We can replace it with non-trigonometry by observing that triangles ABC and ACD are similar, meaning... 19 replies | 193 view(s) • April 23rd, 2018, 12:55 I like Serena replied to a thread Right triangle. in Geometry Let's make a drawing: \begin{tikzpicture} %preamble \usetikzlibrary{arrows} \def\a{sqrt(17)} \def\b{sqrt(68)} \def\c{sqrt(85)} ... 19 replies | 193 view(s) • April 23rd, 2018, 09:13 I like Serena replied to a thread Special Relativity in Physics Indeed, the distance travelled is v\Delta t' = 240\text{ lightyear}. So we have \Delta t' = \frac{240\text{ lightyear}}{v} and \Delta t =... 11 replies | 135 view(s) • April 22nd, 2018, 17:50 mathmari replied to a thread Banach space in Analysis I understand!! Thank you very much!! (Smile) 6 replies | 87 view(s) • April 22nd, 2018, 12:30 Let me throw in another XKCD: I do hope that no one drops our people table! (Worried) 19 replies | 522 view(s) • April 22nd, 2018, 11:32 I like Serena replied to a thread Optics: Lenses in Physics Normally the object and image are at different sides of the lense, meaning their distances add up. However, it is possible that we have a virtual... 6 replies | 89 view(s) • April 22nd, 2018, 10:17 MarkFL replied to a thread Olinguito in Introductions Hello, and welcome to MHB! (Wave) 8 replies | 106 view(s) • April 22nd, 2018, 09:31 Good! (Happy) We have:$$\epsilon = -\d \Phi t = -\d {(B_i A_i + B_o A_o)} t\approx -A_i\d {B_i} t  where $A_i$ is the cross sectional area...
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• April 22nd, 2018, 09:17
mathmari replied to a thread Banach space in Analysis
I got stuck right now. I haven't really understood that part. (Thinking) I thought that the Cauchy sequence should be in $Y$ but its limit...
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• April 22nd, 2018, 08:59
I like Serena replied to a thread Banach space in Analysis
$C(\mathbb{R})$ means that $\forall t\in \mathbb R: \lim_{h\to 0} f(t+h)=f(t)$. Will that be true if $f$ is the absolute converging limit of a...
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• April 22nd, 2018, 08:47
How about not letting ourselves be stopped by that, but just do the math?
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• April 22nd, 2018, 08:45
I like Serena replied to a thread Optics: Lenses in Physics
Good! It means that we have the set of equations: \begin{cases} d_o+d_i=2.4 \text{ m} \\ f = 0.55 \text{ m}\\ \frac 1{d_o} + \frac 1 {d_i} = \frac...
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• April 22nd, 2018, 08:29
We have $\varepsilon = -\d \Phi t$. This applies to all of the magnetic field that passes through the loop. And that includes the magnetic field that...
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• April 22nd, 2018, 08:05
mathmari replied to a thread Banach space in Analysis
Great!! (Yes) How could we justify that $C(\mathbb{R})$ is a Banach space? (Wondering) Do we have to show that $\frac{\sin^2\left... 6 replies | 87 view(s) • April 22nd, 2018, 07:58 I like Serena replied to a thread Banach space in Analysis Hey mathmari!! (Wave) What you have is correct. So$X$is indeed complete and therefore a Banach space. And$Y\$ has a Cauchy sequence that shows...
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# Dimitrios Bachtis (Swansea)

• This event has passed.

# Dimitrios Bachtis (Swansea)

## April 28 @ 4:00 pm - 5:00 pm UTC+0

Quantum field-theoretic machine learning

The precise equivalence between discretized Euclidean field theories and Markov random fields, as established by the Hammersley-Clifford theorem, opens up the opportunity to investigate machine learning from the perspective of quantum field theory. In this talk I will discuss a variety of interconnected topics: Markov properties for quantum fields, the derivation of machine learning algorithms and of neural networks from the $\phi^{4}$ scalar field theory and the minimization of distance metrics between probability distributions. I will then conclude by presenting applications pertinent to the quantum field-theoretic machine learning algorithms.

## Details

Date:
April 28
Time:
4:00 pm - 5:00 pm UTC+0
Event Categories:
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# LocalDate plus() method in Java

An immutable copy of a LocalDate where the required duration is added to it can be obtained using the plus() method in the LocalDate class in Java. This method requires two parameters i.e. the duration to be added and the TemporalUnit of the duration. Also, it returns the LocalDate object with the required duration added to it.

A program that demonstrates this is given as follows −

## Example

Live Demo

import java.time.*;
import java.time.temporal.*;
public class Demo {
public static void main(String[] args) {
LocalDate ld = LocalDate.parse("2019-02-15");
System.out.println("The LocalDate is: " + ld);
System.out.println("LocalDate after adding 10 days is: " + ld.plus(10, ChronoUnit.DAYS));
}
}

## Output

The LocalDate is: 2019-02-15
LocalDate after adding 10 days is: 2019-02-25

Now let us understand the above program.

First the LocalDate is displayed. Then an immutable copy of the LocalDate where 10 days are added to it is obtained using the plus() method and this is displayed. A code snippet that demonstrates this is as follows −

LocalDate ld = LocalDate.parse("2019-02-15");
System.out.println("The LocalDate is: " + ld);
System.out.println("LocalDate after adding 10 days is: " + ld.plus(10, ChronoUnit.DAYS));
karthikeya Boyini

I love programming (: That's all I know


text:
I recently introduced the Decimal Number System, the one we are used as humans.

As I said in that post, as humans we commonly have 10 fingers and we can count up to 10, hence the popularity of that system in our history.

The Binary Number System is the second most important system for our species, as it led the electronics and computer revolution.

In electronics, we have 2 states: 0 or 1. There’s 0 volts, or there’s 5 (or 9, 12, whatever). A gate is open, or it’s closed.

It’s either one or the another.

The digit in the binary number system is called bit.

As the decimal number system, also the binary number system is positional.

We sum each digit in the binary number system mutiplied by the power of 2 depending on their position, starting at position 0 from the right.

Given that:

$2^0$ equals to 1

$2^1$ equals to 2

$2^2$ equals to 4

$2^3$ equals to 8, and so on..

We can represent numbers using a series of bits:

1 can be represented as $1\times2^0$

10 can be represented as $1\times2^1 + 0\times2^0$

111 can be represented as $1\times2^2 + 1\times2^1 + 1\times2^0$

Leading zeros in a number can be dropped, or added if needed, because they do not mean anything on the left of the top left 1: 110 can be represented a 0110 or 00000110 if needed. It holds the same exact meaning, because as the system above explained, we are simply multiplying a power of 2 times zero.

Using binary numbers we can represent any kind of number in the decimal number system.

We need to have an adequate number of digits to represent enough numbers. If we want to have 16 numbers, so we can count from 0 to 15, we need 4 digits (bits). With 5 bits we can count 32 numbers. 32 bits will give us 4,294,967,296 possible numbers.

64 bits will give us 9,223,372,036,854,775,807 possible numbers. It grows pretty quickly.

Here is a simple conversion table for the first 4 digits, which we can generate using just 2 bits:

Decimal number Binary number
0 00
1 01
2 10
3 11

Here is a simple conversion table for the first 8 digits:

Decimal number Binary number
0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111

If you notice, I repeated the above sequence, adding 1 instead of 0 in the series from 4 to 7.

Here is a simple conversion table for the first 16 digits:

Decimal number Binary number
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111

Again, I repeated the sequence we used to get the first 8 numbers, prepended 0 to the first set 0-7 and prepended 1 to 8-15.

I’ll soon talk about performing operations like sum and division with binary numbers, the hexadecimal numbers system, how to convert from binary to decimal and to hexadecimal, without looking at tables like these, and vice versa.

⤴⤴⤴


text:
## Elementary Algebra

It takes 2$\frac{1}{4}$ seconds to download 1 song, which is equal to $\frac{9}{4}$ seconds. To download 16 songs, we multiply the time needed for 1 song by 16. So, $\frac{9}{4}$ $\times$ 16 = $\frac{9 \times 16}{4}$ Factor the numerator and the denominator: $\frac{9\ \times\ 4\ \times 4}{4}$ = Cancel the common factors in the numerator and the denominator to obtain: 9 $\times$ 4 = 36 seconds


text:
# NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions

NCERT In-text Question Page No. 286
Question 1.
In Step 2 of the construction using ruler and compasses, what would happen if we take the length of radius to be smaller than half the length of $$\overline{\mathrm{AB}}$$ ?
In case we take the radius smaller than half of the length of $$\overline{\mathrm{AB}}$$, the arcs will not intersect each other at two points P and Q.

NCERT In-text Question Page No. 289
Question 1.
In Step 2 above, what would happen if we take radius to be smaller than half the length BC?
If we take radius to be smaller than half of BC, the arcs drawn with centres B and C will not intersect each other.

NCERT In-text Question Page No. 290
Question 1.
How will your construct at 15° angle?
Steps of construction:
Step I: Construct an angle ∠ABC of 60°.
Step II: Bisect ∠ABC to get an angle of 30°
i. e. ∠ABD = 30°.

Step III: Bisect ∠ABD, such that $$\overline{\mathrm{BE}}$$ is bisector of ∠ABD.
Thus, ∠ABD = $$\frac { 1 }{ 2 }$$(30°) = 15°.

NCERT In-text Question Page No. 291
Question 1.
How will you construct a 150° angle?
Steps of construction:
Step I: Draw a line I and mark a point O on it.
Step II: With centre O and a convenient radius, draw an arc intersecting I at A.
Step III: With the same radius and centre at A, draw an arc to cut the first arc at B.
Step IV: Again with the same radius and centre at B, draw another arc to intersect the
first arc at C.

Step V : Once again with the same radius and centre at C, draw an arc to cut the first are at D.
Step VI: Now, bisect ∠COD, such that ∠COE = ∠EOD = 30°.
Step VII: Since, 150° = 120° + 30°, therefore ∠AOC + ∠COE = ∠AOE. Thus, ∠AOE is the required angle whose measure is 150°.

Question 2.
How will you construct a 45° angle?
Step II: Draw OR, the angle bisector of ∠POQ such that $$\frac { 1 }{ 2 }$$– [∠POQ] = $$\frac { 1 }{ 2 }$$ (90°) = 45° or ∠POR = 45°


text:
### Session S6: Turbulence: Mixing III

3:05 PM–5:15 PM, Tuesday, November 22, 2011
Room: 309

Chair: P.K. Yeung, Georgia Institute of Technology

Abstract ID: BAPS.2011.DFD.S6.9

### Abstract: S6.00009 : The emergence of correlations from distant sources in turbulent flows

4:49 PM–5:02 PM

Preview Abstract MathJax On | Off   Abstract

#### Authors:

Mihkel Kree
(IRPHE, Aix Marseille Universite)

J\'er\^ome Duplat
(Universite Joseph Fourier, Grenoble)

Emmanuel Villermaux
(Aix Marseille Universite)

A scalar field injected from a point source in a turbulent flow resolves into a set of elongated sheet-like structures whose concentration probability distribution $P(C)$ is well understood in terms of the interplay between stretching histories of fluid particles, and molecular diffusion. Additionally, neighboring sheets may diffusively merge and evolve further in an undistinguishable fashion. We report here on experiments using two distinct dyes (Fluorescein and Rhodamine in water) injected form sources separated by a variable distance $s$ in a turbulent jet. This method allows to separate the total concentration field $C$ into its subparts $C_1$ and $C_2$ originating from each sources. Joint probability distributions $Q(C_1,C_2)$ and correlations between the fields can thus be measured. We evidence this way the crossover scale above which the fields are and remain independent in the flow.

To cite this abstract, use the following reference: http://meetings.aps.org/link/BAPS.2011.DFD.S6.9


text:
# Publications

## 2022

1. Zarei, F., Lee, J., Mackenzie, R., & Le Men, V. (2022). Evaluation of the uniformity of sound-masking systems in an open-plan office. Applied Acoustics, 186, 108464. https://doi.org/10.1016/j.apacoust.2021.108464

## 2021

1. Nik-Bakht, M., Lee, J., & Dehkordi, S. H. (2021). BIM-based reverberation time analysis. Journal of Information Technology in Construction (ITcon), 26(3), 28–38. https://doi.org/10.36680/j.itcon.2021.003
2. Lee, J., Mackenzie, R., Le Men, V., Gariépy, F., & Zarei, F. (2021). The effect of sound masking on employees’ acoustic comfort and performance in open-plan offices in Canada. Internoise 2021, 6. https://doi.org/10.3397/IN-2021-3215
3. Mackenzie, R., Zarei, F., Le Men, V., & Lee, J. (2021). Spatial uniformity tolerances for sound masking systems in open-plan offices. Internoise 2021, 7. https://doi.org/10.3397/IN-2021-3199
4. Zhang, S., & Lee, J. (2021). Diffuseness Quantification in a Reverberation Chamber and Its Variation with Fine-Resolution Measurements. Buildings, 11(11), 16. https://doi.org/10.3390/buildings11110519

## 2020

1. Lee, J., Zarei, F., Mackenzie, R., & Le Men, V. (2020). The spatial uniformity of an electronic sound masking system in an open-plan space. The Journal of the Acoustical Society of America, 148, 2440–2440. https://doi.org/10.1121/1.5146732
2. Hadavi, S., & Lee, J. (2020). Calculation of activity noise levels in classrooms by using a Gaussian mixture model. Internoise 2020, 261, 1779–1784. https://www.ingentaconnect.com/contentone/ince/incecp/2020/00000261/00000005/art00093
3. Dabirian, S., Han, S. H., & Lee, J. (2020). Stochastic-based noise exposure assessment in modular and off-site construction. Journal of Cleaner Production, 244, 118758. https://doi.org/10.1016/j.jclepro.2019.118758
4. Lee, J., & Wang, L. M. (2020). Investigating multidimensional characteristics of noise signals with tones from building mechanical systems and their effects on annoyance. The Journal of the Acoustical Society of America, 147(1), 108–124. https://doi.org/10.1121/10.0000487

## 2019

1. Zhang, S., & Lee, J. (2019). Diffuseness quantification of a reverberation chamber and uncertainty with fine-resolution measurements. 26th International Congress on Sound and Vibration, 7.
2. Erfani, K., Mahabadipour, S., Nik-Bakht, M., & Li, J. (2019). BIM-based Simulation for Analysis of Reverberation Time. IBPSA Building Simulation 2019. http://www.ibpsa.org/proceedings/BS2019/BS2019_211379.pdf
3. Hadavi, S., & Lee, J. (2019). A Survey of The Unoccupied Acoustic Conditions of Active Learning Classrooms in Montreal. Canadian Acoustics, 47(1), 81–86. https://jcaa.caa-aca.ca/index.php/jcaa/article/view/3252

## 2018

1. Dabirian, S., Han, S., & Lee, J. (2018). Identification of modular construction activity noise levels by using k-means clustering. Internoise 2018, 258, 4319–4325. https://www.ingentaconnect.com/contentone/ince/incecp/2018/00000258/00000003/art00035#expand/collapse
2. Dabirian, S., Lee, J., & Han, S. (2018). Noise exposure assessment of a modular construction manufacturing factory. The Journal of the Acoustical Society of America, 144, 1756–1756. https://doi.org/10.1121/1.5067778
3. Jha, D. D., Lee, J., & Zaheeruddin, M. (2018). Characterization, analysis, and noise control measures of a mechanical room. The Journal of the Acoustical Society of America, 144, 1790–1790. https://doi.org/10.1121/1.5067902
4. Erfani, K., Mahabadipour, S., Lee, J., & Nik-Bakht, M. (2018). Compatibility study between building information modeling and acoustic simulation software. The Journal of the Acoustical Society of America, 144, 1918–1919. https://doi.org/10.1121/1.5068403
5. Hadavi, S., & Lee, J. (2018). Acoustic conditions for students’ engagement in active learning classrooms. The Journal of the Acoustical Society of America, 144, 1894–1894. https://doi.org/10.1121/1.5068303
6. Hadavi, S., & Lee, J. (2018). The new acoustic design challenges in active learning classrooms. Internoise 2018, 6.
7. Lee, J., & Wang, L. M. (2018). Development of a model to predict the likelihood of complaints due to assorted tone-in-noise combinations. The Journal of the Acoustical Society of America, 143(5), 2697–2707. https://doi.org/10.1121/1.5036731

## 2017

1. Lee, J., & Wang, L. M. (2017). Uncertainty in tone quantification methods of background noise for enclosed spaces. The Journal of the Acoustical Society of America, 141, 3501–3501.
2. Lee, J., & Zaheeruddin, M. (2017). Acoustics Specialization for Building Engineers. Proceedings of the Acoustics Week in Canada 2017, 45, 100–101. https://jcaa.caa-aca.ca/index.php/jcaa/article/view/3083
3. Lee, J., Francis, J. M., & Wang, L. M. (2017). How tonality and loudness of noise relate to annoyance and task performance. Noise Control Engineering Journal, 65(2), 71–82. https://doi.org/10.3397/1/376427


text:
• Swagato K Ray

Articles written in Proceedings – Mathematical Sciences

• Cowling-price theorem and characterization of heat kernel on symmetric spaces

We extend the uncertainty principle, the Cowling-Price theorem, on noncompact Riemannian symmetric spacesX. We establish a characterization of the heat kernel of the Laplace-Beltrami operator onX from integral estimates of the Cowling-Price type.

• Mixed Norm Estimate for Radon Transform on Weighted $L^p$ Spaces

We will discuss about the mapping property of Radon transform on $L^p$ spaces with power weight. It will be shown that the Pitt’s inequality together with the weighted version of Hardy–Littlewood–Sobolev lemma imply weighted inequality for the Radon transform.

• # Proceedings – Mathematical Sciences

Current Issue
Volume 129 | Issue 5
November 2019

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019


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Main content

# Radicals and rational exponents | Lesson

## What are radicals and rational exponents, and how frequently do they appear on the test?

Exponential expressions are algebraic expressions with a coefficient, one or more variables, and one or more exponents. For example, in the expression start color #7854ab, 3, end color #7854ab, start color #ca337c, x, end color #ca337c, start superscript, start color #208170, 4, end color #208170, end superscript:
• start color #7854ab, 3, end color #7854ab is the coefficient.
• start color #ca337c, x, end color #ca337c is the base.
• start color #208170, 4, end color #208170 is the exponent.
In start color #7854ab, 3, end color #7854ab, start color #ca337c, x, end color #ca337c, start superscript, start color #208170, 4, end color #208170, end superscript, start color #7854ab, 3, end color #7854ab is multiplied by start color #ca337c, x, end color #ca337c start color #208170, 4, end color #208170 times:
start color #7854ab, 3, end color #7854ab, start color #ca337c, x, end color #ca337c, start superscript, start color #208170, 4, end color #208170, end superscript, equals, start color #7854ab, 3, dot, end color #7854ab, start color #208170, left parenthesis, end color #208170, start color #ca337c, x, end color #ca337c, start color #208170, dot, end color #208170, start color #ca337c, x, end color #ca337c, start color #208170, dot, end color #208170, start color #ca337c, x, end color #ca337c, start color #208170, dot, end color #208170, start color #ca337c, x, end color #ca337c, start color #208170, right parenthesis, end color #208170
An expression can also be raised to an exponent. For example, for left parenthesis, start color #ca337c, 3, x, end color #ca337c, right parenthesis, start superscript, start color #208170, 4, end color #208170, end superscript, the expression start color #ca337c, 3, x, end color #ca337c is multiplied by itself start color #208170, 4, end color #208170 times:
left parenthesis, start color #ca337c, 3, x, end color #ca337c, right parenthesis, start superscript, start color #208170, 4, end color #208170, end superscript, equals, start color #ca337c, 3, x, end color #ca337c, start color #208170, dot, end color #208170, start color #ca337c, 3, x, end color #ca337c, start color #208170, dot, end color #208170, start color #ca337c, 3, x, end color #ca337c, start color #208170, dot, end color #208170, start color #ca337c, 3, x, end color #ca337c, equals, start color #7854ab, 81, end color #7854ab, start color #ca337c, x, end color #ca337c, start superscript, start color #208170, 4, end color #208170, end superscript
Notice how 3, x, start superscript, 4, end superscript, does not equal, left parenthesis, 3, x, right parenthesis, start superscript, 4, end superscript !
Rational exponents refer to exponents that are/can be represented as fractions: start fraction, 1, divided by, 2, end fraction, 3, and minus, start fraction, 2, divided by, 3, end fraction are all considered rational exponents. Radicals are another way to write rational exponents. For example, x, start superscript, start superscript, start fraction, 1, divided by, 2, end fraction, end superscript, end superscript and square root of, x, end square root are equivalent.
In this lesson, we'll:
1. Review the rules of exponent operations with integer exponents
2. Apply the rules of exponent operations to rational exponents
3. Make connections between equivalent rational and radical expressions
On your official SAT, you'll likely see 1 question that tests your ability to work with radicals and rational exponents. All questions on quadratics, polynomials, and solving radical equations also require a basic understanding of exponents and radicals.
You can learn anything. Let's do this!

## What are the rules of exponent operations?

### Powers of products & quotients (integer exponents)

Khan Academy video wrapper
Powers of products & quotients (integer exponents)See video transcript

### The rules of exponent operation

#### Adding and subtracting exponential expressions

When adding and subtracting exponential expressions, we're essentially combining like terms. That means we can only combine exponential expressions with both the same base and the same exponent.
\begin{aligned} ax^n\pm bx^n &= (a\pm b )x^n \end{aligned}

#### Multiplying and dividing exponential expressions

When multiplying two exponential expressions with the same base, we keep the base the same, multiply the coefficients, and add the exponents. Similarly, when dividing two exponential expressions with the same base, we keep the base the same and subtract the exponents.
\begin{aligned} ax^m \cdot bx^n &= ab\cdot x^{m+n} \\\\ \dfrac{ax^m}{bx^n} &= \dfrac{a}{b}\cdot x^{m-n} \end{aligned}
When multiplying or dividing exponential expressions with the same exponent but different bases, we multiply or divide the bases and keep the exponents the same.
\begin{aligned} x^n\cdot y^n &= (xy)^n \\\\ \dfrac{x^n}{y^n} &= \left(\dfrac{x}{y}\right)^n \end{aligned}

#### Raising an exponential expression to an exponent and change of base

When raising an exponential expression to an exponent, raise the coefficient of the expression to the exponent, keep the base the same, and multiply the two exponents.
left parenthesis, a, x, start superscript, m, end superscript, right parenthesis, start superscript, n, end superscript, equals, a, start superscript, n, end superscript, dot, x, start superscript, m, n, end superscript
When the bases are numbers, we can use a similar rule to change the base of an exponential expression.
left parenthesis, a, start superscript, b, end superscript, right parenthesis, start superscript, n, end superscript, equals, a, start superscript, b, n, end superscript
This is useful for questions with multiple terms that need to be written in the same base.

#### Negative exponents

A base raised to a negative exponent is equivalent to 1 divided by the base raised to the
of the exponent.
x, start superscript, minus, n, end superscript, equals, start fraction, 1, divided by, x, start superscript, n, end superscript, end fraction

#### Zero exponent

A nonzero base raised to an exponent of 0 is equal to 1.
x, start superscript, 0, end superscript, equals, 1, comma, x, does not equal, 0

### How do the rules of exponent operations apply to rational exponents?

Every rule that applies to integer exponents also applies to rational exponents.

### Try it!

try: divide two rational expressions
In order to divide 12, x, start superscript, start superscript, start fraction, 5, divided by, 2, end fraction, end superscript, end superscript by 3, x, start superscript, start superscript, start fraction, 1, divided by, 2, end fraction, end superscript, end superscript, we
the coefficients and
the exponents of x.
start fraction, 12, x, start superscript, start superscript, start fraction, 5, divided by, 2, end fraction, end superscript, end superscript, divided by, 3, x, start superscript, start superscript, start fraction, 1, divided by, 2, end fraction, end superscript, end superscript, end fraction, equals

Try: raise an exponential expression to an exponent
To calculate left parenthesis, 2, y, start superscript, start superscript, start fraction, 4, divided by, 3, end fraction, end superscript, end superscript, right parenthesis, cubed, we
and
the exponents start fraction, 4, divided by, 3, end fraction and 3.
left parenthesis, 2, y, start superscript, start superscript, start fraction, 4, divided by, 3, end fraction, end superscript, end superscript, right parenthesis, cubed, equals

## How are radicals and fractional exponents related?

### Rewriting roots as rational exponents

Khan Academy video wrapper
Rewriting roots as rational exponentsSee video transcript

### Roots and rational exponents

Squares and square roots are inverse operations: they "undo" each other. For example, if we take the square root of 3 squared, we get square root of, 3, squared, end square root, equals, 3.
The reason for this becomes more apparent when we rewrite square root as a fractional exponent: square root of, x, end square root, equals, x, start superscript, start superscript, start fraction, 1, divided by, 2, end fraction, end superscript, end superscript, and square root of, 3, squared, end square root, equals, left parenthesis, 3, squared, right parenthesis, start superscript, start superscript, start fraction, 1, divided by, 2, end fraction, end superscript, end superscript, equals, 3, start superscript, 1, end superscript.
When rewriting a radical expression as a fractional exponent, any exponent under the radical symbol (square root of, empty space, end square root) becomes the numerator of the fractional exponent, and the value to the left of the radical symbol (e.g., root, start index, start color #7854ab, 3, end color #7854ab, end index) becomes the denominator of the fractional exponent. Square root is equivalent to root, start index, 2, end index.
root, start index, n, end index, equals, x, start superscript, start superscript, start fraction, m, divided by, n, end fraction, end superscript, end superscript
All of the rules that apply to exponential expressions with integer exponents also apply to exponential expressions with fractional exponents. Similarly, for radical expressions:
\begin{aligned} \sqrt[n]{x} \cdot \sqrt[n]{y} &= \sqrt[n]{xy} \\\\ \dfrac{\sqrt[n]{x}}{\sqrt[n]{y}} &= \sqrt[n]{\dfrac{x}{y}} \end{aligned}
When working with radical expressions with the same radical, we can choose whether to convert to fractional exponents first or multiply what's under the radical symbol first to our advantage.

### Try it!

Try: determine equivalent expressions
Determine whether each of the radical expressions below is equivalent to x, start superscript, start superscript, start fraction, 3, divided by, 2, end fraction, end superscript, end superscript, y, start superscript, start superscript, start fraction, 1, divided by, 3, end fraction, end superscript, end superscript.
Equivalent $\sqrt{x^3}\cdot\sqrt[3]{y}$square root of, x, cubed, end square root, dot, cube root of, y, end cube root $\sqrt{xy}$square root of, x, y, end square root $\sqrt{x^3y}$square root of, x, cubed, y, end square root $\sqrt[6]{x^9y^{2}}$root, start index, 6, end index

## Your turn!

Practice: multiply rational expressions
Which of the following is equivalent to 2, x, cubed, dot, 3, x, start superscript, 5, end superscript ?
Choose 1 answer:

Practice: change bases
If a, start superscript, start superscript, start fraction, b, divided by, 2, end fraction, end superscript, end superscript, equals, 25 for positive integers a and b, what is one possible value of b ?

Practice: raise to a negative exponent
If n, start superscript, start superscript, minus, start fraction, 1, divided by, 3, end fraction, end superscript, end superscript, equals, x, where n, is greater than, 0, what is n in terms of x ?
Choose 1 answer:

Practice: simplify radical expressions
start fraction, cube root of, 8, x, start superscript, 8, end superscript, y, start superscript, 6, end superscript, end cube root, divided by, square root of, 4, x, squared, y, start superscript, 6, end superscript, end square root, end fraction
Which of the following is equivalent to the expression above?
Choose 1 answer:

## Things to remember

Adding and subtracting exponential expressions:
\begin{aligned} ax^n\pm bx^n &= (a\pm b )x^n \end{aligned}
Multiplying and dividing exponential expressions:
\begin{aligned} ax^m \cdot bx^n &= ab\cdot x^{m+n} \\\\ \dfrac{ax^m}{bx^n} &= \dfrac{a}{b}\cdot x^{m-n} \\\\ x^n\cdot y^n &= (xy)^n \\\\ \dfrac{x^n}{y^n} &= \left(\dfrac{x}{y}\right)^n \end{aligned}
Raising an exponential expression to an exponent and change of base:
\begin{aligned} \left(ax^m\right)^n &= a^n \cdot x^{mn} \\\\ (a^b)^n &=a^{bn} \end{aligned}
Negative exponent:
x, start superscript, minus, n, end superscript, equals, start fraction, 1, divided by, x, start superscript, n, end superscript, end fraction
Zero exponent:
x, start superscript, 0, end superscript, equals, 1, comma, x, does not equal, 0
All of the rules that apply to exponential expressions with integer exponents also apply to exponential expressions with fractional exponents.
\begin{aligned} \sqrt[n]{x^m} &= x^{^{\scriptsize \dfrac{m}{n}}} \\\\ \sqrt[n]{x} \cdot \sqrt[n]{y} &= \sqrt[n]{xy} \\\\ \dfrac{\sqrt[n]{x}}{\sqrt[n]{y}} &= \sqrt[n]{\dfrac{x}{y}} \end{aligned}

## Want to join the conversation?

• I think there is some problem in Multiplying and dividing exponential expressions. The base are not same. Even there is a problem in (your turn!). It also have wrong problem. According to rules base should be same then only you can solve.
(0 votes)
• In school you might have learned that exponents with different bases can't mix, just like fractions with different denominators can't either, but for some reason or the other, the SAT likes throwing you problems with exponents that have different bases. The trick to the vast majority of these is that most of the bases are powers of each other, allowing you to apply your exponent rules to get them to the same base.
We know that 2^3^4 is the same thing as (2^3)^4 because of the way that the order of operations works, and is the same thing as 2^(3*4) because of our exponent rules. If we simplify the parentheses, we get that:
8^4 = 2^12
And we have different bases. If you have to convert them to the same base, always look at the factors of each base. We know that 8 is 2^3, so we can apply that power to a power rule in reverse:
8^4 = 2^12
(2^3)^4 = 2^12
2^(3*4) = 2^12
2^12 = 2^12
(11 votes)
• I think there may be a mistake, but I could be wrong. In a question it gives the options to 'cube 2' and 'square 3'. Shouldn't it be the other way? It was the question where the answer was 8y^4.
(1 vote)
• For me, the question doesn't talk about squaring 3, just multiplying 3 times the existing exponent on y. The explanation looked correct to me. We distribute the power to all factors, so:
(2y^(4/3))^3 = 2^3 * (y^(4/3))^3
= 8 * y^(4/3 * 3) = 8y^4
(3 votes)
• Can someone please take the time to talk to me and explain how these problems work? Like in a google meeting or something?
(2 votes)
• In the last problem, I believe that x and y must be positive (> 0) for the expressions to be equivalent. Technically, the sqrt(x^2*y^6) does not equal x*y^3, but it does equal the abs_val(x*y^3). For example, if x=-1 and y=1, the original expression and the expression in the answer are not equal.

I don't think this technical detail needs to be described in this lesson, but the problem should specify that x and y are positive to avoid this technical issue.
(1 vote)
• I have a question how would I solve a problem like (49 to the 1/4 power ) then -2 power
(1 vote)
• There is a question on the SAT radical and rational exponents practice that I don't understand very well. The question goes like this: 2^203 + 2^204, which of the following is equal to the value above?
It says there in the explanation that the expression can be rewritten with the following fact: x^y=x × x^y-1
Using the fact it says it should be 3(2)^203 = 2^203 +2^204=1(2)^203+2(2)^203=1+2(2)203=3(2)^203.
I understand the last two steps , but not how we got there.
(1 vote)


text:
# Does it exist a way how to find the ways to get from one point to another when certain points must be avoided in a grid?

The problem is as follows:

The diagram from below shows a grid of $$6\times 6$$. How many different ways can you get from $$A$$ to $$B$$ without going through any of the highlighted points?

The alternatives given are as follows:

$$\begin{array}{ll} 1.&\textrm{265 ways}\\ 2.&\textrm{365 ways}\\ 3.&\textrm{395 ways}\\ 4.&\textrm{405 ways}\\ \end{array}$$

Does it exist a way to simplify this problem?. How exactly can I find the method to solve this?. There isn't any indication of which sort of path should be taken. Hence there could be tons of ways and I'm stuck on it. Can someone help me here?.

It would help a lot to me to include some sort of diagram or drawing to justify a resonable method to solve this.

I'm assuming the paths must go right and down only.

You can count the paths by using the inclusion-exclusion principle. Without considering the points that must be avoided, there are $$\binom{6+6}{6}$$ paths. Now subtract the $$\binom{4+1}{1}\binom{2+5}{5}$$ paths that visit the first bad point, the $$\binom{2+4}{4}\binom{4+2}{2}$$ paths that visit the second bad point, and the $$\binom{4+5}{5}\binom{2+1}{1}$$ paths that visit the third bad point. Then add back in the paths that visit two bad points. No paths visit all three bad points.

\begin{align} &\binom{12}{6} -\left(\binom{5}{1}\binom{7}{5} +\binom{6}{4}\binom{6}{2}+\binom{9}{5}\binom{3}{1}\right) +\left(\binom{5}{1}\binom{4}{4}\binom{3}{1}+\binom{6}{4}\binom{3}{1}\binom{3}{1}\right)\\ &=924-(5\cdot 21+15\cdot 15+ 126\cdot 3)+(5\cdot 1\cdot 3+15\cdot 3\cdot 3)\\ &=924-(105+225+378)+(15+105)\\ &=924-708+120\\ &= {\color{red}{366}} \end{align}

Here's an alternative solution that uses a Pascal-type recursion. Let $$p(i,j)$$ be the number of good paths that start from $$A=(0,0)$$ and reach point $$(i,j)$$. We want to compute $$p(6,6)$$. By conditioning on the last step into $$(i,j)$$, we find that $$p(i,j)$$ satisfies the following recursion: $$p(i,j)= \begin{cases} 0 &\text{if (i,j) is a bad point}\\ 1 &\text{if i=0 or j=0}\\ p(i-1,j)+p(i,j-1) &\text{otherwise} \end{cases}$$ The resulting values of $$p(i,j)$$ are: $$\begin{matrix} i\backslash j &0 &1 &2 &3 &4 &5 &6 \\ 0 &1 &1 &1 &1 &1 &1 &1 \\ 1 &1 &2 &3 &4 &0 &1 &2 \\ 2 &1 &3 &6 &10 &10 &11 &13 \\ 3 &1 &4 &10 &20 &30 &41 &54 \\ 4 &1 &5 &0 &20 &50 &91 &145 \\ 5 &1 &6 &6 &26 &0 &91 &236 \\ 6 &1 &7 &13 &39 &39 &130 &{\color{red}{366}} \end{matrix}$$ So $$p(6,6)=366$$.

• @RobPratt The answer is $365$ but I have no idea how to get there. – Chris Steinbeck Bell Mar 21 '20 at 2:07
• Besides the inclusion-exclusion argument, I also listed them explicitly and got 366. – RobPratt Mar 21 '20 at 2:17
• Let's suppose what if I don't want to resort to binomials. Does it exist a way to solve this by brute force or counting them using Pascal's logic?. – Chris Steinbeck Bell Mar 21 '20 at 2:21
• I have no idea how did the author got to $365$?. Could it be that is a typo or he's not accounting for something. – Chris Steinbeck Bell Mar 21 '20 at 2:21
• @ChrisSteinbeckBell: The Pascal logic for that goes this way: we are calculating the number of paths through the lower two dots. To get to the first dot you have to make $6$ moves, of which are $4$ are down and $2$ are right. You can do that in $6 \choose 4$ ways because you can choose any $4$ of the $6$ moves to be down and the others are right. Then to get to the next dot you have to move $3$ moves of which $1$ is down and to get to $B$ you again make $3$ moves of which $1$ is down. The choices are independent, so you multiply them. – Ross Millikan Mar 21 '20 at 3:11

Denote the points to be avoided by $$1$$, $$2$$, $$3$$, in the order Left, Top, Bottom.

Number of paths through $$1 = \binom{6}{2}\cdot \binom{6}{2}$$

Number of paths through $$2= \binom{5}{1}\cdot\binom{7}{2}$$

Number of paths though $$3= \binom{9}{4}\cdot \binom{3}{1}$$

Number of paths through $$1$$ and $$3= \binom{6}{2}\cdot \binom{3}{1}\cdot \binom{3}{1}$$

Number of paths through $$2$$ and $$3 = \binom{5}{1}\cdot \binom{3}{1}$$

Number of paths through $$1$$ and $$2 = 0$$.

Now we have to substract from the number of paths from $$A$$ to $$B = \binom{12}{6}$$ the first three amounts and add the next three amounts, which gets $$366$$ ( we use the inclusion-exclusion).

• Where did you used the inclusion-exclusion principle?. This part I'm not getting it. Is it where the part you mentioned subtracting A to B?. Why is it the number of paths through $1=\binom{6}{2} \cdot \binom{6}{2}$? Why is it that? – Chris Steinbeck Bell Mar 21 '20 at 2:57
• @Chris Steinbeck Bell: First, do you understand why the number of paths from $A$ to $B$ is $\binom{12}{6}$ ? There are 6 vertical steps and 6 horizontal steps. Out of $12$ steps, you have to choose which ones will be the horizontal ones, so $12$ choose $6$ ways. – orangeskid Mar 21 '20 at 3:01
• Probably this is the reason why I requested the inclusion of a drawing thus I could understand this better. I can see there are seven vertical lines and seven horizontal lines considering A and B. I don't know where do you get the 6 vertical steps and the 6 horizontal steps. why is it 6 ways from those 12?. I'm stuck there. – Chris Steinbeck Bell Mar 21 '20 at 3:14
• @Chris Steinbeck Bell: Check out this source: youtube.com/… also en.wikipedia.org/wiki/Lattice_path – orangeskid Mar 21 '20 at 3:24

I believe you are meant to always move either to the right or down along the grid. Without that restriction there would be infinitely many paths.

To get from $$A$$, which is $$(0,0)$$, to $$B$$, which is $$(6,6)$$, you need to take six steps down and six steps to the right. (I'm ignoring the restriction to avoid the three marked points for now.) It is true that there are seven horizontal and seven vertical lines in the figure, as you mention in a comment, but the steps are the segments between vertices, not the vertices themselves, which is why only six steps down are needed instead of seven.

One possible path from $$A$$ to $$B$$ is $$RRDDRDRRDDDR$$. Another is $$RDRDRDRDRDRD$$. A third is $$RRRRRRDDDDDD$$. In fact, any "word" consisting of six $$R$$s and six $$D$$s corresponds to a path. So the problem is reduced to counting words with six $$R$$s and six $$D$$s. Such a word is completely specified by stating where the $$R$$s are (or alternatively stating where the $$D$$s are). Any set of six elements chosen from $$1,2,3,4,5,6,7,8,9,10,11,12$$ is a set of possible $$R$$ positions. For the first possible path mentioned above, the set of $$R$$ positions is $$\{1,2,5,7,8,12\}$$. For the second it is $$\{1,3,5,7,9,11\}$$, and for the third it is $$\{1,2,3,4,5,6\}$$. There are $$\binom{12}{6}$$ ways of forming such a set.

Now that we know how many paths there are from $$A$$ to $$B$$, we want to subtract paths that are invalid in that they do pass through one of the marked points. There are, for example, $$\binom{5}{4}\binom{7}{2}$$ paths that pass through the marked point in the second row. The first binomial coefficient is the number of ways to get from $$A$$ to the marked point; the second is the number of ways to get from the marked point to $$B$$.

You can compute the number of paths passing through each of the other two marked points by similar reasoning. You will see, however, that some paths have been subtracted twice, so you will need to add these back. That is, you need to use the principle of inclusion-exclusion. One set of paths that needs to be added back is the set of paths that pass through both $$(2,4)$$ and $$(4,5)$$. They were subtracted once because they pass through $$(2,4)$$, and they were subtracted a second time because the pass through $$(4,5)$$. The number that needs to be added back to compensate for the double subtraction is $$\binom{6}{2}\binom{3}{2}\binom{3}{2}$$, which is the number of ways of going from $$(0,0)$$ to $$(2,4)$$, then from $$(2,4)$$ to $$(4,5)$$, and then from $$(4,5)$$ to $$(6,6)$$.

Added: Rob Pratt has given a beautiful solution using a Pascal's-triangle-like recurrence. I think it worth pointing out that Pascal's triangle is a table of binomial coefficients, so the binomial coefficients are still there in the background when you use that method. A formula for the result of applying the recurrence can be obtained by combining a sequence of arrays, each of which is a shifted Pascal's triangle multiplied by one or more binomial coefficients.

If no points are required to be avoided, the relevant portion of Pascal's triangle is $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 3 & 6 & 10 & 15 & 21 & 28\\ 1 & 4 & 10 & 20 & 35 & 56 & 84\\ 1 & 5 & 15 & 35 & 70 & 126 & 210\\ 1 & 6 & 21 & 56 & \color{red}{126} & 252 & 462\\ 1 & 7 & 28 & 84 & 210 & 462 & 924 \end{array}$$ where, if rows and columns are both labeled $$0$$ through $$6$$, the entry in row $$i$$, column $$j$$ is $$\binom{i+j}{i}$$. To eliminate paths that pass through the marked point in the second-to-last row, we need to subtract the array $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 126 & 126 & 126\\ 0 & 0 & 0 & 0 & 126 & 252 & 378 \end{array}$$ whose row $$i$$, column $$j$$ entry is $$\binom{5+4}{5}\binom{i-5+j-4}{i-5}=126\binom{i-5+j-4}{i-5}$$ for $$i\ge5$$, $$j\ge4$$. This leaves $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 3 & 6 & 10 & 15 & 21 & 28\\ 1 & 4 & 10 & 20 & 35 & 56 & 84\\ 1 & 5 & \color{red}{15} & 35 & 70 & 126 & 210\\ 1 & 6 & 21 & 56 & 0 & 126 & 336\\ 1 & 7 & 28 & 84 & 84 & 210 & 546 \end{array}$$ Similarly, to eliminate paths that pass through the marked point in Row $$4$$, Column $$2$$, subtract the array $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 15 & 15 & 15 & 15 & 15\\ 0 & 0 & 15 & 30 & 45 & 60 & 75\\ 0 & 0 & 15 & 45 & 90 & 150 & 225 \end{array}$$ whose row $$i$$, column $$j$$ entry is $$\binom{4+2}{4}\binom{i-4+j-2}{i-4}=15\binom{i-4+j-2}{i-4}$$ for $$i\ge4$$, $$j\ge2$$. This leaves $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 3 & 6 & 10 & 15 & 21 & 28\\ 1 & 4 & 10 & 20 & 35 & 56 & 84\\ 1 & 5 & 0 & 20 & 55 & 111 & 195\\ 1 & 6 & 6 & 26 & \color{red}{-45} & 66 & 261\\ 1 & 7 & 13 & 39 & -6 & 60 & 321 \end{array}$$ At this point, paths that pass through both the marked point in Row $$5$$ and the one in Row $$4$$ have been subtracted twice, which explains the negative entries. To add these back, add the array $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 45 & 45 & 45\\ 0 & 0 & 0 & 0 & 45 & 90 & 135 \end{array}$$ whose row $$i$$, column $$j$$ entry is $$\binom{4+2}{4}\binom{1+2}{1}\binom{i-5+j-4}{i-5}=45\binom{i-5+j-4}{i-5}$$ for $$i\ge5$$, $$j\ge4$$. This leaves $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & \color{red}{5} & 6 & 7\\ 1 & 3 & 6 & 10 & 15 & 21 & 28\\ 1 & 4 & 10 & 20 & 35 & 56 & 84\\ 1 & 5 & 0 & 20 & 55 & 111 & 195\\ 1 & 6 & 6 & 26 & 0 & 111 & 306\\ 1 & 7 & 13 & 39 & 39 & 150 & 456 \end{array}$$ A similar subtraction followed by addition are needed to eliminate paths that pass through the marked point in Row $$1$$, Column $$4$$. Subtracting $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 5 & 5 & 5\\ 0 & 0 & 0 & 0 & 5 & 10 & 15\\ 0 & 0 & 0 & 0 & 5 & 15 & 30\\ 0 & 0 & 0 & 0 & 5 & 20 & 50\\ 0 & 0 & 0 & 0 & 5 & 25 & 75\\ 0 & 0 & 0 & 0 & 5 & 30 & 105 \end{array}$$ whose entries are $$\binom{1+4}{1}\binom{i-1+j-4}{i-1}=5\binom{i-1+j-4}{i-1}$$ for $$i\ge1$$, $$j\ge4$$, gives $$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 & 0 & 1 & 2\\ 1 & 3 & 6 & 10 & 10 & 11 & 13\\ 1 & 4 & 10 & 20 & 30 & 41 & 54\\ 1 & 5 & 0 & 20 & 50 & 91 & 145\\ 1 & 6 & 6 & 26 & \color{red}{-5} & 86 & 231\\ 1 & 7 & 13 & 39 & 34 & 120 & 351 \end{array}$$ To eliminate double subtraction of paths the pass through both $$(1,4)$$ and $$(5,4)$$ add $$\begin{array}{lllllll} 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 5 & 5 & 5\\ 0 & 0 & 0 & 0 & 5 & 10 & 15 \end{array}$$ whose entries are $$\binom{1+4}{1}\binom{4+0}{4}\binom{i-5+j-4}{i-5}=5\binom{i-5+j-4}{i-5}$$ for $$i\ge5$$, $$j\ge4$$. This gives the array in Rob Pratt's answer.

Assembling all these arrays to get a non-recursive formula for the entries in the final array is equivalent to the binomial coefficient method with inclusion-exclusion.

• I'm not familiar with the use of combinatorics, but I do understand how to calculate $\binom{12}{6}$ but I don't know how you got it that binomial. Perhaps does it exist another way which doesn't require this?. I would see it better if there is some justification visually of how to account for those ways?. And I'm thinking that the allowed paths are what you say. But there isn't an indication of that. – Chris Steinbeck Bell Mar 21 '20 at 1:50
• I have to confess that when I apply the method of my answer I don't manage to match any of the given alternatives, although I do come very close to one of them. I may, of course have made an arithmetic error somewhere. – Will Orrick Mar 21 '20 at 1:50
• Gee this doesn't give me much confidence though. – Chris Steinbeck Bell Mar 21 '20 at 1:51
• Let $R$ indicate a right move and $D$ a down move. One possible path is $RRDDRDRRDDDR$. In fact, any "word" consisting of six $R$s and six $D$s corresponds to a path. So the problem is reduced to counting words with six $R$s and six $D$s. This is where $\binom{12}{6}$ comes from. (You have to chose the positions of the six $R$s from twelve possible positions.) The number of paths that pass through the marked point in Row 2 is found by splitting the path into two subpaths at that point. – Will Orrick Mar 21 '20 at 1:53
• Well, I'm confident of the method, and have checked the arithmetic a few times. Are you 100% sure there's no typo in the answer choices? – Will Orrick Mar 21 '20 at 1:55


text:
# Add Caching to a Data Access Layer

Dynamic web pages are great; you can adapt the resulting page to your user, show other user's activity, offer different products to your customers based on their navigation history, and so on. But the more dynamic a website is, the more database queries you'll probably need to perform. Unfortunately, these database queries consume the largest portion of your running time.

In this tutorial, I will demonstrate a way to improve performance, without running extra unnecessary queries. We'll develop a query caching system for our data layer with small programming and deployment cost.

### 1. The Data Access Layer

Adding a caching layer transparently to an application is often difficult because of the internal design. With object oriented languages (like PHP 5) it is a lot easier, but it still can be complicated by poor design.

In this tutorial, we set our starting point in an application that performs all its database access through a centralized class from which all data models inherit the basic database access methods. The skeleton for this starting class looks like this:

Let's implement it step by step. First, the constructor that will use the PDO library to interface with the database:

We connect to the database using the PDO library. For the database credentials I use a static class named "app_AppConfig" that centralizes the application's configuration information.

To store the database connection, we use a static attribute ($DB). We use a static attribute in order to share the same connection with all the instances of "model_Model", and, because of that, the connection code is protected with an if (we don't want to connect more than once). In the last line of the constructor we set the exception error model for PDO. In this model, for every error the PDO finds, it throws an exception (class PDOException) instead of returning error values. This is a matter of taste, but the rest of the code can be kept cleaner with the exceptional model, which is good for this tutorial. Executing queries can be very complex, but in this class we have taken a simple approach with a single doStatement() method: This method executes the query, and returns an associative array with the entire result set (if any). Note that we are using the static connection (self::$DB). Note, also, that this method is protected. This is because we don't want the user to execute arbitrary queries. Instead of that we will provide concrete models to the user. We will see this later, but before let's implement the last method:

The "model_Model" class is a very simple but convenient class for data layering. Although it's simple (it can be enhanced with advanced features like prepared statements if you want), it does the basic stuff for us.

To complete the configuration part of our application, let's write the "app_Config" static class:

As stated before, we will provide concrete models to access the database. As a little example, we will use this simple schema: a documents table and an inverted index to search whether a document contains a given word or not:

From the basic data access class (model_Model), we derive as many classes as needed by the data design of our application. In this example, we can derive those two self-explanatory classes:

Those derived models is where we add the public information. Using them is extremely simple:

The result for this example might look similar to that (obviously it depends on your actual data):

What we have written is shown in the next UML class diagram:

### 2. Planning our Caching Scheme

When things start to collapse in your database server, it is time to take break and consider optimizing the data layer. After having optimized your queries, adding the proper indexes, etc., the second move is to try to avoid unnecessary queries: why make the same request to the database on every user request, if this data hardly changes?

With a well-planned and well-decoupled class organization, we can add an extra layer to our application almost with no programming cost. In this case, we are going to extend the "model_Model" class to add transparent caching to our database layer.

#### The Caching Basics

Since we know that we need a caching system, let's focus on that particular problem and, once sorted out, we will integrate it in our data model. For now, we won't think in terms of SQL queries. It's easy to abstract a little and build a general enough scheme.

The simplest caching scheme consist of [key,data] pairs, where the key identifies the actual data we want to store. This schema is not new, in fact, it is analogous to PHP's associative arrays, and we use it all the time.

So we will need a way to store a pair, to read it, and to delete it. That's enough to build our interface for cache helpers:

The interface is quite easy: the get method gets a value, given its identifying key, the put method sets (or updates) the value for a given key, and the delete method deletes it.

With this interface in mind, it's time to implement our first real caching module. But before doing it, we will choose the data storage method.

#### The Underlying Storage System

The decision to build a common interface (like cache_CacheHelper) for caching helpers will allow us to implement them nearly on top of every storage. But on top on what storage system? There are a lot of them we can use: shared memory, files, memcached servers or even SQLite databases.

Often underestimated, DBM files are perfect for our caching system, and we are going to use them in this tutorial.

DBM files work naively on (key,data) pairs, and do it very fast due to its internal B-tree organization. They also do the access control for us: we don't need to worry about blocking the cache before writing (like we will have to do on other storage systems); DBM does it for us.

DBM files are not driven by expensive servers, they do their work inside a lightweight library on the client side accessing locally to the actual file that stores the data. In fact they actually are a family of file formats, all of them with the same basic API for (key,data) access. Some of them allow repeated keys, others are constant and don't allow writes after closing the file for the first time (cdb), etc. You can read more about that on http://www.php.net/manual/en/dba.requirements.php

Nearly every UNIX system install one type or more of these libraries (probably Berkeley DB or GNU dbm). For this example, we will use "db4" format (Sleepycat DB4 format: http://www.sleepycat.com). I have found that this library is often preinstalled, but you can use whichever library you want (except cdb, of course: we want to write on the file). In fact you could move this decision into the "app_AppConfig" class and adapt it for every project you do.

With PHP, we have two alternatives to deal with DBM files: the "dba" extension (http://php.net/manual/en/book.dba.php) or the "PEAR::DBA" module (http://pear.php.net/package/DBA). We will use the "dba" extension, which probably you already have installed in your system.

#### Wait a minute, we are dealing with SQL and result sets!

DBM files work with strings for key and values, but our problem is to store SQL result sets (that can vary in structure quite a lot). How could we manage to convert them from one world to the other?

Well, for keys, it is very easy because the actual SQL query string identifies a set of data very well. We can use the MD5 digest of the query string to shorten the key. For values, it is trickier, but here your allies are the serialize() / unserialize() PHP functions, which can be used to convert from arrays to string and vice versa.

We will see how all this works in the next section.

### 3. Static Caching

In our first example, we will deal with the easiest way to perform caching: caching for static values. We will write a class called "cache_DBM" implementing the interface "cache_CacheHelper", just like that:

This class is very easy: a mapping between our interface and dba functions. In the constructor, the given file is opened,
and the returned handler is stored in the object in order to use it in the other methods.

A simple example of use:

Below, you'll find what we have done here expressed as an UML class diagram:

Now let's add the caching system to our data model. We could have changed the "model_Model" class in order to add caching to each of its derived classes. But, if we had done so, we would have lost the flexibility to assign the caching characteristic only to specific models, and I think this is an important part of our job.

So we will create another class, called "model_StaticCache", which will extend "model_Model" and will add caching functionality. Let's start with the skeleton:

In the constructor, we first call the parent constructor in order to connect to the database. Then, we create and store, statically, a "cache_DBM" object (if not created before elsewhere). We store one instance for every derived class name because we are using one DBM file for every one of them. For that purpose, we use the static array "$cache". To determine in which directory we have to write the cache files, we have used again the application's configuration class: "app_AppConfig". And now: the doStatement() method. The logic for this method is: convert the SQL statement to a valid key, search the key in the cache, if found return the value. If not found, execute it in the database, store the result and return it: There are two more things worth noting. First, we are using the MD5 of the query as the key. In fact, it is not necessary, because the underlying DBM library accepts keys of arbitrary size, but it seems better to shorten the key anyway. If you are using prepared statements, remember to concatenate the actual values to the query string to create the key! Once the "model_StaticCache" is created, modifying a concrete model for its use is trivial, you only need to change its "extends" clause in the class declaration: And that's all, the magic is done! The "model_Document" will perform only one query for every document to retrieve. But we can do it better. ### 4. Caching Expiration In our first approach, once a query is stored in the cache, it remains valid forever until two things occur: we delete its key explicitly, or we unlink the DBM file. However this approach is only valid for a few data models of our application: the static data (like menu options and this kind of things). The normal data in our application is likely to be more dynamic than that. Think about a table containing the products we sell in our web page. It is not likely to change every minute, but there is the chance that this data will change (by adding new products, changing selling prices, etc.). We need a way to implement caching, but have a way to react to changes in data. One approach to this problem is to set an expiration time to the data stored in the cache. When we store new data in the cache, we set a window of time in which this data will be valid. After that time, the data will be read from the database again and stored into the cache for another period of time. As before, we can create another derived class from "model_Model" with this functionality. This time, we will call it "model_ExpiringCache". The skeleton is similar to "model_StaticCache": In this class we have introduced a new attribute:$expiration. This one will store the configured time window for valid data. We set this value in the constructor, the rest of the constructor is the same as in "model_StaticCache":

The bulk of the job comes in the doStatement. The DBM files have no internal way to control expiration of data, so we must implement our own. We'll do it by storing arrays, like this one:

This kind of array is what we serialize, and store into the cache. The "time," key is the modification time of the data in the cache, and the "data" is the actual data we want to store. On read time, if we find that the key exists, we compare the creation time stored with the current time and return the data if not expired.

If the key doesn't exist or is expired, we continue executing the query and storing the new result set in the cache before returning it.

Simple!

Now let's convert the "model_Index" to a model with expiring cache. As it happens, with "model_Documents," we only need to modify the class declaration and change the "extends" clause:

About the expiration time... some considerations must be made. We use a constant expiration time (1 hour = 3,600 seconds), for the sake of simplicity, and because we don't want to modify the rest of our code. But, we can easily modify it in a lot of ways to allow us to use different expiration times, one for each model. Afterward we will see how.

The class diagram for all our job is as follows:

### 5. Different Expiration

In every project, I am sure you will have different expiration time for nearly every model: from a couple of minutes to hours, or even days.

If only we could have a different expiration time for every model, it would be perfect... but, wait! We can do it easily!

The most direct approach is to add an argument to the constructor, so the new constructor for "model_ExpiringCache" will be this one:

Then, if we want a model with a 1 day expiration time (1 day = 24 hours = 1,440 minutes = 86,400 seconds), we can accomplish it this way:

And that's all. However, the drawback is that we must modify all the data models.

Another way of doing it is to delegate the task to the "app_AppConfig":

And then add the call to this new method on the "model_ExpiringCache" constructor, like this:

This latest method allows us to do fancy things, like use different expiration values for production or development environments in a more centralized way. Anyway, you can choose yours.

In UML, the total project looks like this:

### 6. Some Caveats

There are some queries that cannot be cached. The most evident ones are modifying queries like INSERT, DELETE or UPDATE. These queries must arrive to the database server.

But even with SELECT queries, there are some circumstances in which a caching system can creaet problems. Take a look at a query like this one:

This query selects randomly 10 banners for the "home" zone of our website. This is intended to generate movement in the banners shown in our home, but if we cache this query, the user will not see any movement at all, until the cached data expires.

The rand() function is not deterministic (as it is not now() or others); so it will return a different value on every execution. If we cache it, we will freeze only one of those results for all the caching period, and therefore breaking the functionality.

But with a simple re-factoring, we can obtain the benefits of caching and show pseudo-randomness:

What we are doing here is to cache fifty different random banners configurations, and select them randomly. The 50 SELECT's will look like this:

We have added a constant condition to the select, which has no cost to the database server but renders 50 different keys for the caching system. A user will need to load the page fifty times to see all the banner's different configurations; so the dynamic effect is achieved. The cost is fifty queries to the database to fetch the cache.

### 7. A Benchmark

What benefits can we expect from our new caching system?

First, it must be said that, in raw performance, sometimes our new implementation will run slower than database queries, specially with very simple, well-optimized queries. But for those queries with joins, our DBM cache will run faster.

However, the problem we solved is not raw performance. You will never have a spare database server for your tests in production. You'll probably have a server with high workloads. In this situation, even the fastest query can run slowly, but with our caching scheme, we are not even using the server, and, in fact, we are reducing its workload. So the real performance increase will come in the form of more petitions per second served.

In a website that I am currently developing, I have done a simple benchmark to understand the benefits of caching. The server is modest: it runs Ubuntu 8.10 running on top of an AMD Athlon 64 X2 5600+, with 2 GB of RAM and an old PATA hard disk. The system runs Apahce and MySQL 5.0, that comes with the Ubuntu distribution without any tuning.

The test was to run Apache's benchmark program (ab) with 1, 5 and 10 concurrent clients loading a page 1,000 times from my development website. The actual page was a product detail that has no less than 20 queries: menu contents, product details, recommended products, banners, etc.

The results without cache were 4.35 p/s for 1 client, 8.25 for 5 clients, and 8.29 for 10 clients. With caching (different expiration), the results were 25.55 p/s with 1 client, 49.01 for 5 clients, and 48.74 for 10 clients.

### Final Thoughts

I've shown you an easy way to insert caching into your data model. Of course, there are a plethora of alternatives, but this one is but one choice that you have.

We have used local DBM files to store the data, but there are even faster alternatives that you might consider exploring. Some ideas for the future: using APC's apc_store() functions as underlying storage system, shared memory for the really critical data, using memcached, etc.

I hope you have enjoyed this tutorial as much as I did writing it. Happy caching!


text:
5

# Prove thie following lemia: Lerma Let {Ta}i sequence of real nurnbers We define C) new sequences {E,k_] and {O-}=1 Vn â‚¬ 2+ I Vn â‚¬ 2 Tin-1IF the scquene...

## Question

###### Prove thie following lemia: Lerma Let {Ta}i sequence of real nurnbers We define C) new sequences {E,k_] and {O-}=1 Vn â‚¬ 2+ I Vn â‚¬ 2 Tin-1IF the scquenees (E,} =1and {O};=1 #re both convergent tothe same litnit, THEN the scquence {Ia}i 1 also convergent: Suygeston: Use the definition o Iitit,

Prove thie following lemia: Lerma Let {Ta}i sequence of real nurnbers We define C) new sequences {E,k_] and {O-}=1 Vn â‚¬ 2+ I Vn â‚¬ 2 Tin-1 IF the scquenees (E,} =1and {O};=1 #re both convergent tothe same litnit, THEN the scquence {Ia}i 1 also convergent: Suygeston: Use the definition o Iitit,

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current Altempt in progress Identify the following compound(s)expected to be themost soluble in butane; CH;CHzCHzCH3? H H H H H A-C-C-C-C-c-H H H H H H H H H H H-C-C-C-C-0-H H H H H 4 H H-C-C-0-C-C-H A-d&_8-a Iland IlI...
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text:
# Thread: 243.13.01.18 motion of a particle

1. $$\tiny{243.13.01.18}$$
$\textsf{The following equations each describe the motion of a particle.}\\$
$\textsf{For which path is the particle's velocity vector always orthogonal to its acceleration vector?}$

\begin{align*} \displaystyle (1) r(t)&=t^8i+t^5j\\ (2) r(t)&=\cos(8t)i+\sin(2t)j\\ (3) r(t)&=ti+t^3j\\ (4) r(t)&=\cos{(10t)}i+\sin(10t)j\\ \end{align*}

I presume this would mean a constant value?

2.

3. How do we know if two vectors are orthogonal?

I would suppose perpendicular to each other
But how would that be determined from the selection.

5. Originally Posted by karush
I would suppose perpendicular to each other
But how would that be determined from the selection.
Yes, perpendicular, normal, orthogonal all mean the same thing. Is there a vector formula involving the angle between two vectors? If two vectors in the plane are orthogonal, what is the projection of one onto the other?

6. More simply, if two vectors are orthogonal, what does that say about the dot product of one vector with the other? With velocity vector t2i+ t8j, the acceleration vector is 2ti+ 8t7j. Now, how do you determine if they are orthogonal?

7. Originally Posted by karush
$\textit{In the notation of the dot product,}\\$
$\textit{the angle between two vectors$u$and$v$is:}$
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right]
\end{align*}
So then if $u=t^2i+ t^8j$ and $v=2ti+ 8t7j$ then
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{(t^2i+ t^8j)\cdot (2ti+ 8t^7j)}{|t^2i+ t^8j||2ti+ 8t^7j|} \right]
\end{align*}
We know:

$\displaystyle \vec{a}\cdot\vec{b}=\left|\vec{a}\right|\left|\vec{b}\right|\cos(\theta)$

If $\displaystyle \theta=\frac{\pi}{2}$, then:

$\displaystyle \vec{a}\cdot\vec{b}=0$

Or (in the case of two-dimensional vectors):

$\displaystyle a_1b_1+a_2b_2=0$

The book answer to this was (4)

\begin{align*} \displaystyle
(1) r(t)&=t^8i+t^5j\\
(2) r(t)&=\cos(8t)i+\sin(2t)j\\
(3) r(t)&=ti+t^3j\\
(4) r(t)&=\cos{(10t)}i+\sin(10t)j\\
\end{align*}

so
\begin{align*} \displaystyle
r(t)&=\cos{(10t)}\textbf{i}+\sin(10t)\textbf{j}\\
r'(t)&=10\cos(10t)\textbf{j}-10\sin(10t)\textbf{i}
\end{align*}

9. Originally Posted by karush
The book answer to this was (4)

\begin{align*} \displaystyle
(1) r(t)&=t^8i+t^5j\\
(2) r(t)&=\cos(8t)i+\sin(2t)j\\
(3) r(t)&=ti+t^3j\\
(4) r(t)&=\cos{(10t)}i+\sin(10t)j\\
\end{align*}
Yes, that's what I got as well, but can you demonstrate this is the case?

\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{u\cdot v}{|u||v|} \right]\\
&=\cos^{-1}\left[
\frac{(\cos{(10t)}\textbf{i}+\sin(10t)\textbf{j}\cdot (-10\sin(10t)\textbf{i}+10\cos(10t)\textbf{j})}
{|\cos{(10t)}\textbf{i}+\sin(10t)\textbf{j}||-10\sin(10t)\textbf{i}+10\cos(10t)\textbf{j}|}
\right]
\end{align*}
hopefully before I plow into it

- - - Updated - - -

ok I don't see how you do a dot product on this?

11. I would just set:

$\displaystyle \vec{r}'(t)\cdot\vec{r}''(t)=0$

Using the dot-product formula:

$\displaystyle \vec{a}\cdot\vec{b}=\sum_{k=1}^{n}\left(a_kb_k\right)$


text:
5

# The Coefficient of Variation is an example of a noise to signalratio statistic? True or False?...

## Question

###### The Coefficient of Variation is an example of a noise to signalratio statistic? True or False?

The Coefficient of Variation is an example of a noise to signal ratio statistic? True or False?

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# Classical transport theory (plasmapy.formulary.braginskii)¶

Functions to calculate classical transport coefficients.

## Introduction¶

Classical transport theory is derived by using kinetic theory to close the plasma two-fluid (electron and ion fluid) equations in the collisional limit. The first complete model in this form was done by S. I. Braginskii 1.

This module uses fitting functions from literature (1, 2, 3, 4, 5 and the next section) to calculate the transport coefficients, which are the resistivity, thermoelectric conductivity, thermal conductivity, and viscosity.

Keep in mind the following assumptions under which the transport equations are derived:

1. The plasma is fully ionized, only consisting of ions and electrons. Neutral atoms are neglected.

2. Turbulent transport does not dominate.

3. The velocity distribution is close to Maxwellian. This implies:

1. Collisional mean free path ≪ gradient scale length along field.

4. The plasma is highly collisional: collisional frequency ≫ gyrofrequency.

When classical transport is not valid, e.g. due to the presence of strong gradients or turbulent transport, the transport is significantly increased by these other effects. Thus classical transport often serves as a lower bound on the losses / transport encountered in a plasma.

## Transport Variables¶

For documentation on the individual transport variables, please take the following links to documentation of methods of ClassicalTransport.

## Using the module¶

Given that many of the transport variables share a lot of the same computation and many are often needed to be calculated simultaneously, this module provides a ClassicalTransport class that can be initialized once with all of the variables necessary for calculation. It then provides all of the functionality as methods (please refer to its documentation).

If you only wish to calculate a single transport variable (or if just don’t like object oriented interfaces), we have also provided wrapper functions in the main module namespace that use ClassicalTransport under the hood (see below, in the Functions section).

Warning

The API for this package is not yet stable.

## Classical transport models¶

In this section, we present a broad overview of classical transport models implemented within this module.

### Braginskii 1¶

The original Braginskii treatment as presented in the highly cited review paper from 1965. Coefficients are found from expansion of the kinetic equation in Laguerre polynomials, truncated at the second term in their series expansion ($$k = 2$$). This theory allows for arbitrary Hall parameter and include results for Z = 1, 2, 3, 4, and infinity (the case of Lorentz gas completely stripped of electrons, and the stationary ion approximation).

### Spitzer-Harm 23¶

These coefficients were obtained from a numerical solution of the Fokker- Planck equation. They give one of the earliest and most accurate (in the Fokker-Planck sense) results for electron transport in simple plasma. They principally apply in the unmagnetized / parallel field case, although for resistivity Spitzer also calculated a famous result for a strong perpendicular magnetic field. Results are for Z = 1, 2, 4, 16, and infinity (Lorentz gas / stationary ion approximation).

### Epperlein-Haines 4¶

Not yet implemented.

### Ji-Held 5¶

This is a modern treatment of the classical transport problem that has been carried out with laudable care. It allows for arbitrary hall parameter and arbitrary $$Z$$ for all coefficients. Similar to the Epperlein-Haines model, it corrects some known inaccuracies in the original Braginskii results, notably the asymptotic behavior of alpha-cross and beta_perp as Hall → +infinity. It also studies effects of electron collisions in the ion terms, which all other treatments have not. To neglect electron-electron collisions, leave $$μ = 0$$. To consider them, specify mu and theta.

References

1(1,2,3)

Braginskii, S. I. “Transport processes in a plasma.” Reviews of plasma physics 1 (1965): 205. (1965)

2(1,2)

Spitzer Jr, Lyman, and Richard Härm. “Transport phenomena in a completely ionized gas.” Physical Review 89.5 (1953): 977. (1953)

3(1,2)

Physics of Fully Ionized Gases, L. Spitzer (1962)

4(1,2)

Epperlein, E. M., and M. G. Haines. “Plasma transport coefficients in a magnetic field by direct numerical solution of the Fokker–Planck equation.” The Physics of fluids 29.4 (1986): 1029-1041.

5(1,2)

Ji, Jeong-Young, and Eric D. Held. “Closure and transport theory for high-collisionality electron-ion plasmas.” Physics of Plasmas 20.4 (2013): 042114.

### Classes¶

 ClassicalTransport(T_e, n_e, T_i, n_i, ion, m_i) Classical transport coefficients (e.g.

### Functions¶

 electron_thermal_conductivity(T_e, n_e, T_i, …) Calculate the thermal conductivity for electrons. electron_viscosity(T_e, n_e, T_i, n_i, ion) Calculate the electron viscosity. ion_thermal_conductivity(T_e, n_e, T_i, n_i, ion) Calculate the thermal conductivity for ions. ion_viscosity(T_e, n_e, T_i, n_i, ion[, …]) Calculate the ion viscosity. resistivity(T_e, n_e, T_i, n_i, ion[, m_i, …]) Calculate the resistivity. thermoelectric_conductivity(T_e, n_e, T_i, …) Calculate the thermoelectric conductivity.