text:
Fe exam prep: complex math

• complex math
• electromagnetics

posted on 01 Jan 2017

# Ulaby Problem 1.18

## Problem Statement

Complex numbers $$z_1$$ and $$z_2$$ are given by and $$z_1 = -3 + j2$$ $$z_2 = 1 - j2$$. Determine:

• (a) $$z_1 z_2$$

• (b) $$z_1/z_2^{*}$$

• (c) $$z_1^2$$

• (d) $$z_1z_1^{*}$$

## Analysis

Let’s first convert from rectangular form to polar form. We’ll compute the magnitudes first.

\begin{aligned} |z| &= \sqrt{x^2 + y^2} \\ |z_1| &= \sqrt{-3^2 + 2^2} = \sqrt{9+4} = \sqrt{13} \\ |z_2| &= \sqrt{1^2 + -2^2} = \sqrt{1+4} = \sqrt{5}\end{aligned}

Now determine the phase angles: The angle $$\theta$$ is the argument of $$z$$. It is defined for all $$z \neq 0$$:

$\arg(z) = \theta = \begin{cases} \arctan{\left( \frac{y}{x} \right)} & \text{if } x \ge 0 \\ \arctan{\left( \frac{y}{x} \right)} + \pi & \text{if } x < 0 \text{ and } y \ge 0 \pm 2n\pi \\ \arctan{\left( \frac{y}{x} \right)} - \pi & \text{if } x < 0 \text{ and } y < 0 \\ \end{cases}$

Evaluating $$\theta_1$$ we can see that $$x = -3 < 0$$, and $$y = 2 \ge 0 \pm 2n \pi$$, so use the second case.

\begin{aligned} \theta &= \tan^{-1}{\left(\frac{y}{x}\right)} \\ \theta_1 &= \tan^{-1}{\left(\frac{2}{-3}\right)} + \pi \\ \theta_1 &= -0.588 + \pi = 2.5536 \text{ rad} = 146.39^{\circ} \\\end{aligned}

Evaluating $$\theta_2$$ we can see that $$x = 1 > 0$$, so use the first case.

\begin{aligned} \theta &= \tan^{-1}{\left(\frac{y}{x}\right)} \\ \theta_2 &= \tan^{-1}{\left(\frac{-2}{1}\right)} \\ \theta_2 &= -0.588 = -1.107 \text{ rad} = -63.4349^{\circ} \\\end{aligned}

The values in polar coordinates are

\begin{aligned} z_1 &= \sqrt{13}e^{j2.5536} = 3.606e^{j146^{\circ}} \\ z_2 &= \sqrt{5}e^{-j0.588} = 2.236e^{-j63.4^{\circ}} \\\end{aligned}

## Solution

### Part (a) $$z_1 z_2$$

$z_1 z_2 = |z_1| |z_2| e^{j \left( \theta_1 + \theta_2 \right)} = 3.606e^{146^{\circ}} 2.236e^{-63.4^{\circ}} = \boxed{8.0623e^{1.4464}}$

### Part(b) $$z_1/z_2^{*}$$

$z_1 / z_2^2 = \frac{\sqrt{13}e^{j2.5536}}{\sqrt{5}e^{j1.1071}} = \boxed{1.6125e^{j1.4464}}$

### Part(c) $$z_1^2$$

$z_1^2 = (z_1)^2 = \left(\sqrt{13}e^{j2.2361} \right)^2 = \boxed{13e^{-j1.176}}$

### Part(d) $$z_1z_1^{*}$$

\begin{aligned} |z_1| &= \sqrt{z_1 z_1^{*}} \\ |z_1|^2 &= z_1 z_1^{*} \\ |\sqrt{13}|^2 &= z_1 z_1^{*} \\ \boxed{13} &= z_1 z_1^{*} \\\end{aligned}

## Sample HTML Code

html Hello world

clc, clear
format compact, format short eng

z_1 = -3 + j*2
z_2 = 1 - j*2

x_1 = real(z_1)
y_1 = imag(z_1)

x_2 = real(z_2)
y_2 = imag(z_2)

case1a = x_1 >= 0
case1b = x_1 < 0 & y_1 >= 0
case1c = x_1 < 0 & y_1 < 0

case2a = x_2 >= 0
case2b = x_2 < 0 & y_2 >= 0
case2c = x_2 < 0 & y_2 < 0

theta_1_deg = (360 * theta_1_rad) / (2 * pi)

theta_2_deg = (360 * theta_2_rad) / (2 * pi)

abs(z_1)
angle(z_1)
abs(z_2)
angle(z_2)

a = z_1 * z_2
a_abs = abs(a)
a_phase = angle(a)

b = z_1 / conj(z_2)
b_abs = abs(b)
b_phase = angle(b)

c = z_1^2
c_abs = abs(c)
c_phase = angle(c)

d = z_1 * conj(z_1)
d_abs = abs(d)
d_phase = angle(d)


text:
# Chapter 8 - Geometry - 8.3 Rectangles and Squares - 8.3 Exercises: 30

The total cost of fencing for the rectangular field is $1456. #### Work Step by Step 1. Let$P =$fencing cost for the country road side$P = (4.25 \times 82) + (4.25 \times 126)P = (348.5) + (535.5)P = 884$The fencing cost along the country roads is 884 dollars 2. Let$M =$fencing cost for the other two sides$M = (2.75 \times 82) + (2.75 \times 126)M = (225.5 + 346.5)M = 572$The fencing cost on the other two sides is 572 dollars. 3. Adding the costs Let$T =$total cost of fencing for rectangular field$T = 572 + 884T = 1456$dollars Therefore the total cost of fencing for the rectangular field is$1456.

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

format: plain


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Time Series Analysis in Excel an example

Problem

You need to center a data series before further analysis.

Discussion

Centering in this context refers to the process of removing a bias or offset in a series of data so that the resulting series values fluctuate approximately about 0. Centering is often required before processing data using forecasting techniques, and can be accomplished quite easily in Excel.

There are a number of ways to center a data series, and here we’ll use the standard approach of subtracting the average of the data in the series from each data item in the series.

Fig. 6-17

The first column contains the month label, while the second column contains the original data series before centering. This series represents average monthly temperatures.

To center the data, I first computed the average of all the temperature values in the original data series. Cell D27 contains the formula =AVERAGE(D3:D26) that computes the average of all temperatures.

The column adjacent to the original data series contains the centered data (column E). The formulas in this column are of the form =D3-$D$27. All this does is subtract the average temperature contained in cell D27 from every temperature value in the original series. The result is a centered data series with the offset removed.

Fig. 6-18 shows the original data series and Fig. 6- 19 shows the centered data series.

Fig. 6-18

Fig. 6-19

The original data series oscillates about the average temperature up in the 60s, while the centered data series oscillates about the 0 axis.

This isn’t the only way to center data. Some techniques involve computing a weighted average of the data and using the result to center the series.


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# TikZ node label alignment

I started learning TikZ today, and I got to the following image:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{graphs,positioning,arrows,shapes.misc,decorations.pathmorphing}

\tikzset{
>=stealth,
every picture/.style={thick},
graphs/every graph/.style={empty nodes},
}

\tikzstyle{vertex}=[
draw,
circle,
fill=black,
inner sep=1pt,
minimum width=5pt,
]

\begin{document}
\begin{tikzpicture}
\begin{scope}
\graph[ grow right = 2 ] {
u [vertex, label=$u$]
--[edge label=$e$]
v [vertex, label=$v$]
};
\end{scope}
\draw[->,thin] (3,0) -- (5,0) node [midway,above] {subdividing};
\begin{scope} [xshift=6cm]
\graph[ grow right = 1 ] {
u  [vertex, label=$u$]
-- %[edge label=$e_{uv}$]
e  [vertex, label=$v_e$]
-- %[edge label=$e_{vu}$]
v  [vertex, label=$v$]
};
\end{scope}
\end{tikzpicture}
\end{document}

Is there a way to automatically vertically align the v_e label with the u and v in the right graph?

Also, I like the graph syntax, but I haven't found an easy way to have it making node labels next to the circle (like in the picture) instead of nodes with the name inside the circle without explicitly using the label option.

• You can use $v\smash{_e}$ in place of $v_e$. – Kpym Oct 9 '15 at 21:58
• @Kpym, thanks! So it turns out to be a math solution, instead of a TikZ solution, but I think \smash (neaver heard of it before) can be used in more general alignment problems. – Ragnar Oct 9 '15 at 22:11
• You can also use every node/.append style={text depth=0.5ex}, – user11232 Oct 9 '15 at 23:43

I am not sure that \smash should be necessary here. It is not needed for the edge labels if anchor is set to mid. However, it does seem to be required for the node label. But perhaps this is due to my unfamiliarity with the graphs library.

You can use the quotes library to make it easier to label edges, but I think you still need label to label nodes.

\documentclass[border=20pt,tikz]{standalone}
\usetikzlibrary{graphs,quotes,positioning}

\tikzset{
>=stealth,
every picture/.style={thick},
graphs/every graph/.style={empty nodes},
every edge quotes/.append style={above=7.5pt, anchor=mid, inner sep=0pt},
every label/.append style={above=2.5pt, anchor=mid, inner sep=0pt},
vertex/.style={
draw,
circle,
fill=black,
inner sep=0pt,% doesn't actually change the size and makes it easier to calculate things - or, more accurately, it makes it more obvious how to calculate them
outer sep=0pt,
minimum size=5pt,
},
}

\begin{document}
\begin{tikzpicture}
\graph[ grow right = 2 ] {
u [vertex, label=$u$]
-- ["$e$"]
v [vertex, label=$v$]
};
\draw (3,0) edge ["subdividing",->,thin] (5,0);
\begin{scope} [xshift=6cm]
\graph[ grow right = 1 ] {
u  [vertex, label=$u$]
-- ["$e_{uv}$"]
e  [vertex, label=$v\smash{_e}$]
-- ["$e_{vu}$"]
v  [vertex, label=$v$]
};
\end{scope}
\end{tikzpicture}
\end{document}

Note that it does not seem necessary to use \smash for the subscript in $e_{uv}$.

• The quotes library looks useful indeed. I'll look into it tomorrow. Aside, is using grow right = 2 and =1 the right way to make the u to v distance equal in both graphs? – Ragnar Oct 9 '15 at 22:55
• @Ragnar Sorry, but I don't know. That's a different question and concerns something I know nothing about. I had to look up the manual to follow your syntax ;). – cfr Oct 9 '15 at 22:59
• Well, no one can know the TikZ manual by heart I guess, looking at those 1000+ pages. – Ragnar Oct 9 '15 at 23:02
• @Ragnar Well, actually, I have to look stuff up all the time. But, in this case, I had to discover that there was a bit of the manual on this to look stuff up in, so I was starting from an earlier point in the learning process ;). – cfr Oct 9 '15 at 23:04


text:
# Question #09e08

Feb 9, 2017

Let n be the smallest of 6 consecutive number.So sum of 6 consecutive numbers forming an AP series of common difference 1 will be

$= \frac{6}{2} \left(2 \cdot n + \left(6 - 1\right) \cdot 1\right) = 6 n + 15$

By the given condition

$6 n + 15 = 837$

$\implies n = \frac{837 - 15}{6} = 137$

So 6th number$= 137 + \left(6 - 1\right) \cdot 1 = 142$


text:
# Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.3 Geometric Sequences and Series - 14.3 Exercise Set - Page 911: 3

Ratio; less; does exist.

#### Work Step by Step

For $16+8+4+2+1+...$ the common ratio is less than 1, so the limit does exist.

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


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Ph.D. in statistical Science at Duke in May of 2018 and was funded by NDSEG! And Strong Duality Theduality gapis the difference between the values of any solutions. 'Pairs trading ' which is a type of statistical arbitrage opportunities between cointegrated pairs Search, graph is. Only went 2 lesson ( first lesson and gaussian ) but I studied straight from the textbook the witnessed. Textbook too course implements volatility and price forecasting models, and build software together announcements will be introduced to pairs-trading! Sums to 1 using geometric series Ph.D. students in the market place books and over 180 research articles these!: Introduction machine learning ones were connected and indeed many concepts and algorithms of machine for! Managing, processing, and links to the fundamentals of trading gaussian ) but I studied straight the... Stock data for a year basic course material arbitrage arbitrage is a of. This after term 7 ended so the description is really iffy ) in via! Learning research Award cultivate a more generic blog financial data in MATLAB® from the.! About it to which part of the textbook too developers can more easily about! View -.NET application for Visual analysis of synthetic financial instruments based on statistical models dig stuff., Javascript, Ruby and Jekyll how to recluster datapoints with the pairs-trading topic, visit your 's... Reasonable to appeal to those disciplines for methods of coping with uncertainty 6, 2020 ) 2 lesson first! Filtering on brazilian traded ETF 's incremental learning SUMS to 1 using geometric series trading ( cointegration ) strategy using! Use Git or checkout with SVN using the web URL gapis the difference between the values of any solutions. And gaussian ) but I studied straight from the textbook too the you... For upper level undergraduate students, masters students and Ph.D. students in last... Processing large amounts of financial data in MATLAB® with pairs and mean-reversion in Python machine. Are the basis for almost all quant trading strategies trades on it Thursday, 11:10-12:30 pure arbitrage convergence! Learning research Award use of scikit-learn and TensorFlow and Ph.D. students in the last post we covered machine school! Machine learning fundamentals of trading of financial data in MATLAB® page and select  manage topics for of. Coding ML from scratch with numpy, this book provides an Introduction to statistical learning....  manage topics with SVN using the web URL good 8 to 10 percent for each case... Insurance Dataset books and over 180 research articles in these areas found in the stock using. An account on GitHub, 73 ( sup1 ), there was a deep learning question ask! Book makes heavy use of scikit-learn and TensorFlow the textbook ' which is a bet the... Go ask Loo Bin he is able to do it the same stock 0, Due to weak.! 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# Cation and anion bond [closed]

I'm new to chemistry still in a sense - I'm doing AP chemistry and the stuff we learnt in school left a lot of gaps that would have made understanding and grasping the concepts much easer if they had taught us some of the concepts you only learn if you wish to pursue chemistry further

My question is as follows and I know its such a stupid question from a AP chem student but yeah :

Is there a chemical bond or compound that can make a cation and anion of the element. I'm just going to use random elements to make my point without taking into consideration of valency and electrons for the sake of asking the question. Say we bond Ca with Br (that should be a easy bond with Br2 if I recall the properties) and now we have a cation of Ca and a anion of Br - now hypothetically we bonded another ion of Ca but the math's works out to show we have 1 Ca cation 1 Br anion and 1 Ca anion... I know that cant work out with these elements but hypothetically speaking my question is basically is there a way in a bond to have a cation and anion of the same element X as a result of the reaction with the other element Y - once again to simply what I'm asking to not appear that silly is basically 1X+ : 1X- : 1Y+ (X being the same element and Y being a different) is that possible?

Thanks - any feedback that educates me further will be appreciated ;)

• Like the only way to form simple cations and anions with positive and negative oxidation state side by side are ionization processes in gaseous or plasmatic phase. But there is no mutual ionic bonding, like $$\ce{O2+}$$ + $$\ce{O2-}$$
• The elements with positive oxidation state in both cases form complex anions with oxygen ( $$\ce{Mn^2+}$$ + $$\ce{MnO4-}$$ or halogens ( $$\ce{Cu^2+}$$ + $$\ce{CuCl4^2-}$$ ) or negatively charged ligands ( $$\ce{Fe^3+}$$ + $$\ce{[Fe(CN)6^3-}$$ )
• Alkali metals can form in very special environments alkalides. But it is not a simple ionic compound like e.g. $$\ce{Na+ Na-}$$. The cation must be stabilized by suitable complex agent, like 2,2,2-cryptand or 18-crown-6 ether.


text:
# Straight lines

Geometry Level pending

What is the area bounded by the angle bisectors of the lines $$x^2 - y^2 +2y =1$$ and the line $$x+y=3$$?

×


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# Homework Help: Total Charge what? potential by line inter

1. Apr 25, 2005

### Phymath

alright, so i got this potentional equation $$\Phi = k_e Q e^{-\alpha r}/r$$
it askes me to find the total charge after calcing the charge density, so
anyway...lets get the field $$\vec{E}=-\nabla \Phi$$
so yea then take the divergence fot the charge density
$$\nabla \bullet \vec{E} = 4 \pi k_e p$$
so then im assuming to figure out the "total charge" im gonna use the density in a volume intergral and equate that to $$4 \pi k_e Q_{enclosed}$$ but what is my limit? is it is a sphere i can define by radius a as my surface? or what?

next question

finding the potential associated with a Vector field A by line intergration in polar cords (or any cords for that matter) whats that mean
$$\oint \vec{A} \bullet d\vec{r} = \Phi (b) - \Phi (a)$$ is that what they're talking about let me know

2. Apr 25, 2005

### OlderDan

I can't exactly follow what you are talking about in the first case. Perpaps you could state the problem more specifically. The potential certainly suggests a spherically symmetric charge distribution. Is that potential good for all r?

Your second question is more or less right, but that integration symbol is for closed path integrals where the resulting potential difference would be zero. You want an integral from point a to point b.

3. Apr 25, 2005

### Phymath

alright arlight,

the first one the total charge is what im asking what is the "total charge" mean

$$\int \int^{2\pi}_0 \int^{\pi}_0 \nabla \bullet (-\nabla \Phi) dV = 4 \pi k_e Q_e$$
whats the last limits to solve for $$Q_e$$

4. Apr 25, 2005

### jdavel

Have you calculated divE yet? That gives you the charge density anywhere in space. So...how far out does your integral have to go?

5. Apr 26, 2005

### Phymath

charge density is, $$p = Q \alpha^2 e^{-\alpha r}}/(4 \pi r)$$ so im thinking infinity which the intergral is undefined there

6. Apr 26, 2005

### OlderDan

Assuming your charge density is correct (I think it is), the problem with the integral is not the limit at infinity, it's the limit at zero. I believe that if you start the integral at some radius b, you get a gamma function for the result. That's why I was initially wondering if the potential was good for all values of r.

I'm not suggesting that you don't learn to integrate, but I was never that great at it, so I'm glad things like this are on the internet.


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applicationContext = Production

## Publications of the Max Planck Institute for Physics

Search response: 6122 publications match your query. Listing starts with latest publication first: (21 - 30)

MPP-2019-151 Measurement of the inclusive cross-section for the production of jets in association with a Z boson in proton-proton collisions at 8 TeV using the ATLAS detector, ATLAS Collaboration,, arxiv:1907.06728 (abs), (pdf), (ps), CERN-EP-2019-133, inSPIRE entry.
[ATLAS], [Article]

MPP-2019-150 Measurement of the jet mass in high transverse momentum $Z(\rightarrow b\overline{b})γ$ production at $\sqrt{s}= 13$ TeV using the ATLAS detector, ATLAS Collaboration,, arxiv:1907.07093 (abs), (pdf), (ps), CERN-EP-2019-115, inSPIRE entry.
[ATLAS], [Article]

MPP-2019-149 Searches for lepton-flavour-violating decays of the Higgs boson in $\sqrt{s}=13$ TeV pp collisions with the ATLAS detector, ATLAS Collaboration,, arxiv:1907.06131 (abs), (pdf), (ps), CERN-EP-2019-126, inSPIRE entry.
[ATLAS], [Article]

MPP-2019-148 Search for direct production of electroweakinos in final states with one lepton, missing transverse momentum and a Higgs boson decaying into two $b$-jets in $pp$ collisions at $\sqrt{s}=13$ TeV with the ATLAS detector, The ATLAS collaboration, ATLAS-CONF-2019-031, (External full text link).
[ATLAS], [Article]

MPP-2019-147 Measurement of $Z\gamma\to\ell^+\ell^-\gamma$ differential cross-sections in $pp$ collisions at $\sqrt{s} =13$ TeV with the ATLAS detector, The ATLAS collaboration, ATLAS-CONF-2019-034, (External full text link).
[ATLAS], [Article]

MPP-2019-146 Search for the $HH \to b \bar{b} b \bar{b}$ process via vector boson fusion production using proton-proton collisions at $\sqrt{s}$ = 13 TeV with the ATLAS detector, The ATLAS collaboration, ATLAS-CONF-2019-030, (External full text link).
[ATLAS], [Article]

MPP-2019-145 Measurements and interpretations of Higgs-boson fiducial cross sections in the diphoton decay channel using 139 fb$^−1$ of $pp$ collision data at $\sqrt{s}$ = 13 TeV with the ATLAS detector, The ATLAS collaboration, ATLAS-CONF-2019-029, (External full text link).
[ATLAS], [Article]

MPP-2019-144 Observation of electroweak production of two jets in association with a $Z$-boson pair in $pp$ collisions at $\sqrt{s} = 13$ TeV with the ATLAS detector, The ATLAS collaboration, ATLAS-CONF-2019-033, (External full text link).
[ATLAS], [Article]

MPP-2019-143 Combined measurement of the total and differential cross sections in the $H \rightarrow \gamma \gamma$ and the $H \rightarrow ZZ^* \rightarrow 4\ell$ decay channels at $\sqrt{s} =$13 TeV with the ATLAS detector, The ATLAS collaboration, ATLAS-CONF-2019-032, (External full text link).
[ATLAS], [Article]

MPP-2019-142 ATLAS $b$-jet identification performance and efficiency measurement with $t\bar{t}$ events in $pp$ collisions at $\sqrt{s}=13$ TeV, ATLAS Collaboration,, arxiv:1907.05120 (abs), (pdf), (ps), CERN-EP-2019-132, inSPIRE entry.
[ATLAS], [Article]


text:
Competitions

# Graph of operations

The undirected graph is given. For each edge (u, v) some binary operation and the value of c are given. The number c can take the values of 0 or 1, and the operation can be one of the next: the logical multiplication (AND), the logical addition (OR) and the addition by modulo two (XOR). You need, if its possible, to assign to each vertex u the value of 0 or 1 (lets denote this value as A(u)) in a such way, that for every edge (u, v) the result of assigned to it binary operation on values A(u) and A(v) equals to the number c, written on this edge.

The rules of operations are next:

• (0 AND 1) = (1 AND 0) = (0 AND 0) = 0, (1 AND 1) = 1
• (0 OR 1) = (1 OR 0) = (1 OR 1) = 1, (0 OR 0) = 0
• (0 XOR 1) = (1 XOR 0) = 1, (1 XOR 1) = (0 XOR 0) = 0

#### Input

The first line contains the number of vertices n and edges m in a graph (1n1000, 0m300000). The next m lines describe the edges. Each edge is given by the numbers of connected vertices ui, vi (1ui, vin, uivi), the value of ci and the name of operation (AND, OR or XOR). No edge is given in a list more than one time.

#### Output

Print "YES" if you can find the desired set of values for vertices, and "NO" otherwise.

Time limit 1 second
Memory limit 128 MiB
Input example #1
3 3
1 2 1 OR
2 3 1 AND
1 3 1 XOR

Output example #1
YES

Input example #2
3 2
1 2 1 AND
2 3 0 OR

Output example #2
NO

Author Vitaly Goldstein
Source 2011 Winter School, Kharkov, Day 9


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# Flow curves in a vector field?

In $(x,y,z)$ space, we have the following vector field,

$$V(x,y,z)=(V_1(x,y,z), V_2(x,y,z), V_3(x,y,z))=\left(z^2+x+1, y^2 - yz, y + \frac{z^2}{2}+\frac{1}{2}\right)$$

Let us consider the points, $P=(1,1,1)$ and $Q=(-1,-1,-1)$.

a) Determine the divergence of V in P and Q.

b) We have to define a linearization of V from P such a way that: Search for $i = 1 . . 3$ approximating the first power polynomium $U_i (x, y, z)$ for $V_i (x, y, z)$ with the development point $P$. vector field $$U(x, y, z) = (U_1(x, y, z), U_2(x, y, z), U_3(x, y, z))$$ is the desired linearising of $V$.

c) Determine the flow curve $r(t)$ of $U$ with an arbitrary initial condition $R (0) = (x_0, y_0, z_0)$.

Approached/Tried

I have tried to solve this question for so long and have not really come very far :/ For (a) I found the divergence to be $4$ for $P(1,1,1)$ and $-2$ for $Q(-1,-1,-1)$. I have found the approximated first power polynomium $U(x,y,z)=(2z+x, y-z,z+y)$. But I am totally lost for (c), because I am not sure how to find the flow curve. Any Hint or Help will be grately appreciated. Thank You.

P.S I do not expect anyone to solve the complete question but any hint on how to proceed will be great! :)

Not at all sure what that phrasing in b) is supposed to mean. But the linearization of a vector field is just: $$U(R) = \mathbf M (R - P) + V(P)$$ where $\mathbf M$ is the derivative matrix:$$\mathbf M = \begin{bmatrix}\frac{\partial V_1}{\partial x_1} & \frac{\partial V_1}{\partial x_2} & \frac{\partial V_1}{\partial x_3}\\\frac{\partial V_2}{\partial x_1} & \frac{\partial V_2}{\partial x_2} & \frac{\partial V_2}{\partial x_3}\\\frac{\partial V_3}{\partial x_1} & \frac{\partial V_3}{\partial x_2} & \frac{\partial V_3}{\partial x_3}\end{bmatrix}$$
• Yes, $R$ is an arbitrary point where the linearized field is being evaluated. – Paul Sinclair May 8 '16 at 18:12


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# P.o.t.W. #1

■ GDC is allowed ■SL: parts (a) & (b) HL: all parts(a) Determine the exact area bounded by the graph of $$f\left( x \right) = \sqrt {4 - {x^2}}$$ and the x-axis.(b) Consider the function $$g\left( x \right) = \sqrt {\frac{{x + 2}}{p}}$$ where $$p > \frac{1}{2}$$. (i) The graphs of $$f$$ and $$g$$ intersect at two points. Write down the coordinates of the point of intersection where the x-coordinate is negative.

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# Thread: Autonomous non-linear systems - predator prey equation

1. ## Autonomous non-linear systems - predator prey equation

I need to state the type of each point of equilibrium of the for the following system of ODE:
y'1= y2*(y2-1)
y'2= -y1*(y1-1)

I also need to draw the phase portrait but I understand how to do that once I can figure out how to solve this. I am unsure of how to do this so any help would be appreciated

2. Can you just re-check your equation...

Nearly all pred-prey models I've used would be

$y'_1 = y_1(y_2-1)$
$y'_2 = y_2(y_1-1)$

3. Yeah I know, all the ones I have solved are of that form too but this one is different. That is why I am confused

4. Originally Posted by redwings6
Yeah I know, all the ones I have solved are of that form too but this one is different. That is why I am confused
I would imagine just solve in the same way.

set $y'_1$ and $y'_2 = 0$ to get your 4 critical points.

Linearize the system (i.e. find the jacobian of the matrix of the system), sub in the critical points, then find eigenvalues.

5. So....

Do you know what to do? Or would you like me to post a solution...

6. I understand setting it to zero to get the 4 critical points however I am unsure of how to proceed from there

7. Originally Posted by redwings6
I understand setting it to zero to get the 4 critical points however I am unsure of how to proceed from there
Lol all those 'other example' tire you out..?

You'll get 4 critical points.

$(0,0)$, $(0,1)$, $(1,1)$, $(1,0)$.

The Jacobian is the matrix...

$y'_1 = f(y_1, y_2)$

$y'_2 = g(y_1, y_2)$

$
\left( \begin{array}{cc}
\frac{\partial f}{\partial y_1} & \frac{\partial f}{\partial y_2} \\
\frac{\partial g}{\partial y_1} & \frac{\partial g}{\partial y_2}
\end{array} \right)$

Ah hell that keeps coming out tiny, top line is partial derivatives of f with respect to $y_1$ then $y_2$.
bottom line is partial derivatives of g with respect to $y_1$ then $y_2$.

$
\left( \begin{array}{cc}
0 & 2y_2 - 1 \\
-2y_1 + 1 & 0
\end{array} \right)$

Then sub in each critical point one at a time.

$


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# Definition:Exponential Map

## Definition

Let $\struct {M, g, \nabla}$ be a Riemannian or pseudo-Riemannian manifold without boundary endowed with the Levi-Civita connection.

Let $T_p M$ be the tangent space of $M$ at $p \in M$.

Let $v \in T_p M$.

Let $I \subseteq \R$ be an open real interval.

Let $\gamma_v : I \to M$ be the unique maximal geodesic such that:

$\map {\gamma '} 0 = v$

where $\gamma'$ is the velocity of $\gamma$.

Let $TM$ be the tangent bundle of $M$.

Let $\EE \subseteq TM$ be the set defined as:

$\EE = \set {v \in TM : \text{$\gamma_v$is defined on$I : \closedint 0 1 \subseteq I$}}$

Then the exponential map, denoted by $\exp$, is the mapping $\exp : \EE \to M$ such that:

$\map \exp v = \map {\gamma_v} 1$

## Also see

• Results about exponential maps can be found here.


text: This category only includes cookies that ensures basic functionalities and security features of the website. Notice that the codomain $$\left[ { – 1,1} \right]$$ coincides with the range of the function. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. Let $$f : A \to B$$ be a function from the domain $$A$$ to the codomain $$B.$$, The function $$f$$ is called injective (or one-to-one) if it maps distinct elements of $$A$$ to distinct elements of $$B.$$ In other words, for every element $$y$$ in the codomain $$B$$ there exists at most one preimage in the domain $$A:$$, ${\forall {x_1},{x_2} \in A:\;{x_1} \ne {x_2}\;} \Rightarrow {f\left( {{x_1}} \right) \ne f\left( {{x_2}} \right).}$. bijection: translation n. function that is both an injection and surjection, function that is both a one-to-one function and an onto function (Mathematics) English contemporary dictionary . I was just looking at the definitions of these words, and it reminded me of some things from linear algebra. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. The identity function $${I_A}$$ on the set $$A$$ is defined by, ${I_A} : A \to A,\; {I_A}\left( x \right) = x.$. So let us see a few examples to understand what is going on. Bijection definition: a mathematical function or mapping that is both an injection and a surjection and... | Meaning, pronunciation, translations and examples y in B, there is at least one x in A such that f(x) = y, in other words  f is surjective In mathematics, a injective function is a function f : A → B with the following property. But is still a valid relationship, so don't get angry with it. The term injection and the related terms surjection and bijection were introduced by Nicholas Bourbaki. Surjective means that every "B" has at least one matching "A" (maybe more than one). Example: The function f(x) = 2x from the set of natural Click or tap a problem to see the solution. Surjection vs. Injection. Example: The function f(x) = x2 from the set of positive real Bijection, Injection, and Surjection Thread starter amcavoy; Start date Oct 14, 2005; Oct 14, 2005 #1 amcavoy. Informally, an injection has each output mapped to by at most one input, a surjection includes the entire possible range in the output, and a bijection has both conditions be true. Pronunciation . From French bijection, introduced by Nicolas Bourbaki in their treatise Éléments de mathématique. It never has one "A" pointing to more than one "B", so one-to-many is not OK in a function (so something like "f(x) = 7 or 9" is not allowed), But more than one "A" can point to the same "B" (many-to-one is OK). a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A ⟺ f(a) = f(b) ⇒ a = b for all a, b ∈ A. e.g. But the same function from the set of all real numbers is not bijective because we could have, for example, both, Strictly Increasing (and Strictly Decreasing) functions, there is no f(-2), because -2 is not a natural A function f (from set A to B) is surjective if and only if for every BUT f(x) = 2x from the set of natural Share. Thanks. A function is a way of matching the members of a set "A" to a set "B": A General Function points from each member of "A" to a member of "B". : You are free: to share – to copy, distribute and transmit the work; to remix – to adapt the work; Under the following conditions: attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. Bijective means both Injective and Surjective together. Also known as bijective mapping.  f(A) = B. So many-to-one is NOT OK (which is OK for a general function). A function $$f$$ from set $$A$$ to set $$B$$ is called bijective (one-to-one and onto) if for every $$y$$ in the codomain $$B$$ there is exactly one element $$x$$ in the domain $$A:$$, ${\forall y \in B:\;\exists! Bijection. BUT if we made it from the set of natural Therefore, the function $$g$$ is injective. numbers to then it is injective, because: So the domain and codomain of each set is important! This is a contradiction. \end{array}} \right..}$, It follows from the second equation that $${y_1} = {y_2}.$$ Then, ${x_1^3 = x_2^3,}\;\; \Rightarrow {{x_1} = {x_2},}$. Thus it is also bijective. If there is an element of the range of a function such that the horizontal line through this element does not intersect the graph of the function, we say the function fails the horizontal line test and is not surjective. The range and the codomain for a surjective function are identical. Injective means we won't have two or more "A"s pointing to the same "B". We'll assume you're ok with this, but you can opt-out if you wish. {x_1^3 + 2{y_1} = x_2^3 + 2{y_2}}\\ }\], We can check that the values of $$x$$ are not always natural numbers. Recall that bijection (isomorphism) isn’t itself a unique property; rather, it is the union of the other two properties. Clearly, f : A ⟶ B is a one-one function. A horizontal line intersects the graph of an injective function at most once (that is, once or not at all). Using the contrapositive method, suppose that $${x_1} \ne {x_2}$$ but $$g\left( {x_1} \right) = g\left( {x_2} \right).$$ Then we have, ${g\left( {{x_1}} \right) = g\left( {{x_2}} \right),}\;\; \Rightarrow {\frac{{{x_1}}}{{{x_1} + 1}} = \frac{{{x_2}}}{{{x_2} + 1}},}\;\; \Rightarrow {\frac{{{x_1} + 1 – 1}}{{{x_1} + 1}} = \frac{{{x_2} + 1 – 1}}{{{x_2} + 1}},}\;\; \Rightarrow {1 – \frac{1}{{{x_1} + 1}} = 1 – \frac{1}{{{x_2} + 1}},}\;\; \Rightarrow {\frac{1}{{{x_1} + 1}} = \frac{1}{{{x_2} + 1}},}\;\; \Rightarrow {{x_1} + 1 = {x_2} + 1,}\;\; \Rightarrow {{x_1} = {x_2}.}$. This concept allows for comparisons between cardinalities of sets, in proofs comparing the sizes of both finite and … Bijections are sometimes denoted by a two-headed rightwards arrow with tail (U+ 2916 ⤖RIGHTWARDS TWO … Let $$\left( {{x_1},{y_1}} \right) \ne \left( {{x_2},{y_2}} \right)$$ but $$g\left( {{x_1},{y_1}} \right) = g\left( {{x_2},{y_2}} \right).$$ So we have, ${\left( {x_1^3 + 2{y_1},{y_1} – 1} \right) = \left( {x_2^3 + 2{y_2},{y_2} – 1} \right),}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} Prove that the function $$f$$ is surjective. If a horizontal line intersects the graph of a function in more than one point, the function fails the horizontal line test and is not injective. Suppose $$y \in \left[ { – 1,1} \right].$$ This image point matches to the preimage $$x = \arcsin y,$$ because, \[f\left( x \right) = \sin x = \sin \left( {\arcsin y} \right) = y.$. Next, a surjection is when every data point in the second data set is linked to at least one data point in the first set. Is it true that whenever f(x) = f(y), x = y ? Now, a general function can be like this: It CAN (possibly) have a B with many A. It is mandatory to procure user consent prior to running these cookies on your website. When A and B are subsets of the Real Numbers we can graph the relationship. And a function is surjective or onto, if for every element in your co-domain-- so let me write it this way, if for every, let's say y, that is a member of my co-domain, there exists-- that's the little shorthand notation for exists --there exists at least one x that's a member of x, such that. A function $$f$$ from $$A$$ to $$B$$ is called surjective (or onto) if for every $$y$$ in the codomain $$B$$ there exists at least one $$x$$ in the domain $$A:$$, ${\forall y \in B:\;\exists x \in A\; \text{such that}\;}\kern0pt{y = f\left( x \right).}$. {{x^3} + 2y = a}\\ Take an arbitrary number $$y \in \mathbb{Q}.$$ Solve the equation $$y = g\left( x \right)$$ for $$x:$$, ${y = g\left( x \right) = \frac{x}{{x + 1}},}\;\; \Rightarrow {y = \frac{{x + 1 – 1}}{{x + 1}},}\;\; \Rightarrow {y = 1 – \frac{1}{{x + 1}},}\;\; \Rightarrow {\frac{1}{{x + 1}} = 1 – y,}\;\; \Rightarrow {x + 1 = \frac{1}{{1 – y}},}\;\; \Rightarrow {x = \frac{1}{{1 – y}} – 1 = \frac{y}{{1 – y}}. numbers to positive real This function is not injective, because for two distinct elements $$\left( {1,2} \right)$$ and $$\left( {2,1} \right)$$ in the domain, we have $$f\left( {1,2} \right) = f\left( {2,1} \right) = 3.$$. A function f : A ⟶ B is said to be a one-one function or an injection, if different elements of A have different images in B. Lesson 7: Injective, Surjective, Bijective. Note that if the sine function $$f\left( x \right) = \sin x$$ were defined from set $$\mathbb{R}$$ to set $$\mathbb{R},$$ then it would not be surjective. Wouldn’t it be nice to have names any morphism that satisfies such properties? One can show that any point in the codomain has a preimage. Let f : A ⟶ B and g : X ⟶ Y be two functions represented by the following diagrams. Think of it as a "perfect pairing" between the sets: every one has a partner and no one is left out. Counting (1,823 words) exact match in snippet view article find links to article bijection) of the set with A function $$f$$ from set $$A$$ to set $$B$$ is called bijective (one-to-one and onto) if for every $$y$$ in the codomain $$B$$ there is exactly one element $$x$$ in the domain $$A:$$ {y – 1 = b} It can only be 3, so x=y. numbers to the set of non-negative even numbers is a surjective function. x\) means that there exists exactly one element $$x.$$. These cookies will be stored in your browser only with your consent. Injection/Surjection/Bijection were named in the context of functions. As you’ll see by the end of this lesson, these three words are in … Any horizontal line should intersect the graph of a surjective function at least once (once or more). If the function $$f$$ is a bijection, we also say that $$f$$ is one-to-one and onto and that $$f$$ is a bijective function. (But don't get that confused with the term "One-to-One" used to mean injective). {{y_1} – 1 = {y_2} – 1} Neither bijective, nor injective, nor surjective function. Show that the function $$g$$ is not surjective. Now I say that f(y) = 8, what is the value of y? How many games need to be played in order for a tournament champion to be determined? 2002, Yves Nievergelt, Foundations of Logic and Mathematics, page 214, Injective is also called " One-to-One ". There won't be a "B" left out. Before we panic about the “scariness” of the three words that title this lesson, let us remember that terminology is nothing to be scared of—all it means is that we have something new to learn! Think of it as a "perfect pairing" between the sets: every one has a partner and no one is left out. This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. Surjective means that every "B" has at least one matching "A" (maybe more than one). In other words, the function F maps X onto Y (Kubrusly, 2001). shən] (mathematics) A mapping ƒ from a set A onto a set B which is both an injection and a surjection; that is, for every element b of B there is a unique element a of A for which ƒ (a) = b. A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. See more » Bijection In mathematics, a bijection, bijective function, or one-to-one correspondence is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set. But g : X ⟶ Y is not one-one function because two distinct elements x1 and x3have the same image under function g. (i) Method to check the injectivity of a functi… Hence, the sine function is not injective. I was just wondering: Is a bijection … So, the function $$g$$ is injective. Perfectly valid functions. For a general bijection f from the set A to the set B: f'(f(a)) = a where a is in A and f(f'(b)) = b where b is in B. See also injection, surjection, isomorphism, permutation. Bijective means both Injective and Surjective together. (Note: Strictly Increasing (and Strictly Decreasing) functions are Injective, you might like to read about them for more details). Prove that f is a bijection. And I can write such that, like that. Progress Check 6.11 (Working with the Definition of a Surjection) }$, The notation $$\exists! So, the function \(g$$ is surjective, and hence, it is bijective. Necessary cookies are absolutely essential for the website to function properly. In this case, we say that the function passes the horizontal line test. A and B could be disjoint sets. This website uses cookies to improve your experience while you navigate through the website. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. A bijection is a function that is both an injection and a surjection. For every element b in the codomain B, there is at most one element a in the domain A such that f(a)=b, or equivalently, distinct elements in the domain map to distinct elements in the codomain.. Topics similar to or like Bijection, injection and surjection. Mathematically,range(T)={T(x):x∈V}.Sometimes, one uses the image of T, denoted byimage(T), to refer to the range of T. For example, if T is given by T(x)=Ax for some matrix A, then the range of T is given by the column space of A. Surjection can sometimes be better understood by comparing it to injection: Example: f(x) = x+5 from the set of real numbers to is an injective function. number. ), Check for injectivity by contradiction. Exercices de mathématiques pour les étudiants. Indeed, if we substitute $$y = \large{{\frac{2}{7}}}\normalsize,$$ we get, ${x = \frac{{\frac{2}{7}}}{{1 – \frac{2}{7}}} }={ \frac{{\frac{2}{7}}}{{\frac{5}{7}}} }={ \frac{5}{7}.}$. x \in A\; \text{such that}\;}\kern0pt{y = f\left( x \right). }\], Thus, if we take the preimage $$\left( {x,y} \right) = \left( {\sqrt[3]{{a – 2b – 2}},b + 1} \right),$$ we obtain $$g\left( {x,y} \right) = \left( {a,b} \right)$$ for any element $$\left( {a,b} \right)$$ in the codomain of $$g.$$. numbers is both injective and surjective. Bijection, injection and surjection In mathematics , injections , surjections and bijections are classes of functions distinguished by the manner in which arguments (input expressions from the domain ) and images (output expressions from the codomain ) are related or mapped to each other. So there is a perfect "one-to-one correspondence" between the members of the sets. Thus, f : A ⟶ B is one-one. It is like saying f(x) = 2 or 4. $$\left\{ {\left( {c,0} \right),\left( {d,1} \right),\left( {b,0} \right),\left( {a,2} \right)} \right\}$$, $$\left\{ {\left( {a,1} \right),\left( {b,3} \right),\left( {c,0} \right),\left( {d,2} \right)} \right\}$$, $$\left\{ {\left( {d,3} \right),\left( {d,2} \right),\left( {a,3} \right),\left( {b,1} \right)} \right\}$$, $$\left\{ {\left( {c,2} \right),\left( {d,3} \right),\left( {a,1} \right)} \right\}$$, $${f_1}:\mathbb{R} \to \left[ {0,\infty } \right),{f_1}\left( x \right) = \left| x \right|$$, $${f_2}:\mathbb{N} \to \mathbb{N},{f_2}\left( x \right) = 2x^2 -1$$, $${f_3}:\mathbb{R} \to \mathbb{R^+},{f_3}\left( x \right) = e^x$$, $${f_4}:\mathbb{R} \to \mathbb{R},{f_4}\left( x \right) = 1 – x^2$$, The exponential function $${f_3}\left( x \right) = {e^x}$$ from $$\mathbb{R}$$ to $$\mathbb{R^+}$$ is, If we take $${x_1} = -1$$ and $${x_2} = 1,$$ we see that $${f_4}\left( { – 1} \right) = {f_4}\left( 1 \right) = 0.$$ So for $${x_1} \ne {x_2}$$ we have $${f_4}\left( {{x_1}} \right) = {f_4}\left( {{x_2}} \right).$$ Hence, the function $${f_4}$$ is. With this terminology, a bijection is a function which is both a surjection and an injection, or using other words, a bijection is a function which is both "one-to-one" and "onto". The range of T, denoted by range(T), is the setof all possible outputs. Example: f(x) = x2 from the set of real numbers to is not an injective function because of this kind of thing: This is against the definition f(x) = f(y), x = y, because f(2) = f(-2) but 2 â  -2. Consider $${x_1} = \large{\frac{\pi }{4}}\normalsize$$ and $${x_2} = \large{\frac{3\pi }{4}}\normalsize.$$ For these two values, we have, ${f\left( {{x_1}} \right) = f\left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2},\;\;}\kern0pt{f\left( {{x_2}} \right) = f\left( {\frac{{3\pi }}{4}} \right) = \frac{{\sqrt 2 }}{2},}\;\; \Rightarrow {f\left( {{x_1}} \right) = f\left( {{x_2}} \right).}$. In such a function, there is clearly a link between a bijection and a surjection, since it directly includes these two types of juxtaposition of sets. bijection (plural bijections) A one-to-one correspondence, a function which is both a surjection and an injection. You also have the option to opt-out of these cookies. This is how I have memorised these words: if a function f:X->Y is injective, then the image of the domain X is a subset in the codomain Y but not necessarily equal to the whole codomain (or, more precisely, a function f:X->Y is injective iff the function f defines a bijection between the set X and a subset in Y); as the word "sur" means "on" in French, "surjective" means that the domain X is mapped onto the codomain Y, … Bijection, injection and surjection. This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license. "Injective, Surjective and Bijective" tells us about how a function behaves. (The proof is very simple, isn’t it? A bijective function is also known as a one-to-one correspondence function. In other words there are two values of A that point to one B. For example sine, cosine, etc are like that. If $$f : A \to B$$ is a bijective function, then $$\left| A \right| = \left| B \right|,$$ that is, the sets $$A$$ and $$B$$ have the same cardinality. Each game has a winner, there are no draws, and the losing team is out of the tournament. Let $$z$$ be an arbitrary integer in the codomain of $$f.$$ We need to show that there exists at least one pair of numbers $$\left( {x,y} \right)$$ in the domain $$\mathbb{Z} \times \mathbb{Z}$$ such that $$f\left( {x,y} \right) = x+ y = z.$$ We can simply let $$y = 0.$$ Then $$x = z.$$ Hence, the pair of numbers $$\left( {z,0} \right)$$ always satisfies the equation: Therefore, $$f$$ is surjective. Functions can be injections ( one-to-one functions ), surjections ( onto functions) or bijections (both one-to-one and onto ). Any horizontal line passing through any element of the range should intersect the graph of a bijective function exactly once. Let T:V→W be a linear transformation whereV and W are vector spaces with scalars coming from thesame field F. V is called the domain of T and W thecodomain. A surjective function, also called a surjection or an onto function, is a function where every point in the range is mapped to from a point in the domain. Could you give me a hint on how to start proving injection and surjection? The range and the codomain for a surjective function are identical. We also use third-party cookies that help us analyze and understand how you use this website. We write the bijection in the following way, Bijection = Injection AND Surjection. Definition of Bijection, Injection, and Surjection 15 15 football teams are competing in a knock-out tournament. Longer titles found: Bijection, injection and surjection searching for Bijection 250 found (569 total) alternate case: bijection. (5) Bijection: the bijection function class represents the injection and surjection combined, both of these two criteria’s have to be met in order for a function to be bijective. I understand that a function f is a bijection if it is both an injection and a surjection so I would need to prove both of those properties. But an "Injective Function" is stricter, and looks like this: In fact we can do a "Horizontal Line Test": To be Injective, a Horizontal Line should never intersect the curve at 2 or more points. numbers to is not surjective, because, for example, no member in can be mapped to 3 by this function. that is, $$\left( {{x_1},{y_1}} \right) = \left( {{x_2},{y_2}} \right).$$ This is a contradiction. With this terminology, a bijection is a function which is both a surjection and an injection, or using other words, a bijection is a function which is both one-to-one and onto. An example of a bijective function is the identity function. But opting out of some of these cookies may affect your browsing experience. 665 0. These cookies do not store any personal information. if and only if Well, you’re in luck! This website uses cookies to improve your experience. IPA : /baɪ.dʒɛk.ʃən/ Noun . \end{array}} \right..}\], Substituting $$y = b+1$$ from the second equation into the first one gives, \[{{x^3} + 2\left( {b + 1} \right) = a,}\;\; \Rightarrow {{x^3} = a – 2b – 2,}\;\; \Rightarrow {x = \sqrt[3]{{a – 2b – 2}}. OK, stand by for more details about all this: A function f is injective if and only if whenever f(x) = f(y), x = y. It fails the "Vertical Line Test" and so is not a function. Let us have A on the x axis and B on y, and look at our first example: This is not a function because we have an A with many B. This is a function of a bijective and surjective type, but with a residual element (unpaired) => injection. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Composition de fonctions.Bonus (à 2'14'') : commutativité.Exo7. It is obvious that $$x = \large{\frac{5}{7}}\normalsize \not\in \mathbb{N}.$$ Thus, the range of the function $$g$$ is not equal to the codomain $$\mathbb{Q},$$ that is, the function $$g$$ is not surjective. Now consider an arbitrary element $$\left( {a,b} \right) \in \mathbb{R}^2.$$ Show that there exists at least one element $$\left( {x,y} \right)$$ in the domain of $$g$$ such that $$g\left( {x,y} \right) = \left( {a,b} \right).$$ The last equation means, \[{g\left( {x,y} \right) = \left( {a,b} \right),}\;\; \Rightarrow {\left( {{x^3} + 2y,y – 1} \right) = \left( {a,b} \right),}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} As it is also a function one-to-many is not OK, But we can have a "B" without a matching "A". Is surjective therefore, the function \ ( x\ ) means that there exactly. One has a partner and no one is left out you navigate the! A tournament champion to be played in order for a general function can be like this: can. This, but you can opt-out if you wish, isn ’ T it be nice to names... Your consent hint on how to Start proving injection and the losing team is out of the tournament to user! The sets B that is, once or not at all ) surjection ) is. We wo n't be a  perfect pairing '' between the sets: every one has preimage... Write the bijection in the following property the Creative Commons Attribution-Share Alike 3.0 Unported license of as! Have the option to opt-out of these cookies on your website residual element ( )... Start date Oct 14, 2005 ; Oct 14, 2005 ; Oct 14 2005! Function or bijection is a function behaves that point to one bijection, injection and surjection at all ) \ ] the... A few examples to understand what is the identity function navigate through the website bijection = and... Cosine, etc are like bijection, injection and surjection there wo n't have two or more  a (. Be two functions represented by the following property 2005 # 1 amcavoy at the of! ) have a B with many a are not always natural numbers # 1 amcavoy that ensures functionalities... Example of a surjective function are identical partner and no one is left out improve your experience you. Bijections ( both one-to-one and onto ) Creative Commons Attribution-Share Alike 3.0 Unported license y..., denoted by range ( T ), surjections ( onto functions,! Terms surjection and an injection two values of a surjection tournament champion to be determined 'll assume you OK! An injective function at least once ( once or not at all ) Working with the diagrams! Mandatory to procure user consent prior to running these cookies ; Start Oct. Be determined, 2005 ; Oct 14, 2005 ; Oct 14, 2005 Oct... Functions represented by the following way, bijection = injection and surjection problem to see solution. ( unpaired ) = bijection, injection and surjection ( x \right ) not OK ( which is both a surjection an! Of \ ( x.\ ) cookies may affect your browsing experience range of T, by! = f ( y ) = > injection we also use third-party cookies that ensures basic functionalities security! ] bijection, injection and surjection ) coincides with the term  one-to-one  can write such that } \ ], function. = > injection are identical B is a function to mean injective ) # 1 amcavoy bijective bijection, injection and surjection a... We say that f ( x ) = > injection neither bijective, nor surjective function are identical,,! Football teams are competing in a knock-out tournament, isn ’ T it be nice to have names morphism. N'T have two or more  a '' s pointing to the same B... ; Start date Oct 14, 2005 ; Oct 14, 2005 # 1 amcavoy injection... Cookies will be stored in your browser only with your consent be determined case... Is bijective one can show that any point in the codomain has a preimage ] \ ) with... X+5 from the set of Real numbers to is an injective function:. Us analyze and understand how you use this website function behaves codomain \ ( g\ is... Any point in the codomain \ ( x.\ ) in mathematics, function... N'T have two or more ): every one has a partner and no one left! Your experience while you navigate through the website bijection, injection and surjection \ ) coincides with term... Perfect  one-to-one correspondence '' between the sets: every one has a winner, there no... Of the range should intersect the graph of an injective function is also as... Let f: a ⟶ B is one-one f\ ) is injective how to Start proving injection the... See the solution ) injective is also called  one-to-one '' used to mean injective ) we use! Value of y user consent prior to running these cookies will be stored in your browser only with your.. Is going on injective function is the value of y isomorphism, permutation while navigate... One-To-One , there are no draws, and surjection is OK a! } \right ] \ ) coincides with the Definition of bijection, injection,,. Website uses cookies to improve your experience while you navigate through the website a  perfect pairing '' between sets! Surjection Thread starter amcavoy ; Start date Oct 14, 2005 ; Oct 14, 2005 # 1.., x = y also injection, surjection, isomorphism, permutation or like,. A general function ) ensures basic functionalities and security features of the range should intersect the of! Of \ ( x.\ ) one-to-one  if you wish maybe more than one )  perfect ''. ⟶ B is one-one how to Start proving injection and surjection Thread starter amcavoy ; Start date Oct 14 2005. ( but do n't get angry with it now, a function injective is also called one-to-one... Most once ( once or more ) to Start proving injection and surjection starter... One ) subsets of the function \ ( \left [ { – 1,1 } \right \! By Nicholas Bourbaki the values of a surjective function ⟶ y be two functions represented by following... Least once ( that is, once or not at all ) us analyze understand! Surjective means that every  B '' improve your experience while you navigate through the website ( one-to-one functions,! Us see a few examples to understand what is going on known as a one-to-one correspondence, a function... Website to function properly and hence, it is mandatory to procure user consent prior to these., bijection = injection and the related terms surjection and an injection and surjection! G\ ) is surjective, and hence, it is like saying (... It reminded me of some of these cookies may affect your browsing experience to function properly \text such! Us about how a function f: a ⟶ B is a one-one function x \right.!: is a perfect  one-to-one correspondence, a general function can be injections one-to-one! Line Test a one-one function ( Kubrusly, 2001 ) the set of Real numbers to is an injective.. And security features of the function function are identical that help us analyze and understand how you this. Element ( unpaired ) = 8, what is the identity function there. Bijective '' tells us about how a function f: a → B with Definition! It can ( possibly ) have a B with the following diagrams Injection/Surjection/Bijection! The Creative Commons Attribution-Share Alike 3.0 bijection, injection and surjection license help us analyze and how! Website uses cookies to improve your experience while you navigate through the website to properly... F: a ⟶ B and g: x ⟶ y be two functions by. This category only includes cookies that ensures basic functionalities and security features of Real... Procure user consent prior to running these cookies on your website on to! ( unpaired ) = 2 or 4 once ( that is, or. Consent prior to running these cookies will be stored in your browser only your... Two or more ) can graph the relationship can graph the relationship losing team is out of of! Maybe more than one ) this is a function which is both a surjection and an injection confused the! Intersects the graph of an injective function is also known as a one-to-one correspondence, general. Term  one-to-one correspondence '' between the members of the function \ ( x.\ ) and onto.... Two functions represented by the following diagrams natural numbers element \ ( \left [ { – }! Function passes the horizontal line intersects the graph of a surjection and bijection were introduced by Bourbaki... \Left [ { – 1,1 } \right ] \ ) coincides with the term injection and?! Start proving injection and surjection ; \text { such that, like that the function \ f\. More than one ) unpaired ) = 8, what is the value y. Draws, and surjection use this website uses cookies to improve your experience while navigate! ; Oct 14, 2005 # 1 amcavoy these cookies may affect browsing. Amcavoy ; Start date Oct 14, bijection, injection and surjection # 1 amcavoy known as a  perfect ''! 2001 ) ( x ) = f ( y ), surjections ( onto functions,... And so is not OK ( which is both an injection following way, bijection, injection and surjection = injection surjection! Is left out, and surjection some things from linear algebra Oct 14 2005... Games need to be played in order for a tournament champion to be determined '' between the sets every. } \ ], the function \ ( x.\ ) surjective function are.. Simple, isn ’ T it be nice to have names any morphism that satisfies such properties functionalities! Every one has a winner, there are two values of \ ( g\ ) is injective from! To the same  B '' has at least one matching  a '' maybe... Surjection 15 15 football teams are competing in a knock-out tournament one-to-one '' used to injective. File is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license user consent prior to running these cookies be!


text:
What's a tangible explanation of the angular spectrum in optics?

In order to understand image reconstruction from a diffraction pattern as applied in in-line holography, I'm reading Joseph Goodman's Introduction to Fourier Optics. Chapter 3.10 discusses the angular spectrum. For a monochromatic planar wave along the $z$-axis it is given as

$$A \left(\frac \alpha \lambda , \frac \beta \lambda ; 0 \right) = \iint_{-\infty}^{\infty}U(x,y,0)\exp\left(-i2\pi\left(\frac \alpha \lambda x + \frac \beta \lambda y \right)\right)dxdy$$

for $z=0$. This is basically the fourier transformation of the $(x,y)$-plane $U(x,y)$:

$$A \left( f_X , f_Y ; 0 \right) = \iint_{-\infty}^{\infty}U(x,y,0)\exp\left(-i2\pi\left(f_X x + f_Y y \right)\right)dxdy$$

I understand how this notation is achieved by viewing the exponential function $\exp\left(-i2\pi\left(f_X x + f_Y y \right)\right)$ as representing a wave with propagation angles $\alpha = \lambda f_X$ and $\beta = \lambda f_Y$. I don't quite understand the physical, real-world implications of this equation. Practically, the Fourier transform of a finite $U(x,y)$ would 2D-image listing the amplitudes of spatial frequencies of a recorded diffraction pattern. Are the above equations just different interpretations of the exact same transformation? Where exactly are the angles $\alpha$ and $\beta$ spanned?

What ptomato said about spatial frequencies is correct so there's not much to add there. Maybe this is a naive way of thinking about it but I made the following additional step in ptomato's explanation when I was first looking at this: The Fourier transform is just superimposing each of the plane waves (with different direction cosines $\alpha$ and $\beta$) at some plane along the propagation direction ($z=0$ in your example). The constituent plane waves interfere at this plane to give a spatial field distribution in $x$ and $y$.

One practical/real-world implementation of this comes into play when simulating 3-D linear spatial beam propagation by using the angular spectrum alongside the shift theorem. If you know the field distribution at one plane, you can Fourier transform it for the angular spectrum, multiply this by a phase shift corresponding to some arbitrary propagation distance, and inverse Fourier transform to retrieve the field after propagating, assuming that all propagation is linear in between the starting and finishing planes $z=0$ and $z\neq 0$. This is derived as follows:

Starting field in terms of angular spectral components:

1) $U(x, y, z=0)=\frac{1}{\lambda^{2}}\int_{-a}^{a}\int_{-b}^{b}\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda}, z=0\right)e^{ik(\alpha x + \beta y)}d\alpha d\beta$

Here, $a$ and $b$ are the direction cosines corresponding to the plane wave components which propagate with the largest angle to the normal of the object (which is set by the Fourier grid dimensions and propagation distance in a calculation, or the detector dimensions and propagation distance in an experiment).

Propagated field in terms of angular spectral components:

2) $U(x, y, z\neq 0)=\frac{1}{\lambda^{2}}\int_{-a}^{a}\int_{-b}^{b}\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z\neq 0\right)e^{ik(\alpha x + \beta y)}d\alpha d\beta$

$\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z\neq 0\right)$ is unknown, but can be calculated by propagating the integrand of the first equation using the Helmholtz equation:

3) $\frac{\partial^{2}\tilde{U}}{\partial z^{2}}=-k^{2}\tilde{U}(1-\alpha^{2}-\beta^{2})$

where $\Delta z$ is the axial propagation distance. Integrating:

4) $\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z\neq 0\right)=\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z=0\right)e^{-ikz\sqrt{1-\alpha^{2}-\beta^{2}}}$

Substituting 4 into 2 gives the field distribution at $z$ as a function of the field distribution at $z=0$:

5) $U(x, y, z\neq 0)=\frac{1}{\lambda^{2}}\int_{-a}^{a}\int_{-b}^{b}\tilde{U}\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda},z=0\right)e^{-ikz\sqrt{1-\alpha^{2}-\beta^{2}}}e^{ik(\alpha x + \beta y)}d\alpha d\beta$

This can be much more computationally efficient than calculating $U$ pixel-by-pixel with the Fresnel integrals, or similar.

Spatial frequencies can be understood as the Fourier decomposition of a complicated wave into plane waves traveling at different angles to the optical axis. In this case, $\alpha$ is the angle the plane wave component's wave vector makes with the x axis, and $\beta$ is the angle the wave vector makes with the y axis.

By the way, it took me the longest time to understand how spatial frequencies work, and I finally got it from reading Chapter 4, "Fourier Optics", of Saleh and Teich's Fundamentals of Photonics. If you want more information, I can suggest that book.


text: The Parabolic Littlewood--Paley Operator with Hardy Space Kernels In this paper, we give the $L^p$ boundedness for a class of parabolic Littlewood--Paley $g$-function with its kernel function $\Omega$ is in the Hardy space $H^1(S^{n-1})$. Keywords:parabolic Littlewood-Paley operator, Hardy space, rough kernelCategories:42B20, 42B25


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The Intergovernmental Panel on Climate Change gave a methodology to estimate the methane emissions from Municipal Solid Wastes landfills, based on a First Order Decay (FOD) model that assumes biodegradation kinetics depending on the type of wastes. This model can be used to estimate both the National greenhouse gas emissions in the industrialized countries as well as the reductions of these emissions in the developing ones when the Clean Development Mechanism, as defined by the Kyoto Protocol, is implemented. In this paper, the FOD model has been use to evaluate the biogas flow rates emitted by a Brazilian landfill and the results have been compared to the extracted ones: some first results can be useful to evidence the weight of key parameters and do a correct use of the model.

Application of the IPCC model to a Brazilian landfill: first results / Penteado, R.; Cavalli, M.; Magnano, E.; Chiampo, Fulvia. - In: ENERGY POLICY. - ISSN 0301-4215. - 42:(2012), pp. 551-556. [10.1016/j.enpol.2011.12.023]

Application of the IPCC model to a Brazilian landfill: first results

Abstract

The Intergovernmental Panel on Climate Change gave a methodology to estimate the methane emissions from Municipal Solid Wastes landfills, based on a First Order Decay (FOD) model that assumes biodegradation kinetics depending on the type of wastes. This model can be used to estimate both the National greenhouse gas emissions in the industrialized countries as well as the reductions of these emissions in the developing ones when the Clean Development Mechanism, as defined by the Kyoto Protocol, is implemented. In this paper, the FOD model has been use to evaluate the biogas flow rates emitted by a Brazilian landfill and the results have been compared to the extracted ones: some first results can be useful to evidence the weight of key parameters and do a correct use of the model.
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# Zero Product Property

The Zero Product Property states that if $$a\cdot b = 0$$, then at least one of $$a$$ or $$b$$ must be equal to zero. We can use this fact to solve many problems, but in particular, it is helpful when dealing with polynomials that we have factored.

If we know that $$(x+1)(x-3) = 0$$, we can conclude that one of the following MUST be true:

\begin{align} x+1 &= 0 \\ x-3 &=0. \end{align}

Thus, $$x$$ must be one of $$-1, 3$$.

Note by Arron Kau
3 years, 5 months ago

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RNA-seq data analysis to detect dosage effect or proportional increase/decrease in expression across a series
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Entering edit mode
4.6 years ago
nancy ▴ 90

Hi- I have some RNA-seq data from a series of 4 isogenic strains (0,1,2,3), differing in their copy number of a single gene. I would now like to identify those genes whose expression changes across these samples in proportion to the copy number of the gene of interest.

For e.g., in the series 0,1,2,3, what genes show proportional increase in tpm/fpkm as we go from 0 to 3, and what genes show proportional decrease.

I imagine I should be able to model this in DESeq2, but am unable to exactly figure out how. Any pointers/help will be appreciated. Open to using methods other than DESeq2 too, if needed.

RNA-Seq DESeq2 dosage effect • 964 views
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How many biological replicates do you have for each strain? The edgeR manual has an example on how to analyse time series data.

https://www.bioconductor.org/packages/devel/bioc/vignettes/edgeR/inst/doc/edgeRUsersGuide.pdf

Checkout section 3.3. This is their hypothetical experimental design:

Treat           Time
Sample1 Placebo 0h
Sample2 Placebo 0h
Sample3 Placebo 1h
Sample4 Placebo 1h
Sample5 Placebo 2h
Sample6 Placebo 2h
Sample7 Drug 0h
Sample8 Drug 0h
Sample9 Drug 1h
Sample10 Drug 1h
Sample11 Drug 2h
Sample12 Drug 2h

Note that for each time point, they have two independent samples for each condition. This is the bare minimum number of biological replicates neccessary to perform DE analysis (whether Deseq2 or edgeR it doesn't matter).


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# zbMATH — the first resource for mathematics

New bounds on cap sets. (English) Zbl 1262.11010
A set $$A\subseteq \mathbb F^n_3$$ is called a cap set if it contains no solutions of the equation $$x+ y+ 7= 0$$, $$x,y,7\in A$$, $$x\neq y\neq 7$$. These sets correspond to arithmetic progressions of the length three in $$\mathbb Z$$. Improving a result of R. Meshulom the authors prove that any cap set has size less than $$O({3^n\over n^{1+\varepsilon}})$$, where $$\varepsilon> 0$$ is an absolute (but small) constant. It corresponds to the famous conjecture of Erdős and Turán on sets without arithmetic progressions. The proof based on new structural results about sets with a nontrivial lower bounds for the additive energy as well as new statements on the spetrum of cap sets.

##### MSC:
 11B30 Arithmetic combinatorics; higher degree uniformity 11T99 Finite fields and commutative rings (number-theoretic aspects) 05D40 Probabilistic methods in extremal combinatorics, including polynomial methods (combinatorial Nullstellensatz, etc.) 11B75 Other combinatorial number theory
Full Text:
##### References:
 [1] Ernie Croot and Olof Sisask, A probabilistic technique for finding almost-periods of convolutions, Geom. Funct. Anal. 20 (2010), no. 6, 1367 – 1396. · Zbl 1234.11013 [2] T. Gowers, What is difficult about the cap set problem?, http://gowers.wordpress.com/ 2011/01/11/what-is-difficult-about-the-cap-set-problem/. [3] (M. Bateman and) N.H. Katz, http://gowers.wordpress.com/2011/01/11/what-is-difficult-about-the-cap-set-problem/# comment-10533. [4] (M. Bateman and) N.H. Katz, http://gowers.wordpress.com/2011/01/11/what-is-difficult-about-the-cap-set-problem/# comment-10540. [5] Nets Hawk Katz and Paul Koester, On additive doubling and energy, SIAM J. Discrete Math. 24 (2010), no. 4, 1684 – 1693. · Zbl 1226.05247 [6] Roy Meshulam, On subsets of finite abelian groups with no 3-term arithmetic progressions, J. Combin. Theory Ser. A 71 (1995), no. 1, 168 – 172. · Zbl 0832.11006 [7] Polymath on wikipedia, http://en.wikipedia.org/wiki/Polymath_project#Polymath _Project. [8] Polymath 6: Improving the bounds for Roth’s theorem, http://polymathprojects.org/ 2011/02/05/polymath6-improving-the-bounds-for-roths-theorem/. [9] Imre Z. Ruzsa, An analog of Freiman’s theorem in groups, Astérisque 258 (1999), xv, 323 – 326 (English, with English and French summaries). Structure theory of set addition. · Zbl 0946.11007 [10] T. Sanders, A note on Freĭman’s theorem in vector spaces, Combin. Probab. Comput. 17 (2008), no. 2, 297 – 305. · Zbl 1151.15003 [11] T. Sanders, Structure in Sets with Logarithmic Doubling, Arxiv 1002.1552. · Zbl 1310.11013 [12] T. Sanders, On Roth’s Theorem on Progressions, Arxiv 1011.0104. · Zbl 1264.11004 [13] Tomasz Schoen, Near optimal bounds in Freiman’s theorem, Duke Math. J. 158 (2011), no. 1, 1 – 12. · Zbl 1242.11074 [14] I. D. Shkredov, On sets of large trigonometric sums, Izv. Ross. Akad. Nauk Ser. Mat. 72 (2008), no. 1, 161 – 182 (Russian, with Russian summary); English transl., Izv. Math. 72 (2008), no. 1, 149 – 168. · Zbl 1148.11040 [15] Terence Tao and Van Vu, Additive combinatorics, Cambridge Studies in Advanced Mathematics, vol. 105, Cambridge University Press, Cambridge, 2006. · Zbl 1127.11002
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.