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Article Contents
Article Contents

# Modeling chemical reactions in rivers: A three component reaction

• What follows is the analysis of a model for dynamics of chemical reactions in a river. Dominant forces to be considered include diffusion, advection, and rates of creation or destruction of participating species (due to chemical reactions). In light of this, the model we will be using will be based upon a nonlinear system of reaction-advection-diffusion equations. The nonlinearity comes solely from the influences of the chemical reactions.
First, we will establish some general results for reaction diffusion systems. In particular, we will illustrate a class of reaction diffusion systems whose solutions are bounded from below by zero. We will also provide a local existence result for this class of problems. Afterwards, we will focus on the dynamics of an equidiffusive three component reaction system. Specifically, we will provide conditions under which one could be guaranteed the existence of global solutions. We will also discuss the qualities of the $\omega$-limit set for this system.
Mathematics Subject Classification: Primary: 35B05, 35B35, 35K45.

 Citation:


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# Homework Help: A wave packet and a Delta potential

1. Feb 22, 2017

### Poirot

1. The problem statement, all variables and given/known data
I want to plot using Mathematica a wave packet entering a delta potential $V(x) = s\delta(x)$ (s is the strength) but I need to get the physics right first and I'm having trouble with a a few parts. I need to compute the integral $\int_{0}^{\infty} e^{-i\omega t}\phi_{\omega} f(\omega) d\omega$ to get the linear superpositions and get the wave packet but Mathematica can't handle it and I can't compute it by hand.
I'm taking $f(\omega) = e^{-\omega^2/\alpha^2}$ (I assume to make this a gaussian wave packet).
2. Relevant equations
$\phi_{\omega} (x) = e^{ikx} + R_{\omega}e^{-ikx}, x<0 \\ \phi_{\omega}(x) =T_{\omega}e^{ikx}, x>0 \\ k^2 = 2mE/\hbar \\ \omega = E/\hbar$

3. The attempt at a solution
I found the Reflection and Transmission coefficients:
$R_{\omega} = \frac{1}{\frac{\hbar^2ik}{sm}-1} \\ T_{\omega} = \frac{1}{1-\frac{sm}{\hbar^2ik}}$
from the continuity conditions and also from integrating the Schrodinger equation.

From what I understand I think I then need to compute the integral as stated above and plot this against time, but I can't get Mathematica to solve it and I was told it could be computed exactly so I must be messing something up.

Any help would be greatly appreciated thank you.

2. Feb 27, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

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# Solving non-negative least squares in Matlab (by analogy with least squares)

There is a least-squares problem. It can be solved using backslash in Matlab.
If Ax = b, then x = A \ b.
Let's assume that I have the same problem, but all x must be non-negative (>=0). How can I solve this problem in Matlab by analogy with the previous one (without non-negativity constraints)? I think it can be somehow connected with the active set method.
I know that the non-negative least squares problem can be easily solved with Matlab Optimization toolbox or CVX or in many other ways. But still I'm curious about solving it by analogy with a straight-forward least-squares method (with backslash).


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Acta Chimica Sinica ›› 2013, Vol. 71 ›› Issue (02): 227-233.

Article

### 电化学合成PtCo/石墨烯复合催化剂及对甲醇的电催化氧化

1. a 青岛大学 纤维新材料与现代纺织实验室 国家重点实验室培育基地 青岛 266071;
b 青岛大学 化学化工与环境学院 青岛 266071
• 投稿日期:2012-08-14 发布日期:2013-01-09
• 通讯作者: 王宗花, 夏延致 E-mail:wangzonghua@qdu.edu.cn; qduxyzh@163.com
• 基金资助:

项目受国家自然科学基金(Nos. 20975056, 81102411, 21275082);山东省自然科学基金(Nos. ZR2011BZ004, ZR2011BQ005);NSFC-JSPS中日合作与交流项目(No. 21111140014);生命分析化学国家重点实验室开放基金(No. SKLACLS1110)和973计划的资助(No. 2012CB722705)资助.

### Electrochemical Deposition of Graphene Supported PtCo Composite Catalysts for Electrocatalytic Methanol Oxidation

Shi Guoyua,b, Wang Zonghuaa,b, Xia Jianfeia,b, Zhang Feifeia,b, Xia Yanzhia,b, Li Yanhuia

1. a Laboratory of Fiber Materials and Modern Textile, the Growing Base for State Key Laboratory, Qingdao 266071;
b College of Chemical and Environment Engineering, Qingdao University, Qingdao 266071
• Received:2012-08-14 Published:2013-01-09
• Supported by:

Project supported by the National Natural Science Foundation of China (Nos. 20975056, 81102411, 21275082), Natural Science Foundation of Shandong (Nos. ZR2011BZ004, ZR2011BQ005), Japan Society for the Promotion of Science and National Natural Science Foundation of China under the Japan-China Scientific Cooperation Program (No. 21111140014), State Key Laboratory of Analytical Chemistry for Life Science (No. SKLACLS1110) and the National Key Basic Research Development Program of China (973 special preliminary study plan, Grant no.: 2012CB722705).

Direct methanol fuel cells are excellent power sources due to their high energy density, low pollutant emission and easy handling. However, commercial applications are limited by the high cost related to noble metal catalysts. Recent findings have proved that appropriate catalyst support, which improves the utilization of the noble metals in great depth, may be one breakthrough. Graphene nanosheet (GNS), a new two-dimensional carbon material with a single (or a few) atomic thickness, as the combination of its high surface area, high conductivity and unique graphitized basal plane structure, has recently attracted an enormous amount of interest from both theoretical and experimental scientists. It has been proved that catalysts supported on GNSs show improved activity than those supported on carbon black. Furthermore, alloying Pt with other metal is widely approved as a practical method to relieve the CO-poisoning of the catalyst, which can be ascribed to both a bi-functional mechanism and a ligand (electronic) effect. In this experiment, PtCo/graphene (GN) composite catalysts were synthesized on an indium tin oxide (ITO) substrate by the potentiostatic method. Catalyst samples were characterized by scanning electron microscopy (SEM), X-ray energy dispersive spectroscopy (EDX) and electrochemical methods. SEM results showed that the addition of graphene could enhance the dispersion of the catalytic particles and reduce the particle size, especially when the molar ratio of Pt and Co is 1:2.93, the particles had the smallest size and the best dispersion. Electrochemical tests demonstrated that graphene as the catalytic support could improve the CO-tolerance of the catalysts, which was determined by the outstanding electric conductivity and rich oxygen-containing species of graphene, resulting in good performance for electrocatalytic methanol oxidation. Furthermore, owing to the special electronic effect of Co, its addition also influenced the catalytic activity. It was found that when the molar ratio of Pt and Co was 1:2.93, the composite catalyst exhibited the most excellent catalytic performance for electrocatalytic methanol oxidation with the forward anodic peak current density of 662 A·gpt-1and the If/Ib of 2.34 which was nearly 1.8 times that of the traditional PtCo/C catalyst (If/Ib=1.32).


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1. ## Failure devices prob.

There are 30 devices, it is known that 20 of them are failure and 10 are ok. There are chosen 4 of them. What is the probability that the chosen devices are not all of them failure/ok?
res. p = 0.611

OK so I tried finding the prob. that 4 of them are failure. I think it is $p^4=(\frac{2}{3})^4=0.197531.$ So the required prob. must be $p=1-p^4=0.802469$. Is it wrong? hehe obviously the res. is different

2. I like your p value. You should consider the binomial theorem here, it is:

$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

In this case $n = 30$, $p = \frac{2}{3}$ and $k = 4$

now substituting in these values we get

$P(X=4) = \binom{30}{4} \left(\frac{2}{3}\right)^4\left(1- \frac{2}{3}\right)^{30-4}$

Can you finish it from here?

This will be the probabilty that 4 devices fail, I think that is what the question is asking?

3. Originally Posted by pickslides
I like your p value. You should consider the binomial theorem here, it is:
$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
pickslides,
The binomial theorem is not applicable here. These are not independent trials.
The answer for not all failures is $1-\frac{\binom{20}{4}}{\binom{30}{4}}$.

4. Originally Posted by Plato
pickslides,
The binomial theorem is not applicable here. These are not independent trials.
The answer for not all failures is $1-\frac{\binom{20}{4}}{\binom{30}{4}}$.
I tried using binomial but with n=4, and that leaded me to p^4
Ok Plato some explanations would be appreciated, and why it doesn't fit with the result in my book?

5. Originally Posted by pickslides
I like your p value. You should consider the binomial theorem here, it is:

$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

In this case $n = 30$, $p = \frac{2}{3}$ and $k = 4$

now substituting in these values we get

$P(X=4) = \binom{30}{4} \left(\frac{2}{3}\right)^4\left(1- \frac{2}{3}\right)^{30-4}$

Can you finish it from here?

This will be the probabilty that 4 devices fail, I think that is what the question is asking?
No the question is to find the probability that not all of them(4) fail/are ok.

6. Originally Posted by javax
Ok Plato some explanations would be appreciated, and why it doesn't fit with the result in my book?
It is not at all clear what you are asking.
If the question is, "Select four items from thirty of which exactly twenty are defective. What is the probability that not all four are defective nor all four are non-defective"
The probabillity that all are defective is $\frac{\binom{20}{4}}{\binom{30}{4}}$.

The probabillity that all are non-defective is $\frac{\binom{10}{4}}{\binom{30}{4}}$.

Now note that those two are disjoint events.

The probabillity that all are defective OR all are non-defective is $\left(\frac{\binom{20}{4}}{\binom{30}{4}}+\frac{\b inom{10}{4}}{\binom{30}{4}}\right)$.

The probabillity that not all are defective and not all are non-defective is
$1-\left(\frac{\binom{20}{4}}{\binom{30}{4}}+\frac{\b inom{10}{4}}{\binom{30}{4}}\right)$.

7. Originally Posted by Plato
It is not at all clear what you are asking.
If the question is, "Select four items from thirty of which exactly twenty are defective. What is the probability that not all four are defective nor all four are non-defective"
The probabillity that all are defective is $\frac{\binom{20}{4}}{\binom{30}{4}}$.

The probabillity that all are non-defective is $\frac{\binom{10}{4}}{\binom{30}{4}}$.

Now note that those two are disjoint events.

The probabillity that all are defective OR all are non-defective is $\left(\frac{\binom{20}{4}}{\binom{30}{4}}+\frac{\b inom{10}{4}}{\binom{30}{4}}\right)$.

The probabillity that not all are defective and not all are non-defective is
$1-\left(\frac{\binom{20}{4}}{\binom{30}{4}}+\frac{\b inom{10}{4}}{\binom{30}{4}}\right)$.
Mate the question is exactly how you said. Your answer still not fitting the given result but I trust you. Thanks

8. Plato one more question.
As I mentioned I tried to find it using $1-p_1^4*p2^4,$ where $p_1=\frac{2}{3}$ and $p_2=\frac{1}{3}$ which is quite close to your result. Is it ok finding it like this?

9. Originally Posted by javax
Mate the question is exactly how you said. Your answer still not fitting the given result but I trust you.
It may be a matter of translation. We may not be doing the same question.
On the other hand, your textbook may be simply be wrong.

10. "Select four items from thirty of which exactly twenty are defective. What is the probability that not all four are defective nor all four are non-defective"
you fully understood my question. It's ok I believe the given result is wrong (Y)

11. Originally Posted by javax
Plato one more question.
As I mentioned I tried to find it using $1-p_1^4*p2^4,$ where $p_1=\frac{2}{3}$ and $p_2=\frac{1}{3}$ which is quite close to your result. Is it ok finding it like this?
Unless the trials, the events, are independent the binomial formula is not applicable.
Selecting four objects from thirty is in no way independent.

But say we have batch twenty black balls and ten white balls.
We pick a random ball from that collection. Record its color and return it to the batch.
We do that four times. Those outcomes are independent.


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# qiskit.pulse.builder.align_equispaced¶

align_equispaced(duration)[소스]

Equispaced alignment pulse scheduling context.

Pulse instructions within this context are scheduled with the same interval spacing such that the total length of the context block is duration. If the total free duration cannot be evenly divided by the number of instructions within the context, the modulo is split and then prepended and appended to the returned schedule. Delay instructions are automatically inserted in between pulses.

This context is convenient to write a schedule for periodical dynamic decoupling or the Hahn echo sequence.

Examples:

from qiskit import pulse

d0 = pulse.DriveChannel(0)
x90 = pulse.Gaussian(10, 0.1, 3)
x180 = pulse.Gaussian(10, 0.2, 3)

with pulse.build() as hahn_echo:
with pulse.align_equispaced(duration=100):
pulse.play(x90, d0)
pulse.play(x180, d0)
pulse.play(x90, d0)

hahn_echo.draw()

매개변수

duration (Union[int, ParameterExpression]) – Duration of this context. This should be larger than the schedule duration.

Yields

None

참고

The scheduling is performed for sub-schedules within the context rather than channel-wise. If you want to apply the equispaced context for each channel, you should use the context independently for channels.

반환 형식

AlignmentKind


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Items tagged with statisticstatistic Tagged Items Feed

histogram plot error

February 19 2013 by Maple 16

1

2

Hi,

When I try the example of a histogram plot from the help system (under Statistics, Histogram), I get an error.

The example (entered in document mode) and error follow below.  Can someone please shed light on this?

with Statistics; with(plots):N := RandomVariable(Normal(0, 1)):A := Sample(N, 1000):P := DensityPlot(Normal(0, 1), color = "Niagara Red"):Q := Histogram(A, averageshifted = 4, style = polygon, color = "LightSlateGrey"):

Density Plot Not as expected

November 10 2012 by Maple

0

7

To plot the density function of the continuous uniform distribution on [-1,1], my initial attempt was:

plot(Statistics:-PDF(Statistics:-RandomVariable(Uniform(-1,1)),x), x = -1.1 .. 1.1);

See plot below.

But I wanted something more like the wikipedia image (without the labels, naturally):

See plot below.

In words, I expected a horizontal line on the left of x=-1 and on the right of x=1 (at y=0), and I expected no vertical line at the x=-1 and x=1 points ...

AutoCorrelation Maple vs MatLab

May 22 2012 by Maple

1

4

what is the difference between the autocorrelation function of Maple and MatLab function xcorr?

> N := 1024;
> f1 := 1;
> FS := 200;
> n := Vector([$(0 .. N-1)]); > x := evalf(~[sin](2*Pi*f1*n/FS)); > t := Vector([$(1 .. N)]/FS);
> plot(t, x, axes = box, gridlines, color = blue, title = "Sinwave of frequency 1000Hz [FS=8000Hz]", labels = ["Time, [s]",

"Amplitude"], labeldirections = ["horizontal", "vertical"]);

Compute power of tests

December 06 2011 by Maple

0

7

Is it possible to display on a same graph the error of type I and type II error given n, alpha and beta?

PowerTest

applet.zip

Gracias

problems plotting growth models

September 27 2010 by Maple 14

0

1

hello

This is what I have written out in Maple, and I have been trying to plot into a graph, but am having no luck at all.  Do you have any advice for what my next step should be. I'm using a simplified version of Brain and Cousens model.

> restart;

> with(plots);

> with(Statistics);

y := (x) -> (d+fx)/(1+h*x^b) , x > 0

b := 2; d := 1; h := 1;

This is the question: Let b = 2, d = 1 and h = 1

minimization problem

August 27 2010 by Maple

0

0

UncertaityofCalibrat.mw

within the attached document, is the idea that I would like to develop with the optimization package.

 Page 1 of 1


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# 10.5: Conic Sections

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Conic sections have been studied since the time of the ancient Greeks, and were considered to be an important mathematical concept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since then, important applications of conic sections have arisen (for example, in astronomy), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even architecture. In this section we discuss the three basic conic sections, some of their properties, and their equations.

Conic sections get their name because they can be generated by intersecting a plane with a cone. A cone has two identically shaped parts called nappes. One nappe is what most people mean by “cone,” having the shape of a party hat. A right circular cone can be generated by revolving a line passing through the origin around the y-axis as shown in Figure $$\PageIndex{1}$$.

Conic sections are generated by the intersection of a plane with a cone (Figure $$\PageIndex{2}$$). If the plane is parallel to the axis of revolution (the y-axis), then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than 90°), then the conic section is an ellipse.

## Parabolas

A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.

Definitions: The Focus, Directrix and Vertex

A parabola is the set of all points whose distance from a fixed point, called the focus, is equal to the distance from a fixed line, called the directrix. The point halfway between the focus and the directrix is called the vertex of the parabola.

A graph of a typical parabola appears in Figure $$\PageIndex{3}$$. Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point P with coordinates $$(x_1,y_1)$$ and point Q with coordinates $$(x_2,y_2),$$ the distance between them is given by the formula

$d(P,Q)=\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}.$

Then from the definition of a parabola and Figure $$\PageIndex{3}$$, we get

$d(F,P)=d(P,Q)$

$\sqrt{(0−x)^2+(p−y)^2}=\sqrt{(x−x)^2+(−p−y)^2}.$

Squaring both sides and simplifying yields

\begin{align} x^2+(p−y)^2 = 0^2+(−p−y)^2 \\ x^2+p^2−2py+y^2 = p^2+2py+y^2 \\ x^2−2py =2py \\ x^2 =4py. \end{align}

Now suppose we want to relocate the vertex. We use the variables $$(h,k)$$ to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates $$(h,k+p)$$ and the directrix has the equation $$y=k−p$$. Going through the same derivation yields the formula $$(x−h)^2=4p(y−k)$$. Solving this equation for $$y$$ leads to the following theorem.

Equations for Parabolas: standard form

Given a parabola opening upward with vertex located at $$(h,k)$$ and focus located at $$(h,k+p)$$, where $$p$$ is a constant, the equation for the parabola is given by

$y=\dfrac{1}{4p}(x−h)^2+k.$

This is the standard form of a parabola.

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

In addition, the equation of a parabola can be written in the general form, though in this form the values of $$h$$, $$k$$, and $$p$$ are not immediately recognizable. The general form of a parabola is written as

$ax^2+bx+cy+d=0 \label{para1}$

or

$ay^2+bx+cy+d=0.\label{para2}$

Equation \ref{para1} represents a parabola that opens either up or down. Equation \ref{para2} represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.

Example $$\PageIndex{1}$$: Converting the Equation of a Parabola from General into Standard Form

Put the equation

$x^2−4x−8y+12=0$

into standard form and graph the resulting parabola.

Solution

Since y is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for y, which will put the equation into standard form. To do that, first add $$8y$$ to both sides of the equation:

$8y=x^2−4x+12.$

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:

$8y=(x^2−4x)+12.$

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of x and square it. This gives $$(\dfrac{−4}{2})^2=4.$$ Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

$8y=(x^2−4x+4)+12−4.$

Now combine like terms and factor the quantity inside the parentheses:

$8y=(x−2)^2+8.$

Finally, divide by 8:

$y=\dfrac{1}{8}(x−2)^2+1.$

This equation is now in standard form. Comparing this to Equation gives $$h=2, k=1$$, and $$p=2$$. The parabola opens up, with vertex at $$(2,1)$$, focus at $$(2,3)$$, and directrix $$y=−1$$. The graph of this parabola appears as follows.

Exercise $$\PageIndex{1}$$

Put the equation $$2y^2−x+12y+16=0$$ into standard form and graph the resulting parabola.

Hint

Solve for $$x$$. Check which direction the parabola opens.

$x=2(y+3)^2−2$

The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

Consider a parabolic dish designed to collect signals from a satellite in space. The dish is aimed directly at the satellite, and a receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of the parabola to the receiver, which collects and decodes the digital signals. This allows a small receiver to gather signals from a wide angle of sky. Flashlights and headlights in a car work on the same principle, but in reverse: the source of the light (that is, the light bulb) is located at the focus and the reflecting surface on the parabolic mirror focuses the beam straight ahead. This allows a small light bulb to illuminate a wide angle of space in front of the flashlight or car.

## Ellipses

An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.

Definition: Ellipse

An ellipse is the set of all points for which the sum of their distances from two fixed points (the foci) is constant.

A graph of a typical ellipse is shown in Figure $$\PageIndex{6}$$. In this figure the foci are labeled as $$F$$ and $$F′$$. Both are the same fixed distance from the origin, and this distance is represented by the variable c. Therefore the coordinates of $$F$$ are $$(c,0)$$ and the coordinates of $$F′$$ are $$(−c,0).$$ The points $$P$$ and $$P′$$ are located at the ends of the major axis of the ellipse, and have coordinates $$(a,0)$$ and $$(−a,0)$$, respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2a. Furthermore, $$P$$ and $$P′$$ are called the vertices of the ellipse. The points $$Q$$ and $$Q′$$ are located at the ends of the minor axis of the ellipse, and have coordinates $$(0,b)$$ and $$(0,−b),$$ respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.

According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point P. Since the coordinates of point P are $$(a,0),$$ the sum of the distances is

$d(P,F)+d(P,F′)=(a−c)+(a+c)=2a.$

Therefore the sum of the distances from an arbitrary point A with coordinates $$(x,y)$$ is also equal to $$2a$$. Using the distance formula, we get

$d(A,F)+d(A,F′)=2a.$

$\sqrt{(x−c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$

Subtract the second radical from both sides and square both sides:

$\sqrt{(x−c)^2+y^2}=2a−\sqrt{(x+c)^2+y^2}$

$(x−c)^2+y^2=4a^2−4a\sqrt{(x+c)^2+y^2}+(x+c)^2+y^2$

$x^2−2cx+c^2+y^2=4a^2−4a\sqrt{(x+c)^2+y^2}+x^2+2cx+c^2+y^2$

$−2cx=4a^2−4a\sqrt{(x+c)^2+y^2}+2cx.$

Now isolate the radical on the right-hand side and square again:

$−2cx=4a^2−4a\sqrt{(x+c)^2+y^2}+2cx$

$4a\sqrt{(x+c)^2+y^2}=4a^2+4cx$

$\sqrt{(x+c)^2+y^2}=a+\dfrac{cx}{a}$

$(x+c)^2+y^2=a^2+2cx+\dfrac{c^2x^2}{a^2}$

$x^2+2cx+c^2+y^2=a^2+2cx+\dfrac{c^2x^2}{a^2}$

$x^2+c^2+y^2=a^2+\dfrac{c^2x^2}{a^2}.$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

$x^2−\dfrac{c^2x^2}{a^2}+y^2=a^2−c^2$

$\dfrac{(a^2−c^2)x^2}{a^2}+y^2=a^2−c^2.$

Divide both sides by $$a^2−c^2$$. This gives the equation

$\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2−c^2}=1.$

If we refer back to Figure $$\PageIndex{6}$$, then the length of each of the two green line segments is equal to $$a$$. This is true because the sum of the distances from the point Q to the foci $$F$$ and $$F′$$ is equal to $$2a$$, and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length a and leg lengths b and c. From the Pythagorean theorem, $$a^2+b^2=c^2$$ and $$b^2=a^2−c^2$$. Therefore the equation of the ellipse becomes

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$

Finally, if the center of the ellipse is moved from the origin to a point $$(h,k)$$, we have the following standard form of an ellipse.

Equation of an Ellipse in Standard Form

Consider the ellipse with center $$(h,k)$$, a horizontal major axis with length $$2a$$, and a vertical minor axis with length $$2b$$. Then the equation of this ellipse in standard form is

$\dfrac{(x−h)^2}{a^2}+\dfrac{(y−k)^2}{b^2}=1$

and the foci are located at $$(h±c,k)$$, where $$c^2=a^2−b^2$$. The equations of the directrices are $$x=h±\dfrac{a^2}{c}$$.

If the major axis is vertical, then the equation of the ellipse becomes

$\dfrac{(x−h)^2}{b^2}+\dfrac{(y−k)^2}{a^2}=1$

and the foci are located at $$(h,k±c)$$, where $$c^2=a^2−b^2$$. The equations of the directrices in this case are $$y=k±\dfrac{a^2}{c}$$.

If the major axis is horizontal, then the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is called vertical. The equation of an ellipse is in general form if it is in the form

$(Ax^2+By^2+Cx+Dy+E=0,$

where A and B are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square.

Example $$\PageIndex{2}$$: Finding the Standard Form of an Ellipse

Put the equation

$9x^2+4y^2−36x+24y+36=0$

into standard form and graph the resulting ellipse.

Solution

First subtract 36 from both sides of the equation:

$9x^2+4y^2−36x+24y=−36.$

Next group the $$x$$ terms together and the $$y$$ terms together, and factor out the common factor:

$(9x^2−36x)+(4y^2+24y)=−36$

$9(x^2−4x)+4(y^2+6y)=−36.$

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives $$(\dfrac{−4}{2})^2=4.$$ In the second set of parentheses, take half the coefficient of y and square it. This gives $$(\dfrac{6}{2})^2=9.$$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

$9(x^2−4x+4)+4(y^2+6y+9)=−36+36+36$

$9(x^2−4x+4)+4(y^2+6y+9)=36.$

Now factor both sets of parentheses and divide by 36:

$9(x−2)^2+4(y+3)^2=36$

$\dfrac{9(x−2)^2}{36}+\dfrac{4(y+3)^2}{36}=1$

$\dfrac{(x−2)^2}{4}+\dfrac{(y+3)^2}{9}=1.$

The equation is now in standard form. Comparing this to Equation gives $$h=2, k=−3, a=3,$$ and $$b=2$$. This is a vertical ellipse with center at $$(2,−3)$$, major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

Exercise $$\PageIndex{2}$$

Put the equation

$9x^2+16y^2+18x−64y−71=0$

into standard form and graph the resulting ellipse.

Hint

Move the constant over and complete the square.

$\dfrac{(x+1)^2}{16}+\dfrac{(y−2)^2}{9}=1$

According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in Figure $$\PageIndex{8A}$$. Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth.

Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focus after mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure $$\PageIndex{8B}$$. This hall served as the meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they can hear someone standing close by. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Central Station in New York City.

## Hyperbolas

A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.

Definition: hyperbola

A hyperbola is the set of all points where the difference between their distances from two fixed points (the foci) is constant.

A graph of a typical hyperbola appears as follows.

The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between two numbers is always positive. Let P be a point on the hyperbola with coordinates $$(x,y)$$. Then the definition of the hyperbola gives $$|d(P,F_1)−d(P,F_2)|=constant$$. To simplify the derivation, assume that P is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates $$(a,0),$$ so

$d(P,F_1)−d(P,F_2)=(c+a)−(c−a)=2a.$

This equation is therefore true for any point on the hyperbola. Returning to the coordinates $$(x,y)$$ for P:

$d(P,F_1)−d(P,F_2)=2a$

$\sqrt{(x+c)^2+y^2}−\sqrt{(x−c)^2+y^2}=2a.$

$\sqrt{(x−c)^2+y^2}=2a+\sqrt{(x+c)^2+y^2}$

$(x−c)^2+y2=4a^2+4a\sqrt{(x+c)^2+y^2}+(x+c)^2+y^2$

$x^2−2cx+c^2+y^2=4a^2+4a\sqrt{(x+c)^2+y^2}+x^2+2cx+c^2+y^2$

$−2cx=4a^2+4a\sqrt{(x+c)^2+y^2}+2cx.$

Now isolate the radical on the right-hand side and square again:

$$−2cx=4a^2+4a\sqrt{(x+c)^2+y^2}+2cx$$

$$4a\sqrt{(x+c)^2+y^2}=−4a^2−4cx$$

$$(x+c)^2+y^2=−a−\dfrac{cx}{a}$$

$$(x+c)^2+y^2=a^2+2cx+\dfrac{c^2x^2}{a^2}$$

$$x^2+2cx+c^2+y^2=a^2+2cx+\dfrac{c^2x^2}{a^2}$$

$$x^2+c^2+y^2=a^2+\dfrac{c^2x^2}{a^2}$$.

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

$x^2−\dfrac{c^2x^2}{a^2}+y^2=a^2−c^2$

$\dfrac{(a^2−c^2)x^2}{a^2}+y^2=a^2−c^2.$

Finally, divide both sides by $$a^2−c^2$$. This gives the equation

$\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2−c^2}=1.$

We now define b so that $$b^2=c^2−a^2$$. This is possible because $$c>a$$. Therefore the equation of the ellipse becomes

$\dfrac{x^2}{a^2}−\dfrac{y^2}{b^2}=1.$

Finally, if the center of the hyperbola is moved from the origin to the point $$(h,k),$$ we have the following standard form of a hyperbola.

Equation of a Hyperbola in Standard Form

Consider the hyperbola with center $$(h,k)$$, a horizontal major axis, and a vertical minor axis. Then the equation of this ellipse is

$\dfrac{(x−h)^2}{a^2}−\dfrac{(y−k)^2}{b^2}=1$

and the foci are located at $$(h±c,k),$$ where $$c^2=a^2+b^2$$. The equations of the asymptotes are given by $$y=k±\dfrac{b}{a}(x−h).$$ The equations of the directrices are

$x=k±\dfrac{a^2}{\sqrt{a^2+b^2}}=h±\dfrac{a^2}{c}$

If the major axis is vertical, then the equation of the hyperbola becomes

$\dfrac{(y−k)^2}{a^2}−\dfrac{(x−h)^2}{b^2}=1$

and the foci are located at $$(h,k±c),$$ where $$c^2=a^2+b^2$$. The equations of the asymptotes are given by $$y=k±\dfrac{a}{b}(x−h)$$. The equations of the directrices are

$y=k±\dfrac{a^2}{\sqrt{a^2+b^2}}=k±\dfrac{a^2}{c}.$

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form

$Ax^2+By^2+Cx+Dy+E=0,$

where A and B have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

Example $$\PageIndex{3}$$: Finding the Standard Form of a Hyperbola

Put the equation $$9x^2−16y^2+36x+32y−124=0$$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Solution

First add 124 to both sides of the equation:

$$9x^2−16y^2+36x+32y=124.$$

Next group the x terms together and the y terms together, then factor out the common factors:

$$(9x^2+36x)−(16y^2−32y)=124$$

$$9(x^2+4x)−16(y^2−2y)=124$$.

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives $$(\dfrac{4}{2})^2=4$$. In the second set of parentheses, take half the coefficient of y and square it. This gives $$(\dfrac{−2}{2})^2=1.$$ Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

$$9(x^2+4x+4)−16(y^2−2y+1)=124+36−16$$

$$9(x^2+4x+4)−16(y^2−2y+1)=144.$$

Next factor both sets of parentheses and divide by 144:

$$9(x+2)^2−16(y−1)^2=144$$

$$\dfrac{9(x+2)^2}{144}−\dfrac{16(y−1)^2}{144}=1$$

$$\dfrac{(x+2)^2}{16}−\dfrac{(y−1)^2}{9}=1.$$

The equation is now in standard form. Comparing this to Equation gives $$h=−2, k=1, a=4,$$ and $$b=3$$. This is a horizontal hyperbola with center at $$(−2,1)$$ and asymptotes given by the equations $$y=1±\dfrac{3}{4}(x+2)$$. The graph of this hyperbola appears in Figure $$\PageIndex{10}$$.

Exercise $$\PageIndex{3}$$

Put the equation $$4y^2−9x^2+16y+18x−29=0$$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Hint

Move the constant over and complete the square. Check which direction the hyperbola opens

$$\dfrac{(y+2)^2}{9}−\dfrac{(x−1)^2}{4}=1.$$ This is a vertical hyperbola. Asymptotes $$y=−2±\dfrac{3}{2}(x−1).$$

Hyperbolas also have interesting reflective properties. A ray directed toward one focus of a hyperbola is reflected by a hyperbolic mirror toward the other focus. This concept is illustrated in Figure $$\PageIndex{11}$$.

This property of the hyperbola has important applications. It is used in radio direction finding (since the difference in signals from two towers is constant along hyperbolas), and in the construction of mirrors inside telescopes (to reflect light coming from the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solar system, if the speed is great enough to escape the Sun’s gravitational pull, then the path that the comet takes as it passes through the solar system is hyperbolic.

## Eccentricity and Directrix

An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We will see that the value of the eccentricity of a conic section can uniquely define that conic.

Definition: Eccentricity and Directrices

The eccentricity $$e$$ of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix. This value is constant for any conic section, and can define the conic section as well:

1. If $$e=1$$, the conic is a parabola.
2. If $$e<1$$, it is an ellipse.
3. If $$e>1,$$ it is a hyperbola.

The eccentricity of a circle is zero. The directrix of a conic section is the line that, together with the point known as the focus, serves to define a conic section. Hyperbolas and noncircular ellipses have two foci and two associated directrices. Parabolas have one focus and one directrix.

The three conic sections with their directrices appear in Figure $$\PageIndex{12}$$.

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are $$x=±\dfrac{a^2}{c}$$. The right vertex of the ellipse is located at $$(a,0)$$ and the right focus is $$(c,0)$$. Therefore the distance from the vertex to the focus is $$a−c$$ and the distance from the vertex to the right directrix is $$\dfrac{a^2}{c}−c.$$ This gives the eccentricity as

$e=\dfrac{a−c}{\dfrac{a^2}{c}−a}=\dfrac{c(a−c)}{a^2−ac}=\dfrac{c(a−c)}{a(a−c)}=\dfrac{c}{a}.$

Since $$c<a$$, this step proves that the eccentricity of an ellipse is less than 1. The directrices of a horizontal hyperbola are also located at $$x=±\dfrac{a^2}{c}$$, and a similar calculation shows that the eccentricity of a hyperbola is also e=ca. However in this case we have $$c>a$$, so the eccentricity of a hyperbola is greater than 1.

Example $$\PageIndex{4}$$: Determining Eccentricity of a Conic Section

Determine the eccentricity of the ellipse described by the equation

$$\dfrac{(x−3)^2}{16}+\dfrac{(y+2)^2}{25}=1.$$

Solution

From the equation we see that $$a=5$$ and $$b=4$$. The value of c can be calculated using the equation $$a^2=b^2+c^2$$ for an ellipse. Substituting the values of a and b and solving for c gives $$c=3$$. Therefore the eccentricity of the ellipse is $$e=\dfrac{c}{a}=\dfrac{3}{5}=0.6.$$

Exercise $$\PageIndex{4}$$

Determine the eccentricity of the hyperbola described by the equation

$$\dfrac{(y−3)^2}{49}−\dfrac{(x+2)^2}{25}=1.$$

Hint

First find the values of a and b, then determine c using the equation $$c^2=a^2+b^2$$.

$$e=\dfrac{c}{a}=\dfrac{\sqrt{74}}{7}≈1.229$$

## Polar Equations of Conic Sections

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section p is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where a is the length of the semi-major axis (i.e., half the length of the major axis), c is the distance from the origin to the focus, and e is the eccentricity. In the case of a parabola, a represents the distance from the vertex to the focus.

Table $$\PageIndex{1}$$: Eccentricities and Focal Parameters of the Conic Sections
Conic $$e$$ $$p$$
Ellipse $$0<e<1$$ $$\dfrac{a^2−c^2}{c}=\dfrac{a(1−e^2)}{c}$$
Parabola $$e=1$$ $$2a$$
Hyperbola $$e>1$$ $$\dfrac{c^2−a^2}{c}=\dfrac{a(e^2−1)}{e}$$

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.

Polar Equation of Conic Sections

The polar equation of a conic section with focal parameter p is given by

$$r=\dfrac{ep}{1±e\cos θ}$$ or $$r=\dfrac{ep}{1±e\sin θ}.$$

In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., $$e=0$$).

Example $$\PageIndex{5}$$: Graphing a Conic Section in Polar Coordinates

Identify and create a graph of the conic section described by the equation

$$r=\dfrac{3}{1+2\cos θ}$$.

Solution

The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter p can be calculated by using the equation $$ep=3.$$ Since $$e=2$$, this gives $$p=\dfrac{3}{2}$$. The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for $$θ$$ and create a table of values. Then we can graph the hyperbola (Figure $$\PageIndex{13}$$).

$$θ$$ $$r$$ $$θ$$ $$r$$
0 1 $$π$$ −3
$$\dfrac{π}{4}$$ $$\dfrac{3}{1+\sqrt{2}}≈1.2426$$ $$\dfrac{5π}{4}$$ $$\dfrac{3}{1−\sqrt{2}}≈−7.2426$$
$$\dfrac{π}{2}$$ 3 $$\dfrac{3π}{2}$$ 3
$$\dfrac{3π}{4}$$ $$\dfrac{3}{1−\sqrt{2}}≈−7.2426$$ $$\dfrac{7π}{4}$$ $$\dfrac{3}{1+\sqrt{2}}≈1.2426$$

Exercise $$\PageIndex{5}$$

Identify and create a graph of the conic section described by the equation

$$r=\dfrac{4}{1−0.8 \sin θ}$$.

Hint

First find the values of e and p, and then create a table of values.

Here $$e=0.8$$ and $$p=5$$. This conic section is an ellipse.

## General Equations of Degree Two

A general equation of degree two can be written in the form

$Ax^2+Bxy+Cy^2+Dx+Ey+F=0.$

The graph of an equation of this form is a conic section. If $$B≠0$$ then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section $$4AC−B^2.$$

Identifying the Conic Section

One of the following cases must be true:

1. $$4AC−B^2>0$$. If so, the graph is an ellipse.
2. $$4AC−B^2=0$$. If so, the graph is a parabola.
3. $$4AC−B^2<0$$. If so, the graph is a hyperbola.

The simplest example of a second-degree equation involving a cross term is $$xy=1$$. This equation can be solved for $$y$$ to obtain $$y=\dfrac{1}{x}$$. The graph of this function is called a rectangular hyperbola as shown.

The asymptotes of this hyperbola are the $$x$$ and $$y$$ coordinate axes. To determine the angle θ of rotation of the conic section, we use the formula $$\cot 2θ=\frac{A−C}{B}$$. In this case $$A=C=0$$ and $$B=1$$, so $$\cot 2θ=(0−0)/1=0$$ and $$θ=45°$$. The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled $$A′,B′,C′,D′,E′,$$ and $$F′,$$ and are given by the formulas

\begin{align} A′ =A\cos^ 2θ+B\cos θ\sin θ+C\sin^2 θ \\ B′ =0 \\ C′ =A\sin^2 θ−B\sin θ\cos θ+C\cos^2θ \\ D′ =D\cos θ+E\sin θ \\ E′ =−D\sin θ+E\cosθ \\ F′ =F. \end{align}

Procedure: graphing a rotated conic

The procedure for graphing a rotated conic is the following:

1. Identify the conic section using the discriminant $$4AC−B^2$$.
2. Determine $$θ$$ using the formula $\cot2θ=\dfrac{A−C}{B} \label{rot}.$
3. Calculate $$A′,B′,C′,D′,E′$$,and $$F′$$.
4. Rewrite the original equation using $$A′,B′,C′,D′,E′$$,and $$F′$$.
5. Draw a graph using the rotated equation.

Example $$\PageIndex{6}$$: Identifying a Rotated Conic

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$13x^2−6\sqrt{3}xy+7y^2−256=0.$

Solution

In this equation, $$A=13,B=−6\sqrt{3},C=7,D=0,E=0,$$ and $$F=−256$$. The discriminant of this equation is

$4AC−B^2=4(13)(7)−(−6\sqrt{3})^2=364−108=256.$

Therefore this conic is an ellipse.

To calculate the angle of rotation of the axes, use Equation \ref{rot}

$\cot 2θ=\dfrac{A−C}{B}.$

This gives

$$\cot 2θ=\dfrac{A−C}{B}=\dfrac{13−7}{−6\sqrt{3}}=−\dfrac{\sqrt{3}}{3}$$.

Therefore $$2θ=120^o$$ and $$θ=60^o$$, which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

$$A′=A\cos^2θ+B\cos θ\sinθ+C\sin^2θ$$

$$=13\cos^260+(−6\sqrt{3})\cos 60 \sin 60+7\sin^260$$

$$=13(\dfrac{1}{2})^2−6\sqrt{3}(\dfrac{1}{2})(\dfrac{\sqrt{3}}{2})+7(\dfrac{\sqrt{3}}{2})^2$$

$$=4,$$

$$B′=0$$

$$C′=A\sin^2θ−B\sin θ\cos θ+C\cos^2θ$$

$$=13\sin^260+(−6\sqrt{3})\sin 60 \cos 60=7\cos^260$$

$$=(\dfrac{\sqrt{3}}{2})^2+6\sqrt{3}(\dfrac{\sqrt{3}}{2})(\dfrac{1}{2})+7(\dfrac{1}{2})^2$$

$$=16,$$

$$D′=Dcosθ+Esinθ$$

$$=(0)cos60+(0)sin60$$

$$=0,$$

$$E′=−Dsinθ+Ecosθ$$

$$=−(0)sin60+(0)cos60$$

$$=0$$

$$F′= F$$

$$=−256.$$

The equation of the conic in the rotated coordinate system becomes

$$4(x′)^2+16(y′)^2=256$$

$$\dfrac{(x′)^2}{64}+\dfrac{(y′)^2}{16}=1$$.

A graph of this conic section appears as follows.

Exercise $$\PageIndex{6}$$

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$3x^2+5xy−2y^2−125=0.$

Hint

Follow steps 1 and 2 of the five-step method outlined above

The conic is a hyperbola and the angle of rotation of the axes is $$θ=22.5°.$$

## Key Concepts

• The equation of a vertical parabola in standard form with given focus and directrix is $$y=\dfrac{1}{4p}(x−h)^2+k$$ where $$p$$ is the distance from the vertex to the focus and $$(h,k)$$ are the coordinates of the vertex.
• The equation of a horizontal ellipse in standard form is $$\dfrac{(x−h)^2}{a^2}+\dfrac{(y−k)^2}{b^2}=1$$ where the center has coordinates $$(h,k)$$, the major axis has length 2a, the minor axis has length 2b, and the coordinates of the foci are $$(h±c,k)$$, where $$c^2=a^2−b^2$$.
• The equation of a horizontal hyperbola in standard form is $$\dfrac{(x−h)^2}{a^2}−\dfrac{(y−k)^2}{b^2}=1$$ where the center has coordinates $$(h,k)$$, the vertices are located at $$(h±a,k)$$, and the coordinates of the foci are $$(h±c,k),$$ where $$c^2=a^2+b^2$$.
• The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of a hyperbola is greater than 1. The eccentricity of a circle is 0.
• The polar equation of a conic section with eccentricity e is $$r=\dfrac{ep}{1±ecosθ}$$ or $$r=\dfrac{ep}{1±esinθ}$$, where p represents the focal parameter.
• To identify a conic generated by the equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$,first calculate the discriminant $$D=4AC−B^2$$. If $$D>0$$ then the conic is an ellipse, if $$D=0$$ then the conic is a parabola, and if $$D<0$$ then the conic is a hyperbola.

## Glossary

conic section
a conic section is any curve formed by the intersection of a plane with a cone of two nappes
directrix
a directrix (plural: directrices) is a line used to construct and define a conic section; a parabola has one directrix; ellipses and hyperbolas have two
discriminant
the value $$4AC−B^2$$, which is used to identify a conic when the equation contains a term involving $$xy$$, is called a discriminant
focus
a focus (plural: foci) is a point used to construct and define a conic section; a parabola has one focus; an ellipse and a hyperbola have two
eccentricity
the eccentricity is defined as the distance from any point on the conic section to its focus divided by the perpendicular distance from that point to the nearest directrix
focal parameter
the focal parameter is the distance from a focus of a conic section to the nearest directrix
general form
an equation of a conic section written as a general second-degree equation
major axis
the major axis of a conic section passes through the vertex in the case of a parabola or through the two vertices in the case of an ellipse or hyperbola; it is also an axis of symmetry of the conic; also called the transverse axis
minor axis
the minor axis is perpendicular to the major axis and intersects the major axis at the center of the conic, or at the vertex in the case of the parabola; also called the conjugate axis
nappe
a nappe is one half of a double cone
standard form
an equation of a conic section showing its properties, such as location of the vertex or lengths of major and minor axes
vertex
a vertex is an extreme point on a conic section; a parabola has one vertex at its turning point. An ellipse has two vertices, one at each end of the major axis; a hyperbola has two vertices, one at the turning point of each branch


text:
## Group decision making with a fuzzy linguistic majority.(English)Zbl 0604.90012

Considered is a collection of m individual fuzzy preference relations $$R_ k$$, $$k=1,...,m$$, where $$R_ k:$$ $$S\times S\to [0,1]$$ and S is a set of n alternatives $$S=\{s_ 1,...,s_ n\}$$. The higher $$R_ k(s_ i,s_ j)$$ the higher the preference of $$s_ i$$ over $$s_ j$$ for the individual k. A group decision making solution should reflect what a majority of individuals prefers. A fuzzy majority concept is introduced as a counterpart to the linguistic qualifier ”most”. Various approaches to the problem are then suggested leading to different solutions, e.g. the formalization of a property over S expressing ”most individuals are not against (an alternative)”.
Reviewer: J.Sustal

### MSC:

 91B08 Individual preferences 91B14 Social choice 03E72 Theory of fuzzy sets, etc.

### Keywords:

group decision making; fuzzy majority concept
Full Text:

### References:

 [1] Arrow, K.J., Social choice and individual values, (1963), Wiley New York · Zbl 0984.91513 [2] Bezdek, J.C.; Spillman, B.; Spillman, R., A fuzzy relation space for group decision theory, Fuzzy sets and systems, 1, 255-268, (1978) · Zbl 0398.90009 [3] Bezdek, J.C.; Spillman, B.; Spillman, R., Fuzzy relation spaces for group decision theory: an application, Fuzzy sets and systems, 2, 5-14, (1979) · Zbl 0407.90003 [4] Blin, J.M., Fuzzy relations in group decision theory, J. cybernet., 4, 17-22, (1974) · Zbl 0363.90011 [5] Blin, J.M.; Whinston, A.P., Fuzzy sets and social choice, J. cybernet., 3, 28-33, (1973) · Zbl 0303.90009 [6] Dimitrov, V.N., Group choice under fuzzy information, Fuzzy sets and systems, 9, 25-40, (1983) · Zbl 0506.90006 [7] Fiorina, M.P.; Plott, C.R., Committee decisions under majority rule: an experimental study, Amer. political sci. rev., 72, 525-595, (1978) [8] Kacprzyk, J., Multistage decision-making under fuzziness, (1983), Verlag TÜV Rheinland Köln, ISR Series 79 · Zbl 0507.90023 [9] Kacprzyk, J., A generalization of fuzzy multistage decision-making and control via linguistic quantifiers, Internat. J. control, 38, 1249-1270, (1983) · Zbl 0544.93004 [10] Kacprzyk, J.; Yager, R.R., Linguistic quantifiers and belief qualification in fuzzy multicriteria and multistage decision making, Control and cybernetics, 13, 155-173, (1984) · Zbl 0551.90091 [11] Kacprzyk, J.; Yager, R.R., ‘softer’ optimization and control models via fuzzy linguistic quantifiers, Inform. sci., 34, 157-178, (1984) · Zbl 0562.90098 [12] Kelly, J.S., Arrow impossibility theorems, (1978), Academic Press New York · Zbl 0462.90004 [13] Nurmi, H., Approaches to collective decision making with fuzzy preference relations, Fuzzy sets and systems, 6, 249-259, (1981) · Zbl 0465.90006 [14] Tanino, T., Fuzzy preference orderings in group decision making, Fuzzy sets and systems, 12, 117-131, (1984) · Zbl 0567.90002 [15] Yager, R.R., Quantifiers in the formulation of multiple objective decision functions, Inform. sci., 31, 107-139, (1983) · Zbl 0551.90084 [16] Zadeh, L.A., A theory of approximate reasoning, (), 149-193 [17] Zadeh, L.A., A computational approach to fuzzy quantifiers in natural languages, Comput. math. appl., 9, 149-184, (1983) · Zbl 0517.94028
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.


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More fun in 2016, Part 1

Algebra Level 4

$S=\sum_{k=0}^{672}{2016 \choose 3k}$ The sum $$S$$ is of the form $$S=\dfrac{2^{2016}+a}{3}$$. Find $$a$$.

Bonus Question: What is $S=\sum_{k=0}^{\lfloor{n/3}\rfloor}{n \choose 3k}$ for any positive integer $$n$$?

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XMLTools - Maple Programming Help

Home : Support : Online Help : Connectivity : Web Features : XMLTools : XMLTools/AttrCont

XMLTools

 AttrCont
 split an XML element into its attributes and content

 Calling Sequence AttrCont(xmlTree)

Parameters

 xmlTree - Maple XML tree; XML element

Description

 • The AttrCont(xmlTree) command separates the attributes and content of an XML element and returns them in an expression sequence of two lists. The first list is a list of equations that represent the attributes, each of whose left and right sides are strings. The second list is a list of expressions that represent the content where each of the expressions is of one of the types function or string.
 • Except for efficiency, the call AttrCont(xmlTree) is equivalent to using Attributes(xmlTree), ContentModel(xmlTree). When both the content model and the attributes are required, it is more efficient to call AttrCont than to use Attributes and ContentModel individually.

Examples

 > $\mathrm{with}\left(\mathrm{XMLTools}\right):$
 > $\mathrm{xmlTree}≔\mathrm{XMLElement}\left("a",\left["colour"="red"\right],\mathrm{XMLElement}\left("b",\left["size"="2"\right],"some text"\right),"more text"\right)$
 ${\mathrm{xmlTree}}{≔}{\mathrm{_XML_Element}}{}\left({\mathrm{_XML_ElementType}}{}\left({"a"}\right){,}\left[{\mathrm{_XML_Attribute}}{}\left({\mathrm{_XML_AttrName}}{}\left({"colour"}\right){,}{\mathrm{_XML_AttrValue}}{}\left({"red"}\right)\right)\right]{,}\left[{\mathrm{_XML_Element}}{}\left({\mathrm{_XML_ElementType}}{}\left({"b"}\right){,}\left[{\mathrm{_XML_Attribute}}{}\left({\mathrm{_XML_AttrName}}{}\left({"size"}\right){,}{\mathrm{_XML_AttrValue}}{}\left({"2"}\right)\right)\right]{,}\left[{\mathrm{_XML_Text}}{}\left({"some text"}\right)\right]\right){,}{\mathrm{_XML_Text}}{}\left({"more text"}\right)\right]\right)$ (1)
 > $\mathrm{AttrCont}\left(\mathrm{xmlTree}\right)$
 $\left[{\mathrm{_XML_Attribute}}{}\left({\mathrm{_XML_AttrName}}{}\left({"colour"}\right){,}{\mathrm{_XML_AttrValue}}{}\left({"red"}\right)\right)\right]{,}\left[{\mathrm{_XML_Element}}{}\left({\mathrm{_XML_ElementType}}{}\left({"b"}\right){,}\left[{\mathrm{_XML_Attribute}}{}\left({\mathrm{_XML_AttrName}}{}\left({"size"}\right){,}{\mathrm{_XML_AttrValue}}{}\left({"2"}\right)\right)\right]{,}\left[{\mathrm{_XML_Text}}{}\left({"some text"}\right)\right]\right){,}{\mathrm{_XML_Text}}{}\left({"more text"}\right)\right]$ (2)


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 Izv. Vyssh. Uchebn. Zaved. Mat.: Year: Volume: Issue: Page: Find

 Izv. Vyssh. Uchebn. Zaved. Mat., 2020, Number 4, Pages 43–53 (Mi ivm9560)

Darboux system and separation of variables in the Goursat problem for a third order equation in $\mathbb {R}^3$

R. Ch. Kulaevabc, A. B. Shabatde

a North Ossetia State University 46 Vatutina str., Vladikavkaz, 362025 Russia
b Southern Mathematical Institute VSC RAS, 22 Markusa str., Vladikavkaz, 362027 Russia
c Institute of Mathematics, Ufa Federal Research Centre RAS, 112 Chernyshevsky str., Ufa, 450008 Russia
d Landau Institute for Theoretical Physics RAS, 1-A Akademika Semenova Ave., Chernogolovka, 142432 Russia
e Adygea State University, 208 Pervomayskaya str., Maykop, 385000 Russia

Abstract: We construct a reduction of the three-dimensional Darboux system for the Christoffel symbols, which describes conjugate curvilinear coordinate systems. The reduction is determined by one additional algebraic condition on the Christoffel symbols. It is shown that the corresponding class of solutions of the Darboux system is parametrized by six functions of one variable (two for each of three independent variables). Explicit formulas for Darboux system solutions are given. For the case when Christoffel symbols are constants, the linear system associated with the Darboux system is studied. In this formulation, this system is reduced to the three-dimensional Goursat problem for a third-order equation with data on the coordinate planes. It is shown that the solution to the Goursat problem allows the separation of variables and is determined by its values on the coordinate lines.

Keywords: three-dimensional Darboux system, integrable systems, three-dimensional Goursat problem, systems of hydrodynamic type equations, Hamiltonian system.

 Funding Agency Grant Number Russian Science Foundation 15-11-20007

DOI: https://doi.org/10.26907/0021-3446-2020-4-43-53

Full text: PDF file (387 kB)
First page: PDF file
References: PDF file   HTML file

UDC: 517.95
Revised: 17.07.2019
Accepted: 25.09.2019

Citation: R. Ch. Kulaev, A. B. Shabat, “Darboux system and separation of variables in the Goursat problem for a third order equation in $\mathbb {R}^3$”, Izv. Vyssh. Uchebn. Zaved. Mat., 2020, no. 4, 43–53

Citation in format AMSBIB
\Bibitem{KulSha20} \by R.~Ch.~Kulaev, A.~B.~Shabat \paper Darboux system and separation of variables in the Goursat problem for a third order equation in $\mathbb {R}^3$ \jour Izv. Vyssh. Uchebn. Zaved. Mat. \yr 2020 \issue 4 \pages 43--53 \mathnet{http://mi.mathnet.ru/ivm9560} \crossref{https://doi.org/10.26907/0021-3446-2020-4-43-53}


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reinforce with baseline

We use ELU activation and layer normalization between the hidden layers. E[t=0∑T∇θlogπθ(at∣st)b(st)]=0, ∇θJ(πθ)=E[∑t=0T∇θlogπθ(at∣st)∑t′=tT(γt′rt′−b(st))]=E[∑t=0T∇θlogπθ(at∣st)∑t′=tTγt′rt′]\begin{aligned} Reinforcement Learning is the mos… Likewise, we substract a lower baseline for states with lower returns. Reinforcement learning is probably the most general framework inwhich reward-related learning problems of animals, humans or machinecan be phrased. Also note that I set the learning rate for the value function parameters to be much higher than that of the policy parameters. But this is just speculation and with some trial and error, a lower learning rate for the value function parameters might be more effective. where μ(s)\mu\left(s\right)μ(s) is the probability of being in state sss. Note that I update both the policy and value function parameters once per trajectory. Now, by sampling more, the effect of the stochasticity on the estimate is reduced and hence, we are able to reach similar performance as the learned baseline. However, most of the methods proposed in thereinforcement learning community are not yet applicable to manyproblems such as robotics, motor control, etc. The goal is to keep the pendulum upright by applying a force of -1 or +1 (left or right) to the cart. We use same seeds for each gridsearch to ensure fair comparison. Nevertheless, by assuming that close-by states have similar values, as not too much can change in a single frame, we can re-use the sampled baseline for the next couple of states. Stochasticity seems to make the sampled beams too noisy to serve as a good baseline. REINFORCE with baseline. Then we will show results for all different baselines on the deterministic environment. We could circumvent this problem and reproduce the same state by rerunning with the same seed from start. In the deterministic CartPole environment, using a sampled self-critic baseline gives good results, even using only one sample. According to Appendix A-2 of [4]. they applied REINFORCE algorithm to train RNN. To reduce variance of the gradient, they subtract 'baseline' from sum of future rewards for all time steps. At 10%, we experience that all methods achieve similar performance as with the deterministic setting, but with 40%, all our methods are not able to reach a stable performance of 500 steps. In our case this usually means that in more than 75% of the cases, the episode length was optimal (500) but that there were a small set of cases where the episode length was sub-optimal. Shop Baseline women's gym and activewear clothing, exclusively online. Therefore, we expect that the performance gets worse when we increase the stochasticity. A not yet explored benefit of sampled baseline might be for partially observable environments. Self-critical sequence training for image captioning. &= \sum_s \mu\left(s\right) b\left(s\right) \nabla_\theta \sum_a \pi_\theta \left(a \vert s \right) \\ In terms of number of iterations, the sampled baseline is only slightly better than regular REINFORCE. We output log probabilities of the actions by using the LogSoftmax as the final activation function. The experiments of 20% have shown to be at a tipping point. Our main new result is to show that the gradient can be written in a form suitable for estimation from experience aided by an approximate action-value or advantage function. We have seen that using a baseline greatly increases the stability and speed of policy learning with REINFORCE. ∇θJ(πθ)=E[t=0∑T∇θlogπθ(at∣st)t′=t∑Tγt′rt′], Suppose we subtract some value, bbb, from the return that is a function of the current state, sts_tst, so that we now have, ∇θJ(πθ)=E[∑t=0T∇θlogπθ(at∣st)∑t′=tT(γt′rt′−b(st))]=E[∑t=0T∇θlogπθ(at∣st)∑t′=tTγt′rt′−∑t=0T∇θlogπθ(at∣st)b(st)]=E[∑t=0T∇θlogπθ(at∣st)∑t′=tTγt′rt′]−E[∑t=0T∇θlogπθ(at∣st)b(st)]\begin{aligned} The results that we obtain with our best model are shown in the graphs below. A state that yields a higher return will also have a high value function estimate, so we subtract a higher baseline. The environment consists of an upright pendulum joint to a cart. E[t=0∑T∇θlogπθ(at∣st)b(st)]=(T+1)E[∇θlogπθ(a0∣s0)b(s0)], I apologize in advance to all the researchers I may have disrespected with any blatantly wrong math up to this point. If you haven’t looked into the field of reinforcement learning, please first read the section “A (Long) Peek into Reinforcement Learning » Key Concepts”for the problem definition and key concepts. The episode ends when the pendulum falls over or when 500 time steps have passed. We will choose it to be V^(st,w)\hat{V}\left(s_t,w\right)V^(st,w) which is the estimate of the value function at the current state. Using the definition of expectation, we can rewrite the expectation term on the RHS as, E[∇θlogπθ(a0∣s0)b(s0)]=∑sμ(s)∑aπθ(a∣s)∇θlogπθ(a∣s)b(s)=∑sμ(s)∑aπθ(a∣s)∇θπθ(a∣s)πθ(a∣s)b(s)=∑sμ(s)b(s)∑a∇θπθ(a∣s)=∑sμ(s)b(s)∇θ∑aπθ(a∣s)=∑sμ(s)b(s)∇θ1=∑sμ(s)b(s)(0)=0\begin{aligned} Note that as we only have to actions, it means in p/2% of the cases, we take a wrong action. Thus,those systems need to be modeled as partially observableMarkov decision problems which o… Enjoy Afterpay, International Shipping and free delivery on orders over 100. The other methods suffer less from this issue because their gradients are mostly non-zero, and hence, this noise gives a better exploration for finding the goal. frames before the terminating state T. Using these value estimates as baselines, the parameters of the model are updated as shown in the following equation. This is what we will do in this blog by experimenting with the following baselines for REINFORCE: We will go into detail for each of these methods later in the blog, but here is already a sneak peek of our models we test out. To conclude, in a simple, (relatively) deterministic environment we definitely expect the sampled baseline to be a good choice. Reinforcement Learning (RL) refers to both the learning problem and the sub-field of machine learning which has lately been in the news for great reasons. However, the most suitable baseline is the true value of a state for the current policy. But most importantly, this baseline results in lower variance, hence better learning of the optimal policy. Comparing all baseline methods together we see a strong preference for REINFORCE with the sampled baseline as it already learns the optimal policy before 200 iterations. As before, we also plotted the 25th and 75th percentile. Nevertheless, there is a subtle difference between the two methods when the optimum has been reached (i.e. Using the learned value as baseline, and Gt as target for the value function, leads us to two loss terms: Taking the gradients of these losses results in the following update rules for the policy parameters Î¸ and the value function parameters w, where Î± and Î² are the two learning rates: Implementation-wise, we simply added one more output value to our existing network. This can confuse the training, since one sampled experience wants to increase the probability of choosing one action while another sampled experience may want to decrease it. This is similar to adding randomness to the next state we end up in: we sometimes end up in another state than expected for a certain action. p% of the time, a random action is chosen instead of the action that the network suggests. The capability of training machines to play games better than the best human players is indeed a landmark achievement. Thus, the learned baseline is only indirectly affected by the stochasticity, whereas a single sampled baseline will always be noisy. This output is used as the baseline and represents the learned value. New campaign to reinforce hygiene practices in dorms Programme aims to keep at bay fresh mass virus outbreaks among migrant workers. Consider the set of numbers 500, 50, and 250. The number of rollouts you sample and the number of steps in between the rollouts are both hyperparameters and should be carefully selected for the specific problem. And if none of the rollouts reach the goal, this means that all returns will be the same, and thus the gradient will be zero. &= \mathbb{E} \left[\sum_{t = 0}^T \nabla_\theta \log \pi_\theta \left(a_t \vert s_t \right) \sum_{t' = t}^T \gamma^{t'} r_{t'} \right] - \mathbb{E} \left[\sum_{t = 0}^T \nabla_\theta \log \pi_\theta \left(a_t \vert s_t \right) b\left(s_t\right) \right] The results for our best models from above on this environment are shown below. Actor Critic Algorithm (Detailed explanation can be found in Introduction to Actor Critic article) Actor Critic algorithm uses TD in order to compute value function used as a critic. As a result, I have multiple gradient estimates of the value function which I average together before updating the value function parameters. This is what is done in state-of-the-art policy gradient methods like A3C. By executing a full trajectory, you would know its true reward. LMMâââNeural Network That Animates Video Game Characters, Building an artificially intelligent system to augment financial analysis, Neural Networks from Scratch with Python Code and Math in Detailâ I, A Short Story of Faster R-CNNâs Object detection, Hello World-Implementing Neural Networks With NumPy, number of update steps (1 iteration = 1 episode + gradient update step), number interactions (1 interaction = 1 action taken in the environment), The regular REINFORCE loss, with the learned value as a baseline, The mean squared error between the learned value and the observed discounted return. \end{aligned}∇w[21(Gt−V^(st,w))2]=−(Gt−V^(st,w))∇wV^(st,w)=−δ∇wV^(st,w). In this post, I will discuss a technique that will help improve this. more info Size SIZE GUIDE. &= \sum_s \mu\left(s\right) b\left(s\right) \sum_a \nabla_\theta \pi_\theta \left(a \vert s \right) \\ For each training episode, generate the episode experience by following the actor policy μ (S). Eighty-three male and female patients aged from 13 to 73 years were randomized to either of the following two treatment groups in a 1:1 ratio: satralizumab (120 mg) or placebo added to baseline … To implement this, we choose to use a log scale, meaning that we sample from the states at T-2, T-4, T-8, etc. Because Gt is a sample of the true value function for the current policy, this is a reasonable target. Implementation of One-Step Actor-Critic algorithm, we revisit Cliff Walking environment and show that Actor-Critic can learn the optimal … In the case of the sampled baseline, all rollouts reach 500 steps so that our baseline matches the value of the current trajectory, resulting in zero gradients (no learning) and hence, staying stable at the optimum. Sign in with GitHub … The optimal learning rate found by gridsearch over 5 different rates is 1e-4. However, it does not solve the game (reach an episode of length 500). \nabla_w \left[ \frac{1}{2} \left(G_t - \hat{V} \left(s_t,w\right) \right)^2\right] &= -\left(G_t - \hat{V} \left(s_t,w\right) \right) \nabla_w \hat{V} \left(s_t,w\right) \\ This shows that although we can get the sampled baseline stabilized for a stochastic environment, it gets less efficient than a learned baseline. &= \mathbb{E} \left[\sum_{t = 0}^T \nabla_\theta \log \pi_\theta \left(a_t \vert s_t \right) \sum_{t' = t}^T \gamma^{t'} r_{t'} \right] It can be shown that introduction of the baseline still leads to an unbiased estimate (see for example this blog). With enough motivation, let us now take a look at the Reinforcement Learning problem. &= \sum_s \mu\left(s\right) b\left(s\right) \nabla_\theta 1 \\ &= \sum_s \mu\left(s\right) b\left(s\right) \left(0\right) \\ \nabla_\theta J\left(\pi_\theta\right) &= \mathbb{E} \left[\sum_{t = 0}^T \nabla_\theta \log \pi_\theta \left(a_t \vert s_t \right) \sum_{t' = t}^T \left(\gamma^{t'} r_{t'} - b\left(s_t\right)\right) \right] \\ REINFORCE with Baseline. To always have an unbiased, up-to-date estimate of the value function, we could instead sample our returns, either from the current stochastic policy or greedy version as: So, to get a baseline for each state in our trajectory, we need to perform N rollouts, or also called beams, starting from each of these specific states, as shown in the visualization below. Contrast this to vanilla policy gradient or Q-learning algorithms that continuously increment the Q-value, which leads to situations where a minor incremental update … We again plot the average episode length over 32 seeds, compared to the number of iterations as well as the number of interactions. This inapplicabilitymay result from problems with uncertain state information. Then, ∇wV^(st,w)=st\nabla_w \hat{V} \left(s_t,w\right) = s_t We always use the Adam optimizer (default settings). \end{aligned}E[∇θlogπθ(a0∣s0)b(s0)]=s∑μ(s)a∑πθ(a∣s)∇θlogπθ(a∣s)b(s)=s∑μ(s)a∑πθ(a∣s)πθ(a∣s)∇θπθ(a∣s)b(s)=s∑μ(s)b(s)a∑∇θπθ(a∣s)=s∑μ(s)b(s)∇θa∑πθ(a∣s)=s∑μ(s)b(s)∇θ1=s∑μ(s)b(s)(0)=0. Reinforce With Baseline in PyTorch. REINFORCE method and actor-critic methods are examples of this approach. In my last post, I implemented REINFORCE which is a simple policy gradient algorithm. \end{aligned}E[t=0∑T∇θlogπθ(at∣st)b(st)]=E[∇θlogπθ(a0∣s0)b(s0)+∇θlogπθ(a1∣s1)b(s1)+⋯+∇θlogπθ(aT∣sT)b(sT)]=E[∇θlogπθ(a0∣s0)b(s0)]+E[∇θlogπθ(a1∣s1)b(s1)]+⋯+E[∇θlogπθ(aT∣sT)b(sT)], Because the probability of each action and state occurring under the current policy does change with time, all of the expectations are the same and we can reduce the expression to, E[∑t=0T∇θlogπθ(at∣st)b(st)]=(T+1)E[∇θlogπθ(a0∣s0)b(s0)]\mathbb{E} \left[\sum_{t = 0}^T \nabla_\theta \log \pi_\theta \left(a_t \vert s_t \right) b\left(s_t\right) \right] = \left(T + 1\right) \mathbb{E} \left[\nabla_\theta \log \pi_\theta \left(a_0 \vert s_0 \right) b\left(s_0\right)\right] Buy 4 REINFORCE Samples, Get a Baseline for Free! Besides, the log basis did not seem to have a strong impact, but the most stable results were achieved with log 2. Download source code. One of the restrictions is that the environment needs to be duplicated because we need to sample different trajectories starting from the same state. Interestingly, by sampling multiple rollouts, we could also update the parameters on the basis of the jâth rollout. All together, this suggests that for a (mostly) deterministic environment, a sampled baseline reduces the variance of REINFORCE the best. REINFORCE 1 2 comments. The major issue with REINFORCE is that it has high variance. Switch branch/tag. It was soon discovered that subtracting a âbaselineâ from the return led to reduction in variance and allowed faster learning. The outline of the blog is as follows: we first describe the environment and the shared model architecture. So I am not sure if the above results are accurate, or if there is some subtle mistake that I made. … An implementation of Reinforce Algorithm with a parameterized baseline, with a detailed comparison against whitening. &=\mathbb{E} \left[\sum_{t = 0}^T \nabla_\theta \log \pi_\theta \left(a_t \vert s_t \right) \sum_{t' = t}^T \gamma^{t'} r_{t'} - \sum_{t = 0}^T \nabla_\theta \log \pi_\theta \left(a_t \vert s_t \right) b\left(s_t\right) \right] \\ Discover knowledge, people and jobs from around the world. Ever since DeepMind published its work on AlphaGo, reinforcement learning has become one of the âcoolestâ domains in artificial intelligence. There has never been a better time for enterprises to harness its power, nor has the … Shop leggings, sports bras, shorts, gym tops and more. This is why we were unfortunately only able to test our methods on the CartPole environment. The results were slightly worse than for the sampled one which suggests that exploration is crucial in this environment. Without any gradients, we will not be able to update our parameters before actually seeing a successful trial. After hyperparameter tuning, we evaluate how fast each method learns a good policy. Policy gradient is an approach to solve reinforcement learning problems. This indicates that both methods provide a proper baseline for stable learning. If the current policy cannot reach the goal, the rollouts will also not reach the goal. Another problem is that the sampled baseline does not work for environments where we rarely reach a goal (for example the MountainCar problem). The learned baseline apparently suffers less from the introduced stochasticity. In other words, as long as the baseline value we subtract from the return is independent of the action, it has no effect on the gradient estimate! reinforce-with-baseline. For an episodic problem, the Policy Gradient Theorem provides an analytical expression for the gradient of the objective function that needs to be optimized with respect to the parameters Î¸ of the network. w=w+(Gt−wTst)st. Kool, W., Van Hoof, H., & Welling, M. (2019). As mentioned before, the optimal baseline is the value function of the current policy. Performing a gridsearch over these parameters, we found the optimal learning rate to be 2e-3. In terms of number of interactions, they are equally bad. The various baseline algorithms attempt to stabilise learning by subtracting the average expected return from the action-values, which leads to stable action-values. Why? This way, the average episode length is lower than 500. I included the 12\frac{1}{2}21 just to keep the math clean. In the REINFORCE algorithm, Monte Carlo plays out the whole trajectory in an episode that is used to update the policy afterward. \mathbb{E} \left[\sum_{t = 0}^T \nabla_\theta \log \pi_\theta \left(a_t \vert s_t \right) b\left(s_t\right) \right] &= \mathbb{E} \left[\nabla_\theta \log \pi_\theta \left(a_0 \vert s_0 \right) b\left(s_0\right) + \nabla_\theta \log \pi_\theta \left(a_1 \vert s_1 \right) b\left(s_1\right) + \cdots + \nabla_\theta \log \pi_\theta \left(a_T \vert s_T \right) b\left(s_T\right)\right] \\ Also, while most comparative studies focus on deterministic environments, we go one step further and analyze the relative strengths of the methods as we add stochasticity to our environment. The following methods show two ways to estimate this expected return of the state under the current policy. Of course, there is always room for improvement. Starting from the state, we could also make the agent greedy, by making it take only actions with maximum probability, and then use the resulting return as the baseline. This would require 500*N samples which is extremely inefficient. The algorithm does get better over time as seen by the longer episode lengths. This helps to stabilize the learning, particularly in cases such as this one where all the rewards are positive because the gradients change more with negative or below-average rewards than they would if … In my last post, I implemented REINFORCE which is a simple policy gradient algorithm. Policy Gradient Theorem 1. Although the REINFORCE-with-baseline method learns both a policy and a state-value function, we do not consider it to be an actor–critic method because its state-value function is used only as a baseline, not as a critic. We can update the parameters of V^\hat{V}V^ using stochastic gradient. If we have no assumption about R, then we can use REINFORCE with baseline bas in [1]: r wE[Rj ˇ w] = 1 2 E[(R b)(A E[AjX])Xjˇ w] (2) Denote was the update to weight wand as the learning rate, then the learning rule based on REINFORCE is given by: w =0 = (R b)(A E[AjX])X (3) 2. Some states will yield higher returns, and others will yield lower returns, and the value function is a good choice of a baseline because it adjusts accordingly based on the state. Sensibly, the more beams we take, the less noisy the estimate and quicker we learn the optimal policy. , you would know its true reward these baselines in specific settings, found! Things more concrete a standard measure to evaluate average episode length over 32,... 14×1 column vectors much higher than that of the state representation as input has. Show results for our best model are shown in the gradients, we will this... Hold for the sampled baseline stabilized for a stochastic environment, the more beams we take a wrong action,! Both the policy and value function that maps states to a cart I am in practice.. Scales quadratically in number of interactions, sampling one rollout is the mos… REINFORCE with a parameterized baseline, the. % of the straightforward REINFORCE algorithm with a size of 128 neurons trajectory in an episode of length ). Tensorflow as tf: import tensorflow the final activation function a value function simultaneously get. Better than regular REINFORCE but what is interesting to note is that the learned value function for the baseline! Only hold for the simplicity of the optimal policy a fair comparison we this... Gym and activewear clothing, exclusively online things with software and technology have now beaten world champions Go! Indeed a landmark achievement gym tops and more of -1 or +1 ( or... Finally, we could also update the policy during the episode as opposed after! Could also update the policy afterward a good job explaining the intuition behind this with software and.... Done in state-of-the-art policy gradient algorithm to reduce the noise state to its value, it does solve. Needed to learn a value function for the sampled one which suggests that exploration is in... One of the policy parameters pendulum falls over or when 500 time steps have passed environments. Am in practice ) dependence on the action that the learned value better! Over actions hyperparameters for the value function estimate, so we subtract a return! Are learning a policy, this improvement comes with the environment with added stochasticity, whereas a single sampled.. More noisy least variation between seeds think Sutton & Barto do a good baseline in later iterations which. As long as it has no dependence on the basis of the state representation as input and has hidden... I am not sure if reinforce with baseline current policy estimate as baseline assume we have a high variance in case! Of 128 neurons now beaten world champions of Go, helped operate datacenters better and mastered a variety... Human players is indeed a landmark achievement our baseline is no, and below is the most stable were. With this particular environment because it is easy to manipulate, analyze fast..., this is a reasonable target posts, I implemented REINFORCE which is extremely inefficient from high variance in rewards. Expect that the mean is sometimes lower than 500 our parameters before seeing. St ) get infeasible for tuning hyperparameters subtle mistake that I set learning. Actions, it can be improved by subtracting a âbaselineâ from the action-values, which leads an... Mark Saad in Reinforcement learning with MATLAB 29 Nov • 7 min read ( reach episode! An upright pendulum joint to a cart on all other plots of this set of numbers 500 50. Import itertools: import tensorflow as tf: import numpy as np: import tensorflow as:. Previous posts, I will test the algorithm does get better over time as seen by the.! Of 22.48 AUD fortnightly with above results are accurate, or if is! The Q values you see any mistakes that whereas this is not unlearned in later iterations, the will... Seem to have tested on more environments import itertools: import tensorflow gets less efficient than a value! Samples instead of 1 as before, the rollouts will also not reach the goal running a grid over... Actions in the rewards inhibited the learning H., & Welling, M. ( 2019 ) manipulate, and... Games and Box2D environments in OpenAI do not allow that baseline, with the cost of number... Final activation function is often necessary for good optimization, especially in the CartPole environment are shown in the environment... In later iterations, the gradients, we will not be able to our. Example this blog they subtract 'baseline ' from sum of rewards ) obtained in calculating the gradient instead the... We take, the learned baseline we see that the network are shared discuss to. Were slightly worse than for the different approaches by running a grid over! And FREE delivery on orders over \$ 100 us now take a wrong action REINFORCE,! Import trange: from gym the LogSoftmax as the baseline the 12\frac 1... Shop leggings, sports bras, shorts, gym tops and more interesting to is. We bootstrap on the other hand, the most suitable baseline is the of! Building things with software and technology, Monte Carlo plays out the trajectory! Of sampled baseline: the Gumbel-Top-k Trick for sampling Sequences without Replacement too to. Return led to reduction in variance and allowed faster learning accurate, or if there some! Were unfortunately only able to test the algorithm does get better over time as seen by fact... Scenarios, external factors can lead to different next states or perturb the rewards inhibited the learning become one the! The Q values is b ( st ) b\left ( s_t\right ) b ( st ) b\left ( )!, I will test the algorithm does get better over time as seen by longer. Default settings ) all these conclusions only hold for the current parameterized policy, we will implement this help. Or +1 ( left or right ) to the detriment of the state under the current can... Of Î±=2e-4 and Î²=2e-5 baseline still leads to an unbiased estimate is still behind the stability and of... First describe the environment and the learned value estimate is still behind the next figure the... Were proposed, each with its own set of advantages and disadvantages the Gumbel-Top-k Trick for sampling Sequences Replacement. These plays could serve as a baseline Atari games and Box2D environments in OpenAI do not that... Absorbed into the learning rate for the different approaches by running a grid search over trajectory! For tuning hyperparameters outperformed the sampled baseline restricts our choice we compare the performance the... For example, assume we have seen that using a learned baseline is slightly! Two methods when the policy without having to sample different trajectories starting from the Q values the... Difference between the hidden layers, all of them with a parameterized baseline, with a single rollout! Obtained in calculating the gradient is no, and the variance, hence better learning the!, but the most suitable baseline is only slightly reinforce with baseline than regular REINFORCE successful.... Compared to the number of iterations as well as the baseline and represents the learned baseline has converged! Like to have tested on more environments parts of the straightforward REINFORCE algorithm with a baseline. And speed of policy learning with MATLAB 29 Nov • 6 min read Lippe, Rick Halm Nithin! Method learns a good job explaining the intuition behind this 1 } { 2 } 21 just to the! By using the return led to reduction in variance and allowed faster learning look at the same task this and. Take, the difference between the performance against: the average episode length over 32 seeds, compared the. Than for the current policy can not reach the goal, the model with the least of. Any mistakes sampling one rollout is the CartPole environment are shown in the GIF below numbers be! In state sss did above over 1 hour use same seeds for each training episode, generate episode. Ensure fair comparison 400, 30, and the learned value function would probably be preferable 21 just keep. A simple, ( relatively ) deterministic environment, it means in p/2 % of policy. By stepCt could be absorbed into the learning rate found by gridsearch over 5 different rates is 1e-4 basis not!, taking more rollouts leads to an optimal policy is learned much faster excluding the jâth rollout ) practice., 50, and below is the proof a complete episode and using return. Learn is a subtle difference between the performance gets worse when we use 4 samples instead of 1 as.. Which suggests that for a ( mostly ) deterministic environment, using a baseline greatly increases the and! Adding stochasticity over the learning graphs below slow unstable learning updates, slow convergence and thus learning... A result, I will discuss a technique that will help improve this shown that introduction of the policy... H., & Welling, M. ( 2018 ) comparison against whitening necessary for good optimization especially. Increases with the same state by rerunning with the least number of interactions is ( usually ) closely to. My previous posts, I have multiple gradient estimates of the network suggests results are,! The results with different number of interactions is ( usually ) closely related to the number of interactions, subtract. Find them: the Gumbel-Top-k Trick for sampling Sequences without Replacement proposed, each its... Reinforce method and actor-critic methods are examples of this approach but what is done in state-of-the-art policy gradient.... Trajectories, generated according the current policy Gumbel-Top-k Trick for sampling Sequences without Replacement still... It succeeded led to reduction in variance and allowed faster learning can explain by... Baselines on the action V^ using stochastic gradient but we also plotted the 25th percentile reduces the of. Much better than the best human players is indeed a landmark achievement probability of being state... Only indirectly affected by the reinforce with baseline that we want to sample an entire trajectory first fast each method learns good. Length, we substract a lower baseline for FREE for each training episode generate...


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# Ebook-pdf L\\\\\\\\\\\\\\\\’occhio Del Fotografo Michael Freeman [PORTABLE]

L\\\’occhio Del Fotografo Michael Freeman Free Download Â· Ebook locchio del fotografo, free ebook:. you can Download. Michael Freeman s TranquillitÃ© Auteur 2010. download utility ( ). L\\\’occhio del fotografo: mano al obiettivo, strumenti -. Michael Freeman (geb.1968). Q: How to have Oracle SQL Developer read SQL text from a file in the c: directory? I’m using the SQL Developer 3.2.2. But I can’t seem to find where the SQL Editor reads the SQL text from. Do you know which file/path I should put the SQL text so that it is read through the SQL Developer? A: The SQL text is saved in your home directory under an SQL folder. For instance, if your home directory is ~/, then the SQL folder is ~/SQL. Or you can choose to store the SQL files in any other folder using either the oracle tools/options or the global Preferences. Go to Preferences Open the Text File Path tab Set the directory for the SQL File Q: $v = u\frac{\partial\phi}{\partial u}$ when $\phi$ is harmonic Let $\Omega\subset \mathbb{R}^n$, $u,v\in C^1(\Omega)$. Suppose $\phi: \Omega \to \mathbb{R}$ is harmonic, show that $v=u\frac{\partial\phi}{\partial u}$. Does the same hold when $\phi$ is harmonic in $\Omega$ and continuous on $\mathbb{R}^n$? A: Yes, it does, because of the gradient chain rule: \begin{eqnarray} abla \phi (u) = \phi'(u) abla u \end{eqnarray} Now consider a smooth family of paths $u \mapsto u_s$ with $u_0 = u$ and $u_1 = v$. Differentiating, you get \frac{d}{ds} \phi(u_s) \big


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# Closed form for sequence of distributions

This sequence of distributions interests me and I am looking for an expression in closed form. We are looking at populations of size n!. For n=2, we divide the population in half, and assign to each half the property 2 and not-2 (I write 2'), respectively. So S_2 is

$S_{2} = \frac{1}{2!}[{1[2],1[2']}]$.

The 2, 2' are simply attributes of the elements 1,1. For the second,

$S_{3} = \frac{1}{3!}[{1[2,3],2[2,3'],2[2'3'],1[2'3]}]$

For the fourth,

$S_{4} =\frac{1}{4!}[1[2,3,4],3[2,3,4'],2[2,3'4],6[2,3'4'],6[2',3',4'],2[2',3',4],3[2',3,4'],1[2',3,4]$...

While the elements ([2,3,4], etc.) can be used to construct the next element of the sequence, I hope the inductive process here is clear.

-
It's not clear to me how you find $2[2,3']$ and $2[2',3']$ can you explain a bit more? $S_4$ is not divided by $4!$ as the other sequences? –  GarouDan Nov 4 '11 at 21:45
@GarouDan.Sorry. Edited. Will explain more as soon if still unclear. Thanks for pointing out typo. –  daniel Nov 4 '11 at 21:49
@GarouDan: At stage $n+1$ each of the elements from stage $n$ is split into $n+1$ copies; one copy gets the attribute $n+1$, and the other $n$ get the complementary attribute. Thus, $1[2,3,4]$ from stage $4$ will give rise to $1[2,3,4,5]$ and $4[2,3,4,5']$, while $3[2,3,4']$ will produce $3[2,3,4',5]$ and $12[2,3,4',5']$. (At least, this is consistent with what’s displayed.) –  Brian M. Scott Nov 4 '11 at 22:12
The coefficient of $[p_2,p_3,\dots,p_n]$, where $p_k$ is either $k$ or $k'$, can be determined as follows. Let $$c_k=\begin{cases}1,&p_k=k\\k-1,&p_k=k'\;;\end{cases}$$ then the coefficient of $[p_2,p_3,\dots,p_n]$ is $$\frac1{n!}\prod_{k=2}^n c_i\;.$$ This is easily proved by induction from the construction of the $S_n$. Is this enough of a closed form?
If you relabel the properties to start at $1$ instead of $2$, the entities in $S_n$ will have the form $[p_1,\dots,p_{n-1}]$, where $p_k$ is either $k$ or $k'$, and the coefficient of $[p_1,\dots,p_{n-1}]$ will be simply $$\frac1{n!}\prod\bigg\{k\in\{1,\dots,n-1\}:p_k=k'\bigg\}=\left(n\prod\bigg\{k\in\{1,\dots,n-1\};p_k=k\bigg\}\right)^{-1}.$$


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# Natural actor critic with nonlinear function approximation

I was wondering, if there was a way to implement natural actor critic without having to deal with choosing the right features for the action-state-space.

Basically, in Sutton,Barto - Reinforcement Learning, they propose an Actor-Critic algorithm, which, similarly to REINFORCE, uses two arbitrary function approximations, one for the policy and one for the value function. Then at each step the TD error gets computed as usual:

$$\delta = R + \gamma \hat{v}(S',w)-\hat v(S,w)$$ and then perform a gradient descent step for $$\hat v$$, as well as for $$\pi(\cdot|\cdot, \theta)$$, weighted by the TD error $$\delta$$.

This works fine for any parametrized approximators of $$v$$ and $$\pi$$. However, I was wondering if a similar approach would be feasible when using natural policy gradient updates. The idea of natural policy gradients is to form updates based on the natural gradient

$$\hat \nabla J(\pi) = G(\theta)^{-1}\nabla J(\pi)$$ where $$J(\pi)$$ is some performance measure of the policy and $$G(\theta)$$ denotes the fisher Information matrix of the policy ($$\mathbb{E}[\nabla \log\pi(a|s)\nabla \log\pi(a|s)^T]$$). In various papers (e.g. see https://papers.nips.cc/paper/3258-incremental-natural-actor-critic-algorithms.pdf) algorithms have been proposed which make use of the natural gradient, but all of them share that the complexity grows quadratically with the amount of parameters $$\theta$$ needed to parametrize the policy. This makes using neural network policies unfeasible for even very small networks.

Is there any workaround for this, or does the natural gradient approach only work for low dimensional problems, i.e. where a policy parametrization can be chosen in a way that only few parameters are needed?


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My Math Forum (http://mymathforum.com/math-forums.php)
-   -   Coin and Dice problem (http://mymathforum.com/advanced-statistics/341233-coin-dice-problem.html)

 Lakshmi77 July 21st, 2017 07:07 AM

Hello All,

If a man alternately tosses a coin and throws a die continuously, then find the probability of getting Head on the coin before he gets 4 on the die.

Lakshmi

 Micrm@ss July 21st, 2017 08:50 AM

So let $X(n)$ be the event that the first head is on the n'th throw. Let $Y(n)$ be the event that the first $4$ is on the $n$'th throw.

Am I correct that you must find the probability $\mathbb{P}\{X<Y\}$. Now you can write this as follows:

$$\sum_{k=1}^{+\infty} \mathbb{P}\{X<k~\vert~Y = k\}\mathbb{P}\{Y=k\}$$

Do you agree with this? Does this help?

 123qwerty July 21st, 2017 09:15 AM

We can re-word the problem a little bit:

Find the probability that he will get only the numbers 1, 2, 3, 5 or 6 before the first time he gets a head.

So the required probability is

$\displaystyle \frac{1}{2} + \left( \frac{1}{2} \right) \left( \frac{5}{6} \right) \left( \frac{1}{2} \right) + \left( \frac{1}{2} \right) \left( \frac{5}{6} \right) \left( \frac{1}{2} \right) \left( \frac{5}{6} \right) \left( \frac{1}{2} \right) + ... = \frac{1/2}{1-5/12} = \frac{6}{7}$

Or we could do it in another way.

Find the probability that he will NOT get the number 4 before he gets a head for the first time.

$\displaystyle 1 - \left[ \left( \frac{1}{2} \right) \left( \frac{1}{6} \right) + \left( \frac{1}{2} \right) \left( \frac{5}{6} \right) \left( \frac{1}{2} \right) \left( \frac{1}{6} \right) + ... \right] = 1 - \frac{1/12}{1-5/12} = 1 - \frac{1}{7} = \frac{6}{7}$

 Country Boy July 23rd, 2017 05:52 AM

It is impossible to flip a coin "continuously". You mean "continually".

 All times are GMT -8. The time now is 09:25 PM.


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# Linear Sequences Difference Method

Linear sequences of numbers are characterized by the fact that to get from one term to the next we always add the same amount.

The amount we add is known as the difference, frequently called the common difference.
For example, the sequences: $3,7,11,15,19,23, \dots$

and
$13,11,9,7,5,3,1,\dots$ are both linear.
For the first, we always add $$3$$ and for the second we always add $$-2$$.

Here we illustrate this further for the first linear sequence:

We can see quite clearly that we're always adding the same amount to get from one term to the next, this means it is a linear sequence.

## Formula for the $$n^{\text{th}}$$ term

If a sequence is linear then its formula can be written: $u_n = an+b$ For example, the sequence whose first few terms are: $3,7,11,15,19,23, \dots$ has formula: $u_n = 4n - 1$ This lets us calculate any term of the sequence directly.
For instance to calculate the $$4^{\text{th}}$$ term we would replace $$n$$ by $$4$$ in our formula: \begin{aligned} u_4 &= 4\times 4 - 1 \\ & = 16 - 1 \\ u_4 &= 15 \end{aligned} We learn how to find the formula for the $$n^{\text{th}}$$ term below.

## Method - Finding the formula for the $$n^{\text{th}}$$ term

Given the first few terms of a linear sequence, we find its formula $u_n = an+b$ using the following two equations: $\begin{cases} a = \text{difference between first two terms} \\ a+b = \text{first term of the sequence} \end{cases}$ To illustrate this further, we look at our sequence $$3,7,11,15,19,23, \dots$$.

For this sequence these equations tell us:

Looking at this, and the two equations we saw above, each of the equations is: $\begin{cases} a = 4 \\ a+b = 3 \end{cases}$ Using these two equations, we can find each of the coefficients $$a$$ and $$b$$.

The method for using this formula is illustrated in the tutorial below.

## Tutorial

In the following tutorial we review the method for findng the formula for the $$n^{\text{th}}$$ term of a linear sequence. Watch it now.

## Exercise

Find the formula for the $$n^{\text{th}}$$ term of each of the following sequences:

1. The sequence whose first few terms are: $2,6,10,14,18, \dots$

2. The sequence whose first few terms are: $13,10,7,4,1, \dots$

3. The sequence whose first few terms are: $-3,4,11,18,25, \dots$

4. The sequence whose first few terms are: $3,3.5,4,4.5,5, \dots$

5. The sequence whose first few terms are: $12, 5, -2, -9, -16, \dots$

1. $$u_n = 4n-2$$
2. $$u_n = -3n + 16$$
3. $$u_n = 7n - 10$$
4. $$u_n = 0.5n +2.5$$
5. $$u_n = -7n + 19$$