text:
# Temperature jump

The temperature jump method is a technique used in chemical kinetics for the measurement of very rapid reaction rates. It is one of a class of chemical relaxation methods pioneered by the German physical chemist Manfred Eigen in the 1950s. In these methods, a reacting system initially at equilibrium is perturbed rapidly and then observed as it relaxes back to equilibrium.[1][2][3] In the case of temperature jump, the perturbation involves rapid heating which changes the value of the equilibrium constant, followed by relaxation to equilibrium at the new temperature.

The heating usually involves discharging of a capacitor (in the kV range) through a small volume (< 1 mL) of a conducting solution containing the molecule/reaction to be studied. In some versions of the apparatus used, the solution is heated instead by the output of a pulsed laser which emits in the near infra-red. When laser heating is employed, the solution need not be conducting. In both cases, the temperature of the solution is caused to rise by a small amount in microseconds (or less in the case of laser heating). This allows the study of the shift in equilibrium of reactions that equilibrate in milliseconds (or microseconds with laser temperature jump), these changes most commonly being observed using absorption spectroscopy or fluorescence spectroscopy. Due to the small volumes involved the temperature of the solution returns to that of its surroundings in minutes.[4]

The fractional extent of the reaction (i.e. the percentage change in concentration of a measurable species) depends on the molar enthalpy change (ΔH°) between the reactants and products and the equilibrium position. If K is the equilibrium constant and dT is the change in temperature then the enthalpy change is given by the Van 't Hoff equation:

${\displaystyle {\Delta H^{o}}={RT^{2}}.{\frac {d\ln K}{dT}}}$

where R is the universal gas constant and T is the absolute temperature. When a single step in a reaction is perturbed in a temperature jump experiment, the reaction follows a single exponential decay function with time constant ${\displaystyle (\tau )}$ equal to a function of the forward (ka) and reverse (kb) rate constants. For the perturbation of a simple equilibrium ${\displaystyle {\ce {A <=> B}}}$ which is first order in both directions, the reciprocal of the time constant equals the sum of the two rate constants[2]

${\displaystyle 1/\tau =k_{a}+k_{b}}$

The two rate constants can be determined from the values of ${\displaystyle (\tau )}$ and the equilibrium constant :${\displaystyle K=k_{a}/k_{b}}$, yielding two equations for two unknowns.

In more complex reaction networks, when multiple reaction steps are perturbed, then the reciprocal time constants are given by the eigenvalues of the characteristic rate equations. The ability to observe intermediate steps in a reaction pathway is one of the attractive features of this technology.[5]

Related chemical relaxation methods include pressure jump,[6][3] electric field jump[6] and pH jump.[3][7]

## References

1. ^ Laidler K.J. and Meiser J.H., Physical Chemistry (Benjamin/Cummings 1982) p.362, ISBN 0-8053-5682-7
2. ^ a b Atkins P. and de Paula J. Atkins' Physical Chemistry (8th ed., W.H.Freeman 2006) p.805, ISBN 0-7167-8759-8
3. ^ a b c Steinfeld J.I., Francisco J.S. and Hase W.L., Chemical Kinetics and Dynamics (2nd ed., Prentice-Hall 1998) p.140-3, ISBN 0-13-737123-3
4. ^ Daniel L. Purich, R. Donald Allison (1999). Handbook of biochemical kinetics. Academic Press. ISBN 0-12-568048-1.
5. ^ Urbanke, C; Wray, J. (2001). "A fluorescence temperature-jump study of conformational transitions in myosin subfragment 1" (PDF). Biochemical Journal. 358 (Pt 1): 165–173. doi:10.1042/0264-6021:3580165. PMC . PMID 11485564.
6. ^ a b Espenson J.H. Chemical Kinetics and Reaction Mechanisms (2nd ed., McGraw-Hill 2002) p.256 ISBN 0-07-288362-6
7. ^ Gutman, M. and Huppert, D.J. (1979) Biochem. Biophys. Methods 1, 9-19

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# 🤔 Image Processing¶

Whether you are into Facebook, Instagram, or Snapchat you are probably familiar with all kinds of image filters and manipulations. It turns out that lots of the filters you can apply to images are pretty fun to code. Hopefully you have read the Nested Iteration and Image Processing section to get yourself ready for this project. In any case you will probably want to open up that section in another tab so you can refer to the image module functions that are provided.

You can use the following image using the name “golden_gate.png” or you can use any image you choose by using the full URL to the image. For example: http://reputablejournal.com/images/ComputerHistory/TeleType.png will use an automatically reduced size image of a picture the author took in the computer history museum.

## Basic Filters¶

To start, we’ll try some pixel by pixel filters. We have a few for you to try here, but feel free to experiment on your own as well. You really can’t go wrong here so let your imagination run wild.

The first thing to try is to create a grayscale version of a color image. Grayscale is not quite black and white where each pixel would be “all on” or “all off” but rather a grayscale image is one in which the the red, green, and blue components of each pixel are all the same and in the range from 0 to 255. Your first task is to figure out how to turn RGB values for each pixel which are likely different into three values for RGB that are the same. There are several different ways you could do this, and there’s not really a clear right or wrong, so just try the first thing that occurs to you and it will probably look pretty good.

Now that you have a grayscale image, try turning it into a black and white image by setting a threshold value for your gray_value. That is, if gray_value is less than your threshold make r,g,b all 0. If its more make r,g,b all 255.

Here is another pretty standard filter for photos called “Sepia tone” It will remind you of the old-west photographer style images. The formula to convert a photo into sepia tone is as follows:

newR = (R × 0.393 + G × 0.769 + B × 0.189)
newG = (R × 0.349 + G × 0.686 + B × 0.168)
newB = (R × 0.272 + G × 0.534 + B × 0.131)

And finally here’s an activecode place for you to go wild. Try making everything neon. Take away all of the green, double the blue. whatever you can think of. If you find something cool you can come back to it and try it on some different images.

## Rotating, Scaling and Blending¶

In addition to filters, another really common thing to do with images is to crop, resize, and rotate them. We will start with rotating , moving on to resizing and then we will apply the cropping operation to combine multiple images into one by taking parts of two or more images and adding them into a final image. Note that for this group of exercises we will not change the original image in place. Instead we’ll make a new Empty image and move the pixels from the original image into the appropriate place in the new image.

Lets start by rotating an image by 90 degrees in the clockwise direction. This is an easy one to get wrong as your initial thought might be to just take a pixel from position x, y and put it at position y, x. In fact this is easy to try so you should do that first to see why it is not quite correct. To get this one right you might want to work through a small example to understand the correct pattern. Thats good problem solving practice and really helps in this case.

Now lets make an image larger. We’ll begin by enlarging the image by the same amount in both the width and the height. This preserves a property of the image known as its aspect ratio. You should think about this before you start as how you solve this particular problem will make a huge difference in the complexity of your code.

If get this one mastered then think about how you might enlarge the image by different factors in height and width. You can make yourself look tall and thin (with an odd shaped head). This is optional so if you decide to do it you can write it as an enhancement to the code in the activecode window above.

If you tried to enlarge an image really big you would notice that it starts to look like an 80’s vintage video game. That is the image will get really blocky. Later on in this project we’ll introduce the idea of smoothing an image which can soften this blocky effect.

Once you have conquered enlarging an image its time to take on reducing an image. The key to this is to start simple. Don’t try to invent the perfect solution to this problem before you solve a simple version. What I mean by that is that in order to shrink an image, the ideal solution would be to summarize the colors contained in a block of pixels down to one. But one way to do that summary is to simply pick one pixel to be the representative for the whole group. If you get that strategy working then you might think about more advanced statistical techniques such as using the median of the color values or taking an average of all of the color values in a block of pixels.

For our final project from this section lets take parts from two different images and glue them into a new image. If your art department has a green screen this is a fun chance to put yourself into a scene of some kind. If not, its still fun to take parts of two images and blend them together. You can blend two images by averaging their pixel values. Of course if you prefer to have one image be “on top” of another image then you can just replace the pixel values of the bottom image with the top.

Challenge: Can you figure out how to rotate your image by an arbitrary angle? Here’s a diagram that will give you a pretty big hint, but remember that in the diagram the x and y coordinates grow up and to the right with 0,0 in the lower left. However, in your image 0,0 is in the upper left corner and x and y grow down and to the right. Also you’ll have to be really careful about how you size your resulting image to make sure you have room for your rotated image.

## Image Kernels for Machine Learning¶

This is definitely a more advanced section, but if you are comfortable with all of the exercises up to now, you are going to like this.

### Cleaning up noise¶

Here is a “friend” of mine in a photo taken long ago. Its been in a box in the closet for years, gathering dust, getting crushed by books and generally aging as old photos tend to do. I recently scanned it to add it to my digital collection. But I’m not too happy with the result.

noisyman.png

Your job is to digitally restore my friend and make him look like new. How are you going to do that? Well, what do we have to work with? If you look at the image, most of the speckles are just one pixel that is out of whack caused by dust on the picture, or a small scratch. Clearly that pixel value is incorrect with respect to the pixels surrounding it. So we need to fix that.

Your first inclination would be to find the bad pixels and fix only those, but there is an even easier solution for us. We can simply pretend that all pixels need to be fixed. There are two strategies we can use:

1. Replace every pixel with the average of the 8 pixels around it.

2. Replace every pixel with the median pixel value of the 8 pixels around it.

This strategy should work pretty well as the “bad” pixels tend to be close to 0 or 255 whereas the good pixels are in more in the middle.

To find the neighbors we will use some nested loops where we calculate the range of the loops based on the current pixel location. For example if we are trying to fix the pixels at row 11 and column 23 then we would want to look at the all the pixel values between row 10, column 22 and row 12 and column 24. This process of iterating over the neighbors of a pixel is called a kernel and is widely used in image processing.

One word of caution before you dive into this, is that there is literally an “edge case” and a “corner case” that you need to worry about or your program will crash. That is the pixels around the edge do not have 8 neighbors. We can deal with this the hard way or the easy way. The hard way is to add some conditionals to your program to detect these edges and respond by dealing with a different number of neighbors.

The easy way to deal with this is to make the tradeoff that the pixels at the edge of the image are fine as they are, and we can start fixing our image at row 1, column 1 and stop 1 column from the right and 1 row from the bottom. Now there are no special cases to worry about and you probably won’t even notice the difference.

One super clever strategy is to use the max and min functions to figure out the correct neighbor indexes. You might try to figure this out if you are really a perfectionist.

### Smoothing¶

This exercises is really a remix of the last problem and a return to our image enlargement problem, and we can fix the blocky nature of the enlarged image by replacing each pixel with the average of its neighbors.

### Edge detection¶

the Sobel kernel has two parts to it one to calculate the gradient, that is how the darkness of the image is changing from left to right and another to measure how the darkness of the image is changing from top to bottom.

$\begin{split} G_x = \left[ {\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 0 & -2 \\ 1 & 0 & -1 \\ \end{array} } \right]\end{split}$
$\begin{split} G_y = \left[ {\begin{array}{ccc} 1 & 2 & 1 \\ 0 & 0 & 0 \\ -1 & -2 & -1 \\ \end{array} } \right]\end{split}$

You apply each of the kernels to the neighboring pixels by multiplying the neighbors by the value in the small matrix. Then we combine the x and y gradients using

$$G = \sqrt{G_x^2 + G_y^2}$$

This definitely gives you a taste of why image processing requires so much computational power. its going to take a while for our Python in the browser to work its way over all of the pixels doing all of this computation for each one. Its also why this one is last as it can be really time consuming and frustrating to debug something.

Post Project Questions

During this project I was primarily in my...
• Comfort Zone
• Learning Zone
• Panic Zone
Completing this project took...
• Very little time
• A reasonable amount of time
• More time than is reasonable
Based on my own interests and needs, the things taught in this project...
• Don't seem worth learning
• May be worth learning
• Are definitely worth learning
For me to master the things taught in this project feels...
• Definitely within reach
• Within reach if I try my hardest
• Out of reach no matter how hard I try


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
8 초 512 MB 0 0 0 0.000%

## 문제

In a secret military base, a new communication technology is being tested. For the experiment, m antennas were constructed inside.

The terrain around the base is perfectly flat, and the base, seen from above, is a convex polygon. The boundary of the polygon is a wall that protects the base from intruders, as well as blocks the radio waves from leaving the base to be possibly intercepted by foreign agents.

Unfortunately, some construction works are required in the facility, and two of the polygon’s walls must be torn down. This creates a security risk: if two spies are placed outside the base in such a way that two of the antennas lie on the line between them, and there is no wall blocking this line, then the spies may listen to the communication between those two antennas.

Your goal is, for some possible scenarios of removal of two walls, to determine the number of pairs of antennas which are compromised in the way described above.

The picture above corresponds to the first case of the example input from the “Example” section. In this case, the base is a pentagon with four antennas, denoted by little crosses. All the lines between pairs of antennas are also shown.

## 입력

The first line of input contains the number of test cases z (1 ≤ z ≤ 200 000). The test cases follow, each one in the following format:

The first line of a test case contains an integer n (3 ≤ n ≤ 10) – the number of vertices of the polygon. The next n lines contain two integers – the coordinates of the vertices, presented clockwise. The vertices are numbered 0, 1, . . . , n − 1 in order in which they appear.

The next line contains an integer m (2 ≤ m ≤ 50 000) – the number of antennas inside the base – and the m following lines contain the coordinates of the antennas.

The next line contains another integer q (1 ≤ q ≤ 10) – the number of scenarios to consider. The last q lines describe scenarios – the i-th line contains two integers ai, bi (0 ≤ ai < bi ≤ n−1). Such a pair denotes removing the walls ai and bi and requires to compute the number of distinct lines that go through some two antennas and do not cross neither the segment between the vertices ai and ai + 1 nor the segment between bi and (bi + 1) mod n.

All coordinates are integers whose absolute values do not exceed 109. In any single testcase, all points of the input are distinct and no three of them are collinear.

Every test case, including the first, is preceded by a single empty line.

The sum of all m values in all test cases does not exceed 300 000.

## 출력

For every testcase output, in separate lines, the answers to all given scenarios.

## 예제 입력 1

2

5
0 0
0 5
3 7
6 5
6 0
4
1 2
1 3
5 2
5 3
3
0 3
1 4
1 2

4
-1 -1
-1 1
2 1
2 -1
2
0 0
1 0
6
0 1
0 2
0 3
1 2
1 3
2 3

## 예제 출력 1

4 1 0
0 1 0 0 0 0


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# Mathematics Colloquia and Seminars

We develop a new multi-scale framework flexible enough to solve a number of problems involving embedding random sequences into random sequences. Grimmett, Liggett and Richthammer asked whether there exists an increasing M-Lipschitz embedding from one i.i.d. Bernoulli sequences into an independent copy with positive probability. We give a positive answer for large enough M. A closely related problem is to show that two independent Poisson processes on $\mathbb{R}$ are roughly isometric (or quasi-isometric). Our approach also applies in this case answering a conjecture of Szegedy and of Peled. Our theorem also gives a new proof to Winkler's compatible sequences problem. This is joint work with Allan Sly.


text:
Turning on a nuclear briefcase with the smallest possible number of keystrokes

Mathematics Asked by Delta Account on August 14, 2020

On the front panel of the "nuclear briefcase" there are $$12$$ buttons. Each button controls its own switch: pressing it toggles it from ON to OFF and back. The initial position of the switches is unknown. The nuclear case triggers an inaudible (ultrasonic) frequency alarm when at least eight switches are in the ON position.

Find the shortest way using as few keystrokes as possible to ensure that the suitcase will sound an alarm.

I tried to do this with examples, but actually I’ve not idea how can I determine which button to press.

(I improved my previous answer, and now I think it corresponds to Delta Account's answer, but maybe in a bit more detail.)

Consider the following list of $$16$$ switch states where 0 corresponds to "in its original position" and 1 corresponds to the opposite:

000000000 0 0 0
000000000 0 0 1
000000000 0 1 1
000000000 0 1 0
000000000 1 1 0
000000000 1 1 1
000000000 1 0 1
000000000 1 0 0
111111111 1 0 0
111111111 1 0 1
111111111 1 1 1
111111111 1 1 0
111111111 0 1 0
111111111 0 1 1
111111111 0 0 1
111111111 0 0 0

(The spacing is there to help the pattern be clear, but doesn't mean anything.)

The "all switches are ON" position corresponds to some unknown sequence of twelve 0s and 1s. Whatever it is, it is four steps away from one of the states above. To see this, take an arbitrary state $$x_1x_2x_3dots x_{12}$$. Then replace the first $$9$$ switches by whichever of 0 and 1 occurs more often among them, changing at most 4 switches, and you'll get one of the states above.

To go through the states above in order, we need $$9 + 14cdot1 = 23$$ steps: there is one step where we press $$9$$ buttons to get from one state to the next, and fourteen steps where we just need to press $$1$$ button.

Answered by Misha Lavrov on August 14, 2020

let's divide 12 buttons into two groups of 9 buttons and 3 buttons. consider a group of 9 buttons there may be 5 switched on switches and in this case we need to switch three switches so that we have a total of 8 switched on switches. in a group of three buttons, we can consider all possible combinations and in this case we will definitely have 8 switches on. the number of combinations in a group with three buttons will be $$2^3-1=7$$ , since initially there is already a combination . now if, in our original group with nine buttons, five of these switches were not on but off, then we need 9 more moves to turn all the switches and then we will have 5 switched on switches, then again all the combinations to the second group with three buttons and then we will certainly have 8 switched on switches and the total number of moves is $$7+9+7=23$$

Answered by Delta Account on August 14, 2020

UPDATE: A better answer with $$47$$ moves. (Credit: inspired by comment of @WW1 in main thread.)

Paint the buttons $$2$$ red, $$1$$ green, $$9$$ blue.

Using a gray code or similar, you can have the $$2$$ red buttons go through all $$4$$ states with $$3$$ moves.

Lemma: If you press green, then all the blues once each, then green again, then at some point there are $$ge 6$$ ONs among these $$10$$.

Proof: After pressing green and all the blues, the $$10$$ bits will be negated. Either the initial state has $$6+$$ ON, or the final state has $$6+$$ ON, or both states have $$5$$ ON. So we only need to worry about this last case.

• If green initial state is OFF, then among the initial blue there were already $$5$$ ON. When you press green the first time you already got $$6$$ ON simultaneously.

• If green initial state is ON, then before pressing green the second time, green was OFF and there are $$5$$ blues ON. The second green press will be the $$6$$th ON. $$~~~~~~~~square$$

So if you do the red buttons as an outer loop and the other buttons as an inner loop, you need only $$4 times (11 + 1) = 48$$ moves.

As Ross Milikan pointed out, this can be slightly optimized by omitting the first gray-code move, for $$11 + 1 + 11 + 1 + 11 + 1 + 11 = 47$$ moves.

Not a provably optimal answer, but a method that takes only $$80 ll 2^8 = 256$$ moves.

Paint the last $$3$$ buttons red. Using a gray code or similar, you can have them go through all $$8$$ possibilities in $$8$$ moves.

For the first $$9$$ buttons, if you press each once then either the initial state or the final (bitwise negated) state will have $$5$$ or more switches ON.

So if you do the red buttons as an outer loop and the other buttons as an inner loop, you need only $$8 times (9 + 1) = 80$$ moves.

Answered by antkam on August 14, 2020

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text:
# How can we know "by inspection" the location of maximum deflection of a simply supported beam with a point load?

I am working the following problem:

The simply supported beam shown in Fig. 12-12a is subjected to the concentrated force P. Determine the maximum deflection of the beam. EI is constant.

The text later says:

By inspection of the elastic curve, Fig. 12-12b, the maximum deflection occurs at D, somewhere within region AB. Here the slope must be zero.

Why does the author say that, by inspection, the maximum deflection occurs at D? How do we know that? He didn't show any work or explain why it should be at D and not somewhere else.

If I consider the maximum slope to occur at a point 2 m from A, then I will take EI(dv2/dx2) = 0, then my answer is, -2((x2)^2) + 12x2 -44/3 = 0, then x = 4.29 m ....

here's the full question . in hree , we can notice that there are 2 sets of slope equation that we can use . Which is equation 5 & 7 . In equation 5 , we will get 1.633 as in the working . ( the author use by 'inspection' the max deflection occur at region AB)

However , when as @Jmac stated , we dont know where is the position where the max deflection is located , how can we use equation 5 to solve ?

why We dont have to consider equation , which involve region DC ??? in

solving equation 7 = 0 , i have x = 5.23 , 3 and 0.763 , which is correct ?

• No, it says the maximum deflection occurs at D. Therefore the slope at D is zero. The slope must be zero by the "common-sense" argument that if it was not zero, a point close to D would have a bigger deflection than D. Or you can make a rigorous argument using theorems that are proved in a calculus course. Dec 3 '16 at 16:51
• $D$ is defined as the point where the maximum deflection occurs. Notice that the location of $D$ is defined as "somewhere within $AB$". I agree that using the term "by inspection" is odd, though.
– Wasabi
Dec 4 '16 at 2:06
• For point loading, I think the max deflection only occurs at the point of loading when it's loaded at the centre (and both ends are supported). Consider that the place you are loading, it's closer to one support than the other. That means that it's closer to a point resiting the load, so the maximum deflection ocurs somewhere else. The case of being loaded in the middle is a special case because you are the same distance from both supports, so the maximum deflection is at the point loading.
– JMac
Dec 5 '16 at 10:48
• @kelvinmacks Not necessarily at "another side"; just further from the support. Consider if you were to place the load just to the left of point C. It couldn't really deflect the beam that much; there's a support right beside it. It would bend the beam a lot though; so the maximum deflection would be somewhere to the left of the load. The load is causing bending along with displacement. This bending is what makes the beam deform more further to the left.
– JMac
Jan 4 '17 at 11:57
• @kelvinmacks I honestly don't really get where you're getting lost there. They use equation 5 to solve for the location of D (by finding where the slope is 0).
– JMac
Jan 4 '17 at 14:14

Engineers, by design, are lazy. If you are looking at a beam, where you can see that under a given load it deflects most at a certain point, then you can say that it deflects most at that point. It's like when you have symmetry conditions to only have to figure out half of the numbers for a beam, and extrapolating.

Textbook writers say 'by inspection' when they can use their engineering intuition and what they believe to be common sense in order to save labor.

But if you only had the beam, some forces, and we knew that the beam would bend downwards, bottom out, and start to bend back up to the other end, it's intuitive that if you rolled a marble from either end of the beam, it would roll down and settle on the lowest point, where it had a max deflection and a slope of zero.

Now, if there was a beam with multiple forces pointing up and down, with the beam looking like a ride at Disney Land, then that would be a different story. Then, you would have to go through more calculations and really get that engineering intuition into action.

But most often, engineering intuition is knowing how to use the minimum computation for the required results. The author simply said, there is only one global and local minimum, and said, "close enough for government work."


text:
## Evidence for s-channel single top quark production in $p\bar{p}$ collisions at sqrt(s) = 1.96 TeV    [PDF]

D0 Collaboration
We present measurements of the cross sections for the two main production modes of single top quarks in $p\bar{p}$ collisions at $\sqrt{s} = 1.96$\;TeV in the Run II data collected with the D0 detector at the Fermilab Tevatron Collider, corresponding to an integrated luminosity of 9.7\;fb$^{-1}$. The s-channel cross section is measured to be $\sigma({\ppbar}{\rargap}tb+X) = 1.10^{+0.33}_{-0.31}$\;pb with no assumptions on the value of the t-channel cross section. Similarly, the t-channel cross section is measured to be $\sigma({\ppbar}{\rargap}tqb+X) = 3.07^{+0.54}_{-0.49}$\;pb. We also measure the $s+t$ combined cross section as $\sigma({\ppbar}{\rargap}tb+X, tqb+X) = 4.11^{+0.60}_{-0.55}$\;pb and set a lower limit on the CKM matrix element $|V_{tb}| > 0.92 at 95% C.L., assuming$m_t=172.5$\;GeV. The probability to measure a cross section for the s channel at the observed value or higher in the absence of signal is$1.0\times 10^{-4}\$, corresponding to a significance of 3.7 standard deviations.
View original: http://arxiv.org/abs/1307.0731


text:
# Package FiniteFields

In a Mathematica session, I evaluate:

Needs["FiniteFields"];
fld = GF[2];

Now, for example, I'd like to compute $a^3+a^5$ for $a \in{Z_2}$. The result I'd like to have is $0$, because in $Z_2$, $a^n=a$ and $a+a=0$.

I tried:

fld[{a^3}] + fld[{a^2}]

But Mathematica gives me:

Subscript[{Mod[a^2 + a^3, 2]}, 2]

How can I simplify $a^3+a^2$ in $Z_2$ to get zero?

-
My advice conflicts in some ways with that of @m_goldberg. I do not claim that advice was bad-- for some purposes the FiniteFields` package might in fact be what's needed. I only advocate avoiding it when there are alternatives, as it is unmaintained and moreover sometimes will do, in technical parlance, "weird things". – Daniel Lichtblau May 23 '14 at 0:09


text:
# How to calculate the equilibrium constant of acetic acid-hydroxide ion equilibrium?

I am trying to calculate the equilibrium constant of the following equilibrium:

$$\ce{CH3COOH (aq) + OH- (aq) <=> CH3COO- (aq) + H2O (l)}$$

I am aware that this reaction essentially goes to completion (the position of the equilibrium lies far to the right). However, I still would like to calculate the equilibrium constant.

I know that $K_\text{eq} = K_\mathrm{a}(\ce{CH3COOH}) \times K_\mathrm{b}(\ce{OH-)} / K_\mathrm{w}$ (I have derived this from the definitions $K_\mathrm{a}$ and $K_\mathrm{b}$). I know that $K_\mathrm{a}(\ce{CH3COOH}) = 1.8\times10^{-5}$ and that $K_\mathrm{w} = 10^{-14}$. However, I am struggling to determine what $K_\mathrm{b}(\ce{OH-})$ is.

I learnt that $K_\mathrm{b}$ is called the "base ionization constant." However, $\ce{OH-}$ is a base, but it is already an ion! So, how can something that is already an ion, ionize? According to this webpage, $\ce{OH-}$ has a $K_\mathrm{b}$ of 1.0. However, what is the equilibrium that led to this number? Is it $\ce{OH- + H2O <=> H2O + OH-}$?

Moreover, if the $K_\mathrm{b}$ of $\ce{OH-}$ is 1.0, is $\ce{OH-}$ considered to be a weak base? Generally, strong bases have a $K_\mathrm{b}$ that is much greater than 1.0. On the other hand, if bases that contain hydroxide ions, such as sodium hydroxide and potassium hydroxide, are strong bases, shouldn't the hydroxide ion itself also be a strong base?

• Interesting question! Though in the third step you've written $K_\text{eq} = K_\mathrm{a}(\ce{CH3COOH}) \times K_\mathrm{b}(\ce{OH-)} / K_\mathrm{w}$ and said that "I have derived this from the definitions $K_\mathrm{a}$ and $K_\mathrm{b}$" That means you know the definition of $K_\mathrm{b}$ (base dissociation constnat) for $\ce{OH-}$ ion, don't you? If not, then how did you insert it into that formula in the first place? – Gaurang Tandon Jun 7 '18 at 3:59
• Good point. I shouldn't have made that assumption. I have used the equation Ka(HA) * Kb(B) = Keq * Kw in the past for acid base reactions of the form HA + B <=> BH+ + A-, and I assumed that it would work for the acetic acid - hydroxide ion equilibrium. – user62238 Jun 8 '18 at 3:16

Firstly, to derive this equilibrium constant ($K_\text{eq}$), you don't need to bother about $K_\mathrm{b}$ of $\ce{OH-}$. There is a a far simpler way of doing it.
So from the equation, $\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}$, you can easily write, $$K_{\text{eq}} = \frac{\ce{[CH_3COO^-]}}{\ce{[CH_3COOH][OH-]}} = \frac{\ce{[CH_3COO^-]}[\ce{H^+}]}{\ce{[CH_3COOH]}} \times \frac{1}{\ce{[H^+][OH^-]}}= \frac{K_{\mathrm{a}}(\mathrm{CH_3COOH})}{K_\mathrm{w}}$$ So, here no $K_\mathrm{b}$ of $\ce{OH^-}$ arises and therefore we don't have to bother about it.
Secondly, if we want to find out $K_\mathrm{b}$ of $\ce{OH^-}$, At $\pu{25^o C}$, $\mathrm{p}K_\mathrm{w} = 14$, and from the equation of dissociation of water, $\ce{H2O(l) <=> H+(aq) + OH-(aq)}$, we can consider water as a weak acid, and $K_\mathrm{a} = \ce{[H^+][OH^-]} = K_\mathrm{w}$, so, $\mathrm{p}K_\mathrm{a}= \mathrm{p}K_\mathrm{w}$, and from the relation $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}$ (note that in the relation the definitions are that $\mathrm{p}K_\mathrm{a}$ is the acid dissociation constant of a weak acid and $\mathrm{p}K_\mathrm{b}$ is the base dissociation constant of its conjugate base). The conjugate base of water is $\ce{OH-}$, so $\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{a} = 0$ as, both are same for water. So, $K_\mathrm{w}= 10^{-\mathrm{p}K_\mathrm{a}} = 10^0 =1.$ Now, you can see why $K_\mathrm{b}$ of $\ce{OH-}$ is 1 at $\pu{25^o C}$
Thirdly, yes, $\ce{OH^-}$ can't be said as a strong base. It's a moderately strong base. And, the concept that $\ce{NaOH}$ or $\ce{KOH}$ being a strong base should imply $\ce{OH^-}$ should be strong is a totally wrong concept. There is no logic behind it. Those are ionic compounds and completely dissociates and, those are Arrhenius bases. But to become a base, $\ce{OH^-}$ should accept a proton, and hence its Brønsted base, and there can't be a comparison as such. If you have a weak Brønsted acid, its conjugate base must be strong and vice versa. But here the cases are completely different, and there can't be any direct implication.
• Not at all. If they have to behave as Bronsted-Lowry Base, the whole molecule should undergo reaction like $$\ce{NaOH + H^+ <=> NaOH2^+}$$ Does such equilibrium ever exist ??? – Soumik Das Jun 8 '18 at 8:49


text:
# Monsoons and Mumbai

Every year towards the end of May, the city of Mumbai begins to change. People start to prepare for the coming of the Monsoon, which is itself derived from the word “mausam” in Arabic which roughly translates as a season. Perhaps in no other place than Mumbai, the might of monsoon is felt by such a large and varied section of people. All the residents, in all the areas of Mumbai, irrespective of their class do prepare for the annual coming of the monsoon. In Mumbai, everywhere you would see temporary structures being erected, plastic sheets of many colours and the most prominent blue tarpaulins being used to cover myriad of houses. It seems that the city itself is getting ready to greet the incoming monsoon. The various preparations done by people of all classes reminds one of festive preparation that families do. Some buy waterproofing for their bodies, some for their homes and shops and some for their vehicles. Some, who are very rich, even buy waterproofing for their pets. Government bodies like the BMC clean up the gutters and nallahs, which were freshwater rivers once, hard to believe when you look at their present states. You can see piles of cleaned up garbage and debris stacked neatly alongside the gutters manholes and nallahs. They say it helps in preventing floods. So, every year, like a seasonal change, you will see people cleaning them, and stacking the filth accumulated in these water bodies, both above the ground and underground in neat stacks.

Business establishments erect skeletons of bamboos and timber around the areas which are not covered. This happens in the case of hotels which have open-air seating areas. Then on these skeletons, they apply sheaths of water-proofing materials. At the end of it, the area is ready for use even in the heaviest rains. Shops which would sell, clothes and other stuff in other seasons, have umbrellas in their display. Even the street and traffic signal vendors start selling them. The umbrellas come in a variety of sizes, colours and designs. Males mostly carry black ones, while the females carry ones with a variety of colours. The umbrella is the protection of the commons against the rain. In the crowded local trains and buses, it is not seen kindly if you wear a raincoat and enter and transfer your wetness to others in the process. Umbrellas, on the other hand, can be folded and kept inside your bag. These are the kind of umbrellas, which are compact are the ones which are most common. The price range generally indicates the number of seasons that an umbrella will last. Also, the cobblers, who double up as umbrella repairers can be seen fixing broken ribs, handles and mechanisms of old umbrellas. Then there are the large ones, which you need to carry in hand, for they are larger than any bags you would usually carry. I myself carry one of the largest sizes available. Because when it really pours, the compact ones, though easy to pack and carry, aren’t going to protect you from getting wet in the Mumbai monsoon.

Then there is the rain footwear. The idea is that in the rains leather (even faux leather) will get damaged. Hence one should wear something made of rubber or plastic which is not affected by the rains. All the branded footwear companies have a stock of rainwear, which people buy as a preparation for the rain. But this is not for the real elite, they will wear their suedes even in the rains. From the by-lanes of Kurla, there emerge cheap replicas of the designer rainwear which are sold at less than 10th of the prices of the originals. They are sold outside the stations, on the streets, and in the shops. Each year, the designs, patterns and colours change depending on what is in vogue in the market.

People also buy protection for their mobile phones and wallets. Just before the rains pavement sellers, who sell a plethora of mobile trinkets and accessories, also start with selling waterproofing for the mobiles. Same is the case with the bags that working people and school children carry. Those not well off use covers of plastic to cover the valuables inside their bags. Good quality plastics are always in demand for such things. But now with the ban, we don’t know what will happen. Others purchase rain covers for their bags. Even with all these precautions mobiles and stuff inside the bags do get wet. And they get damaged.

They say when you are already anticipating something, the shock value isn’t that much as you would expect. But in the case of Mumbai rains, it is not so. Amongst the places I have lived in, the longest has been in Mumbai. Even then, I consider myself alien to this city, an outsider. I have seen and explored parts of it, yet I do not consider myself as a Mumbaikar. There are two primary reasons for this. The first is that even though after staying for more than a decade, I don’t use public transport for my daily commute. Neither the train nor the bus I use regularly. The two major forms of transport in Mumbai. So far, I have not stayed very far away from my place of work. Hence I do not have to suffer [safar?] daily travel. The other reason is that I have not yet made myself at home with the rains. It is not that I do not enjoy the rains, I do, but only for the first few weeks. Then it becomes torture. A melancholy if you will. The sunless skies for weeks on end are depressing enough, and then you have to prepare for the wetness. No matter how hard you try, the rains will get you. There hasn’t been a season where I haven’t got completely drenched. I have given up.

In case, I am caught unawares, without umbrella or any other protection. I just let the rain do its thing. I don’t fight it. But the city itself is fully prepared for it. Unless it is a literal flood, the city continues its routine as if, nothing has happened. All the supply lines will be working as they are supposed to. Waterlogging will produce delays in traffic and trains, but that is about it. If half the amount of rain that lashes Mumbai every season pours into any other metro it will come to a standstill. But Mumbai has made sure that the services operate despite this amount of rain. It is because of anticipation and will of the people to work.

The area of Mumbai is about $600 \, km^{2}$, and the average rainfall in Mumbai is about $2200 \,mm$. This essentially means that we have rainfall equivalent to about $2.2 \, m$ water present on the entire surface of about $600 \, km^{2}$. So the total volume of water that Mumbai receives each year is

$V_{Rain} = 600 \times 10^{6} \, m^{2} \cdot 2.2 \, m = 1.3 \times 10^{9} \, m^{3}$

As a first approximation we can consider the rain drops to be of uniform size, and the diameter of the drop is about $1 \,mm = 10^{-3} m$. So the volume of the drop is about
$V_{Drop} = \frac{4}{3} \,\pi \,(0.5 \times 10^{-3} m)^{3} = 5.2 \times 10^{-10} \,m^{3}$

Hence in each year the number of raindrops on Mumbai would be

$N = \frac{V_{Rain}}{V_{Drop}} = 2.5 \times 10^{18}$

As a better approximation we can take into account the fact that not all raindrops have the same size. We can then make a distribution of the raindrops according to their size. We can have a distribution of the raindrops according to their size as follows:

Type of
Shower
Diameter
of drops
(mm)
Percentage Volume $m^{3}$
Drizzle 0.5 15 $6.5 \times 10 ^{-11}$
Normal 1 70 $5.2 \times 10 ^{-10}$
Thunder 3 15 $8.1\times 10 ^{-9}$

So the total volume of our rail gets distributed according to the above table. Now we calculate the number of drops for each type of shower:

$V_{Drizzle} = \frac{15}{100} \times 1.3 \times 10^{9} \, m^{3} = 1.9 \times 10^{8} \, m^{3}$

Hence the number of drops in drizzle are:

$N_{Drizzle} = \frac{1.9 \times 10^{8} \, m^{3} }{6.5 \times 10 ^{-11}} = 2.9\times 10^{18}$

Similarly for normal shower we get

$V_{Normal} = \frac{70}{100} \times 1.3 \times 10^{9} \, m^{3} = 9.1 \times 10^{8} \, m^{3}$

Hence the number of drops in normal shower are:

$N_{Normal} = \frac{9.1 \times 10^{8} \, m^{3} }{5.2 \times 10 ^{-10}} = 1.75\times 10^{18}$

Similarly for thunder shower we get

$V_{Thunder} = \frac{15}{100} \times 1.3 \times 10^{9} \, m^{3} = 1.9 \times 10^{8} \, m^{3}$

Hence the number of drops in normal shower are:

$N_{Thunder} = \frac{9.1 \times 10^{8} \, m^{3} }{8.1 \times 10 ^{-9}} = 1.1\times 10^{17}$

So if we add all these up we get the total number of drops:

$N = N_{Thunder} + N_{Normal} + N_{Drizzle} = 1.1\times 10^{17}+ 1.75\times 10^{18}+ 2.9\times 10^{18} = 4.7\times 10^{18}$

This is not very different from our first rough estimate.

# Just for fun or how to invite readers to immerse in your book

These problems are for fun. I never meant them to be taken too seriously. Some you will find easy enough to answer. Others are enormously difficult, and grown men and women make their livings trying to answer them. But even these tough ones are for fun. I am not so interested in how many you can answer as I am in getting you to worry over them.

What I mainly want to show here is that physics is not something that has to be done in a physics building. Physics and physics problems are in the real, everyday world that we live, work, love, and die in. And I hope that this book will capture you enough that you begin to find your own flying circus of physics in your own world. If you start thinking about physics when you are cooking, flying, or just lazing next to a stream, then I will feel the book was worthwhile. Please let me know what physics you do find, along with any corrections or comments on the book. However, please take all this as being just for fun.

From Preface of Jearl Walkers The Flying Circus of Physics

# Anonymity and community in the age of RTI

(Draft under work.)

The year 2005 was a significant event in the history of India, as it
saw the introduction of bill for Right to Information. Under this
act any person could request data from government departments
pertaining to issues of interest. This was seen as first step
towards transparency of government of India, and indeed it was. The
RTI act was a weapon in hands of activists, who could now get the
required information officially. What was important also was that
the RTI query. The information gathered from the various RTI queries
which individuals across the country empowered the citizens about
their rights regarding many issues. Also equally importantly RTI was
a tool for exposing the corruption running rampant in various
government departments. A RTI query could expose the irregularities
and bypassing of rules and regulations, leading to
corruption. There are many famous cases for their expose of big
names and massive corruption which were brought to light. (Some
examples?)

Some people became what are now known as RTI activists. The RTI
activists were whistleblowers of the RTI age. These activists
exposed many scams at local and national levels. Such expose make
the people involved very uncomfortable. And as it happens in many
cases the people involved are very powerful and would not stop at
anything to attain their goals. In many such cases the RTI activists
were “easy” targets. By using the four-fold approach of /saam/,
/daam/, /dand/ and finally /bhed/ those in trouble try to stop the RTI
activists. Unfortunately in many cases, it turns out the activists
were killed as a result of their whistleblowing acts. (Give
examples)

What the RTI activists in various parts lack is a safety net, by
this I mean they lack support from people with similar
interests. They also lack, in a sense, a feeling of community who
will stand by them in case of such bad incidents. What can be done
for them? What kind of mechanism can result in their protection?
There was a proposed whistleblower bill, which if enacted will
provide security to such people. But the bill was not passed. Is
there any other option? Is there a way in which people can still ask
for information under RTI, without revealing their identity, so that
they cannot be “targetted” as whistleblowers. Is there a way in
which despite being anonymous there is a sense of community among
the activists of the RTI? It seems there can be an alternative way in
which we the people can provide a sense of anonymity and at the same
time that of a community to the RTI activists. What follows is
such a proposal which will try to cover these fundamental issues.

The proposal is to setup a front, which for lack of a name I would
call a /collective/ for now, which will mask those requesting RTI
from various government departments. The activists can send their
queries to the government departments through this collective. This
collective will in a sense create a buffer between activists and the
respective government departments, and associated problematic
elements. Since the RTI queries will be sent through this
collective, there will be sense of anonymity for the RTI activists,
as their names wouldn’t come under spotlight as in the earlier
case. One can also have a way in which the activist need not reveal
their identity to collective, that is we should also allow for
anonymous requests for queries. Some may point out that this might
be abused, but we have to give in this possibility hoping that pros
will outnumber the cons. Thus at no point we would have any data
regarding the identity of the applicant, hence we will not be able
to reveal it, when asked. This as will be pointed out can be abused,
but Thus the collective will form a mask for all the RTI
requests. Of coure people who want their name displayed for their
work, will be proud public faces of the collective. It would be much
more difficult to subdue or silence or attack the collective as it
would not be single point contact, but rather a distributed system.

So what do the activists gain by submitting to the collective,
apart from the anonymity. The collective will have a policy that
all the data that it receives from these RTI queries, will be put
up in public domain. This will in turn create a public archive of
information which will be accessible to anyone. Such a archive will
address many issues, like redundancy in filing of RTI queries, or
making the future RTI queries much more pointed and making material
available for researchers.

Such is a very rough proposal for the formation of the
collective. Suggestions and refinements in the proposal and
possible way of execution of this collective are needed.


text:
Power Series - Differential Equation (check my answer)

1. Dec 7, 2009

vucollegeguy

Using the power series method to solve the differential equation
y'+xy=0 when y(0)=1

Write the solution in the form of a power series and then recognize what function it represents.
************************************

$$\sum$$(-1)k*[(x2k)/(2k)*k!]

Is my answer correct?
Is it written in the form of a power series?
What function does it represent?

Thanks to anyone that helps!

2. Dec 7, 2009

Dick

Yes, it's correct. And yes, it's a power series. It looks like exp(f(x)) to me. Can you guess what f(x) might be?

3. Dec 8, 2009

vucollegeguy

I'd guess and say that f(x)=cos(x).
So the function would be f(x)=x2cos(x).

Please correct me if I am wrong.
Thank you for helping thus far.

4. Dec 8, 2009

HallsofIvy

Staff Emeritus
Not quite. Your sum reduces to
$$\sum(-1)^k \frac{\left(\frac{x^2}{2}\right)^2}{k!}$$
which is a "cosine" but it is
$$cos(\frac{x}{2})$$

$$x^2cos(x)$$
would be
$$x^2\sum(-1)^k \frac{x^k}{k!}= \sum (-1)^k \frac{x^{k+2}}{k!}$$

5. Dec 8, 2009

vucollegeguy

Ok -got it.
A big "THANKKK YOUUU!" to all for all of your help!


text:
Comment
Share
Q)

# A carpenter bought a piece of wood that was 0.7 metres long. Then she sawed 0.1 metres off the end. How long is the piece of wood now?

( A ) 0.8 m
( B ) 0.9 m
( C ) 0.6 m
( D ) 0.5 m

Comment
A)
Length of the wood = 0.7 m
Length of the wood sawed = 0.1 m
\begin{align*} \text {Length of the wood left} &= {0.7 - 0.1} \\&= 0.6 m \end{align*}


text:
# Outer space (mathematics)

In the mathematical subject of geometric group theory, the Culler–Vogtmann Outer space or just Outer space of a free groupFn is a topological space consisting of the so-called “marked metric graph structures” of volume 1 on Fn. The Outer space, denoted Xn or CVn, comes equipped with a natural action of the group of outer automorphismsOut(Fn) of Fn. The Outer space was introduced in a 1986 paper[1] of Marc Culler and Karen Vogtmann, and it serves as a free group analog of the Teichmüller space of a hyperbolic surface. Outer space is used to study homology and cohomology groups of Out(Fn) and to obtain information about algebraic, geometric and dynamical properties of Out(Fn), of its subgroups and individual outer automorphisms of Fn. The space Xn can also be thought of as the set of Fn-equivariant isometry types of minimal free discrete isometric actions of Fn on Fn on R-treesT such that the quotient metric graph T/Fn has volume 1.

## . . . Outer space (mathematics) . . .

The Outer space

${displaystyle X_{n}}$

was introduced in a 1986 paper[1] of Marc Culler and Karen Vogtmann, inspired by analogy with the Teichmüller space of a hyperbolic surface. They showed that the natural action of

${displaystyle operatorname {Out} (F_{n})}$

on

${displaystyle X_{n}}$

is properly discontinuous, and that

${displaystyle X_{n}}$

is contractible.

In the same paper Culler and Vogtmann constructed an embedding, via the translation length functions discussed below, of

${displaystyle X_{n}}$

into the infinite-dimensional projective space

${displaystyle mathbb {P} ^{,{mathcal {C}}}=mathbb {R} ^{mathcal {C}}!-!{0}/mathbb {R} _{>0}}$

, where

${displaystyle {mathcal {C}}}$

is the set of nontrivial conjugacy classes of elements of

${displaystyle F_{n}}$

. They also proved that the closure

${displaystyle {overline {X}}_{n}}$

of

${displaystyle X_{n}}$

in

${displaystyle mathbb {P} ^{,{mathcal {C}}}}$

is compact.

Later a combination of the results of Cohen and Lustig[2] and of Bestvina and Feighn[3] identified (see Section 1.3 of [4]) the space

${displaystyle {overline {X}}_{n}}$

with the space

${displaystyle {overline {CV}}_{n}}$

of projective classes of “very small” minimal isometric actions of

${displaystyle F_{n}}$

on

${displaystyle mathbb {R} }$


text:
Stockfuse provides an interactive, graphical view to show your profits and losses on each of your holdings over time:

• Go to your Portfolio Dashboard;
• Click on the Analytics tab and select P&L Attribution in the dropdown menu.

By default, the chart shows the daily P&L for every stock you’ve traded on over the past month; however, the control strip allows you to customize the chart in many different ways. For example, if you want to know the cumulative P&L on everything you’ve traded since the beginning of time, simply click on the Daily button and switch it to Cumulative, and then change the horizon to All Time.

This feature is currently available only on the web.


text:
# Multinomial¶

class paddle.distribution. Multinomial ( total_count, probs ) [source]

Multinomial distribution parameterized by total_count and probs.

In probability theory, the multinomial distribution is a generalization of the binomial distribution, it models the probability of counts for each side of a k-sided die rolled n times. When k is 2 and n is 1, the multinomial is the bernoulli distribution, when k is 2 and n is grater than 1, it is the binomial distribution, when k is grater than 2 and n is 1, it is the categorical distribution.

The probability mass function (PMF) for multinomial is

$f(x_1, ..., x_k; n, p_1,...,p_k) = \frac{n!}{x_1!...x_k!}p_1^{x_1}...p_k^{x_k}$

where, $$n$$ is number of trials, k is the number of categories, $$p_i$$ denote probability of a trial falling into each category, $${\textstyle \sum_{i=1}^{k}p_i=1}, p_i \ge 0$$, and $$x_i$$ denote count of each category.

Parameters
• total_count (int) – Number of trials.

• probs (Tensor) – Probability of a trial falling into each category. Last axis of probs indexes over categories, other axes index over batches. Probs value should between [0, 1], and sum to 1 along last axis. If the value over 1, it will be normalized to sum to 1 along the last axis.

Examples:

import paddle

print(multinomial.sample((2, 3)))
# Tensor(shape=[2, 3, 3], dtype=float32, place=Place(gpu:0), stop_gradient=True,
#        [[[1., 4., 5.],
#          [0., 2., 8.],
#          [2., 4., 4.]],

#         [[1., 6., 3.],
#          [3., 3., 4.],
#          [3., 4., 3.]]])

probs ( value )

Probability density/mass function.

Note

This method will be deprecated in the future, please use prob instead.

property mean

mean of multinomial distribuion.

Returns

mean value.

Return type

Tensor

property variance

variance of multinomial distribution.

Returns

variance value.

Return type

Tensor

prob ( value )

probability mass function evaluated at value.

Parameters

value (Tensor) – value to be evaluated.

Returns

probability of value.

Return type

Tensor

log_prob ( value )

probability mass function evaluated at value

Parameters

value (Tensor) – value to be evaluated.

Returns

probability of value.

Return type

Tensor

sample ( shape=() )

draw sample data from multinomial distribution

Parameters

sample_shape (tuple, optional) – [description]. Defaults to ().

entropy ( )

entropy of multinomial distribution

Returns

entropy value

Return type

Tensor

property batch_shape

Returns batch shape of distribution

Returns

batch shape

Return type

Sequence[int]

property event_shape

Returns event shape of distribution

Returns

event shape

Return type

Sequence[int]

kl_divergence ( other ) [source]

The KL-divergence between self distributions and other.

rsample ( shape=() )

reparameterized sample


text:
# LightsOut - A homework framework for teaching CS to non-engineers

This provides a fill-in-the-blanks homework framework to implement the game "LightsOut".

LightsOut is a game with a grid of lights. When you touch a light, it toggles it’s state; it ALSO toggles the state of the lights adjacent to it (in a cross pattern). The goal is to turn all the lights off.

In addition to the normal game operation, each cell will have a number of uses/clicks until it is disabled. The goal is not readability/extensibility in the scaffolding section as students will not concern themselves with reading it; rather the focus is on making it hard for students to get lost while they try to learn arrays and custom types.

• Two distinct parts, solution and scaffolding. Students are supposed to fill in the solution and should have no reason to touch or look at the scaffolding
• Solution should be able to be found sequentially without forced gaps where code is unable to be run
• Whole project must be in one file (file management complexity)
• UI independant (no need to use Visual Studio's UI editor & helps with distribution)
• Actual solution to the LightsOut game must use arrays and structures
• Scaffolding should be resilient to simple typos
• Comments should be clear but not verbose (jargon tends to get tuned out)
• Scaffolding should not hinder performance to the point of inconvenience
• Length. While this scaffolding implementation turned out fairly long as more features were added, shorter would be better as it simply distracts less from the assignment
• Assignment will be paired with a handout, however I'd like the comments to be nearly sufficient

Visual Basic.NET code follows:

Option Strict On
Option Explicit On
Public Class Form1
' The board is square so this should be a perfect square number. (it will be rounded down to the nearest perfect square)
Const sizeOfBoard As Integer = 25
Dim sizeOfBoardEdge As Integer = CInt(Math.Floor(Math.Sqrt(sizeOfBoard)))

' PART 1: Make a custom structure with two variables inside. A Boolean named boolLightOn and an Integer named intClicksRemaining
' Put your structure declaration here ↓

' Put your structure declaration here ↑
' When you make your custom structure, change the type of this array. It should end with:    As YourStructureName
Private boardArray(sizeOfBoard - 1) As Boolean

Me.Text = "Lights Out!"
' If you want to setup the board (maybe randomize it?) you can do that here
' PART 1: Write a loop that makes all the lights start ON (Right now they start off because Booleans are false by default)

RedrawBoard() ' This will draw the board for the first time
End Sub

Public Sub ButtonGotClicked(ByVal sender As Object, ByVal e As System.EventArgs)
' You can use intButtonID to get the number of the button clicked
Dim intButtonID As Integer = CInt(CType(sender, Button).Tag)
' Since each row is sizeOfBoardEdge long, you can divide and round down to find the row number
Dim intRowCoordinate As Integer = intButtonID \ sizeOfBoardEdge
' Since each row is sizeOfBoardEdge long, you can divide and get the remainder to find the column number
Dim intColumnCoordinate As Integer = intButtonID Mod sizeOfBoardEdge

' ============================================================================================================
' Your code goes here in this area

' PART 1: When you add your custom structure edit the line below to use boolLightOn
boardArray(intButtonID) = Not boardArray(intButtonID) ' This updates boardArray to change the button you clicked
' PART 2: When you click a button, you should subtract 1 from the clicks remaining.

' PART 2: When you click a button it should toggle the buttons adjacent to it in a cross shape

' You can delete this when you understand what is going on
MessageBox.Show("You clicked button number: " + Convert.ToString(intButtonID) +
" Coordinate: (" + Convert.ToString(intRowCoordinate) + ", " + Convert.ToString(intColumnCoordinate) + ")")

' Leave everything below this line alone
' ============================================================================================================
RedrawBoard() ' Now that you have updated boardArray we have to redraw the board
End Sub

' You don't need to look at this function. Just call RedrawBoard() when you need to update the display
Sub RedrawBoard()
Me.Size = New Size(sizeOfBoardEdge * 66, sizeOfBoardEdge * 66)
Me.ClientSize = New Size(sizeOfBoardEdge * 66, sizeOfBoardEdge * 66)
Dim Btns(sizeOfBoardEdge * sizeOfBoardEdge - 1) As Button
Dim table As TableLayoutPanel = New TableLayoutPanel
Dim type = boardArray.GetType()
Dim stateList As List(Of Boolean) = New List(Of Boolean)
Dim scoreList As List(Of Integer) = New List(Of Integer)
Console.WriteLine("Type was: " + type.ToString)
Console.WriteLine(type.IsValueType)
Console.WriteLine(type.IsPrimitive)
Console.WriteLine(type.Namespace.StartsWith("System"))
Console.WriteLine(type.IsEnum)
table.Size = Me.ClientSize

For i As Integer = 0 To Btns.Length - 1
Btns(i) = New Button
Btns(i).Anchor = AnchorStyles.Bottom Or AnchorStyles.Left Or AnchorStyles.Right Or AnchorStyles.Top
If Not type.IsValueType And Not type.IsPrimitive And Not type.Namespace.StartsWith("System") And Not type.IsEnum Then
' This is for boardArray As Tile
Console.WriteLine("Struct: " + (i).ToString)
Dim item = boardArray(i)
Try
Dim clicksRemaining = CInt(CallByName(item, "intClicksRemaining", CallType.Get, Nothing))
Btns(i).Text = (clicksRemaining).ToString
Console.WriteLine("    intClicksRemaining: " + (clicksRemaining).ToString)
Catch ex As MissingMemberException
MessageBox.Show("Your custom struct did not contain a Integer named intClicksRemaining. Array index: " + (i).ToString, "Error", MessageBoxButtons.OK, MessageBoxIcon.Stop)
End
End Try
Try
Dim lightOn = CallByName(item, "boolLightOn", CallType.Get, Nothing)
Dim enabled = CBool(lightOn)
Btns(i).ForeColor = If(enabled, Color.Black, Color.White)
Btns(i).BackColor = If(enabled, Color.White, Color.Black)
Console.WriteLine("    boolLightOn: " + (enabled).ToString)
Catch ex As MissingMemberException
MessageBox.Show("Your custom struct did not contain a Boolean named boolLightOn. Array index: " + (i).ToString, "Error", MessageBoxButtons.OK, MessageBoxIcon.Stop)
End
End Try
Else
' This is for boardArray As Boolean
Dim enabled = CBool(CObj(boardArray(i)))
Btns(i).BackColor = If(enabled, Color.White, Color.Black)
End If
Btns(i).Tag = i
Btns(i).Size = New Size(60, 60)
table.Controls.Add(Btns(i), i Mod sizeOfBoardEdge, i \ sizeOfBoardEdge)
Next
Me.Controls.Clear()

Dim wonTheGame As Boolean = Not stateList.Contains(True)
Dim lostTheGame As Boolean = True
For Each x In scoreList
If x <> 0 Then
lostTheGame = False
Exit For
End If
Next

If lostTheGame Then
For Each btn In Btns
btn.Enabled = False
Next
MessageBox.Show("Sorry you ran out of moves.", "You lost", MessageBoxButtons.OK, MessageBoxIcon.Exclamation)
End
End If
If wonTheGame Then
Dim score = 0
For Each btn In Btns
btn.Enabled = False
Try
Dim clicksRemaining = CInt(CallByName(btn, "intClicksRemaining", CallType.Get, Nothing))
score += clicksRemaining
Catch ex As Exception
End Try
Next
MessageBox.Show("WINNER: Your score was: " + score.ToString, "You won", MessageBoxButtons.OK, MessageBoxIcon.Exclamation)
End If
End Sub

Sub DiscoParty(btns() As Button)
While (True)
For Each btn In btns
btn.BackColor = getRandomColor()
Next
End While
End Sub

Function getRandomColor() As Color
Return Color.FromArgb(getRandomColorChannel(), getRandomColorChannel(), getRandomColorChannel())
End Function

' Generate Random number in range 0 ... 255
' https://msdn.microsoft.com/en-us/library/f7s023d2(v=vs.90).aspx
Function getRandomColorChannel() As Integer
Const upperbound = 255
Const lowerbound = 0
Return CInt(Math.Floor((upperbound - lowerbound + 1) * Rnd())) + lowerbound
End Function
End Class

Questions:

• Does the solution fit the requirements?
• Where can this be improved to provide a better experience to a non-savvy intro student?
• Where can the performance be improved? Some of the students run in VMs (as do I) and some run on extremely outdated hardware.
• Win/lose checking code definitely smells, cleaner solution?
• Can any of the comments be reworded to be clearer?
• (minor) General suggestions to naming, design, and refactoring?

Obviously, please do not solve the actual homework.


text:
# Revision history [back]

You should definitely add a code snippet...

However, if I'm correctly understanding you have a problem with the following code:

class MyNode : public rclcpp::Node
{
MyNode() : Node("my_node")
{


text:
Math Help - Standard Deviation on Results of Stock Trades

1. Standard Deviation on Results of Stock Trades

I have calculated the expected value and standard deviation for a trading strategy for the financial markets, but think that my standard deviation is too low.

Assume the following:
A trading strategy has the folllowing possible outcome:
probability of winning trade = 50%, win will be $1,500 probability of losing trade = 50%, loss will be -$1,000

EV = 0.5 x $1,500 + 0.5 x -$1,000 = $750 + -$500 = $250 Is the standard deviation equal to$883.88? I used the =STDEVA formula in MS Excel i.e.=STDEVA( 750, -500) - is this correct?

If I then put the trade on 100 times will my EV and SD be as follows?
EV = $250 x 100 =$25,000
SD = sqrt(100) x $883.88 =$8,839

2. Originally Posted by george12345
I have calculated the expected value and standard deviation for a trading strategy for the financial markets, but think that my standard deviation is too low.

Assume the following:
A trading strategy has the folllowing possible outcome:
probability of winning trade = 50%, win will be $1,500 probability of losing trade = 50%, loss will be -$1,000

EV = 0.5 x $1,500 + 0.5 x -$1,000 = $750 + -$500 = $250 Is the standard deviation equal to$883.88? I used the =STDEVA formula in MS Excel i.e.=STDEVA( 750, -500) - is this correct?

If I then put the trade on 100 times will my EV and SD be as follows?
EV = $250 x 100 =$25,000
SD = sqrt(100) x $883.88 =$8,839
I agree with your expected value but not the standard deviation.

E(x)=$250 E(x^2)=1500^2 x 0.5 + (-1000)^2 x 0.5 = 1625000 Var(x) = 1625000 -250^2 = 1562500 SD =$ 1250

The last part is correct after you modify the SD.

3. hi mathaddict, thanks for your quick response. Your explanation is very helpful and it also makes sense. After one trial the SD should be $1,250 (the difference between -$500 and +$750). I couldn't make sense of the meaning of my$883.88 figure!

Thanks again!

4. Originally Posted by george12345
hi mathaddict, thanks for your quick response. Your explanation is very helpful and it also makes sense. After one trial the SD should be $1,250 (the difference between -$500 and +$750). I couldn't make sense of the meaning of my$883.88 figure!

Thanks again!
That's the SD of -500 and 750.

What you are asked to calculate is the SD of the grouped data

X ............... P(X=x)

-1000 ............... 0.5

1500 ................ 0.5

Of course, you are welcome.


text:
If you think that the regressors or the errors are likely to be uncorrelated within a potential group, then there is no need to cluster within that group. I found myself writing a long-winded answer to a question on StatsExchange about the difference between using fixed effects and clustered errors when running linear regressions on panel data. (using Stata). To learn more, see our tips on writing great answers. “Can save energy” denotes the degree to which the respondent is confident about her ability to reduce energy, consumption - 0 (not at all) to 10 (completely confident). I have 19 countries over 17 years. work” contains the reported frequency of being involved in work for, voluntary/charitable organizations in the last 12 months, ranging from 1 (never) to 6 (at least once a week), and “Supports fuel tax” captures the support for, a tax on fuels, ranging from 1 (strongly against) to 5 (strongly in favor). Does software exist to automatically validate an argument? For example, errors may be clustered by country and by city, or errors may be clustered by country and by year. Note #2: While these various methods yield identical coefficients, the standard errors may differ when Stataâs cluster option is used. But, to conclude, Iâm not criticizing their choice of clustered standard errors for their example. For example, looking at outcomes for students nested in classrooms, there are clear factors that all students within a class are exposed to that vary between classes (e.g. But, it is in theory possible to have $\mathbf{E} \left[ \hat{u}_{jc} \hat{u}_{kc} \right] < 0$ - i.e. Heteroscedasticity robust standard errors with the PLM package. I have a cross-sectional dataset with M&A acquirers and study performance. It is counterproductive to read very long text books during an MSc program. Notice that where $j = k$ in the inner two summations, we'll be multiplying a residual by itself, so that will always be positive. As one of the comments to your question noted, clustering at the larger level is more conservative (and likely better). Then there is no need to adjust the standard errors for clustering at all, even if clustering would change the standard errors. She therefore assigns teachers in "treated" classrooms to try this new technique, while leaving "control" classrooms unaffected. It seems to me (and Iâm about as formally untrained in statistics as one can getâitâs a long story) that both OLS with clustered standard errors and hierarchical modeling are intended to address the same problem: the correlation of residuals within a cluster (be it a state, in some of your research, or a country, as in my research). But fixed effects do not affect the covariances between residuals, which is solved by clustered standard errors. As they say, the innocent (i.e. Then model errors for individuals in the same region may be correlated, while model errors for individuals in different regions are assumed to be uncorrelat ed. sport, culture, union, professional, consumer, humanitarian, environmental, religious, political, science and other) is reported, and 0 otherwise. I know that this is the criterium for the choice of the cluster level. Can anyone please explain me the need > then to cluster the standard errors at the firm level? Note that if each observation had its own cluster, then the cluster estimator would be identical to the HW estimator. When analyzing her results, she may want to keep the data at the student level (for example, to control for student-level obsâ¦ By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. But, to the extent that we have $\mathbf{E} \left[ \hat{u}_{jc} \hat{u}_{kc} \right] \neq 0$ for $j \neq k$ within the cluster, then the two will differ. Focus on the middle term (the three nested summations) in the cluster estimator. One leading example of âclustered errorsâ is in dividual-level cross-section data with clustering on geographical region, such as village or state. If the errors within a cluster are indeed independent of each other, than we should have, by definition of that independence, $\mathbf{E} \left[ \hat{u}_{jc} \hat{u}_{kc} \right] = 0$ for $j \neq k$. Thus, we'll always have the standard sum of $\hat{u}_{jc}^2 x_{jc}x'_{jc}$ like we do in the basic HW estimator. *** < 0.01, ** p< 0.05, * p< 0.1 Note: Exogenous controls include whether a cadet is black or Hispanic, GPA, SAT math and verbal scores, cadet leadership score, cadet ï¬tness aptitude, and recruited NCAA athlete. Table A.5: Regression estimates for pro-environmental outcomes II, significant at the 1%, 5%, and 10%-level, respectively. How to decide on the clustering of standard errors? Clustered Standard Errors 1. Now, compare this to the HW robust variance estimator, that does not account for clustering: $$\widehat{\text{var}} \left( \hat{\beta} \right)_{HW} = (X'X)^{-1} \left[ \sum_{i=1}^n \hat{u}_i^2 x_ix_i' \right] (X'X)^{-1}$$. Course Hero is not sponsored or endorsed by any college or university. All regions are part of a country (~12 countries). Trust = 5 (average trust in the ESS) and Trust = 9 (largest point estimate in a regression of life satisfaction on trust). White standard errors (with no clustering) had a simulation standard deviation of 1.4%, and single-clustered standard errors had simulation standard deviations of 2.6%, whether clustering was done by firm or time. > > Different assumptions are involved with dummies vs. clustering. Related. Robust standard errors account for heteroskedasticity in a modelâs unexplained variation. Thanks for contributing an answer to Cross Validated! Making statements based on opinion; back them up with references or personal experience. Clustered standard errors belong to these type of standard errors. the residuals from two different regions in a country), then we'll have, at least in expectation, $\mathbf{E} \left[ \hat{u}_{jc} \hat{u}_{kc} \right] = 0$ for $j \neq k$. “Supports energy subs.” measures. Thanks for your response! site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. “CC resp.” is the degree to which reducing climate change is felt as a personal responsibility, - 0 (not at all) to 10 (a great deal). âSupports energy subs.â measures the support for subsidies on renewable energy to reduce climate change - â¦ It is meant to help people who have looked at Mitch Petersen's Programming Advice page, but want to use SAS instead of Stata.. Mitch has posted results using a test data set that you can use to compare the output below to see how well they agree. It only takes a minute to sign up. This preview shows page 42 - 44 out of 103 pages. For example, suppose that an educational researcher wants to discover whether a new teaching technique improves student test scores. $\endgroup$ â paqmo May 21 '17 at 15:50 That is, if the amount of variation in the outcome variable is correlated with the explanatory variables, robust standard errors can take this correlation into account. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. @paqmo do you mean that if you cluster at the regional level the standard errors will be larger? MathJax reference. How would you approach this? the environmental organization (rather than being a passive member only). “Char. Standard Errors are clustered at the tactical ofï¬cer level. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Standard errors clustered at the country level in parentheses Envir org member. “p(Trust 9 = Trust 5)” denotes the p-value of an F-test for equality of coefficients for Trust = 5 (average. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. here are a couple of solutions: Clustered standard errors at the group level; Clustered bootstrap (re-sample groups, not individual observations) and they indicate that it is essential that for panel data, OLS standard errors be corrected for clustering on the individual. The coef_test function from clubSandwich can then be used to test the hypothesis that changing the minimum legal drinking age has no effect on motor vehicle deaths in this cohort (i.e., $$H_0: \delta = 0$$).The usual way to test this is to cluster the standard errors by state, calculate the robust Wald statistic, and compare that to a standard normal reference distribution. That's probably the case, just wanted to be sure. Clustered standard errors are often useful when treatment is assigned at the level of a cluster instead of at the individual level. Typically, the motivation given for the clustering adjustments is that unobserved components in outcomes for units within clusters are correlated. What would be a good way to decide on this? Why does air pressure decrease with altitude? Standard errors clustered at the country level in parentheses. The difference is in the degrees-of-freedom adjustment. Serially Correlated Errors Learn more about Bowling Alone: The Collapse and Revival of American Community with Course Hero's FREE study guides and Since the errors are unobserved and a characteristic of the underlying population, there is no straight forward trick to determine the level to cluster. Grouped Errors Across Individuals 3. A: The author should cluster at the most aggregated level where the residual could be correlated. See Bertrand, Duflo, and Mullainathan for a longer discussion of this. Why does using \biggl \biggl not throw an error? In that case, those terms will add to the variance of your estimator, but that is the appropriate thing in such a situation, since if your observations and errors are correlated, then you don't actually have as many independent observations as your simple sample size would indicate. Adjusting for Clustered Standard Errors. There's no formal test that will tell you at which level to cluster. 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Comfortable with  clustered standard errors country level '' classrooms to try this new technique, while leaving  control classrooms... Writing great answers when Stataâs cluster option is used component of statistical inference account... Strategies by first explaining what it is essential that for panel data, OLS errors. Are correlated what should drive the approach while leaving  control '' classrooms to try this new,... Decide on the middle term ( the three nested summations ) in substitutions in the when! $C$ clusters, each containing $N_c$ observations at the individual level space... Explaining what it is common to report standard errors clustered at the larger is. Correlation across entities ( firm or industry or country ) 44 out of 103 pages I am already adding and! The environment - 1 ( not at all, even if clustering would change the errors!, but focuses on having Donated to such an organization policy and cookie policy 's formal. 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text:
# Why do the owner of the house and the governor accede to the demands of the missionary?

Question from Rain by Somerset Maugham

##### Answers 1

The quartermaster bargains for a reduction in price,

“I tell you shes a friend of mine, Jo,” said the quartermaster. “She cant pay more than a dollar, and you`ve sure got to take her for that.”

I think the trader was happy to rent the rooms for the right price.


text:
# 8.3 Polar coordinates  (Page 5/8)

 Page 5 / 8

How are the polar axes different from the x - and y -axes of the Cartesian plane?

Explain how polar coordinates are graphed.

Determine $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ for the point, then move $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ units from the pole to plot the point. If $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is negative, move $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ units from the pole in the opposite direction but along the same angle. The point is a distance of $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ away from the origin at an angle of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ from the polar axis.

How are the points $\text{\hspace{0.17em}}\left(3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ related?

Explain why the points $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3,-\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ are the same.

The point $\text{\hspace{0.17em}}\left(-3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ has a positive angle but a negative radius and is plotted by moving to an angle of $\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$ and then moving 3 units in the negative direction. This places the point 3 units down the negative y -axis. The point $\text{\hspace{0.17em}}\left(3,-\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ has a negative angle and a positive radius and is plotted by first moving to an angle of $\text{\hspace{0.17em}}-\frac{\pi }{2}\text{\hspace{0.17em}}$ and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y -axis.

## Algebraic

For the following exercises, convert the given polar coordinates to Cartesian coordinates with $\text{\hspace{0.17em}}r>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}0\le \theta \le 2\pi .\text{\hspace{0.17em}}$ Remember to consider the quadrant in which the given point is located when determining $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ for the point.

$\left(7,\frac{7\pi }{6}\right)$

$\left(5,\pi \right)$

$\left(-5,0\right)$

$\left(6,-\frac{\pi }{4}\right)$

$\left(-3,\frac{\pi }{6}\right)$

$\left(-\frac{3\sqrt{3}}{2},-\frac{3}{2}\right)$

$\left(4,\frac{7\pi }{4}\right)$

For the following exercises, convert the given Cartesian coordinates to polar coordinates with $\text{\hspace{0.17em}}r>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le \theta <2\pi .\text{\hspace{0.17em}}$ Remember to consider the quadrant in which the given point is located.

$\left(4,2\right)$

$\left(-4,6\right)$

$\left(3,-5\right)$

$\left(\sqrt{34},5.253\right)$

$\left(-10,-13\right)$

$\left(8,8\right)$

$\left(8\sqrt{2},\frac{\pi }{4}\right)$

For the following exercises, convert the given Cartesian equation to a polar equation.

$x=3$

$y=4$

$r=4\mathrm{csc}\theta$

$y=4{x}^{2}$

$y=2{x}^{4}$

$r=\sqrt[3]{\frac{sin\theta }{2co{s}^{4}\theta }}$

${x}^{2}+{y}^{2}=4y$

${x}^{2}+{y}^{2}=3x$

$r=3\mathrm{cos}\theta$

${x}^{2}-{y}^{2}=x$

${x}^{2}-{y}^{2}=3y$

$r=\frac{3\mathrm{sin}\theta }{\mathrm{cos}\left(2\theta \right)}$

${x}^{2}+{y}^{2}=9$

${x}^{2}=9y$

$r=\frac{9\mathrm{sin}\theta }{{\mathrm{cos}}^{2}\theta }$

${y}^{2}=9x$

$9xy=1$

$r=\sqrt{\frac{1}{9\mathrm{cos}\theta \mathrm{sin}\theta }}$

For the following exercises, convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented.

$r=3\mathrm{sin}\text{\hspace{0.17em}}\theta$

$r=4\mathrm{cos}\text{\hspace{0.17em}}\theta$

${x}^{2}+{y}^{2}=4x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\frac{{\left(x-2\right)}^{2}}{4}+\frac{{y}^{2}}{4}=1;$ circle

$r=\frac{4}{\mathrm{sin}\text{\hspace{0.17em}}\theta +7\mathrm{cos}\text{\hspace{0.17em}}\theta }$

$r=\frac{6}{\mathrm{cos}\text{\hspace{0.17em}}\theta +3\mathrm{sin}\text{\hspace{0.17em}}\theta }$

$3y+x=6;\text{\hspace{0.17em}}$ line

$r=2\mathrm{sec}\text{\hspace{0.17em}}\theta$

$r=3\mathrm{csc}\text{\hspace{0.17em}}\theta$

$y=3;\text{\hspace{0.17em}}$ line

$r=\sqrt{r\mathrm{cos}\text{\hspace{0.17em}}\theta +2}$

${r}^{2}=4\mathrm{sec}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}\theta$

$xy=4;\text{\hspace{0.17em}}$ hyperbola

$r=4$

${r}^{2}=4$

${x}^{2}+{y}^{2}=4;\text{\hspace{0.17em}}$ circle

$r=\frac{1}{4\mathrm{cos}\text{\hspace{0.17em}}\theta -3\mathrm{sin}\text{\hspace{0.17em}}\theta }$

$r=\frac{3}{\mathrm{cos}\text{\hspace{0.17em}}\theta -5\mathrm{sin}\text{\hspace{0.17em}}\theta }$

$x-5y=3;\text{\hspace{0.17em}}$ line

## Graphical

For the following exercises, find the polar coordinates of the point.

$\left(3,\frac{3\pi }{4}\right)$

$\left(5,\pi \right)$

For the following exercises, plot the points.

$\left(-2,\frac{\pi }{3}\right)$

$\left(-1,-\frac{\pi }{2}\right)$

$\left(3.5,\frac{7\pi }{4}\right)$

$\left(-4,\frac{\pi }{3}\right)$

$\left(5,\frac{\pi }{2}\right)$

$\left(4,\frac{-5\pi }{4}\right)$

$\left(3,\frac{5\pi }{6}\right)$

$\left(-1.5,\frac{7\pi }{6}\right)$

$\left(-2,\frac{\pi }{4}\right)$

$\left(1,\frac{3\pi }{2}\right)$

For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis.

$5x-y=6$

$r=\frac{6}{5\mathrm{cos}\theta -\mathrm{sin}\theta }$

$2x+7y=-3$

${x}^{2}+{\left(y-1\right)}^{2}=1$

$r=2\mathrm{sin}\theta$

${\left(x+2\right)}^{2}+{\left(y+3\right)}^{2}=13$

$x=2$

$r=\frac{2}{\mathrm{cos}\theta }$

${x}^{2}+{y}^{2}=5y$

${x}^{2}+{y}^{2}=3x$

$r=3\mathrm{cos}\theta$

For the following exercises, convert the equation from polar to rectangular form and graph on the rectangular plane.

$r=6$

$r=-4$

${x}^{2}+{y}^{2}=16$

$\theta =-\frac{2\pi }{3}$

$\theta =\frac{\pi }{4}$

$y=x$

$r=\mathrm{sec}\text{\hspace{0.17em}}\theta$

$r=-10\mathrm{sin}\text{\hspace{0.17em}}\theta$

${x}^{2}+{\left(y+5\right)}^{2}=25$

$r=3\mathrm{cos}\text{\hspace{0.17em}}\theta$

## Technology

Use a graphing calculator to find the rectangular coordinates of $\text{\hspace{0.17em}}\left(2,-\frac{\pi }{5}\right).\text{\hspace{0.17em}}$ Round to the nearest thousandth.

$\left(1.618,-1.176\right)$

Use a graphing calculator to find the rectangular coordinates of $\text{\hspace{0.17em}}\left(-3,\frac{3\pi }{7}\right).\text{\hspace{0.17em}}$ Round to the nearest thousandth.

Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(-7,8\right)\text{\hspace{0.17em}}$ in degrees. Round to the nearest thousandth.

$\left(10.630,131.186°\right)$

Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(3,-4\right)\text{\hspace{0.17em}}$ in degrees. Round to the nearest hundredth.

Use a graphing calculator to find the polar coordinates of $\text{\hspace{0.17em}}\left(-2,0\right)\text{\hspace{0.17em}}$ in radians. Round to the nearest hundredth.

$\text{\hspace{0.17em}}\left(2,3.14\right)or\left(2,\pi \right)\text{\hspace{0.17em}}$

## Extensions

Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{sec}\text{\hspace{0.17em}}\theta ;a>0.$

Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{sec}\text{\hspace{0.17em}}\theta ;a<0.$

A vertical line with $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units left of the y -axis.

Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{csc}\text{\hspace{0.17em}}\theta ;a>0.$

Describe the graph of $\text{\hspace{0.17em}}r=a\mathrm{csc}\text{\hspace{0.17em}}\theta ;a<0.$

A horizontal line with $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units below the x -axis.

What polar equations will give an oblique line?

For the following exercise, graph the polar inequality.

$r<4$

$0\le \theta \le \frac{\pi }{4}$

$\theta =\frac{\pi }{4},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}2$

$\theta =\frac{\pi }{4},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}\ge -3$

$0\le \theta \le \frac{\pi }{3},\text{\hspace{0.17em}}r\text{\hspace{0.17em}}<\text{\hspace{0.17em}}2$

$\frac{-\pi }{6}<\theta \le \frac{\pi }{3},-3

what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
can I see the picture
How would you find if a radical function is one to one?