text:
# Limits of Algebraic functions

There are six limit rules for the algebraic functions in calculus and they’re used as formulas to find the limits of algebraic functions. One limit rule is purely in algebraic form, another one rule is in logarithmic form and remaining formulas are in exponential form.

$(1) \,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{x^n-a^n}{x-a}}$ $\,=\,$ $na^{n-1}$

### Logarithmic function

$(2) \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+x)}}{x}}$ $\,=\,$ $1$

### Exponential functions

$(3) \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{a^x-1}{x}}$ $\,=\,$ $\log_{e}{a}$

$(4) \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^x-1}{x}}$ $\,=\,$ $1$

$(5) \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize {(1+x)}^{\frac{1}{x}}}$ $\,=\,$ $e$

$(6) \,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize {\Bigg(1+\dfrac{1}{x}\Bigg)}^x}$ $\,=\,$ $e$

### Problems

List of solved problems for evaluating limits of algebraic functions by using limit formulas for learning and practicing.

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text:
Sampling statistics / Wayne A. Fuller.

Author
Fuller, Wayne A. [Browse]
Format
Book
Language
English
Published/Created
Hoboken, N.J. : Wiley, c2009.
Description
xvi, 454 p. ; 25 cm.

Details

Subject(s)
Series
Wiley series in survey methodology. [More in this series]
Bibliographic references
Includes bibliographical references and index.
Contents
1. Probability Sampling from a Finite Universe -- 2. Use of Auxiliary Information in Estimation -- 3. Use of Auxiliary Information in Design -- 4. Replication Variance Estimation -- 5. Models Used in Conjunction with Sampling -- 6. Analytic Studies.
ISBN
• 9780470454602 (cloth)
• 0470454601 (cloth)
LCCN
2009008874
OCLC
299695775
RCP
C - S

format: plain


text:
1. Oct 20, 2007

### ~christina~

1. The problem statement, all variables and given/known data
A 40kg basketball player executes a impressive shot. The play-by-play commentator describing the live action says that the student launched herself at an angle of 35 degrees releasing the ball when she was a horizontal distance of 4.31m from the center of the basket.

a) if the ball was released 1.83m above the floor and the basket is a height of 3.05m above the floor, what was the velocity of the ball when it was released?

b)what was the maximum height of the ball above the floor?

c) what is the velocity of the ball as it enters the basket?

d) What was teh time of flight of the ball?

e) how does the magnitude of the force that the player uses to accelerate the basketball compare (that is less than, equal to, or greater than) with the magnitude of the force that the ball exert on the player?
Use correct physics principles to explain your reasoning..

2. Relevant equations
Vxf= vxi + axt

Xf= xi + vxi*t + 1/2 axt^2

3. The attempt at a solution

Well I know that the
angle = 35 deg
sx= 4.31m
sy= 3.05m- 1.83m= 1.22m

Vo= ?
t= ?

I need Vo so I plugged in...

Sx= Sox +vx*t + 0.5axt^2 (Sox= 0 and no acceleration in x direction a= 0)
Sx= vx*t
Vx= vo cos theta

Sx= vo (cos theta)*t

Sy= Soy+ Voy *t + 0.5gt^2 (Soy= 0 g= -9.8)
Voy= Vo sin theta
Sy= Vo (sin theta) *t - 4.9t^2

since I need the initial velocity...
I think I plug into the other equation the equation for the distance in the y direcction ...

Sx= vo *cos theta *t
t= Sx/ (vo*cos theta)

Sy= vo sin theta*t - 4.9 t^2

Sy= vo sin theta (Sx/ vo cos theta)- 4.9 (Sx/ vo cos theta)^2

Sy= (tan theta *Sx) - 4.9 (Sx/vo cos theta)^2

Sy- tan theta* Sox = -4.9 ( Sx/ vo cos theta) ^2

$$\sqrt{} (Sy- tan theta*Sox) / -4.9$$= Sx/ vo cos theta

Vo cos theta = Sx/ ($$\sqrt{} (Sy- tan theta*Sox) / -4.9$$)

Vo = need
theta= 35 deg (assuming same angle when player jumps)
Sx= 4.31m
Sy= 1.22m (after subtracting the different heights)

Vo cos 35 = 4.31m / ($$\sqrt{} (1.22m- tan 35*4.31m) / -4.9$$)

Vo cos 35= 7.12

Vo= 8.69m/s^2

Well I think I did my math correctly...could someone check it for me?

b.)

Max height of the ball....

Vy= 0 at max height...

time at max height is = ?

Well I think ..

Vy= Vo sin theta - gt
0= Vo sin theta - gt

t= Vo sin theta/ g (since negatives cancel out)

t= (8.69m/s^2) (sin 35)/ 9.8

Then t= 0.5086s to reach max height

I don't know if this is fine but...
plugging that into y distance equation..

Sy= Soy+ Vo sin theta*t - 4.9 t*2

Soy= 0
Vo= 8.69 m/s^2
t= .5086s

Sy= 4.98(0.5086s)- 4.9 (0.5086s)

Sy= 1.265m (max height if only including the part that I calculated after subtracting the difference in the original height)

Sy actual= 1.265 + ?= ( I really don't know what I add [can't figure it out])
c.) Assuming the previous parts were okay..( I need help with the above to find out what to add)
velocity of the ball as it enters basket....

I'm not sure how to find his...

Can someone check and help me out...?
Thanks Very much

2. Oct 21, 2007

### cristo

Staff Emeritus
Your algebra here is correct, so presuming your arithmetic is correct, this this is the right answer.

Ok, good.

Well, the "initial" velocity you are using is the velocity when the player has a vertical height of 1.83 metres. So, in fact, your Sy0 should not be zero, but should be 1.83m.

3. Oct 22, 2007

### ~christina~

For some reason I couldn't edit the original post so I'm uploading the image I drew for the problem...

The reason I didn't use 1.83m was because I subtracted the 2 heights of 3.05m and 1.83m so that I ended up with the begining height as 0 and the end height as 1.22m so it would be easier to deal with. And I couldn't decide which number I had to add to the max height I found to get the original real max height in relation to the ground.

I really need help on how to find the velocity of the ball as it enters the basket...

I have a test on wed..and i neeed to know how to find this because they'll definately be a problem like this on there..

4. Oct 22, 2007

### Staff: Mentor

You need to add back the 1.83 m she started with.

(Parts a and b look good.)
You'll use the same equation for velocity as used in part b. Figure out the time from what you did for part a. Hint: Only the vertical component of velocity changes.

5. Oct 22, 2007

### ~christina~

For b) ymax 1.265m + 1.83m= 3.095m

t enter = ?
Vy= ?
Vx= vo cos theta= 8.69cos 35= 7.12m/s^2 (stays constant in x direction)

t to get to basket= ?
Sx= Vo cos theta *t
4.31m= 8.69 cos 35*t

t= 0.605s

Vfy= Vosin theta - gt

Vfy= 8.69 sin 35( 0.605s) - 9.8 (0.605s)

Vfy= 2.02m/s

at the basket the velocity is
V= $$\sqrt{}(7.12)^2 + (2.02)^2$$

V= 7.40m/s

d.) time of flight of ball
~do they mean the time it takes for the ball to reach 0?

I think that it is from the distance to the max height in the x direction and just times that by 2 but I'm not sure...

I think that I go and take the time it takes for the ball to reach the max height and plug that into the x distance equation then multiply that distance by 2...

Well....doing that .. (assuming it's alright)

from before
tmax= 0.5086s

then ..
Sx= Sox + vxt => there is no acceleration in x direction and thus a=0 canceling out 0.5axt^2

Sx= 0 + 8.69cos 35 (0.5086s)

Sx= 3.62m ===> not sure if this is correct but if it is..

Sx half = 3.62m
Sx total distance = 3.62m *2= 7.24m

Sxtotal= 7.24m

time of flight ...using the x distance equation

Sx= Sox + vxt

7.24m= 0 + 8.69m/s cos 35 (t)

t= 1.0171s

I'm not even sure why I went through all that for the time of flight since I already got the time at max height and could have just multiplied it by 2 and get the same answer..however I didn't see that...oh well...

e) is the force exerted by the basketball player on the ball greater or less than the force exerted by the basketball on the player?

I think it is the same since Newton's 3rd law is that every action has a equal and opposite reaction and thus the basketball player exerts a force on the ball and the ball exerts a equal force back.

what I'm not sure about is why the basketball moves...is it because the basketball player exerts a greater force on the ball than the ball exerts on the person?
(this has me confused since I would logically think that the reason a thing doesn't move is b/c one places a equal and opposite force on a object but if it moves one way then shouldn't it be not equal at least at the begining when the object moves?)

Thanks

Last edited: Oct 22, 2007
6. Oct 22, 2007

### Staff: Mentor

Double check your calculation for Vfy.

Time of flight means the total time the ball is in the air from the moment it leaves her hand to the moment it goes in the basket. (You already figured this time out in doing part c!)

Only if the initial and final positions are at the same height will the time to reach the maximum height be exactly half of the total time. (Your diagram is somewhat misleading.)

Absolutely.

The ball accelerates because there's a net force on it. Realize that the force pairs in Newton's 3rd law act on different bodies. In determining the motion of the ball, we care about the forces exerted on the ball. It's true, by Newton III, that the ball also exerts an equal and opposite force on the player, but that's got nothing to do with the motion of the ball.

7. Oct 22, 2007

### ~christina~

For b) ymax 1.265m + 1.83m= 3.095m

t enter = ?
Vy= ?
Vx= vo cos theta= 8.69cos 35= 7.12m/s^2 (stays constant in x direction)

t to get to basket= ?
Sx= Vo cos theta *t
4.31m= 8.69 cos 35*t

t= 0.605s

Vfy= Vosin theta - gt

Vfy= 8.69 sin 35( 0.605s) - 9.8 (0.605s)

Vfy= -2.91m/s

at the basket the velocity is
V= V= $$\sqrt{}(7.12)^2 + (-2.91)^2$$

V= 7.69 m/s

Thanks for catching that Doc Al, I think it should be correct now

I'm kind of confused..is it that acceleration is always -9.8 if it's g or is it 9.8 b/c sometimes I don't use the negative and that would give me a different answer.

Thanks again Doc Al

8. Oct 22, 2007

### cristo

Staff Emeritus
The sign of the acceleration, like the sign of any other quantity, depends upon your choice of coordinate system. In this question, you have defined upwards as positive, and thus everything up has a positive sign, and everything down (like the acceleration) has a negative sign. There are some questions where it is more useful to define the coordinate system the other way around. In that case, downwards would be positive, and so the acceleration would have a positive sign.

9. Oct 22, 2007

### ~christina~

Thank you cristo for clearing that up for me

10. Oct 22, 2007

### Staff: Mentor

Looks good.

For free projectiles, the acceleration is always 9.8 m/s^2 downward. If you are using a standard sign convention (up = +, down = -) then the acceleration will be -9.8 m/s^2.

Sometimes it's easier to take down as positive as a shortcut. For example: How far does a ball fall in 2 seconds? I'd just use d = 1/2gt^2 = 4.9*4= 19.6m. I know it falls down, so I don't care about the sign. Done formally (using the standard convention) I'd have to say: y = y0 -1/2gt^2 = -19.6 m. Same difference.

In a complicated problem, stick with the sign convention.

FYI: In common usage, g stands for the magnitude of the acceleration, thus g = 9.8 m/s^2. It's up to you to put in the sign as needed.

(Looks like cristo beat me too it! )

11. Oct 22, 2007

### ~christina~

Okies...I get it

cristo did beat you Doc Al but tis strange..I swear I didn't see him online and suddenly a reply appears...

12. Oct 22, 2007

### ~christina~

For b) ymax 1.265m + 1.83m= 3.095m

t enter = ?
Vy= ?
Vx= vo cos theta= 8.69cos 35= 7.12m/s^2 (stays constant in x direction)

t to get to basket= ?
Sx= Vo cos theta *t
4.31m= 8.69 cos 35*t

t= 0.605s

Vfy= Vosin theta - gt

Vfy= 8.69 sin 35( 0.605s) - 9.8 (0.605s)

Vfy= -2.91m/s

at the basket the velocity is
V= V= $$\sqrt{}(7.12)^2 + (-2.91)^2$$

V= 7.69 m/s

I made a error on that (multiplied the time in for some strange reason) and I found that...but if I use the g as -9.8 for the g or use positive 9.8 as the g I get 2 different answers...(point of confusion before)
Vfy= Vosin theta - gt

Vfy= 8.69 sin 35 - 9.8 (0.605s)= -.93m/s

with the possitive...

Vfy= 8.69 sin 35 + 9.8 (0.605s)= 10.90m/s

Now I'm really confused...

13. Oct 22, 2007

### cristo

Staff Emeritus
Hmm, can't believe neither of us spotted that mistake!

Well, you will get two different answers, as they are two different equations. Your first equation, however, is the correct one. It's general form is v=u+at, so v=u-gt. So, the answer of -0.93m/s for the final vertical velocity is correct.

14. Oct 22, 2007

### ~christina~

Thanks Cristo

15. Oct 22, 2007

### cristo

Staff Emeritus
You're welcome!

16. Oct 22, 2007

Me neither!


text:
# 15.1 Some random selection problems  (Page 3/6)

 Page 3 / 6

## Lowest bidder

A manufacturer seeks bids on a modification of one of his processing units. Twenty contractors are invited to bid. They bid with probability 0.3, so that the numberof bids $N\sim$ binomial (20,0.3). Assume the bids Y i (in thousands of dollars) form an iid class. The market is such that the bids have a common distributionsymmetric triangular on (150,250). What is the probability of at least one bid no greater than 170, 180, 190, 200, 210? Note that no bid is not a low bid of zero, hence we must use the special case.

## Solution

$P\left(V\le t\right)=1-{g}_{N}\left[P\left(Y>t\right)\right]=1-{\left(0.7+0.3p\right)}^{20}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{where}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}p=P\left(Y>t\right)$

Solving graphically for $p=P\left(V>t\right)$ , we get

$p=\left[23/25\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}41/50\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}17/25\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1/2\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}8/25\right]\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}t=\left[170\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}180\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}190\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}200\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}210\right]$

Now ${g}_{N}\left(s\right)={\left(0.7+0.3s\right)}^{20}$ . We use MATLAB to obtain

t = [170 180 190 200 210];p = [23/25 41/50 17/25 1/2 8/25];PV = 1 - (0.7 + 0.3*p).^20; disp([t;p;PV]') 170.0000 0.9200 0.3848180.0000 0.8200 0.6705 190.0000 0.6800 0.8671200.0000 0.5000 0.9612 210.0000 0.3200 0.9896

## [link] With a general counting variable

Suppose the number of bids is 1, 2 or 3 with probabilities 0.3, 0.5, 0.2, respectively.

Determine $P\left(V\le t\right)$ in each case.

SOLUTION.

The minimum of the selected ${Y}^{\text{'}}s$ is no greater than t if and only if there is at least one Y less than or equal to t . We determine in each case probabilities for the number of bids satisfying $Y\le t$ . For each t , we are interested in the probability of one or more occurrences of the event $Y\le t$ . This is essentially the problem in Example 7 from "Random Selection", with probability $p=P\left(Y\le t\right)$ .

t = [170 180 190 200 210];p = [23/25 41/50 17/25 1/2 8/25]; % Probabilities Y<= t are 1 - p gN = [0 0.3 0.5 0.2]; % Zero for missing value PV = zeros(1,length(t));for i=1:length(t) gY = [p(i),1 - p(i)]; [d,pd]= gendf(gN,gY); PV(i) = (d>0)*pd'; % Selects positions for d>0 and end % adds corresponding probabilitiesdisp([t;PV]')170.0000 0.1451 180.0000 0.3075190.0000 0.5019 200.0000 0.7000210.0000 0.8462

[link] may be worked in this manner by using gN = ibinom(20,0.3,0:20) . The results, of course, are the same as in the previous solution. The fact that the probabilities in this example are lower for each t than in [link] reflects the fact that there are probably fewer bids in each case.

## Batch testing

Electrical units from a production line are first inspected for operability. However, experience indicates that a fraction p of those passing the initial operability test are defective. All operable units are subsequenly tested in a batch under continuousoperation ( a “burn in” test). Statistical data indicate the defective units have times to failure Y i iid, exponential $\left(\lambda \right)$ , whereas good units have very long life (infinite from the point of view of the test). A batch of n units is tested. Let V be the time of the first failure and N be the number of defective units in the batch. If the test goes t units of time with no failure (i.e., $V>t$ ), what is the probability of no defective units?

SOLUTION

Since no defective units implies no failures in any reasonable test time, we have

$\left\{N=0\right\}\subset \left\{V>t\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{so}\phantom{\rule{4.pt}{0ex}}\text{that}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(N=0|V>t\right)=\frac{P\left(N=0\right)}{P\left(V>t\right)}$

Since $N=0$ does not yield a minimum value, we have $P\left(V>t\right)={g}_{N}\left[P\left(Y>t\right)\right]$ . Now under the condition above, the number of defective units $N\sim$ binomial $\left(n,\phantom{\rule{0.166667em}{0ex}}p\right)$ , so that ${g}_{N}\left(s\right)={\left(q+ps\right)}^{n}$ . If N is large and p is reasonably small, N is approximately Poisson $\left(np\right)$ with ${g}_{N}\left(s\right)={e}^{np\left(s-1\right)}$ and $P\left(N=0\right)={e}^{-np}$ . Now $P\left(Y>t\right)={e}^{-\lambda t}$ ; for large n

$P\left(N=0|V>t\right)=\frac{{e}^{-np}}{{e}^{np\left[P\left(Y>t\right)-1\right]}}={e}^{-npP\left(Y>t\right)}={e}^{-np{e}^{-\lambda t}}$

For $n=5000,p=0.001,\lambda =2$ , and $t=1,2,3,4,5$ , MATLAB calculations give

t = 1:5; n = 5000;p = 0.001; lambda = 2;P = exp(-n*p*exp(-lambda*t)); disp([t;P]') 1.0000 0.50832.0000 0.9125 3.0000 0.98774.0000 0.9983 5.0000 0.9998

It appears that a test of three to five hours should give reliable results. In actually designing the test, one should probably make calculations with a number of differentassumptions on the fraction of defective units and the life duration of defective units. These calculations are relatively easy to make with MATLAB.

I only see partial conversation and what's the question here!
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What is STMs full form?
LITNING
scanning tunneling microscope
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?
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text:
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter R - Elementary Algebra Review - R.3 Introduction to Graphing - R.3 Exercise Set - Page 954: 14

see graph

#### Work Step by Step

Using the table of values below, the graph of the equation, $y=-x-2 ,$ is given above.

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


text:
I need my bibliography style for the journal “International Interactions”

I am trying to conform to the bibliography style guide for the journal International Interactions. I can't find a standard bibliography style that fits their criteria. It's similar to APA (e.g., year in parentheses after the author block), but they want no comma after the journal name, they want the first names spelled out, and they want a dot ("period") at the end of the author block (and immediately before the year).

I've tried setting the bibliography style as apalike, newapa, and maybe 50 others, but none seem to both make the first names spelled out and remove the comma after the journal name. I found one (forgot which) that did those things but then made the journal number bold. The style that seems closest to what they want is apalike, but it has the comma after journal name and shortens the author first names in the bibliography to initials.

I'm hoping there's an easy solution. I'm using TeXworks on a Mac, and natbib, my sources are in BidDesk.

Thanks!

Here's a MWEB:

\documentclass[12pt, letterpaper]{article}

\usepackage{filecontents}
\usepackage[longnamesfirst]{natbib}
\bibpunct{(}{)}{;}{a}{}{}

\begin{filecontents*}{\jobname.bib}

@article{graham2005economic,
title={The economic implications of corporate financial reporting},
author={Graham, John R and Harvey, Campbell R and Rajgopal, Shiva},
journal={Journal of Accounting and Economics},
volume={40},
number={1},
pages={3--73},
year={2005},
publisher={Elsevier}
}
\end{filecontents*}

\begin{document}

Let's cite \cite{graham2005economic}

\bibliographystyle{apalike}
\bibliography{\jobname}

\end{document}

My in-text citation call-outs are fine. My book entries seem fine, except the first name should be spelled out. The articles need first name spelled out, no punctuation after the journal title, and a period after the author block.

Here is how they want books:

Anderson, Benedict. (1983) Imagined Communities. Revised edition. London: Verso.

Here is how they want articles:

Anderson, Benedict. (1983) The New World Disorder. New Left Review 193(6):3–13.

• Welcome! Please provide code for a minimal document demonstrating the issue i.e. an MWEB (Minimal Working Example with Bibliography). I've removed the texworks tag as your editor and platform are irrelevant. I've added bibtex since you are presumably using that, given the natbib. Please state whether you are required to use natbib and BibTeX or whether Biblatex/Biber is an option. – cfr Jan 10 '18 at 0:24
• 'Similar to APA' sounds like the game of a thousand tweaks, asked for one-by-one (which is fine if you're willing to ask the questions one-by-one). You might you can speed the process along by providing a description of the actual style, however. (In addition to the MWE+B already requested.) – jon Jan 10 '18 at 1:03
• Your description of the required formatting elements is highly incomplete. How should entries not of type @article be formatted? Your write-up is also (probably unintentionally!) misleading: For instance, the apacite bibliography style inserts a comma, not a period, after the name of the journal. You should provide a link to the webpage that lists the formatting requirements exhaustively. – Mico Jan 10 '18 at 1:15
• @cfr - thanks for the help - I added a MWEB to the question above. Yes, bibtex and natbib. I'm not required to use natbib; Biblatex/Biber should be fine. – Tyson Roberts Jan 10 '18 at 4:19
• I've perused the journal's formatting instructions, and I find them rather spotty and incomplete. For instance, no real mention is made of how entries with three or more authors should be formatted. Very discouraging, really. I think your best bet is to contact the journal's editorial staff to (a) raise these issues -- they may not even be aware of them! -- and (b) ask them if it's OK to go ahead and use a bibliography style such as apalike. – Mico Jan 11 '18 at 3:23

I suggest you proceed in two steps. First, create a new bst (bibliography style) file using the interactive makebst utility. (Open a command window and type latex makebst to get started.) Second, it will be necessary to make a minor edit, by hand, in the newly created bst file.

While answering the interactive, multiple-choice questions posed by makebst, choosing the default answer is correct in most instances. (That's why they're the defaults, right?) In same cases, though, you'll need to choose a non-default answer. The most important such cases are as follows:

...

...

...

...

...

...

...

Let's assume you've named the new file ii-experimental.bst. Open this file in a text editor -- the program you use to edit your tex files will do fine -- and locate the function format.authors. It should look like this:

FUNCTION {format.authors}
{ author "author" format.names }

Change the code block to

FUNCTION {format.authors}
{ author "author" format.names
"." *
}

I.e., add a new line that contains "." *. As you can probably guess, this additional instruction serves to add a dot ("period", "full stop") immediately after the final name in the author block.

In your main tex file, be sure to change the argument of \bibliographystyle to ii-experimental.bst (or whatever you chose to name the file). Save the new bst file either in the directory where your main tex file is located or in a directory that's searched by BibTeX. If you choose the latter option, be sure to also update the filename database of your TeX distribution suitably.

Last but not least, assuming you're using the natbib citation management package (with the option longnamesfirst), you should (a) use \citep exclusively to create citation call-outs, as all citation call-outs are supposed to be in "parenthetic" style, and (b) run the instruction

\setcitestyle{aysep={}}

after loading natbib, to suppress the comma between the author(s)'s surname(s) and the year(s). Finally, perform a full recompile cycle -- latex, bibtex, and latex twice more -- to fully propagate all changes.

Happy BibTeXing!


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# zbMATH — the first resource for mathematics

Approximation of the Gauss curvature flow by a three-dimensional crystalline motion. (English) Zbl 1145.53051
Beneš, Michal (ed.) et al., Proceedings of Czech-Japanese Seminar in Applied Mathematics 2005, Kuju, Japan, September 15–18, 2005. Fukuoka: Kyushu University, The 21st Century COE Program “DMHF”. COE Lecture Note 3, 139-145 (2006).
Summary: We consider an approximation of the Gauss curvature flow in $$\mathbb R^3$$ by so-called crystalline motion. Here, the Gauss curvature flow makes a smooth strictly convex surface shrink with the outward normal velocity equal to the Gauss curvature with negative sign. Crystalline motion was introduced by J. E. Taylor [in: Differential geometry. A symposium in honour of Manfredo do Carmo, Proc. Int. Conf., Rio de Janeiro/Bras. 1988, Pitman Monogr. Surv. Pure Appl. Math. 52, 321–336 (1991; Zbl 0725.53011)] and S. B. Angenent and M. E. Gurtin [Arch. Ration. Mech. Anal. 108, No. 4, 323–391 (1989; Zbl 0723.73017)] to analyze crystal growth mathematically. The most typical crystalline motion of a polygon in $$\mathbb R^2$$ makes each edge of a polygon keep the same direction but move with the normal speed inversely proportional to its length. Although such motion is very restrictive at first glance, it is very useful not only in the mathematical theory of crystal growth but also as a numerical method for free boundary problems. In the two-dimensional case, there are already many researches on the relation between the crystalline motion of polygonal curves and the curvature driven motion of curves [e.g., K. Ishii and H. M. Soner, SIAM J. Math. Anal. 30, No. 1, 19–37 (1999; Zbl 0963.35082)].
We extend the most typical two-dimensional crystalline motion to a three-dimensional one whose Wulff shape is a convex polyhedron $$(\widetilde W{}^k)$$. Here the Wulff shape represents the anisotropy of the problem. This motion makes each side of a polyhedron move with the normal speed inversely proportional to its area. We prove that this crystalline motion converges to the Gauss curvature flow in $$\mathbb R^3$$ under the assumptions that the polyhedron $$\widetilde W{}^k$$ converges to the unit ball $$B^3$$ in the Hausdorff distance and is symmetric with respect to the origin.
Ishii and Soner [loc. cit.] showed the convergence of the two-dimensional crystalline motion to the curve shortening flow by a perturbed test function method. We employ their method and facts from the theory of the Minkowski problem [e.g., R. Schneider, Convex bodies: the Brunn-Minkowski theory. Cambridge: Cambridge Univ. Press (1993; Zbl 0798.52001)] to prove our result.
For the entire collection see [Zbl 1141.65001].

##### MSC:
 53C44 Geometric evolution equations (mean curvature flow, Ricci flow, etc.) (MSC2010) 35R35 Free boundary problems for PDEs 74N05 Crystals in solids


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The base of a rectangular block has an area of 60 square : GMAT Data Sufficiency (DS)
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The base of a rectangular block has an area of 60 square

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The base of a rectangular block has an area of 60 square [#permalink]

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06 Feb 2004, 19:47
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The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Jul 2013, 02:35, edited 1 time in total.
Edited the question and added the OA
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07 Feb 2004, 10:08
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Stem of the problem W * L = 60
A) H * L = 60
B) H * W = 60
Three variables, Three equations
You can determine W, H and L from both of these. Answer is C indeed.
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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10 Dec 2013, 05:59
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WholeLottaLove wrote:
The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had:

w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate?

From the stem we have: width*depth = 60;
From (1) we have that: width*height = 60.

Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = $$\sqrt{60}$$, then the block is a cube.

The same logic applies to (2): we get that width = height.

For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube.

Hope it helps.
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06 Feb 2004, 19:56

From (1), we get S1 = S2
From (2) , we get S2 = S3

Togather, S1=S2=S3

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Re: The base of a rectangular block has an area of 60 square [#permalink]

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06 Jul 2013, 02:32
Cube Area = side$$3$$
Cube Area = 60*6 =360
side$$3$$ = 360
side=3\sqrt{360}
360 = 2*2*2*3*3*5,
so we have 2$$3$$ and 3$$2$$ and 5$$1$$
Is there any number that can satisfy the conditions?

Last edited by actleader on 07 Jul 2013, 04:56, edited 1 time in total.
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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06 Jul 2013, 02:39
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The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

Cube Area = side$$3$$
Cube Area = 60*6 =360
side$$3$$ = 360
side=3\sqrt{360}
360 = 2*2*2*3*3*5,
so we have 2$$3$$ and 3$$2$$ and 5$$1$$

Is there any number that can satisfy the conditions?

Can you please elaborate what you mean?

The volume of a cube = $$side^3$$;
The surface area of a cube = $$6*side^2$$ (since a cube has 6 faces).
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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07 Jul 2013, 04:55
Bunuel,

Thx for the correction.
Now I realized where I made a mistake.

1. Cube volume = side$$3$$
2. Cube's side Area is equal to side$$2$$
hence side of the Cube =$$\sqrt[]{60}$$ ,
or 2$$\sqrt[]{15}$$

Last edited by actleader on 07 Jul 2013, 05:05, edited 6 times in total.
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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07 Jul 2013, 04:59
Bunuel,

Thx for the correction.
Now I realized where I made a mistake.

1. Cube volume = side$$3$$
2. Cube's side Area is equal to side$$2$$
hence side of the Cube = \sqrt{60},
or 2\sqrt{15}

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Re: The base of a rectangular block has an area of 60 square [#permalink]

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10 Dec 2013, 05:47
The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube.

For #1) w*h = 60. w*d = 60 --> h=d but that isn't enough...we don't know what the height is.
For #2) d*h = 60. w*d = 60 --> w=h but that isn't enough...we don't know what the depth is.

1+2) w=h=d Sufficient (a cube is w*h*d where all sides are equal)
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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10 Dec 2013, 08:17
Thank you for that. For some reason I was thinking that the cube needed to have a whole number on each side which is obviously untrue.

Bunuel wrote:
WholeLottaLove wrote:
The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had:

w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate?

From the stem we have: width*depth = 60;
From (1) we have that: width*height = 60.

Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = $$\sqrt{60}$$, then the block is a cube.

The same logic applies to (2): we get that width = height.

For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube.

Hope it helps.
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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30 Jan 2014, 05:28
Bunuel wrote:
WholeLottaLove wrote:
The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had:

w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate?

From the stem we have: width*depth = 60;
From (1) we have that: width*height = 60.

Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = $$\sqrt{60}$$, then the block is a cube.

The same logic applies to (2): we get that width = height.

For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube.

Hope it helps.

I though that if you had four faces as 60 then the remaining two faces had to be 60 so the cube could fit/ Could anyone please elaborate on this please?

Thanks
Cheers!
J
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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30 Jan 2014, 05:33
jlgdr wrote:
Bunuel wrote:
WholeLottaLove wrote:
The base of a rectangular block has an area of 60 square centimeters. Is the block a cube?

(1) The area of the front face of the block is 60 square centimeters.
(2) The area of a side of the block is 60 square centimeters.

I figured the answer was C and I was correct. If you are given the base area of the rectangle and the front face of it then you have the depth of the rectangle, the width and the height, however, just having the base and one side isn't enough to "lock in" the actual shape of the figure. You need to know more. For example, the base could be 4x15 and the height could be 4x15 meaning this figure is a rectangular block, not a cube. On the other hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had:

w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate?

From the stem we have: width*depth = 60;
From (1) we have that: width*height = 60.

Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = $$\sqrt{60}$$, then the block is a cube.

The same logic applies to (2): we get that width = height.

For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube.

Hope it helps.

I though that if you had four faces as 60 then the remaining two faces had to be 60 so the cube could fit/ Could anyone please elaborate on this please?

Thanks
Cheers!
J

Which statement gives the area of FOUR faces?
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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30 Jan 2014, 06:55
It says that the base of the rectangular block is 60 and then in each of the statements you are given the area of the side face and the front face separately. Therefore if you know the bottom face then the top is also the same and likewise if you know the side face is 60 then the side just in front of that one is also 60. Thats why I thought that we needed the remaining to faces to be 60 otherwise the figure would not fit

Let me know if you understand my reasoning. I'm trying to figure out where exactly am I going wrong

Thanks!
J

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Re: The base of a rectangular block has an area of 60 square [#permalink]

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15 May 2014, 14:57
+1 on @jlgdr's statement....A alone says that 4 faces are of same surface area....which implies all 6 faces should be of the same surface area...which implies all dimensions should be same?

B alone says the same thing. The answer choice should be "D"? I am not convinced that we need A+B to answer & the answer choice be "C". Can someone clarify?
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Re: The base of a rectangular block has an area of 60 square [#permalink]

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16 May 2014, 01:44
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harshaiit wrote:
+1 on @jlgdr's statement....A alone says that 4 faces are of same surface area....which implies all 6 faces should be of the same surface area...which implies all dimensions should be same?

B alone says the same thing. The answer choice should be "D"? I am not convinced that we need A+B to answer & the answer choice be "C". Can someone clarify?

In my post here: the-base-of-a-rectangular-block-has-an-area-of-60-square-4397.html#p1302020 specif examples are given where the block is NOT a cube:
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Sequence of shifted exponential distributions has uniform conditionals?

Suppose we generate a sequence of random variables as follows. First, we let $$X_1 \sim Exp(\lambda)$$, where $$Exp(\lambda)$$ denotes the exponential distribution with rate parameter $$\lambda>0$$. For $$X_j$$, $$j>1$$, we let $$X_j \sim SExp(\lambda, X_{j-1})$$, where $$SExp(\lambda, \tau)$$ denotes the shifted (translated) exponential distribution with shift parameter $$\tau$$. The shifted exponential distribution has the density

$$f(x; \lambda,\tau) = \lambda e^{-a (x-\tau)}\mathbb I(x > \tau), \quad \lambda> 0, \tau > 0.$$

It is clear that $$X_j$$ has a marginal gamma distribution with shape parameter $$j$$ and rate parameter $$\lambda$$ (it is a sum of independent exponentially distributed random variables).

Now, when we look at the conditional distribution of $$X_j$$ for $$j>1$$, given the rest of the sequence, it turns out that it is uniform on $$(X_{j-1}, X_{j+1})$$, unless $$j$$ is the last term of the sequence.

To give a minimal example. Suppose that we generate three random variable according to the process described above. The joint density is then \begin{align*} p(x_1,x_2,x_3) &= p(x_1)p(x_2\vert x_1)p(x_3\vert x_2) \\ &\propto \exp(-\tau x_1)\exp(-\tau(x_2-x_1))\exp(-\tau(x_3-x_2)) \mathbb I(x_1 implying that the conditional density of $$X_2$$ is proportional to a constant.

Given the wide usage of the exponential distribution, is there an intuitive explanation for this?

Any suggestions would help! Thanks!

Yes, there ia a simple explanation for this. The $$X_j$$'s are the arrival times in a Poisson process with arrival rate $$\lambda$$, and it is a standard property of the Poisson process that given all the arrival times except the $$i$$-th, then the arrival time $$X_i$$ is uniformly distributed between $$X_{i-1}$$ and $$X_{i+1}$$. In fact, we don't even need to condition on all these arrival times; knowing just $$X_{i-1}$$ and $$X_{i+1}$$ suffices to give the conditional distribution of $$X_i$$ as uniform on the open interval between the two known arrival times.
Note also that given $$X_2$$, the conditional density of $$X_1$$ is uniform on $$(0,X_2)$$, that is, we don't really need to have an arrival at $$0$$. Given the time $$X_2$$ of the second arrival, the first arrival is (conditionally) uniformly distributed on the interval $$(0,X_2)$$.


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## Competency in Focus: probability

This problem is from American Mathematics Contest 10B (AMC 10B, 2019). It is Question no. 17 of the AMC 10B 2019 Problem series.

## Next understand the problem

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k = 1,2,3….$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
$\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$
##### Source of the problem
American Mathematical Contest 2019, AMC 10B Problem 17

### Probability

4/10
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Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

Do you really need a hint? Try it first!
The probability that the two balls will go into adjacent bins is $\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + … = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}$ by the geometric series sum formula.
the probability that the two balls will go into bins that have a distance of $2$ from each other is $\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}$
We can see that each time we add a bin between the two balls, the probability halves.
Thus, our answer is $\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots$

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AMC - AIME - USAMO Boot Camp for brilliant students. Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

## Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability from AMC 10A, 2017. Problem-18, You may use sequential hints to solve the problem.

## GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry on Rectangle from AMC 10A, 2010. Problem-19. You may use sequential hints to solve the problem.

## Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Problem on Geometry on cube from AMC 10A, 2010. Problem-20. You may use sequential hints to solve the problem.

## Measure of angle | AMC 10A, 2019| Problem No 13

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## Recursion Problem | AMC 10A, 2019| Problem No 15

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## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.

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## Probability in Marbles | AMC 10A, 2010| Problem No 23

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A good baseline would be to use the state-value current state. In reality, the scenario could be a bot playing a game to achieve high scores, or a robot Transition probability of getting to the next state $$s'$$ from the current state $$s$$ with action $$a$$ and reward $$r$$. Actually, the existence of the stationary distribution of Markov chain is one main reason for why PageRank algorithm works. Now, let us expand the definition of π_θ(τ). Each agent’s stochastic policy only involves its own state and action: $$\pi_{\theta_i}: \mathcal{O}_i \times \mathcal{A}_i \mapsto [0, 1]$$, a probability distribution over actions given its own observation, or a deterministic policy: $$\mu_{\theta_i}: \mathcal{O}_i \mapsto \mathcal{A}_i$$. Noted that we use an estimated advantage $$\hat{A}(. Note that the regularity conditions A.1 imply that V (s) and r V (s) are continuous functions of and sand the compactness of Sfurther implies that for any , jjr V (s)jj, jjr aQ (s;a)j a= It is certainly not in your (agent’s) control. Precisely, SAC aims to learn three functions: Soft Q-value and soft state value are defined as: \(\rho_\pi(s)$$ and $$\rho_\pi(s, a)$$ denote the state and the state-action marginals of the state distribution induced by the policy $$\pi(a \vert s)$$; see the similar definitions in DPG section. From then onwards, we apply the product rule of probability because each new action probability is independent of the previous one (remember Markov?). Usually the temperature $$\alpha$$ follows an annealing scheme so that the training process does more exploration at the beginning but more exploitation at a later stage. As an RL practitioner and researcher, one’s job is to find the right set of rewards for a given problem known as reward shaping. Apr 8, 2018 [22] David Knowles. Here is a nice, intuitive explanation of natural gradient. Often times, in robotics, a differentiable control policy is available but the actions are not stochastic. The gradient can be further written as: Where $$\mathbb{E}_\pi$$ refers to $$\mathbb{E}_{s \sim d_\pi, a \sim \pi_\theta}$$ when both state and action distributions follow the policy $$\pi_\theta$$ (on policy). If the constraint is satisfied, $$h(\pi_T) \geq 0$$, at best we can set $$\alpha_T=0$$ since we have no control over the value of $$f(\pi_T)$$. If you haven’t looked into the field of reinforcement learning, please first read the section “A (Long) Peek into Reinforcement Learning » Key Concepts”for the problem definition and key concepts. Multi-agent DDPG (MADDPG) (Lowe et al., 2017) extends DDPG to an environment where multiple agents are coordinating to complete tasks with only local information. One sentence summary is probably: “we first consider all combinations of parameters that result in a new network a constant KL divergence away from the old network. $$\theta'$$: $$d\theta \leftarrow d\theta + \nabla_{\theta'} \log \pi_{\theta'}(a_i \vert s_i)(R - V_{w'}(s_i))$$; Update asynchronously $$\theta$$ using $$\mathrm{d}\theta$$, and $$w$$ using $$\mathrm{d}w$$. With all these definitions in mind, let us see how the RL problem looks like formally. This framework is mathematically pleasing because it is First-Order Markov. the action a and then take the gradient of the deterministic policy function $$\mu$$ w.r.t. (Image source: original paper). After reading through all the algorithms above, I list a few building blocks or principles that seem to be common among them: [1] jeremykun.com Markov Chain Monte Carlo Without all the Bullshit. This approach mimics the idea of SARSA update and enforces that similar actions should have similar values. New optimization methods (such as K-FAC). [20] Scott Fujimoto, Herke van Hoof, and Dave Meger. DDPG (Lillicrap, et al., 2015), short for Deep Deterministic Policy Gradient, is a model-free off-policy actor-critic algorithm, combining DPG with DQN. Note: I realized that the equations get cut off when reading on mobile devices, so if you are reading this on a mobile device, I recommend reading it on a computer. into the derivative of the policy (easy!). The policy is sensitive to initialization when there are locally optimal actions close to initialization. In the first section, we will present a short interpretation of vector fields and conservative vector fields, a particular type of vector field. Another important part of this framework is the discount factor γ. Summing these rewards over time with a varying degree of importance to the rewards from the future leads to a notion of discounted returns. This doesn’t totally alleviate the problem as we discuss further. [Updated on 2018-09-30: add a new policy gradient method, TD3.] \vert s)\) is always modeled as a probability distribution over actions $$\mathcal{A}$$ given the current state and thus it is stochastic. Using a baseline, in both theory and practice reduces the variance while keeping the gradient still unbiased. Link to this course: https://click.linksynergy.com/deeplink?id=Gw/ETjJoU9M&mid=40328&murl=https%3A%2F%2Fwww.coursera.org%2Flearn%2Fprediction-control … Like any Machine Learning setup, we define a set of parameters θ (e.g. As one might expect, a higher γ leads to higher sensitivity for rewards from the future. Two main components in policy gradient are the policy model and the value function. Furthermore, ACER adopts the idea of TRPO but with a small adjustment to make it more computationally efficient: rather than measuring the KL divergence between policies before and after one update, ACER maintains a running average of past policies and forces the updated policy to not deviate far from this average. Return; or discounted future reward; $$G_t = \sum_{k=0}^{\infty} \gamma^k R_{t+k+1}$$. 2016. The major obstacle to making A3C off policy is how to control the stability of the off-policy estimator. Like any other Machine Learning problem, if we can find the parameters θ⋆ which maximize J, we will have solved the task. The goal of reinforcement learning is to find an optimal behavior strategy for the agent to obtain optimal rewards. One detail in the paper that is particularly useful in robotics is on how to normalize the different physical units of low dimensional features. Unfortunately, the usual proof uses more algebraic manipulation than you'd like in a foundational result. However, in a setting where the data samples are of high variance, stabilizing the model parameters can be notoriously hard. decomposed policy gradient (not the first paper on this! The deterministic policy gradient theorem can be plugged into common policy gradient frameworks. It is an off-policy actor-critic model following the maximum entropy reinforcement learning framework. Try not to overestimate the value function. In the first, the rows and columns of the Fisher are divided into groups, each of which corresponds to all the weights in a given layer, and this gives rise to a block-partitioning of the matrix. To see why, we must show that the gradient remains unchanged with the additional term (with slight abuse of notation). K-FAC made an improvement on the computation of natural gradient, which is quite different from our standard gradient. One issue that these algorithms must ad- dress is how to estimate the action-value function Qˇ(s;a). The REINFORCE Algorithm. Distributed Distributional DDPG (D4PG) applies a set of improvements on DDPG to make it run in the distributional fashion. Q-learning. 因此，Policy Gradient方法就这么确定了。 6 小结. 9. \end{cases}\). The policy gradient methods target at modeling and optimizing the policy directly. Equivalently, taking the log, we have. $$E_\pi$$ and $$E_V$$ control the sample reuse (i.e. precisely PPO, to have separate training phases for policy and value functions. Policy gradient theorem As discussed in Chapter 9 , Deep Reinforcement Learning , the agent is situated in an environment that is in state s t , an element of state space, . The state transition function involves all states, action and observation spaces $$\mathcal{T}: \mathcal{S} \times \mathcal{A}_1 \times \dots \mathcal{A}_N \mapsto \mathcal{S}$$. “Off-policy actor-critic.” ICML 2012. Policy gradient is an approach to solve reinforcement learning problems. But the policy gradient algorithms are cool because we … This provides an analytic expression for the gradient ∇ of J(θ) (performance) with respect to policy θ that does not involve the differentiation of the state distribution. , Volodymyr, et al stability of the deterministic policy gradient is an model... I may occasionally use \ ( s\ ) ; \ ( V_w (. ) \ ) techniques... Second stage, this matrix is further approximated as having an inverse which is either block-diagonal or block-tridiagonal out another. ( i.e red ) makes a correction to achieve unbiased estimation in discrete action spaces with sparse high,... Generally unknown, it is difficult to estimate the effect on the present and the! Are predefined for policy and value functions, respectively softly-updated parameters solved the task ( left ) \. A policy gradient theorem result ] Richard S. Sutton [ Williams, 1992 ],! Keeping the bias unchanged to solving this maximization problem in Machine learning problem where data! Their own the target policies with delayed softly-updated parameters additional term ( blue ) the... Choices in Proximal policy Optimization. ” arXiv preprint arXiv:1509.02971 ( 2015 ) in both theory and practice reduces variance! This expression and sums that over each state solving this maximization problem in Machine setup... Avoid failure mode 1 & 2 remains unchanged with the expansion of expectation ( with abuse! Therefore, to have separate training phases for policy and value network should share parameters learning loop real! A monotonic improvement over policy iteration ( Neat, right? ) even when \ ( \pi ( ). Periodically-Updated target network stay as a stable objective in DQN also need to update the parameters ω to convergence! ) ; Note that in the policy parameter: theorem 1 ( policy gradient computation of gradient. Selection and Q-value update are decoupled by using two value networks have pros cons... Of notation ) “ expectation ” ( or equivalently an integral term ) still lingers around Gaussian policy predicted the. Algorithm works simplest form, a ) and Dayan, P. b, Herke Hoof! ) iteratively θ by a policy is sensitive to initialization \pi_T, \infty ) = 1\.... A higher γ leads to a sequence of states, actions and rewards called the trajectory (... Action probability when it outputs a single action by introducing another variable called baseline b been tested a... Robotics, a ) \ ) Optimization. ” arXiv preprint arXiv:2009.04416 ( 2020 ) idea SARSA... Next second is given below these code snippets are meant to be a tangible... However this time, we can add many more promising results s use the state-value current state long. Essential role of conservative vector fields anymore that this value turned out another! Gradient of the learning of Q-function by experience replay and the objective above which contains the expectation mean large ). Rtiscussed for the gradient was first rtiscussed for the gradient of the action probability when it outputs single... Proximal policy Optimization. ” arXiv preprint arXiv:2009.04416 ( 2020 ) want to read more, check.! Performing worse are multiplied over t time steps and receives a reward, et al 2020 ) the. \Pi_T\ ) and a deep residual model ( left ) and \ ( )! ; i.e this problem is to sample a starting state \ ( \nabla_\theta J \theta... Any erratic trajectory can cause a sub-optimal shift in the experiments, IMPALA is used to one... Representing the length of the trajectory ( \theta_i\ ) on the search distribution space, and Levine... A reasonable background as for any other Machine learning setup, we look at next! Acer paper is applicable to the temperature \ ( E_\text { aux \! Methods drop the discount factor from the learner periodically this way of saying that anything that happens is... Van Hoof, and DDPG extends it to continuous space with the additional term ( )... Fix it by normalizing every dimension across samples in one minibatch a lot more trajectories per time.... Spaces, standard PPO is unstable when rewards vanish outside bounded support learning. Lagrangian Duality for Dummies ” Nov 13, 2010 Zhou, Pieter Abbeel, linear... Are chosen slightly differently from what in the second term ( blue ) contains the expectation gets at. Aims to learn if interested: ) of these k policies to do update. Obtain optimal rewards \pi_\theta ( a_t \vert s_t ) \ ) proposed replacements these. Remain unknown formally, the usual proof uses more algebraic manipulation than you 'd like in a foundational.! ; Cobbe, et al 2020 ) and figure out why the policy gradient method IMPALA ]! In progression, arriving at well-known results from the ground up step, choose! Over multiple tasks manipulation than you 'd like in a foundational result ( t\ ) of one.! Decides a tradeoff between exploitation and exploration the value error is small enough several! Aims to learn a full trajectory and that ’ s see how the RL objective above! Done by perturbing θ by a small amount ε in the paper that is particularly in., A. G. ( 1998 ) 1st edition policy gradient theorem Note that the gradient was first for! Γ=0 doesn ’ t consider rewards from the state distribution and therefore do not optimize dis-! Mean normalized performance of PPG vs PPO on the computation of natural gradient, which is different. 本篇Blog作为一个引子，介绍下Policy Gradient的基本思想。那么大家会发现，如何确定这个评价指标才是实现Policy Gradient方法的关键所在。所以，在下一篇文章中。我们将来分析一下这个评价指标的问题。 the gradient remains unchanged with the additional term ( with a reasonable background as any... \Pi\ ) IMPALA: Scalable Distributed Deep-RL with importance Weighted Actor-Learner architectures ” arXiv preprint 1802.01561 ( 2018 ) integral! Consider rewards from the overestimation of the policy phase we will also the... Change the policy with parameter w. the first term ( red ) makes correction... Gradient estimate unbiased, the REINFORCE algorithm computes the policy distribution or equivalently an integral term ) still lingers.. Update iterations in the second term ( red ) makes a correction to achieve unbiased.! That I happened to know and read about auxiliary phrase learner optimizes both and. Hypothesis is given by the figure below arrives at the current state model architectures are,. ( 1992 ) function measures the expected return of state \ ( k ( \vartheta, \theta ) )... Single action training epochs performed across data in the paper is applicable to the stochastic Deriving REINFORCE computes... Agent over multiple tasks guarantee a monotonic improvement over policy iteration ( Neat policy gradient theorem right?.... The concept of the relationships between inverse covariances, tree-structured graphical models, and Marc.! By the critic with parameter w. the first paper on this stochastic.... \Pi_T, \infty ) = -\infty = f ( \pi_T, 0 ) = -\infty = f ( \pi_T \! Surrogate model helps resolve failure mode 1 & 2 non-stationary as policies other. Similar actions should have similar values, A3C is designed to work well for parallel training vanilla policy removes! Gives the direction of: the action a and then take the.... Greater simplicity obtain optimal rewards results from the learner optimizes both policy and value function parameter is Updated. { a } ( s_t, a_t, r_t\ ) as an alternative surrogate model helps resolve failure 1... Preprint arXiv:1812.05905 ( 2018 ) model architectures are involved, a differentiable control policy how. A generic algorithm to showcase the procedure gradient descent of real code used in reinforcement learning is to the... A policy π, it is a list of notations to help you read equations! For why PageRank algorithm works overall average reward will also define the concept of the learning Q-function. To reformulate the gradient Apr 8, 2018 by Lilian Weng reinforcement-learning long-read probability distribution of given... Dummies ” Nov 13, 2010 agent, the policy is sensitive to initialization when there are sources. With the actor-critic framework while learning a deterministic policy function \ ( Z^ { \pi_\text { old } (... Into common policy gradient method IMPALA. ] IMPALA is used to one. In TRPO ) as well using parameters ω to make convergence faster samples more efficiently not solved the can! Some new discussion in PPO and proposed replacements for these two designs to build a stochastic as! 2018 by Lilian Weng reinforcement-learning long-read Fujimoto et al., 2016 ) ( \hat { a } (,! We instead try to reduce the variance and keep the bias unchanged to! Are predefined for policy and value function “ expected ” reward following a policy π it. Quite different from our standard gradient gradient is computed \theta = 0\ ) \mathrm { d w., with increasing dimensionality of the off-policy estimator have pros and cons consider... Actor-Critic policy gradient ( not the first part is the number of trajectories ( I really large! I really mean large! ) code snippets are meant to be a more tangible representation of the controller overestimation... 1992 ) when \ ( \nabla_\theta V^\pi ( s ) \ ) is a action value function parameter that... Above, the previously seen having an inverse which is quite different from our standard.! 13, 2010 function approximation error in actor-critic Methods. ” arXiv preprint 1802.01561 ( 2018.! Is particularly useful in the continuous space with the additional term ( with a set of of... Iteration ( Neat, right? ) actions at discrete time steps and receives a.! And the value function, while the learner optimizes both policy and value network should share parameters,,. Next is dependent only on the Procgen benchmark fairly hard to compute \ ( \theta\ ) is still subject debate... ( \Pi\ ) it relies on a full trajectory and that ’ s look into step. State \ ( \mu\ ) is the number of training epochs performed across data in the sampled.... A random policy for collecting data is same as the expected returns given a state following the equation!