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Physics Help Forum Why is the product of a scalar and a vector, a vector?
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 Apr 22nd 2018, 12:12 AM #1 Senior Member     Join Date: Feb 2017 Posts: 202 Why is the product of a scalar and a vector, a vector? Refer the title.
 Apr 22nd 2018, 02:00 AM #2 Senior Member   Join Date: Aug 2010 Posts: 347 The first thing one would need to know, to answer the general question "Why is A a B" is the definition of "A"! You are asking "why is the product of a scalar and a vector a vector" so I would ask "what is the definition of the product of a scalar and a vector?" benit13 likes this.
 Apr 23rd 2018, 02:11 AM #3 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 207 As the operations are currently defined for vectors, scalar multiplication is distributive, similar to the following algebra: $\displaystyle 5(x + y) = 5x + 5y$ Similarly: $\displaystyle 5(\hat{i} + \hat{j}) = 5\hat{i} + 5 \hat{j}$ A scalar changes the magnitude of the vector, but not its direction if the coordinate system has a set of orthogonal basis vectors Last edited by benit13; Apr 23rd 2018 at 02:14 AM.
Apr 23rd 2018, 04:23 AM   #4
Physics Team

Join Date: Apr 2009
Location: Boston's North Shore
Posts: 1,557
 Originally Posted by avito009 Refer the title.
It seems to me that I always do my very best thinking when I'm on "the throne."

What does multiplication by a number mean to you? S'pose I wrote 3 x 5 dollars. I'm sure you'd know the answer to that

3x5 dollars = 5 dollars + 5 dollars + 5 dollars = 15 dollars

. Extend that to vectors. What should 5 x A = 5A mean? It means

5A = A + A + A + A + A

In other words you're taking a vector and multiplying it to get a larger or smaller vector in the same direction.

Last edited by topsquark; Apr 23rd 2018 at 05:18 AM.

 Apr 23rd 2018, 04:27 AM #5 Senior Member     Join Date: Feb 2017 Posts: 202 Eucledian space. Now if we take magnitude of a vector it is >=0. Which menas magnitude cant be negative. So a scalar is a quantity having only magnitude. Which means its positive. This holds only in Euclidean space. So the line in x axis where -1 and + 2 is taken if -1 is taken to be a vector then it signifies 1 in the backward direction and +2 means 2 in forward direction. So here -1 is a vector and |1| is a scalar. Since scalar cant have a negative direction. But in case of vector space. What study is vector space? Its linear algebra. So in linear algebra a scalar can have a negative value. Please correct me if I am wrong.
Apr 23rd 2018, 05:24 AM   #6

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,434
 Originally Posted by avito009 Now if we take magnitude of a vector it is >=0. Which menas magnitude cant be negative. So a scalar is a quantity having only magnitude. Which means its positive. This holds only in Euclidean space. So the line in x axis where -1 and + 2 is taken if -1 is taken to be a vector then it signifies 1 in the backward direction and +2 means 2 in forward direction. So here -1 is a vector and |1| is a scalar. Since scalar cant have a negative direction. But in case of vector space. What study is vector space? Its linear algebra. So in linear algebra a scalar can have a negative value. Please correct me if I am wrong.
It all depends on where you are getting your scalars from. If we take the scalar to be a real number then we find that it can be negative, zero, or positive. But we can also take that scalar to be a complex number, and complex numbers are not positive or negative though they can, of course, be zero. The point is that the space of scalar values we are working with does not have to be related to the vector space in question.

This is all much more general than you are looking at. Given any field F and any vector space V we can construct an "F-vector space." In the example you are talking about F is a real number and V is the "usual" vector space on the plane.

-Dan
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Apr 23rd 2018, 05:30 AM   #7
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 Originally Posted by avito009 Now if we take magnitude of a vector it is >=0. Which menas magnitude cant be negative. So a scalar is a quantity having only magnitude. Which means its positive.
A scalar can be negative. All the negative sign means is that the new vector points in the opposite direction as the original.

And i you ever study relativity you'll see vectors whose magnitude itself is negative.

 Originally Posted by avito009 But in case of vector space. What study is vector space? Its linear algebra. So in linear algebra a scalar can have a negative value.
Although the elements of a vector space are called "vectors" in actuality the can mean a lot of things such as matrices, quantum states, etc.

 Apr 23rd 2018, 05:42 AM #8 Senior Member     Join Date: Jun 2016 Location: England Posts: 586 A vector has magnitude and direction. A scalar has only magnitude. You can change the magnitude of a Vector by multiplying it by a Scalar. You can change the Direction of a Vector by multiplying it by another Vector . (Note that, unless the multiplying vector has a length of one, the magnitude will also change) Watch out for the difference between the of two vectors and the of two vectors. __________________ ~\o/~
 Apr 23rd 2018, 08:29 AM #9 Senior Member     Join Date: Feb 2017 Posts: 202 Clarification. Let me clarify why I have doubt that a scalar cant be negative. One of the definitions of scalar would be this: Lets say we have an acceleration of -5 m/s (Deceleration actually). Now whats its scalar? Its |5| which is 5. Notice that the negative sign is dropped. So this made me think that a scalar is always positive in Euclidean space.
Apr 23rd 2018, 03:28 PM   #10

Join Date: Apr 2008
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Posts: 2,434
 Originally Posted by avito009 Let me clarify why I have doubt that a scalar cant be negative. One of the definitions of scalar would be this: Lets say we have an acceleration of -5 m/s (Deceleration actually). Now whats its scalar? Its |5| which is 5. Notice that the negative sign is dropped. So this made me think that a scalar is always positive in Euclidean space.
You are making a number of threads which can make it hard to collate everything.

A vector cannot be positive or negative... It is not a number. If there is a negative involved then it must be from the scalar.

Let's get out of 1-D for a moment. Consider the vector $\displaystyle \textbf{v} = 5 \hat{i} - 2 \hat{j} + 3 \hat{k}$. Is this positive or negative? There is nothing to determine it one way or another so we can't say that it is either.

By the way, you really need to work on your units. m/s is the unit for a speed, m/s^2 is the unit of an acceleration.

-Dan

PS I'm sorry, I keep finding things to add. You say that your vector -5 has a scalar |-5| = 5. This is the length or size or more properly "norm" of the vector. It is a scalar but a vector does not have "a scalar" in this sense. There can be many properties of a vector that are real numbers.
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Last edited by topsquark; Apr 23rd 2018 at 03:32 PM.

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# Foci of a general conic equation

The general equation of a conic is $A x^2 + B x y + C y^2 + D x + E y + F = 0$. At Wikipedia, there is an equation for the eccentricity, based on ABCDEF.

Is there a similar equation for getting the foci or directrix for a general ellipse, parabola, hyperbola from ABCDEF? Please assume that a non-degenerate form of the equation is given.

-

You'll first want to check either the discriminant or the eccentricity of your conic before proceeding to use any expression(s) for the foci; the central conics have two foci while the parabola only has one.

For the central conics, it is known that the two foci are at a distance $a\epsilon$ from the center, where $\epsilon$ is the eccentricity and $a$ is the semimajor axis for an ellipse, and the semitransverse axis for a hyperbola.

To simplify things, you'd first want to perform a translation on your central conic, such that the conic's center is now at the origin, and the standard equation becomes

$$\alpha x^2+\beta xy+\gamma y^2=1$$

with that, you can use the formula

$$a=\sqrt{\frac2{\alpha+\gamma+\mathrm{sgn}(\alpha-\gamma)\sqrt{\alpha^2+\beta^2+\gamma^2-2\alpha\gamma}}}$$

along with the eccentricity formula (like the one here) and the formula for the slope of the major/transverse axis to figure out the coordinates of your foci.

The parabolic case is a bit tricky, and I'll leave that to another answerer.

-

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## Happy Pi Day, 2013

Today, March 14, is one of the days that is sometimes celebrated as “Pi Day”, in honor of the best-known irrational and transcendental number, the ratio of the circumference of a circle to its diameter, usually written as the Greek letter π (pi).  The date, 3/14, is chosen because the approximate value of π is 3.14159265…   Legend has it that the value was named π because pi is the first letter of the Greek word “περίμετρος”, meaning perimeter.

The New Scientist reports that this year, to observe Pi Day, Professor Marcus du Sautoy of the University of Oxford is sponsoring Pi Day Live, a project to “crowd source” the calculation of π (pi).  The value has, of course, alreay been calculated to trillions of decimal places; because it is an irrational number, it cannot be represented exactly by any finite decimal number.  (Pi is transcendental, also, of course.)  Pi Day Live is suggesting some relatively easy methods of getting an approximate value for π, including Buffon’s Needle.  I mentioned Buffon’s Needle in a Pi Day post back in 2010.  The New Scientist headline calls it an “ancient” method, which I think is a bit over the top for something described in the 18th century.

That earlier Pi Day post also tells a related story, of the Indiana state legislature’s attempt to set the value of pi by law, one of the all-time great accomplishments of legislative lunacy.

Finally, take a thought today for 134th anniversary of the birth of Albert Einstein.

##### Update Thursday, 14 March, 15:35 EDT

I’ve just noticed that there is a rendering error (at least in Firefox) on the “Find Pi” page I linked above.  The equation for the estimated value of pi is a bit garbled (where it reads 2L\over xp; the correct equation (using the “Find Pi” variable names) is:

$\pi = \dfrac {2 L}{x p}$

I’ve dropped the site a note with the correction.


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[SOLVED]Convergence 3

dwsmith

Well-known member
$\sum\limits_{n = 1}^{\infty}\left(\sqrt{1 + n^2} - n\right)$

$$\sqrt{1 + n^2} - n = \frac{1}{\sqrt{1 + n^2} + n}$$
Now what?

Chris L T521

Well-known member
Staff member
$\sum\limits_{n = 1}^{\infty}\left(\sqrt{1 + n^2} - n\right)$

$$\sqrt{1 + n^2} - n = \frac{1}{\sqrt{1 + n^2} + n}$$
Now what?
$\frac{1}{\sqrt{1+n^2}+n}\sim \frac{1}{n}$

So it makes sense to compare this to $\dfrac{1}{n}$. I'd suggest using the limit comparison test to do this and show that

$\lim_{n\to\infty}\dfrac{\dfrac{1}{n}}{\dfrac{1}{ \sqrt{1+n^2} +n}}\rightarrow L$

If the limit converges (i.e. $L<\infty$), both terms have the same behavior (in this case, the limit should converge, implying that both series diverge).

I hope this helps!

dwsmith

Well-known member
$\frac{1}{\sqrt{1+n^2}+n}\sim \frac{1}{n}$

So it makes sense to compare this to $\dfrac{1}{n}$. I'd suggest using the limit comparison test to do this and show that

$\lim_{n\to\infty}\dfrac{\dfrac{1}{n}}{\dfrac{1}{ \sqrt{1+n^2} +n}}\rightarrow L$

If the limit converges (i.e. $L<\infty$), both terms have the same behavior (in this case, the limit should converge, implying that both series diverge).

I hope this helps!
Wouldn't it be easier than to say that $\frac{1}{n} > \frac{1}{\sqrt{1+n^2}+n}$ and the Riemann Zeta function only converges for the $\text{Re}(\sigma) >1$. Since $\text{Re} (\sigma) =1$, it diverges.

Chris L T521

Well-known member
Staff member
Wouldn't it be easier than to say that $\frac{1}{n} > \frac{1}{\sqrt{1+n^2}+n}$ and the Riemann Zeta function only converges for the $\text{Re}(\sigma) >1$. Since $\text{Re} (\sigma) =1$, it diverges.
I'm not that confident with working with Zeta functions. However, I must say that in order to show divergence, you need to find a general term that is smaller than the term of the series you're looking at; that is, if you're comparing two series $\sum a_n$ and $\sum b_n$ and $a_n\leq b_n$, then $\sum b_n$ diverges if $\sum a_n$ does. So, if you want to use zeta functions, you want to find some multiple $k$ of the summand term in $\zeta(1)$ such that $\dfrac{k}{n}<\dfrac{1}{\sqrt{1+n^2}+n}$.

It's going to diverge no matter what, but you need to be careful in showing how it does so.


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# A Central Limit Theorem for Wasserstein type distances between two different laws

Abstract : This article is dedicated to the estimation of Wasserstein distances and Wasserstein costs between two distinct continuous distributions $F$ and $G$ on $\mathbb R$. The estimator is based on the order statistics of (possibly dependent) samples of $F$ resp. $G$. We prove the consistency and the asymptotic normality of our estimators. \\
Keywords :
Document type :
Preprints, Working Papers, ...
Domain :

Cited literature [17 references]

https://hal.archives-ouvertes.fr/hal-01624786
Contributor : Thierry Klein <>
Submitted on : Wednesday, November 29, 2017 - 10:09:36 AM
Last modification on : Tuesday, February 11, 2020 - 1:58:03 PM

### Files

Wasserstein_HALrev1.pdf
Files produced by the author(s)

### Identifiers

• HAL Id : hal-01624786, version 2

### Citation

Philippe Berthet, Jean-Claude Fort, Thierry Klein. A Central Limit Theorem for Wasserstein type distances between two different laws. 2017. ⟨hal-01624786v2⟩

Record views


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## Pages

### A Bag of Marbles - Probability

There are four red, three yellow and five blue marbles in a bag. What is the probability of selecting a blue marble?

There are a total of 4 + 3 + 5 = 12 marbles in the bag. Out of these, five a blue. Thus the number of favorable outcomes is 5 (since selecting any one of the five blue marbles gives the favorable outcome) and the total number of possible outcomes is 12 (since any one of the twelve marbles can be chosen). Applying the formula for theoretical probability,

P("blue marble") = "Number of favorable outcomes"/"Total number of possible outcomes"
P("blue marble") = 5/12

Thus the probability of selecting a marble that is blue is 5/12.

From the same bag mentioned above, you pick one marble, replace it in the bag, and then pick another marble. What is the probability that both marbles will be red?

First find the probability of selecting a red marble from the bag. There are a total of 12 marbles, thus the total number of possible outcomes is 12. Further, there are four red marbles in the bag. Thus there are four favorable outcomes.

P("red") = "Number of favorable outcomes"/"Total number of possible outcomes" = 4/12 = 1/3

After selecting the first marble and then replacing it in the bag, you select another marble. Since the first marble was replaced, therefore the total number and number of red marbles is same. Thus the probability of selecting a red marble in the second attempt is the same as that in the first attempt, 1/3.

Now find the combined probability of both marbles being red.

P("both red") = P("first red and second red")

The word 'and' in the above statement implies that since the two events of selecting the marbles are independent events, their combined probabilities is the product of their individual probabilities. Thus,

P("both red") = 1/3 * 1/3 = 1/9

Thus the probability of selecting two red marbles by replacing the first marble in the bag is 1/9.

In the same question as above, the second marble is selected without replacing the first marble in the bag. Then find the probability of getting two red marbles.

This question is different from the above one in that it does not replace the first marble in the bag. Thus, the probability of getting a red marble in the second attempt changes.

The probability of the first marble selected being red is the same as before, 1/3, since total number of marbles and number of red marbles is the same.

P("first red") = 1/3

After selecting the first marble, 11 marbles are left in the bag. Further, if we assume that the first marble selected was red, then 3 red marbles are left in the bag. The probability of getting a red marble from the bag now is

P("second red") = "Number of red marbles left in the bag"/"Total number of marbles left in the bag" = 3/11

Now the combined probability of the first marble being red and the second marble being red is the product of their individual probabilities since both are independent events.

P("first red and second red") = P("first red") * P("second red")
 = 1/3 * 3/11 = 1/11

Thus, the probability of getting two red marbles from the bag without replacing the first one is 1/11.


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# Maclaurin Series of Common functions

?Refer to Wiki: List of Maclaurin series of some common functions

Maclaurin series of these common functions are very useful, which we really want to memorize.

## Example

Solve:

• This Maclaurin series has a clear pattern of cos(x) Maclaurin series:

• In this case, it?s:

• So we could convert the series back to the trig function:

Solve:

## Example

Solve:

• Looks quite a complicated series, but we look at the factorial dominators, we found that resembles the pattern of the sin(x)’s maclaurin series.
• By comparing to the sin(x) maclaurin series, we notice that x = ?/2, and all the rest are the same.
• So rewrite the series to the function:

• And we get the result is 1.


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Problems

# Transformation

Lets take some positive integer n. We shall change it nest way: if number is even, divide it by 2, if odd, add 1 to it. After several such changes we always get the number 1. For example, from number 11 we get 12, then 6, 3, 4, 2 and at last 1. So in order to get 1 from 11 we need to make 6 changes.

Given a positive integer, find the number of its changes to get 1.

#### Input

One positive integer n (1n109).

#### Output

Print the number of changes for n to get 1.

Time limit 1 second
Memory limit 128 MiB
Input example #1
11

Output example #1
6


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Greedy algorithm, $Z^1$ case

I.A. Kurkova, M.V. Menshikov

1997, v.3, Issue 2, 243-259

ABSTRACT

We consider a single server queueing system with stations in all integer points of the real line. The customers arrival streams at the different stations are independent Poisson processes. The service times of customers are mutually independent and exponentially distributed. The server serves each station exhaustively, i.e. till the station is empty. The next station to be served, is selected using the greedy algorithm: the server goes to the neighbouring station with the maximum number of customers. We study the trajectories of the server and his asymptotic position, as time tends to infinity.

Keywords: queueing system,Poisson stream,greedy server,Markov chain,transience


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# Velocity time graph problem?.

May 13, 2018

See below

#### Explanation:

Sketch graph:

We are asked the show that the shaded region $20 \le t \le 26$

$s = 0.375 {t}^{2} - 13 t + 202$

Area under graph from $t = 20 \to t = 26$ is the area of the trapezium.

$\frac{1}{2} \left(2 + 6.5\right) \cdot 6 = 25.5 m$

Using:

$s = 0.375 {t}^{2} - 13 t + 202$

Distance from $t = 20 \to t = 26$

$s = 0.375 {\left(26\right)}^{2} - 13 \left(26\right) + 202 - \left(0.375 {\left(20\right)}^{2} - 13 \left(20\right) + 202\right)$

$\setminus \setminus \setminus \setminus \setminus = 25.5 m$

Shown.


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# Solve for x, y

## Main Question or Discussion Point

is there any way to solve the following set of equations algebraically

$$(x+1)^y = a$$

$$x^y = b$$

yes there is...now you try it, and tell us how far you get

cough. how many more you have there?

i take back what i said...what on earth is that equation :)

Hmmm... it appears that if there is such y that $$b^(1/y) = a^(1/y) - 1$$, any x will satisfy these equations?

Hmmm... it appears that if there is such y that $$b^(1/y) = a^(1/y) - 1$$, any x will satisfy these equations?
maybe you mean if there is such y that $$b^{\frac{1}{y}} = a^{\frac{1}{y}} - 1$$, any x will satisfy these equations?

then it also appears that if there is such x that $$\frac{\ln a}{\ln (x+1)} = \frac{\ln b}{\ln x}$$, any y will satisfy these equations.

cough. how many more you have there?
lol. i have no more. i got the equation in the thread you mention from
(x+1)y = 216
xy = 125

Last edited:
hmm. no it's me who takes back what I said. "any" x will not do.


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# Base API: Making the self hosted version

In a previous post I announced that Base API can now be self-hosted.

In this post I will show how I managed to compile the same application that is running as a SaaS software to be able to run as a single binary (with 90% of the same functionality).

The goal was to have one code-base for both versions, to keep them in sync feature wise and to make maintaining easier.

## Stack

Currently Base API is running in two places:

• On Heroku - the backend application is compiled on deploy and the binary is running in a dyno. The only thing needed there is the Crystal Buildpack

• On Netlify - the frontend of the application is built on Netlify and served as a Progressive Web App

## Self-Hosted vs. SaaS

The SaaS version has some features that are not needed in the self-hosted version:

• authentication and user management - since it's a service people are able to sign up, login, etc...

• limits on services - rate-limiting and email sending limits

• error reporting - runtime errors are reported to Sentry

Also the Self-Hosted version contains it's own features:

• status of the server - a dashboard that lists information about the server

• user and password authentication for the interface

## Crystal

When I started to research I had some options to handle the two different versions:

• use an environment variable at runtime - this is easiest version to go with, the downside is that all of the application is compiled into the binary and by using the right environment variable the users could host their SaaS version which is not desired.

• use a compile time flag - this is the more complicated version, it has the upside of the binary only containing the code for the self-hosted version, but it's harder to test and compile.

I went with the compile time flag which can be used in macros like this:

{% if flag?(:standalone) %}
# This will be in self-hosted version
{% else %}
# This will be in the SaaS version
{% end %}

So in every case there is something present in one version and not the other version a macro like that needed to be added.

At the end I just need to compile with the flag:

crystal build src/base.cr -D standalone

and Crystal would do it's magic.

## Mint

Since Mint does not have a macro system, the only way to go is with environment variables.

To check which version we are in I created a computed property on the main store:

store Application {
get isStandalone : Bool {
!String.isEmpty(@STANDALONE)
}

...
}

And everywhere I needed to check I just used that property for deciding what to do, for example on one of the routes:

/download {
if (Application.isStandalone) {
Window.navigate("/")
} else {
sequence {
Application.initialize()
}
}
}

You can provide the environment variables using a .env file when building:

mint build -e .env.standalone

## Putting them together

Crystal has a nice shard for exactly this purpose: https://github.com/schovi/baked_file_system it allows you to "bake" files into the compiled binary.

The frontend after compiling, its baked into the final binary and served through a custom HTTP handler.

The static binary is compiled in a Docker container on Alpine linux.

## Licensing

I needed to figure out how to limit people for using the software without permission, and for this I used Public-key cryptography.

There is a key pair - the public key is built into the binary and the private key is supplied to the server.

Every 10 minutes or so the self-hosted version checks if it can keep running or not (stops if any of the steps fail):

• it sends a request to the server with a unique id and a license key (which represents the user)

• the server then checks if the license key is valid or not

• if it's not valid then sends an error

• if it's valid then sings the payload with its private key and sends the payload back with along the signature

• the self-hosted server checks the signatures validity

This way I can check the validity of the license and if it's invalid the self-hosted server will just stop, also it is very hard to circumvent since it needs an active internet connection and the requests are random every time.

## UX

So now I had the binary but it was not easy to use so I added a CLI to do migrations and start the server. The configuration can be read from a YAML file or from environment variables. Also logging needed to be unified and configurable.

It took around two days to do the modifications, if you have any questions let me know!

Nice post, Szikszai. I've never heard of Crystal or Mint but checked both of them out. I presume you've gone with Crystal for C performance with a ruby-like syntax?

Mint looks like an interesting project of yours too. What made you want to start working on something like this? How would you compare it to React?

Thanks!

Yes I've gone with Crystal because I used Ruby and love the syntax.

I started Mint after I learned Elm and got frustrated by it for a couple of reasons, and wanted to same developer experience without constraints. It is actually uses React under the hood (kind of like a platform) but it has all of the features of third party libraries like stores and routing and things like that.

Maybe I'll write a post about it here soon :)

Interesting. I had a look at the Mint site (which looks great, btw) but I'm not that skilled up on any JS libs or frameworks. Perhaps a post would help someone like me :)

I'm sure it's something @drenther would be curious about, though.


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# Functions Functions for Modeling

Strategy for presentation:
I thought we would lead the participants through this material, so that there is a common basis for later.

## What are the inputs?

Models should be intended for a purpose. So it matters what is the input and what is the output, since you typically want to transform a particular quantity into another type of quantity.
Snow Tree Cricket (from Chapter 1 of Hughes-Hallett). See Math 135 notes from 25 Jan. 2012

## Difference between traditional math and computer notation for functions.

$$y = mx + b$$, $$f(x) = mx + b$$, $$\int x dx$$, $$\sin(x)$$, $$\sqrt{x}$$, and so on.

## Parameters.

Where do parameters get introduced in Stewart? Why would we ever write $$\sin x$$?

## Specific Functions for Modeling

### Straight-Line functions $$f(x) = mx + b$$

#### Parameterization

Students already know a lot about these: slope and intercept. Emphasize that these are parameters. The parameters specify a particular function out of the whole set of functions.

Other parameterizations (e.g. point-slope, two-points, point-angle) can be converted to slope-intercept.

#### Applications

This is the general purpose modeling function. All simple relationships can and are represented this way.
1. World record swim times versus year. swim100m.csv
2. Wage versus education. cps.csv
3. Natural gas usage versus temperature. utilities.csv
See more details on these examples

### Exponentials

#### Proportional versus constant change

1. Atmosphere pressure, water pressure
2. Population of Mexico (Math 135 27 Jan 2012)
3. Exponential models of data Income and housing variables

#### Parameterization of exponentials

1. Half-life, interest rate, exponential time constant. Which is best depends on what information you have.
2. Conversion: Rule of 72.

#### Models

1. Half-lives of meds. How often should you take the pill to keep the drug level within 10% over the long run?
2. Activity on doubling and half-lives
3. Sums of exponentials. Why is cooling not a simple exponential? Fitting Stan's data

### Logarithms

#### Activities

1. Log axes — how to interpret. Example of log axes. Activity: Plot intervals on these scale to indicate absolute and relative precision on both log and linear axes, for instance, 10% precision on measurements of 1 and of 100 on a log axis.

#### Modeling project

1. Prices versus log prices in the stock market. Which generates a better simulation of stock variation: a random walk on $$\Delta$$ prices or a random walk on $$\Delta$$ log prices?

### Splines

1. Squeezing Orange Juice … measuring the cumulative. (Make a video of the scale when squeezing orange juice.)

### Power Laws

1. Random walks. How does the typical distance scale? Explanation in terms of random vectors being perpendicular. (Later application: Explaining the power law: integration of the square distance compared to integration of the absolute distance)
2. Chest circumference versus weight, tree height and thickness. (Later example in multivariate: relationships in the internal combustion engine)

### Sines and Cosines

#### Parameterization

Amplitude, time lag, offset.

#### Models

1. Day length as a function of the day of the year. (For later example, as a function of longitude.) Where does the model break down (near the arctic circle)?
2. Tone and period. Model a musical scale. How does the period change from note to note? Model a chord.
3. Modeling tide data. Example on Hawaiian tides
4. Kepler and the planets. (See kepler.Rnw)


text:
MathSciNet bibliographic data MR1646211 (2000j:57074) 57S17 (57R85) Wang, Yanying; Wu, Zhende; Ma, Kai \$(Z\sb 2)\sp k\$$(Z\sb 2)\sp k$-actions with fixed point set of constant codimension \$2\sp k+1\$$2\sp k+1$. Proc. Amer. Math. Soc. 128 (2000), no. 5, 1515–1521. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# Find axis to minimize moment of inertia

1. Nov 14, 2006

### vu10758

Two masses (M1 and M2) are connected by a thin massless rod of length L.

a) Determine the location of an axis perpendicular to the rod) which minimizes the moment of inertia of the system.

The answer is "with M1 at the origin x = M2*L/(M1+M2)

I know that I = Sum of ( M1*r1^2)
However, how do we know that M1 is at origin? I know that M1 + M2 is the total mass of the system. I know L is the length between the two points. But I don't know why the answer is above. I think I am missing something in understanding this but I don't know what.

b) Verify that the moment of inertia about this axis can be obtained by determine I about an axis through M2 and then using the parallel-axis theorem.

I know that I = I_cm + Mh^2 , where h is the distance between the axis, M is the total mass (M1+M2). Is the distance between the axis L? How do I use this to verify the moment of inertia.

2. Nov 14, 2006

### physics girl phd

for part a)
Hint --
How would you calculate the inertia of the system if the system was rotated about some known position -- say a distance "x" down the rod from the mass M1?

3. Nov 14, 2006

### vu10758

Wouldn't I use the formula I = Mx^2 then where x is the distance down from the rod?

4. Nov 14, 2006

### physics girl phd

Not quite -- what do you mean by "M"... are both masses contributing?

5. Nov 14, 2006

### FunkyDwarf

Ok theres two ways you can do this. First is intuitively. The moment of intertia is effectively a measure of mass for rotational motion. So if you have your axis straight through the middle of the rod, you can picture just through everyday experience that its going to require a fair bit of effort or torque in order to spin it. If you have the rod through one of the masses, the torque required to spin that object is virtually nil (t = fd and if d is some infintesimle radius...) And the only major force drain is the other mass at the max length. As im sure you'll see from the answer you have there, if the masses are equal its best to have it in the middle, if ones larger its best to have the axis closer to that one, or through it.

This can also be solved mathematically. You have written one definition of I but theres another, more useful (for me anyway) term.

$$I = \int r^2dm$$
Now this more general form is more for complex objects and so your sumation one is actually better for this point particle problem, but its a good idea to keep the above one in mind. Your moment of inertia is going to be

$$I = r^2M1 + (L-r)^2M2$$

right? From your calculus classes you know that if you want to minimize a function you can differentiate it and equate to zero. Bobs your uncle :)

-G

6. Nov 14, 2006

### vu10758

For part 2,

I know that I = I_cm + mh^2, where h is the distance between the axes. So is h, L-r in this case? How can I get I about an axis through M2.

Is it

I = M2*L^2 + M1*(L-r)^2 ?

7. Nov 15, 2006

### civil_dude

It is my recollection that the min. I is that axis which is the same as the centor of mass. Am I right?


text:
Vector Inverse is Negative Vector

Theorem

Let $\left({\mathbf V, +, \circ}\right)_{\mathbb F}$ be a vector space over $\mathbb F$, as defined by the vector space axioms.

Then:

$\forall \mathbf v, -\mathbf v \in \mathbf V: -\mathbf v = -1_{\mathbb F} \circ \mathbf v$

Proof

Utilizing the vector space axioms:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \mathbf v + \left({-1_{\mathbb F} \circ \mathbf v }\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({1_{\mathbb F} \circ \mathbf v}\right) + \left({-1_{\mathbb F} \circ \mathbf v }\right)$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \left({1_{\mathbb F} - 1_{\mathbb F} }\right) \circ \mathbf v$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle 0 \circ \mathbf v$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \mathbf 0$$ $$\displaystyle$$ $$\displaystyle$$ Vector Scaled by Zero is Zero Vector

... so $-1_{\mathbb F} \circ \mathbf v$ is an additive inverse of $\mathbf v$.

From Vector Inverse Unique, $-1_{\mathbb F} \circ \mathbf v = -\mathbf v$.

$\blacksquare$


text:
# lowest energy conformation of butan-2,4-diol

Do the OH groups want to be furthest apart? or the CH3 groups?

I'm trying to determine the lowest energy conformation - and I don't know which group has the bigger steric effect?

• Consider H-bonding. Mar 16 '17 at 7:59

I presume it is in the gauche form that its energy is minimun because of possible hydrogen bonding between the hydroxy groups

In butan-2-ol you have one alcohol, which is indeed smaller than the methyl group.

• sorry i've just realised that I meant a diol. Mar 13 '17 at 10:59


text:
# Can someone explain rigorously Taleb's criticism of Nate Silver's election forecasting?

Taleb makes the claim in this paper (and others) that there exists some sort of bound on the variance of a binary forecast such that if a forecaster's binary predictions exceed the bounds on variance there exists a method to arbitrage him. Would someone familiar with the dispute (and perhaps various other papers) be able to explain exactly and technically what he means?

• Can you provide a link? I think I know what the discussion is about, but without a link, I may be making assumptions regarding the discussion. Aug 9 '19 at 22:34
• Yes. I think his main point revolves around the volatility of Silver's forecasts being too high. He claims that this would expose Silver to arbitrage if he were to trade on his forecasts. I do not understand how he can know this. Is there a rigorous proof or is it just Taleb's "feeling" that the vol of Silver's forecast is too high.
– roz
Aug 11 '19 at 2:29

hope I am not too late to the party.

tl;dr Taleb's paper draws incorrect conclusions from a set of wrong assumptions. In practice, the movements of the forecast at 538 are very much in line with what can be defined an "arbitrage free prediction" based on a binary option model.

The gist of the article is summarized in its incipit.

A standard result in quantitative finance is that when the volatility of the underlying security increases, arbitrage pressures push the corresponding binary option to trade closer to 50%, and become less variable over the remaining time to expiration. Counterintuitively, the higher the uncertainty of the underlying security, the lower the volatility of the binary option. This effect should hold in all domains where a binary price is produced – yet we observe severe violations of these principles in many areas where binary forecasts are made, in particular those concerning the U.S. presidential election of 2016.

There are quite a number of fallacies in the above statements:

Taleb's assumptions:

1. Any forecast of a dichotomous outcome is equivalent to the pricing of a binary option.
2. Increase in the underlying's volatility pushes a binary option price to trade close to 50% of its payoff. Incorrect
3. Increase in the underlying's volatility reduces the volatility of the option price itself. Incorrect

Taleb's Conclusions:

Any forecasts of binary outcomes should behave according to principles 1 and 2. Therefore any prediction that - at the same time - a) claims high uncertainty and that b) fluctuates heavily, goes against no-arbitrage principle, or put otherwise: 538's predictions is rubbish. For the sake of clarity, this is not a valid claim!

Analysis of assumption no.1 Assume we are happy with 1, i.e. we agree to model the an election forecast as a binary option. From a financial perspective, the underlying S can be interpreted as the "consensus" for the two candidates, being the strike K the amount of consensus required to "exercise" the option (i.e. to claim the elections). Tbh, one could argue on the appropriateness of the choice of a binary option since the "consensus variable" S that has a lower bound (zero, as prices cannot go negative). Let us pass on that critique assuming that it is not a big issue when S and K are sufficiently distant from 0. Side note - we will be working with an European Call Option (as elections are normally "callable" only at expiry, i.e. only after election day has passed - wink wink).

Analysis of assumption no.2 We hereby simulate how the price of a binary option with the above mentioned features behaves with changes in volatility. We examine the outcome for ATM, ITM, and OTM options, for the sake of completeness. Regardless from the moneyness of the option, the increase in volatility pushes the option value towards 0. This invalidates assumption no.2

• The price of an OTM binary option is 0% of the payoff for very small and for very large volatilities. There is a sweet volatility spot where option value is maximized, which can be interpreted as the level of volatility that maximises the chances of S > K at maturity. Away from that sweep spot the increasing chances that the S < K at maturity push option value to 0.

• The price of an ATM binary option is exactly 50% of its payoff for small volatilities then decays to 0. The higher the volatility the higher the chances that the S < K at maturity.

• The price of an ITM binary option is 100% of the payoff for small volatilities then decays to 0. The higher the volatility the higher the chances that the S < K at maturity.

A more extensive discussion on binary option vega can be found here: How does volatility affect the price of binary options?

Analysis of assumption no.3 We hereby simulate how the price of a binary option fluctuates as it approaches maturity, based on the volatility of the underlying process S. We examine here the case of an ATM Option, outcome for OTM and ITM are however aligned. Simulation covers one year time period prior to the exercise. Our methodology requires that the same binary option is re-priced daily based on a) the TTM (time-to-maturity) and on b) the value of the underlying discrete random process S which is assumed to behave as a random walk. Perfect knowledge of the underlying process is assumed, i.e. the vol of the binary option is perfectly aligned to the true volatility volatility of S.

The increase in volatility pushes the option value towards 0 for larger TTM but the process converges (i.e the forecast) converges closer to expiry. Regardless from the volatility of the underlying process, the price of the binary option never converges to 50% This invalidates assumption no.3

There is an interesting post on Quora that highlights how Taleb "proves" his point by means of a chart (Fig 3. in Taleb's paper) generated with a incorrect Mathematica code. I am not sure if he purposefully uses that chart to deceive a reader already confused by his math or if this is a genuine - though misguided - attempt of his to numerically prove the theory. Either way, it is not correct.

I feel there is quite some confusion revolving this paper and its commentaries, partially due to the cryptic and unapologetic style of Taleb as well as to the fact that many opinions on the issue are imo excessively polarized by the pov of his follower/detractors.

For the sake of transparency: all the analysis above are based on the implementation of binary option pricer from https://www.quantlib.org/ and the script used for the analysis is available for scrutiny and corrections at my github.

• That was quite a nice read on a Wednesday morning, thanks. Nov 11 '20 at 8:10
• How did you model $S$? If you used Geometric Brownian Motion, then with discretization, as volatility increases, there could be paths that are absorbed into 0 thus the result that all OTM/ATM/ITM prices converge to 0. However 0 should not be an absorbing level in this case. If GBM, it would be nice to have the same results but with Reflected Brownian Motion. Nov 11 '20 at 9:37
• Upvoted -- this should (in my opinion) be the accepted answer. Very nicely done! Nov 11 '20 at 13:53
• Are you aware that Nassim Taleb does not model the forecast probability as a binary option price (with an underlying following GBM)? The decay of price as volatility increases you are showing is well understood. Taleb uses a Bachelier-type model where the underlying follows an arithmetic Brownian motion. In that case rather than $N(d_2) \to 0$ as $\sigma \to \infty$ we have $N\left( \frac{S_0-K}{\sigma \sqrt{T}}\right) \to 0.5$. I assume zero drift here to illustrate.
– RRL
Nov 14 '20 at 20:25
• If you are going to criticize Taleb in a public forum, make sure to have all your ducks in a row. I would also highly recommend against casting aspersions by insinuating that a paper which acknowledges Bruno Dupire, Peter Carr, et al. is propagating "fallacies" and that Taleb is either "misguided" or intentionally trying to "deceive a reader." -- incorrect analysis by you here notwithstanding.
– RRL
Nov 14 '20 at 23:26

Taleb argues that under uncertainty, election forecasts should be seen as a Binary option. A similar thought is presented by De Finetti's principle that probability should be treated like a two-way "choice" price. Therefore, under high levels of volatility, forecast should not have extreme variation across time (equivalently, the price of the binary option should not change significantly even if polls reveal a large change in the dynamics among candidates). Under high levels of uncertainty, the price of the binary option converges to 0.5. Therefore, the probability of winning the elections in a two-candidate environment should converge to 0.50. A quick look on Silver's forecasts, shows high volatility across time. For instance, Trump's probability of winning the elections ranges roughly from 0.15 to 0.45

For more details on Taleb's no-arbitrage argument, one should check his recent publication and a response to this publication:

1. Taleb, Nassim Nicholas. "Election predictions as martingales: an arbitrage approach." Quantitative Finance 18.1 (2018): 1-5.

2. https://www.tandfonline.com/doi/citedby/10.1080/14697688.2019.1639802?scroll=top&needAccess=true

• I understand that as volatility increases the price of the binary approaches 0.5. I also understand the mapping between binary bets, choice prices, and probabilities. What I do not understand is what exactly the arbitrage process is. Taleb makes a claim that Silver can be arbitraged via a Dutch book sort of procedure. But I do not see how. I have read the papers you linked and either they do not explain it or I am just not understanding it. Do you know how?
– roz
Aug 12 '19 at 13:34
• Buying the binary when its price is very low and selling it when its price is very high based on the assumption that vol is very high is not an arbitrage; you are just trading vol. So if this is the process that Taleb is referring to when he says arbitrage, I think he is not correct.
– roz
Aug 12 '19 at 13:38
• @roz The key point is that the prices should reflect proper updating, which Taleb defines as martingale valuation-based revision. Otherwise, the pricing is systematically wrong and theoretically can be exploited by Dutch booking. e.g. one casual commentator suggests that when the 538 forecast for a candidate goes up one day, it is more likely to go up the next day as well. Clearly the "underlying volatility of the asset" (as Taleb would say) is not being properly priced in at any stage. Jan 14 '21 at 22:40


text:
# Weeks manifold

In mathematics, the Weeks manifold, sometimes called the Fomenko–Matveev–Weeks manifold, is a closed hyperbolic 3-manifold obtained by (5, 2) and (5, 1) Dehn surgeries on the Whitehead link. It has volume approximately equal to 0.9427... and Gabai, Meyerhoff & Milley (2009) showed that it has the smallest volume of any closed orientable hyperbolic 3-manifold. The manifold was independently discovered by Weeks (1985) and Matveev & Fomenko (1988).

## Volume

Since the Weeks manifold is an arithmetic hyperbolic 3-manifold, its volume can be computed using its arithmetic data and a formula due to A. Borel:

${\displaystyle {\frac {3\cdot 23^{3/2}\zeta _{k}(2)}{4\pi ^{4}}},}$

where k is the number field generated by θ satisfying θ 3 − θ + 1 = 0 and ζ k is the Dedekind zeta function of k (Ted Chinburg, Eduardo Friedman & Kerry N. Jones et al. 2001)

## Related manifolds

The cusped hyperbolic 3-manifold obtained by (5, 1) Dehn surgery on the Whitehead link is the so-called sibling manifold, or sister, of the figure-eight knot complement. The figure eight knot's complement and its sibling have the smallest volume of any orientable, cusped hyperbolic 3-manifold. Thus the Weeks manifold can be obtained by hyperbolic Dehn surgery on one of the two smallest orientable cusped hyperbolic 3-manifolds.


text:
# Power needed for saturated steam compression...

1. Aug 21, 2016

### pranj5

Want to discuss a matter here. If steam remains saturated during compression, how to calculate the power consumption for that by using steam table.
Suppose, saturated steam has been compressed from pressure level A to pressure level B and it remained saturated during and after compression. We have steam table before us and we know the gross enthalpy of steam a level A and level B.
Now, to calculate the power consumption with 100% efficiency, do we deduct gross enthalpy of steam at level A from gross enthalpy of steam at level B or there is some other method to be applied here?

2. Aug 21, 2016

### Staff: Mentor

Some clarification questions:

Are we talking about a continuous flow compressor, with steam entering one end and leaving through the other, or are we talking about compression in a piston-cylinder (closed system) arrangement?

Are we talking about a mixture of saturated steam and liquid water, or are we talking about pure saturated steam at pressure levels A and B?

Chet

3. Aug 21, 2016

### pranj5

This is pure saturated steam and it's a continuous process. Saturated steam enters the system and being compressed and leaves at higher pressure. But, by some special means, the steam remains saturated during the process and also leaves as saturated steam.

4. Aug 22, 2016

### Staff: Mentor

The power required is going to depend on the specific details of your proprietary process. It is not just the change in enthalpy. For example, to maintain the steam 100% quality throughout the compressor, the only way I can think of for accomplishing this is to remove heat from the system in a very precise way along the length of the compressor (so that the steam does not become superheated). For such a compressor, the open system version of the first law of thermodynamics would reduce to: $$\dot{Q}-\dot{W_S}+\dot{m}(h_A-h_B)=0$$where $\dot{Q}$ is the rate of adding heat, $\dot{W_S}$ is the rate at which the system is doing shaft work on the surroundings, $\dot{m}$ is the mass rate of flow of steam through the system, and h is the enthalpy per unit mass of the steam. The rate of doing shaft work on the surroundings is related to the power P by:$$P=-\dot{W_S}$$. Therefore,$$P=-\dot{Q}+\dot{m}(h_B-h_A)$$ So the power will be equal to the mass flow rate times the change in enthalpy per unit mass minus the rate at which heat is being added to the system. But, since we will be removing heat from the system, $\dot{Q}$ will be negative, and the power requirement will be higher than just that associated with the enthalpy change.

If you would like me to continue this analysis to show how the steam tables can be used to precisely quantify the power requirements for this particular scheme, please let me know.

5. Aug 22, 2016

### pranj5

Well, I like to explain your maths above in this way.
If the systems will be a 100% efficient, then Q will be zero here. But, like all man made machinery, this system too is inefficient upto a degree and Q is the extra amount of work that will be done by compressor other than increasing the enthalpy hB to hA. This additional amount of heat has to be removed from the system to keep the steam saturated. Am I right?
At present, many steam compressor manufacturers try to control temperature by spraying water and I have already mentioned that the steam will be in contact with water during compression. Therefore, can we conclude that this extra amount of heat will be converted into some extra steam?

6. Aug 22, 2016

### Staff: Mentor

The part about removing heat to keep the steam saturated is correct. The rest is mostly not correct. It is still possible to operate the compressor reversibly, even if there is heat removed or added. The reversible operation assumes that there is no dissipation of mechanical energy or thermal energy (i.e., no inefficiency of this kind)

I don't quite understand this question. Are you saying that you are considering spraying liquid water into the compressor along the length of the compressor in order to make sure that the flowing steam remains saturated at 100% quality? Are you saying that you would like me to analyze such a process quantitatively?

7. Aug 22, 2016

### pranj5

If there is no inefficiency of any kind, then where this extra heat will do? As far as I can understand the process, for a 100% efficient system all the work will be added to the steam and increase its enthalpy. But, unlike other compressors where steam is superheated from the very beginning and behave as an ideal gas, in this process the steam will remain saturated and at the end, the work done can be calculated by subtracting the gross enthalpy of State B from State A. Here in the process, the steam compression always follow the saturation curve.
Kindly just tell me what the extra heat will do in such a scenario. What I want to know is nothing but the power consumption for steam compression if the steam remains saturated during the compression. I am not saying that I am spraying water to keep the steam saturated, but just want to say that the steam will be in contact with water during compression and therefore remain saturated. How? That's a different question. Just keep it in mind that the steam is saturated during the process.

8. Aug 22, 2016

### Staff: Mentor

So you're saying that you do have a saturated mixture of steam and liquid water that you are processing in the compressor? Is that correct?

9. Aug 22, 2016

### pranj5

Yes. You can say that. But, it's not just wet steam. The amount of water here is much more than the amount of steam. I hope you have remembered that in such a scenario, a part of steam will dissolve in water and raise its enthalpy and the steam will always remain saturated.

10. Aug 22, 2016

### Staff: Mentor

So it's mostly liquid water with a smaller fraction of steam (by mass, not volume)?

11. Aug 22, 2016

### pranj5

Yes, you can say that.

12. Aug 22, 2016

### Staff: Mentor

So, what are the starting and ending conditions of the process?

13. Aug 22, 2016

### pranj5

That has already been stated at the starting of this thread. There is a specific volume of saturated steam here that has been compressed from level B to level A and before, during and after the compression; the steam remained saturated and not superheated.

14. Aug 22, 2016

### Staff: Mentor

Problem Statement:

There is a saturated mixture of steam and liquid water being processed through a continuous adiabatic reversible compressor at $\dot{m}$ kg/sec. The inlet mixture (State A) is at pressure $p_A$ and water vapor mass fraction $x_A$. The mixture at the outlet (State B) is saturated at pressure $p_B$. The outlet mass fraction of water vapor $x_B$ is, as yet, unknown. The objective is to determine the mechanical power required to run the compressor. Since the compressor is operating adiabatically and reversibly, the entropy per unit mass of the outlet stream $s_B$ is equal to the entropy per unit mass of the inlet stream $s_A$.

Step 1: Determine the entropy per unit mass and the enthalpy per unit mass of the inlet stream in State A
a. From the saturated steam tables, at pressure $p_A$, determine the following:
• $h_{LA}$ = enthalpy per unit mass of liquid water
• $h_{VA}$ = enthalpy per unit mass of water vapor
• $s_{LA}$ = entropy per unit mass of liquid water
• $s_{VA}$ = entropy per unit mass of water vapor
b. Using these data, calculate the enthalpy per unit mass $h_A$ and the entropy per unit mass $s_A$ of the entering mixture in State A
$$h_A=h_{LA}+(h_{VA}-h_{LA})x_A$$
$$s_A=s_{LA}+(s_{VA}-s_{LA})x_A$$

Step 2: Determine the mass fraction of water vapor in the outlet stream (State B) from the compressor $x_B$
a. From the saturated steam tables, at pressure $p_B$, determine the following:
• $h_{LB}$ = enthalpy per unit mass of liquid water
• $h_{VB}$ = enthalpy per unit mass of water vapor
• $s_{LB}$ = entropy per unit mass of liquid water
• $s_{VB}$ = entropy per unit mass of water vapor
b. Since $s_B=s_A$, using these data, calculate the exit water vapor mass fraction $x_B$ from:
$$x_B=\frac{s_A-s_{LB}}{s_{VB}-s_{LB}}$$

Step 3 Calculate the enthalpy per unit mass of the mixture in outlet State B $h_B$
$$h_B=h_{LB}+(h_{VB}-h_{LB})x_B$$

Step 4 Calculate the mechanical power that must be supplied to the compressor
$$P=\dot{m}(h_B-h_A)$$

15. Aug 22, 2016

### Staff: Mentor

No you didn't: I'm asking for the actual numbers. What pressure, in pascals or bar, are states A and B?

Last edited: Aug 23, 2016
16. Aug 23, 2016

### pranj5

@Chestermiller, many many thanks for your valuable contribution. I will answer your questions in a few days. But, I want to be assured about a few points. In Step 4, you have considered the power consumption to be equal to the difference in gross enthalpy of steam a Pressure level B and Pressure level A multiplied by steam flow. Kindly just tell me that do you agree to that if the steam remain saturated before, during and after compression; then the power consumption would be just the same amount as per the formula in step 4. It's a very very important assumption, at least for me.

17. Aug 23, 2016

### Staff: Mentor

Yes, provided you calculate the enthalpy per unit mass in States A and B according to the equations I presented in Steps 1 and 3, and the compression operates ideally (i.e., adiabatic and reversibly).

18. Aug 23, 2016

### pranj5

In case, the compression isn't idea, what will happen to the excess amount of energy. Do this add to the system as a whole and evaporate a little more amount of water?

19. Aug 23, 2016

### Staff: Mentor

Yes the final temperature will be higher, and you will have to do more work to compress it.

20. Aug 24, 2016

### pranj5

The answer isn't clear. If the inefficient compression will evaporate more water, that means adding some more steam to the system. But how that means higher temperature and more power to compress? Though it can be easily understood that this extra amount of steam will increase the density and pressure. But, if the pressure level is the goal, then it means reaching to the desired pressure level before time. In short, the extra steam will help in increasing the pressure and if there is special valve available and will open when the pressure inside will reach a level, then it would be opened earlier.


text:
CGAL 5.5 - 2D Arrangements
CGAL::Arr_geodesic_arc_on_sphere_traits_2< Kernel, X, Y >::Construct_curve_2 Class Reference

#include <CGAL/Arr_geodesic_arc_on_sphere_traits_2.h>

## Definition

Construction functor of geodesic arcs.

Is Model Of:

Assignable

CopyConstructible

AdaptableUnaryFunction

AdaptableBinaryFunction

AdaptableTernaryFunction

## Types

typedef Arr_geodesic_arc_on_sphere_traits_2< Kernel, X, Y >::Point_2 Point_2

typedef Arr_geodesic_arc_on_sphere_traits_2< Kernel, X, Y >::Curve_2 result_type

typedef Kernel::Direction_3 Direction_3

typedef Direction_3 argument_type

## Operations

X_monotone_curve_2 operator() (const Direction_3 &normal)
Construct a full great circle from a normal to a plane. More...

Curve_2 operator() (const Point_2 &p, const Point_2 &q)
Construct the minor geodesic arc from two endpoints. More...

Curve_2 operator() (const Point_2 &p, const Point_2 &q, const Direction_3 &normal)
Construct a geodesic arc from two endpoints and a normal to the plane containing the arc. More...

## ◆ operator()() [1/3]

template<typename Kernel , typename X , typename Y >
 X_monotone_curve_2 CGAL::Arr_geodesic_arc_on_sphere_traits_2< Kernel, X, Y >::Construct_curve_2::operator() ( const Direction_3 & normal )

Construct a full great circle from a normal to a plane.

Parameters
 normal the normal to the plane containing the great circle.

## ◆ operator()() [2/3]

template<typename Kernel , typename X , typename Y >
 Curve_2 CGAL::Arr_geodesic_arc_on_sphere_traits_2< Kernel, X, Y >::Construct_curve_2::operator() ( const Point_2 & p, const Point_2 & q )

Construct the minor geodesic arc from two endpoints.

The minor arc is the one with the smaller angle among the two geodesic arcs with the given endpoints.

1. Find out whether the arc is x-monotone.
2. If it is x-monotone,
1. Find out whether it is vertical, and
2. whether the target is larger than the source (directed right).

An arc is vertical, iff

1. one of its endpoint direction pierces a pole, or
2. the projections of the endpoint directions onto the xy-plane coincide.
Parameters
 [in] p the first endpoint. [in] q the second endpoint.
Precondition
p and q must not coincide.
p and q cannot be antipodal.

## ◆ operator()() [3/3]

template<typename Kernel , typename X , typename Y >
 Curve_2 CGAL::Arr_geodesic_arc_on_sphere_traits_2< Kernel, X, Y >::Construct_curve_2::operator() ( const Point_2 & p, const Point_2 & q, const Direction_3 & normal )

Construct a geodesic arc from two endpoints and a normal to the plane containing the arc.

The two endpoints determine the plane. The normal determines the orientation of the plane and the final arc (whether its the minor arc or the major arc). The right-hand rule can be used to select the appropriate normal.

Parameters
 [in] p the first endpoint. [in] q the second endpoint. [in] normal the normal to the oriented plane containing the arc.
Precondition
Both endpoints lie on the given oriented plane.


text:
# Performance of ID's checking between Access and Excel

I want to increase the performance of this code:

Dim...
...
...
...
...

Set objExcel = CreateObject("Excel.Application")
objExcel.Visible = True

Set rs = db.OpenRecordset("Select DataPointVID from My_ValidReport", dbOpenSnapshot)

Do Until rs.EOF = True
For Each ws In wb.Worksheets
Set ValidMeldeposition = ws.UsedRange.Cells.Find(What:=rs!DataPointVID)
If Not (ValidMeldeposition Is Nothing) Then ValidMeldeposition.Interior.ColorIndex = 43
Next ws
rs.MoveNext
Loop

j = 0
j = j + 1
Dokumentenpfad = CurrentProject.Path & "\Dateiname" & Format(Date, " dd.mm.yyyy") & " (" & j & ")" & ".xlsx"
Loop

wb.SaveCopyAs (NeuerReport)
MsgBox "Your file was saved here: " & vbCrLf & "'" & NeuerReport & "'"
objExcel.Visible = True

Set objWorkbook = Nothing
Set objExcel = Nothing

End Sub

The problem is that the Excel file contains over 70 Worksheets and every Sheet includes ID's which are checked by the code and also by the Access-Query which includes the ID's.

• This sounds like a really fun use case for a 3 dimensional array. I hope you dont mind getting some grey hairs LOL Nov 21 '19 at 20:13
• what? :D I don't know how i can solve this :D never.. :D Nov 22 '19 at 8:55


text:
235 views
for (i=n ; i>0 ; i--)
{
for ( j=1;j<n ; j=j*2)
{
for ( k=0;k<j ;k++)
{

}

}

}

time complexity ??
how to find of inner and innermost loop?

edited | 235 views
0
**********
+1
$O\left ( nlog^{3}n \right )$
As first for loop goes upto n iterations
2nd for loop goes upto log n iterations
3rd for loop goes upto $\left ( log n \right )^{2}$ iterations
+1
I'm getting O($n^{2}$). Is it correct?
+1
@hemant yes you are corrrect.I made a mistake , solved now
0
I am getting O($n^{2} log n$)
+1
Why so much discussion on this? Outer loop is independent. Multiply O(n) at last. Analyze inner two step by step, you will get O(n).... now multiply. You get O(n^2).  This is an easy question. I guess some people are misreading the second loop. i.e     for ( j=1;j<n ; j=j*2)   ... It is  j<n ....not j<i
0
And as k is depedent on j, so only iteration of k will be counted  and iteration of j will not count
0
@Ahwaan @srestha

Kindly check my comment on the answer to this question. I hope the explanation  makes sense.
0
@srestha Iteration for j is already counted when we take one by one values of j and say what happens to k....   @rishi yes O(n sq)  and it was in GP too.

Outermost loop will run for $n$ times , will not effect $2^{nd}$ and $3^{rd}$ loop.

outermost loop will run for $n$times

$j=1\rightarrow k=1$

$j=2 \rightarrow k=1,2$

$j=4\rightarrow k=1,2,3,4$

$j=8\rightarrow k=1,2,3,4,5,6,7,8$

$\vdots \vdots$

$j=n\rightarrow k=1,2,3,4,5,6,7,8.....2^h$

k will run -:

3rd loop$=2^0+2^1+2^2+.....2^h$

$=2^{h+1}-1$

$2^h=n\Rightarrow h=\log n$

3rd loop$=2^{\log n +1}-1=n^{\log 2} \times 2-1\approx n$

Outermost loop will run $n$times

Overall time complexity=$\Theta (n^2)$

selected
0
Okay, so iterations of 3rd loop (innermost) covers the iteration of 2 nd loop.

And the outer most loop has no relation with any of inner most loop. That is time complexity = n * timeComplexity of inner loops.

Is it ?
0
yes $n*n$
0
@sourav

I have got some problem in ur ans

Say, i=1 -> j=1  -> k=0,1

i=2 ->j=1,2 -> k=0,1

k=0,1,2

i=3 ->j=1,2 -> k=0,1

k=0,1,2

i=4 ->j=1,2,4-> k=0,1

k=0,1,2

k=0,1,2,3,4

So, j will not run only 1,2,4............

but it will run like 1,2,2,4..............
0

@rishi @sourav

for i loop each n times

inner loops (logn + n) ~= n

therfore n2

am i right in my approach

0

@Sreshtha :first of all you do not bother about outermost loop,they have no any relation with 2nd and 3rd loop.

Now see carefully the innermost loop -:

for ( k=0;k<j ;k++)

let us write the above snippet as-:

for ( k=1;k<=j ;k++)

Now see k is running for every value of j.

j=1,k will run 1 times

j=2, k will run 2 times

j=4, k will run 4 times

j=8,k will run 8 times

.....

see the pattern

$1+2+4+8+...2^h$

0
But outer most loop also take a part for calculating inner two loops

and that is why k will take log n complexity and not n
0
there is no statement ,then how you calculating complexity
0
yes some time it depend on statement

But not always statement matters

As statement not given, we may think , it is statement independent
+2
@Srestha

Why are you doing like this?

--------------------------------------------------------

Say, i=1 -> j=1  -> k=0,1

i=2 ->j=1,2 -> k=0,1

k=0,1,2

i=3 ->j=1,2 -> k=0,1

k=0,1,2

i=4 ->j=1,2,4-> k=0,1

k=0,1,2

k=0,1,2,3,4

So, j will not run only 1,2,4............

but it will run like 1,2,2,4..............
--------------------------------------------------------

'j' is independent of 'i'. j <n is there and not j<i.
0

@A_i_s_h,

Lets forget about the outer loop for a while.

for ( j=1;j<n ; j=j*2){

for ( k=0;k<j ;k++){

}

}
 Number of Outer Loop Iteration 0th 1st 2nd 3rd . . . xth iteration value of j 1 2 4 8 .. . n Number of times inner loop runs 1 time 2 times 4 times 8 times . .. $2^{x}$ times

Now, we are concerned only about the number of times inner loop runs, which is a GP of powers of 2. = $2^{x+1}-1$

Now how is x nd n related ? $2^{x} = n$

Therefore, $2^{x+1}-1$ = 2n - 1 = O(n)

Now bring in the outer loop, which runs n times irrespective of any of the inner loop. Therefore, net time complexity  = O($n^{2}$)

0

@rishi

2x+1−1 = 2n - 1 = O(n)

can u elaborate this part

0
$2^{x} = n$

$2^{x+1}-1 = (2*2^{x})-1 = (2*n)-1$
0
tanku :)

@rishi

+1 vote


text:
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# Solve the Following Quadratic Equations by Factorization: 1/(X-1)(X-2)+1/(X-2)(X-3)+1/(X-3)(X-4)=1/6 - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

#### Question

Solve the following quadratic equations by factorization:

1/((x-1)(x-2))+1/((x-2)(x-3))+1/((x-3)(x-4))=1/6

#### Solution

We have been given,

1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)=1/6

((x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2))/((x-1)(x-4)(x-2)(x-3))=1/6

(3(x^2-5x+6))/((x^2-5x+4)(x^2-5x+6))=1/6

18 = x2 - 5x + 4

x2 - 5x - 14 = 0

x2 - 7x + 2x - 14 = 0

x(x - 7) + 2(x - 7) = 0

(x + 2)(x - 7) = 0

Therefore,

x + 2 = 0

x = -2

or,

x - 7 = 0

x = 7

Hence, x = -2 or x = 7.

Is there an error in this question or solution?

#### APPEARS IN

Solution Solve the Following Quadratic Equations by Factorization: 1/(X-1)(X-2)+1/(X-2)(X-3)+1/(X-3)(X-4)=1/6 Concept: Solutions of Quadratic Equations by Factorization.
S


text:
An easy proof of the uncountability of bijections on natural numbers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:03:02Z http://mathoverflow.net/feeds/question/29475 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29475/an-easy-proof-of-the-uncountability-of-bijections-on-natural-numbers An easy proof of the uncountability of bijections on natural numbers? Abhishek Parab 2010-06-25T06:47:53Z 2012-11-27T17:16:01Z
The proof that I have in mind is as follows -

$\text{Gal }(\overline{\mathbb Q}/\mathbb Q)$ is a proper uncountable subgroup of the group of permutations on countably many symbols, hence the latter is uncountable. .

But it needs a lot of jargon from topology and algebra. Is there a neat proof like Cantor's diagonal argument?
 http://mathoverflow.net/questions/29475/an-easy-proof-of-the-uncountability-of-bijections-on-natural-numbers/29476#29476 Answer by Ewan Delanoy for An easy proof of the uncountability of bijections on natural numbers? Ewan Delanoy 2010-06-25T07:07:19Z 2010-06-25T07:07:19Z
Indeed, Cantor's argument is readily adapted. Let $\pi_1,\pi_2,\pi_3, \ldots$ be any countable sequence of permutations of $\mathbb N$ ; let us show that this sequence does not exhaust all permutations, by constructing a permutation $\pi$ different from all the $\pi_i$. We first define $\pi$ on the even integers inductively, then define $\pi$ on the odd integers.

Let $X$ be any subset of $\mathbb N$ such that both $X$ and ${\mathbb N} \setminus X$ are infinite (e.g. the even integers, the prime numbers ...).
 Set $\pi(0)$ to be an integer in $X$ different from $\pi_1(0)$. Set $\pi(2)$ to be an integer in $X$ not in $\lbrace \pi(0),\pi_2(2)\rbrace$. Set $\pi(4)$ to be an integer in $X$ not in $\lbrace \pi(0),\pi(2)\pi_3(4)\rbrace$. Continuing like this inductively, we define an injection from the even integers to $X$, such that $\pi(2k-2) \neq \pi_k(2k-2)$ for any $k$ (and hence $\pi \neq \pi_k$).

Finally, the set A=${\mathbb N} \setminus \pi(2\mathbb N)$ is countably infinite ; setting $\pi(2k-1)=$ the $k$-th element of $A$ finishes the proof.
 http://mathoverflow.net/questions/29475/an-easy-proof-of-the-uncountability-of-bijections-on-natural-numbers/29488#29488 Answer by villemoes for An easy proof of the uncountability of bijections on natural numbers? villemoes 2010-06-25T09:30:21Z 2010-06-25T09:30:21Z
The fact that any conditionally convergent series [and that such exists] can be rearranged to converge to any given real number $x$ proves that there is an injection $P$ from the reals to the permutations of $\mathbb{N}$. The usual proof (pick positive terms until we exceed $x$, pick negative terms until we "deceed" $x$, repeat) is constructive; if we use a series consisting of easily computable rational numbers (the alternating harmonic series being the canonical example) and a real number $x$ for which we can decide $q<x$ or $q>x$ or $q=x$ for every rational number $q$, we may compute $P(x)(n)$ for every $n$.
 http://mathoverflow.net/questions/29475/an-easy-proof-of-the-uncountability-of-bijections-on-natural-numbers/29500#29500 Answer by Carl Mummert for An easy proof of the uncountability of bijections on natural numbers? Carl Mummert 2010-06-25T11:32:50Z 2010-06-25T13:36:10Z
I'm not sure what counts as an "easy" proof, but the following seems easy to me. Make a subtree of $\mathbb{N}^{<\mathbb{N}}$ by declaring that a sequence $\sigma$ is in the subtree if (1) $\sigma$ is injective and (2) whenever $\sigma(2n+2)$ is defined, it is equal to the least natural number not in the range of $\sigma \upharpoonright (2n+2)$. Then any path through this tree is a bijection of $\mathbb{N}$, and the tree is perfect because any sequence of even length has infinitely many immediate extensions in the tree (any number not already in the range is a candidate). So the tree has continuum-many paths.
 http://mathoverflow.net/questions/29475/an-easy-proof-of-the-uncountability-of-bijections-on-natural-numbers/29517#29517 Answer by Qiaochu Yuan for An easy proof of the uncountability of bijections on natural numbers? Qiaochu Yuan 2010-06-25T15:06:04Z 2010-06-25T22:08:51Z
Bijections of the natural numbers are equinumerous with functions $\mathbb{N} \to \mathbb{N}$, which are equinumerous with continued fraction expansions of positive irrational numbers.

Edit: Here's one proof of the first assertion. Clearly there are at most as many bijections $\mathbb{N} \to \mathbb{N}$ as functions $\mathbb{N} \to \mathbb{N}$. On the other hand, for any function $f : \mathbb{N} \to \mathbb{N}$ there exists a bijection with $f(i)$ cycles of length $i$. Now apply Cantor-Schroeder-Bernstein.
 http://mathoverflow.net/questions/29475/an-easy-proof-of-the-uncountability-of-bijections-on-natural-numbers/29541#29541 Answer by Pete L. Clark for An easy proof of the uncountability of bijections on natural numbers? Pete L. Clark 2010-06-25T20:02:33Z 2010-06-25T20:02:33Z
Consider the following map $\Phi$ from bijections of the natural numbers to subsets of the natural numbers: to each bijection $\sigma$, associate its fixed point set $S(\sigma) = \{x \in \mathbb{N} \ | \ \sigma(x) = x \}$.

It is easy to see that this map is almost surjective: the only subsets which are not in the image are those whose complement consists of a single element. Indeed, a subset of the natural numbers admits a fixed point free permutation iff it does not consists of a single element. In particular, the complement of the image of $\Phi$ is countable. Since the codomain of $\Phi$ is uncountable, so therefore is the image of $\Phi$.

This shows that the set of bijections is uncountable. By the usual Schroder-Bernstein argument, it follows that it has continuum cardinality.
 http://mathoverflow.net/questions/29475/an-easy-proof-of-the-uncountability-of-bijections-on-natural-numbers/114674#114674 Answer by Adam Kapelner for An easy proof of the uncountability of bijections on natural numbers? Adam Kapelner 2012-11-27T17:16:01Z 2012-11-27T17:16:01Z
I think there's an easier proof but I can't quite do every step.

Imagine the bijection from $f : A \longrightarrow A$ where $A = \{1,2,3\}$. 1 can be mapped to 3 numbers, 2 can be mapped to the remaining 2 numbers and 3 can be mapped to 1 number. Hence the size of the size of the bijection is the factorial of the size of the set being mapped, ie $|A|!$.

Thus, the size of the bijection $f : \mathbb{N} \longrightarrow \mathbb{N}$ should be $|\mathbb{N}|!$ which is clearly bigger than $2^{|\mathbb{N}|}$, hence it is uncountable.

Can you let me know if this suffices?


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## California releases summary of Advanced Clean Car Rules: GHG, ZEV, LEV III and Clean Fuel Outlets

##### 16 November 2011

The California Air Resources Board (ARB) released a summary of its proposed Advanced Clean Cars (ACC) regulation package (earlier post), which establishes stringent new emissions standards for light duty vehicles from 2015 through 2025. Further documents will be released on 7 December. The Advanced Clean Cars package combines four different regulations: light-duty vehicle greenhouse gas limits; the Zero Emission Vehicle (ZEV) mandate; the next phase of the Low Emission Vehicle (LEV-III) program; and a new Clean Fuels Outlet regulation.

Separately, the US DOT and EPA released their joint proposal to set stronger fuel economy and greenhouse gas pollution standards for Model Year 2017-2025 passenger cars and light trucks. (Earlier post.) The California and Federal greenhouse gas emissions standards are aligned; California will accept compliance with the Federal standards as meeting state requirements.

The Advanced Clean Cars package of regulations is designed to deliver:

• A 47% reduction of greenhouse gas emissions by 2025, compared to today’s levels;

• A further 75% reduction in smog-forming emissions by 2025;

• One in seven new cars sold in 2025 (15.4%) be a zero-emission or plug-in hybrid vehicle;

• A total of 1.4 million zero-emission and plug-in hybrid vehicles on the road in California by 2025;

• A reduction of 40 million tons of greenhouse gases in 2025, the equivalent of taking eight million cars off the road; and,

• A savings of $5 Billion in operating costs in 2025 for California drivers. This will rise to$10 Billion in 2030 when more advanced cars are on the road.

The proposed Advanced Clean Car package has been in development over the past three years and comprises four separate but related regulations:

• Greenhouse gas standard for cars and light trucks, model years 2017-2025 This regulation builds on California’s first-in-the-nation greenhouse gas standard that was later adopted by the federal government as part of a national program in 2009. The current proposal to strengthen the greenhouse gas standard was developed in tandem with the federal government over the past three years, toward accommodating manufacturers’ desire for a single national program within California’s separate motor vehicle emission control program. It is designed to parallel the national greenhouse gas and fuel economy standards rule the U.S. EPA and the Department of Transport are releasing today under their coordinated rulemaking.

The proposed new standard drops greenhouse gas emissions to 166 grams per mile, a reduction of 47% from current levels. This will be achieved through existing technologies (such as hybrid cars), the use of stronger and lighter materials, and more efficient drivetrains and engines.

• LEV III. California will need to reduce smog-forming pollution by an additional 75% from 2016 levels to help meet more stringent Federal air quality standards expected in the next few years. Since California continues to have the nation’s worst air quality, and has more than 26 million cars on the road, it is necessary to further reduce smog-forming pollution from cars. This regulation will drive the development of the cleanest cars yet that use today’s diesel, gasoline-powered, or typical gas-electric hybrid internal combustion engines.

In comments at the Los Angeles Auto Show, ARB Chairman Mary Nichols said that California LEV III will be harmonized with the upcoming US EPA Tier 3 regulations.

• Zero Emission Vehicle Regulation. This regulation builds on the program in place since 1990 and is designed to rapidly increase ZEV production to early commercial volumes, establishing a sustainable and growing market for these advanced technology vehicles. This will place California on a path to reduce greenhouse gas emissions by 80% by 2050, a goal adopted by many nations and believed necessary to stabilize climate temperature.

ARB analysis demonstrates that the ZEV regulation is required to put 1.4 million ZEVs on the road by 2025 (15.4 percent of new vehicle sales in that year) in order to be on track to reach the 2050 greenhouse gas reduction goal. ARB says that a transitional model—the plug-in hybrid car—will play a significant role over the next twenty years but by mid-century, 87% of cars on the road will need to be full zero-emission vehicles to achieve climate goals.

• Clean Fuel Outlets. This regulation is designed to address the pending commercialization of zero-emission hydrogen fuel cell vehicles by requiring the construction of hydrogen fueling stations. Construction of the new stations will provide a convenient fueling infrastructure, first within the major air basins but ultimately throughout the state.

### Comments

Some Good, Some bad, Some just absolutely Stupid.

Good: Tightened toxic emissions to the final ever levels, just pure clean AIR, which we already do.

Just Bad: Tightening CAFE mileage requirements on a ridiculous schedule.

Just Stupid: CO2 regulations on a continent that emits ZERO net CO2 already, and 21st Century Science shows is just a scientific nullity and all the pseudo-scientific Truth of leftist Lysenko-ism.

I suspect long before these go into effect, California will go bankrupt, and organizations like CARB will be RIFFed.

Clean electric heating, cooking, vehicles, trains, off-grid clean distributed energy production, energy efficiency programs etc etc have nothing to do with California's and USA's potential bankruptcy. It is common sense evolution that may return more wealth and buying power to the majority.

USA (and many other countries) are going bankrupt because 1% (and even fewer tomorrow) are pilfering 90% of the national wealth. The other 99% are getting poorer, can't maintain their acquired lifestyle and can no longer afford to buy what the 1% is over-producing and pushing on them.

Surpluses are piling up. Factories are closing. The middle class is going down to the lower class level and buying less and less. The lower class is going for food stamps and is not buying. Total unemployment is going up another step every month.

The anti-snowball effect is progressively melting down the national economy. Governments are going deeper in debt trying to give a sense of economic security. Sooner or latter, people will realize that less than 1% cannot continue to be millionaires and billionaires for extended periods and that the other 99% will become as poor as the serf of the middle ages.

We may find ourselves right back where we were 500 years ago.

"requiring the construction of H2 fueling stations". This makes no sense to me.
And "requiring" 15.4% of new cars sold be ZEV by 2025. Well, let's hope it's more than that(without "requiring" anything).

damn....One hundred + (100) years ago, ICE vehicles required many thousand new gas stations. It made NO SENSE but they well built!!!!.

Now is the time to move on to other energy sources and leave wood, coal and oil/NG burning behind.

The world will have to electrify. Resistance is futile.

Harvey, if H2 vehicles really are a viable reality, i'm all for "priming the pump" and building some public stations at tax payer expense.
But i can't see spending that money if these vehicles are decades off.
100 yrs ago the govt didn't have to mandate the building of gas stations. Entrepreneurs saw it as a way to make money and spent their own dollars.
There is nobody more ready to move onto something new. But i see electrical outlets everywhere...no need to mandate their construction.

danm....in the last 100+ years the dirty oil/gas industry received XXXX$B in subsidies, hand outs and oil wars cost. In all fairness, clean H2O and EVs industry should get many times more. Of course, private industry will also contribute xxxB$ because it will be beneficial to do so.

The comments to this entry are closed.


text:
main-content

## Über dieses Buch

Where do solutions go, and how do they behave en route? These are two of the major questions addressed by the qualita tive theory of differential equations. The purpose of this book is to answer these questions for certain classes of equa tions by recourse to the framework of semidynamical systems (or topological dynamics as it is sometimes called). This approach makes it possible to treat a seemingly broad range of equations from nonautonomous ordinary differential equa tions and partial differential equations to stochastic differ ential equations. The methods are not limited to the examples presented here, though. The basic idea is this: Embed some representation of the solutions of the equation (and perhaps the equation itself) in an appropriate function space. This space serves as the phase space for the semidynamical system. The phase map must be chosen so as to generate solutions to the equation from an initial value. In most instances it is necessary to provide a "weak" topology on the phase space. Typically the space is infinite dimensional. These considerations motivate the requirement to study semidynamical systems in non locally compact spaces. Our objective here is to present only those results needed for the kinds of applications one is likely to encounter in differen tial equations. Additional properties and extensions of ab stract semidynamical systems are left as exercises. The power of the semidynamical framework makes it possible to character- Preface ize the asymptotic behavior of the solutions of such a wide class of equations.

## Inhaltsverzeichnis

### Chapter I. Basic Definitions and Properties

Abstract
After the appropriate definitions in Section 2 we present in Section 3 a simple example of a semidynamical system in an infinite dimensional space. The example arises in conjunction with a Poisson process. The objective of Section 4 is to extend semidynamical systems to the negative time domain. This results in a classification of the maximal time domains of semidynamical systems. In Section 5 we characterize critical and periodic motions and prove an extension theorem for periodic motions. Examples are presented which distinguish semidynamical system behavior from its precursor, dynamical systems. Section 6 is devoted to both an algebraic and a topological classification of positive orbits. The chapter is closed following a reparametrization theorem. It permits us to treat global semidynamical systems throughout the book. This results in a considerable savings as regards the theory without losing the applicability of the results to many important examples.
Stephen H. Saperstone

### Chapter II. Invariance, Limit Sets, and Stability

Abstract
The main concerns of this chapter are two-fold. Firstly, where do positive motions go as t → ∞, and secondly, what can we say about the behavior of the resulting limiting orbits?
Stephen H. Saperstone

### Chapter III. Motions in Metric Space

Abstract
Stability and attraction for sets as defined in Chapter II only yield information on the behavior of motions near a positively invariant set. Much detail is obscured by only considering the positive orbit or its hull. As Sell points out in [1], stability and attraction of sets are too crude to give much information about the behavior of motions within a set. A finer tool is needed here. In part, this can be accomplished by endowing the phase space with a metric or uniform structure. Also, the consideration of almost periodic motions requires completeness of the phase space. This added structure allows us to answer the question posed in the opening paragraph of Section 4 of Chapter II; namely, when does a semidynamical system extend (uniquely) to a dynamical system? Moreover, in this setting we can complete- the classification of compact positively minimal sets into the closure of recurrent, uniformly recurrent, almost periodic, periodic, and critical motions.
Stephen H. Saperstone

### Chapter IV. Nonautonomous Ordinary Differential Equations

Abstract
The solutions of the autonomous ordinary differential equation
$$\dot{x} = f(x)$$
(1.1)
(where ẋ stands for dx/dt) give rise to a semidynamical (even dynamical) system on IRd provided f: W → IRd is continuous on the open subset W ⊂ IRd and the solutions of Equation (1.1) through any point (x0,t0) ∈ W × IR are uniquely defined and remain in W for all time. In fact, if Φ(x0;t) denotes the solution of Equation (1.1) through (x0,0) evaluated at time t ∈ IR+, it can be verified that (W,Φ) is a semidynamical system.
Stephen H. Saperstone

### Chapter V. Semidynamical Systems in Banach Space

Abstract
Material fundamental to the existence and qualitative behavior of partial differential equations and differential delay equations (to name just two areas) are developed in this chapter. The general formulation is an evolution equation in a Banach space. The work of Crandall and Liggett on the nonlinear version of the Hille — Yosida — Phillips theorem for linear semigroups has spawned an elegant analysis of nonlinear evolution equations in Banach spaces. One special feature of the semigroup generation theorem (linear or nonlinear) is that we obtain a representation of the solutions to du/dt + Au = 0 in terms of the operator A. The classical approach was to establish the existence, uniqueness, and continuous dependence of the solutions of the particular partial differential equation, for example, and then demonstrate that the solutions generate a semigroup. This was essentially the approach we also took in Chapter IV.
Stephen H. Saperstone

### Chapter VI. Functional Differential Equations

Abstract
A distinguishing feature of ordinary differential equations is that the future behavior of solutions depends only upon the present (initial) values of the solution. Numerous physical, economic, biological, and social systems, though, exhibit hereditary dependence. That is, the future state of the system depends not only upon the present state, but also upon past states. Models of such systems must take into account this hereditary effect. We illustrate this with some examples.
Stephen H. Saperstone

### Chapter VII. Stochastic Dynamical Systems

Abstract
We turn to a semidynamical system which is generated by a Markov process. Here again we obtain, in general, a non-differentiable system.
Stephen H. Saperstone

### Chapter VIII. Weak Semidynamical Systems and Processes

Abstract
Many of the important properties of semidynamical systems which were developed in the first three chapters can essentially be obtained with a weaker continuity axiom (Definition 2.1(iii) of Chapter I); namely, assume that π(x,t) is only continuous in x ∈ X. In particular, we still obtain weak in-variance of compact positive limit sets. In addition, if the continuity in x is uniform with respect to t ∈ IR+, then (X,π) extends to a weak dynamical system on the positive limit sets. Moreover, the positive limit sets will then be minimal with respect to this flow. Finally, we will still be able to show, as in Chapter III, that the positive limit sets are equi-almost periodic.
Stephen H. Saperstone

### Backmatter

Weitere Informationen


text:
# How to calculate the area of a parabolic dish

I feel like this should be really easy, but I'm not sure if I'm doing it correctly so I'm going to give it a go here, and if I'm not very good at maths (I'm not) then you can hopefully correct me!

Suppose I want to paint the surface of a parabolic dish, how would I calculate that area?

To make it slightly more clear, this type of thing:

Where a cross section of it would be some function:

$y = a x^2\quad$ (1)

from $y = 0$ (what I will call the vertex) to some point $y=h$ at the opening of the dish.

I thought then that the radius at a certain height would be:

$r = \sqrt{\frac{y}{a}}\quad$ (2)

and therefore the radius at the top, $d$, would be given by the equation:

$d = \sqrt{\frac{h}{a}}$

or in other words:

$a = \frac{h}{d^2}$

So I rewrote (1) and (2) using that:

$y = \frac{h}{d^2} x^2 \quad$ (3)

$r = \sqrt{\frac{d^2y}{h}}\quad$ (4)

Now I can find the areas of small strips around the dish, and integrate. And this is where I get a bit stuck. I initially did the area of rectangles, but of course the strips (if you unfold them) are actually like this:

(of course they're not exactly like that because that's a cone and we have a dish, but they're similar in how they're both not rectangles for almost the same reason, if that makes sense)

And I have no idea how to express that area mathematically...

• The practical answer is to measure the diameter of the opening of the dish $d$, compute the area of a circle with that diameter $\frac \pi 4 d^2$, and buy a little more paint than that. The true area won't be too much more. – Ross Millikan Jun 27 '15 at 15:29

The formula for the surface area of a curve rotated about the $y$-axis (see the Theorem of Pappus) is \begin{align} \int_0^r\overbrace{\vphantom{\sqrt{y'^2}}2\pi x}^{\text{length of strip}}\overbrace{\sqrt{1+y'^2}\,\mathrm{d}x}^{\text{width of strip}} &=\int_0^r2\pi x\sqrt{1+4a^2x^2}\,\mathrm{d}x\\ &=\frac{\pi}{4a^2}\int_0^{4a^2r^2}\sqrt{1+t}\,\mathrm{d}t\\[6pt] &=\frac{\pi}{6a^2}\left(\sqrt{1+4a^2r^2}^3-1\right)\tag{1} \end{align} where $t=4a^2x^2$. You can use $h=ar^2$ to get $$\frac{\pi r}{6h^2}\left(\sqrt{r^2+4h^2}^3-r^3\right)\tag{2}$$

• As $a\to0$, the expression in $(1)$ tends to $\pi r^2$. – robjohn Jun 28 '15 at 1:46

As you describe it, the origin is at the center of the dish, $x$ is radial, and $y$ is up the tower. If you know the radius of the dish $r$ and the height of the edge $h$, you have $h=ar^2, a=\frac h{r^2}$ Now consider a small piece of the curve from $x$ to $x+dx$. The area swept out as you rotate this part of the curve is the arc length between $x$ and $x+dx$ times $2 \pi x$. The element of arc length is $ds=\sqrt{dx^2+dy^2}=dx\sqrt{1+(\frac {dy}{dx})^2}$ So you want $\int_0^r dx2 \pi x\sqrt{1+(\frac {dy}{dx})^2}=\int_0^r dx2 \pi x\sqrt{1+(2ax)^2}$

Added: Alpha integrates this to $\frac {\pi (4a^2r^2+1)^{3/2}}{6a^2}$, which should reduce to $\pi r^2$ in the limit $a \to 0$, but I can't get it to do so. The integral above does reduce that way nicely.

• I specifically didn't want it in terms of $a$ which was why I eliminated $a$ in my equations. $h$ and $r$ are measurable you see, so I wanted the area in terms of those. But never mind, we could just substitute back in. – Jamie Twells Jun 27 '15 at 15:50
• Actually, could you expand on why $2\pi x \mathrm{d}s$ is the area? Why is it not more complicated than that? I mean, imagine a rectangle with dimensions $2 \pi x$ and $\mathrm{d}s$. If you fold that rectangle around to meet, then lie it on the dish, it leaves a V shape where the bottom corners meet but the top corners have a little gap, surely? – Jamie Twells Jun 27 '15 at 16:00
• So does this eventually give $A = \frac{\pi}{hd}(2hx^2+d^2)^\frac{3}{2}$ where $d$ is radius at the opening (widest radius) and $h$ is the height – Jamie Twells Jun 27 '15 at 16:57
• For your second comment, the overlap area is second order in differentials-it is proportional to $ds^2$ so doesn't contribute. For the integral, Alpha gets $\frac {2 \pi (1+ar^2)^{3/2}}{3a}$. I haven't checked if that matches your solution. You shouldn't have an $x$ in your result. For comparison, the area of a spherical cap is $2 \pi h R$, where $R$ is the radius of the sphere. Your shape is not very different. To get $R$ you have $R^2=r^2+(R-h)^2$ – Ross Millikan Jun 27 '15 at 19:45
• yes sorry should have been $d^2$ not $x^2$ – Jamie Twells Jun 27 '15 at 19:48