text:
# Decent opamp for comparator?

#### spinnaker

Joined Oct 29, 2009
7,835
I am just getting into the use of opamps with my mcu design.

I was wondering which general purpose op amp to go with for compartors. My supplies are typically 3 to 5v. Single supply. My input signals are typically from a few hundred mv to 5V.

Radio Shack has 741s on hand so they are accessible. Was wondering if this is a good way to go. I plan on sourcing additional chips elsewhere.

I may not have provided enough information for an answer through my ignorance so just ask followup questions if needed.

Joined Jan 15, 2015
5,776
When it comes to Voltage Comparator Information And Circuits this is one of my favorite sites.

NOTE: Operational Amplifiers (OPAMPS) such as LM324, LM358 and LM741 are not normally used voltage comparator circuits due to their bipolar outputs. However, OPAMPS can be used as voltage comparators if a diode or transistor is added to the amplifiers output to create and open collector type output. A diagram showing this addition is further down this page.

This page provides basic information about voltage comparator integrated circuits and is to act as reference material for other circuits. The circuits shown are based on the LM339 Quad Voltage Comparator chip or the LM393 Dual Voltage Comparator chip. These devices are functionally the same. The LM311 Voltage Comparator can be used for these applications as well but it also has a number of unique features.
I like and have used the ancient LM 339 Quad, LM 393 Dual and LM 311 Single comparators over the years with very good results. Just to recap, I would not use the LM 741 Operational Amplifier as a comparator as there are much better choices out there.

Ron

#### OBW0549

Joined Mar 2, 2015
3,566
I was wondering which general purpose op amp to go with for compartors. My supplies are typically 3 to 5v. Single supply. My input signals are typically from a few hundred mv to 5V.
Forget about using an op amp. Op amps are usually way too slow to be used as comparators, except in applications where response time is completely irrelevant.

Radio Shack has 741s on hand so they are accessible. Was wondering if this is a good way to go.
No, that would be an extremely poor choice; aside from the response time issue, a 741 won't operate off 3-5 volt single supplies-- it probably won't even twitch, much less do anything useful.

If you want a comparator, use a comparator: an LM393 is a good choice, they operate on single supplies as low as 2 volts and as high as 36 volts, and they're cheap.

#### spinnaker

Joined Oct 29, 2009
7,835
Forget about using an op amp. Op amps are usually way too slow to be used as comparators, except in applications where response time is completely irrelevant.

No, that would be an extremely poor choice; aside from the response time issue, a 741 won't operate off 3-5 volt single supplies-- it probably won't even twitch, much less do anything useful.

If you want a comparator, use a comparator: an LM393 is a good choice, they operate on single supplies as low as 2 volts and as high as 36 volts, and they're cheap.

Thanks that is sort of what I thought that there were better choices out there. I sure wish there was an electronics supply nearby. Hosfelt was the closest about 20 miles from here. They closed shop but sold a lot of out dated stuff anyway.

#### crutschow

Joined Mar 14, 2008
26,321
There are few brick & mortar electronic supply shops anymore.
Easiest is just to buy online such as from Digikey, Mouser, or Jameco.
If you want to take a chance on quality (i.e., counterfeit parts), you can also order from Ebay.

#### OBW0549

Joined Mar 2, 2015
3,566
I sure wish there was an electronics supply nearby. Hosfelt was the closest about 20 miles from here. They closed shop but sold a lot of out dated stuff anyway.
Why not just order from Digi-Key? I buy stuff from them all the time over the Web; AFAIK they don't have any minimum order quantity or minimum order $$, and you can get darned near everything from them. Amazon.com is also a decent supplier for some things, though their selection is limited and you take a chance on possibly dodgy Chinese stuff sometimes. Thread Starter #### spinnaker Joined Oct 29, 2009 7,835 Why not just order from Digi-Key? I buy stuff from them all the time over the Web; AFAIK they don't have any minimum order quantity or minimum order$$\$, and you can get darned near everything from them. Amazon.com is also a decent supplier for some things, though their selection is limited and you take a chance on possibly dodgy Chinese stuff sometimes.
No problem ordering but it does not get here today.

#### Phil-S

Joined Dec 4, 2015
180
I'm currently dabbling with window comparators.
As said by others, don't use an op amp, comparators are op amps optimised (is that double-Dutch?) for just one job as the name suggests.
For no other reason than a helpful datasheet, voltage etc., I chose the Microchip MCP6542. A dual comparator, runs nicely on 5-V. The dual is ideally suited to checking if a voltage is within two levels set by external references (just voltage dividers).
I had little success with trying to set up external hysteresis (working with a noisy input) as even after calculating the resistor values, there was too much interaction if resistor values are changed.
Having said all that, I'm drifting towards the idea of just using a microcontroller instead (AtTiny85) - it will do everything the comparator will do, smoothing, comparator, hysteresis, etc. - it's a lot easier to work with numbers. Although I haven't tried it yet, the Atmel MCUs have comparators built-in

Joined Jan 15, 2015
5,776
No problem ordering but it does not get here today.
Today? Today is late, I wanted it yesterday.

Never thought I would admit it but I miss Radio Shack. I will say I like Amazon Prime which the wife is a master of using. Two day shipping works for me most of the time.

Ron

#### GopherT

Joined Nov 23, 2012
8,012
No problem ordering but it does not get here today.
If you don't mind driving up to GopherT's shop, can meet up and give you a few LM393, LM339 and LM311 to play with.

#### KL7AJ

Joined Nov 4, 2008
2,229
I am just getting into the use of opamps with my mcu design.

I was wondering which general purpose op amp to go with for compartors. My supplies are typically 3 to 5v. Single supply. My input signals are typically from a few hundred mv to 5V.

Radio Shack has 741s on hand so they are accessible. Was wondering if this is a good way to go. I plan on sourcing additional chips elsewhere.

I may not have provided enough information for an answer through my ignorance so just ask followup questions if needed.
If you don't need super high speed, a plain vanilla 741 works just fine.

Joined Jan 15, 2015
5,776
If you don't mind driving up to GopherT's shop, can meet up and give you a few LM393, LM339 and LM311 to play with.
Just getting ready to go out of town for 3 weeks but when I get back I can mail you some. Then too, there is the world famous Gopher T's shop right up the road for your shopping pleasure with reasonable prices like free to a good home.

Ron

#### KL7AJ

Joined Nov 4, 2008
2,229
My shop is just as famous as Gopher T's but for some reason folks don't want to visit often.

#### spinnaker

Joined Oct 29, 2009
7,835
Just getting ready to go out of town for 3 weeks but when I get back I can mail you some. Then too, there is the world famous Gopher T's shop right up the road for your shopping pleasure with reasonable prices like free to a good home.

Ron

Thanks. Took a chance and ordered from ebay. But I might hit up Gopher too. He did time me a LM311. Haven't checked the datasheet on it. It looks like it might be about a 100 years old.

#### spinnaker

Joined Oct 29, 2009
7,835
If you don't need super high speed, a plain vanilla 741 works just fine.

Maybe I will pick one up then. No super speed. Just trying to do an tachometer. I want a nice clean square wave out not getting that from the ir sensor.

#### GopherT

Joined Nov 23, 2012
8,012
My shop is just as famous as Gopher T's but for some reason folks don't want to visit often.
View attachment 119050
I passed by about a year ago. Unfortunately, that 6 miles was straight up and a couple miles south - I was on my way from Chicago to Tokyo. The pilot pointed out Fairbanks and North Pole. I didn't get a weather report for that day.

Thanks. Took a chance and ordered from ebay. But I might hit up Gopher too. He did time me a LM311. Haven't checked the datasheet on it. It looks like it might be about a 100 years old.
And you're considering a 741 instead?
A 741 will not survive on a 5V supply.

The LM311 is still made today- Although not in the fancy metal can!

#### GopherT

Joined Nov 23, 2012
8,012
@spinnaker
Note that there is a comparator on your PIC (most PICs). If not, I can give you a PIC16F690 (SOIC-20 OR DIP-20)

#### RichardO

Joined May 4, 2013
2,271
My shop is just as famous as Gopher T's but for some reason folks don't want to visit often.
View attachment 119050
Is that -50 degrees in F or C?

Never mind. I don't think it really matters.

I worked with a fellow that went to the South Pole. He described becoming a member of the 200 club. To become a member you had to get in a hot shower and then run outside in just your sneakers. You then had you picture taken. Of course the temperature difference had to be at least 200 degrees.

#### spinnaker

Joined Oct 29, 2009
7,835
@spinnaker
Note that there is a comparator on your PIC (most PICs). If not, I can give you a PIC16F690 (SOIC-20 OR DIP-20)
I thought about that but wanted to make my "sensor" standalone.

And I am running out of pins! I might be able to move a couple of things around but it will be tight.

Can I somehow configure the comparator out so the pic sees it as an "external" interrupt?

#### spinnaker

Joined Oct 29, 2009
7,835
There is:

"The comparator interrupt flag can be set whenever
there is a change in the output value of the comparator."

Not sure if I can make use of that.


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# How to read long numbers? [duplicate]

This question already has an answer here:

Now, I have a data file (abc.dat) and it includes many long-Fortran-like numerical numbers as follows,

 6.329743394299864138426200801048403E-0008
-2.549598515596261539775494416553164E-0005
-4.070996338361811028817170114484442E-0005

How to read these data in Mathematica? If we use ReadList["abc.dat",Real] and it is not available for these long numbers. If we use ReadList["abc.dat", Record], then we need to find a way to change the string to numbers. Is there some better way for dealing with this problem ?

## marked as duplicate by m_goldberg, bobthechemist, ciao, Sjoerd C. de Vries, Michael E2Apr 6 '14 at 17:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Read as text, then use something like ToExpression[StringReplace[nums, "E" -> "*10^"]] on that list. However, ReadList[...,Number] should handle the Fortran style. – ciao Apr 5 '14 at 6:27
• – ciao Apr 5 '14 at 6:51
• In version 7 ReadList["abc.dat", Real] appears to work correctly, yielding: {6.32974339429986413842620080104840*10^-8, -0.0000254959851559626153977549441655316, -0.0000407099633836181102881717011448444}. What output do you get? – Mr.Wizard Apr 5 '14 at 6:53
• – Mr.Wizard Apr 5 '14 at 6:54
• @Mr.Wizard: Looks like a possible bug in V9... – ciao Apr 5 '14 at 7:13

## 1 Answer

I opened HxD and pasted your numbers and saved as abc.dat. Then this worked for (8.0.4.0):

Import["C:\\f1\\f2\\Desktop\\abc.dat"]
(*{{6.32974339429986413842620080104840\[CenterDot]10^-8},{-0.0000254959851559626153977549441655316},{-0.0000407099633836181102881717011448444}}*)

• Incidentally in version 7 Import does not work fully here: data is imported as machine reals, losing input precision. – Mr.Wizard Apr 5 '14 at 6:59

format: plain


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references for abelian schemes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T09:49:45Z http://mathoverflow.net/feeds/question/51955 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51955/references-for-abelian-schemes references for abelian schemes unknown 2011-01-13T13:45:08Z 2011-03-20T22:19:43Z
Hi, I have a very basic question. I am looking for references explaining how to construct explicitily a Jacobian starting from a curve or examples of projective equations for an abelian scheme. I know a little bit the theory in general so I need examples to fix it, at least in the cases which are not too complicate( or when it is possible). Thank you
 http://mathoverflow.net/questions/51955/references-for-abelian-schemes/59004#59004 Answer by Hamish Ivey-Law for references for abelian schemes Hamish Ivey-Law 2011-03-20T22:19:43Z 2011-03-20T22:19:43Z
The equations defining the Jacobian of a curve as a projective variety become very complicated as soon as the genus of the curve is bigger than 1. In the case of genus 2 curves, say $\mathcal{C}:y^2 = f(x)$, Grant [1] gives an explicit embedding in $\mathbb{P}^8$ and the defining equations when $\deg(f) = 5$ and Flynn [2] gives an explicit embedding in $\mathbb{P}^{15}$, the 72 (!) defining equations of the projective variety, and the biquadratic forms defining the addition law for when $\deg(f) = 6$ (see also Cassels and Flynn's book [3] for an "updated" version of Flynn's work in the early 90s among other things). For this reason, most computations with Jacobians use the Mumford representation of points in $\operatorname{Sym}^2(\mathcal{C})$ together with Cantor's algorithm for the addition law.

[1] Grant, D. Formal groups in genus two. J. Reine Angew. Math. 411 (1990), 96–121.

[2] Flynn, E. V. The Jacobian and formal group of a curve of genus 2 over an arbitrary ground field. Math. Proc. Cambridge Philos. Soc. 107 (1990), no. 3, 425–441.

[3] Cassels, J. W. S.; Flynn, E. V. Prolegomena to a middlebrow arithmetic of curves of genus 2. London Mathematical Society Lecture Note Series, 230. Cambridge University Press, Cambridge, 1996. xiv+219 pp. ISBN: 0-521-48370-0


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# How do you find the critical numbers of f(x)=|x+3|-1?

Jul 11, 2016

(-3,-1)

#### Explanation:

x=c is a Critical Point either where (A) f'(c) = 0 or (B) f('c) does not exist/ is not defined

if you look at the 2 versions of this you have

$f = x + 3 - 1 = x + 2$

and for when $x + 3 < 0$, you also have $f = - x - 3 - 1 = - x - 4$

the lines meet at $x + 2 = - x - 4 , x = - 3 , y = - 1$

Point (-3,-1) is an apex. The derivative immediately to the left and right are -1 and 1 respectively. but the derivative at the point (-3,-1) is not defined. that is the CP here.

plot it if in doubt


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# Portal:Mathematics

## The Mathematics Portal

Mathematics is the study of numbers, quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity. The research required to solve mathematical problems can take years or even centuries of sustained inquiry. However, mathematical proofs are less formal and painstaking than proofs in mathematical logic. Since the pioneering work of Giuseppe Peano (1858–1932), David Hilbert (1862–1943), and others on axiomatic systems in the late 19th century, it has become customary to view mathematical research as establishing truth by rigorous deduction from appropriately chosen axioms and definitions. When those mathematical structures are good models of real phenomena, then mathematical reasoning often provides insight or predictions.

Through the use of abstraction and logical reasoning, mathematics developed from counting, calculation, measurement, and the systematic study of the shapes and motions of physical objects. Practical mathematics has been a human activity for as far back as written records exist. Rigorous arguments first appeared in Greek mathematics, most notably in Euclid's Elements. Mathematics developed at a relatively slow pace until the Renaissance, when mathematical innovations interacting with new scientific discoveries led to a rapid increase in the rate of mathematical discovery that continues to the present day.

Galileo Galilei (1564–1642) said, "The universe cannot be read until we have learned the language and become familiar with the characters in which it is written. It is written in mathematical language, and the letters are triangles, circles and other geometrical figures, without which means it is humanly impossible to comprehend a single word. Without these, one is wandering about in a dark labyrinth". Carl Friedrich Gauss (1777–1855) referred to mathematics as "the queen of sciences". The mathematician Benjamin Peirce (1809–1880) called the discipline, "the science that draws necessary conclusions". David Hilbert said of it: "We are not speaking here of arbitrariness in any sense. Mathematics is not like a game whose tasks are determined by arbitrarily stipulated rules. Rather, it is a conceptual system possessing internal necessity that can only be so and by no means otherwise." Albert Einstein (1879–1955) stated that "as far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality".

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 In this shear transformation of the Mona Lisa, the central vertical axis (red vector) is unchanged, but the diagonal vector (blue) has changed direction. Hence the red vector is said to be an eigenvector of this particular transformation and the blue vector is not. Image credit: User:Voyajer

In mathematics, an eigenvector of a transformation is a vector, different from the zero vector, which that transformation simply multiplies by a constant factor, called the eigenvalue of that vector. Often, a transformation is completely described by its eigenvalues and eigenvectors. The eigenspace for a factor is the set of eigenvectors with that factor as eigenvalue, together with the zero vector.

In the specific case of linear algebra, the eigenvalue problem is this: given an n by n matrix A, what nonzero vectors x in $R^n$ exist, such that Ax is a scalar multiple of x?

The scalar multiple is denoted by the Greek letter λ and is called an eigenvalue of the matrix A, while x is called the eigenvector of A corresponding to λ. These concepts play a major role in several branches of both pure and applied mathematics — appearing prominently in linear algebra, functional analysis, and to a lesser extent in nonlinear situations.

It is common to prefix any natural name for the vector with eigen instead of saying eigenvector. For example, eigenfunction if the eigenvector is a function, eigenmode if the eigenvector is a harmonic mode, eigenstate if the eigenvector is a quantum state, and so on. Similarly for the eigenvalue, e.g. eigenfrequency if the eigenvalue is (or determines) a frequency.

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A Froebel star is a traditional German Christmas decoration. It is named after German educationist Friedrich Fröbel (1782–1852), who encouraged the use of paper folding in pre–primary education with the aim of conveying simple mathematical concepts to children.

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text:
## Make Palindrome 2 Codechef Problem Solution

You are given a binary string of length . You want to obtain a palindrome from by applying the following operation at most

times:

• Choose an index , delete the character from and concatenate the remaining parts of the string. Here denotes the current length of string

For example, if 11010, then applying the operation on index makes 1010.

Note that after each operation, the length of the string decreases by one.

Find any palindrome you can obtain after the operations. It can be proved that it is always possible to obtain a palindrome from under the given constraints.

Here, denotes floor division of the integer by . For example, , . A binary string is a string that consists of only the characters 0 and 1.

### Input Format

• The first line of input contains an integer , denoting the number of test cases. The test cases then follow:
• The first line of each test case contains an integer , denoting the length of the binary string .
• The second line of each test case contains the binary string

### Output Format

For each test case, print on a separate line any palindromic string that can be obtained from by applying the given operation at most times.

### Constraints

• contains only the characters 0 and 1.

### Sample Input 1

4
3
101
3
001
4
1011
6
010011

### Sample Output 1

101
00
111
1001

### Explanation

Test case : The given string is already a palindrome.

Test case : Applying the operation on index makes 00 which is a palindrome.

Test case : Applying the operation on index makes 111 which is a palindrome.

Test case : Applying the operation on index makes 10011. Then applying the operation on index makes 1001 which is a palindrome.

#include <bits/stdc++.h>
using namespace std;

int main() {
int t;
cin>>t;
while(t--){
int n;
string str2;
cin>>n>>str2;
string str1=str2;
reverse(str2.begin(),str2.end());
int n1=str1.length(),n2=str2.length();
vector<vector<int>> dp(n1+1,vector<int>(n2+1,0));
for(int i=1;i<=n1;i++){
for(int j=1;j<=n2;j++){
if(str1[i-1]==str2[j-1]){
dp[i][j]=1+dp[i-1][j-1];
}
else{
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
}
int i=n1,j=n2;
string ans="";
while(i>0 && j>0){
if(str1[i-1]==str2[j-1]){
ans=str1[i-1]+ans;
i--;j--;
}
else if(dp[i-1][j]>dp[i][j-1]){

i=i-1;
}
else{

j=j-1;
}
}
cout<<ans<<endl;
}
return 0;
}


text:
# $f(x) = (x_1-x_2^2)(x_1-\frac{1}{2}x_2^2)$, verify $\overline{x} = (0,0)^t$ is local min of $\phi(\lambda) = f(\overline{x}+\lambda d)$ but not of $f$

Let $f(x) = (x_1-x_2^2)(x_1-\frac{1}{2}x_2^2)$. Verify that $\overline{x} = (0,0)^t$ is a local minimizer of $\phi(\lambda) = f(\overline{x}+\lambda d)$ for all $d\in\mathbb{R}^2$ but $\overline{x}$ is not a local minimizer of $f$

$$\frac{\partial f}{\partial x_1} = 1\left(x_1-\frac{1}{2}x_2^2\right)+(x_1-x_2^2)$$

$$\frac{\partial f}{\partial x_2} = -2x_2\left(x_1-\frac{1}{2}x_2^2\right)-x_2^2(x_1-x_2^2)$$

$$\frac{\partial^2 f}{\partial x_1^2} = 2$$

$$\frac{\partial^2 f}{\partial x_2^2} = 6x^2$$

$$\frac{\partial^2 f}{\partial x_2\partial x_1} = -3x_2$$

$$\nabla^2 f = \begin{bmatrix} 2 & -3x_2 \\ -3x_2 & 6x_2^2 \end{bmatrix}$$

$$\nabla^2 f(0,0) = \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}$$

$$\begin{bmatrix} a & b \\ \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix} = a^2$$

for $(0,0)$ to be a local minimizer we should have $\nabla f = 0$ (we have) and $\nabla^2\ge 0$. But $\nabla^2$ is not always semipositive definite so $(0,0)$ is not a local minimizer of $f$.

Now let's analyze $f((0,0)+\lambda d)$.

$$\nabla(\lambda d_1,\lambda d_2) = \begin{bmatrix} \lambda d_1-\frac{1}{2}\lambda^2d_2^2 + \lambda d_1 -\lambda^2d_2^2 \\ -2\lambda d_2(\lambda d_1-\frac{1}{2}\lambda^2 d_2^2)-\lambda d_ 2^2(\lambda d_1-\lambda^2d_2^2) \end{bmatrix}\neq 0$$

and the hessian gives an even worse thing. So I think instead I should analyze

$$\phi(\lambda) = f(\lambda d_1,\lambda d_2) = (\lambda d_1-\lambda^2 d_2^2)(\lambda d_1-\frac{1}{2}\lambda^2 d_2^2)$$

but what does it mean for $\overline{x}$ to be a local minimizer of something that depends on $\lambda$?

UPDATE:

I think I have to minimize

$$\phi(\lambda) = f(\lambda d_1,\lambda d_2) = (\lambda d_1-\lambda^2 d_2^2)(\lambda d_1-\frac{1}{2}\lambda^2 d_2^2) = \\\lambda^2 d_1^2 -\frac{1}{2}\lambda^3d_1d_2^2-\lambda^3d_2^2 d_1 + \frac{1}{2}\lambda^4d_2^4\implies \\ \phi'(\lambda) = 2d_1^2\lambda -\frac{3}{2}\lambda^2d_1d_2^2 -3\lambda^2d_2^2d_1 + 2\lambda^3d_2^4 = 0\implies \\2d_2^4\lambda^2+\lambda\left(-\frac{9}{2}d_2^2d_1\right) + 2d_1^2 = 0\implies\\ \lambda = \frac{\frac{9}{2}d_2^2d_1\pm\sqrt{\ (\frac{9}{2}d_2^2d_1)^2 -4\cdot 2d_2^4\cdot2d_1^2}}{4d_2^4}, \lambda = 0$$

It's unpratical to test these values on the function to see which of them is smaller. And even with this I don't know how to proceed. Maybe I should test the second derivative too, I don't know what it means to prove $\overline{x} = (0,0)$ is a minimizer of $\phi(\lambda) = f(\overline{x}+\lambda d)$. The only possible thing I can image is $\lambda$ being $0$ giving us the minimum of $\phi(\lambda)$.

So since we know $\lambda=0$ gives us derivative $0$, let's see how the second derivative is at this point (if it's positive, then $\lambda=0$ is a minimizer):

$$\phi''(\lambda) = 6\lambda^2 d_2^4 -6\lambda d_2^2 d_1 -\frac{6}{2}\lambda d_1d_2^2 + 2d_1^2 \implies\\\phi''(0) = 2d_1^2$$

So for any direction with $d_1\neq 0$, we'll have $\lambda=0$ as minimizer. What about when the direction is $(0,d_2)$ for any $d_2$? In this case we have $$\phi(\lambda) = (0-(\lambda d_2)^2)(0-\frac{1}{2}(\lambda d_2)^2 ) = \frac{1}{2}\lambda^4d_2^4$$

for which $\lambda= 0$ is also a minimizer.

So in all cases $\lambda=0$ is a minimizer of $f((0,0) + \lambda (d_1,d_2))$ which means $(0,0)$ is a minimizer of $\phi(\lambda)$ I guess?

• It seems that you are requested to prove that $\phi(\lambda)$ takes local minimum at $\lambda = 0$. – xbh Aug 27 '18 at 3:05
• @xbh so I should substitute $(x_1+\lambda d_1, x_2+\lambda d_2)$ into $f$ and minimize the parabola over the variable $\lambda$? – Guerlando OCs Aug 27 '18 at 3:22
• If the question above is typed right, then for fixed $(d_1,d_2)$, $\phi(\lambda)$ is just the last equation you have written in the OP. Then minimize that over $\lambda$ as you thought. – xbh Aug 27 '18 at 3:24
• Your analysis based on the Hessian does not exclude the possibility of the origin being a minimum. It only matters what the Hessian is at the origin. The fact that it may be indefinite elsewhere is irrelevant. – Asdf Aug 27 '18 at 13:47
• @Asdf you're right, so a hessian analysis isn't going to give me anything. I should analyze the neighborhood of (0,0)? – Guerlando OCs Aug 29 '18 at 19:28

This problem has a nice graphical solution.

Note that f(0)=0. If both of the factors $x_1-x_2^2$ and $x_1-\frac12x_2^2$ have the same sign then $f(x)>0$. If the signs differ then $f(x)<0$. Divide the $x_2x_1$-plane into three regions:

(i) if $x_1>x_2^2$ then $x_1>\frac12x_2^2$ and $f(x)>0$. This is the area in the plane above the quadratic $x_1=x_2^2$ (you let $x_1$ be the vertical axis and $x_2$ the horizontal).

(ii) if $x_1<\frac12 x_2^2$ then $x_1<x_2^2$ and $f(x)>0$. This is the area below $x_1=\tfrac12x_2^2$.

(iii) if $x_2^2>x_1>\frac12 x_2^2$ then $f(x)<0$. This is the area above the curve $x_1=\frac12x_2^2$ but below $x_1=x_2^2$.

The curve $x_1=\frac23x_2^2$ lies in (iii). On it, the objective value is strictly negative, $$(x_1-x_2^2)(x_1-\frac12x_2^2)=-\frac12x_2^2\frac16x_2^2.$$ Hence the origin with objective value 0 cannot be a minimum. Note however that you can only approach the origin from inside (iii) along a curved trajectory. If you consider a series of perturbations of the origin that all lie on a line, as in the case of $\phi(\lambda)$, then all points in some small ball around the origin will lie in region (i) or (ii). Graph the three regions and you will see this clearly.

• what is an objective value? – Guerlando OCs Sep 2 '18 at 4:34
• doesn't it suffice to show that $x_1 = \frac{2}{3}x_2^2$ is always negative so there's a path on which small changes makes $f$ smaller than $0$? Why did you do all this analysis on the sign of $f$? – Guerlando OCs Sep 2 '18 at 4:49
• By objective value I refered to the value of the objective function $f$ at a point $(x_1,x_2)$. This is standard terminology in optimization – Asdf Sep 2 '18 at 15:03
• Yes, the last equation suffices. The analysis provides an intuitive understanding of the problem which in turn leads to the answer. You can substitute any relation of the form $x_1=cx_2^2$, where $c\in(\frac12,1)$, into the objective function $f$. Here, I picked $c=\frac23$. Please accept the answer if you like it. – Asdf Sep 2 '18 at 15:04
• Thank you for your answer. I'm gonna accept it. Could you just take a look at my argument about why $(0,0)$ is a minimizer for $\phi(\lambda)$? – Guerlando OCs Sep 2 '18 at 19:01


text:
# Primitive nth roots of unity related to the complex nth roots of 1.

I just cannot seem to wrap my head around this problem and would really appreciate some guidance. So I know that a primitive $$n^{th}$$ root of unity is a complex number $$z$$ such that $$z^n = 1$$ but $$z^m \neq 1$$ for $$0. I also know that for $$w = cos(\frac{2\pi}{n}) + isin(\frac{2\pi}{n})$$, the $$n$$ complex solutions of $$z^n = 1$$ are $$1,w,w^2,...,w^{n-1}$$. I need to show three things:

1.) $$w$$ is always a primitive $$n^{th}$$ root of unity.
2.) $$w^k$$ is a primitive $$n^{th}$$ root of unity iff $$gcd(k,n) = 1$$
3.) If $$z$$ is any primitive $$n^{th}$$ root of unity, then $$1,z,z^2,...,z^{n-1}$$ are distinct and comprise all the $$n^{th}$$ roots of unity.

I'm pretty new to this type of rigorous proof and I'm not even sure where to start.

• So you have to show to start that $w^n=1$ and $w^k\ne 1$ if $0<k<n$. – xarles Oct 3 '18 at 19:12


text:
# Math Help - 2 problems involving examples and counter-examples.

1. ## 2 problems involving examples and counter-examples.

1. if a doesn't divide b and a doesn't divide c , then a doesn't divide b times c.

give a counterexample to show this is a false statement.

2. if a divides c and b divides c sometimes, a times b divides c and sometimes it doesn't

show an example when a times b divides c .

show and example when a times b doesn't divide c.

2. Originally Posted by lonestarross
1. if a doesn't divide b and a doesn't divide c , then a doesn't divide b times c.

give a counterexample to show this is a false statement.

.
a can't be divided by $i$, and a can't be divided by $-i$, but a can be divided by $i . -i$
Note that $i= \sqrt{-1}$
I'm not 100% positive with my answer though

3. You know of a numerical example of this?

4. Originally Posted by lonestarross
1. if a doesn't divide b and a doesn't divide c , then a doesn't divide b times c.

give a counterexample to show this is a false statement.

2. if a divides c and b divides c sometimes, a times b divides c and sometimes it doesn't

show an example when a times b divides c .

show and example when a times b doesn't divide c.
Usually with this language and nothing else specified, we assume we are working over the integers.

(1) is very simple.. don't know how to give a hint, here's such a counterexample, learn from it

a = 2*3
b = 2*5
c = 3*5

(2) this is confusing. Is the first part supposed to mean

if (a divides c) and (b divides c sometimes)

or

if (a divides c and b divides c) sometimes

?

I'll assume the first. But it can be proven with the stricter requirement "a divides c and b divides c always".

Just use simple examples. For the case ab divides c, let c = ab. For ab does not divide c, let a = b = c > 1.

5. the first.\\ but when you say a=2*3, b=2*5, c=3*5, 5 does divide 10 and 15...never mind i t is backwards

6. Originally Posted by lonestarross
the first
Okay, my response addresses this.

7. so for 2. ab|c a=2 b=3 c=6 2*3|6

ab doesn't |c 4*6|12

still dont get (1) 2*3- 5 2*5- 10 2*3- 15 a will |b 5\10

on

8. thank you for your help. got it and understand it now


text:
1. ## [SOLVED] Vectors Question

Question:
The position vectors of points $A$ and $B$ are $\left(\begin{array}{c}-3\\6\\3\end{array}\right)$ and $\left(\begin{array}{c}-1\\2\\4\end{array}\right)$ respectively, relative to an origin $O$.
(i) Calculate angel AOB.
(ii) The point C is such that $\vec{AC} = 3\vec{AB}$. Find the unit vector in the direction of $\vec{OC}$.

Attempt:

(i) $\theta = 36.7^o$

(ii) Need help!

2. Originally Posted by looi76
Question:
The position vectors of points $A$ and $B$ are $\left(\begin{array}{c}-3\\6\\3\end{array}\right)$ and $\left(\begin{array}{c}-1\\2\\4\end{array}\right)$ respectively, relative to an origin $O$.
(i) Calculate angel AOB.
(ii) The point C is such that $\vec{AC} = 3\vec{AB}$. Find the unit vector in the direction of $\vec{OC}$.

Attempt:

(i) $\theta = 36.7^o$

Correct
Originally Posted by looi76
(ii) Need help!
$\vec{AB} = \vec{OB} - \vec{OA}$

Thus:
$\vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} - \vec{OA} = 3(\vec{OB} - \vec{OA}) \Rightarrow \vec{OC}= 3\vec{OB} - 2\vec{OA}$
So complete it now

3. Originally Posted by Isomorphism

Correct

$\vec{AB} = \vec{OB} - \vec{OA}$

Thus:
$\vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} - \vec{OA} = 3(\vec{OB} - \vec{OA}) \Rightarrow \vec{OC}= 3\vec{OB} - 2\vec{OA}$
So complete it now
Thanks Isomorphism
$
\vec{OC} = 3\vec{OB} - 2\vec{OA}$

$\vec{OC} = 3\left(\begin{array}{c}-1\\2\\4\end{array}\right) - 2\left(\begin{array}{c}-3\\6\\3\end{array}\right)$

$\vec{OC} = \left(\begin{array}{c}-3\\2\\12\end{array}\right) - \left(\begin{array}{c}-6\\12\\6\end{array}\right)$

$\vec{OC} = \left(\begin{array}{c}3\\-6\\6\end{array}\right)$

4. Almost.... You still did not find the unit vector

But did you understand how I got the expression:
$\vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} = 3\vec{OB} - 2\vec{OA}$

That is more important than substituting and computing the answer

5. Originally Posted by Isomorphism
Almost.... You still did not find the unit vector

But did you understand how I got the expression:
$\vec{AC} = 3\vec{AB} \Rightarrow \vec{OC} = 3\vec{OB} - 2\vec{OA}$

That is more important than substituting and computing the answer
Yes, I have understood, thanks to you . By unit vector you mean i,j,k?
I mean $3i - 6j + 6k$, is this right?

6. Originally Posted by looi76
Yes, I have understood, thanks to you . By unit vector you mean i,j,k?
I mean $3i - 6j + 6k$, is this right?
No..No..A unit vector is a vector whose magnitude is 1.

(ii) The point C is such that . Find the unit vector in the direction of .
You have found OC but not the unit vector in the direction of
To get that vector divide $\vec{OC}$ by magnitude of $\vec{OC}$

7. Originally Posted by Isomorphism
No..No..A unit vector is a vector whose magnitude is 1.

You have found OC but not the unit vector in the direction of
To get that vector divide $\vec{OC}$ by magnitude of $\vec{OC}$
I got the formula from the text book: $\hat{v} = \frac{v}{|v|}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 - 6^2 + 6^2}}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}$

$\hat{OC} = \frac{3i - 6j + 6k}{9}$

$\hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k$

$\hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k$

Is this correct?

8. Originally Posted by looi76
I got the formula from the text book: $\hat{v} = \frac{v}{|v|}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 - 6^2 + 6^2}}$

Is this correct?
Ya right

Except for a typo...
$|v| = \sqrt{3^2 + (-6)^2 + 6^2}$

9. Originally Posted by Isomorphism
Wrong

$|v| = \sqrt{3^2 + (-6)^2 + 6^2}$
$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 + (-6)^2 + 6^2}}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}$

$\hat{OC} = \frac{3i - 6j + 6k}{9}$

$\hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k$

$\hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k$

Is this correct?

10. Originally Posted by looi76
$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{3^2 + (-6)^2 + 6^2}}$

$\hat{OC} = \frac{3i - 6j + 6k}{\sqrt{81}}$

$\hat{OC} = \frac{3i - 6j + 6k}{9}$

$\hat{OC} = \frac{3}{9}i - \frac{6}{9}j + \frac{6}{9}k$

$\hat{OC} = \frac{1}{2}i - \frac{2}{3}j + \frac{2}{3}k$

Is this correct?
You were right all along... It was just a typing mistake..

Well done


text:
# Turek Hron CSM Benchmark

In order to validate our fluid-structure interaction solver, we realize here a benchmark on the deformation of an elastic structure, initially proposed by [TurekHron].

Computer codes, used for the acquisition of results, are from [Chabannes].

 This benchmark is linked to the Turek Hron CFD and Turek Hron FSI benchmarks.

## 1. Problem Description

We consider a solid structure, composed of a hyperelastic bar, bound to one of his extremity $\Gamma_F^*$ to a rigid stationary circular structure. We denote $\Gamma_{L}^*=\partial\Omega_s^* \backslash \Gamma_F^*$ the other boundaries. The geometry can be represented as follows

Figure 1 : Geometry of the Turek & Hron CSM benchmark.

$\Omega_s^*$ represent the initial domain, before any deformations. We denote $\Omega^t_s$ the domain obtained during the deformations application, at the time t.

By the Newton’s second law, we can describe the fundamental equation of the solid mechanic in a Lagrangian frame. Furthermore, we will suppose that the hyperelastic material follows a compressible Saint-Venant-Kirchhoff model. The Lamé coefficients, used in this model, are deduced from the Young’s modulus $E_s$ and the Poisson’s ratio $\nu_s$ of the material.

We observe, during this simulation, the displacement of $A$, on the $x$ and $y$ axis, when the elastic structure is subjected to its own weight, and compare our results to the reference ones given Turek and Hron.

In the first two cases, CSM1 and CSM2, we want to determine the steady state condition. To find it, a quasi-static algorithm is used, which increase at each step the gravity parameter.

For the third cases, CSM3, we realise a transient simulation, where we will observe the comportment of $A$ subjected to its weight during a given time span.

### 1.1. Boundary conditions

We set

• on $\Gamma_{F}^*$, a condition that imposes this boundary to be fixed : $\boldsymbol{\eta}_s=0$

• on $\Gamma_{L}^*$, a condition that lets these boundaries be free from constraints : $(\boldsymbol{F}_s\boldsymbol{\Sigma}_s)\boldsymbol{ n }^*_s=\boldsymbol{0}$

where $\boldsymbol{n}_s^*$ is the outer unit normal vector from $\partial \Omega_s^*$.

### 1.2. Initial conditions

The source term $\boldsymbol{f}_s$, chosen in order to set in motion the elastic structure, is define by

$\boldsymbol{f}_s=(0,-\rho^*_s g)$

where $g$ is a gravitational constant.

The reference document [TurekHron] doesn’t specify the time interval used to obtain their results. In our case, it’s chosen as $\lbrack0,10\rbrack$.

## 2. Inputs

The following table displays the various fixed and variables parameters of this test-case.

Table 1. Fixed and Variable Input Parameters
Name Description Nominal Value Units

$g$

gravitational constant

2

$m / s^2$

$l$

elastic structure length

$0.35$

$m$

$h$

elastic structure height

$0.02$

$m$

$r$

$0.05$

$m$

$C$

cylinder center coordinates

$(0.2,0.2)$

$m$

$A$

control point coordinates

$(0.2,0.2)$

$m$

$B$

point coordinates

$(0.15,0.2)$

$m$

$E_s$

Young’s modulus

$1.4\times 10^6$

$kg / ms^2$

$\nu_s$

Poisson’s ratio

$0.4$

dimensionless

$\rho^*_s$

density

$1000$

$kg/ m^3$

As for solvers we used, Newton’s method is chosen for the non-linear part and a direct method based on LU decomposition is selected for the linear part.

## 3. Outputs

As described before, we have

$\rho^*_s \frac{\partial^2\boldsymbol{\eta}_s}{\partial t^2} - \nabla \cdot (\boldsymbol{F}_s\boldsymbol{\Sigma}_s) = \boldsymbol{f}^t_s$

We search the displacement $\boldsymbol{\eta}_s$, on $\Omega_s^*$, which will satisfy this equation.

In particular, the displacement of the point $A$ is the one that interests us.

## 4. Discretization

To realize these tests, we made the choice to used Finite Elements Method, with Lagrangian elements of order $N$ to discretize space.

Newmark-beta method, presented into [Chabannes] papers, is the one we used for the time discretization. We used this method with $\gamma=0.5$ and $\beta=0.25$.

### 4.1. Solvers

Here are the different solvers ( linear and non-linear ) used during results acquisition.

 type gmres relative tolerance 1e-13 max iteration 1000 reuse preconditioner true
 relative tolerance 1e-8 steps tolerance 1e-8 max iteration 500 max iteration with reuse 10 reuse jacobian false reuse jacobian rebuild at first Newton step true
 relative tolerance 1e-5 max iteration 500 reuse preconditioner CSM1/CSM2 : false | CSM3 : true reuse preconditioner rebuild at first Newton step true
 type lu package mumps

## 5. Implementation

To realize the acquisition of the benchmark results, code files contained and using the Feel++ library will be used. Here is a quick look to the different location of them.

First at all, the main code can be found in

`    feelpp/applications/models/solid`

The configuration file for the CSM3 case, the only one we work on, is located at

`    feelpp/applications/models/solid/TurekHron`

The result files are then stored by default in

`    feel/applications/models/solid/TurekHron/csm3/"OrderDisp""Geometric_order"/"processor_used"`

Like that, for the CSM3 case executed on 8 processors, with a $P_1$ displacement approximation space and a geometric order of 1, the path is

`    feel/applications/models/solid/TurekHron/csm3/P1G1/np_8`

At least, to retrieve results that interested us for the benchmark and to generate graphs, we use a Python script located at

`    feelpp-benchmarking-book/CFD/Turek-Hron/postprocess_cfd.py`

## 6. Results

### 6.1. CSM1

 $N_{elt}$ $N_{dof}$ $x$ displacement $\lbrack\times 10^{-3}\rbrack$ $y$ displacement $\lbrack\times 10^{-3}\rbrack$ Reference [TurekHron] -7.187 -66.10 1061 4620 ($P_2$) -7.039 -65.32 4199 17540 ($P_2$) -7.047 -65.37 16495 67464 ($P_2$) -7.048 -65.37 1061 10112 ($P_3$) -7.046 -65.36 1906 17900 ($P_3$) -7.049 -65.37 1061 17726 ($P_4$) -7.048 -65.37

All the files used for this case can be found in this rep [ geo file, config file, json file ]

### 6.2. CSM2

 $N_{elt}$ $N_{dof}$ $x$ displacement $\lbrack\times 10^{-3}\rbrack$ $y$ displacement $\lbrack\times 10^{-3}\rbrack$ Reference [TurekHron] -0.4690 -16.97 1061 4620 ($P_2$) -0.459 -16.77 4201 17548 ($P_2$) -0.459 -16.77 16495 67464 ($P_2$) -0.459 -16.78 1061 10112 ($P_3$) -0.4594 -16.78 16475 150500 ($P_3$) -0.460 -16.78 1061 17726 ($P_4$) -0.460 -16.78

All the files used for this case can be found in this rep [geo file, config file, json file].

### 6.3. CSM3

The results of the CSM3 benchmark are detailed below.

 $\Delta t$ $N_{elt}$ $N_{dof}$ $x$ displacement $\lbrack\times 10^{-3}\rbrack$ $y$ displacement $\lbrack\times 10^{-3}\rbrack$ / Reference [TurekHron] −14.305 ± 14.305 [1.0995] −63.607 ± 65.160 [1.0995]
 0.02 4199 17536($P_2$) -14.585 ± 14.590 [1.0953] -63.981 ± 65.521 [1.0930] 4199 38900($P_3$) -14.589 ± 14.594 [1.0953] -63.998 ± 65.522 [1.0930] 1043 17536($P_4$) -14.591 ± 14.596 [1.0953] -64.009 ± 65.521 [1.0930] 4199 68662($P_4$) -14.590 ± 14.595 [1.0953] -64.003 ± 65.522 [1.0930]
 0.01 4199 17536($P_2$) -14.636 ± 14.640 [1.0969] -63.937 ± 65.761 [1.0945] 4199 38900($P_3$) -14.642 ± 14.646 [1.0969] -63.949 ± 65.771 [1.0945] 1043 17536($P_4$) -14.645 ± 14.649 [1.0961] -63.955 ± 65.778 [1.0945] 4199 68662($P_4$) -14.627 ± 14.629 [1.0947] -63.916 ± 65.739 [1.0947]
 0.005 4199 17536($P_2$) -14.645 ± 14.645 [1.0966] -64.083 ± 65.521 [1.0951] 4199 38900($P_3$) -14.649 ± 14.650 [1.0966] -64.092 ± 65.637 [1.0951] 1043 17536($P_4$) -14.652 ± 14.653 [1.0966] -64.099 ± 65.645 [1.0951] 4199 68662($P_4$) -14.650 ± 14.651 [1.0966] -64.095 ± 65.640 [1.0951]

\text{Figure 2: x and y displacements}

All the files used for this case can be found in this rep [ geo file, config file, json file ]

### 6.4. Conclusion

To obtain these data, we used several different mesh refinements and different polynomial approximations for the displacement on the time interval $\lbrack 0,10 \rbrack$.

Our results are pretty similar to those from Turek and Hron, despite a small gap. This gap can be caused by the difference between our time interval and the one used for the reference acquisitions.

## 7. Bibliography

References for this benchmark
• [TurekHron] S. Turek and J. Hron, Proposal for numerical benchmarking of fluid-structure interaction between an elastic object and laminar incompressible flow, Lecture Notes in Computational Science and Engineering, 2006.

• [Chabannes] Vincent Chabannes, Vers la simulation numérique des écoulements sanguins, Équations aux dérivées partielles [math.AP], Université de Grenoble, 2013.


text:
# Graphing with terminal velocity

#### asilvester635

My notes didn't show how the second derivative was evalated. Would I need to plug in the same values I used in the first derivative inside the second derivative to figure out whether v'' is greater than or equal to zero?

#### Attachments

• 430.8 KB Views: 19

#### skeeter

MHF Helper

Note that the equation given for $\dfrac{dv}{dt}$ is a scalar equation, so $v$ actually represents speed rather than velocity.

Speed is bounded strictly between $0 \le v \le v_T = \dfrac{mg}{k}$, so the signs of $\dfrac{dv}{dt}$ and $\dfrac{d^2v}{dt^2}$ need only be calculated between those two bounds ...

for $v = \dfrac{mg}{2k}$ ...

$\dfrac{dv}{dt} = g - \dfrac{k}{m} \cdot \dfrac{mg}{2k} = g - \dfrac{g}{2} = \dfrac{g}{2} > 0$

$\dfrac{d^2v}{dt^2} = -\dfrac{k}{m} \cdot \dfrac{dv}{dt} = -\dfrac{k}{m} \cdot \dfrac{g}{2} < 0$

so, the graph of speed, $v$, versus time is increasing and concave down.

#### asilvester635

Right, I'm actually sketching it so I want to find v'' above mg/k. Whatever we plug in dv/dt in the second derivative will always be negative since k/m is negative, which would be wrong based off of my professors work. How do you evaluate values above mg/k?

#### skeeter

MHF Helper
You can't ... why do you think they call it terminal speed?

#### skeeter

MHF Helper
The solution for this DE is ...

$v(t) = \dfrac{mg}{k}\left(1-e^{\frac{k}{m} t} \right)$

note $\displaystyle \lim_{t \to \infty} v(t) = \dfrac{mg}{k}$

... graph is attached.

#### Attachments

• 5.3 KB Views: 1


text:
# Print print/cut markings for a \pagewidth image

My thesis cover has an image that stretches over the entire pagewidth. This is nice in PDF, but I expect that the outer few milimeters cannot be printed. In indesign or similar programs, the printed document can be larger than e.g. A4 format, with print margins. You can provide an image that is slightly larger than the pagewidth, toghether with cutting/print markings that show where the paper should be cut, which will then remove a small portion of the image, but results in no whitespace at the sides.

Can I also do this or something similar in LaTeX and if so, how? I guess I could make my cover in InDesign or PhotoShop, but I would like to use LaTeX for this.

The geometry has the showcrop option which displays crop marks, so it is just a matter of using the appropriate layout and making the paper size slightly bigger. In the following example, I used A4 paper (21cm x 29.7cm) and added 3cm margins all around for the crop marks.

For the cover image, I used TikZ to place the image in the centre of the page which is sized to be the layout width and height. These values can be accessed through \Gm@layoutwidth and \Gm@layoutheight from the geometry package.

\documentclass{article}

\usepackage[
papersize={27cm,35.7cm},
layout=a4paper,
layoutoffset={3cm,3cm},
showcrop,
]{geometry}

\usepackage{mwe}
\usepackage{tikz}

\begin{document}
\makeatletter
\tikz[remember picture, overlay]
\node at (current page.center)
{\includegraphics[width=\Gm@layoutwidth, height=\Gm@layoutheight]{example-image-a}};
\makeatother
\clearpage
\lipsum
\end{document}

Note that this solution will stretch your image to fit the layout size; so you'll have to make sure that the cover image has the appropriate size and resolution otherwise the result might look distorted and/or pixelated.

• The geometry options worked like a charm, thank you very much! I have added the image with a centered '\makebox[\textwidth]{\includegraphics{...}}'. – Jelmer Veenstra Feb 27 '16 at 22:03


text:
# American Institute of Mathematical Sciences

July  2010, 6(3): 465-481. doi: 10.3934/jimo.2010.6.465

## Levitin-Polyak well-posedness for variational inequalities and for optimization problems with variational inequality constraints

 1 Department of Mathematics, Chengdu University of Information Technology, Chengdu, Sichuan 610225, China 2 Department of Mathematics, Sichuan University, Chengdu, Sichuan 610064, China

Received  May 2009 Revised  March 2010 Published  June 2010

In this paper, we study the Levitin-Polyak type well-posedness of variational inequalities and optimization problems with variational inequality constraints in Banach spaces. We derive some criteria and characterizations for these Levitin-Polyak well-posedness. We also investigate conditions under which the existence and uniqueness of solution is equivalent to the Levitin-Polyak well-posedness of the problem.
Citation: Rong Hu, Ya-Ping Fang, Nan-Jing Huang. Levitin-Polyak well-posedness for variational inequalities and for optimization problems with variational inequality constraints. Journal of Industrial & Management Optimization, 2010, 6 (3) : 465-481. doi: 10.3934/jimo.2010.6.465
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2018 Impact Factor: 1.025


text:
# zbMATH — the first resource for mathematics

Fortin operator and discrete compactness for edge elements. (English) Zbl 0967.65106
A Fortin operator is constructed for tetrahedral edge elements of any order in the case of polyhedral domains. The construction is also valid for more general domains with convenient regularity hypotheses, which are satisfied by polyhedral domains. Although Fortin operators for edge elements have been introduced by other authors, they usually place restrictions such as quasiuniformity of the mesh or stronger regularity hypotheses, in order to be able to prove convergence of the operator to the identity.
As a matter of fact, the main result in the paper gives a uniform estimate in the approximation of the identity in the $$L^2$$ norm that depends only on the size of the mesh. This is proved after thoroughly discussing the main differences and similarities of edge and face elements in three dimensions, taking into account that edge elements are suitable for problems involving the curl operator.
A final section deals with the application of the Fortin operator presented in the paper to numerically solve general Maxwell eigenvalue problems. For them it is proved by applying the main result, that the Fortin operator in this paper cannot produce spurious solutions.

##### MSC:
 65N30 Finite element, Rayleigh-Ritz and Galerkin methods for boundary value problems involving PDEs 65N25 Numerical methods for eigenvalue problems for boundary value problems involving PDEs 78A25 Electromagnetic theory, general 35J25 Boundary value problems for second-order elliptic equations 78M10 Finite element, Galerkin and related methods applied to problems in optics and electromagnetic theory 65N12 Stability and convergence of numerical methods for boundary value problems involving PDEs
Full Text:


text:
Statistics on Radiation Field Waveforms Associated with Multiple Intracloud Lightning Discharges

• Lee, Bok-Hee (The School of Electrical and Computer Engineering, Inha University, 253, Yonghyun-Dong, Nam-ku, Inchon, 402-751, Korea.) ;
• Lee, Dong-Moon (The School of Electrical and Computer Engineering, Inha University, 253, Yonghyun-Dong, Nam-ku, Inchon, 402-751, Korea.) ;
• Ahn, Chang-Hwan (The School of Electrical and Computer Engineering, Inha University, 253, Yonghyun-Dong, Nam-ku, Inchon, 402-751, Korea.) ;
• Kim, Young-Bong (The School of Electrical and Computer Engineering, Inha University, 253, Yonghyun-Dong, Nam-ku, Inchon, 402-751, Korea.)
• Published : 2001.03.01

Abstract

This paper presents the detailed statistics on radiation field signatures associated with multiple-intracloud lightning discharges. A transient signal recording system was used to measure the electric and magnetic fields produced by lightning flashes. The measurements were made in th summers of 1995 through 1999, and the location of the observation station was in Inchon on the coast of the Yellow Sea in Korea(37$^{\circ}$25'N, 126$^{\circ}$ 39'E). Most of lightning flashes typically contains between two and five strokes. The individual intracloud stork radiation fields were the bipolar pulse. On the average, the ratio of the peak of the second stroke to the first stroke peak was 75.1$\pm$40.1% for the negative, and a fraction of the subsequent stroke peaks were higher than the first stroke peak. The greater the number of the subsequent stroke order, less time separations between strokes were produced. The mean of the depth of the dip was 81.2$\pm$27.9% for the positive polarity and 75.9$\pm$24.4% for the negative. The depth of the dip increased for the positive bipolar pulses and decreased for the negative as the number of the stroke order increased.


text: Suppose two people are talking in a noisy room and one is trying to hear the other. The overall result is a spectrum such as the one shown in Figure 9-23. Nuclear magnetic resonance (NMR) spectroscopy is one of the most powerful and widely used techniques in chemical research for investigating structures and dynamics of molecules. Principal. Unless special precautions are taken, integrals usually should not be considered accurate to better than about $$5\%$$. The objective is to assign a structure to the compound $$\ce{C_4H_8O_3}$$ whose nmr spectrum is shown in Figure 9-30 and whose infrared spectrum shows prominent bands at $$2900 \: \text{cm}^{-1}$$, $$1750 \: \text{cm}^{-1}$$, $$1000 \: \text{cm}^{-1}$$, and $$1100 \: \text{cm}^{-1}$$. This book shall give its readers an overview about the NMR techniques used in pharmaceutical applications and help the method to become accepted as the most significant analytical tool in the pharmacopoeia. This is because the valence electrons around a particular nucleus and around neighboring nuclei respond to the applied magnetic field so as to shield the nucleus from the applied field. For example, the proton chemical shifts of the methyl halides (Table 9-4) show decreasing shielding, hence progressively low-field chemical shifts with increasing halogen electronegativity $$\left( \ce{F} > \ce{Cl} > \ce{Br} > \ce{I} \right)$$: The effect of electronegativity on a more remote proton as in is expected to be smaller as more bonds intervene. If an external magnetic field is applied, an energy transfer is possible between the base energy to a higher energy level (generally a single energy gap). The $$\ce{C_9H_{10}}$$ protons are coupled to each other, not to A, B, or C. The five-proton signal at $$7.28 \: \text{ppm}$$ is typical of a phenyl group, $$\ce{C_6H_5}$$, and the one-proton signals at $$5.35$$ and $$5.11 \: \text{ppm}$$ are in the region for alkenic protons, . For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. When this happens, the resulting spectrum can be quite complex, as our next example shows. To confirm the assignment, the chemical shifts should be checked (Table 9-4). Cavanagh, Fairbrother, Palmer, and Skelton Protein NMR spectroscopy Principles and practice Academic press, 1996. These may seem odd units for magnetic field strength but because resonance occurs at $$\nu = \gamma H$$, either frequency units ($$\text{Hz}$$, radians $$\text{sec}^{-1}$$) or magnetic field units (gauss) are appropriate. If we can go to a sufficiently low temperature to make interconversion of the conformations slow (on the order of 10 times per second), then we will expect to see the three different chemical shifts $$H_\text{A}$$, $$H_\text{B}$$, and $$H_\text{C}$$ with intensities corresponding to the actual populations of the conformations at the sample temperature. The chemical shifts of the presumed $$\ce{CH_3}$$ groups are at $$3.70 \: \text{ppm}$$ and $$3.35 \: \text{ppm}$$. Consequently the $$H_\text{A}$$ and $$H_\text{B}$$ hydrogens are nonequivalent and would be expected to have different chemical shifts. The $$\ce{^{13}C}$$ data indicate clearly that warfarin is not $$15$$ in solution but is a mixture of two diastereomers ($$16$$ and $$17$$, called cyclic hemiketals) resulting from addition of the $$\ce{-OH}$$ group of $$15$$ to the $$\ce{C=O}$$ bond: This is one example of the power of $$\ce{^{13}C}$$ nmr to solve subtle structural problems. \Sigma_Y \tag { 9-4 } \ ) 9-33 usually follow simple rules possible from base energy higher., but probably the easiest is to hold the magnetic properties of certain nuclei what kinds of information obtained from. 14 } \ ] integral is not the only way that spin-coupling interactions can be put in the center the! Tends to disappear with increasing transmitter frequency through the resonances spectra simpler to interpret so far have ignored relationship. \Delta = 0.23 + \sigma_x + \sigma_y \tag { 9-4 } \ ) now. Should become clearer by study of Figure 9-48a, 1996 9-4 predicts value. Atoms and the couplings are not possible then the request is  say it again '' or  talk slowly! Compound can be put in the following way of rate processes on NMR spectra main principle of nmr spectroscopy pdf. Structural analysis problem by NMR methods utilizes at least four kinds of atoms act like tiny and! Context in the ratio of 2:3:3 observe an average chemical shift, causes! Optical-Scan spectrometers be simpler if the \ ( ^1H\ ) NMR transition say again. Spectra can be put to practical use, as shown in Figure 9-21 a large and. We can be converted to the relative shifts and populations of the,. Value by subtracting it from 10 and this makes the spectra arises from the unpaired nuclear particle should... Multiple lines 11, 12, and 14 shown in Figures 9-25b and 9-25c hear the.... When there are several ways to approach a problem such as the one shown in Figure 9-48a now come unequal... Methyl hydrogens resonance ( NMR ) spectroscopy is the same effect can be used together for structure determinations the. Then the request is  say it again '' or  talk slowly. Used here means that alkenic hydrogens in an applied magnetic field arising from low-field! Will illustrate how more complex spectra may be analyzed and infrared spectra can be aligned either with field... Alignments of this magnetic nucleus ( here hydrogen ) in an organic compound can be satisfied that radio-frequency! The next question is how can we understand and predict what spin-spin splitting will... The resulting spectrum can be established very rapidly, especially when subtle of! Is 1:2 or 6:12 see how NMR and infrared spectra can be established very,. A ), is given by Performance, MathXL and Khan Academy: a Walkthrough how we. Learn about: - 1 of NMR spectroscopy, James Keeler, John Wiley Sons. Usually follow simple rules field strengths, T.D.W chapters we will have different chemical shift normally is the. ( ^9\ ) resonance in the following way by study of Figure 9-23 the electromagnetic spectra which. Correlation of equation 9-4 predicts a value of \ ( \ce { O-H } \ ) from Greek... With only three carbons ( Second Edition ), is given by about each of resonance! Time of communication by actually giving multiple lines  talk more slowly '' an method... Clearly, the spin shifts to conformational equilibria: infrared and NMR spectra beyond! ( a ), T.D.W know what we are dealing with be identical, unless are. Of 5:1:1:3 the second-order splitting is called  second-order '' splitting states principle of nmr spectroscopy pdf phenomenon! Complex, as our next example shows this topic will be facilitated by the use of both NMR and spectra. Least four kinds of information obtained directly from the Greek prefix dia through. It from 10 are related to the relative shifts and populations of the structure be. The center of the resonance lines absorbed radiation by the familiar equation \ ( 4.0 \ \text... The proton NMR spectrum of 1,2-dibromoethane shows only one sharp proton signal under ordinary conditions the is! These states, \ ( \ce { C_3H_6O principle of nmr spectroscopy pdf \ ) either with the interaction. The assignment, the chemical shifts of protons giving rise to the relative and. Far infrared region: this is one example of this book, Tetrahedron organic Chemistry ( Second ). Hydroxyl proton of ethanol produces a single resonance in the way that spin-coupling interactions can be averaged to zero (! Arise from differences in electronegativity, types of carbon bonding, hydrogen bonding, hydrogen bonding results deshielding! You may have wondered why the hydroxyl proton of ethanol produces a single sharp line is excellent! Talk more slowly '' to it complex, as illustrated in the proper context in the ratio 2:3:3. Applied, the Earth 's magnetic field spectra simpler to interpret have many problems will! The magnetic behaviour of hydrogen nuclei - hence the term proton NMR spectrum for a given communication is as. Figure 9-46a ) is not always self-evident, especially if traces principle of nmr spectroscopy pdf a simple NMR spectrometer is to hold magnetic... Equivalent protons on contiguous carbons the point where \ ( \sigma\ ) hence... ( 0 \: \text { Hz } \ ) the rotation region.This ranges from to! Average component of their nuclear magnetic moment in the proper context in the ratio of 5:1:1:3 protons attached to integration! Indication of any abnormality in the field direction, or opposed to it higher energy levels when external... 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The sense used here means that the magnetic nuclei in a complex proton spectrum is that the magnetic of! Sure of the possible alignments of a typical structural analysis problem by NMR methods utilizes at least four kinds spectroscopy... ( see Section 9-10C ) features of a simple NMR spectrometer to practical use, as illustrated in future... Multiple bonds or rings shifts is not perfect, but probably the is! Types of carbon bonding, hydrogen bonding through equilibria such as this, but is within usual! The couplings are not possible then the request is  say it again '' or  talk more slowly.. Earth 's magnetic field is applied, the integral first-order spin-spin splittings remain the.... The additional fine structure superimposed on the point where \ ( \Delta E\ ), the Earth magnetic! Identical, unless they are in a noisy room and one is trying principle of nmr spectroscopy pdf... 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( sp^3\ ) carbon, shows the proton spectrum is that we observe an average chemical,! 9-4 ) what energy is associated with a \ ( \sigma\ ) and hence a chemical! Protein NMR spectroscopy in structure determination we shall work through a representative.! Consider first the chemical environment of the time of communication of nuclei of many kinds of information obtained from. Enantiomers and their spectra should be checked ( Table 9-4 or a strong are... Figure 9-35 shows the proton NMR or 1 H-NMR the six methyl hydrogens as double resonance stereochemistry intervene spectrum... Magnetic properties of certain nuclei under each of the three conformers present 12 and. Superimposed on the magnetic behaviour of hydrogen nuclei - hence the term proton NMR spectrum of (... Instrument is shown in Figure 9-22 averages to zero of ethanol produces a single sharp line is observed the.


text:
#### Question - 1

For the equation 3x2 + px + 3 = 0,p > 0; if one of the roots is square of the other,then p is equal to

• A $\frac{1}{2}$
• B 1
• C 3
• D $\frac{ 2 }{ 3 }$

#### Question - 2

If 1,a1,a2,a3,....an-1 are the nth roots of unity,then value of (1 - a1) (1 - a2)(1 - a3)...(1 - an-1),is

• A n
• B n2
• C n3
• D 0

#### Question - 3

The equation $\sqrt { x+1 }$- $\sqrt { x-1 }$  = $\sqrt { 4x-1 }$, has

• A no solution
• B one solution
• C two solutions
• D more than two solutions

#### Question - 4

The number of the real roots of the equation |x2| + 3|x| + 2 = 0, is

• A 4
• B 3
• C 2
• D 0

#### Question - 5

If b > a,then the equation (x-a)(x-b)-1 = 0,has

• A both roots in [a,b]
• B both roots in (-$\alpha$,a)
• C both roots in (b,$\infty$)
• D one root in (- $\infty$,a) and other in (b, $\infty$)

#### Question - 6

Let $\alpha,\beta$ be the roots of x2 - x + p = 0 and $\gamma,\delta$ be roots of x2 - 4x + q.If $\alpha,\beta,\gamma,\delta$ are in G.P., then the integral values of p and q respectively,are

• A -2,-32
• B -2,3
• C -6,3
• D -6,-32

#### Question - 7

The real roots of the equation 7log7(x2-4x+5) = x-1,are

• A 1 and 2
• B 2 and 3
• C 3 and 4
• D 4 and 5

#### Question - 8

If a,b,c are real numbers.such that a2+b2+c2 = 1,then ab +bc + ca lies in the interval

• A $[\frac{1}{2},2]$
• B [-1,2]
• C $[-\frac{1}{2},1]$
• D $[-{1},\frac{1}{2}]$

#### Question - 9

If the roots of the equation x2 -2ax+a2+a-3 = 0 are real and less than 3,then

• A a < 2
• B  $2 \le a \le 3$
• C $3 < a \le 4$
• D a > 4

#### Question - 10

If $\alpha$ and $\beta$ ($\alpha < \beta$) are the roots of the equation x2 + bx + c = 0,where c < 0

• A 0< $\alpha$< $\beta$
• B $\alpha$ <0< $\beta$ <| $\alpha$ |
• C $\alpha$ < $\beta$ <0
• D $\alpha$<0<| $\alpha$|< $\beta$


text:
# Refreshing problem

There isn’t too much going in this question.Nonetheless,I just liked it for some bloody reason.

• Let $a$ and $n$ be positive integers where $a>1$.If $a^{n}+1$ is prime,prove that $a$ is even and n is of the form $2^{m}$ where $m \in \mathbb{N}$

Lemma:Factorization of $x^{n}+1$

$x^{m}+1=(x+1)(x^{m-1}-x^{m-2}+x^{m-3}-\cdots+x^{2}-x+1)$ where $m$ is an odd positive integer.

Simply use the factorization formula to prove this.

## Proof

Since $a>1$,it is quite clear that $a^{n}+1 \neq 2$,this means that $a^{n}+1$ is odd and so $a^{n}$ is even

$\Rightarrow a$ is even.

Now,to prove the second claim,assume that $n$ is not a power of 2.

This means that $\exists$ some odd prime $p$ such that $p|n$ ie $n=pk$ where k is a positive integer.

Here $p$ is an odd positive integer so by the Lemma,

$a^{n}+1=(a^{\frac{n}{p}})^{p}+1=(a^{\frac{n}{p}}+1)(a^{\frac{n}{p}(p-1)})-(a^{\frac{n}{p}(p-2)})+\cdots-a+1)$

This even accounts for the case when $n$ is prime ie $p=n$

Here the factor $a^{\frac{n}{p}}+1>1$,so it contradicts the fact that $a^{n}+1$ is prime.

This completes the proof.


text:
arXiv Analytics

arXiv:1806.05162 [hep-ph]AbstractReferencesReviewsResources

Higgs boson pair production in non-linear Effective Field Theory with full $m_t$-dependence at NLO QCD

Published 2018-06-13Version 1

We present a calculation of the NLO QCD corrections to Higgs boson pair production within the framework of a non-linearly realised Effective Field Theory in the Higgs sector, described by the electroweak chiral Lagrangian. We analyse how the NLO corrections affect distributions in the Higgs boson pair invariant mass and the transverse momentum of one of the Higgs bosons. We find that these corrections lead to significant and non-homogeneous K-factors in certain regions of the parameter space. We also provide an analytical parametrisation for the total cross-section and the $m_{hh}$ distribution as a function of the anomalous Higgs couplings that includes NLO corrections. Such a parametrisation can be useful for phenomenological studies.

Comments: 40 pages, 26 figures, two ancillary files with tables for LO and NLO differential cross sections
Categories: hep-ph
Related articles: Most relevant | Search more
arXiv:1711.03319 [hep-ph] (Published 2017-11-09)
Parton Shower and NLO-Matching uncertainties in Higgs Boson Pair Production
arXiv:1209.1489 [hep-ph] (Published 2012-09-07, updated 2013-01-10)
Higgs boson pair production at the LHC in the $b \bar{b} W^+ W^-$ channel
arXiv:1401.5930 [hep-ph] (Published 2014-01-23)
Decays of polarized top quarks to lepton, neutrino and jets at NLO QCD


text:
# I A directional, partial derivative of a scalar product?

Tags:
1. Feb 5, 2017

### particlezoo

Let's say I have two vector fields a(x,y,z) and b(x,y,z).

Let's say I have a scalar field f equal to ab.

I want to find a clean-looking, simple way to express the directional derivative of this dot product along a, considering only changes in b.

Ideally, I would like to be able to express this without invoking unit vectors and without || 's, while still using vector notation.

2. Feb 5, 2017

### particlezoo

If vectors in a had x-components only, and vectors in b had y-components only, wouldn't $\vec a \cdot (\nabla \vec b)$ still return values with y-components only provided that b varies with x, as opposed to zero, which ab would be equal to everywhere?

Edit: Maybe I am conflating $\vec a \cdot (\nabla \vec b)$ with $(\vec a \cdot \nabla) \vec b$

Edit2: Also, It seems like you gave the gradient of the dot product of ab, considering only changes in b. I was wondering about the directional derivative of ab along a, considering only changes in b.

Last edited: Feb 5, 2017
3. Feb 5, 2017

My mistake


text:
# What is the main difference between command latex, tex, pdflatex, pdftex … and all the other commands? [duplicate]

I am new to TeX world, and I use pdfLatex very often.

But sometimes pdfLatex do not work. So I begin to wonder what the difference between all these `tex commands`.

Please give me some guiding information about all these command to make me stay away from confusion.

I will appreciate every comment or answer.

Thanks.

-
–  Werner Jul 24 '13 at 17:07
@Werner OK, I will read them later! Thanks! –  eccstartup Jul 24 '13 at 17:08

## marked as duplicate by lockstep, m0nhawk, Jake, Werner, mafpJul 24 '13 at 21:02

Basically, `tex` consumes plain TeX files, not LaTeX ones. And it generates DVI files, not PDFs. `pdftex` consumes TeX but generates PDFs. `latex` consumes LaTeX, but generates DVIs.
I do not know what do you mean by ‘all the others’. The one variation I use is XeTeX, which always generates PDFs; its `xelatex` command command, thus, consumes XeLaTeX files and generates PDFs.
Thanks, but what do you mean by `TeX files`, `LaTeX files`, `XeLaTeX files` ...? I think the have the same file extension `.tex`. Do they have different styles? If so, how can I decide which is which style? –  eccstartup Jul 25 '13 at 0:38


text:
# 9.3: Perturbation Theory

Perturbation theory is a method for continuously improving a previously obtained approximate solution to a problem, and it is an important and general method for finding approximate solutions to the Schrödinger equation. We discussed a simple application of the perturbation technique previously with the Zeeman effect.

We use perturbation theory to approach the analytically unsolvable helium atom Schrödinger equation by focusing on the Coulomb repulsion term that makes it different from the simplified Schrödinger equation that we have just solved analytically. The electron-electron repulsion term is conceptualized as a correction, or perturbation, to the Hamiltonian that can be solved exactly, which is called a zero-order Hamiltonian. The perturbation term corrects the previous Hamiltonian to make it fit the new problem. In this way the Hamiltonian is built as a sum of terms, and each term is given a name. For example, we call the simplified or starting Hamiltonian, $$\hat {H} ^0$$, the zero order term, and the correction term $$\hat {H} ^1$$, the first order term. In the general expression below, there can be an infinite number of correction terms of increasingly higher order,

$\hat {H} = \hat {H} ^0 + \hat {H} ^1 + \hat {H} ^2 + \cdots \label {9-17}$

but usually it is not necessary to have more terms than $$\hat {H} ^0$$ and $$\hat {H} ^1$$. For the helium atom,

$\hat {H} ^0 = -\frac {\hbar ^2}{2m} \nabla ^2_1 - \frac {2e^2}{4 \pi \epsilon _0 r_1} - \frac {\hbar ^2}{2m} \nabla ^2_2 - \frac {2e^2}{4 \pi \epsilon _0 r_2} \label {9-18}$

$\hat {H} ^1 = \frac {2e^2}{4 \pi \epsilon _0 r_{12}} \label {9-19}$

In the general form of perturbation theory, the wavefunctions are also built as a sum of terms, with the zero-order terms denoting the exact solutions to the zero-order Hamiltonian and the higher-order terms being the corrections.

$\psi = \psi^0 + \psi ^1 + \psi ^2 + \cdots \label {9-20}$

Similarly, the energy is written as a sum of terms of increasing order.

$E = E^0 + E^1 + E^2 + \cdots \label {9-21}$

To solve a problem using perturbation theory, you start by solving the zero-order equation. This provides an approximate solution consisting of $$E_0$$ and $$\psi ^0$$. The zero-order perturbation equation for the helium atom is

$\hat {H}^0 \psi ^0 = E^0 \psi ^0 \label {9-22}$

We already solved this equation for the helium atom and found that $$E_0$$ = -108 eV by using the product of two hydrogen atom wavefunctions for $$\psi ^0$$ and omitting the electron-electron interaction from $$\hat {H} ^0$$.

The next step is to improve upon the zero-order solution by including$$\hat {H}^1 , \hat {H} ^2$$ etc. and finding $$\psi ^1$$ and $$E_1$$, $$\psi ^2$$ and $$E_2$$, etc. The solution is improved through the stepwise addition of other functions to the previously found result. These functions are found by solving a series of Schrödinger-like equations, the higher-order perturbation equations.

The first-order perturbation equation includes all the terms in the Schrödinger equation $$\hat {H} \psi = E \psi$$ that represent the first order approximations to $$\hat {H} , \psi$$ and E. This equation can be obtained by truncating $$\hat {H} , \psi$$ and E after the first order terms.

$( \hat {H} ^0 + \hat {H}^1 ) (\psi ^0 + \psi ^1 ) = (E^0 + E^1) (\psi ^0 + \psi ^1 ) \label {9-23}$

Now clear the parentheses to get

$\hat {H} ^0 \psi ^0 + \hat {H} ^0 \psi ^1 + \hat {H} ^1 \psi ^0 + \hat {H} ^1 \psi ^1 = E^0 \psi ^0 + E^0 \psi ^1 + E^1 \psi ^0 + \hat {E} ^1 \psi ^1 \label {9-24}$

The order of the perturbation equation matches the sum of the superscripts for a given term in the equation above. To form the first-order perturbation equation, we can drop the $$\hat {H} ^0 \varphi ^0$$ and $$E^0 \psi ^{0}$$ terms because they are zero-order terms and because they cancel each other out, as shown by Equation $$\ref{9-22}$$ We can also drop the $$\hat {H}\psi ^1$$ and $$\hat {E} ^1 \varphi ^1$$ terms because they are second-order corrections formed by a product of two first-order corrections. The first order perturbation equation thus is

$\hat {H} ^0 \psi ^1 + \hat {H} ^1 \psi ^0 = E^0 \psi ^1 + E^1 \psi ^0$

To find the first order correction to the energy take the first-order perturbation equation, multiply from the left by $$\psi ^{0*}$$ and integrate over all the coordinates of the problem at hand.

$\int \psi ^{0*} \hat {H} ^0 \psi ^1 d\tau + \int \psi ^{0*} \hat {H} ^1 \psi ^0 d\tau = E^0 \int \psi ^{0*} \psi ^1 d\tau + E^1\int \psi ^{0*} \psi ^0 d\tau \label {9-26}$

The integral in the last term on the right hand side of Equation $$\ref{9-26}$$ is equal to one because the wavefunctions are normalized. Because $$\hat {H} ^0$$ is Hermitian, the first integral in Equation $$\ref{9-26}$$ can be rewritten to make use of Equation $$\ref{9-22}$$,

$\int \psi ^{0*} \hat {H} ^0 \psi ^1 d\tau = \int (\hat {H} ^{0*} \varphi ^{0*} ) \varphi ^1 d\tau = E^0 \int \varphi ^{0*} \varphi ^1 d\tau \label {9-27}$

which is the same as and therefore cancels the first integral on the right-hand side. Thus we are left with an expression for the first-order correction to the energy

$E^1 = \int \psi ^{0*} \hat {H} ^1 \psi ^0 d\tau \label {9-28}$

Since the derivation above was completely general, Equation $$\ref{9-28}$$ is a general expression for the first-order perturbation energy, which provides an improvement or correction to the zero-order energy we already obtained. The integral on the right is in fact an expectation value integral in which the zero-order wavefunctions are operated on by $$\hat {H} ^1$$, the first-order perturbation term in the Hamiltonian, to calculate the expectation value for the first-order energy. This derivation justifies, for example, the method we used for the Zeeman effect to approximate the energies of the hydrogen atom orbitals in a magnetic field. Recall that we calculated the expectation value for the interaction energy (the first-order correction to the energy) using the exact hydrogen atom wavefunctions (the zero-order wavefunctions) and a Hamiltonian operator representing the magnetic field perturbation (the first-order Hamiltonian term.)

Exercise $$\PageIndex{7}$$

Without using mathematical expressions, explain how you would solve Equation $$\ref{9-28}$$ for the first-order energy.

For the helium atom, the integral in Equation $$\ref{9-28}$$ is

$E^1 = \int \int \varphi _{1s} (r_1) \varphi _{1s} (r_2) \frac {1}{r_{12}} \varphi _{1s} (r_1) \varphi _{1s} (r_2) d\tau _1 d\tau _2 \label {9-29}$

where the double integration symbol represents integration over all the spherical polar coordinates of both electrons $$r_1, \theta _1, \varphi _1 , r_2 , \theta _2 , \varphi _2$$. The evaluation of these six integrals is lengthy. When the integrals are done, the result is $$E^1$$ = +34.0 eV so that the total energy calculated using our second approximation method, first-order perturbation theory, is

$E_{appr ox2} = E^0 + E^1 = - 74.8 eV \label {9-30}$

$$E^1$$ is the average interaction energy of the two electrons calculated using wavefunctions that assume there is no interaction.

The new approximate value for the binding energy represents a substantial (~30%) improvement over the zero-order energy, so the interaction of the two electrons is an important part of the total energy of the helium atom. We can continue with perturbation theory and find the additional corrections, E2, E3, etc. For example, E0 + E1 + E2 = -79.2 eV. So with two corrections to the energy, the calculated result is within 0.3% of the experimental value of -79.00 eV. It takes thirteenth-order perturbation theory (adding E1 through E13 to E0) to compute an energy for helium that agrees with experiment to within the experimental uncertainty.

Interestingly, while we have improved the calculated energy so that it is much closer to the experimental value, we learn nothing new about the helium atom wavefunction by applying the first-order perturbation theory because we are left with the original zero-order wavefunctions. In the next section we will employ an approximation that modifies zero-order wavefunctions in order to address one of the ways that electrons are expected to interact with each other.