text:
# 5.4.2 Waves on a string: d’Alembert’s solution

The displacement $w(x,t)$ of an ideal string with tension $T$ and mass per unit length $m$ satisfies the differential equation

$$\frac{\partial^2w}{\partial x^2}= \frac{1}{c^2} \frac{\partial^2w}{\partial t^2}\tag{1}$$

where $c=\sqrt{T/m}$ is the wave speed, from eq. (5) of section 3.1.1. We can make a change of variables: define

$$p=x+ct,\mathrm{~~~~}q=x-ct. \tag{2}$$

Now use the chain rule repeatedly:

$$\frac{\partial w}{\partial x}=\frac{\partial w}{\partial p}\frac{\partial p}{\partial x}+\frac{\partial w}{\partial q}\frac{\partial q}{\partial x}$$

$$=\frac{\partial w}{\partial p}+\frac{\partial w}{\partial q}$$

so

$$\frac{\partial^2 w}{\partial x^2}=\frac{\partial^2 w}{\partial p^2}+2\frac{\partial^2 w}{\partial p \partial q}+\frac{\partial^2 w}{\partial q^2} . \tag{3}$$

Similarly

$$\frac{\partial w}{\partial t}=c \frac{\partial w}{\partial p} -c \frac{\partial w}{\partial q}$$

so that

$$\frac{\partial^2 w}{\partial t^2}= c^2 \left[ \frac{\partial^2 w}{\partial p^2} – \frac{\partial^2 w}{\partial p \partial q} \right]-c^2 \left[ \frac{\partial^2 w}{\partial p \partial q} -\frac{\partial^2 w}{\partial q^2} \right] . \tag{4}$$

Substituting eqs. (3) and (4) into eq. (1) thus gives

$$\frac{\partial^2 w}{\partial p^2}+2\frac{\partial^2 w}{\partial p \partial q}+\frac{\partial^2 w}{\partial q^2} = \frac{\partial^2 w}{\partial p^2}-2\frac{\partial^2 w}{\partial p \partial q}+\frac{\partial^2 w}{\partial q^2}$$

which reduces to

$$4\frac{\partial^2 w}{\partial p \partial q}=0.\tag{5}$$

This equation can be integrated immediately. But we have to remember that these are partial derivatives, not ordinary derivatives. So instead of picking up arbitrary constants of integration, we pick up arbitrary functions. For example, from the equation

$$\frac{\partial w}{\partial p}=0$$

we would deduce that $w$ could be any arbitrary function of $q$. So the general solution of eq. (5) is

$$w(p,q) = f(p) +g(q)$$

where $f$ and $g$ are arbitrary functions. The final result is that the free motion of the string must be of the form

$$w(x,t) = f(x+ct) + g(x-ct),\tag{6}$$

in other words the sum of a wave of any shape travelling leftwards at speed $c$ and another wave of any shape travelling rightwards at speed $c$.

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text:
Thread: Can someone help me with this

1. Question on MIPS, clock cycle & clock per instruction

Hi everyone

I am doing COmputer Organisation, I cant find any hints to solve this question.Can someone help me?

TWO programs, X and Y were executed in THREE different machines, C1, C2 and C3 with the following outcome of execution times in second respectively; 8, 5; 6, 4; 3, 7. Calculate and determine:

a) Relative performance of the machines for each program.
any hint/formula for the next questions, can;t find anything related to this..thank you for your kind help & support
b) MIPS of C1, C2 and C3 given the clock rate 250 MHz, 180 MHz and 150 MHz respectively
c) Clock cycle for program X in C1
d) Clock per instruction for program Y on C2
e) Which machine is the fastest for each of the program?

Appreciate all your kind help & support.thank you & regards

2. Originally Posted by anderson
Hi everyone

I am doing COmputer Organisation, I cant find any hints to solve this question.Can someone help me?

TWO programs, X and Y were executed in THREE different machines, C1, C2 and C3 with the following outcome of execution times in second respectively; 8, 5; 6, 4; 3, 7. Calculate and determine:

a) Relative performance of the machines for each program.
b) MIPS of C1, C2 and C3 given the clock rate 250 MHz, 180 MHz and 150 MHz respectively
c) Clock cycle for program X in C1
d) Clock per instruction for program Y on C2
e) Which machine is the fastest for each of the program?

Appreciate all your kind help & support.thank you & regards
a) Chose one of the machines for a reference, say C1, then the relative performance on program i on machine j is:

$\tau_{i,j}= \dfrac{t_{i,1}}{t_{i,j}}$

Now what difficulties are you having with the others?

CB

3. Hi everyone
I am doing COmputer Organisation, I cant find any hints to solve this question.Can someone help me?
TWO programs, X and Y were executed in THREE different machines, C1, C2 and C3 with the following outcome of execution times in second respectively; 8, 5; 6, 4; 3, 7. Calculate and determine:
a) Relative performance of the machines for each program.
$\frac{8}{5}$, $\frac{6}{4}$, $\frac{3}{7}$
any hint/formula for the next questions, can;t find anything related to this..thank you for your kind help & support
b) MIPS of C1, C2 and C3 given the clock rate 250 MHz, 180 MHz and 150 MHz respectively
c) Clock cycle for program X in C1
d) Clock per instruction for program Y on C2
e) Which machine is the fastest for each of the program?
Appreciate all your kind help & support.thank you & regards

4. Originally Posted by anderson
Hi everyone
I am doing COmputer Organisation, I cant find any hints to solve this question.Can someone help me?
TWO programs, X and Y were executed in THREE different machines, C1, C2 and C3 with the following outcome of execution times in second respectively; 8, 5; 6, 4; 3, 7. Calculate and determine:
a) Relative performance of the machines for each program.
$\frac{8}{5}$, $\frac{6}{4}$, $\frac{3}{7}$
No that is not what you were told, relative to machine 1 the performances on the two programs are: $1,1;\frac{8}{6}, \frac{4}{5}; \frac{8}{3}, \frac{4}{7}$

CB

5. Hi
Thank you for replying, i think there is some mistake in your answer.THis is what i did so far, need your help to comment & advise.Thank you for allkind help & support.

1.) Relative performance:
Program X: 8/4/3 = 0.667
Program Y : 5/3/7 = 0.238
2) MIPS of C1,C2,
MIPS of C1: 250M/8= 31.25MIPS
MIPSfor C2:180M/6 = 30MIPS
MIPS for C3: 150M/3=50MIPS
3) Clock cycle for program X
Clock cycle time for
C1= 1/250M = 4ns
CPU execution time for a program=CPU clock cycles for a program x clock cycle time
For program A in C1
8seconds = Clock cycles x 4ns
Clock cycle for program X in C1 = 8/4ns = 2 x10^9
4) Clock per instruction for program Y on C2
Clock cycle time for
C2= 1/180M = 5.56ns
Clock per instruction = clock cycle time/number of instruction
= 5.56ns/180M = 3.086x10^-17
5)?

6. Originally Posted by anderson
Hi
Thank you for replying, i think there is some mistake in your answer.THis is what i did so far, need your help to comment & advise.Thank you for allkind help & support.

1.) Relative performance:
Program X: 8/4/3 = 0.667
Program Y : 5/3/7 = 0.238
2) MIPS of C1,C2,
MIPS of C1: 250M/8= 31.25MIPS
MIPSfor C2:180M/6 = 30MIPS
MIPS for C3: 150M/3=50MIPS
3) Clock cycle for program X
Clock cycle time for
C1= 1/250M = 4ns
CPU execution time for a program=CPU clock cycles for a program x clock cycle time
For program A in C1
8seconds = Clock cycles x 4ns
Clock cycle for program X in C1 = 8/4ns = 2 x10^9
4) Clock per instruction for program Y on C2
Clock cycle time for
C2= 1/180M = 5.56ns
Clock per instruction = clock cycle time/number of instruction
= 5.56ns/180M = 3.086x10^-17
5)?
Thry explaining what you think you are doing

CB


text:
#### Volume 11, issue 4 (2011)

 Recent Issues
 The Journal About the Journal Editorial Board Editorial Interests Subscriptions Submission Guidelines Submission Page Policies for Authors Ethics Statement ISSN (electronic): 1472-2739 ISSN (print): 1472-2747 Author Index To Appear Other MSP Journals
$\mathrm{SO}(3)$ homology of graphs and links

### Benjamin Cooper, Matt Hogancamp and Vyacheslav Krushkal

Algebraic & Geometric Topology 11 (2011) 2137–2166
##### Abstract

The $SO\left(3\right)$ Kauffman polynomial and the chromatic polynomial of planar graphs are categorified by a unique extension of the Khovanov homology framework. Many structural observations and computations of homologies of knots and spin networks are included.

##### Keywords
categorification, Khovanov homology, Kauffmann polynomial, chromatic polynomial, spin network
##### Mathematical Subject Classification 2010
Primary: 57M27
Secondary: 57M25, 05C10, 05C31


text:
# Why does my gameobject fly of into the air when the screen is touched?

I am encountering a peculiar problem when updating a piece of code that translates my gameobject left and right based on mobile touch input.

To provide context, the platform upon which the gameobject, which is a sphere in my case, translates, is this:

The platform is 10 units wide, with the sphere in the center.

In its original form, the code looks like this:

void FixedUpdate () {
if(Input.touchCount == 1) {
float touchXComponent = Input.GetTouch(0).deltaPosition.x;
transform.Translate(Vector3.right * Time.smoothDeltaTime * touchXComponent * 0.5f);
}
}

It provides the desired left and right movement, but it has a tendency to allow the ball to leave the platform, and then start moving up and down in the air with each further touch.

To combat that, I tried editing the code as follows:

void FixedUpdate(){
if (Input.touchCount == 1){
float touchXComponent = Input.GetTouch(0).deltaPosition.x;
transform.Translate((Vector3.right * Time.smoothDeltaTime * touchXComponent * 0.5f).x, transform.position.y, transform.position.z);
}
}

This, however, causes the ball to fly of the screen the second it is touched, leaving me with this error in the console:

Assertion failed: Invalid worldAABB. Object is too large or too far away from the origin.

I would have thought that the update I made to the code would keep the ball stationary on the y and z-axes, while allowing it to translate freely on the x-axis. Why is that not the case?

Thank you.

• Does the sphere have a Rigidbody? In many cases, mixing rigid body physics simulation and direct transform manipulation can cause hiccups and unexpected behaviours. So if you're using physics you generally want to do all your work physically and stay hands-off with the transform component. – DMGregory Jun 27 '18 at 10:02
• @DMGregory Thank you. I'll try using something like AddForce instead. On another note, were the updates to my other question (about getting smooth movement like current popular games) any good? – Jamie Corkhill Jun 27 '18 at 10:04
• If other users are upvoting or answering your question, that's a sign they find it clear and high-quality. If it's downvoted or idling without activity, that's usually a sign it could benefit from clarifying edits. You don't need me to give a personalized rating — the reception of the site's community is what matters. – DMGregory Jun 27 '18 at 10:13

As this doc says :

If relativeTo is left out or set to Space.Self the movement is applied relative to the transform's local axes.

Which means you're actually translating your object by it's own position every update. To avoid that just change your call to :

transform.Translate((Vector3.right * Time.smoothDeltaTime * touchXComponent * 0.5f).x, 0, 0);

• Thank you. After making the changes, if you drag the screen, the ball still sometimes lifts into the air. – Jamie Corkhill Jun 27 '18 at 8:35
• For this problem, you should freeze Y position of your object. Either by using rigidbody.constraints, or by setting the Y position to a constant every update. – Sasugasm Jun 27 '18 at 8:44
• I wanted to do that, but there are other game elements where the ball will be falling, in which case I do want its Y value to change. I need to figure a way to translate the object independent of it y-axis transform. Also, I can't upvote your answer because I only have 14 rep. – Jamie Corkhill Jun 27 '18 at 8:47
• I'd add a collider to your object so that it will collide with the plane, let gravity make it fall when there is a hole and add a constraint to freeze X axis. If the X axis matters, you may may clamp it's position for every update between two constant values. – Sasugasm Jun 27 '18 at 8:51
• Thank you. That makes sense. Can I ask you one last question? Is the method I'm using the best way to get smooth movement from touch input? I was thinking something involving Vector3.Lerp() or some other easing function would be better, but I'm unsure how to go about that. – Jamie Corkhill Jun 27 '18 at 8:54

From what I understand, you want the ball to move on the X axis only, relative to world coordinates. The current method you're using will do it in the local coordinates (relative to the object).

Since Transform.Translate method already gives you this functionality, you can just change your line from

transform.Translate((Vector3.right * Time.smoothDeltaTime * touchXComponent * 0.5f).x, transform.position.y, transform.position.z);

to

transform.Translate((Vector3.right * Time.smoothDeltaTime * touchXComponent * 0.5f).x, transform.position.y, transform.position.z, Space.World);

and it should work in theory (I don't have access to Unity right now).


text:
# Attempting to understand relativity and time dilation

Tags:
1. Dec 2, 2015

### Prierin

1. The problem statement, all variables and given/known data
t = t0/(1-v2/c2)1/2

2. Relevant equations
t = 10/(1-.95c^2/c^2)1/2

3. The attempt at a solution

The provided solutiojn to the example given above is:

t = 10/(1- (.95c)2/c2)1/2

t = 10/(1- .952)1/2

t = 10/ .312

t = 32

Unfortunately, no matter what I do the answer I continuously come up with is 51.28. I've tried four different approaches and come up with the same EXACT answer each time without fail. I am not so bold as to say the example above is incorrect as this level of math isn't my forte, so I am stuck. Can anyone explain to me WHY the answer would be 32 rather than 51.28?

I am not familiar with the 1/2 at the end of the formula and I suspect that may be what is throwing my numbers off. I am nto clever enough, it seems, to get my calculator to understand that portion correctly.

(On a side note, I have also seen this forumal written as t=t0/(1-v^2/c^2)^1/2. Not sure which is correct.)

2. Dec 2, 2015

### andrewkirk

That's what's correct, and your use of the other one is probably what's causing you problems.

3. Dec 2, 2015

### Prierin

OK, that's good to know, thanks. However. even with that I am still coming up with the same result of 51.28.

if c= 299792458 and .95c = 284802835.1

10/(1-284802835.1^2/299792458^2)^1/2 = 51.28205128205128205128

The example answer, however, is 32, so I am still confused as how they reached that conclusion.

EDIT: I just attempted to edit the formula somewhat and FINALLY gor 32 as an answer by using .5 rather than 1/2.

10/(1-284802835.1^2 / 299792458^2)^.5 = 32.02 I had a feeing the 1/2 was being interpreted incorrectly by the calculator as a division rather than a fraction... still, I'll need confirmation from someone who knows the math better than I before I celebrate

Last edited: Dec 2, 2015
4. Dec 2, 2015

### Ray Vickson

The correct formula is
$$t = \frac{t_0}{\sqrt{ 1 - v^2/c^2} }= \frac{t_0}{(1 - v^2/c^2)^{1/2} }$$
If you use that you will get 32.03. The "1/2" is not a fraction or anything like it; it is an exponent (= the "1/2"th power = square root).

5. Dec 2, 2015

### Prierin

BEAUTIFUL! I knew I was doing something wrong.

Now, I have calculated the formula both ways: 10/(sqrt(1-284802835.1^2/299792458^2)) & 10/(1-284802835.1^2/299792458^2)^.5 and have come up with the same EXACT answer of 32.02563076101742669665 (or 32.03) so now I am a happy chappie!

Thanks and kudos!

Now if I can just fogure out how to save values in the calculator LOL

6. Dec 2, 2015

### andrewkirk

It wasn't being interpreted incorrectly, it was being punched in incorrectly.
The standard rule for precedence of operators (ie which arithmetic operations get done before others) is that exponentiation is done before multiplication and division, which are done before addition and subtraction.
So if you punched in

10/(1-284802835.1^2/299792458^2)^1/2

then that means

10/((1-284802835.1^2/299792458^2)^1)/2

which is

$$\bigg[\frac{10}{(1-\frac{(284802835.1)^2}{(299792458)^2})^1}\bigg]\div 2$$

which is not what you want.

7. Dec 2, 2015

### Prierin

I meant to say, the calculator wasn't misinterpreting - I was. LOL

It makes sense now. As I pointed out, if I use ^.5 the answer comes out correctly so now I can breathe a little easier. I didn't *kneed* to know any of this math, but I have a curious mind and like to understand how things work...


text:
Chapter 30 Atomic Physics

# 246 30.8 Quantum Numbers and Rules

### Summary

• Define quantum number.
• Calculate angle of angular momentum vector with an axis.
• Define spin quantum number.

Physical characteristics that are quantized—such as energy, charge, and angular momentum—are of such importance that names and symbols are given to them. The values of quantized entities are expressed in terms of quantum numbers, and the rules governing them are of the utmost importance in determining what nature is and does. This section covers some of the more important quantum numbers and rules—all of which apply in chemistry, material science, and far beyond the realm of atomic physics, where they were first discovered. Once again, we see how physics makes discoveries which enable other fields to grow.

The energy states of bound systems are quantized, because the particle wavelength can fit into the bounds of the system in only certain ways. This was elaborated for the hydrogen atom, for which the allowed energies are expressed as $\boldsymbol{E_n \propto 1/n^2}$, where $\boldsymbol{n = 1, 2, 3, \cdots }$. We define $\boldsymbol{n}$ to be the principal quantum number that labels the basic states of a system. The lowest-energy state has $\boldsymbol{n = 1}$, the first excited state has
$\boldsymbol{n = 2}$, and so on. Thus the allowed values for the principal quantum number are

$\boldsymbol{n = 1, 2, 3, \cdots}$

This is more than just a numbering scheme, since the energy of the system, such as the hydrogen atom, can be expressed as some function of $\boldsymbol{n}$, as can other characteristics (such as the orbital radii of the hydrogen atom).

The fact that the magnitude of angular momentum is quantized was first recognized by Bohr in relation to the hydrogen atom; it is now known to be true in general. With the development of quantum mechanics, it was found that the magnitude of angular momentum $\boldsymbol{L}$ can have only the values

$\boldsymbol{L = \sqrt{l (l + 1)}}$$\boldsymbol{\frac{h}{2 \pi}}$$\;\;\;\; \boldsymbol{(l = 0, 1, 2, \cdots , n - 1)}$

where $\boldsymbol{l}$ is defined to be the angular momentum quantum number. The rule for $\boldsymbol{l}$ in atoms is given in the parentheses. Given $\boldsymbol{n}$, the value of $\boldsymbol{l}$ can be any integer from zero up to $\boldsymbol{n - 1}$. For example, if $\boldsymbol{n = 4}$, then $\boldsymbol{l}$ can be 0, 1, 2, or 3.

Note that for $\boldsymbol{n = 1}$, $\boldsymbol{l}$ can only be zero. This means that the ground-state angular momentum for hydrogen is actually zero, not $\boldsymbol{h/2 \pi}$ as Bohr proposed. The picture of circular orbits is not valid, because there would be angular momentum for any circular orbit. A more valid picture is the cloud of probability shown for the ground state of hydrogen in Chapter 30.6 Figure 3. The electron actually spends time in and near the nucleus. The reason the electron does not remain in the nucleus is related to Heisenberg’s uncertainty principle—the electron’s energy would have to be much too large to be confined to the small space of the nucleus. Now the first excited state of hydrogen has $\boldsymbol{n = 2}$, so that $\boldsymbol{l}$ can be either 0 or 1, according to the rule in $\boldsymbol{L = \sqrt{l (l + 1)} \frac{h}{2 \pi}}$. Similarly, for $\boldsymbol{n = 3}$, $\boldsymbol{l}$ can be 0, 1, or 2. It is often most convenient to state the value of $\boldsymbol{l}$, a simple integer, rather than calculating the value of $\boldsymbol{L}$ from $\boldsymbol{L = \sqrt{l(l+1)} \frac{h}{2 \pi}}$. For example, for $\boldsymbol{l = 2}$, we see that

$\boldsymbol{L = \sqrt{2 (2+1)}}$$\boldsymbol{\frac{h}{2 \pi}}$$\boldsymbol{= \sqrt{6}}$$\boldsymbol{\frac{h}{2 \pi}}$$\boldsymbol{= 0.390h = 2.58 \times 10^{-34} \;\textbf{J} \cdot \textbf{s}}$

It is much simpler to state $\boldsymbol{l = 2}$.

As recognized in the Zeeman effect, the direction of angular momentum is quantized. We now know this is true in all circumstances. It is found that the component of angular momentum along one direction in space, usually called the $\boldsymbol{z}$ -axis, can have only certain values of $\boldsymbol{L_z}$. The direction in space must be related to something physical, such as the direction of the magnetic field at that location. This is an aspect of relativity. Direction has no meaning if there is nothing that varies with direction, as does magnetic force. The allowed values of $\boldsymbol{L_z}$ are

$\boldsymbol{L_z = m_l}$$\boldsymbol{\frac{h}{2 \pi}}$$\;\;\;\boldsymbol{(m_l = -l, -l +1, \cdots , -1, 0, 1, \cdots l - 1,l)}$

where $\boldsymbol{L_z}$ is the $\boldsymbol{z}$ -component of the angular momentum and $\boldsymbol{m_l}$ is the angular momentum projection quantum number. The rule in parentheses for the values of $\boldsymbol{m_l}$ is that it can range from $\boldsymbol{-l}$ to $\boldsymbol{l}$ in steps of one. For example, if $\boldsymbol{l = 2}$, then $\boldsymbol{m_l}$ can have the five values –2, –1, 0, 1, and 2. Each $\boldsymbol{m_l}$ corresponds to a different energy in the presence of a magnetic field, so that they are related to the splitting of spectral lines into discrete parts, as discussed in the preceding section. If the $\boldsymbol{z}$ -component of angular momentum can have only certain values, then the angular momentum can have only certain directions, as illustrated in Figure 1.

### Example 1: What Are the Allowed Directions?

Calculate the angles that the angular momentum vector $\boldsymbol{L}$ can make with the $\boldsymbol{z}$ -axis for $\boldsymbol{l = 1}$, as illustrated in Figure 1.

Strategy

Figure 1 represents the vectors $\boldsymbol{L}$ and $\boldsymbol{L_z}$ as usual, with arrows proportional to their magnitudes and pointing in the correct directions. $\boldsymbol{L}$ and $\boldsymbol{L_z}$ form a right triangle, with $\boldsymbol{L}$ being the hypotenuse and $\boldsymbol{L_z}$ the adjacent side. This means that the ratio of $\boldsymbol{L_z}$ to $\boldsymbol{L}$ is the cosine of the angle of interest. We can find $\boldsymbol{L}$ and $\boldsymbol{L_z}$ using $\boldsymbol{L = \sqrt{l (l+1)}\frac{h}{2\pi}}$.

Solution

We are given $\boldsymbol{l = 1}$, so that $\boldsymbol{m_l}$ can be +1, 0, or −1. Thus $\boldsymbol{L}$ has the value given by $\boldsymbol{L = \sqrt{l (l+1)} \frac{h}{2 \pi}}$.

$\boldsymbol{L =}$$\boldsymbol{\frac{\sqrt{l (l+1)}h}{2 \pi}}$$\boldsymbol{=}$$\boldsymbol{\frac{\sqrt{2}h}{2 \pi}}$

$\boldsymbol{L_z}$ can have three values, given by $\boldsymbol{L_z = m_l \frac{h}{2 \pi}}$.

$\boldsymbol{L_z = m_l}$$\boldsymbol{\frac{h}{2 \pi}}$$\boldsymbol{=}$$\boldsymbol{ \{ }$ $\begin{array}{cccc} \boldsymbol{\frac{h}{2 \pi},} & \boldsymbol{m_l} & \boldsymbol{=} & \boldsymbol{+1} \\[1em] \boldsymbol{0,} & \boldsymbol{m_l} & \boldsymbol{=} & \boldsymbol{0} \\[1em] \boldsymbol{- \frac{h}{2 \pi},} & \boldsymbol{m_l} & \boldsymbol{=} & \boldsymbol{-1} \end{array}$

As can be seen in Figure 1, $\boldsymbol{\textbf{cos} \;\theta = \textbf{L}_z / \textbf{L}}$ and so for $\boldsymbol{m_l = +1}$, we have

$\boldsymbol{\textbf{cos} \;\theta _1=}$$\boldsymbol{\frac{L_Z}{L}}$$\boldsymbol{=}$$\boldsymbol{\frac{\frac{h}{2 \pi}}{\frac{2h}{2 \pi}}}$$\boldsymbol{=}$$\boldsymbol{\frac{1}{\sqrt{2}}}$$\boldsymbol{= 0.707}$

Thus,

$\boldsymbol{\theta _1 = \textbf{cos}^{-1} 0.707 = 45.0 ^{\circ}}$.

Similarly, for $\boldsymbol{m_l = 0}$, we find $\boldsymbol{\textbf{cos} \theta _2 = 0}$; thus,

$\boldsymbol{\theta _2 = \textbf{cos}^{-1} 0 = 90.0 ^{\circ}}$.

And for $\boldsymbol{m_l = -1}$,

$\boldsymbol{\textbf{cos} \theta _3 =}$$\boldsymbol{\frac{L_Z}{L}}$$\boldsymbol{=}$$\boldsymbol{\frac{- \frac{h}{2 \pi}}{\frac{2h}{2 \pi}}}$$\boldsymbol{=}$$\boldsymbol{- \frac{1}{\sqrt{2}}}$$\boldsymbol{= -0.707}$,

so that

$\boldsymbol{\theta _3 = \textbf{cos}^{-1} (-0.707) = 135.0^{\circ}}$.

Discussion

The angles are consistent with the figure. Only the angle relative to the $\boldsymbol{z}$ -axis is quantized. $\boldsymbol{L}$ can point in any direction as long as it makes the proper angle with the $\boldsymbol{z}$ -axis. Thus the angular momentum vectors lie on cones as illustrated. This behavior is not observed on the large scale. To see how the correspondence principle holds here, consider that the smallest angle ($\boldsymbol{\theta _1}$ in the example) is for the maximum value of $\boldsymbol{m_l = 0}$, namely $\boldsymbol{m_l = l}$. For that smallest angle,

$\boldsymbol{\textbf{cos} \theta =}$$\frac{L_z}{L}$$\boldsymbol{=}$$\boldsymbol{\frac{l}{\sqrt{l (l+1)}}}$,

which approaches 1 as $\boldsymbol{l}$ becomes very large. If $\boldsymbol{\textbf{cos} \theta = 1}$, then $\boldsymbol{\theta =0 ^{\circ}}$. Furthermore, for large $\boldsymbol{l}$, there are many values of $\boldsymbol{m_l}$, so that all angles become possible as $\boldsymbol{l}$ gets very large.

# Intrinsic Spin Angular Momentum Is Quantized in Magnitude and Direction

There are two more quantum numbers of immediate concern. Both were first discovered for electrons in conjunction with fine structure in atomic spectra. It is now well established that electrons and other fundamental particles have intrinsic spin, roughly analogous to a planet spinning on its axis. This spin is a fundamental characteristic of particles, and only one magnitude of intrinsic spin is allowed for a given type of particle. Intrinsic angular momentum is quantized independently of orbital angular momentum. Additionally, the direction of the spin is also quantized. It has been found that the magnitude of the intrinsic (internal) spin angular momentum, $\boldsymbol{S}$, of an electron is given by

$\boldsymbol{S = \sqrt{s (s+1)}}$$\boldsymbol{\frac{h}{2 \pi}}$$\;\;\; \boldsymbol{(s= 1/2 \;\textbf{for electrons})}$,

where $\boldsymbol{s}$ is defined to be the spin quantum number. This is very similar to the quantization of $\boldsymbol{L}$ given in $\boldsymbol{L = \sqrt{l (l + 1)} \frac{h}{2 \pi}}$, except that the only value allowed for $\boldsymbol{s}$ for electrons is 1/2.

The direction of intrinsic spin is quantized, just as is the direction of orbital angular momentum. The direction of spin angular momentum along one direction in space, again called the $\boldsymbol{z}$ -axis, can have only the values

$\boldsymbol{S_z = m_s}$$\boldsymbol{\frac{h}{2 \pi}}$$\;\;\; \boldsymbol{(}$$\boldsymbol{m_s = -}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{,+}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{)}$

for electrons. $\boldsymbol{S_z}$ is the $\boldsymbol{z}$ -component of spin angular momentum and $\boldsymbol{m_s}$ is the spin projection quantum number. For electrons, $\boldsymbol{s}$ can only be 1/2, and $\boldsymbol{m_s}$ can be either +1/2 or –1/2. Spin projection $\boldsymbol{m_s = +1/2}$ is referred to as spin up, whereas $\boldsymbol{m_s = -1/2}$\ is called spin down. These are illustrated in Chapter 30.7 Figure 5.

### Intrinsic Spin

In later chapters, we will see that intrinsic spin is a characteristic of all subatomic particles. For some particles $\boldsymbol{s}$ is half-integral, whereas for others $\boldsymbol{s}$ is integral—there are crucial differences between half-integral spin particles and integral spin particles. Protons and neutrons, like electrons, have $\boldsymbol{s = 1/2}$, whereas photons have $\boldsymbol{s = 1}$, and other particles called pions have $\boldsymbol{s = 0}$, and so on.

To summarize, the state of a system, such as the precise nature of an electron in an atom, is determined by its particular quantum numbers. These are expressed in the form n, l,ml,msn, l,ml,ms —see Table 1 For electrons in atoms, the principal quantum number can have the values $\boldsymbol{n=1, 2, 3, \cdots}$ . Once $\boldsymbol{n}$ is known, the values of the angular momentum quantum number are limited to $\boldsymbol{l = 1, 2, 3, \cdots , n - 1}$ . For a given value of $\boldsymbol{l}$, the angular momentum projection quantum number can have only the values $\boldsymbol{m_l = -l,-l + 1, \cdots , -1, 0, 1, \cdots ,l - 1,l}$ . Electron spin is independent of $\boldsymbol{n}$, $\boldsymbol{l}$, and $\boldsymbol{m_l}$, always having $\boldsymbol{s=1/2}$. The spin projection quantum number can have two values, $\boldsymbol{m_s=1/2}$ or $\boldsymbol{-1/2}$.

Name Symbol Allowed values
Principal quantum number $\boldsymbol{n}$ $\boldsymbol{1, 2, 3, \cdots}$
Angular momentum $\boldsymbol{l}$ $\boldsymbol{0, 1, 2, \cdots n- 1}$
Angular momentum projection $\boldsymbol{m_l}$ $\boldsymbol{-l, -l + 1, \cdots , -1, 0, 1, \cdots ,l-1,l \; (\textbf{or} \; 0, \pm 1, \pm 2, \cdots , \pm l)}$
Spin1 $\boldsymbol{s}$

$\boldsymbol{1/2 \; (\textbf{electrons})}$

Spin projection $\boldsymbol{m_s}$ $\boldsymbol{-1/2, +1/2}$
Table 1: Atomic Quantum Numbers

Figure 2 shows several hydrogen states corresponding to different sets of quantum numbers. Note that these clouds of probability are the locations of electrons as determined by making repeated measurements—each measurement finds the electron in a definite location, with a greater chance of finding the electron in some places rather than others. With repeated measurements, the pattern of probability shown in the figure emerges. The clouds of probability do not look like nor do they correspond to classical orbits. The uncertainty principle actually prevents us and nature from knowing how the electron gets from one place to another, and so an orbit really does not exist as such. Nature on a small scale is again much different from that on the large scale.

We will see that the quantum numbers discussed in this section are valid for a broad range of particles and other systems, such as nuclei. Some quantum numbers, such as intrinsic spin, are related to fundamental classifications of subatomic particles, and they obey laws that will give us further insight into the substructure of matter and its interactions.

### PhET Explorations: Stern-Gerlach Experiment

The classic Stern-Gerlach Experiment shows that atoms have a property called spin. Spin is a kind of intrinsic angular momentum, which has no classical counterpart. When the z-component of the spin is measured, one always gets one of two values: spin up or spin down.

# Section Summary

• Quantum numbers are used to express the allowed values of quantized entities. The principal quantum number $\boldsymbol{n}$ labels the basic states of a system and is given by
$\boldsymbol{n = 1, 2, 3, \cdots}$
• The magnitude of angular momentum is given by
$\boldsymbol{L = \sqrt{l(l+1)}}$$\boldsymbol{\frac{h}{2 \pi}}$$\boldsymbol{(l=0, 1, 2, \cdots , n-1)}$

where $\boldsymbol{l}$ is the angular momentum quantum number. The direction of angular momentum is quantized, in that its component along an axis defined by a magnetic field, called the $\boldsymbol{z}$ -axis is given by

$\boldsymbol{L_z = m_l}$$\boldsymbol{\frac{h}{2 \pi}}$$\;\;\;\boldsymbol{(m_l = -l, -l+1, \cdots , 1, 0, 1, \cdots l-1,l)}$

where $\boldsymbol{L_z}$ is the $\boldsymbol{z}$ -component of the angular momentum and $\boldsymbol{m_l}$ is the angular momentum projection quantum number. Similarly, the electron’s intrinsic spin angular momentum $\boldsymbol{S}$ is given by

$\boldsymbol{S= \sqrt{s (s+1)}}$$\boldsymbol{\frac{h}{2 \pi}}$$\;\;\;\boldsymbol{(s = 1/2 \;\textbf{for electrons})}$

$\boldsymbol{s}$ is defined to be the spin quantum number. Finally, the direction of the electron’s spin along the $\boldsymbol{z}$ -axis is given by

$\boldsymbol{S_z = m_s}$$\boldsymbol{\frac{h}{2 \pi}}$$\boldsymbol{(}$$\boldsymbol{m_s = -}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{,+}$$\boldsymbol{\frac{1}{2}}$$\boldsymbol{)}$,

where $\boldsymbol{S_z}$ is the $\boldsymbol{z}$ -component of spin angular momentum and $\boldsymbol{m_s}$ is the spin projection quantum number. Spin projection $\boldsymbol{m_s = +1/2}$ is referred to as spin up, whereas $\boldsymbol{m_s= -1/2}$ is called spin down. Table 1 summarizes the atomic quantum numbers and their allowed values.

### Conceptual Questions

1: Define the quantum numbers $\boldsymbol{n}$, $\boldsymbol{l}$, $\boldsymbol{m_l}$, $\boldsymbol{s}$ , $\boldsymbol{m_s}$.

2: For a given value of $\boldsymbol{n}$, what are the allowed values of $\boldsymbol{l}$?

3: For a given value of $\boldsymbol{l}$, what are the allowed values of $\boldsymbol{m_l}$? What are the allowed values of $\boldsymbol{m_l}$ for a given value of $\boldsymbol{n}$? Give an example in each case.

4: List all the possible values of $\boldsymbol{s}$ and $\boldsymbol{m_s}$ for an electron. Are there particles for which these values are different? The same?

### Problem Exercises

1: If an atom has an electron in the $\boldsymbol{n=5}$ state with $\boldsymbol{m_l=3}$, what are the possible values of $\boldsymbol{l}$?

2: An atom has an electron with $\boldsymbol{m_l = 2}$. What is the smallest value of $\boldsymbol{n}$ for this electron?

3: What are the possible values of $\boldsymbol{m_l}$ for an electron in the $\boldsymbol{n = 4}$ state?

4: What, if any, constraints does a value of $\boldsymbol{m_l=1}$ place on the other quantum numbers for an electron in an atom?

5: (a) Calculate the magnitude of the angular momentum for an $\boldsymbol{l = 1}$ electron. (b) Compare your answer to the value Bohr proposed for the n=1n=1 size 12{n=1} {} state.

6: (a) What is the magnitude of the angular momentum for an $\boldsymbol{l=1}$ electron? (b) Calculate the magnitude of the electron’s spin angular momentum. (c) What is the ratio of these angular momenta?

7: Repeat Chapter 30.8 Exercise 6 for $\boldsymbol{l=3}$.

8: (a) How many angles can $\boldsymbol{L}$ make with the $\boldsymbol{z}$ -axis for an $\boldsymbol{l = 2}$ electron? (b) Calculate the value of the smallest angle.

9: What angles can the spin $\boldsymbol{S}$ of an electron make with the $\boldsymbol{z}$ -axis?

## Footnotes

1. 1 The spin quantum number s is usually not stated, since it is always 1/2 for electrons

## Glossary

quantum numbers
the values of quantized entities, such as energy and angular momentum
angular momentum quantum number
a quantum number associated with the angular momentum of electrons
spin quantum number
the quantum number that parameterizes the intrinsic angular momentum (or spin angular momentum, or simply spin) of a given particle
spin projection quantum number
quantum number that can be used to calculate the intrinsic electron angular momentum along the $\boldsymbol{z}$ -axis
z-component of spin angular momentum
component of intrinsic electron spin along the $\boldsymbol{z}$ -axis
magnitude of the intrinsic (internal) spin angular momentum
given by $\boldsymbol{S = \sqrt{s (s+1)} \frac{h}{2 \pi}}$
z-component of the angular momentum
component of orbital angular momentum of electron along the $\boldsymbol{z}$ -axis

### Solutions

Problem Exercises

1: $\boldsymbol{l = 4, 3}$ are possible since $\boldsymbol{l 3: [latex]\boldsymbol{n = 4 \Rightarrow l = 3, 2, 1, 0 \Rightarrow m_1 = \pm 3, \pm 2, \pm 1, 0}$ are possible.

5: (a) $\boldsymbol{1.49 \times 10^{-34} \;\textbf{J} \cdot \textbf{s}}$

(b) $\boldsymbol{1.06 \times 10^{-34} \;\textbf{J} \cdot \textbf{s}}$

7: (a) $\boldsymbol{3.66 \times 10^{-34} \;\textbf{J} \cdot \textbf{s}}$

(b) $\boldsymbol{s = 9.13 \times 10^{-35} \;\textbf{J} \cdot \textbf{s}}$

(c) $\boldsymbol{\frac{L}{S} = \frac{\sqrt{12}}{\sqrt{3/4}} = 4}$

9: $\boldsymbol{\theta = 54.7^{\circ} , 125.3^{\circ}}$


text:
# Thread: Taylor Series and shifting indices

1. ## Taylor Series and shifting indices

For the truncated form of the Taylor series, the Nth Taylor polynomial is given in our notes as

$T_{N}(x) = \sum_{n = 0}^N \frac{f^{(n)}(0)}{n!} \cdot x^{n}$

But then later, he defines $T_{3}(x)$ for $f(x)=\sin x$

as $x-\frac{x^{3}}{6}$

Should it not be expanded to three terms, or is the notation signifying that it should be expanded until n, rather than N, reaches 3?

Also, I have a question about shifting indices. Our notes give this equality

$\sum_{n = 1}^\infty \frac{(-1)^{n}4^{n}}{n!} = \sum_{n = 1}^\infty \frac{(-4)^{n}}{n!} = \sum_{n = 0}^\infty \frac{(-4)^{n}}{n!} - 1 = e^{-4} - 1 ,$

but is it not:

$\sum_{n = 1}^\infty \frac{(-4)^{n}}{n!} = \sum_{n = 0}^\infty \frac{(-4)^{n}}{n!} - \frac{4}{n+1}?$

Thanks

2. For the first, notice that all even-numbered terms cancel since $\sin(0)=0$. For the second, you're just adding the term corresponding to $n=0$, I don't really know what you did to get to the other expression.

3. Oh ok I see now, his notes just said to take 1 away from n in the index and increase it in the summand, not very clear to be honest! Thank you for that


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# Question about GLMM in R

We have a dataset containing drug seizures every month from 2014 - 2018 in Ohio (60 months). For each drug seizure, we have the following variables in a dataframe I’m calling df, with some 130,000 rows.

• Date = month/year it was seized,
• Weight = the weight of the seizure in grams,
• County = the county in Ohio where the seizure occurred, the chemical composition (only used to create the variable Have_Fentanyl),
• Have_Fentanyl = 1 if fentanyl is present and 0 otherwise, LabCaseNo = lab case identifier

If two drugs were seized in the same bust, they have the same lab case identifier. We also have deaths due to drug overdose every month/year, in a column with a zillion repetitions (because 60 values spread out over 130,000 rows).

We are trying to fit a generalized linear mixed model to predict deaths from the other variables. Because seizures that happened as part of the same bust are not independent, we have one level for seizures, one for busts, one for county, and one for month/year. In R, we fit this model via the following code

library(lme4)
summary(glmer(Death ~ log(Weight) + Have_Fentanyl
+  (1 | LabCaseNo) + (1 | County) + Date,
data = df[complete.cases(df\$Death),],
control = glmerControl(optimizer = "bobyqa"),
family = "poisson"))

(1) As the response variable is count data, we are using Poisson residuals. How would we modify this code to make this quasi-Poisson or negative binomial?

When we run this code we get an error that says Model is nearly unidentifiable: large eigenvalue ratio - Rescale variables? When we start dropping variables like Date and County we get other error messages like Model failed to converge with max|grad| = 2.64655 (tol = 0.001, component 1) Model is nearly unidentifiable: very large eigenvalue - Rescale variables?

We get these errors for every choice of optimizer (“bobyqa” above) that we have tried. The source we learned about the GLMM model from also had “failed to converge” errors, and ignored them.

(2) Do we need to worry about any of these errors?

(3) Why are we getting these errors? Should we use a different GLMM command?

We’ve already taken the log of our weight variable. All the other variables are categorical. When we run the model on only 100 data points we don’t get errors, so it might have to do with the number of rows.

Lastly, we are a bit confused about which variables should be “fixed” effects and which “random” effects. We’ve read other posts on Cross Validated, blogs, course webpages, and the writings of Gelman. I had envisioned a four level model with a level for time, one for county, and one for LabCaseNo, but I don’t actually know how to fit that model with glmer or lmer. If anyone has a tip please share!

• What's the format of the variables such as county, date, labcaseno, etc? Are they factors? Are you able to fit a glm without random effects? How do you tie an overdose to an individual bust? – jsk Jul 16 at 2:49
• County is categorical, with 88 levels. Date is categorical with 60 levels, but glmer is able to fit it as quantitative too (1-60), and LabCaseNo is categorical with many, many levels (probably about 40,000). When we tried glmer without random effects, the number of dummy variables breaks the model. We also can't tie individual overdoses to individual busts. Only deaths (per month and per county) is available. – David White Jul 16 at 3:41
• Labcaseno with that many levels will probably break glmer as well. If you only have deaths per month and county, then how can you predict deaths from labcaseno? – jsk Jul 16 at 5:24
• @jsk I think you are hitting on the main problem. My method had been to make a column, Deaths, containing the number of deaths in the month when the drug seizure happened. But, because we can easily have 2,000 seizures in a month, that death value is repeated 2,000 times. The answer below says this is what's breaking glmer. – David White Jul 16 at 14:45

I think there are two problems going on here. First, I'll hazard a guess that your model is far too complex—all of these nestings are probably the source of the errors. Second, and more importantly, you are fitting different data to the same outcome variables many times over.

Your outcome (please correct me if I am wrong here) is drug overdose deaths in Ohio during a given month, so you have 60 outcome datapoints. Also, while these deaths per month are technically count data, if they are "big enough" you could reasonably ditch the poisson assumption.

I think a better strategy might be to try a little feature engineering to make some new variables that are more informative.

For example, why not transform your variables weight, county and date from their current form (130,000 rows) into something that summarizes the total amount of drugs seized across the state in a particular month, like total_grams_of_drugs_siezed? (forgive my verbose variable names). And maybe sum up all of your Have_Fentanyl instances during a given month into a new variable total_fentanyl_seizures?

Then you could have something like deaths ~ total_grams + total_fentanyl_seizures in a plain, boring old lm(). That's a good start. If you wanted to get fancy, you could apply some longitudinal techniques (a GAM with a s(time) term, maybe?) but I'd be worried that you're actually regressing out something you care about—yes, drug overdose deaths are changing over time, but don't you think they might be changing because there are more fentanyl-laced drugs?

A final note, another type of variable that you might want to try constructing would be a "time-lagged" variable—drug deaths in October might be predicted by drug deaths in September, for example.

• Thanks for your answer. We did both of these suggestions earlier in our paper. We began with a "plain, boring old lm()" with total_fentanyl_seizures, and were advised not to use total_grams because one high weight seizure should not be the same as a bunch of low weight seizures (indeed, low weight seizures tend to be deadlier, because of how fentanyl is mixed into the drug supply). In the next section, we fit an ARIMA model with deaths and total_fentanyl_seizures. We wanted the GLMM as an omnibus model to finish the paper. Can you suggest ways to correct the 2 problems and still use a GLMM? – David White Jul 16 at 3:39
• I agree with John Davis - you shouldn't be repeating your outcome over and over again within one model. Some further feature engineering of your "plain, boring old lm()" should fix this problem. For example, you could create a new variable called "number of low weight seizures". Use your expert knowledge to decide on appropriate predictors and thresholds when defining them. You just need to make sure you're clear about how each predictor was defined, and give good explanations for how/why you defined them as such, when reporting your results. – rw2 Jul 16 at 13:17
• @rw2, thanks for your feedback. Since we've already demonstrated the importance of weight in an earlier section, I might end up just deleting this GLMM section. – David White Jul 16 at 14:48


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# Math Help - Simultaneous Equation

1. ## Simultaneous Equation

Hi there, I've got a simultaneous equation that I am not sure of how to solve. I have looked at substitution, however I don't know how to reduce one variable to a single without making the rest of the equation look strange. Basically I don't see how to either eliminate one variable, or get one variable to a coefficient of 1.
Here they are

90y + 49x= $1746 62y+43x =$1286

If you could show me how to either eliminate one variable, or reduce one variable to a coefficient of 1, I'd be able to take the rest from there.

Thankyou.

2. ## Re: Simultaneous Equation

Originally Posted by Tren301
Hi there, I've got a simultaneous equation that I am not sure of how to solve. I have looked at substitution, however I don't know how to reduce one variable to a single without making the rest of the equation look strange. Basically I don't see how to either eliminate one variable, or get one variable to a coefficient of 1.
Here they are

90y + 49x= $1746 62y+43x =$1286

If you could show me how to either eliminate one variable, or reduce one variable to a coefficient of 1, I'd be able to take the rest from there.

Thankyou.
To eliminate one of the variables, they need to have the same coefficients. So what is the Lowest Common Multiple of 90 and 62 (if you want to eliminate the y's), or the Lowest Common Multiple of 49 and 43 (if you want to eliminate the x's)?

3. ## Re: Simultaneous Equation

Originally Posted by Tren301
90y + 49x= $1746 [1] 62y+43x =$1286 [2]
Divide [1] by 90 and [2] by 62:
y + 49x/90 = 1746/90 [1]
y + 43x/62 = 1286/62 [2]
Subtract 'em to get:
208x/1395 = 208/155
Solve for x; then for y : OK?


text:
# Thread: induction inequality

1. ## induction inequality

I have to prove that $2^n < n! for all n>4$ using induction.

So,um I started like this,

$p(n) : 2^n < n !$
so $p(n+1) : 2^(n+1) < (n+1)!$
=> $2^n * 2 < (n+1) * n!$
=> $2^n < (n+1)/2 * n!
$

so um idk what to do next.How do I prove that this prove holds for all n>4

2. Originally Posted by NidhiS

I have to prove that $2^n < n! for all n>4$ using induction.

So,um I started like this,

$p(n) : 2^n < n !$
so $p(n+1) : 2^(n+1) < (n+1)!$
=> $2^n * 2 < (n+1) * n!$
=> $2^n < (n+1)/2 * n!
$

so um idk what to do next.How do I prove that this prove holds for all n>4
Compare the factors at both sides:

$2^n * 2 < (n+1) * n!~\implies~\underbrace{2^n3}$


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# Romberg Integration

Romberg Integration is used to approximate $$\int_2^3 \! f(x) \, \mathrm{d} x$$ If $f(2)=.51342, f(3)=.36788, R_{3,1}=.43687 , R_{3,3}=.43662$ Find $f(2.5)$

I am quite confused how to proceed since I don't even know what the function is. The only peice of information that may be helpful is $$R_{1,1}= { h \over 2}(f(2)+f(3))={1 \over 2}(.51342+.36788)=.44065$$
Maybe putting this into a piece-wise function may be helpful

-
There must probably be one of the $R(n,0)$ of the form $$\frac h2 \left( f(2) + f(2.5) + f(3) \right)$$ To be honest I didn't like Romberg integration when I learned it and I don't feel like diving back into this ugly recurrence relation... –  Patrick Da Silva Oct 14 '12 at 3:41
You really want to try to backtrack the recurrence relation to find a linear combination of $f(2)$, $f(2.5)$ and $f(3)$ using $R_{3,3}$ and $R_{3,1}$. I'm going to try that now. –  Patrick Da Silva Oct 14 '12 at 4:26
The problem seems to be similar to the one, which is solved in this exam sheet. –  Martin Sleziak Oct 14 '12 at 6:15
@math101: If you've found the solution, you could write it up as an answer for posterity and accept it so the question doesn't remain unanswered. –  joriki Oct 14 '12 at 7:00
@joriki I have a few questions that have not been answered. When I have the time I will answer them so that others can have the answer –  math101 Oct 14 '12 at 14:32

$R_{1,1}$
$R_{2,1},R_{2,2}$
$R_{3,1}, R_{3,2}, R_{3,3}$
We need to use the recursive formulas. The first formula is simple and easy: $$R_{1,1}={h \over 2}( f(a)+f(b))={1 \over 2}(.51342+.36788)=.44065$$ $$R_{3,3}={16R_{3,2}-R_{2,2} \over 15}=.43662$$ $$R_{3,2}={4R_{3,1}-R_{2,1} \over 3}={4(.43687)-R_{2,1} \over 3}=.58249-{ 1 \over 3}R_{2,1}$$ $$R_{2,2}={4R_{2,1}-R_{1,1} \over 3}={4r_{2,1}-.44065\over 3}={4 \over 3}R_{2,1}-.14688$$
Now we have to subsitute $R_{3,2}$ and $R_{2,2}$ into $R_{3,3}$ $$R_{3,3}={1 \over 15}(16(.58249-{1 \over 3}R_{2,1})-({4 \over 3}R_{2,1}-.14688)=.43662$$ Now that we only have one variable its simple to solve and $R_{2,1}=.43761$ Now we can solve $R_{2,1}$ $$R_{2,1}={ 1\over 2}R_{1,1}+h_1f(a+h_2)={ 1\over 2}R_{1,1}+h_1f(2.5)$$ $$f(2.5)={2R_{2,1}-R_{1,1} \over 2h}={2(.43761)-(.44065) \over 2(.5)}=.43457$$ So we can conclude that $f(2.5)=.43457$


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# Propagation of error using a range of values

The power law function for gamma-ray burst is defined for amplitude ($$A$$), energy ($$E$$), pivot energy ($$E_{piv}$$) (fixed) and index $$\lambda$$ as

$$f(E; A, \lambda) = N(E) = A*\left(\dfrac{E}{E_{piv}}\right)^\lambda$$

Where we use a range of energy for model fitting of GRB. The fitting gives us a single value for all variables except for energy (which is the input), and error propagation for the said formula can be calculated by partial differentiating it with respect to it's variables. What I'm confused at is that what value will I give to $$E$$ when I'll evaluate the error expression

$$err_{pl} = \displaystyle\sqrt(\left(\frac{E}{E_{piv}}\right)^{2\lambda}s_A^2 + \frac{A^2\lambda^2\left(E/E_{piv}\right)^{2\lambda-2}s_E^2}{E_{piv}^2} + A^2\left(\frac{E}{E_{piv}}\right)^{2\lambda}\log\left(\frac{E}{E_{piv}}\right)^2s_{\lambda}^2)$$

• Have you tried plotting the error versus E? – Ed V Jul 8 at 13:34
• okay, If I got your point correct yes I can plot errors versus E as an envelope over the fitted model since the error is symmetrical so it doesn't change over the entire energy range – Syed Ali Jul 8 at 14:24
• Just to make sure I understand the situation: when you do the nonlinear curve fitting, of your real data to your model (top equation), I assume you know and enter the pivot energy and hold it constant in the fit. The fit returns estimates of A and lambda and their standard errors. Then your error equation gives you the error as a function of E. If the error was independent of E, then you could use any reasonable E. If my understanding is correct, no need to respond! Contiinued in next comment. – Ed V Jul 8 at 16:54
• You may find these papers helpful: "A simple, all-purpose nonlinear algorithm for univariate calibration," by J. Tellinghuisen, Analyst 125, 1045-1048 (2000), and "Statistical Error Propagation," by J. Tellinghuisen, J. Phys. Chem. A 105, 3917-3921 (2001). – Ed V Jul 8 at 16:56


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# woo.fem¶

## Tetra¶

class woo.fem.Tetra

Tetra (triangle in 3d) particle.

canonicalizeVertexOrder(self: woo.fem.Tetra) → None

Order vertices so that signed volume is positive.

getCentroid(self: woo.fem.Tetra) → minieigen11.Vector3

Return centroid of the tetrahedron

getVolume(self: woo.fem.Tetra) → float

Return volume of the tetrahedron.

static make(vertices, fixed=True, wire=True, color=None, highlight=False, mat=<function defaultMaterial>, visible=True, mask=3, __class=<class 'woo.fem.Tetra'>)

Create tetrahedral particle

### Tet4¶

class woo.fem.Tet4

4-node linear interpolation tetrahedron element with best-fit co-rotated coordinates.

node(= None)

Local coordinate system

[type: None, read-only in python]

refPos(= MatrixX())

Reference nodal positions in local coordinates

[type: None, read-only in python]

uXyz(= VectorX())

Nodal displacements in local coordinates

[type: None, read-only in python]

KK(= MatrixX())

Stiffness matrix

[type: None, not shown in the UI, read-only in python]

EB(= MatrixX())

$$E B$$ matrix, used to compute stresses from displacements.

[type: None, not shown in the UI, read-only in python]

ensureStiffnessMatrix(self: woo.fem.Tet4, young: float, nu: float) → None

Ensure that stiffness matrix is initialized; internally also sets reference configuration. The young parameter should match woo.dem.ElastMat.young attached to the particle.

getStressTensor(self: woo.fem.Tet4) → minieigen11.Matrix3
reset(self: woo.fem.Tet4) → None
setRefConf(self: woo.fem.Tet4) → None

Set the current configuration as the reference one

update(self: woo.fem.Tet4) → None

Update current configuration; creates reference configuration if not existing

## Membrane¶

class woo.fem.Membrane

Facet as triangular element, with 2 translational and 2 rotational degrees of freedom in each node.

Since positions of nodes determine the element’s plane, the $$z$$ degrees of freedom have zero displacements and are condensed away from the KKdkt matrix (force reaction, however, is nonzero in that direction, so the matrix is not square).

Drilling rotations can be computed, but are ignored; this can lead to instability in some cases – wobbly rotation of nodes which does not decrease due to non-viscous damping.

The element is assumed to be under plane-stress conditions.

Mass of the element is lumped in to nodes, but this is not automatized in any way; it is your responsibility to assign proper values of DemData.mass and DemData.inertia.

node(= <Node @ 0x2ce57d0, at (0, 0, 0)>)

Local coordinate system

[type: None, read-only in python]

refRot(= [])

Rotation applied to nodes to obtain the local coordinate system, computed in the reference configuration. If this array is empty, it means that reference configuration has not yet been evaluated.

[type: None, read-only in python]

refPos(= Vector6(0, 0, 0, 0, 0, 0))

Nodal coordinates in the local coordinate system, in the reference configuration

[type: None, read-only in python]

uXy(= Vector6(0, 0, 0, 0, 0, 0))

Nodal displacements, stored as ux0, uy0, ux1, uy1, ux1, uy2.

[type: None, read-only in python]

surfLoad(= 0.0)

Normal load applied to this facet (positive in the direction of the local normal); this value is multiplied by the current facet’s area and equally distributed to nodes.

[type: None, unit: Pa]

phiXy(= Vector6(0, 0, 0, 0, 0, 0))

Nodal rotations, only including in-plane rotations (drilling DOF not yet implemented)

[type: None, read-only in python]

KKcst(= MatrixX())

Stiffness matrix of the element (assembled from the reference configuration when needed for the first time)

[type: None]

KKdkt(= MatrixX())

Bending stiffness matrix of the element (assembled from the reference configuration when needed for the first time).

[type: None]

enableStress(= False)

Set to evaluate EBcst and DBdkt when stiffness matricess are being computed. After than, using sigCST and sigDKT will return stresses.

[type: None]

EBcst(= MatrixX())

CST displacement-stress matrix, for computation of stress tensor (see stressCst).

[type: None, read-only in python]

DBdkt(= MatrixX())

DKT displacement-stress matrix, for computation of stress tensor (see stressDkt.

Warning

This matrix is not computed correctly, therefore also stressDkt returns garbage.

[type: None, read-only in python]

noWarnExcessRot(= False)

Set to disable warning about excessive in-plane rotation. Only do this if you know what you’re doing.

[type: None]

reset(self: woo.fem.Membrane) → None

Reset reference configuration; this forces using the current config as reference when update is called again.

setRefConf(self: woo.fem.Membrane) → None

Set the current configuration as the reference one.

stressCst(self: woo.fem.Membrane, glob: bool=False) → object

Return CST stresses (product of EBcst and uXy), provided that EBcst was computed previously by setting enableStress when building stiffness matrices. The value returned is either $$(\sigma_x,\sigma_y,\sigma_{xy})$$ (local stresses), or Matrix3 representing stress tensor in global coordinates (with glob=True).

stressDkt(self: woo.fem.Membrane) → minieigen11.Vector6

Return Vector6 of DKT stresses (product of DBdkt and phiXy), see stressCst for conditions; additionaly, bending must have been enabled.

Warning

This function returns nonsense currently and must be fixed!

update(self: woo.fem.Membrane, dt: float, rotIncr: bool=False) → None

Update current configuration; create reference configuration if it does not exist.

## Bo1_Tetra_Aabb¶

class woo.fem.Bo1_Tetra_Aabb

Creates/updates an Aabb of a Tetra.

## In2_Tet4_ElastMat¶

class woo.fem.In2_Tet4_ElastMat

Apply contact forces and compute internal response of a Tet4.

contacts(= False)

Apply contact forces to Tetra nodes (not yet implemented)

[type: None, read-only in python]

nu(= 0.25)

Poisson’s ratio used for assembling the $$E$$ matrix (Young’s modulus is taken from ElastMat). Will be moved to the material class at some point.

[type: None]

## In2_Membrane_ElastMat¶

class woo.fem.In2_Membrane_ElastMat

Apply contact forces and compute internal response of a Membrane. Forces are distributed according to barycentric coordinates when bending is enabled; otherwise forces are distributed equally (thirds) to all nodes, to avoid contacts punching through the mesh which has no bending resistance. This can be overridden by setting applyBary, in which case forces will be always applied weighted by barycentric coords.

Note

If your particles are made of ~:obj:woo.dem.FrictMat, use In2_Membrane_FrictMat instead, if you run into ambiguous dipatch errors.

contacts(= True)

Apply contact forces to facet’s nodes (FIXME: very simply distributed in thirds now)

[type: None]

nu(= 0.25)

Poisson’s ratio used for assembling the $$E$$ matrix (Young’s modulus is taken from ElastMat). Will be moved to the material class at some point.

[type: None]

thickness(= nan)

Thickness for CST stiffness computation; if NaN, try to use the double of Facet.halfThick.

[type: None]

bendThickness(= nan)

Thickness for DKT stiffness computation; if NaN, use thickness.

[type: None]

bending(= False)

Consider also bending stiffness of elements (DKT)

[type: None]

applyBary(= False)

Distribute force according to barycentric coordinate of the contact point; this is done normally with bending enabled, this forces the same also for particles without bending.

[type: None]

rotIncr(= False)

Compute nodal rotation incrementally (by integration of angular velocities) rather than by subtracting from reference rotations (the advantage of incremental is that it is numerically stable even for huge rotations, but perhaps less precise).

[type: None]

### In2_Membrane_FrictMat¶

class woo.fem.In2_Membrane_FrictMat

Workaround for current dispatching mechanism limitations so that membrane with woo.dem.FrictMat is not matched by woo.dem.In2_Facet.

Tip

Ask questions or report issues at github.


text:
• Create Account

Banner advertising on our site currently available from just $5! ### #ActualOrymus3 Posted 05 July 2013 - 01:30 PM So tying this all together, would this persistent world of sorts bother you? Let's say it is not forced, and customers can be generated in your own little world, but if you 'connect' you have access to not only viewing other people's creations and work, but being a part of a 'living and breathing' world of customers who simply want to have a good time and have different needs. Would there be a geographical element to the game? (travelling, etc.) In this vein, I would imagine that two competing similar businesses could exist if only because a portion of the users wouldn't bother going the distance to the "best" of these places and settle for the closest one... this is a decision a lot of us do on a daily basis Is this a problem of perhaps I'm worried players will become bored waiting for games to end? Do you think this can be avoided by giving the players enough to do other than simply working on the courts, then watching players for ever and ever play on them? What if they have a 'perfect' business? Would you simply watch the games and earn money so that you could spend it on, well, nothing besides perhaps some internet fame (or building another 'location') for procuring 10000 professional players in your corporation's life time? We still watch football don't we? Besides, I've seen fully asynchronous games where you get to play "a turn" once per week, and they still do good. If your concept is sound, and if you have a subset of users in mind, some will survive these limitations for sure... If you can make the outcome of matches played uncertain, keep a level of competition, and make it very hard to break ahead, games will remain fun to watch, and will provide entertainment while money comes in. For the rest of the post, I'm affraid that you've been very unclear about the exact nature of your concept, perhaps because even some of it remains unclear to you. I don't come out of this lenghty reading with a precise understanding of what you're trying to achieve, which makes it hard for me to comment efficiently on your questions... My appreciation is that you're currently having "coldfeet" and not making the decisions you should be making. Merging two gameplays? sure. Scare of losing playerbase as a result? there's actually no avoiding this, and if you try to please everyone, your game will suck for everyone. Decide on a persona, a subset of users, and stick to that plan, rather than try to pick up on large ensembles such as casual and simulation players. Pick "Jonnhy", the guy that works at a software company 50h/week, two kids, a wife, has a 10 years old car parked in the garage, and earn approx 40k$/year. This guys doesn't have much time, but he wants to make the most out of it, but he wants to appreciate his time, so he's probably not going to try min/max your game. If you provide him with a clever experience with some self-expression, he might just as well jump in.

Why this persona? (you could definitely build another). Over the last couple of years, we've focused so much on Soccer-mums, that I don't think soccer-mums have anymore time left to play games. They're so busy on Farmville they can't even raise their kids. Might as well focus on somewhat busy adults with limited gaming time and build an experience suited for them. These guys are having a rough time following up with the amount of time it takes to play games like WoW, and always being far behind.

### #1Orymus3

Posted 05 July 2013 - 01:25 PM

So tying this all together, would this persistent world of sorts bother you? Let's say it is not forced, and customers can be generated in your own little world, but if you 'connect' you have access to not only viewing other people's creations and work, but being a part of a 'living and breathing' world of customers who simply want to have a good time and have different needs.

Would there be a geographical element to the game? (travelling, etc.)

In this vein, I would imagine that two competing similar businesses could exist if only because a portion of the users wouldn't bother going the distance to the "best" of these places and settle for the closest one... this is a decision a lot of us do on a daily basis

Is this a problem of perhaps I'm worried players will become bored waiting for games to end? Do you think this can be avoided by giving the players enough to do other than simply working on the courts, then watching players for ever and ever play on them? What if they have a 'perfect' business? Would you simply watch the games and earn money so that you could spend it on, well, nothing besides perhaps some internet fame (or building another 'location') for procuring 10000 professional players in your corporation's life time?

We still watch football don't we? Besides, I've seen fully asynchronous games where you get to play "a turn" once per week, and they still do good.

If your concept is sound, and if you have a subset of users in mind, some will survive these limitations for sure...

If you can make the outcome of matches played uncertain, keep a level of competition, and make it very hard to break ahead, games will remain fun to watch, and will provide entertainment while money comes in.

For the rest of the post, I'm affraid that you've been very unclear about the exact nature of your concept, perhaps because even some of it remains unclear to you. I don't come out of this lenghty reading with a precise understanding of what you're trying to achieve, which makes it hard for me to comment efficiently on your questions...

PARTNERS


text:
## Solving an extension of the 2D wave equation to 2n dimensions

The 2D wave equation $$left(frac{partial^2}{partial^2 t} – frac{partial^2}{partial^2 z}right)f(t,z) = 0$$ can be written after changing variables to $$x = t – z$$, $$y = t+ z$$ as $$frac{partial}{partial x}frac{partial}{partial y}f(x,y) = 0,$$ with general solution $$f(x,y) = g(x) + h(y),$$ where $$g$$ and $$h$$ are arbitrary functions.

I want to solve a similar (looking) PDE for $$(x,y) in mathbb{R}^{n}timesmathbb{R}^{n},$$ which is given by $$frac{partial}{partial x}cdotfrac{partial}{partial y}f(x,y) = 0,$$ where the dot product is with the Euclidean metric.

Is it possible to find a general solution to this equation in terms of some arbitrary functions?
I asked Mathematica but unfortunately it didn’t know how to solve this one. It can solve the 2D version fine.

## linear algebra – Inequality for dimensions of eigenspaces

I want to proof that for an endomorphism $$f$$ over vector space $$V$$ with eigenvalue $$lambda$$ and
$$V = U oplus V/U$$ with $$f(U) subseteq U$$ the following holds:
$$dim E(f,lambda) leq dim E(f_U,lambda) + dim E(f_{V/U}, lambda)$$
I managed to proof that the inequality holds for the algebraic multiplicity of $$lambda$$ via splitting the characteristic polynomial of $$f$$. However I can’t find a way to connect that to the geometric multiplicity. I would appreciate any hint.

## calculus and analysis – Problems with improper integrals in higher dimensions

Calculating numerically certain energy integrals in three and four dimensions related to
the Riesz potential
and the capacity, I try

``````NIntegrate( 1/Sqrt((x - p)^2 + (y - q)^2 + (z - r)^2), {x, y, z} (Element)
Ball({0, 0, 0}, 1), {p, q, r} (Element) Ball({0, 0, 0}, 1),
AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 25,
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0})
``````

`0`

and a warning message
“NIntegrate::moptxn: The option SymbolicProcessing of the method FiniteElement is not one of {Method,MeshOptions}.”
As far as I understand it, this means that the `NIntegrate` command does not accept the set of the integration
in the form `{x, y, z} (Element) Ball({0, 0, 0}, 1), {p, q, r} (Element) Ball({0, 0, 0}, 1)`. However, if it is so,
the input, not `0`, should be returned.

My next try is

`````` NIntegrate( 1/((x - p)^2 + (y - q)^2 + (z - r)^2)^(1/2), {x, -1, 1}, {y, -Sqrt(1 - x^2), Sqrt(1 - x^2)},
{z, -Sqrt(1 - x^2 - y^2), Sqrt(1 - x^2 - y^2)}, {p, -1, 1}, {q, -Sqrt(1 - p^2), Sqrt(1 - p^2)},
{r, -Sqrt(1 - p^2 - q^2), Sqrt(1 - p^2 - q^2)}, AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 25,
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0})
``````

`-1244.482640558337558417913`

and a warning
“NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 54 recursive bisections
in z near {x,y,z,p,q,r} = <<1>>. NIntegrate obtained -1244.482640558337558417913
and 3534.518334552443660338233`25. for the integral and error estimates.”
A similar issue with

``````NIntegrate(1/((x - p)^2 + (y - q)^2 + (z - r)^2)^(1/2)*
Boole(x^2 + y^2 + z^2 <= 1 && p^2 + q^2 + r^2 <= 1), {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, {p, -1, 1},
{q, -1, 1}, {r, -1, 1}, AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 25,
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0})
``````

`-1203.034524853306755966531`

I wonder negative numbers since the integrand is positive.
Then i switch to MonteCarlo methods.

``````NIntegrate( 1/((x - p)^2 + (y - q)^2 + (z - r)^2)^(1/2)*
Boole(x^2 + y^2 + z^2 <= 1 && p^2 + q^2 + r^2 <= 1), {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, {p, -1, 1},
{q, -1, 1}, {r, -1, 1}, AccuracyGoal -> 2, PrecisionGoal -> 2, WorkingPrecision -> 25,
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0},Method -> "QuasiMonteCarlo")
``````

`21.12327556039856680489716`

and a warning “NIntegrate::maxp: The integral failed to converge after 50000 integrand evaluations.
NIntegrate obtained 21.12327556039856680489716`25. and 0.2251832352724191939747851`25. for the integral and error estimates.”
A more or less reliable result is obtained by

``````NIntegrate( 1/((x - p)^2 + (y - q)^2 + (z - r)^2)^(1/2)*
Boole(x^2 + y^2 + z^2 <= 1 && p^2 + q^2 + r^2 <= 1), {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, {p, -1, 1},
{q, -1, 1}, {r, -1, 1},  AccuracyGoal -> 2, PrecisionGoal -> 2, WorkingPrecision -> 25,
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0}, Method -> "AdaptiveMonteCarlo")
``````

`20.32729987338035891791629`

without any warning. Unfortunately, ` AccuracyGoal -> 3, PrecisionGoal -> 3` is not achieved.
Also this works in eight dimensions:

``````NIntegrate(1/((x - p)^2 + (y - q)^2 + (z - r)^2 + (w - s)^2)^((4 - 2)/2)*
Boole(x^2 + y^2 + z^2 + w^2 <= 1 &&  p^2 + q^2 + r^2 + s^2 <= 1), {x, -1, 1}, {y, -1, 1}, {z,-1, 1},
{w, -1, 1}, {p, -1, 1}, {q, -1, 1}, {r, -1, 1}, {s, -1, 1}, AccuracyGoal -> 2, PrecisionGoal -> 2,
WorkingPrecision -> 25,Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 + (w - s)^2 == 0},
``````

`25.78510573458365881830859`

BTW, the `Exclusions` option works in the above: compare with `24.68762920929857902438239` witout it.

The questions arise: what are other methods to calculate such integrals?
is a three-digit result possible with MonteCarlo methods?

## In what dimensions can I print a photo shot on Canon EOS 2000D

In what dimensions can I print a photo shot on Canon EOS 2000D at full resolution and save the photo as JPEG in a 3:2 aspect ratio? Let’s assume I want to print it with 600 dpi.

## parameters – Different FourierParameters for different dimensions of discrete Fourier transform

This question asks for a way to implement different FourierParameters for different dimensions of a discrete Fourier transform. The question received no answers, but a workaround was mentioned in the comments for the particular case considered by the OP.

I am faced with essentially the same problem, but in my case I would like to use non-integer values for FourierParameters, for which the aforementioned workaround doesn’t help. Given that several years have passed since the linked question was posed, I wonder if there might now be a better solution to the problem of assigning different FourierParameters to different dimensions of a Fourier transform?

## How to add missing dimensions to all images to avoid cls?

My sites have lots of external images without dimensions, and that’s causing layout shift.
I need a code which automatically add missing width and height to all images.
I tried 1 plugin Specify Missing Image Dimensions but site become too slow after activating this plugin.
Thanks

## Can I use Google Analytics 4 custom dimensions in the Analysis Hub?

In the Analysis Hub:

Can I use a custom dimension I’ve previously added?

For example:

I’ll send the following events and I would like to do analysis based on one of its parameters:

``````gtag("event", "my_custom_event", {
my_custom_dimension: "SOME_VALUE"
});
``````

## complexity – Finding closest point in N dimensions in reasonable time (O(log(n) ?)

Is it possible to find the closest point (or k closest points) to any arbitrary point in a set of n points (in dimension N)?

When I write closest point, I mean smallest Euclidean distance. I’m looking for an algorithm for multiple searches, where the preprocessing time for the n points is negligible?

Is it possible to get the closest point(s) in a reasonable time (O(log(n)) for example with a method similar to binary search).

Any reference is welcome.


text:
You may often find a table or figure is too big to fit into one column when your article has two columns. Use this to insert a figure (same with table) and it will save you:

\begin{figure*}
% figure here
\end{figure*}

Note the star (*) appended after figure? That's the trick.


text:
# Stellar Dynamical Models for 797 z ∼ 0.8 Galaxies from LEGA-C

Houdt, Josha van and Wel, Arjen van der and Bezanson, Rachel and Franx, Marijn and D'Eugenio, Francesco and Barisic, Ivana and Bell, Eric F. and Gallazzi, Anna and Graaff, Anna de and Maseda, Michael V. and Pacifici, Camilla and Sande, Jesse van de and Sobral, David and Straatman, Caroline and Wu, Po-Feng (2021) Stellar Dynamical Models for 797 z ∼ 0.8 Galaxies from LEGA-C. The Astrophysical Journal, 923 (1). ISSN 0004-637X

Text (2108.08142v1)
2108.08142v1.pdf - Accepted Version
Restricted to Repository staff only until 7 December 2022.

## Abstract

We present spatially resolved stellar kinematics for 797 $z=0.6-1$ galaxies selected from the LEGA-C survey and construct axisymmetric Jeans models to quantify their dynamical mass and degree of rotational support. The survey is $K_s$-band selected, irrespective of color or morphological type, and allows for a first assessment of the stellar dynamical structure of the general $L^*$ galaxy population at large lookback time. Using light profiles from Hubble Space Telescope imaging as a tracer, our approach corrects for observational effects (seeing convolution and slit geometry), and uses well-informed priors on inclination, anisotropy and a non-luminous mass component. Tabulated data include total mass estimates in a series of spherical apertures (1, 5, and 10 kpc; 1$\times$ and 2$\times$\re), as well as rotational velocities, velocity dispersions and anisotropy. We show that almost all star-forming galaxies and $\sim$50\% of quiescent galaxies are rotation-dominated, with deprojected $V/\sigma\sim1-2$. Revealing the complexity in galaxy evolution, we find that the most massive star-forming galaxies are among the most rotation-dominated, and the most massive quiescent galaxies among the least rotation-dominated galaxies. These measurements set a new benchmark for studying galaxy evolution, using stellar dynamical structure for galaxies at large lookback time. Together with the additional information on stellar population properties from the LEGA-C spectra, the dynamical mass and $V/\sigma$ measurements presented here create new avenues for studying galaxy evolution at large lookback time.

Item Type:
Journal Article
Journal or Publication Title:
The Astrophysical Journal
This is an author-created, un-copyedited version of an article accepted for publication/published in The Astrophysical Journal. IOP Publishing Ltd is not responsible for any errors or omissions in this version of the manuscript or any version derived from it. The Version of Record is available online at doi:10.3847/1538-4357/ac1f29
Uncontrolled Keywords:
/dk/atira/pure/subjectarea/asjc/1900/1912
Subjects:
Departments:
ID Code:
161951
Deposited By:
Deposited On:
09 Nov 2021 12:10
Refereed?:
Yes
Published?:
Published


text:
# Why is eigendecomposition $V \Lambda V^{-1}$ not $V^{-1} \Lambda V$

Say you have a linear transformation matrix $A$. In the basis of eigenvectors, this transformation simply becomes a scaling, represented by the diagonal matrix of eigenvalues.

Thus, intuitively the transformation A can be decomposed into the following:

1. Transform into the basis of eigenvectors (using the transformation matrix $V$, where the eigenvectors form the columns)
2. Apply the scaling.
3. Transform back.

This would seem to correspond to $V^{-1} \Lambda V$, where the standard notation of matrices being applied on the left and vectors on the right holds.

Yet everywhere I always see the formula as $V \Lambda V^{-1}$. Why isn't my intuition correct?

• You intuition would be correct if you calculate the left eigenvector. People tend to like the right eigenvectors more, so that is the "convention". By the way, the equation as it stands is incomplete as you should explain what $\Lambda$ and $V$ are in terms of the eigenvalues and eigenvectors. – Fabian May 9 '17 at 4:51
• $Y=V^{-1}$. Now how does it change. what looks more pleasing to you? – dineshdileep May 9 '17 at 4:51
• @Fabian What is meant by right vs. left eigenvalues? In the equation $A \vec{v} = \lambda \vec{v}$ would $\lambda$ be a right or left eigenvalue? – 1110101001 May 9 '17 at 4:53
• That looks like a right eigenvector to me. The left eigenvector is $\vec{w} A = \lambda \vec{w}$ – Fabian May 9 '17 at 4:55
• @Fabian Hm alright after reading up a bit it seems a left eigenvector would then be $\vec{v} A = \lambda \vec{v}$. But I'm still missing some key intuition here. The right eigenvector reads naturally as applying the linear transformation A to $\vec{v}$ and finding that the result is scaled by $\lambda$. Yet for some reason this "natural" interpretation seems to go awry when following the line of reasoning in the opening post. How would you even interpret $\vec{v} A$ — applying a vector on a matrix?? – 1110101001 May 9 '17 at 5:00

It all depends on what you define your coordinate transformation matrix $$V$$ to be; obviously if you replace it by the inverse matrix (which carries the same information) then the two possible formulae for the diagonalisation are interchanged. Now typically people take $$V$$ to be matrix whose columns contain the coordinates of a chosen basis of eigenvectors, the coordinates being expressed of course in terms of the basis for which the matrix $$A$$ was originally expressed. And it is a sad fact of life that multiplying by that matrix will perform the coordinate transformation in the opposite sense, in other words convert a vector expressed in coordinates on the basis of eigenvectors to its expression in the original basis. Think of it: if you apply $$V$$ to a standard basis vector, the result is a column of $$V$$, and therefore will express an eigenvector (one whose coordinates with respect to the eigenvector basis are given by that standard basis vector) in coordinates with respect to the original basis.

The formula was generated from the equation $AV=V\Lambda$ which is a compact way of presentation a set of formulas $Av_i=\lambda_i{v_i}$ for eigenvectors.
In this case matrix $\Lambda$ is scaling column vectors ${v_i}$ grouped in the matrix $V=[v_1 \ \ v_2 \ \dots \ \ v_n]$.


text:
# Frequency response using fourier

i need to find the frequency response of the following equations:

$$h(t) = e^{-3t} u(t)$$ $$x(t) = 1+\cos\left(\frac{4\pi}{3}t\right)$$

find $y(t)$

However i am quite confused on how to do this, please explain and show every step so i can learn

• Is this homework? – Phonon Nov 20 '13 at 4:01
• kinda, he asked us to solve it but we dont have to turn it in however i want to learn how to do it – user998316 Nov 20 '13 at 19:09
• Equations don't have frequency responses; systems do. You need to say what, if any, is the relationship between the desired output $y(t)$ and the givens $x(t)$ and $h(t)$. Please edit to add this information if your instructor gave it to you, and if he did not, ask your instructor for this information. -1 for now, pending edits. – Dilip Sarwate Dec 21 '13 at 19:54

## 2 Answers

You have to calculate $H(f)$, the Fourier transform of $h$. Similarly, you have to calculate $X(f)$ which is FT of $x$. Then, recall that, convolution in time domain is multiplication in frequency domain. Using this property, you will get $Y(f)$. Take inverse Fourier transform.

Since this is probably a HW, I will not give further details, the general way is like that. You can manage algebra.

Hint #1: look up the definition of signals and systems. "Calculating the frequency response" and "finding y(t)" which presumably is x convolved with h are two different things.

Hint #2: This looks like a typo. It's most likely $h(t)=e^{-3t}u(t)$


text:
# Incompatibility of Nexus Style Class and mhchem

I was very impressed with the nexus style class on the latex showcase (see here). Unfortunately when I tried to implement it, I needed to load the mhchem package and when doing so, the tabs at the bottom fail to render properly:

# When Loading mhchem package:

As you can see, not only is the page footer the wrong colour, it is also lacking the tabs.

Here is my not working code:

% !TeX program = lualatex
\documentclass[11pt,oneside]{book}

%%%%%%%%%%%%% Geometry
\usepackage[a4paper,left=2.5cm,right=2.5cm, bottom=2.5cm,top=2.5cm]{geometry}

%%%%%%%%%%%%%%% Les paquets

\usepackage[english]{babel}
\usepackage[palette=munch]{nexus}

%%%%%%%%%%%%%%%% hyperref
\usepackage{lipsum}
\usepackage[verbose]{hyperref}
\hypersetup{
}
\usepackage{mhchem} % <----- MHCHEM, COMMENT THIS LINE AND COMPILE TWICE TO SEE PROPER LAYOUT

\begin{document}

\include{titlepage}

\tableofcontents

\chapter{This is a Chapter}
\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-3]

\subsection{This is a subsection}
\lipsum[1-2]

\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-1]
\subsection{This is a subsection}
\lipsum[1-2]

\chapter{This is a Chapter}
\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-3]

\subsection{This is a subsection}
\lipsum[1-2]

\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-1]
\subsection{This is a subsection}
\lipsum[1-2]

%%%%%%%%%%%%%%%%%%%%%%%

\chapter{This is a Chapter}
\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-3]

\subsection{This is a subsection}
\lipsum[1-2]

\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-1]
\subsection{This is a subsection}
\lipsum[1-2]

\chapter{This is a Chapter}
\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-3]

\subsection{This is a subsection}
\lipsum[1-2]

\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-1]
\subsection{This is a subsection}
\lipsum[1-2]

%%%%%%
\begin{appendices}
\chapter{This is a Chapter}
\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-3]

\subsection{This is a subsection}
\lipsum[1-2]

\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-1]
\subsection{This is a subsection}
\lipsum[1-2]

\chapter{This is a Chapter}
\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-3]

\subsection{This is a subsection}
\lipsum[1-2]

\section{This is a section}
\subsection{This is a subsection}
\lipsum[1-1]
\subsection{This is a subsection}
\lipsum[1-2]

\end{appendices}

\end{document}

The relevant class files can be found from the creator's github repository available here. One will notice that the above code is simply the test.tex with \usepackage{mhchem} added in which causes the error.

# Things I have tried

I have tried following some suggestions from this answer and added \mhchemoptions{text-greek=default, math-greek=default} to the line after \usepackage{mhchem} but alas to no avail. It did not solve my problem. I am not sure what is causing this issue

your style loads assoccnt. mhchem loads calc. As the documentation of assoccnt says:


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# For a combinatorial proof of a symmetric identity

In my paper Supercongruences involving dual sequences [Finite Fields Appl. 46(2017), 179-216], I gave a new symmetric identity which states that if $x+y=-1$ then $$\sum_{k=0}^n(-1)^k\binom xk^2\binom{y+z}{n-k}=\sum_{k=0}^n(-1)^k\binom yk^2\binom{x+z}{n-k}.$$ (See (1.17) of the paper.)

QUESTION: Is there a combinatorial proof of the above symmetric identity?


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 Dal'nevost. Mat. Zh.: Year: Volume: Issue: Page: Find

 Dal'nevost. Mat. Zh., 2015, Volume 15, Number 2, Pages 133–155 (Mi dvmg305)

On regular systems of algebraic $p$-adic numbers of arbitrary degree in small cylinders

N. V. Budarinaa, F. Götzeb

a Institute for Applied Mathematics, Khabarovsk Division, Far-Eastern Branch of the Russian Academy of Sciences, 680000 Khabarovsk, Russia, Dzerzhinsky st., 54
b Faculty of Mathematics, University of Bielefeld, P. O. Box 10 01 31, 33501 Bielefeld, Germany

Abstract: In this paper we prove that for any sufficiently large $Q\in{\mathbb N}$ there exist cylinders $K\subset{\mathbb Q}_p$ with Haar measure $\mu(K)\le \frac{1}{2}Q^{-1}$ which do not contain algebraic $p$-adic numbers $\alpha$ of degree $\deg\alpha=n$ and height $H(\alpha)\le Q$. The main result establishes in any cylinder $K$, $\mu(K)>c_1Q^{-1}$, $c_1>c_0(n)$, the existence of at least $c_{3}Q^{n+1}\mu(K)$ algebraic $p$-adic numbers $\alpha\in K$ of degree $n$ and $H(\alpha)\le Q$.

Key words: integer polynomials, algebraic $p$-adic numbers, regular system, Haar measure

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UDC: 511.42
MSC: Primary 11K60; Secondary 11J61, 11J83
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Citation: N. V. Budarina, F. Götze, “On regular systems of algebraic $p$-adic numbers of arbitrary degree in small cylinders”, Dal'nevost. Mat. Zh., 15:2 (2015), 133–155

Citation in format AMSBIB
\Bibitem{BudGot15} \by N.~V.~Budarina, F.~G\"otze \paper On regular systems of algebraic $p$-adic numbers of arbitrary degree in small cylinders \jour Dal'nevost. Mat. Zh. \yr 2015 \vol 15 \issue 2 \pages 133--155 \mathnet{http://mi.mathnet.ru/dvmg305} \elib{http://elibrary.ru/item.asp?id=25058090}


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# scipy.special.expit¶

scipy.special.expit(x) = <ufunc 'expit'>

Expit ufunc for ndarrays.

The expit function, also known as the logistic function, is defined as expit(x) = 1/(1+exp(-x)). It is the inverse of the logit function.

Parameters: x : ndarray The ndarray to apply expit to element-wise. out : ndarray An ndarray of the same shape as x. Its entries are expit of the corresponding entry of x.

Notes

As a ufunc expit takes a number of optional keyword arguments. For more information see ufuncs

New in version 0.10.0.

#### Previous topic

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#### • Class 11 Physics Demo

Explore Related Concepts

# How to Simplify Negative Fractions

How to Simplify Negative Fractions?

In Mathematics, we know that whole numbers can be negative or positive. Similarly, fractions can also be positive or negative.

Negative Fractions
The negative fractions are handled in the same way as we handle negative whole numbers.
The rules for mathematical operations on negative whole numbers such as addition, subtraction, multiplication and division also apply to fractions.
The negative fractions can also be calculated using number line.

Points to be noted in a Negative Fraction

•    A negative sign in front of a fraction implies that either the numerator or the denominator is negative.
•    The fraction that has negative numbers in both the numerator and the denominator can be simplified.
After simplifying this negative fraction, the two minus signs will get cancelled and we get the same numbers as a positive fraction.

Simplifications of Negative Fractions

The negative fractions can be simplified using the mathematical operators. Let us see some methods of simplification of negative fractions using mathematical operations:

CASE 1:

1               5           4               2
-   __    +    __   =     __  =      __

6               6              6             3

In the above example, as the signs of the fractions are different, the two fractions are subtracted and the sign of the highest number is applied in the result.
As 5 is greater than 1 and the difference of 5 and 1 is 4, we get +4 divided by the denominator 6 as the result.

CASE 2:
If both the numbers to be added are negative, then the result will be the total of the two fractions preceded by a negative sign.

CASE 3:
In a mathematical problem where a negative fraction is to be subtracted from a positive fraction, minus of a negative fraction will become positive and so the two numbers will be added and the result will be a positive fraction being the sum of the two fractions given.

CASE 4:
In case of subtraction of two fractions in which the first fraction is a negative fraction and the second is a positive fraction, then the result will be the addition of two fractions preceded by a negative sign.

CASE 5:
In case of multiplication, product of one positive fraction and one negative fraction will result in a negative fraction. But the product of two negative fractions results in a positive fraction as “minus into minus is plus”.

CASE 6:
In case of division of fractions, if one fraction is positive and the other fraction is negative, then the resultant fraction will be negative. A negative fraction divided by another negative fraction results in a positive fraction.

An important point to be considered in simplifying negative fractions is that once the result is obtained, the numerator and the denominator should be reduced in its value by dividing it with a common factor, as we saw in CASE 1.
This should be repeated until there is no common factor between the numerator and the denominator.


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# duodecimal

The duodecimal system is a positional number system with base 12, using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A and B.

The duodecimal comes fro Latin.  There is also a less used synonym duodecadic mixed of a Latin begin and a Greek end and still a purely Greek dodecadic.

Some divisibility tests in duodecimal are:

$n$ is divisible by 2 if its least significant digit is 0, 2, 4, 6, 8 or A.

$n$ is divisible by 3 if its least significant digit is 0, 3, 6 or 9.

$n$ is divisible by 4 if its least significant digit is 0, 4 or 8.

$n$ is divisible by 6 if its least significant digit is 0 or 6.

$n$ is divisible by 11 if it has digital root B.

$n$ is of course divisible by 12 if it ends in a 0.

$n$ is divisible by 13 if the difference of the odd placed digits and the even place digits of $n$ is a multiple of 13.

Title duodecimal Duodecimal 2013-03-22 16:20:55 2013-03-22 16:20:55 PrimeFan (13766) PrimeFan (13766) 5 PrimeFan (13766) Definition msc 11A63


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# Euler's Formula

Euler's formula, the most remarkable formula in mathematics, according to Richard Feynman and many others, states that for any real number $$x$$,

$e^{ix} = \cos x + i \sin x$

In this section we will derive Euler's formula to get a better understanding of how Euler's formula can be used to describe rotations in space and time, eventually including the Lorentz transformation.

##### Skip if you prefer...
The derivation of Euler's formula may be a little complicated if you've not touched calculus in a while. Feel free to skip this section for now and go straight to the next section, coming back later as you need.

## Finding a function whose derivative is the function itself

Euler's formula comes about by first considering a function of $$x$$, for which the derivative of the function for any value of $$x$$ has the same value as function itself. That is:

$\frac{d}{dx}f(x) = f(x)$

If we start with a guess of $$f(x) = 1 + x$$ then we see it is not quite right as

$\frac{d}{dx}(1 + x) = 1 \ne (1 + x)$

but we can improve it by adding another term

$f(x) = 1 + x + \frac{x^2}{2}$

which is a little closer in that

$\frac{d}{dx}(1 + x + \frac{x^2}{2}) = 1 + x \ne f(x)$

but still not the same. So we improve it again with

$f(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{2\times 3}$

and so on, until we end up with an infinite sum of terms:

$f(x) = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum_{k=0}^{\infty} \frac{x^k}{k!}$

We now have a function satisfying $$\frac{d}{dx} f(x) = f(x)$$.

##### The derivative equals the function
Take a couple of minutes to see if you can show that the derivative of our function: $$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$ is itself.

Solution

We can work out the derivative of each term in the infinite sum of terms,

\begin{aligned} \frac{d}{dx} f(x) &= \frac{d}{dx} \sum_{k=0}^{\infty} \frac{x^k}{k!} \\ &= \sum_{k=0}^{\infty} \frac{kx^{k-1}}{k!} \\ \end{aligned}

so we can completely remove the term for $$k=0$$ (as zero times anything is still zero), and continue:

\begin{aligned} \frac{d}{dx} f(x) &= \sum_{k=1}^{\infty} k\frac{x^{k-1}}{k!} \\ &= \sum_{k=1}^{\infty} \frac{x^{k-1}}{(k - 1)!} \\ &= \sum_{k=0}^{\infty} \frac{x^k}{k!} \\ &= f(x) \end{aligned}

Done!

Great, but where does $$e$$ fit in?

## Understanding what an exponent is

We learn in high school that $$3^2 \times 3 = 3^3$$ and $$2^2 \times 2 \times 2 = 2^4$$ and later that $$x^2 \times x^2 = x^4$$ etc., but the general definition of exponentiation is that when multiplying two numbers with the same base ($$3^2 \times 3^1$$), the powers (or exponents) are added ($$3^{2+1} = 3^3$$):

$b^m \times b^n = b^{m + n}$

## Our function is an exponential

If we take our function which satisfies the property $$\frac{d}{dx}f(x) = f(x)$$ we can show that it behaves just like an exponent should, in that:

$$f(m) \times f(n) = f(m+n)$$

To show this requires quite a bit of re-arranging:

\begin{aligned} f(m) \times f(n) &= \sum_{k=0}^{\infty} \frac{m^k}{k!} \times \sum_{k=0}^{\infty} \frac{n^k}{k!} \\ &= (\frac{m^0}{0!} + \frac{m^1}{1!} + \frac{m^2}{2!} + \frac{m^3}{3!} + ...) \times (\frac{n^0}{0!} + \frac{n^1}{1!} + \frac{n^2}{2!} + \frac{n^3}{3!} + ...) \\ &= 1 + (\frac{m^1}{1!} + \frac{n^1}{1!}) + (\frac{m^2}{2!} + \frac{m^1 n^1}{1! 1!} + \frac{n^2}{2!}) + (\frac{m^3}{3!} + \frac{m^2 n^1}{2! 1!} + \frac{m^1 n^2}{1! 2!} + \frac{n^3}{3!}) ... \\ &= 1 + (\frac{m^1}{1!} + \frac{n^1}{1!}) + (\frac{m^2}{2!} + \frac{2}{2}\frac{m^1 n^1}{1! 1!} + \frac{n^2}{2!}) + (\frac{m^3}{3!} + \frac{3}{3}\frac{m^2 n^1}{2! 1!} + \frac{3}{3}\frac{m^1 n^2}{1! 2!} + \frac{n^3}{3!}) ... \\ &= \sum_{k=0}^{\infty} \frac{(m + n)^k}{k!} \\ &= f(m + n) \end{aligned}

Also similar to an exponent, we can show for our function $$f(x)$$, that $$f(0) = 1$$.

##### Identity of f(0)
Show that $$f(0) = 1$$ for our function: $$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$

Solution:

Substituting $$x=0$$,

\begin{aligned} f(0) &= \sum_{k=0}^{\infty} \frac{0^k}{k!} \\ &= \frac{0^0}{0!} + \frac{0^1}{1!} + \frac{0^2}{2!} + ... \\ &= 1 + 0 + 0 + ... \\ &= 1 \end{aligned}

Done!

Given that our function $$f(x)$$ behaves just like an exponent should behave, it must be equivalent to some number to the power of $$x$$. We can find that number by working out $$f(1)$$.

##### Find f(1)
Show that $$f(1) \approx 2.717$$

Solution:

Evaluating,

\begin{aligned} f(1) &= \sum_{k=0}^{\infty} \frac{1!}{k!} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \\ &= \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... \\ &= 1 + 1 + 0.5 + 0.1667 + 0.0417 + 0.0083... \\ &= 2.717 \end{aligned}

The exact value (if we summed to infinity) is assigned a special name in maths, e.

So,

$e^1 = f(1) = \sum_{k=0}^{\infty} \frac{1}{k!} \approx 2.71828...$

Now any value of our function can be evaluated simply as a power of $$e$$:

$f(3) = e^3 = 2.71828... ^3 = 20.08534...$

## Derivatives of exponentials

What if we need to work with other exponents of $$e$$, such as $$e^{3x}$$? How do we work out the derivative - how fast $$e^{3x}$$ is changing as $$x$$ changes? Your memory may (or may not) tell you that:

$\frac{d}{dx} e^{3x} = 3e^{3x}$

or more generally, if $$g$$ is another function of $$x$$, that

$\frac{d}{dx} e^{g} = \frac{dg}{dx}e^{g}$

This is all we need for our tooling here, though we can also see why this is the case given the chain rule for derivatives which says that:

$\frac{d}{dx}f(g) = \frac{dg}{dx} \times \frac{d}{d g}f(g)$

together with the special property of our function $$f(x) = e^x$$ that the derivative at any point is equal to the function itself, $$\frac{d}{dx}f(x) = f(x)$$.

## Euler's formula

Now we have enough background to appreciate the beauty of Euler's formula. Euler, like Roger Cotes before him, noticed that if he evaluated this exponent function with a special type of value, an "imaginary" value whose square is negative, the result is a combination of the trigonometric functions $$\cos$$ and $$sin$$. That is, if we define $$i$$ such that $$i^2 = -1$$, and evaluate $$e^{ix}$$, it conveniently splits into two sums, one of the odd powers and one of the even powers:

\begin{aligned} e^{ix} &= \sum_{k=0}^{\infty} \frac{(ix)^k}{k!} \\ &= \frac{(ix)^0}{0!} + \frac{(ix)^1}{1!} + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} ... \\ &= (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} ...) + i(\frac{x^1}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} ...) \\ &= \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{2k!} + i \sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!} \\ &= \cos(x) + i\sin(x) \end{aligned}

That is, the two separate summations are the actual definitions of $$\cos(x)$$ and $$\sin(x)$$.

Notice that, with our introduction to Geometric Algebra, we did not need to invent an imaginary unit $$i$$ here, as we have already seen that the product of basis vectors $$\mathbf{xy}$$ has the same property that $$(\mathbf{xy})^2 = -1$$, but we will pick up this thread in Euler's formula and Geometric Algebra next.

## Derivatives of cos and sin

It's worth noting that we can calculate the derivative of $$e^{ix}$$ and as a result work out the standard derivations of $$\cos$$ and $$sin$$ that we may have memorized in school or otherwise:

\begin{aligned} \frac{d}{dx}e^{ix} &= ie^{ix} \\ &= i\cos(x) + i^2\sin(x) \\ &= -\sin(x) + i\cos(x) \end{aligned}

and so comparing the real parts of $$e^{ix}$$ and $$\frac{d}{dx}e^{ix}$$ show that

$\frac{d}{dx}\cos(x) = -\sin(x)$

while comparing the imaginary parts show that

$\frac{d}{dx}\sin(x) = \cos(x)$