text:
# Homework Help: Rotation about CM - Can I use Moment of Inertia?

1. Mar 26, 2012

### RoshanBBQ

Rotation about CM -- Can I use Moment of Inertia?

1. The problem statement, all variables and given/known data
I have a program I coded to simulate the movement of a some point masses connected to each other rigidly. Each mass is the same. I am trying to code in the correct equations for rotation, but I am having some difficulty with a nonzero product of inertia.

3. The attempt at a solution
I have the x, y, and z coordinates to these masses where x y and z are n by 1 matrices (where I have n masses).

I calculated the center of mass as:
xcm = average(x)
ycm = average(y)
zcm = average(z)

Then I defined new coordinates, xx yy and zz:
xx = x - xcm
yy = y - ycm
zz = z- zcm

The following Wikipedia article then says, "With respect to a coordinate frame whose origin coincides with the body's center of mass, they can be expressed in matrix form as:"

and then it shows that the rotation depends only on the moments of inertia.
http://en.wikipedia.org/wiki/Newton–Euler_equations

However, I then compute the product of inertia:
sum(yy.*zz), which does not equal 0!

where .* means element-wise multiplication.

What is going on here? Am I computing something incorrectly, is Wikipedia wrong, or am I just misunderstanding something?

Last edited: Mar 26, 2012
2. Mar 27, 2012

### Andrew Mason

Re: Rotation about CM -- Can I use Moment of Inertia?

This is the first problem you have. This is not the definition of the centre of mass. You have to multiply the mass elements by their displacements from an origin, sum them and then divide by the total mass. That gives you the displacement of the centre of mass from the origin.

The second problem you have is with the moment of inertia. The moment of inertia relates the torque to the angular acceleration produced by that torque: $\vec{\tau} = I\vec{\alpha}$

An unconstrained rigid body will rotate about an axis through the centre of mass of the body. You have to then find the moment of inertia of the body about that axis. This can be a difficult calculation and depends on the geometry of the object. See, for example: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mi

You will have to have a solid understanding of rotational motion before you can write your program.

AM

3. Mar 28, 2012

### RoshanBBQ

Re: Rotation about CM -- Can I use Moment of Inertia?

Since all masses are the same, the center of mass is the average of the coordinates.

Let us assume we have n point masses of mass m located on the x-axis with coordinates x_1, x_2, ..., x_n.
$$M = \sum_{k=1}^n m=m\sum_{k=1}^n 1=mn$$
$$x_{cm} = \frac{1}{M}\sum_{k=1}^n x_n m=\frac{m}{M}\sum_{k=1}^n x_n=\frac{m}{nm}\sum_{k=1}^n x_n=\frac{1}{n}\sum_{k=1}^n x_n$$

The result is known as the arithmetic average of the coordinates.

Also, the moment of inertia calculation is not difficult as you claimed it can be since I am dealing with point masses. It is simply the sum of squared coordinates scaled by m.

4. Mar 30, 2012

### Andrew Mason

Re: Rotation about CM -- Can I use Moment of Inertia?

Ok.
So would the moment of inertia about the axis of rotation (z axis) not be:

$$m\sum_{i=1}^{n} |x_i-x_{cm}|^2 + |y_i-y_{cm}|^2$$?

AM

5. Mar 30, 2012

### D H

Staff Emeritus
Re: Rotation about CM -- Can I use Moment of Inertia?

That wikipedia article is poorly written, which results in you misunderstanding something.

That first equation in that wikipedia article essentially says that translational and rotational dynamics are decoupled in the special case in which the origin of the frame describing the rigid body has its origin at the body's center of mass. It does not say that "rotation depends only on the moments of inertia." The Jc in that wiki article's equation is the inertia tensor, not the moments of inertia.

To understand this equation, you need to know a bit about what I'll call body frames and structure frames. A structure frame has its origin at some pre-determined point on the rigid body and has some pre-determined set of orthogonal axes that are based on the structure. A body frame is also based on the rigid body, but has its origin at the center of mass of the body.

For example, an aircraft structure frame often has the origin in front of and well below the nose of the aircraft. The x-hat axis points toward the rear of the aircraft along the aircraft's longitudinal axis, the the z-hat axis points up (away from the wheels), and the y-hat axis completes a right-handed coordinate system. This makes the x and z coordinates of any point inside the aircraft positive. A typical aircraft body frame has its origin at the aircraft center of mass, with x-hat and z-hat pointing in the opposite directions as the structural x-hat and z-hat axes. Thus the body frame's x-hat axis points toward the nose, y-hat points out the right wing, and z-hat points down.

These body axes almost certainly are not the principal axes of the aircraft. The principal axes are convenient for simplifying the equations of motion but are quite useless in reality. Think of a plane with sumo wrestlers seated on the left side, a bunch of school kids seated on the right, and a somewhat unbalanced cargo below. The principal axes vary from flight to flight, and even within flight as the plane burns fuel.

You can define a structure frame and a body frame for your set of rigidly connected point masses. Just pick some orthogonal axes that make sense. Working with the center of mass (a body frame) is a lot easier than some other point as the origin (a structure frame) because translational and rotational dynamics decouple with this choice. Translational motion is described by good old $\vec F_{\text{ext}} = d{\vec p}/dt = m \ddot{\vec x}_{\text{cm}}$. Rotational dynamics is a bit hairier, but a lot less hairy than using something other than the center of mass as your origin.

For a number of reasons, the angular velocity is typically expressed as angular velocity of the body frame with respect to inertial but expressed in body frame coordinates. One reason is tradition: This goes way back to Euler. Another is that the inertial tensor is a time-varying beast when expressed in inertial coordinates but it is constant (assuming a true constant mass rigid body) in the body frame. Yet another is that sensor readings are in some frame whose orientation is fixed with respect to structure / body.

This representation of angular velocity in body frame coordinates means that the rotational equations of motion are not the simple rotational equivalent of F=dp/dt=ma, $\vec {\tau}_{\text{ext}} =\mathbf J \dot{\vec{\omega}}$, where J is the inertia tensor. The relation F=dp/dt does have a rotational equivalent; it is
$$\vec {\tau}_{\text{ext}} = \left(\frac{d\vec L}{dt}\right)_I$$
where that subscript "I" on the derivative means "from the perspective of an inertial observer."

The time derivative of any vector quantity $\vec u$ from the perspective of an inertial (non-rotating) observer and a rotating observer are related via
$$\left(\frac{d\vec q}{dt}\right)_I = \left(\frac{d\vec q}{dt}\right)_R + \vec{\omega}\times \vec q$$
where the subscript "R" on the derivative on the right hand side means "from the perspective of a rotating observer."

The angular momentum as expressed in body frame coordinates is $\vec L = \mathbf J \vec{\omega}$. Using the rotational equivalent of F=dp/dt and the expression that relates inertial and rotating time derivatives, the rotational equations of motion become
$$\vec {\tau}_{\text{ext,body}} = \mathbf J \frac{d\vec{\omega}}{dt} + \vec{\omega} \times \left(\mathbf J \vec{\omega}\right)$$
Here $\vec {\tau}_{\text{ext,body}}$ are the real external torques on the body expressed in body frame coordinates (rather than inertial).

One final note: The above gives the angular velocity in body frame coordinates. It does not give the orientation of the body frame with respect to inertial. You cannot simply integrate angular velocity. Do you need to know the orientation of your body, and if so, do you know how to compute that?

6. Mar 30, 2012

### RoshanBBQ

Re: Rotation about CM -- Can I use Moment of Inertia?

Yes, that is how I computed it. I used xx yy and zz, which are relative to CM.

That makes much more sense. Wikipedia calls J "moment of inertia about the center of mass." They should call it the inertia tensor if it is the inertia tensor. I was previously using an inertia tensor in my program when I was using a structure frame (But at the time, I didn't know that meant a coupling between the two types of motion. When I figured it out, I promptly switched to a body frame and threw away my tensor stuff because of this Wikipedia article). You should really edit this Wikipedia article's definition of J if you have the knowledge to do it confidently.

As for your question, I do not think I need to know the angle with respect to the inertial observer. I am modeling a boat with 10 states in my state space: x y z (relative to starting point), their derivatives, roll and yaw, and their derivatives. I am assuming pitch is always zero to simplify the controls problem of directing the boat's motion.

So basically, my code outputs this:
Code (Text):
a = tensor\torque.'; % inv(tensor)*torque.'
where a is the instantaneous angular acceleration of pitch, roll, and yaw about the CM. "torque" is a 1 by 3 vector (which is why I transpose it during its interaction with the inertia tensor) defined as
$$\vec {\tau}_{\text{ext,body}} - \vec{\omega} \times \left(\mathbf J \vec{\omega}\right)$$

I then have the output go into 2 integrators in a Simulink model. Roll/Yaw and dRoll/Dyaw are fed back into the function for use in computing buoyancy forces etc. as well as torque in the above equation.

I have one question for now: Let us assume I have a 3x3 inertia tensor, T, relative to CM for a certain orientation I call "0 degrees from x, y, and z". Is there a way to transform T into T' where T' is the inertia tensor for that same object rotated a, b, and c degrees from x, y, and z respectively? That would save a lot of computation. As of now, I simply recompute the inertia tensor each time (and I have 100,000s of point masses). The code is called 100s of times, so it would be nice to cut off a few ms. As of now, the MATLAB code runs .5 seconds the first time (when it computes a lot of stuff, most burdensome being the x,y, and z coordinates of the hull). Subsequent calls to the function use persistent variables to keep many results, taking only .1 seconds. To take out the tensor calculation would probably cut .03 seconds or so since it is 100,000s of multiplications, sums, and squares. The simulink finally compiles the MATLAB code into C++ code, so it is hard to tell how fast that executes.

PS
Before I made this thread, I tried to do some serious self-studying on the topic. Take a look at that misleading image I uploaded from a book I was using. It seemed all sources agreed you could use only moments of inertia to analyze motion about the CM (wikipedia, books they cite, the book i quote, etc.), but failed to stress you would still need the principal axes for that frame of reference. The book I quote does, however, mention the idea about principal axes. I was left to assume one of two things: I could analyze that motion with that equation about the CM GIVEN I HAVE CHOSEN AXES THAT WERE PRINCIPAL or I could analyze motion about the CM BECAUSE THOSE AXES ARE ALWAYS PRINCIPAL AXES. That is just not a nice ambiguity. You cleared it up well though -- apparently it was the former. I am doing this for an engineering introduction to controls class. I am not well-versed even in 1-d physics, so I have had to do a lot of thinking and self-studying to formulate my 3-d model.

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7. Apr 2, 2012

### D H

Staff Emeritus
Re: Rotation about CM -- Can I use Moment of Inertia?

RoshanBBQ,

I am a bit troubled by this last post of yours. What exactly are you trying to do here? What problem are you trying to solve? You talked in the initial post about rigid bodies, but now you are talking about thousands (hundreds of thousands) of point masses! So which is it?

In a rigid body formalism, you have essentially twelve independent variables that describe each body. With hundreds of thousands of point masses, each with six degrees of freedom (position and velocity), you have on the order of a million independent variables! Whoa! A mesh of hundreds of thousands points would stress even the best finite element analysis package. You might want to rethink what you are doing.

Compared to your million or so degree of freedom system, a rigid body has but twelve degrees of freedom:
- Three for the position of some selected point on the rigid body,
- Three for the time derivative (velocity) of that point,
- Three for the orientation of the body or structure frame, and
- Three for the rotation rate (angular velocity) of the body frame.

There are reasons to use a finite elements approach. Real bodies are deformable rather than rigid. With a rigid body formalism you are forgoing looking at what happens inside the body. This is obviously a disadvantage, but the advantage of a rigid body formalism is that the number of independent variables shrinks by orders and orders of magnitude. There's no reason to use a finite elements approach if the object of interest is fairly rigid and you don't care about behaviors within the body.

The issues that need to be addressed in a rigid body formalism are
- How to physically describe the body,
- How to represent the body's dynamical state,
- How to formulate the equations of motion, and
- How to integrate those equations of motion with respect to time.

How to physically describe the body
A point mass has but one physical characteristic, its mass. A rigid body has mass plus an inertia tensor. The inertia tensor of a rigid body is constant in a frame fixed with respect to the body. Assuming the body truly is rigid, you need to calculate the inertia tensor but once.

Here's where you can make use of your hundreds of thousands of point masses. Calculate the inertia tensor for this system of masses. You'll need to transform to the body frame if the frame in which the locations of the point masses are represented is not the body frame. Suppose you calculate the inertia tensor in some frame A, but you want it in frame B. Let T be the transformation matrix that transforms a vector as represented in frame A to the representation of that same vector in frame B. The inertia tensor transforms according to
$$\mathbf I_B = \mathbf T\, \mathbf I_A\,\mathbf T^T$$

How to represent state
Rigid bodies have a translational and rotational state. The translational and rotational equations of motion decouple if the translational state is represented as the position and velocity of the center of mass of the rigid body. There are reasons to a structure frame whose origin is not the center of mass, but I recommend that you start with a center of mass body frame. The equations of motion for the translational state such a frame are fairly simple: It's just Fext=mbodya. It is rotational state that is the tough nut to crack.

One way to represent rotational state is via the transformation matrix. This is in a sense the canonical form. After all, the mathematics of rotation in Euclidean three space is the group SO(3), where SO is short for special orthogonal (matrices). There is an issue with this representation scheme: Too many numbers. A 3x3 matrix has nine elements, but rotation has but three degrees of freedom. This will lead to challenges when it comes to integrating the equations of motion. There are ways around this. The problem is not intractable.

F=ma is a second order differential equation. The rotational equations of motion are also second order. Some rotational analog of velocity is needed here. The obvious choice is angular velocity. One advantage of using angular velocity is that it is not over-specified. Angular velocity comprises three parameters, and there are three degrees of freedom. For a number of reasons, angular velocity is typically represented as angular velocity of the rotating body with respect to inertial but represented in body frame coordinates.

How to formulate the equations of motion
The translational equations of motion are simply F=ma for a center of mass representation. The tricky issue is rotational state. The equations of motion for angular velocity is given in post #5. That leaves rotation itself. If you represent the orientation of your rigid body by the transformation matrix T from inertial to body coordinates, the time derivative of this matrix is
$$\dot{\mathbf T} = -\mathbf X(\vec{\omega}) \mathbf T$$
where $\vec{\omega}$ is the angular velocity as represented in body frame coordinates and $\mathbf X(\vec{\omega})$ is the cross product matrix generated from the vector $\vec{\omega}$. If you instead use the transformation from body to inertial, this becomes
$$\dot{\mathbf T} = \mathbf T \mathbf X(\vec{\omega})$$

How to integrate those equations of motion with respect to time
Position, velocity, and angular velocity are conceptually easy. Just use some numerical integration technique. Once again it is rotation that is tricky. Use a typical numerical integration technique on either of the above derivative equations and you will get some matrix that is not orthogonal. One way to get around this is to force the matrix to be orthogonal. Pick some row and normalize (make that row a unit vector). Next, remove the projection of that vector from some second row and normalize. Finally, remove the projections of both of these unit vectors from the third row and normalize. If you use this approach it is best to alternate your choices of this first, second, and third row with each integration step.

Another approach is to use a better integration technique. Google the term "geometric integration". Yet another approach is to use a better representation of orientation. There are many different ways to represent SO(3) that are better suited to numerical integration. Some of these alternative approaches are unit quaternions, Rogrigues parameters, and modified Rogrigues parameters.


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## Electronic Communications in Probability

### Balanced random and Toeplitz matrices

#### Abstract

Except for the Toeplitz and Hankel matrices, the common patterned matrices for which the limiting spectral distribution (LSD) are known to exist share a common property–the number of times each random variable appears in the matrix is (more or less) the same across the variables. Thus it seems natural to ask what happens to the spectrum of the Toeplitz and Hankel matrices when each entry is scaled by the square root of the number of times that entry appears in the matrix instead of the uniform scaling by $n^{−1/2}$. We show that the LSD of these balanced matrices exist and derive integral formulae for the moments of the limit distribution. Curiously, it is not clear if these moments define a unique distribution

#### Article information

Source
Electron. Commun. Probab., Volume 15 (2010), paper no. 14, 134-148.

Dates
Accepted: 27 April 2010
First available in Project Euclid: 6 June 2016

Permanent link to this document
https://projecteuclid.org/euclid.ecp/1465243957

Digital Object Identifier
doi:10.1214/ECP.v15-1537

Mathematical Reviews number (MathSciNet)
MR2643593

Zentralblatt MATH identifier
1225.60014

Rights

#### Citation

Basak, Aniran; Bose, Arup. Balanced random and Toeplitz matrices. Electron. Commun. Probab. 15 (2010), paper no. 14, 134--148. doi:10.1214/ECP.v15-1537. https://projecteuclid.org/euclid.ecp/1465243957

#### References

• Bai, Z. D. (1999). Methodologies in spectral analysis of large dimensional random matrices, a review. Statistica Sinica 9, 611–677 (with discussions).
• Bai, Z. D. and Silverstein, J. (2006). Spectral Analysis of Large Dimensional Random Matrices. Science Press, Beijing.
• Basak, Anirban (2009). Large dimensional random matrices. M.Stat. Project Report, May 2009. Indian Statitstical Institute, Kolkata.
• Bhamidi, Shankar; Evans, Steven N. and Sen, Arnab (2009). Spectra of large random Trees. Available at http://front.math.ucdavis.edu/0903.3589. %
• Bhattacharya, R. N.; Ranga Rao, R. Normal approximation and asymptotic expansions.Wiley Series in Probability and Mathematical Statistics.John Wiley & Sons, New York-London-Sydney, 1976. xiv+274 pp.
• Bose, Arup; Gangopadhyay, Sreela and Sen, Arnab (2009). Limiting spectral distribution of $XX^{\prime}$ matrices. Annales de l'Institut Henri Poincaré. To appear. Currently available at http://imstat.org/aihp/accepted.html.
• Bose, Arup; Sen, Arnab. Another look at the moment method for large dimensional random matrices. Electron. J. Probab. 13 (2008), no. 21, 588–628.
• Bryc, Włodzimierz; Dembo, Amir; Jiang, Tiefeng. Spectral measure of large random Hankel, Markov and Toeplitz matrices. Ann. Probab. 34 (2006), no. 1, 1–38.
• Hammond, Christopher; Miller, Steven J. Distribution of eigenvalues for the ensemble of real symmetric Toeplitz matrices. J. Theoret. Probab. 18 (2005), no. 3, 537–566.

format: plain


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# Lines CSV file (*.csv)¶

Lines CSV file (*.csv) is a text file to import or export [Lines].

Table 47 shows the list of columns in CSV file, and List 22 shows an example, respectively.

The encoding of Lines CSV file is UTF-8.

Table 47 The list of columns in Lines CSV file
Column number Name Description
1 lid Line ID. The value should be the same for the vertices in the same line.
2 vid Vertex ID. For vertices in the same line, the values should be 1, 2, …
3 x X coordinate
4 y Y coordinate
5 name Name of line
6 value Value of line
List 22 Example of Lines CSV file
lid,vid,x,y,name,value
1,1,-1.93,0.117,Line4,4
1,2,-2.191,1.632,Line4,4
1,3,-2.096,3.005,Line4,4
1,4,-0.77,4.047,Line4,4
1,5,0.722,4.355,Line4,4
1,6,2.758,4.094,Line4,4
2,1,2.498,-1.422,Line3,3
2,2,3.469,-2.89,Line3,3
2,3,4.321,-4.69,Line3,3
2,4,2.285,-5.305,Line3,3
2,5,0.248,-4.642,Line3,3
2,6,-0.675,-3.766,Line3,3
2,7,-1.031,-2.985,Line3,3
3,1,-0.746,2.224,Line2,2
3,2,2.048,2.911,Line2,2
3,3,3.232,1.774,Line2,2
3,4,4.416,-0.594,Line2,2
3,5,4.534,-2.346,Line2,2
3,6,4.392,-2.866,Line2,2
4,1,-0.949,1.167,Line1,1
4,2,-0.869,-0.571,Line1,1
4,3,-0.174,-1.867,Line1,1
4,4,1.266,-3.482,Line1,1
4,5,2.569,-3.814,Line1,1
4,6,2.829,-3.979,Line1,1


text:
# Plot 3 variables in view similiar to ArrayPlot

Is there any function to plot vectors in form $(x,y,z)$ where first coordinate is X value, second- Y value and color of square represents third coordinate? It should look like ArrayPlot.

Here is a sample of my data (http://pastebin.com/nHprReDu) and the results I obtain using ListDensityPlot:

ListDensityPlot will deal with data in that format. An interpolation order of zero will reproduce the "blocky" ArrayPlot output as well.

data = Flatten[Table[{x, y, Sin[x] + Sin[y]}, {x, -3 Pi, 3 Pi}, {y, -3 Pi, 3 Pi}], 1];
ListDensityPlot[data, InterpolationOrder -> 0]

Thank you for posting your data. Unfortunately, the data you have is not regularly sampled, hence the irregular shapes in the ListDensityPlot. However, you can interpolate and resample if you want "square" blocks. Below, data is your data from pastebin.

if = Interpolation[data];
ListDensityPlot[
Flatten[Table[{x, y, if[x, y]}, {x, 1, 23, 2}, {y, 1, 14}], 1],
InterpolationOrder -> 0
]

• Thank you for idea, but in my case it returns not only squares, what's weird because values are different – pupilx Jan 26 '17 at 22:16
• @pupilx You should add your data and the results in the question, so we can compare results. – MarcoB Jan 26 '17 at 22:18
• Please see my edit – pupilx Jan 26 '17 at 22:31
• @pupilx Thank you for the data. I have amended my answer. Take a look at the new version. – MarcoB Jan 26 '17 at 22:41
• Thank you, it helps me a lot. The most important for me is not to lose information if value < 1, because then I need to construct true/false matrix , will it be ok? – pupilx Jan 26 '17 at 22:47


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# Differential geometric approach to quantum mechanics

+ 6 like - 0 dislike
25846 views

Plenty of books/papers have been written about differential geometry in relation with general relativity, string theory, classical/quantum/gauge field theory and classical mechanics (Mathematical Methods of Classical Mechanics by V. I. Arnold comes to mind). I was wondering if people have investigated non-relativistic quantum mechanics using differential geometry and if this approach has been fruitful? Any good resources/books that discuss this method would be greatly appreciated.

edited May 12, 2014

Yeah, the approach basically uses Kahler manifolds --- for a review of this "geometric description", see Ashtekar and Schilling "Geometrical Formulation of Quantum Mechanics" arXiv:gr-qc/9706069.

That looks interesting, thanks!

For a geometrical perspective on quantum information theory, and just quantum theory and quantum states in general, look up The Geometry of Quantum States.

+ 5 like - 0 dislike

I do not understand well the question. Are you asking whether the issue of (first) quantization on manifolds has been investigated? If this is the question the answer is positive. Several researchers focused on that problem. The point is that the standard procedure based on Stone-von Neumann theorem generally does not hold for manifolds different form ${\mathbb R}^n$. The algebra of elementary observables (that generated by position and momentum in $\mathbb R^n$ is very difficult to define). Naive approaches face the technical problem of  symmetric operators which should represent observables but are not (essentially) self adjoint.  The situation becomes simpler when the space is homogeneous, i.e. when there is a (at least topological) group acting transitively on the space. The action can be defined either  in terms of isometries, conformal transformations  or diffeomorphisms. In this case  there are procedures, in particular due to Isham and Landsman and collaborators (Letters in Mathemattical Physics 20:11-18, 1990, Nuclear Physics B365 (1991) 121-160) leading to  a *-algebra of essentially self adjoint operators defining observables in a suitable Hilbert space associated with the manifold. These procedures include (generalizations of) anyons theory and Aharonov Bohm quantization as well as the standard quantization procedure in $\mathbb R^n$.

answered May 13, 2014 by (2,025 points)
edited May 13, 2014

Thanks! I have only recently starting to recognize how important differential geometry appears to be in physics, and I'm getting more and more interested in it. I am basically trying to understand in what level of details we can describe quantum mechanics using differential geometry. So "(first) quantization on manifolds" sounds really interested, although I will be lying if I say I understood most of your message ;) (this is due to my lack of knowledge). But if I have some spare time then I will try to look up some of the terms you have mentioned.

+ 5 like - 0 dislike

I'd say the answer to this is clearly: geometric quantization.

I'd argue that this is the theory of quantization, everything else is an approximation to it (in particular the popular algebraic deformation quantization is an approximation to full non-perturbative geometric quantization). And it is all based on differential geometry -- on higher differential geometry, actually, if one includes quantization of local field theory -- I have some technical details on this here.

For an exposition of how differential geometry and Lie theory yield the theory of quantization see here.

answered May 13, 2014 by (5,900 points)

Thanks! I've saved your pdf file on my computer and hope to be able to tackle it after I have learnt more differential geometry. I'm currently reading "Geometry, Topology and Physics" by Mikio Nakahara, but I have the feeling I may need to buy a more advanced book on diff. geometry before I can better appreciate your paper ;).

@Hunter, my remark on field theory was for completeness, to drive home the point that this really eventually gives the full story, not just some fragment. But I gather you want to and probably should start with just quantum mechanics for the time being, hence with just standard geometric quantization. A fairly comprehensive and commented list of literature on that is here. That lists introductions, and textbooks, and original articles and recent developments. Dig around a bit and see what suits your needs.

@UrsSchreiber Thanks, that is indeed exactly what I was looking for!

+ 3 like - 0 dislike

An excellent book is Marsden and Ratiu, Introduction to mechanics of symmetry .

It exposes the symplectic geometry underlying much of classical mechanics and even some of quantum mechanics.

answered May 23, 2014 by (13,199 points)
edited May 23, 2014

Nice, and it is free as well! Thanks

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text:
# Problem: Consider the reaction 8H2S(g) + 4O2 (g) → 8H2O(g) + S8 (g) Δ[H2S]/Δt = -0.031 M/s Part AFind Δ[O2]/Δt. Express your to two significant figure and include the appropriate units. Part BFind Δ[H2O]/Δt. Express your answer to two significant figures and include the appropriate units. Part CFind Δ[S8]/Δt. Express your answer to two significant figure and include the appropriate units. Part D Find the rate of the reaction. Express your answer to two significant figure and include the appropriate units.

🤓 Based on our data, we think this question is relevant for Professor Sheng's class at FGCU.

###### Problem Details

Consider the reaction

8H2S(g) + 4O2 (g) → 8H2O(g) + S8 (g)

Δ[H2S]/Δt = -0.031 M/s

Part A

Find Δ[O2]/Δt.

Express your to two significant figure and include the appropriate units.

Part B

Find Δ[H2O]/Δt.

Express your answer to two significant figures and include the appropriate units.

Part C

Find Δ[S8]/Δt.

Express your answer to two significant figure and include the appropriate units.

Part D

Find the rate of the reaction. Express your answer to two significant figure and include the appropriate units.


text:
maths > statistics-basics

Standard Random Experiments

what you'll learn...

overview

In this page, the following standard random experiments are explained.

•  tossing a coin

•  rolling a dice

•  picking a ball of a color from a box

In each of these, the statistical methods to predict the outcome of an experiment requires large effort and small variations in data can cause errors in the prediction. So the possible outcomes are theorized as having equal chances and the probability of an outcome is defined.

recap

In the topic Probability of an Event in Random Experiments, we have studied the following.

Dropping a ball is not a random experiment. If a ball is dropped $10$$10$ times, it will always hit the floor.

Tossing a coin is a random experiment.

Let us say, when the coin was tossed once, it landed 'heads' up. The result does not indicate in anyway that the result will be 'heads' whenever the coin is tossed. It can either be 'heads' or 'tails'. That is why tossing-a-coin is called a random experiment.

That is, a random experiment has a number of possible outcomes. A random experiment, when performed, results in one of the possible outcomes.

In this chapter, the following random experiments are considered

•  tossing a coin : A coin with 'heads' or 'tails' is flipped. The face seen on top is the outcome of the experiment.

•  rolling a dice : A dice has $6$$6$ faces marked with numbers $1$$1$ to $6$$6$. The dice lands with one of the faces up. The outcome is the number on the face that is facing up.

•  picking a ball from a box : A box has a number of red and blue balls. A person picks a ball without looking into the box. The color of the ball is the outcome of the experiment.

tossing a coin

Consider tossing a coin. From the knowledge we have gained in statistics, the right way to predict the outcome, is conduct the experiment a large number of times and collect data. Based on the data one can predict. An example is illustrated below.

Considering tossing a coin. We have conducted the experiment $40$$40$ times and collected the data. The prediction goes like the following

•  If the coin is tossed $80$$80$ times, the outcome 'heads' will occur $40$$40$ times

•  If the coin is tossed $20$$20$ times, the outcome 'heads' will occur $10$$10$ times

•  If the coin is tossed $1$$1$ time, the outcome 'heads' will be predicted as $20$$20$ of $40$$40$ times chance.

This is the fundamental part in probability and this is explained in detail below.

variations

Considering tossing a coin. The figure shows examples of conducting the experiment many times. There is small variation in the data.

It is noted that, in random experiments, the data will have errors due to variations.

Since, the data has errors due to variations, the data is not used to define the probability of the outcome. Instead, the two outcomes are theorized to be having equal chance and the data is used to verify if that assumption is correct.

data or theory

To find the probability of a possible outcome in a random experiment, two possible methods are used.

•  repeat the experiment a large number of times such that any error due to variations is negligibly low.

•  theoretically figure-out that the possible outcomes have equal chance and calculate the probability accordingly.

Probability = number of outcomes in the desired event / number of outcomes in the sample-space.

large number of trials

Considering tossing a coin.

The figure shows examples of conducting the experiment many times. There is small variation in the data.

Theoretically, the following is derived.

•  the sample space is 'heads' and 'tails'

•  The two outcomes in the sample space have equal chance

•  It is expected that if the experiment is conducted large number of trials, then the difference between number of 'heads' and 'tails' will be negligibly small.

•  Therefore, probability of 'heads' is $\frac{1}{2}$$\frac{1}{2}$ and for the 'tails' is $\frac{1}{2}$$\frac{1}{2}$

honest

A coin that adheres to this theorization of equal probability is called an honest coin.

It is assumed that coins used in random experiments are honest coins. If that assumption is not valid, then it is explicitly given that the coin is biased.

rolling a dice

Consider the random-experiment: Rolling a dice. Without conducting the experiment and looking at the data that has errors because of variations, we conclude the following.

•  the sample space is $1$$1$, $2$$2$, $3$$3$, $4$$4$, $5$$5$, and $6$$6$

•  all the six outcomes in the sample space have equal chance.

•  so the probability of $1$$1$ is $1/6$$1 / 6$ and so on for other events.

A dice adhering to this theorization of equal probability is called an honest dice.

picking a color

Consider the random experiment: Picking a ball at random from a box of 3 red balls and 2 blue balls. Without conducting the experiment and looking at the data that has errors because of variations, we conclude the following

•  The sample space is 'red1', 'red2', 'red3', 'blue1', and 'blue2' color balls.

•  The five balls have equal chance.

•  The desired outcome or the event is a red ball. That can be one of $3$$3$ in the sample space.

•  The probability of picking a red ball = count of possible outcomes in desired event / count of possible outcomes in the sample space
Probability of picking a red ball = $3/5$$3 / 5$

summary

Honest random experiment : Random experiments, when they are conducted, result in variation in outcome probabilities depending on various parameters. A honest random experiment is the hypothetical experiment that exhibits outcome probabilities equal to the theorized outcome probabilities.

A coin that adheres to the theorization of equal probability between heads and tails is called an honest coin.

A dice that adheres to the theorization of equal probability between $6$$6$ faces is called an honest dice.

Outline


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# Singular Valued Decomposition: Nearly SVD

LINALG-IDAQ9X

If $A=\begin{bmatrix} -5 & 0 & 0 \\\ 0 & 3 & 0\end{bmatrix}$ then $A=U\Sigma V^{T}$ for $U=\begin{bmatrix} 0 & 1 \\\ 1 & 0\end{bmatrix}$, $\Sigma=\begin{bmatrix} 3 & 0 & 0 \\\ 0 & 5 & 0\end{bmatrix}$, and $V=\begin{bmatrix} 0 & 1& 0 \\\ -1 & 0 & 0 \\\ 0 & 0 & 1\end{bmatrix}$.

Why is the equation $A=U\Sigma V^{T}$ not a singular value decomposition of $A$?

A

The entries of $A$ must be non-zero.

B

The eigenvalues of $A^{T}A$ are 9 and 25.

C

The diagonal entries of $\Sigma$ must be arranged in descending order starting with the upper left entry.

D

The matrix $A$ is not a square matrix.

E

The columns of $U$ are not eigenvectors of $A^{T}A$


text:
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# Compare the Modal Ages of Two Groups of Students Appearing for an Entrance Test: - CBSE Class 10 - Mathematics

#### Question

Compare the modal ages of two groups of students appearing for an entrance test:

 Age (in years): 16-18 18-20 20-22 22-24 24-26 Group A: 50 78 46 28 23 Group B: 54 89 40 25 17

#### Solution

 Age (in years) Group ‘A’ Group ‘B’ 16–18 50 54 18–20 78 89 20–22 46 40 22–24 28 25 24–25 23 17

For group “A”

The maximum frequency is 78 so the modal class is 18–20.

Therefore,

l = 18

h = 2

f = 78

f1 = 50

f2 = 46

Mode rArr=l+(f-f1)/(2f-f1-f2)xxh

=18+(78-50)/(156-50-46)xx2

=18+28/60xx2

=18+14/15

= 18 + 0.93

= 18.93

For group “B”

The maximum frequency 89 so modal class 18–20.

Therefore,

l = 18

h = 2

f = 89

f1 = 54

f2 = 40

Mode rArr=l+(f-f1)/(2f-f1-f2)xxh

=18+(89-54)/(178-54-40)xx2

=18+35/84xx2

=18+35/42

=18+5/6

= 18 + 0.83

= 18.83

Thus, the modal age of group A is 18.93 years whereas the modal age of group B is 18.83 years.

Is there an error in this question or solution?

#### APPEARS IN

Solution Compare the Modal Ages of Two Groups of Students Appearing for an Entrance Test: Concept: Mode of Grouped Data.
S


text:
Browse Questions

# A circular disc of radius $'R'$ is removed from a bigger circular disc of radius $'2R'$ such that the circumference of the disc touch. The center of mass of the new disc is at a distance $'\alpha R'$ from the center of the bigger disc. The value of $\alpha'$ is

$\begin {array} {1 1} (a)\;\frac{1}{2} & \quad (b)\;\frac{1}{3} \\ (c)\;\frac{1}{4} & \quad (d)\;\frac{1}{6} \end {array}$

$(b)\;\frac{1}{3}$

–1 vote


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Is this proof that massless objects cannot be charged?

In the realm of pre relativistic physics.

$$\vec{p}=m\vec{v}$$

$$F= \frac{dp}{dt}= m\vec{a}$$

If there exists an electric field in space, the force experienced by it would be $$F= q\vec{E}$$

Applying newtons laws:

$$q\vec{E} = m \vec{a}$$

This an equation stating a relation between the force, and the acceleration of an object.

Given $$m=0$$

It follows that $$q\vec{E} = 0$$

Since we assumed in the beginning that there exists and electric field.

$$q=0$$

Thus a massless object must have no charge. [Or newtons laws don't work for massless objects even in pre relativistic physics]

My question is: is my reasoning correct? And is there a generalisation of this to the relativistic equations.

• Related/possible duplicates: physics.stackexchange.com/q/7905/50583 and its linked questions Jun 27 at 18:11
• A massless object has to always travel at the speed of light, so using non-relativistic mechanics would be completely invalid. Jun 27 at 18:12
• It's a little unfair, since you start off by assuming you can use the kinematic equations that govern the motion of a classical particle with mass. You then do some manipulations and find what seems to be a contradiction when you set the mass to zero. Because your starting place seems inappropriate it calls your conclusions into question.
– hft
Jun 27 at 18:48
• I wouldn't call it a contradiction. yes a is undefined, but q=0 is an inevitability, not a contradiction. But yes, m=0 does seemingly violate newtons law when finding acceleration atleast. But yes, I already knew newtonian mechanics wouldn't be correct Jun 27 at 19:26
• A "proof" requires axioms. Here your axioms apparently include "Newton's laws hold when interpreted according to the rules of algebra for finite real numbers". If so then it is valid, because $m=0$ does then imply $ma=0$. But physics is often done under what are in effect other axioms of algebra, e.g. distribution theory, so your choice is not necessarily the only reasonable one. Jun 28 at 20:40

As was pointed out in the comments above, one has to use relativistic mechanics to talk meaningfully about massless particles; you can't just write $$F = ma$$ and expect it to work. And, indeed, it turns out that we can come up with a reasonable question of motion for a charged massless particle if we use the machinery of relativistic dynamics.

The equation of motion of a massive charged relativistic particle is $$\frac{d p^\mu}{d\tau} = \frac{q}{m} F^{\mu}{}_\nu p^\nu, \tag{1}$$ where $$m$$ is the particle's rest mass, $$q$$ is its charge, $$p^\mu$$ is its four-momentum, $$F^{\mu \nu}$$ is the field strength tensor, and $$\tau$$ is the proper time along the (massive) particle's world-line. In particular, $$\tau$$ serves mainly as a parameter that traces out the particle's world-line through spacetime.

If we multiply both sides by $$m$$ and introduce a new parameter $$\lambda = \tau/m$$, it turns out that the above equation is equivalent to $$\frac{d p^\mu}{d \lambda} = q F^\mu {}_\nu p^\nu. \tag{2}$$ What's more, under this parameterization we have $$\frac{d x^\mu}{d \lambda} = m \frac{d x^\mu}{d \tau} = p^\mu. \tag{3}$$

The new parametrization (2) has a perfectly well-behaved limit as $$m \to 0$$; in fact, this is a conventional parametrization for the worldlines of massless particles. Defining our $$x$$-direction to be the direction of the field, we have $$F^{01} = - F^{10} = E_x$$, and so the equations of motion are \begin{align*} \dot{p}^t &= q E_x p^x & \dot{p}^x &= q E_x p^t & \dot{p}^y = \dot{p}^z = 0 \tag{4} \end{align*} where dots denote differentiation with respect to $$\lambda$$. This can then be solved (in principle) for the four-momentum $$p^\mu$$ as a function of $$\lambda$$. If desired, one can then find the trajectories $$t(\lambda)$$, $$x(\lambda)$$, etc. in parametric form, and (in principle) invert the first of these to obtain $$x(t)$$, $$y(t)$$ and $$z(t)$$.

This means that the energy ($$p^t$$) and $$x$$-momentum ($$p^x$$) of a massless charged particle will increase as it travels. But you can also show from the equations of motion (4) (try it!) that the quantity $$p_\mu p^\mu = -(p^t)^2 + (p^x)^2 + (p^y)^2 + (p^z)^2$$ is constant with respect to $$\lambda$$. So these equations ensure that a massless charged particle (with $$p_\mu p^\mu = - m^2 = 0$$) stays massless as it travels; and that means that it always travels at the speed of light, even as its energy and momentum change.

• Massless charged particles have very interesting properties then! I haven't learned tensors yet, however atleast with what I know, I am confused how conservation of energy\momentum applies here. How can the momentum of the particle increase if the speed doesn't? I only know of field momentums contribution to the total momentum, outside of the formula $p= \gamma m_{0}\vec{v}$, which I presume wouldn't apply here. What form does this momentum take? Obviously you state p in the eq of motion but intuatively what "stores" this momentum. [If that is a better phrasing] Jun 27 at 21:31
• I think a massless charged sphere is impossible because the field energy contributes some non-zero mass and the material of the sphere cannot have negative mass to compensate this. Also, in the limit of a pointlike entity the field energy blows up unless the charge goes to zero. Jun 27 at 23:35
• @AndrewSteane: Your last objection would also rule out massive charged particles, not just massless ones; it's a well-known problem with point charges in classical electrodynamics. Also, note that the fields have momentum, not just mass-energy, and so it's entirely possible that when we take all of that into account we still have $m^2 = E^2 - p^2 = 0$ for the system. I'll admit I haven't done those calculations, though. Jun 28 at 2:37
• I think there is a certain similarity to gravitational forces because while a photon doesn't have "rest mass", all existing photons do have mass, and experience "forces" in "gravitational fields" potentially increasing their energy and momentum -- much like a charge would in an electric field -- even though they don't get any faster. Jun 28 at 9:34
• @hft: Your summary is correct. Jun 30 at 11:47

I disagree with your conclusion that $$qE=0$$ if $$m=0$$. My interpretation is that in such a case the acceleration is infinite, so the product is well defined. It is only natural to think that in Newtonian mechanics a massless object will experience infinite acceleration if it experiences any finite force. For a given force, the smaller the mass you use, the larger the acceleration you get.

Note about massless pulleys and similar situations: Given what I said above, how can a pulley of zero mass have a finite angular acceleration? Should it not be infinite? Well, the acceleration needs to be infinite only if the force is finite. For zero force, the acceleration is arbitrary. So there is no conflict: massless pulleys can have any acceleration (determined by the assumptions/constraints of the problem, like rope does not slip), as far as they feel or make no finite force. But all this simplifications or lack of rigour works because they mathematically work in the limit. So real physical problems in which the pulley has very low mass behave as if the mass of the pulley were zero and the acceleration given by the problem constraints. When you take the limit of mass goes to zero, you can chose that the force is proportional to the mass, so the less mass the less force. In the limit of zero mass, the force will be zero but the acceleration will remain finite.

• Mathematically the conclusion that qE=0 is correct. Why do you dissagree? $$qE=ma$$ $$\frac{qE}{m}= a$$ $$\frac{qE}{0} = a$$ therefore acceleration is undefined not infinite we aren't dealing with limits, the product of m and a is defined, it is 0. Y=mx when x=0, y=0. Jun 27 at 20:27
• $$Y=mx$$ $$[ x=0]$$ then $$y = m*0=0$$ you can't then say "$\frac{y}{x}= m$", thus m is infinite. m can be anything. There are infinitely many m's that satisfy those conditions, hence its undefined. But what you can say for certain, is that when x=0 -> y=0. Which the the same for m=0 -> qE=0 Jun 27 at 20:34
• The most you can say is that the acceleration is undefined, not that a charged massless particle cannot exist. In such a case you would need to complete your theory. One way is to propose that there are no charged massless particles. Another is to propose that such particles have infinite acceleration. Only experiment can decide in such a case. The experiment says that Newtonian mechanics is wrong in many other ways, so... Jun 27 at 21:22
• Then why constrained systems containing multiple massless pulleys in newtonian physics have finite accelerations? Jun 28 at 4:26
• @Shiva yes, I sense something that could be problematic there. Let me refresh my memory reading some problems and let me go back. Jun 28 at 6:41

I think a massless electron traveling at $$c$$ under classical Maxwell's Eq. would be problematic, but:

The Standard Model says electrons were massless and charged before Spontaneous Symmetry Breaking, but there wasn't enough time to cause problems.

Also: gluons are massless and carry color/anti-color charge. Of course, they're confined so they never get out to cause problems.

The correct answer is your parenthetical possibility that "Newton's laws don't work for massless objects in nonrelativistic physics." This has nothing to do with electricity or electric charge: you can't have a meaningful massless particle in Newtonian physics at all, because in that case $$F = ma$$ would imply that any net force would induce an infinite acceleration, which is difficult to make mathematically rigorous.

We sometimes talk about "massless pulleys" in the context of Newtonian physics, but these aren't actually dynamical objects - they're just an idealization representing the fact that we semi-magically require that the tensions in both ropes coming out of the pulley must be equal. (In fact, when we talk about a "massless pulley", we usually actually mean that the pulley has a negligible moment of inertia, not mass.)

Whether a charged massless particle could exist in relativistic physics is a more interesting and complicated question, but that is out of the scope of your question.

As described in Michael Seifert's answer, the question requires relativity, not Newtonian mechanics. As explained in the comments by Andrew Steane and Michael Seifert, the question also requires quantum mechanics. So we're in the realm of quantum field theory.

Massless charged particles are problematic in QFT, because then we would get pair production at arbitrarily low energies. This would manifest itself in ordinary blackbody radiation, for example, and that's not what we observe. There are various other issues, but this is the most obvious. It's one of the reasons that Penrose's CCC theory, at least in its original version, was totally non-viable.


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# Supremum is not necessarily Greatest Element

## Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $T$ admit a supremum in $S$.

Then the supremum of $T$ in $S$ is not necessarily the greatest element of $T$.

## Proof

Consider the subset $T$ of the set of real numbers $\R$:

$T := \set {x \in \R: 1 \le x < 2}$

The number $2$ cannot be the greatest element of $T$ as $2 \notin T$.

However, $2$ is the supremum of $T$ in $S$.

Indeed, by definition:

$\forall x \in T: x < 2$

So, let $x < 2$.

Then consider $y = \dfrac x 1 + \dfrac 2 1 = \dfrac {x + 2} 2$.

We have that $x < y < 2$ by Mediant is Between.

Thus $y \in T$ but $y > x$ and so $x$ cannot be the greatest element of $T$.

Neither can $y$ be the supremum of $T$ in $S$.

The conclusion is that there is no greatest element of $T$.

Hence the result.

$\blacksquare$


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C Specification

If the pNext chain of VkBindImageMemoryInfo includes a VkBindImageMemorySwapchainInfoKHR structure, then that structure includes a swapchain handle and image index indicating that the image will be bound to memory from that swapchain.

The VkBindImageMemorySwapchainInfoKHR structure is defined as:

typedef struct VkBindImageMemorySwapchainInfoKHR {
VkStructureType    sType;
const void*        pNext;
VkSwapchainKHR     swapchain;
uint32_t           imageIndex;
} VkBindImageMemorySwapchainInfoKHR;

Members

• sType is the type of this structure.

• pNext is NULL or a pointer to an extension-specific structure.

• swapchain is VK_NULL_HANDLE or a swapchain handle.

• imageIndex is an image index within swapchain.

Description

If swapchain is not NULL, the swapchain and imageIndex are used to determine the memory that the image is bound to, instead of memory and memoryOffset.

Memory can be bound to a swapchain and use the pDeviceIndices or pSplitInstanceBindRegions members of VkBindImageMemoryDeviceGroupInfo.

Valid Usage
• imageIndex must be less than the number of images in swapchain

Valid Usage (Implicit)
• sType must be VK_STRUCTURE_TYPE_BIND_IMAGE_MEMORY_SWAPCHAIN_INFO_KHR

• swapchain must be a valid VkSwapchainKHR handle

Host Synchronization
• Host access to swapchain must be externally synchronized


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# Instrumental Variables

Instrumental variables are an incredibly powerful for dealing with unobserved heterogenity within the context of regression but the language used to define them is mind bending. Typically, you hear something along the lines of “an instrumental variable is a variable that is correlated with x but uncorrelated with the outcome except through x.” At this point examples are listed — taxes on smoking likely effect health only through their actions on smoking — or the author drops right into the math stats.

I like math stats (when I am not getting a grade for it at least!) and will work through it. But at some point, I want to play with a simulation of the process. However, books like Mostly Harmless Econometerics just throw a bunch of nonsense in terms of emperical studies at you. Sure, the OLS and IV estimates are different… but I don't know which is right for sure in an empirical setting. Simulations are great for this (the lack of simulations is one of my biggest issues with that book. Right after the use of Stata.).

I turned to Google and did several searches and the only simple simulation that I could find was done using Stata. At least it was a simulation… but I wanted more. Since I couldn't find any examples using R, I play around until I got it and then write up the simulation to see if it helps anyone.

## Overview

Suppose that you have a continuous variable $$y$$ with the known mean response function

$E(y) = \beta_0 + \beta_1 x + \beta_2 c$

and further that $$x$$ and $$c$$ are correlated with each other. If you knew $$c$$, estimating $$\beta_1$$ would be easy by simply running lm(y ~ x + c). You would complete the regression, throw together a few diagnostics and call it a day.

But often we don't observe $$c$$ in our data. $$c$$ could be any number of things, such as treatment practices at a hospital or unmeasured differences between patients, but it is in the direct casual path of $$y$$ and you don't know it. If you instead do the regression lm(y ~ x), you will not get an accurate estimate of $$\beta_0$$. If you think about the model fit in that call it looks like

$y_i = \beta_0 + \beta_1 x_i + \epsilon_i$ but in this case $\epsilon_i = f(c_i, \eta_i)$ where $$\eta_i$$ is white noise centered on zero. As long as that function is linear, the expectation of $$\epsilon$$ will be some multiple of $$c$$. The result is that $$x$$ and $$\epsilon_i$$ are correlated. The estimates of $$\beta_1$$ will nessecarly be wrong. Interested parties can read more here and see just how wrong it will be.

Suppose there is a third variable $$z$$, $$x$$ can be expressed as some function of $$z$$, $$z$$ has no effect on $$y$$ except through $$x$$ (therefore $$z$$ is independent of any other variable that effects $$y$$). Then we have the equations

$E(x) = \alpha_0 + \alpha_1 x^* + \alpha_2 z$ and $E(y) = \beta_0 + beta_1 x + f(c)$

where $$x^*$$ is some latent part of $$x$$ and $$c$$ is still unobserved. Looking at the first equation and realizing we don't know what $$x^*$$ is the resulting model will have two regression coefficients. The intercept will be $$\alpha_0 + \alpha_1 E(x^*)$$ which is not correlated with $$f(c)$$ as and $$z$$ is independent of $$f(c)$$ by assumption.

If we take the fitted values of $$x$$ from the first equation and plug it into the second equation, the $$x$$ term is no independent of $$f(c)$$. This independence means that we can produce consistent estimates of $$\beta_1$$ when we replace $$x$$ with the fitted values of $$x$$ from equation one. When the estimator used is OLS, this method is called two-stage least squares (2SLS).

## Simulations

We need to start by generating values of $$x$$ and $$c$$ that are correlated. The easiest way to do this is to use the function mvrnorm() from MASS. Lets start by generating a 1000x2 random matrix with $$\rho = 0.5$$.

library(MASS)
# we are really generating x* and c and using a common variance
xStarAndC <- mvrnorm(1000, c(20, 15), matrix(c(1, 0.5, 0.5, 1), 2, 2))
xStar <- xStarAndC[, 1]
c <- xStarAndC[, 2]

and lets generate $$z$$ and make the observed $$x$$

z <- rnorm(1000)
x <- xStar + z

And now generate the response variable

# using 1 makes it easy to estimate how 'wrong' an estimator is and toss
# some noise on y
y <- 1 + x + c + rnorm(1000, 0, 0.5)

Lets check to make sure that everything worked so far:

cor(x, c)

## [1] 0.3027

cor(z, c)

## [1] -0.02993

We have a moderate correlation between $$x$$ and $$c$$ and it appears that $$z$$ and $$c$$ are uncorrelated. So far, so good.

Lets estimate the full model (the model that we would estimate if we knew $$c$$)

lm(y ~ x + c)

##
## Call:
## lm(formula = y ~ x + c)
##
## Coefficients:
## (Intercept)            x            c
##       1.716        0.987        0.970

Not bad 0.9868 is basically 1. Now what would happen if we ignored the fact that $$c$$ was unknown and estimated anyways?

lm(y ~ x)

##
## Call:
## lm(formula = y ~ x)
##
## Coefficients:
## (Intercept)            x
##        12.0          1.2

Yikes! The complete model definitely works but when we ignore the heterogeneity caused by $$c$$ we get terrible estimates. The estimate was 120% of the true effect of $$x$$. That isn't good.

What happens when we break the correlation between the “x part” and the unobserved $$c$$ by using the instrument and 2SLS?

xHat <- lm(x ~ z)\$fitted.values
lm(y ~ xHat)

##
## Call:
## lm(formula = y ~ xHat)
##
## Coefficients:
## (Intercept)         xHat
##      16.895        0.954

That is a pretty good estimate, only off by 4.6% of the true value of 1. Pretty nifty that we are able to accurately estimate the effect of $$x$$ using the instrument despite the unobserved value of $$c$$.

Its worth noting now that if you want to do this in R, you probably do not want to do 2SLS as I have done it here. The parameter estimates are correct but the errors are not. Using the method I have done above, you get standard errors from the second equation residuals whereas you want them from the first. The package AER includes a function ivreg() which will take care of this for you.

If you look at the equations, two things probably jump out at you. First, if $$x$$ does a poor job predicting $$x$$ (i.e., $$z$$ is a weak instrument), then it is unlikely that IV and 2SLS will provide you with meaningful results. In fact, it is possible that the IV estimates will be more inaccurate than the naive OLS estimator. A rule of thumb is the first stage (i.e., the regression of x on the instrument and any covariates) should have an F statistic of at least 10.

Second, if $$x$$ and $$c$$ are only weakly correlated, the OLS and IV estimates should converge. As $$x$$ and $$c$$ become more correlated, however, the estimates move apart quickly and with even mild correlation ($$\rho \sim 0.2$$) there can be significant differences in the estimates.

The dynamic graph below allows you to explore the effects of altering these parameters on the resulting OLS and IV estimates. The solid black line is the true value, the green line is the estimate using the full model (if you knew $$c$$), red is the IV estimate and blue is using the OLS estimator.


text:
# Introduction

Calculus allows to work with situations where a quantity is continuously changing and its rate of change isn't necessarily constant.

It applies to all situations where one quantity changes smoothly with respect to another quantity.

Basic calculus splits into

• differential calculus
• integral calculus

# 1 What is differentiation?

## 1.1 Graphs of relationships

If a graph that represents the relationship between two quantities is curved, then there's no single gradient value that applies to the whole graph. Instead, the graph has a gradient at each point on the graph.

As with straight-line, the gradient at each point is the rate of change of the quantity on the vertical axis with respect to the quantity on the horizontal axis.

## 1.2 Gradients of curved graphs

The gradient of straight line is a measure of how steep it is and is calculated as $$\text{gradient} = \frac{\text{rise}}{\text{run}}$$ Vertical lines don't have gradient. Hence:

The gradient of the straight line through the points $(x_1, y_1)$ and $(x_2, y_2)$, where $x_1 \ne x_2$, is given by $$\text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$$

A straight-line graph that

• Slopes down from left to right has a negative gradient.
• Is horizontal has a gradient of zero.
• Slopes up from left to right has a positive gradient.

On the other hand curved graphs, as $y = x^2$ don't have a single gradient value. Instead, any particular point on it can be chosen and gradient can be found for the graph at that given point.

The straight line that 'just touches' the graph at a given point is called the tangent.

The gradient of a graph at a particular point is the gradient of the tangent to the graph at that point.

If a graph has a 'sharp corner', like $y = |x|$ then it has no gradient at that point. If a graph has a 'break' at a point (the discontinuity), then the graph has no tangent at that point, hence no gradient. If a graph has a vertical tangent at a point, then the graph has no gradient at it.

## 1.3 Gradients of the graph of $y = x^2$

$y = x^2$ has a gradient at every point since it has a tangent, which is not vertical, at every point.

To find a gradient of a point on $y = x^2$ one may start by finding an approximate value by choosing a second point on the graph that is fairly close to first one. Then a gradient of this approximation can be calculated in usual way.

The closer second point is to the first one, the better the approximation is.

E.g. for point $(1, 1)$ and second point $(1.1, 1.21)$, the approximate value for the gradient is $$\frac{1.21 - 1}{1.1 - 1} = \frac{0.21}{0.1} = 2.1$$ Second point can lie on both sides of first one.

Following table for $(1, 1)$ and various second points can be devised

$x$-coord of second point0.80.90.990.9991.0011.011.11.2

That can be confirmed algebraically as follows. Consider $h$ denoting the increase in the $x$-coordinate from $(1, 1)$ to the second point. The value of $h$ can't be zero. The $x$-coordinate of second point is $1 + h$. The $y$-coordinate of second point is $(1 + h)^2$.

The gradient of the line that passes through both points is

\begin{align*} \text{gradient} &= \frac{(1 + h)^2 - 1}{(1 + h) - 1} \\ &= \frac{1 + 2h + h^2 - 1}{1 + h - 1} \\ &= \frac{2h + h^2}{h} \\ &= \frac{h(2 + h)}{h} \\ &= 2 + h \end{align*}

As the value of $h$ gets closer and closer to zero, the gradient of the line $2 + h$ gets closer and closer to 2. Hence the gradient of the graph at $(1, 1)$ must bey exactly 2.

Expressed mathematically as

$2 + h$ tends to $2$ as $h$ tends to $0$

or

the limit of $2 + h$ as $h$ tends to zero is $2$.

More generic formula can be devised as

\begin{align*} \text{gradient} &= \frac{(x + h)^2 - x^2}{(x + h) - x} \\ &= \frac{x^2 + 2xh + h^2 - x^2}{x + h - x} \\ &= \frac{2xh + h^2}{h} \\ &= \frac{h(2x + h)}{h} \\ &= 2x + h \end{align*}

As the second point gets closer and closer to the original point, the value of $h$ gets closer and closer to $0$ and the gradient gets closer and closer to $2x$. So the gradient of the graph of $y = x^2$ at the point whose $x$-coordinate is denoted by $x$ is given by the formula $$\text{gradient} = 2x$$

## 1.4 Derivatives

Consider a function $f$ and input value $x$. The point on the graph of $f$ that corresponds to $x$ is $(x, f(x))$. If the graph of $f$ has a gradient at that point, then we say that $f$ is differentiable at $x$.

It's convenient to think about various gradients of a function $f$ as defining a new function that is related to $f$. The rule of the new function is:

If the input value is $x$, then the output value is the gradient of the graph of $f$ at the point $(x, f(x))$.

This new function is called the derivative (or derived function) of the function $f$ and is denoted by $f'$.

The domain of $f'$ consists of all the values at which $f$ is differentiable.

Differentiation is the process of finding the derivative of a given function $f$.

The value of the derivative of a function $f$ at a particular input value $x$ is called the derivative of $f$ at $x$.

### Differentiation from first principle

Suppose $f$ and $x$ in the domain of $f$ such that $f$ is differentiable at $x$ (the graph of $f$ has a gradient at $(x, f(x))$). Consider second point with the offset of $h$, then:

The gradient of the line that passes through the two points is given by $$\text{gradient} = \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}$$ This expression is known as the difference quotient for the function $f$ at the value $x$.

As the value of $h$ gets closer and closer to zero, the value of the difference quotient gets closer and closer to the gradient of the graph at the point $(x, f(x))$, that is $f'(x)$.

To find the formula for $f'(x)$ I need to find, in terms of $x$, the limit of the difference quotient as $h$ tends to zero. In other words, I need to consider what happens to the difference quotient for $f$ at $x$, as $h$ gets closer and closer to zero.

Saying that a function $f$ is differentiable at a particular value of $x$ is the same as saying that the difference quotient for $f$ at $x$ tends to a limit as $h$ tends to zero. It must tend to the same limit for positive values of $h$ as for negative values.

For any function $f$, the derivative $f'$ of $f$ is given by the equation $$f'(x) = \lim_{h \to 0} \frac{ f(x + h) - f(x) }{ h }$$ for each value of $x$ in the domain of $f$ for which this limit exists.

### Leibniz notation

The notation with $f'$ denoting the derivative of a function $f$ is called Lagrange notation or prime notation. Leibniz notation is other extensively used notation.

In general, Lagrange notation is used when thinking in terms of a variable, and a function of this variable. Leibniz notation is used when thinking more of the relationship between two variables.

Consider $y = x^2$, formula for its gradient is $\text{gradient} = 2x$. In Leibniz notation this equation is written as $$\frac{\text{d}y}{\text{d}x} = 2x$$ Often referred to as the derivative of $y$ with respect to $x$.

It's important to mention that

• The notation is not a fraction
• $\text{d}$ doesn't have any meaning outside of the notation
• $\text{d}$ is not a factor

The usefulness of the Leibniz notation can be demonstrated as follows

The emphasis is on the relationship between the variables $x$ and $y$, rather than on the idea of $f$ as a function of $x$. Change in the $x$- and $y$-coordinate is denoted by $\delta x$ and $\delta y$ respectively. $\delta$ (delta) indicates 'a small change in'. So the $x$- and $y$-coordinates of second point are $x + \delta x$ and $y + \delta y$ respectively.

The gradient of the line that passes through he two points is then $$\frac{\text{rise}}{\text{run}} = \frac{\delta y}{\delta x}$$

The gradient is the limit of $\delta y / \delta x$ as the second point gets closer and closer to the first point.

### Functions whose domains include endpoints

Any function whose domain includes an endpoint isn't differentiable at this endpoint. E.g. $f(x) = x^{3/2}$ has domain $[0, \infty)$:

The endpoint at $0$ (at the origin) can be approached from right but can't from left. Hence the graph doesn't have tangent at the origin, it's not differentiable at $x = 0$. But the graph has a 'gradient on the right' at origin, namely $0$.

So we can say that the function $f(x) = x^{3/2}$ is right-differentiable at $x=0$, and that its right derivative at 0 is 0.

### Summary of derivatives

The derivative (or derived function) of a function $f$ is the function $f'$ such that $$f'(x) = \text{gradient of the graph of } f~ \text{at the point } (x, f(x))$$ The domain of $f'$ consists of the values in the domain of $f$ at which $f$ is differentiable (that is, the $x$-values that give points at which the gradient exists).

If $y = f(x)$, then $f'(x)$ is also denoted by $\frac{\text{d}y}{\text{d}x}$.

The derivative $f'$ is given by the equation $$f'(x) = \lim_{h \to 0} \frac{ f(x + h) - f(x) }{ h }$$ The procedure of using this equation to find a formula for the derivative $f'$ is called differentiation from first principles.

# 2 Finding derivatives of simple functions

## 2.1 Derivatives of power functions

Any functions of the form $f(x) = x^n$, where $n$ is a real number, is called a power function.

Previously seen formulas for the derivatives of three power functions are: $$\frac{\text{d}}{\text{d}x}(x^2) = 2x, ~~~ \frac{\text{d}}{\text{d}x}(x^3) = 3x^2, ~~~ \frac{\text{d}}{\text{d}x}(x^4) = 4x^3$$ They all follow the same pattern:

### Derivative of a power function

For any number $n$, $$\frac{\text{d}}{\text{d}x}(x^n) = nx^{n -1}$$ This formula holds for all values of $n$, including negative and fractional values.

$f(x) = x$ has derivative $f'(x) = 1 \times x^0$; that is$f'(x) = 1$. $f(x) = 1$ (same as $f(x) = x^0$) has derivative $f'(x) = 0$.

## 2.2 Constant multiple rule

If any graph is scaled vertically by a particular factor, then its gradient at any particular $x$-value is multiplied by this factor.

Scaling a graph by a factor of 3:

### Constant multiple rule (Lagrange notation)

If the function $k$ is given by $k(x) = af(x)$, where $f$ is a function and $a$ is a constant, then $$k'(x) = af'(x)$$ for all values of $x$ at which $f$ is differentiable.

### Constant multiple rule (Leibniz notation)

If $y = au$, where $u$ is a function of $x$ and $a$ is a constant, then $$\frac{ \text{d}y }{ \text{d}x } = a\frac{ \text{d}u }{ \text{d}x }$$ for all values of $x$ at which $u$ is differentiable.

### Derivative of a constant function

If $a$ is a constant, then $$\frac{\text{d}}{\text{d}x}(a) = 0$$

## 2.3 Sum rule

When two functions are added, then the $y$-coordinates of the points on the two graphs are added.

If I add two functions, then the gradient of the sum function at any particular $x$-value is the sum of the gradients of the two original functions at that $x$-value. E.g.:

### Sum rule (Lagrange notation)

If $k(x) = f(x) + g(x)$, where $f$ and $g$ are functions, then $$k'(x) = f'(x) + g'(x)$$ for all values of $x$ at which both $f$ and $g$ are differentiable.

### Sum rule (Leibniz notation)

If $y = u + v$, where $u$ and $v$ are functions of $x$, then $$\frac{ \text{d}y }{ \text{d}x } = \frac{ \text{d}u }{ \text{d}x } + \frac{ \text{d}v }{ \text{d}x }$$

Every polynomial function (with domain $\mathbb{R}$) is differentiable at every value of $x$.

# 3 Rates of change

The idea of a gradient as a rate of change applies to curved graphs. However the rate of change is 'instantaneous' value.

Another way to think about the derivative $\text{d}y/\text{d}x$ is that it is the rate of change of $y$ with respect to $x$.

## 3.1 Displacement and velocity

### Relationship between displacement and velocity

Suppose that an object is moving along a straight line. If $t$ is the time that has elapsed since some chosen point in time, $s$ is the displacement of the object from some chosen reference point, and $v$ is the velocity of the object, then $$v = \frac{\text{d}s}{\text{d}t}$$ (Time, displacement and velocity can be measured in any suitable units, as long as they're consistent).

# 4 Finding where functions are increasing, decreasing or stationary

Because the derivative of a function gives the gradient at each point on the graph of the function, it gives information about the shape of the graph.

## 4.1 Increasing/decreasing criterion

### Functions increasing or decreasing on an interval

A function $f$ is increasing on the interval $I$ if for all values $x_1$ and $x_2$ in $I$ such that $x_1 < x_2$ $$f(x_1) < f(x_2)$$ A function $f$ is decreasing on the interval $I$ if for all values $x_1$ and $x_2$ in $I$ such that $x_1 < x_2$ $$f(x_1) > f(x_2)$$ (The interval $I$ must be part of the domain of $f$.)

### Increasing/decreasing criterion

If $f'(x)$ is positive for all $x$ in an interval $I$, then $f$ is increasing on $I$. If $f'(x)$ is negative for all $x$ in an interval $I$, then $f$ is decreasing on $I$.

## Stationary points

The gradient of $f(x) = 2x^3 - 3x^2 - 36x$ is

• positive on the interval $(-\infty, -2)$
• negative on the interval $(-2, 3)$
• positive on the interval $(3, \infty)$

Between these open intervals, there are points at which the gradient of the graph is zero. Such a point is called a stationary point.

At the left-hand stationary point the value taken by the function is larger than at any other point nearby, hence the function has a local maximum at this point. Similarly the function has a local minimum a the right-hand point.

Turning point is a point where a function has a local maximum or local minimum.

A stationary point of functions as: when the graph of a functions is constantly decreasing or increasing, is called a horizontal point of inflection.

### Strategy to fund the stationary points of a function $f$

Solve the equation $f'(x) = 0$

### First derivative test (for determining the nature of a stationary point of a function $f$)

If there are open intervals immediately to the left and right of a stationary point such that

• $f'(x)$ is positive on the left interval and negative on the right interval, then the stationary point is a local maximum
• $f'(x)$ is negative on the left interval and positive on the right interval, then the stationary point is a local minimum
• $f'(x)$ is positive on both intervals or negative on both intervals, then the stationary point is a horizontal point of inflection.

### Strategy to apply the first derivative test by choosing sample points

1. Choose two points (that is, two $x$-values) fairly close to the stationary point, one on each side.
2. Check that the function is differentiable at all points between the chosen points and the stationary point, and that there are no other stationary points between the chosen points and the stationary point.
3. Find the value of the derivative of the function at the two chosen points.
• If the derivative is positive at the left chosen point and negative at the right chosen point, then the stationary point is a local maximum.
• If the derivative is negative at the left chosen point and positive at the right chosen point, then the stationary point is a local minimum.
• If the derivative is positive at both chosen points or negative at both chosen points, then the stationary point is a horizontal point of inflection.

## 4.3 Uses of finding stationary points

### Properties of graphs of cubic functions

The graph of every cubic function has the following properties.

• There are two, one or no stationary points
• Apart from at any stationary points and in the interval between them if there are two,
• the graph slopes up from left to right if the coefficient of $x^3$ in the rule of the function is positive
• the graph slopes down from left to right if the coefficient of $x^3$ is negative.
• If there are two stationary points, then one is a local maximum and the other is a local minimum, and the graph slopes the other way in the interval between them.
• If there is one stationary point, then it is a horizontal point of inflection.
• There are three, two or one $x$-intercepts.
• There is one $y$-intercept.
• The graph tends to plus or minus infinity for large positive and large negative values of $x$.

### Finding the greatest and least values of a function an interval of the form $[a, b]$

(This strategy is valid when the function is continuous on the interval, and differentiable at all values in the interval except possibly the endpoints.)

1. Find the stationary points of the function.
2. Find the values of the function at any stationary points inside the interval, and at the endpoints of the interval.
3. Find the greatest or least of the function values found.

# 5 Differentiating twice

## 5.1 Second derivative

The derivative of a function is itself a function, sometimes called the derived function. Meaning it can itself be differentiable. The second derivative (or second derived function) is the function that is obtained by differentiating a function twice.

In Lagrange notation, the second derivative of a function $f$ is denoted by $f''$.

In Leibniz notation, the second derivative of $y$ with respect to $x$ is denoted by: $$\frac{d^2y}{dx^2}$$ With following reasoning: The second derivative can be written as $$\frac{d}{dx}(\frac{d}{dx}y)$$ Which was historically abbreviated as $$(\frac{d}{dx})^2y~,\text{and then to}~~ \frac{d^2y}{dx^2}$$

The domain of the second derivative of a function consists of all the values of $x$ at which its first derivative is differentiable. It's said that the original function is twice-differentiable at such values of $x$.

Third, forth, Nth derivative can be taken in same way.

The derivative of a function is sometimes called its first derivative. Hence the word 'first' in 'first derivative test'.

Derivative of a function tells the gradient of the graph of the function. So the second derivative tells the gradient of the graph of the first derivative. Hence the second derivative also gives an information about the shape of the graph of the original function.

E.g. suppose an interval on which the second derivative of a particular function is positive. By the increasing/decreasing criterion, this means that the first derivative is increasing on that interval. In other words, the gradient of the graph of the original function is increasing on that interval.

Meaning that the shape of the graph of the original function on the interval must be something like:

All of these are sections of the graphs with increasing gradient.

A section of a graph shaped like one of the diagrams above is said to be concave up. A graph is concave up on an interval if the tangents to the graph on that interval lie below the graph:

When there is an interval on which the second derivative of a particular function is negative then, by the increasing/decreasing criterion, the first derivative is decreasing on that interval. Meaning the shape of the graph of the original function on the interval must be something like:

A section of a graph shaped like this is said to be concave down. A graph is concave down on an interval if the tangents to the graph on that interval lie above the graph:

Sometimes, the words convex and concave are used instead of concave up and concave down, respectively.

A point where a graph changes from concave up to concave down or vice versa is called a point of inflection. E.g. $f(x) = x^3$ has a point of inflection at the origin:

If the gradient of a graph at a point of inflection is zero, then the point is a horizontal point of inflection. Otherwise the point is a slant point of inflection.

### Concave up/concave down criterion

If $f''(x)$ is positive for all $x$ in an interval $I$, then $f$ is concave up on $I$. If $f''(x)$ is negative for all $x$ in an interval $I$, then $f$ is concave down on $I$.

### Second derivative test (for determining the nature of a stationary point)

If, at a stationary point of a function, the value of the second derivative of the function is

• negative, then the stationary point is a local maximum
• positive, then the stationary point is a local minimum

When the value of the second derivative at a stationary point is zero then I can't use the second derivative test to determine the nature of the stationary point.

## 5.2 Rates of change of rates of change

### Relationship between displacement, velocity and acceleration

Suppose that an object is moving in a straight line. If $t$ is the time that has elapsed since some chosen point in time, and $s$, $v$ and $a$ are the displacement, velocity and acceleration of the object, respectively, then $$v = \frac{\text{d}s}{\text{d}t}, ~~ a = \frac{\text{d}v}{\text{d}t}, ~~ a = \frac{\text{d}^2s}{\text{d}t^2}$$


text:
# Must the Minkowski sum of a Borel set and a *closed* ball be Borel?

Let $A$ be a Borel set in $\mathbb{R}^n$. Must then $A + B(0,1)$ be Borel? Here $B(0,1)$ is the closed ball centered at $0$ of radius $1$.

I know that Erdos and Stone gave an example of a compact set (it is Cantor) and a $G_\delta$-set, whose Minkowski's sum is not Borel. But can we have an example with one of them being a closed ball?

-
Also at MathOverflow –  David Moews Feb 9 at 8:37
Could you give the reference for the Erdos-Stone result? –  Nate Eldredge Feb 9 at 13:30
@NateEldredge: P. Erdős and A. H. Stone, On the sum of two Borel sets, Proc. Amer. Math. Soc. 25 (1970), 304-306. –  Martin Feb 9 at 14:34
@AndresCaicedo Thanks for the link. The answer by Tapio Rajala applies to $n\ge 3$ only. What if $n=2$? (The case $n=1$ has an easy affirmative answer). –  user53153 Feb 9 at 19:27