text:
# Scipy Two-point Boundary value Problem

Nonlinear ODE Statement I would like to use scipy to solve the following:

u'' + (u')^2 = sin(x)
u(0)=0,
u(1)=1

where u = u(x).

Approach I am looking at the following link

https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.integrate.solve_bvp.html

Problem - Not sure about variable wrt x or How to input BC's

The example given in the link does not include a term wrt x, in particular not sure how to include

sin(x)

from the right hand side of my problem, nor the

u(1)=1

Work Done

In the following I try to modify the example from the above link to conform to my ODE above

   from scipy.integrate import *
import numpy as np

from numpy import *
def fun(x, y):
return np.vstack((y[1],np.sin(x) -(y[0])**2))
def bc(ya, yb):
return np.array([ya[0], yb[0]-1])
x = np.linspace(0, 1, 5)
y_a = np.zeros((2, x.size))
y_b = np.zeros((2, x.size))
y_b[0] = 3

from scipy.integrate import solve_bvp
res_a = solve_bvp(fun, bc, x, y_a)
res_b = solve_bvp(fun, bc, x, y_b)
x_plot = np.linspace(0, 1, 100)
y_plot_a = res_a.sol(x_plot)[0]
y_plot_b = res_b.sol(x_plot)[0]
import matplotlib.pyplot as plt
plt.plot(x_plot, y_plot_a, label='y_a')
plt.plot(x_plot, y_plot_b, label='y_b')
plt.legend()
plt.xlabel("x")
plt.ylabel("y")
plt.show()

• Could you add what you actually find questionable about your result? Apart from not using the correct DE as per answer. You also should check res.success or print res.message, as it is not unlikely that something goes wrong with the convergence of the BVP solver. Also explore the tol parameter, the default is rather high. – LutzL Jan 6 at 23:14

I think in NumPy, if y is a matrix, y[0] gets you the 0-th row, but it usually makes the code clearer if you use y[0,:].
The primary problem I see is that in fun you're defining the ODE $$u''+u^2=\sin x,$$ using y[0,:] instead of y[1,:].


text:
How to effectively solve One dimensional Motion Problem

This tour is given to give the feel of the whole chapters plus the type of questions and ways to tackle One-dimensional motion/linear motion problems. This is quite beneficial for anybody studying One-dimensional Motion.

Description:
One dimensional Motion is the study of the motion along a straight line. Complete Study material has been provided at the below link
motion in one dimension

Most Important Points to remember about One dimension Motion
1. Distance and displacement are not the same things. Distance is scalar while displacement is a vector quantity. Distance is never zero in a round trip while the displacement will be zero. Example When a person moves in a circle and return to it normal position, its displacement is zero while the distance traveled is the circumference of the circle
2. Speed is calculated over distance while velocity over displacement. So Average speed in a round trip will never be zero while average velocity will be zero
3. The magnitude of Instantaneous velocity is equal to the instantaneous speed
4. When a particle moves with constant velocity, its average velocity, its instantaneous velocity and its speed are all equal
5. Acceleration is defined as the change in velocity per unit time. A body moving with constant speed but with varying direction will have acceleration as the velocity is changing.

Most Effective way to solve a one-dimensional motion/linear motion  problems

1. First, visualize the question
2. Take starting point as origin as and take one direction as positive and others as negative. This is required as we will be dealing with Vector quantities
3. Write down what is given in the question and what is required
4. If it is uniform motion, then you can utilize the One-dimensional motion equation. It the motion is varying and the relation is given for that motion. Then we can utilize one-dimensional motion derivative equation to find out the solutions
5. Calculate as require

Formula’s of linear motion problems are
v=u+at
s=ut+(1/2)at2
v2=u2+2as
$x=\int v dt$
$v=\int a dt$
v=(dx/dt)
a=dv/dt=(d2x/dt2)
a=vdv/dx

Example -1
A bus starts at Station A from rest with uniform acceleration 2m/sec2. The bus moves along a straight line
1. Find the distance moved by bus in 10 sec?
2. At what time, its velocity becomes 20m/sec?
3 How much time it will take to cover a distance of 1.6km

Solution:
Now first step to attempt such question is to visualize the whole process. Here the bus is moving along a straight line and with uniform acceleration
Now what we have
Initial velocity=0
Acceleration=2m/sec2

Now since it is uniform motion we can use given motion formula’s in use
v=u+at
s=ut+(1/2)at2
v2=u2+2as

1. distance (s)=?
time(t)=10 sec

So here the most suitable equation is
s=ut+(1/2)at2
Substituting given values
s=(0)10 +(1/2)2(10)2=100 m

1. velocity(v)=20 m/s
time(t)=?
So here the most suitable equation is
v=u+at
20=0+2t
or t=10 sec

3.distance(s)=1.6km=1600m
t=?
So here the most suitable equation is
s=ut+(1/2)at2
1600=(1/2)(2)t2
or t=40 sec

Example -2
A object is moving along an straight line.The motion of that object is described by
x=at+bt2+ct3
where a,b,c are constants and x is in meters and t is in sec.
1. Find the displacement at t=1 sec
2. Find the velocity at t=0 and t=1 sec
3. Find the acceleration at t=0 and t=1 sec

Solution:
Now first step to attempt such question is to visualize the whole process.Here the object is moving along a straight line and its motion is described by the given equation
Now we
x=at+bt2+ct3
Now since its motion is described by the given equation,following formula will be useful in determining the values
$x=\int v dt$
$v=\int a dt$
v=(dx/dt)
a=dv/dt=(d2x/dt2)
a=vdv/dx

1. x=? t=1sec

Here by substituting t=1 in given equation we get the answer
x=a(1)+b(1)2+c(1)3
x=a+b+c m

2.v=? t=0,v=? t=1
Here we are having the displacement equation,so first we need to find out the velocity equation
So here the most suitable formula is
v=(dx/dt)
or v=d(at+bt2+ct3)/dt
or v=a+2bt+3ct2
Substituing t=0 we get
v=a m/s
Substituing t=1 we get
v=a+2b+3c m/s

3 a=? t=0,a=? t=1
Now we are having the velocity equation,we need to first find the acceleration equation.
So here the most suitable formula is
a=(dv/dt)
or a=d(a+2bt+3ct2)/dt
a=2b+6ct
Substituting t=0 we get
a=2b m/s2
Substituting t=1 we get
a=2b+6c m/s2

Some other  good questions for your Assignments

a) The displacement of the body x(in meters) varies with time t (in sec) as
x=(-2/3)t2 +16t+2
find following
a. what is the velocity at t=0,t=1
b. what is the acceleration at t=0
c. what is the displacement at t=0
d .what will the displacement when it comes to rest
e .How much time it takes to come to rest.

b). A man runs at a speed at 4 m/s to overtake a standing bus. When he is 6 m behind the door at t=0, the bus moves forward and continues with a constant acceleration of 1.2 m/s2
find the following
a. how long does it take for the man to gain the door
b if in the beginning he is 10m behind the door, will be running at the same speed ever catch up the bus?

Important Graph in One-dimensional Motion

A) Position Time Graph

This is a  plot of Position and time of an object moving along the straight line. We can find out velocity, speed, distance, and displacement.  We can determine acceleration sign from it but cannot determine the value of it

i) Velocity is given by the Slope of the Position time Graph. Velocity is positive if the slope is positive  and Velocity is negative if the slope is negative

ii) The numerical value of the Slope without sign gives the speed of the object

iii) Displacement can be calculated using position at the time interval

iv) Displacement-time graph is same as Position Time graph.Distance time can be drawn from position-time graph by the taking the mirror of the image of the position-time graph from the point zero velocity onwards

iv) The slope of the chord drawn to displacement time or distance-time graph gives the average velocity or average speed  over the time interval to which the chords corresponds to

v) If the graph is concave up, acceleration is + positive and If the graph is concave down, acceleration is -ve

B) Velocity Time graph

i) The area under the velocity-time curve provides the displacement

ii) The slope of the curve provides the acceleration

C) Acceleration -time Curve

i) The area under the curve provides the change in velocity

Related Articles

Motion in straight line practice problems for class 11

how to study physics problems

Solving physics problems

Describing Motion and Rest

Kinematics Good conceptual problems

Acceleration formula Explained with Examples

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## Saturday, 12 February 2022

### Completely Incomplete

If there's one thing above all others I've learned over decades of lay study of science, it's that the universe really doesn't do boxes. Chimps in couture love 'em, though. We can't help ourselves. We go blithely about sticking things in containers, impressing ourselves immensely with our ability to classify, constrain and label.

To be fair, they can be incredibly useful. A good box with a well-defined boundary can... But hang on, maybe I need to talk about what I mean by 'boxes'.

A box is a boundary. A binary; in/out; yes/no; 1/0. It's a definition. In fact, 'definition' is pretty much the entirety of what I'm trying to convey here, and I could just as easily write the same post using that term, but picture a physical box anyway, as thinking of it as a constraint will have some value

In the jargon of the mathematician, we can most closely relate it to 'set' while, in the jargon of the physicist, we'd more closely relate it to 'system'. In biology, the nearest term is 'clade' (any species and all its progeny). In reality, what they are is patterns we perceive and the stories we tell ourselves about them. Any time you see an obviously important noun in what follows, think 'box'.

This suite of pattern-seeking and categorisation skills is, as with many of our abilities, traceable to our deep evolutionary history, and is a critical component of our survival strategy as a species. The ability to spot a pattern and stick it into a box deprived more than one predator of an entrée, and is the essential foundation of the modern world. They're key to developing good intuitions, allowing us to categorise variables and garner expertise*. We have to be a little careful though, because these boxes don't only constrain the things we put in them, they also massively constrain the way we think about the things we've put in them.

Still, there's one box of particular interest, and I'm going to label it 'what I know'. I'm not referring here to my specific example, this is a Platonic 'what I know' box, the essential 'what I know', of which each of us has their own perception. This box is, for each the wellspring from which our intuitions derive, and it's an incredible tool, but there are some problems. The first is a problem in specifying boxes. Let's start with a story:

Pierre-Simon de Laplace was a French polymath - some cite him as a prime candidate for 'the last man who knew everything' - whose contributions to the sum of human knowledge were vast and broad-reaching, spanning mathematics, physics, engineering, probability and statistics... he was the Dave Grohl of modern science.

He also famously said something we would recognise as our classical picture of how the world works. In 1814 after reading Newton's masterwork, he remarked if one could grasp the positions and momenta of all the particles in the universe, coupled with knowledge of all the forces and the wit to analyse them robustly, one could "embrace in a single formula the movements of the greatest bodies of the universe and those of the tiniest atom; for such an intellect nothing would be uncertain and the future just like the past would be present before its eyes."[1]

This was probably the first properly scientific expression of what we'd now recognise as 'causal determinism'. There's a problem lurking in there, though (there are many, but this is the most relevant for our purpose here), and it's all to do with specifying a box, this one labelled 'system'.

In classical mechanics, the configuration of a system is completely specified with two pieces of information - position and momentum ($x$ and $p$ respectively)*. If you know these two quantities, you can wind forward and backward in time and make accurate predictions about everywhere it's been and everywhere it will go and when. This is extremely intuitive, which is hardly surprising because, miraculous as it is, the macroscopic world behaves very differently to the real world, to such a degree as to have taken several colossal revolutions in thought merely to furnish us with the tools required to notice the real world was there!

In quantum mechanics, it isn't possible to completely specify a system, if we're operating on this definition of 'specification'. Why? Because we can't know both the position and momentum of a particle at the same time. Know, you say? In fact, it doesn't even make sense to talk about them existing at the same time.

There's a wonderful analogy, originally formulated by Bertrand Russell for a completely unrelated purpose, which I decided to modify slightly here for reasons I hope will become apparent later, as I begin to tie the disparate threads of this post together.

You can liken the differences between classical mechanics and quantum mechanics to the difference between shoes and socks. We'll take an observable quantity - angular momentum about the $x$ axis - and another - angular momentum about the $y$ axis. We'll call these quantities $l$ and $r$ respectively.

Look at a pair of shoes. It's pretty simple to label them $l$ and $r$. They have intrinsic $l$ and $r$ -ness. Their 'chirality' (handedness) is an objective fact about them. One fits the left foot, the other fits the right foot.

What about socks? They're a little trickier. They have no intrinsic chirality. In fact, it doesn't make any sense to talk about their chirality until an observation is made. In physics, it means we need an interaction. Of course, we have very specific ways to interact with socks, and their chirality comes about only as a result of these interactions. The interaction changes them and, once the interaction is complete, they lose their chirality and go back to being non-quantities. This is directly analogous to measuring the position of a 'particle'. As soon as your 'measurement' is complete, you have no idea of where it is or how fast it's moving. Of course, if you have sweaty feet and don't change your socks often, they lose elasticity and fail to return to shape, so the analogy falls apart, much like your socks.

Yes, yes. I know. I always stretch analogies too far. This is, however, how our $x$ and $p$ provably and measurably behave in the real world**. Quantum entities behave far more like socks than they do shoes. One could argue, in fact, shoeness is an emergent property arising from the behaviour of socks. I won't go further here, except to note the existence of better analogies, some of which I've employed in more complete posts about quantum theory. Links in the usual place.

Suffice it to say for succinctness quantum mechanics requires a different definition for 'specification'. We have to modify our expectations of what we can 'know'. Since we can't specify $x$ and $p$, we have to find something else to work with. I'm not going to discuss the wavefunction ($\Psi$) here, not least because I have plans to deal with it in some considerable detail in future posts. What matters is having a way to completely specify a system so it can be evolved in time and space in a way which allows us to make predictions.

"But this is all physics, what about mathematics?" say you. Damn, I was hoping you wouldn't notice. Ah, well. Here goes, then.

What the above should demonstrate reasonably well is this; assumptions are dangerous to the progress of knowledge. One might argue, in fact, the most formidable barrier to knowledge is the certainty we already possess it. This is what our assumptions really represent unless we learn to treat them with care. Not for nothing is the central aim of philosophy 'how to identify presupposition failure', as defined by one professor of philosophy of my acquaintance.

My friend Phil Scott, mathematician and computer scientist, wrote a wonderful piece here about the foundations of mathematics as they shifted from geometry to set theory. He tells us:

An exodus from geometry occurred in the 18th century, and was completed when the geometric foundation itself fell into total crisis. Remember the four axioms from Euclid I gave above? I missed out the fifth, not simply because it is an overly complex axiom, but because its status as axiom was in question from the early commentaries of Euclid. It is an axiom governing parallel lines, but perhaps the most familiar of its consequences is that the angles of a triangle add up to 180 degrees.
With the axiom’s status in doubt, several mathematicians tried to show its necessity by showing that its denial would be commitment to absurdity. In the late 18th century, they began exploring strange and speculative worlds in which the angles of a triangle do not sum to 180 degrees.

And, as a result...

The solid foundations of geometry were replaced with a fluid space of infinite geometries, a shifting sand that was no place to build the rest of mathematics.

The point is, absent Euclid's axioms, mathematics had no axiomatic foundation. This is a bit of a problem for an axiomatic-deductive system of logic, as it can't be difficult to appreciate. Enter set theory.

Of course, there's more to the story than a simple cape-enshrouded leap onto the stage to save the day, as this glib comment might imply. I recommend reading Phil's delightful piece, as it provides some critical context and is a thing of profound beauty.

In a nutshell, set theory is the idea things can be gathered into groupings called 'sets' based on shared characteristics. It's a source of some amusement to me the word 'set' has the largest set of discrete definitions of any word in the English language, at something like 430 entirely different meanings. You might say the set S of all statements defining the word 'set' has 430 members (and none of them pay the greens fees, the bastards...)

Set theory owes much of its existence to Georg Cantor. He was wrestling with the infinite and, in particular, the notion there could be different "sizes" of infinity. This might seem odd, but it's not incredibly difficult to see with an example. Take two sets of numbers, $\mathbb{Z}$ and $\mathbb{R}$. I can tell you all sorts of things about them but the most interesting is how large they are. I chose these particular terms for a reason; they're conventionally used to define two sets of numbers.

$\mathbb{Z}$ is the integers. The integers are all the whole numbers from minus infinity to infinity ($-\infty,\infty$). it's conventional at this point in any explanation to show a plot on a line to make this intuitive, so let's do that. Here's a snippet of the number line for set $\mathbb{Z}$:

No real surprises to be had there. Now let's look at an example of the set $\mathbb{R}$:

As we should be able to see, the set $\mathbb{R}$ contains not only all the integers, but also all the numbers between them. It contains all the rational numbers (numbers expressible as the ratio between two numbers, a.k.a. fractions), all the irrational numbers (numbers not expressible as the ratio between two numbers. Pi ($\pi$) is an example, but any number whose decimal expansion is infinite is irrational). In fact, for any two lines you choose at random, there is an infinite number of numbers between them. $\mathbb{R}$ is the real numbers.

To a non-mathematician, the idea of different-sized infinities sounds insane, even after more than a century of trans-finite mathematics, but just these two core examples show it must be true. It's intuitively obvious, even to a non-mathematician, the sets $\mathbb{Z}$ and $\mathbb{R}$ are not the same size, despite both being infinite. The former set is quite obviously a subset of the latter, so they can't possibly be the same size.

This was a problem, and we needed a way to treat it mathematically. Cantor had a solution, though. He introduced the notion of 'cardinality of infinites'. As the name might suggest, what he introduced was a way of counting with infinities of different sizes. The cardinal numbers are the 'counting' numbers, non-negative integers, or 'natural numbers' ($\mathbb{N}$), to use the proper term.

The cardinality of a set is simply the number of members. This is defined in terms of bijection - essentially direct comparison to other sets. In the jargon, two sets have the same cardinality if there is a direct one-to-one mapping, one to the other. For example, the little pigs in a popular children's story and the blind mice in a nursery rhyme have the same cardinality; 3. There's a direct bijection between the mice and the pigs in these sets.

Another important term to introduce is 'countable'. This is a source of some confusion, because infinities are uncountable under any vernacular definition of 'countable'. The meaning here is very specific, though. An infinity is countable if and only if there is direct one-to-one mapping with the natural numbers. It's an in-principle countability, rather than a practical one; a countability of orderliness, if you like.

The cardinality of the set $\mathbb{N}$ of all natural numbers is denoted Aleph-nought ($\aleph_0$). This is the smallest infinite cardinality. The set $\mathbb{Z}$ is also $\aleph_0$, as it has a direct one-to-one mapping with $\mathbb{N}$. This might seem counter-intuitive, but Cantor proved the cardinality of any unbounded subset of $\mathbb{N}$ is the same as the cardinality of $\mathbb{N}$ itself.

What about $\mathbb{R}$? You might well ask...

Unsurprisingly, there is no one-to-one mapping between $\mathbb{R}$ and $\mathbb{N}$. $\mathbb{R}$ has a larger cardinality, $\aleph_1$. In fact, the cardinality of the reals has a special place and a special designation $\mathfrak{c}$. Its calculated cardinality is in fact $2^{\aleph_0}$ but, due to a hypothesis forwarded by Cantor, this is taken as equivalent to $\aleph_1$. This 'continuum hypothesis' is as follows: there is no set whose cardinality lies between that of the natural numbers and that of the reals. This hypothesis hasn't been proven, but it has some important consequences. It was the first of the problems forwarded by David Hilbert when laying out his vision of how mathematics should progress in the coming century. Hilbert's vision will have some import, so we'll come back to it shortly.

So we have our arithmetic foundation, and all seems well with the world, but there's a problem lurking in there. It's not immediately obvious, but it's a biggie, and it has a lot to do with the fact, other than some bits of additional notation, this set theory is constructed in natural language. Regulars here will already be hearing cognitive klaxons sounding, because we've looked in excruciating detail at the pitfalls of natural language, where misunderstandings can arise, where ambiguity creeps in...

By Mad Hatter, of course, I mean the great Bertrand Russell, upon whom the illustration of the Mad Hatter in Carroll's early editions is allegedly (and credibly) based†.

What if, he reasoned, you constructed a set of all sets not members of themselves? This set would have to be a member, but then it would be a member of itself, so it couldn't be, which meant it had to be included... and so on. Is your head exploding yet?

Well, this little paradox comprehensively exploded set theory as it stood, leaving arithmetic once again without a foundation. We needed something consistent - free of contradictions. We needed something complete.

This was the meat of the second of Hilbert's problems, namely to prove the axioms of arithmetic consistent.

The obvious route to progress was to reconstruct set theory. It had many incredibly useful features. We just had to be careful about our definitions to remain free of paradox. One way to do this is to formalise it, to remove the potential ambiguities of natural language. Russell did some work on this himself, founding 'type' theory, still used in some settings today (such as Church's $\lambda$ calculus) Ernst Zermelo constructed a system which, once some inconsistencies were ironed out with input from Abraham Fraenkel, is now pretty much universally recognised as the foundation of arithmetic.

We're not entirely out of the woods yet, though. There are still some potential issues lurking, even with Zermelo-Fraenkel (ZF) theorem. First, there's an axiom associated with ZF theorem historically seen as controversial. It's known as the 'axiom of choice' (AC), and it was constructed by Zermelo in his proof of his 'well-ordering' theorem, the idea every set can be well-ordered (in context, this is in fact logically equivalent to AC). In a nutshell, it says it's possible to select at least one item from a collection of boxes each containing at least one item, even if the number of boxes is infinite. For the most part, this is fairly uncontroversial.

We can talk again about shoes and socks here and, in fact, this is the original setting of this analogy. If your boxes contain shoes, the assertion some rule can be applied to selecting one from each box is trivial, to the extent one would think having an axiom saying you can make a selection might seem a bit silly. But what about socks?

Well, with socks, we don't actually have any means of distinguishing between them. Here's where the axiom of choice comes into play. In essence, the axiom of choice simply asserts the existence of some applicable criterion, regardless of whether we actually have access to any such criterion.

Extending this a little to the notion of cardinality, think about how you'd go about ordering the set $\mathbb{R}$. Bearing in mind it includes all decimal expansions, all fractions and every possible number on the number line, and between any two numbers there is an infinite number of other numbers, and you begin to see why imposing the axiom of choice is useful, and also its relationship to the well-ordering theorem. This is also why the set $\mathbb{R}$ is uncountable and, in fact, what it means for an infinite set to be uncountable.

You can see why this axiom might make some mathematicians uneasy. At this point, it's unclear whether AC is entirely free of paradox, but ZFC (Zermelo-Fraenkel theorem with the axiom of choice) is widely accepted as the foundation of arithmetic. We can breathe easy.

Or can we? let's look again at what we needed...

We needed something consistent - free of contradictions. We needed something complete.

So, is ZFC consistent? It appears to be. Is it complete? What do we mean by complete anyway?

Here's a definition: A set of axioms is syntactically complete if any statement rendered in the language of the set can be proved or disproved with the axioms of the set alone.

Let's take an example we've met before. Euclidean geometry.

We can think of Euclid's postulates as a set of axioms.

1. To draw a straight line from any point to any point.

2. To produce a finite straight line continuously in a straight line.

3. To describe a circle with any centre and distance.

4. That all right angles are equal to one another.

5. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

We know the fifth postulate to be problematic. We can construct - and demonstrate the existence in the real world of - geometries in which this postulate simply doesn't hold; non-Euclidean geometries, in other words. General Relativity, for example, describes just such a geometry for spacetime. We can traverse a right-angled triangle whose internal angles do not sum to 180 degrees; whose internal angles are, in fact, all right. Start at the North Pole, walk one mile South, turn 90 degrees and walk one mile East, then turn North and walk for one mile. You've just traversed a right-angled triangle with three right angles and three identical hypotenuses††. Eat that, Pythagoras!

Of course, if we remove the fifth postulate, Euclid's axioms are incomplete, because the fifth postulate is a statement expressed in the language of the set which can neither be proved nor disproved from the remaining axioms alone.

So what about ZFC? Are there any statements expressed in its language which can't be proved or disproved from the axioms of ZFC alone?

There are, and we've already met one; the continuum hypothesis. In fact, the proof the continuum hypothesis could be neither proven nor disproven with the axioms of ZFC came in two stages. The proof it couldn't be proven came in 1963 from Paul Cohen, who further proved both the continuum hypothesis and the axiom of choice were in fact independent of ZF theory. The proof it couldn't be disproven (again, along with the axiom of choice) came earlier in 1940 from somebody who I'm gently wending my way towards, because he's going to drop a bomb any minute.

So Russell cracked on with type theory, and set about resolving Hilbert's second problem. He constructed, along with Alfred North Whitehead, a three volume work, the Principia Mathematica, a complete, ground-up complete and consistent axiomatisation and formalisation of mathematics, published between 1910 and 1913. At least, that's what it said on the tin (maybe; we'll come back to this). There was a villain lurking in the wings, and his name was uncertainty.

What we've encountered above are iterations of Gödel's First Incompleteness Theorem.

Properly stated:

If a (logical or axiomatic formal) system is omega-consistent, it cannot be syntactically complete.

We have to be careful here, because Gödel's Incompleteness Theorems are the subject of an awful lot of bullshit. Indeed, my first foray into learning about this fascinating area of thought stemmed from Phil Scott's gentle correction of my misunderstanding of it. The most important factor to be kept front and centre is domain of applicability and, in Gödel's theorems, it is tightly defined. All else aside, omega-consistency is a property only applicable to statements about natural numbers.

Gödel's Second Incompleteness Theorem is an extension of the first, and it states:

The consistency of axioms cannot be proved within their own system.

In other words, there is a fundamental barrier to the existence of a complete and consistent set of axioms for mathematics. We can now imagine exhausting the limits of human imagination, yet we're prohibited from ever standing in a place from which we can say our mathematics is complete and consistent, because there are limits to where we can stand - without even considering the notion of whether such a place might be one of  pair of conjugate variables, meaning to even speak of its existence makes no sense.

I'm reasonably confident there are things I've gotten wrong here. All else aside, the domain of applicability of Gödel's Incompleteness Theorems is very narrow, and it's important not to apply this too broadly. Moreover, some of what I said about Russell and Whitehead quite possibly doesn't withstand rigorous scrutiny. Some mathematicians argue compellingly at least part of the motivation in constructing Principia was to show the expectation of completeness and consistency unreasonable. I've certainly been very suggestive in my use of language, and led you to a way of thinking about some boxes, and it's motivated by what comes next, as we go back to the very beginning. I want to look more closely at the box we labelled 'what I know', and cast an eye over it from the perspective of the one category Rumsfeld didn't invoke in his exposition of the fundamental tautologies of epistemology in the header; unknown knowns.

OK, that's going to cause some mental backflips, so let me add some stability. We each have this box we label 'what I know'. It's the place from which all our intuitions derive. As we develop expertise in a given domain, every new variable we identify and place in this box, and every solution to those variables we derive, all fuel our intuition and drive us toward better, more considered conclusions. This is, in fact, what it means to have expertise; to understand the variables in a given domain to a high degree. This box drives our thinking enormously.

But then we interface with the boxes of other people, people who... know more. This is what I'm referring to when I talk about unknown knowns. The things we know that I don't know. The higher degree of understanding the variables. And here's the problem, of course, because my 'what I know' box for any given domain undoubtedly contains inferences in it which, because I lack some understanding of critical variables, seem perfectly consistent to me, because I literally can't think outside the box.

To invoke a favourite example, our collective 'what I know' box contained conclusions and inferences which stood the test of time, right until they didn't. Newton's Law of Universal Gravitation was the final word on gravity - even knowing the precession of Mercury's perihelion didn't match predictions - right until we discovered it was wrong. The discrepancy was tiny, amounting to only 43 seconds of arc per century, and many hypotheses were erected to explain it (this is the source of the planet Vulcan, in a sense). Einstein eventually showed it was just wrong, a reasonable approximation in a limited domain, but what did he really do? He took Newton's 'what I know' box and picked it apart. He identified two things which, although absolute and immutable in Newton's schema and fundamental to the propagation of gravity, were in fact variables, and interdependent variables at that. He showed Newton's box was inconsistent, and the inconsistency arose from not properly treating critical variables, confounders with a demonstrable impact on observables. He brought new expertise to the table, along with better intuitions and a considerably larger collective box.

And there's a lesson in thought here for us all. Sometimes, in order to understand the inconsistency in the conclusions you draw, you have to be able to think outside the box. All our conclusions seem perfectly consistent to us at all times. It could hardly be otherwise since, if we could spot the inconsistency, we wouldn't draw the conclusions. Sometimes, we have to acknowledge the unknown knowns, and recognise others have greater understanding of the variables; greater expertise; a bigger box. It should also pose a question we should always be prepared to ask ourselves when our intuition tells us something the experts in a given field accept as true is wrong; what variables are they aware of that I'm not? What do they know that I don't?

But we know that, right? Completely.

[1] A Philosophical Essay on Probabilities - Laplace 1814

Gödel's Theorem: An Incomplete Guide to Its Use and Abuse - Torkel Franzen. This is a truly stupendous book, a reasonably comprehensive exposition of Gödel's Incompleteness Theorems that deconstructs a lot of the nonsense. I can't recommend this highly enough.

Definitions and Axioms - A wonderful potted history of mathematics by Phil Scott.

Very Able - Why expertise is fundamentally about recognising and understanding variables. A fun story from the history of astrophysics.

Where Do You Draw the Line - How expertise changes your perception and improves intuition. A thought experiment.

Well, Blow Me Down! - When even really well-developed intuitions fail.

The Certainty of Uncertainty - Heisenberg's Uncertainty Principle and its implications.

Give Us A Wave! - Waves, waves everywhere. Waves in quantum mechanics, light and sound.

Did You See That? - Consciousness and the observer effect - socks and shoes in the real world. Why observation is about interaction, not consciousness.

The Map is not the Terrain - Semantics and the pitfalls of natural language.

*Momentum is a vector quantity, which means it contains information about both magnitude and direction. Properly, it's a composite quantity, the product of mass and velocity ($p=mv$).

**And not just particles, either. Quantum effects have been measured in very large objects, including a 20 kilo mirror in a hypersensitive pendulum experiment.

†Russell and Lewis Carroll were friends, both mathematicians and logicians. It is even plausible to suggest Russell's Paradox is one of the factors motivating Carroll's absurdist fiction, as a means of railing against absurdities arising in mathematical logic.

††In fact, this isn't true of Earth's surface because of a complication arising from physics. Because of the interplay of forces (pseudoforces, more accurately) the Earth is slightly oblate; it's wider around the equator than it is around the poles. Because of this, lines of latitude are longer at higher latitudes, thus the two sides of such a right triangle running North and South would actually be slightly longer than the East-West line. The right-angled equilateral describe here only applies on a perfect sphere.


text:
Re: Three questions

RE "subclassing" records, I have found a way to do it which doesn't cause TLC to enumerate any sets. The approach has some downsides, but it's nice enough to be useful, so I'm posting it here:

Is(r, types) ==

\A type \in types :

/\ DOMAIN type \subseteq DOMAIN r

/\ \A f \in DOMAIN type: r[f] \in type[f]

(other variations are possible, such as taking a set of "types" rather than one)

And then:

id     |-> Nat]

[data |-> A]}

TypeOK == Is(data, DataT)

The downside is that it requires abandoning the [f1 : s1, f2 : s2] syntax, and that handling sequences requires another definition.

Ron

On Wednesday, October 28, 2015 at 1:33:30 PM UTC+2, Ron Pressler wrote:
Thanks, Stephan!

2. Right... So simple... <facepalm>

1 + 3: Sure, I was just wondering if there perhaps was a better way I missed. Future improvements, perhaps?

Ron

On Wednesday, October 28, 2015 at 1:14:32 PM UTC+2, Ron Pressler wrote:
Hi.

1. "Subclassing"/composing records

Is there a way to represent record composition (or subclassing) -- e.g. to have a header reused by several message record "types" (i.e. sets of the form [field1 : ..., field2: ...]) -- that's efficient in TLC? I've tried:

LOCAL Fields(type) == DOMAIN CHOOSE x \in type : TRUE

Project(r, type) == [f \in Fields(type) |-> r[f]]

IsInstance(r, types) == \A type \in types : Project(r, type) \in type

\* Same as TLC's @@ operator

Build(r1, r2) == [f \in DOMAIN r1 \cup DOMAIN r2 |-> IF f \in DOMAIN r2 THEN r2[f] ELSE r1[f]]

But aside from Build (TLC's @@), everything else requires an element in the set to get the domain (found by CHOOSE : TRUE), and performing CHOOSE requires enumerating the set, which is something I'd like to avoid.

2. Transition (action) invariants

TLC allows checking invariants in the form of state predicates, or liveness properties in the form of temporal formulas that may specify actions or subactions (e.g. <><A>_var). Can TLC check "action invariants", e.g.  \A x \in {1,2,3} : ~A(k)   ? My intention is to check whether the action (or subaction) A(x) for those particular values of x is ever taken. I don't mean checking if A(x) is ENABLED, but actually taken (i.e., the primed variables are taken into account)

3. Can the _expression_ { <<x, y>> \in { <<x, y>> : x \in 1..3, y \in 2..7 } : y <= x } be written more concisely? I've tried { <<x, y>>  : x \in 1..3, y \in 2..7 : y <= x }, but that doesn't work.

Ron


text:
Discussion in 'Homework Help' started by JimmyB, Jun 19, 2010.

1. ### JimmyB Thread Starter Member

Jun 1, 2010
38
0
Hi is anyone able to help me with this question??

draw to scale 3 phasors representing the following voltages, taking e1 as reference phasor

e1:= 80 sin ωt volts

e2= 60 cos ωt volts

e3 = 100 sin (ωt-∏/3) volts

I think i understand this part, is it 1 sin wave and 1 cosine wave and 1 lagging sin wave on one 2∏ cycle chart,

the part im not sure is it then asks to sum these 3 voltages and express it in the form Em sin (ωt ± θ), i think i can sum the 2 sin waves though i'm unsure, it is the cosine i'm unsure with...

is anyone able to help please

jim

Last edited: Jun 20, 2010
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Consider the phasor e2 cosine as leading the reference e1 sine phasor by 90° [∏/2 radians]. You've typed e2 twice but I assume the 3rd phasor is e3. Phasor e3 lags e1 by ∏/3 [60°].

Use phasor addition to find the sum e1+e2+e3

$e_{tot}=e_{1}+e_{2}+e_{3}=80\angle 0^{o}+60\angle 90^{o}+100\angle -60^{o}$

Then convert the resultant phasor back to the time-domain equivalent.

Last edited: Jun 19, 2010
3. ### JimmyB Thread Starter Member

Jun 1, 2010
38
0
sorry it was ment to be e3.. edited now...

Thanks tnk, does that mean convert into rectangular add the votages then back into polar form..

Am I right about the drawing?? all 3 waves are drawn on one 2∏ cycle..

thanks

jim

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
If you are lucky enough to have a calculator that will do everything (including addition) in polar form then no - you don't have to convert. Otherwise you will probably have to go to rectangular form to do the addition and then back to polar.

Are you sure you know what a phasor diagram is? I'm not sure of your meaning in - "all 3 waves on a 2∏ cycle". You may be thinking of the time based waveform diagram - which is not a phasor diagram.

5. ### JimmyB Thread Starter Member

Jun 1, 2010
38
0
are Phasor diagrams straight lines?

or can I draw 2 sin waves and 1 cosine wave superimposed on one graph..

If they are staright lines do i have e1 horizontal mag 80, e2 same origin but vertical mag 60, e3 same origin mag 100 -60° from e1.. is that what is ment by phasor diagram.. or should e2 start from the end of e1, and e3 start from the end of e2.. I know that is what you should do to add the phasors..

thanks

jim

Last edited: Jun 20, 2010
6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Something like this .....

File size:
3.4 KB
Views:
37

Jun 1, 2010
38
0
Thanks tnk


text:
Dept. of Mathematics

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• (2017-12-03)
Given $\alpha>0$, the $\alpha$-Grushin plane is $\mathbb{R}^2$ equipped with the sub-Riemannian metric generated by the vector fields $X = \partial_1$ and $Y = |x_1|^{\alpha} \partial_2$. It is a standard example in ...

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In this thesis we prove the existence of Jordan Decomposition in $D_{G/k}$, the ring of invariant differential operators on a semisimple algebraic group over a field of positive characteristic, and its corollaries. In ...

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We study two different problems: generic behavior of a measure preserving transformation and extending partial isometries of a compact metric space. In Chapter $1$, we consider a result of Del Junco--Lema\'nczyk [\ref{DL_B}] ...

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The equivariant 𝔼∞G operad has the property that 𝔼∞G(n) is the total space for the G-equivariant universal principal Σn bundle. There is a forgetful functor from 𝔼∞G-algebras to 𝔼∞-algebras, where 𝔼∞ is the classic ...

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In this thesis, we study certain aspects of Levy processes and their applications. In the first part of this thesis, we study the applications of Levy processes in actuarial mathematics. Our topics are closely related to ...

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E_req {eddington} R Documentation

## Determine the number of additional rides required to achieve a specified Eddington number

### Description

Determine the number of additional rides required to achieve a specified Eddington number.

### Usage

E_req(rides, candidate)

### Arguments

 rides A vector of mileage, where each element represents a single day. candidate The Eddington number to test for.

### Value

An integer vector of length 1. Returns 0L if E is already achieved.

### See Also

E_cum, E_next, E_num, E_sat

[Package eddington version 2.1.1 Index]


text:
Lecture 18 - Numerical Differentiation

Presentation on theme: "Lecture 18 - Numerical Differentiation"— Presentation transcript:

Lecture 18 - Numerical Differentiation
CVEN 302 July 19, 2002

Lecture’s Goals Finite Difference Schemes
Taylor Series Expansion for Differentiation Basic Numerical Integration Trapezoidal Rule Simpson’s Rule Midpoint Gaussian Quadrature

Calculus - Numerical Methods
Differentiation Integration v t dv dt v t a t y

Numerical Differentiation
Estimate the derivatives (slope, curvature, etc.) of a function by using the function values at only a set of discrete points Ordinary differential equation (ODE) Partial differential equation (PDE)

Numerical Differentiation
Represent the function by Taylor polynomials or Lagrange interpolation Evaluate the derivatives of the interpolation polynomial at selected nodal points

Numerical Differentiation
A Taylor series or Lagrange interpolation of points can be used to find the derivatives. The Taylor series expansion is defined as:

First Derivative at a Point
i i i i i+2

Numerical Differentiation
Use the Taylor series expansion to represent three points about single location:

Numerical Differentiation
Assume that the data points are “equally spaced” and the equations can be written as:

Forward Differentiation
For a forward first derivative, subtract eqn[2] from eqn[1]: Rearrange the equation:

Forward Differentiation
As the Dx gets smaller the error will get smaller The error is defined as:

Backward Differentiation
Subtract eqn[3] from eqn[2]: The error is defined as:

Central Differentiation
Subtract eqn[3] from eqn[1]: The error is defined as:

Differential Error Notice that the errors of the forward and backward 1st derivative of the equations have an error of the order of O(Dx) and the central differentiation has an error of order O(Dx2). The central difference has an better accuracy and lower error that the others. This can be improved by using more terms to model the first derivative.

Numerical Differentiation
If you want to improve the accuracy and decrease the error you will need to eliminate terms : = =

Higher Order Errors in Differentiation
The terms become : The terms become A=-3, B= 4 and C=-1

Higher Order 1st Derivative
Parabolic curve i i i i i+2 Forward difference Backward difference

Higher Order Derivatives
To find higher derivatives, use the Taylor series expansions of term and eliminate the terms from the sum of equations. To improve the error in the problem add additional terms.

2nd Derivative of the Function
It will require three terms to get a central 2nd derivative of discrete set of data. = # = =

2nd Order Central Difference
The terms become : The terms become A=1,B=-2 and C=1. Therefore

Lagrange Differentiation
Another form of differentiation is to use the Lagrange interpolation between three points. The values can be determine for unevenly spaced points. Given:

Lagrange Differentiation
Differentiate the Lagrange interpolation Assume a constant spacing

Lagrange Differentiation
Differentiate the Lagrange interpolation Various locations

Lagrange Differentiation
To find a higher order derivative from the Lagrange interpolation for a three point Lagrange Take the derivative

Partial Derivatives Straightforward extension of one-dimensional formula (i, j+2) (i-1, j+1) (i, j+1) (i+1, j+1) (i-2, j) (i-1, j) (i, j) (i+1, j) (i+2, j) (i-1, j-1) (i, j-1) (i+1, j-1) (i, j-2)

(i, j+2) (i-1, j+1) (i, j+1) (i+1, j+1) (i-2, j) (i-1, j) (i, j) (i+1, j) (i+2, j) (i-1, j-1) (i, j-1) (i+1, j-1) (i, j-2)

Partial Derivatives Laplacian Operator j+1 j j-1 i-1 i i+1

Partial Derivatives Mixed Derivative j+1 j j-1 i-1 i i+1

Bi-harmonic Operator j+2 j+1 j j-1 j-2 i-2 i-1 i+1 i+2 i

Richardson Extrapolation
This technique uses the concept of variable grid sizes to reduce the error. The technique uses a simple method for eliminating the error. Consider a second order central difference technique. Write the equation in the form:

Richardson Extrapolation
The central difference can be defined as Write the equation with different grid sizes

Richardson Extrapolation
Expand the terms:

Richardson Extrapolation
Multiply eqn [2] by 4 and subtract eqn [1] from it.

Richardson Extrapolation
The equation can be rewritten as: It can be rewritten in the form

Richardson Extrapolation
The technique can be extrapolated to include the higher order error elimination by using a finer grid.

Richardson Extrapolation Example
The function is given: Find the first derivative at x=1.25 using a central difference scheme and Dh = The exact solution:

Richardson Extrapolation Example
The data points are: The derivatives using central difference

Richardson Extrapolation Example
The results of the central difference scheme are: The Richardson Extrapolation uses these results to find a better solution

Summary Finite Difference Techniques Error Calculation Taylor Series
Lagrange Polynomials Error Calculation

Summary Finite Difference Techniques Forward Difference Scheme
Backward Difference Scheme Central Difference Scheme

Homework Check the Homework webpage


text:
# Finding the Equations of a Circle provided a point and the radius

So I tried googling the exact question and I never found the solution. This is homework so I really don't want to know the answer but how to arrive at the answer.

The question that was given was "What are the equations of the circles tangent to the x-axis of radius 4 and whose center is at (-2,y)?"

From the given and a little graphing, I found that (-2,0) would be the point where the circle is tangent to the x-axis. Am I correct?

if I am correct, then I have 2 given pieces of information which is the radius of 4 and a point as well as the x-value of the center.

With all the equations my professor given me, I still don't know how to solve to get the center. Can someone point me to the right direction?

thanks.

• Hint: $(x-a)^2+(y-b)^2=r^2$ Is a circle radius $r$ centred at $(a,b)$ – Karl Feb 20 '16 at 13:56
• If the $x$ axis is a tangent then there are 2 possible circles. One of which is centred at $(-2,4)$ what is the other? – Karl Feb 20 '16 at 14:01

Let's use $(a,b)$ to denote the coordinates of the center of the circle.

The circle has radius $4$ and it's tangent to the $x$-axis, so its center must be $4$ units above or below the $x$-axis. This means that $b=4$ or $b=-4$.

You're also told (not very clearly) that the center coordinates have the form $(-2,y)$, for some $y$. This means that $a=-2$.

So, now we know that the center is at $(a,b) = (-2,4)$, or $(a,b) = (-2,-4)$, and we know the radius is $4$, so you can probably figure out the equations of the two circles.

• No problem upvoted – Karl Feb 20 '16 at 14:12

Hint:

You have correctly found the point $(-2,0)$ where the circle is tangent to the $x$ axis. Now you know the radius $4$ and you also know that the radius is orthogonal to the tangent. So, where can be the center of the circle?


text:
### Recent Submissions

• #### Inflammation and physical dysfunction: responses to moderate intensity exercise in chronic kidney disease

Background: People with chronic kidney disease (CKD) experience skeletal muscle wasting, reduced levels of physical function and performance, and chronic systemic inflammation. While it is known that a relationship exists between inflammation and muscle wasting, the association between inflammation and physical function or performance in CKD has not been well studied. Exercise has anti-inflammatory effects, but little is known regarding the effect of moderate intensity exercise. This study aimed to (i) compare systemic and intramuscular inflammation between CKD stage G3b-5 and non-CKD controls; (ii) establish whether a relationship exists between physical performance, exercise capacity and inflammation in CKD; (iii) determine changes in systemic and intramuscular inflammation following 12 weeks of exercise; and (iv) investigate whether improving inflammatory status via training contributes to improvements in physical performance and muscle mass. Methods: This is a secondary analysis of previously collected data. CKD patients stages G3b-5 (n = 84, n = 43 males) and non-CKD controls (n = 26, n = 17 males) underwent tests of physical performance, exercise capacity, muscle strength and muscle size. In addition, a subgroup of CKD participants underwent 12 weeks of exercise training, randomized to aerobic (AE, n = 21) or combined (CE, n = 20) training. Plasma and intramuscular inflammation and myostatin were measured at rest and following exercise. Results: Tumour necrosis factor-α was negatively associated with lower $^{^{^{.}}}{\rm V}$O2Peak (P = 0.01), Rectus femoris-cross sectional area (P = 0.002) and incremental shuttle walk test performance (P < 0.001). Interleukin-6 was negatively associated with sit-to-stand 60 performances (P = 0.006) and hand grip strength (P = 0.001). Unaccustomed exercise created an intramuscular inflammatory response that was attenuated following 12 weeks of training. Exercise training did not reduce systemic inflammation, but AE training did significantly reduce mature myostatin levels (P = 0.02). Changes in inflammation were not associated with changes in physical performance. Conclusions: Systemic inflammation may contribute to reduced physical function in CKD. Twelve weeks of exercise training was unable to reduce the level of chronic systemic inflammation in these patients, but did reduce plasma myostatin concentrations. Further research is required to further investigate this.


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# Continuous image of connected set is connected: Proof

In De La Fuente's Mathematical Methods and Models for Economists, the following is said:

Let $f:X\to Y$ be a continuous mapping between two metric spaces. If C is a connected subset of $X$, then $f(C)$ is connected.

The proof goes as in Rudin's Principles, and I cannot understand exactly what Rudin also does not explain:

Suppose $f(C)$ is not connected. Then $f(C)=P\cup Q$, where $P$ and $Q$ are nonempty, separated subsets of $Y$, that is,

$clP\cap Q = \emptyset$ and $P\cap clQ = \emptyset$

Let $$A = C\cap f^{-1}(P) \\ B = C\cap f^{-1}(Q)$$ and notice that then $$C = A\cup B$$ where neither $A$ nor $B$ is empty, and $$f(A)=P \\ f(B) = Q$$

The proof goes on, but this is where De La Fuente loses me. I can clearly see that $f(A)\subseteq P$, but not that $P\subseteq f(A)$. Any thoughts?

Thanks!

• I have added (proof-explanation) tag - see the tag-info - since your question seems to be about this specific proof (rather than asking for any proof of the fact mentioned in the title). – Martin Sleziak Sep 4 '17 at 22:15

Let $p\in P$. Then as $P\subset f(C)$, we know that $f(c)=p$ for some $c\in C$. Can you go from there?

• Yes! Thanks a lot! – Miguel Santana Sep 5 '17 at 8:42

The proof has to be properly aligned with the OP's definition of connectedness:

Definition: A subspace $C$ of a topological space $X$ is connected if and only if for any non-trivial partition $A \cup B$ of $C$,

$\qquad \overline{A} \cap B \ne \emptyset\; \text{ OR }\; A \cap \overline{B} \ne \emptyset$.

Proposition 1: A function $f: X \to Y$ is continuous $\text{ iff for every } A \subset X \text{, } f(\overline{A})\subseteq \overline{f(A)}$.

The above proposition and some general properties about sets and functions should be enough to construct the desired proof.


text:
# I Complex integral path

1. Jun 27, 2017

### Silviu

Hello! If I have a real integral between $-\infty$ and $+\infty$ and the function to be integrated is holomorphic in the whole complex plane except for a finite number of points on the real line does it matter how I make the path around the poles on the real line? I.e. if I integrate on the semicircle in the upper plane and I have a pole at 0 let's say, do I get the same result if I go around the pole above or below the real axis? Thank you!

2. Jun 27, 2017

### FactChecker

The sign will change because one path is clockwise around the pole and the other is counter-clockwise.

3. Jun 27, 2017

### Silviu

Do you mean the sign of the semicircle around the pole (whose radius t=you take to go to 0 in the end)? But you also have to consider that in one case the pole is inside the contour (so you must use Residue theorem) and in the other is outside (SO by Cauchy theorem the integral around everything would be 0), right?

4. Jun 27, 2017

### strangerep

You'll get the same result IF you do both integrals properly.

In the first case, the pole is outside the contour, so no residue contribution. But there's potentially a contribution from the small semicircle over the pole, traversed clockwise.

In the second case, the pole is inside the contour so there's a residue contribution. But there's also a contribution from the small semicircle under the pole, this time traversed anticlockwise.

You should end up with the same result either way.

5. Jun 28, 2017

### FactChecker

Yes. Sorry. My post was too brief. @strangerep gave a better answer.


text:
# 9. Green’s Reciprocation Theorem

## What It Is

One simple theorem George Green published in his 1828 paper is his Reciprocation Theorem. (This is Jackson's term, Wikipedia calls it reciprocity.) It seems almost trivial, but often leads to surprising results with very little effort.  Here it is:

Consider two different volume and surface charge distributions, in the same identical geometry.

Charge distribution A has volume charge density ${\rho }_{A}\left(\stackrel{\to }{r}\right)$ and boundary surface charge densities ${\sigma }_{A}\left(\stackrel{\to }{r}\right),$ generating electrostatic potential ${\phi }_{A}\left(\stackrel{\to }{r}\right).$

In that same space (meaning with the same surfaces) charge distribution B, with densities ${\rho }_{B}$, ${\sigma }_{B}$ would give potential ${\phi }_{B}\left(\stackrel{\to }{r}\right).$

Then the Reciprocation Theorem is:

$\underset{V}{\int }{\rho }_{A}{\phi }_{B}dV+\underset{S}{\int }{\sigma }_{A}{\phi }_{B}da=\underset{V}{\int }{\rho }_{B}{\phi }_{A}dV+\underset{S}{\int }{\sigma }_{B}{\phi }_{A}da.$

In words:  If the A charge densities were at rest in the B potential, the total potential energy would be the same as that of the B charge densities held at rest in the A potential.

To prove it, we’ll just consider volume charges (the surface charges could be taken as volume charges in a thin layer limit anyway).

Then

${\phi }_{B}\left(\stackrel{\to }{r}\right)=\frac{1}{4\pi {\epsilon }_{0}}\underset{V}{\int }\frac{{\rho }_{B}\left({\stackrel{\to }{r}}^{\prime }\right)}{\left|\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }\right|}$

and

$\underset{V}{\int }{\rho }_{A}\left(\stackrel{\to }{r}\right){\phi }_{B}\left(\stackrel{\to }{r}\right){d}^{3}r=\frac{1}{4\pi {\epsilon }_{0}}\underset{V}{\int }\frac{{\rho }_{A}\left(\stackrel{\to }{r}\right){\rho }_{B}\left({\stackrel{\to }{r}}^{\prime }\right)}{\left|\stackrel{\to }{r}-{\stackrel{\to }{r}}^{\prime }\right|}{d}^{3}r{d}^{3}{r}^{\prime }.$

But this is symmetric in A, B, proving the theorem: it’s that simple!

## Earnshaw’s Theorem from the Reciprocation Theorem

Earnshaw’s theorem is often stated simply as: In a charge-free region, the potential cannot have a maximum or minimum. (Because locally the field in all directions would have to point all inwards or all outwards, so nonzero divergence, meaning there must be charge present.)

A more informative version is: in a charge-free region, the potential at a point is the same as the average potential on any spherical surface centered at that point, provided only that the spherical surface is itself within the charge-free region.

To prove this using the Reciprocation Theorem, take:

System A:  the existing setup, with charge distribution ${\rho }_{A}\left(\stackrel{\to }{r}\right)$ all outside the region of interest, generating potential ${\phi }_{A}\left(\stackrel{\to }{r}\right).$

System B:  a spherical surface, radius $R,$ with uniform surface charge density $\sigma$, the total surface charge being $4\pi {R}^{2}\sigma =-Q,$ and a point charge $+Q$ at the center of the sphere.

The spherical surface is taken to be in the charge-free region of system A.

The B system, the charged sphere plus the equal but negative central point charge, gives zero potential outside the sphere (which is where all the A system charge is), so

$\int {\rho }_{A}\left(\stackrel{\to }{r}\right)\text{ }{\phi }_{B}\left(\stackrel{\to }{r}\right)=0.$

Therefore from the theorem $\int {\phi }_{A}\left(\stackrel{\to }{r}\right){\rho }_{B}\left(\stackrel{\to }{r}\right)\text{ }\text{\hspace{0.17em}}=0,$ and taking the center of the sphere as the origin for convenience, this becomes, integrating over the area of the spherical surface (and cancelling out $Q$ from both sides):

$\frac{1}{4\pi {R}^{2}}\int {\phi }_{A}\left(\stackrel{\to }{r}\right){R}^{2}d\Omega \text{\hspace{0.17em}}\text{ }={\phi }_{A}\left(0\right).$

Note:  if this looks too easy, there are many more difficult proofs on the web.

Exercise:  Suppose we take a new system B: a sphere centered at the origin with surface charge density proportional to the coordinate $z.$ We replace the center point charge with a point dipole, such that the potential from this system is zero outside the sphere. System A is as before: what does the Reciprocation Theorem tell you this time? Hint: you can think of that surface charge as two equally-sized solid spheres of charge of opposite sign, their centers a very small distance apart.

## Revisiting an Earlier Result Using the Reciprocation Theorem

We established in the Electrostatics II lecture that if different parts of a spherical surface are held at different potentials, the consequent potential at any point P in space outside the sphere can be found by integrating over the spherical surface with a weighting factor equal to the density of induced charge on a perfectly conducting grounded sphere with a unit charge at the point P.

In fact, this result follows immediately from the reciprocation theorem!

For

$\underset{V}{\int }{\rho }_{A}{\phi }_{B}dV+\underset{S}{\int }{\sigma }_{A}{\phi }_{B}da=\underset{V}{\int }{\rho }_{B}{\phi }_{A}dV+\underset{S}{\int }{\sigma }_{B}{\phi }_{A}da,$

consider two systems A, B with identical geometry: just a sphere plus one point outside it.

System A is our spherical surface divided into parts held at different potentials, labeled by the variable potential ${\phi }_{A}\left({\stackrel{\to }{r}}^{\prime }\right),$ ${\stackrel{\to }{r}}^{\prime }$ being any point on the surface of the sphere, and system A also includes the point P, position $\stackrel{\to }{r}$ say, at which we want to find the potential from the charge distribution on the spherical surface (but there is no volume charge in A).

System B is the geometrically identical sphere, but now a fully connected conducting surface, grounded and so at zero potential, and now there is a unit charge at the point P, that is, ${\rho }_{B}\left(\stackrel{\to }{r}\right)=\delta \left(\stackrel{\to }{r}-{\stackrel{\to }{r}}_{\text{P}}\right).$

The left-hand side of the equation above is identically zero: ${\rho }_{A}=0$ in the volume, ${\phi }_{B}=0$ on the surface.

The right-hand side, using ${\rho }_{B}\left(\stackrel{\to }{r}\right)=\delta \left(\stackrel{\to }{r}-{\stackrel{\to }{r}}_{\text{P}}\right),$ gives ${\phi }_{A}\left(\stackrel{\to }{r}\right)=-\underset{S}{\int }{\sigma }_{B}\left({\stackrel{\to }{r}}^{\prime }\right){\phi }_{A}\left({\stackrel{\to }{r}}^{\prime }\right)da,$ where ${\sigma }_{B}\left({\stackrel{\to }{r}}^{\prime }\right)$ would be the surface charge density induced at ${\stackrel{\to }{r}}^{\prime }$ on a grounded conducting sphere by unit charge at $\stackrel{\to }{r}.$

Note also that this proof generalizes from a conducting sphere to any closed conducting surface.

## Symmetry of the Dirichlet Green's Function

We’ve shown the Dirichlet Green’s function is symmetric

${G}_{D}\left(\stackrel{\to }{r},{\stackrel{\to }{r}}^{\prime }\right)={G}_{D}\left({\stackrel{\to }{r}}^{\prime },\stackrel{\to }{r}\right).$

This also follows easily from the Reciprocation Theorem: take two systems A, B having the same set of grounded conducting surfaces, one with a single unit charge at $\stackrel{\to }{r},$ the other with a single unit charge at ${\stackrel{\to }{r}}^{\prime }$. Now, by definition, ${G}_{D}\left(\stackrel{\to }{r},{\stackrel{\to }{r}}^{\prime }\right)$ is the potential at $\stackrel{\to }{r}$ from the single unit charge at ${\stackrel{\to }{r}}^{\prime }$ plus the charges induced on the grounded surfaces, and vice versa.  Symmetry follows from The Theorem. (The result is certainly not intuitively obvious: think of an odd shaped conductor, say a sphere but with a tall thin conical "mountain" somewhere. Now put $\stackrel{\to }{r}$ just above the mountain peak, ${\stackrel{\to }{r}}^{\prime }$ above the plane on the other side.)

Exercise:  Write this out explicitly, in terms of ${\rho }_{A},{\phi }_{A},{\rho }_{B},{\phi }_{B}.$


text:
# Mathematical Sciences Research Institute

Home » Free Probability and Non-commutative Banach Spaces

# Workshop

Free Probability and Non-commutative Banach Spaces January 22, 2001 - January 27, 2001
 Registration Deadline: January 27, 2001 about 22 years ago October 22, 2000 over 22 years ago
Parent Program: Operator Algebras
Organizers P. Biane, D. Shlyakhtenko, R. Speicher, D. Voiculescu, E. Effros, E. Kirchberg, V. Paulsen, G. Pisier, Z-J. Ruan and A. Sinclair
Speaker(s)

## Show List of Speakers

Description
As part of the full-year 2000-2001 program on Operator Algebras, MSRI will host a one-week workshop on Free Probability and Non-Commutative Banach Spaces, January 22-26, 2001. The Free Probability section of the workshop will cover several aspects of the subject: applications to von Neumann algebras and C*-algebras of free product type, connections with random matrix theory, free stochastic processes and free stochastic integration, combinatorial approach via noncrossing partitions, free entropy. Probable participants in the Free Probability section include: G. Ben Arous, H. Bercovici, P. Biane, M. Bozejko, M. Anshelevich, T. Cabanal - Duvillard, K. Dykema, L. Ge, D. Gaboriau, A. Guionnet, U. Haagerup, K. Johannsen, A. Nica, F. Radulescu, D. Shlyakhtenko, R. Simion, R. Speicher, M. Stefan, S. Thorbjoernsen, D.Voiculescu, P.Zinn - Justin. The Non-commutative Banach Space section will cover the central concepts of the recently developed theory of operator spaces such as: exactness, local reflexivity and injectivity with applications to C* tensor products, operator algebras and operator modules. The non-commutative Lp-spaces, which play an important role in this theory, provide many points of contact with free probability. Probable participants in the Noncommutative Banach Spaces section include: W. Arveson, D. Blecher, E.G. Effros, U. Haagerup, M. Junge, E. Kirchberg, C. Le Merdy, J. Kraus, B. Magajna, N. Ozawa, T. Oikhberg, V. Paulsen, G. Pisier, C. Pop, N. Randrianantoanina, H.P. Rosenthal, Z.-J. Ruan, A. Sinclair, R. Smith, C. Webster
Keywords and Mathematics Subject Classification (MSC)
Primary Mathematics Subject Classification No Primary AMS MSC
Secondary Mathematics Subject Classification No Secondary AMS MSC
Funding & Logistics

## Show Funding

To apply for funding, you must register by the funding application deadline displayed above.

Students, recent Ph.D.'s, women, and members of underrepresented minorities are particularly encouraged to apply. Funding awards are typically made 6 weeks before the workshop begins. Requests received after the funding deadline are considered only if additional funds become available.

## Show Lodging

MSRI does not hire an outside company to make hotel reservations for our workshop participants, or share the names and email addresses of our participants with an outside party. If you are contacted by a business that claims to represent MSRI and offers to book a hotel room for you, it is likely a scam. Please do not accept their services.

MSRI has preferred rates at the Hotel Shattuck Plaza, depending on room availability. Guests can call the hotel's main line at 510-845-7300 and ask for the MSRI- Mathematical Science Research Institute discount. To book online visit this page (the MSRI rate will automatically be applied).

MSRI has preferred rates at the Graduate Berkeley, depending on room availability. Reservations may be made by calling 510-845-8981. When making reservations, guests must request the MSRI preferred rate. Enter in the Promo Code MSRI123 (this code is not case sensitive).

MSRI has preferred rates at the Berkeley Lab Guest House, depending on room availability. Reservations may be made by calling 510-495-8000 or directly on their website. Select "Affiliated with the Space Sciences Lab, Lawrence Hall of Science or MSRI." When prompted for your UC Contact/Host, please list Chris Marshall (coord@msri.org).

MSRI has a preferred rates at Easton Hall and Gibbs Hall, depending on room availability. Guests can call the Reservations line at 510-204-0732 and ask for the MSRI- Mathematical Science Research Inst. rate. To book online visit this page, select "Request a Reservation" choose the dates you would like to stay and enter the code MSRI (this code is not case sensitive).

Additional lodging options may be found on our short term housing page.

## Show Visa/Immigration

Schedule, Notes/Handouts & Videos
Show Schedule, Notes/Handouts & Videos
Jan 22, 2001
Monday
 08:00 AM - 05:00 PM On finite representability of operator spaces T. Oikhberg 08:00 AM - 05:00 PM C* Spaces E. Kirchberg 08:00 AM - 05:00 PM Variations on injectivity V. Paulsen 08:00 AM - 05:00 PM Completely contractive projections on non-commutative L^1 spaces P.W. Ng 09:30 AM - 10:50 AM Applications of free entropy Liming Ge 10:50 AM - 11:50 AM Large deviations for Gaussian matrices and free entropies Thierry Cabanal-Duvillard 11:50 AM - 02:00 PM Some properties of free group subfactors Marius Stefan 02:00 PM - 03:00 PM Non-commutative Lp spaces and their connections with operator algebras Zhang-Jin Ruan 02:00 PM - 03:00 PM Approximation properties for crossed products Roger Smith
Jan 23, 2001
Tuesday
 09:30 AM - 10:00 AM Multipliers of H1 and power bounded operators G. Pisier 10:00 AM - 11:00 AM The length of a Gamma factor Erik Christensen 11:00 AM - 12:00 PM Curvature and Dirac operator of a commuting d-tuple William Arveson (University of California, Berkeley) 12:00 PM - 02:00 PM Subspaces of non-commutative Lorentz-spaces Narcisse Randrianantoanina 02:00 PM - 04:20 PM Free probability analogue of the Wasserstein distance on trace state space Dan Virgil Voiculescu (University of California, Berkeley) 02:00 PM - 04:20 PM The enumeration of planar maps, links and knots P. Zinn-Justin 04:20 PM - 05:20 PM R-cyclic families of matrices in free probability Alexandru Nica
Jan 24, 2001
Wednesday
 09:30 AM - 10:50 AM Free cummulants, matrix orbits and characters of symmetric groups Philippe Biane (École Normale Supérieure) 10:50 AM - 11:50 AM A characterization of freeness by a factorization of operator-valued cummulants Roland Speicher 11:50 AM - 02:00 PM Ito formula for free Levy processes Michael Anshelevich 02:00 PM - 04:00 PM TBA Timur Oikhberg 02:00 PM - 04:00 PM On the local theory of noncommutative Lp spaces Marius Junge 04:00 PM - 04:30 PM Approximation of hyperfinite martingales by finite dimensional martingales Magdalena Musat 04:30 PM - 05:00 PM Semi-Transitive Operator Algebras H. Rosenthal 05:00 PM - 05:30 PM Singular masas in factors from hyperbolic groups Alistair Sinclair (University of California, Berkeley) 05:30 PM - 06:30 PM Normal modules and maps Bojan Magajna
Jan 25, 2001
Thursday
 09:30 AM - 11:00 AM TBA Eberhard Kirchberg (Humboldt-Universität) 11:00 AM - 12:00 PM Multiplier operator algebras and M-structure David Blecher 12:00 PM - 01:30 PM An operator Hilbert space without the operator approximation property Alvaro Arias 01:30 PM - 02:00 PM $$\ell^2$$-invariants for groups and equivalence relations Damien Gaboriau 02:00 PM - 03:50 PM Convex sets of moments in von Neumann algebras Florin Radulescu 03:50 PM - 04:30 PM Strong and weak limit theorems for free random variables Hari Bercovici 04:30 PM - 05:30 PM Self decomposability and Levy processes S. Thorbjoernsen
Jan 26, 2001
Friday
 09:30 AM - 10:50 AM On free Araki-Woods factors Dimitri Shlyakhtenko (University of California, Los Angeles) 10:50 AM - 11:50 AM Decomposability of Voiculescu's circular operator Kenneth Dykema 11:50 AM - 02:00 PM Invariant subspaces of R-diagonal operators Piotr Sniady 02:00 PM - 03:15 PM TBA Vern Paulsen (University of Texas) 03:15 PM - 03:50 PM TBA Ping Ng 03:50 PM - 04:10 PM H1 has no completely unconditional basis E. Ricard
Jan 27, 2001
Saturday
 09:30 AM - 10:35 AM Operator space properties of decomposable operators Christian Le Merdy 10:35 AM - 11:40 AM On the lifting problem for separable C*-algebras Narutaka Ozawa 11:40 AM - 02:00 PM One-sided M Structure in operator spaces Vrej Zarikian 02:00 PM - 04:00 PM From lacunary sets on groups to free and boolean probability with applications to operator spaces Marek Bozejko 02:00 PM - 04:00 PM Monotonic independence for noncommutative random variables Naofumi Muraki (Iwate Prefectural University) 04:00 PM - 05:00 PM Cummulants, lattice paths and orthogonal polynomials Franz Lehner


text:
# rhomboid

This command draws a rhomboid in the geometry.

## Inline mode usage

rhomboid -h
Displays the help file that summarizes the parameters for this command.
rhomboid –n <name> -p <x1> <y1> <z1> <x2> <y2> <x3> <y3>
Draws a rhomboid in the geometry given three of its corner points are defined by the user.

## Interactive mode usage

Invocation
rhomboid
Parameters
• First point.
• Second point.
• Third point.

The fourth point of the rhomboid will be calculated by adding the difference between the second point and the first point to the third point.

## Example

> rhomboid
First point [x y z]: 1 1 0
Second point [x y]: 3 2
Third point [x y]: 2 5
>


text:
# Adjusting Time Step for Discrete Models

Developing Models for Kalman Filters

In this installment we consider one final model transformation, one that modifies the time step that your model uses.

When in the process of developing a new model, you need to pick carefully the time step for your measurement experiments, so that relevant system behaviors can be effectively observed to produce a robust numerical representation.

• You want the time step short. Abundant samples means better resolution, redundancy that supports better noise filtering, reduced aliasing hazards, and easier conversions between digital and analog signals.
• You want the time step long. Longer steps make response characteristics easier to identify, yielding better numerical precision and better model accuracy.

The "rule of thumb" of least 10 samples per cycle for the highest meaningful frequency in your system response suggests a reasonable compromise. However, that choice is likely to be overruled later because of practical restrictions. For example, after you develop your model using a 200 samples per second sample rate, you might be forced to apply the model using a 60 updates per second digital timing clock compatible with existing PID control equipment — for better or worse. (And you will want to study how much worse...)

## Motivation: sampling exponential physical responses

We have previously discussed how a stable linear system responds to disturbances: with an exponentially decaying response. States influence each other as this progresses, so all of the states show a combination of the characteristic exponential responses. These responses are always in proportion to the values of the state variables. The exponential characteristic remains the same, regardless of what happened in past history to put the system into its current state.

### single state case

Suppose that the system is stable and has a single state variable. Its impulse response decreases from an initial value of 1, until eventually settling to zero. The rate of decay in the state variable value can be characterized by a rate constant `λ` such that

`    e-λT = D   `

where D is the state value after the time interval `T` has passed, and scalar parameter `λ` indicates the rate of decay. We can use a logarithm function to solve for the rate constant parameter.

`    log( e-λ·T) =  log(D)   →  λ = -log(D) / T `

### multiple state case

For a two-variable pair associated with a complex eigenvalue of the state transition matrix, the associated settling for both variables is bounded by the same exponential decay rate.

All other multiple-state cases are just different combinations of the previous two cases, always with exponential settling in all of the terms involved in the mix, just with different proportions and rates.

There some obvious special cases.

1. Zero time step case. States do not have any opportunity to change during a degenerate "instantaneous" time step. The state transition matrix to represent this is the Identity Matrix.
2. Infinite time step case. Given enough time, all states settle fully to zero. The state transition matrix to represent this limiting case is the Zero Matrix.
3. Existing model case. A known model provides a state transition matrix for one particular chosen time step.
4. Multiple-step special case. The state transition matrix at a time interval that is twice as long can be obtained by applying the original state transition operation twice. This idea can be extended for additional integer number of time steps.

## The matrix logarithm

The time step used in the model doesn't make any difference to the actual system. The system will show its exponentially decaying impulse responses, and your choice of time steps only determines which positions along those decay curves you wish to represent.

Starting from the identity matrix of a zero-time transition, as the step-time interval is allowed to increase, more exponential decays occurs, and more state interaction occurs, eventually reaching the known model transition matrix at time interval `T`. Is there a way to interpolate along this trajectory to evaluate the state transitions that occur at arbitrary time intervals shorter than `T`?

As it turns out, there is a matrix-valued exponential decay function that does this, with a matrix-valued decay rate `Λ`.

`    e-Λ·T = A `

At least for the special cases of non-pathological state transition matrices, a "matrix logarithm" function exists that can calculate the `Λ` parameter matrix.[1] Knowing Λ, you can then substitute any value you want for `T` to obtain the state transition matrix for that time interval. General conditions for existence and uniqueness of the matrix logarithm are complicated, and you can research these if you wish. The following series approximation for the "principal matrix logarithm" is typically easier to evaluate than closed-form formulas.

$X\phantom{\rule{1.0em}{0ex}}\stackrel{\scriptscriptstyle\mathrm{def}}{=}\phantom{\rule{1.0em}{0ex}}A-I$ $\mathrm{log}\left(A\right)\phantom{\rule{1.0em}{0ex}}=\phantom{\rule{1.0em}{0ex}}-\sum _{k=1}^{\infty }\phantom{\rule{0.5em}{0ex}}\frac{{\phantom{\rule{1.0em}{0ex}}X}^{k+1}}{k}$

This can be reversed. Given a matrix logarithm, the value of the matrix can be reconstructed using a series expansion for a matrix exponential function.[2]

$\mathrm{exp}\left(\Lambda \right)\phantom{\rule{1.0em}{0ex}}=\phantom{\rule{1.0em}{0ex}}\sum _{k=0}^{\infty }\phantom{\rule{0.5em}{0ex}}\frac{{\Lambda }^{k}}{\mathrm{k!}}$

## The test

As usual, we don't fully believe it until we see it...

Suppose that a state transition dynamic model has been developed to work with a 1/200 second sample rate. This is the state transition matrix as it exists for the initial 0.005 second time step.

``` A0p005 =
0.98644   0.01000  -0.04000
0.00000   0.94544  -0.16244
0.00000   0.16244   0.80144
```

We can apply the series approximation formula to obtain a good estimate for the principal matrix logarithm.

```log0p005 = zeros(3,3);
X = eye(3,3)-A0p005;
Xpow = eye(3,3);

for term=1:6
Xpow = Xpow*X;
log0p005 = log0p005 + Xpow*(1/term);
end
log0p005 = -log0p005```
```log0p005 =
-0.01365   0.01411  -0.04343
-0.00000  -0.03993  -0.18428
-0.00000   0.18428  -0.20329
```

I found that six terms was sufficient for adequate convergence of the power series formula. Did this really produce the matrix logarithm as intended? Check by calculating the power series expression for a matrix exponential function.

```exp0p005 = zeros(3,3);
X = eye(3,3);
for  term=1:6
exp0p005 = exp0p005 + X;
X = (X * log0p005) / term;
end
```
```A0p005 =
0.98644   0.01000  -0.04000
0.00000   0.94544  -0.16244
0.00000   0.16244   0.80144

exp0p005 =
0.98644   0.01000  -0.04000
0.00000   0.94544  -0.16244
0.00000   0.16244   0.80144
```

So far so good! The matrix exponential successfully "reversed" the matrix logarithm to restore the original matrix.

For example, the updating interval is 0.005 seconds for the initial sampling rate. The system must operate (for better or worse) at 60 samples per second with a new sampling interval of 0.01666667 seconds. We can get this with three steps at the original sample interval of 0.005 seconds, followed by a 0.00166667 sample fractional step. That final fraction is 1/3 of one sample interval at the original rate. The exponent of that fractional step is obtained simply by multiplying the full-step matrix logarithm by 1/3.

```log0p00166 =
-0.004551   0.004705  -0.014478
-0.000000  -0.013309  -0.061428
-0.000000   0.061428  -0.067763
```

The corresponding fractional state transition matrix is then calculated by evaluating the matrix exponential.

```A0p00166 =
0.99546   0.00423  -0.01410
0.00000   0.98495  -0.05896
0.00000   0.05896   0.93269
```

So now build the new transition matrix from a cascade of three of the initial transition steps followed by the fractional step.

```A0p0166 = A0p005 * A0p005 * A0p005 * A0p00166 =
0.95551   0.00801  -0.10704
0.00000   0.74068  -0.38917
0.00000   0.38917   0.39569
```

Just for comparison, let's approach this in a different way. As it happens, the new 0.01666667 second step could also be obtained by applying 1/3 step transitions 10 times, each fractional transition spanning 0.00166667 seconds. Let's try doing it that way.

``` A0p0166 = eye(3,3);
for term=1:10
A0p0166 = A0p0166 * exp0p00166;
end
```
```A0p0166 =
0.95551   0.00801  -0.10704
0.00000   0.74069  -0.38918
0.00000   0.38918   0.39569
```

The two different calculations produce equivalent results.

## A loose end

Don't you hate loose ends? In this case, the inconvenient detail is that, so far, we have the time-adjusted the state transition matrix, but not the other matrices for the dynamic model.

The `C` matrix bases its reports on the current values of the system state. It makes absolutely no difference what sort of state transition steps were used to reach any given system state. Ergo: no adjustment required!

But the `B` matrix? Not so lucky. Consider carefully what it means. Suppose that a "unit impulse" signal `u` is applied to the system in a zero state. The impulse does something to displace the system in a way that we cannot know, but results show up at the next sample instant as a new state vector equal to `B`.

That is how we model the system at least. For the real linear system, after the impulse takes effect, there is a short piece of exponential decay, just like any other passage of time, until the target sampling interval is reached. This suggests a close relationship between the input response and the state transition response.

There is another way that the system state could arrive at the value `B` other than being driven there by an impulse. The system could have started in some state such that over the one sample time interval the exponential state decay takes it naturally to equal the value `B`. In fact, once past the instantaneous initial impulse, there is no difference. This idea motivates the following scheme for adjusting the `B` matrix for a new time step.

1. Let `AT1` be the state transition matrix for the model at the original time step `T1`.
2. Let the corresponding input coupling matrix be `BT1`.
3. Let `AT2` be the state transition matrix for the new time step `T2`, as determined using the methods from the rest of this installment.

Then the equivalent input coupling matrix for the new time step `T2` is then given by

`     BT2 = `AT2 AT1-1 BT1` `

If you are not fully convinced, you already know how to set up your system model and improve the modified input coupling matrix using the LMS method. Or, become a master at linear systems mathematics, and recalculate everything exactly according to theory.

## A retrospective

Let's briefly review how far we have come. If you have followed this series from the beginning:

1. You can apply Linear Least Squares or Total Least Squares methods to fit the parameters of a linear state transition model.
2. You can compare how your model predictions perform compared to actual system outputs by simulation with actual input data.
3. You can restructure models that require multiple steps of past history into models requiring only one step of past history and multiple state variables using shift register techniques.
4. You can transform ARMA models developed using statistical methods into equivalent state transition models.
5. You can improve your models numerically, using LMS gradient learning techniques.
6. You can apply scaling, transposition, restructuring, and eigenvalue transformations to change the form of the state transition matrix without changing its performance.
7. You can use correlation analysis to check whether your model captures all of the important behaviors of the system and to deduce an appropriate model order.
8. You can adjust your model order by learning new response characteristics or removing insignificant ones.
9. You can adjust your model so that it can be used at a different time step than the one you used to collect your I/O test data sets.

All of these techniques take an adaptive systems approach to modeling. That is: assume that your system model needs to be improved to make it match the real system more closely. That is surely the case when you start with no model at all!

When you reach the point that your state transition equations give rather credible but imperfect tracking of the real system behavior, it is time to change gears and consider the Kalman Filter approach. We know that the model results are going to be imperfect, because the real system has noise disturbances that the model does not and cannot know about. The random disturbances propagate in the state variables to send the system along a different trajectory than the model predicts.

We can't be completely sure at any individual point what the noise effects are, but over a longer period, we can start to see systematic trajectory changes. When we see those, we can be increasingly confident that they resulted from random effects, and we can take action to restore accurate tracking. We begin this new direction next time.

Footnotes:

[1] You can find out more about Wikipedia at Matrix Logarithm, Andreas Ambrosioy, 2013-03-21, available at the planetmath.org site.

[2] The Wikipedia article on Matrix Exponentials.

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