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# Tag Info

2

A tensile design for a space elevator cable requires there to be a counterweight some distance beyond the point of geostationary orbit. The counterweight would be hurled with centrifugal force and would keep the entire cable taut. The cable must not only support a payload, but also must be a tether for the counterweight, and provide for centripetal force ...

1

If you work through the numbers, you will find that all of the air on earth has a mass that is less than 1 millionth the mass of the earth. You can't get more than a tiny fraction of that air close to your falling object, so the effects of wind and other disturbances would FAR outweigh any effects due to gravity, because G is sooooooo small.

1

If you consider that gravity is weak compared to the electromagnetic force because $G \approx 6.67 \times 10^{-11} Nm^2 kg^{-2}$ and $k_e \approx 8,987 \times10^9 N m^2 C^{-2}$ it would require very small distances in order for the gravitational force to be effective, but at this distances the electromagnetic force would be several times higher, ...

1

The object of making the silicon sphere was not to define the kilogram by the mass of sphere, but redefine Avogadro's constant in terms of the number of silicon atoms. Then use the fixed Avogadro's constant to fix the kilogram. Avogadro's constant was defined as the number of atoms in 12 grams of carbon atoms. You can read up on the Avogadro project here ...

1

Con: There are 3 naturally occurring isotopes of silicon, so isotopic purification would be required after elemental purification. I'm unsure about how hard it is to maintain pure silicon. Isotopically, beryllium would be better (100% $^9$Be, naturally), but I don't know about its reactivity, either. Carbon has two naturally occurring isotopes, but we know ...

0

The analogy between how a massive object bends a rubber sheet and how sources of mass-energy-momentum bend spacetime is a poor one. It's essentially comparing the mechanical deformation of a two dimensional sheet in space to the curvature of four dimensional (3+1) spacetime, which have a mathematical similarity, but not a physical one. In Newton's theory of ...

0

If you take away mass in Newtonian mechanics, then any force leads to an infinite response. The spring force then promises an infinite negative-feedback response to any deviation from equilibrium, and depending on how you take the limits involved, you either get infinite sinusoidal motion with a period of 0, or perfect rigidity with no motion and a period of ...

1

The general theory of relativity predicts that kinetic energy will contribute to gravitational mass. Here is a paper that explores the gravitational effect of kinetically energetic particles within a system: http://arxiv.org/PS_cache/gr-qc/pdf/9909/9909014v1.pdf. Here is an interesting article by Frank Helle on the production of gravity by relativistic ...

2

there are many experiments that put different contraints on the neutrino masses. Here is a good collection from the particle data group.

0

A Non-Relativistic Interpretation of Relativistic Results Million years ago, when self-studying special relativity, an exercise was created in order to understand non-relativistically the relativistic result that a force could not accelerate a particle to speed greater than $\:c\:$. The exercise, the Figure and the solution are already in LaTeX and are ...

0

Particles that have no mass, as photons, have momentum based on their energy, more especificaly on the relation $$p=\frac{E}{c}$$ Where $p$ is the momentum, $E$ is the energy and $c$ is the speed of light. You can think of it intuitively by noticing Einstein's famous equation, $E=mc^2$, and substituting it on the classical momentum formula $p=mv$, using ...

0

By inertia I assume you mean momentum. The momentum is related to the energy of the object by: $$E^2 = p^2c^2 + m^2c^4$$ and to the velocity by: $$p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}$$ The momentum does indeed tend to infinity as $v \rightarrow c$, but note that it will never reach an infinite value because no massive object can travel at the ...

1

The equation $E = m^2c^2 + p^2 c^2$ is restricted to Special Relativity. However, in classical physics we have $$\vec{F} = m \vec{a},$$ and $$\vec{F} = m \vec{g},$$ whence $$m \vec{a} = m \vec{g}.$$ This can be written as $$m \big( \vec{a} - \vec{g}\big) = \vec{0}.$$ From a mathematical point of view we have $$\big( m = 0 \big) \vee \big( \vec{a} - ... 0 For a friction less drop we simply use Newtons law:$$\frac{dv}{dt} = g$$and solving for v gives v(t) = gt. Introducing the friction component into the differential equation, I make the approximation that the drag is laminar:$$ \frac{dv}{dt} = g - \frac{\gamma}{m} v$$The solution of this is: v(t) = \frac{mg}{\gamma} - \frac{mg}{\gamma} ... 1 Lets just say were solving for any specific speed. Not the terminal velocity. I'm still not sure what to do. It might help to start with the speed an object attains after falling a short distance, ignoring air resistance and assuming the effect of gravity is constant. If we fall from Mount Everest (about 8000 meters) to sea level, we can treat gravity ... 1 Okay, so here's the basic physics: it comes down to differential equations. Big objects in air tend to see v^2 drag. Plugging this into F = m a, this means that something which is falling straight downward with speed v will obey the nonlinear differential equation:$$\frac {dv}{dt} = \frac 1\lambda v^2 - g$$for some length \lambda. The terminal ... 1 Mass is a Lorentz invariant quantity! The relativistic mass is not the real mass, it is is just called relativistic "mass" for obvious reasons. This term is abandoned by most textbooks, as it often causes this confusion. -2 It's quite wrong to assert that relativistic mass is obsolete. Gary Oas of Standford university did a study of a sample of the SR and GR textbooks. Of the ones published between 1995 and 2005 there were 60% of them which use relativistic mass. In this video, Alan H. Guth of MIT describes why the mass of a photon is not zero. ... 0 When you breath in helium, you gain mass, volume, and buoyancy, but lose density and weight. 1 The mechanism for "giving mass" to elementary bosons and fermions is different. With bosons, it is related to the gauge symmetry (SU(3)_c \times SU(2)_L \times U(1)_Y) which is partially broken (and become SU(3)_c \times U(1)_{em}). The unbroken part imposes its associated bosons (gluons and photon) to be massless to respect this symmetry. With ... -1 I'm wondering if there is any explanation for why bosons(specifically gauge bosons) can be massless (photon and gluon) but we don't see any fundamental massless fermions. It's because a fermion is a "body", and because "the mass of a body is a measure of its energy content". See Einstein's E=mc² paper. He talks about a body and an electron here. "The ... 2 So, unfortunately, if a box contains a certain amount of kinetic/potential energy stored within it, that energy appears as a mass to the outside world. This is actually the basis for using the so-called "mass defect" (change in mass) in radioactive reactions for detecting the energy released: One over-large Uranium nucleus becomes some smaller parts with ... 0 As in Ben Crowell's Answer, the concept of "Relativistic Mass" is not wrong, but it is awkward. There are several things a loose usage of the word "mass" could imply, all different and thus it becomes a strong convention to talk about the meaning of the word "mass" that is Lorentz invariant - namely the rest mass, which is the square Minkowski "norm" of the ... 3 Suppose your object is a sphere with a radius r and mass m. The aerodynamic drag on a sphere is given by:$$ F_{drag} = \tfrac{1}{2}C_d \rho \,\pi r^2 \,v^2 \tag{1} $$where \rho is the density of the air and C_d is the drag coefficient. The drag coefficient varies with speed (the NASA article I linked shows how C_d changes with speed) but over a ... -1 Look at the paragraph "gravity and the photon" in this link: In the relativistic framework, i.e. large velocities, any energy is also a relativistic mass: For the photon this means the following equation: m is the relativistic mass of a photon with energy h*nu. Gravity attracts relativistic mass, and the photon has one. Read the link further to ... 0 Imagine 3 objects. One is a flat piece of paper. The second is an identical piece of paper rolled into a ball. The third object is the exact same shape and size as the rolled up piece of paper but it's made of iron. If you drop the flat paper and the rolled up paper at the same time, the rolled up paper hits first because it has less air resistance due to ... 1 You misunderstand. All objects have some escape velocity, which is the velocity needed for anything (photons or matter) to escape from that object's gravitational field. And that's not the velocity it needs to maintain under some sort of constant thrust, but the initial velocity it needs to, shall we say, coast away from the object. For a black hole that ... 1 So what I understand from reading the other answers is this: Here on earth: 1 Kilogram of lettuce: Mass-> 1 Kg, Weight-> 1 Kgf In another hypotetical planet where gravity is half of earth: 1 Kilogram of lettuce: Mass-> 1 Kg, Weight-> 0.5 Kgf Since there is no practical easy way to measure mass, in everyday life we use the kilogram as a unit of weight ... 0 The object's center of mass will fall straight downward, accelerated at g. All objects accelerate at the same rate in a vacuum, regardless of their weight, so neither end will accelerate faster than the other. That said, with air resistance, both ends of the object will feel the same drag force, so the end of the object with less mass will feel a greater ... 2 Gravity, like all cause-effect relationships, propagates at a maximum speed of c; indeed from the Einstein field equations small amplitude (linear limit) gravitational waves travel at precisely c. A good idea for what is going on comes from an approximation of General Relativity called Gravitoelectromagnetism. This makes an approximate analogy between ... 1 Is it possible to decrease the mass of the object? Perhaps surprisingly the answer is yes. All you need to do is it drop it. Then some of the object's mass-energy, which we call potential energy, is converted into kinetic energy, which ends up getting dissipated. You're then left with a mass deficit. The mass of the object is reduced. It is known ... 1 We know how real numbers (in the mathematical sense) behave. There is no a priori reason however to assume that masses of objects behave as real numbers. The proposed experiment can be seen as a check to see if masses actually can be modeled by (a subset of the) real numbers. 2 What is mass of elementary particles? It is the "length" of the four vector (p_x,p_y,-_z,E), for complex systems it is called rest mass. As the length of three dimensional vectors is not additive ( think adding two opposite momenta), rest masses are not additive, vector algebra has to be used. The invariant mass of your two gammas must be larger than the ... 3 "Rest mass" is probably a bit more abstract in modern science than what you seem to be thinking. It is simply this: if you can put yourself into an inertial frame such that a body is at rest relative to you, then that body's rest mass is defined as that body's total energy measured in this particular inertial frame - multiplied by c^2, if you want to ... 1 Where does mass come from in pair production? From the energy-momentum of the gamma photons. Think of photon momentum as resistance to change-in-motion for a wave propagating linearly at c. You've maybe heard of electron spin and the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular ... -2 This question goes back to the question that Feynman asked his father about why a ball in a wagon rolled to the back, when he pulled on the wagon. His father replied, "That, nobody knows." In other words, you're asking what relativistic momentum is. We don't even know what non-relativistic momentum is. Sure, we know the effects, we measure them, I've ... 2 Your equations (1) and (2) are dimensionally equivalent. Take your equation (1):$$ F = BI\ell $$The dimensions I are CT^{-1}, so the right hand side are [B]\,CT^{-1}L - I've left the dimensions of B unspecified for now and just written them as [B]. Now take your equation (2):$$ F = BQv $$which has the dimensions [B]\,CLT^{-1}. If we set ... 2 Since you have not specified the "real world" size of the ship, let's take a 74-gun ship of the line https://en.wikipedia.org/wiki/Seventy-four_(ship) as the desired type, firing a 36-pound cannon. The bore on such a cannon https://en.wikipedia.org/wiki/36-pounder_long_gun was about 175 mm, with a shot weight of about 39 lb. Shrinking this cannon to a bore ... 1 Short: No Because: Assume material SG ~= 7 Assume ~= non-sonic shot. Say 500 m/S based on this superb reference R = 0.05mm = 5E-5m E= 1/2 mV^2 m kg = 4/3.Pi.R^3 x sg x 1000 kg/m^3 = 4/3x 3.14 x (5E-5)^3 x 1000 x 7 = 3.E-9 kg At say V = 500 m/s E = 0.5 x 3.5E-9 x 500^2 ~~= 500 micro Joule. (438 uJ calc) A VERY solid teaspoon weighs 50 gram. ... 1 I think the content of his statement is that "the assignment of mass to an object as defined using the conservation of momentum experiment is a transitive property". Not all relations are transitive properties (for instance, "Wolves eat Deer, Deer eat grass, but wolves don't eat grass"). For instance, if we had defined "the mass of an object" as a weight ... 1 Who can with certainty fathom the mind of a Feynman? :-) However, it sounds like he is saying that there is no reason to decide with certainty that Gold, Copper and Aluminum do not have properties which affect how they behave when accelerated by a force. An example that has some correspondence is carrying out the same experiment in air. Mount 1 kg spheres ... 2 As wikipedia says about the second law, As Newton's second law is only valid for constant-mass systems, mass can be taken outside the differentiation operator by the constant factor rule in differentiation This is expanded in the article about variable mass systems. You derived the formula$$ F = m\dot v + \dot m v.  where $\dot v \equiv ... 0 So,$F = \frac{dp}{dt} = \frac{d}{dt}(mv)=m\frac{dv}{dt}+v\frac{dm}{dt} = m\frac{dv}{dt} =\frac{dp}{dv} \times \frac{dv}{dt} = \frac{dp}{dt} = F $nope, no problem. Note that$\frac{dm}{dt}$is assumed zero because the situation described by the equation requires that the mass is constant, and hence time invariant. This is of course inapplicable if somehow ... 0 Even without gravity, momentum conservation will still hold. If you elastically scatter an unknown mass$m$with an initial (known) velocity$v$against a known mass at rest, for instance we can take the SI standard of$1$Kg, then from the resulting measured velocities you should be able to find$m$. Thus Yes, there will still be a mass. 0 While defining "matter" we refer to "mass", but the definition of "mass" refers back to "matter". So isn't this wrong? What will be the right definitions? Circular definitions are indeed undesirable, because they talk a lot but they tell little. The solution is to break the circle by removing the requirement to define the word and refer to experience ... 4 The word "matter" needs to be used with a great deal of care. In science, the word has definitely passed its use-by date. Some problems with the precise usage of the word are listed on the Wikipedia page for Matter and I quote: ...As such, there is no single universally agreed scientific meaning of the word "matter". Scientifically, the term "mass" is ... 1 Assuming constant total energy,$E^2 = (m_0^2c^2)^2 + (pc)^2 = \gamma^2 m_0^2c^4$leads to the equation$E^2 = \frac{m^2c^6}{c^2-v^2}$and then$m = \frac{E}{c^3} \sqrt{c^2 - v^2}$, which also yields the quarter of an ellipse:$v^2 +\frac{c^6}{E^2}m^2 = c^2$, producing the following graph (ignore the number): where the vertical axis is rest mass which ... 0 In my textbook the definition of matter and mass is as such: Matter: Any thing that occupies space and has mass. Mass: The amount of matter contained in a body. Why are the two definitions interlinked? Because it's not a very good textbook. Take a look at Einstein's E=mc² paper, and note this: "The mass of a body is a measure of its energy-content". It ... 0 These aren't good definitions. They do help students grasp the concept, but certain students, such as you, will find it silly that one definition depends on the other. Anyway, one definition depending on the other doesn't necessarily mean that a definition is flawed. In mathematics many functions are defined over itself, yet the definition is effective in ... 5 There's no one-to-one relationship. With zero rest mass, a particle must always be observed to move at$c$. A particle with nonzero rest mass, on the other hand, can move at any speed in$[0,\,c)\$ (note the closed-open interval). At the risk of putting words in your mouth, I think I can recall the exact same question in my mind and it went something like ...

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##### Tautology

Show that each of these conditional statements is a tautology :

a) (p ∧ q) → p

b) p → (p ∨ q)

c) ¬p → (p → q)

d) (p ∧ q) → (p → q)

e) ¬(p → q) → p

f ) ¬(p → q) → ¬q

Things you need to Understand

Point 1

A compound proposition that is always true, no matter what the truth values of the propositional variables that occur in it, is called a tautology.

Point 2

Observe the truth table for implication (→):

 p q p  → q F F T F T T T F F T T T

Checkpoint : The given compound proposition might fail to be a tautology when the RHS of (→) is False, and LHS is True.

Method 1

Step 1: Make RHS False

Step 2: Try to make LHS True

Step 3: If Step 1 possible , then cannot be tautology

else, given compound proposition is tautology.

a) (p ∧ q) → p

Step 1: Make RHS False

p = F

Step 2: Try to make LHS True

LHS = p ∧ q = F ∧ q = F

∴  LHS → RHS = T

Hence, Tautology

b) p → (p ∨ q)

Step 1: Make RHS False

RHS = (p ∨ q)

for LHS to be False , both p = F and q = F

Step 2: Try to make LHS True

LHS = p = (∵ for RHS we made p = F)

∴ LHS → RHS = T

Hence, Tautology

c) ¬p → (p → q)

Step 1: Make RHS False

RHS = (p → q)

for RHS to be False, p = T and q = F

Step 2: Try to make LHS True

LHS = ¬p = ¬ = F

LHS → RHS = T

Hence, Tautology.

d) (p ∧ q) → (p → q)

Step 1: Make RHS False

RHS = p → q

for RHS to be False, p = T and q = F.

Step 2: Try to make LHS True

LHS =  (p ∧ q) = T  ∧ F = F

LHS → RHS = T

Hence, Tautology.

e) ¬(p → q) → p

Step 1: Make RHS False

RHS = p

for RHS to be False , p = F

Step 2: Try to make LHS True

LHS =  ¬(p → q)  = ¬(F → q) = ¬ (T) = F

∴  LHS → RHS = T

Hence, Tautology

f) ¬(p → q) → ¬q

Step 1: Make RHS False

RHS = ¬q

for RHS to be False,  q= T

Step 2: Try to make LHS True

LHS = ¬(p → q) = ¬(p → T) = ¬ (T) = F

LHS → RHS

Hence, Tautology.

format: plain


text:
# Plus Two Maths Model Question Paper 3

## Kerala Plus Two Maths Model Question Paper 3

Time : 2 1/2 Hours
Cool off time : 15 Minutes
Maximum : 80 Score

General Instructions to Candidates :

• There is a ‘Cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘Cool off time’ to get familiar with questions and to plan your answers.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

QUESTIONS

Question 1 to 7 carry 3 scores each. Answer any six questions only

Question 1.
a. Let f : R → R be a function defined by f (x) = x3 + 5. Then f1 (x) is
i. (x+5)1/3
ii. (x-5)1/3
iii. (5-x)1/3
iv. 5-x

b. Let * be a binary operation defined on Q
a*b = a-b + ab. Check whether
i. It is commutative?
ii. Is * associative ?

Question 2.

Question 3.

Question 4.

Question 5.
Prove that the function f given by f (x)= log sin x is strictly increasing on ( 0,$$\frac { \pi }{ 2 }$$)

Question 6.

Question 7.

Question 8 to 17 carry 4 scores each. Answer any eight questions only

Question 8.
a. Show that the relation R in set of real numbers defined as R = {(a,b): a < b2} is neither reflexive nor symmetric not transitive.
b. Show that the operation * on Q, defined by a*b = a+b-ab is commutative, and ex-its and identity elements find it.

Question 9.
a. The principal value of the expression cos-1 cos (680) is …………..

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.
a. Find the distance between the planes x-y + z-5 = 0 and 2x-2y + 2z = 0.
b. Write the vector equation corresponding to Cartesian equation of a line

Question 16.
Find the shortest distance between the lines

Question 17.

Question 18 to 25 carry 6 scores each. Answer any 5 questions only

Question 18.

Question 19.
a. Use differential to approximate (0.999)1/10
b. A window is in the form of rectangle sur-mounted by a semicircular opening. The total perimeter of the windows is 1 Om. Find the dimensions of the window to admit maximum light through the whole opening.

Question 20.
Find the area lying above x axis and included between the circle x2 + y2 = 8x and inside of the parabola y2 = 4x. Also draw a neat diagram.

Question 21.
Evaluate:

Question 22.
Minimize and maximize Z = 5x + 10 y subject to x + 2y < 120, x + y > 60, x – 2y > 0, x, y > 0
a. Draw the feasible region
b. Find the comer points
c. Find the maximum and minimum profit.

Question 23.
a. Find the distance of the point (-1, -5, -10) from the point of intersection of the line

Question 24.
a. Two numbers are selected at random (with-out replacement) from the Pt six positive integers.

Let X denote the larger of the two numbers obtained. Find E(X) and Var(X)

b. A card from a pack of 52 cards is lost from the remaining X cards of the pack, two cards are drawn and are found to be both spades. Find the probability of the lost card being a spade.

a. ii
b. a * b = a – b + ab
b*a = b- a + ba = b – a + ab
∴ * is not commutative.
(a,b) * c = d * c = d- c + dc
= a- b + ab – c + ac – bc + abc
= a- b + ab – c + ac – bc + abc
= a – b – c + ab – bc + ca + abc
a * (b * c) = a * d = a- d + ad
= a – (b – c + be) + a (b – c + be)
= a- b + c – bc + ab – ac + abc
= a – b + c ab – be + ca + abc
∴ * is not associative.

We are giving that

a. R = {(a,b): a < b2}
Relation R is defined in the set of real numbers.
i. Reflexive
Consider a ∈ R
If a ∈ R ⇒ a < a2 which is false (a, a) ∈ R
R is not reflexive.

ii. Symmetric
Let a,b ∈ R and
(a,b) ∈ R ⇒ a < b2 and b < a2,
which is false ⇒ (a,b) ∈ R, but (b,a) ∈ R
∴ R is not symmetric.

iii. Transitive
Let a, b, c ∈ R

The given equation of the circle x2 + y2 = 8x can be expressed as (x – 4)2 + y2 = 16. Thus, the centre of the circle is (4,0) and radius is 4. Its intersection with the parabola y2 = 4x gives
x2 + 4x = 8x
or
x2 – 4x = 0
or
x(x-4) = 0
or
x = 0, x = 4
Thus, the point of intersection of these two curves are 0 (0,0) and P (4,4) above the x-axis.

From the above the required area of the region OPQCO included between these two curves above x-axis is
= (area of the region OCPO)
+ (area of the region PCQP)

a. The feasible region determine! by the constraints,
x + 2y < 120, x + y > 60, x – 2y > 0,
x > 0 and y > 0
is as follows.

b. The comer points of the feasible are region are A(60,0), C(60,30) and D (40,20).
The values of Z at these comer points are as follows.

 Corner point Z=5x + 10y A (60,0) 300 → Minimum B (120,0) 600 → Maximum C (60,30) 600 → Maximum D (40,20) 600

a. The minimum value of Z is 300 at (60,0) and the maximum value of Z is 600 at all the points on the line segment joining (120,0) and (60,30)

a. The two positive integers can be sele-cted from the fist six positive integers without replacement in 6 x 5 = 30 ways.

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2,3,4,5 or 6.

For X=2, the possible observations are (1,1)and(2,1)
∴ P (x = 2) = $$\frac { 2 }{ 30 }$$ = $$\frac { 1 }{ 15 }$$

For X = 3 the possible observations are (1,3), (2,3), (3,1) and (3,2).
∴ p (x = 3) = $$\frac { 4 }{ 30 }$$ = $$\frac { 2 }{ 15 }$$

For x = 4 the possible observations are

(1,4), (2,4), (3,4), (4,3), (4,2) and (4,1).
∴ p (x = 4) = $$\frac { 6 }{ 30 }$$ = $$\frac { 1 }{ 5 }$$

For X = 5, the possible observations are (1.5) , (2,5), (3,5), (4,5), (5,4), (5,3), (5,2) and (5,1).
∴ p (x = 5) = $$\frac { 8 }{ 30 }$$ = $$\frac { 4 }{ 15 }$$

For X = 6, the possible observations are (1.6), (2,6), (3,6), (4,6), (5,6), (6,4), (6,3), (6,2) and (6,1)
∴ p (x = 6) = $$\frac { 10 }{ 30 }$$ = $$\frac { 1 }{ 3 }$$

Therefore, the required probability distribution is as follows.

b. Let E and E, be the respective events of choosing a spade card and a card which is not spade. Out of 52 cards, 13 cards are spade and 39 cards are not spade.


text:
# Query about SDG (Synthetic Differential Geometry)

(Edited 10/17/17): With the hope of obtaining informed responses on the following intriguing remark of Marta Bunge on the status of Synthetic Differential Geometry, I have added a third question to the original two and expanded Bunge's quote to provide further context.

In her "A Personal tribute to Bill Lawvere", presented at Union College in January 2013, Marta Bunge made the following observation about synthetic differential geometry (SDG):

The basic idea of Synthetic Differential Geometry, in the form of the Kock-Lawvere axiom, requires, for a topos $\mathcal{E}$ with a ring object $\textrm{R}$ in it, that the subobject $\textrm{D}$ of $\textrm{R}$, consisting of those elements of square zero, be tiny and representing of tangent vectors at $0$ of arrows from $\textrm{R}$ to $\textrm{R}$. During the period 1981-88, I devoted myself almost totally to SDG, involving students and collaborators (Murray Heggie, Patrice Sawyer, Eduardo Dubuc, Felipe Gago) and participating in the workshops organized by Anders Kock at Aarhus, as well as in related special meetings. Lawvere’s intuition of the role of atoms (or “tiny objects”) in developing a simple form of Analysis going back to the ideas of Newton and Leibniz, and in the same spirit as in the work of André Weil, was both simple and attractive. In my work with my student Felipe Gago on a synthetic theory of smooth mappings, we used two additional axioms (Bunge-Dubuc 1987) to SDG, to wit, the representability of germs of smooth mappings by the sub object $\Delta = \neg \neg \{0\}$ of $\textrm{R}$, required to be tiny, and the existence and uniqueness of solutions of ordinary differential equations. However, no well adapted model of SDG is known at present to satisfy both of these axioms. This open problem is, in my view, pivotal for further progress in this fascinating area, which includes a synthetic proof of Mather’s theorem on the equivalence of locally stable and infinitesimally stable germs of smooth mappings (Bunge-Gago 1988), as well as Morse theory, developed synthetically in the thesis work of Felipe Gago at McGill.

I have three questions regarding this remark.

(i) Is the problem of the existence of a well adapted model of SDG that satisfies the above-stated two additional axioms still open?

(ii) Assuming it is, how widely shared is her view on this matter in the SDG research community? That is, how widely is it held by members of the community that further progress in SDG is contingent on the existence of such a well adapted model?

(iii) (New Question): Assuming the two additional assumptions are natural, which they appear to be, are there any cogent arguments opposed to her view?

• Marta Bunge I would have to agree that in the SDG community her views are widely accepted. She was in Los Caracas last Sunday and spoke with a lot of us for about 6 hours straight. She is a very smart lady and we have great respect for her. I don't know of any cogent arguments opposed to her view or should I say that I've never heard of any. Oct 17 '17 at 20:43
• Winston, thank you for your helpful report. In Los Caracas, did she explicitly reiterate the view expressed in the quote? Moreover, did she assert that the existence question is still open? Oct 17 '17 at 20:56
• I did not get to stay as I was called in but several people were talking about how long they spoke with her. I wish I would have been able to speak with her alone so I could pick her brain a little. I was mostly star struck before I left. Oct 17 '17 at 21:03
• Progress in what direction -- i.e. how might one expect better models to help with Mather's theorem or Morse theory or other areas? Oct 17 '17 at 21:17
• Matt, I suspect she is referring to results she and others have established using the additional assumptions. See e.g. Synthetic Aspects of $C^\infty$-mappings II: Mather’s Theorem…”, J. Pure App Alg 55 (1988). Oct 17 '17 at 21:45

In a paper by Marta Bunge and Eduardo Dubuc. "Local concepts in SDG and germ representability" (1987) certain axioms were laid down towards a synthetic theory of differential topology based on logical infinitesimal notions given by Jacques Penon in his 1985 Universite Paris VII thesis.

One of them was Postulate WAII on $\Delta$-integration of vector fields, where $\Delta = \neg \neg \{0\}$ is a subobject of $R$ (the ring of line type in the Kock-Lawvere axioms) that represents germs of mappings from $R$ to $R$ by one of the basic axioms of what was to become SDT. As shown therein, this postulate is equivalent to the existence and uniqueness of (local) solutions to ODE within SDG.

On the basis of Postulate WAII, along with further axioms for a theory now called SDT (Synthetic Differential Topology) an extension of SDG, Marta Bunge and her McGill University student Felipe Gago in "Synthetic aspects of $C^{\infty}$-mappings II : Mather's theorem for infinitesimally represented germs" (1988) proved the theorem of the title. Felipe Gago in "Morse theory for infinitesimally represented germs" (1988) developed Morse theory which was part of his 1988 McGill University thesis. In her 1999 thesis, Ana Maria San Luis, a student of Felipe Gago at Sgo de Compostela, gave an alternative proof of Mather's Theorem (without the so-called Preparation Theorem) still using Postulate WAII.

Subsequently, Eduardo Dubuc, in "Germ representability and local integration of vector fields in a well adapted model of SDG" (1980) gave proofs of the validity of the basic axiom of germ representability as well as of Postulate WAII both in the topos $(G, R)$ with $G$ the topos of sheaves on the opposite of the category $B$ of finitely generated $C^{\infty}$-rings determined by a local (or germ determined) ideal, known as the Dubuc topos. Unfortunately, an error in the validity of the uniqueness part of the proof of Postulate WAII in the Dubuc topos $G$ given therein was found by Michael Makkai in discussions with Gonzalo Reyes.

This meant therefore that there was at that time no known well adapted model of all of the axioms and postulates used in the work of Bunge-Gago-San Luis on a synthetic theory of smooth mappings and their singularities. Hence the remarks made by myself concerning the need to find a suitable well adapted model of SDT to validate our work. At the time this was indeed an open question but it is no longer one. As shown by myself at the Octoberfest, Ottawa, October 31-November 1, 2015, in a talk "Synthetic Theory of Stable Mappings and their Singularities" (whose slides have been posted in my Research Gate page), the uniqueness part of Postulate WAII (called Postulate VIII therein) is in fact valid in the Dubuc topos $G$, which is then a well adapted model of SDT (as well as of SDG).

This result is now included in Chapter 12 ("$G$ as a WAM of SDT") of a forthcoming book by Marta Bunge, Felipe Gago and Ana San Luis, "Synthetic Differential Topology", Cambridge University Press, to appear in 2018. In Chapter 12, proofs of the validity of all (general and special) axioms and postulates used in our work are given with references to their various sources. Further developments of differential topology might need additional axioms and postulates, but it seems reasonable now (as it was so before the error was found!) to expect that those too will be shown valid in $(G, R)$.

An interesting characterization of well adapted models of SDG (and so also of SDT) involving the Dedekind reals in a topos was given in a 1980 paper by Marta Bunge and Eduardo Dubuc "Archimedian local $C^{\infty}$-rings and models of SDG".


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# File:Uniform curl.svg

 Description A plot of the vector field $F(x,y)=y\boldsymbol{\hat{x}}-x\boldsymbol{\hat{y}}$. Self-made with Mathematica and MS Paint. Converted to SVG by 03:00, 21 July 2007 (UTC). Updated to reflect Mathematica 7.0 syntax and formatting Feb 17, 2011 (UTC). Date 2007-05-23 (first version); 2007-05-23 (last version) Source Originally from en.wikipedia; description page is/was here. Author Original uploader was Loodog at en.wikipedia Permission( Reusing this file) CC-BY-2.5; Released under the GNU Free Documentation License.

## Licensing

Loodog at en.wikipedia, the copyright holder of this work, hereby publishes it under the following licenses:
 Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled GNU Free Documentation License.


text:
Journal of Prime Research in Mathematics

# Weighted homogeneous polynomials with isomorphic milnor algebras

Imran Ahmed
Department of Mathematics, COMSATS Institute of Information Technology, Lahore, Pakistan.

$$^{1}$$Corresponding Author: drimranahmed@ciitlahore.edu.pk.

### Abstract

We recall first some basic facts on weighted homogeneous functions and filtrations in the ring A of formal power series. We introduce next their analogues for weighted homogeneous diffeomorphisms and vector fields. We show that the Milnor algebra is a complete invariant for the classification of weighted homogeneous polynomials with respect to right-equivalence, i.e. change of coordinates in the source and target by diffeomorphism.

#### Keywords:

Milnor algebra, right-equivalence, weighted homogeneous polynomial.


text:
# Homework Help: Diatomic molecule Hamiltonian

1. Apr 27, 2012

### Andurien

1. The problem statement, all variables and given/known data
I have to find the hamiltonian for a diatomic molecule, where the molecule can only rotate and translate and we supose that potencial energy doesn't change.

2. Relevant equations

3. The attempt at a solution

Okey so I used Spherical coordinate system such as the kinetic energy of the molecule is
$$T=\frac{1}{2}(m\dot{r}^2+I(\dot{\phi}^2+I(\dot{θ})^2)=L$$

To find the Hamiltonian i've considered:

$$P_r=m\dot{r} \Rightarrow \dot{r}=\frac{P_r}{m}$$

And so on for the other momentum so the final solution for the hamiltonian is:

$$H=\frac{1}{2}(\frac{P_r^2}{m}+\frac{P_\phi^2}{I}+ \frac{P_θ^2}{I})$$

BUT the correct solution given by my professor is:
$$H=\frac{1}{2}(\frac{P_r^2}{m}+\frac{P_\phi^2}{I \sin^2\theta}+ \frac{P_θ^2}{I})$$

So I don't know why is there a sin^2\theta factor.

Last edited: Apr 27, 2012
2. Apr 27, 2012

### vela

Staff Emeritus
The problem lies in where you started. How'd you come up with this?

3. Apr 27, 2012

### Andurien

Yep I already solved the problem. You were right Vela i started wrong. I divided the Hamiltonian between the lagrangian for the translation kinetic energy and the rotation kinetic energy so:
$$T_t=\frac{1}{2}M(\dot{x}^2+\dot{y}^2+\dot{z}^2)= \frac{1}{2M} (P_x^2+P_y^2+P_z^2)$$

And then the rotation one:
$$T_{rot}=\frac{1}{2}Mr^2[\dot{\theta^2}+\dot{\phi}^2\sin^2(\theta)]$$

And then just like before i get the hamiltonian.

Thank you for the response and sorry for my english ;)


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 0 0 0 0.000%

## 문제

Killbots is a simple game of evading killer robots. Who created the robots and why they have been programmed to destroy, no one knows. All that is know is that the robots are numerous and their sole objective is to destroy everything on their way. Fortunately, their creator focused on quantity rather than quality and as a result the robots are severely lacking in intelligence. Killbots is a turn-based game, played on a rectangular grid. Each cell can contain one of the following:

• Hero (@) - The hero is the unfortunate soul stuck in a room with a bunch of killler robots. In the original Killbots game, the player controls the hero by moving it around the grid. But in this problem, it is assumed that the hero stays at his place and does not move at all. Instead, it can destroy any robot trying to come to its cell.
• Robot (+) - Robots are mechanical thugs desiring on to crush the hero with the metallic girth. They have never heard of the "Three Laws of Robotics" and would probably crush anyone who tried to explain them. Each robot will take a single step toward the hero on every turn. If the robot is in the same row or the same column as the hero, it will step directly towards the robot. Otherwise, it will take the diagonal step that makes it as closer to the hero.
• Fastbot (#) - Fastbots are a much speedier version of the basic robot. For each move a robot makes, they take two steps. Fortunately, their increased mobility was not paired with increased intelligence; they often only succeed in destroying themselves twice as quickly as their slower brethren.
• Junkheap (*) - When two or more robot or fastbots collide, they are destroyed and the resulting debris produces a junkheap. Robots or fastbots blindly ignore any junkheaps in their path and are destroyed upon crashing into them.

The following figure is an example of how the game works. The cell E5 is where the hero stands. The cells containing junkheap are marked with '*'. Robots and fastbots are shown with '+' and '#' respectively. If we start from the grid on the left, the robots at B5 an E4 take direct steps directly towards the hero (to the right and down respectively). The latter is destroyed by the hero. The one at B8 takes an up-right step to C7. The robot at H6 takes an up-left step to G5 where it crashes into junkheap and is destroyed. The robot at F7 toes to E6 (where it will be destroyed in the next step by the fastbot initially at G8. Note that all robots and fastbots move simultaneously so at the same time F7 moves to E6, the fastbot at G8 move to F7, without collision. After a turn that all robots a fastbots move, the fastbots move another step while the robots stay. So, the fastbot at F7 moves to E6 crashing into the robot staying there. They are both destroyed leaving junkheap at E6. The game continue until all robots and fastbots are destroyed either by the hero, or crashing into other robots or junkheap.

The goal of this problem is count the number of robots and fastbots destroyed by the hero.

There are multiple test-case in the input each specifying a game grid. Each test case contains several lines of the same length. Each line describes a row of the grid, and each character of the line is a cell of the grid. Empty cells are marked by '.', and non-empty cells are marked by the special characters described in the problem description. Af each test-case, there is a line in the input containing a single '\$' character. You may assume there is one and only one cell in the input containing '@'. Each side of the grid has at least one and at most 1000 cells. ## 출력 For each test case, write a single line containing the number of fastbots and robots destroyed by the hero when the game ends (after all robots and fastbots are destroyed). ## 예제 입력 1 ...*.... .......# .....*.. ....+... [email protected]*. .......+ .*...+.. .+....#.$
...
[email protected]
.*.
.+.
\$

## 예제 출력 1

4
1


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# Polyharmonic spline

Polyharmonic splines are used for function approximation and data interpolation. They are very useful for interpolating and fitting scattered data in many dimensions. A special case is the thin plate spline.[1][2]

## Definition

A polyharmonic spline is a linear combination of polyharmonic radial basis functions (RBFs) denoted by ${\displaystyle \phi }$ plus a polynomial term:

${\displaystyle f(\mathbf {x} )=\sum _{i=1}^{N}w_{i}\phi (|\mathbf {x} -\mathbf {c} _{i}|)+\mathbf {v} ^{\textrm {T}}{\begin{bmatrix}1\\\mathbf {x} \end{bmatrix}}}$

(1)

where

Polyharmonic basis functions
• ${\displaystyle \mathbf {x} =[x_{1}\ x_{2}\ \cdots \ x_{d}]^{\textrm {T}}}$ (${\displaystyle {\textrm {T}}}$ denotes matrix transpose, meaning ${\displaystyle \mathbf {x} }$ is a column vector) is a real-valued vector of ${\displaystyle d}$ independent variables,
• ${\displaystyle \mathbf {c} _{i}=[c_{i,1}\ c_{i,2}\ \cdots \ c_{i,d}]^{\textrm {T}}}$ are ${\displaystyle N}$ vectors of the same size as ${\displaystyle \mathbf {x} }$ (often called centers) that the curve or surface must interpolate,
• ${\displaystyle \mathbf {w} =[w_{1}\ w_{2}\ \cdots \ w_{N}]^{\textrm {T}}}$ are the ${\displaystyle N}$ weights of the RBFs,
• ${\displaystyle \mathbf {v} =[v_{1}\ v_{2}\ \cdots \ v_{d+1}]^{\textrm {T}}}$ are the ${\displaystyle d+1}$ weights of the polynomial.

The polynomial with the coefficients ${\displaystyle \mathbf {v} }$ improves fitting accuracy for polyharmonic smoothing splines and also improves extrapolation away from the centers ${\displaystyle \mathbf {c} _{i}.}$ See figure below for comparison of splines with polynomial term and without polynomial term.

The polyharmonic RBFs are of the form:

${\displaystyle {\begin{matrix}\phi (r)={\begin{cases}r^{k}&{\mbox{with }}k=1,3,5,\dots ,\\r^{k}\ln(r)&{\mbox{with }}k=2,4,6,\dots \end{cases}}\\[5mm]r=|\mathbf {x} -\mathbf {c} _{i}|={\sqrt {(\mathbf {x} -\mathbf {c} _{i})^{T}\,(\mathbf {x} -\mathbf {c} _{i})}}.\end{matrix}}}$

Other values of the exponent ${\displaystyle k}$ are not useful (such as ${\displaystyle \phi (r)=r^{2}}$), because a solution of the interpolation problem might not exist. To avoid problems at ${\displaystyle r=0}$ (since ${\displaystyle \log(0)=-\infty }$), the polyharmonic RBFs with the natural logarithm might be implemented as:

${\displaystyle \phi (r)={\begin{cases}r^{k-1}\ln(r^{r})&{\mbox{for }}r<1\\r^{k}\ln(r)&{\mbox{for }}r\geq 1\end{cases}}.}$

The weights ${\displaystyle w_{i}}$ and ${\displaystyle v_{j}}$ are determined such that the function interpolates ${\displaystyle N}$ given points ${\displaystyle (\mathbf {c} _{i},f_{i})}$ (for ${\displaystyle i=1,2,\dots ,N}$) and fulfills the ${\displaystyle d+1}$ orthogonality conditions

${\displaystyle \sum _{i=1}^{N}w_{i}=0,\;\;\sum _{i=1}^{N}w_{i}\mathbf {c} _{i}=\mathbf {0} .}$

All together, these constraints are equivalent to the symmetric linear system of equations

${\displaystyle {\begin{bmatrix}A&B\\B^{\textrm {T}}&0\end{bmatrix}}\;{\begin{bmatrix}\mathbf {w} \\\mathbf {v} \end{bmatrix}}\;=\;{\begin{bmatrix}\mathbf {f} \\\mathbf {0} \end{bmatrix}}\;\;\;\;}$

(2)

where

${\displaystyle A_{i,j}=\phi (|\mathbf {c} _{i}-\mathbf {c} _{j}|),\quad B={\begin{bmatrix}1&1&\cdots &1\\\mathbf {c} _{1}&\mathbf {c} _{2}&\cdots &\mathbf {c} _{N}\end{bmatrix}}^{\textrm {T}},\quad \mathbf {f} =[f_{1},f_{2},\cdots ,f_{N}]^{\textrm {T}}.}$

In order for this system of equations to have a unique solution, ${\displaystyle B}$ must be full rank. ${\displaystyle B}$ is full rank for very mild conditions on the input data. For example, in two dimensions, three centers forming a non-degenerate triangle ensure that ${\displaystyle B}$ is full rank, and in three dimensions, four centers forming a non-degenerate tetrahedron ensure that B is full rank. As explained later, the linear transformation resulting from the restriction of the domain of the linear transformation ${\displaystyle A}$ to the null space of ${\displaystyle \textstyle {B^{\textrm {T}}}}$ is positive definite. This means that if ${\displaystyle B}$ is full rank, the system of equations (2) always has a unique solution and it can be solved using the Cholesky decomposition after a suitable transformation. The computed weights allow evaluation of the spline for any ${\displaystyle \mathbf {x} \in \mathbb {R} ^{d}}$ using equation (1). Many practical details of implementing and using polyharmonic splines are explained in Fasshauer.[3] In Iske[4] polyharmonic splines are treated as special cases of other multiresolution methods in scattered data modelling.

## Reason for the name "polyharmonic"

A polyharmonic equation is a partial differential equation of the form ${\displaystyle \Delta ^{m}f=0}$ for any natural number ${\displaystyle m}$. For example, the biharmonic equation is ${\displaystyle \Delta ^{2}f=0}$ and the triharmonic equation is ${\displaystyle \Delta ^{3}f=0}$. All the polyharmonic radial basis functions are solutions of a polyharmonic equation (or more accurately, a modified polyharmonic equation with a Dirac delta function on the right hand side instead of 0). For example, the thin plate radial basis function is a solution of the modified 2-dimensional biharmonic equation.[5] Applying the 2D Laplace operator (${\displaystyle \Delta =\partial _{xx}+\partial _{yy}}$) to the thin plate radial basis function ${\displaystyle f_{\text{tp}}(x,y)=(x^{2}+y^{2})\log {\sqrt {x^{2}+y^{2}}}}$ either by hand or using a computer algebra system shows that ${\displaystyle \Delta f_{\text{tp}}=4+4\log r}$. Applying the Laplace operator to ${\displaystyle \Delta f_{\text{tp}}}$ (this is ${\displaystyle \Delta ^{2}f_{\text{tp}}}$) yields 0. But 0 is not exactly correct. To see this, replace ${\displaystyle r^{2}=x^{2}+y^{2}}$ with ${\displaystyle \rho ^{2}=x^{2}+y^{2}+h^{2}}$ (where ${\displaystyle h}$ is some small number tending to 0). The Laplace operator applied to ${\displaystyle 4\log \rho }$ yields ${\displaystyle \Delta ^{2}f_{\text{tp}}=8h^{2}/\rho ^{4}}$. For ${\displaystyle (x,y)=(0,0),}$ the right hand side of this equation approaches infinity as ${\displaystyle h}$ approaches 0. For any other ${\displaystyle (x,y)}$, the right hand side approaches 0 as ${\displaystyle h}$ approaches 0. This indicates that the right hand side is a Dirac delta function. A computer algebra system will show that

${\displaystyle \lim _{h\to 0}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }8h^{2}/(x^{2}+y^{2}+h^{2})^{2}\,dx\,dy=8\pi .}$

So the thin plate radial basis function is a solution of the equation ${\displaystyle \Delta ^{2}f_{\text{tp}}=8\pi \delta (x,y)}$.

Applying the 3D Laplacian (${\displaystyle \Delta =\partial _{xx}+\partial _{yy}+\partial _{zz}}$) to the biharmonic RBF ${\displaystyle f_{\text{bi}}(x,y,z)={\sqrt {x^{2}+y^{2}+z^{2}}}}$ yields ${\displaystyle \Delta f_{\text{bi}}=2/r}$ and applying the 3D ${\displaystyle \Delta ^{2}}$ operator to the triharmonic RBF ${\displaystyle f_{\text{tri}}(x,y,z)=(x^{2}+y^{2}+z^{2})^{3/2}}$ yields ${\displaystyle \Delta ^{2}f_{\text{tri}}=24/r}$. Letting ${\displaystyle \rho ^{2}=x^{2}+y^{2}+z^{2}+h^{2}}$ and computing ${\displaystyle \Delta (1/\rho )=-3h^{2}/\rho ^{5}}$ again indicates that the right hand side of the PDEs for the biharmonic and triharmonic RBFs are Dirac delta functions. Since

${\displaystyle \lim _{h\to 0}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }-3h^{2}/(x^{2}+y^{2}+z^{2}+h^{2})^{5/2}\,dx\,dy\,dz=-4\pi ,}$

the exact PDEs satisfied by the biharmonic and triharmonic RBFs are ${\displaystyle \Delta ^{2}f_{\text{bi}}=-8\pi \delta (x,y,z)}$ and ${\displaystyle \Delta ^{3}f_{\text{tri}}=-96\pi \delta (x,y,z)}$.

## Polyharmonic smoothing splines

Polyharmonic splines minimize

${\displaystyle \sum _{i=1}^{N}(f(\mathbf {c} _{i})-f_{i})^{2}+\lambda \int \limits _{{\mathcal {B}}\subset \mathbb {R} ^{d}}|\nabla ^{m}f|^{2}\,d\mathbf {x} }$

(3)

where ${\displaystyle {\mathcal {B}}}$ is some box in ${\displaystyle \mathbb {R} ^{d}}$ containing a neighborhood of all the centers, ${\displaystyle \lambda }$ is some positive constant, and ${\displaystyle \nabla ^{m}f}$ is the vector of all ${\displaystyle m}$th order partial derivatives of ${\displaystyle f.}$ For example, in 2D ${\displaystyle \nabla ^{1}f=(f_{x}\ f_{y})}$ and ${\displaystyle \nabla ^{2}f=(f_{xx}\ f_{xy}\ f_{yx}\ f_{yy})}$ and in 3D ${\displaystyle \nabla ^{2}f=(f_{xx}\ f_{xy}\ f_{xz}\ f_{yx}\ f_{yy}\ f_{yz}\ f_{zx}\ f_{zy}\ f_{zz})}$. In 2D ${\displaystyle |\nabla ^{2}f|^{2}=f_{xx}^{2}+2f_{xy}^{2}+f_{yy}^{2},}$ making the integral the simplified thin plate energy functional.

To show that polyharmonic splines minimize equation (3), the fitting term must be transformed into an integral using the definition of the Dirac delta function:

${\displaystyle \sum _{i=1}^{N}(f(\mathbf {c} _{i})-f_{i})^{2}=\int \limits _{\mathcal {B}}\sum _{i=1}^{N}(f(\mathbf {x} )-f_{i})^{2}\delta (\mathbf {x} -\mathbf {c} _{i})\,d\mathbf {x} .}$

So equation (3) can be written as the functional

${\displaystyle J[f]=\int \limits _{\mathcal {B}}F(\mathbf {x} ,f,\partial ^{\alpha _{1}}f,\partial ^{\alpha _{2}}f,\dots \partial ^{\alpha _{n}}f)\,d\mathbf {x} =\int \limits _{\mathcal {B}}\left[\sum _{i=1}^{N}(f(\mathbf {x} )-f_{i})^{2}\delta (\mathbf {x} -\mathbf {c} _{i})+\lambda |\nabla ^{m}f|^{2}\right]\,d\mathbf {x} .}$

where ${\displaystyle \alpha _{i}}$ is a multi-index that ranges over all partial derivatives of order ${\displaystyle m}$ for ${\displaystyle \mathbb {R} ^{d}.}$ In order to apply the Euler-Lagrange equation for a single function of multiple variables and higher order derivatives, the quantities

${\displaystyle {\partial F \over \partial f}=2\sum _{i=1}^{N}(f(\mathbf {x} )-f_{i})\delta (\mathbf {x} -x_{i})}$

and

${\displaystyle \sum _{i=1}^{m^{2}}\partial ^{\alpha _{i}}{\partial F \over \partial (\partial ^{\alpha _{i}}f)}=2\lambda \Delta ^{m}f}$

are needed. Inserting these quantities into the E-L equation shows that

${\displaystyle \left(\sum _{i=1}^{N}(f(\mathbf {x} )-f_{i})\delta (\mathbf {x} -\mathbf {c} _{i})\right)+(-1)^{m}\lambda \Delta ^{m}f=0.}$

(4)

A weak solution ${\displaystyle f(\mathbf {x} )}$ of (4) satisfies

${\displaystyle \int \limits _{\mathcal {B}}\left(\sum _{i=1}^{N}(f(\mathbf {x} )-f_{i})\delta (\mathbf {x} -\mathbf {c} _{i})\right)g(\mathbf {x} )+(-1)^{m}\lambda (\Delta ^{m}f)g(\mathbf {x} )\,d\mathbf {x} =0}$

( 5)

for all smooth test functions ${\displaystyle g}$ that vanish outside of ${\displaystyle {\mathcal {B}}.}$ A weak solution of equation (4) will still minimize (3) while getting rid of the delta function through integration.[6]

Let ${\displaystyle f}$ be a polyharmonic spline as defined by equation (1). The following calculations will show that ${\displaystyle f}$ satisfies (5). Applying the ${\displaystyle \Delta ^{m}}$ operator to equation (1) yields

${\displaystyle \Delta ^{m}f=\sum _{i=1}^{M}w_{i}C_{m,d}\delta (\mathbf {x} -\mathbf {c} _{i})}$

where ${\displaystyle C_{2,2}=8\pi ,}$ ${\displaystyle C_{2,3}=-8\pi ,}$ and ${\displaystyle C_{3,3}=-96\pi .}$ So (5) is equivalent to

{\displaystyle {\begin{aligned}\int \limits _{\mathcal {B}}&\sum _{i=1}^{N}\delta (\mathbf {x} -\mathbf {c} _{i})(f(\mathbf {x} )-f_{i}+(-1)^{m}\lambda C_{m,d}w_{i})g(\mathbf {x} )\,d\mathbf {x} \\&=\sum _{i=1}^{N}(f(\mathbf {c} _{i})-f_{i}+(-1)^{m}\lambda C_{m,d}w_{i})g(\mathbf {c} _{i})\\&=0.\end{aligned}}}

(6)

The only possible solution to (6) for all test functions ${\displaystyle g}$ is

${\displaystyle f(\mathbf {c} _{j})-f_{j}+(-1)^{m}\lambda C_{m,d}w_{j}=0\quad {\text{for}}\ j=1,2,\dots ,N}$

(7)

(which implies interpolation if ${\displaystyle \lambda =0}$). Combining the definition of ${\displaystyle f}$ in equation (1) with equation (7) results in almost the same linear system as equation (2) except that the matrix ${\displaystyle A}$ is replaced with ${\displaystyle A+(-1)^{m}C_{m,d}\lambda I}$ where ${\displaystyle I}$ is the ${\displaystyle N\times N}$ identity matrix. For example, for the 3D triharmonic RBFs, ${\displaystyle A}$ is replaced with ${\displaystyle A+96\pi \lambda I.}$

In (2), the bottom half of the system of equations (${\displaystyle B^{\textrm {T}}\mathbf {w} =0}$) is given without explanation. The explanation first requires deriving a simplified form of ${\displaystyle \textstyle {\int _{\mathcal {B}}|\nabla ^{m}f|^{2}\,d\mathbf {x} }}$ when ${\displaystyle {\mathcal {B}}}$ is all of ${\displaystyle \mathbb {R} ^{d}.}$

First, require that ${\displaystyle \textstyle {\sum _{i=1}^{N}w_{i}=0.}}$ This ensures that all derivatives of order ${\displaystyle m}$ and higher of ${\displaystyle \textstyle {f(\mathbf {x} )=\sum _{i=1}^{N}w_{i}\phi (|\mathbf {x} -\mathbf {c} _{i}|)}}$ vanish at infinity. For example, let ${\displaystyle m=3}$ and ${\displaystyle d=3}$ and ${\displaystyle \phi }$ be the triharmonic RBF. Then ${\displaystyle \phi _{zzy}=3y(x^{2}+y^{2})/(x^{2}+y^{2}+z^{2})^{3/2}}$ (considering ${\displaystyle \phi }$ as a mapping from ${\displaystyle \mathbb {R} ^{3}}$ to ${\displaystyle \mathbb {R} }$). For a given center ${\displaystyle \mathbf {P} =(P_{1},P_{2},P_{3}),}$

${\displaystyle \phi _{zzy}(\mathbf {x} -\mathbf {P} )={\frac {3(y-P_{2})((y-P_{2})^{2}+(x-P_{1})^{2})}{((x-P_{1})^{2}+(y-P_{2})^{2}+(z-P_{3})^{2})^{3/2}}}.}$

On a line ${\displaystyle \mathbf {x} =\mathbf {a} +t\mathbf {b} }$ for arbitrary point ${\displaystyle \mathbf {a} }$ and unit vector ${\displaystyle \mathbf {b} ,}$

${\displaystyle \phi _{zzy}(\mathbf {x} -\mathbf {P} )={\frac {3(a_{2}+b_{2}t-P_{2})((a_{2}+b_{2}t-P_{2})^{2}+(a_{1}+b_{1}t-P_{1})^{2})}{((a_{1}+b_{1}t-P_{1})^{2}+(a_{2}+b_{2}t-P_{2})^{2}+(a_{3}+b_{3}t-P_{3})^{2})^{3/2}}}.}$

Dividing both numerator and denominator of this by ${\displaystyle t^{3}}$ shows that ${\displaystyle \textstyle {\lim _{t\to \infty }\phi _{zyy}(\mathbf {x} -\mathbf {P} )=3b_{2}(b_{2}^{2}+b_{1}^{2})/(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})^{3/2}},}$ a quantity independent of the center ${\displaystyle \mathbf {P} .}$ So on the given line,

${\displaystyle \lim _{t\rightarrow \infty }f_{zyy}(\mathbf {x} )=\lim _{t\rightarrow \infty }\sum _{i=1}^{N}w_{i}\phi _{zyy}(\mathbf {x} -\mathbf {c} _{i})=\left(\sum _{i=1}^{N}w_{i}\right)3b_{2}(b_{2}^{2}+b_{1}^{2})/(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})^{3/2}=0.}$

It is not quite enough to require that ${\displaystyle \textstyle {\sum _{i=1}^{N}w_{i}=0,}}$ because in what follows it is necessary for ${\displaystyle f_{\alpha }g_{\beta }}$ to vanish at infinity, where ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are multi-indices such that ${\displaystyle |\alpha |+|\beta |=2m-1.}$ For triharmonic ${\displaystyle \phi ,}$ ${\displaystyle w_{i}u_{j}\phi _{\alpha }(\mathbf {x} -\mathbf {c} _{i})\phi _{\beta }(\mathbf {x} -\mathbf {d} _{j})}$ (where ${\displaystyle u_{j}}$ and ${\displaystyle \mathbf {d} _{j}}$ are the weights and centers of ${\displaystyle g}$) is always a sum of total degree 5 polynomials in ${\displaystyle x,}$ ${\displaystyle y,}$ and ${\displaystyle z}$ divided by the square root of a total degree 8 polynomial. Consider the behavior of these terms on the line ${\displaystyle \mathbf {x} =\mathbf {a} +t\mathbf {b} }$ as ${\displaystyle t}$ approaches infinity. The numerator is a degree 5 polynomial in ${\displaystyle t.}$ Dividing numerator and denominator by ${\displaystyle t^{4}}$ leaves the degree 4 and 5 terms in the numerator and a function of ${\displaystyle \mathbf {b} }$ only in the denominator. A degree 5 term divided by ${\displaystyle t^{4}}$ is a product of five ${\displaystyle b}$ coordinates and ${\displaystyle t.}$ The ${\displaystyle \textstyle {\sum w=0}}$ (and ${\displaystyle \textstyle {\sum u=0}}$) constraint makes this vanish everywhere on the line. A degree 4 term divided by ${\displaystyle t^{4}}$ is either a product of four ${\displaystyle b}$ coordinates and an ${\displaystyle a}$ coordinate or a product of four ${\displaystyle b}$ coordinates and a single ${\displaystyle c_{i}}$ or ${\displaystyle d_{j}}$ coordinate. The ${\displaystyle \textstyle {\sum w=0}}$ constraint makes the first type of term vanish everywhere on the line. The additional constraints ${\displaystyle \textstyle {\sum _{i=1}^{N}w_{i}\mathbf {c} _{i}=0}}$ will make the second type of term vanish.

Now define the inner product of two functions ${\displaystyle f,g:\mathbb {R} ^{d}\rightarrow \mathbb {R} }$ defined as a linear combination of polyharmonic RBFs ${\displaystyle \phi _{m,d}}$ with ${\displaystyle \textstyle {\sum w=0}}$ and ${\displaystyle \textstyle {\sum w\mathbf {c} }=0}$ as

${\displaystyle =\int \limits _{\mathbb {R} ^{d}}(\nabla ^{m}f)\cdot (\nabla ^{m}g)\,d\mathbf {x} .}$

Integration by parts shows that

${\displaystyle =(-1)^{m}\int \limits _{\mathbb {R} ^{d}}f(\Delta ^{m}g)\,d\mathbf {x} .}$

(8)

For example, let ${\displaystyle m=2}$ and ${\displaystyle d=2.}$ Then

${\displaystyle =\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }(f_{xx}g_{xx}+2f_{xy}g_{xy}+f_{yy}g_{yy})\,dx\,dy.}$

(9)

Integrating the first term of this by parts once yields

${\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f_{xx}g_{xx}\,dx\,dy=\int _{-\infty }^{\infty }f_{x}g_{xx}{\big |}_{-\infty }^{\infty }\,dy-\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f_{x}g_{xxx}\,dx\,dy=-\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f_{x}g_{xxx}\,dx\,dy}$

since ${\displaystyle f_{x}g_{xx}}$ vanishes at infinity. Integrating by parts again results in ${\displaystyle \textstyle {\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }fg_{xxxx}\,dx\,dy.}}$

So integrating by parts twice for each term of (9) yields

${\displaystyle =\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(g_{xxxx}+2g_{xxyy}+g_{yyyy})\,dx\,dy=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(\Delta ^{2}g)\,dx\,dy.}$

Since ${\displaystyle \textstyle {(\Delta ^{m}f)(\mathbf {x} )=\sum _{i=1}^{N}w_{i}C_{m,d}\delta (\mathbf {x-\mathbf {c} _{i}} ),}}$ (8) shows that

{\displaystyle {\begin{aligned}&=(-1)^{m}\int \limits _{\mathbb {R} ^{d}}f(\mathbf {x} )\sum _{i=1}^{N}w_{i}(-1)^{m}C_{m,d}\delta (\mathbf {x-\mathbf {c} _{i}} )\,d\mathbf {x} =(-1)^{m}C_{m,d}\sum _{i=1}^{N}w_{i}f(\mathbf {c} _{i})\\&=(-1)^{m}C_{m,d}\sum _{i=1}^{N}\sum _{j=1}^{N}w_{i}w_{j}\phi (\mathbf {c} _{i}-\mathbf {c} _{j})=(-1)^{m}C_{m,d}\mathbf {w} ^{\textrm {T}}A\mathbf {w} .\end{aligned}}}

So if ${\displaystyle \textstyle {\sum w=0}}$ and ${\displaystyle \textstyle {\sum w\mathbf {c} =0},}$

${\displaystyle \int \limits _{\mathbb {R} ^{d}}|\nabla ^{m}f|^{2}\,d\mathbf {x} =(-1)^{m}C_{m,d}\mathbf {w} ^{\textrm {T}}A\mathbf {w} .}$

(10)

Now the origin of the constraints ${\displaystyle B^{\textrm {T}}\mathbf {w} =0}$ can be explained. Here ${\displaystyle B}$ is a generalization of the ${\displaystyle B}$ defined above to possibly include monomials up to degree ${\displaystyle m-1.}$ In other words,

${\displaystyle B={\begin{bmatrix}1&1&\dots &1\\\mathbf {c} _{1}&\mathbf {c} _{2}&\dots &\mathbf {c} _{N}\\\vdots &\vdots &\dots &\vdots \\\mathbf {c} _{1}^{m-1}&\mathbf {c} _{2}^{m-1}&\dots &\mathbf {c} _{N}^{m-1}\end{bmatrix}}^{\textrm {T}}}$

where ${\displaystyle \mathbf {c} _{i}^{j}}$ is a column vector of all degree ${\displaystyle j}$ monomials of the coordinates of ${\displaystyle \mathbf {c} _{i}.}$ The top half of (2) is equivalent to ${\displaystyle A\mathbf {w} +B\mathbf {v} -\mathbf {f} =0.}$ So to obtain a smoothing spline, one should minimize the scalar field ${\displaystyle F:\mathbb {R} ^{N+d+1}\rightarrow \mathbb {R} }$ defined by

${\displaystyle F(\mathbf {w} ,\mathbf {v} )=|A\mathbf {w} +B\mathbf {v} -\mathbf {f} |^{2}+\lambda C\mathbf {w} ^{\textrm {T}}A\mathbf {w} .}$

The equations

${\displaystyle {\frac {\partial F}{\partial w_{i}}}=2A_{i*}(A\mathbf {w} +B\mathbf {v} -\mathbf {f} )+2\lambda CA_{i*}\mathbf {w} =0\quad {\textrm {for}}\ i=1,2,\dots ,N}$

and

${\displaystyle {\frac {\partial F}{\partial v_{i}}}=2B_{i*}^{\textrm {T}}(A\mathbf {w} +B\mathbf {v} -\mathbf {f} )=0\quad {\textrm {for}}\ i=1,2,\dots ,d+1}$

(where ${\displaystyle A_{i*}}$ denotes row ${\displaystyle i}$ of ${\displaystyle A}$) are equivalent to the two systems of linear equations ${\displaystyle A(A\mathbf {w} +B\mathbf {v} -\mathbf {f} +\lambda C\mathbf {w} )=0}$ and ${\displaystyle B^{\textrm {T}}(A\mathbf {w} +B\mathbf {v} -\mathbf {f} )=0.}$ Since ${\displaystyle A}$ is invertible, the first system is equivalent to ${\displaystyle A\mathbf {w} +B\mathbf {v} -\mathbf {f} +\lambda C\mathbf {w} =0.}$ So the first system implies the second system is equivalent to ${\displaystyle B^{\textrm {T}}\mathbf {w} =0.}$ Just as in the previous smoothing spline coefficient derivation, the top half of (2) becomes ${\displaystyle (A+\lambda CI)\mathbf {w} +B\mathbf {v} =\mathbf {f} .}$

This derivation of the polyharmonic smoothing spline equation system did not assume the constraints necessary to guarantee that ${\displaystyle \textstyle {\int _{{\mathcal {\mathbb {R} }}^{d}}|\nabla ^{m}f|^{2}\,d\mathbf {x} }=Cw^{\textrm {T}}Aw.}$ But the constraints necessary to guarantee this, ${\displaystyle \textstyle {\sum w=0}}$ and ${\displaystyle \textstyle {\sum w\mathbf {c} =0},}$ are a subset of ${\displaystyle B^{\textrm {T}}w=0}$ which is true for the critical point ${\displaystyle w}$ of ${\displaystyle F.}$ So ${\displaystyle \textstyle {\int _{{\mathcal {\mathbb {R} }}^{d}}|\nabla ^{m}f|^{2}\,d\mathbf {x} }=Cw^{\textrm {T}}Aw}$ is true for the ${\displaystyle f}$ formed from the solution of the polyharmonic smoothing spline equation system. Because the integral is positive for all ${\displaystyle w\neq 0,}$ the linear transformation resulting from the restriction of the domain of linear transformation ${\displaystyle A}$ to ${\displaystyle w}$ such that ${\displaystyle B^{T}w=0}$ must be positive definite. This fact enables transforming the polyharmonic smoothing spline equation system to a symmetric positive definite system of equations that can be solved twice as fast using the Cholesky decomposition.[5]

## Examples

The next figure shows the interpolation through four points (marked by "circles") using different types of polyharmonic splines. The "curvature" of the interpolated curves grows with the order of the spline and the extrapolation at the left boundary (x < 0) is reasonable. The figure also includes the radial basis functions phi = exp(-r2) which gives a good interpolation as well. Finally, the figure includes also the non-polyharmonic spline phi = r2 to demonstrate, that this radial basis function is not able to pass through the predefined points (the linear equation has no solution and is solved in a least squares sense).

Interpolation with different polyharmonic splines that shall pass the 4 predefined points marked by a circle (the interpolation with phi = r2 is not useful, since the linear equation system of the interpolation problem has no solution; it is solved in a least squares sense, but then does not pass the centers)

The next figure shows the same interpolation as in the first figure, with the only exception that the points to be interpolated are scaled by a factor of 100 (and the case phi = r2 is no longer included). Since phi = (scale*r)k = (scalek)*rk, the factor (scalek) can be extracted from matrix A of the linear equation system and therefore the solution is not influenced by the scaling. This is different for the logarithmic form of the spline, although the scaling has not much influence. This analysis is reflected in the figure, where the interpolation shows not much differences. Note, for other radial basis functions, such as phi = exp(-k*r2) with k=1, the interpolation is no longer reasonable and it would be necessary to adapt k.

The same interpolation as in the first figure, but the points to be interpolated are scaled by 100

The next figure shows the same interpolation as in the first figure, with the only exception that the polynomial term of the function is not taken into account (and the case phi = r2 is no longer included). As can be seen from the figure, the extrapolation for x < 0 is no longer as "natural" as in the first figure for some of the basis functions. This indicates, that the polynomial term is useful if extrapolation occurs.

The same interpolation as in the first figure, but without the polynomial term

## Discussion

The main advantage of polyharmonic spline interpolation is that usually very good interpolation results are obtained for scattered data without performing any "tuning", so automatic interpolation is feasible. This is not the case for other radial basis functions. For example, the Gaussian function ${\displaystyle e^{-k\cdot r^{2}}}$ needs to be tuned, so that ${\displaystyle k}$ is selected according to the underlying grid of the independent variables. If this grid is non-uniform, a proper selection of ${\displaystyle k}$ to achieve a good interpolation result is difficult or impossible.

• To determine the weights, a dense linear system of equations must be solved. Solving a dense linear system becomes impractical if the dimension ${\displaystyle N}$ is large, since the memory required is ${\displaystyle O(N^{2})}$ and the number of operations required is ${\displaystyle O(N^{3}).}$
• Evaluating the computed polyharmonic spline function at ${\displaystyle M}$ data points requires ${\displaystyle O(MN)}$ operations. In many applications (image processing is an example), ${\displaystyle M}$ is much larger than ${\displaystyle N,}$ and if both numbers are large, this is not practical.

Recently, methods have been developed to overcome the aforementioned difficulties. For example Beatson et al.[7] present a method to interpolate polyharmonic splines at one point in 3 dimensions in ${\displaystyle O(\log N)}$ operations instead of ${\displaystyle O(N)}$ operations.

## References

1. ^ R.L. Harder and R.N. Desmarais: Interpolation using surface splines. Journal of Aircraft, 1972, Issue 2, pp. 189-191
2. ^ J. Duchon: Splines minimizing rotation-invariant semi-norms in Sobolev spaces. Constructive Theory of Functions of Several Variables, W. Schempp and K. Zeller (eds), Springer, Berlin, pp.85-100
3. ^ G.F. Fasshauer G.F.: Meshfree Approximation Methods with MATLAB. World Scientific Publishing Company, 2007, ISPN-10: 9812706348
4. ^ A. Iske: Multiresolution Methods in Scattered Data Modelling, Lecture Notes in Computational Science and Engineering, 2004, Vol. 37, ISBN 3-540-20479-2, Springer-Verlag, Heidelberg.
5. ^ a b Powell, M. J. D. (1993). "Some algorithms for thin plate spline interpolation to functions of two variables" (PDF). Cambridge University Dept. of Applied Mathematics and Theoretical Physics technical report. Retrieved January 7, 2016.
6. ^ Evans, Lawrence (1998). Partial Differential Equations. Providence: American Mathematical Society. pp. 450–452. ISBN 0-8218-0772-2.
7. ^ R.K. Beatson, M.J.D. Powell, and A.M. Tan: Fast evaluation of polyharmonic splines in three dimensions. IMA Journal of Numerical Analysis, 2007, 27, pp. 427-450.


text:
Introduction

This vignette decribes how to solve a TSP using ompr.

Wikipedia gives the following definition:

The travelling salesman problem (TSP) asks the following question: Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the origin city?

Also that Wikipedia article is a good starting point if you want to know more about the topic.

With the basic definition you have a set of verticies (cities) and a set of edges (connection between cities). Each edge has an associated distance $$d > 0$$. That distance could be travel time, distance in km or the monetary cost associated with traveling from one city to another. Restrictions on the distances lead to special cases of the problem. For example the metric-TSP requires that the triangle inequality holds for all triples of edges.

In this vignette we will construct a TSP with random points within an Euclidean space.

Setting

First let us import some librarys

library(knitr)
library(dplyr)
library(ggplot2)

The number of cities:

n <- 10

Boundary of our Euclidean space:

# from 0 to ...
max_x <- 500
max_y <- 500

Some random cities:

set.seed(123456)
cities <- data.frame(id = 1:n, x = runif(n, max = max_x), y = runif(n, max = max_y))
ggplot(cities, aes(x, y)) +
geom_point()

Now the distance matrix

distance <- as.matrix(dist(select(cities, x, y), diag = TRUE, upper = TRUE))

Model formulation

There are essential two prominent ways to model a TSP as a MILP. One is to formulate the full model using the Miller–Tucker–Zemlin (MTZ) formulation and the other option is to use the so-called sub-tour elimination constraints .1

The first formulation is fairly compact (quadratic many constraints and variables) but is not suitable anymore when n gets larger. The second formulation has exponential many constraints at most, but can solve larger TSPs due to the better LP relaxation. The idea of the latter approach is add constraints to the model during the solution process as soon as a solution was found that contains a sub-tour. For solution strategies like this solvers usually offer callbacks that let’s you modify the model during the the branch-and-cut process - this is however not currently supported by ompr.

Therefor we will use the MTZ formulation and solve a fairly small TSP.

Import ompr.

library(ompr)
model <- MIPModel() %>%

# we create a variable that is 1 iff we travel from city i to j
add_variable(x[i, j], i = 1:n, j = 1:n,
type = "integer", lb = 0, ub = 1) %>%

# a helper variable for the MTZ formulation of the tsp
add_variable(u[i], i = 1:n, lb = 1, ub = n) %>%

# minimize travel distance
set_objective(sum_expr(distance[i, j] * x[i, j], i = 1:n, j = 1:n), "min") %>%

# you cannot go to the same city
set_bounds(x[i, i], ub = 0, i = 1:n) %>%

# leave each city
add_constraint(sum_expr(x[i, j], j = 1:n) == 1, i = 1:n) %>%

# visit each city
add_constraint(sum_expr(x[i, j], i = 1:n) == 1, j = 1:n) %>%

# ensure no subtours (arc constraints)
add_constraint(u[i] >= 2, i = 2:n) %>%
add_constraint(u[i] - u[j] + 1 <= (n - 1) * (1 - x[i, j]), i = 2:n, j = 2:n)
model
## Mixed integer linear optimization problem
## Variables:
##   Continuous: 10
##   Integer: 100
##   Binary: 0
## Model sense: minimize
## Constraints: 110

Results

This model can now be solved by one of the many solver libraries. Here we will use GLPK.

library(ompr.roi)
library(ROI.plugin.glpk)
result <- solve_model(model, with_ROI(solver = "glpk", verbose = TRUE))
## <SOLVER MSG>  ----
## GLPK Simplex Optimizer, v4.63
## 110 rows, 110 columns, 434 non-zeros
##       0: obj =   0.000000000e+00 inf =   2.900e+01 (29)
##      33: obj =   2.118421165e+03 inf =   5.551e-17 (0)
## *    62: obj =   1.450925576e+03 inf =   3.997e-15 (0)
## OPTIMAL LP SOLUTION FOUND
## GLPK Integer Optimizer, v4.63
## 110 rows, 110 columns, 434 non-zeros
## 100 integer variables, 90 of which are binary
## Integer optimization begins...
## +    62: mip =     not found yet >=              -inf        (1; 0)
## +    79: >>>>>   1.471334313e+03 >=   1.450925576e+03   1.4% (4; 0)
## +   103: >>>>>   1.469860058e+03 >=   1.450925576e+03   1.3% (3; 1)
## +   119: mip =   1.469860058e+03 >=     tree is empty   0.0% (0; 7)
## INTEGER OPTIMAL SOLUTION FOUND
## <!SOLVER MSG> ----

To extract the solution we can use get_solution method that will return a data.frame which we can further be used with tidyverse packages.

solution <- get_solution(result, x[i, j]) %>%
filter(value > 0)
kable(head(solution, 3))
variable i j value
x 7 1 1
x 1 2 1
x 5 3 1

Now we need to link back the indexes in our model with the actual cities.

paths <- select(solution, i, j) %>%
rename(from = i, to = j) %>%
mutate(trip_id = row_number()) %>%
tidyr::gather(property, idx_val, from:to) %>%
mutate(idx_val = as.integer(idx_val)) %>%
inner_join(cities, by = c("idx_val" = "id"))
kable(head(arrange(paths, trip_id), 4))
trip_id property idx_val x y
1 from 7 267.4290 439.3217
1 to 1 398.8922 398.9946
2 from 1 398.8922 398.9946
2 to 2 376.7825 296.8970

And plot it:

ggplot(cities, aes(x, y)) +
geom_point() +
geom_line(data = paths, aes(group = trip_id)) +
ggtitle(paste0("Optimal route with cost: ", round(objective_value(result), 2)))

Feedback

Do you have any questions, ideas, comments? Or did you find a mistake? Let’s discuss on Github.


text:
# Chapter 7: Logic of Behavior: Sheaves, Toposes, Languages

« Previous | Next »

## Lecture Videos

 Session 13 Chapter 7: Logic of Behavior: Sheaves, Toposes, Languages Part 1 by Dr. David I. Spivak
 Session 14 Chapter 7: Logic of Behavior: Sheaves, Toposes, Languages Part 2 by Dr. Brendan Fong

Internal links in the chapter file below are non-functional. For working links, open the file for the full textbook: An Invitation to Applied Category Theory: Seven Sketches in Compositionality (PDF - 2.6MB).

7.1 How can we prove our machine is safe?

7.2 The category Set as an exemplar topos

7.2.1 Set-like properties enjoyed by any topos

7.2.2 The subobject classifier

7.2.3 Logic in the topos Set

7.3 Sheaves

7.3.1 Presheaves

7.3.2 Topological spaces

7.3.3 Sheaves on topological spaces

7.4 Toposes

7.4.1 The subobject classifier $$\Omega$$ in a sheaf topos

7.4.2 Logic in a sheaf topos

7.4.3 Predicates

7.4.4 Quantification

7.4.5 Modalities

7.4.6 Type theories and semantics

7.5 A topos of behavior types

7.5.1 The interval domain

7.5.2 Sheaves on $$\mathbb{IR}$$

7.5.3 Safety proofs in temporal logic


text:
# No/Infinitely Many Square Roots of 2 by 2 Matrices

## Problem 512

(a) Prove that the matrix $A=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ does not have a square root.
Namely, show that there is no complex matrix $B$ such that $B^2=A$.

(b) Prove that the $2\times 2$ identity matrix $I$ has infinitely many distinct square root matrices.

## Proof.

### (a) The matrix $A$ does not have a square root.

Let $B=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$ be the matrix such that $B^2=A$.
Since
\begin{align*}
B^2=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}
=\begin{bmatrix}
a^2+bc & ab+bd\\
ca+dc & cb+d^2
\end{bmatrix}
=\begin{bmatrix}
a^2+bc & (a+d)b\\
(a+d)c & cb+d^2
\end{bmatrix},
\end{align*}
it follows from $B^2=A$ that
$\begin{bmatrix} a^2+bc & (a+d)b\\ (a+d)c & cb+d^2 \end{bmatrix}=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}.$

Comparing entries we obtain four equations
\begin{align*}
a^2+bc&=0 \tag{1}\\
(a+d)b&=1\tag{2}\\
(a+d)c&=0\tag{3}\\
cb+d^2&=0.\tag{4}
\end{align*}

Equation (3) gives $a+d=0$ or $c=0$.
If $a+d=0$, then equation (2) becomes $0=1$. This is impossible and thus $c=0$.
Since $c=0$, equations (1) and (4) yield that $a=d=0$.
However, inserting these into (2) gives $0=1$.
Hence there is no solution satisfying these equations.

Therefore, there is no square root of the matrix $A$.

### (b) The identity matrix has infinitely many square roots

Let
$B=\begin{bmatrix} 1 & r\\ 0& -1 \end{bmatrix},$ where $r$ be an arbitrary real numbers.

Then we compute directly and obtain
\begin{align*}
B^2=\begin{bmatrix}
1 & r\\
0& -1
\end{bmatrix}\begin{bmatrix}
1 & r\\
0& -1
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}=I.
\end{align*}

Hence $B$ is a square root of the identity matrix $I$.
Since $r$ is an arbitrary real numbers, there are infinitely many square roots of $I$.

## Related Question.

Problem.
Prove that a positive definite matrix has a unique positive definite square root.

For a solution of this problem, see the post
A Positive Definite Matrix Has a Unique Positive Definite Square Root

Using the numbers appearing in $\pi=3.1415926535897932384626433832795028841971693993751058209749\dots$ we construct the matrix \[A=\begin{bmatrix} 3 & 14 &1592& 65358\\ 97932& 38462643& 38& 32\\...


text:
# Return evenly spaced numbers on a log scale and set the base in Numpy

NumpyServer Side ProgrammingProgramming

To return evenly spaced numbers on a log scale, use the numpy.logspace() method in Python Numpy. The 1st parameter is the "start" i.e. the start of the sequence. The 2nd parameter is the "end" i.e. the end of the sequence. The 3rd parameter is the "num" i.e. the number of samples to generate. Default is 50. The 4th parameter is the "base" i.e. the base of the log space. The step size between the elements in ln(samples) / ln(base) (or log_base(samples)) is uniform.

In linear space, the sequence starts at base ** start (base to the power of start) and ends with base ** stop (see endpoint below). The start is the base ** start is the starting value of the sequence. The stop is the base ** stop is the final value of the sequence, unless endpoint is False. In that case, num + 1 values are spaced over the interval in log-space, of which all but the last are returned. The base of the log space. The step size between the elements in ln(samples) / ln(base) (or log_base(samples)) is uniform. Default is 10.0.

The axis in the result to store the samples. Relevant only if start or stop are array-like. By default (0), the samples will be along a new axis inserted at the beginning. Use -1 to get an axis at the end.

## Steps

At first, import the required library −

import numpy as np

To return evenly spaced numbers on a log scale, use the numpy.logspace() method. The 1st parameter is the "start" i.e. the start of the sequence −

arr = np.logspace(10.0, 20.0, num = 3, base = 2.0)
print("Array...\n", arr)

Get the array type −

print("\nType...\n", arr.dtype)

Get the dimensions of the Array −

print("\nDimensions...\n",arr.ndim)

Get the shape of the Array −

print("\nShape...\n",arr.shape)

Get the number of elements −

print("\nNumber of elements...\n",arr.size)

## Example

import numpy as np
# To return evenly spaced numbers on a log scale, use the numpy.logspace() method in Python Numpy
# The 1st parameter is the "start" i.e. the start of the sequence
# The 2nd parameter is the "end" i.e. the end of the sequence
# The 3rd parameter is the "num" i.e the number of samples to generate. Default is 50.
# The 4th parameter is the "base" i.e. the base of the log space.
# The step size between the elements in ln(samples) / ln(base) (or log_base(samples)) is uniform. Default is 10.0.
arr = np.logspace(10.0, 20.0, num = 3, base = 2.0)
print("Array...\n", arr)

# Get the array type
print("\nType...\n", arr.dtype)

# Get the dimensions of the Array
print("\nDimensions...\n",arr.ndim)

# Get the shape of the Array
print("\nShape...\n",arr.shape)

# Get the number of elements
print("\nNumber of elements...\n",arr.size)

## Output

Array...
[1.024000e+03 3.276800e+04 1.048576e+06]

Type...
float64

Dimensions...
1

Shape...
(3,)

Number of elements...
3
Updated on 16-Feb-2022 10:04:24


text:
Lemma 91.12.3. Let $A$ be a ring. Let $P = A[E]$ be a polynomial ring. Set $I = (e; e \in E) \subset P$. The maps $\text{Tor}_ i^ P(A, I^{n + 1}) \to \text{Tor}_ i^ P(A, I^ n)$ are zero for all $i$ and $n$.

Proof. Denote $x_ e \in P$ the variable corresponding to $e \in E$. A free resolution of $A$ over $P$ is given by the Koszul complex $K_\bullet$ on the $x_ e$. Here $K_ i$ has basis given by wedges $e_1 \wedge \ldots \wedge e_ i$, $e_1, \ldots , e_ i \in E$ and $d(e) = x_ e$. Thus $K_\bullet \otimes _ P I^ n = I^ nK_\bullet$ computes $\text{Tor}_ i^ P(A, I^ n)$. Observe that everything is graded with $\deg (x_ e) = 1$, $\deg (e) = 1$, and $\deg (a) = 0$ for $a \in A$. Suppose $\xi \in I^{n + 1}K_ i$ is a cocycle homogeneous of degree $m$. Note that $m \geq i + 1 + n$. Then $\xi = \text{d}\eta$ for some $\eta \in K_{i + 1}$ as $K_\bullet$ is exact in degrees $> 0$. (The case $i = 0$ is left to the reader.) Now $\deg (\eta ) = m \geq i + 1 + n$. Hence writing $\eta$ in terms of the basis we see the coordinates are in $I^ n$. Thus $\xi$ maps to zero in the homology of $I^ nK_\bullet$ as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
## anonymous 5 years ago if x+1/x =8, find the x^2 + x^-2

1. anonymous

Is it ${x+1 \over x}=8?$

2. anonymous

no, it is x+(1/x)

3. anonymous

(x + 1/x)^2 = x^2 + 2 + 1/x^2

4. anonymous

It is further = x^2 + 2 + x^-2

5. anonymous

finally (x + 1/x)^2 - 2 = x^2 + x^-2 a

6. anonymous

$x+{1 \over x}=8 \implies x^2+2+x^{-2}=16 \implies x^2+x^{-2}=14$

7. anonymous

You hv x + 1/x = 8 Squaring both sides (x + 1/x)^2 = 8^2 = 16 S0, 16 - 2 = x^2 + x^-2 so 14 is the answer

8. anonymous

Sorry not 16, it's 64. So the final answer is 62.

9. anonymous

the answer I have are: a) 60, b)58, c)62, d)0, e)non So the answer must be e)

10. anonymous

Correct Anwar!!!

11. anonymous

12. anonymous

option c)

13. anonymous

$x+{1 \over x}=8 \implies (x+{1 \over x})^2=8^2 \implies x^2+2+x^{-2}=64$ $\implies x^2+x^{-2}=62$

14. anonymous

Here is the corrected version You hv x + 1/x = 8 Squaring both sides (x + 1/x)^2 = 8^2 = 64 S0, 64 - 2 = x^2 + x^-2 so 62 is the answer

15. anonymous

16. anonymous

yes, thanks guys its clear now

17. anonymous

You're welcome!


text:
“Be kind, for everyone you meet is fighting a hard battle” - Often attributed to Plato but likely from Ian McLaren (pseudonym of Reverend John Watson)

## Tuesday, March 31, 2009

### Horsepower, fuel efficiency, and thermodynamic efficiciency

I received a comment to my post on the Moller Skycar Volantor suggesting that I "look up the Wankel rotary engine (Mazda RX8) - 1.3L = 200+ HP." I'm not completely sure what the commenter was getting at, but I think he or she was saying that it's possible to get the power claimed by Moller in a package of the type he claims to have it in. Let me be clear (quoting our President): I agree. There's no question in my mind that that's possible. I didn't discount that possibility in my post.

What I did say isn't possible is to develop that amount of power and still achieve 20 miles per gallon of fuel at the speed claimed. That is, the combination of airspeed, fuel economy, and engine power claimed is not realistic. I'd like to delve into this in a little more detail. An internal combustion engine works by taking some combustible fuel into an enclosed space and igniting it. This results in a large temperature rise, and the consequent increase in pressure is used to push down a piston, turn a rotor, or turn a turbine thus converting the potential energy in the chemical bonds of the fuel into mechanical energy and using it to do work.

Any given fuel has a fixed amount of energy available for release by oxidation for any given mass, volume, or number of "moles" of substance. In a perfect world, unencumbered by the second law of thermodynamics, we could harness 100% of this energy to do useful work. Even in such a perfect world, no more could be had. And once the time over which this chemical potential energy is converted by oxidation into internal energy or "heat" is noted, we can divide the energy in the quantity of fuel (energy available is the same as work that can be done) by the time, we have work divided by time, or power.

For example, my Piper Saratoga PA32R-301T will burn about 18 gallons of fuel per hour to go about 170 knots. A "knot" is one nautical mile per hour, and a nautical mile is 6080 feet or about 1.152 statute miles. So, the plane burns 18 gallons in an hour to go 170*1.152 or 195.6 miles. Thus it's achieving 10.9 m.p.g., but more to the point of this post, it's burning 18 gallons per hour. There are a variety of sites with somewhat differing values for the energy density of the 100LL avgas I burn in the Saratoga (see here and here for example) so I'll use an average of 32.6 megajoules/liter or 123.4 (easy to remember) megajoules/gallon.

So, I'm releasing 18*123.4 megajoules/hour of chemical potential energy. Since an hour is 3600 seconds that's 18*123.4/3600 megajoules per second or 617,000 joules per second. By definition, a joule/second is a watt so this is 617 kilowatts or 827 horsepower. This is interesting, my pilot's operating handbook says that the power setting producing this fuel flow is 70% of the 300 maximum continuous horsepower available, or 210 horsepower. Thus, I'm using (210/827)*100% or 25.4% of the heat released by burning the avgas. This is pretty close to the type of efficiency we've come to expect of internal combustion engines in typical applications.

The maximum possible efficiency allowed by the laws of thermodynamics for a "heat engine" was determined in the 19th Century beginning with the work of Sadi Carnot and is determined solely by the temperature of the working fluid (fuel/air mixture as it oxidizes in this case) and the cold reservoir (the atmosphere) and for reasonable temperatures, as stated in my post on the Moller Skycar, is about 82%. For a working internal combustion engine, the maximum efficiency is related to the compression ratio and, for a typical compression ratio of 10.5:1, works out to be about 61%. Various factors involving friction, the operating points of the engine, inefficiencies in the fuel delivery and exhaust, and many other factors cause the actual efficiency to be as low as it is. These considerations apply equally to the Wankel or rotary engines in the Moller Skycar.

But, if the rate of fuel burn in volume/time and the type of fuel are known (or can be calculated) then the maximum available power can be calculated. An estimate of engine efficiency can then give the power available to turn wheels, turn a propellor, turn a ducted fan, etc. This is what I did in the Moller Skycar post, using the miles per hour divided by miles per gallon to give gallons per hour and thus the heat energy available. This enabled me to demonstrate that the claims made by Moller for the Skycar are not feasible.

## Sunday, March 29, 2009

### "Exponentially" really does mean something

In my never ending quest to understand what people believe about the world, and specifically about energy and fuel economy, I run across repeated misunderstanding of the word "exponentially." This or that "increases exponentially." Now some things do, in fact, increase exponentially. Money in a bank account at a fixed rate of interest comes to mind. The population of bacteria in a petri dish or humans on a finite planet (for a while) are other examples.

But it means more than "gets big quickly." People who should know better, or at least should learn better before using it this way frequently use it to mean this. An example is an article at the Planet Green web site entitled "Become a "Hypermiler," Save Even More Gas" by one Collin Dunn of Corvallis, OR. As usual, driving more slowly is first on his list. OK, I agree completely, but then he states that "The amount of drag your vehicle generates increases exponentially with each increase in speed; that is, driving a little faster generates a lot more drag, which requires more gas to overcome."

Well, it's just some guy writing an article for a Cable Channel web site, right? But he got this gem from the Toyota Open Road Blog where they go into detail to precisely state their error: "The amount of drag your vehicle generates is not linear – it does not increase at the same rate as your vehicle’s speed does. Instead, drag is more or less proportional to the square of speed. It increases exponentially." NO, NO, NO!

"More or less proportional to the square of speed" is correct. That is, if you know the drag at some speed, and want the drag at some other speed, take the ratio of the second speed to the first, square it, multiply the drag at the first speed by the squared ratio and there's your approximate drag at the second speed. For example, doubling the speed would approximately increase aerodynamic drag by a factor of four. This is NOT exponential, the mathematically astute refer to it as a power function. The variable (speed) is the base, the power (2 in this case) is the exponent. The exponent is fixed.

An exponential function has a fixed base, and the variable is the exponent. Now, depending on the ranges of the variables and the fixed numbers and the proportionality constants, the exponential function may be smaller than the power function at some values of the variable, but the exponential function is always ultimately larger for large values of the variable. So while something that increases exponentially really does get large very fast, at least after a while, it's not true that anything that increases quickly at any point increases exponentially. This most assuredly does include aerodynamic drag.

And don't even get me started on "mega."

## Thursday, March 26, 2009

### A potpourri of cluelessness

In my reading of various blogs, news articles, forum posts, etc. I've encountered a large number of writings indicative of the strange and distorted ideas people have about the subjects of this blog, that is, of fuel economy, energy use, physics, and life. I've decided to periodically quote some of them along with, where it's relevant, my own comments.

The following is from Physics Forums where the question was "Does RPM affect gas mileage?" An answer was:

"Assume that you travel a set distance at the same speed, first in 3rd gear, then in 4th. Can you see that the engine will turn more times in the lower gear. Each revolution of the engine will "consume" the same amount of air/fuel mixture. Therefore you must consume more air/fuel mixture at the lower gear.

Now a lot of modern engines are getting smarter about feeding fuel so, the assumption of a constant a/f mixture may not be valid. With intelligent fuel metering the millage difference may be small.

The biggest difference to fuel consumption is your rate of acceleration. If you like to feel some acceleration and take pride in your ability to get to 60mph (100kph) then you will see improvement by playing "old lady" for a while.

You do more work when doing 5s to 60 vs 10s to 60. This increased work MUST be reflected in fuel consumption."

Well. The fact is that RPM will affect gas mileage but this guy's argument is full of errors. He (I assume it's "he") correctly states that the lower gear will require more revolutions to travel a given distance than a higher gear, but then claims that each revolution will consume the same amount of fuel/air mixture regardless of the gear. This is false. The lower gear will travel a lesser distance and therefore do less work for each revolution, thus requiring less throttle since the gear ratio will give a greater mechanical advantage. But you'll need to apply this lesser throttle through more revolutions of the engine. Now, due to throttling and frictional losses, you'll be less efficient in the lower gear but it's certainly not true that "Each revolution of the engine will "consume" the same amount of air/fuel mixture." And this was true before cars included computerized controls, it's only a matter of the throttle position.

Next, "Integral" (his screen name, complete with integral symbol avatar) is completely off base with his concept of acceleration and work. Going from 0 to 60 adds the same amount of kinetic energy to the car, and thus takes the same amount of work, regardless of whether it's done in 5 seconds or 10. It's true that taking 10 seconds is more fuel efficient, since the energy is added over a longer distance but Integral confuses work with power. And this is from a Physics Forum where he has enough posts to be a mentor. Further, someone points out his errors and he stridently defends them, even insulting the corrector.

Moving on, "Josh," in reply to a tip at Daily Fuel Economy Tip that suggested minimizing use of electrical accessories said: "This is simply not true …. and alternator is not operated on a clutch therefore it spins at the same speed no matter what is being used in the car. It is not on a clutch like the AC that just kicks in when the compressor comes on."

Of course, this is false. Raising the electrical load causes an increased magnetic field in the alternator field coil, resulting in a larger torque to be overcome by the alternator drive belt, thereby using more fuel.

These are merely misunderstandings of fundamental physical principles and don't reach the heights of looniness achieved by some of the more "outro" world wide web denizens. I'll cover more of each type in subsequent posts.

## Saturday, March 21, 2009

### The Moller Skycar

Since I was a little boy, there has been talk of flying cars. And since I was a little boy, Paul Moller has been a short few years away from going into production on such a vehicle. He still is, though now it's called the Skycar Volantor. It's reminiscent of the wag's remark about fusion energy: "fusion is the energy source of the future, and always will be."

But what about the M400 Skycar? It's "specifications" can be found here. What a fine way to get around! 275 m.p.h. cruise at better than 20 m.p.g. It's stated that the production model will employ eight rotary engines rated at 150 h.p. per engine burning any of a variety of fuels, but ethanol is suggested. This is the power required for the vertical take off capability. At this site, there are a variety of videos, including a hover test of the M400, it apparently will get off the ground. There are also some specifications wherein it's indicated that the "nominal continuous power" is 720 horsepower and the fuel is ethanol. This would be running at 60% power and makes sense. It's not stated whether the engines are turbocharged, but they must be since the operational ceiling is stated to be 36,000 feet (!).

So let's take a look at a vehicle utilizing internal combustion engines to get 20 m.p.g. (the specs. say ">20" so this should be conservative) at 275 m.p.h. This means it's burning 13.75 gallons of ethanol per hour. OK, we find here that ethanol has an energy density of 6,100 watt hours/liter, or about 83,100,000 joules/gallon. So the total available energy in 13.75 gallons of ethanol is 1,143,000,000 joules. Burning this in one hour or 3600 seconds at 100% efficiency will produce 317,395 watts or 426 horsepower. This means the Skycar's engines are using fuel with an efficiency of 169%. I rather doubt it.

Let's suppose, then, that the figures come from burning gasoline. Gasoline has a significantly higher energy density, and would mean the claim is only an efficiency of 106%. Still higher than the figures one typically sees for an internal combustion engine. All right, suppose that the 720 horsepower only applies to the "top speed," listed as 360 m.p.h. Now, with gasoline, we're looking at an efficiency of 81%. Well, at least it's no longer in the category of the "over unity" nut cases, but I've never seen a real engine with such a specification and, unless it's operating at a very high temperature and dumping into a very cold reservoir, thermodynamics won't allow it. If a heat engine is operating from 1500 K into a reservoir of 273 K (about 2240 degrees fahrenheit into 32 degrees fahrenheit) the absolute theoretical maximum efficiency is 81.8%. No real engine comes close.

Is 720 horsepower sufficient to produce a speed of 360 m.p.h.? It's likely that it is. The Piper Meridian, for example, utilizes a Pratt & Whitney PT6A-42A turboshaft engine running at 500 horsepower to cruise at 260 knots, or about 300 m.p.h. It burns on the order of 40 gallons of Jet A fuel each hour to do it though. Speed available goes up approximately with the cube root of power, so 720 horsepower should be able to give an increase of about 12.9% over the Meridian, all else being equal. This would be about 339 m.p.h. Maybe the Skycar is a little aerodynamically cleaner, maybe its ducted fans are slightly more efficient than the Meridian's propeller. I'll call it plausible.

So if the Skycar were to be a real product, what are its claimed advantages? Well, it's driveable at low speed from your garage to an approved takeoff location where it can take off vertically. It's being designed, ultimately, to be fully automated, no pilot intervention necessary, and thus able to be used by those with no flying skills of any kind. You were previously able to buy a place in line for a production Skycar at a cost for delivery of about $400K to$1M (according to some dated web sites) depending on where in line you were. You could put only part of that down to reserve your place. It seems that that's no longer the case though.

Dr. Moller is now stating that, in limited production, the M400 Skycar Volantor will sell for about $500,000 and be available in "about three years." In mass production (i.e., when everyone you know is buying one) they'll sell for$60,000 to $80,000. If you'd rather own the company than the Skycar, it's sold over the counter as MLER.OB on the OTCBB ("Over the Counter Bulletin Board"). You can peruse the financials here. Note the negative book value. No wonder they're not taking deposits. Its market capitalization values the company at$8.27M, with 54% owned by insiders and 5% holders. Pretty thin. You just can't beat the laws of thermodynamics.


text:
# Calculation of Laplacian operator in Riemannian manifold

Let $u$ be non-negative smooth subharmonic function on a complete Riemannian manifold $M$. Take $v=u^{p/2}$ with $p>1$. Then in this paper

Yau, Shing-Tung, Some function-theoretic properties of complete Riemannian manifold and their applications to geometry, Indiana Univ. Math. J. 25, 659-670 (1976); erratum ibid. 31, 607 (1982). ZBL0335.53041,

(more precisely formula (2.19) on the 5th page), it is given that

$$v\Delta v=\frac{p}{2}vu^{p/2-1}\Delta u+\frac{p}{2}(\frac{p}{2}-1)vu^{p/2-2}|du|^2.$$