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# Line Islands in a Word Search

My third word search related challenge in a row. :)

## Challenge:

#### Brief explanation of what a word search is:

In a word search you'll be given a grid of letters and a list of words. The idea is to cross off the words from the list in the grid. The words can be in eight different directions: horizontally from left-to-right or right-to-left; vertically from top-to-bottom or bottom-to-top; diagonally from the topleft-to-bottomright or bottomright-to-topleft; or anti-diagonally from the topright-to-bottomleft or bottomleft-to-topright.

#### Actual challenge:

Given a list of coordinates indicating the crossed out words within a grid (in any reasonable format), output how many line-islands there are.

We've had challenges involving finding the amount of islands in a matrix. But in this case it's different: just looking at the grid, words could be right next to each other, but would still be two separated line-islands. E.g. In the following partially solved word search, we have four separated line-islands, even though the letters of the words are right next to each other.

This partially solved word search above would result in an output of 4.

If 'word' FG was FGH like this:

The output would have been 3 instead, because the FGH+NKH now form a single connected line-island.

In this challenge we'll only be taking the coordinates of the words in a grid as input, and output the amount of line-islands.

## Challenge rules:

• You're allowed to take the input in any reasonable format. It could be pair of coordinates per word to indicate their start/end positions (e.g. [[[0,0],[0,2]],[[1,1],[1,3]],[[3,1],[1,3]],[[3,2],[2,3]]] for the second example above); it could be a full list of coordinates per word (e.g. [[[0,0],[0,1],[0,2]],[[1,1],[1,2],[1,3]],[[3,1],[2,2],[1,3]],[[3,2],[2,3]]]); a list of bit-mask matrices per word (e.g. [[[1,1,1,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]],[[0,0,0,0],[0,1,1,1],[0,0,0,0],[0,0,0,0]],[[0,0,0,0],[0,0,0,1],[0,0,1,0],[0,1,0,0]],[[0,0,0,0],[0,0,0,0],[0,0,0,1],[0,0,1,0]]]); etc. If you're unsure if a possible input-format is valid, leave a comment down below.
• All 'words' are guaranteed to have at least two letters.
• You can assume the input won't be empty.
• You can optionally take the dimensions of the grid as additional input.
• Note that coordinates of 'words' are not guaranteed to be in the same positions to still form a single line-island! E.g. this grid would result in 1 because the lines cross, but neither 'word' share the same letter-positions:

## General rules:

• This is , so the shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.

## Test cases:

• Input as start/ending coordinates of the 'words': [[[0,0],[0,2]],[[1,1],[1,2]],[[3,1],[1,3]],[[3,2],[2,3]]]
• Input as coordinates of the complete 'words': [[[0,0],[0,1],[0,2]],[[1,1],[1,2]],[[3,1],[2,2],[1,3]],[[3,2],[2,3]]]
• Input as bit-mask matrices of the 'words': [[[1,1,1,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]],[[0,0,0,0],[0,1,1,0],[0,0,0,0],[0,0,0,0]],[[0,0,0,0],[0,0,0,1],[0,0,1,0],[0,1,0,0]],[[0,0,0,0],[0,0,0,0],[0,0,0,1],[0,0,1,0]]]

Output: 4

• Input as start/ending coordinates of the 'words': [[[0,0],[0,2]],[[1,1],[1,3]],[[3,1],[1,3]],[[3,2],[2,3]]]
• Input as coordinates of the complete 'words': [[[0,0],[0,1],[0,2]],[[1,1],[1,2],[1,3]],[[3,1],[2,2],[1,3]],[[3,2],[2,3]]]
• Input as bit-mask matrices of the 'words': [[[1,1,1,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]],[[0,0,0,0],[0,1,1,1],[0,0,0,0],[0,0,0,0]],[[0,0,0,0],[0,0,0,1],[0,0,1,0],[0,1,0,0]],[[0,0,0,0],[0,0,0,0],[0,0,0,1],[0,0,1,0]]]

Output: 3

• Input as start/ending coordinates of the 'words': [[[0,0],[1,1]],[[0,1],[1,0]]]
• Input as coordinates of the complete 'words': [[[0,0],[1,1]],[[0,1],[1,0]]]
• Input as bit-mask matrices of the 'words': [[[1,0],[0,1]],[[0,1],[1,0]]]

Output: 1

• Input as start/ending coordinates of the 'words': [[[1,9],[8,9]],[[8,6],[1,6]],[[1,4],[4,7]],[[9,0],[0,9]],[[1,0],[6,0]],[[6,1],[3,4]],[[9,4],[9,9]],[[0,1],[0,8]],[[9,3],[1,3]],[[0,0],[9,9]]]
• Input as coordinates of the complete 'words': [[[1,9],[2,9],[3,9],[4,9],[5,9],[6,9],[7,9],[8,9]],[[8,6],[7,6],[6,6],[5,6],[4,6],[3,6],[2,6],[1,6]],[[1,4],[2,5],[3,6],[4,7]],[[9,0],[8,1],[7,2],[6,3],[5,4],[4,5],[3,6],[2,7],[1,8],[0,9]],[[1,0],[2,0],[3,0],[4,0],[5,0],[6,0]],[[6,1],[5,2],[4,3],[3,4]],[[9,4],[9,5],[9,6],[9,7],[9,8],[9,9]],[[0,1],[0,2],[0,3],[0,4],[0,5],[0,6],[0,7],[0,8]],[[9,3],[8,3],[7,3],[6,3],[5,3],[4,3],[3,3],[2,3],[1,3]],[[0,0],[1,1],[2,2],[3,3],[4,4],[5,5],[6,6],[7,7],[8,8],[9,9]]]
• Input as bit-mask matrices of the 'words': pastebin

Output: 4

• Input as start/ending coordinates of the 'words': [[[8,6],[1,6]],[[1,4],[4,7]],[[9,0],[0,9]],[[6,1],[3,4]],[[9,4],[9,9]],[[9,3],[1,3]],[[0,0],[9,9]]]
• Input as coordinates of the complete 'words': [[[8,6],[7,6],[6,6],[5,6],[4,6],[3,6],[2,6],[1,6]],[[1,4],[2,5],[3,6],[4,7]],[[9,0],[8,1],[7,2],[6,3],[5,4],[4,5],[3,6],[2,7],[1,8],[0,9]],[[6,1],[5,2],[4,3],[3,4]],[[9,4],[9,5],[9,6],[9,7],[9,8],[9,9]],[[9,3],[8,3],[7,3],[6,3],[5,3],[4,3],[3,3],[2,3],[1,3]],[[0,0],[1,1],[2,2],[3,3],[4,4],[5,5],[6,6],[7,7],[8,8],[9,9]]]
• Input as bit-mask matrices of the 'words': pastebin

Output: 1

• Can I take the board dimensions as a 2nd argument? Mar 15 at 3:10
• @Jonah "You can optionally take the dimensions of the grid as additional input." It was already one of the rules, so yes. :) Mar 15 at 7:33
• Woops. Thsnks Kevin Mar 15 at 12:52

# Jelly,  22  20 bytes

żSH$ƝẎ)ṭfƇ@ẎQʋ€ÐLQL  A monadic Link that accepts a list of lists of pairs of integers - the full list of coordinates for each word - and yields the island count. Try it online! (Two islands each being two words crossing at cell corners, where equal-island words are non-adjacent in the input list.) Or see the test-suite. ### How? Needs updating: ṭfƇ@ẎṢQ - Link 1, add crossing words: Coordinates, (all) Word Coordinates @ - with swapped arguments: Ƈ - filter keep (those Word Coordinate lists) for which: f - filter keep (i.e. contain some of the same entries) ṭ - tack to Coordinates Ẏ - tighten Ṣ - sort Q - deduplicate żSH$ƝẎ)ç€ÐLQL - Main Link: Word Coordinates
)        - for each list in Word Coordinates:
\$Ɲ          -   for neighbours:
S             -     sum (vectorises)
H            -     halve
ż              - Word Coordinates zip-with those "intermediate coordinates"
Ẏ         - tighten
   - use as both arguments of:
ÐL    -   loop until no change with:
ç€      -     call Link 1 for each
Q  - deduplicate
L - length

# Python3, 560 bytes:

lambda w:len({str(i[0])for i in g(w)})
S=lambda a:[(t:=(a[1][0]-a[0][0])/([1,(k:=a[1][1]-a[0][1])][k!=0])*(k>0)),[a[0][1],t*-1*a[0][1]+a[0][0]][k!=0]]
M=lambda j,k:[(x:=(k[1]-j[1])/(j[0]-k[0])),j[0]*x+j[1]][::-1]
N=lambda s,p:min(t:=[s[0][0],s[1][0]])<=p[0]<=max(t)and min(t:=[s[0][1],s[1][1]])<=p[1]<=max(t)
V=lambda a,b:(j:=S(a))[0]!=(k:=S(b))[0]and N(a,m:=M(j,k))and N(b,m)
def g(w):
Q,S=[[i]for i in w],[]
while Q:
if(v:=Q.pop(0))[-1]not in S:
if[]==(j:=[i for i in w if i not in S and V(v[-1],i)]):yield v
else:Q=[v+[i]for i in j]+Q
S+=v[-1:]

Try it online!

• t*-1* can be -t* :) Mar 15 at 7:45

# Wolfram Language (Mathematica), 90 bytes

Length@ConnectedComponents@Graph[#->If[#⋂#2=={},##]&@@@Tuples[Total/@#~Tuples~2&/@#,2]]&

Try it online!

A simple set intersection between the lists of coordinates would detect almost all the cases except the 'phantom intersection' of two diagonal 'words'.
In this special case, these phantom intersecting diagonals, have both 2 coordinates that add up to a same one.

+---------+---------+
|A        |B        |
|  x   y  | x+1  y  |
|         |         |
+---------X---------+         A coords.          B coords.
|B        |A        |    (x,y) + (x+1,y+1) = (x,y+1) + (x+1,y)
|  x  y+1 | x+1 y+1 |
|         |         |
+---------+---------+

Unfortunately a test for this would not detect the standard intersection of two words that actually have one common coordinate. As such my first implementation combined two test:

• the intersection between the plain lists of coordinates #
• the intersection between the lists of the sums of consecutive coordinates Partition[#,2,1] (this also assumed the lists would always be in order).

But then I thought why not sum every pair of coordinates (multisets) of a 'word'?
Also pushed by the fact that Tuples[#,2] is shorter than Partition[#,2,1]

## Technicalities

The only doubt: can there be false positives?
That is, can two 'words' that do not share any coordinate be mapped into 'words' that do share some coordinates?

To sum all the coordinates in a 'word' with themselves is essentially a x2 scaling done in a redundant fashion.

So, for each mapped coordinate of a 'word' $$\(x_i+x_j,y_i+y_j)\$$ we can infer its preimage: a pair of coordinates of 'word' that yield the mapped one. Hence if two 'words' share a mapped coordinate they also have to share the relative pair of 'core coordinates'.

This pair of 'core coordinates' (P,b) are the two coordinates around the possibly 'phantom point' (@) $$\(\frac{x_i+x_j}{2},\frac{y_i+y_j}{2})\$$

+---------+---------+   +---------+---------+   +---------+---------+   +---------+---------+    +---------+---------+
|BBBBBBBBB|         |   |PPPPPPPPP|         |   |         |         |   |PPPPPPPPP|         |    |         |PPPPPPPPP|
|BBBB@BBBB|         |   |PPPPPPPPP|         |   |         |         |   |PPPPPPPPP|         |    |         |PPPPPPPPP|
|BBBBBBBBB|         |   |PPPPPPPPP|         |   |         |         |   |PPPPPPPPP|         |    |         |PPPPPPPPP|
+---------+---------+ , +----@----+---------+ , +---------+---------+ , +---------@---------+ OR +---------@---------+
|         |         |   |bbbbbbbbb|         |   |PPPPPPPPP|bbbbbbbbb|   |         |bbbbbbbbb|    |bbbbbbbbb|         |
|         |         |   |bbbbbbbbb|         |   |PPPPPPPPP@bbbbbbbbb|   |         |bbbbbbbbb|    |bbbbbbbbb|         |
|         |         |   |bbbbbbbbb|         |   |PPPPPPPPP|bbbbbbbbb|   |         |bbbbbbbbb|    |bbbbbbbbb|         |
+---------+---------+   +---------+---------+   +---------+---------+   +---------+---------+    +---------+---------+
`


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# Physics 204B (Spring 2019) SALEH & < Problem Set 01 Problem 1 1 of 5> Review...

###### Question:

Physics 204B (Spring 2019) SALEH & < Problem Set 01 Problem 1 1 of 5> Review You may want to review (Pages 613-616) Compute the strength of the net electric force on charge A in the figure (Figure 1). Assume that g Express your answer to two significant figures and include the appropriate units 1.4 nC and g -5.2 nC FA- 1 Value! Units Submit Part B What is the direction of the net electric force on charge A? D right Figure 1 of 1 down gi 92 up O left 1 em 1 cm

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# How could you edit a specific group policy using a batch file

Im working on over 700 computers in a school district and have written a small program that i intend to write to a cd. The program is set to autorun when the disk is inserted and prompt the screen resolution of the computer and what computer the building is in (the different school buildings). Afterwards the program will run a batch file that copies a default desktop from the disk and into the windows\web\wallpaper directory. It also replaces other files that have been customized for the school district.

To finish changing the theme of the computer, i need to have the file make a few edits to the group policy and the registry. How would i be able to use the program to makes these changes? Would it all be written into the batch or would the batch have to initiate another file (like a registry file)?

-

All the group policy editor does is set registry keys. If you can identify what keys are being set for the policy you want, you can use reg.exe to set those keys.

reg.exe add HKCU\Software\path\to\regkey\ /v valuename /d newvalue

-

Registry will work for the first user, but after a new user logs in, GP will change those settings for the new user.

To make deployment gp, do this: Set one machine's GP to the way you want it. Once you are done, goto C:\windows\System32\GroupPolicy and copy the contents. (note: this is a hidden file). On the next machines you want, just paste back in the file.

If you do this in a batch, you will have to run the batch as administrator to touch the C:\windows\system32 folder. (UAC, the bane IT)

-


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# Why are mass and energy related by the speed of light?

1. Oct 12, 2010

### Kyxha

Im a 3rd year physics major. I understand the equation E=mc$$^{2}$$.
But i don't understand why mass and energy are related by the number c.

How are mass, energy, and the speed of light so inextricably related that
we can describe their relationship with one simple equation?

2. Oct 12, 2010

### Sleuth

The point should be that they are not simply related... they are EXACTLY the very same thing....
Mass IS energy, and energy IS mass...

3. Oct 12, 2010

### Kyxha

Okay, they are each other.
Why are the forms of pure energy and pure mass related by c^2?

4. Oct 12, 2010

### tom.stoer

Einstein introduced the idea that the speed of light - c - is identical in all reference frames. From this idea one can derive the relation between space and time, i.e. "spacetime"; one can show that the four-vector (ct,x) has an "invariant length c²t² - x² which is identical in all reference frames. In addition one can show that structurally the same applies to energy and momentum: the four-vector (E,cp) has an "invariant length E² -c²p² = (mc²)² which is identical in all reference frames; it's value (mc²)² is the invariant mass (squared) i.e.the rest mass (squared).

So the relation via rest mass m and energy E = mc² is basically derived from the requirement of universal c.

5. Oct 12, 2010

### diazona

EDIT: looks like tom.stoer just said the same thing I did but more concisely.

It has to do with the energy-momentum four-vector.
$$p^\mu = \begin{pmatrix}E/c \\ p_x \\ p_y \\ p_z\end{pmatrix}$$
(The c there is just a unit conversion thing) You know, of course, that if you change from one inertial reference frame to another, the values of E and p that you get for any particular particle (assuming the particle is not accelerating) will change; for example, you might have both px and E decrease.

One of the insights of special relativity is that those changes in E and p are analogous to a rotation. Think about rotating a vector in 3D space. As an example, suppose the original vector is (1,0,0), then when you rotate it, its x component will decrease and its y and/or z components will increase. The energy-momentum four-vector acts similarly, except that instead of one component increasing when another decreases, both the energy and momentum increase or decrease together. Mathematically, though, both cases look very similar.

Anyway, Einstein realized that if you just plug in the usual expressions for energy and momentum,
$$\begin{pmatrix}\frac{1}{2}mv^2/c \\ mv_x \\ mv_y \\ mv_z\end{pmatrix}$$
and look at how that four-vector changes as you move from one reference frame to another, it doesn't correspond to a rotation. For example, if you go from a reference frame in which a particle is moving to another one in which it is at rest, the vector goes from having nonzero elements to being all zeros, which means its norm goes from nonzero to zero - but a rotation can't change the norm.

In order to make a vector that transforms according to the rules of a generalized rotation (i.e. invariant norm), you need to ensure that not all of the components of the vector go to zero at the same time. Now, if you have a particle at rest, clearly its momentum will be zero. That means its energy has to be nonzero. If you were to calculate what the norm of the vector has to be to keep it invariant, you'd find out that you need to add $mc$ to the energy component, which means that every particle has a "rest energy" of $mc^2$ in addition to whatever kinetic energy it may have. That's where the formula $E = mc^2$ comes from: it's the inherent energy that a particle has just by virtue of being a massive particle.

Now, you may wonder, why c specifically? That also has to do with the idea of shifts in reference frame being like rotations. When you perform one of these "rotations" by speeding up or slowing down, you shift time into space or vice-versa. Since time and space are measured in different units, we need a conversion factor to convert between them. The conversion factor will obviously have to have units of speed. Now, one of the postulates of relativity is that the there is a particular speed which is invariant across all reference frames, i.e. no matter how much you speed up or slow down, something traveling at this invariant speed will never appear to go any faster or slower relative to you. It's that speed that appears in $E = mc^2$, or in the full version, $E^2 = p^2c^2 + m^2c^4$. It just happens that the first reason anyone had to suspect that that speed was special was that it was the speed of light, and that's why we call it "the speed of light".

6. Oct 12, 2010

### enotstrebor

Though the question may have some ambiquity do to interpretation of E=mc^2 equation (today it is a invariant/rest mass only relationship), it is clearly applicable to the invariant rest mass and thus any answer to the question must explicitly apply to the invariant/rest mass to energy relationship.

For the invariant mass energy relationship (E=mc^2), why is energy and mass related by c? As the question applies to the invariant mass energy, the answer can not have anything to do with the energy momentum four-vector.

For the invariant mass energy relationship (E=mc^2), why is energy and mass related by c? As the question applies to the invariant mass energy, the answer can not have anything to do with reference frames.

Yes that is true, but why for every particle is the rest mass and rest energy related by the square of a value that is the same as the value of the speed of light.

A good physics theory should not only mathematically relate values, but explain why the values are related. As far as I know, the SM does not indicate any reason why rest mass and rest energy are related this way.

7. Oct 12, 2010

### diazona

What?! That doesn't make sense at all. Just because the equation deals with rest mass does not mean a derivation based on four-momentum or Lorentz transformations is invalid.

8. Oct 12, 2010

### enotstrebor

If mass is energy and energy is mass then this statement requires the photon to have mass.

Clearly the photon is energy but it has no mass!

The particle mass energy form is an energy form where (by todays view) the measured energy value is dependent on the reference frame but whose mass (characterized by resistance to change in velocity within a given reference frame) never goes to zero in any reference frame.

Photon (particle) energy form is an energy form whose energy value is frame dependent and if the reference frame is moving in the same direction as the photon, but ahead of it, as this reference frame approaches the speed of light, the measured value goes toward zero.

So it is clear that the mass form of energy does not have the properties of the photon form of energy. Thus the above statement is incorrect.

9. Oct 12, 2010

### enotstrebor

It is not a derivation question, it is a fundamental question. One that goes below the mathematics, and asks what is the fundamental nature of particle mass energy.

As the E in E=mc^2 can be photon energy, let me rephrase the question using the particle mass energy and the equivalent photon energy. Note experimentally 2 equivalent rest mass particles, the electron and the anti-electron (positron) can annihilate to produce two photons. I will reference this specific case (the answer to the question must be universally true so it must be true it this specific case also) with the condition that both electron and anti-electron are at rest at time of annihilation (even if not possible to physically do this experiment, the present theory allows us to consider this possibility).

As the two particles are of equal mass and as conservation of momentum requires the two resultant photons must be of equal energy and momentum and must head out in opposite directions (net momentum 0), the rest mass relationship is 2 m c^2 = 2 E_ph

Thus for, lets say the anti-electron (or electron) the single particle's mass is converted to a single photon where the relationship between the single particle's rest mass has the equivalent energy of a photon given by the relationship m c^2 = E_ph.

What is the physics that allows the mass energy form to be at rest and resist a change to velocity and what is the substantive physics difference (not the mathematical difference) of the anti-particle that interacts with the particle nature that causes the transform of the mass form of energy to the photon energy form and why does this energy form always and immediately accelerate to the velocity c and what is the source nature of the photon acceleration which unlike mass energy occurs without requiring any energy input and without any energy loss?

Once you have answered this fundamental question, which requires a deeper understanding of the source of nature (why nature is this way) not just a mathematical form that indicates how nature behaves, then one can answer the question of why the relationship E=mc^2 exists in the first place rather than how relationships are known to behave.

For example string theory attempts to give an underlying foundation (source of behavior) to the present theory, the SM, that models the behavior of particles.

Many/most believe the model of the behavior of the particle is a model of the particle itself.

Some, including some famous physicists do not. If you are interested, read Roger Penrose's book, The Road to Reality. It is both a mathematically informative and a philosophically informative.

10. Oct 13, 2010

### tom.stoer

The italic sentences are not correct. We tried to explain above what happens (perhaps the explanation is not clear for you).

Lorentz transformations are relations for general reference frames. The name "rest mass" is missleading; it should read "invariant mass" b/c it has the same value in all reference frames. The Lorentz transformations automatically generate a relation like E = mc² were c is a numerical constant (speed) used to construct the Lorentz transformations (in SR: speed of light). That's why these relations clearly have something to do with reference frames and Lorentz transformations.

No.

A good physical theory makes some basic assumptions (a small set of axioms and principles) and derives falsifiable predictions. So the "why" is encoded in the axioms and principles. This is what SR does very efficiently. One basic principle is the universal speed of light.

The SM is based on relativistic QFT and therefore automatically has this property E = mc². So the SM can't say anything about the "why", it must simply reproduce this relation as a consistency check.

This is wrong

As I said, the relation E² = mc² applies to the rest mass or invariant mass only. Therefore this relation says that the photon is massless (zero rest mass) which is correct. In general the relation E² - (cp)² = (mc²)² which is modified for photons: E² - (cp)² = 0

Last edited: Oct 13, 2010
11. Oct 13, 2010

### tom.stoer

No.

Both SM and ST do neither question theory of relativity, nor do they add any additional structure which could explain these relations.

They may provide mechanism to generate rest mass like spontaneous symmetry breaking etc., but they do not explain E = mc² in a more fundamental sense. The SM uses SR for its construction is therefore by construction not able to explain its own basics. The same applies to string theory: String theory uses a target space with certain space-time symmetries for its construction is therefore not able to explain the symmetry structure of spacetime.

12. Oct 13, 2010

### Sleuth

It may seem stupid to you, but I think that special relativity does not explain WHY.
The only possible WHY is that for dimensional reasons, if you want to obtain an energy from a mass you ha to multiply it for a velocity squared, and the ONLY universal constant in general relativity with the right dimensions is just c...

13. Oct 13, 2010

### DrDu

I would argue like this: For a massive particle, the energy in the rest frame is M.
Then $$E=\sqrt{M^2+p^2c^2}$$. At low speeds, we get $$E\approx M \left(1+\frac{1}{2} \frac{p^2c^2}{M^2}\right)=M+\frac{1}{2} \frac{ p^2c^2}{M}$$. If we want the second term to have the non-relativistic form $$\frac{1}{2} \frac{p^2}{m}$$, then the constant M has to be chosen as $$mc^2$$.

Last edited: Oct 13, 2010
14. Oct 13, 2010

### enotstrebor

I have never seen a derivation of E=mc^2 from the Lorentz transformation which results in the invariance of E - p^2c^2. Can you show it or give a reference.

I do know Einstein derived the relationship without using the Lorentz transform and one can go to Wiki Mass-energy equivalence and see E=mc^2 derived independent of a Lorentz transform.

Thus, as the relationship E=mc^2 can be derived independently and without using the Lorentz transform, I maintain my stand and statement that the answer to the original question is also independent of the Lorentz transform, (i.e. has nothing to do with Lorentzian transforms or four-vectors).

15. Oct 14, 2010

### tom.stoer

Einstein's derivation was not directly based on the Lorentz trf., instead he found the Lorentz transformation properties; so we know that his derivation and the Lorentz transformation are deeply related. Length contraction and time dilation are nothing else but special cases cases of the Lorentz transformation.

The Lorentz transformation itself follows from various reasons; a physical reason is the Lorentz-covariance of Maxwell’s equations which can be checked both theoretically and experimentally; in addition one finds from Maxwell’s theory a constant speed c of el.-mag. waves.

In the following we need the Lorentz transformation in momentum space. It is structurally identical to the trf. for space-time itself. Einstein gave a derivation for E and p, but the Lorentz symmetry can be seen independently via transforming Maxwell’s equation into momentum space.

The idea is rather simple:

Take the energy-momentum four-vector (in the following c=1)

$$p^\mu = (E, \vec{p})$$

It transforms under a Lorentz-transformation as

$$p^\mu \to (p^\prime)^\mu = \Lambda ^\mu_\nu p^\nu$$

A Lorentz transformation is exactly the transformation that leaves the four-vector "length" invariant:

$$p^\mu p_\mu = (p^\prime)^\mu (p^\prime)_\mu$$

Calculating the invariant length one finds

$$p^\mu p_\mu = E^2 - p^2$$

The reason why the same trf. applies in position and in momentum space is that the expression

$$p^\mu x_\mu$$

must be invariant as well.

As you are familiar with Einstein's derivation of the expressions for E(v) and p(v) you can simply check that

$$p^\mu p_\mu = E^2 - p^2 = m^2$$

holds for all v, that means of all reference frames. Use

$$\gamma(v) = \frac{1}{\sqrt{1-v^2}}$$

$$E(v) = \gamma(v) m$$

$$p(v) = \gamma(v) mv$$

and do the calculation explicitly.

This is all you need. You have the relation between E, p, v and m and you have an invariant "length" m. All what remains is to find out what this m means physically. The simplest way is to make a Taylor expansion for small velocities. You get

$$E(v) = \gamma(v) m = m + \frac{m}{2}v^2 + \ldots$$

That means that you must identify m with the rest mass of the particle. So rest energy and rest mass are related via

$$E = m$$

Introducing the omitted factors of c again into the equations results in

$$E = mc^2$$

So rest mass and rest energy are related via this famous equation.

Your statement that the derivation can be done w/o using the Lorentz transformation is not really true. The Lorentz transformation is somehow hidden in all these expressions, but in principle it is the underlying mathematical structure.

Last edited: Oct 14, 2010
16. Oct 14, 2010

### DrDu

Nowadays, rest mass is defined as a Casimir invariant of "massive representations" of the Poicare algebra. So Lorentz invariance lies at the very heart of the definition of mass.

17. Oct 14, 2010

### tom.stoer

Exactly

18. Oct 14, 2010

### Staff: Mentor

Since mass has units of kg and energy has units of kg m²/s² it is clear that the conversion factor must have units of m²/s². The only combination of universal constants with those units is c².

19. Oct 15, 2010

### enotstrebor

If you would actually go and look on wiki at the evidence you are given, rather than just insisting you are right, you might find there are other avenues you haven't considered.

20. Oct 15, 2010

### tom.stoer

could you give an explanation of the ideas from on wiki?
do you have a link? there are many sites on wiki


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# Understand Disturbance with Data - Flooding and Erosion - Earth analytics python course module

Welcome to the first lesson in the Understand Disturbance with Data - Flooding and Erosion module. This module uses time series data to explore the impacts of a flood. Learn how to use Google Earth imagery, NOAA precipitation data and USGS stream flow data to explore the 2013 Colorado floods.

In this lesson, you will learn to use the time series function in Google Earth to explore changes in the landscape associated with the 2013 Colorado floods.

## Learning Objectives

At the end of this activity, you will be able to:

• Use the timeline function in Google Earth to view time series imagery data.
• Identify and describe a driver and an impact of a flood event.

## What You Need

You will need to download and install Google Earth on your computer and then download the .kml file below.

In early September 2013, a slow moving cold air front moved through Colorado, USA intersecting with a warm, humid air front. The clash between the cold and warm airs fronts yielded heavy rain. This rain, combined with a drought conditions, Colorado soil conditions and other factors yielded devastating flooding across the Front Range in Colorado, USA.

## Use Imagery to Detect Change

Spatially located (georeferenced) imagery, collected using satellites and airplanes provides a powerful visual record of landscape changes. Google Earth, has a time series feature that allows you to look at imagery of the earth, across time in certain areas of the Earth.

You will use this feature to look at the landscape in Boulder, Colorado both before and after the floods.

### How to View Time Series Imagery in Google Earth

• Double click on the .kmz file that you downloaded above. It should open in Google Earth.

Data Tip: If the .kmz file doesn’t automatically open when you double click on it, try to Open Google Earth, go to File –> Open in Google Earth. Finally, navigate to the location of your downloaded file (~Documents/data/co-flood/locations) and open it.

• Once you have the .kmz file open, notice it is listed in the Temporary Places section of the places window. It should automatically zoom you into to an area in North Boulder, Colorado. If it doesn’t double click on the text Locations of Significant Damage.
• Click on the show historical imagery button in Google Earth
• When you click on the show historical imagery, a slider will appear in the upper LEFT hand corner of your window. Scroll back and forth through time to get used to the slider
• Finally, double click on one of the thumbtacks from Locations of Significant Damage. Scroll to 10/2012 and then to 10/2013. Do you see any differences?
• Check out the other thumbtack. What differences do you see?

## Activity: What Changes Do You See?

This activity is a required part of your homework. If you are in class however, you can complete this during the class period. As a group, discuss the following questions.

• What differences do you see in the landscape between 2012 and 2013?
• For each difference: What do you think caused that difference?
• For each difference: How can you quantitatively record the difference?
• For each CAUSE listed above, could you somehow quantitatively record the “size” or impact of the cause?
• Was the cause - caused by something else (i.e. did something else DRIVE the cause)?

Now, go to the Canvas Discussion page. In the wk1-google-earth-activity create a new post that includes:

1. atleast 3 differences or changes that your group noticed in the imagery
2. what data you could use to quantify each different that you noticed
3. what might have caused that change in the landscape?
4. How would you measure that driver that caused the change in the landscape

Consider organizing your response in a table like the one below. Do not use the example below in your Canvas Discussion page post UNLESS you can come up with another way to quantify how many (or the degree to which) trees were lost! Also note the example provided below is using manually measured data. You may know of other ways to measure tree loss!

An example table that shows what you might post in the Canvas Discussion page:

DifferenceData you can use to quantify the changeWhat drove the changeHow could you quantify that driver
Trees are missing after the floodhuman field survey of treesLarge boulders swept downstream by flood waterschanges in terrain pre-post flood

When you have completed your post, as an individual, find atleast 2 other posts that you like and comment on them.


text:
# How To Count Shuffle permutations

Let $n\in \mathbb{N}$ and $S(n)$ the permutation group on $\{1,\ldots,n\}$.

For any $p,q\in \mathbb{N}$ with $p+q=n$, the set $Sh(p,q)\subset S(n)$ is the set of all permutations $\tau$ such that $\tau(1)< \cdots \tau(p)$ and $\tau(p+1)<\cdots <\tau(n)$, usually called the set of $(p,q)$-shuffles.

More generally for any $k;p_1,\ldots,p_k\in\mathbb{N}$ with $p_1+\cdots +p_k=n$, the set of $(p_1,\ldots,p_k)$-shuffles $Sh(p_1,\ldots,p_k)\subset S(n)$ is the set of all permutations $\tau$, such that $\tau(1)< \cdots \tau(p_1)$, $\tau(p_1+1)<\cdots <\tau(p_2)$ (up to) $\tau(n-p_k+1)<\cdots <\tau(n)$.

====

Now the question is: HowTo calculate the cardinality $|Sh(p_1,\ldots,p_k)|$?

I know that $|Sh(p,q)|=\binom{p+q}{q}$ but just as a fact. From empirical tests I would 'guess' $|Sh(p_1,\ldots,p_k)|=\binom{n}{p1,\ldots,p_k}$, where the expression on the left means the multinomial coefficient, but I'm more interested in deriving such an appropriate expression.

Choose $p$ numbers from the set $1..n$ and there is a unique shuffle such that $\tau(1)..\tau(p)$ to have those $p$ values and the other $q$ fill in the remaining places. The multinomial coefficinet $n\choose p_1...p_k$ is the number of ways to partition $1..n$ into subsets of cardinality $p_i$, and each subset can again be ordered each such partition is a unique shuffle.


text:
# Calculus of Variations-Hanging String

Calculus of Variations--Hanging String

## Homework Statement

A uniform string of length 2 meters hangs from two supports of the same height, 1 meter
apart. By minimizing the potential energy of the string, find the equation describing the
curve it forms and, in particular, find the vertical distance between the supports and the
lowest point of the string.

## Homework Equations

U = λ g $\int y(x)dx$
0≤ x ≤ 1
S = $\int\sqrt{dx^2+dy^2}$dt
Euler-Lagrange Equation

## The Attempt at a Solution

Been stuck on this one for awhile, tried solving naively by just solving the lagrange equation for potential energy and that obviously got me an answer of U = 0. I also tried to solve the arc length equation for y(x) and I got y'[x] = √3, so that is obviously wrong. I'm not sure how to add the constraints of the supports into the equation. Im probably going to have to parameterize y and x in terms of some other variable.

I have a feeling that the shape is going to be a cycloid...just having a hard time proving it.


text:
## Journal of the Mathematical Society of Japan

### Special values of the spectral zeta functions for locally symmetric Riemannian manifolds

Yasufumi HASHIMOTO

#### Abstract

In this paper, we establish the formulas expressing the special values of the spectral zeta function $\zeta_{\rm\Delta}(n)$ of the Laplacian $\rm\Delta$ on some locally symmetric Riemannian manifold $\Gamma\backslash G/K$ in terms of the coefficients of the Laurent expansion of the corresponding Selberg zeta function. As an application, we give a numerical estimation of the first eigenvalue of $\rm\Delta$ by computing the values $\zeta_{\rm\Delta}(n)$ numerically, when $\Gamma\backslash G/K$ is a Riemann surface with $\Gamma$ being the quaternion group.

#### Article information

Source
J. Math. Soc. Japan, Volume 57, Number 1 (2005), 217-232.

Dates
First available in Project Euclid: 13 October 2006

https://projecteuclid.org/euclid.jmsj/1160745823

Digital Object Identifier
doi:10.2969/jmsj/1160745823

Mathematical Reviews number (MathSciNet)
MR2114730

Zentralblatt MATH identifier
1084.11052

#### Citation

HASHIMOTO, Yasufumi. Special values of the spectral zeta functions for locally symmetric Riemannian manifolds. J. Math. Soc. Japan 57 (2005), no. 1, 217--232. doi:10.2969/jmsj/1160745823. https://projecteuclid.org/euclid.jmsj/1160745823

#### References

• T. Arakawa, S. Koyama and M. Nakasuji, Arithmetic forms of Selberg zeta functions with applications to prime geodesic theorem, Proc. Japan Acad. Ser. A Math. Sci., 78 (2002), 120–125.
• A. A. Bytsenko, F. L. Williams, Asymptotics of the heat kernel on rank-1 locally symmetric spaces, J. Phys. A, 32 (1999), 5773–5779.
• R. Gangolli, Zeta functions of Selberg's type for compact space forms of symmetric spaces of rank one, Illinois J. Math., 21 (1977), 1–41.
• R. Gangolli and G. Warner, Zeta functions of Selberg's type for some noncompact quotients of symmetric spaces of rank one, Nagoya Math. J., 78(1980), 1–44.
• Y. Hashimoto, The Euler-Selberg constants for nonuniform lattices of rank one symmetric spaces, Kyushu J. Math., 57 (2003), 347–370.
• Y. Hashimoto, Y. Iijima, N. Kurokawa and M. Wakayama, Euler's constants for the Selberg and the Dedekind zeta functions, to appear in Bull. Belg. Math. Soc. Simon Stevin.
• D. Hejhal, The Selberg trace formula of $PSL(2,\bR)$ I, II, Lecture Notes in Math., 548, 1001, Springer, 1976, 1983.
• D. Hejhal, A classical approach to a well-known spectral correspondence on quaternion groups, Number theory, New York, 1983–84, 127–196; Lecture Notes in Math., 1135, Springer, Berlin, 1985.
• H. H. Kim, Functoriality for the exterior square of $GL_4$ and the symmetric fourth of $GL_2$ (with Appendix 1 by D. Ramakrishnan and Appendix 2 by Kim and P. Sarnak), J. Amer. Math. Soc., 16 (2003), 139–183.
• H. H. Kim and F. Shahidi, Cuspidality of symmetric powers with applications, Duke Math. J., 112 (2002), 177–197.
• N. Kurokawa, Gamma factors and Plancherel measures, Proc. Japan Acad. Ser. A Math. Sci., 68 (1992), 256–260.
• W. Luo, Z. Rudnick and P. Sarnak, On Selberg's eigenvalue conjecture, Geom. Funct. Anal., 5 (1995), 387–401.
• R. Miatello, On the Plancherel measure for linear Lie groups of rank one, Manuscripta Math., 29 (1979), 249–276.
• B. Randol, On the analytic continuation of the Minakshisundaram-Pleijel zeta function for compact Riemann surfaces, Trans. Amer. Math. Soc., 201 (1975), 241–246.
• R. J. Schoof, Quadratic fields and factorization, Computational methods in number theory, Part II, Math. Centre Tracts, 155 (1982), 235–286.
• A. Selberg, On the estimation of Fourier coefficients of modular forms, Proc. Sympos. Pure Math., 8 Amer. Math. Soc., Providence, 1965, 1–15.
• H. Shimizu, Hokei kansu. I. (Japanese) [Automorphic functions. I], Second edition, Iwanami Shoten Kiso Sūgaku [Iwanami Lectures on Fundamental Mathematics], 8, Daisū [Algebra], Iwanami Shoten, Tokyo, 1984.
• F. Steiner, On Selberg's zeta function for compact Riemann surfaces, Phys. Lett. B, 188 (1987), 447–454.
• A. Voros, Spectral function, special function and the Selberg zeta function, Comm. Math. Phys., 110 (1987), 439–465.
• H. Wada, A table of ideal class numbers of real quadratic fields, Sophia Kokyuroku in Mathematics, 10 (1981).
• F. L. Williams, Meromorphic continuation of Minakshisundaram-Pleijel series for semisimple Lie groups, Pacific J. Math., 182 (1998), 137–156.


text:
Publicité ▼

## définition - Metric_(mathematics)

voir la définition de Wikipedia

Wikipedia

# Metric (mathematics)

In mathematics, a metric or distance function is a function which defines a distance between elements of a set. A set with a metric is called a metric space. A metric induces a topology on a set but not all topologies can be generated by a metric. A topological space whose topology can be described by a metric is called metrizable.

In differential geometry, the word "metric" is also used to refer to a structure defined only on a differentiable manifold which is more properly termed a metric tensor (or Riemannian or pseudo-Riemannian metric).

## Definition

A metric on a set X is a function (called the distance function or simply distance)

d : X × XR

(where R is the set of real numbers). For all x, y, z in X, this function is required to satisfy the following conditions:

1. d(x, y) ≥ 0     (non-negativity, or separation axiom)
2. d(x, y) = 0   if and only if   x = y     (identity of indiscernibles, or coincidence axiom)
3. d(x, y) = d(y, x)     (symmetry)
4. d(x, z) ≤ d(x, y) + d(y, z)     (subadditivity / triangle inequality).

Note that condition 1 and 2 together produce positive definiteness. Also, observe that the first condition is implied by the others.

A metric is called an ultrametric if it satisfies the following stronger version of the triangle inequality where points can never fall 'between' other points:

For all x, y, z in X, d(x, z) ≤ max(d(x, y), d(y, z))

A metric d on X is called intrinsic if any two points x and y in X can be joined by a curve with length arbitrarily close to d(x, y).

For sets on which an addition + : X × XX is defined, d is called a translation invariant metric if

d(x, y) = d(x + a, y + a)

for all x, y and a in X.

## Notes

These conditions express intuitive notions about the concept of distance. For example, that the distance between distinct points is positive and the distance from x to y is the same as the distance from y to x. The triangle inequality means that the distance from x to z via y is at least as great as from x to z directly. Euclid in his work stated that the shortest distance between two points is a line; that was the triangle inequality for his geometry.

If a modification of the triangle inequality

4*. d(x, z) ≤ d(z, y) + d(y, x)

is used in the definition then property 1 follows straight from property 4*. Properties 2 and 4* give property 3 which in turn gives property 4.

## Examples

$d(x,y)=\sum_{n=1}^\infty \frac{1}{2^n} \frac{p_n(x-y)}{1+p_n(x-y)}$
is a metric defining the same topology. (One can replace $\frac{1}{2^n}$ by any summable sequence $(a_n)$ of strictly positive numbers.)

## Equivalence of metrics

For a given set X, two metrics d1 and d2 are called topologically equivalent (uniformly equivalent) if the identity mapping

id: (X,d1) → (X,d2)

For example, if $d$ is a metric, then $\min (d, 1)$ and ${d \over 1+d}$ are metrics equivalent to $d.$

## Metrics on vector spaces

Norms on vector spaces are equivalent to certain metrics, namely homogeneous, translation-invariant ones. In other words, every norm determines a metric, and some metrics determine a norm.

Given a normed vector space $(X, \|\cdot\|)$ we can define a metric on X by

$d(x,y) := \| x-y\|$.

The metric d is said to be induced by the norm $\|\cdot\|$.

Conversely if a metric d on a vector space X satisfies the properties

• $d(x,y) = d(x+a,y+a)$ (translation invariance)
• $d(\alpha x, \alpha y) = |\alpha| d(x,y)$ (homogeneity)

then we can define a norm on X by

$\|x\| := d(x,0)$

Similarly, a seminorm induces a pseudometric (see below), and a homogeneous, translation invariant pseudometric induces a seminorm.

## Generalized metrics

There are numerous ways of relaxing the axioms of metrics, giving rise to various notions of generalized metric spaces. These generalizations can also be combined. The terminology used to describe them is not completely standardized. Most notably, in functional analysis pseudometrics often come from seminorms on vector spaces, and so it is natural to call them "semimetrics". This conflicts with the use of the term in topology.

### Extended metrics

Some authors allow the distance function d to attain the value ∞, i.e. distances are non-negative numbers on the extended real number line. Such a function is called an extended metric. Every extended metric can be transformed to a finite metric such that the metric spaces are equivalent as far as notions of topology (such as continuity or convergence) are concerned. This can be done using a subadditive monotonically increasing bounded function which is zero at zero, e.g. d′(x, y) = d(x, y) / (1 + d(x, y)) or d′′(x, y) = min(1, d(x, y))).

The requirement that the metric take values in [0,∞) can even be relaxed to consider metrics with values in other directed sets. The reformulation of the axioms in this case leads to the construction of uniform spaces: topological spaces with an abstract structure enabling one to compare the local topologies of different points.

### Pseudometrics

A pseudometric on X is a function d : X × XR which satisfies the axioms for a metric, except that instead of the second (identity of indiscernibles) only d(x,x)=0 for all x is required. In other words, the axioms for a pseudometric are:

1. d(x, y) ≥ 0
2. d(x, x) = 0 (but possibly $d(x,y)=0$ for some distinct values $x\ne y$.)
3. d(x, y) = d(y, x)
4. d(x, z) ≤ d(x, y) + d(y, z).

This is the most common generalization of metrics.[citation needed] In some contexts, pseudometrics are referred to as semimetrics because of their relation to seminorms.

### Quasimetrics

Occasionally, a quasimetric is defined as a function that satisfies all axioms for a metric with the possible exception of symmetry:[1]

1. d(x, y) ≥ 0
2. d(x, y) = 0   if and only if   x = y
3. d(x, y) = d(y, x) (dropped)
4. d(x, z) ≤ d(x, y) + d(y, z).

If d is a quasimetric on X, a metric d' on X can be formed by taking

d'(x, y) = 12(d(x, y) + d(y, x)).

Quasimetrics are common in real life. For example, given a set X of mountain villages, the typical walking times between elements of X form a quasimetric because travel up hill takes longer than travel down hill. Another example is a taxicab geometry topology having one-way streets, where a path from point A to point B comprises a different set of streets than a path from B to A. Nevertheless, this notion is rarely used in mathematics, and its name is not entirely standardized.[2]

A quasimetric on the reals can be defined by setting

d(x, y) = yx if yx, and
d(x, y) = 1 otherwise.

The topological space underlying this quasimetric space is the Sorgenfrey line.

### Semimetrics

A semimetric on X is a function d : X × XR that satisfies the first three axioms, but not necessarily the triangle inequality:

1. d(x, y) ≥ 0
2. d(x, y) = 0   if and only if   x = y
3. d(x, y) = d(y, x)

Some authors work with a weaker form of the triangle inequality, such as:

d(x, z) ≤ ρ (d(x, y) + d(y, z))     (ρ-relaxed triangle inequality)
d(x, z) ≤ ρ max(d(x, y), d(y, z))     (ρ-inframetric inequality).

The ρ-inframetric inequality implies the ρ-relaxed triangle inequality (assuming the first axiom), and the ρ-relaxed triangle inequality implies the 2ρ-inframetric inequality. Semimetrics satisfying these equivalent conditions have sometimes been referred to as "quasimetrics",[3] "nearmetrics"[4] or inframetrics.[5]

The ρ-inframetric inequalities were introduced to model round-trip delay times in the internet.[5] The triangle inequality implies the 2-inframetric inequality, and the ultrametric inequality is exactly the 1-inframetric inequality.

### Premetrics

Relaxing the last three axioms leads to the notion of a premetric, i.e. a function satisfying the following conditions:

1. d(x, y) ≥ 0
2. d(x, x) = 0

This is not a standard term. Sometimes it is used to refer to other generalizations of metrics such as pseudosemimetrics[6] or pseudometrics;[7] in translations of Russian books it sometimes appears as "prametric".[8]

Any premetric gives rise to a topology as follows. For a positive real r, the open r-ball centred at a point p is defined as

Br(p) = { x | d(x, p) < r }.

A set is open if for any point p in the set there is an r-ball centred at p which is contained in the set. In general, the open r-balls themselves need not be open sets with respect to this topology. In fact, the interior of an r-ball may be empty. Thus every premetric space is a topological space, and in fact a sequential space.

As for metrics, the distance between two sets A and B, is defined as

d(A, B) = infxA, yB d(x, y).

This defines a premetric on the power set of a premetric space. If we start with a (pseudosemi-)metric space, we get a pseudosemimetric, i.e. a symmetric premetric. Any premetric gives rise to a preclosure operator cl as follows:

cl(A) = { x | d(x, A) = 0 }.

### Pseudoquasimetrics

The presyllables pseudo-, quasi- and semi- can also be combined, e.g., a pseudoquasimetric (sometimes called hemimetric) relaxes both the indiscernibility axiom and the symmetry axiom and is simply a premetric satisfying the triangle inequality. For pseudoquasimetric spaces the open r-balls form a basis of open sets. A very basic example of a pseudoquasimetric space is the set {0,1} with the premetric given by d(0,1) = 1 and d(1,0) = 0. The associated topological space is the Sierpiński space.

Sets equipped with an extended pseudoquasimetric were studied by William Lawvere as "generalized metric spaces".[9][10] From a categorical point of view, the extended pseudometric spaces and the extended pseudoquasimetric spaces, along with their corresponding nonexpansive maps, are the best behaved of the metric space categories. One can take arbitrary products and coproducts and form quotient objects within the given category. If one drops "extended", one can only take finite products and coproducts. If one drops "pseudo", one cannot take quotients. Approach spaces are a generalization of metric spaces that maintains these good categorical properties.

### Important cases of generalized metrics

In differential geometry, one considers metric tensors, which can be thought of as "infinitesimal" metric functions. They are defined as inner products on the tangent space with an appropriate differentiability requirement. While these are not metric functions as defined in this article, they induce metric functions by integration. A manifold with a metric tensor is called a Riemannian manifold. If one drops the positive definiteness requirement of inner product spaces, then one obtains a pseudo-Riemannian metric tensor, which integrates to a pseudo-semimetric. These are used in the geometric study of the theory of relativity, where the tensor is also called the "invariant distance".

## Notes

1. ^ E.g. Steen & Seebach (1995).
2. ^ Rolewicz, Stefan (1987), Functional Analysis and Control Theory: Linear Systems, Springer, ISBN 90-277-2186-6, OCLC 13064804  This book calls them "semimetrics", which is unfortunate because the same term is frequently used for two other generalizations of metrics.
3. ^ Xia, Q. (2009), "The Geodesic Problem in Quasimetric Spaces", Journal of Geometric Analysis 19 (2): 452–479, DOI:10.1007/s12220-008-9065-4
4. ^ Qinglan Xia (2008), "The geodesic problem in nearmetric spaces", Journal of Geometric Analysis: Volume , Issue (009), Page 19 (2): 452–479, arXiv:0807.3377.
5. ^ a b * Fraigniaud, P.; Lebhar, E.; Viennot, L. (2008), The Inframetric Model for the Internet, "2008 IEEE INFOCOM - The 27th Conference on Computer Communications", IEEE INFOCOM 2008. the 27th Conference on Computer Communications: 1085–1093, DOI:10.1109/INFOCOM.2008.163, ISBN 978-1-4244-2026-1, retrieved 2009-04-17 .
6. ^ Buldygin, V.V.; Kozachenko, I.U.V. (2000), Metric characterization of random variables and random processes .
7. ^ Khelemskiĭ (2006), Lectures and exercises on functional analysis .
8. ^ Arkhangel'skii & Pontryagin (1990). Aldrovandi, R.; Pereira, J.G. (1995), An introduction to geometrical physics .
9. ^ Lawvere, F.W. (2002) [1973], Metric spaces, generalised logic, and closed categories, Reprints in Theory and Applications of Categories, 1, pp. 1–37 .
10. ^ Vickers, Steven (2005), "Localic completion of generalized metric spaces I", Theory and Applications of Categories 14: 328–356

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# scipy.signal.lsim2¶

scipy.signal.lsim2(system, U=None, T=None, X0=None, **kwargs)[source]

Simulate output of a continuous-time linear system, by using the ODE solver scipy.integrate.odeint.

Parameters
systeman instance of the lti class or a tuple describing the system.

The following gives the number of elements in the tuple and the interpretation:

• 1: (instance of lti)

• 2: (num, den)

• 3: (zeros, poles, gain)

• 4: (A, B, C, D)

Uarray_like (1D or 2D), optional

An input array describing the input at each time T. Linear interpolation is used between given times. If there are multiple inputs, then each column of the rank-2 array represents an input. If U is not given, the input is assumed to be zero.

Tarray_like (1D or 2D), optional

The time steps at which the input is defined and at which the output is desired. The default is 101 evenly spaced points on the interval [0,10.0].

X0array_like (1D), optional

The initial condition of the state vector. If X0 is not given, the initial conditions are assumed to be 0.

kwargsdict

Additional keyword arguments are passed on to the function odeint. See the notes below for more details.

Returns
T1D ndarray

The time values for the output.

youtndarray

The response of the system.

xoutndarray

The time-evolution of the state-vector.

Notes

This function uses scipy.integrate.odeint to solve the system’s differential equations. Additional keyword arguments given to lsim2 are passed on to odeint. See the documentation for scipy.integrate.odeint for the full list of arguments.

If (num, den) is passed in for system, coefficients for both the numerator and denominator should be specified in descending exponent order (e.g. s^2 + 3s + 5 would be represented as [1, 3, 5]).

#### Previous topic

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text:
Lemma 10.103.8. Let $R$ be a Noetherian local Cohen-Macaulay ring of dimension $d$. Let $0 \to K \to R^{\oplus n} \to M \to 0$ be an exact sequence of $R$-modules. Then either $M = 0$, or $\text{depth}(K) > \text{depth}(M)$, or $\text{depth}(K) = \text{depth}(M) = d$.

Proof. If $d = 0$, then every nonzero $R$-module has depth $0$ and the lemma is true. Assume $d > 0$. Then $\text{depth}(K) > 0$ as $K$ is a submodule of a module of depth $> 0$. Hence the lemma holds if $\text{depth}(M) = 0$. Assume both $\text{depth}(M) > 0$ and $d > 0$. Then we choose $x \in \mathfrak m$ which is a nonzerodivisor on $M$ and on $R$. Then $x$ is a nonzerodivisor on $M$ and on $K$ and it follows by an easy diagram chase that $0 \to K/xK \to (R/xR)^{\oplus n} \to M/xM \to 0$ is exact. Using Lemmas 10.71.7 and 10.103.2 we find the result follows from the result for $K/xK$ over $R/xR$ which has smaller dimension. $\square$

Comment #2966 by Dario Weißmann on

So what if the depth of $R$ is zero?

Also if $M$ is the zero module then the conventions in section 10.71 say it has infinite depth.

Comment #2967 by Dario Weißmann on

Then every (nonzero) module has depth zero, right? So maybe the first line of the proof should be "If $\text{depth}(M)=0$ or $\text{depth}(R)=0$ then the lemma is clear."

Comment #3092 by on

This is an annoying but important lemma. I tried to fix it so it is actually true! Hope I succeeded this time. The fix is here.

There are also:

• 7 comment(s) on Section 10.103: Cohen-Macaulay rings

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text:
# Impossibility for Byzantine Generals Problem where $n \leq 3m$

In the paper "Reaching Agreement in the Presence of Faults", M. Pease et al. proved that there is no protocol (of some kind) to solve the problem for $n \leq 3m$, where $n$ stands for the number of generals and $m$ stands for the number of traitors. The key of their proof this the impossibility of the case $n=3,m=1$. However, the method they used does not look like an information theoretic proof. Thus, it seems like that their result is not "impossibility of arbitrary protocol".

My question: Is there a infomation-theory-based proof for the case $n=3,m=1$? More formally, is there a proof or counterexample for the proposition "there does not exist any kind of protocol, which solves Byzantine generals problem where $n=3,m=1$"?

Note: The typical protocol $\mathrm{SM}(m)$ (it works for arbitrary $n,m$) suggested by L. Lamport et al. is NOT a suitable counterexample, because it needs a signature mechanism, which is NOT perfect reliable in the sense of information theory, if we assume that traitors have infinity computing resources.

• What is an "information-theory-based proof"? Why do you expect such a proof to exist in this case? – Yuval Filmus Jun 2 '17 at 6:19
• To @YuvalFilmus: I think that (maybe it is a false belief) the original proof of impossibility states that "there is no protocol in a given form for $n \leq 3m$", which is other than "there is no protocol for $n \leq 3m$". So I am seeking a proof (or counterexample) for the latter. Note that the typical protocol $\mathrm{SM}(m)$ (it works for arbitrary $n,m$) suggested by L. Lamport et al. is NOT a suitable counterexample, because it needs a signature mechanism, which is NOT perfect reliable in the sense of information theory, if we assume that traitors have infinity computing resources. – Lwins Jun 2 '17 at 6:41
• So what you're really asking is, what result exactly is proved in the paper "Reaching Agreement in the Presence of Faults". Presumably this is explained in the paper. – Yuval Filmus Jun 2 '17 at 6:44
• To @YuvalFilmus: Not exactly probably. As I mentioned above, I am seeking a proof (or counterexample) for the proposition "there is no protocol for $n \leq 3m$". – Lwins Jun 2 '17 at 6:47

In the synchronous model of communication, there are $n$ agents which share a clock. In each round of communication, each agent sends an arbitrary message to each other agent, and then receives the message sent her by every other agent.

A protocol for byzantine agreement on $n$ agents supporting $m$ byzantine agents is a communication protocol for the agents satisfying the following properties:

• Each agent receives an input bit.
• The agents all start talking at time 0.
• There are at most $m$ byzantine agents, whose behavior is arbitrary.
• The other agents follow the protocol.
• The protocol always terminates (this means that the non-byzantine agents always reach a special "terminate" state of the protocol, and then stop talking forever), with a return value which is also a bit.
• The return values must all agree (note that this only applies to the non-byzantine agents).
• If all input bits are the same, then the return values have to be the same bit.

The impossibility result states that such a protocol exists if and only if $n > 3m$.

There is a different model in which an agent can sign a message, and this signature cannot be tampered with. In this model (which I will not specify formally) the problem can be solved for any $n,m$.

One of the difficulties in the area of distributed systems is the complicated nature of the computation model. If you want to understand the meaning of impossibility results, you have to familiarize yourself with these models in full detail (even more detail than the rather informal treatment in this answer).

• Thanks for your patient explanation : ) And I have the last question: how could I find the proof of the impossibility on synchronous communication model? The proof in paper "Reaching Agreement in the Presence of Faults" seems not to be the one. Could you suggest some literature (say, paper or book) highly related to communication model with Byzantine nodes? – Lwins Jun 2 '17 at 7:22
• I am only aware of one proof, and it is the one in the paper. For the same proof in a slightly different model, see lecture notes of Chiu Yuen Koo. – Yuval Filmus Jun 2 '17 at 7:22
• Here is one textbook: Attiya and Welch, Distributed Computing: Fundamentals, Simulations, and Advanced Topics, second edition, Wiley, 2004. – Yuval Filmus Jun 2 '17 at 7:24


text:
# Re: multicol package

> I am trying to set up a book (8"x10"
> - about 400 pages) with mostly 2 columns.
> I want the chapter titles and section headings to go across
> the page with the
> body text in 2 columsn.
>
> I have set it up as follows:
>
> Chapter Title
> \begin{multicols}{2}
>
> (body text)
>
> \end{multicols}
>
> or
>
> \begin{multicols}{2}
>
> (body text)
> \end{multicols}
>
> It seems to work OK until I'm about half way through the
> book, then when I try
> to view using DVI or PDF, I get an error message "Dimension
> too large."
>
> Is there a limit as to how many times you can changee from
> one to two columns?
> Is there a limit to how many words this setup can handle?
> Any help would be
> appreciated.
>
>
The errors is produced for another cause.
No limits in the words.
I have in several times "Dimension too large" but, for now, I can pinpoint the
origin. I not use multicol but two columns option in koma-script style.
Marcelo

____________________________________________________________________________________
¡Viví la mejor experiencia en la web!
Descargá gratis el nuevo Internet Explorer 8


text:
# $A$ an absolutely flat ring $\Rightarrow$ $S^{-1}A$ is absolutely flat

I was doing some exercises in the book of Atiyah / MacDonald on Commutative Algebra, and I'm a little "stuck" with number 3.10 (i):

If $A$ is an absolutely flat ring and $S\subseteq A$ a multiplicatively closed subset, then the localization $S^{-1}A$ is absolutely flat.

I wanted to use a criterion shown earlier: $A$ absolutely flat $\Leftrightarrow$ every principal ideal of $A$ is idempotent.

So if $\frac{a}{s}\in S^{-1}A$, my plan was to show that there exists a $\frac{b}{t}\in S^{-1}A$ with $\frac{b}{t}\cdot(\frac{a}{s})^2=\frac{a}{s}$. From the absolute flatness of $A$, we know that there are $x,y\in A$ with $xa^2=a$ and $ys^2=s$. So choosing $b:=x$, $t:=y$ one would be done, but I don't see why $y$ should be in $S$? I guess this has to be "obvious" in some way, and I'm just being stupid!?

Then I tried to google it, but all I found was a different solution (here), and I'd like to ask one question regarding this solution, too, but I don't know if this is the appropriate place to do so? Since it's the same topic, I hope so, but if not, please correct me and I'll edit this post.

Let $P$ be an arbitrary $S^{-1}A$-module. The author says there is an obvious isomorphism $S^{-1}P\cong S^{-1}A\otimes_A P\to P$ of $S^{-1}A$-modules. The first isomorphy is clear to me (and proven in the book), but I don't get the second one. I believe the maps should look like this:

$\varphi:S^{-1}A\otimes_A P\to P, \varphi(\frac{1}{s}\otimes p)=\frac{1}{s}p$, and

$\psi:P\to S^{-1}A\otimes_A P, \psi(p)=1\otimes p$.

Then $\varphi\circ\psi=1$. As for $\psi\circ\varphi$, let $\frac{1}{s}\otimes p\in S^{-1}A\otimes_A P$, then $\psi(\varphi(\frac{1}{s}\otimes p))=\psi(\frac{1}{s}p)=1\otimes\frac{1}{s}p$, and then I don't know why I can switch over the $\frac{1}{s}$. I know that there is an $S^{-1}A$-module structure on the tensor product, but I thought it was of the form $\frac{a}{s}\cdot(\frac{1}{t}\otimes p)=\frac{a}{st}\otimes p$. It has to have something to do with the $S^{-1}A$-module structure on $P$. I guess it should be easier to show $S^{-1}M\cong M$ directly, I just had the idea to do it this way, but didn't try it out yet. I am going to do this now, but the question above still remains. It has to be a simple thing I overlook again, something I did not fully understand yet, and I hope you can help me understand it!

Thanks for your help in advance!

## 2 Answers

Hint: for any $S^{-1} A$-module $M$, $x,y \in M$ and $s \in S$, $sx = y \iff x = \frac{1}{s} y$.

• So since $s(1\otimes\frac{1}{s}p)=1\otimes p$, $1\otimes\frac{1}{s}p=\frac{1}{s}(1\otimes p)=\frac{1}{s}\otimes p$, correct? Thank you very much for your help! How can I mark this as "answered" by the way? Feb 6, 2012 at 13:14
• @Rand al'Thor: yes, you got it. (And it seems that you found the check mark as well...) Feb 6, 2012 at 13:17

I am too late with the answer, but I am doing the same exercise now.

Is it possible to use the following argument, since for any $a\in A$ we may find $x\in A$ such that $a = a^2x$ we may try the following trick, take any $a/s\in S^{-1}A$

$$\frac{xs}{1}\left(\frac{a}{s}\right)^2 = \frac{a^2sx}{s^2}=\frac{a^2x}{s}=\frac{a}{s},$$ where of course $xa^2=a$, note that $xs/1$ is a correct element of $S^{-1}A$. It seems that this argument shows that every principal ideal of $S^{-1}A$ is idempotent, doesn't it?


text:
# Diffs in Unit Tests

While working on my refactorable settings code extension, I needed to verify that the output of the tool matched some sample output from the original settings code generation tool. Unfortunately, Assert.AreEqual on two really long strings of generated code doesn't give you a whole lot of helpful information about exactly which characters don't match (especially if non-printing characters are the problem). A little googling turned up an amazingly useful solution from Phil Haacked - a set of classes which display a character-by-character diff of the two strings.

I actually ended up using this slightly modified version by Sina Iravanian. If no one else has put this into a nuget package by the next time I need it, I'm going to have to do it myself; it's too incredibly useful to only be available for cut-and-paste.

Comment


text:
# How can I find the modulus of the angular acceleration?

#### Chemist116

The problem is as follows:

A pulley starts spinning from rest a rotation with constant angular acceleration. After $5\,s$ a point in its periphery
has an instant acceleration which makes a $53^{\circ}$ angle with its linear speed. Find the modulus of the angular acceleration (in $\frac{rad}{s^{2}}$) of the pulley.

The given alternatives in my book are as follows:

$\begin{array}{ll} 1.&0.5\,\frac{m}{s}\\ 2.&0.53\,\frac{m}{s}\\ 3.&0.053\,\frac{m}{s}\\ 4.&0.106\,\frac{m}{s}\\ 5.&1.06\,\frac{m}{s}\\ \end{array}$

For this particular problem. I'm lost as how should I use the given information of the instant acceleration and the linear speed. How should I put those vectors?. Which sort of equation should I use?.

The only equation which comes to my mind for the angular acceleration is how it is related to the tangential acceleration as:

$a_{t}=\alpha \times r$

But in this case there is no radius.

Thus I believe it has something to do with vectors but I can't really find exactly how to use that information. Can somebody help me here?.

#### DarnItJimImAnEngineer

The thing to bear in mind here is that you have two components of the acceleration,
$\vec{a} = \alpha r \hat{e}_t - \omega^2 r \hat{e}_r$ (plus two more terms that are zero if the radius is constant). In this problem, you don't know the radius, but you do know the angle (and therefore the ratio) between the tangential and centripetal accelerations. (Note $\hat{e}_r$ is centrifugal acceleration, away from the centre, not that it matters in this case.)
Then replace $\omega$ with $\alpha t$. This gives you everything you need.

BTW, either you or the book made a mistake with the units in the answers. They should be in $\frac{rad}{s^2}$.

#### Chemist116

The thing to bear in mind here is that you have two components of the acceleration,
$\vec{a} = \alpha r \hat{e}_t - \omega^2 r \hat{e}_r$ (plus two more terms that are zero if the radius is constant). In this problem, you don't know the radius, but you do know the angle (and therefore the ratio) between the tangential and centripetal accelerations. (Note $\hat{e}_r$ is centrifugal acceleration, away from the centre, not that it matters in this case.)
Then replace $\omega$ with $\alpha t$. This gives you everything you need.

BTW, either you or the book made a mistake with the units in the answers. They should be in $\frac{rad}{s^2}$.
Yes. I'm sorry for that. I wrote this question in a rush. It should had been as you correctly indicated. In other words as $\frac{rad}{s^{2}}$.

I'm confused at why the acceleration at the given instant is it a subtraction from these two accelerations, the tangential from the centripetal?

Wouldn't it be the sum?. Or could I be looking at those vectors differently from you? Or what it was intended to say with the negative sign is that just the centripetal is pointing inwards?.

I'm posting a sketch from what I'm understanding by reading your words. Please tell me if I did missunderstood what you meant.

If solely guided by what you mentioned then the tangent function can be used to relate both.

$\tan 53^{\circ}=\frac{\omega ^ 2 r}{\alpha\times r}$

$\frac{4}{3}=\frac{\omega ^ 2}{\alpha}$

Thus:

$\alpha=\frac{3 \omega ^2}{4}$

But I'm not given $\omega$:

Since I calculated it from rest:

$\omega_{f}=\omega_{0}+\alpha \left(5\right)$

$\omega_{f}=5\alpha$

By introducing this in the above equation:

$\alpha=\frac{3 \left(5\alpha\right) ^2}{4}$

Then:

$\alpha=\frac{4}{25 \times 3}\approx 0.0533$

But in this situation I'm not considering the sign you mentioned. Although by comparing this answer to the answers sheet in my book it checks.

But again I'd like you could check if what I'm understanding is correct. :unsure:

#### DarnItJimImAnEngineer

It is the vector sum. The negative sign, just as you said, is because the $\hat{e}_r$ unit vector points (by convention) away from the centre, while the acceleration is towards the centre.

1 person

#### Chemist116

It is the vector sum. The negative sign, just as you said, is because the $\hat{e}_r$ unit vector points (by convention) away from the centre, while the acceleration is towards the centre.
So is my drawing or the way how I understood what you meant correct?.

#### skeeter

Math Team
reference the diagram ...

$a:a_c:a_T = 5:4:3$

$\omega = \alpha \cdot t \implies \omega(5) = 5\alpha$

$a_c = r \cdot \omega^2 \implies a_c(5) = r \cdot 25 \alpha^2$

$a_T = r \cdot \alpha$, a constant

$\dfrac{a_c}{a_T} = \dfrac{4}{3} = \dfrac{r \cdot 25\alpha^2}{r \cdot \alpha} = 25 \alpha$

$\alpha = \dfrac{4}{75} \approx 0.053 \, rad/s^2$

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text:
# 800 Followers problem!

Number Theory Level 4

$100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$

Let $$N$$ denote the number above. Find the remainder when $$N$$ is divided by $$1000$$.

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# Spin-dependent constraints on blind spots for thermal singlino-higgsino dark matter with(out) light singlets

Journal of High Energy Physics, Jul 2017

The LUX experiment has recently set very strong constraints on spin-independent interactions of WIMP with nuclei. These null results can be accommodated in NMSSM provided that the effective spin-independent coupling of the LSP to nucleons is suppressed. We investigate thermal relic abundance of singlino-higgsino LSP in these so-called spin-independent blind spots and derive current constraints and prospects for direct detection of spin-dependent interactions of the LSP with nuclei providing strong constraints on parameter space. We show that if the Higgs boson is the only light scalar the new LUX constraints set a lower bound on the LSP mass of about 300 GeV except for a small range around the half of Z 0 boson masses where resonant annihilation via Z 0 exchange dominates. XENON1T will probe entire range of LSP masses except for a tiny Z 0-resonant region that may be tested by the LZ experiment. These conclusions apply to general singlet-doublet dark matter annihilating dominantly to $t\overline{t}$. Presence of light singlet (pseudo)scalars generically relaxes the constraints because new LSP (resonant and non-resonant) annihilation channels become important. Even away from resonant regions, the lower limit on the LSP mass from LUX is relaxed to about 250 GeV while XENON1T may not be sensitive to the LSP masses above about 400 GeV.

This is a preview of a remote PDF: https://link.springer.com/content/pdf/10.1007%2FJHEP07%282017%29050.pdf

Marcin Badziak, Marek Olechowski, Paweł Szczerbiak. Spin-dependent constraints on blind spots for thermal singlino-higgsino dark matter with(out) light singlets, Journal of High Energy Physics, 2017, 1-31, DOI: 10.1007/JHEP07(2017)050


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2. Economics
3. you produce coco cola beverages you estimate that the price...

# Question: you produce coco cola beverages you estimate that the price...

###### Question details

You produce Coco cola beverages. You estimate that the price elasticity of demand for your product is 2.7 (in absolute value). Coco cola currently sells for $2.50 per 20-ounce can. Some legislators are considering placing a$1.00 per can tax on your product.(the answer should be a few words)

1-The DWL will be relatively

2-Tax revenue will be relatively

3-Given the price elasticity of demand, your burden of the tax is likely to be relatively


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Welcome to mathimatikoi.org forum; Enjoy your visit here.

True or false

Groups, Rings, Domains, Modules, etc, Galois theory
Papapetros Vaggelis
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

True or false

Examine if the following is true.

If $\displaystyle{p}$ is a prime number and$\displaystyle{\left(G,\cdot\right)}$ is a group

such that $\displaystyle{o(G)=p^3}$ and $\displaystyle{x^p=1\,,\forall\,x\in G}$ then

$\displaystyle{\left(G,\cdot\right)}$ is abelian.


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makePriceN4Function: Price Function N. 4 In smoof: Single and Multi-Objective Optimization Test Functions

Description

Fourth function by Price. The implementation is based on the defintion

f(\mathbf{x}) = (2\mathbf{x}_1^3 \mathbf{x}_2 - \mathbf{x}_2^3)^2 + (6\mathbf{x}_1 - \mathbf{x}_2^2 + \mathbf{x}_2)^2

subject to \mathbf{x}_i \in [-500, 500].

Usage

 1 makePriceN4Function()

Value

[smoof_single_objective_function]

References

W. L. Price, A Controlled Random Search Procedure for Global Optimisation, Computer journal, vol. 20, no. 4, pp. 367-370, 1977.

makePriceN1Function, makePriceN2Function

smoof documentation built on May 20, 2017, 5:35 a.m.

Search within the smoof package
Search all R packages, documentation and source code


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# 59

## Eliezer_Yudkowsky

From Robyn Dawes's Rational Choice in an Uncertain World:

Post-hoc fitting of evidence to hypothesis was involved in a most grievous chapter in United States history: the internment of Japanese-Americans at the beginning of the Second World War.  When California governor Earl Warren testified before a congressional hearing in San Francisco on February 21, 1942, a questioner pointed out that there had been no sabotage or any other type of espionage by the Japanese-Americans up to that time.  Warren responded, "I take the view that this lack [of subversive activity] is the most ominous sign in our whole situation. It convinces me more than perhaps any other factor that the sabotage we are to get, the Fifth Column activities are to get, are timed just like Pearl Harbor was timed... I believe we are just being lulled into a false sense of security."

Consider Warren's argument from a Bayesian perspective.  When we see evidence, hypotheses that assigned a higher likelihood to that evidence, gain probability at the expense of hypotheses that assigned a lower likelihood to the evidence.  This is a phenomenon of relative likelihoods and relative probabilities.  You can assign a high likelihood to the evidence and still lose probability mass to some other hypothesis, if that other hypothesis assigns a likelihood that is even higher.

Warren seems to be arguing that, given that we see no sabotage, this confirms that a Fifth Column exists.  You could argue that a Fifth Column might delay its sabotage.  But the likelihood is still higher that the absence of a Fifth Column would perform an absence of sabotage.

Let E stand for the observation of sabotage, H1 for the hypothesis of a Japanese-American Fifth Column, and H2 for the hypothesis that no Fifth Column exists.  Whatever the likelihood that a Fifth Column would do no sabotage, the probability P(E|H1), it cannot be as large as the likelihood that no Fifth Column does no sabotage, the probability P(E|H2).  So observing a lack of sabotage increases the probability that no Fifth Column exists.

A lack of sabotage doesn't prove that no Fifth Column exists.  Absence of proof is not proof of absence.  In logic, A->B, "A implies B", is not equivalent to ~A->~B, "not-A implies not-B".

But in probability theory, absence of evidence is always evidence of absence.   If E is a binary event and P(H|E) > P(H), "seeing E increases the probability of H"; then P(H|~E) < P(H), "failure to observe E decreases the probability of H".  P(H) is a weighted mix of P(H|E) and P(H|~E), and necessarily lies between the two.  If any of this sounds at all confusing, see An Intuitive Explanation of Bayesian Reasoning.

Under the vast majority of real-life circumstances, a cause may not reliably produce signs of itself, but the absence of the cause is even less likely to produce the signs.  The absence of an observation may be strong evidence of absence or very weak evidence of absence, depending on how likely the cause is to produce the observation.  The absence of an observation that is only weakly permitted (even if the alternative hypothesis does not allow it at all), is very weak evidence of absence (though it is evidence nonetheless).  This is the fallacy of "gaps in the fossil record"—fossils form only rarely; it is futile to trumpet the absence of a weakly permitted observation when many strong positive observations have already been recorded.  But if there are no positive observations at all, it is time to worry; hence the Fermi Paradox.

Your strength as a rationalist is your ability to be more confused by fiction than by reality; if you are equally good at explaining any outcome you have zero knowledge.  The strength of a model is not what it can explain, but what it can't, for only prohibitions constrain anticipation.  If you don't notice when your model makes the evidence unlikely, you might as well have no model, and also you might as well have no evidence; no brain and no eyes.

Part of the sequence Mysterious Answers to Mysterious Questions

Next post: "Conservation of Expected Evidence"

Previous post: "Your Strength as a Rationalist"


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# Math 323

Chien-Lin Su

# Lecture 1 - 2018/01/10

• Midterm is on Mar 14th

## Chapter 1 - Probability Introduction

### 1.1. Definitions

Experiment - process by which an observation is made

• Ex dice tossing is an experiment since we observe the number appearing on the other side

• Random Experiment

Random experiment - TODO

Sample space - S or Ω - collection of every possible outcome

Event - A, B, C - partial collection of sample space

Let P(A) = probability of A ∀A, A ∈ Ω P(A) has the following axioms:

• $P(A) \ge 0$
• $P(S) = 1$ (unit measure / normed)
• if $i \ne j$ and $A_i \cap A_j = \emptyset$, $P(A_i \cup A_j) = P(A_i) + P(A_j)$

# Lecture 2 - 2018/01/12

Limit of a sequence of sets

## Limit of Sequence of Sets

Non-decreasing sequence - sequence of $A_i$ such that $A_j \subseteq A_k$ if $i < k$

Non-increasing sequence - sequence of $A_i$ such that $A_j \supseteq A_k$ if $i > k$

Example 1

Note that the sequence is non-decreasing.

$\limitinfty = \cupinfty = \braces{x \mid 1 < x < 2}$

Note that $x < 2$ is open

Example 2

$A_k = \braces{x \mid 1 < x \le x + \dfrac{1}{k}}$

Note that the sequence is non-increasing.

$\limitinfty = \capinfty = \braces{x \mid 1 < x \le 1} = \emptyset$

Note that $x \le 1$ is still closed.

## 1.4 Partition & Inequalities

Partition - sequence of mutually exclusive sets which together form the whole. More formally:

Let $\braces{A_i}_{i = 1}^{\infty (k)}$ be a sequence of sets

$A_i \le \Omega, \forall i$

If $A_i \cap A_j = \emptyset, \forall i \ne j$ and $\cupinftyup{\infty (k)} = \Omega$
then $\braces{A_i}_{i=1}^{\infty (k)}$ is a partition of sample space $\Omega$

Note - if $B$ is any subset of $\Omega$ & $\braces{A_1, …, A_k}$ is a partition of $\Omega \Rightarrow B = \cupinftyup{k} (A_i \cap B)$

### Theorem 1. Kolmogorov Probability Measure

Let $F$ be a set that collects possible sets from $\Omega$

P: $F \rightarrow R^+$ is called a probability measure if

a. $P(A) \ge 0 \quad \forall A \in F \mathcomment{\text{non-negative}}$
b. $P(\Omega) = 1$
c. $P(\cupinfty) = \Sigma_{i=1}^{\infty} P(A_i), A_i \cap A_j = \emptyset \quad \forall i \ne j$

Some extensions:

1. $P(\phi) = 0$
2. $P(\cupinftyup{k}) = \Sigma_{i=1}^{k} P(A_i), A_i \cap A_j = \emptyset \quad \forall i \ne j$
3. $P(A^C) = 1 - P(A)$
4. $A \subseteq B \Rightarrow P(A) \le P(B)$
5. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
6. $P(A_1 \cup A_2 \cup A_3) = P(A_1) + P(A_2) + P(A_3) - P(A_1 \cap A_2) - P(A_2 \cap A_3) - P(A_1 \cap A_3) + P(A_1 \cap A_2 \cap A_3)$

Proof

Claim: $P(\phi) = 0$

$\because \Omega = \Omega \cup \phi$

$\therefore P(\Omega) = P(\Omega \cup \phi)$

$= P(\Omega) + P(\phi) (\therefore \Omega \cap \phi = \phi)$

$= 1+ P(\phi)$

$\therefore P(\phi) = 0$

Claim: $P(A^C) = 1 - P(A)$

$\because \Omega = A \cup A^C$

$\therefore P(\Omega) = P(A \cup A^C)$

$= P(A) + P(A^C) — (A \cap A^C = \phi)$

$\therefore 1 = P(A) + P(A^C)$

$\therefore P(A^C) = 1 - P(A)$

Claim: $A \subseteq B \Rightarrow P(A) \le P(B)$

$P(B) = P(A \cup (A^C \cap B))$

$= P(A) + P(A^C \cap B) — (A \cap (A^C \cap B) = \emptyset)$

$\ge P(A)$

Claim: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\because A \cup B = A \cup (A^C \cap B)$

$\therefore P(A \cup B) = P(A \cup (A^C \cap B)) — (A \cap (A^C \cap B) = \emptyset)$

$= P(A) + P(A^C \cap B)$

Note: $P(B) = P((A \cap B) \cup (A^C \cap B))$

$= P(A \cap B) + P(A^C \cap B)$

$\Rightarrow P(A^C \cap B) = P(B) - P(A \cap B)$

//todo include other stuff

1.5 Conditional Probability and Baye’s Rule

Def 1 The conditional probability of an event A, given that an event B has occurred, is equal to

$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} if P(B) > 0$

Remark 1:

Here A is the event whose uncertainty we want to update, and B is the evidence we observe (or want to treat as given

We call P(A) the prior probability of A, and P(A | B) the posterior probability of A.)

Prior - before updating based on evidence

Posterior - after updating based on evidence

Remark 2:

$P(A \mid B)$ is a probability measure

Check 1:

$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} > 0$

Since $P(B) > 0$ and $P(A \cap B) > 0$

Check 2:

$P(\Omega \mid B) = \dfrac{P(\Omega \cap B)}{P(B)} = \dfrac{P(B)}{P(B)} = 1$

Check 3:

see check 3


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# Complexity Parameter in Decision Tree

As the complexity parameter is calculated? What is the meaning of it?

From what I read, the cp is a value at which the tree makes divisions in the nodes until the reduction in the relative error is less than a certain value.

There are places I read that say the CP affects only the growth of the tree and others say that interferes with pruning too. For min appears that it interferes only in growth but not sure.

I am using rpart () package to create trees, in the case of the classification tree exists missclassification rate to evaluate the ratings, but in the case of regression is not anything to evaluate the predictions beyond the MSE?

• Have a look at this stats.stackexchange.com/questions/117908/… – Simone Oct 31 '15 at 13:58
• @Simone I've read it, but I did not understand well if it goes for regression tree also – user72621 Oct 31 '15 at 14:09

This is answered in this rpart resource. From p. 25:

For regression models (see next section) the scaled cp has a very direct interpretation: if any split does not increase the overall $R^2$ of the model by at least cp (where $R^2$ is the usual linear-models definition) then that split is decreed to be, a priori, not worth pursuing. The program does not split said branch any further, and saves considerable computational effort.

That same page gives this formula for how the cp parameter affects calculation of a tree's risk:

$$R_{cp}(T) ≡ R(T) + cp ∗ |T| ∗ R(T_1)$$

($T_1$ here is a tree with no splits, $|T|$ the splits in the tree. The full formal definition of risk is outside the scope of your question, but for reference the definition is on p. 4.)

• But the error associated with the cp would be cross-validation error? – user72621 Nov 2 '15 at 19:58
• I'm not sure what you mean by "the error associated with the cp," but based on this quote from the same page, cp doesn't appear to directly involve CV error: "The default value of .01 has been reasonably successful at ‘pre-pruning’ trees so that the cross-validation step need only remove 1 or 2 layers, but it sometimes over prunes, particularly for large data sets." More, section 4.2 mentions setting the complexity parameter via CV. – Sean Easter Nov 2 '15 at 21:57
• The $cp$ is involved both in the growth of the tree as in pruning then? – user72621 Nov 2 '15 at 22:35
• It doesn't appear that way to me. It seems the complexity parameter stops the tree from growing, though one can think of that as a sort of preemptive pruning. I'm looking for an explicit description of how that package uses the risk formula in deciding whether to split, but I can't yet find it. – Sean Easter Nov 2 '15 at 22:44
• One thing that can not find is how do you measure the quality of predictions in regression tree in the classification tree there is missclassification rate at each terminal node, but there is no option $extra=3$ in regression tree using $method="anova"$ in $rpart()$ – user72621 Nov 2 '15 at 22:48