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Moduli of Double Epw-sextics (Memoirs of the American by Kieran G. O'grady

The writer reports the GIT quotient of the symplectic grassmannian parametrizing lagrangian subspaces of $\bigwedge^3{\mathbb C}^6$ modulo the usual motion of $\mathrm{SL}_6$, name it $\mathfrak{M}$. it is a compactification of the moduli area of delicate double EPW-sextics and as a result birational to the moduli area of HK $4$-folds of variety $K3^{[2]}$ polarized by way of a divisor of sq. $2$ for the Beauville-Bogomolov quadratic shape. the writer will verify the good issues. His paintings bears a powerful analogy with the paintings of Voisin, Laza and Looijenga on moduli and classes of cubic $4$-folds.

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Moduli of Double Epw-sextics (Memoirs of the American Mathematical Society)

The writer reviews the GIT quotient of the symplectic grassmannian parametrizing lagrangian subspaces of $\bigwedge^3{\mathbb C}^6$ modulo the average motion of $\mathrm{SL}_6$, name it $\mathfrak{M}$. this can be a compactification of the moduli house of tender double EPW-sextics and as a result birational to the moduli house of HK $4$-folds of style $K3^{[2]}$ polarized by way of a divisor of sq. $2$ for the Beauville-Bogomolov quadratic shape.

Additional info for Moduli of Double Epw-sextics (Memoirs of the American Mathematical Society)

Example text

Let U ⊂ V be complementary to W . 12) ( 2 W) ∧ U ⊕ W ∧ ( ∼ U ) −→ EW . 13) EW = ([v0 ] ∧ W0 ∧ U ⊕ [v0 ] ∧ ( 2 U )) ⊕ (( 2 W0 ) ∧ U ⊕ W0 ∧ ( U )). ) Given the above decomposition the scheme CW,A ∩ (P(W ) \ P(W0 )) is described as the degeneracy locus of a family of quadratic forms. One identiﬁes the family of quadratic forms with {(q A + q w )}w∈W0 and the claim follows. Licensed to Tulane Univ. 78. org/publications/ebooks/terms 42 3. 3. 5). 14) ZW0 ,A := V (q A ) ∩ Gr(2, V0 )W0 ⊂ P( 2 V0 / W0 ).

A ∈ Σ∞ ) or A ∈ Σ[2]. By deﬁnition we may assume that A ∈ BFX for X one of A, A∨ , . . , F2 , or A ∈ XFN3 , where F is the basis {v0 , . . , v5 } of V . 2. It remains to consider A ∈ (BFC1 ∪ BFE1 ∪ BFE ∨ ∪ BFF2 ∪ XFN3 ). 1) one easily checks the following: 1 3 2 If A ∈ (BFC1 ∪ BFE1 ∪ BFE ∨ ) then V02 ⊂ A and dim(A ∩ ( V02 ∧ V )) ≥ 3, if 1 A ∈ (BFF2 ∪ XFN3 ) there exists a 3-dimensional subspace W ⊂ V03 containing V01 such that 3 W ⊂ A and dim(A ∩ ( 2 W ∧ V )) ≥ 3. 5 for details. 6. The GIT-boundary Let M ⊂ M be the (open) subset parametrizing PGL(V )-orbits of stable points; the GIT-boundary of M is ∂M := (M \ Mst ).

3), with weights in decreasing order: e0 > e1 > . . > es . 3) dX = (d0 , d1 , . . , d[(s−1)/2] ). 1) that contains BFX (or 3 XN3 ) . Let SFX ⊂ LG( V ) be the set of A which are λX -split of type dX . 4 gives the following: Licensed to Tulane Univ. 78. org/publications/ebooks/terms 30 2. 1. Every point of BX is represented by a point of SFX and every point of XN3 is represented by a point of SFN3 . Next let F be the basis of V obtained by reading the vectors in F in reverse order: F := {v5 , v4 , v3 , v2 , v1 , v0 }.


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# Converting a Simple Spin-half Bra-ket State into its Matrix Representation in another Basis

## Prompt

Write the spin-half state $\vert + \rangle$ in the matrix representation of the $\vert \pm \rangle_y$ basis.

This SWBQ

## Wrap Up

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# Choose the correct answer from the given four options:The $4^{\text {th }}$ term from the end of the AP: $-11,-8,-5, \ldots, 49$ is(A) 37(B) 40(C) 43(D) 58

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Given:

Given A.P. is $-11,-8,-5, \ldots, 49$.

To do:

We have to find the 4th term from the end of the given arithmetic progression.

Solution:

In the given A.P.,

$a_1=-11, a_2=-8, a_3=-5$

First term $a_1 = a= -11$, last term $l = 49$

Common difference $d = a_2-a_1 = -8 - (-11) = -8+11=3$

We know that,

nth term from the end is given by $l - (n - 1 ) d$.

Therefore,

4th term from the end $= 49 - (4 - 1) \times (3)$

$= 49 - 3 \times 3$

$= 49 -9$

$= 40$.

The 4th term from the end of the given A.P. is $40$.

Updated on 10-Oct-2022 13:27:27


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# HP Forums

Full Version: Triangular number AND sum of first m factorials
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
153 (my favorite number) is both a triangular number (the sum of the integers 1 through $$n$$; in this case $$n=17$$) as well as the sum of the factorials $$1!$$ through $$m!$$ (in this case $$m=5$$).

The first three natural numbers which have both of those properties are 1, 3 (both trivial) and 153. Find the next number in this sequence. For extra credit, find the mathematical relationship between $$n$$ and $$m$$ for all members of this sequence (which apparently is not yet in OEIS).

001 LBL A
002 DEC X
003 ENTER
004 INC X
005 LBL 00
006 INC X
007 RCL* Y
008 DSE Y
009 GTO 00
010 PSE 10
011 STO+ X
012 +/-
013 1
014 ENTER
015 ENTER
016 R/\
017 SLVQ
018 X<>Y
019 FP
020 PSE 10
021 RCL L
022 RTN

2 A -> 3 -> 0 -> 2
5 A -> 153 -> 0 -> 17

The next m should be greater than 31, but I can’t find it using the wp34s, at least not with help of this simple program…

Edited to fix a typo pointed out by Dieter below.
(01-09-2018 04:31 PM)Joe Horn Wrote: [ -> ]153 (my favorite number) is both a triangular number (the sum of the integers 1 through $$n$$; in this case $$n=17$$) as well as the sum of the factorials $$1!$$ through $$m!$$ (in this case $$m=5$$)

It's also a narcissistic number:

153 = 1^3 + 5^3 + 3^3

V.
.
(01-09-2018 08:53 PM)Gerson W. Barbosa Wrote: [ -> ]001 LBL A
002 DEC X
003 ENTER
004 INC C

?!? – increment register C ?
Is this possibly supposed to mean INC X ?

(01-09-2018 08:53 PM)Gerson W. Barbosa Wrote: [ -> ]005 LBL 00
006 INX X

Ah, here is the missing X from line 004, and the "C" from there goes here. ;-)

Dieter
(01-09-2018 10:19 PM)Dieter Wrote: [ -> ]Ah, here is the missing X from line 004, and the "C" from there goes here. ;-)

:-)

Duly swapped. Thanks!

Gerson.
A quick test of sums of factorials up to 69! finds no triangle numbers larger than 153. If one does exist, it has to have well over 100 digits. I may try later on the emulator.

John
Tn&Sf:

« { } 3 ROT
FOR n n Sfac 8 * 1 + ISPF? { √ 1 - 2 / + n I→R + } { DROP } IFTE
NEXT
»

Sfac:

« DUP 1 - 2
FOR m 1 + m * -1
STEP 1 +
»

ISPF?

« 1
»

7 Tn&Sf ->

{ '(√73-1)/2' 3. '(√265-1)/2' 4. 17 5. '(√6985-1)/2' 6. '(√47305-1)/2' 7. }

49G or 50g in exact mode

ISPF? (IsPerfectSquare?) has yet to be implemented. It should return 1 when the argument is a perfect square and 0 otherwise. Then the output would be a list of n and m pairs, separated by dots. Just in case someone wants to try.
(01-10-2018 04:03 AM)Gerson W. Barbosa Wrote: [ -> ]ISPF? (IsPerfectSquare?) has yet to be implemented. It should return 1 when the argument is a perfect square and 0 otherwise.

The HP 50g LongFloat library contains a function like the one you're looking for. It's called ZSqrt, and it returns IP(sqrt(x)) to level 2, and a 0 or 1 to level 1, with 1 meaning that x was a perfect square. It works on integers of any length.

Gerald H also posted a program HERE which seems to do essentially the same thing. It returns IP(sqrt(x)) to level 2 of the stack, and on level 1 it leaves a SysRPL TRUE if x was a perfect square, otherwise a FALSE. That's the same idea as HP's FPTR2 ^ZSQRT, but Gerald's program is more accurate; see Gerald's posting for evidence of HP's function's inaccuracy.

BTW, LongFloat's ZSqrt function gets the same result as Gerald's program when given the example in Gerald's posting, so I surmise that ZSqrt is trustworthy.
I've got a proof that there are only three such numbers.

Consider the last pair of digits in $$\sum_1^n i!$$, from n=9 onwards these never change because subsequent factorial terms will always have a factor of 100 present. These digits are '13'.

Note that n is triangular iff 8n+1 is a perfect square. For the sum of factorials to be triangular, the last two digits must therefore be '05'. Checking all possibilities shows that there are no square numbers that end '05'.

Thus, numbers of the desired form must have n < 9. Checking all cases reveals that only 1, 3 and 153 have the desired properties.

Pauli
Paul: Now THAT is beautiful! Thank you! I can stop my futile hunt now.
Thanks Don't let my proof stop your hunt, you'll be able to wile away many hours looking...

I hadn't realised that all square numbers that end in '5' actually end in '25'. I must have seen this before but never noticed or remembered it.

Pauli
A really great thread Joe et. al. Thank you all.

John
(01-11-2018 10:21 AM)Paul Dale Wrote: [ -> ]Don't let my proof stop your hunt, you'll be able to wile away many hours looking...

At least I can do it a little more efficiently now :-)

100

« { } SWAP 0 1 ROT 1 SWAP
FOR m m * SWAP OVER + ROT
OVER 8 * 1 + ZSqrt
{ 1 - 2 / + m I→R + }
{ DROP }
IFTE
SWAP ROT
NEXT
DROP2
»

EVAL

-->    { 1 1. 2 2. 17 5. }
(about 17 seconds on the real 50g)

ZSqrt from the LongFloat Library

Yes, that's a consequence of the ever growing number of trailing zeros in factorials and the properties of perfect squares.

Gerson.
(01-11-2018 10:21 AM)Paul Dale Wrote: [ -> ]Thanks Don't let my proof stop your hunt, you'll be able to wile away many hours looking...

I hadn't realised that all square numbers that end in '5' actually end in '25'. I must have seen this before but never noticed or remembered it.

Pauli

I figured it was hopeless since I tested all sums of factorials up to 1000 (over 2500 digits) with no triangle numbers found. Took almost 20 min. on the emulator.

An interesting and educational thread indeed!

John
(01-11-2018 06:29 PM)Gerson W. Barbosa Wrote: [ -> ]Yes, that's a consequence of the ever growing number of trailing zeros in factorials and the properties of perfect squares.

Gerson.

Note, however, that due to the first four numbers, all of the sums of factorials end in 3.
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• :


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# Can I make the water boil simply by spinning it in a glass?

Correct me if I'm wrong, I think the pressure in a fluid reduces when the speed increases(The airplane rises because the air above the airfoil moves faster than the air below it).

Next, looking at the air above the surface of the water which is in a spinning glass, I'm wondering if this relative motion decreases the pressure in the air just above the water surface... If yes, does the boiling point of the water reduce due to the loweing of this air pressure over the surface ? (A layer of air just above the water surface spins with the water, so there will be a relative motion between this layer of air and the air above it.)

• related to title reddit.com/r/askscience/comments/3bin9p/… Mar 31 '17 at 5:42
• With brisk enough stirring, one gets cavitation. It's not exactly boiling, but does create a bit of a vapor bubble. Mar 31 '17 at 9:25
• xkcd already did this one Mar 31 '17 at 9:44
• The wing in your example does not rise because air moves faster than the air below it. That image is extremely misleading — a wing only provides lift when it diverts air downward, and that image shows no change in the motion of the air. Now, in a real wing, the air on top will flow faster than the air underneath, but isn't what causes the wing to generate lift. Apr 1 '17 at 3:05

## No, mostly

You mostly can't boil water by spinning the glass. "Mostly" because some weird stuff is possible under extreme conditions like in a rotary evaporator; in such cases, whether or not there's "boiling" starts to become an issue of semantics.

### That explanation of aerodynamic lift is a common misconception

First, to correct a misconception:

Correct me if I'm wrong, I think the pressure in a fluid reduces when the speed increases(The airplane rises because the air above the airfoil moves faster than the air below it).

This statement is a common misconception supported even by some authoritative sources. However the logic behind it presumes that equilibrium behavior applies in dynamic scenarios which doesn't hold; more in the comments below.

"What really allows airplanes to fly?" has some excellent discussion on airfoils.

### Spinning water in a glass can decrease pressure

If you spin a glass of water fast enough, you can get a vortex going in the center. This vortex is sorta like a tornado, with lower pressure in the center and higher pressure at the boundaries, against the glass.

In principle, if you do this fast enough, you could lower the inner pressure down to the static boiling point. Checking a phase diagram for water, it looks like water can boil at room temperature if we drop the pressure down to a just few percent of normal atmospheric pressure.

### Would additional evaporation be due to "boiling"?

In a classical sense, boiling is when a material in the liquid state reaches the point where it'll start turning into its vapor state with appreciable stability throughout its volume. Non-boiling liquids can still turn into vapor, but they usually do so at their boundaries, in which case we call it "evaporation" instead of "boiling".

The distinction between rapid evaporation and boiling kinda breaks down under extreme conditions since the classical sense in which we defined those terms no longer applies, but I think that most people would find the term "boiling" to be misleading in this case.

• Comments are not for extended discussion; this conversation has been moved to chat.
– rob
Apr 1 '17 at 2:26

No, you cannot make water boil simply by spinning it in a glass. It's impossible. The pressure inside the water will not decrease, but increase. Even at the water surface.

### The situation

Assume there is a glass cylinder, which is closed at the bottom and initially open at the top. It is partially filled with water and spinning at a very high velocity (a very, very, very high velocity if you like). The water is dragged along with the glass cylinder and at some point spins with the same velocity as the glass, generating a vortex.

At some velocity (depending on the length of the glass and the amount of water), the water will start to spill out on top of the glass. To prevent this, we place either a ring on top, leaving a hole in the center, or we close the glass completely. This does not make a big difference.

There are two phases involved: Water and air. Let's first look at ...

### The water.

The pressure distribution inside the water can be calculated using e.g. the idea of potential energy. There effectively are two pressure gradients which are superposed:

1. the usual hydrostatic pressure gradient due to Earth's gravity from top to bottom (roughly 1 bar per 10 m),

2. a pressure gradient due to the rotation from the center line outwards.

The following formula can be derived for the pressure $p$ inside the water: $$p = p_0 - \rho g (z-z_0) + \tfrac 12 \rho \omega^2 r^2$$ ($z$ vertical coordinate pointing upward, $r$ radial coordinate pointing outward, $\rho$ fluid density, $g$ gravity of Earth, $\omega$ angular velocity, $p_0=p(z=z_0,r=0)$).

Note that the assumptions that result in this pressure distribution are only a spinning tube in Earth's gravity field, filled with incompressible fluid. No assumption has been made yet on the position of the boundary between water and air.

This means that the pressure is increasing with increasing $r$. At the boundary between water and air, the pressure is lowest, while at the inner wall of the cylindrical part of the glass, the pressure is highest. Due to Earth's gravity, the highest pressure is reached at the bottom of the glass.

Important statement: The pressure inside the water is increasing when moving from the inside (surface) to the outside (glass).

Now let's look at ...

### The air.

For the air that spins inside the cylinder, the same principle holds true as for the water. Since air is compressible, the above equation cannot be used directly for air, but a similar statement can be made:

In the air, too, the pressure increases when we move from the inside (rotation axis) to the outside (water surface).

This by the way also holds true for tornados: The outside pressure is greater than the inside pressure. For tornados, the outside pressure is set by the surrounding atmospheric pressure.

One big difference is of course the density: $\rho_{\mathrm{water}} \approx 1000 \mathrm{kg/m^3}$, $\rho_{\mathrm{air}} \approx 1.2 \mathrm{kg/m^3}$. This results in a much smaller pressure gradient inside the air when compared to the pressure gradient in the water. But for very, very high velocities, the pressure gradient in air can become non-negligible.

Now we need ...

### The Euler–Cauchy stress principle.

This is one of the fundamental assumption of continuum mechanics. It states that the stress vector is continuous. For the water and the air in our glass this means that the pressure at the water-air boundary is equal inside the water and the air. Put in other words: The pressure between water and air is continuous and does not jump.

Now we finally need the ...

### Reference pressure.

This is of course der springende Punkt, as we say in German ("the jumping point" - the important point).

If the tube is partially open, the pressure will equalize at the opening. Since the tube is spinning really, really fast, there is a pressure gradient inside the air as well. This will result in air flowing into the tube at the rotation axis, and air flowing out of the tube at the outer rim of the opening. So we get a secondary flow here. This secondary flow leads to a decreased pressure through the Bernoulli equation (see below), but the pressure at the opening cannot drop below atmospheric pressure, because otherwise the secondary flow would stop.

For a stationary rotation, this secondary flow is stationary. Because the air pressure near the rotation axis must be smaller than the air pressure at the water surface, air must flow in near the rotation axis and exit the tube near the outer rim of the opening (assuming we do not consume or generate matter inside the tube).

Therefore, the pressure at the outer part of the opening must be above atmospheric pressure.

Since the pressure gradient from rotation axis to water surface extends throughout the whole length of the tube, the secondary flow also extends through the whole length of the tube. In order to keep going, the pressure at the water surface at the bottom of the tube must therefore be slightly above the pressure at the outer rim of the opening.

Therefore, the pressure at the entire water surface must be above atmospheric pressure.

The closed tube is easier. There is no secondary flow, just a pressure increase towards the water. Since the amount of air inside the tube has not changed, and the pressure at the water surface is above the pressure at the tube center, the pressure at the water surface must be above atmospheric pressure. The pressure at the tube center is below atmospheric pressure.

In total, all you can get is a pressure increase at the water-air boundary as @Peter.A.Schneider pointed out in his comment. This is independent on how fast the glass spins.

For an open glass, the secondary flow will by the way result in water being evaporated, which will lead to a cooling of the water. So instead of boiling it is more likely that the water will freeze.

### Bernoulli equation

Finally, the original posts states:

I think the pressure in a fluid reduces when the speed increases

A pressure decrease in a moving fluid is relative to the fluid at rest. You are referring to the momentum conservation equation for an inviscid fluids, aka Bernoulli equation: $$\tfrac12 (v_2^2-v_1^2) + \frac 1 \rho (p_2-p_1)=0$$ (here without the gravitation term). This is valid along a streamline only. It is not a general principle that the pressure decreases in fast moving fluids.

In the spinning glass, the streamlines are circles concentric to the spin axis. Along these streamlines, neither velocity nor pressure changes. In short, using Bernoulli's equation on your spinning glass does not tell you anything on the pressure in the water.

• I think that people are looking at this question with too many simplifying assumptions. For example, consider an extreme case in which a cup is continuously spun until it creates a vortex in the local environment and generates a literal tornado around it. Then pressure drop, right? Not that someone's likely to do that, just "impossible"'s going too far. That said, I agree that it'd take a lot of contriving to get there. =P Anyway, there're a few more routes toward getting pressure drops; it's definitely possible.
– Nat
Mar 31 '17 at 14:13
• @Nat, I extended my answer. Yes, the tornado can be generated. No, this can never lead to a pressure decrease at the surface of the water. It can lead to a pressure decrease in the center of the tornado, but there is no water there to boil. Mar 31 '17 at 18:32
• @Nat No, the pressure would not drop. Why would it? In weather, the updraft creates a local pressure drop. There is no updraft because air is spinning circularly in a glass, I think. Mar 31 '17 at 19:34
• Actually, if you have a concave wall and stir really fast, wouldn't the water heat up from compression? I know it's not very compressible, but it's not completely incompressible either. Mar 31 '17 at 19:35
• Hm... looking at the phase diagram one can see that stirring fast enough always creates ice, which I think is the opposite of what the OP wanted ;-). Mar 31 '17 at 19:44

You are trying to use the Bernoulli Principle were it is not applicable! If we temporarily ignore the radial force created by the spinning, we would have the water spinning at some angular velocity $w$. Because of drag, the air above it would also spin. However, the fastest angular velocity that the air could attain, would be the same as the water. For the Bernoulli Principle to apply, the air would have to spin faster than the water, which can't happen - under the given conditions!

If a liquid is starting to boil I'd usually apply that to the whole liquid not just a part of it.

If you just make a part of it evaporate due to extremely low pressure in one part that does not make the whole liquid boil.

• irrelevant and nonstandard Mar 31 '17 at 12:02
• It does make sense to me. Why is it not relevant @CarlWitthoft Mar 31 '17 at 14:06

Here is a link saturated vapour pressure

There is plot in this link and I attached here.

So if we want the water to boil, we need bring either pressure or temperature to the curve. I think the above discussion is assuming the temperature is room temperature e.g. 20-25 degrees C. So the focus is how to bring the pressure down by fast stirring. This is also the reasoning in the question. I think making your stirring spoon having a shape of airfoil can reduce local pressure down to the vapor pressure if you spin fast. You will get local evaporation (cold though).

It could be thought in another way, fast stirring can increase the temperature to 100 degrees and thus boil the water. The stirring does not to be necessary in one direction. The goal is to let the work dissipated to heat by viscosity.

• If you use an airfoil to stir the water, you can (at least in theory) create cavitation bubbles, that's true. However, OP asks: "does the boiling point of the water reduce due to the loweing of this air pressure over the surface". So he's not asking for boiling inside the water, but specifically for boiling at the water surface. And yes, dissipating energy into the water can obviously raise the temperature to the boiling point. This would probably take a while, though, and is no pressure effect. :) Mar 31 '17 at 20:24


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# What does $\mathbb{P}(X_1=x_1,\ldots,X_n=x_n,\sum_{i=1}^n X_i = t)$ actually mean?

Let $X = (X_1,X_2,\ldots,X_n)$ be a vector of iid Bernoulli random variables with parameter $\theta$.
We want to show that $T(X):= \sum_{i=1}^n X_i$ is a sufficient statistic for $\theta$.

This is what they did:

$$P\left(X=x\mid \sum_{i=1}^n X_i = t\right) = \frac{P(X_1 = x_1,X_2=x_2,\ldots,X_n=x_n,\sum_{i=1}^n X_i = t)}{P(\sum_{i=1}^n X_i=t)} \\ =\frac{\theta^{\sum_{i=1}^n x_i}(1-\theta)^{n-\sum_{i=1}^n x_i}}{\binom{n}t \theta^t (1-\theta)^{n-t}} = \frac{1}{\binom{n}t}.$$

What I don't understand is from the first equality to the second, it just looks like they used the fact that
$$P(X_1=x_1,\ldots,X_n=x_n) = P(X_1=x_1)\ldots P(X_n=x_n) = \theta^{\sum_{i=1}^n x_i} (1-\theta)^{n-\sum_{i=1}^n x_i}$$ ... so it seems that $$P(X_1=x_1,\ldots,X_n=x_n) = P(X_1=x_1)\ldots P(X_n=x_n) = P(X_1=x_1,\ldots,X_n=x_n) = P(X_1=x_1,\ldots ,X_n=x_n,\sum_{i=1}^n X_i = t)$$
Can someone please explain this? Shouldn't there be another factor of $$P\left(\sum_{i=1}^n X_i = t\right)$$
by independence as well?

• You should have used indicator functions in the second inequality to see that $\{(X_1,\cdots,X_n)\}\cap\{\sum X_i=t\}$ is $(X_1,\cdots,X_n)$ if $\sum X_i=t$ and $0$ otherwise. – StubbornAtom Mar 1 '18 at 3:43
• That's right. It looks like they were a bit sloppy here and failed to state that they require $\sum x_i = t$. – Ben Mar 1 '18 at 4:00

Generally the notation there is deceptive: the numerator in the second line does involve exponents of $\sum_i x_i$ and $n-=\sum_i x_i$, but these are supposed to be assumed to be $t$ and $n-t$ respectively.

The probability of a given configuration which has sum $t$ is $\theta^t (1-\theta)^{n-t}$: you have $t$ factors of $\theta$ and $n-t$ factors of $(1-\theta)$ in the multiplication of the probabilities. The probability of having sum $t$ in total is given by the binomial PMF. As you know, each of these configurations with a particular sum has the same probability, so the binomial PMF is just ${n \choose t}$ times the probability of each configuration. Thus the probability of each configuration cancels out, and so the conditional probability can be written in the usual "number of ways for X to happen divided by number of elements in sample space" form.

The main point of this calculation from the point of view of statistical inference is that once you know the sum, you cannot discern any more information about $\theta$ by inspecting the actual configuration, because conditioned on the value of the sum, the probabilities of each possible configuration are all the same.

By $$P\left(X_1=x_1,\ldots ,X_n=x_n,\sum_iX_i=t\right)$$ they do mean what it looks like: the probability that $X_1=x_1$ and $X_2=x_2,$ etc and that also $\sum_iX_i=t.$ And they do write the correct expression in the numerator except that there should be a constraint that $\sum_i x_i = t.$ So we can replace the occurrences of $\sum_i x_i$ with $t,$ which is how they get to the final expression (which should also have a constraint on it). To be explicit, we have $$P\left(X_1=x_1,\ldots ,X_n=x_n,\sum_iX_i=t\right) = \theta^{\sum_i x_i}(1-\theta)^{n-\sum_i x_i}1_{\sum_i x_i=t}= \theta^t(1-\theta)^{n-t}1_{\sum_i x_i=t}.$$

The reasoning behind this expression is that of course if $\sum_i X_i = t$ we are going to have zero unless $\sum_i x_i = t.$ And on the other hand, if we do meet that constraint, then the $\sum_i X_i=t$ is redundant, so we just have the same thing as $P(X_1=x_1,\ldots,X_n=x_n).$

First to show sufficient statistics you want to show the probability you're calculating doesn't depend on $\theta$. This goal is achieved in the end, as you can see in the end. To answer your question, you can see that $t$ is redundant since if you have all values of $x$ then their sum is also know. So $P(X=x,\sum X=t)=P(X=x)$.

The statement $X_1 = x_1,X_2=x_2,\ldots,X_n=x_n,\sum_{i=1}^n X_i = t$ is redundant and means the same thing as $X_1 = x_1,X_2=x_2,\ldots,X_n=x_n.$ And then you use independence of the $n$ random variables to write that as a product: $$\Pr(X_1=x_1) = \theta^{x_1} (1-\theta)^{1-x_1} = \begin{cases} \theta & \text{if } x_1=1, \\ 1-\theta & \text{if } x_1=0. \end{cases}$$ Then multiply and get $\theta^{x_1+\cdots+x_n} (1-\theta)^{n-(x_1+\cdots+x_n)}.$


text:
🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Oh no! Our educators are currently working hard solving this question. In the meantime, our AI Tutor recommends this similar expert step-by-step video covering the same topics. Numerade Educator ### Problem 105 Easy Difficulty # Fission, the process that supplies energy in nuclear power plants, occurs when a heavy nucleus issplit into two medium-sized nuclei. One such reaction occurs when a neutron colliding witha$^{235} \mathrm{U}$(uranium) nucleus splits that nucleus into a$^{141} \mathrm{Ba}$(barium) nucleus and a$^{92} \mathrm{Kr}$(krypton) nucleus. In this reaction, two neutrons also are split off from the original$^{235} \mathrm{U}$Before the collision, the arrangement is as shown in Fig. 8.49$\mathrm{a}$. After the collision, the$^{141} \mathrm{Ba}$nucleus is moving in the$+z$-direction and the$^{92} \mathrm{Kr}$nucleus in the$-z$-direction. The three neutrons are moving in the$x y$-plane, as shown in Fig. 8.49$\mathrm{b}$. If the incoming neutron has an initial velocity of magnitude$3.0 \times 10^{3} \mathrm{m} / \mathrm{s}$and a final velocity of magnitude$20 \times 10^{3} \mathrm{m} / \mathrm{s}$in the directions shown, what are the speeds of the other two neutrons, and what can you say about the speeds of the 141$\mathrm{Ba}$and$^{92} \mathrm{Kr}$nuclei? (The mass of the 14$\mathrm{Ba}$nucleus is approximately$23 \times 10^{-25} \mathrm{kg},$and the mass of$^{92} \mathrm{Kr}$is about$1.5 \times 10^{-25} \mathrm{kg} . )$### Answer ##$=1.5 v_{\mathrm{Be}}\$

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text:
40m QIL Cryo_Lab CTN SUS_Lab TCS_Lab OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop
 PSL Not logged in
 Wed Jan 30 15:07:32 2019, awade, anchal, HowTo, FSS, Shot noise in FSS loops? Fri Feb 1 14:59:01 2019, awade, anchal, HowTo, FSS, Noise tallies Fri Feb 1 20:20:11 2019, awade, HowTo, FSS, Modifying North FSS EOM path Tue Feb 5 11:40:02 2019, awade, HowTo, FSS, Modifying South FSS boost Tue Feb 5 20:39:52 2019, anchal, Summary, FSS, Complete model of TTFSS box with TF and crossover analysis Wed Feb 6 19:13:06 2019, anchal, Summary, FSS, Complete model of TTFSS box with TF and crossover analysis Wed Feb 6 21:30:24 2019, awade, Summary, FSS, Complete model of TTFSS box with TF and crossover analysis Thu Feb 7 10:33:27 2019, anchal, Summary, FSS, Complete model of TTFSS box with TF and crossover analysis
Message ID: 2301     Entry time: Fri Feb 1 14:59:01 2019     In reply to: 2300     Reply to this: 2302
 Author: awade, anchal Type: HowTo Category: FSS Subject: Noise tallies

We've hit a flat (white) noise floor at around 70 mHz/rtHz.  Time to tally all the unaccounted for noises.

Photo detector sensor contributions

Tally of photodetector sensor contribution to final white noise floor
Noise Output measured noise Input referred noise Freq. Equivalent Noise
BN detector (Koji 27 MHz, G=1.27 kΩ) Dark 9 nA/rtHz 12.0 pW/rtHz 48.0 nrad/rtHz * f or 48 µHz/rtHz @ 1 kHz
BN detector (Koji 27 MHz) Shot noise (est from DC power 0.5 mW) - 13.7 pW/rtHz 54.8 nrad/rtHz *f or 54.8 µHz/rtHz @1 kHz
FSS North PD (36 MHz, G=2.7 kΩ) Dark (measured) 14.7 nV/rtHz out of mixer, 26.7 nV/rtHz at detector 13.2 pW/rtHz 3 mHz/rtHz
FSS North PD (36 MHz, G=2.7 kΩ) Shot (est from DC power 1.2 mW, gamma=0.3, vis=0.5) - 15.6 pW/rtHz 2.9 mHz/rtHz
FSS South PD (36 MHz, G=1.5 kΩ) Dark (measured) 16.1nV/rtHz out of mixer, 26.0 nV/rtHz at detector 14.5 pW/rtHz 5.9 mHz/rtHz
FSS North PD (36 MHz, G=1.5 kΩ) Shot (est from DC power 1.2 mW, gamma=0.3, vis=0.5) - 15.6 pW/rtHz 2.9 mHz/rtHz
Total - - 7.8mHz/rtHz

Here the FSS resonant detector dark noises were measured using a Lv13 ZX05-1MHW+ mixer with a 4th order LP + 50 Ω terminator on a SR785.

Frequency equivalent noise is just input referred noise divided by the PDH discriminator value. The PDH discriminator value is computer assuming gamma=0.3 rad modulation depth, visibility of 0.5, Finesse of 15000, cavity length 3.8 cm.  This gives 4.4 nV/Hz for the slope of the error signal about resonance.

BN detector input referred noise is just input referred power divided by 1/2 Vpp of the beat note.  For 0 dBm beat note, then input equivalent power is (dividing through 1.27 kΩ TI gain) works out to 0.25 mW/rad.

So the upshot  is that there is too much FSS RFPD noise to see Brownian noise clearly but is not our current dominant noise source. We are missing about 69.6 mHz/rtHz.

Also noise contributions at the summing stage in FSS

I'm including this because it was a question we thought about and dismissed.  There is a 100kΩ - 1 µF - 100 kΩ  LP filter from the FSS offset channel into U3 on the servo board (R8-C5-R9, in D040105-C). There is an estimated 0.4 pA/rtHz of thermal current noise there. Also the AD829 has a current noise of 1.5 pA/rtHz  (op amp U3).  These are additive and indistinguishable from sensor noise.  The mixer output equivalent voltage noise is just the equivalent voltage across R3 and R9 in the FSS RF board (LIGO-D0901894): i.e. 0.4 pA/rtHzx146 Ω = 58.4 pV/rtHz at mixer output.  Dividing back through the mixer conversion, transformer coupler and PD gains (see PSL: 2242 for values and links) we come to some input referred noises in units of optical power tabulated below.

 Noise Output measured noise Input referred noise Freq. Equivalent Noise Offset LP filter Johnson noise R8-C5-R9, D040105-C) 0.4 pA/rtHz 0.05 pA/rtHz 12 µH/rtHz U3 current noise (D040105-C) 1.5 pA/rtHz 0.2 pA/rtHz 45 µH/rtHz

So electronic noise, at least from the point of summing the error signal with offsets is negligible.  We might want to revisit this later, but most points of noise injection into the FSS loop will be suppressed by the closed loop function and will be small.

Laser frequency noise

The current state of the FSS loops is ~180 kHz UGF for north and ~70 kHz UGF for south (woeful).  South is worst because the RF PD has only half the TI gain and we are at maximum gain for the FSS variable gain stages.  The whole point of these loops is to suppress laser frequency noise.  Their current performance is just not good enough.

If we assume that the laser noise is modeled as

$\dpi{100} S_f^\textrm{Laser} =\frac{1\textrm{Hz}}{f}\times10^4 \quad\textrm{[Hz}/\sqrt{\textrm{Hz}}\textrm]$

and that we need a clearance of 10 dB at the 100 Hz point where we expect to see Brownian noise (10 mHz/rtHz @ 100 Hz), then we need to get to 1 mHz/rtHz or a a factor of 1e5 1/(1+G) suppression. This means that its very likely that our current limitation is actually FSS loop bandwidth.

Here is what we need to do.

Naively if we assume 1/f loop shape, that means that we would need a UGF of 10 MHz to reach 1e5 gain by 100 Hz.  Nope.  Of course we have some loop shaping in the lower frequency reaches of the PZT path of the FSS.  There is a boost pole-zero pair installed in the common path (PZT servo path U7) that uses R29 of 5.6 kΩ and capacitors of either 4.7 µF or 6.8 µF depending on the box.  This would put the corners between 6.0Hz and 4.1 Hz, too low (don't know why so low). If we move the boost up to 1 kHz or higher then we would be able to reach 1 mHz/rtHz by 100Hz with a overall path open loop gain crossing 0dB at 1 MHz.

Right now we cannot turn up the gain of the loops much more than 200 kHz.  We can see from looking at the EOM control signals that above a certain amount of OLG the EOM is taking too much of the actuation load and rails.  Basically the the cut off frequency for the EOM servo is not aggressive enough and the EOM is working too hard in the frequency band below 10 kHz where the laser PZT should be doing the heavy lifting.  I outlined some of the changes that need to be done in PSL:1902.  This is based on Matt Evan's modifications of the MIT Squeezer FSS system.

MIT's FSS boxes are a few versions on from our FSS boxes but the main points remain:

• R19 24.9k to 2k (lower first zero from 1.94 kHz to 24.1 kHz)
• C23 1uF to 33nF (more AC coupling, move first zero from 145 Hz to 4.38 kHz)
• C24 1uF to 100nF (more AC coupling, move second pole from 122 Hz to 1.22 kHz)

In addition to this we want to modify the boost stage by removing the ~5 µF chunkers and trying something more like 28 nF or more.  I think what we need here is a metal film cap (no ceramic) and the question is whether setting the corner so high will be an issue with it railing between locks if there is some kind of offsetting there.  Maybe this isn't an issue.

These are relatively strait forward changes providing we have the appropriate capacitors in stock. It seems like boosting FSS UGF by offloading EOM path LF onto the PZT and making the boost good is the most obvious way to squash this noise and make fast progress.  Lets try and hit the 1 mHz/rtHz target with a 1 MHz FSS UGF effort.

Efforts on characterizing thermal sensors need to wrapped up but we want to get to our actual noise source.  The more recent PID improvements on the cavity temperature control (PSL:2295) have gotten us to a point where we have usable 1kHz/V VCO measurements which is as good as we need for now. Maybe a shift in effort from thermal control to laser stabilization is the best use  of our time.

I've attached a block diagram of the FSS controls below.  It shows at a glance the current configuration of the EOM, PZT and slow laser controls.

Edit awade Sat Feb 2 00:44:47 2019: Revised down FSS shot noise estimates based on correct sidebands, visibility and modulation depth values (from Tara's thesis).

 Attachment 1: CTN_FSS_south_blockdiagram.pdf  48 kB  Uploaded Fri Feb 1 17:11:17 2019
ELOG V3.1.3-


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## Cryptology ePrint Archive: Report 2011/271

Practical Key-recovery For All Possible Parameters of SFLASH

Charles Bouillaguet and Pierre-Alain Fouque and Gilles Macario-Rat

Abstract: In this paper we present a new practical key-recovery attack on the SFLASH signature scheme. SFLASH is a derivative of the older $C^*$ encryption and signature scheme that was broken in 1995 by Patarin. In SFLASH, the public key is truncated, and this simple countermeasure prevents Patarin's attack. The scheme is well-known for having been considered secure and selected in 2004 by the NESSIE project of the European Union to be standardized.

However, SFLASH was practically broken in 2007 by Dubois, Fouque, Stern and Shamir. Their attack breaks the original (and most relevant) parameters, but does not apply when more than half of the public key is truncated. It is therefore possible to choose parameters such that SFLASH is not broken by the existing attacks, although it is less efficient.

We show a key-recovery attack that breaks the full range of parameters in practice, as soon as the information-theoretically required amount of information is available from the public-key. The attack uses new cryptanalytic tools, most notably pencils of matrices and quadratic forms.

Category / Keywords: public-key cryptography / SFLASH, multivariate cryptography, practical cryptanalysis, key-recovery


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# Ex.6.2 Q7 Triangles Solution - NCERT Maths Class 10

Go back to  'Ex.6.2'

## Question

Using Theorem $$6.1$$, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

## Text Solution

Reasoning:

We know that theorem $$6.1$$ states that “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio ($$BPT$$)”.

Steps:

In $$\Delta ABC$$$$, D$$ is the midpoint of $$AB$$

Therefore,

$AD = BD$

$\frac{AD}{BD}=1$

Now,

\begin{align}DE&||BC \\ \Rightarrow\frac{AE}{EC}&=\frac{AD}{BD}\,\,\,\text{[Theorem} \,6.1] \\ \Rightarrow \frac{AE}{EC}&=1 \end{align}

Hence $$,E$$ is the midpoint of $$AC.$$

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# De Morgan's Laws (Logic)

This proof is about De Morgan's Laws in the context of propositional logic. For other uses, see De Morgan's Laws.

## Theorem

### Disjunction of Negations

#### Formulation 1

$\neg p \lor \neg q \dashv \vdash \neg \left({p \land q}\right)$

#### Formulation 2

$\vdash \left({\neg p \lor \neg q}\right) \iff \left({\neg \left({p \land q}\right)}\right)$

### Conjunction of Negations

#### Formulation 1

$\neg p \land \neg q \dashv \vdash \neg \left({p \lor q}\right)$

#### Formulation 2

$\vdash \left({\neg p \land \neg q}\right) \iff \left({\neg \left({p \lor q}\right)}\right)$

### Conjunction

#### Formulation 1

$p \land q \dashv \vdash \neg \left({\neg p \lor \neg q}\right)$

#### Formulation 2

$\vdash \left({p \land q}\right) \iff \left({\neg \left({\neg p \lor \neg q}\right)}\right)$

### Disjunction

#### Formulation 1

$p \lor q \dashv \vdash \neg \left({\neg p \land \neg q}\right)$

#### Formulation 2

$\vdash \left({p \lor q}\right) \iff \left({\neg \left({\neg p \land \neg q}\right)}\right)$

## The Intuitionist Perspective

Note that this:

$\neg p \land \neg q \dashv \vdash \neg \left({p \lor q}\right)$

can be proved in both directions without resorting to the LEM.

All the others:

$\neg p \lor \neg q \vdash \neg \left({p \land q}\right)$

$p \land q \vdash \neg \left({\neg p \lor \neg q}\right)$

$p \lor q \vdash \neg \left({\neg p \land \neg q}\right)$

are not reversible in intuitionistic logic.


text:
# Geometric Measure Theory On Euclidean Spaces

Dear all,

I have recently found the following discussion: Different Measures On R2 regarding different boundary measures on $\mathbb{R}^n$ .

Is there any example in the opposite direction?

i.e. an example of a set $A \subseteq \mathbb{R} ^n$ such that $Leb^+ (A) > H^{n-1} (\partial A)$ ?

-

Take $[0,1]\times \{0\} \subset \mathbb{R}^2$.

-
Dear @Tapio: Thanks for your answer, but I don't think I understand it: We have $H^1(\partial [0,1]\times0 )=H^1([0,1]\times0 ) = 1$ . But I can't understand how to calculate the minkowski's content of this thing- $Leb ( [0,1] \times 0 ) = 0$ since we are in $\mathbb{R}^2$ . But what is the Lebesgue measure of an $\epsilon$ extension of this thing? Thanks ! –  Jeremy Young Jul 23 '12 at 6:57
@Jeremy, $\mathop{Leb}^+=2$. –  Anton Petrunin Jul 23 '12 at 7:53
@Anton: Can you please explain this calculation? Thanks! –  Jeremy Young Jul 23 '12 at 11:24
@Jeremy: An $\epsilon$ extension of that set is very simple, and the computation of its area is completely elementary. Just draw a picture of it. –  Mark Meckes Jul 23 '12 at 12:14
@Mark: I agree that an $\epsilon$ extension of this set is pretty simple to calculate in the $\mathbb{R}$ case (and it's equal to $2\epsilon$ there indeed, which gives $Leb^+ =2$ as you said. But since we're looking at the $\mathbb{R}^2$ case, the $\epsilon$ extension we are looking at is actually some kind of an ellipse, which I have no idea how to calculate its volume. I hope I made my misunderstandings clear now. Thanks a lot ! –  Jeremy Young Jul 23 '12 at 15:57
$A=\mathbb{Z}\subset \mathbb R^2$. In this case $\partial\mathbb{Z}= \mathbb{Z}$ and $\mathop{Leb}^+(\mathbb Z)=\infty$ and $H^1(\mathbb Z)=0$.
You can get a bounded example of the same type. Take a countable nowhere dense set $A$ in the unit disc such that the $\varepsilon$-neighborhood of $A$ contains a disc of radius $\sqrt[3]{\varepsilon}$.
But $\partial \mathbb{Q} = \mathbb{R}$. Perhaps you mean $\mathop{Leb}^+(\mathbb Z)=\infty$ and $H^1(\mathbb Z)=0$? –  Tapio Rajala Jul 22 '12 at 19:44
Thanks @Anton, but I was thinking about this question in the context of boundary measures...In this case, since $\partial \mathbb{Z}$ is empty , it doesn't give me an answer. Thanks anyway! –  Jeremy Young Jul 23 '12 at 6:52
@Jeremy, $\partial \mathbb Z=\mathbb Z$. –  Anton Petrunin Jul 23 '12 at 7:46


text:
#### Volume 20, issue 5 (2020)

 Recent Issues
 The Journal About the Journal Editorial Board Editorial Interests Subscriptions Submission Guidelines Submission Page Policies for Authors Ethics Statement ISSN (electronic): 1472-2739 ISSN (print): 1472-2747 Author Index To Appear Other MSP Journals
An unstable change of rings for Morava $E$–theory

### Robert Thompson

Algebraic & Geometric Topology 20 (2020) 2145–2176
##### Abstract

The Bousfield–Kan (or unstable Adams) spectral sequence can be constructed for various homology theories, such as Brown–Peterson homology theory $\mathit{BP}\phantom{\rule{-0.17em}{0ex}}$, Johnson–Wilson theory $E\left(n\right)$ or Morava $E$–theory ${E}_{n}$. For nice spaces the ${E}_{2}$–term is given by Ext in a category of unstable comodules. We establish an unstable Morava change of rings isomorphism between ${Ext}_{{\mathsc{𝒰}}_{{\Gamma }_{\phantom{\rule{-0.17em}{0ex}}B}}}\left(B,M\right)$ and ${Ext}_{{\mathsc{𝒰}}_{{E}_{n\ast }{E}_{n}}∕{I}_{n}}\left({E}_{n\ast }∕{I}_{n},{E}_{n\ast }{\otimes }_{{\mathit{BP}}_{\ast }}M\right)$, where $\left(B,{\Gamma }_{\phantom{\rule{-0.17em}{0ex}}B}\right)$ denotes the Hopf algebroid $\left({v}_{n}^{-1}{\mathit{BP}}_{\ast }∕{I}_{n},{v}_{n}^{-1}{\mathit{BP}}_{\ast }\mathit{BP}∕{I}_{n}\right)$. We show that the latter groups can be interpreted as $Ext$ in the category of continuous modules over the profinite monoid of endomorphisms of the Honda formal group law. By comparing this with the cohomology of the Morava stabilizer group we obtain an unstable Morava vanishing theorem when $p-1\nmid n$

##### Keywords
unstable Adams spectral sequence, Morava changes of rings theorem
##### Mathematical Subject Classification 2010
Primary: 55N20, 55Q51, 55T15


text:
MAT244--2020F > Chapter 2

Help for section 2.4, questions 14-15

(1/1)

kavinkandiah:
Sorry, I think I asked my question in the wrong folder!!! Here is my question:

Hi, I'm having a lot of trouble with questions 14 and 15. I know how to find the solution, but I don't know how to do the rest, can someone please help??

Zhiyue Yu:
Hi, here is the full steps of the question, it's just plugging in the points, hope that is helpful!

Victor Ivrii:
Almost done, but you need also define interval, on which solution exists; it must include $0$


text:
### Symmetry, Integrability and Geometry: Methods and Applications (SIGMA)

SIGMA 13 (2017), 029, 14 pages      arXiv:1612.03674      https://doi.org/10.3842/SIGMA.2017.029

### Isomonodromy for the Degenerate Fifth Painlevé Equation

Primitivo B. Acosta-Humánez a, Marius van der Put b and Jaap Top b
a) Universidad Simón Bolívar, Barranquilla, Colombia
b) University of Groningen, Groningen, The Netherlands

Received December 12, 2016, in final form May 01, 2017; Published online May 09, 2017

Abstract
This is a sequel to papers by the last two authors making the Riemann-Hilbert correspondence and isomonodromy explicit. For the degenerate fifth Painlevé equation, the moduli spaces for connections and for monodromy are explicitly computed. It is proven that the extended Riemann-Hilbert morphism is an isomorphism. As a consequence these equations have the Painlevé property and the Okamoto-Painlevé space is identified with a moduli space of connections. Using MAPLE computations, one obtains formulas for the degenerate fifth Painlevé equation, for the Bäcklund transformations.

Key words: moduli space for linear connections; irregular singularities; Stokes matrices; monodromy spaces; isomonodromic deformations; Painlevé equations.

pdf (385 kb)   tex (23 kb)

References

1. Chekhov L., Mazzocco M., Rubtsov V., Painlevé monodromy manifolds, decorated character varieties and cluster algebras, Int. Math. Res. Not., to appear, arXiv:1511.03851.
2. Fokas A.S., Its A.R., Kapaev A.A., Novokshenov V.Yu., Painlevé transcendents. The Riemann-Hilbert approach, Mathematical Surveys and Monographs, Vol. 128, Amer. Math. Soc., Providence, RI, 2006.
3. Gromak V.I., On the theory of Painlevé's equations, Differential Equations 11 (1975), 285-287.
4. Inaba M., Moduli of parabolic connections on curves and the Riemann-Hilbert correspondence, J. Algebraic Geom. 22 (2013), 407-480, math.AG/0602004.
5. Inaba M., Iwasaki K., Saito M.-H., Dynamics of the sixth Painlevé equation, in Théories asymptotiques et équations de Painlevé, Sémin. Congr., Vol. 14, Soc. Math. France, Paris, 2006, 103-167.
6. Inaba M., Iwasaki K., Saito M.-H., Moduli of stable parabolic connections, Riemann-Hilbert correspondence and geometry of Painlevé equation of type VI. I, Publ. Res. Inst. Math. Sci. 42 (2006), 987-1089, math.AG/0309342.
7. Inaba M., Iwasaki K., Saito M.-H., Moduli of stable parabolic connections, Riemann-Hilbert correspondence and geometry of Painlevé equation of type VI. II, in Moduli Spaces and Arithmetic Geometry, Adv. Stud. Pure Math., Vol. 45, Math. Soc. Japan, Tokyo, 2006, 387-432, math.AG/0605025.
8. Jimbo M., Miwa T., Ueno K., Monodromy preserving deformation of linear ordinary differential equations with rational coefficients. I. General theory and $\tau$-function, Phys. D 2 (1981), 306-352.
9. Jimbo M., Miwa T., Monodromy preserving deformation of linear ordinary differential equations with rational coefficients. II, Phys. D 2 (1981), 407-448.
10. Ohyama Y., Kawamuko H., Sakai H., Okamoto K., Studies on the Painlevé equations. V. Third Painlevé equations of special type ${\rm P}_{\rm III}(D_7)$ and ${\rm P}_{\rm III}(D_8)$, J. Math. Sci. Univ. Tokyo 13 (2006), 145-204.
11. Ohyama Y., Okumura S., R. Fuchs' problem of the Painlevé equations from the first to the fifth, in Algebraic and Geometric Aspects of Integrable Systems and Random Matrices, Contemp. Math., Vol. 593, Amer. Math. Soc., Providence, RI, 2013, 163-178, math.CA/0512243.
12. Okamoto K., Sur les feuilletages associés aux équations du second ordre à points critiques fixes de P. Painlevé. Espaces des conditions initiales, Japan. J. Math. 5 (1979), 1-79.
13. Okamoto K., Isomonodromic deformation and Painlevé equations, and the Garnier system, J. Fac. Sci. Univ. Tokyo Sect. IA Math. 33 (1986), 575-618.
14. Okamoto K., Studies on the Painlevé equations. III. Second and fourth Painlevé equations, ${\rm P}_{{\rm II}}$ and ${\rm P}_{{\rm IV}}$, Math. Ann. 275 (1986), 221-255.
15. Okamoto K., Studies on the Painlevé equations. IV. Third Painlevé equation ${\rm P}_{{\rm III}}$, Funkcial. Ekvac. 30 (1987), 305-332.
16. Okamoto K., The Hamiltonians associated to the Painlevé equations, in The Painlevé property, CRM Ser. Math. Phys., Springer, New York, 1999, 735-787.
17. van der Put M., Families of linear differential equations related to the second Painlevé equation, in Algebraic Methods in Dynamical Systems, Banach Center Publ., Vol. 94, Polish Acad. Sci. Inst. Math., Warsaw, 2011, 247-262.
18. van der Put M., Families of linear differential equations and the Painlevé equations, in Geometric and Differential Galois Theories, Sémin. Congr., Vol. 27, Soc. Math. France, Paris, 2013, 207-224.
19. van der Put M., Saito M.-H., Moduli spaces for linear differential equations and the Painlevé equations, Ann. Inst. Fourier (Grenoble) 59 (2009), 2611-2667, arXiv:0902.1702.
20. van der Put M., Singer M.F., Galois theory of linear differential equations, Grundlehren der Mathematischen Wissenschaften, Vol. 328, Springer-Verlag, Berlin, 2003.
21. van der Put M., Top J., A Riemann-Hilbert approach to Painlevé IV, J. Nonlinear Math. Phys. 20 (2013), suppl. 1, 165-177, arXiv:1207.4335.
22. van der Put M., Top J., Geometric aspects of the Painlevé equations ${\rm PIII}(\rm D_6)$ and ${\rm PIII}(\rm D_7)$, SIGMA 10 (2014), 050, 24 pages, arXiv:1207.4023.
23. Saito M.-H., Takebe T., Classification of Okamoto-Painlevé pairs, Kobe J. Math. 19 (2002), 21-50, math.AG/0006028.
24. Saito M.-H., Takebe T., Terajima H., Deformation of Okamoto-Painlevé pairs and Painlevé equations, J. Algebraic Geom. 11 (2002), 311-362, math.AG/0006026.
25. Saito M.-H., Terajima H., Nodal curves and Riccati solutions of Painlevé equations, J. Math. Kyoto Univ. 44 (2004), 529-568, math.AG/0201225.
26. Witte N.S., New transformations for Painlevé's third transcendent, Proc. Amer. Math. Soc. 132 (2004), 1649-1658, math.CA/0210019.


text:
# Help with differential equation

1. Jan 3, 2012

### PhyPsy

1. The problem statement, all variables and given/known data
$\frac{dr}{\sqrt{r^2- 2mr}}= \frac{d\rho}{\rho}$
Solve for r.

2. Relevant equations

3. The attempt at a solution
For those familiar with GR, this is an attempt to come up with the isotropic form of the Schwarzschild solution to the field equations, but I'm having trouble with the math, so I just need someone to check my math. The solution I am supposed to get is $r=\rho(1+ \frac{1}{2}m/\rho)^2$, but I am coming up with $r=\rho(\frac{1}{2}+ m/\rho)^2$.

After integration, the equation I came up with is $2\ln(\sqrt{r-2m}+ \sqrt{r})= \ln\rho$
$e^{2\ln(\sqrt{r-2m}+ \sqrt{r})}= e^{\ln\rho}$
$e^{\ln(\sqrt{r-2m}+ \sqrt{r})+ \ln(\sqrt{r-2m}+ \sqrt{r})}= e^{\ln\rho}$
$e^{\ln(\sqrt{r-2m}+ \sqrt{r})}e^{\ln(\sqrt{r-2m}+ \sqrt{r})}= e^{\ln\rho}$
$(\sqrt{r-2m}+ \sqrt{r})^2= \rho$
$r- 2m+ 2\sqrt{r^2- 2mr}+ r= \rho$

I moved all terms on the left side of the equal sign to the right except for the square root term and squared both sides.
$4(r^2- 2mr)= (\rho+ 2m- 2r)^2= \rho^2+ 4m\rho- 4r\rho+ 4m^2- 8mr+ 4r^2$

The terms $4r^2$ and $-8mr$ cancelled, and then I moved the term $-4r\rho$ to the left side of the equal sign.
$4r\rho= \rho^2+ 4m\rho+ 4m^2$
$r\rho= (\frac{1}{2}\rho+ m)^2$
$r= \rho^2(\frac{1}{2}+ m/\rho)^2/\rho= \rho(\frac{1}{2}+ m/\rho)^2$

Does anyone see a problem?

2. Jan 3, 2012

### Dick

I see one. The integration will give you $2\ln(\sqrt{r-2m}+ \sqrt{r})= \ln\rho +C$. There's an undetermined constant, C. I don't think you can assume it's zero. I think a choice of that constant will give you the books answer. Don't you have some way to determine the constant?

3. Jan 4, 2012

### PhyPsy

Doing the math in my head, the constant would have to be $\ln(\rho-m)$. The constant from integration of the left side of the equation allows for the rho term. I risk leaving the topic of this forum by asking this, but every book and article I have seen uses the specific transformation $r=\rho(1+\frac{1}{2}m/\rho)^2$ when deriving the isotropic form of the Schwarzschild solution. Why could I not use 0 as the constant of integration and have the transformation be $r=\rho(\frac{1}{2}+m/\rho)^2$? It seems to me that also leads to a legitimate isotropic construct of the Schwarzschild solution.

4. Jan 4, 2012

### Dick

$\ln(\rho-m)$ isn't a constant. You can't add that. You generally have some boundary conditions you want your coordinates to satisfy. In the books solution for large $\rho, r$ you have $\rho \approx r$. Your version wouldn't have that.

5. Jan 4, 2012

### Staff: Mentor

you could try differentiating your answer to see what factors got changed then go back thru your steps to see if you properly accounted for them.

in undergrad we once did an eperiment to measure the speed of light via intereference and a rotating magnetic and got 1/2 the seepd of light. We never figured out why and the prof couldn't find anything wrong with what we did. After many years, I realized we may not have factered in that the mirror was double sided which meant it effectively spun at twice the speed. Oh well, that's why I went into programming that and the money even though physics was way more interesting.

6. Jan 4, 2012

### quZz

how did you get this? Seems you should have had
$$\ln(\sqrt{(r-m)^2-m^2} + r-m) = \ln(C \rho)$$
choosing $C=2$ you get your answer

7. Jan 4, 2012

### PhyPsy

Bah, you're right, since $\rho$ is a function of r, I can't use it as a constant. My bad.

Ha ha, I'm actually in IT myself, though I am not a hardcore programmer; I do some here and there. Physics is a very interesting subject to me, but a career in it, on the other hand, doesn't appeal to me too much.

Well, to be honest, I cheated and used Wolfram to compute the integral for the left side. I thought it might be a bad idea, but I looked into using substitution and integration by parts and couldn't get anywhere using those techniques. Yes, differentiating your solution does indeed return $\frac{dr}{\sqrt{r^2-2mr}}$. I am curious to know how you came up with that.

Also, I was not aware you would get a constant multiplied by $\rho$ on the right side. I don't remember learning about that in calculus, but it has been 7 years since I took it, so maybe I just completely forgot.

8. Jan 4, 2012

### quZz

Ok, first you notice that $r^2-2mr = (r-m)^2-m^2$ and then you substitute $x\equiv r-m$, and $dr=dx$. The integral then becomes
$$\int\frac{dx}{\sqrt{x^2-m^2}} = \ln(\sqrt{x^2-m^2}+x)$$
Second, on the right hand side you have
$$\int\frac{d\rho}{\rho}=\ln\rho + C_1$$
Let's write $C_1=\ln C$ so that we have pure $\ln$s on both sides
$$\ln\rho + C_1 = \ln(C\rho)$$


text:
# arxivststuff from arxiv that you should probably bookmark

## Boosting with Structural Sparsity: A Differential Inclusion Approach

Abstract · Apr 16, 2017 23:58 ·

bregman consistency jiechao xinwei inclusions split boosting lbi lasso chendi stat-ml

### Arxiv Abstract

• Chendi Huang
• Xinwei Sun
• Jiechao Xiong
• Yuan Yao

Boosting as gradient descent algorithms is one popular method in machine learning. In this paper a novel Boosting-type algorithm is proposed based on restricted gradient descent with structural sparsity control whose underlying dynamics are governed by differential inclusions. In particular, we present an iterative regularization path with structural sparsity where the parameter is sparse under some linear transforms, based on variable splitting and the Linearized Bregman Iteration. Hence it is called \emph{Split LBI}. Despite its simplicity, Split LBI outperforms the popular generalized Lasso in both theory and experiments. A theory of path consistency is presented that equipped with a proper early stopping, Split LBI may achieve model selection consistency under a family of Irrepresentable Conditions which can be weaker than the necessary and sufficient condition for generalized Lasso. Furthermore, some $\ell_2$ error bounds are also given at the minimax optimal rates. The utility and benefit of the algorithm are illustrated by several applications including image denoising, partial order ranking of sport teams, and world university grouping with crowdsourced ranking data.

Read the paper (pdf) »


text:
# Air Column Resonance

#### Purpose

To create standing waves in an air column with variable length.

#### Objectives

• To create standing wave patterns in an open-end tube and to determine their wavelength
• To learn how to record and organize experimental data using spreadsheets

#### Equipment

• Plastic cylindrical tube connected to water container
• Tuning fork, mallet, a pencil

#### Theory

The velocity of a  wave with frequency f and wavelength $\lambda$ is given by

$v = \lambda f\;\;\;\;(\text{Eq.1})$

For a sound wave in air, the velocity depends on the temperature as

$v = 331+0.6T\;\;\;\;(\text{Eq.2})$

For standing wave patterns to be occur in a tube with one open and one closed end, the length of the tube L and the wavelength $\lambda$ must satisfy the following condition

$L = m \left(\frac{\lambda}{4}\right) \;\text{where }\; m=1,3, 5,\ldots \;\;\;\;(\text{Eq.3})$

In the experiment, the length of the tube is determined by the water level, y. Thus, the wavelength of the waves producing the standing wave pattern will be :

$\lambda = \frac{4 y}{m}\;m=1,3,5,7,\ldots\;\;\;\;(\text{Eq.4})$

#### Pre-lab Activity

The water level in a vertical glass tube 1.00 m long can be adjusted to any position in the tube. A tuning fork vibrating at 1024 Hz is held just over the open top end of the tube, in order to set up a standing wave of sound in the air-filled portion of the tube. Considering room temperature (20oC), at what positions of the water level will sound from the fork set up resonance in the air column inside the tube?

#### Experimental Setting

Here is step-by-step detailed instructions on how to set up the lab.

###### Procedure

Fill in the tube with water. Using the mallet, set the tuning fork to vibrate in its fundamental frequencies. Move the water container up and down and listen for the position where resonance occurs. Mark them in pencil on the outside surface of the tube.

• Measure the room temperature and calculate the speed of sound in the room using Eq. 2.
• Calculate the expected wavelength for the 512 Hz and 1024 Hz fork by using Eq.1
 T (oC) v (m/s) wavelength (m) 512 Hz 1024 Hz

#### Things To Watch For!

###### Don’t spill water!

Use a pitcher and fill in the tube-container system with water. Be careful so that water does not spill. As you move the water container up and down, water level in the container and the tube will always be the same (water seeks its level). This is the way you can control the length of the air column in the plastic tube.

###### The standing waves are in the AIR-filled portion of the tube!

The standing waves are created in the air space between the opening of the tube and the water level. The resonance (standing waves) will occur only when the water is at such a level that the length of the air column L meets the conditions in Eq.3

###### Listen for the resonance of the fundamental frequency produced by the fork.

The sound becomes distinctively loud when the conditions of Eq.3 are met. There will be positions that you will hear fainter sound – those might be the positions for standing waves due to the higher harmonics.  Generally, if you are unsure whether the sound is loud or not, it is most likely not the resonance you are looking for.

#### Data

Record your data in a data table.

 512 Hz 1024 Hz y1  in (m) y2  in (m) y3  in (m) y4  in (m)

#### Results

Using your measurements above, determine the wavelength for each fork by using Eq. 4. Each position y should yield the same value of wavelength – so use your measurements to determine those values, and average it. The average value is your best result for the wavelength.

Check your value for the wavelength against what it should be, and calculate the percent difference between the two.


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# Optimizing CAL Report Hadoop MapReduce Jobs

#### eBay's Central Application Logging system (CAL) collects log data from all kinds of applications. The summary reports for log data are created using Hadoop MapReduce jobs. This article discusses our experiences optimizing these jobs.

eBay's Central Application Logging (CAL) collects log data from all kind of applications to provide a log viewer and summary report service. Application owners can use logging summary reports to find out the following:

• Percentile of time spent on each transaction
• Service transaction calls
• DB operations for an application

CAL uses Hadoop MapReduce jobs to generate reports for each kind of application.

This article discusses about our experiences optimizing Hadoop jobs. We hope to provide developers some insight to the identification and solutions/options to optimize and resolve Hadoop MapReduce (MR) problems, including the reduced consumption of resources.

### Problem statement

There are three issues that need to be addressed:

• Data Set—The volume of CAL logs is increasing 70% YoY. Also, the CAL upstream applications log patterns are different. Some are database operation intensive and some contain complex nested transactions, and the log volume of each application varies widely.
• Computing Resource—CAL uses a shared Hadoop Cluster that can only use 19% capacity to run jobs during the hours of 23:00-8:00 GMT-7. Additionally, jobs running more than 1 hour are killed. A current job takes ~50% resources of all clusters to complete.
• Success Rate—The success rate is 92.5% for current MapReduce jobs. Also, some jobs take up to 6 hours to complete.

### MapReduce jobs

Before we go into the solution, let’s briefly visit the Hadoop MapReduce phases.

There are five phases in the CAL Hadoop MR job: CombineFileInputFormat, Mapper, Combiner, Partitioner, and Reducer.

Figure 1: Hadoop MapReduce Job Phases

In CombineFileInputFormat, small files are combined into a split, and the split is assigned to a mapper to be processed. The split is a logical representation of the data stored in file blocks on Hadoop.

Mapper transforms input records into intermediate records, which are then processed later in the reducer.

The intermediate records are transferred from the mapper node to the reducer node through the network. If the intermediate record set is too large, the effort would be expensive. The combiner is a semi-reducer in MR that preprocesses intermediate results and delivers them to the reducer.

The partition assigns intermediate results to different reducers, according to the key of the intermediate results. Bad practices can result in data skew in the reducer.

Reducer reduces a set of intermediate values that share a key to values you would expect.

The duration of a Hadoop job depends on the slowest map task and the slowest reduce task.

Given:

• $T_{Map}$ stand for Map task duration
• $Count_{Map}$ stand for Map task number
• $T_{Reduce}$ stand for Reduce task duration
• $Count_{Reduce}$ stand for Map task number

Then $T$, which is the MR job duration, could be expressed as:

$$T = \max_{Count_{Map}}T_{Map} + \max_{Count_{Reduce}}T_{Reduce}$$

Map time $T_{Map}$ or reduce time $T_{Map}$ is proportional to the input record number $N_{input}$ and its output record number $N_{output}$.

$$T_{Map} = { \theta * (\alpha_{0} * N_{input}+ (\alpha_{1} * N_{output}) + T_{GC}}$$

$$T_{Reduce} = { \theta * (\alpha_{0} * N_{input}+ (\alpha_{1} * N_{output}) + T_{GC}}$$

Also, the computation complexity $\theta$ has an impact on the Hadoop job duration. For some abnormal jobs, garbage collection (GC) time $T_{GC}$ should be take into consideration.

In order to control the Hadoop job duration, we need to work on three aspects:

1. GC, which can reduce $T_{GC}$
2. Data skew, aimed at reducing max $T_{Map}$ and max $T_{Reduce}$
3. Computation, which is about reducing $\theta$ in the expression

#### Resource usage

Given:

• $Size_{Map}$ stands for the Mapper container memory size in an MR job
• $Size_{Reduce}$ is the container memory size for reducer
• $Size_{AM}$ stands for the application master container memory size in an MR job
• $Count_{MAP}$ stands for the number of mappers in each MR job
• $Count_{Reduce}$ is for the reducer

Then

• Total memory resource usage as R should be related to container size and container holding time, which can be expressed as:

$$R = \sum_{Count_{Map}}Size_{Map}* T_{Map} + \sum_{Count_{Reduce}}Size_{Reduce}* T_{Reduce} + T_{job} * Size_{AM}$$

Therefore, to reduce resource usage, we need to do the following:

1. Reduce Map or Reduce task number: $Count_{Map}$
2. Decrease Map or Reduce task container size: $Size_{Map}$, $Size_{Reduce}$
3. Optimize time cost: $T_{Map}$, $T_{Reduce}$, $T_{job}$

### Solution

#### Garbage collection problem

Garbage collection is the standout problem in CAL. The reason for job failure usually is “GC overhead” or “Out of Memory.” Even for successful jobs, GC times are high. At the same time, jobs are using the CPU of the Hadoop node during GC.

##### Garbage collection in Mapper

In CAL, the logging event is represented as a CAL record. A CAL transaction is a logical concept that is composed of multiple records.

A transaction logs events related to a user request. Each transaction might involve other transactions in response to its user request. That is, transactions are organized in a tree-like hierarchical structure. Each transaction only knows its parent transaction, but the child transaction wants to know its root transaction. However, a transaction is not always processed before its children. Hence, all transaction information needs to be kept for traceability. And for any transaction, we have no idea when its children's transactions have been totally received. We need keep all those transactions in order to give the upcoming transaction information about its parent transaction or its children's transactions.

To summarize, Figure 2 shows a typical tree-like hierarchy of transactions. Transaction F is the root transaction. It will involve transaction B and G to handle a user request. If we already have transaction F, then transaction B, and finally transaction C, we can’t help C find its root transaction F until the transaction D comes in. Transaction B should always be kept in memory so that its child transaction A can be reached, because B has no idea how many children it has, same as other transactions.

Figure 2: Transaction Hierarchy

Unfortunately, this situation will cause out of memory (OOM). A transaction should have a time window property. If the time window is t, and the start timestamp of transaction is ts, all child transactions should happen before ts + t.

In my experiment, the time window was set at 5 min. We verified this assumption with logging data from 12 applications with the largest log volume. By discarding transactions that are out of this time window, 10 of 12 applications could promise almost 100% accuracy. At the same time, we set a whitelist for the remaining two applications, disabling this function for them.

The time window reduces the Hadoop job duration effectively and increases success rate from 93% to 97%.

Figure 3: Transaction and Metrics

Previously, mapper would read all records for that transaction in memory, then extract useful metrics once we got entire transaction, as shown in Figure 3. However, we can get metrics while reading records; whole transaction information is not required.

The combiner could reduce data transfers between the mapper and reducer tasks. But, when the output of mapper is even larger, GC time would be high because of sorting. So, we pre-aggregate in mapper to decrease output intermediate record numbers.

##### Garbage collection in reducer

Reducer was facing a GC issue similar to the Mapper's issue.

In MR, metrics are extracted from raw logs with its timestamp. Time granularity for a CAL report is 15 minutes, so the timestamp could round up to 4 timestamp values in an hour. The reducer of a CAL report job outputs two kinds of files—a base file that is used to store metrics in a 15-minute time granularity and an aggregate file that is used to store metrics in 1 hour time granularity.

In Reducer, the input record is sorted according to its key. The previous key format is on the left in Figure 4. Aggregation keeps records until records for last timestamp comes in. We changed the key of input of reducer to the format on the right in Figure 4. Memory usage in reducer was reduced effectively.

Optimization shouldn’t change the output of a job, even data order. In the base file, a record is sorted by timestamp. So we write records for 15-minutes granularity into different temporary files according to its timestamp, and merge them together when reducer has completed.

Figure 4: Aggregation in Reducer

This solution solves failure in Reducer and makes Reducer scalable for increasing input.

#### Data skew

Another obvious problem is data skew. While checking the map task and reduce task in a Hadoop Job, we learned that a map task executed time differences from 3s to more than 1 hour. The reduce task experiences a similar problem.

According to the formula, assigning even input records to mapper or reducer, max $T_{Map}$and max $T_{Reduce}$can be minimized.

For data skew in Mapper, CombineFileInputFormat can help.

In previous CAL MR jobs, CombineFileInputFormat was not enabled. But now, combining log files from same application and setting the max input file size of each mapper to 256MB in CombineFileInputFormat of MR job decreases the mapper task number to around half of what was before.

For data skew in Reducer, Partition takes over. In a CAL report, there are two concepts, report name and metrics name. For each kind of report, there are multiple metrics. Previously, the partition strategy used a hash of the report name. Now, it’s changed to a hash of the report name and metrics name. Data skew is much better.

#### Repetitive computation

In the expression in the Hadoop Job duration section, the job duration should be proportional to the input record number. In my experiment, there are two data sets. Data set A is 20MB and data set B is 100MB. An MR job with input set A takes 90 minutes to complete. A job with B as input takes only 8 minutes. To figure out what is going on, we need to look into the log data.

In a CAL log, there are two types of logs: SQL and general logs. The general log is an event log. The SQL log is related to DB operations like SQL statements. The general log might reference the SQL log. Parsing the SQL log is more time consuming.

Accordingly, we counted the SQL logs in data sets A and B. They are almost equal. Counting the general log that referenced SQL log provides some idea; the number for B is much larger than A.

The implementation suffers some repetitive computation on the reference SQL log part. Each SQL would be parsed every time, if referenced. To solve this issue, we cached the parsed result of the SQL log and reused it once referenced. The job for A can now complete in 4 minutes.

#### Resource usage

After the optimization of memory for mapper, reducer, the data skew problem, and the repetitive computation issue, we also need to also look at resource usage.

The action item to reduce resource usage includes:

1. Combining small files to reduce mapper number using CombineFileInputFormat.
2. Reducing memory usage.
3. Optimizing time cost.

Item 1 was addressed in the Data skew section. The solution for item 2 was addressed in the GC problem section. Item 3 was addressed in the repetitive computation chapter. With all these items addressed, the usage of Hadoop cluster decreased from about 50% to less than 19%.

### Job verification and monitoring

Optimization work should always care about data quality. The result for a user shouldn't be changed. Before we do this work, the verification plan should come first. We can use a metric with the idempotent property for evaluation.

Another important issue is monitoring—bringing the KPI we care about, the success rate, resource usage, and job duration into a dashboard to observe trends during optimization.

### Optimization results

Before optimization, a CAL report job required 50% resources in the Hadoop cluster to complete. Now, it can be completed with 19% capacity. Figure 5 shows the memory resource usage on Hadoop Cluster. The left side shows the situation before optimization, the right is the current situation.

Figure 5: Queue Resource Usage in Hadoop Eagle

We can save more than 50% relative computing resources, which equates to nearly ~200 Hadoop nodes in the Hadoop Cluster.

Figure 6: Compute Usage in Hadoop Eagle

The success rate has increased from 92.5% to about 99.9%. A failure rate of 00.1% is due to jobs running more than 1 hour. These jobs are killed because they need more complex computing.

95% of the Hadoop jobs now have a running time of about 40 minutes, down from 90 minutes before optimization.

Figure 7: CAL Report MR Job Running Time Trend

Because the CAL log volume increases 70% YoY, the volume will continue to increase 70% each year. We need to use CombineFileInputFormat and Partition to make MR scale.

### Conclusion

Data volumes increase with each passing day. Hadoop MR is a straightforward solution for big data analysis. Coupled with increasing computing resources to handle data volume increases, optimization can be lead to major cost savings. These optimizations for MR jobs have helped save more than 50% resources than before and have increased success rates to 99.9%.


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## Stop writing Regular Expressions. Express them with Verbal Expressions.

From the post:

GitHub user jehna has fashioned a runaway hit with his unique way of constructing difficult regular expressions.

VerbalExpressions turns the often-obscure-and-tricky-to-type regular expression operators into descriptive, chainable functions. The result of this is quite astounding. Here’s the example URL tester from the README:

// Create an example of how to test for correctly formed URLs
var tester = VerEx()
.startOfLine()
.then( "http" )
.maybe( "s" )
.then( "://" )
.maybe( "www." )
.anythingBut( " " )
.endOfLine();

You can think of other regex languages that are more concise, but can you think of one as easy to teach to new users?

I think there is a lesson here for hand editing of topic maps.

Such as discovering what commons terms particular people would use for topic map constructs.

Those terms could become their topic map authoring language, with a translation script that casts their expression into some briefer notation.

Yes?

### 3 Responses to “Stop writing Regular Expressions. Express them with Verbal Expressions.”

1. clemp says:

Sounds like a great idea but wouldn’t casting something into a briefer language mean the user is still working at the lowest semantic level just using a different vocabulary: like assembly code instead of machine language? Maybe the Domain Specific Language could be at a higher level of abstraction so that is could be expanded into multiple constructs that represent higher level structures: like the Topic map equivalent of C, Python, or SQL.

However, the first step would be to create a catalog of common topic map structures or patterns. It seems like such a catalog could eventually enable automated or computer assisted construction of topic maps to supplement hand-editing of topic maps. Hand-editing is a necessary first step but it does not scale well. Imagine how few applications there would be now if everything had to be coded in assembler. Or how few databases would there be now if everyone had to build their own out of B-Trees.

2. Patrick Durusau says:

I was thinking more along the lines of higher level languages, DSL for sure.

Think of it as being like auto-completion. When a lawyer says, “…the party of the first part,” an association is created with a topic for the party of the first part and as other players are named, they are added to the association. Roles could be composed of their obligations in the contract (we say seller/buyer as a shorthand for those). All the parts of a number of transactions, mortgages for example, are fairly well known. And there are a lot of them.

That’s not the unconstrained authoring case but I think the unconstrained case is vastly overrated. There is a lot of mileage in fairly routine and constrained cases.

+1! on hand editing. I think we need assistive interfaces suggest solutions and/or provide information to allow you to do the hard cases.

3. clemp says:

That sounds like a great idea. I agree about unconstrained authoring. The flexibility is great for experts but it usually comes at the cost of making it harder to do the routine things that I’d prefer to let the computer do for me.


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# Prove that the lim (x->1) x^2 + 2 is NOT equal to 2.999?

1. Oct 19, 2004

### bezgin

How can we prove that the lim (x->1) x^2 + 2 is NOT equal to 2.999? (example I made up right now) At the end of each proof we find a relation between epsilon and delta. What does it mean?

2. Oct 19, 2004

### MiGUi

You cant take back the 2 as a constant, and the limit of x² will be 1

3. Oct 19, 2004

### Galileo

Since
$$\lim_{x\rightarrow 1}x^2+2=2.999 \iff \lim_{x\rightarrow 1}x^2=0.999$$
we can work with the righthanded expression. (it's easier).
We'll have to find an $\epsilon>0$ such that for any $\delta>0$ we have $|x-1|<\delta$ AND $|x^2-0.999|>\epsilon$

EDIT: Removed 'proof', because of errors.

Last edited: Oct 19, 2004
4. Oct 19, 2004

### jatin9_99

my friend it's real simple; as the 2 won't be effected by the limit, so only part affected by limit is x^2 when x->1

although u are write it might go to 2.999 when it approach from left, but u need to consider another fact when limit approach from right than it will be 3.00001 that's how we prove that because u don't specify as limit apporaches from right or left

5. Oct 20, 2004

### spacetime

Proving that $$\lim_{x\rightarrow a} f(x) = b$$ means that it should be possible, by choosing a $$\delta$$ , to make the make the difference in values of f(x) and b smaller than any positive number $$\epsilon$$, for all the values of $$|x-a|$$ less than $$\delta$$.

For 3, you can show this to be true. But for 2.999, although you can choose a number for which the difference is smaller than any positive number, but this will not be true for all the $$|x-a|$$ smaller than that chosen number.

That is the reason why 2.999 isn't the limit of the function.

spacetime
www.geocities.com/physics_all/index.html

6. Oct 21, 2004

### bezgin

These replies make think, once again, what does it mean to be "close"?
Can't we find a relation between epsilon and delta by doing the operations abs(x^2 + 2 - 2.999) < epsilon and abs(x-1) < delta?
I'm really stuck with this issue for two weeks and think I will not understand the concept of the formal definition of a limit. I've tried nearly anything on the web :(

7. Oct 21, 2004

### arildno

Sure, here's how:
$$|x^{2}+2-2.9991|=|(x-1)(x+1)+r|,r=0.0009$$
1) Assume x>1:
Then r+(x-1)(x+1)>r>0, so:
$$|(x-1)(x+1)+r|>r$$.
Choose in this case, $$\epsilon<r$$
2)Assume x<1.
Then, you can find $$\delta$$, so that $$(1-x)(x+1)<\frac{r}{2}$$
Hence, by the triangle inequality, we have:
$$|x^{2}+2-2.9991|=|(x-1)(x+1)+r|\geq(r-(1-x)(x+1))\geq(r-\frac{r}{2})=\frac{r}{2}$$

Hence, the choice $$\epsilon<\frac{r}{2}$$ is always out of reach, so 2.9991 cannot be the limit.

Last edited: Oct 21, 2004


text:
If i define a function

$$F(t)=f􏰀(x(t),y(t))$$

with

$$x(t)=x_0+∆x * t + ∆^2x * t^2$$ $$y(t)=y_0+∆y * t + ∆^2y * t^2$$

$∆$ is the slope $dx/dt$ and $∆^2$ is the 2nd derivative $d^2x/dt^2$ of x (or y)

I know that for just F(t) = f(x(t)) the expansion around ($t$,$t_k$) looks like $$F(t) = f(x_0) + f_x\ {∆x}\ (t-t_k) + [f_x * ∆^2x + \frac{1}{2} f_{xx}(∆x)^2](t-t_k)^2$$

How does the taylor expansion up to order 2 look like?

If you expand around $(x_0,y_0)$ (I.e. expand around $t=0$) then you would have $F(t)=f(x_0,y_0)+f_x(x_0,y_0)(\Delta xt+\Delta^2xt^2)+f_{xy}(x_0,y_0)(\Delta xt+\Delta^2xt^2)(\Delta yt+\Delta^2yt^2)$ (plus other terms in $y$ which I will leave to you, check out multi variable Taylor theorem on wikipedia.


text:
# Math Help - inverse trigonometric functions

1. ## inverse trigonometric functions

Help needed on these two problems -

1. Prove that

tan^(-1) x = sin^(-1)( x / sqrt(1 + x^2))

2. Find the exact value of -
$\cos\left( \tan^{-1}(\frac {5}{12}) + \cot^{-1}(\frac {5}{12})\right)
$

2. Originally Posted by darkangel
Help needed on these two problems -

1. Prove that

tan^(-1) x = sin^(-1)( x / sqrt(1 + x^2))

2. Find the exact value of -
$\cos\left( \tan^{-1}(\frac {5}{12}) + \cot^{-1}(\frac {5}{12})\right)$
for the second one rewrite it as $cos\bigg(arctan\bigg(\frac{5}{12}\bigg)+arctan\big g(\frac{12}{5}\bigg)\bigg)$..then split it apart using $cos(x+y)=cos(x)\cdot{cos(y)}+sin(x)\cdot{sin(y)}$ and use the fact that $cos(arctan(x))=\frac{1}{\sqrt{1+x^2}}$ to finish

3. The second one should be easy.

$tan^{-1}(x)+cot^{-1}(x)=\frac{\pi}{2}$

4. 1. Set $y = \sin^{-1} \frac{x}{\sqrt{1+x^{2}}}$. This implies:

$\sin y = \frac{x}{\sqrt{1+x^{2}}}$

Using the identity: $\sin^{2} x + \cos^{2} x = 1 \Rightarrow \cos x = \sqrt{1 - \sin^{2} x}$

So, find an expression of $\cos y$.

Then, find tan y using the relation: $\tan y = \frac{\sin y}{\cos y}$ and solve for y.

5. aite Thanks for everyone's help

6. Originally Posted by galactus
$tan^{-1}(x)+cot^{-1}(x)=\frac{\pi}{2}$
Just a quick note here.

Observe that $\left( {\tan ^{ - 1} x} \right)' + \left( {\cot ^{ - 1} x} \right)' = 0,$ so $\tan ^{ - 1} x - \frac{\pi }
{4} + \cot ^{ - 1} x - \frac{\pi }
{4} = 0,$
and the result follows.

7. woah ! thanks Krizalid !!