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Next: Potential due to uniform Up: Newtonian gravity Previous: Potential due to uniform

Potential outside uniform spheroid

Let us now calculate the gravitational potential generated outside a spheroid of uniform mass density and mean radius . A spheroid is the solid body produced by rotating an ellipse about a major or a minor axis. Let the axis of rotation coincide with the -axis, and let the outer boundary of the spheroid satisfy

 (3.56)

where is termed the ellipticity. In fact, the radius of the spheroid at the poles (i.e., along the rotation axis) is , whereas the radius at the equator (i.e., in the bisecting plane perpendicular to the axis) is . Hence,

 (3.57)

Let us assume that , so that the spheroid is very close to being a sphere. If then the spheroid is slightly squashed along its symmetry axis and is termed oblate. Likewise, if then the spheroid is slightly elongated along its axis and is termed prolate. (See Figure 3.1.) Of course, if then the spheroid reduces to a sphere.

Now, according to Equation (3.45) and (3.46), the gravitational potential generated outside an axially symmetric mass distribution can be written

 (3.58)

where is the total mass of the distribution, and

 (3.59)

Here, the integral is taken over the whole cross section of the distribution in - space.

It follows that for a uniform spheroid, for which ,

 (3.60)

Hence,

 (3.61)

giving

 (3.62)

to first order in . It is thus clear, from Equation (3.42), that, to first order in , the only nonzero are

 (3.63) and (3.64)

Thus, the gravitational potential outside a uniform spheroid of total mass , mean radius , and ellipticity is

 (3.65)

By analogy with the preceding analysis, the gravitational potential outside a general (i.e., axisymmetric, but not necessarily uniform) spheroidal mass distribution of mass , mean radius , and ellipticity (where ) can be written

 (3.66)

where . In particular, the gravitational potential on the surface of the spheroid is

 (3.67)

which yields

 (3.68)

where use has been made of Equation (3.56). For the case of a uniform-density spheroid, for which [see Equation (3.64)], the preceding expression simplifies to

 (3.69)

Consider a self-gravitating spheroid of mass , mean radius , and ellipticity , such as a star or a planet. Assuming, for the sake of simplicity, that the spheroid is composed of uniform-density incompressible fluid, it follows that the gravitational potential on its surface is given by Equation (3.69). However, the condition for an equilibrium state is that the potential be constant over the surface. If this is not the case then there will be gravitational forces acting tangential to the surface. Such forces cannot be balanced by internal fluid pressure, which only acts normal to the surface. Hence, from Equation (3.69), it is clear that the condition for equilibrium is . In other words, the equilibrium configuration of a uniform-density, self-gravitating, fluid, mass distribution is a sphere. Deviations from this configuration can only be caused by forces in addition to self-gravity and internal fluid pressure; for instance, internal tensile forces, centrifugal forces due to rotation, and tidal forces due to orbiting masses. (See Chapter 6.) The same is true for a self-gravitating mass distribution of non-uniform density.

We can estimate how small a rocky asteroid, say, needs to be before its material strength is sufficient to allow it to retain a significantly non-spherical shape. The typical density of rocky asteroids in the solar system is . Moreover, the critical pressure above which the rock out of which such asteroids are composed ceases to act like a rigid material, and instead deforms and flows like a liquid, is (de Pater and Lissauer 2010). We must compare this critical pressure with the pressure at the center of the asteroid. Assuming, for the sake of simplicity, that the asteroid is roughly spherical, of radius , and of uniform density , the central pressure is

 (3.70)

where is the gravitational acceleration at radius . [See Equation (3.53).] This result is a simple generalization of the well-known formula for the pressure a depth below the surface of a fluid. It follows that

 (3.71)

If then the internal pressure in the asteroid is not sufficiently high to cause its constituent rock to deform like a liquid. Such an asteroid can therefore retain a significantly non-spherical shape. On the other hand, if then the internal pressure is large enough to render the asteroid fluidlike. Such an asteroid cannot withstand the tendency of self-gravity to make it adopt a spherical shape. The same applies to any rocky body in the solar system. The condition is equivalent to , where

 (3.72)

It follows that only a rocky body whose radius is significantly less than about --for instance, the two moons of Mars, Phobos (see Figure 3.2) and Deimos--can retain a highly non-spherical shape. On the other hand, a rocky body whose radius is significantly greater than about --for instance, the asteroid Ceres (see Figure 3.3) and the Earth's moon--is forced by gravity to be essentially spherical.

According to Equations (3.58) and (3.63) , the gravitational potential outside a general axisymmetric body of mass and mean radius can be written

 (3.73)

where the are (or smaller) dimensionless parameters that depend on the exact shape of the body. However, a long way from the body (i.e., ), the right-hand side of the preceding expression is clearly dominated by the first term inside the round brackets, so that

 (3.74)

However, this is just the gravitational potential of a point particle, located at the center of the body, whose mass is equal to that of the body. This suggests that the gravitational interaction between two irregularly shaped bodies in the solar system can be approximated as the interaction of two point masses, provided that the distance between the bodies is much larger than the sum of their radii.

Next: Potential due to uniform Up: Newtonian gravity Previous: Potential due to uniform
Richard Fitzpatrick 2016-03-31


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# If the zeroes of the polynomial x3 - 3x2 + x + 1 are a - b, a and a + b respectively, then the values of 'a' and 'b' are respectively.

This question was previously asked in
DSSSB TGT Maths Female Subject Concerned - 18 Nov 2018 Shift 3
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1. $$1, ±\sqrt{2}$$
2. $$3, -\sqrt2$$
3. $$-1, \sqrt2$$
4. $$1, ± \sqrt{\dfrac{3}{2}}$$

Option 1 : $$1, ±\sqrt{2}$$

## Detailed Solution

Given:

Zeroes of the polynomial = (a - b), a and (a + b)

Formula used:

Sum of zeroes = –B/A

Product of zeroes = -D/A

Where,

A = Coefficient of x3

B = Coefficient of x2

C = Coefficient of x

Calculations:

Using the given polynomial

Sum of zeroes = (a - b) + a + (a + b) = 3

⇒ 3a = 3

⇒ a = 1

Product of zeroes = (a – b) × a × (a + b) = -1

⇒ (1 - b) × 1 × (1 + b) = -1

⇒ 1 - b2 = -1

⇒ b2 = 2

⇒ b = ±√2

∴ The values of 'a' and 'b' are 1 and ±√2

format: plain


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# least square problem

Let $1<p<\infty$.We define the space: $L_{V}^{p}(-1,1)=\left \{ f:(-1,1)\rightarrow \mathbb{R}:\int_{-1}^{1}\left | f(x) \right |^{p}V(x)dx<\infty \right \}$

We define the norm: $\left \| f \right \|_{L_{V}^{p}}=(\int_{-1}^{1}\left | f(x) \right |^{p}V(x)dx)^{\frac{1}{p}}$

Consider $W\subset L_{V}^{p}(-1,1)$ to be a finite dimensional subspace. For a given $f$ in $L_{V}^{p}(-1,1)$, we define the minimizer $m$ in $W$ such that: $\left \| f(x)-m(x) \right \|_{L_{V}^{p}}= min\left \| f(x)-q(x) \right \|_{L_{V}^{p}}$ for all $q$ in $W$.

I need to show that $m$ satisfies the following:

$\int_{-1}^{1}\left | f(x)-m(x) \right |^{p-2}(f(x)-m(x))q(x)V(x)dx=0$ for all $q\in W$

I have no idea how to prove the above identity based on the information given in the problem. Any ideas?

-

Hint: Since $m$ is the minimizer of $\|f - m\|_p$, for any $q \in W$, $$\frac{d}{dt} \|f - m - tq\|_p^p = 0$$ (why?) Then try and evaluate this expression.

-
Can you, please, explain why is the fact that $m$ is a minimizer implies $\frac{d}{dt}\left \| f-m-tq \right \|_{p}^{p}=0$ ? –  M.Krov Dec 2 '12 at 3:19
For any $t \in \mathbb{R}$, $m + tq$ is in $W$, so you know $||f - m||_p \leq ||f - m - tq||_p$. Some tools from calculus (of a single real variable) can help here. –  Pot Dec 2 '12 at 12:28


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# Geometry/Algebra Question

My text states the following areas of two similar triangles: $$A_1 = \frac{1}{2}h_1b$$

$$A_2=\frac{1}{2}h_2e$$

$$\frac{A_1}{A_2}=\frac{\frac{1}{2}h_1b}{\frac{1}{2}h_2e}=\left(\frac{h_1}{h_2}\right) \left(\frac{b}{e}\right)=\left(\frac{b}{e}\right) \left(\frac{b}{e}\right)=\left(\frac{b^2}{e^2}\right)$$

I'm not following how they're getting:

$$\left(\frac{h_1}{h_2}\right) \left(\frac{b}{e}\right)=\left(\frac{b}{e}\right) \left(\frac{b}{e}\right)$$

I fear it's something painfully obvious.

$$\frac{h_1}{h_2}=\frac{b}{e}$$


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# Alkali metal acts as strong reducing agents explain.

Verified
145.8k+ views
Hint: Reducing agent is an element or compound which loses an electron to an electron recipient in a redox chemical reaction. A reducing agent thus gets oxidized. The reducing agents "reduce" the oxidizing agents. For an element to be a strong reducing agent it should be able to get oxidised easily which means it should lose an electron very easily.

Alkali metals belong to group 1 of the periodic table.
Reducing agents are those compounds or elements which can reduce others and get oxidized themselves.
Alkali Metals act as strong reducing agents, this is because alkali metals have only one valence electron in their valence shell which they can easily lose and attain nearest noble gas configuration and become more stable than they were earlier. Thus they lose an electron and oxidize themselves while reducing others, hence they are called reducing agents.
This makes them a strong reducing agent.Their general outermost electronic configuration is $n{{s}^{1}}$. This means that they have only one electron in the outermost shell.So they can very easily lose this electron to other electron-accepting species.In this way, they themselves get oxidised then and hence act as a reducing agent. Consider the reaction:
$2Na+C{{l}_{2}}\to 2NaCl$
Here, the $Na$ metal is getting oxidised and$Cl$ is getting reduced. $Na$ metal loses one electron and forms$N{{a}^{+}}$ and $Cl$accepts one electron forming .$C{{l}^{-}}$ Thus, here $Na$ metal itself gets oxidised and $Cl$reduces the atom. Hence, $Na$ metal acts as a strong reducing agent.

Note: Group 1 metals in the Modern / long-form periodic tables are called Alkali Metals.
All the alkali metals are mentioned below:
Lithium $\left[ Li \right]$
Sodium $\left[ Na \right]$
Potassium $\left[ k \right]$
Rubidium $\left[ Rb \right]$
Caesium $\left[ Cs \right]$
Francium $\left[ Fr \right]$


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ah many thanks ! I understand better now ( change quite  a bit since 2003 :-)
I indeed can see the classes .

are these new renderers already mapped into wx, or should we use them as such , directly  ?
any sample /  test available , out of chance ?

Many thanks for your patience, anyway :-)

Luc

--------------
Luc
Envoyé avec Sparrow

Le lundi 25 mars 2013 à 17:07, Henk-Jan van Tuyl a écrit :

On Mon, 25 Mar 2013 10:16:00 +0100, Luc TAESCH <luc.taesch@gmail.com>
wrote:

… but just for 10 mn as I cannot find these on the current 0.90.3. ;-< ??

I see you are pointing to an old lib, 0.12.7. Currently I cannot find
the Graphics-UI-WXCore-WxcClassesAL
file, nor any containing GraphicsRenderer

I guess there have been some refactoring since 0.12.7.

could you help pointing at the files ?

or am i getting the wrong source ?
I got the source from

The functions you are looking for, are defined in file
. This file is automatically created when you run
cabal install
or


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# Congruences involving binomial coefficients and Lucas sequences - Mathematics > Number Theory

Congruences involving binomial coefficients and Lucas sequences - Mathematics > Number Theory - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Abstract: In this paper we obtain some congruences involving central binomialcoefficients and Lucas sequences. For example, we show that if p>5 is a primethen $\sum {k=0}^{p-1}F k*binom2k,k-12^k$ is congruent to 0,1,-1 modulo paccording as p=1,4 mod 5, p=13,17 mod 30, and p=7,23 mod 30 respectively,where {F n} is the Fibonacci sequence. We also raise several conjectures.

Autor: Zhi-Wei Sun

Fuente: https://arxiv.org/


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# Proof of white noise process

I have the process $$y_t=e_t+ae_{t-1}e_{t-2}$$ where $$e_t$$ is iid with mean of $$0$$ and variance $$\sigma^2$$

How do I go about mathematically proving that this is a white noise process?

• I'm not sure how to interpret the expression. Is it $y_t=e_t+ae_{t-1}e_{t-2}$ where $e_t \stackrel{iid}{\sim} N(0,\sigma^2)$? Perhaps you mean the MA(2) process $y_t=e_t+q_1e_{t-1}+q_2e_{t-2}$ ? – Peter Leopold Sep 16 '19 at 5:00

Clearly, the process is zero-mean. Let's check its autocovariance $$k\geq 1$$: \begin{align}E[y_ty_{t-k}]&=E[(e_t+ae_{t-1}e_{t-2})(e_{t-k}+ae_{t-1-k}e_{t-2-k})]\\&=E[e_te_{t-k}]+aE[e_te_{t-1-k}e_{t-2-k}]+aE[e_{t-k}e_{t-1}e_{t-2}]+a^2E[e_{t-1}e_{t-2}e_{t-1-k}e_{t-2-k}]\end{align} Here, $$E[e_te_{t-k}]=E[e_t]E[e_{t-k}]=0$$, similarly for $$E[e_te_{t-1-k}e_{t-2-k}]$$. And, for $$E[e_{t-k}e_{t-1}e_{t-2}]$$, even if $$k=1$$ or $$2$$, one of the terms will leave the expectation and make the expression $$0$$. For the last term, the expectation is non-zero only if $$k=0$$. So, we have non-zero autocorrelation, $$r_y(k)$$, only if $$k=0$$, and the process is zero-mean. Then, this is a white process. Note that you can find the $$r_y(0)$$ as $$1+a^2$$ from the above equation.


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CW complex for infinite holed torus? (Surface of infinite genus)

1. Mar 19, 2013

dumbQuestion

I am just trying to figure out how to make a CW complex for this. For the n-genus orientable manifold (connect sum of n-tori) I feel like a lot of things make sense, fundamental group, CW complex, etc. But in the infinite case, things seem to fall apart. For example, I can not figure out how the fundamental group is a free group. I was hoping to figure this out by first looking at the CW complex of this surface, but I'm not sure I can picture it.

IN a finite case, I just have a single 0 cell (1 vertex), 2n 1-cells, and a single 2 cell. BUt does this hold at the infinite case? If not what's an alternate way to visualize it?


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# STCONN in $O(n)$ time?

This is a very basic question on $$s$$-$$t$$-connectivity in directed graphs. As a baseline, using DFS (or BFS), one can solve the problem on a graph $$G=(V,E)$$ in $$O(n+m)$$ time and $$O(n)$$ space, where $$n=|V|$$ and $$m=|E|$$.

However, I have seen mentioned in several places that DFS/BFS let you solve the problem in $$O(n)$$ time and space. For instance:

Two basic results on the complexity of $$\mathsf{STCONN}$$ exist for over 20 years. The first follows from the discovery of the very efficient algorithms for graph traversal, BFS (Breadth-First Search) and DFS (Depth-First Search), which use only linear time and space in $$n$$, the number of vertices.

Theorem 4. $$\mathsf{STCONN}\in\mathrm{TISP}(n,n)$$.

from [1]. How does that work? (Sorry if this question is trivial...) I would have thought maybe this followed from taking $$n$$ as the input size (not the number of vertices), but the text before does contradict this interpretation.

[1] Avi Wigderson. The Complexity of Graph Connectivity. MFCS 1992: 112-132

• @RyanWilliams $s$-$t$ graph connectivity requires an explicit labeling of $s$ and $t$, so strictly speaking it is not a graph property and the standard set of graph evasiveness results need not (directly) apply. As a trivial example, consider the still-monotone modification where we also require the $s$-$t$ path length to equal $1$. That said, the usual adversarial approach, in which the edge existence oracle keeps answering "no" to all queries until doing so would induce an $s$-$t$ cut, can still be used to derive an $\Omega(n^2)$ lower bound in the edge query model. – Yonatan N Mar 22 at 18:17
• When given as a standard-form adjacency matrix with $s$ at the first index and $t$ in the last, the problem requires us to read $\Omega(n^2)$ entry queries even under the additional promise that the top left and bottom right $(n/2)$-by-$(n/2)$ blocks are all $1$: to prove that $s$ and $t$ are disconnected, one would need to ascertain that the other $\Omega(n^2)$ entries of the matrix are all $0$. – Yonatan N Mar 22 at 22:59
Per the section quoted in the comments, it appears that one of the cited documents takes input in the form of an adjacency matrix. In this scenario, taking $$s$$ and $$t$$ to correspond to the first and last indices respectively, we can prove that $$\Omega(n^2)$$ bits of the matrix may need to be read to decide STCONN even under the promise that the $$(n/2)\times(n/2)$$ square blocks in the top-left and bottom-right are all ones (with the necessary adjustments for odd $$n$$).
Indeed, we have some $$\Omega(n^2)$$ entries unaccounted for in the matrix, and there is an $$s$$-$$t$$ path iff any one of them is a $$1$$. So in the worst case, we might need to read all of these $$\Omega(n^2)$$ bits to prove $$s$$-$$t$$-non-connectivity (or half of them if the graph is known to be undirected, which doesn't change the order of growth). We can turn this into a lower bound against randomized algorithms by considering the case that with 50% probability we either add or don't add in a $$1$$ uniformly at random: we must read some $$\Omega(\varepsilon n^2)$$ bits to gain a $$\varepsilon$$ advantage over a random coin flip.
Graphically, this corresponds to the case where $$s$$ and $$t$$ each lie within disjoint cliques of size $$n/2$$, and we wish to determine if there is at least one edge joining these cliques. One can also do the same thing with any other pair of connected structures of order $$\Omega(n)$$ each.


text:
# Find the cube root of 9, using the Newton's method.

Find the cube root of 9, using the Newton's method.
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text:
## Pseudoforces and Geometry (or) BigBang and BlackHoles

### February 24, 2006

In mechanics, the concept of a pseudo-force is a useful one. It allows you to work in frames which are non-inertial and apply there the usual laws of Newton. But, one thing about pseudoforces which is rarely discussed, is the fact that pseudoforces tell you something about the geometry of the space around you.

One has to wait till one learns general relativity to recognise this fact. And even then, it is quite likely that this fact is missed in the rain of tensor indices and four vector algebra :)

But the relation between pseudoforces and geometry is a simple one, independent of relativity or gravity. And this relation between pseudoforces and geometry is better discussed first in the familiar Newtonian framework.

Sidenote : Once you understand how pseudoforces affect geometry, the relation between geometry and gravity- the core idea of GR is self-evident. According to Einstein’s equivalence principle, gravitational forces are indistinguishable from pseudoforces and if pseudoforces affect geometry, gravitational forces should do so too !

Well, the beauty of the equivalence principle is that by identifying gravitational forces with pseudoforces it connects a lot of seemingly unrelated facts into a beautiful theory. All this factsabout pseudoforces, geometry and gravity fall into place in General relativity which combines all of them by identifying gravitational forces with pseudoforces and making matter affect geometry via gravitational/pseudo-forces

Now, coming back – how do we see that pseudoforces affect geometry, and yeah why haven’t we seen it before when we learnt mechanics ?

I’ll answer the second question first. We haven’t seen it before, because the usual examples chosen for non-inertial frames – accelerating/rotating frames have such a familiar geometry that we tend to overlook the effect of pseudoforces. The geometry of space is same as the usual High-school geometry which applies to inertial frames , so there is not much new to talk about here. Once we realise this, the question how do we see the effect of pseudoforces on geometry has a straightforward answer : Think about more general non-inertial frames !

Sidenote : By the way, if you take Einstein’s relativity into account, you would see that this is not true – you can prove that there is no non-inertial frame which has the same geometry as the highschool geometry. In particular, in accelerating and rotating frames we would see two additional geometric effects- gravitational redshift and Sagnac effect due to pseudoforces. That indeed is the reason why in relativity, you are forced to acknowledge the effect of pseudoforces on geometry, whether you like it or not.

Now, what do I mean by non-inertial frames more general than accelerating/rotating frames ? Consider, for example, what I call an “expanding” frame – a frame of observers who when seen from an inertial frame are expanding radially outwards from the origin.The frame is defined by the usual convention that according to these set of observers, they are all at rest with respect to each other. If you think about it a bit, you will see why this frame is different from the usual non-inertial frames .

In the case of accelerating/rotating frames, the corresponding observers are in a rigid-body like motion. In particular, this means that the distances between the observers does not change. But, for an expanding frame, the distance between the observers increases with time. It is very simple to convince oneself of this fact- just think of two diametrically opposite observers. The distance between them increases with time. And since distances are absolute in Newtonian framework, these observers will actually see the distance between them increasing, though they are at “rest” with respect to each other (by definition) !

I like this particular example very much, since it brings into a sharp focus our conception of what do we mean by being at rest . If you say that being at rest with respect to some frame means not moving, then these two observers are not moving in the “expanding” frame – they are at rest due to the very way the “expanding” frame has been defined. And the distance between the two observers can still increase, though neither of them is “moving” in this sense ! So, how do the expanding frame’s observers interpret this weird thing they observe ? There is only one way out – in that frame, the space itself is expanding ! Though they are at rest, the space is continuously “created” in between them resulting in the increase of distance.

But, what causes this “creation of space” in this non-inertial frame ? What is that entity which is responsible for the creation of space ? Well, that is easily answered. Remember this golden rule about pseudoforces- “Whatever weird is happening in non-inertial frame, can eventually be blamed on the pseudo forces” ! So, it is the pseudoforce fields that create space.

You can convince yourself by calculating the pseudoforces due to this expansion and by showing that the expanding observers can predict the rate of creation of space by measuring the pseudoforces in this frame. So here we have the first example of effect of pseudoforces on geometry- there are some pseudoforce fields which can create space and increase the distance between the two objects which are at rest. In fact, they can do the reverse thing too, they can “eat up” or “swallow” the space in between two objects at rest.

So in this sense, non-inertial frames are like the wonderlands of Lewis Caroll. If you’ve read “Alice in the Wonderland” you would remember this scene in which the red queen and Alice start running

The most curious part of the thing was, that the trees and the other things round them never changed their places at all : however fast they went, they never seemed to pass anything. …
Alice looked round her in great surprise. Why, I do believe we’ve been under this tree the whole time! Everything’s just as it was!’

Of course it is,’ said the Queen, what would you have it?’
Well, in our country,’ said Alice, still panting a little, you’d generally get to somewhere else — if you ran very fast for a long time, as we’ve been doing.’

A slow sort of country!’ said the Queen. `Now, here, you see, it takes all the running you can do, to keep in the same place.

If you want to get somewhere else, you must run at least twice as fast as that!’

Expanding-frame is a bit like that. You’ve to run in this frame to be at the same distance from something at rest in this frame !

And once you realise this, you are on your way to understand the philosophy behind GR- that space is not a quite thing sitting silently allowing all other things to move. It is not a dead-silent stage for the other actors. The stage called space itself is an actor in the drama. It can be created and eaten up, squished and stretched by the pseudoforces.

In fact, there are only two more steps from here to GR. First, is the realisation that inertial frames need not exist- a spacetime where there is no frame in which pseudoforces vanish everywhere is said to be a “curved” spacetime. So, this realisation can be rephrased as the fact that the spacetime can be “curved”. This terminology comes from the fact that for example, a two-dimensional being over a sphere cannot establish a frame over the sphere in which all the pseudoforces vanish. So, this is one way of defining what do you mean by a curved space-time – you are in a curved spacetime if you cannot establish a single inertial frame everywhere.

The universe we live in is such a space. There are pseudoforce fields that extend all over the universe creating new space – this is what we mean by the expansion of the universe. But unlike the example before, this expansion cannot be gotten rid of. The pseudoforce fields cannot be transformed away everywhere by a change of frame since there are no global inertial frames. And in fact, you can use this to tell about the past of our universe- in the past when all this space was not yet created by the pseudoforce fields, the universe was small and hot, and this is what the BigBang theory is all about.

And this should clear the confusing question “where did BigBang happen ?” – you see when bigbang happened there was only one place where it could happen – the rest of the space was simply not there !

And then there are people who think everything in the universe expands – you know from our measuring rods to earth-sun distance to distance between the galaxies. Then they get confused about how we can see the expansion if everything expands. Well, I just tell them – think of this expanding frame and imagine what those non-inertial guys would see. Equivalence principle assures you that what we see in an expanding universe would be very similar.(The only difference is that the pseudoforces cannot be wished away bya frame choice- the efffect of pseudoforces on geometry are very similar.). It is immediately obvious that the measuring rods these people carry would not expand. True there will be a tendency to expand- there will be stresses developed on the rods due to the pseudoforces for example ! But, they won’t expand appreciably unless the pseudoforces are really strong .

Second, is of course the equivalence principle- There is no way to distinguish gravitational forces from the pseudoforces in a general spacetime. So, they are just two different names of the samething. This in turn, means that matter can produce pseudoforce fields which can then eat up space. How cool !

This gives rise to really cool objects called Black Holes ! What happens here is that the matter produces a pseudoforce field which eats up space at a large rate. In fact, it eats up space so fast that invariably any two things inside a blackhole are brought near each other. The rate of eating of space is so high that you can’t even run away ! Remember – there is a limit to how fast you can run since you cannot run faster than light. There is no escape from the interior of a blackhole because eventually the pseudoforces will eat up the space between any two objects and everything will be crushed together. Like the expanding universe case, these pseudoforces which create blackholes cannot be transformed away or gotten rid of by changing frames.

So, in GR you have this picture of space which is active and alive , with birth and death ! The space is like a deep sea on which all ships stand. When the sea is calm and serene, the explorers aboard can delude themselves that they are on the land – That there is a ‘ground’ beneath their feet, that their life is lived over a rock-steady stage. Oh explorers! how far would you delude yourselves ? When the sea comes to her full glory, When her waters drench your clothes, When the platform on which you stand begins to shake, would you still mainatain that you are on land ? Would you still dismiss its dynamical nature ? That, in a way, is the challenge of gravity to anyone who seeks to understand her.

P.S. : Do have a look at this article titled Albert Einstein’s Theory of Relativity- In Words of Four Letters or Less . It talks about the same topic but in a different(and probably better) way.

### 11 Responses to “Pseudoforces and Geometry (or) BigBang and BlackHoles”

1. bharath Says:

give me some example where I can see a pseudo force. Suppose a body is moving faling from a mountain. I observe it from a frame of reference that itself is falling at a lower rate of acceleration. then what should I see? what is difference if it is not accelerating.

the article seems little too abstract. it would be nice to know the objects about which you are talking. at least for the non experts.

2. nayagam Says:

Hi Bharath,

give me some example where I can see a pseudo force.

Well, Pseudoforces are there everywhere ! Say, you are sitting in a bus/car and say the driver applies a brake causing the moving vehicle to stop suddenly – Imagine what would it feel like ? It would feel like something is pushing you – you would feel a force. But, if you just look around, there is NOTHING really there pushing you ! It is as if there is some invisible ghost which is producing this force – hence, the name “pseudo-force”(literally false-force or fake-force). If you will, it is the same “ghostly” force which you feel when you are sitting on a vehicle which takes a sharp turn.

It is an amazing fact about our world and it is a real weird thing, if you come to think of it ! (In fact, historically, Mach was among the first to remark on how weird this whole thing is. And,Einstein was motivated a lot by Mach’s ideas when he was putting together general relativity.)

3. bharath Says:

thats the D’Alembert principle! does inertial and non-inertial frames mean stationary and moving frames of reference? like a frame stuck to a billboard as opposed to one stuck to a moving car?

then what is the change in geometry due to this pseudo force? I can see that the world appear to behave (move) differently when we go from a stationary to a moving frame of reference. is that it?

when you mean an expanding frame, you attach a distance measure to the frame which changes as the frame expands, so that two points are always at a contant distance with respect to the frame. but for an observer on the bill-board (say) the points will be receeding from each other? if this understanding is right, then why is it natural (any intuitive reasons?) to choose a distance measure depending on the frame of reference.

I think I am slowly getting to the hard part of the blog now. thanks for keeping up the patience.

4. nayagam Says:

Suppose a body is moving faling from a mountain. I observe it from a frame of reference that itself is falling at a lower rate of acceleration. then what should I see? what is difference if it is not accelerating.

Ha ! Great! This question in fact goes into the heart of Equivalence principle and the crucial difference between how Newton and Einstein look at the same thing! I didn’t want to go into this in this post, since equivalence principle demands a separate post altogether. But, now that you’ve asked, let me try to explain as briefly as possible.

If you don’t mind, can I make it a ball dropped from a tall tower ? Say, let us start with the case when you are standing at rest relative to earth’s surface. Ok, now since the days of Galileo, we have known what happens to a ball dropped from a tower on earth. As it comes down, it moves more and more fast- it accelerates. And that too at a steady-rate of g=9.8 metres/second-squared, i.e, for every single second the ball spends falling, the speed of the ball increases by 9.8 metres/second.This tells us immediately via Newton’s laws that there is a force acting on the ball. (In fact, we know the name of this force-“Weight” :) )Now, the million dollar question is this -Is this a pseudoforce ?

Now, that’s a tricky question. In fact, it’s so tricky that Newton answered it wrongly ! Newton thought that this is not a pseudoforce, because there is earth’s gravitational field to take the blame for applying force on the ball. So, this is supposed to be “real” force and not a “fake”-force.

Now, enter Einstein ! Einstein brought with him what is called the equivalence principle – there is no way to distinguish between the “real” forces produced due to gravitation and the “fake” pseudo-forces, so they are one and the same. Another way of putting it is as follows- You always measure only the sum of the “real”+”fake” forces, and there is no experiment you can design which can separate them(according to Equivalence principle) – so this very division of forces into gravitational forces vs pseudoforces is meaningless according to Einstein ! So, under this kind of philosophy, what is acting on the ball can justifiably called pseudoforce.

And the story is same for each kind of frame you may care to imagine. Newton goes to every frame, looks at the force and almost arbitrarily divides them into “real¨ vs “fake” whereas Einstein looks at the whole force and calls it the one and the only one gravitational pseudoforce.

5. bharath Says:

loved the alice in wonderland example. so the expading frame is like rubber like frame thats getting stetched due to pseudo forces (‘invisible forces’)

okie. so what is spacetime? As I understood it, the frame of referece only changes the way space gets represented. time appears to be a common denominator with no contraint on it.

6. bharath Says:

I will let you in peace and let you blog this in your leisure. so you can do justice to it, and it might also benefit others. :) what say?

I pestered you because your blog starts with topics I understood decently, but then I couldn’t quite see how it all worked out as you describe in your blog.

7. nayagam Says:

thats the D’Alembert principle!

D’Alembert’s principle ? Oh, I am kind of getting now the way you are thinking about this. But, I’ll rather keep D’Alembert out of this- I am afraid we will really go far from this topic, if we start talking about analytical mechanics !

does inertial and non-inertial frames mean stationary and moving frames of reference? like a frame stuck to a billboard as opposed to one stuck to a moving car?

Oh!no! sorry ! I had taken for granted the definitions of frames(inertial and non-inertial) were wellknown – my mistake !

You know,I am resisting very hard my temptation to putforth the standard mathematical definitions of these objects. The advantage of math is of course that things can be defined very precisely and unambiguously.But, the disadvantage is that quite some mathematical agility is required to relate new math to the old already learnt physics.

Ok, let me give it a try ! By a frame I mean basically a set of observers who observe everything- events happening always and everywhere. These observers are by convention at rest with respect to each other.(Further, I prefer giving to each observer a set of three spatial vectors of unit length orthogonal to each other-e_x,e_y and e_z and again by convention all e_x’s are parallel to each other , all e_y’s are parallel to each other etc.)

If this frame is such that (according to the conventions of rest/spatial parallelism of the frame) there are no pseudoforces acting on any particle, such a frame is called a (global)inertial frame. As I’ve already mentioned in the article, if a spacetime does not allow such a frame, then the spacetime is said to be curved. A non-inertial frame is just a frame which is not an inertial frame.

8. nayagam Says:

then what is the change in geometry due to this pseudo force? I can see that the world appear to behave (move) differently when we go from a stationary to a moving frame of reference. is that it?

Almost, but not quite. Notice that, if you go from one frame to a second frame which is moving with a uniform velocity relative to first,you won’t see any weird thing. In fact, the principle of relativity demands that you SHOULD NOT see anything weird.

But, in the case of expanding frames, something weird does happen according to that frame’s observers. Those observers think that they are at rest with respect to each other,but, still, the distances between them change ! They are forced to blame the dynamics of space itself to explain why distances between them increase. Further, as I was telling in the post they are forced to blame the pseudoforces for this dynamics of space.

9. bharath Says:

got the inertial and non-inertial stuff. I think its coming clear to me now. thanks!

I knew only two kinds of mechanics. one that worked out through force interactions of Newton, the other through invariance principle of Lagrange and Hamilton. mmm.

a frame of reference is just a origin paiared with its coordinate system, right? I think I get that ok.

10. nayagam Says:

when you mean an expanding frame, you attach a distance measure to the frame which changes as the frame expands, so that two points are always at a contant distance with respect to the frame. but for an observer on the bill-board (say) the points will be receeding from each other? if this understanding is right, then why is it natural (any intuitive reasons?) to choose a distance measure depending on the frame of reference.

Hmm, well, there is only one distant-measure/metric for the whole spacetime.Only the components of the metric are different in different frames. But, hey wait, I thought I didn’t mention metric at all in this post ! Strike that out ! Actually I don’t follow your line of thought here. May be you want me to explain how metric fits into all this…

so the expading frame is like rubber like frame thats getting stetched due to pseudo forces (’invisible forces’)

well, I’m still not sure I am conveying to you what I want to convey.Let me give another try. The observers are radially expanding as seen from the inertial frame.They think they are at rest relative to each other, but they see that distances between them is increasing.According to inertial observer, this is just plain motion away from each other- No big deal! But, since the non-inertial observers are adamant on the fact that they are at rest and not moving, the non-inertial observers are FORCED to blame whatever they are seeing on the dynamics of space-time due to pseudoforces. that is briefly my argument. And by the way, pseudoforces are not anymore invisible than ordinary forces. What is invisible (in Newton’s mechanics)is the thing which produces these pseudoforces(the so called absolute space).

okie. so what is spacetime? As I understood it, the frame of referece only changes the way space gets represented. time appears to be a common denominator with no contraint on it.

Well, We owe our concept of spacetime to Minkowski. The main idea is that time is like one more dimension. I wrote a post some time ago
A time-like what ? which you might like to see along with 2-3 posts following it. There I am trying to show how relativity naturally leads to the concept of a spacetime.

11. nayagam Says:

a frame of reference is just a origin paiared with its coordinate system, right? I think I get that ok.

I rather think of frame as a set of orthonormal basis vectors at each event.What you are calling as a “frame” is more usually called a “co-ordinate system” or better “a co-ordinate chart” over some domain.

These two concepts can often be related. For examle, consider a two dimensional being on a sphere. Your “frame” would mean establishing a \theta,\ph” co-ordinates all-over the sphere(except possibly the poles). My “Frame” would mean the two unit vectors e_theta,e_phi at each point (except poles).


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# Interesting construction problems

Here are some interesting construction problems that I stumbled upon.

In the following problems, constructions can only be done using an unmarked straightedge and compass.

I came across the first problem accidentally while watching my friend fold origami. However, it might have been posed long ago by some mathematician. The third problem is easier than the others.

1. Given a line segment of length $$l$$ and any positive integer $$n$$. Show that, using straightedge and compass, it is always possible to divide the line into $$n$$ equal segments, no matter what $$n$$ is.

2. Given a line segment of length $$\sqrt{2}+\sqrt{3}+\sqrt{5}$$, is it possible to construct a line segment of length $$1$$?

3. Given triangle ABC, construct the circumcircle, and the incircle.

4. Given the perpendicular from A and two medians from A, B onto BC, AC respectively, reconstruct triangle ABC.

Note by Joel Tan
3 years, 4 months ago

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2: Follows directly from the definition of a constructible number.

Constructively, show that $$\sqrt{n}$$ is always constructible when $$n \in \mathbb{Z}^{+}$$ and use compass equivalence theorem to add them up.

3: Let $$O$$ be the circumcenter of $$\Delta ABC$$. It's pretty obvious that $$\Delta AOB$$ is isoceles, and that the angle bisector of $$\angle AOB$$ is the perpendicular bisector of $$AB$$; hence, by similar argument for the other sides, $$O$$ is the intersection of the perpendicular bisectors of the sides of $$\Delta ABC$$. Circumcircle construction is just a corollary.

Let $$O^{\prime}$$ be the incenter of $$\Delta ABC$$. Let the shortest distance from the incenter to $$AB$$ intersect the latter at $$X$$; do similarly for $$AC$$, with the intersection $$Y$$. $$O^{\prime}X \equiv O^{\prime}Y$$ (radii), and $$AX \equiv AY$$ (convergent tangents). It's easy to see the angle bisector of $$\angle ABC$$ passes through the incenter; thus, by similar argument for the other angles, $$O^{\prime}$$ is the intersection of the angle bisectors of the interior angles of $$\Delta ABC$$. Again, incircle is just a corollary (albeit a slightly more complicated one).

- 3 years ago

I'm new to compass and straightedge, so I'm sorry if I used any theorems incorrectly.

As for the incircle corollary, my method is to pick any side of $$\Delta ABC$$ and construct any circle with center $$O^{\prime}$$ that intersects that side. The median of the two intersections on the side will be a point on the circumference of the incircle.

- 3 years ago


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# Integral in an old math book

1. Mar 14, 2017

### TheoEndre

1. The problem statement, all variables and given/known data
Hi.
Can anyone here solve this integral for me:
$\int\frac{d^3y}{\operatorname dx^3}\;y\;dx$

2. Relevant equations
I haven't seen such integral before, it is included in an old math book
and it should be solved using "Integration by Parts" technique because it is an exercise of it.
3. The attempt at a solution
I don't no from where to start the solution so I don't know what to write here.

Thank you very much!

2. Mar 14, 2017

### Biker

I too started integration, Here in PhysicsForums we don't give direct answers. I will try to walk you through though.

First it says integrate it with respect to dx.
Second, In most of integration by parts questions. The best way to solve it is to just do the integration by parts a lot of times see if you arrive at a pattern.

Do you know the table method? this should make this easier for you. If not just do the normal way and see if you can reach something you can make use of.

3. Mar 14, 2017

### doktorwho

You say that you don't know where to start but since it is an integration by parts problem you should know the procedure of it's doing. From the integral you choose your $u$ and you $v'$ which are the parts you seek. Here is a hint that will help you, usually for $u$ we choose functions that we don't know or are hard to integrate and for $v'$ we choose the function we know or are easy to integrate.
An example to clarify would be $\int x \ln x \, dx$ where you are bound to take $u=\ln x$ and $v'=x$ since the first you know to differentiate and the second to integrate. Try to find whats best to fit the given description for your problem :) .

4. Mar 14, 2017

### TheoEndre

Thank you very much for your help.
I've done a lot of integration by parts problems, and I know the table method but still I don't think I'm solving correctly in this one.

I've this so far:
I took $y$ as $D$ and I took $\frac{d^3y}{\operatorname dx^3}$ as $I$ and made it to:
$y\;\frac{d^2y}{\operatorname dx^2}\;-\;\int\frac{dy}{\operatorname dx}\;\frac{d^2y}{\operatorname dx^2}\;dx$

5. Mar 14, 2017

### TheoEndre

I didn't mean I don't know the integration by parts thing, but I mean I don't know what to take as $u$ and what as $dv$ in this problem.

6. Mar 14, 2017

### Biker

Excellent, Now notice the second part.
If I do to it integration by parts a couple of more times just as you did with the first one, What will I get?
Note: Notice what we are trying to find $\int\frac{d^3y}{\operatorname dx^3}\;y\;dx$

you will find that on both side we have the same integration.

7. Mar 14, 2017

### doktorwho

This is correct. Now may i suggest a different method of continuation than what Byker suggested. Look at the second integral. Can you introduce a substitution that would clarify everything without the need of further by parts integration?

8. Mar 14, 2017

### TheoEndre

$\frac{d^2y}{\operatorname dx^2}\;$ is the derivative of $\frac{dy}{\operatorname dx}$?
Are you hinting for $udu$ substitution?

9. Mar 14, 2017

### doktorwho

Yes.

10. Mar 14, 2017

### TheoEndre

I just want to clarify something:
$\frac{d^2y}{\operatorname dx^2}\;$ is the derivative of $\frac{dy}{\operatorname dx}$ but not $\frac{d^2y}{\operatorname dx^2}\;dx$ right?
I thought about making $\frac{dy}{\operatorname dx}\;\frac{d^2y}{\operatorname dx^2}\;$ as $dv$ and $dx$ as $u$
$\begin{array}{l}dv=\frac{dy}{\operatorname dx}\;\frac{d^2y}{\operatorname dx^2}\\v=\frac12\left(\frac{dy}{dx}\right)^2\\u=dx\\du=0\end{array}$
so:
$y\;\frac{d^2y}{\operatorname dx^2}\;-\;\frac12\left(\frac{dy}{dx}\right)^2dx\;-\;\int0$
What do you think guys?

11. Mar 14, 2017

### Biker

That is not how substitution work, neither integration by parts. I thought I should lead you the way through integration by parts only so you can practice.

When you do integration by parts, You do it to the functions inside the integral not to dx. You can think of dx as just a notation to tell you what you are integrating with respect to.

You have two choices, Either continue integration by parts as you did in the first step or make a substitution and integrate (Which is easier).

Here: https://www.mathsisfun.com/calculus/integration-by-substitution.html

12. Mar 14, 2017

### TheoEndre

Sorry for that mistake, I had the idea that $dx$ is something more than a notation.

$y\;\frac{d^2y}{\operatorname dx^2}\;-\;\frac12\left(\frac{dy}{dx}\right)^2+\;c$
I hope I didn't mess it up this time.

13. Mar 14, 2017

### Biker

Excellent, good job :D.

Integration means infinite sum of f(x) dx. But you can't just treat dx as part of what you are integrating. you can substitute with du in a sense (The reverse of the Chain rule).

14. Mar 14, 2017

### TheoEndre

Thank you very very much!
I'm glad that I got the help from great man like you Biker, and nice to meet you :)
and thank you doktorwho for your help, you gave me a really great hint. and nice to meet you too :)
Sorry for the inconvenience and sorry for my language (English is not my native language)
not to forget, I completed the solution in the two methods you gave me and I understand it very much!
thank you again!
see you later


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# 21+ Hasse Diagram In Set Theory PNG

21+ Hasse Diagram In Set Theory
PNG
. Remarkably, this hasse diagram is. A hasse diagram simply represents a nite partially ordered set, in the form of a drawing of its transitive reduction.

My question is, is there any nice ways of visualizing a tree of height $\omega ^2$ as a hasse diagram like this? In order theory, a hasse diagram is a type of mathematical diagram used to represent a finite partially ordered set, in the form of a drawing of its transitive reduction. Chan, hubert, a parameterized algorithm for upward planarity testing, proc.

### Understanding comparability in a partially ordered sets with hasse diagrams.

These are examples of hasse diagrams. Hasse diagrams are graphical renderings of partially ordered sets shown via the cover relation of the partially ordered sets with an implied you can use this hasse diagram template to change the theme, colors, background or restructure it based on your own data sets. Hasse diagrams can be described as the transitive reduction as an abstract directed acyclic graph. It results in the following diagram:


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# Solve a system of differential equations in a loop

Posted 8 months ago
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 Hello, What I try to do is. Solve a system of differential equations and obtain a set of values. Compare these values, in this case when it becomes >1, then change accordingly one initial value (with an If ) and repeat. Let's say I would like to do this five times. The code I use is the following. Αpparently, I do not implement the If command correctly in the code.I hope this makes sense.Thank you for your effort.Best regards,A.PS. I have also attached the file.  \[CapitalGamma] = 5/3; c4 = 6; smax = 50; Rinit = 1.05; \[Gamma]1 = 0.8; \[Gamma]2 = 0.9; t = Flatten@Table[i, {i, 1, 5, 1}]; \[Gamma]uinit = (\[Gamma]1 + \[Gamma]2)/2; Sol1 = Quiet[ Catch[NDSolveValue[{D1[s] == 2 (-1 + R[s]) R[ s] (-(-1 + \[CapitalGamma]) \[Gamma]u[s] + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-2 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[s]^( 2 - \[CapitalGamma]) + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-1 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^-\[CapitalGamma]), N1[s] == -(1 + \[Gamma]u[s]^2) (1 - \[CapitalGamma] + c4 (-1 + R[s])^(1/2 - \[CapitalGamma]/2) R[s]^(-(3/2) (-1 + \[CapitalGamma])) (2 + 4 R[s] (-1 + \[CapitalGamma]) - 3 \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[s]^( 1 - \[CapitalGamma])), R'[s] == D1[s]/Sqrt[D1[s]^2 + N1[s]^2], Derivative[1][\[Gamma]u][s] == N1[s]/Sqrt[D1[s]^2 + N1[s]^2], R[0] == Rinit, \[Gamma]u[0] == \[Gamma]uinit, WhenEvent[N1[s]* D1[s] == 0, Throw[{#, s, Abs[N1[s]/D1[s]] , R[s]}]; "Stop Integration"]}, {R, \[Gamma]u}, {s, 0, smax}]], NSDolveValue::noout] & /@ t j = Select[Sol1, #[[3]] > 1 &][[All, {1, 2, 3, 4}]]; j[[All, 3]]; If[j[[3, 3]] > 1, \[Gamma]2 = \[Gamma]uinit] If[j[[3, 3]] < 1, \[Gamma]1 = \[Gamma]uinit]  Attachments:
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Posted 8 months ago
 Agapi,There is no need to loop. WhenEvent has the ability to change states when events happen and continue the integration. Look at the example in the documentation of WhenEvent. The first example is exactly what you need. I did not dive into your code to see what state you are checking in the If statement but I'll assume its [Gamma]u. You would do something like WhenEvent[N1[s]* D1[s] == 0,If[\[Gamma]u > 1, \[Gamma]2 = \[Gamma]uinit,\[Gamma]1 = \[Gamma]uinit]] Although I really do not understand what you are doing so you will have to change the action above to fit your situation. The documentation examples are good and you can execute multiple statements inside a list in the WhenEvent.RegardsNeil
Posted 8 months ago
 Thank you very much for your input. I have changed my code a bit using what you proposed and with the help of the documentations. What i can't do is to make the code start over a new integration using the new starting value of γuinit that meets the If condition. \[CapitalGamma] = 5/3; c4 = 6; smax = 2; Rinit = 1.05; \[Gamma]1 = 0.8; \[Gamma]2 = 0.81; \[Gamma]uinit = (\[Gamma]1 + \[Gamma]2)/2; {sol1, values} = Reap@NDSolve[{D1[s] == 2 (-1 + R[s]) R[ s] (-(-1 + \[CapitalGamma]) \[Gamma]u[s] + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-2 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[s]^( 2 - \[CapitalGamma]) + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-1 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^-\[CapitalGamma]), N1[s] == -(1 + \[Gamma]u[s]^2) (1 - \[CapitalGamma] + c4 (-1 + R[s])^(1/2 - \[CapitalGamma]/2) R[s]^(-(3/2) (-1 + \[CapitalGamma])) (2 + 4 R[s] (-1 + \[CapitalGamma]) - 3 \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[s]^( 1 - \[CapitalGamma])), R'[s] == D1[s]/Sqrt[D1[s]^2 + N1[s]^2], Derivative[1][\[Gamma]u][s] == N1[s]/Sqrt[D1[s]^2 + N1[s]^2], R[0] == Rinit, \[Gamma]u[0] == \[Gamma]uinit, WhenEvent[ {D1[s] == 0, N1[s] == 0} , Sow[{s, D1[s], N1[s], Abs[N1[s]/D1[s]]}]; smax = s; If[Abs[N1[s]/D1[s]] > 1, {\[Gamma]uinit = \[Gamma]2, Print[\[Gamma]uinit]}]; If[Abs[N1[s]/D1[s]] < 1, {\[Gamma]uinit = \[Gamma]1, Print[\[Gamma]uinit]}]]}, {R, \[Gamma]u}, {s, 0, smax}]; ParametricPlot[Evaluate[{R[s], \[Gamma]u[s]} /. sol1], {s, 0, smax}, AspectRatio -> 1, PlotStyle -> ColorData["GreenPinkTones"], AxesLabel -> {"R", "\[Gamma]u"}]
Posted 8 months ago
 Your bug was that you included the line smax = s in the WhenEvent[]. By doing this you forced the integration to stop just when you wanted the If statement to alter your constants.This works fine: \[CapitalGamma] = 5/3; c4 = 6; smax = 2; Rinit = 1.05; \[Gamma]1 = 0.8; \[Gamma]2 = 0.81; \[Gamma]uinit = (\[Gamma]1 + \[Gamma]2)/2; {sol1, values} = Reap@NDSolve[{D1[s] == 2 (-1 + R[s]) R[ s] (-(-1 + \[CapitalGamma]) \[Gamma]u[s] + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-2 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^(2 - \[CapitalGamma]) + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-1 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^-\[CapitalGamma]), N1[s] == -(1 + \[Gamma]u[s]^2) (1 - \[CapitalGamma] + c4 (-1 + R[s])^(1/2 - \[CapitalGamma]/2) R[ s]^(-(3/2) (-1 + \[CapitalGamma])) (2 + 4 R[s] (-1 + \[CapitalGamma]) - 3 \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^(1 - \[CapitalGamma])), R'[s] == D1[s]/Sqrt[D1[s]^2 + N1[s]^2], Derivative[1][\[Gamma]u][s] == N1[s]/Sqrt[D1[s]^2 + N1[s]^2], R[0] == Rinit, \[Gamma]u[0] == \[Gamma]uinit, WhenEvent[{D1[s] == 0, N1[s] == 0}, {Sow[{s, D1[s], N1[s], Abs[N1[s]/D1[s]]}], If[Abs[N1[s]/D1[s]] > 1, \[Gamma]uinit = \[Gamma]2; Print[\[Gamma]uinit]]; If[Abs[N1[s]/D1[s]] < 1, \[Gamma]uinit = \[Gamma]1; Print[\[Gamma]uinit]]}]}, {R, \[Gamma]u}, {s, 0, smax}]; Regards,Neil
Posted 8 months ago
 Thank you again for your help, but this is not doing what I was expecting. I think I did not explain my goal correctly. I want the integration to stop the time the condition D[1]==0,N1[s]==0 is met. This value i name smax. Then it should go back at the beginning with a new initial value of γuinit and start the integration all over (s=0) again. The visual result should be a different line close to the original that it will stop at a new smax, when the same condition is met. Rinse and repeat let's say n times.
Posted 8 months ago
 Its going to be something like this: (if I understand what you want) \[CapitalGamma] = 5/3; c4 = 6; smax = 2; Rinit = 1.05; \[Gamma]1 = 0.8; \[Gamma]2 = 0.81; \[Gamma]uinit = (\[Gamma]1 + \[Gamma]2)/2; sols = Table[ Reap@NDSolve[ smax = 2; {D1[s] == 2 (-1 + R[s]) R[ s] (-(-1 + \[CapitalGamma]) \[Gamma]u[s] + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-2 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^(2 - \[CapitalGamma]) + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-1 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^-\[CapitalGamma]), N1[s] == -(1 + \[Gamma]u[s]^2) (1 - \[CapitalGamma] + c4 (-1 + R[s])^(1/2 - \[CapitalGamma]/2) R[ s]^(-(3/2) (-1 + \[CapitalGamma])) (2 + 4 R[s] (-1 + \[CapitalGamma]) - 3 \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^(1 - \[CapitalGamma])), R'[s] == D1[s]/Sqrt[D1[s]^2 + N1[s]^2], Derivative[1][\[Gamma]u][s] == N1[s]/Sqrt[D1[s]^2 + N1[s]^2], R[0] == Rinit, \[Gamma]u[0] == \[Gamma]uinit, WhenEvent[{D1[s] == 0, N1[s] == 0}, Sow[{s, D1[s], N1[s], Abs[N1[s]/D1[s]]}]; smax = s; If[Abs[N1[s]/D1[s]] > 1, {\[Gamma]uinit = \[Gamma]2, Print[\[Gamma]uinit]}]; If[Abs[N1[s]/D1[s]] < 1, {\[Gamma]uinit = \[Gamma]1, Print[\[Gamma]uinit]}]]}, {R, \[Gamma]u}, {s, 0, smax}], 5]; data = Map[({R[s], \[Gamma]u[s]} /. #) &, sols[[All, 1, 1]]]; ParametricPlot[data, {s, 0, smax}, AspectRatio -> 1, AxesLabel -> {"R", "\[Gamma]u"}] The idea is to make a list of solutions and then plot the list of results.Regards,Neil
 Τhank you for the help Neil.The list and the plot works as it should. The "If" doesn't do what I want it to do. The idea is to change the γuinit to (γ1+(γ1+γ2)/2)/2 inside the first If or to (γ2+(γ1+γ2)/2)/2 inside the second, in order to bisect the initial interval of γ1,γ2. I tried the Piecewise also instead of if ,after reading some posts, but did not work either. I used the last line in order to visual see the results, but this only shows the result of the last integration. Is it possible to see it for the previous integrations? Any ideas would be really appreciated. Regards, Agapi \[CapitalGamma] = 5/3; c4 = 6; smax = 2; Rinit = 1.05; \[Gamma]1 = 0.8; \[Gamma]2 = 0.9; \[Gamma]uinit = (\[Gamma]1 + \[Gamma]2)/2; sol1 = Table[ Reap@NDSolve[{D1[s] == 2 (-1 + R[s]) R[ s] (-(-1 + \[CapitalGamma]) \[Gamma]u[s] + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-2 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[s]^( 2 - \[CapitalGamma]) + c4 ((-1 + R[s]) R[s]^3)^((1 - \[CapitalGamma])/ 2) (-1 + \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[ s]^-\[CapitalGamma]), N1[s] == -(1 + \[Gamma]u[s]^2) (1 - \[CapitalGamma] + c4 (-1 + R[s])^(1/2 - \[CapitalGamma]/2) R[s]^(-(3/2) (-1 + \[CapitalGamma])) (2 + 4 R[s] (-1 + \[CapitalGamma]) - 3 \[CapitalGamma]) \[CapitalGamma] \[Gamma]u[s]^( 1 - \[CapitalGamma])), R'[s] == D1[s]/Sqrt[D1[s]^2 + N1[s]^2], Derivative[1][\[Gamma]u][s] == N1[s]/Sqrt[D1[s]^2 + N1[s]^2], R[0] == Rinit, \[Gamma]u[0] == \[Gamma]uinit, WhenEvent[ {D1[s] == 0, N1[s] == 0} , j = Sow[{s, N1[s], D1[s], Abs[N1[s]/D1[s]]}]; smax = s; If[ Abs[N1[s]/D1[s]] > 1, {\[Gamma]uinit = (\[Gamma]1 + \[Gamma]uinit)/2, Print[\[Gamma]uinit]}]; If[ Abs[N1[s]/D1[s]] < 1, {\[Gamma]uinit = (\[Gamma]2 + \[Gamma]uinit)/2, Print[\[Gamma]uinit]}]]}, {R, \[Gamma]u}, {s, 0, smax}], 5]; data = Map[({R[s], \[Gamma]u[s]} /. #) &, sol1[[All, 1, 1]]]; ParametricPlot[data, {s, 0, smax}, AspectRatio -> 1, AxesLabel -> {"R", "\[Gamma]u"}] j[[All]] & /@ sol1


text:
#### Vol. 1, No. 1, 2022

 Recent Issues Volume 1, Issue 1
 The Journal About the Journal Editorial Board Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN (electronic): 2832-904X ISSN (print): 2832-9058 To Appear Other MSP Journals
Henselianity in NIP $\mathbb{F}_p$-algebras

### Will Johnson

Vol. 1 (2022), No. 1, 115–128
DOI: 10.2140/mt.2022.1.115
##### Abstract

We prove an assortment of results on (commutative and unital) NIP rings, especially ${\mathbb{𝔽}}_{p}$-algebras. Let $R$ be a NIP ring. Then every prime ideal or radical ideal of $R$ is externally definable, and every localization ${S}^{-1}R$ is NIP. Suppose $R$ is additionally an ${\mathbb{𝔽}}_{p}$-algebra. Then $R$ is a finite product of henselian local rings. Suppose in addition that $R$ is integral. Then $R$ is a henselian local domain, whose prime ideals are linearly ordered by inclusion. Suppose in addition that the residue field $R∕\mathfrak{𝔪}$ is infinite. Then the Artin–Schreier map $R\to R$ is surjective (generalizing the theorem of Kaplan, Scanlon, and Wagner for fields).

##### Keywords
NIP, henselian rings
Primary: 03C60


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# Frequency response

In signal processing, the frequency response of an analog or digital equipment or software is some representation, usually numeric, graphical, or mathematical, of how that equipment or software impacts frequencies.

Since a frequency is fully defined by its amplitude and phase, the frequency response typically describes how the equipment, software (or any system) impacts the amplitude and phase of a set of frequencies. When describing only amplitudes, the impact is called the magnitude response of the system. When describing only phase, the impact is called the phase response of the system. Additional detail and examples of magnitude responses and phase responses is provided in their respective topics.

## Frequency response and transfer functions

In the general case, when the system is defined with a transfer function, the frequency response (both magnitude response and phase response) can be computed from the transfer function. For example, given the Z transform transfer function H(z) of a linear time-invariant systems, the magnitude of H(z), |H(z)|, z = A e-j ω, when evaluated at A = 1 (on the unit circle), produces the magnitude response of the system

$$|H(e^{-j\,\omega})|=\sqrt{Re(H(e^{-j\,\omega}))^2+Im(H(e^{-j\,\omega}))^2}$$

where Re and Im are the real and imaginary parts of H. The phase response of the system is

$$\Phi(e^{-j\,\omega})=atan2(Im(H(e^{-j\,\omega})),Re(H(e^{-j\,\omega})))$$

Similarly, the magnitude response of a system defined with its Laplace transform transfer function H(s), s = σ + j ω, evaluated at σ = 0, is |H(j ω)| and its phase response is arg(H(j ω)).

## Example: Frequency response of a finite impulse response filter

A finite impulse response filter a(k) of length N computes the output signal y(k) from the input signal x(k) with the formula

$$y(k)=\sum_{n=0}^{N-1} a(n) \, x(k-n)$$

The Z transform of both sides of the equation above produces

$$Z(y(k))=Z(\sum_{n=0}^{N-1} a(n) \, x(k-n))=Z(x(k)) \sum_{n=0}^{N-1} a(n) \, z^{-n}$$

and the transfer function of the filter is

$$H(z)=\frac{Z(y(k))}{Z(x(k))}=\sum_{k=0}^{N-1} a(k) \, z^{-k}$$

Thus, the magnitude response of the filter at the frequency f is

$$|H(f)|=\sqrt{(\sum_{k=0}^{N-1} a(k) \, cos(\frac{2\pi\,f\,k}{N}))^2+(\sum_{k=0}^{N-1} a(k) \, sin(\frac{2\pi\,f\,k}{N}))^2}$$

If the filter is one with coefficients that are symmetric around the middle, it can be shown that the phase response is as follows (see Phase response).

$$\Phi(f)=\frac{2\pi\,f}{N} \frac{N-1}{2}$$

## Example: frequency response of headphones

Occasionally, you may see the frequency response of headphones (or, say, speakers) listed as (20 Hz, 20 KHz) for example. This is a particularly simplistic representation of the frequency response and it means that the headphones are able to reproduce frequencies between 20 Hz and 20 KHz. Thus, if an input signal contains frequencies in this interval it will be reproduced by the headphones. You would not know, however, how well these headphones will do for specific frequencies. It may be that a bass at 300 Hz is reproduced, but at lower amplitude, which means that the bass will be heard through the headphones, but not very well.

### Filtered HTML

• Freelinking helps you easily create HTML links. Links take the form of [[indicator:target|Title]]. By default (no indicator): Click to view a local node.


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By using properties of determinants, show that

Question:

By using properties of determinants, show that

$\left|\begin{array}{lll}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$

Solution:

$\Delta=\left|\begin{array}{lll}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we have:

$\Delta=\left|\begin{array}{lll}1+x+x^{2} & 1+x+x^{2} & 1+x+x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$

$=\left(1+x+x^{2}\right)\left|\begin{array}{lll}1 & 1 & 1 \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$

Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$, we have:

$\Delta=\left(1+x+x^{2}\right)\left|\begin{array}{lll}1 & 0 & 0 \\ x^{2} & 1-x^{2} & x-x^{2} \\ x & x^{2}-x & 1-x\end{array}\right|$

$=\left(1+x+x^{2}\right)(1-x)(1-x)\left|\begin{array}{lll}1 & 0 & 0 \\ x^{2} & 1+x & x \\ x & -x & 1\end{array}\right|$

$=\left(1-x^{3}\right)(1-x)\left|\begin{array}{lll}1 & 0 & 0 \\ x^{2} & 1+x & x \\ x & -x & 1\end{array}\right|$

Expanding along R1, we have:

\begin{aligned} \Delta &=\left(1-x^{3}\right)(1-x)(1)\left|\begin{array}{ll}1+x & x \\ -x & 1\end{array}\right| \\ &=\left(1-x^{3}\right)(1-x)\left(1+x+x^{2}\right) \\ &=\left(1-x^{3}\right)\left(1-x^{3}\right) \\ &=\left(1-x^{3}\right)^{2} \end{aligned}

Hence, the given result is proved.


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### How to Claim a Computational Feat

Clémence Chevignard, Rémi Géraud-Stewart, Antoine Houssais, David Naccache, and Edmond de Roffignac

##### Abstract

Consider some user buying software or hardware from a provider. The provider claims to have subjected this product to a number of tests, ensuring that the system operates nominally. How can the user check this claim without running all the tests anew? The problem is similar to checking a mathematical conjecture. Many authors report having checked a conjecture $C(x)=\mbox{True}$ for all $x$ in some large set or interval $U$. How can mathematicians challenge this claim without performing all the expensive computations again? This article describes a non-interactive protocol in which the prover provides (a digest of) the computational trace resulting from processing $x$, for randomly chosen $x \in U$. With appropriate care, this information can be used by the verifier to determine how likely it is that the prover actually checked $C(x)$ over $U$. Unlike traditional'' interactive proof and probabilistically-checkable proof systems, the protocol is not limited to restricted complexity classes, nor does it require an expensive transformation of programs being executed into circuits or ad-hoc languages. The flip side is that it is restricted to checking assertions that we dub \emph{refutation-precious}'': expected to always hold true, and such that the benefit resulting from reporting a counterexample far outweighs the cost of computing $C(x)$ over all of $U$.

Available format(s)
Category
Cryptographic protocols
Publication info
Preprint. MINOR revision.
Keywords
proof of workhashing
Contact author(s)
david naccache @ ens fr
History
Short URL
https://ia.cr/2021/1554

CC BY

BibTeX

@misc{cryptoeprint:2021/1554,
author = {Clémence Chevignard and Rémi Géraud-Stewart and Antoine Houssais and David Naccache and Edmond de Roffignac},
title = {How to Claim a Computational Feat},
howpublished = {Cryptology ePrint Archive, Paper 2021/1554},
year = {2021},
note = {\url{https://eprint.iacr.org/2021/1554}},
url = {https://eprint.iacr.org/2021/1554}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# Inversion (in combinatorics)

Jump to: navigation, search

Inversion may refer to:

• An inversion of a permutation $\pi$ on the ordered set $\{1,2,\ldots,n\}$ is a pair $i < j$ such that $\pi(i) > \pi(j)$
• A transposition, a permutation that exchanges two elements
• A derangement, a permutation with no fixed points
How to Cite This Entry:
Inversion (in combinatorics). Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Inversion_(in_combinatorics)&oldid=39878


text:
# Compute in a Sentence  🔊

Definition of Compute

to calculate or work out

Examples of Compute in a sentence

In order to compute the correct numbers, Leanna needed a calculator.  🔊

One of his key responsibilities at work is to estimate and compute total prices.  🔊

My first task was to compute the cost of the water filtration system.  🔊

Her major mistake was to compute the problem without working the numbers in parentheses first.  🔊

To compute the problem unassisted proved difficult even to the mathematician.  🔊

# Other words in the Money and Business category:

Most Searched Words (with Video)


text:
Sharing reading notes.

I recently started a reading notes sharing page. For now, it is hosted on this Github repository. It should really be a hosted wiki — let me know if you want to help on that!

My aim is to share my thoughts on books I’m reading these days, and get your own thoughts on books you’ve enjoyed. The more we are reading the same books, the more we can share, and the more profound our insights.

Looking forward to your collaboration!

Machine Learning: a definition

Hi everyone, welcome back. Tonight I was having dinner with some engineer friends (sadly not a plat en sauce) and the subject of Machine Learning popped up.

The discussion quickly almost turned religious, between believers and non believers, those that thought it’s over hyped, those who just use it as a tool and those who research it like myself. All in all, it made me feel that even engineers have a very vague sense of what exactly is Machine Learning. I’ve noticed people in general often mix it up with other concepts that either include it such as AI or that it includes such as the Supervised Learning paradigm. Since it’s been a long time that I’ve been wanting to address this issue, here I go.

The aim of this article is to explain ML assuming that you already have some general mathematical background, and have (very) basic knowledge of optimization i.e. you know what argmin is. I also assume that you can read equations.

My definition of Machine Learning is the following:

Choosing a task and learning a statistical model from data that will perform well on this task.

This definition seems extremely vague, so I’ll explain each part in more detail. I’m intentionally leaving out complications due to taking into account randomness in the data. Of course we need this both in theory and in practice.

Choosing a task

Examples of tasks are recognizing handwritten digits, translating sentences from a language to another, identifying people in pictures, forecasting the weather or stock prices, grasping an object with a robot arm… Basically any clear-cut problem is a task. This part doesn’t suppose an actual mathematical expression of the task. On the other hand, it does require to be very precise as we’ll see in the following example.

A model that will perform well

To define what perform well means, you often want to define a loss or cost function for your task, that given your data and model, i.e. the specific function of your data you’re considering, outputs a measure of its performance. A simple example: suppose my (very basic) task is to always output the number 0. A good metric could be the distance of my function output to 0. Then it’s obvious that if f and g are the constant functions respectively equal to 42 and 0.2, g performs better on this task than f, for every distance function other than the 0-1 distance. All these different distance functions yield a different metric for the same task.
The problem is then to minimize your loss function over the class of functions that you will choose in the next step.

A statistical model

Choosing a class of functions that take as inputs the data you want to use for the task, i.e. often a high-dimensional vector, and outputs a real number or another vector. A class of functions is a group of functions that share a certain structure, with different parameters. An example of this is the class of linear functions that can we written : $f(X) = AX + b$, where matrix $A$ and vector $b$ are the parameters.

Learning the model

This here is the actual way that you will choose your function that minimizes your loss over your data in the class of functions that you chose. This is the learning and what makes ML so compelling to me, and makes it treated as almost magical in the media.

This is also what made the success of deep learning, where in the parameter space, the overall loss is differentiable with regard to the parameters, allowing us to simply glide towards a (near) optimal solution using Stochastic Gradient Descent.

Who does what?

ML to me is described by these 4 steps. Both researchers and engineers usually focus on a subset of these steps. For example, for a well-known task such as classification (i.e. for each data point, outputting the category it should belong to), or regression (for each data point, predicting the value of something), both the task and the loss function are well known. Most work can then focus on choosing the model family and how to compute the best model of the family. Engineers on the other hand will often have a hard time modelling their task in a nice loss function, and then apply known model families and optimization schemes.

Now for an example

Let’s consider the task (step 1) of recognizing hand written digits, supposing that we know the ground truth for a subset of our data — we’ll call this subset our training set. Our input is going to be a 784 (ie. 28×28)-vector representing an image. Each of its dimensions corresponds to a pixel in the image. Here is a sample image for a 1:

Now that we identified our task, we want to model it using a loss function (step 2). We want our output to tell us if the sample is a 0, 1, 2… or 9. An intuitive way to represent this is to have a 0-1 error function : if we predict the correct label, our loss is 0 and 1 if we make a mistake. Machine Learners rarely use this in practice because the discrete nature of this loss make the 4th step, the learning, harder.

A common and smart way to represent a very similar task is to have our function output a vector of probabilities over the 10 classes, i.e. a positive vector of size 10, with coordinates that sum to 1. Our new proxy task is predicting how likely a sample is to belong to a certain class, given labelled training data. We’ll call this vector $\hat{y}$ as opposed to $y$ which is the correct probability distribution, where $y_i = 1$ if and only if $i$ is the correct class of our data point. We need a metric to tell if $\hat{y}$ is far from $y$. Cross entropy is the go-to method for this. Its expression for each data sample is:

$L(\hat{y}, y) = -\sum_i y_i log(\hat{y_i})$

Note that if we did not know the $y_i$, it would be a different task, and would thus require a different loss function.

In practice, we’ll minimize the empirical average of this loss over our training data.

At this point, we want to find our model class, i.e. the structure in our functions f. An easy example is using a one-hidden-layer neural network, with a sigmoid activation function $\sigma$. f can be written as:

$f(x) = W_1 \sigma (W_0 X + b_0) + b_1$

So our model is the set of functions that can be written this way, for any weights $W_0, W_1, b_0, b_1$.

Finally comes the learning. We want to learn what function of our class minimizes our loss. This is equivalent to finding the weights that minimize our loss, given the preceding structure. Mathematically, our minimization problem is:

$\min_{W_0, W_1, b_0, b_1} - \sum_{i,j} y_i log(W_1 \sigma (W_0 X_j+ b_0) + b_1)$

where the sum over $i$ refers to the different classes (0, 1…9) in our problem and the sum over $j$ refers to the different samples in our training set.

NB for you who are familiar with optimization: This problem is somewhat nice, because the loss function is differentiable with respect to the weights, although it is not convex in the weights. This explains why people use Stochastic Gradient Descent, an algorithm that guarantees us to find a local minimum of our loss and the corresponding weights.

Once we find the optimal weights, we have our candidate, and have finished our learning! For this particular digit recognition dataset, common classification error runs around 5% on a validation set. The classification error used here is the intuitive 0-1 loss that we defined earlier.

A validation set is a set similar to the training set, meaning that we also know the correct labels $y_i$. The difference with the training set is that we never fed it to the learning algorithm. This means our model’s parameters were not optimized for outputting correct results for the validation set. The hope is that the parameters we learned generalize enough to work well on the validation set. We want this so that we are confident our model will perform well on any new data we feed to it, assuming this data was generated in the same manner as the training data. More on this in a future post.

Conclusion

I went over what makes a problem a Machine Learning problem: the 4 phases of Choosing a Task, Modelling the Task, Choosing a Model Class, and Learning i.e. Optimizing over the Model Class. I hope the example helped to clear this up. We also saw that by trying to solve a specific task, we may have to define a new task that is a good proxy for our original task. In the end though, we always validate our approach on the first task on held-out data, using the loss function that we designed for it, here the 0-1 error loss.

I hope this helps making Machine Learning less of a buzz word and more understandable! I’m looking forward to your comments, don’t hesitate to leave one below.

Going further

The Bible of Optimization, basic and advanced:
Boyd, S.; and Vandenberghe, L. (2004), Convex Optimization. Cambridge University Press, New York, NY, USA.

The Bible of Deep Learning:
Goodfellow, I.; Bengio Y.; and Courville, A. (2016), Deep Learning. MIT Press

The Plat en Sauce Abstraction

Hi all! Here is my first post on food! I hope you enjoy it and am looking forward to your comments! Today, I’ll focus on the French Plat en sauce concept.

I found an old cooking book in one of my parents’ shelves a few months ago and finally got around to checking it out and trying some things in it. The book is called Le Répertoire de la Cuisine (1914 for the first edition) by Théodore Gringoire and Louis Saulnier. The Répertoire is a major guide/recipe book for people who already know some technical French cuisine vocabulary – but nowadays, you can just Google them and find out all about what monder, singer or chemiser mean. I won’t reproduce the recipe from the book here – you can probably find a PDF on internet or buy it on Amazon. I will add my own recipes in the following days, tune back in soon!

I moved to the US a few months ago,  and started missing the homey family French cuisine tastes. In my mind, nothing is more characteristic of this than a Blanquette and other plats en sauce : sauce entrées where meat and veggies are slowly cooked with herbs and wine.  I read and tried a few different recipes from the Répertoire inspired from Estouffade (traditionnelle and à la provençale) and Blanquette.

My maths/CS background made me realize that these can all be abstracted to a generic Plat en sauce, which most of you experienced cooks probably already know. Couscous, Coq au vin, Bœuf bourguignon, Fricassée or even Curry implement this. They’re all pretty cheap and healthy dishes, and the abstraction allows for creation and variation depending on your current tastes and desires. Here is what makes up a good plat en sauce in my mind, with the usual French interpretations of each category.

• Meat: beef stew or equivalent for other animals such as veal or lamb.
• Fonds:  stock – traditionally either white if made from poultry, veal or vegetables or brown if made from red meat
• Wine: same color as the fonds, i.e. red for dark fonds and white for light fonds.
• Vegetables: usually onions, carrots and mushrooms at least – any other root can do, e.g. parsnip for some tartness
• Carbs: usually rice or steamed potatoes, the former being my favorite to have a donburi-style Plat en sauce where the rice gets soaked in the sauce.
• Spices, herbs and various seasonings: bouquet garni (thyme, rosemary and sage) is the most frequent for French dishes, but Ras El-Hanout would be its couscous equivalent. I also consider the egg/heavy cream in Blanquette in this category.
• Optional: a roux to make the sauce thicker.  Always add this at the very end.

Onions have two uses in these recipes: sliced onions, usually in quarters, are for eating, and whole onions are spiked with cloves to flavor the broth and to facilitate removing the cloves in one go before serving the dish. This allows not to end up with a violent clove surprise while eating.

Roux: mix melted butter and flour (around half and half) and cook till the mixture is the color you want. Darker gives a more nutty taste, but doesn’t thicken the sauce as much. Always mix with a few ladles of sauce first before pouring in the main pot to avoid lumps.

I’ll progressively add my recipes for different implementations of Plat en sauce as soon as I’ve tweaked them well enough, come back soon for more! Check my Recipes page for the up to date version.

Let me know if this helped you make one of these dishes or if you tried deriving a new plat en sauce from the abstraction! Here are some ideas that I’d like to try:

• a vegetarian one, using tofu and vegetable stock
• using ichiban-dashi (kombu seaweed and fish stock) for an umami-tasting base


text:
# Math Help - separation of variabes,new constant help needed

1. ## separation of variabes,new constant help needed

I have separated the variables of dy/dx +2ycot2x = 0
By taking -1/2y to the left i ended u with

y = 1/2Asin2x i. 1/2Acosec2x where A in a constant, my question in since the mark scheme gives 1/Acosec2x can i make a new constant B from the 2A which would represent the right answer in the mark scheme?

If not i believe there may be something i'm missing at the stage
-0.5ln2y = 0.5lnsin2x + c which would get me the answer. I have an exam tommorow and really worried if someone couldhe me here this would be great thanks!!

2. Originally Posted by i_zz_y_ill
I have separated the variables of dy/dx +2ycot2x = 0
$y'+2y\cot 2x = 0$ thus $\frac{y'}{y} = - 2\tan 2x$.
Now you can use seperation of variables.

3. Yeah thanks but that didnt realli answer my question, can i vreate a new constant in such a circumstance. Your wright i should have separated it that way as is more convenient but i may not always do it that way. Unless I am wrong in my method please tell me.

4. Originally Posted by i_zz_y_ill
I have separated the variables of dy/dx +2ycot2x = 0
By taking -1/2y to the left i ended u with

y = 1/2Asin2x i. 1/2Acosec2x where A in a constant, my question in since the mark scheme gives 1/Acosec2x can i make a new constant B from the 2A which would represent the right answer in the mark scheme?

If not i believe there may be something i'm missing at the stage
-0.5ln2y = 0.5lnsin2x + c which would get me the answer. I have an exam tommorow and really worried if someone couldhe me here this would be great thanks!!
The entire post is very hard to understand. Could you please reword it, so that we can understand it clearly?

If I understand correctly you are asking if A is a constant then is B = 2A another constant? Then, yes you are right.


text: For a given record, the Confidence Interval gives the mean value estimation with 95% probability. An example data set having three independent variables and single dependent variable is used to build a multivariate regression model and in the later section of the article, R-code is provided to model the example data set. This option can take on values of 1 up to N, where N is the number of input variables. Solution: Anything to the left of this line signifies a better prediction, and anything to the right signifies a worse prediction. This also creates a baseline interaction term of BMW:Mileage, which is not specifically included in the model but comes into play by setting Jaguar and Porche equal to 0: $\hat{Price} = b_0 + b_1 * Mileage + b_2 * Porche + b_3 * Jaguar + b_4 Mileage*Jaguar + b_5 Mileage*Porche.$. Equal variances across explanatory variable: Check the residuals plot for fan-shaped patterns. Data Mining. The Prediction Interval takes into account possible future deviations of the predicted response from the mean. Matrix Formulation of Linear Regression 3. If this procedure is selected, FOUT is enabled. XLMiner computes DFFits using the following computation, y_hat_i = i-th fitted value from full model, y_hat_i(-i) = i-th fitted value from model not including i-th observation, sigma(-i) = estimated error variance of model not including i-th observation, h_i = leverage of i-th point (i.e. Its delivery manager wants to find out if there’s a relationship between the monthly charges of a customer and the tenure of the customer. DFFits provides information on how the fitted model would change if a point was not included in the model. We show below how we can obtain one of these $$p$$-values (for CarTypeJaguar) in R directly: We, therefore, have sufficient evidence to reject the null hypothesis for Mileage and the intercept on Porche compared to the intercept on BMW (which is also significant), assuming the other terms are in the model. Alternative hypothesis: At least one of the coefficients on the parameters (including interaction terms) of the least squares regression modeling price as a function of mileage and car type are nonzero. See the following Model Predictors table example with three excluded predictors: Opening Theatre, Genre_Romantic, and Studio_IRS. For the given lines of regression 3X–2Y=5and X–4Y=7. Multiple linear regression is a method we can use to understand the relationship between two or more explanatory variables and a response variable. Import the relevant libraries and load the data In order to shown the informative statistics, we use the describe() command as shown in figure. Solve via QR Decomposition 6. To do so, we need to incorporate interaction terms on the dummy variables of Porche and Jaguar with Mileage. R-Squared: Adjusted R-Squared values. 5.1. The probabilistic model that includes more than one independent variable is called multiple regression models. Does this same conjecture hold for so called “luxury cars”: Porches, Jaguars, and BMWs? The model describes a plane in the three-dimensional space of , and . The closer the curve is to the top-left corner of the graph (the smaller the area above the curve), the better the performance of the model. For a variable to come into the regression, the statistic's value must be greater than the value for FIN (default = 3.84). Linear Regression Dataset 4. Interest Rate 2. XLMiner displays The Total sum of squared errors summaries for both the Training and Validation Sets on the MLR_Output worksheet. This tutorial shares four different examples of when linear regression is used in real life. Stepwise selection is similar to Forward selection except that at each stage, XLMiner considers dropping variables that are not statistically significant. The decile-wise lift curve is drawn as the decile number versus the cumulative actual output variable value divided by the decile's mean output variable value. A simple linear regression equation for this would be $$\hat{Price} = b_0 + b_1 * Mileage$$. When this is selected, the covariance ratios are displayed in the output. Multiple Linear Regression is one of the regression methods and falls under predictive mining techniques. The value for FIN must be greater than the value for FOUT. This option can take on values of 1 up to N, where N is the number of input variables. The process is fast and easy to learn. Now we define the dependent and independent variables. In this example, we see that the area above the curve in both data sets, or the AOC, is fairly small, which indicates that this model is a good fit to the data. There are many hypothesis tests to run here. The residual plot does show a bit of a fan-shaped patter from left to right, but it is not drastic. Multivariate Regression Model. Let $$x_1 = [1, 3, 4, 7, 9, 9]$$ ... Really what is happening here is the same concept as for multiple linear regression, the equation of a plane is being estimated. (3.2) may often still be analyzed by multiple linear regression techniques. Backward Elimination in which variables are eliminated one at a time, starting with the least significant. Then the data set(s) are sorted using the predicted output variable value. To compute coefficient estimates for a model with a constant term (intercept), include a column of ones in the matrix X. Best Subsets where searches of all combinations of variables are performed to observe which combination has the best fit. After the model is built using the Training Set, the model is used to score on the Training Set and the Validation Set (if one exists). The following model is a multiple linear regression model with two predictor variables, and . Noah can only work 20 hours a week. The greater the area between the lift curve and the baseline, the better the model. Solution: Solving the two regression equations we get mean values of X and Y . It’s hard to tell exactly whether the slopes will also be statistically significantly different when looking at just the scatterplot. Example. Mileage of used cars is often thought of as a good predictor of sale prices of used cars. How can he find this information? Summary statistics (to the above right) show the residual degrees of freedom (#observations - #predictors), the R-squared value, a standard deviation type measure for the model (i.e., has a chi-square distribution), and the Residual Sum of Squares error. Recall that these sample coefficients are actually random variables that will vary as different samples are (theoretically, would be) collected. Problem Statement . This point is sometimes referred to as the perfect classification. However, since there are several independent variables in multiple linear analysis, there is another mandatory condition for the model: Non-collinearity: Independent variables should show a minimum of correlation with each other.


text:
# I Green's function for the wave equation

#### dRic2

Gold Member
Hi, I'm reading "Wave Physics" by S. Nettel and in chapter 3 he introduces the Green's function for the 1-dimensional wave equation. Using the separation of variables method he restricts his attention to the spatial component only. Let $u(x)$ be the spatial solution to the wave equation and $G(x|x_0)$ the Green's function. The author now states the following:

Let us now suppose that it is $u(x)$ that is specified at an end point. Then we must choose $G(x|x_o)$ to vanish at that end point.
I don't really understand this connection. Can someone help me ?

Thanks
Ric

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#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
You are looking to write down the solution on the form
$$u(x) = \int G(x|x_0) \kappa(x_0) dx_0 + f(x)$$
where $\kappa$ is the inhomogeneity in your wave equation and $f(x)$ solves the homogeneous wave equation and satisfies your boundary conditions. If $G(x|x_0)$ does not vanish at the boundary, then the integral will generally not be zero there and your function will not satisfy the boundary conditions.

#### dRic2

Gold Member
In the book I'm reading there is a slightly different version of the solution:

(forget about the $\frac 1 {\rho}$)

BTW I also do not understand your answer. If, for example, $k$ vanishes at the end point the integral could also be zero and the boundary condition would be verified.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
If, for example, $k$ vanishes at the end point the integral could also be zero and the boundary condition would be verified.
No, this is incorrect. The point is to evaluate $x$ at the boundary, not $x_0$, which is an integration variable.

#### dRic2

Gold Member
Sorry, dumb mistake. Now I get your argument, thanks.

"Green's function for the wave equation"

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text:
## anonymous 3 years ago A 26.3 kg block (m1) is on a horizontal surface, connected to a 5.70 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.057 m and a moment of inertia I=0.090 kgm2. A force F = 192.9 N acts on m1 at an angle theta = 30.3°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

1. anonymous

2. Qasim_Ahmed

look buddy i dont know the absolute solution for i dont recall the equation but i'll tell you what to do .. first you should analyse F to calculate the force component that acts on the system.. in this case this component equals $F * \cos \theta$ then you can calculate the m1's acceleration by newtons 2nd low $F = m_{1} a$ now comes my erased part of my memory ! you can write down the equation of inertia and put in your variables to find the up going acceleration i know i didn't made lot of help .. so please forgive me .. i have an important exam tomorrow in Plasma Physics and i am so distracted .. Good Luck

3. anonymous

Thanks Ahmed! Let me try this way and see. Also, good luck on your exam :)

4. anonymous

this may help


text:
# Math Help - Roots of oolynomial equations

1. ## Roots of oolynomial equations

Sorry if this is in the wrong section. I wasn't sure where it went!
If you have an equation:
$
x^2 + 4x + 7 = 0
$

Which has roots
$\alpha and \beta$
And you want an equation that has roots
$
\alpha + 2\beta and \beta + 2\alpha
$

$let u = \alpha + 2\beta$
Im going to get alpha in terms of beta and u...and if i sub this into the equation im going to still have beta :S...
I know you can do it via sum of roots = -b/a etc but im told in this example to use substitution -,-.

Thanks.

2. Originally Posted by AshleyT
Sorry if this is in the wrong section. I wasn't sure where it went!
If you have an equation:
$
x^2 + 4x + 7 = 0
$

Which has roots
$\alpha and \beta$
And you want an equation that has roots
$
\alpha + 2\beta and \beta + 2\alpha
$

$let u = \alpha + 2\beta$
Im going to get alpha in terms of beta and u...and if i sub this into the equation im going to still have beta :S...
I know you can do it via sum of roots = -b/a etc but im told in this example to use substitution -,-.

Thanks.
You don't, you use the information that if $a$ and $b$ are roots of $x^2+ux+v$ then $u=a+b$ and $v=ab$.

CB

3. Originally Posted by CaptainBlack
You don't, you use the information that if $a$ and $b$ are roots of $x^2+ux+v$ then $u=a+b$ and $v=ab$.

CB
Yeah , but annoyingly im not allowed to use that method in this question ..I have to use a substitution method.