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# Speed of Light Always The Same?

I know this has been beaten to death in this forum and other places -> The problem is that I have not come across an explanation which I can understand!!

I understand the most basic things of time dilation. And it makes sense to me.

However, I cannot grasp the concept of how light always travels at roughly the same speed in a vacuum.

If a spaceship travelling at 99% of the speed of light is racing after a photon of light, how is it that the photon is moving away from the spaceship at 100% of the speed of light...?
Shouldn't the spaceship be only a small distance from the photon which it is chasing?

It seems though that this is wrong. And that the photon should be moving away from the spaceship at c... meaning the photon would be extremely far away from the spaceship?!

I have read many explanations, but none do it for me >.<
Can someone explain this in a very simple manner?

excuse my stupidity on the matter..

HallsofIvy
Homework Helper
I know this has been beaten to death in this forum and other places -> The problem is that I have not come across an explanation which I can understand!!

I understand the most basic things of time dilation. And it makes sense to me.

However, I cannot grasp the concept of how light always travels at roughly the same speed in a vacuum.
Not "roughly" the same speed- exactly the same speed!

If a spaceship travelling at 99% of the speed of light is racing after a photon of light, how is it that the photon is moving away from the spaceship at 100% of the speed of light...?
Shouldn't the spaceship be only a small distance from the photon which it is chasing?
No, the photon moves away from the space ship at 100% the speed of light. That would be true in clasical physics. In classical physics, a stationary person would say that the photon is moving, relative to him, at 199% the speed of light because he would use the "speed addition formula" u+ v. But according to relativity (and verified by experiment) for very high speeds the correct formula is
$$\frac{u+ v}{1+ \frac{uv}{c^2}}$$
In your example u= c. That formula gives
$$\frac{v+ c}{1+ \frac{vc}{c^2}}= \frac{v+c} {\frac{c+v}{c}}= c$$

That works out that way for any speed, not just .99c. That's why the speed of light, relative to anyone, no matter how fast, the spaceship that originates the light is moving relative to them, is c.

It seems though that this is wrong. And that the photon should be moving away from the spaceship at c... meaning the photon would be extremely far away from the spaceship?!

I have read many explanations, but none do it for me >.<
Can someone explain this in a very simple manner?

excuse my stupidity on the matter..

Last edited by a moderator:
Hmm sorry, I don't understand the mathematics.

I accept what you told me is true :) But it doesn't help me explaining how it is true... >.<

For example, I don't know why the spaceship would see the photon far away from them

But someone not moving and watching would see the photon close to the spaceship?

Or the other way around? hmm... I don't understand. Surely the photon is either close or far away from the spaceship... but not both?

What if we froze time and the spectator walks up to the photon... where would it be then?! close to the spaceship or far away?

(if you have a problem with freezing time, let's say we have a simulation of the universe in a computer. And we froze that instead.) :p

JesseM

Different observers measure the speed of anything using rulers and clocks which are at rest relative to themselves, and synchronized in their own frame using the Einstein clock synchronization convention. But the rulers of different observers don't measure distance the same way (each observer measures rulers that are moving relative to themselves to be shrunk by a factor of sqrt(1 - v^2/c^2) relative to their own ruler/clock system, the effect known as length contraction), the clocks of different observers don't measure time the same way (each observer measures clocks that are moving relative to themselves to have the time between ticks expanded by a factor of 1/sqrt(1 - v^2/c^2), the effect known as time dilation), and clocks that are synchronized in their own frame will be measured to be out-of-sync in other frames (clocks that are synchronized and have a distance x between them in their own rest frame will be out-of-sync by vx/c^2 in another frame where they are moving at speed v, due to what's called the relativity of simultaneity). So, it's because of all these differences that two observers can each think light is moving at c relative to themselves.

For a numerical example of how all these effects come together to ensure both observers measure a single light ray to move at c, see my post #6 on this thread.

you are confused for the same reason that all newbies get confused. 'relativity of simultaneity'. once you get the idea its not as hard as it sounds.

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So, it's because of all these differences that two observers can each think light is moving at c relative to themselves.

This wording is going to lead to cos-like discussion about what is "real" and "true". It's not just a case of each of the observers thinking that light is moving at c relative to themselves, light is moving at c relative to themselves.

The scenario raised here is almost precisely the one which got me interested in this area and I am not sure that someone who has approached from the OP's starting point will immediately appreciate the "relativity of simultaneity" approach.

The visualisation which I came up with to explain the phenomenon worked quite well for me, but months of effort have shown it is difficult to explain to others.

cheers,

neopolitan

the best advice I can give you is to learn to draw a spacetime diagram.

JesseM
This wording is going to lead to cos-like discussion about what is "real" and "true". It's not just a case of each of the observers thinking that light is moving at c relative to themselves, light is moving at c relative to themselves.
Well, it's only moving at c relative to themselves in a particular choice of coordinate system. Coordinate systems are human inventions, are they not? That's why I said "thinks", although I could equally well have said something like "measures using their own ruler/clock system constructed in the particular way defined by Einstein".
neopolitan said:
The scenario raised here is almost precisely the one which got me interested in this area and I am not sure that someone who has approached from the OP's starting point will immediately appreciate the "relativity of simultaneity" approach.
The relativity of simultaneity was only part of my answer, the more intuitive difference is that they measure length differently because of length contraction, and time differently because of time dilation (as well as because of the relativity of simultaneity if they are measuring the one-way speed of light, although you can dispense with simultaneity issues altogether if one of them is measuring the two-way speed of light and therefore needs only one clock), and each one defines "speed" in terms of their own distance/time measurements.

Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!

sylas
Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!

Not at two places at once... rather that for any fixed event (place and instant) the distances and times to other events will depend on the observer.

It's not that the photon is in two places at once, but that the single place-and-instant is at many different distances at once, depending on how you look at it.

Fredrik
Staff Emeritus
Gold Member
you are confused for the same reason that all newbies get confused. 'relativity of simultaneity'. once you get the idea its not as hard as it sounds.
the best advice I can give you is to learn to draw a spacetime diagram.
I agree with these comments, but I would like to add that even though it's not as hard as it sounds (mathematically), it's even more counterintuitive than it sounds. (When you hear about it the first time, you aren't even aware of some of the more bizarre consequences of the invariance of the speed of light). That's why spacetime diagrams are so useful. The only way to develop some sort of intuition for these things is to starting thinking in terms of spacetime diagrams.

Fredrik
Staff Emeritus
Gold Member
Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!
A photon is the worst example you can pick, because photons aren't really localized in space. (That's a very difficult quantum mechanical result that I don't fully understand myself). The detection of a photon is a better example, because that's an event. It's something that happens at a specific time at a specific place. The set of all events is called "spacetime". All observers agree about what events are a part of spacetime. They just disagree about what coordinates to assign to the events. Each observer uses his/her own coordinate system, which is just a function that assigns four numbers to each event.

Well, it's only moving at c relative to themselves in a particular choice of coordinate system.

Can you demonstrate a coordinate system in which c is not moving at c relative to an observer? (Stress is on the relative and note I write c, not the figure which is a consequence of the standard metre and the standard second, so c in terms of fathoms per fortnight or furlongs per full moon is still c.)

cheers,

neopolitan

A photon is the worst example you can pick, because photons aren't really localized in space.

The OP specifically had a photon based scenario.

cheers,

neopolitan

DaveC426913
Gold Member
Hmm, but it seems as if the same photon is in two or more places at once... depending on how many observers there are?!
I don't know - to me it's far simpler than everyone else is trying to say.

The observer on Earth watches a photon leave the spaceship and travel 186,000 miles in one second. The spacehip in that time has moved 99% of that distance, and so has moved 184.2 thousand miles. (The photon is only 1.86 miles ahead of the spaceship)

The observer in the spaceship is time dilated by, say a factor of 10*. i.e. compared to his Earth counterpart, he's moving ten times as slow.

So, the photon is now 1.8 miles in front of his ship - but only 0.01 seconds has passed in his spaceship. He measures the photon's speed as 1.8miles per 0.01 seconds or 186,000mph.

(Alternately, he waits for his appropriate one second, then measures the photon to be 186,000 miles beyond his ship. However, on Earth, ten seconds has passed, so they see the photon 1,860,000 miles in front of the ship.)

*numbers in this example are picked for their clarity of explanation, not for their accuracy

I need help with this as well; I'm trying to get my head around it and its so much... fun and frustration at the same time, fun because its so interesting, frustrating because I don't understand it as much as I wish too.

Here is my similar problem:

A spacecraft is moving at 80% the speed of light (0.8c) from planet X, a radio station sends a light beam parallel to the spacecraft.
What would the observer on planet X and the spacecraft pilot see in regards to 'watching' this light beam move towards the spacecraft.

In both situations light would obviously move at 1c, but how does this work for the spacecraft and observer... damn Counter-Intuitive addition of velocities... where does Time dilation and the Length Contraction come into effect, and how.

regards

I don't know - to me it's far simpler than everyone else is trying to say.

The observer on Earth watches a photon leave the spaceship and travel 186,000 miles in one second. The spacehip in that time has moved 99% of that distance, and so has moved 184.2 thousand miles. (The photon is only 1.86 miles ahead of the spaceship)

The observer in the spaceship is time dilated by, say a factor of 10*. i.e. compared to his Earth counterpart, he's moving ten times as slow.

So, the photon is now 1.8 miles in front of his ship - but only 0.01 seconds has passed in his spaceship. He measures the photon's speed as 1.8miles per 0.01 seconds or 186,000mph.

(Alternately, he waits for his appropriate one second, then measures the photon to be 186,000 miles beyond his ship. However, on Earth, ten seconds has passed, so they see the photon 1,860,000 miles in front of the ship.)

*numbers in this example are picked for their clarity of explanation, not for their accuracy

due to relativity of simultaneity he will measure distances differently too. not just time (the length of an object is the distance between the front and back at one simultaneous moment)

this is why learning to draw spacetime diagrams is so important

Fredrik
Staff Emeritus
Gold Member
The OP specifically had a photon based scenario.
I guess I'm just nitpicking the word "photon". The OP's question is fine if we interpret "photon" as "massless classical particle", which is probably what he had in mind anyway. It's a convenient way to think about light in SR. (The word "photon" is actually defined by a quantum field theory which says that photons aren't localized).

JesseM
Can you demonstrate a coordinate system in which c is not moving at c relative to an observer?
Do you mean a coordinate system in which light is not moving at c (or even at a constant speed) relative to the observer? If so you could pick something like Rindler coordinates which are commonly used for an observer experiencing constant proper acceleration (also discussed here). But really you have total freedom in defining non-inertial coordinate systems any way you like, for example if x,y,z,t represent the coordinates of an inertial coordinate system I could define a non-inertial coordinate system in terms of any transformation I can think of, a simple one would be:

x' = x - at^2
y' = y
z' = z
t' = t

But you could also do a crazy one like:

x' = x * sin(t/t0)
y' = y - t*c
z' = pi*z - 1008
t' = t + 3x/c + 7y/c + 0.5z/c

JesseM
I don't know - to me it's far simpler than everyone else is trying to say.

The observer on Earth watches a photon leave the spaceship and travel 186,000 miles in one second. The spacehip in that time has moved 99% of that distance, and so has moved 184.2 thousand miles. (The photon is only 1.86 miles ahead of the spaceship)

The observer in the spaceship is time dilated by, say a factor of 10*. i.e. compared to his Earth counterpart, he's moving ten times as slow.

So, the photon is now 1.8 miles in front of his ship - but only 0.01 seconds has passed in his spaceship. He measures the photon's speed as 1.8miles per 0.01 seconds or 186,000mph.

(Alternately, he waits for his appropriate one second, then measures the photon to be 186,000 miles beyond his ship. However, on Earth, ten seconds has passed, so they see the photon 1,860,000 miles in front of the ship.)

*numbers in this example are picked for their clarity of explanation, not for their accuracy
But by picking fake numbers you've actually given a misleading explanation here--in reality you can't explain the fact that both observers measure the photon to move at c using only time dilation as you suggest, you also need to take into account length contraction and the relativity of simultaneity.

JesseM
I need help with this as well; I'm trying to get my head around it and its so much... fun and frustration at the same time, fun because its so interesting, frustrating because I don't understand it as much as I wish too.

Here is my similar problem:

A spacecraft is moving at 80% the speed of light (0.8c) from planet X, a radio station sends a light beam parallel to the spacecraft.
What would the observer on planet X and the spacecraft pilot see in regards to 'watching' this light beam move towards the spacecraft.

In both situations light would obviously move at 1c, but how does this work for the spacecraft and observer... damn Counter-Intuitive addition of velocities... where does Time dilation and the Length Contraction come into effect, and how.

regards
In the planet X rest frame:
Suppose as the spacecraft is leaving planet X, they synchronize clocks, so that the planet X clock reads t=0 and the spacecraft clock reads t'=0 at the moment it leaves planet X. Now suppose that planet X sends the signal 75 years later at t=75 on the planet X clock, and that at the same moment in this frame, the spacecraft is passing another planet, planet Y, which is at rest relative to planet X and 60 light years away in this frame (since the spacecraft was traveling at 0.8c in this frame, naturally 75 years after it left planet X it would be at a distance of 75*0.8=60 light-years from planet X). There is a clock at planet Y which is synchronized with the on at planet X in this frame, so it also reads t=75 when the spacecraft passes it. In this frame the spacecraft clock is slowed down by a factor of sqrt(1 - 0.8^2) = 0.6 due to time dilation, so it only reads t' = 75*0.6 = 45 when it passes planet Y.

The spacecraft starts out 60 light years away from planet X when the signal is sent at t=75, so after 300 years have passed in this frame the spacecraft will have traveled an additional 300*0.8c = 240 light years away, so 300 years later at t=375 years, it'll be 60 + 240 = 300 light-years away from planet X. And naturally if we assume the signal travels at c = 1 light year/year in this frame, the signal will also be 300 light-years away from planet X after 300 years have passed from the time it was sent, so t=375 years must be the time in the planet X frame that the signal reaches the craft. However, because of the time dilation factor, at t=375 the spacecraft clock only reads 375*0.6 = 225 years, so t'=225 years must be the time on the spacecraft's clock when the signal reaches it.

In the spacecraft frame:
All frames agree on local events, so it must be true in this frame too that the planet Y clock reads t=75 at the moment the spacecraft is passing it, at which time the spacecraft clock reads t'=45, as we figured out before. However, in this frame the planet X clock does not read t=75 at this moment, due to the relativity of simultaneity. The way the relativity of simultaneity works is that if two clocks are synchronized and a distance L apart in their own rest frame, then in a frame where they are moving at speed v parallel to the axis between them, the clock in the rear will show a time that's ahead of the clock at the front by vL/c^2. So, in the spacecraft's frame at the moment it's passing planet Y, the planet X clock must be behind by (0.8c)*(60 light-years)/c^2 = 48 years, so the planet X clock reads t=75-48=27 years at the moment the spacecraft is passing planet Y in the spacecraft frame. Since the planet X clock is slowed down by a factor of 0.6 in this frame, we know the planet X clock will take an additional 48/0.6 = 80 years to tick forward to t=75, i.e. the planet X clock will read 75 at t'=45+80=125 in the spacecraft frame. And again, all frames agree on local events, so it must be true in this frame as well that the signal was sent when the planet X clock read t=75, at t'=125 in the spacecraft frame.

Meanwhile, because of length contraction, if the distance between planet X and planet Y was 60 light years in their own rest frame, then in the spacecraft rest frame the distance is shrunk by a factor of sqrt(1 - 0.8^2) = 0.6, so the distance between planet X and planet Y is only 60*0.6 = 36 light-years in the spacecraft rest frame. So at t'=45 when the spacecraft was next to planet Y, planet X was 36 light-years away; and since planet X is moving at 0.8c away from the spacecarft in this frame, that means 80 years later at t'=125 when planet X was sending the signal in this frame, planet X must have been an additional 80*0.8=64 light-years away, for a total of 64+36=100 light-years away from the spacecraft at the moment the signal was sent at t'=125. So, under the assumption that light travels at c in this frame too, the signal will reach the spacecraft 100 years later at t'=225.

By the way, another way to approach this problem would be to use the Lorentz transformation, which says that if an arbitrary event has coordinates x,t in one frame, then in a second frame moving at v along the first frame's x-axis (and with the two frames having a common origin, so x=0,t=0 in the first frame coincides with x'=0,t'=0 in the second), this same event will have coordinates:

x'=gamma*(x - vt)
t' =gamma*(t - vx/c^2)

with gamma = 1/sqrt(1 - v^2/c^2)

In this problem v=0.8c so gamma=1/0.6=1.666..., which means for example that if the event of the signal being sent had coordinates (x=0, t=75) in the planet X frame, in the spacecraft frame the event of the signal being sent has coordinates:

x'=1.666...*(0 - 0.8*75) = -100
t' =1.666...*(75 - 0.8*0) = 125

And if the event of the signal reaching the spacecraft happened at (x=300, t=375) in the planet X frame, in the spacecraft frame this event has coordinates:

x'=1.666...*(300 - 0.8*375) = 0
t'=1.666...*(375 - 0.8*300) = 225

So, you can see that this agrees with the previous calculations, and it's a little simpler to figure out although if you use the Lorentz transform you don't get to think about the different contributions of time dilation, length contraction and the relativity of simultaneity.

read einsten's "special theory of relativity",trust me ,all your doubts will vanish.

Thanks JesseM, this really helped in understanding it all, and the semi-complex nature of special relativity. Definitely printing that out for keeps in helping my understanding.

regards

Do you mean a coordinate system in which light is not moving at c (or even at a constant speed) relative to the observer? If so you could pick something like Rindler coordinates which are commonly used for an observer experiencing constant proper acceleration (also discussed here). But really you have total freedom in defining non-inertial coordinate systems any way you like, for example if x,y,z,t represent the coordinates of an inertial coordinate system I could define a non-inertial coordinate system in terms of any transformation I can think of, a simple one would be:

x' = x - at^2
y' = y
z' = z
t' = t

But you could also do a crazy one like:

x' = x * sin(t/t0)
y' = y - t*c
z' = pi*z - 1008
t' = t + 3x/c + 7y/c + 0.5z/c

Yes, the first c should have been "it" from your quote and "it" was light.

As for your coordinate system, you just jumped out of SR and the question was posed in the context of SR (look at the OP's original question).

Let me rephrase, can you demonstrate a coordinate system in which light is not moving at c relative to an inertial observer?

It doesn't make sense to have a non-inertial coordinate system if the observer is inertial, but I think you will find that even if a non-inertial coordinate system is used then, in that non-inertial coodinate system, the speed of light relative to the observer will be c. So, the choice of inertial coordinate system is not the culprit in your example, but rather non-inertia.

And not being inertial is not a human invention.

cheers,

neopolitan


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# Dealing with grouped data in a reggression

Suppose I have the following database where I have, for each age, the average years of experience and income of the group and the number of individuals observed for such age (but not the income/experience of each individual):

----------------------------------------------------------------------------
| Average income | Age | Average years of experience | number of individuals
----------------------------------------------------------------------------
|     105.40     | 18  |            2.1              |          23
----------------------------------------------------------------------------
|     205.50     | 19  |            3.4              |          12
----------------------------------------------------------------------------
|     465.40     | 20  |            4.5              |          33
----------------------------------------------------------------------------
|     678.40     | 21  |            5.1              |          57
----------------------------------------------------------------------------
|     815.40     | 22  |            6.6              |          11
----------------------------------------------------------------------------
|     902.56     | 23  |            6.7              |          20
----------------------------------------------------------------------------
|     997.13     | 24  |            7.3              |          13

I want to get a regression [not necessarily a linear one] of the form:

$(Avg\_income) = \beta_0 + \beta_1 * (age) + \beta_2 * (Avg\_years\_xp)$

or

$(Income) = \beta_0 + \beta_1 * (age) + \beta_2 * (Years\_xp)$

The first approach I thought of was ignoring the "number of individuals" column and doing the regression as if I had only 7 observations, but this seems to be a waste of information, right?

The second was replicating each row by "number of individuals", but is this approach really correct? Also, I'd like to have this work in a more generalized way such that I can work a non-integer "number of individuals" (ok, doesn't make any sense in this example, but it does in my real problem).

Obs.: I'm planning doing a Bayesian inference, but Frequentist approaches are welcomed too. R syntax is good too.

• averages of more points will tend to be more precise. If the variance of individual observations is constant, the variance of averages will be proportional to $1/n_i$. You need to weight the averages so the more precise ones get more weight. – Glen_b Jan 31 '14 at 11:20
• @Glen_b thanks, do you know how one can use weights in other regression models like generalized linear models that don't have a variance parameter? (maybe this is subject to another question...) – random_user Feb 2 '14 at 23:59
• GLM functions will usually let you supply a vector of weights or some equivalent. – Glen_b Feb 3 '14 at 0:01

format: plain


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Yeah, you need longer variable names. I suggest:

pi_thing instead of p (since pi may cause confusion).

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slicer4ever    6769

int l,dijk=di+nx*(dj+oy*dk),dijkl,dijkr,ima=step_div(dk-ez,nz);
int qj=dj+step_int(-ima*byz*ysp),qjdiv=step_div(qj-ey,ny);
int qi=di+step_int((-ima*bxz-qjdiv*bxy)*xsp),qidiv=step_div(qi,nx);
int fi=qi-qidiv*nx,fj=qj-qjdiv*ny,fijk=fi+nx*(fj+oy*(dk-ima*nz)),fijk2;
double disy=ima*byz+qjdiv*by,switchy=(dj-ey)*boxy-ima*byz-qjdiv*by;
double disx=ima*bxz+qjdiv*bxy+qidiv*bx,switchx=di*boxx-ima*bxz-qjdiv*bxy-qidiv*bx;
double switchx2,disxl,disxr,disx2,disxr2;

/*
so you have created qj, qi, and more badly named
crap at the top of your function... again.
*/

O man, that's just....I don't even know. is their actually comma separated code in that? and a random variable doing...nothing on the 4th line?
that student needs to get a talking to.

I think with things like visual assist x, and code-completion tabbing. i've pretty much stopped doing this practice(although i never did it very much anyway). I like to use full names for things most of the time, just because of the fact that it looks nicer at the end of the day.

edit: that's actually making me a bit sick looking at it, i'm ganna go lie down for a bit now. Edited by slicer4ever

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int l;

is especially nice, especially if you use a font which makes l and 1 look very similar.

EDIT: Obviously it should be called lambda

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Nercury    812

a random variable doing...nothing on the 4th line?

Well to be fair I have hand-picked the worst from this code, and this is just a beginning of the method.

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Oh sure it is more compact. And that is NOT a good thing. Of COURSE I can figure out what it all means eventually, I am not THAT stupid. But it is not the point.

I can't and I'm not that stupid (OK, I am, but still). That thing is just plain unreadable, I have an easier time reading blobs from hex editors... I'd kick immediately anybody who came up with code like that. The only place where I'd tolerate something like that is for stuff like 4k demoscene compos or something similar (and where the 4k limit applies on the source code, e.g. javascript demos and such).

Also oh god that last link... WHY (?°?°)?? ???

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Bacterius    13165
remap(ai,aj,ak,ci,cj,ck,x,y,z,ijk);
vc.find_voronoi_cell(x,y,z,ci,cj,ck,ijk,w,mrs);
This code has to be generated by a machine. Or he did a search & replace at the very end. No sane human can make sense of this.

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int l,dijk=di+nx*(dj+oy*dk),dijkl,dijkr,ima=step_div(dk-ez,nz);
int qj=dj+step_int(-ima*byz*ysp),qjdiv=step_div(qj-ey,ny);
int qi=di+step_int((-ima*bxz-qjdiv*bxy)*xsp),qidiv=step_div(qi,nx);
int fi=qi-qidiv*nx,fj=qj-qjdiv*ny,fijk=fi+nx*(fj+oy*(dk-ima*nz)),fijk2;
double disy=ima*byz+qjdiv*by,switchy=(dj-ey)*boxy-ima*byz-qjdiv*by;
double disx=ima*bxz+qjdiv*bxy+qidiv*bx,switchx=di*boxx-ima*bxz-qjdiv*bxy-qidiv*bx;
double switchx2,disxl,disxr,disx2,disxr2;

/*
so you have created qj, qi, and more badly named
crap at the top of your function... again.
*/

What the?  What human could just look at that and know exactly what is going on?  If I had to read crap like that every day I'd just quit programming.  I don't think it'd be worth the pain every day.

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alh420    5995

Makes me remember my best friend in highschool.

I don't think I ever say him use a variable name with more then 4 characters. mostly 3.

I guess it looked good together with the embedded asm he was writing.

He also didn't like to use circuit boards when building his electronics, because they were a waste of time and space.

Instead he build his circuits with all air-connections, in 3D... (yes, they did work, perfectly)

He's now pretty far into his academic career :P

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slicer4ever    6769

struct Vertex
{
int index;
float x;
float y;
float z;
Vertex* next;
};
{
while(p)
{
if(p->index == index)
return p->x;
p = p->next;
}
return 0;
}
{
while(p)
{
if(p->index == index)
return p->y;
p = p->next;
}
return 0;
}
{
while(p)
{
if(p->index == index)
return p->z;
p = p->next;
}
return 0;
}
struct Face
{
int index;
int i0;
int i1;
int i2;
Face* next;
};
{
while(f)
{
if(f->index == index)
return f->i0;
f = f->next;
}
return 0;
}
{
while(f)
{
if(f->index == index)
return f->i1;
f = f->next;
}
return 0;
}
{
while(f)
{
if(f->index == index)
return f->i2;
f = f->next;
}
return 0;
}
void drawMesh(Face* f, Vertex* v, int n)
{
glBegin(GL_TRIANGLES);
for(int i=0;i<n;++i)
{
glVertex3f(getX(getV0(f,i)), getY(getV0(f,i)), getZ(getV0(f,i)));
glVertex3f(getX(getV1(f,i)), getY(getV1(f,i)), getZ(getV1(f,i)));
glVertex3f(getX(getV2(f,i)), getY(getV2(f,i)), getZ(getV2(f,i)));
}
glEnd();
}

o god, that's horrid, and so incredibly inefficient, I don't even know.

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RobTheBloke    2553

o god, that's horrid, and so incredibly inefficient, I don't even know.

It was all like that, but his exam marking was his worst offence. He'd often fail 50% of a year group, and barely pass the rest. All because he had absolutely no clue. If the students had me as a second marker, they were lucky. Some other 2nd markers weren't as diligant though :(

Why do you need to name your class fields as c, cc, d, e? What the hell are you trying to obfuscate?

Most lecturers are generally pretty level headed, and do try to keep their code clean and readable. The more research based they are however, the more chance there is for strangeness to creep in. Academic research can be quite an isolating experience, so you don't often have people around to bounce ideas off, which does tend to encourage people to slip into some really bizarre habits.

There is also a tendancy for some senior academics to stick their name on top any paper from any student they are supervising. This experience can make a junior researcher quite paranoid, and overly protective of their work. Once they've got a Phd, often this paranoia doesn't leave the academic, but simply becomes much more engrained. Maintaining secrecy, and deliberately obfuscating code, is not uncommon in the academic world as a result.....

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I also imagine that for those with a heavy maths background they will try to name program variables as if they were doing so in an equation out of custom (since in equations everything is a letter and maybe a subscript).

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Bacterius    13165

I also imagine that for those with a heavy maths background they will try to name program variables as if they were doing so in an equation out of custom (since in equations everything is a letter and maybe a subscript).

True, but in general equation variables are named sensibly. I have trouble conceiving an equation mapping to this:

step_int((-ima*bxz-qjdiv*bxy)*xsp)

Nah it just looks like someone trying to use 1-letter variables, running out, moving on to 2-letter variables, running out, and so on.

Reasonable variable names are descriptive stuff like "delta", "x_prime", "du", and so on. But having multiple variables called "disx", "disy", "disxl", "disxr", "disx2", "disxr2" is unreasonable. It's insane. You have to look up their meaning to invoke them when needed, unless you know the code by heart (which should not be the case for even remotely self-documenting code). It might be good for a code-and-forget kind of project or library, but good luck maintaining this.

PS: or excessive precomputation, maybe.

Edited by Bacterius

Khatharr    8812

"du"?

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Bacterius    13165

"du"?

Change in the variable "u". For a physics library I can see that coming up and people would know what it is referring to (if the "u" variable is clear, of course). Maybe "dt" is a better example (change in time, and the ubiquitous update(double dt) function signature)

Edited by Bacterius

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Khatharr    8812

Oh, you mean delta_u.

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Bacterius    13165

Oh, you mean delta_u.

Nah, too long, "du" is used everywhere in mathematics as notation so it's a perfect candidate as a variable name. On the other hand, more complicated rates of change could use better variable names, "d3udv3" doesn't work out so well

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Krohm    5031

I fully endorse this product and/or service.

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Tom KQT    1704

I wouldn't say that using p as a name for verticle position variable is a coding horror, especially if it is some particle class with position (p), velocity (v) and acceleration (a), where the meaning is quite obvious. I personaly would use at least pos, vel, acc, but probably not the full text. Acceleration is so long to write...

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ZebraQuake    131

Even "p" as the position of a particle is ambiguous - it could easily be momentum (which is denoted as "p"). I've always despised single-letter variables (Volume? Velocity? Voltage? Acceleration? Amperes? Area?) even when dealing with mathematics. Walking into programming where it's [i]expected[/i] that variable names be descriptive with such behavior is, in my opinion, inexcusable. I mean, "vel" is already a massive improvement over "v" (likewise with "pos" and such) -_-

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Azaral    467

Yes, I hate trying to read code where they use short hand variable names.

The human brain is wired to instantly recognize words it knows. You read number all over the place. When you see number even if it is not in the correct order, you will be able to recognize it as such.

Writing out words instead of short handing them makes the code SOOO much easier to read.

"Aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it deosn't mttaer in waht oredr the ltteers in a wrod are, the olny iprmoatnt tihng is taht the frist and lsat ltteers be at the rghit pclae. The rset can be a toatl mses and you can sitll raed it wouthit porbelm. Tihs is bcuseae the huamn mnid deos not raed ervey lteter by istlef, but the wrod as a wlohe."

That right there should be evidence enough for people to use FULL AND MEANING FULL words for their code. Is it really that much more effort to type a few extra letters.

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Bacterius    13165

That right there should be evidence enough for people to use FULL AND MEANING FULL words for their code. Is it really that much more effort to type a few extra letters.

Better yet - now we can randomize variable names without impeding code readability, as lnog as the frsit and lsat cahracetrs are in the rghit palce

That said there is an anti-argument to this - ridiculously long variable and function names lead to extremely long code lines (think business java code with getCustomerFormAndIDFileSecureTransactionDBConcurrent() methods) which are difficult to read, not to mention print. So a balance needs to be achieved somehow.

Edited by Bacterius

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Azaral    467

That right there should be evidence enough for people to use FULL AND MEANING FULL words for their code. Is it really that much more effort to type a few extra letters.

Better yet - now we can randomize variable names without impeding code readability, as lnog as the frsit and lsat cahracetrs are in the rghit palce

That said there is an anti-argument to this - ridiculously long variable and function names lead to extremely long code lines (think business java code with getCustomerFormAndIDFileSecureTransactionDBConcurrent() methods) which are difficult to read, not to mention print. So a balance needs to be achieved somehow.

Heh. They COULD be random, if you're happy with your code not compiling

And yeah, obviously if its TOO long then you've got a problem too.

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## Superfluid phases of ultracold Fermi gases on a checkerboard superlattice    [PDF]

M. Iskin
We analyze the ground-state phase diagram of two-component Fermi gases loaded into a two-dimensional checkerboard superlattice, i.e. a double-well optical lattice, potential within the BCS mean-field theory. We show that, by coupling the two s-wave sublattice superfluid order parameters, a checkerboard potential gives rise to a Hamiltonian that has the form of a two-band superfluidity with three (two intraband and an interband) nonlocal order parameters. We study the evolution of these order parameters as a function of particle filling, interaction strength and checkerboard potential, and find that the system always prefers the 0-phase solutions, i.e. the phase difference between sublattice order parameters is 0, but never the $\pi$-phase one. In addition, we find that the ground-state of the system undergo a superfluid-insulator quantum phase transition at half fillings beyond a critical checkerboard potential $C$, the threshold of which is precisely determined by the magnitude of the order parameter at $C = 0$.
View original: http://arxiv.org/abs/1304.8111


text:
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# A rectangular-shaped carpet remnant that measures x feet by

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A rectangular-shaped carpet remnant that measures x feet by [#permalink]

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28 Jun 2012, 14:09
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A rectangular-shaped carpet remnant that measures x feet by y feet is priced at $50. What is the cost of the carpet, in dollars per square yard? (9 square feet = 1 square yard) A. 50xy B. 450xy C. xy/9 D. xy/50 E. 450/(xy) [Reveal] Spoiler: OA _________________ Kudos if you like the post! Failing to plan is planning to fail. Intern Joined: 14 Apr 2012 Posts: 11 Re: A rectangular-shaped carpet remnant that measures [#permalink] ### Show Tags 28 Jun 2012, 16:37 xy sq ft =$50
1 sq ft = $50/xy multiplying by 9 on both side 9 sq ft =$450/xy
or 1 sq yard = $450/xy Hence E. Manager Status: Rising GMAT Star Joined: 05 Jun 2012 Posts: 131 Location: Philippines Concentration: General Management, Finance GPA: 3.22 WE: Corporate Finance (Consulting) Re: A rectangular-shaped carpet remnant that measures [#permalink] ### Show Tags 28 Jun 2012, 17:41 3 This post received KUDOS Stiv wrote: A rectangular-shaped carpet remnant that measures x feet by y feet is priced at$50. What is the cost of the carpet, in dollars per square yard? (9 square feet = 1 square yard)
1)$$50xy$$
2) $$450xy$$
3) $$\frac {xy}{9}$$
4)$$\frac {xy}{50}$$
5) $$\frac {450}{xy}$$

x feet
y feet
total price = $50 find: cost of carpet ($/yd^2)

now the trap here would be for you to convert x feet and y feet using the given conversion in the question.

see, the conversion states that 9 square feet = 1 yard -->> if you analyze this statement it means square feet! SQUARE! this means that two units have already been multiplied (hope you're catching my drift)

this means that we should multiply x by y first, we have xy (area in square feet; this makes the conversion consistent)

(xy square feet) x (1 square yard)/(9 square feet) = xy square yard / 9 -->> see we want to cancel out square feet in numerator and square feet in denominator

now we want to get the cost of carpet per square yard, this means we just divide the total cost by what we got earlier; we have..

50/(xy/9) = 50 * 9 / xy = 450/xy

I think the main takeaway here is the trap that I said earlier. You have to think like the test maker. That being said, this question is very GMAT-esque, and I'm not surprised since the source is GMATPrep.

Anyway, can I get a kudos?

Thanks!
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Re: A rectangular-shaped carpet remnant that measures [#permalink]

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28 Jun 2012, 22:58
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Stiv wrote:
A rectangular-shaped carpet remnant that measures x feet by y feet is priced at $50. What is the cost of the carpet, in dollars per square yard? (9 square feet = 1 square yard) 1)$$50xy$$ 2) $$450xy$$ 3) $$\frac {xy}{9}$$ 4)$$\frac {xy}{50}$$ 5) $$\frac {450}{xy}$$ Area of the carpet = xy square feet Cost per square foot =$50/xy square feet

Cost per square yard = $50/(xy/9) =$450/xy square yard

To learn how to quickly convert units, check: http://www.veritasprep.com/blog/2011/01 ... nd-ounces/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 11 Oct 2016 Posts: 1 Re: A rectangular-shaped carpet remnant that measures x feet by [#permalink] ### Show Tags 13 Oct 2016, 02:22 Could you elaborate the cost? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7934 Location: Pune, India Re: A rectangular-shaped carpet remnant that measures x feet by [#permalink] ### Show Tags 13 Oct 2016, 04:08 LavetaLocklin wrote: Could you elaborate the cost? Cost =$50

Area of rug = x feet * y feet = xy square feet

We need cost in dollars per square yard

We are given that 9 square feet = 1 square yard
1 square foot = 1/9 square yard
So xy square feet = (1/9)*xy square yard = xy/9 square yard

Cost in dollars/square yard = $50 / (xy/9) square yard = (450/xy) dollars per square yard _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A rectangular-shaped carpet remnant that measures x feet by [#permalink]

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14 Oct 2016, 04:45
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Stiv wrote:
A rectangular-shaped carpet remnant that measures x feet by y feet is priced at $50. What is the cost of the carpet, in dollars per square yard? (9 square feet = 1 square yard) A. 50xy B. 450xy C. xy/9 D. xy/50 E. 450/(xy) We are given that a rectangular-shaped carpet remnant that measures x feet by y feet is priced at$50. We need to determine the cost, in dollars per square yard.

The area of the carpet is xy square feet and we have to convert that value to square yards.

(xy square feet) x (1 square yard/9 square feet) = xy/9 square yards

The cost of the carpet, in dollars per square yard, is 50/(xy/9) = 450/xy

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Re: A rectangular-shaped carpet remnant that measures x feet by [#permalink]

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Re: A rectangular-shaped carpet remnant that measures x feet by   [#permalink] 13 Jan 2018, 22:23
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# How to sort all chinese references before English ones

I'm writing my phd thesis, which is in Chinese. It's required to sort the all Chinese references before English ones in the Section of reference.

However, it seems that there is not any existing sorting options that can support this.

I tried to use \usepackage[backend = biber, style=alphabetic, sorting=nty]{biblatex}. All Chinese references were listed behind English ones.

Is there any way that can sort all the Chinese references before all the English ones?

• Welcome to TeX.SX! Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – user31729 May 22 '14 at 4:50
• You should specify a a sortkey for each of the Chinese references so they are placed before the English ones. You may have to sort them manually, but this would only have to happen once. – Werner May 22 '14 at 4:53
• It could be interesting to see whether a custom sorting scheme with the language field before sortname would do the trick. (You'd need the language field present in the entries of course, but that is the same amount of work as adding presort.) – equaeghe Oct 18 '17 at 14:26

A simple solution would be to use the presort field. Unlike sortkey, presort would be able to sort into broad categories, leaving the sorting within the categories to biblatex, instead of having to sort every entry individually (as with sortorder).

biblatex.def defines the nyt sorting style as:

\DeclareSortingScheme{nyt}{
\sort{
\field{presort}
}
\sort[final]{
\field{sortkey}
}
\sort{
\field{sortname}
\field{author}
\field{editor}
\field{translator}
\field{sorttitle}
\field{title}
}
......

So, by adding a presort = 1 for all Chinese entries in your .bib file and a presort = 2 for all English entries, they will be sorted by language first, then proceed to be sorted by the regular nyt/name,year,title scheme after that as usual.

A very simple solution might be the following: - edit your *.bbl file - mark and cut all english entries - paste them after the chinese entries - save this file as your new *.bbl - and run xelarex (or ist lua?) again.


text:
# Vector calculus problem

f = 2xy in the x dir, (x^2 - z^2) in the y dir, -3xz^2 in the z dir.
particle travles from (0,0,0) to (2,1,3) along the segments (0,0,0) ->(0,1,0)->(2,1,0)->(2,1,3)

the integral is F dot product with dl.
i cant figure this out.

do i dot product each part individually and evaluate? i mean dot the x component of F and dl and evaluate them on an integral?

Galileo
Homework Helper
You have to cut your integral into 3 parts for the three line segments:

$$\int_{(0,0,0)}^{(0,1,0)}\vec F \cdot d\vec l+\int_{(0,1,0)}^{(2,1,0)}\vec F \cdot d\vec l+\int_{(2,1,0)}^{(2,1,3)}\vec F \cdot d\vec l$$

Since each of the the line segments are parallel to either the x,y or x-axis, getting $d\vec l$ is very simple.

Galileo said:
You have to cut your integral into 3 parts for the three line segments:

$$\int_{(0,0,0)}^{(0,1,0)}\vec F \cdot d\vec l+\int_{(0,1,0)}^{(2,1,0)}\vec F \cdot d\vec l+\int_{(2,1,0)}^{(2,1,3)}\vec F \cdot d\vec l$$

Since each of the the line segments are parallel to either the x,y or x-axis, getting $d\vec l$ is very simple.

so the first integral would be from (0,0,0) to (0,1,0), and i would be evaluating 2xy*dx? thats the dot product of Fx and dl sub x. what do i do with the y?
i tried it this way earlier and i get a final answer after evaluating the three integrals of -22. the book says -50.

Galileo
Homework Helper
Note the direction. If you go from (0,0,0) to (0,1,0) in a straight line, then you only have the y-component to worry about (not the x!)
So the first integral is:

$$\int_0^1 F_y(0,y,0)dy=0$$
since the y component of $\vec f$ is zero along this line.
Do the same for the 2nd and 3rd integral. The second yields 4, the third is -54.

ive got another problem, my electro-magnetics prof didnt get a chance to do any examples on this stuff. D = 2 Ro z^2 in the Ro dir, and Ro cos^2 theta in the z dir.
find closed surface integral of D dot Ds.
0<Ro<5
-1<z<1
0<theta<2pi

is this the same thing or is it different?

this is the same problem but this time they want me to evaulate it on the straight line from (0,0,0) to (2,1,3).
everything i can find on the net wants me to parametrize it, and the book doesnt say anything about that. so im sure theres an easier way to do it.

the book gives the answer of -39.5

if i make z = 3/2 x and y = 1/2 x and substitute those into the integral of F dot dl i end up with an answer of -27.66667. the way im doing it makes sense from the things ive seen on the net. but the book gives an answer of -39.5. ??


text:
# Work Word Problems

Related Topics:
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This video is for the redesigned SAT which is for you if you are taking the SAT in March 2016 and beyond.

Calculator: Not Allowed

$$\frac{1}{x} + \frac{2}{x} = \frac{1}{5}$$

Anise needs to complete a printing job using both of the printers in her office. One of the printers is twice as fast as the other, and together the printers can complete the job in 5 hours. The equation above represents the situation described. Which of the following describes what the $$\frac{1}{x}$$ expression represents in this equation?

A) The time, in hours, that it takes the slower printer to complete the printing job alone
B) The portion of the job that the slower printer would complete in one hour
C) The portion of the job that the faster printer would complete in two hours
D) The time, in hours, that it takes the slower printer to complete $$\frac{1}{5}$$ of the printing job

The correct answer is B) The portion of the job that the slower printer would complete in one hour.

The quickest method would be to recognize that the given equation is similar to the formula used for work problems:

If we let x be the time taken by the slower printer then the time taken by the faster printer would be $$\frac{1}{2}x$$ since it is twice as fast. Substitute that into the work formula and we get
$$\frac{1}{x} + \frac{2}{x} = \frac{1}{5}$$
More examples of work problems using algebra

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text:
missashleyn Group Title Can someone help me please? A wooden pyramid, 12 inches tall, has a square base. A carpenter increases the dimensions of the wooden pyramid by a factor of 5 and makes a larger pyramid with the new dimensions. Describe in complete sentences the ratio of the volumes of the two pyramids. one year ago one year ago

1. Dido525

Do you know what the volume of a pyramid is?

2. missashleyn

1/3 (b)(h)?

3. Dido525

Yep. So in the first case, what would the volume be? You to to find the ration you divide the new value by the old value right?

4. Dido525

To find the ratio*

5. missashleyn

Well I know the height, but what's the base?

6. Dido525

No need. You got $v=\frac{ 1 }{ 3 }A*h$

7. Dido525

h=12.

8. Dido525

Therefore: $v=\frac{ 1 }{3 } * A *12$ = $V=4A$

9. Dido525

Now we are given the ara of the base was increased by a factor of 5. In other words, it was multiplied by 5.

10. missashleyn

So, 20?

11. Dido525

Ratios mean you divide, not multiply.

12. missashleyn

Now I'm confused :(

13. Dido525

14. missashleyn

I thought you said it was multiplied by 5?

15. Dido525

Haha, yeah for the new volume. But when you find a ration you divide the new value by the old value.

16. Dido525

ratio*

17. missashleyn

So... what is the ratio of the volumes of the 2 pyramids? o.o

18. Dido525

Lol, I don't want to give you an answer :P . I would rather you understand it.

19. Dido525

Allright better explanation.

20. Dido525

It's a 1/5 ratio right?

21. Dido525

Because we scale the pyramid by a factor of 5.

22. Dido525

1:5 ratio sorry.

23. Dido525

Ignore all the other parts.

24. missashleyn

I'm seriously so confused it's not even funny! Lol okay so the dimensions are increased by a facor of 5. So you're saying the ratio of the 2 pyramids is 1:5?

25. Dido525

Ahh but remember, since this is volume, we cube that 5.

26. Dido525

1/5^3

27. Dido525

or 1:125 which can also be 1/125.

28. missashleyn

My head is starting to hurt..so I cube the ratio 1:5? WHAT? Lol you have to break it down for me all at once. I don't do step by step :b

29. Dido525

Allright.

30. Dido525

The scale factor of the Pyramid is 1:5 right? This can be re-written as 1/5 . So the ratio would be 1:5 but since we are dealing with volumes, We have to cube that ratio or in other words, 1/5^3 or 1/125. We can say the New pyramid is 125x more greater than the old pyramid.

31. missashleyn

Is that the answer or are you waiting for me to answer? Lol I don't know how to do this.

32. Dido525

What don't you understand?

33. Dido525

I want to try and help :c .

34. missashleyn

I'm confused on the entire thing lol, I mean I'd like to understand how to get the right answer so I can fully explain my answer in complete sentences but at this point my head is starting to hurt and I just want to finish it. Lol

35. dumbcow

initial dimensions; height = 12 let the square base have dimensions of 1 so initial volume = 4 new dimensions, multiply lengths by 5 height = 60 square base have sides of 5 .... so area of 25 new volume = (1/3)(25)(60) = 500 volume ratio = 500/4 = 125 does that help?

36. dumbcow

in general, the rule is that the volume ratio = (length ratio)^3 --> 5^3 = 125

37. missashleyn

Yes! Thank you both for helping me.. sorry I get confused so easily! I appreciate you both! :)

38. dumbcow

:)


text:
## Friday, July 1, 2016

### Transiting LHS 6343C is a T Class Brown Dwarf

Benchmark Transiting Brown Dwarf LHS 6343 C: Spitzer Secondary Eclipse Observations Yield Brightness Temperature and mid-T Spectral Class

Authors:

Montet et al

Abstract:

There are no field brown dwarf analogs with measured masses, radii, and luminosities, precluding our ability to connect the population of transiting brown dwarfs with measurable masses and radii and field brown dwarfs with measurable luminosities and atmospheric properties. LHS 6343 C, a weakly-irradiated brown dwarf transiting one member of an M+M binary in the Kepler field, provides the first opportunity to probe the atmosphere of a non-inflated brown dwarf with a measured mass and radius. Here, we analyze four Spitzer observations of secondary eclipses of LHS 6343 C behind LHS 6343 A. Jointly fitting the eclipses with a Gaussian process noise model of the instrumental systematics, we measure eclipse depths of 1.06 \pm 0.21 ppt at 3.6 microns and 2.09 \pm 0.08 ppt at 4.5 microns, corresponding to brightness temperatures of 1026 \pm 57 K and 1249 \pm 36 K, respectively. We then apply brown dwarf evolutionary models to infer a bolometric luminosity log(L_star / L_sun) = -5.16 \pm 0.04. Given the known physical properties of the brown dwarf and the two M dwarfs in the LHS 6343 C system, these depths are consistent with models of a 1100 K T dwarf at an age of 5 Gyr and empirical observations of field T5-6 dwarfs with temperatures of 1070 \pm 130 K. We investigate the possibility that the orbit of LHS 6343 C has been altered by the Kozai-Lidov mechanism and propose additional astrometric or Rossiter-McLaughlin measurements of the system to probe the dynamical history of the system.


text:
# Convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)}$

Study the convergence of the following series: $$\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0$$

Applying d'Alembert criterion I have that $$\lim_{n\to\infty} \frac{n^\alpha \ln^\beta(n)}{(n+1)^\alpha \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1$$ so the nature of the series is inconclusive.

If $\alpha = \beta = 0$, then the series diverges, since $\sum_{n=2}^\infty 1 = \infty$.

Should I study the rest of the cases (i.e. if $\alpha = 0, \beta > 0$ the root test and the ratio test are also inconclusive). What is the best form to study the series?.

-

You can apply Cauchy's condensation test. Moving to $$\sum_{n = 2}^{\infty} \frac{2^n}{2^{n\alpha} n^{\beta}} = \sum_{n = 2}^{\infty} \left( 2^{1-\alpha} \right)^n \frac{1}{n^{\beta}},$$ the series surrenders to the ratio test when $\alpha \neq 1$. When $\alpha=1$, this is a well-known series.
Sorry for commenting in an old post but I think that by the ratio test if $\alpha > 1$ the series converges. –  user50554 Jan 28 '13 at 18:37
Yeah, which is consistent with the ratio test as $\frac{\left(2^{1-\alpha}\right)^{n+1}\frac{1}{\left(n+1\right)^{\beta}}}{\left(2^{1-\alpha}\right)^{n}\frac{1}{n^{\beta}}} = 2^{1 - \alpha}\left(\frac{n}{n+1}\right)^{\beta} \to 2^{1 - \alpha} < 1$ if $\alpha > 1$. –  levap Jan 28 '13 at 22:19
I think you should use the integral test. In order to establish the convergence/divergence of $$\int_2^{\infty} \frac{dx}{x^a\log^b{x}}$$ I suggest you to use the substitution $x=e^t$ which transforms the integral into $$\int_{\log 2}^{\infty} \frac{dt}{e^{(a-1)t}t^b}$$ which is easier. I think you should be able to conclude by yourself, now.


text:
# string¶

The string type is used for strings of text. It may contain Unicode characters.

In Python, “string” is analogous to the unicode type on Python 2.x, and the str type on Python 3.x.
In Ruby, “string” is analogous to the String type.
{ "type": "string" }

"This is a string"

Unicode characters:

"Déjà vu"

""

"42"

42

## Length¶

The length of a string can be constrained using the minLength and maxLength keywords. For both keywords, the value must be a non-negative number.

{
"type": "string",
"minLength": 2,
"maxLength": 3
}

"A"

"AB"

"ABC"

"ABCD"

## Regular Expressions¶

The pattern keyword is used to restrict a string to a particular regular expression. The regular expression syntax is the one defined in JavaScript (ECMA 262 specifically). See Regular Expressions for more information.

Note

When defining the regular expressions, it’s important to note that the string is considered valid if the expression matches anywhere within the string. For example, the regular expression "p" will match any string with a p in it, such as "apple" not just a string that is simply "p". Therefore, it is usually less confusing, as a matter of course, to surround the regular expression in ^...$, for example, "^p$", unless there is a good reason not to do so.

The following example matches a simple North American telephone number with an optional area code:

{
"type": "string",
"pattern": "^(\$$[0-9]{3}\$$)?[0-9]{3}-[0-9]{4}\$"
}

"555-1212"

"(888)555-1212"

"(888)555-1212 ext. 532"

"(800)FLOWERS"

## Format¶

The format keyword allows for basic semantic validation on certain kinds of string values that are commonly used. This allows values to be constrained beyond what the other tools in JSON Schema, including Regular Expressions can do.

Note

JSON Schema implementations are not required to implement this part of the specification, and many of them do not.

There is a bias toward networking-related formats in the JSON Schema specification, most likely due to its heritage in web technologies. However, custom formats may also be used, as long as the parties exchanging the JSON documents also exchange information about the custom format types. A JSON Schema validator will ignore any format type that it does not understand.

### Built-in formats¶

The following is the list of formats specified in the JSON Schema specification.


text: dc.contributor.advisor McIntire, Larry V. Zaidi, Tabish Naseem 2009-06-04T00:45:47Z 2009-06-04T00:45:47Z 1996 https://hdl.handle.net/1911/14039 We have developed a transient computational model of mural thrombogenesis, simulating whole blood flow in real time in a parallel plate flow chamber. Finite element computational methods were utilized to solve governing partial differential equations of mass conservation and fluid flow. The spatial and temporal variation of concentrations of four platelet agonists known to be synthesized by or released from activated platelets (ADP, vWF, thromboxane A$\sb2$ and thrombin) were computed. The embolizing shear, tensile stresses and torque acting on the aggregates due to fluid forces caused by blood flow were also quantitated. The maximum thrombin concentrations (occurring in stagnating fluid pockets proximal to the thrombi) at the end of two minute blood perfusion are more than 800 fold larger compared to those required for irreversible platelet aggregation in vitro (1-3 nM) under all blood flow conditions. While ADP and vWF might only be important at the initial stages of thrombogenesis and at lower wall shear rates, TxA$\sb2$ concentrations by themselves are not sufficient to cause platelet activation. All fluid stresses increase with wall shear rates and are independent of time. The torque on platelet thrombi increases with time and wall shear rate. 90 p. application/pdf eng Chemical engineeringBiomedical engineering Dynamic mathematical modeling of mural thrombogenesis over type I collagen Thesis Text Chemical and Biomolecular Engineering Engineering Rice University Masters Master of Science Zaidi, Tabish Naseem. "Dynamic mathematical modeling of mural thrombogenesis over type I collagen." (1996) Master’s Thesis, Rice University. https://hdl.handle.net/1911/14039.


text:
m
• E

F Nous contacter

0

Multi angle

H 1 $L^2$-cohomology and the theory of weights

Auteurs : Saper, Leslie (Auteur de la Conférence)
CIRM (Editeur )

Résumé : The intersection cohomology of a complex projective variety $X$ agrees with the usual cohomology if $X$ is smooth and satisfies Poincare duality even if $X$ is singular. It has been proven in various contexts (and conjectured in more) that the intersection cohomology may be represented by the $L^2$- cohomology of a Kähler metric defined on the smooth locus of $X$. The various proofs, though different, often depend on a notion of weight which manifests itself either through representation theory, Hodge theory, or metrical decay. In this talk we discuss the relations between these notions of weight and report on new work in this direction.

Codes MSC :
14F43 - Other algebro-geometric (co)homologies
55N33 - Intersection homology and cohomology

 Informations sur la Vidéo Langue : Anglais Date de publication : 01/07/16 Date de captation : 14/06/16 Collection : Research talks ; Algebraic and Complex Geometry ; Topology Format : MP4 Durée : 00:59:04 Domaine : Topology ; Algebraic & Complex Geometry Audience : Chercheurs ; Doctorants , Post - Doctorants Download : https://videos.cirm-math.fr/2016-06-14_Saper.mp4 Informations sur la rencontre Nom de la rencontre : Analysis, geometry and topology of stratified spaces / Analyse, géométrie et topologie des espaces stratifiésOrganisateurs de la rencontre : Mazzeo, Rafe ; Leichtnam, Eric ; Piazza, PaoloDates : 13/06/16 - 17/06/16 Année de la rencontre : 2016 URL Congrès : http://conferences.cirm-math.fr/1422.htmlCitation Data DOI : 10.24350/CIRM.V.19000903 Cite this video as: Saper, Leslie (2016). $L^2$-cohomology and the theory of weights. CIRM. Audiovisual resource. doi:10.24350/CIRM.V.19000903 URI : http://dx.doi.org/10.24350/CIRM.V.19000903

### Voir aussi

Bibliographie

Ressources Electroniques (Depuis le CIRM)

Books & Print journals

Recherche avancée

0
Z


text:
Free Version
Easy

# Archimedes

GRE-CULYH3

Select ONE entry for each blank from the corresponding column of choices. Fill ALL BLANKS in the way that best completes the text.

One of the greatest mathematicians of

, Archimedes

the ratio of the radius to the circumference.


text:
# Effects of contaminants and trophic cascade regulation on food chain stability: application to cadmium soil pollution on small mammals-raptor systems

Type
Publication
In Ecological Modelling. 382:33-42
Date

## Abstract

Within the current context of global change, ecosystems face simultaneously multiple stressors such as environmental pollution, excessive enrichment, and additional disturbances affecting population dynamics. Numerous theoretical and empirical studies explored the transfer of contaminants by food ingestion between trophic levels, and how contaminant effects on survival and fecundity may change regime shifts of classical predator-prey dynamics. However, the extent to which those contaminants may influence the cascade effect of external stressors, as a change in resources (bottom-up cascade) or a variation in predator abundance (top-down cascade) is still poorly understood. We develop a data-driven model to explore how soil contamination modulate the food chain stability and resilience to changes in prey nutrient and in the apex predator mortality rate. We particularly focus on the ecotoxicological impact of the trace metal, cadmium, on a widespread raptor, the barn owl (Tyto alba), feeding on several prey distinct by their trophic positions: herbivores (Microtus spp), omnivores (wood mouse Apodemus sylvaticus and bank vole Myodes glareolus) and insectivores (shrews: Crocidura spp and Sorex spp). Our model reveals the alternative steady states in population dynamics and the occurrence and position of regime shifts where a subtle change in conditions causes a sudden shift in the ecological system. Based on mathematical modelling and bifurcation analysis, the results show for instance that under toxicity threshold, where no population decline is observed, the contaminant weakens food-chain resilience. Then at higher contamination, the toxic effects on predator releases the top-down control over prey that may increase. This range of chemical stress overturns the paradox of enrichment, a central concept in trophic cascade theory. The transition phase at the highest contamination, where the whole community collapse, exhibits multiple patterns, from smooth to abrupt, depending on external stressors and the prey population. Thus, this work provides a methodology to identify ecological traits of preys that are critical for transferring adverse effects of contaminants across the whole community.

## Keywords

Critical transitions, Contaminant transfer, Food chain, Ordinary differential equations

### Citation in bibtex

  @Article{Baudrot2018EcoMod,
author    = {Baudrot, Virgile and Fritsch, Cl{\'e}mentine and Perasso, Antoine and Banerjee, Malay and Raoul, Francis},
title     = {Effects of contaminants and trophic cascade regulation on food chain stability: Application to cadmium soil pollution on small mammals--Raptor systems},
journal   = {Ecological Modelling},
year      = {2018},
volume    = {382},
pages     = {33--42},
doi       = {10.1016/j.ecolmodel.2018.05.002},
publisher = {Elsevier}
}


text:
# Block Matrices and Matrix multiplication

Suppose i have a vector space $V=\mathbb{F}^{3\times2}$ of $3\times2$ matrices and a vector space $W=\mathbb{F}^{2\times2}$ of $2\times2$ matrices over the field $\mathbb{F}$.

Any Linear transformation $T: V\to W$ would take a $3\times2$ matrix and transform it into a $2\times2$.

I tried to compare this with the fact that multiplying a $2\times3$ matrix by a $3\times2$ matrix also results in a $2\times2$ matrix.

But obviously this $2\times3$ matrix is not the matrix representation of $T$.

To better understand, i defined a standard basis for $V$ and made some algebra resulting in this fact :
http://postimg.org/image/uhh8e02fx/

Where the elements of the $6\times1$ matrix are the vectors from the standard basis of $\mathbb{F}^{3\times2}$ after being transformed by $T$.

By all of this, i found that there are at least two ways to implement matrix multiplication to result in a $2\times2$ matrix.

My question :

Is the ( $1\times6$ times $6\times1$ ) matrix multiplication the only one that is equivalent to the ($2\times3$ times $3\times2$) matrix multiplication ?

I can't find detailed information about matrix multplication, even on linear algebra books ( even wikipedia http://en.wikipedia.org/wiki/Matrix_multiplication doesn't mention the equivalent matrix multiplications i found out empirically ).

Is there any topic name, site, resource, book that can teach much more than the algorithm approach (row dot column ) of matrix multiplication ?

-
I didn't understand your explicit question, but here's a comment about $T$ itself: $T$ is a linear transformation from a six-dimensional vector space $V$ to a four-dimensional vector space $W$. Therefore if you choose a basis for $V$ and a basis for $W$, the corresponding matrix representation of $T$ will be a $4\times6$ matrix. –  Greg Martin Apr 23 '13 at 7:45

Indeed, the space of linear maps from $F^{3\times 2}$ to $F^{2\times 2}$ is $24$-dimensional ($6\cdot 4=24$). Some of them are realized as matrix multiplication $M\mapsto AM$ where $A\in F^{2\times 3}$. However, the matrix $A$ has only $6$ entires, so the linear spaces of all such transformations is $6$-dimensional: it is a proper subspace of $L(F^{3\times 2},F^{2\times 2})$.
However, every linear map from $F^{3\times 2}$ to $F^{2\times 2}$ can be written as a finite sum of two-sided multiplication maps $M\mapsto AMB$. Indeed, $24$ terms suffice: $$M\mapsto \sum_{j=1}^{24} A_j MB_j$$ where each matrix $A_j$ and $B_j$ has one nonzero entry. To see why, observe that by placing $1$ at one entry of $A$ and one entry of $B$ one can move any given entry of $M$ into any desired position in $AMB$.


text:
# Row, Column and Null Space

We have now seen what a vector space is and how to determine if a given set forms a basis of that vector space. We now turn our attention to finding a basis a vector space spanned by a given set of vectors. One such way to do this is to turn the vector into either the rows or columns of a matrix, then find the row or column space of that matrix. That is what we will focus on in this post.

## Matrix

We let
\begin{align*}
\begin{bmatrix}
1 & 2 & 4 & 66 \\
3 & 6 & 1 & 11 \\
7 & 14 & 6 & 88
\end{bmatrix}
\end{align*}
with this matrix we will rank of the matrix, and the bases of the row, column and null space.

### Row Reduce

In order to find all of these, we will need to begin by row reducing the matrix. If you would like to review row reducing, I would suggest seeing the posts Solving a System of Equations, Inverse Matrix or Finding a Determinant (Part 2). Since we have seen how to do this, we will instead using a calculator to find that this matrix rew reduces to the matrix
\begin{align*}
\begin{bmatrix}
1 & 2 & 0 & -2 \\
0 & 0 & 1 & 7 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}

### Rank, row space, column space

Now that we have the row reduced echelon form of the matrix, we are ready to provide our solution. In particular, we find the rank of $$A$$ by counting the left most 1s in the matrix. That is, if we look at the first row, the left most nonzero number is in the first column and is a 1. In the second row, the left most nonzero number is in the third column and is a 1. However, there are no nonzero numbers in the third row.

Because we have two rows with leading 1s, the rank$$(A)=2$$.

Furthermore, the resulting rows with left most 1s will form a basis of the row space of $$A$$. That is, the basis of the row space of $$A$$ is the set $$\left\{(1,2,0,-2),(0,0,1,7)\right\}$$.

Also, the columns that have left most ones tell us which vectors to use as the basis for the column space. That is, we should use the original column vectors of $$A$$ that are in the columns with left most 1s. Therefore, the basis of the column space of $$A$$ is the set $$\left\{(1,3,7),(4,1,6)\right\}$$.

### Null space

In order to find the null space of $$A$$, we have to find solutions to $$A\mathbf{x}=\mathbf{0}$$. While doing this, we will notice that columns that do not correspond to left most 1s will be free variable. Therefore, we will let $$x_{2}=s$$ and $$x_{4}=t$$ be the free variables.

Using the first row, we see that $$x_{1}+2x_{2}-2x_{4}=0$$ in $$A\mathbf{x}=\mathbf{0}$$, so $$x_{1}=-2x_{2}+2x_{4}$$, since $$x_{2}=s$$ and $$x_{4}=t$$, we then get $$x_{1}=-2s+2t$$. Then, doing the same thing for the second row, we find that $$x_{3}-7x_{4}=0$$, so $$x_{3}=7x_{4}$$. Hence $$x_{3}=7t$$.

This gives us that the vector $$\mathbf{x}$$ is in the null space of $$A$$ if
\begin{align*}
\mathbf{x}=\begin{bmatrix} -2s+2t \\ s \\7t \\ t\end{bmatrix}
\end{align*}
for some $$s,t \in \mathbb{R}$$. Therefore, if we want to find a basis, we find the vectors that correspond to $$s=1,t=0$$ and $$s=0,t=1$$. This gives us that the basis of the null space is the set $$\left\{(-2,1,0,0),(2,0,1,1)\right\}$$.

## Conclusion

As always, I hope you learned something and had some fun along the way. If you did, make sure to share this on Social Media and subscribe to our YouTube channel.

## 1 thought on “Row, Column and Null Space”

This site uses Akismet to reduce spam. Learn how your comment data is processed.


text:
# The order of $-i$ in the multiplicative group of $4^{th}$ roots of unity is

$\begin{array}{1 1}(1)4&(2)3\\(3)2&(4)1\end{array}$

The roots of fourth roots of unity are
$1.-1,i,-i$
$(-i)^4=i^4=1$
Order of $(-i) =4$
Hence 1 is the correct answer.


text:
Study Guide

# Basic Algebra - Evaluating Algebraic Expressions

## Evaluating Algebraic Expressions

Expressions are made of variables, or letters that take the place of unknown numbers. But what if we know the numbers? We can take out the variable, replace it with the number, and do the math. This is called evaluating an algebraic expression.

If we have the expression 3x + 1 and we find out that x = 10, we can evaluate it. The expression means "three times whatever x is, plus one."

We rewrite the expression, replacing the x with 10 in parentheses, and do the math.

3x + 1 =
3(10) + 1 =
30 + 1 =
31

### Sample Problem

Evaluate 4xy + 3x if x = 8 and y = 5.

We take out all the variables and put in the numbers in parentheses. Replace x with 8 and swap out y with 5.

4xy + 3x =
4(8)(5) + 3(8) =
160 + 24 = 184

Don't Forget: we'll need our faithful friend PEMDAS (a.k.a. the Order of Operations) to make sure we do the math in the right order.


text:
Browse Questions

# Given $x$ is positive, the values of $f(x)=-3 \cos \sqrt {3+x+x^2}$ lie in the interval .............. .

Can you answer this question?


text:
Categories

## Reaction rate mixing sulfuric acid with formic acid

Mixing sulfuric acid $\text{H}_2\text{SO}_4$ with formic acid $\text{HCOOH}$ yields the following reaction $\text{H}_2\text{SO}_4 + \text{HCOOH} \rightarrow \text{CO}+\text{H}_2\text{SO}_4 \bullet\text{H}_2\text{O}$ Let’s pretend I have a $2$L solution made up of $96$% sulfuric acid and a $1$L solution made up of $65$% formic acid. If my computations are right, this should yield $960$ml of pure sulfuric acid and […]

Categories

## Craft updater still spinning after at least half an hour–no error message

I hadn’t updated site in a while now. Checked Craft’s control panel and something like 12 updates were available and i clicked update. “Updating Composer dependencies (this may take a minute)…” has been on display for over half an hour, maybe an hour, and the activity spinner just keeps spinning. No error report or any […]

Categories

## Nautilus “Connect to Server” is much slower than the terminal’s SSH

In both machines I usually operate on, one at home and another at work, Nautilus seens to work pretty fast when it comes to my local files. However, as soon as I connect it using the SSH option on Nautilus, it becomes really painfully slow, taking 2-5 seconds to open a folder, while a terminal […]

Categories

## Nautilus “Connect to Server” is much slower than the terminal’s SSH

In both machines I usually operate on, one at home and another at work, Nautilus seens to work pretty fast when it comes to my local files. However, as soon as I connect it using the SSH option on Nautilus, it becomes really painfully slow, taking 2-5 seconds to open a folder, while a terminal […]


text:
# The positive aspects of smoothness in Banach lattices

The positive aspects of smoothness in Banach lattices Let X be a Banach lattice, and let $${x\in X{\setminus}\{0\}}$$ . We study the structure of the set Grad(x), of all supporting functionals of x. If X is a Dedekind σ-complete Banach lattice, there is an isometry from Grad(x) onto Grad(|x|); hence the elements x and |x| are smooth simultaneously. And if, additionally, X* is strictly monotone then Grad(|x|) consists of positive functionals. As a by-product of our results we obtain that an arbitrary Banach lattice X is strictly monotone whenever its dual X* is smooth. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Positivity Springer Journals

# The positive aspects of smoothness in Banach lattices

, Volume 17 (2) – Feb 26, 2012
7 pages

Loading next page...

/lp/springer_journal/the-positive-aspects-of-smoothness-in-banach-lattices-0A0UeTTGOU
Publisher
Springer Journals
Copyright
Copyright © 2012 by The Author(s)
Subject
Mathematics; Fourier Analysis; Operator Theory; Potential Theory; Calculus of Variations and Optimal Control; Optimization; Econometrics
ISSN
1385-1292
eISSN
1572-9281
D.O.I.
10.1007/s11117-012-0163-y
Publisher site
See Article on Publisher Site

### Abstract

Let X be a Banach lattice, and let $${x\in X{\setminus}\{0\}}$$ . We study the structure of the set Grad(x), of all supporting functionals of x. If X is a Dedekind σ-complete Banach lattice, there is an isometry from Grad(x) onto Grad(|x|); hence the elements x and |x| are smooth simultaneously. And if, additionally, X* is strictly monotone then Grad(|x|) consists of positive functionals. As a by-product of our results we obtain that an arbitrary Banach lattice X is strictly monotone whenever its dual X* is smooth.

### Journal

PositivitySpringer Journals

Published: Feb 26, 2012

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text:
Model description

MaltBERTa is a large pre-trained language model trained on Maltese texts. It was trained from scratch using the RoBERTa architecture. It was developed as part of the MaCoCu project. The main developer is Rik van Noord from the University of Groningen.

MaltBERTa was trained on 3.2GB of text, which is equal to 439M tokens. It was trained for 100,000 steps with a batch size of 1,024.

The training and fine-tuning procedures are described in detail on our Github repo.

How to use

from transformers import AutoTokenizer, AutoModel, TFAutoModel

tokenizer = AutoTokenizer.from_pretrained("RVN/MaltBERTa")
model = AutoModel.from_pretrained("RVN/MaltBERTa") # PyTorch
model = TFAutoModel.from_pretrained("RVN/MaltBERTa") # Tensorflow

Data

For training, we used all Maltese data that was present in the MaCoCu, Oscar and mc4 corpora. After de-duplicating the data, we were left with a total of 3.2GB of text. We ran experiments with only training on data that came from the .mt domain in Oscar and mc4, but got better performance by incorporating all data.

Benchmark performance

We tested the performance of MaltBERTa on the UPOS and XPOS benchmark of the Universal Dependencies project. Moreover, we test on a Google Translated version of the COPA data set (see our Github repo for details). We compare performance to the strong multi-lingual models XLMR-base and XLMR-large, though note that Maltese was not one of the training languages for those models. We also compare to the recently introduced Maltese language models BERTu, mBERTu and our own MaltBERTa. For details regarding the fine-tuning procedure you can checkout our Github.

Scores are averages of three runs for UPOS/XPOS and 10 runs for COPA. We use the same hyperparameter settings for all models for UPOS/XPOS, while for COPA we optimize on the dev set.

UPOS UPOS XPOS XPOS COPA
Dev Test Dev Test Test
XLM-R-base 93.6 93.2 93.4 93.2 52.2
XLM-R-large 94.9 94.4 95.1 94.7 54.0
BERTu 97.5 97.6 95.7 95.8 55.6
mBERTu 97.7 97.8 97.9 98.1 52.6
MaltBERTa 95.7 95.8 96.1 96.0 53.7

Acknowledgements

Research supported with Cloud TPUs from Google's TPU Research Cloud (TRC). The authors received funding from the European Union’s Connecting Europe Facility 2014- 2020 - CEF Telecom, under Grant Agreement No.INEA/CEF/ICT/A2020/2278341 (MaCoCu).

Citation

If you use this model, please cite the following paper:

@inproceedings{non-etal-2022-macocu,
title = "{M}a{C}o{C}u: Massive collection and curation of monolingual and bilingual data: focus on under-resourced languages",
author = "Ba{\~n}{\'o}n, Marta  and
Espl{\a}-Gomis, Miquel  and
Garc{\'\i}a-Romero, Cristian  and
Kuzman, Taja  and
Ljube{\v{s}}i{\'c}, Nikola  and
van Noord, Rik  and
Sempere, Leopoldo Pla  and
Ram{\'\i}rez-S{\'a}nchez, Gema  and
Rupnik, Peter  and
Suchomel, V{\'\i}t  and
Toral, Antonio  and
van der Werff, Tobias  and
Zaragoza, Jaume",
booktitle = "Proceedings of the 23rd Annual Conference of the European Association for Machine Translation",
month = jun,
year = "2022",
`