[QUOTE=JohnnyMo1;46377075]Fourier transform best transform[/QUOTE]
i know, [I]i have to do them for homework[/I]
[QUOTE=JohnnyMo1;46377075]Fourier transform best transform[/QUOTE]
That's not how you spell Legendre
So my group is doing a presentation for my math class, we asked our teacher if we could present another topic other than what he gave us.
What's the application of the Prime Numbers in general and Prime Spiral to be specified?
[QUOTE=PopLot;46381191]That's not how you spell Legendre[/QUOTE]
Legendre transformations are for nerds who like Hamiltonian mechanics
Lagrange 4 lyfe
Hey, i probably shouldn't be asking this as a Calc 2 student, but here it goes.
Lets say i have this polynomial: [img]http://latex.codecogs.com/gif.latex?4x%5E2+4x+65[/img]
I know how to get it into this form: [img]http://latex.codecogs.com/gif.latex?4%28x+%5Cfrac%7B1%7D%7B2%7D%29%5E2+64[/img]
but, what's the process when i want this form: [img]http://latex.codecogs.com/gif.latex?%282x+1%29%5E2+64[/img], when starting from the original?
or should i just distribute the root of the factored multiple? (distribute sqrt(4) into (x+(1/2))^2)
You're pretty close in your second line. You simply have to put the 4 inside the square and you should have it.
[QUOTE=_Axel;46392209]You're pretty close in your second line. You simply have to put the 4 inside the square and you should have it.[/QUOTE]
yeah, that's what i was thinking. I was just worried i was getting there in a wrong way.
I heard somebody talk something about me
I need help figuring out a certain probability, my math is not really that advanced.
We didn't even start learning probability in school yet.
So if you could help that'd be nice.
Lets say that I have 2 pairs of custom imaginary dice (4), they each have 37 sides.
considering all possible combinations, what are the odds of me getting the same combo on the pairs?
i.e what are the odds of me getting this combo: (AF) (AF) - (AF) (AF) or (Z8) (Z8) - (Z8) (Z8) etc..
I'm not sure if i'm thinking about this properly but if I am thinking properly then lets say that I have a limit of 50 tries to roll the 2 pairs of dice and get this combo, how do my chances change with each try?
do they increase or decrease or stay the same?, this is really interesting so if you could explain that be awesome.
Thank you, if you need me to clarify anything I'll be sure to do it.
Do any of these qualify as "the same combo"?
(AF)(AF) - (Z8)(Z8)
(AF)(Z8) - (AF)(Z8)
[QUOTE=Krinkels;46394853]Do any of these qualify as "the same combo"?
(AF)(AF) - (Z8)(Z8)
(AF)(Z8) - (AF)(Z8)[/QUOTE]
No they don't, they all have to be the same
And you're throwing 2 pairs of cubes (4 cubes total)
Also each pair is thrown separately and not together, so first you throw the first pair and then the second (I'm not sure if this matters)
[QUOTE=Aus;46394485]I need help figuring out a certain probability, my math is not really that advanced.
We didn't even start learning probability in school yet.
So if you could help that'd be nice.
Lets say that I have 2 pairs of custom imaginary dice (4), they each have 37 sides.
considering all possible combinations, what are the odds of me getting the same combo on the pairs?
i.e what are the odds of me getting this combo: (AF) (AF) - (AF) (AF) or (Z8) (Z8) - (Z8) (Z8) etc..
I'm not sure if i'm thinking about this properly but if I am thinking properly then lets say that I have a limit of 50 tries to roll the 2 pairs of dice and get this combo, how do my chances change with each try?
do they increase or decrease or stay the same?, this is really interesting so if you could explain that be awesome.
Thank you, if you need me to clarify anything I'll be sure to do it.[/QUOTE]
First, denote the probability of an event A occurring by P(A).
In this case throwing two pairs is the same as throwing all four at once.
The probability of rolling four of a kind with four thirty-seven sided dice is 37*(1/37)^4.
How did I get this? First fix a number k between one and thirty seven.
The outcome of each roll is not affected by the outcome of the others. This is known as independence. If two events are independent, the probability that they both happen is the product of the probabilities that each happens. In symbols:
Independence means that P(A and B) = P(A)P(B).
The event where 'all four rolls are k' is the same as the event where 'the first three dice rolled k and so did the fourth one'. These events are independent, so:
P(four rolls are k) = P(first three rolls are k)P(last roll is k)
Similarly we can interpret 'three rolls are k' as 'these two dice rolled k and so did the third one'.
P(four rolls are k) = P(first two rolls are k)P(third roll is k)P(last roll is k)
And again,
P(four rolls are k) = P(first roll is k)P(second roll is k)P(third roll is k)P(last roll is k)
The probability that any roll is k is 1/37. The probability that you roll four ks is (1/37)^4.
Now, if two events are mutually exclusive then the probability that either happens is the sum of the probabilities that each happens. In symbols:
If A and B are mutually exclusive, P(A or B) = P(A) + P(B)
The event where you roll four of one number and the event where you roll four of another are mutually exclusive.
P('four ones are rolled' or 'four twos are rolled' or 'four threes are rolled' etc.) = P('four ones are rolled') + P('four twos are rolled') + ... + P('four thirtysevens are rolled')
It was already established that the probability that four of anything was rolled is (1/37)^4. So, since there are thirty-seven terms in that sum, the probability that four of a kind is rolled is 37(1/37)^4 = 1/(37^3).
The probability that you roll four of a kind on any given try is the same. The probability that you roll four of a kind within the first fifty tries is higher, though. To find what it is exactly, we need the following formula:
P(A) + P(not A) = 1
That is to say, it is certain that either A will happen or A will not happen for any event A. You can find the probability that you do not roll four of a kind in fifty tries fairly easily using this formula:
P('four of a kind is not rolled within fifty tries') = (1-P('four of a kind is rolled on the first try')(1-P('four of a kind is rolled on the second try')...(1-P('four of a kind is rolled on the fiftieth try')) = (1-(1/37)^3)^50
Now you can find the probability of rolling four of a kind in fifty tries by applying the last formula again:
P('four of a kind is rolled within fifty tries') = 1 - (1-1/(37^3))^50
As you would expect, the probability decreases as the number of tries decreases, and increases when the number of tries increases. At fifty, it is extremely close to 1.
[QUOTE=Krinkels;46395618]First, denote the probability of an event A occurring by P(A).
In this case throwing two pairs is the same as throwing all four at once.
The probability of rolling four of a kind with four thirty-seven sided dice is 37*(1/37)^4.
How did I get this? First fix a number k between one and thirty seven.
The outcome of each roll is not affected by the outcome of the others. This is known as independence. If two events are independent, the probability that they both happen is the product of the probabilities that each happens. In symbols:
Independence means that P(A and B) = P(A)P(B).
The event where 'all four rolls are k' is the same as the event where 'the first three dice rolled k and so did the fourth one'. These events are independent, so:
P(four rolls are k) = P(first three rolls are k)P(last roll is k)
Similarly we can interpret 'three rolls are k' as 'these two dice rolled k and so did the third one'.
P(four rolls are k) = P(first two rolls are k)P(third roll is k)P(last roll is k)
And again,
P(four rolls are k) = P(first roll is k)P(second roll is k)P(third roll is k)P(last roll is k)
The probability that any roll is k is 1/37. The probability that you roll four ks is (1/37)^4.
Now, if two events are mutually exclusive then the probability that either happens is the sum of the probabilities that each happens. In symbols:
If A and B are mutually exclusive, P(A or B) = P(A) + P(B)
The event where you roll four of one number and the event where you roll four of another are mutually exclusive.
P('four ones are rolled' or 'four twos are rolled' or 'four threes are rolled' etc.) = P('four ones are rolled') + P('four twos are rolled') + ... + P('four thirtysevens are rolled')
It was already established that the probability that four of anything was rolled is (1/37)^4. So, since there are thirty-seven terms in that sum, the probability that four of a kind is rolled is 37(1/37)^4 = 1/(37^3).
The probability that you roll four of a kind on any given try is the same. The probability that you roll four of a kind within the first fifty tries is higher, though. To find what it is exactly, we need the following formula:
P(A) + P(not A) = 1
That is to say, it is certain that either A will happen or A will not happen for any event A. You can find the probability that you do not roll four of a kind in fifty tries fairly easily using this formula:
P('four of a kind is not rolled within fifty tries') = (1-P('four of a kind is rolled on the first try')(1-P('four of a kind is rolled on the second try')...(1-P('four of a kind is rolled on the fiftieth try')) = (1-(1/37)^3)^50
Now you can find the probability of rolling four of a kind in fifty tries by applying the last formula again:
P('four of a kind is rolled within fifty tries') = 1 - (1-1/(37^3))^50
As you would expect, the probability decreases as the number of tries decreases, and increases when the number of tries increases. At fifty, it is extremely close to 1.[/QUOTE]
Thank you very much for this long and full explanation I understand it a lot more now.
[highlight](User was permabanned for this post ("Alt" - Swebonny))[/highlight]
[QUOTE=JohnnyMo1;46383634]Legendre transformations are for nerds who like Hamiltonian mechanics
Lagrange 4 lyfe[/QUOTE]
lol ok hotshot. try to solve those coupled Lagrangian equations directly. i dare you
[QUOTE=Aus;46395936]Thank you very much for this long and full explanation I understand it a lot more now.
[highlight](User was permabanned for this post ("Alt" - Swebonny))[/highlight][/QUOTE]
RIP
When I wondered, how long mathematicians lived, I checked and looks like they lived quite long. Maybe is it attributed to constant use of the brains, thus being "active"?
[QUOTE=Fourier;46397974]When I wondered, how long mathematicians lived, I checked and looks like they lived quite long. Maybe is it attributed to constant use of the brains, thus being "active"?[/QUOTE]
[url=http://en.wikipedia.org/wiki/%C3%89variste_Galois]Except when they tried to be active with their bodies and not with their brains.[/url]
[QUOTE=PopLot;46397504]lol ok hotshot. try to solve those coupled Lagrangian equations directly. i dare you[/QUOTE]
we solve systems of 10 coupled nonlinear second-order PDEs all the time in general relativity because [I]that's just how einstein rolled[/I]
[QUOTE=Fourier;46397974]When I wondered, how long mathematicians lived, I checked and looks like they lived quite long. Maybe is it attributed to constant use of the brains, thus being "active"?[/QUOTE]
I'd say it's more due to the fact that smart people tend to do less stupid shit that's likely to get them killed. Also famous academics were probably mostly upper class, and thus were much more well fed/sheltered/doctored.
[QUOTE=PopLot;46398060][url=http://en.wikipedia.org/wiki/%C3%89variste_Galois]Except when they tried to be active with their bodies and not with their brains.[/url][/QUOTE]
that boy got some bad luck :(.
[editline]3rd November 2014[/editline]
[QUOTE=chonks;46400472]I'd say it's more due to the fact that smart people tend to do less stupid shit that's likely to get them killed. Also famous academics were probably mostly upper class, and thus were much more well fed/sheltered/doctored.[/QUOTE]
Yeah, just wanted to state that we mathematicians are smart in non obvious way.
[QUOTE=JohnnyMo1;46398461]we solve systems of 10 coupled nonlinear second-order PDEs all the time in general relativity because [I]that's just how einstein rolled[/I][/QUOTE]
Proof or gtfo.
That sounds badass though. Can't wait until I have time to go through my GR textbook.
[QUOTE=PopLot;46402005]Proof or gtfo.
That sounds badass though. Can't wait until I have time to go through my GR textbook.[/QUOTE]
I'm trying to find it in Carroll but my professor may have done it differently, but we proved Birkhoff's theorem (i.e. Schwarzschild is the unique spherically symmetric vacuum solution) by directly solving Einstein's equations and making it actually doable by appealing to symmetry properties.
Which GR textbook do you have?
[editline]3rd November 2014[/editline]
But that's what Einstein's field equation is: a set of 10 (independent) coupled non-linear PDEs. Any spacetime satisfying the EFE is a solution to them.
[QUOTE=JohnnyMo1;46403188]I'm trying to find it in Carroll but my professor may have done it differently, but we proved Birkhoff's theorem (i.e. Schwarzschild is the unique spherically symmetric vacuum solution) by directly solving Einstein's equations and making it actually doable by appealing to symmetry properties.
Which GR textbook do you have?
[editline]3rd November 2014[/editline]
But that's what Einstein's field equation is: a set of 10 (independent) coupled non-linear PDEs. Any spacetime satisfying the EFE is a solution to them.[/QUOTE]
[I]A First Course in General Relativity[/I] by Schutz. Sadly the discussion of Birkhoff's Theorem is pretty truncated and points to texts like Misner for a more general treatment. I'll take your word for it.
[sp] But Hamilton > Lagrange [/sp]
I've been trying to gain an intuition of quaternions and while doing so I think I figured out math itself. I was reading about quaternion algebra and I was wondering why all math seemed so arbitrary, and the answer is that it is a bit arbitrary, although it has a method to it. Hamilton designed his quaternion algebra to interact with quaternions algebraically and he assigned properties to the algebra so that the operations on quaternions actually made sense.
With that said, I'm a total math noob. I only have a high school level of math but I want to intuitively understand more advanced concepts like quaternions. I've been starting with Unity and they use quaternions instead of Euler angles and me being me, I wanted to figure out how they actually work, even if I already know why they're used and that I can use them even without even knowing how they work. How can I figure out and gain an intuition on quaternions? What material should I look up?
quaternions are for nerds, real ballers use octonians
[editline]4th November 2014[/editline]
haha jk anyone with any sense stops at complex numbers
[editline]4th November 2014[/editline]
[QUOTE=PopLot;46404707][I]A First Course in General Relativity[/I] by Schutz. Sadly the discussion of Birkhoff's Theorem is pretty truncated and points to texts like Misner for a more general treatment. I'll take your word for it.
[sp] But Hamilton > Lagrange [/sp][/QUOTE]
Hamilton kinda sucks for field theory though.
[QUOTE=JohnnyMo1;46410723]quaternions are for nerds, real ballers use octonians
[editline]4th November 2014[/editline]
haha jk anyone with any sense stops at complex numbers
[editline]4th November 2014[/editline]
Hamilton kinda sucks for field theory though.[/QUOTE]
[I]You[/I] kinda suck for field theory though.
hey, I'm having to look at the fractional quantum Hall effect for a short literature review and I've stumbled upon the Laughlin ground state for a collective excitation of electrons
[URL]http://journals.aps.org/rmp/pdf/10.1103/RevModPhys.71.863[/URL]
at the bottom left of page 869, equation 21. can anyone tell me what the angular brackets j<k> mean in the context of the pi product here? I've never seen anything like that before
Physics feels so unrigorous sometimes. I feel like half the problems in physics textbooks could be phrased as "Show that if you ignore just the right amount of subtlety, you can derive this expression."
[QUOTE=Turnips5;46415782]hey, I'm having to look at the fractional quantum Hall effect for a short literature review and I've stumbled upon the Laughlin ground state for a collective excitation of electrons
[URL]http://journals.aps.org/rmp/pdf/10.1103/RevModPhys.71.863[/URL]
at the bottom left of page 869, equation 21. can anyone tell me what the angular brackets j<k> mean in the context of the pi product here? I've never seen anything like that before[/QUOTE]
I have no clue. There's no context. That's really annoying.
I've encountered something slightly similar in a course that dealt with the Ising Model and phase changes and whatnot.
What it meant there was (I think, it's a while ago) that <i,j> denoted a sum over every unique pair of indices. The reason for this is that it had to do with the interaction energy, and then you obviously don't want to include terms where the spin interacts with itself (i,i), or because the sum could have a term with index (i,j) and another one (j,i), which would result in counting a pair twice.
Also note that in the Pi-product it would be obligatory, because if any two indices are the same then the entire wave function becomes zero.
Anyway, that's the context where I saw a similar notation, maybe I'm completely off.
[QUOTE=Fourier;46397974]When I wondered, how long mathematicians lived, I checked and looks like they lived quite long. Maybe is it attributed to constant use of the brains, thus being "active"?[/QUOTE]
the guy who did a lot of work on infinity spent most of his adult life in a mental institution
[url]http://en.wikipedia.org/wiki/Georg_Cantor[/url]
mathematicians didn't all live long lives however,
[quote]Évariste Galois (French: [evaʁist ɡaˈlwa]) (25 October 1811 – 31 May 1832) was a French mathematician born in Bourg-la-Reine. While still in his teens, he was able to determine a necessary and sufficient condition for a polynomial to be solvable by radicals, thereby solving a 350 years-standing problem. His work laid the foundations for Galois theory and group theory, two major branches of abstract algebra, and the subfield of Galois connections. He was the first to use the word "group" (French: groupe) as a technical term in mathematics to represent a group of permutations. A radical Republican during the monarchy of Louis Philippe in France, he died from wounds suffered in a given gun duel under questionable circumstances related to Stéphanie-Félicie Poterin du Motel[1] at the age of 20.[/quote]
[editline]5th November 2014[/editline]
ah crap beaten to the punch
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