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OmnipotentEntity wrote:
EDIT: I figured out my mistake. I didn't substitute correctly and I dropped a critical factor of 1/2 in equation 14. now my result agrees with yours.
FWIW, here's my solution. I essentially kept playing on Wolfram Alpha until I found an identity involving the polynomials (an+bn+cn). The identity I found is:
(a+b+c)⁴ = 6(a⁴+b⁴+c⁴) - 8(a³+b³+c³)(a+b+c) + 6(a²+b²+c²)(a+b+c)² - 3(a²+b²+c²)²
The proof can be found visiting here.
Now, simply substitute everything:
1 = 6x - 8*3 + 6*2 - 3*4 => 6x = 25 => x = 25/6
After doing this I wonder if there is some closed formula for identities like this involving higher powers, or more terms. But this discussion would make this post much longer, so it's better to leave it out.
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I did not solve it by writing the polynomial equation, so I cannot review your method.
But I am sure that the answer (which I got in many different ways) is 25/6, or in decimals 4.16666... (which is incidentally approximately 4, which is what the sequence would suggest :D)
So, I think you should see if you did not miss something...
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Here's a relative simple problem I came up with today. It does involve a bit of calculation, though.
Let a, b, c be three numbers satisfying
a + b + c = 1
a² + b² + c² = 2
a³ + b³ + c³ = 3
What is the value of a⁴+b⁴+c⁴?
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FractalFusion wrote:
Riddler Express this week sounds interesting:
You purchase a new clock but are dismayed to realize that both of its hands are identical. At first, it seems it’s going to be impossible to tell the time because you don’t know which hand is for the minutes and which is for the hours.
However, you realize you don’t need to know which is which for every time — for example, when it’s 12:30, the minute hand will be exactly on the 6 and the hour hand will be halfway between the 12 and the 1. It can’t be the other way around because if the hour hand were exactly on 6, the minute hand would have to exactly on 12, which it’s not. So you know what time it is.
How many times during the day will you not be able to tell the time?
I found a very elegant solution for this problem using complex numbers!
If you measure the angle of each hand from the 12 marker, making it positive when it turns clockwise, the angular velocity of the minutes pointer is 12 times larger than the angular velocity of the hours pointer.
Because of that, the angle of the minutes hand will be (up to possible multiples of 2pi) 12 times the angle of the hours hand. It's impossible to tell the time if the angle of the hours hand also turns out to be 12 times the angle of the minutes hand, excluding, of course, the places where the pointers are at the same angle.
So, given two angles ah and am for hours and minutes, respectively, you can define two complex numbers:
zh=exp(i*ah)
zm=exp(i*am)
Now, we will always have zm = (zh)12.
At an ambiguous time, we'll have zh = (zm)12
That's the same as saying (zh)144=zh or (zh)143 = 1
That happens when the hours angle is a multiple of 2pi/143, so in a twelve hour interval, that condition will happen 143 times. Now, we need to look at where in these places the two pointers will be at the same angle, which gives just (zh)12=zh or (zh)11 = 1.
Since 143/11 = 13, we want only 12/13 of the 143 times. So, in a twelve hour period, we have 12/13*143 = 12*11 = 132, and of course 2*132=264 during the day.
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Ferret Warlord wrote:
For the most part, if we allow the range to extend into complex numbers, is the absolute value thing even necessary?
The short answer is, you can't extend it to complex numbers.
The long answer is, integration assumes some properties about the topology of the space you are working on, and the topology of the space in R is totally different from that of R² and that of the complex numbers C.
For example, in R², a space can be path connected and not simply connected, while in R that's impossible (all path connected spaces in R are intervals, which are always simply connected). There are other differences, too, like a function with arguments in R that has a derivative is always differentiable, while it's possible for a function with domain in R² to have derivatives in all directions and not be differentiable.
In the space of complex numbers, a function having a derivative (with respect to complex numbers) is a notion even stronger than differentiability. For example, if a complex function is differentiable in a simply connected space, any integral around a closed loop gives exactly zero!
So, as a rule you cannot simply assume that what you found for an ODE will extend to a complex-valued function. In the case we were discussing, involving logarithms, it gets even uglier because the complex logarithm has branch cuts, and it's not simple to do analysis there.
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FractalFusion wrote:
(Technically the ln functions should all have absolute value signs, though it doesn't matter for ln(cos) (since -cos is just a shifted cos) and ln(cosh) (since cosh>=1).)
To be honest, I've never been a fan of using absolute value in logs, it's standard practice, so I do it, but I think it gives a wrong idea of what's happening.
When you have an ODE like xy' = 1, for example, what's really happening is that at x=0 you have a singular point, and you actually have two possible solutions, ln x, for x>0 and ln(-x) for x<0. This does not violate the uniqueness theorem for ODE's, so there's no problem in that.
This is the same problem you pointed out at the missing solution, thanks for bringing it up. The function being integrated, 1/(x²-b²) is singular at x = +-b, so you should expect two different solutions getting separated at x=b and x=-b.
When people say that the solution of xy' = 1 is y = ln |x|, x != 0, they actually give the impression that you can solve an ODE and find a function that's defined on a set of points that's not an interval, which is something that makes no sense, in light of all the theorems. So, yeah, although I use absolute values all the time, I'd be happier if people wrote the functions, with their correct intervals :)
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Ferret Warlord wrote:
I am looking for two functions, f(x) and g(x), such that
f2 - g2 = 1
(f')2 + (g')2 = 1
All I can figure is that lim f' = lim g' = √(2)/2 as x-> inf, but that's only on an intuitive level, nothing rigorous.
If it helps, f(0) = 1, g(0) = 0
Make the change of variables: f(x) = cosh(u(x)), g(x) = sinh(u(x))
This way, the first equation is automatically satisfied. Now, for the second:
(sinh² u + cosh² u)(u')² = 1
Now, if you use hyperbolic identities, you find that the first factor is just cosh(2u)
Taking the square root and separating:
sqrt(cosh 2u)du = dx
So, if you could integrate sqrt(cosh 2u), you could invert this equation and write the answer. The problem is that the function is not elementary, so you'd need to use special functions. If you just want the asymptotic behavior at infinity, it's simpler. Remember that:
cosh 2u = (exp(2u) + exp(-2u))/2
Now, if u is very large, the exp(2u) term dominates, so we can approximate the equation:
exp(u)du = sqrt(2)dx => exp(u) = sqrt(2)x
Notice that I dropped the integration constant because it's negligible in the limit where both u and x go to infinity.
Now, we remember our expressions for f and g:
f = cosh(u(x)), g = sinh(u(x))
But notice that, as u->infty, we have cosh(u) ~ sinh(u) ~ exp(u), so
f ~ g ~ exp(u)/2 = x*sqrt(2)/2
So, we see that, for large x, the asymptotic expansion of f and g is just x*sqrt(2)/2. If you derive it, it's clear that we should have f' ~ g' ~ sqrt(2)/2
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Warp wrote:
Time for an actual challenge for a change.
In his last couple of videos, the youtuber blackpenredpen has tackled the question of finding a non-zero function y such that
y'⋅y'' = y'''
(iow. the derivative of y times the double-derivative of y is equal to the triple-derivative of y.)
It seemed surprisingly non-trivial.
There's a trick to solve ODE's like this. First, call f the derivative of y. You have
f⋅f' = f''
Now, let us call the argument of the function x. It's a second order ODE, but since it does not depend explicitly on x, you can always make a change of variables that will convert it to first order.
Set u = df/dx. Apply the chain-rule and get f'' = du/dx = du/df * df/dx = u*du/df
Now plug everything:
uf = u du/df => f df = du => f² + c = 2u
Now substitute u and separate:
dx/2 = df/(f²+c)
Now you integrate. I'll assume c is positive. If c is negative, I'll just state what needs to be changed later. Make c = b². Then:
x/2 = arctan(f/b)/b - k/b
f = b*tan(bx/2 + k)
Finally, y is simply the integral of f. The integral of tan(x) is just -ln(cos(x)), so:
y = -2 * ln(cos(bx/2+k)) + a
That has three free parameters (as expected from a third order ODE).
To include the solution y = a, notice that if you take b = 0, then f=0, and y = a.
Now, if you want to check, knowing that the derivative of tan(x) is sec²(x), and that the derivative of sec(x) is sec(x)tan(x):
y' = b*tan(bx/2+k),
y'' = b²/2 * sec²(bx/2+k),
y''' = b³/2 * sec²(bx/2+k)tan(bx/2+k)
So, if you multiply the first two you do get the third one, as expected.
If c is negative, you have to do c = -b², and since the integral of 1/(x²-b²) is -arctanh(x/b)/b, you get:
f = -b tanh(bx/2+k),
and
y = -2 * ln(cosh(bx/2+k)) + a
You can again check the solution, knowing that the derivative of tanh(x) is sech²(x) and that of sech(x) is -sech(x)tanh(x). The derivative of the hyperbolic trigonometric functions are always the analogues of the normal trigonometric ones, with the difference of some minus signs, and I always have to look up a table to know which sign it is xD. Anyway:
y' = -b*tanh(bx/2+k),
y'' = -b²/2 * sech²(bx/2+k),
y''' = b³/2 * sech²(bx/2+k)tanh(bx/2+k)
So, again, if you multiply the first two you get the third. These, with the constant function y = a, are all possible solutions.
EDIT: Spoke too soon! The equation at the beginning:
uf = u du/df,
also has a solution if u = 0!
That means y'' = 0, which has the solution
y(x) = ax + b,
which is another possible family of solutions. In this case, the second and third derivatives vanish, and the differential equation is, thus, always satisfied.
EDIT 2:
And, I missed another. If c = 0, we have:
dx/2 = df/f² => x/2 = -1/f - k/2 => f = -2/(x+k) => y = -2*ln |x+k| + a. To verify:
y' = -2/(x+k)
y'' = 2/(x+k)²
y''' = -4/(x+k)³
Multiplying the first two you get the third. So, the entire set of solutions should be:
y = -2 * ln(cos(bx/2+k)) + a
y = -2 * ln(cosh(bx/2+k)) + a
y = -2*ln |x+k| + a
y = ax + b
It's the first time I've seen so many different solutions to an ODE!
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Easy. Arrange all cards from 1 to 9 in a 3x3 magic square. There is exactly one 3x3 magic square (excluding reflections and rotations). You pick three cards that add to 15 iff you fill up a row, column, or diagonal of the magic square. So, the game in question is simply tic-tac-toe. Since with perfect play, tic-tac-toe is a draw, this game is a draw also.
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Warp wrote:
p4wn3r wrote:
The problem is that not everyone trusts charitable organizations.
Some people, including some speedrunners, have presented criticism of the PCF in particular. But perhaps this is not the place to discuss that...
I have taken a look at it in the past and, for what it's worth, I have seen nothing concrete to conclude PCF does not do what it states. Their finances look typical for a charity that works with research, which usually has large employee expenses.
The issue I have is that oncology is perhaps the wealthiest sector in the research industry, and also one that does not have a good reputation. If you look at RetractionWatch, for example, you'll see that every two months a prominent oncologist gets exposed for falsifying research. Remember that dude that got arrested some time ago, Martin Shkreli? He made millions of dollars in the stock market betting that new drug companies would fail, because it's common practice to hype up the research and the company not getting a viable product and failing three years after being launched.
So, yeah, I'm 99% sure that PCF does exactly what it states in its website. If you have a good reason to donate to them, well, that's up to the person. Like any grant, though, their foundation will retain the right to the inventions if the researchers do their job correctly, and they just get university professors to evaluate proposals. By simply donating to a university, you'd get the same thing while skipping their board of directors. And by investing on a company, for example, you at least have a chance of getting the money back.
Now, please don't take from this that I dislike non-profits. Disclosure: I am actually developing an app for one, and I am charging less than I would for a for-profit company, and I think it's a very nice project. But during the meetings, everything I do is very similar to a typical company, these foundations have many positions that are all about money, and people should take all of this into account when supporting a non-profit.
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Hello, dwango. While I understand your personal and religious motivations to charitable contributions, the feedback I wish to give you is the following:
1) Your life comes first. No matter what your goals with TASbot are, it's safe to say that if you get steamrolled by debt you certainly won't be able to accomplish them. Do not be afraid to cut down costs, even if it results in a sub-optimal experience, if your life is requiring you to do it. The community will understand.
2) I have to agree with keylie that you are putting yourself in an unnecessary bind by stating that you'll place 10% at a charity. The problem is that not everyone trusts charitable organizations. Most are tax exempt and some receive millions in government contracts, there are many expenses in employees and fundraising activities which have little to do with their stated purpose, and many have endowment funds where they actually can invest the money you give to them in a random company before finally spending it on their real activities. The more detailed you are with the destination of the donations, the more likely you are to receive the money. So, use the money to fund your expenses. If people want to give it to GDQ, encourage them to donate during the event.
3) Consider seeking sponsorships or asking GDQ to fund your costs. You are, after all, working for them. Maybe this is related to what I said in (2), but GDQ being a charitable event does not mean that there are a lot of people there making money. Twitch is sponsoring hoping to get publicity, the organizers and the hotel hosting are also charging fees. It's simply fair that your costs get compensated by someone, since you are providing content to them.
4) Finally, if your vision is more long-term (50+ years), and you are willing to keep doing this, maybe it's a good idea to make a foundation yourself to ask for donations? I have the feeling that once you formalize in an institution everything you told us here, it will be easier to get people to contribute and join, especially if more issues happen to you and you need to enlist other people to help.
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Nach wrote:
I wouldn't call releasing a lot of encodes in a single day by several encoders/publishers an "egregious cases of abuse" . Common sense would tell me there's no problem at all. Encoders/publishers are doing what they should be doing. The outcome as described in the staff thread was probably a surprise to all involved, and possibly even unintended by those among them who had malicious intent.
You are consistently missing the point and, given how much you do this, I cannot imagine that this is not intentional on your part.
The assertion that the release of a lot of encodes in a single day was abuse is not even an inference from the evidence. The person in question admitted to it. Every property has its social purpose. If the rules of your site do not say publishers should not do this, the problem is with the rules, not with the people who find this behavior disturbing.
Certainly, there might have been occasions where publishers uploaded too much and annoyed users. Very nice that you curbed that, but the suggestion that accidentally uploading too much is somehow similar to flooding the channel with the admitted intention of reducing a particular movie's popularity is not only silly, but offensive to your publishers who had never done such thing before.
And, also, doubling down on something is never a reasonable solution. If you double down on things that are doomed to failure, all that achieves is giving the illusion of gains for a short while with catastrophic losses occurring eventually. Doubling down when things go wrong only increases the problems.
If you don't understand this, and by the history of these forums, I actually think that you don't understand, it's starting to look reasonable to doubt whether you are in a position to decide what is "threaded gruefood" at all!
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Memory wrote:
Now while this may not seem directly related to the comments at first glance, the fact is that Spikestuff talks about TASVideos drama to so many people and makes a spectacle of it. Because he does this, it's easily possible for somebody else to have jumped in on whatever he was making fun of at the time and created one of the comments previously listed.
I really tried to keep away from this stuff, but unfortunately this is just silly. So you're suggesting that, two years ago, someone who does not like Spikestuff created lots of sockpuppet accounts with the intention of following him everywhere agreeing with everything that he says and attacking people he doesn't like with the intention of, two years later, getting him falsely accused of creating sockpuppets so that he could be demoted? Seriously?
Also, by the way, in the thread of the movie that sparked all this, I suggested that making votes in submission threads non-anonymous would help identify abuse, which was answered by complaining about me and grue'ing the posts. Interestingly, we're now discussing the possibility of YouTube comments coming from fake accounts exactly because youtube comments are not anonymous!
I understand that you might be upset about confidential information leaking, but the problems discussed in the OP are far worse, and mentioning minor leaks only dilutes the issue. I am sorry, but I don't feel the rest of the staff are handling this matter properly. By the way, if Bruno contacted you, why are we seeing this posted in the public forums before something was concluded internally anyway? It seems to me you are being far too conivent and damaging the credibility of the site.
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Nice! Another way to find the limit for small is to look at the Taylor series for log(1+x). If x-> 0, you can use only the linear term and ln(1+x)~x
Anyway, the reason it takes much less time than expected is because of the power of exponential growth. The exponential function grows really fast, but to see its power you have to use time. We can see from the formula. If we call n=C/A, the formula has ln(1+nx), so to make use of exponential growth you want nx to be large enough so that you cannot approximate the logarithm, or else you have linear growth.
The Taylor series for ln(1+nx) stops converging for nx = 1. For typical monthly returns of 1% , it only makes sense if the amount of deposits you make are of the order of 100. Few people have the discipline to do that, that's why they don't see the returns.
The S&P 500 usually grows 11% per year. By reinvesting dividends and renting ETF shares for short sellers it's usually possible to get 1% monthly. People get put-off because the stock market is volatile and they fear losing money. However, if you look at time scales of around 3-5 years the stock market grows remarkably stable.
It's because of formulas like this that I think there's little sense I'm trying to accumulate money with wages. At best, if you don't get fired and are promoted regularly, your salary will grow linearly with time, which means your capital increases quadratically, which is a lot slower than exponentially.
Instead, what people often do is take mortgage loans. Here the interest in these loans is around 1.5 % monthly. The difference now is that the exponential growth is working against them, and the natural consequence is that they unconsciouslyrics work the entire life to pay the bank...
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In a math test that would be perfect. However, I think most people in the forum were not trained in these methods. Try to deduce/explain the formula with high school math. Discuss some limits of it and explain the puzzling thing. With no gains per month it would be 500 months, almost 42 years. How does 1% make it 15 years, almost one third of the time? I think it's nice to discuss this!
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Finance time!
Suppose you have an amount of capital C you want to reach. Your plan is to do so by investing an amount A periodically. Suppose that, by the time you have to deposit this amount A, the capital you already had increased by x. That means, if you had M in your account, after a time unit passes, you'll have M*(1+x). Find the amount of deposits n of value A in order to reach the goal C. Prove that the formula tends to C/A when x->0
In a situation closer to real life. Suppose you want to reach 1 million dollars by investing $2000 every month. Suppose the capital invested increases by 1% every month. After approximately how much time will you become a millionaire?
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There are two ways to derive E=mc2, or the more general E=gamma*mc2, where gamma, the Lorentz factor, becomes 1 if the particle is at rest, v=0. The one that was used by Einstein, and is actually what's used to test relativity everyday in particle colliders, is to look at what happens in collisions.
It turns out that if you require that a theory conserves energy and momentum, and analyze the result of a collision from different frames where space and time transform according to the postulates of relativity, you find that the only consistent choice of energy is E=gamma*mc2 plus an unobservable constant.
In particular, if the masses of the particles are different in initial and final states (a particle "becomes" another with different mass), the mass energy converts to kinetic energy and vice-versa.
Actually, you are right that we can measure energy in terms of mass. If you include quantum mechanics, you can also measure length and time in terms of energy. This is done in particle physics all the time.
An example: if you want to produce a Higgs boson, with mass of 126 GeV by colliding an electron and a positron, their kinetic energies should be at least 63 GeV (you can neglect the electron mass energy because it's much smaller than that) so that the sum can produce the Higgs at rest.
The second way to deduce is more mathematical but you can get the answer with less calculation. One way to derive physical theories is to use the principle of least action, which states a particle's path should minimize a given quantity. It turns out that, from the postulates of relativity, you can deduce that a free particle maximizes its proper time when moving.
Then, using techniques from analytical mechanics you find, using a few derivatives that, a particle that maximizes proper time necessarily has an energy of gamma*mc2.
Finally, the c that appears in the formula is the velocity that all inertial frames agree. That means, if an object is traveling at c in a given inertial frame, it necessarily travels at the same speed at all other inertial frames. That uniquely fixes c, so the velocity in the formula cannot be any value.
Of course, nothing stops you from creating a theory where c is replaced by the velocity of sound, for example. You would get a consistent theory where energy is proportional to the square of the speed of sound. However, you would get bizarre results which are not observed experimentally, like massive things unable to break the speed of sound.
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Here's an example from Khan Academy: https://www.khanacademy.org/math/geometry-home/geometry-angles/geometry-angles-between-lines/v/proof-corresponding-angle-equivalence-implies-parallel-lines
In the video, the "proof" given is that if you suppose the lines are not parallel, you get a triangle and the sum of its angles being 180 degrees causes a contradiction.
The problem is that to prove the sum of internal angles in a triangle is 180 degrees you have to draw parallel lines, show that some of them are equal using the corresponding angle theorem and see they sum to 180. So, the argument used is completely circular.
Euclid knew this, that's why he computes the sum of internal angles much later in the Elements. Perhaps we should start teaching geometry using Stoicheia again? Preferably in greek so that people don't introduce errors translating? :P
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Since we're talking about circularity of arguments, here's something that I see that almost every school textbook gets wrong. More specifically, it's the corresponding angle theorem, which states that the corresponding angles in two parallel lines cut by a straight line.
Pretty much every textbook I've seen either just states it or uses completely circular arguments to prove it. In fact, the only rigorous treatment I've seen of it is in Euclid's Elements!
Of course, proofs in geometry can be quite different depending on the postulates you use, but it's still a "hard" proof that I think textbooks should not omit, because it's a key point in Euclidean geometry.
In any case, I found Euclid's proof ingenious. First, he proves that an exterior angle of a triangle is always greater than the two other internal ones. Then, he uses that to show that the sum of two angles of a triangle is always less than 180 degrees.
Then, he uses this fact to demonstrate that when corresponding angles are equal, the lines are parallel, it's remarkable that this is true even without the parallel postulate!
The corresponding angle theorem is the converse of this statement, and Euclid's parallel postulate is written in a form very convenient to prove this theorem, which Euclid then uses to derive a proof by contradiction.
I think this is actually another case of textbooks handwaiving some things and leaving interesting mathematics behind, in this case the necessity of the parallel postulate to prove fundamental statements in geometry, something that was known very well by the Greeks!
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It's very difficult to answer questions like "What's the meaning of this entity?" objectively. The issue is that this question boils down to what you consider fundamental and what are derived statements in mathematics, and there are many ways of exchanging fundamental and non-fundamental roles so that math remains equivalent, and it all reduces to personal opinion.
If what you consider fundamental is geometry, then certainly, sine and cosine of real numbers have a much clearer meaning. For the complex function, if the argument is in the imaginary axis, it's related to the hyperbolic sine and cosine, and I think it should possible to come up with more complicated geometric constructions to represent arguments with both real and imaginary parts.
If analysis is more fundamental, though, there's little difference. You can argue that the "true" functions are the complex ones and you are just limiting yourself to the reals because geometry is more familiar.
Although rigorously, it can be questioned why being analytic is so important. One property of the sine is that it's periodic, but the analytic continuation is not periodic in the imaginary axis like it is in the real one. You can come up with complex functions that are periodic in both axes, but they will always have singularities at some points. It turns out that the only analytic function that's periodic in areas of the complex plane is the constant function.
That's a very important lesson when you want to generalize something. You cannot keep all the properties of the thing you're trying to generalize. The only function with all the properties of sin(x) for real x is, of course, sin(x) for real x. The very idea of generalizing something means that you're taking one property as more important than the others and maintaining it, while not bothering if the rest holds or not.
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The principle behind Euler's formula making sense for complex arguments is analytic continuation, you can look it up to understand it.
The main idea is: there's no a priori answer to an expression like sin(i). There is an infinite amount of functions from the complex numbers to the complex numbers that match sin(x) in the real line. To see this, just pick any function that matches sin(x) on the reals and add one that is 0 for every real number but does not vanish in imaginary ones. These two functions will be different, but both are extentions of sin(x) to the complex plane.
However, if you want this extension to be differentiable, then it's a completely different story. Calculus with complex numbers is much more different than with the reals. While it's completely possible that a real function is differentiable once but not twice or more, in the complex plane if it has only the first derivative, that means it will have infinitely many derivatives and that its Taylor series will converge for some small enough radius.
The consequence of this is that requiring a function to be complex differentiable is very restrictive. In particular, it implies that there's only one function that extends sin(x) and is differentiable (or analytic, which in complex numbers is the same thing). To prove Euler's formula you only need to look at the Taylor series of e^x, sin(x) and cos(x). If you treat this series as a sum of complex numbers instead of reals, you will see that it converges just as well, and the function you obtain from it is guaranteed to be the analytic continuation of them, so you can safely interpret Euler's formula as a relation between complex numbers.
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FractalFusion wrote:
So it does turn out that there is a nice factorization of the sextic into cubics in general, even though it's not over the integers (except when 4a-7 is a square).
By the way, the (a-2-a sqrt(4a-7)) should be over 2; that is, (a-2-a sqrt(4a-7))/2.
Corrected. Thanks for taking a look!
FractalFusion wrote:
p4wn3r wrote:
Therefore, the Galois group should be generated by two cyclic Z3 groups, and another Z2 group that switches 3 roots from one cycle to the other.
This is probably beside the point, but finding Galois groups is quite complicated and I can't be certain that the Galois group has to be Z3×Z3×Z2 (unless there's something I'm missing). Yes it's true that, if we look at the problem purely as a permutation of 6 roots (that is, the maximal subgroup of S6 satisfying the cycle conditions), then it is Z3×Z3×Z2. But Galois groups have additional properties because they describe automorphisms over Q.
Perhaps I should have been more accurate and have said that the Galois group is a subgroup of Z2 x Z3 x Z3. I know that the root permutation approach has many limitations, because it ignores the subtleties of the field being extended. Another case where Z2 would not appear is if a was in a quadratic number field instead of a rational, it's possible that sqrt(4a-7) is denested, so does not generate a new field. The point is that, taking these subtleties into account does not help to understand why the factorization is possible.
If you want to understand why you should try to factor the sextic in this problem and not a general one, the reason is that, the problem is formulated in a way that lets you see that in the worst case the Galois group is abelian, and thus solvable, unlike a general sextic.
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I used a different sign convention to solve it, but I could factor it in the general case. The system is
x^2 = y + a, y^2 = z + a, z^2 = x + a
We have x = ((x^2-a)^2-a)^2-a or
x^8 - 4ax^6 + 2(3a^2-a)x^4 - 4(a^3-a^2)x^2 - x + a^4 - 2a^3 + a^2 - a = 0
Now, as Fractal said, because of the ambiguity of the square root sign, the roots of the equation are related to all possible patterns of signs in the infinite nested radical. If x=y=z then this reduces to the quadratic x^2-x-a = 0, which should factor the previous polynomial. The quotient is
x^6 + x^5 + (1-3a)x^4 + (1-2a)x^3 + (1-3a+3a^2)x^2 + (1-2a+a^2)x + 1 - a + 2a^2 - a^3 = 0
Now, the roots of this are part of a "limit cycle" corresponding to the -+- and +-+ patterns. If we're going to use field theory, we can look at the Galois group by seeing which changes of the roots preserve the equations of the limit cycle. If x,y,z are in a cycle, then the mappings (x,y,z)->(y,z,x)->(z,x,y) preserve the relations, but arbitrary permutations should not. Therefore, the Galois group should be generated by two cyclic Z3 groups, and another Z2 group that switches 3 roots from one cycle to the other. That means the sextic should factorize into two cubics after solving a quadratic.
(Whenever I look at an identity from Ramanujan, I think I am seeing something from a higher intelligence. It's amazing how he could see the factorization without any formal education in algebra. The only explanation I have for this is that he somehow developed an intuition of Galois theory after solving lots of algebraic equations)
In any case, once you guess the factorization into cubics, it's easier. Set x+y+z=u. Summing the equations and using
(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx),
we find xy + yz + zx = (u^2-u-3a)/2
Now change the equations a bit:
x^2 y = ay + y^2, x^2 z = az + yz
y^2 z = az + z^2, y^2 x = ax + zx
z^2 x = ax + x^2, z^2 y = ay + xy
Sum everything and use
(x+y+z)(xy+yz+zx) = x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y + 3xyz
You should get xyz = (u^3 - 2u^2 - (1-7a)u - 3a)/6
Therefore x,y,z are roots of the cubic
x^3 - ux^2 + x(u^2-u-3a)/2 - (u^3-2u^2-(1-7a)u -3a)/6 = 0
Now write the same thing for u -> v, and when you multiply the two cubics, you should get the sextic! The coefficient of x^5 gives
u+v=-1
The coefficient of x^4 gives a triviality when substituting u+v=-1. The coefficients of x^3 give
uv = 2 - a
Solving the quadratic, the factorization magically works and you get the polynomial
x^3 + x^2(1+sqrt(4a-7))/2 - x(2a+1-sqrt(4a-7))/2 + (a-2-a sqrt(4a-7))/2 = 0
The other is found by flipping the sign of the square root. Now you can solve this by depressing the cubic and matching with a trigonometric relation. This is boring, the root that matters is, when
tan(p) = (2sqrt(4a-7)-1)/3sqrt(3),
x = -(1+sqrt(4a-7))/6 - 2sqrt(4a-sqrt(4a-7))sin(p/3)/3
Now, amazingly, when a = 8, sqrt(4a-7) = 5 and
tan(p) = sqrt(3)!
So, p = pi/3 and
x = -1 -2sqrt(3)sin(pi/9)
With my sign convention, it appears negative. To prove that
x = -sqrt(8-sqrt(8+sqrt(8-...)))
You can use numerics or just plug the solutions in the system, which you can check with trigonometric identities.
(1+2sqrt(3)sin(pi/9))^2 = 8 - (1+2sqrt(3)sin(2pi/9))
(1+2sqrt(3)sin(2pi/9)^2 = 8 + (-1+2sqrt(3)sin(4pi/9))
(-1+2sqrt(3)sin(4pi/9))^2 = 8 - (1+2sqrt(3)sin(pi/9))
Nesting this, you can convince yourself that the pattern is -+-,-+-,... Although there's of course the question of convergence that I didn't bother to check...
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One of my friends sent me a problem that's insane. Prove that
sqrt(8-sqrt(8+sqrt(8-...))) = 1 + 2sqrt(3)sin(pi/9)
The pattern of signs on the infinitely nested radical is -,+,-,-,+,-,-,+,-,...
This problem is from legendary indian mathematician Srinivasa Ramanujan. It takes a lot of calculations, it might be a good idea to use a computer algebra system to help. Even though I had seen a lot of problems like this, it took me a lot of time to see the idea. At first I thought my solution was a brute-force way, but then I saw it was Ramanujan's intended approach.
Hint: The obvious way to deal with infinite radicals is to realize that taking away one period of the iteration does not change the limit. Then:
x = sqrt(8-sqrt(8+sqrt(8-x)))
The problem is that by denesting you get an equation of degree 8, which has no obvious solution. Suppose that the pattern was instead +,+,+,+,+,...
Then it's not necessary to use three radicals, and you can do instead
x = sqrt(a+x)
And you get a quadratic, which obviously has a spurious solution. Try to understand what this spurious solution actually means. Now generalize, consider you have a system:
x^2 = y + a, y^2 = z + a, z^2 = x + a
Apparently this computes sqrt(a+sqrt(a+sqrt(a+...))), but think about the spurious solutions. What would they compute? Can you use that information to factor the degree 8 polynomial that comes up when you denest everything?
Also, look at the value sin(pi/9). It would come up by trisecting pi/3 with a trigonometric relation, which is a cubic equation. Can you make a cubic appear in the octic polynomial?
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Warp wrote:
It appears that even university math professors can sometimes miss the obvious, because it's quite trivial to see that numbers of that form can never be prime.
I can tell you how unsurprising this situation is. There are many "research" departments where all they do is learn and teach people how to run some software packages and publish whatever it spits out without having any idea of what they're doing. They're essentially technicians who specialized in some very specific software, but universities keep hyping them up to get money.
About probable primes, it's actually more sane than other fields. Some of these tests would actually be rigorous if the Riemann hypothesis is proven true, so you can treat some of them as conditional proofs of primality.
There are many other fields where what the program is doing involves lots of approximations and when they get the right answer no one knows if it's because of the theory or the approximation somehow canceled the error. Many times there are no precise instructions on how to repeat the calculation and the paper is nothing more than advertisement.
Some fields are just plain bogus. When someone publishes that their program calculated a value that was observed in an experiment you can be sure he just ran 30 different codes and published the one that worked. There are some pretty dumb cases where some people announced a measurement, then later said it was wrong and put a new value. And meanwhile people always managed to come up with codes that "calculated" both values! That said, I have seen more awareness from the media about these issues and these "junk science" things are getting a hit in funding, which is a good thing.
