Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
On that note, sin(x) and cos(x) have a quite clear meaning when the parameter is a real number. However, when the parameter is a complex number, do they have a clear meaning or interpretation? What are they representing?
Player (42)
Joined: 12/27/2008
Posts: 873
Location: Germany
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.
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
In one of his latest videos, blackpenredpen states that you can't use L'Hopital's rule in order to solve lim(x->0) sin(x)/x because "that would be circular reasoning". I didn't really understand why.
Editor, Expert player (2080)
Joined: 6/15/2005
Posts: 3284
Warp wrote:
In one of his latest videos, blackpenredpen states that you can't use L'Hopital's rule in order to solve lim(x->0) sin(x)/x because "that would be circular reasoning". I didn't really understand why.
If all you want is the answer to "What is lim(x->0) sin(x)/x?", then sure, there is nothing whatsoever that stops you from using L'Hopital's rule to evaluate it. The reason he talks about "circular reasoning" is because when someone brings up lim(x->0) sin(x)/x, the question isn't so much "What is lim(x->0) sin(x)/x?" so much as "Why exactly is lim(x->0) sin(x)/x=1?" If you bring up L'Hopital's rule, the implied question then becomes "Why exactly is the derivative of sin(x) equal to cos(x)?". And to prove that the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x), you first need to show that lim(x->0) sin(x)/x=1 and lim(x->0) (cos(x)-1)/x=0, without using L'Hopital's rule.
Player (42)
Joined: 12/27/2008
Posts: 873
Location: Germany
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!
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
p4wn3r wrote:
Pretty much every textbook I've seen either just states it or uses completely circular arguments to prove it.
Could you give an example of such a circular argument used in textbooks, out of curiosity?
Player (42)
Joined: 12/27/2008
Posts: 873
Location: Germany
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
Player (36)
Joined: 9/11/2004
Posts: 2631
A simple way to prove that a triangle's internal angles are 180 degrees to simply use the fact that the external angles sum to 360 around any closed convex polyhedra, and the internal angle + the external angle is 180, and there's three such angles, 3*180 - 360 = 180. As far as I know, this doesn't use parallel lines at all. The sum of external angles can be proven by observing that in order to walk around a polyhedra you need to end up facing the same direction you began in. And the external angle is the change in heading at each corner. Whereas the fact that an internal angle + an external angle comes from the definition of an external angle directly. Actually, upon reflection it does, because it assumes that the heading direction does not change between beginning a side and ending it, which is rather antithetical to the notion of a straight line, but proving it requires constructing parallel lines. I'm leaving it just in case someone else had this idea, to help explain why it's wrong.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
OmnipotentEntity wrote:
the internal angle + the external angle is 180
Isn't the internal angle + the external angle at a vertex 360 degrees?
arflech
He/Him
Joined: 5/3/2008
Posts: 1120
i imgur com/QiCaaH8 png
Player (42)
Joined: 12/27/2008
Posts: 873
Location: Germany
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?
Player (36)
Joined: 9/11/2004
Posts: 2631
This is just a F/A analysis right? If then you just need to solve: 500 = ((1 + i)^N - 1)/i where i = 0.01, for N. Which gives 180.07 months, or 15 years.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Player (42)
Joined: 12/27/2008
Posts: 873
Location: Germany
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!
Editor, Expert player (2080)
Joined: 6/15/2005
Posts: 3284
p4wn3r wrote:
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
If you make n deposits of value A with interest rate x, then the value in your account will be: which simplifies to (*) Setting equal to C (the goal) and solving for n, we get: (**) For the example that you gave, plugging in C=1000000, A=2000, x=0.01 gives approximately 180.07 months, so it takes 181 months (~15.08 years) to reach the goal. But that's not your actual question. The actual question is to prove that the formula tends to C/A when x->0. We could use L'Hopital's rule on (**). Or instead we could just see that (*) approaches An as x->0. This is because ((1+x)^n - 1) / (1+x - 1) is precisely the value of the derivative of z^n at z=1, by definition. This gives An=C, so it follows that n=C/A in the case x->0.
OmnipotentEntity wrote:
This is just a F/A analysis right?
What is F/A analysis? Never heard of it before.
Player (42)
Joined: 12/27/2008
Posts: 873
Location: Germany
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...
Player (36)
Joined: 9/11/2004
Posts: 2631
FractalFusion wrote:
What is F/A analysis? Never heard of it before.
Took a class called "Engineering Economy." It had financial formulas. That one was called F/A, which as far as I've been able to gather stands for "Future Value given Annualized Payment."
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Editor, Skilled player (1348)
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
I'm still interested in the problem JXQ suggested here over 1 year ago. You have a d-dimensional cube of side n divided into side 1 cubes. That is, a "nxnxnx"... cube. The question is how many lines do you need to 'pierce' every one of the n^d small cubes. Each line can pierce at most d(n-1) + 1 cubes. So, we easily know that the amount of lines is at least the ceiling of (n^d)/(d(n-1)+1). When n > 2, it's easy to find examples where the minimum amount of lines is bigger than that. That's because in order to a line to pierce all d(n-1) + 1 cubes it has to pass by some 'central' cubes, thus if every line pierced this amount of cubes there would be a lot of overlap. When n = 2, though, this is not a problem, because there are no 'central' cubes. I conjectures ceil[(2^d)/(d+1)] is a sufficient amount of lines to pierce every small cube of a d-dimensional side 2 cube. I've verified it up to d = 6. d = 4 d = 5 d = 6 Cubes with the same color are pierced by the same line. I think it's easy to understand how I set up the dimensions in those drawings. Now d = 7 is a very interesting case because (2^d)/(d+1) = 16 is an integer. So if we want a 16-line solution, every line needs to pass by 8 cubes, and no overlap is allowed. This is a potential counter example to what I conjectured, but I still feel it's possible. The challenge can also be thought as creating 16 sets of length-7 binary sequences. Each set must have 8 sequences and 2 consecutive sequences inside a set can only differ at 1 position. The goal is to cover all 128 possible binary sequences with our 16*8 = also 128 sequences. Edit: I forgot to mention something in the reformulation of the problem - inside each set, the digit each pair of sequences differs on must be different. That is, a valid set of sequences would be {0000000, 0010000, 0010001, 0010101, 0110101, 1110101, 1110111, 1111111}.
My YouTube channel: https://www.youtube.com/channel/UCVoUfT49xN9TU-gDMHv57sw Projects: SMW 96 exit. SDW any%, with Amaraticando. SMA2 SMW small only Kaizo Mario World 3
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
In the latest Numberphile video the guest needs to shuffle a deck of cards, but in such a manner that not only is their order randomized, but also a random amount of them is face up and the rest face down. This got me thinking how that could be done with a physical deck of cards, to assure as high-quality randomness as possible. There exist very good techniques to just shuffle the deck with all cards face down (like the techniques used by professional poker dealers), but what if you indeed had to have a random amount of cards in the deck, chosen at random, be face up? (In the video in question the suit and number of the cards didn't matter for what she was trying to demonstrate, but let's assume that in our scenario we want them randomized as well.) I suppose that one way would be to shuffle the deck with all cards eg. face down, then using some external method get a fair random number between 0 and 52, and turn that amount of cards face up, and then shuffle again. Of course this would require an external way of getting an evenly-distributed random number between 0 and 52. What if we don't want to require such a thing? Can it be done by physically manipulating the cards alone, without external devices to give us random numbers? (And obviously we want the number of cards that are turned face up to be as evenly random as possible. For this we cannot just rely on the human selecting a random amount of cards. At least not directly. Humans are notoriously bad at generating evenly-distributed random numbers.)
Player (36)
Joined: 9/11/2004
Posts: 2631
Wouldn't just shuffling the standard way, but flip one half of the deck each time, work?
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Patashu
He/Him
Joined: 10/2/2005
Posts: 4045
The initial question doesn't specify a specific probability distribution that the random amount of face up/face down cards should satisfy, does that matter or is any distribution ok?
My Chiptune music, made in Famitracker: http://soundcloud.com/patashu My twitch. I stream mostly shmups & rhythm games http://twitch.tv/patashu My youtube, again shmups and rhythm games and misc stuff: http://youtube.com/user/patashu
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
OmnipotentEntity wrote:
Wouldn't just shuffling the standard way, but flip one half of the deck each time, work?
Then you would get half of the cards face up, not a random amount of them.
Masterjun
He/Him
Site Developer, Expert player (2047)
Joined: 10/12/2010
Posts: 1185
Location: Germany
Can't you just shuffle normally, but then afterwards drop each card from a sufficiently high elevation? Instead of flipping a coin you're flipping the cards. Every configuration of cards would have the same probability. This means that having 26 face up cards out of 52 is much more likely than having 0 face up cards. Just like how rolling two dice will give you a sum of 7 more often than 2. If you want an even distribution on the number of cards being face up, the easiest way would probably be to add a 53th card, shuffle the deck (all face down), use the extra card to divide the deck in a top and a bottom half (and remove the extra card), turn around the top half, and then shuffle again. Edit: If you don't have a 53th card, then of course you can just use the randomness of a shuffled deck, for example: Use eight cards from 1 to 8 (1 being Ace), shuffle, take a card, subtract 1 then multiply by 8. Reshuffle the eight cards, subtract 1 and then simply add the number to the previous result. This gives you an equal distribution from 0 to 63. If you get a number above 52 just redo the process. The rest should be trivial.
Warning: Might glitch to credits I will finish this ACE soon as possible (or will I?)
Player (36)
Joined: 9/11/2004
Posts: 2631
Warp wrote:
OmnipotentEntity wrote:
Wouldn't just shuffling the standard way, but flip one half of the deck each time, work?
Then you would get half of the cards face up, not a random amount of them.
Mathematically, yes. However practically, there is some variation between the deck halves, and the deck halves may not actually be halves. So you'll get a normal distribution of face up and face down cards, where each card has a 50/50 chance of being face up (but just because each probability is 50/50 doesn't mean you'll get exactly 26 face up cards.) Were you looking for a uniform distribution on the number of face up cards? You didn't specify.
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
OmnipotentEntity wrote:
Were you looking for a uniform distribution on the number of face up cards? You didn't specify.
Well, I did talk about "an evenly-distributed random number between 0 and 52"
Player (36)
Joined: 9/11/2004
Posts: 2631
Warp wrote:
OmnipotentEntity wrote:
Were you looking for a uniform distribution on the number of face up cards? You didn't specify.
Well, I did talk about "an evenly-distributed random number between 0 and 52"
Ah, so you did. I apologize for missing that. Well, you can generate a random number between 1 and 64 using two distinguishable d8. Then if the number is between 1 and 52 flip that many cards over and shuffle normally. (If not regenerate the random number.) (EDIT: This seems to be equivalent to what Masterjun wrote.)
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.