Euler's formula is traditionally written as:
eix = cos(x) + i sin(x)
However, due to the cyclic nature of the trigonometric functions, shouldn't it be more precisely:
eix = cos(x + 2πm) + i sin(x + 2πn) , m,n∈ℤ
Why isn't it written like that?
I'm not sure I understand what you mean by "more precisely", both formulas you provided are correct. Certainly, the second formula is more general than the first, since it reproduces the first formula when m=n=0.
There are many ways that you could rewrite the formula using properties of trigonometric functions. For example:
eix = 1 - 2sin2(x/2) + 2i sin(x/2)cos(x/2)
The way you write the formula depends mostly on your intentions in your proof/exposition. If the periodicity of sine and cosine is important, for example if you want to study the analytic properties of eix, by all means make it explicit.
Most of the time the formula is introduced, people are only interested in how to evaluate expressions with a complex exponent, and for this case, the first identity suffices. But again, it's an aesthetic choice, both formulas are correct.
For the abovementioned reason, ii has infinitely many values.
At first I thought if it's only a property of imaginary numbers that a closed-form expression using them may have more than one value, but then I remembered that x1/2 has two values even for real values of x.
Is there any closed form expression that has infinitely many values using only real numbers? (I don't consider equations to be included in these closed-form expressions.)
The concept which you are arriving is this: https://en.wikipedia.org/wiki/Branch_point
The explanation from complex analysis for multiple definitions of ii, and also of the complex square root x1/2 is that it is impossible to define the complex function log(z) continuous everywhere. Any definition of log(z) must have a branch cut, where you find discontinuities.
The theory behind this is very rich. For example, if instead of functions you consider multifunctions, you can make the image continuous and end up with a Riemann surface. The uniformization theorem guarantees that these surfaces must have constant curvature, so up to topological complications, there are only three types of Riemann surfaces, which correspond to the three possible geometries: Euclidean, elliptic and hyperbolic.
To answer my own question, eg. arctan(1) has infinitely many values...
Which got me thinking... Wouldn't by this same logic 0/0 have infinitely many values?
After all, what does "arctan(1)" mean? It means "what is the value for which the tan of that value gives 1?" There are infinitely many such values.
Likewise "0/0" means "what is the value which, when multiplied by 0, gives 0?" Likewise there are infinitely many such values.
Arctan is normally defined to take a value between -pi/2 and pi/2. It can also be defined in a number of other ways, such as the integral of 1/(1+x^2), which remove the ambiguity. In principle you could choose any other principal branch and you'd be fine as long as you applied it consistently, but it seems a lot more natural to take the branch covering x=0 as the principal branch.
0/0 is different. It really is undefined. There is no real concept of branches of that "function" because its solutions are infinitesimally close together.
After all, what does "arctan(1)" mean? It means "what is the value for which the tan of that value gives 1?" There are infinitely many such values.
Arctan is a function, and the definition of a function requires a domain and an image, and that for every value in the domain, there is exactly one value assigned to it in the image set. The remark that there is an infinite number of values whose tangent is 1 only proves that the algebraic property "what is the value for which the tan of that value gives 1?" is not sufficient to define a function.
In practice, we impose further restrictions to make the function easier to work with. One convenient choice is to make arctan a continuous function, and like thatguy said, if you use the definition of arctan(x) "what is the value between -pi/2 and pi/2 for which the tan of that value gives x?", then there is no ambiguity, and the function is continuous on the entire real line, and it's also an odd function, which might help to carry out some simplifications.
Warp wrote:
Likewise "0/0" means "what is the value which, when multiplied by 0, gives 0?" Likewise there are infinitely many such values.
Division must have the property that only one value satisfies the algebraic requirement. Otherwise, it is not a function, and this remark again only points out that division by 0 cannot be a function.
The main difference with respect to arctan is that, in 0/0, continuity cannot help you to define this operation at all. That's because we can prove that if, in some limit x->a, we have continuous functions f(x) -> 0 and g(x) -> 0, then this says nothing about the limit f(x)/g(x) when x -> a. It can be 0, infinity, any other value, or not exist at all.
Arctan is a function, and the definition of a function requires a domain and an image, and that for every value in the domain, there is exactly one value assigned to it in the image set.
In the strict sense, a well-defined function associates one, and only one, output to any particular input. The term "multivalued function" is, therefore, a misnomer because functions are single-valued.
and
Wikipedia wrote:
We can treat arctan as a single-valued function by restricting the domain of tan x to −π/2 < x < π/2 – a domain over which tan x is monotonically increasing. Thus, the range of arctan(x) becomes −π/2 < y < π/2.
I'm not usually all that good at FiveThirtyEight's The Riddler puzzles since I usually give up before working out an answer. This week, however, I think I got both of the answers right. The "Riddler Express" puzzle, a game of hot potato, took longer for me to solve than the "Riddler Classic" puzzle, determining whether a bathroom is occupied. I should note that the bathroom puzzle has a somewhat ambiguous problem statement. Nothing is said about whether or how the employees' behavior is influenced by the status of the sign. I assumed that all employees, regardless of their habits, knock or attempt to open the door, then leave to find another bathroom if it is occupied. Basically, I assumed that the bathroom always returned to a vacant state before anyone even attempts to open the door again. It is also plausible that people might wait to enter if the sign says "occupied" or semi-forgetful and conscientious employees would switch the sign to "occupied" if it says vacant while the bathroom is occupied. While I didn't perform any analysis beyond what I assumed of the problem statement, my method generalizes pretty easily to other employee behaviors.
So keeping in mind that there are about 14 hours until the submission deadline (might want to wait before replying), what results did you get?
Riddler Express: It's one of those questions where either you get it or you don't.
Every child has the same chance (1/30) of winning. At some point, the potato is first held by either of the two people beside you; let's call them A and B, and without loss of generality, A has the potato; thus B never had it. To win, the potato must be passed from A to B the long way around (i.e. not through you). This applies to every child regardless of how the potato was passed previously; therefore every child has the same chance of winning.
Riddler Classic: I assume likewise as Bobo the King did: whoever attempts to use the washroom leaves immediately if the bathroom is occupied (i.e. the only person modifying the door sign is the current user of the bathroom).
First, in the long run, the state of the door sign when vacant tends to probability 1/2 for both "occupied" and "vacant" (this is because only 2/3 of the users affect the sign: half of them set it to "occupied" and half of them set it to "vacant").
Given that, we can calculate the probability of cases. I call user behavior T1 if they ignore the sign, T2 if they set it to "occupied" but not "vacant" at the end, and T3 if they set it to "occupied", then "vacant" at the end. Each behavior occurs with equal probability (1/3).
occupied, user is T1, sign="occupied": 1/2 * 1/3 * 1/2 = 1/12 (because of the long run probability for the sign)
occupied, user is T2, sign="occupied": 1/2 * 1/3 * 1 = 1/6
occupied, user is T3, sign="occupied": 1/2 * 1/3 * 1 = 1/6
occupied, sign="occupied": 5/12
occupied, sign="vacant": 1/2 - 5/12 = 1/12
vacant, last user was T1, sign="occupied": 1/2 * 1/3 * 1/2 = 1/12 (because of the long run probability for the sign)
vacant, last user was T2, sign="occupied": 1/2 * 1/3 * 1 = 1/6
vacant, last user was T3, sign="occupied": 1/2 * 1/3 * 0 = 0
vacant, sign="occupied": 1/4
vacant, sign="vacant": 1/2 - 1/4 = 1/4
sign="occupied": 5/12 + 1/4 = 2/3
sign="vacant": 1/3
occupied given sign="occupied": (5/12) / (2/3) = 5/8
vacant given sign="vacant": (1/4) / (1/3) = 3/4
Riddler Express: It's one of those questions where either you get it or you don't.
Every child has the same chance (1/30) of winning. At some point, the potato is first held by either of the two people beside you; let's call them A and B, and without loss of generality, A has the potato; thus B never had it. To win, the potato must be passed from A to B the long way around (i.e. not through you). This applies to every child regardless of how the potato was passed previously; therefore every child has the same chance of winning.
Oooh, so sorry! The correct answer is 1/29. The initial holder of the potato has zero probability of winning. :P
Kidding aside, that was the same logic I used to solve the problem and I'm glad to see someone smarter than me agrees.
FractalFusion wrote:
Riddler Classic: I assume likewise as Bobo the King did: whoever attempts to use the washroom leaves immediately if the bathroom is occupied (i.e. the only person modifying the door sign is the current user of the bathroom).
First, in the long run, the state of the door sign when vacant tends to probability 1/2 for both "occupied" and "vacant" (this is because only 2/3 of the users affect the sign: half of them set it to "occupied" and half of them set it to "vacant").
Given that, we can calculate the probability of cases. I call user behavior T1 if they ignore the sign, T2 if they set it to "occupied" but not "vacant" at the end, and T3 if they set it to "occupied", then "vacant" at the end. Each behavior occurs with equal probability (1/3).
occupied, user is T1, sign="occupied": 1/2 * 1/3 * 1/2 = 1/12 (because of the long run probability for the sign)
occupied, user is T2, sign="occupied": 1/2 * 1/3 * 1 = 1/6
occupied, user is T3, sign="occupied": 1/2 * 1/3 * 1 = 1/6
occupied, sign="occupied": 5/12
occupied, sign="vacant": 1/2 - 5/12 = 1/12
vacant, last user was T1, sign="occupied": 1/2 * 1/3 * 1/2 = 1/12 (because of the long run probability for the sign)
vacant, last user was T2, sign="occupied": 1/2 * 1/3 * 1 = 1/6
vacant, last user was T3, sign="occupied": 1/2 * 1/3 * 0 = 0
vacant, sign="occupied": 1/4
vacant, sign="vacant": 1/2 - 1/4 = 1/4
sign="occupied": 5/12 + 1/4 = 2/3
sign="vacant": 1/3
occupied given sign="occupied": (5/12) / (2/3) = 5/8
vacant given sign="vacant": (1/4) / (1/3) = 3/4
I'm not sure if your work here matches what you did on paper, but I opted to model the problem explicitly as a Markov chain. I drew up a directed graph showing the likelihoods of transitioning from/to each of the six allowed states. What's especially nifty is that a symmetry is apparent, reducing the effective number of variables by half. Draw the graph and you'll find that:
P(sign="vacant" & occupant=full forgetful) = P(sign="occupied" & occupant=full forgetful)
P(sign="occupied" & occupant =half forgetful) = P(sign="occupied" & occupant=conscientious)
P(sign="vacant" & occupant=null) = P(sign="occupied" & occupant=null)
So I used that graph to draw up a 6x6 sparse matrix. It can be shown that the matrix corresponding to a Markov chain has eigenvalue 1, so I just plugged in that eigenvalue and went to work solving for the three distinct probabilities. My answers agree with yours.
If the behavior of the employees is dependent on the state of the sign, it's fairly easy to modify the graph and corresponding matrix. The only bad thing about it is that it would likely break the symmetry of the problem as I outlined it.
Oooh, so sorry! The correct answer is 1/29. The initial holder of the potato has zero probability of winning. :P
Based on the statement of the question, I'm getting 1/30, if I'm understanding this right.
Riddler Express wrote:
A class of 30 children is playing a game where they all stand in a circle along with their teacher. The teacher is holding two things: a coin and a potato. The game progresses like this: The teacher tosses the coin. Whoever holds the potato passes it to the left if the coin comes up heads and to the right if the coin comes up tails. The game ends when every child except one has held the potato, and the one who hasn’t is declared the winner.
By my understanding, the circle is 31 people large, and none of the 30 children start with the potato, because the teacher does.
Oooh, so sorry! The correct answer is 1/29. The initial holder of the potato has zero probability of winning. :P
Based on the statement of the question, I'm getting 1/30, if I'm understanding this right.
Riddler Express wrote:
A class of 30 children is playing a game where they all stand in a circle along with their teacher. The teacher is holding two things: a coin and a potato. The game progresses like this: The teacher tosses the coin. Whoever holds the potato passes it to the left if the coin comes up heads and to the right if the coin comes up tails. The game ends when every child except one has held the potato, and the one who hasn’t is declared the winner.
By my understanding, the circle is 31 people large, and none of the 30 children start with the potato, because the teacher does.
Huh. Well, that's annoying. I could have sworn that an earlier version of the problem statement included the sentence, "The teacher begins by giving the potato to a random student," in which case the only nontrivial way to interpret the challenge is the 29 remaining students' probabilities based on their positions relative to the initial potato-holder. I can't find that sentence anymore, so either I dreamed it up or the page was edited.
I just checked out a cached version from August 4th and... no mention of passing the potato to a random student initially. Damn it, looks like I wasted a submission.
The youtuber blackpenredpen recently uploaded a video where he calculates on a whiteboard the result of
(7 + sqrt(50))1/3 + (7 - sqrt(50))1/3
It has a real solution, and you don't need to calculate any square or cube roots. Of course if you are lazy you could just use a calculator, or watch the video. However, if you want to challenge yourself, maybe try deducing the answer by hand. I think the solution was interesting.
A surd of the form a+b*sqrt(2) has the cube (a^2+6b^2)a+(3a^2+2b^2)b*sqrt(2), so if the cube roots of 7+5sqrt(2) and 7-5sqrt(2) are surds, you can find them by solving a pair of cubic equations (I'm starting with the + case):
(a^2+6b^2)a=7
(3a^2+2b^2)b=5
From the first equation, 2b^2=7/(3a)-(1/3)a^2, so substituting into the second,
(3a^2+7/(3a)-(1/3)a^2)b=5, so
(8a^3+7)b=15a, so b=15a/(8a^3+7).
Then squaring and re-substituting,
450a^2/(8a^3+7)^2=7/(3a)-(1/3)a^2, and then
1350/(8a^3+7)^2=7/a^3-1
1350a^3=(7-a^3)(8a^3+7)^2
1350a^3=(7-a^3)(64a^6+112a^3+49)
1350a^3=343+735a^3+336a^6-64a^9
64a^9-336a^6+615a^3-343=0
This is a cubic in a^3, and testing potential rational roots yields the root a^3=1, from which a=1.
Using that root, b=15/(8+7)=1; indeed, (1+sqrt(2))^3=1+3sqrt(2)+3*2+2sqrt(2)=7+5sqrt(2).
This suggests that maybe the algebraic cube root of 7-5sqrt(2) is 1-sqrt(2): (1-sqrt(2))^3=1-3sqrt(2)+3*2-2sqrt(2)=7-5sqrt(2).
Therefore, the given expression simplifies to 2.
Hm, I haven't seen the word "surd" used a lot. It looks weird.
Anyway, before I look at the video, I'll post two other methods:
Method 1 (along the same lines as arflech):
Let x=(7+sqrt(50))1/3, y=(7-sqrt(50))1/3. We want to find z=x+y.
Now
z3 = x3+3x2y+3xy2+y3 = x3+y3+3xy(x+y) = x3+y3+3xyz
= 7+sqrt(50) + 7-sqrt(50) + 3(7+sqrt(50))(7-sqrt(50))z = 14 - 3z.
So (7+sqrt(50))1/3+(7-sqrt(50))1/3 is a solution to z3+3z-14=0. We see that 2 is a root of z3+3z-14, and it factors as (z-2)(z2+2z+7). Since z2+2z+7 only has complex roots, then 2 is the only real root of z3+3z-14, and so (7+sqrt(50))1/3+(7-sqrt(50))1/3=2.
Method 2 (if you are familiar with the Pell equation x2-2y2=±1):
Note that 7+sqrt(50)=7+5sqrt(2)=(1+sqrt(2))3. (This is recognizable if you already know that 7+5sqrt(2) is the third power of the fundamental solution x+y*sqrt(2) to x2-2y2=±1.) Similarly 7-sqrt(50)=7-5sqrt(2)=(1-sqrt(2))3. Then we immediately get:
(7+sqrt(50))1/3 + (7-sqrt(50))1/3 = 1+sqrt(2) + 1-sqrt(2) = 2.
Edit: Video uses Method 1.
I was previously familiar with this week's Riddler express puzzle, making it especially easy. If you haven't seen it before, you can find the answer by plugging it into the OEIS, but you wouldn't cheat, right?
Anyway, the second riddle is, as expected, tougher. Oddly enough, the question is even a little bit of a hint for the first riddle. I was able to get it in maybe 90 minutes of work, most of which was spent on algebra. I'm confident in my answer, which I'll post sometime around Sunday evening. Did anyone else get it? I'm looking at you, FractalFusion.
Edit: Mixed up express and classic. I was familiar with the Riddler Express puzzle, not with the classic puzzle.
Did anyone else get it? I'm looking at you, FractalFusion.
How the sequence is defined (borrowing from Riddler):
The sequence has the property that, if you replace every 3 with 3332 and 2 with 332, the resulting sequence will be the same.
Let x be the long-term proportion of 2s in the sequence. Replace every 3 with 3332 and 2 with 332. If we look at only the 2s being replaced with 332, the proportion of 2s in the new sequence is 1/3, whereas if we look at only the 3s being replaced with 3332, the proportion of 2s in the new sequence is 1/4. So the true proportion of 2s in the new sequence is x*(1/3) + (1-x)*(1/4). Since this replacement does not change the sequence:
x = x*(1/3) + (1-x)*(1/4)
x = (1/12)*x + 1/4
(11/12)*x = 1/4
x = 3/11
So the proportion of 2s in the sequence is 3/11, and the proportion of 3s in the sequence is 8/11. The ratio of 3s to 2s is 8/11 : 3/11, or 8:3.Edit: This post is wrong. See http://tasvideos.org/forum/viewtopic.php?p=457755#457755 for the correction.
Did anyone else get it? I'm looking at you, FractalFusion.
How the sequence is defined (borrowing from Riddler):
The sequence has the property that, if you replace every 3 with 3332 and 2 with 332, the resulting sequence will be the same.
Let x be the long-term proportion of 2s in the sequence. Replace every 3 with 3332 and 2 with 332. If we look at only the 2s being replaced with 332, the proportion of 2s in the new sequence is 1/3, whereas if we look at only the 3s being replaced with 3332, the proportion of 2s in the new sequence is 1/4. So the true proportion of 2s in the new sequence is x*(1/3) + (1-x)*(1/4). Since this replacement does not change the sequence:
x = x*(1/3) + (1-x)*(1/4)
x = (1/12)*x + 1/4
(11/12)*x = 1/4
x = 3/11
So the proportion of 2s in the sequence is 3/11, and the proportion of 3s in the sequence is 8/11. The ratio of 3s to 2s is 8/11 : 3/11, or 8:3.
I got something different. Let me see if I can find a flaw in your reasoning...
We'll construct a sequence with the porportion of 2s you derived:
33323332332
That's eleven digits, three of which are 2s. Much later in the sequence, we'll find:
33323332333233233323332333233233323332332
That's 41 digits, 11 of which are 2s, a slightly lower proportion of 2s. This suggests that the ratio you derived is not a fixed point.
I believe your error lies where you say, "So the true proportion of 2s in the new sequence is x*(1/3) + (1-x)*(1/4)." Because the strings 3332 and 332 are not of equal length, you can't just average the ratios together to find the the combined ratio. At least I think that's correct.
The sequence has the property that, if you replace every 3 with 3332 and 2 with 332, the resulting sequence will be the same.
