A Neat Little Math Tidbit

[AetherFox]

While fiddling around with my graphing calculator today, (during math class) I discovered that apparently all lines with the formula Y= X+Asin(X) where A can be any real number, all intersect for the first time in the first quadrant at the ordered pair (Pi, Pi)!!
Pretty wacky stuff!
Anyway, I'm sure there is a perfectly logical exlepanation for this, and I'd love to hear it, but I'm just throwing out my (amazing? probably not) discovery to anyone who cares.
:smallbiggrin:

@V, Umm, Well, hmmmmmmm.......
I still don't see why....
Aww, apparently this isn't as cool as I thought it was.......:smallfrown:

[Tirian]

It's because sin(pi) = 0. Not really revolutionary, but it is neat when math jazzes you like that.

[afroakuma]

Y=X+sin(x) and Y=X+2sin(X) give me a single intercept at (0,0)

I see no other intercept.

Did I do something wrong?

(TI-83 Plus)

[AetherFox]

Oh, yes, sorry.
I'm in radians.

[Tirian]

@V, Umm, Well, hmmmmmmm.......
I still don't see why....
Aww, apparently this isn't as cool as I thought it was.......:smallfrown:

Let A be any real number, and k be any integer. Then k*pi + A*sin(k*pi) = k*pi, so all of the curves will meet at (k*pi, k*pi) for all integral values of k.

[golentan]

Alrighty, how many of you can figure this one out. It's one of my personal favorites!

X^X^X^X^X^X^X... add infinitum = 2

What does X equal and how do you know?

I'll tell you when you get it right.

[RS14]

x^x^x^x^x^x^...=2
So x^x^x^x^x^x...=x^2=2
So x=(2)^(1/2)

Edit:
I've seen similar problems before, but usually involving an infinite fraction (a/(1+a/(1+a/...)))=c, for some c.

[golentan]

Yep, 100% correctomundo. The fun thing is that it stumps most math/physics majors, but programmers tend to solve in ten seconds flat. I'm sort of disappointed: I was hoping for some wild flails in the dark before anyone got it.

[RS14]

Yep, 100% correctomundo. The fun thing is that it stumps most math/physics majors, but programmers tend to solve in ten seconds flat. I'm sort of disappointed: I was hoping for some wild flails in the dark before anyone got it.
Well, for the record, I'm a math major, but have gone to math competitions for the last couple of years.

Since I got that one too quick, here is another to chew on:

You are standing in a strange amusement park. You can choose to take any one of three slides. Two slides return you to your current location, and disorient you so that all three slides are indistinguishable. The third slide leads out of the park. Thus, each time you slide, you have a 1/3 chance of leaving the park, no matter how many times you have already slid. If you slide until you leave the park, what is the average number of times you slide?

[Collin152]

Should I be concerned that this slide somehow deposits me at its starting point?

[JaxGaret]

I'm going to say 1 and five-sixths.

Probably wrong.

EDIT: Yep, totally wrong. I completely misinterpreted the question.

[golentan]

Having gone back and consulted on the matter of series, I have to say the answer is 3. Please don't make me translate the work I did over to the forum, as I solved in all generality and that's a lot of letters and symbols.

Edit: And yes, I realize how pathetic that is given the simplistic relation of probability of an escape to the answer I got, but I still don't want to write it out.

[RS14]

Having gone back and consulted on the matter of series, I have to say the answer is 3.
This is what I get.
Summary:
S=a/(1-r), where S is the sum of an infinite geometric sequence, a is the first term, and r is the ratio between terms. r must have absolute value less than 1.
The average is 1/3+2*1/3*(2/3)^1+3*1/3*(2/3)^2...
Consider just 1/3+1/3*2/3*1/3*(2/3)^2...
This equals 1, by the general formula for a infinite series.
Then the full expression equals 1+1*2/3+1*(2/3)^2..., which equals 3, by the same formula.

[golentan]

Edit: Removed due to embarassing comments showing author's ignorance. Sorry guys.

P.P.S.: To make up for it: it's stupid, and it's illegal, but it is the classic "proof" that 1 = 2
Given 1 = a and a = b
a^2 = ab
(a^2) - (b^2) = ab - (b^2)
(a - b)(a + b) = b(a - b)
a + b = b
a + a = a
2a = a
2 = 1

It also does 1 = 0
a + b = b
a = 0
1 = 0

Oh, how I wish that this worked.

[Saeveo]

You are standing in a strange amusement park. You can choose to take any one of three slides. Two slides return you to your current location, and disorient you so that all three slides are indistinguishable. The third slide leads out of the park. Thus, each time you slide, you have a 1/3 chance of leaving the park, no matter how many times you have already slid. If you slide until you leave the park, what is the average number of times you slide?

It's just a Markov chain, yeah?

[ghost_warlock]

Threads like this make me sad because I fail at math. I want to participate, but my brain just doesn't work right. :smallfrown:

*goes off to listen to Jonathan Coulton's song (http://www.youtube.com/watch?v=gEw8xpb1aRA) and dream*

[First Speaker]

Should I be concerned that this slide somehow deposits me at its starting point?

AARGH IT'S BREAKING THERMODYNAMICS HELP MEEE!

(Well, admittedly, there might be an external source of energy, but that's not how slides work.)

[AgentPaper]

2

If you slide 3 times, on average you will be out of the park once out of those three. Statistically, an even amount of people will be sent out on the first, second, and third times. Average of 1, 2 and 3 is 2.

AARGH IT'S BREAKING THERMODYNAMICS HELP MEEE!

(Well, admittedly, there might be an external source of energy, but that's not how slides work.)

The way I imagine it, you're standing in front of three ladders to the slides, and have to climb up. The whole thing is in a cylinder room, so you can't see where the slides go. All the slides come in between the ladders, and look the same. A third slide comes in from the outside, which is how you get in in the first place.

Heh, you could make a math themed amusement park. Most of it would fly over peoples heads. (And most even over people like me. The calculus classes! They do nothing!)

[Thufir]

2

If you slide 3 times, on average you will be out of the park once out of those three. Statistically, an even amount of people will be sent out on the first, second, and third times. Average of 1, 2 and 3 is 2.

Your reasoning is flawed. The chance of leaving the park is 1/3 every time, so you can take more than 3 attempts, thus taking the average of 1, 2 and 3 makes no sense as a method. I assume the answer is 3 as other people stated earlier, but I can't be bothered to check.

[Khanderas]

Your reasoning is flawed. The chance of leaving the park is 1/3 every time, so you can take more than 3 attempts, thus taking the average of 1, 2 and 3 makes no sense as a method. I assume the answer is 3 as other people stated earlier, but I can't be bothered to check.
Exactly.
First slide 33% chance of getting out. If you dont get out, slide again.
Second slide 33% chance to get out. If not, slide again.
thrid try, 33% chance to get out, fail and slide again.
Fourth try 33% chance to get out, fail and slide again.
and so on.
It is always 33% chance since you dont know what slide you took the last time(s).

How is this old classic then ?

3 travelling salesmen get to a hotel and check in for the night. The room they get costs 25 €. Each salesman gives a 10 € note to the boy who helped them carry their bags up to their room. The kid runs down and gets 5 1 € notes.
5 € is hard to split up among 3 people, so each salesman takes one € back and the kid keeps 2€ as a tip.

So each salesman paid 10 - 1 = 9 €
The kid kept 2 €.

3 x 9 + 2 = ....... 29 € ?

Where did the last € go ?

[paddyfool]

3 travelling salesmen get to a hotel and check in for the night. The room they get costs 25 €. Each salesman gives a 10 € note to the boy who helped them carry their bags up to their room. The kid runs down and gets 5 1 € notes.
5 € is hard to split up among 3 people, so each salesman takes one € back and the kid keeps 2€ as a tip.

So each salesman paid 10 - 1 = 9 €
The kid kept 2 €.

3 x 9 + 2 = ....... 29 € ?

Where did the last € go ?

That's a very old chestnut (much older than the Euro), and the answer is very simple - you don't add two, since the 2 came out of the 3*9. What happened in effect was that each salesman paid about 8 1/3 € towards the hotel bill, and 2/3 of a € towards the tip, getting one back from their ten.

Here's another old chestnut to go with that one:



On the final phaze of a given gameshow, the participant is always shown three doors, and is told there is a prize behind one of them. On being asked to guess which door the prize lies behind, the participant points to one. The host then opens one of the other doors to show that the prize is not behind that door, and gives the participant the choice of sticking to the door they guessed at first, or changing their guess to the third door. Should the participant stick or switch, and what difference does it make?

[Tirian]

You are standing in a strange amusement park. You can choose to take any one of three slides. Two slides return you to your current location, and disorient you so that all three slides are indistinguishable. The third slide leads out of the park. Thus, each time you slide, you have a 1/3 chance of leaving the park, no matter how many times you have already slid. If you slide until you leave the park, what is the average number of times you slide?

Let x be the expected number of rides that you take. If you initially take the correct slide, then it only requires one slide. Otherwise, you have taken a ride and are still in the same situation, so that you now expect it to take x+1 rides. Therefore x = 1/3(1) + 2/3(x+1), or x = 3.

Here's a similar question. A scientist creates a single cell of a new species of bacteria. The bacteria has a 2/3 chance of surviving until it (asexually) reproduces, and if it does there is a 50% chance that it fissions into two parts and 50% that it will split into three parts. Each of the daughter cells is identical to the original, and there is no possible potential of overpopulation, or any other external risk (including the destruction of Earth or the universe). What is the probability that the species will eventually become extinct?

[Khanderas]

That's a very old chestnut (much older than the Euro)
Well yeah, thats why I called it a classic.
I needed to call the currency something and € seemed to work.


Let x be the expected number of rides that you take. If you initially take the correct slide, then it only requires one slide. Otherwise, you have taken a ride and are still in the same situation, so that you now expect it to take x+1 rides. Therefore x = 1/3(1) + 2/3(x+1), or x = 3.
I had to think about this one a few times, even went as to ask what the heck you were thinking, but I got it now. Good cause I had no good setup myself to crack it. Understand the numbers but I would never have come up with them on my own.

[Dallas-Dakota]

With all these math threads, maybe one of you should make a general maths thread?:smallconfused:

[The Rose Dragon]

On the final phaze of a given gameshow, the participant is always shown three doors, and is told there is a prize behind one of them. On being asked to guess which door the prize lies behind, the participant points to one. The host then opens one of the other doors to show that the prize is not behind that door, and gives the participant the choice of sticking to the door they guessed at first, or changing their guess to the third door. Should the participant stick or switch, and what difference does it make?

You should switch to double your chances.

Easiest question ever.

[Dallas-Dakota]

You should switch to double your chances.

Easiest question ever.

Hmm, it would not double you changes.
Instead of a 33.3-% chance it turns to a 50% chance of getting the right one.

[JaxGaret]

Rose Dragon was correct.

Here (http://en.wikipedia.org/wiki/Monty_hall_problem) is the reasoning why.

[Dallas-Dakota]

Rose Dragon was correct.

Here (http://en.wikipedia.org/wiki/Monty_hall_problem) is the reasoning why.
Blegh.

*leaves icky maths alone*

[Tirian]

I had to think about this one a few times, even went as to ask what the heck you were thinking, but I got it now. Good cause I had no good setup myself to crack it. Understand the numbers but I would never have come up with them on my own.

Yeah, mathematical expectation is tricky when you first see it, and being an elementary Markov chain it's even a little more wrinkly because of the x+1 "trick". It is a part of probability theory, which you'll run into if you get off the calculus train in college and take a course in introductory discrete math, combinatorics, or statistics.

Actually, the proof that I gave has much more general application in statistics, which is that if you repeat an experiment whose probability of success is p, then you would expect that it would take 1/p trials before you get your first success. This is a useful factoid to know if you want to take on another fun problem, the coupon-collector's problem:

A box of cereal contains one of eight toys, all equally distributed. On average, how many boxes would you need to buy to have at least one of each toy?

[AgentPaper]

Blegh.

*leaves icky maths alone*

After thinking it over a bit, I finally understand the problem. The main issue is that most people assume that the host opened one of the doors at random. If he did, then switching would have no effect. However, he always chooses a door that didn't win. That's what makes the probability change. There's a 2/3 chance that a car is behind one of those doors, and if you switch, you get that same 2/3 chance, since the only wrong answer among that 2/3 probability has been removed.

[evisiron]

*reads through first few posts*

So this is what its like when a non-gamer listens to DnD players talking...

[Allysian]

Two words, holy crap...

Edit: also remember that i'm a seventh grader, but still, you people are intellectual gods!

[JaxGaret]

you people are intellectual gods!

Nah. It's just maths.

[Thufir]

you people are intellectual gods!

Yes. Yes I am.

[InaVegt]

You are standing in a strange amusement park. You can choose to take any one of three slides. Two slides return you to your current location, and disorient you so that all three slides are indistinguishable. The third slide leads out of the park. Thus, each time you slide, you have a 1/3 chance of leaving the park, no matter how many times you have already slid. If you slide until you leave the park, what is the average number of times you slide?

Note, I've not looked at other answers beforehand

As I mark the slide I take before sliding it, I have a limit of three times.

The chance of me getting out exactly the first time is 1/3.
The chance of me getting out exactly the second time is 2/3 * 1/2, or 1/3
The chance of me getting out exactly the third time is 1/3 as well, as there are no other options left.

Now, we multiply the chance each event happens by the time it is, for 1/3 * 1 + 1/3 * 2 + 1/3 * 3 = 1/3 + 2/3 + 3/3 = 2.

As such, the average number of slides is 2.

If we take the literal question, I'm not good enough at math.

[golentan]

I'm pretty sure marking the slide doesn't work.

New problem:You and your significant other go to a dinner party. The table is a big, round picnic style table that seats ten, and seating is at random (order and actual location both). What is the chance you'll get to play footsies with the person you came with? What if you are seated 3rd and they're seated 7vth?

This should be fun.

[Jorkens]

In the spirit of the original post, here's a reall nice elementary(ish) theorem:

sum over all n of ( 1 / n^2 ) = product over all primes p of ( 1 / (1 - 1/p^2 ) )

If you can see why this is true you're doing well...

[Silence]

Reading through this thread and only being in precal makes my head hurt... I barely understood half.

[Tirian]

Reading through this thread and only being in precal makes my head hurt... I barely understood half.

Yeah, the rotten thing about math education is that there is so much fun and awesome stuff out there that people could understand, but they don't learn it because it isn't on what I like to call "the death march to differential equations". Logic, graph theory, combinatorics, elementary statistics, decision theory, finite state automata: all neat stuff that you'll pretty much only touch if you actually major in math.

[AgentPaper]

I'm pretty sure marking the slide doesn't work.

New problem:You and your significant other go to a dinner party. The table is a big, round picnic style table that seats ten, and seating is at random (order and actual location both). What is the chance you'll get to play footsies with the person you came with? What if you are seated 3rd and they're seated 7vth?

This should be fun.

100%. Or is there something stopping you from doing it if you're not right next to each other or directly opposite? If you're doing it at all, you're doing it in the center of the table, and you can reach that from anywhere, or not at all.

[golentan]

Well, it's virtually impossible to get your feet to people not sitting directly next to you at a ten seat picnic table. It has a wide diameter and lots of thick supports underneath. But I guess you could try, as long as you wouldn't be embarrassed if you got the wrong person by mistake. The question was meant to be how likely are you to be sitting next to each other, before I cutesied it up.