Interesting math probleem thread

Word problems belong on

Do all of the non-trivial zeros of the reimann zeta function lie on the critical line?

Oh, my sorry, WORD problem. Guess I'll go over to Dunce.

3 loads. And no.

Fuck you

OP here, this is correct. I've got the official solution, it's quite confusing, but if you concentrate enough you'll understand it.

...

I've got. A solution with 5, but can't prove its the smallest.
You drop 1/3 of your gas at 166miles, and repeat it 3 times. On the fourth run you fill the truck at 166 miles and go on from there another 166 miles and drop 1/3 of the tank and return. Now you'll have 188miles worth of gas at the 166th mile post and the same amount at 332 miles. With a full tank you can get across because you'll have full tank at 332 miles.

You'll have 166 miles with of gas*

Worth* damn autocorrect

Both are wrong, according to my notes

During a lecture all of the students fell asleep exactly once. For each pair of students there was some moment when both were asleep at the same time ( students A and B were asleep at the same time at some point, so were A and C and so were C and B and so on). Prove that at some point all of the students were sleeping simultaneously.

x>y

A = x-B or y-B
B = x-A or y-A

B knows that if A = x-B then A knows B = B or y-x+B
if A = y-B then A knows B = x-y+B or B

So if y-x+B =< 0 then B knows A = y-B
If B =< x-y then B will answer yes on his first turn

If B > x-y then B will answer no

So if B = y-A then y-A > x-y and A < 2y-x
If A >= 2y-x then A knows B = x-A and says yes on his second turn

A knows that if B = x-A then B knows A = A or y-x+A
if B = y-A then B knows A = x-y+A or A

So if y-x+A =< 0 then A knows B = y-A
If A =< x-y then A will answer yes on his second turn

If A > x-y then A will answer no

So if A = y-B then y-B > x-y and B < 2y-x
If B >= 2y-x then B knows A = x-B and says yes on his second turn

If A = x-B < 2y-x then B > 2x-2y
If B =< 2x-2y then B knows A = y-B and says yes on his second turn

If B = y-A > 2x-2y then A < 3y-2x

If A = x-B >= 3y-2x then B =< 3x-3y

etc.

So the bounds of A and B which result in "yes" on the nth turn are:

1 =< B =< x-y
2y-x =< A =< x-y
2y-x =< B =< 2x-2y
3y-2x =< A =< 2x-2y
3y-2x =< B =< 3x-3y
etc.

[math] \arctan(1)+\arctan(2)+\arctan(3) = \pi [/math]

You flip a coin repeatedly. What is the probability that the first instance of two consecutive heads appears before the first instance of three consecutive tails?

Why is this formality necessary?

Both boys are presented with two choices. One goes first. If he's wrong, the other choice must be correct, and the second boy should guess the other number, by process of elimination.

They don't guess the numbers out loud, they just say if they can.

The power tower [math]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdot^{\cdot^{\cdot}}}}}} [/math] converges.

Find the supremum of all [math]x[/math] such that [math]x^{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}} [/math] converges.

What kind of question is this, you haven't even implied the set you are working on in which x would lie.

e.g. "Let A be a set of real numbers" ...

Supremum obviously implies reals.

The supremum of a subset S of a partially ordered set T is the least element in T that is greater than or equal to all elements of S, if such an element exists. Consequently, the supremum is also referred to as the least upper bound (or LUB).

No supremum does not always imply reals. You can have the complex plane with a supremum on its set. BUT SURE MAN SUREEEEE

e

Supremum implies totally ordered set.

thank you. Holy shit.

Holy shit you autists. Of course supremum doesn't necessitate the reals, but what other domain could have possibly been a reasonable interpretation of the set over which is working? There is literally no reasonable interpretation but [math]\mathbb{R}[/math]; any other set in which power towers converge (implying completeness) and with suprema (implying a natural ordering) would be stupid and contrived.

Natural guess, but no.

It's [math]e^{1/e}[/math]. Formulate it as [math]y_0 = 1[/math] and [math]y_n = x^{y_{n - 1}}[/math] and use analytical methods from there.

Not the guy who posted the problem but I was really hoping someone would post a full proof that even a retard could understand.

Which came first: the chicken or the road?

To elaborate, a fixed point [math]y = lim_{n \rightarrow \infty}{y_n}[/math] only exists if [math]y = x^y[/math] for some [math]y[/math] given a fixed [math]x[/math] because [math]\{y_n\}[/math] is strictly increasing. Calculus I shows that there is one value of [math]x[/math] for which [math]f(y) = x^y[/math] is tangent to [math]f(y) = y[/math], namely [math]x = e^{1/e}[/math]. For smaller values of [math]x[/math], the functions crossover multiple times (twice). A more rigorous proof involves rewriting [math]y[/math] in terms of the Lambert W function, restricting it to the real line, and using other fixed-point theorems. I'm too lazy to write it out, though. More detailed sketches of both proofs are available online.

>would be stupid and contrived.
Any moreso than the reals?

>Any moreso than the reals?

You had to go there...

>And so on

So the teacher is going to keep asking kid A and B for an answer, ad infinitum?

>A, do you know the answer?
>No
>B, do you know the answer?
>No
>A, do you know the answer?
>No
>B, do you know the answer?
>No
etc...

That proof is more complicated than it needs to be. You definitely don't need to invoke the Lambert W function

What total ordering are you putting on C exactly?

Proofs that C is not an ordered field are instantly searchable.

Let F be a bijection from C to R. Then all you need to do is use F for the usual order. In other words, were a and b are complex numbers

a < b if and only if F(a) < F(b)

easy

And that total ordering is completely useless. Congratulations.

Why?
To me it looks pretty good.

Because for example, you lose ability to say if x and y are both greater than 0, then xy >0. (You can't multiply anymore and retain order.)

I'm guessing if you understood proofs you would have just read one, so here's an example.
Let F(i)=1, F(-1)=-1, F(0)=(0), and F be a bijection between C and R, existence of such a bijection exists between C and R is obvious. But i>0 since F(i)=1>0=F(0), but i^2

OP what book is this from? Need to step my math game up.

>Imposing a total ordering on C necessarily take away your ability to use its most desirable property, namely its arithmetic.

But that is useless for the problem they were talking about.

What you would have to do is simply compute all of the complex numbers c that satisfy the property of its infinite exponential converging. Then you would compute F(c) for all these numbers and then you would look for the F(c) that is the biggest of them all and that would be your supremum for that problem.

Your trolling is sad.

Written out proof is attached.

>Your trolling is sad.

But I am not trolling. I am not arguing against any of the proposed answers. I am just saying that this problem could be solved in any number of dimensions my nigga because all we need to do is apply the order of the real numbers to any 'bigger' set.

He's going to ask them about the same numbers forever, I presume.

1 load goes up to 500/1 miles.
2 loads go up to 500/3 + 500/1 = 633.33... miles.
3 loads go up to 500/5 + 500/3 + 500/1 = 733.33... miles.
4 loads go up to 500/7 + 500/5 + 500/3 + 500/1 = 838.095238095238... miles.
And so on. There's a general formula for this but I'm on a phone now. And according to the formula there is no limit to the max distance travelled, but there are diminishing returns for each extra load.

For the record, the answer for 800 miles is 4 loads, if it isn't already obvious.

>piss easy induction question
OP asked for interesting problems m8.

Sorry don't have it anymore and I don't recall the author(s) but the title was "Mathematical miniatures". It was a book for studying for IMO (international mathematics Olympiad). Only high school stuff but at the highest level.

Well yes but eventually one of them will say "yes", and soon will the other. It seems impossible, but it's not

1. truck drops 300miles of gas at 100miles, returns
2. Truck fills up at 100 miles, drops 300miles of gas at 200 miles, return to 100, fills, returns
(Now there's gas for300 at 200 miles and 100 at 100 miles)
3. Truck fills itself up at 100 and at 200, then drops 300 miles worth of gas at 300 miles, fills itself up at 200, and returns
(Now there's 300 of gas at 300 miles)
4. Truck crosses the desert

If you're not trolling then you're just an idiot, for multiple reasons.

If you want to talk about the convergence of any sequence, and tetration, the nested exponentiation here, is a a question about the convergence of a sequence, then you have to have a topology on your set.

To ask, in C, what z^w is, or in any set what exponentiation means you arithmetic, for C you need to use C's arithmetic. The arithmetic of C forces a topology on you that is the unique topological field(ring) structure of C.

The order you've imposed generates a different topology on C. It's the order topology.

So then for C, or indeed any 'bigger' (fucking retarded) set, when you answer this question, which topology is the sequence converging in?

If the answer is the order topology, then that convergence will almost surely not make sense because you won't have arithmetic. Certainly in C in won't make any sense without arithmetic.

So what happens if we use the arithmetic of C to determine convergence, then impose a bijection based order? Depending on the bijection I choose, there might not be a supremum for my set. That is the bijection you chose might have mapped the infinite set complex numbers converging under tetration to an unbounded set in R. So then the question becomes unanswerable.

So let's for the sake of argument say you only pick bijections which are mapping the numbers which converge under tetration to a bounded set in R, then there will be a supremum necessarily. However at this point, even under these conditions, given any complex number, z, there exists a bijection that induces an order which forces z to be the supremum in question.

So in the two scenarios possible, either the supremum isn't defined, or it's meaninglessly defined, because the set of numbers which converge under tetration is independent of your chosen order. Your supremum is meaningless.

>dimensions

Seriously, don't post about math until you learn some.