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/sci/ - Science & Math

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3403606 No.3403606 [Reply] [Original] [archived.moe]

mathematics thread?
mathematics thread.

>> No.3403609

my erdos number is 2 u mad bro

>> No.3403613

the lowest erdos number that's possible on /sci/ is probably 4. anyone who says otherwise is obviously a mental patient visiting 4chan on the hospital computer.

>> No.3403635

bump 4 erdos

>> No.3403653

pics or it didnt happen

>> No.3403666

I'm grothendieck. sup

>> No.3403667
File: 174 KB, 506x1063, apocalypse.png [View same] [iqdb] [saucenao] [google] [report]

>> No.3403677

pic of u with fields medal or it didn't happen
and datestamp

>> No.3403681 [DELETED] 

Erdos number 4

Kevin Bacon number 3

u jelly?

>> No.3403684

step one: picture a flat washer O (circle with circle removed from the middle). find the area of the washer with only a single straight line measurement.

step two: do the same thing with finding the volume of a sphere that's had a sphere taken out from the middle.

>> No.3403688 [DELETED] 

morphy number 6

u jelly?

>> No.3403692
File: 80 KB, 301x294, dsggh.png [View same] [iqdb] [saucenao] [google] [report]

ITT: Godtier individuals who arent fucking retarded.

Needs more euler though..

>> No.3403696

draw tangent to smaller (removed) circle. call the point of tangency A.
call the points where it hits the larger circle B and C.
measure AB.

>> No.3403704
File: 2 KB, 202x61, euler1.jpg [View same] [iqdb] [saucenao] [google] [report]

can you generalise this to n dimensions?
for example in 2 dimensions we have v-e=0
for n=3 we have v-e+f=2

>> No.3403706


derive pi

>> No.3403707

What is the proof for that?

>> No.3403713

pi is old skool man. tau is cooler!!

>> No.3403719

call the small radius r and the big radius R
the area we're trying to find is pi (R^2-r^2)
now make a right triangle with legs r, AB and hypotenuse R
by Pythagoras AB^2 = (R^2-r^2)
area= pi (R^2 - r^2) = (AB^2) pi

>> No.3403732


sexy as fuck. this is why i love math.

btw whats your major

>> No.3403734

x, y > 1
1. let m^m = x^y; solve for m
2. let n^n = y^x; solve for n
3. replace x with m and y with n
4. repeat unless x = y
Does this algorithm always end?

>> No.3403738

not in college yet... in high school
I just do a lot of math
and im OP btw

>> No.3403747

cool proof but that doesn't work for the sphere!
volume is 4/3 pi (R^3 - r^3)
AB^3 <> (R^3 - r^3)

>> No.3403750


good job man. you have a bright future ahead of you. Any math books you can recommend for not so mathematically gifted community college student?

>> No.3403751

let y=1 and x=4
it wont end... y is always 1 and x keeps on decreasing

>> No.3403757

get The Act and Craft of Problem Solving by Paul Zeitz
it is the best book every written. It changed my life.
there is no math book out there that is even close.

>> No.3403767

since you know the area of the washer that you get with the intersection of a plane and the sphere (the plane goes through the center), you can use pappus theorem to get the volume.

>> No.3403773

>x, y > 1

>> No.3403776

prove that the product of four consecutive integers cannot be a perfect square

>> No.3403789

check the number of 2's in the prime factorization

>> No.3403802


>> No.3403875

someone solve this

>> No.3403910

problem by erdos:

prove that in any subset of n+1 elements of {1,2,3...2n} there exist an a and b in the subset such that a divides b.

>> No.3403951
File: 17 KB, 344x400, euler1.jpg [View same] [iqdb] [saucenao] [google] [report]

bumping with euler

>> No.3403973

>high schooler on /sci/
>actually smart and not retarded

this is as amazing as a girl on /b/ during 2004

>> No.3403978

prove that the power set of any set A has cardinality larger than lAl.

>> No.3404021

OP here again... i think ive seen something similar before...

if the set is finite, its trivial since 2^n>n
otherwise, suppose there exists a bijection f.
now consider the set S (a subset of A) such that for all s in S f(s) does not contain s.
now consider f^-1(S)
it can neither be in S nor in S complement
so were done.

>> No.3404040


wow kid I bet you no more about math then most engineers here

>> No.3404047


wow kid I bet you know more about math then most engineers here

>> No.3404053

Assume the opposite is true, i.e. that there is a subset without a and b such that a divides b. Let x be such a subset. Let f(x) be the smallest number in x. Let y be a subset such that f(y) is maximal among all f(x), i.e. there is no valid subset z with f(z) > f(y). 2f(y) is not an element of y, since otherwise a=f(y) and b=2f(y) would disqualify y. Since any number that nontrivially divides 2f(y) is smaller than or equal to f(y), the only such number that is an element of y is f(y). Hence, the set y with f(y) replaced by 2f(y), say y', also contains no a and b such that a divides b. However, f(y') > f(y) since f(y) is not an element of y'. Since we assumed that y was chosen such that f(y) was maximal, this is a contradiction.

>> No.3404082

engineers don't do real math

>> No.3404083

I got a math number theory puzzle for you guys:

prove that 3N^4+2N^3-3N^2-2N is divisible by 24 for any integer where N is greater than 0.

It is rather difficult so don't be too ashamed if you can't do it.

>> No.3404102

just check all residues mod 3
and then all residues mod 8.

there might be a prettier solution... but this works too

>> No.3404120


nice solution
here's mine:

we partition the set into subsets in the form {a, 2a, 4a, 8a...} for odd a
we will have n sets (one for every odd number<=2n)
clearly if we have 2 elements in the same subset then one of them divides the other, and since we have n subsets and n+1 elements then we are done.

im OP again btw

and send more problems /sci/fags

>> No.3404134

A proof by simple induction is straightforward. But I'm guessing that's not what you had in mind?

>> No.3404147

(N - 1)N(N + 1)(3N + 2)

The first two guarantee the result is even.
The first three guarantee the result is divisible by 3.
Therefore the result is divisible by 6.
If N is odd, the first and the third guarantee it's divisible by 4.
If N is even, the second and the fourth guarantee it's divisible by 4.
So, it's divisible by 24 .

>> No.3404154

someone's gotta solve this

>> No.3404162

Prove that for any sequence of integers a1, a2, ... , an (not necessarily distinct), there is some nonempty subsequence whose sum is a multiple of n.

>> No.3404175


OP here again... send harder problems...
consider the sequence of sums (mod n)
a1, a1+a2, a1+a2+a3, a1+a2+a3+a4...
we have n such sums, meaning either there is a sum that is 0 mod n or there are two sums S1 and S2 with S1=S2 (mod n).
If there is a sum that's 0 mod n we're done.
Otherwise, we take S2-S1 to get a sum that's divisible by n.

for example if we have
a1+a2+a3+a4+a5+a6=m (mod n)
a1+a2+a3+a4=m (mod n)
then we have
a5+a6=0 (mod n)

>> No.3404186

Prove that any even integer > 2 can be expressed as a sum of 2 primes

>> No.3404187

Here's a differential equation I've been having trouble with recently:
<span class="math">
Any suggestions would be greatly appreciated.

>> No.3404195

Haha OP I bet you can't solve this!
> Generic unsolved problem I just found in Wikipedia.

>> No.3404214

f(x) = k ln(x)

>> No.3404216

...Wrote 'me own Tan function just today...
Testing so far looks like (+/-) 0.1% Accuracy =)

...I'll Call it 1% just to be safe ;)
Anyone wanna look?

>> No.3404240

OP again...
>be in school
>history class
>write sums for goldbach for 2<n<=1000
>continued doing that for 45 mins everyday until n=15000
>fuck history

>> No.3404274

Let A_n be the matrix filled in left to right, top to bottom with integers 1 to n^2 in order. Show that if det(A_n) = -2 that n=2

>> No.3404279

That's nice... I'm looking for x(t).

>> No.3404311

send harder problems
so subtract the last column from two other columns
these two columns are multiples of each other.

>> No.3404325

Uses two less accurate Approximations(?) to make one good(?)-ish one...

...Its still not perfect though =(
Half a degree out probably not close enough eh....

But here you go...
Using Side a / side b // angle A
(Assumes side a>=b // Haven't tested using a<b)
A1 = [a / ( a + b)] * 90
A2 = [ a*a / (a*a + b*b)] *90

Angle A = (A1 * 2 + A2) / 3

Pretty simple hey?

>> No.3404350

Wait... nevermind. I read that completely wrong, as d^2y/dx^2 = -k / x^2

>> No.3404377

It's all good... it's just like 3a.m. here, so I'm a bit crabby. I tried this:
<span class="math">
which makes it separable, but by the time you work it out it is ridiculously complicated (not to mention the 2nd time derivative is NOT -k/x^2).

>> No.3404379

Show for integers a,b a<b that if a^b = b^a that a=2 and b=4

>> No.3404388

Haha, I'm on EST too. And don't mind me, I've never taken differential equations.

>> No.3404414

theres -4, -2...
but except that
if you solve it then u get that (x,y)= (1 + 1/a)^a, (1 + 1/a)^(a+1)
for some a
and if x,y are rational then a is rational
and that clearly clearly can never be an integer

>> No.3404450
File: 38 KB, 800x600, mangel1.jpg [View same] [iqdb] [saucenao] [google] [report]

send moar problems

>> No.3404479

This one still needs a solution:
halp pl0x

>> No.3404492

sorry can't do much calculus...
the only calculus i know is from deriving myself and reading wiki...

>> No.3404502

If you're in highschool shouldn't you be doing Calculus by now? I recently passed AP Calc (just finished highschool), it was hella fun. Free university credit for scoring a 5 on the AP exam.

>> No.3404513

well i do know the stuff from AP calc BC, and i took the AP and got a 5. They don't really teach much there.
and i know some multivariable stuff...
but i mean i don't know differential equations or anything...

>> No.3404535

Wait you know matrices in high school? Your educational system is like fucking awesome...

>> No.3404541

nah... i taught myself... my school system is as shitty as all other ones...

>> No.3404549

Matrices are simple, why wouldn't they teach them in highschool? I was taught them here in highschool.

>> No.3404550


Try this:

x''(t) = -k/x(t)²
x''(t)*x'(t) = -k*x'(t)/x(t)²
x'(t)² = 2*k/x(t) + A (A is a constant)
dx/dt = sqrt(2*k/x + A)
dx/sqrt(2*k/x + A) = t

Then you gotta integrate this shit, too lazy to do it myself

>> No.3404555

Well they seem simple now, but I certainly wasn't taught about them in high school. I don't live in the USA, maybe that's why.

>> No.3404577

Yup, same here. We did Gaussian elimination on, like, 8-9th grade.

>> No.3404587

My erdos is 4.

>> No.3404639

...In true trial & error style, should I be putting Pi in there?

Angle A = (A1 * 2.14 + A2) / 3.14...

...Seems to even out its accuracy a bit =)
0.3 Degree's out on angle 67.5 (gives 67.8 instead of 68 // wants to be 67.5 really)

...Is this awesome or am i stupid?
=D either way....

>> No.3404660

bumping for more problems!!!!

>> No.3404723

What is the sum of every possible combination of 3 numbers 1 through N multiplied by eachother? In terms of N.

A.K.A. if N=3, the sum is 1*2*3=6
if N=4, the sum is 1*2*3+1*2*4+1*3*4+2*3*4=50

if this is too difficult, you can try the sum of every combination of two (aka if N=3, 1*2+2*3+1*3)

>> No.3404735

you use induction... it trivializes it...

>> No.3405076 [DELETED] 

erdos number 5
u jelly?

>> No.3405088

Yes, it can. Look for Gauß-Bonnet theorem.

>> No.3405109

bumpin for interest

>> No.3405127

anyone have a relatively simple proof showing for the irrationality of pi?

>> No.3405151

Basically it's
<span class="math">\sum_{i=1}^{n}\sum_{j=i+1}^{n}\sum_{k=j+1}^{n}ijk = \sum_{i=1}^{n}i\sum_{j=i+1}^{n}j(\frac{n(n+1)}{2}-\frac{j(j+1)}{2}) [/spoiler]

and you calculate this shit and so on

I tried to find its exact expression but holy shit that takes too long. It will be like N^5 or something.

>> No.3405616


Ok I solved it. If you start with y > x > e, it won't terminate in finite time. Nice problem - where'd you get it?

>> No.3405633

I am confused as to how you want it generalised.
Take R^n, add a point to get a n dimensional sphere. Triangulate using simplicial complexes and take the homology complex. Taking dimensions and adding along the chain gives you what you want I think.

>> No.3405690



Say y > x. We have m^m = x^y, so y > m > x, because z^z is an increasing function (for argument z > 1, which we're assuming here), and x^x < x^y < y^y. Same goes for n. So, at each step of the algorithm, the numbers get closer together.

Say we start at y_0 > x_0, with x_0 > e (you'll see why soon). If it finishes after finite steps, then let's call the last values before it terminates (x,y), with y > x. That it terminates means x^y = y^x = m^m.

Let A = m log m = y log x = x log y. Since m > x, then if A > m, we have A > x. If m > e, then A = m log m > m, so let's suppose that x_0 > e => x > e => m>e => A > m => A > x.

From y log x = A, we can write y = A / log x, and substituting this we have A = x log y = x log [A / log x]. One solution to this is x = y = m, or A = x log x, but that's not the one we're looking for, because we need one with y > x. So, are there any others (with A > x)?

Let's differentiate x log[ A / log x] - A with respect to A. The derivative is x/A - 1, and since we're restricting to A > x, it's negative. So, there can't be multiple zeroes on A > x, which means there's only the one.

So, if x_0 > e, it won't finish. QED

Question: Does the gap between x and y shrink to zero asymptotically? I think so, but I don't know how to show it

>> No.3405759

>implying Erdos ever actually touched a woman.

>> No.3405765

>implying this is important

>> No.3405785

I did matrices in high school. (British further maths)

>> No.3405786

Erdos had 13 children, 5 of which outlived him.

>> No.3405787

>implying op's pic is irrelevant to the discussion

>> No.3405794

lolwut? you got some sauce there, chief?

>> No.3405809 [DELETED] 
File: 35 KB, 600x480, Facepalm.jpg [View same] [iqdb] [saucenao] [google] [report]

>mfw I'm stoned and I read and wrote euler instead of erdos.

>> No.3405839


Canada fag here
You learn of matrices and things along those lines depending on which grade 12. math course you take.
You learn it in the simpler one knows as applied.

>> No.3405897

Wait what? Most people in the US math system learn matrices a little bit in 8th/9th grade, then cover them more in depth in 10th/11th grade

>> No.3405944

Erdős never married and had no children.


>> No.3405954

he had some problem where his genitals hurt if he got hard...
but who needs sex when u got math and meth?

>> No.3406051
File: 32 KB, 740x308, purity.png [View same] [iqdb] [saucenao] [google] [report]

moar math!

>> No.3406117

Oh yeah, the fucker was on amphetamines, he was challenged to a 500$ bet that he we addicted, and couldn't do a month without them. Motherfucker won the bet and then resumed the use.

>> No.3406122

erdos afterwards told the dude who bet him that he put his math back a month...
something about how he couldn't get ideas without it...

>> No.3406232

I made it up, thinking about how to find the exponential mean of two numbers. Someone in a different thread suggested I just solve log log m + log log m = log log x + log log y, which made sense and avoids ambiguity, but I still thought the original problem was interesting.

>> No.3406855


>> No.3408181


>> No.3408198

mine is 0...that's right...I'm not dead....

>> No.3408225

there are 3 people, one always speaks the truth, one always lies, and one guesses randomly. How can you find out which is which with only 3 questions?

>> No.3408299

Yes, I know how to do this.
V - E + F
Okay, Let D(n) be the number of n-1 dimensional 'pieces', so D(1) = V, D(2) = E, D(3) = F.

For an N dimensional shape the number
D(1) - D(2) + D(3) - D(4) + D(5) - ... + (-1)^N*D(N) is a constant. Not sure which one though, maybe N-1 or something.

>> No.3408306

Whoops, that constant looks like it might be 2N - 2.

>> No.3408313

aw god damnit, the formula is wrong too:
D(1) - D(2) + D(3) - D(4) + D(5) - ... + (-1)^(N-1)*D(N)
is the right one.

>> No.3408384

find the infinite sum
of the reciprocals of primes squared

>> No.3408415

prove fermat's last for specific exponents.

>> No.3408450

>implying this is possible

>> No.3408475

it is very possible

>> No.3408490

find all positive integer solutions to

>> No.3408498

a^2 + b^2 = c^2.
solve for A.

>> No.3408543

so we know the sum of reciprocals of squares is pi^2/6=a
and the sum were trying to find is B
and this is a geometric series we can easily solve...

>> No.3408559

a = sqrt(c^2 - b^2)

>> No.3408576
File: 10 KB, 230x311, JH_Poincare.jpg [View same] [iqdb] [saucenao] [google] [report]

Sup, bitches? Where's my boy Perelman?

>> No.3409446

/sci/ needs moar math

>> No.3409461

these aren't solved yet

>> No.3409489

Flip a coin repeatedly. You "win" if more than 2/3rds of the flips are heads. What is the probability that you win?

>> No.3409518

You win as soon as the total heads > (2/3) number of flips?

>> No.3409524


Oh, apologies for the ambiguity. Yes, the moment 2/3rd of your flips are heads, you win. If you get a head on your first flip, you win.

>> No.3409545

A part a one inch long needle is placed randomly on 2x2 surface. If another one inch needle is placed in the same manner, what is the probability that the two needles overlap?

>> No.3409553

> or >= ?
>=, then you win before the first flip.

>> No.3409597

Let X and Y be metric spaces and f: X -> Y be a map:
The map f is proper if its inverse image of a bounded set is bounded.

The map f is (uniformly) bornologous if for every R>0 there exists S>0 such that if d(x,y) < R then d(f(x), f(y)) < S for all x,y in X.

The map f is coarse if f is proper and bornologous.

Two maps g,h from a set X to a metric space Y are close if d(g(x), h(x)) is uniformly bounded in X.

Check that two maps f,g are close if and only if they arise from a coarse map X x [0,1] -> Y by evaluation at 0 and 1.

>> No.3409707


>> No.3409874
File: 55 KB, 654x1105, CoinGame python program.png [View same] [iqdb] [saucenao] [google] [report]

so i did an experiment. it looks like you'll win about 69% of the time. sounds good to me!

>> No.3409914

You'd be amazed how many people are utterly convinced the answer's 100%. /sci/'s had a couple shitstorms over this one. Josef was in the 100% camp for a while, I think.

>> No.3409935

The exact answer, by the way, is (5 - sqrt(5)) / 4.

>> No.3410659


Proof? I got the same answer, but did it in a hamfisted way.

>> No.3410662

But the game doesn't specify when you lose.

>> No.3410667

is there a way to prove that math isn't just made up bullshit like the rules of chess?

>> No.3410671

>prove that the product of four consecutive integers cannot be a perfect square

<span class="math">0 x 1 x 2 x 3 = 0^2[/spoiler]

>> No.3410683
File: 29 KB, 479x358, yessir.jpg [View same] [iqdb] [saucenao] [google] [report]


When you don't win, of course.

>> No.3410696



Nope. That doesn't imply that our Mathematical System is flawed, but we can't prove that it's not.

>> No.3410699

postin' in a cool thread

>> No.3410729 [DELETED] 

Me and my friends had this little math going going where you get three of the same positive integral and the common operators as well as the square root, the factorial and parentheses and have to arrive at 3. It's my turn at 11 and I'm now trying to prove it's not possible. To reduce the cases to a finite number, I'm trying to eliminate the square root and the factorial, as they are the only ones that can be used infinitely. What I'm asking now is: Is there any natural n greater than 1 so that sqrt(n!) is a natural number again?
I remember reducing the problem to 'Is there a prime k so that there are no primes p so that k<p<=2k', but don't recall the exact procedure.

>> No.3410748

Me and my friends had this little math thing going going where you get three of the same positive integer and the common operators as well as the square root, the factorial and parentheses and have to arrive at 3. It's my turn at 11 and I'm now trying to prove it's not possible. To reduce the cases to a finite number, I'm trying to eliminate the square root and the factorial, as they are the only ones that can be used infinitely. What I'm asking now is: Is there any natural n greater than 1 so that sqrt(n!) is a natural number again?
I remember reducing the problem to 'Is there a prime k so that there are no primes p so that k<p<=2k', but don't recall the exact procedure.

>> No.3410787

Found this gem in a philosophy book.

"Nevertheless, there are many unexpected connections in mathematics that are difficult to explain. For example, pi turns up in all kinds of quite unrelated places, such as the solution to Buffon's needle problem [infinite set of parallel lines one unit apart, one unit needle dropped, what's the probability of the needle landing on a line?]... The answer to this problem turns out to be [...], but why pi should turn up in this context is a complete mystery. At a deep level, then, there remains something mysterious about mathematics. "

>> No.3410921

Not sure why that's so hard to explain, since the probability of it crossing a line depends on the angle between the needle and the lines. You take the integral over the possible angles and there you are. At least I guess that'd be how it works

>> No.3411415
File: 8 KB, 458x339, circles.png [View same] [iqdb] [saucenao] [google] [report]


Yep, just take the expected value of the sine function over the interval (0, 1/2pi).

For a more interesting problem, replace the parallel lines with an infinite set of 1 unit radius circles tangent to 4 other circles.

>> No.3411476

mystery in math = you don't understand the material

>> No.3411805

You lose if you flip forever and never get up to the 2/3 ratio.

I never figured out the proof, although I've seen at least one posted. It involved recurrence relations, I think.

>> No.3411817

Pi shows up there because you put it in there. It'd be like an Object-oriented programmer saying it's a problem to skeptics that object oriented programming languages are so perfect for describing things. It's because real math is Turing complete, and every language is just as productive, though less effective in various carious, for solving problems.

>> No.3413313

positive integers was implied

>> No.3413344

you need an even number of each prime
so say the biggest prime you have is p
now to have it again you need 2p,
but there is another prime between p and 2p, a contradiction that p is the largest prime you have...

OP here btw and holy shit the threads still alive :D

>> No.3413349
File: 60 KB, 533x594, watt.jpg [View same] [iqdb] [saucenao] [google] [report]


>6 hours since last post

>thread is 'alive'


>> No.3413364

when i got here it was front pg
cause of >>3413313

>> No.3413558

if the first flip is heads u win, if not then use recursion (so then if it happens again after that tails you need 2 heads in a row to win, if u get HT or TH after then u do recursion again, if TT then recursion a bit differently)
this trivializes it

>> No.3413579

number the people 1,2,3, then ask if the number assigned to random+1=the number assigned to true (mod 3)
the rest is simple

>> No.3413594

consider solution with smallest a
but a,b,c all divisible by 3, so you can get a smaller a by dividing all variables by 3, a contradiction

>> No.3413623

and why is b divisible by 3 exactly?

>> No.3413638

for all n, n^2=0 or 1 (mod 3)
if b is 1 then there are no solutions since
1+0 or 1=0 (mod 3) cant happen

>> No.3413667


>> No.3413770

bumping a win thread

>> No.3413804


>> No.3413832

>implying this is possible

>> No.3413833

Prove that P=NP

>> No.3413877

... yeah

>> No.3413916

<span class="math">\sum_{n=0}^{+\infty}\frac{1}{2^{\frac{n(n+1)}{2}}} [/spoiler]

>> No.3413928

I may sound stupid but why is n² = 2 (mod 3) impossible?

>> No.3413994
File: 189 KB, 1000x753, 1299738168630.png [View same] [iqdb] [saucenao] [google] [report]


Hey , this looks like Gibbs phase rule...
any reason for that ? or it's just a coincidence ?


>> No.3414937

Yeah, that was probably the way I reasoned. Any idea on how to solve the actual problem? I doubt there is a prime with those properties, but I don't have much of a clue about how primes work.

>> No.3414948

this got me a 2.7/10 on physical chemistry.

>> No.3415033

We need more threads like this on /sci/ good work mentalmen.

>> No.3415140

I have just finished a degree in maths, so don't take me to be a crank but I don't accept the solution to the Monty-Hall problem.

Here's why. Instead of doors, have boxes with gold or coal.
After he opens the first coal box, you close your eyes and the two that are left are moved around so you don't know which is which. now you choose one, and have a half chance to get the gold. If you choose the same door as you did before, you STILL have a half chance to win.

So staying and switching give the same result, half a chance to win.

If anyone can explain why this is wrong please do because it really pisses me off.

>> No.3415352

well, 0^2=0, 1^2=1, 2^2=1 (mod 3)

>> No.3415362

experiment... take 3 cups, put a coin under one of them, get someone to play and do like 100 rounds. it should give you the intuition.

>> No.3415376

its well known that there's always a prime between n and 2n.

erdos (i think) wrote this poem about the theorem:

ill tell you once
and ill tell you again
there is always a prime
between n and 2n

>> No.3415382

i think he meant in a closed form...
as in no infinite sum...

>> No.3415383

It's called Bertrand's postulate.

>> No.3415385

well, if you know which one you chose earlier it doesn't matter that you shuffled them...

>> No.3415390

toasting in an epic bread
/sci/ needs more threads like this one.

>> No.3415394

Yeah, I meant they are indistinguishable obviously.

>> No.3415396

Try it. There are only 3 numbers mod 3. square them

>> No.3415405

When you first guess you have a 1/3 chance. Then the host basically lets you swap that door, for the other two. Of those two at least one will have a goat and he reveals it. If you switch you get 2 doors.

>> No.3415749

when you pick randomly then if you pick the one you had before there's a 1/3 chance, if not then 2/3.
there's a 1/2 chance you will pick either one, and 1/2(1/3)+1/2(2/3)=1/2
so the probability is 1/2, but if you pick the one you didn't have its 2/3

>> No.3415771

how do you prove pi is irrational?

>> No.3415775
File: 7 KB, 248x169, chrislanganreaction.jpg [View same] [iqdb] [saucenao] [google] [report]

I can prove God exists using binary logic. Umad, mathfags?

>> No.3415916

bump for moar problems

>> No.3415953
File: 66 KB, 550x733, 1310103261968.jpg [View same] [iqdb] [saucenao] [google] [report]

Are the mods even deleting math threads now? I was posting in the pure math thread just a few minutes ago but now it's vanished. It had some discussions relating to stock market speculation, I wonder if that's what caused it to be zapped?

>> No.3416002

they probably deleted it for a different reason

>> No.3416006
File: 200 KB, 595x425, broomrapefl.gif [View same] [iqdb] [saucenao] [google] [report]

You have a set S of 12 variables. Each is a real number. One variable has a value that is not equal to one; the other eleven variables all have a value of one.

You do not know which variable is the unusual one, nor do you know if it is greater or less than the others.

Define a function f: P -> R,
where P is the set of all subsets of S, and R is the set of real numbers. f(X) is the sum of all elements in X.

You have an oracle, which knows the value of each variable. However, you can only communicate with the oracle in the following way: You can present the oracle with two disjoint subsets of S, denoted X and Y, and the oracle will tell you which of the following is true:
f(X) > f(Y)
f(X) < f(Y)
f(X) = f(Y)

You may communicate with the oracle in this way three times. From this, you must deduce which variable has a unique value.

Note: This is entirely possible. If you try to prove that it is not, I will call you a troll.

Pic is broomrape.

>> No.3416051

Like the naked girl in the first post.

>> No.3416065

first question, you take variables 1,2,3,4 and 5,6,7,8.
if they are the same, then you know one of 9,10,11,12 is the unusual one, so you compare 1,2 and 9,10
if 1,2,3,4, and 5,6,7,8 are different, suppose WLOG that the sum of variables 1,2,3,4 is greater. now compare 9,10,11,12 with 1,2,5,6.
if they are different and WLOG 9,10,11,12 is larger, you know 5 or 6 is the different one.
if they are the same then one of 3,4,7,8 is different. so compare 4,7 with 9,10.

>> No.3416114

What if:
You compare 1,2,3,4 against 5,6,7,8 and 1,2,3,4 is greater.
Then you compare 1,2,5,6 against 9,10,11,12 and they are the same.
Then you compare 4,7 with 9,10 and find that they are the same.
Then the unique variable must be either 3 or 8 but you are out of tries!

I don't think your solution works in all cases.

>> No.3416266

you should compare on the second step 9,10,11,12,8 and 3,4,5,6,7

>> No.3416330

moar liek this

>> No.3416403

Has anyone got a good way of remebering the basic graph transformations? (Stretchs, translations)
I'm 20 and still have to stop and think when i'm doing them

>> No.3416407
File: 44 KB, 468x564, 1295250563872.jpg [View same] [iqdb] [saucenao] [google] [report]


ConspiracyCat is pleased with your answer

>> No.3416430

you mean the names of the transformations?

>> No.3416440

i dont think so(unless they have formal names i dont know), the stuff you learn when you're about 15/16ish. Like f(x+2) is a horizontal shift of a graph 2 units to the right.
Mostly the reflections are what trip me up

>> No.3416450

Thanks, I had no idea.

>> No.3416462

oh... i learned that when i was very young (6-7) so its intuitive for me.
but this is kind of how I think about it.
the trick is usually to find where its f(0)

so if you have f(ax+b), (i wont do anything more difficult than linear transformations cause its similar) then this is the same as
f(a(x+b/a)), which means you dilate by a with the center (f(0)) at -b/a.
so you put f(0) at x=-b/a and then you just put the graph, but on your axis, instead of intervals of a, you put intervals of 1.

hope this makes sense...

>> No.3416534

someone solve this

>> No.3416545

Find the coordinates of stationary points and nature of them of y=(2x-3)^6

i dy/dx this and i got 12(2x-3).

i solved for x and got 3/2 but when i sub it back in after i d2y/dx^2 i get 0 which means its a point of inflection ntot a minimum like in the answer.s any help pls?

>> No.3416553

shouldnt the first derivative be 12(2x-3)^5?

>> No.3416555

if the 2nd derivative is 0 it doesn't imply its a point of inflection. you have to check the 3rd derivative, and if it is 0 it may still be a min.

>> No.3416559

I'm having trouble interpreting OP's image.

Does it mean
>Have sexual intercourse with women and prove theorems


>Disregard women and prove theorems.

>> No.3416560

sorry i meant that

are you sure? ive never heard of finding the 3rd dy/dx never used it before...

>> No.3416564


the latter

>> No.3416565

erdos (guy in pic) had some problem and he couldn't have sex...
so he did math instead
and meth

>> No.3416571

take y=x^6 at x=0.
the 2nd derivative is 0, but it clearly is a minimum.

>> No.3416581


>> No.3416594

A cylindrical hole is drilled straight through the center of a solid sphere. The length of hole in the sphere (i.e. of the remaining empty cylinder) is 6 units. What is the volume of the remaining solid object (i.e. sphere less hole)?

>> No.3416596

just take lim z-->0 of f(3/2+z).
since this is always greater than f(3/2) then its a min

>> No.3416608

reword the problem please. I don't get what you're asking.

>> No.3416658

this is simple with calculus.
any discrete way to do it?

>> No.3416664

I like this one.

There are n balls rolling along a line in one direction and k balls rolling along the same line in the opposite direction. The speeds of the balls in the first group and in the second group are equal. Initially the two groups of balls are separated from one another and at some point the balls start colliding. The collisions are assumed to be elastic. How many collisions will there be?

By the way, if you are ever bored or want to waste several hours:


>> No.3416673

this is really simple... just think about the collisions as the balls passing through each other...
so each one of the n hits each of the k 1 time so its just nk

>> No.3416675

picture a scenario in which there are 100 doors instead of 3, with 1 having a car behind it and 99 having goats behind them.
You choose your initial door, and Monty opens 98 doors, showing you the goats behind them. Would you consider it beneficial to switch to the only other remaining door?

>> No.3416680

By your logic, wouldn't it be min(n,k)?

>> No.3416687

no. try k=1. clearly there wont be just 1 collision.
k=1 n=2 is easy to visualize.

>> No.3416691

If I closed my eyes and forgot which door I chose and choose randomly I have a half chance of getting the right door, right?

Flipping a coin to choose whether you stick or not gives you a half chance of being right no?

>> No.3416696


explained here:


>> No.3416701

There's a simple problem I've been having trouble with. I'm new to combinatorics, but I still feel a little dumb.

There are 45 white balls and 5 black balls distributed uniformly at random into 10 bins such that each bin has exactly 5 balls.
How many ways are there for the first two bins to have all white balls?

Answer 1: There are 45 white balls and 10 being picked from them, so nCr(45,10)

Answer 2: The first 10 balls are fixed, and there are 35 white balls and 5 black ones left, so it's nCr(40,5)

Obviously, one of these is wrong. Can someone help me out?

>> No.3416707

No, that's the thing. Think about your chances of picking the door with the car vs your chances of picking a door with a goat in the first place.

>> No.3416710


Why is there a 1/3 chance if I pick randomly the one I chose at first. The previous decision has no bearing on this one. Now there are only two things to choose from and I choose one at random. The fact that earlier I chose something has had no effect on where the car is and so cannot effect the probability of which door it is behind.

>> No.3416723

suppose the balls are distinguishable.
so number all the balls so that 1-45 are white and 46-50 are black.
now in bin 1 we chose 5 white ones so 45C5
in bin 2 we choose 5 whites, but there are only 40 left since 5 are in bin 1, so 40C5
in the next bin anything is fine, and there are 40 balls left, so 40C5
next one is 35C5
and so on
so its


if they are indestinguishable, its just how many different ways can you put the 5 black balls in 8 bins, which you can do cases:
case 1: all 5 balls in same bin
case 2: 4 balls in 1 bin, 1 ball in another
and so on

>> No.3416732

suppose you pick the one you chose 1st
now we know you picked it first (we are supposing that)
and then the probability is 1/3.

>> No.3416741

Your previous decision does, in fact, alter the situation.

In the beginning, you have a 1/3 chance of choosing the car. Monty opens a goat door. Now if you switch, you lose, and if you stay, you win.

You have a 2/3 chance of choosing a goat initially. Monty opens the one other goat door. Now if you switch you win, and if you stay you lose.

Therefore, your chance of winning by switching is 2/3 and your chance of winning by staying is 1/3.

>> No.3416754
File: 44 KB, 1024x768, monty.png [View same] [iqdb] [saucenao] [google] [report]

Here's a handy diagram I drew for a thread a few weeks ago

>> No.3416771

Thank you. I accept the reasoning and will never bring it up again.

My problem was this: I can get 50% chance of winning by flipping a coin to switch or not.

Answer is: You gett 66% chance of winning by always swapping.

I'm an idiot. Thank you /sci/

>> No.3416800


>> No.3416830

Hard to know if you've seen these before but this is always a good one:

In a room there are some people who shake hands with each other. No one shakes hands with themselves and some pairs of people don't shake hands.

Show that an even number of people shook an odd number of hands?

>> No.3416846

easy... the sum of the number of handshakes of everyone must be even (since each handshake involves 2 people), and if an odd number shook an odd number of times then the sum is odd.

>> No.3416855


>> No.3416888

nah you're not an idiot this 'monty hall problem' is famous and alot of very intelligent people get it wrong at first.

>> No.3416899

Two players play a game where on their respective turns, each player says a number between 1 and 10 higher than the current score, which starts at 0. Whoever says 100 first wins. Does the first player have a winning strategy?

>> No.3416917

yes, he picks 1, and then picks 11-(what his opponent said last turn).
that way first player says all numbers in the form 11n+1, and 100=11*9+1

i'm disappointed /sci/... post harder stuff

>> No.3416989
File: 35 KB, 612x500, Augustin-Louis Putin.jpg [View same] [iqdb] [saucenao] [google] [report]

I'll just leave this here.

>> No.3417009

2 people shake hands once. Then the number of handshakes is 1. ??

>> No.3417025

but we count the sum of the number of handshakes of each person.
person 1: 1 handshake
person 2: 1 handshake
1+1=2 which is even

>> No.3417035

Cauchy's proofs for the Cauchy Schwarz inequality and the AM-GM inequality are two of the most beautiful discrete mathematical gems.

>> No.3417039

Here's a hard one: Show that the probability of two positive integers chosen at random being relatively prime is 6/(pi^2)

>> No.3417045


Isn't the proof for the Cauchy Schwarz inequality just tiresome algebraic manipulation? I've never seen a different proof.

>> No.3417064


by random integers i believe you mean random between 1 and n as n goes to infinity...
the probability that they are both divisible by some prime p is 1/p^2, and we need to have that they are not both divisible by any prime.
from here its simple... (i solved the calculations earlier here: >>3408384 >>3408543)

>> No.3417066

the prettiest one is this:
(a_1x+b_1)^2+(a_2x+b_2)^2+...+(a_nx+b_n)^2= f(x)
now since this is always >=0, its descriminant is <=0

>> No.3417080


you can also prove it with vector dot products, and how cos(x) is between 1 and -1
but the proof with the quadratic is prettier

>> No.3417108

yes 1,2,3...., i said "positive integers".

You've skipped too many steps, how did you show that the prob that they are both divisible by some prime p is 1/p^2?

>> No.3417122

well, if we want them to be divisible by p,
the prob that the first one is is 1/p
same for the second
so we multiply 1/p*1/p=1/p^2

>> No.3417132

and what i meant was that you cant pick a random integer... you can only pick a random integer in an interval
so we have to define the interval for our random variable N to be between 1 and n, and then we can take lim n-->infinity

>> No.3417190

actually its easier to do the calculation like this:
the product over primes of (1-1/p^2)=product 1/(1/(1-1/p^2))= 1/(the infinite sum 1/1^2+1/2^2+1/3^2...)=6/pi^2

>> No.3417209

Science & Math?
In my /sci/?


>> No.3417218

shit thread is shit

>> No.3417239

win thread is win

>> No.3417447


>> No.3417466
File: 42 KB, 300x300, madmen.jpg [View same] [iqdb] [saucenao] [google] [report]

>recursion a bit differently
>this trivializes it

>> No.3417477

it does...
then theres some algebra and shit... you can let the candyass engineers do that

>> No.3417500
File: 16 KB, 251x141, AAA.png [View same] [iqdb] [saucenao] [google] [report]


(+100 Internetz for a good hint as well)

>> No.3417548

check them dubs

>> No.3417578

Betty and Sam are both given a natural number less than twenty.

Betty is instructed to subtract two from her number, divide the result in half, and then add two to the new result.

Sam is instructed to divide her number in half, add two to the result, and divide the new result in half.

Before carrying out their instructions, Betty and Sam secretly swap numbers. Their results should have been the same, but instead they differ by six.

What numbers were given to Betty and Sam?

>> No.3417723


Betty has the number 8 and Sam has the number 16.

<span class="math"> \frac{b-2}{2}+2=\frac{b+2}{2} [/spoiler]
<span class="math"> \frac{\frac{s}{2}+2}{2}=\frac{s+4}{4} [/spoiler]
<span class="math"> \frac{b+2}{2}=\frac{s+4}{4} [/spoiler]
<span class="math"> 2(b+2)=2b+4=s+4 [/spoiler]
<span class="math"> 2b=s [/spoiler]

Now when you switch them the difference is 6, so:

<span class="math"> \left | \frac{s+2}{2}-\frac{b+4}{4} \right |=\left | \frac{2s-b}{4} \right |=6 [/spoiler]
<span class="math"> \left | 2s-b \right |=24 [/spoiler]
<span class="math"> \left | 2(2b)-b \right |=\left | 3b \right |=24 [/spoiler]
<span class="math"> \left | b \right |=8 [/spoiler]
<span class="math"> b=\pm8 [/spoiler]

Because b<20 this gives b = 8 and therefore s = 16. You can plug those numbers back into the original equations and see that it satisfies everything.

>> No.3417733

I keep seeing this problem on this board. Does it have a canonical name?

>> No.3417782


the math looks correct, but my question is this:

since they differ by six, why is b=8 and s=16. that's a difference of 8. what's the deal? =|

>> No.3417797


The problem doesn't state that the numbers Betty and Sam get differ by six. It says that after they switch numbers with each other and do the appropriate calculations the resulting numbers differ by six.

>> No.3417847

Let n be any finite sequence of digits abc...k, show that there exists an n such that 2^n = abc...k....m

ex: pick the sequence 3, then 2^5 = 32, so n=5 works.

>> No.3417876

sorry mistake:
>Let abc...k be any finite sequence of digits,

>> No.3417918


Still not totally sure what it is you're asking.

>> No.3417940

If you give me any finite sequence of digits, say 1531513553634, then i can always find some integer n such that 2^n = 1531513553634(somemore digits)

like if you picked 409, I'd say n=12 since 2^12 = 4096

show that i can always do this.

>> No.3417949

It emerged in my research.

>> No.3417960

So, for any natural number X, X*10^a + b = 2^n for some n.

>> No.3418177


>> No.3418377

how do you display math nicely here in the forum?

>> No.3418385


put it in [m ath] [/m ath] tags (without the spaces).

>> No.3418955
File: 104 KB, 233x350, 1251689670442.png [View same] [iqdb] [saucenao] [google] [report]


>> No.3419974

captcha: godel

>> No.3420192

therefore pi

>> No.3420219

why are you using vartheta and varphi?

>> No.3420229

somehow looks like cauchy-riemann-eqn ^^

>> No.3420231

help with this please

4x over (x-6) times 2 over (x+6)

>> No.3420256
File: 180 KB, 828x1720, geometry.png [View same] [iqdb] [saucenao] [google] [report]

This thread seriously lacks geometry.

>> No.3420268

all the problems in your post are easy as fuck

>> No.3420272

Feel free to post some problems that are of higher difficulty so that others like yourself may enjoy solving them. :)

>> No.3420279

prove that if 2 internal angle bisectors of a triangle are congruent, then the triangle is isosceles.

>> No.3420283

prove that all euclidean constructions with a compass and straightedge can be done with only a compass.

>> No.3420287

wait, this thread is actually 4 days old. I'm proud, /sci/

>> No.3420299

OP wins an internet for starting this

>> No.3420304

In a triangle ABC, prove <div class="math">tan(A)tan(B)tan(C) = tan(A) + tan(B) + tan(C)</div>.

>> No.3420344

(tan A+ tan B)/(1-tan A tanB)= (tan 2PI- tan C)/(1+tan 2PI tan C) = -tanC
(tan A+ tan B)= -tan C(1-tan A tan B)

>> No.3420354

I've never even paid attention to that.
wtf? this shows tanA+tanB+tanC=tanA*tanB

>> No.3420358

Look at that last line again. Multiply tan(C) through both terms of the parenthesis.

>> No.3420362

oh I'm sorry, I must be really high today! my bad!

>> No.3420370

tg(A+B) = tg(Pi-C)
given the fact that tg(Pi - x) = -tgx, we have
tg(Pi-C) = -tgC
which means that:
tg(A+B) = (tgA+tgB)/(1-tgA.tgB) = -tgC

<=> tgA + tgB = -tgC (1-tgA.tgB)
<=> tgA + tgB + tgC = tgA.tgB.tgC

>> No.3420487

Suppose <span class="math"> k [/spoiler] runners having distinct constant speeds start at a common point and run laps on a circular track with circumference 1. Then for any given runner, is there a time at which that runner is distanced at least <span class="math"> \frac{1}{k} [/spoiler] (along the track) away from every other runner?

>> No.3420577

Could you help me out on #23 then?

>> No.3420579

Shit, I mean the second #23, the one with 2 semicircles.

>> No.3420622

notice that the answers don't depend on the position of the center of the circles.
If the point is at the center of the square, you have a single circle (the two halves are the same)
Therefore, you have a circle with R=1
The surface is then: <span class="math">\pi[/spoiler]

>> No.3420627
File: 37 KB, 478x384, s.jpg [View same] [iqdb] [saucenao] [google] [report]

The answer is B.

Refer to picture and what I have labeled as s.

The radius of the small circle is s/2.

The radius of the big circle is s*root(2)/2.

The area of the shaded region is 3s^2*pi/8.

The sum of the radii is 2.

s/2 + s*root(2)/2 = 2

Solve for s and substitute in 3s^2*pi/8 for the area.

It's messy, I'm sure there's an easier way to do it, but it works.

>> No.3420635

The problem explicitly states they are different sized semicircles.

>> No.3420642

the problem doesn't prevent you from taking a "limit". (the surface is a continuous fonction, since it is constant, according to the possible solutions)
I can then assume that the point is at the exact middle of the diagonal, can't I? (I'm really asking, maybe I'm wrong but this is the reason that makes me think I'm not)

>> No.3420656

No, you can't assume that. The answer is pi when the centers match the center of the square, but not otherwise. Assume the radii are distinct and calculate it.

>> No.3420657

>>3420627 plus I think there is a mistake: the radius of the small circle isn't s/2! (what you refer at in the drawing isn't a diameter)

>> No.3420676

so you deny that the sum of the two disctinct sufaces is a continuous function? I can't agree :/

>> No.3420693
File: 45 KB, 478x384, s.jpg [View same] [iqdb] [saucenao] [google] [report]

What? Yes it is. Draw the rest of the square, clown.

>> No.3420699

It's not a function dude. There is only one possible situation where there are two semicircles of different radius inside a square. You can't just pick any point on the diagonal and draw semicircles in there. It's a specific point. There's no limit to take because the function is not continuous for there are infinite points on the diagonal where such semicircles cannot be constructed.

>> No.3420700 [DELETED] 

oh, you're right.
However, tell me:
you say "solve for s"
Which means that the way you reason allows a single value for s. What about the surface for a different value of s?

>> No.3420719

oh right, I didn't pay attention to that. Thanks for highlighting it!

>> No.3420777

Why is combinatorics very difficult? I found those geometry questions fair but problems with combinations and anything discrete leave me with a blank stare.

>> No.3420792

People need to learn how to count. A lot of people lack conditional probability skills, and results in probability are sometimes counter-intuitive.

"How the hell am I supposed to count ALL those things?"

Such as the Iraq.

>> No.3421049

Calculate sqrt(2 + sqrt(2 + sqrt(2 + ... )))

>> No.3421057

Let that number be equal to x.

Square both sides.

x^2 = 2 + x


x = 2 or x = -1.

2 works.

>> No.3421305

another challenge for you guys:
show in two different ways that
<span class="math"> \lim_{n\to +\infty} n^2 \int_0^{1/n} xf(x) \,\mathrm{d}x = \frac{f(0)}{2} [/spoiler]
(one of them is obviously a change of variable, but I'm struggling with finding a second method)

>> No.3421332

actually, people who have these exercises to do don't have access to series expansion (which works too, before integrating), so if someone has a more elementary solution, it is welcome!

>> No.3421385

In pentagon ABCDE we have the following:
AB || CE
AC || DE
AD || BC
AE || BD
BE || CD

Prove pentagon ABCDE is a regular pentagon.

>> No.3421438

well phi (1.61...) in nested radical form is sqrt(1 + sqrt(1 + sqrt(1 + ... )))

>> No.3421442

This thread is inspiring me to open up my Calculus book and stop dicking around.

>> No.3421647

Thanks a lot man, this thread is quite inspiring. Although I'm not bad in math (for my age and location) all these people make me realize that I really don't know shit, and that I need to learn math asap.


>> No.3421663


>> No.3421721


>> No.3422741


I spent about 10minutes looking at this.

If you work out the derivatives (ie quotient rule on the left, product rule on the right) and then square and sum both of those equations you'll get an equation without any cos(f) or sin(f) that looks like

(df/dy)^2 = [cos^2(x) + sin^2(x)*(df/dx)^2]/sin^4(x)

I don't have much knowledge of PDE's and it doesn't seem to be solvable by separation of variables.

gl bro

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