/sci/ - Science & Math

[Anonymous No.8249063]

ITT: Equations, results and proofs that blow your mind

[Anonymous No.8249076]

what the heck is this?

[Anonymous No.8249140]

>>8249063
wut

[Anonymous No.8249215]

>>8249063
>> observe that for all n > 1 and all m > 1, mn is positive
>> observe that 0(1) is 0
>> observe that 0 + something positive is positive
>> observe that (1)(2) is 2
>> observe that 2 + something positive > 2
>> observe that you are a giant retard OP who misunderstands the theorem

[Anonymous No.8249451]

>>8249063
>writing your m's and n's like that
K Y S
Y
S

[math]\mathcal{FA\,M}[/math]

[Anonymous No.8249467]

>>8249063
Can someone explain?

(1)(0) = 0
0 + (2)(1) = 2
0 + 2 + (3)(2) = 8
0 + 2 + 8 + (4)(3) = 22
... = infinity?

How is this not infinity, and in fact 1/12?

[Anonymous No.8249565]

>>8249467
OP is just being a dummy

[Anonymous No.8249574]

>>8249467
Cesaro convergent series :v

[Anonymous No.8249579]

3+7=4+6

[Anonymous No.8249956]

>>8249215
0 is not positive

[Anonymous No.8250038]

>>8249579
WAIT.....do that again...

[Anonymous No.8250080]

how do I do this
Please help a retard in need

sqrt(x+7)+sqrt(x)=7
It's part of irrational equation chapter
I see that squaring 7 on its own doesn't give the right number, and I put this into a program showing that x=9
How do I do this.

[Anonymous No.8250086]

Imagine you would not be able to recognize your cock in POV; Than you would saw tons of your porn with observer would be different; How would you percieve that?

[Anonymous No.8250089]

>>8249063
Surely it would be 1/144?

[Anonymous No.8250090]

>>8249467
Overfloats!? :D

I dont know really; I cannot.

I prefer playing generals to math/.

[Anonymous No.8250091]

>>8250090
Po slovensky by som povedal len ze sa niekto snazi vypocitat obsah kruznice.

[Anonymous No.8250104]

>>8250080
multiply both sides by sqrt(x+7) - sqrt(x)
this yields x + 7 - x = 7sqrt(x+7) - 7sqrt(x)
so sqrt(x+7) + sqrt(x) = 7sqrt(x+7) - 7sqrt(x)
or 8 sqrt(x) = 6sqrt(x+7) <=> 64x = 36x + 252
=> x = 252/28 = 9

[Anonymous No.8250112]

>>8250104
thanks alot
All I wonder now is why I have to multiply sqrt(x+7) MINUS sqrt(x)

I know the sign changes when you put it on the other side, but why is it not -sqrt(x+7) - sqrt(x)?

[Anonymous No.8250121]

>>8250112
because (a+b)(a-b) = a^2 - b^2

[Anonymous No.8251294]

>>8249467
He's trying to meme off of convergence theory that iirc states that all integers between 1 and infinite add up to 1/12 if you make some retard assumptions

[Anonymous No.8251297]

>>8249956
you cant read

[Anonymous No.8251301]

>>8251294
no
thats never what it was
its the reinmann zeta function which is complex valued, that alone should tell you it is not the same as summing all the natural numbers, which clearly diverges

[Anonymous No.8251734]

>>8251294
Nope. Summing the natural numbers never totals to -1/12 under any consistent assumptions.

However, the Reimann Zeta function Z(s), which happens to equal the sum of 1/(1^s) + 1/(2^s) + 1/(3^s) ... for all s with real part greater than 1, can be shown by complex analysis to equal -1/12 for s=-1.

However, the Riemann Zeta function at s=-1 is *not* in fact the sum of all natural numbers; it is a different function with a different meaning, due to it actually being defined on the complex plane instead of the real numbers, which have somewhat different structure. However, if you find yourself having to sum all the natural numbers for a calculation in which the answer should obviously be finite, consider that the equations may actually be asking you to evaluate the Reimann Zeta function. This comes up sometimes in the kind of informal, messy math physicists do.

[Anonymous No.8251750]

only perelman could recognize this

[Anonymous No.8251755]

I really like the proof that the conic sections are what they are through the application of dandelin spheres. Picture shows the case for the ellipse.

[Anonymous No.8251770]

>>8251750
>fake quote
>promoting some stupid shit
>promoting some stupid kid
eh...

[Anonymous No.8251810]

pi = 4(the infinite sum of ((-1)^k)/(2k+1) in which k starts at 0

e= the infinite sum of 1 / k! in which k starts at 0

phi (the golden ratio) = (1 + the square root of 5) / 2

[Anonymous No.8251893]

>>8250112
bra it's got nothing to do with 'putting it on the other side', you simply multiply LHS by the conjugate (minus version) so that you can rid it of the square roots, because as >>8250121 said, it forms the difference of two squares factorisation

[Anonymous No.8252307]

>>8249215
>Observe

faggot

[Anonymous No.8253073]

I'd love to see a derivation of the result in OP. Summing divergent series is my fetish. Here's something pretty mind blowing: the product of all prime numbers is 4π^2.

http://cds.cern.ch/record/630829/files/sis-2003-264.pdf

[Anonymous No.8253135]

>>8251297
slw;'WIhnddn;kj zag4Bnbn\]pf

[Anonymous No.8253515]

Anyone wanna take a crack at this?