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/sci/ - Science & Math

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3284510 No.3284510 [Reply] [Original]

I think I have the potential to be the next Ramanujan.
Test me.
(Obviously not ridiculously technical questions such as the Riemann hypothesis, difficult but elementary number theory is my specialty).

>> No.3284520

Ramanujan was an Indian who only had access to a few old math books at a library. What was amazing is the limited knowledge he had access to. By being on the internet, you do not fit those qualifications.

>> No.3284524

Prove the following:

Every simply connected, closed 3-manifold is homeomorphic to the 3-sphere.

>> No.3284557
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3284557

Skwirmer is it? alright then kiddo, ill give ye something the consida

Give me a solution to this here problem is it?

Supposin a1, a2, . . . , an is a list of n numbers with the following propertays:
The sum of those there n numbers is 500. The sum of the smallest three of them there numbers is 48. The sum of the largest two of the numbers is 35.
now skipper repetition is allowed in them there series

>> No.3284555
File: 227 KB, 815x1222, cutey_Emma_gelb.jpg [View same] [iqdb] [saucenao] [google]
3284555

Write a sentence which, like thin one, only includes numbers represented in base 10.

>> No.3284567

Prove there is no largest Mersenne Prime.

>> No.3284571

Prove the following:

Every even integer greater than 2 can be expressed as the sum of two primes.

>> No.3284578

solve for x:

y = x + 1

>> No.3284640 [DELETED] 

>>3284571
>>3284567
>>3284557
>>3284555
>>3284524

well, fuck. nigga, you got a lot of work to do if you want to be the next Ramanujan.

>> No.3285330

>>3284578
Fuck you need to get to those books NOW!

>> No.3285342 [DELETED] 
File: 4 KB, 210x229, 1307490330596.png [View same] [iqdb] [saucenao] [google]
3285342

>>3284571
>mfw you posted that

>> No.3285352

>>3284571
express 3 as the sum of 2 primes

>> No.3285368

>>3285352
>implying 3 is prime

Its factors are 1.5 and 2 dumbshit.

>> No.3285373

prove that a set of integers contains the integers it contains

>> No.3285381

>>3285368
2/10

>> No.3285378

>>3285352
3 is not an even integer numbfuck

>> No.3285376

I'm really tempted posting my logarithm problem again.

>> No.3285397

Prove that every prime of the form 4n+1 can be written as the sum of two squares. ex. 5 = 1^2 + 2^2

I'm obviously talking about integers only.

>> No.3285407

>>3285376

Fuck your logarithm problem.
I saw it more than once and I hate the fact that I'm not able to solve it.

>> No.3285414

>>3285407
I would post coolface now if your answer didn't apply to myself as well.

>> No.3285417

>>3285376
post it

>> No.3285429
File: 34 KB, 310x310, 1289354062878.jpg [View same] [iqdb] [saucenao] [google]
3285429

>>3284510
>I think I['m]. . .the next Ramanujan.
>Test me.
>elementary number theory is my specialty

>> No.3285431
File: 105 KB, 2700x1300, nested logarithms.png [View same] [iqdb] [saucenao] [google]
3285431

>>3285417
You just took the red pill dude

>> No.3285436

>>3285431
Haven't added the limit discussion we've had yesterday yet, the latex is probably a week old. (Not that I've made any progress worth mentioning since then)

>> No.3285471

>>3285431
damn you, i still haven't stopped thinking about this

>> No.3285484

>>3285431
>why.jpeg

>> No.3285506

>>3285484
Because Joseph is an underage failure who hands out red pills on the internet

>> No.3285512

>>3285431
i cant even fully comprehend the problem and its still melting my mind

>> No.3285516

>>3285431

The answer is "yes, there is a way to make the index continuous"

next question.

>> No.3285520

>>3285431
I've actually done this before for sine and cosine, but the process should be the same.

Protips for great success:
1. http://en.wikipedia.org/wiki/Function_iteration
2. f^(n+1)(x) = f(f^n(x))
3. Taylor series about the fixed point of the function.

I'll leave you guys to figure out the rest. What I have in mind is not an analytical solution, but you can approximate fractional and imaginary iteration counts to arbitrary precision.

>> No.3285526

>>3285512
Just read the abstract, it's pretty straight forward.
<span class="math">\ln(x)[/spoiler] is "apply the ln once to x".
<span class="math">\ln(\ln(x))[/spoiler]is "apply the ln to x twice"
Is there a function that applies the ln half a time (i.e. applying that function twice would result in the ln)? What about a function you have to apply <span class="math">\pi[/spoiler] times to get the ln?

>> No.3285555

>>3285520 What I have in mind is not an analytical solution, but you can approximate fractional and imaginary iteration counts to arbitrary precision.
Blaaah. You can of course fit some function through there, but who's to say it's the right one. The actual function might even diverge for <span class="math">k=\frac12,\frac32,\ldots[/spoiler]. Some arbitrary interpolation does by no means satisfy things like <span class="math">nlog_k\circ nlog_l=nlog_{k+l}[/spoiler] etc.

From a mathematical point of view the first step would be to show existence, but that seems far out of my range, so I'm trying to prove existence by example. It's very likely to be unsuccessful though.

>> No.3285564

>>3285555
>Some arbitrary interpolation does by no means satisfy things like nlogknlogl=nlogk+l etc.
The way I did it preserved that property. I didn't interpolate; I solved a bunch of recurrence relations.

>> No.3285572

>>3285564
Then I'm interested in how you did it for sin/cos. Hooray, new input!

>> No.3285606

>>3285520
logarithms don't have a fixed point on the real number, so fixed point studies seem a bit arbitrary.
But another possible route of studie is to find a way to define the property for any rational number, then argue by density to define a continuous map on R.

>> No.3285610

>>3285606
I'm considering <span class="math">\ln(1+x)[/spoiler], which luckily has a fixed point at 0.

>> No.3285627

No one wants to ask simple questions, so here's one:

<span class="math">\int_D{x^2+y^2+1}dxdy \; \; \; D=\{(x,y)/x^2+y^2\leq 1\}[/spoiler]

Just give the step necessary to transform this into a simple calculation

>> No.3285635

>>3285572
Let me see if I can find the mathematica notebook I was toying with...

If I don't, then the easiest way to explain it is this:
<span class="math">f^{n+1}(x) = f(f^n(x))[/spoiler]
So you differentiate for x, and then solve the recurrence relation for n. <span class="math">f^{n}(x) = x[/spoiler] if x is the fixed point (0 for sin and ln(x+1)). The recurrence relation you solved for is the first derivative in the Taylor series. Differentiate again and solve the recurrence relation again for the second term, etc. etc. It comes down to a computational clusterfuck because you use the chain rule, and higher derivatives of the chain rule end up being a combinatorial mess.

I may not have explained it the best way, but in the end you get a Taylor series that's basically a function of two variables: f(x,n), where n is the number of times the function is applied. You've gotta construct the Taylor series around the fixed point of the function or it doesn't work.

>> No.3285667

>>3285635
I don't understand. I've got the derivative of the nested logarithm, it's
<div class="math">\frac\partial{\partial x}\mathrm{nlog}_k(x) = \prod_{j=0}^{k-1}\frac1{1+\mathrm{nlog}_{k-1}(x)}</div>... where nlog is a nested ln(1+x).
Is that what you meant by "solve the recurrence equation"?

However, I don't understand how this relates to the Taylor expansion around some value of k.

>> No.3285670

>>3285667
the k-1 in the fraction should be a j

>> No.3285691

Alright Ramanujan,
Let us start simply:

>Given 7 distinct positive integers that add up to 100, prove that some three of them
add up to at least 50.

>> No.3285735

>>3285667
I didn't do it that way. I meant differentiate <span class="math">f^{n+1}(x) = f(f^{n}(x))[/spoiler] for x, replace x with the fixed point of the funciton (i.e. zero for ln(x+1)) and then solve for a closed form solution in n. I'm not sure how else to explain it...

I wish I could find my old high school notebooks. I was jacked up on amphetamine 24/7 and this is one of the problems I attacked vigorously. I know I ended up deriving a solution for any function that was smooth and had a fixed point, but it's been a while. If I ever find it, I'll email it to you.

>> No.3285754

>>3285735
Uh?
<span class="math">\partial_xf^{n+1}(x)=\frac1{1+f^n(x)}\partial_xf^n(x)[/spoiler].
Insert fixed point <span class="math">f^n(0)=0[/spoiler], <span class="math">\partial_xf^n(0)=1[/spoiler], result: <span class="math">\partial_xf^{n+1}(0)=\frac111=1[/spoiler]. Uhm ...

>> No.3285764

>>3285691

Does it have to be exactly 3 of the int?

>> No.3285766

>Hey guys, I think I might be a genius. I have absolutely nothing new to contribute but I do have access to Wolfram Alpha. Quiz my copy/pasting abilities.

>> No.3285777

>>3285764
That is what the question says, yes.
Also, 2 is false and 4 is trivial.

>> No.3285781

>>3285777
what if one of the
integers is 50?

>> No.3285785

>>3285777

Seems a pointless case check, although possibly there's a way to speed it up. If I needed to result I'm sure I'd do the work, but it doesn't excite me in any way.

>> No.3285790

>>3285754
Yeah, I've fucked up the explanation. Let me sit down and derive it again...

>> No.3285797

Didn't this guy focus on infinite series and stuff?

>> No.3285804

>>3285781
That is perfectly fine. Although, we require a solution for the general case.
>>3285785
I am sorry that this problem does not interest you, I find it quite elegant. Checking cases naively can overcomplicate things. There is a simple solution. Hint: Arrange the integers in ascending order.

>> No.3285807

q1: sec^2(x)+5 = 7

q2: 12cos^2(x)+6cos(x) = 6

q3: 2cos(x)+sqrt(2) = 0

q4: -12cos^2(x)-14cos(x)=4

q5: sec^2(x)+3=4

and finally

q6: -2cos(-x) = 4cos(x) - 4
prove it.

>> No.3285831

>>3285804

You start doing high enough level maths when you're not guaranteed a 'nice' solution exists you don't really stop to look for a two line solution when a 10 line solution is apparent.

>> No.3285839

>>3285804

Ps. you don't find the problem elegant you find the solution elegant

>> No.3285860

@OP: this is not the place (if you are not a troll). I refer you to www.artofproblemsolving.com: establish yourself there.

>> No.3285879

>>3285839
Sure.

>> No.3285886

>>3285879
u mad bro?