Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: Yeong on March 02, 2011, 05:06:16 pm
-
Well, mah favorite math law is L'Hospital's Law.
-
Taylor Series. Just the fact that we can take practically any function and write it as a polynomial is mind-blowing. Not to mention you can use them to prove e^(ix)+1=0, which ties for second for me along with 1+2+3+4...=-1/12
-
Both of those are among my favorites, but since I study sequences (both infinite and finite) in my spare time, I think I gotta love things like the Riemann Zeta function... it's also yet another reason to love those Taylor Series ♥
-
1+2+3+4...=-1/12
Remind me again how this can be even remotely possible? There seems to be no way to get that to work O.O
-
Is it 1+2+3+4+... or is it 1-2+3-4+5-6...? I know one of them was proven to be -1/12 and the other to be -1/4
-
My question is, how can 1+2+3+4... ever end up being less than 1?? You are starting with 1 and only adding positive numbers, how can you get negative? It seems mathematically impossible O.o
-
Euler's formula forever. It's actually true.
-
it seems that Eulers formula only deals with trig and complex numbers ??? Can i see some sort of mathematical proof or explanation?
-
I think you use Ramanujan sums? Yeah, these funky summation systems (Abel, Cesaro, Euler, etc) give you results like 1+1+1+1+... = -1/2, 1-1+1-1+1-... = 1/2, and 1-2+3-4+5-6+... = 1/4.
Let's add a geometry theorem: Given a hexagon that has an incircle, its opposite diagonals meet at a single point. (Brianchon)
-
It's here: http://en.wikipedia.org/wiki/Euler's_formula#Proofs It makes no sense, but apparently it makes sense O.o
-
Okay, after doing some reading, it was not fully explained in the original post that it is not a traditional sum, but in fact a Ramanujan Sum, which is *not* the same :P
-
That is because even math is theoretical :D
But think of this... when working with mod systems, negatives are really positive. So, say you are working in mod 8... you would count {0,1,2,3,...,6,7,0,1...}...
Now also consider the numbers that are not infinity, but by performing a mathematical operation, you still get that number (so n+a=n, n*a=n, et cetera).
Now say you work in mod r... -1 really equals r-1. Taking all of this into account, working with mod (r) where r is equal to one of the before mentioned "infinite" numbers, the very largest number is in fact -1. No matter how high you count, it has been proven you will never reach these numbers, but in a mod system, -1 means you are just 1 away from that number!
Anywho, that has nothing to do with the proof that 1+2+3+4... is less than 1, but it is still interesting, yes?
-
1+2+3+4... does not equal -1/12. It eauls -1/12R where R is the symbol denoting a Ramunujan Sum :P
-
∭∞-∞(√(x²+y²+z²)e-(x²+y²+z²)))dxdydz =2ㅠ
The reason why is very fun to figure out, but should be rather obvious to some people ;)
-
(d/dx)ln x = 1/x
I know this is basic stuff, but without this, there's no complex d/dx ing! XD
-
I like Fermat's Last Theorem, as well as the Trapezoid Rule
-
∭∞-∞(√(x²+y²+z²)e-(x²+y²+z²)))dxdydz =2ㅠ
The reason why is very fun to figure out, but should be rather obvious to some people ;)
I didn't even know there was a triple-integral character O.O
Oh, and vertical integration (for APUSH nerds):
∫
f(x)
dx
-
AP US history ???
-
1+2+3+4...=-1/12
Remind me again how this can be even remotely possible? There seems to be no way to get that to work O.O
It seems that you have forgotten that when numbers exceed 32,767 they return to -32,768, therefore allowing a sequence of 1+2+3+4.... to be less than 1. ;)
-
Well, it's the same basic idea as those proofs that show 0=1: when you deal with infinite numbers in the wrong way, weird things happen :D
AP US history ???
http://en.wikipedia.org/wiki/Vertical_integration
Like Carnegie's monopoly over the steel industry.
-
I like the fact that from that you can seemingly prove infinity equals negative one half
1+1=1+1+1+1+1+1+1+1+... is an infinity of ones added together so it equals 1*infinity
1+1+1+1+1+1+1+1+1+1+...=-1/2 divide both sides by one and you get infinity equals -1/2
-
You didn't do your order of operations ;)
-
Distributive law. :P
jk, they're all great. ;)
-
I love how Euler's identity includes all the principal mathematical constants in the same formula.
(http://upload.wikimedia.org/math/9/e/9/9e9a547076c6820b95e439dd1a5d6a32.png)
-
(http://img16.imageshack.us/img16/3020/butterflyeffect.png) (http://img16.imageshack.us/i/butterflyeffect.png/)
Buttefly Effect! LOL I don't know what the equation is about, but I love the concept behind the Butterfly Effect
-
ln (-1)
------ = Chuck Norris
0
:P
-
ln(-1)=(pi)(i)
-
Formula for finding prime numbers.
(k + 2)(1 −
[wz + h + j − q]2 −
[(gk + 2g + k + 1)(h + j) + h − z]2 −
[16(k + 1)3(k + 2)(n + 1)2 + 1 − f2]2 −
[2n + p + q + z − e]2 −
[e3(e + 2)(a + 1)2 + 1 − o2]2 −
[(a2 − 1)y2 + 1 − x2]2 −
[16r2y4(a2 − 1) + 1 − u2]2 −
[n + l + v − y]2 −
[(a2 − 1)l2 + 1 − m2]2 −
[ai + k + 1 − l − i]2 −
[((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2]2 −
[p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m]2 −
[q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x]2 −
[z + pl(a − p) + t(2ap − p2 − 1) − pm]2)
> 0
Strangely enough I made a program on calc and computer that use this formula but I never got it to work. :(
-
I don't know who discovered this equation, but I call it the "mosquito curve" and I found it by accident. (This is in polar graphing)
r = sin(cos(tan(theta)))
-
Formula for finding prime numbers.
(k + 2)(1 −
[wz + h + j − q]2 −
[(gk + 2g + k + 1)(h + j) + h − z]2 −
[16(k + 1)3(k + 2)(n + 1)2 + 1 − f2]2 −
[2n + p + q + z − e]2 −
[e3(e + 2)(a + 1)2 + 1 − o2]2 −
[(a2 − 1)y2 + 1 − x2]2 −
[16r2y4(a2 − 1) + 1 − u2]2 −
[n + l + v − y]2 −
[(a2 − 1)l2 + 1 − m2]2 −
[ai + k + 1 − l − i]2 −
[((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2]2 −
[p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m]2 −
[q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x]2 −
[z + pl(a − p) + t(2ap − p2 − 1) − pm]2)
> 0
Strangely enough I made a program on calc and computer that use this formula but I never got it to work. :(
If I remember correctly, k is prime iff there exists a Diophantine solution on the other variables. (aka: practically, this formula is "computationally useless" :P) You'd need to solve for the other 25 or so integer variables such that the inequality holds; not an easy task :P
-
So brute forcing it like I was trying to do won't work then. :D
-
I don't know who discovered this equation, but I call it the "mosquito curve" and I found it by accident. (This is in polar graphing)
r = sin(cos(tan(theta)))
Lol, I do that too XD But r=sin(sin(sin(tan(θ is better.
Another fun thing to do is Y=sin(cos(tan(X, change it to Dot mode, then set the window range to (-50,50) and (-1.25,0.75). It's a nearly perfect butterfly O.O
-
I don't know who discovered this equation, but I call it the "mosquito curve" and I found it by accident. (This is in polar graphing)
r = sin(cos(tan(theta)))
Lol, I do that too XD But r=sin(sin(sin(tan(θ is better.
Another fun thing to do is Y=sin(cos(tan(X, change it to Dot mode, then set the window range to (-50,50) and (-1.25,0.75). It's a nearly perfect butterfly O.O
"Nearly" is a good word, because there is an actual "butterfly curve." (It's called that, too.) Although I forgot the equation for the butterfly curve.
-
Here's an interesting prime generating algorithm and a nice programming challenge at the same time (figure out how this works!):
(17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1)
Start with the number n=2. Take the one leftmost that, when multiplied to n, gives an integer, and multiply that number to n and replace. Repeat process indefinitely.
The primes will be the exponents of all powers of two after the initialization step :D
-
I don't understand that at all, can you give a better example? You start with n=2, but then what leftmost one? And when you say "when multiplied to n", do you mean the leftmost one to the nth power? And why do you "multiplied to n" to get an integer, and then "multiply that number to n". I don't understand what you're trying to say at all, please give an example.
-
So,
(17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1)
Start with the number n=2. Take the one leftmost that, when multiplied to n, gives an integer, and multiply that number to n and replace. Repeat process indefinitely.
2*(15/2) is the first fraction that gives me an integer when multiplied, Store 15 to n.
15*(55/1) = 825
825*(29/33) = 725
725*(77/29)=1925
1925*(13/11)=2275
2275*(17/91)=425
425*(78/85)=390
390*(11/13)=330
330*(29/33)=290
290*(77/29)=770
Repeating this process gives you
910, 170, 156, 132, 116, 308, 364, 68, 4. 4=2^2. Prime #1 found!
-
Ok, I see (actually, wouldn't 55/1 be an integer when multiplied? 55/1*2 = 110
-
You pick the leftmost fraction that can give you an integer. Since 15/2 is to the left of 55/1, it has precedence.
-
Xn+Yn=Zn..... ;D ;D ;D
Fermats Last Theorem!!!!!!!!!!!!!!!
Okay fine i'm not just showing off i know about it because its actually not a very interesting equation.
I also respect something much easier:1/12+1/22+1/32+....................=PIE2/6.
Its called the Basel (harry potter) series but it was first proven by Yuler.
-
*Euler.
Oiler refers to either a various hockey teams, a profession, a ship, or a football team.
-
I don't know who discovered this equation, but I call it the "mosquito curve" and I found it by accident. (This is in polar graphing)
r = sin(cos(tan(theta)))
Lol, I do that too XD But r=sin(sin(sin(tan(θ is better.
Another fun thing to do is Y=sin(cos(tan(X, change it to Dot mode, then set the window range to (-50,50) and (-1.25,0.75). It's a nearly perfect butterfly O.O
Is this a dog with wings?
r=tan(cos(cos(sin(theta^2)^2)^2)^3
-
I don't know who discovered this equation, but I call it the "mosquito curve" and I found it by accident. (This is in polar graphing)
r = sin(cos(tan(theta)))
Lol, I do that too XD But r=sin(sin(sin(tan(θ is better.
Another fun thing to do is Y=sin(cos(tan(X, change it to Dot mode, then set the window range to (-50,50) and (-1.25,0.75). It's a nearly perfect butterfly O.O
Is this a dog with wings?
r=tan(cos(cos(sin(theta^2)^2)^2)^3
Oh the things we do when we're bored :P
The dog doesn't seem to work with me, though. Any particular window settings?
-
I don't know who discovered this equation, but I call it the "mosquito curve" and I found it by accident. (This is in polar graphing)
r = sin(cos(tan(theta)))
Lol, I do that too XD But r=sin(sin(sin(tan(θ is better.
Another fun thing to do is Y=sin(cos(tan(X, change it to Dot mode, then set the window range to (-50,50) and (-1.25,0.75). It's a nearly perfect butterfly O.O
Is this a dog with wings?
r=tan(cos(cos(sin(theta^2)^2)^2)^3
Oh the things we do when we're bored :P
The dog doesn't seem to work with me, though. Any particular window settings?
No, I just hit ZStandard and it comes up.
It might just be a side view of Hot Dog's mosquito though :P
-
Euler's.