Favorite Equation: Quadratic Formula - Solving for X

  • Thread starter rock4christ
  • Start date
In summary: To me it seems that Euler's identity is a trivial instance of Euler's formula. Euler's identity is a special case of Euler's formula, yes. But it's not trivial at all, and it's not just a "special case" in the sense that it's not as important or something - it's just as important, and the fact that it's a special case makes it more beautiful, IMO.
  • #71
Z being the integers is certainly not just an american thing. And in fact it's primarily a german thing as Z is for 'Zahlen' (numbers). Before english, german was the official language of science and many words have their origin in german. For instance, what we call fields in english are 'Körper' in german, meaning "body" and the french word for field is 'corp', meaning "body" as well. 'Ring' is the litteral german translation of 'Zalhring', a term first coined by Hilbert according to Wiki.

An example from physics: the partition function, conventionally noted Z, stands for 'Zugstansum' (probably spelt wrong) meaning 'sum over all states'.
 
Last edited:
Mathematics news on Phys.org
  • #72
I want a stand world wide one too! I don't care, Z or J, though i think Z looks better in BlackBold :)
 
  • #73
[tex]- \frac{\hbar^2}{2m}\nabla^2\Psi +V\Psi = i\hbar\frac{\partial\Psi}{\partial t}
[/tex]
 
Last edited:
  • #74
Nice. Very nice. You do Schrodinger proud.
 
  • #75
Gib Z said:
Nice. Very nice. You do Schrodinger proud.

How can he be proud if he is dead?
 
  • #76
>.<"

Have you observed him dead? Hes in a superposition of many states, one of them in which he is alive and proud! :P
 
  • #77
great answer, Gib Z.
 
  • #78
[tex]e^{i\theta}=\cos\theta + i \sin\theta[/tex]

This has already been mentioned but I thought it was the coolest thing ever when using series to show it.
 
Last edited:
  • #79
Laplace L(f)=integral(e^(-sx))(f)dx
 
  • #80
Us physics guys chiming in

[itex]G_{\mu \nu}=T_{\mu\nu}[/itex]

(screw the 8pi)
 
  • #81
[tex]\int_{-\infty}^{+\infty} e^{-x^{2}} = \sqrt{\pi}[/tex]
 
Last edited:
  • #82
[tex]\frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2}[/tex]

[tex]\frac{\pi^4}{90} = \sum_{n=1}^{\infty} \frac{1}{n^4}[/tex]

Discovered by Euler, who continued to study the Riemann Zeta-function,


[tex]\varsigma(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}[/tex]

And, as uart tells us on page 1: [tex] \prod_{n=1}^\infty \frac{1}{1-(1/p_n)^a} = \sum_{n=1}^\infty 1/n^a[/tex]
 
Last edited:
  • #83
d_leet said:
My point was more or less that there are algebraic system/structures(I'm not sure which would be the correct term) where 1+1 does not necessarily equal 2, well even this might be incorrect because in the group Z2 1+1 does at least belong to the equivalence class of 2.. namely [1]+[1]=[2]=[0], where [a] represents the equivalence class of a, and a relates b if and only if 2 divides a-b. I'm not completely sure of the correctness of any this at the moment because I'm tired and it still is fairly new to me so if anyone would care to make a correction please feel free, however, I believe my point still stands that there are algebraic systems where 1+1 is not necessarily equal to 2.
There are algebraic systems where there is a 1 and a + but no 2, and hence in such systems "1 + 1 = 2" is not true (but only because it has no meaning; there is no use for a symbol '2' and hence there is none). But any algebraic system with a 1, a +, and a 2 will have 1 + 1 = 2. [Unless someone purposefully makes a senseless interpretation of the symbols]
 
  • #84
theperthvan said:
Yeah, it obviously does. But WHAT if it didn't?
Are you calling Hilbert an "it"? :) No listen, the guy said "we defined 1 + 1 to be 2" so we did it, not an "it". So what if we didn't define it that way? Then everything would be the same as it is today, we'd just be using different symbols.
 
  • #85
Not too sure what you're saying.
If you mean that we, mathematicians, human beings, whatever, decided that we will make 1+1=2, then good. Have a lolly.
Even if it is defined differently, the fact still remains that if Bob has an apple and Sally gives him another apple, Bob now has two (whatever two means) apples, which is what I meant. Not really talking about how it is written or what base is being used or any definition stuff, just the concept.
 
  • #86
arunma said:
If this were one of the physics forums, I'd probably cite Maxwell's Equations. But since this is the world of math, I'm definitely going to have to go with the Fundamental Theorem of Calculs. But Uart's example was also interesting...



This, I must admit, is pretty awesome. I wonder how it's derived, especially since there's no obvious formula for calculating the nth prime.

This formula is even better. If n=2 then is equals PI^2/6
 
  • #87
lim Re(zeta(1 + ix)) = gamma
x to 0

It's awesome because it involves the coolest function ever, and the coolest constant ever.

I tried to write it with Latex, how do I make it appear in my post?
 
  • #88
I would have to say "I = 1,000,000,000"

where I represents me, and 1,000,000,000 represents the amount of money in my bank account.
 
  • #89
[tex]\int_\Omega \mathrm {d}\omega = \oint_{\partial \Omega} \omega[/tex]

Stokes Theorem, so useful all the time...
 
  • #90
The story about L'Hopital's rule being work for hire done by Bernoulli because L'Hopital wanted to have something named after him, which now all first year calculus students hear about, is awesome.
 
  • #91
y=x-sin(x)

(solve for x)

because after 20 years of scratching my head, some smarty said he was going to re-invent math to solve the problem.

I warned him...

:devil:

(I finally found a "mathematician" that could explain the silliness to me)
 
  • #92
OmCheeto said:
y=x-sin(x)

(solve for x)

because after 20 years of scratching my head, some smarty said he was going to re-invent math to solve the problem.

I warned him...

:devil:

(I finally found a "mathematician" that could explain the silliness to me)

But that's easy to solve. [tex]x=Om(y)[/tex], where Om(y) is defined as the inverse of x-sin(x)
 
  • #93
Char. Limit said:
But that's easy to solve. [tex]x=Om(y)[/tex], where Om(y) is defined as the inverse of x-sin(x)

Gulp. Did you know for the last 15 years I've been offering a $100 to anyone who could either solve the equation, or explain why it could not be solved.

I've never mentioned that at this forum as:
a. There are just way too many smart people here
b. It's a sign of a crackpot

I don't know why people can offer a million dollars for such things(Millennium Prize), but I get a bad label for doing such things.

Maybe I should have called it the "OmCheeto Prize"?
 
  • #94
Well, I believe you know me on Facebook, so you probably have my address. I expect my $100 within two weeks.
 
  • #95
I think I have two favorites, first, the infamous Fourier Series:

[itex]f(x) = A_o + \sum_{n=1}^{\infty} A_n\cos{\frac{n\pi x}{L}} + \sum_{n=1}^{\infty} B_n\sin{\frac{n\pi x}{L}}[/itex]

which I think is one of the more interesting ideas in all of mathematics, and obviously one of the more applicable mathematical tools we use in everyday life. Joseph Fourier was truly brilliant to think along these lines (every function can be represented as an infinite series of sine and cosine, well, when you do a Fourier Transform anyway..), though I don't exactly know how much exactly he contributed to the theory of Fourier Series, I'm giving him the benefit of the doubt of total creativity :DSecond, I always liked the simple weighted average:

[itex]\bar{x} = \sum_{i=1}^{n} P_i x_i[/itex]

I guess simply because it's extremely useful and just aesthetic to me, it's just always been on of my favorites, from quantum theory (expectation values) to statistical mechanics (with partition functions, Boltzmanm factors, etc.) it just always takes a conceptual center stage.
 
  • #96
Ah, here we go:

[tex]\lim_{x\to0}\zeta(1 + ix)=\gamma[/tex]
 
  • #97
Mike_Bson said:
Ah, here we go:

[tex]\lim_{x\to0}\zeta(1 + ix)=\gamma[/tex]

But I get something different...

Actually, after the addition of [tex]\frac{i}{x}[/tex], I get the E-M constant.
 
  • #98
Mine is definitely v=v0+at
 
  • #99
Distance along a curve is my favorite.
[tex]\int \sqrt{1-({{dy} \over {dx}})^2}dx[/tex]
 
  • #100
I'm not sure what my favorite is, and I would probably keep switching favorite formula anyway.
I kind of like this one: [tex]\pi=\lim_{n\rightarrow\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt 2}}}}}_{n}[/tex]
I like it because it goes quicker towards pi than what many other formulas do, but also because I managed to prove it 2 days ago, using regular 2n-gons.
 
  • #101
PhilosophyofPhysics said:
[tex]e^{i\theta}=\cos\theta + i \sin\theta[/tex]

This has already been mentioned but I thought it was the coolest thing ever when using series to show it.

For all of you talking about Euler's formula, i'd recommend looking at:
http://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/

Which gives what to me is a very good explanation, kinda breaks the wonder in a way but deepens the understanding.

I don't have a favorite formula, though the one posted by yayness is very interesting!
 
  • #102
Polar Inertial Momentum Inequality
 
  • #103
djosey said:
I don't have a favorite formula, though the one posted by yayness is very interesting!

It is, but it's not recommended if you want to calculate really large amounts of decimals in π. Even though it goes quickly towards π, you need to calculate a lot more decimals in [tex]\sqrt 2[/tex] than what the number of correct decimals in π will be.
Let's say you calculate k decimals in [tex]\sqrt 2[/tex], then you'll have k/2 correct decimals in [tex]\sqrt{2+\sqrt 2}[/tex] and k/4 correct decimals in [tex]\sqrt{2+\sqrt{2+\sqrt 2}}[/tex], and then k/8, k/16 and so on. I still like the formula though. It is simple and easy to remember.
 
  • #104
[tex]n! \approx\sqrt{2 \pi n}(\frac{n}{e})^{n}[/tex]

Stirling's Approximation. Haha.
 
  • #105
My favorite equation is

[tex]e^\pi-\pi=20[/tex]

It's a good way to check whether your computer is experiencing rounding errors :biggrin:

http://xkcd.com/217/
 
Last edited:

Similar threads

Replies
19
Views
2K
Replies
8
Views
2K
Replies
2
Views
1K
Replies
11
Views
2K
Replies
1
Views
2K
Replies
6
Views
2K
Replies
36
Views
5K
Back
Top