Deja Vu

Most Elegant Math Equations Ever

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Sam Bridge: Where the hell does x^2+10 cross the x axis? I don't see any "i" values on the y-axis.

 

You're right, and you shouldn't. smile.png Imaginary and complex numbers actually hold crucial application in the real world. If you haven't been formally introduced to the topic yet, then it certainly can sound confusing and senseless. Before even thinking about Euler's identity, I would suggest starting a thread on imaginary and complex numbers (as not to stray the discussion here).

 

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[math]\displaystyle\int_{a}^{b} f(t)\;dt=F(b)-F(a)[/math]

 

The fundamental theorem of calculus, which relates differentiation and definite integration.

 

[math]\Gamma(z)=\displaystyle\int_{0}^{\infty} t^{z-1} e^{-t}\;dt[/math]

 

The gamma function, though beyond my understanding as a student, seems quite beautiful by its definition via an improper integral. Shifting its argument equates it to the factorial, and both are ubiquitous in math.

 

[math]e^{i\tau}=1[/math]

 

An identity from Euler's formula. Here, [math]\tau[/math] denotes the constant ratio of a Euclidean circle's circumference to its radius (isolated from [math]\pi[/math] to emphasize its geometric definition). Values of [math]\cos(x)+i\sin(x)[/math] parametrize a unit circle in the complex plane. [math]\tau[/math], being the circumference-radius ratio, is equivalent to a full circle as an angular measure. Thus, it helps to show the meaning behind complex exponentials, where this ratio "completes a full turn" in the complex unit circle.

 

[math]e=\displaystyle\lim_{x\to\pm\infty}\dfrac{^2{(x+1)}}{^2{x}}-\dfrac{^2{x}}{^2{(x-1)}}[/math]

 

Just think this one's really cool. It's a symmetric limit for a symmetric complex-valued function, that gives [math]e[/math], and involves tetration! Also, it's simply attractive from an aesthetic point of view. Just look how elegant it is.

Edited by Amaton

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1 + 1 = 2

 

It looks simple, and easy, but it emphasizes the importance of what are and are not additive. Consider this example from my high school days —

 

A driver wants to average 60 mph for two laps on a race track that's one mile around. The first lap is completed at 30 mph. Considering that the driver can change the car's speed instantaneously. What speed must the car go on the second lap.

 

We found students typically blurting out answers in the following order —

— 90 mph, thinking that speeds are additive { (30+90)/2=60 }, and they foresee the arithmetic mean. Wrong.

— 120 mph, thinking that the speeds are multiplicative { √(30∙120)=60 }, and they foresee the geometric mean. Wrong.

— Hmm.....

 

The answer involves the fact that, in this situation, only the distances and the times are additive —

 

v = Σd/Σt

 

So, for the first lap —

 

d1 = 1 mile and t1 = 2 minutes so that

 

v1 = 1 mile / 2 minutes = 30 mph

 

For the second lap —

 

d2 = 1 mile and t2 = (to be determined) so that

 

2 miles / 2 + t2 minutes = 60 mph = (by definition) 2 miles / 2 minutes

 

Thus, 2 + t2 minutes = 2 minutes, making t2 = 0 minutes

 

And v2 = 1 mile / 0 minutes = mph

 

As with the hare catching up to the tortoise, its time has expired, but it must still cover the distance.

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Imagine a planet with a tunnel drilled from a point on the surface through to the centre and all the way to the antipodean point at the other end. Then an object dropped into the tunnel and affected only by the planet’s gravitation will oscillate in the tunnel in simple harmonic motion with a period given by

[latex]T=\sqrt{\frac{3\pi}{\rho G}}[/latex]


where [latex]G[/latex] is the gravitational constant and [latex]\rho[/latex] the density of the planet. What I find elegant about this formula is that the period of oscillation does not depend on the size or the mass of the planet – only on its density. cool.png

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Imagine a planet with a tunnel drilled from a point on the surface through to the centre and all the way to the antipodean point at the other end. Then an object dropped into the tunnel and affected only by the planet’s gravitation will oscillate in the tunnel in simple harmonic motion with a period given by

 

[latex]T=\sqrt{\frac{3\pi}{\rho G}}[/latex]

where [latex]G[/latex] is the gravitational constant and [latex]\rho[/latex] the density of the planet. What I find elegant about this formula is that the period of oscillation does not depend on the size or the mass of the planet – only on its density. cool.png

 

Interestingly a satellite orbiting very close to the Earth's surface has the same period smile.png .

 

Also, that equation assumes the Earth is of uniform density. For a more accurate discussion about the falling through a hole in the Earth, see DH's post in this thread: http://www.scienceforums.net/topic/64539-earth-hole/

Edited by elfmotat

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Imagine a planet with a tunnel drilled from a point on the surface through to the centre and all the way to the antipodean point at the other end. Then an object dropped into the tunnel and affected only by the planet’s gravitation will oscillate in the tunnel in simple harmonic motion with a period given by

 

[latex]T=\sqrt{\frac{3\pi}{\rho G}}[/latex]

where [latex]G[/latex] is the gravitational constant and [latex]\rho[/latex] the density of the planet. What I find elegant about this formula is that the period of oscillation does not depend on the size or the mass of the planet – only on its density. cool.png

 

Also time does not depend on route of tunnel - doesn't have to go via centre.

 

Would echo idea of checking DH's posts

 

 

Echo

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