27 Feb '10 19:40

Do you guys have a favorite equation?

Mine is e^(i*pi)+1=0.

Mine is e^(i*pi)+1=0.

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01 Mar '10 08:101 editOne of the most interesting equations, and one of the most famous ones too, I believe, is this:

a^n + b^n = c^n, where a,b,c is an integer >0 and n is an integer >2.

Fermat got interested in it, and myriads of mathematicians thereafter.

One of the matematicians working with this equation is Sophie Germain (1776 - 1831) who has one of the most interesting life stories. She was a lady matematician in the area no women worked before. There is even a set of integers that is named after her.

The problem is now solved: There are no a,b,c, and n that can solve the equation. This is not interesting. What's interesting IMHO is that a seemingly simple equation is so difficult to deal with. Not many mathematicians can understand the proof. But in its simpler form it is understandable for amateur mathematicians. Like the case n=4 and some more.

Simon Sing brought the attention to this equation in his book "Femat's Last Theroem".- Joined
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podunk, PA02 Mar '10 02:551 edit

A mathematical identity is one that remains constant regardless of the values of the variables in it.*Originally posted by talvtal***An identity of what?**

ex.

(sin(A))^2 + (cos(A))^2 = 1

no matter what value you substitue for "A" the equation is true.

although now I'm slightly confused because there are no variables in e^(pi*i) = -1

?

Edit: ok actually its just a special case (x=pi) of

e^(ix) = cos(x) + i*sin(x)- Joined
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podunk, PA02 Mar '10 03:21

And while im at it*Originally posted by joe shmo***A mathematical identity is one that remains constant regardless of the values of the variables in it.**

ex.

(sin(A))^2 + (cos(A))^2 = 1

no matter what value you substitue for "A" the equation is true.

although now I'm slightly confused because there are no variables in e^(pi*i) = -1

?

Edit: ok actually its just a special case (x=pi) of

e^(ix) = cos(x) + i*sin(x)

I don't have a favorite yet, but one i'm currently looking at

y" + (k^2)y = 0

its soluion relates to Eulers identity.- Joined
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