Tosh5457 said:
Stokes[/PLAIN] theorem and
residue theorem. They're so important that they contain much of the information of vector calculus and complex analysis, respectively.
Stokes' Theorem is extremely elegant and very important. It has a lot of consequences such as Green's theorem, the divergence theorem, the fundamental theorem of algebra, Brouwer's fixed point theorem, the invariance of domain theorem, a lot of complex analysis theorems, etc. It's definitely one of the most important theorems out there. Too bad that most books don't give a nice derivation of the theorem.
The residue theorem is also very nice. Complex analysis has a lot of beautiful theorems. For exampel, the Cauchy integral formula is very nice too:
http://en.wikipedia.org/wiki/Cauchy's_integral_formula
BOAS said:
I don't know if it has a nice name, but it was the first theorem I worked through largely on my own (had done others in class), and the feeling of satisfaction was great.
"The sum of the first n positive integers is \frac{n(n+1)}{2}"
This was the theorem that Gauss discovered when he was about 5 years old:
This is attributed to an early school lesson when the teacher thought he would keep the class busy whilst he popped out for something. He set the test of adding all the whole numbers from 1-100. By the time he reached the door, Gauss had the answer.Gauss imagined the problem as 1 + 2 + 3 +...+98 + 99 + 100, but then he wrote the numbers underneath but in reverse order. 100 +99 + 98...+3 + 2 + 1. So each 100 pairs of vertical numbers added up to 101 so the total was 10100 but this is twice the true answer as each number is included twice. The total is therefore 5050. This lead to the general formula that the sum of consecutive numbers from 1 to n is n(n+ 1) ÷ 2.
Another very nice proof of this result is geometrical:
http://www.mathsisfun.com/algebra/triangular-numbers.html
The generalization of this result would be: what is the sum
1^2 + 2^2 + 3^2 + ... + n^2
or in general, what is the sum of 1^k + 2^k + 3^k + ... + n^k
This is not at all obvious. The results for the first few ##k## are:
\begin{eqnarray*}<br />
1 + 2 + 3 + 4 + 5 + ... + n & = & \frac{1}{2}n^2 + \frac{1}{2}n\\<br />
1^2 + 2^2 + 3^2 + 4^2 + 5^2 + ... + n^2 & = & \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n\\<br />
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + ... + n^3 & = & \frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2\\<br />
1^4 + 2^4 + 3^4 + 4^4 + 5^4 + ... + n^4 & = & \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3 - \frac{1}{30}n\\<br />
1^5 + 2^5 + 3^5 + 4^5 + 5^5 + ... + n^5 & = & \frac{1}{6}n^6 + \frac{1}{2}n^5 + \frac{5}{12}n^5 - \frac{1}{12}n^2\\<br />
1^6 + 2^6 + 3^6 + 4^6 + 5^6 + ... + n^6 & = & \frac{1}{7}n^7 + \frac{1}{2}n^6 + \frac{1}{2}n^5 - \frac{1}{6}n^3 + \frac{1}{42}n\\<br />
1^7 + 2^7 + 3^7 + 4^7 + 5^7 + ... + n^7 & = & \frac{1}{8}n^8 + \frac{1}{2}n^7 + \frac{7}{12}n^6 - \frac{7}{24}n^4 + \frac{1}{12}n^2\\<br />
1^8 + 2^8 + 3^8 + 4^8 + 5^8 + ... + n^8 & = & \frac{1}{9}n^9 + \frac{1}{2}n^8 + \frac{2}{3}n^7 - \frac{7}{15}n^5 + \frac{2}{9}n^3 - \frac{1}{30}n\\<br />
1^9 + 2^9 + 3^9 + 4^9 + 5^9 + ...+ n^9 & = & \frac{1}{10}n^{10} + \frac{1}{2}n^9 + \frac{3}{4}n^8 - \frac{7}{10}n^6 + \frac{1}{2}n^4 - \frac{3}{20}n^2\\<br />
1^{10} + 2^{10} + 3^{10} + 4^{10}+ 5^{10} + ... + n^{10} & =& \frac{1}{11}n^{11} + \frac{1}{2}n^{10} + \frac{5}{6}n^9 - n^7 + n^5 - \frac{1}{12}n^3 + \frac{5}{66}n<br />
\end{eqnarray*}
The question is to find a pattern for the general case. We can see some parts of the pattern, but finding the general case is not at all easy. If you want a tough challenge, you can try it.
Here is the solution:
http://en.wikipedia.org/wiki/Faulhaber's_formula
A very nice derivation of the general formula can be found in the following intriguing document which attempts to generalize calculus to discrete situations:
https://www.cs.purdue.edu/homes/dgleich/publications/Gleich 2005 - finite calculus.pdf