# Morse theory and a scalar field equation on compact surfaces by Malchiodi A.

By Malchiodi A.

Best mathematics books

The Mathematics of Paul Erdos II (Algorithms and Combinatorics 14)

This can be the main complete survey of the mathematical lifetime of the mythical Paul Erd? s, probably the most flexible and prolific mathematicians of our time. For the 1st time, the entire major parts of Erd? s' examine are lined in one venture. due to overwhelming reaction from the mathematical group, the venture now occupies over 900 pages, prepared into volumes.

Additional info for Morse theory and a scalar field equation on compact surfaces

Sample text

Thus, by (MP), the number of ways to arrange any r elements from N n in a row is given by n(n - l)(n - 2 ) . . (n - r + 1). For convenience, let us call an arrangement of any r elements from N n in a row, an r-permutation of N n , and denote by P™ the number of r-permutations of N n . Thus, we have P r " = n(n - l)(n - 2 ) . . 2. 1) for P™ looks a bit long. We shall make it more concise by introducing the following useful notation. Given a positive integer n, define n! to be the product of the n consecutive integers n, n — 1 , .

4 • 3 • 2 • 1 = 24. " is read "n factorial". By convention, we define 0! = 1. Using the "factorial" notation, we now have P r n = n(n - 1 ) . . (n - r + 1) n{n — 1 ) . . (n — r + l)(n — r)(n — r — 1 ) . . 2 • 1 _ n\ (n — r)(n — r — 1 ) . . 3) When n = 4 and r = 3, we obtain 4! r 3 — (4 - 3)! P4 - 4! 1! 4• 3• 2• 1 = 4 • 3 • 2 = 24, 1 which agrees with what we found before. 3) is valid when 0 < r < n. Consider two extreme cases: when r = 0 and r = n respectively. 3), pn __ n! = ^ = 1 (n - 0)!

1. ,n n choices n-\ choices 1 n-(r-2) n-0-l) choices choices n—2 choices ... 1 rth Counting 20 We wish to choose r elements from {1, 2 , . . ,n} to fill the r spaces, where the ordering of elements matters. There are n choices for the 1st space. After fixing one in the 1st space, there are n — 1 choices remaining for the 2nd space. After fixing one in the 2nd space, there are n — 2 choices left for the 3rd space, and so on. After fixing one in the (r — l)th space, there are n — (r — 1) choices left for the rth space.