Mathematics 2

If a is not a positive integer, then the summation 1^a+...+n^a is not a polynomial.

2013-11-13 18:04:12 補充：
We can prove that 1^a+...+n^a ~ O(n^(a+1)) for a is not -1.

2013-11-13 19:22:29 補充：
Case 1: a in N
Let a function sequence {B(n,x), n=0,1,...} be defined by ∫_[x,x+1] B(n,t)dt=x^n,then
B(0,x)=1, B(1,x)=x-1/2, B(2,x)=x^2-x+1/6,... and
B(n,x) is a polynomial of degree n and the leading coefficient of B(n, x) is 1.
(Namely, B(n,x)= x^n+ c1 x^(n-1) + c2 x^(n-2) + ...+ c(n) )
So, B(0,x), B(1,x), ..., B(n,x) is a basis of the space {p(x) | degree(p) < n+1},
then x^n = B(n,x)+c1 B(n-1,x) + c2 B(n-2, x) + ...+cn B(0,x)
1^a+2^a +...+n^a = ∫_[1~2] B(a,t)dt+∫_[2~3] B(a,t)dt+ ...+∫_[n~n+1] B(a,t) dt
=∫_[1~n+1] B(a, t) dt
=∫_[1~n+1] [t^a + c1 t^(a-1)+...] dt
= [t^(a+1) /(a+1) + ... ] for t=1~n+1
= [(n+1)^(a+1) /(a+1)+...]
= [n^(a+1) /(a+1)+...] is a poly. of deg. a+1 and its leading coef. is 1/(a+1).

Case 2: a > -1
Consider lim(n→∞) (1^a+2^a+...+n^a)/n^(a+1)
=lim(n→∞) ∑_[k=1~n] (1/n) (k/n)^a
=∫_[0~1] x^a dx (Riemann's sum)
= 1/(a+1)
so 1^a+2^a+...+n^a ~ O(n^(a+1)) (big O)

Please help another one of mine posted at the same time as this one titled with Mathematics 1.
Thanks really.
Simon YAU.

師姐果然不簡單，連 Landau notation 都應付自如。

將來 asymptotic 的問題都可以請教她。。。