我以目前知道的方法給你看看,為了清楚寫得有點繁瑣.
你只要知道
(i) Sturm's theorem .
☆ Sturm's theorem is a symbolic procedure to determine the number of
distinct real roots of a real polynomial .
the recurrence relations of Laguerre polynomials Ln :
(ii) (n+1) L (x) = (2n+1-x) L (x) - n L (x)
n+1 n n-1
(iii) x L'(x) = n L (x) - n L (x)
n n n-1
i
i n x
(iv) L (x) = Σ (-1) ( ) ───
n 0≦i≦n n-i i!
等要證明原命題時,要用(i)時後,要注意選擇的區間 .(i)+(ii)+(iii)+(iv)立即得到
你的問題答案.
先介紹 Sturm's theorem :
" We restricted our domain to an interval (a,b)"
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A finite sequence of real polynomials
P (x) , P (x) , ............ , P (x) --------------(*) (n≧2)
0 1 n
of decreasing degree is a Sturm chain if
1) If P (x) = 0 then sign(P (x)) = sign( P'(x) ) . (a<x<b)
0 1 0
2) If P (x) = 0 ( 0<i<n) , sign(P (x)) = (-1)* sign(P (x) )
i i+1 i-1
3) P (x) is a constant function .
n
( note sign(a) = 1 or -1 or 0 , 實數a 是 "正"或"負"或 "0" )
Let σ(ξ) denote the number of sign changes in the sequence
P (ξ) , P (ξ) , ............ P (ξ)
0 1 n
( ex : + + 0 - + , σ(ξ) = 2 )
Sturm's theorem :
Given a Sturm chain ,
if c , d are not roots of P (x) and a < c < d < b , then :
0
the number of distinct roots of P (x) in the open interval (c,d) is
0
σ(c) ─ σ(d)
注意這裡是允許 P (x) is non-square-free .
0
證明不難 , 你可以依照 WIKI 寫下的"大略(並不完整)"的證明 , 寫出完整的證明 .
但要注意的是它是先證 P (x) is square-free 再推到 non-square-free case.
0
WIKI 網址
http://en.wikipedia.org/wiki/Sturm_chain
你可以舉些簡單的多項式P (x) 試看看 , 網址裡有給出其他 P (x) 的一種取法
0 i
(0<i≦n)
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n階Laguerre多項式為
1 x n n -x d
L (x) = ─ * e D ( x * e ) , D : differential operator ──
n n! dx
有這些工具之後便能簡單證出 "n階Laguerre多項式有 n 個不同正實根"
(n-i) n
Let P (x) = (-1) L (x) !!! A = { P (x) }
i (n-i) i i=0
由 (iii) 條件得知,在區間 (0,∞) , A 滿足 (1)
由 (ii) 條件得知,A 滿足 (2) ,
由 P (x) = 1 , A 滿足 (3) , ( 在此要使用Sturm's theorem只能在 (0,∞) )
n
So A is Sturm chain .
Fixed n , choose z > 0 large enough so that for all i (0≦i≦n)
sign(P (z)) = sign ( the leading coefficients of P (x) )
i i
n n-1
( Note : if Q(x) = a x + a x + ..... + a then
n n-1 0
a is called of the leading coefficients of Q(x) . )
n
Choose z > 0 small enough so that for all i (0≦i≦n)
0
sign(P (z )) = sign ( P (0) )
i 0 i
(n-i) n-i
(這裡這樣取法合理的的原因 , 因為 P (0) = (-1) L (0) = (-1) )
i n
P (z) , P (z) , ........ , P (z)
0 1 n
i
(n-i) (-1) n
sign (P (z)) = sign ( (-1) * ─── ) = sign ( (-1) )
i i!
so σ(z) = 0 !!!
P (z ) , P (z ) , ........ , P (z ) => ..+ - + - ..+
0 0 1 0 n 0
σ(z ) = n !!! , σ(z ) -σ(z) = n
0 0
n
Hence P (x) = (-1) L (x) has n distinct roots in (z ,z )
0 n 0
( 0 < z < z < ∞ )
0
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有什麼地方錯誤麻煩告知我一下 !
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