Ln(x)=0恰有n個實根

Proof:
Part 1: Prove the Rodrigues's formula
Proof Legendre polynomials are the polynomials which satisfied by the differential equation , (1 - x2) P"n(x) -2x P'n(x) + n(n + 1) Pn(x) = 0.
Consider the polynomials Gn(x) = dn/dxn (x2 - 1)n. When differentiated n times, it becomes a polynomial of order n consisting of either all odd or all even powers of x, as n is odd or even. The coefficient of the highest power of x is 2n(2n-1)(2n-2)...(n+1), and the first two polynomials are 1 and 2x. If G(x) is substituted in the recurrence relation for the Legendre polynomials, it is found to satisfy it. If we divide G(x) by the constant 2nn!, then the first two polynomials are 1 and x. Therefore, Pn(x) = (1/2nn!) dn/dxn (x2 - 1)n.
Part 2: Legendre polynomials are orthogonal with respect to the L2 inner product on the interval −1 ≤ x ≤ 1:
Proof: Since Legendre differential equation can be viewed as a Sturm–Liouville problem. Accroding Sturm-Liouville theory. The normalized eigenfunctions form an orthonormal basis.
Part 3 : Given an orthogonal polynomial sequence Pi(x), any nth-degree polynomial S(x) can be expanded in terms of P0,P1,...Pn. That is, there are coefficients α0,α1,...αnsuch that

S(x)= ΣαiPi(x)
Proof by mathematical induction. Chooseαnα0so that the xn term of S (x) matches that of αnPn(x). Then S (x)-αnPn(x) is an (n − 1)th-degree polynomial. Continue downward.
Part 4: Given an orthogonal polynomial sequence, each of its polynomials is orthogonal to any polynomial of strictly lower degree.
Proof: Given n, any polynomial of degree n − 1 or lower can be expanded in terms of P0,P1,...Pn-1. Pnis orthogonal to each of them.

2009-01-19 16:19:08 補充：
Part 5: Each polynomial in an orthogonal sequence has all n of its roots real, distinct, and strictly inside the interval of orthogonality.

2009-01-19 16:20:33 補充：
Let m be the number of places where the sign of Pn changes inside the interval of orthogonality, and let x1,x2,...xmbe those points. Each of those points is a root of Pn. By the fundamental theorem of algebra, m ≤ n.

2009-01-19 16:20:37 補充：
Now m might be strictly less than n if some roots of Pn are complex, or not inside the interval of orthogonality, or not distinct. We will show that m = n.

2009-01-19 16:21:13 補充：
Let S(x)=Π(x-xj )

This is an mth-degree polynomial that changes sign at each of the xj, the same way that Pn(x) does. S(x)Pn(x) is therefore strictly positive, or strictly negative, everywhere except at the xj.

2009-01-19 16:21:17 補充：
S(x)Pn(x)W(x) is also strictly positive or strictly negative except at the xj and possibly the end points.

2009-01-19 16:21:35 補充：
Therefore, , the integral of this, is nonzero. But, Pn is orthogonal to any polynomial of lower degree, so the degree of S must be n.

2009-01-19 16:22:03 補充：
Part 6: Conclusion

We have shown that any polynomial in an orthogonal sequence has n distinct real roots. Since Legendre polynomials is an orthogonal polynomial sequences. Legendre polynomial has n distinct real roots.

2009-01-19 16:22:07 補充：
On the other hand, the formula of Legendre polynomial is (1/2nn!) dn/dxn (x2 - 1)n. So Ln(x)= 0 should have n distinct real roots.

我比較好奇的是，除了利用正交性有沒有別的方法可以證明

那是否只要用到 f(x) 的 degree 為 2n, n 階導數後的 degree 剩下 n 就可以得出 n 個根的結論?

2009-01-19 09:24:21 補充：
准許用歸納法嗎? 如果能做到時.

Yes! 您應可作答!
這是後續應用問題的重要依據!

2009-01-19 09:29:14 補充：
任何方法都可以啊! 注意是n個實根就好

2009-01-21 18:58:04 補充：
利用正交性質證明n個相異實根,已經是很方便了,別的方法我也不知吔!

Legendre Polynomial...