integration2

是否限制f(x)為2次多項式,否則f(x)會有無限多可能!

2010-02-22 16:08:14 補充：
There are infinite fn. satisfies your condition. Give you one of them.
Let f(x)=1 + A cos πx+ B cos(2πx)
1. ∫[0~1] f(x)dx=1
2. ∫[0~1] xf(x)dx= 1/2 - 2/π^2 A = α, then A is known constant.
3. ∫[0~1] x^2 f(x)dx=1/3-2/π^2 A+0.75/π^2 B=α^2, then B can be obtained.

2010-02-22 21:24:00 補充：
we can choose A, B such that
1/2 - 2/π^2 A = α
1/3- 2/π^2 A+0.75/π^2 B=α^2, then B can be obtained.
then∫[0~1] f(x)dx=1
∫[0~1] xf(x)dx=α
∫[0~1] x^2 f(x)dx=α^2
so, f(x) satisfies the given 3 condition.

2010-02-23 00:10:53 補充：
更正:1/3- 2/π^2 A+0.5/π^2 B=α^2, 則
f(x) conti. on [0,1] and satisfies given 3 conditions.

2010-02-23 00:35:41 補充：
Ooo: R+ , then there is no solution.
∫[0~1] f(x)dx=1, ∫[0~1] xf(x)dx=α, ∫[0~1] x^2 f(x)dx=α^2
then ∫[0~1] f(x)(x-α)^2 dx=∫[0~1] x^2 f(x)dx -2α∫[0~1] f(x)xdx+α^2∫[0~1] f(x)dx=0
∫[0~1] f(x)(x-α)^2dx=0 then f(x)=0 contradits ∫[0~1] f(x)dx=1
so, "No solution!"