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integration2
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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!"
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see~~see~~see~~see~~see~~ 圖片參考：http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2010-02-21-19-41-12.png 更新: f(x)=1 + A cos πx+ B cos(2πx) is not a solution Hint: f(x) satisfies all the THREE conditions (at the same time) 更新 2: um.... see the ALL the REQUIREMENTS of f... 更新 3: f : [0,1] -> |R+最佳解答:
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!"
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