Question

确定常数a,b,使得∫[0,1] [f(x)-(a+bx)]^2 dx最小?

Answer 1

f(a,b)=∫[0,1] [f(x)-(a+bx)]^2 dx
令f'a=-2∫[0,1][f(x)-(a+bx)] dx=0
0= ∫[0,1] [f(x)-(a+bx)] dx=∫[0,1] f(x)dx-a-b/2
a+b/2=∫[0,1] f(x)dx (1)
f'b=2∫[0,1] [f(x)-(a+bx)](-x) dx=0
0=∫[0,1] xf(x)dx-a/2-b/3
a/2+b/3=∫[0,1] xf(x)dx (2)

由 (1) (2)解出a,b即可

Answer 2

I(a,b)=∫[0,1] [f(x)-(a+bx)]^2 dx;
Ia(a,b)=-2∫[0,1] [f(x)-(a+bx)] dx=0,
Ib(a,b)=-2∫[0,1] [f(x)-(a+bx)]x dx=0;
2a+b=2∫[0,1] f(x)dx,
3a+2b=6∫[0,1] xf(x)dx;
a=4∫[0,1] f(x)dx-6∫[0,1]x f(x)dx,
b=-6∫[0,1] f(x)dx+12∫[0,1] xf(x)dx.