Linear Algebra - Inner Product Spaces

Question

The inner product on vector space of continuous functions on interval [a, b] is defined such that

< f,g> := integrate[f(x)g(x) dx] from a to b

Consider two functions x and x3. Which one is closer to x2 on the interval [0; 1]?

Answer 1

Ashley,

for sure the inner product of any two functions, say x1 and x2 ALSO equals

|x1|*|x2|*cos (angle between x1 and x2).

A larger cosine means a smaller angle (cos b =1 means b=0)

So you not only need to integrate both x1*x2 and separately x2*x3

(and find value from a to b for each)

you will need to divide the first by |x1|*|x2| and the second by |x2|*|x3|

if a was 0 and b=1 then |x1| is the norm of x1

but generally it the square root of the integral from a to b of |x1|²

same with |x3| and |x2|, integral rom a to b of the function absolute value

squared. once all is computed see which cosine is bigger and those are

(the two with the larger cosine) are closest.

Answer 2

The norm in that space is given by the quantity ||f-g||=sqrt{<f-g,f-g>}. So the only thing you have to do is calculate the integrals <x-x^2,x-x^2> and <x^3-x^2,x^3-x^2> and see which one is the smallest.