Compute llg-fll^2 in C(1,3) Real Linear Space

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Homework Statement


In the real linear space C(1, 3) with inner product (f, g) = intergal (1 to 3) (f(x)g(x))dx, let f(x) = 1/x. Knowing that g = (1/2)log3 is the constant polynomial g that is nearest to f. Compute llg-fll2 for this g.

Homework Equations

The Attempt at a Solution


I devised llg-fll = llgll - llfll = sqrt(g, g) - sqrt(f,f). Therefore, llg-fll2 = (g, g) - (f, f)
Using the equations, this equals integral(1 to 3)(1/2log3)2dx - integral (1 to 3)(1/x)2dx
Simplifying I get (1/2log3)2x evaluated from 1 to 3 +1/x evaluated from 1 to 3 = log23-2/3

My book says the answer is supposed to be 2/3 - 1/2log23 but I do not get this. Where am I going wrong?
 
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Cassi said:

Homework Statement


In the real linear space C(1, 3) with inner product (f, g) = intergal (1 to 3) (f(x)g(x))dx, let f(x) = 1/x. Knowing that g = (1/2)log3 is the constant polynomial g that is nearest to f. Compute llg-fll2 for this g.

Homework Equations

The Attempt at a Solution


I devised llg-fll = llgll - llfll = sqrt(g, g) - sqrt(f,f).

This is wrong. \|g - f\|^2 = (g,g) - 2(f,g) + (f,f). Hence \|g - f\| = \sqrt{(g,g) - 2(f,g) + (f,f)}. However you should in finding g have needed to calculate \|g - f\|^2 = \|f - g\|^2 = \int_1^3 (f(t) - g(t))^2\,dt anyway, so you can simply substitute the found g.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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