Pure Maths Question 2...

you still do not give the answer of question 1 !

b(i)

∫ [f(x)-Σa(i)φi(x)]^2 dx
=∫ [f(x)^2] dx -∫ 2f(x)Σa(i)φi(x) dx+∫[Σa(i)φi(x)]^2 dx

The first term is equal to the first term of the expression
The second term is equal to the third term of the expression

Consider +∫[Σa(i)φi(x)]^2 dx

which is equal to ∫[Σa(i)^2φi(x)^2+Σa(i)a(j)φi(x)φj(x)] dx
i not equal to j

We know that if i,j both greater than 0, thenφi(x)φj(x)=0
if i=0 but j not equal to 0, then
or j=0 but i not equal to 0
Also φi(x)φj(x)=0

Finally
∫[Σa(i)^2φi(x)^2+Σa(i)a(j)φi(x)φj(x)] dx
=∫[Σa(i)^2φi(x)^2] dx

if i=0, then ∫[Σa(0)^2φ0(x)^2] dx=a(0)^2
if i is odd (i=2m-1), then ∫[Σa(i)^2φi(x)^2] dx=a(i)^2(1/π)∫cos^2(mx) dx=a(i)^2
if i is even (i=2m), then ∫[Σa(i)^2φi(x)^2] dx=a(i)^2(1/π)∫sin^2(mx) dx=a(i)^2

So Our third term is equal to the second term of the expression

l