Posted by mathstudent on .
Find the least squares approximation of x over the interval [0,1] by a polynomial of the form a + b*e^x

The polynomial produces an output space with two linearly independent basis vectors: u1 = 1, u2 = e^x
I believe these are the steps to solve the problem
1) Select a valid inner product that makes steps 23 simple.
2) Compute two orthonormal basis vectors (g1, g2) from the linearly independent basis vectors (u1, u2)
3) Calculate the projection of f(x) = x onto the output space of the polynomial represented by (g1, g2) by
<x, g1>*g1 + <x, g2>*g2
I'm not sure I'm picking a good inner product, because the numbers aren't very clean.
I pick for an inner product
<f,g> = Definite integral over [0,1] of f(x)*g(x) dx
Via GramSchmidt:
Linearly Independent Vector u1 = 1
Orthogonal Vector v1 = 1
Orthonormal vector g1 = 1
Linearly Independent Vector u2 = e^x
Orthogonal Vector v2 = e^x  e + 1
Orthonormal vector g2 = (e^x  e + 1)/sqrt(1/2*(e5)*(e+1)
Projection of f(x)=x onto {g1,g2}=
<f,g1>*g1 + <f,g2>*g2
= 1/2 + 1/4(e^xe1)/(e5)
Which can be rewritten in the a + b*e^x form as:
(1/2  (e+1)/(e5)) + 1/(4(e5)) * e^x
This answer isn't close to the book answer:
1/2 + 1/(e1) * e^x
Where did I go wrong?

math 
Count Iblis,
Linearly Independent Vector u2 = e^x
Orthogonal Vector v2 = e^x  e + 1
v2^2 = Integral from zero to 1 of
[e^x  e + 1]^2 dx =
Integral from zero to 1 of
[e^(2x) + 2 (1e)e^(x) + (1e)^2] dx =
1/2 (e^2 1)  (e1)^2 =
(e1)[1/2 (e+1)  (e1)] =
1/2(e1)(3  e)
So, g2 should be:
g2 = (e^x  e + 1)/sqrt(1/2*(e1)*(3e)
<f,g2>*g2 =
2(e^x  e + 1)/[(e1)(3e)] Integral from zero to 1 of x (e^x  e + 1) dx
Integral from zero to 1 of e^(p x) dx =
1/p [e^p  1]
Differentiate both sides w.r.t. p:
Integral from zero to 1 of x e^(p x) dx =
1/p^2 [e^p  1] + 1/p e^p
For p = 1 this is:
Integral from zero to 1 of x e^x dx = 1
We thus have:
<f,g2>*g2 =
2(e^x  e + 1)/[(e1)(3e)] Integral from zero to 1 of x (e^x  e + 1) dx =
2(e^x  e + 1)/[(e1)(3e)] *
[1 + 1/2(1e)] =
2(e^x  e + 1)/[(e1)(3e)] *
[3/2  1/2 e] =
(e^x  e + 1)/(e1) =
e^x/(e1) 1
The projection is 1/2 plus this which is:
e^x/(e1)  1/2 
math 
mathstudent,
Thanks so much for working that out.
In hindsight, I did the problem right except that I made a mistake in calculating <v2,v2>