Posted by **mathstudent** on Friday, January 18, 2008 at 12:13am.

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

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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 2-3 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 Gram-Schmidt:

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*(e-5)*(e+1)

Projection of f(x)=x onto {g1,g2}=

<f,g1>*g1 + <f,g2>*g2

= 1/2 + 1/4(e^x-e-1)/(e-5)

Which can be rewritten in the a + b*e^x form as:

(1/2 - (e+1)/(e-5)) + 1/(4(e-5)) * e^x

This answer isn't close to the book answer:

-1/2 + 1/(e-1) * e^x

Where did I go wrong?

- math -
**Count Iblis**, Friday, January 18, 2008 at 9:09am
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 (1-e)e^(x) + (1-e)^2] dx =

1/2 (e^2 -1) - (e-1)^2 =

(e-1)[1/2 (e+1) - (e-1)] =

1/2(e-1)(3 - e)

So, g2 should be:

g2 = (e^x - e + 1)/sqrt(1/2*(e-1)*(3-e)

<f,g2>*g2 =

2(e^x - e + 1)/[(e-1)(3-e)] 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)/[(e-1)(3-e)] Integral from zero to 1 of x (e^x - e + 1) dx =

2(e^x - e + 1)/[(e-1)(3-e)] *

[1 + 1/2(1-e)] =

2(e^x - e + 1)/[(e-1)(3-e)] *

[3/2 - 1/2 e] =

(e^x - e + 1)/(e-1) =

e^x/(e-1) -1

The projection is 1/2 plus this which is:

e^x/(e-1) - 1/2

- math -
**mathstudent**, Friday, January 18, 2008 at 5:35pm
Thanks so much for working that out.

In hindsight, I did the problem right except that I made a mistake in calculating <v2,v2>

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