Thursday
March 30, 2017

Post a New Question

Posted by 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 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 - ,

    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 - ,

    Thanks so much for working that out.

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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question