Question

Question:Let \(V\) be the space \(C\left( { - 1,1} \right)\) with the inner product of Example 7. Find an orthogonal basis for the subspace spanned by the polynomials 1, \(t\), and \({t^2}\). The polynomials in this basis are calledLegendre polynomials.

Short answer

The orthogonal basis is \(\left\{ {1,\,t,\,3{t^2} - 1} \right\}\).

Step-by-step solution

Use the given information

If the pair of vectors\(\left\langle {f,g} \right\rangle \)belong to vector space \(V\) defined in \(C\left( { - 1,1} \right)\), then their inner product is given by \(\left\langle {f,g} \right\rangle = \int_{ - 1}^1 {f\left( t \right)g\left( t \right)dt} \).

It is given that \(1,\,t,\,{t^2}\) are orthogonal, then \(\left\langle {1,t} \right\rangle \) must be 0, that is\(\int_{ - 1}^1 {tdt} = 0\).

Use Gram Schmidt process

According to the Gram Schmidt process, the third element in the orthogonal basis span\(\left\{ {1,t,{t^2}} \right\}\)can be defined as\({t^2} - \frac{{\left\langle {{t^2},1} \right\rangle }}{{\left\langle {1,1} \right\rangle }}1 - \frac{{\left\langle {{t^2},1} \right\rangle }}{{\left\langle {t,t} \right\rangle }}t\).

Find\(\left\langle {{t^2},1} \right\rangle ,\,\left\langle {1,1} \right\rangle ,\,\left\langle {t,t} \right\rangle \), as follows:

\(\begin{aligned}{}\left\langle {t,t} \right\rangle &= \int_{ - 1}^1 {{t^2}dt} \\ &= \left( {\frac{{{t^3}}}{3}} \right)_{ - 1}^1\\ &= \left( {\frac{1}{3} - \left( { - \frac{1}{3}} \right)} \right)\\ &= \frac{2}{3}\end{aligned}\)

\(\begin{aligned}{}\left\langle {1,1} \right\rangle &= \int_{ - 1}^1 {1dt} \\ &= \left( t \right)_{ - 1}^1\\ &= \left( {1 - \left( { - 1} \right)} \right)\\ &= 2\end{aligned}\)

\(\begin{aligned}{}\left\langle {{t^2},t} \right\rangle &= \int_{ - 1}^1 {{t^3}dt} \\ &= \left( {\frac{{{t^4}}}{4}} \right)_{ - 1}^1\\& = \left( {\frac{1}{4} - \frac{1}{4}} \right)\\ &= 0\end{aligned}\)

Simplify for Gram Schmidt inequality

Plug the above obtained values in \({t^2} - \frac{{\left\langle {{t^2},1} \right\rangle }}{{\left\langle {1,1} \right\rangle }}1 - \frac{{\left\langle {{t^2},1} \right\rangle }}{{\left\langle {t,t} \right\rangle }}t\) and simplify as follows:

\(\begin{aligned}{}{t^2} - \frac{{\left\langle {{t^2},1} \right\rangle }}{{\left\langle {1,1} \right\rangle }}1 - \frac{{\left\langle {{t^2},t} \right\rangle }}{{\left\langle {t,t} \right\rangle }}t &= {t^2} - \frac{{2/3}}{2}1 - \left( 0 \right)t\\ &= {t^2} - \frac{1}{3}\end{aligned}\)

The third element in the orthogonal basis for subspace spanned by the polynomials \(1,\,t,\,{t^2}\)can be obtained by scaling\({t^2} - \frac{1}{3}\)by 3, to get it as\(3{t^2} - 1\).

Thus, the orthogonal basis is \(\left\{ {1,\,t,\,3{t^2} - 1} \right\}\).