Project e^x using the monomial and legendre basis

Question

Project $e^x$ onto $P_2$ with the $L^2$ inner product on $[-1,1]$ using:

a) The monomial basis {$1, x, x^2$}.
b) The Legendre basis {$1,x,\frac{1}{2}(3x^2 - 1)$}.

Answer 1

Let, $$ A_{nm} = \int_{-1}^1 x^n x^m dx, $$

write,

$$ e^x = \sum_{n=0}^2 c_n x^n + (\text{ higher order terms}) $$

our goal is to use the inner product formulas to determine the $c_n$'s. Multiply both sides by $x^p$ and integrate from $[-1,1]$ (this is the inner product with $x^p$).

$$\int_{-1}^1 x^p e^x dx= \int_{-1}^1 x^p \sum_{n=0}^2 c_n x^n dx + (\text{ higher order terms})$$

$$\int_{-1}^1 x^p e^x dx= \sum_{n=0}^2 \int_{-1}^1 x^p x^n dx c_n + (\text{ higher order terms})$$

$$\int_{-1}^1 x^p e^x dx= \sum_{n=0}^2 A_{pn} c_n + (\text{ higher order terms})$$

Define $E_p = \int_{-1}^1 x^p e^x$ and we have,

$$ E_p = \sum_{n=0}^2 A_{pn} c_n + (\text{ higher order terms}),$$

If we neglect the higher order terms we then have the following equation,

$$ E_p = \sum_{n=0}^2 A_{pn} c_n,$$

which can be written in matrix form,

$$ \boxed{E = A c}$$

The vector $E$ and the matrix $A$ are computed from integrals. The vector $c$ contains the unknown coefficients. You can solve for $c$ using ordinary matrix methods (I would do a Cholesky decomposition).

Once you know your function in terms of the monomials you can convert this to a representation in terms of Legendre polynomials using the following procedure.

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Suppose we have,

$$ f(x) = 1 + 2 x + 3 x^2 $$

we want to write $f$ as a linear combination of Legendre polynomials. We start wiht the highest order term ($x^2$) and write it in terms of $P_2$.

$$P_2 = \frac12 ( 3 x^2 -1 ) \Leftrightarrow x^2 = \frac13(2P_2+1)$$

$$ f(x) = 1 + 2 x + 3 \frac13(2P_2+1) $$

$$ f(x) = 2 + 2 x + 2P_2 $$

$$ f(x) = 1 + 2 x + 2P_2 $$

$$ f(x) = 1 P_0 + 2 P_1 + 2P_2 $$

Then the projection of $f(x)$ onto $P_2$ is the coefficient $2$.

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For part $(b)$ equation will be

$$e^x = \sum_{n=0}^2 c_n P_n(x) + (\text{higher order terms})$$

Multiply both sides by $P_m(x)$ and integrate over the interval $[-1,1]$.

$$e^x P_m(x) = \sum_{n=0}^2 c_n P_n(x) P_m(x) + (\text{higher order terms})$$

$$\int_{-1}^1 e^x P_m(x) \mathrm{d}x = \int_{-1}^1 \sum_{n=0}^2 c_n P_n(x) P_m(x) \mathrm{d}x + (\text{higher order terms})$$

$$\color{blue}{\int_{-1}^1 e^x P_m(x) \mathrm{d}x} = \sum_{n=0}^2 c_n \color{blue}{\int_{-1}^1 P_n(x) P_m(x) \mathrm{d}x} + (\text{higher order terms})$$

We will label the integral on the left as $E_m$ and the integral on the right as $A_{nm}$.

$$\color{blue}{E_m} = \sum_{n=0}^2 c_n \color{blue}{A_{nm}} + (\text{higher order terms})$$

Lets neglect the higher order terms.

$$E_m = \sum_{n=0}^2 c_n A_{nm} $$

This is equivalent to the matrix equation,

$$ E = A c$$