How orthogonality of Legendre's polynomial is helpful in calculating coefficients?

Question

This question is in context of Legendre's polynomials. I don't understand following phrase:

One of the main characteristic to be taken into account is the “Orthogonality”. This property of certain polynomials gives the advantage of calculating coefficients without using systems of equations, as for example in the least mean squares technique.

Answer 1

To illustrate the problem related to curve fit, suppose that we want to approximate $e^x$ by polynomials over the range $(-1\leq x\leq 1)$.

So, for degree $1$, we shall minimize $$F_1=\int_{-1}^1(a+bx-e^x)^2\,dx$$ This will lead to $$a=\frac{e^2-1}{2 e}\qquad , \qquad b=\frac 3e$$

for degree $2$, we shall minimize $$F_2=\int_{-1}^1(a+bx+cx^2-e^x)^2\,dx$$ This will lead to $$a=\frac{3 \left(11-e^2\right)}{4 e}\qquad , \qquad b=\frac 3e\qquad , \qquad c=\frac{15 \left(e^2-7\right)}{4 e}$$ You can notice that coefficient $a$ changed value.

Let us do the same using Legendre polynomials (we shall not repeat the linear case since $P_0(x)=1$ and $P_1(x)=x$. For degree $2$, we shall minimize $$G_2=\int_{-1}^1(aP_0(x)+bP_1(x)+cP_2(x)-e^x)^2\,dx$$ This will lead to $$a=\frac{e^2-1}{2 e}\qquad , \qquad b=\frac 3e\qquad , \qquad c=\frac{5 \left(e^2-7\right)}{2 e}$$ Doing it again for degree $3$, minimizing $$G_3=\int_{-1}^1(aP_0(x)+bP_1(x)+cP_2(x)+dP_3(x)-e^x)^2\,dx$$ and get $$a=\frac{e^2-1}{2 e}\qquad , \qquad b=\frac 3e\qquad , \qquad c=\frac{5 \left(e^2-7\right)}{2 e}\qquad , \qquad d=\frac{7 \left(37-5 e^2\right)}{2 e}$$

As you see, we can increase the number of Legendre polynomials, the coefficients do not change; this is because of their orthogonality.

This does not change the results $(F_k=G_k)$ but when adding one extra term in the basis of orthogonal polynomials, we know that previous coefficients remain unchanged (this is not the case with simple powers). Then, their obvious interest in this area.

Edit

You should probably be very interested by this paper.

Answer 2

That is a general property of orthogonal basis. Let $\{p_{k}\}$ be a set of orthogonal polynomials on $[0,1]$ and suppose $f$ is finite linear combination of $p_{k}$'s, such as $f(x)=\sum_{j=1}^{n}a_{j}p_{j}(x)$. Then by orthogonality, we can compute coefficients by \begin{align} \int_{0}^{1}f(x)p_{i}(x)dx=\sum_{j=1}^{n}a_{j}\int_{0}^{1}p_{j}(x)p_{i}(x)dx=a_{i}\int_{0}^{1}p_{i}^{2}(x)dx\Rightarrow a_{i}=\frac{\int_{0}^{1}f(x)p_{i}(x)dx}{\int_{0}^{1}p_{i}(x)^{2}dx} \end{align}