91Ó°ÊÓ

When a series of process is used to convert a set of linearly independent vectors into orthonormal vectors set, then the process is known as Gram-Schmidt technique (or process). The space of orthonormal vectors set would be same as the original set.

An inner product, which varies from space to space, defines orthonormality. This issue involves an inner product on an interval, which means that the integral's limits are contained inside the provided range.

Here idea is to first take n^th polynomial of the form $x^n$and remove all other components (Gram-Schmidt procedure) and after doing so, normalize it. Construct a first vector (first polynomial). Let us denote it by:

$L_1=1$

No need to check orthogonality to others, since this is first vector, there is nothing to compare it to! However, check whether it is normalized.

$$\begin{gathered} {∫}_{-1}^1{\left(L_1\right)}^{2}dx \\ ={∫}_{-1}^{1}1dx \\ =x{|}_{-1}^1 \\ =2 \end{gathered}$$

It is not normalized in the terms of given inner product! Therefore, to normalize it, modify $L_1$by adding a constant denote the normalized polynomials with tilde:

$$\begin{gathered} \widetilde{L_1}=\frac{1}{\sqrt{2}}{L}_1 \\ =\frac{1}{\sqrt{2}} \end{gathered}$$

The goal of the problem is to construct the next polynomial. Use Gram-Schmidt procedure, first remove the$~L_1$component from the next polynomial x to obtain $L_2$which is orthogonal to $L_1$.

Now one can easily conclude that the latter integral vanishes since here integration is done on an odd function on an even interval. Again, normalize it:

$$\begin{gathered} {∫}_{-1}^1{\left(L_2\right)}^{2}dx \\ ={∫}_{-1}^1{x}^{2}dx \\ ={\overline{)\frac{1}{3}{x}^3}}_{-1}^1 \\ =\frac{2}{3} \\ \mathrm{Therefore},\mathrm{for}\mathrm{the}\mathrm{latter}\mathrm{integral}\mathrm{to}\mathrm{be}\mathrm{equal}\mathrm{to}1,\mathrm{multiply}\mathrm{the}{\mathrm{L}}_2\mathrm{as}\mathrm{follows}: \\ \stackrel{□}{{\mathrm{L}}_2}=\sqrt{\frac{3}{2}}{\mathrm{L}}_2 \\ =\sqrt{\frac{3}{2}}\mathrm{x} \end{gathered}$$

To construct the third vector (polynomial), subtract the $~L_1$and $~L_2$components from $x^2$

Once again, normalize it, and see that the norm of the $L_3$vector is:

$$\begin{gathered} {∫}_{-1}^1{\left(L_3\right)}^{2}dx \\ ={∫}_{-1}^1{\left(x^2\frac{1}{3}\right)}^{2}dx \\ =\frac{8}{45} \end{gathered}$$

The integral is simply evaluated by expanding the square brackets and integrating polynomials. Therefore, the normalized version of the vector is:

$\stackrel{â–¡}{L_3}=\sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)$

Finally, to obtain the fourth vector, start by subtracting the other already obtained components from $x^3$,as the Gram-Schmidt procedure instructs.

And proceeding in the same manner as previously, normalize the polynomial:

$$\begin{gathered} {∫}_{-1}^1{\left(L_4\right)}^{2}dx \\ ={∫}_{-1}^1{\left(x-\frac{3}{5}x\right)}^{2}dx \\ =\frac{8}{175} \end{gathered}$$

Which can be simply evaluated by integrating polynomials. These steps have been skipped due to simplicity of the latter. Therefore, the final normalized polynomial is given by:

localid="1656315849650" $\widetilde{L_4}=\sqrt{\frac{175}{8}}\left(x^3-\frac{3}{5}x\right)$