Question

Consider the linear space Pof all polynomials, with inner product

$\mathbf{(}\mathbf{f}\mathbf{,}\mathbf{g}\mathbf{)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathit{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathit{d}\mathit{t}$

For three polynomials f, g, and hwe are given the following inner products:

For example,$\left(f,f\right)\mathbf{=}\mathbf{4}$andlocalid="1660796365768" $(g,h)=\left(h,g\right)=3$

a.Find $\left(f,g+h\right)$

b.Findlocalid="1660796420390" $||g+h||$

c.Find ${\mathbf{proj}}_{\mathbf{E}}\mathbf{h}$, where E=span $\left(f,g\right)$. Express your solution as linear combinations of fand g.

d.Find an orthonormal basis of span $\left(f,g,h\right)$.Express the functions in your basis as linear combinations of

f,g, and h.

Short answer

(a) The value of$\left(f,g+h\right)$ is 8.

(b) The value of${||g+h||}^2$ is 57.

(c) the projection in linear combination is $2f+3g$.

(d) The orthonormal basis is $\frac{f}{2},g,\frac{h-2f-3g}{5}$.

Step-by-step solution

Consider for part (a).

Given is that P is the set of all polynomials, with the inner product

$(\mathrm{f},\mathrm{g})={\int }_0^{1}f\left(\mathrm{t}\right)g\left(\mathrm{t}\right)dt$

a) So,

$$\begin{gathered} \left(f,g+h\right)=\left(f,g\right)+\left(f,h\right) \\ =0+8 \\ =8 \end{gathered}$$

Hence, the value of$\left(f,g+h\right)$ is 8.

Consider for part (b).

b)

$$\begin{gathered} {||g+h||}^2=\left(g+h,g+h\right) \\ =\left(g,g\right)+\left(g,h\right)+\left(h,g\right)+\left(h,h\right) \\ =1+3+3+50 \\ =57 \end{gathered}$$

Hence, the value of${||g+h||}^2$ is 57.

Consider for part (c).

(c)

First of all observe that {f,g} is an orthogonal set as $\left(f,g\right)=0$. By normalizing it we will get an orthonormal basis for E.

$||f||=2,||g||=1$

Thus, an orthonormal basis will be $\left\{\frac{f}{2},g\right\}$.

$$\begin{gathered} pro{j}_{E}h=\left\{\frac{f}{2},h\right\}\frac{f}{2}+\left(g,h\right)g \\ =\frac{1}{2}\left(f,h\right)\frac{f}{2}+\left(g,h\right)g \\ =\frac{1}{2}\times 8\times \frac{f}{2}+3g \\ =2f+3g \end{gathered}$$

Hence, the projection in linear combination is $2f+3g$.

Consider for part (d).

(d)

Observe that from the last part we have an orthonormal set $\left\{\frac{f}{2},g\right\}$.

It will apply the Gram-Schmidt to get an orthonormal vector to the space span $\left\{\frac{f}{2},g\right\}$.

$$\begin{gathered} \stackrel{r}{h}=\frac{h-\left({\displaystyle \frac{f}{2}},g\right){\displaystyle \frac{f}{2}}-\left(g,h\right)g}{||h-\left({\displaystyle \frac{f}{2}},g\right){\displaystyle \frac{f}{2}}-\left(g,h\right)g||} \\ =\frac{h-pro{j}_{E}h}{||h-pro{j}_{E}h||} \\ =\frac{h-2f-3g}{||h-2f-3g||}.....\left(1\right) \end{gathered}$$

In order to compute (1), it needs to compute $||h-2f-3g||$. Consider,

$$\begin{gathered} ||h-2f-3g||=\left(h-2f-3g,h-2f-3g\right) \\ =\left(h,h\right)-2\left(h,f\right)-3\left(h,g\right)-2\left(f,h\right)+4\left(f,f\right)+6\left(f,g\right)-3\left(g,h\right)+6\left(g,f\right)+9\left(g,g\right) \\ =50-16-9-16-+!6+0-9+0+9 \\ =25 \end{gathered}$$

Hence, the orthonormal basis will be $\frac{f}{2},g,\frac{h-2f-3g}{5}$.