91影视

Consider the set of all polynomial such that and .

The set $\left\{1,t,t_2,鈥�,t\right\}$is linear independent set ofV if there exist constant${\mathit{a}}_{\mathbf{i}}\mathbf{鈭�}\mathit{R}$such that${\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}{\mathit{t}}_{\mathbf{1}}\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\mathit{t}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{a}}_{\mathbf{n}}{\mathit{t}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$where ${\mathbf{a}}_{\mathbf{0}}\mathbf{=}{\mathbf{a}}_{\mathbf{1}}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{=}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$.

Any polynomial $f\left(t\right)鈭�P_3$is defined as follows.

$f\left(t\right)=b_0+b_{1}t+b_2{t}^2+b_3{t}^3$

Substitute the value 1 for t in the equation$f\left(t\right)=b_0+b_{1}t+b_2{t}^2+b_3{t}^3$as follows.

$$\begin{gathered} f\left(t\right)=b_0+b_{1}t+b_2{t}^2+b_3{t}^3 \\ f\left(1\right)=b_0+b_1+b_2+b_3 \end{gathered}$$

As $f\left(1\right)=0$, substitute the value 0 for $f\left(1\right)$in the equation $f\left(1\right)=b_0+b_1+b_2+b_3$as follows.

$\begin{array}{l}f\left(1\right)=b_0+b_1+b_2+b_3\\ b_1=-\left(b_0+b_2+b_3\right)\end{array}$

Take integration both side in the equation $f\left(t\right)=b_0+b_{1}t+b_2{t}^2+b_3{t}^3$as follows.

role="math" localid="1659407877530" $$\begin{gathered} f\left(t\right)=b_0+b_{i}t+b_2{t}^2+b_3{t}^3 \\ {鈭�}_{-1}^{1}f\left(t\right)dt={鈭�}_{-1}^1{b}_0+b_{1}t+b_2{t}^2+b_2{t}^{3}dt \\ ={\left\{b_{0}t+b_1\frac{t^2}{2}+b_2\frac{t^3}{3}+b_3\frac{t^4}{4}\right\}}_{-1}^1 \\ {鈭�}_{-1}^{1}f\left(t\right)dt=b_0+b_1\frac{1}{2}+b_2\frac{1}{3}+b_3\frac{1}{4}+b_0-b_1\frac{1}{2}+b_2\frac{1}{3}-b_3\frac{1}{4} \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {鈭�}_{-1}^{1}f\left(t\right)dt=b_0+b_2\frac{1}{3}+b_0+b_2\frac{1}{3} \\ {鈭�}_{-1}^{1}f\left(t\right)dt=2b_0+b_2\frac{2}{3} \end{gathered}$$

As ${鈭�}_{-1}^{1}f\left(t\right)dt=0$, substitute the value 0 for${鈭�}_{-1}^{1}f\left(t\right)t$in the equation${鈭�}_1^{1}f\left(t\right)t=2b_b+b_2\frac{2}{3}$as follows.

$$\begin{gathered} {鈭�}_{-1}^{1}f\left(t\right)=2b_0+b_2\frac{2}{3} \\ 0=2b_0+b_2\frac{2}{3} \\ {b}_0=-b_2\frac{1}{3} \end{gathered}$$

Substitute the value $-b_2\frac{1}{3}$for $b_0$in the equation$b_1=-(b_0+b_2+b_3)$as follows.

$$\begin{gathered} b_{-1}=-\left(b_0+b_2+b_3\right) \\ =-\left(-b_2\frac{2}{3}+b_2+b_3\right) \\ =-\left(b_2\frac{2}{3}+b_3\right) \\ {b}_1=-b_2\frac{2}{3}-b_3 \end{gathered}$$

Finally, substitute the value $-b_2\frac{2}{3}-b_3$for $b_1$in the equation $f\left(t\right)=b_0+b_{1}t+b_2{t}^2+b_3{t}^3$as follows.

$$\begin{gathered} f\left(t\right)=b_0+b_{1}t+b_2{t}^2+b_3{t}^3 \\ f\left(t\right)=b_0+\left(-b_2\frac{2}{3}-b_3\right)t+b_2{t}^2+b_3{t}^3 \\ f\left(t\right)=b_0+b_2\left(t^2-\frac{2}{3}\right)+b_3(t^3-t) \end{gathered}$$

From the equation, the set $\left\{1\left(t^2-\frac{2}{3}t\right),\left(t^3-t\right)\right\}$span $P_3$if $f\left(1\right)=0$and${鈭�}_1^{1}f\left(t\right)dt=0$.

Compare the equations$b_0+b_2\left(t^2-\frac{2}{3}t\right)+b_3\left(t^3-t\right)=0$ both side as follows.

$\begin{array}{l}b+b_2\left(t^2-\frac{2}{3}t\right))+b_3(t^3-t)=0\\ b_0+b_2\left(t^2-\frac{2}{3}t\right)+b_3(t^3-t)=\left(0\right)+\left(0\right)\left(t^2-\frac{2}{3}t\right)+\left(0\right)t^3-t)\end{array}$

By the definition of linear independence, the subset$\left\{1,\left(t^2-\frac{2}{3}t\right),\left(t^3-t\right)\right\}$ is L.I.

Therefore, the set of all polynomial$P_3$ such that$f\left(1\right)=0$ and${鈭�}_1^{1}f\left(t\right)dt=0$ has dimension 3 and spanned by $span\left\{1,\left(t^2-\frac{2}{3}t\right),\left(t^3-t\right)\right\}$.