91影视

Consider a polynomial $f\left(t\right)=6$.

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

Assume there exist the constant $b_i$such that $b_1\left(t-3\right)+b_2{\left(t-3\right)}^2+b_3{\left(t-3\right)}^3=6$.

Simplify the equation$b_1\left(t-3\right)+b_2{\left(t-3\right)}^2+b_3{\left(t-3\right)}^3=6$ as follows.

$$\begin{gathered} b_1\left(t-3\right)+b_2{\left(t-3\right)}^2+b_3{\left(t-3\right)}^3=6 \\ {b}_{1}t-3b_1+b_2\left(t^2+9-6t\right)+b_3\left(t^3-27-9t^2+27t\right)=6 \\ {b}_{1}t-3b_1+b_2{t}^2+9b_2-6t{b}_2+b_3{t}^3-27b_3-9t^2{b}_3+27t{b}_3=6 \\ \left(-3b_1+9b_2-27b_3\right)+\left(-6b_2+b_1+27b_3\right)t+\left(-9b_3+b_2\right)t^2+b_3{t}^3=6 \end{gathered}$$

Compare the equation$\left(-3b_1+9b_2-27b_3\right)+\left(-6b_2+b_1+27b_3\right)t+\left(-9b_3+b_2\right)t^2+b_3{t}^3=6$ both side as follows.

$\begin{array}{rcl}\left(-3b_1+9b_2-27b_3\right)+\left(-6b_2+b_1+27b_3\right)t+\left(-9b_3+b_2\right)t^2+b_3{t}^3& =& 6+\left(0\right)t+\left(0\right)t^2+\left(0\right)t^3\\ -9b_3+b_2& =& 0\\ -6b_2+b_1+27b_3& =& 0\\ b_3& =& 0\\ -3b_1+9b_2-27b_3& =& 6\\ & & \end{array}$

Substitute the value for$b_3$ in the equation $-9b_3+b_2=0$as follows.

$\begin{array}{rcl}-9b_3+b_2& =& 0\\ -9\left(0\right)+b_2& =& 0\\ b_2& =& 0\\ & & \end{array}$

Substitute the values 0 for $b_3$and 0 for$b_2$ in the equation$-6b_2+b_1+27b_3=0$ as follows.

$\begin{array}{rcl}-6b_2+b_1+27b_3& =& 0\\ -6\left(0\right)+b_{1A}+27\left(0\right)& =& 0\\ b_1& =& 0\\ & & \end{array}$

Substitute the values 0 for $b_2$,0 for $b_3$,0 for $b_1$in the equation $-3b_1+9b_2-27b_3=6$as follows.

$\begin{array}{rcl}-3b_1+9b_2-27b_3& =& 6\\ -3\left(0\right)+9\left(0\right)-27\left(0\right)& =& 6\\ 6& =& 0\\ & & \end{array}$

The above condition is not possible.

Therefore, every polynomial of degree cannot be express as a linear combination of the polynomial $\left(t-3\right)$, ${\left(t-3\right)}^2$and${\left(t-3\right)}^3$

Hence, the statement is false.