Consider these Values: A=32, B=15 and C=17. The set ot first four Hermite polynomials is. S={1,2t,-2+4t^2,-12t+8t^3}

Given:
Consider the values ${\displaystyle A=32,B=15\text{and}C=17}$
The set of first four Hermite polynomials is S={${\displaystyle 1,2t,-2+4t^2,-12t+8t^3}$}
Calculation:
a) dim(${\displaystyle P_3}$)=4
The standard basis for ${\displaystyle P_3}$={${\displaystyle 1,t,t^2,t^3}$}
For, ${\displaystyle c_1\left(1\right)+c_2\left(t\right)+c_3(-2+4t^2)+c_4(-12t+8t^3)=0}$
${\displaystyle (c_1-2c_3)+(c_2-12c_4)t+4t^2{c}_3+8t^3{c}_4=0+0t+0t^2+0t^3}$
${\displaystyle 8t^3{c}_4=0t^3\Rightarrow c_4=0}$
${\displaystyle 4t^2{c}_3=0t^2\Rightarrow c_3=0}$
${\displaystyle (c_1-2c_3)=0\Rightarrow c_1-2\left(0\right)=0\Rightarrow c_1=0}$
${\displaystyle (c_2-12c_4)t=0t\Rightarrow c_2-12\left(0\right)=0\Rightarrow c_2=0}$
then ${\displaystyle c_1=c_2=c_3=c_4=0}$
Answer (a):
${\displaystyle c_1=c_2=c_3=c_4=0}$
The polynomials are Linearly Independent and hence they form a basis for ${\displaystyle P_3}$
b) ${\displaystyle P\left(t\right)=-1+At+8t^2+B{t}^3=-1+32t+8t^2+15t^3}$
${\displaystyle (c_1-2c_3)+(c_2-12c_4)t+4t^2{c}_3+8t^3{c}_4=-1+32t+8t^2+15t^3}$
${\displaystyle 8t^3{c}_4=15t^3\Rightarrow c_4=\frac{15}{8}}$

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