 # 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} allhvasstH 2021-09-05 Answered

Consider these Values: $A=32,B=15\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}C=17$
The set ot first four Hermite polynomials is
S={$1,2t,-2+4{t}^{2},-12t+8{t}^{3}$}. These polynomials have a wide variety of applications in physics and engineering.
1) Show that the set S of first four Hermite polynomials form a basis for ${P}_{3}$
2) Find the coordinate vector of the polynomial $p\left(t\right)=-1+At+8{t}^{2}+B{t}^{3}$ relative to S

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Given:
Consider the values $A=32,B=15\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}C=17$
The set of first four Hermite polynomials is S={$1,2t,-2+4{t}^{2},-12t+8{t}^{3}$}
Calculation:
a) dim(${P}_{3}$)=4
The standard basis for ${P}_{3}$={$1,t,{t}^{2},{t}^{3}$}
For, ${c}_{1}\left(1\right)+{c}_{2}\left(t\right)+{c}_{3}\left(-2+4{t}^{2}\right)+{c}_{4}\left(-12t+8{t}^{3}\right)=0$
$\left({c}_{1}-2{c}_{3}\right)+\left({c}_{2}-12{c}_{4}\right)t+4{t}^{2}{c}_{3}+8{t}^{3}{c}_{4}=0+0t+0{t}^{2}+0{t}^{3}$
$8{t}^{3}{c}_{4}=0{t}^{3}⇒{c}_{4}=0$
$4{t}^{2}{c}_{3}=0{t}^{2}⇒{c}_{3}=0$
$\left({c}_{1}-2{c}_{3}\right)=0⇒{c}_{1}-2\left(0\right)=0⇒{c}_{1}=0$
$\left({c}_{2}-12{c}_{4}\right)t=0t⇒{c}_{2}-12\left(0\right)=0⇒{c}_{2}=0$
then ${c}_{1}={c}_{2}={c}_{3}={c}_{4}=0$
${c}_{1}={c}_{2}={c}_{3}={c}_{4}=0$
The polynomials are Linearly Independent and hence they form a basis for ${P}_{3}$
b) $P\left(t\right)=-1+At+8{t}^{2}+B{t}^{3}=-1+32t+8{t}^{2}+15{t}^{3}$
$\left({c}_{1}-2{c}_{3}\right)+\left({c}_{2}-12{c}_{4}\right)t+4{t}^{2}{c}_{3}+8{t}^{3}{c}_{4}=-1+32t+8{t}^{2}+15{t}^{3}$
$8{t}^{3}{c}_{4}=15{t}^{3}⇒{c}_{4}=\frac{15}{8}$