vittorecostao1

2022-06-19

Showing that the matrix transformation $T(f)=x\ast {f}^{\prime}(x)+{f}^{\u2033}(x)$ is linear

Layla Love

Beginner2022-06-20Added 29 answers

By the definition of T, you have for every function h in your domain of T, that

$T(h)=x\cdot {h}^{\prime}(x)+{h}^{\u2033}(x)$

If now $h=f+g$, we get

$T(f+g)=x\cdot (f+g{)}^{\prime}(x)+(f+g{)}^{\u2033}(x),$which equals your calculated result, which moreover used the linearity of differentiation.

$T(h)=x\cdot {h}^{\prime}(x)+{h}^{\u2033}(x)$

If now $h=f+g$, we get

$T(f+g)=x\cdot (f+g{)}^{\prime}(x)+(f+g{)}^{\u2033}(x),$which equals your calculated result, which moreover used the linearity of differentiation.