Question

Give an interval over which the set of two functions${\mathit{f}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$and${\mathit{f}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{x}\mathbf{\left|}\mathit{x}\mathbf{\right|}$is linearly independent. Then give aninterval over which the set consisting of${\mathit{f}}_{\mathbf{1}}$and${\mathit{f}}_{\mathbf{2}}$is linearlydependent.

Short answer

The set of functions is linearly independent on the interval$\text{(-}\infty \text{,}\infty \text{)}$ , and the set of functions is linearly dependent on the interval $\text{(-}\infty \text{,0)}$or $(0,\infty )$.

Step-by-step solution

Step 1:Definecriterion for linearity independent solution.

If and only if $\text{W}\left({\text{y}}_{\text{1}}{\text{,y}}_{\text{2}}\right)\ne \text{0}$for every X in the interval, the set of two solutions ${\text{y}}_{\text{1}}$and ${\text{y}}_{\text{2}}$is linearly independent on the interval .

Find an interval over which the set is linearly dependent.

The${\text{f}}_{\text{2}}\text{(x)=x|x|}$function can be divided into two components.

${\text{f}}_{\text{2}}\text{(x)=}\{\begin{array}{ll}{\text{-x}}^{\text{2}}& \text{for(-}\infty \text{,0)}\\ {\text{x}}^{\text{2}}& \text{for(0,}\infty \text{)}\end{array}$

Because${\text{f}}_{\text{2}}\text{(x)}$in the interval$\text{(-}\infty \text{,}\infty \text{)}$changes its sign at$\text{x=0}$, neither${\text{f}}_{\text{1}}{\text{(x)=x}}^{\text{2}}$nor${\text{f}}_{\text{2}}\text{(x)=x|x|}$can be expressed as a linear combination of the other, we infer that the set of functions is linearly independent on the interval.$\text{(-}\infty \text{,}\infty \text{)}$

The Wronskian for the interval$\text{(-}\infty \text{,0)}$is,

$\begin{array}{c}\text{W}\left({\text{f}}_{\text{1}}{\text{(x),f}}_{\text{2}}\text{(x)}\right)\text{=W}\left({\text{x}}^{\text{2}}{\text{,-x}}^{\text{2}}\right)\\ \text{=}\left|\begin{array}{ll}{\text{x}}^{\text{2}}& {\text{-x}}^{\text{2}}\\ \text{2x}& \text{-2x}\end{array}\right|\\ {\text{=x}}^{\text{2}}\text{×(-2x)-2x×}\left({\text{-x}}^{\text{2}}\right)\\ {\text{=-2x}}^{\text{3}}{\text{+2x}}^{\text{3}}\text{=0}\end{array}$

Similarly, the Wronskian for the interval $(0,\infty )$is $\text{W}\left({\text{f}}_{\text{1}}{\text{(x),f}}_{\text{2}}\text{(x)}\right)\text{=W}\left({\text{x}}^{\text{2}}{\text{,x}}^{\text{2}}\right)\text{=0}$. Since $\text{W}\left({\text{f}}_{\text{1}}{\text{(x),f}}_{\text{2}}\text{(x)}\right)\text{=0}$for both intervals, then the set of functions is linearly dependent on the interval $\text{(-}\infty \text{,0)}$or $(0,\infty )$.