Question

In Problems$8\text{to}15,\text{use}\left(8.5\right)$ show that the given functions are linearly independent.

$\text{14.}{e}^{ix},e^{-ix}$

Short answer

It has been shown that the functions $e^{ix},e^{-ix}$are linearly independent for all values of $x$

Step-by-step solution

 Step 1: Definition of linearly independent functions 

The functions $f_1\left(x\right),f_2\left(x\right),\dots ,f_n\left(x\right)$are linearly independent if the determinant

localid="1657470388333" $W=\left|\begin{array}{cccc}{f}_1\left(x\right)& f_2\left(x\right)& \cdots & f_n\left(x\right)\\ f_1^{\mathrm{\prime }}\left(x\right)& f_2^{\mathrm{\prime }}\left(x\right)& \cdots & f_n^{\mathrm{\prime }}\left(x\right)\\ ⋮& ⋮& \ddots & ⋮\\ f_1^{(n-1)}\left(x\right)& f_2^{(n-1)}\left(x\right)& \cdots & f_n^{(n-1)}\left(x\right)\end{array}\right|\ne 0$

Here, W is called the Wronksian of functions.

 Step 2: Use the Wronksian to show that the given functions are linearly independent.

Find the derivatives of the function $f_1\left(x\right)=e^{ix}\text{of order}1$

$\begin{array}{r}{f}_1\left(x\right)=e^{ix}\\ f_1^{\mathrm{\prime }}\left(x\right)=i{e}^{ix}\end{array}$

Find the derivatives of the function $f_2\left(x\right)=e^{-ix}\text{of order}1$

$\begin{array}{r}{f}_2\left(x\right)={\mathrm{e}}^{-ix}\\ f_2^{\mathrm{\prime }}\left(x\right)=-{\mathrm{e}}^{-ix}\end{array}$

Substitute the derivatives in the Wronksian formula and simplify as follows:

$\begin{array}{r}W=\left|\begin{array}{cc}{e}^{ix}& e^{-ix}\\ i{e}^{ix}& -i{e}^{-ix}\end{array}\right|\\ =\left(e^{ix}\times i{e}^{-ix}\right)-\left(e^{-ix}\times i{e}^{ix}\right)\\ =-i-i\\ =-2i\end{array}$

Since , $W\ne 0\text{, therefore,}{\mathrm{e}}^{ix},{\mathrm{e}}^{-ix}$are the linearly independent functions.