Question

In Problems 15–22 Determine whether the given set of functions is linearly independent on the interval $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$ .

${\mathbf{f}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathbf{f}}_{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathbf{f}}_{\mathbf{3}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{cos}}^{\mathbf{2}}\mathbf{x}$

Short answer

The set of function is linearly dependent.

Step-by-step solution

Use wronskian and drive the equation

We have the set of functions

${\mathrm{f}}_1\left(\mathrm{x}\right)=\cos 2\mathrm{x},{\mathrm{f}}_2\left(\mathrm{x}\right)=1,{\mathrm{f}}_3\left(\mathrm{x}\right)={\cos }^2\mathrm{x}$

And we have to determine if this set is linearly independent on the interval $(-\infty ,\infty )$or not as the following technique:

First, we have to obtain the wronskian of this set of functions as

$\begin{array}{rcl}\mathrm{W}\left({\mathrm{f}}_1,{\mathrm{f}}_2,{\mathrm{f}}_3\right)& =& \left|\begin{array}{ccc}{\mathrm{f}}_1& {\mathrm{f}}_2& {\mathrm{f}}_3\\ {\mathrm{f}\text{'}}_1& {\mathrm{f}\text{'}}_2& {\mathrm{f}\text{'}}_3\\ {\mathrm{f}}_1\text{'}\text{'}& {\mathrm{f}}_2\text{'}\text{'}& {\mathrm{f}}_3\text{'}\text{'}\end{array}\right|\\ & =& \left|\begin{array}{ccc}\cos 2\mathrm{x}& 1& {\cos }^2\mathrm{x}\\ -2\sin 2\mathrm{x}& 0& -2\mathrm{cosx}\sin \mathrm{x}\\ -4\cos 2\mathrm{x}& 0& 2{\sin }^2\mathrm{x}-2{\cos }^2\mathrm{x}\end{array}\right|\end{array}$

Second, we have to find the determinate of this matrix as

$\begin{array}{rcl}\mathrm{W}\left(\cos 2\mathrm{x},1,{\cos }^2\mathrm{x}\right)& =& 0-\left(1\right)\times \left|\begin{array}{cc}-2\sin 2\mathrm{x}& -2\mathrm{cosx}\sin \mathrm{x}\\ -4\cos 2\mathrm{x}& 2{\sin }^2\mathrm{x}-2{\cos }^2\mathrm{x}\end{array}\right|+0\\ & =& -\left[(-2\sin 2\mathrm{x})\left(2{\sin }^2\mathrm{x}-2{\cos }^2\mathrm{x}\right)-(-2\cos \mathrm{x}\sin \mathrm{x})(-4\cos 2\mathrm{x})\right]\\ & =& 4\sin 2\mathrm{x}{\sin }^2\mathrm{x}-4\sin 2\mathrm{x}{\cos }^2\mathrm{x}+8\sin \mathrm{x}\cos \mathrm{x}\cos 2\mathrm{x}\\ & =& -4\sin 2\mathrm{x}\left({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}\right)+4\cos 2\mathrm{x}(2\sin \mathrm{x}\cos \mathrm{x})······\left(1\right)\\ & & \end{array}$

Third, we know that

$$\begin{gathered} {\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}=\cos 2\mathrm{x}······\left(2\right) \\ 2\sin \mathrm{x}\cos \mathrm{x}=\sin 2\mathrm{x}······\left(3\right) \end{gathered}$$

Then if we substitute the equations (2) and (3) into equation (1), we can have

role="math" localid="1663827649393" $\begin{array}{rcl}\mathrm{W}\left(\cos 2\mathrm{x},1,{\cos }^2\mathrm{x}\right)& =& -4\sin 2\mathrm{x}(\cos 2\mathrm{x})+4\cos 2\mathrm{x}(\sin 2\mathrm{x})\\ & =& -4\sin 2\mathrm{x}\cos 2\mathrm{x}+4\sin 2\mathrm{x}\cos 2\mathrm{x}\\ & =& 0\end{array}$

Since the value of the determinant is zero and the Wronskian of the set of functions is zero,

Therefore, the set of functions is linearly dependent.