# Checking Linear Independence. Using Wronskian vs. Using Definition

1. Feb 10, 2013

### mef51

1. The problem statement, all variables and given/known data
Is the set $$\{cos(x), cos(2x)\}$$ linearly independent?

2. Relevant equations

Definition: Linear Independence
A set of functions is linearly dependent on a ≤ x ≤ b if there exists constants not all zero
such that a linear combination of the functions in the set are equal to zero.

Definition: Wronskian
http://en.wikipedia.org/wiki/Wronskian

Theorem
If the Wronskian of a set of n functions defined on the interval a ≤ x ≤ b is nonzero for at least one point then the set of functions is linearly independent there.

3. The attempt at a solution

Let's say I'm using the interval [-∞, ∞]. First, I'll use the definition.

Consider
$$a*cos(x) + b*cos(2x)$$
Now, pick x = 0, a = 1, b = 1
$$1*cos(0) - 1*cos(0) = 0$$
Since a ≠ 0 and b≠ 0, I conclude from the definition that the functions are linearly dependent.

Now, I'll use the Wronskian.

$$W(cos(x), cos(2x)) = \left| \begin{array}{cc} cos(x) & cos(2x) \\ -sin(x) & -2sin(2x) \end{array} \right| = -2sin(2x)cos(x) + sin(x)cos(2x)$$

Pick x = ∏/4. Then,

$$W = -2sin(\frac{\pi}{2})cos(\frac{\pi}{4}) + sin(\frac{\pi}{4})cos(\frac{\pi}{2}) = \frac{-2}{\sqrt{2}} ≠ 0$$

So, by the Theorem above, since the Wronskian is nonzero, I conclude that the functions are linearly independent.

A contradiction. What in flying flip went wrong?

Last edited: Feb 10, 2013
2. Feb 10, 2013

### mef51

Hey got it!

I misinterpreted the definition of linear dependence.

The constants need to be non-zero for all x on the interval. I just chose one x.