Testing Linear Independence of 3 Functions

[CalcYouLater]

Homework Statement

I have five sets of three functions that I need to test for linear independence. I know that I could write them as a linear combination of scalar multiples of the functions, set it equal to 0 and see if I can solve for the scalar multiples. To clarify I mean:

Functions:
$$cos(x),cos(2x),cos(3x)$$

$$a_{1}cos(x)+a_{2}cos(2x)+a_{3}cos(3x)=0$$

Since the only solutions to that equation are a_1=a_2=a_3=0, the functions are linearly independent.

I also know that I could use the Wronskian of the functions, but this is time consuming and can lead to meaningless information.

Does anyone know of a more efficient method to determine if 3 functions are linearly independent or dependent?

 

[vela]

Those are the only two methods that come to mind. How did you go about showing a₁=a₂=a₃=0?

 

[CalcYouLater]

Hi. Thanks for responding. I just went back over my work, and it looks like I basically just made an assumption. Thanks for calling my attention to it. I first attempted to find the determinant of a 3X3 matrix containing the cosine functions listed above and their derivatives. If the determinant was not equal to 0, then I would say that they are linearly independent. I began to work out the expression, but it was a nightmare. So, I started thinking about what a graph of the function looked like. I figured that since each one had a different frequency, none of them could be written as a linear combination of one or more of the others. Now, that line of reasoning doesn't seem so airtight. Is it possible to have an interference in a wave such that the "new" wave has a different frequency than the "old" wave?

 

[vela]

Sort of. If you sum two sine waves, you get a beat, but you don't actually have any new frequency components.

http://hyperphysics.phy-astr.gsu.edu/hbase/sound/beat.html#c3

To solve for a_n, try multiplying the equation by cos nx and integrating from -pi to pi.

 

[fzero]

The Wronskian is probably the easiest rigorous method to use and it probably what your prof is expecting. There are other ways to show the result (Fourier analysis techniques for example) but they rely on a lot of results that you probably aren't meant to assume for your course.

 

[CalcYouLater]

Of course, I should have thought about orthogonality. Their inner product with cos nx is non-zero (provided n=1,2,3 for cos(x),cos(2x),cos(3x)). Thanks for the insight, I think this will make things faster.

 

[vela]

You could also evaluate the Wronskian for various values of x and show there is at least one x where it's not equal to zero.

 

[CalcYouLater]

I was thinking in terms of solving the Wronskian analytically. That's why I thought it would be a nightmare, I was thinking about possible trig substitutions and I was getting over-whelmed. Plugging in values of x makes sense. My expression for the Wronskian shows clearly there are more than a few choices here. This method seems like it will work well when the functions are all in different families.

 

[Office_Shredder]

Worst case scenario, just start plugging values of x into your expression
$$a_{1}cos(x)+a_{2}cos(2x)+a_{3}cos(3x)=0$$

For example, if we plug in x=0 we get
$$a_1 + a_2 + a_3=0$$.

if we plug in x=pi/2 we get

$$-a_2=0$$

And if x=pi/3:

$$\frac{a_1}{2}-\frac{a_2}{2}-a_3=0$$

Use these to solve for the only possible choice of $$a_1,a_2,a_3$$