I was doing a problem where the goal was to find whether two functions: $$f(x) = \sin(2x) ,~~~~~ \text{and}~~~~~~ g(x) = \cos(2x)$$ are linearly independent or not using the wronskian.

The problem is simple enough, and after evaluating the wronskian I got a result of $~-2~$. Therefore, the implication is that the two functions are linearly independent for any value of $~x~$. And the way we've defined linear dependence is the following:

If we can find some $~c_1~$ and $~c_2~$ constants such that $$c_1~f(x) + c_2~g(x) = 0~,$$ where $~c_1~$ and $~c_2~$ can't both be zero, then $~f(x)~$ and $~g(x)~$ are linearly dependent.

However, with this definition, I can easily think of a counter example to the result of our wronskian: If I pick $~c_1 = 1~$ and $~c_2 = -1~$, then at $~x = \frac{\pi}{8}~$, we get that $~\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} = 0~$.

So clearly, $~f(x)~$ and $~g(x)~$ are linearly dependent at $~x = \frac{\pi}{8}~$ and, in fact, any $~x~$ such that $~x = \frac{\pi}{8} +2 ~\pi~n~$

This contradicts the result of our wronskian, which implies that $~f(x)~$ and $~g(x)~$ are linearly independent for any $~x~$.

Did I miss something?

Maybe I'm interpreting some definition the wrong way.

  • 1
    $\begingroup$ For any $x \in \mathbb{R}$, $f(x)$ and $g(x)$ are linearly dependent. If $f$ and $g$ have non-zero Wronskian, then $f$ and $g$ are linearly independent functions. That is, if $a, b \in \mathbb{R}$ and $af + bg = 0$, then $a = b = 0$. $\endgroup$ Jul 12 '15 at 22:40

In order for $f(x)$ and $g(x)$ to be linearly dependent functions, there have to be constants $a$ and $b$ such that $$ af(x)+bg(x)=0 $$ for all $x$. You have found constants that work for one specific value of $x$, but they won't work for other values.

Notice that I haven't said anything about the Wronskian yet. There are two relevant facts about the Wronskian:

Fact 1: If the Wronskian of $f_1,f_2,\dots,f_n$ is nonzero at any point, then $f_1,\dots,f_n$ are linearly independent.

Fact 2: If $f_1,f_2,\dots,f_n$ are solutions to a nonsingular $n$th-order differential equation, then their Wronskian is nonzero at every point if they're linearly independent, and zero at every point if they're linearly dependent.

What this means is, if you're given any $n$ functions, there are four possible options:

  1. The Wronskian never vanishes, and those functions could all be linearly independent solutions to the same nonsingular $n$th-order ODE.
  2. The Wronskian vanishes in some places and not in others. If you got the functions as solutions to an $n$th-order ODE, either that ODE has singularities at the points where the Wronskian vanishes, or you screwed up somehow.
  3. The Wronskian vanishes everywhere, but the functions are linearly independent anyway. In practice, this doesn't happen very often.
  4. The functions are linearly dependent. Their Wronskian vanishes everywhere.

The functions $\sin 2x$ and $\cos 2x$ are examples of option 1; whatever functions you computed to have Wronskian $4x^2$ would be examples of option 2 (that is, any ODE which has them as solutions must be singular at $0$).

  • $\begingroup$ Right. This actually confused forever when I told linear/ diff eq. It should probably be mentioned in textbooks. $\endgroup$
    – Elliot G
    Jul 12 '15 at 22:43
  • $\begingroup$ On that note you can use a far simpler example. The functions 1 and x are independent but clearly are the same at x=1. Also, 1 and sin x even have infinite solutions, but are clearly independent. $\endgroup$
    – Elliot G
    Jul 12 '15 at 22:48
  • $\begingroup$ Hmmm, I see your point. I believe what threw me off is that I saw an example where the wronskian gave us 4x^2 and we had to mention that the functions were linearly independent except for when x=0. $\endgroup$
    – user203373
    Jul 12 '15 at 22:57
  • $\begingroup$ Well, there's the problem: that example is wrong, or at least misleading. Two functions (defined on some interval) whose Wronskian exists and is not everywhere $0$ in the interval are linearly independent. $\endgroup$ Jul 13 '15 at 0:13
  • $\begingroup$ @FrenchToastCrunch: See my update. $\endgroup$
    – Micah
    Jul 13 '15 at 0:23

Without using Wronskian: $$ a\sin(2x)+b\cos(2x)=\sqrt{a^2+b^2}\sin(2x+\delta)=0,\quad\forall x\in[x_1,x_2] $$ then $\sqrt{a^2+b^2}=0$ since $\sin$ cannot be identical zero on an interval. It makes $a=b=0$, thus, independent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy