TITLE: The derivatives of the tangent generate an infinite set of linearly independent elements
QUESTION [5 upvotes]: The method of undetermined coefficients requires, that the input of the ODE be a function that returns to some linear combination of itself if you take the derivative enough times.
Tangent is offered as an example of a function that fails this condition. The derivatives of the tangent produce functions containing higher and higher powers of $\tan^n x$. My books says they are independent, or rather, that new independent terms will always be produced by taking the derivative.
I tried to check this idea with the wronskian of $\tan^n x$ and $\tan^m x$, but I got a function that is often zero, 
$(m-n)\tan^{n+m-1}x\sec^2 x$
so it is inconclusive. How can I simply prove independence?
Also, why is the tangent so badly behaved compared to sin and cosine?  I don't know if there is a succinct answer to that, but I thought I'd ask anyway.

REPLY [3 votes]: One way to see this is that $\mathrm{tan}$ is meromorphic but not analytic. At $\pi/2$ the tangent function has a pole of order $1$ and hence has a laurent expansion around  $\pi/2$ of the form $a_{-1}(x-\pi/2)^{-1} + O(1)$. It follows 
$$tan^{(n)}(x) =  n!(-1)^{n-1}a_{-1}(x-\pi/2)^{-n} + O((x-\pi/2)^{-n+1}).$$
There is obviously no linear relation among these Laurent series. Your book's assertion follows.