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.

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    $\begingroup$ Sine and cosine are a linear combination of exponentials, and tangent isn't. (Polynomials times) linear combinations of exponentials are precisely the functions with the property you're looking at, and it's not hard to verify that $\tan x$ isn't a function of this form (for example because it has poles). $\endgroup$ Jun 28 '11 at 18:58
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    $\begingroup$ It is also not hard to see that the functions $\tan^n x$ are linearly independent. Consider a linear dependence $P(\tan x) = 0$ where $P$ is a polynomial. Since $\tan x$ takes infinitely many values, one of them is not a root of $P$; contradiction. $\endgroup$ Jun 28 '11 at 19:05

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.


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