I was bored in my Algebra 2 class and wanted to try to illustrate the tan(x) function as an infinite summation of rational functions.
I recalled that the solution to a rational function's numerator is equivalent to the roots of the function. The denominator's solution is the vertical asymptote(s). Since tan(x) has infinite roots and asymptotes, if I expressed it as a rational function it'd be infinitely long. My teacher suggested just using an infinite summation of rational functions, so I did that instead.
The first thing I did was set up the numerator and denominator so that they could each equal the corresponding roots and asymptotes; since it's (sorta? kinda? maybe) an even function, I just used variations of $x\pm \pi$ for both sides. Given that all roots of tan(x) are whole multiples of pi and all asymptotes are odd multiples of pi/2, I came up with the following equation (I hate mathjax so I'm just gonna draw it):
Where n is any integer from negative infinity to infinity, my idea is that the end product of this is tan(x). Is this even remotely correct or did I leave something out? Thanks!
(I do recognize that the horizontal asymptote is apparently 1 with this equation, but when I plot a section of this in desmos it seems to pass through y=1 anyway. There might be something going on that I could simplify in terms of rewriting an "x^2" somewhere, or maybe I'm just trippin.)
Best Answer
Your idea is good, but the series does not converge.
Actually, from the Mittag-Leffler's theorem (somewhat advanced complex analysis), one has the expansion $$\tan x = \sum_{n=0}^\infty \frac{8x}{(2n+1)^2\pi^2 - 4x^2}.$$ (See https://en.wikipedia.org/wiki/Mittag-Leffler%27s_theorem .) Now the quadratic dependence of $n$ in the denominator ensure convergence! Note that the above series correctly diverges at $x = (n+1/2)\pi$, and is zero at $x=n\pi$.