# Error bound for Taylor approximation to $\tan(x)$

I'm looking to bound the error for the Taylor series of $\tan(x)$ so that I will know how many terms I need to go out to get a desired precision. I've already searched and came across this, but the bound provided in that answer doesn't make much sense.

In particular, for $f(x):=\tan(x)$, I want to first expand around $x=0$ then approximate $\tan(1)$ to a precision of $N$ digits (note that $\tan$ is analytic for $x\in (-\pi/2,\pi/2)$). After using the Lagrange error bound from elementary calculus I have $$R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(1-0)^{n+1}$$ for $c\in (0,1)$. We want $|R_n|\leq 10^{-N}$. The problem is that I can't find a good way to bound the term $f^{(n+1)}(c)$, as higher-order derivatives of $\tan$ are polynomials in $\tan(x)$ and have a very nasty closed form.

The first answer in the link I referred to had some $\frac{c}{(n+1)!}$ bound on it, but because of the way $c$ is defined, after using Mathematica I see that $\frac{c}{(n+1)!}\rightarrow \infty$ for $n\rightarrow \infty$, which doesn't make sense since increasing the number of terms should decrease the error.

In the above, $R_n\rightarrow \infty$ as $n\rightarrow \infty$ as well. I'm not sure if I'm just overlooking something--am I just doing something wrong? Could anybody give me any suggestions?

This is on a complex analysis assignment by the way, but it seems that complex-analytic methods aren't required here.

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Doing Limit[c/(n + 1)!, n -> Infinity, Assumptions -> 0 < c < 1] MMA provides the right answer. – Pragabhava Oct 3 '12 at 4:15
The problem is that $c$ is not a constant, but an expression that depends on $n$: in the other thread, $c=(\pi/4)^{n+1}tan^{(n+1)}(\pi/4)$. Edit: sorry for the notation abuse, but the $c\in (0,1)$ above is not the same as the $c$ referred to in the other thread. – MathStudent1208 Oct 3 '12 at 4:19
That doesn't make sence at all. $\frac{\pi}{4} < 1$ and $\tan(\frac{\pi}{4}) = 1$, so $c_{n+1}<c_n<...<\frac{\pi}{4}$. Still, Limit[(\[Pi]/4)^(n + 1) Tan[\[Pi]/4]^(n + 1)/(n + 1)!, n -> Infinity] gives the correct answer in MMA 8. – Pragabhava Oct 3 '12 at 4:35
$\tan^{(n+1)}$ denotes the $n+1$-st derivative, and not power, of $\tan$... – MathStudent1208 Oct 3 '12 at 4:50

There is a good reason this is on a complex analysis assignment rather than a calculus one. The easiest way to bound the derivative term is to use $$f^{(n)}(z) = \frac{n!}{2\pi i} \int_{\gamma_r} \frac{ f(t) }{(t-z)^{n+1}} dt$$

which comes from differentiating Cauchy's Integral formula. Here $\gamma_r$ is the circle of radius $r$ centered at $z.$

Don't evaluate the integral exactly, just make rough bounds. We have $z\in (0,1)$ and $\gamma_r$ is the circle of radius $r$ around $z.$ $$| f^{(n)} (z)| = \frac{n!}{2\pi} \int_{\gamma_r} \frac{ |\tan t|}{|t-z|^{n+1}} dt \leq \frac{n!M_r}{r^n}$$

where $M_r$ is the maximum of $|\tan z|$ on the contour.

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Great approach! +1 – Pragabhava Oct 3 '12 at 4:22
I've tried using the Cauchy estimate before actually, but that didn't work out well. I'll try this and see how it goes... – MathStudent1208 Oct 3 '12 at 4:50
So we have $R_n=\frac{1}{2\pi i}\int_\gamma \frac{\tan(z)}{(z-c)^{n+2}}dt$, and the integrand has a pole of order $n+2$ at $z=c$. Computing this residue would make me differentiate $\tan(z)$ $n+1$ times. Is there a better way to go about this? – MathStudent1208 Oct 3 '12 at 5:12
@MathStudent1208 Please see my edited answer. – Ragib Zaman Oct 3 '12 at 5:20
Hi Ragib, thanks for your update. This is the same as using Cauchy's estimate, which I tried before. In particular, with the bound above, we have $|R_n|\leq \frac{M_r}{r^{n+1}}$. But since $\tan$ is only analytic on $(-\pi/2,\pi/2)$ and $z\in (0,1)$, if we take $z>\pi/2-1$ we will have $|r|<1$, as the largest circle around $z$ must be contained in the strip of convergence. In that case, since $|r|<1$, it appears that $|R_n|\rightarrow \infty$ as $n\rightarrow \infty$, which is puzzling because $n\rightarrow \infty$ should make the approximation better. – MathStudent1208 Oct 3 '12 at 6:42