# Maclaurin polynomial of tan(x)

The method used to find the Maclaurin polynomial of sin(x), cos(x), and $e^x$ requires finding several derivatives of the function. However, you can only take a couple derivatives of tan(x) before it becomes unbearable to calculate.

Is there a relatively easy way to find the Maclaurin polynomial of tan(x)?

I considered using tan(x)=sin(x)/cos(x) somehow, but I couldn't figure out how.

• A recurrence formula for the coefficients was obtained here. Commented Nov 25, 2015 at 22:35

Long division of series.

$$\matrix{ & x + \frac{x^3}{3} + \frac{2 x^5}{15} + \dots \cr 1 - \frac{x^2}{2} + \frac{x^4}{24} + \ldots & ) \overline{x - \frac{x^3}{6} + \frac{x^5}{120} + \dots}\cr & x - \frac{x^3}{2} + \frac{x^5}{24} + \dots\cr & --------\cr & \frac{x^3}{3} - \frac{x^5}{30} + \dots\cr & \frac{x^3}{3} - \frac{x^5}{6} + \dots\cr & ------\cr &\frac{2 x^5}{15} + \dots \cr &\frac{2 x^5}{15} + \dots \cr & ----}$$

Starting with $y=\tan x$ you can generate successive derivatives quite easily by getting rid of terms involving $x$ as follows:

$$y'=\sec^2 x=1+y^2$$ $$\Rightarrow y''=2yy'$$ $$\Rightarrow y'''=2(y')^2+2yy''$$ $$\Rightarrow y^{(4)}=4y'y''+2y'y''+2yy'''=6y'y''+2yy'''$$

And so on...

Tangent's derivatives are not cyclic. If you differentiate $e^x$, $\cos(x)$, $\sin(x)$ and even $\sinh(x)$ and $\cosh(x)$ enough times, you'll eventually end up with the original function providing a nice formula for the Taylor coefficients. A way to state this periodicity is to say that they all satisfy a differential equation of the form $y^{(n)}=y$ for some $n$. This property means that their Taylor Series will have some kind of periodic relation among the coefficients.

Not so with tangent.

Instead, tangent satisfies the differential equation $y'=1+y^2$. This is because $\frac{d}{dx}\tan(x) = \sec^2(x) = 1+\tan^2(x)$, via the Pythagorean Theorem. This can be used to compute specific values for the coefficients. But the general form of the Taylor Expansion is

$$\tan(x) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{2n(2n-1)!}x^{2n-1}$$

where the $B_n$ are the Bernoulli Numbers, which are defined to be the Taylor Series coefficients of $\frac{x}{e^x-1}$. I'll let ProofWiki provide the proof of this formula.

• It seems like the ProofWiki proof is not there anymore... While I'm struggling to find a proof for this
Commented Oct 6, 2020 at 14:55
• ok, a nice proof is here math.stackexchange.com/questions/2098941/…
$\tan x=\displaystyle\sum_{n=0}^\infty A_{2n+1}\cdot\frac{x^{2n+1}}{(2n+1)!}~,~$ where $A_{2n+1}$ are the so-called “zag-” or tangent numbers.
They go hand-in-hand with their complementary counterparts $A_{2n}$ , commonly called “zig-”
or secant numbers : $~\sec x=\displaystyle\sum_{n=0}^\infty A_{2n}\cdot\frac{x^{2n}}{(2n)!}~.~$ The entire sequence $A_n$ is called “zigzag-” or