Does the Taylor series of sinx divided by the Taylor series of cosx forms the Taylor series of tanx?

Question

The question is in the title

if not, I'd like to know why it doesn't work

Thanks in advance.

Answer 1

It works, indeed but it is slightly tedious $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}+O\left(x^{11}\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320}-\frac{x^{10}}{3628 800}+O\left(x^{11}\right)$$ $$\tan(x)=\frac{\sin(x)}{\cos(x)}=\frac{x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}+O\left(x^{11}\right)}{1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320}-\frac{x^{10}}{3628 800}+O\left(x^{11}\right)}$$ Now, long division to get $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+\frac{17 x^7}{315}+\frac{62 x^9}{2835}+O\left(x^{11}\right)$$

Answer 2

The ratio does not write as a Taylor series directly. To do so, the product and composition rules are used. Recognizing the Taylor series $\frac{1}{1-u} = 1 + u + \dots$, one writes \begin{equation} \begin{aligned} \frac{\sin x}{\cos x} &= \left(x - \frac{x^3}{6} + \dots\right) \frac{1}{1 - \frac{x^2}{2} + \dots}\\ &= \left(x - \frac{x^3}{6} + \dots\right)\left(1 + \frac{x^2}{2} + \dots\right) \\ &= x + \frac{x^3}{3} + \dots \, , \end{aligned} \end{equation} which is the Taylor series of $\tan$.

An important point is convergence. The Taylor series of $\sin$ and $\cos$ are convergent all over $\mathbb{R}$, whereas the Taylor series of $\tan$ is convergent over $]-\pi/2,\pi/2[$.

Note: another technique to obtain the Taylor series of $\tan$ consists in exploiting the fact that $\tan$ is an odd function, and the relationship with the derivative $\tan' = 1 + \tan^2$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

A recursive approach.

$$ \sin\pars{x} \equiv \sum_{n = 0}^{\infty}s_{n}x^{2n + 1}\,,\qquad \cos\pars{x} \equiv \sum_{n = 0}^{\infty}c_{n}x^{2n}\,,\qquad \tan\pars{x} \equiv \sum_{n = 0}^{\infty}t_{n}x^{2n + 1} $$ where $\ds{s_{n} = {\pars{-1}^{n} \over \pars{2n + 1}!}}$ and $\ds{c_{n} = {\pars{-1}^{n} \over \pars{2n}!}}$

---

\begin{align} s_{n} & = \bracks{x^{2n + 1}}\cos\pars{x}\tan\pars{x} = \bracks{x^{2n + 1}}\sum_{i = 0}^{\infty}c_{i}x^{2i} \sum_{j = 0}^{\infty}t_{j}x^{2j + 1} \\[5mm] & = \bracks{x^{2n + 1}}\sum_{i = 0}^{\infty}\sum_{j = 0}^{\infty} c_{i}t_{j}\sum_{k = 0}^{\infty}\delta_{2i + 2j + 1,2k + 1}\,\,x^{2k + 1} \\[5mm] & = \bracks{x^{2n + 1}}\sum_{k = 0}^{\infty}x^{2k + 1} \bracks{\sum_{j = 0}^{\infty}c_{k - j}t_{j} \sum_{i = 0}^{\infty}\delta_{i,k - j}}\implies \bbx{\ds{\sum_{j = 0}^{n}c_{n - j}t_{j} = s_{n}}} \end{align}

From the last expression I'll get a recursive relation for $\ds{\braces{t_{j}}}$. Namely,

$$ \left\{\begin{array}{rcl} \ds{t_{0}} & \ds{=} & \ds{1} \\ \ds{t_{n}} & \ds{=} & \ds{s_{n} - \sum_{j = 0}^{n - 1}c_{n - j}t_{j} = \pars{-1}^{n}\bracks{{1 \over \pars{2n + 1}!} - \sum_{j = 0}^{n - 1} {\pars{-1}^{\,j} \over \pars{2n - 2j}!}\,t_{j}}\,,\qquad n \geq 1} \end{array}\right. $$