Working out $\tan x$ using sin and cos expansion

Question

Using only the series expansions $\sin x = x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...$ and $\cos x = 1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...$ Find the series expansions of the $\tan x$ function up to the $x^5$ term.

So it is: $$ \frac {x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...} {1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...} $$

However many times I retry, trying long division results :

$$ \require{enclose} \begin{array}{rll} Q(x) \\[-3pt] 1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ... \enclose{longdiv}{x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...}\kern-.2ex \\[-3pt] \underline{x-\dfrac{x^3} {2!} + \dfrac{x^5} {4!}\phantom{0000}} \\[-3pt] \dfrac{-x^3} {3!} + \dfrac{x^3} {2!}+ \dfrac{x^5} {5!}-\dfrac{x^5} {4!} \phantom{00}&& \\[-3pt] \underline{\phantom{0}\dfrac{-x^3} {3!} + \dfrac{x^3} {2!}- \dfrac{x^5} {2!2!}+ \dfrac{x^5} {3!2!}} && \\[-3pt] \dfrac{x^5} {5!} - \dfrac{x^5} {4!} + \dfrac{x^5} {2!2!}-\dfrac{x^5} {3!2!}&& \\[-3pt] \end{array} $$

Sorry for the horrible format. but this is the best i can do..

$$ Q(x) = x + (\frac {x^3}{2!} - \frac {x^3}{3!}) + (\dfrac {x^5}{5!} - \dfrac {x^5}{4!} + \dfrac {x^5}{2!2!} - \dfrac {x^5}{3!2!}) + ... $$

Which results the coefficients of the $x^5$ term $\dfrac 1 {20}$.
This is clearly a wrong value. I can't seem to get $\dfrac 2 {15}$ instead.
Where did I get wrong?

Many thanks in advance.

Answer 1

You may exploit the fact that $\tan x$ is an odd function, hence in a neighbourhood of the origin: $$ \tan x=\sum_{n\geq 0} a_{2n+1}\, x^{2n+1}\tag{1} $$ as well as: $$ \frac{d}{dx}\tan x=\frac{1}{\cos^2 x}=1+\color{purple}{\tan^2 x} = \sum_{n\geq 0}(2n+1)\,a_{2n+1}\,x^{2n} \tag{2}$$ from which it follows that $a_1=1$ and:

$$ a_{2n+1} = \frac{1}{2n+1}\sum_{k=1}^{n}a_{2k-1}a_{2n-2k+1}\tag{3}$$

so the Taylor coefficients of the tangent function can be computed with a rather simple recursion. $(3)$ leads to: $$ a_3 = \frac{1}{3}a_1^2 = \frac{1}{3},\quad a_5=\frac{2}{5}a_1 a_3=\frac{2}{15}.\tag{4}$$ No polynomial divisions involved. Pretty nice, don't you think?

Answer 2

The following way may be easier.

Since $\tan x$ is an odd function, it can be represented as $\tan x=ax+bx^3+cx^5+\cdots.$

Since $\sin x=\cos x\tan x$, one has $$x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)\left(ax+bx^3+cx^5+\cdots\right).$$

Comparing the coefficients gives you $$1=a$$ $$-\frac{1}{3!}=b-\frac{a}{2!}$$ $$\frac{1}{5!}=c-\frac{b}{2!}+\frac{a}{4!}.$$

Answer 3

You get that the coefficient for $x^5$ is $\dfrac {1}{5!} - \dfrac {1}{4!} + \dfrac {1}{2!2!} - \dfrac {1}{3!2!}$, which is indeed $\dfrac2{15}$. So, no need to change method.