Write the first four terms of the power series in x
Write the first four terms of the power series in $x$ for $\tan x$.
Explanation of Method
Since $\tan x=\sin x/\cos x$ has no elementary closed-form coefficient formula, the series is built by writing $\tan x=\sum a_{2k+1}x^{2k+1}$ (odd powers only, by symmetry), multiplying by the known $\cos x$ series, and matching coefficients against the known $\sin x$ series order by order.
- Write $\tan x\cdot\cos x=\sin x$ with unknown coefficients $a_1,a_3,a_5,a_7,\ldots$.
- Match coefficients of each power of $x$ successively.
- Solve for $a_1,a_3,a_5,a_7$ in turn.
Step 1 · Set up the matching equation
With $\cos x=1-\frac{x^2}2+\frac{x^4}{24}-\cdots$ and $\sin x=x-\frac{x^3}6+\frac{x^5}{120}-\cdots$, write $\tan x=a_1x+a_3x^3+a_5x^5+a_7x^7+\cdots$ and require $(\cos x)(\tan x)=\sin x$.
Step 2 · Match coefficients
$x^1$: $a_1=1$. $\quad x^3$: $a_3-\frac{a_1}2=-\frac16\Rightarrow a_3=\frac13$. $\quad x^5$: $a_5-\frac{a_3}2+\frac{a_1}{24}=\frac1{120}\Rightarrow a_5=\frac2{15}$.
Step 3 · Continue to the fourth term
$x^7$: $a_7-\frac{a_5}2+\frac{a_3}{24}-\frac{a_1}{720}=-\frac1{5040}\Rightarrow a_7=\frac{17}{315}$.
Summary Table
| Property | Value |
|---|---|
| First four terms | $x,\ \frac{x^3}3,\ \frac{2x^5}{15},\ \frac{17x^7}{315}$ |
| Coefficients (excl. $x$-power) | $1,\frac13,\frac2{15},\frac{17}{315}$ |
| Radius of convergence | $\pi/2$ |
Additional Notes
- All coefficients are positive, unlike $\sin x$ or $\cos x$ — a reflection of $\tan x$'s monotonic increase between its poles.
- Long division of two convergent series (here $\sin x$ by $\cos x$) is valid precisely because $\cos x\ne0$ near the origin; the resulting series' radius of convergence, $\pi/2$, is exactly the distance to the nearest zero of $\cos x$.
- These same coefficients appear (scaled) in the general formula involving Bernoulli numbers: $\tan x=\sum_{n=1}^\infty\frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}x^{2n-1}$.