The Tangent Series From a Differential Identity

Why Use a Differential Identity?

Tangent can be written as a quotient of sine and cosine, so its series could be obtained by dividing their Maclaurin series. There is another route: tangent satisfies a simple differential equation. Equating coefficients in that equation produces a recurrence that determines its series one coefficient at a time.

This method illustrates a useful principle in Taylor-series construction: a differential identity can encode relationships among the coefficients of a function’s series. The recurrence gives the terms, while the singularities of the function explain how far the resulting series converges.

Key idea: If $y=\tan x$, then $y'=1+y^2$ and $y(0)=0$. Substituting a power series for $y$ into this identity determines its coefficients.

Deriving the Coefficient Recurrence

Near $x=0$, cosine is nonzero, so $\tan x=\sin x/\cos x$ is analytic there and has a Maclaurin series. Write that series as

y(x)=\tan x=\sum_{n=0}^{\infty}a_nx^n.

The derivative of tangent is $\sec^2x$, and the identity $\sec^2x=1+\tan^2x$ gives $y'=1+y^2$. Also, $y(0)=\tan 0=0$, so $a_0=0$. Differentiating the series and multiplying it by itself gives

y'(x)=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n, \qquad y(x)^2=\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n}a_ka_{n-k}\right)x^n.

The inner sum is the coefficient convolution for a product of power series. Comparing the coefficients of $x^n$ in $y'=1+y^2$ yields the recurrence.

Coefficient recurrence: For $n\geq 0$, $$ (n+1)a_{n+1}=\delta_{n0}+\sum_{k=0}^{n}a_ka_{n-k}, \qquad a_0=0, $$ where $\delta_{n0}=1$ when $n=0$ and $\delta_{n0}=0$ when $n\geq1$. The extra term accounts for the constant $1$ in $1+y^2$.

At each step, the recurrence gives $a_{n+1}$ using coefficients already found. For $n=0$, it gives $a_1$; for $n=1$, it gives $a_2$, and so on.

Worked Example: Generate the Series Through the Fifth Power

Start with $a_0=0$ and apply the recurrence for $n=0,1,2,3,4$.

For $n=0$,

$$ a_1=1+a_0a_0=1. $$

For $n=1$,

2a_2=a_0a_1+a_1a_0=0, \qquad a_2=0.

For $n=2$,

3a_3=a_0a_2+a_1a_1+a_2a_0=0+1+0=1, \qquad a_3=\frac{1}{3}.

For $n=3$,

4a_4=a_0a_3+a_1a_2+a_2a_1+a_3a_0=0, \qquad a_4=0.

For $n=4$,

5a_5=a_0a_4+a_1a_3+a_2a_2+a_3a_1+a_4a_0 =0+\frac13+0+\frac13+0=\frac23, \qquad a_5=\frac{2}{15}.

Thus the terms through degree five are

\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\cdots.

Check the calculation by substituting the truncated polynomial $P(x)=x+x^3/3+2x^5/15$ into the differential identity and comparing coefficients through degree four:

P'(x)=1+x^2+\frac23x^4, \qquad 1+P(x)^2=1+x^2+\frac23x^4+\text{terms of degree at least }6.

The two sides agree through degree four, as required for the coefficients calculated.

Why Only Odd Powers Appear

Tangent is an odd function: $\tan(-x)=-\tan x$. An analytic odd function has zero coefficients for every even power. The recurrence also reflects this pattern: after $a_0=0$, its first coefficients $a_2,a_4,\ldots$ vanish.

Here is the recurrence argument. Suppose the even coefficients below $a_{2m}$ have already been shown to be zero. Use the recurrence with $n=2m-1$ to calculate $a_{2m}$. Every product $a_ka_{2m-1-k}$ has indices whose sum is odd, so one index is even. If that even index is zero, its coefficient is $a_0=0$; if it is positive and less than $2m$, its coefficient vanishes by the induction assumption. Therefore every product in the sum is zero, and $2m\,a_{2m}=0$. Hence $a_{2m}=0$.

Consequence: The Maclaurin series of tangent contains only odd powers: $$ \tan x=a_1x+a_3x^3+a_5x^5+a_7x^7+\cdots. $$

Worked Example: Continue Through the Ninth Power

Using the coefficients already found, calculate $a_7$ with $n=6$. The only nonzero terms in the convolution are those with index pairs $(1,5)$, $(3,3)$, and $(5,1)$:

7a_7=a_1a_5+a_3a_3+a_5a_1 =\frac{2}{15}+\frac{1}{9}+\frac{2}{15} =\frac{6}{45}+\frac{5}{45}+\frac{6}{45} =\frac{17}{45}.

Therefore $a_7=17/315$. Next, use $n=8$ to find $a_9$. The contributing index pairs are $(1,7)$, $(3,5)$, $(5,3)$, and $(7,1)$:

9a_9 =a_1a_7+a_3a_5+a_5a_3+a_7a_1 =\frac{17}{315}+\frac{2}{45}+\frac{2}{45}+\frac{17}{315} =\frac{34}{315}+\frac{28}{315} =\frac{62}{315}.

Thus $a_9=62/2835$, and the series through degree nine is

\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15} +\frac{17x^7}{315}+\frac{62x^9}{2835}+\cdots.

For a coefficient check, the coefficient of $x^6$ in $y^2$ is $a_1a_5+a_3^2+a_5a_1=17/45$, matching $7a_7$. The coefficient of $x^8$ in $y^2$ is $62/315$, matching $9a_9$. These are exactly the coefficient equations required by $y'=1+y^2$.

Locating the Nearest Singularities

The recurrence constructs coefficients, but does not by itself tell us the interval of convergence. For that, consider tangent as a complex function, using $\tan z=\sin z/\cos z$. Its possible singularities occur where $\cos z=0$, unless the numerator vanishes there too.

Write $z=u+iv$, where $u,v$ are real. The exponential formula for cosine shows that $\cos z=0$ exactly when $e^{2iz}=-1$. Since the modulus of $e^{2iz}$ is $e^{-2v}$, this equation requires $e^{-2v}=1$, and therefore $v=0$. For real $u$, $e^{2iu}=-1$ precisely when $2u=(2k+1)\pi$, with $k\in\mathbb Z$. Thus the zeros of cosine are

z=\frac{\pi}{2}+k\pi,\qquad k\in\mathbb Z.

At each such point, sine is nonzero, so tangent has a pole there. The nearest poles to $0$ are $z=\pi/2$ and $z=-\pi/2$, each at distance $\pi/2$.

Radius of convergence: The Maclaurin series for $\tan x$ has radius of convergence $R=\pi/2$, and hence converges for $|x|<\pi/2$.

The reason is that a function analytic in a disk has a Taylor series converging throughout that disk, and the radius is limited by the nearest singularity. Here, tangent is analytic for $|z|<\pi/2$, since cosine has no zeros in that disk. Its poles at $z=\pm\pi/2$ prevent the Taylor series from having a larger radius: if it did, the series would define an analytic function near those poles, contradicting the fact that it agrees with tangent within its disk of convergence. This identifies the radius; endpoint convergence is a separate question and is not needed to locate it.

Worked Example: Find the Nearest Pole

To locate the zeros of cosine, solve $e^{2iz}=-1$. For $z=u+iv$, the modulus equation gives

|e^{2iz}|=e^{-2v}=|-1|=1, \qquad v=0.

With $z=u$ real, the phase equation is $e^{2iu}=-1$, so

2u=(2k+1)\pi, \qquad z=u=\frac{(2k+1)\pi}{2}.

The choices $k=0$ and $k=-1$ give $z=\pi/2$ and $z=-\pi/2$, whose distances from $0$ are both $\pi/2$. For example, $\sin(\pi/2)=1$, so the numerator does not cancel the zero of cosine there. The nearest singularities are therefore poles at distance $\pi/2$, confirming the radius of convergence.

A Common Coefficient Mistake

When comparing coefficients, it is easy to forget the constant $1$ in the differential equation. For instance, an incorrect calculation at $n=0$ might use only the constant coefficient of $y^2$:

$$ a_1= a_0a_0=0. $$

This would incorrectly remove the linear term. The right-hand side is $1+y^2$, not just $y^2$. At $n=0$, the correct equation is

$$ a_1=1+a_0a_0=1. $$

A second safeguard is to keep the derivative index straight: the coefficient of $x^n$ in $y'$ is $(n+1)a_{n+1}$, not $na_n$. Keeping both details in view prevents an incorrect recurrence from the first step.

Why the Method Matters

The differential identity links each new coefficient to products of earlier ones, turning a function equation into an algebraic construction. The recurrence is especially useful when a function’s differential equation is simpler than the quotient or composition needed to represent it. It also offers a built-in check: coefficients computed at each stage must satisfy the corresponding coefficient equation.

For tangent, the two parts of the construction answer different questions. The recurrence gives the actual coefficients, while the nearest poles determine the radius within which the series represents the function. A correct list of coefficients does not alone establish a convergence interval; that conclusion depends on the function’s analytic behavior.

Check Your Understanding

Use the recurrence and the singularity argument to answer these questions.

Why does the recurrence include a separate $\delta_{n0}$ term?
Use the recurrence to calculate $a_3$ and $a_5$.
Why do the even-power coefficients vanish?
Where are the zeros of cosine in the complex plane?
Why do the poles at $z=\pm\pi/2$ imply that the Maclaurin series cannot have radius greater than $\pi/2$?

The Tangent Series From a Differential Identity

What You'll Learn

Why Use a Differential Identity?

Deriving the Coefficient Recurrence

Worked Example: Generate the Series Through the Fifth Power

Worked Example: Generate the Series Through the Fifth Power

Why Only Odd Powers Appear

Worked Example: Continue Through the Ninth Power

Worked Example: Continue Through the Ninth Power

Locating the Nearest Singularities

Worked Example: Find the Nearest Pole

Worked Example: Find the Nearest Pole

A Common Coefficient Mistake

Why the Method Matters

Check Your Understanding

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