The Sine Series

Defining Sine by a Power Series

The exponential series introduced a function by specifying its coefficients and then studying the resulting power series. We now use a related series, with alternating signs and odd powers, to define the sine function. Its factorial denominators ensure convergence even when the input is large, while its pattern of powers will give the function a useful symmetry.

x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}.

Definition: For $x\in\mathbb{R}$, define $$ S(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}, $$ once convergence has been established. We write $\sin x=S(x)$. The power series on the right is called the sine series.

The definition is not yet complete until convergence is checked for every real $x$. As with the Exponential Series, the ratio of consecutive absolute terms gives a direct way to control the tail. The index matters here: the exponent and factorial both increase by two from one term to the next.

Convergence for Every Real Input

Theorem (Absolute Convergence of the Sine Series): The sine series converges absolutely for every real $x$. Its radius of convergence is infinite.

Proof. Fix $x\ne0$ and set $u_n=|x|^{2n+1}/(2n+1)!$. Each $u_n$ is positive, and

\frac{u_{n+1}}{u_n} =\frac{|x|^{2n+3}}{(2n+3)!}\frac{(2n+1)!}{|x|^{2n+1}} =\frac{|x|^2}{(2n+2)(2n+3)}.

For this fixed $x$, the last expression tends to zero as $n$ tends to infinity. Choose an index $N$ such that $u_{n+1}\leq u_n/2$ for every $n\geq N$. Repeated application gives $u_{N+j}\leq u_N(1/2)^j$ for every integer $j\geq0$. Therefore,

\sum_{n=N}^{\infty}u_n \leq u_N\sum_{j=0}^{\infty}\left(\frac12\right)^j =2u_N<\infty.

The finitely many terms preceding $u_N$ do not affect convergence, so the series of absolute values converges. When $x=0$, every term is zero, and convergence is immediate. Thus the sine series converges absolutely for every real input. It converges at every real distance from its center $0$, so its radius of convergence is infinite. $\square$

This also justifies defining $S(x)$ for all real $x$. By the Uniform Convergence Inside the Radius Theorem and the Term-by-Term Differentiation of a Power Series Theorem, the series may be differentiated on the real line. The derivative produces a companion series with even powers. We introduce that series only to state the derivative formulas; its further properties belong to the study of the cosine series.

Derivatives and the Sine Differential Equation

Definition: For $x\in\mathbb{R}$, let $$ C(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}. $$ This companion power series also has infinite radius of convergence.

The infinite radius follows from the same ratio argument: the ratio of consecutive absolute terms, for $x\ne0$, is $x^2/((2n+1)(2n+2))$, which tends to zero. Term-by-term differentiation is therefore valid for both series.

Theorem (Derivative Relations for the Sine Series): The functions $S$ and $C$ are differentiable on $\mathbb{R}$, and $$ S'(x)=C(x),\qquad C'(x)=-S(x). $$ In particular, $S$ is twice differentiable and satisfies $S''(x)=-S(x)$ for every real $x$.

Proof. Differentiate the series for $S$ term by term:

S'(x) =\sum_{n=0}^{\infty}(-1)^n\frac{(2n+1)x^{2n}}{(2n+1)!} =\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} =C(x).

For the derivative of $C$, the constant term contributes zero. For each remaining term, differentiation gives

C'(x) =\sum_{n=1}^{\infty}(-1)^n\frac{x^{2n-1}}{(2n-1)!} =-\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!} =-S(x),

where the second equality uses $k=n-1$, so that $(-1)^n=-(-1)^k$. The differentiation theorems apply because both power series have infinite radius of convergence. Since $S'=C$ and $C'=-S$, differentiating the first identity gives $S''=C'=-S$, as claimed. $\square$

The differential equation is a consequence of the coefficients: differentiating twice shifts the odd powers back to odd powers, and the alternating signs change the series to its negative. The series also immediately gives the initial values $S(0)=0$ and $S'(0)=C(0)=1$.

Parity and Estimates Near Zero

Every power in the sine series is odd. Replacing $x$ by $-x$ therefore changes the sign of each term. Because the series converges absolutely, its value may be computed term by term, giving the symmetry below.

Proposition (Oddness of the Sine Series): For every real $x$, $S(-x)=-S(x)$.

Proof. For each $n\geq0$, $(-x)^{2n+1}=-x^{2n+1}$. Thus the $n$th term at $-x$ is the negative of the $n$th term at $x$. Summing the convergent series gives $S(-x)=-S(x)$. $\square$

For small positive inputs, the terms of the sine series decrease in absolute value. Indeed, if $0<x\leq1$, the ratio of consecutive absolute terms is

\frac{x^{2n+3}/(2n+3)!}{x^{2n+1}/(2n+1)!} =\frac{x^2}{(2n+2)(2n+3)} \leq\frac{1}{6}<1.

The Alternating Series Test and its remainder bound then place the sum between consecutive partial sums. Keeping the first two terms gives $x-x^3/3!\leq S(x)\leq x$. In particular, $x-x^3/6=x(1-x^2/6)>0$ for $0<x\leq1$. This establishes positivity near zero directly from the series, without assuming it from a graph or a geometric definition.

Worked Example: Bounding the Series at Two

At $x=2$, the first three terms, through index $n=2$, sum to

2-\frac{2^3}{3!}+\frac{2^5}{5!} =2-\frac{4}{3}+\frac{4}{15} =\frac{14}{15}.

To bound the remaining tail, put $u_n=2^{2n+1}/(2n+1)!$. Its first omitted term is $u_3=2^7/7!=8/315$, and for every $n\geq3$,

\frac{u_{n+1}}{u_n} =\frac{4}{(2n+2)(2n+3)} \leq\frac{4}{8\cdot9} =\frac{1}{18}.

Repeated use of this ratio bound shows that the absolute tail is at most a geometric sum:

\sum_{n=3}^{\infty}u_n \leq\frac{8}{315}\sum_{j=0}^{\infty}\left(\frac{1}{18}\right)^j =\frac{8}{315}\cdot\frac{18}{17} =\frac{16}{595}.

Consequently, the error after the three displayed terms has absolute value at most $16/595$. The estimate is obtained from the series itself and does not require evaluating its sum in another way.

Worked Example: Proving Positivity at One-Half

For $x=1/2$, the first two terms of the sine series sum to

\frac12-\frac{(1/2)^3}{3!} =\frac12-\frac{1}{48} =\frac{23}{48}.

The absolute terms decrease, since their consecutive ratio is at most $1/6$ when $0<x\leq1$. By the alternating-series bounds, the full sum lies between this two-term partial sum and the first partial sum:

\frac{23}{48}\leq S\left(\frac12\right)\leq\frac12.

Since $23/48>0$, this proves $S(1/2)>0$. It also gives a concrete enclosure: the value lies within $1/48$ of $1/2$.

Worked Example: Reading the Initial Derivatives from the Series

The derivative relations give $S(0)=0$, $S'(0)=C(0)=1$, and $S''(0)=-S(0)=0$. Differentiating $S''=-S$ once more gives $S'''=-S'$, so $S'''(0)=-1$. Thus the series determines the first derivatives at the origin directly:

S(0)=0,\qquad S'(0)=1,\qquad S''(0)=0,\qquad S'''(0)=-1.

In particular, the definition of the derivative at zero gives $$ \lim_{x\to0}\frac{S(x)-S(0)}{x}=S'(0)=1. $$ Since $S(0)=0$, this says $S(x)/x\to1$ as $x\to0$. The function therefore has slope one at the origin, in agreement with the leading term $x$ in its defining series.

What the Series Reveals

The sine series supplies several useful facts before any further identities are considered. Absolute convergence makes the function well defined on the whole real line. Term-by-term differentiation gives the companion series and the differential equation $S''=-S$. Oddness follows from the powers, while alternating-series estimates show that $S(x)$ is positive for $0<x\leq1$ and close to $x$ there.

A common pitfall is to use an alternating-series estimate for every real input without checking that the absolute terms decrease. For $0<x\leq1$, the ratio calculation verifies this condition at every index. For larger inputs, the terms need not decrease from the beginning, even though the series still converges absolutely. The geometric tail argument, which only needs the ratio to be sufficiently small eventually, is the appropriate convergence tool in that case.

Another important distinction is between convergence and an estimate for a particular input. Infinite radius of convergence guarantees that the series converges, but a numerical error bound still requires a calculation, such as a geometric tail estimate or a valid alternating-series remainder bound. Keeping those arguments separate makes it clear which conclusions hold for all real inputs and which use an additional restriction.

Takeaway: The sine series converges absolutely for every real input, is odd, and satisfies $S'=C$ and $S''=-S$. Near zero, alternating-series bounds give $x-x^3/6\leq S(x)\leq x$ for $0<x\leq1$.

Check Your Understanding

Use the convergence proof, derivative relations, and alternating-series estimates to answer the following questions.

What is the ratio of consecutive absolute terms in the sine series at a fixed nonzero input $x$?
Why does convergence for every real input imply that the sine series has infinite radius of convergence?
How does reindexing the derivative series for $C$ produce $-S$?
Which condition on the absolute terms allows the alternating-series bounds to be used for $0<x\leq1$?
What does $S'(0)=1$ imply about the limit of $S(x)/x$ as $x$ tends to zero?

What You'll Learn