The Exponential Series

From a Power Series to the Exponential Function

The previous tutorial proved when integration may be performed term by term. The same power-series framework also lets us define an important function directly from its coefficients. In this tutorial, the coefficients are the reciprocals of the factorials. Their rapid decrease makes the series converge at every real input, and differentiating the series will reproduce the original series.

For a nonnegative integer $n$, write $n!=1\cdot2\cdots n$, with $0!=1$. We study the series

\sum_{n=0}^{\infty}\frac{x^n}{n!} =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.

Its sum will be denoted by $E(x)$. Once its basic properties are established, we will use the standard notation $\exp(x)$ for the same function and set $e=\exp(1)$. The series definition comes first: convergence and the properties we need will follow from arguments about the series itself.

Definition: For each real number $x$, define $$ E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}, $$ provided the series converges. We will prove that it converges absolutely for every real $x$, so this definition applies on all of $\mathbb{R}$. We write $\exp(x)=E(x)$ and $e=E(1)$.

Convergence at Every Real Input

Fix $x\ne0$ and consider the absolute values $u_n=|x|^n/n!$. Their successive ratio is

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

As $n$ increases, this ratio tends to zero. In particular, there is 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 nonnegative integer $j$. The tail is therefore bounded by a convergent geometric series. If $x=0$, all terms after the constant term are zero. Thus the series converges absolutely for every real $x$, and its radius of convergence is infinite.

By the Uniform Convergence Inside the Radius Theorem, the series converges uniformly on every bounded closed interval. This permits us to apply the Term-by-Term Differentiation of a Power Series Theorem throughout the real line. The key consequence is that the function defined by the series satisfies a simple differential identity.

Theorem (The Exponential Series Is Its Own Derivative): The function $E$ is differentiable on $\mathbb{R}$, and $$ E'(x)=E(x)\qquad\text{for every }x\in\mathbb{R}. $$

Proof. The coefficient of $x^n$ in the defining series is $c_n=1/n!$. Its radius of convergence is infinite, as just proved. The Term-by-Term Differentiation of a Power Series Theorem therefore gives

E'(x) =\sum_{n=1}^{\infty}\frac{n}{n!}x^{n-1} =\sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!} =\sum_{k=0}^{\infty}\frac{x^k}{k!} =E(x).

For the third expression, the index was changed by setting $k=n-1$. This identity holds for every real $x$, since both power series have infinite radius of convergence. $\square$

The reindexing is the essential calculation: differentiating $x^n/n!$ gives $x^{n-1}/(n-1)!$, exactly the term with index $n-1$ in the original series. This self-reproduction under differentiation is one reason the factorial coefficients are especially useful.

The Multiplication Law

The derivative identity determines a strong relation between values of $E$ at different inputs. The proof uses the product rule and the Mean Value Theorem: a differentiable function whose derivative is zero on an interval must be constant there. First apply this idea to $E(x)E(-x)$. By the chain rule and the derivative identity,

\frac{d}{dx}\bigl(E(x)E(-x)\bigr) =E(x)E(-x)-E(x)E(-x) =0.

This product is constant on $\mathbb{R}$, and its value at $x=0$ is $E(0)E(0)=1$. Thus $E(x)E(-x)=1$. For a fixed real $y$, the same reasoning applied to $E(x+y)E(-x)$ gives the full multiplication law.

Theorem (Multiplication Law for the Exponential Series): For all real $x$ and $y$, $$ E(x+y)=E(x)E(y). $$ Moreover, $E(x)>0$ for every real $x$.

Proof. The function $G(x)=E(x)E(-x)$ is differentiable, and the product and chain rules give

$$ G'(x)=E(x)E(-x)-E(x)E(-x)=0. $$

For any two real inputs, the Mean Value Theorem applied on the interval between them shows that $G$ has the same value at both inputs. Hence $G$ is constant. Since $E(0)=1$, we have

$$ E(x)E(-x)=G(x)=G(0)=E(0)E(0)=1. $$

Now fix $y$ and define $H(x)=E(x+y)E(-x)$. Differentiating gives

$$ H'(x)=E(x+y)E(-x)-E(x+y)E(-x)=0. $$

Again the Mean Value Theorem implies that $H$ is constant. Evaluating at $x=0$ yields $H(x)=H(0)=E(y)$, so $E(x+y)E(-x)=E(y)$. Multiplying both sides by $E(x)$, and using $E(x)E(-x)=1$, proves $E(x+y)=E(x)E(y)$.

The identity $E(x)E(-x)=1$ also shows that $E(x)\ne0$ for every $x$. Applying the multiplication law with both inputs equal to $x/2$ gives

E(x)=E\left(\frac{x}{2}+\frac{x}{2}\right) =E\left(\frac{x}{2}\right)^2>0.

The final inequality holds because $E(x/2)$ is real and nonzero, so its square is strictly positive. This proves both claims. $\square$

In particular, $E(-x)=1/E(x)$: the series at the negative input gives the reciprocal of its value at the positive input. Positivity also combines with $E'=E$ to show that $E$ is strictly increasing. Indeed, the Mean Value Theorem applied to $E$ on any interval $[u,v]$ with $u<v$ gives $E(v)-E(u)=E'(c)(v-u)>0$ for some $c\in(u,v)$.

Worked Examples

Worked Example: Estimating the Value at One

The value at one is $$ e=E(1)=\sum_{n=0}^{\infty}\frac{1}{n!}. $$ The partial sum through $n=4$ is

1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24} =\frac{24+24+12+4+1}{24} =\frac{65}{24}.

For $u_n=1/n!$, the ratio $u_{n+1}/u_n=1/(n+1)$ is at most $1/6$ for $n\geq5$. Consequently, the omitted tail beginning with $u_5=1/120$ is bounded by a geometric series:

\sum_{n=5}^{\infty}\frac{1}{n!} \leq \frac{1}{120}\sum_{j=0}^{\infty}\left(\frac{1}{6}\right)^j =\frac{1}{120}\cdot\frac{1}{1-1/6} =\frac{1}{100}.

The tail is positive, so these calculations give the useful enclosure $$ \frac{65}{24}<e\leq\frac{65}{24}+\frac{1}{100}. $$ The estimate comes directly from the defining series and a geometric bound; it does not require knowing a closed form for the sum.

Worked Example: Evaluating at a Negative Input

Use the multiplication law with $x=1$ and $y=-1$. Since $E(0)=1$, $$ 1=E(0)=E(1+(-1))=E(1)E(-1)=eE(-1). $$ We have already proved $e=E(1)>0$, so division by $e$ is valid and $$ E(-1)=\frac{1}{e}. $$ More generally, the same calculation with $x=t$ and $y=-t$ gives $E(-t)=1/E(t)$ for every real $t$. The nonzero condition needed for this reciprocal is guaranteed by the multiplication law, rather than assumed.

Worked Example: Translating an Input by Two

Apply the multiplication law to $x$ and $2$: $$ E(x+2)=E(x)E(2). $$ Applying it again to $1+1$ gives $$ E(2)=E(1+1)=E(1)E(1)=e^2. $$ Therefore $E(x+2)=e^2E(x)$ for every real $x$. For example, at $x=-2$ this reads $E(0)=e^2E(-2)$, so $E(-2)=1/e^2$, in agreement with the reciprocal identity. The translation formula obtains values at shifted inputs from a single value of the function.

Why These Properties Matter

The series definition supplies more than a way to compute approximations. Absolute convergence everywhere allows the same power series to define a function on the whole real line; term-by-term differentiation identifies its derivative; and the derivative identity yields the multiplication law. From that law follow reciprocal values, positivity, and strict increase. These properties are often used together, so it is useful to keep track of which conclusion rests on which earlier step.

A common pitfall is to treat the multiplication law as an immediate consequence of multiplying two infinite series and collecting terms. Such a calculation requires justification for the rearrangement of infinitely many terms. Here it was proved instead from the derivative identity using only the product rule and the Mean Value Theorem. This route also made the nonvanishing needed for reciprocal values explicit.

Takeaway: The series $E(x)=\sum_{n=0}^{\infty}x^n/n!$ converges for every real $x$, satisfies $E'=E$, and obeys $E(x+y)=E(x)E(y)$. It is strictly positive and increasing, with $E(-x)=1/E(x)$.

Check Your Understanding

Use the convergence argument, derivative identity, and multiplication law to answer the following questions.

For fixed nonzero $x$, what is the ratio of successive absolute terms in the exponential series?
Why does the differentiated series have the same terms as the original series after an index change?
How does the Mean Value Theorem show that $E(x)E(-x)$ is constant?
Use the multiplication law to express $E(-3)$ in terms of $E(3)$.
Why does $E(x)=E(x/2)^2$ imply strict positivity rather than merely nonnegativity?

What You'll Learn