Representation of Functions by Series
Imagine students is trying to describe a complicated curve using only simpler pieces 📈. In Calculus 2, one of the most powerful ideas is that some functions can be written as infinite sums called power series. This lesson explains how functions can be represented by series, why that matters, and how to recognize when a series really matches a function.
What it means to represent a function by a series
A power series is an infinite sum of the form
$$\sum_{n=0}^{\infty} a_n (x-c)^n$$
where $a_n$ is the coefficient sequence, $c$ is the center, and $x$ is the variable. When a function $f(x)$ can be written in this form for values of $x$ near $c$, we say the function is represented by a series.
The big idea is that a complicated function can sometimes be rebuilt from an infinite polynomial-like expression. This is useful because polynomials are easier to manipulate, differentiate, integrate, and approximate. For example, the function $e^x$ can be represented by the series
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
for every real number $x$. That means the exponential function is not just close to this series — it is exactly equal to it wherever the series converges.
This topic is part of the broader study of power series because it connects three major ideas: finding a series, checking where it converges, and using the series as a function rule. ✅
Why function representations by series are useful
Representing a function by a series gives a practical way to work with functions that are hard to handle directly. In science and engineering, this shows up when a calculator or computer uses an approximation of a function instead of the exact formula. A series can also reveal hidden patterns in a function.
For example, the function
$$\frac{1}{1-x}$$
has the geometric series representation
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$
for $|x|<1$. This is a huge deal because the left side looks like a single fraction, but the right side shows it as an infinite sum. If students needs an approximation near $x=0$, the series gives a quick answer:
$$\frac{1}{1-0.2} = 1 + 0.2 + 0.2^2 + 0.2^3 + \cdots$$
The first few terms already give a good estimate.
Another reason series representations matter is that they let us define or study functions by local behavior near a center point $c$. Many functions can be matched to a power series that looks like a “zoomed-in” polynomial model around that point.
How to build a series representation from a known series
A common strategy is to start with a familiar series and transform it. The most important base series in Calculus 2 is the geometric series:
$$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r} \quad \text{for } |r|<1$$
By substituting a new expression for $r$, we can represent many functions.
For example, if we replace $r$ with $x^2$, then
$$\frac{1}{1-x^2} = \sum_{n=0}^{\infty} x^{2n} \quad \text{for } |x|<1$$
If we replace $r$ with $-x$, then
$$\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n \quad \text{for } |x|<1$$
These examples show an important skill: changing variables inside a known series to create a new representation. This is a standard Calculus 2 procedure and a very common test question.
You can also shift the center. For instance,
$$\frac{1}{1-(x-2)} = \sum_{n=0}^{\infty} (x-2)^n$$
for $|x-2|<1$. Here the series is centered at $c=2$ instead of $c=0$.
Differentiation and integration help create new representations
Once a power series is known, calculus operations can produce more series. If
$$f(x) = \sum_{n=0}^{\infty} a_n (x-c)^n$$
inside its interval of convergence, then it can be differentiated term by term:
$$f'(x) = \sum_{n=1}^{\infty} n a_n (x-c)^{n-1}$$
It can also be integrated term by term:
$$\int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-c)^{n+1}$$
These rules are extremely useful for representing new functions.
For example, start with
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \quad \text{for } |x|<1$$
Differentiate both sides:
$$\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}$$
This gives a series representation for $\frac{1}{(1-x)^2}$.
Now integrate the geometric series:
$$\int \frac{1}{1-x}\,dx = \int \sum_{n=0}^{\infty} x^n\,dx$$
which becomes
$$-\ln|1-x| = C + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$$
Choosing $C=0$ so the left and right sides match at $x=0$, we get
$$\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n} \quad \text{for } |x|<1$$
This is a major representation in Calculus 2 and is used often in approximation and theoretical problems. ✨
Recognizing common function series
Some functions appear often enough that it helps to know their standard series representations.
The exponential function:
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
The sine function:
$$\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
The cosine function:
$$\cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$$
The inverse tangent function:
$$\arctan x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \quad \text{for } |x|\le 1$$
These formulas are not just memorization facts. They are examples of how a function can be described by a series, and each one has a specific interval of convergence. For example, the series for $\arctan x$ converges at $x=\pm 1$, but the series for $e^x$ converges for all real $x$.
If students sees a function like $\frac{1}{1+x^2}$, a useful move is to rewrite it as
$$\frac{1}{1-(-x^2)}$$
then use the geometric series:
$$\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n} \quad \text{for } |x|<1$$
This is a clear example of representing a function by a series without needing a new theorem.
Interval of convergence matters
A function representation by series is only valid where the series converges. The radius of convergence $R$ and interval of convergence determine where the representation is actually true.
For a power series
$$\sum_{n=0}^{\infty} a_n (x-c)^n$$
there is always a number $R \ge 0$ such that the series converges when
$$|x-c|<R$$
and diverges when
$$|x-c|>R$$
The endpoints $x=c\pm R$ must be checked separately.
This matters because a series representation is not automatically valid everywhere. For example,
$$\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$
is valid on $(-1,1]$, not on all real numbers. The endpoint $x=1$ works, but $x=-1$ does not.
So whenever students represents a function by a series, the final answer should include the interval where the equality is true. That interval is part of the function representation itself.
A step-by-step example
Suppose we want a series for
$$f(x) = \frac{x}{1+x^2}$$
First, rewrite the denominator using the geometric form:
$$\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n} \quad \text{for } |x|<1$$
Now multiply both sides by $x$:
$$\frac{x}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n+1} \quad \text{for } |x|<1$$
This is the series representation of the function. The function is now expressed as an infinite polynomial in odd powers of $x$.
This kind of problem tests several Calculus 2 skills at once: recognizing a known series, rewriting the function, and keeping track of the interval of convergence. It also shows how series can be used to model functions in a simpler form.
Conclusion
Representing a function by a series means writing the function as an infinite sum that matches it on some interval. This idea is central to Power Series because it combines algebra, limits, and calculus into one tool. students should remember that a valid function representation requires both the correct series and the correct interval of convergence. When a function can be represented by a series, it becomes easier to differentiate, integrate, and approximate. That is why series representations are one of the most important ideas in Calculus 2. 🚀
Study Notes
- A power series has the form $\sum_{n=0}^{\infty} a_n (x-c)^n$.
- A function is represented by a series when the function equals the series on the interval where the series converges.
- The geometric series is the main starter series: $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ for $|r|<1$.
- Common transformations include substitution, shifting the center, differentiating, and integrating.
- Term-by-term differentiation gives $\sum_{n=1}^{\infty} n a_n (x-c)^{n-1}$ inside the interval of convergence.
- Term-by-term integration gives $C + \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-c)^{n+1}$ inside the interval of convergence.
- Important standard series include $e^x$, $\sin x$, $\cos x$, $\arctan x$, and $\ln(1+x)$.
- The radius of convergence $R$ tells where the series works inside $|x-c|<R$.
- Endpoints must be checked separately because convergence there is not automatic.
- A correct series representation must always include the domain where it is valid.