🌊 3. Fourier Series for Functions with Period 2π

Note: References (2.2.x) are explained in 📐 2. Fourier Series General Concepts.

⚙️ 3.1 Formal treatment — computing the coefficients

Suppose the function \( f(x) \), with period \( 2\pi \), is represented by the series

\[f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty }(a_{n}\cos(nx)+b_{n}\sin(nx))\quad(3.1.1 )\]

Then the coefficients can formally be determined as follows.

• 📏 \( a_{0} \) Finding the constant term \( a_{0} \)

If we integrate both sides of (3.1.1) over one period \( 2\pi \), and assume that the integral \( \int_{}^{} \) and the sum \( \sum_{}^{} \) may be interchanged, we obtain

\[ \int_{-\pi}^{\pi}f(x)dx=\frac{a_{0}}{2}
\underbrace{
\int_{-\pi}^{\pi}dx}_{=2\pi\text{ (Hint 1)}}
+\sum_{n=1}^{\infty }\left[ a_{n}
\underbrace{
\int_{-\pi}^{\pi}\cos(nx)dx}_\text{= 0     (See Hint 2)}
+b_{n}
\underbrace{
\int_{-\pi}^{\pi}\sin(nx)dx}_{\substack{\text{= 0   odd integrand}\\\text{over symmetric limits}}}
\right] \]

Hint 1: \( \int_{-\pi}^{\pi}1dx= (\pi - (-\pi))=2\pi \)

Hint 2:

• Positive and negative parts cancel over a full period. \( \int_{-\pi}^{\pi}\cos(kx)\,dx = 0 \quad (k \ne 0) \)

As a result: \[ a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx\quad(3.1.2 ) \]

• ✖️ ∫ cos Finding the Cosine Coefficients

If we multiply both sides of (3.1.1) by \( \cos(mx) \) and then integrate both sides over one period \( 2\pi \), assuming again that \( \int_{}^{} \) and \(\sum_{}^{} \) may be interchanged, we obtain

\[ \int_{-\pi}^{\pi}f(x)cos(mx)dx =\frac{a_{0}}{2}
\underbrace{
\int_{-\pi}^{\pi}\cos(mx)dx}_{\text{=0  (See Hint 2)}}
+\sum_{n=1}^{\infty }\left[ a_{n}
\underbrace{
\int_{-\pi}^{\pi}\cos(nx)\cos(mx)dx}_{\substack{
(2.2.3) \quad
\begin{cases}
0 & n \neq m \\
\pi & n=m>0 \\
2\pi & n=m=0
\end{cases}}
}
+b_{n}
\underbrace{
\int_{-\pi}^{\pi} \sin(nx)\cos(mx)dx }_\text{= 0     (2.2.1)}
\right] \]

On the right-hand side only the second term remains. For this term the condition \( (n=m=0) \) is not possible because the summation starts at \( (n=1) \). During the summation all terms with \( (n \neq m) \) are 0 and disappear. The only remaining term is \( (n = m \gt 0) \).

Therefore we obtain

\[ a_{m}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(mx)dx\quad(3.1.3 ) \]

• ✖️ ∫ sin Finding the Sine Coefficients

If we multiply both sides of (3.1.1) by \( \sin(mx) \) and then integrate both sides over one full period \( 2\pi \), we can isolate the coefficient \(b_{m} \).

Assuming that we are allowed to swap the sum and the integral, we obtain:

\[ \int_{-\pi}^{\pi}f(x)sin(mx)dx =\frac{a_{0}}{2}
\underbrace{
\int_{-\pi}^{\pi}\sin(mx)dx}_{\text{=0  (See Hint 3)}}
+\sum_{n=1}^{\infty }\left[ a_{n}
\underbrace{
\int_{-\pi}^{\pi}\cos(nx)\sin(mx)dx}
_\text{= 0     (2.2.1)
}
+b_{n}
\underbrace{
\int_{-\pi}^{\pi} \sin(nx)\sin(mx)dx }
_{\substack{
(2.2.2) \quad
\begin{cases}
0 & n \neq m \\
\pi & n=m>0 \\
0 & n=m=0
\end{cases}}
}
\right] \]

Hint 3:

• \( \sin(mx) \) is an odd function:

\[ \sin(-mx)=-\sin(mx) \]

• For any odd function:

\[ \int_{-a}^{a} f(x)\,dx = 0 \]

• So the positive area cancels the negative area.

On the right-hand side only the third term remains. For this term the condition \( (n = m = 0) \) is not possible because the summation starts at \( (n=1) \). During the summation all terms with \( (n \neq m) \) are 0 and disappear. The only remaining term is \( (n = m \gt 0) \).

Therefore we obtain:

\[ b_{m}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x) \sin(mx) dx\quad(3.1.4) \]

📘 Definition — Fourier Series

This definition bundles the above knowledge.

Let \( f(x) \) be a function with period \( 2\pi \).

• The Fourier series associated with \( f(x) \) is then defined as:

\[ \frac{a_{0}}{2}+ \sum_{n=1}^{\infty}\left( a_{n} \cos(nx) + b_{n} \sin(nx) \right) \quad(3.1.1) \]

• The Cosine Fourier coefficients are computed using the following formula:

\[ a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx \quad(n=0,1,2,\cdots ) \quad(3.1.5) \]

• The Sine Fourier coefficients are computed using the following formula:

\[ b_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx\quad(n=1,2,3,\cdots ) \quad(3.1.6) \]

💡 Remarks (Important!)

1️⃣ Why do we use \( \frac{a_0}{2} \) instead of \( a_{0} \) in the Fourier series equation?

In a Fourier series, we write:

\[f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty }(a_{n}\cos(nx)+b_{n}\sin(nx))\quad(3.1.1 )\]

At first glance, it might seem more natural to just write \( a_{0} \).

So why do we divide it by 2?

Step 1 — What does \( a_{0} \) represent?

The coefficient is defined as:

\[ a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx \]

But the average value of a function over \( (-\pi,\pi) \)is:

\[ \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx \]

So: \( a_{0} \) = 2 × (average value)

👉 This means \( a_{0} \) is twice as large as the constant we actually need.

Step 2 — Fixing the size

To get the correct constant term, we divide by 2:

\[ \frac{a_{0}}{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx \]

✅ Now it equals the true average of the Fourier function.

Step 3 — Why not just change the formula instead?

We could redefine \( a_{0} \), but then:

• the formula for \( a_{n} \) would be different for \( n=0 \) and \( n\ge 0 \)
• we would lose one clean, consistent formula:

\[ a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx\quad\text{(works for all }n\ge 0) \]

Using \( \frac{a_0}{2} \) keeps everything uniform.

Step 4 — Why is \( n=0 \) special?

• \( \cos(0x)=1 \to \) gives a constant term
• \( \sin(0x)=0 \to \)gives nothing

So:

• the constant part comes only from cosine
• we pull it out of the summation by starting at \( n=1 \) and write it separately as \( \frac{a_0}{2} \)

\[f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty }(a_{n}\cos(nx)+b_{n}\sin(nx))\quad(3.1.1 )\]

Additional advantage:

• The formula for the coefficients are kept consistent for all n — including \( n = 0 \).

The cosine coefficients are defined as:

\[ a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx \quad(n=0,1,2,\cdots ) \quad(3.1.5) \]

Now look at what happens when \( n = 0 \):

• \( \cos(0x) = 1 \)
• So:

\[ a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx\quad(3.1.2 ) \]

This gives the average value (up to a factor) of the Fourier function.

Bottom line: Using \( \frac{a_0}{2} \):

• keeps the formulas symmetric
• avoids treating \( a_{0} \) differently
• lets one single formula for \( a_{n} \) work for all \( n\ge 0 \)

One sentence intuition:

• We use \( \frac{a_0}{2} \) so the constant term fits neatly into the same pattern as all the other cosine terms, without special exceptions.

2️⃣ A note on rigor (important but subtle)

The method we used is not fully rigorous yet.

We assumed that:

• A Fourier series representation of \( f(x) \) existed (3.1.1).
• We are allowed to swap summation and integration during the calculation of the coefficients.

These assumptions are not automatically true.

👉 Therefore, in a complete theory, we must specify conditions on \( f(x) \) that guarantee:

• the Fourier series exists
• the series actually represents the function

This is one of the most difficult parts of Fourier theory, so we usually state the result as a theorem called the Dirichlet Conditions (often without proof at first).

🧠 3.2 Theorem: Dirichlet Conditions