Recall that we defined the partial sum to be
$$
\begin{align*}
S_Nf(x)=\sum_{n=-N}^{N}\widehat f(n)e^{inx}.
\end{align*}
$$
and recall that our goal is to understand when will this sum coverage to \(f(x)\). Recall that the Fourier coefficient is
$$
\begin{align*}
\widehat{f}(n) &= \frac{1}{2\pi}\int_{-\pi}^{\pi} f(y)e^{-iny}\ dy
\end{align*}
$$
If we plug in the coefficient back in the partial sum, we get
$$
\begin{align*}
S_Nf(x) = \sum_{n=-N}^{N} \left( \frac{1}{2\pi}\int_0^{2\pi}f(y)e^{-iny}\,dy \right)e^{inx}.
\end{align*}
$$
Note that \(e^{inx}\) does not depend on \(y\). Hence move it inside the integral.
$$
\begin{align*}
S_Nf(x) = \frac1{2\pi}\sum_{n=-N}^{N} \int_0^{2\pi} e^{inx}f(y)e^{-iny}\,dy. \tag{1}
\end{align*}
$$
Then note that in general, for finitely many integrable functions \(g_n(y)\),
$$
\begin{align*}
\sum_{n=-N}^{N}\int_0^{2\pi}g_n(y)\,dy = \int_0^{2\pi}\sum_{n=-N}^{N}g_n(y)\,dy.
\end{align*}
$$
Hence swap the finite sum and integral in (1) to get
$$
\begin{align*}
S_Nf(x) = \frac1{2\pi}\int_0^{2\pi} \sum_{n=-N}^{N} e^{inx}f(y)e^{-iny}\,dy.
\end{align*}
$$
But note that \(f(y)\) doesn’t depend on the sum so we can take it outside
$$
\begin{align*}
S_Nf(x) = \frac1{2\pi}\int_0^{2\pi} f(y)\sum_{n=-N}^{N}e^{inx}e^{-iny}\,dy.
\end{align*}
$$
Now combine the exponents to get
$$
\begin{align*}
S_Nf(x) = \frac1{2\pi}\int_0^{2\pi} f(y)\sum_{n=-N}^{N}e^{in(x-y)}\,dy. \tag{2}
\end{align*}
$$
Now define the Dirichlet Kernel to be
$$
\begin{align*}
\boxed{
D_N(t):=\sum_{n=-N}^{N}e^{int}.}
\end{align*}
$$
In (2), we have \(t = x - y\). So
$$
\begin{align*}
\boxed{
S_Nf(x) = \frac1{2\pi}\int_0^{2\pi}f(y)D_N(x-y)\,dy.
}
\end{align*}
$$
So the Fourier partial sum is a convolution of \(f\) with the Dirichlet kernel stated in the following lemma
Lemma
The Fourier Partial sum can be written as
\[S_Nf(x) = \frac1{2\pi}\int_0^{2\pi}f(y)D_N(x-y)\,dy.\]
where
\[
D_N(t):=\sum_{n=-N}^{N}e^{int} = \frac{\sin\left((N+\tfrac12)t\right)}{\sin(t/2)}
\]
where \(D\) is the Dirichlet Kernel
Proof
We now show the closed form. Expand the Dirichlet Kernel to see that
$$
\begin{align*}
D_N(t) = e^{-iNt}+e^{-i(N-1)t}+\cdots+1+\cdots+e^{i(N-1)t}+e^{iNt}.
\end{align*}
$$
and let \(\omega=e^{it}\). Hence
$$
\begin{align*}
D_N(t)=\sum_{n=-N}^{N}\omega^n =\omega^{-N}+\omega^{-N+1}+\cdots + \omega^N.
\end{align*}
$$
Factoring out \(\omega^{-N}\),
$$
\begin{align*}
D_N(t) = \omega^{-N}\left(1+\omega+\omega^2+\cdots+\omega^{2N}\right).
\end{align*}
$$
Therefore, \(D_N(t)\) is a finite geometric sum. Hence
$$
\begin{align*}
1+\omega+\omega^2+\cdots+\omega^{2N} = \frac{\omega^{2N+1}-1}{\omega-1},
\qquad \omega\ne1.
\end{align*}
$$
and therefore
$$
\begin{align*}
D_N(t)=\omega^{-N}\frac{\omega^{2N+1}-1}{\omega-1}.
\end{align*}
$$
Now multiply \(\omega^{-N}\) into the numerator
$$
\begin{align*}
D_N(t) &= \frac{\omega^{-N}\omega^{2N+1}-\omega^{-N}}{\omega-1} \\
&=\frac{\omega^{N+1}-\omega^{-N}}{\omega-1}
\end{align*}
$$
factor \(\omega^{1/2}\) from both numerator and denominator. So in the numerator we will have
$$
\begin{align*}
\omega^{N+1} = \omega^{1/2}\omega^{N+1-\frac12} =
\omega^{1/2}\omega^{N+\frac12}.
\end{align*}
$$
If we substitute
$$
\begin{align*}
D_N(t)=\frac{\omega^{1/2}\left(\omega^{N+\frac12}-\omega^{-N-\frac12}\right)}{\omega^{1/2}\left(\omega^{1/2}-\omega^{-1/2}\right)}.
\end{align*}
$$
Cancel the common factor:
$$
\begin{align*}
D_N(t)= \frac{\omega^{N+\frac12}-\omega^{-(N+\frac12)}}{\omega^{1/2}-\omega^{-1/2}}.
\end{align*}
$$
Now substitute back \(\omega=e^{it}\):
$$
\begin{align*}
D_N(t)= \frac{e^{i(N+\frac12)t}-e^{-i(N+\frac12)t}}{e^{it/2}-e^{-it/2}}.
\end{align*}
$$
Now recall that
$$
\begin{align*}
e^{in\theta}-e^{-in\theta}=2i\sin(n\theta).
\end{align*}
$$
Set \(\theta = (N+\frac12)t\) for the numerator and \(\theta = t/2\) in the denominator so
$$
\begin{align*}
D_N(t) &= \frac{2i\sin((N+\frac12)t)}{2i\sin(t/2)} \\
&= \frac{\sin((N+\frac12)t)}{\sin(t/2)}.
\end{align*}
$$
Now recall that we excluded \(\omega = 1\). Since \(\omega=e^{it}\),
$$
\begin{align*}
\omega=1 \quad\Longleftrightarrow\quad t=2\pi k,\qquad k\in\mathbb Z.
\end{align*}
$$
So when \(t = 2\pi k\), let’s go back to the original definition:
$$
\begin{align*}
D_N(t)=\sum_{n=-N}^{N}e^{int}.
\end{align*}
$$
At \(t=2\pi k\), every term is
$$
\begin{align*}
e^{in(2\pi k)}=1.
\end{align*}
$$
There are \(2N+1\) terms, so \(D_N(2\pi k)=2N+1\). If we summarize everything, we get
$$
\begin{align*}
D_N(t)=
\begin{cases}
\displaystyle \frac{\sin\left((N+\tfrac12)t\right)}{\sin(t/2)},
& t\notin 2\pi\mathbb Z,\\[8pt]
2N+1,
& t\in 2\pi\mathbb Z.
\end{cases}
\end{align*}
$$
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi