TITLE: Connection between Fourier transform and Taylor series
QUESTION [149 upvotes]: Both Fourier transform and Taylor series are means to represent functions in a different form. 
What is the connection between these two? Is there a way to get from one to the other (and back again)? Is there an overall, connecting (geometric?) intuition?

REPLY [11 votes]: Taylor series at $t=0$ of some function $\textrm{f}(t)$ is defined as 
$$  \textrm{f}\left(t\right) =\sum_{j=0}^{\infty}
  {
   h_j\cdotp\frac{d^{j}}{dt^{j}}\textrm{f}(0)\cdotp t^{j}
  }
$$
where $ h_j=1/{j!}$ and $\frac{d^0}{dt^0}\textrm{f}\left(t\right)=\textrm{f}\left(t\right)$
Fourier series is defined as
$$
\textrm{f}\left(t\right) =    
  \sum_{n=1}^{\infty}
  { \left(
   a_n\cdot\cos \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)+
   b_n\cdot\sin \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)
   \right)
  }
$$ 
 with coefficients:
 $$
 \begin{align}
  a_n&=\frac{2}{T}\cdotp\int_{t_1}^{t_2}{\textrm{f}(t)\cdotp\cos\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)\,dt}\\
  b_n&=\frac{2}{T}\cdotp\int_{t_1}^{t_2}{\textrm{f}(t)\cdotp\sin\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)\,dt}
 \end{align}
$$
For full-wave function $t_1=-{T}/{2}$ and $t_2=+{T}/{2}$ for any positive period $T$.
Let find Taylor series of cosine and sine functions:
$$
\begin{align}
  \cos\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)&=\sum_{k=0}^{\infty}{\frac{(-1)^k}{\left(2\cdotp k \right)!}
   \cdotp \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)^{2\cdotp k}}\\
  \sin\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)&=\sum_{k=0}^{\infty}{\frac{(-1)^k}{\left(2\cdotp k + 1\right)!}
   \cdotp \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)^{\left(2\cdotp k +1\right)}}
  \end{align}
$$
and substitute this expansion to Fourier coefficients:
$$
\begin{align}
  a_n&=\frac{2}{T}\cdotp
  \int_{t_1}^{t_2} 
  {
   \underbrace
   { 
    \left(
    \sum_{j=0}^{\infty}
    {
     h_j\cdotp\frac{d^{j}}{dt^{j}}\textrm{f}(0)\cdotp t^{j}
    }
    \right)\cdotp 
    \left(\sum_{k=0}^{\infty}
    {
     \frac{(-1)^k}{\left(2\cdotp k \right)!}
     \cdotp \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)^{2\cdotp k}
    }\right)
   }_{\mbox{$\textrm{Tc}(t)$}}
   \,dt
  }\\
  b_n&=\frac{2}{T}\cdotp
  \int_{t_1}^{t_2}
  { 
   \underbrace
   {
    \left(
    \sum_{j=0}^{\infty}
    {
     h_j\cdotp\frac{d^{j}}{dt^{j}}\textrm{f}(0)\cdotp t^{j}
    }
    \right)
    \cdotp 
    \left(\sum_{k=0}^{\infty}
    {\frac{(-1)^k}{\left(2\cdotp k + 1\right)!}
     \cdotp \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)^{\left(2\cdotp k +1\right)}
    }
    \right)
   }_{\mbox{$\textrm{Ts}(t)$}}
   \,dt
  }
  \end{align}
$$
Now consider $\textrm{Tc}(t)$:
For some first indicies $j$ and $k$ brackets can be disclosured and terms multiplied in sequence as shown:
$$
 \begin{align*}
    \textrm{Tc}(t)&=&\textrm{f}(0)+\frac{d}{dt}\textrm{f}(0)\cdotp t + \left(-2\cdotp\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{2}\cdotp\frac{d^2}{dt^2}\textrm{f}(0)\right)\cdotp t^{2}+\\&&
    +\left(-2\cdotp\frac{d}{dt}\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{6}\cdotp\frac{d^3}{dt^3}\textrm{f}(0)\right)\cdotp t^{3}+\\&&
    +\left(\frac{2}{3}\cdotp\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{4}-\frac{d^2}{dt^2}\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{24}\cdotp\frac{d^4}{dt^4}\textrm{f}(0)\right)\cdotp t^{4}+\dots
 \end{align*}
$$
Now integrate this function for each term: 
$$
\begin{align*}
  \int_{t_1}^{t_2}{\textrm{Tc}(t)\,dt}&=&\textrm{f}(0)\left(t_2-t_1\right)+\frac{1}{2}\cdotp\frac{d}{dt}\textrm{f}(0)\cdotp\left(t_2^2-t_1^2\right) + \\&& + \frac{1}{3}\cdot\left(-2\cdotp\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{2}\cdotp\frac{d^2}{dt^2}\textrm{f}(0)\right)\cdotp\left(t_2^3-t_1^3\right) +\\&&
  +\frac{1}{4}\cdotp\left(-2\cdotp\frac{d}{dt}\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{6}\cdotp\frac{d^3}{dt^3}\textrm{f}(0)\right)\cdotp\left(t_2^4-t_1^4\right)+\\&&
  +\frac{1}{5}\left(\frac{2}{3}\cdotp\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{4}-\frac{d^2}{dt^2}\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{24}\cdotp\frac{d^4}{dt^4}\textrm{f}(0)\right)\cdotp\left(t_2^5-t_1^5\right)+\dots
  \end{align*}
$$
Collect coefficients  at previous integral above $\frac{d^i}{dt^i}\textrm{f}(0)$:
$$
   \begin{align*}
    \int_{t_1}^{t_2}{\textrm{Tc}(t)\,dt}&=&\textrm{f}(0)\cdotp\left( \left(t_2-t_1\right)-
  \frac{2}{3}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^3-t_1^3\right)+\frac{2}{15}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^5\cdotp\left(t_2^5-t_1^5\right)+\dots\right) +
  \\&& 
  +\frac{d}{dt}\textrm{f}(0)\cdotp\left(\frac{1}{2}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^2-t_1^2\right)-
  \frac{1}{2}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^4-t_1^4\right)+\frac{1}{9}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^4\cdotp\left(t_2^6-t_1^6\right)+\dots\right)+
  \\&&
  +\frac{d^2}{dt^2}\textrm{f}(0)\cdotp\left(\frac{1}{6}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^3-t_1^3\right)-
  \frac{1}{5}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^5-t_1^5\right)+\frac{1}{21}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^4\cdotp\left(t_2^7-t_1^7\right)+\dots\right)+
  \\&&
  +\frac{d^3}{dt^3}\textrm{f}(0)\cdotp\left(\frac{1}{24}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^4-t_1^4\right)-
  \frac{1}{18}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^6-t_1^6\right)+\frac{1}{72}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^4\cdotp\left(t_2^8-t_1^8\right)+\dots\right)+\dots    
   \end{align*}
$$
Now it is easy to recognize sequences at brackets (rhs expression is multiplied by $2/T$).
 For $\textrm{f}(0)$:
$$
\begin{align*}
\left(t_2-t_1\right)-
    \frac{2}{3}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^3-t_1^3\right)+\dots &:&  \frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{  1\cdot\left(2\cdot i+1 \right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot  \left(t_2^ \left(2\cdot i+1 \right) -t_1^ \left(2\cdot i+1 \right)\right) }{T^ \left(2\cdot i+1 \right)}
\end{align*}
$$
For $\frac{d}{dt}\textrm{f}(0)$:
$$
\begin{align*}
\frac{1}{2}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^2-t_1^2\right)-\frac{1}{2}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^4-t_1^4\right)+\dots &:&\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ 1\cdot \left(2\cdot i+2 \right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot  \left(t_2^ \left(2\cdot i+2 \right) -t_1^ \left(2\cdot i+2 \right)\right) }{T^ \left(2\cdot i+1 \right)}
\end{align*}
$$
For $\frac{d^2}{dt^2}\textrm{f}(0)$:
$$
\begin{align*}
\frac{1}{6}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^3-t_1^3\right)-\frac{1}{5}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^5-t_1^5\right)+\dots &:&\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{  1\cdot2\cdot \left(2\cdot i+3 \right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot  \left(t_2^ \left(2\cdot i+3 \right) -t_1^ \left(2\cdot i+3 \right)\right) }{T^ \left(2\cdot i+1 \right)}
\end{align*}
$$
For $\frac{d^3}{dt^3}\textrm{f}(0)$:
$$
\begin{align*}
\frac{1}{24}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^4-t_1^4\right)-
  \frac{1}{18}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^6-t_1^6\right)+\dots&:&\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ 1\cdot 2\cdot 3\cdot\left(2\cdot i+4 \right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot  \left(t_2^ \left(2\cdot i+4 \right) -t_1^ \left(2\cdot i+4 \right)\right) }{T^ \left(2\cdot i+1 \right)}
\end{align*}
$$
and so on. 
Finally overall sequence for $\frac{d^m}{dt^m}$ is computed as:
$$ 
\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ m!\cdot\left(1+m+2\cdot i\right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot  \left(t_2^ \left(2\cdot i+m+1 \right) -t_1^ \left(2\cdot i+m+1 \right)\right) }{T^ \left(2\cdot i+1 \right)}
$$
Now we can find sum using CAS:
$$
\textrm{Ct}(n,m)=\sum_{i=0}^{\infty}{\left(\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ m!\cdot\left(1+m+2\cdot i\right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot  \left(t_2^ \left(2\cdot i+m+1 \right) -t_1^ \left(2\cdot i+m+1 \right)\right) }{T^ \left(2\cdot i+1 \right)}\right)}
$$
and $\textrm{Ct}(n,m)$ becomes quite complex expression containing Lommel's or hypergeometric functions.
In particular, when $m=0$ function becomes
$$
\textrm{Ct}(n,0)=\frac{\sin\left(\frac{2\pi n\cdot t_2}{T})\right)-\sin\left(\frac{2\pi n \cdot t_1}{T}\right)}{\pi n}
$$
for $m=1$:
$$
 \textrm{Ct}(n,1)= \frac
{2\pi n
 \left( 
    \sin\left(\frac{2\pi n\cdot t_2}{T})\right)\cdot t_2-\sin\left(\frac{2\pi    n\cdot t_1}{T}\right)\cdot t_1
  \right)
+T\cdot
\left( 
\cos\left(\frac{2\pi n\cdot t_2}{T})\right)-\cos\left(\frac{2\pi n\cdot t_1}{T}\right)
\right)
}
{ 2\cdot(\pi n)^2 }
$$
and so on.
So we can write expression for $a_n$:
$$
a_n=\sum_{m=0}^{\infty}{\frac{d^m}{dt^m}\textrm{f}(0)\cdot\textrm{Ct}(n,m)}
$$
or
$$
a_n=\sum_{m=0}^{\infty}{\frac{1}{m!}\cdot\frac{d^m}{dt^m}\textrm{f}(0)\cdot\left(\sum_{i=0}^{\infty}{\left(\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ \left(1+m+2\cdot i\right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot  \left(t_2^ \left(2\cdot i+m+1 \right) -t_1^ \left(2\cdot i+m+1 \right)\right) }{T^ \left(2\cdot i+1 \right)}\right)}\right)}
$$
We can easy see that $\frac{1}{m!}$ is Taylor's coefficient, thus relationship between the Fourier coefficients and the coefficient in the Taylor expansion using special functions is established.
In the particular case of full wave function $t_1=-{T}/{2}$ and $t_2=+{T}/{2}$ we can write for $\textrm{Ct}(n,m)$ more simple closed form:
$$
\frac{
\left(-1 \right)^n T^m  
\left(
  \left(\pi n\right)^ \left(-m-\frac{1}{2} \right)\left(\textrm{I}\cdot\textrm{L}_{\textrm{S1}}\left(m+\frac{3}{2}, \frac{1}{2}, -\pi n \right)-\textrm{L}_{\textrm{S1}}\left(m+\frac{3}{2}, \frac{1}{2}, \pi n \right)\right)+1+\left(-1 \right)^m \right)
}
{2^ {m}\cdot\left(m+1 \right)!}
$$
where $\textrm{L}_{\textrm{S1}}(\mu,\nu,z)=\textrm{s}_{\mu,\nu}(z)$ is first Lommel function and $\textrm{I}=(-1)^{2^{-1}}$ (Complex $a_n$ coefficient). 
For example let consider parabolic signal with period $T$: $\textrm{g}(t)=A\cdot t^2+B\cdot t + C$.
Coefficients $a_n$ can be found using Fourier formula:
$$
a_n=\frac{2}{T}\cdotp\int_{-T/2}^{T/2}{\left(A\cdot t^2+B\cdot t + C\right)\cdotp\cos\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)\,dt} = A\cdot\left(\frac{T}{\pi n}\right)^2\cdot(-1)^n
$$
For $\textrm{g}(t)$ function: $\textrm{g}'(t)=2A\cdot t + B$ , $\textrm{g}''(t)=2A$, derivatives greater than two is zero. So we can use $\textrm{Ct}(n,m)$ for $m=0,1,2$. It is easy to check if $t_1=-{T}/{2}$ and $t_2=+{T}/{2}$ then $\textrm{Ct}(n,0) = 0,\textrm{Ct}(n,1)=0$ we have zero values for odd and non zero for even values, in particular, whenfor $m=2$:
$$
\textrm{Ct}(n,2) = \frac{1}{2}\left(\frac{T}{\pi n}\right)^2\cdot(-1)^n
$$ 
for $m=4$ as example $\textrm{Ct}(n,4) = \frac{1}{48}\left(\frac{T}{\pi n}\right)^4\cdot\left((\pi n)^2-6\right)\cdot(-1)^n
$.
Finally we can obtain for example for $m$ up to 4:
$$
\begin{align*}
a_n=\sum_{m=0}^{4}{\frac{d^m}{dt^m}\textrm{g}(0)\cdot\textrm{Ct}(n,m)}=&\\
=\left(A\cdot 0^2+B\cdot 0 + C\right)\cdot 0+\left(2A\cdot 0+B\right)\cdot 0+\frac{1}{2}\left(\frac{T}{\pi n}\right)^2\cdot(-1)^n \cdot 2A+0\cdot0+...\cdot 0=\\
=A\cdot\left(\frac{T}{\pi n}\right)^2\cdot(-1)^n 
\end{align*}
$$
the same result using Fourier integral. There is interesting  result for non-integer harmonic for $\textrm{g}(t)$ is:
$$
a_n=\frac{1}{120}A\cdot T^2 \cdot (-1)^{n+1} \left({}_{2}\textrm{
F}_1\left( 1, 3; \frac{7}{2}; -\left(\frac{\pi n}{2}\right)^2\right)\cdot(\pi n)^2-20\right) 
$$
where ${}_{2}\textrm{
F}_1\left(a,b;c;z\right)$ is hypergeometric function.
So we can plot coefficients calculated from Fourier integral and for this special result (for $T=1,\,A=1$):
Cosine series. Red circles and red line is Fourier cosine coefficients (for real part of non-integer power of -1), solid blue line is real part of obtained expression with hypergeometric function and dash green line is imaginary. For integer $n$ imaginary part is zero and $a_n$ is real. 
Similar expressions can be obtained for the sine $b_n$ series.