Csdrhrt

A holomorphic function in an annulus containing the unit circle has a Laurent series about zero which generalizes the Taylor series of a holomorphic function in a neighborhood of zero. When restricted to the unit circle, this Laurent series gives a Fourier series of the corresponding periodic function. (This explains the connection between the Cauchy integral formula and the integral defining the coefficients of a Fourier series.)

But it's worth mentioning that the Fourier transform is much more general than this and applies in a broad range of contexts. I don't know that there's a short, simple answer to this question.

Edit: I guess it's also worth talking about intuition. One intuition for the Taylor series of a function $f(x)$ at at a point is that its coefficients describe the displacement, velocity, acceleration, jerk, and so forth of a particle which is at location $f(t)$ at time $t$. And one intuition for the Fourier series of a periodic function $f(x)$ is that it describes the decomposition of $f(x)$ into pure tones of various frequencies. In other words, a periodic function is like a chord, and its Fourier series describes the notes in that chord.

(The connection between the two provided by the Cauchy integral formula is therefore quite remarkable; one takes an integral of $f$ over the unit circle and it tells you information about the behavior of $f$ at the origin. But this is more a magic property of holomorphic functions than anything else. One intuition to have here is that a holomorphic function describes, for example, the flow of some ideal fluid, and integrating over the circle gives you information about "sources" and "sinks" of that flow within the circle.)

There is a big difference between the Taylor series and Fourier transform. The Taylor series is a local approximation, while the Fourier transform uses information over a range of the variable.

The theorem Qiaochu mentions is very important in complex analysis and is one indication of how restrictive having a derivative in the complex plane is on functions.

There is an analogy, more direct for fourier series. Both Fourier series and Taylor series are decompositions of a function $f(x)$, which is represented as a linear combination of a (countable) set of functions. The function is then fully specified by a sequence of coefficients, instead of by its values $f(x)$ for each $x$. In this sense, both can be called a transform
$f(x) leftrightarrow { a_0, a_1, ...}$

For the Taylor series (around 0, for simplicity), the set of functions is ${1, x , x^2, x^3...}$. For the Fourier series is ${1, sin(omega x), cos(omega x), sin(2 omega x), cos(2 omega x) ...}$.

Actually the Fourier series is one the many transformations that uses an orthonomal basis of functions. It is shown that, in that case, the coefficients are obtained by "projecting" $f(x)$ onto each basis function, which amounts to an inner product, which (in the real scalar case) amounts to an integral. This implies that the coefficients depends on a global property of the function (over the full "period" of the function).

The Taylor series (which does not use a orthonormal basis) is conceptually very different, in that the coeffients depends only in local properties of the function, i.e., its behaviour in a neighbourhood (its derivatives).

Taylor series at $t=0$ of some function $textrm{f}(t)$ is defined as

$$ textrm{f}left(tright) =sum_{j=0}^{infty}
{
h_jcdotpfrac{d^{j}}{dt^{j}}textrm{f}(0)cdotp t^{j}
}
$$
where $ h_j=1/{j!}$ and $frac{d^0}{dt^0}textrm{f}left(tright)=textrm{f}left(tright)$
Fourier series is defined as
$$
textrm{f}left(tright) =
sum_{n=1}^{infty}
{ left(
a_ncdotcos left({frac{2picdotp n cdotp t}{T}}right)+
b_ncdotsin left({frac{2picdotp n cdotp t}{T}}right)
right)
}
$$
with coefficients:
$$
begin{align}
a_n&=frac{2}{T}cdotpint_{t_1}^{t_2}{textrm{f}(t)cdotpcosleft({frac{2picdotp n cdotp t}{T}}right),dt}\
b_n&=frac{2}{T}cdotpint_{t_1}^{t_2}{textrm{f}(t)cdotpsinleft({frac{2picdotp n cdotp t}{T}}right),dt}
end{align}
$$
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:
$$
begin{align}
cosleft({frac{2picdotp n cdotp t}{T}}right)&=sum_{k=0}^{infty}{frac{(-1)^k}{left(2cdotp k right)!}
cdotp left({frac{2picdotp n cdotp t}{T}}right)^{2cdotp k}}\
sinleft({frac{2picdotp n cdotp t}{T}}right)&=sum_{k=0}^{infty}{frac{(-1)^k}{left(2cdotp k + 1right)!}
cdotp left({frac{2picdotp n cdotp t}{T}}right)^{left(2cdotp k +1right)}}
end{align}
$$
and substitute this expansion to Fourier coefficients:
$$
begin{align}
a_n&=frac{2}{T}cdotp
int_{t_1}^{t_2}
{
underbrace
{
left(
sum_{j=0}^{infty}
{
h_jcdotpfrac{d^{j}}{dt^{j}}textrm{f}(0)cdotp t^{j}
}
right)cdotp
left(sum_{k=0}^{infty}
{
frac{(-1)^k}{left(2cdotp k right)!}
cdotp left({frac{2picdotp n cdotp t}{T}}right)^{2cdotp k}
}right)
}_{mbox{$textrm{Tc}(t)$}}
,dt
}\
b_n&=frac{2}{T}cdotp
int_{t_1}^{t_2}
{
underbrace
{
left(
sum_{j=0}^{infty}
{
h_jcdotpfrac{d^{j}}{dt^{j}}textrm{f}(0)cdotp t^{j}
}
right)
cdotp
left(sum_{k=0}^{infty}
{frac{(-1)^k}{left(2cdotp k + 1right)!}
cdotp left({frac{2picdotp n cdotp t}{T}}right)^{left(2cdotp k +1right)}
}
right)
}_{mbox{$textrm{Ts}(t)$}}
,dt
}
end{align}
$$
Now consider $textrm{Tc}(t)$:
For some first indicies $j$ and $k$ brackets can be disclosured and terms multiplied in sequence as shown:
$$
begin{align*}
textrm{Tc}(t)&=&textrm{f}(0)+frac{d}{dt}textrm{f}(0)cdotp t + left(-2cdotptextrm{f}(0)cdotpleft(frac{picdotp n}{T}right)^{2}+frac{1}{2}cdotpfrac{d^2}{dt^2}textrm{f}(0)right)cdotp t^{2}+\&&
+left(-2cdotpfrac{d}{dt}textrm{f}(0)cdotpleft(frac{picdotp n}{T}right)^{2}+frac{1}{6}cdotpfrac{d^3}{dt^3}textrm{f}(0)right)cdotp t^{3}+\&&
+left(frac{2}{3}cdotptextrm{f}(0)cdotpleft(frac{picdotp n}{T}right)^{4}-frac{d^2}{dt^2}textrm{f}(0)cdotpleft(frac{picdotp n}{T}right)^{2}+frac{1}{24}cdotpfrac{d^4}{dt^4}textrm{f}(0)right)cdotp t^{4}+dots
end{align*}
$$
Now integrate this function for each term:
$$
begin{align*}
int_{t_1}^{t_2}{textrm{Tc}(t),dt}&=&textrm{f}(0)left(t_2-t_1right)+frac{1}{2}cdotpfrac{d}{dt}textrm{f}(0)cdotpleft(t_2^2-t_1^2right) + \&& + frac{1}{3}cdotleft(-2cdotptextrm{f}(0)cdotpleft(frac{picdotp n}{T}right)^{2}+frac{1}{2}cdotpfrac{d^2}{dt^2}textrm{f}(0)right)cdotpleft(t_2^3-t_1^3right) +\&&
+frac{1}{4}cdotpleft(-2cdotpfrac{d}{dt}textrm{f}(0)cdotpleft(frac{picdotp n}{T}right)^{2}+frac{1}{6}cdotpfrac{d^3}{dt^3}textrm{f}(0)right)cdotpleft(t_2^4-t_1^4right)+\&&
+frac{1}{5}left(frac{2}{3}cdotptextrm{f}(0)cdotpleft(frac{picdotp n}{T}right)^{4}-frac{d^2}{dt^2}textrm{f}(0)cdotpleft(frac{picdotp n}{T}right)^{2}+frac{1}{24}cdotpfrac{d^4}{dt^4}textrm{f}(0)right)cdotpleft(t_2^5-t_1^5right)+dots
end{align*}
$$
Collect coefficients at previous integral above $frac{d^i}{dt^i}textrm{f}(0)$:
$$
begin{align*}
int_{t_1}^{t_2}{textrm{Tc}(t),dt}&=&textrm{f}(0)cdotpleft( left(t_2-t_1right)-
frac{2}{3}cdotpleft(frac{picdotp n}{T}right)^2cdotpleft(t_2^3-t_1^3right)+frac{2}{15}cdotpleft(frac{picdotp n}{T}right)^5cdotpleft(t_2^5-t_1^5right)+dotsright) +
\&&
+frac{d}{dt}textrm{f}(0)cdotpleft(frac{1}{2}cdotpfrac{picdotp n}{T}cdotp left(t_2^2-t_1^2right)-
frac{1}{2}cdotpleft(frac{picdotp n}{T}right)^2cdotpleft(t_2^4-t_1^4right)+frac{1}{9}cdotpleft(frac{picdotp n}{T}right)^4cdotpleft(t_2^6-t_1^6right)+dotsright)+
\&&
+frac{d^2}{dt^2}textrm{f}(0)cdotpleft(frac{1}{6}cdotpfrac{picdotp n}{T}cdotp left(t_2^3-t_1^3right)-
frac{1}{5}cdotpleft(frac{picdotp n}{T}right)^2cdotpleft(t_2^5-t_1^5right)+frac{1}{21}cdotpleft(frac{picdotp n}{T}right)^4cdotpleft(t_2^7-t_1^7right)+dotsright)+
\&&
+frac{d^3}{dt^3}textrm{f}(0)cdotpleft(frac{1}{24}cdotpfrac{picdotp n}{T}cdotp left(t_2^4-t_1^4right)-
frac{1}{18}cdotpleft(frac{picdotp n}{T}right)^2cdotpleft(t_2^6-t_1^6right)+frac{1}{72}cdotpleft(frac{picdotp n}{T}right)^4cdotpleft(t_2^8-t_1^8right)+dotsright)+dots
end{align*}
$$

Now it is easy to recognize sequences at brackets (rhs expression is multiplied by $2/T$).
For $textrm{f}(0)$:
$$
begin{align*}
left(t_2-t_1right)-
frac{2}{3}cdotpleft(frac{picdotp n}{T}right)^2cdotpleft(t_2^3-t_1^3right)+dots &:& frac{left(-1 right)^icdot 2^ left(2cdot i+1 right)cdot n^ left(2cdot i right)}{ 1cdotleft(2cdot i+1 right)cdot left(2cdot i right)! }cdot frac {pi^ left(2cdot i right)cdot left(t_2^ left(2cdot i+1 right) -t_1^ left(2cdot i+1 right)right) }{T^ left(2cdot i+1 right)}
end{align*}
$$
For $frac{d}{dt}textrm{f}(0)$:
$$
begin{align*}
frac{1}{2}cdotpfrac{picdotp n}{T}cdotp left(t_2^2-t_1^2right)-frac{1}{2}cdotpleft(frac{picdotp n}{T}right)^2cdotpleft(t_2^4-t_1^4right)+dots &:&frac{left(-1 right)^icdot 2^ left(2cdot i+1 right)cdot n^ left(2cdot i right)}{ 1cdot left(2cdot i+2 right)cdot left(2cdot i right)! }cdot frac {pi^ left(2cdot i right)cdot left(t_2^ left(2cdot i+2 right) -t_1^ left(2cdot i+2 right)right) }{T^ left(2cdot i+1 right)}
end{align*}
$$
For $frac{d^2}{dt^2}textrm{f}(0)$:
$$
begin{align*}
frac{1}{6}cdotpfrac{picdotp n}{T}cdotp left(t_2^3-t_1^3right)-frac{1}{5}cdotpleft(frac{picdotp n}{T}right)^2cdotpleft(t_2^5-t_1^5right)+dots &:&frac{left(-1 right)^icdot 2^ left(2cdot i+1 right)cdot n^ left(2cdot i right)}{ 1cdot2cdot left(2cdot i+3 right)cdot left(2cdot i right)! }cdot frac {pi^ left(2cdot i right)cdot left(t_2^ left(2cdot i+3 right) -t_1^ left(2cdot i+3 right)right) }{T^ left(2cdot i+1 right)}
end{align*}
$$
For $frac{d^3}{dt^3}textrm{f}(0)$:
$$
begin{align*}
frac{1}{24}cdotpfrac{picdotp n}{T}cdotp left(t_2^4-t_1^4right)-
frac{1}{18}cdotpleft(frac{picdotp n}{T}right)^2cdotpleft(t_2^6-t_1^6right)+dots&:&frac{left(-1 right)^icdot 2^ left(2cdot i+1 right)cdot n^ left(2cdot i right)}{ 1cdot 2cdot 3cdotleft(2cdot i+4 right)cdot left(2cdot i right)! }cdot frac {pi^ left(2cdot i right)cdot left(t_2^ left(2cdot i+4 right) -t_1^ left(2cdot i+4 right)right) }{T^ left(2cdot i+1 right)}
end{align*}
$$
and so on.
Finally overall sequence for $frac{d^m}{dt^m}$ is computed as:
$$
frac{left(-1 right)^icdot 2^ left(2cdot i+1 right)cdot n^ left(2cdot i right)}{ m!cdotleft(1+m+2cdot iright)cdot left(2cdot i right)! }cdot frac {pi^ left(2cdot i right)cdot left(t_2^ left(2cdot i+m+1 right) -t_1^ left(2cdot i+m+1 right)right) }{T^ left(2cdot i+1 right)}
$$
Now we can find sum using CAS:
$$
textrm{Ct}(n,m)=sum_{i=0}^{infty}{left(frac{left(-1 right)^icdot 2^ left(2cdot i+1 right)cdot n^ left(2cdot i right)}{ m!cdotleft(1+m+2cdot iright)cdot left(2cdot i right)! }cdot frac {pi^ left(2cdot i right)cdot left(t_2^ left(2cdot i+m+1 right) -t_1^ left(2cdot i+m+1 right)right) }{T^ left(2cdot i+1 right)}right)}
$$
and $textrm{Ct}(n,m)$ becomes quite complex expression containing Lommel's or hypergeometric functions.
In particular, when $m=0$ function becomes
$$
textrm{Ct}(n,0)=frac{sinleft(frac{2pi ncdot t_2}{T})right)-sinleft(frac{2pi n cdot t_1}{T}right)}{pi n}
$$
for $m=1$:
$$
textrm{Ct}(n,1)= frac
{2pi n
left(
sinleft(frac{2pi ncdot t_2}{T})right)cdot t_2-sinleft(frac{2pi ncdot t_1}{T}right)cdot t_1
right)
+Tcdot
left(
cosleft(frac{2pi ncdot t_2}{T})right)-cosleft(frac{2pi ncdot t_1}{T}right)
right)
}
{ 2cdot(pi n)^2 }
$$

and so on.

So we can write expression for $a_n$:
$$
a_n=sum_{m=0}^{infty}{frac{d^m}{dt^m}textrm{f}(0)cdottextrm{Ct}(n,m)}
$$
or
$$
a_n=sum_{m=0}^{infty}{frac{1}{m!}cdotfrac{d^m}{dt^m}textrm{f}(0)cdotleft(sum_{i=0}^{infty}{left(frac{left(-1 right)^icdot 2^ left(2cdot i+1 right)cdot n^ left(2cdot i right)}{ left(1+m+2cdot iright)cdot left(2cdot i right)! }cdot frac {pi^ left(2cdot i right)cdot left(t_2^ left(2cdot i+m+1 right) -t_1^ left(2cdot i+m+1 right)right) }{T^ left(2cdot i+1 right)}right)}right)}
$$
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:
$$
frac{
left(-1 right)^n T^m
left(
left(pi nright)^ left(-m-frac{1}{2} right)left(textrm{I}cdottextrm{L}_{textrm{S1}}left(m+frac{3}{2}, frac{1}{2}, -pi n right)-textrm{L}_{textrm{S1}}left(m+frac{3}{2}, frac{1}{2}, pi n right)right)+1+left(-1 right)^m right)
}
{2^ {m}cdotleft(m+1 right)!}
$$
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 $T$: $textrm{g}(t)=Acdot t^2+Bcdot t + C$.
Coefficients $a_n$ can be found using Fourier formula:
$$
a_n=frac{2}{T}cdotpint_{-T/2}^{T/2}{left(Acdot t^2+Bcdot t + Cright)cdotpcosleft({frac{2picdotp n cdotp t}{T}}right),dt} = Acdotleft(frac{T}{pi n}right)^2cdot(-1)^n
$$
For $textrm{g}(t)$ function: $textrm{g}'(t)=2Acdot t + B$ , $textrm{g}''(t)=2A$, 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 $m=2$:

$$
textrm{Ct}(n,2) = frac{1}{2}left(frac{T}{pi n}right)^2cdot(-1)^n
$$
for $m=4$ as example $textrm{Ct}(n,4) = frac{1}{48}left(frac{T}{pi n}right)^4cdotleft((pi n)^2-6right)cdot(-1)^n
$.
Finally we can obtain for example for $m$ up to 4:

$$
begin{align*}
a_n=sum_{m=0}^{4}{frac{d^m}{dt^m}textrm{g}(0)cdottextrm{Ct}(n,m)}=&\
=left(Acdot 0^2+Bcdot 0 + Cright)cdot 0+left(2Acdot 0+Bright)cdot 0+frac{1}{2}left(frac{T}{pi n}right)^2cdot(-1)^n cdot 2A+0cdot0+...cdot 0=\
=Acdotleft(frac{T}{pi n}right)^2cdot(-1)^n
end{align*}
$$
the same result using Fourier integral. There is interesting result for non-integer harmonic for $textrm{g}(t)$ is:
$$
a_n=frac{1}{120}Acdot T^2 cdot (-1)^{n+1} left({}_{2}textrm{
F}_1left( 1, 3; frac{7}{2}; -left(frac{pi n}{2}right)^2right)cdot(pi n)^2-20right)
$$
where ${}_{2}textrm{
F}_1left(a,b;c;zright)$ 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.

I think that the missing link that connects the Fourier transform to the Taylor series expansion is Euler's formula, $e^{jmath x}=cos(x) +jmath sin(x)$. This celebrated formula establishes a relationship between trigonometric functions of real entities and exponential functions of complex (i.e. imaginary) entities. In doing so, it establishes a fundamental connection with the Fourier transform, which is essentially trigonometric in nature. In fact, the Fourier transform for a function $f(t)$ is, by definition, a Fourier series of $f(t)$ interpreted as $trightarrow infty$, and a Fourier series is, by definition, a linear summation of sin and cos functions. The Taylor Series comes into play in the derivation of Euler's formula.

As previously mentioned, Euler's formula states $e^{jmath x}=cos(x) +jmath sin(x)$. Hence, it's acceptable to conceptually superimpose the conventional (x, y) unit circle and the real-complex plane, as they both portray the polar Eulerian expression on the continuous interval from $0$ to $2pi$. We therefore, evaluating Euler's formula at $x=2pi$, arrive at
$$e^{2pi*jmath theta } = {cos(}2pitheta)+j*{sin}(2pitheta)$$
This expression is intrinsically intertwined with the nature of a Fourier transform because Fourier transforms aim to convert a function from the time domain to the frequency domain. To do so, we decompose periodic functions into simple, linear sums of sin and cos, and allow this new function to approach infinity.

So, where does the Taylor series fit into all this? We use the Taylor series expansion (or more precisely, the McLauren series expansion) to obtain Euler's formula. Here's the proof (or derivative, depending on your perspective):

Proposition: For any complex number $z$, where $z=x+jmath y=x+sqrt{-1}*y$, i.e. $x=Re {z } $ and $y=Im {z } $, it can be said that, for real values $x$,

$$ e^z=e^{jpi} = cos(x)+jmath sin(x)$$

Lemma 1: The Taylor series expansion of an infinitely differentiable function $f(x)$ in the neighborhood $a$ is defined
$$f(x)=sum_{k=0}^{infty} frac{f^k(a)}{k!}(x-a)^k$$
We'll say that we're working in the neighborhood $a$ surrounding $theta$. So, when $theta = 0$, we get a special case of the Taylor series, called the McLauren series. The expansion of the Taylor (or McLauren) series for $sin(x)$ and $cos(x)$ are
$$ sin (theta)=sum_{k=0}^{infty} frac{d^{k} sin(theta)}{dtheta^k} mid _{theta=0} = sum_{k=0}^{infty}frac{theta^k}{k!} = theta - frac{theta^3}{3!} + frac{theta^5}{5!} - frac{theta^7}{7!} ...$$
$$ cos (theta)=sum_{k=0}^{infty} frac{d^{k} cos(theta)}{dtheta^k} mid _{theta=0} = sum_{k=0}^{infty}frac{theta^k}{k!} = 1 - frac{theta^2}{2!} + frac{theta^4}{4!} - frac{theta^6}{6!} ...$$
This is not a surprise, as we know that sine and cosine are odd and even functions, respectively. We've now accounted for the trigonometric portions of $e^z$, but have not yet addressed the complex component.

Lemma 2: We apply the previously defined Taylor (more specifically McLauren) series expansion to the function $e^z$ when $z=jmath x$, and we get
$$e^{jmath x} =sum_{k=0}^{infty} frac{d^{k} e^{jmath x}}{dtheta^k} mid _{theta=0} = sum_{k=0}^{infty}frac{theta^k}{k!} = 1 + jmath x -
$$

Comparing the right-hand terms from the Taylor series expansions of $ cos (theta)$, $sin (theta)$, and $e^{jmath theta}$ performed in Lemmas 1 and 2, we see that summing the expansions from Lemma 1, we arrive, by the procedure I'm too lazy to type out in LaTeX but not too lazy to explain via my good friends Google and krotz, at
$e^{jmath x}=cos(x) +jmath sin(x)$. [Q.E.D.]

Now that we've proven Euler's formula, we can evaluate it at $x=2pi$ to acquire

$$e^{2pi*jmath theta } = {cos(}2pitheta)+j*{sin}(2pitheta)$$

Due to the characteristics of the sin and cos functions, it is possible to use simple integration to recover the amplitude of each sin and cos wave represented in a Fourier transform (similar to the reverse of the above proof). In an overwhelming majority of cases, it's highly useful to select Euler's formula as the function to integrate over. Because Euler's formula and the Fourier transform are both (at least, in part) fundamentally trigonometric in nature, the use of Euler's formula greatly simplifies most of the real portions of the Fourier analyses. Additionally, for the complex case, frequencies can be represented inverse-temporally using a combination of Euler's formula and the Fourier series expansion.

So, it's a bit messy and convoluted (etymologically, not integrally), but it really boils down to the fact that the Taylor (or McLauren) series, the Fourier series and transform, and Euler's formula all relate a trigonometrically
The differences between the three arise by nature of application. Taylor series are used to represent functions as infinite sums of their derivatives. Fourier series and transforms are used in linear systems &/or differential equations to convert signals or DEs from the time to frequency domain. Euler's formula is used to relate trigonometric and complex exponential (complexponential?!) funcions, and is also a formula that, when evaluated at $x=pi$, yields Euler's identity, $e^{jmath pi}+1=0$, an equation so austerely eloquent and aesthetically arousing that I'd be down to stare at all day.