What is the relationship between Fourier transformation and Fourier series? Is there any connection between Fourier transformation of a function and its Fourier series of the function? I only know the formula to find Fourier transformation and to find Fourier coefficients to find the corresponding Fourier series.
 A: I like to think of both Fourier series and Fourier transforms as decomposing functions against spectral measures on $\mathbb{S}^1$ and $\mathbb{R}$, respectively. 
A Fourier series takes a function $f$ on $\mathbb{S}^1$ and decomposes it as a sum of projections onto eigenspaces. Writing $e^{i\lambda x} = e_\lambda(x)$ and the $L^2$ inner product as $(\cdot |\cdot )$, 
$$ f(x) = \sum_{\pm\lambda\in2\pi\mathbb{Z}} \bigg( \int f(\xi)e^{-i\lambda\xi}d\xi \bigg)e_\lambda(x) = \sum_{\pm\lambda\in 2\pi\mathbb{Z}}(f|e_\lambda)e_\lambda. $$
Here, $\lambda$ ranges over the spectrum of the Laplace operator. For each $\lambda$ in the spectrum, $f$ is projected onto the eigenspace (spanned by $e_{\pm\lambda}(x)$). The original function $f$ is recovered by summing over all projections.
The same idea illustrates the Fourier transform. The inversion formula indicates
$$ f(x) = \int_\mathbb{R} \hat{f}(\lambda) e^{i\langle x,\lambda\rangle}\ d\lambda = \int_\mathbb{R} \bigg( \int_\mathbb{R} f(y)e^{-i\langle \lambda,y\rangle}\ dy \bigg)e_\lambda(x)\ d\lambda\ ``=" \int_\mathbb{R} (f|e_\lambda) e_\lambda(x) d\lambda$$
Morally, the Fourier transform $\hat{f}$ evaluated at a single frequency $\lambda$ is the projection of $f$ onto the $\lambda$-eigenspace of the Laplace operator in an appropriate function space. The Fourier inversion formula indicates how one recovers the original function by "summing" (integrating) over all the projections of $f$ onto all the eigenspaces.
The general theory here is that of the spectral measure. A spectral measure is useful (aside from having an unbearably awesome name) because it provides a formal unifying framework for studying this kind of decomposition associated to operators on Hilbert spaces. It's a measure on $\mathbb{C}$ which takes values not in the extended complex plane, but in projections on a Hilbert space $H$: $\mu$ takes a Borel set $\sigma\subset\mathbb{C}$ and maps it to an orthogonal projection $\mu(\sigma):H\to H$. In our examples, the projections map a function to the integral of its projections onto each $\lambda$ eigenspace, for each $\lambda\in \sigma$.
Now we see that this perspective unifies both Fourier series and Fourier transforms: both are given by spectral measures associated to the Laplace operator on the spaces $\mathbb{S}^1$ and $\mathbb{R}$. The spectrum of the Laplacian on $\mathbb{S}^1$ is discrete, so the integral is a sum of projections; the spectrum of the Laplacian on $\mathbb{R}$ is continuous, so the integral is an actual integral of projections.
There's some subtle analysis I'm leaving out of this -- questions of convergence, inner products, and so on. I learned (am learning) about spectral measures from Reed & Simon, Methods of Modern Mathematical Physics (book 1). Rudin in Functional Analysis (ch 12, 13) calls this a "spectral decomposition." I have not checked, but would be surprised to learn the topic is not covered in Taylor's magnum opus Partial Differential Equations.
A: Given a locally compact abelian group $G$, one can define the character group of $G$ as the group of continuous homomorphisms $G \to S^1$. (It should actually land in $\mathbf C^\times$, but for the purpose at hand, this is good enough.)
The character group of the circle $S^1$ is isomorphic with $\mathbf Z$ (the characters are $\chi_n : \theta \mapsto e^{2\pi i n\theta}$).
On the other hand, the character group of $\mathbf R$ is isomorphic with $\mathbf R$ itself. The characters are $\chi_t : s \mapsto e^{2\pi i st}$.
It is a general principle that the characters of a locally compact group form a "basis" for the space of "nice-enough" functions on the group. Thus, periodic functions (i.e. functions on the circle) have a decomposition as sums of $e^{2\pi i n\theta}$ (Fourier series), whereas functions on $\mathbf R$ have a decomposition as Fourier integrals (inverse Fourier transform of their Fourier transform).
I'm sweeping hundreds of years of analysis under the rug (not that I know all of it), but this is the general idea.
A: One can (at least on an informal level) think of Fourier transforms as a kind of limit of Fourier series.
If $f$ is a function on $\mathbb R$, we can restrict it to the interval $[N,-N]$,
where it has a Fourier series, namely
$$f(x) = \sum_{n = -\infty}^{\infty} a_n  e^{\frac{\pi i n x}{N}} ,$$
where $$a_n = \frac{1}{2N}\int_{-N}^N f(x) e^{\frac{-\pi i n x}{N}}dx
= \frac{1}{2N}\int_{-N}^N  f(x)e^{-ix y} dx,$$ where $y = \pi n/N$.
This is valid for $x \in [-N,N].$
So $$f(x) = \sum_{n = - \infty}^{\infty} e^{ixy} \cdot \int_{-N}^N
f(x) e^{-ixy} dx \cdot \frac{\Delta y}{2\pi},$$ where $y = \pi n/N$, and $\Delta y = \pi/N$
(again, valid for $x \in [-N,N]$).
Letting $N \to \infty$, defining $$\hat{f}(y) = \int_{-\infty}^{\infty}
f(x)e^{-ixy} dy,$$ and thinking of the sum as a Riemann sum, we find that
$$f(x) = \frac{1}{2\pi}  
\int_{-\infty}^{\infty} \hat{f}(y)e^{ixy} dy,$$
now valid for all $x \in (-\infty,\infty).$
This is the representation of $f$ as the integral of its Fourier transform.
(The above is informal, since I didn't carefully discuss convergence issues.  It is a very traditional and well-known intuition, though, and can be made rigorous in various contexts; e.g. my memory is that Wiener proved his Tauberian theorem by passing from the Fourier series to the Fourier transform context using this sort of argument.)
A: The series and the transform are related. The series occurs when we transform a periodic signal. 
Informally speaking, the transform of a periodic function looks like a comb and the spikes in that comb correspond to the terms of the series.
The transforms of nonperiodic functions are not like this. They contain a continuum of frequencies, not just discrete multiples.
