I understand that integral transforms are generalisations of the dot product of functions that could be interpreted as infinite dimensional vectors.
The most significant advantage then is that differentiation and integration become multiplication and division.
For intuition purposes I am trying to construct an example which shows that similar effects can be achieved with finite dimensional vectors (e.g. with 10 elements).
So I would start with two vectors and calculate the dot product of them. After that I want do multiplication/division on this resultant term and re-transform it to show that this resulted in the equivalent of differentiation/integration of one of the original vectors.
Obviously for this to work you must use a vector with which you calculate the dot product that is equivalent to the exponential function, so I am thinking along the lines of $2^{-st}$ for that vector. Additionally e.g. differentiation could be understood as differences of the consecutive elements of the respective vector.
My question
Something is not working out correctly so that I have difficulties creating this intuitive example - do you understand what I am trying to achieve and if yes could you help me?
I'll give a simple example of the $\mathcal{Z}$-transform to see if this is what you mean. Assume you have a sequence $a_n$, $n=0,1,\ldots,N$ and you compute its $\mathcal{Z}$-transform:
$$A(z)=\sum_{n=0}^Na_nz^{-n}$$
This is obviously a dot product if you define two vectors, one with elements $a_n$ and the other with elements $z^{-n}$. Take now as an example a function $B(z)$ defined by
$$B(z)=1-z^{-1}$$
If you multiply $A(z)$ with $B(z)$ you get
$$C(z)=A(z)B(z)=\sum_{n=0}^Na_nz^{-n}-\sum_{n=0}^Na_{n}z^{-(n+1)}=\\ =\sum_{n=0}^Na_nz^{-n}-\sum_{n=1}^{N+1}a_{n-1}z^{-n}= a_0+\sum_{n=1}^N(a_n-a_{n-1})z^{-n}+a_Nz^{-(N+1)}$$
Transforming back gives
$$c_n=\begin{cases}a_0,&n=0\\a_n-a_{n-1},&0<n\le N\\a_N,&n=N+1\end{cases}$$
So, apart from the first and last value, the new sequence is the sequence of first order differences of the original sequence. Of course, other choices of $B(z)$ are possible and useful.