I'm interested in finding a recursion or simple representation for a "Hadamard product" of two power series.

The Hadamard Product

The Hadamard product is defined on generating functions $f(x)$ and $f(y)$ as $f(z)$ below:

$f(x) = \sum_{i=0}^\infty{c_i x^i}$

$f(y) = \sum_{i=0}^\infty{d_i y^i}$

$f(z) = \sum_{i=0}^\infty{c_i \cdot d_i z^i}$

My Particular Interest

Now suppose that we work on the series above with only the first N terms as possibly nonzero, and the rest will be considered to be zero.

Take the coefficients of $f(x)$ to be the reciprocals of the first $N$ naturals, e.g.

$f(x) = \sum_{i=1}^N{\frac{1}{i} x^{i+1}}$

Now suppose that $f(y)$ is in the form of an expression, i.e.

$f(y) = \frac{1-(2y)^{N-1}}{1 - 2y}$ represents the first N terms of the form $(2y)^N$


$f(y) = \frac{1-(n+1)x^N+Nx^{N+1}}{(x-1)^2}$ represents the first N naturals in series

or some similar function...

My Question

Can we find a simple expression, such as a recursion or formula, to represent the "Hadamard product" of these new series?

We could simply write out the $N$ terms in the resulting function $f(z)$. However, I'm wondering if there is a more compact way of representing this. I'm particularly interested when $f(x)$ is the series of reciprocals mentioned above. However, I am interested in finding this result for almost any $f(y)$. Is there a way to do this?

I'd like to consider both the cases when coefficients of $f(y)$ take on real values and when it is complex-valued.


It's $z \int_0^z \frac{f(y)}{y^2} dy$ after you remove the constant and linear terms.

  • $\begingroup$ Qiaochu: Thanks. I'm hoping that there may be an alternative formulation as a recursion, perhaps. I'm attempting to find a recursion to basically evalutate that integral. I'm sorry, I should have made that clear. Do you think a recursion or an alternate expression without integration is possible, or is that begging the question? $\endgroup$ – Matt Groff Nov 3 '10 at 18:01
  • $\begingroup$ @Matt: no, not really. It simply is the integral, and computing it is as hard as computing the integral. $\endgroup$ – Qiaochu Yuan Nov 3 '10 at 19:02
  • $\begingroup$ I guess I'm just frustrated. I have no reason to expect an alternate method of computation, but it seems that this follows a pattern that we can somehow find an alternate method for. I guess I'm just trying to get a better feeling for integration; it seems if we can find a limit for the appropriate sequence we may find an "easier", but equivalent, alternative. $\endgroup$ – Matt Groff Nov 3 '10 at 19:24
  • $\begingroup$ @Matt: I still don't really understand what you're trying to do. As far as I can tell it is something like using generating function techniques as a shortcut for computing Boolean functions. It doesn't work like that. Generating function techniques help you organize structured sequences but they're not much help on arbitrary sequences, especially if you want to compute Hadamard products. You can't just escape the difficulty of working with Boolean functions like that. $\endgroup$ – Qiaochu Yuan Nov 3 '10 at 19:33
  • $\begingroup$ I had an initial intuition that generating functions have properties of a Turing machine (a computer) and there still may be a way to tap into this potential. I had success in organizing strings of bits into a function that has a constant size (yet the strings can be arbitrarily large, and the pattern can be different.) I'm very stubborn at just laying this idea to rest, although I probably should branch out in my interests... But you're right, eventually my methods all depend on Hadamard products. I'm hoping I can find a way to avoid this. I had success when limiting the function... $\endgroup$ – Matt Groff Nov 3 '10 at 20:04

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