1
$\begingroup$

How to compute $$\sum_{j=1}^{K}\frac{P(j)}{Q(j)}\exp(2\pi ija) $$ where $\left|a\right|<1,\ K\in Z$,$\frac{P(j)}{Q(j)}$-rational function.Roots $Q(j)$ are known, complex.$$$$ In my case $$Q(j)=(1+j^{2q})(j^{2p}+(j-A)^{2p})$$ with integer $q,p>0$ and real $A>=0$.

$$P(j)=(j-A)^{2p}$$ or $$P(j)=(j)^{2p}$$ or $$P(j)=(j)^{2}(j-A)^{2p}$$

  • 0
    Numerically?$\,$2011-05-02
  • 0
    Why $|a|<1$? But $K$ should be positive, presumably?2011-05-02
  • 0
    J.M not numerically,but but some approximation mistake is ok2011-05-02
  • 0
    Didier Piau, K is positive, integer. $\left|a\right|<1$ in my case. I don't know if it is helpful2011-05-02
  • 0
    My idea is to approximate the sum with integral, use partial fraction decomposition, and try to express it with exponential integrals. Don't know if its going to work2011-05-02
  • 0
    So you want an (approximate?) analytic expression for your sum, then?2011-05-02
  • 0
    yes. At least approximate expression2011-05-02
  • 0
    Are the roots of $Q(j)$ distributed everywhere in the complex plane or are the in the upper half-plane, or ...?2011-05-02
  • 0
    everywhere, but they have a nice form (I have edited my question)2011-05-02
  • 0
    What is $x$ doing in the definition of $Q(j)$?2011-05-02
  • 1
    @Katja: If $Q(j)$ is so simple, maybe $P(j)$ can also be provided in the question? And yes: what is $x$?2011-05-02
  • 0
    Your rational function certainly will have a complicated partial fraction decomposition...2011-05-03
  • 0
    I can deal with partial fraction decomposition. But it seems that approximation by integral doesn't work :(2011-05-04
  • 0
    This type of problem is dealt with in Melzak "Concrete Mathematics" chapter 3 section 5. It does involve a partial fraction expansion and a couple of tricks to get to a digamma/dilogarithm form.2016-08-20

0 Answers 0