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 Developing contour summation JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 12/13/2013, 11:40 PM (This post was last modified: 12/20/2013, 06:44 PM by JmsNxn.) Well I have been working on formally justifying this and I have found a way. It's actually a lot like what you said would happen. But it's still quite magical Suppose f is analytic in the open region G save a finite number of singularities or poles, further more that there exists an antisum for f in G, F such that it is analytic on G save a finite number of poles and singularities and if $z,z+1 \in G$ then $f(z+1) + F(z) = F(z+1)$ Then, quite beautifully, if $\gamma$ is some differentiable arc in G, the following two definitions are equivalent: if $\gamma:[a,b]\to G$ then $t_0 = a < t_1 < t_2 <... such that for all $\epsilon>0$ there exists an N such that $t_j - t_{j-1} < \epsilon$, Then letting $\gamma_j = \gamma(t_j)$ $\sum_{\gamma} f(z) \bigtriangledown z = \lim_{N\to \infty} \sum_{j=1}^{N} \sum_{\gamma_{j-1}}^{\gamma_j} f(z) \bigtriangledown z$ Where this equals: $\sum_{\gamma} f(z) \bigtriangledown z = \lim_{N\to\infty} \sum_{j=1}^{N} F(\gamma_j) - F(\gamma_{j-1})$ But by the Mean value theorem $F(\gamma_j) - F(\gamma_{j-1}) = F'(\gamma(m_j)) \gamma'(m_j) (t_j - t_{j-1})$ where $t_{j-1} < m_j < t_j$ But this converges to the Riemann integral, since $F$ is holomorphic on $\gamma$ and $\gamma$ is differentiable on $[a,b]$: Therefore! $\sum_\gamma f(z) \bigtriangledown z = \int_\gamma F'(z) dz$ What does this mean? We can reduce summation to integration! This formula holds for real line calculations. So residues do occur. In fact! if C is some circle around zero in G, then $\sum_C \log(z/z-1) \bigtriangledown z = 2\pi i$ I've also managed to work out that: $\sum_C \int_{\zeta-1}^{\zeta} \frac{f(w)}{w-z}\,dw \bigtriangledown \zeta = 2\pi i f(z)$ so long as z is in the interior of the contour. Now to avoid confusion we call these discrete residues. $\ln(z)$ has discrete residues at all negative integers, each equaling $-2\pi i$ This is because $\frac{d}{dz}\sum \ln (z) \bigtriangledown z = \frac{d}{dz}\ln(\Gamma(z+1)) = \psi(z+1)$ and $\psi$ has poles with residue -2 pi i at negative integers. Now unfortunately I can't know what to call these points where discrete residues occur. It's a little discouraging, but it's not all because of poles. I think it has more to do with branch cuts. For example: $\sum_C \frac{\bigtriangledown z}{z^n} = 0$ which gives us some odd behaviour. I'm currently working on trying to reduce these results to give me a way of calculating infinite sums using discrete residues. one function I can trivially show it will work on is $f(z) = \frac{e^{i z}}{1+2z^2} - \frac{e^{i(z-1)}}{1+2z^2 -4z+2}$ We will calculate $\sum_{j=-\infty}^{\infty} f(j)$ knowing that $F(z) = \frac{e^{iz}}{1+2z^2}$ Take the semi circle $C_R$ consisting of an arc of radius R,A_R and the real line from [-R,R], then $f$ is analytic on $C_R$ and furthermore $\sum_{C_R} f(z) \,\bigtriangledown z = \sum_{j=-R}^{R} f(j) + \sum_{A_R} f(z)\,\bigtriangledown z$ but it is easy to show: $|\sum_{\gamma}f(z)\,\bigtriangledown z| \le L(\gamma) sup_{z \in \gamma} |F'(z)|$ therefore, since $L(A_R) = \pi R$ and $g(z) = F'(z) = \frac{d}{dz} \frac{e^{iz}}{1+2z^2} = \frac{ie^{iz}}{1+2z^2} - \frac{4z e^{iz}}{(1+2z^2)^2}$ it follows that $L(A_R) sup_{z \in A_R} F'(z) \to 0$ as $R \to \infty$ Therefore we are left with only calculating the residues of the poles within the contour, which happen at $z=i/\sqrt{2}$. $\sum_{j=-\infty}^{\infty} f(j) = 2 \pi i Res_{z=i/\sqrt{2}}\{F'(z)\}$ with a little calculation $Res_{z=i/\sqrt{2}} F'(z) = 0$ $\sum_{j=-\infty}^{\infty} \frac{e^{ij}}{1+2j^2} - \frac{e^{i(j-1)}}{1+2j^2 -4j + 2} =0$ But using more trivial methods this sum can be easily shown to be zero. It just happens to have a trivial antisum. Calculating antisums into closed form expressions is quite difficult, and so finding discrete residues is a little difficult. I'm working on trying to make this more effable. I'm trying to rigorize my methods of evaluating the antisum using the mellin transform and I've only been able to let it be defined on some strip a < Re z < b You can begin to see the beautiful applications this should have. I'm still working on investigating this intensely, trying to see if I can find some interesting sums. It's just proving a little laborious to fully justify all this. « Next Oldest | Next Newest »

 Messages In This Thread Developing contour summation - by JmsNxn - 09/15/2013, 07:33 PM RE: Developing contour summation - by JmsNxn - 09/15/2013, 08:40 PM RE: Developing contour summation - by tommy1729 - 09/23/2013, 08:50 PM RE: Developing contour summation - by JmsNxn - 12/13/2013, 11:40 PM

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