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 Developing contour summation JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 09/15/2013, 07:33 PM Well I've been working a lot on continuum sums/indefinite sums and I've found a way to perhaps recover a chain rule, and with it the concept of a contour summation. It is a linear substitution law for indefinite summation. I think I can recover a full chain rule by using luinear approximations but I am not quite sure. The rule is very simple to state and I can give a concrete reason for its result. Being quick I'll state it: I: $\sum_0^{a b} f(y) \Delta y= \sum_0^a \sum_0^{b-1} f(x + ta)\Delta t \Delta x$ A good argument for this is a few examples in integers, lets take the identity function and show it works. $\sum_0^{nm} y \Delta y = \sum_{j=1}^{nm} j =1 + 2 + 3 +....+ nm -1 + nm$ But this is equal to two sums across n and m: $\sum_{j=1}^n \sum_{k=0}^{m-1} j + kn$ which follows from the division algorithm. This substitution holds for complex numbers because the right hand side of I behaves like the left hand side when adding $\sum_{nm}^{n(m+1)}$ or $\sum_{nm}^{(n+1)m}$ This may not seem like a big rule but it allows us to define a chain rule for linear substitutions, and what more gives us a glimmer of hope at finding a regular chain rule so that we may find a form of substitution for continuum sums. A neat thing I've been investigating alongside this is contour summation. Imagining C is some contour in complex plane parameterized by $\phi: [a,b] \to \mathbb{C}$ then the contour sum is defined as: $\sum_C f(z) \Delta z = \lim_{n\to \infty} \sum_{j=0}^{n-1} \sum_{\phi(a + \frac{jb}{n})}^{\phi(a+\frac{(j+1)b}{n})} f(t) \Delta t$ Given if f is analytic and has an antidifference on C, then the result depends only on the end points (like contour integration). Closed contours are zero if there is an antidifference. The result does not depend on the parameterization. And I have a belief that there are non-analytic functions where there is a type of residue according to closed contours around singularities. This requires more research however. But using our definition of the contour sum and our linear substitution there is a rule for given for circles. I wont write it all out but it gives me the thought that we can find some neat closed form formula without limits for closed contour summation. More on this as I progress. I was just wondering if anyone sees any quick applications for the linear substitution rule JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 09/15/2013, 08:40 PM (This post was last modified: 09/15/2013, 08:41 PM by JmsNxn.) We can try the result for monomials, and by induction: $\sum_0^{a^n}f(z)\Delta z = \sum_0^{a}\sum_0^{a-1} .... \sum_0^{a-1} f(x + (t_1 + t_2 + ...+t_{n-1})a) \Delta t_1 \,\Delta t_2\,...\Delta t_{n-1} \Delta x$ This leaves us wondering what an addition formula would be--given it could generate substitution of polynomials. tommy1729 Ultimate Fellow Posts: 1,365 Threads: 333 Joined: Feb 2009 09/23/2013, 08:50 PM The only contour summation I believe in is the riemann sum of a contour integral. A few reasons 1) If the sum is over finite terms , the value DOES depend on the path. 2) If the sum is over infinite terms , the set MUST be dense on the path ( such as gaussian rational approximations ). 3) If the sum is over a non fixed amount of terms , it is continuum summation. Or conditional continuum summation. 4) the arguments above apply to all types of numbers and dimensions ( real complex bicomplex etc ) I tried to counter my own arguments for about a week but failed, hence Im confident that contour summation does not exist ? I once considered uncountable sums , but they turned out to be equivalent to contour integration too. So I doubt the consistancy of the idea of contour summation. regards tommy1729 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 »

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