what is summation in the sense of a principal value? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:18:27Z http://mathoverflow.net/feeds/question/16066 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16066/what-is-summation-in-the-sense-of-a-principal-value what is summation in the sense of a principal value? vilvarin 2010-02-22T16:19:32Z 2010-02-23T11:49:09Z <p>In one paper I saw this equality:</p> <p>$$\sum_{\eta=-\infty}^{\infty}\frac{z}{(z+\eta)}=\pi z\cot(\pi z)$$ which is the same as $$\sum_{\eta=-\infty}^{\infty}\frac{1}{(z+\eta)}=\pi \cot(\pi z)$$ where summation is understood in the sense of a principal value. What does it mean?</p> <p>In another paper I found the next expression:</p> <p>$$\frac{\exp(2\pi iaz)}{\exp(2\pi iz)-1}=\frac{1}{2\pi i}\sum_{n=-\infty}^{\infty}\frac{\exp(2\pi ina)}{z-n}$$ for $a=0$ it is equivalent to $$\frac{1}{\exp(2\pi iz)-1}=\frac{1}{2\pi i}\sum_{n=-\infty}^{\infty}\frac{1}{z+n}$$ which is not exactly the same expression like in the first case. $$\sum_{n=-\infty}^{\infty}\frac{1}{z+n}=\pi Cot[\pi z]-i\pi$$</p> <p>Where is my mistake? </p> <p>If the second formula is wrong, what is the correct formula for the second case? $$\sum_{n=-\infty}^{\infty}\frac{\exp(2\pi ina)}{z+n}=?$$</p> http://mathoverflow.net/questions/16066/what-is-summation-in-the-sense-of-a-principal-value/16067#16067 Answer by L Spice for what is summation in the sense of a principal value? L Spice 2010-02-22T16:24:10Z 2010-02-22T17:11:10Z <p>A principal-value sum (or integral) is usually one in which unconditional summation (or integration) does not converge, so one needs to sum in a particular way to achieve convergence. I suspect that, in this case, the necessary summation is symmetric, so that we consider <code>$\lim_{N \to \infty} \sum_{n = -N}^{n = N} f(n)$</code> instead of <code>$\sum_{n = 1}^\infty f(-n) + \sum_{n = 0}^\infty f(n)$</code>.</p> <p><strike>It's not quite clear to me what your issue is with the two formul&aelig; you mention. Since you are summing different functions ($1/(z + n)$ versus $z/(z + n)$), it is no surprise that the answers are different. What am I missing?</strike> (Sorry, I did not notice that you had already factored out the $z$.)</p> http://mathoverflow.net/questions/16066/what-is-summation-in-the-sense-of-a-principal-value/16072#16072 Answer by Harald Hanche-Olsen for what is summation in the sense of a principal value? Harald Hanche-Olsen 2010-02-22T17:19:04Z 2010-02-22T17:19:04Z <p>See the answer of L Spice for the principal value bit.</p> <p>For the second bit, the formula from the second paper is rather suspect. For example, $a=1/2$ produces the divergent sum (even in the principal value sense) <code>$$\sum_{n=-\infty}^\infty \frac{(-1)^n}{z-n}.$$</code> And for your case $a=0$, $z=1/2$ yields <code>$\sum(z-n)^{-1}=0$</code> by symmetry, so the formula cannot be right then either.</p>