Are these two functions equal? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:29:40Z http://mathoverflow.net/feeds/question/71643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71643/are-these-two-functions-equal Are these two functions equal? Anixx 2011-07-30T08:28:38Z 2012-03-09T18:13:54Z <p>The question here is sparked by the discussion inside <a href="http://mathoverflow.net/questions/41011/what-is-the-indefinite-sum-of-tanx/42903#42903" rel="nofollow">this question</a> about indefinite sum(antidifference) of tan(x).</p> <p>A proposed solution was a function $$f_1(x)=ix-\psi _{e^{2 i}}^{(0)}\left(x+\frac{\pi }{2}\right)+C$$ The function involved is the <a href="http://mathworld.wolfram.com/q-PolygammaFunction.html" rel="nofollow">q-digamma function</a>. Symbolically it resolves to be antidifference of tan(x), one can check it by following this link to Wolfran Alpha's result: <a href="http://tiny.cc/60mmf" rel="nofollow">http://tiny.cc/60mmf</a></p> <p>But it turned out that neither Wolfram Alpha, nor any other software is able to evaluate the q-polygamma function with $q=e^{2 i}$ numerically. Attempting to evaluate it manually also turned out to be too difficult.</p> <p>A second solution seemed to be easier to evaluate. It was a series that converged absolutely and also has been proven to be an antidifference of tan(x):</p> <p>$$f_2(x)=-\sum _{k=1}^{\infty } \left( \psi \left(k \pi -\frac{\pi }{2}+1-x\right)+\psi \left(k \pi -\frac{\pi }{2}+x\right) \right.$$ $$\qquad \left. -\psi \left(k \pi -\frac{\pi }{2}+1\right)-\psi \left(k \pi -\frac{\pi }{2}\right)\right)+C$$</p> <p>This function has a fancy graphic (see the initial discussion). Since it is known that a function can have several antidifferences which differ between each other by a 1-periodic function, it is interesting, whether the both functions $f_1(x)$ and $f_2(x)$ are equal or what is the difference between them (of course if the first function actually can be evaluated numerically).</p>