Convergence of Newton series for sin ax - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:40:32Z http://mathoverflow.net/feeds/question/99166 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99166/convergence-of-newton-series-for-sin-ax Convergence of Newton series for sin ax Anixx 2012-06-09T09:34:07Z 2012-06-10T01:39:58Z <p>Let's define half discrete-analytic function as a function whose Newton series converges to that function for each $x>0$:</p> <p>$$f(x)=\sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)=\sum_{m=0}^{\infty} \binom {x}m \sum_{k=0}^m\binom mk(-1)^{m-k}f(k)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f(k)}{(x-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(x-k) k!(n-k)!}}$$</p> <p>Let's define weak discrete-analytic a function a function whose bi-directional Newton expansion converges to that function at any real $x$:</p> <p>$$f(x)=\lim_{n\to\infty}\frac{\sum _{k=-n}^n \frac{(-1)^k f(k)}{(x-k) (k+n)! (n-k)!}}{\sum _{k=-n}^n \frac{(-1)^k}{(x-k) (k+n)! (n-k)!}}$$</p> <p>It seems that the function $\sin x$ is both half discrete-analytic and weak discrete-analytic. But the function $\sin \frac{\pi x}2$ is only weak discrete-analytic.</p> <p>So what is the maximum $a$ such that $\sin ax$ is half discrete-analytic?</p> <p>I also interested to know at which $a$ $\sin ax$ is weak discrete-analytic. For example, is $\sin 3x$ weak discrete-analytic or not?</p> <p>This question is motivated by my old search for a natural fractional integration and integration constant (integral analog of Ramanjuan sum). I want to find a natural generalization of Newton series to functions whose Newton series normally diverges. Once that accomplished it would be possible to find Newton expansions for consecutive derivatives of a function and by analytically continuing them into negative domain, get natural integral. Unfortunately finding natural fractional integral of even such simple function as $\sin x$ requires building Newton expansion for $\sin \frac{\pi x}2$ which diverges.</p> http://mathoverflow.net/questions/99166/convergence-of-newton-series-for-sin-ax/99176#99176 Answer by Gerald Edgar for Convergence of Newton series for sin ax Gerald Edgar 2012-06-09T13:20:22Z 2012-06-09T13:27:47Z <p><strong>half-discrete analytic</strong></p> <p>First do the formal calculation, then discuss its validity. \begin{align} &amp;\sum_{m = 0}^{\infty} \binom{x}{m} \sum_{k = 0}^{m} \binom{m}{k} (-1)^{(m - k)} \operatorname{sin} (a k) \cr &amp;=\sum_{m = 0}^{\infty} \frac{-i}{2}\binom{x}{m} \biggl(\bigl(\operatorname{e} ^{i a} - 1\bigr)^{m} - (-1)^{m} \Bigl(\operatorname{e} ^{(-ia)}-1\Bigr)^{m}\biggr) \cr &amp;=-\frac{i}{2}\biggl(\sum_{m = 0}^{\infty}\binom{x}{m} \bigl(\operatorname{e} ^{i a} - 1\bigr)^{m} - \sum_{m = 0}^{\infty} \binom{x}{m} \Bigl(\operatorname{e} ^{(-ia)}-1\Bigr)^{m}\biggr) \cr &amp;= -\frac{i}{2}\Biggl(\left(\operatorname{e}^{ia}\right)^x - \left(\operatorname{e}^{-ia}\right)^x\Biggr) \cr &amp;= -\frac{i}{2}\left(\operatorname{e}^{iax} - \operatorname{e}^{-iax}\right) = \sin(ax) \end{align} There are two places where the computation may be invalid. First, convergence of the binomial series, we require $$\left|\operatorname{e}^{ia}-1\right| &lt; 1, \qquad \left|\operatorname{e}^{-ia}-1\right| &lt; 1$$ Second, we need to do this: $$\left(\operatorname{e}^{ia}\right)^x = \operatorname{e}^{iax},$$ where that $x$ exponent is the sum from the binomial series.</p> <p>Both of these will be OK for $-\pi/3 &lt; a &lt; \pi/3$.</p> http://mathoverflow.net/questions/99166/convergence-of-newton-series-for-sin-ax/99209#99209 Answer by Gerald Edgar for Convergence of Newton series for sin ax Gerald Edgar 2012-06-10T01:39:58Z 2012-06-10T01:39:58Z <p><strong>weak discrete-analytic</strong></p> <p>Closed form here involves some ${}_2F_1$ functions, so I cannot provide proofs. Numerically, though, it seems that $\sin(ax)$ is weak discrete-analytic for $a$ up to some value just above $3$. Maybe $\pi$, I guess. </p> <p>In this diagram we have $\sin(ax)$ in blue (for $x=5.678$); and 5000 terms for the Newton series in red. They agree for $a$ up to about $3.1$ or $3.2$...</p> <p><img src="http://i47.tinypic.com/21lo7yh.jpg" alt="alt text"></p>