User richard hevener - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:57:22Z http://mathoverflow.net/feeds/user/16839 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84097/divergence-of-dirichlet-series Divergence of Dirichlet series Richard Hevener 2011-12-22T15:24:27Z 2012-12-06T00:12:02Z <p>Suppose $s$ is a complex number with $\Re(s) \in (0,1]$ and ${a_n}$ is a complex sequence converging to $a \neq 0$. Must the Dirichlet series $$\sum_{n=1}^\infty\frac{a_n}{n^s}$$ diverge?</p> <p>I asked this question on math.stackexchange on 2011-12-13. Here is the link: <a href="http://math.stackexchange.com/questions/91218/divergence-of-dirichlet-series" rel="nofollow">http://math.stackexchange.com/questions/91218/divergence-of-dirichlet-series</a>.</p> <p>For reference, put $s = r + it$. It is easy to see that the above series must converge if $r > 1$ and must diverge if $r \leq 0$ or if $s = 1$. On math.stackexchange I was able to prove divergence for $0 &lt; r &lt; 1$, using a result from Knopp. This leaves the question open for $s = 1 + it$, where $t \neq 0$.</p> http://mathoverflow.net/questions/84097/divergence-of-dirichlet-series/84556#84556 Answer by Richard Hevener for Divergence of Dirichlet series Richard Hevener 2011-12-29T23:53:23Z 2012-12-06T00:12:02Z <p>I had a devil of a time figuring out the previous answer, but I now believe it to be fundamentally sound. Based on that answer, I am providing another one with a number of gaps filled in.</p> <p>First note that we can assume WLOG that $a = 1$. Under this assumption we will show that either the real or the imaginary part of $\sum_{n=1}^\infty \frac {a_n} {n^s}$ diverges, where $s = 1 + it$ and $t \neq 0$. We also assume WLOG that $t > 0$. (Otherwise, replace $t$ by $-t$ in appropriate places below.) Put $a_n = u_n + iv_n$, where $u_n \rightarrow 1$ and $v_n \rightarrow 0$. Note that $$\Re(\frac {a_n} {n^s}) = (u_n \cos (t \log (n)) + v_n \sin (t \log (n))) / n$$ and $$\Im(\frac {a_n} {n^s})= (-u_n \sin (t \log (n)) + v_n \cos (t \log (n))) / n.$$</p> <p>There must either be infinitely many values of n for which $|\cos (t \log (n))| > 1/2$ or infinitely many values of n for which $|\sin (t \log (n))| > 1/2$ . We assume the former. (Otherwise, switch the roles of sin and cos, and work with the imaginary part of the series instead of the real part; the necessary changes are fairly straightforward.) Likewise, WLOG we assume that there are inf. many values of n for which $\cos (t \log (n)) > 1/2$, as opposed to $&lt; -1/2$. (Otherwise, replace cos by -cos and $\Re$ by $-\Re$ in appropriate places below.)</p> <p>Now $\cos(x)$ is uniformly continuous on the real line. Select $\delta > 0$ such that if $|x - y| &lt; \delta$, then $|\cos(x) - \cos(y)| &lt; 1/4$ . We can also assume that $\delta &lt; t$.</p> <p>To prove divergence under the above assumptions, we will show that the sequence of partial sums of the real part of our series is not Cauchy. Let $M$ be a positive integer.</p> <p>First find $M_1 > M$ such that for all $n > M_1$, $u_n > 1/2$ and $|v_n| &lt; 1/16$. Choose $N$ an integer such that $N > \max(M_1, 2t / \delta)$ and $\cos (t \log (N)) > 1/2$ . Now $N \delta / 2t > 1$, whence $N \delta / t - N \delta / 2t > 1$. Thus, we can select $K$ an integer with $1 &lt; N \delta / 2t &lt; K &lt; N \delta / t$ . Let $n$ be any integer with $n > N$. From calculus we know that $$t(\log (n) - \log (N)) &lt; t(n - N) / N.$$<br> Observe that if $N &lt; n \leq N + K$ , then $$t(\log (n) - \log (N)) &lt; tK / N &lt; \delta,$$ whence $$|\cos(t \log (n)) - \cos(t \log (N))| &lt; 1/4.$$ Therefore, \eqalign{&amp;u_n \cos (t \log (n)) + v_n \sin (t \log (n))\geq u_n \cos (t \log (n)) - |v_n \sin (t \log (n))|\cr&amp;\qquad\gt \cos (t \log (n)) / 2 - 1/16\cr&amp;\qquad\geq (\cos (t \log (N)) - |\cos (t \log (n)) - \cos (t \log (N))|) / 2 - 1/16\cr&amp;\qquad\gt (1/2 - 1/4) / 2 - 1/16 = 1/16.\cr}</p> <p>It follows that \eqalign{\sum_{n=N+1}^{N+K} \Re(\frac {a_n} {n^s}) &amp;\gt \sum_{n=N+1}^{N+K} 1 / 16n \gt K / (16(N + K))= 1 / (16(N / K + 1))\cr&amp; \gt 1 / (16(2t / \delta + 1))= \delta / (16(2t + \delta)) \gt \delta / 48t.\cr}<br> Thus, the sequence of partial sums of the real part of our series is not Cauchy.</p> http://mathoverflow.net/questions/65623/divergence-of-the-dirichlet-series-for-the-riemann-zeta-function-for-re-s-1-im/82764#82764 Answer by Richard Hevener for Divergence of the Dirichlet series for the Riemann zeta function for Re s = 1, Im s <> 0 Richard Hevener 2011-12-06T03:05:11Z 2011-12-06T20:23:51Z <p>I wanted to correct a couple of the inequalities in David Speyer's answer and comments.</p> <p>I believe one in the answer should read $\left| \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right| \leq |s|/(2n^{{Re}(s)+1})$. (This has now been corrected by Mr. Speyer, omitting the "2" in the denominator, which is fine.)</p> <p>Also, I believe the one in the following comment should read $|f'(x) (x - \lfloor x \rfloor)| \leq |s|/|x|^{{Re}(s)+1}$.</p> http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial/71702#71702 Answer by Richard Hevener for If $f$ is infinitely differentiable then $f$ coincides with a polynomial Richard Hevener 2011-07-31T02:39:00Z 2011-11-01T03:52:16Z <p>In Andrey Gogolev's answer the following two assertions appear:</p> <p>"It is clear that $X$ is a non-empty . . . set" and "Now consider any maximal interval $(c,e) \subset ((a,b) - X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$."</p> <p>These are true, but perhaps not transparently obvious. In attempting to fill the gaps, I developed a variation of the proof which requires neither the observation that $X$ has no isolated points nor any argument about degrees of polynomials. Here is my adaptation, borrowed freely from Gogolev:</p> <p>I use the symbol "$\bot$" for "contradiction."</p> <p>Define $I = [0,1]$ and $X = \{x \in I: \forall (a,b) \ni x: f|_{(a,b) \cap I} \; is \; not \; a \; polynomial\}$ .</p> <p>We first establish the following:</p> <p>Lemma: Suppose $[c,d] \subset I$ is an interval on which $f$ coincides with a polynomial $p$. Then there exists a maximal subinterval $[cm,dm]$ having the properties $[c,d] \subset [cm,dm] \subset I$ and $f = p$ on $[cm,dm]$. Furthermore, $cm \in X \cup \{0\}$ and $dm \in X \cup \{1\}$.</p> <p>Proof: Let $cm$ = LUB $\{x: f(x) \neq p(x)\} \cup \{0\}$ and $dm$ = GLB $\{x: f(x) \neq p(x)\} \cup \{1\}$. It is clear that $[cm,dm]$ is maximal. Supppose that $cm \not \in X$ and $cm \neq 0$. Then we can find another interval $(u,v)$ with $cm \in (u,v) \subset I$ on which $f$ coincides with a polynomial $q$. But on $[cm,v]$ we have $f = p = q$, whence $f = p$ on $[u,dm]$. Since $u &lt; cm$, we see that $[cm,dm]$ is not maximal ($\bot$). Therefore, $cm \in X$ or $cm = 0$. Likewise, $dm \in X$ or $dm = 1$.</p> <p>Now we begin the proof-by-$\bot$ of the main result. Suppose that $f$ is not a polynomial on $I$.</p> <p>If $X = \emptyset$, we begin with any $[c,d]$, and the lemma tells us that $cm = 0$ and $dm = 1$, so $f$ is a polynomial on $I$ ($\bot$). Thus, $X \neq \emptyset$. Now define $S_n = \{x: f^{(n)}(x) = 0\}$. $X$ and $S_n$ are clearly closed. Applying the Baire category theorem to the covering ${X \cap S_n}$ of the complete metric space $X$, we get that there exists an interval $(a,b)$ such that $(a,b) \cap X \neq \emptyset$ and $(a,b) \cap X \subset S_n$ for some $n$. (It is important here that $S_n$ is closed.)</p> <p>Put $J = (a,b) \cap I$, and let $a1$ and $b1$ be the left and right end-points of $J$. (Observe that it is possible that $a1 = 0$ or $b1 = 1$, so J may not be open.) If $J \subset S_n$, then $f$ is a polynomial on $J$, whence $(a,b) \cap X = (a,b) \cap I \cap X = J \cap X = \emptyset$ ($\bot$). Thus, we can choose a point $t \in J - S_n$. Now $t \not \in X$, since $(a,b) \cap X \subset S_n$. Therefore, we can find an interval $(c,d) \ni t$ such that $f$ coincides with a polynomial $p$ on $(c,d) \cap I$. Furthermore, $f = p$ on the closure of $(c,d) \cap I$, which is an interval of the form $[c1,d1] \subset I$. Apply the lemma to $[c1,d1]$ to obtain a maximal interval $[cm,dm]$ having the stated properties. Since $t \not \in S_n$ and considering $p$, we see that $cm \not \in S_n$. Suppose $cm > a1$. Then we have $a \le a1 &lt; cm \le c1 \le t &lt; b$, so $cm \in (a,b)$. From the lemma, $cm \in X$, since $cm > a1 \ge 0$. Thus, $cm \in (a,b) \cap X \subset S_n$ ($\bot$). Therefore, $cm \le a1$. Likewise, $dm \ge b1$. Thus, $f$ is a polynomial on $J \subset [a1,b1] \subset [cm,dm]$, whence, as above, $(a,b) \cap X = \emptyset$ ($\bot$). We are at last forced to conclude that $f$ must indeed be a polynomial on $I$.</p> http://mathoverflow.net/questions/78596/power-function-inequality Power function inequality Richard Hevener 2011-10-19T17:32:03Z 2011-10-21T04:24:57Z <p>Let $x$ and $p$ be real numbers with $x \ge 1$ and $p \ge 2$ . Show that $(x - 1)(x + 1)^{p - 1} \ge x^p - 1$ .</p> <p>I recently discovered this result. I am sure it is known, but it is new to me. It is quite easy to prove if $p$ is an integer, even a negative one. I have a proof in the general case above, but it seems overly complicated. Can someone provide a simple demonstration?</p> http://mathoverflow.net/questions/84097/divergence-of-dirichlet-series Comment by Richard Hevener Richard Hevener 2011-12-29T23:39:01Z 2011-12-29T23:39:01Z Yes, the theorem at math.nus.edu is much like the one I used from Knopp. Also, thanks, GH--I have clarified my argument on SE. http://mathoverflow.net/questions/78596/power-function-inequality Comment by Richard Hevener Richard Hevener 2011-10-22T05:23:26Z 2011-10-22T05:23:26Z Thanks for the advice, Gerhard. http://mathoverflow.net/questions/78596/power-function-inequality/78605#78605 Comment by Richard Hevener Richard Hevener 2011-10-20T19:12:49Z 2011-10-20T19:12:49Z Beautiful argument--thanks. As cardinal observed over at math.stackexchange, you don't need Jensen, just convexity. http://mathoverflow.net/questions/78596/power-function-inequality Comment by Richard Hevener Richard Hevener 2011-10-20T19:10:46Z 2011-10-20T19:10:46Z I wanted to explain that I noticed this result while studying an argument of Mazur's that Lp is homeomorphic to L1 for 1 &lt; p &lt; oo. It went unanswered for a week on math.stackexchange before I posted it here.