User john engbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T20:59:33Z http://mathoverflow.net/feeds/user/23578 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99213/stick-knot-questions-simple-but-may-not-be-easy/99326#99326 Answer by John Engbers for Stick knot questions: simple but may not be easy John Engbers 2012-06-12T01:02:15Z 2012-06-12T01:02:15Z <p>A paper by <a href="http://www.calvin.edu/~venema/pdfs/Six-Chains.pdf" rel="nofollow">Gerard Venema and Tom Clark</a> classified stick knots with 6 segments (using the lengths of the segments); they are using chains for their knots.</p> http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/97416#97416 Answer by John Engbers for Examples of interesting false proofs John Engbers 2012-05-19T18:06:12Z 2012-05-19T18:06:12Z <p>In the definition of an equivalence relation $\sim$, the reflexivity of $\sim$ is redundant: Indeed, for any $x$, by the symmetric property we have $x \sim y$ implies $y \sim x$. By transitivity we have $x \sim y$ and $y \sim x$ imply $x \sim x$. Therefore, using only symmetry and transitivity, we obtain reflexivity.</p> http://mathoverflow.net/questions/97132/sum-of-exponential-functions Sum of exponential functions John Engbers 2012-05-16T15:32:51Z 2012-05-17T17:44:03Z <p>Suppose that we have $q$ positive integers $a_1, \ldots, a_q$ satisfying $a_1 \leq \cdots \leq a_q \leq q$. I'm interested in the possible types of behaviors for the function given by $$f(x) = (a_1^{x-1} + \cdots + a_q^{x-1})^{1/x},$$ where $x \in [2,\infty)$. In particular, I'm interested in the behavior at integer values of $x$ in that range, but I think that the continuous version of $f$ might be easier to handle.</p> <p>It isn't hard to see that $\lim_{x \to \infty} f(x) = a_q$. I can show that if there is an $x_0 \in (2,\infty)$ with $f(x_0) > a_q$, then $f$ is decreasing at $x_0$, and furthermore I can show that $f$ is either eventually decreasing, increasing, or constant (this almost entirely comes from the number of $i$ with $a_i = a_q$). </p> <p>Each example $(a_1,\ldots,a_q)$ that I've used also has the nice property that it is either always decreasing, always constant (this in fact only occurs if $q=a_q$ and $a_i = a_q$ for each $i$), or decreasing on $[2,x_0)$ and increasing on $(x_0,\infty)$ for some $x_0$ which depends on the $a_i$. Does anyone know of either a proof or counterexample of this property? Is any other behavior possible? Simply computing the derivative of $f$ doesn't seem to be too helpful, but perhaps I'm missing the correct viewpoint on this derivative.</p> <p><strong>Edit</strong>: added the restriction $a_q \leq q$. Without this restriction, $f(x)$ can be strictly increasing as well, e.g. $a_1=3,a_2=4$. An example of the decreasing/increasing is $a_1=1, a_2=2$.</p> http://mathoverflow.net/questions/99213/stick-knot-questions-simple-but-may-not-be-easy/99326#99326 Comment by John Engbers John Engbers 2012-06-12T01:10:28Z 2012-06-12T01:10:28Z I realize now that you're looking specifically for circuits - my apologies. http://mathoverflow.net/questions/98970/solving-problem-area-of-triangle Comment by John Engbers John Engbers 2012-06-06T17:05:05Z 2012-06-06T17:05:05Z You should try math.stackexchange.com for this type of question. It is not appropriate for MO. http://mathoverflow.net/questions/98583/percolation-on-infinite-percolation-clusters Comment by John Engbers John Engbers 2012-06-01T16:53:12Z 2012-06-01T16:53:12Z Can't you just imagine running percolation on $\mathbb{Z}^d$ twice (independently), and losing edges that get deleted either time? So you're running percolation with parameter $p^2$ (or $pq$)? http://mathoverflow.net/questions/98569/sample-variance-calculation Comment by John Engbers John Engbers 2012-06-01T13:25:18Z 2012-06-01T13:25:18Z This is probably more appropriate for mathstackexchange. I'm guessing it will be closed soon, so you might want to try asking it there. http://mathoverflow.net/questions/97713/the-order-of-the-inverse-image-of-a-subgroup-under-a-surjective-homomorphism Comment by John Engbers John Engbers 2012-05-23T03:02:14Z 2012-05-23T03:02:14Z Re-read point 3) under first isomorphism theorem in wikipedia. In your situation, you get $C$ is isomorphic to $A/\ker f$, which does the trick. http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/95613#95613 Comment by John Engbers John Engbers 2012-05-19T18:28:18Z 2012-05-19T18:28:18Z True story that I witnessed in a US precalculus class: the teacher told the class that $\pi$ was a <i>rational</i> number, since $\pi = C/d$, where $C$ is the circumference of a circle and $d$ is the diameter. Since $\pi$ can be written as a fraction, it is rational. This still makes me cringe to this day. http://mathoverflow.net/questions/97132/sum-of-exponential-functions/97240#97240 Comment by John Engbers John Engbers 2012-05-17T19:24:34Z 2012-05-17T19:24:34Z Thank you Igor - I've got it now. http://mathoverflow.net/questions/97132/sum-of-exponential-functions/97240#97240 Comment by John Engbers John Engbers 2012-05-17T18:44:39Z 2012-05-17T18:44:39Z I seem to be missing something here. Lemma 2 that gives the result, but doesn't this state that the function from $1/x$ to $f = (a_1^{x-1} + \cdots + a_q^{x-1})^{1/x}$ is log convex, and so the function $f$ might not be log convex? For example, plotting $g(x) = \log(1+2^{x-1})/x$ from 1.5 to 20 gives a non-convex function (while plotting $g(1/x)$ does appear convex, which is what the lemma tells us). <a href="http://www.wolframalpha.com/input/?i=Plot+Log%281%2B2" rel="nofollow">wolframalpha.com/input/?i=Plot+Log%281%2B2</a>^%28x-1%29%29%2Fx+for+x%3D1.5+to+20 http://mathoverflow.net/questions/97132/sum-of-exponential-functions Comment by John Engbers John Engbers 2012-05-17T15:41:08Z 2012-05-17T15:41:08Z Thanks for the suggestions --- I've tried using a H\&quot;{o}lder/Jensen/etc inequality, but haven't found the right thing to use them on yet. @Tom My initial use of the power mean theorem was hurt by the sum of the weights, which also get hit by the $x$th root; I shall keep poking around along these lines. @David Unless I'm missing something here, I don't think that the log convex trick will work, since $f$ isn't necessarily convex (e.g. $a_1=1, a_2=2$).