tag:blogger.com,1999:blog-1920088135580776574.post3006536499302917765..comments2013-10-28T08:30:20.487-07:00Comments on Ask Doctor Math: Nothing is Certain but Death and Logarithmsdrmathhttp://www.blogger.com/profile/17936175968300765200noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-1920088135580776574.post-43960334906484324572009-11-03T09:23:46.726-08:002009-11-03T09:23:46.726-08:00Do you plan to resume your blog? It used to be one...Do you plan to resume your blog? It used to be one of my favourites and as I was cleaning up among my RSS feeds, I couldn't just delete it in some vague hope of an article showing up again after a longish break.<br /><br />I guess if you had moved you'd have posted about it. Any chance of having you back?rooksnoreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-54432401312323062392009-10-30T21:56:51.606-07:002009-10-30T21:56:51.606-07:00This comment has been removed by a blog administrator.Delly News Bloghttp://www.blogger.com/profile/13750296861408964040noreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-85888083912118699632009-10-06T03:27:30.393-07:002009-10-06T03:27:30.393-07:00This comment has been removed by a blog administrator.Delly News Bloghttp://www.blogger.com/profile/13750296861408964040noreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-72943629658289693472009-07-04T08:07:52.858-07:002009-07-04T08:07:52.858-07:00dear dr. math,
can you find an equation for this ...dear dr. math,<br /><br />can you find an equation for this pattern?<br /><br />this really gives me a hard time figuring it out.<br /><br />1=0<br />2=1<br />3=4<br />4=15<br />5=64<br />6=325<br />7=1956<br /><br />there... it's like x=???<br /><br />x is the number in the right side of the equal sign...Zyrexhttp://www.blogger.com/profile/09511506851323586188noreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-69280830097834819672009-04-17T06:27:00.000-07:002009-04-17T06:27:00.000-07:00@Richard,Matt: I was quite aware that most readers...<B>@Richard,Matt:</B> I was quite aware that most readers wouldn't be able to read the German article, but thought I share it anyway for anybody who does speak German. So my comment wasn't meant as some way of telling that Dr. Math's explanation isn't good, just that I find the German WP article better this time. »Better« doesn't necessarily imply »bad« for the compared thing.<br /><br /><B>@Matt:</B> In regard to the English article in the WP I share your sentiments. And would extend that statement to a lot of other articles. I'd say that generally the information found in the German WP is more accurate and better presented than in the corresponding English article. The overall quality seems to be better. A notion which Mr. Wales shares. <br /><br /><B>@drmath:</B> For now I must pass on the translation as my time is very limited right now. Especially because I would more or less translate the entire article, because it was an overall observation, which led to the statement above.<br /><br />Greetings,<br />DrizztAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-32630959616126877292009-04-17T04:44:00.000-07:002009-04-17T04:44:00.000-07:00see also: Stigler's law .see also: <A HREF="http://en.wikipedia.org/wiki/Stigler's_law_of_eponymy" REL="nofollow"> Stigler's law </A>.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-39851782899133854472009-04-16T13:56:00.000-07:002009-04-16T13:56:00.000-07:00Thanks for the feedback, everyone!
Drizzt, if som...Thanks for the feedback, everyone!<br /><br />Drizzt, if some part of the German article is particularly good, maybe you could give us a translation? I had to pass a German reading exam to get my doctorate, but I'm pretty rusty now.<br /><br />Matt, thanks for that derivation! I couldn't think of a good way to get from the scale-invariance property to the actual distribution without doing some more complicated math. But your reasoning gives a good estimate in only a few steps!<br /><br />I don't know that quote, but it certainly sounds true. The phenomenon of things being named after the wrong people seems to be very common, to the point of almost being a joke. For example, my all-time favorite theorem, Stokes' Theorem, was named neither for the person who discovered it nor the person who proved it, but for the person who transmitted it from one to the other. <br /><br />I guess it works out well in some contexts, though, because otherwise half of math would be named after Euler and the other half after Cauchy. :)drmathhttp://www.blogger.com/profile/17936175968300765200noreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-52914034178664311432009-04-16T08:34:00.000-07:002009-04-16T08:34:00.000-07:00Oh, PS: I like the line "named (as is often the ca...Oh, PS: I like the line "named (as is often the case) for the second person to discover it".<br /><br />When you wrote this, were you thinking of the quip about theorems being named after the first person after Euler to discover them?<br /><br />(And do you know who said that?)Mattnoreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-6488224221591864192009-04-16T08:29:00.000-07:002009-04-16T08:29:00.000-07:00The German wiki article may be great (no idea, sad...The German wiki article may be great (no idea, sadly, since I don't speak German), but the English one wasn't -- too technical for a lay audience, not technical enough for a technical audience (IMO, obviously). The point of this blog is to convey mathematical ideas in simple terms, so I think Dr M's presentation is pretty good.<br /><br />That said... Dr Math, may I -- as one Dr M to another -- comment on your scale invariance "derivation"? You mention a conversion to half-feet; you could push this example a little further, since it alone gives a good qualitative feel for the distribution. Divide the digits into four groups: those starting with 1, those starting with 2 or 3, those with 4, and the rest. As you showed, the first two are the same size, and both are the same size as the last one. Given how many digits are encompassed by each set (1, 2 & 5, respectively), it's clear that the number of occurrences is decreasing as the digit increases. This means that the remaining set (numbers starting with 4) is probably fairly small, which means that each of the three equal-sized sets are a little less than a third of the total.<br /><br />You can even get some bounds on that "little less than a third" estimate. Let x be the size of the three equal-sized sets. Assuming that the distribution is monotonically decreasing, the size of the set of numbers starting with 4 is smaller than the size of the set of numbers starting with 3 which, in turn is less than x/2 (since the total of those numbers starting with 2 & 3 is x, giving an average of x/2 for each). Similarly, it's also greater than the size of the set of numbers starting with 5 which, in turn, is greater than x/5. So, the size of this "extra" set is somewhere between x/2 and x/5. That seems like a wide range, but... the total then lies between x + x + x/2 + x = (7/2)x and x + x + x/5 + x = (16/5)x. Since the total is fixed, we get that x must be between 2/7 (28.6%) and 5/16 (31.3%) of the total. Not a bad estimate, for such a simple argument, based on a single rescaling!<br /><br />I just thought that was kinda cool.Mattnoreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-59949596057725142332009-04-16T06:23:00.000-07:002009-04-16T06:23:00.000-07:00I enjoyed the discussion of Benford's law presente...I enjoyed the discussion of Benford's law presented here by Dr. Math. It has one advantage over the German Wikipedia article for me, it is written in English. I myself have analyzed large lists of mathematical and scientific constants and other piles of data just to rediscover Benford's law for my own entertainment so the topic is of personal interest.<br /><br />Most of us appreciate your entertaining exploration of math and math concepts. Please keep up the good work.Richardhttp://www.blogger.com/profile/18117573567556136072noreply@blogger.comtag:blogger.com,1999:blog-1920088135580776574.post-23603156904712880982009-04-15T04:46:00.000-07:002009-04-15T04:46:00.000-07:00I must say, that in this particular case I'd like ...I must say, that in this particular case I'd like the explanation of the <A HREF="http://de.wikipedia.org/wiki/Benfordsches_Gesetz" REL="nofollow" TITLE="German Wikipedia on »Benford's Law«">German Wikipedia</A> somewhat better. SO for anybody who likes to read it (and understands German), go ahead and read the article marked as »worth reading«.<br /><br />Sorry to Dr.Math, I really appreciate reading this blog, but this time I found the article over at Wikipedia to be better. ;)<br /><br />Greetings,<br />DrizztAnonymousnoreply@blogger.com