# stevenvh

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bio website stevenvh.net/steven.php location Flanders, Belgium age 54 member for 4 years, 3 months seen Apr 1 at 9:38 profile views 240

That's "Steven" (with the "n" at the end)

"The whole problem with the world is that fools and fanatics are always so certain of themselves, but wiser people so full of doubts." — Bertrand Russell

Product designer, consumer electronics: audio (with Philips), home automation.
Done computer science in a previous life too.

Belbin team roles: Plant and Resource Investigator

Personal values: respect, honesty, pride, modesty, fairness

I yell because I care

favorite candy

 Mar30 comment How to explain Real Big Numbers? @user136774 - That's why I say for most practical purposes. I remember a quote by a professor: "In theory the summation goes to infinity, but in practice infinity is five." When was the last time you needed Graham's Number in a real-life application? Or even 10^20? Dec23 comment I have learned that 1/0 is infinity, why isn't it minus infinity? @millimoose - No, $\infty$ isn't a real number, I never said it was. But you can extend the reals to include $\infty$, that's the hyperreals, and then you can define operations which include reals and $\infty$. (Probably not a ring like I claimed earlier; none of the 4 basic operation has an inverse for $\infty$.) Dec22 comment I have learned that 1/0 is infinity, why isn't it minus infinity? @millimoose - The conclusion is correct inasmuch the limit can't be defined if left and right limit are different. That does not imply that it would be defined if they are equal. And $\infty$ + $\infty$ is $\infty$, by definition. $\infty$ is not a number, but that doesn't mean you can't define a ring for it. And operations with reals, like $\infty$ + r = $\infty$. The fact that $\infty$ + $\infty$ = $\infty$ is the reason why $\infty$ - $\infty$ is not defined; you can't find a result which is consistent with the definition for addition. Dec21 comment I have learned that 1/0 is infinity, why isn't it minus infinity? @millimoose - Sorry, I still don't agree. lim(1/x^2) for x->0 is defined: it's +infinity. That may not be a number in R, but it has a defined symbol, for which a limited number of operations are defined. Like infinity + infinity = infinity, but infinity - infinity is not defined. lim(1/x) on the other hand has no defined value, not in R, nor + or -infinity. That's undefined, which is different from + or -infinity. Dec19 comment I have learned that 1/0 is infinity, why isn't it minus infinity? @millimoose - Not sure that's correct. For instance for 1/x^2 both the left limit and right limit for x -> 0 are +infinity. Then the limit is +infinity, so defined. Dec19 comment I have learned that 1/0 is infinity, why isn't it minus infinity? IEEE-754 doesn't have to give infinity as the result of 1/0; it has a type "NaN" (Not a Number) for these cases. Sep29 comment Resources to learn the meaning of any math symbol @Hagen - I guess Adrián saw it in the document referred to in this answer. Unfortunately this assumes a rather advanced level from the reader, and familiarity with notation, so symbols and operators aren't explained there. Sep26 comment Steps to Re-Learn Mathematic the right way You don't want a "complete guide". Today's mathematics covers so many areas that the complete guide wouldn't fit on your bookshelf, let alone that you would have the means to absorb all of it: that's time, and mental capacity. No offense: nobody understands everything in mathematics. It would be a good idea if you would indicate what application of your maths you have in mind. Aug27 comment Prove $\sqrt{\arctan(x)} = (1/2) \arccos((1-x)/(1+x))$ This statement is not true: Sqrt[ArcTan[x]] === 1/2 ArcCos[(1 - x)/(1 + x)] --> False. Can anybody check if my TeX-ification is correct? Aug27 comment Prove $\sqrt{\arctan(x)} = (1/2) \arccos((1-x)/(1+x))$ This looks like a math problem, nothing to do with Mathematica. Migrate? Jul4 comment If a function is uniformly continuous on $(-\infty,-1]$ and $[-1,\infty)$ is it uniformly continuous on $\mathbb{R}$ @Lierre - I don't think it's a typo. It would be silly to leave out ]-1,1[ and conclude that it's uniformly continuous over $\mathbb{R}$. -1 and -1 mentioned 3 times in the question, and that's also how Henning in the accepted answer interpreted it. I therefore rolled back your edit. Jul4 comment If a function is uniformly continuous on $(-\infty,-1]$ and $[-1,\infty)$ is it uniformly continuous on $\mathbb{R}$ @Lierre - the question is about -1 as endpoints of both intervals, not -1 and 1. Jul4 comment If a function is uniformly continuous on $(-\infty,-1]$ and $[-1,\infty)$ is it uniformly continuous on $\mathbb{R}$ I don't agree. Your argument would be valid if the intervals were half-closed, but both are closed, so the union is uniformly continuous. Jul2 comment What is the most frequent number of edges of Voronoi cells of a large set of random points? Not a useful answer if all you get to see is a table of contents. Jun11 comment How do I scale 3 fractions to 3 natural numbers? Second question: now that I see it, it seems obvious. I just couldn't see how to take $a$ and $b$ apart. Thanks. Jun11 comment How do I scale 3 fractions to 3 natural numbers? I saw that, but misinterpreted. Sorry about that. Jun11 comment How do I scale 3 fractions to 3 natural numbers? Ah, the $x$, $y$ and $z$ are not my natural number result then? Because if they were, their $\gcd$ would also be natural. Jun11 comment How do I scale 3 fractions to 3 natural numbers? "and that r is maximal rational number". Is $r$ rational, or natural? Jun11 comment Missing steps in the calculation of limit? Yes, thanks. +1. It's too early for an accept, there might come other answers. Jun11 comment Missing steps in the calculation of limit? Thanks for the quick reply. But my problem is that in the application frequency is zero, I apply the limit for that case, as I mention in my question.
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