Luboš Motl
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 Jan 15 answered Normal Distribution sample mean and population mean? Jan 15 comment Prove that $f_{n}(x) = \frac{1}{x} \chi _{[\frac 1 n, 2]}$ is not uniformly convergent on $[0,2]$? OK, you should have written this from the beginning, it's important. Jan 14 answered Change of variables from multiple to single Jan 14 comment How many elements set $\{\cos(\frac{2\pi k}{n})| k \in \mathbb Z\}$ has, $n \in \mathbb N$ I am just saying - here and there - that you only need to distinguish according to the remainder of $n$ after division by $2$, i.e. whether $n$ is even or odd, not division by $4$. Jan 14 comment How many elements set $\{\cos(\frac{2\pi k}{n})| k \in \mathbb Z\}$ has, $n \in \mathbb N$ No, I think that there is no "modulo 4" problem here, just "modulo 2". You could get to "modulo 4" if there were an absolute value of the cosine or something like that. Jan 14 answered How many elements set $\{\cos(\frac{2\pi k}{n})| k \in \mathbb Z\}$ has, $n \in \mathbb N$ Dec 30 comment Range of $(x+\sqrt{x})(10-x+\sqrt{10-x})$ Just to be sure, if you draw a graph of $f(x)$, it is a boring quasi-parabolic bump between $x=0$ and $x=10$, having $f=0$ for $x=0$ and $x=10$, the boundaries, and reaching something like $f\approx 52.36$ for $x=5$. Dec 30 comment Range of $(x+\sqrt{x})(10-x+\sqrt{10-x})$ You need to run a few more arguments. First, the function $f(x)$ is only well-defined for $0\leq x \leq 10$. Second, the factor $g(x)$ is an increasing function for $0\leq x \leq 10$ and relatively speaking, it is more quickly increasing (percentage per unit $x$) than it is for a higher $x$, or, equivalently,than $g(10-x)$. So between $0$ and $5$, $f(x)$ inherits the increasing character from $g(x)$. ... You may always verify that there are no other local extrema between $0,10$ (by setting $f'(x)=0$) and you may verify that the second derivative $f''(5)\lt 0$ which proves it is a maximum. Dec 29 answered Range of $(x+\sqrt{x})(10-x+\sqrt{10-x})$ Sep 19 awarded Nice Answer Aug 20 awarded Great Answer May 7 awarded Yearling Mar 1 comment Why does $1+2+3+\cdots = -\frac{1}{12}$? I don't think so - my identity for $(n)+ (n+1)+\dots$ works for an arbitrary fractional $n\in R$, too. The value for $n=1/2$ is as important in superstring theory as the value for $n=0$. Feb 19 comment Why does $1+2+3+\cdots = -\frac{1}{12}$? What is wrong about your calculation is that you are assigning an incorrect value to this $1+1+1+\dots$. In that sum, one must really keep track from which value of $n$ each term $1$ comes from. So the sum $1+1+1+$ starting at $n=1$ is $-1/2$ but if it starts at $n=0$, the sum is $+1/2$, for example. However, no such ambiguity exists for the analogous value of $\zeta(-1)$. Incidentally, the general sum $(n)+(n+1)+(n+2)+\dots$ is equal to $(n-n^2/2) -1/12$. You may check that it is equal to $-1/12$ both for $n=0$ and $n=1$ and it obeys the consistency checks when removing $k$ initial terms, too Feb 19 comment Why does $1+2+3+\cdots = -\frac{1}{12}$? Dear @MarioCarneiro, some sums may be hard or even ambiguous but I assure you that both $0+1+2+3+\dots$ and $1+2+3+4+\dots$ are equal to $-1/12$. The sum honors everything that needs to be honored to be certain that $-1/12$ is the only right finite value that may be attributed to it. Feb 3 awarded Nice Answer Dec 9 awarded Caucus Sep 30 awarded Explainer Aug 2 comment Why does $1+2+3+\cdots = -\frac{1}{12}$? Dear @AxelBoldt, $0+1+2+3+4+\dots$ is also demonstrably and always equal to $-1/12$, and it may be shown by pretty much the same proof. And $1+1=2$. If all entries except for a finite number are zero, then one adds a finite number of terms which always has uncontroversial rules. May 7 awarded Yearling