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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
9
comment Plis, What is the orthogonality conditions for associated legendre polynomials with both two different indexes
If some indices are equal, the results are mathworld.wolfram.com/AssociatedLegendrePolynomial.html
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
Mar
6
comment Why does $1+2+3+\cdots = -\frac{1}{12}$?
Sure, it's a completely analogous sum. Just like $1+2+3+\dots$ quantifies the ground state energy in 1+1 dimensions (of a string), $1+1+1+\dots$ quantifies the ground state's charge (of a system of free fermions, for example). It's really the reason why the two degenerate states $|0\rangle$ and $c_0|0\rangle$ have ghost numbers $\pm 1/2$, for example.