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 Apr 4 comment Exercising divergent summations: $\lim 1-2+4-6+9-12+16-20+\ldots-\ldots$ Dear Gottfried, it's the most convincing proof of the finite value yet. But I still believe that the fact that the arguments are shifted from $s$ to $s-1$ and $s-2$ and you are combining several of them,even though the original sum had a uniform argument $s$, if you get my point, is illegitimate in the zeta regularization. There may be a set of rules in which your would be an allowed value - but maybe those rules would be allow any value. I sort of feel why your calculation would produce physically wrong results in physics. Feb 10 comment Is Z3 a sub-field of R? Kobi, it's not the "same" 0,1,2. In real numbers, 1+2=3, while in Z3, 1+2=0. They behave differently, so they're not the same thing. Feb 7 answered Proving the convergnce of a sequence Feb 6 comment Simplify exponential equation This is almost certainly an equation that can't be solved analytically for general values of $A_1,A_2,c_1,c_2$. Feb 6 answered Finding all permutations which satisfy given condition Feb 6 comment Proving the convergnce of a sequence Have you tried to find the change of the distances from the -3 fixed point, just like you did for +2? One always gets attracted to +2, and repelled from -3, so one ends up with the constant sequence at -3 and the same limit if you start with -3. Otherwise, even for -2.99 or -3.01, you will end up with +2. Feb 6 comment Proving the convergnce of a sequence You should first find the candidate limit because you have to do it, anyway, and it's in fact simpler than to prove the existence of the limit. If there is a limit, what can you say about $a_{n+1}-a_n$ for a large enough $n$? What does it imply? Feb 4 comment Is the prime counting function differentiable? But aside from the naive definition of the derivative, it's very useful to write the jumps in terms of the Dirac delta-function, $\theta'(x)=\delta(x)$, and analytically, $\delta(x)$ may moreover be written via $1/(x-i\epsilon)=i\pi\delta(x)$. The form of the delta-function (i.e.a piece of the OP's derivative) on the left-hand side is analytic in a half-plane and this is helpful in various manipulations with $\pi(x)$. While the "undergrad" statement is that the derivative doesn't exist, the moral answer is that it not only does but it may be very useful. Feb 4 comment How many solution are there to equation $f(x)=f(f(x))$ given the following function? One may also prove that $f(x)=x$ and $f(x)=6-x$ have no solutions outside these two intervals, and all these four solutions are different from each other because $f(x)=x$ and $f(x)=6-x$ could only be obeyed by the same $x$ if $x=6-x$ i.e. $x=3$ but that's clearly not a solution. Feb 4 comment How many solution are there to equation $f(x)=f(f(x))$ given the following function? @N74: to prove it, one really wants to say (and see from the graph, whatever it is, or from the explicit form of the function with $824-(x-3)^{10}$) that the function $f(x)-x$ is monotonic around the points $x=1+\epsilon$ (e.g. in the interval 0.5-1.5) and $x=5-\epsilon$ (in 4.5-5.5), so by monotonicity and continuity, one finds exactly 1 solution of $f(x)=x$ in each interval. Similarly, one finds a pair of solutions to $f(x)=6-x$ in these two intervals which is also OK. The monotonicity follows from the graph visually and from monotonicity of $y=x$. Feb 4 comment How many solution are there to equation $f(x)=f(f(x))$ given the following function? Probably yes, I forgot to add "6-" at one place, and both pairs of solutions seem to be OK. Feb 4 comment How many solution are there to equation $f(x)=f(f(x))$ given the following function? $f(x)=x$ has solutions about $x\sim 1.04$ and $x\sim 4.96$, so these are two solutions of the original problem, too. It's easy to prove by inequalities that the intersection of the curve above with $y=x$ has to exist for $x=1+\epsilon$ and $x=5-\epsilon$. When we add solutions to $f(x)=6-x$ which is also sufficient, here are 4 solutions in total, about 1.043, 1.044, and 6-these two values. But the latter two, from $f(x)=6-x$, don't work. Feb 4 comment For a sequence, why must $\lim _{n→∞} {||x_n||} = ∞$, $\lim _{n→∞} {||x_n||} = 0$, or there exists a convergent subsequence with a nonzero limit? @Jimm, something like that could make the proposition right. But where in the world can you define $||x_n||$ as a supremum of a set that contains not just $x_n$ but also other $x_{m}$? The norm of a vector $x_n$ is just the Pythagorean etc. length of it, isn't it? Feb 4 comment Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ The task was to compute the sum, not to approximate it, and you haven't solved the task correctly. Alternatively, you may say that the difference between +1 and -1 doesn't matter, either. Feb 4 comment Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ It depends for whom. For a mathematician, the difference between these two numbers usually does matter. Feb 4 comment Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ I don't know whether you understood what I wrote. The result is 22.50000000000000000000000, not 22.500002. Feb 4 comment For a sequence, why must $\lim _{n→∞} {||x_n||} = ∞$, $\lim _{n→∞} {||x_n||} = 0$, or there exists a convergent subsequence with a nonzero limit? Dear @Jimm, I agree that my sequence, interpreted e.g. as vectors in $R^1$, is a counterexample, even though I didn't dare to suggest that it could have been so. I think that to show that the proposed lemma is false counts as a solution, too. Feb 4 comment Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ LOL, what about the 0.000002 error? I assure you that the result 22.5 is exact and no such error arises. Feb 4 comment Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ Just a non-essential addition. People normally use arguments in radians, so whenever I write $\cos A^\circ$ or $\exp(iA^\circ)$, I mean $\cos(\pi A/180)$ or $\exp(i\pi A/180)$. Feb 4 comment For a sequence, why must $\lim _{n→∞} {||x_n||} = ∞$, $\lim _{n→∞} {||x_n||} = 0$, or there exists a convergent subsequence with a nonzero limit? The absolute value or norm may also be zero. But that doesn't matter, the absolute value has no impact on my sequence at all. What you're missing is that the typical hard sequences that you must address don't have any limit at all. You're a priori assuming that a sequence always has some limit which is wrong.