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Apr
19
comment Analog clock with same hands - sometimes one can't tell time
I just notice: If taken literally, at all moments the current time cannot be inferred from the position of the hands, since you cannot tell from the hands whether it is am or pm. ;-)
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
… which is the only one not doubled by the symmetry. Therefore the total number of times (before correction for the case of both hands being at the same position) must be odd; however $144$ is clearly an even number.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
While I don't yet know the right solution, I'm now convinced that $144$ is wrong: Why? Well, it's obvious that (using the time scale of the answer) if $t$ is a solution, then $-t$ is also a solution (it just corresponds to the mirror image of the clock's face). Now there are exactly two times where $t\equiv -t (\mod 1)$, namely $t=0$ (midnight/noon) and $t=\frac12$ (6 o'clock). Now at 6 o'clock, there's no question which time it is, since if the minute hand points down, the hour hand must point in between two numbers, so it can't point at 12. OTOH midnight/noon is a solution, …
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
Yes, it should affect the result. Assuming the $144$ is right (I'm still thinking about this one), it means you have to subtract $11$ instead of $1$ from it, giving $133$ times in $12$ hours, or $266$ times per day.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
There are in total 11 times in 12 hours where both hands are in the same position; in each of those exchanging them trivially gets the same time, but you know the exact time.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
If you mean the moving things which point to the digits on the clock face then hand is the right word.
Apr
19
comment Analog clock with same hands - sometimes one can't tell time
With "needles", do you mean the hands?
Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
Indeed, there are only counterexamples to the claim: There are no real numbers $a$, $b$ such that $\log_ab=-\log_ba$.
Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
@Arthur: But $1/x\ne -x$
Apr
19
comment Prove that $\log_a(b)=-\log_b(a)$
@corciacandy: No, $b^{-x} = (a^x)^{-x} = a^{-x^2} \ne a$ unless $x=0$ and $a=1$.
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
Never. See joois answer for a more general proof.
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
BTW, I think generalization to non-positive functions could be done by simply splitting the domain into the parts where $f>0$ and $f<0$ (parts where $f=0$ obviously don't contribute, since they are subset of the $x$ axis which is of measure zero). Obviously $f$ and $-f$ have the same measure.
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
Indeed, it is. +1
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
@Ali: Actually, the bounded domain cannot be absolutely general; if the domain is a non-measurable set, I'm sure the graph on that domain will also be unmeasurable; however if it is unmeasurable, it certainly will not have a measure greater than zero either.
Apr
19
comment Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
@Ali: Well, the proof uses that the sine is bounded, so you can only use it on bounded functions. For unbounded functions (like $1/x$) you'll need other arguments. I think the best way to prove it for the general bounded domain would be to decompose the open interval into countably many closed intervals and use the $\sigma$-additivity of the Lebesgue measure.
Apr
19
answered Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$
Apr
19
comment Error in my proof?
@PVanchinathan: Yes, that's a simple proof that the sequence converges. But the answer is missing such a proof.
Apr
18
comment Error in my proof?
That argumentation only works if you already know that the sequence converges. Otherwise you'll find that $1+2+4+8+\ldots=-1$ because it obviously fulfils the equation $2x+1=x$, and $-1$ is the only solution to that equation.
Apr
18
answered Number Theory - Multiple of $36$ problem
Apr
18
answered Mathematical induction