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I will, but I'll be busy the next two days, so it will take a while.benorin said:Anyhow, could you please take a look and my original post? Thanks @fresh_42
The rest will hopefully follow soon...
I will, but I'll be busy the next two days, so it will take a while.benorin said:Anyhow, could you please take a look and my original post? Thanks @fresh_42
The rest will hopefully follow soon...
... where the latter are Dedekind groups. Thus a Hamilton group is a non abelian Dedekind group.Math_QED said:It surprises me that nobody solved 5 yet.
For those who are interested and too lazy to look up the definition: A Hamilton group is a non-abelian group in which all subgroups are normal.
Why is the center the only possible subgroup of index ##4##?fishturtle1 said:Proof: First we show ##Q_8## is non abelian. We know ##ijk = -1##. Multiplying by ##ji## on the left gives ##k = -ji##. Similarly, multiplying by ##k## on the right gives ##-ij = -k##. So, ##(-j)i = k \neq i(-j)##.
We note that if ##G## is a group and ##H## is a subgroup of ##G## with index ##2##, then ##H## is normal in ##G##. It seems clear that any nontrivial subgroup of ##Q_8## is generated by a single element of ##Q_8##. Any subgroup generated by ##\pm i, \pm j, \pm k## has index ##2## and is therefore normal. The subgroup generated by ##-1## is the center of ##Q_8## and therefore is normal. So, ##Q_8## is Hamiltonian.
Next we note from Wikipedia: Let ##N## be a normal subgroup of ##G## and ##H## a subgroup of ##G##. We say ##G## is the semi direct product written ##N \rtimes H## if ##G = NH##, and these subgroups intersect trivially.
Let ##A, B## be two non trivial subgroups of ##G##. Then ##A \cap B \neq \lbrace 1 \rbrace## since ##-1 \in A \cap B##. And so the only way to write ##Q_8## as a semi direct product is ##Q_8 = 1 \rtimes Q_8##. []
You have to calculate ##P\left(\dfrac{S_n}{n};0.5\right) = P(Z<0.02)## for CLT and ##P\left(\left| \bar{X}-0.48 \right|>0.02\right)\leq \dfrac{0.2496}{n(0.02)^2}## for WLLN.benorin said:I think I've got the first half done ok: CLT.
From the Central Limit Theorem (CLT) we have that ##Z=\tfrac{np - \mu _{p^{\prime}}}{\sigma _{p^{\prime}}}=1.645## where the value ##Z=1.645## comes from the ##95%## confidence level from a Z-table, ##\mu _{p^{\prime}} =np## is the population mean, and ##\sigma _{p^{\prime}} = \sqrt{\tfrac{p(1-p)}{n}}## because it's CLT for a proportion and here ##p=.48## is the probability of heads for the sample of biased coin flips and we get
$$Z= \tfrac{0.48n-0.5n}{\sqrt{ n(0.48)(0.52) }} = 1.645 \Rightarrow n=\left( 1.645 \tfrac{\sqrt{ (0.48)(0.52)}}{0.02} \right) ^{\tfrac{2}{3}} = 12$$
where the ceiling function was applied because ##n## is integer.
Here's where it get fuzzy...
According to the weak law of large numbers, in particular the associated Chebyshev's inequality we have
$$P\left( \left| \bar{x}_n - \mu\right|\geq \epsilon\right)\leq \tfrac{\sigma ^2}{n\epsilon ^2}$$
From the margin of error we have ##\epsilon =.02,## and since it's binomial trials we have that ##\sigma ^2 = np(1-p),## here ##p## is the probability of heads for the sample being ##p=0.48##. Hence
$$P\left( \left| p - \mu\right|\geq .02\right)\leq \tfrac{ n(0.48)(0.52)}{ n(0.02)^2}=0.05$$
since we are testing at the 95% confidence level--and freaking ##n## cancels instead of letting us solve for it: was I supposed to have used ##\sigma _{p^{\prime}} ^2 = \tfrac{p(1-p)}{n}## as in the CLT version? It might be wrong it's been a long while since my last stats course, and I'm kinda stoned so may have gremlins mucking about, speaking of which ##n=12## seems low.
I think the only subgroup of index ##4## is the center ##\lbrace \pm1 \rbrace##. Any subgroup of order ##2## must be generated by an element of order ##2##. There is only one element of order ##2## in ##Q_8##, namely ##-1##, and so there can only be one subgroup of order ##2##. Equivalently, there is only one subgroup of index ##4## which we have found.fresh_42 said:They have ##4## elements ##\{\pm 1, \pm i\}## hence index ##2##. And the center has index ##4##. Other indices are not possible except for ##Q_8## and ##\{1\}##. So the only question is: what are the subgroups of index ##4##? (Or you could simply conjugate elements and calculate what happens, but your idea is nicer.)
You can definitely write it in a more elegant way, because it is troublesome to proofread your solution. The key inequality comes directly from ##(xyz−1)^2>0.## If you multiply the inequality by its denominators and apply it, then things become at least readable. And of course: write down the solution in reverse order!lekh2003 said:I'd be interested in a non-brute force method.
fresh_42 said:You can definitely write it in a more elegant way, because it is troublesome to proofread your solution. The key inequality comes directly from ##(xyz−1)^2>0.##
##\geq ## is sufficient, since we have all inequalities with the possibility of being equal. And ##\geq 0## is obvious. There is another little thing to consider: ##6+\ldots = 4+2+\ldots \leq 4+ xyz +\frac{1}{xyz}.##lekh2003 said:Ahh, that is exactly what I ended up with. So, I did it? Do I need to prove that (xyz-1)^2 is >= 0?
Oh yeah, of course its ##\geq 0##, I'm a fool.fresh_42 said:##\geq ## is sufficient, since we have all inequalities with the possibility of being equal. And ##\geq 0## is obvious.
How is this related?fresh_42 said:There is another little thing to consider:
This is too hand wavy with too many unproven statements. Here is an image along which you (hopefully) formalize your argument:lekh2003 said:Time for another half-assed proof :P I hope I interpreted the problem correctly.
To maximise the area of the tetrahedron would involve maximising the area of the face ABC, which can maximise the lengths of BC, BA, and AC. This would involve taking the limit of the tetrahedron as it approaches a flat tetrahedron where the area of ABC is equal to the area of ABD+BCD+ADC. Any other type of tetrahedron will slowly reduce the area since AD, BD, CD are constant and the more spread out they are, the larger ABC's area is.
Hence, this is simply a problem of finding the area of an equilateral triangle with a distance of 1 from the center to each of the vertices. Using simple trigonometry and law of cosines, you can find the side length of this equilateral triangle. I used a triangle with tau/3 and a side of length 1 on either side. The opposite is ##\sqrt 3##. That's the side length of the equilaterqal triangle. Area of an equilateral triangle is ##\frac{\sqrt 3}{4}s^2##. Hence, the triangle has area ##\frac{3\sqrt 3}{4}##.
Since the surface area of the tetrahedron must be double the area of this equilateral triangle (when flat, this being the limit), then the maximum limit of area is ##\frac{3\sqrt 3}{2}.
Well, since you already got the credit and the solution is in any case a kind of brute force, I think I could simply post my solution:lekh2003 said:Oh yeah, of course its ##\geq 0##, I'm a fool.
How is this related?
You can differentiate ##x \longmapsto \tan(x)=\dfrac{\sin(x)}{\cos(x)}## and use it to differentiate ##x\longmapsto x=(\tan \circ \operatorname{arctan})(x).## This would be a proof by differentiation instead of a direct integration. For a direct integration I'd try to do it with a Weierstraß substitution:Mayhem said:For my own edification, how do I prove that ##\int \frac{1}{x^2+1} dx = \arctan(x) + C##...
This depends on what you take for granted. E.g. if you know the graph of the tangent function then you automatically have the graph of the inverse function by reflection at the diagonal ##y=x##. Another possibility is to use approximation formulas like... and the value of the two limits?
Can arctan(x) be differentiated using the definition of the derivative? Then you could use F'(x) = f(x) as a "proof".fresh_42 said:You can differentiate ##x \longmapsto \tan(x)=\dfrac{\sin(x)}{\cos(x)}## and use it to differentiate ##x\longmapsto x=(\tan \circ \operatorname{arctan})(x).## This would be a proof by differentiation instead of a direct integration. For a direct integration I'd try to do it with a Weierstraß substitution:
https://en.wikipedia.org/wiki/Weierstrass_substitution
This depends on what you take for granted. E.g. if you know the graph of the tangent function then you automatically have the graph of the inverse function by reflection at the diagonal ##y=x##. Another possibility is to use approximation formulas like
$$
\operatorname{arctan} x \approx \begin{cases}
\dfrac{x}{1+0.28x^2} &\text{ if } |x|\leq 1\\
\dfrac{\pi}{2}-\dfrac{x}{x^2+0.28}&\text{ if }x>1\\
-\dfrac{\pi}{2}-\dfrac{x}{x^2+0.28}&\text{ if }x<-1
\end{cases}
$$
It might work if you write it as a power series, but I guess it is quite hard if you only haveMayhem said:Can arctan(x) be differentiated using the definition of the derivative? Then you could use F'(x) = f(x) as a "proof".
I think any starting point is fine as long as the starting point is 1) well defined 2) doesn't result in circular reasoning. Using the fact that ##(\arctan(x))' = 1/x^2 + 1 ## isn't circular, but in practice it is a bad way thinking, as guess-and-check quickly becomes unfeasible.fresh_42 said:It might work if you write it as a power series, but I guess it is quite hard if you only have
$$
\lim_{h \to 0} \dfrac{\operatorname{arctan}(x+h)-\operatorname{arctan}(x)}{h}
$$
The problem with any proof is always: 'Where are we allowed to start?', means what do we know already. If you can show ##F'(x)=f(x)## then you have automatically proven ##\int F'(x)\,dx = f(x)+C##. And even this uses the fundamental theorem of calculus.
My first suggestion should work. ##(\tan x)'=\left(\dfrac{\sin x}{\cos x}\right)'=1+\tan^2 x.## SoMayhem said:I think any starting point is fine as long as the starting point is 1) well defined 2) doesn't result in circular reasoning. Using the fact that ##(\arctan(x))' = 1/x^2 + 1 ## isn't circular, but in practice it is a bad way thinking, as guess-and-check quickly becomes unfeasible.
I'm not super familiar with Weierstrass substitution, but it does seem to yield a useful result.fresh_42 said:My first suggestion should work. ##(\tan x)'=\left(\dfrac{\sin x}{\cos x}\right)'=1+\tan^2 x.## So
\begin{align*}
(id)'(x)=1&=(\tan \circ \operatorname{arctan})'(x)=(\tan y)'\cdot \operatorname{arctan }'(x)\\
&\Longrightarrow \\
\operatorname{arctan}'(x)&=\dfrac{1}{1+\tan^2(y)}=\dfrac{1}{1+\tan^2(\operatorname{arctan}(x))}=\dfrac{1}{1+x^2}
\end{align*}
but this uses ##(\sin)'=\cos\, , \,(\cos)'=-\sin## and ##1=\sin^2+\cos^2## in order to differentiate the tangent function.
I think that the direct way for ##\int\frac{dx}{1+x^2}## with the Weierstraß substitution works as well: ##t:=\tan\left(\dfrac{x}{2}\right)## which yields ##dx=\dfrac{2dt}{1+t^2}.## At least it immediately connects the rational polynomial with the tangent function.