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closed as off-topic by Arthur Fischer Sep 18 at 10:37

  • This question does not appear to be about Mathematics Stack Exchange or the software that powers the Stack Exchange network within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. –  Asaf Karagila Jul 18 '12 at 8:35
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(+1) For thinking outside the (sand)box. –  cardinal Jul 18 '12 at 19:40
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At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! –  Grace Note Oct 5 '12 at 14:45
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To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. –  leo Dec 17 '12 at 18:03
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This "sandbox" is being closed to prevent the creation of new answers. To start a draft, simply edit one of the existing free answers. –  Arthur Fischer Sep 18 at 10:37

17 Answers 17

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@Committingtoaname , sorry, forgot this was here. Is not going so well, can't find a way of clearly proving an important argument. –  Edvin Orlov Oct 11 at 3:46

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Definition A: Given $b\in\Bbb R$ we define a map $m\mapsto b^m$ from $\Bbb Z^{\ge 0}$ to $\Bbb R$ recursively by $b^0:=1,b^{m+1}:=b^m\cdot b.$

Observe that this map is increasing for $b>1,$ decreasing for $0<b<1,$ and constant for $b=1.$

Claim 1: For $b>1,$ the map from Definition A is unbounded above.

Proof: Given $m\in\Bbb Z^{>0},$ let $a_m:=b^m-b^{m-1},$ so that each $a_m>0.$ It is readily shown by induction on $m$ that for all $m\in\Bbb Z^{>0},$ we have $a_{m+1}=a_m\cdot b$ and $$b^m=1+\sum_{k=1}^m a_k,$$ so in particular, $$b^m\ge 1+\sum_{k=1}^m a_1=1+ma_1>ma_1.$$ Take any $y\in\Bbb R.$ By the Archimedean Property of Reals, since $a_1>0,$ then there is some $m\in\Bbb Z^{>0}$ such that $ma_1>y,$ and so $b^m>y.$ Since this holds for all $y\in\Bbb R,$ then the map is unbounded above, as desired. $\Box$

Claim 2: Given $m\in\Bbb Z^{\ge0},$ the map $\Bbb R^{>0}\to\Bbb R^{>0}$ given by $b\mapsto b^m$ is non-decreasing, and is increasing for $m\ne 0$.

Proof: It is readily seen that the map $b\mapsto b^0=1$ is nondecreasing, and that the map $b\mapsto b^1=b$ is increasing. Suppose for some $m>0$ that the map $b\mapsto b^m$ is increasing, and take $b,c\in\Bbb R$ with $0<b<c.$ We know that $b^m<c^m$ by inductive hypothesis. Since $0<b$ and $b^m<c^m,$ then $b^{m+1}:=b^m\cdot b<c^m\cdot b.$ Since $b<c$ and $0<c^m,$ then $c^m\cdot b<c^m\cdot c=:c^{m+1},$ and so $b^{m+1}<c^{m+1},$ as desired. $\Box$


Definition B: Given $c\in\Bbb R^{\ge1}$ and $n\in\Bbb Z^{>0},$ we define $c^{\frac1n}:=\sup\{b\in\Bbb R:b^n\le c\}.$ Note in particular that $1^n=1\le c,$ so $\{b\in\Bbb R:b^n\le c\}$ is non-empty, and by Claim 1 is bounded above, so that $c^{\frac1n}$ is well-defined for $n\in\Bbb Z^{\ge 2}$. Furthermore, observing that $b^1=b$ for any $b\in\Bbb R,$ we have $$c^{\frac11}:=\sup\{b\in\Bbb R:b^1\le c\}=\sup\{b\in\Bbb R:b\le c\}=c=c^1,$$ so that Definitions A and B agree, and $c^{\frac1n}$ is well-defined for all $n\in\Bbb Z^{>0}.$

Claim 3: Given $c\in\Bbb R^{\ge 1}$ and $n\in\Bbb Z^{>0},$ $x=c^{\frac1n}$ is a positive real solution to $x^n=c.$ Moreover, it is the unique such solution by Claim 2.

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@cameronbuie How did this go? –  Committing to a challenge Oct 22 at 0:43
    
@Committingtoachallenge: I've not yet finished. Would you like me to tag you once I've done so? –  Cameron Buie Oct 22 at 2:14
    
@CameronBuie Yes :), what is it for? –  Committing to a challenge Oct 22 at 2:16

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This is a free answer for the use of anyone

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@committingtoaname: no, and I'd actually forgotten this was here. For one, I realized that there were some errors in what is here; moreover, achielle hui's comments in that question both shows the location I was heading and why it's hard to proceed further. So I abandoned this direct route. –  Semiclassical Oct 8 at 2:04

$1. a)$ For which values of $n$ is the graph $K_n$ Eulerian?

We want a cycle that crosses every edge precisely once.

Theorem: A connected graph has an Eulerian cycle iff it has no vertices of odd degree.

Hence it is easy to recognize that for $K_n$, each vertex has $n-1$ degree, so for all odd $n$ we have $K_n$ as an Eulerian graph.


$1.b)$ For which values of $n$ is the graph $K_n$ Hamiltonian?

We want a cycle that visits each vertex once.

From the definition it is rather obvious that all $K_n$ have a hamiltonian cycle for $n\geq 3$ which can easily be visualized if you consider a regular $n-gon$


$1.c)$ For which values of $m$ and $n$ is the graph $K_{m,n}$ Eulerian?

Reapplying the above theorem: A connected graph has an Eulerian cycle iff it has no vertices of odd degree.

We consider the two groups of vertices. If we have $m$ vertices in group $1$ and $n$ vertices of group $2$, we note that every vertex of group $1$ has degree $n$ and every vertex of group $2$ has degree $m$.

Hence $K_{m,n}$ is Eulerian iff $m,n$ are both even.


$1.d)$ For which values of $m$ and $n$ is the graph $K_{m,n}$ Hamiltonian?

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$\def\i{\large\int}\def\xd{\dot{x}}\def\f{\frac}\def\p{\partial}\def\l{\left[}\def\r{\right]}$

$\i_1^2 \f{\xd^2}{t^3} dt$ with $x(1)=2$ and $x(2)=17$

Since this is a fixed end point problem we need to first consider the E-L equation.

E.g.

$$\f{\p f}{\p x}-\f{d}{dt}\l\f{\p f}{\p \xd}\r=0$$ $$0-\f{d}{dt}\l\f{2\xd}{t^3}\r=0$$ $$\f{2\xd}{t^3}=c$$ $$\xd=\f{ct^3}{2}$$ $$\f{dx}{dt}=\f{ct^3}{2}$$ $$x=\f{ct^4}{8}+D$$

Using the end points we can determine these constants.

$$2=\f{c}{8}+D$$ $$17=2c+D$$ $$D=17-2c$$ $$2=\frac{c}{8}+17-2c$$ $$-15=-\frac{15c}{8}$$ $$1=\frac{c}{8}$$ $$c=8,D=1$$

Hence: $x=t^4 + 1$ & $\xd= 4t^3$ $$J=\i_1^2 16t^3 dt=\left.4t^4\right|_1^2=63$$


$\i_0^{\f{\pi}{2}} (x^2 - \xd^2 - 2x\sin t) \, dt$ with $x(0)=1, x(\f{\pi}2 ) =2$

Once again we apply $E-L$ form:

$$\f{\p f}{\p x}-\f{d}{dt}\l\f{\p f}{\p \xd}\r=0$$ $$2x-2\sin t - \f{d}{dt}\l -2\xd\r=0$$ $$2x-2\sin t - 2\ddot{x}=0$$

This is simply a second order ODE that can be solved using superpositioning.

$$2\ddot{x} + 2x = 2\sin t$$

We assume the solution is of the form $$x = Ax \cos t + Bx \sin t + C$$ $$\dot{x} = -Ax\sin t + A\cos t + Bx\cos t + B\sin t$$ $$\ddot{x} = -Ax\cos t - A\sin t - A\sin t + Bx\sin t + B\cos t + B\cos t$$ $$= -Ax\cos t - 2A\sin t + 2B\cos t + Bx\sin t$$ $$-Ax\cos t - 2A\sin t + 2B\cos t + Bx\sin t + 2Ax\cos t + 2Bx \sin t + 2C=2\sin t$$ $$Ax\cos t -2A\sin t +2B\cos t +3Bx\sin t+2C=2\sin t $$ $$\sin t(-2A+3Bx)+\cos t(Ax+2B) + 2C = 2\sin t$$ $$C=0,-2A+3Bx=2,Ax+2B=0$$

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System

$$\dot{x}_1 = 3x_1 + x_2$$ $$\dot{x}_2 = 4x_1 + 3x_2 + u$$

where $|u|\leq 1$

First we will convert this to maxtrix form and compare eigenvalues:

$\def\b{\begin{pmatrix}}\def\e{\end{pmatrix}}\def\d{\dot{x}}\b\d_1 \\ \d_2\e=\b3&1\\4&3\e\b x_1\\x_2 \e+ \b 0 \\u\e$

So we can clearly see that we have eigenvalues calculated from $(3-\lambda)^2 -4=0$

$\lambda^2 -6\lambda +5=0=(\lambda-5)(\lambda-1), \lambda=1,5$

Hence we have a repulsive node.

Now we want to find the eigenvectors:

For $\def\l{\lambda}\l=5. \b-2 &1\\4&-2\e=\b 0\\0\e\implies \underset{\sim}{v}_1=\b1\\2\e$

For $\l=1. \b2 & 1\\4&2\e =\b 0 \\ 0\e\implies \underset{\sim}{v}_2=\b1\\-2\e$

Thus we get the general solution: $A\b1\\2\e e^{5t} + B \b 1 \\-2 \e e^t$

Hamiltonian:$$H= -1 + \psi_1(3x_1+x_2) + \psi_2(4x_1+3x_2 +u)$$ $$= -1 + \psi_1(3x_1+x_2) + \psi_2(4x_1+3x_2) +\psi_2u$$

$$\underset{\sim}{\dot{\psi}}=-A^T\underset{\sim}{\psi}=\b3&4\\1&3\e\underset{\sim}{\psi}$$

$H$ is linear in $u$, $|u|\leq 1$, hence we have $H$ max at $u^*=sgn(\psi_2) = \pm 1$

We now need equilibrium points for $u^*=1$ $$3x_1+x_2=0$$ $$4x_1+3x_2+1=0$$ $$x_1 = \frac15,x_2 = -\frac35$$

And for $u^*=-1$

$$3x_1+x_2=0$$ $$4x_1+3x_2-1=0$$ $$x_1 = -\frac15,x_2 = \frac35$$

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I have trouble with a proof of the following fact

Let $(F,U,\eta,\varepsilon):M\to N$ be a Quillen adjunction, i.e. an adjunction between model categories such that $F$ preserves cofibrations and acyclic cofibrations or, equivalently, $U$ preserves fibrations and acyclic fibrations. Then the total left derived functor $\Bbb LF:HoM\to HoN$ and the total right derived functor $\Bbb RU:HoN→HoM$ exist and form an adjoint pair.

The proof goes as follows

Let $\gamma:M→HoM$ and $\delta:N→HoN$ be the localizations. Since $F$ maps acyclic cofibrations to weak equivalences, $δF$ sends acyclic cofibrations between cofibrant objects to iso's, which implies that it has a left derived functor $\Bbb LF$. Similarly, since $U$ preserves acyclic fibrations, the right derived functor $\Bbb RU$ exists.

So the functors between the homotopy categories exist. We want to obtain an isomorphism $HoN(LFX,Y)\to HoM(X,RUY)$

Assume $X$ is cofibrant and $Y$ is fibrant. From $U$ preserving acyclic fibrations, it follows easily that if $f\simeq g:FX\to Y$, then $Uf\simeq_l Ug$. Since $U$ preserves fibrations and terminal objects, $UY$ is fibrant, hence $Uf\circ\eta\simeq Ug\circ\eta:X\to UY$. This gives a bijection $$\pi N(FX,Y)\cong\pi M(X,UY)$$

We generalize to arbitrary $X$ and $Y$:

On objects we have $\Bbb LF=FQ$ and $\Bbb RU=UR$. We have bijections: $$HoN(\Bbb LFX,Y)\cong HoN(FQX,RY)\cong HoM(QX,URY)\cong HoM(X,\Bbb RUY)$$ The first bijection is induced by $r:Y\to RY$, the last one by $q:QX\to X$. For the middle bijection, note that there is a bijection $$\pi(X,Y)\to HoM(X,Y)\\ [f]\mapsto[RQf]$$ if $X$ is cofibrant and $Y$ is fibrant.

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Theorem: If $f:A\to B, g:B\to C, h:C\to A$ are three mappings such that $fgh=1_A$, $ghf=1_B$ and $hfg=1_C$ then each of $f,g,h$ is a bijection.

Proof:


Observation A: We can see that $f^{-1}=(gh)$, $g^{-1}=(hf)$,$h^{-1}=(fg)$


Claim 1: $fgh$ is injective

Proof: $afgh=bfgh=aff^{-1}=bff^{-1}=a=b$

Hence $fgh$ is injective

Observation B: Injectivity of $ghf$ and $hfg$ follows similarly


Claim 2: $f$ is injective

Proof: If $f$ were not injective, then $afgh\ne bfgh$

Hence $f$ is injective.


Claim 3: $ghf$ is surjective.

Proof: $(\forall b\in A)(\exists b\in C)|bghf=b)$

Since $ghf=1_B,bghf=b$, thus $ghf$ is clearly surjective.


Claim 4: $ghf$ surjective $\implies f$ is surjective.

Proof: We have $ghf$ surjective, meaning $(\forall b\in B)(\exists b\in B)|(bghf=b)$

Hence $bg=c$ gives us $(\forall b\in B)(\exists b\in B)|((ch)f=b)$

and $ch=a$ gives us $(\forall b\in B)(\exists b\in B)|(af=b)$ Which is the definition of $f$ being surjective.


Observation C: $f$ is a bijection and $g$ and $h$ can be proven as above. $\blacksquare$

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