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H: If cumulative probability distributions $(F_n)_n$ converge pointwise to a continuous cdf $F$, then $(F_n)_n$ converges uniformly to $F$.
I have a candidate proof for this result, but the solution given in the solution manual for this exercise seems much more complicated than mine so I wonder if I did something wrong.
I argue:
Let $(F_n)_n$ be a sequence of distributions such that $F_n \implies F$. And suppose $F$ is continuous. We want to show that
$$\sup_x|F_n(x)-F(x)| \rightarrow 0$$
$\text{as }n \rightarrow \infty$.
Let $\varepsilon > 0$, let $M_\varepsilon$ be such that
$$1-F(M_\varepsilon)<\varepsilon\text{ and } F(-M_\varepsilon)<\varepsilon\tag{$\star$}$$
Using the fact that a sequence of cumulative distributions is $\textbf{tight}$ if and only if every subsequential (vague) limit is a cumulative distribution function for a probability measure, we may conclude that $(F_n)_n$ is tight. That is, for arbitrary $\varepsilon$ we may find $M_\varepsilon'$ such that
$$\limsup_{n \rightarrow \infty} F_n(-M_{\varepsilon}')+\big(1-F_n(M_\varepsilon')\big) < \varepsilon.\tag{$\star\star$}$$
Taking the maximum between $M_\varepsilon$ and $M_\varepsilon'$ we may suppose $M_\varepsilon=M_\varepsilon'$.
From $(\star)$ and $(\star\star)$ and the fact that the $F_n$ and $F$ are nondecreasing, it readily follows that there exists $N\in \mathbb{N}$ such that
$$\sup_{n\geq N} \sup_{|x|\geq M_\varepsilon}|F_n(x)-F(x)|<\varepsilon$$
additionally, given that convergence is uniform in the compact interval $[-M_\varepsilon,M_\varepsilon]$, there is an $N'$ such that
$$\sup_{n\geq N'} \sup_{|x|\leq M_\varepsilon}|F_n(x)-F(x)|<\varepsilon.$$
Supposing without loss of generality that $N=N'$, we get that
$$\sup_{n\geq N} \sup_{x\in \mathbb{R}}|F_n(x)-F(x)|<\varepsilon$$
which shows uniform convergence.
Am I making some mistake?
Thank you for any feedback.
AI: Even if your functions are continuous it does not follow that the convergence is uniform on $[-M_{\epsilon},M_{\epsilon}]$.
Note that $F$ is uniformly continuous on this interval. Choose $\delta >0$ such that $|F(x)-F(y)| <\epsilon$ for $|x-y| <\delta$. Choose $N$ such that $\frac 2 N M_{\epsilon} <\delta$ and divide the interval $[-M_{\epsilon},M_{\epsilon}]$ in to $N$ equal parts to get a partition $(x_i)$. Now note that if $$x_{i-1} \leq x \leq x_i$$ then $$F_n(x)-F(x) \leq F_n(x_i)-F(x_i)+ (F(x_i)-F(x))< 2 \epsilon $$ if $n$ is sufficiently large since $F_n(x_i) \to F(x_i)$ for each $i$. Use a similar inequality for $F(x)-F_n(x)$ to finish the proof. |
H: Is it possible to construct a matrix norm that uses minimum instead of maximum over a compact convex set?
I'm reading a paper where the following matrix norm is used:
$$ ||A||_{C, 2} = \max_{x \in C} \|Ax\|_2, $$
where A is a $d \times q $ matrix, $C$ is a compact convex set in $\mathbb{R}^q$, and $\|.\|_2$is a standard Euclidean norm in $\mathbb{R}^d$.
The paper uses this norm to prove some properties of some function $\Phi(A)$ over different possible matrices $A$. I want to extend the result of paper in such a manner that the following quantity arises:
$$ s_1(A) = \min_{x \in C} \|Ax\|_2,$$
that is, I need to operate with minimum instead of maximum in all proofs of that paper.
So, the question is as follows: is it possible to define such a norm that will operate with a minimum instead of maximum?
By itself, $s_1(A)$ is not a norm because it is not sub-additive. The following function is sub-additive:
$$ s_2(A) = \frac{1}{\min_{x \in C} \|Ax\|_2},$$
but - another drawback - it causes that $C$ should not contain $x = 0$. To overcome this, we can define the following function:
$$ s_3(A) = \frac{1}{\min_{x \in C \setminus B_{\varepsilon}(0)} \|Ax\|_2},$$
where $B_{\varepsilon}(0)$ is an open ball centered at $0$ of radius $\varepsilon$. To make the function be defined for $A = 0$, we can modify it as follows:
$$ s_4(A) = \frac{1}{\delta + \min_{x \in C \setminus B_{\varepsilon}(0)} \|Ax\|_2},$$
but, however, none of these functions is absolutely homogeneous.
So, my question is as follows: is it possible to construct the norm of matrix by the analogy with the $\|.\|_{C,2}$ norm but that uses the minimum instead of maximum?
AI: Your $s_3(A)$ can be useful, since it is $\|A^{-1}\|^{-1}$ up to a constant, if $C$ is the closed unit ball. This is because, when $A^{-1}$ is invertible, $$c\|x\|\le\|Ax\|\iff c\|A^{-1}y\|\le\|y\|$$
If $C$ is separated from the origin (by a hyperplane), then all of the functions suffer from not being specific enough, i.e., $s(A)=0$ do not imply $A=0$. Since one of the main uses of norms is to define convergence, this defeats their purpose. This apart from not being homogeneous; for example, $s_4(0)=\frac{1}{\delta}$. |
H: Finding $a$ such that $ \sqrt{\frac32x^2-xy+\frac32y^2}=x\cos a+y\sin a$ has at least one solution other than $(0,0)$
Find all values of the parameter $ a $ from the interval $ [0, 2 \pi) $, for which the equation
$$ \sqrt{\dfrac{3}{2}x^2 - xy + \dfrac{3}{2}y^2} = x \cos a + y \sin a $$
has at least one solution $ (x, y) $ other than $ (0,0) $.
AI: We know that $x\cos a+y\sin a\leq \sqrt{x^2+y^2}$ and so, $$ \sqrt{\dfrac{3}{2}x^2 - xy + \dfrac{3}{2}y^2} \leq\sqrt{x^2+y^2}\implies (x-y)^2\leq0\implies x=y$$
Now its trivial from here imho. |
H: How to prove distance from foci on an ellipse is equal to twice the semi-major axis (for specific ellipse)
Prove that for any point (x,y) on the conic, the sum of the distances to the two foci is always twice the semi-major axis.
I know that this can be proven in general for all ellipses but the practice question specifically asks for this to be proven for $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1. I feel like I'm really close but I've managed to math myself into a corner somehow.
Let the foci ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0) be denoted as F and F'. Let the point on the conic be denoted P(x,y). We are required to show PF + PF' = 2a. In this case, since a = 3, 2a = 6.
PF = $\sqrt{(x-\sqrt{5})^2 + y^2}$ and PF' = $\sqrt{(x+\sqrt{5})^2 + y^2}$
By rearranging the equation for the ellipse, we get y$^2$ = 4 - $\frac{4}{9}$x$^2$.
Substitute this into PF and PF' to get:
PF = $\sqrt{(x-\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 - 2\sqrt{5}x + 9}$ = $\sqrt{(x - \frac{9\sqrt{5}}{5})^2}$ = x - $\frac{9\sqrt{5}}{5}$
PF' = $\sqrt{(x+\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 + 2\sqrt{5}x + 9}$ = $\sqrt{(x + \frac{9\sqrt{5}}{5})^2}$ = x + $\frac{9\sqrt{5}}{5}$
Therefore PF + PF' = 2x
And then I got stuck
AI: An ellipse is a plane curve surrounding two foci, such that for all points on the curve, the sum of the two distances to the foci is a constant. One starts with $\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$ (your question) to arrive at $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $2a$ is any constant (which ends up being the length of the semi-major axis), $b^2=a^2-c^2$, and the foci are $(-c,0),(+c,0)$. Note that $a,b,c\in\mathbb R^+$.
Addendum
Finding the distance of either foci from a point $P(x,y)$ on the ellipse, $$PF=\sqrt{(x-c)^2+y^2}=\sqrt{(x-c)^2+b^2-\frac{b^2}{a^2}x^2}=\sqrt{\frac{c^2}{a^2}x^2-2cx+a^2}=\left\lvert\frac ca x-a\right\rvert=a-ex$$
$$PF^\prime=\sqrt{(x+c)^2+y^2}=\sqrt{(x+c)^2+b^2-\frac{b^2}{a^2}x^2}=\sqrt{\frac{c^2}{a^2}x^2+2cx+a^2}=\left\lvert\frac ca x+a\right\rvert=a+ex$$
since $x\in[-a,+a]$, where $e=\frac ca$. |
H: Let $B$ be a collection of pairwise disjoint intervals $[a_i,b_i)$ where $a_i\in\Bbb R$ and $b_i\in\Bbb Q$. Can $B$ be uncountable?
In an exercise I am trying to solve the following question appeared:
Let $A_i$ denote the following interval: $[a_i,b_i) \subset \mathbb R$, with $a_i \in \mathbb R$ and $b_i \in \mathbb Q$. Let $B := \{A_i \subset \mathbb R$, such that if $\forall A_i,A_j \in B,$ then $A_i \cap A_j = \emptyset$$\}$. Is the set $B$ countable or uncountable?
Without that final restriction it would be easy to prove that this set is uncountable because we would have: $B \sim \mathbb R \times \mathbb Q$, but I don't even know how to approach the problem due to that restriction. How can I solve this?
AI: Just pick a rational number $r_i \in A_i$ for each $i$ and show that $A_i \to r_i$ is injective. |
H: Prove that a polynomial has no rational roots
Let $P(x)$ be an integer polynomial whose leading coefficient is odd. Suppose that $P(0)$
and $P(1)$ are also odd.
Prove that $P(x)$ has no rational roots.
I have been able to prove that there are no integer roots (using the binomial theorem), and I'm stuck.
AI: Let $P(x)$ be an integer polynomial whose leading coefficient is odd. $P(0)$
and $P(1)$ are also odd. Prove that $P(x)$ has no rational roots.
Let the polynomial $P(x)$ have a rational root $\frac ab\Rightarrow P(x)=(bx-a)Q(x)$, where $Q(x)$ is an integer polynomial. $b,a$ have to be odd since the leading coefficient, $P(0)$ are odd respectively. Now $P(1)$ is even since $b-a$ is even. Hence, a contradiction. |
H: Question about cardinalities of sets
Let $A\cap C$ and $B\cap C$ be finite such that $\left |A\cap C \right|\ge \left |B\cap C \right|$. From this, can we conclude that $\left |\neg B\cap C \right|\ge \left |\neg A\cap C \right|$?
My gut feeling tells me that we can. Either $C$ is finite or not. If $C$ is finite, then the conclusion follows trivially. If $C$ is not finite, i.e. it is infinite, then $\left |\neg B\cap C \right|= \left |\neg A\cap C \right|$. Is this reasoning correct?
AI: Assuming that your $\neg B\cap C$ is equivalent to the set difference $C\setminus B$, your reasoning is correct assuming the axiom of choice.
In the absence of the axiom of choice it is consistent that $C$ be an amorphous set, and in that case there is no bijection between $C\setminus B$ and $C\setminus A$ if $|A\cap C|>|B\cap C|$: such a bijection could be extended by a bijection between $B\cap C$ and a proper subset of $A\cap C$ to yield a bijection between $C$ and a proper subset of itself, and an amorphous set admits no such bijection.
However, in that case we can actually show that $|C\setminus A|<|C\setminus B|$. Let $A'=A\cap C$, $B'=B\cap C$, and $D=C\setminus(A'\cup B')$. Since $|A'|>|B'|$, $|A'\setminus B'|>|B'\setminus A'|$, and there is an injection $\varphi:B'\setminus A'\to A'\setminus B'$. Let
$$f:C\setminus A'\to C\setminus B':x\mapsto\begin{cases}
x,&\text{if }x\in D\\
\varphi(x),&\text{if }x\in B'\setminus A'\,;
\end{cases}$$
then $f$ is an injection, so $|C\setminus A|<|C\setminus B|$. |
H: Condition in an inequality
I have an inequality that is reduced to :
$h(1 - 2v) \geq \frac{1-2v}{2}$
I need to find that :
if $v < 1/2$, then $h \geq 1/2 $
if $v > 1/2$, then $h \leq 1/2 $
But I am only able to find that :
$h(1 - 2v) \geq \frac{1-2v}{2}$
$h \geq \frac{1-2v}{(1 - 2v) 2}$
$h \geq \frac{1}{2}$
It may be very simple for someone who is used with mathematics, but I don't see how to solve it. Could you please help me to find these conditions ?
AI: You begin with an inequality
\begin{align}
h\cdot(1-2v) \geq \frac{1-2v}{2}.
\end{align}
Now the change in sign that you are trying to measure comes from when you divide across the inequality by (1-2v). Notice that if $v > \frac{1}{2}$, then $ 1 - 2v < 0$, i.e. is negative: so denoting by $|1-2v|$ the positive part of $1-2v$
\begin{align}
h\cdot(1-2v) \geq \frac{1-2v}{2} \iff -h\cdot |1-2v| \geq -\frac{|1-2v|}{2}.
\end{align}
Rearranging the inequality on the right hand side then gives us
\begin{align}
h\cdot(1-2v) \geq \frac{1-2v}{2} \iff h\cdot |1-2v| \leq \frac{|1-2v|}{2},
\end{align}
after which we can divide across by the positive $|1-2v|$ to obtain that $h \leq \frac{1}{2}$.
Now if instead $v < \frac{1}{2}$, we find that $1 - 2v > 0$ and so we can perform the division straight away to obtain that $h \geq \frac{1}{2}$. |
H: No simple group of order 720
In his Notes on Group Theory, 2019 edition (http://pdvpmtasgaon.edu.in/uploads/dptmaths/AnotesofGroupTheoryByMarkReeder.pdf p. 83 and ff.)
Mark Reeder gives a proof of the non-existence of simple groups of order 720.
P. 83, before the proof, he says : "In the former case, where $n_3(G) = 40$, the normalizer of a Sylow3-subgroup P acts by an involution on P with trivial fixed points, and normalizes every subgroup of P."
A little lower, in the proof of Lemma 10.26, he says :
"If $n_{3}(G) = 40$ then $N_{G}(P)$ contains an element inverting $P$, hence normalizing $Q$."
If I understand it correctly, the reasoning is as follows : if $G$ is a simple group of order 720, if the number of Sylow 3-subgroups of $G$ is 40, then the normalizer $N_{G}(P)$ of a Sylow 3-subgroup $P$ of $G$ has order 18 and is not abelian. So far, so good (the normalizer is nonabelian in view of Burnside's normal complement theorem). M. Reeder seems to find it obvious that this implies that $N_{G}(P)$ is isomorphic either to the dihedral group of order 18 or to the generalized dihedral group constructed on a noncyclic group of order 9. But a nonabelian group $H$ of order 18 can also be isomorphic to the direct product of a group of order 3 with $S_{3}$ and in this case, it is not true that every element of order 2 of $H$ normalizes every subgroup of order 3 of $H$. Thus, for me, the remark of Mark Reeder is not evident.
Mark Reeder gives the following link to a proof by Derek Holt :
http://sci.tech-archive.net/Archive/sci.math/2006-12/msg07456.html
but this link no longer works.
I can prove that $G$ has exactly 10 Sylow 3-subgroups and deduce from this that these Sylow 3-subgroups have trivial pairwise intersections, but my proof is quite long, so, reading M. Reeder, I'm afraid that something is escaping me.
Thus, my question is : can you explain the two sentences of M. Reeder that I quoted above ? Thanks in advance.
By the way, I think that the non-existence of simple groups of order 720 can be proved in the following way. Let us define a colian group as a finite froup G with the following properties :
1° G is simple;
2° the order of G is divisible by 9 and not by 27;
3° the Sylow 3-subgroups of G are in number 10;
4° the Sylow 3-subgroups of G are noncyclic;
5° the Sylow 3-subgroups of G interset pairwise trivially.
The proof given by Cole of the isomorphy of all simple groups of order 360 (or, in any case the variant of this proof given here : https://fr.wikiversity.org/wiki/Th%C3%A9orie_des_groupes/ chapter 35) can easily be extended to the following statements :
1° every simple group of order 360 is colian;
2° every colian group is isomorphic to $A_{6}$;
3° (and thus every simple group of order 360 is isomorphic to $A_{6}$.)
Then we prove that a simple group of order 720 should be colian, and thus should be isomorphic to $A_{6}$, which is absurd since $A_{6}$ has order 360.
Edit 1 (September 18, 2020). There is no problem with this part of Mark Reeder's proof. He proves (lemma 10.16) that if $P$ is an abelian Sylow subgroup of a nonabelian finite simple group $G$, then no non-identity element of $P$ is centralized by $N_{G}(P)$. Thus if $\vert P \vert = 9$, $N_{G}(P)$ cannot be the direct product of a group of order $3$ with a group isomorphic to $S_{3}$.
Edit 2 (September 22, 2020). I think that the end of the proof of lemma 10.26 in M. Reeder's exposition (p. 83-84) can be simplified.
The author assumes that $G$ is a simple group of order 720 and that $Q$ is a subgroup of order 3 of $G$ contained in several Sylow 3-subgroups of $G$ and he needs to draw a contradiction from it.
He proves that $N_{G}(Q)$ has order 72, so $Q$ has exactly 10 conjugates in $G$. Let $X$ denote the set of the conjugates of $Q$ in $G$. Thus, $X$ has cardinality 10 and, as noted by the author, $G$ acts faithfully on $X$ by conjugation. The author proves that the $Q$-orbits in $X$ have sizes 1, 3, 3, 3. Thus, if $t$ is an element of $Q \setminus \{1\}$,
(1) the permutation $M \mapsto tMt^{-1}$ of $X$ has only one fixed point.
The author also notes that, by the $N/C$ theorem, $C_{G}(Q)$ has order 36 or 72.
From here, I would say what follows. Just remember that $C_{G}(Q)$ has even order. That implies that $t$ is the square of an element of order 6. (Choose $a$ of order 2 in $C_{G}(Q)$, then $t$ is the square of $t^{-1}a$ and $t^{-1}a$ has order 6.) Thus $t = u^{2}$, with $u$ of order 6. In view of simplicity of $G$, $u$ acts on $X$ by conjugation as an even permutation of order 6 and thus $t$ acts on $X$ by conjugation as the square of an even permutation of order 6. But an even permutation of order 6 of a set with cardinality 10 has cyclic structure 6-2-1-1, 3-3-2-2 or 3-2-2-1-1-1, thus the square of such a permutation has at least 4 fixed points, which contradicts the result (1) of the author,
If I'm wrong, please say it me.
Edit 3. (October 26, 2020) There is another problem, perhaps more serious, with M. Reeder's proof. See (No simple group of order 720, again).
Edit 4. (March 26, 2023) @Derek Holt. Introducing your proof, you said : "Let me know if it would helpful to include any further details anywhere, or if you can shorten any parts of the proof."
I think your proof is correct, but I can perhaps make some remarks. Here is my first remark. (I will continue if it seems to interest you.)
1° I think that in the proof of Jordan's proposition, it is not necessary to distinguish between the cases $|\Delta \cap g(\Delta)| = 1$ and the other case. Here is a proof without this distinction. It is long, but the reason is perhaps that I try to be exhaustive.
For a finite set $\Omega$ and a subset $E$ of $\Omega$, I will note $Alt_{\Omega}(E)$ the subgroup of $Alt({\Omega})$ formed by the even permutations of $\Omega$ that fix all elements of $\Omega \setminus E$. $Alt_{\Omega}(E)$ is canonically isomorphic to $Alt(E)$. (You note it $Alt(E)$.)
If $\Omega$ is a finite set and $G$ a subgroup of $Sym(\Omega)$, I will say that a subset $E$ of $\Omega$ is richly permutated by $G$ if $Alt_{\Omega}(E)$ is contained in $G$.
The two following lemmas are easy to prove.
Lemma 1. Let $\Omega$ be a finite set, $E$ a subset of $\Omega$ and $\sigma$ an element of $Sym(\Omega)$. Then $Alt_{\Omega}(\sigma(E)) = \sigma Alt_{\Omega}(E) \sigma ^{-1}$.
Lemma 2. Let $\Omega$ be a finite set, $G$ a subgroup of $Sym(\Omega)$, $E$ a subset of $\Omega$ richly permutated by $G$. For every element $g$ of $G$, $g(E)$ is also a subset of $\Omega$ richly permutated by $G$.
Theorem (Jordan). Let $\Omega$ be a finite set and $G$ a primitive subgroup of $Sym(\Omega )$. If $G$ contains a $3$-cycle, then $G$ contains $Alt(\Omega )$ (and is thus equal to $Alt(\Omega )$ or to $Sym(\Omega )$).
Proof. By hypothesis, $G$ contains a $3$-cycle $(x_{1} \ x_{2} \ x_{3})$. Then the subset $\{x_{1}, \ x_{2}, \ x_{3})$ of $\Omega$ is richly permutated by $G$, so there is at least a subset of $\Omega$ that is richly permutated by $G$ and that has cardinality at least $3$ (for example, $\{ x_{1}, \ x_{2}, \ x_{3} \}$ is such a subset). So, among the subsets of $\Omega$ that are richly permutated by $G$ and that have cardinality at least $3$, we can choose one, say $\Delta$, that is maximal for inclusion. We will prove that $\Delta$ is the whole $\Omega$.
Assume, by contradiction, that
(hyp. 1) $\Delta$ is not the whole $\Omega$.
Then, since $| \Delta | > 1$, $\Delta$ is not a trivial block for $G$. Thus, since $G$ is primitive by hypothesis, $\Delta$ is not a block for $G$. So, there is an element $g$ of $G$ such that we can choose
$e_0 \in \Delta \cap g(\Delta )$
and $c_0 \in g(\Delta ) \setminus \Delta$.
For every distinct $a, b$ in $\Delta \setminus \{e_0 \}$, $a, b$ and $e_0$ are three distinct elements of $\Delta$, thus (since $\Delta$ is richly permutated by $G$)
(2) $(a \ b \ e_0 ) \in G$.
On the other hand, $c_0 $ and $e_0 $ are two distinct elements of $g(\Delta )$. Since $|\Delta| \geq 3$, which implies $|g(\Delta )| \geq 3$, we can choose
an element $d_0 $ of $g(\Delta )$ that is distinct from $c_0 $ and $e_0 $.
Since (Lemma 2), $g(\Delta )$ is a subset of $\Omega$ richly permutated by $G$, we have
(3) $(c_0 \ d_0 \ e_0 ) \in G$.
From (2) and (3) results
$(c_0 \ d_0 \ e_0) \ (a \ b \ e_0 ) \ (c_0 \ d_0 \ e_0 )^{-1} \in G$, i.e. (I compose permutations from right to left)
(4) $(a \ b \ c_0 ) \in G$, for every distinct $a$ and $b$ in $\Delta \setminus \{e_0 \}$.
We proved this for every distinct $a$ and $b$ in $\Delta \setminus \{e_0 \}$. Let us prove that is true for every distinct $a$ and $b$ in $\Delta $. We have to prove that (4) is still true if $a$ or $b$ is equal to $e_0 $. In other words, we have to prove the two following theses :
(thesis 5) for every $b$ distinct from $e_0$ in $\Delta $, $( e_0 \ b \ c_0 ) \in G$
(thesis 6) for every $a$ distinct from $e_0$ in $\Delta $, $( a \ e_0 \ c_0 ) \in G$.
Let $b$ be an element distinct from $e_0$ in $\Delta $. Since $|g(\Delta )| \geq 3$, we can choose an element $a_0$ distinct from $b$ in $\Delta \setminus \{ e_{0} \}$. Then, by (2),
(7) $(a_0 \ b \ e_0 ) \in G$.
On the other hand, we have, by (4),
$(a_0 \ b \ c_0 ) \in G$.
With (7), this implies
$(e_0 \ b \ a_0 ) (a_0 \ b \ c_0 ) \in G$, i.e.
(8) $( e_0 \ b \ c_0 ) \in G$,
which proves our thesis (5).
Let now $a$ be an element distinct from $e_0$ in $\Delta $. Since $|g(\Delta )| \geq 3$, we can choose an element $b_0$ distinct from $a$ in $\Delta \setminus \{ e_{0} \}$. Then by (2) et (4),
$(a \ b_0 \ e_0) \ (c_0 \ b_0 \ a) \in G$, i.e.
(9) $(a \ c_0 \ e_0) \in G$.
which proves our thesis (6).
From (4), (8) and (9), it results that for every distinct $a$ and $b$ in $\Delta $, $(a \ b \ c_0 )$ is ìn $G$. Since $\Delta $ is richly permutated by $G$, it proves that every $3$-cycle in $\Delta \cup \{ c_0 \}$ is in $G$ (more rigorously : the canonical image of every such cycle in $Sym(\Omega)$ is in $G$). Since every alternating group is generated by its $3$-cycles, $\Delta \cup \{c_0 \}$ is thus richly permutated by $G$, which contradicts the maximality of $\Delta $. This contradiction shows that our hypothesis (1) is false, so $\Delta$ is the whole $\Omega$, thus $\Omega$ is richly permutated by $G$, which means that $G$ contains $Alt(\Omega)$, which proves Jordan's proposition.
By the way, a primitive group is by definition a transitive group with no non-trivial block. If I'm not wrong, the proof of Jordan's proposition doesn't depend on transitivity, only on the non-existence of non-trivial blocks. So, Jordan's proposition can be stated : "Let $\Omega$ be a finite set and $G$ a subgroup of $Sym(\Omega )$ with no non-trivial block. If $G$ contains a $3$-cycle, then $G$ contains $Alt(\Omega )$ (and is thus equal to $Alt(\Omega )$ or to $Sym(\Omega )$)."
It is not really a strengthening of Jordan's proposition, in the sense that there is no case where the weak hypotheses are satisfied and the strong hypotheses are not, since the " strengthening" of Jordan's proposition shows that transitivity results from the weak hypotheses (since $Alt(\Omega)$ is transitive when $\Omega$ has at least $3$ elements).
AI: If $N_G(P)=S_3\times C_3$ then you should be able to transfer off the $C_3$, which is a quotient. The focal subgroup theorem proves this directly. Or you can use Gruen's first theorem, which for abelian Sylow $p$-subgroups states that $P\cap G'=P\cap N_G(P)'$. |
H: How to evaluate the volume of tetrahedron bounded between coordinate planes and tangent plane?
Find the volume of the tetrahedron in $\mathbb{R}^3$ bounded by the coordinate planes $x =0, y=0, z=0$, and the
tangent plane at the point $(4,5,5)$ to the sphere $(x -3)^2 +(y -3)^2 +(z -3)^2 = 9$.
My attempt: I started with determining the equation of tangent plane which comes out to be $x+2y+2z=24$. This is because direction ratios of normal to sphere at $(4, 5, 5)$ are $2, 4, 4$. So, then equation of tangent plant is given by $2(x-4)+4(y-5)+4(z-5)=0$ which means $x+2y+2z=24$.
The required volume is $$\int _{x=0}^4\int _{y=0}^{12-\frac{x}{2}}\int _{z=0}^{12-y-\frac{x}{2}}\:\:dz\:dy\:dx$$ but this is not giving me the required answer which is $576$. Please help.
AI: Your work seems fine, there is only an issue for the $x$ upper limit
$$V=\int _{x=0}^{24}\int _{y=0}^{12-\frac{x}{2}}\int _{z=0}^{12-y-\frac{x}{2}}\:\:dz\:dy\:dx$$
As noticed in the comments, to check the result we can use that
$$V=\frac13 Sh$$
which in this case leads to
$$V=\frac13 144 \cdot 12=576$$ |
H: Explicit solution to an ODE
Consider the nonlinear ODE
$$y'(t)=\frac{a(t)+b(t)}{a(t)b(t)}b(y(t)), \qquad y(0)=0, \qquad 0<t<1,$$
where $a,b \in C^0([0,1])$ are positive and Lipschitz.
Can I find $y$ explicitly in terms of $a,b$ ?
If $b(t)=b$ is constant, then obviously $y(t)=t+b\int_0^t \frac{1}{a(s)} \, ds$. What about the general case ?
AI: You can rewrite the equation as
$$
\int\frac{dy}{b(y)}=\int\left(\frac{1}{a(t)}+\frac{1}{b(t)}\right)dt
$$
so if you know the primitives of $1/a,1/b$ you can obtain the solution. |
H: Use of Lim Sup in proof, rather than Lim
In my textbook on Advanced Probability it reads
"Definition: $X_n$ converges in probability to X if for all $\epsilon >0$, $\lim_{n\rightarrow \infty} P(|X_n-X| \geq \epsilon) = 0$"
Now in a lemma, we set out to prove that $X_n$ also converges in probability to $X$ if and only if for all $\epsilon > 0$, $\lim_{n\rightarrow \infty} P(|X_n-X| > \epsilon) = 0$.
Now in the proof, proving the "if" direction, they fix $\epsilon > 0$ and reason (without further argument)
$${\lim \sup}_{n \rightarrow \infty} P(|X_n -X| \geq \epsilon) \leq {\lim \sup}_{n \rightarrow \infty}P(|X_n - X| > \epsilon/2) $$
and from here conclude that $\lim_{n\rightarrow \infty} P(|X_n-X| \geq \epsilon) = 0$.
I have two questions:
Why use Lim Sup, rather than ordinary Lim, when the definition only concerns the limit?
Why is this inequality "obviously" true?
AI: The inequality follows from the fact that the event $|X_n-X| \geq \epsilon$ is contained in the event $|X_n-X| >\epsilon/2$. Recall that $A \subseteq B$ implies $P(A) \leq P(B)$.
When you want to prove that $a_n \to 0$ (where $a_n \geq 0$ ) but you don't know that $\lim a_n$ exists you start with $\lim \sup a_n$. But once you show that $\lim \sup a_n=0$ you can conclude that $\lim a_n$ exist and is $0$. |
H: Finding position vector of orthocentre
I wanted to know the position vector of orthocentre of a $\triangle ABC$. Given position vectors of vertices as $A(\mathbf a),\,B(\mathbf b),\,C(\mathbf c)$, can we find a general formula for orthocentre like for centroid it is $\displaystyle G\left(\frac{\mathbf{a+b+c}}{3}\right)$. All help is greatly appreciated.
AI: Signum function: $sgn(x)=\frac{x}{|x|}$. $a,b,c,H$ are all vectors. $A,B,C$ are angles of the triangle.
$$cos(A)=(b-a)·(c-a)⇒tan(A)=sgn((b-a)·(c-a))\sqrt{\left(\frac{1}{(b-a)·(c-a)}\right)^2-1}$$
$$cos(B)=(c-b)·(a-b)⇒tan(B)=sgn((c-b)·(a-b))\sqrt{\left(\frac{1}{(c-b)·(a-b)}\right)^2-1}$$
$$cos(C)=(a-c)·(b-c)⇒tan(C)=sgn((a-c)·(b-c))\sqrt{\left(\frac{1}{(a-c)·(b-c)}\right)^2-1}$$
$$H=\frac{atan(A)+btan(B)+ctan(C)}{tan(A)+tan(B)+tan(C)}$$ |
H: Continuous Images of Arc Connected spaces
Arc Connected: $X$ is arc connected if for any $x,y\in X$, $\exists$ homeomorphism $f:I\to X$ such that $f(0) = x, f(1) = y.$
If $g:X\to Y$ is a continuous surjective function and $X$ is arc connected, is $Y$ arc connected too?
I don't think so, but haven't been able to find a proof/counterexample in a book. Also, Wikipedia says that $Y$ should be arc connected.
My Reasoning: As in path-connected, the function $g \circ f$ is a continuous map from $I$ to $Y$, but this need not be a homeomorphism.
AI: No. An image of arc connected (in your definition $f$ should be a homeomorphism onto image, not just homeomorphism) space does not have to be arc connected.
Consider $[0,1]$ with the Euclidean topology. Now put $x\sim y$ iff $x=y=1$ or $x,y\in [0,1)$. So we collapse $[0,1)$ to a point. Finally take the quotient map $q:[0,1]\to[0,1]/\sim$. Note that the quotient is not arc connected.
The same works with any arc connected space and a relationship that produces at most countable (greater than $1$) number of equivalence classes. |
H: If $Y\subseteq X:=\prod_{i\in I}X_i$ then there exist $Y_i\subseteq X_i$ for each $i\in I$ such that $Y=\prod_{i\in I}Y_i$
Statement
If $Y\subseteq X:=\prod_{i\in I}X_i$ then there exist $Y_i\subseteq X_i$ for each $i\in I$ such that $Y=\prod_{i\in I}Y_i$.
Defining $Y_i:=\pi_i[Y]$ for each $i\in I$ then clearly $Y\subseteq\pi^{-1}_i[Y_i]$ for each $i\in I$ and so $Y\subseteq\bigcap_{i\in I}\pi^{-1}_i[Y_i]$ but un fortunately I don't be able to prove the other inclusion. So could someone help me, please?
AI: This is false. For example, take $I = \{i, j\}$, $X_i = \{0, 1\}$, $X_j = \{a, b\}$. Then $X = X_i \times X_j = \{(0, a), (0, b), (1, a), (1, b)\}$. If we take $Y = \{(0, a), (1, b)\}$ then $Y$ is not a cartesian product. |
H: Dirac Measure (weak limit)
I am wondering why Dirac measure is weak limit of the function?
AI: because for all $\varphi \in \mathcal C_c^\infty (\mathbb R^n)$, $$\lim_{r\to 0}\int_{\mathbb R^n}\varphi (x)f_r(x)\,\mathrm d x=\varphi (0)=\left<\varphi ,\delta _0\right>.$$ |
H: Making sense of linear transformations under change of basis
Let $T: V \rightarrow V$ be a linear transformation, where $V$ is some $n$-dimensional space. Let $A, B$ be two ordered bases for $V$. Let $T_A$ and $T_B$ represent the matrix representations of $T$ with respect to $A$ and $B$ respectively. Let $x_A$ and $x_B$ denote the representations of an arbitrary vector $x \in V$ with respect to $A$ and $B$ respectively. Let $M$ be the change-of-basis matrix such that $x_A = Mx_B$.
By definition, we have $T(x_A) = T_Ax_A$ and $T(x_B) = T_Bx_B$.
Consider this proof that $T_B = M^{-1}T_AM:$
We have, $$\begin{aligned} T_Bx_B &= (T(x_A))_B \\&= (T_Ax_A)_B \\&= M^{-1}T_Ax_A \\&= M^{-1}T_AMx_B\end{aligned}$$
and hence, $$\begin{aligned} T_B = M^{-1}T_AM \end{aligned}$$
Is this proof correct? I'm having trouble understanding the first step: why is $T_Bx_B = (T(x_A))_B$? Also, in the third step, we implicitly assume $(T_Ax_A)_A = T_Ax_A$. Why is this true?
AI: I suggest to proceed as follows, we have that $y_A=T_Ax_A $, $y_B=T_Bx_B $ and since $y_A=My_B \iff y_B=M^{-1}y_A$ we have
$$y_B=T_Bx_B $$
$$M^{-1}y_A=T_BM^{-1}x_A $$
$$y_A=MT_BM^{-1}x_A $$
that is
$$\boxed{T_A=MT_BM^{-1}\iff T_B = M^{-1}T_AM }$$ |
H: Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + 2$ = $(x - 1)(x - 2)$
and I can see $(x - 1)^2 \equiv 1$ $($mod $x - 2)$ . We also have :-
$$\frac{(x - 1)^{100}}{(x - 1)(x - 2)} = \frac{(x - 1)^{99}}{(x - 2)}.$$
We have :- $(x - 1)^{98} \equiv 1$ $($mod $x - 2).$ $\rightarrow (x - 1)^{99} \equiv (x - 1)$ $($mod $x - 2)$. Now for the case of $(x - 2)^{200}$ we have :-
$$\frac{(x - 2)^{200}}{(x - 1)(x - 2)} = \frac{(x - 2)^{199}}{(x - 1)}.$$
We have :- $(x - 2) \equiv (-1)$ $($mod $x - 1)$ $\rightarrow (x - 2)^{199} \equiv (-1)$ $($mod $x - 1)$.
Adding all these up we have :- $(x - 1)^{100} + (x - 2)^{200} \equiv (x - 2)$ $($mod $x² - 3x + 2)$ .
On checking my answer with wolfram alpha , I found the remainder to be $1$, so I messed up in some step .
Can anyone help me?
AI: You are right that $(x-1)^{98}\equiv1\pmod{(x-2)}$. But that implies
$$(x-1)^{100}\equiv(x-1)^2=x(x-2)+1\equiv1\pmod{x-2}.$$
More naively, as
$$x-1\equiv1\pmod{x-2}$$
then
$$(x-1)^{100}\equiv1^{100}=1\pmod{x-2}.$$
Similarly,
$$x-2\equiv-1\pmod{x-1}$$
and
$$(x-2)^{200}\equiv(-1)^{200}=1\pmod{x-1}.$$
So $(x-1)^{100}+(x-2)^{200}$ is congruent to $1$ modulo both $x-1$ and $x-2$,
and so also modulo $(x-1)(x-2)$. |
H: Given a function $f(x)$ that is define by $f(x-1)$, by knowing $f(0)$ is it possible to rewrite $f(x)$ without using $f(x-1)$
Let a function $f(x)$ that is written using the function itself. Something like Fibonacci sequence $f(x)=f(x-2)+f(x-1)$. Now given enough result of $f(x)$ (in the example of Fibonacci sequence, $f(1)$ and $f(2)$), is it possible to rewrite the function so that it doesn’t require calling itself again, but only uses the input $x$ and the pre-given results (again for Fibonacci sequence it can be written as $f(x)=\frac{\phi^x-(1-\phi)^x}{\sqrt 5}$? Is it possible to prove that any of these self-calling function could be rewritten?
AI: Let me start by saying that expressions such as recursive definition of the Fibonacci sequence you mentioned are usually called recurrence relations. From what I gathered, you ask the question of whether or not such recurrence relations always admit closed form solutions. Indeed this is not the case as explained e.g. here, however for large classes of recurrence relations it can be shown that closed form solutions exist. One particular such class is linear recurrence relations with constant coefficients like the Fibonacci recurrence where one can use linear algebra to obtain a closed form solution by defining a linear function $T$ which given some vector consisting of $k$ values of the sequence described by the recurrence and some initial values (which is what you call $f(0)$) produces another such vector containing elements of the aforementioned sequence further along the index. One can then find a formula for the $n$th sequence element by repeatedly applying $T$ to the initial values, which really boils down to multiplying a matrix $A_T$ representing $T$ to the vector of initial values a bunch. A closed form solution for the powers of $A_T$ needed for this can be obtained using Eigendecompositions. |
H: Moving from log points to percentage points
I'm trying to understand the formula to move from log points to percentage points. I know the same question has already been asked here: How to interpret the difference in log points
and I can follow PaulB's answer easily until the taylor expansion, is the last step that I have troubles understanding. Could anyone please help me claryfying that? It seems to me like the "-1" should be part of the log, but it's clearly not correct.
Any help would be highly appreciated
AI: We have that when x is "small" (let try with calculator)
$$\log(1+x)\approx x$$
therefore for small pecentage $\frac{\%\Delta}{100}$
$$\log\left(1+\frac{\%\Delta}{100}\right) \approx \frac{\%\Delta}{100}$$
For the last step, by exponentiation
$$\log(x)-\text{log}(y)=\log\left(\frac{\%\Delta}{100}+1\right)$$
$$e^{\log(x)-\text{log}(y)}=e^{\log\left(\frac{\%\Delta}{100}+1\right)}=\frac{\%\Delta}{100}+1$$
$$\frac{\%\Delta}{100}=e^{\log(x)-\text{log}(y)}-1$$ |
H: $M^{2\times 2}(\mathbb{Z})$ has identity $e_1$, but a certain subring $R$ has another identity $e_2$ and $e_1 \notin R$.
I'm a bit confused. Consider the following situation:
We know $M^{2\times 2}(\mathbb{Z})$ has an identity for $\cdot$ which is $$e_1 =\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}.$$
Now consider the subring $$A = \left\{\begin{pmatrix}
x & 0 \\
0 & 0 \\
\end{pmatrix}\mid x \in \mathbb{Z}\right\},$$
which has identity element:$$e_2 =\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}.$$
In my book on algebra, it says that if $1\in R$ and $S\subset R$ is a subring of $R$, then $1\in S$ needs to be met. But in this example, this is not the case. Can someone help with my confusion?
AI: The definition of "ring" varies in whether it includes 1 (and even commutativity of multiplication). Generally, those who hold 1 isn't required for a ring will not require a subring to have the same 1. It's a matter of differing conventions.
On the other hand, if this occurs in the same book that says subrings must include the same 1, that's an issue. |
H: Metric Space Question: is $H(x)$ in this neighborhood?
Let the metric $d$ be defined as
$$
d(f,g) =\sup_{x\in[0,1]}|f(x)-g(x)|,
$$
and let
$$
H(x) = \begin{cases} 0 \text{ if } x \leq \frac{1}{2}\\ 1 \text { if } x > \frac{1}{2} \end{cases}.
$$
Is $f(x) = x$ in $B_\frac{1}{2}(H)$ ?
My answer. No, because
$$
d(H(x),f(x)) = \sup_{x\in[0,1]}|f(x)-H(x)| = \frac{1}{2}.
$$
Therefore, $f(x) \not \in B_\frac{1}{2}(H)$
I am not sure if my answer is correct, and I found that it is hard to visualize this metric. Can someone helps me on this?
AI: $d(H,f) \ge |H(\frac12)- f(\frac12)| = |0 - \frac12|= \frac12 $ as the sup of a set is an upper bound for it. So $f \notin B_{\frac12}(H)$ as claimed. Well done. |
H: Does multivariate polynomial over a finite field always have a solution (in the field)?
Let $K = F_{p^e}$ be a finite field. Say I have a single polynomial $f \in K[x_1,\ldots, x_n]$ of degree $d$.
Under what conditions on $n$ and $d$ can I claim that a root to $f$ always exists? In other words, do there exist polynomials over finite fields with an arbitrary number of variables and bounded degree that have no roots in the field?
If a root exists, can I compute it efficiently?
AI: You can't do it with $n$ and $d$ alone, unless $d=1$ (which is a hyperplane and so is easy to find all solutions).
For example, we know $\mathbb{F}_{p^{2e}}$ is a degree 2 extension of $\mathbb{F}_{p^e}$, so there is an irreducible quadratic $x^2+ax+b$ in $\mathbb{F}_{p^e}[x]$. Consequently,
$$
(a_1x_1+\dots+a_nx_n)^2+a(a_1x_1+\dots+a_nx_n)+b=0
$$
has no solutions for all $(a_1,\dots,a_n)\in\mathbb{F}_{p^e}^n-\{(0,\dots,0\}$. Similar construction gives no $d>1$ can guarantees the existence of a zero regardless of how large $n$ is.
Note that this doesn't contradict Chavelley-Warning theorem, since the number of zeros, $0$, is divisible by $p$. |
H: Algorithm to generate insecure random numbers
I would like an algorithm which can generate a list of random, uniformly distributed floating point numbers from a given seed, ideally also being able to specify the number of decimal places.
The use case if for randomly generating datasets for education, so does not need to be secure. What would a good algorithm be for this? Ideally one that is easy to understand and easy to write in a programming language.
AI: Use a LCG https://en.wikipedia.org/wiki/Linear_congruential_generator to generate a uniform integer and rescale it as you wish. |
H: If $\frac{a}{b}$ is irreducible, then the quotient of the product of any $2$ factors of $a$ and any $2$ factors of $b$ are irreducible.
$a,b\in \mathbb{Z}$
Factors of $a$: $a_1,a_2,...,a_n$.
Factors of $b$: $b_1,b_2,...,b_m$
Prove that if $\frac{a}{b}$ is irreducible, then
$\frac{a_ia_j}{b_kb_l}$ is irreducible for all $i,j,k,l$.
I proved the case where $i=j, k=l$ using the fact that the square root of any non-perfect square is irrational.
I can't prove the more general case though.
EDIT: not just prime factors, just all the integer factors. E.g. 12: 1,2,3,4,6,12
AI: Alternative approach : proof by contradiction.
Suppose that $\frac{a_i a_j}{b_k b_l}$ is not irreducible.
Then $\exists \;$ prime $\;p \;\ni p|(a_i \times a_j)$ and
$p|(b_k \times b_l) \Rightarrow$
$p|a$ and $p|b \Rightarrow \frac{a}{b}$ is not irreducible. |
H: $6\times 6$ grid problem
[Edited to be consistent with the version I proposed in an answer below, which the OP agreed contained the essence of the problem.--John Hughes]
Some months ago, one friend proposed me a problem that I still do not find the solution. That is:
"You have a $6\times 6$ grid ($36$ squares) with an integer in each cell. There are $5$ different operators: you can add one to each cell any $n\times n$ square inside this $6\times 6$ grid (where $n=2,3,4,5,6$).
Can you, using these operators as many times as you want, make all numbers of this grid be multiples of $3$?"
The first thought it comes to me was to put all numbers modulo $3$. Then, find some coloration such that using any operator there is an invariant modulo $3$, but I did not find any. Later, I thought that using an specific combination of operators I can add some quantity to one square, so it implies that this is possible. Again, I did not find any successful combination.
I surrender, but if anyone can tell me your solution (because I do not want to tell my friend I do not solve it hehe).
AI: Assuming my interpretation of the question is correct, the answer is "no, you cannot always make the grid consist of multiples of $3$.
The proof involves a little linear algebra. We'll do everything mod 3.
The space of all "problems" is a 36-dimensional space, with a basis given by the 36 matrices each of which has a "1" in exactly one location (so that $E_{3, 5}$ is all zeroes except for a $1$ in the 3rd row, fifth column).
If we have a grid $A$, an "operation" amounts to adding a matrix $Q$ to $A$, where $Q$ is all zeroes except for a $k \times k$ block of $1$s, where $k = 2, 3, 4, 5,$ or $6$. So "solving" a puzzle $A$ amounts to finding a sequence $Q_1, Q_2, \ldots, Q_n$ where $$ A + Q_1 + Q_2 + \ldots + Q_n = 0.$$ Letting $R_i = -Q_i$, this amounts to saying $$A = R_1 + \ldots + R_n.$$
To rephrase the previous paragraph, the puzzle is "Given any matrix $A$, is there a linear combination of the 'operation/ matrices which equals $A$?" (Note that mod 3, $-Q_1 = 2Q_i$, so each $R_i$ is in fact a linear combination of "operation" matrices.) Said even better, is the span of all operation matrices, mod 3, equal to the 36-dimensional space of `problems'?
As it happens, I wrote a little matlab program to generate all the operation matrices, and convert each one to a 36-vector (by reading off the columns one after another). That gave me $55$ vectors in a 36 dimensional space. And then I computed the rank of this $36 \times 55$ matrix, and it turns out to be $35$ (alas).
It takes a little thinking to see that the fact that the dimension of the column space over the reals is 35 means that the dimension of the columns space over the integers mod 3 cannot be more than 35, but that's actually true. Hence there's some "problem" matrix $A$ that cannot be represented by a linear combination (mod 3) of "operation" matrices. |
H: Combinatorics calculation
I am trying to solve a problem and stuck at an intermediate step. Let $s_M$ be average of elements of a set $M\subset N$, $|N|=n$. Find an average of all $s_M$.
I got result as :
(Sum of all set elements)*(1),where (1) is given below :
$$ \tag{1}
{n-1 \choose 0}/1+
{n-1 \choose 1}/2+
{n-1 \choose 2}/3+\ldots+
{n-1 \choose n-1}/n
$$
But given result is :
((sum of all set elements)$(2^n-1))/n$................(2)
I know C(n,0)+C(n,1)...C(n,n)=($2^n$).
(1) and (2) gives same result.But how can we go from (1) to (2)?
AI: They just find a closed form for your sum. Let $$f(x):= {n-1 \choose 0}x/1+
{n-1 \choose 1}x^2/2+
{n-1 \choose 2}x^3/3+\ldots+
{n-1 \choose n-1}x^{n}/n
$$
You need to find $f(1)$.
Then \begin{align} f'(x)&={n-1 \choose 0}+
{n-1 \choose 1}x+
{n-1 \choose 2}x^2+\ldots+
{n-1 \choose n-1}x^{n-1} \\
&= (1+x)^{n-1}
\end{align}
So \begin{align}f(x) &= \int (1+x)^{n-1}\,dx
= {(1+x)^{n}\over n}+c
\end{align}
Since $f(0)=0$ we have $c=-{1\over n}$
Now put $x=1$ and you are done: $f(1) = {2^{n}-1\over n}$ |
H: a probabilistic series limit?
Could anyone tell me $\lim_{n\to \infty}\frac{1}{n+1} \sum\limits_{j=0}^{n}x(j)=?$ If I am given that $x(j)=1$ with probability $p$ and $x(j)=0$ with probability $q$, $p+q=1$
Thanks for helping.
AI: You cannot find the limit in general but if $x(j)$'s are indepedent then SLLN can be applied. The limit is the mean value of $x(1)$ which is $p$. |
H: Solving $\frac{dy}{dx}=\sin(10x+6y)$. Why doesn't my approach work?
I do not want the solution to this question. I want to know why we cannot apply what I did.
A curve through origin satisfies $\frac{dy}{dx}=\sin(10x+6y)$. Find it.
My method:
Let $10x+6y=t$
This gives $\frac{dt}{dx}-6\sin t=10$
Integrating factor is -6 and I solved it.
Correct method:
After assuming $t$, $\frac{dt}{6\sin t+10}=dx$ and then followed by integrating it. This proves to be very tedious with to many substitutions.
We can assume $1$ multiplied with $-6$ as $x^0$.
AI: $$\frac{dt}{dx}-6\sin t=10$$
For the integrating factor method you need to have a DE of the form:
$$y'+\alpha (x)y=\beta (x)$$
This is not the case. The DE is not linear.
$$\frac{dt}{dx}-6\sin t=10$$
The DE is separable. So method 2 is correct. |
H: Is this proof of $C[0,1]$ and $C[a,b]$ being isometric correct?
From the book Introductory Functional Analysis with Applications-Kreyszig:
Let $C[a,b]$ be the metric space of continuous, real valued functions defined on $[a,b]\subset \mathbb{R}$ with the metric $d(x,y)=\max_{t\in[a,b]}|x(t)-y(t)|$. Show that for any choice of $a,b\in \mathbb{R}$ with $a<b$, $C[0,1]$ and $C[a,b]$ are isometric.
By definition, two metric spaces are isometric if there exists an bijective isometry between the spaces, in other words a bijective mapping $T:X\to Y$ such that for all $x,y\in X$: $d_{X}(x,y)=d_{Y}(Tx,Ty)$.
My attempt: Between $[0,1],[a,b]\subset \mathbb{R}$ there exists an bijective mapping $f:[b,a]\to[0,1]$ defined by $f(x)=\frac{x-a}{b-a}$ whos inverse is $f^{-1}(x)=x(b-a)+a$. We define $T:C[0,1]\to C[a,b]$ by $Tx(t)=x(f(t))$. I claim that $T$ is a bijective isometry:
Injective: Suppose that $Tx=Ty$ then by the coincidence axiom $$d(Tx,Ty)=\max_{t\in [a,b]}|x(f(t))-y(f(t))|=0.$$ since $f$ is a bijection this implies that $$\max_{t\in[0,1]}|x(t)-y(t)|=0$$
that is $x=y$ and $T$ is thus injective.
Surjective: Assume that $y\in C[a,b]$ then $y(f^{-1}(t))\in C[0,1]$ and $T(y(f^{-1}(t)))=y(t)$ so $T$ is surjective and hence bijective.
Distance preserving: Let us suppose that $x,y \in C[0,1]$ and the maximum between $x$ and $y$ occurs at $t_{0}\in[0,1]$ then $f^{-1}(t_{0})=t_{0}(b-a)+a\in [a,b]$ and since for every $t\in [a,b]: f(t)\in[0,1]$
$$d(Tx,Ty)=\max_{t\in[a,b]}|x(f(t))-y(f(t))|=|x(f(f^{-1}(t_{0})))-y(f(f^{-1}(t_{0})))|=d(x,y)$$
Therefore we conclude that $T$ is an isometry between $C[0,1]$ and $C[a,b]$ and thus they are isometric spaces.
Question: is this correct?
PS. Any comments would also be helpful!
AI: Yes, it is correct. Note also that $T$ is a linear map, so we can really identify $C([a,b])$ with $C([0,1]$) as Banach spaces! One says that $C([a,b])$ and $C([0,1])$ are isometrically isomorphic Banach spaces. |
H: Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that these are the only solutions. Also I'm not sure that squaring three times is a good approach or not.
AI: Let $\sqrt[4]{x}=a$ and $\sqrt[4]{x+1}=b$.
Thus, $a\geq0,$ $b\geq1$, $b^4-a^4=1$ and $$a+b=\sqrt[4]{a^4+b^4}$$ or
$$(a+b)^4=a^4+b^4$$ or $$2ab(2a^2+3ab+2b^2)=0,$$ which gives $$ab=0.$$
Can you end it now. |
H: Probability of winning a ticket with a red dot
Question:
You have obtained some interesting information about the local lottery. There was a malfunction at the printer that accidentally marked a bunch of tickets with a red dot. This malfunction disproportionately affected winning lottery tickets. In total $40\%$ of winning tickets were marked with a red dot, while only $20\%$ of losing tickets were marked with a red dot. You have a probability of $\frac{3}{39}$ of winning the lottery.
You have found a ticket marked with a red dot. What is the probability that this is a winning ticket?
What i have done is the following:
A= probability of a winning ticket.
B= probability of having a red dot.
P(A|B)= P(A intersection B) / P(B)
P(B)= P(red dot| winning ticket)$\times$ P(winning ticket) + P(red dot| losing ticket)$\times$ P(losing ticket)
$0.4 \times \frac{3}{39} + 0.20 \times \frac{12}{13}= 0.215$
P(A intersection B)= P(A|B)$\times$P(B)= $0.4\times 0.215=0.086$
So P(A|B)= $\frac{0.086}{0.215}=0.4$
I'm not sure if I'm doing this right. Maybe someone can give feedback.
AI: Out of $39$ tickets, $3$ are winning tickets and $36$ are losing tickets.
So, number of winning tickets with red dot = $3\times 0.4$
Number of losing tickets with red dot = $36\times 0.2$
You need to find probability of winning of a ticket with red dot = $\dfrac{3\times 0.4}{3\times 0.4+36\times 0.2} = \dfrac{1}{7}$ |
H: Infinite product limit and estimate
I came across this product series in research and need to understand and estimate it. It appears to be unbound, but what would be the law?
$$\lim_{n \to \infty} \prod_{k=1}^{n}\frac1{(1 - \frac1{2k+1})}$$
Many thanks if you know the answer.
AI: The first thing I would do is write the product term to accurately reflect the fact that multiplication is taking place over odds:
$$\lim_{n \to \infty} \prod_{k=1}^{n}\frac{2k+1}{2k}=\lim_{n \to \infty} \prod_{k=1}^{n}\left(1+\frac{1}{2k}\right)$$
Regarding convergence, a useful theorem says if $a_k>0$, then $\sum a_k$ and $\prod (1+a_k)$ converge or diverge together: not to the same value, but one is finite if and only if the other is. So we can study
$$\frac{1}{2}
\sum _{k=1}^{\infty} \frac{1}{k},
$$a well-known series known to be divergent called the harmonic series. Your intuition is correct, i.e. the product is unbounded (does not converge). |
H: If $X_n$ converges to $X$ in probability, then for $f$ continuous, then $f(X_n)$ converges in probability to $f(X)$
The following is an exercise in the book Measure Theory and Probability, by Athreya and Lahire.
Let $X_n$ converge to $X$ in probability. If $f$ is continuous, then $f(X_n)$ converges in probability to $f(X)$
This seems like a simple exercise, but I haven’t been able to solve it. This is what I have tried so for.
Assuming that $f$ is uniformly continuous, then for
$$\epsilon > 0 \ \exists \delta >0 : \mid X_n(w) - X(w) \mid < \delta \implies \mid f(X_n(w)) - f(X(w)) \mid < \epsilon
$$
Hence, we know that $P(\mid X_n - X\mid < \delta ) \leq P(\mid f(X_n) - f(X) \mid < \epsilon )$, and since $X_n \rightarrow_p X$, then
$$
1 = \lim_{n \to \infty } P(\mid X_n - X\mid < \delta ) \leq \lim_{n\to \infty}P(\mid f(X_n) - f(X) \mid < \epsilon )
$$
Now, is the above solution correct? And how does one prove if $f$ is not uniformly continuous?
AI: Theorem: Let $(X_n)$ be a sequence of random variables. Then $X_n \to X$ in probability, iff for every subsequence $(n_k)$ there exists sub-subsequence $(n_{k_m})$ such that $X_{n_{k_m}} \to X$ almost surely.
Firstly lemma: If $X_n \to X$ in probability, then there exists subsequence $(n_k)$ such that $X_{n_k} \to X$ almost surely.
Proof of lemma: By convergence in probability, we have a subsequence $(n_k)$ such that $\mathbb P(|X_{n_k} - X| \ge \frac{1}{k^2}) \le \frac{1}{k^2}$. Hence $\sum_{k=1}^\infty \mathbb P(|X_{n_k} - X| \ge \frac{1}{k^2}) < \infty$ so by borel cantelli, almost surely we have $|X_{n_k} - X| < \frac{1}{k^2}$ starting from some $k>K$, so $X_{n_k} \to X$ almost surely.
Proof of theorem.
By Lemma we have => direction (since for any subsequence $(n_k)$ we have $X_{n_k} \to X$ in probability, too).
"<=" Assume contrary, that there $X_n \not \to X$ in probability. Hence by definition, there exists $\varepsilon >0, \delta >0$ and subsequence $(n_k)$ such that $\mathbb P(|X_{n_k} - X| > \varepsilon) > \delta$ for every $k \in \mathbb N$. But from that subsequence $(n_k)$ we cannot choose any sub-subsequence $(n_{k_m})$ such that $X_{n_{k_m}} \to X$ almost surely (because $X_{n_{k_m}} \not \to X$ in probability and this is necessary condition).
Having theorem we can proceed as follows in your question:
Let $f$ be continuous and $X_n \to X$ in probability. We want to show that $Y_n := f(X_n) \to f(X) =: Y$ in probability. Take any subsequence $(n_k)$. We know that there exists sub-subsequence $(n_{k_m})$ such that $X_{n_{k_m}} \to X$ almost surely. Hence $Y_{n_{k_m}} \to Y$ almost surely (cause continuous functions of pointwise convergent sequences are pointwise convergent). Since $(n_k)$ was arbitrary, by theorem we know that $Y_n \to Y$ in probability. |
H: Why does definition of the inverse of a matrix involves having $AB=I=BA$?
So, I was reviewing the first course in Linear Algebra which I took and got curious about the reason behind defining the inverse of a matrix in the following way (from Wikipedia):
In linear algebra, an $n$-by-$n$ square matrix $A$ is called invertible (also nonsingular or nondegenerate) if there exists an $n$-by-$n$ square matrix $B$ such that
$$
AB=BA=I
$$
Now, I had an exercise to prove that if $AB=I$, then $BA=I$. Then, what is the reason to put both the equalities in the definition? Is that somewhat traditional or is it because of some specific reason which I'm not aware of?
I'd be happy if someone could help me out.
Thanks in advance!
AI: We want to require $AB=I$ and $BA=I$ for any linear operators $A$ and $B$. The second is redundant for finite-dimensional spaces but not in general.
Say $V$ is the space of all one-sided sequences $x=(x_1,x_2,\dots)$. Define $A,B:V\to V$ by $$Ax=(x_2,x_3,\dots),$$ $$Bx=(0,x_1,x_2,\dots).$$Then $AB=I$ but $BA\ne I$.
So: We need the condition for the infinite-dimensional case, so the reason it's included in the definition in the finite-dimensional case is so the definition is the same in every vector space. |
H: Evalution of a function where $t = x + \frac{1}{x}$
Consider a function $$y=(x^3+\frac{1}{x^3})-6(x^2+\frac{1}{x^2})+3(x+\frac{1}{x})$$ defined for real $x>0$. Letting $t=x+\frac{1}{x}$ gives: $$y=t^3-6t^2+12$$
Here it holds that $$t=x+\frac{1}{x}\geq2$$
My question is: how do I know that $t=x+\frac{1}{x}\geq2$ ?
I want to know how to get to this point without previouly knowing that $t=x+\frac{1}{x}\geq2$
AI: Because by AM-GM $$x+\frac{1}{x}\geq2\sqrt{x\cdot\frac{1}{x}}=2.$$
Your calculation of $y$ is right:
$$y=t^3-3t-6(t^2-2)+3t=t^3-6t^2+12$$ and you got it without using $t\geq2$. |
H: How to Find Solutions to a Multivariate Polynomial System
I have a system of polynomials, where the first one is a multivariate linear polynomial, but the rest are univariate quadratic polynomials. How would I solve such a system (finding one or all solutions, or showing there are no solutions)? For example,
$$17x+16y-5z-67=0 \\ x^2+3x-5=0 \\ 4y^2-7y-4=0 \\ z^2-6z-3=0$$
AI: The system has no solution. This can be seen by computing a Groebner basis, for example. But also a direct approach is possible. We can compute $x,y,z$ from the second, third and last equation (two solutions each) and then substitute it into the first one. Even if we would replace the first equation by
$$
17x+16y-5z-a=0
$$
where $a$ is an integer, there is no solution. |
H: Some parts of the proof of downward Löwenheim–Skolem theorem I need to clarify
Downward Löwenheim–Skolem theorem states that, for every signature $\sigma$ of a first order language, every infinite $\sigma$-structure $\mathscr M$ with domain $M$ and every infinite cardinal number $\kappa \ge \vert\sigma\vert$, there is a $\sigma$-structure $\mathscr N$ with domain $N$, such that $\vert N\vert = κ$ and such that if $κ < \vert M\vert$, then $\mathscr N$ is an elementary substructure of $\mathscr M$.
So, here is a sketch of the proof, from Wikipedia:
For each first-order $\sigma$-formula $\phi(y,x_1,…,x_n)$, the axiom of choice implies the existence of a function $f_\phi:M^n\rightarrow M$ such that, for all $a_1,…,a_n\in M$, either $\mathscr M\vDash \phi \bigl(f_\phi(a_1,a_2,…,a_n),a_1,a_2,…,a_n \bigl)$ or $\mathscr M\vDash\lnot\exists y\phi(y,a_1,a_2,…,a_n)$.
The family of functions $f_\phi$
gives rise to a preclosure operator $F$ on the power set of $M$, $F(A) = \lbrace f
_\phi(a_1,a_2,…,a_n)\in M\ \vert\ \phi \in \sigma\ ;\ a_1,a_2,...,a_n \in A\rbrace$, for $A\subseteq M$.
Iterating $F$ countably many times results in a closure operator $F_\omega$. Taking an arbitrary subset $A\subseteq M$ such that $\vert A\vert =\kappa$, and having defined $N=F^\omega (A)$, one can see that also $\vert N\vert = \kappa$. Then, $\mathscr N$ is an elementary substructure of $\mathscr M$, by the Tarski–Vaught test.
What about formulas $\phi$ with only one variable? I guess then, functions $f_\phi$ are of zero arity, and thus, they are constants, is that right?
It is not clear to me, how $A\subseteq F(A)$, as is should, in order for $F$ to be a preclosure operator. In the definition of $F(A)$, only elements of M that satisfy some $\sigma$-formula are selected by the $f_\phi$'s. But, $A$, on the other hand, is an arbitrary subset of $M$ with cardinality $\kappa$. So, it may contains elements that do not satisfy any $\sigma$-formula, is that right? If it does, then, how do these elements get in $F(A)$, since they are not going to be selected by any $f_\phi$?
"Iterating $F$ countably many times", actually means that $F^\omega (A)= \lbrace a\in F^m(A)\ \vert\ m\in\Bbb N \rbrace$, is that right?
AI: For $(1)$ and $(3)$: yes, that's right.
For $(2)$: you're right that elements of $A$ need not be definable, and so we're not going to be able to get them in $F(A)$ via one-variable formulas (a la point $(1)$). However, we can get around this: consider the formula $$\varphi_{id}:\equiv y=x_1.$$ The corresponding $f_{\varphi_{id}}:M\rightarrow M$ is just the identity map, and so we do indeed get $F(A)\supseteq A$. |
H: Integration of Sign(x)
I'm currently trying to create a function for a coding project. It takes a function of the form $ax^{b}$ and integrates it up until a value $c$ beyond which the value is 0.
From this I played around in desmos and came up with the following function:
$$
\frac{1}{2}ax^b(1-\operatorname{sgn}(x-c))
$$
When I put the integral of this into WolframAlpha I get:
$$
-\frac{ax^{b+1}\left(\operatorname{sgn}\left(c-x\right)^{2}-4\operatorname{sgn}\left(c-x\right)-5\right)}{8\left(b+1\right)}
$$
When I type this into desmos however an odd thing happens. When $x > c$ the value of the integral should be constant, but non-zero. In this scenario it is constant but zero... Performing the calculations myself I can understand why the value is zero from the function supplied. Is this just a bug with wolfram or is my understanding of what the output should be incorrect?
Note: I am aware that i could use:
g(x) => x>c ? f(c) : f(x)
but it feels incomplete.
AI: Wolfram Alpha is calculating an indefinite integral--that is to say, an antiderivative of the function you provided. Such an expression is not unique and only needs to satisfy the property that its derivative equals the input function.
If instead you want $$F(x) = \int_{t=-\infty}^x f(t) \, dt, \tag{1}$$ so that $F$ is continuous, then you need to state this. For instance,
Integrate[Piecewise[{{0, t < 0}, {a t^b, 0 <= t <= c}, {0, t > c}}],
{t, -Infinity, x}, Assumptions -> {c > 0, b > 0, a > 0, Element[x, Reals]}]
Note that your expression $$f(x; a, b, c) = \frac{1}{2}a x^b (1 - \operatorname{sign}(x - c)) \tag{2}$$ lacks a condition to handle the case $x < 0$, in which case $(1)$ should read $$F(x) = \int_{t=0}^x f(t) \, dt. \tag{3}$$ Such an expression can be integrated properly in Mathematica with the command
Integrate[a t^b/2 (1 - Sign[t - c]), {t, 0, x},
Assumptions -> {c > 0, b > 0, a > 0, 0 <= x < Infinity}] |
H: derivative of multivariable recursive function
Given a recursive function
$$
f(x,y,z) = f(h(x),g(y,z),z)
$$
I want to get the derivative of the function to $z$
$$
{d\over dz } f(x,y,z) = ?
$$
My guess is
$$
{d \over dz}f(x,y,z) = f'(h(x),g(y,z),z) g'(y,z) {dy \over dz}
$$
But I'm not sure if I'm right, especially the last part $dy \over dz$ because the $y$ for the next call of $f$ is $g(y,z)$ and it is affected by $z$.
Am I wrong?
AI: You need to use the chain rule properly:
if $\phi(x, y, z) = f(h(x), g(y, z))$, we have:
$\frac{d \phi}{dz}(x, y, z) = \frac{\partial f}{\partial g}(h(x), g(y, z)) \frac{\partial g}{\partial z}(y, z)$. |
H: Prove $\sum_{k=1}^{\infty} \frac{{(-1)}^n}{k^2} \sum_{j=0}^{\infty} \frac{{(-1)}^j}{2k+j+1}=-\frac{\pi^2}{12}\ln{2}+\pi C-\frac{33}{16} \zeta(3)$
Prove $$\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2} \sum_{j=0}^{\infty} \frac{{(-1)}^j}{2k+j+1}=-\frac{\pi^2}{12}\ln{2}+\pi C-\frac{33}{16} \zeta(3)$$
where C is catalan's constant.
Wolfram Alpha confirms that the sums converge to approximately the right side. Wolfram Alpha also evaluates the first sum in terms of Hurwitz lerch transcendent or digamma function but how do I then evaluate the outer sum with either of these functions.
Original question is $$\int_0^1 \frac{\text{Li}_2(-x^2)}{1+x} \; \mathrm{d}x$$
and I've got it to the double sum here by writing Li as its series form and forming a geometric series with $\frac{1}{1+x}$.
Any tips or suggestions? maybe other approach to the integral?
Edit: Integration by parts may work better?
$$\ln{(1+x)}\text{Li}_2(-x^2) \bigg \rvert_0^1 + 2\int_0^1 \frac{\ln{(1+x)}\ln{(1+x^2)}}{x} \; \mathrm{d}x$$
Wolfram says that second integral is $\pi C -\frac{33 \zeta(3)}{16}$ which is very good here but I don't know how to evaluate that integral.
$$\int_0^1 \frac{2\ln{(1+x)}\ln{(1+x^2)}}{x} \; \mathrm{d}x=\int_0^1 \frac{\ln^2{(1+x)(1+x^2)}}{x} \; \mathrm{d}x-\int_0^1 \frac{\ln^2{(1+x)}}{x} \; \mathrm{d}x - \int_0^1 \frac{\ln^2{(1+x^2)}}{x} \; \mathrm{d}x$$
Last integral is 0
$$\int_0^1 \frac{2\ln{(1+x)}\ln{(1+x^2)}}{x} \; \mathrm{d}x=\int_0^1 \frac{\ln^2{(1+x)(1+x^2)}}{x} \; \mathrm{d}x-\int_0^1 \frac{\ln^2{(1+x)}}{x} \; \mathrm{d}x$$
AI: We have that
$$\sum_{j=0}^{\infty} \frac{{(-1)}^j}{2k+j+1}=\sum_{j=1}^{\infty} \frac{{(-1)}^{j+1}}{j}-\sum_{j=1}^{2k} \frac{{(-1)}^{j+1}}{j}=\ln 2-\sum_{j=1}^{2k} \frac{{(-1)}^{j+1}}{j}$$
then
$$\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2} \sum_{j=0}^{\infty} \frac{{(-1)}^j}{2k+j+1}
=\ln 2\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2}-\sum_{k=1}^{\infty} \frac{{(-1)}^n}{k^2}\sum_{j=1}^{2k} \frac{{(-1)}^{j+1}}{j}$$
with
$$\ln 2\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2}=-\frac{\pi^2}{12}\ln{2}$$
and using the results indicated here and here
$$\sum_{k=1}^{\infty} \frac{{(-1)}^k}{k^2}\sum_{j=1}^{2k} \frac{{(-1)}^{j+1}}{j}=\sum_{k=1}^{\infty} \frac{{(-1)}^kH_{2k}}{k^2}-\sum_{k=1}^{\infty} \frac{{(-1)}^k H_k}{k^2}= \frac{23}{16} \zeta (3) - \pi \mathbf{G}+\frac{5}{8}\zeta(3)$$ |
H: What's meant by the number of "distinct $C^k$ differential structures" other than the amount of distinct maximal atlases?
When reading the Wiki page on differential structures, I'm struck by the exceptional case of $R = 4$.
However, the definition of differential structure leaves me nonplussed, as it seems to just be another name for "maximal atlas" in the context of https://i.stack.imgur.com/MBce6.png, which says: "There is an "essentially unique" smooth structure for any topological manifold of dimension smaller than 4."
When they say "smooth structure" do they simply mean a unique maximal atlas (C-infinity) or is there more to it?
Thanks so much for helping with the clarification.
AI: You are on the right track, but not quite there yet.
First, a smooth structure is indeed nothing more than a maximal $C^\infty$ atlas.
However, to say that the smooth structure is "essentially unique" does not mean that there exists a unique maximal $C^\infty$ atlas. Such a strong uniqueness property is always false: for any smooth manifold $M$ there exists a non-smooth homeomorphism $h : M \to M$, and using $h$ you can then define a different maximal $C^\infty$ atlas by transport of structure, which means replacing your given atlas $\{\phi_i : U_i \to \mathbb R^m\}$ with the new atlas $\{\phi_i \circ h^{-1} : h(U_i) \to \mathbb R^m \}$. For example, even the standard structure on $\mathbb R$ can be changed by transport of structure using the non-smooth homeomorphism $h(x) = \sqrt[3]{x}$.
Instead, for the smooth structure to be "essentially unique" usually means that it is unique up to transport of structure: for any two maximal $C^\infty$ atlases $\{\phi_i : U_i \to \mathbb R^m\}$ and $\{\psi_j : V_j \to \mathbb R^m\}$ there exists a homeomorphism $h : M \to M$, and a bijection between the index sets, such that if $i \leftrightarrow j$ correspond under that bijection then $V_j = h(U_i)$ and $\psi_j = \phi_i \circ h^{-1}$. This, of course, is equivalent to the statement that $h$ is a diffeomorphism from the structure $\{\phi_i : U_ \to \mathbb R^m\}$ to the structure $\{\psi_j : V_j \to \mathbb R^m\}$. |
H: Why does $I(\overline{S})=I(S)$?
Let $S\subset \operatorname{Spec}A$, where $A$ is a commutative ring with $1$. Define $I(S)$ to be the set of functions vanishing on $S$. In other words, $I(S)=\bigcap_{P\in S}P\subset A$. Why is it true that $I(\overline{S})=I(S)$? Here $\overline{S}$ denotes the Zariski closure of $S$.
One inclusion is clear to me. Namely, since $I(\cdot)$ is inclusion reversing, we have $I(\overline{S})\subset I(S)$.
However, why is the reverse inclusion true?
AI: $T\subseteq \operatorname{Spec}A$ is a closed set iff it is of the form
$$
V(J) = \{P \in \operatorname{Spec}A\mid J\subseteq P\}
$$
for some ideal $J\subseteq A$. The closure $\overline S$ is the intersection of all such sets that contain $S$. And we have $S\subseteq V(J)$ iff $J\subseteq \bigcap S=I(S)$.
Now note that for any set $\{I_i\}_i$ of ideals in $A$, we have
$$
\bigcap_iV(I_i) = V\left(\sum_iI_i\right)
$$
So we are after the sum of all ideals that contain the intersection of $S$. But the intersection of $S$ is an ideal. So really, the closure of $S$ is just
$$
V\left(I(S)\right) = \{R\in\operatorname{Spec}A\mid{}I(S)\subseteq R\}
$$
From here, the conclusion follows immediately. |
H: Integrate over a set $B = \left\{ (x,y) \in \Bbb R^2: 2\leq x \leq y \leq 6 \right\}$.
How do I integrate over the following set?
$$B = \left\{ (x,y) \in \Bbb R^2: 2\leq x \leq y \leq 6 \right\}$$
This may seem trivial but I really am not sure how to find the bounds. I thought since $x \leq y$ it would indicate an area under a function $y = x$ where $x \in [2,6]$ and so from the graph I could determine $y \in [0, x]$ but I don't think this is correct since $2 \leq y$ so the bounds for $y$ should be $y \in [2,x]$ but I still am not sure whether this is correct. And also isn't it possible to rewrite the set so that I was able to see the bounds straight away?
AI: The condition $B = \left\{ (x,y) \in \Bbb R^2: 2\leq x \leq y \leq 6 \right\}$ ie equivalent to
$2 \leq y \leq 6$
$2\leq x \leq y $
then make a sketch of the lines
$y=2$, $y=6$
$x=2$, $x=y$
and the region $B$ is given by the the triangle between the lines $x=2$, $y=6$ and $x=y$. |
H: Checking Presentations in GAP
If I have the following presentation for $A_5$ $$\langle x,y,z\mid x^3 = y^3= z^3 =(xy)^2=(xz)^2= (yz)^2= 1\rangle$$ with subgroup $$ H = \left\langle {x,y} \right\rangle$$ and let GAP apply coset enumeration to my generators and relations, as with the code below, is there a command I can use to check whether this presentation is indeed for $A_5$?
<free group on the generators [ x, y, z ]>
gap> x:=F.x;
x
gap> y:=F.y;
y
gap> z:=F.z;
z
gap> rels:=[x^3,y^3,z^3,(x*y)^2,(x*z)^2,(y*z)^2];
[ x^3, y^3, z^3, (x*y)^2, (x*z)^2, (y*z)^2 ]
gap> G:=F/rels;
<fp group on the generators [ x, y, z ]>
gap> gens:=GeneratorsOfGroup(G);
[ x, y, z ]
gap> xG:=gens[1];
x
gap> yG:=gens[2];
y
gap> zG:=gens[3];
z
gap> H:=Subgroup(G,[xG,yG]);
Group([ x, y ])
gap> ct:=CosetTable(G,H);
[ [ 1, 3, 4, 2, 5 ], [ 1, 4, 2, 3, 5 ], [ 1, 3, 5, 4, 2 ],
[ 1, 5, 2, 4, 3 ], [ 2, 3, 1, 4, 5 ], [ 3, 1, 2, 4, 5 ] ]
gap> Display(TransposedMat(ct));
[ [ 1, 1, 1, 1, 2, 3 ],
[ 3, 4, 3, 5, 3, 1 ],
[ 4, 2, 5, 2, 1, 2 ],
[ 2, 3, 4, 4, 4, 4 ],
[ 5, 5, 2, 3, 5, 5 ] ]
I'm asking because I'm doing research where I will enter candidate presentations into GAP and check if the presentation is equal to a certain alternating group.
AI: Yes.
Use IdGroup(G);. If G is indeed (a presentation for a group isomorphic to) $A_5$, the output of this command is [60, 5]. |
H: Exercise with maximum and minimum between real numbers
Let $a, b, \alpha, \beta>1$ be four real numbers. Consider
$$\max\lbrace a^{-\alpha}, b^{-\beta}\rbrace.$$
I am looking for the right quantity C such that
$$\max\lbrace a^{-\alpha}, b^{-\beta}\rbrace\cdot C =1.$$
I guess that it is $\min\lbrace a^{\alpha}, b^{\beta}\rbrace$, but I don’t know how to prove it.
Could anyone please tell me if it is true and help me to prove it?
Thank you in advance!
AI: Let $u$ and $v$ be positive real numbers. Then $\max(u,v)^{-1}=\min(u^{-1},v^{-1})$.
This is since if say $u<v$ then $\max(u,v)=v$ and $\min(u^{-1},v^{-1})=v^{-1}$
etc. |
H: absolute convergence of the series, $\sum_{n=1}^\infty \frac{nz^{n-1}\{(1+n^{-1})^n-1\}}{(z^n-1)\{z^n-(1+n^{-1})^n\}}$
We need to prove the absolute convergence of the series, $\sum_{n=1}^\infty \frac{nz^{n-1}\{(1+n^{-1})^n-1\}}{(z^n-1)\{z^n-(1+n^{-1})^n\}}$.
Since
\begin{align*}
\sum_{n=1}^\infty \frac{nz^{n-1}\{(1+n^{-1})^n-1\}}{(z^n-1)\{z^n-(1+n^{-1})^n\}}=\sum_{n=1}^\infty \frac{n\{(1+n^{-1})^n-1\}}{(z-\frac{1}{z^{n-1}})\{z^n-(1+n^{-1})^n\}}\\
=\sum_{n=1}^\infty \frac{n\{(1+n^{-1})^n-1\}}{z^{n+1}-(1+(1+n^{-1})^n)z+\frac{(1+n^{-1})^n}{z^{n-1}}},
\end{align*}
So if $|z|>1$, then the series $\approx \sum_{n=1}^\infty \frac{n\{(1+n^{-1})^n-1\}}{z^{n+1}}$, whose radius of absoulute convergence is 1, and so the series converges absolutely;
if $|z|<1$, then the series $\approx \sum_{n=1}^\infty \frac{n\{(1+n^{-1})^n-1\} z^{n-1}}{(1+n^{-1})^n}$, whose radius of absoulute convergence is 1, and so the series converges absolutely.
But the problem states that when $z=(1+\frac{1}{m})e^{2k\pi i/m}$ where $k=0,1,\dots, m-1; m=1,2,3,\dots$, we have $|z|>1$ and the series doesn't converge absolutely, which contradicts the above proof. Why?
AI: In the original expression
$$\sum_{n=1}^\infty \frac{nz^{n-1}\{(1+n^{-1})^n-1\}}{(z^n-1)\color{red}{\{z^n-(1+n^{-1})^n\}}}$$
we need
$$z^n-(1+n^{-1})^n \neq 0 \implies z^n \neq \left(1+\frac1n\right)^n \implies z\neq \left(1+\frac{1}{m}\right)e^{2k\pi i/m}$$
for $m=1,\ldots,m-1$ with $m=1,2,\ldots$. |
H: If $K$ is compact then $K\cap Y$ is compact in $Y$ too for any closed $Y\subseteq X$
Definition
A subspace $K$ of a topological space $X$ is compact if every its open cover has a finite subcover.
Lemma
If $X$ is compact then any its closed subspace is compact too.
Proof. Omitted.
Theorem
If $K$ is compact and closed then $K\cap Y$ is compact in $Y$ too for any closed $Y\subseteq X$
Proof. If $K$ is compact and $Y$ is closed then $K\cap Y$ is closed in $K$ so that by previous lemma it is compact too in $K$. Now if $\mathcal{T}$ is the topology of $X$ then we observe that any $A\in\mathcal{T}$ is such that $$(A\cap K)\cap(K\cap Y)=A\cap(K\cap Y)=(A\cap Y)\cap(K\cap Y)$$ so that we conclude $\mathcal{T}|_K|_{K\cap Y}=\mathcal{T}|_{K\cap Y}=\mathcal{T}|_Y|_{K\cap Y}$ and so the theorem holds.
So I ask if the proof of the last theorem is correct and if it is not correct I ask how to prove it: perhaps could it be that the statement is false?
AI: Let $\Omega$ be a topological space and $A$ a subspace of $\Omega$. Then if $F$ is a subset of $A$, $F$ is compact in $A$ iff $F$ is compact in $\Omega$.
Now using the lemma $K\cap Y$ is a compact subset of $K$. The iff condition implies that $K\cap Y$ is compact in $X$ and again the iff condition implies that $K\cap Y$ is compact in $Y$ |
H: Finding the image of a matrix given two examples of transformations
I was given the following question:
Let $e_1=\left(\begin{matrix}1\\0\\\end{matrix}\right)$ and $e_2=\left(\begin{matrix}0\\1\\\end{matrix}\right)$, $y_1=\left(\begin{matrix}3\\5\\\end{matrix}\right)$ and $y_2=\left(\begin{matrix}-1\\8\\\end{matrix}\right)$. Let $T:\mathbb{R^2}\rightarrow\mathbb{R^2}$ be a linear transformation that maps $e_1$ to $y_1$ and $e_2$ to $y_2$. Find the image of $\left(\begin{matrix}4\\-3\\\end{matrix}\right)$.
It seems that I am supposed to find the matrix that results in the given transformations, and then apply that to the given vector. Is there a method to doing this? Am I misunderstanding the problem?
AI: Yes, your thought on how to approach it is correct. Every linear transformation is uniquely determined by its action on a basis. In this case, suppose the matrix of the transformation is $A = \left( \begin{array}{cc} a & b \\ c & d \end{array}\right)$. So you know
$$
\left( \begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{c} 1 \\ 0 \end{array}\right) = \left
(\begin{array}{c} 3 \\ 5 \end{array}\right)
$$
Does that tell you anything about the matrix $A$? |
H: How to calculate $ \int_0^\infty \exp(-\frac{a^2}{x^2}-x^2)~\mathrm{d}x $
suppose $a>0$, how to integrate:
$$
\int_0^\infty e^{-a^2/x^2}e^{-x^2}~\mathrm{d}x
$$
AI: In general,
\begin{align}
\int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right) dx
&= \int_{-\infty}^{0}f\left(x-\frac{1}{x}\right) dx+ \int_{0}^{\infty}f\left(x-\frac{1}{x}\right) dx\\
& \overset{t=-\frac1x} = \int^{\infty}_{0}f\left(t-\frac{1}{t}\right) \frac {dt}{t^2}+ \int^{0}_{-\infty}f\left(t-\frac{1}{t}\right) \frac {dt}{t^2}\\
&= \frac12\int_{-\infty}^{0}f\left(t-\frac{1}{t}\right)\left( 1+\frac1{t^2}\right)dt + \frac12\int_{0}^{\infty}f\left(t-\frac{1}{t}\right)\left( 1+\frac1{t^2}\right)dt\\
& \overset{x=t-\frac1t } =\int_{-\infty}^{\infty}f(x)dx
\end{align}
Thus,
$$
\int_0^\infty e^{-a^2/x^2}e^{-x^2}dx
\overset{ t=\frac x{\sqrt a}}=\frac12\sqrt{a}e^{-2a }\int_{-\infty}^\infty e^{-a(t-\frac1t)^2 }dt\\
= \frac12\sqrt{a}e^{-2a }\int_{-\infty}^\infty e^{-a t^2 }dt
= \frac12\sqrt{a}e^{-2a } \cdot \sqrt{\frac\pi a}= \frac12\sqrt{\pi}e^{-2a }
$$ |
H: Finding the distribution of a sequence of random variables
I'm having some trouble with this problem:
Let $(X_n)_{n\ge1}$ a sequence of random variables that for all ${n\ge1}$, $X_n$ follows an exponential law with parameter $1/n$. Let $Y_n = X_n - \left\lfloor X_n\right\rfloor$.
$(Y_n)_{n\ge1}$ converges in distribution to which law?
I know for sure it is a Beta distribution, but I don't know the parameters. How should I go about it?
AI: You can try to calculate cumulative distribution function of $Y_n$. Indeed, $Y_n$ takes values $(0,1)$ almost surely. Hence for $t \in (0,1)$ we have:
$$ \mathbb P(Y_n \le t) = \mathbb P(X_n \in (k,k+t] ; k \in \mathbb N) = \sum_{k=0}^\infty \int_k^{k+t} \frac{1}{n}\exp(-\frac{1}{n}x)dx = \sum_{k=0}^\infty -\exp(-\frac{1}{n}x) \Big |_{x=k}^{x=t+k} = \sum_{k=0}^\infty \exp(-\frac{k}{n}) - \exp(-\frac{k}{n} - \frac{t}{n}) = \sum_{k=0}^\infty \exp(-\frac{k}{n})(1-\exp(-\frac{t}{n}) = (1-\exp(-\frac{t}{n}))\sum_{k=0}^\infty (\exp(-\frac{1}{n}))^k $$
So that $$F_{Y_n}(t) = \begin{cases} 0 & t \le 0 \\ \frac{1-\exp(-\frac{t}{n})}{1-\exp(-\frac{1}{n})} & t \in (0,1) \\ 1 & t \ge 1 \end{cases}$$
As $n \to \infty$ we have $$F_{Y_n}(t) \to \begin{cases} 0 & t \le 0 \\ t & t \in (0,1) \\ 1 & t \ge 1 \end{cases}$$
So that $Y_n \Rightarrow Y$, where $Y$ is uniformly distributed on $(0,1)$
Calculation of limit: Note that $\exp(x) = 1 + x + o(x)$, hence
$$ \frac{1-\exp(-\frac{t}{n})}{1-\exp(-\frac{1}{n})} = \frac{\frac{t}{n} + o(\frac{1}{n})}{\frac{1}{n} + o(\frac{1}{n})} = \frac{t + o(1)}{1 + o(1)} \to t$$ |
H: Example of an omitted type taken from Hodges' book
So when reading section 6.2 of Hodges' A shorter model theory I came across this example:
Now, what does it mean precisely "by symmetry"? What is the argument he is trying to make?
I tried to prove on my own that for each $s\subseteq\omega$ there exists a countable model of $T$ that omits $\Phi_s$, but the only "straightforward" way I came up to do it is by constructing explicit models with $\omega$ as the domain, starting from a "ground" model in which every element (natural number) satisfy the $i$-th relation $P_i$ according to the $i$-th digit of its binary expansion and then adjusting it according to the type I want to omit.
But of course I have the feeling that there is a more succinct and elegant way of proving this. Any hint?
Thanks!
AI: The symmetry Hodges is referring to is the symmetry between $P_i$ and $\lnot P_i$ in the axioms of $T$, for all $i\in \omega$.
Let $M$ be a model of $T$. Now pick some $X\subseteq \omega$, and define a structure $M_X$ by starting with $M$ and, for each $i\in X$, replacing the interpretation $P_i^M$ of the relation symbol $P_i$ by its complement $M\setminus P_i^M$.
Then $M_X\models T$ "by symmetry". For each axiom $\varphi\in T$, there is a corresponding axiom $\varphi_X\in T$, obtained by replacing $P_i$ by $\lnot P_i$ and replacing $\lnot P_i$ by $P_i$, for each $i\in X$. Since $M\models \varphi_X$, it follows that $M_X\models \varphi$.
Now if we want to omit a given type $\Phi_s$, we pick an arbitrary countable model $M$ and note that $M$ must omit some type $\Phi_t$. Letting $X = s\triangle t$ (the symmetric difference of $s$ and $t$), we have $M_X\models T$ and $M_X$ omits $\Phi_s$. |
H: Let $a,b \in \Bbb C$, show that $\sum_{k=0}^{n-1} |a+w^kb|=\sum_{k=0}^{n-1} |b+w^ka|$ with $w=\exp\bigg(\dfrac{2i\pi}{n}\bigg)$
Let $a,b \in \Bbb C$, and let's denote $w=\exp\bigg(\dfrac{2i\pi}{n}\bigg)$
Show that for every $n\ge 2$ $$\sum_{k=0}^{n-1} |a+w^kb|=\sum_{k=0}^{n-1} |b+w^ka|$$
$\bullet~$ My attempts:
I tried to compare the two modules $|a+w^kb|,|b+w^ka|$ for each $k$ and I noticed that $|w|^k=1$ could help me figure it out through some algebraic manipulation but I can't find it. It's easy to show that $$|w^kb-w^ka|=|b-a|$$ But I can't bring each term to the other side because of the module.
Any help is greatly appreciated!
AI: Note that $\lvert a+w^kb\rvert=\lvert w^k(b+aw^{n-k})\rvert=\lvert b+aw^{n-k}\rvert$ and so when you sum over $k$ from $0$ to $n-1$, $k$ and $n-k$ hit the same set of values (modulo $n$, which is all that matters for the exponent of an $n$th root of unity). |
H: Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ for a natural number $ n \geq 2$?
Under what conditions does $ \ (a+b)^{n}=a^{n}+b^{n}$ holds for a natural number $ n \geq 2$?
My attempt at solving:
Using $(a+b)^2=a^2+2ab+b^2$; if $(a+b)^2=a^2+b^2$, $2ab=0$ therefore $a$ and/or $b$ must be $0$.
If $a$ and/or $b$ is $0$ then $a^2$ and/or $b^2$ will be $0$.
Therefore $(a+b)^2$ can never equal $a^2+b^2$.
AI: For even $n$ you are right. For odd $n$ we have also $a+b=0$. |
H: Difference between $\det(A) \neq 0$, the columns are linearly independent, and the rows are linearly independent?
The question is such that,
$A$ is an $m\times n$ matrix.
$x$ is an $n$ vector, $b$ is a $m$ vector.
We want the condition that ensures the existence of a solution for $$Ax=b$$
The options had,
$\det(A) \neq 0 $
the columns of $A$ are linearly independent
the rows of $A$ are linearly independent
So far from what I have studied from Anton's Linear Algebra, all these 3 options are equivalent and seem correct. I cant seem to figure out difference between them and also of course the answer for this question. I see that we can reject the option $1.$ on the basis that $A$ is not a square matrix and thus its determinant does not exist. But I can't figure out anything beyond this.
AI: They are equivalent for square matrices.
For non-square matrices #1 makes no sense and #2 and #3 are different (in fact it is impossible for #2 and #3 to both be true for a non-square matrix).
Regarding the correct condition for existence of a solution to $Ax=b$ for any $b$, this requires that row operations will not produce a row of zeros in the un-augmented $A$. Otherwise, if $b$ is chosen appropriately, the augmented $A$ will reduce to have a row of the form $0,0,\dots,0,c$ where $c \neq 0$, which encodes the inconsistent equation $0=c$. This means the rows of $A$ must be linearly independent.
One can test the rows of $A$ for linear independence by doing elimination; one can test the columns for linear independence by doing elimination to $A^T$. Of course you really only have to do one or the other, considering the properties of the rank of a matrix.
Note that on the reverse side, the condition for uniqueness of a solution if one exists is that the columns are linearly independent. This is actually quite intuitive: given $x \neq y$ such that $Ax=b,Ay=b$ you have $A(x-y)=0$, and now you have a nontrivial linear combination of columns of $A$ adding up to zero. |
H: Finding sum of expressions involving coefficients of terms in the expansion $(1+x+x^2)^n$
We take:
$$(1+x+x^2)^n=a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_{2n}x^{2n}$$
and we need to find the values of the expressions:
$$i)a_1+a_4+a_7+a_{10}+\cdots$$
$$ii)a_0-a_2+a_4-a_6+\cdots$$
I have solved similar expressions for eg.
$$1) a_0+a_2+a_4+\cdots$$
$$2) a_1+a_3+a_5+\cdots$$
$$3) a_0+a_3+a_6+\cdots$$
by using simple substitutions like $x=1, x=-1, x=\omega, x=\omega^2$ etc. but in these two expressions I'm completely stumped as I've tried using combinations of previous substitutions to create the kinds of expressions in part $i)$ and $ii)$ but was unsuccessful.
Any help would be appreciated.
AI: Let $f(x) = (1 + x + x^2)^n$, and let $g(x) = f(x)/x$.
For 1, consider $g(1) + g(\omega) + g(\omega^2)$, where $\omega = e^{2 \pi i/3}$.
For 2, consider $f(i) + f(-i)$. |
H: Proof of the following equality with vectors
Let $\{v_1,v_2,\dots,v_n\}$ be an orthogonal set in $V$, and let $a_1,a_2,\dots,a_n$ be scalars. Prove that
$$\left\Vert \sum_{i=1}^na_iv_i \right\Vert^2=\sum_{i=1}^n|a_i|^2\Vert v_i\Vert^2$$
Here's what I've tried, but I don't know if it is correct: $$\left\Vert \sum_{i=1}^na_iv_i \right\Vert^2=\left<\sum_{i=1}^n a_iv_i, \sum_{i=1}^n a_iv_i\right>=\sum_{i=1}^na_i\left<v_i, \sum_{i=1}^na_iv_i \right>=\sum_{i=1}^na_i\left(\sum_{i=1}^n\overline{a_i} <v_i,v_i>\right)=\sum_{i=1}^na_i\left(\sum_{i=1}^n\overline{a_i}\ \Vert v_i\Vert^2 \right)$$
I think that when proving this, I should also relate to the fact that the vectors are orthogonal.
AI: I think the only mistake in your work is the fact that the two summations are linked.
Instead, you should have
\begin{align}
\left\| \sum^n_{j=1} a_j v_j\right\|^2 =&\ \left\langle \sum^n_{j=1} a_j v_j, \sum^n_{i=1} a_i v_i\right\rangle\\
=&\ \sum^N_{i=1}\sum^N_{j=1} \bar a_ja_i\langle v_j, v_i\rangle \\
=&\ \sum^N_{i=1}\sum^N_{j=1} \bar a_ja_i \delta_{i, j}\|v_i\|^2 = \sum^N_{j=1} |a_j|^2\|v_i\|^2.
\end{align} |
H: Wrong proof: in a ring $R$ such that $r^n=r$ for every $r\in R$, there are no non-trivial ideals.
The claim is
Let $R$ be a commutative ring with an identity element $1\not=0$ with the property that for every $r\in R$ there is an $n\geq 2$ such that $r^n=r$. Then there are no non-trivial ideals.
My proof which I'm convinced must be wrong:
Let $r\in R/\{0\}$. Since $r^{n-1}r=r=rr^{n-1}$, and $1$ is unique, we must have $r^{n-1}=1$. Now let us look at the ideal $(r)=\{a\cdot r\mid a\in R\}$ of $R$. Since $r^{n-2}\in R$, then $r^{n-2}\cdot r= r^{n-1}=1\in(r)$. This implies that $(r) = R$.
Now let $I$ be an ideal of $R$ and suppose $I\not=\{0\}$. Then there is an element $r\in I$. We will show that $(r)\subseteq I$. Indeed, take $g\in(r)$. Then there is an element $a\in R$ such that $g = a\cdot r$. Since $I$ is an ideal, $g\in I$. Hence $(r) = R\subseteq I$, but this implies that $I = R$.
I am convinced this claim is wrong (and hence the proof must be too) since I am asked to solve an exercise which cannot be solved if this proposition is true.
AI: How do you know that $R$ is an integral domain? We have $(r^{n-1}-1)r = 0,$ but that doesn't mean either $r = 0$ or $r^{n-1} = 1.$ You have only shown the result is true if $R$ is an integral domain. |
H: Incomplete information of player’s choice in Prisoner’s Dilemma
What happens if the players in a prisoner’s dilemma or stag hunt game don’t always have control over their choices?
Instead of deciding to cooperate or defect the players have to draw from a deck. There are an equal number of cooperate and defect cards and each player picks two. They must each play one card. 50% of the time (they have both cards) they can make their own decision about what to do. But the other times it is not their choice since they have two of the same cards.
Does this do anything to the optimal strategy? Does it just take longer to come to equilibrium as you try to guess if they really meant to defect or they were forced to?
AI: The strategy does not change. The key point is from either player's perspective it doesn't matter if the other player intended to defect or was forced to, it is always better to defect regardless of the choice of the other player or how they made the desision. |
H: Conditions under which a series converges to another.
In a proof I'm reading they seem to be using a claim like this:
Let $(a_{s,n})_{s,n}$ and $(b_n)_n$ be sequences of real numbers. Suppose $\sum_n b_n$ converges and $\sum_n a_{s,n}$ converges for each fixed $s$. Suppose further that for fixed $n$ we have $a_{s,n} \rightarrow b_n$ as $s \rightarrow \infty$. Then we may conclude that
$$\sum_n |a_{s,n}-b_n| \rightarrow 0$$as $s \rightarrow \infty$.
In particular $$\Big|\sum_n a_{s,n} - \sum_n b_n \Big| \rightarrow 0$$ as $s \rightarrow \infty$,
Is it true? How can I prove it?
AI: This is not true. Consider $b_n$ a summable sequence and define $a_{s,n}= b_n$ for $s\neq n$ and $a_{n,n}=b_n+1$.
Then $\sum_na_{s,n}$ is convergent and $a_{s,n}\to b_n$ as $s\to \infty$. But, for any $s$,
$$\sum_n |a_{s,n}-b_n|=\sum_n a_{s,n}-b_n= \sum_n a_{s,n}-\sum_nb_n= \left|\sum_n a_{s,n}-\sum_nb_n\right| = 1$$ |
H: show that $v_n \leq 2u_n$
let $(u_n)_{n \geq 1}, \, (v_n)_{n \geq 1}$ such that :
$$\forall n\geq1, \,\,\,v_n \leq \frac{u_n}{(1-u_n)^2}$$
and $u_n \to 0$, prove that for $n$ large enough we have $v_n \leq 2u_n$.
now intuitively speaking I know why this would be the case, as $(1-u_n)^2$ would be close to $1$ and thus less than $2$, but how would a formal proof of this look like ? follow up question : what is the mathematical definition of "large enough" ?
AI: Since $u_n \to 0,$ there exists $N$ such that $|u_n| \le 1/4$ for all $n \ge N.$ Then whenever $n \ge N, u_n \le 1/4 \Rightarrow (1-u_n)^2 \ge (3/4)^2 = 9/16 \Rightarrow v_n \le \frac{16}{9} u_n \le^* 2u_n.$
Large enough means exactly what I just used: A statement holds for "$n$ large enough" if there exists $N$ such that it holds for all $n \ge N.$
*The statetement you are trying to prove is false unless we assume $\{u_n\}$ is a non-negative sequence. Else, let $u_n = -1/n, v_n = \frac{u_n}{(1-u_n)^2} = -\frac{n}{(1+n)^2}.$ The inequality $v_n \le 2u_n$ transforms to $\frac{n^2}{(n+1)^2} \ge 2,$ which is actually false for all $n.$ |
H: Is this proof of Bernoulli’s inequality correct?
I have to prove $(1+x)^{n} >1+nx$ for $n=2,3,4.... $and $x>-1$ and $x$ isnt 0. There are a lot of proofs of this but l want to know if this one works. If not, can u show where my reasoning is weak.
If $x>-1$ then $1+x>0$
Hence $ (1+x)^{n}>0$ for $n>=2$ and $x>-1$ . For $1+nx$ . I can show that $1+nx=0$ for $n=2,x=-.5$
Since $nx+1>0$ isn't true for a single pair (n,x) in the list of all pairs, the statement that it is true for all pairs of (n,x) is also false.
So $(1+x)^{n}>1+nx$ for $n=2,3,4.... $and $x>-1$.
AI: Note that your proof is not valid, since we are not trying to disprove anything - so a counterexample approach won't work. "It's not true for a single pair, so it's not true for all pairs" is the incorrect reasoning. That's like saying "every car I've seen today has been red, so every car is red". It doesn't work.
Instead, you could prove the statement by induction. Take arbitrary ${x>-1}$ and ${n \in \{2,3,4,5,6...\}}$. Then for ${n=2}$, the statement is clear
$${(1+x)^2 = 1 + 2x + x^2 \geq 1 + 2x}$$
(since ${x^2\geq 0}$ for any ${x}$). Hence the statement ${(1+x)^n \geq 1+nx}$ is true for ${n=2}$. Now assume that for some ${n=k}$ where ${k \in \{2,3,4,...\}}$ that
$${(1+x)^k \geq 1+kx}$$
Then notice
$${(1+x)^{k+1}=(1+x)(1+x)^k\geq (1+x)(1+kx)=1+kx+x+kx^2 = 1+(k+1)x + kx^2}$$
And again since ${kx^2}$ will be ${\geq 0}$ we have
$${1+(k+1)x + kx^2 \geq 1+(k+1)x}$$
So
$${(1+x)^{k+1}\geq 1+(k+1)x}$$
Since we have shown that the statement is true for ${n=k+1}$ if it's true for ${n=k}$, and we have shown it's true for ${n=2}$ we have the statement is true also for ${n=3,4,5....}$ and so by the Principle of Mathematical Induction we now have that
$${(1+x)^{n}\geq 1+nx}$$
for ${n \in \{2,3,4,5,6,...\}}$ and ${x>-1}$, as required. |
H: Proof of $\sum_{k=1}^n \left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$
Proof of $\displaystyle\sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$
The following proof is from a book, however, there is something that I don't quite understand
for $k\geq 2$ we have:
(1): $\displaystyle\frac{1}{k^2}\le \frac{1}{k\left(k-1\right)}$
so
(2):$\displaystyle \sum _{k=1}^n\left(\frac{1}{k^2}\right)\le \sum _{k=2}^n\left(\frac{1}{k\left(k-1\right)}\right)$
(3):$\displaystyle \sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:1+\left(1-\frac{1}{n}\right)$
(4):$\displaystyle\sum_{k=1}^n\left(\frac{1}{k^2}\right)\le \:\:2-\frac{1}{n}$
I don't understand why in line (2), on the right, we start the sum at $k=2$ while
on the left, we start on $k=1$. Why start at $k=2$?
AI: Indeed we have
$$\begin{aligned}\sum _{k=1}^n\left(\frac{1}{k^2}\right)&=1+\sum _{k=2}^n\left(\frac{1}{k^2}\right) \\
&\le 1+\sum _{k=2}^n\left(\frac{1}{k\left(k-1\right)}\right) \\
&=1+\sum _{k=2}^n\left(\frac1{k-1}-\frac1k\right) \\
&=2-\frac1n\end{aligned}$$ |
H: Why does $V(I(S))=\overline{S}$?
Let $S\subset\operatorname{Spec}A$, where $A$ is a commutative ring with $1$. I am having trouble seeing why $V(I(S))=\overline{S}$, where $\overline{S}$ is the Zariski closure of $S$.
My attempt is as follows. It is not hard to see that $S\subset V(I(S))$: $V(I(S))$ is the set of all prime ideals of $A$ containing $I(S)$, and $I(S)$ is the intersection of all elements (prime ideals) of $S$. Since every element of $S$ contains $I(S)$, it follows that $S\subset V(I(S))$. This implies that $\overline{S}\subset V(I(S))$.
The issue I'm having is proving the reverse inclusion. Suppose $V(J)$ is any closed set containing $S$. Then $I(V(J))\subset I(S)$ since $I(\cdot)$ is inclusion reversing. I'm not sure where to go from here or even if this is the correct line of thought to have. What am I missing?
AI: The idea of hitting $V(J)$ with $I$ is a good idea and can probably be made to work, but I think it's more straightforward to reason directly with $V(J)$ and $J$. It's a matter of unwinding the definitions completely:
Suppose $V(J)\supset S$ as you said. What is $V(J)$? It is the set of all primes containing $J$, where $J$ is some ideal in $A$. What does $S\subset V(J)$ tell us about the relationship between $J$ and the prime ideals that are members of $S$? Well, $\mathfrak{p}\in V(J)$ means $\mathfrak{p}\supset J$, by definition. So $S\subset V(J)$ means that for all $\mathfrak{p}\in S$, we have $J\subset \mathfrak{p}$. Thus, intersecting over all these primes, we get
$$J\subset \bigcap_{\mathfrak{p}\in S} \mathfrak{p}.$$
Recalling that $I(S)$ is nothing but the right side of this, we have
$$J\subset I(S).$$
But then this means that any prime ideal containing $I(S)$ also contains $J$! In other words, any member of $V(I(S))$ is also a member of $V(J)$! This is the assertion that
$$V(I(S))\subset V(J).$$
So any closed set containing $S$ actually contains $V(I(S))$, and it follows that $V(I(S))\subset \overline S$, as desired.
(To me it feels like saying the same sentence over and over again in slightly different language until it becomes the desired statement.) |
H: Suppose that $f$ is surjective and relation preserving. Then $\mathcal{R}$ is reflexive iff $\mathcal{S}$ is reflexive.
This is a problem about relations from Proofs and Fundamentals by Ethan D. Bloch that I’m having some doubts and I would really appreciate if you could guide me.
The problem starts with the following definition:
Definition: Let $A,B$ be sets, and let $\mathcal{R}, \mathcal{S}$ be relations on $A$ and $B$, respectively. Let $f: A \rightarrow B$ be a map. We say that $f$ is relation preserving if and only if $xRy$ iff $f(x)Sf(y)$ for all $x, y \in A$.
After this we have the following result:
Result: Suppose that $f$ is surjective and relation preserving. Then $\mathcal{R}$ is reflexive, symmetric or transitive iff $\mathcal{S}$ is reflexive, symmetric or transitive (respectively).
I attempt to show the reflexive part. And I have the following:
Proof: $\implies$. Suppose that $\mathcal{R}$ is reflexive. By definition, we have that $aRa$ for all $a \in A$. Since $f$ is relation preserving, it follows that $f(a)Sf(a)$. Since $f$ is surjective, we know that for all $b \in B$, there existis an element $a \in A$ such that $b = f(a)$. From this we know that $f(a)Sf(a)$ for all $f(a) \in B$. Hence, by definition, $\mathcal{S}$ is reflexive.
$\Longleftarrow$. Suppose that $\mathcal{S}$ is reflexive. By definition, we have that $bSb$ for all $b \in B$.
Although, I don’t know what the next step should be. How can I deduce that $\mathcal{R}$ is reflexive?
Also, will the proof for the other two properties be similar to this one?
Thank you for your attention!
AI: Suppose that $\mathcal{S}$ is reflexive, let $a\in A$, and let $b=f(a)$. Since $f$ is relation-preserving, $a\,\mathcal{R}\,a$ iff $f(a)\,\mathcal{S}\,f(a)$, i.e., iff $b\,\mathcal{S}\,b$. And $\mathcal{S}$ is reflexive, so $b\,\mathcal{S}\,b$, and therefore $a\,\mathcal{R}\,a$. Thus, $\mathcal{R}$ is reflexive.
Yes, the other two properties can be proved in much the same fashion. |
H: Pathwise Connectification of Spaces
For any space $X$, let $Y=X\times I$, and topologize $Y$ by defining basic neighbourhoods of $(x,y)$ as -
$(x,y), y\neq0: U_{(x,y)} = \{x\}\times B_\epsilon(y)$
$(x,0): U_{(x,0)} = \{(x',z):z'\in U, 0\leq z < \epsilon_z \},$ where $U$ is a neighbourhood of $x$ in $X$, and $\epsilon_z > 0\ $ $\forall z\in U$
Let $X^*$ be the quotient space of $Y$ on identifying all points $(x,1), x\in X$.
Then, show that $X^*$ is path connected, and that every continuous function $f:X\to Z$ can be extended to continuous $F:X^*\to Z$, where $Z$ is path connected.
I've been able to show that $X^*$ is connected. However, I've failed to show the extension property. Any help would be appreciated!
AI: HINT: Let $p$ be the point at the ‘top’. Pick any base point $x_0\in X$, and let $z_0=f(x_0)$. Let $F(\langle x_0,y\rangle=z_0$ for $y\in[0,1)$, and let $F(p)=z_0$ as well. For each $x\in X\setminus\{x_0\}$ there is a path in $Z$ from $f(x)$ to $z_0$; define $F$ on $\{x\}\times[0,1)$ so that it follows that path. To show continuity, use the fact that the sets $\{x\}\times(0,1)$ in $X^*$ are pairwise disjoint and open. |
H: What is the opposite of "coprime integers"?
What do you call two integers that are not relatively prime? In my language, there is a clear term for that, but I can't seem to find one in English.
AI: There isn't a standard term for that notion.
"Integers with a nontrivial common factor" is probably the best you can do without symbols. |
H: When does the inequality hold?
I am trying to find a condition on $c$ such that the below inequality holds true
$$ \frac{1 - e^{-st}}{st} - \frac{1}{st+c} > 0 $$
where $s$, $c$ and $t$ are greater than $0$. I tried simpyfing it and got $c > (c + st) e^{-st}$, but I am not sure what do next.
AI: We have
$$\frac{1 - e^{-st}}{st} - \frac{1}{st+c} > 0 \iff e^{-st}<1-\frac{st}{st+c}=\frac{c}{st+c}$$
$$\iff e^{st}>1+\frac{st}{c}$$
which is always true for $c\ge 1$.
To prove that, let consider $f(x)=e^x-1-\frac x c$ with $f(0)=0$ and show that
for $c\ge 1 \implies \forall x>0\: f(x) >0 $
for $0<c< 1 \implies \exists x_0>0 \:f(x_0) =0$ |
H: Integral of second-order derivative
How can I perform the integration of second-order T ($\int \partial^2T=0$), so that I can arrive at equation 5.85, where T is a variable of $\xi$ and $\eta$?
Here is what I get:
$$ \int \partial^2T=0 $$
$$ T\partial + C = 0 $$
and I'm not sure what to do with the $T\partial$. Doesn't seem right at all.
AI: Start with what you are given:$$\frac{\partial}{\partial\xi}\left(\frac{\partial}{\partial\eta }T(\xi,\eta)\right) = 0$$
Integrate once with respect to $\xi$:
$$\frac{\partial}{\partial\eta }T(\xi,\eta) = C_1(\eta)$$
where $C_1$ is an arbitrary function of $\eta$ (since it vanishes if you apply $\frac{\partial}{\partial\xi}$).
Then integrate once with respect to $\eta$:
$$T(\xi,\eta) = \tilde{ C_1}(\eta) + C_2(\xi)$$
where $C_2$ is an arbitrary function of $\xi$ (since it vanishes if you apply $\frac{\partial}{\partial\eta}$) and $\tilde{C_1}$ is an antiderivative of $C_1$.
Now just call $\tilde{\psi} = \tilde{C_1}$ and $\tilde{\varphi}=C_2$. |
H: $\lambda_{\max}(XDX^T)$ smaller than $\lambda_{\max}(XX^T)$?
$X\in\mathbb{R}^{n\times d}$ and $D$ is a $d$-dimensional diagonal matrix. All elements on the diagonal of $D$ are in $[0,1]$. I am wondering whether the largest eigenvalue $\lambda_{\max}(XDX^T)$ of $XDX^T$ is smaller or equal to $\lambda_{\max}(XX^T)$. Intuitively, this is true and when $D=I_n$, we will have $\lambda_{\max}(XDX^T) = \lambda_{\max}(XX^T)$. But I cannot prove it or disapprove it. Please help!!!
AI: It is indeed true that $\lambda_{\max}(XDX^T) \leq \lambda_{\max}(XX^T)$. In particular, we note that for any positive semidefinite definite matrices $A,B$, we have $\lambda_{\max}(A + B) \geq \lambda_{\max}(A)$. Thus, we have
$$
\lambda_{\max}(X^TX) = \lambda_{\max}(X^TDX + X^T(I - D)X) \geq \lambda_{\max}(X^TDX).
$$ |
H: Prove $A$ is dense in $C([0,1]\times[0,1])$
Given
$$
A=\left\{\sum^n_{k=0}f_k(x)g_k(y) : \ n \in \mathbb{Z}^+, \ f_k, g_k\in C[0,1]\right\}.
$$
I am trying to use the Stone-Weierstrass Theorem to prove that $A$ is dense in $C([0,1]\times[0,1])$.
It is easy to see that $A$ is an algebra. I know $A$ vanishes nowhere, but I am confused about how to prove $A$ separates points. I know according to the Weierstrass Approximation Theorem, for all $\epsilon>0$ there exists a polynomial $p$ such that $|p(x,y)-f(x,y)|<\epsilon$, when $f\in C([0,1]\times[0,1])$, and I know polynomials separate points, and I have proved that all polynomials are in $A$.
My question is, how to prove $A$ separates points?
AI: Hint: Given two distinct points $(x_1,y_1),(x_2,y_2) \in [0,1]\times[0,1]$, try to give an example of a function of the form $F(x,y) = f(x)g(y)$ for which
$$
F(x_1,y_1) \neq F(x_2,y_2).
$$
You might find it helpful to separately handle the case where $x_1 \neq x_2$ and the case where $y_1 \neq y_2$. Note that our function $F$ is allowed to depend on the choice of points. |
H: $a_n$ is convergent.
Let $\{a_n\}$ be a bounded sequence of real numbers and $a_{n+1}\geq a_n - 2^{-n}$. Prove that $a_n $ is convergent.
My attempt: Suppose $a_n $ is not convergent then $\limsup a_n \neq \liminf a_n$. Let $\{x_n\}$ and $\{y_n\}$ converges to limsup and liminf. Then for some $x_p=a_{n_p}$ and $y_q=a_{n_q}$ we have $x_p-y_q>\epsilon$. where $n_q>n_p$. Also from given condition $x_p-y_q < \frac{1}{2^{n_p}}$. This is a contradiction for bigger $n_p$.
Is this correct?
Please help!
AI: Hint: Let $b_n=a_n-2^{1-n}$. Then the given condition amounts to $b_{n+1}\ge b_n$. |
H: Finding the volume of a rectangular prism using only surface area
Three surfaces of a rectangular prism are 25 cm squared, 18 cm squared, and 8 cm squared. What is its volume?
Can someone please explain how to solve the problem without using guess and check? The book where I found this problem said the answer is 60 cm squared but I don't know how to get to that answer.
Thanks.
AI: Hint: The volume of a rectangular prism is $abc$, where $a,b,c$ are the lengths in the three dimensions. You have $ab = 25$, $ac=18$, $bc=8$. What happens if you multiply these equations together? |
H: Is the summation $\sum_{i=1}^{n}\frac1{i} \binom{n}{i}$ possible?
I want to compute the following sum:
$$
\sum\limits_{i=1}^{n} \frac{{n\choose{i}}}{i}
$$
What I have done so far:
We know that $$(1+x)^n=\sum\limits_{r=0}^{n} {n\choose{r}}x^r$$
so, $$\frac{(1+x)^n-1}{x}=\sum\limits_{i=1}^{n} {{n\choose{i}}}x^{i-1}$$
therefore, upon integration we get,
$$\int\limits_{0}^{1}\frac{(1+x)^n-1}{x}dx=\sum\limits_{i=1}^{n} \frac{{{n\choose{i}}}}{i}$$
I cannot get any further with the LHS of the above equation.
Primary questions to be addressed:
Is such an integration possible (why so)?
Are there any other approximations for the sum?
AI: To expand a little on @Sangchul Lee's comment - after the substitution of ${y=1+x}$ the integral becomes
$${\Rightarrow \int_{1}^{2}\frac{y^n-1}{y-1}dy}$$
(This substitution isn't "necessary" for the next part, however it makes it a bit clearer). Now we can use a special factoring formula:
$${a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})}$$
To get
$${\int_{1}^{2}\frac{(y-1)(y^{n-1} + y^{n-2} + ... + y + 1)}{(y-1)}dy=\int_{1}^{2}y^{n-1} + y^{n-2} + ... + y + 1dy}$$
Evaluating that last integral gives us the sum
$${\Rightarrow \sum_{k=1}^{n}\frac{2^{k}-1}{k}}$$
So overall you have that
$${\sum_{k=1}^{n}\frac{{n\choose k}}{k}=\sum_{k=1}^{n}\frac{2^k-1}{k}}$$
Other than that though, I'm not sure if there's any more useful form. WolframAlpha gives some very nasty looking closed forms involving special functions. |
H: Laurent series of $\sin(-\frac{1}{z^2})$ radii of convergence
I am calculating radii of convergence of series for function:
$$
f(z)=\sin(-\frac{1}{z^2})
$$
I started with Taylor expansion for $\sin$ and then inserted $-\frac{1}{z^2}$. I got:
$$
-\frac{1}{z^2}+\frac{1}{3!}\frac{1}{z^6}-\frac{1}{5!}\frac{1}{z^{10}}+\frac{1}{7!}\frac{1}{z^{14}}...=\sum_{n=0} ^{\infty} (-1)^n \frac{1}{(2n-1)!}z^{4n-2}
$$
Is this even right?
But furthermore, I have to determine the radii of convergence?
I tried to use root formula ($\limsup \sqrt[n] {a_n}$), I tried to calculate it with L'Hospital's rule, but I don't know the derivative of faculteta... Any idea, how to continue?
AI: You can certainly insert $\frac{-1}{z^2}$ in $\sin$ if $z\neq0$ and you did it correctly but the series is
$\sum_{n=0}^{\infty} (-1)^{n+1} \frac{z^{-4n-2}}{(2n+1)!}$.
This case is easier to use the quotient rule:
$|\frac{a_{n+1}}{a_{n}}|=\frac{(2n+1)!}{(2n+3)!}|z^{-4}|\rightarrow0$ when $n\rightarrow\infty$, Then $\sin(\frac{-1}{z^2})$ converges for any $z\in\mathbb{C}$ except $z=0$. You could also notice that $\sin$ converges for any $z\in\mathbb{C}$, then in the domain of $g(z)=\frac{-1}{z^2}$ you have that $\sin(g(z))$ converges. |
H: why is R matrix inversion and transposition the same, but matrix Q has different inversion and transposition results
$R ={\begin{bmatrix}0.9697253054707993 & 0.04804422035332832 & -0.2394255308445735\\-0.01069682073773017 & 0.9878712527451343 & 0.1549063137056192\\ 0.2439639521643922 & -0.1476554803940568&0.9584776727467001\end{bmatrix}}$
$R.inv() = \underbrace {\begin{bmatrix}0.9697253054707994 & -0.01069682073773021 & 0.2439639521643922\\0.04804422035332829 & 0.9878712527451345 & -0.1476554803940568\\ -0.2394255308445735 & 0.1549063137056193 &0.9584776727467004\end{bmatrix}}_\text{inverse R}$
$R.t() = \underbrace {\begin{bmatrix}0.9697253054707994 & -0.01069682073773021 & 0.2439639521643922\\0.04804422035332829 & 0.9878712527451345 & -0.1476554803940568\\ -0.2394255308445735 & 0.1549063137056193 &0.9584776727467004\end{bmatrix}}_\text{transpose R}$
above inverse R = R transposed
consider Q:
$Q = {\begin{bmatrix}2 & 3 & 1\\5 & 4 & 3\\2 &3 &6\end{bmatrix}}$
$Q.inv() =\underbrace {\begin{bmatrix}-0.4285714285714285 & 0.4285714285714285 & -0.1428571428571428 \\0.6857142857142857 & -0.2857142857142857 & 0.02857142857142857\\-0.2 &-0 &0.2\end{bmatrix}}_\text{Q inverted}$
$Q.t() = \underbrace {\begin{bmatrix}2 & 5 & 2\\3 & 4 & 3\\1 &3 &6\end{bmatrix}}_\text{Q transposed}$
inverse Q $\neq$ Q transposed
Why isn't Q.inv() = Q.t() as with R?
C++ are same types well:
Mat Q is of type 64FC1 and should be accessed with Mat.at(y,x)
Mat R is of type 64FC1 and should be accessed with Mat.at(y,x)
AI: The matrix $R$ is an orthogonal matrix, whereas the matrix $Q$ is not. Note that the matrices $R$ and $Q$ are not similar by the usual definition of matrix similarity. |
H: k-partite Subgraph
I'm just working on a problem, but can only show, that the statement is true for $k=2$.
Let $G$ be a graph with $E(G)$ edges and $k \ge 2$. Show, that there is a $k$-partite subgraph $G*$ of $G$, so that $E(G*) \ge \frac{k-1}{k} E(G)$.
For $k = 2$, I solved the problem by induction over the number of vertices of $G$.
Thanks a lot
AI: I think induction works for any $k$.
Fix $k$, induct over the order $n$ of the graph.
If $k\leq n$, this is trivial, each vertex can be in its own partite set.
So assume it's true for all graphs of order $n$, and let $G$ be a graph of order $n+1$.
Pick a vertex $v$ of $G$. The graph $G-v$ has order $n$, so there's a $k$-partite subgraph $(G-v)^*$ of $G$ such that $|E((G-v)*)| \geq \frac{k-1}{k}|E(G-v)|$. We may assume without loss of generality that $(G-v)^*$ contains every vertex of $G-v$.
Let $V_1$, $V_2$, $\dots$, $V_k$ be the partite sets of $(G-v)^*$.
Choose the partite set $V_i$ that minimises $|N(v)\cap V_i|$ (so pick a partite set that contains the fewest neighbours of $v$), and notice that $|N(v)\cap V_i| \leq \frac{1}{k}|N(v)|$. Thus there are at least $\frac{k-1}{k}|N(v)|$ edges from $v$ to $G-V_i$.
The edges of $G$ that are not in $G-v$ are exactly those incident with $v$, so create $G^*$ from $(G-v)^*$ by adding $v$, and all the edges between $v$ and $(G-V_i)$.
$G^*$ is $k$-partite, since you can keep the same partite sets, and just put vertex $v$ into the partite set $V_i$. |
H: Find the smallest positive real $x$ such that $\lfloor{x^2}\rfloor - x\lfloor{x}\rfloor = 6$
Problem
Find the smallest positive real $x$ such that $\lfloor{x^2}\rfloor - x\lfloor{x}\rfloor = 6$.
What I've done
I set $m = \lfloor x \rfloor$ and $n = \{x\}$. Then I proceeded as below:
$\lfloor (m+n)^2 \rfloor -(m+n)m = 6$
Since $m$ is a non-negative integer, thus $\lfloor (m+n)^2 \rfloor = \lfloor m^2 \rfloor + \lfloor 2mn + n^2 \rfloor = m^2+ \lfloor 2mn + n^2 \rfloor$.
Then I got $\lfloor 2mn+n^2 \rfloor -mn = 6$
Where I'm stuck
I don't know how to go on.
AI: First note that if $m=0$ then the equation becomes $0=6,$ which is absurd, so $m>0$.
Then, note $$k = mn = \lfloor 2mn+n^2 \rfloor - 6$$ is an integer and $k/m = n < 1,$ so $k=mn < m$.
Then the last equation just states $k=6$ and since $6=k<m$ the smallest $m$ can be is $7$.
We conclude $$x = m + n = m + \frac{k}{m} = 7 + \frac67$$ |
H: Sheaf associated to a locally free presheaf of modules
Just checking here. It is true isn't it, that the sheaf associated to a locally free presheaf of $O_X$ modules (I suppose there is an example that is not a sheaf??) over a scheme $X$ is a locally free sheaf of $O_X$ modules? If the associated sheaf is coherent then this just textbook (e.g. Hartshorne Exercise II.5.7b), but otherwise I still would suppose this is true simply by definition of locally free presheaf (i.e. free when restricted to an open set)
AI: Suppose $\mathcal{F}$ is a presheaf of $\mathcal{O}_X-$modules that is locally free. This means that there exists an open cover $\{U_\alpha\}$ such that there exists $\varphi_\alpha:\mathcal{F}|_{U_\alpha}\xrightarrow{\sim}\bigoplus_{I}\mathcal{O}_{U_\alpha}$ for each $\alpha$. Such a map is an isomorphism on stalks. We know that $\mathcal{F}|_{U_\alpha}$ can be regarded as a presheaf on $U_\alpha$, and then $\mathcal{F}|_{U_\alpha}^+\cong \mathcal{F}^+|_{U_\alpha}$. By the universal property of the sheafification, the isomorphism $\varphi_\alpha:\mathcal{F}|_{U_\alpha}\to \bigoplus_I \mathcal{O}_{U_\alpha}$ induces a map $\mathcal{F}^+|_{U_\alpha}\to \bigoplus_I \mathcal{O}_{U_\alpha}.$ This map is an isomorphism: it can be checked on stalks but we know that the stalk maps were isomorphisms to begin with since $\varphi_\alpha$ was an isomorphism. So, $\mathcal{F}^+|_{U_\alpha}\cong \bigoplus_I\mathcal{O}_{U_\alpha}$.
Actually, this wasn't strictly necessary. It suffices to note that since $\mathcal{F}|_{U_\alpha}$ is isomorphic to a sheaf, it is itself a sheaf on $U_\alpha$. Hence, $\mathcal{F}^+|_{U_\alpha}\cong \mathcal{F}|_{U_\alpha}\cong \bigoplus_I \mathcal{O}_{U_\alpha}.$
By the way, the tensor product of sheaves needs to be sheafified to get a sheaf in general, so I think you can construct an example of a locally free presheaf that is not a sheaf by tensoring a pair of locally free sheaves. |
H: Uniform random variables question
Let U and V be independent random variables, both uniformly distributed on [0, 1]. Find
the probability that the quadratic equation $x^
2 + 2Ux + V = 0$ has two real solutions.
My solution:
The probability of two real solutions is the probability that $4U^2 - 4V > 0$.
$$
P(4U^2 - 4V > 0) = P(U^2>V)\\
=\int_{0}^{1} P(U^2>V|V=k)f_V(k)dk\\
=\int_{0}^{1} P(U>\sqrt{k})f_V(k)dk\\
=\int_{0}^{1} (1-\sqrt{k})f_V(k)dk\\
=\int_{0}^{1} (1-\sqrt{k}) dk
=\frac{1}{3}
$$
Does anyone see any problems with this?
AI: A simpler way to compute $\Pr[U^2 > V]$ is to observe that in the unit square $(U,V) \in [0,1]^2$, the region for which $U^2 > V$ is satisfied is given by $$0 \le V < U^2 \le 1,$$ hence $$\Pr[U^2 > V] = \int_{u=0}^1 \int_{v=0}^{u^2} f_{U,V}(u,v) \, dv \, du = \int_{u=0}^1 u^2 \, du = \frac{1}{3}.$$ This is possible because the joint density of $U, V$ is uniform on the unit square. |
H: The set $\{P \mid d(P,A) = k\cdot d(P,B)\}$ always represents a circle.
I'm trying to demonstrate that, given a real number $k$, with $k>0$ and $k\neq 1$, the set $\{P \mid d(P,A) = k\cdot d(P,B)\}$ always represents a circle.
I simply gave the coordinates $A=(m,n)$, $B=(p,q)$ and $P=(x,y)$ and put into $d(P,A) = k \cdot d(P,B)$ and found the equation: $$(1-k^2)x^2 + (1-k^2)y^2 + (2k^2p-2m)x + (2k^2q-2n)y + (m^2+n^2-k^2p^2-k^2q^2)=0$$ that, written as $Ex^2 + Ey^2 + Fx + Gy + H = 0$ is a circle, with the respectives $E,F,G,H$.
I'm stuck at this point. The material shows that this is enough proof, but I'm not convinced because some questions before at the same list, I proved that $Ex^2 + Ey^2 + Fx + Gy + H = 0$ is a circle if, and only if, $E \neq 0$ and $F^2+G^2-4EH>0$.
As $k\neq1$, there's no problem with $E \neq 0$. But how to prove thaat $F^2+G^2-4EH>0$? The equations simply don't match... I tried some substitutions, but I'm ending with a lot of variables, with no relation to each other... I can't find a way around that.
Any insights?
AI: Without the loss of generality you may assume that $A=(0,0)$ and $B=(b,0)$ (you can shift and rotate the axis without changing the geometry), then the equation you get is much simpler.
$$x^2+y^2=k^2[(x-b)^2+y^2] \implies (1-k^2)x^2+(1-k^2)y^2+2bk^2x-b^2k^2=0.$$
Now you have
$$E=1-k^2, F=2bk^2, G=0, H=-b^2k^2.$$
The condition $F^2+G^2-4EH>0$ can be easily verified. |
H: How to divide with exponentiation?
Let's say I wanted to multiply but couldn't actually use the multiply operation. I could do this:
$$(a+b)^2 = a^2 + 2ab + b^2 \implies ab = \frac{(a+b)^2 - a^2 - b^2}{2} $$
Now, logically it should be possible to reverse this process, I should be able to divide two numbers without using the division operation. But I have no clue how to do that. I actually need this for something very niche and also I think it's an interesting problem
EDIT: Okay, you're right I didn't include enough information.
Only operate on Natural numbers
Only use addition, subtraction, exponentiation, root (no logs, no mult/div)
As @WhatsUp put it nicely, the goal is to express the division function $N \times N \rightarrow N,(a,b) \mapsto a/b$ in terms of polynomials of a and b
Remainder can be discarded or kept
AI: Despite all the close votes, I think this could be an interesting problem if some more details are given.
As @tomasz said in the comment, the important thing is what operations are allowed.
My interpretation: in a field $K$, one wants to express the division function $(\cdot, \cdot): K \times K^\times \rightarrow K, (a, b) \mapsto a/b$ in terms of polynomials of $a$ and $b$.
In an infinite field, one may argue that this is not possible: if $a/b = h(a, b)$ holds for some polynomial $h$ and all $a, b\in K\times K^\times$, then we would have $a = bh(a, b)$, and putting $a = 1$ gives $1 = bh(1, b)$ for all $b \in K^\times$.
This cannot happen, as the polynomial $bh(1, b) - 1$, viewed as polynomial of $b$, has non-zero constant term and hence is non-zero, so it can only have finitely many roots.
In a finite field, the solution is simple, as $b^{-1} = b^{q - 2}$, where $q$ is the cardinality of $K$. |
H: Under what conditions will the covariance matrix be identical to the correlation matrix?
Under what conditions will the covariance matrix be identical to the correlation matrix?
I have been looking everywhere but no webpage or book seems to answer my question.
I just want to know when could this situation happen, and what that means for the variables.
Thanks
AI: If all the variables $X_1 \ldots X_n$ are variance=1 - that is - they have the unit scale then $n\times n$ covariance matrix will be (in theory) identical to the correlation matrix. Note - they don't have be normally distributed for this to hold. That said, numerical differences due to the difference in algorithms ( i.e. formulas) may arise - so you definitely don't want to be doing $Cov == Corr$. The differences depend on how degenerate your system is and how much "divide-by-near-zero" error you accumulate. |
H: Show $\pi$ is the orthogonal projection of $W$ iff $\|\pi(u)\| \leq \|u\|$ for all $u \in V$
Let $V$ be an inner product space and $W \subseteq V$ a finite-dimensional subspace of $V$. Let $\pi \in \mathcal L(V,V)$ a projection with $W$ as image. Show that $\pi$ is the orthogonal projection $\operatorname{pr}_W$ of $W$ if and only if $\|\pi(u)\| \leq \|u\|$ for all $u \in V$.
I prove the first part $(\Rightarrow)$, but I can't figure out how to start the second part $(\Leftarrow)$.
AI: Suppose $w \in W^\perp$ such that $\pi(w) \neq 0$ and consider $u = tw + \pi(w)$ for real scalar $t$. Then $$\|u\|^2 = t^2\|w\|^2 + \|\pi(w)\|^2 \\ \|\pi(u)\|^2 = (t+1)^2\|\pi(w)\|^2$$ Then solve for $t$ such that $\|\pi(u)\| > \|u\|$ |
H: Would my proof of induction be accepted in an intro Abstract Algebra Course. Self-studying and New to proofs.
Hello I'm self studying and I'm also new to proofs and would like to know whether my proof is rigorous enough for a first course in Abstract Algebra.
I'm asked to proof Induction of the second kind which states that:
Suppose P(n) is a statement about the positive integers and c is some fixed positive integer. Assume
i) P(c) is true
ii) for every $m > c $, if P(k) is true for all k such that $c \leq k < m $ , then P(m) is true
Then P(n) is true for all $n \geq c $
The proof also has to use the well-ordering principle which states that:
Every nonempty subset of the positive integers has a smallest element.
My proof:
First let $M = \{x\mid x \in N \land x>c \land P(x) \text{ is false} \}$
Now we assume M is nonempty. Then by the well-ordering principle there exists a smallest element of M which we'll call $y.$ We know that $P(n)$ for all $c \leq n < y $ is true thus $P(y)$ is true by ii. This is a contradiction thus $M$ is the empty set which means that $P(n)$ is true for all $n \geq c $
I'm self-studying and do not know whether my proof would be acceptable for a intro abstract algebra class. So my questions are:
Is my proof correct?
and if it's not correct, why not and how would I prove it then?
and if it's correcct, is there anything you would change?
Thanks in advance.
AI: As said in the other answer, the proof is almost correct, however I would "fix" it differently.
Namely, nothing is wrong with your set $M$ the way it is, but this only lets you conclude that $P(n)$ is true for $n\gt c$. Now (and that is the missing step, however trivial), you use (i), which says $P(c)$ is true too, to conclude $P(n)$ is true for $n\ge c$. |
H: Let $L/K$ be a finite Galois extension and $\alpha\in L\setminus K$. Then there exists $h\in G$ with prime power order not fixing $\alpha$.
Let $L/K$ be a finite Galois extension of fields and let $G=\text{Gal }L/K$. Let $\alpha\in L$ with $\alpha\notin K$. Show that there exists $h\in G$ with $h$ of prime power order not fixing $\alpha$.
I struggle to get started. An earlier part of the problem asks us to show that there exists $g\in G$ with $g(\alpha)\neq \alpha$, but this is trivial. I am not sure how to use this. One strategy I considered involved raising $g$ to some power $m$ to make $g^m$ have prime power order, but I am not sure how to show that such $g^m$ need not fix $\alpha$. I also considered inducting on the number of prime factors of the order of $g$, but no clear opportunity arises to apply the inductive hypothesis.
AI: Suppose $g$ does not fix $\alpha$ and that $g$ has order $r = p^at$ where $t$ is not divisible by $p$. Choose integers $c, d$ such that $cp^a+dt=1$. If both $g^{p^a}$ and $g^t$ fix $\alpha$, then so do $(g^{p^a})^c$ and $(g^t)^d$. But then $(g^{p^a})^c\cdot (g^t)^d = g^{cp^a+dt} = g$ also fixes $\alpha$. Now use induction on the number of prime factors of $r$. |
H: Law of large numbers question
Let $a, b, p \in (0, 1)$. What is the distribution of the sum of $n$ independent Bernoulli random variables with parameter $p$? By considering this sum and applying the weak law of large numbers, identify the limit
$$
\lim_{n \to \infty} \sum_{r \in \mathbb{N}:an<r<bn} \binom{n}{r} p^r(1 − p)^{n−r}
$$
in the cases (i) $p < a$; (ii) $a < p < b$; (iii) $b < p$.
I feel like the answers should be
(i) $0$,
(ii) $1$,
(iii) $0$,
but I don't know how to answer this question rigorously.
AI: WLLN says that the probability that the sample mean of $n$ Bernoulli variables deviates from $p$ by more than any $\epsilon > 0$ goes to zero as $n \to \infty$. This means the probability that a binomial random variable deviates from $np$ by more than $n \epsilon$ goes to zero as $n \to \infty$.
Now take $0<\epsilon<\min \{ |p-a|,|p-b| \}$. Then in case (i) and (iii), $(an,bn)$ doesn't intersect with $((p-\epsilon)n,(p+\epsilon)n)$, thus it is contained in the complement, so
$$P(X \in (an,bn)) \leq P(X \not \in ((p-\epsilon)n,(p+\epsilon)n)) \to 0.$$
In case (ii), instead $(an,bn)$ contains all of $((p-\epsilon)n,(p+\epsilon)n)$ so you have
$$P(X \in (an,bn)) \geq P(X \in ((p-\epsilon)n,(p+\epsilon)n)) \to 1.$$
It may help to draw a number line to help see the set relationships going on here. |
H: Some Counterexamples on Connectedness
There are abundant counterexamples in literature of the $2$ statements -
$X$ is Path Connected $\implies$ $X$ is Locally Path Connected
$X$ is Arc Connected $\implies$ $X$ is Locally Arc Connected
In all of the counterexamples I've found, they hold as the space is Path/Arc Connected, but is not Locally Connected (for example, the Extended Topologist's Sine Curve and the Closed Infinite Broom).
So, I wish to ask - are there counterexamples of the above $2$ statements, if we also assume that $X$ is Locally Connected? How about Locally Path Connected for Statement $2$?
AI: Let $Y=[0,1]\times[0,1]$ with the lexicographic order topology. For each $x\in[0,1]$ let $I_x$ and $I^x$ be copies of $[0,1]$ with its usual topology, and for each $t\in[0,1]$ let $t_x$ and $t^x$ be the copies of $t$ in $I_x$ and $I^x$, respectively. For $x\in[0,1]$ identify $\langle x,0\rangle\in Y$ with $0_x\in I_x$ and $\langle x,1\rangle\in Y$ with $0^x\in I^x$. Then identify all of the points $1_x$ and $1^x$ to a single point $p$ to get the space $X$.
Informally, we attach a ‘sticker’ in the form of a copy of the closed unit interval to each point on the bottom and top edges of the lexicographically ordered square, and we identify the free ends of the stickers.
Then $X$ is path connected and locally connected, but it is not locally path connected at any of the points $\langle x,0\rangle\sim 0^x$ or $\langle x,1\rangle\sim 0^x$. |
H: Little $o$ notation in the proof of central limit theorem.
In the proof im reading for the CLT, they seem to be using the following claim:
If $h(t)=o(t^2)$ then if $$g(n)\stackrel{\text{def}}{=}h\Big(\frac{t}{\sigma\sqrt{n}}\Big)$$ we have that $g(n)=o(\frac{1}{n})$.
More explicitly, I have a function $h$ that satisfies $$\lim_{t \rightarrow \infty} \frac{h(t)}{t^2} = 0$$
(and I seem to know nothing else about $h$) and then they are telling me that
$$\lim_{n \rightarrow \infty} h\Big(\frac{t}{\sigma\sqrt{n}}\Big) n = 0 .$$
But I don't see how I can say anything about $\lim_{x \rightarrow 0} h(x)$ just from knowing $h(t)=o(t^2)$.
Is the claim true? How can I show it?
AI: They likely meant to say $h(t)\in o(t^2)$ as $t\to0$ rather than $t\to\infty$, since otherwise the claim is false: take $h(t)=t$ (so that $h(t)\in o(t^2)$ as $t\to\infty$), then $g(n) = \frac t{\sigma\sqrt{n}}$ is not in $o(\frac1n)$ as $n\to\infty$ since $\lim_{n\to\infty}\frac{tn}{\sigma\sqrt n}=\infty$.
However, if $h(t)\in o(t^2)$ as $t\to0$, then the claim is indeed true since
$$
\lim_{n\to\infty}h\left(\frac t{\sigma\sqrt n}\right)n = \frac{t^2}{\sigma^2}\lim_{s\to0}\frac{h(s)}{s^2} = 0
$$
using the change of variables $s = \frac t{\sigma\sqrt n}$, showing that $g(n)\in o(\frac1n)$ as $n\to\infty$. |
H: Can any integer be expressed as sums of powers of three?
I heard a long time ago that any integer can be expressed as sums (or differences) of powers of three, using each power only once. Examples:
$5=9-3-1$
$6=9-3$
$22=27-9+3+1$
etc.
To my surprise, I couldn't find anything about this on the Internet. So..
Is this true? (So far I couldn't find any integer I can't express like this)
Is there a proof to it?
(Kind of a side question, assuming this is true): Is there a formula that gives the coefficients (-1, 0 or 1) for each power for a given integer?
Thanks
AI: This system is well-known and is called balanced ternary. It generalizes to all odd bases, but instead of the digits $-1,0$, and $1$, for base $2n+1$ you have the digits $0,\pm 1,\ldots,\pm n$; you can read a bit more here (with further generalizations in the rest of that article) and in this question and answer. |
H: Spivak's Calculus: chapter 2, problem 18(c)
In Spivak's calculus book, I cannot understand the solution proposed for question (c) of problem 18 in chapter 2:
Prove that $\sqrt{2}+\sqrt[3]{2}$ is irrational. Hint: Start by
working out the first 6 powers of this number.
Working out the powers is quite easy:
$(2^\frac{3}{6}+2^\frac{2}{6})^0 = 1$
$(2^\frac{3}{6}+2^\frac{2}{6})^1 = 2^\frac{3}{6} + 2^\frac{2}{6}$
$(2^\frac{3}{6}+2^\frac{2}{6})^2 = 2^\frac{6}{6} + 2 \cdot 2^\frac{3}{6} \cdot 2^\frac{2}{6} + 2^\frac{4}{6} = \\
2 \cdot 2^\frac{0}{6} + 2^\frac{4}{6} + 2 \cdot 2^\frac{5}{6}$
$(2^\frac{3}{6}+2^\frac{2}{6})^3 =
2^\frac{9}{6} + 3 \cdot 2^\frac{6}{6} \cdot 2^\frac{2}{6} + 3 \cdot 2^\frac{3}{6} \cdot 2^\frac{4}{6} + 2^\frac{6}{6} = \\
2 \cdot 2^\frac{0}{6} + 6 \cdot 2^\frac{1}{6} + 6 \cdot 2^\frac{2}{6} + 2 \cdot 2^\frac{3}{6}
$
$(2^\frac{3}{6}+2^\frac{2}{6})^4 = 2^\frac{12}{6} + 4 \cdot 2^\frac{9}{6} \cdot 2^\frac{2}{6} + 6 \cdot 2^\frac{6}{6} \cdot 2^\frac{4}{6} + 4 \cdot 2^\frac{3}{6} \cdot 2^\frac{6}{6} + 2^\frac{8}{6} = \\
4 \cdot 2^\frac{0}{6} + 2 \cdot 2^\frac{2}{6} + 8 \cdot 2^\frac{3}{6} + 12 \cdot 2^\frac{4}{6} + 8 \cdot 2^\frac{5}{6}
$
$(2^\frac{3}{6}+2^\frac{2}{6})^5 = 2^\frac{15}{6} + 5 \cdot 2^\frac{12}{6} \cdot 2^\frac{2}{6} + 10 \cdot 2^\frac{9}{6} \cdot 2^\frac{4}{6} + 10 \cdot 2^\frac{6}{6} \cdot 2^\frac{6}{6} + 5 \cdot 2^\frac{3}{6} \cdot 2^\frac{8}{6} + 2^\frac{10}{6} = \\
40 \cdot 2^\frac{0}{6} + 40 \cdot 2^\frac{1}{6} + 20 \cdot 2^\frac{2}{6} + 4\cdot2^\frac{3}{6} + 2 \cdot 2^\frac{4}{6} + 10 \cdot 2^\frac{5}{6}$
$(2^\frac{3}{6}+2^\frac{2}{6})^6 = 2^\frac{18}{6} + 6 \cdot 2^\frac{15}{6} \cdot 2^\frac{2}{6} + 15 \cdot 2^\frac{12}{6} \cdot 2^\frac{4}{6} + 20 \cdot 2^\frac{9}{6} \cdot 2^\frac{6}{6} + 15 \cdot 2^\frac{6}{6} \cdot 2^\frac{8}{6} + 6 \cdot 2^\frac{3}{6} \cdot 2^\frac{10}{6} + 2^\frac{12}{6} = \\
12 \cdot 2^\frac{0}{6} + 24 \cdot 2^\frac{1}{6} + 60 \cdot 2^\frac{2}{6} + 80 \cdot 2^\frac{3}{6} + 60 \cdot 2^\frac{4}{6} + 24 \cdot 2^\frac{5}{6}
$
I am at a loss as to how this can help towards the solution... The first question in problem 18 is asking to prove the "rational root theorem" but I don't see how I can combine that with this hint to solve the problem.
UPDATE: following Gerry Myerson's advice I create a polynomial $c_0 \cdot 2^\frac{0}{6} + c_1 \cdot 2^\frac{1}{6} + c_2 \cdot 2^\frac{2}{6} + c_3 \cdot 2^\frac{3}{6} + c_4 \cdot 2^\frac{4}{6} + c_5 \cdot 2^\frac{5}{6}$ where each $c_i$ multiplies the coeffficients of the respective powers I found above for $x^0, \dots, x^6$.
So each upper-case $C_i$ is the sum of the coefficients from each of the powers $x_n$. For example $C_5 = 24 + 10c_5 + 8c_4 + 2c_2$. The expansion of this polynomial is:
$
\begin{aligned}
C_0 \cdot 2^\frac{0}{6} + C_1 \cdot 2^\frac{1}{6} + C_2 \cdot 2^\frac{2}{6} + C_3 \cdot 2^\frac{3}{6} + C_4 \cdot 2^\frac{4}{6} + C_5 \cdot 2^\frac{5}{6} &&= \\
[ 12 + 40 \cdot c_5 + 4 \cdot c_4 + 2 \cdot c_3 + 2 \cdot c_2 + c_0] \cdot 2^\frac{0}{6} &+ &\\
[ 24 + 40 \cdot c_5 + 6 \cdot c_3] \cdot 2^\frac{1}{6} &+ & \\
[ 60 + 20 \cdot c_5 + 2 \cdot c_4 + 6 \cdot c_3 + 1 \cdot c_1] \cdot 2^\frac{2}{6} & + & \\
[ 80 + 4 \cdot c_5 + 8 \cdot c_4 + 2 \cdot c_3 + 1\cdot c_1 ] \cdot 2^\frac{3}{6}& + & \\
[ 60 + 2 \cdot c_5 + 12 \cdot c_4 + 1 \cdot c_2 ] \cdot 2^\frac{4}{6} & + & \\
[ 24 + 10 \cdot c_5 + 8 \cdot c_4 + 2 \cdot c_2 ] \cdot 2^\frac{5}{6} & & \\
\end{aligned}
$
So as Danny Pak - Keung Chan explained, in order to use the rational root theorem I need to set each of these $C_i$ equal to zero and solve the system of equations:
$
\begin{aligned}
12 + 40 \cdot c_5 + 4 \cdot c_4 + 2 \cdot c_3 + 2 \cdot c_2 + c_0 = 0&\\
24 + 40 \cdot c_5 + 6 \cdot c_3 = 0 \\
60 + 20 \cdot c_5 + 2 \cdot c_4 + 6 \cdot c_3 + 1 \cdot c_1 = 0 \\
80 + 4 \cdot c_5 + 8 \cdot c_4 + 2 \cdot c_3 + 1\cdot c_1 = 0 \\
60 + 2 \cdot c_5 + 12 \cdot c_4 + 1 \cdot c_2 = 0\\
24 + 10 \cdot c_5 + 8 \cdot c_4 + 2 \cdot c_2 = 0 \\
\end{aligned}
$
Now this will hopefully yield integer solutions which I can replace into the polynomial and thus apply the rational root theorem (knowing that the value $\sqrt{2} + \sqrt[3]{2}$ is a root) which means it is either an integer or it is irrational.
At that point I guess I will just need to prove (easily verifying inequalities) that $\sqrt{2} + \sqrt[3]{2}$ is not an integer therefore it must be irrational.
This is a LONG way to solve this problem.
Update 2
I had GNU Octave solve the equations and got:
$c_0=-4. c_1=-24, c_2=12, c_3=-4, c_4=-6, c_5=0$
So our polynomial is:
$x^6 + 0x^5 -6x^4 -4x^3 +12x^2 -24x^1 -4$
This really does match the accepted answer's "minimal polynomial"
So now as I mentioned one can use part (a) of the problem (integral root theorem) to claim that the roots are either integral or irrational.
At this point it is easy to check that:
$1.4 < \sqrt{2} < 1.5$ and $1.2 < \sqrt[3]{2} < 1.3$ so adding we have:
$2.6 < \sqrt{2} + \sqrt[3]{2} < 2.8$
Therefore $\sqrt{2} + \sqrt[3]{2}$ is not an integer so by (a) it must be irrational and the long proof has come to its end.
AI: Hint: I guess the idea is to Try to write $(\sqrt{2}+\sqrt[3]{2})^6$ in terms of $\sqrt{2}+\sqrt[3]{2}$ then we get a something like
$$(\sqrt{2}+\sqrt[3]{2})^6 = p(\sqrt{2}+\sqrt[3]{2})$$ where $p(x)$ is a polynomial then using the rational zeros theorem we can show that $p(x)-x^6$ has no rational roots but $p(\sqrt{2}+\sqrt[3]{2})-(\sqrt{2}+\sqrt[3]{2})^6=0$. Thus we get that $\sqrt{2}+\sqrt[3]{2}$ is irrational.
To check your work see the link:
https://www.wolframalpha.com/input/?i=minimal+polynomial+2%5E%281%2F2%29+%2B+2%5E%281%2F3%29
Since you still looking for the answer and you want to use your work I will start to write what you did using $b=\sqrt[6]{2}$ and $a=\sqrt{2}+\sqrt[3]{2}$ the your work
$$a^2=2+2b^5+b^4$$
$$a^3=2+6b+6b^2+2b^3$$
And so on the eliminates the $b$ powers do you want me to continue? |
H: Appropriateness of Poisson distribution for low number of trials with the probability = 0.5 of success
I'm working on the following problem from Ross "A First Course in Probability" (9th edition):
People enter a gambling casino at a rate of 1 every 2 minutes. (a)
What is the probability that no one enters between 12:00 and 12:05?
(b) What is the probability that at least 4 people enter the casino
during that time?
(problem 63, chapter 4, page 178)
The official solution manual approaches the problem as a Poisson process, but why is it appropriate? I feel like for a small number of trials (n = 5) and high probability (p = 0.5) the Binomial distribution would be more fit. The answer is quite different depending on what distribution we use:
Binomial(n = 5, p=0.5): $ P(X=0) = 0.03125; P(X \geq 4) = 0.1875 $
Poisson($\lambda = 2.5$): $ P(X = 0) = 0.082; P(X \geq 4) = 0.242 $
AI: It's not just five trials; it's infinitely many trials with an infinitely small probability of success on each trial. Or, if you like, one can approximate it by dividing the dividing every minute into $10\,000$ small intervals, each with probability $1/20\,000$ of someone entering the casino, so it's $\operatorname{Binomial}(2.5\times10\,000,\,\,\, 1/20\,000).$ And an even closer approximation to the Poisson distribution (but not very much closer since you're extremely close already) is achieved by dividing each minute into $100\,000$ parts, each with probability $1/200\,000$ of a customer entering. |
H: When evaluating the limit of $f(x, y)$ as $(x, y)$ approaches $(x_0, y_0)$, should we consider only those $(x, y)$ in the domain of $f$?
When evaluating the limit of $f(x, y)$ as $(x, y)$ approaches $(x_0, y_0)$, we should or should not consider only those $(x, y)$ in the domain of $f(x, y)$ ? I am confused by different practices of Calculus textbooks. Have anyone searched and found some authoritative opinion ?
Thomas Calculus 14e §14.2 Example 2 (Page 802-803) $\lim_{(x, y) \to (0, 0)} \frac{x^2 - x y}{\sqrt{x} - \sqrt{y}}$ considers only those $(x, y)$ in the domain. The authors' answer ($\mathbf{0}$) is the same as the answer by WolframAlpha . See textbook page 802 and textbook page 803 .
Larson Calculus 10e §13.2 Exercise 27 (Page 887) $\lim_{(x, y) \to (0, 0)} \frac{x - y}{\sqrt{x} - \sqrt{y}}$ considers NOT only those $(x, y)$ in the domain. The authors' answer (DNE) is NOT the same as the answer by WolframAlpha ($\mathbf{0}$). See textbook page 887 and solution manual page 1268 .
AI: Larson defines the (potential) limit of a function $f$ only at points $P$ where $f$ is defined in a punctured open neighborhood of $P$.
That explains the answer from the answer key.
But it's a minority point of view.
Note that Larson is still only considering points $(x,y)$ approaching $(x_0,y_0)$ where the points $(x,y)$ are in the domain of $f$, but the author is only allowing consideration of limits at points $(x_0,y_0)$ where $f$ is defined in a punctured open neighborhood of $(x_0,y_0)$.
As evidence for the "minority point of view" claim, the following texts define the (potential) limit of $f$ at a point $P$, assuming only that $P$ is such that every punctured neighborhood of $P$ contains points of the domain of $f$.
Edwards & Penney -- Calculus - Early Trancendentals, 7th Ed (2007)
Stewart -- Calculus - Early Transcendentals, 6th Ed (2008)
Thomas & others -- Calculus, 11th Ed (2004) |
H: Why there is no suspension axiom for homology ? and why there is no excision axiom for cohomology theory?
Here are the axioms of reduced cohomology theory as given to me in the lecture:
1- $\tilde{H}^n(-;G): J_{*} \rightarrow Ab_{*}$ is a contravariant functor.
2- $\tilde{H}^n(X;G) \cong \tilde{H}^{n+1}(\sum X;G).$
3- Homotopy Axiom. homotopic maps induce the same map in cohomology.
4- Exactness. cofibre sequence induced a LES.
5- Dimension Axiom:
$$\tilde{H}^k(S^n ; \mathbb{Z}) = \mathbb{Z}, \text{ if } k = n \text{ and } \tilde{H}^k(S^n ; \mathbb{Z})= 0 \text{ if } k \neq n. $$
My professor was depending on the book " Modern classical homotopy theory " by Jeffery Strom.
While the homology theory axioms from Rotman book(on pg.231) "Introduction to algebraic topology " are as below :
My questions are:
1-Why there is no suspension axiom for homology or what is its equivalence? and why there is no excision axiom for cohomology theory or what is its equivalence?
2-If I transformed the dimension axiom given by my professor to homology theory, I do not understand how the statement I obtained is the same as the statement mentioned in Rotman. Here is the statement I obtained for homology dimension axiom:
$$\tilde{H}_k(S^n ; \mathbb{Z}) = \mathbb{Z}, \text{ if } k = n \text{ and } \tilde{H}^k(S^n ; \mathbb{Z})= 0 \text{ if } k \neq n. $$
Could anyone explain to me how they are equivalent, please?
AI: I work with reduced homology here. Keep in mind that it is a straightforward consequence of definitions that for $A$ nonempty $\widetilde H_*(X, A) = H_*(X,A)$.
The fact that $\widetilde H_{k+1}(\Sigma Y; \Bbb Z) = \widetilde H_k(Y; \Bbb Z)$ follows immediately from:
the homotopy axiom, which shows that because the cone $CY$ is contractible, we have $\widetilde H_*(CY) = 0$; this is also used to show that $\widetilde H_*(\Sigma Y, CY) \cong \widetilde H_*(\Sigma Y, *)$.
the exactness axiom, applied to the pair $(CY, Y)$, which shows that $\partial: H_{k+1}(CY, Y) = \widetilde H_{k+1}(CY, Y) \to \widetilde H_k(Y)$ is an isomorphism for all $k$;
the excision axiom applied to $X = \Sigma Y, A = CY$, and $U = \Sigma Y \setminus CY$, which provides an isomorphism $$\widetilde H_*(\Sigma Y) = \widetilde H_*(\Sigma Y, *) \cong \widetilde H_*(\Sigma Y, CY) \cong \widetilde H_*(CY, Y).$$
Putting this together this gives your desired "axiom". |
H: Solving for matrix system using least squares quadratic
So I'm given the following coordinates below and I'm asked to set up a matrix system to solve for the least squares expressions.
I have the first question right, and I have matrix A of the second question correct. I'm a little stumped on how I would find the matrix for b though? I know b generally represents the y coordinates, but since there is a 2nd degree I'm not exactly sure what to do. Am I supposed to substitute coordinates into some sort of formula? I would appreciate any help!
AI: If I understand the question, I believe the $y$ values will be the same as in the first problem, i.e. $b=-9,8, \ldots$. The change is only on the left hand side with the $\beta$ and $x$ part of the problem.
I hope this helps. |
H: What does this mathematical expression mean?
I am reading a natural language processing paper and I came across this expression. I don't know what it means. Especially the unif part.
$$m_i \sim \operatorname{unif}\{1,n\}\text{ for } i = 1 \text{ to } k$$
AI: It means you have $k$ random variables, $m_1, m_2, \ldots , m_k$, each of which was chosen from a uniform distribution with extreme values $1$ and $n$. It is fundamentally ambiguous as to whether the distribution is discrete or continuous. (Check your source.)
Example: if $k = 3$ and $n = 5$, you could have...
...in the continuous case:
$m_1 = 2.938, m_2 = 1.155, m_3 = 4.076$.
... and in discrete case:
$m_1 = 4, m_2 = 5, m_3 = 1$.
Given your source is about natural language processing having (discrete) words, I suspect (but don't know) that the latter is the case. |
H: Derivative of a function of two variables
Let $f:\mathbb R^2\rightarrow \mathbb R $ be defined by
$f(x,y)=\exp(-\frac{1}{x^2+y^2})$ if $(x,y)\neq(0,0)$ and $f(0,0)=0$. Check whether $f$ is differentiate at (0,0) or not.
I have checked that $f$ is continuous at (0,0) and both partial derivatives of $f$ exists at (0,0) and both equals to 0. But how do I check the differentiability?
AI: Hint: you should check if
$$\frac{\exp(-\frac{1}{x^2+y^2})}{\sqrt{x^2+y^2}} \to 0$$
when $$(x,y) \to (0,0)$$
For this most simple, imho, is consider functions $z=x^2+y^2$ and $t=\frac{\exp(-\frac{1}{z})}{\sqrt{z}}$ and take limit on their composition. |
H: Defining derivative of powers of $x$
We know that derivative of $x^n$ is $nx^{(n-1)}$ if $n$ is an integer.
My question is how do we define derivative of $x^r$ is $r$ is an irrational number. For example what is the derivative of $x^\sqrt2$ or $x^\pi$?
AI: We define $x^r$ as $x^r = e^{r \log x}$ so that
$$\begin{aligned}
\frac{d}{dx} x^r &= \frac{d}{dx} e^{r \log x} \\&= \frac{r}{x} e^{r \log x} \\&= \frac{r}{x} x^r \\&= r x^{r-1}.
\end{aligned}$$ |
H: Why does AM>GM when applied on functions gives the absolute minima.
In some cases we use the relation AM>GM to find the minima for example take $f(x)=x+\frac1x$ $[x\gt 0]$ using the result AM>GM we can find the minima as $2$.It is the same minima which we get if we use the methods of derivative.
But why do we get the same minima why not say we get $f(x)>1$.This statement ($f(x)>1$) is not false but $1$ is just not the absolute minima.Can we prove that the relation AM>GM will always give the absolute minima in the given domain? (Note-I am not asking for the proof of result AM>GM.)
Hence as we are getting the absolute minima using the result AM>GM I conclude that there is some kind of relation between applying the result $AM>GM$ and using the methods of derivative.Is there any intuitive way to understand that relation?
AI: Using AM-GM not always gives an extreme value. We need also to save the case of the equality occurring. In AM-GM it happens for equality case of all variables, which not always good.
Also, under AM-GM we have convexity of $\ln$, which has a relation with a second derivative.
Also, there are very many methods to find an extreme value with derivatives and these methods have no any relation with AM-GM.
For example.
Let we need to find a maximal value of $(a^2-ab+b^2)(a^2-ac+c^2)(b^2-bc+c^2)$ by AM-GM, where $a$, $b$ and $c$ are non-negatives such that $a+b+c=3$.
We see that $(a,b,c)=(2,1,0)$ gives a value $12$.
We'll prove that $12$ is a maximal value.
We can not use AM-GM here in the following form. $$\prod_{cyc}(a^2-ab+b^2)\leq\left(\frac{\sum\limits_{cyc}(a^2-ab+b^2)}{3}\right)^3$$ because it does not save the case of equality occurring.
For $(a,b,c)=(2,1,0)$ we obtain:
$$a^2-ab+b^2=3,$$$$a^2-ac+c^2=4$$ and $$b^2-bc+c^2=1,$$ which is not good because we need $$a^2-ab+b^2=a^2-ac+b^2=b^2-bc+c^2,$$ which is wrong and it says that we got a wrong inequality:
$$\left(\frac{\sum\limits_{cyc}(a^2-ab+b^2)}{3}\right)^3\leq12,$$ which easy to understand after checking $(a,b,c)=(2,1,0).$
By the way, the following solution saves the case of equality occurring.
Let $a\geq b\geq c$.
Thus, $$(a^2-ab+b^2)(a^2-ac+c^2)(b^2-bc-c^2)\leq(a^2-ab+b^2)a^2b^2=$$
$$=((a+b)^2-3ab)a^2b^2\leq(9-3ab)a^2b^2=12(3-ab)\left(\frac{ab}{2}\right)^2\leq$$
$$\leq12\left(\frac{3-ab+\frac{ab}{2}+\frac{ab}{2}}{3}\right)^3=12.$$
Id est, we solved this problem by AM-GM.
Actually, using AM-GM was possible because for $(a,b,c)=(2,1,0)$ we have $$3-ab=\frac{ab}{2}=\frac{ab}{2}.$$
We can use the Lagrange Multipliers method here (id est, we can use derivatives),
but it not so easy here. |
H: Can a non-inner automorphism map every subgroup to its conjugate?
Let $G$ be a finite non-cyclic group. Can a non-inner automorphism map every subgroup to its conjugate? Namely, can there be a non-inner automorphism $\alpha$ that, for every $H\le G$, there exists some $g$ in $G$ such that $\alpha(H)=H^g$?
AI: Yes, the dihedral group of order 10 has this property. Its subgroup structure is very simple: $D_{10}$, $\{e\}$, the rotation subgroup, and the 5 subgroups generated by a flip. Any automorphism fixes the first three, so those are done, and shuffles the last five, and those 5 subgroups are all conjugate to each other.
All we need now is to show that there is a non-inner automorphism, but this is easy; the inner automorphisms always send a rotation to its inverse (or fix it) so we only need an automorphism which doesn't do that. Let the generators be $\sigma,\tau$, rotation and flip, and consider the automorphism defined on the rotations by $\sigma \mapsto \sigma^2$ and fixing $\tau$. |
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