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H: Equivalent metrics give the same topology
Definition
If $d_1$ and $d_2$ are distance in $X$ then we say that they are equivalent iff and only if there exist $\alpha,\beta>0$ such that
$$
\alpha\cdot d_1(x,y)\le d_2(x,y)\le\beta\cdot d_1(x,y)
$$
for any $x,y\in X$.
Theorem
Two metric are equivalent if and only if they give the same topology
If $d_1$ and $d_2$ are equivalent then
$$
\alpha\cdot d_1(x,y)\le d_2(x,y)\le\beta\cdot d_1(x,y),\,\,\text{and}\,\,\frac{1}\beta d_2(x,y)\le d_1(x,y)\le\frac{1}\alpha\cdot d_2(x,y)
$$
and so using this inequality I tried to prove that for any $d_1$-ball there exist a $d_2$-ball that is contained in the first and vice versa. Unfortuately I don't be able to do this and then I don't be able to prove that if two metric give the same topology they are equivalent. So could someone help me, please?
AI: Hint: If $d_1$ and $d_2$ are equivalent, can you find relationships between $d_1$-balls and $d_2$-balls?
The converse is false: The discrete metric and the Euclidean metric on $\Bbb{Z}$ give the same topology (the discrete one) yet they are not equivalent. |
H: Why doesn't $\sum_{n=0}^\infty2^n=-1$?
Now of course I'm not stranger to the fact that adding finite (and in many cases - infinite) amount of positive numbers always yeilds a positive number, but in many cases, often the finite limit isn't equivalent to the strange nature of infinity. It seems that mathematics tends to prefer that if $\sum_{n=0}^\infty2^n$ were to converge, it likes
$$\sum_{n=0}^\infty2^n=-1$$
This can be seen in many "proofs" like forcing the geometric series formula:
$$\sum_{n=0}^\infty r^n=\frac{1}{1-r}\implies\sum_{n=0}^\infty2^n=\frac{1}{1-2}=-1$$
Or the digit function (which returns the digit in the $b^n$ column of $x$ in base $b$):
$$D_b^n(x)=\lfloor\frac{x}{b^n}\rfloor-b\lfloor\frac{x}{b^{n+1}}\rfloor$$
(E.g; $D_{10}^{-2}(\pi)=4$)
For $-1$:
$$D_2^n(-1)=1 \forall n>0$$
$$D_2^n(-1)=0 \forall n\leq0$$
Hence it could be seen that $-1_{(10)}=\bar1.0_{(2)}$, which implies $$\sum_{n=0}^\infty2^n=-1$$
Besides conventions for convergence, what exactly is stopping this sum from being equal to $-1$? Would such an equivalence lead to a contradiction?
Computationally concerned, using $-1_{(10)}=\bar1.0_{(2)}$ as a binary expansion holds all the expected arithmetic properties like $\bar1.0_{(2)}\times\bar1.0_{(2)}=\bar01.0_{(2)}$
AI: If $$\sum_{n=0}^\infty 2^n$$ converges, then it converges to $-1$
is a true statement.
So is the statement
If $$\sum_{n=0}^\infty 2^n$$ converges, then it converges to $42$.
And so is the statement
If $$\sum_{n=0}^\infty 2^n$$ converges, then the Earth is flat.
All three statements are true because $A\implies B$ is always true if $A$ is false. The simple fact is that the expression $$\sum_{n=0}^\infty 2^n$$ denotes the sum of a divergent series, and it therefore cannot be used in an equality because it is undefined.
The expression $$\sum_{n=0}^\infty a_n$$ is simply shorthand for $$\lim_{N\to\infty} \sum_{n=0}^Na_n.$$
If you plug in $a_n=2^n$ into that expression, you see that what you are really asking is why
$$\lim_{N\to\infty} \sum_{n=0}^N 2^n=-1$$ is not true. And it is not true because the right-hand side of the equality is a real number, while the left hand side is an undefined expression. |
H: Is this series $\sum_{n=0}^{\infty} 2^{(-1)^n - n} = 3 $ or $ \approx 3$
Is this series $\sum_{n=0}^{\infty} 2^{(-1)^n - n} = 3 $ or $ \approx 3$
Hello all.
In Ross' elementary Analysis Chapter 14 there is an example of the above series.
It simply says it converges.
I have computed that it converges to 3.
The way I did it is that I thought:
$$2 + \frac{1}{4} + \frac{1}{2} + + \frac{1}{16} + + \frac{1}{8} + + \frac{1}{64} + + \frac{1}{32}....$$
can be broken up to $$ 2 + \sum_{n=1}^{\infty} (\frac{1}{4})^n + \frac{1}{2}\sum_{n=0}^{\infty} (\frac{1}{4})^n$$
Which is 2 + 2 geometric series that I computed to $\frac{1}{3} $ and $ \frac{2}{3}$ respectively getting exactly 3.
Using the formula I learned first term divided by 1 - ratio.
Now wolfram alpha says this is only approximately 3.
Did I go wrong somewhere above?
Or could it be that there is some rule that makes me not be able to give an exact result?
I apologize if this is very basic. I have been thinking about this since yesterday.
Thanks a lot!
AI: Wolfram alpha is not wrong. The result is approximately 3.
You are also not wrong. The result is exactly 3.
Where you are wrong is when you use the word "only". You say:
wolfram alpha says this is only approximately 3.
This implies that you think that the number $3$ is not "approximately" equal to itself. In fact, it is. $3\approx 3$ is a true statement, so a better way of putting it would be:
wolfram alpha says only that this is approximately $3$
In other words, WA is not wrong, it's just not telling the entire story. It is saying that the sum is approximately $3$ (which means that it is either exactly $3$ or a number close to, but not equal to $3$), when in fact, it could state the logically stronger claim that the sum is exactly $3$. |
H: Convergence of $\sqrt{1-\sqrt{{1}-\sqrt{{1}-\sqrt{{1-\sqrt{{1}}}...}}}}$
Recently, as is evident from many of my recent questions, I have been very interested in nested radicals. Recently I attempted to investigate the following infinite nested radical and arrived at a strange, counter-intuitive result.
$$\sqrt{1-\sqrt{{1}-\sqrt{{1}-\sqrt{{1-\sqrt{{1...}}}}}}}$$
At first glance, it would seem that the value of the nested radical must be either $0$ or $1$. However, if we make the assumption that a single value can be assigned to it, ie it converges, we arrive at a different value for it:
Let
$$x=\sqrt{1-\sqrt{{1}-\sqrt{{1}-\sqrt{{1-\sqrt{{1...}}}}}}}\implies x=\sqrt{1-x}\implies x^2=1-x \implies x^2+x-1=0$$
Using the quadratic formula, we get $$x=\frac{-1+\sqrt5}{2}$$
as $x$ is necessarily positive. Can this result be true? Or is my assumption that allowed me to label the radical as $x$ and assume it converges incorrect?
AI: Whenever we find $\ldots$ in a formula we must be clear what they are meaning. My interpreation of a formula like
$$\sqrt{1-\sqrt{1-\sqrt{1-\sqrt{1-\ldots}}}}$$
is that this is the limit of the sequence
$$x_1=\sqrt{1}\\
x_2=\sqrt{1-\sqrt{1}}\\
x_3=\sqrt{1-\sqrt{1-\sqrt{1}}}\\
\ldots\\
x_{n+1}=\sqrt{1-x_n}$$
If this is the meaning of this expression then we have
$$x_1=1\\
x_2=0\\
x_3=1\\
x_4=0\\
x_5=1\\
x_6=0\\
\ldots
$$
and this sequence does not converge at all.
Your calculation says:
If $x_n$ converges, then the limit will be $\frac{-1+\sqrt5}{2}$ or $\frac{-1-\sqrt5}{2}$
This is true. But $x_n$ does not converge, so this result is useless. |
H: How can I prove that the following are happening: $\ln\Big(1+\frac{1}{x}\Big)=\frac{1}{x}+o\Big(\frac{1}{x}\Big)$?
How can I prove that the following are happening ($x\to\infty$): $\ln\Big(1+\frac{1}{x}\Big)=\frac{1}{x}+o\Big(\frac{1}{x}\Big)$ and $\Big(1+\frac{1}{x}\Big)^{p}=1+\frac{p}{x}+o\Big(\frac{1}{x}\Big)$, where o is the notation for Little-o.
I thought it could be shown directly with the definition that 2 functions are asymptotically equivalent ($\lim_{x\to x_0}\vert\frac{f(x)}{g(x)}\vert$=0), but I'm not sure.
AI: Using Taylor's formula, you know that
$$
\log(1+y) = y + o(y), \quad (y \to 0)
$$
Setting $y=1/x$ gives you
$$
\log\left(1+\frac 1x\right) = \frac 1x + o\left(\frac 1x\right), \quad (x\to +\infty)
$$
The same can be accomplished in the second example:
$$
(1+y)^p = 1 + p y + o(y) (\textrm{ as }y \to 0) \Rightarrow \left(1+\frac 1x\right)^p = 1+ \frac px + o\left(\frac 1x \right) (\textrm{ as } x\to +\infty)
$$ |
H: Does the Riemannian distance function obey the Leibniz rule?
Let $M$ be a Riemannian manifold, and let $
\alpha,\beta:[0,\delta) \to M$ be two smooth paths satisfying $\alpha(t) \neq \beta(t)$ for every $t$.
Suppose that for every $t$ there is a unique length-minimizing geodesic from $\alpha(t)$ to $\beta(t)$. This implies that the Riemannian distance function $d:M \times M \to \mathbb R$ is differentiable at the point $\left(
\alpha(t),\beta(t)\right)$.
Now, for a given (fixed) point $p \in M$, denote by $d_p:M \to \mathbb R$ the distance function from $p$, that is $d_p(q)=d(p,q)$. Its differential at a point $q \in M$ is denoted by $d(d_p)_q:T_qM \to \mathbb R$.
Question:
Does
$$\frac{d}{{dt}}d\left(\alpha(t),\beta(t)\right)=d(d_{\alpha(t)})_{\beta(t)}(\dot \beta(t))+d(d_{\beta(t)})_{\alpha(t)}(\dot \alpha(t))$$
hold?
This is a sort of "Leibniz-rule"...I tried to play with the formula described here, using exponential maps and initial velocities of geodesics, but so far without success.
This Leibniz-rule clearly holds for the case of $\mathbb R^n$, endowed with the standard Euclidean metric.
AI: Actually, this isn't related the Riemannian geometry at all. The result follows from the following general assertion:
Let $f:M \times M \to \mathbb R$ be smooth. Then
$$
\frac{d}{{dt}}f\left(\alpha(t),\beta(t)\right)=d(f_{\alpha(t)}^L)_{\beta(t)}(\dot \beta(t))+d(f_{\beta(t)}^R)_{\alpha(t)}(\dot \alpha(t)),
$$
where $f_{p}^L$ and $f_{q}^R$ are the functions $M \to \mathbb R$ obtained by fixing the first and second variables to be $p,q$ respectively.
This is easily proved via observing the splitting of $T(M \times M)$ as a direct sum. |
H: Is it possible to calculate $x$ and $y$?
I am struggling with this.
If $p=\frac{(x+y)}{2}$ and $q=\frac{y}{x}$ and you know the values for $p$ and $q$, can you calculate what $x$ and $y$ are?
I tried $p=3.5$ and $q=2.5$ as an example. Given this $p$ and $q$, is the only possible solution $x=2$ and $y=5$? Or are there other solutions?
Thank you.
AI: $y = q x$ so your first equation becomes $p = \frac{1+q}{2} x$, therefore (assuming $q \ne -1$) the only solution is
$$ x = \frac{2p}{1+q},\ y = \frac{2qp}{1+q}$$
where we need $p \ne 0$.
Second case: $1+q = 0$. Then if $p \ne 0$, there is no solution, while if $p = 0$ the first equation becomes $0=0$ the solutions are $y=-x$, $x=$ anything except $0$. |
H: dimension of intersection of subspaces, one of which of dimension $n-1$
Let $H$ be a linear subspace of dimension $n-1$ of a linear space $V$ of dimension $n$.
Let $W$ be some subspace of $V$.
Show one of the following holds:
$W \subseteq H$ or
$\dim(W \cap H) = \dim(W) - 1$
This makes a lot of sense to me, as if there is some element in w that isn't in a largest subspace of V, then one of its dimensions can't be spanned by any of the vectors in the basis of $H$, thus its dimension is at most $\dim(W) - 1$.
It is intuitive, but I can't come up with a proof...
please help
AI: Hint: use the Grassman formula for dimension of $H + W$
$$\dim (H+W)+\dim (H\cap W)=\dim (W) + \dim (H)$$
If $W \subset H$ then $H + W=H$, otherwise it is all the ambient space $V$.
Edit: suppose $H+W=V$, then $\dim (H+W)=n$ and by Grassman formula:
$$n + \dim (H\cap W)=\dim (W) + \dim (H)=\dim (W) + n-1$$ |
H: Understanding notation and meaning of uniform convergence of a power series
I am trying to understand correctly the idea of uniform convergence in power series, and in general the notation for series of functions.
When we define a series of functions, we are defining a 'new' object, in which the terms of the series are functions rather than real values. Then we define this new object's value in terms of limits of sequences of functions. All of this is analogous to defining what a series of real values if, and we write:
$$\sum_{n=1}^{\infty}f_n = f$$
and we specify whether we mean uniformly or pointwise.
Here the series is a summation of functions, and its value is precisely a function. This seems to be quite commonly written:
$$\sum_{n=1}^{\infty}f_n(x) = f(x)$$
taken to mean the exact same thing as the previous equality (including whether it is pointwise or uniformly convergent to $f$). This notation sometimes becomes a bit confusing for me, because dependent on context we either mean $f(x)$ is a function itself, or $f(x)$ is the real value image of some parameter $x$ under $f$.
Say I define a power series and write $\sum_{n=1}^{\infty}a_n x^n = f(x)$ In this case, the symbol $x$ refers to a parameter, and $f(x)$ is simply the limit of the series of real values whenever $x$ lies in the radius of convergence. This means that the power series IS the function $f$, across the radius of convergence.
My question is then to clarify the exact meaning and notation when we say a power series is uniformly convergent on an interval. We have that $\sum_{n=1}^{\infty}a_n x^n = f(x)$ (in defining the power series), and I have often seen the uniform convergence written as $\sum_{n=1}^{\infty}a_n x^n = f(x)$ (uniformly), where sometimes the '(uniformly)' is dropped. I am hoping to clarify my understanding of these two identical notations.
Am I correct in saying that the first notation (defining the power series as a function), is information about the convergence of real valued series within the radius of convergence? That is I treat $x$ as some particular but unspecified value, and so the series is NOT a series of functions, but of real values? I.e. this allows me to create a well defined function $f$, whose domain is the radius of convergence.
In the second notation, is it correct to say that now $f_n(x)$ is being treated not as a real value on some parameter $x$, but instead to say that it IS a function? I.e. $f_n(x) = f_n$, where $f_n$ is defined by $f_n(x) = a_n x^n$? That is to say, the second notation asserts that the series of functions $\sum_{n=1}^{\infty}f_n$ is uniformly convergent to the function $f$ (which was defined above).
Ultimately, is asserting that a 'power series is uniformly convergent' equivalent to saying that a series of functions (given by $\sum_{n=1}^{\infty}f_n$) uniformly converges onto the power series itself?
AI: Yes, $\sum_{n=1}^\infty f_n= f$ is a sum of functions and is defined by saying that, for a specific value of x, f(x) is the value of the numeric sum $\sum_{n=1}^\infty f_n(x)$.
"And we specify whether we mean uniformly or pointwise". Not quite! Convergence of a sequence of series of functions is always "pointwise". It may then "converge uniformly" or not. The definition of convergence of "$\lim_{n\to\infty} f_n= f$" is that the numerical sequence "$\lim_{n\to\infty} f_n(x)$" converges to "$f(x)$" for every x in the domains of all the $f_n$. That is "pointwise" convergence- it converges at every "point"- every x-value.
We can then determine whether or not that convergence is also uniform.
Determining "pointwise" convergence is exactly the same as for numerical sequence. A sequence of functions, $f_n$, converges pointwise to f if and only if the numerical sequence, $f_n(x)$, converges to f(x) for every x in the domain. That is, if for a specific x, given $\epsilon> 0$ there exist N such that if n> N then $|f_n(x)- f(x)|< \epsilon$. This is done at every value of x- given an $\epsilon$, what N will work may be different for different x values.
The sequence of functions converges uniformly (on some interval) if, in addition to converging pointwise, given $\epsilon$ there exist an N that will work for all x in that interval.
Notice I was talking about sequences here. A series, $\sum_{n= 1}^\infty f$ converges to f if and only if the sequence of finite sums, $\sum{n=1}^N f_n$ converges to f so the same distinction between "pointwise" and "uniform" convergence applies. A series, if it converges, always converges "pointwise". It them may or may not converge "uniformly". |
H: If an abelian group has subgroups of orders $m$ and $n$, respectively, then it has a subgroup whose order is $\operatorname{lcm}(m,n)$.
I'm solving Exercise 2.5.26 from Topics in Algebra by Herstein. Could you please verify if my attempt is fine or contains logical mistakes?
Let $G$ be an abelian group and $H,K$ its subgroups of orders $m$ and $n$ respectively. Then $G$ has a subgroup of order $\operatorname{lcm}(m,n)$.
My attempt: First, we prove a simple lemma
Lemma: Let $G$ be an abelian group and $H,K$ its subgroups such that $o(H)= m$, $o(K|)= n$, and $H \cap K = \{1\}$. Then the $o(H \lor K) = mn$ where $H \lor K$ is the smallest subgroup containing both $H$ and $K$.
Proof: Because $G$ is abelian, $H \lor K =\{hk \mid h \in H \text{ and } k \in K\}$. Let $a,b \in H$ and $x,y \in K$ such that $ax = by$. Then $a^{-1}b = xy^{-1} \in H \cap K$ and thus $a^{-1}b =1$. Consequently, $a=b$ and $x=y$. The result then follows.
Come back to our general case, let $\operatorname{lcm}(m,n) = p_1^{n_1} \cdots p_k^{n_k}$ where $p_1,\ldots,p_k$ are prime numbers. For each $i$, we pick $C_i$ from the set $\{C \in \{H, K\} \mid p_i^{n_i} \text{ divides } o(C)\}$. By below Sylow theorem for abelian groups (in which I use first isomorphism theorem to give a simple proof here),
Let $G$ be an abelian group, $p$ a prime number, and $n$ a natural number. If $o(G)$ is divisible by $p^n$, then $G$ has a subgroup of order $p^n$.
we have $G$ has a subgroup $G_i$ with $o(G_i) = p_i^{n_i}$ for each $i$. Notice that $G_i \cap G_j = \{1\}$ for all $i \neq j$. Then the result follows by applying our Lemma repeatedly.
Update: Thanks to @aschepler for pointing out my typo. The correct version should be
we have $\color{blue}{C_i}$ has a subgroup $G_i$ with $o(G_i) = p_i^{n_i}$ for each $i$. Notice that $G_i \cap G_j = \{1\}$ for all $i \neq j$. Then the result follows by applying our Lemma repeatedly.
AI: $\DeclareMathOperator{\lcm}{lcm}$One possibility is to use the formula
$$
o(H K)
=
\frac{o(H) \cdot o(K)}{o(H \cap K)},
$$
then note that since $H \cap K \le H, K$, by Lagrange
$$
o(H \cap K) \mid o(H), o(K),
$$
so that
$$
o(H \cap K) \mid \gcd(o(H), o(K)),
$$
and finally recall that $\gcd(x, y) \cdot \lcm(x, y) = x y$.
With this you obtain that $G$ has a subgroup of order a multiple of the lcm, so you will have to know that the converse of Lagrange's theorem holds in a finite abelian subgroup. |
H: Is there a general form for the relation between $df(x)$ and $dx$?
Let's say that $f(x)$ is a function and $dx$ is a slight change in the input, $x$. It has been used here as standard calculus notation and hence, $dx \rightarrow 0$. $df$, or $df(x)$ is again, by standard calculus notation, the change in the value of $f(x)$ with the slight change in $x$ equal to $dx$ i.e. $df = f(x+dx) - f(x)$.
Now, I want to know if there is some general form for $df$ in terms of $dx$.
I have stated some examples below to make it clearer.
$$\text{If }f(x) = x^2 \implies df = (2x\cdot dx) + (dx)^2$$
$$\text{If }f(x) = x^3 \implies df = 3x(dx^2) + 3x^2(dx) + (dx)^3$$
$$\text{If }f(x) = \sin(x) \implies df = (\cos(x))\cdot dx$$
$$\text{If }f(x) = \cos(x) \implies df = (-\sin(x))\cdot dx$$
As far as I've noticed, for most of the functions, $df = a(dx) + b(dx)^2 + ... + k(dx)^n$, where $a \neq 0$. But, can there exist such a function that $df$ includes a constant term?
Let's assume that such a function exists. Let's call it $f_2$. Now, let's say that $df_2 = c + a(dx)$, where $c \in \Bbb R$. Now, how would we evaluate $f_2'(x)$ in this situation?
$f_2'(x) = \dfrac{df_2}{dx} = \dfrac{a+c(dx)}{dx} = \dfrac{a}{dx} + c$ as $dx \rightarrow 0$. So, does that mean that $f_2'(x) = \infty$?
Edit (more on the intentions for posting this question)
While deriving the power rule in one of his videos, Grant Sanderson (3Blue1Brown) considers $(df)(dg)$ as negligible but that didn't satisfy me much. So, while deriving $\dfrac{dx^2}{dx}$, I first simplified the RHS for the LHS being $\dfrac{dx^2}{dx}$ and the right hand side being $\dfrac{2x(dx)+(dx)^2}{dx} = 2x+dx$ as $dx\rightarrow 0$, which makes it equal to $2x$. Now, I thought of using the same for deriving $\dfrac{d((f(x)⋅g(x))}{dx}$ when it has been deduced that $d(f(x)⋅g(x)) = f(x)(dg) + g(x)(df) + (df)(dg)$.
Now, if I somehow, establish that $dx$ is a part of every single term in the expansion of $df$, then I can just take it out as a common part and then cancel it with the $dx$ in the denominator to have at least a single term without any power of $dx$. All the terms other than that term would then, collapse as they would all have $(dx)^n$ in them for $n \geq 1$ and as $dx \rightarrow 0$, those terms would approach $0$ as well. This would help me get a clearer view of the derivation and of differentiation itself.
Thanks!
AI: As I commented on a previous post of yours, one should keep in mind that writing expressions such as $\frac{a+c\cdot dx}{dx}$ really is taking notation too seriously, since the derivative is not a fraction and so on.
However, with the right interpretation one can kind of make sense of (at least some of) such manipulations. In this view, then the equation
$$df=a+c(dx)$$
(where $a$ is not an infinitesimal) doesn't make sense because on the left hand side you have an infinitesimal, and on the right hand side you have a nonzero constant (which cannot be infinitesimal).
Edit. Regarding the question of whether $df$ is always of the form $a dx + b (dx)^2+\dots+(dx)^n$.
As I mentioned before, Grant makes a series of informal but intuitive statements. One could choose different methods to formalize these informal statements, and the answer to the form of $df$ really depends on which method you use. For example, in synthetic analysis we always have $(dx)^2=0$, and hence $(dx)^n=0$ for $n>1$.
There is also something called nonstandard analysis, which is a system that also can be used to formalize the intuitive statements made by Grant. I'm not familiar with that formalism, though.
And of course there is the usual framework that uses limits to formalize calculus, and in this context the expressions $df$ and $dx$ by themselves don't make any sense.
Edit. knowing more about the motivation behind the question.
Since you want to know this in order to "cancel a factor $dx$ with the $dx$ in the denominator" you should know that, in all treatments, let it be the standard limit-based calculations, synthetic analysis or nonstandard analysis, every function which has a derivative can be said to "admit a differential which has at least one factor of $dx$ in every term" (this needs to be made rigourous in each of the frameworks, though). In this vague sense, the answer to your question would be "yes" for every differentiable function, and if this is not the case then the function is not differentiable.
For example, here is a calculation that shows the case you mention, as it happens in the usual limit-based framework.
Let $f:\mathbb R\to\mathbb R$ be defined by $f(x)=1$ for $x\neq 0$ and $f(0)=0$. Graphically, the situation is as follows:
Then for $x=0$ and any $h\neq 0$ you have
$$f(x+h)-f(x)=f(h)-f(0)=1-0=1$$
which you could interpret as telling you that "$df=1$ at $x=1$". What happens, then? Well, then the function is not differentiable at $x=0$, so the derivative does not exist. |
H: proof of uniform convergence of sequence of functions
Let be $f_n$ a sequence of functions with $f_n:\mathbb{R}\supset[\alpha, \beta] \to \mathbb{R}$ and $f_n$ continuous and bounded for all $n\in \mathbb{N}$. Further, $f_n$ converges point-wisely: $\lim\limits_{n\to\infty}f_n(x) = f(x)$, where $f(x)$ is also continuous and bounded for all $x\in[\alpha, \beta]$.
Can I conclude that $f_n$ converges uniformly?
I tried to construct a finite covering of $[\alpha,\beta]$ which should have delivered a $n_0$ such that for all $n>n_0$ the condition of uniform convergence holds. However, it didn't work...
May be it is not possible with further assumptions. In that case I would be interested in which further assumptions I need to prove uniform convergence?
AI: No, that is not true. Take, for instance, $f_n\colon[0,1]\longrightarrow\Bbb R$ defined by $f_n(x)=nx(1-x)^n$. Each $f_n$ is continuous and bounded and $(f_n)_{n\in\Bbb N}$ converges uniflmly to the null function. But the convergence is not uniform, since$$(\forall n\in\Bbb N):f\left(\frac1{n+1}\right)=\left(\frac n{n+1}\right)^{n+1}\text{ and }\lim_{n\to\infty}\left(\frac n{n+1}\right)^{n+1}=\frac1e\ne0.$$However, if, for each $x\in[\alpha,\beta]$, the sequence $(f_n(x))_{n\in\Bbb N}$ is increasing (or if, for each $x\in[\alpha,\beta]$, the sequence $(f_n(x))_{n\in\Bbb N}$ is decreasing), then, yes, the convergence is uniform; that's Dini's theorem. |
H: Finding the Constant Term of a Polynomial
How the coefficient of $t^{m+q}$ in the product $g'h'$ is
$$g_0h_{m+q}+g_1h_{m+q-1}+ \cdots +g_mh_q+\cdots g_{m+q}h_0 ?$$
For example, if $g_2=x^2+1, h_3=x^3+x^2+1$, then what is the value of $g_0h_{m+q}$? Plz explain reason in general for other terms.
The source of the problem -
AI: You have to decide the value of $m+q$. The given expression sums up the different ways to get the exponent of $t$ to be $m+q$. The first term comes from multiplying the constant term in $g$, which is $g_0$ by the $t^{m+q}$ term in $h$. The second comes from multiplying the linear term in $g$, which is $g_1t$ by the corresponding term in $h$, which is $h_{m+q-1}t^{m+q-1}$ and so on.
In your example, $g_0=1, g_1=0, g_2=1, h_0=1,h_2=1,h_3=1$ |
H: Why doesn’t this work: proof for the lower bound of the size $\sigma(k)$ of the smallest tournament with the Schütte property $S_k$?
Definitions:
Tournament and Dominating set
In a tournament $T$, for $u,v \in V(T)$, an edge “$u \rightarrow v$” means $u$ defeats $v$.
We say $T$ has the Schütte property $S_{k}$ if it has no dominating set of size at most $k$
Let $\sigma(k)$ be the minimum number of vertices in a tournament with the property $S_{k}$
I need to show that $\sigma(k) = \Omega(k2^k)$, i.e. any tournament on $k2^k$ vertices would have a dominating set of size at most $k$ . I’m trying a probabilistic argument below, but it turns out to be valid for any positive bound, and I can’t see where it went wrong.
Proof:
For every pair $x,y \in V(T)$, choose $x \rightarrow y$ or $y \rightarrow x$ uniformly at random.
Given a set $S \in\left(\begin{array}{c}{[k2^k]} \\ k\end{array}\right),$ let $E_{S}$ be the event that $S$ is a dominating set.
For $S$ to dominate a fixed vertex $v$, cannot have all edges $v \rightarrow S$
$k$ edges, chosen independently $\Rightarrow$ probability is $1-2^{-k}$
This must be true for all vertices in $V(T) \backslash S$
Edges again independent $\Rightarrow \mathbb{P}\left(E_{S}\right)=\left(1-2^{-k}\right)^{k2^k - k}$
Consider $\sum_{S} \mathbb{P}\left(E_{S}\right)$. If this quantity is positive, then at least one of the $\mathbb{P}(E_S)$ is positive.
But indeed $\sum_{S} \mathbb{P}\left(E_{S}\right)= {{k2^k} \choose k}\left(1-2^{-k}\right)^{k2^k - k} > 0$.
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Edit: I’m still stuck, but have come up with several ideas. Would like to hear if any of those could be developed further.
$\ $
Proof attempt 1: I’ve computed that, in a tournament on $[n]$ vertices, the total out-degree of all $k$-sets is $\frac{1}{2} {n \choose k} k(n-k)$, and so the average out-degree of the $k$-sets is $\frac{1}{2}k(n-k)$.
Assume for a contradiction that a tournament $T$ on $k2^k$ doesn’t have a dominating set of size at most $k$, and let $S$ be a $k$-set of maximum out-degree, and $v$ is a vertex in $V(T) \setminus S$ that dominates $S$.
By the above, the out-degree of $S$ must be at least $\frac{1}{2}k(n-k)$, and I’m trying to see if another $k$-set with larger out-degree exists, hence contradiction.
$\ $
Proof attempt 2: I’m trying a constructive proof to find a dominating set of size $k$ in the tournament $T$, with $V(T) = k2^k$, by repeated applications of the pigeon-hole principle.
Since the average out-degree of a vertex is $\frac{1}{2}(k2^k -1)$, for a vertex $v_1$ of maximum out-degree, the size of the set of vertices that dominates $v_1$ must be at most $k2^k - \frac{1}{2}(k2^k -1) - 1 = \frac{1}{2}(k2^k -1)$. From this set of vertices, we can find a vertex $v_2$ of maximum out-degree (w.r.t. this set of vertices), and so on.
I was hoping to construct a sequence of dominating set this way, but it turns out that I’d need to pick more than $k$ vertices before the size of the remaining vertices becomes less than $1$.
AI: You've shown that in a random tournament, $\sum_{S} \mathbb{P}\left(E_{S}\right) > 0$. (Incidentally, looking at the sum is unnecessary to see that $\mathbb{P}(E_S)$ is positive: you have a formula for $\mathbb{P}(E_S)$.)
When looking at an arbitrary tournament, you're not allowed to say that the probability of having edges $v \to s$ for all $s \in S$ is $2^{-k}$. Once you've chosen $S$, those edges are what they are. Your only possible source of randomness is in the choice of $S$.
In general, my advice when writing a probabilistic proof would be to begin by saying what exactly you're choosing at random, and from what distribution. |
H: Group actions, faithful, transitive
I have some questions on group actions, which occured from this problem:
We have the group $G=\operatorname{GL}_m(\mathbb{R})\times \operatorname{GL}_n(\mathbb{R})$ and the set $X=\operatorname{Mat}_{m\times n}(\mathbb{R})$
The group action is given as follows
$G\times X\to X$, $((P,Q),A)=PAQ^{-1}$.
As described in the text my questions are: How do you find the orbits of the group action in this specific case, and how to show that this actions is faithful or not.
I struggle a little bit with the notation/definition of group actions, as made clear in the text, and feel the need of some explanation.
Thanks in advance for helping me out.
This is indeed a group action as
$1X=((\mathbb{1}_m,\mathbb{1}_n), X)=\mathbb{1}_mX\mathbb{1}_n=X$
For $P=(P_1, P_2)$ and $Q=(Q_1, Q_2)$ in $G$ we have $PQ=(P_1Q_1, P_2Q_2)$ and
$(PQ)A=P(QA)$ is immediate to show.
I am asked to describe the orbits of the group action.
In general the orbit of an element $x\in X$ is the set $Gx=\{gx|g\in G\}$.
Also it is obvious that the orbit of the null matrix $0$ is $G0=\{0\}$.
Since the orbits are pairwise disjoint and the union is $X$ we can already conclude that the action is not transitive.
But I struggle to figure out what the other orbits are.
I guess that every other orbit is equal, so we have $GA=GB$ for $A\neq B$ and $A,B$ are not the null matrix.
But I can not proof it yet, as I do not know when you are able to manipulate a matrix $A$ into a matrix $B$ by muliplication with matrices of the general linear group, but I am convinced that this is possible.
I looked in a beginners book about linear algebra, but I did not find something necessary. Also this looks a little bit overly complicated.
However, my main question is about this figuring out if the group action is faithful.
The lecture notes read as follows:
Is $G$ a group and $\varphi$ a homomorphism $\varphi: G\to \operatorname{Sym}(X)$, we say that $G$ acts on $X$ and write $\varphi: G\times X\to X$.
I do not understand.
I know how to get the homomorphism $\varphi: G\to\operatorname{Sym}(X)$, but the group action (for example the one I am working with) is not a homomorphism. In general $X$ is just given as a set, and not with a structure.
To show that a group action is faithful we have to show that $\ker\varphi=\{1\}$.
But with which describtion do I work now? Besides that the kernel is only defined for a homomorphism.
Can I conclude from the group action given, or do I have to transform it into $\varphi: G\to \operatorname{Sym}(X)$?
Do you have to use both describtions side by side?
AI: Edit: I didn't notice that $m$ and $n$ were different.
Let's deal with the case $n=m$ first, as it gives you the right idea for the general case.
Since $P$ and $Q$ are invertible matrices, and you are running over all of them, you may replace $P$ by $PQ$. Then you are looking at $PQAQ^{-1}$. You might as well replace $QAQ^{-1}$ by $A$ because in the end you are likely to want a description that is independent of basis anyway.
So now, you want to find some invariant that distinguishes matrices apart that is not affected by multiplication by an invertible matrix. Once you have done that, then you want to decide if any two matrices with the invariant are related by multiplying on the left and right by invertible matrices, or equivalently as I have shown, a change of basis and a single multiplication.
First, what is the orbit containing $I$?
So once you look at the case $m=n$, you see that you want to use the one matrix to obtain a normal form of some sort, and then you can take transposes to push the 'adjustment' matrix to the other side and take normal forms again. The obvious one to do in general is Gaussian elimination, as that is more or less the only thing we can do with arbitrary matrices. |
H: Spiral equation
Considering concentric arcs, of equal developed length, whose start point is aligned:
I am looking for the equation of the spiral passing through the end points.
Some help to solve this problem will be welcome!
Edit: The result
AI: In polar coordinates, every arc starts at $\theta=0$ and ends at $\theta=L/r$, where $L$ is the length of each arc and $r$ is the radius for respective arc. So this is the equation:
$$\theta=L/r.$$
In Cartesian coordinates:
$$(x,y) = \left(r\cdot\cos\frac Lr,\, r\cdot\sin\frac Lr\right)$$
for $0 < r < \infty.$
The spiral is called hyperbolic spiral, or a reciproke spiral – see my post Does the spiral Theta = L/R have a name? and the answer to it. |
H: Given $\frac{z_1}{2z_2}+\frac{2z_2}{z_1} = i$ and $0, z_1, z_2$ form two triangles with $A, B$ the least angles of each. Find $\cot A +\cot B$
Question: If $z_1$ and $z_2$ are two complex numbers satisfying
$\frac{z_1}{2z_2}+\frac{2z_2}{z_1} = i$
and $0, z_1, z_2$ form two non-similar triangles. $A, B$ are the least angles in the two triangles, then $\cot A +\cot B$ equals:
I tried solving the first equation by trying to complete the square but to no avail. Then I tried taking $\frac{z_1}{z_2}$ as another variable $z$, hoping to use the rotation method but I couldn't figure out what to do with it. I think that I'm missing something, but even if I calculate $z_1$ and $z_2$ then would it not be insufficient to find a condition for minimum values of the cotangents of the angles?
AI: Write the equation $\frac{z_1}{2z_2}+\frac{2z_2}{z_1} = i$ as $(\frac{z_1}{2z_2})^2 -i \frac{z_1}{2z_2} +1 =0 $ and solve to obtain
$$\left(\frac{z_1}{z_2}\right)_{1,2}=(\sqrt5+1)e^{i\frac\pi2},\> (\sqrt5-1)e^{i\frac{3\pi}2}$$
Thus, the two are right triangles with $|z_1|>|z_2|$ and
$$\cot A + \cot B= \left|\left(\frac {z_2}{z_1}\right)_1\right|+ \left|\left(\frac {z_2}{z_1}\right)_2\right|=
{\sqrt5+1}+ {\sqrt5-1}=2{\sqrt5}
$$ |
H: Notation for "and"
Consider the real numbers $a$ and $b$. I would like to express in math notation, that $c = 2$ if $a > 0$ and $b > 0$. Can I use the wedge symbol here like
$c = 2 \quad \text{ if }a>0 \wedge b>0$
or is this wrong usage of the wedge symbol? Is there a better way to express this?
AI: It depends on you. Anyway in my opinion "and" is more elegant choice in this case. |
H: Proving that, given a surjective linear map, a set is a generator of a vector space
I know how to prove that, given $L: V \rightarrow W$ to be an injective linear map and $\{v_1, v_2, ..., v_n\}$ to be a linearly independent set in $V$, $\{L(v_1), L(v_2),...,L(v_n)\}$ is a linearly independent set in $W$
Proof:
We're given that $\{v_1, v_2, ..., v_n\}$ is a linearly independent set in $V$. Thus we have (let $a_i \in \Re$ where $1 \leq i \leq n$)
$$a_1 v_1 + a_2 v_2 \ + \ ... + \ a_n v_n =0$$
Where $v_i \in V$
Thus we know that $a_i = 0$
We want to show that $\{L(v_1), L(v_2),...,L(v_n)\}$ is a linearly independent set in $W$. Let us start off by writing
$$a_1 L(v_1) + a_2 L(v_2) \ + \ ... + \ a_n L(v_n) =0$$
Where $L(v_i) \in W$
As we're given $L$ to be linear
$$a_1 L(v_1) + a_2 L(v_2) \ + \ ... + \ a_n L(v_n) = L(a_1 v_1 + a_2 v_2 \ + \ ... + \ a_n v_n) = 0$$
As we're given $L$ to be injective (if $f(a) = f(a') \Rightarrow a=a'$) and we know that $0=L(0)$ (which can be easily proven using linearity) we get
$$L(a_1 v_1 + a_2 v_2 \ + \ ... + \ a_n v_n) = L(0) \Rightarrow a_1 v_1 + a_2 v_2 \ + \ ... + \ a_n v_n =0$$
Thus $a_i = 0$ and $\{L(v_1), L(v_2),...,L(v_n)\}$ is linearly independent in $W$
QED.
OK, I gave the above info because is related to what I'd like to ask.
My issue is how to prove that, given $L: V \rightarrow W$ to be a surjective linear map and $\{v_1, v_2, ..., v_n\}$ to be a generator of $V$, $\{L(v_1), L(v_2),...,L(v_n)\}$ is a generator of $W$
My try:
We're given that $\{v_1, v_2, ..., v_n\}$ is a generator of $V$. Thus we have (let $v \in V$ and $b_i \in \Re$ where $1 \leq i \leq n$)
$$V = \text{span} (v_1, v_2, ..., v_n)$$
And then
$$v = b_1 v_1 + b_2 v_2 \ + \ ... + \ b_n v_n$$
But I do not know how to show that
$$L(v) = b_1 L(v_1) + b_2 L(v_2) \ + \ ... + \ b_n L(v_n)$$
Based on linearity and surjectivity (where $L(v) \in W$).
AI: Let $w\in W$. Since $L$ is surjective, there is some $v\in V$ such that $L(v)=w$. By your observation, we may write $v=b_1v_1+\cdots + b_nv_n$ for some scalars $b_1,\dots, b_n$. Hence
$$ w=L(v)=L(b_1v_1+\cdots + b_nv_n)=b_1L(v_1)+\cdots + b_nL(v_n).$$
So $W=\mathrm{span}\{L(v_1),\dots,L(v_n)\}$. |
H: Nature of the series $\sum_{n=1}^\infty u_n$ where $u_{n+1} = \int_{0}^{u_{n}} \cos(x)^{n}dx$
Can you find the nature of the series $\sum_{n=1}^\infty$ given by $u_{n+1} = \int_{0}^{u_{n}} \cos^n(x)dx$?
You can show $u_{n}$ is convergent and the limit is 0. However it seems more difficult to find an equivalent to $u_{n}$ in order to study the series.
The series seems divergent and $u_{n}$ evolves like $\frac{1}{n}$ but I am unable to find a proof.
Any help ?
AI: If $n$ is odd and $x\geq 0$ then $\int_{0}^{x}\cos^n(t)\,dt$ is non-negative and bounded by $1$, so up to index-shifting we may assume that $u_0\in(0,1]$. We have that $\{u_n\}_{n\geq 0}$ is decreasing and positive, so it is convergent. Decreasing since
$$ u_{n+1} = \int_{0}^{u_n}\cos(t)^n\,dt < \int_{0}^{u_n}1\,dt = u_n.$$
Let us assume that $\lim_{n\to +\infty} u_n=L> 0$. Then for any $\varepsilon >0$ we have
$$ L' = \int_{0}^{L''}\cos(t)^n\,dt $$
for any sufficiently large $n$, with $L'$ and $L''$ being at most $\varepsilon$-apart from $L$. Since $\cos(t)^n$ is pointwise convergent to zero on $(0,1)$, by the monotonic/dominated convergence theorem $L$ must be zero.
Since $u_n\to 0$,
$$ u_{n+1}\sim\int_{0}^{u_n}e^{-nt^2/2}\,dt \sim u_n-\frac{n}{6}u_n^3 $$
such that $u_n\sim\frac{\sqrt{6}}{n}$ from the solution of the separable ODE
$$ f'(x) = -\frac{x}{6}f(x)^3 $$
and $\color{red}{\sum_{k=1}^{n}u_n\sim \sqrt{6}\log(n)}$ as $n\to +\infty$. |
H: If a complex analytic function is injective on a dense subset of an open connected domain, is it injective on the whole domain?
Let $D$ be an open connected domain inside $\mathbb C$ and let $g: D\rightarrow \mathbb C$ be an analytic function. Suppose there is a dense subset of $D$ on which $g$ is injective. Does $g$ have to be injective on all of $D$?
This type of property is certainly false in general for continuous or real differentiable maps. For example, one can take $h(x)=(\cos x, \sin x)$ on $[0,2\pi]$. It's less clear to me in the complex analytic case. Though, I still think it is false in general.
AI: As stated it is not true since one can take $f(z)=z^2$ on the unit disc and take the set $A$ which consists of points with rational coordinates in the right half of the disc $\Re z >0$ and points with irrational coordinates in the left half $\Re z <0$; obviously $A$ is dense and $f$ is $1-1$ on $A$
It is true if the dense set of injectivity $A$ is not "sparse" - for example, if for any open subset $U$ the measure of $U-A$ is zero.
first, it is clearly locally true in the sense that $f'(z) \ne 0, z \in D$ since near a critical point $f'(w)=0$ the local form of the function, $f(z)=f(w)+a(z-w)^{n+1}+O(z-w)^{n+2}, a \ne 0$ shows it cannot be injective but it is $n+1 \to 1$ where $n \ge 1$ is the order of the critical point, so in particular, there is a small open neighbourhood $U$ of $f(w)$ for which $f^{-1}(U)$ contains $n+1$ disjoint open subsets sent by $f$ into the same open set $W$ - the non-sparse hypothesis leads to a contradiction since if $U_1,U_2$ are two such disjoint open sets $f(U_1 \cap A) \cap f(U_2 \cap A)$ has full measure in $f(U_1)=f(U_2)=W$ so they cannot be disjoint
But now assuming $f(z)-c$ has two distinct zeroes $w_1,w_2$, using again the local form of $f$ at $w_1,w_2$, it follows that $f$ sends disjoint small neighbourhoods of $w_1,w_2$ into the same small neighborhood of $c$ again contradicting the hypothesis as above |
H: Is $R[x]/(g) \cong R[x]^{n-1}$ as groups, where $g$ is monic of degree $n$?
I am trying a problem that asks if it is true that $R[x]/(g) \cong R[x]^{(n-1)}$ (as groups under addition), when $g$ is a monic polynomial of degree $n$, and where $(g)$ is the ideal generated by $g$. Note, $R[x]^{(n)} = \{f \in R[x] : \deg(f) \leq n\}$ and $R$ is a commutative ring.
I know if $g = x^n$, this statement holds intuitively. Though, if $g = x^n + x + 1$ or any other polynomial with more than one term, I'm not so sure.
My hunch is that it is true and I would like to try and apply the first isomorphism theorem, but I cannot find a group homomorphism with kernel $(g)$, given that I believe $(g)$ can be irreducible.
I would appreciate any help on this!
AI: The map $R[x]^{(n-1)}\to R[x]/(g)$ that is induced by the inclusion is injective and onto. |
H: Proof of $\exists!A \ \forall B \ (A \cup B = B) $
This is example 3.6.2 of How to Prove It by Velleman, and the proof he gives is the following:
proof
Existence: Clearly $∀ B ( \varnothing ∪ B = B )$, so ∅ has the required property. Uniqueness: Suppose $∀ B ( C ∪ B = B )$ and $∀ B ( D ∪ B = B )$. Applying the first of these assumptions to $D$ we see that $C ∪ D = D$ , and applying the second to $C$ we get $D ∪ C = C$. But clearly $C ∪ D = D ∪ C$ , so $C = D $.
Is something wrong with this proof? I understand the reasoning and why it should be valid, but isn't it also true that $\forall B \ (B\cup B = B)$? If the proof is valid, then that means that $B = \varnothing$, but that's clearly not right.
The other possibility I can see is that writing $B\cup B$ is somehow invalid, but going by the definition of set union I don't see why this would be a problem.
AI: Proved is that a unique $A$ exists such that for every set $B$ we have $A\cup B=B$ and secondly that $A=\varnothing$.
In mathematical notation:$$\exists!A\forall B(A\cup B)=B\text{ and }\forall B(\varnothing\cup B)=B\tag1$$
Yes, it is true that $B\cup B=B$ for every $B$ but that does not lead to the conclusion that $B=\varnothing$.
It would lead to that conclusion if it was proved that for every set $B$ a unique set $A$ exists such that $A\cup B= B$ and secondly that $A=\varnothing$.
In mathematical notation:$$\forall B\exists!A(A\cup B)=B\text{ and }\forall B(\varnothing\cup B)=B\tag2$$
But that has not been proved, and also cannot be proved.
This simply because the first part of statement $(2)$ is false. |
H: Does ${f(x)=\ln(e^{x^2})}$ reduce to ${x^2\ln(e)}$ or ${2x\ln(e)}$?
I'm confused with the expression ${f(x) = \ln(e^{x^2})}$.I know the rule ${\log_a(x^p) = p\log_a(x)}$. So does the given expression reduce to ${x^2\ln(e)}$ or ${2x\ln(e)}$?
AI: It depends whether by ${e^{x^2}}$ you mean
$${e^{(x^2)}}$$
or
$${(e^{x})^2}$$
If you intend to say the first, then the answer reduces to ${x^2\ln(e)=x^2}$
, and by convention by ${e^{x^2}}$ we usually are referring to ${e^{(x^2)}}$.
If you are referring to the second, the answer indeed reduces to ${2x\ln(e)=2x}$.
In any case - if you want to clear any ambiguity in expressions - always put brackets. But like I said, convention would dictate that ${e^{x^2}=e^{(x^2)}}$ |
H: Let $G$ be a finite nilpotent group and $G'$ its commutator subgroup. Show that if $G/G'$ is cyclic then $G$ is cyclic.
So I thought the cleanest way to do this was to simply prove $G' = 1$ since if $G$ is cyclic $G' = 1$ and then $G \cong G/G'$, but I got no where with this.
My next idea was since $G$ is nilpotent I know it's the direct product of normal Sylow subgroups which commute with one another, so it suffices to show that each Sylow subgroup is cyclic. If $P$ is a Sylow $p$ subgroup then $PG'/G'$ is cyclic so by the second isomorphism theorem so is $P/P\cap G'$. So if $P$ intersects $G'$ trivial y then $P$ is cyclic. But I don't know what to do with the case where $P\cap G' \neq 1$.
AI: Let $$G=\gamma_0(G)>[G,G]=\gamma_1(G)>...>\gamma_n(G)=1$$ be the lower central series of $G$. We know that $G/\gamma_1(G)$ is cyclic. Let $k>1$ be the maximal number such that $G/\gamma_k(G)$ is cyclic. If $k=n$, we are done. So suppose $k<n$. Then $G_k/G_{k+1}$ is central in $G/G_{k+1}$. Since $G/G_k\equiv (G/G_{k+1})/(G_k/G_{k+1})$ is cyclic, we have that $G/G_{k+1}$ is abelian, so $\gamma_1(G)\le \gamma_{k+1}(G)$, a contradiction. QED |
H: Exists $t^*\in \mathbb{R}$ such that $y(t^*)=-1$?
$y'=y^2-3y+2, y(0) = \frac{3}{2}$
Exists $t^*\in \mathbb{R}$ such that $y(t^*)=-1$?.
How to prove without solving the ode? Any hint?
AI: Your equation admits two constant solutions, $y=1$ and $y=2$. By uniqueness, any other solution cannot intersect these constant solutions. Since $y(0) \in (1, 2)$, then you must have $1 < y(t) < 2$ for every $t$. |
H: Topological proof for the unsolvability of the quintic
Sorry, but in order to ask the question, you will have to view this video
http://drorbn.net/dbnvp/AKT-140314.php.
Here a topological proof for the unsolvability of the quintic is given, based on ideas of Vladimir Arnold. It's amazing because it does not require Galois Theory! And the proof takes only about 15 minutes in the video. (min 6 - min 22)
Here is my question: At min 7:10 Dror Bar-Natan removes all coefficients which lead to multiple roots, and says something like "It's a codimension 1 thing that we remove".
This is not quite clear to me. What does that mean?
Unfortunately I have not found an exposition of this proof, which is written down in a paper or a book, and which is as clear and easy as the exposition is this this video. If you have a reference here, please let me know!
Here are some more references I have found:
https://www.youtube.com/watch?v=zeRXVL6qPk4
https://www.maths.ed.ac.uk/~v1ranick/papers/abel.pdf
https://projecteuclid.org/download/pdf_1/euclid.tmna/1471875703
https://web.williams.edu/Mathematics/lg5/394/ArnoldQuintic.pdf
https://arxiv.org/pdf/1908.00972.pdf
Mathematical Omnibus, Dmitry Fuchs, Lecture 5
AI: When you have a submanifold of dimension $k$ inside an (ambient) manifold of dimension $n$, we define its codimension as $n-k$. Now you know that the discriminant as a function on $\mathbb{C}^6$ is given by some function (choose your favourite formula, mine is the Vandermonde determinant), but any formula you use should be smooth, and it gives you a function $\mathbb{C}^6 \to \mathbb{C}$. The "bad" points are where the discriminant is $0$ and you can check that $0$ is a regular value, so $f^{-1}(0)$ is something $5$-(complex) dimensional i.e. something codimension $1$. As for why it didn't matter, I can't really say anything besides I didn't see any issue in throwing out these points in the ensuing discussion (which is probably what Bar-Natan meant you'll see it doesn't matter). |
H: Assert the range of a binomial coefficient divided by power of a number
I found this question in a previous year post graduate entrance exam for mathematics.The question was
What is the range of
\begin{equation*}
\frac{200 \choose 100}{4^{100}}
\end{equation*}
The choices were
\begin{align*}
[\frac{3}{4}, 1) && \text{or} &&(0, \frac{1}{2}) && \text{or} && [1, \infty) && \text{or} && [\frac{1}{2}, \frac{3}{4})
\end{align*}
AI: Hint:
Compare $\binom{200}{100}$ with $\sum_{k=0}^{200}\binom{200}k$.
This in the understanding that: $$\sum_{k=0}^{200}\binom{200}k=2^{200}=4^{100}$$ |
H: Linearized system for $ \begin{cases} \frac{d}{dt} x_1 = -x_1 + x_2 \\ \frac{d}{dt} x_2 = x_1 - x_2^3 \end{cases} $ is not resting at rest point?
Assume there is the dynamical system
$$
\begin{align}
\frac{d}{dt} x_1 &= -x_1 + x_2 \\
\frac{d}{dt} x_2 &= x_1 - x_2^3
\end{align}
$$
The system is at rest at the point $(x_1, x_2) = (1, 1)$ and the point is stable. At this point of course
$$
\begin{align}
\frac{d}{dt} x_1 &= 0 \\
\frac{d}{dt} x_2 &= 0
\end{align}
$$
I want to investigate the rest point more and so I use the linear model from the Taylor series at the rest point:
$$
\frac{d}{dt}x = \begin{pmatrix} -1 & 1 \\ 1 & -3 \end{pmatrix}x
$$
I want to simulate both nonlinear and linear model. But something is strange. At the rest point I have:
$$
\frac{d}{dt}x = \begin{pmatrix} -1 & 1 \\ 1 & -3 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ -2 \end{pmatrix}
$$
So although the nonlinear model is at rest at $(1, 1)$ the linear model is not at rest there! So when I simulate both systems they are very different even at the start and even if the start point is very near to the rest point. Look:
The red $x_2$ trajectory is going even in the wrong direction at the start. What is the cause of the problem? Shouldn't the linear system approximate the nonlinear system at least when it starts near the rest point?
AI: The linearized model should read $\frac{dx}{dt}=J(x_0) \cdot (x-x_0)$, so in your case $\frac{dx}{dt}=\begin{pmatrix} -1 & 1 \\ 1 & -3 \end{pmatrix} \begin{pmatrix} x_1 - 1 \\ x_2 - 1 \end{pmatrix}$. This is the correct first order Taylor expansion of $F(x)=\begin{pmatrix} -x_1 + x_2 \\ x_1 - x_2^3 \end{pmatrix}$ at $(1,1)$. |
H: Is it possible to construct a continuous and bijective map from $\mathbb{R}^n$ to $[0,1]$?
Let $U$ be a non-trivial finite-dimensional vector space over $\mathbb R.$ I am trying to use a bijective and continuous map $f: U \to [0,1]$ and $d(x,y)=|f(x)-f(y)|$ to prove that there exist a metric on $U$ that makes $U$ compact. However, I couldn't find such continuous and bijective map: $f:U\to[0,1] \text{ (or $[0,1]^n$).}$ Is there any example? Or is there any other way to prove there exists a metric on $U$ that makes $U$ compact?
Edited: Thank you for all of your comments. I just started to learn compactness these days so I am not very good at some of the concepts. Now I understand that there is no need to construct a continuous map to prove the compactness. I also know that there does not exist a norm on U which makes U compact. My question is: how to prove there does exist a metric on U which makes U compact?
AI: To answer the question in the title: No for $n>1$.
If $f:\mathbb R^n\to[0,1]$ is continuous and surjective then $f^{-1}([0,\frac12))$ is a proper clopen subset of $f^{-1}([0,1]\setminus\frac12)$. That means $f^{-1}([0,1]\setminus\frac12)$ is disconnected. But $\mathbb R^n$ minus a single point is connected, so $f$ must not be injective. |
H: What is the Fourier transform of $|x|$?
I am trying to find the Fourier transform of $|x|$ in the sense of distributions in its simplest form. Here is what I have done so far:
Let
$$f(x)=|x|=\lim_{a\rightarrow 0}\frac{1-e^{-a|x|}}{a},$$
then the Fourier transform is given by
$$\begin{aligned}
\hat{f}(\xi)&=\int_{-\infty}^\infty f(x)e^{-2\pi i x \xi}dx \\
&=\lim_{a\rightarrow 0}\frac{1}{a}\left(\delta(\xi)-\frac{2a}{a^2+4\pi^2\xi^2}\right).
\end{aligned}$$
Using the identity (see here),
$$\delta(\xi)=\lim_{a\rightarrow 0}\frac{1}{\pi}\frac{a}{a^2+\xi^2},$$
we know that
$$2\pi\delta(2\pi\xi)=\lim_{a\rightarrow0}\frac{2a}{a^2+4\pi^2\xi^2}.$$
Hence, using the identity,
$$\delta(b x)=\frac{1}{|b|}\delta(x),$$
we know that
$$\hat{f}(\xi)\stackrel{?}{=}\lim_{a\rightarrow0}\frac{1}{a}[\delta(\xi) - \delta(\xi)].$$
This doesn't seem right... Can you see where I have gone wrong and do you know how to calculate $\hat{f}(\xi)$ in its simplest form?
AI: So, a way to compute it is to write $|x| = x\mathop{\mathrm{sign}}(x)$. By definition, we have
$$
\langle \mathcal{F}(|x|),\varphi\rangle = \langle |x|,\mathcal{F}(\varphi)\rangle = \langle x\mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi)\rangle
$$
Since $x∈ C^\infty$, we can then write
$$
\langle x\mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi)\rangle = \langle \mathop{\mathrm{sign}}(x),x\,\mathcal{F}(\varphi)\rangle = \frac{1}{2i\pi}\langle \mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi')\rangle
$$
where I used the formula for the Fourier transform of a derivative. Now, by definition again, and then using the fact that $\mathcal{F}(\mathop{\mathrm{sign}}(x)) = 1/{i\pi} \,\mathrm{P}(\tfrac{1}{x})$ (the principal value of $1/x$) we get
$$
\frac{1}{2i\pi}\langle \mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi')\rangle = \frac{1}{2i\pi}\langle \mathcal{F}(\mathop{\mathrm{sign}}(x)),\varphi'\rangle
\\
= \frac{-1}{2\pi^2}\langle \mathrm{P}(\tfrac{1}{x}),\varphi'\rangle = \frac{1}{2\pi^2}\langle \mathrm{P}(\tfrac{1}{x})',\varphi\rangle
$$
so that
$$
\mathcal{F}(|x|) = \frac{1}{2\pi^2} \mathrm{P}(\tfrac{1}{x})' = \frac{-1}{2\pi^2} \mathrm{P}(\tfrac{1}{x^2})
$$
where $\mathrm{P}(\tfrac{1}{x^2})$ is the Hadamard finite part of $\tfrac{1}{x^2}$. Away from $0$, we can thus say that
$$
\mathcal{F}(|x|) = \frac{-1}{2\pi^2x^2}
$$
(if I did not make mistakes in the constants and signs ...) |
H: How to graph the following type of functions and discuss its differentiability:
I am having trouble in graphing a particular type of functions where a function is divided piecewise, and for some pieces we have to be draw maximum part and for some we have to draw minimum part, for example,
Graph $g(x)$: $$
\text { Let } f(x)=x^{2}-2|x| \text { and } g(x)=\left\{\begin{array}{ccc}
\min \{f(t)\}, & -2 \leq t< 0, & -2 \leq x<0 \\
\max \{f(t)\}, & 0 \leq t \leq 2, & 0 \leq x \leq 2 \\
f(x) & x>2
\end{array}\right.
$$
Can anyone explain the concepts behind graphing such type of functions?
PS: Please provide a source for theory of graphing such type of functions.
AI: The first step is find $\min\{f(t)\}$ on $-2 \leq t < 0$ and $\max\{f(t)\}$ on $0 \leq t \leq 2$. Let's call these quantities $m$ and $M$ respectively. Among other methods, let's graph $f$ to find $m$ and $M$ (since the question is interested in graphing). Below is a graph of $f$ (I used Desmos for the graphing):
We quickly see $m = -1$ and $M = 0$. Using this information, the definition of $g(x)$ becomes:
$$
g(x) = \begin{cases} -1 & -2 \leq x < 0, \\
0 & 0 \leq x \leq 2, \\
f(x) & x > 2.
\end{cases}
$$
This definition is easier to visualize (we have eliminated the parameter $t$), and graphing this, we have:
(sorry if the blue lines are a bit hard to see) |
H: A finite group $G$ is called an $N$-group if the normalizer $N_G(P)$ of every non-identity p-subgroup $P$ of $G$ is solvable.
Prove that if $G$ is an $N$-group, then either (i) $G$ is solvable, or (ii) $G$ has a unique minimal normal subgroup $K$, the factor group $G/K$ is solvable, and $K$ is simple.
Suppose that $G$ is an $N$-group and $G$ is not solvable. Then we know that $G$ can not have a normal $P$ group. Also since for any non trivial normal subgroup $N$, let $P$ be a sylow subgroup of $N$, then by the Fratini Argument $G = N_G(P)N$. So by the second isomorphism theorem $G/N = N_G(P)N/N \cong N_G(P)/N \cap N_G(P)$. So since $N_G(P)$ is solvable and quotients of solvable groups are solvable $N_G(P)/N \cap N_G(P)$ is solvable, and hence $G/N$ is solvable.
So it suffices to show that if $G$ is an $N$-group and $G$ is not solvable that $G$ has a minimal normal subgroup. Since $G$ is not solvable we know that if we keep taking the commutator subgroup of the previous commutator subgroup we eventually get for some integer $n$ that $G^{n} = G^{n+ 1} \neq 1$ where $G^n$ is the $n^{th}$ commutator subgroup. Now we have that $G^n$ is characteristic and hence normal.
So I am convinced that I want to let $K$ be a minimal normal subgroup of $G^{n}$, since $
AI: So here is a full proof to complete the question.
Let $M$ be a minimal normal subgroup of $G$. Then $M$ is the direct product of isomorphic simple groups $S$. If $S$ is cyclic then $M$ is a $p$-subgroup such that $N_G(M)=G$, so $G$ is soluble. Thus $S$ is non-abelian and $M$ is the direct product of $n$ copies of $S$. If $Q$ is a Sylow $p$-subgroup of $S$ for some $p\mid |S|$ then $n-1$ copies of $S$ appear in $N_G(Q)$. Since $N_G(Q)$ is soluble though, this means $n=1$. Thus $M$ is non-abelian simple.
Finally, by a Frattini argument $G=M\cdot N_G(Q)$, where $Q\in\mathrm{Syl}_p(M)$. Since $N_G(Q)$ is soluble by assumption, and $G/M\cong N_G(Q)/N_M(Q)$, we see that $G/M$ is soluble. |
H: Right adjoint to the forgetful functor $\text{Ob}$
Let $\text{Ob}:\textbf{Cat}\rightarrow\textbf{Set}$ be the forgetful functor mapping a small category to its set of objects. Consider the functor $R:\mathbf{Set}\rightarrow\textbf{Cat}$ mapping a set $X$ to the category having $X$ as its set of objects and a single morphism between each pair of objects. I am trying to show that $R$ is right adjoint to $\text{Ob}$.
For $x,y\in X$, should the single morphism between $x$ and $y$ in $R(X)$ be unique? Or, can I use the same morphism for each pair of objects in $R(X)$?–Does it matter?
Edit:
Giving unique arrows would complicate the matter because I would have to specify the composition for each pair; whereas, in giving a constant morphism $*$, I can merely specify $*\circ*:=*$. Right?
AI: All that matters is that the hom-set $R(X)(x, y)$ is a singleton for each $x, y \in X$. Identities and composition are therefore trivial, because there's only a single choice. It doesn't matter exactly what the singleton sets are (e.g. whether they are equal or not). You should try to avoid thinking of sets up to equality, and instead consider them only up to isomorphism. |
H: Let $\lambda$ be a real eigenvalue of matrix $AB$. Prove that $|\lambda| > 1$.
Let $A$ and $B$ be real symmetric matrices with all eigenvalues strictly greater than 1. Let
$\lambda$ be a real eigenvalue of matrix $AB$. Prove that $|\lambda| > 1$.
My solution:
Let $a$ and $b$ be eigenvalues of $A$ and $B$ corresponding the eigenvectors $y$ and $x$, respectively.
Looking at the following dot product:
$$\langle ABx,y \rangle = \langle Bx,A^Ty \rangle=\langle Bx,Ay \rangle = \langle bx,ay \rangle=ab\langle x,y \rangle=\langle abx,y \rangle$$
we get $$(AB)x=(ab)x$$ Therefore, $\lambda := ab$ is an eigenvalue of $AB$. Since $a>1$ and $b>1$, it follows that $\lambda > 1$
However, it doesn't seem ok as the problem was actually asking to prove $|\lambda|>1$. Indeed, $\lambda > 1 \implies |\lambda|>1$, but then the problem wouldn't write $|\lambda|$ in my opinion.
The given solution:
The transforms given by $A$ and $B$ strictly increase the length of every nonzero vector, this
can be seen easily on a basis where the matrix is diagonal with entries greater than $1$ in the diagonal.
Hence their product $AB$ also strictly increases the length of any nonzero vector, and therefore its real
eigenvalues are all greater than $1$ or less than $-1$.
Any help is appreciated.
AI: $AB$ is similar to $A^{1/2}BA^{1/2}$. Hence its eigenvalues are positive.
Full solution:
Since $A$ and $B$ are positive definite, $AB$ is similar to the positive definite matrix $A^{1/2}BA^{1/2}$. Hence $AB$ has a positive spectrum.
Furthermore, since $A$ and $B$ are unitarily diagonalisable and their eigenvalues are greater than $1$, we have $\|Ax\|_2,\|Bx\|_2>\|x\|_2$ for all nonzero vector $x$. It follows that $\|ABx\|_2=\|A(Bx)\|_2>\|Bx\|_2>\|x\|_2$ for all nonzero $x$. As $AB$ has a positive spectrum, the eigenvalues of $AB$ must be positive numbers greater than $1$.
Alternative solution (that uses the induced $2$-norm for matrices). Since $A,B\succ I$, we have $0\prec A^{-1},B^{-1}\prec I$ and $\|A^{-1/2}B^{-1}A^{-1/2}\|_2\le\|A^{-1/2}\|_2^2\|B^{-1}\|_2=\|A^{-1}\|_2\|B^{-1}\|_2<1$. Hence $0\prec A^{-1/2}B^{-1}A^{-1/2}\prec I$ and $A^{1/2}BA^{1/2}\succ I$. Since $A^{1/2}BA^{1/2}$ is similar to $AB$, all eigenvalues of $AB$ are positive numbers greater than $1$. |
H: Evaluating an improper integral - issues taking the cubic root of a negative number
Problem:
Evaluate the following integral.
$$ \int_{-1}^{-1} \frac{dx}{x^\frac{2}{3}} $$
Answer:
This integral includes the point $x = 0$ which results in a division by $0$. To get around this difficulty, we break the integral into two integrals.
\begin{align*}
\int_{-1}^{-1} \frac{dx}{x^\frac{2}{3}} &= \int_{-1}^{0} \frac{dx}{x^\frac{2}{3}} + \int_{0}^{1} \frac{dx}{x^\frac{2}{3}} \\
\int_{-1}^{0} \frac{dx}{x^\frac{2}{3}} &= \int_{-1}^{0} x^{-\frac{2}{3}} \,\,\, dx \\
\int_{-1}^{0} \frac{dx}{x^\frac{2}{3}} &= 3x^{\frac{1}{3}} \Big|_{-1}^0
= \lim_{x \to 0} 3x^{\frac{1}{3}} - \lim_{x \to -1} 3x^{\frac{1}{3}} \\
\lim_{x \to 0} 3x^{\frac{1}{3}} &= 0 \\
\end{align*}
Am I right so far? I do not know how to evaluate the following limit:
$$ \lim_{x \to -1} 3x^{\frac{1}{3}} $$
The problem is taking the cube root of a negative number.
AI: Cube root of a real number $p$ is the unique real number $q$ such that $q^3=p$. Therefore, $x^{\frac{1}{3}} \to -1$ as $x \to -1$ because $(-1)^3=-1$ and so $(-1)^{\frac{1}{3}}=-1$. |
H: Integral of a product of Bessel functions of the first kind
I want to do this integral $H(\rho)=\int_{0}^{\infty} J_1(2 \pi Lr)J_0(2\pi \rho r)dr$, where $J_1$ and $J_0$ are Bessel functions of the first kind and $L\in \mathbb{R}$ is a constant, so I tried to do this in the Mathematica, but he failed. When I tried to put some value to $L$ and $\rho$, the software calculate numerically, so I ploted $H(\rho)$ for a fixed $L$ and the result of the plot is a function like $rect(x/L)$, such that
\begin{equation}
{\displaystyle \operatorname {rect} (t)=\left\{{\begin{array}{rl}0,&{\text{if }}|t|>{\frac {1}{2}}\\{\frac {1}{2}},&{\text{if }}|t|={\frac {1}{2}}\\1,&{\text{if }}|t|<{\frac {1}{2}}.\end{array}}\right.}
\end{equation}
I'm not sure about this result for $H(\rho)$, so I searched in the internet and didn't find none property for solve this integral, I don't know if, in fact $H(\rho)=rect(x/L)$ or something of this type. Someone knows if this result is correct? This integral have an analytic solution?
AI: The general relationship you need is Eqn. 10.22.63 in the Digital Library of Mathematical Functions:
$$
\int_{0}^{\infty}J_{\mu}\left(ax\right)J_{\mu-1}\left(bx\right)\mathrm{d}x=%
\begin{cases}b^{\mu-1}a^{-\mu},&0<b<a,\\
(2b)^{-1},&b=a(>0),\\
0,&0<a<b,\end{cases}
$$
assuming $\Re(\mu) > 0$.
In your case, $a = 2 \pi L$, $b = 2 \pi \rho$, and $\mu = 1$. The result you found above is what you'd get if you set $a = 1$ and $b = t$.
While the given equation seems to require positive $a$ and $b$, I believe you can extend the result to all real non-zero $a$ and $b$ by changing the variable of integration $t \to - t$ and/or using the property that $J_n(-x) = (-1)^n J_n(x)$. |
H: Connectives in George Tourlakis' Mathematical Logic
In page 10 of Mathematical Logic, Tourlakis says that "Readers who have done some elementary course in logic, or in the context of a programming course, may have learned that ¬, ∨ are the only connectives one really needs since the rest can be expressed in terms of these two."
Capture from the book
How can we express the other connectives (→, ∧, ≡) using only ¬ and ∨?
Capture from the book
AI: The following logical equivalences (or equations in Boolean algebra) can be checked in a variety of ways:
\begin{align}
P \wedge Q & = \neg (\neg P \vee \neg Q) \\
P \to Q & = \neg P \vee Q \\
P \leftrightarrow Q & = (P \to Q) \wedge (Q \to P) \\
& = (\neg P \vee Q) \wedge (\neg Q \vee P) \\
& = \neg\bigl( \neg (\neg P \vee Q) \vee \neg (\neg Q \to P)\bigr)
\end{align}
The first, for example, can be considered as using De Morgan's Law, but could also be verified using truth tables. |
H: Inverting product of non-square matrices?
I am working on an Optimization problem and I need to show that
$$AB(B^TAB)^{-1} B^T = I_n$$
where $A$ is $n \times n$ and invertible, $B$ is $n \times k$ with rank $k$, and $k \le n$.
If $B$ is square, then this is a simple calcuation. For $B$ non-square, I have tried the following way (which I'm pretty sure is supposed to work, but there must be a slight flaw in my reasoning somewhere): Since $B$ represents an injection, it has a left inverse; therefore there must be a matrix $D$ such that $DB = I_k$ (and thus $B^T D^T = I_k$). Now, we claim that
$$(B^TAB)^{-1} = DA^{-1}D^T.$$
To prove this, we would like to show that
$$(B^TAB) (DA^{-1}D^T) = I_k.$$
However, the multiplication doesn't simplify, since $D$ is a left inverse, not a right inverse.
Any help is greatly appreciated. Thank you!
AI: If $k < n$ this is impossible, as the rank of the left side can't be more than $k$ (i.e. $B$ has rank at most $k$, and $\text{rank}(UV) \le \min(\text{rank}(U),\text{rank}(V))$. |
H: Let $L\in End(V)$ with $L(V)=W$. Then $Tr(L)=Tr(L|_W)$
Let $V$ be a finite dimensional vector space and $W$ a subspace. Let $L$ be an endomorphism with image is $W$. Then $Tr(L)=Tr(L|_W)$ where $L|_W$ denotes the restriction of $L$ to $W$.
I am unsure what definition of trace I should use to prove this, it seems "obvious" if the trace is simply the sum of the diagonal elements of a matrix. But I am not sure if we are allowed to restrict linear maps when using such a definition.
One idea is: $L|_W$ is an isomorphism of $W$, this gives is a nice basis of $W$ which we can extend to a basis of $V$ such that the matrix of $L$ is 0 in the last $n-m$ collums and scalars along the first m diagonals. Then maybe $L|_W$ is given as matrix just by getting rid of the last $n-m$ columns and then we have the result.
Edit:
Specifically I am confused with how to calculate the trace of the restriction of a map. I know that the trace is the sum of the diagonals of the matrix that represents the map, and the this does not depend on the basis that we choose. But how do we do this when we restrict our map to a subspace? It cannot be that we still sum over the diagonals. Suppose that was the case then we have $0=Tr(L|_{Ker(L)}=Tr(L)\neq 0$.
AI: This is a special case of the general result that if $W$ is an invariant subspace for $L \in \operatorname{End}(V)$ (i.e. $L(W) \subseteq W$), then we can form a restriction map $L {\mid}_W \in \operatorname{End}(W)$ and an induced map $\bar L \in \operatorname{End}(V / W)$; and:
$$\operatorname{Tr}(L) = \operatorname{Tr}(L {\mid}_W) + \operatorname{Tr}(\bar L).$$
In the particular case given here, we certainly have that $L(W) \subseteq L(V) \subseteq W$, so $W$ is an invariant subspace for $L$. Also, the induced map $\bar L$ is equal to 0 in this case since $\bar L(x + W) = (Lx) + W = 0 + W$ for every $x \in V$. Therefore, $\operatorname{Tr}(\bar L) = 0$, and the desired result follows.
The idea for the proof of the general result is: start with a basis $\beta_W = \{ x_1, \ldots, x_m \}$ of $W$, and extend to a basis $\beta_V = \{ x_1, \ldots, x_m, x_{m+1}, \ldots, x_n \}$ of $V$. Then in $[L]_{\beta_V}$, the upper left $m \times m$ submatrix is equal to $[L {\mid}_W]_{\beta_W}$, and the lower right $(n-m) \times (n-m)$ submatrix is equal to $[\bar L]_{\beta_{V / W}}$ where $\beta_{V / W} := \{ x_{m+1} + W, \ldots, x_n + W \}$ is a basis for $V / W$. |
H: Connected and Hausdorff topological space whose topology is stable under countable intersection,
We know that the converging sequences of a discrete space are the stationary sequences.
I am looking for two examples for spaces (not empty or reduced to a singleton)
connected and Hausdorff topological space where the converging sequences are the stationary sequences.
connected and Hausdorff topological space whose topology is stable under countable intersection.
AI: There is an example in Arvind K. Misra, A topological view of P-spaces, General Topology and its Applications, Volume 2, Issue 4, December 1972, 349-362. It starts with the space $E_0$ that he constructs in Example $\bf{3.1}$, a Hausdorff $P$-space (i.e., one in which $G_\delta$-sets are open) with two points $a$ and $b$ that cannot be separated by a continuous function. In Example $\bf{5.3}$ he recursively constructs from $E_0$ spaces $E_n$ for $n\in\omega$ in such a way that $E_n$ is embedded in $E_{n+1}$ and then defines $E_\omega$ to be the direct limit of the sequence $\langle E_n:n\in\omega\rangle$. (The topology on $E_\omega$ is the final topology determined by the embeddings.) $E_\omega$ is a Hausdorff $P$-space on which every real-valued continuous function is constant, so it is connected.
In any $P$-space every convergent sequence is eventually constant. |
H: let $\{a_n\} \downarrow 0$, show that $\sum_1^\infty a_n$ converge if and only if $\sum_1^\infty 2^na_{2^n}$ converge
let $\{a_n\} \downarrow 0$, show that $\sum_1^\infty a_n$ converge if and only if $\sum_1^\infty 2^na_{2^n}$ converge.
If $\{a_n\}$ is a non-increasing sequence, then this is Cauchy condensation test for infinite series.
But the question says $\{a_n\}$ tends to $0$ from the right, and a sequence does not need to be non-increasing to tend to $0$, for example, $1/2, 1, 1/3, 1/2, 1/4, 1/3, \dots$ tends to $0$.
So how should I prove this statement in this case?
AI: The notation $\{a_n\} \downarrow 0$ usually means that the sequence is non-increasing (and tends to zero).
The Cauchy condensation test does not work without the monotony condition: If you define $a_n = 0$ if $n = 2^k$ for some $k$, and $a_n = 1/n$ otherwise, then $\sum_{n=1}^\infty 2^na_{2^n}$ converges, but $\sum_{n=1}^\infty a_n$ does not converge. |
H: Solution to autonomous differential equation with locally lipscitz function
As I was learning about the following theorem and its proof from the book Nonlinear Systems by H. K. Khalil, I encountered a difficulty in grasping some parts of the proof.
Theorem: Consider the scalar autonomous differential equation
\begin{equation}
\dot{y}=-\alpha(y),\ y(t_0)=y_0,\tag{1}
\end{equation}
where $\alpha$ is a locally Lipschitz class $\kappa$ function defined on $[0,a)$. For all $0\leq{y_0}<a$, this equation has a unique solution $y(t)$ defined for all $t\geq{t_0}$. Moreover,
\begin{equation}
y(t)=\sigma(y_0,t-t_0),\tag{2}
\end{equation}
where $\sigma$ is a class $\kappa\ell$ function defined on $[0,a)\times[0,\infty)$.
The proof goes as follows.
Since $\alpha(.)$ is locally Lipschitz, the equation (1) has a unique solution $\forall\ {y_0}\geq{0}$. Because $\dot{y}(t)<0$ whenever $y(t)>0$, the solution has the property that $y(t)\leq{y_0}$ forall $t\geq{t_0}$. By integration we have,
\begin{equation}
-\int_{y_0}^{y} \dfrac{dx}{\alpha(x)}= \int_{t_0}^{t} d\tau.
\end{equation}
Let b be any positive number less than $a$ and define $\eta(y)=-\int_{b}^{y}\dfrac{dx}{\alpha(x)}$. The function $\eta(y)$ is strictly decreasing differentiable function on $(0,a)$. Moreover, $\lim_{y\to{0}}\eta(y)=\infty$. This limit follows from two facts.
First, the solution of the differential equation $y(t)\to{0}$ as $t\to\infty$, since $\dot{y}(t)<0$ whenever $y(t)>0$.
Second, the limit $y(t)\to{0}$ can happen only asymptotically as $t\to\infty$; it cannot happen in finite time due to the uniqueness of the solution.
Here I do not quite understand the second fact (in italics) how the uniqueness of solution ensures that $y(t)$ goes to $0$ asymptotically as $t\to\infty$.
Any hints on this are greatly appreciated.
AI: That's not what it's saying. It's saying $y(t) \to 0$ can't happen in finite time, i.e. there can't be a solution $Y(t)$ of the differential equation with $Y(t_0) = y_0$ and $Y(t_1) = 0$ for some $t_1 > t_0$.
Suppose that did happen.
Note that $y(t) = 0$ is also a solution of the differential equation, because part of the definition of class $\kappa$ is $\alpha(0)=0$. So this would contradict the Existence and Uniqueness Theorem, as there would be two different solutions $Y$ and $0$ having the same value at $t_1$. |
H: Evaluating $\sum\limits_{i=\lceil \frac{n}{2}\rceil}^\infty\binom{2i}{n}\frac{1}{2^i}$
Let $a_n = \sum\limits_{i=\lceil \frac{n}{2}\rceil}^\infty\binom{2i}{n}\frac{1}{2^i}$. Prove that $a_{n+1} = 2a_n + a_{n-1}$.
I have tried considering the derivatives of $\frac{1}{1-x^2}=1+x^2+x^4+...$, and although this may work, it certainly does not seem like it would be the best way to go about solving this problem.
AI: First, as Steven Stadnicki's question comment states, you can also get the same result as what you're asking for by starting "from $i = 0$" because "$\binom{a}{b} = 0$ when $a \lt b$" (note $\binom{a}{b}$ means the number of ways to choose $b$ items from among $a$ items, but there are $0$ ways to do this if $b \gt a$).
Next, for positive natural numbers $n$ and $k$, Pascal's rule states
$$\binom{n}{k} = \binom{n - 1}{k} + \binom{n - 1}{k - 1} \tag{1}\label{eq1A}$$
With positive natural numbers $i$ and $n$, this gives
$$\binom{2i}{n + 1} = \binom{2i - 1}{n + 1} + \binom{2i - 1}{n} \tag{2}\label{eq2A}$$
Using \eqref{eq1A} on both terms of the RHS of \eqref{eq2A} gives
$$\binom{2i - 1}{n + 1} = \binom{2i - 2}{n + 1} + \binom{2i - 2}{n} \tag{3}\label{eq3A}$$
$$\binom{2i - 1}{n} = \binom{2i - 2}{n} + \binom{2i - 2}{n - 1} \tag{4}\label{eq4A}$$
Substituting these into \eqref{eq2A}, multiplying both sides by $\frac{1}{2^{i}}$, summing from $i = 1$ to $\infty$, changing the summation indices on the RHS and making a few algebraic manipulations, you end up with
$$\begin{equation}\begin{aligned}
\binom{2i}{n + 1} & = \binom{2i - 2}{n + 1} + 2\binom{2i - 2}{n} + \binom{2i - 2}{n - 1} \\
\binom{2i}{n + 1}\frac{1}{2^i} & = \frac{1}{2}\binom{2i - 2}{n + 1}\frac{1}{2^{i-1}} + \binom{2i - 2}{n}\frac{1}{2^{i-1}} + \frac{1}{2}\binom{2i - 2}{n - 1}\frac{1}{2^{i-1}} \\
\sum_{i=1}^{\infty}\binom{2i}{n + 1}\frac{1}{2^i} & = \frac{1}{2}\sum_{i=1}^{\infty}\binom{2(i - 1)}{n + 1}\frac{1}{2^{i-1}} + \sum_{i=1}^{\infty}\binom{2(i - 1)}{n}\frac{1}{2^{i-1}} + \frac{1}{2}\sum_{i=1}^{\infty}\binom{2(i - 1)}{n - 1}\frac{1}{2^{i-1}} \\
a_{n+1} & = \frac{1}{2}\sum_{i=0}^{\infty}\binom{2i}{n + 1}\frac{1}{2^{i}} + \sum_{i=0}^{\infty}\binom{2i}{n}\frac{1}{2^{i}} + \frac{1}{2}\sum_{i=0}^{\infty}\binom{2i}{n - 1}\frac{1}{2^{i}} \\
a_{n+1} & = \left(\frac{1}{2}\right)a_{n+1} + a_{n} + \left(\frac{1}{2}\right)a_{n-1} \\
\left(\frac{1}{2}\right)a_{n+1} & = a_{n} + \left(\frac{1}{2}\right)a_{n-1} \\
a_{n+1} & = 2a_n + a_{n-1}
\end{aligned}\end{equation}$$ |
H: Prove/disprove that $ \mathbb{R}^2 / $~ is hausdorff, when: $(x_1,x_2) $~$(y_1,y_2)$ if there is $t>0$ such that $x_2 = tx_1 $ and $ty_2 = y_1$
We look at the following equivalence relation on $\mathbb{R}^2$:
$(x_1,x_2) $~$(y_1,y_2)$ iff there exists $t>0$ such that $x_2 = tx_1 $ and $ty_2 = y_1$
The task is to prove/disprove: $Y = \mathbb{R}^2 / $~ is Hausdorff.
Well, I am not sure how exactly to approach this problem. I tried to prove this stright forward but had no lack.
I also thought of maybe finding a homeomorphism from $Y$ to a Hausdorff space, which will prove it since homeomorphims, preserves Hausdorff. But I couldn't find such of a homeomorphism.
Help would be appreciated.
AI: For the purposes of visualization, it may help you to think of this space as the quotient by a group action: let $(0,\infty)$ act on $\mathbb{R}^2$ according to $t\cdot(x,y) = (tx,y/t)$. The orbits of this action are mainly hyperbolas, as shown in the plot below, though there are also five other "exceptional" orbits (what are they?).
Although it isn't actually necessary, things probably look a little more standard if we work with the "logarithm" of this action, i.e. convert it into into an action of the additive group $\mathbb{R}$ so that the orbits of the action are integral curves of the vector field associated to the action. In this form, the $\mathbb{R}$ action is the linear action $t \cdot (x,y) = (e^t x , e^{-t}y)$ and, taking $\frac{d}{dt}|_{t=0}$, the generating vector field is the linear vector field, $X(x,y) = (x,-y)$, whence the plot shown above.
I think by staring at this picture, you will probably convince yourself quite quickly that the quotient should not be Hausdorff. Any neighbourhoods (saturated with respect to the equivalence relation or not) of the two disjoint orbits consisting of the positive and negative $x$-axis must overlap. I encourage you to fill in the details yourself. |
H: Prove a metric space is totally bounded
Let $X = 2^{\mathbb{Z+}}$ be the space of binary sequences $(x_k)_{k\ge1}$ with each $x_k \in \{0,1\}.$ Define a metric on $X$ by $d(x,y) = \sum^\infty_{k=1} |x_k−y_k|/2^k.$ I am trying to use the definition of a totally bounded space but I haven't found the $\varepsilon$-net of $X.$ My question is how to find the $\varepsilon$-net so that we can prove that $X$ is totally bounded?
AI: If I'm understanding the question correctly, what you're looking for is, for each $\varepsilon>0,$ a finite set of points in your space such that every point in the space is within a distance $\varepsilon$ of some point in that finite set.
Find the smallest positive integer $k$ such that $2^{-k}<\varepsilon.$ Then consider the set of all sequences of the following form: $$x_1, x_2, x_3, \ldots,x_k,\,\underbrace{0, 0, 0, 0, 0, \ldots\ldots}$$
There are only $2^k$ of these, a finite number. And every point is within $\varepsilon$ of one of these. |
H: Definition of ring of dual numbers
In an exercise from Vakil's algebraic geometry notes, he asks us to describe the set $\rm Spec\space k[\epsilon]/(\epsilon^2)$, where $k$ is a field. A comment from this question gives a solution, but it's under the assumption that $\epsilon$ is transcendental, and so the assumption may be made that $k[\epsilon]$ is a PID. However, I am not sure why we can assume this. In the question, Vakil says "you should think of $\epsilon$ as a very small number, so small that its square is 0 (although itself is not zero)". If $\epsilon$ is any number we want, they we can certainly choose a value for which $k[\epsilon]$ is not a PID. Am I missing something?
AI: The precise definition is $k[\epsilon]=k[X]/(X^{2})$, where $\epsilon = X \bmod X^2$. Then $\epsilon^2=0$.
$k[\epsilon]$ is a PIR because it's a homomorphic image of $k[X]$, but it's not a domain because $\epsilon$ is a zero divisor. |
H: The Grothendieck-Serre Correspondence : Obstructions to the construction of the cyclic group of order 8 as a Galois group?
I am reading this note called as The Grothendieck-Serre Correspondence by Leila Schneps where this quote occurs:
the author still recalls Serre’s unexpected reaction of spontaneous delight upon being shown a very modest lemma on obstructions to the construction of the cyclic group of order 8 as a Galois group, simply because he had never spotted it himself.
I am quite confused. I read the statement as:
obstruction to the [construction of cyclic group of order 8] as a Galois group
We can create a cyclic group of order 8: $(\mathbb Z/8\mathbb Z, +, 0)$. So I'm not sure what the "galois group obstruction is". Is the above quote to be read as:
obstruction to the [construction of cyclic group of order 8 as a Galois group] ?
I am still confused, because I thought the field $\mathbb Q(\zeta_8)/\mathbb Q$ has galois group $\mathbb Z/8\mathbb Z$ (where $\zeta_8$ is the $8$-th root of unity), from Kummer theory?
So what obstruction is the quote referring to which delighted Serre?
AI: This is more of a long comment than an answer (I believe I found an exact reference to the obstruction, but my summary of it might not be correct).
One guess is the obstruction discussed in "On Cyclic Field Extensions of Degree 8" (a paper written by the author of the article you cite, and which also mentions Serre).
The correct statement of the problem is: there is no "versal" polynomial for $\mathbb Z/8\mathbb Z$ over $\mathbb Q$, a polynomial with generic coefficients whose Galois group over its coefficient field is that group, and whose specializations give rise to all the polynomials with that galois group (see the article for more details).
The references therein explain the obstruction. I only skimmed things quickly and might be badly misunderstanding what is going on, but I think the idea is that if there was such a polynomial existed then you could find a specialization unramified and inert over $2$, but that contradicts the fact that every extension of $\mathbb Q$ with Galois group $\mathbb Z/8\mathbb Z$ such that 2 doesn’t split is ramified over $2$. Hopefully someone more knowledgeable can jump in. |
H: Are chain complexes chain equivalent to free ones?
Given a chain complex $A_\bullet\in\mathrm{Ch(\mathbf{Ab})}$, are there exist some chain complex $A'_\bullet\in\mathrm{Ch(\mathbf{Ab})}$ which is chain equivalent to $A_\bullet$ such that $A'_p$ are all free abelian groups?
AI: I'm answering my own question.
No. A counter example is $A_\bullet := A_0\to A_1\to A_2 := 0\to\mathbf{Z}/2\mathbf{Z}\to 0$.
Assume there exists a chain complex $A'_\bullet$ of free abelian group with $\varphi:A_\bullet\to A'_\bullet, \psi:A'_\bullet\to A_\bullet$ and a homotopy $\alpha$ between $\psi\circ\varphi$ and $\mathrm{id}_{A_\bullet}$.
It lead to a contradiction Because
$\phi_1:\mathbf{Z}/2\mathbf{Z}\to B$ is a null map because $B$ is free and
the only $\alpha_0$ and $\alpha_1$ are null maps. |
H: Question about step in Ahlfors' proof of Cauchy's inequality in complex analysis.
I'm reading Ahlfors' Complex Analysis. In this book, he provides the following proof of Cauchy's inequality. Using $|a -b|^2 = |a|^2 + |b|^2 - 2 \Re\left(a\overline{b}\right)$ he establishes the following
$$0\le \sum_{k=1}^n \bigr\lvert a_k - \lambda \overline{b_k}\bigr\rvert^2 = \sum_{k=1}^n |a_k|^2 + \underbrace{|\lambda|^2\sum_{k=1}^n |b_k|^2}_{a)} - 2 \underbrace{\Re\left(\overline{\lambda}\sum_{k=1}^n a_kb_k\right)}_{b)}$$
where $\lambda$ is some arbitrary complex number. He then proceeds to take the particular value of $\lambda$ to be
$$ \lambda =\frac{\sum_{j=1}^n a_jb_j}{\sum_{j=1}^n |b_j|^2} $$
and using this, he says that after simplifications you obtain the following:
$$ \sum_{k=1}^n |a_k|^2 - \frac{\Bigr\lvert\sum_{k=1}^n a_kb_k\Bigr\rvert^2}{\sum_{k=1}^n |b_k|^2}\ge0$$
which proves Cauchy's inequality.
I wanted to expand this and check this result for myself. I separated the problem into $2$ parts:
$\textbf{a)}$ For $|\lambda|^2\sum_{k=1}^n |b_k|^2$ I got the follwing:
$$
|\lambda|^2\sum_{k=1}^n |b_k|^2 = \frac{\Bigr\lvert\sum_{j=1}^n a_jb_j\Bigr\rvert^2}{\Bigr\lvert\sum_{j=1}^n |b_j|^2\Bigr\rvert^2}\sum_{k=1}^n |b_k|^2 = \frac{\Bigr\lvert\sum_{j=1}^n a_jb_j\Bigr\rvert^2}{\left(\sum_{j=1}^n |b_j|^2\right)^2}\sum_{k=1}^n |b_k|^2 = \frac{\Bigr\lvert\sum_{k=1}^n a_kb_k\Bigr\rvert^2}{\sum_{k=1}^n |b_k|^2}
$$
Here I use the property $\bigr\lvert|x|^2+|y|^2\bigr\rvert = |x|^2 + |y|^2$. I believe this is justified because $|x|^2 + |y|^2\in \mathbb{R}$, that $|x|^2 + |y|^2\ge 0 + 0 \ge 0$, and that the modulus of a positive real number is the real number itself. I think the reasoning is correct, but I'm not completely sure.
$\textbf{b)}$ For $\Re\left(\overline{\lambda}\sum_{k=1}^n a_kb_k\right)$ I got the following:
$$
\Re\left(\overline{\lambda}\sum_{k=1}^n a_kb_k\right) = \Re\left(\frac{\overline{\sum_{j=1}^n a_jb_j}}{\overline{\sum_{j=1}^n |b_j|^2}}\sum_{k=1}^n a_kb_k\right) = \Re\left(\frac{\sum_{j=1}^n \overline{a_jb_j}}{\sum_{j=1}^n |b_j|^2}\sum_{k=1}^n a_kb_k\right) =\frac{\Re\left( \sum_{j=1}^n\sum_{k=1}^n\overline{a_jb_j}a_kb_k\right)}{ \sum_{j=1}^n |b_j|^2}
$$
Where I used the fact that $\overline{x} = x$ for $x \in \mathbb{R}$, and also that $\Re\left(\frac{x}{c} + i \frac{y}{c}\right) = \frac{x}{c} = \frac{\Re(x + iy)}{c}$. And here is where I ran into trouble.
I know that for the specific case where $j=k$ I can make a simplification using the fact that $z \overline{z} = |z|^2$, but this still leaves the other cases where $j \neq k$, and I don't know how I could find the real part of these terms.
I also tried using the fact that $\Re(z) \le |z|$ and that $|a +b| \le |a| + |b|$. Using this I got that
$$
\frac{\Re\left( \sum_{j=1}^n\sum_{k=1}^n\overline{a_jb_j}a_kb_k\right)}{ \sum_{j=1}^n |b_j|^2} \le \frac{\Bigr\lvert \sum_{j=1}^n\sum_{k=1}^n\overline{a_jb_j}a_kb_k\Bigr\rvert}{ \sum_{j=1}^n |b_j|^2} \le \frac{ \sum_{j=1}^n\sum_{k=1}^n|a_jb_j|\cdot|a_kb_k|}{ \sum_{j=1}^n |b_j|^2}\le \frac{ \sum_{s=1}^n|a_sb_s|^2}{ \sum_{j=1}^n |b_j|^2}
$$
where for the last inequality I just did $\sum_{j=k} + \sum_{j\neq k} \le \sum_{j=k}$. But even with this, this doesn't give me a result that simplifies to the desired conclusion.
I don't if there's a step I'm doing wrong or of there's something I'm missing, but I can't seem to get to the inequality I want to arrive at. Can anyone tell if I'm on the right track? Thank you!
AI: Hint: With the choice of $\lambda=\frac {\sum_{j=1}^n a_jb_j}{\sum_{j=1}^n|b_j|^2},$ $$\bar{\lambda}\sum_{k=1}^na_kb_k$$ is already real and equals $$\frac{|\sum_{j=1}^n a_j b_j|^2}{\sum
_{j=1}^n|b_j|^2}.$$ |
H: Is the determinant a tensor?
I was reading Schutz's book on General Relativity. In it, he says that a(n) $M \choose N$ tensor is a linear function of $M$ one-forms and $N$ vectors into the real numbers.
So does that mean the determinant of an $n \times n$ matrix is a $0 \choose n$ tensor because it is a function that maps the $n$ column vectors of the matrix to a real number (the value of the determinant)?
But then, the determinant also maps the $n$ column vectors of the matrix to the same real number (the value of the determinant).
So would the tensor representation of the determinant be different if you choose the map for the column vectors than the map for the row vectors?
AI: Yes, in fact the determinant $\det:(\mathbb R^n)^n\to \mathbb R$ is (up to constant multiple) the only alternating $n$-multilinear map (i.e. alternating $n$-covariant tensor, see John Lee, Introduction to
Riemannian
Manifolds, 2nd edition (2018), pages 400 and following). See for instance this question for a proof of this fact. |
H: Injection from $(K\otimes_{\mathbb{Z}}\hat{\mathbb{Z}})^*\to \prod_p(K\otimes_{\mathbb{Z}}\mathbb{Z}_p)^*$?
Suppose $K$ is a number field. The projections $\hat{\Bbb{Z}} = \prod_p\Bbb{Z}_p\to\Bbb{Z}_p$ give a map
$$(K\otimes_{\Bbb{Z}}\hat{\Bbb{Z}})^*\to \prod_p(K\otimes_{\Bbb{Z}}\Bbb{Z}_p)^*.$$
I want to show that this map is injective. It is clear to me that if $x\otimes (y_p)\in K\otimes_{\Bbb{Z}}\hat{\Bbb{Z}}$ is a unit, then it maps to $\prod_p 1\otimes 1$ if and only if $x=1$ and each $y_p=1$. But surely there are units that are not single tensors, and that is where I don't know how to proceed; I have very little intuition about either tensor products or $\hat{\Bbb{Z}}$.
I would prefer some explanation of the concepts, or of the problem I'm having, to a solution to my specific question.
AI: In fact, more is true: the ring homomorphism $K\otimes_{\Bbb Z}\hat{\Bbb Z} \rightarrow \prod_p(K \otimes_\Bbb Z \Bbb Z_p)$ is injective.
To see this, we first look at the case $K = \Bbb Q$. It is easy to see that every element of $\hat{\Bbb Q} = \Bbb Q\otimes_{\Bbb Z} \hat{\Bbb Z}$ can be written as a pure tensor (because a "common denominator" can be found for any finite sum of pure tensors), thus you should be able to understand this special case.
For general $K$, we fix an isomorphism $K \simeq \Bbb Q^d$ as $\Bbb Q$-vector spaces. We then have $$K\otimes_{\Bbb Z}\hat{\Bbb Z} \simeq K\otimes_{\Bbb Q}\Bbb Q \otimes_{\Bbb Z} \hat{\Bbb Z} \simeq \hat{\Bbb Q}^d$$ and also $$K\otimes_{\Bbb Z} \Bbb Z_p \simeq K\otimes_{\Bbb Q}\Bbb Q \otimes_{\Bbb Z}\Bbb Z_p \simeq \Bbb Q_p^d.$$
Together with the projection maps $K\otimes_{\Bbb Z} \hat{\Bbb Z}\rightarrow K\otimes_{\Bbb Z}\Bbb Z_p$ and $\hat{\Bbb Q}^d \rightarrow \Bbb Q_p^d$, we get a commutative square. The injectivity then follows from the above special case.
There is actually a more explicit description of $K\otimes_Z \hat{\Bbb Z}$. It is equal to the following set: $$\{(x_p)\in\prod_p (K\otimes_{\Bbb Z}\Bbb Z_p): x_p\in \mathcal O_K\otimes_{\Bbb Z}\Bbb Z_p, \text{almost all } p\},$$ where "almost all" means "all but finitely many". This is sometimes called a "restricted tensor product".
The group of units $(K \otimes_{\Bbb Z}\hat{\Bbb Z})^\times$ can be described similarly, by adding $^\times$ everywhere.
You may read the wiki page on adele rings or many books on the subject for more information. |
H: What are some ("small") Riesel numbers without a covering set?
In the thread Does every Sierpinski number have a finite congruence covering? some examples of proven Sierpiński numbers that seem to have no full covering sets, are given. So it seems natural to ask if the same type of examples have been found for Riesel numbers, i.e. fixed odd positive numbers $k$ with the property that $k\cdot 2^n-1, n\in\mathbb{N}$ produces no primes.
AI: (Answering my own question.)
Indeed, explicit examples of such Riesel numbers are known. We are talking about Riesel numbers $k$ where algebraic factorizations account for some subset of the exponents $n$, and a partial covering takes care of the remaining cases.
The interesting aspect of such numbers $k$, is that $k\cdot 2^n - 1$ is composite for all $n\in\mathbb{N}$ (proven), yet the least prime factor of $k\cdot 2^n - 1$ seems to be unbounded as $n$ tends to infinity. See thread linked in the question above.
The algebraic factorization is maybe a bit simpler in the Riesel case ($-1$, as opposed to the $+1$ in Sierpiński). For every $a>1$, we have the high-school factorization of integer polynomials:
$$x^a - 1 = (x-1)(x^{a-1}+x^{a-2}+\dots+x^2+x+1)$$
So if we pick a $k$ that is a perfect $a$th power, $k=\ell^a$, then for all $n$ divisible by $a$ we have that:
$$k\cdot 2^n - 1 = \ell^a\cdot 2^{ma} - 1 = (\ell\cdot 2^m)^a - 1$$
is algebraically factorizable from the above formula (take $\ell\cdot 2^m = x$). So we only have to cover the $n$ that are not divisible by $a$ with some "partial covering" of fixed primes.
So it seems most natural to start with perfect squares ($a=2$). And in fact this has been done in a published paper, namely Michael Filaseta, Carrie Finch, and Mark Kozek: On powers associated with Sierpiński numbers, Riesel numbers and Polignac's conjecture, Journal of Number Theory, Volume 128, Issue 7, July 2008. In it, they use the Riesel number:
$$k=3896845303873881175159314620808887046066972469809^2$$
and so all the even exponents $n$ in the expression $k\cdot 2^n-1$ are covered by the factorization, and the odd exponents are covered by the prime set:
$$\{ 7, 31, 127, 17, 151, 257, 41, 337, 241, 113, 65537, 71, 97, 673, 1321, 641, 6700417, 281, 14449, 29191 \}$$
for which the modulus (least common multiple of the orders of 2 modulo the primes) is $6720$.
I am not sure how Filaseta et al. came up with this covering set, but it may not be so easy to improve on as you might think at first glance. The problem with square Riesel numbers is that the excellent primes $3$ and $5$ (of orders 2 and 4, respectively) cannot be used to cover any positions (that are not already covered by the algebraic factorization). So without those two primes in the set, which normally cover 75% of all positions, and with other primes "lost" as well (see Sierpiński thread linked in question), the covering set becomes big, and so the least $k=\ell^2$ (found with the Chinese remainder theorem) becomes big as well.
While I have not come up with a better set than Filaseta et al., I did find smaller $\ell$ values that work with the same covering. The best of them being:
$$k=1956890318063807573129004316753152275056499243^2$$
However, I expect smaller square Riesel numbers (with the property that no full covering seems to exist) can be found. Has anyone beaten my example here? New records wanted!
But when I thought about this, in connection to the linked thread, it occurred to me that smaller Riesel numbers of this type should be possible by going from squares to cubes, i.e. $a=3$. I already knew a partial covering for cubes from the other thread, and its $k$ (a Sierpiński) was smaller. I discussed this with user "Gelly", and he found for me the cube:
$$k=1469583304447640330447613742^3$$
To prove it is Riesel, all $n$ (from expression $k\cdot 2^n-1$) divisible by 3 are handled by the factorization, and for the remaining, use the covering:
$$\{ 3, 5, 127, 17, 43, 257, 29, 113, 5419, 15790321, 449, 2689, 2017 \}$$
of modulus $672$. And note that this, just like the other examples mentioned, seems to be a $k$ that does not admit a full covering set. When $n$ is $39 \pmod{672}$ or $375 \pmod{672}$, we conjecture that no finite covering suffices. So we think that the least prime factor of $1469583304447640330447613742^3\cdot 2^{672m+39}-1$ (and the same with exponent $672m+375$) is unbounded.
Again, I think this cube can be "improved", i.e. a smaller example found. Has this been done by someone already? References or new small/minimal examples are very welcome.
Conclusion: Perfect powers such as $1956890318063807573129004316753152275056499243^2$ (found by me, with the covering set from the linked paper) and $1469583304447640330447613742^3$ (found by user "Gelly") are proven Riesel numbers for which it seems that no covering sets exist. |
H: Understanding the $\alpha$ existence in the definition of the boundary map in case of simplicial homology and its absence in singular homology.
Here is the boundary map in case of simplicial homology(AT pg.105):
And here is the boundary map in case of singular homology(AT pg.108):
My question is:
I know that the difference between the 2 homologies is that in the first case we are using simplicial complexes as basis but in the second case we are using continuous maps as basis. My question is, why $\alpha$ exists in the definition of the boundary map in case of simplicial homology and it is absent in singular homology definition?
Could anyone help me answer this question, please?
AI: These pictures are from Hatcher yes? You can actually find the explanation there, but I'll try explain it a little more clearly than he does.
So, a $\Delta$-complex structure on $X$ is defined as a collection of continuous maps $\sigma_\alpha:\Delta^n \to X$ subject to various conditions. Then $\Delta_n(X)$ are the free abelian groups generated by these maps. And from the example that is given in Hatcher after the definition of a $\Delta$-complex is given, we see that for a torus this can be given by $6$ such maps, $2$ from $\Delta^2$, $3$ from $\Delta^1$ and $1$ from $\Delta^0$. So for the torus $\Delta_0(X)$ is generated by $1$ map, and so on. Therefore when you talk about the boundary map and how it acts on the basis of $\Delta_n(X)$ it makes sense to include the index $\alpha$.
On the other hand each $C_n(X)$ is generated by all continuous maps $\sigma:\Delta^n\to X$. There are no other conditions on these maps, and there are likely uncountably many of them, and therefore no natural way to index them.
Besides that the boundary maps are basically the same, as mentioned in the comments below the question. The appearance of an index in the first definition is just cosmetic; there to remind you that you're defining it for an easily indexed basis. |
H: Is this a valid strategy to find group new group automorphisms given you already know some?
It is well known that Aut($G$), the group of all automorphisms of a group $G$, is a group under function composition. Lets say that you know $n$ of the automorphisms (plus the identity isomorphism) on $G$, hence you have
$$H=\{id, \phi_1, \phi_2,\dots,\phi_n\}$$
Lets also say we do not know if $H = \text{Aut}(G)$ yet. I thought of the following strategy to come up with new Automorphisms:
Check if $H$ is a group. If not, take some $\phi_i,\phi_j \in H$ and find $\phi_i \cdot \phi_j$. If $\phi_i \cdot \phi_j$ is not in $H$, we have found another automorphism on $G$; add it to $H$. If it is not a new automorphism, pick new $\phi_i, \phi_j$. Repeat this process until $H$ is a group.
My question is, is this even a good strategy? One clear problem is that if we arrive at a subgroup of $\text{Aut}(G)$, we are obviously not going to generate any more automorphisms of $G$. Is there any way to optimize this strategy? Another thought I had was to take
$$\phi_i^2, \phi^3,\dots,id$$
to find new automorphisms instead of composing $2$ randomly chosen ones, but I am not sure if this is better or worse.
AI: Stated somewhat differently: suppose you have some elements $h_1, \ldots, h_n$ of a finite group.
Start with $S = \{h_1, \ldots, h_n\}$, and mark them all.
Repeat while $S$ has some marked elements:
... Let $a$ be the first marked element $a$ of $S$.
.. Place into $S$ all elements $ab$ with $b \in S$ that are not already in $S$, and mark them.
.. Remove the mark from $a$.
When done, $S$ will be the subgroup of your group generated by $h_1, \ldots, h_n$. |
H: Why does this not satisfy the conditions of a metric?
Suppose we would like to define a metric on New York City, let t : NYC × NYC → R+ is
a function that measures the time it takes to travel between two points in New York City.
Why doesn’t t satisfy the criteria of being a metric?
I know that a metric must satisfy triangle inequality, non-negativity, and symmetry. I feel like this satisfies all of these conditions. Why is it not a metric, though? Thanks for the help!
AI: First and most important, it’s not well-defined: the time needed to get from one point in New York City to another depends on the time, date, and means of transportation. Secondly, it need not be symmetric even at a particular time and date using a particular means of transportation. |
H: $X_{1},X_{2},X_{3}\overset{i.i.d.}{\sim}N(0,1)$, find m.g.f. of $Y=X_{1}X_{2}+X_{1}X_{3}+X_{2}X_{3}$
I tried this
$X_{1}X_{2}+X_{1}X_{3}+X_{2}X_{3}=X_{1}(X_{2}+X_{3})+\frac{1}{4}(X_{2}+X_{3})^{2}-\frac{1}{4}(X_{2}-X_{3})^{2}$
$U=X_{2}+X_{3}\sim N(0,2)$
$\psi_{X_{1}(X_{2}+X_{3})}(t)=\psi_{X_{1}U}(t)=\frac{1}{\sqrt{2}\cdot 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{x_{1}ut}e^{-\frac{1}{2}(x_{1}^{2}+\frac{u^{2}}{2})}\, dx_{1}\, du$
$=\frac{1}{\sqrt{2}\cdot 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(2(x_{1}-ut)^{2}-2u^{2}t^{2}+u^{2})}\, dx_{1}\, du$
$V=(x_{1}-ut)\qquad dv=dx_{1}\\
=\frac{1}{\sqrt{2}\cdot 2\pi}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(1-2t^{2})u^{2}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}v^{2}}\, dv\, du\\$
$=\frac{1}{\sqrt{\pi}\cdot 2}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(1-2t^{2})u^{2}}\, du$
$ w=\sqrt{\frac{1-2t^{2}}{2}}u,\qquad \sqrt{\frac{2}{1-2t^{2}}}\, dw=du\\
\psi_{XU}(t)=\frac{1}{\sqrt{1-2t^{2}}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}w^{2}}\, dw=\frac{1}{\sqrt{1-2t^{2}}}$
$Z=X_{2}-X_{3}\sim N(0,2)$
$\psi_{\frac{1}{4}(X_{1}+X_{2})^{2}}(t)=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{\frac{1}{4}z^{2}t}e^{-\frac{1}{4}z^{2}}\, dz$
$=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(z^{2}-z^{2}t)}\, dz=\frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{4}(1-t)z^{2}}\, dz\\$
$v=\sqrt{1-t}z,\quad \frac{1}{\sqrt{1-t}}\, dv=dz\\
\psi_{\frac{1}{4}(X_{1}+X_{2})^{2}}(t)=\frac{1}{\sqrt{1-t}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{4\pi}}e^{-\frac{1}{4}z^{2}}\,dz=\frac{1}{\sqrt{1-t}}$
$\psi_{\frac{1}{4}(X_{1}+X_{2})^{2}-\frac{1}{4}(X_{2}-X_{3})^{2}}(t)=\frac{1}{\sqrt{1-t}}\cdot \frac{1}{\sqrt{1-(-t)}}=\frac{1}{\sqrt{(1-t)(1+t)}}=\frac{1}{\sqrt{1-t^{2}}}$
$\psi_{X_{1}(X_{2}+X_{3})+\frac{1}{4}(X_{2}+X_{3})^{2}-\frac{1}{4}(X_{2}-X_{3})^{2}}(t)=\frac{1}{\sqrt{1-2t^{2}}}\cdot \frac{1}{\sqrt{1-t^{2}}}=\frac{1}{\sqrt{1-2t^{2}}\sqrt{1-t^{2}}}$
But I have learned that this is incorrect. What mistakes did I make?
AI: The the three terms in the first line are not independent. How can you just multiply the MGF's to get the MGF of their sum? MGF of $X+Y$ is the product of the MGF's of $X$ and $Y$ if $X$ and $Y$ are independent, not in general. |
H: Why random variables is a function? It seems that it violates the definition of function.
I am watching Lecture 5 of MIT 6.041 Probabilistic Systems Analysis and Applied Probability. In the lecture, Professor said Random Variable is a function that maps elements from the sample space to a real number. For example, if the sample space is the students in the classroom, a random variable $x$ can be the height of the students.
However, I remember that a function is defined as $\forall a \in A \exists! b \in B((a,b) \in F)$. That means for all inputs, there is one unique output. In the example above, we may have two students with the same height. Will that violate the rule of function? Or Professor just used the word function loosely?
AI: It is not a violation of the definition of a function when two unequal inputs return equal outputs. Rather, the rule is that a single input cannot return more than one output!
For example, when one first is introduced to functions, one is mostly paying attention to things like $f(x) =x^2$, which defines a function from the set of real numbers to the set of real numbers. In this case, there are two "students", say the inputs $x=1$ and $x=-1$, which have the same "height", both inputs give the output of $1$. This is not a problem and $f(x)=x^2$ is definitely a function. The same is true in the context of random variables. Your reading of the definition is just slightly off, is all. |
H: Finding the volume when a parabola is rotated about the line $y = 4$.
Problem:
Find the volume generated by revolving the region bounded below by the
parabola $y = 3x^2 + 1$ and above by the line $y = 4$ about the line $y = 4$.
Answer:
Let $V$ be the volume we are trying to find. We want to find the points where the two curves intersect. Hence,
set up the following equation:
$$ 3x^2 + 1 = 4$$.
From this equation, we find two solutions: $ x = \pm 1 $
\begin{align*}
du &= -dx \\
V &= \pi \int_{-1}^{1} (4 - (3x^2 + 1))^2 \,\,\, dx = \pi \int_{-1}^{1} (3 - 3x^2)^2 \,\,\, dx \\
V &= 9 \pi \int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \int_{-1}^{1} x^4 - 2x^2 + 1 \,\,\, dx \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{x^5}{5} - \frac{2x^3}{3} + x \Big|_{-1}^{1} \\
%
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &=
\frac{^5}{5} - \frac{2(1)^3}{3} + 1 - \left( \frac{(-1)^5}{5} - \frac{2(-1)^3}{3} - 1 \right) \\
%
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{1}{5} - \frac{2}{3} + 1 - \left( -\frac{1}{5} + \frac{2}{3} - 1 \right) \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{1}{5} - \frac{2}{3} + 1 + \frac{1}{5} - \frac{2}{3} + 1 \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= 2 - \frac{4}{3} + \frac{2}{5} = 2 - \frac{20}{15} + \frac{6}{15} \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{16}{15} \\
V &= 9 \pi \left( \frac{16}{15} \right) \\
V &= \frac{ 48 \pi }{5}
%
\end{align*}
The book's answer is:
$$ \frac{144 \pi}{15} $$
I claim my answer is right. That is, the book failed to simplify its answer. Am I missing something? Please comment.
AI: $$\frac{144 \pi}{15}$$
dividing the numerator and denominator by three:
$$\frac{48 \pi}{5}$$
In the future, if you don't feel like simplifying, you can just divide the two answers and see if it lines up
$$\frac{144 \pi}{15} \approx 30.1593$$
$$\frac{48 \pi}{5} \approx 30.1593$$
So, yes, you're right. |
H: $f:[0,1]\rightarrow[0,1]$, measurable, and $\int_{[0,1]}f(x)dx=y\implies m\{x:f(x)>\frac{y}{2}\}\geq\frac{y}{2}$.
Question: Suppose $f:[0,1]\rightarrow[0,1]$ is a measurable function such that $\int_0^1f(x)dx=y$. Prove that $m\{x:f(x)>\frac{y}{2}\}\geq\frac{y}{2}$.
My thoughts: Since we have $\int_0^1f(x)dx=y\implies \frac{1}{2}\int_0^1f(x)dx=\frac{y}{2}$. Now, I was hoping to be able to split the integral bounds and consider $\int f$ over $\{x:f(x)>\frac{y}{2}\}$ and over $\{x:f(x)\leq\frac{y}{2}\}$, and somehow use Markov's inequality in there somewhere. I think I have something off though and I am not sure if I can do what I am saying, or even if it is correct. Any suggestions, ideas, etc. are greatly appreciated! Thank you.
AI: Suppose $A = \{x:f(x)>\frac{y}{2}\}$ and $m(A)<\frac{y}{2}$. Also lets denote $B = \{x:f(x)\leqslant\frac{y}{2}\}$. Now we have
$$\int\limits_{0}^{1} = \int\limits_{A}+ \int\limits_{B}$$
For first we use that $f \leqslant 1$ and $m(A)<\frac{y}{2}$. For second we have estimation $\leqslant \frac{y}{2}$. So, sum cannot be $y$. |
H: Showing that $\|f\|_{\infty}\leq \liminf_{p\to \infty}\|f\|_p$.
i'm trying to prove the next problem, and I wanted to know if my answer is correct.
Problem:
Let $(\Omega,\mathcal{F},\mu)$ be a $\sigma$-finite measurable space.
If $f\in L^p$ for all $p\in [1,\infty)$, show that
$$\|f\|_{\infty} \leq \liminf_{p\to \infty}\|f\|_p.$$
Solution:
Suppose $0<\mu (\Omega)\leq \infty$.
Let $0\leq M\leq\|f\|_{\infty}$, and $A=\{x\in \omega:|f(x)|>M\}$, then $\mu (A)>0$ and $\liminf_{p\to \infty} \mu(A)^{1/p}=1$. Then,
$$\liminf_{p\to \infty} \|f\|_p \geq M \liminf_{p\to \infty} \mu(A)^{1/p}=M,$$
for all $M\in [0,\|f\|_{\infty})$. Hence, $\|f\|_{\infty}\leq \liminf_{p\to \infty}\|f\|_p$.
AI: You need to define $A$ differently. The current definition yields $\mu(A)=0$. You should let $\epsilon >0$ and then define $A = \{x \in \Omega \mid |f(x)| > M- \epsilon\}$. Then you have, by definition of the essential supremum, $\mu(A)>0$. Now it could happen that $\mu(A)=\infty$, so $\liminf\limits_{p \to \infty} \mu(A)^{1/p}=\infty$... However by sigma finiteness there must be a set $B$ with $\infty > \mu(B)>0$ and $\mu(A \cap B) >0$. Now substitute $A$ by $A \cap B$ and the rest of your proof works. |
H: Smoothly varying finite dimensional vector subspaces tracing out infinite dimensional locus
Is it possible that a smoothly varying finite dimensional vector subspaces tracing out infinite dimensional locus?
More precisely, Let $E$ be a vertor bundle on a (compact) smooth manifold $M$ and $T_t:\Gamma(M,E)\to \Gamma(M,E)$ is a smooth family of linear operators, where $\Gamma(M,E)$ is the space of sections. $T_t$ is smooth on $t\in \mathbb{R}$ means that if $\{\sigma_i\}$ is a local frame then the coefficients of each $T_t\sigma_i$ with respect to the frame $\{\sigma_i\}$ is smooth in $t$.
Now if $V\subset \Gamma(M,E)$ is a linear subspace of finite dimension.
Is it possible that
$$
\dim span_{\mathbb{R}}\left(\bigcup_{t\in\mathbb{R}}\{T_t(V)\}\right)=\infty~?
$$
Is it possible that for any sufficiently small $\varepsilon>0$,
$$
\dim~ span_{\mathbb{R}}\left(\bigcup_{|t|<\varepsilon}\{T_t(V)\}\right)=\infty~?
$$
A typical example of this question is when $T_t=e^{-f(t)}de^{f(t)}$ ($f(t)$ is a smooth function on $M$ which depends smoothly on $t$), $E=\bigwedge^\bullet T_M^*$ is the exterior bundle and $V=\mathcal{H}^\bullet(M)$ is the space of harmonic forms.
Added:If the above possiblities are possible, I want to know whether $\dim\ker(T_t\mid_V)$ is upper semicontinuous on $t$ for small $t$? That is,
$$
\dim\ker(T_0\mid_V)\geq \dim\ker(T_t\mid_V),\quad |t|<\varepsilon~.
$$
AI: It's absolutely possible. Consider the trivial bundle $\mathbb{R}\times\mathbb{R}$ and the operator defined by $T_tf(x)=\cos(tx)f(x)$. Let $V$ be the subspace of constant functions. Since $\{\cos(tx):t\in[0,\infty)\}$ is linearly independent, the union $\bigcup_{t\in I}T_t(V)$ is infinite dimensional for any open $I\subset\mathbb{R}$ |
H: Find minimum value of $\frac{\sec^4 \alpha}{\tan^2 \beta}+\frac{\sec^4 \beta}{\tan^2 \alpha}$
Find minimum value of $\frac{\sec^4 \alpha}{\tan^2 \beta}+\frac{\sec^4 \beta}{\tan^2 \alpha}$
I know this question has already answered here Then minimum value of $\frac{\sec^4 \alpha}{\tan^2 \beta}+\frac{\sec^4 \beta}{\tan^2 \alpha}$ but my answer is coming different.
i applied AM-GM directly to two fractions and by changing terms of sec and tan into sin and cos and simplifying a little we get that
$$\frac{\sec^4 \alpha}{\tan^2 \beta}+\frac{\sec^4 \beta}{\tan^2 \alpha} \geq\frac2{\cos\alpha \cos\beta \sin\alpha \sin\beta}\geq2$$
but minimum value is coming 8 ???
AI: $$\frac{2}{\cos\alpha\cos\beta\sin\alpha\sin\beta} = \frac{8}{\sin2\alpha\sin2\beta} \geq 8$$
Now why did the previous inequality only give 2 whereas when we use this we get 8? Basically, $\alpha$ and $\beta$ are independent of each other, hence we can minimise the second expression by putting $\alpha = \beta = \frac{\pi}{4}$
On the other hand, in the expression you reduced to, both $\sin\alpha$ and $\cos\alpha$ cannot simultaneously be $1$, hence the product will actually have a different maximum value (which is $\frac{1}{2}$) |
H: On Hypergeometric Series and OEIS Sequence
I have been searching an integer sequence in OEIS. The sequence is the following: OEIS A321234 (https://oeis.org/A321234) . So far, so good. However, this sequence is the denominator of a Hypergeometric Series, the following one:
$${}_3 F_2([1/2, 1, 1], [3/2, 3/2], x).$$
The problem is: I do not even know what those kind of series are. Do not even know what this notation means. Could someone recomend me any book references so as to understand it better? I have read Wiki's page, but it seems not enough.
Thanks a lot
AI: $$\, _3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};x\right)$$ is one of the many hypergeometric functions (google for that).
They are very special functions corresponding to infinite sums. For this one, the first terms of its expansion are
$$1+\frac{2 x}{9}+\frac{8 x^2}{75}+\frac{16 x^3}{245}+\frac{128 x^4}{2835}+\frac{256
x^5}{7623}+\frac{1024 x^6}{39039}+\frac{2048 x^7}{96525}+\frac{32768
x^8}{1859715}+O\left(x^{9}\right)$$
You would find the numerators in sequence $A046161$ |
H: Fundamental Group of $\mathbb{RP}^n$
I was tring to culculate the fundamental group of $\mathbb{RP}^n$ with VAN KAMPEN to have a better understanding on how to use this theorem.
$\left(\mathbb{RP}^n:=S^n/ \sim \left((x_0,\cdots,x_n) \sim(-x_0,\cdots,-x_n)\right)\right)$
By consider the $A=\left\{[(x_0,\cdots, x_n)] \in \mathbb{RP}^n|-1<x_0<1\right\}$, $B=\left\{[(x_0,\cdots, x_n)] \in \mathbb{RP}^n|x_0>0\right\}$.
In case $n=2$
Obviously, we have that the fundamental group of $B$ is $1$ and of $A$ is $\mathbb{Z}$ and consider the element of fundamental groub of $A$ which is $a$. We have that $a^2$ is in the intersection of $A$ and $B$. Thus, $a^2$ shoud be $1$ in fundamental group of $\mathbb{RP}^2$.
Therefore the fundamental group of $\mathbb{RP}^2$ shoud be $\mathbb{Z}/2$.
Then I consider the $n>2$ case, I realise that the fundamental group of $B$ is still $1$. And $A$ is the $\mathbb{RP}^{n-1}$. The fundamental group of $A\cap B$ shoud be 1 when $n>2$, because the $A\cap B$ is $S^{n-1}$.
I am not sure if I could use the Van Kampen directly to say that the fundamental group of $A\cup B$ is the same as $A$ because the fundamental group of both $A$ and $A\cap B$ is $1$.
AI: Your result for $n=2$ is correct, but the process seems to be unclear to me. In general, we can proceed inductively to find the fundamental group.
The real projective space $\Bbb{RP}^n$ is homeomorphic to $D^n/{(x\sim-x)}$ where $x\in\partial D^n\approx S^{n-1}$. Now, let $U$ be a smaller open $n$-ball $\subset D^n$, then $\pi_1(U)=1$. Let $V$ be an enlargement of $D^n\setminus A$ (open), then $V\simeq S^{n-1}/(x\sim-x)$ by deformation retraction. You may find that $V\approx \Bbb{RP}^{n-1}$ because the most typical definition for real projective spaces is $S^n/(x\sim-x)$.
By Seifert Van-Kampen's Theorem, we conclude that $\pi_1(\Bbb{RP}^n,x_0)\cong \pi_1(U)*\pi_1(V)/N$, but $\pi_1(U\cap V)=1\implies N=e$ (for the reason that $U\cap V\approx S^n, n\ge 2$ is simply connected), so this reduces to $\pi_1(\Bbb{RP}^n)\cong\pi_1(V)$. Because $V\simeq\Bbb{RP}^{n-1}$, $\pi_1(\Bbb{RP}^{n})\cong\pi_1(\Bbb{RP}^{n-1})$. Inductively, it should be $\Bbb{Z}/2$.
However, the case is different when $n=1,2$ because $\Bbb{RP}^1\approx S^1$ and hence has a fundamental group isomorphic to $\Bbb{Z}$ and $\Bbb{RP}^2$ doesn't follow the induction due to the reason that $\pi_1(U\cap V)\cong \Bbb{Z}$.
In sum,
$$\pi_1(\Bbb{RP}^n)\cong
\begin{cases}
\Bbb{Z} & n=1\\
\Bbb{Z}/2 & n\ge2
\end{cases}$$
Remark: I explored the last statement made by Ted Shifrin in details. |
H: the definition of invertible sheaf on a functorial scheme in category theory
We define a functorial scheme as in "Two functorial definitions of schemes".A invertible sheaf is important and I'm interested in category theory, so I hope to define invertible sheaf in category theory like a scheme. However, the definition of a invertible sheaf(or a locally free sheaf) require a ringed space. Therefore to define a invertible sheaf for a functorial scheme seems to be difficult.
My question is: Can we define invertible sheaf on a scheme in category theory?
Thanks in advance.
AI: I presume you already know how to define a sheaf of groups/rings/modules on a scheme-defined-as-a-sheaf. If not then you will have to start there. There are basically only two more ingredients needed to define invertible sheaves:
The structure sheaf of a scheme is represented by the scheme $O = \operatorname{Spec} \mathbb{Z} [x]$. This is a ring object in the category of schemes, so for every scheme $X$, the set of morphisms $X \to O$ has a natural ring structure. This defines a sheaf of rings on the category of all schemes, but precomposing it with the forgetful functor gives you a sheaf $O_X$ on the category of schemes over $X$ (or on the small Zariski site of $X$ – take your pick).
An invertible sheaf $M$ on $X$ is a sheaf of $O_X$-modules that is locally isomorphic to $O_X$. Locally isomorphic means there is a cover of $X$ consisting of morphisms $U \to X$ such that pulling back $M$ along the morphism yields an $O_U$-module isomorphic to $O_U$. (Strictly speaking this is the definition of a locally free sheaf of rank 1... but as you know, they are the same thing as invertible sheaves.)
Notice that the above makes equal sense for schemes-defined-as-sheaves and schemes-defined-as-ringed-spaces. |
H: A question on locally integrable function on $\mathbb{R}^n$
I am currently doing some practice problems for the analysis qual. I have some thoughts on the following problem, and it would be great if someone could see if I am on the right track:
Problem:
Let $O\subset \mathbb{R}^n$ be an open set, and let $f\in L_{loc}^1(\mathbb{R}^n)$ (the set of locally integrable functions on $\mathbb{R}^n$). Assume that for all $\phi \in C_c(O)$ (the set of continuous functions with compact support in $O$), we have $$\int_Of\phi dm = 0,$$ where $dm$ stands for Lebesgue measure. Prove that $f(x) = 0$ for a.e. $x\in O$.
My thoughts are as follows: I want to show by contradiction that if there exists $f \in L_{loc}^1(\mathbb{R}^n)$ s.t. there is a Lebesgue measurable set $E$ with $m(E) > 0$ and $|f| > 0$ on $E$, then one can find some $\phi \in C_c(O)$ s.t. $$\int_Of\phi dm \ne 0.$$ Since $E$ is with positive measure, then by inner regularity of Lebesgue measure we can find a nonempty compact set $K$ s.t. $K \subset U$. Then by Urysohn lemma, there exists a nonnegative function $\phi \in C_c(O)$ s.t. $\phi|_K = 1$ and $\phi$ vanishes outside $O$. Now I claim that $\phi$ does the job: Note that $$\int_O|f|\phi = \int_E|f|\phi \ge \int_K|f|\phi = \int_K|f| > 0,$$ where the last inequality is by the fact that $K\subset E$. But now I don't find a good way to conclude that $$\int_Of\phi dm \ne 0.$$
Could anyone help me fill in this gap or point out which place goes wrong here? Much appreciated in advance!
AI: Hint:
there are compacts sets $K_n$ such that $K_n\subset\operatorname{Int}(K_{n+1})$ and $O=\bigcup_nK_n$.
You may try to show that $f=0$ in $K_n$. Any measurable subset $E$ of $K_n$ can be approximated in $L_1$ by a sequence of continuous functions with support in $\operatorname{Int}(K_{n+1})$ and which are uniformly bounded. Then by dominated convergence $\int f\mathbb{1}_E=0$.
This implies that $f=0$ in $K_n$ for each $K_n$. |
H: Why does $\frac{n!n^x}{(x+1)_n}=\left(\frac{n}{n+1}\right)^x\prod_{j=1}^{n}\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^x$
Why does $$\frac{n!n^x}{(x+1)_n}=\left(\frac{n}{n+1}\right)^x\prod_{j=1}^{n}\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^x$$ where the subscript n is the rising factorial in the left denominator
my attempt:
the index n in the product indicates the indicates the term $\left(1+\frac{x}{j}\right)^{-1}$ may be some series of infinite geometric sums from the rising factorial but how?
This is page 2 in Andrews, Askey, Roy Special Functions.
AI: $$\prod_{j=1}^n\left(\frac{j+1}j\right)^x$$ telescopes and equals $(n+1)^x$. Multiplying
by
$$\left(\frac n{n+1}\right)^x$$
gives $n^x$.
$$\prod_{j=1}^n\left(1+\frac xj\right)^{-1}=\frac{n!}{(x+1)(x+2)\cdots(x+n)}
=\frac{n!}{(x+1)^{(n)}}$$
etc. (I prefer $x^{(n)}$ and $x_{(n)}$ for the rising and falling factorials resp.) |
H: Discrete Probability: alice and mary take a math exam
So I have a problem here, I know the answer but I have no idea how to solve it. This is one of a sample problem given to us by our prof. Can someone please help me out how to figure this problem out?
Alice and Mary take a math exam. The probability of passing this exam for Alice and Mary is 2/3 and 3/5, respectively. What is the probability that at least one of them will pass the exam?
AI: Fisrt of all we have to assume independence (understood hypothesis)
Let's set the events
"0= not pass"
"1=pass"
And let's analyze all the possible probabilities
$$\mathbb{P}[0;0]=\frac{1}{3}\times \frac{2}{5}=\frac{2}{15}$$
$$\mathbb{P}[0;1]=\frac{1}{3}\times \frac{3}{5}=\frac{3}{15}$$
$$\mathbb{P}[1;0]=\frac{2}{3}\times \frac{2}{5}=\frac{4}{15}$$
$$\mathbb{P}[1;1]=\frac{2}{3}\times \frac{3}{5}=\frac{6}{15}$$
As a verification
$\frac{2}{15}+\frac{3}{15}+\frac{4}{15}+\frac{6}{15}=1$
Now you can answer to any question they ask you, as you know the probability of every elementary event of the event space |
H: Finding $E \in \mathcal A \otimes \mathcal B$ such that $E \neq E^y \times E_x,$ for some $x \in X, y \in Y.$
Let $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be two measure spaces. What has been stated in my book is that $\mathcal A \times \mathcal B$ may not necessarily be a $\sigma$-algebra of subsets of the product space $X \times Y$ but it's a semi-algebra of subsets of $X \times Y.$ Let $\mathcal A \otimes \mathcal B$ be the $\sigma$-algebra of subsets of $X \times Y$ generated by $\mathcal A \times \mathcal B.$ Then the following theorem holds $:$
Theorem $:$ For every $E \in \mathcal A \otimes \mathcal B,$ for every $x \in X$ and for every $y \in Y$ $$E_x \in \mathcal B\ \ \ \text {and}\ \ \ E^y \in \mathcal A$$
where $E_x$ and $E^y$ are the $x$-section and $y$-section of $E$ respectively defined by \begin{align*} E_x : & = \left \{y \in Y\ |\ (x,y) \in E \right \} \\ E^y : & = \left \{x \in X\ |\ (x,y) \in E \right \} \end{align*}
Does the above theorem imply that $E = E^y \times E_x\ $? I don't think so. For otherwise $\mathcal A \otimes \mathcal B = \mathcal A \times \mathcal B,$ which is not necessarily true. Can anybody help me finding one such example where $E \in \mathcal A \otimes \mathcal B$ but $E \neq E^y \times E_x\ $?
Thanks in advance.
AI: Take both spaces to be the real line with Borel sigma algebra. Consider the open unit disk $E =\{(x,y): x^{2}+y^{2} <1\}$. Then $E_0=[-1,1]=E^{0}$ and $E_0 \times E^{0}=[-1,1] \times [-1,1] \neq E$. |
H: Problem with $A.M.\geq G.M.$ Inequality
Question:
If $x^2+y^2=1$, prove that $-{\sqrt2}\leq x+y \leq\sqrt2$
My approach:
$$\frac{x^2+y^2}{2}\geq\sqrt{x^2y^2}$$
$$ \frac12\geq xy$$
$$\frac{-1}{\sqrt2}\leq\sqrt{xy}\leq \frac{1}{\sqrt2} $$
Now how do I proceed from here?
AI: Proceeding from your approach, you had
$$2xy \leq 1$$
Adding $x^2 + y^2$ to both sides,
$$\implies x^2+2xy+y^2 \leq 1 + x^2 + y^2$$
$$\implies (x+y)^2 \leq 1 + 1$$
And you're done. |
H: Yes/ No Is $X$ is homeomorphics to $Y$?
$X=\{(x,y) \in \mathbb{R^2}: x^2+y^2=1 \}$ and $Y=\{(x,y) \in \mathbb{R^2}: x^2+y^2=1 \} \cup \{(x,y) \in \mathbb{R^2}: (x-2)^2+y^2=1 \} $ be the subspaces of $\mathbb{R}^2$
Now my question is that
Is $X$ is homeomorphics to $Y$ ?
My attempt : Yes
Both $X$and $Y$ are connected and compact
Now by using the theorem connected subspaces of connected sets is connected
So $X$ is homeomorphics to $Y$
Is its true ?
AI: No. The set $X$ is a circle. It is connected and it remains connected if you remove any point from it.
But $Y$ is the union of two circles, with a common point, which is $(1,0)$. If you remove that point from $Y$, what you get is disconnected.
So, $X$ and $Y$ are not homeomorphic. |
H: Proving that $\mathbb{Z}[i]/\langle 2+3i\rangle $ is a finite field
Prove that $\mathbb{Z}[i]/\langle 2+3i\rangle $ is a finite field.
Hi. I can't try a few steps in the next solution
$$\mathbb{Z}[i]/\langle 2+3i\rangle \simeq \mathbb{Z}[x]/\langle 1+x^2,2+3x\rangle$$ and $9(1+x^2)+(2-3x)(2+3x)=13$ then $13\in \langle 1+x^2,2+3x\rangle $. Thus $$\mathbb{Z}[x]/\langle 1+x^2,2+3x\rangle=\mathbb{Z}/\langle 13,1+x^2,2+3x\rangle \\
\simeq \mathbb{Z}_{13}[x]/\langle 1+x^2,2+3x\rangle \simeq \mathbb{Z}_{13}$$
The last isomorphism induced by $x\mapsto 8$ (check $\langle 1+x^2,2+3x\rangle=\langle x-8\rangle $ in $\mathbb{Z}_{13}[x]$)
Therefore $\mathbb{Z}[i]/\langle 2+3i\rangle \simeq \mathbb{Z}_{13}$ finite field.
Question 1. Why $\mathbb{Z}[i]/\langle 2+3i\rangle\simeq \mathbb{Z}[x]/\langle1+x^2,2+3x\rangle$?
I have this: Let $$f:\mathbb{Z}[x]\to \mathbb{Z}[i]/\langle 2+3i\rangle $$ with $f(p(x))=p(i)+\langle 2+3i\rangle $ homomorphism with $\ker(f)=\langle 1+x^2,2+3x\rangle $ then $$\mathbb{Z}[x]/\langle 1+x^2,2+3x\rangle \simeq \mathbb{Z}[i]/\langle 2+3i\rangle $$It is correct?
Question 2. Why $\mathbb{Z}/\langle 13,1+x^2,2+3x\rangle \simeq \mathbb{Z}_{13}[x]/\langle 1+x^2,2+3x\rangle $?
Question 3. Why $\langle 1+x^2,2+3x\rangle=\langle x-8\rangle$?
AI: Question 1: We have $\Bbb Z[i]\simeq \Bbb Z[x]/\langle x^2+1\rangle$. And the third isomorphism theorem says that when we are dividing out by first $x^2+1$, and then $2+3x$, we are allowed to divide out but both of them simultaneously.
Question 2: Again justified by the third isomorphism theorem, dividing out by $13$ before the other generators.
Question 3: Here we have
$$
9(2+3x)=18+27x=-8+x
$$
So $\langle x^2+1,2+3x\rangle$ contains $x-8$. Now also note that
$$
3(x-8)=3x-24=3x+2\\
(x-8)^2+3(x-8)=x^2-16x+64 +3x-24=x^2+1
$$
So $\langle x-8\rangle$ contains both $x^2+1$ and $2+3x$. Since each ideal contains the generators of the other ideal, the two ideals must be equal. |
H: What is $T(p(a x^2+b))$ when $T(p(x))=x^2p'(x)$
Let $T$ be linear transformation on $\mathbb{P} \rightarrow \mathbb{P}$ that $T(p(x))=x^2p'(x)$.
1. Then, what is the value of $T(p(ax^2+b))$?
I think it could be $x^2p'(ax^2+b)$ or $2ax^3p'(ax^2+b)$ or even something else.
2. What would be the result when $T$ is on $\mathbb{D} \rightarrow \mathbb{D}$? ($\mathbb{D}$ is set of differentiable functions)
AI: $T(p(ax^{2}+b))=x^{2} \frac d {dx} (p(ax^{2}+b))=x^{2}p'(ax^{2}+b) (2ax)$ by Chain Rule. The answer is same for $\mathbb D$. |
H: A question for a measurable function $g$ on a finite measure space such that $fg\in L^p$ for all $f\in L^p$
Let $\mu$ be a finite positive measure on $X$ and let $1\leq p<\infty$. Suppose that $g:X\to \Bbb R$ satisfies $fg\in L^p$ whenever $f\in L^p(\mu)$. I want to show that $||g||_\infty =\sup \{||fg||_p:f\in L^p(\mu)$ such that $||f||_p=1\}<\infty$.
The inequality $||g||_\infty \geq \sup \{||fg||_p:f\in L^p(\mu)$ such that $||f||_p=1\}$ is obvious. So it is left to show the converse inequality and that the value is finite, but I have no idea here. Any hints?
AI: Hint: If $\frac 1{\mu (E)} \int_E |g| d\mu\leq M$ for every set $E$ of positive measure then $|g| \leq M$ a.e. so $\|g\|_{\infty} \leq M$.
Now let $M=\sup \{\|fg\|_p: \|f\|_p=1\}$. Take $f=\frac 1 {\mu (E)^{1/p}} \chi_E$. Then $\|f\|_p=1$. By Holder's inequality we have
$\frac 1{\mu (E)} \int_E |g| d\mu\leq \frac 1{\mu (E)} \|f(\mu (E)^{1/p}) g\|_p \|I_E\|^{q}$ where $\frac 1 p+\frac 1 q=1$. The desired equality now follows.
To show that $\|g\|_{\infty} <\infty$ consider $f =\sum a_n\chi_{n \leq |g| <n+1}$ where $a_n$ are positive numbers. Let $c_n=\mu \{x: n \leq |g(x)| <n+1\}$. The hypothesis implies that $\sum a_n^{p} n^{p}c_n <\infty$ whenever $\sum a_n^{p} c_n <\infty$. I leave it to you to show that this can happen only when $c_n=0$ for $n$ sufficiently large. Of course this would then show that $g$ is essentially bounded. |
H: Maximum of function abs
Let function $f(x)=|2x^3-15x+m-5|+9x$ for $x\in\left[0,3\right]$ and $m\in R$. Given that $\max f(x) =60$ with $x\in\left[0,3\right]$, find $m$.
I know how to solve this kind of problem for $g(x)=|2x^3−15x+m−5|$. However, the $+9x$ is confusing me.
AI: Case 1:
Let $2x^3-15x+m-5\gt0$ at the point where maximum occurs.
Then our $f(x)$ becomes $2x^3-6x+m-5$
note that $f(x)$ decreases for $(0,1)$ and then increases for $x\gt1$ so the maximum of $f(x)$ is at $x=3$(as $x\in\left[0,3\right]$)
Plugging in the $f(3)=60$ we get $m = 29$.
We can put in $x=3$ and $m = 29$ in $2x^3-6x+m-5$ and verify that it is positive.
Case 2 :
Let $2x^3-15x+m-5\lt0$ at the point where maximum occurs
Then our $f(x)$ becomes $-2x^3+24x-m+5$
This function increases in $(0,2)$ and then decreases for $x\gt2$ so maximum for $f(x)$ is at $x=2$
Putting $f(2)=60$ we get $m=-23$.
We can put in $x=2$ and $m = -23$ in $-2x^3+24x-m+5$ and verify that it is negative.
So we get $m= \left({29\ \ and\ -23}\right)$ |
H: $\lim\limits_{R\to0^+}\int\limits_{x^2+y^2\le R^2}e^{-x^2}\cos(y)dxdy=?$
$$\lim\limits_{R\to0^+}\int\limits_{x^2+y^2\le R^2}e^{-x^2}\cos(y)dxdy=?$$
First I want to show $f(x,y)=e^{-x^2}\cos(y)$ doesn't go crazy at $(0,0)$ otherwise it is already clearly continuous and bounded.
So $$|e^{-x^2}\cos(y)|\le e^{-x^2}\to 1$$ when $$x^2+y^2\to 0$$
So Now finite and bounded integrand's integral has to go 0 because region vanishes, but how to properly show it?
AI: The function $|e^{-x^2}\cos(y)|$ is bounded by $1$ from above for any $x$ and $y$. So the absolute value of your integral is less than or equal to the area of the circle $x^2+y^2\leq R^2$ times $1$. |
H: Does the spiral Theta = L/R have a name?
Note: I intentionally left the equation in the title in plain text instead of MathJax, so it is searchable.
Here is a spiral's equation in polar coordinates:
$$\theta=L/r,$$
and in Cartesian coordinates:
$$(x,y) = \left(r\cdot\cos\frac Lr,\, r\cdot\sin\frac Lr\right)$$
for $0 < r < \infty$ and some positive constant $L$.
It appeared here, at Math SE, as an answer to the Spiral equation question.
Does this curve have a name?
AI: As @YvesDaoust said in a comment it is called a hyperbolic spiral, or a reciproke spiral as the circle inversion of an Archimedean spiral
Wikipedia has a couple of images, depending on whether you look at one or two arms of the hyperbola underlying $r=\frac a \varphi$ |
H: Every $T_1$, $C_2$, regular space is normal
By my attempt, I know that if $X$ is $T_1$, $C_2$, and regular, then it is metrizable. So is every metrizable space also normal? this seems correct but not sure how to see it.
AI: Is every metrizable space normal? Yes.
Let $(X,d)$ be a metric space and $P$, $Q$ be two nonempty disjoint closed subsets of $X$.
First note that if $Y$ is a closed subset of $X$, then for $x \in X$, $d(x,Y)=0$ if and only if $x \in Y$.
For each $p \in P$, set $B_p$ denote the open ball centered at $p$ with radius $\frac{d(p,Q)}{2}\neq 0$ by the above result.
Similarly, for each $q \in Q$, define $B_q$ to be the open ball centered at $q$ with radius $\frac{d(q,P)}{2}\neq 0$.
Now prove that the sets $U=\bigcup_{p\in P}B_p$ and $V=\bigcup_{q\in Q}B_q$ are the required open neighborhoods of the closed sets $P$ and $Q$ respectively. |
H: What is the difference between ${3 \choose 2}$ and ${3 \choose 1}{2 \choose 1}$?
While choosing 2 from 3, we do ${3 \choose 2}$. But what would happen if we do ${3 \choose 1}{2 \choose 1}$? [Choosing 1 from 3 and again 1 from the remaining two)?
I know the latter is incorrect but can anyone give me conceptual view of how that's wrong and what would I be doing if I did the latter?
AI: ${3 \choose 2}$ chooses 2 objects out of three without paying attention to the order in which they were chosen. It only matters which objects are chosen.
${3 \choose 1}{2 \choose 1}$ chooses 2 objects sequentially, so the order does matter.
Example: the three objects are ABC.
The first method counts the number of sets of two that can be chosen: $\{A,B\},\{A,C\},\{B,C\}$.
The second method counts the number of 2 letter words that can be built from different letters: AB, BA, AC, CA, BC, CB. |
H: Probability question about picking $2$ types of balls out of $3$
I need help with this question:
A bag contains $2$ red balls, $6$ blue balls and $7$ green balls. Victoria draws $2$ balls out of
the bag. What is the probability that she gets a red ball and a blue ball?
I can figure out the probability of picking $1$ ball ($\frac{2}{15}$,$\frac{2}{5}$,$\frac{7}{15}$ respectively). But I am stuck a finding out the probability of 2.
Any help would be appreciated.
AI: Your probabilities are correct to draw one ball.
To draw two balls $B_1$ and $B_2$, you multiply the probability of drawing $B_1$ with the probability of drawing $B_2$ after drawing $B_1$ (only $14$ balls are remaining).
Then, how many ways can you draw a red and a blue ball? You can draw a red first, and then a blue, and you can also draw a blue first, then a red.
So you have to calculate both probabilities (red,blue) and (blue,red) and sum them. This is your result. |
H: Two conditional expectations equal almost everywhere
Suppose $X$ is a continuous random variable. If $\mathbb{E}[X\,|\,\mathcal{F}]=\mathbb{E}[X\,|\,\mathcal{G}]$ almost everywhere for two sub-sigma algebra $\mathcal{F}$ and $\mathcal{G}$, does this imply $\mathcal{F}$ and $\mathcal{G}$ are set theoretically identical?
AI: No. If $X,Y,Z$ are independent, $\mathcal F=\sigma (Y)$ and $\mathcal G =\sigma (Z)$ then $\mathbb E( X|\mathcal F)=\mathbb E(X|\mathcal G)=\mathbb EX$. |
H: Evaluate $ \cos a \cos 2 a \cos 3 a \cdots \cos 999 a $ where $a=\frac{2 \pi}{1999}$
Evaluate
$
\cos a \cos 2 a \cos 3 a \cdots \cos 999 a
$
where $a=\frac{2 \pi}{1999}$
I know this question has already been answered many times but my doubt is different
Solution: Let $P$ denote the desired product, and let
$
Q=\sin a \sin 2 a \sin 3 a \cdots \sin 999 a
$
Then
$
\begin{aligned}
2^{999} P Q=&(2 \sin a \cos a)(2 \sin 2 a \cos 2 a) \cdots(2 \sin 999 a \cos 999 a) \\
=& \sin 2 a \sin 4 a \cdots \sin 1998 a \\
=&(\sin 2 a \sin 4 a \cdots \sin 998 a)[-\sin (2 \pi-1000 a)] \\
& \cdot[-\sin (2 \pi-1002 a)] \cdots[-\sin (2 \pi-1998 a)] \\
=& (\sin 2 a \sin 4 a \cdots \sin 998 a) \sin 999 a \sin 997 a \cdots \sin a=Q
\end{aligned}
$
how they got this last step from previous one ???
AI: To reach penultimate step, use $\sin(2\pi-\theta)=-\sin\theta$, and to go from penultimate to ultimate step, use
$$2\pi-1000a=2\pi-1000\cdot\frac{2\pi}{1999}=2\pi\left(1-\frac{1000}{1999}\right)=2\pi\frac{999}{1999}=999a$$
Also, there comes out $(-1)^{999-500+1}$ if one counts only even numbers from $1000$ to $1998$. |
H: Positive semi-definite real matrix with unit diagonal
Give an example of a $n\times n$ positive semi-definite real matrix $M\in \mathbb{R}^{n \times n}$, such that the following two conditions hold:
the eigenvalues $\lambda_1, \dots, \lambda_n$ of $M$ are $\lambda_i \leq 1$ for all $i\in [n]$;
the diagonal entries are $m_{i, i} = 1$, for all $i \in [n]$.
Is it possible to define any such matrix $M$ with the additional property that $\det (M) = 0$?
AI: No.
Since we have a symmetric PSD matrix we have the following,
$$Tr(M) = \sum\limits_{i=1}^n \lambda_i$$
and
$$\det(M) = \prod\limits_{i=1}^n \lambda_i.$$
By assumption, $Tr(M) = \sum\limits_{i=1}^nm_{i,i}=\sum\limits_{i=1}^n 1= n$. Thus, $\sum\limits_{i=1}^n\lambda_i = Tr(M) = n$. Since, for each $i\in[n]$, $0\leq \lambda_i\leq 1$, we have that $\lambda_i=1$ for each $i\in[n]$. Then, the determinant is necessarily $1$ since
$$\det(M) = \prod\limits_{i=1}^n\lambda_i = \prod\limits_{i=1}^n 1 = 1.$$ |
H: Is $\sin(\frac{1}{|z|})$ holomorphic on $\Bbb C-\{0\}$?
$f(z)=\sin(\frac{1}{|z|})$, $z\in \Bbb C-\{0\}$.
$$\frac{\partial f}{\partial\bar{z}}=\frac{\partial \sin(\frac{1}{|z|})}{\partial\bar{z}}=\frac{\partial}{\partial \bar{z}}\sin(\frac{1}{(z\bar z)^{1/2}})=\cos(\frac{1}{|z|})(2^{-1}z)(z\bar{z})^{\frac{-3}{2}} \neq 0$$
So, $f(z)$ is not holomorphic on $\Bbb C-\{0\}$.
Am I correct?
AI: Yes, that is correct. Another way of proving it is this: if $z=x+yi$, with $x,y\in\Bbb R$; then write $f(x+yi)$ as $u(x,y)+v(x,y)i$. Then $v$ is the null function. So, if $f$ was holomorphic, you would have (by the Cauchy-Riemann equations) $u_x=u_y=0$ and s $u$ would be constant. But then $f$ itself would be constant. But it is not. |
H: Checking compactness of $[0,1]$ using the definition
By definition a set is compact if every open covering has a finite subcover.
I made an open covering of $[0,1]$ by taking a fixed $\epsilon= \dfrac1{10,00000}$ (or even more small) radius neighborhood around each point in $[0,1$]. But I could not find its finite subcover. By Heine-Borel theorem it's clear that it will be compact. But, I was trying to use definition but could not find its finite subcover.
AI: The size of $\varepsilon$ doesn't matter. Take the intervals $(-\varepsilon,\varepsilon)$, $(-\varepsilon/2,3\varepsilon/2)$, $(0,2\varepsilon)$, and so on… |
H: Closed-form expression for $\prod_{n=0}^{\infty}\frac{(4n+3)^{1/(4n+3)}}{(4n+5)^{1/(4n+5)}}$?
I have recently come across this infinite product, and I was wondering what methods I could use to express the product in closed-form (if it is even possible):
$$\prod_{n=0}^{\infty}\dfrac{(4n+3)^{1/(4n+3)}}{(4n+5)^{1/(4n+5)}}=\dfrac{3^{1/3}}{5^{1/5}}\cdot \dfrac{7^{1/7}}{9^{1/9}}\cdot\dfrac{11^{1/11}}{13^{1/13}}\cdot\dfrac{15^{1/15}}{17^{1/17}}\cdot\dotsb$$
Thanks in advance!
AI: $$\exp\sum_{n\geq 0}\left(\frac{\log(4n+3)}{4n+3}-\frac{\log(4n+5)}{4n+5}\right)=\exp\left[-\sum_{n\geq 1}\frac{\chi_4(n)\log(n)}{n}\right]$$
equals
$$\lim_{s\to 1^+}\exp\left[\frac{d}{ds}\sum_{n\geq 1}\frac{\chi_4(n)}{n^s}\right]=\exp\beta'(1)$$
with $\beta$ being Dirichlet's beta function. According to equation (18) the closed form is
$$ \exp\,\left[\frac{\pi}{4}\left(\gamma+\log\frac{4\pi^3}{\Gamma\left(\frac{1}{4}\right)^4}\right)\right].$$
A derivation can be found here. |
H: Are homology groups of a chain complex isomorphic to that of free chain complex?
Given a chain complex $A_\bullet\in\mathrm{Ch(\mathbf{Ab})}$, are there exist some chain complex $A'_\bullet\in\mathrm{Ch(\mathbf{Ab})}$ whose homology groups are all isomorphic to that of $A_\bullet$ such that $A'_p$ are all free abelian groups?
I first thought of stronger condition:
Are chain complexes chain equivalent to free ones?
but that was false, so I'm wondering if this weaker condition holds.
AI: I think this boils down to "given a sequence of Abelian groups, is there a free chain complex having these as homology groups?".
The answer to this is yes. Let $A_n$ be your $n$-th Abelian group. Then it has
a free presentation
$$0\to F_n'\to F_n\to A_n\to0$$
which is an exact sequence with $F_n$ and $F_n'$ free. Then the homology of the complex
$$\cdots\to0\to F_n'\to F_n\to0\to\cdots\tag{*}$$
is just $A_n$ in one position. Now take the direct sum of the (*) over all $n$. |
H: Can one obtain from the following diagram that the map $f_3$ is injective?
Let $A_i$ and $B_i$ be $R$-modules $(i=1,2,3)$.
If in the diagram
each map is $R$-linear, the rows are exact, both squares commute, and $f_1, f_2, \alpha_1, \beta_1$ are injective, is it possible to prove that $f_3$ is injective?
AI: Let $A_1=B_3=0$, $A_2=A_3=B_1=B_2=R$ and $f_2$ the identity map (the other $f_i$
have to be zero maps). Then $\ker f_3=A_3=R$. |
H: Karush-Kuhn-Tucker in Quadratic Program
I read an paper on Quadratic Programming:
Paul A. Jensen, Jonathan F. Bard: Operations Research Models and Methods Nonlinear Programming Methods.S2 Quadratic Programming Available here: https://www.me.utexas.edu/~jensen/ORMM/supplements/methods/nlpmethod/S2_quadratic.pdf
It defines the Quadratic Program as:
$Minimize \;\; f(x)=cx+\frac{1}{2}x^{T}Qx$
$Subject\, to\;\; Ax\leq b\;\; and\;\; x\geq 0$
On page 2 it states that the Karush-Kuhn-Tucker conditions for this are:
(JB.1) $\frac{\partial L}{\partial x_{j}}\geq 0,\; \; j = 1,...,n\; \; \; \; \; \; c+x^{T}Q+\mu A\geq0$
(JB.2) $\frac{\partial L}{\partial \mu_{i}}\leq 0,\; \; i = 1,...,m\; \; \; \; \; \; Ax-b\leq0$
(JB.3) $x_{j}\frac{\partial L}{\partial x_{j}}=0,\; \; j = 1,...,n\; \; \; \; \; \; x^{T}\left ( c^{T}+Qx+A^{T}\mu \right )=0$
(JB.4) $\mu_{i}g_{i}(x)=0,\;\;i=1,...,m \;\;\;\;\;\; \mu\left ( Ax-b \right )=0$
(JB.5) $x_{j}\geq 0,\;\; j=1,...,n \;\;\;\;\;\; x\geq 0$
(JB.6) $\mu_{i}\geq 0,\;\; i=1,...,m \;\;\;\;\;\; \mu\geq 0$
As far as I know the Karush-Kuhn-Tucker (in general, not in the special case of Quadratic Programming) are these:
(KKT.Stationarity) $\triangledown f(x) + \sum_{i=1}^{m} \mu_{i} \triangledown g_{i}(x) + \sum_{j=1}^{l} \lambda_{j} \triangledown h_{j}(x) = 0$
(KKT.PrimalFeasibility) $g_{i}(x)\leq 0,\;\; i=1,...,m \;\; and \;\; h_{j}(x)= 0,\;\; j=1,...,n$
(KKT.DualFeasibility) $\mu_{i}\geq 0,\;\; i=1,...,m$
(KKT.ComplementarySlackness) $\sum_{i=1}^{m} \mu_{i}g_{i}(x)=0$
I see how the following are related:
(JB.2) and (JB.5) <=> (KKT.PrimalFeasibility) with $g(x)=\begin{bmatrix} A & 0 \\ 0 & -I \end{bmatrix} x - \begin{bmatrix} b \\ 0 \end{bmatrix}$ and $h(x)=0$
(JB.6) <=> (KKT.DualFeasibility)
(JB.4) <=> (KKT.ComplementarySlackness)
However, I do not understand the connection between:
(JB.1), (JB.3) on the one hand and (KKT.Stationarity) on the other hand
Could you please explain me, how these follow from each other?
AI: The authors didn't mention the multipliers (say $\lambda \ge 0$) for $x \ge $ 0 explicitly. Usually, you would have:
$c+x^{T}Q+\mu A - \lambda = 0$ and $\lambda \ge 0$ and $x_j \lambda_j = 0$
but this indeed implies
$c+x^{T}Q+\mu A \ge 0$ and $x^T (c+x^{T}Q+\mu A) = 0$
when you eliminate $\lambda = c+x^{T}Q+\mu A$. |
H: Solve $\lfloor \ln x \rfloor \gt \ln \lfloor x\rfloor$
The question requires finding all real values of $x$ for which $$\lfloor \ln x\rfloor \gt \ln\lfloor x\rfloor $$ To start off, one could note that $$\lfloor \ln x \rfloor =\begin{cases} 0,& x\in[1,e) \\ 1,& x\in[e,e^2) \\ 2, &x\in [e^2,e^3) \\ 3,& x\in [e^3,e^4) \\ \vdots \end{cases}$$ and
$$\ln\lfloor x\rfloor =\begin{cases} 0, &x\in[1,2) \\ \ln 2, &x\in [2,3) \\ \ln 3,& x\in[3,4) \\ \ln 4,& x\in[4,5) \\ \vdots \end{cases}$$Although, from here it is not exactly clear to me how I can proceed. It appeas to me that there are infinitely many intervals of $x$ for which this inequality is true, but how can I find a generalized form of such an interval? E.g. something of the form $x\in \big(f(k), g(k)\big)$ for $k\in\mathbb N$ ?
AI: You have a solution in every interval $[k, k+1), k \in \mathbb N$ which contains a power of $e.$
To prove this, consider breaking up $[k, k+1)$ as $[k, e^\alpha) \cup[e^\alpha, k+1).$
In the first interval, $\lfloor{\ln x}\rfloor = \alpha -1, $ while $\ln\lfloor{ x}\rfloor = \ln k$.
It is easy enough to see that the inequality doesn't hold in this region (Using the fact that $k > e^{\alpha - 1}$)
For the second interval, $\lfloor{\ln x}\rfloor = \alpha, $ while $\ln\lfloor{ x}\rfloor = \ln k$.
Obviously, $\alpha > \ln k$ so the inequality holds.
Hence, your solution set is of the form: $[e^\alpha, \lceil{e^\alpha}\rceil) \quad\forall \alpha \in \mathbb N$ |
H: Showing that for some group it is abelian iff $x • (y • x ^{−1} ) = y$
I have to show that for some group it is abelian iff $x • (y • x^{−1}) = y$.
This is what I did:
Starting with the given statement $x • (y • x^{−1}) = y$ which implies $x • (y • x^{−1}) • x= y • x$. Since it is a group associativity can be used giving $(x • y) • (x^{−1} • x) = y • x$ implying $x • y = y • x$. So since the group commutes it can be implied to be abelian.
Is this sufficient? Thank you
AI: No, it is not. All you did was to prove that if that condition holds, then the group is Abelian. But you also have to prove it in the other direction. That is not difficult, though: if the group is Abelian, then$$x\bullet(y\bullet x^{-1})=x\bullet(x^{-1}\bullet y)=(x\bullet x^{-1})\bullet y=y.$$ |
H: If $x$ is a local minimum of $f$ restricted to an open subset, why can we conclude it is a minimum of $f$ as well?
Let $(E,\tau)$ be a topological space and $f:E\to\mathbb R$. We say that $x\in E$ is a local minimum of $f$ if $$f(x)\le f(y)\;\;\;\text{for all }y\in N\tag1$$ for some open neighborhood $N$ of $x$.
Let $\Omega\subseteq E$ and $x$ be a local minimum of $\left.f\right|_\Omega$. If $\Omega$ is open, why can we conclude that $x$ is a local miminum of $f$?
AI: If $x \in \Omega$ is a local minimum of $\left.f\right|_\Omega$, then there is an open neigborhood $N \subset \Omega$ of $x$ such that $f(x) \le f(y) $ for all $y \in N.$
Since $\Omega$ is open, $N$ is open in $E$, and the result follows. |
H: Probability one random variable is less than another random variable but higher than the same other random variable with a factor
Consider the two independent random variables $X$ and $Y$. Assume both are uniformly distributed on $[0,1]$. I want to calculate the probability that $X$ lies between some "low" linear transformation of $Y$, say $\delta_1Y$, and some "higher" linear transformation of $Y$, say $\delta_2Y$, where $\delta_1<\delta_2\Leftrightarrow\delta_1Y<\delta_2Y$. Further we know that $\delta_1>0$ and $\delta_2\geq1$. Thus, the probability I want to calculate is $\mathbb{P}(\delta_1Y<X<\delta_2Y)$.
I know that the probability that $X$ is less than $Y$ can be calculated by:
$$\mathbb{P}[X<Y]=\int_{y=0}^{1}\int_{x=0}^{y}P_{X}(x)P_{Y}(y)dxdy$$
where $P_X(x)$ is the PDF of $X$ and $P_Y(y)$ is the PDF of $Y$. A simple solution would be if you simply could calculate the probability as:
$$\mathbb{P}[\delta_{1}Y<X<\delta_{2}Y]=\mathbb{P}[\delta_{1}Y<X]\cdot\mathbb{P}[X<\delta_{2}Y]$$
But since this implicitly would assume that the two probabilities are independent, I'm pretty sure this is not a viable option. Does anyone have an idea how to solve this?
AI: You could start with:
$$\begin{aligned}P\left(\delta_{1}Y<X<\delta_{2}Y\right) & =\mathbb{E}\mathbf{1}_{\delta_{1}Y<X<\delta_{2}Y}\\
& =\int_{0}^{1}\int_{0}^{1}\mathbf{1}_{\delta_{1}y<x<\delta_{2}y}dxdy\\
& =\int_{0}^{1}\int_{\max\left(0,\delta_{1}y\right)}^{\min\left(1,\delta_{2}y\right)}dxdy\\
& =\int_{0}^{1}\min\left(1,\delta_{2}y\right)-\max\left(0,\delta_{1}y\right)dy
\end{aligned}
$$
If you already know that $\delta_1\geq0$ then $\max(0,\delta_1y)=\delta_1y$ and things become a bit easyer.
Go on with discerning cases. |
H: What is the function $E(x)$?
When reading Problems in Calculus of One Variable (a translated Russian book), I came across unfamiliar notation "$E(x)$". It is neither expected value nor $\exp(x)$. Here is a picture of the function used in context, which I hope someone can deduce what it means from
$$\int\limits_0^x E(x)\mathrm d x=\frac{E(x)(E(x)-1)}{2}+E(x)[x-E(x)]$$
It is not defined in the book, nor specific to context, and also used in multiple instances.
AI: I think it is the floor, $E(x) = \lfloor x\rfloor$. If $x$ is an integer, then you find the expression for the sum of integers, and if $x$ is not an integer, you add the missing part of the rectangle.
In French, the notation $E(x)$ is used for the floor since it is called "partie entière". |
H: Show that $\lim_n a_n = 0$ implies $\lim_n \frac{\sum_{m=1}^n m a_m}{\sum_{m=1}^n m} = 0$
Let ${a_n}$ be a sequence in $\mathbb{R}$. I want to show that
$$
\lim_n a_n = 0 \implies \lim_n \frac{\sum_{m=1}^n m a_m}{\sum_{m=1}^n m} = 0.
$$
My attempt is like this:
Fix any $\varepsilon>0$. There exists $N$ such that $|a_n|<\varepsilon$ for all $n>N$. Then,
$$
\begin{align*}
\left|\frac{x^{1}+2x^{2}+\cdots+mx^{m}}{1+2+\cdots+m}\right| & =\left|\frac{x^{1}+\cdots+Mx^{M}+\cdots+mx^{m}}{1+\cdots+M\cdots+m}\right|\\
& <\left|\frac{x^{1}+\cdots+Mx^{M}+\varepsilon((M+1)+\cdots+m)}{1+\cdots+M\cdots+m}\right|\\
& \le\left|\frac{x^{1}+\cdots+Mx^{M}}{1+\cdots+M\cdots+m}\right|+\varepsilon\left|\frac{(M+1)+\cdots+m}{1+\cdots+M\cdots+m}\right|\\
& <\left|\frac{x^{1}+\cdots+Mx^{M}}{1+\cdots+M\cdots+m}\right|+\varepsilon
\end{align*}.
$$
However I get the feeling that this is going nowhere.
AI: Everything is fine by now. Fix any $\varepsilon > 0$. Then, there exists $N$ such that $|a_m| \le \varepsilon$ for $m \ge N$.
Now you have:
$$ |\frac{\sum_{k=1}^n ka_k}{\sum_{k=1}^n k}| = |\frac{\sum_{k=1}^N ka_k}{\sum_{k=1}^n k} + \frac{\sum_{k=N+1}^n ka_k}{\sum_{k=1}^n k}| \le \frac{\sum_{k=1}^N |ka_k|}{\sum_{k=1}^n k} + \varepsilon $$
Now, choose $M$ so that for $n \ge M$ you have $\frac{\sum_{k=1}^N |ka_k|}{\sum_{k=1}^n k} \le \varepsilon$ (you can find such $M$, because the numerator is fixed, and $\sum_{k=1}^n k \to \infty$ as $n \to \infty$.)
Hence, for any $\varepsilon > 0$, you found $K = \max\{N,M\}$ such that for $n \ge K$, you have:
$$ | \frac{\sum_{k=1}^n ka_k}{\sum_{k=1}^n k} | \le 2\varepsilon $$ |
H: Solve the following multiple integral
Let
$I = \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 + x_2 + x_3 - x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4 $
Then $I$ equals
$ (a)\ \frac{1}{2} \\
(b)\ \frac{1}{3} \\
(c)\ \frac{1}{4} \\
(d)\ 1 $
I've tried to solve this and I also could do the whole integration down to the last variable but the process is tremendously lengthy and I have a term $log7$ remaining in my solution. I don't know where I've made a calculative mistake but I just want some hint to solve this problem in a easier and a compact way.
Also I've thought about change of variable transformations but couldn't think of any suitable one.
AI: By symmetry we can see that $$\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 + x_2 + x_3 - x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4= \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 + x_2 - x_3 +x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4= \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 -x_2 + x_3 +x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4= \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{-x_1 + x_2 + x_3 +x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4$$
So $$I=\frac 14 \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \frac{x_1 + x_2 + x_3 - x_4}{x_1 + x_2 + x_3 + x_4} + \frac{x_1 + x_2 - x_3 +x_4}{x_1 + x_2 + x_3 + x_4} + \frac{x_1 -x_2 + x_3 +x_4}{x_1 + x_2 + x_3 + x_4} \frac{-x_1 + x_2 + x_3 +x_4}{x_1 + x_2 + x_3 + x_4} dx_1 dx_2 dx_3 dx_4= \frac 14 \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} 2 dx_1 dx_2 dx_3 dx_4=\frac 12.$$ |
H: Raising Indices of a Conformally Transformed Metric
This is a bit of a silly question, but if a conformally transformed metric is given by
$g_{ij} = A^4 h_{ij}$,
where $A$ is a function of the spatial and time coordinates, and if one raises the indices does one accordingly have
$g^{ij} = A^4 h^{ij}$?
AI: $g^{ij}$ is typically defined as the pointwise inverse of the metric, $g^{ij}g_{jk}=\delta^i_k$. As such, if you have a conformally related metric $g_{ij}=A^4h_{ij}$, the inverse would have the reciprocal of the conformal factor $g^{ij}=A^{-4}h^{ij}$. |
H: Why is the spectrum of a shift operator the closed unit disk?
Consider the following text from Murphy's: "$C^*$-algebras and operator theory":
In example 2.3.2, why is $\sigma(u) = \Bbb{D}$ (= the closed unit disk)?
I can see that $\sigma(u) \subseteq \Bbb{D}$ and $\sigma(u^*) = \Bbb{D}.$
Thanks in advance!
AI: $$ \lambda \in \sigma(u) \iff \overline{\lambda} \in \sigma( u^*).$$ |
H: Evaluate the limit $\lim_{x\to 0} \frac{\ln |1+x^3|}{\sin^3 x}$
I tried this with L’Hopital and got $1$ as the answer. I don’t know if it can be applied considering the presence of the modulus function, but the answer is right. But I want a solution without using that rule, and I don’t know how to start this. Can I get some insight into this?
Edit: MY ATTEMPT
$$\lim_{x\to 0} \frac{\ln |1+x^3|}{|1+x^3|}\lim_{x\to 0} \frac{|1+x^3|}{\sin^3x}$$
$$=\lim _{x\to 0} \frac{x^3|\frac{1}{x^3}+1|}{\sin^3x}$$
AI: It comes down to two known limits as follows:
$$\lim_{x\to 0} \frac{\ln |1+x^3|}{\sin^3 x}=\lim_{x\to 0} \frac{\ln |1+x^3|}{x^3}\frac{x^3}{\sin^3 x}=1\cdot 1=1$$ |
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