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H: Why the digit Sum of a non negative number(n) is always the remainder after division by 9 , until n becomes only one digit?
lets assume for $689$ : ,
$$689 = (6+ 8 + 9 ) = 23$$
$$23 = (2 + 3) = 5$$
We can get this way:
$$689 \!\!\mod \;9 = 5$$
I found a approach like this:
$$10 = (9 * 1) + 1 $$
$$100 = (9 * 11) + 1 $$
this way :
$$689 = 9 + (8*10) + (6*100)$$
$$689 = 9 + (8*(9+1)) + (6*(9 * 11 + 1))$$
$$689 = (9+8+6) + 9 * ((8 * 1) + (6 * 11))$$
But I couldn't figure out out how to proceed forward after this?
AI: (Hint) :- Use modular arithmetic .
Note that $10^n \equiv 1\ (\textrm{mod}\ 9)$ for any positive integer $n$ and every number can be uniquely represented as $10^na_1 + 10^{n - 1}a_2 + ... + a_n$ where $a_1,a_2,...,a_n$ are positive integers from the set $[1,2,3,4,5,6,7,8,9]$ .
You have actually done the same thing in the case of $689$ . Since you got :-
$$(9 + 8 + 6) + 9 * ((8*1) + (6*11))$$
We get :- $689 \equiv (9 + 8 + 6)\ (\textrm{mod}\ 9)$
=> $(9 + 8 + 6) \equiv 5\ (\textrm{mod}\ 9)$
, and that's the same dealing with the remainder of the sum of digits after division by $9$ .
Edit :- As mentioned by @Toby Mak, the remainder found from the division by $9$ is called it's digital root . |
H: Convergent Sequences in Extremally Disconnected Hausdorff Spaces
It's written in Willard (15G.3) that the only convergent sequences in a Hausdorff Extremally Disconnected space are the eventually constant sequences. However, it has not provided a proof.
I've tried to derive this myself, but am unable to do so. There's another post on Math Stackexchange about this problem, but a solution wasn't presented there. So, any help in proving this is appreciated!
AI: Let $(x_n)$ be a sequence in $X$ such that $x_n \to p \in X$.
Assume it is not eventually constant. Then in particular $x_n \ne p$ for infinitely many $n$, i.e. there exists a subsequence $(x_{n _k})$ such that $y_k = x_{n _k} \ne p$ for all $k$. Let us inductively construct a subsequence $(y_{k_r})$ of $(y_k)$ and a sequence of pairwise disjoint open sets $U_r$ such that $y_{k_r} \in U_r$ and $p \notin \overline U_r$.
For $r=1$ we take $k_1 = 1$. There exist disjoint open neigborhoods $U_1$ of $y_1$ and $V_1$ of $p$. Thus $p \notin \overline U_1$. For the induction step observe that $W_r = X \setminus \bigcup_{i=1}^r \overline U_i$ is an open neigborhood of $p$. Since $y_k \to p$, we find $k_{r+1}$ such that $y_{k_{r+1}} \in W$. There exist disjoint open subsets $U_{r+1}$ and $V_{r+1}$ of $W_r$ such that $y_{k_{r+1}} \in U_{r+1}$ and $p \in V_{r+1}$. These subsets are also open in $X$. Clearly $p \notin \overline U_{r+1}$ and $U_{r+1} \cap \bigcup_{i=1}^r \overline U_i = \emptyset$ which shows that $U_1,\ldots, U_{r+1}$ are pairwise disjoint.
Let $U = \bigcup_{i=1}^\infty U_{2i}$ which is open. Thus also $\overline U$ is open. Since $y_{k_{2i}} \in U$ and $y_{k_{2i}} \to p$, we have $p \in \overline U$. Therefore $y_{k_r} \in \overline U$ for $r \ge R$. Now let $r \ge R$ be odd. Then $U_r \cap U = \emptyset$, thus $U_r \cap \overline U = \emptyset$ because $U \subset X \setminus U_r$. We conclude that $y_{k_r} \notin \overline U$. This is the desired contradiction. |
H: Integral definition of curl
My book, Mathematical Methods for Physics and Engineering - K. F. Riley,
explains that the 'Integral definition of curl' is:-
$$\nabla\times a = \lim_{V\to0}(\frac{1}{V}\int_S dS\times a) $$
and then, after a few pages, it mentions:-
$$\int_S dS \times \nabla\phi = \int_C\phi dr$$
Now, if we take $a = \nabla\phi$ in the first equation, then $\int_S dS\times a$ becomes $0$ (because the curl of the gradient of anything is $0$). But this contradicts with the second equation.
So, what is the mistake here?
AI: Presumably, in the first definition $S$ is a closed surface (i.e. a surface with no boundary), and $V$ is the volume enclosed by the surface.
In the second definition, $S$ is not necessarily a closed surface. $S$ may have a boundary, and the boundary of $S$ is the curve $C$.
But in the special case where $S$ is closed, then $C$ is an empty curve, so $\int_C \phi dr = 0$. There is therefore no contradiction. |
H: Is it true that $\sum_{i=0}^{n-1} 2^i$ divides $\sum_{i=n}^{2n-1} 2^i$?
Some investigation suggested to me that
$$\sum_{i=0}^{n-1} 2^i \Bigg| \sum_{i=n}^{2n-1} 2^i$$
The point being that binary numbers in the form exclusively of an even number of $1$s appear to be divisible by the number represented by just half that number of $1$s.
Is this the case, and if so why is that please?
AI: Note you have
$$\begin{equation}\begin{aligned}
\sum_{i=n}^{2n-1} 2^i & = 2^n + 2^{n+1} + \ldots + 2^{2n-1} \\
& = 2^n + 2^n(2) + \ldots + 2^n(2^{n-1}) \\
& = 2^n(1 + 2 + \ldots + 2^{n-1}) \\
& = \sum_{i=0}^{n-1} 2^n(2^i) \\
& = 2^n\left(\sum_{i=0}^{n-1} 2^i\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$ |
H: What's the name for sequences sitting between geometric and arithmetic, i.e. whose recurrence relation is of the form $ax+b$?
What's the name for sequences sitting between geometric and arithmetic?
E.g. let $x_{n+1}=ax_n+b$
I can't find a general name for these.
These sequences may also be a Lucas Sequence but that's a pretty convoluted way of defining these.
AI: I think the term you are looking for is Linear recurrence Relations |
H: Easy way to check that real part of $e^{-\frac{1}{z^{4}}}$ is harmonic.
Let $z=x+i y$ and $$ u(x,y) = Re \left(e^{-\frac{1}{z^{4}}}\right) ,~ \text{for}~ (x,y) \ne (0,0)$$ and $0$ otherwise. Then is there any short way to check that $u$ satisfies Laplace equation ?
I can apply brute force to check that given function is harmonic in $\mathbb{R}^{2}$, but I think there may be some easy way to check it is harmonic.
AI: That function is holomorphic and the real part of a holomorphic functions is always harmonic. |
H: Why is $\operatorname{Im}{\frac{1-e^{iny}}{-2i\sin{\frac{y}{2}}}}=\frac{1}{2\sin{\frac{y}{2}}}\operatorname{Re}(1-e^{iny})$?
I am studying Fourier analysis and had a question involving the following equality:
$$\operatorname{Im}{\frac{1-e^{iny}}{-2i\sin{\frac{y}{2}}}}=\frac{1}{2\sin{\frac{y}{2}}}\operatorname{Re}(1-e^{iny})$$
I can see that $\operatorname{Im}{\frac{1}{-2i\sin{\frac{y}{2}}}}=\frac{1}{-2\sin{\frac{y}{2}}}$, but I cannot get to the RHS of the above equation. What am I missing?
AI: We have $\frac{1}{-2i\sin{\frac{y}{2}}}=\frac{i}{2\sin{\frac{y}{2}}}$ as $i^2=-1$.
Then $\frac{1-e^{iny}}{-2i\sin{\frac{y}{2}}}=\frac{1-e^{iny}}{2\sin{\frac{y}{2}}}i=\frac{i-i(\cos(ny)+i\sin(ny))}{2\sin{\frac{y}{2}}}=\frac{\sin(ny)+i(1-\cos(ny))}{2\sin{\frac{y}{2}}}$ and
Re($1-e^{iny})=1-\cos(ny)$. |
H: Divergence of $\sum_{n=1}^{\infty}\prod_{k=1}^n q_k$ for some enumeration $(q_n)_{n}$ of $\mathbb{Q}\cap (0,1)$
Given an enumeration $(q_n)_{n}$ of $\mathbb{Q}\cap (0,1)$, let us consider the series
$$\sum_{n=1}^{\infty}\prod_{k=1}^n q_k.$$
Find an enumeration such that the series is convergent.
Find an enumeration such that the series is divergent.
is rather easy: take any enumeration $(a_n)_{n}$ of $\mathbb{Q}\cap (0,1/2]$ and any enumeration $(b_n)_{n}$ of $\mathbb{Q}\cap (1/2,1)$. By letting $(q_n)_n=a_1,b_1,a_2,b_2,\dots$, it follows that
$$\sum_{n=1}^{\infty}\prod_{k=1}^n q_k< 2\sum_{n=1}^{\infty}\frac{1}{2^n}=2.$$
seems to be more challenging and I did not solve it so far. Any hints?
I read about this problem a few years ago, but I can't remember the source. If someone find it please let me know!
Bonus question 1'. Is there any enumeration $(q_n)_{n}$ of $\mathbb{Q}^+$ such that the series is convergent?
AI: Fix an enumeration of $X = \mathbb{Q}\cap(0,1)$. Using this, we construct an ordered list $\mathtt{L}$ by the following algorithm:
set $\mathtt{L}_0 = () $.
For each $k$ in the list $\mathbb{N}_{1} = (1,2,3,\dots)$,
Pick the first element $q$ of the set $X \setminus \mathtt{L}_{k-1}$ under the prescribed enumeration.
Pick $N_k \in \mathbb{N}_{1}$ such that $(N_k - \left|\mathtt{L}_{k-1}\right|) \left( q \prod_{r \in \mathtt{L}_{k-1}} r \right) > 1 $.
Write $m = N_k - \left|\mathtt{L}_{k-1}\right| - 1$ and pick $s_1, \dots, s_m \in X\setminus(\mathtt{L}_{k-1}\cup\{q\})$ such that
$$ (N_k - \left|\mathtt{L}_{k-1}\right|) \left( s_1 \dots s_m q \prod_{r \in \mathtt{L}_{k-1}} r \right) > 1 \tag{*} $$
Set $\mathtt{L}_k$ as the concatenation of $\mathtt{L}_{k-1}$ and $(s_1, \dots, s_m, q)$. (Note: We have $\left|\mathtt{L}_{k}\right| = N_k$.)
This algorithm clearly exhausts all the elements of $X$. Moreover, it defines a list $\mathtt{L}$ as the 'limit' of $\mathtt{L}_k$'s such that $\mathtt{L}$ enumerates $X$. Finally, if we write $\mathtt{L}=(q_n)_{n\geq 1}$, then by $\text{(*)}$,
$$ \sum_{n=1}^{\infty} \prod_{j=1}^{n} q_j
\geq \sum_{k=1}^{\infty} (\left|\mathtt{L}_{k}\right| - \left|\mathtt{L}_{k-1}\right|) \prod_{j=1}^{\left|\mathtt{L}_k\right|} q_j > \sum_{k=1}^{\infty}1 = \infty. $$
Answer to Bonus Question. Fix an enumeration of $X=\mathbb{Q}\cap(0,\infty)$. Then run the following algorithm:
Set $\mathtt{L}_0 = ()$.
For each $n$ in $(1,2,3,\dots)$,
Pick the first element $b_n$ of $X\setminus\mathtt{L}_{n-1}$ under the prescribed order.
Pick an element $a_n$ of $X\setminus(\mathtt{L}_{n-1}\cup\{b_n\})$ such that $a_n \leq \frac{1}{2}$ and $a_n b_n \leq \frac{1}{4}$.
Set $\mathtt{L}_n$ as the concatenation of $\mathtt{L}_{n-1}$ and $(a_n, b_n)$.
Now let $\mathtt{L}$ be the 'limit' of $\mathtt{L}_n$ as $n\to\infty$. (Regarding $\mathtt{L}_n$'s as partial functions from $\mathbb{N}_1$ to $X$, this amounts to taking the union of all $\mathtt{L}_n$'s.) By the construction, it is clear that $\mathtt{L}$ is an enumeration of $X$. Moreover, if we write $\mathtt{L}_n=(q_n)_{n\geq1}$, then the construction tells that
$$ \prod_{k=1}^{n} q_k \leq 2^{-n} \qquad \text{for all} \quad n \geq 1, $$
and so, the series $ \sum_{n=1}^{\infty} \prod_{k=1}^{n} q_k $ converges. |
H: Prove that set of permutation on $3$ elements is not isomorphic to $(\Bbb Z_6,+)$.
Prove that set of permutation on $3$ elements is not isomorphic to $(\Bbb Z_6,+ )$ (the group with $+$ on $\Bbb Z_6$).
Hello everyone,
I tried to build a function and show that associativity doesnt work because of the permutation but I don't know how to write the proof, I tried using table or examples.
Can someone help please ?
Thank you
AI: The group $S_3$ of permutations of $3$ elements is not abelian.
For instance, consider the transpositions $\tau=(1\,2)$, $\tau'=(2\,3)$. Then
\begin{align}
&&\tau\tau'&=(1\,2\,3)&\qquad& (\text{denote it }\gamma)\\
&\text{whereas }\qquad\qquad
&\tau'\tau&=(1\,3\,2)=\gamma^{-1}
\end{align} |
H: System of congurences and the Chinese Remainder Theorem
I have the following system of congruences:
\begin{align*}
x &\equiv 1 \pmod{3} \\
x &\equiv 4 \pmod{5} \\
x &\equiv 6 \pmod{7}
\end{align*}
I tried solving this using the Chinese remainder theorem as follows:
We have that $N = 3 \cdot 5 \cdot 7 = 105$ and $N_1=35, N_2=21, N_3=15$.
From this, we get the following
\begin{align*}
35x_1 &\equiv 1 \pmod{3} \\
21x_2 &\equiv 1 \pmod{5} \\
15x_3 &\equiv 1 \pmod{7}
\end{align*}
and this will result in
\begin{align*}
2x_1 &\equiv 1 \pmod{3} \\
x_2 &\equiv 1 \pmod{5} \\
x_3 &\equiv 1 \pmod{7}
\end{align*}
so from CRT $x =x_1N_1b_1 + x_2N_2b_2 + x_3N_3b_3 = 2 \cdot 35 \cdot3 + 1 \cdot 21 \cdot 5 + 1 \cdot 15 \cdot7 = 420 $.
However $420$ doesn't seem to satisfy the given system, what would be the problem here?
AI: From the Chinese remainder theorem:
$$x = x_1N_1b_1 + x_2N_2b_2 + x_3N_3b_3$$
$$= 2 \cdot 35 \cdot \color{red}{1} + 1 \cdot 21 \cdot \color{red}{4} + 1 \cdot 15 \cdot \color{red}{6} = 244$$
The general solution is when $x \equiv 244 \pmod {\text{lcm}(3,5,7)} \Rightarrow x \equiv 244 \pmod {105} \equiv 34 \pmod {105}$, or when $x = 105k + 34$ when $k$ is an integer.
This problem is the exact problem that appears in the Brilliant article on the Chinese remainder theorem. Their method involves rewriting the largest congruence, $x \equiv 6 \pmod 7$ in the form $7j+6$, then substituting the expression into the next largest congruence, so that $7j + 6 \equiv 4 \pmod 5$. Solving this congruence gives $j \equiv 4 \pmod 5$. Repeating this process gives $j = 5k +4$, and substituting into the equation for $j$ gives $x = 7(5k + 4) + 6 = 35k + 34$. $35k + 34 \equiv 1 \pmod 3$ results in $k \equiv 0 \pmod 3$. Therefore $k = 3l$ and $x = 35(3l) + 34$, so $x$ is in the form $105k + 34$. |
H: Physics and Riemann hypothesis
I was reading the article "Quantum physics sheds light on Riemann hypothesis" from Bristol University (http://www.bristol.ac.uk/maths/research/highlights/riemann-hypothesis/) and stopped here:
From a conference in 1996 in Seattle, aimed at fostering collaboration between physicists and number theorists, came early evidence of correlation between the arrangement of the Riemann zeroes and the energy levels of quantum chaotic systems. If this were true it would prove the Riemann hypothesis.
I think this would not prove the Riemann hypothesis, since mathematics is a formal science, as opposed to natural sciences, and is independent of our universe. It can only go the other way around, i.e., mathematics implies that something in our universe is true (not our universe implies something in mathematics is true).
Still, I'd like to ask whether the mathematical community agrees or disagrees with the above article.
AI: There are two possible interpretations.
One interpretation is that a correspondence between the zeroes of the Riemann zeta function and the energy levels of certain quantum systems may suggest a new theoretical approach that leads to a logical proof (or disproof) of the Riemann hypothesis. This is possible, but the proof/disproof would still lie within the field of mathematics, even if the method was suggested by physical considerations.
A second interpretation is that the experimental measurement of the energy levels of certain quantum systems could add to the weight of heuristic evidence for (or against) the Riemann hypothesis. This is very unlikely, since the hypothesis has already been verified by numerical calculations for at least the first $10^{13}$ zeroes of the Riemann zeta function (and for some ranges of values beyond that). I cannot see how any physical experiment could match either the volume or the precision of those existing numerical results.
A key sentence in the press-release style summary that you linked to is this:
We don't really know why RMT methods work in calculating the moments of the Riemann zeta function.
If we don't (yet) understand the approach then it isn't (yet) mathematics. |
H: Why does the irreflexivity of $L^1$ follow from $(L^1)'' \subsetneq L^1$?
My question is about the proof that $l^1$ is irreflexive. I have seen multiple proofs ($L^1$ and $L^{\infty}$ are not reflexive, Dual of $l^\infty$ is not $l^1$ and more), and all proofs stop after showing that there exists $f \in (l^\infty)' \cong (l^1)''$ that is not representable as
\begin{equation} f(x) = \sum_{i \in \mathbb{N}} a_i x_i \quad \text{with} \quad a \in l^1 \end{equation}
like the Hahn-Banach extension of $\lim_{n \to \infty} x_n$ on $c \subseteq l^\infty$.
With the isometric, linear injection $X \to X'', x \mapsto (\phi \mapsto \phi(x))$ we get that $l^1 \subsetneq (l^1)''$, but why does it follow from this that $l^1 \not\cong (l^1)''$? After all, there are spaces that are isometrically isomorphic to proper subspaces of themselves, like $l^1 \cong X := \{ x \in l^1 | x_1 = 0 \} \subsetneq l^1$ by
\begin{equation}
\phi: l^1 \to X, \quad (x_1, x_2, x_3, ...) \mapsto (0, x_1, x_2, ...)
\end{equation}
AI: By definition a normed linear space $X$ is reflexive if the canonical injection from $X$ to the second dual is surjective.
For the definition see https://en.wikipedia.org/wiki/Reflexive_space |
H: Finding coefficients in expansions
Show that the coefficient of $x^{−12}$ in the expansion of
$$\left(x^4−\frac{1}{x^2}\right)^5\left(x−\frac{1}{x}\right)^6$$
is $−15$, and calculate the coefficient of $x^2$. Hence, or otherwise, calculate the coefficients of $x^4$ and $x^{38}$ in the expansion of $$(x^2−1)^{11}(x^4+x^2+1)^5.$$
The first part of this exercise is solved easily considering "partitions", i.e. what can be added to get the exponents and turns out to be $-15$ and $215$ respectively. I was able to solve the second part of this exercise easily using multinomial theorem. However, this exercise is given before multinomial theorem is introduced, and I would really like to know what the "Hence" was gunning for. Once I factor out to get$$x^{21}\left(x-\frac{1}{x}\right)^{11}\left(x^2+\frac{1}{x^2}+1\right)^5,$$I can see similarities with the first question. I can even deduce that since $4-21=-17$, $38-21=17$, and inspecting signs, that the coefficients are equal in magnitude and opposite in sign. Please explain how I can use the first part of the question to proceed, rather than going "otherwise"/multinomial theorem.
AI: \begin{align*}
(x^2 - 1)^{11}(x^4 + x^2 + 1)^5
&=(x^2 - 1)^{5}(x^4 + x^2 + 1)^5(x^2 - 1)^{6} \\
&=(x^6 - 1)^5(x^2 - 1)^6 \\
&=x^{16}\left(x^4 - \frac{1}{x^2}\right)^5\left(x - \frac{1}{x}\right)^6
\end{align*}
Thus, the coefficient of $x^4$ in the new equation is the same as the coefficient of $x^{-12}$ in the original equation, which is $-15$.
And like you mentioned, the coefficient of $x^4$ and $x^{42 - 4} = x^{38}$ should be the same but opposite in sign. So the coefficient of $x^{38}$ is $15$. |
H: Determine if a statement can haapen
let $ V= M_{10}\left(F\right) $ a vector space over some field $ \mathbb{F} $
I have to determine if its possible that exists $ A\in M_{10}\left(F\right) $
such that $ M_{10}\left(F\right)=span\left\{ A^{i}:0\leq i\leq100\right\} $
I guess it cant be. But I cant prove why. I tried to separate
the cases - if $ A $ is invertible and if it is not. But still, cant find anything that would help me decide.
I noticed that there are 101 matrices in the set $ \left\{ A^{i}:0\leq i\leq100\right\} $ so one of them has to be linear independent in the rest of them. But I cant see how to continue
Thanks in advance
AI: The answer is indeed no. This is trivial if we can use the Cayley Hamilton theorem.
Note that each element of the span of $\{A^i : 0 \leq i \leq 100\}$ can be written in the form $p(A)$ for some polynomial $p$. However, the characteristic polynomial $\chi(x)$ (which has degree $n = 10$) satisfies $\chi(A) = 0$.
Note that any polynomial $p(x)$ can be written in the form
$$
p(x) = q(x) \chi(x) + r(x)
$$
with $r$ of degree at most $n-1$ via polynomial division. This means that
$$
p(A) = q(A)\chi(A) + r(A) = r(A).
$$
Because $r$ has degree at most $n-1$, $p(A) = r(A)$ is a linear combination of the elements $I,A^1,\dots,A^{n-1}$.
So, the span of $\{A^i : 0 \leq i \leq 100\}$ is at most $n$-dimensional, which means it cannot contain all of $V$.
Without Cayley-Hamilton:
Case 1: $A$ is a multiple of $I$; in this case it is clear that the powers of $A$ do not span $V$.
Case 2: $A$ is not a multiple of $I$. In this case, there exists at least one matrix $B$ for which $AB \neq BA$. It follows that $B$ is not an element of the span of $\left\{ A^{i}:0\leq i\leq100\right\}$. |
H: Alternate methods to prove $(1+a)(1+b)(1+c)(1+d) \geq 16$ if $abcd =1$.
I found this question some time ago in an Elementary Olympiad book:
If $a, b, c, d$ are positive integers such that $abcd =1$, then prove that $(1+a)(1+b)(1+c)(1+d) \geq 16$.
Evidently this was a direct consequence of Hölder's inequality, so I merely gave the following proof.
$$(1^4+(a^{\frac{1}{4}})^4)(1^4+(b^{\frac{1}{4}})^4)(1^4+(c^{\frac{1}{4}})^4)(1^4+(d^{\frac{1}{4}})^4) \geq (1+(abcd)^{\frac{1}{4}})^4 $$
$$(1+a)(1+b)(1+c)(1+d) \geq (1+1)^4$$
$$(1+a)(1+b)(1+c)(1+d) \geq 16$$
QED.
However, I was requested by my instructor to try the problem again using only the AM-GM inequality or the Cauchy-Schwarz inequality, because it is supposedly solvable using only those two.
At this point I expanded the expression believing that all the terms so formed would be of the degree $4$, and would, when taken together in groups, yield a power of $abcd$ on the GM side, which could then be added.
Doing so, however, I only get the weaker result that $(1+a)(1+b)(1+c)(1+d) \geq 14$.
I would appreciate a hint for this tantalizing problem.
AI: Maybe it means the following way.
By AM-GM
$$\prod_{cyc}(1+a)=$$
$$=1+(a+b+c+d)+(ab+ac+ad+bc+bd+cd)+$$
$$+(abc+abd+acd+bcd)+abcd\geq$$
$$\geq1+4+6+4+1=16.$$ |
H: If $\lim_{x\to 0} \frac{x^2 \sin (bx)}{ax-\sin x}=1$, then find $a,b$
$$\lim_{x\to 0} \frac{x^2 \frac{\sin bx}{bx} (bx)}{x(\frac{ax}{x}-\frac{\sin x}{x})}$$
$$=\lim_{x\to 0} \frac{bx^2 \frac{\sin bx}{bx}}{a-\frac{\sin x}{x}}$$
$$=0$$
Which is obviously wrong. I think I am suffering from lack of conceptual clarity in limits, so what exactly is going wrong in this question?
AI: $$L=\lim_{x\to 0} \frac{x^2\sin bx}{ax-\sin x}=\lim_{x\to 0} \frac{x^2(bx-b^3x^3/6+...)}{ax-x+x^3/6}=\lim_{x\to 0} \frac{bx^3-b^5 x^5/6}{ax-x+x^3/6}$$
If $L=1$, then $a=1,b=1/6.$ |
H: Meaning of a phrase from Zorich II
Taken this excerpt directly from Zorich, Mathematical Analysis II:
My question is simply this: I'd like to understand what the author means by the last sentence highlighted in red, which seems to me to seem to be in contrast with the first (always highlighted in red).
AI: It's just the notation that's unfortunate here. I haven't seen this done with Riemann integrals, but with Lebesgue integrals, the vector space of Lebesgue integrable functions on $E$ is denoted $\mathcal L(E)$, while the quotient space in which almost everywhere equal functions are identified is denoted $L(E)$. I'd suggest doing the same here: Denote the space of Riemann integrable functions as $\mathcal R(E)$, and the quotient space in question as $R(E)$, so they are easily distinguished.
Now what Zorich says is essentially just that while the integral is linear as a function $\mathcal R(E)\longrightarrow\mathbb R$, it is also linear as a function $R(E)\longrightarrow\mathbb R$, where $[f]\mapsto\int_E[f]\mathrm dV:=\int_E f\mathrm dV$. Here, $[f]$ is the equivalence class of $f$ in the quotient space. |
H: When are the zeros of $x^T A x$ equal to kernel of $A$?
I am wondering when the set of solutions to $x^T Ax = 0$ is equal to the kernel of the matrix $A$. Is there a general answer to this?
It seems to be true in the positive-definite case:
If $x^TAx = 0$ then $x$ must be $0$ (otherwise $x^TAx > 0$), so $Ax = 0$.
On the other hand, if $Ax = 0$ then $x^T Ax = x^T(Ax) = x^T \cdot 0 = 0$.
Similarly for the negative-definite case.
AI: For a symmetric semi-definite matrix it still holds, as you see from diagonalising the matrix. But even for $A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ it is no longer true. |
H: morphism is epic?
If we have three morphisms f, g and h between objects of a category.
Suppose gf = h. If g and h are epic, can we conclude that so is f ? Any help would be appreciated!
AI: Hint:
In category set let the domain and codomain of $h=g\circ f$ be a singleton.
Further let the codomain of $f$ have more than one element.
What can be said about being epic of $f,g,h$? |
H: Number of Relations that satisfy a condition
This is a multiple choice question from my Text Book
Let $A=\{1,2,3\}$. The no. of relations containing $(1,2)$ and $(1,3)$ which are reflexive and Symmetric but not transitive is
(A) $1$
(B) $2$
(C) $3$
(D) $4$
My Approach:
$A=\{1,2,3\}$
Relation $R$ must contain $(1,2)$ and $(1,3)$
For $R$ to be Reflexive, it must contain $(2,2)$ and $(1,1)$
For $R$ to be symmetric, it must contain $(2,1)$ and $(3,1)$
For $R$ to not to be Transitive, it must not contain $(2,3)$ and $(3,2)$
\therefore, $R=\{(1,2),(1,3),(2,2),(1,1),(3,1),(2,1)\}$
Anyother addition to $R$ will not satisfy the stated condition.
Hence, option $A$ is correct
Am I right?
[Edit:
R contains $(3,3)$ as well]
AI: You must also have $(3,3)$ for the relation to be reflexive. Because you have $(2,1)$ and $(1,3)$ you would need $(2,3)$ for the relation to be transitive. Similarly, you need $(3,2)$. If either or both of these are missing, the relation is not transitive, so there would be $3$ choices, but the requirement of symmetry says they have to both be missing and there is only one choice. |
H: Question about Schur's Lemma
Maybe a very dumb question about Schur's Lemma, but somehow this confuses me.
Let $G, H$ be groups. A function $f: G\rightarrow H$ is called group isomorphism, if $f(g\cdot h) = f(g)*f(h)$, and $f$ is bijective.
Let $(D_1,V_1)$ and $(D_2,V_2)$ be irreducible representations of a group $G$, and $A \in L(V_1 \rightarrow V_2)$ a linear function. If $A D_1(g) = D_2(g)A \hspace{5mm} \forall g \in G$, than A is either the $0-$ map or an isomorphism.
So in the proof of Schur's Lemma, (for example here) it is shown, that A has to be either the $0-$function or bijective. But what is the argument to say that it is an bijective group homomorphism (= group isomorphism)? Do I have a wrong understanding of isomorphism?
AI: There are many types of homomorphisms/isomorphisms in mathematics. Here the meaning is not an isomorphism of groups, but an isomorphism of representations. Given two representations $(V_1,\pi_1), (V_2,\pi_2)$ an isomorphism is an invertible linear map $A:V_1\to V_2$ such that $A(g.v)=g.(Av)$ for all $g\in G, v\in V_1$. |
H: Undertstanding Cup products
I am trying to understand cup products from the following notes and am having some difficulty understanding theorem which in a way defines cup product for cohomology: https://www.math.ucla.edu/~sharifi/groupcoh.pdf#theorem.1.9.5
What does it mean to say that the map 'cup products'
$$H^i(G, A)\otimes _{\mathbb Z} H^j(G,B) \xrightarrow{\smile} H^{i+j}(G, A \otimes _{\mathbb Z}B) $$
is "natural on $A$ and $B$" ?
And how do we get the commutative diagram on the top of page 37 ( https://www.math.ucla.edu/~sharifi/groupcoh.pdf#page.37 ) ?
I believe it has something to do with the 'naturality'.
Any help is appreciated and feel free to give any reference.
AI: It means naturality in the sense of category theory. Saying the cup product is natural in $A$ means that for any $B$ and any map $f: A \to A'$ you get induced maps $H^i(G, A)\otimes_{\mathbb{Z}} H^j(G, B) \to H^i(G, A')\otimes_{\mathbb{Z}} H^j(G, B)$ and $H^{i+j}(G, A\otimes_\mathbb{Z} B) \to H^{i+j}(G, A' \otimes_{\mathbb{Z}} B)$, and the square formed by these maps and the cup product commutes.
$\delta$ here is the connecting homomorphism arising from whichever short exact sequence is under consideration. The vertical maps in the diagram at the top of p.37 are $\delta \otimes 1$ on the left, and $\delta$ - arising from $0 \to A \otimes B \to A \Uparrow \otimes B \to A^* \otimes B \to 0$ (1.9.4), on the right. Commutativity is precisely the identity (ii) of 1.9.5. |
H: Evaluating $\int_0^1dx\int_x^{\frac{1}{x}}\frac{y\,dy}{(1+xy)^2(1+y^2)}$
Evaluate the integral (change of order of integration may be useful):$$\int_0^1dx\int_x^{\frac{1}{x}}\frac{y\,dy}{(1+xy)^2(1+y^2)}$$
I searched the region, I got this graph
But, can't identify the region exactly. In fact, the lower bound of y and upper bound does not match to a region. I can be wrong. Please give any hints or you can add a solution.
AI: The region of interest is bounded between the y-axis, the line $y=x$, and the line $y=\frac{1}{x}$, where $0<x<1$. Maybe this image will help you visualize it better:
To change the order of integration, first notice that we need to split the region into two along the line $y=1$ because of the shape of the region.
Consider the first region $0 \leq y \leq 1$ and $0 \leq x \leq y$ and the second region $1 \leq y < \infty$ and $0 \leq x \leq \frac{1}{y}$. These two regions together are equal to the original region.
Rewrite your integral as follows:
$$\int_0^1 \int_x^{\frac{1}{x}} \frac{y}{{(1+xy)}^2{(1+y^2)}} \; dy \; dx=\int_0^1 \int_0^{y} \frac{y}{{(1+xy)}^2{(1+y^2)}} \; dx \; dy + \int_1^{\infty} \int_0^{\frac{1}{y}} \frac{y}{{(1+xy)}^2{(1+y^2)}} \; dx \; dy$$
$$=\int_0^1 -\frac{1}{{(1+y^2)}^2} + \frac{1}{1+y^2}\; dy+ \frac{1}{2} \int_1^{\infty} \frac{1}{1+y^2} \; dy$$
$$=-\frac{\pi}{8}-\frac{1}{4}+\frac{\pi}{4}+\frac{\pi}{8}=\boxed{\frac{\pi}{4}-\frac{1}{4}}$$ |
H: Proving uniqueness of antipodes in Hopf algebras
Let $(H,\mu,\nu,\Delta,\epsilon)$ be a Bialgebra where H is the vector space, $\mu, \nu$ are the product and unit whilst $\Delta, \epsilon$ are the coproduct and counit. Now, for $f,g \in end(H)$ define $f@g \in end(H)$ by $f@g=\mu(f \otimes g)\Delta(x)=\Sigma_{(x)}f(x')g(x'')$ (via Sweedler notation).
An element $S \in end(H)$ is called an antipode if
$S@id_H=id_H@S=\nu\circ\epsilon$
If a Bialgebra has an antipode, then it is unique. To see this, suppose $S,T$ are antipodes for the bialgebra $H$. Then we have:
$S = S@(\nu\epsilon)=S@(id_H@T)=(S@id_H)@T=(\nu\epsilon)@T=T$
Can somebody explain to me the first equality? Why do we get $S = S@(\nu\epsilon)$?
AI: We know that an antipode for a bialgebra $H$ is a linear map $S:H \rightarrow H$ which is a convolution inverse of $id_H$.
We have to remember that the unit for the convolution product is
$$1_{\text{End}(H)}=\nu \circ\epsilon $$ Proof: $$f*(\nu\epsilon) = \mu \circ(f \otimes \nu\epsilon) \circ \Delta = \mu \circ (A \otimes \nu) \circ (f \otimes \mathbb{k})\circ (C \otimes \epsilon) \circ \Delta = \\= r_A \circ (f \otimes \mathbb{k}) \circ(r_C)^{-1} = f$$ (if you miss the second equality you can read a complete explanation here)
So, $\forall f \in \text{End}(H)$, the following holds (because $f*id = f$ in every algebraic context): $$f * 1_{\text{End}(H)}= f$$
Now, take $S \in \text{End}(H)$ and write the same thing:
$$S = S * 1_{\text{End}(H)}$$
but $1_{\text{End}(H)}=\nu \circ\epsilon $, so
$$S = S *\nu \epsilon$$ |
H: If $\lim_{x\to 0} \frac{1+a\cos 2x + b\cos 4x}{x^4}$ exists for all $x\in\mathbb R$ and is equal to $c$, find $\lfloor a^{-1} +b^{-1} + c^{-1}\rfloor$
$$\lim_{x\to 0} \frac{1+a(1-\frac{4x^2}{2!} + \frac{16x^4}{4!}-\cdots)+b(1-\frac{16x^2}{2!} + \frac{256x^4}{4!}-\cdots)}{x^4}$$
$$=\lim_{x\to 0} \frac{ (1+a+b) -\frac{x^2}{2!} (4a+16b) + \frac{x^4}{4!} (16a+256b)}{x^4}$$
For limit to exist, $1+a+b=0$ and $4a+16b=0$
So $a=-\frac 43$ and $b=\frac 13$
Now $$c=\lim_{x\to 0} \frac{\frac{x^4}{4!} (16a+256b)\cdots}{x^4}$$
$$c=\frac{16a+256b}{24}$$
$$c=\frac{2a}{3}+\frac{32}{3}b=\frac{24}{9}$$
Then $$\lfloor a^{-1} + b^{-1} + c^{-1} \rfloor = \left\lfloor 3-\frac 34 + \frac{9}{24}\right\rfloor = \lfloor 2.625\rfloor=2$$
The given answer is $8.$
What is wrong in my solution?
AI: $$L=\lim_{x\to 0} \frac{1+a\cos 2x + b\cos 4x}{x^4}$$
$$L=\lim_{x\to 0} \frac{1+a(1-\frac{4x^2}{2!} + \frac{16x^{4}}{4!}\cdots)+b(1-\frac{16x^2}{2!} + \frac{256x^4}{4!}\cdots)}{x^4}$$
$$\implies L=\lim_{x\to 0} \frac{(1+a+b)+(-2a-8b)x^2+(2a/3+32b/3)x^4}{x^4}$$
for the limit to exist $(1+a+b)=0, (-2a-8b)=0, (2a/3+32b/3=c$ then
$a=-4/3,b=1/3, c=8/3; [1/a+1/b+1/c]=[21/8]=2.
OP's answer is right. |
H: Why use limit laws to verify continuity instead of direct substitution?
My textbook (Calculus Early Transcendentals, 8th edition, by James Stewart ) asks to verify a function is continuous at a point using the definition of continuity and the limit laws. However, why would the text explicitly state to use the limit laws to verify continuity? I'm assuming this implies that direct substitution shouldn't be used and looking at solutions online confirms my suspicions as they all use limit laws as well. However, using direct substitution is much faster than using the limit laws and both methods achieve the same result.
For example, show that $f(x) = (x+2x^3)^4$ is continuous at $x=-1$
By definition of continuity, we're trying to show that $\lim_{x\to-1}(x + 2x^3)^4 =f(-1)$
Show that $f(-1)$ exists
$$f(-1)=(1+2(-1)^3)^4 = 81$$
Now for the limit
Using the limit laws:
$$\lim_{x\to-1}(x + 2x^3)^4$$
$$=[\lim_{x\to-1}x + 2x^3]^4$$
$$=[\lim_{x\to-1}x + \lim_{x\to-1}2x^3]^4$$
$$=[\lim_{x\to-1}x + 2\lim_{x\to-1}x^3]^4$$
$$=[-1 + 2(-1)^3]^4 = 81$$
Using direct substitution:
$$\lim_{x\to-1}(x + 2x^3)^4 = (1+2(-1)^3)^4=81$$
In either method we reach the same result that $\lim_{x\to-1}(x + 2x^3)^4 = 81$ which verifies that $f(x)$ is continuous at $x=-1$ since $\lim_{x\to-1}(x + 2x^3)^4 =f(-1)$
Perhaps I'm missing some important connection between continuity and limits as to why limit laws are used instead of direction substitution to verify continuity?
AI: Maybe it'll help to break down very strictly what we are doing with the limit laws. Firstly,
$${(x + 2x^3)^4 = (x+2x^3)(x+2x^3)(x+2x^3)(x+2x^3)}$$
Now, if we have that the limit
$${\lim_{x\to -1}(x+2x^3)}$$
exists, we can say that
$${\lim_{x\to -1}(x+2x^3)^4 = (\lim_{x\to -1}(x + 2x^3))(\lim_{x\to -1}(x + 2x^3))(\lim_{x\to -1}(x + 2x^3))(\lim_{x\to -1}(x + 2x^3))=(\lim_{x\to -1}(x + 2x^3))^4}$$
this is because the limit of the product is the product of the limits (provided the individual limits exist). This is indeed a limit law.
We know that both ${x}$ and ${2x^3}$ will be continuous functions, and so will their sum. Hence we can say that
$${\lim_{x\to -1}x+2x^3=-1+2(-1)^3=-3}$$
Notice we can only directly substitute ${x=-1}$ because we know ${x+2x^3}$ is continuous. This step becomes invalid if we did not know the continuity of ${x+2x^3}$. And now since the individual limit of ${x+2x^3}$ exists, we have that
$${\lim_{x\to -1}(x+2x^3)^4 = (-3)^4 = 81}$$
And ${f(x)=(x+2x^3)^4}$ also has ${f(-1)=81}$, we can conclude ${f(x)}$ is continuous at ${x=-1}$.
If you just directly plug in the number ${-1}$ into
$${(x+2x^3)^4}$$
and claiming this is
$${\lim_{x\to -1}(x+2x^3)^4}$$
you are assuming that ${(x+2x^3)^4}$ is continuous. So you cannot use this to prove continuity, as this is circular reasoning, as also stated by @JMoravitz
Edit: another way to show continuity is to notice that if ${f(x) = x^4}$ and ${g(x) = x + 2x^3}$, notice that your function is just
$${(x+2x^3)^4 = f(g(x))}$$
Both ${f,g}$ are continuous, and the composition of two continuous functions is continuous - so ${(x+2x^3)^4}$ is continuous |
H: How to quantify which expression for a given mathematical quantity converges the fastest?
Which converges faster to $e$: $\lim_{ n \to \infty} \sum_{k=0}^{n} \frac{1}{k!}$ or $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$?
To check this, I thought of taking ratio of the two and taking limit as n goes to infinity,
Now the question is,
$$ \lim_{ n \to \infty} \dfrac{\displaystyle \sum_{k=0}^{n} \frac{1}{k!}}{ \left(1 + \frac{1}{n}\right)^n} =L$$
Now, what is 'L'? I think it is one. So, this idea seems inconclusive.
AI: By the binomial theorem, $$\left(1 + \frac{1}{n}\right)^n = \sum_{k = 0}^n \frac{{n \choose k}}{n^k} = \sum_{k = 0}^n \frac{1}{k!} \frac{n(n-1)\cdots(n-k+1)}{n^k} < \sum_{k = 0}^n \frac{1}{k!} < e$$
so $\displaystyle\sum_{k = 0}^n \frac{1}{k!}$ converges "faster". |
H: How to calculate the limit of sequence
The question is calculate limit of sequence
$$
\lim_{n \to \infty}
\frac{\left(2\,\sqrt[\Large n]{\, n\,}\, -
\,\sqrt[\Large n]{\, 2\,}\right)^{n}}{n^2}
$$
I'm trying to simplify the equation, like divide $\,\sqrt[\Large n]{\, n\,}\,$, but can't get more. I drew the continuous function plot, which shows value tends to $0.4$.
Do any method show more details about this question ？.
AI: Note that we have
$$\begin{align}
\left(2\sqrt[n]{n}-\sqrt[n]{2}\right)^n&=\left(2e^{\frac1n \log(n)}-e^{\frac1n\log(2)}\right)^n\\\\
&=\left(1+\frac1n\log(n^2/2)+O\left(\frac{\log^2(n)}{n^2}\right)\right)^n\\\\
&=e^{n\log\left(1+\frac1n\log(n^2/2)+O\left(\frac{\log^2(n)}{n^2}\right)\right)}\\\\
&=e^{n\left(\frac1n\log(n^2/2)+O\left(\frac{\log^2(n)}{n^2}\right)\right)}\\\\
&=\frac{n^2}{2}+O\left(n\log^2(n)\right)\tag1
\end{align}.$$
Upon dividing $(1)$ by $n^2$ and letting $n\to \infty,$ we find
$$\lim_{n\to\infty }\frac{\left(2\sqrt[n]{n}-\sqrt[n]{2}\right)^n}{n^2}=\frac12$$ |
H: Are $\forall x\forall y(p(x,y)\leftrightarrow p(y,x))$ and $\forall x\forall y p(x,y) \leftrightarrow \forall y\forall x p(x,y)$ tautologies?
$$\forall x\forall y( p(x,y) \leftrightarrow p(y,x) )$$
I don't know if this formula is a tautology or not.
I think the order inside the predicates is not important and therefore, it's a tautology.
Thanks for your help!
What about this one ?
$$\forall x\forall y p(x,y) \leftrightarrow \forall y\forall x p(x,y).$$
The order in the all-quantifier isn't important too, so i think this one is a tautology too, right ?
AI: $$\forall x\forall y( p(x,y) \leftrightarrow p(y,x) )$$
Consider the interpretation $\mathcal{I}(P) = \{\langle x, y \rangle: x \text{ loves }y\}$. Sadly enough, just because you love someone doesn't always mean they loves you back: We can easily think of a structure where John loves Peter (i.e. $\langle John, Peter \rangle \in \mathcal{I}(P)$) but Peter doesn't love John ($\langle Peter, John \rangle \not \in \mathcal{I}(P)$). Thus $P(x,y)$ does not imply $P(y,x)$. Since we found is at least one interpretation that makes the formula false, it is not tautological.
The interpretation of a predicate is a set of tuples. Tuples are ordered: $\langle a, b \rangle \neq \langle b, a\rangle$. Since the order of the objects in the interpretation of the predicate matters, so does the order in which we write the terms down in the expression.
$$\forall x\forall y p(x,y) \leftrightarrow \forall y\forall x p(x,y).$$
The order in the all-quantifier isn't important
A bit sloppy, but yes: In general, $\forall x \forall y \phi$ is logically equvivalent to $\forall y \forall x \phi$; that is, we can commute quantifiers of the same type. So yes, this formula indeed is tautological. |
H: Let $G=\mathbb{Z}\times\mathbb{Z}$ and $H=\{(a,b)\in\mathbb{Z}\times\mathbb{Z}: 8\mid a+b\}$. What is the index $[G:H]$?
Let $G=\mathbb{Z}\times\mathbb{Z}$ and $H=\{(a,b)\in\mathbb{Z}\times\mathbb{Z}: 8\mid a+b\}$. What is the index $[G:H]$?
From a separate exercise part of this problem we are given $H\lhd G$ and that $G/H\cong\mathbb{Z}_8.$
Considering the definition of $H$ I believe $|H|=8n$ (Since there are $8n$ entries that satisfy the condition).
How would I find $|G|$? Would this just be $\infty$?
AI: By definition, when we have $H\lhd G$ such that $G/H$ is finite, then $[G:H]:=|G/H|.$ In this case, then, $[G:H]=|\Bbb Z_8|=8.$ |
H: Construct compact subset of a set
In the context of uniform continuity in our analysis course we often need compact sets or compact subsets to prove certain properties. I am not sure if I have fully understood how to construct a compact subset and why we are allowed to do this. May be someone can give me an explanation to the following issue.
Let's assume that $M\subseteq \mathbb{R}^n$ is an open set, with $M\neq\emptyset$ . As $M$ is open there exists a neighbourhood for an arbitrary point $m\in M$: $U_{\delta}(m):= \{x\in M~|~\Vert x-m\Vert<\delta\}\subset M$, where $\delta >0$. Now, I simply take a $\delta_0$ with $0<\delta_0<\delta$ and define a compact subset by $C:=\{x\in M~|~\Vert x-m\Vert\leq\delta_0\}\subset M$. Is this correct?
Is it possible to construct a compact subset in this way when we don't have furhter information on $M$ (open, not-open, closed, not-closed)?
AI: If $M$, yes, what you did is correct: the set $C$ that you have difined is compact.
If there is no assumption about $M$, then it may well happen that the only compact subsets of $M$ are the finite ones (note that every finite subset of $\Bbb R^n$ is compact). That's the case is, for instance, $M=\Bbb Q^n$. |
H: how to find eigenvalues of $T_n:R^n\to R^n$ where $T_n (x)=(0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1})$, what happen if $n\to\infty$
how to find eigenvalues of $T_n:R^n\to R^n$ where $T_n (x)=(0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1})$ what happen if $n\to\infty$, well by definition i do $T_n(x)=\lambda x$ hence
$(0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1})=(\lambda x_1,\ldots,\lambda x_n)$ imply
$0=\lambda x_1=\lambda^2 x_2=2\lambda^3 x_3=\ldots =(n-1)!\lambda^n x_n $ so i think that $\lambda$ have to be zero right? and when $n\to\infty$ the operator $T_n$ go to null operator right? i will apreciate any hint please
AI: It is clear that $0$ is an eigenvalue because $T_n(x) = 0$ iff $x_1 = \ldots = x_{n-1} = 0$. Thus all $(0,\ldots,0,x_n)$ with $x_n \ne 0$ are eigenvectors.
No $\lambda \ne 0$ can be an eigenvalue because then $0=\lambda x_1=\lambda^2 x_2=2\lambda^3 x_3=\ldots =(n-1)!\lambda^n x_n $ implies that $x = 0$.
An alternative way to see this is decribed in Thomas Andrews' comment. We have $(T_n)^n = 0$. If $T_n$ has an eigenvalue $\lambda \ne 0$, then we can take an eigenvector $x \ne 0$ and get $(T_n)^n(x) = \lambda^n x \ne 0$.
Edited:
It is not clear what is meant by $n \to \infty$. There are many possible interpretations, here is one:
Let $\mathbb R^\infty$ denote the vector space of all real sequences $(x_n)$. Define $T_\infty : \mathbb R^\infty \to \mathbb R^\infty, T_\infty(x_1,x_2,\ldots) = (0,x_1,\frac{x_2}{2},\ldots,\frac{x_{n-1}}{n-1},\ldots)$. This map does not have eigenvalues. In fact, it is injective which means that $0$ is no eigenvalue. No $\lambda \ne 0$ can be an eigenvalue because $T_\infty(x) = \lambda x$ implies $0=(n-1)!\lambda^n x_n $ for all $n \in \mathbb N$. This means $x = 0$. |
H: If $M$ is compact subset of a metric space $X$, then it is bounded.
I have proved this theorem using open cover definition. I also read that this result can be proved using sequence definition. That is, if a set $M$ is compact then every sequence in $M$ has a convergent subsequence.
Using this definition, book showed that $M$ is bounded.
$Proof:$
Let $M$ be unbounded set , then there is a sequence $(y_n)$ which is unbounded, $ie.$ $d(y_n,b)>n$ where $b$ is fixed element. Now this sequence cannot have a convergent subsequence since convergent sequence will be bounded.
I have a doubt in the last statement since unbounded sequence can have a convergent subsequence and bounded subsequence. Please tell me what this last line actually mean.
AI: Yes, the proof is not perfect, since an unbounded sequence can indeed have a convergent subsequence. However, it is correct that such a sequence as the one mentioned in that proof cannot have a convergent subsequence. That's because not only that sequence is unbounded as every subsequence of that sequence is unbounded too. In fact, if $(y_{n_k})_{k\in\Bbb N}$ is a subsequence of $(y_n)_{n\in\Bbb N}$, then, for each $k\in\Bbb N$, $d(y_{n_k},b)>n_k$. So, since $\lim_{k\to\infty}n_k=\infty$, the sequence $(y_{n_k})_{k\in\Bbb N}$ is indeed unbounded. |
H: let $|S| = 3$. Prove $(P(S),\Delta)$ is not isomorphic to $(\mathbb{Z_8},+)$.
Hi everyone I hope somone can help
I have some set $S$ with $3$ elements $P(S)$ is the power set.I need to prove that $(P(S),\Delta)$ is not isomorphic to $(\mathbb{Z_8},+)$. I tried building a function to show that associativity doesn't work with no success .
any idea ?
Thank you very much
AI: $(\mathbb Z_8,+)=C_8$ is cyclic. If it would be isomorphic to $(\mathcal P(S), \Delta)$, $(\mathcal P(S), \Delta)$ would be generated by a unique element. That can’t be as $A \Delta A= \emptyset$ for all $A \in \mathcal P(S)$. |
H: Characterization of noetherian modules via short exact sequences (understanding a step in the proof)
I am currently dealing with this result from William Stein's Algebraic Number Theory notes, and I also underlined the part of which I am not sure yet:
I know that $g(M_0)$ is trivial since we have a short exact sequence and $M_0$ lies in the image of $f$. So it is $g(M'/M_0) \simeq g(M')$. The latter is a submodule of $N$. But how can I make the transition to $M'/M_0$ (I only showed that its image is isomorphic to some submodule of $N$)?
I think it is probably just some small detail but I am currently at loss.
AI: Restrict $g$ to a map $g':M'\to N$. The kernel of $g$ is $M_0$
so by the First Isomorphism Theorem $g'$ sets up an isomorphism $M'/M_0\to g'(M')=g(M')$, a submodule of $N$. |
H: How do you show that $\mathbb{Z}^2$ is a closed set in $\mathbb{R}^2$?
In $(\mathbb{R}, \tau_{st})$, we can write $\mathbb{R} \setminus \mathbb{Z} = \bigcup_{n \in \mathbb{Z}} (n,n+1)$, and hence $\mathbb{R} \setminus \mathbb{Z}$ is an open set. Thus the complement, $\mathbb{Z}$, is closed.
In $(\mathbb{R}^2, \tau_{st})$ I am not sure how to write an expression for $\mathbb{R}^2 \setminus \mathbb{Z}^2$ in terms of open balls. So the first part of my question is whether we can write this in a 'neat' expression like above?
I can try to write $\mathbb{R}^2 \setminus \mathbb{Z}^2 = \mathbb{R}^2 \setminus \bigcup_{n,m \in \mathbb{Z}} \big\{(n,m)\big\}$. So knowing that $\big\{(n,n)\big\}$ is a closed set, we have the complement of the union of closed sets. But this is an infinite union, so I can't actually conclude that $\mathbb{Z}^2 = \bigcup_{n,m \in \mathbb{Z}} \big\{(n,m)\big\}$ is actually closed in this manner.
I know that we can say $\mathbb{Z}^2$ is closed by arguing its set of limit points is empty, but is there any other way I can show this? In particular, is there a way to show (analogous to the $\mathbb{R}^1$ case above) that $\mathbb{R}^2 \setminus \mathbb{Z}^2$ is closed, without relying on an argument by limit points? I.e. Can I show the set as an explicit construction of union of open balls?
AI: $\mathbb{R}^2 \setminus \mathbb{Z}^2$ is the union of the open balls centered on points both having irrational coordinates and radius small enough to not intersect $\mathbb{Z}^2$.
That makes a lot of balls... but it works! |
H: Suppose $A$, $B$, and $C$ are sets. Prove that $C\subseteq A\Delta B$ iff $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$.
Not a duplicate of
Suppose $A$, $B$, and $C$ are sets. Prove that $C ⊆ A △ B$ iff $C ⊆ A ∪ B$ and $A ∩ B ∩ C = ∅$.
Suppose $A, B$, and C are sets. Prove that $C\subset A\Delta B \Leftrightarrow C \subset A \cup B$ and $A \cap B \cap C = \emptyset $
Set theory: Prove that $C \subseteq A \Delta B \iff C \subseteq A \cup B \wedge A \cap B \cap C = \emptyset$
This is exercise $3.5.21$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $A$, $B$, and $C$ are sets. Prove that $C\subseteq A\Delta B$ iff $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$.
Here is my proof:
$(\rightarrow)$ Suppose $C\subseteq A\Delta B$.
$(1)$ Let $x$ be an arbitrary element of $C$. From $C\subseteq A\Delta B$ and $x\in C$, $x\in A\Delta B$. Now we consider two cases.
Case $1.$ Suppose $x\in A\setminus B$. Ergo $x\in A\cup B$.
Case $2.$ Suppose $x\in B\setminus A$. Ergo $x\in A\cup B$.
Since the above cases are exhaustive, $x\in A\cup B$. Thus if $x\in C$ then $x\in A\cup B$. Since $x$ is arbitrary, $\forall x(x\in C\rightarrow x\in A\cup B)$ and so $C\subseteq A\cup B$. Therefore if $C\subseteq A\Delta B$ then $C\subseteq A\cup B$.
$(2)$ Suppose $A\cap B\cap C\neq\emptyset$. So we can choose some $x_0$ such that $x_0\in A$, $x_0\in B$, and $x_0\in C$. From $C\subseteq A\Delta B$ and $x_0\in C$, $x_0\in A\Delta B$. Now we consider two cases.
Case $1.$ Suppose $x_0\in A\setminus B$. Ergo $x_0\notin B$ which contradicts $x_0\in B$ and so it must be the case that $A\cap B\cap C=\emptyset$.
Case $2.$ Suppose $x_0\in B\setminus A$. Ergo $x_0\notin A$ which contradicts $x_0\in A$ and so it must be the case that $A\cap B\cap C=\emptyset$.
Since the above cases are exhaustive, $A\cap B\cap C=\emptyset$. Therefore if $C\subseteq A\Delta B$ then $A\cap B\cap C=\emptyset$.
From parts $(1)$ and $(2)$ we can conclude that if $C\subseteq A\Delta B$ then $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$.
$(\leftarrow)$ Suppose $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$. Let $x$ be an arbitrary element of $C$. From $C\subseteq A\cup B$ and $x\in C$, $x\in A\cup B$. Now we consider two cases.
Case $1.$ Suppose $x\in A$. Now we consider two cases.
Case $1.1.$ Suppose $x\in A\setminus B$. Ergo $x\in A\Delta B$.
Case $1.2.$ Suppose $x\notin A\setminus B$ and so $x\notin A$ or $x\in B$. Now we consider two cases.
Case $1.2.1.$ Suppose $x\notin A$ which is a contradiction.
Case $1.2.2.$ Suppose $x\in B$ which is a contradiction since $A\cap B\cap C=\emptyset$.
Since cases $1.2.1$ and $1.2.2$ lead to a contradiction then case $1.2$ leads to a contradiction. From case $1.1$ or case $1.2$ we can conclude $x\in A\Delta B$.
Case $2.$ Suppose $x\in B$ and a similar argument shows $x\in A\Delta B$.
Since case $1$ and case $2$ are exhaustive, $x\in A\Delta B$. Thus if $x\in C$ then $x\in A\Delta B$. Since $x$ is arbitrary, $\forall x(x\in C\rightarrow x\in A\Delta B)$ and so $C\subseteq A\Delta B$. Therefore if $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$ then $C\subseteq A\Delta B$.
From $(\rightarrow)$ and $(\leftarrow)$ we can conclude $C\subseteq A\Delta B$ iff $C\subseteq A\cup B$ and $A\cap B\cap C=\emptyset$. $Q.E.D.$
Is my proof valid$?$ Is my proof unnecessarily redundant or every step is needed$?$
Thanks for your attention.
AI: Your proof is correct. Here is a proof that avoids any mention of specific elements (following the theme of my answer to one of your previous questions). The key statements we use are the following:
(a) If $X$ and $Y$ are sets then $X \subseteq Y$ iff $X \setminus Y = \emptyset$.
(b) If $X$ and $Y$ are sets then $X \cup Y = \emptyset$ iff $X = \emptyset$ and $Y = \emptyset$.
(We discussed both of these before, so let's not reprove them!)
Now, in this problem we care about when $C \subseteq A \Delta B$. So, guided by property (a), we should examine $C \setminus (A\Delta B)$. Use axioms of set operations (e.g., De Morgan etc) to prove:
$$
C \setminus (A\Delta B) = \big(C \setminus (A\cup B)\big) \cup \big(A \cap B \cap C\big)\tag{1}
$$
I have hidden the proof of $(1)$ at the bottom of this answer; but try it yourself first. It's also a sensible thing to say out loud: $A \Delta B$ is the set of elements that are in either $A$ or $B$, but not both. So being in $C \setminus (A \Delta B)$ is the same as either being in $C$ and not in $A$ or $B$, or being in $C$ and in both $A$ and $B$.
Once you have $(1)$, the rest is very straightforward.
\begin{align}
C \subseteq A \Delta B &\iff C \setminus (A \Delta B) = \emptyset \tag{using (a)} \\
&\iff \big(C \setminus (A\cup B)\big) \cup \big(A \cap B \cap C\big) = \emptyset \tag{using (1)}\\
&\iff C \setminus (A \cup B) = \emptyset \text{ and } A \cap B \cap C = \emptyset \tag{using (b)}\\
&\iff C \subseteq A\cup B \text{ and } A\cap B\cap C = \emptyset \tag{using (a)}
\end{align}
Proof of $(1)$:
Recall that $$A \Delta B = (A \cup B) \setminus (A \cap B) = (A \cup B) \cap \neg(A \cap B)\tag{2}$$
So \begin{align}C \setminus (A \Delta B) &= C\cap \neg\big((A \cup B)\cap \neg (A \cap B)\big) \tag{by (2)} \\ &= C \cap \big(\neg (A \cup B) \cup (A \cap B)\big) \tag{De Morgan} \\ &= \big(C \cap \neg (A \cup B)\big) \cup \big(C \cap (A \cap B)\big) \tag{distributivity} \\ &= \big(C \setminus (A \cup B)\big) \cup \big(A \cap B \cap C\big)\end{align}
In the last line we used the definition of set difference on the left side, and associativity/commutativity of intersection on the right side. |
H: $g_n = m(E_n)^{\frac{-1}{q}}\chi_{E_n}$. Show $\int fg_n \rightarrow 0$
Let $f\in L^p(\mathbb{R})$. Let $E_n$ be measurable sets of finite measure, where $\lim_{n \to \infty}m(E_n)=0$. Let $p,q$ be such that $1<p,q<\infty$ and $\frac{1}{p}+\frac{1}{q} = 1$. And define $g_n = m(E_n)^{\frac{-1}{q}}\chi_{E_n}$. Show $\lim_{n\to\infty}\int f g_n = 0$.
I can't figure out how to get this one off the ground. I initially thought to use Holder's inequality and then take the limit of $||f||_p||g_n||_q$, but $||g_n||_q = 1$, so that's not getting me anywhere.
Any thoughts would be greatly appreciated.
Thanks in advance.
AI: Hint: Hölder gives you
$$\left\vert \int fg_n \right\vert
\leq \int \vert fg_n \vert
\leq \Vert f \chi_{E_n}\Vert_p \cdot \Vert g_n \Vert_q = \Vert f \chi_{E_n}\Vert_p,$$
which will allow you to conclude what you want. |
H: The multinomial coefficient and number of unique words
Let's define $C$ as the number of possible sequences of length $N$ we can form using symbols $A_i$ from an alphabet $A = \{A_1,\ldots , A_M\}.$
To my knowledge, we always have $C = M^{N+1}$.
For example, the total count of numbers containing 3 digits that we can form using decimal digits is $10^4$ = 1000, which would be the cardinality of $\{000, 001, ... 999\}$.
Or for example, the total count of binary numbers we can build with 3 bits is $2^4$ = 16, which would be the cardinality of $\{000, ..., 111\}$.
Separately, I have read that the multinomial coefficient ${n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}$ can be understood as:
The number of distinct ways to permute a multiset of n elements, and $k_i$ are the multiplicities of each of the distinct elements.
How are these problems and quantities related? Are they one and the same for some choices of $k_i$, $n$, $N$ and $M$?
Put another way, can $C$ be written as a multnomial coefficient?
AI: Suppose that you have an alphabet $A=\{a_1,\ldots,a_m\}$ of $m$ symbols. Using this alphabet we can form $m^n$ distinct sequences of length $n$ (not $m^{n+1}$). Now suppose that $$k_1+k_2+\ldots+k_m=n\;;$$ the multinomial coefficient $\binom{n}{k_1,\ldots,k_m}$ is the number of distinct sequences of length $n$ that contain exactly $k_1$ instances of $a_1$, $k_2$ instances of $a_2$, and so on. The relationship between the two is then straightforward:
$$m^n=\sum_{\substack{\langle k_1,\ldots,k_m\rangle\in\Bbb N^m\\k_1+\ldots+k_m=n}}\binom{n}{k_1,\ldots,k_m}\;:$$
the sum is taken over all possible counts of the different symbols. |
H: Why is $\sum_{i=0}^{t-1}r^i \leq r^t$?
On page 348 of this article by Siegelmann and Sontag, there's a claim that depends on the following inequality:
$$\sum_{i=0}^{t-1}r^i \leq r^t$$
(The claim appears after a line consisting only of "Therefore,". I'm using $r$ for $LW$ in the text. The same claim can also be found on page 70 of Siegelmann's book.)
I don't understand why the inequality must hold. The only constraint for $r$ implied in the text, as far as I can see, is that it must be non-negative. I know that the inequality is equivalent to
$$\frac{1-r^t}{1-r} \leq r^t$$
and I have been playing around with algebraic manipulations of this second inequality, but I have not seen how to justify it. I think there is some simple algebraic trick that I am missing.
AI: Looking into the article, the definition of both $L$ and $W$ yields $L> 2$ and $W\ge 1$. Therefore, you should add the assumption $r = LW> 2$.
This directly leads to $r-1>1$ and therefore
$$ \sum_{i=0}^{t-1}r^i = \frac{r^t-1}{r-1} < r^t-1< r^t. $$ |
H: In a Hopf algebra $H$, unit $\circ$ counit = Identity?
Let $(H,\mu,\nu,\Delta,\epsilon,S)$ be a Hopf algebra where the convolution is denoted by $*$.
Then $S=S*\nu\epsilon$ and thus $\nu\epsilon = 1_{End(H)}$. This would imply that $\epsilon$ is injective and $\nu$ is surjective. I am surprised that this mysterious $\epsilon$ is injective. Can somebody please help me to understand why that is?
For more context, check my previous post:Proving uniqueness of antipodes in Hopf algebras
AI: The counit is never injective and the unit is never surjective, unless $H$ is of dimension $1$. It's the other way around: the unit is always injective while the counit is always surjective. After all (by the axioms of bialgebra) $\epsilon\circ\nu=id_K$ where $K$ is the underlying field while $id_K(x)=x$ is the identity function.
I think that the confusion comes from the fact that "$1_{End(H)}$" in the other question is not the identity function $id_H(x)=x$. Moreover $1_{End(H)}(x)\neq x$ for at least one $x\in H$, unless $H$ is of dimension $1$. Actually the dimension (over $K$) of the image of $1_{End(H)}$ is $1$, regardless of $H$.
The $1_{End(H)}$ endomorphism is the identity but with respect to convolution, not composition. In fact $1_{End(H)}=\nu\circ\epsilon$ is the definition of $1_{End(H)}$. Which then is proven to be the neutral element of the convolution operator. |
H: Asymptotics of Recursive Bound
Suppose that we knew that for $n>N$, we have $$F(n) \le F(n-g(n)) +h(n)$$ for some well behaved functions $g,h$. (for a concrete example, let’s say $N=1,g(n)=n^\alpha, h(n)=n^\beta$ where $\alpha,\beta \in (0,1))$)
I was wondering then how we can get an asymptotic upper bound of $F$. Would this be an easy job for Riemann-Stieltjes integration? (I am not very knowledgeable about this form of integration, so if this is the case, working through my concrete example would be appreciated)
AI: Here's an approach that works well if you don't care about the leading constant. For simplicity assume $F$ is an increasing function, although one could probably do the technicalities to reduce to this case.
Suppose we have $F(n) \le F(n-n^\alpha) + n^\beta$. Starting from $F(N)$, let's iterate this inequality as many times as it takes so that the argument of $F$ in the upper bound is at most $N/2$. Each time we use the inequality, the argument is decreased by $n^\alpha$ for whatever our current value of $n$ is; since that value (by construction) is always at least $N/2$, we decrease the argument of $F$ by at least $(N/2)^\alpha$. Therefore the argument will go below $N/2$ in at most $(N/2)^{1-\alpha}$ iterations.
At each iteration, the upper bound increases by $n^\beta$ for the current value of $n$; since these values are all at most $N$, the upper bound increases by at most $N^\beta$ each time. The result of this iteration is the new recursive upper bound
$$
F(N) \le F(N/2) + (N/2)^{1-\alpha}N^\beta = F(N/2) + 2^{\alpha-1}N^{\beta+1-\alpha}.
$$
Now we iterate this inequality for $N$, $N/2$, $N/4$, and so on down to $N=1$ or wherever we want to stop. The resulting upper bound is a geometric series
$$
F(N) \le F(1) + 2^{\alpha-1} \big( N^{\beta+1-\alpha} + (N/2)^{\beta+1-\alpha} + (N/4)^{\beta+1-\alpha} + \cdots \big);
$$
notice that the exponent $\beta+1-\alpha$ is greater than $1$, so the first term is the dominant term (and we might as well bound the finite geometric series by the infinite geometric series it's the start of, whence the $\cdots$). The result is
$$
F(N) \le F(1) + 2^{\alpha-1} \frac{N^{\beta+1-\alpha}}{1-2^{\alpha-1-\beta}}.
$$
To get tighter and tighter upper bounds, one can reduce the $2$ in $N/2$ to a smaller and smaller constant; but the order of magnitude of this upper bound is probably correct. The limiting case is perhaps a differential equation? (If one assumes that $F(N)$ is bounded by a power of $N$, this could probably be made rigorous.) I'm not sure that I see a way to use Riemann–Stieltjes integration. |
H: Basic Geometry: Partitions and Intersections
Once more unto the breach, dear friends, once more!
So I'm currently working on a problem which I have somehow been able to simplify to the point where if I can simply prove that if three lines each partition a compact subset of $\mathbb{R}^2$ into two pairs of equal parts, then the portion of the compact set contained in the "triangle of intersections" between these lines must be zero.
To illustrate, consider this figure:
Where we know that
$$ \begin{align}
A & = D_1 + D_2 \\\\
B_1 + B_2 & = C_1 + C_2 \\\\
B_2 & = C_1 + D_1 \\\\
C_2 & = B_1 + D_1 \\\\
A + C_1 & = B_2 + D_2 \\\\
A + B_1 & = C_2 + D_2
\end{align} $$
If I can just prove that $D_1 = 0$, then everything would work. Alas, after hours of banging my head against the wall (and also doing some computations, mind you), I am unable to get it to work.
Thus, I appeal to you.
Looking forward to your answers.
AI: Your equations can be solved for $A,B_1, B_2$ in terms of the $C$'s and $D$'s:
$$ \eqalign{A &= D_1 + D_2\cr B_1 &= C_2 - D_1\cr
B_2 &= C_1 + D_1}$$
You'll want $C_2 \ge D_1$ so $B_1 \ge 0$, but otherwise $C_1, \ldots, D_2$ can be arbitrary nonnegative numbers. |
H: Order of statements in the Delta-Epsilon Limit definition
So the definition of Limit I see is
$$\lim_{x \to a}f(x) = L$$ means:
for all $\epsilon >0$, there exists a $\delta >0$ such that $$0<|x - a| < \delta \Rightarrow |f(x) - L| < \epsilon $$
I was wondering if modifying the definition to: Limit exists when for all $\delta >0,$ there exists a $\epsilon > 0$ such that
$0<|f(x) - L| < \epsilon $ $\Rightarrow |x - a| < \delta$
would cause any problem.
Because to me it seems like this definition should also work except it does seem a lot harder when it comes to the actual proving part.
Can this flipped version of the limit definition also work? Why did mathematicians define in this order?
AI: Huge problems.
Basic issue. We hone in on the function by honing in on the image of the function by honing in on the domain. We can't hone in on the domain by honing in on the function because the function needn't be invert-- it may fail the horizontal line test-- and we have point in the image close together that are miles apart in the domain.
Example a constant function. Or the periodic $\sin$ function where $|f(x) - L| < \epsilon$ would not mean $|x-a| < \delta$ as $x$ may be multiples of $2\pi$ distant for $a$. So $|f(x)-L| < \epsilon \not \implies |x-a| < \delta$.
... Now that counter-intuitive aspect that confuses every student (except the students who lie) is that it seems if we are honing in one the domain ($\delta$) to hone in on the image, should we start with the $\delta$??? Doesn't starting with the $\epsilon$ seem backwards.
And .... yes, it seems that way, but if you draw enough pictures and practice enough you will see that we must work backwards by establishing "acceptable error" range in the final image and finding the intial input range to produce the acceptable error. That's the only way it'll work. The wording sound backwards... but it's how it must be.
We start with the "acceptable error" range in the end result, to retrofit and finish with ant initial input range.
That's why it's $\epsilon$ first, then $\delta$ and not the other way around.
And if you are confused, rest assured every student before you either was equally confused at one time or are is lying about it. |
H: When does $[a][b]=[ab]$ hold, where [] is an equivalence class?
Let $A$ and $B$ be subsets of the integers. Define $A \boxtimes B $ = $\{ ab : a \in A,\ b \in B\}$. I want to know what properties an equivalence relation ~ must have such that $[a]\boxtimes[b] = [ab]$. I have already proved that this works for a congruence relation mod m, and trivially, for the relation =. but I am unsure where to begin pinpointing the specific properties that an equivalence relation must have such that this property holds. Proofs of such properties would also be helpful.
AI: I doubt there is any nice characterization. But here's another example: the equivalence relationed defined as $m\sim n$ if the powers of $2$ dividing $m$ and $n$ are the same.
In general, if $f$ is any monoid homomorphism from the domain of your set to another set (so that both sets have multiplications and $f(mn)=f(m)f(n)$), then the equivalence relation defined as $m\sim n$ if and only if $f(m)=f(n)$ will have this property. Both your examples have this form: $f(n)=n\mod m$ in the first instance, and $f(n)=n$ in the second instances. The example above also has this form, where the codomain is the multiplicative monoid $\{1,2,4,8,16,\dots\}$. |
H: Are all vector fields gradients of functions?
Just what the title says. I wanted to know if all vector fields are gradients of functions?
AI: If the function is one dimensional then yes. Take partial derivative of each variable to get the gradient.
If the function has higher dimensions with more than 1 variable then you obtain a matrix or tensor.
For example, take derivate of the gradient and you may obtain the Jacobian matrix which is not a vector.
Here is gradient of gradient What is the gradient of a gradient? |
H: Seeming contradiction in switching order of limits problem
I had to solve the following problem:
Find $0\neq f \in \mathcal{C}[-1,2]$ such that $\int_{-1}^2x^{2n}f(x) \,dx = 0$ for all $0\leq n \in \mathbb{Z}$.
Now, I have indeed managed to find such a function, and verified my answer with the solutions. But then I was thinking about the problem some more and thought to do the following and obtained a contradiction.
Take such an $f\in \mathcal{C}[-1,2]$. Note that $\lim\limits_{n\rightarrow \infty}\int_{-1}^2x^{2n}f(x) = 0$. By Weierstrass-Approximation Theorem, I can find a sequence of polynomials $p_k \in \mathcal{C}[-1,2], \ p_k = a_0+\cdots+a_{m_k}x^{m_k}$ such that $p_k \rightarrow f$ uniformly. Then we have,
$$
\begin{align}
\lim\limits_{n\rightarrow \infty}\int_{-1}^2x^{2n}f(x)\, dx &= \lim\limits_{n\rightarrow \infty}\int_{-1}^2x^{2n}\lim\limits_{k\rightarrow\infty}p_k(x)\, dx\\
&=\lim\limits_{n\rightarrow \infty}\lim\limits_{k\rightarrow\infty} \int_{-1}^2x^{2n}p_k(x) \, dx \quad \text{by uniform convergence}\\
&=\lim\limits_{k\rightarrow \infty}\lim\limits_{n\rightarrow\infty} \int_{-1}^2x^{2n}p_k(x) \, dx \quad \text{again by uniform convergence}\\
&= \lim\limits_{k\rightarrow \infty}\lim\limits_{n\rightarrow\infty} \Big(\frac{a_{m_k}}{2n+m_k+1} \ 2^{2n+m_k+1}+\cdots \Big)\\
&= \pm\infty
\end{align}
$$
Of course, this is a contradiction, but I can't see where I went wrong. I think it might be where I switched limits, but I am unsure.
AI: The problem comes from the fact that you cannot perform the following switch
$$ \lim_{n\to\infty}\int_{-1}^2 x^{2n}f(x)dx \not= \int_{-1}^2 \lim_{n\to\infty} x^{2n}f(x)dx$$
This is because unless $f = 0$, you will have that $x^{2n}f(x)$ will converge pointwise to a discontinuous function. Namely the function $$ g(x) = \begin{cases} f(-1) \hspace{4mm}\text{if}\hspace{4mm} x = -1\\
0 \hspace{13mm}\text{if}\hspace{4mm} x\in (-1,1)\cup\{x:f(x)=0 \}\\
f(1) \hspace{8mm}\text{if}\hspace{4mm} x = 1\\
\infty \hspace{11mm}\text{if}\hspace{4mm} x\in (1,2]\backslash\{x:f(x)=0 \}\\
\end{cases}$$
Further, you would never be able to switch the order of integration because the Riemann integral would not exist. But you could switch the order if you know that $f\geq 0$ and if you are working with the Lebesgue integral instead by using Monotone Convergence Theorem on $[1,2]$ and Dominated Convergence Theorem on $[-1,1].$ |
H: What is E[X-Y]?
I was given X~N(1,3) and Y~N(5,7) and for E[X+Y]= E[X]+E[Y], I just added the 1 and 5 and got the answer correct, but when I subtracted it for E[X-Y] I didn't get the correct answer.
I was lost thinking if E[X-Y] = E[X]-E[Y] does not exist. Can I get help on what E[X-Y] means?
AI: Expectation is a linear operator. Just rewrite your expression as $Z=X+(-1)\cdot Y$, hence the expectation of $Z, \mathbf{E}Z = \mathbf{E}X + (-1) \cdot \mathbf{E}Y = 1+(-1)\cdot 5=-4 $ |
H: Connection between words and sets of solutions to integral equations
Consider any nonnegative integral solution $x_1, x_2,\ldots, x_n$ of the equation $x_1 + x_2 + \ldots + x_n = m.$ For each $i = 1,2,\ldots,n$, let $y_i = x_1 + x_2 + \ldots + x_i$. Then $0 \le y_1 \le y_2 \le \ldots \le y_n = m$. Conversely, suppose $(y_1, y_2, \ldots, y_n)$ is any $n$-tuple of nonnegative integers such that $0 \le y_1 \le y_2 \le \ldots \le y_n = m$, and let $x_1 = y_1$, and $x_i = y_i - y_{i- 1}$ for all integers $i = 1,2,\ldots,n.$ Then $x_1 + x_2 + \ldots + x_n = y_1 + (y_2 - y_1) + (y_3 -y_2) + \ldots + (y_n - y_{n -1}) = y_n= m$, and so $x_1 + x_2 + \ldots + x_n = m.$ $\color{red}{\text{Consequently}}$, the number of nonnegative integral solutions of the equation $x_1 + x_2 + \ldots + x_n = m$ is the same as the number of $n$-tuples of nonnegative integers $(y_1, y_2, \ldots, y_n)$ such that $0 \le y_1 \le y_2 \le \ldots \le y_n = m$.
I don't see how the conclusion $\color{red}{\text{bolded in red}}$ above follows from the preceding steps. Is there some sort of unspoken/implicit bijection somewhere in the text above?
To make sure the question is not too short, I'll add the way I personally understand the connection between integral solutions to equations like the one above and words:
Questions:
$1.$ What's the number of words of length $n$ sourced out of $\{|,*\}$ where the $m$ letters are reserved for *s?
$2.$ How many integral solutions are there for $\sum_{i=1}^nx_i=m?$
Answers:
I'll do a special case where $n = 4, m = 7$ which is similar to the general case.
$1$. Consider the following example word: *|***|**|*. To count words of this kind, we simply choose seven places out of ten for stars which can be done in $\binom{10}{7} = \binom{7 + 4 - 1}{7} = \left(\!\!{4\choose 7}\!\!\right)$ ways. The last equality is by definition.
$2.$ Consider the word *|***|**|* again. The letter "|" naturally divides the word into four substrings. If we let each of these substrings stand for a variable and the number of *s -- for variable values, then this problem is exactly the same as the one above.
AI: There is an explicit bijection there: it’s between the set $S_0$ of solutions in non-negative integers to the equation $$x_1+\ldots+x_n=m\tag{1}$$ and the set $S_1$ of $n$-tuples $\langle y_1,\ldots,y_n\rangle$ of integers such that $0\le y_1\le\ldots\le y_n=m$. The map $\varphi:S_0\to S_1$ takes a solution $x_1,\ldots,x_n$ to $(1)$ to the $n$-tuple $\langle y_1,\ldots,y_n\rangle$, where $y_k=x_1+\ldots+x_k$ for $k=1,\ldots,n$.
For instance, if $n=4$ and $m=6$, the solution $2+1+0+3=6$ to $(1)$ would be taken to the $4$-tuple $\langle 2,3,3,6\rangle$
This is clearly a bijection, because it has an inverse, which is also described in the opening paragraph: given an $n$-tuple $\langle y_1,\ldots,y_n\rangle\in S_1$, we can reconstruct the solution to $(1)$ that corresponds to it, namely, $x_1=y_1$, $x_2=y_2-y_1$, and in general $x_k=y_k-y_{k-1}$ for $k=2,\ldots,n$. |
H: Hilbert space of multivariate random variable
From this lecture notes (Page 7, Example 4), the space of univariate random variables with finite variance is a Hilbert space, when the inner product is chosen as $\langle X, Y \rangle = \mathrm{Cov} (X, Y)$. I'm asking for answers or references for the following two questions:
(1) How to generalize the prescribed idea of constructing Hilbert space to multi-variate random variables?
(2) Can we define a Hilbert space of random variables, with inner product as $\langle X, Y \rangle = \mathbb{E} (X^T Y)$?
AI: For a multivariate random variable $X = (X_1,\ldots X_d)$, we normally define the expected value of $X$ to be defined to be $\mathbb{E}(X) = (\mathbb{E}(X_1),\ldots \mathbb{E}(X_d))$ and say that a multivariate random variable $X$ has finite variance if $\mathbb{E}[|X - \mathbb{E}[X]|^2]<\infty$ where $|\cdot|$ is the standard $l^2$ norm on $\mathbb{R}^d$. You can then extend your inner product for $d=1$ to be given by $\left< X,Y\right> = \mathbb{E}[(X-\mathbb{E}[X])\cdot (Y-\mathbb{E}[Y])]$, where the $\cdot$ is represents your traditional dot product. This gives the same definition for your definition in $d=1$ but is defined for arbitrary $d$. This function is in fact an inner product and inherits its properties from the expected value operator. This said, I am not sure if there are other inner products on multivariate random variables to reduce to your case if $d=1$.
That's exactly how the Hilbert space is defined normally for multivariate random variables. |
H: Question about ring isomorphism out of a quotient ring
Suppose $f:R\rightarrow S$ is a map of commutative rings with $1$. Further suppose that $I$ is an ideal of $R$ contained in the kernel of $f$. This induces a map on the quotient $f':R/I\rightarrow S$. Suppose further that $f'$ is an isomorphism.
Is it true then that $I=\ker f$? Certainly $I\subset \ker f $ by I am not sure how to see the opposite inclusion.
AI: In fact, we can say more: given a ring morphism $f : R\to S$ and an ideal $I\subseteq\ker f,$ we have $\ker f = I$ if and only if $f' : R/I\to S$ is injective.
Suppose we have $f : R\to S$ and an ideal $I\subseteq\ker f$ such that the induced map $f' : R/I\to S$ is injective. Then if $x\in\ker f,$ it follows that $f'(x + I) = 0.$ Injectivity of $f'$ implies that $x+I = I,$ so that $x\in I.$ Thus, $\ker f = I.$
Conversely, if $I = \ker f,$ then the first isomorphism theorem tells us that $f' : R/\ker f\to S$ is an isomorphism onto the image of $S,$ and is in particular injective. |
H: How to calculate series sum $\sum_{n=1}^{\infty}\frac{\sin(kn)}{n}=\frac{\pi-k}{2}$
According to Wikipedia: $$\sum_{n=1}^{\infty}\frac{\sin(kn)}{n}=\frac{\pi-k}{2}, 0<k<2\pi$$
How do I prove this? What if $k \ge 2\pi$?
AI: Take the function $\;f(x):=\cfrac{\pi-x}2\;$ on $\;[0,\pi]\;$ and extend it to an odd periodic function on $\;(-\pi,\pi]\;$ , so that in fact the extension is
$$F(x)=\begin{cases}f(x)=\cfrac{\pi-x}2,&x\in[0,\pi)\\{}\\
-f(-x)=-\cfrac{\pi+x}2,&x\in(-\pi,0)\end{cases}\;\;\;\;,\;\;\;F(x+2\pi)=F(x)\;$$
We can thus write the Fourier series of $\;F\;$ which will only contain sines as $\;F\;$ is an odd function, so $\;a_n=0\;$ for all $\;n=0,1,2,...\;$ , and:
$$b_n=\frac1\pi\int_{-\pi}^\pi F(x)\sin nx\,dx=\frac2\pi\int_0^\pi \frac{\pi-x}2\sin nx\,dx=\int_0^\pi\sin nx\,dx-\frac1\pi\int_0^\pi x\sin nx\,dx=$$
$$=\left.-\frac1n\cos nx\right|_0^\pi+\left.\frac{x\cos nx}\pi\right|_0^\pi+\frac\pi x\int_0^\pi\cos nx\,dx=-\frac{(-1)^n}n+\frac1n+\frac{(-1)^n}n=\frac1n\implies$$
$$\frac{F(x^+)+F(x^-)}2=\sum_{n=1}^\infty\frac{\sin nx}n$$
By Dirichlet's Theorem since $\;F\;$ is piecewise smooth (=continuously differentiable)
Thus,at any continuity point in $\;[-\pi,\pi)\;$, namely: at any $\;x\;$ in this interval different from zero, we have
$$\frac{\pi-x}2=\sum_{n=1}^\infty\frac{\sin nx}n\;,\;\;\text{with periodicity}\;\;2\pi$$
For example, evaluate the above for $\;x=\cfrac\pi2\;$ and get a rather important and very nice infinite sum. |
H: Showing that the field of fractions of $\mathbb{Z}[\sqrt{d}]$ is $\mathbb{Q}[\sqrt{d}].$
Let $d\in\mathbb{Z}$ be an integer that it not a square. Let $\sqrt{d}\in \mathbb{C}$ be a square root of $d$. Let $\mathbb{Z}[\sqrt{d}]:=\left\{a+b\sqrt{d}:a,b\in\mathbb{Z}\right\}$. Let $F$ be its field of fractions. Show that $F$ can be identified with $\mathbb{Q}[\sqrt{d}]:=\left\{a+b\sqrt{d}:a,b\in\mathbb{Q}\right\}$.
My solution. With $f:F\to \mathbb{Q}[\sqrt{d}]$ by $f((a+b\sqrt{d},a'+b'\sqrt{d}))=\frac{a+b\sqrt{d}}{a'+b'\sqrt{d}}=\frac{aa'-bb'd}{a'^2-b'^2d}+\frac{a'b-ab'}{a'^2-b'^2d}\sqrt{d}$. It must be shown that $f$ is a ring isomorphism and bijective, but it is too tedious.
Is there a simpler way to prove that $F$ and $\mathbb{Q}[\sqrt{d}]$ are isomorphic?
AI: Nice question. Your approach is fine, and you're right that it's tedious. Here's another one. Let $R$ be an integral domain and $F$ a field together with a map $R \to F$. We can show that $F$ is the field of fractions of $R$ by showing that it is the smallest field containing $R$.
Now it is obvious that $\mathbb{Q}[\sqrt{d}]$ contains $\mathbb{Z}[\sqrt{d}]$, and it is obvious that any field containing $\mathbb{Z}[\sqrt{d}]$ has to contain $\mathbb{Q}[\sqrt{d}]$, since elements of the form $m$ and $n \sqrt{d}$ have to be invertible. So the only thing to check is that $\mathbb{Q}[\sqrt{d}]$ is a field, i.e. you have to prove that non-zero elements have inverses. |
H: When Should I Use Symbols in a Proof?
The question I'm about to ask might sound weird, I hope i can deliver the idea.
I have noticed that in some mathematics books (especially English ones) the proofs are written in words and symbols are used only when necessary, example: for all, implies, there exists, if and only if... etc, but in other books symbols are used more, example: $\implies$, $\forall$, $\exists$, $\iff$... etc
** So is there a rule that should be followed here? Or is it a choice that the writer makes? Will it be familiar if the proofs are mostly written with symbols?
I am asking this question because, right now, I am translating a book to English, and the writer of the original book barely used words in proofs, he only used connectors like: Therefore, Hence, Thus... etc. I do not want to translate over 300 pages for nothing.
** So should I keep the original proofs? Or should I reformulate them?
Sorry for the long question, but I am really confused here, I would be grateful if you help me.
Thank you all in advance.
AI: Keep the original style. You will respect the author's intent, and, more importantly, you will avoid introducing errors due to possible misunderstandings or clumsy rewrites. As a bonus, you don't have to translate the symbols, they are international. |
H: Difference between "measure" and "metric"
I was reading about mathematical structure and came across the distinction of metric and measure as follows:
A measure: intervals along the real line have a specific length, which can be extended to the Lebesgue measure on many of its subsets.
A metric: there is a notion of distance between points.
Question: Isn't metric a super category of measure? So a measure is a form of a metric?
AI: No, a measure is not a kind of metric.
In the contexts you quote, a measure tells you the size of a set. A metric tells you the distance between two points. |
H: Expected value of dice game
You throw a die. Each time you get a 4,5, or 6, you get the value on the face of the die. When you get a 1, 2, or 3, you quit the game (but keep any winnings). What's the expected value of this game?
Here's how I'm approaching this. I tried to first write an equation for the EV, as so:
$EV = (1/2)\cdot (0) + (1/2) \cdot (5+EV)$ - however, this assumes that you don't get to keep your winnings; how do I account for the winnings?
Next, I tried approaching it like this, where I sum up all of the possibility; you can get a $\{1,2,3\}$ on your first roll, second roll, third role, etc. where the EV of a non-$\{1,2,3\}$ roll is $5$.
$(1/2)(0) + (1/2)(1/2)(5) + (1/2)(1/2)(1/2)(10) + (1/2)(1/2)(1/2)(1/2)\cdot 15 ....$ how do I find a closed for expression for this series (it looks both like an arithmetic and geometric series)
AI: Both approaches look correct! The $5+{}$in the first equation does account for the winnings you get to keep.
The series you found in the second case is
$$
\frac52 \sum_{k=1}^\infty \bigg( \frac12 \bigg)^k k.
$$
This is indeed a notch more complicated than a simple geometric series, but we can still handle it! Starting from
$$
\sum_{k=0}^\infty x^k = \frac1{1-x},
$$
valid for $|x|<1$, we can take the derivative of both sides to get
$$
\sum_{k=0}^\infty kx^{k-1} = \frac1{(1-x)^2},
$$
which implies
$$
c\sum_{k=1}^\infty kx^k = \frac{cx}{(1-x)^2}.
$$
This is exactly the series you want, with $c=\frac52$ and $x=\frac12$, and therefore your series equals
$$
\frac{(5/2)(1/2)}{(1-1/2)^2} = 5,
$$
which is the correct expected value. |
H: Finite index subgroups of $SL(2,\mathbb Z)$
A complete classification of genus $0$ congruence subgroups of $SL(2,\mathbb Z)$ has been carried out by A. Sebbar [1]. They fall into 33 conjugacy classes with index divisible by $6$. I was wondering if dropping the requirement of a congruence subgroup has also been studied. To be concrete, is there a classification of ALL index 6 subgroups of $SL(2,\mathbb Z)$? I would be very grateful for recommendations on the literature. Thanks!
[1] A. Sebbar, Classification of Torsion-Free Genus Zero Congruence Groups, http://www.jstor.org/stable/2668773
AI: Did you try GAP? It should give you all index 6 subgroups. The group has presentation $\langle a,b \mid a^6=b^4=1, a^3=b^2\rangle.$ In general classification of all finite index subgroups of that group is a hopeless task. There are too many of them. For example every finite group generated by an element of order two and an element of order 3 is a quotient of that group. The kernel is then of finite index. And there are lots of such finite groups.
But one can try to apply something similar to Stallings core graphs that work for the free groups. |
H: How may I teach myself pure mathematics "from scratch"?
I'm in my 19's and I keen on becoming a mathematician yet currently I can't fund going to university. I am to work as a customer service representative in order to linger. I harness mornings to study maths.
How may I teach myself pure mathematics "from scratch"?
I reckon I don't have the skills to start this major either: Syllabus (best books in your minds, courses,etc).
AI: You could look for videos on Coursera and on YouTube. MIT OpenCourseWare is pretty good and free on YouTube. They will advice you the books. |
H: How do I know if I should divide or subtract cases to remove over counting in combinatorics?
Sometimes I do questions where I divide my answer by a factor to remove excess counting and sometimes do subtraction. Now, I wonder, would both methods be equivalent? Because I definitely think we can just pluck out unfavourable cases from the total cases by subtraction instead of dividing. However is the converse true could we plug out cases that we subtract by division?
And finally as a bonus, are there other methods of case removal?
Examples: 1. 30 people in room hand shake each other, find the number of handshakes done
2. ways to p ick three items from a set of 'n' and put into three slots ( you divide by 3! on this one)
AI: It depends on whether your count is too high because of a known quantity or a known factor. Some examples might clarify.
Ten people like strawberry ice cream, twenty like chocolate, and five like both. How many people are there in total? An estimate would be be $10 + 20$ (strawberry plus chocolate), but this overcounts by a known quantity (the five who like both). You would correct this overcount with subtraction: $10 + 20 - 5$.
There are ten people in a room and each one shakes hands with each other person. How many handshakes take place? An estimate would be $10 \cdot 9$ (each of the ten people needs to shake hands with nine other people), but this overcounts by a known factor (every handshake is getting counted twice). You would correct this with division: $\frac{1}{2} \cdot 10 \cdot 9$. |
H: Cauchy problem and global solution
Let $y(x)$ the unique solution of the Cauchy problem
\begin{cases}
y'(x)=\sqrt{\ln(1+y(x)^2)}
\\
y(0)=y_0
\end{cases}
i) Show that, for $y_0>0$, $y(x)$ is strictly increasing and convex.
ii) Show that $y(x)$ is globally defined.
For i), I note that $$\ln(1+y(x)^2) = 0 \text{ iff } y(x)=0$$
So, if $y(0)=0$, then $y(x)=0$ is the unique solution.
If $y_0>0$, then in a right neighbourhood of $x=0$ I have that $y'(x)>0$. Also, by uniqueness, I can't cross the stationary solution $y(x)=0$, and therefore $$y'(x)=\ln(\sqrt{1+y^2})>0$$ for every $x>0$. Since this is the rhs of the equation, this proves that the function is strictly increasing.
To show convexity, I just computed $$y''(x)= \frac{y y'(x)}{1+y(x)^2} >0 $$ since $y'(x)>0$ and $y(x)>0$ by uniqueness.
ii) Here I don't know how to argue formally. It is clear from a picture that $\lim_{x \rightarrow \infty} y(x)=+\infty$, but I don't know how to show that $y(x)$ is globally defined.
AI: Let $f(x,y) := \sqrt{\log(1 + y^2)}$ be the RH side of your ODE. Since $\log$ is concave, you have $\log(1+t) \leq t$ for every $t >-1$ therefore:
$\log(1 + y^2) \leq y^2\quad \Rightarrow\quad f(x,y) \leq \sqrt{y^2} = |y|$
and $f(x,y)$ is sublinear in $y$ (uniformly w.r.t. $x$).
Now classical global existence theorems apply and ii follows. |
H: $\rho(f,g)=\int_E \min(1,|f-g|)dm$. Prove that $f_n$ converges to $f$ in measure if and only if $\rho(f_n,f)\rightarrow 0$ as $n\rightarrow\infty$
Question: Suppose $m$ is a finitemeasure on a measurable space $E$. Define $\rho(f,g)=\int_E \min(1,|f-g|)dm$. Prove that $f_n$ converges to $f$ in measure if and only if $\rho(f_n,f)\rightarrow 0$ as $n\rightarrow\infty$.
My Thoughts: For the backwards direction, if $\rho(f_n,f)\rightarrow 0$ as $n\rightarrow 0$, then we have that $\lim_{n\rightarrow\infty}\int_E\min(1,|f_n-f|)dm=\lim_{n\rightarrow\infty}\int_E|f_n-f|dm=0$, and so we have uniform convergence, which implies in measure convergence. For the forward direction, I am not really sure if I should consider cases, that is, when $1$ is the minimum and then when $f_n-f$ is the minimum and show that if $1$ is the minimum then the statement can't be true, and so $f_n-f$ must be the minimum and then show that the only way we get in measure convergence is if that integral equals $0$... but I am not quite sure how to do that. Any thoughts, suggestions, etc. are greatly appreciated! Thank you.
AI: In the backwards direction you don't have uniform convergence, only that $\int_E \min(1,|f_n-f|)\,dm \to 0$. Also, the conclusion $$\int_E \min(1,|f_n-f|)\,dm \to 0 \implies \int_E |f_n-f|\,dm \to 0$$ is not entirely spelled out in your question.
Anyway, for $0 < \varepsilon < 1$ you can use Markov's inequality to obtain
$$m(|f_n-f| > \varepsilon) =m\Big(\min(1,|f_n-f|) > \varepsilon\Big)\le \frac1\varepsilon \int_E\min(1,|f_n-f|)\,dm \xrightarrow{n\to\infty} 0.$$
Conversely, assume that $f_n \to f$ in measure $m$ and let $0 < \varepsilon <1$. Pick $n_0 \in \Bbb{N}$ such that for $n \ge n_0$ we have $$m\left(|f_n-f| >\frac\varepsilon{2m(E)}\right) < \frac\varepsilon2.$$
Then for all $n \ge n_0$ we have
\begin{align}
\int_E \min(1,|f_n-f|)\,dm &\le \int_{|f_n-f| \le \frac\varepsilon{2m(E)}} |f_n-f|\,dm + \int_{|f_n-f| > \frac\varepsilon{2m(E)}} 1\,dm\\
&\le \frac\varepsilon{2m(E)}m\left(|f_n-f| \le\frac\varepsilon{2m(E)}\right) + m\left(|f_n-f| >\frac\varepsilon{2m(E)}\right)\\
&\le \frac\varepsilon2 + \frac\varepsilon2\\
&=\varepsilon
\end{align}
so we conclude $\int_E \min(1,|f_n-f|)\,dm \to 0$. |
H: Trace Inequality for difference of positive definite matrices
Prove that for positive definite matrices $A$ and $B$ where $A - B$ is also positive definite, show
$$2Tr((A-B)^{1/2}) + Tr(A^{-1/2}B) \leq 2Tr(A^{1/2})$$
My attempt so far:
We know that $A - B$ positive definite $\implies Tr(A - B) \geq 0 \implies Tr((A-B)^{1/2}(A-B)^{1/2}) \geq 0$
I'm not sure whether I'm on the right track, or where to go from here. Any direction or solution would be appreciated.
AI: Let $Y=B^{1/2},\,Z=(A-B)^{1/2}$ and $X=A^{1/2}=(Y^2+Z^2)^{1/2}$. Then $X,Y,Z\succ0$ and
\begin{aligned}
&\operatorname{tr}\left(2A^{1/2}-2(A-B)^{1/2}-A^{-1/2}B\right)\\
&=\operatorname{tr}\left(2X-2Z-X^{-1}Y^2\right)\\
&=\operatorname{tr}\left(2X-2Z-X^{-1}(X^2-Z^2)\right)\\
&=\operatorname{tr}\left(X-2Z+X^{-1}Z^2\right)\\
&=\left\|X^{1/2}-X^{-1/2}Z\right\|_F^2\ge0.
\end{aligned} |
H: Is a set of full measure also dense?
So I found this post about subsets of measure $1$ in $[0,1]$ being necessarily dense. I was wondering if this holds true for more general measure spaces, i.e. are sets of full measure necessarily dense?
I asked a friend this question and we think we have a counterexample, although we don't know if this counts as a properly defined measure space.
Just take $[0,1]$ again but with every singleton defined to be an open set. Then the irrational points have measure 1, but any rational point is open and doesn't intersect the irrationals, so the subset of irrationals in the interval is not dense. Is this allowed?
AI: Yes, your counter-example is allowed. The topology you are describing is the Discrete Topology on $[0,1]$. In this topology, any subset is both open and closed, so in particular the closure of any subset is itself, hence the only dense subset is the entire space. As you note, there are, however, proper subsets of full measure.
For the type of conclusion you want, the measure on your space should have something to do with the topology on your space. To this end, let $X$ be any topological space. The Borel algebra $\mathcal{B}(X)$ is the $\sigma$-algebra on $X$ generated by all the open sets. A measure $\mu$ on $\mathcal{B}(X)$ is called a Borel measure.
Claim: For a topological space $X$ and a Borel measure $\mu$ on $X$, the properties "any Borel set of full measure is dense" and "any non-empty open set has positive measure" are equivalent.
Proof:"$\Rightarrow:$" Assume $U\neq\emptyset$ is an open subset of $X$ and $\mu(U)=0$. Then $\mu(X\setminus(X\setminus U))=\mu(U)=0$, so $X\setminus U$ has full measure. Since $U$ is open, $X\setminus U$ is closed, hence $\overline{X\setminus U}=X\setminus U$, but $X\setminus U\neq X$, since $U\neq\emptyset$, so $X\setminus U$ is not dense in $X$. The claim follows by contrapositive.
"$\Leftarrow:$" Assume $A\subseteq X$ is Borel and not dense in $X$, i.e. $\overline{A}\neq X$. Then $X\setminus\overline{A}\neq\emptyset$. Since $\overline{A}$ is closed, $X\setminus\overline{A}$ is open, so, by hypothesis, $\mu(X\setminus\overline{A})>0$. By definition, $\overline{A}$ does not have full measure. The claim follows by contrapositive.
Note that the second property is also known as strict positivity. In particular, this property holds for every Haar measure, such as the Lebesgue measure on any $\mathbb{R}^n$. |
H: How to compute the solutions of $d\alpha = \omega$ for a given exact form $\omega$?
Let $\omega \in \Omega^k(\mathbb R^n)$ be an exact form. How can I compute all the forms $\alpha \in \Omega^{k-1}(\mathbb R^n)$ such that $\omega = d\alpha$? I am mostly interested in the case $k = n$, but an answer for general $k$ would be greatly appreciated.
(Assume all forms to be smooth. I'm not interested in any merely $C^r$ nonsense.)
AI: Let $k \geq 2$, $\omega \in \Omega^k(\Bbb{R}^n)$ be an exact form, and denote $\text{prim}(\omega) := \{\beta \in \Omega^{k-1}(\Bbb{R}^n)| \, \, d\beta = \omega\}$; this is the set of primitives of the form $\omega$. As long as you can find one form $\alpha$ such that $d\alpha = \omega$, then we can show that
\begin{align}
\text{prim}(\omega) = \{\alpha + d\phi| \, \, \phi\in \Omega^{k-2}(\Bbb{R}^n)\} \tag{$*$}
\end{align}
In other words the set of primitives is precisely all the forms which differ from $\alpha$ by an exact form.
To prove this, note that in $(*)$, the inclusion $\supseteq$ is clear because $d^2 = 0$. For the inclusion $\subseteq$, suppose $\beta \in \text{prim}(\omega)$. Then, $d\beta = \omega = d \alpha$. Hence, $d(\beta-\alpha) = 0$. By Poincare's lemma (since $\Bbb{R}^n$ is star-shaped with respect to the origin) $\beta-\alpha$ is exact. Hence, $\beta = \alpha + d\phi$ for some $\phi\in \Omega^{k-2}(\Bbb{R}^n)$.
Now, of course, if $k =1$, we have to interpret this slightly differently; in this case, $d(\beta- \alpha) = 0$ (where $\beta,\alpha$ are $0$-forms, i.e real-valued functions) implies that $\beta - \alpha = \text{constant function}$
As for finding a particular primitive $\alpha$, there is an integral expression for it. This is the analogue of finding primitives in single variable calculus: given a form $f\,dx \in \Omega^1(\Bbb{R})$, we define the function $F(x) := \int_0^x f(u)\, du$, then $dF = f\, dx$ (by the fundamental theorem of calculus), though there may not be a simpler way to express the integral in terms of nice elementary functions. Every other primitive differs from $F$ by a constant.
Anyway, here's an explicit expression (taken from Spivak's Calculus on Manifolds, from the Proof of Poincare's Lemma, with modified notation), we first express the form $\omega$ as
\begin{align}
\omega &= \sum_{I}\omega_{i_1\dots i_k} \, dx^{i_1} \wedge \dots \wedge dx^{i_k},
\end{align}
where the sum on $I = (i_1, \dots, i_k)$ is being taken over all injective subsets of $\{1, \dots, n\}$, of length $k$ (for example, take them to be all increasing), and then we define the form $\alpha$ pointwise as
\begin{align}
\alpha(x) := \sum_{I}\sum_{\mu=1}^k (-1)^{\mu-1} \left(\int_0^1 t^{k-1}\omega_{i_1\dots i_k}(tx)\, dt\right)\cdot x^{i_{\mu}} \, (dx^{i_1} \wedge \dots \widehat{dx^{i_{\mu}}}\wedge \dots \wedge dx^{i_k})(x)
\end{align}
(be careful to distinguish the point $x\in \Bbb{R}^n$ from the form $dx^i$). A pretty tedious calculation, which I don't feel like doing will show that $d\alpha = \omega$ (because $d\omega = 0$). |
H: Fractional part of a real number: questions
I was reading this question Evaluate the following integral $ \int_1^{\infty} \frac{\lbrace x\rbrace-\frac{1}2}{x} dx$ and I have seen that the user have used the mantissa of a real number or the fractional part.
I know that the mantissa of $x\in\Bbb R$ is defined by $\text{mant}(x)=x-\lfloor x\rfloor$ and $0\leq \text{mant}(x)<1$. I add the different symbologies.
$$\text{mant}(x)=\begin{cases}
\{x\}& \\
\text{frac}(x) &\\
M(x)
\end{cases}
$$
the mantissa of a real number is connected to the logarithms and in what way?
If I have $x=-0.43$, $\text{mant}(x)= -0.43-\lfloor (-0.43)\rfloor=\color{red}{ -0.43-(-1)\cdot \lfloor(+0.43)\rfloor}$? i.e. can I bring out the minus sign from if $x\in\Bbb R$, $\lfloor (-x)\rfloor \iff -\lfloor x\rfloor$? Is there a specific property?
What is the relevance of this mantissa function?
AI: A short answer, there is much more to say about this topic.
$2$: You can bring a minus outside of a floor, but when you do this, the floor becomes a ceiling. In other words,
$$
\lfloor -0.43\rfloor = -1
$$
since the floor function means to round down. On the other hand,
$$
\lfloor -0.43\rfloor = -\lceil 0.43\rceil=-1.
$$
Note that the fractional part of a negative number is sometimes defined in other ways (such as rounding towards zero).
$1$ and $3$: One big application of the mantissa is in how numbers are stored on a computer. There are various ways to do this (and tricks which I am omitting). But, a number is stored as a mantissa $m$ and an exponent $e$. From this pair, the corresponding number is $m2^e$ (on a real computer the representation is slightly different, but I'm just trying to go for the idea). This also shows where logarithms come into play, the exponent $e$ is a (rounded) logarithm of the desired number.
Example (using base $10$, instead of base $2$, which a computer would use): The number $1,578$ could be written as
$$
0.1578\cdot 10^4.
$$
The $0.1578$ would be the mantissa and the $4$ is the exponent. Note that $4\leq\log_{10}1578<5$, a connection to logarithms.
Aside: On a real computer, the representation could be closer to
$$
1.578\cdot 10^3,
$$
where only the mantissa $0.578$ would be stored (on a computer, there is only one leading digit since binary has only numbers, $0$ and $1$. |
H: Show that $M:= \{x\in \mathbb{R}^n~|~ \Vert x-a\Vert \leq \delta\}$ is closed
Let's consider the following problem:
We want to show that $M:= \{x\in \mathbb{R}^n~|~ \Vert x-a\Vert \leq \delta\}\neq\emptyset$ is closed. According to our definition:
A set $M$ is closed iff $M$ contains all its limit points.
Let be $b$ a limit point of $M$. Then we find a point $x\in M$ with $x \neq b$ in every neigbourhood. So for every $\epsilon>0$ the open ball $U_{\epsilon}(b):=\{x\in M~|~ \Vert x-b\Vert < \epsilon\}$ is non empty and contains a point $x\neq b$. Then we let $\epsilon \to 0$ and simply conclude: $\Vert b-a\Vert = \Vert b-x+x-a\Vert \leq \Vert b-x \Vert+\Vert x-a\Vert\leq 0 + \delta$ which shows that $b\in M$. Hence, $M$ is closed as $b$ was arbitrarily chosen.
Actually this seems to me like a typical analysis technique but I am not sure if I have a wrong intuition in my mind or maybe the reasoning above is wrong?
In my mind $\epsilon$ will never reach $0$ it will only be very very close to it so that $\Vert b-x \Vert$ also will never become $0$. So following this logic the inequality above must be wrong?! May be someone can help me with this confusion.
AI: I don't agree that we can say “we let $\varepsilon\to0$”. I would do that proof as follows: take $\varepsilon>0$. There is some $x\in M$ such that $\|b-x_\varepsilon\|<\varepsilon$. So,$$\|b-a\|\leqslant\|b-x_\varepsilon\|+\|x_\varepsilon-a\|<\delta+\varepsilon.$$Since this occurs for each $\varepsilon>0$, then$$\|b-a\|\leqslant\lim_{\varepsilon\to0}\delta+\varepsilon=\delta.$$Note that at no point I took $\varepsilon=0$. |
H: What is the pdf for a jointly uniform distribution inside a triangle?
I have a triangle bounded by $0 \leq x, y \leq 1$ and $x + y \geq 1.5$. I'm told that points are uniformly distributed within this triangle.
I am wondering how I can find the pdf? Is it simply solving for the following?
$$
\int_{0.5}^1 \int_{0.5}^1 f(x,y) dx dy = 1
$$
$f(x,y) = 4$ satisfies this equation, but is it unique?
AI: By definition uniform density in a region has a constant value equal to the reciprocal Of the area of the region (and $0$ outside the region) . In this case the region is the triangle formed by the points $(1,1), (1,0.5)$ and $(0.5,1)$. So the density is $\frac 1 A$ for all points in this triangle where $A$ is the area. I will let you calculate the area.
Note that the limits for the integral are not the ones you have written. You should get $\int_{0.5}^{1}\int_{1.5-y}^{1} f(x,y) dxdy$. |
H: Existence of two sequences
Does there exists two nonnegative sequences
$\{a_{n, m} \} $ and $\{r_m\} $
such that
(1). $a_{n, m}\in (\frac{1}{N_o},1)$ for some integer $N_o$ for all $n, m$
(2). $\sum_{n=1}^{\infty}a^2_{n, m}=r_m$
and
$\sum_{m=1}^{\infty}r_{ m} $ converges.
If yes, can we find a family of such sequences?
AI: No. If $\sum_{n=1}^{\infty}a^2_{n, m}$ converges, $a_{n,m}\to 0$ as $n\to \infty$ and thus it cannot be bounded from below. |
H: Can basis vectors contain more components than the dimension of the spanned vector space?
For example, can I say $\left(1,0,1\right)$ and $\left(0,1,1\right)$ form the basis for a plane in $\mathbb{R}^{3}$?
AI: The answer to your question is yes. The dimension of the subspace (the plane) is $2$, but since it is a subspace of $\mathbb{R}^{3}$, one must have $3$ components that describe the basis vectors which span it, or any other subspace of $\mathbb{R}^{3}$.
I hope this helps. |
H: Asking for a hint: Convex sets and inner product spaces
I'm trying to solve the following question:
Suppose $V$ is an inner product space and $B$ is the open unit ball in $V$ (thus
$B = \{ f \in V : \Vert f \Vert < 1 \}$ ). Prove that if $U$ is a subset of $V$ such that
$B \subset U \subset \bar{B}$, then $U$ is convex. Here $\bar{B}$ is the closure of $B$.
Can I get a hint? I have tried to use the fact that it's an inner product space and played around with Cauchy Schwartz inequality, but to no avail. I'm completely stuck.
AI: Let $x,y \in U$ and $0<t<1$. Then $\|tx+(1-t)y\|\leq t\|x\|+(1-t)\|y\| \leq t+(1-t)=1$. If we have strict inequality then $tx+(1-t)y \in B \subset U$. If equality holds then $tx$ and $(1-t)y$ are nonnegative multitples of each other: $tx =s(1-t)y$ for some $s \geq 0$. So $x =ry$ for some $r \geq 0$. The case where $\|x\|<1$ or $\|y\|<1$ is trivial. If you assume that $\|x\|=\|y\|=1$ the we get $r=1$ and $x=y$. This completes the proof.
I have used the fact that $\|x+y\|=\|x\|+\|y\|$ implies $x =ay$ for some $a \geq 0$ (or $y=ax$ for some $a \geq 0$). This is true in any inner product space. |
H: Show that $\int_0^1\frac{1}{|f(x)-x_0|}dx$ is unbounded
Let $f$ be a Lebesgue measurable functinon on such that $f:[0,1]\rightarrow [0,1]$. Prove that for any $M$ there exists $x_0\in[0,1]$ such that
$$
\int_0^1\frac{1}{|f(x)-x_0|}dx\geq M.
$$
My attempt:
There exists a sequence of step functions $f_n$ such that $f_n$ increasing to $f$ pointwise and
$$
\int_0^1f_n(x)dx=\int_0^1f(x)dx.
$$
I am not sure whether we can prove that
$$
\lim\int_0^1\frac{1}{|f_n(x)-x_0|}dx=\int_0^1\frac{1}{|f(x)-x_0|}dx
$$
since the denominator is not bounded. If above is true, then we can get the desired result.
AI: Prove by contradiction. If the integral is bounded by some $M$ for all $x_0$ then $\int_0^{1}\int_0^{1} \frac 1 {|f(x)-y|} dxdy \leq M <\infty$. By Fubini/Tonelli Theorem we can write this as $\int_0^{1}\int_0^{1} \frac 1 {|f(x)-y|} dydx \leq M$ This is a contradiction because the inside integral is $\infty$ for every $x$. |
H: Why is $\det \Phi =\det \Psi =1 $.
I found this in Partial Differential Equations, Evans, page 627.
I would to like understand why Evans claims that $\det \Phi =\det \Psi=1$. Here, $y_i=x_i:= \Phi^{i}$, $i=1,\ldots,n-1$ and $y_n=x_n -\gamma(x_1, \ldots,x_{n-1})$ ($y=\Phi(x)$).
The inverse transform is similarly defined: $x_i=y_i$ when $i=1,\ldots,n-1$, $x_n=y_n + \gamma(x_1, \ldots,x_{n-1}) $.($x=\Psi(y)$). The function $\gamma :\mathbb R^{n-1}\to \mathbb R$ is of class $C^2$. The book says $\det \Phi =\det \Psi=1$.
I tried with $n=2$ as follows:
The transformation $T$ defined on $\mathbb R^2$ by $y_1=x_1$ and $y_2=x_2 -\gamma(x_1)$. I chose the basis $e_1,e_2$ of $R^2$ so $[T(e_1)]_{e_1,e_2}=(1,-\gamma(1))$ and $[T(e_2)]_{e_1,e_2}=(0,1-\gamma(0))$, so when I do the determinant of the matrix that represents $T$ I obtain $1-\gamma(0)$, which is different from $1$.
Please, can somebody help me or give me some hint? Thank you. Here is a poor quality screenshot (sorry):
https://i.stack.imgur.com/tzRqf.jpg
AI: Its not clear what the determinant of $\Phi(x)\in\mathbb R^n$ should be. Instead it should be that the determinant of $\nabla\Phi(x) \in \mathbb R^{n\times n}$ is $1$. This has been corrected in the Second Edition:
The reason the determinant of $\nabla\Phi$ is identically one is because the matrix is lower triangular with all diagonal entries equal to 1. I write it out below in the special case of dimension 2:
$$ \nabla\Phi(x) = \begin{pmatrix}\partial \Phi_1/\partial x_1 & \partial \Phi_1/\partial x_2 \\ \partial \Phi_2/\partial x_1 & \partial \Phi_2/\partial x_2 \end{pmatrix}(x) = \begin{pmatrix} 1 & 0 \\ -\gamma'(x_1) & 1\end{pmatrix}.$$
The determinant of the gradient of the inverse map follows. |
H: Generated $\sigma$-algebras identity (unions)
I'm trying to prove
$$\sigma (\mathcal{C}_1 \cup \mathcal{C}_2) = \sigma (\sigma( \mathcal{C}_1 ) \cup \sigma (\mathcal{C}_2))$$
for non-empty $\mathcal{C}_1,\mathcal{C}_2 \in \mathscr{P}(\Omega)$.
So I know that $\mathcal{C}_1 \cup \mathcal{C}_2 \subseteq \sigma( \mathcal{C}_1 ) \cup \sigma (\mathcal{C}_2)$ and therefore
$$\sigma (\mathcal{C}_1 \cup \mathcal{C}_2) \subseteq \sigma (\sigma( \mathcal{C}_1 ) \cup \sigma (\mathcal{C}_2)),$$
But I still don't know why the other contention is true.
As always, thanks in advance.
AI: You'll need the fact that, if $\mathcal A$ is a $\sigma$-algebra, then the $\sigma$-algebra it generates is itself: $\sigma(\mathcal A)=\mathcal A$. We'll also need monotonicity: If $\mathcal C\subseteq\mathcal D$ then $\sigma(\mathcal C)\subseteq\sigma(\mathcal D)$. Armed with these observations, we can proceed as follows.
Since $\mathcal C_1\subseteq\mathcal C_1\cup\mathcal C_2$, we have $\sigma(\mathcal C_1)\subseteq\sigma(\mathcal C_1\cup\mathcal C_2)$, and similarly $\sigma(\mathcal C_2)\subseteq\sigma(\mathcal C_1\cup\mathcal C_2)$. So $\sigma(\mathcal C_1)\cup\sigma(\mathcal C_2)\subseteq\sigma(\mathcal C_1\cup\mathcal C_2)$. By monotonicity, we get $\sigma(\sigma(\mathcal C_1)\cup\sigma(\mathcal C_2))\subseteq\sigma(\sigma(\mathcal C_1\cup\mathcal C_2))$. Finally, our initial observation simplifies the right side of this to just $\sigma(\mathcal C_1\cup\mathcal C_2)$, as required. |
H: Prove that $ba^k + ab^k \leq a^{k+1} + b^{k + 1}$ for $a, b \geq 0$, $k \in \mathbb N$
I'm in the middle of a proof by induction that for all $n \in \mathbb N$, $\left(\frac{a + b}{2}\right)^n \leq \frac{a^n + b^n}{2}$ for all nonnegative real numbers $a, b$. I've reached a point in my proof where I want to show that $ba^k + ab^k \leq a^{k + 1} + b^{k + 1}$, and I'm stuck here.
I attempted to prove this by showing a more general statement that $c + d \leq \frac{a}{b}c + \frac{b}{a}d$ for all $a, b > 0$ and $c, d \geq 0$, but I quickly realized that I can't prove it if it isn't true (whoops). That was my only idea; now I'm stuck. Well, not really -- my only other idea was to prove this statement itself by induction, but I'm convinced there's another way to show it without having a nested proof by induction. What must I consider to discover this route?
Thank you very much in advance.
AI: Hint: For $a,b\geq 0,$ and $k\in{\mathbb N},$ $$(a-b)(a^k-b^k)\geq 0.$$ |
H: Finding Work done by the Force Field $\vec{F}$
I need to find the work required to move a certain block from point A to point B.
$\vec{F}(x,y)=2y^{3/2}\ \textbf{i}+3x\sqrt{y} \ \textbf{j}$, where point A is $(1,1)$ and point B is $(2,9)$. I do know how to solve this problem, but I thought of another way to solve it.
The first method I used to solve this problem:
$P=2y^{3/2}$ and $Q=3x\sqrt{y}$
Because the force field $\vec{F}$ has to be conservative, $\frac{\partial P}{\partial x}=2y^{3/2}$ and $\frac{\partial Q}{\partial y}=3x\sqrt{y}$.
I integrated $\frac{\partial f}{\partial x}$ and got $f(x,y)=2xy^{3/2}+h(y)$. I differentiated both sides with respect to $y$ and got $\frac{\partial f}{\partial y}=3y\sqrt{x}+\frac{\partial h(y)}{\partial y}$.
I get that $\frac{\partial h(y)}{\partial y}=0$, which means that $h(y)=c$. If we assume that $c=0$, $f(x,y)=2xy^{3/2}$.
Using the FTC of Line Integrals, I get that $$\int\limits_C \nabla f\cdot d\vec{r}=f\left(\vec{r}\left(b\right)\right)-f\left(\vec{r}\left(a\right)\right)$$. And found work as $$W=\int\limits_C \nabla f\cdot d\vec{r}=f\left(2,9\right)-f\left(1,1\right) = 2\cdot2\cdot9^{3/2}-2\cdot1\cdot1^{3/2} = 106$$
The second method that I thought of was using parametric equations. Finding a parametric equation that passes through point A and B, through time $t = 0$ to $t = 1$, I get $x=1+t$ and $y=8t+1$. I use a different way to evaluate the FTC Line Integral: $$\int\limits_C \vec{F}\cdot d\vec{r} = \int_0^1 \begin{bmatrix}
2y^{3/2} \\
3x\sqrt{y} \\
\end{bmatrix} \cdot \frac{d}{dt}\left(\begin{bmatrix}
1+t \\
1+8t \\
\end{bmatrix}\right) dt = \int_0^1 \begin{bmatrix}
2y^{3/2} \\
3x\sqrt{y} \\
\end{bmatrix} \cdot \begin{bmatrix}
1 \\
8 \\
\end{bmatrix} dt = \int_0^1 2y^{3/2}+24x\sqrt{y} \ dt$$.
I substitute $x=1+t$ and $y=8t+1$ into the integral and evaluate the following integral. $$\int_0^1 {2\left(8t+1\right)^{3/2}+24\cdot\left(t+1\right)\cdot\sqrt{8t+1}} \ dt = 106$$, thus giving the same answer.
I want to know why the ways of solving the problem results in the same solution, but the first way of solving the problem requires me to assume that $c=0$? Am I missing something that allows me to safely assume that the constant of integration is 0?
AI: In fact it doesn’t matter what constant of integration $c$ you have chosen, if you realize in your last step of method 1, the constant cancel out of each other when you subtract the potential.
Physically you may just treat the constant as some sort of reference value, and it is immaterial of which reference value you have chosen. |
H: Let $D$ be a dense subset of a banach space $X$. Show that any $x$ can be written as a sum of elements of $D$ with a certain conditon.
Let $D$ be an everywhere dense subset of a Banach space $B$ with norm $\|\cdot\|$. Show that any $x\in B$ can be written as the sum of series $$x=\sum_{k=1}^\infty x_k\,,$$ where $x_k\in D$ and $\|x_k\|\le 3\cdot 2^{-k}\|x\|$ for every $k\ge 1$.
I am not really sure how to do this problem. It feels like i have to define $x_n$ inductively but I could not figure out the specifics. Any hints or solutions appreciated. Perhaps the trick $\sum y_n-y_{n-1}=y_n
$ could come in handy, but again I could not figure it out as I did not know how to force the bound onto $y_n-y_{n-1}$
AI: Hint: you want $x_1$ to be close to $x$, which can be done. And then you want $x_2$ close to $x - x_1$...
EDIT:
Actually you can get it with $\|x_k\| \le 2^{1-k} \|x\|$.
We may assume $x \ne 0$, otherwise take all $x_k = 0$.
Lemma: Given $x \ne 0$ and $\epsilon > 0$, there is $y \in D$ with $\|y\| \le \|x\|$ and $0 < \|y - x\| < \epsilon$.
Proof: take $y$ in the intersection of $D$ with the open ball of radius $\epsilon/2$ about $r x$ where $r = 1 - \epsilon/(2 \|x\|)$.
I will inductively choose $x_k$ such that
$$0 < \left \|x - \sum_{j=1}^{k} x_j \right\| \le 2^{-k} \|x\|\ \text{and}\ \|x_k\| \le 2^{1-k} \|x\| $$
This can be done using the above lemma, rewriting the first
inequality above as $$0 < \left\|x_k - \left(x - \sum_{j=1}^{k-1} x_j\right)\right\| \le 2^{-k} \|x\| $$
and noticing that $$2^{1-k} \|x| \ge \|x - \sum_{j=1}^{k-1} x_j\|$$ |
H: propositions of Interior and closure
Let $(X,{\tau})$ a topological space, i have to show:
$A{\bigcup}B={X} \Rightarrow \text{cl } A{\bigcup}\text{int } B={X}$
$A{\bigcap}B=\varnothing \Rightarrow \text{cl }A{\bigcap}\text{int }B={\varnothing}$
Well, my proof of 2) is this:
$x\in (\text{cl }A{\bigcap}\text{int }B) {\Rightarrow}$ $x\in \text{cl }A$ and $x\in \text{int }B$
${\hspace{3.3cm}}$ $\Rightarrow$ $(\exists U\in \mathcal{N}(x),U\subset B)$ and $(\forall V\in \mathcal{N}(x), V{\bigcap}A \neq \varnothing)$
Here $\mathcal{N}(x)$ denote neighborhood system of $x$, so, $V{\bigcap}A {\subset} A$ and by hypotesis $A{\bigcap}B={\varnothing}$, i can conclude:
$(V{\bigcap}A){\bigcap} B\subset A{\bigcap}B \subset \varnothing$
Then $x\in\varnothing$. Conversely, it's always true.
For the 1) I have this part of the proof:
$x\in (\text{cl }A{\bigcup}\text{int }B) {\Rightarrow}$ $x\in \text{cl }A$ or $x\in \text{int }B$
${\hspace{3.3cm}}$ $\Rightarrow$ $(\exists U\in \mathcal{N}(x),U\subset B)$ or $(\forall V\in \mathcal{N}(x), V{\bigcap}A \neq \varnothing)$
(And this analysis is similar to the first proof) but I have trouble by the other hand. Can you give some advice or hint to do this? Do you think I'm right on my proofs? Thank you very much.
AI: $\newcommand{\cl}{\operatorname{cl}}$ $\newcommand{\int}{\operatorname{int}}$Your proof of (2) starts off fine, but it goes a little astray at the end. (It would also be clearer if you used more words and fewer symbols.)
Suppose that $x\cl A\cap\int B$. Then $x\in\cl A$, and $x\in\int B$. Since $x\in\int B$, there is a $U\in\mathcal{N}(x)$ such that $U\subseteq B$, where $\mathcal{N}(x)$ is the nbhd system at $x$, and since $x\in\cl A$, $V\cap A\ne\varnothing$ for each $V\in\mathcal{N}(x)$. In particular, then, $U\cap A\ne\varnothing$, so $$\varnothing\ne U\cap A\subseteq B\cap A=\varnothing\;,$$ which is absurd. It follows that there is no such $x$ and hence that $\cl A\cap\int B=\varnothing$.
For (1) it’s probably easiest to prove the contrapositive: show that if $\cl A\cup\int B\ne X$, then $A\cup B\ne X$. Suppose that $x\in X\setminus(\cl A\cup\int B)$. Then $x\in X\setminus\cl A$, so there is a $U\in\mathcal{N}(x)$ such that $U\cap A=\varnothing$. And $x\in X\setminus\int B$, so $x\notin\int B$, and therefore $U\nsubseteq B$. At this point the proof is almost finished; can you complete it from here? |
H: 2010 USAMO #5:Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for integers $m$ and $n$, then $m - n$ is divisible by $p$.
Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let $S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q(q+1)(q+2)}
$
Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for coprime integers $m$ and $n$, then $m - n$ is divisible by $p$.
My Progress till now: $$2S_q = 2\sum_{x=1}^{\frac{q+1}{3}} \frac{1}{(3x-1)(3x)(3x+1)} = \sum_{x=1}^{\frac{p-1}{2}} \left[\frac{1}{3x(3x-1)}-\frac{1}{3x(3x+1)}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}} \left[ \frac{1}{3x-1} - \frac{2}{3x} +\frac{1}{3x+1}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}}\left[ \frac{1}{3x-1} + \frac{1}{3x} +\frac{1}{3x+1}\right] - \sum_{x=1}^{\frac{p-1}{2}} \frac{1}{x} $$
With the help of @user10354138 , I have got $\frac{1}{p} - 2S_q = \frac{1}{p} + \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}$
But then I am stuck.
Please give me some hints rather than a solution.
Thanks in advance.
PS: I didn't post it in AOPS, because there we don't get any guidance.
AI: (Original) Hint: You are almost there with the simplification. Note that you are summing over $\frac1n$ from $n=2$ to $\frac{3p-1}2$ in the first. So
$$
2S_q+1=\sum_{n=(p+1)/2}^{(3p-1)/2}\frac1n
$$
If you tweak the RHS slightly, you would be summing over $\frac1n$ as $n$ runs through representative of each of the nonzero residue classes mod $p$. So ...
Addendum (2020-07-29): As discussed in the comments,
\begin{align*}
\frac1p-2S_q-1&=-\left(\sum_{n=(p+1)/2}^{p-1}\frac1n+\sum_{n=p+1}^{p+(p-1)/2}\frac1n\right)\\
&=-\sum_{i=1}^{(p-1)/2}\left(\frac1{p-i}+\frac1{p+i}\right)
\end{align*}
and now
$$
\frac1{p-i}+\frac1{p+i}=\frac{p}{(p-i)(p+i)}
$$
so the numerators are divisible by $p$ and the denominators are not. So putting everything over a common denominator, we see
$$
\frac{m-n}{n}=-\sum_{i=1}^{(p-1)/2}\frac{p}{(p-i)(p+i)}=\frac{p\times \text{some integer}}{\text{some integer not divisible by }p}.
$$
That is, every representation of $\frac{m-n}{n}$ must have more factors of $p$ in the numerator than in the denominator, hence $m-n$ is divisible by $p$. |
H: $K$-topology of the real line and quotient topology.
Let $K$ be the set $\{1,1/2,1/3,\cdots,1/n,\cdots\}$, which is the set of reciprocal of all positive integers. The $K$-topology on $\mathbb R$ is defined as generated by the usual open intervals $(a,b)$ and also $(a,b)-K$. For the text below allow me to write $\mathbb R_K$ for the $K$-topology.
The question is: Let $Y$ be the quotient space obtained from $\mathbb R_K$ by collapsing the set $K$ to a point; let $p:\mathbb R_K\to Y$ be the quotient map. Show that $Y$ satisfies the $T_1$ axiom, but is not Hausdorff.
I let $p(1/n)=b$ for all $n\in \mathbb Z_+$, it also map other real numbers into itself. From here I'm understanding that $Y=(\mathbb R\setminus K)\cup\{b\}$, where $b$ can be an element of $K$, or can not.
I can show that $Y$ satisfies $T_1$-axiom. Another problem want me to show that $Y$ is not a Hausdorff space, so I need to find two different point $x_1,x_2$ in $Y$ so that every pair of their respective neighborhood $U_1,U_2$ has nontrivial intersection, right?
My work is as follow: If $b\neq 0$, then we can consider the two points $b,0$ in $Y$. If $U$ is a neighborhood of $0$, then there is $r>0$ such that $(-r,r)\subset U$. If $V$ is a neighborhood of $b$, this means $p^{-1}(V)$ is open in $\mathbb R_K$. But this means $p^{-1}(V)$ contains $K$, so it contains every open set $(1/n-r_n,1/n+r_n)$ for some appropriate $r_n$. Maps this back to $Y$ we see $V$ contains these neighborhoods $(1/n-r_n,1/n+r_n)\setminus\{1/n\}$. Therefore, for $n$ large enough, $1/n$ would smaller than $r$, this makes $U$ and $V$ intersects, then we are done.
However, if $b=0$ then I've no idea how to choose the other point. I think I've messed up some concept about the quotient space.
From: Munkres General Topology, Chapter 22 Question 6(a).
AI: Formally the quotient space $Y$ is actually the set
$$\big\{\{x\}:x\in\Bbb R\setminus K\big\}\cup\{K\}\;;$$
its points are the sets $\{x\}$ for $x\in\Bbb R\setminus K$ and the set $K$. However, in this case there is no real harm in thinking of the quotient as $X=(\Bbb R\setminus K)\cup\{b\}$, where each $x\in\Bbb R\setminus K$ corresponds to the point $\{x\}$ in the quotient, and the point $b$ of $X$ corresponds to the point $K$ of the quotient; you just have to make sure that you get the topology right, which you did when you described a nbhd of $b$. And just as the point $K$ of the quotient is different from each of the points $\{x\}$ for $x\in\Bbb R\setminus K$ — in particular, $K\ne\{0\}$ — so your point $b$ is different from each of the other points of $X$, and in particular, it is not $0$: $b\ne 0$. And as you correctly showed, $0$ and $b$ do not have disjoint open nbhds in $X$, so $X$ is not Hausdorff, and neither is $Y$. |
H: Let $A$ and $B$ be $n\times n$ matrices. Let $\operatorname{rank}(A)=s$ and $\operatorname{rank}(B)=t$. Then rank of $A+B\ge\cdots$
$\newcommand{\rank}{\operatorname{rank}}$Suppose $A$ and $B$ are $n\times n$ matrices. Let $\rank(A)=s$ and $\rank(B)=t$. Then rank of $A+B$ is at least ..............
My attempt
$\rank(A)=s$ $\implies$ $A$ has $s$ linearly independent $n\times 1$ column vectors. Let it be $\{x_1,x_2,\ldots,x_s\}$
$\rank(B)=t$ $\implies$ $B$ has $t$ linearly independent $n\times 1$ column vectors. Let it be $\{y_1,y_2,\ldots,y_t\}$
How do I complete the question?
AI: There is no general low bound. If $B=-A$, then $A+B=0$ with rank $0$ independently of the rank of $A$. For different $s,t$ you can probably expect $rank(A+B)\ge |s-t|$. To prove (or disprove) it, you can assume that, say, $A$ is in the RREF. |
H: Arnold on Homogeneous Linear Equations
In this section, Arnold formulates the equation that we know as the homogeneous linear equation. My question is why the term $\partial a/\partial X$ vanishes when linearisation is done?
Completing the steps for linearisation around the point $(p, 0)$, we get
$$
\begin{align}
\frac{\partial Y}{\partial X}\Big|_{(p,0)} &= a(X,Y)\Big|_{(p,0)} \\
&= a(p, 0) + \frac{\partial a}{\partial X}\Big|_{(p,0)}(X-p) + \frac{\partial a}{\partial Y}\Big|_{(p,0)}(Y) \\
&= \frac{\partial a}{\partial X}\Big|_{(p,0)}(X-p) + f(X)\Big|_{(p,0)}Y
\end{align}
$$
because $a(p, 0) = 0$ and re-writing $\frac{\partial a}{\partial Y} = f(X)$
Now, why is $\frac{\partial a}{\partial X} = 0$? How does $a(X+T, Y) = a(X, T)$ translate to this fact?
Using the limit definition of differentiation, we have
$$
\lim _{\Delta X \to 0} \frac{a(X+\Delta X, 0) - a(X, 0)}{\Delta X}
$$
but we can't just say that $a(X+\Delta X) = 0$ right? (because it only happens at periods of T)
AI: I realised $a(X + \Delta X) = 0$ and $a(X, 0) = 0$ simply because of the fact that the $a(X, 0) = 0$ is given. The fact $a(X+T, Y) = a(X, Y)$ should have no utility here. |
H: Isomorphism in UFD
Is this statement true:
Let $R$ be a UFD and $p,q\in R$ are irreducible which are not associated. Prove that for all $n,m\in\mathbb N$:
$$R/(p^mq^n)\cong R/(p^m)\times R/(q^n).$$
I don't think that the chinese remainder theorem works here, however, it is easy to construct a map $f$ from $R$ to $R/(p^m)\times R/(q^n)$ with $\ker f=(p^mq^n)$. The only problem is to prove the map is onto.
I'll appreciate any help.
AI: @JCAA probably gave the best answer to this question in comments. However if you want further elaboration read on:
Let $R=\mathbb{R}[x,y]$. Then $R$ is a UFD, and $x,y$ are irreducible, and not associates. If some element of $w\in R$ mapped to $(0,1)$ in $R/(x)\times R/(y)$ then $$w=ax=1+by$$ for some $a,b\in R$. Then we would have $ax-by=1$, which is impossible as $(x),(y)$ are not coprime.
Edit:
The above just proves that your map is not an isomorphism. To show that the rings are not isomorphic, note that $R/(x)\times R/(y)$ contains the idempotent $(1,0)$, which is non-trivial (not $(0,0)$ or $(1,1)$). To see that $R/(xy)$ has no such idempotent, note that elements of $R/(xy)$ may be uniquely written in the form $\alpha=c+xp(x)+yq(y)$, where $p,q$ are polynomials over $\mathbb{R}$ and $c\in \mathbb{R}$. Let $$\alpha^2=c'+xp'(x)+yq'(y),$$ for a constant $c'$ and polynomials $p',q'$.
Then if $\alpha^2=\alpha$ we have $\deg(q')=\deg(q)$, and $\deg(p')=\deg(p)$. Thus $q=p=0$ and $\alpha=0$ or $1$. |
H: Finding the positive integers that can be written in the form $x^2+xy+5y^2$
I'm working on a homework problem that asks which positive integers can be written in the form $x^2+xy+5y^2$. An example was given on how to find all positive integers that can be written as the difference of two squares (i.e. $x^2-y^2$), but it was a proof by cases that doesn't seem to carry over to more generalized versions of the form $ax^2+bx+cy^2$.
Is there any method to approaching these kinds of problems?
AI: See this text. It describes precisely which numbers are represented by a binary quadratic form (Proposition 4.1). In your case the discriminant is $-19$, so a number $n$ is represented by your form iff $-19$ is a square modulo $4n$. You can conclude by using the Gauss reciprocity law. |
H: Probability that a King of hearts will be chosen, given there is a king
I am currently stuck on this question and how it would be worked out.
I currently have the probaility that at least one king will be 0.341
And that the probability of drawing a King of hearts is 51 c 1 / 52 c 5 [editted! I typed it wrong 51 c 4 / 52 c 5]
But the following question is stumping me a bit,
A hand of five cards is dealt from a normal pack of 52 cards. Find the probability that the hand will contain the king of hearts, given that there is at least one king.
Thanks,
AI: HINT: There are $\binom{48}5$ hands with no king, so there are $\binom{52}5-\binom{48}5$ hands with at least one king. There are $\binom{51}4$ hands that contain the king of hearts. Can you put those pieces together to get the desired probability? |
H: The solutions of a linear differential equation!
I can't see how if a differential equation is linear then if it attains a complex solution, then the complex conjugate of the solution is a solution also?
Thanks is advance.
Modification: the linear differential equation has to be real too (from the comments).
AI: As others have said, this is only true if your coefficients are real, e.g. $y’ = iy$ won’t have this property. So suppose your equation is of the form
$$\sum_{k = 0}^n A_k(x) y^{(k)} = 0 ,$$
and $y$ is a solution. First, I claim that $(\overline{y}) ‘ = \overline{(y’)}$. This should be straightforward to check by breaking $y$ into real and imaginary parts. By induction, this means $(\overline{y})^{(k)} = \overline{(y^{(k)})}$. Thus
\begin{align*}
\sum_{k = 0}^n A_k(x) (\overline{y})^{(k)} & = \sum_{k = 0}^n A_k(x) \overline{ (y^{(k)}) } \\
& = \sum_{k = 0}^n \overline{A_k(x)} \overline{ (y^{(k)}) } \\
& = \sum_{k = 0}^n \overline{ A_k(x) (y^{(k)}) } \\
& = \overline{ \sum_{k = 0}^n A_k(x) (y^{(k)}) } \\
& = \overline{0} \\
& = 0.
\end{align*} |
H: Book recommendations for Euclidean/Non-Euclidean Geometry
Request for Book Recommendations:
Background introduction
Disclaimers: If the tone below is a little arrogant I apologize beforehand, but I'm being very specific here because I want to make
sure that I will be learning and accessing the right material, and
also that which is more tailored to my interests
So this is the first time I'm using the forum here to ask a question, although I've visited this site a couple of times already. But I do notice similar sounding questions have been asked several times, so I want to be highly specific so as to separate this question from other similar questions, so that it does not get tagged as a duplicate or lacks specificity to the point it is vague. What I want is a personalized recommendation, (not a generalized recommendation), tailored to my level of knowledge and interest.
First off, I am an incoming freshman going to college, and am majoring in physics, however I am interested in eventually double majoring in mathematics as well, and one area of mathematics I would really like to choose as an area of study is geometry.
Specifically, I'm searching for a recommendation in Euclidean geometry/Non-Euclidean Geometry, whether it is a book, a pdf, or a website tutorial. I do not want an book with an axiomatic treatment style for right now. It would be highly helpful if the book were more problem oriented, teaching specific techniques in solving geometry problems That being said, the book should not just be a collection of problems without any exposition or explanations. Hopefully the book contains a complete treatise that connects concepts and techniques, while also managing workable problems
Preferred style and difficulty of the material
For example, some geometric techniques I want to learn about include homotheties, spiral similarity, inversion, projective transformations(I know this lies out the scope somewhat), complete quadrilaterals.
There's much more, but in general I want a somewhat comprehensive, encyclopedic text on the different theorems and techniques(without being pedantic and overtly wordy, and having enough exercises)
For reference two books whose chapters catalog and encompass what I am looking for, but are inaccessible for my current level are Evan Chen's "Euclidean Geometry in Mathematical Olympiads",
and Dan Pedoe's "A comprehensive course in geometry". In EGMO, the problems are clearly meant for those who go to IMO and math olympiads, which is not what I'm looking for, while in Dan Pedoe's book, the exercises are scant and few.
A note about the difficulty of the books: The geometry taught in high school is boring, SAT style-dry and does not vary in both concepts and problem type. Obviously I am not asking for a book introducing me to the extreme basics in Euclidean geometry.
That being said, I'm still a novice in comparison to many of the seasoned Geometers out there(or those who have studied geometry in depth)
I have read AOPS Introduction to geometry, which I think is closer to the level I want for getting to solve a variety of geometry problems while also using different techniques, the book's difficulty level is actually quite easy, but then again I don't want an International Mathematical Olympiad difficulty style type of book, but not a book where everything is spoon-fed to you either.
Please do not recommend the following:
BOOKS NOT TO RECOMMEND
I recognize some of the texts included below are good texts in themselves, but for my present purpose, I do not want to get mired in the actual theory behind what a geometry/geometries is, as well as the axiomatic foundations of geometry. I want to learn math, not what the math behind really means at this point.
"Elementary Geometry from an Advanced Standpoint" Edwin Moise
"Advanced Euclidean Geometry" Robert Hartshorne
Euclid's elements(Seriously? Why would anyone recommend this book to a beginner? Just as no-one would recommend Newton's principia to someone who is first learning physics either)
EGMO Evan Chen (See above)
"Geometry-A comprehensive Course" Dan Pedoe
"An Introduction to Geometry" Richard Ruscyzyk (Already read the book, good for beginners but only focuses on some of the more basic techniques)
"Geometry revisited" Coxeter (Too few problems, there is little to be gained from reading it imo)
"Geometry Unbound" Kiran S. Kedlaya
"PROBLEMS IN PLANE AND SOLID GEOMETRY" Viktor Prasolov (This is almost all problems and no exposition at all)
Also-no Olympiad style books, unless the difficulty is manageable.
Further questions
One final question to those who have studied the aforementioned and more in geometry: I do want to eventually learn projective geometry and differential geometry, are there any specific course recommendations, or areas I have to learn first?(With regard to geometry, topology, and abstract algebra, since I have some knowledge of analysis) Thanks-
AI: It looks like you've looked around for books and found a lot that are not to your taste, or are at the wrong level of difficulty.
Have you browsed through archive.org? If you type the search term "Geometry" into the Internet Archive search box you get hundreds, maybe thousands of books on geometry. Browse away until you find the right level and mix of exercises.
Geometry as a subject was a much bigger part of the math curriculum during the 19th century than it was in the 20th century. Archive.org has a wealth of 19th century texts, and I've found that many of them are full of exercises. So I'd recommend that you look through some of those. The great thing about math is that it doesn't become obsolete, and everything you're likely to learn about Euclidean geometry at this stage was well known in the 19th century.
You can of course narrow down your search terms to focus on euclidean, solid, or projective geometry. |
H: If $(x^2 - x - 1)^n = a_{2n}x^{2n} + a_{2n - 1}x^{2n - 1} + \cdots + a_2x^2 + a_1x + a_0$, then find the value of $a_0 + a_2 + a_4 + \cdots + a_{2n}$
So here is the Problem :-
If $(x^2 - x - 1)^n = a_{2n}x^{2n} + a_{2n - 1}x^{2n - 1} + \cdots + a_2x^2 + a_1x + a_0$, then find the value of $a_0 + a_2 + a_4 + \cdots + a_{2n}$ .
I don't know how to factorise $(x^2 - x - 1)^n$ and I was only able to solve this by substituting $n = 1$, where that would give $$(x² - x - 1) = a_2x^2 + a_1x + a_0.$$
This immediately gives the answer as $0$ . But I want a solution for $n$ itself, but not substituting $n$ as any number. Can anyone give me an idea of it ?
AI: Substituting $x=1$ into the equality we have
$(-1)^n=a_0+a_1+a_2+a_3+\ldots+a_{2n-1}+a_{2n}$,
Substituting $x=-1$ into the equality we have
$1=a_0-a_1+a_2-a_3+\ldots-a_{2n-1}+a_{2n}$,
Adding these two, we get
$2(a_0+a_2+\ldots+a_{2n-2}+a_{2n})=(-1)^n+1$
Therefore,
$a_0+a_2+\ldots+a_{2n-2}+a_{2n}=\frac{(-1)^n+1}{2} =\begin{cases}
1, & \text{if $n$ is even} \\
0, & \text{if $n$ is odd}
\end{cases} $ |
H: What is the value of $\int_{0}^{2\pi}f(e^{it}) \cos t \, dt$ if $f$ is analytic?
If $f(z)$ is an analytic function, then find the value of the integration
$$\int_{0}^{2\pi}f(e^{it}) \cos t \, dt.$$
My Work:
Taking $e^{it} = z$ the integrand becomes of the form
$$i\frac{f(z)\operatorname{Re(z)}}{z}$$
on the simple closed contour $|z|=1$. But how can I proceed from here? Please help.
AI: Hint: If you write $\cos t$ as $\frac {e^{it}+e^{-it}} 2$ you will see that the integral is $\frac 1 {2i}\int_{\gamma} \frac {f(z){(1+z^{2})}} {z^{2}}$. You can evaluate this using Cauchy's Integral Formula or the Residue Theorem. |
H: Continuity of a function $f: X\to \mathbb Z$
If you have a set $X$ which is equipped with a topology: what is required for the function $f: X\to \mathbb Z$ to be continuous?
The fact that $X$ has a topology should certainly be helpful. But $\mathbb Z$ is a discrete space, so I am not totally sure how to establish continuity.
I need to establish that every open set $A\subset \mathbb Z$ has an open pre-image $f^{-1}(A)$ in $X$. But what are open sets in $\mathbb Z$, and how could they be linked to the topology in $X$? Does the topology on $X$ imply, via $f$, a topology on $f(X)\subset\mathbb Z$, and hence (somehow) on all of $\mathbb Z$?
AI: Every subset of $\Bbb Z$ is open, so the preimage of every subset of $\Bbb Z$ must be open in $X$ in order for $f$ to be continuous. In particular, let $U_n=f^{-1}[\{n\}]$ for each $n\in\Bbb Z$: in order for $f$ to be continuous, it is necessary and sufficient that all of the sets $U_n$ be open. Of course, since each $U_n=X\setminus\bigcup_{k\in\Bbb Z\setminus\{n\}}U_k$, each $U_n$ is also closed in $X$. In other words, $X$ must have a partition into countably many clopen (i.e., both closed and open) subsets, and $f$ must send each of these clopen sets to an integer.
A non-trivial example is given by letting $X=\Bbb Z\times\Bbb R$ with the product topology and letting $f$ be the projection of $X$ to the first factor:
$$f:X\to\Bbb Z:\langle n,x\rangle\mapsto n\;.$$
Then $f$ is continuous, because each of the sets $\{n\}\times\Bbb R$ is clopen in $X$. |
H: When is a closed ball inside another closed ball?
I was solving a question and I came upon a statement which I can't really prove. I know that this is indeed true when our metric space is $\mathbb{R}^2$ with the euclidean metric.
Let $(X,d),$ be a metric space. Take $\overline{B}_r(x)$ be the closed ball around a point $x\in X$ of radius $r.$ Take another point $x'\in \overline{B}_r(x).$ Suppose $\textrm{dist}(x',\partial \overline{B}_r(x))=k,$ where $\textrm{dist}(x,A)=\inf\limits_{y\in A} d(x,y), A\subseteq X.$ Take $r'\leq k,$ and consider the closed ball $\overline{B}_{r'}(x').$ Then we have $\overline{B}_{r'}(x')\subseteq \overline{B}_r(x).$
I am wondering if there exists a metric/metric space in which this property does not satisfy...
I tried using the triangle inequality but I am not getting anything useful, and I am starting to wonder if this property only follows when the ball $\overline{B}_r(x)$ is path-connected, as then we can apply the property, though I am not sure if this holds in all spaces:
For a point $x$ in a closed ball, the shortest path from $x$ to the boundary of the disc is the path from $x,$ along the radius of the disc (that passes through $x)$, to the boundary circle.
Any help regarding this would be much appreciated!
AI: There is, as other have noticed, the $0$-$1$ distance, where $\partial\overline B_{r}(x)=\emptyset$ for all $r$, thus incurring at least in definitory issues. For another example, we can observe that, by Fermat-Wiles, in the metric space $(\Bbb Q^2,d_3)$ with $d_3(x,y)=\sqrt[3]{\lvert x_1-y_1\rvert^3+\lvert x_2-y_2\rvert^3}$ it holds that $\partial\overline B_1(0)=\{(1,0),(0,1),(-1,0),(0,-1)\}$, and therefore, by simple geometric considerations, there are several $x\in B_1(x)$ and $r<d\left(x,\partial \overline B_1(0)\right)$ such that $B_{r}(x)\nsubseteq \overline B_1(0)$.
This can be adapted to a a path-connected counterexample by considering the metric space $X=\{x\in\Bbb R^2\,:\, \lVert x\rVert_3\ne 1\lor x\in\Bbb Q^2\}$ with the distance induced by the ambient space $(\Bbb R^2,\lVert \bullet\rVert_3)$. Then, the same considerations as before on $\overline B_1(0)$ hold, and that specific ball is path-connected. Unfortunately, I cannot think of a complete counterexample, or of a counterexample where all balls are path-connected, and I don't know if they are possible. |
H: Does there exist a closed set that is not semi-open?
A subset $A$ of a topological space $(X,T)$ is said to be semi-open if there exist an open set $B \in ( X,T)$ such that $B \subseteq A \subseteq \overline B$.
Now my question is that
Give an example of a closed set that is not necessarily a semi-open set.
My attempt : I think about closed interval $A=[0,1]$ and $A= \mathbb{R}$ all are satisfied the semi-open properties.
I think such kind example doesn't exist.
AI: Like quangtu123 said, every closed subset with empty interior does the trick. If you want something in $\Bbb{R}$ which is not just a singleton or something similar, I would suggest looking at the Canot set. |
H: Solving a system of coupled recurrence relations
I am required to solve the below system of recurrence relations:$$\begin{cases}a_n-a_{n-4}=t_n-t_{n-3}\\a_n-a_{n-1}=25t_{n-1}-t_{n-3}\end{cases}$$As you can see, I can't isolate $a_i$ or $t_i$. I would like to know if a non-constant closed-form solution exists and if yes, how to obtain it.
AI: Hint:
$$a_n - a_{n-4}=(a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + (a_{n-2} - a_{n-3}) + (a_{n-3} - a_{n-4})$$
Here, the LHS and each summand on the RHS can be expressed via $t$ using one of your two equations, giving a recurrence relation in just $t$. |
H: Solving definite integrals with periodic integrand
So, the question is:
If $f(2-x) = f(2+x)$ and $f(4-x) = f(4+x)$ where $f(x)$ is a function for which
$$\int_{0}^{2}f(x)dx =5$$ ,
then prove that :
$$\int_{0}^{50}f(x)dx =125$$
$$\int_{-4}^{46}f(x)dx =125$$
$$\int_{2}^{52}f(x)dx =125$$
also comment whether $\int_{1}^{51}f(x)dx =125$ is true or false.
My approach:
since $f(2-x) = f(2+x)$ , replace $x$ by $x+2$
and we get $f(x) = f(4+x)$ hence period of
$f(x)$ is $4$
now,
$$\int_{0}^{50}f(x)dx = \int_{0}^{48}f(x)dx +\int_{0}^{2}f(x)dx $$
hence
$$\int_{0}^{50}f(x)dx =12* \int_{0}^{4}f(x)dx +\int_{0}^{2}f(x)dx $$
which becomes
$$\int_{0}^{50}f(x)dx =12*( \int_{0}^{2}f(x)dx+\int_{0}^{2}f(x+4)dx) +5 $$
so,we get $$\int_{0}^{50}f(x)dx =125$$
similarly for
$$\int_{-4}^{46}f(x)dx $$,substitute $x+4=t$
therefore,
$$\int_{-4}^{46}f(x)dx =\int_{0}^{50}f(t)dt =125 $$
however I am not able to prove on the final part and comment on whether $\int_{1}^{51}f(x)dx =125$ is true or false.Kindly help me out.
AI: Hint: $\int_2^{52} f(x)dx=\int_0^{52} f(x)dx -\int_0^{2} f(x)dx=(13)\int_0^{4} f(x)dx-5$. |
H: Are $\operatorname{Spec}\overline{\mathbb{Q}}[x]$ and $\operatorname{Spec}\mathbb{Z}$ homeomorphic?
Let $\overline{\mathbb{Q}}$ be the algebraic closure of the field $\mathbb{Q}$.
I know that $\operatorname{Spec}\mathbb{Z}=\{0\}\cup\{(p): p\text{ is prime}\}$ and the closed points of $\operatorname{Spec}\mathbb{Z}$ is $(p)$. The closure of $\{0\}$ is $\operatorname{Spec}\mathbb{Z}$.
And I have proved that $\overline{\mathbb{Q}}$ is countable. But I am not sure about how does $\operatorname{Spec}\overline{\mathbb{Q}}[x]$ look like. I was thinking if I could find a bijection of closed point and non-closed point of these two sets, then I can say they are homeomorphic in Zariski topology. Is this right?
AI: The prime ideals of $\overline{\Bbb Q}[x]$ are $\{0\}$ and $(x-a)$
for $a\in\overline{\Bbb Q}$. So the spectrum consists of countably infinitely many
closed points and one dense point, just as the spectrum of $\Bbb Z$ has.
These spectra are therefore homeomorphic. |
H: Binomial exponent to be a Martingale process
Let $0 < p < 1$. Let $\{X_n\}_{n\in\mathbb N}$ be a sequence of independent random variables such that
$$P (X_n = 1) = p, P (X_n = −1) = 1 − p$$
for all $n \in \mathbb N$. For each $n\in\mathbb N$, set
$$S_n := X_1 + X_2 + · · · + X_n$$
for the filtration $F_n := \sigma(X_i,1 \leqslant i \leqslant n).$
Now I define the following stochastic process $\{M_n\}$ by
$$M_n := (\frac{1-p}{p})^{S_n},\forall n \in\mathbb N,$$
I want to show that $M_n$ is an $F_n$-martingale.
I have tried to prove the following,
$$E=[M_{n+1}|F_n]=M_n. $$
However, by inserting $M_n$, I don't get the desired result.
AI: $E(\frac {1-p} p)^{S_{n+1}}|\mathcal F_n)=(\frac {1-p} p)^{S_n}E(\frac {1-p} p)^{X_{n+1}}=(\frac {1-p} p)^{S_n} [(\frac {1-p} p) p+\frac p {1-p} (1-p)]=(\frac {1-p} p)^{S_n}$ |
H: evaluate the limit of $\frac {x}{|x|^s}$ as $s<1$ and $x$ goes to $0$
Evaluate $lim_{x\to 0} \frac {x}{|x|^s}$ for $s<1$ and $x\in \mathbb R^n$. For $n = 1$, I think this limit is equal to $0$. However I am trying to evaluate it when $x$ is an vector in arbitrary dimension. Does this limit exist?
AI: Hint: $\displaystyle\left\|\frac x{\sqrt{\|x\|}}\right\|=\sqrt{\|x\|}$. |
H: determining the quotient group in Mayer-Vietoris sequence
I am having trouble to determine the quotient group in the following Mayer-Vietoris sequence.
I know this problem in Hatcher exists here but my question is not to have a solution (because I do have one). I want to understand how we come up with it.
X is the space obtained by attaching a Mobius strip M to the projective plane $RP_2$ by attaching its boundary to the copy of $S_1= RP_1 \subset RP_2$. I was able to find out using Mayer-vietoris sequence that $H_1(X)=(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}) / Im \phi$ where $\phi(c)=(a,-2b)$. So $H_1(X)=\langle a,b\rangle / \langle 2a,a-2b \rangle $.
Up to here, I understand all the steps. Now this is what I am struggling with: How can we determine $H_1(X)$? I know the answer should be $\mathbb{Z_4}$ but I am having trouble to see that.
PS: I saw in one of the sites at the internet this: $\langle a,b \rangle / \langle 2a,a-2b \rangle =\langle a+2b,b \rangle /\langle a+2b,-4b \rangle \simeq\mathbb{Z_4}$ but I don't understand why at all. I understand that this is an algebraic question but I would rather put the context to it as well.
AI: I don't understand the explanation you've quoted as well. However $a-2b$ in the denominator means means that $[a]=2[b]$ in the quotient. Thus the $a$ generator is redundant and we can replace it with $2b$. This gives us the following:
$$\langle a,b\rangle/\langle 2a,a-2b\rangle\simeq\langle 2b,b\rangle/\langle 4b,0\rangle =\langle b\rangle/\langle 4b\rangle\simeq\mathbb{Z}_4$$
Note that this is not a very precise and formal proof. More like an intuition. The first isomorphism requires a bit of work with free (abelian) groups. |
H: Is $f(x)=\left.\begin{cases}x\,\text{sgn}(\sin\frac{1}{x})&\text{if $x\neq0$}\\0&\text{if $x=0$}\end{cases}\right\}$ Riemann integrable?
For $x\in[-1,1]$, let
$$
f(x)=
\begin{cases}
x\,\operatorname{sgn}(\sin\frac{1}{x}), &\text{if $x\neq0$} \\
0, &\text{if $x=0$}
\end{cases}
$$
where $\text{sgn}$ denotes the signum function. Then:
$f$ is continuous on $[-1,1]$
$f$ is not differentiable at any point of $[-1,1]$
$f$ is Riemann integrable on $[-1,1]$.
The set of points of discontinuity of $f$ in $[-1,1]$ is finite.
I found this question in a previous year entrance paper. The answer given is option $3$.
Now, I know that $f$ is discontinuous at $x=0$, and that $f$ is differentiable
at many points in $[-1,1]$. But what about options $3$ and $4$? Aren't they both essentially the same?
AI: For each $0 < \epsilon <1$, $f$ is piecewise differentiable on $[-1,-\epsilon] \cup [\epsilon , 1]$ and therefore Riemann integrable on those intervals. Also $f$ is discontinuous at all points of the infinite set $S=\{1/k\pi \mid k \in \mathbb Z\}$.
This makes 1., 2. and 4. claims false.
Finally $f$ is bounded on $[-1,1]$. So given the above, $f$ is Riemann integrable on $[-1,1]$ and 3. is correct.
Recall that a map $f$ that is bounded on $[a,b]$ and Riemann integrable on all $[c,b]$ with $a<c<b$ is Riemann integrable on $[a,b]$. |
H: counterexample completeness of metric spaces
I'm looking for a counterexample. I already proved that if $(X,d_X)$ and $(Y,d_Y)$ are metric spaces and $X$ is complete and $f:X\to Y$ is a bijection that $Y$ is complete if $f^{-1}$ is uniform continuous. I know that if $f^{-1}$ isn't uniform continuous $Y$ isn't complete. I just can't find a counterexample.
I tried to look for $f:\mathbb{R}\to\mathbb{Q}$ or something but it's never right. I found online that $f:\mathbb{R}\to(0,1):x\mapsto1/(1+2^{-x})$ is a counterexample but i don't really see how. Can someone maybe give an easier example or explain me why this is a great counterexample?
Edit: i'm looking for a function with f is uniform continuous and f^-1 is normal continuous, where X is complete but Y isn't
AI: $f(x)=\frac x {1+|x|}$ is a homeomorphism from $\mathbb R$ onto $(-1,1)$. The domain is complete but the range is not.
Alternatively use the homeomorphism $\arctan x$ from $\mathbb R$ onto $(-\pi /2 ,\pi /2)$.
These two functions are uniformly continuous bijections. |
H: Problem with sequences and cofinite topology
Let $X$ be an infinite Topological Space with Cofinite Topology.
Let $\lbrace x_{n}\rbrace\subseteq X$ be a sequence and let $a$, $b\in X$ be two points such that $x_{n}=a$ and $x_{n}=b$ for endless values of $n$.
Prove that $\lbrace x_{n}\rbrace$ does not converge.
Honestly, I don't know how to proceed. I though something about the fact that $X$ is not $T_{2}$, or I though to work on the definition of limit in Topological Spaces... but I don't know how to start. Can anyone help me?
AI: Suppose that $a \neq b$ and that $L$ is the limit of the sequence $\{x_n\}$.
If $L \notin \{a,b\}$, $U = X \setminus \{a,b\}$ is an open subset of the cofinite topology that contains $L$ but cannot contain all but a finite number of $\{x_n\}$ knowing the hypothesis on the sequence.
And if $L \in \{a, b \}$, say $L=a$ for example, then $V = X \setminus\{b\}$ is an open subset that contains $L$ and cannot contain all but a finite number of $\{x_n\}$. |
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