text stringlengths 83 79.5k |
|---|
H: Is the set of all rationals with denominators less than $10^6$ closed in $\mathbb{R}$?
I'm wondering if the set of all rationals with denominators less than $10^6$ is closed in the real number system. I think it is not, so that's what I've been trying to prove. I've tried looking at its complement and showing that it is open, but I didn't progress much. My last resort is to show that there is a sequence of rationals in my set converging to an irrational number (or to a rational whose denominator is greater than $10^6$). The problem is, how do I ensure that all simplified fractions in my sequence will have denominator less than $10^6$?
AI: Hint: Let $N=\operatorname{lcm}\{1,2,3,\ldots,10^6\}$, and let our set be
$$S=\{\tfrac{a}{b}\in\mathbb{Q}\mid 1\leq b\leq 10^6\}.$$
Consider the homeomorphism $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=Nx$. Then $f(S)\subseteq\mathbb{Z}$. |
H: A problem about invariant subspaces
I need help with the following problem:
Let $H\subseteq\mathbb R^3$ be a plane with cartesian equation $x=y,$ and let $r$ be the straight line generated by $(1,1,2)$.
Find an endomorphism $\phi$ of $\mathbb R^3$ such that $\phi(\mathbb R^3)=H$ and $\phi^2(\mathbb R^3)=r$.
Is there a standard way to choose such endomorphism $\phi$?
AI: Extend $v_1 = (1,1,2)$ to a basis $\{v_1,v_2\}$ of $H$ and then to a basis $\{v_1, v_2, v_3\}$ of $\mathbb{R}^3$.
Take the endomorphism $\phi$ which maps $v_3$ to $v_2$, $v_2$ and $v_1$ to $v_1$. |
H: Irreducible polynomial not attaining squares over finite field
Is it possible to construct an irreducible polynomial $f$ over $\mathbb{F}_{q}$ such that $f(x)$ is a non-square for any $x \in \mathbb{F}_{q}$?
I can prove the existence of irreducible polynomials (Euclid's argument), and I can construct polynomials with no square values (for example by Lagrange interpolation through non-squares), but satisfying these 2 conditions feels difficult.
Motivation: I am trying to construct an hyper-elliptic curve $y^2 = f(x)$ with no rational points.
EDIT: Can you construct such an $f$ that will not be constant on the ground field $\mathbb{F}_{q}$?
AI: One way is with Artin-Schreier polynomials $f(x)=x^q-x+a$ with $a$ a non-square. |
H: $f$ has an essential singularity in $z_0$. What about $1/f$?
Let $\Omega$ be a non-empty, open subset of $\mathbb C$. Consider an holomorphic function $f:\Omega \setminus \{z_0\} \to \mathbb C$ and suppose we know $z_0$ is an essential singularity of $f$.
I am wondering what we can say about the function $\tilde{f}:=\frac{1}{f}$ and its singularity in $z_0$. Do you know any theorem that answers to this question?
Actually, I can't prove anything, since I do not know the answer: I've studied some examples. For instance, if you take $f(z)=e^{\frac{1}{z}}$ then $\tilde{f}$ will still have an essential singularity, hasn't it?
On the other side, if we take $f(z)=\sin(\frac{1}{z})$ then I think that $z_0=0$ becomes a limit point of poles for $\tilde{f}$ (so we can't classify it, because it isn't an isolated singularity).
Wha do you think? Do you know any useful theorem concerning this? Thank you in advance.
AI: Your two examples essentially span all the possibilities. By the Big Picard Theorem, we know that $f$ assumes all but one value in $\Bbb{C}$ infinitely often, in any neighborhood of $z_0$.
If the missed value is $0$, then $\frac{1}{f}$ has an isolated singularity at $z_0$, and it must clearly be essential (since otherwise $f$ itself would have a pole or removable singularity).
If the missed value is not $0$, or if there is no missed value, then $\frac{1}{f}$ will have a sequence of poles that converges to $z_0$, and hence the singularity at $z_0$ will not be isolated. So, thinking of $\frac{1}{f}$ as a holomorphic function on its largest possible domain (which will omit some sequence of points converging to $z_0$), the singularity at $z_0$ is technically unclassifiable.
On the other hand, even in this case you can regard $\frac{1}{f}$ as a meromorphic function from $\Omega \backslash \{z_0\}$ to the extended complex plane $\hat{\Bbb{C}}$, and when considered as such it will have an essential singularity at $z_0$. This is a natural enough thing to do that I suspect most people would be happy just saying "$\frac{1}{f}$ has an essential singularity at $z_0$." |
H: Prove that an odd Collatz sequence will at some point have two consecutive even numbers
Is there an existent proof for this? It would be an important step to proving the conjecture in any case.
AI: Write the initial odd number in binary notation. The binary representation ends with a number of 1s, preceded by a 0 -- say 'abcd01111'. Now imagine computing two steps ahead in the sequence to $\frac{2x+x+1}{2}$ in binary:
abcd01111
+ abcd011110
+ 1
------------
opqrs01110 divided by 2 is opqrs0111
When the carries have settled you find that this next step has one less digit 1 at the end. Proceed by induction; after finitely many steps you have used up all of the 1s, and the $\frac{2x+x+1}{2}$ is actually even:
abcd01
+ abcd010
+ 1
---------
opqrs00 divided by 2 is opqrs0
at which point two consecutive halvings will happen.
(This could all have been written in a more mathematically dignified form by observing that for every $n\in\mathbb N$ there is exactly one $k\ge 0$ such that $n\equiv 2^k-1 \pmod{2^{k+1}}$. If $k>0$, then the next step in the sequence will be $2^{k+1}+2^k-2\equiv 2(2^{k-1}-1) \pmod{2^k+1}$ and the one after that is $2^{k-1}-1\pmod{2^k}$ -- in other words the two steps have made $k$ decrease by one. But that is not as instructive a way to phrase it, I think).
This argument also implies that the ratio between the starting number and maximum number in the sequence can be arbitrarily large. |
H: Fat Tail / Large Kurtosis Discrete Distributions?
All,
I'm wondering if there are any notable, basic discrete probability distributions with "fat/heavy tails" or a large kurtosis? I know the Geometric Distribution's excess kurtosis approaches 6, but I can't find any that are larger.
Are there any discrete distributions where the kurtosis ( or other higher normalized moment ) is infinite / undefined?
( By basic, I guess I mean something one could easily implement with game tokens like dice or cards. )
AI: For a "fat tail" you might take
probability mass function $p(x) = \zeta(s)^{-1}/x^s$ for positive integers $x$, where $s > 1$. This has finite variance for $s > 3$, and finite kurtosis for $s > 5$. As $s \to 5+$ the excess kurtosis is, according to Maple,
$ \dfrac{8100 \zeta(5)}{8100 \zeta(3) \zeta(5) - \pi^8} (s - 5)^{-1} + O(1)$, where
$8100 \zeta(5)/(8100 \zeta(3) \zeta(5) - \pi^8) \approx 13.82154183$.
For something you can implement easily with a fair coin, you might try this. Let $N$ be the number of flips of the coin until the first heads, and $X = r^N$.
For the kurtosis to be finite, you need $r < 2^{1/4}$. According to Maple, the excess
kurtosis is $${\frac {-13\,{r}^{7}+16\,{r}^{5}+2\,{r}^{4}+8\,{r}^{3}+16\,{r}^{2
}-16}{2 \left(2- {r}^{4}\right) \left(2- {r}^{3} \right) }}$$
For example, if $r=1.1$ the excess kurtosis is approximately 24.22623314. |
H: Significant figures with a plus-minus
A question on my homework asks me to give the amount of significant figures of $2900±100$. Would this be one, two, or both?
AI: Well, if it had been $2900\pm 50$ then there would clearly have been 2 significant figures, namely the 2 and 9. But how do we account for the fact that the error interval is wider here?
A principled answer would be that there are now $\log_{10}(\frac{2900}{2\cdot 100}) \approx 1.16$ digits of precision (the denominator here is the width of the interval specified, thus the factor of 2). However, in practice the "number of significant figures" are usually supposed to be an integer, and the convention is such that "10.01" and "99.99" both count as "four significant figures", so we should round the $1.16$ upwards to $2$ to get the (probably) expected answer. |
H: Differentiation from first principles of specific form.
I've been posed a question in which I'm to differentiate with respect to $x$ a function of the form $(x+a)^k$. I've successfully completed (matches the book's answer) the question by using the chain rule, however I cannot achieve the same result using the definition of the derivative.
I would like a worked example of differentiation from first principles of a function of the form $(x+a)^k$.
AI: For concreteness, let's first take $k=3$.
We use the factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$.
Here we have $(x+a+h)^3-(x+a)^3=h((x+a+h)^2+(x+a+h)(x+a)+(x+a)^2)$.
The difference quotient is then
$$\frac{(x+a+h)^3-(x+a)^3}{h}=((x+a+h)^2+(x+a+h)(x+a)+(x+a)^2).$$
As $h$ goes to zero, this becomes $3(x+a)^2$, as desired.
How does this generalize? We have similar factorizations for all $k$:
$$x^k-y^k=(x-y)(x^k+x^{k-1}y+\cdots xy^{k-1}+ y^k)$$
where there are $k$ terms in the second parenthesis. This factorization is easy to check, because the terms all telescope and cancel.
So what I did for $k=3$ can be repeated for general $k$, and when $h$ goes to zero you get $k$ terms of $(x+a)^{k-1}$ added together, which is exactly the same derivative you get using the chain rule. |
H: Finding Asymptotes of Hyperbolas
To find a asymptote its either b2/a2 or a2/b2 depending on the way the equation is written.
With the problem
$$\frac{(x+1)^2}{16} - \frac{(y-2)^2}{9} = 1$$
The solutions the sheet I have is giving me is $3/4x - 3/4$ and $3/4 x + 5/4$
I thought it was just supposed to be $\pm 3/4x$.
AI: For the hyperbola $$\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1$$
The asymptotes are $y - k = \pm \dfrac{b}{a}(x - h)$.
You could leave your answer as $y - 2 = \pm \dfrac{3}{4}(x + 1)$, or write two separate equations.
Edit... If you do write separate equations, you'll have
$y - 2 = \dfrac{3}{4}(x + 1)$ and $y - 2 = - \dfrac{3}{4}(x + 1)$, which are, in "slope-intercept form":
$y = \dfrac{3}{4} x + \dfrac{11}{4}$ and $y = - \dfrac{3}{4}x + \dfrac{5}{4}$ |
H: Prove that there exists analytic $f$ such that $f(z) = 1/\bar{z}$ on the boundary
I'm doing some self-study in complex analysis, and came to the following question:
Let $D(a,1) \subset \mathbb{C}$ be the disk of radius $1$ with center at $a \in \mathbb{C}$, and let $\partial D(a,1)$ be the boundary of $D(a,1)$. Prove that $|a| <1$ if and only if there exists a function $f$ analytic on $D(a,1)$ and continuous up to $\partial D(a,1)$ such that $f(z) = 1/\bar{z}$ for $z \in \partial D(a,1)$.
I don't know what area of the theory to try to apply here. I know that $|a| <1$ iff $\; 0 \in D(a,1)$, and I know that the function $1/\bar{z}$ is nowhere analytic. However, I can't see how to get to the desired conclusion.
Thanks.
AI: If you know about harmonic functions, then you can use the maximum principle on the real and imaginary parts of $f(z) - {1 / \bar{z}}$ to show that if $a > 1$, then if such a $f(z)$ existed then $f(z)$ would have to be ${1 / \bar{z}}$ for all $z$.
If $a < 1$, I'd use an explicit construction like Henning Makholm suggests. |
H: What are some examples of non-identity bijections $f: X \to X$ such that $f^{-1} = f$
One example I can think of is $f: \mathbb{Z_2} \to \mathbb{Z_2}$ given by $f(1) = 0$ and $f(0) = 1$.
AI: In a sense, every such bijection is going to look the same. If $a,b \in X$, then we will either have things that look like $f(a) = a$ or $f(a) = b, f(b) = a$ (which I'm going to refer to as a single 'transposition.'
But this gives us infinitely many examples to choose from, even just in $\mathbb{Z}$. You might let $f$ be the identity on every element except, say, $1$ and $5$, such that $f(1) = 5, f(5) = 1$. Or you can have as many transpositions as you'd like. |
H: Exercise on compact $G_\delta$ sets
I'm having trouble proving an exercise in Folland's book on real analysis.
Problem: Consider a locally compact Hausdorff space $X$. If $K\subset X$ is a compact $G_\delta$ set, then show there exists a $f\in C_c(X, [0,1])$ with $K=f^{-1}(\{1\})$.
We can write $K=\cap_1^\infty U_i$, where the $U_i$ are open.
My thought was to use Urysohn's lemma to find functions $f_i$ which are 1 on $K$ and $0$ outside of $U_i$, but I don't see how to use them to get the desired function. If we take the limit, I think we just get the characteristic function of $K$.
I apologize if this is something simple. It has been a while since I've done point-set topology.
AI: As you have said, we can use Urysohn's lemma for compact sets to construct a sequence of functions $f_i$ such that $f_i$ equals $1$ in $K$ and $0$ outside $U_i$.
Furthermore, $X$ is locally compact, so there is an open neighbourhood $U$ of $K$ whose closure is compact. We can then assume without loss of generality that $U_i\subseteq U$
Then we can put $f=\sum_i2^{-i} f_i$. Clearly, $f^{-1}[\{1\}]=K$.
Moreover, $f$ is the uniform limit of continuous functions (because $f_i$ are bounded by $1$), so it is continuous, and its support is contained in $U$, so $f$ is the function you seek. |
H: Understanding this summation identity
I'm currently reading a book in which part of the solution to the problem involve this identity:
$$\sum_{j=i+1}^{n}j = \sum_{j=1}^{n}j-\sum_{j=1}^{i}j$$
Which I cannot derive myself. The only thing I can do with it is this:
$$\sum_{j=i+1}^{n}j = \sum_{j=1}^{n}j+i = \sum_{j=1}^{n}j + \sum_{j=1}^{i}i$$
Which seems to me completely useless.
Any help in understanding this (as I am unaccustomed to summation manipulation in general) would be greatly appreciated.
I know it's related to " Calculate integer summation when lower bound is a variable " but I still don't see the why.
AI: We have
$$\sum_{j=1}^{i}j+\sum_{j=i+1}^n j=\sum_{j=1}^n j.\tag{$1$}$$
The result you are looking for follows by subtraction.
If $(1)$ seems unclear, let us take particular numbers, say $i=7$ and $n=19$.
We have
$$\sum_{j=1}^{i}j =\sum_{j=1}^7j=1+2+\cdots+7$$
and
$$\sum_{j=i+1}^{n}j =\sum_{j=8}^{19} j=8+9+\cdots +19.$$
If you add them, you get
$$1+2+\cdots+7+8+9+\cdots +19.$$
This is equal to
$$\sum_{j=1}^{19} j.$$
Remark: Exactly the same argument shows that if $a_1,a_2,\dots$ is any sequence, then
$$\sum_{j=i+1}^n a_j=\sum_{j=1}^n a_j -\sum_{j=1}^i a_j.$$ |
H: An increasing sequence $a_i \in \mathbb{N}$ such that $a_{j}-a_{i}\mid a_{j}-1$ for all $i
How to prove or disprove that there exists a set of natural numbers
$$a_{1}<a_{2}<\cdots$$ with
the property that, for every $i<j$, $$a_{j}-a_{i}\mid a_{j}-1\quad ?$$
I think it doesn't work for very big $a_{j}$'s and very small $a_{i}$'s.
AI: There is no such sequence.
Suppose by contradiction that we have such a sequence, and for any $j>i$ we have $a_j-a_i\vert a_j-1$. Then either $a_j=1$ or $a_i=1$ or $a_j>2(a_j-a_i)$ (because $a_j=1+k(a_j-a_i)$ for some $k\geq 0$, the three cases corresponding to $k=0,1$ and $k\geq 2$).
The first two cases are only possible for small $i$, so we can discard them (we can assume that the sequence starts with arbitrarily large numbers by cutting off the beginning).
The last one gives us $2a_i>a_j$ with any sufficiently large fixed $i$ and all $j>i$. But of course $a_{i+a_i}\geq 2a_i$, so we have a contradiction. |
H: How many significant digits should be retained mid-calculation?
I always feel paranoid when dealing with long series of calculations on paper. How many significant digits should one use to reduce to negligible the probability of getting a wrong final result in a modestly long series of calculations? (I know that the answer to my question depends on the setting. If it helps, I'm a college student; but I would be interested in learning anything on the matter.)
AI: If on a test you are not allowed to use a calculator, the calculations, if done properly, do not involve much work with decimals. You just have to resist the urge to convert fractions to decimals until, possibly, the end.
If you are using a calculator, learn how to use the memory feature present on most scientific and financial calculators. This has many advantages. For one thing, it means you do not have to rekey intermediate results. Keying mistakes are a frequent source of error. You are also likely to save a fair bit of time.
It is particularly dangerous to round off interest rates, when the period you are talking about involves several years. Even a smallish error in the interest rate, when compounded for a long enough period, can lead to substantial dollar differences. Another example of the magic of compound interest!
Remark: An additional subtlety is that calculators calculate internally to higher precision than the precision they display. So even if you key in exactly what the calculator shows, you are potentially losing some of the accuracy that the calculator is capable of giving. |
H: If $E$ and $F$ are subfields of a finite field $K$ and $E\cong F$, prove that $E = F$
If $E$ and $F$ are subfields of a finite field $K$ and $E\cong F$, prove that $E = F$.
A finite field is a simple extension of each of its subfields and $\mathbb{Z}_p$ is a subfield of every finite field. Hence $E\cong \mathbb{Z}_p(u)$ and $F\cong \mathbb{Z}_p(v)$ for some $u,v\in K$. Proving that $u = v$ given $\mathbb{Z}_p(u)\cong \mathbb{Z}_p(v)$ may be stronger than I need though, since $u$ and $v$ could be different generators for the same set.
Can anyone help me with this? Many thanks.
Edit: This is a Galois Theory free zone.
AI: The isomorphism in particular implies that the orders of $E$ and $F$ are equal. That's all we need. Let us assume that you know that a finite field with $q=p^k$ elements is a splitting field for the polynomial $x^q-x=0$. Thus each of $E$ and $F$ consists of the roots (in $K$) of $x^q-x=0$. That implies that $E=F$. |
H: Is there a concrete description of the ideal $I$ such that $\mathbb{Q}[x]/I\cong\mathbb{Q}[\sqrt{2}+\sqrt{3}]$?
I know that in general if $R[u]$ is the ring obtained by adjoining an element $u$ to a ring $R$, then $R[u]\cong R[x]/I$ for some ideal $I$ such that $I\cap R=\{0\}$.
In a particular instance, I'm working with $u=\sqrt{2}+\sqrt{3}$, so I'm wondering if there is a concrete description of an ideal $I$, say generated by some polynomials of $\mathbb{Q}[x]$ or something like that, such that $\mathbb{Q}[x]/I\cong\mathbb{Q}[u]$.
It's clear there is a surjective homomorphism from $\mathbb{Q}[x]\to\mathbb{Q}[u]$, so I would take the kernel, which is just the set of all polynomials with $u$ a root. Is this ideal generated by anything easy to write down? Or is that the best description I can give?
AI: Let $p(x)$ be the minimimal polynomial of $\sqrt{2}+\sqrt{3}$. Then $I=(p(x))$. According to Wolfram|Alpha, $p(x)=x^4-10x^2+1$. |
H: Why is a strict $p$-ring whose residue ring is a field necessarily local?
Let $A$ be a strict $p$-ring. Recall that this means $A$ is $p$-adically separated and complete, $p:A\rightarrow A$ is injective, and $A/pA$ is a perfect $\mathbf{F}_p$-algebra. If $A/pA$ is a field, then $A$ is known to be a discrete valuation ring. The only way I have seen this proved is to use the actual construction of the $p$-Witt ring $W(A/pA)$, and I'm wondering if there is a way to prove this using only the definition of a strict $p$-ring.
In general, a ring $B$ is a discrete valuation ring if and only if it is local, max-adically separated, and its maximal ideal is principal and non-nilpotent (this is basically Proposition 2 in Serre's Local Fields, although in place of the separated hypothesis, he has a Noetherian hypothesis). If $A$ is a strict $p$-ring with $A/pA$ a field, then $pA$ is maximal, and non-nilpotent by definition. Also, $A$ is $p$-adically separated by definition. But I can't think of a simple way to prove that $A$ is local, i.e., that $pA$ is the unique maximal ideal of $A$. I suppose one could work out the universal formulas which define the ring operations on such a ring and use them to characterize the non-units as precisely the elements of $pA$, but I was hoping there might be a way to avoid this.
The reason I expect such a thing might be possible is that Serre claims that "a complete discrete valuation ring, absolutely unramified, with perfect residue field $k$, is nothing other than a strict $p$-ring with residue ring $k$." He doesn't give any argument for why a strict $p$-ring whose residue ring is a field is local. This leads me to believe I'm missing a potentially obvious argument.
I apologize if this question doesn't seemed well-defined. The ring operations on a strict $p$-ring can be derived from the definitions (starting by proving the existence of a multiplicative system of representatives, then getting series expansions, etc.), so, proving locality from this, strictly speaking, is just using the definitions, but I would say there's considerable work involved (or at least considerable messiness). Basically I'm trying to avoid writing down the polynomials that give the ring structure.
AI: If $x\not\in pA$, then there is some $y$ such that $xy\equiv1$(mod $p$). Then $x$ is invertible mod $p^n$ for any $n$ since $(xy)^{p^n}\equiv 1$(mod $p^n$). Thus $x$ is invertible. |
H: proof for $ (\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A}\cdot\vec{C})\vec{B}-(\vec{B}\cdot\vec{C})\vec{A}$
this formula just pop up in textbook I'm reading without any explanation
$ (\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A}\cdot\vec{C})\vec{B}-(\vec{B}\cdot\vec{C})\vec{A}$
I did some "vector arithmetic" using the determinant method
but I'm not getting the answer to agreed with above formula.
I'm wondering if anyone saw this formula before and know the proof of it?
the final result that i get is
$b_{1}(a_{3}c_{3}+a_{2}c_{2})-a_{1}(b_{2}c_{2}+b_{3}c_{3})i$
$b_{2}(a_{3}c_{3}+a_{1}c_{1})-a_{2}(b_{3}c_{3}+b_{1}c_{1})j$
$b_{3}(a_{2}c_{2}+a_{1}c_{1})-a_{3}(b_{2}c_{2}+b_{1}c_{1})k$
But I failed to see any correlation for $(\vec{A}\cdot\vec{C})$ and $(\vec{B}\cdot\vec{C})$ part...
AI: Let $a=\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and likewise for $b,c.$ The first component of $a\times (b \times c)$ is seen by applying the definition to be $a_2(b_1c_2 - b_2c_1) - a_3(b_3c_1-b_1c_3).$ With some algebra, this is seen to be $(a_2c_2+a_3c_3)b_1 - (a_2b_2+a_3b_3)c_1.$ The tricky transformation to apply here is simulataneously adding and subtracting the quantity $a_1b_1c_1.$ This allows us now to write the first component as $(a_1c_1 + a_2c_2 +a_3c_3)b_1 - (a_1b_1 + a_2b_2 + a_3b_3)c_1.$ Applying similar arguments to the second and third coordinates, we see that $a\times (b\times c) = \begin{pmatrix}(a_1c_1 + a_2c_2 +a_3c_3)b_1 - (a_1b_1 + a_2b_2 + a_3b_3)c_1 \\ (a_1c_1 + a_2c_2 +a_3c_3)b_2 - (a_1b_1 + a_2b_2 + a_3b_3)c_2 \\ (a_1c_1 + a_2c_2 +a_3c_3)b_3 - (a_1b_1 + a_2b_2 + a_3b_3)c_3 \end{pmatrix} = (a\cdot c)b - (a\cdot b)c$ |
H: Method to solve $xx'-x=f(t)$
I would like to resolve this differential equation:
$xx'-x=f(t)$
any suggestions (or any online texts on similar differential equation) please?
Thanks.
AI: This is an 'Abel equation of the second kind in the canonical form'. |
H: Continuous function from $(0,1)$ onto $[0,1]$
While revising, I came across this question(s):
A) Is there a continuous function from $(0,1)$ onto $[0,1]$?
B) Is there a continuous one-to-one function from $(0,1)$ onto $[0,1]$?
(clarification: one-to-one is taken as a synonym for injective)
I figured the answer to A is yes, with $\frac{1}{2}\sin(4\pi x)+\frac{1}{2}$ as an example.
The answer to part B is no, but what is the reason?
Sincere thanks for any help.
AI: There isn't such a function. If $f$ is such a function, then $f$ is monotonic, and $f^{-1}$ too. But a monotonic function defined on an interval is continuous iff its range is an interval. So we have that $f$ is an homeomorphism from $(0,1)$ to $[0,1]$, which is impossible since one is compact and the other is non compact. |
H: Why would anti-symmetric (0,2) tensor be traceless?
As it is, why would anti-symmetric (0,2) tensor be traceless? Is it because trace should allow any variable for its indices?
AI: Let $A_{ij}$ be (the entries of) a totally antisymmetric tensor. Then, its trace is
$$g^{j i} A_{i j} = g^{i j} A_{i j} = -g^{i j} A_{j i} = - g^{j i} A_{i j} = 0$$
because $g^{i j}$ is (the entries of) a totally symmetric tensor. |
H: What's $T\left(n\right)$?
If $T\left( n \right) = 8T\left( n-1 \right) - 15T\left( n-2 \right); T\left(1\right) = 1; T\left( 2 \right) = 4$,
What's $T\left(n\right)$ ?
I use this method:
Let $c(T(n) - aT(n-1)) = T(n-1) - aT(n-2)$
from $T(n) = 8T(n-1) - 15T(n-2)$, we can get $\begin{cases}c = \frac{1}{3}\\a = 5\end{cases}$ and $\begin{cases}c = \frac{1}{5}\\a = 3\end{cases}$,
then, we get $\frac{T(n)-5T(n-1)}{T(n-1)-5T(n-2)} = 3$,
so, we reach the answer: $T(n) = \frac{3^{n-1}+5^{n-1}}{2}$.
Sorry for my poor English.
AI: This is second order recurrence equation. Since $T(n)=8T(n-1)-15T(n-2)$, we first solve the quadratic equation: $x^2=8x-15$. Its roots are $3$ and $5$. Therefore, $T(n)$ must be in the form:
$$T(n)=a\cdot 3^n+b\cdot 5^n, n\geq 1$$for some constants $a$ and $b$.
To find $a$ and $b$, since $1=T(1)=3a+5b$ and $4=T(2)=9a+25b$, we get $a=1/6$ and $b=1/10$, i.e.
$$T(n)=\frac{1}{6}\cdot 3^n+\frac{1}{10}\cdot 5^n, n\geq 1$$ |
H: Mode for unique numbers
I'm trying to construct a rough statistical software and I'm confused about the following:
What is the mode if all numbers are unique?
What is the mode if 2 numbers have same (highest) frequency?
What I feel:
Should output nothing.
Should output the average of the 2.
Just as reference:
a = [1 2 3 4 5]
mode(a) %Prints 1
a = [2 2 3 3]
mode(a) %Prints 2
in Octave
But I'm not sure how it's supposed to be done.
AI: In general, a data set can have multiple modes. If you must return a single value, pick any one of them. (Octave seems to pick the least one.) But don't pick something that isn't a mode. So don't use your second idea: when asked for the mode of $[1,1,2,5,5]$, the value $3$ is not a sensible answer. |
H: How to go about evaluating summation of something?
I am new to math and algorithms, I am trying to understand how does this equation:
$$|N| \le \sum\limits_{i = 1}^{h} 2^{i-1} $$
Come to a conclusion of:
$$|N| \le 2^h - 1$$ and then
$$h \ge log(|N|+1)$$
This is the evaluation of a height of a binary tree and N is the total number of elements in a binary tree.
PS: I am new to math.stackexchange. Please recommend some better tags for this question.
AI: We can see that the sum of a geometric progression: $\sum_{a\le k\lt b}{c^{k}}=\frac{c^{b}-c^{a}}{c-1}$, where $c\not=1$. Therefore, manipulating the range of the sum gives:
$$\sum_{1\le k \lt h+1}{2^{k-1}}=\sum_{0\le k \lt h}{2^{k}}=\frac{2^{h}-1}{2-1}=2^{h}-1,$$
Which is the answer you are looking for. In regards to the next part of your question, given:
$$\left|N\right|\le2^{k}-1 \implies |N|+1\le2^{k}$$
Taking $\log_{10}$ of both sides:
$$\log_{10}{(|N|+1)} \le k\log_{10}{(2)}$$
However, as $\log_{10}(2)\lt 1$:
$$\log_{10}{(|N|+1)}\le k$$
Which is the inequality you desired.
As mentioned by André Nicolas in his comments below, I shall explain a little more about the usage of the most common logarithms and the different notations regarding them.
In computer science, by far the most common form of logarithm is $\log_{2}$, usually written as $\lg$ or $\operatorname{lb}$ in CS literature. This form of logarithm is also known as the "Binary Logarithm" in some literature, and is the solution to the equation:
$$2^{x}=a\implies x=\log_{2}{a}=\lg{a}=\operatorname{lb}{a}$$
A simple example use of the binary logarithm is in determining the number of bits required to represent some integer, $n$, which we can find by evaluating:
$$\left\lfloor\lg{n}\right\rfloor+1,$$
Where $\left\lfloor x \right\rfloor$ is the greatest integer less than or equal to $x$.
The next most common logarithm in computer science is $\log_{10}$ also just written $\log$. This is also the most common form of logarithm in fields such as engineering, and is called the "Common Logarithm". These are solutions to equations of the form:
$$10^{x}=a \implies x=\log_{10}{a}=\log{a}$$
Finally, whilst not commonly used in computer science (but often used in other disciplines such as maths and physics), we have $\log_{e}$, usually written $\ln$, and called the "Natural Logarithm". This logarithm has properties of particular importance in maths, and you will see many examples of its use whilst browsing this site. It provides solutions to equations of the form:
$$e^{x}=a\implies x=\log_{e}{a}=\ln{a}$$
There are various rules which we can use to manipulate logarithms (above I have used the rule that $\log_{b}{a^{n}}=n\log_{b}{a}$, for all bases $b$), you can find many of these rules here and here. |
H: What does it mean to be: "at least logarithmic"?
I am going through some CS basics and the documents says in places that:
We need the run time to be at least logarithmic.
What does that mean?
By def, log means:
The logarithm of a number is the exponent by which another fixed
value, the base, has to be raised to produce that number.
What does it truly mean when we say we want a run time to be logarithmic?
AI: Your question has two answers (although one is considerably more likely).
Firstly, the author could mean that the running time must be slower or as slow as some $C\log{n}$ (as pointed out by anon in the comments), i.e:
$$f(n)\ge C\log{n}\implies f(n)\in\mathcal{\Omega}{(\log{n})}$$
This means that given some set of $n$ inputs, the algorithm will complete no faster than $C\log{n}$, for some arbitrary constant $C\in\mathbb{R}^{+}$.
However, as you have stated it is a piece of CS literature, what is more likely is that the author intended that the algorithm would complete faster or as fast as some $C\log{n}$, i.e:
$$f(n)\le C\log{n} \implies f(n)\in\mathcal{O}{(\log{n})}$$
This means that the algorithm, given a set of $n$ inputs as before, will complete no slower than $C\log{n}$, for some arbitrary constant $C\in\mathbb{R}^{+}$.
Bear in mind that we often use asymptotic notation (e.g. $f(n)\in\mathcal{\Omega}(\log{n})$, $f(n)\in\mathcal{O}(n^{2})$) in computer science to approximate the running time of algorithms. Also note that $f(n)\in\mathcal{O}(n)$ could also be written as $f(n)=\mathcal{O}(n)$ (indeed this notation is more common in CS literature).
Hope this helps clear things up! |
H: How to learn from proofs?
Recently I finished my 4-year undergraduate studies in mathematics. During the four years, I met all kinds of proofs. Some of them are friendly: they either show you a basic skill in one field or give you a better understanding of concepts and theorems.
However, there are many proofs which seem not so friendly: the only feeling I have after reading them is "how can one come up with that", "how can such a long proof be constructed" or "why does it look so confusing". What's worse, most of those hard proofs are of those important or famous theorems. All that I can do with these hard proofs is work hard on reciting them, leading me to forget them after exams and learn nothing from them. This makes me very frustrated.
After failing to find the methodology behind those proofs, I thought, "OK, I may still apply the same skill to other problems." But again, I failed. Those skills look so complicated and sometimes they look problem-specific. And there are so many of them. I just don't know when to apply which one. Also, I simply can't remember all of them.
So my questions are: How to learn from those hard proofs? What can we learn from them? What if the skill is problem-specific? (How do I find the methodology behind them?)
I need your advice. Thank you!
P.S. Threre are a lot of examples. I list only four below.
Proof of Sylow Theorem in Algebra
Proof of Theroem 3.4 in Stein's Real Analysis.
Theorem 3.4 If $F$ is of bounded variation on $[a,b]$, then $F$ is differentiable almost everywhere.
Proof of Schauder fixed point theorem in functional analysis.
Proof of open mapping theorem in functional analysis.
AI: One important thing about proofs is that you will never be able to appreciate them, and therefore to learn from them, if you are not capable of reading the statement to be proved with a sceptical attitude, and to try to imagine it is untrue:
What's this nonsense they are claiming, it cannot be true!
Certainly it must be possible to satisfy the hypotheses without being obliged to accept the conclusion!
Once you have some mental idea of what a counterexample to the statement would look like, you can interpret the proof as an argument that systematically talks this idea out of your head, convincing you that it really is not possible to ever come up with such a counterexample.
Then you will have acquired a feeling of what the proof is really about, and you will be far more likely to retain it, and to come up with similar arguments when you need to prove something yourself.
But if you take a docile attitude and accept the statement to be proved from the onset, you will never be able to understand what all this reasoning was needed for in the first place. |
H: How to identify symmetric positive definite matrices?
I'm working on a project, implementing Successive over-relaxation (SOR) method (http://en.wikipedia.org/wiki/Successive_over-relaxation) using Python. SOR can only apply if given matrix is,
symmetric positive-definite (SPD)
OR
strictly or irreducibly diagonally dominant.
So I want identify that given matrix is SPD or not.I found two articles about this.
1.Java doc (http://www.codezealot.org/opensource/org.codezealot.matrix/docs/org/codezealot/matrix/Matrix.html#isPositiveDefinite())
If a Matrix ANxN is symmetric, then the Matrix is positive definite if
For all i ≤ N, ai,i > 0 and
For all i ≤ N, ai,i > ∑ ai,j, for all j ≤ N, where j ≠ i
2.Numerical Analysis for Engineering (https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/04LinearAlgebra/posdef/)
A symmetric matrix is positive definite if:
all the diagonal entries are positive, and
each diagonal entry is greater than the sum of the absolute values of all other entries in the corresponding row/column.
These articles says different about the second property.
So,
- What is the correct one?
- Is there any other SPD properties I can use?
- Any suggestions are also welcome.
Thank you in advance. Sorry for my bad English.
AI: The second criteria in both conditions are roughly restatements of Gershgorin's Circle Theorem. The JavaDoc statement is wrong in two regards: (1) it should include absolute values inside the summation $\sum_{j\neq i}|a_{i,j}|$ and (2) the summation should be run through $j \leq N$. As stated by @Marc, the condition is sufficient but not necessary. |
H: Product of two symmetry groups
Do there exist 2-sylow subgroups of $S_4\times S_3$ that are normal?
Do there exist 3-sylow subgroups of $S_4\times S_3$ that are normal?
Thank you for helping!
AI: Fact $1$: Suppose $G$ and $H$ are finite groups and $P_G$ and $P_H$ are Sylow $p$-subgroups of $G$ and $H$, respectively. Then $P_G \times P_H$ is a $p$-Sylow subgroup of $G \times H$.
Fact $2$: $A \times B \trianglelefteq G \times H$ if and only if $A \trianglelefteq G$ and $B \trianglelefteq H$.
Fact $3$: If there is at least one Sylow $p$-subgroup that is not normal in $G$, then $G$ has no normal Sylow $p$-subgroup.
Fact $4$: $S_4$ does not have a normal Sylow $3$-subgroup and $S_3$ does not have a normal Sylow $2$-subgroup.
From this you can conclude that $S_4 \times S_3$ does not have a normal Sylow $2$-subgroup or a normal Sylow $3$-subgroup. |
H: Polynomial over characteristic two finite field of odd degree with certain image
I am trying to construct a polynomial $f \in \mathbb{F}_{2^k}$ of odd degree, such that $\forall x \in \mathbb{F}_{2^k} \exists \alpha \in \mathbb{F}_{2^k}$ such that $f(x)=\alpha ^2 -\alpha$.
Working with some normal basis $\{ \alpha ^{2^i} | 0 \le i \le k-1 \}$, the image of $x \to x^2 -x$ is $\{ \sum c_i \alpha ^{2^i} | \sum c_i = 0, c_i \in \mathbb{F}_{2} \}$. In other words, we require $f(\mathbb{F}_{2^k}) \subseteq Im(t^2-t)$, and this image is a subspace of dimension $k-1$ of the field.
It is easy to construct such $f$ of even degree: $f(x)=x^{2^{i+1}} - x^{2^i}$, for any $i$.
EDIT: I managed to find an example, but I don't like it much: taking $f(x) = x^{2^k + 1} - x$. For every $x \in \mathbb{F}_{2^k}$, $f(x)=x^2-x$.
AI: Another way to state your condition is that, for every $\beta \in \mathbf{F}_{2^k}$, $f(\beta)$ has trace 0 over $\mathbf{F}_2$.
For $k=5$, an example is $f(x) = x^9 - x^5$. To see that this has trace 0 for any $\beta \in \mathbf{F}_{32}$:
$$ \begin{align*} \mathop{Tr}(\beta^9 - \beta^5)
&=
\mathop{Tr}(\beta^9) - \mathop{Tr}(\beta^5)
\\&=
(\beta^9 + \beta^{18} + \beta^{36} + \beta^{72} + \beta^{144})
- (\beta^5 + \beta^{10} + \beta^{20} + \beta^{40} + \beta^{80})
\\&=
(\beta^9 + \beta^{18} + \beta^{5} + \beta^{10} + \beta^{20})
- (\beta^5 + \beta^{10} + \beta^{20} + \beta^{9} + \beta^{18})
\\&= 0 \end{align*}
$$
In general, the condition means that
$$ f(\beta) + f(\beta)^2 + \cdots + f(\beta)^{2^{k-1}} = 0 $$
for every $\beta \in \mathbf{F}_{2^k}$, or equivalently,
$$ f(x) + f^\sigma(x^2) + f^{\sigma^2}(x^4) + \cdots + f^{\sigma^{k-1}}(x^{2^{k-1}}) \equiv 0 \pmod{ x^{2^k} - x} $$
where $f^\sigma$ is the polynomial formed by conjugating each coefficient of $f$ by the action $u \mapsto u^2$.
We can split the powers of $x$ into orbits under the action $x \mapsto x^2 \pmod{x^{2^k} - x}$ -- i.e. writing
$$ f(x) = u + h(x) (x^{2^k} - x) + \sum_m g_m(x^m) $$
where $u$ is a constant with trace 0, and each $g$ has the form
$$g(x) = \sum_{j=0}^{k-1} c_j x^{2^j}.$$
Each $g(x)$ must satisfy the constraint, and any choice of valid $g's$ gives a valid $f$. So,
$$0 \equiv \sum_{i=0}^{k-1} g^{\sigma^i}(x^{2^i})
\equiv \sum_{i=0}^{k-1} \sum_{j=0}^{k-1} c_j^{2^i} x^{2^{i+j}}
= \sum_{j=0}^{k-1} x^{2^{j}} \sum_{i=0}^{k-1} c_{j-i}^{2^{i}}
\pmod{x^{2^k} - x}
$$
(there was a change of variable $j \mapsto j-i$) where the index on $c$ is taken modulo $k$. Each coefficient gives the same condition:
$$\sum_{i=0}^{k-1} c_{k-1-i}^{2^i} = 0$$
Now, how can we use this to find a polynomial of small, odd degree? Consider $k=5$. One of the orbits of powers of x is:
$$ x^5, x^{10}, x^{20}, x^{40} \equiv x^9, x^{80} \equiv x^{18} $$
By setting the coefficients on $x^9$ and a smaller term to $1$ and the rest to $0$ will satisfy the equation. This is the example at the top of my answer. |
H: Idea behind factoring an operator?
Suppose if we have an operator $\partial_t^2-\partial_x^2 $ what does it mean to factorise this operator to write it as $(\partial_t-\partial_x) (\partial_t+\partial x)$ When does it actually make sense and why ?
AI: In the abstract sense, the decomposition $x^2-y^2=(x+y)(x-y)$ is true in any ring where $x$ and $y$ commute (in fact, if and only if they commute).
For sufficiently nice (smooth) functions, differentiation is commutative, that is, the result of derivation depends on the degrees of derivation and not the order in which we apply them, so the differential operators on a set of smooth ($C^\infty$) functions (or abstract power series or other such objects) form a commutative ring with composition, so the operation makes perfect sense in that case in a quite general way.
However, we only need $\partial_x\partial_t=\partial_t\partial_x$, and for that we only need $C^2$. Of course, differential operators on $C^2$ do not form a ring (since higher order derivations may not make sense), but the substitution still is correct for the same reasons. You can look at the differentiations as polynomials of degree 2 with variables $\partial_\textrm{something}$.
For some less smooth functions it might not make sense. |
H: Compute Angle Between Quaternions (in Matlab)
I am working on a project where I have many quaternion attitude vectors, and I want to find the 'precision' of these quaternions with respect to each-other.
Without being an expert in this type of thing, my first thought is to find the angle between each (normalized) quaternion, and then find the RMS of that angle. That will give a measure of the precision of our attitude measurements. I was going to use a simple dot product to get this angle.
I don't know that this is a good solution though. Don't know if the dot product will necessarily work between quaternions though.
Note that just want a scalar representation of the pointing precision, and that the differences will on the order of arcseconds. I had also thought to convert to Euler Angles and then just use the vector part of that the Euler Angles.
Doing the math in MATLAB, but really I'm just looking for the theory behind it.
Thanks for the help!
AI: For small differences the Euclidian distance between the two vectors is sufficient. To get an exact answer you would have to use the following process. Assume your quaternions $x$ and $y$ are represented as $x = [x_0, x_1, x_2, x_3]$ and $y = [y_0, y_1, y_2, y_3]$ and that they are unit quaternions. Then let $\operatorname{inv}()$ denote the inverse of a quaternion which for unit quaternions is equivalent to the conjugate (i.e. $ \operatorname{inv}(x)=\operatorname{conj}(x) = [x_0, -x_1, -x_2, -x_3]$). Then the quantity that captures the true difference is $z = x*\operatorname{conj}(y)$. We can then recover the angle using $\theta = 2\arccos(z_0)$. Then $\theta$ gives you an angle by which the two quaternions differ. |
H: If $ u \in W^{2,3} ( \Omega ) $ then $u \in L^3 ( \Omega )$?
If $ u \in W^{2,3} ( \Omega ) $ then $u \in L^3 ( \Omega )$ ?
In wikipedia, the definition of Sobolev space is $$ W^{k,p} ( \Omega) = \{ u \in L^p ( \Omega)\mid D^{\alpha} u \in L^p , | \alpha| \leqslant k \},$$ where $ \Omega $ is an open set in $\mathbb R^n$. So I think it's trivial, is this right?
AI: The set-builder notation you see causes $W^{k,p}(\Omega)$ to consist of functions in $L^p(\Omega)$. |
H: What's the difference between a curve and a graph?
My book states "... if $S$ is a level curve and $C$ is a curve in $S$ passing through a point $a$ ..."
What is a curve in $S$? Is it simply a subset of $S$?
From what I comprehend, a graph is usually a ordered set of the form
$ X = \{(a,f(a))\ |\ a \in \mathrm{dom}(f)\}$
AI: By "curve in $S$" your book means a restriction of $S$ which still represents a curve. For example, if we consider the curve described by
$$
(x,y) = (\sin\theta, \cos\theta)
$$
for $\theta \in [0, 2\pi]$, then an example of a curve "inside" this curve would be
$$
(x,y) = (\sin\theta, \cos\theta)
$$
for $\theta \in [0, \pi/2]$.
Probably the reason your book worded it this way is because it doesn't want to include every such restriction. For instance, restricting the domain to $[0,2\pi ] \cap \mathbb{Q}$ gives a subset of the curve, but not a "subcurve" (this terminology is not used as far as I know, but I didn't know what else to call it.)
Note that in introductory calculus courses often times there is no distinction made between a curve, the range of a curve, the graph of a curve, and the parametric representations of a curve. This is often times a source of confusion (as it may have contributed here), but the distinction should be properly handled in a more rigorous analysis setting.
You are correct that the graph of a curve is given by $\{(a,f(a)) | a \in \mathrm{dom}(f)\}$ |
H: Quick question about comparing functions.
Suppose I have functions f and g; is there any theorem or technique that I can use to know if $f(x) \leq g(x)$ or if $f(x) \geq g(x)$ or neither in a given interval?
AI: I don't know if this is helpful and it doesn't solve your general problem, but as an example: if you have sat two real valued (differentiable) functions $f$ and $g$ defined on the same say closed interval $[a, b]$ of the real numbers, you could for example note that say $f(a) > g(a)$ for the left endpoint of the interval and then note that the derivative of $f$ is greater than or equal to the derivative of $g$ on the whole interval. If this is all so, then you would be able to conclude that $f(x) > g(x)$ for all x in the interval. But again, this approach would only apply to a small set of functions.
Another idea could be to (as above) first note that $f$ is greater than $g$ at some point in the interval and then show that $f(x) = g(x)$ does not have any solutions. Then $f$ clearly would be greater than $g$ on the whole interval. |
H: Summations manipulation: is this one right?
I've got a summation like this:
$\sum_{l=1}^L \sum_{i=1}^I p_l c_l^i = \sum_{l=1}^L \sum_{i=1}^I p_l w_l^i$
Is it right to bring $ p_l $ out of the symbol $ \sum_{i=1}^I $ such that:
$\sum_{l=1}^L p_l \sum_{i=1}^I c_l^i = \sum_{l=1}^L p_l \sum_{i=1}^I w_l^i$ ?
Thank you in advance
AI: Yes that is correct. Since the $p_l$ does not depend on $i$, you can bring it out. For a fixed $l$ the $p_l$ is just a common factor of all the terms in the sum $\sum_{i=1}^I$ where $i$ varies. |
H: Exact DE with inital value
Solve the given initial value problem and determine at least approximately where the solution is valid $(2x - y) dx + (2y- x) dy = 0, y(1) = 3 $.
My Solution: Let $M=2x-y \Rightarrow M_y=-1$ and let $N=2y-x\Rightarrow N_x=-1$, therefore the given DE is exact. Let $\Psi_x=M=2x-y\Rightarrow \Psi = x^2-xy+h(y)\Rightarrow \Psi_y=-x+h'(y)\Rightarrow h'(y)=\Psi_y+x=2y-x+x=2y \Rightarrow \Psi=x^2-xy+2y=C$, initial value condition given is $y(1)=3 \Rightarrow C=(1)^2-(1)(3)+2(3)=4 \Rightarrow x^2-xy+2y=4\Rightarrow y=\frac{4-x^2}{2-x}=2+x$, but the given answer is $y = [x+\sqrt{28-3x^2}]/2, |x|<\sqrt{28/3} $
AI: You have made a small error,
$h'(y) = 2y$ tells us that $h(y) = y^2+K$, so you should have gotten
$$\Psi = x^2-xy+y^2=C.$$
I think you know how to solve the problem from here. |
H: Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.
I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$
I am trying to go ahead of this step.
AI: Let $x_n=\big(1+\frac{1}{n}\big)^n$ be the $n^\text{th}$ element of the sequence. Then
$$
\begin{align}
\frac{x_{n+1}}{x_{n}}
&=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^{n}}
\\&=\bigg(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)
\\&=\bigg(1-\frac{1}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)
\\&≥^*\bigg(1-\frac{n}{(n+1)^2}\bigg)\bigg(1+\frac{1}{n+1}\bigg)
\\&>\bigg(1-\frac{n+1}{(n+1)^2}\bigg)\bigg(1+\frac{1}{n+1}\bigg)
\\&=1
\end{align}$$
The $\ge^*$ step follows from Bernoulli's inequality, which says that $(1+x)^m\ge 1+mx $ whenever $m\in \mathbb N$ and $x\ge -1$. |
H: Parametrise curve by angle and convex curves
Can one parametrise any closed curve by the angle its tangent makes to the $x$-axis? I seem to remember that this is only possible for convex curves. Could anyone tell me why, please?
Also is necessarily true that for convex curves, the mean curvature is always positive? I can see why since the mean curvature is kinda like a second derivative and for convex functions the second derivative is positive..
AI: A parametric representation of a curve $\gamma\subset{\mathbb R}^2$ is an at least continuous function
$${\bf f}:\quad t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)\ .$$
A priori the independent variable $t$ is not required to have any geometrical interpretation and can be thought of as "time".
It is a fact of life that any curve $\gamma$, which is a static or "drawn" geometrical object, has many equivalent parametric representations, among them some where the independent variable has a geometric meaning connected with the curve; e.g. it could be the $x$-coordinate of the running point, the arc length along the curve, measured from a starting point, etc. One such "geometrical" variable could be the changing tangential angle $\theta={\rm arg}(\dot x,\dot y)$, but it is not suitable for all curves. It is definitely necessary that the above function ${\bf f}$ is well defined, and this means that to any $\theta\in{\mathbb R}$ (or $\in{\mathbb R}/(2\pi)$) should correspond at most one point of the curve having this tangential angle $\theta$. Such is the case for strictly convex curves, but not for curves containing pieces of lines or for a sinusoid $x\mapsto(x,\sin x)$ $\ (-\infty<x<\infty)$. |
H: Spherical shell enclosing the Earth's surface
What is the minimum thickness of a spherical shell that encloses the Earth's surface?
The two spheres are concentric, the outer sphere encloses all the surface (to the highest mountain), and the inner sphere excludes all the surface (down to the deepest ocean trench).
Yes, I know this is not strictly a mathematics question, but it is mathematical in some extended sense. Of course, just subtracting the highest mountain peak from the deepest ocean
trench ignores the oblate spheroid shape of the Earth. I am curious to learn how close the Earth is to a sphere, and the shell thickness divided by, say, the inner radius, is one measure of that closeness. Thanks!
AI: Three notes:
The size of the equatorial bulge is 42.72 km.
It is a generally-accepted fact that the farthest point from the Earth's center is the peak of Chimborazo, in Ecuador. This is very close to the equator, and 6310 m above sea level.
The deepest point in the Arctic Ocean is 5450 m below sea level. I'm guessing this is probably the closest point to the Earth's center -- obviously it's closer than anything Antarctic, and the bulge is so large relative to the size of anything geographical that I suspect the greater depth of any trenches would be beaten out by their greater distance from the poles. If I'm wrong about this, the obvious candidate to check would be the Aleutian Trench (which has a maximum depth of 7679 m).
Adding those first three numbers would give 54.48 km. Assuming I'm right that these are the sites we should be measuring between, this is definitely an overestimate (since Chimborazo isn't on the equator and the deepest point isn't on the pole), but it seems like a reasonable starting point. Unfortunately Wikipedia doesn't give the latitude of any of the oceanic sites we care about, which means you'd have to dig a little deeper to get any kind of exact value... |
H: Compute $\lim_{n\to\infty} \frac{\int_{0}^{1} f(x) \sin^{2n} (2 \pi x) \space dx}{\int_{0}^{1} e^{x^2}\sin^{2n} (2 \pi x) \space dx}$
Suppose that $f$ is continuous on $[0, 1]$. Then calculate the following limit:
$$\lim_{n\to\infty} \frac{\displaystyle\int_{0}^{1} f(x) \sin^{2n} (2 \pi x) \space dx}{\displaystyle\int_{0}^{1} e^{x^2}\sin^{2n} (2 \pi x) \space dx}$$
What should i start with? (high school problem)
AI: For every continuous function $g$ defined on $(-1,1)$ and every $\delta$ in $[0,1]$, define
$$
I_n(g)=\int_{-1}^1g(x)\cos^{2n}\left(\frac\pi2 x\right)\,\mathrm dx,\qquad
J_n(\delta)=\int_{-\delta}^{+\delta}\cos^{2n}\left(\frac\pi2 x\right)\,\mathrm dx.
$$
Fix a continuous function $g$ defined on $(-1,1)$. Since $g$ is continuous, there exists some finite $c$ such that $|g|\leqslant c$ uniformly on $(-1,1)$. Since $g$ is continuous at $0$, for every $(a,b)$ such that $a\lt g(0)\lt b$, there exists some $\delta$ in $[0,1]$ such that $a\lt g\lt b$ uniformly on $(-\delta,\delta)$. Hence,
$$
aJ_n(\delta)-c(J_n(1)-J_n(\delta))\leqslant I_n(g)\leqslant bJ_n(\delta)+c(J_n(1)-J_n(\delta)).
$$
Here is a basic but crucial fact:
For every $\delta$ in $[0,1]$, $J_n(1)-J_n(\delta)\ll J_n(\delta)$ when $n\to\infty$.
Now, fix some continuous functions $g$ and $h$ defined on $(-1,1)$ such that $h(0)\ne0$, say, $h(0)\gt0$. Assume without loss of generality that $g\gt0$ everywhere (if necessary, add to $g$ a large multiple of $h(0)$). Then, for every positive $(a,b,a',b')$ such that $a\lt g(0)\lt b$ and $a'\lt h(0)\lt b'$,
$$
\frac{a}{b'}\leqslant\liminf\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}\leqslant\limsup\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}\leqslant\frac{b}{a'}.
$$
Hence,
$$
\lim\limits_{n\to\infty}\frac{I_n(g)}{I_n(h)}=\frac{g(0)}{h(0)}.
$$
To solve the question asked, apply the result above to the functions $g$ and $h$ defined on $(-1,1)$ by $g(x)=f(\frac14(x+1))+f(\frac14(x+3))$ and $h(x)=e(\frac14(x+1))+e(\frac14(x+3))$ with $e(t)=\mathrm e^{t^2}$.
Note: To prove the basic but crucial fact mentioned above, one can show that $J_n(\delta)\sim2/\sqrt{\pi n}$ when $n\to\infty$, for every $\delta$ in $(0,1]$. |
H: P, Q, R, S four points lie in a plane and PQ = PR = QR = PS then how many possible values of angle QSR can exist?
P, Q, R, S four points lie in a plane and PQ = PR = QR = PS then how many possible values of angle QSR can exist?
I think 2 values because PQRS is either a square or rhombus.
AI: Let $r = PQ = PR = QR = PS$. The triangle $PQR$ is equilateral with side length $r$ and $S$ is some point on the circle with center $P$ and radius $r$ (this circle also passes through $Q$ and $R$). The angle $QSR$ is therefore half of $QPR$, i.e., $30$ degrees, when $S$ is on the big arc $QR$ of the circle, and is $180$ minus that, i.e., $150$ degrees, when $S$ is on the small arc $QR$ of the circle. |
H: linear transformation matrix
Any help on this linear transformation question is very much appreciated.
Let $V$ denote the real vector space $R^2$ and $\psi : V \rightarrow V$ be a real linear transformation such
that $\psi ((1, 0)) = (11, 8)$ and $\psi ((0, 1)) = (4, 3)$. Express the image $\psi ((x, y))$ of $(x, y)$ in terms
of $x$ and $y$.
Assume that $w_1 = (4, 5)$ and $w_2 = (9,11)$ form an ordered basis $B$ for $V$ . Working from the
denition determine the matrix $M^B_B (\psi)$ with respect to the basis $B$.
does $ M^B_B(\psi)=
\begin{bmatrix}
-469 & -1048 \\
388 & 867
\end{bmatrix}$ ?
thanks in advance for any help
AI: Denote $E$ the standard basis $(1,0)$, $(0,1)$. By definition, the matrix $M^E_E(\psi)$ (using your notation) is given by
$$M^E_E(\psi)=
\begin{bmatrix}
11 & 4 \\
8 & 3
\end{bmatrix}.$$
Now, the transition matrix from $B$ to $E$ is given by
$$M^E_B(id)=
\begin{bmatrix}
4 & 9 \\
5 & 11
\end{bmatrix}.$$
Hence the transition matrix from $E$ to $B$ is given by
$$M^B_E(id)=
\begin{bmatrix}
4 & 9 \\
5 & 11
\end{bmatrix}^{-1}=\frac{1}{-1}\begin{bmatrix}
11 & -9 \\
-5 & 4
\end{bmatrix}=\begin{bmatrix}
-11 & 9 \\
5 & -4
\end{bmatrix}.$$
Combining all these, the matrix $M^B_B(\psi)$ is given by
$$M^B_B(\psi)=M^B_E(id)\cdot M^E_E(\psi)\cdot M^E_B(id)=\begin{bmatrix}
-11 & 9 \\
5 & -4
\end{bmatrix}\cdot\begin{bmatrix}
11 & 4 \\
8 & 3
\end{bmatrix}\cdot\begin{bmatrix}
4 & 9 \\
5 & 11
\end{bmatrix}=...$$ |
H: Proof that every normed vector space is a topological vector space
The topology induced by the norm of a normed vector space is such that the space is a topological vector space.
Can you tell me if my proof is correct? Of course we have to show that addition and scalar multiplication are continuous with respect to the product topology (induced by the norm).
(1) To show that $(x,y) \mapsto x + y$ is continuous let $\varepsilon > 0$. I can show that the norm $\|\cdot\|_{V \times V}: V \times V \to \mathbb R$ defined as $\|(x,y) - (x_0, y_0) \| = \|x-x_0\| + \|y_0 - y\|$ induces the same topology as the product topology on $V \times V$. Hence we can choose $\delta = \varepsilon$ to get $$ \| (x+y) - (x_0+y_0)\| \leq \|x-x_0\| + \|y-y_0\| < \delta = \varepsilon$$
(2) To show that $V \times K \to V$, $(v, \alpha) \mapsto \alpha v$ is continuous at $(v,\alpha)$, observe that $$\| \alpha v - \beta w\| = \| \alpha v - \beta w + \alpha w - \alpha w\| = \|\alpha(v-w) + (\alpha - \beta) w\| \leq |\alpha| \|v-w\| + |\alpha - \beta| \|w\|$$
Hence $\| \alpha v - \beta w\| < \varepsilon$ if $\|v-w\| < \frac{\varepsilon}{2 |\alpha|}$ and $|\alpha - \beta| < \frac{\varepsilon}{2 \|w\|}$. Unfortunately, the second inequality depends on $w$. How do I make it independent of $w$? Thanks.
AI: The first point is fine. For the second, fix $(v_0,\alpha_0)\in V\times K$ and $\varepsilon >0$. We have to find $\delta>0$ such that if $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$ then $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$. We have
\begin{align}
\lVert \alpha_0v_0-\alpha v\rVert&\leq \lVert \alpha_0v_0-\alpha v_0\rVert+
\lVert \alpha v_0-\alpha v\rVert\\
&=|\alpha_0-\alpha|\lVert v_0\rVert+|\alpha|\lVert v-v_0\rVert\\
&\leq |\alpha_0-\alpha|(\lVert v_0\rVert+\lVert v-v_0\rVert)+|\alpha_0|\lVert v-v_0\rVert.
\end{align}
We take $\delta$ such that $\delta^2+\delta(\lVert v_0\rVert+|\alpha_0|)\leq \varepsilon$ (which is possible).
In this case, $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$ when $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$. |
H: $(a+b)^\beta \leq a^\beta +b^\beta$ for $a,b\geq0$ and $0\leq\beta\leq1$
It seems that $(a+b)^\beta \leq a^\beta +b^\beta$ for $a,b\geq0$ and $0\leq\beta\leq1$. However, I could not prove this nor the same result for a general concave and increasing function (for which it might not hold). If the inequality is true, does it follow from some general inequality or is there some other simple proof?
AI: When $a=b=0$, the result is obvious. Assume that $(a,b)\ne(0,0)$. Then, $t=a/(a+b)$ is in $[0,1]$ hence $t^{\beta}\geqslant t$ because $\beta\leqslant1$. Likewise, $1-t=b/(a+b)$ is in $[0,1]$ hence $(1-t)^{\beta}\geqslant 1-t$. Summing these two inequalities yields $t^{\beta}+(1-t)^{\beta}\geqslant 1$, which is your result. |
H: Maximum-likelihood estimation for continuous random variable with unknown parameter
Let $X$ be a random variable with the unknown parameter $\lambda$ and the following pdf
$$f(t)=2\lambda t\cdot\mathrm e^{-\lambda t^2}\cdot\textbf{1}_{[0,\infty)}(t)$$
where $\textbf{1}_A(x)$ is an indicator function with
$$\textbf{1}_A(x)=\begin{cases}1,&\text{if }x\in A,\\0,&\text{else.}\end{cases}$$
Let $\vec x=(x_1,\ldots,x_n)$ be a sample of $X$. Determine the maximum-likelihood estimator $\widehat{\lambda}$ such that the following is true for the likelihood-function $\mathcal L(\vec x;\lambda)$:
$$\forall \lambda\;:\;\mathcal L(\vec x;\lambda)\leq \mathcal L(\vec x;\widehat\lambda)$$
For the sake of simplicity my first thoughts were to get the log-likelihood this way:
$$\mathcal L(\vec x;\lambda)=\prod\limits_{i=1}^nf(x_i)\implies \ln(\mathcal L(\vec x;\lambda))=\sum\limits_{i=1}^n\ln(f(x_i))$$
This is the point where I'm stuck: i don't know how to compute the derivative to maximize the function $$\frac{\mathrm d \ln(\mathcal L(\vec x;\lambda))}{\mathrm d\lambda}\overset{!}{=}0.$$
Any hints on how to derive the sum would be appreciated.
AI: We have
$$L(\vec x,\lambda)=\prod_{j=1}^n(2\lambda x_j)\cdot e^{-\lambda x_j^2}=2^n\left(\prod_{k=1}^nx_k\right)\lambda ^n\exp\left(-\lambda \lVert x\rVert^2\right)\chi_{x_j\geq 0\forall j},$$
hence assuming the $x_j> 0$.
$$\log L(\vec x,\lambda)=n\log 2+\sum_{j=1}^n\log x_j+n\log\lambda-\lambda\lVert x\rVert^2.$$
Now taking the derivative with respect to $\lambda$, we get
$$\partial_{\lambda}\log L(\vec x,\lambda)=\frac n{\lambda}-\lVert x\rVert^2.$$
I let you finish the computation. |
H: Check if a relation is a partial or total order and find minimum, maximum, minimal and maximal elements
Given $\mathbb Z^-=\{x\in \mathbb Z:x<0\}$ and $T = \mathbb Z^-\times \mathbb N$, let the binary relation $\odot$ be defined as follows:
$$\begin{aligned} (a,b) \odot (c,d) \Longleftrightarrow a \leq c \land b \mid d \end{aligned}$$
Check if the operation $\odot$ sets order, in particular total order and also find minimum, maximum, maximal and minimal elements in $(T, \odot)$ if they exists.
In order for $\odot$ to be a partial order, its reflexivity, antisymmetry and transitivity has to be proved.
Reflexivity
It can be easily shown that $\forall (a,b) \in T$
$$\begin{aligned}(a,b) \odot (a,b) \Longleftrightarrow a \leq a \land b \mid b\end{aligned}$$
so this relation is reflexive.
Antisymmetry
In order for this property to be true the following condition must be valid $\forall (a,b),(c,d) \in T$:
$$\begin{aligned} (a,b) \odot (c,d) \land (c,d) \odot (a,b) \Rightarrow (a,b) = (c,d) \end{aligned}$$
The antisymmetry doesn't apply to this relation because while the following is always true:
$$\begin{aligned} a \leq c \land c \leq a \Rightarrow a = c \end{aligned}$$
we cannot state the same for the following:
$$\begin{aligned} b \mid d \land d \mid b \Rightarrow b = d\end{aligned}$$
because as $b \neq 0 \land d = 0$ (or similarly $b = 0 \land d \neq 0)$ then
$$\begin{aligned}\exists \alpha \in \mathbb Z : \alpha b = d \end{aligned}$$
$$\begin{aligned}\exists \beta \in \mathbb Z : \beta d = b \end{aligned}$$
whereas $\alpha = 0$ but $\nexists \beta \in \mathbb Z : \beta 0 = d$. Hence this relation isn't antisymmetric.
Transitivity
We need to prove
$$\begin{aligned} (a,b) \odot (c,d) \land (c,d) \odot (e,f) \Rightarrow (a,b) \odot (e,f) \end{aligned}$$
it's valid the following:
$$\begin{aligned} a \leq c \land c \leq e \Rightarrow a \leq e \end{aligned}$$
the following is valid as well
$$\begin{aligned} a \mid c \land c \mid f \Rightarrow a \mid f \end{aligned}$$
as $\exists x \in \mathbb Z : ax = c$ and $\exists y \in \mathbb Z : yc = f$
it's safe to say $yxa=f$.
Conclusion: this relation isn't partial order, so in particular isn't a total order.
Is everything ok with my exercise or did I do anything wrong? When it comes to look for maximum, minimum and minimal or maximal elements I feel lost. So I think there's no maximum or minimum because both $\mathbb Z^-$ and $\mathbb N$ are inifinite. Apparently I don't have a clue on how to spot minimal or maximal elements for $T$. Will you please help me out with that?
AI: Actually, it is a partial order. Given natural numbers $a,b$, we can conclude from $a\vert b\wedge b\vert a$ that $a=b.$ Your counterexample doesn't work, because, while everything divides $0$, $0$ only divides itself.
A maximum element would be some $(a,b)\in T$ such that for all $(c,d)\in T$, $c\leq a\wedge d\vert b$. This would require that $a$ be the greatest element of $\Bbb Z^-$ and that $b$ be a natural number that all natural numbers divide. Can you think of what that would be? Note that if there is a maximum element, then it is the only maximal element.
A minimum element would be some $(a,b)\in T$ such that for all $(c,d)\in T$, $c\geq a\wedge b\vert d$. Then $a$ would have to be the least element of $\Bbb Z^-$, which is not possible.
Given any $(a,b)$, we have that $(a-1,b)\odot(a,b)$, so there is always a different element preceding it. Thus, there are no minimal elements, either.
To determine whether or not $\odot$ totally orders $T$, we must determine whether for distinct $(a,b),(c,d)\in T$ we need have $(a,b)\odot(c,d)$ or $(c,d)\odot(a,b)$. It shouldn't be difficult to see that this need not hold (I leave it to you to find an example), so that $\odot$ does not totally order $T$, after all.
Edit: Well done in showing that it isn't a total order, and in finding the maximum element. It is in fact maximum (not just maximal) since for each $(c,d)\in T$ we have $c\leq-1$ and $d\vert 0$. Partial orders may or may not have a maximum or a minimum element, and may or may not have maximal or minimal elements. For an example of a partial order with a maximum element, take any set $A$, any $b\notin A$ and define a partial order $\precsim$ on $A\cup\{b\}$ by $c\precsim c$ for all $c$, and $a\precsim b$ for all $a\in A$. Then $b$ is the maximum element, and each $a\in A$ is incomparable to the others, so minimal, but not minimum if $A$ has multiple elements.
To elaborate on the antisymmetry of your example, let's suppose that $m,n\in\Bbb N$ such that $m\vert n\wedge n\vert m$. Hence, $$m=nv\wedge n=mw$$ for some $v,w\in\Bbb N$. If $m=0$, then so is $n$. If $m\neq 0$, then since $m=nv=mvw$, then $vw=1$, which is only possible for $v=w=1$ since $v,w\in\Bbb N$. Thus, again, $m=n$. Does that clear things up? |
H: Calculating new rotation matrix with its derivative given
I've got a skew-symmetric matrix representing gyroscope measurements, say $\Omega = [p,q,r]^T$, with $p$, $q$, $r$ being the angular velocities around $X$, $Y$ and $Z$ axes. I know my system's dynamics is:
$\dot{R} = R \Omega_\times$
with $\Omega_\times$ being a skew-symmetric matrix built on $\Omega$. How do I integrate $R$ to obtain new rotation matrix from its derivative? I already have the derivative, so just the integration process seems to overwhelm me. Simple addition of $R_{new} = R + \dot{R}$ violates $SO(3)$ group's constraints ($det(R) \neq 1$).
Thanks for any help.
AI: The general solution is $R(t)=R_0\mathrm e^{t\Omega_\times}$, where $R_0$ is any rotation matrix. You can rotate to coordinates in which $\Omega_\times$ takes the form
$$\Omega_\times=\pmatrix{0&\omega&0\\-\omega&0&0\\0&0&0}\;,$$
which leads to
$$\mathrm e^{t\Omega_\times}=\pmatrix{\cos\omega t&\sin\omega t&0\\-\sin\omega t&\cos\omega t&0\\0&0&1}\;.$$ |
H: From an equality to a comparison
Let $X$ be a set. Let $0$ be an element of $X$.
For a function $P$ defined on tuples of $n$ elements of the set $X$ we know (for every tuples $f$ and $g$ each having $n$ elements)
$$\forall i \in n : ( f_i \neq 0 \wedge g_i \neq 0 ) \wedge P f = P
g \Rightarrow f = g.$$
Let $X$ be also a poset with least element $0$.
Under which additional conditions can we prove:
$$\forall i \in n : ( f_i \neq 0 \wedge g_i \neq 0 ) \wedge P f \le P
g \Rightarrow \forall i\in n: f_i \le g_i?$$
It is a practical task to prove it, don't be afraid to assume additional conditions. Maybe we should require that $X$ is a (semi)lattice?
AI: There is no answer to this question. Consider for example when a reverse of the above formula holds:
$$\forall i \in n : ( f_i \neq 0 \wedge g_i \neq 0 ) \wedge P f \le P
g \Rightarrow \forall i\in n: f_i \ge g_i.$$
This case can't be distinguished from what is wanted in the question, without explicitly specifying a partial order for the image of $P$. |
H: If $p, q$, and $r$ are relatively primes, then there exist integers $x$, $y$, and $z$ such that $px + qy + rz = 1$
True/False
If $p, q$, and $r$ are relatively primes, then there exist integers $x, y$, and $z$ such that $px + qy + rz = 1$
NOTE: $p, q$, and $r$ are positive primes.
AI: Let's take the more general question: if $a$, $b$, and $c$ are integers and they are relatively prime, i.e., $\gcd(a,b,c)=1$, but not necessarily pairwise relatively prime, then there exist integers $x,y,z$ such that $ax+by+cz=1$.
Indeed, $\gcd(a,b,c) = \gcd(\gcd(a,b),c)$. Let $d=\gcd(a,b)$; then there exist integers $m$ and $n$ such that $am+bn=d$. And since $\gcd(d,c)=1$, there exist integers $t$ and $z$ such that $dt + cz = 1$. Now substituting the value of $d$, let $x=mt$ and $y=zn$ to get
$$1 = dt + cz = (am+bn)t + cz = a(mt) + b(nt) + cz = ax + by + cz.$$ |
H: Stuck on space curves for vector valued functions
I'm working through the James Stewart Calculus text to prep for school. I'm stuck at this particular point.
How would you sketch the graph for the parametric equations:
$x = \cos t$, $y = \sin t$, and $z = \sin 5t$? I understand that if it were the case that $z=t$, I'd merely get a helix around the $z$-axis, as $x$ and $y$ form an ellipse. However, I cannot make the leap to solve more exotic problems such as the problem posed or even the case when $z = \ln(t)$.
Some help and a push in the right direction would be appreciated.
AI: Here is an animation I made that might help. The left is a plot of $(\cos(t),\sin(t),\sin(5t))$ and the right is a plot of $\sin(5t)$.
For the case of $(\cos(t),\sin(t),\ln(t))$, here is the corresponding animation:
As a sanity check, note that in each animation, you can see that the point on the circle makes its first full revolution as $t=2\pi\approx 6.28$.
Mathematica code for my (and anyone else's) future reference:
size = 1.5
slices = 150
Slice[t_,z_] := {Show[ParametricPlot3D[{Cos[2 Pi*s], Sin[2 Pi*s], z},
{s, 0, 1}, PlotRange -> {{-size, size}, {-size, size}, {-size, size}}],
Graphics3D[{PointSize[Large], Point[{Cos[t], Sin[t], z}]}]],
Show[Plot[Sin[5 s], {s, 0, 2 Pi}, Ticks -> {{0, 2 Pi/5, 4 Pi/5, 6 Pi/5,
8 Pi/5, 2 Pi}}], Graphics[{PointSize[Large], Point[{t, Sin[5 t]}]}]]}
NewSlice[t_,z_] := {Show[ParametricPlot3D[{Cos[2 Pi*s], Sin[2 Pi*s], z},
{s, 0, 1}, PlotRange -> {{-size, size}, {-size, size}, {-2, 2}}],
Graphics3D[{PointSize[Large], Point[{Cos[t], Sin[t], z}]}]],
Show[Plot[Log[s], {s, 0.5, 8}, PlotRange -> {{0, 8}, {-1, 2}},
AspectRatio -> 1/2], Graphics[{PointSize[Large], Point[{t, Log[t]}]}]]}
Export["sin.gif", Table[Slice[2 Pi*t/slices, Sin[5*2 Pi*t/slices]], {t,
0,slices}], "DisplayDurations" -> 0.15]
Export["ln.gif",Table[NewSlice[t, Log[t]], {t, 0.5, 7.5, 7/slices}],
"DisplayDurations" -> 0.15] |
H: If $A^n$ has a free subset of $n+1$ elements, it has an infinite free subset.
I want to prove the following:
Let $A$ be a ring and $n$ a natural number. If the left $A$-module $A^n$ contains a free subset of $n+1$ elements, then $A^n$ already contains an infinite free subset.
Since we can embed $A^{n+1}$ into $A^n$, we can embed $A^{n+2}$ into $A^{n+1}$, and so on. So $A^n$ contains finite free subsets of all sizes, but I really have no clue how to construct an infinite free subset from that.
Can someone help me?
AI: As you have said, we can embed $A^m$ for any $m$. In particular, there is an embedding of $A^{2n}=A^n\oplus A^n$. To clarify a bit, write the embedding as $A_0\oplus A_0'$, both $A_0$ and $A_0'$ submodules of $A^n$ isomorphic to $A^n$.
Then we can inductively construct a sequence: given $A_j,A_j'$ each isomorphic to $A^n$, we can embed $A^{2n}$ into $A_j'$ as $A_{j+1}\oplus A_{j+1}'$. Then $A_\infty=\bigoplus_j A_j$ is free of rank $\aleph_0$. |
H: Show that the limit of functions is continuous
Let $f_n$ be a sequence of not necessarily continuous functions $\mathbb{R} \rightarrow \mathbb{R}$ such that $f_n(x_n) \rightarrow f(x)$ whenever $x_n \rightarrow x$. Show that f is continuous.
What I am trying to do is to show that whenever we have $x \in \mathbb{R}$ and $x_n \rightarrow x$, then $f(x_n) \rightarrow f(x)$, when $n \rightarrow \infty$.
These types of things are usually showed by using the triangle inequality. I know I can make $|f(x) - f_n(x_n)|$ as small as possible by choosing a big enough n. I can also make $|f(x) - f_n(x)|$ as small as possible. But I am not able to combine these to prove that $|f(x) - f(x_n)|$ can be made as small as possible.
AI: First note that the hypothesis implies that the $f_n$ converge pointwise to $f$. To see this, consider the constant sequence $\langle x_n:n\in\Bbb N\rangle$ where $x_n=x$ for each $n\in\Bbb N$: $$\langle f_n(x):n\in\Bbb N\rangle=\langle f_n(x_n):n\in\Bbb N\rangle\to f(x)\;.$$
Now suppose that $f$ is not continuous at $x$, and let $\langle x_n:n\in\Bbb N\rangle\to x$ be such that $\langle f(x_n):n\in\Bbb N\rangle$ does not converge to $f(x)$. Then there is an $\epsilon>0$ such that $|f(x_n)-f(x)|\ge\epsilon$ for infinitely many $n\in\Bbb N$, so you can find a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ such that $|f(x_{n_k})-f(x)|\ge\epsilon$ for every $k\in\Bbb N$. Since $\langle x_{n_k}:k\in\Bbb N\rangle\to x$, you might as well assume from the start that you have a sequence $\langle x_n:n\in\Bbb N\rangle$ and an $\epsilon>0$ such that $\langle x_n:n\in\Bbb N\rangle\to x$ and $|f(x_n)-f(x)|\ge\epsilon$ for all $n\in\Bbb N$.
By hypothesis $\langle f_n(x_n):n\in\Bbb N\rangle\to f(x)$. Choose $n_0\in\Bbb N$ so that $|f_n(x_0)-f(x_0)|<\epsilon/2$ for all $n\ge n_0$; we can do this, since the $f_n$’s converge pointwise to $f$. Now choose $n_1>n_0$ so that $|f_n(x_1)-f(x_1)|<\epsilon/2$ for all $n\ge n_1$. Continue in this way to construct an increasing sequence $\langle n_k:k\in\Bbb N\rangle$ such that $|f_n(x_k)-f(x_k)|<\epsilon/2$ for all $n\ge n_k$.
Now form a new sequence $\langle y_n:n\in\Bbb N\rangle$ as follows:
$$y_n=\begin{cases}
x_0,&\text{if }n\le n_0\\
x_k,&\text{if }n_{k-1}<n\le n_k\text{ for some }k\ge 1\;.
\end{cases}$$
It’s not hard to see that $\langle y_n:n\in\Bbb N\rangle\to x$: it’s just the sequence $\langle x_n:n\in\Bbb N\rangle$ with each term repeated some finite number of times. Note that $y_{n_k}=x_k$ for every $k\in\Bbb N$. Thus, for each $k\in\Bbb N$ we have $f_{n_k}(y_{n_k})=f_{n_k}(x_k)$, which by the choice of $n_k$ implies that $|f_{n_k}(y_{n_k})-f(y_{n_k})|<\epsilon/2$.
Now $\langle f_n(y_n):n\in\Bbb N\rangle\to f(x)$, so $\langle f_{n_k}(y_{n_k}):k\in\Bbb N\rangle\to f(x)$, and there must be a $k\in\Bbb N$ such that $|f_{n_k}(y_{n_k})-f(x)|<\epsilon/2$. But then
$$|f(y_{n_k})-f(x)|\le|f(y_{n_k})-f_{n_k}(y_{n_k})|+|f_{n_k}(y_{n_k})-f(x)|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon\;,$$
which is a contradiction: $|f(y_{n_k})-f(x)|=|f(x_k)-f(x)|\ge\epsilon$.
Thus, $f$ must in fact be continuous at $x$. |
H: Closed form solution of Fibonacci-like sequence
Could someone please tell me the closed form solution of the equation below.
$$F(n) = 2F(n-1) + 2F(n-2)$$
$$F(1) = 1$$
$$F(2) = 3$$
Is there any way it can be easily deduced if the closed form solution of Fibonacci is known?
AI: Any of the standard methods for solving such recurrences will work. In particular, whatever method you would use to get the Binet formula for the Fibonacci numbers will work here, once you establish initial conditions. If you set $F(0)=0$ and $F(1)=1$, as with the Fibonacci numbers, the closed form is
$$F(n)=\frac{(1+\sqrt3)^n-(1-\sqrt3)^n}{2\sqrt3}\;;$$
I don’t see any way to derive this directly from the corresponding closed form for the Fibonacci numbers, however.
By the way, with those initial values the sequence is OEIS A002605.
Added: The general solution is $$F(n)=A(1+\sqrt3)^n+B(1-\sqrt3)^n\;;\tag{1}$$ Argon’s answer already shows you one of the standard methods of obtaining this. To find $A$ and $B$ for a given set of initial conditions, just substitute the known values of $n$ in $(1)$. If you want $F(1)=1$, you must have $$1=F(1)=A(1+\sqrt3)^1+B(1-\sqrt3)^1\;,$$ or $A+B+\sqrt3(A-B)=1$. To get $F(2)=3$, you must have $$\begin{align*}3&=F(2)=A(1+\sqrt3)^2+B(1-\sqrt3)^2\\
&=A(4+2\sqrt3)+B(4-2\sqrt3)\;,
\end{align*}$$
or $4(A+B)+2\sqrt3(A-B)=3$. You now have the system
$$\left\{\begin{align*}&A+B+\sqrt3(A-B)=1\\
&4(A+B)+2\sqrt3(A-B)=3\;.
\end{align*}\right.$$
Multiply the first equation by $2$ and subtract from the second to get $2(A+B)=1$, and multiply the first equation by $4$ and subtract the second from it to get $2\sqrt3(A-B)=1$. Then you have the simple system $$\left\{\begin{align*}&A+B=\frac12\\&A-B=\frac1{2\sqrt3}\;,\end{align*}\right.$$ which you should have no trouble solving for $A$ and $B$. |
H: Pythagorian quadruples
From my work on hyperelliptic equations I found how to get infinitely many solutions of the equation $a^4+b^4+c^2=d^4$. I call these solutions harmonic:
$$\begin{array}{rcccccl}
1^4 &+& 2^4 &+& 8^2 &=& 3^4\\
2^4 &+& 3^4 &+& 48^2 &=& 7^4\\
3^4 &+& 4^4 &+& 168^2 &=& 13^4\\
4^4 &+& 5^4 &+& 440^2 &=& 21^4
\end{array}$$
and so on. All numbers natural.
Does anyone know if there are infinite non harmonic solutions of this equation?
AI: EDIT
There seems to be a lot of non-harmonic solutions.
Here are couple of one parameter family of non-harmonic solutions.
$b = a(a-1)$, $d = a^2 - a + 1$ and $c = (a-1)(2a^2 - a +1)$ which relies on the identity $$a^4 + \left( a(a-1)\right)^4 + \left((a-1)(2a^2-a+1) \right)^2 = (a^2 - a + 1)^2$$ You could scale these up appropriately i.e. $$\left(ka,ka(a-1),k^2(a-1)(2a^2 - a + 1),k \left(a^2-a+1 \right) \right),$$ to get other solutions.
$b = a(a+1), d = a^2 + a + 1$ and $c = (a+1)(2a^2+a+1)$ which relies on the identity $$a^4 + \left( a(a+1)\right)^4 + \left((a+1)(2a^2+a+1) \right)^2 = (a^2 + a + 1)^2$$ You could again scale these up appropriately i.e. $$\left(ka,ka(a+1),k^2(a+1)(2a^2 + a + 1),k \left(a^2+a+1 \right) \right),$$ to get other solutions.
You could also take your harmonic solution $(a,a+1,a(a+1)(a^2 + a + 2),a^2+a+1)$ and scale appropriately, i.e. $$\left(ka,k(a+1),k^2a(a+1)(a^2 + a + 2),k \left(a^2+a+1 \right) \right),$$ to get other solutions.
Yes. Below is a one parameter family of infinite solutions.
$b = a+1$, $d = a^2+a+1$, $c = (a^2+a+1)^2-1$.
$$b^4 = (a+1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1$$
$$d^4 = (a^2 + a +1)^4 = 1+4 a+10 a^2+16 a^3+19 a^4+16 a^5+10 a^6+4 a^7+a^8$$
Hence,
\begin{align}
d^4 - b^4 - a^4 & = 4 a^2+12 a^3+17 a^4+16 a^5+10 a^6+4 a^7+a^8\\
& = a^2 \left(4 +12 a+17 a^2+16 a^3+10 a^4+4 a^5+a^6 \right)\\
& = a^2 (a+1)^2 (a^2 + a + 2)^2
\end{align}
Hence, choose $c = a(a+1)(a^2+a+2) = (a^2 + a + 1 -1)(a^2 + a + 1 +1) = \left( \left(a^2 + a + 1 \right)^2 -1 \right)$ |
H: To show sum of residues of $f(z)$ over all poles is $0$
Let $p(z)$ and $q(z)$ be relatively prime polynomials with complex co-efficients so that $deg(q(z))\ge deg(p(z))+2$ and let $f(z)=p(z)/q(z)$. We need to show that the sum of residues of $f(z)$ over all poles is $0$
Well, I tried like this:
by Residue theorem: If $f$ is analytic in a domain except for isolated singularities at $a_1,\dots a_k$ then for any closed contour $\gamma\in D$ on which none of the points $a_k$ lie, we have $$\frac{1}{2\pi i}\int_{\gamma}f(z)dz=\sum_{1}^{k}n(\gamma;a_k)Res[f(z);a_k]$$
as $p$ and $q$ are relatively prime to each other we have $r,s$ such that $p(z)r(z)+q(z)s(z)=1$
$$\frac{1}{2\pi i}\int_{\gamma}f(z)dz=\sum_{1}^{k}Res[f(z);a_k ]$$
$$\frac{1}{2\pi i}\int_{\gamma}\frac{p(z)}{q(z)}dz=\sum_{1}^{k}n(\gamma;a_k)Res[f(z);a_k]$$
Now, I am confused where to use the given facts, should I replacing $p(z)$ from the relatively prime condition? and how to implement the given degree condition? thank you for help
AI: Hint: Let the contour be a circle of radius $R$ (large enough to contain all the poles). Let $R\to\infty$. What happens to the integral? |
H: Is there easier way to calculate the limit of this function?
$$
\lim_{K\rightarrow\infty}\frac{(1-\epsilon)^K}{1+(1-\epsilon)^K}\frac{\sum_{i=1}^{\frac{K-1}{2}}\left(\begin{array}{l}
K \\
i \end{array}\right)\left[\left(\frac{2\epsilon-\epsilon^2}{(1-\epsilon)^2}\right)^i-\left(\frac{\epsilon}{1-\epsilon}\right)^i\right]}{(1-\epsilon)^K\left[1+\sum_{i=1}^{\frac{K-1}{2}}\left(\begin{array}{l}
K \\
i \end{array}\right)\left(\frac{2\epsilon-\epsilon^2}{(1-\epsilon)^2}\right)^i\right]+\left[1+\sum_{i=1}^{\frac{K-1}{2}}\left(\begin{array}{l}
K \\
i \end{array}\right)\left(\frac{\epsilon}{1-\epsilon}\right)^i\right]}=?
$$
where $1>\epsilon>0$. I wonder especially when $\epsilon\rightarrow 0$ but $\epsilon\neq 0$
I even dont know if this is a difficult or an easy question for a mathematician. If you could comment on this matter, I will be happy.
AI: An idea which leads directly to the solution is to write the sums involved as (multiples of) probabilities of events involving some binomial random variables $X_K$ and $Y_K$ with respective parameters $(K,\eta)$ and $(K,\epsilon)$, where $\eta=2\epsilon-\epsilon^2$. To wit, the ratio of interest $R_K$ can be rewritten as
$$
R_K=\frac{(1-\epsilon)^K}{1+(1-\epsilon)^K}\cdot\frac{A_K-B_K}{(1-\epsilon)^K\cdot A_K+B_K},
$$
with
$$
A_K=\sum_{i=0}^{\frac{K-1}{2}}{K\choose i}\left(\frac{\eta}{1-\eta}\right)^i=(1-\eta)^{-K}\cdot\mathrm P(X_K\leqslant\tfrac12(K-1)),
$$
and
$$
B_K=\sum_{i=0}^{\frac{K-1}{2}}{K\choose i}\left(\frac{\epsilon}{1-\epsilon}\right)^i=(1-\epsilon)^{-K}\cdot\mathrm P(Y_K\leqslant\tfrac12(K-1)).
$$
If $\eta\lt\frac12$, then $\epsilon\lt\frac12$ and, by the (weak) law of large numbers, $\mathrm P(X_K\leqslant\tfrac12(K-1))\to1$ and $\mathrm P(Y_K\leqslant\tfrac12(K-1))\to1$ when $K\to\infty$.
Thus,
$$
(1-\epsilon)^{2K}\cdot A_K=(1-\eta)^{K}\cdot A_K\to1,\qquad
(1-\epsilon)^{K}\cdot B_K\to1.
$$
Finally, for every $\epsilon$ such that $\eta=2\epsilon-\epsilon^2\lt\frac12$, that is, for every $\epsilon\lt1-\frac{\sqrt2}2=0.393$,
$$
\lim\limits_{K\to\infty}R_K=\tfrac12.
$$ |
H: Polynomial-related manipulation
My question is:
Factorize: $$x^{11} + x^{10} + x^9 + \cdots + x + 1$$
Any help to solve this question would be greatly appreciated.
AI: $$
\begin{align}
& {}\quad (x^{11} + x^{10}) + (x^9 + x^8)+(x^7+x^6)+(x^5+x^4)+(x^3+x^2 )+( x + 1)\\[8pt]
& =x^{10}(x+1)+x^8(x+1)+x^6(x+1)+x^4(x+1)+x^2(x+1)+(x+1)\\[8pt]
& =(x+1)(x^{10}+x^8+x^6+x^4+x^2+1)\\[8pt]
& =(x+1)(x^8(x^2+1)+x^4(x^2+1)+x^2+1)\\[8pt]
& =(x+1)((x^2+1)(x^8+x^4+1))\\[8pt]
& =(x+1)(x^2+1)(x^4+1-x^2)(x^4+1+x^2)\\[8pt]
& =(x+1)(x^2+1)(x^4+1-x^2)(x^2+1-x)(x^2+1+x)
\end{align}
$$ |
H: Splitting field and subextension
Definition: Let $K/F$ be a field extension and let $p(x)\in F[x]$,
we say that $K$ is splitting field of $p$ over $F$ if $p$ splits
in $K$ and $K$ is generated by $p$'s roots; i.e. if $a_{0},...,a_{n}\in K$
are the roots of $p$ then $K=F(a_{0},...a_{n})$.
What I am trying to understand is this: in my lecture notes it is
written that if $K/E$,$E/F$ are field extensions then $K$ is splitting
field of $p$ over $F$ iff $K$ is splitting field of $p$ over $E$.
If I assume $K$ is splitting field of $p$ over $F$ then
$$\begin{align*}K=F(a_{0,}...,a_{n})\subset E(a_{0,}...,a_{n})\subset K &\implies F(a_{0,}...,a_{n})=E(a_{0,}...,a_{n})\\
&\implies K=E(a_{0,}...,a_{n}).
\end{align*}$$
Can someone please help with the other direction ? help is appreciated!
AI: This is false. Let $F=\mathbb{Q}$, let $E=\mathbb{Q}(\sqrt{2})$, let $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$, and let $p=x^2-3\in F[x]$. Then the splitting field for $p$ over $E$ is $K$, but the splitting field for $p$ over $F$ is $\mathbb{Q}(\sqrt{3})\subsetneq K$.
Let's say that all fields under discussion live in an algebraically closed field $L$. Letting $M$ be the unique splitting field for $p$ over $F$ inside $L$, then the splitting field for $p$ over $E$ inside $L$ is equal to $M$ if and only if $ME=M$, which is the case if and only if $E\subseteq M$.
In other words, you'll get the same splitting field for $p$ over $F$ and over $E$ if and only if $E$ were already isomorphic to a subfield of the splitting field for $p$ over $F$. When $E$ does not have that property, there is "extra stuff" in $E$ (for example, $\sqrt{2}\in E$ in the example) that will need to also be contained in the splitting field for $p$ over $E$. |
H: Representation of Cyclic Group over Finite Field
The post Irreducible representations of a cyclic group over a field of prime order discusses the irreducible representations of a cyclic group of order $N$ over a finite field $\mathbb{F}_p$ where $N$ does not divide $p$.
Where can I find information about the irreducible representations in the case where $p$ does divide $N$?
(I'm interested in this because I'm wondering if, given a finite dimensional vector space $\mathbb{F}_{p}^{M}$ if there are examples of particularly simple invertible linear operators $T$ such that $T$ does not preserve a subspace. Since the vector space contains finitely many elements it seems that this is the same thing as an irreducible representation of a cyclic group.
However, based on the post I read above, where p does not divide N, it seems like T is just multiplication by a primitive element that generates the extension $\mathbb{F}_{q^{M}}$.)
AI: The question of how to classify all finite-dimensional representations of $C_n$ over an arbitrary field $F$ can be studied using the structure theorem for finitely-generated modules over a principal ideal domain, in this case $F[x]$. The structure theorem asserts that any finitely-generated module is uniquely a finite direct sum of modules of the form $F[x]/p(x)^r$ where $p \in F[x]$ is irreducible and $r$ is a non-negative integer.
If $T$ is an operator acting on $F^k$ for some $n$, then $F^k$ becomes a finitely-generated module over $F[x]$ with $x$ acting by $T$. $T$ gives a representation of the cyclic group $C_n$ if and only if $T^n = 1$, in which case the summands $F[x]/q(x)^r$ in the decomposition of $F^k$ must have the property that $q(x)^r | x^n - 1$.
If $F$ has characteristic $0$ or has characteristic $p$ and $p \nmid n$, then $x^n - 1$ is separable over $F$, hence $r \le 1$ and $F^k$ is a direct sum of irreducible representations, all of which are of the form $F[T]/q(T)$ where $q$ is an irreducible factor of $x^n - 1$ over $F$.
If $F$ has characteristic $p$ and $p | n$, then writing $n = p^s m$ where $p \nmid m$ we have
$$x^n - 1 = (x^m - 1)^{p^s}$$
from which it follows that $r \le p^s$ (but now it is possible to have $r > 1$). If $r > 1$, then the corresponding representation $F[T]/q(T)^r$ is indecomposable and not irreducible, where $q$ is an irreducible factor of $x^m - 1$ over $F$. The irreducible representations occur precisely when $r = 1$. In other words,
The irreducible representations of $C_{p^s m}$, where $p \nmid m$, over a field of characteristic $p$ all factor through the quotient $C_{p^s m} \to C_m$.
One can also see this more directly as follows. If $V$ is an irreducible representation of $C_{p^s m}$ over a field of characteristic $p$ and $T : V \to V$ is the action of a generator, then
$$T^{p^s m} - 1 = (T^m - 1)^{p^s} = 0.$$
Thus $T^m - 1$ is an intertwining operator which is not invertible, so by Schur's lemma it is equal to zero. |
H: integration of a continuous function $f(x) $ and $xf(x)$ is zero
Possible Duplicate:
Prove that $\exists a<b$ s.t. $f(a)=f(b)=0$ when $\int_0^1f(x)dx=\int_0^1xf(x)dx=0$
Suppose that $f:[0,1]\to \mathbb{R}$ is continuous, and that $$\int_{0}^{1} f(x)=\int_{0}^{1} xf(x)=0.$$
How does one prove that $f$ has at least two distinct zeroes in $[0,1]$?
Well, if not say $\forall a,b\in [0,1] \ni a<b$, but $f(a)\neq f(b), f(a)>0$ then there will be a neighborhood of $a$ say $(a-\epsilon,a+\epsilon)$ where $f(x)>0$ and hence the integral will not be equal to $0$, but I don't know where I am using the other integral condition. am I wrong anywhere in my proof? please help.
AI: Define $F(x):= \int_0^xf(t)\,dt$ for $x \in [0,1]$. Then the second integral tells us
$$
\begin{aligned}
0 = \int_0^1xf(x)\,dx \, & = \, xF(x)\Bigr|_0^1-\int_0^1F(x)\,dx
\\
& = 1\,F(1) - 0\,F(0) - \int_0^1F(x)\,dx
\\
& = \int_0^1f(x)dx - \int_0^1F(x)\,dx
\\
& = -\int_0^1F(x)\,dx.
\end{aligned}
$$
By the mean value theorem and continuity of $F$, which follows from continuity of $f$, this tells us that there is $c \in (0,1)$ such that $F(c)=0$. That is,
$$
\int_0^cf(x)\,dx = 0.
$$
It follows that
$$
\int_c^1f(x)\,dx = \int_0^1f(x)\,dx-\int_0^cf(x)\,dx = 0 - 0 = 0
$$
as well.
We then apply the mean value theorem two more times, using continuity of $f$, for the intervals $[0,c]$ and $[c,1]$ to find $a \in (0,c)$ and $b \in (c,1)$ respectively such that $f(a)=f(b)=0$. |
H: Partition of Unity question
I am starting to read the book "Differential Forms in Algebraic Topology" by Bott and Tu.
In the proof of the exactness of the Mayer - Vietoris sequence (Proposition 2.3, page 22 - 23) a partition of unity $\{\rho_U,\rho_V\}$ subordinate to an open cover of two open sets $U,V$ is applied to a function $f$ which is defined on the intersection $U \cap V$.
Then the authors emphasize that $\rho_V \,f$ is a function on $U$. I struggle to see why this is true, I thought $\rho_V$ has support contained in $V$, and so from the picture in the book I have the impression that the function $\rho_U\,f$ is a function with support in $U$. I am aware this must be false, the authors even stress the fact that one has to multiply by the partition function of the other open set! Any help to understand this would be great, many thanks !!
AI: The support (Wikipedia article) of $\rho_V$ is contained in $V$, so that $\rho_V(x)=0$ for any $x\notin V$. Thus, even though $f$ is only defined on $U\cap V$, we can extend $\rho_Vf$ to $U\cap V^c$ by declaring $\rho_Vf(x)=0$ for all $x\in U\cap V^c$ (this extending of the domain of the function is being done implicitly by the authors). This defines $\rho_Vf$ on all of $U=(U\cap V)\cup (U\cap V^c)$. |
H: Different Lifts of the Same Function
I'm just learning algebraic topology and have hit a problem I can't do. Lets say we have two topological spaces $X$ and $Y$ where $X$ is connected, and a continuous function $f:X \to Y$. Let $p:Y^\prime\to Y$ be a covering map of $Y$. Say $f_1^\prime$ and $f_2^\prime$ are two lifts of $f$. I want to prove that if $f_1^\prime(x_0)=f_2^\prime(x_0)$ for some $x_0$ then $f_1^\prime(x)=f_2^\prime(x)$ for all $x$. Could someone point me in the right direction?
AI: I think we need some more hypotheses. Taking $X = \{0, 1\}$ will already give a lot of counterexamples. So I will assume that $X$ is connected. The idea is to show that the subset of $X$ on which your lifts agree is open and closed.
So take $x \in X$, and an evenly covered neighborhood $U$ of $f(x)$. Let $U_1'$ and $U_2'$ be the slices above $U$ containing $f_1'(x)$ and $f_2'(x)$, respectively. Argue that there is an open neighborhood $V$ of $x$ such that $f_1'(V) \subset U_1'$ and $f_2'(V) \subset U_2'$, and that the two lifts either agree or disagree on all of $V$. Let me know if I should say more! |
H: Construction of special $\omega_1$-Aronszajn tree
Problem from Kunen II.40:
The definition is the following: An $\omega_1$-Aronszajn tree $T$ called special iff $T$ is the union of $\omega$ antichains.
Need to prove that $T$ is special iff there is a map $f: T \rightarrow \mathbb{Q}$ such that for $x,y \in T, x < y \rightarrow f(x) < f(y)$, and show that a special Aronszajn tree exist.
The hint is to construct $T$ and $f$ simultaneously by induction.
It seems like I should somehow "pack" the antichains (equivalence classes?), to achieve a map from a "large" tree to a relatively small set. Couldn't advance any furher than that, though.
Any help?
Thanks in advance.
AI: Suppose $T$ is a countable union of antichains. We are going to construct a map $g: T\to 2^\omega$ so that range of $g$ is countable and it is strictly increasing (with respect to the lexicographical ordering on $2^\omega$). Pick a function $f: T\to \omega$ so that $f^{-1}(n)$ is an antichain for all $n$. For $t\in T$, define $g(t)=x$ by: $x(n)=1$ if and only if $n\leq f(t)$ and $\{ s\in T: s\leq t\}\cap f^{-1}(n)\ne\varnothing$. It is easy to verify that $g$ is as required.
To construct a special A-tree. Consider the subtree of the A-tree constructed in Kunen's book which consists of nodes of successor height. |
H: Is the integral closure of a Henselian DVR $A$ in a finite extension of its field of fractions finite over $A$?
This question is related to the one here: A question related to Krull-Akizuki theorem
In the answers to that question, some examples are given of a discrete valuation ring $A$ and a finite (necessarily inseparable) extension of its field of fractions in which the integral closure of $A$ is not finitely generated as an $A$-module. My question is whether there are similar counterexamples with $A$ Henselian (it's possible that the examples given there are Henselian, but I do not understand them well enough to know for sure).
If $A$ is a complete discrete valuation ring, then the integral closure of $A$ in any finite extension of its field of fractions is finite over $A$. For $A$ just Henselian, one can at least say that the integral closure of $A$ in a finite extension of $\mathrm{Frac}(A)$ is a Henselian discrete valuation ring. In the separable case, one knows the integral closure is a principal ideal domain which is finite over $A$, and, since $A$ is Henselian, this implies that the integral closure is a finite product of local rings, hence is itself local. For the inseparable case, the reference I know is Neukirch's book on algebraic number theory. He proves that the valuation on $\mathrm{Frac}(A)$ extends uniquely to any finite extension (with valuation ring the integral closure of $A$), and it is visible from the formula for the extended valuation that it is again discrete.
Maybe an example can be obtained by taking the Henselization of an example as in the question referenced above.
EDIT: I just noticed that Exercise 1 of section II.4 of Serre's Local Fields states that, if every finite purely inseparable extension of $\mathrm{Frac}(A)$ has integral closure finite over $A$, then the completion of $\mathrm{Frac}(A)$ is separable over $\mathrm{Frac}(A)$. So I guess a counterexample would have to be a non-excellent Henselian DVR.
AI: Let $A$ be a DVR with field of fractions $K$ such that for some purely inseparable finite extension $L$, the integral closure $B$ of $A$ in $L$ is not finite over $A$ (so $A$ is non-excellent). Then as in your guess, the henselization $A^h$ of $A$ is not excellent.
First $A^h\otimes_A B$ is the inductive limit of $A'\otimes_A B$ with $A'$ étale over $A$ so that $A'\otimes_A B$ is normal, thus $A^h\otimes_A B$ itself is normal. As $L$ is purely inseparable over $K$ and $K^h:=\mathrm{Frac}(A^h)$ is algebraic separable over $K$, $K^h\otimes_K L$ is a field and it is the field of fractions of $A^h\otimes_A B$. Therefore $A^h\otimes_A B$ is the integral closure of $K^h$ in $K^h\otimes_K L$.
It remains to show that $A^h\otimes_A B$ is not finite over $A^h$. Suppose the contrary. A system of generators involves finitely many elements of $B$. So there exists a finitely generated sub-$A$-module $M$ of $B$ such that $A^h\otimes_A M\to A^h\otimes_A B$ is surjective. So $A^h\otimes_A (B/M)=0$. By the faithful flatness of $A^h$ over $A$, we get $B=M$ is finite over $A$. Contradiction. |
H: Percentages Issue
I am having a problem with the following question could you guys tell me what I am doing wrong?
When the tires of a taxicab are under-inflated, the cab's odometer will read $10\%$ over the true mileage. If the odometer of a cab with under-inflated tires read $m$ miles, what is the actual distance driven? (answer = $10m/11$)
Here is how I am doing it:
$$ m = m - \frac{10m}{100}$$
$$ m = \frac{9m}{10}$$
Could anyone tell me what I am doing wrong ?
AI: Let $x$ be the true mileage; then $$m=x+\frac{10x}{100}=x+\frac{x}{10}=\frac{10x+x}{10}=\frac{11x}{10}\;,$$ so $$x=\frac{10m}{11}\;.$$
What you wrote clearly cannot be right: $m$ cannot be nine-tenths of $m$ unless $m=0$. You seem to have tried to use $m$ to represent both the odometer reading and the true mileage; since those are different, this cannot work. |
H: Am I allowed to realize one object twice within one set-theory?
Say I consider a set theory with the Axioms of Extensionality and the Axiom of Pairing.
As I understand it, stating the axiom allows me to make a definition like
$$(a,b):=\{\{a\},\{a,b\}\}$$
and work with that $(a,b)$ in the context of my theory. Pairing says "it exists" (I can write it down with my language) and Extensionality says the abstract idea of it is unique as a set.
Is that way of thinking correct? Is that the purpose?
Because (if I know $a$ and $b$ exists and since I know what set brackets are) in a way I feel the set $\{\{a\},\{a,b\}\}$ existed already before the existence of a pair was guaranteed by the axiom - the possibility of nesting of sets as for the definition seems to be apriori to me, I asked a related question here.
Secondly, since there are more set-constructions of the ordered pair, like say
$$(a,b)':=\{b,\{a,b\}\}$$
as an alternative, I wonder:
Am I allowed to realize the ordered pair twice in one theory?
Then I could for example put ordered pairs as elements of ordered pairs of the second type and so on.
Is there really only one realization of the ordered pair in say ZFC or are there in fact all thinkable versions in the theory and we just choose one if we prove stuff about the abstract thing (which implies that the statements are true for all models)?
Or another idea: Should I view the whole thing in a way that I only define the thing using a concrete relization so that I can prove stuff about the "actual" abstract object, which is really only implicitly postulated to exist in the axiom.
If that point of view ist true then I don't really see what the real difference of two realizations can be.
AI: I think I wrote this as an answer to one of your previous questions and then I deleted it (I think Henning wrote another answer incorporating the point I was making).
We hardly make actual use of the properties of an ordered pair beyond the fact that it is actually a collection of two elements which may not be distinct and the order does matter.
Much like the proofs in real analysis do not depend on how you interpret the real numbers within a model of ZFC, but rather on the properties of the structure, a similar thing can be said here.
What we write when we write a proof is more of a schema for a proof. We consider abstract (non-pure set) objects and we say "plug the definition here, and insert the definition there". Ordered pairs make an excellent example as they appear almost everywhere. However there are only a few places where you actually care for the contents of the interpretation of the ordered pair.
Most of the time you care about the fact that an ordered pair allows you to distinguish between the two elements, even if they are the same (e.g. $\langle a,a\rangle$). As long as you have a way of telling which is the left coordinate and which is the right there is no real danger in replacing the definition.
However you should remember, again, that every time you are using an instance of the replacement axiom schema in contexts of ordered pairs then the axiom you are using may be different; but the logic behind the proof remains the same. |
H: Product norm on infinite product space
Today I proved that if $V$ is a normed space with norm $\|\cdot\|$ then I can define a norm on $V \times V$ that induces the same topology as the product topology as follows: $\| (v,w) \|_{V \times V} = \|v\| + \|w\|$.
I think I can do the same for an infinite product $V^{\mathbb N}$ by defining $\|(v_n)\|_{\mathbb N} = \sum_{n=0}^\infty \frac{1}{2^n} \|v_n\|$ and I proved it using the proof of the case $V \times V$ and changing some minor things.
Can you confirm that this result is correct? Thanks.
AI: Proposition. If $X_i, i \in \mathbb{N}$ is any sequence of nonzero topological vector spaces, then the product topology on $X = \prod_i X_i$ is not normable.
Proof. Suppose to the contrary there is a norm $\|\cdot\|$ on $X$ which induces the product topology. Let $B \subset X$ be the open unit ball of $\|\cdot\|$. $B$ is open in the product topology and contains 0, so we can find a basic open neighborhood of 0 which is contained in $B$. That is, there is a set of the form $U = U_1 \times U_2 \times \dots \times U_n \times X_{n+1} \times \cdots$, where $U_i \subset X_i$ is open and nonempty, such that $0 \in U \subset B$. Choose any nonzero $x_{n+1} \in X_{n+1}$ and set $x = (0,\dots, 0, x_{n+1}, 0, \dots)$. Then for any $t \in \mathbb{R}$ we have $tx \in U \subset B$. If we take $t = 2/\|x\|$, we have $\|tx\| = 2$ which is absurd since $B$ is the unit ball.
In short, the problem is that every open set of the product topology contains a line, and balls of a norm do not. |
H: Lying-over theorem without Axiom of Choice
This question is motivated by this and this.
Can the following proposition be proved without Axiom of Choice?
Proposition:
Let $k$ be a field.
Let $A$ and $B$ be commutative algebras without zero-divisors which are finitely generated over $k$. Suppose that $A$ is a subring of $B$ and $B$ is integral over $A$.
Let $P$ be a prime ideal of $A$.
Then there exists a prime ideal $Q$ of $B$ such that $P = A \cap Q$.
AI: First we show
Lemma If $A\to B$ is a finite homomorphism of rings with $A$ local and $B\ne 0$, then the maximal ideal $P$ of $A$ is the pre-image of a maximal ideal $Q$ of $B$.
Proof. By Nakayama's lemma, $PB\ne B$. The quotient $B/PB$ is a finite $k$-algebra (where $k$ is the field $A/P$) and is no zero. The set of proper ideals of $B/PB$ is non-empty and has an element of maximal $k$-vector space dimension. The latter is then a maximal ideal, hence equal to $Q/PB$ for some maximal ideal $Q$ of $B$ containing $P$. The pre-image $P'$ of $Q$ is maximal because $A/P'$ is contained in $B/Q$ and the later is finite over $A/P'$. So $P'=P$.
Now we prove your proposition. As $B$ is a finitely generated $A$-algebra, $B$ integral over $A$ implies that $B$ is finite over $A$. Hence $A_P\to A_P\otimes_A B$ is finite with $A_P\otimes_A B\ne 0$. By the above lemma, $PA_P$ is the pre-image of a maximal ideal of $A_P\otimes_A B$. The existence of $Q$ as desired follows from standard arguments on localizations.
This been said, I agree with the second part of Martin Brandenburg's comment. |
H: Set of finite subsets as vector space: Double dual?
Some problem I've found while thinking about duals of vector spaces:
Be $S$ an arbitrary set. Denote by $F(S)$ the set of finite subsets of $S$, and by $P(S)$ its power set.
Now it is easy to see that $F(S)$ forms a vector space over $\mathbb{Z}/2\mathbb{Z}$ if you define vector addition as symmetric difference and multiplication with scalar by the simple relations $0A=\emptyset$ and $1A=A$. A particular basis of that vector space is $\{\{s\}: s\in S\}$.
From that, it is also easy to construct the dual space $F(S)^*$: Its members are simply given by the elements of $P(S)$ with the following application rule: If $\alpha\in P(S)$ and $v\in F(S)$, then $$\alpha(v) = \begin{cases}1 & \text{if $\alpha\cap v$ has an odd number of elements}\\0 & \text{if $\alpha\cap v$ has an even number of elements}\end{cases}$$ (or simply, express the number of elements in $\alpha\cup v$ in $\mathbb{Z}/2\mathbb{Z}$. It is also not hard to show that vector addition in $F(S)^*$ is again the symmetric difference. So one can say that, given the application rule above, $F(S)^* = P(S)$ Note that for finite $S$, $F(S)=P(S)$, while for infinite $S$, $F(S)$ is smaller than $P(S)$.
So far, so good. However, what about the double-dual $F(S)^{**} = P(S)^*$? Since for finite $S$, $P(S)=F(S)$, one would conclude that the very same construction works again. However for infinite $S$, it cannot work, because you cannot say whether an infinite set has an even or odd number of elements. Also, $P(S)$ already contains all subsets of $S$, and as far as I understand, $P(S)^*$ should then be larger than $P(S)$.
Therefore my question: Does there exist a simple representation of $F(S)^{**}=P(S)^*$, and if so, what does it look like? Ideally one should easily see the equivalence to $F(S)$ in the case of finite $S$.
AI: As a vector space, $F(S)$ is a vector space over $\mathbb{F}_2$ with basis indexed by $S$ itself. So it has dimension $|S|$. It is isomorphic to the direct sum of $|S|$ many copies of $\mathbb{F}_2$, and so the dual space is isomorphic to the direct product of $|S|$ many copies of $\mathbb{F}_2$.
As a vector space, $P(S)$ is the direct product of $|S|$ many copies of $\mathbb{F}_2$, so it is indeed isomorphic to $F(S)^*$, and so $F(S)^{**} = P(S)^*$.
This just follows from the universal property: a linear map
$$f\colon \bigoplus_{s\in S}(\mathbb{F}_2)_s\to \mathbb{F}_2$$
is completely determined by what happens to the basis. Specifying what happens to each basis element $1_s$ is equivalent to specifying an element of $\mathop{\prod}\limits_{s\in S}\mathbb{F}_2$, where the map $f$ corresponds to the tuple $(f(1_s))_s$. Conversely, any element $\mathbf{f}$ of the product defines a function $\mathop{\oplus}\limits_{s\in S}(\mathbb{F}_2)_s\to\mathbb{F}_2$ by mapping the basis element $1_s$ to $\mathbf{f}_s$.
Once you've identified $F(S)^*$ with $P(S)$, it follows that $F(S)^{**}$ can be identified with $P(S)^*$. However, in order to make a similar correspondence you would need to start with a basis for $P(S)$ (so as to express it as a direct sum); this is difficult to do (you need AC, I believe, to guarantee the existence of a basis).
Note that when $S$ is infinite, it is difficult to even express $F(S)^{**}$, let alone make it correspond to something "obvious". It takes a bit of cardinal arithmetic to even show that it cannot be isomorphic to $F(S)$.
(Yes, if $V$ is an infinite dimensional vector space, then $V^*$ is a vector space of dimension strictly larger than $V$: see for example this answer; in particular, if $S$ is infinite, then $P(S)$ is an infinite dimensional vector space of dimension strictly larger than $|S|$, and so $P(S)^*$ is itself of dimension strictly larger than $\dim(P(S))$. ) |
H: Angles of a quadrilateral from a ratio.
I cant seem to find the angles here any suggestions ?
A quadrilateral has angles in the ratio 1:2:3 and a fourth angle that is 31 degrees larger than the smallest angle.What is the difference in degree between the middle two angles ?
AI: The angles in the quadrilateral sum to $360^\circ$. Let the smallest angle be $x^\circ$. Solve for $x$ using $x+2x+3x+x+31=360$. |
H: support of a differential form on manifold
In the book "Differential forms in Algebraic Topology" by Bott and Tu, the support of a differential form $\omega$ on a manifold $M$ is defined to be "the smallest closed set $Z$ so that $\omega$ restricted to $Z$ is not $0$." (page 24).
I am a little confused, suppose we let $M = \mathbb{R}$ with the trivial atlas $\{ \mathbb{R}, \text{Id} \}$ and consider the $0-$form $\omega = x$. Then any non - zero point would constitute a set on which the restriction of $\omega$ is non - zero. But I expect the authors want the support to be the smallest closed set containing all the points at which $\omega$ is non - zero. Where is my misunderstanding ? Lots of thanks for help!
AI: Your expectation is the correct definition: the support of a differential form $\omega \in \Omega^k(M)$ is the set
$$\mathrm{supp}(\omega) = \overline{\{p \in M : \omega_p \not\equiv 0\}}.$$
I looked in my copy of Bott and Tu and indeed they defined it incorrectly. Perhaps a better way to phrase their sentence would be "the support of $\omega$ is the smallest closed set $Z$ so that $\omega$ restricted to any point in the interior of $Z$ is not identically $0$," or, as Micah suggests in the comments, "the support of $\omega$ is the smallest closed set $Z$ such that $\omega$ restricted to $M \setminus Z$ is identically $0$." |
H: Proving $\frac{1-q}{q} - \frac{1-q^{x}}{xq^{x}} \geq 0 $
I am trying to find a nice way to verify that whenever $q \in (0,1)$ and $x \in (0,1)$, then $$\frac{1-q}{q} - \frac{1-q^{x}}{xq^{x}} \geq 0 .$$
AI: Take the derivative with respect to $q$ to get $$\frac{q^{1-x} - 1}{q^2}$$ which is negative for $q,x \in (0,1)$. This means that $\frac{1-q}{q} - \frac{1-q^x}{xq^x}$ is monotone decreasing in $q$ (for a chosen $x$), so it's enough to verify that the identity holds in $q=1$, which it does. |
H: Prove pseudoprime $N$ and base a must be relatively prime.
I would really appreciate some hints. Sorry if this is too easy. Thanks sincerely. Also, technically this is not homework but it is a problem from a textbook. Prove:
$a^{N-1} \not \equiv 1($mod$ \ N)$ if the gcd$(a,N)>1$. Where $a,N \in \mathbb{Z}$ and $N \geq 1$.
I have tried by contradiction as follows:
Suppose $a^{N-1} \equiv 1 $ mod$(N)$. Then,
$$ \begin{align}
a^{N} & \equiv a \ \text{mod}(N) \iff \\ cN & = a^{N} - a, \ \text{for some} \ c \in \mathbb{Z}
\end{align} $$
But I can't seem to get a contradiction. It looks like I could try two things here 1) trying to the right side as $a(a^{N-1}-1)$ or rewriting the equation as $a = \lambda a+ \mu N$ for some $\lambda,\mu \in \mathbb{Z}$.
Similarly, I have tried beginning with gcd$(a,N)>1$, so
$$
\begin{align}
\text{gcd}(a,N)= d & = \lambda a+ \mu N \ \text{for some} \ \lambda,\mu \in \mathbb{Z} \\ \mu N & = d - \lambda a \iff \\
N& | d - \lambda a \iff \\
\lambda a & \equiv d \ (\text{mod}N)
\end{align} $$
Which looks like a rabbit trail.
AI: Suppose that $\gcd(a,N)=d>1$. Then $d\mid a$ and $d\mid N$. Because $d\mid a$, we have $d\mid a^s$ for any $s\geq 1$. Thus $d\mid \gcd(a^s,N)$ for any $s\geq 1$, so that for any integers $x$ and $y$,
$$d\mid xa^s+yN.$$
But the statement that $a^{N-1}\equiv 1\bmod N$ is just that there is an integer $y$ such that
$$a^{N-1}+yN=1,$$
so when $N\geq 2$ we would have to have $d\mid 1$, which is a contradiction. |
H: what is the physical importance of unitary group
What would be the physical importance of unitary group? By physical, I mean geometric intuiton and the usages in physics.
Also, how would special unitary group be used?
AI: In quantum mechanics, the time evolution of states is unitary. Also, many important symmetry operations are unitary as well (there are also some anti-unitary symmetry operations, most notably time reversal).
The special unitary group can be used instead for time evolution because the global phase doesn't matter (actually states are described not by vectors, but by rays in some Hilbert space; if you change the phase of your state vector, you are still in the same ray).
Also note that the gauge transformations from the gauge theories are local unitaries; you apply a different one in each spacetime point (which means that the actual group isn't one-parameter but infinitely-many parameters). That's why the $U(1)$ symmetry is relevant there: You definitely do $not$ get the same state if you do a local $U(1)$ transformation. |
H: Upper half plane is complete with the Lobatchevski metric
How do I show that the Upper half plane is complete with the Lobatchevski metric? I tried to use the fact that $M$ is complete iff the lengh of any divegert curve is unbounded,but did not get any results.thanks.
AI: Here's one possible approach:
If a Riemannian manifold is homogeneous (meaning for any pair of points there is an isometry moving one to the other), then it is complete.
The upper half plane with Lobatchevski metric is homogeneous.
To prove 1, argue as follows: Pick a point $p$. Then for some $\epsilon > 0$, the exponential map is defined on all vectors of length less than $\epsilon$. By homogeneity, this $\epsilon$ works at all points. Intuitively, this means that from any point and in any direction, a geodesic is allowed to flow a least a distance $\epsilon$. This, in turn, easily implies all geodesics are defined for all time.
To prove 2, recall the metric is $ds^2 = \frac{1}{y^2}(dx^2 + dy^2)$. Now, show that if $T_a(x,y) = (a+x,y)$, then $T$ is an isometry. This implies we can move any point to one of the form $(0,y)$. Next, show the map $D_\lambda(x,y) = (\lambda x, \lambda y)$ is an isometry for $\lambda > 0$.
Putting these together shows the hyperbolic plane is homogeneous: To move $(x,y)$ to $(x',y')$, move $(x,y)$ to $(0,y)$ using $T_{-x}$, then use $D_{y'/y}$ to move $(0,y)$ to $(0,y')$, then use $T_{x'}$ to move $(0,y')$ to $(x',y')$. |
H: Prove that if $R$ is von Neumann regular and $P$ a prime ideal, then $P$ is maximal
Let $R$ be a commutative ring with $1\neq 0$. $R$ is said to be von Neumann regular if for all $a\in R$, there is some $x\in R$ such that $a^2x=a.$ Prove that if $R$ is von Neumann regular and $P$ a prime ideal, then $P$ is maximal.
My idea: We know that $P$ is a prime ideal and $R$ is a commutative ring, so $R/P$ is an integral domain. If we can show that $R/P$ is a field, then $P$ is maximal. Further, every finite integral domain is a field, although I am not sure it will be helpful here.
Any suggestions/comments/answers are welcome. Thanks.
AI: You're on the right track! For any prime ideal $P\in R$ and $a\notin P$, we have $a^2x=a$ in $R$ for some $x\in R$, so that in $R/P$, we have $\bar{a}^2\bar{x}=\bar{a}$ (where $\bar{s}$ means the equivalence class $s+P$). Do you see how to proceed?
Rest of solution (mouse over to reveal):
Rewriting, we have$$\bar{a}^2\bar{x}-\bar{a}=\bar{a}(\bar{a}\bar{x}-\bar{1})=\bar{0}.$$Because $a\notin P$, we have $\bar{a}\neq\bar{0}$, so that because $R/P$ is an integral domain, we can conclude $\bar{a}\bar{x}-\bar{1}=0$. Thus any $\bar{a}\neq\bar{0}$ in $R/P$ has an inverse, so $R/P$ is a field. |
H: An integral domain whose every prime ideal is principal is a PID
Does anyone has a simple proof of the following fact:
An integral domain whose every prime ideal is principal is a principal ideal domain (PID).
AI: Here is a proof followed by conceptual elaboration, from my 2008/11/9 Ask an Algebraist post.
Let R be an integral domain. Let every prime ideal in R be principal.
Prove that R is a principal ideal domain (PID)
Below I present a simpler way to view the proof, and some references.
First let's recall one well-known proof, as presented by P.L. Clark (edited):
Proof$_{\,1}$ $\ $ Suppose not. Then the set of all nonprincipal ideals is nonempty.
Let $\{I_i\}$ be a chain of nonprincipal ideals and put $\,I = \cup_i I_i.\,$ If $I = (x)$ then $x \in I_i$ for some $i,$ so $I = (x) \subset I_i$ implies $I = I_i$ is principal, contradiction.
Thus by Zorn's Lemma there is an ideal $I$ which is maximal with respect to the
property of not being principal. As is so often the case for ideals maximal
with respect to some property or other, we can show that I must be prime.
Indeed, suppose that $ab \in I$ but neither $a$ nor $b$ lies in I. Then the ideal
$J = (I,a)$ is strictly larger than $I,$ so principal: say $J = (c).$
$I:a := \{r \in R\ :\ ra \in I\}$ is an ideal containing $I$ and $b,$ so strictly larger
than $I$ and thus principal: say $I:a = (d).$ Let $i \in I,$ so $i = uc.$ Now $u(c) \subset I$ so $ua \in I$ so $u \in I:a.$ Thus we may write $u = vd$ and $i = vcd.$ This shows $I \subset (cd).\ $ Conversely, $d \in I:a$ implies $da \in I$ so $d(I,a) = dJ \subset I$ so $cd \in I.$ Therefore $I = (cd)$ is principal, contradiction. $\ \ $ QED
We show that the second part of the proof is just an ideal theoretic
version of a well-known fact about integers. Namely suppose that the
integer $i>1$ isn't prime. Then, by definition, there are integers $\,a,b\,$
such that $\,i\mid ab,\ \,i\nmid a,b.\,$ But this immediately yields a
proper factorization of $\,i,\,$ namely $\,i = c\, (i\!:\!c),$ where $c = (i,a).\,$ Hence: $ $ not prime $\Rightarrow$ reducible (or: irreducible $\Rightarrow$ prime). $ $
A similar constructive proof works much more generally, namely
Lemma $ $ If ideal $I\ne 1$ satisfies: ideal $\,J \supset I \Rightarrow J\,|\,I\,$ then $I$ not prime $\Rightarrow I\,$ reducible (properly).
Proof $\ $ $I$ not prime $\Rightarrow$ exists $\,a,b \not\in I\,$ and $\,ab \in I.$
$\ A := (I,a)\supset I \Rightarrow A\mid I,\,$ say $\,I = AB;$ wlog we may assume $\,b \in B\,$
since $A(B,b) = AB\,$ via $Ab = (I,a)b \subset I = AB.$ The factors $A,B$
are proper: $A = (I,a),\, a \not\in I;\,\ B \supset (I,b),\, b \not\in I.\quad$ QED
Note that the contains $\Rightarrow$ divides hypothesis: $J\supset I \Rightarrow J\,|\,I\,$
is trivially true for principal ideals $J$ (hence proof$_{\,1}$),
and also holds true for all ideals in a Dedekind domain.
Generally such ideals J are called multiplication ideals.
Rings whose ideals satisfy this property are known as
multiplication rings and their study goes back to Krull.
The OP's problem is well-known: it is Exercise $1\!-\!1\!-\!10\ p.8$
in Kaplansky: Commutative Rings, namely:
(M. Isaacs) In a ring R let $I$ be maximal among non-principal
ideals. Prove that $I$ is prime. (Hint: adapt the proof of Theorem 7.
We have $(I,a) = (c).$ This time take $J =$ all $x$ with $xc \in I.$ Since
$J \supset (I,b),\ J$ is principal. Argue that $I = Jc$ and so is principal.)
For generalizations of such Kaplansky-style Zorn Lemma arguments see the papers referenced in my post here.
Below is an interesting reference on multiplication rings.
Mott, Joe Leonard. Equivalent conditions for a ring to be a multiplication ring.
Canad. J. Math. 16 1964 429--434. MR 29:119 13.20 (16.00)
If "ring" is taken to mean a commutative ring with identity and a
multiplication ring is a "ring" in which, when A and B are ideals with
A $\subset$ B, there is an ideal C such that A = BC , then it is shown that
the following statements are equivalent.
(I) R is a multiplication ring;
(II) if P is a prime ideal of R containing the ideal A, then there is an
ideal C such that A = PC;
(III) R is a ring in which the following three
conditions are valid:
$\qquad$ (a) every ideal is equal to the intersection of its
isolated primary components;
$\qquad$ (b) every primary ideal is a power of its
radical;
$\qquad$ (c) if P is a minimal prime of B and n is the least positive
integer such that $\rm P^n$ is an isolated primary component of B , and if
$\rm P^n \ne P^{n+1},$ then P does not contain the intersection of the remaining
isolated primary components of B . (Here an isolated P-primary component of
A is the intersection of all P-primary ideals that contain A .)
Reviewed by H. T. Muhly |
H: Find the radius of the smallest circle in terms of the side of the square.
The side length of the square is $a$. Two different quadrants are inscribed in it as follows. A small red circle is also inscribed between them.
Find the radius of the smallest circle (red one) in terms of '$a$' only.
AI: $r=\frac{3^2-2^2\sqrt{2}}{7^2} a$.
Take $a=1$ to simplify. To find $r$, find the solution to the following that has $(x,y) \in [0,1]^2$:
$$x^2+y^2 = (1+r)^2\\
(x-1)^2+(y-1)^2 = (\sqrt{2}-1+r)^2\\
x = 1-r .$$
$x,y$ represent the center of the smaller circle, $r$ is the radius. There are two solutions, only one has $(x,y) \in [0,1]^2$. Since I took $a=1$ to simplify, I need to scale the answer by $a$, which gives the answer above. |
H: Every prime ideal is either zero or maximal in a PID.
$(1)$ Let $R$ be a commutative ring with $1\neq 0.$ If $R$ is a PID, show that every prime ideal is either zero or maximal.
In many books I have found the proof of the above statement where they show that
(2)Let $R$ be a commutative ring with $1\neq 0.$ If $R$ is a PID, then every nonzero prime ideal is maximal.
I have revised the question now, how can prove that $(1)$ is true using $(2)$. I am of the opinion that both questions are similar and that's why I am asking this question if I am mistaken then please explain why these two are different. Thanks.
AI: In your revised question, you ask us to show that Every nonzero prime ideal is maximal in a nontrivial PID $\implies$ Every prime ideal is either zero or maximal in a nontrivial PID.
This is trivial. Take a prime ideal. If it's zero, that's fine. If it's not, then the assumed statement says exactly that it's maximal. There's nothing left to show.
Showing the other direction is as trivial, too.
EDIT
The OP asks in a comment:
Sometimes the trivial things are difficult to understand! You said, "If it's zero, that's fine." and my question is that: why is it fine? Everyone says it is trivial but I cannot wrap my brain around it. Can you elaborate and explain it with a microscopic lens?
The statement we want to show is Every prime ideal is either zero or maximal in a nontrivial PID. In other words, if we take a prime ideal, we have to show that it is either zero, or that it is maximal.
If it is not zero, then by assumption we know that it is maximal (this is the statement from (2) ).
If it is zero, than we don't care. Why don't we care? Because we wanted to show that prime ideals are either zero or maximal. So we have shown that all nonzero prime ideals are maximal, and the zero ideal is in fact the zero ideal. That's why I can say that "The zero ideal is zero, and that's fine." There is nothing to prove about that case. |
H: Motivation for adjoint operators in finite dimensional inner-product-spaces
Given a finite dimensional inner-product-space $(V,\langle\;,\rangle)$ and an endomorphism $A\in\mathrm{End}(V)$ we can define its adjoint $A^*$ as the only endomorphism such that $\langle Ax, y\rangle=\langle x, A^*y\rangle $ for all $x,y\in V$.
While all of this lets us prove some stuff about unitary, normal and hermitian matrices, I'd like to know if there's some other motivation behind its introduction (Is there a geometric interpretation? Any other algebraic remark?)
AI: Here is an algebraic approach to adjoint operators. Let us strip away the existence of an inner product and instead take two vector spaces $V$ and $W$. Furthermore, let $V^*$ and $W^*$ be the linear duals of $V$ and $W$, that is, the collection of linear maps $V\to k$ and $W\to k$, where $k$ is the base field. If you're working over $\mathbb R$ or $\mathbb C$, or some other topological field, you might want to work with continuous linear maps between topological vector spaces.
Given a linear operator $A: V\to W$, we can define a dual map $A^*: W^* \to V^*$ by $(A^*(\phi))(v)=\phi(A(v))$. It is straight forward to verify that this gives a well defined linear map between the vector spaces. This dual map is the adjoint of $A$. For most sensible choices of dual topologies, this map should also be continuous.
The question is, how does this relate to what you are doing with inner products? Giving an inner product on $V$ is the same as giving an isomorphism between $V$ and $V^*$ as follows:
Given an inner product, $\langle x, y \rangle$, we can define an isomorphism $V\to V^*$ via $x\mapsto \langle x, - \rangle$. This will be an isomorphism by nondegeneracy. Similarly, given an isomorphism $\phi:V\to V^*$, we can define an inner product by $\langle x,y\rangle =\phi(x)(y)$. The "inner products" coming from isomorphisms will not in general be symmetric, and so they are better called bilinear forms, but we don't need to concern ourselves with this difference.
So let $\langle x,y \rangle$ be an inner product on $V$, and let $\varphi$ be the corresponding isomorphism $\varphi:V\to V^*$ defined above. Then given $A:V\to V$, we have a dual map $A^*:V^* \to V^*$. However, we can use our isomorphism to define a different dual map (also denoted $A^*$, but which we will denote by $A^{\dagger}$ to prevent confusion) by $A^{\dagger}(v)=\varphi^{-1}(A^*\phi(v))$. This is the adjoint that you are using.
Let us see why. In what follows, $x\in V, f\in V^*$. Note that $\langle x, \varphi^{-1} f \rangle = f(x)$ and so we have
$$ \langle Ax, \varphi^{-1}f \rangle = f(Ax)=(A^*f)(x)=\langle x, \varphi^{-1}(A^* f) \rangle $$
Now, let $y=\varphi^{-1}f$ so that $\varphi(y)=f$ Then we can rewrite the first and last terms of the above equality as
$$\langle Ax, y \rangle = \langle x, \varphi^{-1}(A^* \phi(y)) \rangle = \langle x, A^{\dagger}y \rangle $$ |
H: Are there "variables/unknowns" for operations?
We use letters for unknowns/variables:
$x^2=4$
Are there variables/unknowns for operations too?
$8 \star 7 $
With the $\star $ being any operation.
AI: Of course. For example, the Cayley-Hamilton theorem states that, if $a_nx^n+\cdots+a_1x+a_0$ is the characteristic polynomial of a linear operator $M$, then $M$ is a root of $a_nX^n+\cdots+a_1X+a_0$ where $X$ is a variable representing a linear operator (often called a matrix). A less common example (but probably more in the spirit of your question) is the Eckmann-Hilton argument, which shows that any two binary operators $\cdot$ and $\star$ which satisfy certain conditions are equivalent. |
H: $p$ is a $3$-digit prime, then there always exist $p$ consecutive composite numbers.
True/False
If $p$ is a $3$-digit prime, then there always exist $p$ consecutive composite numbers.
How to approach this?
AI: $P_k =(s + 1)! + k$ for $k = 2$ to $(s + 1)$
are $s$ consecutive positive integers
as $P_k$ is always divisible by $k$, hence the statement is true |
H: Constructions of small set with big difference set
Does anyone know any constructions of a small set with a big difference set? Mathematically speaking:
Let $A\subseteq \mathbb{Z}$, such that $A-A=\mathbb{Z}_n$. Please give a sequence $(A_n)_{n\in \mathbb{N}}$ such that $|A_n|$ is small in terms of $n$.
AI: Here is an example which you can easily generalize: 0,1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100. It gives the best possible order $\sqrt{n}$. |
H: How can I properly isolate the variables for this differential equation?
I'm attempting to solve a problem involving this differential equation:
$$\frac{dy}{dx} = x^2y^2 + x^2 - y^2 - 1$$
Because this is a separable differential equation, I tend to split the equation between variables $x$ and $y$, and integrate both sites; but I can't seem to get them separated, due to the $x^2y^2$.
$$\frac{dy}{dx} + y^2 = x^2y^2 + x^2 - 1$$
$$ dy + y^2 = x^2y^2\space dx + x^2\space dx - dx$$
Now the only $y$ variable left to get on the opposite side is the $y^2$ that's trapped in $x^2y^2$, but extracting $y^2$ would leave a $y^2$ variable in all the other terms. That's sort of the problem.
$$\frac{dy + y^2}{y^2} = x^2 dx + \frac{x^2}{y^2} dx - \frac{dx}{y^2}$$
$$\frac{dy}{y^2} + 1 = x^2 dx + \frac{x^2}{y^2} dx - \frac{dx}{y^2}$$
And now, if I want the $y$ variables out of the left side, I end up to the previous equation.
$$ dy + y^2 = x^2y^2\space dx + x^2\space dx - dx$$
It's a really bad loop I'm in right now. How can I get around this problem, and get each variable to its seperate side for integration?
AI: You can factor the right side as $(y^2+1)(x^2-1)$ |
H: how to differentiate a single vector
I am trying to understand the basic rules of vector differentiation.
In a non-scalar expression where x is a column vector, is it valid to differentiate with respect to the kth element of x ?
For example, what would be the result of the following?
$ \partial/\partial x_k (x^T) $
$ \partial/\partial x_k (x) $
AI: Indeed, if you consider a vector $x$ to be a function of its entries. Then we have
$$\frac{\partial}{\partial x_k}x = \begin{pmatrix} 0\\ \vdots\\ 1\\ 0\\ \vdots\\ 0\end{pmatrix} \text{ and }\frac{\partial}{\partial x_k}x^T = \begin{pmatrix} 0, \ldots, 1, 0, \ldots,0\end{pmatrix}$$
where the $1$ is in the $k^{th}$ spot. More generally, there is a notion of derivative for functions between any two Banach spaces (which includes $\mathbb R^n$ for all $n$), under which the derivative of a function $f$ at a point $a$ is the unique linear function $A$ such that $$\lim\limits_{x\to a} \frac{\|f(x)-f(a)-A(x-a)\|}{\|x-a\|}=0$$
and this agrees with the notion of a derivative from Calculus I because when $f$ is a function from $\mathbb R$ to $\mathbb R$ we get a linear function $cx$, and the constant $c$ is precisely the derivative (in the Calc I sense) of $f$ at $a$. |
H: Question in do Carmo's book Riemannian geometry
This is a question on Do Carmo's book "Riemannian Geometry" (question 7 from chapter 7):
Let $f:M\to \bar{M}$ be a diffeomorphism beetwen two riemannian manifolds. Suppose $\bar{M}$ complete and that there is $c>0$ such that: $$|v|\geq c|df_pv|$$ for every $p\in M$ and $v\in T_pM$. Prove that $M$ is complete.
I think one approach could be using the Hopf-Rinow Theorem, because this question is in a chapter about this theorem.Thanks.
AI: Let $\{x_n\}$ be a Cauchy sequence in $M$. Lets see that $\{x_n\}$ is convergent on $M$. This shows $M$ is a complete metric space. By Hopf-Rinow's theorem, $M$ is complete, in geodesic sense.
Claim: $d_{\bar{M}}(y_m,y_n)\leqslant \frac{1}{c}d_M(x_m,x_m)$
Proof: given a curve $\gamma$ in $M$, we have
$$\ell(\gamma)=\int|\gamma'|dt\geqslant c\int|df_{\gamma}\gamma'|=c\cdot \ell(f(\gamma))$$
Once $f$ is a diffeomorfism, the differentiable curves of $M$ and $\bar{M}$ are in bijection. So, considering the curves joinning $x_m$ to $x_n$ (and its images joinning $y_m$ to $y_n$), we can take infimum and it leads us to the result.
Claim: $y_n=f(x_n)$ is Cauchy in $\bar{M}$.
Proof: it follows directly from the claim above.
Hence, once $\bar{M}$ is complete, we have $y_n\rightarrow y$ in $\bar{M}$. But $f$ is a diffeomorphism. Then $x_n=f^{-1}(y_n)\rightarrow f^{-1}(y)$ in $M$. |
H: $f(x)$ is irreducible if and only if $f(x)$ does not have a root in $\mathbb{Z}/\mathbb{2Z}.$
Let $f(x)$ be a polynomial in $(\mathbb{Z}/\mathbb{2Z})[x]$ of degree $2$ or $3$. Prove that $f(x)$ is irreducible if and only if $f(x)$ does not have a root in $\mathbb{Z}/\mathbb{2Z}.$
I know that $f(x)$ is irreducible if and only if $F[x]/(f(x))$ is a field.
Any suggestions/hints will be appreciated.
AI: Having a root in the field of coefficients is equivalent to having a linear factor. If a polynomial of degree 2 or 3 factors in a non-trivial way, then at least one of the factors is linear. |
H: In a field $F=\{0,1,x\}$, $x + x = 1$ and $x\cdot x = 1$
Looking for some pointers on how to approach this problem:
Let $F$ be a field consisting of exactly three elements $0$, $1$, $x$. Prove that
$x + x = 1$ and that $x x = 1$.
AI: Hint 1: We know that $1x=x$ and $0x=0$. But $x$ must have a multiplicative inverse, since $x\neq 0$. So the multiplicative inverse has to be <fill in the blank>
Hint 2. Note that $1+x$ must be either $0$, $1$, or $x$. Can it be $1$? Then we would have $1+x=1$, which implies <fill in the blank>. Can it be $x$? What does that leave for $x+x$? |
H: Solve the differential equation $y'=|x|$, $y(-1)=2$
Given the differential equation,
$y'=|x|$, $y(-1)=2$
I believe I understand how to solve the implicit solution but have questions about using the initial condition to solve the explicit solution. I have outlined my solution below in case there is a mistake I have overseen.
I rewrite the differential equation piecewise and solve each piece independently.
$y'=x$ for $x\ge0$
$y'=-x$ for $x<0$
This yields,
$y_1=\frac{1}{2}x^2+c_1$ for $x\ge0$
$y_2=-\frac{1}{2}x^2+c_2$ for $x<0$
I use the initial condition to solve for $c_2$ and get,
$y_2=-\frac{1}{2}x^2+\frac{5}{2}$ for $x<0$
Does this mean that the solution is only $y_2$? I am thinking that if possible we want to find $c_1$ such that the solution is continuous and differentiable for the largest possible interval of definition. If $c_1=c_2$ then the piecewise solution is continuous and differentiable for all reals and satisfies the initial condition.
Thank you in advance for your time and assistance.
AI: Those constants aren't independent (there's really only one). $|x|$ is continuous, so the derivative $y'$(and therefore $y$ itself) must also be continuous. So you really only have the one function, defined over all $\mathbb{R}$:
$$y=-\frac 1 2 x^2 + \frac 5 2, x<0$$
$$y=\frac 1 2 x^2 + \frac 5 2, x\geq 0$$
This passes through $(-1,2)$ and solves the equation, like you said. In general the interval on which the solution is defined is taken to be the largest possible such interval. In this case that's the entire real number line. |
H: Proving a metric space to be compact
I have the following metric space: The set $X$ of all sequences with members from the set $\{1,2,\ldots, n\}$, together with the metric $$d(x,y)=\frac{1}{\min\{j\in\mathbb{N}:x_j\ne y_j\}}.$$ I wish to prove two things about this space:
1) $X$ is compact.
2) If $T:X\to X$ is defined by $Tx_n=x_{n+1}$ then $T$ is continuous.
Its many years since I studied topology so if someone can help me even a little bit I will be very grateful. Thanks.
AI: Hint for 1: Note that a metric space is compact iff every sequence has a convergent sub-sequence. Let $(y_k)$ be a sequence in $X$. Since there are only finitely many choices for the first term of an element of $(y_k)$, there is some sub-sequence $(y_k^1)$ of $(y_k)$ such that all elements have the same first term. Since there are only finitely many choices for the second term, there is some sub-sequence $(y_k^2)$ of $(y_k^1)$ such that all elements have the same first term. Continue in this manner, and show that the sub-sequence $(y_k^k)$ (often called the diagonal sub-sequence) is convergent.
Hint for 2: Let $(x_j),(y_j)\in X$. Note that $\min\{j\in \mathbb N: x_j\neq y_j\}=\min\{j\in \mathbb N: x_{j+1}\neq y_{j+1}\}+1$, so $$\frac{1}{d((x_j),(y_j))}=\frac{1}{d(T(x_j),T(y_j))}+1$$
and do some rearranging. |
H: Find functions family satisfying $ \lim_{n\to\infty} n \int_0^1 x^n f(x) = f(1)$
I wonder what kind of functions satisfy
$$ \lim_{n\to\infty} n \int_0^1 x^n f(x) = f(1)$$
I suppose all functions must be continuous.
AI: Your equation can be rewritten as
$$\lim_{n \to \infty} (n+1) \int_0^1 x^n (f(x) - f(1))\ dx = 0 $$
(I'd rather use $n+1$ than $n$, because $(n+1) \int_0^1 x^n \ dx = 1$)
Note that as $n \to \infty$, $(n+1) x^n \to 0$ uniformly on $[0,1-\delta]$ for any $\delta > 0$, implying that $(n+1) \int_0^{1-\delta} x^n (f(x) - f(1))\ dx \to 0$ for any $f$ that is integrable on $[0,1]$.
On the other hand, if $f$ is continuous from the left at $1$, for any $\epsilon > 0$ there is $\delta > 0$ such that $|f(x) - f(1)| < \epsilon$ for $x \in [1-\delta, 1]$, and then
$$\left|(n+1) \int_{1-\delta}^1 x^n (f(x) - f(1))\ dx\right| < \epsilon (n+1) \int_{1-\delta}^1 x^n\ dx < \epsilon$$
So your equation is true for functions that are integrable on $[0,1]$ and continuous from the left at $1$.
However, this is only a sufficient condition, which certainly could be weakened. I doubt that there is a simple necessary and sufficient condition.
Note also that $(n+1) x^n < -1/(e x \ln(x))$, so the equation is also true if $(f(x) - f(1))/(x \ln(x))$ is integrable on $[1-\delta,1]$ for some $\delta > 0$. For an example that is not continuous from the left at $1$, take the indicator function of the union of the intervals $J_k = [1-2^{-k}-3^{-k}, 1-2^{-k}]$ for positive integers $k$. |
H: Torque calculation, to achieve clean spin+tumble
Here's a pencil-like robotic spaceship carrying an experiment, it is a solid mass 100m long, 100 inches thick and weighs 1000kg. We're in deep solar space 100au above the sun.
Assume we can apply any platonic torque to the object.
Notice the global unchanging XYZ axis shown.
If the body is rotating around it's own long axis, we'll call that "spinning".
If the body is rotating only around the global Z axis, we'll call that "tumbling".
To be clear (i): if it is "tumbling", then the "spin" would not always
be around the global X axis. "spin" is around the long axis of the
object: that axis would change globally as the object moves in other
ways.
To be clear (ii): what we describe as "tumbling" would take place
only in the global YX plane, with no movement to-or-fro in the Z
direction.
Now:
what we want the spacecraft to do is spin and tumble cleanly, at the same time.
To be clear: we want the want the angular momentum to be the sum of two items. A constant component along the global z axis. And a
rotating component which is perpendicular to the z axis. (Indeed,
along the body length for clarity.)
Of course ... that's absolutely impossible because of gyroscopic effects.
If you give it two naive torques to spin it, and, make it tumble, in fact there will always be a "wobble" in the third global axis (ie: seen from overhead it will rotate slightly back and fore in the y axis).
So - what torque would you have apply
over time
to get it "spinning" and "tumbling" cleanly with no wobble in the Y axis?
In a word:
what is the torque you must apply over time to force a body, to have two angular momentums, one causing the body to rotate at h HZ along
the Z axis, and the other causing the body to rotate at i HZ along an
axis which is rotating at h HZ (ie, in sync with the first-mentioned
rotation) in the XY plane, with no other motion or wobble?
I imagine the answer is something quite simple like "oh you must apply sin(time) torque here" or some other cohesive solution.
This would seem to be an everyday issue to engineers and the like, so I imagine there is some well-explored solution. But I couldn't find it.
An interesting follow up question:
Joriki has kindly pointed out that the solution is a torque where: the axis of the torque initially points along the global y axis, and then, the axis of the torque rotates in the global XY plane matching perpendicularly the z-rotation of the object.
However I ran this in a simulator and it doesn't work. If you set the z-rotation of the torque axis to some fixed value, it just twists the object somewhat randomly. If you set the z-rotation of the torque axis to zero and slowly increase the z-rotation of the torque axis, again it largely just twists it around randomly.
It occurs to me that the formula to get the object to do the behaviour described, would have to, in fact, take in to account the current angular momentum of the object at any moment. Perhaps? It is very confusing to see how you would start the motion or describe maintaining said movement. Both problems - starting or maintaining - seem astoundingly difficult.
AI: The torque is the time derivative of the angular momentum. If I understand correctly what you mean by "spinning and tumbling cleanly with no wobble", you want the angular momentum to be the sum of a constant component along the $z$ axis and a rotating component perpendicular to the $z$ axis, along the body axis. The time derivative of this sum is a vector perpendicular to both the $z$ axis and the body axis, which rotates in sync with the body. In your snapshot of the motion, it would point along the $y$ axis. |
H: Check if $(\mathbb Z_7, \odot)$ is an abelian group, issue in finding inverse element
Take $\mathbb Z_7$ and the operation $\odot$ defined on it as follows $\forall a,b \in \mathbb Z_7$:
$$\begin{aligned} a \odot b=a+b+3\end{aligned}$$
Check if $(\mathbb Z_7, \odot)$ is a group and in particular if it is an abelian group.
Associativity
It can be easily proved that $\odot$ is associative:
$$\begin{aligned} (a \odot b) \odot c = a \odot (b \odot c)\end{aligned}$$
$$\begin{aligned}(a + b + 3) \odot c = a \odot (b+c+3)\end{aligned}$$
$$\begin{aligned} (a + b + 3) + c +3 = a + (b+c+3)+3 \end{aligned}$$
$$\begin{aligned} a + b + c +6 = a +b+c+6 \end{aligned}$$
Commutativity
$\odot$ is commutative:
$$\begin{aligned} a \odot b = b \odot a \end{aligned}$$
$$\begin{aligned} a + b + 3 = b +a+3 \end{aligned}$$
Identity element
There is an identity element for $(\odot, \mathbb Z_7)$
$$\begin{aligned} a \odot \mathbb 1_{\mathbb Z_7} = a \end{aligned}$$
$$\begin{aligned} a + e + 3 = a \end{aligned}$$
$$\begin{aligned} e + 3 = 0 \end{aligned}$$
$$\begin{aligned} e = 4 \end{aligned}$$
and similarly if we repeat the calculation for $\mathbb 1_{\mathbb Z_7} \odot a$, still $e=4$
What I am struggling with is the inverse element, how to check if there is one. I know that it should be:
$$\begin{aligned} a \odot a^{-1} = e \\ a + a^{-1} + 3 = e \\ a + a^{-1} + 3 = 4\end{aligned}$$
Is it acceptable to state $a^{-1} = a+1$ (and similarly for $a^{-1} \odot a = e$)? If so, have I succeeded in proving $(\odot, \mathbb Z_{7})$ to be an abelian group?
AI: The inverse is $a^{-1}=1-a$, as $a\odot (1-a)=a+(1-a)+3=4=e$. |
H: Is the adjoint representation of SO(4) self-dual?
The adjoint representation (over the complex numbers) of SO(4) is 6-dimensional. Is this representation self-dual?
Other than the adjoint representation and its dual, are there other irreducible 6-dimensional representations of SO(4) over the complex numbers?
AI: The adjoint representation of any semisimple Lie group $G$ is self-dual. This follows from the fact that the Killing form gives a nondegenerate $G$-invariant bilinear map $\mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ for such $G$ (and we can then extend scalars to $\mathbb{C}$ to get a corresponding such map for $\mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C}$).
The adjoint representation of $\text{SO}(4)$ is also not irreducible (and this is the only $n \ge 2$ for which this is true). This follows from the fact that $\mathfrak{so}(4) \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$ as a Lie algebra and so the adjoint representation of $\mathfrak{so}(4)$ on itself is a direct sum of two $3$-dimensional representations.
Edit: In fact $\text{SO}(4)$ admits no irreducible $6$-dimensional representations. Its universal cover is $\text{Spin}(4) \cong \text{SU}(2) \times \text{SU}(2)$ which admits two irreducible $6$-dimensional representations given by $V \otimes S^2(V)$ and $S^2(V) \otimes V$ where $V$ is the defining representation of $\text{SU}(2)$. The kernel of the covering map $\text{Spin}(4) \to \text{SO}(4)$ is generated by $(-1, -1)$ and this acts nontrivially in both of the above representations, so neither descends to a representation of $\text{SO}(4)$. |
H: Hölderian path connectedness
Let $X$ be a complete metric space and $\alpha \in (0,1)$. Suppose that for every $x,y \in X$ there exists $z \in X$ s.t.
$$
d(x,z) \le \frac{1}{2^\alpha}d(x,y), \qquad d(y,z) \le \frac{1}{2^\alpha}d(x,y).
$$
Then $X$ is Hölderian path-connected, i.e. for every $x,y \in X$ we can find a $\alpha$-Hölderian path $\gamma \colon [0,1] \to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$.
Have you got any ideas on how to solve this problem? To be honest, I do not know how to start. I'm stumped for the hypothesis of completness... How can we use it?
I thank you in advance for any useful idea.
AI: Some steps
Construct, inductively, for each integer $n$, $2^n+1$ points $z_{n,1},\dots,z_{n,2^n+1}$ such that $d(z_{n,k+1},z_{n,k})\leq 2^{-n\alpha}$ and $z_{n,1}=x$, $z_{n,2^n+1}=y$.
Define $f(k2^{-n})=z_{n,k}$. Now we shall extend it in a Hölder continuous map.
Let $t\in [0,1]$. We can find a sequence $\{t_j\}$ of the form $\{k_j2^{-n_j}\}$ which converges to $t$. Since $f$ satisfies a Hölder continuity condition, the sequence $f(t_k)$ is Cauchy in $X$.
Define $\gamma(t):=\lim_{k\to +\infty}f(t_k)$, checking that it's well-defined and $\alpha$-Hölderian. |
H: Given that $x=\dfrac 1y$, show that $∫\frac {dx}{x \sqrt{(x^2-1)}} = -∫\frac {dy}{\sqrt{1-y^2}}$
Given that $x=\dfrac 1y$, show that $\displaystyle \int \frac 1{x\sqrt{x^2-1}}\,dx = -\int \frac 1{\sqrt{1-y^2}}\,dy$
Have no idea how to prove it.
here is a link to wolframalpha showing how to integrate the left side.
AI: Substitute $x=1/y$ and $dx/dy=-1/y^2$ to get $\int y^2/(\sqrt{1-y^2}) (-1/y^2)dy$ |
H: Dirichlet problem, uniqueness & counterexample
Help me please with this example:
Let's $q(x)$ be a continuous and constant sign function in $[0,1]$;
uniqueness of solution of Dirichlet problem to equation: $u''+qu=0$ depends on sign of $q$.
Prove the theorem of uniqueness in cases then it's true and give counterexample in case then it's not true.
Thanks a lot!
AI: For the uniqueness part, other than using maximum principle as Thomas said, you can also prove in this way: For $u$ satisfying $$\tag{1}u''+qu=0$$ with $q\leq 0$ and $u(0)=u(1)=0$, we must have $u\equiv 0$. To see this, multiply $(1)$ by $u$ and integrate it over $[0,1]$, we get
$$\int_0^1uu''+\int_0^1qu^2=0.$$
Integration by parts, we have
$$\big[uu'\big]_0^1-\int_0^1(u')^2+\int_0^1qu^2=0.$$
Since $u(0)=u(1)=0$ by assumption, we have
$$-\int_0^1(u')^2+\int_0^1qu^2=0.$$Left hand side is nonpositive since $q\leq 0$, which implies that $u'\equiv 0$, i.e. $u\equiv c$ for some constant $c$. Since $u(0)=u(1)=0$, $c=0$, i.e. $u\equiv 0$ as required. |
H: Show that $f'' > 0$, $\lim_{x \to b^-} = \infty$ implies that $\lim_{x \to b^-} f'(x) = \infty$
Let $f$ be a continuous function on $[a,b)$, $f$ twice differentiable in $(a,b)$ so that $f''(x)>0$ for each $x \in (a,b)$. Prove that if $$\lim_{x\to b-}f(x) =\infty $$ then
$$ \lim _{x\to b-}f'(x)=\infty $$
AI: Suppose $|f'|$ is bounded around $b$. Then for any $x$ and $y$ close to $b$,
$$
|f(x)-f(y)| \leq \left( \sup |f'| \right) |x-y|,
$$
and therefore $\lim_{x \to b-} f(x)$ exists and is finite. This contradiction implies that $|f'|$ is unbounded around $b$. But $f'$ is monotonically increasing, and therefore $f'(x) \to +\infty$ as $x \to b^{-}$. |
H: Why isn't GL system of provability logic reflexive?
Formula $\square p \rightarrow p$ (axiom T; corresponding to reflexive modal frames) is interpreted as "if p is provable, then p", or more precisely: for all realizations (all substitutions for $p$), $PA \vdash Bew(\ulcorner p \urcorner) \rightarrow p$ (PA is system whose provability we discuss, i.e. some standard first-order axiomatization of Peano Arithmetic).
Now if we forget the fact that T combined with Gödel-Löb axiom produces a contradiction-proving disaster, what I'm wondering is what is intuitively wrong with T? Because to me it seems perfectly obvious that if we have a proof of $p$, then we also have $p$, since $p$ is just the last step of the mentioned proof. How can we have a proof (in PA) of something that doesn't hold (in PA)? Since proof is nothing more than coded array of sentences, we can simply follow its steps and get to $p$.
Even $\square \bot \rightarrow \bot$ seems ok: if we can prove $\bot$, then $\bot$ does hold in this system (it's not T's fault that we're dealing with inconsistent system).
I'm new to provability logic (just started reading The Logic of Provability) so I guess I'm overlooking something obvious. Thanks!
AI: This is an area where it is very easy to get oneself confused from lack of precision. You write ${\it Bew}(\ulcorner p\urcorner)$ without specifying which axioms $p$ is to be proved from, and that makes an important difference.
Let's first consider
$$ T_0(p) \equiv {\it Bew}_{PA}(\ulcorner p \urcorner)\to p$$
There's nothing wrong with this; if we believe in Peano Arithmetic at all, it must be true in the standard model, and so in particular $PA+T_0$ is consistent. (In general I will tacitly assume that $PA$ is true and consistent).
On the other hand, individual instances of $T_0$ are not necessarily theorems of $PA$. Because of Gödel-Rosser's incompleteness theorem, there is a model of $PA$ in which ${\it Bew}_{PA}(\ulcorner \bot\urcorner)$ is true. But that means that in that model $T_0(\bot)$ is false -- so it cannot be a theorem.
In fact, Löb's theorem shows that $T_0(p)$ is a theorem of $PA$ if and only if $p$ itself is.
Since $T_0$ is consistent with $PA$ (and even true), we can choose to adopt it as an axiom. We may then consider
$$ T_1 \equiv {\it Bew}_{PA+T_0}(\ulcorner p\urcorner)\to p$$
The story of $T_1$ is then much the same as that of $T_0$: We can convince ourselves that it is consistent with $PA+T_0$, in fact true in the standard model, but not a theorem of $PA+T_0$. And we can then continue by induction to
$$ T_n \equiv {\it Bew}_{PA+T_0+T_1+\cdots+T_{n-1} }(\ulcorner p\urcorner) \to p$$
for any finite $n$.
However, bad things happen if we attemt to "go to the limit" and consider
$$ T_\infty(p) \equiv {\it Bew}_{PA+T_\infty}(\ulcorner p\urcorner)\to p $$
Note that instead of a limiting process we could have arrived at $T_\infty$ directly by being sloppy about the precise meaning of $\Box$ in the pithy formulation "take $\Box p \to p$ as an axiom".
First, it isn't immediately clear that $T_\infty$ is well-defined at all, but I think, without having checked the details, that one of the recursion theorems can produce a formula that meets the above specification. So let's put that objection aside.
It is a much more serious problem that $T_\infty$ directly implies that $PA+T_\infty$ is consistent! That means that $PA+T_\infty$ can prove its own consistency and (Gödel-Rosser again) this is an offense punishable by inconsistency. So $T_\infty$ is not consistent with $PA$.
If we unfold the construction in the proof of Gödel-Rosser, we would get a concrete $p_0$ such that $PA \vdash \neg T_\infty(p_0)$. But that is not very enligthening, so let's instead consider intuitively why our intuition that $T_n$ must be true doesn't work for $T_\infty$.
Suppose we have a proof of some sentence $p$ from $PA+T_\infty$; we wish to convince ourselves that then $p$ is indeed true. The proof is a certain finite sequence of formulas; we can check one by one that each of these formulas is true, given that all of the previous ones. Except, that is, if one of the formulas is an instance of $T_\infty$ and is justified by "that is an axiom". Since the truth of $T_\infty$ was what we were trying to argue for, we cannot just assume it at that point. And it won't work to attempt to justify it by induction either, because there is no guarantee that the proof the "inner" $T_\infty$ is applied to is shorter or simpler than the outer one. So we might find ourselves in an infinite ascent there.
In the first version of this answer I wrote $T_\omega$ instead of $T_\infty$, but arguably $T_\infty$ isn't really the limit of the $T_n$s. In fact, we can continue the $T_n$ sequence transfinitely to
$$ T_\omega \equiv {\it Bew}_{PA+T_0+T_1+\cdots+T_n+\cdots }(\ulcorner p\urcorner) \to p$$
and indeed
$$ T_\alpha \equiv {\it Bew}_{PA+\{T_\beta\mid\beta < \alpha\} }(\ulcorner p\urcorner) \to p$$
for as many ordinals as we can describe in $PA$ in the first place. Neither of these have the circularity problem that plague $T_\infty$. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.