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H: Prove by induction that $3\mid n^3 - n$
Prove by induction that $3\mid n^3 - n$.
I'm having an argument with my professor whether my exam was right or not. Before I sign a formal complain to get a review on my exam, I'd like to be sure it's correct. My answer:
Proof by induction:
Proposition: $\forall n \in\mathbb{ N}$ (belongs to natural numbers), $3 | (n^3 - n)$;
Basic Step: $P(0) = 0^3 - 0$ is divisible by 3.
Inductive Step:
$p(n+1) = (n+1)^3 - (n+1)$
$= n^3 + 3n^2 + 3n + 1 - n - 1$
$= n^3 + 3n^2 + 2n$
$= n^3 + 3n^2 + 2n - (n^3 - n)$
$= 3n^2 + 3n$
By induction, $p(n+1)$ is true if:
$3\mid n$ and $3\mid m$ and $3 | n - m$
to prove that, assume that $n$ is $3p$ and $m$ is $3q$, therefore $3 | 3p - 3q$.
My professor wanted:
$= (n^3 -n) + 3n^2 + (2n + n)$
AI: It is clear to me that your logic is correct. However your solution is not well-written and as a result is confusing to read. For example, you have written
$$n^3+3n^2+2n = n^3+3n^2+2n - (n^3-n)$$
(This is the equality between your 3rd and 4th lines.) This is false. It is clear to me that what you mean is that, since $p(n+1)=n^3+3n^2+2n$, then $p(n+1)-p(n)=n^3+3n^2+2n-(n^3-n)$. However, this is not what you wrote.
Thus, while the substance of your answer is essentially correct, the form in which it is being communicated leaves something to be desired. There is no objective standard regarding how such a situation should be graded; it is up to the professor.
This is not mathematical advice, but I suggest you approach your professor again rather than lodging a formal complaint, because the case is not sufficiently clear-cut to be adjudicated by outside parties. It will help you have the kind of conversation you want with them if you approach that conversation with the attitude that you know you had the right idea, but that you didn't communicate it well and would like to learn how to do it better, but that you want to make sure that they understand that you did have the right idea. |
H: Convergence to zero of integral
A simple question: Let $f: \mathbb{R} \to \mathbb{R}$ be a function with $\int_{-\infty}^{\infty} |f(z)| dz < \infty$. Is it then the case that $\lim_{h \to \infty} \int_{|f(z)| > h} f(z)dz = 0$ ?
Thanks!
AI: It's enough to prove the statement when $f \geq 0$.
We know
$$\int f(z)dz = \sum_{i=0}^{\infty} \int{f(z)1_{i \leq f(z) < i+1} dz} < \infty$$
therefore the remainder of the series
$$\sum_{i=h}^{\infty} \int{f(z)1_{i \leq f(z) < i+1}}=\int_{|f(z)|\geq h}f(z) dz$$ goes to zero.
Edit: Another possibility is to write the limit as $\lim_{h \to \infty} \int f(z) 1_{|f(z)| \geq h} dz$ and use dominated convergence. |
H: A small question: In vectors, is true that |w|+|v|=|w+v|?
A small question: In vectors, is true that |w|+|v|=|w+v|?
Thanks in advance for any help!
AI: Not. Let $v$ be a nonzero vector, and $w=-v$. Then $|w|=|v|>0$ and $|w+v|=|0|=0$. |
H: What are the coordinates of the vertices of a regular tetrahedron, relative to its centroid?
I am trying to draw an equilateral/regular tetrahedron in Processing (subset of Java), so I have to define 4 triangles that meet at the 4 vertices. I have been able to find the coordinates for the vertices assuming 1 of the vertices is at the origin.
However, the tetrahedron rotates, and I would like its rotation to be about its centroid. This where I get stuck, I need to define the 4 planes of the tetrahedron when the centroid is at the origin.
I hope my question is clear. :/ Thanks in advance.
AI: According to this Wikipedia article, if the origin is at the centre, then you can take the vertices to be
$$(\pm 1, 0, -1/\sqrt{2}),\,\,\,\,\, (0, \pm 1, 1/\sqrt{2}).$$
The article does also give some alternatives such as $(1, 1, 1, ), (1, -1, -1), \dots $ which might be easier to work with. |
H: Integral over a Jordan curve
I've got the task:
Let $C$ be a Jordan curve, piecewise smooth. Compute $\int_C \frac{e^z}{z(1-z)^3} dz$, if:
a)$z_0 = 0$ is inside $C$, and $z_1 = 1$ outside.
b) both $z_0$ and $z_1$ are outside of $C$.
Well, how to work with it? If those points were inside, it would look like Cauchy integral theorem, but if one of them/ both are outside...? Then what can I do? May you help? I have no idea, even how to begin, which theorems I should use, nothing...
AI: For part $a$, only $z_0$ lies inside, so let $f(z)=\frac{e^z}{(1-z)^3}$. The integral is now
$$\int_C\frac{f(z)}{z}dz$$ where we can use the Cauchy Integral Formula.
Use the same logic for part $b$. |
H: Show that $\mu\{x\in X : f(x)=\infty\}=0$.
I'm working through some problems in a first course on Measure Spaces.
The problem is:
Let $(X,\Sigma, \mu)$ be a measure space and let $f:X \to[0,\infty]$ be a measurable function such that $\int_Xfd\mu\lt\infty$. Show that $\mu\left(\{x\in X : f(x)=\infty\}\right)=0$.
The solution goes as:
Let $E=\{x\in X: f(x)=\infty\}$. Then E is measurable. Suppose $\mu(E)\gt0$. Then for each $n \in \Bbb{N}$,$$f\ge f\chi_E \ge n\chi_E .\qquad(1)$$
So,$$\infty\gt\int_Xfd\mu\ge \int_X n\chi_E d\mu=n\mu(E).\qquad $$.
But $$\lim_{n \to \infty}n\mu(E)=\infty$$ which is a contradiction. Thus $\mu(E)=0$
My question is:
i) I can't understand how the step denoted by (1) is made, particularly the reasoning behind $f\chi_E\ge n\chi_E$. Would someone please enlighten me as to how?
Many thanks in advance.
AI: $E$ is defined as the set on which $f(x)= \infty$. Do you agree that $\infty > n$ for all $n \in \mathbb{N}$? There's not much more to it. Multiplying $f$ by the characteristic function of $E$ is what allows you to get a function that is identical to $f$ on $E$ but vanishes everywhere else. |
H: How to prove Schwarz inequality for Hermitian forms?
I'm trying to do something like the proof of the Schwarz inequality for inner product.
If $h(y,y)\neq 0$, then we can take $\alpha=-h(x,y)/h(y,y)$ and calculate $h(x+\alpha y,x+\alpha y)$ which is a nonnegative number. The desired conclusion follows of this calculation.
If $h(y,y)=0=h(x,y)$, then the inequality is trivial.
My question is: how do we know that the case $h(y,y)=0$ and $h(x,y)\neq 0$ is not possible? When we are working with an inner product, this case is not possible because, by definition, $\langle y,y\rangle=0\Rightarrow y=0\Rightarrow\langle x,y\rangle=0$. But we don't have this condition for hermitian forms.
Thanks.
AI: If $h(y,y) = 0$ but $h(x,y) \ne 0$, then taking $\alpha = - t h(x,y)$ with $t > 0$
we would have
$h(x + \alpha y, x + \alpha y) = h(x, x) - 2 t |h(x,y)|^2$,
which would be negative for sufficiently large $t$. |
H: What is the value of sin(arcsin(4))?
In this case arcsin() is the unrestricted sin inverse function. I know that it is either undefined or has the value of 4. It could be undefined because arcsin() has only a doman of -1...1 and 4 is out of the domain. On the other hand, it could be that since they are inverses the intermediary result does not matter and they will cancel to get back 4. Which one is it?
AI: If you are working with real numbers, $\arcsin(4)$ doesn't exist.
Note that the formula
$$\sin(\arcsin(x))=x$$
only holds over the domain of $\arcsin(x)$ which is $x \in [-1,1]$.
Things are different when you work with complex numbers though.... |
H: What's the derivative of $\frac{\cos((n+1)\theta)-\cos((n-1)\theta)}{2^n},\theta=\frac{x\pi}{n}$ for $x \in \{0,...,n-1\}$?
What's the derivative of $\dfrac{\cos((n+1)\theta)-\cos((n-1)\theta)}{2^n}$,
where $\theta=\dfrac{x\pi}{n}$ for $x \in \{0,...,n-1\}$ and $n$ is fixed?
So I've tried using the chain rule, and I got:
$a)$
$$\left( (n+1)\theta \right)'=\left( \pi x-\frac{x\pi}{n} \right)'=\pi(1-\frac{1}{n})$$
So because of that:
$$\left(\dfrac{\cos((n+1)\theta)-\cos((n-1)\theta)}{2^n}\right)'\stackrel{a)}{=}\dfrac{1}{2^n}\left( \pi(1-\frac{1}{n}) \sin(n-1)\theta- \pi(1-\frac{1}{n})\sin(n+1)\theta \right)$$
But my textbook says, that it's
$$\frac{(n+1)\sin((n+1)\theta)}{\sin\theta } - \frac{(n-1)\sin((n-1)\theta)}{\sin\theta }$$
How come?
AI: The book is obviously wrong. I think you are also wrong. The correct answer is $$ \dfrac {n \sin x \cos (nx) + \cos x \sin (nx)}{2^{n-1}}. $$ |
H: Transitive, symmetric $R$ such that for all $x$, there is $y$ such that $xRy$, is an equivalence relation
I'm stuck at one particular task I'm working on.
Here is the task:
Let R be a transitive and symmetrical relation on $S$. Assume that for all
$x \in S$ there is a $y \in S$ so that $xRy$. Prove that $R$ is an equivalence
relation.
How can I prove that $R$ is an equivalence relation?
I would appreciate any help.
Thank you.
AI: All you need to do here is make the case that $R$ must be reflexive. Since for every $a \in S$ there is $b,$ such that $aRb,$ then by symmetry $bRa$ and by transitivity $aRa.$ Since this holds for every $a\in S$, the relation is necessarily reflexive, and hence, is an equivalence relation. |
H: The sequence 1,11,111,... and the prime factorization of its elements
I have been recently investigating the sequence 1,11,111,... I found, contrary to my initial preconception, that the elements of the sequence can have a very interesting multiplicative structure. There are for example elements of the sequence that are divisible by primes like 7 or 2003.
What I am interested in is for what numbers, other than 2 and 5 can we say that they divide no element of the sequence?
AI: In fact, every number coprime with $10$ (that is, those that aren't integer multiples of $2$ and/or $5$) divides some element of that sequence. See this question.
On the other hand, it is immediately clear that no even number or integer multiple of $5$ can divide an element of that sequence. |
H: What would qualify as a valid reason to believe there is a closed form?
I noticed that almost every non-homework-level integral posted on this site prompts somebody to ask "Do you have any reason to believe there is a closed form?" (some recent examples here and here)
I actually never fully understood either meaning or purpose of such questions.
What would qualify as a valid reason to believe there is a closed form?
Why should a lack of such reason inhibit an interest in an integral?
Personally, I hold the following view at this subject: we should always believe there is a closed form. Even if eventually somebody manages to prove there is no closed form in a certain class of functions, we might introduce a new special function or constant to represent a value of the integral/product/sum in question, study its properties, find other integrals that can be represented using it, discover interesting connections with other functions, etc. This happened many times in the past (think of $\operatorname{erf}(x)$, $\Gamma(x)$, $\zeta(x)$, $\operatorname{Li}_s(x)$, $K$, $A$, etc) and always helped to move math forward.
AI: The question of whether a given integral has a closed form does not have a consistent answer from one practitioner to the next. In my experience, however, there's a plausibility that there is one when the integration limits include a branch point or an otherwise integrable singularity.
But what is a closed form? That said, we can debate until we turn blue as to what constitutes a closed form. In my humble opinion, a closed form implies a means of computing the value of the integral that results in fewer operations that simply computing the integral by some numerical scheme. Knowing when this condition has been satisfied takes a certain level of familiarity with the special functions we use to express our integration results.
For example, erf is considered a closed form, and rightly so, because we have amazingly accurate schemes to compute the error function using very few operations. These schemes are much more often then not much more efficient than any numerical integration scheme, so we may call it a closed form.
On the other hand, consider the following answer to a question posed yesterday:
$$-{\frac {\,{\mbox{$_3$F$_2$}(1/6,1/2,1/2;\,7/6,3/2;\,1)}\Gamma
\left( 5/6 \right) \Gamma \left( 2/3 \right) -{\pi }^{3/2}}{6 \Gamma
\left( 5/6 \right) \Gamma \left( 2/3 \right) }} $$
Is this considered a closed form? It depends. Is there a scheme that computes such a hypergeometric faster than the original (double) integral, which has an integrable singularity? As I am not very familiar with such hypergeometrics, I'd have a hard time answering that question. My rule of thumb is that, when I see any hypergeometric higher than $_2F_1$ in any result, I do not consider it closed form because it looks so...icky. People who are better versed than me in such matters may feel free to disagree with me.
(NB The actual answer to the question, by the way, is $\pi/24$, which is a whole other can of worms.)
Of course, we could go on about the notion of computability in general, but I'd rather just stick with the notion of tolerance and counting operations.
In any case, I hope these observations won't temper the OP's enthusiasm with working with integrals like the ones he poses and solves here, but will further challenge him to provide more beautiful results that can be useful as well. |
H: Show that these two Eigenvalues are positive
In the course of a rather long calculation, I got two Eigenvalues, namely
$$
\lambda_{1,2}=\frac{1}{2}(2+x^2+u\pm\sqrt{u^2-2ux^2+x^4+4x^2}).
$$
I have to show, that these two Eigenvalues are positive. (Here x is out of a bounded domain $\Omega\subset\mathbb{R}^2$ and $u\in C^2(\Omega)\cap C^1(\overline{\Omega})$ is a non-negative function.)
Could you please help me to show, tha $\lambda_{1,2}$ are positive? I do not have an idea, how to show that.
AI: You need to just show that $2+x^2+u > \sqrt{u^2-2ux^2+x^4+4x^2}$
$$\iff (2+x^2+u)^2 > u^2-2ux^2+x^4+4x^2$$
$$\iff 4+4 u+u^2+4 x^2+2 u x^2+x^4 > u^2 - 2 ux^2 + x^4 + 4 x^2$$
$$\iff 4 + 4 u + 4 u x^2 > 0$$
which is true as $u \ge 0$. |
H: Constant distribution in the factorial 2n!
a very simple question: how does 2 distribute in the factorial $2n!$ ?
$2((n)(n-1)(n-2))$
would be treating the expression as $2(n!)$, which is different from:
$(2n)(2n-1)(2n-2)$
which is different still from:
$(2(n))(2(n-1))(2(n-2))$
Thanks.
AI: Your first expression is $2(n!)$, your second is $(2n)!$ and the third is $(2n)!!$ The fist two possibilities give a good reason to use parentheses-they are quite different. I think $(2n)!$ would be the default reading if there are no parentheses, but I would look for confirmation in whatever I was reading. |
H: $\delta$ proof of a limit at a point
problem: Evaluate the following limits or show they do not exist.
I'm stuck on trying to show $\lim_{x \to 1} \frac{x}{x-1} (x\neq 1)$ does not exist.
I proved that $\lim_{x \to 1^+} \frac{x}{x-1} (x\neq 1)=\infty$
Proof:
Let $\alpha\in\mathbb{R}$ such that $\alpha>0$. Let $\delta=\frac{1}{\alpha}$ such that if $0<x-1<\delta\implies \frac{1}{x-1}>\delta=\alpha$. Since $x>1$ it follows that $\frac{x}{x-1}>\frac{1}{x-1}$. Thus $\frac{x}{x-1}>\frac{1}{x-1}>\alpha$
Now I want to show $\lim_{x \to 1^-} \frac{x}{x-1} (x\neq 1)=-\infty$
Proof:
Let $\alpha\in\mathbb{R}$ and let $\delta= \frac{1}{1+|\alpha|}$ such that if $0<1-x<\frac{1}{1+|\alpha|}\implies x-1>-\frac{1}{1+|\alpha|}\implies \frac{1}{x-1}<-(1+|\alpha|)<\alpha$. Thus $\lim_{x \to 1^-} \frac{x}{x-1} (x\neq 1)=-\infty$
Conclusion: Since $\lim_{x \to 1^-} \frac{x}{x-1} \neq \lim_{x \to 1^+} \frac{x}{x-1}\implies \lim_{x \to 1} \frac{x}{x-1}$ DNE
Would this be correct?
AI: Edit: At this point, it looks pretty much fine, though it doesn't matter that $\frac1{x-1}<\alpha.$ Rather, since $\frac1{x-1}<-(1+|\alpha|),$ then we have $$\frac{x}{x-1}=\frac{(x-1)+1}{x-1}=1+\frac1{x-1}<1+-(1+|\alpha|)=-|\alpha|\le\alpha,$$ and so we're done. |
H: What is the value of $ (a+b) $ where $ a\log_{1971}3 + b\log_{1971}{73} = 2012 $
I have two integers which are a and b . They satisfy the following equation which is $ a\log_{1971}3 + b\log_{1971}{73} = 2012 $ . I want to know the value of $ (a+b) $ .
I have tried to solve this question and got the following equation . Then I cant proceed .
$ 1971^{2012} = 3^a73^b $
What can I do ?
AI: We know that $1971=3^3\cdot73$. Since $1971$ is raised to the $2012^{th}$ power, we can easily see what $a$ and $b$ are. |
H: What's the name of $x^x$?
I know that
$$f{(x)} = a^x$$
is called exponential function and
$$f{(x)} = x^a $$
is the power function.
But what is the name of $f{(x)} = x^x$?
AI: As far as I'm aware, it doesn't have a nontrivial one. It is its own name. |
H: $\text{SL}(2, \mathbb Z)$ homework question
I'm trying to show that if $G = \langle \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}, \begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}\rangle \subset \text{SL}(2, \mathbb Z)$, then $-\text{Id} \in G$. So far I have described $G_1 = \langle \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}\rangle$ and $G_2 = \langle \begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix}\rangle$ and then tried to express $-\text{Id}$ as a composition of elements of $G_1$ and $G_2$ but it didn't work. I'm not even convinced thought that's the right strategy. Maybe there's a more theoretic argument that I'm missing? Can somebody please give me a hint how to proceed? Thanks.
AI: Hints:
Well, to begin with:
$$S^{-1}=\begin{pmatrix}1&1\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}1&-1\\0&\;\;1\end{pmatrix}$$
$$T^{-1}=\begin{pmatrix}1&0\\1&1\end{pmatrix}^{-1}=\begin{pmatrix}\;\;1&0\\-1&1\end{pmatrix}$$
$$ST^{-1}=\begin{pmatrix}\;\;0&1\\-1&1\end{pmatrix}\;,\;\;\left(ST^{-1}\right)^2=\begin{pmatrix}-1&1\\-1&0\end{pmatrix}\;\ldots\ldots$$ |
H: Is this graph a graceful tree?
Suppose we have a graph $G=(V,E)$ where $V=\{0,1,\ldots,n\}$ and $E$ consists of $n$ edges
in such a way that the set of absolute differences $\{|i-j||ij\in E\}$ is exactly the set
$\{1,\ldots,n\}$. In other words: if $G$ is a tree, it is a graceful tree.
But can we be sure that $G$ indeed is a tree? Why can't there be any cycles?
I think $G$ must be a tree, but I am missing the right approach to prove it. Any ideas for a proof? Counterexamples?
AI: $G$ doesn't have to be a tree. For example let $n=4$ and let the edges be $02, 03, 23, 04$. |
H: Measuring water puzzle
Given two unmarked jugs, one which holds 7 liters,
and another which holds 11 liters,
an unlimited supply of water, and no need to conserve,
how do you measure exactly 6 liters?
I would also like to know whether there is a single solution , many solution or no solution?
Can it measure any amount?
AI: Multiples of 7: 7,14,21,28,35...
Multiples of 11: 11,22,33.
We observe that the difference 6 appears between 28 and 22
So you keep filling the 7 liter jug and emptying it into the 11 liter jug
After 4 fillings you will be left with 6 liters in the 7-liter jug
Edit:adding 2nd solution.
Likewise you can search for a multiple of 11 that is 6 more than a multiple of 7.
55=5*11, 49=7*7
So you can fill the 11 liter jug 5 times and empty it in the 7 liter jug. In the end you will be left with 6 liters in the 11-liter jug |
H: Is there a relationship between SVD and similarity transformation?
Is there a relationship between SVD and similarity transformation? I mean, in articles I read about the SVD method, I came across the equation $A=USV^T$ which seems to me like similarity transformation $A^\prime=QAQ^T$. Is that right? How are these two concepts connected?
AI: Not directly. An SVD of $A$ is related to the orthogonal (unitary in $\mathbb{C}$) similarity diagonalization of a positive definite matrix $A^TA$ (or $AA^T$), because
$$A^TA = (U \Sigma V^T)^T (U \Sigma V^T) = V \Sigma^T \Sigma V^T.$$
If $A$ is a square matrix, then
$$A^TA = V \Sigma^2 V^T.$$
Other than that, (also if $A$ is a square matrix!) you can put
$$A = U \Sigma V^T = U (\Sigma V^T U) U^T,$$
which is a similarity of $A$ and $\Sigma V^T U$, which is a product of a diagonal and orthogonal matrix, but I don't see how would that be useful. Similarly, you could also do
$$A = U \Sigma V^T = V (V^T U \Sigma) V^T.$$
Of course, if $A$ is positive semidefinite, then you can choose $U$ and $V$ such that $U = V$, so the SVD is the same as the Eigenvalue decomposition, but this is just a special case. |
H: Riddle that doesn't make sense and makes sense at the same time
How many letters are there to the CORRECT answer to this question?
And if you do get the answer please explain how you got it.
AI: Four.
$${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$$
In Spanish: Cinco.
In Turkish: Dört.
In Russian: три.
In response to the edit:
Parsing the question is not hard. The question seems nonsensical because it means almost nothing in daily life. But if you examine the structure, it becomes clear.
The question begins with "how many," which indicates a numerical answer. The question then asks for a number of letters in an answer. Understanding that the answer is a textual response, then the number of letters referred to must be the total number of letters in the response.
Since any numerical answer can be an answer, we then examine possible answers for correctness. Since there is an implied self-reference, we conclude that the only way for an answer to be correct is if the number of letters in the response is the same as the value of the response.
Hence, four... in English. |
H: Algebra Math Contest Question
A novel has 6 chapters. As usual, starting from the first chapter begins on a new page. The last chapter is the longest and the page numbers of its pages add up to 2010: How
many pages are there in the first 5 chapter?
Let the pages be numbered
$$1,\dots,X_{1}; X_{1}+1,\dots,X_{2}; X_{2}+1,\dots,X_{3}; X_{3}+1,\dots,X_{4}; X_{4}+1,\dots,X_{5}; X_{5}+1,\dots,X_{6}$$
Sum of the last chapter pages $= [(X_{6}+X_{5}+1)(X_{6}-X_{5})]/2 = 2010$
(Note sum of arithmetic series with $a_{1} = X_{5}+1$ and $a_{n} = X_{6}$)
Now $2\cdot 2010 = 4020 = 67\cdot 60 = 134\cdot 30 = 268\cdot 15$
Thus $X_{6}+X_{5}+1 = 67$ and $X_{6}-X_{5} =60$, which gives $X_{6} = 63$, and $X_{5} = 3$. This is not possible as there are five chapters.
Next $X_{6}+X_{5}+1 = 134$ and $X_{6}-X_{5} = 30$ gives non-integer. So this is not possible.
Next $X_{6}+X_{5}+1 = 268$ and $X_{6}-X_{5} = 15$ gives $X_{6} = 141$, and $X_{5} = 126$. This seems plausible but it does not pass the test that chapter 6 is the longest as you could have a longer chapter when $X_{5} = 126$. Any other product of factors would still not hold the condition of the "Longest Chapter". What seems to to be answer for this?
AI: You missed some factorizations. For example
$$4020=201*20$$
$$X_{6}+X_{5}+1 = 201 \,;\, X_6-X_5=20 \Rightarrow X_6=110, X_5=90$$
And I think, there are few more you missed.
Note
$$4020= 2^2*3*5*67$$
has $(2+1)(1+1)(1+1)(11)=24$ divisors, thus there are exactly $24/2$ ways of writing it as a product of two integers. Anyhow, since the two factors must have different parity, only $(1+1)(1+1)(1+1)=8$ ways [namely one factor has to be $4$ times a divisor of $3*5*67$] are relevant for this problem.
Note To solve the problem, you need to check all the $8$ cases. Note that in a case, if you get $ X_6-X_5 < \frac{X_5}{5}$ this tells you that it is impossible for the last chapter to be the longest.
You are hoping to get your answer by elimination. If you can prove that 7 cases are impossible, since the problem is telling you there is one possibility by elimination it has to be the remaining case... |
H: Brownian motion is almost surely not differentiable everywhere
Could anyone point out the difference between the statement of the following theorems:
1) For any $t\ge0$, Brownian motion is almost surely not differentiable at $t$.
2) Almost surely, the sample paths of Brownian motion are not Lipschitz continuous at any point. In particular, the sample of Brownian motion are almost surely not differentiable at any point.
The author of the book argues that to show 2) a more refined argument is needed. I'm a bit confused about the difference between the above statements. Could someone explain this?
AI: Since there are uncountably many $t$'s, it is not automatic that something that is almost surely true for any given $t$ is almost surely true for all $t$.
For example, consider a Poisson process $N(t)$. For each $t$, $N(t)$ almost surely doesn't have a jump at $t$. On the other hand, with probability $1$ the process does have a jump somewhere. |
H: Is the following legal: $\frac{d}{d \theta}\cos(n\theta)=\frac{d \cos n\theta}{d \cos \theta}\frac{d \cos \theta}{d \theta}$?
Is the following application of chain rule legal? $$\dfrac{d}{d \theta}\cos(n\theta)=\dfrac{d \cos n\theta}{d \cos \theta}\dfrac{d \cos \theta}{d \theta}$$
I need to find out how $\cos n\theta$ changes in accordance to $\cos \theta$, so can I do it by dividing the left side by the second part of the right side? I'm just not used to seeing 'entering' the result like I do here when I put an $n$ inside $\cos \theta$.
AI: It's stated in a somewhat bizarre way, but basically it's true. A better way of saying it: $\cos(n\theta) = T_n(\cos(\theta))$ (where $T_n$ happens to be a Chebyshev polynomial, assuming $n$ is a nonnegative integer), and then
$$\dfrac{d}{d\theta} \cos(n\theta) = T_n'(\cos(\theta)) \dfrac{d}{d\theta} \cos(\theta)$$ |
H: Calculus finding critical points
Find the Critical Points for this equation:
$$x e^x = 0$$
I know its going to be either 0 or 1 but how do I get there? Thanks
AI: The equation
$xe^x = 0 \tag{1}$
has exactly one solution, $x = 0$, since $e^x \ne 0$ for all $x$. If one defines a critical point in the usual sense to be a point where the derivative of a function is zero, then one needs to consider a function for sense to be made; in this case I suppose we should set
$y(x) = xe^x; \tag{2}$
then
$y'(x) = e^x + xe^x \tag{3}$
and if we want $x$ such that $y'(x) = 0$, we must have
$e^x + xe^x = 0, \tag{4}$
implying
$x = -1. \tag{5}$
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: Question on Galois Connections
I have been learning about Galois Connections, and came across Steve Roman's book "Lattices and Ordered Sets." Two questions:
There is a theorem which says that if $(\Pi,\Omega)$ is a Galois connection on $(P,Q)$, where we write $p^*=\Pi(p)$ and $q'=\Omega (q)$, then $$p^{*'*}=p^* \text{ and } q^{'*'}=q'.$$ However, then it claims that from this fact it follows that the composite maps $(\;)^{*'}$ and $(\;)^{'*}$ are closure operators on $P$ and $Q$, respectively. Isn't it better to say that they are closure operators on $Q'$ and $P^{*}$, respectively? (hopefully the notation is clear). It seems to me there is no guarantee they are closure operators on all of $P$ or $Q$, right? Or am I missing something form the definition of closure operator?
An exercise in the same section asks to prove that, if $P$ and $Q$ are lattices, then De Morgan's laws hold in Cl$(P)$ and Cl$(Q)$.
I am not sure how to begin with this problem. I don't see how the fact that $P$ and $Q$ have to be lattices is relevant.
I have a feeling that I will have to go back and forth between $P$ and $Q$ in some way, but I do not know how to relate the lattice operations to the maps in the Galois Connection, nor do I know how to reconcile the fact that all you really know is that $p\leq p^{*'}$ (no equality).
Any ideas or hints as to how to get started?
AI: This asterisk and apostrophe stuff is a little annoying to type, so I'm going to switch to $F:P\to Q$ and $G:Q\to P$.
So, $GF$ is a closure operator on $P$ in the sense that it is nondecreasing ($p\leq GF(p)$) and idempotent $GF(GF(p))=GF(p)$. Something similar applies for $FG$ on $Q$. The idea is that $GF$ extends elements to a "closed" element containing the original element. The equality $FGF=F$ allows you to conclude that $GF$ is an idempotent operator on $P$, and similarly the other equality allows you to conclude that $FG$ is idempotent on $Q$.
Restricting $GF$ to operate only on $G(Q)$ would be boring because all of the elements of $G(Q)$ are "already closed": $GF(G(q))=G(q)$ for all $q\in Q$! So $GF$ doesn't actually expand those elements at all. It's more interesting in cases where $GF(p)>p$, where the nonclosed element $p$ became closed after $GF$ is applied.
For $2)$, DeMorgan's laws are statements about greatest lower bounds and least upper bounds. You need the posets to be lattices so that GLB's and LUB's are defined in the first place. Otherwise you are trying to prove things about things that don't exist :)
Added
When I get to the DeMorgan's Laws part, I have a feeling you might have been meant to work with antitone connections all along. So for this part, let $F$ and $G$ be order reversing. $FG$ and $GF$ still have the idempotence and nondecreasingness they had before. Let $a,b\in F(P)$.
In particular, $FG(a)=a$ and $FG(b)=b$. Actually, $F(P)$ is closed under meets and joins. Here is how to show it for joins: I will leave meets up to you. Since $G\underline{(a\vee b)}\leq \underline{G(a\vee b)}$, it follows that $F(\underline{G(a\vee b)})\leq \underline{a\vee b}$. But this implies $FG(a\vee b)\leq a\vee b$, and the other containment is taken care of by the fact $FG$ is nondecreasing. Thus $FG(a\vee b)=a\vee b$.
One of DeMorgan's Laws on $F(P)$ would now look like this: $G(a\vee b)=G(a)\wedge G(b)$. Let's establish this particular one.
You have that $a,b\leq a\vee b$, which immediately yields $G(a\vee b)\leq G(a), G(b)$. By definition of the greatest lower bound, we have
$$G(a\vee b)\leq G(a)\wedge G(b)\leq G(a),G(b).$$
Applying $F$ reverses these:
$$a\vee b\geq F(G(a)\wedge G(b))\geq a,b$$
(Notice I used the fact that $FG$ is the identity on $F(P)$ to delete some unnecessary $FG$'s.)
But then by definition of the least upper bound, $a\vee b= F(G(a)\wedge G(b))$. Another application of $G$ on both sides yields that $G(a\vee b)=GF(G(a)\wedge G(b))=G(a)\wedge G(b)$.
I'm going to encourage you to work out $G(a\wedge b)=G(a)\vee G(b)$ on your own. You could reprove everything for $F$ too, but your teacher might rather want to see you appeal to similarity instead of the extra text :) |
H: If two limits don't exist, are we sure the limit of sum doesn't exist?
This is a true/false question.
If the limits $$\lim_ {x\to a}\ f(x)\quad \text{ and }\quad \lim_ {x\to a}\ g(x)$$ don't exist, limits $$\lim_ {x\to a}\ (f(x)+g(x))\quad \text{ and }\quad \lim_ {x\to a}\ (f(x)g(x))$$ don't exist also.
My idea is, that, because $$\lim_ {x\to a}\ (f(x)+g(x))=\lim_ {x\to a}\ f(x)+\lim_ {x\to a} g(x)$$ then the limits doesn't exist, correct?
AI: $\mathbb{FALSE}$.
Let $f=1/(x-a)$ and $g=-1/(x-a)$
Your product hypothesis is false too;
Let $f=1/\sin(1/(x-a))$ and $g=\sin(1/(x-a))$ |
H: Fibonacci Proof: Prove that $\frac{F_n-F_{n+16}}{7}$ is always an odd integer.
Im stuck on this question, i have to prove that $$\frac{F_n-F_{n+16}}{7}$$ is always an odd integer. I tried induction to do this but i just can't see how to prove it.
thanks for any help
AI: I suspect that, instead, you are merely supposed to show that $$\frac{F_n-F_{n+16}}7$$ is an integer for all $n$. As discussed in the comments above, the integer needn't be odd.
I leave the two base cases to you (we need two of them for an induction proof).
Now, let's suppose that $$\frac{F_n-F_{n+16}}7$$ and $$\frac{F_{n+1}-F_{(n+1)+16}}7$$ are both integers for some $n$. Note, then, that
$$\begin{align*}
\frac{F_{n+2}-F_{(n+2)+16}}7&=\frac{(F_n+F_{n+1})-(F_{n+16}+F_{(n+1)+16})}7\\\\
&=\frac{F_n-F_{n+16}}7+\frac{F_{n+1}-F_{(n+1)+16}}7\;.
\end{align*}$$ |
H: Eigenvector and eigenvalue for differential operator
My friends and I have been struggling with the following problem, and don't understand how to do it. We have gotten several different answers, but none of them make sense. Can you help us?
Problem statement: Let $V$ be the vector space of real-coefficient polynomials of degree at most $3$. Let $D:V \rightarrow V$ be the differential operator; $D(p(x))=\frac{d}{dx}p(x)$. Give an example of an eigenvector for $D$. What is the corresponding eigenvalue?
We ended up getting that $\frac {d}{dx}p(x)=\lambda p(x)$, so that $p(x)=\frac{x^2}{2}$. Is this correct?
Thanks.
AI: You can restrict $D$ to the finite dimensional subspace $V_n$ of polynomials of degree $\leq n=\deg p.$ Then $D^{n+1}=0$ on $V_n.$ Hence the only eigenvalue of $D$ is $0$ and the eigenvectors a constant polynomials. |
H: Expected value on non-normalised PDFs
Suppose the following is known:
$$\int{g(x)dx}>\int{g'(x)dx}$$
Considering that $g=kf$ and $g'=k'f'$ where $f$ and $f$ are probability distributions on $X\in[0,1]$. Is the following true:
$$E_g(x)> E_{g'}(x)$$
More rigorously:
$$\int{xg(x)dx}>\int{xg'(x)dx}$$
AI: That is not true. Simple example: $g(x)=\frac{7}{12}$ and $g'(x)=x$. |
H: Combined Cumulative Probabilities of Success
If you undertake an iterative action 10 times and each time the probability of that action failing is 4% each time, how do you calculate the probability that the outcome of all 10 actions will be successful (meaning, each of the 10 action must succeed)? The result I am getting is 66.5%
Now, how do you calculate the probability that if this iterative 10-step action were taken twice at the same time by different actors at least ONE of the actors would succeed?
AI: The probability of success of single actor is approximately $0.665$. The probability they both fail the task is approximately $(1-0.665)^2$. The probability at least one will succeed is... |
H: If a and b are matrices then what does a|b stand for?
Given $a =(1, 2, 3)^T$ and $b = (5, 8, 4)^T$
what is $C=(a|b)$ mean?
is this a given $b$? or a over $b$? how to work out $C$?
AI: It means augmented. So in this case:
$$
\mathbf{C}=\mathbf{a}|\mathbf{b}=\big[\begin{array}{c|c}\mathbf{a}&\mathbf{b}\end{array}\big]=\left[\begin{array}{c|c}1 & 5 \\ 2 & 8 \\ 3 & 4 \end{array}\right].
$$ |
H: An implication involving filters
Let $z$ be a filter on a set $U$ and $I$, $J$ be sets of filters on $U$.
Is the following implication always true?
$$(\forall K \in z\, \exists x \in I \cup J : x \supseteq \mathop\uparrow\!\! K) \Rightarrow (\forall K
\in z\, \exists x \in I : x \supseteq \mathop\uparrow\!\! K) \vee (\forall K \in z\, \exists x \in J : x
\supseteq \mathop\uparrow\!\! K)$$
where $\mathop\uparrow\!\! K$ is the principal filter generated by a set $K$.
If it is not always true (counterexample?), which additional conditions may we require to make it provable?
I need this for my own research.
AI: The assertion is true. It’s equivalent to the simpler assertion
$$\forall K\in z\,\exists x\in I\cup J(K\in x)\to\Big(\forall K\in z\,\exists x\in I(K\in x)\lor\forall K\in z\,\exists x\in J(K\in x)\Big)\;.$$
Suppose that it fails for some $z,I$, and $J$. Then there are $K_I,K_J\in z$, $x_I\in I$, and $x_J\in J$ such that $K_I\in x_I$, $K_J\in x_J$, $\forall x\in I(K_J\notin x)$, and $\forall x\in J(K_I\notin x)$. Let $K=K_I\cap K_J\in z$; there is an $x\in I\cup J$ such that $K\in x$. Without loss of generality $x\in I$. But then $K_J\supseteq K\in x\in I$, so $K_J\in x\in I$, contradicting the choice of $K_J$. |
H: $G/H$ is a Factor Group. Are its subgroups isomorphic to other factor groups?
Given a group $G$, let $H$ be a normal subgroup of $G$. Suppose $A$ is a subgroup of $G/H$. Does there exist a normal subgroup $K$ of $G$ such that $G/K$ is isomorphic to $A$?
AI: No. Several posters have already given the correct formulation of the type of statement you are after. I will give you a counterexample to your statement as is.
Recall that $A_5$, the alternating group on $5$ symbols, is simple (a similar argument will work with any other simple group). $A_5$ is a quotient of $A_5\times A_5$ (just mod out by the first factor). Now we can take a non-trivial subgroup of $A_5$ such as $A_4$. There is no way to realize $A_4$ as a quotient of $A_5\times A_5$.
To see this, suppose we have a homomorphism from $A_5\times A_5$ to $A_4$. Then we have the composition $A_5\to A_5\times A_5 \to A_4$, where the first homomorphism is given by $x\mapsto (x,id)$. Then, since $A_5$ is simple and $A_5$ is larger than $A_4$, our homomorphism $A_5\to A_4$ sends everything to the identity. Thus anything of the form $(x,id)\in A_5\times A_5$ maps to $0$. But we can say the same thing (for the same reasons) about elements of the form $(id,y)$. Since every element $(x,y)$ in $A_5$ can be expressed as a product $(x,id)\cdot (id,y)$, every homomorphism from $A_5\times A_5$ to $A_4$ is thus trivial, and thus there is clearly no surjective homomorphism, as you require. |
H: Differentiation: chain rule over multiple arguments
I am wondering whether the following generally holds:
$$ \frac{ \delta f(g(x), h(x))} {\delta x} = \frac{ \delta f(g(x), h(x))} {\delta g(x)} *\frac{ \delta g(x)} {\delta x} + \frac{ \delta f(g(x), h(x))} {\delta h(x)} *\frac{ \delta h(x)} {\delta x} $$
If so, what is the name of this rule, or of which existing rule is it a special case?
AI: Yes. It is the chain rule for multivariable calculus.
If you put $p(x) = (g(x),h(x))$, you can write the multivariable chain rule as
$$\frac{df}{dx} = \nabla f(p) \cdot p'(x)$$
to more closely resemble the one-variable chain rule. |
H: Cyclic subgroups dealing with prime numbers Question
Let G be a group with |G| = pq. (p and q are both prime).
How can we prove that every proper subgroup of G is cyclic?
AI: Hints:
1) Lagrange's Theorem
2) A group of order a prime a number is always cyclic . |
H: Image of a nilpotent Lie algebra
Suppose $L_1$ and $L_2$ are two Lie algebras, and that $f: L_1\to L_2$ is a Lie algebra homomorphism. If $L_1$ is nilpotent, does it follow that $f(L_1)$ is nilpotent?
Remark 1. The corresponding statement for groups is true. In other words, homomorphic image of a nilpotent group is nilpotent.
Remark 2. If $L_1$ is solvable Lie algebra, then $f(L_1)$ is solvable. This is also well-known.
AI: Unless I'm missing something, isn't it obvious? No need to quote a book. Take any sequence $f(x_1), f(x_2), ..., f(x_r)$ in $f(L_1)$, then if the sequence is large enough $ad_{x_1} ... ad_{x_r}$ is zero on any $x \in L_1$, and since $f$ is a Lie algebra morphism this implis that $ad_{f(x_1)} ... ad_{f(x_r)}$ is zero on $f(x)$. This is the definition of nilpotency. |
H: Prove that symmetric closure of R $h_{sym}(R) = R \cup R^T$
I have a question about proving statements of the form in which the given is the union of two sets. Well first of all let me just write how I tried to prove the statement:
To show that $h_{sym}(R) = R \cup R^T$, you can show that $h_{sym}(R) \subseteq R \cup R^T$ and $R \cup R^T \subseteq h_{sym}(R)$.
Assume $(x,y) \in h_{sym}(R)$. Since by definition $h_{sym} R = \bigcap \{S \subseteq A \times A : $ S is symmetric and $R \subseteq S \}$, it follows that $R\subseteq h_{sym}(R)$.
Now my question is, is this sufficient to prove that $(x,y)\in R \cup R^T$? Basically what $R \cup R^T$ means is that for any elements $(x,y)$, $(x,y)\in R \vee (x,y)\in R^T$, so it should be sufficient to prove that (x,y) is an element of one of the sets of the union, correct?
For the other direction:
Assume $(x,y)\in R\cup R^T$, then there are two cases to be looked at:
Case 1: Let $(x,y)\in R$, since $R\subseteq h_{sym}(R)$, it follows that $(x,y)\in h_{sym}(R)$.
Case 2: Let $(x,y)\in R^T$, since by definition $R\subseteq h_{sym}(R)$, it follows that $(x,y)\in R\cap R^T$, since $h_{sym}(R)$ is symmetric, it follows that $(x,y)\in h_{sym}(R)$.
As you can probably tell I'm still new to writing proofs and would very much appreciate an answer to my question and a brief feedback on whether the proof is correct (I have my doubts about Case 2 being correct) and in general what could be done better.
AI: Your argument that $h_{sym}(R)\subseteq R\cup R^T$ simply doesn’t work. This is actually the harder direction. Here’s one way to do it.
Let $\langle x,y\rangle\in h_{sym}(R)=\bigcap\{S\subseteq A\times A:S\text{ is symmetric and }R\subseteq S\}$. Suppose that $\langle x,y\rangle\notin R\cup R^T$; then $\langle x,y\rangle\notin R$, and $\langle x,y\rangle\notin R^T$, so $\langle y,x\rangle\notin R$. Let $$T=h_{sym}(R)\setminus\{\langle x,y\rangle,\langle y,x\rangle\}\;.$$ Now get a contradiction by showing that $T$ is a symmetric relation on $A$, and $$R\subseteq T\subsetneqq h_{sym}(R)\;.$$ Conclude that $\langle x,y\rangle\in R\cup R^T$ after all.
However, there’s an easier way, using the fact that if $\mathscr{S}$ is any collection of sets, and $S\in\mathscr{S}$, then $S\supseteq\bigcap\mathscr{S}$:
Observe that $R\cup R^T$ is symmetric and contains $R$, so that $$R\cup R^T\in\{S\subseteq A\times A:S\text{ is symmetric and }R\subseteq S\}\;;$$ this immediately implies that $$R\cup R^T\supseteq\bigcap\{S\subseteq A\times A:S\text{ is symmetric and }R\subseteq S\}\;.$$
The second case of your argument for the opposite inclusion is also flawed. You have $\langle x,y\rangle\in R^T$, and you want to conclude that $\langle x,y\rangle\in h_{sym}(R)$. The fact that $R\subseteq h_{sym}(R)$ does not imply that $\langle x,y\rangle\in R$, which is what you’d need in order to justify your claim that $\langle x,y\rangle\in R\cap R^T$. However, if $\langle x,y\rangle\in R^T$, then $\langle y,x\rangle\in R\subseteq h_{sym}(R)$, and $h_{sym}(R)$ is symmetric, so $\langle x,y\rangle\in h_{sym}(R)$, as desired. |
H: induction proof of a determinant $n \times n$
I have to proof the following property:
Can somebody help my with a few steps for n=n+1?
Thanks in advance. Cheers.
AI: Note that the matrix can be written as
$$(a-1)I_n + \vec{1} \vec{1}^T$$
Now we will make use of the powerful Sylvester determinant theorem, we have
$$\det\left(I_n + uv^T \right) = 1 + v^Tu$$
Hence,
$$\det\left((a-1)I_n + \vec{1} \vec{1}^T \right) = (a-1)^n \det \left(I_n + \dfrac{\vec{1} \vec{1}^T}{a-1}\right) = (a-1)^n\left(1+\dfrac{ \vec{1}^T\vec{1}}{a-1}\right) = (a-1)^{n-1}(a+n-1)$$ |
H: Looking for math software to play around with?
I am doing some research and I want to use software like Scilab or MATLAB to plug in values for a, for example, $2\times 2$ matrix and see how the linear combination of vectors' direction and magnitude are changed when different numbers (or eigenvalues, etc). are manipulated. However I tried scilab and it is extremely hard and I don't know how to write matrices in scilab. In fact, what is the best software to do this in?
I can quickly clarify any unclearances.
I FORGOT TO ADD I HAVE A MACBOOK.
AI: IMHO you really could learn a lot from Mathematica, language is pretty simple and very connected to algebraic manipulations, also with very few knowledge of the language (even plain and simple English) you can do most of the computational stuff from calculus, linear algebra, complex analysis and discrete mathematics. |
H: Let p be an odd prime number, and a be an integer. Prove that if a^2 ≡ 1 (mod p), then a ≡ ±1 (mod p).
Not sure how to approach this. I can't apply FlT since p is not given. Can I just raise both sides to 0.5? Doesn't seem proper.
AI: If
$$a^2 = 1\mod p,$$ then $$a^2-1=0\mod p.$$ That means $$(a+1)(a-1)=0\mod p.$$ Therefore, $p|(a+1)(a-1)$. Since $p$ is prime, what does that tell you about $a+1$ and $a-1$? |
H: In a ring if a product is invertible with one factor is not a divisor of zero then both factors are invertible
Let $(A,+,.)$ be a ring, and $a$ et $b$ two elements of $A$ such that $ab$ is invertible and $b$ is not a divisor of zero. I want to show that $a$ and $b$ are invertible.
My try: since $ab$ is invertible there exists $x\in A$ such that $(ab)x=1$ and $x(ab)=1$ by associativity of $"."$, we have that $a(bx)=1$ so to show that $a$ is invertible it remains to show that $(bx)a=1$ but i don't see how to do it. Thanks for your help!!
AI: From $xab=1$ you have $bxab=b$, hence $(bxa-1)b=0$ and as $b$ is not a zero divisor $bxa=1$. Since also $abx=1$, we see that $bx$ is a (two-sided) inverse of $a$ and $xa$ is a (two-sided) inverse of $b$. |
H: Open balls connecting two points of connected set
Let $U\subseteq\mathbb{R}^n$ be an open, connected set, and let $a,b\in U$. Is it always true that there exists open balls $B_1,\ldots,B_k\subseteq U$ such that $a\in B_1,b\in B_k$, and $B_i\cap B_{i+1}\neq\emptyset$?
I think it should be true, but it's not clear how this will follow from the definition of a connected set (cannot be partitioned into two nonempty disjoint open sets.)
AI: Pick $a\in U$ and let $A$ be the set of points for which suc a ball chain to $a$ exists.
Then $A$ is open as from any reachable point $x$ we may reach all points in a ball around $x$.
But $A$ is also (relatively) closed because if $x$ is not reachable, then also points sufficiently close to $x$ are not reachable (namely if $B_r(x)\subseteq U$, then this holds for all points in $B_{\frac r2}(x)$).
The only nonempty clopen subset of a connected space is the whoile space, voila.
This is essentially the usual argument to show path-connectedness of connected opne subsets of $\mathbb R^2$, but here applied directly to "ball-chain" connectedness. |
H: Arithmetic and Geometric Mean Inequality
Use the AM - GM inequality (no other method is acceptable), to prove that for all positive integers $n$:
$$\left(1 +\dfrac{1}{n}\right)^n \leq \left(1 + \dfrac{1}{n+1}\right)^{n+1}$$
I see that it is increasing... Don't know how to keep going.
AI: One way is to rewrite the inequality as:
$$\sqrt[n+1]{\left(1+\frac{1}{n}\right)^n\cdot 1}\le \frac{n\left(1+\frac{1}{n}\right)+1}{n+1},$$
and note that this is precisely AM-GM for $\left(1+\frac{1}{n}\right)$ taken $n$ times and $1.$ |
H: Number of elements in $\{ x^{p^n}: x \in K \} \subseteq K$ for a field $K$
I have another question about ring theory.
Let $K$ b a field, with $p>0$ its characteristic. Define the set $T \:= \ \{ x \in K: x^{p^n}=x \} $. I had to prove that $|T| \leq p^n$.
Research effort
We may assume that $p$ is prime, because otherwise the ring would have zero divisors. The exercise stated that $T$ itself is a field. I tried several things to prove the statement above.
First of all $x \in T \ \iff \ x^{p^n}=x \ \iff \ x(x^{p^2-1}-1)=0 \ \iff \ x=0 \ \ \text{or} \ \ x^{p^2-1}=1$.
We see now if $T^*$ would be cyclic, $|T^*| = |\langle g\rangle| \leq p^n-1$, so that $|T| \leq p^n$.
My syllabus states the following:
Let $R$ be a domain, and let $G$ be a finite subgroup of $R^*$, than $G$ is cyclic.
This means that it's enough to show that $T$ is a finite set, but I failed to do so.
Could you provide me a hint please? It would be appreciated
AI: A polynomial of degree $n$ in a field $F$ has at most $n$ roots. |
H: Given a theorem can it always be reduced logically to the axioms?
It's probably a silly question but I’ve been carrying this one since infancy so i might as well ask it already.
let ($p \implies q$) be a theorem where $p$ is the hypotheses and $q$ is the conclusion. If stated in logical symbols can it always be reduced to the axioms by logical manipulation?
Does that mean that if different rules of logic are chosen (say quantum logic) one cannot prove theorems the "normal" (language and common sense) way since intuition fails?
AI: This question is quite natural, as the notion of "rigorous proof" largely depends on the context in which it is mentioned.
The short answer is Yes. Every mathematical proof, if correct, can be formulated as a derivation starting from axioms (usually of ZFC), and using basic deduction rules.
In practice, it is of course infeasible, so what we call a "proof" is a description containing enough information to convey the ideas necessary to build this rigorous proof. This is (almost) never done, but should be always possible.
Therefore, what is sufficient to constitute a "proof" highly depends on the interlocutors exchanging it, because they have to be able to fill the gaps with their knowledge. For instance, a proof of most highschool problems is reduced to "trivial" for professional mathematicians.
In the end, the answer is "yes", but you are wrong with your last sentence: a "human" proof is transformed into a "logical" one not by logical manipulation, but by filling in a lot of gaps, that are omitted in human formulations.
If, as you mention, different rules for logic are chosen, the proofs need to be compatible with these new rules, i.e. extendable into logical proofs following these rules. |
H: distance it's attained on closed and finite dimensional subspaces of a Banach space
Let $X$ be a Banach space. And let $F\subset X$ be a closed and linear subspace (in particular is Banach). I want to prove the following:
Let $d(x,F)=\displaystyle \inf_{y\in F} ||x-y||$. Is it true that there exist $y \in F$ such that $d(x)=||x-y||$?
What if $F$ it's also finite dimensional?
AI: In general, there need not exist a $y \in F$ that realises the distance. If $F$ is finite-dimensional, or if $X$ is reflexive, then there always exists such a $y$.
In the case of finite-dimensional $F$, the intersection of the closed balls around $x$ and $F$ is compact, since $F$ is locally compact. Thus the intersection
$$\bigcap_{r > d(x,F)} F\cap \overline{B_r(x)}$$
is not empty, since all finite intersections are non-empty. All elements of that intersection realise the distance.
If $X$ is reflexive, the closed balls are weakly compact, and the same argument of compactness guarantees that the above intersection is non-empty. |
H: isomorphism between two factor groups
We know that $A$ and $B$ are subgroups of $G$. Moreover, $B$ is a normal group of $G$. I've proved that if $BA=AB$ then $AB$ is a subgroup of $G$.
Then I have to prove that factor group $A/(A \cap B)$ and $BA/B$ are isomorphic.
I have shown that there are two natural homomorphisms $\pi: A \rightarrow BA/B$ and $\phi: A \rightarrow A/(A \cap B)$.
Could you give me a suggestion how to finish the proof?
AI: Hint: It suffices to show that $\pi$ is onto and that $\ker(\pi)=A\cap B$ then the conclusion follows from the first isomorphism theorem. |
H: determinant of the following matrix
I have to find the determinant of the following $n\times n$ matrix in terms of $n$:
$\begin{bmatrix} 2&1&0&0&&&&\\ 1&2&1&0&&&& \\ 0&1&2&1&&&&\\ 0&0&1&2&&&& \\ &&&&\ddots \\&&&&&2&1\\&&&&&1&2\end{bmatrix}$
I found a recursion. Let $d_n$ be the determinant of the $n\times n$ matrix, then $d_{n+1}=2d_n-d_{n-1}$, this can be easily seen. From here I cannot find a explicit formula for the sequence. I need hints. Thanks
AI: If $d_{n+1}=2d_n−d_{n−1}$ then $d_{n+1}-d_n=d_n−d_{n−1}$ so you have an arithmetic progression. So you just need to calculate the first two terms to know the rest.
$|2|=2$ and $\begin{vmatrix} 2&1\\ 1&2\end{vmatrix} =2^2-1^2=3$ which tell you the starting point and the rate of increase |
H: What is the variance of this random variable?
A clerk drops $n$ matching pairs of letters and envelopes. He then places the letters into
the envelopes in a random order. Let $X$ be the number of correctly matched pairs. Find the
variance of $X$.
AI: Let $Y_i=1$ if the $i$-th letter is in the right envelope, and let $Y_i=0$ otherwise. Then
$X=Y_1+Y_2+\cdots+Y_n$.
Now $E(X)$ is easy to find using the linearity of expectation. It remains to find $E(X^2)$.
Square $Y_1+Y_2+\cdots+Y_n$. We get $\sum Y_i^2$ plus a bunch of "mixed" terms. If we can find the expectation of each, we will be finished.
Calculation of $E(Y_i^2)$ is easy, since $Y_i^2=Y_i$. To find $E(Y_iY_j)$, find the probability that $Y_i=1$ and $Y_j=1$. |
H: Expressing an oscillator as a series of ODEs
Consider an oscillator satisfying the initial value problem $u''+w^2u=0$, where $u(0)=u_0$, $u'(0)=v_0$. Let $x_1 = u$, $x_2=u'$, and transform the equations given into the form $x' = Ax, x(0)$. Then using $$\exp(At) = I + \sum_{n=1}^\infty {A^nt^n\over n!}$$ show that $$\exp(At)= I\cos(wt) + A({\sin(wt)\over w})$$
I've gotten as far as substituting $x_1$ and $x_2$ for $u$, but I am not sure what to do next. I have $x_2'+ w^2x_1=0$. Any help you can give will, as always, be greatly appreciated.
AI: We are given the second order system:
$$\tag 1 u''+ \omega^2~u=0$$
We have:
$x_1 = u$
$x_1' = u' = x_2$
$x_2' = u'' = -\omega^2 u = -\omega^2 x_1$
We can now write the system in matrix form as $x'= Ax$, which yields:
$$x'(t) = \begin{bmatrix}x_1' \\ x_2'\end{bmatrix} = Ax = \begin{bmatrix} 0& 1 \\ -\omega^2 & 0\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}$$
If we solve for the eigenvalues of $|A-\lambda I| = 0$, we have:
$$\lambda_{1,2} = \pm i ~ \omega$$
These are complex conjugate eigenvalues with zero real part, so what does that tell us about the solution? Here is a phase portrait for $\omega = 1$:
Can you continue it from here?
If we write the matrix exponential, we have:
$$e^{At} = \begin{bmatrix} \cos(\omega t) & \dfrac{\sin(\omega t)}{\omega} \\ -\omega \sin(\omega t) & \cos(\omega t)\end{bmatrix}$$
Now, look at this form and the hint in your problem and what do you notice? However, I left the details for you to fill in as it is very important for you to understand! |
H: interior boundaries and closures
Let $X = \{a,b,c,d,e\}$ where the $5$ elements are all different. Let $T = \bigl\{X, \emptyset,\{a\},\{a,b\},\{a,c,d\},\{a,b,c,d\},\{a,b,e\}\bigr\}$. Is $T$ a topology on $X$? List all the closed subsets of $X$. Determine the closures of $\{a\},\{b\},$ and $\{a,e\}.$ Also find the interior of those three sets and their boundaries.
Here is what I know:
I know $T$ is a topology on $X$.
I know the closed subsets of $X$ are $X, \emptyset, \{b,c,d,e\},\{c,d,e\},\{b,e\},\{e\},\{c,d\}$.
What is the interior of $\{b\}$ though? I know for the closure the complement has to be open. I was also thinking that the the boundary would be the closure minus the interior. Can someone please show how they would get the interior? I am still not sure about the working.
AI: Let's look at one example to see how this works. Take $\{a,e\}$. The interior is the union of all open sets inside of it, so perusing the list of open sets, the only such open sets are $\emptyset$ and $\{a\}$. So the interior is $\{a\}$. The closure is the intersection of all closed sets containing $\{a,e\}$. Looking at your list of closed sets, which I am assuming is accurate, the only one containing $\{a,e\}$ is $X$ itself. Thus the closure is all of $X$. Finally the boundary is defined to be the closure minus the interior. So in this case, it is $X\setminus\{a\}=\{b,c,d,e\}$.You can do the other sets similarly. |
H: Proof that a ODE system admits symmetric solutions
I have a ODE system of the form $\dot{x} = f(x)$ with $x \in \mathbb{R}^3$. Now it is claimed that if $(x_1,x_2,x_3)$ is a solution to the system that also $(-x_1,-x_2,-x_3)$ is a solution.
How can I show that this holds, I'm namely not sure how to deal with this? I first thought that I would just show that $-f(x) = f(-x)$ but when computing I saw I wasn't getting anywhere.
ps. If you are wondering why I did not post the ODE system itself, it is because I want to try it for myself.
-edit-
Since I am not really getting response the ODE about which it is about it is
$$\dot{x}_1 = -a_1 x_1 + b_1 x_2 \\
\dot{x}_2 = -a_2 x_2 + b_2 x_1 x_3 \\
\dot{x}_3 = -a_3(x_3 - \lambda) - b_3 x_1 x_2
$$
-edit 2-
A friend of mine told me that if I would calculate the fixed points of the system with $(-x_1,-x_2,-x_3)$ that they give the same fixed points as $(x_1,x_2,x_3)$ and that it would imply that the solutions are the same? But I cannot see how this would imply that.
-edit 3- apparently there was a error in the exercise that I had. I had to proof that if $(x_1,x_2,x_3)$ then $(-x_1,-x_2,x_3)$ was also a solution and not $-x_3$. Subsituting then these variables then in the system delivers the same state space system.
AI: If $x$ and $-x$ are both solutions, $\dot{x} = f(x)$ while $\dfrac{d}{dt}(-x) = -\dot{x} = f(-x)$, so it must be true that $f(-x) = -f(x)$. |
H: Frobenius Norm with Unitary Operators
For something I'm working on, I have a matrix $A$ with another matrix $U$ which is unitary ($U^*U = I$), and I'm trying to show that, for the Frobenius norm, $\|A\| =\|UA\|$. Now, I can do this pretty easily if an inner product space exists. For example, $\|A\| = \sqrt{\langle A,A\rangle}$ and $\|UA\| = \sqrt{\langle UA,UA\rangle} = \sqrt{\langle A,U^*UA\rangle} = \sqrt{\langle A,A\rangle} = \|A\|$. However, I'm not sure if I evoke the inner product space if I'm just told that the Frobenius norm exists. Is this the appropriate approach?
AI: By definition of Frobenius norm of a matrix $A\in\mathbb{C}^{n\times n}$,
$$
\| A\|_F:=\sqrt{\mbox{trace}(A^*A)}=\ldots =\sqrt{\sum_{j=i}^n\sum_{i=1}^n\overline{A_{ji}}\cdot A_{ij}}
$$
This norm is a norm defined by an inner product $\langle \cdot , \cdot \rangle$ (i.e. $\|A\|_F:=\sqrt[2]{\langle A , A\rangle}$ ). In this case, for $A=(A_{\,\alpha\,\beta})_{n\times n}\in\mathbb{C}^{n\times n}$ and $B=(B_{\,u\,v})_{n\times n}\in\mathbb{C}^{n\times n}$ we have
\begin{align}
\langle A, B\rangle_F := & \mbox{trace}\bigg(A^* B\bigg)\\
\end{align}
Remenber that, for all $X=(X_{\,i\,j} )_{n\times n}\in\mathbb{C}^{n\times n}$, $\mbox{trace}(X)=\sum_{\ell=1}^{n}X_{\ell\ell}$ and
\begin{align}
C^*\cdot D=&(C_{ij})_{n\times n}^*\cdot(D_{uv})_{n\times n} \\
=&(\overline{C_{ji}})_{n\times n}\cdot(D_{uv})_{n\times n} \qquad \mbox{ definition of $*$ } \\
=& \bigg(\sum_{k=1}^{n}C_{jk}\cdot D_{kv} \bigg)_{n\times n}\qquad \mbox{ rule of matrix product }
\end{align}
Now, your question
\begin{align}
\| A\|:=&\sqrt{\mbox{trace}(A^*A)}\\
=&\sqrt{\mbox{trace}(A^*IA)}\\
= & \sqrt{\mbox{trace}(A^*[U^*U]A)}\\
= & \sqrt{\mbox{trace}([A^*U^*][UA])}\\
= & \sqrt{\mbox{trace}([UA]^*[UA])}\\
=& \| UA\|
\end{align} |
H: Probability that an independent exponential random variable is the least of three
Let $Y_1, Y_2, Y_3$ be independent exponentially distributed random variables, with parameters $\lambda_1, \lambda_2, \lambda_3$ respectively. Why is it the case that:
$P(Y_1=min(Y_1,Y_2,Y_3))=\frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}$?
I just came across this fact but I'm not sure where it comes from. I do know that if we define $Y=min(Y_1,Y_2),$ then $Y$ is an exponential random variable with parameter $\lambda_1+\lambda_2$. Is this fact related to the one above?
Thanks
AI: Not sure what the intuition is, but it's a straightforward integral to work out.
$$
\begin{aligned}\int_0^\infty \int_{0}^\infty& \int_{0}^\infty [x_1 \le x_2] [x_1 \le x_3] \lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1 -\lambda_2 x_2-\lambda_3 x_3}dx_3 dx_2 dx_1\\
&=
\int_0^\infty \int_{x_1}^\infty \int_{x_1}^\infty \lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1 -\lambda_2 x_2-\lambda_3 x_3}dx_3 dx_2 dx_1\\
&=\int_{0}^\infty \lambda_1 e^{-(\lambda_1 +\lambda_2 +\lambda_3) x_1}dx_1
\\&=\frac{\lambda_1}{\lambda_1 +\lambda_2 +\lambda_3}\end{aligned}$$
(This is using the fact that expectation of the indicator function of an event equals the probability of that event.) |
H: integral of f(x), where f(x) is infinite at one point while zero at other parts
A function f(x) has the value of zero expect one point, where the value is infinite. Does the integral of f(x) equal 0? Or any other values? Thanks a lot.
AI: Assuming that you're talking about Lebesgue integration (since we cannot take the Riemann integral of an unbounded function), then the integral is $0$. This is because the function is equal to the zero function except at a single point, and a singleton is a set of measure $0$, so we ignore what happens there.
Now, for example, the Dirac delta function can be loosely thought of as a function that is zero everywhere on the real line, except at $0,$ where it is infinite, and has an integral over the whole real line of $1,$ but again, this is just a rough idea. It is more accurately defined as a measure/probability distribution. |
H: Optimization Word Problem: Hard time setting up the objective function
So the biggest problem that I have with these questions is to establish an objective function. It really irks me that I can do all the steps after setting up the equation, but without it, Im obviously stuck. So for this word problem:
A 10 inch piece of wire will be cut, and the 2 pieces will be formed into an equilateral triangle and a square. Find the dimensions of the two figures that yield a minimum total area.
So its quite obvious they are looking for a minimum function but how do I set up an objective function to work with, get the derivative, take the critical points, evaluate them and the whole nine yards?
AI: You are looking for minimum area, so let's start with that. $A_{square} = s^2$, $A_{triangle} = {1 \over 2} bh$. Now, this is an equilateral triangle, so with a little trig, we get $h= \sqrt{b^2 -{b\over 2}^2}$. Now we calculate circumference. $C_{square} = 4s$, $C_{triangle} = 3b$. So put it all together and we get:
$$A_{total} = s^2+{1 \over 2}b\sqrt{b^2-{b\over 2}^2}$$
with $$4s + 3b = 10$$ |
H: Absolute Value Proof: if $-a \leq x \leq a$, then $|x| \leq a$.
I want to prove the following proposition:
If $-a \leq x \leq a$, then $|x| \leq a$, where $x,a \in \mathbb{R}$.
Here's my proof:
By trichotomy, there are two possibilities: either $x \geq 0$, or $x<0$. If $x \geq 0$, then $|x| = x$, so $|x| \leq a$. If $x<0$, then $|x| = -x$. We note that $-x > a$ iff $x < -a$, but this is impossible because $x \geq -a$. It follows that $-x \leq a$, i.e. $|x| \leq a$.
I think it works. I'd be pleased if someone could verify my work - or suggest a more effective way of proving the proposition. Thanks!
AI: It looks good!
As one alternative, for the $x<0$ part, we have $x=-|x|,$ so $-a\le-|x|,$ and so $|x|\le a.$
As another alternative, note that $x-a\le 0$ and $0\le x+a$,so $x^2-a^2\le0(x-a)=0,$ so $x^2\le a^2.$ Taking the square root on both sides preserves the inequality, since $t\mapsto\sqrt t$ is an increasing function on $[0,\infty),$ and squares of real numbers are non-negative, so we have $$\sqrt{x^2}\le\sqrt{a^2}.$$ Note that by assumption, $-a\le a,$ so $0\le 2a,$ so $0\le a,$ and so $$\sqrt{x^2}\le a.$$ Finally, if you know (or can prove) that $|x|=\sqrt{x^2}$ for all $x\in\Bbb R,$ then we have $$|x|\le a,$$ as desired. |
H: Mathematical constants
Am I mistaken, or is there a mistake on the mathematical constants Wikipedia page that describes the Conic constant / Schwarzschild constant in terms of Napier's constant?
AI: Yes, it appears that $e$, which is the eccentricity of a conic section, has been
misinterpreted as ${\rm e} = 2.71828\ldots$. |
H: Deriving a formula in terms of slopes of an angle between two oblique lines
Two oblique lines intersect at the origin, both passing through the first quadrant. Let the angle in the first quadrant between the lines be $\theta$. Derive an appropriate formula for finding this angle in terms of the slope of the lines.
So I'm not sure what to do to derive a formula.
AI: Hint: the slopes are the tangents of the angles to the $x$ axis.
Added: here is a picture with 10 and 30 degree lines. The angle between them is 20 degrees |
H: If $T$ is linear and maps $\mathbb{R}^3$ to $\mathbb{R}^2$ where $T(a_1,a_2,a_3)=(a_1+a_2,a_3)$ is $N(T)=\{(a_1,a_1,0)\}$?
I am trying to find the basis. Please don't give me the basis because it's for a graded assignment. Thanks in advance.
AI: Since $T:\Bbb R^3\to\Bbb R^2,$ then $$N(T)=\{(x,y,z)\in\Bbb R^3:T(x,y,z)=(0,0)\}.$$ Using the specific definition of $T$, we have $$N(T)=\{(x,y,z)\in\Bbb R^3:(x+y,z)=(0,0)\},$$ which is the same as $$N(T)=\{(x,y,z)\in\Bbb R^3:x+y=0,z=0\}.$$ Do you see why this is different from what you have? |
H: Why $(\alpha-1)^{-1}\le u^2$ where $u$ is a fundamental unit in $\mathbb{Z}[\alpha]$ and $\alpha=2^{1/3}$?
Given $\alpha = 2^{1/3},$ I want to show that $\beta = (\alpha-1)^{-1}$ is a unit in $\mathbb{Z}[\alpha]$ and is between 1 and $u^2$, where $u$ is a fundamental unit in $\mathbb{Z}[\alpha]$. I see why its a unit and why 1 is a lower bound of $\beta$, but I don't see how to get $\beta \le u^2$. Any help would be appreciated.
AI: I take it that you don't yet know that $1 + \alpha + \alpha^2 = \beta$ is the fundamental unit of ${\mathbb Z}[\alpha]$, but that you're busy establishing this.
Presumably, you have some theorem that gives an estimate on the absolute value of the fundamental unit. For instance
Theorem (Artin's Inequality). Let $K$ be a real cubic field and let $v \in {\mathcal O}_K^*$ with $v > 1$. Then $|\text{disc}(K)| \leq 4v^3 + 24$.
In your case, applying this to $K = {\mathbb Q}(\alpha)$ and the fundamental unit $u$ of ${\mathbb Z}[\alpha]$ gives $108 \leq 4 u^3 + 24$, so $u^2 \geq (\sqrt[3]{21})^2 \approx 7.61$. Now $\beta \approx 3.85$, so $1 < \beta < u^2$. |
H: Basic reductions
I'm trying to learn about reductions and I came across this example in my book:
Let the "merge" of two languages L1,L2$\subset${0,1}* be:
L1$\bot$L2 = {x0 | x$\in$L1}$\cup${y1|y$\in$L2}
I think that either original language is many-one reducible to L1$\bot$L2, but I don't quite know how I would prove it.
AI: Let $f:\{0,1\}^*\to\{0,1\}^*:x\mapsto x0$; then for any $x\in\{0,1\}^*$ we have $x\in L_1$ if and only if $f(x)\in L_1\perp L_2$, so it’s true that $L_1\le_m L_1\perp L_2$. A very similar argument shows that $L_2\le_m L_1\perp L_2$ as well.
Of course you do have to show that $x\in L_1$ if and only if $f(x)\in L_1\perp L_2$, but this isn’t hard. If $x\in L_1$, then by definition $f(x)\in L_1\perp L_2$, and if $f(x)=\in L_1\perp L_2$, then either $f(x)=y0$ for some $y\in L_1$, or $f(x)=y1$ for some $y\in L_2$. Since anything of the form $f(x)$ ends in $0$, it must be that $f(x)=y0$ for some $y\in L_1$, and clearly this implies that $x=y\in L_1$. |
H: $\frac{x^4 - x^3 + ax^2 + bx + c}{x^3 + 2x^2 - 3x + 1}$, remainder $3x^2 - 2x + 1$. Find $(a + b)c$.
Given the polynomials
$P(x) = x^4 - x^3 + ax^2 + bx + c\\
Q(x) = x^3 + 2x^2 - 3x + 1\\
R(x) = 3x^2 - 2x + 1$
such that $P(x) = D(x)Q(x) + R(x)$, find $(a + b)c$.
I would normally apply little Bézout's theorem here and get a system of three equations which I would solve for $a$, $b$ and $c$, but I can't find the roots of $Q(x)$, nor factor it for that matter. Please help.
AI: Start dividing $Q(x)$ into $P(x)$.
$$P(x)-xQ(x)=-3x^3+(a+3)x^2+(b-1)x+c\;,$$ getting rid of the fourth power, so $D(x)=x-3$. Now
$$D(x)Q(x)=(x-3)(x^3+2x^2-3x+1)=x^4-x^3-9x^2+10x-3$$
(assuming no mental algebraic errors on my part), and you can equate coefficients in
$$R(x)=P(x)-D(x)Q(x)$$
to find $a,b$, and $c$. |
H: Does $\sum_{n=2}^\infty (n\ln n)^{-1}$ diverge?
Is $\sum_{n=2}^\infty (n\ln n)^{-1}=\infty$ ?
This seems like elementary calculus, but I can't figure this out. Can anyone supply a hint?
AI: Hint.
$$\sum_{n=2^N}^{2^{N+1}-1} \frac{1}{n\ln n}\ge\frac{2^N}{(N+1)2^{N+1}}=\frac{1}{2(N+1)}$$ |
H: Cutting a rectangle into squares
A carpenter has a rectangular board, $x$ feet long and $y$ feet wide, with total area $n = xy
$square feet. The board is to be divided into n squares (each 1 foot x 1 foot) by successively
cutting a rectangle into two smaller rectangles whose sides are each a whole number of feet.
Using strong induction, show that exactly $n -1$ cuts are needed, no matter how the carpenter decides to perform the cuts.
Any help please? I was thinking of maybe trying to write down the possible number of arrangements he can do it but I think it is pointless since it states no matter how it is done.
AI: Suppose the result is true for all $m\lt n$. We show the result is true for $n$. One cut divides the board into two rectangles of area $a$ and $b$ where $a+b=n$, and $a,b\lt n$. By the induction hypothesis, it takes exactly $a-1$ cuts to split the board of area $a$ into $1\times 1$'s, and exactly $b-1$ to split the board of area $b$ into $1\times 1$'s, for a total of $1+(a-1)+(b-1)=n-1$. |
H: Euler Method - Why is my intuition incorrect?
$$\dfrac{dA}{dt} = 0.5 \times A \times (1 - \dfrac{A}{100}) - 10$$ with $A(0) = 70$ and we want to use Euler's method to get an approximate value for $A(10)$, with a step size of 1.
So the answer sheet says you basically have to use $\text{Ans} + 0.5 \times \text{Ans} \times (1-\dfrac{\text{Ans}}{100}) - 10$ with the first $\text{Ans}$ being $70$, and then of course repeat 10 times.
But I'm wondering, doesn't this actually give you $\dfrac{dA(10)}{dt}$? How is this a correct method?
AI: The Euler method does not give you $\frac {dA}{dt}$. You give it a formula for $\frac {dA}{dt}$, such as the one in your question. Then from any given point, like your start of $(0,70)$ it puts a straight line through the point with slope $\frac {dA}{dt}$ of that point. From your expression, $\frac {dA}{dt}|_{(0,70)}=0.5$ so we step one unit in $t$ at a slope of $0.5$, giving the $A$ value of the next point as $70+0.5\cdot 1=70.5$. Now we are at $(70.5,1)$, we calculate $\frac {dA}{dt}$ at this point and take another step along the $t$ axis, and so on until we get to $t=10$ |
H: Find a Subgroup of $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$ Isomorphic to $\mathbb{Z}_9 \oplus \mathbb{Z}_4$
This is what I have so far:
$$\mathbb{Z}_{12} = \{0, 1, 2,\cdots, 11\}$$
$$\mathbb{Z}_{18} = \{0, 1, 2,\cdots, 17\}$$
$$\mathbb{Z}_9 = \{0, 1, 2,\cdots, 8\}$$
$$\mathbb{Z}_4 = \{0, 1, 2, 3\}$$
Using that fact that isomorphisms must contain the same number of elements, we must find a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ with the same number of elements as in $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$.
Now, let's determine the number of elements in $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$:
$$|\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}| = \text{lcm}(|\mathbb{Z}_{9}|, |\mathbb{Z}_{4}|) = \text{lcm}(|9|, |4|) = 36$$
Next, let's find a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ with order $36$. First, $\langle 3 \rangle$ in $\mathbb{Z}_{12}$ gives us the group $\{0, 3, 6, 9\}$ which has order $4$. Second, $\langle 2\rangle$ in $\mathbb{Z}_{18}$ gives us the group $\{0, 2, 4, 6, 8, 10, 12, 14, 16\}$ which has order $9$. Then, the group generated by $(3, 2)$ has order $36$ (because $\text{lcm}(9,4) = 36$)
My question is, does this properly show that there exists a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ that is isomorphic to $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$? Or do I have to show that the mapping is one-to-one, onto, and preserves the group operation? If so, what is the mapping function to show one-to-one, onto, and operation preservation?
Thanks!
AI: Hint: Is there an element in $\mathbb{Z}_{12}$ that doesn't generate all of $\mathbb{Z}_{12}$ (hence is a factor of $12$) but generates $\mathbb{Z}_{9}$ or $\mathbb{Z}_4$? Then is there an element in $\mathbb{Z}_{18}$ not relatively prime to $18$ that generates $\mathbb{Z}_9$ or $\mathbb{Z}_4$?
Call these two elements $a$ and $b$. Then look at the element $a \oplus b \in \mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$. Create an 'obvious' mapping from this element into $\mathbb{Z}_9\oplus \mathbb{Z}_4$. Prove that this mapping must be $1:1$, onto, and preserves the group operation.
But makes things easy on yourself! Focus on the element $a \oplus b$, it should generate $\mathbb{Z}_9 \oplus \mathbb{Z}_4$ (why?) then those three things are easy to prove. (again, think about why).
EDIT: I'll expand a bit more on my hint then. You already looked at a few elements in $\mathbb{Z}_{12}$. Look closely at the subgroup $\langle 3 \rangle=\{0,3,6,9\}$. Could we map that to $\mathbb{Z}_4$ with a mapping that preserves the operation, say $\varphi$? Now look at $\langle 2 \rangle =\{0,2,4,6,8,10,12,14,16\}\subset \mathbb{Z}_{18}$. Can we take those elements and map it to $\mathbb{Z}_9$, say the mapping is $\theta$, in a way that preserves the operation? Now since clearly $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18} \cong \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$, would the 'mapping'
$$\psi((a,b))=\varphi(a) \oplus \theta(b)$$
for $(a,b) \in \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ be an isomorphism to $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$?
FINAL EDIT: With your last comment you essentially have the idea. It is clear that $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18} \cong \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ (just take the map that sends $(a,b)$ to $(b,a)$). So we just need to find a subgroup of $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ isomorphic to $\mathbb{Z}_9 \oplus \mathbb{Z}_4$. Your idea is right. We make a mapping $\varphi$ that takes each element in $\mathbb{Z}_{18}$ mod $2$ and a mapping $\theta$ that takes each element in $\mathbb{Z}_{12}$ mod $3$. Then show that $\varphi \oplus \theta$ is an isomorphism. Now here is where you need to be careful. You said it would be an isomorphism of $\mathbb{Z}_4 \oplus \mathbb{Z}_9$ to $\mathbb{Z}_{18} \oplus \mathbb{Z}_12$. This is wrong. It's an isomorphism from $\mathbb{Z}_{4} \oplus \mathbb{Z}_9$ to $\langle 2 \rangle \oplus \langle 3 \rangle$ in $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ (so a subgroup of that group). Since this group is isomorphic to $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$, we would be done since it would be clear what the subgroup was. But there is an easier way to do this.
Let's look at the subgroup $\langle 2 \rangle \oplus \langle 3 \rangle$ of $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$. Let's create a mapping $\varphi$ of $\langle 2 \rangle$ in $\mathbb{Z}_{18}$ into $\mathbb{Z}_9$. Notice that $1$ generates $\mathbb{Z}_9$. Moreover, $\langle 2 \rangle$ generates the subgroup (itself) in $\mathbb{Z}_{18}$. So just map $2 \in \mathbb{Z}_{18}$ to $1 \in \mathbb{Z}_{9}$. Similarly, we can map $3 \in \mathbb{Z}_{12}$ to $1 \in \mathbb{Z}_4$. Since we are mapping generators to generators, it is clear that we preserve the operations, it is $1:1$, and onto. So we have the required isomorphism. But all the same, if this shorter way isn't as clear to you then you can can always show it by the preceding paragraph (which was what you stated in your comment, I just cleaned up a bit of the language). Hope that helped finally clear things up. |
H: Bayesian Network - unclear homework example
I am not sure if it is me or the example:
A doctor gives a patient a drug dependent on their age and gender. The patient has a probability to recover depending on whether s/he receives the drug, how old s/he is and which gender the patient has. Additionally it is known that age and gender are conditional independent if nothing else is known from the patient.
i) Draw the Bayesian Network which describes the situation.
ii) How does the factorized probability distribution look like?
iii) Write down the formula to compute the probability that a patient recovers given that you know if s/he gets the drug. Write down the formula using only probabilities which are part of the factorized probability distribution.
I think I could handle task ii) and iii) if someone would show me how the network looks like.
What is not clear to me is the give fact that "A doctor gives a patient a drug dependent on their age and gender" and "patient has a probability to recover depending on whether s/he receives the drug, how old s/he is and which gender the patient has".
Could someone help me with that?
AI: i)
Let $R$ represent the patient recovering, $D$ represent the drug being administered, $A$ represent the patient's age, and $G$ represent the patient's gender. Lastly $X\rightarrow Y$ means that there's an arrow from $X$ to $Y$ in the Bayesian network. Now let's break down the problem.
A doctor gives a patient a drug dependent on their age and gender.
This translates to $A\rightarrow D$ and $G\rightarrow D$.
The patient has a probability to recover depending on whether s/he receives the drug, how old s/he is and which gender the patient has.
This translates to $D\rightarrow R$, $A\rightarrow R$, and $G\rightarrow R$.
Additionally it is known that age and gender are conditional independent if nothing else is known from the patient.
This doesn't add any new arrows and translates to $P(A,G)=P(A)P(G)$. However, if we knew $D$ and let's say $A$, we might be able to infer insight on the value of $G$ (and maybe infer $A$ if we knew $G$ and $D$). Same story is we replace $D$ with $R$.
ii)
This isn't the only way to factorize the joint distribution (there's a more elegant way that uses the conditional independence of $A$ and $G$), but this way will help us with part iii),
\begin{align*}
P(A,G,R,D)
&=P(A|G,R,D)P(G,R,D) \\\
&=P(A|G,R,D)P(G|R,D)P(R,D) \\\
&=P(A|G,R,D)P(G|R,D)P(R|D)P(D) \\\
\end{align*}
On the other hand, they may be looking for a form which conforms more clearly to the arrow relations like,
\begin{align*}
P(R,D,A,G)
&=P(R|D,A,G)P(D,A,G) \\\
&=P(R|D,A,G)P(D|A,G)P(A,G) \\\
&=P(R|D,A,G)P(D|A,G)P(A)P(G) \\\
\end{align*}
iii)
Using the fully factored form in the first part of ii) we can solve for $P(R|D)$,
$$
P(R|D) = \frac{P(A,G,R,D)}{P(A|G,R,D)P(G|R,D)P(D)}
$$ |
H: Simplify the equation, containing radicals
I'm having trouble simplifying this equation, while keeping it as an exact expression.
$x=\sqrt{28800-14400\sqrt{3}}$
I'm looking for the steps to change that above equation into $120\sqrt{2-\sqrt{3}}$
AI: $x=\sqrt{28800-14400\sqrt{3}}\\\implies x=\sqrt{14400(2-\sqrt{3})}\\\implies \color{blue}{{x=120\sqrt{2-\sqrt{3}}}}$. |
H: Something about the proof of Urysohn's lemma
Let $X$ be a normal space and $F,G\subseteq X$ closed sets such that $F\cap G=\emptyset$.
In the proof of Urysohn's lemma, we proved that if $D=\{q_n:n=0,1,...\}=\mathbb{Q}\cap [0,1]$ with $q_0=0$ and $q_1=1$, then there exists a family of open sets $\mathcal{U}=\{U_{q_n}:n=0,1,...\}$ such that $F\subseteq U_0$, $U_1=X\setminus G$ and if $r<s$ with $r,s\in D$ then $Cl(U_r)\subseteq U_s$.
Now, we define $f(x)= \inf \left\{{r\in D:x\in U_r}\right\} $ if $x\in X\setminus G$ and $f(x)=1$ if $x\in G$.
I don't understand why if $x\in X\setminus G$ and $b\in (0,1)$, then $f(x)>b$ if and only if there exists $r\in D$ such that $r>b$ and $x\notin U_r$.
Can someone explain this to me?
Thanks!
Edit: I meant to say $q_0=0$. I already edited it.
AI: If $f(x)>b$, there is $r\in D$ such that $f(x)> r > b$. By definition of $f$, $x\notin U_r$. On the other hand, if such an $r$ exists and $x\in U_p$ for some $p\in D$, then $p>r$ (or $x\in U_r$). Then $f(x)\geq r >b$. |
H: If $[K : \mathbb{Q}] = 2$, then $K = \mathbb{Q}(\sqrt{d})$
This isn't for homework, but I would just a like a hint please. The question asks
If $K$ is an extension field of $\mathbb{Q}$ and $[K : \mathbb{Q}] = 2$ (the dimension of $K$ over $\mathbb{Q}$), then $K = \mathbb{Q}(\sqrt{d})$ for some square-free integer $d$.
I started by considering the linearly independent set $\{ 1 \}$ in $K$. Now since $[K : \mathbb{Q}] = 2$, I can extend this set to a basis $\{ 1, v \}$ of $K$ over $\mathbb{Q}$, where $v \notin \text{span} \{ 1 \}$. I see that $v^2 \in \text{span} \{ 1, v \}$, so that $v^2 = a_0 + a_1 v$ for some $a_0, a_1 \in \mathbb{Q}$. Now $a_0 \neq 0$ for otherwise $a_1 = v$ and $v \in \text{span} \{ 1 \}$. However, I'm not sure how to conclude that $v$ is a square-free integer. I feel like I am on the right track (hopefully), and would greatly appreciate a hint please!
AI: Remember that you can change basis freely - even if $\nu$ isn't a square-free integer, as a solution of a quadratic equation with coefficients in $\mathbb{Q}$ it can be written in the form $\nu=p+q\sqrt{d}$ for some $d\in\mathbb{Z}$ (check this!), and obviously any square factor in $d$ can be pulled out into $q$. Then $\mathbb{Q}(\nu) = \mathbb{Q}(\sqrt{d})$ (check this too!) and you're done. |
H: Squeeze Theorem on this Sine Function
$$\lim_{x \to \infty}\frac{\sin 2x}{4x}$$
My book says:
Start by examinig the numerator of the given function, $\sin 2x$.
The $\sin$ function has a minimum absolute value of $0$ and a maximum absolute value of $1$.
Thus, the range of the absolute value of $\sin 2x$ is:
$$0 \leq |\sin 2x| \leq 1.$$
Divide each part of the inequality by $4x$:
$$0 \leq |\frac{\sin 2x}{4x}| \leq \frac{1}{4x}.$$
My question is:
1) Why do we use these absolute values? Why not squeeze $\sin 2x$ between $-1$ and $1$? Isn't $[-1,1]$ the range of the $\sin$ function? Why are we considering the range of the absolute value?
2) Can we squeeze the whole function $\frac{\sin 2x}{4x}$ between two values, why does it concentrate on just the numerator, $\sin 2x$?
Thank you.
AI: Succint answers to your questions:
1)
When having limits that oscilate between negative and positive values, it is a common technique to use absolute values.
Yes you can do so!
Yes it is.
This is the same question as the first one
2)
Because $\sin2x$ is problematic to deal with, but $\frac{1}{x}$ is not. |
H: When is first order induction valid?
Assume we know $\forall x(P(x))$ is true in a model of Peano arithmetic (PA). Does this mean we can prove $\forall x(P(x))$ using induction? If not, why not? If $P(x)$ is true for all $x$ then $P(0) \land \forall x(P(x) \rightarrow P(S(x)))$ is a true statement.
What if $\forall x(P(x))$ is false in some other model of PA?
I want to emphasize the "if not, why not" part. What more is needed to prove $\forall x(P(x))$ if satisfying the induction axiom schema is not enough.
I think I figured out my confusion. When we say "$\forall x P(x)$" is provable by induction what we mean is we can prove $P(0) \land \forall x(P(x) \rightarrow P(Sx))$ from the other axioms of PA and we can then use modus ponens and the induction axiom $(P(0) \land \forall x(P(x) \rightarrow P(Sx))) \rightarrow \forall x P(x)$ to deduce $\forall x P(x)$. Assume we can prove $\forall x P(x)$ from the other axioms of PA. This does not mean we can prove $\forall x P(x)$ "using induction". We would still have to prove $P(0) \land \forall x(P(x) \rightarrow P(Sx))$ before we could use modus ponens. Of course, we don't need induction to prove $\forall x P(x)$ if it can be derived from the other axioms.
The only way the induction axiom can be false is when $P(0) \land \forall x(P(x) \rightarrow P(Sx))$ is true and $\forall x P(x)$ is false. The easiest way for this to happen is when a proper subset of a model is closed under successor.
Let $P(x)$ = "x does not encode a proof of 0=1". If PA is consistent then there is a model of PA + $\exists x \neg P(x)$. Such a model is $\omega$ - inconsistent. The standard natural numbers are a proper subset of this model and are closed under successor. Even though $P(0) \land \forall x(P(x) \rightarrow P(Sx))$ is true for the standard natural numbers, $\forall x P(x)$ is false. It would seem a model of PA + $\exists x \neg P(x)$ must fail to satisfy induction. We get around this by talking about "definable" subsets of natural numbers. If PA is consistent then the standard natural numbers can't be definable in the language of PA.
AI: There are sentences of the form you describe that are neither provable nor refutable in first-order PA. If $\varphi$ is such a sentence, it will be true in some model of PA, but unprovable from the axioms of PA. One can arrange for $\phi(0)$ to be a theorem. So the induction step will not be a theorem. |
H: Why is $E[u] = \int_{\Omega}\frac12 |\nabla u|^2 + \frac14u^4 dx$ a convex functional?
In our PDE class, we examined the functional
$$E[u] = \int_{\Omega}\frac12 |\nabla u|^2 + \frac14u^4 dx.$$
My professor claimed that this functional was convex. When I take the derivative of the functional at $u$ in the $(v-u)$ direction, I get
$$\partial_{tt} E[u+t(v-u)] = \int_{\Omega}|\nabla (v-u)|^2 dx + \int_{\Omega}u^2(v-u)^2\\+2t\int_{\Omega}3u(v-u)^3 dx + 3t^2 \int_{\Omega}(v-u)^4.$$
If it is convex, I should be able to show that this positive for all $t>0$. But the third term is giving me trouble. What am I missing?
AI: Here's how I'd write it:
$$\partial_{t} E[u+t(v-u)] = \int_{\Omega}\left[(\nabla u + t\nabla (v-u))\cdot \nabla (v-u) + (u+t(v-u))^3(v-u) \right] dx.$$
$$\partial_{tt} E[u+t(v-u)] = \int_{\Omega} \left[ |\nabla (v-u)|^2 + 3(u+t(v-u))^2(v-u)^2 \right] dx.$$
In your equation, there should be a 3 in front of the second term, which allows you to complete the square. |
H: Limit $\lim\limits_{n\to\infty}((n-1)!)^{\frac{1}{n}}$
I try to find the $\lim_{n \to \infty} ((n - 1)!)^{\frac{1}{n}}$ using Stolz-Cesaro convergece theorem and it goes like this: $L = \lim_{n \to \infty} \frac{n!}{(n-1)!} = \infty$. Am I right?
AI: Hint
$$\ln \left( \sqrt[n]{(n - 1)!}\right)=\frac{\ln(1)+\ln(2)+..\ln(n-1)}{n} \,.$$
Now if you use S-C you get exactly what you said.
I guess you used a consequence of SC, which says:
C: If $a_n >0$ and $\lim_n \frac{a_n}{a_{n-1}}$ exists and is equal to $l$ then
$$\lim_n \sqrt[n]{a_n}=l$$ |
H: Prove that $f$ is a bounded function on $S.$
The question is this.
Let$(f_n)$ be a sequence of bounded functions on a set $S$, and suppose that $f_n \rightarrow f$ uniformly on $S$. Prove that $f$ is a bounded function on $S$.
My work is below.
Proof.
Since $(f_n)$ is bounded sequence of functions, we know that $|f_n(x)| < M,\forall x \in S$ and for some real number $M.$ Also, we know that $\forall \epsilon > 0, \exists N \ s.t\ n > N \Rightarrow |f_n(x) - f(x)| < \epsilon, \forall x\in S$ and $\forall n > N.$ Then,
$$
\begin{eqnarray}
-\epsilon&<& f_n(x) - f(x)&<& \epsilon \\
-\epsilon - f_n(x) &<& f(x) &<&\epsilon - f_n(x)
\end{eqnarray}
$$
Since we know that $-f_n(x) \leq M$ and $-M \leq -f_n(x),$ we have
$$-\epsilon - M \leq -\epsilon - f_n(x) < f(x) <\epsilon - f_n(x)\leq \epsilon + M.$$
Setting $\epsilon = 1$, we have $|f(x)| \leq M+1, \forall x \in S. \square$
Is this valid??? My text book says some other thing.
AI: Your proof is almost right, unfortunately almost right means wrong.
The problem with your proof is that when you say that $|f_n(x)| < M,\forall x \in S$, the $M$ depends on $n$, different $n$'s lead to different $M$'s.
But you are on the right track, all you have to do is to change the order of the steps. You know that $\epsilon =1$ is what will work, so start by picking $\epsilon =1$, pick then a good $n$ and only last pick $M$ for that chosen $n$.
Corrected version
Pick $\epsilon =1$. Then $\exists N$ s.t for all $n > N$ we have
$$ \left|f_n(x) - f(x) \right| < 1, \forall x\in S \,.$$
Pick some fixed $n >N$.
Since $f_n$ is bounded there exists some $M$ such that
$$ |f_n(x)| < M,\forall x \in S \,.$$
Then, by the triangle inequality we have
$$|f(x) \leq |f(x)-f_n(x)|+|f_n(x) < 1+M \forall x \in S \,.$$
\square
As you see, you had all the right ideas, you just did the steps in the wrong order :) |
H: How to read this mathematical expression?
I am very much interested in machine learning. I would like to do research in this subject. But presently the mathematical language used in this subject is hard for me. Here is an expression in wiki page on Support vector machine :
Given some training data $\mathcal{D}$ , a set of n points of the form
How to read this?
AI: That means : $\cal D $ has $n$ éléments $z_1,\cdots,z_n$ having the form: $$z_i=(x_{i,1},x_{i,2},\cdots,x_{i,p},1)$$ or $$z_i=(x_{i,1},x_{i,2},\cdots,x_{i,p},-1)$$ where $x_{i,1},x_{i,2},\cdots,x_{i,p}$ are real numbers.
If we want, we can represent this by a matrix :
$$A_{\cal D}=\begin{pmatrix} x_{11}& x_{12}&\cdots& x_{1p}& \varepsilon_1 \\ x_{21}& x_{22}&\cdots& x_{2p}& \varepsilon_2 \\ x_{31}& x_{32}&\cdots& x_{3p}& \varepsilon_3 \\ \vdots &\vdots&\cdots&\vdots& \vdots \\ x_{n1}& x_{n2}&\cdots& x_{np}& \varepsilon_n \\ \end{pmatrix} $$
Where $\varepsilon_i=1$ or $-1$ for all $i \in \{1,..,n\}$
We can now see elements of $\cal D$ as lines of the matrix $A_{\cal D}$. |
H: Using Lagrange Multipliers with Constraints of a Line and a Parabola
Find the absolute maximum and minimum of $f(x,y)= y^2-2xy+x^3-x$ on the region bounded by the curve $y=x^2$ and the line $y=4$. You must use Lagrange Multipliers to study the function on the curve $y=x^2$.
I'm unsure how to approach this because $y=4$ is given. Is this a trick question?
AI: You need to look for points within the region and on the boundary of the region which will be candidates for the absolute maximum and minimum of $f$. In the interior of the region, you should look for critical points by finding what points $P$ give $\nabla f(P) = 0$. However, you also have to check the values of $f$ on the boundary. In this case the boundary above will be the line $y=4$, while the boundary below will be $y=x^2$. You need to check both boundaries! One method to check for the possible candidates for maxima and minima which are on the boundary is to use the method of Lagrange multipliers. For example, if you want to check for extreme points of $f$ on $y=x^2$, you can set $g(x,y) = y-x^2$ and apply the Lagrange multiplier method giving you the system of equations
$$
y=x^2 \\
\nabla f = \lambda \nabla g
$$
Make sure to pay attention to the fact that the boundary $y=x^2$ only considers points with $-2 \leq x \leq 2$, since otherwise you've gone past your other boundary $y=4$.
On the other hand, to look for extreme points (max/min points) of $f$ on $y=4$, you can plug $4$ in for $y$ into the equation for $f$ and decide which values of $x$ between $-2 \leq x \leq 2$ make $f$ the largest and smallest.
Once you have made a list of all possible candidates for max and min (the critical points on the interior of the region, and those found on the boundary), you then decide what the absolute max and min are for $f$ by simply checking the value of $f$ evaluated at each of those points. |
H: Contractibility of topological spaces associated to posets
Suppose $\mathcal{P}$ is a partially ordered set. To $\mathcal{P}$ we can associate a simplicial complex $K(\mathcal{P})$ whose $n$-simplices are the chains of length $n+1$ in $\mathcal{P}$. Since subchains of chains are again chains, it follows that $K(\mathcal{P})$ is indeed a simplicial complex.
We can also define $K(\mathcal{P})$ to be the clique complex of the graph $G(\mathcal{P})$ whose vertices are the elements of $\mathcal{P}$, with edges between all pairs of vertices $x,y\in\mathcal{P}$ with $x<y$.
To each simplicial complex $K$, we have the associated topological space $|K|$, also referred to as the geometric realization of $K$.
Now if $\mathcal{Q}$ is a subposet of $\mathcal{P}$, then $|K(\mathcal{Q})|$ is a subspace of $|K(\mathcal{P})|$. In general, subspaces of contractible spaces need not be contractible, but in this specific case, does contractibility of $|K(\mathcal{P})|$ imply that of $|K(\mathcal{Q})|$?
I'm still trying to develop intuition for concepts such as contractibility and the geometric realization of a simplicial complex, so any answers that help me think about these concepts are much appreciated. Perhaps we can state contractibility in terms of the graph $G(\mathcal{P})$, and translate this question to a problem in graph theory.
AI: No.
Observe that $|K(\mathcal{P})|$ is contractible whenever $\mathcal{P}$ has a minimum element, since you can contract everything to that vertex.
So let $\mathcal{Q}$ be any poset such that $|K(\mathcal{Q})|$ is not contractible, and let $\mathcal{P}$ be the poset obtained from $\mathcal{Q}$ by adding a minimum element. Then $|K(\mathcal{P})|$ is contractible, but $|K(\mathcal{Q})|$ is not.
For example, let $\mathcal{Q} = \{1,2,3,4\}$ with order
$$
1<3,\quad 1<4,\quad 2<3,\quad\text{and}\quad 2<4,
$$
and let $\mathcal{P} = \{0,1,2,3,4\}$, where $0<q$ for all $q\in\mathcal{Q}$. Then $|K(\mathcal{Q})|$ is a square of edges, and $|K(\mathcal{P})|$ is the cone $|K(\mathcal{Q})|$, i.e. the four triangular faces of a square pyramid. |
H: Why divide numerator and denominator by highest power in a rational function?
$$\lim_{x \to \infty}\frac{x^2+3}{x^3+1} \;\;\;\;\;\;\;\; \textrm{or}\;\;\;\;\;\;\;\; \lim_{x \to \infty}\frac{10x^3}{2x^3+3x^2+6x}$$
Why do we proceed by first dividing each term in the numerator and denominator by the highest power of $x$ in the denominator?
In the first function, we would divide each term by $x^3$, and in the second function, we would divide each term by $x^3$.
Why divide by $x$ with the highest power in the denominator when determining the limit?
Thank you.
AI: The rules describing the relation of limits with the arithmetic operations fail when we have the difference, or the quotient of to functions that tend to (+)infinity. But when their limits are zero then it only fails for the quotient. If an expression is only formed by additions (and subtractions) and quotients. By turning parts of this expression from tending to infinity to tending to zero, you are improving the chances of the rules to work. Moreover, by dividing by one of the terms already in the expression you ensure you are no going to run into the quotient of two functions that tend to zero.
The technique then only requires to have an expression that is formed by adding, subtracting, multiplying and dividing terms that tend to infinity. After writing the expression as one quotient of two sums of terms that tend to infinity; if the term that tends to infinity faster can be identified, then the technique will work.
For example: $\lim_{x\rightarrow\infty}\frac{2^x+3^x}{e^x-x}$ can be computed using this same technique. |
H: Minimizing and maximizing $ax+by+cz$ for unit vectors $(x,y,z)$?
Suppose you have a unit vector $(x,y,z)$ and you want to minimize or maximize some linear relation $ax+by+cz$. Of course, one could do this with lagrance multipliers.
Is there an alternative way to do this just with linear algebra in general? If so, why does it work?
I'm curious because earlier I had to do a homework problem minimizing $x+2y+3z$ on the unit sphere.
AI: This can be done using just linear algebra.
The expression you want to maximize ($ax + by +cz$) is equivalent to the dot product $(a, b , c) \cdot (x, y, z)$
From the identity $A\cdot B = |A||B|\cos\theta$, where $\theta$ is the angle between $A$ and $B$, and the fact that $\cos\theta$ has a maximum whene $\theta = 0$, we then know that, to maximize $(a, b , c) \cdot (x, y, z)$, we need the vector $(x,y,z)$ to point in the direction of of $(a,b,c)$.
The rest is just a scalar multiplication. |
H: estimate the average speed
the parametric equation of x=t*cos(t), y=t*sin(t)
By calculating the position at t = 2.5 and t=2.51, estimate the speed at t=2.5.
AI: Let $p(t) = t(\cos t, \sin t)$, then the velocity at $t=2.5$ is approximated by $p'(2.5) \approx \frac{1}{2.51-2.5}(p(2.51)-p(2.5))$, and the corresponding speed by $\|\frac{1}{2.51-2.5}(p(2.51)-p(2.5))\|$.
Substituting numbers: $p'(2.5) \approx (-2.2932, -1.4199)$, and $\|p'(2.5) \|\approx 2.6972$.
Using derivatives, we have $p'(t) = (\cos t, \sin t)+t(-\sin t, \cos t)$, which
gives $p'(2.5) \approx ( -2.2973, -1.4044)$ and $\|p'(2.5)\| \approx 2.6926$, so the finite difference approximation is fairly close. |
H: Using Chebychev's Inequality to find a lower bound
A fair coin is flipped 100 times. Find the lower bound, based on Chebyshev's bound, on the probability of the event that the number of heads observed is in {45,46,..,55}. Compare this to the exact probability 0.73.
So n = 100. p = 0.5. E(X) = 50. Standard deviation = sqrt(100*.5*.5) = 5
45 - 55 is within 1 standard deviation of the mean 50. So I get that k = 1, 1 - 1/k = 1 - 1/1 = 0.
P(|X-E(X)| < 5) = P(|X-E(X)| < 5σ/5) = P(|X-E(X)| < σ)
Then flip it around so we're looking at things outside of 45-55.
P(|X-E(X)| > σ) <= σ^2/σ^2
P(|X-E(X)| > σ) <= 1
How is this useful. I don't get how to use Chebyshev's when k = 1. Isn't this just telling me the probability is <= 1?
Not sure where to go from here...
AI: The Chebyshev Inequality does not necessarily give good bounds!
However, the Inequality was slightly misused. I believe it says that the probability that $|X-\mu| \ge k\sigma$ is $\le \frac{1}{k^2}$. In our case, we are interested in $k\sigma=6$. |
H: Providing an example function
Give an example of a function $f: \mathbb{R}^2 \to \mathbb{R}$ for which the function $f(tx,ty)$, for $t \in \mathbb{R}$, has a local minimum at $t = 0$ for all $(x,y) \in \mathbb{R}^2$, but $(0,0)$ is not a local minimum of $f$.
AI: Let $g:\mathbb R \to \mathbb R$ be a function such that $g(0) = 0$ is a local minimum and $g(1)<0$. Define $f:\mathbb R^2 \to \mathbb R$ by
$$f(x, y) = g(kx) \text{ if } x=ky,\ k\in \mathbb N$$
and $f(x, y) = x^2 + y^2$ otherwise. Then $f$ satisfies the condition and $(0,0)$ is not a local minimum: for any $\epsilon >0$, there is $k \in \mathbb N$ such that $1/k^2 + 1/k^4<\epsilon^2$. Thus
$$\bigg(\frac{1}{k},\frac{1}{k^2} \bigg) \in B_\epsilon\big((0,0)\big)\text{ and } f(\frac{1}{k}, \frac{1}{k^2}) = g(1)<0 = f(0,0)\ .$$ |
H: Prove that $1+a+a^2+\cdots+a^n=(1-a^{n+1})/(1-a)$.
I have problem. Prove this using Mathematical Induction. I am a newbie in Mathematics. Please help me.
$$1+a+a^2+\cdots+a^n = \frac{1-a^{n+1}}{1-a}$$
This is my way for get the proof
Basic Induction:
$$ p(1)= a^1 = 1-a^1+1+1/1-a$$
$$ = 1-a^3/1-a$$
Really I don't understand this case.
AI: The basic induction should be for n=0, then $1 = \frac{1-a}{1-a} = 1$. Now assume it's true for $n=k$ and prove it for $n=k+1$.
So, $p(k+1) = 1+a+a^2+...+a^k +a^{k+1} = \frac{1-a^{k+1}}{1-a} + a^{k+1} = \frac{1-a^{k+1}+a^{k+1} - a^{k+2}}{1-a} = \frac{1-a^{k+2}}{1-a}$ which is exactly what we want |
H: Differential equation in $\mathbb{R}^n$
Given the following ODE for $x$ in $\mathbb{R^n}$
$$ \textbf{x'}(t)=\frac{4\textbf{x}}{|\textbf{x}|}$$
How can I solve it given that $\textbf{x}$ is a function of time? The magnitude of $\textbf{x}$ is really confusing me. I know it should end up being some sort of exponential, right? Thank you for the help/suggestions :)
AI: The equation is only defined for $x \neq 0$.
If $x_0 \neq 0$, then the right hand side satisfies a Lipschitz condition and so a unique local solution is well defined.
It is easy to check that $x(t) = (1+4 \frac{ t }{\|x_0\|})x_0$ is a solution, hence it is the solution.
This is not surprising, the ODE says that the velocity has magnitude 4 in the direction of the current position. Hence it reduces to a one dimensional equation along the ray $\{\alpha x_0\}$. |
H: Polynomial equation should be easy to solve for $c$ given $c \ne 1$?
So I am reviewing questions for a probability exam tomorrow, and have paused on the following
Let $X$ be such that $\mathbf{P}(X=1) = p = 1 - \mathbf{P}(X = -1)$. Find $c \ne 1$ such that $\mathbf{E}(c^X) = 1$
For sake of time, I'll spare rewriting the work which leads to the following conclusion that the equation to solve for $c$ is (given $c \ne 1$),
$$\mathrm{E}[c^X]=1=c^{-1}(1-p)+ cp$$
This is surely easy to solve, but I am nearing my final 1/2 hour for today and tomorrow is the exam. At the moment, I just don't see or recall how to solve this.
We are expected to solve these kinds of equations by hand (i.e. no wolframalpha.com and arguably we shouldn't have to rely on it for solving equations like these)
My next step to solve for $c$:
$$1=c^{-1}(1-p)+ cp \implies \frac{1}{c}-\frac{p}{c}+cp - 1 = 0$$
What next? This looks to be almost a chance to solve with the Quadratic Equation,
$$
c^{-1}(1-p) +cp - 1 = 0
$$
Assuming it's valid to multiply both sides by the variable $c$ (I don't remember whether it is), I get
$$
\tag{1} c^2p - 1 + 1 - p = 0 \implies c^2 p - p = 0
$$
Now, one question I have so far is, have I made a mistake anywhere in this work up to here?. If my work is correct, then I am having trouble seeing whether my answer is equivalent to my wolframalpha's answer:
$$\tag{2} c = \frac{1}{p} - 1, p \ne 0, p \ne 1, c \ne 1$$
If my work is incorrect, please point out the mistakes. But if my work is correct and my answer is indeed equivalent to wolframalpha's answer, then what are the next algebraic transformations I need to get from (1) to (2)? It should be obvious to see for sure, but I'm at the end of my day, and not seeing how.
AI: There is a mistake in the simplification. When you multiplied through by $c$, you should have obtained $c^2p-c+(1-p)=0$. This is a quadratic equation in $c$, and can be solved using the Quadratic Formula.
In fact, the quadratic factors nicely, as $(c-1)(pc-(1-p))$. |
H: How do I find the middle(1/2), 1/3, 1/4, etc, of a line?
Similar to this question: How to calculate the middle of a line? where it's explained how to find the middle of a line (x,y), so that's half the line 1/2, but I also need to find one third of the line, one fourth, and so on.
I tried dividing by 4 instead of 2 to get 1/4 but it didn't seem to work:
1/2 (works):
$x = \dfrac{x_1 + x_2}{2}$ $y = \dfrac{y_1 + y_2}{2}$
1/4 (doesn't work):
$x = \dfrac{x_1 + x_2}{4}$ $y = \dfrac{y_1 + y_2}{4}$
AI: Think about the formula for the midpoint of a line as
$$x = \frac{1}{2}x_{1} + \frac{1}{2}x_{2}.$$
We find the midpoint by taking equal contributions from either end of the line.
If we want to find a point which is one third of the way along the line we should use the following
$$x = \frac{2}{3}x_{1} + \frac{1}{3}x_{2}$$
where we take uneven contributions from the two endpoints.
The point one quarter of the way along the line can be found using
$$x = \frac{3}{4}x_{1} + \frac{1}{4}x_{2}.$$
The y-coordinates are found in a similar manner. |
H: Let $G$ be finite group and $H
Let $G$ be finite group and $H<G$. For every proper subgroup $K$, $[G:H]\leq[H:K]$.
I want to prove $H$ is normal subgroup.
I fixed $K:=g^{-1}Hg$ but this doesn't work. Can somebody advise me?
AI: Why do you try other ways?
Assume to contrary, there exists element $g$ such that $H^g\neq H$. By assumption, we have $[G:H]\leq[H:H^g\cap H]$. Thus we can conclude that $|G|\leq|HH^g|$ and so $G=HH^g$.
It is not hard to see that it is contradiction because there exist elements $h_1,H_2\in H$ such that $g=h_1g^{-1}h_2g$. Hence $g\in H$. |
H: differentiability and linear operators
I could give suggestions for this question.
$f:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is differentiable in $a$, then for each $ h \in \mathbb{R}^{n}$ exists a linear transformation $B(h):\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ such that $h \rightarrow B(h)$ is continuous in $h=0$ and $f(a+h)-f(a)=B(h).h$
I have difficulties to define the operator $B(h):\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$
thank you very much for any suggestions.
AI: Since $f$ is differentiable at $x=a$, we can write $f(a+h) = f(a)+Df(a)h + r_a(h)$, where $r_a$ is $o(h)$.
Define $B(0) = Df(a)$ and $B(h) = Df(a) + r_a(h)\frac{h^T}{\|h\|^2}$ for $h \neq 0$, then we have $B(h)h = Df(a)+ r_a(h) = f(a+h)-f(a)$.
Since $\|B(h)-B(0)\| \le \frac{\|r_a(h)\|}{\|h\|}$, we see that $\lim_{h \to 0} B(h)=B(0)$, and so $B$ is continuous at $h=0$.
Note: If $f$ was $C^1$ we could write $f(a+h)-f(a) = \int_0^1 DF(a+th)h dt$,
and letting $B(h) = \int_0^1 DF(a+th) dt$ would do the trick. |
H: Bijection from $\mathbb N^\mathbb N$
Let's say I'm trying to find a bijection from $\mathbb N^\mathbb N$, i.e., the set of all functions from $\mathbb N$ to $\mathbb N$, to some other set, say an open interval $(a,b)∈R$. What do I need to establish, to say there can exist a bijection? I do understand that I need to prove it's both injective and surjective, but how do you do it with a set of functions as the domain?
Let's say I'm trying to see if the injective part is true. So, I need to find a function that takes every possible sequence of natural numbers as input, and associates each of them with one real number in the interval $(a,b)$. But what does it mean to have a sequence of natural numbers as the input of a function? And even if that makes sense, what sort of functions can satisfy such a purpose? I hope I sound sane.
AI: A function from $\mathbb N$ to $\mathbb N$ would be of the form
$$f(1)=n_1, f(2)=n_2, f(3)=n_3, f(4)=n_4, f(5)=n_5, \ldots$$ for particular values of $n_1,n_2,n_3,n_4,n_5,\ldots$, so producing a sequence of natural numbers. You can then regard the function and the sequence as equivalent.
A function from these to the real numbers is much as you might expect, so for example the continued fraction:
$$g(n_1,n_2,n_3,n_4,n_5,\ldots)= n_1 + \cfrac{1}{n_2 + \cfrac{1}{n_3 + \cfrac{1}{n_4 + \cfrac{1}{n_5+\cdots} } } }.$$
To show this is injective (which it is in this case) you need to show that each different sequence produces a different real number
To show this is surjective (which it is not quite in this case - try finding a rational) you need to show that each real number is produced by some sequence |
H: $a_1=3$ and $a_{n+1}=\dfrac{2a_n}{3}+\dfrac{4}{3a_n^2}$. Show that $4^{1/3} \le a_n$for all $1\le n$.
$a_1=3$ and $a_{n+1}=\dfrac{2a_n}{3}+\dfrac{4}{3a_n^2}$
By considering the function $f(x)=\dfrac{2x}{3}+\dfrac{4}{3x^2}$, show that $4^{1/3} \le a_n$ for all $1\le n$.
AI: $f'(x)=\dfrac{2}{3}-\dfrac{8}{3x^3}=0 \implies x=4^{\frac{1}{3}}$ it is easy to verify this is min point when$x>0$ ,so $f_{min}=4^{\frac{1}{3}}$
$a_n$ is belong to $f(x)$ so $a_n \ge 4^{\frac{1}{3}}$ |
H: Conditioning a CTMC on the future on a Yule pure birth process
I need to solve a problem where I am asked to calculate $M=P(X(0)=2|X(1)=3,X(2)=4,X(3)=5)$ in a Yule pure birth process where $\lambda=1$, so $\lambda_n=\lambda.n=n$ and $\mu_i=0$ (the death rate is 0) for a CTMC: Continuous Time Markov Chain.
The first thing that comes to mind is to reformulate the problem using $P(B|A)=P(A,B)/P(A)$ where making:
$A:\{X(1)=3,X(2)=4,X(3)=5\}$
$B:\{X(0)=2\}$
I get:
$M=\frac{P(X(0)=2,X(1)=3,X(2)=4,X(3)=5)}{P(X(1)=3,X(2)=4,X(3)=5)}$
and decomposing the terms and using the markovian property:
$M=\frac{P(X(3)=5|X(2)=4)P(X(2)=4|X(1)=3)P(X(1)=3|X(0)=2)}{P(X(3)=5|X(2)=4)P(X(2)=4|X(1)=3)}=P(X(1)=3|X(0)=2)$
However this doesn't seem right to me, and I don't know where I'm making the mistake. Suppose I solve the problem using Bayes rule: $P(A|B)P(B)=P(B|A)P(A)$
then $M=\frac{P(X(1)=3,X(2)=4,X(3)=5|X(0)=2)}{P(X(1)=3,X(2)=4,X(3)=5)}P(X(0)=2)$,
Here I can simplify the numerator and denominator again and get in the most reduced form:
$M=P(X(1)=3|X(0)=2)P(X(0)=2)$.
As you can see the P(X(0)=2) term popped up which is new. This answer makes more sense to me, but the problem is how am I supposed to calculate P(X(0)=2)? I'm assuming I can get it by computing the stationary distribution with $\pi.G=0$, and get $\pi_j=\pi_2$. However given that $\pi_j \rightarrow P_{ij}(t)$, when $t \rightarrow \infty$; $P_{ij}(t)\rightarrow 0$, given that $P_{ij}(t)=\binom{j-1}{i-1}e^{-i\lambda t}(1-e^{-\lambda t})^{j-1}, j\geq i$.
I can't seem to find the mistake, my math says that the M should be zero, but my intuition suspects that this isn't the case.
AI: This is a strange question, because usually in a CTMC the value at $t=0$ is specified.
The transition rates don't determine the probabilities by themselves, you have
to say how the process starts out.
You could get the result in terms of the probabilities $P(X(0)=n)$: I get
$$ P(X(0)=2 | X(1)=3, X(2)=4, X(3) = 5) = \dfrac{P(X(0)=2) P_{23}(1)}{P(X(0)=0) P_{03}(1) + P(X(0)=1) P_{13}(1) + P(X(0)=2) P_{23}(1) + P(X(0)=3) P_{33}(1)}$$
You are correct that for a pure birth process with $\lambda$ bounded below, there is no stationary distribution: the population just keeps increasing indefinitely. |
H: Suppose $2^p-1$ is a Mersenne prime. Does $p|2^p-2$?
So, I'm wondering if, given a Mersenne prime, $2^p-1$ whether or not $p$ also divides $2^p-2$. From the wikipedia page on Mersenne prime, it seems like it. For example, $7 = 2^3-1$ is a Mersenne prime and 6 is divisible by 3. Similarly, $8191 = 2^{13}$ and 8190 is divisible by 13.
AI: Since any prime $p$ with $\gcd(2,p)=1$ divides $2^{p-1}-1$ (Little Fermat, Euler) it does also divide $ 2^p-2 = 2(2^{p-1}-1)$ . The reference to the property of Mersenne-primes is a hidden hint: this is the remark, which defines $p$ being element of the set of primes: $2^p-1$ being a Mersenneprime implies $ p \in \mathbb P$ so we are talking about a prime $p$ here ... |
H: cal average speed
the following table represents the number of miles traveled per day by a hiker on his week-long excursion. Find the hiker's average speed in miles per day in the latter half of the week(days 4-7)
Day 0 1 2 3 4 5 6 7
Miles 0 15 10 12 7 12 18 21
AI: To compute the average speed per day you just need to sum all of the miles walked during those days and then divide by the number of days. In your case this amounts to
$$ \frac{7+12+18+21}{4} = \frac{58}{4} = 14,5 $$
Hence your hiker walked, during the last half of the week, with an average speed of $14,5$ miles per day. |
H: $Seq (\mathbb{N})$ of all finite sequences of elements of $\mathbb{N}$ is countable.
I have few questions to the proof of this claim that is illustrated in a textbook.
$Seq (\mathbb{N})$ of all finite sequences of elements of $\mathbb{N}$ is countable.
Proof: Since $Seq(\mathbb{N}) = \bigcup^{\infty} _{n=0} \mathbb{N}^n,$ it suffices to produce a sequence $\left \langle {a_n | n \in \mathbb{N}}\right \rangle $ of enumerations of $\mathbb{N}^n,$ i.e. for each $n \in \mathbb{N}, a_n =\left \langle {a_n(k)| k \in \mathbb{N}}\right \rangle $ is an infinite sequence, and $\mathbb{N}^n = \{a_n(k) | k \in \mathbb{N}\}$
Let $g$ be a bijection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}.$ Define $a_1(i) = \left \langle {i}\right \rangle, \forall i \in \mathbb{N}; a_{n+1}(i) = \left \langle {b_0,...,b_{n-1},i_2}\right \rangle, $ where $g(i) = (i_1,i_2)$ and $\left \langle {b_0,...,b_{n-1}}\right \rangle = a_n(i_1).$ Hence $a_n$ is onto $\mathbb{N}^n.$
My questions: This means $a_2(i) = \left \langle {i_1, i_2}\right \rangle,$ where $g(i) =(i_1,i_2).$ Then $a_3(i) = \left \langle {i_1,i_2,i_2}\right \rangle $ $,a_4(i) =\left \langle {i_1,i_2,i_2,i_2}\right \rangle $ and $a_n(i) =\left \langle {i_1,i_2,...,i_2}\right \rangle$ etc? If so, given $f = \left \langle {f(0),...,f(n-1)}\right \rangle \in \mathbb{N}^n,$ I don't understand how could $f $ be expressed as $a_n(i),$ for some $i \in \mathbb{N}?$
Appreciate any clear instruction. Thank you.
AI: Let $\pi_i$ be a projection of the $i$-th coordinate, and set $g_i = \pi_i \circ g$, that is
\begin{align}
g_1(i) &= i_1, \\
g_2(i) &= i_2,
\end{align}
then your sequence looks like this:
\begin{align}
a_1(k) &= \Big\langle k \Big\rangle \\
a_2(k) &= \Big\langle g_1(k), g_2(k) \Big\rangle \\
a_3(k) &= \Big\langle (g_1\circ g_1)(k), (g_2\circ g_1)(k), g_2(k) \Big\rangle \\
a_4(k) &= \Big\langle (g_1\circ g_1\circ g_1)(k), (g_2\circ g_1\circ g_1)(k), (g_2\circ g_1)(k), g_2(k) \Big\rangle
\end{align}
and so on.
A more intuitive (by my subjective opinion) argument on the cardinality of $\mathrm{Seq}(\mathbb{N})$ would be the following function:
$$f(x_1,x_2,\ldots, x_n) = -1+\prod_{i=1}^n p_i^{x_i+1}$$
where $p_i$ is the $i$-th prime number. In other words, any number gives you a finite sequence of numbers by its prime numbers decomposition, and any sequence can be encoded as show above (be aware that this is not a bijection). The $-1$'s and $+1$'s are to deal with $0 \in \mathbb{N}$. However, appropriate proof would need to use some basic results from number theory and in result might be a bit more complex than the one you have.
I hope this helps $\ddot\smile$ |
H: Comparing different topologies
I have to solve the following:
For every function $f\in C[0,1]$, $\varepsilon>0$ and every finite set $A$, set $U(f,A,\varepsilon)$ is defined by $U(f,A,\varepsilon)=\{g\in C[0,1]:\forall x\in A\; |f(x)-g(x)|<\varepsilon\}$ and family of those sets forms a base of some topology $\mathcal{T}$ on $C[0,1]$. If $U$ is topology induced by metric $d_\infty$ and $M$ topology induced by $d_1$, compare topologies $\mathcal{T}$, $\mathcal{U}$, $\mathcal{M}$.
(By $d_\infty$ and $d_1$ we denote $d_\infty(f,g)=\max_{x\in[0,1]}|f(x)-g(x)|$ and $d_1(f,g)=\int_0^1 |f(x)-g(x)|dx$). So, set $U(f,A,\varepsilon)$ contains continuous functions whose graphs pass through
the finite set of $\varepsilon$-balls (whose centers are points from $A$). We have to check two properties of base for topology $\mathcal{T}$ (first, sets $U(f,A,\varepsilon)$ cover $C[0,1]$ and the second one, with intersection of two base sets, that I'm having problem with). Also, I have got that $\mathcal{M}\subset\mathcal{U}$ (it is easy to check that $d_1(f,g)\leq d_\infty(f,g)$ ) and those topologies are not equivalent, and $\mathcal{U}\subset\mathcal{T}$. But, what happens with $\mathcal{M}$ and $\mathcal{T}$? Also, are topologies $\mathcal{T}$ and $\mathcal{U}$ equivalent?
Detailed explanations are welcome. Thanks in advance.
AI: A partial answer that should help you a little - Comparing $\mathcal{T}$ and $\mathcal{M}$ :
a) Consider $V:= U(0,\{1\}, 1/2) \in \mathcal{T}$, I claim that $V \notin \mathcal{M}$ :
Define $f \in V$ to be the constant function $f \equiv 1/4$. Now I claim that for any $\delta > 0$, there is a $g \in C[0,1]$ such that
$$
d_1(f,g) < \delta \text{ and } g \notin V
$$
For this you can think of a picture of a function $g$ such that
$$
g(1) = 1, \text{ and } g(x) = 1/4\quad\forall x < 1-\delta
$$
and $g$ describes a thin triangle between $x = 1-\delta$ and $x=1$. Then,
$$
d_1(f,g) = \int_{1-\delta}^1 |g(x) - 1/4|dx \leq \int_{1-\delta}^1 (1-1/4)dx = \frac{3\delta}{4} < \delta
$$
Hence, $d_1(f,g) < \delta$. However,
$$
|g(1) - 0| = 1 > 1/2 \Rightarrow g \notin V
$$
b) Consider $W := \{g \in C[0,1] : \int |g(x)|dx < 1\} \in \mathcal{M}$, I claim that $W \notin \mathcal{T}$ :
Choose $f \equiv 1/2$. For any finite subset $A \subset [0,1]$ and any $\delta > 0$, we can construct a function $g \in C[0,1]$ such that
$$
g(x) = 1/2 \quad\forall x \in A, \text{ but } \int |g(x)|dx > 1
$$
In fact, just take the smallest $x_0 \in A$, and build a really large triangle from $(0,2)$ to $(x_0,1/2)$. Now let $g$ be the hypotenuse of that triangle for $x \leq x_0$ and $g(x) = 1/2$ for all $x > x_0$.
Hence, $\mathcal{T}$ and $\mathcal{M}$ are not related by inclusion. |
H: Order of a product in an abelian group.
Suppose $G$ is a finite abelian group and has two element $a$ and $b$, such that $\circ(a)=m$ and $\circ(b)=n$ and $lcm(m,n)\neq m,n$. Is it true that $\circ(ab)=lcm(m,n)$? Thanks in advance.
AI: The statement is not true. In $\mathbb{Z}/30 \mathbb{Z}$, $o(3) = 10$ and $o(5) = 6$, but lcm$(6,10) = 30$ and $o(3+5) = o(8) = 15$. |
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