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H: Random variables X, G are have functional relationship G=g(X). How does g relate the graphs of their distributions?
More specifically, as an educational tool I want to prepare a slideshow showing (in 2-D) the graph of $F_X$ transforming into the graph of $F_G$. (I think it can be done in 4 steps (e.g., graphs on $R$, then $X$, then $G$, and back up to $R$) and am looking for rigor behind the visuals.)
The thought process:
A random variable, X, is defined as a particular type of function from a probability space to the reals, its image the support of its cdf, FX. Thus, “X” denotes a function and, in a supporting role, a set. Given random variable, G=g(X), transform the graph of FX to that of FG in R2 as follows:
1.{(r,FX(r)): rϵR} to {(x,FX(x)): xϵX}: We simply highlight the sub-graph for X.
2.{(x,FX(x)): xϵX} to {(g(x),FX(x)): xϵX}: For each (x,FX(x)) plot (g(x),FX(x)).
3.{(g(x),FX(x)): xϵX} to {(g,FG(g)): gϵG}: We arrive at the sub-graph for G by “sorting” the (g(x),FX(x)) in place vertically by the size of g(x): the resulting 2nd coordinate in (g,FG(g)) being the “sum” over all the F(x)-F(x-) for those (g(x),FX(x)) for which g(x)≤g .
4.{(g,FG(g)): gϵG} to {(r,FG(r)): rϵR}: As G is the support, if rϵR and not in G then FG(r)=0 or FG(r)=FG(g_r) where g_r is the largest value in G less than r.
At issue in particular is the rigor/validity of Step 3.
AI: If $g$ is nondecreasing, $F_G(g(x)) = P(g(X) \le g(x)) = P(X \le x) = F_X(x)$.
If $g$ is nonincreasing, $F_G(g(x)) = P(g(X) \le g(x)) = P(X \ge x) = 1 - F_X(x-)$.
Other cases are more complicated. |
H: $s(x)$ is a arc length function, find $s'(x)$
Here is the problem is my textbook:
Suppose $s(x)$ is the arc length function for the curve $y=\sin x$ taking
$(0,1)$ as the starting point. Find $s’(x)$.
According to arc length formula, I have :
$$ L = \int_0^1 \sqrt{1+\left((\sin x)'\right)^2}dx = \int_0^1 \sqrt{1+\cos^2x}dx$$
But this integral is hard to solve (and When I calculate in Maple, I cannot have the exactly result.) So, I cannot find $s(x)$ to calculate $s’(x)$
Thanks :)
AI: You're given $$s(x):=\int_0^x\sqrt{1+\cos^2t}\,dt$$and since this Riemann integral exists (the function in the integral is continuous and thus integrable in the inverval $\,[0,1]\,$), the primitive $\,G(x)\,$ of the function is a derivable function of the upper limit, so if $$s(x)=\int_0^x\sqrt{1+\cos^2t}\,dt=G(x)-G(0)\Longrightarrow s'(x)=G'(x)=\sqrt{1+\cos^2x}$$By the way, the point$\,(0,1)\,$ is NOT on the curve $\,y=\sin x\,$ , so the function might be $\,\displaystyle{y=\sin\left(\frac{\pi}{2}+ x\right)}\,$ or else you meant something else. |
H: Given pairwise distances of $N$ data points and find the minimal dimenion of space can fit the data
Given a set $D$ consist of all pairwise distance of $N$ unknown dimension points.
e.g. If there is 3 points, ${x_a,x_b,x_c}$
$$D=\{||x_a-x_b||,||x_b-x_c||,||x_a-x_c||\}$$
How can I find the minimum dimension of space which can place all these $N$ points on it and satisfy the distance constraints?
AI: For this topic I can recommend the pretty old paper "Discussion of a set of point in terms of their mutual distances" by Gale Young and A.S. Householder. Although it is from 1938 and uses not "up-to-date" mathematical language it is pretty accessible. By the way: The answer lies in computing the rank of one specific matrix and you can calculate a constellation of points which produce the desired distances by one SVD. |
H: Calculate alpha from $\alpha + \sin(\alpha)$ = K
Sorry for the dumb question, but I'm not involved in math.
I need to reverse the following formula, to calculate $\alpha$:
$$a = b(\alpha + \sin \alpha)/c$$
So I have:
$$(\alpha + \sin \alpha)=ac/b = K$$
Since $a$, $b$, $c$ are constant, I put equal to $K$.
$\alpha$ is measured in radians. I need to find the value of $\alpha$ (in radians or degree).
Thanks to all!!
AI: There is no "closed-form" solution to this equation. You can use numerical methods to solve it for any given value of $K$. If $K$ is small, you can use a series:
$$\alpha = {\frac {1}{2}}K+{\frac {1}{96}}{K}^{3}+{\frac {1}{1920}}{K}^{5}+{\frac {43}{1290240}}{K}^{7}+{\frac {223}{92897280}}{K}^{9} + \ldots $$
The error in the approximation using the terms above is less than about $2 \times 10^{-7}$ for $-1 \le K \le 1$. |
H: Question about a recurrence
In a syllabus of mine, they try to find a closed form of the following recurrence relation
$$\begin{align*}
T(2k) &\leq 3T(k) + ck & k \geq 1\\
T(1) &= 1
\end{align*}$$
The method I usually use to find the closed form of a recurrence is expand it a few times and try to find a pattern. Then I verify that pattern using induction.
In my syllabus they only show the verification part, by using $T(2^l) \leq c(3^l - 2^l)$ as hypothesis, where $c$ is chosen such $T(2)\leq c$ and $l \geq 1$.
So, I tried expanding the recurrence relation to see where I could find that pattern, but I don't get anything close to it.
For example:
$$\begin{align*}T(2^l) &\leq 3(3T(2^{l-2})+c2^{l-2})+c2^{l-1}\\
&= 3^2T(2^{l-2})+3c2^{l-2}+c2^{l-1}\\
&= 3^2T(2^{l-2})+5c2^{l-2}\\
&\leq 3^2(3T(2^{l-3})+c2^{l-3})+5c2^{l-2}\\
&= 3^3T(2^{l-3})+3^2c2^{l-3}+10c2^{l-3}
\end{align*}$$
So, my guess is, since they multiply $3^l$ by $c$, you can subtract $c2^l$ and still remain bigger than $T(2^l)$. But, while expanding, I doubt I would have come up with that.
So my question: is there perhaps a method that I can use to find this hypothesis by expanding, and without to many `magic' manipulations to get to that hypothesis?
AI: Hint: Consider $S(k)=T(2^k)+c\cdot2^k$. Show that $S(k+1)\leqslant 3S(k)$. Compute $S(0)$. Deduce that $S(k)\leqslant 3^k(1+c)$ for every $k\geqslant0$ and, finally, that $T(2^k)\leqslant 3^k+c\cdot(3^k-2^k)$ for every $k\geqslant0$. |
H: How do I solve this problem?
Exercise: If $a+2b=125$ and $b+c=348$, find out $2a+7b+3c$. Here $a$, $b$, $c$ are natural numbers.
The answer is: $2a+7b+3c = 1294$
I tried but just can't figure out how to get to this answer. I have a lot of exercises similar to this one but don't know how to aproach them. Can anyone write the steps in order to get to the answer above? Also it would be great if you can write in a general way so I can apply it to other exercises similar to this.
AI: If you multiply the first equation of the following system by $2$ and the second equation by $3$ you get an equivalent system.
$$\left\{
\begin{array}{c}
a+2b=125 \\
b+c=348
\end{array}
\right. \overset{\times 2}{\underset{\times 3}{\Leftrightarrow }}\left\{
\begin{array}{c}
2a+4b=250 \\
3b+3c=1044
\end{array}
\right. $$
Now you can add the two equations ... |
H: Winning strategy
Is there a winning strategy for player one or two in the following scenario:
The game begins with the number
2012. In one turn, a player can subtract from the current number any natural number
less than or equal to it that is a power of 2. The player who reaches 0 wins.
AI: It seems a player loses if the number is a multiple of 3 on his/her turn. Indeed, in that situation it is impossible to subtract a power of 2 and obtain a difference that is again a multiple of 3. On the other hand, if the number is not a multiple of 3, then the player can force the difference to become a multiple of 3 by playing either 1 or 2.
In particular, the first player wins in this game, where the starting number 2012 is not a multiple of 3. He/she starts by playing 2 (or 512 if he/she wants to go a little faster) and then just makes sure to make the difference a multiple of 3 at the end of all his/her turns. |
H: Least value for addition
We know that $$0\leq a \leq b \leq c\leq d\leq e\,\,\text{ and}\,\, a + b + c + d + e = 100$$. What would be the least
possible value of $\,\,a + c + e\,\,$ ?
I apologize for poor syntax.
AI: $$2(a+c+e) =a+a+c+c+e+e \geq a+b+c+d+e =100$$
With equality if and only if $a=0$, $b=c$ and $d=e$. |
H: Zeroing the carrier measure of an exponential family
I'm trying to derive the general process of changing variables so that an exponential family has zero carrier measure. Distributions in the exponential family have cdf
$$dF(\mathbf{x}|\boldsymbol\eta) = \exp\left({\boldsymbol\eta \cdot T(\mathbf{x}) - g(\boldsymbol\eta)}\right)\, dh(\mathbf{x}).$$
I guess this is a Lebesgue integral, but I don't understand the notation that well.
I would like to find the function $z$ so that
$$dF(z(\mathbf{y})|\boldsymbol\eta) = \exp\left({\boldsymbol\eta \cdot T(z(\mathbf{y})) - g(\boldsymbol\eta)}\right)\, dh(z(\mathbf{y})).$$
and I want the function $h$ to disappear. I want the it to be the standard Lebesgue measure.
So, for the poisson distribution, $h(x) = \frac{1}{x!}$. What is $z$? Is it just $h^{-1}$?
AI: Ok, so if I understood correctly (and adding some stuff of my own to your comment), we have that $\,f:I\to\mathbb{N}\,$ , where $\,I\subset\mathbb{R}\,$ is open or at least contains some open interval within it (otherwise we won't be able to define its derivative at no point).
But if we're talking of $\,f'(x)\,$ then the function is continuos to the discrete space $\,\mathbb{N}\,$, and such functions are as boring as a friday's evening spent with an insurance agent: they are constant functions and thus their derivative is zero, so unless you want to impose some further conditions (say, different topologies for the reals and/or the naturals, or some other definition domain and range of $\,f\,$), I can't see how can we make some other sense of this question |
H: Name for a discrete ordered set with finite subranges
Is there a term for a set that is:
discrete
totally ordered
has finite subranges (not a technical term), i.e. for any a and b, $\{x|a<x<b\}$ is finite
At first glance, it seems this implies that the set is either finite or isomorphic (in some sense) with the integers. Is there a counterexample? What if condition 3 is relaxed? (For example, if the subrange is finite only for certain a and b.)
AI: I am not sure if a such a linear ordering has a name.
However they are not all order isomorphic to finite linear ordering or $\mathbb{Z}$. For example the order type $\omega$ also has this property. As well as the order type $\omega^*$, which is the backward $\omega$.
The finite case is fine. Suppose that $L$ is an infinite linear order. Let $S$ denote the successor relations. Define $x \equiv y$ if there exists a natural number $n$ such that $S^n(x) = y \vee S^n(y) = x$. In order to satisfy these properties, you can see that there can only be one equivalence class. Now ask if the equivalence class has a least element or a greater element. Note that it is impossible to have both and satisfy the three properties above. If it has a least element then you have $\omega$, if it has greater element you have $\omega^*$, if it has neither you have $\mathbb{Z}$.
I think the above sketches the possible linear ordering with the above property. |
H: Can monotone classes be finite?
I am new to measure theory and real analysis and am trying to double check my understanding of monotone classes.
My question:
Can monotone classes be finite? (It is not clear to me whether the idea of increasing or decreasing sets refers to STRICTLY increasing or decreasing sets.)
A related question:
Is any subset of a monotone class itself a monotone class? (The reason I ask is that I do now know the answer to the previous question.)
Thanks in advance.
AI: The definition of monotone class refers to not-necessarily-strictly increasing or decreasing sequences of sets.
An example of a finite monotone class of a set $X$ is $\{\varnothing\}$.
An example of a subset of a monotone class that is not itself a monotone class is the subset
$$\{\text{finite subsets of }\mathbb{N}\}\subset\mathcal{P}(\mathbb{N}).$$
The power set of $\mathbb{N}$, $\mathcal{P}(\mathbb{N})$, is certainly a monotone class, but the increasing sequence of sets
$$\varnothing\subset\{1\}\subset\{1,2\}\subset\cdots$$
each of which lies in the set $\{\text{finite subsets of }\mathbb{N}\}$, has a union of $\mathbb{N}$, which is not in that set. |
H: Solving $\frac{dy}{dx} = xy^2$
This problem appears to be pretty simple to me but my book gets a different answer.
$$\frac{dy}{dx} = xy^2$$
For when y is not 0
$$\frac{dy}{y^2} = x \, dx$$
$$\int \frac{dy}{y^2} = \int x \, dx$$
$$\frac{-1}{y^1} = \frac{x^2}{2}$$
$$\frac{-2}{x^2} = y$$
Is there anything wrong with this solution? It is not what my book gets but it is similar to how they do it in the example.
AI: When you apply indefinite integration, you need to add a "${}+C\,$" (or whatever letter you want) at the end; this is rather infamous for tripping up students. In your problem this gives
$$-\frac{1}{y}=\frac{x^2}{2}+C~~\implies~~ y=\frac{1}{-(x^2/2+C)}=\frac{2}{-x^2-2C}.$$
If we write $K=-2C$ (or again, any old letter), we can write this simply as $\displaystyle\frac{2}{K-x^2}$.
The reason we have "plus a constant" at the end is to prevent erroneous derivations like this:
$$\frac{d}{dx}f(x)=\frac{d}{dx}\big(f(x)+1\big)\implies f(x)=f(x)+1\implies 0=1.$$
In other words, antidifferentiation will only find the antiderivative you want up to addition by an unknown constant. (However with some initial conditions this constant, or constants as it may be in more complicated problems, may be computed exactly.)
Note that it is only necessary to write an add-a-constant to one side of an equation, because for example something like $f(x)+A=g(x)+B$ can be written instead as $f(x)=g(x)+C$ with our constant $C=B-A$. |
H: $1 +1$ is $0$ ?
Possible Duplicate:
-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$?
So:
$$
\begin{align}
1+1 &= 1 + \sqrt{1} \\
&= 1 + \sqrt{1 \times 1} \\
&= 1 + \sqrt{-1 \times -1} \\
&= 1 + \sqrt{-1} \times \sqrt{-1} \\
&= 1 + i \times i \\
&= 1 + (-1) \\
&= 1 - 1\\
&= 0
\end{align}
$$
I can't see anything wrong there, and I can't see anything wrong in $1+1=2$ too. Clearly, $1+1$ is $2$, but I really want to know where is the incorrect part in the above.
AI: $$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$$ is valid only for non-negative real numbers $a$ and $b$. Hence, the error is in the step $$\sqrt{(-1) \times (-1)} = \sqrt{-1} \times \sqrt{-1}$$ |
H: Is the ring of integers in a relative algebraic number field faithfully flat over a ground ring?
Let $L$ be a finite extension of an algebraic number field $K$.
Let $A$ and $B$ be the rings of integers in $K$ and $L$ respectively.
Is $B$ faithfully flat over $A$?
What if $L$ is an infinite algebraic extension of $K$?
AI: A module over a Dedekind domain is flat if and only if it is torsion-free. So any extension of rings $A\rightarrow B$ with $A$ Dedekind and $B$ a domain is flat. In particular, if the extension is integral, it is faithfully flat (by lying-over, as you point out in your comment). |
H: Why is $X_1 + X_2 +\ldots + X_n$ a martingale?
If we have $X_k$ random variables with average $0$ and independent, why is the $\sum_{k=1}^n X_k$ a martingale for the sigma algebra $\mathcal F_n$ generated by $\{X_1,\ldots, X_n\}$?
I basically only have to prove that the expected value of $X_{n+1}$ knowing $\mathcal F_n$ is $0$, but somehow this isn't intuitive to me at all.
Could anyone give me pointers on this?
AI: Just check the definition. Let $S_n:=\sum_{k=1}^nX_k$.
We have that $S_n$ is integrable since so are $X_j$.
Let $n$ an integer. We have by linearity of conditional expectation that
$$E[S_{n+1}\mid\mathcal F_n]=E[S_n+X_{n+1}\mid\mathcal F_n]=E[S_n\mid\mathcal F_n]+
E[X_{n+1}\mid\mathcal F_n].$$
Since $S_n$ is $\mathcal F_n$ measurable, we have $E[S_n\mid\mathcal F_n]=S_n$ and since $X_{n+1}$ is independent of $\mathcal F_n$, we have that $E[X_{n+1}\mid \mathcal F_n]=E[X_{n+1}]=0$. |
H: Friendship theorem and a group of 9 guests
Our task is to prove that there exists 4 strangers OR 4 friends within this group of 9 guests.
Now what's the best way to go about finding this out?
Using the Friendship Theorem? or using the Pigeonhole Principle ?
AI: Here is a counterexample to the claim that nine guests is sufficient to force four mutual friends or four mutual strangers.
Suppose your guests are divided into three disjoint cliques (in both senses of the word). That is, the guests are divided into three groups such that
everyone knows everyone else within their own group and
nobody knows anyone else outside their own group.
By this construction, there cannot be four individuals that all know each other, since at least two of them belong to different groups (and so are strangers). Similarly, there cannot be four individuals that are mutual strangers, since at least two of them belong to the same group (and so are friends). |
H: Expressing $\sin(2x)-8\cos(2x)$ as a single sine function
I am asked as a part of a question to express $\sin(2x)-8\cos(2x)$ as a single sine function.
I know it has something to do with the trigonometric identity $$\sin(a-b)=\sin(a) \cos(b)-\cos(a)\sin(b)$$ but I can't get my head around it because of that $8$ in front of $\cos2x$.
Any tips on how I can move on?
AI: Hint: Our function is
$$\sqrt{65}\left(\frac{1}{\sqrt{65}}\sin 2x -\frac{8}{\sqrt{65}}\cos 2x\right).$$
Let $b$ be an angle whose cosine is $\frac{1}{\sqrt{65}}$ and whose sine is $\frac{8}{\sqrt{65}}$, and use the identity you quoted. |
H: existence of a harmonic function
Let $\Omega\subset\mathbb R^n$ open, not bounded and $n\ge3$. Let $\partial\Omega$ bounded and regular concering the laplace operator. Given a continuous function $\phi:\partial\Omega\rightarrow\mathbb R$ and $\gamma\in\mathbb R$ there exists a harmonic function $u\in C^2(\Omega)\cap C^0(\overline\Omega)$ with $u=g\space\space\text{on}\space\partial\Omega$ and $\lim\limits_{|x|\rightarrow\infty}u(x)=\gamma$
How can you prove the existence?
AI: This is very nontrivial and called the Dirichlet problem.
If you are looking for a solution using complex analysis (well, methods connected to complex analysis), both Ahlfors and Gamelin's textbooks on complex analysis give a proof of existence using subharmonic functions. This is called "Perron's method." I have looked at this closely (I was tested over it) and it gives good insight into exactly how pathological the boundary can be before you can no longer solve the problem.
Another route to existence is given the third volume of Stein and Shakarchi's four volume Princeton Lectures in Analysis series, in the book titled Real Analysis. I have not read it closely, but it might be a useful reference if you are not familiar with complex analysis or subharmonic functions.
You question is a little confusing to me because you say that $\Omega$ is not bounded, but $\partial\Omega$ is. Both references I gave only deal with the case where the region is bounded. However, I have no doubt that armed with the magic search term "Dirichlet Problem," you will be able to find literature about the unbounded case. This is a classic problem, and I'm sure others can give more authoritative references if you are looking for something research-level. |
H: Does the isomorphism $k[x]\otimes_k k[x]\cong k[x,y]$ hold?
I think I am finally beginning to understand tensor products of algebras, and I could use a reality check. If I am understanding correctly, then $k[x]\otimes_k k[x]$ is ring-isomorphic to $k[x,y]$ by the map $x^i\otimes x^j\mapsto x^iy^j$. Is this right?
AI: Yes, that's right. The tensor product over $k$ is the coproduct in the category of commutative $k$-algebras, and the polynomial ring in $n$ variables over $k$ is the free commutative $k$-algebra on $n$ elements. So the coproduct of the free $k$-algebra on one object and itself can only be the free $k$-algebra on two objects by an examination of the functor it represents.
Geometrically, the tensor product is the product in the category of affine schemes over $k$, and the geometric fact being described here is that the product of $\mathbb{A}^1$ with itself is $\mathbb{A}^2$. |
H: Finite groups of functions under function composition
Over the years I have done many questions along the lines of the following:
"Given functions $\phi, \theta$ (usually defined on $\mathbb{R}$ or $\mathbb{C}$, or a suitable subset of $\mathbb{R}$ or $\mathbb{C}$) prove that the collection of all functions obtained from $\theta$ and $\phi$ by function composition form a group G." (Frequently G is $\mathbb{Z}/4 \mathbb{Z}$ or $S_3$.)
A typical example might be functions $\theta:x\mapsto 1-x$ and $\phi:x\mapsto \frac{1}{x}$, (defined on the set of non-zero reals) generating a group isomorphic to $S_3$.
When I have been inventing questions for my students, rather than simply copying examples from previous exam papers, I have sometimes wondered if it is possible to create a similar example with a function $\psi$ which has order 5 in the group - just for a bit of variety! However a crucial restriction is that I need the functions to be simple algebraic functions (something like $x\mapsto \frac{ax+b}{cx+d}$ with $a, b, c, d\in \mathbb{Z}$), which rules out things like rotations of $\mathbb{C}$ through an angle of $2\pi/5$.
My question is then: does anyone know of such a function, or has anyone come across a similar exam question which gives an element of order 5 (or 7 for that matter ...) arising from such a simple type of function?
I have tried investigating the possibilities at various times, and have easily found functions which have order 2, 3, 4, but never one of order 5 or 7.
PS There is no great urgency here, as I have now retired from teaching this sort of stuff.
AI: Depending on exactly what your requirements are, there does not exist such a function. First let's suppose you're working with rational functions $f(z) = \frac{p(z)}{q(z)}$ where $z$ is a complex parameter and $p, q$ have no common roots. In order for $f$ to be invertible by another rational function, it needs to be a bijective function from the Riemann sphere to itself. Now:
If $\deg p > 1$ then $f^{-1}(0)$ consists of more than one point, and
If $\deg q > 1$ then $f^{-1}(\infty)$ consists of more than one point.
It follows that in order for $f$ to be invertible we must have $f(z) = \frac{az + b}{cz + d}$; moreover we need $ad - bc \neq 0$, and this is sufficient.
The resulting group is very well-understood. It is the group of Möbius transformations or, equivalently, the projective special linear group $\text{PSL}_2(\mathbb{C})$, and all of its finite subgroups are known. Remarkably, they are described by Dynkin diagrams (see e.g. this MO question). Explicitly they are:
the finite cyclic groups,
the finite dihedral groups,
the tetrahedral group,
the octahedral group, and
the icosahedral group.
The good news is that there are elements of order $5$. The bad news is that they cannot be realized over $\mathbb{Q}$.
To see this, note that $\text{PSL}_2(\mathbb{C})$ admits a 2-to-1 homomorphism $\text{SL}_2(\mathbb{C}) \to \text{PSL}_2(\mathbb{C})$; thus if $f(z) = \frac{az + b}{cz + d}$ (written so that $ad - bc = 1$) is an element of order $5$ or $7$, its preimages in $\text{SL}_2(\mathbb{C})$, one of which is given by the matrix
$$M = \left[ \begin{array}{cc} a & b \\\ c & d \end{array} \right],$$
must have order $5, 7, 10$ or $14$. Since it has determinant $1$, its eigenvalues must be a primitive root of unity $\zeta_n$ (where $n = 5, 7, 10, 14$) and $\zeta_n^{-1}$. This means that its trace must be
$$\zeta_n + \zeta_n^{-1} = 2 \cos \frac{2 \pi i}{n}$$
but it must also be $a + d$, which is rational. Since $\zeta_n + \zeta_n^{-1}$ is a sum of algebraic integers, it is an algebraic integer, hence is rational if and only if it is an integer. But $2 \cos \frac{2 \pi i}{n}$ is an integer if and only if it's equal to $2, 1, 0, -1, -2$, and this gives $n = 1, 2, 3, 4, 6$. Thus we cannot have $n = 5, 7, 10, 14$ or any number greater than $6$ for that matter.
The above argument is closely related to the crystallographic restriction theorem. |
H: Proving that the magnitude of the sample correlation coefficient is at most $1$
How can you show that the magnitude of the sample correlation coefficient is at most $1$?
The formula is huge, I'm not even sure how to approach this. Can anyone point me in the right direction?
Note that this is the sample correlation coefficient:
$$r_{xy} = \dfrac{\displaystyle \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{(n-1)s_xs_y} = \dfrac{\displaystyle \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\displaystyle \sum_{i=1}^{n} (x_i - \bar{x})^2 \displaystyle \sum_{i=1}^{n} (y_i - \bar{y})^2}}$$
AI: This follows from Cauchy–Schwarz inequality. The Cauchy–Schwarz inequality states that for any two vectors $a$ and $b$ in an inner product space, we have that
$$\lvert \langle a, b \rangle \rvert^2 \leq \lvert \langle a, a \rangle \rvert \lvert \langle b, b \rangle \rvert$$
In your case, the vector $a$ is taken as $a_i = (x_i-\bar{x})$ and the vector $b$ is taken as $b_i = (y_i-\bar{y})$ and the inner product of $a$ and $b$ is taken as $\displaystyle \langle a, b \rangle = \sum_{i=1}^n a_i b_i$. Hence, we get that
$$\displaystyle \langle a, b \rangle = \sum_{i=1}^n a_i b_i = \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})$$
$$\displaystyle \langle a, a \rangle = \sum_{i=1}^n a_i a_i = \sum_{i=1}^n (x_i - \bar{x})^2$$
$$\displaystyle \langle b, b \rangle = \sum_{i=1}^n b_i b_i = \sum_{i=1}^n (y_i - \bar{y})^2$$
Hence, by Cauchy–Schwarz inequality, we get that
$$\left(\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})\right)^2 \leq \left( \sum_{i=1}^n (x_i - \bar{x})^2 \right) \left( \sum_{i=1}^n (y_i - \bar{y})^2\right)$$
Taking the squareroot, we get that
$$\left|\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})\right| \leq \sqrt{\left( \sum_{i=1}^n (x_i - \bar{x})^2 \right) \left( \sum_{i=1}^n (y_i - \bar{y})^2\right)}$$
Hence, we can conclude that
$$|r_{xy}| = \dfrac{\left|\displaystyle \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})\right|}{\displaystyle \sqrt{\left( \sum_{i=1}^n (x_i - \bar{x})^2 \right) \left( \sum_{i=1}^n (y_i - \bar{y})^2\right)}} \leq 1$$
EDIT
Proof of Cauchy Schwarz inequality:
First note that if the vector $b$ is zero, then the inequality is trivially satisfied since both sides are zero. Hence, we can assume that $b \neq 0$. Now look at the component of $a$ orthogonal to $b$ i.e. $$c = a - \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b$$
i.e.
$$a = c + \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b$$
You can check that $c$ is orthogonal to $b$ by computing $$\langle c, b \rangle = \langle a,b \rangle - \dfrac{\langle a, b \rangle}{\langle b, b \rangle} \langle b, b \rangle = \langle a,b \rangle - \langle a,b \rangle = 0$$
You can also check that $\langle c, \alpha b \rangle = 0 = \langle \beta c, b \rangle$.
We now have that
\begin{align}
\langle a,a \rangle & = \left \langle c + \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b, c + \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b \right \rangle\\
& = \langle c,c \rangle + \left \langle c,\dfrac{\langle a, b \rangle}{\langle b, b \rangle} b \right \rangle + \left \langle \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b, c \right \rangle + \left \langle \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b, \dfrac{\langle a, b \rangle}{\langle b, b \rangle} b \right \rangle\\
& = \langle c,c \rangle + \left \lvert \dfrac{\langle a, b \rangle}{\langle b, b \rangle} \right \rvert^2 \langle b, b \rangle = \langle c,c \rangle + \dfrac{\left \lvert \langle a, b \rangle \right \rvert^2}{\langle b, b \rangle}
\end{align}
Now $\langle c,c \rangle \geq 0$. This gives that
$$\langle a,a \rangle \geq \dfrac{\left \lvert \langle a, b \rangle \right \rvert^2}{\langle b, b \rangle}$$
Rearranging, we get what we want, namely
$$\lvert \langle a, b \rangle \rvert^2 \leq \lvert \langle a, a \rangle \rvert \lvert \langle b, b \rangle \rvert$$ |
H: Is the complement of a finite dimensional subspace always closed?
Let $F$ be a finite dimensional subspace of an infinite dimensional Banach space $X$, we know that $F$ is always topologically complemented in $X$, that is, there is always a closed subspace $W$ such that $X=F\oplus W$.
I am thinking about the converse. Suppose $W$ is a subspace of $X$ such that $X=F\oplus W$ for some finite dimensional subspace $F$. Is $W$ necessarily closed?
I guess the answer should be negative but I cannot find such an example. Can somebody give a hint?
Thanks!
AI: If $f:X\to\mathbb C$ is a discontinuous linear functional, then $\ker f$ is not closed. If $v$ is in $X\setminus \ker f$, then $F=\mathbb C v$ and $W=\ker f$ gives a counterexample. ($X$ is the internal direct sum of $F$ and $W$ as vector spaces, but it is not a topological direct sum.) |
H: Basic probability problem
Problem states:
Consider two events $A$ and $B$, with $P(A) = 0.4$ and $Pr(B) = 0.7$. Determine the maximum and the minimum possible values for $P(A \& B)$ and the conditions under which each of these values is attained.
To solve, I considered the event with the lowest probability $A$ to be a subset of the other, so maximum value is attained under that circumstance giving a probability of $0.4$.
But the book states that the minimum is $0.1$, if $P(A \cup B) = 1$.
I don't understand why! Because I thought that the minimum value is get when the two events are disjoint... So the minimum value must be $0$...
AI: Use the fact that
$$P(A\cup B) = P(A) + P(B) - P(A\cap B).$$
This gives you
$$P(A\cup B) = 1.1 - P(A\cap B).$$
Since $P(A\cup B) \le 1$ we have
$$1\ge 1.1 - P(A\cap B),$$
yielding $P(A\cap B)\ge .1$.
If $A\subseteq B$, then $P(A\cap B) = P(A) = .4.$
This represents the largest possible value for $P(A\cap B)$.
so $.1\le P(A\cap B) \le .4$ |
H: Formula to calculate the number of possible positions for $x$ numbers
What formula do I use to calculate the number of possible positions for $x$ numbers?
Let's say I have $3$ people in a race. What are all the possible combinations of the order they can finish in? Let's assume ties are not possible. I heard I use factorial but its been a while since I have used factorials. So I want to verify.
AI: It is easy to enumerate for three people. Lets call the people $x_1,x_2,x_3$. The possible orderings are shown below.
$x_1,x_2,x_3$
$x_1,x_3,x_2$
$x_2,x_1,x_3$
$x_2,x_3,x_1$
$x_3,x_1,x_2$
$x_3,x_2,x_1$
In general, in a $n$ people race, lets denote the people by $x_1,x_2,\ldots,x_n$. The first position can be taken by any one of the $n$ individuals. Hence, there are $n$ options for the first place. Now given that the first position is taken by $1$ individual, for each of the $n$ options for the first position, the second position can be taken by any one of the remaining $n-1$ individuals. Proceeding like this, in general the $k^{th}$ position can be taken by any one of the remaining $n-k+1$ individuals. Hence, the total number of ordering is given by $$n \times (n-1) \times \cdots \times (n-k+1) \times \cdots 2 \times 1$$which is nothing but $n!$.
In you case, $n=3$ gives us $3! = 6$ which matches with our enumeration. |
H: Easy condition for positive definite endomorphism
This problem is taken from Golan's linear algebra book.
Problem: Let $V$ be an inner product space over $\mathbb{R}$ and let $\alpha$ be an endomorphism of $V$. Show that $\alpha$ is positive definite if and only if $\alpha+\alpha^*$ is positive definite.
Definition: An endomorphism $\alpha$ is positive definite if and only if it is selfadjoint and satisfies the condition that $\langle \alpha(v), v\rangle$ is a positive real number for all nonzero $v\in V$.
AI: First, assume $\alpha$ is positive definite. We have $\langle \alpha(v), v\rangle\in \mathbb{R}^+.$, and because $\alpha=\alpha^*$, their sum is self-adjoint.
Now suppose $\alpha+\alpha^*$ is positive definite. We have $\langle (\alpha+\alpha^*)(v), v\rangle \in \mathbb{R}^+$. Also,
$$\langle \alpha(v), v\rangle =\langle v, \alpha^*(v)\rangle =\langle \alpha^*(v), v\rangle.$$
This shows that
$$\langle (\alpha+\alpha^*)(v), v\rangle =\langle \alpha(v), v\rangle+ \langle \alpha^*(v), v\rangle = \langle \alpha(v), v\rangle+\langle \alpha(v), v\rangle$$
is always nonnegative, which shows that $\langle \alpha(v), v\rangle$ is always nonnegative, and it is easy to check that equality only holds when $v=0$. |
H: Divisibility is transitive: $\ a\mid b\mid c\,\Rightarrow\ a\mid c$
As the title says, if a number is divisible by a number, is it always divisible by that number's factors?
An example being that $100$ is divisible by $20$, it is also divisible by $10, 5, 4, 2$ as well?
Does this always apply?
AI: Yes. It is indeed true. The proof also follows immediately. Before looking at the proof, lets us understand what it means to say that $x \in \mathbb{Z}$ divides $y \in \mathbb{Z}$.
We say that $x \in \mathbb{Z}$ divides $y \in \mathbb{Z}$, if there exists $n \in \mathbb{Z}$, such that $ y = x \times n$.
For instance, $6$ divides $-30$, since we have $-5 \in \mathbb{Z}$ such that $-30 = 6 \times (-5)$.
Similarly, $27$ divides $108$, since we have $4 \in \mathbb{Z}$ such that $108 = 27 \times 4$.
Now lets prove your claim.
Claim: If $a$ divides $b$ and $b$ divides $c$, then $a$ divides $c$, where $a,b,c \in \mathbb{Z}$.
Proof:
Since $a$ divides $b$, we have $n_1 \in \mathbb{Z}$ such that $b = a \times n_1$.
Similarly, since $b$ divides $c$, we have $n_2 \in \mathbb{Z}$ such that $c = b \times n_2$.
Making use of the fact that $b = a \times n_1$ in the above equation, we get that $$c = \underbrace{(a \times n_1) \times n_2 = a \times (n_1 \times n_2)}_{\text{By associativity of multiplication}} = a \times n$$ where $n = n_1 \times n_2 \in \mathbb{Z}$.
Hence, $a$ divides $c$. |
H: Proving that $\{0,1\}$ is a field with $1+1:=0$
EDIT: Hopefully question made clearer. Unfortunately this is a question found in analysis book and I do not actually have background on abstract algebra. Sorry for the confusion arisen.
As the title says, I am trying to show that a set of $\{0,1\}$, equipped with the obvious multiplication and with $1+1:=0$, is a field.
I encounter this question after the axioms of addition and multiplication of field in $\mathbb{R}$. I am not sure that how would I approach this question. Should I just check by brute force, e.g. check $x+y = y+x$ for all elements in $\{0,1\}$? How about associative of addition, i.e. $(x+y)+z= x+(y+z)$, how do I find myself such a third element $z$ from $\{0,1\}$?
Thanks
AI: If you have no knowledge of quotient rings such as $\:\mathbb Z/2\mathbb Z\:$ then brute-force verfification of all field axioms may be your only choice. This can be done intelligently, e.g. to verify the distributive law $\rm\:x(y\!+\!z) = xy\!+\!xz\:$ doesn't require checking all $8$ possibilities arising from the two possible values $(0$ or $1)$ for each variable. Instead, note that if $\rm\:x=1\:$ it becomes $\rm\:y\!+\!z = y\!+\!z,\:$ so is true for all $\rm\:y,z.\:$ Else $\rm\:x\ne 1\:\Rightarrow\:x=0\:$ so it becomes $\:0 = 0,\:$ so is true for all $\rm\:y,z.\:$ Therefore the distributive law is true for all $\rm\:x,y,z.\:$ Analogous optimizations exist for the other axioms. |
H: Functions - Set Theory Proof
Let $f$ be a function from $X$ to $Y$, and let $A$, $B$ be subsets (non-proper) of $X$. For each of the following statements, either prove the statement or else give a counter example:
a.) $f(X\setminus A)=Y\setminus f(A)$
b.) $f(X\setminus A) \subseteq Y\setminus f(A)$
c.) $Y\setminus f(A) \subseteq f(X\setminus A)$
d.) $f(A\cup B) = f(A)\cup f(B)$
e.) $f(A) \cap f(B) = f(A \cap B)$
I have an exam tomorrow and have been lagging on the set theory.
Much appreciated.
AI: a) False If $A \subsetneq X$, consider the constant function.
b) False Use $(a)$
c) False. Suppose $f$ is not surjective. Suppose $Y$ contains more than one element and $f$ is a content function.
d) True. Suppose $x \in f(A \cup B)$, then $x = f(y)$ for $y \in A$ or $y \in B$ so $x \in f(A)$ or $x \in f(B)$. Suppose $x \in f(A) \cup f(B)$. Then $x \in f(A)$ or $x \in f(B)$. So there exists a $y \in A$ or a $y \in B$ such that $f(y) = x$ so $x \in f(A \cup B)$.
e) True. You wrote $f(A) \cap f(B) = f(A) \cap f(B)$. Do you mean $f(A) \cap f(B) = f(A \cap B)$??? If you meant the latter, this is false. Suppose $f$ is a constant function and $A$ and $B$ are disjoint. |
H: "Fully correlated" definition
Really sorry to be a noob, but I'm a programmer, not a mathematician, and all of my knowledge about statistics come from this book "Schaum's Outline of Theory and Problems of Probability, Random Variables, and Random Processes".
I'm implementing an UKF for target tracking using C++. Everything went well until an error about covariance matrix of state is not positive definite happened.
After a little research, I found this link Under what circumstance will a covariance matrix be positive semi-definite rather than positive definite? which almost answer everything I need.
Only one thing I don't understand: The answer says "This happens if and only if some linear combination of X is ‘fully correlated’". Can anyone explain for me what does "fully correlated" mean? And example would be great. I have search Google about its definition but there is no luck at all.
AI: I think this means that the relationship is exactly linear and the correlation is 1 or -1. |
H: Is there a distributive law for ideals?
I'm curious if there is some sort of distributive law for ideals.
If $I,J,K$ are ideals in an arbitrary ring, does $I(J+K)=IJ+IK$?
The containment "$\subset$" is pretty clear I think. But the opposite ontainment doesn't feel like it should work. I couldn't work out a counterexample with ideals in $\mathbb{Z}$ however. So does such an equality always hold or not?
AI: Note that if $A$, $B$, and $C$ are ideals, and $B\subseteq C$, then $AB\subseteq AC$; and if $A$ and $B$ are both contained in $C$, then $A+B\subseteq C$.
Since $J\subseteq J+K$, then $IJ\subseteq I(J+K)$. Since $K\subseteq J+K$, then $IK\subseteq I(J+K)$. Therefore, $IJ$ and $IK$ are both contained in $I(J+K)$, so $IJ+IK\subseteq I(J+K)$.
For the converse inclusion, a general element of $I(J+K)$ is of the form
$$\sum a_i(j_i+k_i)$$
with $a_i\in I$, $j_i\in J$, and $k_i\in K$. And we have
$$\sum a_i(j_i+k_i) = \sum\Bigl( a_ij_i + a_ik_i\Bigr) = \left( \sum a_ij_i\right) + \left(\sum a_ik_i\right) \in IJ + IK.$$ |
H: The function $f(x) = \int_0^\infty \frac{x^t}{\Gamma(t+1)} \, dt$
Does anyone know if this function has a name? I came up with it by looking at the power series for $e^z$, changing the summation to an integral, and substituting the gamma function for the factorial function.
AI: Almost tautological, but we can also find that
$$ f(x) = e^{x} \left( 1 - \int_{0}^{1} \frac{\Gamma(s, x)}{\Gamma(s)} \; ds \right),$$
where $\Gamma(s, x) = \int_{x}^{\infty} t^{s-1}e^{-t} \; dt$ is the incomplete gamma function. Indeed,
$$ f'(x) = \int_{0}^{\infty} \frac{s x^{s-1}}{\Gamma(s+1)} \; ds = \int_{0}^{1} \frac{x^{s-1}}{\Gamma(s)} \; ds + f(x)$$
and thus
$$ (f(x)e^{-x})' = \int_{0}^{1} \frac{x^{s-1}e^{-x}}{\Gamma(s)} \; ds.$$
Then integrating both sides with respect to $x$ gives the desired relation.
Now we try to find its asymptotic formula. Since $\Gamma(s,x) / \Gamma(s) \searrow 0$ as $x \to \infty$, we find
$$f(x) = (1 - o(1))e^{x}$$
asymptotically. To obtain a better approximation, note that $1/\Gamma(s) \asymp s$ near $s = 0$. Thus
$$ \begin{align*}
\int_{0}^{1} \frac{\Gamma(s, x)}{\Gamma(s)} \; ds
&\asymp \int_{0}^{1} s\Gamma(s, x) \; ds \\
&= \int_{0}^{1} (\Gamma(s+1,x) - x^{s}e^{-x}) \; ds \\
&= \int_{0}^{1} \int_{x}^{\infty} t^{s}e^{-t} \; dt ds - \frac{x-1}{\log x}e^{-x} \\
&= \int_{x}^{\infty} \left( \frac{t-1}{\log t}\right) e^{-t} \; dt - \frac{x-1}{\log x}e^{-x} \\
&= \int_{x}^{\infty} \left( \frac{t-1}{\log t}\right)' e^{-t} \; dt \\
&\asymp \int_{x}^{\infty} \frac{e^{-t}}{\log t} \; dt \\
&\asymp \frac{e^{-x}}{\log x}.
\end{align*}$$
In particular, we have
$$f(x) = e^{x} - O\left( \frac{1}{\log x} \right).$$ |
H: Can such a function exist?
Denote by $\Sigma$ the collection of all $(S, \succeq)$ wher $S \subset \mathbb{R}$ is compact and $\succeq$ is an arbitrary total order on $S$.
Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ such that for all $(S, \succeq) \in \Sigma$ there exists a compact interval $I$ with the properties that
$f(I) = S$
$x \geq y$ implies $f(x) \succeq f(y)$ for all $x,y \in I$?
If so, how regular can we take $f$ to be? The motivation is that basically, I am trying to construct the analogue of a normal sequence but on $\mathbb{R}$ instead of $\mathbb{N}$.
EDIT: As Brian M. Scott points out, this is not possible if the orderings have no greatest and least elements. However, since adding this assumption doesn't go against the intuition of generalizing normal sequences, I am still interested in the answer if we restrict the various total orders to have minimal and maximal elements.
Thanks in advance.
AI: If I understand correctly what you’re asking, the answer is no. Let $\preceq$ be a linear order on $[0,1]$ having no last element; no compact subset of $\Bbb R$ can be mapped onto $[0,1]$ in such a way that $f(x)\preceq f(y)$ whenever $x\le y$, because every compact subset of $\Bbb R$ has a last element with respect to the usual order. That is, if $I$ is a compact subset of $\Bbb R$, let $u=\max I$, and let $x\in[0,1]$ be such that $f(u)\prec x$; then $x\notin\operatorname{ran}f$, so $f[I]\ne[0,1]$.
Added: Restricting the linear orders to those with first and last elements doesn’t help. Well-order $[0,1]=\{y_\xi:\xi<2^\omega+1\}$. Now let $I$ be any compact interval, say $[a,b]$, and let $f:I\to[0,1]$ be an order-preserving surjection. Clearly $f(b)=y_{2^\omega}$ is the last element of in the well-ordering of $[0,1]$. Let $\langle x_n:n\in\omega\rangle$ be a strictly increasing sequence in $[a,b]$ converging to $b$. Then $\langle f(x_n):n\in\omega\rangle$ is an increasing sequence in the well-ordering of $[0,1]$, and it has a supremum, say $y_\alpha$. Then $y_\alpha<2^\omega$, since $\operatorname{cf}2^\omega>\omega$, and no $y_\beta$ such that $\alpha<\beta<2^\omega$ is in the range of $f$.
Actually, this requires a bit of modification if $f$ is not required to be strictly order-preserving (i.e., a bijection). In that case let $B=f^{-1}[\{y_{2^\omega}\}]$; $B$ must be of the form $(c,b]$ or $[c,b]$ for some $c\in I$. If $B=[c,b]$, replace $b$ by $c$ in the previous paragraph. If $B=(c,b]$, note that nothing between $f(c)$ and $f(b)$ is in the range of $f$. |
H: Why is $a$ and $b$ coprime if $a\equiv 1 \pmod{b}$?
$a$ and $b$ are coprime if their greatest common divisor is 1. How do I conclude that from the fact that $b$ divides $a-1$?
AI: If $b\mid a-1$, then there is an integer $n$ such that $a-1=bn$, and therefore $a-bn=1$. Suppose that $d\mid a$ and $d\mid b$: then $d\mid a-bn$. (Why? If you’re in any doubt, you should write out the details of the argument.) What does that tell you about $d$? |
H: which are positive definite matrix
Given that $A,B$ are positive definite matrix, are they also PD?
$A+B$
$AB$
$A^2 +I$
$ABA^{*}$
$x^TAx>0, x^TBx>0$ so $1$, is correct, could you tell me about the 2, 3,4?
AI: Mex
in 4. we can argue as follows:
$\langle x, ABA^* x\rangle = \langle A^*x, BA^* x\rangle =
\langle B^*A^*x, A^* x\rangle = \langle B Ax, A x\rangle
= \langle B y, y\rangle >0 $,
where $y=Ax$ and in the two last equalities we use that positive matrix are self-adjoint and since they do not have $0$
as eigenvalue then $Ay\neq 0$.
for 3. the argument is similar. |
H: Finding the transformation when given transformation matrix
Lets say, there is a transformation: $T:\Re ^{n}\rightarrow \Re ^{m}$ transforming a vector in $V$ to $W$.
Now the transformation matrix, $A=\begin{bmatrix}
a_{11} & a_{12} &...&a_{1n} \\
a_{21} & a_{22} &...&a_{2n} \\
. & . & .\\
. & . & .\\
. & . & .\\
a_{n1} &a_{n2} &...&a_{n3}\\
\end{bmatrix}$
The basis vectors of $V$ are {$v_{1},v_{2},v_{3}...,v_{n}$} which are all non standard vectors and similarly {$w_{1},w_{2},...,w_{m}$}.
My question is, in the absence of the basis vectors being standard vectors what is the procedure of finding $T$
Soham
AI: I’ll use the example in your comment: $M=\pmatrix{4&2&1\\0&1&3}$, with respect to the bases $$\beta=\{v_1=[1,0,0],v_2=[1,1,0],v_3=[1,1,1]\}$$ and $$\beta\,'=\{w_1=[1,0],w_2=[1,1]\}\;.$$
This transformation takes as input a column vector $$v=\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}$$
representing the linear combination $\alpha_1v_1+\alpha_2v_2+\alpha_3v_3$ of basis vectors in $\beta$ and produces as output the product
$$Mv=\pmatrix{4&2&1\\0&1&3}\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}=\pmatrix{4\alpha_1+2\alpha_2+\alpha_3\\\alpha_2+3\alpha_3}\;.$$
The entries $4\alpha_1+2\alpha_2+\alpha_3$ and $\alpha_2+3\alpha_3$ are then to be interpreted as coefficients of the vectors $w_1$ and $w_2$ in $\beta\,'$:
$$T(v)=(4\alpha_1+2\alpha_2+\alpha_3)w_1+(\alpha_2+3\alpha_3)w_2\;.$$
If you want to know what this looks like in terms of the standard basis $\{[1,0],[0,1]\}$, just multiply out:
$$\begin{align*}
T(v)&=(4\alpha_1+2\alpha_2+\alpha_3)\pmatrix{1\\0}+(\alpha_2+3\alpha_3)\pmatrix{1\\1}\\
&=\pmatrix{4\alpha_1+3\alpha_2+4\alpha_3\\\alpha_2+3\alpha_3}\;.
\end{align*}$$
Note that this is exactly what you get from the product
$$AMv=\pmatrix{1&1\\0&1}\pmatrix{4&2&1\\0&1&3}\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}\;,$$
where $A=\pmatrix{1&1\\0&1}$ is a change-of-basis matrix: it translates a representation in terms of $\beta\,'$ into one in terms of the standard basis. It’s easy to construct this change-of-basis matrix: its columns are just the representations of $w_1$ and $w_2$ in terms of the standard basis.
It follows that if you start with $v$, representing a vector in $\Bbb R^3$ in terms of the basis $\beta$, and multiply it by the matrix $$AM=\pmatrix{4&3&4\\0&1&3}\;,$$ you get $T(v)$ expressed in terms of the standard basis for $\Bbb R^2$. Perhaps, though, you want to be able to input $v$ in terms of the standard basis for $\Bbb R^3$. Then you need another change-of-basis matrix, this time to convert from the standard basis for $\Bbb R^3$ to the basis $\beta$. We already know how to go the other way: to transform from a representation in terms of $\beta$ to one standard coordinates, multiply by the matrix $$B=\pmatrix{1&1&1\\0&1&1\\0&0&1}$$ whose columns are the representations of $w_1,w_2$, and $w_3$ in terms of the standard basis. (In other words, do exactly what we did to get $A$.) If you take the vector $\alpha_1v_1+\alpha_2v_2+\alpha_3v_3$, represented by the matrix $$v=\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}$$ in terms of the basis $\beta$, you can find its representation in terms of the standard basis by multiplying by $B$ to get $$Bv=\pmatrix{1&1&1\\0&1&1\\0&0&1}\pmatrix{\alpha_1\\\alpha_2\\\alpha_3}=\pmatrix{\alpha_1+\alpha_2+\alpha_3\\\alpha_2+\alpha_3\\\alpha_3}\;.$$
Unfortunately, this isn’t quite what we need: we want to start with a vector $v$ in standard coordinates and convert it to one in $\beta$ coordinates so that we can multiply by $AM$ and get $T(v)$ in standard coordinates. That requires changing base from standard to $\beta$; multiplying by $B$ goes in the opposite direction, from $\beta$ coordinates to standard ones. As you might expect, the matrix that does the change of basis in the other direction is $B^{-1}$, which I’ll let you compute for yourself. Once you have it, you can express $T$ in terms of a matrix multiplication that involves standard coordinates on both ends:
$$T(v)=AMB^{-1}v$$
gives $T(v)$ in standard $\Bbb R^2$ coordinates if $v$ is expressed in standard $\Bbb R^3$ coordinates. |
H: Quotient sets and tori
My trek through MacLane and Birkhoff's Algebra has brought me to exercise 1.9.3, which defines the equivalence relation $E$ in $\Bbb R^2$ by $(x, y) E (x', y')$ iff $x-x'$ and $y-y'$ are both integers and asks me to prove that the quotient set $\Bbb R^2/E$ may be described as the set of points on a torus. Sadly, I can't fathom what this question is trying to express, since the quotient set appears to me to be a set of like grids of points that you could connect to make a bunch of $1\times 1$ squares if you wanted. I mean, a torus is three-dimensional last time I checked, so I don't know how you'd go about saying a partition of $\Bbb R^2$ is the same as a subset of $\Bbb R^3$. If someone could help me out in solving this problem, that'd be great. Thanks!
AI: By torus they mean just the surface of the ‘doughnut’, which is indeed two-dimensional.
Note that the relation actually gives you just one $1\times 1$ square: the points $\langle 0.2,0.3\rangle,\langle 1.2,0.3\rangle$, $\langle 5.2,100.3\rangle,\langle-1.8,23.2\rangle$, etc. all get collapsed to a single point.
Added: Let $S=\{\langle x,y\rangle:0\le x,y<1\}$, the unit square in the first quadrant including the left and bottom edges but not the top or right edge. $S$ contains exactly one point from each $E$-equivalent class. You can think of $E$ as cutting up the plane into squares and melting them all together into the single square $S$. Suppose that you start at $\langle 0.5,0.5\rangle$ in $S$ and travel in the positive $x$-direction, through $\langle 0.6,0.5\rangle$, $\langle 0.7,0.5\rangle$, and so on. You approach the point $\langle 1,0.5\rangle$, which isn’t in $S$. But $$\langle 1,0.5\rangle\;E\;\langle 0,0.5\rangle\;,$$ so $\langle 1,0.5\rangle$ and $\langle 0,0.5\rangle$ are the same point of $\Bbb R^2/E$, and $\langle 0,0.5\rangle$ is in $S$. Thus, if you keep going to the right along the line $y=0.5$, you ‘disappear’ off the righthand edge of $S$ and immediately reappear at $\langle 0,0.5\rangle$ on the lefthand edge. (It’s rather like what happened to torpedoes in some of the early arcade games.) For all practical purposes the square has been rolled into a cylinder with the left and right edges glued together.
The same thing happens when you travel vertically: as you ‘disappear’ off the top edge, you pop back in at the bottom. In effect the vertical cylinder has been wrapped around so that its circular rims can be sewn together to make an inner tube $-$ i.e., a torus. |
H: Is it wrong to say $ \sqrt{x} \times \sqrt{x} =\pm x,\forall x \in \mathbb{R}$?
Is it wrong to say $$ \sqrt{x} \times \sqrt{x} =\sqrt{x^2}= \pm x$$
I am quite sure that $\sqrt{(x)^2} = \pm(x)$
But, does $\sqrt{x } \times \sqrt{x} =- (x)$ doesn't holds in $\mathbb{R}$ but if we assume $\mathbb{C}$ it holds right?
AI: Recall that a function returns a unique value. For this reason in the real numbers we have to choose positive or negative values for every number.
However it is vastly more useful to use positive numbers because then we can define composition in a meaningful way, e.g.: $$\sqrt{\sqrt{x}}=\sqrt[4]{x}$$
Had we chosen $\sqrt x<0$ then $\sqrt{\sqrt x}$ would not be a real number, whereas $\sqrt[4]x$ is still a real number (positive or negative).
These are good enough reasons to always require that when taking a root of even order we always choose the positive value, and therefore $\sqrt{x^2}=|x|$. |
H: Notation in Munkres' $\textit{Analysis on Manifolds}$
I am trying to understand Theorem 9.1 of 1991 copy of Munkres' Analysis on Manifolds. I have stated what I don't understand below; there is a heading in bold. This theorem is a precursor to the implicit function theorem and on my copy of the book is on page 73.
Now on page 72 he states the following definition:
Let $A$ be open in $\Bbb{R}^m$; let $f : A \rightarrow \Bbb{R}^n$ be differentiable. Let $f_1,\ldots,f_n$ be the component functions of $f$. We sometimes use the notation
$$Df = \frac{\partial(f_1,\ldots,f_n)}{\partial(x_1,\ldots,x_m)}$$
for the derivative of $f$. On occasion we shorten this to the notation $Df = \partial f /\partial \Bbb{x}$.
This is all good, so now on to theorem 9.1 (which is where my confusion lies).
Theorem 9.1: Let $A$ be open in $\Bbb{R}^{k+n}$; let $f : A \rightarrow \Bbb{R}^n $ be differentiable. Write $f$ in the form $f(\Bbb{x},\Bbb{y})$, for $\Bbb{x} \in \Bbb{R}^k$ and $\Bbb{y} \in \Bbb{R}^n$; then $Df$ has the form
$$Df = \Big[ \partial f/\partial \Bbb{x} \hspace{5mm} \partial f / \partial \Bbb{y}\Big].$$
Suppose there is a differentiable function $g : B \rightarrow \Bbb{R}^n$ defined on an open set $B$ in $\Bbb{R}^k$, such that
$$f(\Bbb{x},g(\Bbb{x})) = 0$$
for all $\Bbb{x} \in B$. Then for $\Bbb{x} \in B$,
$$ \frac{\partial f}{\partial \Bbb{x}}(\Bbb{x},g(\Bbb{x})) + \frac{\partial f}{\partial \Bbb{y}}(\Bbb{x},g(\Bbb{x}))\cdot Dg(\Bbb{x}) = 0.$$
The dot just before $Dg(\Bbb{x})$ means matrix multiplication.
Now the proof of this goes as follows, given $g$, we can define $h : B \rightarrow \Bbb{R}^{k+n}$ by the equation
$$h(\Bbb{x}) = (\Bbb{x},g(\Bbb{x})).$$
The hypotheses of the theorem then imply that the composite function $f(h(\Bbb{X})) = f(\Bbb{x},g(\Bbb{x}))$ is defined and equals zero for all $\Bbb{x} \in B$. The chain rule then implies that
$$\begin{eqnarray*} 0 &=& Df(h(\Bbb{x}))\cdot Dh(\Bbb{x})\\
&=& \Big[\frac{\partial f}{\partial \Bbb{x}}(h(\Bbb{x})) \hspace{4mm} \frac{\partial f}{\partial \Bbb{y}}(h(\Bbb{x})) \Big] \cdot \left[\begin{array}{c} I_k \\ Dg(\Bbb{x}) \end{array}\right]
\end{eqnarray*}.$$
What I don't understand: In the last row above, I get the second matrix on the right hand side, the one involving the identity matrix. However for the first matrix, I can see the notation means that it is formed by concatenating two matrices together, one from $\frac{\partial f}{\partial \Bbb{x}}(h(\Bbb{x}))$ and the other from $\frac{\partial f}{\partial \Bbb{y}}(h(\Bbb{x}))$. My problem now is I don't even no what these matrices look like.
I have tried several ways to interpret them, but keep getting tied up. Also, for the second matrix on the right it is of dimensions
$$(n + k) \times k$$
yes? But if this were so, then how can $Df(h(\Bbb{x}))$ be a map from $\Bbb{R}^{k+n}$ to $\Bbb{R}^n$?
Thanks.
AI: The map is $Df$, which takes a variable in $\Bbb{R}^{k+n}$ and returns a value in $\Bbb{R}^n$, that is to say, it is (because $Df$ is linear) a $n \times (k+n)$ matrix. The point $h(x)$ is the point at which the different partial derivative are evaluated at (when we turn the matrix entries into actual real numbers), it is not "the argument of the linear function".
Note the composition $f \circ h:\Bbb{R}^k \to \Bbb{R}^n$, so we would expect $D(f \circ h)$ to be a $n \times k$ matrix, which it is.
To make an analogy with the 1-dimensional case: when we calculate the slope of a function $f$ at a point $x=a$, we get that the slope is $f'(a)$. A 1-dimensional linear map is a function of the form:
$L(x) = cx$, for some real number $c$, that is a 1x1 matrix. "Which number" we put in the matrix depends on the point "$a$" where we are finding the slope, but $a$ is not what we take "$L$ of".
EDIT: specifically, the matrix $Df$ at the point $X = (x_1,\dots,x_k,\dots,x_{k+n})$ is given by:
$$\begin{bmatrix}\frac{\partial f_1}{\partial x_1}(X)&\dots&\frac{\partial f_1}{\partial x_{k+n}}(X)\\ \vdots&\ddots&\vdots \\\frac{\partial f_n}{\partial x_1}(X)&\dots&\frac{\partial f_n}{\partial x_{k+n}}(X) \end{bmatrix}$$
if we write $X = (x,y)$ for $x \in \Bbb{R}^k, y \in \Bbb{R}^n$, then at the point $h(x) = (x,g(x))$ $Df$ has the form:
$$\begin{bmatrix}\frac{\partial f_1}{\partial x_1}(h(x))&\dots&\frac{\partial f_1}{\partial x_{k+n}}(h(x))\\ \vdots&\ddots&\vdots \\\frac{\partial f_n}{\partial x_1}(h(x))&\dots&\frac{\partial f_n}{\partial x_{k+n}}(h(x)) \end{bmatrix}$$
which we can split into two "blocks":
$$\begin{bmatrix} \frac{\partial f}{\partial x}(h(x))& \frac{\partial f}{\partial y}(h(x)) \end{bmatrix}$$
where the first $k$ columns are the partials with respect to the first $k$ coordinates of $X$, and the last $n$ columns are the partials with respect to the last $n$ coordinates of $X$. |
H: Characterising argmax of uniform distributions
I was thinking about comparisons of uniform random variables of the type $U(0,T)$, when I began to wonder about the argmax. Consider a sequence of parameters $T_1\le T_2\le\ldots T_n$ and corresponding independent random variables $U_1 \sim U(0,T_1)$, $U_2 \sim U(0,T_2)$ and so on. Now define a random variable $M=\arg\max_{0\le i\le n} U_i$. Is it true that $P(M=k)\uparrow k\uparrow$?
I started with $P(M=k)=P(U_k\ge U_1, U_k\ge U_2, \ldots U_k\ge U_{k-1},U_k\ge U_{k+1},\ldots U_k\ge U_n)$.
Here I use: $P(U_i\ge U_j)=\frac{T_i}{2T_j}$ if $T_i\le T_j$ and $P(U_i\ge U_j)=1-\frac{T_j}{2T_i}$ if $T_i\ge T_j$.
Simplifying,
$$P(M=k)=\prod_{i=1}^{k-1}\left(1-\frac{T_i}{2T_k}\right)\times \prod_{j=k+1}^n\left(\frac{T_k}{2T_j}\right)$$
It is here that I am stuck. I tried to use $\frac{P(M=k)}{P(M=k+1)}$, but the expression seems to be unwieldy. Is there a way to see my conjecture to conclusion?
AI: Your formula for $\mathrm P(M=k)$ is wrong because the random variables $U_i$ for $i\ne k$ are not independent.
In fact, $[M=k]=[U_k\gt V_k]$ with $V_k=\max\limits_{i\ne k}U_i$ hence $\mathrm P(M=k)=\mathrm P(V_k\lt U_k)$. Since the random variables $U_k$ and $V_k$ are independent, integrating this identity with respect to the distribution of $U_k$ yields
$$
\mathrm P(M=k)=\frac1{T_k}\int_0^{T_k}\mathrm P(V_k\lt u)\mathrm du=\frac1{T_k}\sum\limits_{i=1}^k\int_{T_{i-1}}^{T_i}\mathrm P(V_k\lt u)\mathrm du,
$$
with the notation $T_0=0$. If $T_{i-1}\lt u\lt T_i$ with $i\leqslant k$, then $U_j\lt u$ with full probability for every $j\leqslant i-1$ hence $[V_k\lt u]=[\forall j\geqslant i,j\ne k,U_j\lt u]$ and
$$
\mathrm P(M=k)=\frac1{T_k}\sum\limits_{i=1}^k\int_{T_{i-1}}^{T_i}\prod\limits_{i\leqslant j\leqslant n,j\ne i}\frac{t}{T_j}\mathrm dt,
$$
that is,
$$
\mathrm P(M=k)=\sum\limits_{i=1}^k\frac{T_i^{n-i+1}-T_{i-1}^{n-i+1}}{(n-i+1)T_iT_{i+1}\cdots T_n}=\sum\limits_{i=1}^k\frac{q_i-q_{i-1}}{n-i+1},
$$
where, for every $0\leqslant i\leqslant n$,
$$
q_i=\frac{T_i^{n-i}}{T_{i+1}T_{i+2}\cdots T_n}.
$$
One sees that $\mathrm P(M=k)\leqslant\mathrm P(M=k+1)$ and that the inequality is strict as soon as $T_i\lt T_{i+1}$ since
$$
q_{i+1}=q_i\left(\frac{T_{i+1}}{T_i}\right)^{n-i}.
$$
A simple construction of $M$ which might help to explain why the result is true is as follows. Consider a random index $1\leqslant I\leqslant n$ whose distribution is characterized by the fact that, for every $1\leqslant i\leqslant n$,
$$
\mathrm P(I\leqslant i)=q_i,\qquad \mathrm P(I=i)=q_i-q_{i-1}.
$$
Then, conditionally on $I$, $M$ is uniformly distributed in the set $\{I,I+1,\ldots,n\}$.
This yields directly the explanation you are after. Choose $1\leqslant k\leqslant n$. Then:
If $i\leqslant k$, then $\mathrm P(M=k\mid I=i)=\mathrm P(M=k+1\mid I=i)$.
For $i=k+1$, $\mathrm P(M=k\mid I=k+1)=0$ while $\mathrm P(M=k+1\mid I=k+1)\gt0$.
If $i\gt k+1$, then $\mathrm P(M=k\mid I=i)=0=\mathrm P(M=k+1\mid I=i)$.
Summing these contributions yields $\mathrm P(M=k)\lt\mathrm P(M=k+1)$. This also yields once again the difference
$$
\mathrm P(M=k+1)-\mathrm P(M=k)=\mathrm P(M=I=k+1)=\frac{\mathrm P(I=k+1)}{n-k}=\frac{T_{k+1}^{n-k}-T_k^{n-k}}{(n-k)T_{k+1}\cdots T_n}.
$$ |
H: Graded Vector Spaces and the Interchange Law
I'm a little confused about how to correctly interchange factors in tensor
products on graded vector spaces.
In particular let $V:= \bigoplus_{n \in \mathbb{N}} V_n$ be a $\mathbb{N}$-graded
vector space and $\bigotimes^k V$ be the graded k-fold tensor product of $V$. Then one basic convention according to the interchange of factors in the tensor product
is
$v_1 \otimes v_2 = (-1)^{|v_1||v_2|}v_2 \otimes v_1$.
But this is ambiguous, isn't it?
For example take $v_1 \otimes v_2 \otimes v_3 \in \bigotimes^3 V$. Then on the one side we could first interchange $v_1$ and $v_2$ to get
$v_1 \otimes v_2 \otimes v_3 = (-1)^{|v_1||v_2|}v_2 \otimes v_1 \otimes v_3$
and then interchange $v_3$ and $v_1$ to get
$v_1 \otimes v_2 \otimes v_3 = (-1)^{|v_1||v_2|+|v_3||v_1|}v_2 \otimes v_3 \otimes v_1$
But on the other side we could first interchange $v_2$ and $v_3$ to get
$v_1 \otimes v_2 \otimes v_3 = (-1)^{|v_2||v_3|}v_1 \otimes v_3 \otimes v_2$
and then interchange $v_1$ and $v_2$ to get
$v_1 \otimes v_2 \otimes v_3 = (-1)^{|v_1||v_2|+|v_2||v_3|}v_2 \otimes v_3 \otimes v_1$
From this we get
$ (-1)^{|v_1||v_2|+|v_3||v_1|}v_2 \otimes v_3 \otimes v_1 = (-1)^{|v_1||v_2|+|v_2||v_3|}v_2 \otimes v_3 \otimes v_1$
and consequently
$(-1)^{|v_3||v_1|} = (-1)^{|v_2||v_3|}$ which is obviously not true in general.
What am I doing wrong here?
AI: The problem is on the line when you say "then interchange $v_1$ and $v_2$ to get"... This kind of manipulation, where you exchange two non-adjacent vectors, isn't covered by the sign rule you gave. What's in the middle matters. |
H: Alternative proof or typo?
In Atiyah-MacDonald, we have the claim that $S^{-1}(I+J) = S^{-1}I + S^{-1}J$ and similarly, $S^{-1}(IJ) = S^{-1}I S^{-1}J$. Here $I,J$ are ideals of a commutative unital ring $R$ and $S$ is a multiplicative subset of $R$.
The proof is omitted with the remark (p.42): "...For sums and products, this follows from (1.18);...."
Where (1.18) is the following exercise on page 10:
I showed $S^{-1}(I+J) = S^{-1}I + S^{-1}J$ and $S^{-1}(IJ) = S^{-1}I S^{-1}J$ showing the usual two inclusions $\supset, \subset$ and the proof is one line.
On the other hand, I don't see where exercise 1.18 comes in here. Could someone show me the proof using exercise 1.18? Or if that's not possible, confirm that this is a typo? Thanks.
AI: If $\iota: R \to S^{-1}R$ is the inclusion morphism, then $S^{-1}(I+J) = (I+J)^e$ for this morphism. |
H: MacNeille completion of a totally ordered set: Dedekind cuts
If $X$ is any partially ordered set with $A\!\subseteq\!X$ and $x\!\in\!X$, define $x\!\leq\!A :\Leftrightarrow \forall a\!\in\!A\!: x\!\leq\!a$ and $A\!\leq\!x :\Leftrightarrow \forall a\!\in\!A\!: a\!\leq\!x$, and denote $A^u\!:=\!\{x\!\in\!X; A\!\leq\!x\}$ and $A^l\!:=\!\{x\!\in\!X; x\!\leq\!A\}$ and $A^\downarrow\!:=\!\{x\!\in\!X; \exists a\!\in\!A\!:x\!\leq\!a\}$ and $A^\uparrow\!:=\!\{x\!\in\!X; \exists a\!\in\!A\!:a\!\leq\!x\}$. The Macneille completion of $X$ is $(\{A^{u\,l}; A\!\subseteq\!X\},\subseteq)$.
If $X$ is any totally ordered set, a cut in $X$ is any pair $(A,B)$ such that $X\!=\!A\!\cup\!B$ is a partition with $A\!=\!A^\downarrow$ and $B\!=\!B^\uparrow$ (hence $A\!\leq\!B$) and $A$ contains no greatest element. A cut $(A,B)$ is proper when $A\!\neq\!\emptyset$ and $B\!\neq\!\emptyset$. The completion by cuts of $X$ is $\{\text{proper cuts in }X\}\!\cup\!\{(\emptyset,X),(X,\emptyset)\}$, where $(A,B)\!\leq\!(A',B'):\Leftrightarrow A\!\subseteq\!A' \Leftrightarrow B\!\supseteq\!B'$.
Wikipedia suggests that the MacNeille completion is the generalization of the completion by cuts. If $X$ is totally ordered, how can I prove that $\{A^{u\,l}; A\!\subseteq\!X\}\:\cong\:\{\text{proper cuts in }X\}$?
I've tried the maps $A^{u\,l}\!\mapsto\!(X\!\setminus\!A^\uparrow,A^\uparrow)$ and $A^{u\,l}\!\mapsto\!(A^{u\,l},X\!\setminus\!A^{u\,l})$, but didn't come far.
AI: The Wikipedia discussion of the relationship between the MacNeille completion and Dedekind cuts makes it appear that you should try $A^{ul}\mapsto\langle A^{ul},A^u\rangle$; all that needs to be shown is that $\langle A^{ul},A^u\rangle$ actually is a cut. Suppose that $a\in A^{ul}\cap A^u$. Then $A\le a\le A^u$, so $a=\sup A$. There are two possibilities.
$a\notin A$: then $\langle A^{ul},A^u\rangle$ is clearly a cut.
$a\in A$: then $\langle A^{ul},A^u\rangle$ fails to be cut (a) because $A^{ul}\cap A^u=\{a\}\ne\varnothing$, and (b) $A^{ul}$ has a greatest element, $a$. Both of these problems are repaired by removing $a$ from $A^{ul}$.
Thus, the correspondence should actually be $A^{ul}\mapsto\langle A^{ul}\setminus A^u,A^u\rangle$.
Added: This gives the expected results for the Dedekind completion of the rationals, for instance. In the setting of arbitrary linear orders, however, one cannot require that the left set of a cut have no largest element. Suppose that $\langle X,\le\rangle$ is a linear order and that $x,y\in X$ are adjacent with $x<y$. The completion should not insert anything between $x$ and $y$, so the cuts corresponding to $x$ and $y$ should be adjacent. The construction given above makes $\big\langle (\leftarrow,x),[x,\to)\big\rangle$ the cut corresponding to $x$ and $\big\langle (\leftarrow,y),[y,\to)\big\rangle=\big\langle (\leftarrow,x],[y,\to)\big\rangle$ the cut corresponding to $y$, which are indeed adjacent. Moreover, they correspond exactly to the scheme for Dedekind cuts in the rationals that makes the cut $\big\langle (\leftarrow,q),[q,\to)\big\rangle$ correspond to the rational $q$.
The only alternative is the scheme that puts an element of $X$ into the left set of its own Dedekind cut; in the rationals this means that no right set has a smallest element, but in the case of my $X$ here it means that the cut corresponding to $x$ will be $\big\langle (\leftarrow,x],(x,\to)\big\rangle=\big\langle (\leftarrow,x],[y,\to)\big\rangle$, and again the halves of the cut will have adjacent endpoints. It simply can’t be avoided.
In particular, it’s possible that $A^{ul}\setminus A^u$ still has a largest element, but if it does, the original linearly ordered set had adjacent elements here. |
H: What is the trellis diagram for a linear block code?
For the convolutional codes there is so-called trellis diagram,
for which the definition is rather clear for me, however in mathematical sense is not.
I have heard that it can be defined for linear block codes also.
Linear codes have very simple mathematical meaning - just the linear subspaces
in $(F_q)^n$.
So I guess the "trellis" should also be something simple.
So what is the trellis for a linear block code ?
AI: This question is studied very extensively in Alex Vardy's chapter in Handbook of Coding Theory. I describe the idea below.
To get a trellis for a linear block code $C$ we need to define something like a time axis for the positions of the individual symbols of all the codewords. So basically an ordering of the symbols: this is the first symbol, this is the second,..., this is the last. Later on we may want to change the ordering, but let's fix one for now.
The trellis collects information about which beginnings of codewords can be matched with which endings in such a way that the end result will be a valid codeword. In the case of a linear code this translates into the following algebraic formulation.
Fix a point $i,0<i<n,$ on the time axis. Consider the so called initially trivial subspace $I_i(C)$ and finally trivial subspace $F_i(C)$ of $C$ defined by
$$
I_i(C)=\{\vec{c}=(c_1,c_2,\ldots,c_n)\in C\mid c_j=0\ \text{for all}\ j\le i\}.
$$
and
$$
F_i(C)=\{\vec{c}=(c_1,c_2,\ldots,c_n)\in C\mid c_j=0\ \text{for all}\ j> i\}.
$$
Let's extend these definitions by declaring that $F_0(C)=I_n(C)=\{0\}$ and
$F_n(C)=I_0(C)=C$.
The key observations are that:
The sum of subspaces $I_i(C)+F_i(C)$ is direct, i.e. they intersect trivially.
If $\vec{c}\in C$, $\vec{c}_I\in I_i(C)$, $\vec{c}_F\in F_i(C)$ are arbitrary,
then the vector $\vec{c}+\vec{c}_I$ is a legal codeword that agrees with $\vec{c}$ up to and including the component $c_i$, and $\vec{c}+\vec{c}_F$ is a legal codeword that agrees with $\vec{c}$ from that point on, i.e. at positions $>i$. Furthermore, all the codewords with these properties are gotten in this way, because if two codewords agree up to and including (resp. from this point on), then their difference is in the initially trivial (resp. finally trivial) subcode.
So we can partition the code $C$ into cosets of the subspace $I_i(C)+F_i(C)$. Next we select a sequence of split points $0=i_0<i_1<i_2<\cdots<i_k=n$. Note that we usually will not use all the points as split points (that is $k<n$). With this data at hand we build a trellis for $C$ as follows. At split point $i_t$ we have the state-space
$$V_t(C)=C/(I_{i_t}(C)\oplus F_{i_t}(C).$$ We observe $V_0(C)$ and $V_k(C)$ are both trivial 0-dimensional spaces. These will give us a unique initial and final state respectively.
The trellis diagram is a directed graph. It has a vertex for each ordered pair $(t,a)$, with
$0\le t\le k$ and $a\in V_t(C).$
For each interval $t,t+1$, we define the subspace $T_t=I_{i_{t+1}}(C)\oplus F_{i_t}(C)$.
We then define a set of edges connecting the vertex $(t,a)$ and the vertex $(t+1,a')$. These are in a bijective correspondence with cosets $\vec{c}+T_t$ such that
$\vec{c}\in a\cap a'$. The number of such edges is thus $|a\cap a'|/|T_t|.$
The edges of the graph are labelled, and the above edge is labelled with the
sequence of symbols $c_{i_t+1},c_{i_t+2},\ldots,c_{i_{t+1}}$. Observe that 1) if $a\cap a'=\emptyset$ there will be no edges from $(t,a)$ to $(t+1,a')$, 2) all the codewords
in the coset $\vec{c}+T_t$ share the same sequence as labels.
What this amounts to is that for each codeword $\vec{c}\in C$ there is a unique path
from $(0,V_0(C))$ to $(k,V_k(C))$ such that at each split point $i_t$ the path passes thru
the vertex $(t,\vec{c}+I_{i_t}(C)\oplus F_{i_t}(C))$, and that it follows the edge
associated with $\vec{c}+T_t$ between those two vertices.
We can read the components of the codeword $\vec{c}$ by concatenating the labels of the edges along the path. Furthermore, it is not to hard to show that all the paths through the trellis represent valid codewords.
We can then decode using the Viterbi algorithm taking advantage of partial codewords shared by many words – exactly as in the case of convolutional codes.
The catch is that the number of either states or edges may easily be prohibitively large. For example, if systematic encoding is used with a $k$-dimensional code, then there are no shared sequences of $k$ first symbols among any codewords (because all those symbols are payload, and can be freely chosen by the user). This means that at such early split points the initially trivial subspaces are all trivial, and consequently the state space may be very large (up to $k$-dimensional, possibly smaller).
We can try and play the game of permuting the symbol positions (using non-systematic encoding). This is kind of a "black art", and it is not guaranteed to reduce the size of the trellis significantly. Therefore this is not a magic tool allowing us to efficiently decode any linear code with the Viterbi algorithm.
An example
Sorry about the residual garbage on the background. The figure shows a trellis of the extended binary linear Hamming code of length 16, rank 11 and minimum distance 4. The split points are at $i_0=0,i_1=4,i_2=8,i_3=12$ and $i_4=16$. The code is defined by the parity check matrix
$$
H=\pmatrix{1111&1111&1111&1111\cr 1111&1111&0000&0000\cr 1111&0000&1111&0000\cr
1100&1100&1100&1100\cr 1010&1010&1010&1010\cr}.
$$
As in all trellises the time axis runs from left to right. The vertices associated with a given split point are stacked on top of each other. Here all the edges are actually double edges, i.e. two parallel edges going from one vertex to another, the labels of the two parallel edges are always bitwise complements of each other. The finally trivial subcodes are spanned by
$$
F_4=\langle 1111 0000 0000 0000\rangle,
$$
$$
F_8=F_4\oplus \langle 0000 1111 0000 0000, 0011 0011 0000 0000, 0101 0101 0000 0000\rangle,
$$
$$
F_{12}=F_8\oplus\langle 0000 0000 1111 0000, 1100 1010 0110 0000, 1010 0110 1100 0000\rangle,
$$
and have respective dimensions 1,4,7. The initially trivial subcodes are
$$
I_{12}=\langle 0000 0000 0000 1111\rangle,
$$
$$
I_8=I_{12}\oplus\langle 0000 0000 1111 0000, 0000 0000 1100 1100, 0000 0000 1010 1010\rangle,
$$
$$
I_4=I_8\oplus\langle 0000 1111 0000 0000, 0000 0110 0101 0011, 0000 0011 0110 0101\rangle,
$$
with complementary dimension such that $\dim I_i\oplus F_i=8$ for all $i\in\{4,8,12\}$.
The words of this code are of the form $(a|b|c|d|)$, where each letter stands for a vector in $\mathbb{F}_2^4$, i.e. a group of 4 bits. The constraints are: $a+b+c+d$ must be either $0000$ or $1111$. The parities of the weights of the groups are either all odd or all even. |
H: About the Wasserstein "metric"
I've just encountered the Wasserstein metric, and it doesn't seem obvious to me why this is in fact a metric on the space of measures of a given metric space $X$. Except for non-negativity and symmetry (which are obvious), I don't know how to proceed.
Do you guys have any advices or links to useful references?
Thanks in advance!
AI: So I assume that what puzzles you are $W_{p}(\mu,\mu)=0$ and the triangle-inequality, where $W_{p}$ denotes the $p$-Wasserstein metric.
Here's some preliminary information. I will denote $\Pi(\mu,\nu)$ the collection of all transference plans from $\mu$ and $\nu$, i.e. $\pi\in\Pi(\mu,\nu)$ iff $\mu$ is the first marginal of $\pi$ and $\nu$ is the second. This can also be expressed in form $\mu=(\mathrm{pr}_{1})_{\#}\pi$ and $\nu=(\mathrm{pr}_{2})_{\#}\pi$, where $\#$ denotes the push-forward. If $(X,d)$ is Polish then for every pair of probability measures $\mu,\nu$ there exists an optimal transference plan $\pi\in\Pi(\mu,\nu)$ so that $W_{p}(\mu,\nu)=\left(\int_{X\times X}d(x,y)^{p}\,d\pi(x,y)\right)^{\frac{1}{p}}$. The proof of this can be found in the book 'Topics in optimal transportation', Cedric Villani, 2003, and the key point consists of noting that $\Pi(\mu,\nu)$ is compact in the weak-convergence of measures (which is shown by using Prokhorov's theorem).
Now to the metric itself.
About the $W_{p}(\mu,\mu)=0$, take the homeomorphism $f:X\to\Delta$ given by $x\mapsto(x,x)$, i.e. $\Delta$ is the "diagonal" of $X\times X$. Then take $\nu:=f_{\#}\mu$ (which is a Borel probability measure on the diagonal $\Delta$) and furthermore define a Borel probability measure $\pi$ on the product space $X\times X$ by setting $\pi(A)=\nu(A\cap\Delta)$ for all Borel sets $A$. Now $\pi$ is a transference plan between $\mu$ to itself (not necessarily optimal!), which is a straight-forward proof, and it vanishes outside the diagonal (i.e. $\pi(\Delta^{c})=0$). Since the diagonal is the zero set of the metric $d$, we conclude that
\begin{equation*}
W_{p}(\mu,\mu)^{p}\leq \int_{X\times X}d(x,y)^{p}\,d\pi(x,y)=\int_{\Delta}d(x,y)^{p}\,d\pi(x,y)+\int_{\Delta^{c}}d(x,y)^{p}\,d\pi(x,y)=0+0=0,
\end{equation*}
whence $W_{p}(\mu,\mu)=0$.
The triangle-inequality uses a so called "Gluing lemma" (also found in Villani's book). It states that if $\mu_{1},\mu_{2},\mu_{3}$ are Borel probability measures on $X$, and $\pi_{1,2}\in\Pi(\mu_{1},\mu_{2})$ and $\pi_{2,3}\in\Pi(\mu_{2},\mu_{3})$ are optimal transference plans, then there exists a Borel probability measure $\mu$ on $X^{3}$ with marginals $\pi_{1,2}$ to the left $X\times X$ and $\pi_{2,3}$ to the right $X\times X$. This measure in a sense glues together $\pi_{1,2}$ and $\pi_{2,3}$. It follows by a simple argument using the marginal properties of each measure that the marginal of $\mu$ to $X\times X$ (the first and third $X$) denoted by $\pi_{1,3}$ is a transference plan in $\Pi(\mu_{1},\mu_{3})$ (not necessarily optimal!) $(*)$. Using minkovski inequality of $L^{p}(X^{3},\mu)$ $(**)$, marginal properties of the measures $(***)$, optimality of $\pi_{1,2}$ and $\pi_{2,3}$ $(****)$, we obtain
\begin{align*}
W_{p}(\mu_{1},\mu_{3}) &\overset{(*)}{\leq} \bigg(\int_{X\times X}d(x,z)^{p}\,d\pi_{1,3}(x,z)\bigg)^{\frac{1}{p}}\overset{(***)}{=}\bigg(\int_{X\times X\times X}d(x,z)^{p}\,d\mu(x,y,z)\bigg)^{\frac{1}{p}} \\
&\leq \bigg(\int_{X\times X\times X}(d(x,y)+d(y,z))^{p}\,d\mu(x,y,z)\bigg)^{\frac{1}{p}} \\
&\overset{(**)}{\leq}\bigg(\int_{X\times X\times X}d(x,y)^{p}\,d\mu(x,y,z)\bigg)^{\frac{1}{p}}+\bigg(\int_{X\times X\times X}d(y,z)^{p}\,d\mu(x,y,z)\bigg)^{\frac{1}{p}} \\
&\overset{(***)}{=}\bigg(\int_{X\times X}d(x,y)^{p}\,d\pi_{1,2}(x,y)\bigg)^{\frac{1}{p}}+\bigg(\int_{X \times X}d(y,z)^{p}\,d\pi_{2,3}(y,z)\bigg)^{\frac{1}{p}} \\
&\overset{(****)}{=}W_{p}(\mu_{1},\mu_{2})+W_{p}(\mu_{2},\mu_{3}).
\end{align*}
So we have the triangle-inequality. |
H: Four color theorem, 3-regular planar graph, Hamiltonian path and spiral chains
Studying the four color problem, I was analyzing all possible 3-regular planar graphs of 12 faces, with the additional restriction that graphs that have one or more faces with less than 5 edges, are not to be considered.
Note: It counts as a face also the surrounding area (infinite) of the graph
Using a Java program that builds all possible graphs of this kind, I saw that all have an Hamiltonian path, and that this path is very simple to compute using an algorithm I am implementing in Java: Cahit spiral chain algorithm.
The algorithm is:
Start from an external vertex
Only to choose the second vertex of the path, move on the external cycle clockwise
For all the other vertices that define the path, always move left (each vertex has three edges, one is the edge I am coming from, for the other two "left" and "right" are referred to the planar representation of the graph)
If moving left, you end up on a already visited vertex, move right
Here is the question:
Is this an obvious observations? Is there a basic theorem that implies the existence of an Hamiltonian path for graphs of this kind (3-regular planar graphs of 12 faces ...)?
AI: Euler's formula states that
$$V - E + F = 2$$
which holds for any finite, connected, planar graph with $V$ vertices, $E$ edges and $F$ faces. Since each vertex has degree $3$, the total degree is $3V$ and the number of edges is half of this, since each edge contributes $2$ to the total degree. So
$$V-\frac{3V}{2}+12 = 2 $$
and therefore:
$$ V = 20$$
The total degree is then $3V=60$ and the number of edges is half of this $E=30$. Since each face is made up of at least $5$ edges, each face is made of exactly $5$ edges. The $3$-regular graph with these properties is the Dodecahedron, which is well-known to be Hamiltonian. Indeed, this is the subject of Hamilton's own Icosian game. |
H: Prove that: $2^a+3^b<3a+4b$
Let be $a, b$ in $(0,1)$ such that $a+b>1$. I need to prove that:
$$2^a+3^b<3a+4b$$
I'm looking for an elementary proof that doesn't resort to the calculus tools.
AI: From the graph of the function $f(x)=2^x$ we see that on interval $(0,1)$ it is bellow the line $y=x+1$ joining the points $(0,f(0))=(0,1)$ and $(1,f(1))=(1,2)$.
Thus we have
$$2^a<1+a$$
for $a\in(0,1)$.
Using similar argument for $3^x$ we get
$$3^b<1+2b$$
for $b\in(0,1)$.
Adding the two inequalities together and using $1<a+b$ we obtain
$$2^a+3^b<2+a+2b<2(a+b)+a+2b=3a+4b.$$ |
H: A question regarding the algebraic closure of a field
I have slight problems understanding a thing about algebraic closures of fields. It seems to me that any algebraic closure $C$ of a field $K$ is a Galois extension, but I read that this is not true. Following are the definitions I use:
An extension $F/K$ is Galois if the fixed field of $Aut_K F$ is $K$.
Note that this definition is equivalent to the definition of F being simultaneously normal and separable over K but I want to use this one as it is this one that I have problems with.
Following is a theorem from my book:
Let $\sigma : K \rightarrow L$ be an isomorphism of fields. If $S$ is a set of polynomials in $K[x]$, $F$ a splitting field of $S$, and $S'$ a set of corresponding polynomials in $L[x]$, $P$ a splitting field of $S'$ then $\sigma$ extends to $F \cong P$.
All right. So my argument is as follows. Take any $u \in C \setminus K$. Then $u$ is algebraic with a nontrivial minimal polynomial, and at least another root $v \in C$. It is elementary knowledge that $K(u) \cong K(v)$. C is an algebraic closure, and therefore splitting field of all irreducible polynomials over $K(u)$ and $K(v)$. Then by the theorem, I should be able to extend the isomorphism to a K-automorphism of $C$ that switches $u,v$. These were chosen arbitrarily in $C$, hence there is a $K$-monomorphism affecting nontrivially every element of $C \setminus K$, hence by the definition, extension is Galois.
Where does this reasoning fail?
AI: The reasoning fails at the last step when you assume $u \neq v$. Consider $K = \mathbb{F}_p(u^p)$.
More generally, if $K$ is not perfect it admits inseparable extensions and the algebraic closure cannot be Galois. The maximal Galois extension is the separable closure. |
H: Puzzle: concatenation is three times product
The numbers A and B have three digits, while C is an odd number with 5 digits. Say you were asked to calculate the integer $(A B) /C$. Instead though you put A and B next to each
other to form a 6-digit number D, and then divided by C. Your answer is now three
times the correct answer. Find A, B and C.
AI: Right so we are trying to solve the equation:
$\frac{3AB}{C} = \frac{1000A + B}{C}$
We can rearrange and factorise this to give:
$(3A - 1)(3B - 1000) = 1000$
Now $3A - 1 \geq 299$ and is an integer factor of $1000$. The only possibilities for this number are $500$ and $1000$. It is easy to see that we must have the first of these, hence $A = \frac{501}{3} = 167$.
Then $3B - 1000 = 2$, giving $B = \frac{1002}{3} = 334$.
To find $C$ remember that it has $5$ digits and is odd. It also divides $AB$.
Now $AB = 55778 = 2\times 167^2$ so that the only possibility for $C$ is $167^2 = 27889$ |
H: Example of an application of a theorem about ideals in rings of fractions in Atiyah-MacDonald
In Atiyah-MacDonald, we have the following theorem (p. 41):
Proposition 3.11.
i) Every ideal in $S^{-1}R$ is an extended ideal.
ii) If $I$ is an ideal in $R$ then $I^{ec} = \bigcup_{s \in S} (I : \langle s \rangle )$. Hence $I^e = (1) = S^{-1}R$ if and only if $I$ meets $S$.
iii) $I = I^{ec}$ if and only if no element of $S$ is a zero-divisor in $R/I$.
iv) The prime ideals of $S^{-1}R$ are in one-to-one correspondence with the prime ideals of $R$ which don't meet $S$.
(I'm omitting point v) of the theorem since my question is about ii),iii) and iv). )
While I am able prove this theorem I'm wondering about how I'll be able to remember it, in particular, statements ii)-iv). Can anyone give me an example of where I'll be using either one of these three statements or all of them?
Thanks.
AI: (too long to be a comment) I don't have a copy of Atiyah-MacDonald in front of me, so I can't currently give exact references. And I wouldn't be surprised if (iii) was mentioned just to give a nice result of (ii) and to facilitate the proof of the (very useful) part (iv).
But when I read (ii)-(iv), the one that jumps out to me in utility is (iv). For example, in proving things about integral extensions, the going-up and going-down theorems, and some forms of Hilbert's Nullstellensatz (all of which are in AM), one uses (iv).
A sort of general pattern will appear. Some rings might be very complicated, so to discern something about their structure we might localize at some prime ideal or another. This is well-behaved, and localizations are often nicer than the original ring itself. This is the pattern used, if I recall correctly, in the proofs of the going-up and going-down theorems in AM.
It might be even better advice to sit down and go through the exercises. AM is known for having many nontrivial results in the exercises, and I'm confident that this proposition will be useful there. |
H: A wrong reasoning about conditional probability
Two of three prisoner A, B and C will be executed, A asks the name of one other than A himself who will be executed. Jailer says that it is B. Merely by asking the question, A reduced the probability that he will be executed from 2/3 to 1/2, regardless of which answer.
What is wrong with the reasoning?
I think the key is jailer answer the question by excluding the situation of A. So the answer just reduced the probability that C will die. A had better ask a question like if B will be executed but I want to ask the accurate reasoning by some formula.
AI: Just to have an answer, and to elaborate on Henning Makholm's comment above:
There is no reasoning given to suggest the probability changes to 1/2 that A will be executed, and evidently, it won't be.
After the jailer names B, there is one other name left on the execution list:
there is a 2/3 chance that it is A, and a 1/3 chance that it is C.
The general lesson to be drawn is that, just because there are two possibilities for
an event (in this case, A or C being the second name on the execution list), they won't necessarily occur with equal probability. |
H: Summation of a series.
I encountered this problem in Physics before i knew about a thing called Taylor Polynomials My problem was that i had to sum this series :
$$\sum^\infty_{n=1}\frac{(-1)^{n+1}}{n}$$
basically $$1,-\frac{1}{2},\frac{1}{3},-\frac{1}{4},\frac{1}{5},-\frac{1}{6},\frac{1}{7}.....$$
So now i know that there is something called a taylor polynomial that says that
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}....$$
So the above summation boils down to $\ln 2$.
What if i never knew the exansion then how would I calculate it?
Earlier I tried solving it like so ,
divide it into two different sets i.e.
$$\text{1 and $\dfrac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13} \ldots$ and $-\dfrac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\ldots$}$$
I said Hey! the first set would contain stuff like, $$\frac{1}{3^n},\frac{1}{5^n},\ldots$$ each of them would probably be reduced to a sum like so $$\sum^\infty_{n=1}\frac1{a^n}=\frac1{a-1}$$ and further become $$\sum^\infty_{a=3}\frac1{a-1}$$ which would subtract all the numbers in the other even set giving 1 as the answer which is wrong .
Where did I go wrong and how could I proceed even without knowing Taylor polynomials
AI: I find Norbert's solution more appealing if you run it backwards.
You're trying to evaluate $$1-{1\over2}+{1\over3}-{1\over4}+\cdots$$ Let $$f(x)=x-{1\over2}x^2+{1\over3}x^3-{1\over4}x^4+\cdots$$ Then we want $f(1)$. So, how can we find a simple formula for $f(x)$? Differentiate it: $$f'(x)=1-x+x^2-x^3+\cdots$$ Recognize this as a geometric series, first term $1$, constant ratio $-x$, so sum is $$f'(x)={1\over1+x}$$ Having differentiated, now antidifferentiate to get $$f(x)=\int{1\over1+x}\,dx=\log(1+x)+C$$ But what is $C$? Well, from the origial formula for $f$, we see $f(0)=0$, so that forces $C=0$, so $f(x)=\log(1+x)$, so $f(1)$, which is what we wanted, is $\log 2$. |
H: Every absolutely continuous function with integrable derivative tends to zero at infinity
I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:
$$\lim_{|x|\rightarrow \infty} f(x)= 0$$
Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?
Thanks in advance.
AI: Because $f'\in L^1$ you know that $\lim_{x\rightarrow \infty} f(x) $ exists (note $|\int_x^y f'| \le \int_x^y|f'| $ and $\int_x^\infty|f'|\rightarrow 0$ if $x\rightarrow\infty$). Since $f\in L^1$ the limit can only be zero, otherwise $\int|f|$ would not exist.
(Of course the reasoning is the same for $x\rightarrow-\infty$) |
H: Proving a language is regular
I know to prove a language is regular, drawing NFA/DFA that satisfies it is a decent way. But what to do in cases like
$$
L=\{ww \mid w \text{ belongs to } \{a,b\}*\}
$$
where we need to find it it is regular or not. Pumping lemma can be used for irregularity but how to justify in a case where it can be regular?
AI: Suppose the language was regular and had a DFA.
After reading, for example, "$\underbrace{aa\ldots a}_nb$" the DFA is in some state, and the identity of this state determines completly what the rest of an input that the machine accepts can be.
But if $n\ne m$, then the possible tails that can follow $\underbrace{aa\ldots a}_nb$ and $\underbrace{aa\ldots a}_m b$ are different sets. That means that the machine must be in different states after having read them.
Now, remember what the F in DFA stands for? |
H: Is there a relationship between the Compactness Theorem and the upward Lowenheim-Skolem Theorem in FOL?
Is there a relationship between the Compactness Theorem and the upward Lowenheim-Skolem Theorem in FOL?
I was thinking of another post of mine "Why accept the axiom of infinity?" when I though, "Well, if someone accepts arbitrary large finite numbers, what stops them from the jumping into the infinite?"
After all, doesn't compactness imply that if you have a theory $T$ with arbitrarily large finite models, then $T$ has arbitrarily large models—infinite or finite? This sounds a lot like the upward Lowenheim-Skolem theorem to me.
Is there a connection?
AI: Well, one might say everything is connected, so it depends on which kind of connections you're looking for.
There certainly is the connection that one can derive upward Löwenheim-Skolem from downward Löwenheim-Skolem and compactness:
Let $T$ be a theory with equality that admits some infinite normal model $M$. For any $\aleph_\beta \ge |T|$ we can construct a model of cardinality $\aleph_\beta$ in two steps. (1) Construct a new theory by adjoining $\aleph_\beta$ new constant letters with axioms that say they are all different. By compactness, the extended theory is consistent and so has a model $N$. Clearly this model must have cardinality at least $\aleph_\beta$, and in particular it is a model of $T$. (2) If $|N|>\aleph_\beta$, use downward Löwenheim-Skolem to cut it down to size as a model of $T$. |
H: does there exist an analytic function such that
If $f$ is analytic in a nbd $\Delta_{\delta}$ of $0$ and $f(z)=-f(-z)\forall z\in\Delta_{\delta} $ Then there exist an analytic function $g\in \Delta_{\delta}$ such that $f(z)=zg(z^2)\forall z\in \Delta_{\delta}$
AI: Expand $f(x)$ at its taylor series you get
$\sum_{i=0}^{\infty} a_i x^i=-\sum_{i=0}^{\infty} a_i (-x)^i=\sum_{i=0}^{\infty} a_i (-1)^{i+1}x^i \Rightarrow \sum_{i=0}^{\infty} a_{2i} x^{2i}=0$ the last equation from analytic continuity and the fact the $k(x)=0$ is analytics means that $a_{2i}=0$ for $i\in \mathbb{N}$ hence $f(x)=\sum_{i=0}^{\infty} a_{2i+1} x^{2i+1}= x\sum_{i=0}^{\infty} a_{2i+1} x^{2i}=xg(x^2)$ and $g(x)$ is analytics because it is a taylor series |
H: Fill in odd combinations in triangle
Can you write a number from 1 to 16 in each of the
triangles, using each number exactly once, such that the
sum of the two numbers in the two cells that share an
edge is always odd?
triangle as such:
http://4.bp.blogspot.com/_PnLYRqe0k9g/SnXtrcvf_nI/AAAAAAAAAKk/RRtqQiVRVqw/s320/Base+4+Triangle+units.png
AI: Took a look at the picture. It appears the answer is no. If the sum of the numbers in adjacent triangles is odd, then one triangle contains an odd number while the other contains an even. Light green triangles are adjacent only to dark green triangles and vice-versa. So if light green triangles contain odd numbers, the dark green triangles contain even numbers. There are equal numbers of odd and even numbers in the range 1 through 16, but there are different numbers of dark and light green triangles.
By the pigeonhole principle, one of the dark green triangles must contain a number of the wrong parity. Therefore, there will be an even sum somewhere. |
H: Why does the Fibonacci Series start with 0, 1?
The Fibonacci Series is based on the principle that the succeeding number is the sum of the previous two numbers. Then how is it logical to start with a 0? Shouldn't it start with 1 directly?
AI: One key number-theoretical reason for starting the sequence $(0,1)$ instead of $(1,1)$ is that it makes the divisibility property of the Fibonacci sequence more straightforward to state; i.e., that $F_k$ divides $F_{nk}$ for any $k,n$. If you start with $F_0=1$ instead of $F_0=0$ then this breaks down (for instance, in that numbering $F_2=2$ but $F_4=5$) and a lot of results have to be presented with indices shifted. This has to do, roughly with the representation of $F_n$ as $\frac{1}{\sqrt{5}}\left(\phi^n-\varphi^n\right)$ (with $\varphi=\frac{1}{\phi} = \phi-1$); the fact that the exponents 'match up' with the index leads to straightforward arguments for the various divisibility properties. |
H: The definition of $f(z)$ being analytic at point $\infty$
Consider this function $f(z) = \frac{1}{1+z}$. We can define $f(\infty) = \lim_{z \rightarrow \infty}{f(z)}$, which is zero for this case. Since $f(\frac{1}{t}) \rightarrow \frac{t}{t+1}$ is analytic at point $t=0$, we can deduce that $f(z)$ is analytic at $z=\infty$.
On the other hand, if we consider the type of singularity at $t=0$ of $f(\frac{1}{t})$. There is singularity because of the existence of $\frac{1}{t}$ but the singularity is removable. So the singularity at $z=\infty$ is removable.
The definition is $f(\infty) = \lim_{z \rightarrow \infty}{f(z)}$. So how we discuss the singularity and think of it as removable if the value of $f(\infty)$ is taken from the limit?
AI: As M Turgeon points out, the whole idea of a removable singularity is that we can "fix" the function at that point so it is equal to the limit, and is then analytic there.
The same idea applies, for example, to the function $f(z)=\dfrac{\sin(z)}{z}$. As it is defined, here, there is a singularity at $z=0$. However, it is removable, since $$\lim_{z\to 0}\frac{\sin z}{z}=1,$$ so we may simply extend the definition of $f$ so that $f(0)=1$, giving us an entire function.
Addendum: Robert Mastragostino brings up a good point, as well. Note that "removable singularity" doesn't (necessarily) mean that we are replacing a given function value with another--indeed, in the example I gave above, $f$ is initially undefined at $z=0$.
An isolated singularity of a function $f$ is a point $z=a$ such that for some $r>0$, we have that $f(z)$ is defined and analytic on $\{z:0<|z-a|<r\}$, but not on $\{z:|z-a|<r\}$. Such a singularity is removable iff there is a function $g$ that agrees with $f$ on $\mathrm{dom}(f)\smallsetminus\{a\}$--note that that doesn't indicate that $a\in\mathrm{dom}(f)$--such that $g(z)$ is defined and analytic on $\{z:|z-a|<r\}$. This may occur by replacing $f(a)$ with $\lim_{z\to a} f(z)$. This may also occur as in the example I gave above, not with replacement, but with extension.
If we are "removing" a singularity at $\infty$, we will generally be extending, rather than replacing a pre-existing function value. A notable exception is with linear fractional transformations $T(z)=\dfrac{az+b}{cz+d}$, where $ad-bc\neq 0$. In such a case, we tend to explicitly define $T(\infty)=\frac{a}{c}$ at the outset. |
H: Conditioning series with positive real numbers
I have to prove that if $\sum_{n=1}^{\infty} a_{n}$ is a convergent series with positive real numbers, then $\sum_{n=1}^{\infty} (a_{n})^\frac{n}{n+1}$ converges. I also wonder if the converse is true. Any suggestion, hint will
be very welcome. Thanks.
AI: if $a_n \geq \frac{1}{2^{n+1}} \Rightarrow a_n^{\frac{n}{n+1}} =\frac{a_n}{a_n^{\frac{1}{n+1}}} \leq 2a_n$ if $a_n \leq \frac{1}{2^{n+1}}$ then $a_n^{\frac{n}{n+1} } \leq \frac{1}{2^n}$. Therefore $a_n^{\frac{n}{n+1} } \leq 2a_n + \frac{1}{2^n}$ and by comparison test you are done. |
H: Creating Unique Values based off Two Sets of Sequential Integers
First off, I apologize if this is the wrong board. I'm a heavy StackOverflow user, and this is technically a programming question (or at least serves programming use), but I find it to be based moreso in math.
I have two sets of sequential integers that relate to each other, for the sake of example, defined as such:
Set A: [1..50]
Set B: [1..150]
I need to generate a third set from a combination of these two sets, with unique values. All integers from Set A must be programmatically paired with all integers from Set B to create Set C, which will contain unique integers. Jumping right into it, I thought "Well, just add the two to together!" I found quickly that such an idea wasn't even close to feasible.
Set A Set B Set C
----- ----- -----
1 1 2
1 2 3 *
1 3 4 *
2 1 3 *
2 2 4 *
2 3 5
*Non-unique Values
Multiplication didn't go so well either...
Set A Set B Set C
----- ----- -----
1 1 1
1 2 2 *
1 3 3
2 1 2 *
2 2 4
2 3 6
*Non-unique Values
So I'm looking for an operator or small formula to put between Set A and Set B to generate a unique, integral Set C.
EDIT
Both sets will have values added to them over time, so I need a solution that could handle an infinitely large Set A and Set B
AI: How about just $ c = 150a + b$? Or $c=150(a-1)+b$ if you want it to be $1$-based.
(From a coding point of view, that's how indexing into packed multidimensional arrays work).
Edit after question was updated: If both of your sets are effectively all of $\mathbb N$, you can use something like
$$ c = \frac{(a+b-1)(a+b-2)}{2} + a $$
which assumes the lowest possible value for $a$ and $b$ is $1$. If the value sets are $0$-based, use
$$ \tag{*} c = \frac{(a+b)(a+b+1)}{2} + a $$
If $a$ and/or $b$ can be negative, first map them to 0-based ones using, for example,
$$ x\mapsto\begin{cases} 2x & \text{if }x\ge 0 \\ -1-2x & \text{if } x < 0\end{cases} $$
and then use $(\text{*})$. |
H: Can we decide a conjecture is decidable without knowing a conjecture is correct or false?
Can we decide a conjecture is decidable without knowing a conjecture is correct or false?
I asked this question because I assume that the millenium prize problem is already to be decidable, otherwise the mathematician would need to consider infinity numbers of cases in order to get the money or the question's logic makes it a fake question.
AI: We need to be very clear and precise, so let me flood you with a bunch of definitions.
If we are talking about "decidable" and "undecidable", we are talking about formal systems.
In a formal system, we have primitive notions (undefined terms and relations/functions among them), axioms, rules of inference, and "grammatical rules" that tell us when a sequence of symbols is a well-formed formula.
A well-formed formula $F$ is provable in the formal system if there is a finite sequence of well-formed formulas such that each term of the sequence is either an axiom, or a consequence of previous formulas in the sequence that follows from the valid rules of inference, and such that the last term of the sequence is $F$.
We say a well-formed formula $F$ is decidable in the formal system if and only if either $F$ is provable, or its negation $\neg F$ is provable. It is undecidable otherwise.
A model of a formal system is a specific interpretation of the primitive terms that makes the axioms true.
A formal system is consistent if and only if it has at least one model.
It is a theorem of first order logic that a well-formed formula $F$ is provable in an axiomatic system if and only if $F$ is true in every model of the system. In particular, $F$ is undecidable if and only if there are models of the system in which $F$ is true, and there are models of the system in which $F$ is false.
Now, notice that while "decidable" and "undecidable" and formal notions: they refer to whether or not a particular formula can be formally deduced from the axiomatic system. By contrast, "true" and "false" are semantic notions: they depend on the meaning (the model) of the terms, and not only on their formal properties (which are given by the axioms and the rules of inference).
In a sense, showing that a formula is undecidable means showing that it is neither true nor false... rather, that it's truth or falsity will depend on the interpretation we give to the terms. A classical example is the Continuum Hypothesis: there does not exist an infinite subset of $\mathbb{R}$ that is not bijectable with either $\mathbb{N}$ or with $\mathbb{R}$. Turns out this is neither true nor false, it is undecidable (assuming Set Theory itself has models): there are models of set theory where it is true, and there are models where it is false. You have one theory in which you can assume it as an axiom, and another theory where you assume its negation as an axiom (in fact, infinitely many theories, one each for what different kinds of such sets there might be).
In a sense, if you already knew whether the conjecture is "true or false" (meaning, true in every model or false in every model), then you've already proven it, it is no longer a conjecture, and it cannot be undecidable. Or else you might be talking about "true in the standard model" (some theories have a 'standard' way of interpreting the terms, such as Peano Arithmetic), or "false in the standard model". In that case, knowing this only tells you that the statement is either provable or formally undecidable. (For example, Goedel proved that there is model of Set Theory in which the Axiom of Choice is true; that showed that it was impossible to disprove the Axiom of Choice, but it did not establish whether the Axiom of Choice was provable or whether it was undecidable; Paul Cohen later showed that there is a model where the Axiom of Choice is false, and that showed that the Axiom of choice was not provable, and so it must be formally undecidable).
For the Millenium Problems, some problems are known to be decidable; others may be undecidable. In some instances, undecidability would establish truth in the standard model. To see an example of this latter situation, think about the Goldbach Conjecture:
Goldbach Conjecture. Every even integer greater than $2$ can be written as a sum of two primes.
Suppose we could show that the Goldbach Conjecture is undecidable in Peano Arithmetic. This would show that it is true in the standard model. Why? Because if there were a counterexample to the Goldbach Conjecture, this could be established in finite time by simply going over all the even numbers and checking all sums of primes smaller than them; this would mean that its falsity in the standard model would be provable, so GC cannot be both undecidable and false-in-the-standard-model. Of course, there are other ways of interpreting the meaning of "number" ("non-standard models") in which GC would be false, but in the standard model it would be true.
The Riemann Hypothesis is one such question: if the Riemann Hypothesis were false, then this could be established by exhibiting a zero and verifying it is a zero (which can be done in finite time). So if the Riemann Hypothesis is formally undecidable, then it must be true in the standard model.
I have no idea what you are going on about with "infinity number of cases"... |
H: A question about Euclidean Domain
This is a problem from Aluffi's book, chapter V 2.17.
"Let $R$ be a Euclidean Domain that is not a field. Prove that there exists a nonzero, nonunit element $c$ in $R$ such that $\forall a \in R$, $\exists q$, $r \in R$ with $a = qc + r$, and either $r = 0$ or $r$ a unit."
Ok, I know that if $c\mid a$ then $r=0$, but if $c\nmid a$, not sure about what to do. I took the classic Euclidean Domain $\mathbb{Z}$ as example, and in $\mathbb{Z}$ I know that $c = 2$ ( also $-2$). Then I tried to generalize this.
I did $c = unit + unit$, but this didn't help and exercise 2.18 showed me that $c$ is not always $unit+unit$. I'm out of ideas, need some help.
Thanks.
AI: If $R$ is not a field, then it has nonunits. Consider the set $S=\{\varphi(a)\mid a\text{ is not a unit, and }a\neq 0\}$, where $\varphi$ is the Euclidean function.
It is a nonempty set of positive integers. By the Least Element Principle, it has a smallest element. Let $c\in R$ be a nonunit, nonzero, such that $\varphi(c)$ is the smallest element of $S$.
Edited. I claim that $c$ satisfies the conditions of the problem. Let $a\in R$. Then we can write $a = qc + r$, with either $r=0$ or $\varphi(r)\lt \varphi(c)$. If $r=0$, we are done. If $r\neq 0$, then $\varphi(r)\lt \varphi(c)$, then $\varphi(r)\notin S$, hence $r$ does not satisfy the condition
$r$ is not a unit, and $r\neq 0$.
Since $r\neq 0$, it follows that $r$ must be a unit.
Thus, for every $a\in R$, there exist $q,r\in R$ such that $a=qc+r$, and either $r=0$ or $r$ is a unit, as desired. |
H: What is the result of sum $\sum\limits_{i=0}^n 2^i$
Possible Duplicate:
the sum of powers of $2$ between $2^0$ and $2^n$
What is the result of
$$2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n\ ?$$
Is there a formula on this? and how to prove the formula?
(It is actually to compute the time complexity of a Fibonacci recursive method.)
AI: Let $S = 2^0 + 2^1 + 2^2 + \cdots + 2^{n}$.
Then $2S = 2^1 + 2^2 + 2^3 + \cdots + 2^{n} + 2^{n+1}$.
Then
$$\begin{align*}
S = 2S - S &= & & 2^1 &+& 2^2 & + & 2^3 & + & 2^4 &+&\cdots &+& 2^{n} &+& 2^{n+1}\\
&& -2^0 -& 2^1 & - & 2^2 & - & 2^3 & - & 2^4 & - & \cdots & - & 2^n
\end{align*}$$
How much is that? |
H: Find volume of region bound by $y=x, y=x^2$ around x-axis
Here is the problem in my textbook:
Find the volume of the solid obtained by rotating the region bounded by
the curves $y=x, y=x^2$ about x-axis.
Here is my solution :
Because equation $x = x^2$ has two roots : $0$ and $1$. we have:
$$ V= \int_0^1{2\pi x(x^2-x)}dx = \frac{\pi}{6}$$
But the solution in my textbook is $\frac{2\pi}{15}$. I think the hole in my solution is : I haven't use the region rotate around x-axis yet. But, I don't know how to use this statement in solution when couting volume.
Thanks :)
AI: Consider the diagram
For your problem, this is equivalent to rotating $R_2$ about the line $y=0$.
Since neither of the curves are touching the axis we are rotating about, we must use washers.
For washers, the volume is given by $$ V = \pi\int^b_a (R(x))^2-(r(x))^2 \ dx$$
Where $R(x)$ is the curve farthest away from our axis of rotation - the top function $y = x$ and $r(x)$ is the curve closest to the axis of rotation - the bottom function $y=x^2.$
As you found the intersecting points, we can set up our integral as
$$V = \pi\int^1_0 (x)^2-(x^2)^2 \ dx$$
$$V = \pi\int^1_0 x^2-x^4 \ dx$$
After integrating and evaluating, you should get $\frac{2\pi}{15}$. |
H: Determining eigenvalues and eigenvectors from a symmetric matrix
Let $A \in \mathbb{R}^{N \times N}$ be symmetric.
a) The respective Eigenvalue $\lambda$ to an approximately defined
Eigenvector $0 \neq x \mathbb \in {R}^n$ from $A$ has to be
calculated, so that $||Ax-\lambda x||^2_2$ is minimal. Specify a
formula to calculate $\lambda$.
This is an introductory class to numerical methods. Can I use here the same formula that I have always used? Considering $Ax=\lambda x$, solving the linear system $(A-\lambda I)x=0$ should give us the (approximated) Eigenvalue $\lambda$, right? Why wouldn't $||Ax-\lambda x||^2_2$ be $0$?
$A$ has furthermore the Eigenvalues $\lambda_1 \leq \lambda_2 \leq ...\leq \lambda_n$. Show that:
b) $\lambda_1 x^T x \leq x^T Ax \leq \lambda_n x^T x$
I think I'm missing something. Can I simply use $Ax=\lambda x$? Thus $x^T Ax$ is the same as $\lambda x^T x$, as $\lambda$ can be any arbitrary Eigenvalue between the smallest ($\lambda_1$) and the largest ($\lambda_n$), it follows that $\lambda_1 x^T x \leq \lambda x^T x \leq \lambda_n x^T x$ is valid.
c) $\lambda_1 = \min_{x \neq 0} \frac{x^T Ax}{x^T x}= \min_{x^T x = 1} x^T Ax$
I'm a little lost here. I would attempt to use the same trick again, thus $\lambda_1 = \min_{x^T x = 1} x^T \lambda x = \min_{x^T x = 1} \lambda x^T x = \min_{x^T x = 1} \lambda$. However I don't think is proving what they want?
d) $\lambda_n = \max_{x \neq 0} \frac{x^T Ax}{x^T x}= \max_{x^T x = 1} x^T Ax$
Analogue to c)
Many thanks in advance!
Edit: Thanks to Martin I know understand better what they want.
For b) and following the tip that I can diagonalize the Matrix, I now have
Let $D=Q^T AQ$. The eigenvalues of A are in the diagonal from $D$. With $y=Q^T x$ it follows that:
$x^T Ax=x^T Q D Q^T x = y^T D y = \sum \lambda_i y_i^2$
and therefore
$\lambda_1 x^T x=\lambda_1 y^T y \leq \sum \lambda_i y_i^2 \leq \lambda_n y^T y = \lambda_n x^T x$
I think I'm very near to being able to answer c) and d) but I'm still missing something. Any help? Thanks!
AI: The point is that solving the system $A-\lambda I=0$ is usually computationally intensive. It might be easy when $A$ is $3\times3$, but there are real life applications where your matrix can be huge. That's why there are lots of numerical methods to approximate eigenvalues, and that's what they are teaching you in your class.
In a), $A$ and $x$ are given, and they want you to find the $\lambda$ that minimizes the formula; with that $\lambda$, you can test whether it really looks like an eigenvalue or not (there are different methods for this, too).
So in a) you want to minimize
$$
||Ax-\lambda x||^2_2=\langle Ax-\lambda x,Ax-\lambda x\rangle=\langle Ax,Ax\rangle - 2\lambda\langle Ax,x\rangle + \lambda^2\langle x,x\rangle.
$$
This is a quadratic with positive degree-two coefficient, so its mininum occurs at "$-b/2a$", i.e. at
$$
\lambda=\frac{2\langle Ax,x\rangle}{2\langle x,x\rangle}=\frac{\langle Ax,x\rangle}{\langle x,x\rangle}.
$$
Edit: for c) and d), you have already proven that
$$
\lambda_1 x^Tx\leq x^TAx\leq \lambda_n x^Tx,
$$
so
$$
\lambda_1 \leq \frac{x^TAx}{x^Tx}\leq \lambda_n.
$$
You achieve the minimum by using an eigenvector for $\lambda_1$, and similarly for $\lambda_n$ and the maximum. This shows the first equality in both c) and d).
For the second equality, if $z^Tz=1$ then
$$
z^TAz=\frac{z^TAz}{z^Tz},
$$
so
$$
\lambda_1\leq z^TAz\leq\lambda_n.
$$
To achieve the equality, let $z$ be an eigenvector for $\lambda_1$; then put $x=z/(z^Tz)^{1/2}$. Then
$$
x^TAx=\frac{z^TAz}{z^Tz}=\lambda_1\,\frac{z^Tz}{z^Tz}=\lambda_1.
$$
And we proceed similarly for $\lambda_n$. |
H: Recurrence telescoping $T(n) = T(n-1) + 1/n$ and $T(n) = T(n-1) + \log n$
I am trying to solve the following recurrence relations using telescoping. How would I go about doing it?
$T(n) = T(n-1) + 1/n$
$T(n) = T(n-1) + \log n$
thanks
AI: In the first case you have $T(n) - T(n-1) = \frac{1}{n}$. If you sum with $n$ going from $2$ to $k$ you get:
$\begin{align*} \big(T(k) - T(k-1)\big) &+ \big(T(k-1) + T((k-2)\big)
+ \big(T(k-2) + T((k-3)\big) + \cdots + \\ &+ \big(T(4) - T(3)\big) +\big(T(3) - T(2)\big) + \big(T(2) - T(1)\big) = T(k) - T(1) = \sum_{m=2}^k \frac{1}{m}\end{align*}$
Notice how every term except $T(k)$ and $T(1)$ cancels out (I added "backwards" to make this more evident). This is the telescoping part. Then we get $T(n) = T(1) - 1 + H_n $, where $H_n$ is the $n$-th harmonic number $\displaystyle \sum_{m=1}^n \frac{1}{m}$ (I substracted $1$ because the first term $\displaystyle \frac{1}{1}$ doesn't appear in $\displaystyle \sum_{m=2}^n \frac{1}{m}$).
Likewise, in the second case you get: $\displaystyle T(n) = T(1) + \sum_{m=2}^n \log m = T(1) + \sum_{m=1}^n \log m$ (this last equality is given by $\log 1 = 0$; it doesn't make any difference but to me it looks nicer). Or you can use Cameron's hint: $T(n) = T(1) + \log n!$, by applying the product rule for logarithms.
The exact solution depends on the value of $T(1)$ in both cases. |
H: Simple question regarding continuous functions on $\mathbb{Q}$
I am a self-studying masochist and I came across an interesting example (to me), that I think will help me with further results. I have little experience writing rigorous proofs, so any explicit help is appreciated.
What I am trying to prove is the following:
Find a measurable indicator function that cannot be approximated by a sequence of continuous functions (demonstrate). For this I believe the "answer" is some form of $1_{\mathbb{Q}}$ on a finite interval. I believe this function is both measurable and not a pointwise limit of continuous functions.
How can I show that if I have a continuous function arbitrarily close to 1 on the rationals, it cannot also be arbitrarily close to 0 on the irrationals? I do know that continuous functions are uniformly continuous on finite intervals, but I cant seem to put all the pieces together.
Any detail would be greatly appreciated.
Thanks!
AI: A somewhat brute-force indirect proof:
Assume that we have a series of continuous functions $(f_n)$ that approximate $1_{\mathbb Q}$ pointwise. For a contradiction we will find an $x$ such that $f_n(x)$ does not converge. This $x$ will be constructed as a point in the intersection of an infinite sequence of closed intervals $I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots$, in the following way:
To begin with, let $I_0=[0,1]$ and $n_0=1$.
Now for each $i\ge 1$, choose a $y_i$ in $I_{i-1}$ such that it is irrational for even $i$ and rational for odd $i$. Since $\lim_n f_n(y_i) = 0$ (or $1$ as appropriate) by assumption, we can choose a large enough $n_i>n_{i-1}$ such that $f_{n_i}(y_i)$ is less than $0.1$ or more than $0.9$. By continuity of $f_{y_i}$ there is then a closed interval around $y_i$ where all values are less than $0.2$ or more than $0.8$. This interval becomes $I_i$; clearly it can be chosen small enough to lie within $I_{i-1}$.
Now, if we take $x$ to be in the intersection of all the intervals (for example it could be the supremum of the lower bounds), then by construction $f_{n_i}(x)$ alternates between being less than $0.2$ or more than $0.8$ according to the parity of $i$ (except for $i=0$). In particular $\lim_i f_{n_i}(x)$ cannot exist, so neither does $\lim_n f_n(x)$, which contradicts the assumption that the $f_n$'s have a pointwise limit.
Later note: Indirect proofs such as this can be difficult to follow, because they start by assuming something that turns out to be impossible, and everything that follows happens in a looking-glass world that doesn't actually exit. That makes it hard to get an intuitive picture of what is going on, because it really isn't going on after all.
Many people, including me, often find it easier to devise an indirect proof than a direct one, but since direct proofs are generally easier to read, in a more polished writeup one ought to try to rewrite he proof such that most of it works in a direct manner and the scope the assume-for-a-contradiction becomes as small as possible. After analyzing which parts of the contradicting assumptions the argument actually uses, we might end up with something like:
Lemma. Let $A$ and $B$ be subsets of $\mathbb R$, and assume that $(f_n)$ is a sequence of continuous functions that converges pointwise to $0$ on $A$ and to $1$ on $B$. If there is an interval $I$ in which $A$ and $B$ are both dense, then $I$ contains a point that is neither in $A$ nor $B$.
Proof. (Almost exactly as above, but with $A$ and $B$ for "irrational" and "rational")
Corollary. There cannot be a sequence of continuous functions that converges pointwise to $1_{\mathbb Q}$. Namely, assume that such as sequence exists. If we set $A=\mathbb R\setminus \mathbb Q$ and $B=\mathbb Q$, then the Lemma concludes that there is a number in $[0,1]$ that is neither rational nor irrational, which is absurd.
Of course, that is only an improvement if there exists some way for the assumptions of the Lemma to be true; otherwise we would still be in looking-glass land. But luckily there is; for example we could let $A$ be the set of dyadic rationals, $B=\mathbb Q\setminus A$, and let $f_n$ have the "correct" value on all fractions with denominators less than $n$ and interpolate linearly between them. (Note that this construction is not necessary for the proof to work; it just provides an example that one can keep in mind while following the steps to the proof). |
H: How to calculate this Frechet derivative
Suppose $F:C^1(\Omega, [0,T]) \to C^1(\Omega, [0,T])$ with $$F(u) = u_t - f(x, t, u, u_x).$$
How do I calculate the Frechet derivative of $F$ at the point $w = f(x,t, 0, 0)t$?
It should be $$F'(w, v) = v_t - \frac{\partial f}{\partial z}\bigg|_{w}v - \frac{\partial f}{\partial p}\bigg|_{w}v_x$$
apparently.
Maybe another day I can do this but forming the difference and then considering another difference to get out the partial derivative is confusing me!
Thanks
AI: If you just want the result and already do know that the functional is Frechet differentiable (or just don't care, which is not a good attitude ;-) you only have to calculate the directional derivative
$$\frac{d}{ds}|_{s=0}F(w+sv) = \frac{d}{ds}|_{s=0}\left\{(w+sv)_t - f(x,t,(w+sv), (w+sv)_x)\right\} = v_t - f_zv -f_p v_x$$
(assuming $f=f(x,t,z,p)$). This reduces the question to a one dimensional differentiation task. |
H: Proving combinator identity KMN=M
Have a problem proving K MN=M
By the K combinator definition
$ (\lambda x y.x) M N $
Parenthesized
$ ((\lambda x. (\lambda y.x)) M) N $
By the principal axiom of lambda calculus
$ (\lambda y.M) N $
Second application of the principal axiom
$ M[y:= N] $ ?
This would give correct result if M is y-free. Apparently there is an error at one of the steps.
AI: When you reduce $(\lambda x.\lambda y.x)M$, the result is $(\lambda y.x)[x:= M]$. This substitution is only defined if$y$ is not free in $M$. Otherwise the $\lambda y$ will need to be $\alpha$-renamed before you can proceed.
Some texts define the substitution function to do this $\alpha$-renaming implicitly, others consider the result of the substitution to be undefined if the variable capture condition is not met. |
H: Pointwise convergence of $1_{(a,b]}$ by a sequence of functions
I am trying to work through an exercise (on my own) out of Resnick's A Probability Path.
One question states the following:
Suppose $-\infty<a\le b<\infty$ and assume we have an indicator function of the form $1_{(a,b]}(x)$. Can this function be approximated by bounded and continuous functions.
Said another way, does there exist a sequence $f_n: 0\le f_n\le 1$ such that $f_n\rightarrow1_{(a,b]}$ pointwise.
Using the hint in the text, my attempted solution was:
Attempt at a solution:
$f_n$ such that:
$f_n(x) = \begin{cases} 0, & x \leq a, x \geq b + \frac1n \\
1, & a + \frac1n < x \leq b
\end{cases}$
and linear otherwise
The problem (or one problem?) is that this doesnt appear continuous to me. I could also use some help in learning how to be more explicit about limits. For example, why does $x>b$ imply $f_n\rightarrow 0$. I "know" this intuitively, but obviously need help explaining it with precision.
Thanks as always!
AI: You are looking at functions which are constant ($0$ or $1$) everywhere with the exception of the intervals $(a, a+1/n)$ and $(b, b+1/n)$. In these intervals you have linear functions with increasing slope which grow from $0$ to $1$ or the other way round.
To prove pointwise (!) convergence just look at:
i) $x\le a$. There your sequence is constantly zero, so it converges to $0$
ii) any $x \in (a, b]$. Choose $N$ such that $a+1/n < x$ for $n\ge N$ . Then $f_n(x) = 1$ for each such $n$, hence $f_n(x)\rightarrow 1$
iii) $x>b$. Choose $N$ such that $x\le b+1/n$ for $n\ge N$. Then $f_n(x)=0$ hence $f_n(x)\rightarrow 0$.
Then you are done. |
H: Inverse Laplace transform with partial fraction
I have the transform below:
$$\frac{(7s+2)(2s-5)}{{s^2}(s-2)}$$
I think this is should be partial fraction to be solved. Can you please help me figure how to consider A, B, C and denominators?
AI: Before doing partial function expansion, we need to make sure that the numerator has a degree smaller than the denominator. If not, we do long division in order to write the fraction as a polynomial and a fraction satisfies the degree condition.
Next, we factor the denominator. For the fraction in the question, both steps are unnecessary as the fraction already satisfies the degree condition and the denominator is already factored.
Now, we consider each term in the denominator:
If it's of the form $(s-a)$, expect $\frac{A}{s-a}$ in the partial fraction expansion.
If it's of the form $(s-a)^n$, expect $\frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + \cdots + \frac{A_n}{(s-a)^n}$.
$(s)$ can be treated as $(s - 0)$.
There are other cases not needed here. See this answer for more information on them.
Given the above, we can write the fraction in the question as:
$$
\frac{(7s+2)(2s-5)}{{s^2}(s-2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s-2}
$$
What's left is plugging in values for $s$ to find $A$, $B$ and $C$. There is an quick way to do it.
Multiply both sides by $(s-2)$ and plug in $s=2$ to get:
$$
\frac{(14 + 2)(4 - 5)}{4} = 0 + 0 + C \Rightarrow C = -4
$$
Multiply both sides by $s^2$ and plug in $s=0$ to get:
$$
\frac{(0+2)(0-5)}{-2} = 0 + B + 0 \Rightarrow B = 5
$$
Plug in $s=1$:
$$
\frac{(7+2)(2-5)}{1\cdot(1-2)} = A + 5 + \frac{-4}{-1} \Rightarrow A = 18
$$
And the fraction becomes:
$$
\frac{(7s+2)(2s-5)}{{s^2}(s-2)} = \frac{18}{s} + \frac{5}{s^2} - \frac{4}{s-2}
$$ |
H: Integral of $\int 2\,\sin^{2}{x}\cos{x}\,dx$
I am asked as a part of a question to integrate $$\int 2\,\sin^{2}{x}\cos{x}\,dx$$
I managed to integrate it using integration by inspection:
$$\begin{align}\text{let } y&=\sin^3 x\\
\frac{dy}{dx}&=3\,\sin^2{x}\cos{x}\\
\text{so }\int 2\,\sin^{2}{x}\cos{x}\,dx&=\frac{2}{3}\sin^3x+c\end{align}$$
However, looking at my notebook the teacher did this:
$$\int -\left(\frac{\cos{3x}-\cos{x}}{2}\right)$$
And arrived to this result:
$$-\frac{1}{6}\sin{3x}+\frac{1}{2}\sin{x}+c$$
I'm pretty sure my answer is correct as well, but I'm curious to find out what how did do rewrite the question in a form we can integrate.
AI: $$2\sin^2(x)\cos(x) = (2\sin(x)\cos(x))\sin(x) = \sin(2x)\sin(x) = \frac{\cos(x) - \cos(3x)}{2}$$
The last step comes from :
$$\cos(A - B) - \cos(A + B) = 2\sin(A)\sin(B)$$ |
H: upper bounds for binomial
I'm trying to calculate the upper bound of the binomial coefficient:
\begin{equation}
\sum\limits_{j=0}^{k} {n\choose j}<\left( \frac{ne}{k} \right)^k
\end{equation}
Using binomial theorem and for $x\ge0$:
\begin{equation}
\sum\limits_{j=0}^{k} {n\choose j}{x^j}\le(1+x)^n
\end{equation}
dividing both sides by $x^k$,we obtain:
$$
\sum\limits_{j=0}^{k} {n\choose j}{\frac{1}{x^{k-j}}}\le\frac{(1+x)^n}{x^k}
$$
For x<1 the term
$\frac{(1+x)^n}{x^k}$
obtain his minimum value at point $x=\frac{k}{n-k}$
I don't understand why... Could you please help. I thank you in advance.
AI: Consider $f(x) = \frac{(1+x)^n}{x^k}$ and apply your favorite calculus-based test for local extrema. |
H: Evaluating: $\lim_{n\to\infty} \int_{0}^{\pi} e^x\cos(nx)\space dx$
Evaluate the limit:
$$\lim_{n\to\infty} \int_{0}^{\pi} e^x\cos(nx)\space dx$$
W|A tells that the limit is $0$, but i'm not sure why is that result or if this is the correct result.
AI: Hint: Integrate by parts, letting $u=e^x$ and $dv=\cos nx \,dx$.
To get an explicit antiderivative, you will have to do two cycles of integration by parts. However, for the limit calculation, one cycle will do, and is in a sense more informative.
Added: We get $du=e^x\,dx$ and can take $v=\frac{1}{n}\sin nx$. Since $uv$ vanishes at both ends, we find that
$$\int_0^\pi e^x \cos nx \,dx=-\frac{1}{n}\int_0^\pi e^x\sin nx \,dx.$$
But $|e^x\sin nx|\le e^x$. |
H: Determining if a language is Recursively Enumerable
Here is a problem from John Hopcroft's "Introduction to Automata Theory" that I'm having a hard time trying to understand.
Exercise 9.2.5:
Let L be recursively enumerable and let Overscript[L, _] be non-RE. Consider the > language:
L' = {0w | w is in L} $\cup$ { 1w | w is not in L }
Can you say for certain whether L' or its complement are recursive, RE, or non-> RE? Justify your answer:
Here is the oficial answer:
"Note that the new language defined in the displayed text should be L'; it is > > different from the given language L, of course. Also, we'll use -L for the > > complement of L in what follows.
Suppose L' were RE. Then we could design a TM M for -L as follows. Given input w, M changes its input to 1w and simulates the hypothetical TM for L'. If that TM > accepts, then w is in -L, so M should accept. If the TM for L' never accepts, > then neither does M. Thus, M would accept exactly -L, which contradicts the fact > that -L is not RE. We conclude that L' is not RE."
and here is what I don't get:
..."Thus, M would accept exactly -L, which contradicts the fact that -L is not RE"....
Why? M is the Turing Machine for -L, so it is supposed to accept -L, right?
If we can construct a Turing Machine for a language, then it is RE. We have constructed a TM for something we know is not RE (-L), based on the TM for the supposedly RE L'. So what? how does this leads to the conclusion that L' is not RE? I'm very confused...
Any help will be much appreciated
Thanks!
AI: The hypothesis of the exercise is that $L$ is RE and $-L$ is not RE. The argument is that if $L'$ were RE, then $-L$ would also be RE, contradicting this hypothesis. |
H: Find Zariski closure of a set
Let $X=\{(x,\sin(x)): x \in \mathbb{A}^{2}\}$. I want to find the closure (with respect Zariski topology) of $X \subseteq \mathbb{A}^{2}$.
OK I've already shown that $X$ is not a closed set. Now consider $cl(X)$ this is a closed subset of $\mathbb{A}^{2}$ so its dimension is $0,1$ or $2$, it is not $0$ because it is not a point. So either $cl(X)$ has dimension $1$ or $2$. I suspect the answer is $2$. If the dimension is $1$ then $X=V(f)$ for some $f \in k[x,y]$ with $f$ irreducible. This implies then that $f(a,\sin(a))=0$ for every $a \in \mathbb{A}^{1}$.
Question: does this implies that $f$ is the zero polynomial?
From this it would follow that the dimension is $2$ so $cl(X)=\mathbb{A}^{2}$.
AI: The Zariski closure is the whole set. Suppose $f(x,\sin(x))=0$ for all $x$. Define $g_y(x)=f(x,y)$ for $y\in [-1,1]$ and note that $g_y$ has infinitely many zeroes, thus must be the zero polynomial. Thus $f$ vanishes on the strip $\mathbb R\times [-1,1]$, so must be the zero polynomial by any number of properties (for example, the fact that polynomials are analytic).
Edit: To clarify, since polynomials are analytic, it suffices to show that $f$ is $0$ on an open set (in the usual metric topology, NOT the Zariski topology) in order to conclude that $f$ is $0$ everywhere. Otherwise, we have some $n$ such that the first $n-1$ derivatives vanish on the set but the $n^{th}$ does not, so is nonzero and of constant sign on some open set, so integrating gives us $f$ is nonzero somewhere on this set. Since $\mathbb R\times [-1,1]$ contains the unit open ball around $(0,0)$, we are done. |
H: using PDE existence to show a map is invertible
I have an existence/uniqueness theorem for the PDE $$u_t = a(x,t)u_{xx} + b(x,t)u_x + c(x,t)u - g(x,t).$$
Now if I have a Gateaux derivative of a map $F$ at a point $p$ satisfying $$DF(p)v = v_t - f_1v_{xx} - f_2v_x - f_3v$$
(the $f_i$ are functions of $(x,t)$) then how can I use my PDE result to say that $DF(p)$ is invertible? I thought I could rearrange to get
$$v_t - f_1v_{xx} - f_2v_x - f_3v - DF(p)v = 0$$
and I want to put the $f_3$ and $DF(p)$ together so that it is in the form of the PDE and I can just quote the existence result which in this cases would tell me that there is a unique $v$ satisfying this equation and hence $DF(p)$ is invertible. But all I know is the the derivative is linear and I can't put the $f_3$ and the $DF(p)$ together.
Thank you
AI: I think that what you've given in your question is just about there: here's an attempt to tidy up. This also needs to be fleshed out in a way that depends on the spaces $V, W$ and on the details of your existence/uniqueness result.
We have $F:V\to W$. Then $DF(p)$ is invertible if it is a bijection from $V$ to $W$. So you need to show two things:
For every $w\in W$, there exists $v\in V$ such that $DF(p)(v)=w$.
If $DF(p)(v_1)=DF(p)(v_2)\in W$, then $v_1=v_2\in V$.
But
$$ DF(p)(v) = v_t-f_1v_{xx}-f_2v_x-f_3v,$$
and (I guess...) your PDE existence and uniqueness result tells you that for every $w=g\in W$ there exists a unique $v\in V$ such that
$$ v_t-f_1v_{xx}-f_2v_x-f_3v = w.$$
Existence gives part 1 above, and uniqueness gives part 2. (The key is that $w$ plays the role of the inhomogeneity $g(x,t)$.) |
H: Expected number of steps/probability in a Markov Chain?
Can anyone give an example of a Markov Chain and how to calculate the expected number of steps to reach a particular state? Or the probability of reaching a particular state after T transitions?
I ask because they seem like powerful concepts to know but I am having a hard time finding good information online that is easy to understand.
AI: The simplest examples come from stochastic matrices. Consider a finite set of possible states. Say that the probability of transitioning from state $i$ to state $j$ is $p_{ij}$. For fixed $i$, these probabilities need to add to $1$, so
$$\sum_j p_{ij} = 1$$
for all $i$. So the matrix $P$ whose entries are $p_{ij}$ needs to be right stochastic, which means that $P$ has non-negative entries and $P 1 = 1$ where $1$ is the vector all of whose entries are $1$.
By considering all the possible ways to transition between two states, you can prove by induction that the probability of transitioning from state $i$ to state $j$ after $n$ transitions is given by $(P^n)_{ij}$. So the problem of computing these probabilities reduces to the problem of computing powers of a matrix. If $P$ is diagonalizable, then this problem in turn reduces to the problem of computing its eigenvalues and eigenvectors.
Computing the expected time to get from state $i$ to state $j$ is a little complicated to explain in general. It will be easier to explain in examples.
Example. Let $0 \le p \le 1$ and let $P$ be the matrix
$$\left[ \begin{array}{cc} 1-p & p \\\ p & 1-p \end{array} \right].$$
Thus there are two states. The probability of changing states is $p$ and the probability of not changing states is $1-p$. $P$ has two eigenvectors:
$$P \left[ \begin{array}{c} 1 \\\ 1 \end{array} \right] = \left[ \begin{array}{c} 1 \\\ 1 \end{array} \right], P \left[ \begin{array}{c} 1 \\\ -1 \end{array} \right] = (1 - 2p) \left[ \begin{array}{c} 1 \\\ -1 \end{array} \right].$$
It follows that
$$P^n \left[ \begin{array}{c} 1 \\\ 1 \end{array} \right] = \left[ \begin{array}{c} 1 \\\ 1 \end{array} \right], P^n \left[ \begin{array}{c} 1 \\\ -1 \end{array} \right] = (1 - 2p)^n \left[ \begin{array}{c} 1 \\\ -1 \end{array} \right]$$
and transforming back to the original basis we find that
$$P^n = \left[ \begin{array}{cc} \frac{1 + (1 - 2p)^n}{2} & \frac{1 - (1 - 2p)^n}{2} \\\ \frac{1 - (1 - 2p)^n}{2} & \frac{1 + (1 - 2p)^n}{2} \end{array} \right].$$
Thus the probability of changing states after $n$ transitions is $\frac{1 - (1 - 2p)^n}{2}$ and the probability of remaining in the same state after $n$ transitions is $\frac{1 + (1 - 2p)^n}{2}$.
The expected number of transitions needed to change states is given by
$$\sum_{n \ge 1} n q_n$$
where $q_n$ is the probability of changing states after $n$ transitions. This requires that we do not change states for $n-1$ transitions and then change states, so
$$q_n = p (1 - p)^{n-1}.$$
Thus we want to compute the sum
$$\sum_{n \ge 1} np (1 - p)^{n-1}.$$
Verify however you want the identity
$$\frac{1}{(1 - z)^2} = 1 + 2z + 3z^2 + ... = \sum_{n \ge 1} nz^{n-1}.$$
This shows that the expected value is
$$\frac{p}{(1 - (1 - p))^2} = \frac{1}{p}.$$
An alternative approach is to use linearity of expectation. To compute the expected time $\mathbb{E}$ to changing states, we observe that with probability $p$ we change states (so we can stop) and with probability $1-p$ we don't (so we have to start all over and add an extra count to the number of transitions). This gives
$$\mathbb{E} = p + (1 - p) (\mathbb{E} + 1).$$
This gives $\mathbb{E} = \frac{1}{p}$ as above. |
H: convergence of $\alpha$-Hölder-continuous functions
Let $\Omega\subset\mathbb R^n$ be compact and $C^{0,\alpha}(\Omega)$ the space of all $\alpha$-Hölder-continuous functions. Define $||u||_{C^{0,\alpha}(\Omega)}:=||u||_{\sup}+\sup\limits_{{x,y\in \Omega\space\&\space x\ne y}}\frac{|u(x)-u(y)|}{|x-y|^\alpha}$ and consider $(C^{0,\alpha}(\Omega),||u||_{C^{0,\alpha}(\Omega)})$ and $\alpha\in]0,1]$ .
How can you prove that for any sequence in bounded closed set of $(C^{0,\alpha}(\Omega),||u||_{C^{0,\alpha}(\Omega)})$ there exists a convergent subsequence (concerning the uniform norm) and it limes is in $(C^{0,\alpha}(\Omega))$?
AI: Hints:
1) Using estimations of norms prove that if set $F$ is bounded in $(C^{0,\alpha}(\Omega),\Vert\cdot\Vert_{C^{0,\alpha}(\Omega)})$ then it is bounded in $(C^{0}(\Omega),\Vert\cdot\Vert_{C^{0}(\Omega)})$
2) If $F$ is bounded in $(C^{0,\alpha}(\Omega),\Vert\cdot\Vert_{C^{0,\alpha}(\Omega)})$ then $$
\exists C>0\quad\forall u\in F\quad \forall x,y\in\Omega\quad |u(x)-u(y)|\leq C|x-y|^\alpha
$$
3) Prove that 2) implies equicontinuity
4) From 1) and 4) you see that $F$ is relatively compact in $(C^{0}(\Omega),\Vert\cdot\Vert_{C^{0}(\Omega)})$.
5) If you get to this paragraph it is remains to prove uniform convergence. Using estimations of norms prove that if sequence $\{u_n:n\in\mathbb{N}\}$ converges in $(C^{0,\alpha}(\Omega),\Vert\cdot\Vert_{C^{0,\alpha}(\Omega)})$, then it converges in $(C^{0}(\Omega),\Vert\cdot\Vert_{C^{0}(\Omega)})$.
6) From statement of paragraph 2) prove that the limit function is $\alpha$-Hölder-continuous. |
H: Leslie Matrix characteristic polynomial
I´m having problems to prove the Leslie matrix characteristic polynomial.
I have to prove that the characteristic polynomial is:
$$
\ λ^{n}-a_{1}λ^{n-1}-a_{2}b_{1}λ^{n-2}-a_{3}b_{1} b_{2}λ^{n-3} - ... -
a_{n}b_{1} b_{2}...b_{n-1}\
$$
I would apreciate some light!
AI: I'm assuming (from Wikipedia) that the matrix is $$\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & ... & a_n \\ b_1 & 0 & 0 & 0 & ... & 0 \\ 0 & b_2 & 0 & 0 & ... & 0 \\0 & 0 & b_3 & 0 & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & b_{n-1} & 0\end{pmatrix}$$
Use induction on $n$. For $n = 1$ the proof is easy.
For $n > 1$ use Laplace expansion on the farthest right column to get
$$\det \begin{pmatrix} t - a_1 & -a_2 & -a_3 & -a_4 & ... & -a_n \\ -b_1 & t & 0 & 0 & ... & 0 \\ 0 & -b_2 & t & 0 & ... & 0 \\0 & 0 & -b_3 & t & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & -b_{n-1} & t\end{pmatrix}$$
= $$t *\det \begin{pmatrix} t - a_1 & -a_2 & -a_3 & -a_4 & ... & -a_{n-1} \\ -b_1 & t & 0 & 0 & ... & 0 \\ 0 & -b_2 & t & 0 & ... & 0 \\0 & 0 & -b_3 & t & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & -b_{n-2} & t\end{pmatrix}$$
$$ + (-1)^n * \det \begin{pmatrix} -b_1 & t & 0 & ... & 0 \\ 0 & -b_2 & t & ... & 0 \\ 0 & 0 & -b_3 & ... & 0 \\ 0 & 0 & 0 & ... & t \\ 0 & 0 & 0 & ... & -b_{n-1} \end{pmatrix}$$ $$= t(t^{n-1} -a_1 t^{n-2} - a_2 b_1 t^{n-2} - ... - a_{n-1}b_1b_2...b_{n-2}) - b_1 b_2...b_{n-1}$$ $$= t^n - a_1 t^{n-1} - a_2 b_1 t^{n-2} - ... - a_n b_1 b_2 ... b_{n-1}$$ as claimed. |
H: Solutions for $P(D)x(t)=0$, $P(x)=(x-x_0)^n$
I am trying to prove the following statement
Let $P(\lambda)=(\lambda-\lambda_{0})^{r}$where $r$ is a positive
integer. Prove that the equation $P(\frac{d}{dt})x(t)=0$ has solutions
$t^{i}e^{\lambda_{0}t},i=0,1,\ldots,r-1$
I thought of three ideas that I am having problems to continue with and I need some help,
Idea 1: Induction, this is clear for $r=1$, and maybe since
$P'(\lambda)=r(\lambda-\lambda_{0})^{r-1}$ then we could use some inductive step ?
Idea 2: Induction again, maybe use $P(\lambda)=(\lambda-\lambda_{0})(\lambda-\lambda_{0})^{r-1}$ and use some inductive step ?
Idea 3: Use the binomial theorem on $P(\lambda)$ and try do this directly
AI: The most direct approach is to multiply it out, one factor at a time. (Note my left hand side fixes a typo in the question)
$$\def\ddt{\frac{d}{d t}}
\begin{align} \left(\ddt - \lambda_0 \right)^r \left( t^i e^{\lambda_0 t} \right)
&= \left(\ddt - \lambda_0 \right)^{r-1} \left(\ddt - \lambda_0 \right)
\left( t^i e^{\lambda_0 t} \right)
\\&= \left(\ddt - \lambda_0 \right)^{r-1}\left( \ddt\left( t^i e^{\lambda_0 t} \right)
- \lambda_0 t^i e^{\lambda_0 t} \right)
\\&= \left(\ddt - \lambda_0 \right)^{r-1}\left( i t^{i-1} e^{\lambda_0 t} + \lambda_0 t^i e^{\lambda_0 t}
- \lambda_0 t^i e^{\lambda_0 t} \right)
\\&= \left(\ddt - \lambda_0 \right)^{r-1}\left( i t^{i-1} e^{\lambda_0 t}\right)
\\&= i \left(\ddt - \lambda_0 \right)^{r-1}\left( t^{i-1} e^{\lambda_0 t}\right)
\end{align}$$
It should be clear how things work from here. |
H: How do I show this formula involving several variables?
This is from Woll's "Functions of Several Variables," but there's no proof.
If $g$ is of class $C^k$ ($k \ge 2$) on a convex open set $U$ about $p$ in $\mathbb{R}^d$, then for each $q \in U$,
$
g(q) = g(p) + \sum_{i=1}^d \frac{\partial g}{\partial r_i} \bigg|_p (r_i(q) - r_i(p)) + \sum_{i,j} (r_i(q) - r_i(p)) (r_j(q) - r_j(p)) \int_0^1 (1-t) \frac{\partial^2g}{\partial r_i \partial r_j} \bigg|_{p + t(q - p)} dt.
$
It looks like Taylor's or mean value theorem. I especially don't understand the integral part.
AI: Consider function
$$
h(t)=g(p+t(q-p))
$$
and its Taylor series with the remainder in the integral form
$$
h(1)=h(0)+h'(0)+\int\limits_{0}^{1}(1-t)h''(t)dt
$$
Now note that
$$
h(0)=g(p)
$$
$$
h'(0)=\sum\limits_{i=1}^d\frac{\partial g}{\partial r_i}\biggl|_p(r_i(p)-r_i(q))
$$
$$
h''(t)=\sum\limits_{i=1}^d\sum\limits_{j=1}^d\frac{\partial^2 g}{\partial r_i\partial r_j}\biggl|_{p+t(q-p)}(r_i(p)-r_i(q))(r_j(p)-r_j(q))
$$ |
H: Does Log-Lipschitz regularity imply Hölder continuity?
A function is Log-Lipschitz if there exists a constant $C$ such that
\begin{equation}
|u(x) - u(y)| \le C|x-y| \log|x-y|
\end{equation}
Is a Log-Lipschitz function $C^{0,\alpha}$ for any $\alpha \in (0,1)
$(Hölder continuous)?
If you need, assume hypothesis. Thank you.
AI: Yes, it is -- assuming that you think and act locally. Think in terms of moduli of continuity $\omega$, i.e., functions such that $|u(x)-u(y)|\le \omega(|x-y|)$. The function $\delta\log (1/\delta)$ is smaller than $\delta^{\alpha}$ ($\alpha<1$) near $0$.
Edit: For $|x-y| < 1$ it is clear that $\log|x-y| < 0$. Commonly this is fixed by adding the modulus: $|\log |x-y||$. In the notation of moduli of continuity the issue is resolved by writing $\log(1/\delta) = - \log(\delta)$, where $\delta$ is small. |
H: Is this an abuse of set-theoretic notation?
The expression is this:
$$\bigcup_{n\in\mathbb{N}}\ \bigcup_{a_0\in\mathbb{Z}}\cdots\bigcup_{a_n\in\mathbb{Z}}\big\{z\in\mathbb{C}:a_0z^n+a_1z^{n-1}+\cdots+a_{n-1}z+a_n=0\big\}.$$
I hope it's clear what this is meant to denote (the set of algebraic numbers), but I'm uneasy about it since the number of unions depends on an element in the first union.
I suspect that the set could be rewritten more concretely as a union indexed by $\mathbb{Z}$-valued sequences that eventually terminate, but I'm not sure how that would look.
AI: $$\bigcup_{n\in\Bbb N}\bigcup_{\langle a_0,\dots,a_n\rangle\in\Bbb Z^{n+1}}\left\{z\in\Bbb C:\sum_{k=0}^na_kz^{n-k}=0\right\}$$
Added: I’d overlooked the fact that what was wanted was actually the set of algebraic numbers, which isn’t quite what’s described here or in the question. As Erick Wong points out in the comments, $\langle 0,\dots,0\rangle$ should be removed from the index set $\Bbb Z^{n+1}$, to avoid including $\{z\in\Bbb C:0=0\}=\Bbb C$. |
H: Proving that $a + b = b + a$ for all $a,b \in\mathbb{R}$
Being interested in the very foundations of mathematics, I'm trying to build a rigorous proof on my own that $a + b = b + a$ for all $\left[a, b\in\mathbb{R}\right] $. Inspired by interesting properties of the complex plane and some researches, I realized that defining multiplication as repeated addition will lead me nowhere (at least, I could not work with it). So, my ideas:
Defining addition $a+b$ as a kind of "walking" to right $\left(b>0\right)$ or to the left $\left(b<0\right)$ a space $b$ from $a$. Adding a number $b$ to a number $a$ (denoted by $a+b$) involves doing the following operation:
Consider the real line $\lambda$ and its origin at $0$. Mark a point $a$, draw another real line $\omega$ above $\lambda$ such what $\omega \parallel \lambda$ and mark a point $b$ on $\omega$. Now, draw a line $\sigma$ such that $\sigma \perp \omega$ and the only point in commom between $\sigma$ and $\omega$ is $b$. Consider the point that $\lambda$ and $\sigma$ have in commom; this point is nicely denoted as $a + b$.
(Note that all my work is based here. Any problems, and my proof goes to trash)
This definition can be used to see the properties of adding two numbers $a$ and $b$, for all $a, b \in\mathbb{R}$.
Using geometric properties may lead us to a rigorous proof (if not, I would like to know the problems of using it).
So, I started:
$a, b \in\mathbb{N}$:
$a+b = \overbrace{\left(1+1+1+\cdots+1\right)}^a + \overbrace{\left(1+1+1+\cdots+1\right)}^b = \overbrace{1+1+1+1+\cdots+1}^{a+b} = \overbrace{\left(1+1+1+\cdots+1\right)}^b + \overbrace{\left(1+1+1+\cdots+1\right)}^a = b + a$
(Implicity, I'm using the fact that $\left(1+1\right)+1 = 1+\left(1+1\right)$, which I do not know how to prove and interpret it as cutting a segment $c$ in two parts -- $a$ and $b$. However, this result can be extended to $\mathbb{Z}$ in the sense that $-a$ $(a > 0)$ is a change; from right to left).
$a, b \in\mathbb{R}$:
Here, we have basically two cases:
$a$ and $b$ are either positive or negative;
$a$ and $b$, where one of them is negative.
Since in my definition $-b, b>0$ means drawing a point $b$ to the left of the real line, there's no big deal interpretating it; subtracting can be interpreted now. So, it starts:
$a + b = c$. However, $c$ can be cut in two parts: $b$ and $a$. Naturally, if $a>c$, then $b<0$ -- many cases can be listed. So, $c = b + a$. But $c = a + b$; it follows that $a + b = b + a$. My questions:
Is there any problem in using my definition of adding two numbers $a$ and $b$, which uses many geometric properties? Is there any way to solve it from informality? Is there anything right here?
Thanks in advance.
AI: First you need to define $\mathbb{R}$ in your construction!
To define $\mathbb{R}$, one way is to go about defining $\mathbb{N}$, then defining $\mathbb{Z}$, then defining $\mathbb{Q}$ and then finally defining $\mathbb{R}$. Once you have these things set up, proving associativity, commutativity of addition over reals essentially boils down to proving associativity, commutativity of addition over natural numbers.
As said earlier, one goes about first defining natural numbers. For instance, $2$ as a natural number is defined as $2_{\mathbb{N}} = \{\emptyset,\{\emptyset\} \}$. We will use the notation that $e$ is $1_{\mathbb{N}}$ and $S(a)$ be the successor function applied to $a \in \mathbb{N}$.
Then we define addition on natural numbers using the successor function. Addition on natural numbers is defined inductively as
$$a +_{\mathbb{N}} e = S(a)$$
$$a +_{\mathbb{N}} S(k) = S(a+k)$$
You can also define $\times_{\mathbb{N}},<_{\mathbb{N}}$ on natural numbers similarly.
Then one defines integers as an equivalence class (using $+_{\mathbb{N}}$) of ordered pairs of naturals i.e. for instance, $2_{\mathbb{Z}} = \{(n+_{\mathbb{N}}2_{\mathbb{N}},n):n \in \mathbb{N}\}$. You can similarly, extend the notion of addition and multiplication of two integers i.e. you can define $a+_{\mathbb{Z}} b$, $a \times_{\mathbb{Z}}b$, $a <_{\mathbb{Z}} b$. Addition, multiplication and ordering of integers are defined as appropriate operations on these set.
Then one moves on to defining rationals as an equivalence class (using $\times_{\mathbb{Z}}$) of ordered pairs of integers. So $2$ as a rational number, $2_{\mathbb{Q}}$ is an equivalence class of ordered pair $$2_{\mathbb{Q}} = \{(a \times_{\mathbb{Z}} 2_{\mathbb{Z}},a):a \in \mathbb{Z}\backslash\{0\}\}$$ Again define $+_{\mathbb{Q}}, \times_{\mathbb{Q}}$, $a <_{\mathbb{Q}} b$. Addition, multiplication and ordering of rationals are defined as appropriate operations on these set.
Finally, a real number is defined as the left Dedekind cut of rationals. i.e. for instance $2$ as a real number is defined as $$2_{\mathbb{R}} = \{q \in \mathbb{Q}: q <_{\mathbb{Q}} 2_{\mathbb{Q}}\}$$
Addition, multiplication and ordering of reals are defined as appropriate operations on these set.
Once you have these things set up, proving associativity, commutativity of addition over reals essentially boils down to proving associativity, commutativity of addition over natural numbers.
Here are proofs of associativity and commutativity in natural numbers using Peano's axiom.
Associativity of addition: $(a+b) + c = a + (b + c )$
Proof:
Let $\mathbb{S}$ be the set of all numbers $c$, such that $ (a+b) + c = a + (b + c )$, $ \forall a,b \in \mathbb{N}$.
We will prove that $ e$ is in the set and whenever $k \in \mathbb{S}$, we have $S(k) \in \mathbb{S}$. Then by invoking Peano’s axiom (viz, the principle mathematical induction), we get that $\mathbb{S} = \mathbb{N}$ and hence $ (a+b) + c = a + (b + c )$, $ \forall a,b \in \mathbb{N}$.
First Step:
Clearly, $ e \in \mathbb{S}$. This is because of the definition of addition.
$ (a+b)+e = S(a+b)$ and $ a + S(b) = S(a+b)$
Hence $ (a+b)+e = a + S(b) = a+ (b+e)$
Second Step:
Assume that the statement is true for some $ k \in \mathbb{S}$.
Therefore, we have $ (a+b)+k = a+(b+k)$.
Now we need to prove, $ (a+b) + S(k) = a+(b+S(k))$.
By definition of addition, we have $ (a+b)+S(k) = S((a+b) + k)$
By induction hypothesis, we have $ (a+b)+k = a+ (b+k)$
By definition of addition, we have $ b + S(k) = S(b+k)$
By definition of addition, we have $ a+S(b+k) = S(a+(b+k))$
Hence, we get,
$ (a + b) + S(k) = S((a+b) + k) = S(a+ (b+k)) = a + S(b+k) = a+ (b + S(k))$
Hence, we get,
$ (a+b) + S(k) = a + (b+S(k))$
Final Step:
So, we have $ e \in \mathbb{S}$. And whenever, $k \in \mathbb{S}$, we have $S(k) \in \mathbb{S}$.
Hence, by principle of mathematical induction, we have that $\mathbb{S} = \mathbb{N}$
i.e. the associativity of addition, viz,
$$(a+b) + c = a + (b+c)$$
Commutativity of addition: $ m + n = n + m$, $ \forall m,n \in \mathbb{N}$.
Proof:
Let $ \mathbb{S}$ be the set of all numbers $ n$, such that $ m + n = n + m$, $ \forall m \in \mathbb{N}$.
We will prove that $ e$ is in the set $ \mathbb{S}$ and whenever $ k \in \mathbb{S}$, we have $ S(k) \in \mathbb{S}$. Then by invoking Peano's axiom (viz, the Principle Mathematical Induction), we state that $ \mathbb{S}=\mathbb{N}$ and hence $ m + n = n + m$, $ \forall m,n \in \mathbb{N}$.
First Step:
We will prove that $ m + e = e + m$ and hence $ e \in \mathbb{S}$.
The line of thought for the proof is as follows:
Let $ \mathbb{S}_1$ be the set of all numbers $ m$, such that $ m + e = e + m$.
We will prove that $ e$ is in the set $ \mathbb{S}_1$ and whenever $ k \in \mathbb{S}_1$, we have $ S(k) \in \mathbb{S}_1$. Then by invoking Peano's axiom (viz, the Principle Mathematical Induction), we state that $ \mathbb{S}_1=\mathbb{N}$ and hence $ m + e = e + m$, $ \forall m \in \mathbb{N}$.
To prove: $ e \in \mathbb{S}_1$
Clearly, $ e + e = e + e$ (We are adding the same elements on both sides)
Assume that $ k \in \mathbb{S}_1$. So we have $ k + e = e + k$.
Now to prove $ S(k)+ e = e + S(k)$.
By the definition of addition, we have $ e + S(k) = S(e + k)$
By our induction step, we have $ e + k = k + e$.
So we have $ S(e+k) = S(k+e)$
Again by definition of addition, we have $ k + e = S(k)$.
Hence, we get $ e + S(k) = S(S(k))$.
Again by definition of addition, $ p + e = S(p)$, which gives us $ S(k) + e = S(S(k))$.
Hence, we get that $ S(k+e) = S(k) + e$.
So we get,
$ e + S(k) = S(e+k) = S(k+e) = S(S(k)) = S(k) + e$.
Hence, assuming that $ k \in \mathbb{S}_1$, we have $ S(k) \in \mathbb{S}_1$.
Hence, by Principle of Mathematical Induction, we have $ m + e = e + m$, $ \forall m \in \mathbb{N}$.
Second Step:
Assume that $ k \in \mathbb{S}$. We need to prove now that $ S(k) \in \mathbb{S}$.
Since $ k \in \mathbb{S}$, we have $ m + k = k + m$.
To prove: $ m + S(k) = S(k) + m$.
Proof:
By definition of addition, we have $ m + S(k) = S(m+k)$.
By induction hypothesis, we have $ m + k = k + m$. Hence, we get $ S(m+k) = S(k+m)$.
By definition of addition, we have $ k + S(m) = S(k+m)$.
Hence, we get $ m + S(k) = S(m+k) = S(k+m) = k + S(m)$.
We are not done yet, since we want to prove, $ m + S(k) = S(k) + m$.
So we are left to prove $ k + S(m) = S(k) + m$.
$S(k) +m = (k+e) + m = k + (e+m) = k + (m+e) = k + S(m)$.
Hence, we get $ m + S(k) = S(k) + m$.
Final Step:
So, we have $ e \in \mathbb{S}$. And whenever $ n \in \mathbb{S}$, we have $ S(n) \in \mathbb{S}$.
Hence, by Principle of Mathematical Induction, we have the commutativity of addition, viz,
$ m + n = n + m$, $ \forall m,n \in \mathbb{N}$.
We might think that associativity is harder/lengthier to prove than commutativity, since associativity is on three elements while commutativity is on two elements.
On the contrary, if you look at the proof, proving associativity turns out to be easier than commutativity.
Note that the definition of addition, viz $m + S(n) = S(m+n)$, incorporates the associativity $m+(n+e) = (m+n)+e$.
For commutativity however, we are changing the roles of $m$ and $n$, (we are changing the "order") and no wonder it is "harder/lengthier" to prove it. |
H: Expectation Values inside absolute value operator
first: are these equality true ?
$$|E[Y]-E[X]|=|E[Y]|-|E[X]|.$$
$$|E[Y]-E[X]|^2=|E[Y]|^2-|E[X]|^2$$
second: what is result of this relation:
$$\sum_{i=1}^{3}p_i.(X_i-\mu)^2=?$$
where the $\mu =\sum_{i=1}^{3}(p_i.X_i)$
AI: "What is the result?" is a bit vague. In some contexts it would be a request for a simplification; Henry's answer seems to treat it as a request for the name of a concept (the variance). Maybe it could also be a request for an alternative expression that might be computationally more tractable. If that last, there's this bit of algebra:
$$
\begin{align}
& {}\quad p_1(X_1-\mu)^2 + p_2(X_2-\mu) ^2+ p_3(X_3-\mu)^2 \\[10pt]
& = p_1(X_1^2-2X_1\mu + \mu^2) + p_2(X_2^2-2X_2\mu + \mu^2) + p_3(X_3^2-2X_3\mu + \mu^2) \\[10pt]
& = p_1 X_1^2 + p_2 X_2^2 + p_3 X_3^2 -2\mu(p_1 X_1+p_2 X_2+p_3 X_3) + (p_1+p_2+p_3)\mu^2 \\[10pt]
& = \left(p_1 X_1^2 + p_2 X_2^2 + p_3 X_3^2\right) -\left(2\mu\cdot \mu\right) + \left(1\cdot\mu^2\right) \\[10pt]
& = \left(p_1 X_1^2 + p_2 X_2^2 + p_3 X_3^2\right) - \mu^2.
\end{align}
$$
Thus, the expected value of the square, minus the square of the expected value equals the variance. |
H: Weird qualification about Cauchy derivative-integral formula
It is known that for a holomorphic function $f$ over $\Omega$, if $C$ is a circle inside $\Omega$, then:
$$f^{(n)}(z) = \frac{n!}{2\pi i}\int_C \frac{f(s)}{(s-z)^{n+1}} \, ds$$
A textbook makes the following claim :
$$f:\mathbb{D}\rightarrow\mathbb{C}$$
where $\mathbb{D}$ is the unit disc, and
$$2f'(0) = \frac{1}{2\pi i}\int_{C_r} \frac{f(s)-f(-s)}{(s-z)^2} \, ds$$
Here $C_r$ is a circle about the origin whose radius is $r$.
The second formula seems to follow easily from the first by setting $n=1$ and summing two identical integrals where a simple change of variable is made to one of the summands. Here is where it gets weird.
The text says that the second formula only holds whenever $0 \lt r \lt 1$. I see no reason for imposing $r \lt 1$. Why would the text go out of its way to make sure $r \neq 1$? I don't see what is so special about this case.
Additional Info:
The statement is made in reference to an exercise problem. The text is Stein & Shakarchi: Princeton Lectures in Analysis II Complex Analysis. You can find the statement on page 65, in Chapter 2, exercise 7 where it says "Hint". At this time, the question and a solution can be found at https://6ba38298-a-62cb3a1a-s-sites.googlegroups.com/site/davegaebler/home-1/solution-manuals/SteinandShakarchiComplexAnalysis.pdf on page 5 of the pdf under "Exercise 7". The author of the solution also goes out of his way to assure $r \neq 1$ and uses a limit argument to compensate.
AI: Note that in the statement of the Cauchy Integral Formula (which share the same necessary conditions as the resulting differentiation formulae), the closure of the disk $D$ must lie completely within the domain of holomorphicity $U.$ Since we don't know that $f$ is holomorphic on an open set containing the entire unit disk, the limit argument has to be used. |
H: the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$
what is the solution set of $\left | \frac{2x - 3}{2x + 3} \right |< 1$ ?
I solved it by first assuming: $-1 < \frac{2x - 3}{2x + 3 } < 1$
ended with: $x > 0 > -3/2$
Is that a correct approach?
And how to derive the solution set from the last inequality?
Is it $(-\frac{3}{2}, \infty)$ or $(0, \infty )$ ?
Thanks in advance.
AI: One thing you can do here is square to get rid of the absolute values and make it easier to solve, obtaining the equivalent statement
$${(2x - 3)^2 \over (2x + 3)^2} < 1$$
The denominator is nonnegative so you can multiply it through, obtaining
$$(2x - 3)^2 < (2x + 3)^2$$
This can be rewritten as
$$4x^2 - 12x + 9 < 4x^2 + 12x + 9$$
This simplifies to just
$$x > 0$$
Note that since there was originally a $(2x + 3)$ in the denominator, if $x = -{3 \over 2}$ were in the above solution we would have had to exclude it. But it wasn't, so we don't have to worry about it. |
H: Norms on inner product space over $\mathbb{R}$
Definition of the problem
Let $\left(E,\left\langle \cdot,\cdot\right\rangle \right)$
be an inner product space over $\mathbb{R}$. Prove that for all $x,y\in E$
we have
$$
\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left\langle x,y\right\rangle \leq\left\Vert x+y\right\Vert \cdot\left\Vert x\right\Vert \left\Vert y\right\Vert .
$$
My efforts
I tried to apply Cauchy-Schwarz inequality:
$$
\left|\left\langle x,y\right\rangle \right|\leq\left\Vert x\right\Vert \left\Vert y\right\Vert \quad\forall x,y\in E,
$$
and since we are in an inner product space over $\mathbb{R}$, we can simplify remove the absolute value from the inner product:
$$
\left\langle x,y\right\rangle \leq\left\Vert x\right\Vert \left\Vert y\right\Vert \quad\forall x,y\in E.
$$
We obtain:
$$
\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\left\langle x,y\right\rangle \leq\left(\left\Vert x\right\Vert +\left\Vert y\right\Vert \right)\cdot\left\Vert x\right\Vert \left\Vert y\right\Vert \quad\forall x,y\in E.
$$
My question
Could you give me a hint/idea on how to solve this problem? Which Lemma/Theorem should I use?
Thank you,
Franck
AI: $\begin{align*} \big( \Vert x+y\Vert \Vert x \Vert \Vert y \Vert \big)^2 &= \Vert x + y \Vert^2 \langle x,x \rangle \langle y,y \rangle = \langle x+y,x+y\rangle \langle x,x \rangle \langle y,y \rangle = \\ &= \big(\langle x,x \rangle + \langle y,y \rangle + 2 \langle x,y \rangle \big) \langle x,x \rangle \langle y,y \rangle = \\ &= \langle x,x \rangle^2\langle y,y \rangle + \langle x,x \rangle \langle y,y \rangle^2 + 2 \langle x,y \rangle \langle x,x \rangle \langle y,y \rangle \geq \\ &\geq \langle x,y \rangle^2\langle x,x \rangle + \langle x,y \rangle^2\langle y,y \rangle + 2\langle x,y \rangle^2 \langle x,x \rangle^\frac{1}{2}\langle y,y \rangle^\frac{1}{2} = \\ &= \big( \langle x,y \rangle \langle x,x \rangle ^\frac{1}{2}+ \langle x,y \rangle \langle y,y \rangle^\frac{1}{2}\big)^2 = \big( (\langle x,x \rangle ^\frac{1}{2}+ \langle y,y \rangle^\frac{1}{2})\langle x,y \rangle \big) ^2 = \\ &= \big( (\Vert x \Vert + \Vert y \Vert)\langle x,y\rangle\big)^2
\end{align*}$
To get to the fourth line, we apply the Cauchy-Schwarz inequality to each summand.
Then, $\big( \Vert x+y\Vert \Vert x \Vert \Vert y \Vert \big)^2 \geq \big( (\Vert x \Vert + \Vert y \Vert)\langle x,y\rangle\big)^2$, and since $\Vert x+y\Vert \Vert x \Vert \Vert y \Vert \geq 0$, we can simply cancel the squares, obtaining $\Vert x+y\Vert \Vert x \Vert \Vert y \Vert \geq \big(\Vert x \Vert + \Vert y \Vert)\langle x,y\rangle$.
Note that we used $\Vert x \Vert = \langle x,x \rangle ^\frac{1}{2}$ and the fact that $\langle x+y,x+y \rangle = \langle x,x+y \rangle + \langle y,x+y\rangle =$ $=\langle x,x \rangle + \langle x,y \rangle + \langle y,x\rangle + \langle y,y \rangle = \langle x,x \rangle + \langle y,y \rangle + 2 \langle x,y \rangle$ since the inner product in a real vector space must be symmetric. |
H: Random variables and sigma field
Given $Z_1,Z_2,\cdots$ i.i.d with $E|Z_i|<\infty$. $\theta$ is an independent r.v. with finite mean and $Y_i=Z_i+\theta$.
If we define $F_n=\sigma(Y_1,\cdots,Y_n), F_\infty=\sigma(\cup_n F_n).$ Do we have $\theta\in F_\infty$?
AI: How about: If $E[Z_i] = m$, then
$$
\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n (Z_i+\theta) = m + \theta
$$
a.e. so $m+\theta$ is $F_\infty$-measurable. So is the constant $m$. |
H: Always a value with uncountably many preimages? (for a continuous real map on the plane)
Let $f$ be a continuous map ${\mathbb R}^2 \to {\mathbb R}$. For $y\in {\mathbb R}$, denote by $P_y$ the preimage set $\lbrace (x_1,x_2) \in {\mathbb R}^2 | f(x_1,x_2)=y \rbrace$.
Is it true that
(1) At least one $P_y$ is uncountable ?
(2) At least one $P_y$ has the same cardinality as $\mathbb R$.
Some easy remarks :
(2) is stronger than (1).
(2) follows from (1) if we assume the GCH.
If there is a point $(x_0,y_0)$ such that the partial derivatives $\frac{\partial f}{\partial x}(x_0,y_0)$ and $\frac{\partial f}{\partial y}(x_0,y_0)$ exist and one of them is nonzero, then
(2) (and hence (1)) follows from the implicit function theorem.
AI: Consider the restriction of $f$ to $\mathbb{R}_x=\mathbb{R} \times \{x\}$ for fixed $x$. If its image is a point, we're done. Otherwise, its image is an interval in $\mathbb{R}$, and that interval contains a subinterval with rational endpoints. Since there are only countably many such intervals, there must be one that's contained in the images of uncountably many $\mathbb{R}_x$; then any point in that interval has uncountable preimage. |
H: Is $A^{q+2}=A^2$ in $M_2(\mathbb{Z}/p\mathbb{Z})$?
I'm wondering, why is it that for $q=(p^2-1)(p^2-p)$, that $A^{q+2}=A^2$ for any $A\in M_2(\mathbb{Z}/p\mathbb{Z})$?
It's not hard to see that $GL_2(\mathbb{Z}/p\mathbb{Z})$ has order $(p^2-1)(p^2-p)$, and so $A^q=1$ if $A\in GL_2(\mathbb{Z}/p\mathbb{Z})$, and so the equation holds in that case. But if $A$ is not invertible, why does the equality still hold?
AI: If $A$ is not invertible, then its characteristic polynomial is either $x^2$ or $x(x-a)$ for some $a\in\mathbb{Z}/p\mathbb{Z}$. In the former case, by the Cayley-Hamilton Theorem we have $A^2 = 0$, hence $A^{q+2}=A^2$. In the latter case, the matrix is similar to a diagonal matrix, with $0$ in one diagonal and $a$ in the other. So, up to conjugation, we have
$$A^{q+2}=\left(\begin{array}{cc}
0 & 0\\
0 & a\end{array}\right)^{q+2} = \left(\begin{array}{cc}
0 & 0\\
0 & a^{q+2}
\end{array}\right).$$
But $a^{p} = a$. Since $q = p^4 -p^3-p^2 + p$, we have
$$a^{q} = \frac{a^{p^4}a^p}{a^{p^3}a^{p^2}} = 1$$
so $a^{q+2} = a^2$, hence $A^{q+2}=A^2$. |
H: What is the probability of two people meeting?
I am trying to figure out a solution to the following problem:
Let there be two groups of people, Group A and Group B. Group A represents x percent (e.g. 1%) of the world's population, and Group B represents y percent (e.g. 2%) of the world's population. What is the probability that a person from Group A will meet a person from Group B? Assume the following things:
The average human being meets z people (e.g. 100,000) in a lifetime.
The world's population is kept at a constant k people.
Everyone in the world was born and will die at the same time.
Disclaimer:
I came up with this question myself, but I'm not a mathematician, so please feel free to clean this up if need be. Also, if there is not enough information in the problem to solve it, add assumptions and please indicate the reasons for adding them. The assumptions I wrote are my attempt at making the problem easier. If they are not necessary, and removing any produces a more accurate answer, then I encourage the removal of them.
AI: Let $a$ be the number of people in group $A$ and $b$ be the number of people in group $B$. Then the chance that you never meet somebody from group $B$ is the chance that all your $z$ acquaintances are outside, $\frac {k-b}k\frac{k-b-1}{k-1} \ldots \frac{k-b-z+1}{k-z+1}=\frac {(k-b)!(k-z)!}{k!(k-b-z)!}$ If the chances that an $A$ individual meets a $B$ individual are independent, the chance they never meet is $\left(\frac {(k-b)!(k-z)!}{k!(k-b-z)!}\right)^a$. Although it is not obvious, this should be symmetric in $a$ and $b$. |
H: Numerically solving 1D Heat
I'm looking at page 10 of http://www4.ncsu.edu/~zhilin/TEACHING/MA402/notes1.pdf
What happened to the boxed term (it doesn't seem to appear in the matrix equation).
(I'm just implementing this and am not familiar with pde/numerical techniques)
AI: Looks like an error in the notes. You do need the boxed term to appear in the equation; each component of the vector on the right-hand side should have an additional $\frac1{\Delta t}u_i^k$ term. |
H: How to prove that the tangent to a circle is perpendicular to the radius drawn to the point of contact?
I've tried drawing a parallel chord to the tangent but then how would you prove that the chord is perpendicular to the radius?
AI: Let $O$ be the centre of the circle, let $\ell$ be a tangent line, and let $P$ be the point of tangency. Suppose that $OP$ is not perpendicular to $\ell$. Draw the line through $O$ which is perpendicular to $\ell$. Then this line meets $\ell$ at a point $Q\ne P$.
Note that $Q$ is outside the circle. Now consider the triangle $OQP$. This is right-angled at $Q$. So $OP$ is the hypotenuse of this triangle, and is therefore bigger than $OQ$. But this is impossible: since $Q$ is outside the circle, we must have $OP\lt OQ$. |
H: Definite Integral (Calculus)
this is a revision problem, not a homework problem. Sincere thanks for help.
Question: Evaluate $\int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}$.
The answer is $\pi/4$, but I am unable to work out the method.
Sincere thanks for help.
AI: Notice that
$$\frac{d}{dx} (\log(\cos x + \sin x)) = \frac{\cos x - \sin x}{\cos x + \sin x}.$$
Now if
$$I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx= \int_0^{\pi/2} 1 - \frac{\sin x}{\cos x + \sin x}dx$$
then we see that
$$2I = \int_0^{\pi/2} 1 + \frac{\cos x - \sin x}{\cos x + \sin x} dx = \pi/2+0.$$ |
H: Maximum value of the function
Find the minimum value of $f(x) = \max\{x^2-4, x , - 1\}$
I am able to do this question by using graph. But if there is any other method please tell me
AI: You can reason as follows. The $-1$ in $\max\{x^2-4,x,-1\}$ guarantees that $f(x)$ is always at least $-1$, so the minimum value of $f(x)$ must be at least $-1$. Is there at least one point $x$ such that $f(x)=-1$? At such a point we would have $x^2-4\le-1$ and $x\le-1$. The question is whether we can have both of these at the same time: is there a point $x\le-1$ such that $x^2-4\le-1$?
The inequality $x^2-4\le-1$ is equivalent to $x^2\le3$, or $-\sqrt3\le x\le\sqrt 3$. Since $-\sqrt3<-1$, we have a whole interval of $x$’s from which to choose: any $x\in[-\sqrt3,-1]$ will have the property that $x^2-4\le-1$ and $x\le -1$ and therefore will satisfy $f(x)=-1$.
Thus, $f(x)\ge-1$ for all $x\in\Bbb R$, and there is at least one $x\in\Bbb R$ such that $f(x)=-1$, so $-1$ must be the minimum value of $f(x)$. |
H: Invert "Gravitational" Force Function or Solve an Intersection
Recall "gravitational"-type force functions, by which I mean anything of the form:
$f(x,y,z) = \frac{k}{((x-x_0)^2+(y-y_0)^2+(z-z_0)^2)^p}, p\in\Re_{>0}, k\in\Re, (x,y,z) \neq(x_0,y_0,z_0)$
(e.g., for gravity, $p=1,k=G m_1 m_2$)
Define a function $g(x,y,z) = f_1(x,y,z) + f_2(x,y,z) + f_3(x,y,z) +\cdots+f_n(x,y,z)$ (a sum of several force functions, presumably each with different $k$ (though probably not $p$).
I have two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ such that $g(x_1,y_1,z_1)<a$ and $g(x_2,y_2,z_2)>a$. My problem is to find the location $(x_3,y_3,z_3)$ such that $g(x_3,y_3,z_3)=a$ where $(x_3,y_3,z_3)$ lies on the straight line between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$. If it helps, I know that exactly two of the xs, ys, and zs are equal.
I'm currently solving this numerically, but it will eventually be solved on an OpenCL kernel where I would very much like a closed form solution.
However, I am having trouble getting such a solution. I'm not even sure it's possible. I tried inverting $g$, but got basically nowhere. Ideas? Thanks.
AI: Let us consider an extremely simplified version of the problem: $p=1$ and just two terms centered at $(0,\bar y_1,0)$ and $(\bar x_2,\bar y_2,0)$, with the line on which the unknown point lies being $y_1 = y_2 = 0$ and $z_1 = z_2 = 0$. Then you want to solve the equation
$$\frac{k_1}{x^2+\bar y_1^2} + \frac{k_2}{(x-\bar x_2)^2+\bar y_2^2} = a,$$
which expands to the fourth-degree polynomial equation in $x$,
$$k_1\big((x-\bar x_2)^2+\bar y_2^2\big) + k_2\big(x^2+\bar y_1^2\big) = a\big(x^2+\bar y_1^2\big)\big((x-\bar x_2)^2+\bar y_2^2\big).$$
Mathematica finds an analytical solution that I cannot reproduce here because its $\LaTeX$ code has 11887 characters in it. If you have more than two terms, of course, you'll get a polynomial equation of degree greater than five, and those are known to have no analytical solutions in general. |
H: Infinitely many primes are of the form $an+b$, but how about $a^n+b$?
A famous theorem of Dirichlet says that infinitely many primes are of the form:$\alpha n+\beta$, but are there infinitely many of the form: $\alpha ^n+\beta$, where $\beta$ is even and $\alpha$ is prime to $\beta$? or of the form $\alpha!+\gamma$, where $\gamma$ is odd?
Out of mere curiosity has this question come, thus any help is greatly appreciated.
AI: There are relatively prime non-trivial $\alpha$ and $\beta$, with $\beta$ even, such that $\alpha^n +\beta$ is not prime if $n \ge 1$. Easy, let $\beta$ have decimal expansion that ends in $4$, and let $\alpha>1$ have decimal expansion that ends in $1$.
A more subtle class of examples is illustrated by $625^n+4$. For this one we use the algebraic identity
$$x^4+4=(x^2-2x+2)(x^2+2x+2)$$
to prove compositeness.
For the factorial question, a necessary condition for primality if $\alpha \gt 1$ is $\gamma=\pm 1$. Unfortunately it is not known whether there are infinitely many primes of the form $n!\pm 1$. |
H: Regular covering implies transitive automorphism group action
We already know the theorem
Theorem
Let $p: (Y,y) \rightarrow (X,x)$ be a covering, with $Y$ connected and $X$ locally path connected, and let $p(y) = x$. If $p_*(\pi_1(Y,y))$ is a normal subgroup of $\pi_1(X,x)$, then $\pi_1(X,x)/p_*(\pi_1(Y,y))$ is isomorphic to Aut(Y/X).
Question
As the situation above, T.F.A.E
(a) the covering is normal (i.e., $p_*(\pi_1(Y,y))$ is a normal subgroup of $\pi_1(X,x)$);
(b) the action of $Aut(X/Y)$ on $p^{-1}(x)$ is transitive;
(c) for every loop $\sigma$ at $x$, if one lifting of $\sigma$ is closed (i.e., still a loop at some $y \in p^{-1}(x)$), then all liftings are closed.
My approach
I already proved (b) $\implies$ (c) and (c) $\implies$ (a). So I need to prove (a) $\implies$ (b), and I know I need to use the theorem above.
So, is that true that for any $y$ and $y' \in p^{-1}(x)$, the there exists $[\sigma] \in \pi_1(X,x)$ such that $y' = [\sigma]y$? (If this is true, then we are done!)
Thank you very much!
AI: Yes, your property is true : $\pi_1(X,x)$ acts transitively on $p^{-1}(x)$ :
First, assume that $Y$ is path connected : let $y,y'\in p^{-1}(x)$, you can find a path $\gamma:y\rightsquigarrow y'$. Then notice that $\gamma$ is the unique lift of $p\circ\gamma:x\rightsquigarrow x$ with starting point y. So $y'=\gamma(1)=y\cdot[p\circ\gamma]$.
Now, notice that $p$ is a local homeomorphism, so $Y$ is connected and locally path connected (since $X$ is), so $Y$ is path connected. |
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