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H: Fast multiplication of orthogonal matrices
Given $A,B\in SO(3)$, direct matrix multiplication computes $C=AB$ with 27 multiplies. The group $SO(3)$ is a $3$-dimensional manifold. This suggests that direct matrix multiplication, which thinks of elements of $SO(3)$ as 9-dimensional, is not optimal. What is the minimal number of multiplies necessary to compute $C$?
AI: Off the top of my head, if you represent elements of $SO(3)$ via lifting them to elements of $SU(2)$, a product can be computed with 8 multiplies. But they're multiplies of complex numbers which are 4 multiplies each for 32 in all.
But, we can use Strassen's algorithm to compute the matrix product in 7 multiplies, and Karatsuba to compute a complex product in 3 multiplies for 21 in all.
Of course, elements of $SU(2)$ aren't general matrices, they have a special form. If you write out the four entries, it's clear you can compute the terms with 4 complex multiplies along with some adds and conjugations, so 12 real multiplies in all if we use Karatsuba or 16 otherwise.
Using Euler-Rodrigues form, we can represent the rotation as 4 real numbers. The stated composition formula requires 16 products. But we can get away with only 10 products. First compute $a_1 a_2$, $b_1 b_2$, $c_1 c_2$, and $d_1 d_2$, then use the idea of Karatsuba to get the rest of the terms in pairs. For example, you can get $a_1b_2 + a_2b_1$ by the single product $(a_1 + b_1)(a_2 + b_2)$ and subtracting off the unwanted terms.
There are more representations listed on wikipedia, but I haven't looked into how much a product would cost in those representations.
Keep in mind that fewer products isn't always better: the adds and other things take time too, and different algorithms behave differently with respect to numerical stability. |
H: Special orthogonal matrix uniquely determined by $n-1 \times n-1$ entries?
For example, consider the specific question: Given $a_{11},a_{12},a_{21},a_{22}$ does that uniquely determine
$A=\begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix}$
where $A\in SO(3)$.
AI: Hint 1: The sum of the squares of each column is $1$.
Hint 2: The third column is $\pm$ the cross product of the first two.
Hint 1 is immediately applicable to $n\times n$ matrices.
There is an $n$-dimensional analog of the cross product that extends Hint 2. |
H: Show that $a^{q-1} \equiv 1 \pmod{pq}$
Assume that $p$ and $q$ are distinct add primes such that $p-1\mid q-1$. If $\gcd(a,pq)=1$ ,show that: $$a^{q-1} \equiv 1 \pmod{pq}$$
I have tried as follows:
$$a^{q-1} \equiv 1 \pmod{q} \quad \text{and} \quad a^{p-1} \equiv 1 \pmod{p}$$
$$\implies a^{(q-1)(p-1)} \equiv 1 \pmod{q} \quad \text{and} \quad a^{(q-1)(p-1)} \equiv 1 \pmod{p}$$
$$\implies a^{(q-1)(p-1)} \equiv 1 \pmod{pq}$$
But then I am stuck - please help.
AI: Remember that you are told that $p-1$ divides $q-1$. Note that you have not yet used this hypothesis. That suggests that you should really try to use it somehow.
Since $p-1$ divides $q-1$, then there exists $k$ such that $q-1 = k(p-1)$. That means that, since $a^{p-1}\equiv 1\pmod{p}$, then
$$1 \equiv 1^k \equiv (a^{p-1})^k \equiv a^{(p-1)k} = a^{q-1}\pmod{p}.$$
So now you have $a^{q-1}\equiv 1\pmod{q}$ and $a^{q-1}\equiv 1\pmod{p}.$ |
H: Understanding Formula for Sampling without Ordering and without Replacement
Alright, so I've been working through a couple combinatorics problems and I'm having trouble understanding the underlying reason for why a formula is written in a certain way. So here's the problem:
Suppose that r flags of different colors are to be displayed on n poles in a row. In how many ways can this be done?
Now this problem was solved by basically treating each flag like a divider and every time you place a flag, you create $n+1$ more choices of position for the next flag to placed which makes total sense. The final answer then is: $n(n+1)(n+2)...(n+r-1)$
What I'm confused about is how to know that $(n+r-1)$ always gives the final number of choices. I drew out a sample case and it does work out but I'm not sure why. This type of formula ending comes up a lot and I really want to understand how it works and when it applies.
AI: You have $r$ flags. There are $n+0$ places to put the first flag.
After putting in the first flag, there are now $n+1$ places for the second flag.
After putting the second flag, there are now $n+2$ places for the third flag.
After putting the third flag, there are now $n+3$ places for the fourth flag.
You see a pattern? After putting the 10th flag, there will be $n+10$ places for the eleventh flag. After putting the 57th flag, there will be $n+57$ places for the $58$th flag.
So, just before you finish, you will have placed the $(r-1)$st flag; that means that there will be $n+(r-1)$ places where you can put the $r$th flag. And once you put the $r$th flag, you are done. There are no more choices to be made. So the last time you needed to count, you had $n+r-1$ ways.
So in total you had $n\times (n+1)\times\cdots \times (n+r-1)$ ways of doing it. |
H: How does he get a perfect swap numerator and denominator.
I'm going through a exercise, in which all the answers are given, but the tutor makes a step and I can't follow at all. A massive jump with no explanation.
Here is the question:
$\lim_{x \to 2} \frac{\frac{1}{2}-\frac{1}{x}}{x-2}$
He then simplifies:
$ \frac{x-2}{2x(x-2)}$
He said he multiplied the entire equation by 2x
How does he know 2x swaps the denominator upto the numerator
Before he gave the simplification I spent perhaps 20 minutes trying to figure something out, and failed, and then he just jumps this massive step
He simplifies because we are finding limits.
Thanks
Joseph G.
AI: If he said that he multiplied the expression by $2x$, he misspoke. He multiplied it by $\frac{2x}{2x}$. Note that $\frac{2x}{2x}=1$, so that it’s entirely permissible to multiply by it, while multiplying by $2x$ would change the value.
He took a small shortcut. I’ll do it the long way first, putting the numerator over a common denominator and simplifying the resulting three-story fraction:
$$\begin{align*}
\frac{\frac12-\frac1x}{x-2}&=\frac{\frac12\cdot\frac{x}x-\frac1x\cdot\frac22}{x-2}\\\\
&=\frac{\frac{x}{2x}-\frac2{2x}}{x-2}\\\\
&=\frac{\frac{x-2}{2x}}{x-2}\\\\
&=\frac{\frac{x-2}{2x}}{x-2}\cdot\frac{2x}{2x}\tag{1}\\\\
&=\frac{x-2}{2x(x-2)}\;.
\end{align*}$$
The tutor merely observed that when the numerator is put over a common denominator, that denominator will be $2x$, and avoided the first few steps of my calculation by essentially going directly to the step marked $(1)$:
$$\begin{align*}
\frac{\frac12-\frac1x}{x-2}&=\frac{\frac12-\frac1x}{x-2}\cdot\frac{2x}{2x}\\\\
&=\frac{\left(\frac{x}{2x}-\frac2{2x}\right)2x}{2x(x-2)}\\\\
&=\frac{x-2}{2x(x-2)}\;.
\end{align*}$$
There’s no swapping of the denominator into the numerator: it just happens that when the numerator is simplified, the result is a fraction whose numerator is the same as the original denominator. |
H: Show that for all $\lambda \geq 1~$ $\frac{\lambda^n}{e^\lambda} < \frac{C}{\lambda^2}$
Show that for any $n \in \mathbb N$ there exists $C_n > 0$ such that for all $\lambda \geq 1$
$$ \frac{\lambda^n}{e^\lambda} < \frac{C_n}{\lambda^2}$$
I can see that both sides of the inequality have a limit of $0$ as $\lambda \rightarrow \infty$ since, on the LHS, repeated application of L'Hôpital's rule will render the $\lambda^n$ term as a constant eventually, while the $e^{\lambda}$ term will remain, and the RHS is obvious.
I can also see that the denominator of the LHS will become large faster than the RHS denominator, but I can't seem to think of anything that will show that the inequality is true for all the smaller intermediate values.
AI: HINT: The inequality $$\frac{\lambda^n}{e^\lambda} < \frac{C}{\lambda^2}$$ is equivalent to the inequality $\lambda^{n+2}e^{-\lambda}<C$. Consider the function $f(x)=x^{n+2}e^{-x}$ on the positive real line; does it have a maximum somewhere? |
H: Giving a hypersurface by higher degree polynomials
Suppose a hypersurface in $\mathbb{P}^n$ is given by an equation $F(Z) = 0$. It is easy to show that polynomials $Z_i F$ ($i = 0,\ldots,n$) give the same hypersurface, but I have trouble demonstrating that $K[Z_0,\ldots,Z_n]/I \cong K[Z_0, \ldots, Z_n]/I'$ where $I = (F)$, $I' = (Z_i F \mid i = 0,\ldots,n)$.
I tried to construct an isomorphism explicitly, but I got stuck: intuitively, $\varphi: G \mapsto \sum_{i=0}^n Z_i G$ seems like it could be right, it is defined correctly because if $G = FH$, then $\varphi(G) = \sum_i Z_i F H \in I'$.
But then I have to show that $\varphi$ is surjective, and I'm stuck: I can't find a way to show that for any polynomial $G$ we have $G \equiv \sum_i Z_i F \: (\mathrm{mod}\ I')$.
Am I on the right track? Is there an easier way to show that $K[Z]/I \cong K[Z]/I'$?
AI: The reason you have trouble showing your algebras are isomorphic is that they are not isomorphic!
For example take $n=1$ and $F=z_0$.
The algebras $K[z_0,z_1]/(z_0)$ and $K[z_0,z_1]/(z_0^2,z_1z_0)$ are not isomorphic because the first one is reduced (meaning its only nilpotent is zero) whereas in the second one the class of $z_0$ is nilpotent.
In scheme theory, a more advanced version of algebraic geometry, one would even say that the locus of zeros of the ideals $(z_0)$ and$(z_0^2,z_1 z_0)$ are different subschemes of $\mathbb P^1_K$, even though their underlying sets are the same, namely the singleton set $\lbrace [0:1]\rbrace$. |
H: What are the zero divisors of $C[0,1]$?
Suppose you have a ring $(C[0,1],+,\cdot,0,1)$ of continuous real valued functions on $[0,1]$, with addition defined as $(f+g)(x)=f(x)+g(x)$ and multiplication defined as $(fg)(x)=f(x)g(x)$. I'm curious what the zero divisors are.
My hunch is that the zero divisors are precisely the functions whose zero set contains an open interval. My thinking is that if $f$ is a function which is at least zero on an open interval $(a,b)$, then there exists some function which is nonzero on $(a,b)$, but zero everywhere else on $[0,1]\setminus(a,b)$. Conversely, if $f$ is not zero on any open interval, then every zero is isolated in a sense. But if $fg=0$ for some $g$, then $g$ is zero everywhere except these isolated points, but continuity would imply that it is also zero at the zeros of $f$, but then $g=0$, so $f$ is not a zero divisor.
I have a hard time stating this formally though, since I'm only studying algebra, and not analysis. Is this intuition correct, and if so, how could it be rigorously expressed?
AI: You’ve the right general idea, but it’s a little more complicated than that: the zero-set of the function $f$ might be a Cantor set, which has no isolated points.
Suppose that $Z=\{x\in[0,1]:f(x)=0\}$ contains no non-empty open interval; $Z$ is closed, so this says that $Z$ is nowhere dense in $[0,1]$. Let $V=[0,1]\setminus Z$: $V$ is a dense open set in $[0,1]$. Now suppose that $fg=0$; clearly we must have $g(x)=0$ for every $x\in V$. But $V$ is dense in $[0,1]$, and $g$ is continuous, so $g(x)=0$ for every $x\in[0,1]$, and $f$ is not a zero-divisor.
Your argument in the other direction is correct: if $Z$ contains a non-empty open interval $(a,b)$, just let
$$g(x)=\begin{cases}
0,&\text{if }x\in[0,1]\setminus(a,b)\\\\
x-a,&\text{if }a<x\le\frac12(a+b)\\\\
b-x,&\text{if }\frac12(a+b)\le x<b\;.
\end{cases}$$
Then $fg=0$, but $g$ is non-zero on $(a,b)$. |
H: How to "stretch" a procedural half-sphere texture on X and/or Y axis
I've implemented an Objective-C function to display the "height" of a half-sphere, with "1.0" being "full-height" and "0.0" being "no-height"
The sphere currently has a few parameters:
Center (x,y: real coordinates)
Radius (r: real value)
These values are currently being drawn in a texture with an independent size, provided that the minimum coordinate is (0,0) and the maximum is (1,1), which could be represented by a mathematical function like f(px, py), where px and py are the points where I want the height of the given half-sphere.
Now, I want to add two more parameters to the sphere, a "x-stretch" and a "y-stretch" (we can call them xs and ys), where I could deform the sphere in the X or Y axis.
The current function is like this (provided that the values of "x", "y" and "r" are known):
$ d(px, py) = \sqrt{(x-px)^2 + (y+py)^2} $ as being the distance from the center to the given point
$ f(px, py) = \sqrt{1 - \frac{d(px, py)}{r}} $ as being the value (from 0.0 to 1.0) of the half-sphere height.
Note that f is only calculated if d is smaller than the radius(r)
I want a way to add xs and ys in the above equations, but I don't really know how to get around that.
If it helps, here's the Objective-C function that "solves" the above problem:
- (CGFloat)sphereValueAtPoint:(NSPoint)point center:(NSPoint)sphereCenter radius:(CGFloat)radius {
CGFloat dist;
dist = sqrt(pow(point.x - sphereCenter.x,2) + pow(point.y - sphereCenter.y, 2));
if (dist > radius)
return(0.0);
return(sqrt(1 - pow(dist/radius, 2)));
}
I've also got a faster, optimized function, but it is not as easy to understand.
EDIT: Note that when xs and ys are both equal to 1.0, no scaling should be done. If either are GREATER than 1.0, the texture is stretched and if they are between 1.0 and 0.0, the texture is compressed. You can assume that xs and ys are always greater than 0.0.
EDIT2: As requested, I'm adding some images for clarification. White is 1.0, Black is 0.0 and the gray is in between (check tooltips).
AI: Ah, what you are doing is computing a height field describing a hemisphere. I would recommend using that phrase in the future, by the way; just calling it a "texture" conveys very little, and makes one think of mapping a textured pattern onto the surface of a sphere.
Your stated formula isn't quite correct, though the code is. Allow me denote the variable point as $(x,y)$ and the center of the hemisphere as $(c_x,c_y)$, as is more usual. The formula for the height of the hemisphere should be
$$f(x,y) = \sqrt{1 - \frac{d(x,y)^2}{r^2}} = \sqrt{1 - \frac{(x-c_x)^2}{r^2} - \frac{(y-c_y)^2}{r^2}}.$$
If you want to scale its size by $s_x$ and $s_y$, all you have to do is
$$f(x,y) = \sqrt{1 - \frac{(x-c_x)^2}{(s_xr)^2} - \frac{(y-c_y)^2}{(s_yr)^2}}.$$
For what it's worth, you're really modelling an ellipsoid with semi-principal axes of length $x_sr$, $y_sr$, and $1$. The implicit equation of an ellipsoid centered at the origin with semi-principal axes of length $a$, $b$, and $c$ is
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.$$
Substitute $a$, $b$, $c$ with their desired values, replace $x$ and $y$ by $x-c_x$ and $y-c_y$ to translate the origin, and solve for $f(x,y) = z$, and you get the formula above. |
H: Arcwise connected part of $\mathbb R^2$
Here's a question that I share:
Show that if $D$ is a countable subset of $\mathbb R^2$ (provided with its usual topology) then $X=\mathbb R^2 \backslash D $ is arcwise connected.
AI: HINT: Not only is $\Bbb R^2\setminus D$ arcwise connected, but you can connect any two points with an arc consisting of at most two straight line segments.
Suppose that $p,q\in\Bbb R^2\setminus D$. There are uncountably many straight lines through $p$, and only countably many of those lines intersect $D$, so there are uncountably many straight lines through $p$ that don’t hit $D$. Similarly, there are uncountably many straight lines through $q$ that don’t hit $D$. Can you finish it from here? |
H: Constructing a holomorphic function with some specific points zero/nonzero
Given $n \in \mathbb{Z}$, is it possible to construct a holomorphic function
$f : \mathbb{C} \rightarrow \mathbb{C}$ such that $f(n) \neq 0$, but
for any integer $m \neq n$ we have $f(m)=0$?
This is actually a homework problem in algebra which I reduced to this statement (in case it is correct).
AI: Here's a hint: $\dfrac{\sin z}{z}$. |
H: If an element has a unique right inverse, is it invertible?
Suppose $u$ is an element of a ring with a right inverse. I'm trying to understand why the following are equivalent.
$u$ has at least two right inverses
$u$ is a left zero divisor
$u$ is not a unit
If $v$ and $w$ are distinct right inverse of $u$, then $u(v-w)=0$, but $v-w\neq 0$, so $u$ is a left zero divisor. It's also clear that if $u$ is a left zero divisor, it cannot be a unit (else I could cancel $u$ from $ub=0$ to see $b=0$).
I'm having a heck of a time seeing why $u$ is not a unit implies $u$ has at least two right inverses. I tried the contrapositive, but saw no good approach. What am I missing?
AI: If $u$ has only one right inverse $v$, then $u(1-vu)=u-(uv)u=0$ hence $u(1-vu+v)=1$ and by uniqueness $1-vu+v=v$, so $1=vu$ and $v$ is a left inverse. Hence $u$ is a unit. |
H: Solving a biquadratic $x^{4}-2x^3 + x^2 - 2x +1 =0$
How do I find the roots of $$x^{4}-2x^3 + x^2 - 2x +1 =0$$
I am not able to find any roots by trial and error.
AI: Divide throughout by $x^2$.
Then you have $x^{2}-2x + 1 - \frac{2}{x} + \frac{1}{x^2} = 0$
You can re-write as $x^{2}+\frac{1}{x^2} - 2(x+\frac{1}{x}) + 1 =0$
Then use $x^{2}+\frac{1}{x^2} = (x+\frac{1}{x})^{2} -2$ and then reduce it to a quadratic by putting $y=x+\frac{1}{x}$. |
H: Find the square root of the polynomial
My question is:
Find the square root of the polynomial-
$$\frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left(\frac{x}y + \frac{y}x\right) + 3$$
AI: Let $t=\dfrac xy +\dfrac yx$.
Then $t^2=\dfrac {x^2}{y^2} +\dfrac{y^2}{x^2}+2$.
Your "polynomial" becomes finally an actual polynomial:
$$t^2-2-2t+3=t^2-2t+1=(t-1)^2=\left(\frac xy+\frac yx -1\right)^2.$$
So, the two square roots are $$\pm \left(\frac xy+\frac yx -1\right).$$ |
H: Angles of triangle inside a cricle
In the figure shown if area of circle with center o is 100pi and CA has length of 6 what is length of AB ?
I looked around on the web and cant seem to get an idea of what the angles AOC and OCA inside the triangle would be. Any suggestions on how I would go about determining them ?
AI: Given that the area of the circle is $100\pi$, its radius $OC$ is $10$. Using Pythagoras Theorem one finds that $OA=8$. From there you may work out $AB$ and use the inverse trigonometric functions to find the angles. |
H: Understanding adjoint functors
To understand adjoint functors I tried to look at an example. Can you tell me if the following is correct?
Before I give the example I'd like to recap the definition: Given two categories $C,D$ and two functors $F: C \to D$ and $G: D \to C$ we say that $F$ and $G$ are adjoint if we can give a natural transformation isomorphism $\eta$ such that for every pair of objects $A \in \text{Obj}(C)$, $B \in \text{Obj}(D)$ and morphisms $f: A \to A^\prime$ in $C$ and $g: B \to B^\prime$ in $D$ the following diagram commutes:
$$
\begin{matrix}
\operatorname{Hom}(FA, B) & \xrightarrow{\eta_{AB}} & \operatorname{Hom}(A, GB) \\
\left\downarrow{\scriptstyle{\operatorname{Hom}(F(f), g)}}\vphantom{\int}\right. & & \left\downarrow{\scriptstyle{\operatorname{Hom}(f, G(g))}}\vphantom{\int}\right.\\
\operatorname{Hom}(FA^\prime, B^\prime)& \xrightarrow{\eta_{A^\prime B^\prime}} & \operatorname{Hom}(A^\prime, GB^\prime)
\end{matrix}
$$
I'm not sure whether $F$ is left adjoint to $G$ or the other way around. Which one is the left adjoint here?
And: is there a better way to display this diagram?
Now the example: We claim that $F = - \otimes_R M$ is the (left?) adjoint of $G = \operatorname{Hom}_R(M, -)$ where $M$ is an $R$-module. To see this we give a natural isomorphism $\eta_{A,B}$ (where $A,B$ are $R$-modules and $C = D = R-\textbf{Mod}$) such that the following diagram commutes:
$$\begin{matrix}\textrm{Hom}(A \otimes M, B)&\xrightarrow{\eta_{AB}}&\operatorname{ Hom}(A, \operatorname{Hom}(M,B))\\
\left\downarrow{\scriptstyle{\textrm{Hom}(f \otimes id_M, g)}}\vphantom{\int}\right.&&\left\downarrow{\scriptstyle{\textrm{Hom}(f, G(g))}}\vphantom{\int}\right.\\
Hom(A' \otimes M, B')&\xrightarrow{\scriptstyle{\eta_{A'B'}}}&\textrm{ Hom}(A^\prime, \operatorname{Hom}(M,B'))\end{matrix}$$
We define $\eta_{AB}$ to be the map $$\eta_{AB}: (f: a \otimes m \mapsto b) \mapsto (g: a \mapsto f(a \otimes -))$$
Then the diagram above commutes. Is this correct?
And is the downarrow map really $\operatorname{Hom}(f \otimes id_M, g)$? I didn't know what else to put there. And did I get the left/right adjointness the correct way around?
AI: Given functors $F:\mathcal{D} \to \mathcal{C}$ and $G:\mathcal{D} \to \mathcal{C}$ with natural bijections $\text{hom}_\mathcal{C}(F(X),Y) \to \text{hom}_\mathcal{D}(X,G(Y))$ we say that $F$ is left adjoint to $G$. Thus tensor product is left adjoint to the Hom functor. I guess this naturally makes sense because in the defining equations the functor $F$ is on the left and $G$ on the right.
In terms of the proof, the map you have written down is correct, but of course one should actually show that everything works; i.e. that your map $\eta_{ab}$ is a bijective and that is is natural (i.e. that the diagram commutes).
It is also fairly standard to write $\text{Hom}_R(f \otimes \text{id}_M,g)$ as $(f \otimes \text{id}_M)^*$
Edit: Please see Bruno's comment below. For a map $f:A \to A'$ and a fixed $B$ it is normal to write $(f \otimes \text{id}_M)^*$. Otherwise $\text{Hom}_R(f \otimes \text{id}_M,g)$ seems to be the correct thing to write. (Note that in your question you need to change the $B$ in the lower left hand corner of the commutative diagram to a $B$') |
H: To show $A\implies B$, is that sufficient to show for all $C$ s.t. $C\implies A$ then $C\implies B$
my question is in the title:
to show $A\implies B$ is it enough to show for any $C$ such that $C\implies A$
we have $C\implies B$?
AI: Yes but that doesn't make it easier since you could choose $C = A$. |
H: Spanning a vector with no zero coordinates
Given a complex square matrix with 1-s on the main diagonal (and arbitrary values elsewhere), do its columns span a vector with no zero coordinate?
Clarification: What I'm asking is, given a complex matrix with 1-s on the main diagonal (and arbitrary values elsewhere), does its column space necessarily contain a vector in which each component is nonzero? Ofcourse, I know not all vectors in the column space are like that, I'm just asking if there is one.
AI: One (final) edit, down toward the bottom.
Take any vector $v$ in the column space. Suppose it has $m$ zeros in it. Take a column that has its diagonal 1 in one of the slots where $v$ has a zero. Adding any non-zero multiple of it to $v$ will create a vector with a non-zero entry in that slot, and there are only finitely many multiples you have to avoid so as not to make any new zeros. So you have reduced the number of zeros by one. Proceed by induction.
This proof fails over a field of non-zero characteristic.
EDIT: It's not just the proof that fails in positive characteristic, it's the result, at least in one case. Over the field of two elements, the matrix $$\pmatrix{1&0&1\cr1&1&0\cr0&1&1\cr}$$ satisfies the hypotheses, but its column space is just $$\lbrace(0,0,0),(1,1,0),(0,1,1),(1,0,1)\rbrace$$ I expect there are examples of other sizes and over other fields but I haven't found them yet.
MORE EDIT: It's cardinality, not characteristic, that matters. The proof in the first paragraph works over any infinite field; there are counterexamples over any finite field. Over the field of 3 elements: $$\pmatrix{1&2&2&0\cr0&1&2&2\cr1&0&1&2\cr1&1&0&1\cr}$$ Over the field of 5 elements, I haven't found any with so much symmetry [but see final edit, below], but here's one just to show it can be done: $$\pmatrix{1&4&4&4&4&0\cr0&1&2&4&3&3\cr1&0&1&3&2&3\cr1&2&0&1&3&4\cr1&3&2&0&1&2\cr1&1&3&0&2&1\cr}$$ Given a field of $q$ elements, here's how to construct a square matrix of order $q+1$ with ones down the diagonal, such that every element of the column space has a zero component:
Let the first column be $v=(1,1,\dots,1,0)$. Let the second column be $w=(0,1,a_1,a_2,a_3,\dots,a_r)$ where $1,a_1,\dots,a_{r-1}$ is any enumeration of the non-zero elements of the field (so $r=q-1$). Now for $k=3,4,\dots,q+1$ let column $k$ be the unique linear combination of $v$ and $w$ that has entry $k$ equal to 1 and entry $k-1$ equal to zero.
FINAL EDIT. Here's a solution for $q=5$ with the kind of symmetry I was hoping for. I reckon this solution makes it clear what to do for any prime $q$, perhaps for any $q$. $$\pmatrix{1&0&4&3&2&1\cr1&1&0&4&3&2\cr1&2&1&0&4&3\cr1&3&2&1&0&4\cr1&4&3&2&1&0\cr0&1&1&1&1&1\cr}$$ |
H: non-trivial common zero of polynomials
The following situation occurs in a proof that I would like to understand: we have polynomials $F_1,\ldots, F_N$ in $k[X_1,\ldots,X_M]$, where $k$ is of characteristic zero and algebraically closed. The polynomials $F_i$ are homogeneous of positive degree. And we have $M>N$.
Now I would like to conclude that these polynomials have a common zero $\neq (0,\ldots,0)$.
(the following seems to be wrong)
By Hilbert's Nullstellensatz it would be enough to show that the ideal $(F_1,\ldots,F_N)$ is proper. This should be easy (probably by looking at dimensions and using $M>N$?), but somehow I don't see a nice argument for that.
AI: Since the $F_i$ are homogeneous of positive degree $(F_1, \ldots, F_N) \subset (X_1, \ldots, X_M) \neq k[X_1,\ldots,X_M]$. So do we need $M > N$? |
H: Deriving the characteristic function for $N(0,2)$
Could someone please help me with an easy derivation of the characteristic function for a $N(0,2)$ distribution? Or a link to somewhere it is done.
AI: Once you know the characteristic function of $N(0,1)$, you can deduce the corresponding for $N(m,\sigma^2)$ for each $m$ and $\sigma$.
Let $\varphi$ the characteristic function of $N(0,1)$. We have
$\varphi(t)=\frac 1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{itx}e^{-x^2/2}dx$. We can take the derivative and find a differential equation satisfied by $\varphi$. Using the initial condition $\varphi(0)=1$ you can completely determine $\varphi$.
An other method is to write $\varphi(t)=\frac 1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-1/2(x-it)^2}dx$ and use a contour integral. |
H: Galois group of $x^6 + 3$ isomorphic to a copy of $S_3$ inside $S_6$
I have seen the the thread here related to the computation of the Galois group of the same polynomial. However, my question is not about the computation itself but about the group presentation of the Galois group. I will explain.
I have determined that the polynomial $x^6 +3 \in \Bbb{Q}[x]$ has Galois group of order 6. The splitting field is $\Bbb{Q}(a)$, where $a$ is a root of $x^6 + 3$. One can take $a = \sqrt[6]{3}\zeta$ where $\zeta = e^{2\pi i/6} = e^{\pi i/6}$.
Now I have determined the rest of the roots to be:
$$\begin{array}{ccccc} \alpha_1 &=& a &&& \alpha_4 = -\alpha_1 \\ \alpha_2 &=& \frac{a^4 + a}{2}= \zeta a &&& \alpha_5 = -\alpha_2 \\ \alpha_3 &=& \frac{a^4 - a}{2} = \zeta^2 a &&& \alpha_6 = - \alpha_3 \end{array}. $$
I have also computed some automorphisms of the Galois group, for example the automorphism $\tau : \alpha_1 \mapsto \alpha_4$ that has order 2, $\sigma : \alpha_1 \mapsto \alpha_2$ that has order 2 and $\rho : \alpha_1 \mapsto \alpha_3$ that has order 3. The presence of two automorphisms of order 2 tells me that the Galois group is isomorphic to $S_3$
However the problem now is if I want to identify my $\tau,\sigma$ and $\rho$ as cycles in $S_6$, I get the cycles $(14)$, $(12)$ and $(132)$. I don't think these cycles lie in the copy of $S_3$ inside of $S_6$; what am I misunderstanding here?
Thanks.
Edit: I made a mistake in the calculations. We actually need $\alpha_1 = a = \sqrt[3]{3}\zeta$ where $\zeta = e^{\pi i/6}$. I did not take a primitive 6-th root of unity earlier. Now if I write $a = \sqrt[6]{3}e^{\pi i/6}$, then $a^3 = \sqrt{3}i$ and so $\frac{1 + a^3}{2} = \frac{1 + \sqrt{3}i}{2} = \zeta^2$. So indeed with the redefined $\zeta$ and $a$, the equations now are
$$\begin{array}{ccccc} \alpha_1 &=& a &&& \alpha_4 = -\alpha_1 \\ \alpha_2 &=& \frac{a^4 + a}{2}= \zeta^2 a &&& \alpha_5 = -\alpha_2 \\ \alpha_3 &=& \frac{a^4 - a}{2} = \zeta^4 a &&& \alpha_6 = - \alpha_3. \end{array} $$
After all this mess, I have got the automorphisms $\sigma = (12)(45)(36)$ and $\gamma = (135)(246)$. We check that $\sigma\gamma = \gamma^2\sigma$. $\sigma\gamma = (12)(36)(45)(135)(246) = (16)(25)(34)$.
$\gamma^2\sigma = (153)(264)(12)(36)(45) = (16)(25)(34)$ so indeed $\sigma\gamma = \gamma^2\sigma$. Hence the Galois group has elements
$$\{1,\gamma,\gamma^2,\sigma,\sigma\gamma, \sigma\gamma^2\} = \{1, (135)(246),(153)(264),(12)(45)(36),(14)(23)(56),(16)(34)(25)\}.$$
AI: By Galois theory we know that there exists an automorphism $\sigma$ with the property
$$\sigma:\alpha_1\mapsto\alpha_2=\frac{\alpha^4+\alpha}2.$$
From this we can deduce that
$$
\sigma(a^3)=\left(\frac{a+a^4}2\right)^3=\frac{a^{12}+3a^9+3a^6+a^3}8.
$$
Here in the numerator $a^{12}=-3a^6$ and $a^9=-3a^3$, so this simplifies to $-a^3$ and hence
$$
\sigma(\zeta_6)=\sigma\left(\frac{1+a^3}2\right)=\frac{1-a^3}2=\zeta_6^{-1}.
$$
Therefore $\sigma(\alpha_2)=\alpha_1$, and by computing the images of the other roots we see that $\sigma$ corresponds to the permutation $(12)(36)(45)$.
Complex conjugation $\rho$ will also be an automorphism of your field, and by plotting the roots on the complex plane we see that $\rho=(16)(25)(34)$.
As products of these we get the other non-trivial automorphisms as permutations of roots:
$$
\rho\sigma=(153)(264),\quad \sigma\rho=(135)(246),
\quad\sigma\rho\sigma=\rho\sigma\rho=(14)(23)(56).
$$
The generators $\sigma$ and $\rho$ satisfy the relations $\sigma^2=\rho^2=(\sigma\rho)^3=1$, so they generate a copy of $S_3$. |
H: abel summable implies convergence
Prove that:
If $\sum c_n$ is Abel summable to $s$ and $c_n=O(\frac{1}{n})$ , then $\sum c_n $ converges to $s$.
"A series of complex number $\sum_{n=0}^{\infty} c_n $ is said to be Abel summable to $s$ if for every $0 \le r <1$ ,the series $A(r)=\sum_{k=0}^{\infty} c_kr^k$ converges ,and $\lim_{r\rightarrow 1^-} A(r)=s$."
AI: The following quote is from the book Jacob Korevaar: Tauberian Theory from Section 7.1: Hardy-Littlewood Tauberians for Abel Summability.
Littlewood [1911] answered Hardy's question whether the condition '$na_n \to 0$' in
Tauber's theorem could be relaxed to boundedness of the sequence $\{na_n\}$.
Theorem 7.1. (Littlewood)
$\sum a_n$ is Abel summable and $|na_n| < C$ $\Rightarrow$ $\sum a_n$ converges.
This 'big O-theorem' for Abel summability is more difficult than the earlier
results. The theorems in this section have attracted much interest and invited many
alternative proofs, frequently with Theorem 7.3 as the first step towards Theorem
7.1. Littlewood's original proof was rather complicated; his key tool was repeated
differentiation; cf. Section 17.
For comments on Littlewood's fundamental article of
1911, see his Collected Papers [1982]; for the history of his discovery, see Littlewood
[1953], A first simple proof for the Theorem was found by Karamata [1930a]; see
Section 11 for his method. A related more direct proof by Wielandt [1952] will be
described in Section 12.
Theorem 7.3. (Hardy and Littlewood) One has the following implication:
$\sum a_n$ is Abel summable and $s_n >-C$ $\Rightarrow$ $\sum a_n$ is Cesaro summable.
The following quote is from the book Boos: Classical and modern methods in summability
H. Tietz and K. Zeller drew my attention to a recent paper (cf. [240]) in
which they give a modification of Wielandt's well-known elegant proof of
the Hardy-Littlewood o-Tauberian theorems for the Abel method. This
is an elementary proof and I decided to use the material of this paper for
the important part of Section 4.4....
The shortest proofs are due to Wielandt [246] and Karamata [124] and, by modifying Wielandt's proof, to Tietz and Zeller [240]. We follow the lines of Tietz and Zeller's
proof which avoids integrals and is based on the Weierstrass approximation
theorem.
References mentioned above:
J. Karamata. Uber die Hardy-Littlewoodsche Umkehrung des Abelschen Stetigkeitssatzes. Math. Z. 32, 319-320 (1930). DOI: 10.1007/BF01194636, GDZ
J. E. Littlewood. The converse of Abel's theorem on power series. Proc. London Math. Soc. (2) 9, 434-448 (1911). doi: 10.1112/plms/s2-9.1.434
H. Tietz and K. Zeller. A unified approach to some Tauberian theorems of Hardy and Littlewood. Acta Comment. Uni. Tartu. Math. 2, 15-18 (1998).
H. Wielandt. Zur Umkehrung des Abelschen Stetigkeitssatzes. Math. Z. 56, 206-207 (1952). DOI: 10.1007/BF01175034, GDZ. |
H: Parametric representation of rectangular form in terms of parameters $\rho$ & $\theta$
I need to represent the cone $z=\sqrt{3x^2+3y^2}$ parametrically in terms of $\rho$ and $\theta$ where $(\rho,\theta,\phi)$ are spherical coordinates.
Attempt. I tried using: $$x=\rho\sin\phi\cos\theta \\y=\rho\sin\phi\sin\theta\\z=\rho\cos\phi$$
and $$\rho^2=x^2+y^2+z^2\\\cos\phi=\frac{z}{\sqrt{x^2+y^2+z^2}}\\\rho^2\sin^2\phi=x^2+y^2$$
I cannot find a way to get rid of $\phi$. Hints please.
This is the graph of it. It is a cone.
AI: If you have $z=\sqrt{3x^2+3y^2}$ and $z^2=3x^2+3y^2$, making the appropriate replacements in $\cos\phi=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$ gives
$$\cos\phi=\dfrac{z}{\sqrt{\frac{z^2}{3}+z^2}}$$
Can you see what to do next? (Hint: your Cartesian equation guarantees that $z$ is always positive, so $\sqrt{z^2}=?$) |
H: Find a maximum of complex function
I am trying to find a simple method that does not use the tools of advanced differential calculus to find following maximum, whose existence is justified by the compactness of the close ball $\Delta$ of $\mathbb C$ and continuity of the function $f:z \mapsto |z^3 + 2iz|$ from $\mathbb C$ to $\mathbb C$
$$ \large { \displaystyle \max_{z \in {\mathbb C},|z| \leq 1} |z^3 +2i z |} $$
Since : $$(\forall z \in \Delta) \quad f(z) \leq 3 $$
is obtained using triangular inequality, we can yet try to find some $z_0 \in {\mathbb C}$ such that $f(z_0)=3$
Does anybody have an idea?
Thanks.
AI: Hint: when do you get equality in the triangle inequality? |
H: Every $k$ vertices in an $k$ - connected graph are contained in a cycle.
Let $G$ be a $k$-connected graph. Meaning, $G$ has no fewer than $k$ vertices, and for every set of $k-1$ or fewer vertices, if we remove them from $G$, the graph stays connected (Of course, $G$ itself is also connected).
I want to prove that for any $k>1$, if $G$ is $k$-connected, then every set of $k$ vertices is contained in a cycle.
I have tried some ways - mainly using induction by removing one of the vertices of the set from the graph, and/or using Menger's theorem to construct the cycle. But I always encounter problems with making sure that the cycle I'm building deosn't have repeating edges etc.
Help would be greatly appreciated :)
Thanks!
AI: Here is a rough argument:
Take a cycle $C$ containing as many of the $k$ designated vertices as possible. If cycle has all the k designated vertices then you are done. Otherwise, By Menger's theorem , you can choose k paths from a missing vertex to the cycle $C$. The end points of the paths in $C$ divide $C$ into $k$ segments. One of the segments does not contain any of the designated vertices. Replace that segment by the two paths connecting the ends of the segment to the missing vertex. |
H: a Function with several periods
A periodic function is given by $ f(x+nT)=f(x) $, with 'n' an integer and T the period.
My question is if we can define a non-constant function with several periods; by that, I mean
$ f(x+T_{i})=f(x) $ with $ i=1,2,3,4,\dots $ a set of different numbers.
For example, a function that satisfies $ f(x+2)=f(x) $, as well as $ f(x+5.6)=f(x) $ and $ f(x+ \sqrt 2) =f(x) $, but $ f(x) $ is NOT a constant.
AI: There are the elliptic functions, which are functions that have two (in general) complex periods. It is a deep theorem of the theory of elliptic functions that the ratio of these two periods is necessarily not real. This was proven by C.G.J. Jacobi in 1835, who also showed that there cannot be a single-valued function of one variable that has more than two periods. |
H: In mean value theorem, does the mean value vary continuously?
Let $f\colon\mathbb R\to\mathbb R$ be continuously differentiable and let's say, for simplicity, that $f(0)=0$. Then by mean value theorem it's
$$f(x)=f'(\xi)\cdot x \,\text{ for some } \xi \in (0, x)$$
What I wondered is: What can we tell about the $\xi$ as we change $x$? My intuition says we should at least be able to find some $\xi\equiv \xi(x)$ that varies continuously with respect to $x$.
Or isn't this necessarily the case?
Thanks for any ideas.
AI: No. Here is a counterexample: Choose $f$ such that
$$f'(x)=\begin{cases} 2x-2 & x \le 1 \\ 0 & 1 \le x \le 2 \\ 2x-4 & x\ge 2\end{cases}$$
so
$$f(x)=\begin{cases} x^2-2x & x\le 1 \\ -1 & 1\le x \le 2 \\ x^2-4x+3 & x\ge 2\end{cases}$$
Then $f(3)=0$, and for $0<x<3$, $f(x)<0$, and $\xi$ has to be chosen to be $\le 1$, and for $x>3$, $f(x)>0$, so $\xi\ge 2$ for these $x$. |
H: Can't understand this solution.
I came across a problem which was already present on the internet.
If an arc with a length of $12\pi$ is $\frac{3}{4}$ of the circumference of the circle, what is the
shortest distance between the endpoints of the arc?
According to a certain site the solution is something like this
$$12\pi\left(\frac{4}{3}\right)=\text{circumference}=16\pi=2\cdot\text{radius}\cdot\pi$$
$$\text{radius}=8$$
$$x^2+y^2=64$$
Let $x=0$ for the first endpoint and let $y=0$ for the other, then find the two points
$(0,8)$ and $(8,0)$. Now find the distance between these two points:
$$d=((0-8)^2+(8-0)^2)^{1/2}=(128)^{1/2}\qquad \text{Ans}$$
I on the other hand decided to take my own approach since I couldnt figure out what happened after the radius
Step 1:
$12\pi = (3/4)$ (Circumference)
Cirum $= 16\pi$ so radius of the circle in question is $8$
Step 2:
Since $16\pi = 360^{\circ}$ so
$12\pi$ is $270^{\circ}$ .
Edited: From the suggestions i got from users here is how i would solve this:
Construct a line from the origin that goes to $270^{\circ}$ which is equal to radius and acts as a base and another line goes from origin to $360^{\circ}$ which acts as a perpendicualr then we calculate the hypotenuse (shortest distance). This definitely makes sense. But what if the question changes and angle is not $90^{\circ}$. I would appreciate it if someone could explain how to solve this using the distance formula as done above without the need of calculating $270^{\circ}$
AI: Draw a picture. By your calculation, the angle subtended at the origin by the arc is $270^\circ$. So going from one endpoint of the arc to the other the short way around, the angle is $360^\circ-270^\circ=90^\circ$.
Then by the Pythagorean Theorem the distance between the endpoints is $\sqrt{8^2+8^2}$, or more simply $8\sqrt{2}$.
Added: Let $A$ and $B$ be on a circle with radius $r$ and centre $O$. There are two arcs joining $A$ and $B$, the "short" one and the "long" one. Let $\theta$ be the angle subtended at $O$ by the short arc. So $\theta=\angle AOB$. We want to calculate $AB$.
Drop a perpendicular from $O$ to $AB$, meeting $AB$ at point $P$. Then $\angle AOP=\theta/2$. We have $\frac{PA}{OA}=\frac{PA}{r}=\sin(\theta/2)$, so $PA=r\sin(\theta/2)$ and therefore
$$AB=2r\sin(\theta/2).$$
Remark: The need to find the chord when one knows the arc is precisely what led to the development that ultimately gave us the modern sine function. It was initially driven by the needs of astronomy (and astrology). The history of trigonometry is very interesting. Many of the trigonometric identities we know and love, and other more obscure ones, were used to speed up "solving triangles," both plane and spherical. |
H: Solving a quadratic Inequality
My question is:
Solve $$9x-14-x^2>0$$
My answer is: $2 < x < 7$
Though I know my answer is right, I want to know in what ways I can solve it and how it can be graphically represented.
Thank you.
AI: Let's rearrange the inequality to get:
$x^2 - 9x + 14 < 0$
i.e.
$(x-7)(x-2) < 0$.
Now the product of two numbers is negative if and only if exactly one of the numbers is negative.
So we have that the inequality is satisfied whenever $x-7 < 0$ or $x-2 < 0$ but not both.
This happens when $2 < x < 7$. |
H: What's the name of this operator?
Let $f,g$ be functions in $C^A$ and $C^B$ respectively.
Let $\boxtimes:C^A \times C^B \to (C\times C)^{A \times B}$ s.t.
$f\boxtimes g(a,b)=(f(a),g(b))$
It seems not the tensor product, nor Cartesian product. Then can we call it direct product? But it seems the term 'direct product' often used on operator between structures.
AI: For the viewpoint of category theory, your map is just $f\times g$ -- it is the image of $f$ and $g$ under the product bifunctor $(-)\times(-)$.
More verbosely, if you compose $f$ and $g$ with the projection maps from $A\times B$, then you get maps $f\circ \pi_1: A\times B \to C$ and $g\circ \pi_2: A\times B \to C$, which factor through $C\times C$ by the universal property of the latter. The mediating morphism is excatly $f\times g$.
If you consider $\mathbf{Set}$ to be a monoidal category by declaring $\times$ to be $\otimes$, then $f\times g$ is indeed the tensor product $f\otimes g$.
On the other hand, in ordinary set theory, we usually identify a function with its graph: $$f=\{\langle x,f(x)\rangle\mid f(x)\text{ is defined}\},$$ and in that sense your $f\boxtimes g$ is of course not the cartesian product of $f$ and $g$. It is closely related though:
$$ f\boxtimes g = \{ \langle\langle a,b\rangle,\langle c,d\rangle\rangle \mid \langle\langle a,c\rangle,\langle b,d\rangle\rangle \in f\times g\}$$
which could be seen as a more vivid justification for Hurkyl's "transpose" terminology. |
H: some uniform continuous functions
We need to find which are uniform continuous (UC) on a) $(0,1)$ and b) $(0,\infty)$.
I have done, could you confirm me, if I am wrong any where?
$\frac{1}{(1-x)}$
$\frac{1}{(2-x)}$
$\sin x$
$\sin(1/x)$
$x^{1/2}$
$x^3$
1) is not UC on a) because limit does not exist when $x\rightarrow 1$, on b) it is discontinous (disco) at $x=1$
2) is UC on a) and it is disco at $x=2$ so not UC on b)
3) is UC on a) and also on b) as we can show by the inequality $|\sin x-\sin y|<|x-y|$
4) is not UC on a) as limit does not exist, and also not UC on b) I am not clear enough for this one.
5) is UC on a) and not UC on b) as derivative is not bounded near $0$
6) is UC on a) and not UC on b) as derivative is not bounded.
AI: I think you need to take another look at (5). If it fails to be UC on $(0,\infty)$ because the derivative is unbounded near $0$, it should fail to be UC on $(0,1)$ for the same reason. If it is UC on $(0,1)$, then the same reasoning should show that it is UC on $(0,\infty)$--since in this case, it is UC on $[1,\infty)$ and is continuous on $(0,\infty)$. I leave it to you to determine which of your arguments is correct.
As for the rest, it looks good, for the most part. One can make an even nicer argument for $\sin x$ being UC, since it is continuous (so UC) on $[0,2\pi]$ and periodic with period $2\pi$, so UC on $\mathbb{R}$, so UC on any subset of $\mathbb{R}$.
Something that may help you to clear up (4): A function $f:E\to\mathbb{R}$ (with $E\subseteq\mathbb{R}$) fails to be uniformly continuous if there is some $\varepsilon>0$ and some sequences $\{x_n\}$ and $\{y_n\}$ of points of $E$ such that $x_n-y_n\to 0$ and such that $|f(x_n)-f(y_n)|\geq\varepsilon$ for all $n$. Consider the points that cause $\sin(1/x)$ to maximize as $x\to 0$ and the points that cause $\sin(1/x)$ to minimize as $x\to 0$, and that should give you your sequences. |
H: Semicontinuity problem
If $A \subset \mathbb{R}^n$, is that claim true?
$$\chi_A \text{ is LSC} \Longleftrightarrow A\text{ is open}$$
And then how can I prove it?
($\chi_A$ is characteristic function : if $x \in A$ than $\chi_A =1$ otherwise zero.)
AI: True. In fact
\begin{equation}
\chi_{A} \mbox{is LSC} \ \Leftrightarrow \ \forall \ t \in \mathbb{R} \ \{x : \chi_A (x) \le t\}\ \mbox{is closed.}
\end{equation}
But
$$
\{x : \chi_A (x) \le t\} = \left\{
\begin{array}{rcl}
\mathbb{R^{n}} &if& t \ge 1\\
A^{c} &if& 0 \le t <1\\
\emptyset &if& 1 < t.\\
\end{array}
\right.
$$
$\Leftrightarrow A$ is open. |
H: a non separable metric space
Let $X$ be a metric space with discrete metric whose points are the positive integers. We have to show $C(X,\mathbb{R})$ is non separable. Well, what I have to do is to show $C(X,\mathbb{R})$ has no countable dense subset. I have no idea how to show that It has no countable as well as dense subset of $C(X,\mathbb{R})$, so far I guess to show it has non dense subset we need to find a sequence of functions $f_n\in C(X,\mathbb{R})$ which has some constant distance to the element of that set. Please, will any one help me to solve the problem?
AI: Hint:
Prove the follwing lemma. If $\{x_i:i\in I\}$ - is an uncountable family in metric space $(M,d)$ such that
$$
\exists \delta>0\quad\forall i\in I\quad\forall j\in I\quad (i\neq j\Longrightarrow d(x_i,x_j)>\delta)
$$
then $(M,d)$ is not separable.
Take a look at binary sequences. |
H: lower enveloper and upper enveloper
Let see $A \subset \mathbb{R}^n$, Define $\overline{f}(x) = \limsup_{y\to x} f(y)$
and $\underline{f}(x) = \liminf_{y\to x} f(y)$
$$\underline{\chi_A} =\chi_{A^o}, \text{ }\overline{\chi_A} =\chi_{A^-} $$
I wanna prove that. How can I approach?
AI: If $x \in A^{\circ}$ there exist $B_\varepsilon(x) \subset A$. Then $\inf \{\chi(x): x \in B_\varepsilon(x) - \{x\} \}= 1 = \chi(x)$. If $x \notin A^{\circ}$ there exist $z \in B_\varepsilon(x)- \{x\}$ such that $x \notin A^{\circ}$ and $\inf\{\chi(x): x \in B_\varepsilon(x)- \{x\} = 0 = \chi(x)$.
Hence $ \underline{\chi_{A}} = \chi_{A^{\circ}}$. |
H: Limit of a function tending to zero.
If $F(t)$ is twice differentiable at $x$ and $$G(h)=\max_{t\in(0,h)}\left[\frac{F'(x+t)-F'(x-t)}{2t}-F''(x)\right],$$ where $x$ is fixed; then how can we show that $\displaystyle\lim_{h\to 0}G(h)=0$.
AI: Hint:
$$\frac{F'(x+t)-F'(x-t)}{t}=\frac{F'(x+t)-F'(x)}{t}+\frac{F'(x-t)-F'(x)}{-t}.$$ |
H: Show that $\frac{(3^{77}-1)}{2}$ is odd and composite
The question given to me is:
Show that $\large\frac{(3^{77}-1)}{2}$ is odd and composite.
We can show that $\forall n\in\mathbb{N}$:
$$3^{n}\equiv\left\{
\begin{array}{l l}
1 & \quad \text{if $n\equiv0\pmod{2}$ }\\
3 & \quad \text{if $n\equiv1\pmod{2}$}\\
\end{array} \right\} \pmod{4}$$
Therefore, we can show that $3^{77}\equiv3\pmod{4}$. Thus, we can determine that $(3^{77}-1)\equiv2\pmod{4}$. Thus, we can show that $\frac{(3^{77}-1)}{2}$ is odd as:
$$\frac{(3^{77}-1)}{2}\equiv\pm1\pmod{4}$$
However, I am unsure how to show that this number is composite. The book I am reading simply states two of the factors, $\frac{(3^{11}-1)}{2}$ and $\frac{(3^{7}-1)}{2}$, but I do not know how the authors discovered these factors.
I'd appreciate any help pointing me in the right direction, thanks.
AI: Hint: $$a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)\,,\,\,\forall n\in\mathbb{N} \wedge \forall a\in\mathbb{R}$$
Added Of course, we also have in this case, applying the above: $$3^{77}-1=\left(3^7\right)^{11}-1=(3^7-1)\left(\left(3^7\right)^{10}+\left(3^7\right)^9+...+3^7+1\right)\,,\,etc.$$ and something similar can be done with $\,3^{77}=\left(3^{11}\right)^7$ |
H: Is $C^1(A)$ a Banach space?
Let $A \subset \mathbb R$ and consider the space $C^1(A)$. I am asked to prove that $( C^1(A), \Vert \cdot \Vert_{C^1(A)})$ is a Banach space, where
$$
\Vert f(x) \Vert_{C^1(A)} = \sup_{x \in A} \vert f(x) \vert + \sup_{x \in A} \vert f'(x) \vert
$$
First question: $A$ should be compact (or at least closed set), shouldn't it?
Secondly, how would you prove this? I've taken a Cauchy sequence, $(f_n)_{n \in \mathbb N} \subseteq C^1(A)$: if I fix $x \in A$, then I obtain two Cauchy sequences $(f_n(x))$ and $(f'_n(x))$ in $\mathbb R$ (?) so they converge to two numbers, $f(x)$ and $f'(x)$. The function $f$ that I obtain is the pointwise limit: how can I prove that this gives me exactly the $C^1(A)$ limit?
I've still one more question: is $( C^1(A), \Vert \cdot \Vert_{\infty})$ still a Banach space? I did not manage to find a counterexample...
Thanks for your help.
AI: If you already know that $C^0$ is a Banach space, and if $f_n$ is Cauchy in $C^1$, then you already do know that $f_n$ and $D_i f_n$ converge uniformly to continuous functions $f, g_i, 1\le i\le n$. You then only have to show that in this case $f$ is differentiable and $D_i f = g_i$. Which theorems do you know about sequences of differentiable functions and the differentiability of the limit?
$(C^1, ||.||_\infty)$ is not a Banach space, consider a sequence of smooth functions converging to $x\mapsto |x|$ on $[-1,1]$ (e.g. polynomials, which are known to be dense in $C^0$).
Since this is homework I deliberately ignore your question about compactness of $A$ and suggest you check corresponing statements for $C^0$ and see whether they carry over. (Note, though, that you usually define differentiability on open sets). |
H: mathematical notation for a logical statement
The proof of the statement below is a homework question, however I did not tag this question as such since I don't need the actual proof: I have already proved the statement wring and don't need a solution; my question here is strictly in regards to the mathematical notation I am using to represent a statement.
Consider the following statement:
"If $a$ and $b$ are integers, then there are integers $m$ and $n$, such that $a=m+n$ and $b=m-n$."
So my question is, will the following be a correct representation of the statement above:
$$(\forall a\in\mathbb{\mathbb{Z}},\forall b\in\mathbb{Z})(\exists m\in\mathbb{Z},\exists n\in\mathbb{Z}: a=m+n, b=m-n)$$
Is this a clear representation or is there a more common way of writing such statement that I don't know about?
Thank you!
AI: Yes, the notation is fine. One slight improvement would be $a=m+n\land b=m-n$.
However you need to understand that mathematical content is simpler when presented in words.
The statement you wrote:
If $a$ and $b$ are integers, then there are integers $m$ and $n$, such that $a=M=n$ and $b=m-n$.
If very clear. On the other hand when parsing (in my head) something of the form:
$$(\forall a\in\mathbb{\mathbb{Z}},\forall b\in\mathbb{Z})(\exists m\in\mathbb{Z},\exists n\in\mathbb{Z}: a=m+n\land b=m-n)$$
I have to step carefully and verify that what is written there makes sense, the result in my mind is exactly the original textual statement. So while it seems "cool" to write things like that (I used to think that in my freshman year) it does not make things clearer. Especially if the original text which you want to translate was very clear to begin with.
To top this off, I suggest you to go to the library and pick a book, or even journal, and see that most of the statements are written in words rather than mathematical notation. This is simply clearer. |
H: Probability; can't understand the maths
For a random variable $x$, define a probability distribution $p[x=n]=c (3^n/n!)$ when $x=0, 1, 2, \dots$ and $p(x)=0$ otherwise. Find the value of $c$.
My professor provided the solution
$$
\sum_{x=0}^\infty \ c\frac{3^n}{n!}=1
$$
so $c\;e^3 = 1$.
I can't understand why the summation has value $1$.
AI: I will change notation a little to make things look more familiar. For $n=0$, $1$, $2$, $3$, and so on, we have
$$P(X=n)=c\frac{3^n}{n!}.$$
The probabilities must add to $1$, so
$$\sum_{n=0}^\infty c\frac{3^n}{n!}=c\left(1+\frac{3}{1!}+\frac{3^2}{2!}+\frac{3^3}{3!}+\frac{3^4}{4!}+\cdots\right)=1.\tag{$1$}$$
Recall from calculus the power series (Taylor series, Maclaurin series) for the exponential function.
$$e^t=\sum_{n=0}^\infty \frac{t^n}{n!}=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots.$$
So (taking $t=3$) we may rewrite $(1)$ as
$$ce^3=1,$$
from which we conclude that $c=\frac{1}{e^3}=e^{-3}$. |
H: Different ways of computing $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$
Possible Duplicate:
On the sequence $x_{n+1} = \sqrt{c+x_n}$
I am wondering how many different solutions one can get to the following question:
Calculate $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$
Post your favorite solution please.
AI: If
$$\phi = \sqrt{1+\sqrt{1+...}}$$
then
$$\phi^2= 1+\sqrt{1+\sqrt{1+...}}=1+\phi$$
hence
$$\phi^2-\phi-1=0$$
$$\phi=\frac{1\pm\sqrt{5}}{2}$$
EDIT:
the sequence $\phi_n=\sqrt{1+\phi_{n-1}}$ is apparently increasing. $\phi_1<2$. Assume $\phi_{k-1}<2$, $k>2$. Then $1+\phi_{k-1}<3\implies \phi_k=\sqrt{1+\phi_{k-1}}<\sqrt{3}<2$, hence the sequence is bounded. |
H: How do I show that this function is always $> 0$
Show that $$f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +
\frac{x^4}{4!} > 0 ~~~ \forall_x \in \mathbb{R}$$
I can show that the first 3 terms are $> 0$ for all $x$:
$(x+1)^2 + 1 > 0$
But, I'm having trouble with the last two terms. I tried to show that the following was true:
$\frac{x^3}{3!} \leq \frac{x^4}{4!}$
$4x^3 \leq x^4$
$4 \leq x$
which is not true for all $x$.
I tried taking the derivative and all that I could ascertain was that the the function became more and more increasing as $x \rightarrow \infty$ and became more and more decreasing as $x \rightarrow -\infty$, but I couldn't seem to prove that there were no roots to go with this property.
AI: Hint:
$$f(x) = \frac{1}{4} + \frac{(x + 3/2)^2}{3} +\frac{x^2(x+2)^2}{24}$$ |
H: Solution to $x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$
Possible Duplicate:
Why does this process, when iterated, tend towards a certain number? (the golden ratio?)
Please post your favorit solution to the following
Compute $x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}$
Thank you
AI: Denote by $x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$. Because the fraction goes on forever, the denominator of the second term in the RHS is identical to $x$. Thus, we have $x=1+1/x$ or the same as $x^2=1+x$. Thus, $x=\frac{\sqrt 5-1}{2}$. |
H: $\ell_1$ distance of a point to a convex polygon
Let us have a set of $n$ points, $x_1, x_2, \ldots, x_n \in \mathbb{R}^d$, that form a convex polytope. And let us have a single point $x \in \mathbb{R}^d$ that is outside of the polytope. How can I compute the $\ell_1$ distance of the point to the polytope?
I wanted to do it similar to $\ell_2$ distance. That is, I check the distance for each line segment. Then I find the $\ell_2$ distance of a line segment to the point and get the minimum among the line segments. It can be done since I can find the smallest distance by using derivation. But, how to do it for $\ell_1$ distance?
AI: This is a linear program. First write the convex polytope as an intersection of linear inequalities and minimize $||x-y||_1$ for for $y$ in the polytope. So in 2D this is $|x_1-y_1|+|x_2-y_2|$. Now introduce two new variables $v_1,v_2$. Then minimize $v1+v2$ subject to $-v_1\leq x_1 - y_1 \leq +v1$, $-v_2\leq x_2 - y_2 \leq +v2$ and $y$ in convex polygon.
Determining the distance from a point to an (unbounded) line is also a linear program. Just minimize $||x-y||_1$ with $ay_1+by_2=c$ for some given $a,b,c$. You don't get such a nice formula for the closest point + distance as in the $l_2$ case. However this still has a derivative (that again isn't a single formula but a piecewise function). So if you program this it involves a few if statements... You go through the possibilities - in German there is the word Fallunterscheidung for that, but i'm not sure if there is a English equivalent.
EDIT: In fact the derivatives of functions involving absolute values may well be one formula, but don't expect them to be continuous everywhere. |
H: Computing: $\lim\limits_{n\to\infty}\left(\prod\limits_{k=1}^{n} \binom{n}{k}\right)^\frac{1}{n}$
I try to compute the following limit:
$$\lim_{n\to\infty}\left(\prod_{k=1}^{n} \binom{n}{k}\right)^\frac{1}{n}$$
I'm interested in finding some reasonable ways of solving the limit. I don't find any easy approach. Any hint/suggestion is very welcome.
AI: All the binomial coefficients except the last one are at least $n$, so the $n$th root is at least $n^{\frac{n-1}{n}}$, so the limit is infinity. |
H: Pigeon principle question: Nine points in a diamond
A diamond (a parallelogram with equal sides) is given, and its sides are 2 cm long. The sharp angels are 60 degrees. If there are nine points inside the diamond, prove that there must be two of them so that the distance between them is at most 1 cm.
Ideas where to start from?
AI: Hint: Split the diamond into eight congruent equilateral triangles. The triangles are your pigeon holes. |
H: Big Oh notation Question in calculus
In my text book, they state the following:
$$\begin{align*}f(x) &= (\frac{1}{x} + \frac{1}{2}) (x-\frac{1}{2}x^2+\frac{1}{3}x^3+O(x^4))-1& ,x \rightarrow 0\\&= 1-\frac{1}{2}x+\frac{1}{3}x^2+\frac{1}{2}x-\frac{1}{4}x^3+O(x^3)-1& ,x \rightarrow 0 \end{align*}$$
However, when I calculate this, I get $1-\frac{1}{2}x+\frac{1}{3}x^2+\frac{1}{2}x-\frac{1}{4}x^3+O(x^3)+\frac{O(x^4)}{2}-1$.
That $O(x^4)$ part disappears I guess, due to the big O notation. However, I cannot figure out why.
Furthermore, a few pages later, they say that $\lim_{x\rightarrow 0} O(x) = 0$. Which I do not really understand, since $O(x)$ defines a set of functions, no?
AI: For your first question, both $O(x^3)$ and $O(x^4)$ are error terms as $x$ approaches zero. Since $x^4$ goes to zero faster than $x^3$ as $x$ goes to zero, the larger error, $O(x^3)$ will subsume the smaller $O(x^4)$.
For your second question, you're correct in interpreting $O(x)$ as a set of functions. In this context, $O(x)$ is the set of all functions $f(x)$ for which $\mid f(x)\mid \le c\mid x\mid$, eventually, for some $c>0$ (which will depend on $f$). The limit $\lim_{x\rightarrow 0} O(x)$ is then interpreted to mean the limit of all such functions $f(x)$ as $x\rightarrow 0$, if it exists. It does in this case, since every $f\in O(x)$ satisfies $\mid f(x)\mid \le c\mid x\mid$, and so has limiting value $0$ as $x\rightarrow 0$. |
H: When was the significance of $i$ first noticed?
Complex analysis is an entire field of mathematics that focuses on the use of the complex constant $i$. When was the significance of $i$, an imaginary number, first noticed?
If I did not know some of the uses of complex analysis, I would likely believe, being the layman that I am, that $i$, as it is not a real number, would be fairly useless, almost like $0/0$. I would have trouble believing it had many uses, because it cannot be used to describe amounts of things like real numbers can (e.g. "I have 5 apples").
Why was any special attention paid to $i$ in the first place?
AI: See this for a history of complex numbers and here for some motivation. Also there's a website on answers and explanations that explain concepts such as imaginary numbers and infinity. |
H: How to evaluate $\lim\limits_{t\to 0} \frac{e^{-1/t}}{t}$?
How can I evaluate
\[
\lim_{t\to 0} \frac{e^{-1/t}}{t}\quad ?
\]
I tried to use L'Hôpital's rule but it didn't help me. Any hints are welcome. Thanks.
AI: Note that as $t \to 0$, $\exp(-1/t)$ tends 'faster' to $0$ than $1/t$ tends to $\infty$. To make this precise, let us proceed as follows.
First note that $\exp(x) \geq 1 + x + \dfrac{x^2}2$ for $x \geq 0$ and $\exp(x) \leq 1 + x + \dfrac{x^2}2$ for $x \leq 0$. For $t \geq 0$,
$$\dfrac{\exp(-1/t)}{t} = \dfrac1{t \exp(1/t)} \leq \dfrac1{t (1 + 1/t + 1/(2t^2))} = \dfrac1{t + 1 + 1/(2t)} = \dfrac{2t}{2t^2 + 2t + 1}$$
Hence, if we let $t \rightarrow 0^+$, we get that
$$0 \leq \lim_{t \rightarrow 0^+} \dfrac{\exp(-1/t)}{t} \leq \lim_{t \rightarrow 0^+} \dfrac{2t}{2t^2 + 2t + 1} = 0$$
Argue similarly, for $t \leq 0$. |
H: Summing the series $\sum u_{n}, u_{n}=\frac{a+n-1}{(a+1)...(a+n)}$
I am trying to sum the series
$$ \sum u_{n}$$
where $$ u_{n}=\frac{a+n-1}{\prod_{j=1}^n (a+j)}$$
$$ a>0$$
We have:
$$ \frac{a+n-1}{\prod_{j=1}^n (a+j)}=\sum_{k=1}^n\frac{b_k}{a+k} $$
$$ b_{k}=\frac{n-k-1}{\prod_{j=1,j\neq k}^n (j-k)}$$
$$ \sum_{n=1}^N u_{n}= \sum_{n=1}^N \sum_{k=1}^n\frac{n-k-1}{(a+k)\prod_{j=1,j\neq k}^n (j-k)}$$
...
AI: Since $(x)_n = \frac{ \Gamma(x+n)}{\Gamma(x)} $ your sum can be written as $$\sum_{k=1}^{\infty} \frac{(a+k-1)\Gamma(a+1)}{\Gamma(a+k+1)} .$$
Note $$\frac{a+k-1}{\Gamma(a+k+1)} = \frac{a+k}{(a+k)\Gamma(a+k)} - \frac{1}{\Gamma(a+k+1)}= \frac{1}{\Gamma(a+k)} - \frac{1}{\Gamma(a+k+1)}.$$
So $$\sum_{k=1}^{\infty} \frac{(a+k-1)\Gamma(a+1)}{\Gamma(a+k+1)} =1.$$ |
H: A particular case of Truesdell's unified theory of special functions
I'm reading through Clifford Truesdell's "An essay toward a unified theory of special functions", Princeton Univ. Press, 1948. All his exposition is based on the functional equation
$$\frac{\partial}{\partial z}\mathrm F(z,\alpha)=\mathrm F(z,\alpha+1)$$
He starts with
We are going to study functions $f (y, \alpha)$ satisfying a functional equation of the type
$$\frac{\partial}{\partial y} f (y, \alpha) = \mathrm A(y, \alpha) f (y, \alpha) + \mathrm B(y, a) f (y, \alpha+1 )$$
Then, we define
$$g\left( {y,\alpha } \right) = f\left( {y,\alpha } \right)\exp \left\{ { - \int\limits_{{y_0}}^y {\mathrm A\left( {v,\alpha } \right)dv} } \right\}$$
We verify that $g$ satisfies
$$\frac{\partial }{{\partial y}}g\left( {y,\alpha } \right) = g\left( {y,\alpha + 1} \right)B\left( {y,\alpha } \right)\exp \left\{ { - \int\limits_{{y_0}}^y {\left[ {A\left( {v,\alpha + 1} \right) - A\left( {v,\alpha } \right)} \right]dv} } \right\}$$
Thus we reduce the equation to
$$\frac{\partial }{{\partial y}}g\left( {y,\alpha } \right) = C\left( {y,\alpha } \right)g\left( {y,\alpha + 1} \right)\tag {1}$$
Now he states
In the case of nearly every special function that I know to satisfy an equation of type $(1)$, the coefficient $C(y, \alpha)$ is factorable, $C(y, \alpha)=Y(y)A( \alpha)$, so we asume
$$\frac{\partial }{{\partial y}}g\left( {y,\alpha } \right) = Y(y)A( \alpha)g\left( {y,\alpha + 1} \right)$$
Now he defines:
$$z:= \int_{y_1}^y Y(v) dv$$
and
$$F(z,\alpha ): = g\left( {y,\alpha } \right)\exp \left\{ {\mathop {\mathrm S}\limits_{{\alpha _0}}^\alpha \log {\text{A}}\left( v \right)\Delta v} \right\}$$
Now this is the operator that is troubling me
$$\mathop {\mathrm S}\limits_{{\alpha _0}}^\alpha h\left( v \right)\Delta v = \mathop {\lim }\limits_{k \to {0^ + }} \left\{ {\int\limits_{{a_0}}^\infty {h\left( v \right){e^{ - kc\left( v \right)}}dv} - \sum\limits_{m = 0}^\infty {h\left( {a + m} \right){e^{ - kc\left( {a + m} \right)}}} } \right\}$$
I can't find any reference to what $c(v)$ is. Is this known operator? What is $c$?
Anyways, I have a simple case I need to transform:
Let $$\mathrm F\left( {x,\alpha } \right) = \int\limits_0^x {{{\left( {\frac{t}{{t + 1}}} \right)}^\alpha }} \frac{{dt}}{t}$$
Then we have the functional equation
$$\frac{\alpha }{x} \mathrm F\left( {x,\alpha } \right) - \frac{\alpha }{x} \mathrm F\left( {x,\alpha + 1} \right) = \frac{\partial }{{\partial x}} \mathrm F\left( {x,\alpha } \right)$$
Following Truesdell's method, I define
$$\mathrm G\left( {x,\alpha } \right) = \frac{{\mathrm F\left( {x,\alpha } \right)}}{{{x^\alpha }}}$$
Then I have the functional equation
$$\frac{\partial }{{\partial x}} \mathrm G\left( {x,\alpha } \right) = - \alpha \mathrm G\left( {x,\alpha + 1} \right)$$
How can I transform it to the $\mathrm F$ equation using Truesdell's method?
The importance of the original $\mathrm F$ I define is that it can be used to show that
$$\log (1+x)=\sum_{n=1}^\infty \frac{1}{n}\left(\frac x {x+1} \right)^n\text{ ; for } x > -\frac 1 2$$
and maybe some other results can be derived. I still have a lot of exposition to read.
AI: The expression with the puzzling $\rm\:c(v)\:$ is Norlund's principal solution of the difference equation $\rm \mathop\Delta\limits_{\alpha}\ \mathop {\mathrm S}\limits_{{\alpha _0}}^\alpha h(v) dv = h(a).\: $ As Truesdell mentions in Appendix II, one can find an exposition of this in Chapter 8 of the classic The Calculus of Finite Differences by Milne-Thomson.
As I have mentioned previously here, Willard Miller showed that Truesdell's method is essentially Lie-theoretic. See his freely available book Lie theory and Special Functions, 1968. There he also shows that, similarly, the Schroedinger-Infeld-Hull ladder / factorization method (a powerful tool widely exploited by physicists to compute eigenvalues, recurrence relations, etc. for solutions of second order ODEs) is essentially equivalent to the representation theory of four local Lie groups. Nowadays it is a special case of Lie-theoretic symmetry methods used for separation of variables in partial differential equations (a major theme in the group-theoretic approach towards a unified theory of special functions). |
H: Automorphism on integers
Is multiplying by a constant m (integer) on group of set of all integers on addition an automorphism?
If so why does the 2nd example in http://en.wikipedia.org/wiki/Automorphism says that the unique non trivial automorphism is negation?
AI: Automorphism is a permutation of a set which respects some structure on the set. What structure? It varies. Automorphism is a general term and does not apply simply to groups, or rings.
In the context of $(\mathbb Z,+)$ as an additive group, we say that $f\colon\mathbb {Z\to Z}$ is an automorphism if:
$f$ is a bijection,
$f(m)+f(n)=f(m+n)$,
$f(0)=0$.
Now suppose that $f$ is an automorphism like that. Well, $f(0)=0$. If $f(1)=1$ then $f$ has to be the identity, because for $n>0$ we have $$f(n)=f(\underbrace{1+\ldots+1}_{n \text{ times}})=\underbrace{f(1)+\ldots+(1)}_{n \text{ times}}=\underbrace{1+\ldots+1}_{n \text{ times}}=n$$ and if $n<0$ then $$0=n+(-n)=f(n)+f(-n)=f(n)+(-n)$$ and therefore $f(n)=n$ as well.
Similarly if $f(1)=-1$ then $f(n)=-n$ for all $n$. So these are two automorphisms.
If $f(1)=n$ for some $n>1$, since $f$ is a bijection we have some $k\in\mathbb Z$ such that $f(k)=1$. We can write, if so, $f(1)=f(k)+\ldots+f(k)=f(k+\ldots+k)$. However $f$ is injective and therefore $k+\ldots+k=1$, which can only happen if $k=1$ and $n=1$. In a similar way we show that if $f(1)$ is negative then it has to be $-1$.
All that is well, but we can have more to say here. Suppose now we take $\mathbb Z$ as an ordered group. This means that now we add the ordering $\leq$ and require it plays nice with the addition, as it does with the standard addition and ordering.
In this case it is easy to show that $f(1)=1$ is the only automorphism possible. Why? Because $0<1$ implies that $f(0)<f(1)$, since now $f$ has to preserve the order too, so the first part shows that you cannot have a group automorphism of $\mathbb Z$ except $n\mapsto\pm n$, and if you want order preservation too you can only have the identity.
If on the other hand, you only want to consider $(\mathbb Z,\leq)$ as an ordered set, then you only wish to preserve $n<m\iff f(n)<f(m)$. In this case, it is not hard to show that for every $k\in\mathbb Z$ the function $f(n)=n+k$ is an automorphism.
To sum up, this is really about the structure you wish to preserve. The standard sets you know, $\mathbb {N,Z,R,Q,C}$ are all sets which have several different structures arising naturally from their properties as we know them. Each of these structures has a different property to preserve, and some would allow us rich automorphism groups, while other structures will limit us to have only a few automorphisms, if not one. |
H: Is there a simple way of arriving at this solution?
Suppose we are given the matrix $$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos(\omega t)& -\sin(\omega t)\\\sin(\omega t)& \cos(\omega t)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$
In other words the new coordinate system is a rotating coordinate system.
Is there a matrix-based way of finding $\begin{pmatrix}\dot{x}\\\dot{y}\end{pmatrix}$ and $\dot{x}^2+\dot{y}^2$ in terms of the new coordinates?
I can, by inspection, see that the latter should be $(\dot{x'}-\omega y')^2+(\dot{y'}+\omega x')^2$ but I would appreciate a way using matrices. Also if such a matrix to "change coordinates" exist, I would appreciate an intuitive explanation of what it is doing.
Many thanks!
AI: If we have $x' = A x$, then $\dot{x'} = \dot{A}x + A \dot{x} = \begin{pmatrix} \dot{A} & A\end{pmatrix}\begin{pmatrix} x \\ \dot{x}\end{pmatrix}$ and thereby $$\sum_k \left(\dot{x_k'}\right)^2 = \left(\dot{x'} \right)^T \left(\dot{x'} \right) = \begin{pmatrix} x \\ \dot{x}\end{pmatrix}^T \begin{pmatrix} \dot{A} & A\end{pmatrix}^T\begin{pmatrix} \dot{A} & A\end{pmatrix} \begin{pmatrix} x \\ \dot{x}\end{pmatrix}$$
$$\sum_k \left(\dot{x_k'}\right)^2 = \begin{pmatrix} x & \dot{x}\end{pmatrix} \begin{pmatrix} \dot{A}^T \dot{A} & \dot{A}^T A \\ A^T \dot{A} & A^TA \end{pmatrix}\begin{pmatrix} x \\ \dot{x}\end{pmatrix}$$
For $k=2$, you get the following.
\begin{align}
\dot{\begin{pmatrix}x'\\y'\end{pmatrix}} & =\dot{\begin{pmatrix}\cos(\omega t)& -\sin(\omega t)\\\sin(\omega t)& \cos(\omega t)\end{pmatrix}}\begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}\cos(\omega t)& -\sin(\omega t)\\\sin(\omega t)& \cos(\omega t)\end{pmatrix} \dot{\begin{pmatrix}x\\y\end{pmatrix}}\\
\begin{pmatrix}\dot{x'}\\\dot{y'}\end{pmatrix} & =\begin{pmatrix}-\sin(\omega t)& -\cos(\omega t)\\\cos(\omega t)& -\sin(\omega t)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}\cos(\omega t)& -\sin(\omega t)\\\sin(\omega t)& \cos(\omega t)\end{pmatrix} \begin{pmatrix}\dot{x}\\\dot{y}\end{pmatrix}\\
\begin{pmatrix}\dot{x'}\\\dot{y'}\end{pmatrix} & =\begin{pmatrix}-\sin(\omega t)& -\cos(\omega t) & \cos(\omega t)& -\sin(\omega t)\\\cos(\omega t)& -\sin(\omega t)&\sin(\omega t)& \cos(\omega t)\end{pmatrix}\begin{pmatrix}x\\y\\\dot{x}\\\dot{y}\end{pmatrix}\\
\left(\dot{x'} \right)^2 + \left(\dot{y'} \right)^2 &= \begin{pmatrix}\dot{x'}&\dot{y'} \end{pmatrix}\begin{pmatrix}\dot{x'}\\\dot{y'} \end{pmatrix}= \begin{pmatrix}\dot{x'}\\\dot{y'}\end{pmatrix}^T\begin{pmatrix}\dot{x'}\\\dot{y'}\end{pmatrix}
\end{align}
Now plugin the previous expression to get the expression in matrix form. |
H: Rotating about a point?
This is part of a program. I am trying to take an image at a given pair of coordinates and move it in a circular path around another given pair of coordinates.
The program is structured so that the image is redrawn 30 times per second, and with each redraw I need to update its position. That is to say, with each redraw I need to take its current coordinates [x,y], and update them to the next position along the circular path relative to the given center point. So I've been trying to determine the equation which would be a function of the present coordinates and the center coordinates and I haven't been able to do it no matter what.
Can someone help me out here? I also realize this should be a function of degrees. Say I move a fixed 5 degrees with each redraw. How do I calculate the new coordinates, relative to a given center?
AI: make a degree variable that holds the value of $\theta$. Every step do $\theta+=5\frac{\pi}{180}$ (to convert from degrees to radians). Then the coordinates of your picture are $$x=x_{center}+radius*\cos(\theta)$$ $$y=y_{center}-radius*\sin(\theta)$$
**minus on the $y$ because computer coordinate systems have $y$ increase downards, and this will produce clockwise rotation with increasing $\theta$ (i.e, angles work in the standard way). |
H: Banach-Tarski theorem without axiom of choice
Is it possible to prove the infamous Banach-Tarski theorem without using the Axiom of Choice?
I have never seen a proof which refutes this claim.
AI: The Banach-Tarski theorem heavily uses non-measurable sets. It is consistent that without the axiom of choice all sets are measurable and therefore the theorem fails in such universe. The paradox, therefore, relies on this axiom.
It is worth noting, though, that the Hahn-Banach theorem is enough to prove it, and there is no need for the full power of the axiom of choice.
More information can be found through here:
Herrlich, H. Axiom of Choice. Lecture Notes in Mathematics, Springer, 2006.
Schechter, E. Handbook of Analysis and Its Foundations. Academic Press, 1997. |
H: Two-sided ideal generated by subset of not commutative ring
Let $R$ be a non-commutative ring, and $S\subset R$ some subset. Let $I_S$ be the smallest two-sided ideal of $R$ such that $I_S\supseteq S$. Is it true that (if $R$ is unital ring) $I_S$ consist only of elements of the form:
$$\sum_{s\in S}\left(\sum_{k=1}^{N_s}x_{k}sy_{k}\right),$$
or (if $R$ isn't unit ring) $I_S$ consist only of elements of the form:
$$\sum_{s\in S}\left(m_{s}s+xs+sy+\sum_{k=1}^{N_s}x_{k}sy_{k}\right),$$
where $N_s\in\mathbb{N}$, $m_{s}\in\mathbb{Z}$ and $x, y, x_{k}, y_{k}\in R$.
Thanks a lot!
AI: Both statements are correct. The simplest way to prove this is to first note that
$$I_S = \bigcap I$$
where the intersection runs over all ideals $I$ of $R$ that contains $S$.
Next, prove that each of the elements in question is contained in any ideal $I$ that contains $S$. This will prove that the set of such elements is contained in the intersection.
Then prove that the set of all such elements is itself an ideal that contains $S$. This will prove that the intersection is contained in the set of all such elements. |
H: Essentially continuous a.e. characteristic function
I'm reading a few papers and trying to understand why one paper is supposed to be a strengthening of the other. The bulk of the papers didn't have to do with what follows, but at one point they imply that an a.e. essentially continuous characteristic function is automatically a characteristic function of a set which is a.e. an open set, and this is stronger. I'll make the definitions rigorous:
We say $x \in \mathbb{R}^n$ is a point of $\textit{essential continuity}$ for $f$ (a function on $\mathbb{R}^n$) if for every $\varepsilon > 0$, there exists a neighborhood $U_{x,\varepsilon}$ of $x$ such that $|f(y)-f(x)| < \varepsilon$ for a.e. $y \in U_{x,\varepsilon}$ (hence the "essentially" continuous). I have two main questions (and one side question which is less important):
Suppose the characteristic function $1_E$ is a.e. essentially continuous, that is, for a.e. $x$, $x$ is a point of essential continuity for $1_E$. Then why must $E$ be a.e. equivalent to an open set?
For this question, I was thinking of letting $\varepsilon < 1$ be fixed, and then the definition of essential continuity implies that for a.e. $x$, there is a neighborhood $U_x$ of $x$ such that for a.e. $y \in U$, we have $|1_E(x)-1_E(x)| < 1$. In particular, this means either both $x$ and $y$ are in $E$, or both are not in $E$. I tried to let my open set $U$ be the union of all the $U_x$ for $x \in E$ which are points of essential continuity, but the problem is there may be points $y$ in each $U_x$ which are not in $E$, and there could be uncountably many $U_x$. Thus the resulting $U$ is not in fact open.
Is there a good example of a set $E$ which is a.e. equivalent to an open set, but $1_E$ is not a.e. essentially continuous?
I don't really know where to start with this one, but I imagine we require a highly disjoint set, like some sort of fat cantor set. This particular set won't work though, since the fat Cantor set, say $K$, is not equivalent to an open set, but perhaps the complement $[0,1] \setminus K$ will work - I'm stuck on this too though.
I'm curious if we can strengthen 1 and 2: with the hypothesis of 1, why must both $E$ and $E^c$ be a.e. equivalent to open sets? And for 2, does there exist a set $E$ a.e. equivalent to an open set such that $E^c$ is also a.e. equivalent to an open set, and $1_E$ is not a.e. essentially continuous?
The motivation for 3 is that I believe the first paper's result can be strengthened from a statement about $E$ being a.e. equivalent to an open set, to a statement about both $E$ and $E^c$ being equivalent to an open set. If so, then in order for the second paper to be a true generalization, point 3 must hold.
Any help is appreciated, thanks.
AI: Re: 1. Taking $\bigcup U_x$ is a good idea, here $x$ should range over essential continuity points that belong to $E$. You have $U_x\subset E$ for all such $x$ (well not really :). Uncountability is not a problem: arbitrary union of open sets is open.
Re: 2. For any open set $E$, the function $1_E$ is not essentially continuous at any point of $\partial E$. Indeed, any neighborhood of a boundary point contains a ball $B\subset E$. So, if the boundary of an open set $E$ has positive measure (e.g., $E$ is the complement of a fat Cantor set), then $1_E$ is not essentially continuous a.e. |
H: Show that $\operatorname{int}(A \cap B)= \operatorname{int}(A) \cap \operatorname{int}(B)$
It's kind of a simple proof (I think) but I´m stuck!
I have to show that $\operatorname{int} (A \cap B)=\operatorname{int} (A) \cap \operatorname{int}(B)$.
(The interior point of the intersection is the intersection of the interior point.)
I thought like this:
Intersection: there's a point that is both in $A$ and $B$, so there is a point $x$, so $\exists ε>0$ such $(x-ε,x+ε) \subset A \cap B$.I don´t know if this is right.
Now $\operatorname{int} (A) \cap \operatorname{int}(B)$, but again with the definition ,there is a point that is in both sets,there's an interior point that is in both sets,an $x$ such $(x-ε,x+ε)\subset A \cap B$. There we have the equality.
I think it may be wrong. Please, I'm confused!
AI: If $x\in\mathrm{int}(A\cap B)$, then there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq A\cap B$. And since $A\cap B\subseteq A$ and $A\cap B\subseteq B$, then...
If $x\in\mathrm{int}(A)\cap\mathrm{int}(B)$, then there exists $\epsilon_1\gt 0$ such that $(x-\epsilon_1,x+\epsilon_1)\subseteq A$, and there exists $\epsilon_2\gt 0$ such that $(x-\epsilon_2,x+\epsilon_2)\subseteq B$. Can you find a single $\epsilon$ that works for both sets? Then what can you say about $(x-\epsilon,x+\epsilon)$? |
H: Find $A \in M_{2}(\mathbb{Z})$ such that $M_{2}(\mathbb {Z})=\{\sum a_{i}A^{i} : a_{i} \in \mathbb{Z}\}$
Question: Does there exist $A \in M_{2}(\mathbb{Z})$ such that every element of $M_{2}(\mathbb{Z})$ can be represented as a linear combination of powers of $A$ with integer coefficients? In other words,
$$\exists A \in M_{2}(\mathbb{Z}) \, \,s.t. M_{2}(\mathbb{Z})=\left.\left\{\sum_{i}a_{i}A^{i} \;\right| \; a_{i} \in \mathbb{Z}, a_j=0\text{ for almost all }j\right\}$$
Motivation: I want to construct a surjective ring homomorphism $\varphi: \mathbb{Z}\left[x\right] \rightarrow M_{2}(\mathbb{Z})$ by letting $\varphi: \displaystyle \sum_{i}a_{i}x^{i} \mapsto \sum_{i}a_{i}A^{i}$. The above property is sufficient for surjectivity.
What I've tried: It's sufficient to find four linear combinations that sum to $\left\{ \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] \right\}$. I haven't been able to find a magic $A$ that does this. For example, powers of $\left[ \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right]$ generate Fibonacci numbers, except that off diagonal elements are the same, so this will never work.
This is not homework. I have a feeling it may be impossible, but I'm not sure why. If for some reason it is impossible over $\mathbb{Z}$, I would be interested to know if it's possible over $\mathbb{Z}_{p}$ for a prime $p$. Thanks for any help.
AI: You cannot construct a surjective morphism $\mathbb{Z}[x] \to \mathcal{M}_2(\mathbb{Z})$ because the former is commutative and the latter is not. No need to appeal to Cayley-Hamilton. This is true when $\mathbb{Z}$ is replaced by any commutative ring. |
H: Non-zero prime ideals in the ring of all algebraic integers
Let $\mathcal{O}$ be the ring of all algebraic integers: elements of $\mathbb{C}$ which occur as zeros of monic polynomials with coefficients in $\mathbb{Z}$.
It is known that $\mathcal{O}$ is a Bezout domain: any finitely generated ideal is a principal ideal.
In addition, $\mathcal{O}$ has no irreducible elements, since any $x \in \mathcal{O}$ which is not a unit can be written as $x = \sqrt{x}\cdot\sqrt{x}$, where $\sqrt{x}$ is also not a unit in $\mathcal{O}$.
My question is:
Does $\mathcal{O}$ have any prime ideal other than $(0)$?
AI: Yes, $\mathcal{O}$ has lots of prime ideals (the axiom of choice is equivalent to every non-unit in any commutative ring being contained in a maximal ideal).
A concrete example is not so easy, but the point is this: for any finite field extension $L/\mathbb{Q}$, pick a prime $\mathfrak{p}_L$ of the ring of integers $L$, and do so compatibly, i.e. if $L \subset L'$ we need $\mathfrak{p}_{L'} \cap L = \mathfrak{p}_L$. The union of all the $\mathfrak{p}_L$ will then be a prime. |
H: gradient flow and what is, for example, $L^2$ gradient?
Am I right that the gradient flow of a functional $E$ is
$$f_t = -\nabla E(f).$$
Solving this for $f$ gives you a minimiser of $E$ in some way?
Here the $\nabla$ denotes the gradient or the first variation or Gateaux derivative or whatever is appropriate.
What is meant by "$L^2$ gradient (flow)" or "$H^{-1}$ gradient (flow)?"
Thanks
AI: If $E$ is a Frechet differentiable functional on a Hilbert space, the gradient of $E$ at $u$ is the element $w$ of the Hilbert space satisfying $E'(u)v = (w,v)$ for all $v$ in the Hilbert space. The $E$ gradient flow starting of $u$ is the solution $\eta(t)$ of the diffential equation $\frac{d}{dt}I(\eta) = -\nabla E(\eta), \eta(0) = u$. Sorry, I don't know how to make the accent mark in Frechet, nor do I know how this works more generally in Banach spaces. If $\eta(t)$ converges to some $w$ as $t\to \infty$, then that $w$ is a critical point of $E$ (not necessarily a minimizer). If someone says ''$L^2$ gradient flow'' or ''$H^{-1}$ gradient flow" they mean that the Hilbert space is $L^2$ or $H^{-1}$, but the gradient flow depends on the functional. |
H: Kaplansky's theorem of infinitely many right inverses in monoids?
There's a theorem of Kaplansky that states that if an element $u$ of a ring has more than one right inverse, then it in fact has infinitely many. I could prove this by assuming $v$ is a right inverse, and then showing that the elements $v+(1-vu)u^n$ are right inverses for all $n$ and distinct.
To see they're distinct, I suppose $v+(1-vu)u^n=v+(1-vu)u^m$ for distinct $n$ and $m$. I suppose $n>m$. Since $u$ is cancellable on the right, this implies $(1-vu)u^{n-m}=1-vu$. Then $(1-vu)u^{n-m-1}u+vu=((1-vu)u^{n-m-1}+v)u=1$, so $u$ has a left inverse, but then $u$ would be a unit, and hence have only one right inverse.
Does the same theorem hold in monoids, or is there some counterexample?
AI: Let $S$ be the semigroup of functions from $\mathbb N=\{z\in \mathbb Z|z\geq 0\}$ to itself, with the composition written traditionally: $(f\circ g)(x)=f(g(x)).$
Let $f\in S$, $f(0)=f(1)=0$ and for $n\geq 2,\,f(n)=n-1.$ Suppose $f\circ g=\operatorname{id}$. Then for $n\geq 1$, we must have $g(n)=n+1$. However, $g(0)$ can be chosen to be either $0$ or $1$ and the equality holds. |
H: question about cofinality and function
In a paper, I want to prove a result that seems to me general.
Let $g:\delta\longrightarrow cf(\lambda)$ where $\delta$ is an ordinal less than $\lambda^+$ and $\lambda$ a cardinal. Suppose that $\forall i<cf(\lambda)$, $g^{-1}[i]$ is not cofinal in $\delta$. Does one have $cf(\lambda)=cf(\delta)$ ?
Remark : I think the notation $g^{-1}[i]$ means the inverse image that is $\{\alpha<\delta : g(\alpha)<i\}$.
We have $\delta=g^{-1}[cf(\lambda)]=\bigcup_{i<cf(\lambda)}g^{-1}[i]=\sup_{i<cf(\lambda)} \delta_i$ where $\delta_i$ is the least upper bound of $g^{-1}[i]$ in $\delta$ so $cf(\delta)\leq cf(\lambda)$. I have some difficulty for the other direction. Could somebody help me ?
Thanks.
AI: Let $\kappa=\operatorname{cf}\delta$, and let $\langle\alpha_\xi:\xi<\kappa\rangle$ be a cofinal sequence in $\delta$. For each $\beta<\operatorname{cf}\lambda$ there is a $\xi(\beta)<\kappa$ such that $g(\alpha_{\xi(\beta)})\ge\beta$, since $g^{-1}[\beta]$ is not cofinal in $\delta$. If $\kappa<\operatorname{cf}\lambda$ there is a $\xi<\kappa$ such that $B=\{\beta<\operatorname{cf}\lambda:\xi(\beta)=\xi\}$ is cofinal in $\operatorname{cf}\lambda$. But then $g(\alpha_\xi)\ge\sup B=\operatorname{cf}\lambda$, which is impossible. Thus $\kappa\ge\operatorname{cf}\lambda$, and since you already have $\kappa\le\operatorname{cf}\lambda$, it follows that $\operatorname{cf}\delta=\kappa=\operatorname{cf}\lambda$. |
H: Represent the following set of points in the XY plane
Represent the following set of points in the XY plane:
{ (x,y) | |x|=1 }
{ (x,y) | |x| is less than or = 1 }
AI: HINT: Note that your set imposes no restriction at all on the $y$-coordinate. If you find a point in the set, every other point with the same $y$-coordinate will also be in the set. Now, what does $\{x\in\Bbb R:|x|\le 1\}$ look like as a subset of the real line?
Added: Your region is the blue stripe in the picture below; it extends infinitely far up and down. |
H: Prove that $\mathcal{W}(\mathbb{R})$ is not metrizable.
A colection $\mathcal{V}$ of open sets in a topological space $X$ is called a Fundamental System of Open Neighborhoods (FSON) of a point $x\in X$ when:
$\forall\ V\in\mathcal{V}$ we have that $x\in V.$
If $A\subset X$ is open set containing $x$ then $\exists\ V\in\mathcal{V}$ such that $V\subset A$.
For example in any metric space the set $\{B(x,\frac{1}{n}); n\in \mathbb{N}\}$ is a FSON of $x$.
Let $\mathcal{W}(\mathbb{R})$ the set of continuous functions $f:\mathbb{R}\to\mathbb{R}$ with a topology defined by the following way: Let $f\in\mathcal{W}(\mathbb{R})$ and a continuous positive function $\varepsilon:\mathbb{R}\to\mathbb{R}^+$ and define the set $B(f,\varepsilon)=\{g\in\mathcal{W}(\mathbb{R}); |g(x)-f(x)|<\varepsilon(x)\ \forall x\in\mathbb{R}\}$ which is a basis for the topology.
Prove that $\mathcal{W}(\mathbb{R})$ is not metrizable.
Hint: Show that $f=0$ doesn't have a countable FSON.
Attempt:
Suppose that we have a countable FSON called $\mathcal{R}_{0}$ of $f=0$ then $\mathcal{R}_{0}=\{A_i\}_{i=1}^{\infty}$. Then for each $A_i$ I can choose $\varepsilon_i$ such that $0\in B(0,\varepsilon_i)=\{g\in\mathcal{W}(\mathbb{R}); |g(x)|<\varepsilon_i(x)\ \forall x\in\mathbb{R}\}\subset A_i$. Then I need to find a positive function $\varphi:\mathbb{R}\to\mathbb{R}^+$ such that for all $i\in\mathbb{N}$ we have that $B(0,\varepsilon_i)\nsubseteq B(0,\varphi)$. If were $\varepsilon_i(x)=\frac{1}{n}\ \forall x \in\mathbb{R}$ I have that $\varphi(x)=\frac{1}{1+x^2}\ \forall x \in\mathbb{R}$ satisfies the condition. But I don't know how to find $\varphi$ in the general case.
AI: For each $n\in\Bbb Z^+$ choose a positive number $r_n<\epsilon_n(n)$. Now construct a continuous $\varphi:\Bbb R\to\Bbb R^+$ such that $\varphi(n)=r_n$ for each $n\in\Bbb Z^+$. (A piecewise linear function is continuous and easy to construct.) Then $B(0,\varphi)$ does what you want. |
H: Building the integers from scratch (and multiplying negative numbers)
Now I understand that what I am about to ask may seem like an incredibly simple question, but I like to try and understand math (especially something as fundamental as this) at the deepest level possible. And for the life of me, I can't shake this feeling I have that something is not quite right.
Let me begin:
First of all, I was reading Terry Tao's discussion about the construction of the standard number system and was very pleased with the way in which $\mathbb{C}$ can be systematically constructed from the natural numbers through a system of homomorphisms (as a direct limit): $$\mathbb{N} \hookrightarrow \mathbb{Z} \hookrightarrow \mathbb{Q} \hookrightarrow \mathbb{R} \hookrightarrow \mathbb{C}$$
In particular, he talks about how the integers can be constructed as the space of equivalence classes of formal differences between natural numbers, where $$[a-b]\sim[c-d ] \iff a+d=b+c$$ with $a,b,c,d\in\mathbb{N}.$ He then goes on to say "with the arithmetic operations extended in a manner consistent with the laws of algebra."
Now, for $(\mathbb{N},+,\cdot)$ addition is straight forward, and $$n\cdot k:=\underbrace{k+\cdots+k}_{n\;\text{times}}$$ is well defined with $n\cdot k=k\cdot n$. Moreover, the distributive law, $n\cdot(m+k)=n\cdot m+n\cdot k$, also holds.
However, when I want to construct $(\mathbb{Z},+,\cdot)$ as above, I run into some trouble. Addition of two equivalence classes seems pretty straightforward, given by $$[a-b] + [c-d] := [ (a+c)-(b+d) ],$$ and is well defined. This makes sense as we simply "add up" the positive and negative amounts together. This definition also behaves nicely with the map $\varphi:\mathbb{N}\hookrightarrow\mathbb{Z}$, given by $n\mapsto[ n-0 ]$; with $\varphi(n+k)=\varphi(n)+\varphi(k)$, where the second addition is the addition of equivalence classes. Moreover, we have commutativity of addition; the existence of an additive identity, namely $[0-0]$; and the existence of additive inverses, $[a-b]+[b-a]\sim[0-0].$
But now, when I try to define the multiplication of classes, I am unsure how to proceed:
If we consider our usual algebraic rules, we get that $$(a-b)\cdot(c-d)=a\cdot c-a\cdot d-b\cdot c+b\cdot d,$$ which might lead us to define the multiplication of two equivalence classes as $$[ a-b ]\cdot[ c-d ]:=[ (a\cdot c+b\cdot d)-(a\cdot c+b\cdot c) ].$$ Now it's easy to check that this is indeed well defined, and obeys the distributive law with the definition of addition we've given above. However I feel that we've simply gone in a circle (logically) as we've assumed a priori that $(-b)\cdot c=-(b\cdot c)$ and $(-b)\cdot(-d)=-(b\cdot d)$.
Now, I am very familiar with the fact that if your set is a ring (with unity) then these results (particularly that negative times negative is positive) come as simply a result of playing around with additive inverses and the distributive laws (see here). However my problem is that we've only gotten this ring structure on our set of equivalence classes by appealing to the ring structure of the integers, which is exactly the thing we are trying to construct from scratch! And I do not want to simply define multiplication in this way "because it works"; it has always been my feeling that results like $(-1)\cdot(-1)=1$ should come as consequences of the structure, and not as properties we impose.
So I'm curious if there is a way around this issue. Is this particular definition of multiplication the only one that:
is well defined?
satisfies the distributive laws?
is associative?
has a multiplicative identity? $$[1-0]\cdot[a-b]=[a-b],\;\text{for all $a,b\in\mathbb{N}$}$$
gives a homomorphism which splits over multiplication? $$\varphi:(\mathbb{N},+,\cdot)\hookrightarrow(\mathbb{Z},+,\cdot),\; n\mapsto[ n-0 ]$$ $$\varphi(n\cdot k)=\varphi(n)\cdot\varphi(k)=[ n-0 ]\cdot[ k-0 ]$$
If our deffiniton of of multiplication has all these properties, then we'll have a ring structure on our set of equivalence classes, and all the familiar properties of the integers will be established. However, it is not immediately clear to me that this is our only option. Can someone shed some light on this?
AI: You've answered your own question. You say you're familiar with the fact that $(-b)c = -bc$ and $b(-d) = -bd$ in a ring. Now, if you write down a definition of multiplication that satisfies all of those properties, you get a ring structure on the set of equivalence classes, and $(-b)c = -bc$ and $b(-d) = -bd$ in a ring...!
Perhaps some extra generality will make this argument seem less circular. You have a rig $R$. This rig has an underlying additive monoid $M = (R, +)$. This abelian monoid has a Grothendieck group $G(M)$, which is the universal group into which $M$ maps. More precisely, there is a forgetful functor $F : \text{Ab} \to \text{CMon}$ from the category of abelian groups to the category of commutative monoids, and the Grothendieck group functor is its left adjoint. $G(M)$ consists of equivalence classes of formal expressions $m - n$ where $m, n \in M$ with addition defined as expected.
Now you want to put a ring structure on $G(M)$ compatible with the multiplication on $R$. It turns out that what you are actually constructing is the universal ring into which $R$ maps; more precisely, there is a forgetful functor from the category of rings to the category of rigs and you are constructing its left adjoint. Well, in any such ring we need
$$(a - b)(c - d) = ac - bc - ad + bd$$
as you well know, so this is the unique possible ring structure on $G(M)$ compatible with the multiplication on $R$ (and distributivity, inverses, etc.) if it is well-defined. This is a general property of adjoint functors (they are unique up to unique isomorphism if they exist). The only thing left is to verify that it actually works.
Another way to say the above is the following. Let $R$ be a rig and let $S$ be a ring, and let $\phi : R \to S$ be any rig homomorphism whatsoever. Then
$$(\phi(a) - \phi(b))(\phi(c) - \phi(d)) = \phi(ac) - \phi(bc) - \phi(ad) + \phi(bd).$$
There is nothing circular about this because $S$ is a ring and the above computation takes place in $S$. |
H: Solving a quadratic?
Lets say you have a quadratic that you factor into root form. To solve for the roots, you let the $y$ value be $0$:
$0 = (x_1-h)(x_2-k)$
Following this you would divide both sides by one of the multipliers.
This would leave you with $x_1-h =0$ and therefore $x_1 =h$ and $x_2-k = 0$ and therefore $x_2 = k$.
However since the product of both the multipliers is zero, one of them has to be zero (the product of two or more numbers can't be zero unless atleast one of them is zero). Therefore, aren't you essentially dividing by zero by dividing by one of the multipliers?
For example, $x_1-h = 0$. To solve for $x_2$, you would do the following:
$0 = (x_1-h)(x_2-k)$
$\frac{0}{x_1-h} = \frac{(x_1-h)(x_2-k)}{(x_1-h)}$
$0 = x_2 - k$
$x_2 = k$
However, since $x_1 -h = 0$, didn't you essentially divide by zero to solve for $x_2$? Wouldn't this classify as undefined behaviour?
AI: I wouldn’t divide by one of the factors. I would simply observe that $(x-h)(x-k)=0$ if and only if at least one of $x-h$ and $x-k$ is $0$: $(x-h)(x-k)=0$ if and only if $x-h=0$ or $x-k=0$, which is the case if and only if $x=h$ or $x=k$. At no stage of the reasoning is any division required.
Added: If you want to go to the trouble of thinking of it in terms of division, you can, but you have to be careful. Start with $(x-h)(x-k)=0$. You can then argue that if $x-h\ne 0$, then you can divide through by it to get $x-k=\frac0{x-h}=0$ and hence $x=k$. If, on the other hand, $x-h=0$, then $x=h$. Thus, either $x=k$ or $x=h$. |
H: Derivative of an implicit function
I am asked to take the derivative of the following equation for $y$:
$$y = x + xe^y$$
However, I get lost. I thought that it would be
$$\begin{align}
& y' = 1 + e^y + xy'e^y\\
& y'(1 - xe^y) = 1 + e^y\\
& y' = \frac{1+e^y}{1-xe^y}
\end{align}$$
However, the text book gives me a different answer.
Can anyone help me with this?
Thank you and sorry if I got any terms wrong, my math studies were not done in English... :)
AI: You can simplify things as follows:
$$y' = \frac{1+e^y}{1-xe^y} = \frac{x+xe^y}{x(1-xe^y)} = \frac{y}{x(1-y+x)}$$
Here in the last step we used $y=x+xe^y$ and $xe^y=y-x$. |
H: Blow-up along an ideal sheaf
Let $k^2=\operatorname{Spec} \; k[x,y]$ where $k$ is an algebraically closed field. Let $\mathcal{I}$ be the ideal sheaf defined by $(x,y)$. Then
$$
Bl_{\mathcal{I}}k^2
$$
is covered by two open charts $\operatorname{Spec} \; k[x, y/x] \cup \operatorname{Spec}\; k[y,x/y]$.
Q1: Why can each chart be described by
$$
\operatorname{Spec} \; k[x,y][t]/(tx-y) \mbox{ and } \operatorname{Spec} \; k[x,y][t]/(ty-x)?
$$
Q2: Isn't $Bl_{\mathcal{I}}k^2=\operatorname{Proj}(\oplus_{i\geq 0} (Rx\oplus Ry)^i t^i)$?
Q3.a: Now let $k^3 =\operatorname{Spec} \; k[x,y,z]$ with $\mathcal{I}$ being defined by $(x,y,z).$ Then
isn't
$$
Bl_{\mathcal{I}}k^3 = \operatorname{Spec} \; k[x,y/x,z/x] \cup \operatorname{Spec}\; k[y,x/y,z/y] \cup \operatorname{Spec}\; k[z,x/z,y/z]?
$$
Q3.b: How can one see that the charts
$$
\operatorname{Spec}\; k[x,y,z][t_1, t_2]/(t_1 x - y, t_2 x-z) $$
$$\operatorname{Spec}\; k[x,y,z][t_1, t_2]/(t_1 y - x, t_2 y-z)
$$
$$\operatorname{Spec}\; k[x,y,z][t_1, t_2]/(t_1 z - x, t_2 z-y)
$$
also cover $Bl_{\mathcal{I}}k^3$?
$$
$$
AI: Let $A=k[x,y]$, then $$k[x,y/x]=k[x,y,y/x]=A[y/x]=A[t]/(y-tx).$$ For the last equality, see the comments below. Similar equalities apply to the other $k[y,x/y]$.
By definition the blowing-up of $k^2$ along the origin should be
$$\operatorname{Proj}(\oplus_{i\geq0}I^i)\longrightarrow k^2,$$
where $I=(x,y)$.
The blowing-up of the affine space $k^n$ along the origin is (see Liu,8.1.13)
$$\operatorname{Proj}(k[t_1,...,t_n][T_1,...,T_n]/(t_iT_j-t_jT_i))\longrightarrow k^n.$$
For your $n=3$ case, we have
$$D_+(T_1)=\operatorname{Spec}(k[t_1,t_2,t_3][\frac{T_2}{T_1},\frac{T_3}{T_1}]/(t_2-t_1\frac{T_2}{T_1},t_3-t_1\frac{T_3}{T_1})),$$
and similar for $D_+(T_2),D_+(T_3)$. By re-writting the symbols carefully these are the three open charts you mentioned exactly. |
H: Get point on ellipse from point and angle
I have the bounds (x from, x to ...) of an ellipse (and thus its radius and center), x and y of a point A that is in the ellipse and an angle. I want to get point B to which the angle points (from point A) and that lies on the ellipse.
This is what I can do:
center_x = (x_from + x_to) / 2
center_y = (y_from + y_to) / 2
radius_x = center_x - x_from
radius_y = center_y - y_from
How to go on?
@All: Thank you very much! The formula works perfectly well!
AI: You need to state as an explicit assumption that the main axes of the ellipse are parallel to the coordinate axes -- if that is not true, knowing the extreme $x$ and $y$ values is not enough to describe the ellipse.
However, if we can make the assumption, the way to proceed is as follows. First compute the lengths of the main semiaxes:
$$ a = \frac12(x_{\rm max}-x_{\rm min}) \qquad b = \frac12(y_{\rm max}-y_{\rm min}) $$
Then the equation for the ellipse is
$$ \tag{*} \left(\frac{x-x_{\rm center}}{a}\right)^2 + \left(\frac{y-y_{\rm center}}{b}\right)^2 = 1$$
Since you also know the angle $\theta$ (between $A$ and your sought point, I assume), you know there's a positive $t$ such that
$$ \tag{**} x= x_A+t\cos\theta \qquad y = y_A+t\sin\theta $$
Substitute these expressions for $x$ and $y$ into $(*)$ and you get a quadratic equation in $t$. Solve that, choosing the positive root (which always exists if $A$ is inside the ellipse), and put the resulting $t$ back into $(**)$. |
H: Proving that $S_n$ has order $n!$
I have been working on this exercise for a while now. It's in B.L. van der Waerden's Algebra (Volume I), page $19$. The exercise is as follows:
The order of the symmetric group $S_n$ is $n!=\prod_{1}^{n}\nu$. (Mathematical induction on $n$.)
I don't comprehend how we can logically use induction here. It seems that the first step would be proving $S_1$ has $1!=1$ elements. This is simply justified: There is only one permutation of $1$, the permutation of $1$ to itself.
The next step would be assuming that $S_n$ has order $n!$. Now here is where I get stuck. How do I use this to show that $S_{n+1}$ has order $(n+1)!$?
Here is my attempt: I am thinking this is because all $n!$ permutations of $S_n$ now have a new element to permutate. For example, if we take one single permutation
$$
p(1,\dots,n)
=
\begin{pmatrix}
1 & 2 & 3 & \dots & n\\
1 & 2 & 3 & \dots & n
\end{pmatrix}
$$
We now have $n$ modifications of this single permutation by adding the symbol $(n+1)$:
\begin{align}
p(1,2,\dots,n,(n+1))&=
\begin{pmatrix}
1 & 2 & \dots & n & (n+1)\\
1 & 2 & \dots & n & (n+1)
\end{pmatrix}\\
p(2,1,\dots,n,(n+1))&=
\begin{pmatrix}
1 & 2 & \dots & n & (n+1)\\
2 & 1 & \dots & n & (n+1)
\end{pmatrix}\\
\vdots\\
p(n,2,\dots,1,(n+1))&=
\begin{pmatrix}
1 & 2 & \dots & n & (n+1)\\
n & 2 & \dots & 1 & (n+1)
\end{pmatrix}\\
p((n+1),2,\dots,n,1)&=
\begin{pmatrix}
1 & 2 & \dots & n & (n+1)\\
(n+1) & 2 & \dots & n & 1
\end{pmatrix}
\end{align}
There are actually $(n+1)$ permutations of that specific form, but we take $p(1,\dots,n)=p(1,\dots,n,(n+1))$ in order to illustrate and prove our original statement. We can make this general equality for all $n!$ permutations: $p(x_1,x_2,\dots,x_n)=p(x_1,x_2,\dots,x_n,x_{n+1})$ where $x_i$ is any symbol of our finite set of $n$ symbols and $x_{n+1}$ is strictly defined as the symbol $(n+1)$.
We can repeat this process for all $n!$ permutations in $S_n$. This gives us $n!n$ permutations. Then, adding in the original $n!$ permutations, we have $n!n+n!=(n+1)n!=(n+1)!$. Consequently, $S_n$ has order $n!$.
How is my reasoning here? Furthermore, is there a more elegant argument? I do not really see my argument here as incorrect, it just seems to lack elegance. My reasoning may well be very incorrect, however. If so, please point it out to me.
AI: You can actually just use a combinatorial argument for this. The permutation group is a bijection from a set of $n$ elements to itself. So look at the first element in the permutation. There are $n$ choices to send that element to. Now for the second element, there are only $n-1$ choices left (because it is a bijection you cannot send two different elements in the domain to the same element in the codomain), and so on until you only have $1$ choice left for the last element. Thus we get $n!$ ways to arrange the permutation. |
H: How does graph theory describe a sequence or line or path of nodes?
I have a dataset of pairs of map coordinates, and I suspect that they could be connected to make a path. However, I'm not sure what the end points are, or if the coordinates actually make a path. I'm using the python networkx library, but it assumes you know a bit of graph theory to know what functions to use.
If I have a non-directed graph that looks like a path (a-b, b-c, c-d, ... y-z), how does graph theory describe the graph? What does it call the 'endpoints' 'a' and 'z'?
AI: I took a look at the documentation. networkx calls a path a path. It calls a the source and z the target of the path. See
http://networkx.lanl.gov/reference/algorithms.shortest_paths.html
or maybe just
http://networkx.lanl.gov/reference/generated/networkx.algorithms.shortest_paths.generic.has_path.html
depending on what exactly you want to do. |
H: Groups where all elements are order 3
I am a student trying to learn some abstract algebra this summer, and I recently proved (as an exercise) that if $G$ is a group where every element has order 2, then $G$ is abelian. I was wondering could we make a similar conclusion about groups where every element has order 3, namely I am asking if $G$ is a group where all elements have order 3, then $G$ is abelian. I think that this is not true, but I cannot think of a counterexample.
The only groups that I can think of which have all elements order 3 are the groups $(\mathbb{Z}/3\mathbb{Z})^n$, but these are abelian. Any help is appreciated. Thanks!
AI: The standard example is the Heisenberg group. Consider the group of all matrices of the form
$$\left(\begin{array}{ccc}
1 & x & y\\
0 & 1 & z\\
0 & 0 & 1
\end{array}\right),$$
where $x,y,z\in\mathbb{Z}/3\mathbb{Z}$. It is not hard to verify that this is a group, that every one of its 27 elements is of exponent $3$, and that it is not abelian. Replacing $\mathbb{Z}/3\mathbb{Z}$ with $\mathbb{Z}/p\mathbb{Z}$ for odd prime $p$ shows that a similar result cannot hold for any prime other than $p=2$.
This is an example of smallest possible order: a finite group in which every element is of exponent $3$ must have order $3^n$ for some $n$ (a consequence of Cauchy's Theorem), and every group of order $3^2$ is abelian.
There is another nonabelian group of order $27$, but in that group there is an element of order $9$:
$$\langle a,b\mid a^9 = b^3 = 1, ba = a^4b\rangle.$$ |
H: Proof that every metric space is homeomorphic to a bounded metric space
I have tried to show that every metric space $(X,d)$ is homeomorphic to a bounded metric space. My book gives the hint to use a metric $d'(x,y)=\mbox{min}\{1,d(x,y)\}$.
If we can show that $d(x,y) \le c_1 \cdot d'(x,y)$ with $c_1$ some positive constant and $d'(x,y) \le c_2 \cdot d(x,y)$ for $c_2$ some positive constant, then the identity map $i:(X,d) \to (X,d')$ is continuous, and also obviously a bijection, thus showing that $(X,d)$ is homeomorphic to $(X,d')$, where $(X,d')$ is bounded, thus giving the desired result.
Suppose $d(x,y)<1$. Then $d'(x,y)=d(x,y)$. If $d(x,y)\ge 1$, then $d'(x,y) \le d(x,y)$. Thus we can set $c_2 = 1$ and $d'(x,y) \le d(x,y)$ for all $x,y \in X$.
But when $d(x,y)>1$ why won't it always be the case that $c_1$ will depend on what $d(x,y)$ is?
AI: You’re working too hard: just show that $d$ and $d'$ generate the same open sets. Remember, a set $U$ is $d$-open if and only if for each $x\in U$ there is an $\epsilon_x>0$ such that $B_d(x,\epsilon_x)\subseteq U$. Once you have that $\epsilon_x$ that’s small enough, you can use any smaller positive $\epsilon$ just as well, so you might as well assume that $\epsilon_x<1$. Can you take it from there? |
H: Evaluate the integral $H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$
$y\geq0$ define $$H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$$ Show that $H$ is a continuous function of $y$ and show $\lim\limits_{y \to +\infty}H(y)=0$.
AI: This is kind of a brute force method where we explicitly find the function $f(a)$.
$$f(a) = \begin{cases} \dfrac{\log(1+a)}{3a} & \text{if }a >0\\ \dfrac13 & \text{if }a=0 \end{cases}$$
This can be obtained as shown below. We have that for $a>0$, $$\dfrac1{x^4+ax} = \dfrac1{x(x^3+a)} = \dfrac1{ax} - \dfrac{x^2}{a(a+x^3)}$$
Hence, $$f(a) = \int_1^{\infty} \dfrac{dx}{x^4+ax} = \int_1^{\infty} \left(\dfrac1{ax} - \dfrac{x^2}{a(a+x^3)} \right) dx = \lim_{R \rightarrow \infty} \int_1^{R} \left(\dfrac1{ax} - \dfrac{x^2}{a(a+x^3)} \right) dx$$
The first integral $$I_1 = \int_1^{R} \dfrac{dx}{ax} = \dfrac{\log(R)}a.$$
The second integral $$I_2 = \dfrac1{3a} \int_1^{R} \dfrac{3x^2dx}{(a+x^3)} = \left. \dfrac1{3a} \log(a+x^3) \right \rvert_{1}^{R} = \dfrac{\log(a+R^3) - \log(a+1)}{3a}$$
Putting these together, we get that
\begin{align}
f(a) & = I_1 - I_2\\
& = \lim_{R \rightarrow \infty} \left(\dfrac{\log(R)}a - \left( \dfrac{\log(a+R^3) - \log(a+1)}{3a}\right) \right)\\
& = \lim_{R \rightarrow \infty} \dfrac{\log(R^3)-\log(a+R^3) + \log(a+1)}{3a}\\
& = \lim_{R \rightarrow \infty} \dfrac{\log \left(\dfrac{R^3}{a+R^3} \right) + \log(a+1)}{3a}\\
& = \dfrac{\log (1) + \log(a+1)}{3a}\\
& = \dfrac{\log(a+1)}{3a}\\
\end{align}
If $a=0$, then $f(0) = \displaystyle \int_1^{\infty} \dfrac{dx}{x^4} = \dfrac13$. Hence, we have that $$f(a) = \begin{cases} \dfrac{\log(1+a)}{3a} & \text{if }a >0\\ \dfrac13 & \text{if }a=0 \end{cases}$$
Clearly, $f$ is a continous function of $a$ for all $a \geq 0$ and $\lim_{a \rightarrow \infty} f(a) = 0$. |
H: Showing inequality for harmonic series.
I want to show that $$\log N<\sum_{n=1}^{N}\frac{1}{n}<1+\log N.$$ But I don't know how to show this.
AI: I think I wrote this up somewhere on this website but anyways here she is again.
From the figure, you can see that the area under the blue-curve is bounded below by the area under the red-curve from $1$ to $\infty$.
The blue-curve takes the value $\frac1{k}$ over an interval $[k,k+1)$
The red-curve is given by $f(x) = \frac1{x}$ where $x \in [1,\infty)$
The green-curve takes the value $\frac1{k+1}$ over an interval $[k,k+1)$
The area under the blue-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k}$ while the area under the red-curve is given by the integral $\displaystyle \int_{1}^{n+1} \frac{dx}{x}$ while the area under the blue-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k+1}$
Hence, we get $\displaystyle \sum_{k=1}^{n} \frac1{k} > \displaystyle \int_{1}^{n+1} \frac{dx}{x} = \log(n+1)$
$\log(n+1)$ diverges as $n \rightarrow \infty$ and hence
$$\lim_{n \rightarrow \infty} \displaystyle \sum_{k=1}^{n} \frac1{k} = + \infty$$
By a similar argument, by comparing the areas under the red curve and the green curve, we get $$\displaystyle \sum_{k=1}^{n} \frac1{k+1} < \displaystyle \int_{1}^{n+1} \frac{dx}{x} = \log(n+1)$$ and hence we can bound $\displaystyle \sum_{k=1}^{n} \frac1{k}$ from above by $1 + \log(n+1)$
Hence, $\forall n$, we have $$\log(n+1) < \displaystyle \sum_{k=1}^{n} \frac1{k} < 1 + \log(n+1)$$
Hence, we get $0 < \displaystyle \sum_{k=1}^{n} \frac1{k} - \log(n+1) < 1$, $\forall n$
Hence, if $a_n = \displaystyle \sum_{k=1}^{n} \frac1{k} - \log(n+1)$ we have that $a_n$ is a monotonically increasing sequence and is bounded.
Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n$ exists. This limit is denoted by $\gamma$ and is called the Euler-Mascheroni constant.
It is not had to show that $\gamma \in (0.5,0.6)$ by looking at the difference in the area of these graphs and summing up the area of these approximate triangles. |
H: Finding the derivative of a function?
Could someone explain how I would find the derivative of the following function? I am completely lost:
$$f(x) = e^{i(x!^{\log x})}$$
AI: For complex $\,z\,$ with $\,\operatorname{Re}(z)>0\,$ , $$\Gamma'(z)=\int_0^\infty t^{z-1}e^{-t}\log t\,dt$$ so $$\left(e^{ix!^{\log x}}\right)'=e^{ix!^{\log x}}\left(ix!^{\log x}\right)'$$ Now, since $\,\displaystyle{x!^{\log x}=e^{\log x\log(x!)}}\,\,and\,\,\log x!=\log\Gamma(x+1)$ , we get$$\left(ie^{\log x\log x!}\right)'=i\left[\frac{\log x!}{x}+\log x\frac{\Gamma'(x+1)}{\Gamma(x+1)}\right]e^{\log x\log x!} $$Now put together all the above and have sweet mathematical nightmares. |
H: How to draw pictures of prime spectra
In Atiyah-MacDonald's Commutative Algebra, they give in Exercise 16 of Chapter 1 the instruction:
Draw pictures of Spec($\mathbb{Z}$), Spec($\mathbb{R}$),
Spec($\mathbb{C}[x]$), Spec($\mathbb{R}[x]$), Spec($\mathbb{Z}[x]$).
What exactly do they have in mind? I can enumerate the prime ideals in each of these rings, but I have little idea of what kinds of pictures one might draw and what advantage a good picture has over a mere enumeration.
AI: Pictures are useful because they're suggestive, and are sometimes better at organizing information.
Sometimes the right picture is a little unclear. For example, when drawing $\mathop{\mathrm{Spec}} \mathbb{C}[x]$, sometimes you might want to draw a line, and imagine each complex number corresponds to some point on this line. Also, you want a splotch off to the side to represent the generic point of the line. This emphasizes the one-dimensionalness and the triviality of its structure as an algebraic curve, and also, it helps us to remember that its ordinary topology doesn't play a role algebraically.
But other times, you would want to draw $\mathop{\mathrm{Spec}} \mathbb{C}[x]$ as the traditional complex plane with its ordinary labeling (again with a splotch to represent the generic point). This picture is useful, for example, if we want to apply results and intuition from complex analysis.
The main thing about having a picture for $\mathop{\mathrm{Spec}} \mathbb{Z}$ is, IMO, really just so that you can then draw a picture of $\mathop{\mathrm{Spec}} \mathbb{Z}[x]$ or to draw a picture of the spectrum of the ring of integers of a number field as a curve with a projection down to $\mathop{\mathrm{Spec}} \mathbb{Z}$. |
H: If $z$ is the unique element such that $uzu=u$, why is $z=u^{-1}$?
I'm trying to figure out why an element $u$ in some ring is invertible with inverse $z$ if any only if
$uzu=u$ and $zu^2z=1$
OR
$uzu=u$ and $z$ is the unique element meeting this condition.
Clearly, both conditions follow if $u$ is a unit with inverse $z$. However, I can't see why either condition implies that $z=u^{-1}$.
I haven't been able to make any decent progress on my own, so does anyone have hints or suggestions on where to go? Thanks.
Edit: From Qiaochu's hint, $zu$ and $uz$ are idempotent. So $(zu)^2=zu$. But $zu$ has right inverse $uz$, so $(zu)^2(uz)=(zu)(uz)\implies zu=1$. The analogous argument for $uz$ shows $uz=1$, so $z=u^{-1}$.
Does anyone have an idea for the second?
AI: For the sake of having an answer:
The strategy is this: since we have $u=uzu$, we also have $0=u(zu-1)=(uz-1)u$. If it can be shown that $u$ is "regular" (in the sense that it is not a nonzero zero-divisor), then we have $zu-1=uz-1=0$, establishing the result.
As per Yuki's comment above, if $u\alpha=0$, then $u(z+\alpha)u=uzu=u$. By uniqueness of $z$, we have $z+\alpha=z$, and so $\alpha=0$. A symmetric argument establishes that if $\alpha u=0$, then $\alpha=0$. Thus, $u$ is regular. |
H: Does $\tan(1/z)$ has a Laurent series convergent?
Well, I have no idea: Does $\tan(1/z)$ have a Laurent series convergent on $0<|z|<R$?
AI: In short, the answer is no. The function $\tan(1/z)$ has poles at $z=\frac{1}{\pi/2+n\pi}$, which means that it is not analytic on the annulas you mentioned for any $R$. |
H: Is this an odd function?
I have this function:
\begin{equation*}
f(x)=%
\begin{cases}
1 &x\in\left[ -\pi,-\pi/2\right[ \\
-1 &x\in\left[ -\pi/2,0\right[\\
1 & x\in\left[ 0,\pi/2\right[ \\
-1 & x\in\left[ \pi/2,\pi\right]\\
\end{cases}
\end{equation*}
First I thought that it was odd, but then I realized that $f(0) = 1 \neq - f(0) = -1$ was true. Does it matter when you calculate an integral and want to use the property of an odd function? For instance can I still deduce that:
\begin{equation*}
\int_{-a}^{a} f(x) dx = 0
\end{equation*}
?
AI: Changing the value of the function in a finite number of points (so that you get an odd function) does not affect the integral, so yes, the integral is zero. |
H: LSC function Problem
$I$ : finite index set, $f_i$ is LSC for each $i \in I$
I wanna prove below. $$\min_i f_i \text{ is LSC}$$
And is there some example of that does not extend to infinite $I$?
AI: We use properties of $\liminf$. First, it's enough to show the result when $I$ has two elements (then use induction on the number of elements of $I$). If $f$ and $g$ are lower semi-continuous, and $x_0\in\Bbb R$, if $x_n\to x_0$ then
\begin{align*}
\liminf_{n\to +\infty}\min\{f(x_n),g(x_n)\}&=\lim_{k\to +\infty}\inf_{n\geq k}\min\{f(x_n),g(x_n)\}\\
&=\lim_{k\to +\infty}\min\{\inf_{n\geq k}f(x_n),\inf_{n\geq k}g(x_n)\}\\
&=\min\{\liminf_{n\to +\infty}f(x_n),\liminf_{n\to +\infty}g(x_n)\}\\
&\geq \min\{f(x),g(x)\}.
\end{align*}
Since it's true for all sequence you can conclude lower semi-continuity.
When $I$ is infinite, it's not true any more. Take $f_i$ the indicator function of $(-\infty,1/i)$. Since this interval is open, $f_i$ is LSC. We have that $\min_{i\in\Bbb N}f_i=\chi_{(-\infty,0]}$, which is not LSC (take $x_0=0$ and $x_n=1/n$). |
H: plotting the following set of points in the XY plane
Represent the following set of points in the XY plane :
$$\{ ( x , y ) \; | \; |x| + |y| = 1 \}$$
What i got:
1) if $x > 0, y > 0 : x = 1 - y$
2) if $x > 0, y < 0 : x = 1 + y$
3) if $x < 0, y > 0 : x = y - 1$
4) if $x < 0, y < 0 : x = -y -1$
Any help to solve this question would be greatly appreciated.
Thank you,
AI: Take each case without the absolute value:$$\,\,\,y=1-x$$$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y=1-(-x)=1+x$$$$-y=1-x$$$$-y=1+x$$so you can see we get four straight lines intersecting each with other two.
Well, now just draw these lines and get your nice... rhomboid (you could know this even before drawing anything from the above equations). |
H: Prove that $(p-1)! \equiv (p-1) \pmod{1+2+3+\cdots+(p-1)}$
Given a prime number $p$ , establish the congruence:
$$(p-1)! \equiv (p-1) \pmod{1+2+3+\cdots+(p-1)}$$
I have proceeded like this:
$$\begin{align*}&(p-1)! \equiv (-1) \pmod{p} \quad \quad \quad \text{by Wilson's Theorem}\\
&(p-1)! \equiv 0 \pmod{\frac{p-1}{2}} \end{align*}$$
Then I know that I have to apply Chinese remainder theorem but I don't have a thorough understanding of it.So please give me an elaborate answer to this question with respect to Chinese remainder theorem.
AI: Note that $1 + 2 + ... + p-1 = \frac{p(p-1)}{2}$ and you can assume that $p > 2$ is odd.
You have a congruence $$ x = (p-1)! \equiv -1 (\bmod p)$$ $$ x = (p-1)! \equiv 0 (\bmod (p-1)/2) $$
Denote $m = p(p-1)/2$ and $m_1 = p , m_2 = (p-1)/2$. Denote $n_i = m/m_i$.
The Chinese Remainder Theorem goes as follows (to our simple case): to solve $x \equiv a_i (\bmod m_i)$ we have to use the fact that $n_i,m_i$ are coprime and thus there are $s_i$ and $r_i$ such that $s_i n_i + r_i m_i = 1$.
In our case you can check that $$ 1 \cdot p - 2 \cdot (p-1)/2 = 1$$ so $n_1 = (p-1)/2 , s_1 = -2$ and $n_2 = p , s_2 = 1$. The general solution is $$x \equiv \sum_{i}a_i s_i n_i \quad (\bmod m)$$ and in our case:
$$ (p-1)! \equiv -1 \cdot (-2)(p-1)/2 + 0 \cdot 1 \cdot p \equiv p-1 \quad (\bmod \frac{p(p-1)}{2})$$
A general layout of the Chinese algorithm can be found in English Wikipedia though the algorithm in the Hebrew Wikipedia is clearer to my taste. |
H: Nature of the series $ \sum u_{n}, u_{n}=n!\prod_{k=1}^n \sin\left(\frac{x}{k}\right) $
Is the series $$ \sum u_{n}$$
$$ u_{n}=n!\prod_{k=1}^n \sin\left(\frac{x}{k}\right)$$
$$ x\in]0,\pi/2] $$
convergent or divergent?
We have:
$$ u_{n}\leq n!\prod_{k=1}^n \frac{x}{k}$$
$$ u_{n}\leq x^n$$
If $0<x<1$ the series is convergent.
$$ u_{n}\geq n! \prod_{k=1}^n \frac{2x}{\pi k}$$
$$ u_{n} \geq \prod_{k=1}^n \frac{2x}{\pi}$$
If $x=\pi/2$, $u_{n}\geq1$, $\sum u_{n}$ is divergent.
What about the case $x\in[1,\pi/2[$ ?
AI: As N.S. shows in his answer, the ratio test quickly resolves the problem for all $x\neq 1$, so for now fix $x=1.$ Then $\displaystyle u_n = \prod_{k=1}^n k\sin(1/k).$
We show that $\displaystyle \lim_{n\to\infty} u_n = L> 0 $ so $\sum u_n$ diverges.
Since $\sin x<x$ we see $u_n<1.$ In order to apply a convergence criterion for infinite products, let us work with $v_n = 1/u_n$ instead. Let us remember that "convergence" of an infinite product demands the partial products tends to a finite non-zero limit. We compute $$v_n=\prod_{k=1}^n \frac{1}{k\sin(1/k)} = \prod_{k=1}^n (1+a_n)$$
where $\displaystyle a_n = \frac{1}{k\sin(1/k)}-1 >0.$ We know that $\displaystyle \prod_{k=1}^n v_n$ converges if and only if $\displaystyle \sum_{k=1}^n a_n$ converges. Using Taylor series for the $\sin$ term and expanding as geometric series we have $$a_n= \frac{1}{1-1/(6k^2) +\mathcal{O}(k^{-3})}-1= \frac{1}{6k^2} + \mathcal{O}\left(\frac{1}{k^3}\right)$$
so we find that $\sum a_n$ converges, so $\sum u_n$ diverges. |
H: Can I change the order of the double integration?
Let $f \in C_0^\infty $, $g \in L^1 $ . Then $$ \int_{\mathbb R^n} \int_{\mathbb R^n} f(x-y)g(y) dy dx = \int_{\mathbb R^n}\int_{\mathbb R^n} f(x-y)g(y) dx dy $$holds? If so, why?
($f,g : \mathbb R^n \to \mathbb R $)
AI: This depends on the finiteness of the integral. Fubini's theorem can be applied:
Let $X$ and $Y$ be complete measure spaces. If
$$\int_{X\times Y} |f(x,y)| \ \mathrm{d}(x,y) < \infty$$
then
$$\int_X \int_Y f(x,y) \ \mathrm{d}y \ \mathrm{d}x = \int_{X\times Y} f(x,y) \ \mathrm{d}(x,y) = \int_Y \int_X f(x,y) \ \mathrm{d}x \ \mathrm{d}y$$
Since $f \in C^\infty_0$ and $g \in L^1$ I think it's safe to say that the integral in question converges. |
H: Continuous function a.e.
I'm not sure these two statement are not same thing.
$$ "f equals a continuous function a.e." & "f is continuous a.e."$$
The concept is too much abstract, so I wanna find some counter examples.
a function $f$ and a continuous function $g$ s.t. $f=g$ a.e
and $f$ is not continuous a.e.
a function $f$ continuous a.e. s.t. there exists no continuous function $g$ with $f=g$ a.e.
In addition I wanna find a Riemann integrable function which has an uncountable set of discontinuities.
Is there some nice example to understand those concepts, note that please.
AI: Function continuous a.e. but not equal a.e. to a continuous function: $f(x) = 0$ if $x < 0$ and $f(x) = 1$ if $x \geq 0$
Function equal a.e. to a continuous function but not continuous a.e.: $f$ = the characteristic function of the rationals.
Riemann integrable function with discontinuities on an uncountable set: $f$ = the characteristic function of the Cantor set in $[0,1]$. Let $C$ be the cantor set. Then $C$ is perfect so it is uncountable and closed. Hence $f$ is continuous on the complement of $C$ because these points are all interior points. But since $C$ contains no open interval, every point of $C$ is a boundary point, so $f$ is discontinuous on $C$. $C$ has measure zero, so $f$ is Riemann integrable. |
H: Equivalent of $ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx, n\rightarrow \infty$
I would like to show that
$$ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx \sim_{n\rightarrow \infty} \frac{1}{n}$$
Using the change of variable $u=x^n$:
$$ I_{n}=\frac{1}{n^2} \int_0^1 \frac{u^{1/n} \ln u}{u^{1/n}-1} \mathrm du=\frac{1}{n^2}\left(\int_0^1 \ln x \mathrm dx+\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \right)=\frac{-1}{n^2}+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx=o(1/n)+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx$$
So I have to show that
$$ \int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \sim_{n \rightarrow \infty} n$$
Could you help me?
AI: $$\frac{x^n}{x-1} = \frac{x^n-1}{x-1} + \frac{1}{x-1}= \left(1+ x+ \cdots + x^{n-1}\right) +\frac{1}{x-1}.$$
So $$I_n = \int^1_0 (1+x+\cdots + x^{n-1}) \log x + \int^1_0 \frac{\log x}{x-1} dx.$$
Integration by parts shows $ \int^1_0 x^k \log x = -1/(k+1)^2$ and expanding $\log$ by Taylor series will show $\displaystyle \int^1_0 \frac{\log x}{x-1} dx = \frac{\pi^2}{6}$ so $$I_n = \sum_{k=n+1}^{\infty} \frac{1}{k^2}.$$
Thus $$nI_n = \frac{1}{n} \sum_{k=n+1}^{\infty} \frac{1}{(k/n)^2} \to \int_1^{\infty} \frac{1}{x^2} dx=1$$
so $I_n \sim 1/n.$
You can skip showing $\displaystyle \int^1_0 \frac{\log x}{x-1} dx = \frac{\pi^2}{6}$ if you expand $x^n/(x-1)$ as a geometric series from the start.
Instead of using Riemann sums we could also have noted that $1/x^2$ is monotonically decreasing and use the well known theorem that if $f$ is monotone then $\displaystyle \int^n_1 f(x) dx \sim \sum_{k=1}^n f(n).$ |
H: Quotient of two free abelian groups of the same rank is finite?
Let $A,B$ be abelian groups such that $B\subseteq A$ and $A,B$ both are free of rank $n$. I want to show that $|A/B|$ is finite, or equivalently that $[A:B
]$ (the index of $B$ in $A$) is finite.
For example, if $A=\mathbb{Z}^n$ and $B=(2\mathbb{Z})^n$ we have that $[A:B]=2^n$ (correct me if I'm wrong).
AI: There are several ways - for example induction on $n$ should work. We can also use the stacked basis theorem, if so desired.
But let's do the following. Select a basis ${\mathcal B}_A =\{a_1,a_2,\ldots,a_n\}$ of $A$.
Let us similarly select a basis ${\mathcal B}_B=\{b_i,b_2,\ldots, b_n\}$ of $B$. As $B\subseteq A$ we get a matrix
$M=(m_{ij})$, with integer entries, uniquely determined by
$$
b_i=\sum_j m_{ij}a_j.
$$
I want to show that the matrix $M$ is non-singular.
Let $V$ be the vector space over $\mathbf{Q}$ with basis ${\mathcal B}_A$. Then ${\mathcal B}_A$
and ${\mathcal B}_B$ are both linearly independent over $\mathbf{Z}$ in $V$, because they were so in $A$. Thus they are both vector space bases of $V$, and $M$ is the change of bases matrix that is then non-singular.
The inverse matrix $M^{-1}$ then has the property that $(\det M)M^{-1}$ also has integer entries (this follows immediately from Cramer's rule). This means that (working inside $V$ again) the basis elements $a_i$ have the property that $(\det M)a_i$ is a linear combination of vectors $b_j$ with integer coefficients. Therefore $(\det M)a_i\in B$. As this holds for all $a_i$, we have shown that
$$
(\det M)A\subseteq B.
$$
The claim follows from this: $A/B$ is a finitely generated abelian group with exponent that is a factor of $|\det M|$, so it must be finite.
A more careful study (using stacked bases theorem) reveals that in fact we have
$$
[A:B]=|\det M|.
$$ |
H: Why $C_0^\infty$ is dense in $L^p$?
Why $C_0^\infty$ is dense in $L^p$?
Would you give me a simple proof or the outline of the proof?
AI: The outline (the proof isn't simple, at least not according to my understanding of simple):
Let $f \in L^p$. Then there is a sequence of simple functions $s_n \in L^p$ that converges pointwise to $f$. Then you show that they converge in norm.
These $s_n$ have finite support. You can therefore apply Lusin's theorem:
(Lusin) Let $\Omega$ be a locally compact Hausdorff space and $\mu$ a Radon measure and $f: \Omega \to \mathbb R$ a $\mu$-measurable function with finite support $E$. Then for every $\delta > 0$ there exists a closed set $K \subset E$ such that $\mu(E \setminus K) < \delta$ and $f$ is continuous on $K$.
to get a sequence of continuous functions.
Then you apply Tietze's theorem:
(Tietze) If $\Omega$ is a locally compact Hausdorff space and $K \subset \Omega$ compact then any $f \in C(K, \mathbb R)$ can be extended to a bounded function in $C(\Omega, \mathbb R)$.
to get a sequence of compactly supported functions.
I did this here for $p=1$.
Once you have $C_c$ dense in $L^p$ you mollify. Which I did here. Hope this helps. |
H: Finding the equation of the holomorphic function from its components
Let $f(z)=u(x,y)+ i v(x,y)$ be a holomorphic function. I often find it difficult to deduce the form of $f$ as a function of $z=x+i y$ only (for example: $f=\sin x \cosh y + i \cos x \sinh y$ cas be written as $f(z)=\sin z$).
However, my book states that given $u$ and $v$ or even $u$ or $v$ only we can deduce the $z$-dependance of $f$ using the following relations:
$f(z)=u(z,0)+i v(z,0)$
$f(z)= 2 u (z/2, -i z/2)$
$f(z)= 2 i v(z/2, -iz/2)$
Can anyone help me understand how they come about?
AI: For a start, it's worth mentioning that you kind of already have $f$ as a function of $z$ only, just using the real part and imaginary part functions a lot. So it's a slightly non-obvious question what we're really trying to achieve here – basically, we want a "neat" expression for $f$, but that's not a terribly mathematical concept. With that in mind, I'll go ahead and see if I can work out what the relations mean.
$f(z) = u(z,0) + iv(z,0)$: This is a bit of a cheat. A priori, $u$ and $v$ need not be capable of taking a complex argument at all. But if they do extend to complex-differentiable functions on $\mathbb C$, then they extend to a function that agrees with $f$ on the real line (because on the real line, the equation is completely straightforward), and it is a general theorem that if two complex-differentiable functions agree on any non-discrete set then they must be equal (as a consequence of the fact that the zeroes of a nonzero complex differentiable function are isolated, so if $f$ and $g$ agree on non-isolated points, then $f-g$ is zero on non-isolated points, so $f - g = 0$, i.e. $f = g$).
The latter two equations do not hold. For example, take $f(z)=1$, so that $u(x,y)=1$ and $v(x,y)=0$. Then $f(0)=1$, but $2u(0,0)=2$ and $2v(0,0)=0$. |
H: Cauchy-Product of non-absolutely convergent series
While grading some basic coursework on analysis, I read an argument, that a Cauchy product of two series that converge but not absolutely can never converge i.e. if $\sum a_n$, $\sum b_n$ converge but not absolutely, the series $\sum c_n$ with $$c_n= \sum_{k=0}^n a_{n-k}b_k$$ diverges.
Although we didn't have any theorem in the course stating something like this, it made me wonder if it was true.
AI: As Davide Giraudo has said in the comments, we can find a counter example by using $a_k = b_k = (-1)^k/k$ for $k\geq 1$ and $a_0=b_0=0.$ In that case we compute $$c_n = \sum_{k=0}^n a_{n-k} b_k = \sum_{k=1}^{n-1} \frac{(-1)^{n-k}}{n-k} \frac{(-1)^k}{k} = (-1)^n \sum_{k=1}^{n-1} \frac{1}{k(n-k)}$$
$$ = \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{n-k+k}{k(n-k)} = \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \left( \frac{1}{k} + \frac{1}{n-k}\right) = 2\frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{1}{k}.$$
We show $\sum c_n$ converges by applying the Leibniz criterion. $c_n \to 0$ is clear, so we need only verify that $d_n = \frac{1}{n} \sum_{k=1}^{n-1}\frac{1}{k} $ is monotonically decreasing for sufficiently large $n.$
We compute $$d_{n+1}- d_n = \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k} - \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{k}= \frac{1}{n(n+1)} - \frac{1}{n(n+1)}\sum_{k=1}^{n-1} \frac{1}{k}.$$
Since $\displaystyle \sum_{k=1}^{n-1} \frac{1}{k} \geq 1$ for all $n\geq 2$ so $d_n$ is indeed monotone. |
H: Riemann Integral Problem
I'm studying Riemann Integral on measure theory class.
There is a function $f : [a,b] \to \mathbb{R}$
that is increasing or decreasing.
Is that function f is Riemann integrable? And if then, what are appropriate step functions?
And is that function continuous a.e.?
AI: Yes, a monotonic function on an interval is Riemann integrable. In fact, one can come up with the step functions naively.
Let $P_n$ denote the partition $\{a, a + \frac{b-a}{n}, a + 2\frac{b-a}{n}, \dots, b\} =: \{x_0, \dots, x_{n}\}$. Then it's easy to see that $x_k - x_{k-1} = \frac{b-a}{n}$. Suppose wlog that $f$ is monotonic decreasing (increasing is more or less the same, or do the same reasoning on $-f$). Then if $M_k$ and $m_k$ denote the maximum and minimum on the $k$th part of the partition, then $M_k = f(x_{k-1})$ and $m_k = f(x_k)$.
Thus
1. (lower sum) $\displaystyle L(f,P_n) = \sum m_k(x_k - x_{k-1}) = \sum f(x_k)\frac{b-a}{n}$ and
2. (upper sum) $\displaystyle U(f,P_n) = \sum M_k(x_k - x_{k-1}) = \sum f(x_{k-1})\frac{b-a}{n}$
So then
$$\begin{align}
U(f,P_n) - L(f,P_n) &= \frac{b-a}{n}\left( \sum_{k = 1}^n f(x_{k-1}) - f(x_k)\right) = \\
&= \frac{b-a}{n} (f(x_0) - f(x_n)) = \frac{b-a}{n}(f(a) - f(b)) \\
&\to 0 \text{ as } (n \to \infty)
\end{align}$$
It's not so hard to show that refinements of these partitions do not change the result.
You also ask whether a monotonic function on $[a,b]$ is continuous a.e. It is, in the sense that there are only a countable number of discontinuities. In particular, you might know that a sum of positive numbers is finite only if it has a countable number of nonzero terms (an easy way to see this - if there are uncountably many, then you can show there are uncountably many rationals as follows: consider the intervals implied by the sum, where each summand gives another interval. Each interval contains a rational.) The function is bounded on $[a,b]$ and monotonic, and so the sum of the 'jumps' at the points of discontinuity is finite. Thus there are at most countably many points of discontinuity. |
H: Geodesics that self-intersect at finitely many points
Notations
$M$ will denote a smooth manifold and $\nabla$ an affine connection on it. A smooth curve $\gamma\colon I \to M$ will be called a geodesic if it is $\nabla$-parallel along itself, that is $\nabla_{\dot{\gamma}(t)}\dot{\gamma}=0$ for every $t \in I$. A geodesic will be said to be maximal if every proper extension of it is not a geodesic.
It is easy to find examples of maximal geodesics which do not self-intersect, like
lines in Euclidean plane, or that intersect in infinitely many points, like great circles on the sphere. On the contrary I cannot find examples of geodesics which self-intersect at finitely many points, like the curve below:
Question Is it possible to determine $M$ and $\nabla$ in such a way that one of the resulting maximal geodesics intersects at finitely many points?
Thank you.
AI: Take a quadrant of the plane and roll it up into a cone by gluing the two edges.
Shown above are the unfolding and the glued manifold, with a single geodesic shown in black; the "seam" runs down the right side of the cone in the right image. In fact, every geodesic that does not pass through the apex intersects itself exactly once.
For a smooth manifold, just round off the top. |
H: If $ f \in C_0^\infty$, then is $f$ uniformly continuous?
If $ f \in C_0^\infty=\{ g: g\in C^\infty, \lim_{|x|\rightarrow \infty}g(x)=0\}$, then is $f$ uniformly continuous on $\mathbb R$?
($ f : \mathbb R \to \mathbb R $)
AI: HINTs
A continuous function on a compact interval is uniformly continuous.
$\lim_{|x| \to \infty} f(x) = 0$ means that $\forall \epsilon...$
Split up the domain to use these two properties. |
H: plotting the following set of points in the XY plane 2 :
Represent the following set of points in the $XY$-plane
$$\left\{ (x,y) \big| (x-|x|)^2 + (y-|y|)^2 \leq 4 \right\}$$
Any help to solve this problem would be greatly appreciated.
Thank you.
AI: As I answered another question of your about absolutes, you have to consider possible cases:
$x>0$ and $y>0$:
In this case, we'll have $0 \leq 4$ which is always true so all the positive values of $x$ and $y$ are correct and a part of our final answer.
$x>0$ and $y<0$:
$$(2y)^2 \leq 4 \implies y^2 \leq 1 \implies -1 \leq y \leq 1$$
So the answer in this quadrant would be $x>0$ and $-1 \leq y \leq 0$
$x<0$ and $y>0$:
$$(2x)^2 \leq 4 \implies x^2 \leq 1 \implies -1 \leq x \leq 1$$
So the answer in this quadrant would be $y>0$ and $-1 \leq x \leq 0$
$x<0$ and $y<0$:
$$4x^2 +4y^2 \leq 4 \implies x^2+y^2 \leq 1 \implies -1 \leq x \leq 1, -\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}$$
To sum up the whole answers, from the first two cases, the union of the two sets is $x>0$ and $\{(y>0) \cup (-1 \leq y \leq 1)\}$ which is $y \geq -1$.
And from the other two cases, $\{(-1 \leq x \leq 0) \cup (-1 \leq x \leq 1)\} $ which is$-1 \leq x \leq 1$ and $\{(y>0) \cup (-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2})\}$ which is $y \geq -\sqrt{1-x^2}$. |
H: How to make derivative operation in matrix space?
\begin{equation}\frac{d}{d\theta}\frac{1}{2}(\theta^TX - y)^2 = 0\end{equation} where, $X$ is $m $ on $ n$ matrix, $y$ is $m$-dimensional vector, $\theta$ is n-dimensional vector.
I can solve this equation, but only intuitively, because of I know solution from lecture I watched:
\begin{equation}
\theta = (X^TX)^{-1}X^Ty
\end{equation}
Could someone solve it step by step and explain transformation?
Below there is my solution, but there are mistakes for sure:
\begin{equation}
\frac{d}{d\theta}\frac{1}{2}(\theta^TX - y)^2 = 0 \\
(\theta^TX - y)X = 0 \\
\theta^TXX - X^Ty = 0 \\
X^T\theta X = X^Ty \\\
\theta = (X^TX)^{-1}X^Ty
\end{equation}
And how I said it's so intuitively. In each line I see something, what doesn't fit to me. Some of transformation I did so I could reach expected solution.
AI: I refuse to answer this question without adding the following comment: one common reason for confusion when it comes to take the derivative of equations from linear algebra is a poor or inconsistent choice of notation. This is the case here, too. If you take the transpose of an $n$- vector you get a $(1,n)$ -matrix. If you apply this from the left to an $(n,m)$ matrix you get $(1, m)$ matrix, an object from which you cannot substract an $m$- vector. The result of this kind of inconsistency in notation are more inconsistencies when you perform operations on these constructions, like taking the derivative. Consequently your computation contains some lines which are not really well defined, eg. if they contain products $XX$ -- you do not assume $X$ to be a square matrix, so this is not defined.
Another choice of notation which can turn inta a trap is using the square notation for the scalar product of a vector with itself, like $v^2 = \langle v, v\rangle = v^T v$ While $v^2$ is common, it is easier to use one of the other forms when it comes to take the derivative.
I'm also not too happy with the notation $\frac{d}{d\theta}$, since $\theta$ is a vector. I prefer something like $D_v f(\theta)$, denoting the directional derivative of $f(\theta)$. Now if $f(\theta) = X\theta$ you just get $D_v f = Xv$. (assuming $X$ is independent of $\theta$).
Combining the comments: what you could look at is either $\langle X\theta - y, X\theta -y\rangle $ or $\langle (\theta^T X)^T - y, (\theta^T X)^T -y\rangle $, depending on how $X$ looks like. I assume you are interested in the first variant. Then, taking the derivative wrt to a vector $v$ you get
$$D_v \frac{1}{2}\langle X\theta-y,X\theta-y\rangle = \frac{1}{2}(\langle Xv,X\theta -y\rangle + \langle X \theta -y, Xv\rangle = \langle Xv,X \theta -y\rangle = \langle v,X^TX\theta-X^Ty\rangle $$
This will be $=0$ for any $v$, so the second factor needs to be $=0$, hence
$$X^TX\theta = X^Ty \Rightarrow \theta = (X^TX)^{-1}X^T y$$
(assuming $X^TX$ is invertible ...) which miraculously is the result you were expecting. |
H: Quick way to check if a polynomial of degree $> 3$ is irreducible?
What's the easiest way to check if a polynomial of degree > 3 is irreducible in $\mathbb{Z}_2[x]$?
I want to find out if $x^7+x^6+1$ is irreducible in $\mathbb{Z}_2[x]$.
If a quadratic polynomial factors, it must be a product of two linear factors, and
if a cubic polynomial factors, then it must be a product of a linear
and a quadratic (or three linear) and it must have a root in $\mathbb{Z}_2[x]$. Easily verified.
If I had a polynomial of degree 5 I would test if I could find any factors by assuming for ex: $(x^3 + Ax^2 + Bx + C) (x^2 + Dx + E)$ and if we find a factor we know that it's reducible.
But if I have degree 7, I would we have to test a lot of combos, $x*x^6$, $x^4*x^3$, $x^3*x^3*x ...$ Is there an easier way to determine if it's irreducible?
AI: There are some methods and tricks. For example here you could by repeated squaring check that the remainder of $x^{128}$ modulo your polynomial is equal to $x$. As the polynomial
$x^{128}+x$ is known to be the product of all the irreducible polynomials of degrees that are factors of seven, it follows that your polynomial is, indeed, irreducible (because it obviously is not a product of linear polynomials). But for this to work so smoothly it is crucial that seven is a prime.
Another thing you could try is to test divisibility by low-dimensional irreducible polynomials. If you play with these enough, you quickly learn that up to degree 3 all the irreducible polynomials of $\mathbb{Z}_2[x]$ are $x$, $x+1$, $x^2+x+1$, $x^3+x+1$ and $x^3+x^2+1$. If your degree 7 polynomial $p(x)=x^7+x^6+1$ were a product of two lower degree polynomials, it would have one irreducible factor that is at most cubic, and hence appear on that list.
We can immediately rule out $x$ and $x+1$ as factors, because $p(x)$ has no zeros in the prime field. We can rule out $x^2+x+1$, because that is a factor of $x^3-1$, and hence also of $x^6-1=x^6+1$. So it can not divide $p(x)$, because then it would have to be a factor of $p(x)-(x^6+1)=x^7$ also, but that is clearly not the case. Eliminating the candidate cubic factors depends on a similar trick (based on the theory of finite fields) that the irreducible cubics are factors of $x^8+x$, and hence also of $x^7+1$. So if either of them divided your $p(x)$, it would also have to be a factor of $p(x)-(x^7+1)=x^6$, which is, again, obviously not the case.
Conclusion. $p(x)$ is irreducible. |
H: Prove that 16, 1156, 111556, 11115556, 1111155556… are squares.
I'm 16 years old, and I'm studying for my exam maths coming this monday. In the chapter "sequences and series", there is this exercise:
Prove that a positive integer formed by $k$ times digit 1, followed by $(k-1)$
times digit 5 and ending on one 6, is the square of an integer.
I'm not a native English speaker, so my translation of the exercise might be a bit crappy. What is says is that 16, 1156, 111556, 11115556, 1111155556, etc are all squares of integers. I'm supposed to prove that. I think my main problem is that I don't see the link between these numbers and sequences.
Of course, we assume we use a decimal numeral system (= base 10)
Can anyone point me in the right direction (or simply prove it, if it is difficult to give a hint without giving the whole evidence). I think it can't be that difficult, since I'm supposed to solve it.
For sure, by using the word "integer", I mean "natural number" ($\in\mathbb{N}$)
Thanks in advance.
As TMM pointed out, the square roots are 4, 34, 334, 3334, 33334, etc...
This row is given by one of the following descriptions:
$t_n = t_{n-1} + 3*10^{n-1}$
$t_n = \lfloor\frac{1}{3}*10^{n}\rfloor + 1$
$t_n = t_{n-1} * 10 - 6$
But, I still don't see any progress in my evidence. A human being can see in these numbers a system and can tell it will be correct for $k$ going to $\infty$. But this isn't enough for a mathematical evidence.
AI: Mark Bennet already suggested looking at the numbers as geometric series, so I'll use a slightly different approach. Instead of writing the squares like that, try writing them as follows:
$$\begin{align} 15&.999\ldots = 16 \\ 1155&.999\ldots = 1156 \\ 111555&.999\ldots = 111556 \\ \vdots\end{align}$$
These numbers can be expressed as a sum of three numbers, as follows:
$$\begin{align} 111111&.111\ldots \\ 444&.444\ldots \\ 0&.444\ldots \\ \hline 111555&.999\ldots \end{align}$$
Since $1/9 = 0.111\ldots$, we get
$$\begin{align} 111111&.111\ldots = \frac{1}{9} \cdot 10^{2k} \\ 444&.444\ldots = \frac{1}{9} \cdot 4 \cdot 10^k \\ 0&.444\ldots = \frac{1}{9} \cdot 4 \\ \hline 111555&.999\ldots = \frac{1}{9} \left(10^{2k} + 4 \cdot 10^k + 4\right). \end{align}$$
But this can be written as a square:
$$\frac{1}{9} \left(10^{2k} + 4 \cdot 10^k + 4\right) = \left(\frac{10^k + 2}{3}\right)^2.$$
Since $10^k + 2$ is always divisible by $3$, this is indeed the square of an integer. |
H: Relation between sides and angles.
Is this phrase safe to consider in general:
Equal sides of a polygon have corresponding equal angles
if not how would you refine or correct it
Example of a corresponding angle would be
Edited:
For example suppose you ignore the fact that the above triangle is an equilateral and treat it as a regular polygon.Then according to the above statement Angle A and Angle B are both equal because Side A corresponds to angle A which is equal to SideB which corresponds to angle B
AI: Construct a pentagon of all equal sides. Physically, out of pieces of straws or something. Note how you can bend it in any which way so that no angles are the same. So the question of "equal sides correspond to equal angles" is true for triangles, can maybe be rephrased in a way that's true for quadrilaterals (since there it forces two pairs of equal angles), but definitely fails to work in general.
A better statement to prove might be the reverse: Do equal angles of a polygon imply equal sides? |
H: For which $n$, $G$ is abelian?
My question is:
For Which natural numbers $n$, a finite group $G$ of order $n$ is an abelian group?
Obviouslyو for $n≤4$ and when $n$ is a prime number, we have $G$ is abelian. Can we consider any other restrictions or conditions for $n$ to have the above statement or the group itself should have certain structure as well? Thanks.
AI: Every group of order $n$ is abelian iff $n$ is a cubefree nilpotent number.
We say that $n$ is a nilpotent number if when we factor $n = p_1^{a_1} \cdots p_r^{a_r}$ we have
$p_i^k \not \equiv 1 \bmod{p_j}$ for all $1 \leq k \leq a_i$.
(adapted from an answer by Pete Clark.) |
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