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H: How to formally justify the existence of a limit with two variables?
Problem: Find the limit of the following functions
a) $\displaystyle \lim_{x \to \infty, \ y\to \infty}$ $\frac{x+y}{x^2 + y^2} $
b) $\displaystyle \lim_{x \to 0,\ y\to 2} \frac{\sin(xy)}{x} $
I know how to make cases when the limit does not exists, but I am having trouble to find the limits when they exists. For example, the limit of the second function is clearly 2, but I couldn't find out how to write it formally (not with the formal definition, but in a way it is correct).
AI: Hints: $$\left|\frac{x+y}{x^2+y^2}\right| \le \frac{|x|}{x^2+y^2} + \frac{|y|}{x^2+y^2} \le \frac1{|x|} + \frac1{|y|}$$
and
$$\frac{\sin xy}{x} = y\cdot \frac{\sin xy}{xy}.$$ |
H: Do the paths intersects? If so where
There are two unidentified objects in the sky. The path of the first object is given by $r(t)=\langle t,-t,1-t\rangle $ and the second object's path is $s(t)= \langle t-3,2t,4t\rangle$
Do the paths intersect? If so where and do the objects collide?
AI: You have two objects with paths $r(t)$ and $s(t)$.
If you ask whether the two paths intersect at some point, then you just want to know if there are values $t_1$ and $t_2$ such that $r(t_1) = s(t_2)$. That is
$$
t_1 = t_2 - 3\\
-t_1 = 2t_2 \\
1 - t_1 = 4t_2
$$
Can you find a $t_1$ and $t_@$ such that this is true? If so, Ttat means that the two object will both pass the point, just not necessarily at the same time.
So want to figure out whether or not the objects collide. That is, you want to figure out whether or not there is one value of $t$ such that $r(t) = s(t)$. That is
$$
\langle t, -t, 1-t\rangle = \langle t - 3, 2t, 4t\rangle
$$
So you ask if this equation has any solution. If it does have a solution, then the two objects with be in the same location at that time, and so they collide. |
H: Prove a relation is asymmetric.
Prove that if a non-empty relation $R$ on $A$ is transitive and irreflexive, then it is asymmetric.
I assume that I need to prove this one by contradiction, but I'm having a hard time wrapping my head around it. If a relation is transitive and irreflexive wouldn't it also be symmetric?
AI: Suppose for a contradiction that we can find $x,y \in A$ such that $(x,y) \in R$ and $(y,x) \in R$. Then by transitivity, $(x,x) \in R$, contradicting irreflexivity.
Edit. By the way, if this still seems at all mysterious, try visualizing it as follows. Consider two points $x$ and $y$, and an arrow going $x \rightarrow y$ and another $y \rightarrow x$. Use transitivity to deduce that there must be a loop at $x$. But irreflexivity basically says: "no loops." |
H: Lambda Calculus: Reduction to Normal Form
I'm working on some problems where I'm supposed to reduce lamda terms to normal form. I'm not sure if I'm doing it right so if someone could let me know, that would be awesome.
$$(\lambda x.\lambda y.x*2+y*3)\; 5 \;4 $$
$$\rightarrow(\lambda y.5*2+y*3) \; 4 $$
$$\rightarrow(5*2+4*3) $$
$$\rightarrow 22$$
And secondly, how does that equation differ from $\lambda x.(\lambda y.x*2+y*3) \; 5$? They just moved the parenthesis over $1\ldots$ does that mean the $λx$ would just stay in the answer? as in:
$$\lambda x.(\lambda y.x*2+y*3) 5$$
$$\rightarrow\lambda x.(x*2+5*3)$$
And would I solve it to be:
$$\rightarrow \lambda x.(x*2+15)$$
Thanks in advance,
Sean
AI: Yes, you are right.
Let $$f = \lambda x.\ \lambda y.\ x\cdot 2 + y \cdot 3,$$
that is, (using different notation) we have $f(x)(y) = 2x+3y$, and so $(f\ 5\ 4) = f(5)(4) = 10+12 = 22$. On the other hand, if you won't take whole $f$ into parentheses, then we arrive at
$$\Big(\lambda x.\ (\lambda y. x \cdot 2 + y \cdot 3)\ 5 \Big)= (\lambda x.\ x \cdot 2+15),$$
if we were to pass $4$ later, then we get $(\lambda x.\ x \cdot 2 + 15)\ 4 = 8+15 = 23$.
The difference is that the scope of $\lambda$ extends to as far right as possible, so the additional $5$ is under the scope of $\lambda x$. Hence, $5$ would be substituted into $y$. However, in the former example, the number $5$ was separated from both lambdas by parentheses, and so it would be substituted into $x$.
I hope this helps $\ddot\smile$ |
H: Inverse of a real matrix plus identity times i
How would you proof that given a real square matrix $A$
then the inverse of the matrix ( $A + i I $) exists?
AI: It won't necessarily exist. For instance, if
$$
A = \left(\begin{array}{cc}0 & 1 \\ -1 & 0 \end{array} \right)
$$
then
$$
A + iI = \left(\begin{array}{cc}i & 1 \\ -1 & i \end{array} \right)
$$
has determinant $i^2 + 1 = 0$.
There are conditions that you can place on $A$ to ensure $A + iI$ is invertible.
HINT: fill in the blank:"$A - \lambda I$ is non-invertible if and only if $\lambda$ is an ____ of $A$". |
H: Is gluing the boundary of $2$ discs removed from a surface also a surface?
Suppose we have a surface $S$. Also, suppose we remove $2$ discs from the surface $S$ and we glue the boundary circles of these two discs together. Is the result a surface?
My beliefis that it is a sphere. But I am not sure how to show it. Can someone help me see that?
AI: You will not get a sphere. If you start with a sphere, then you come up with a cylinder when you delete $2$ discs from it. Now if you glue these two boundaries, you get either a torus or a Klein bottle, depending on how you glue the two boundaries. |
H: How do we explain to students that division by a vector does not make sense?
I have to give a presentation on vector analysis. One of many important things I want to emphasize is that a division by a vector does not make sense.
How do you explain to your students, for example, that division by a vector does not make sense?
Bonus question: Also how do you explain that integration with $dx$ in the denominator does not make sense? Consider the following $\int f(x)\frac{1}{\textrm{d}x}$.
AI: Division should be the inverse of multiplication. What is multiplication for vectors? You could multiply a vector by a scalar obtaining a vector, or we could multiply two vectors to obtain a scalar.
In the first case on could possibly define division... but it would be defined only in the very restrictive case of a vector being a multiple of the other. In the second case (scalar multiplication) division would not be well defined because there are many vectors which multiplied to a given vector give the same result.
As for the second question... you cannot explain all things that do not make sense. You define some things, and that's all. Otherwise you should explain why we don't define untegrals or why we don't define $\sqrt{\int^\partial}$.
Don't let your students think that everything that can be written could possibly make sense... |
H: Determine convergence $\sum_{n=1}^\infty\left(\frac{2 n + 2}{2 n + 3}\right)^{n^2}$
Does $$\sum_{n=1}^\infty\left(\frac{2 n + 2}{2 n + 3}\right)^{n^2}$$ converge?
Hi, I was wondering if anyone knows how to solve this problem? I think I can't use root test... because the result is 1 and it is meaningless. Thank you guys very much.
AI: $$\frac{2n+2}{2n+3}=1-\frac1{2n+3}\leqslant1-\frac1{5n}\leqslant\mathrm e^{-1/(5n)}\implies\left(\frac{2n+2}{2n+3}\right)^{n^2}\leqslant\mathrm e^{-n/5}\leqslant(0.9)^n$$ |
H: Irreducible subsets of a topological space
I found this definition on Hartshorne, Algebraic geometry, page 3...
Definition A nonempty subset $Y$ of a topological space $X$ is irreducible if it cannot be expressed as the union $Y=Y_1\cup Y_2$ of two proper subsets, each one of which is closed in $Y$. The empty set is not considered to be irreducible.
I suppose this definition is made for algebraic sets and Zariski topology, but I was wondering if it could be applied to different contexts...For instance, take $X:=\mathbb{R}^2$ with the standard topology. Let $Y$ be a line, for example the $x$-axis. So $Y$ is reducible (i.e. not irreducible) because for example $Y=(-\infty,1]\cup [0,+\infty)$. First question: does it make sense what i just said?
It seems to me that in the standard topology on $\mathbb{R}^n$ "everything" is reducible in a trivial way, and this seems to me very strange.
So could you provide some examples of irreducible subsets of a not-too-exotic topological space?
A minor question: why is it made explicit the fact that the empty set is not irreducible? I mean, the empty set has no proper subsets, or not!?
AI: In the language of general topology (i.e. not algebraic geometry) such a space is more commonly known as hyperconnected, though either term is acceptable. Intuitively, $\mathbb R$ with its standard topology has far too many open sets to be hyperconnected/irreducible. For a set to be hyperconnected its open sets need to be "large", in the sense that every pair of nonempty open sets needs to have nonempty intersection (this is an equivalent definition to the one Hartshorne uses). Of course, that automatically rules out any Hausdorff spaces (except fairly trivial cases) so examples in terms of "ordinary" things like metric spaces are impossible.
The cofinite topology on any infinite set is an example of a hyperconnected space, and this is probably a better thing to think about intuitively to get a grasp on the Zariski topology. Indeed, the Zariski topology on $\mathbb A^1$ agrees with the cofinite topology, but in larger dimensions the Zariski topology is somewhat larger than that but still fairly small compared to most of the spaces you are used to. |
H: Is Minkowski space not a metric space?
I've just started reading a book on functional analysis, and first definition given there is for a metric and metric space:
Let $\mathfrak{M}$ be an arbitrary set. A function $\rho\colon \mathfrak M\times\mathfrak M\to[0,\infty)$ is called metric if it has the following properties:
1) $\rho(x,y)=0 \iff x=y$ (axiom of identity)
2) $\rho(y,x)=\rho(x,y)\;\forall x,y\in\mathfrak M$ (axiom of symmetry)
3) $\rho(x,y)\le\rho(x,z)+\rho(z,y)\;\forall x,y,z\in\mathfrak M$ (triangle inequality)
The pair $(\mathfrak M,\rho)$ is called metric space.
First and second identities don't make any surprise, I understand them. But what about image of $\rho$ and third inequality? They don't seem to hold in Minkowski space, where if we check interval as candidate for metric, we get $$s^2=t^2-x^2-y^2-z^2,$$
which can be negative and violate triangle inequality (and if take square root, it becomes complex and inequality makes no sense in this case).
So this clearly isn't a metric. But is Minkowski space then not a metric space? What is it then?
AI: Minkowski space is a metric space, but the metric is not the "Minkowski metric". |
H: If a cyclic group has an element of infinite order, how many elements of finite order does it have?
If a cyclic group has an element of infinite order, how many elements of finite order does it have?
I know that the order of the entire group must be infinite, for an element of the group must have an order less than the group order. My first thought was that there are no elements with finite order in this group, however now I'm believing that there are infinitely many elements of finite order, since the group should have infinitely many elements. Can someone explain the solution?
AI: Your first thought was almost right: the only element of finite order is the identity. If $G$ is an infinite cyclic group, then $G=\{g^n:n\in\Bbb Z\}$ for some $g\in G$. If $x\in G$, then $x=g^m$ for some $m\in\Bbb Z^+$. If $m\ne 0$, so that $x$ is not the identity element, then for each $k\in\Bbb Z$ we have $x^k=(g^m)^k=g^{km}$, and $g^{km}=g^0$ if and only if $km=0$ and hence if and only if $k=0$. In other words, no positive power of $x$ is the identity, and therefore $x$ has infinite order. |
H: Some problems about functions.
1- Let $X = \{1,2,3,7,12\} $ and $Y = \{1,15,7,4,20\} $. We use notation $(x,y)$ to denote that the element $x \in X$ is assigned to (or paired with) the element $y \in Y$. For the relations defined below answer the following questions:
Does the relation define a function from $X$ to $Y$?
If it does not define a function, explain why not.
If the relation defines a function, decide whether the function is injective or not and explain why. Also decide whether the function surjective or not and explain why.
$$\begin{align}
(a)&\Big\{ (1,15),(7,7),(3,7),(12,4) \Big\}\\
(b)&\Big\{ (1,1),(3,4),(7,7) \Big\}\\
(c)&\Big\{ (1,15),(3,7),(7,4),(12,20) \Big\}\\
(d)&\Big\{ (1,4),(3,7),(7,1),(1,15) \Big\}
\end{align}$$
2- Let $g \, : X \to Y$ be a function. Suppose $A \subseteq X$, that is $A$ is a subset of or equal to $X$. Suppose $B=g(A) \subseteq Y$. Answer the following questions:
$\begin{align}
(a)&\text{If}\, x \in A, \text{what can you say about}\, g(x)?\\
(b)&\text{If}\, y \in g(A), \,\text{what does this mean?}\\
(c)&\text{If}\, x \in X \, \text{and}\, g(x) \in g(a), \,\text{is it necessarily true that}\, x \in A?
\end{align}$
I just have a few issues.
For 2. I said all of those except (d)[as it is not a function ] are not surjective since there is always one element in Y that is not paired. Is that right?
For 3.
a) g(x)=B?
b) A=x.
c) Is not always true. It is not the case for functions that are not injective.
I am really having doubts about question 3 ( all of it ). Are my answers correct?
Many thx in advance.
AI: (b) is not a function $X\to Y$ either as $12$ is not mapped anywhere. (c) is injective and surjectivity is indeed already not possible because $Y$ has more elements than $X$.
3.(a). No. $g(x)\in B$.
(b) No. $y\in B$ or more interestingly, there exists $x\in A$ with $y=g(x)$.
(c) You are right |
H: Definiteness of A'BA
Let $A$ be a $(k \times n)$ matrix and $B$ a $(k \times k)$ matrix. In that case, is there a general result for the definiteness of the $(n \times n)$ matrix $A'BA$? If not, what if $B$ is known to be positive definite. Can the definiteness of $A'BA$ then be determined?
Best,
Esben
AI: I assume your matrices are real. $A'BA$ has rank at most $k$, so if $k < n$ it certainly can't be positive definite or negative definite. If $B$ is positive semidefinite,
$A'BA$ is positive semidefinite, since for any vector $x$ we have
$$ x' A' B A x = (A x)' B (A x) \ge 0$$ |
H: Help me Verifying that the equation is integrable and finding its solution
How can I verify that the equation is integrable and that find its solution;
$$2y(a-x)dx+[z-y^2+(a-x)^2]dy-ydz=0$$
Honestly, I tried too much, but I got too strange results,thus I couldnt show my efforts here so sorry. Thank you for helping.
AI: Hint:
1-Verify that
$$
curl \left(2y(a-x),\,z-y^2+(a-x)^2,\,-y\right)=(0,0,0)
$$
Actually, it is not true. May be there is a typo. I believe the second component of the vector field is $-[z-y^2+(a-x)^2]$.
2- Find the potential of the vector field $\left(2y(a-x),\,-[z-y^2+(a-x)^2],\,-y\right)$.
edit:
You need a function $f$ such that $\nabla f(x,y,z)=\left(2y(a-x),\,-[z-y^2+(a-x)^2],\,-y\right)$. Then
$$
f(x,y,z)=\int 2y(a-x)\,dx+C(y,z)=-y(a-x)^2+C(y,z)
$$
If we derive in $y$
$$
-[z-y^2+(a-x)^2]= f_y(x,y,z)=-(a-x)^2+C_y(y,z)
$$
then
$$
C(y,z)=-zy+\frac{y^3}{3}+B(z)
$$
and
$$
f(x,y,z)=-y(a-x)^2-zy+\frac{y^3}{3}+B(z)
$$
Now, if we derive in $z$:
$$
-y=f_z(x,y,z)=-y+B'(z)
$$
Then... |
H: Linear program dual
We are trying to find the dual of the following linear program.
$$ \max_x \ 2x_1 \ + x_2 \ \ \ \ -- (1) $$
such that,
$$ x_1 + x_2 \leq 2 \ \ \ \ -- (2)\\ -x_1 - x_2 \leq -4 \ \ \ \ -- (3)\\
x_1 \geq 0 \\ x_2 \geq 0$$
I get the following answer.
$$ \min_\theta \ 4\theta_1 + 2\theta_2 $$
subject to,
$$ -\theta_1 - \theta_2\geq -2 $$
$$ \theta_1 + \theta_2\geq 1 \\ \theta_1 \geq 0 \\ \theta_2 \geq 0$$
Is this correct?
I'm confused because of $(3)$
AI: Following the method described here:
http://www.cs.columbia.edu/coms6998-3/lpprimer.pdf
I get the following answer.
$$ \min_\theta \ 2\theta_1 - 4\theta_2 $$
subject to,
$$ \theta_1 - \theta_2\geq 2 $$
$$ \theta_1 - \theta_2\geq 1 \\ \theta_1 \geq 0 \\ \theta_2 \geq 0$$ |
H: what is a multiplicative group in prime order p?
on pg. 378 section 2 (Overview) it says "We let G be a multiplicative group of prime order p , and g be a generator of G. We let e : G x G --> $G_T$" be a bilinear map.
If somebody could please break each piece of this into smaller parts I would really appreciate it. Here is my attempt:
G is a multiplicative group in order p. Since G is cyclic this means any member of G multiplied by any integer mod p yields identity (1).
g being the generator, means that you always start with g. So you can raise g to a power or you can multiply g by a random integer. But you always have to get 1 mod p for it to be a member of G.
Before looking at binlinear map, I thought I should first read what a linear map is. According to wikipedia, a linear map always yields the same subspace of the input subspaces. So bilinear I think you can end up with something that is not linear (like an elliptic curve?).
My understanding of all this is quite fuzzy and I would appreciate it if someone could explain these to me in simple English or easy-to-understand drawings.
AI: $G$ is not the group of integers mod $p$.
$G$ is a group, that is a set with a composition that obeys the group axioms.
it is multiplicative, that is we agree to use $\cdot$ (and not $+$, say) as symbol for the composition
it s of order $p$, that is its underlying set has $p$ elements
$g$ is a generator, which im plies that $G$ is cyclic
$e\colon G\times G\to G_T$ is bilinear, that is for each $a\in G$, the map $G\to G_T$, $x\mapsto e(a,x)$ is linear and $x\mapsto e(x,a)$ is also linear |
H: Prove d to be a metric
Goodday.
The problem is as follows:
Let $\mathbb{Z}^\mathbb{N}:=\{x:\mathbb{N}\rightarrow \mathbb{Z} \}$.
We define a function $\text{d}:\mathbb{Z}^\mathbb{N} \times
\mathbb{Z}^\mathbb{N} \rightarrow \mathbb{R}$ by the following relation: $-\text{log
d(x,y)} = \text{inf}\{\text{n}\in \mathbb{N}:\text{x(n)}\neq \text{y(n)}\}$
Show that $(\mathbb{Z}^\mathbb{N},d)$ is a metric space.
[inf $\emptyset$ = +$\infty$ and log $0$ = - $\infty$]
I have difficulties proving the triangle inequality for this metric.
Could you give me a hint (no solution if possible)?
Thanks!
AI: Hint: If $n\in\Bbb N$ is such that $\text{x}(n)\ne\text{z}(n),$ then we must have $\text{x}(n)\ne\text{y}(n)$ or $\text{y}(n)\ne\text{z}(n).$ |
H: Understanding syntax for defining a relation.
Let T = {1, 2, 3, 4} and A = T * T. We can define a relation R on A;
(a,b)R(c,d) <=> (a/b)=(c/d)
For example:
(1,2)R(2,4) since (1/2)=(2/4)
Does this mean that ((1,2),(2,4)) ∈ R
or
(1,2) ∈ R and (2,4) ∈ R
AI: It means that ((1,2),(2,4)) ∈ R |
H: Given a circle γ(t) = (cos t, sin t, 0);Is γ an asymptotic curve of the unit sphere with centre in (0, 0, 0)?
I know that the dot product of γ// and N gives the normal curvature andig its 0, then it's called an asymptotic curve.
The equation of the given sphere is x^2 +y^2 + z^2=1---as its an unit sphere with centre at (0,0,0)
But my problem is, here ow can I find out N?i.e.,the unit normal vector on this sphere?
AI: If $(x,y,z)$ is a point on the sphere, then the normal there is also $(x,y,z)$.
In general, if a surface is given as a level set $F(x,y,z) = 0$, then the normal to the surface is the gradient $\nabla F$. |
H: Limit of a sequence true or false statement
Suppose $S(0)=2$, and $$S(n+1)=\dfrac{(S(n))²+2}{2S(n)}.$$ Then $\lim_{n\to\infty} S(n)= \sqrt2$.
So I worked some of the terms out and I get $S(1)=\frac{3}{2}$, $S(2)= \frac{17}{12}$,...
I did the limit of $(x^2+2)/(2x)=∞$ so the statement is false or what am I missing here?
I think $\sqrt{2}$ is the infimum.
AI: If you can show that the sequence converges at all, then letting $L$ be the limit, we take the limit as $n\to\infty$ on both sides of the recurrence $$s_{n+1}=\frac{s_n^2+2}{2s_n},$$ to obtain $$L=\frac{L^2+2}{2L}.$$ Can you take it from there? |
H: conversion of discrete to continuous
Given $N_{j+1}-N_j=kN_j$
How can I substitute some time variable in to make $delta(t)$ small? Meaning change in time.
I need to show $N_j=e^{(j\ln(1+k))}$
How can I rewrite the given in terms of delta t in order to take limit to find derivative ?
AI: Actually this dynamics is solvable, giving $N_j=(1+k)^jN_0$ for every integer $j$, hence $N_j=n(j)$ where $n(t)=(1+k)^tN_0$ for every real number $t$. This is $n(t)=\mathrm e^{at}N_0$ for $a=\ln(1+k)$ hence $n'(t)=a\mathrm e^{at}N_0$, that is, $n'(t)=\ln(1+k)\cdot n(t)$. |
H: Ordinary Differential Equation how to solve this
Hi i need to learn in fast way how to solve ode i still have problem with this.
I need to find $y = y(x)$
having
$y′ = 0$,
$y′′ = 0$,
$y′′′ = 0$,
$y'''' = 0$
I am guessing that $y= e^{rx}$ Then we have some function with degree of 4 but how to solve this. I am not even sure that my guess is correct xD.
I think we can write $a_0y+ a_1y' + a_2y''+a_3y'''+a_4y''' =0$.
It looks maybe simple but i dont knew how to solve this. I probably need to use some high mathematical knowledge. When my knowledge about maths end on simple integrals and etc.
AI: $y' = 0 \iff y = C, C$ constant.
Remember that for y = y(x), y' means the rate of change of y with respect to x. In other words, as you increase or decrease x, how does y change? If y' is 0 it means there's no change in y as you increase or decrease x, which means y has a constant value. To find out the value of C, you'd need further information e.g. an initial condition such as $y(0) = 4$, for which the solution is clearly $y = 4$. |
H: A question about extreme points
If the extreme points of the unit ball of $C[0, 1]$ are $\pm{1}$, where $C[0, 1]$ is the Banach space of all continuous real-valued functions on the unit interval, then what would the extreme points of the unit ball be if we considered all continuous complex-valued functions on the unit interval? Would the extreme points in that case include all complex numbers $z$ such that $|z|= 1$?
AI: The extreme points would be all continuous functions $f$ with $|f(t)| = 1$ for all $t \in [0,1]$. |
H: If the expectation exists and finite, is it always integrable?
If the expectation of a random variable exists and finite, i.e., $EX<\infty$, is it always true that $E |X|<\infty$?
This question arises when I try to prove some stochastic process is a martingale. In the definition of martingale, we need the process to be integrable. But if we already know its expectation and it's finite, do we still have to prove that $E|X|<\infty$?
For example, $W_t^2-t/2$, where $W_t$ is a Wiener process. We know that $E(W_t^2-t/2)=t/2<\infty$, do we have to calculate $E|W_t^2-t/2|$?
AI: Briefly, yes, $E|X|<\infty$. If $E|X| = \infty$, then either $E(X^+)$ or $E(X^-)$ is $\pm\infty$, and $EX$ can't be finite, since $EX = EX^+ + EX^-$. |
H: Special function
I am looking for a function $f:\mathbb{R}\to \mathbb{R}$ with $f(\overline{A})\subset\overline{f(A)}$. $\subset$ means $f(\overline{A})\neq\overline{f(A)}$.
Does anyone has an idea?
AI: Hint: Choose some $y$ that you want to be in $\overline{f(A)}$ but not in $f(\overline{A})$. Now look for a function $f$ that takes values arbitrarily close to $y$ but not $y$ itself. |
H: Using Matlab's bsxfun to subtract matrices of different sizes
My professor gave me some matlab script in which I found the following line:
differences = bsxfun(@minus, A, B);
where A is of size [200,50] and B is of size [1,50]. The documentation and forums didn't give me the answer I'm looking for. My current understanding is that every column of A matrix gets subtracted with B matrix. Is my understanding correct? Currently I don't have matlab installed which makes it a bit problematic to test myself :)
AI: From Octave doc:
The binary singleton expansion function applier performs
broadcasting, that is, applies a binary function F
element-by-element to two array arguments A and B, and expands as
necessary singleton dimensions in either input argument. F is a
function handle, inline function, or string containing the name of
the function to evaluate. The function F must be capable of
accepting two column-vector arguments of equal length, or one
column vector argument and a scalar.
The dimensions of A and B must be equal or singleton. The
singleton dimensions of the arrays will be expanded to the same
dimensionality as the other array.
Sounds like it just expands B from [1,50] to [200,50] in the obvious way (i.e. replicating the first row of B 200 times) to make a matrix C and then performs A - C in this scenario.
Also, off-topic. |
H: Uniqueness of Inverses in Groups Implies Associativity Holds?
I was checking multiplication tables for groups with 4 elements, to see which tables "passed" the group axioms of closure, associativity, identity and inverses. But then I had a question, so hopefully someone can help me with this basic group theory question.
The proof that inverses are unique in a group depends upon the associativity axiom. Let $a$ be an element that has two inverses $b$ and $c$. Then $b = be = b(ac) = (ba)c = ec = c$. Thus the inverse is unique since $b$ must equal $c$.
So my question is this. If we see a multiplication table for a finite group, and we can easily check closure, existence of identity, and existence of inverses, and further we can see that inverses are all unique, does this necessarily imply that the last axiom, associativity, holds? If yes, how?
AI: Not by far. Just try drawing up a multiplication that satisfies identity and existence and uniqueness of inverses. These laws put very few constraints on the multiplication table -- the remaining cells can be filled in at random, and it will be very unlikely that you happen to end up with a group.
Consider, for example, the structure with three elements where each is its own inverse and the inverses are clearly unique:
| e A B
--+-------------
e | e A B
A | A e B
B | B B e
This is not a group, since $Be=BA$ but we cannot cancel the $B$ to conclude $e=A$. |
H: Inverse Trig Derivative
Here's a simple problem that I'm not getting.
y = arcsec (1/x), where o < x < 1.
What is the derivative?
y' = 1 / { x^-1 * (x^(-2) - 1)^(1/2) } * -x^-2
y' = -x / { x^2 * (x^(-2) - 1)^(1/2) }
y' = -1 / { x * (x^(-2) - 1)^(1/2) }
That's what I'm getting.
But the answer should be y' = -1 / {1-x^2}^(1/2)
That's very different.
AI: $$
y =arcsec(\frac{1}{x})\Rightarrow \frac{1}{x}=\sec y=\frac{1}{\cos y}\Rightarrow x=\cos y.
$$
i.e
$$
x(y)=\cos y\Rightarrow x'(y)=-\sin y
$$
$$
x'(y)=\frac{1}{y'(x)}\Rightarrow y'(x)=\frac{1}{x'(y)}
$$
$$
y'(x)=-\frac{1}{\sin y}=-\frac{1}{\sqrt{1-\cos^2 y}}=-\frac{1}{\sqrt{1-x^2}}
$$
$$
y' =arcsec(\frac{1}{x})=-\frac{1}{\sqrt{1-x^2}}
$$ |
H: Nested Cell Property: I'm missing something about this proof
This is a paraphrased version of the beginning of a proof given in The Elements of Real Analysis: Second Edition By Robert G. Bartle, on page 47.
Nested Cell Property. If $n\in\mathbb{\mathbb{N}}$
let $J_{n}$
be a non void closed cell in $\mathbb{R}$
and suppose that that this sequence is nested in the sense that
$$J_{1}\supset J_{2}\supset\cdots\supset J_{n}\supset\cdots$$
Then there exists an element which belongs to all of these cells.
Proof. Suppose that $J_{n}=[a_{n},b_{n}]$
, where $(\forall n\in\mathbb{N})(a_{n}\leqslant b_{n})$
. Thus the set $\{a_{n}\:|\: n\in\mathbb{N}\}$
is bounded above by $b_{1}$
hence the supremum of $\{a_{n}\:|\: n\in\mathbb{N}\}$
exists, let $u=\sup\{a_{n}\:|\: n\in\mathbb{N}\}$
. Then we have that $(\forall n\in\mathbb{N})(a_{n}\leqslant u)$
. The claim is that $u\leqslant b_{n}$
for all $n\in\mathbb{N}$
. Otherwise, $(\exists i\in\mathbb{N})(b_{i}<u)$
. Since $u=\sup\{a_{n}\:|\: n\in\mathbb{N}\}$
there must exist a $a_{k}$
such that $b_{i}<a_{k}$ .
I don't see how we get this last inequality. $a_{k}\leqslant u$
and $b_{i}<u$
, but nothing forces $a_{k}=u$,
so why is it necessary that $b_{i}<a_{k}$? The rest of the proof is clear to me.
AI: For each fixed $m$, $b_m$ is an upper bound for the set $\{a_n:\ n\in \mathbb N\}$. Thus by definition $u\leq b_m$. Or using your line of argument, if $b_i<u$ for some $i$. Let $\epsilon >0$ such that $b_i < u - \epsilon < u$. Then by definition of $u$, there is $a_k$ such that $a_k > u- \epsilon > b_i$. |
H: Why does $(3\sqrt3)^2 = 27$?
How does $(3\sqrt3)^2 = 27$?
I've tried to solve this using binomial expansion and using the FOIL method from which I obtain $9 + 3\sqrt3 +3\sqrt3 + 3$.
it has been a while since I've done this kind of thing so it may be something obvious that I can't see.
AI: Recall that $$(ab)^2 = (a)^2 (b)^2$$
Here, $a = 3$ and $b = \sqrt 3$.
$$(3 \sqrt 3)^2 = (3)^2(\sqrt 3)^2 = 9 \cdot 3 = 27.$$ |
H: Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$
Let $\mathbb{Z}_x$ denote the ring of integers with the operations $\odot, \oplus $ defined as $a \odot b$ = $ a+b -a b$ and $a \oplus b$ = $a+b -1$. Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$.
Not sure how to start.
AI: So we want to find a bijection $f \colon \def\Z{\mathbb Z}\Z \to \Z$ such that
$$ f(a+b) = f(a) \oplus f(b) , \quad f(ab) = f(a) \odot f(b) $$
holds for all $a,b \in \Z$. Let's suppose we have such an $f$ and want to find $f(0)$, for all $a$ we must have
\begin{align*}
f(a) &= f(a+0)\\
&= f(a) \oplus f(0)\\
&= f(a) + f(0) - 1\\
\iff f(0) &= 1
\end{align*}
Using the same idea, we must have
\begin{align*}
f(a) &= f(a\cdot 1)\\
&= f(a) \odot f(1)\\
&= f(a) + f(1) - f(a)f(1)\\
\iff f(1) &= f(a)f(1)
\end{align*}
As $f$ is onto, we must have $f(1) = 0$. Now we can get an idea what to do for a general $n$:
\begin{align*}
f(n+1) &= f(n) \oplus f(1)\\
&= f(n) \oplus 0\\
&= f(n) - 1
\end{align*}
So if there is such an $f$, we must have: $f(1) = 0$, $f(n+1) = f(n)- 1$ for any $n$, so we must have $f(n) = 1-n$ for every $n$. Now check if this $f$ is the desired isomorphism. |
H: Recursively enumerable language of Turing machines
If you have the language $L_{h}=\{M_{i} | (\exists z \in \sum ^{*}) M_{i}\text{ halts on some input } z\}$
where $M_{i}$ are Turing machines, is $L_{h}$ recursively enumerable?
I'm fairly certain it is, but I'm having issues proving that to be the case.
The way I understand RE languages is that they CAN be infinite, which this language most certainly is, but I have no idea where to go from there.
Thanks in advance.
AI: Hint:
First, is the language of all Turing machines enumerable?
Consider a language $L_i = \{M_i \mid M_i \text{ halts on some input }\}$. Its size is at most $1$, how to "enumerate it"?
Run in parallel $L_i$ for all Turing machines, and if some of $L_i$ finishes, output its unique element.
I hope this helps $\ddot\smile$ |
H: sequence of sets with $\limsup A_n = \mathbb N$
Find a sequence of one-point-sets $A_n = \{\ell_n\}$ with $\ell_n\in\mathbb N$ for all $n\in\mathbb N$, such that
$$\limsup_{n\to\infty} A_n=\mathbb N$$
I know the definition of the $\limsup$ of a sequence of sets,
$$\limsup_{n\to\infty} A_n = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$$
I know the sequence has to contain each number of $\mathbb N$ infinitely often, but I'm not able to find a suiting sequence.
AI: $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,\ldots$ |
H: Science Fair Project - Square roots
For a school science fair project, I need four or five square root algorithms to use. Googling gives sites like http://en.wikipedia.org/wiki/Methods_of_computing_square_roots with so many methods that I don't know how to narrow it down to four or five.
I desire the most standard ones, that are simple to program in java. For example, the classic Newton's Iteration and the Bhaskara-Brouncker algorithm (The two i found so far)
Also, I found Square Root Algorithm except I don't know the names of the algorithms used
I need an answer urgently. Thanks in advance.
AI: I'd go for:
1) Newton's Method
2) Midpoint Method
3) Secant Method
4) Taylor Series approximation
They're all similar in spirit in the sense that they all involve the derivative in some way and they're super simple to program because they're just math formulas you can just type in Java. |
H: Is the converse of Stolz Caesaro Lemma true?
Let $(a_n)_n, (b_n)_n$ be two real sequences s.t. $(b_n)_n$ is strictly increasing and unbounded. Prove that $$\lim_{n\rightarrow \infty} \frac {a_n}{b_n}= \lim_{n\rightarrow \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ if limit on the right exists.
I could prove this. But I thinking whether the converse is true, namely: if the limit on the left exists, what about the right limit? Does it exist? Or are there counterexamples?
EDIT: There is this post that answers my question, but does not have much content.
AI: Let $a_n=\sqrt n +(-1)^n$ and $b_n=\sqrt n$. Then the limit on the left is $1$, whereas the general term on the right is
$$ \begin{align}\frac{\sqrt{n+1}+(-1)^{n+1}-\sqrt n-(-1)^n}{\sqrt{n+1}-\sqrt n}&=1-(-1)^n\frac{2}{\sqrt{n+1}-\sqrt n}\\&=1-2(-1)^n(\sqrt{n+1}+\sqrt n)\\&\approx 1\pm4\sqrt n\end{align}$$
and no limit exists on the right. |
H: Equivalence classes for a Relation on a product set.
How do you determine the the equivalence classes for a relation on a product set?
Background:
Let $S=\left\{1,2,3,4\right\}$ and $A=S\times S$. The relation $R$ on $A$ can be defined by
$$\left(a,b\right)R\left(c,d\right) \iff a/b =c/d$$
For example:
$$\left(1,2\right)R\left(2,4\right) \text{ since } 1/2 = 2/4$$
Assuming $R$ is an equivalence relation, what are the equivalence classes for $A/R$?
AI: Just do it one by one: (I use $\sim$ for $R$)
$(1, 1) \sim (2, 2) \sim (3, 3) \sim (4, 4)$
$(1, 2) \sim (2, 4)$
$(1, 3)$
$(1, 4)$
$(2, 1) \sim (4, 2)$
$(2, 3)$
$(3, 1)$
$(3, 2)$
$(3, 4)$
$(4, 1)$
$(4, 3)$ |
H: Subspace of $C^1 [0,1]$
Consider the inner product space of continuously differentiable functions, $C^1 [0,1]$ with inner product:$$\left<f,g\right> =\int_{0}^1f(x)\overline{g(x)}\,dx + \int_{0}^1f'(x)\overline{g'(x)}\,dx $$
Show $\left<f,\cosh\right> = f(1)\sinh(1)$ for any $f$ $\in$ $C^1 [0,1]$
Use part 1. to show the subspace:$$\{f\in C^1 [0,1]:f(1)=0\}$$ is a closed subspace of $C^1 [0,1]$
I can do part 1. but can't get part 2. I think I have to show that every convergent sequence in that subspace has its limit in that subspace. Any help would be greatly appreciated
AI: Hint: Consider $\{\cosh\}^{\perp}$.
$\langle f, \cosh \rangle = 0 \quad \Longleftrightarrow \quad f(1) = 0$. |
H: Number Theory about least common multiple
Let a and b be positive integers and let [a,b] denote the least common multiple of a and b. Show that there exist integers x and y such that
$$ \left(\frac xa\right) + \left(\frac yb\right) = \left(\frac 1 {[a,b]}\right). $$
I need a nudge in the right direction - not sure where to start.
AI: Note that $ab = (a, b)[a, b]$, where $(a, b)$ denotes the greatest common divisor. The rest is given by Euclidean algorithm. |
H: How to find the determinant of this matrix
I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended up with
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1-a & a-1 & 0 & 0 \\
0 & 1-a & a-1 & 1 \\
0 & 0 & 1-a & a-1 \\
\end{bmatrix}
$
I then factored out (1-a) to get this
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 1 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}
$
which will then make my determinant a multipile of $(1-a)^3$:
$\det(A) = (1-a)^3x$
But now here I don't know what to do. I have a feeling that my approach is wrong. Any help or guide please?
AI: To find the determinant you could just do it brute force by "Expansion by Minors" using the first row:
$$\begin{align}
\det \begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix} = a\det
\begin{bmatrix}
a & 1 & 1 \\
1 & a & 1 \\
1 & 1 & a \\
\end{bmatrix} \\- 1\det\begin{bmatrix}
1 & 1 & 1 \\
1 & a & 1 \\
1 & 1 & a \\
\end{bmatrix} \\+ 1\det \begin{bmatrix}
1 & a & 1 \\
1 & 1 & 1 \\
1 & 1 & a \\
\end{bmatrix} \\- 1\det\begin{bmatrix}
1 & a & 1 \\
1 & 1 & a \\
1 & 1 & 1 \\
\end{bmatrix}
\end{align}
$$ |
H: Proving that $\{f \in End(A): \forall a \in A:|a|<\infty \implies f(a)=0\}$ is an ideal
Let $A$ be an abelian group. I need to prove that
$I = \{f \in End(A): f(a)= 0 \ \text{for all $a$ of finite order}\}$,
is an ideal of $\text{End}(A)$. It isn't hard to prove that $I$ is a subgroup of $End(A)$, but it is quite hard to prove that:
$g(x) \cdot f(x)$ and $f(x) \cdot g(x)$ are in $I$ with $f(x)\in I$ and $g(x) \in End(A)$.
I thought that if you multiply $g(x)$ and $f(x)$ and you take an element $a$ that $g(a)*f(a) = g(a)*0 = 0$, so $g(x)f(x)=0$ for $x$ with finite order so $g(x)f(x) \in I$.
Is this proof correct?
AI: If $A$ is an abelian group, the multiplication on $\operatorname{End}(A)$ is the composition. It's customary to write $A$ additively and I'll use this convention.
It's clear that $0\in I$. If $f,g\in I$, then $(f+g)(a)=f(a)+g(a)=0+0=0$ for all $a$ of finite order.
Now, let $f\in I$ and $g$ be an arbitrary endomorphism. Saying $a$ has finite order means that $na=0$ for some integer $n>0$.
Since $ng(a)=g(na)$, we see that if $a$ has finite order, then also $g(a)$ has finite order. Therefore
$$
(fg)(a)=f(g(a))=0
$$
by hypothesis. Proving that $gf\in I$ is even easier: if $a$ has finite order, then
$$
(gf)(a)=g(f(a))=g(0)=0.
$$ |
H: Even function integration problem
Let $f$ be an even function. Show
$$\int_{-a}^af(x)dx=2\int_0^af(x)dx$$
So I thought of breaking it up into two integtrals with one from $-a$ to $0$ and $0$ to $a$. Then I have on the left side
$$\int_{-a}^0f(x)dx$$
Then I thought
$$\int_{-a}^0f(x)dx=-\int_0^{-a}f(x)dx=-\int_0^{-(-a)}f(-x)dx$$ but this seems wrong. Where am I mistaken?
AI: Use the substitution $y=-x, dy=-dx$ in the $\int_{-a}^0 f(x)dx$. This turns it into $\int_0^a f(-y)dy=\int_0^a f(y)dy=\int_0^a f(x)dx$. |
H: Linear Diophantine equation - Find all integer solutions
Using the linear Diphantine equation
121x + 561y = 13200
(a) Find all integer solutions to the equation.
(b) Find all positive integer solutions to the equation.
edit: The answer I have for (a) is an equation:
$x=16800 + 51n$
$y=-3600-11n$
where $n \in \mathbb Z. $
AI: HINT:
First check if the greatest common divisor of 121 and 561 is a multiple of 13200. If so then this equation has a solution, otherwise it doesn't. Also it would be wise to divide by the greatest common divisor, so you'll work with smaller numbers.
And to obtain solutions apply the Euclidean Division Algorith. So you have:
$$561 = 4 \cdot 121 + 77$$
$$121 = 1 \cdot 77 + \cdots$$
Can you spot the pattern? Then going backwards you should be able to obtain one solution and quite easily a closed form of the solution. Then you can check when both solutions are positive.
Also I'm pretty sure there are a lot of nice books and video tutorials about solving a linear Diophantene equation on the internet. This one is quite simple and easy to follow. |
H: example of a topological space such that there exists a sequence that escapes to infinity but has convergent subsequence
Find an example of a topological space such that there exists a sequence that escapes to infinity but has a convergent subsequence
This actually is from exercise 2.15 of introduction to smooth manifolds by John Lee. In that book we knew that if $X$ is a topological manifold, then a sequence escapes to infinity(any compact subspace can only contain finitely many points in the sequence )is equivalent to having no convergent subsequence. However, I think that this might not be true in a general topological space. For example, if $X$ is not metrizable. But I have difficulty finding one.
AI: I found the relevant definition in Lee’s Introduction to Smooth Manifolds at Google Books: a sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ in the space $X$ escapes to infinity if for each compact $K\subseteq X$, $\{n\in\Bbb N:x_n\in K\}$ is finite. In other words, $\sigma$ escapes to infinity if no compact set in $X$ contains infinitely many terms of the sequence.
Let $X$ be a $T_1$ space, and suppose that $\sigma=\langle x_n:n\in\Bbb N\rangle$ has a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ converging to $x\in X$. Let $K=\{x\}\cup\{x_{n_k}:k\in\Bbb N\}$; $K$ is a compact set containing infinitely many terms of $\sigma$, so $\sigma$ does not escape to infinity. Thus, a sequence that escapes to infinity has no convergent subsequence. |
H: Is my answer correct? (And what's the name of the rule?)$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!}$
Want to know if I solved this problem correctly:
$$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!} =\lim_{n \to \infty} \frac{1 \cdot 2 \ldots(n-1)n(n+1)(n+2)(n+3) - 1 \cdot 2 \ldots(n-1)n}{(1 \cdot 2 \ldots(n-1)n(n+1)(n+2))\cdot n}=\lim_{n \to \infty}\frac{(1 \cdot 2 \ldots(n-1) n)\cdot((n+1)(n+2)(n+3)-1)}{(1 \cdot 2 \ldots(n-1)n(n+1)(n+2))\cdot n} =\lim_{n \to \infty} \frac{(n+1)(n+2)(n+3)-1}{(n+1)(n+2)\cdot n} = \frac{1}{1}=1$$
The answer is 1 because we've got $n^3$ in numerator and $n^3 $ in denominator, and the constants are both equals 1 so the limit can be calculated as $\frac{1}{1}$. (Sorry I don't remember that rule. Could you remain me please?)
AI: You can more easily write the expression (cancelling as much as possible) as $$\frac {n+3}n-\frac 1{n(n+1)(n+2)}=1+\frac 3n-\frac 1{n(n+1)(n+2)}$$
Then the limit is very easy to see. |
H: What are non-orthogonal eigenvectors?
Given a symmetric matrix $A$, the maximum of the trace, $Tr(Z^TAZ)$ under the assumption that $Z^TZ=I$ occurs when $Z$ has the eigenvectors of $A$, as $Tr(U^TAU)= \lambda_1 +\lambda_2+...\lambda_ d$ where $Z\in\mathbb{R}^{n\times d}$. I know that the eigen vectors being the solution is as a result of the Courant minimax principle.
q1) Now, I have faintly heard about non-orthogonal eigenvectors, and am very curious to know, why they are called so? Is this because they form a basis of the eigen-space of $A$ and are still non-orthogonal? What is its relation with the Courant Minimax / Courant Fischer characterization, if any?
q2) How does it fit into the trace maximization formulation given above? Especially, if there is no other constraint, when the orthogonality is dropped, doesn't the problem become unbounded or ill-posed? if so, what are these non-orthogonal eigenvectors, optimizing?
q3) If the matrix $A$ was not symmetric or non-normal, then what are its non-orthogonal eigenvectors solving for? or may be is this connected it to an SVD instead in this case?
q4) When i Google, for applications of non-orthogonal eigenvectors, i just find nothing. What are its uses!?
Please restrict your answers to be within the matrix algebra setting as much as possible, unless it strictly requires discussing it through other fields of mathematics.
AI: q1) In the case of repeated eigenvalues, like you say, any eigenvectors in the repeated eigenspace could be chosen even if they are not orthogonal. You could also be hearing about nonorthogonal eigenvectors in the context of nonsymmetric matrices, which generally don't have orthogonal eigenvectors. Having orthogonal eigenvectors is one of the nicest things about symmetric matrices.
q2) If you remove the condition $Z^TZ=I$, you could make the trace as large as you want by choosing any $Z$ where the trace is positive, and scaling $Z \rightarrow c Z$ by a larger and larger constants $c$. If instead the condition is relaxed to have the columns of $Z$ be unit vectors, then the maximum would be where all of the columns of $Z$ are identically equal to the most dominant eigenvector (eigenvector with the largest eigenvalue). If you make the columns of $Z$ all orthonormal, then that is equivalent to the original constraint $Z^TZ=I$.
q3) Not sure what this is asking. They are just the eigenvectors. Perhaps you are asking if there is a variational characterization of eigenvalues of nonsymmetric matrices? I don't know of one but there might be. The singular vectors and values always have such a variational characterization and it is related, but for nonsymmetric matrices the singular value decomposition is a different thing from the eigenvalue decomposition.
q4) Uses of the eigenvalue decomposition, symmetric or not, are hugely useful in more ways than I can post. They can be used when you want to apply the matrix again and again and again without doing the matrix product each time. They characterize the sensitivity of the matrix's action to different inputs - the dominant eigenvector is the most sensitive one, eigenvectors with smaller eigenvalues are correspondingly less sensitive. In basically any application where matrices are, the eigenvalues and eigenvectors are going to represent something important or interesting. In the nonorthogonal case, the minimum angle between eigenvectors characterizes how close the matrix is to being singular, which is important in many nummerical methods. |
H: Limits of a recursively defined sequence
Let $x_1=a$ and define a sequence $\left(x_n\right)$ recursively by:
$x_{n+1} = \dfrac{x_n}{1 + \frac{x_n}{2}}$
For what values of $a$ is it true that $x_n$ approaches $0$?
AI: Whenever the sequence converges (to $L$, say) we must have $L=\frac L{1+\frac L2}$, i.e. $L=0$. So the question is just: For which $a$ does the sequence converge?
If $a>0$ then by induction all $x_n>0$ so that the sequence is strictly decreasing and bounded from below, hence convergent.
If $a=0$ the sequence is constant, hence also convergent.
If $a<-2$ then $x_2>0$ and the sequence converges as in the case $a=x_2>0$.
If $a=-2$ the recursion fails to define $x_2$, i.e. we don ont even have a sequence.
If $-2<a<0$, the sequence starts decreasing as long as $-2<x_n<0$. It cannot remain bounded from below by $-2$ as otherwise it would converge to some number on $[-2,0)$, which is impossible. Thus after finitely many steps we obtain some $x_n<-2$ and from then on convergence as shown above. Or we hit $-2$ exactly and the recursion fails.
To see for which $a$ we hit $-2$, solve the recursion for $x_n$:
$$ x_n = \frac{2x_{n+1}}{1-x_{n+1}}.$$
Consider the sequence $y_n$ given by $y_1=2$ and $y_{n+1}=\frac{2y_n}{1+y_n}$. If and only if $-a=y_n$ for some $n$, our original sequence will hit $-2$ (for we see that $x_k=-y_{n+1-k}$ for $1\le k\le n$).
One readily shows that $y_n=\frac{2^n}{2^n-1}$.
Summary:
If $a=-\frac{2^n}{2^n-1}$ for some $n$, the sequence is only finite.
In all other cases it converges to $0$. |
H: $p\mapsto\Vert f\Vert_{L^p}$ is continuous.
Could someone help me prove the following:
Let $(\Omega,\mathscr{A},\mu)$ be a probability space. Let $f:\Omega\rightarrow\mathbb{R}$ be a non-negative measurable function.
How do I prove that the funtion $N_f:p\in [1,\infty)\mapsto \Vert f\Vert_p\in[0,\infty]$ is continuous?
I already know that this function is non-decreasing. So far, I've tried the following:
Let $p\geq 1$. First suppose that $\Vert f\Vert_p<\infty$. Let $r_n<p$ be a sequence with $r_n\rightarrow p$. Since $|f|^{r_n}<\max (1, |f|^p)$, which is in $L^1(\Omega)$, and $|f|^{r_n}\rightarrow |f|^p$, then by dominated convergence,
we get $\Vert f\Vert^{r_n}_{r_n}\rightarrow\Vert f\Vert_p^p$, hence
$\Vert f\Vert_{r_n}=\exp\left((1/r_n)\log\Vert f\Vert_{r_n}^{r_n}\right)\rightarrow\exp\left((1/p)\log\Vert f\Vert_p^p\right)=\Vert f\Vert_p$. This shows that $N_f$ is left-continuous at p. I don't know how to prove that $N_f$ is right-continuous
If we had $\Vert f\Vert_q<\infty$ for some $q>p$, the same kind of argument would have shown that $N_f$ is continuous at $p$.
Now, suppose $\Vert f\Vert_p=\infty$. Let $r_n\rightarrow p$. Since
$|f|^{r_n}\rightarrow|f|^p$, then, by Fatou-Lebesgue, $\infty=\int|f|^p\leq \liminf\int|f|^{r_n}$, hence $\liminf\Vert f\Vert_{r_n}=\infty$, which shows that $\lim\Vert f\Vert_{r_n}=\infty=\Vert f\Vert_p$.
AI: What you are trying to prove is false, and your problem is exactly at the point where it fails.
Since the function $p \mapsto \|f\|_p$ is non-decreasing, either:
$f \notin \mathbb{L}^\infty$,
or:
$f \in \mathbb{L}^\infty$,
or:
there exists some $p_0 \in [1, \infty]$ such that $f \in \mathbb{L}^p$ for all $p < p_0$ and $f \notin \mathbb{L}^p$ for $p > p_0$, minus the two cases above.
The first case is trivial. You already have solved the second case. For the third case, there are two sub-cases: either $f \in \mathbb{L}^{p_0}$ and $p_0 < + \infty$, or $f \notin \mathbb{L}^{p_0}$ and $p_0 > 1$. You have already solved the case $f \notin \mathbb{L}^{p_0}$, so the only reamining case is $f \in \mathbb{L}^{p_0}$ and $p_0 < + \infty$.
The problem is that, obviously, the function $p \mapsto \|f\|_p$ will be continuous on $[1, p_0]$ and on $(p_0, + \infty]$, but it won't be right-continuous in $p_0$: it jumps from a finite value to $+ \infty$.
The last question is: "Is this case possible"? It turns out it is, or in other words there are counter-examples to the proposition you are trying to prove. I'll let you find one of them.
Hint: to make things easier, take $p_0 = 1$ and $\Omega = [0, 1]$. Your goal is to find a non-negative function $f$ such that $f$ is integrable, but $f^p$ is not integrable for all $p > 1$. |
H: Compute the determinant
a)
| -1 -2 -3 ..... -k | | -1 -2 -3 ..... -k |
| 1 0 3 ..... k | | 0 -2 0 ..... 0 |
| 1 2 0 ..... k | = | 0 0 -3 ..... 0 | =
| ............. k | | ................ |
| 1 2 3 ..... 0 | | 0 0 0 ..... -k |
= (-1)(-2)(-3)...(-k) =
= (-1)^k*1*2*...*k =
= (-1)^k * k!
b)
| 0 0 ..... 0 a_1k | | a_k1 a_k2 ....... a_kk-1 a_kk |
| 0 0 ..... a_2k-1 a_2k | = | ............................. | *(-1)^k =
| .............................. | | 0 0 ....... a_2k-1 a_2k |
| a_k1 a_k2 ..... a_kk-1 a_kk | | 0 0 ....... 0 a_1k |
Please note that _k1, _k2 and all symbols after the _ are in subscript of a (I don't know how to make it).
Any hints how to solve the b) (I'm totally stuck, please help) and is a) correct?
AI: For b), the determinant of a triangular matrix is the product of the diagonals. But b) isn't quite triangular in the sense we understand triangular matrices usually. So, normally the step is to switch the rows around until you get a triangular matrix and every time you do multiply the determinant by -1 but...
A much simpler approach: the determinant is just the product of the diagonals. If your matrix has an odd dimension you multiply that determinant by -1 or else you're fine.
Here are two examples i just typed on wolframalpha. They're basically the same-ish matrix except one is 3x3 and the other is 4x4 and you can look at the results
3x3: http://www.wolframalpha.com/input/?i=determinant+%7B%7B0%2C0%2C1%7D%2C%7B0%2C2%2C3%7D%2C%7B-3%2C1%2C2%7D%7D
4x4: http://www.wolframalpha.com/input/?i=determinant+%7B%7B0%2C0%2C0%2C1%7D%2C%7B0%2C0%2C2%2C3%7D%2C%7B0%2C-3%2C1%2C2%7D%2C%7B1%2C1%2C1%2C1%7D%7D |
H: Prove $\limsup\limits_{n \rightarrow \infty} b_n \leq \limsup\limits_{n \rightarrow \infty} a_n$, given $b_n = \frac{a_1+ \cdots +a_n}{n}$.
I am working on this question for quite a long time. I think I will prove by contradiction, but I can not proceed. Anyone could give a hint? Thanks!
AI: Fix $m$ and pick $n\geqslant m$ large. Then $$\begin{align}
\frac{{{a_1} + {a_2} + \cdots + {a_m} + {a_{m + 1}} + \cdots + {a_n}}}{n} &\leqslant \frac{{{a_1} + {a_2} + \cdots + {a_m} + \left( {n - m} \right)\mathop {\sup }\limits_{k > m} {a_k}}}{n} \cr
&\leqslant \frac{{{a_1} + {a_2} + \cdots + {a_m} }}{n} + \frac{{n - m}}{n}\mathop {\sup }\limits_{k > m} {a_k} \end{align} $$
Now hold $m$ fixed and take $\limsup\limits_{n\to\infty}$. You get $$\limsup_{n\to\infty} \frac{a_1+\cdots+a_n}n\leqslant \sup_{k>m}a_k$$
By the same token you can show that $$\liminf_{n\to\infty} \frac{a_1+\cdots+a_n}n\geqslant \liminf_{n\to\infty }a_n$$ |
H: Derivative of the $\sin(x)$ when $x$ is measured in degrees
So a classic thing to derive in calculus textbooks is something like a statement as follows
Is $\frac{d}{dx}\sin(u)$ the same as the derivative of $\frac{d}{dx}\sin(x)$ where $u$ is an angle measured in degrees and $x$ is measured in radians? and of course the answer is no because of the chain rule.
Except usually this is ambiguously worded as "Is the derivative of $\sin(u)$, where $u$ is measured in degrees, equal to the derivative of $\sin(x)$ where $x$ is the same angle but measured in in radians?"
Then the texts go on to say something like "No and this why we don't work in degrees and instead chose to work in radians, to avoid all the messy constants that come out of taking derivatives." Am I crazy by thinking this is an odd thing to say that will end up confusing students. If your independent variable was an angle measured in degrees, you are probably more interested in it's derivative with respect to degrees not radians, which would infact be equal at the corresponding degrees and radians of an angle. Is my understanding wrong here. Is what the books say fine? I think at minimum they should at least be clear that we are taking the derivative with respect to radians, no?
Note this is not a duplicate of
Derivative of the sine function when the argument is measured in degrees
Even though it is highly related.
AI: The essential problem is writing "$\sin u$, where $u$ is measured in degrees". This doesn't really mean anything. If you want degrees, you need to write them explicitly: $\sin u^\circ$. If you want the sine of an angle, you need to write that: $\sin\angle ABC$ or similar. At some point in geometry or trigonometry class, someone should be teaching that $\vphantom t^\circ=\frac{2\pi}{360}$. |
H: Prove the following determinant identities without expanding the determinants
a)
$$\begin{vmatrix}
\sin^2 x & \cos^2 x & \cos 2x \\
\sin^2 y & \cos^2 y & \cos 2y \\
\sin^2 z & \cos^2 z & \cos 2z \\
\end{vmatrix} = 0;$$
$$\begin{vmatrix}
\sin^2 x & \cos^2 x & \cos^2x-\sin^2x \\
\sin^2 y & \cos^2 y & \cos^2y-\sin^2y \\
\sin^2 z & \cos^2 z & \cos^2z-\sin^2z \\
\end{vmatrix} = 0;$$
$$\begin{vmatrix}
\cos^2 x & \cos^2 x & \cos^2x-\sin^2x \\
\cos^2 y & \cos^2 y & \cos^2y-\sin^2y \\
\cos^2 z & \cos^2 z & \cos^2z-\sin^2z \\
\end{vmatrix} = 0;$$
b)
$$\begin{vmatrix}
1 & a & p+c \\
1 & p & c+a \\
1 & c & a+p \\
\end{vmatrix} = 0,$$
$$\begin{vmatrix}
1 & a & 1\cdot(a+p+c) \\
1 & p & 1\cdot(p+c+a) \\
1 & c & 1\cdot(c+a+p)
\end{vmatrix} =
(a + p + c)\cdot
\begin{vmatrix}
1 & a & 1 \\
1 & p & 1 \\
1 & c & 1
\end{vmatrix} = 0.$$
I am stuck on a), any hints and help is appreciate, and please check if b) is correct.
AI: Remember that $\cos 2x = \cos^2 x - \sin^2 x$. |
H: Uncountable product of separable spaces is separable?
I am asking to what extent this problem is true:
Uncountable product of separable spaces is separable.
I need these statements to be explained if possible:
How to construct a countable subset of uncountable product of separable spaces.
Counterexample for the contradicted case.
AI: A product $X=\prod_{\alpha\in A}X_\alpha$ of non-trivial separable spaces is separable if and only if $|A|\le 2^\omega=\mathfrak{c}$. (By non-trivial I mean that a space does not have the indiscrete topology; for $T_1$ spaces this simply means that it has at least two points.) This is the countable case of the Hewitt-Marczewski-Pondiczery theorem. Hewitt’s original paper can be found here. This result itself is Theorem $16.4$c in Willard’s General Topology; it is also proved here at Ask a Topologist by Henno Brandsma.
Added: Here’s a sketch of the construction of a countable dense subset of $X$ when $|A|\le 2^\omega$. Without loss of generality assume that $|A|=2^\omega$, and identify $A$ with $\Bbb R$. For $\alpha\in\Bbb R$ let $D_\alpha=\{y_\alpha(k):k\in\Bbb N\}$ be a countable dense subset of $X_\alpha$. Let $F=\{q_1,\ldots,q_n\}$ be any finite set of rational numbers with $q_1<\ldots<q_n$, and let $\sigma=\langle m_0,\ldots,m_n\rangle\in\Bbb N^{n+1}$ be any $(n+1)$-tuple of natural numbers. Define a point $x^{F,\sigma}=\langle x_\alpha^{F,\sigma}:\alpha\in\Bbb R\rangle\in X$ as follows:
$$x_\alpha^{F,\sigma}=\begin{cases}
y_\alpha(m_0),&\text{if }\alpha<q_1\\
y_\alpha(m_k),&\text{if }q_k\le\alpha<q_{k+1}\text{ for some }k\in\{1,\ldots,n-1\}\\
y_\alpha(m_n),&\text{if }q_n\le\alpha\;.
\end{cases}$$
Then
$$D=\left\{x^{F,\sigma}:F\subseteq\Bbb Q\text{ is finite and }\sigma\in\Bbb N^{|F|+1}\right\}$$
is a countable dense subset of $X$. |
H: Finding derivative using product rule
When finding the derivative of:
$z=(1+t^2)6^t$
I am working it out to:
$$\frac{dz}{dy} = (1+t^2)(6^t)'+(1)'(t^2)'(6^t)$$
$$=(1+t^2)(ln(6))(6^t)+2t(6^t)$$
$$=6^t((1+t^2)(ln(6)))+2t(6^t)$$
I believe this to be the answer, but correct me if I am wrong. My biggest concern is that I am seeing some sources on the internet saying you need to use the $log$ of the value not the $ln$ even though we have been using the $ln$ value in class.
AI: You answer looks mostly right. Here are a few changes:
$$\begin{align}
\frac{dz}{d\color{red}t} &= (1+t^2)(6^t)'+(\color{red}{1 + }t^2)'(6^t)\\
&=(1+t^2)(\ln(6))(6^t)+2t(6^t)\\
&=6^t(1+t^2)\ln(6)+2t(6^t)\\
&= 6^t[\ln(6) + 2t + \ln(6)t^2].
\end{align}
$$
(Doing the last step might not be necessary.)
Note that you are treating (1+t^2) as a factor, so with the product rule you would have to take the derivative of that factor. That is why you need $(1+t^2)'$.
About your concern about $\log$ and $\ln$. People will disagree on this, but I believe that it is more common to write $\ln$ when we have base $e: \log_e$ and $\log$ when we are talking about base $10$: $\log_{10}$. But granted, this will depend on your textbook/teacher. |
H: An algebraically closed field with characteristic $p>0$
I want to know about an algebraically closed field that is not of characteristic $0$.
I really don't know about infinite fields with characteristic $p$ so I will appreciate your comments.
AI: Any field has an algebraic closure—so the short answer to this is: Just take the algebraic closure of any field of characteristic $p$.
For finite fields, it is possible to describe the algebraic closure fairly explicitly. Let $K=\mathbb{F}_q$, the finite field of order $q$. Then, for any $n\geq 1$, there is a unique (up to isomorphism) field extension $K\to K_n \cong \mathbb{F}_{q^n}$ of degree $n$, with $K_n \subset K_m$ if and only if $n | m$. $K_n$ can be described as the splitting field of any irreducible polynomial of degree $n$ with coefficients in $K$.
Clearly, the algebraic closure of $K$ is just $\overline{K} = \bigcup_n K_n $, which we could also write as $\varinjlim K_n$. |
H: Simplifying the sum of a fraction and an integer under a radical sign
I'm trying to help my little bro, a bit rusty here... Wolfram Alpha is telling me that:
$$
x\sqrt{1+{\frac{x^2}{16-x^2}}}
$$
simplifies to:
$$
4x\sqrt{\frac1{16-x^2}}
$$
I can't for the life of me figure out why. I'm thinking there's a simple rule I'm forgetting about..
AI: Find the common denominator within the radical sign. $$x\sqrt {1 + \frac {x^2}{16 - x^2}} = x\sqrt {\frac {(16 - x^2) + x^2}{16 - x^2}} = x\sqrt{\frac{16}{16- x^2}}=4x \sqrt{\frac{1}{16 - x^2}}$$
In the last step, we simplify $\sqrt{16} = 4$. |
H: Order of the product of two commuting elements with coprime orders in a group.
I want to show that if $g,h\in G$ are group elements with finite coprime orders $m$ and $n$ and $gh=hg$ then the order of $gh$ is $mn$.
We have that
$$
(gh)^{mn}=g^{mn}h^{mn}=1
$$
using $gh=hg$, so $|gh|\leq mn$. On the other hand, $mn=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=\operatorname{lcm}(m,n)$, since $m$ and $n$ are coprime. I am not sure how to proceed, I tried assuming that $|gh|<mn$ and arriving at a contradiction, but it did not take me anywhere.
AI: Let $r$ be such that $(gh)^r = 1$. Then $1 = (gh)^{rm} = g^{rm}h^{rm} = h^{rm}$, and so $n\mid rm$ and thus $n\mid r$. Similarly you can show that $m\mid r$, and combining these results gives $mn\mid r$. |
H: What is meant by 'runs through'?
I'm independently studying abstract algebra for fun (not my forte...) and I'm reading Herstein. He has a question in the chapter on rings:
Let $p$ be an odd prime and let$$\sum_{k=1}^{p-1}\frac{1}{k}=\frac{a}{b}$$ where $a,b\in{\mathbb{Z}}$. Show $p|a$. (Hint: As $a$ runs through $U_p$, so does $a^{-1}$.
$U_p=\{[a]\in\mathbb{Z}_n|(a,n)=1\}$
What does he mean "run through"? I've seen this terminology before and it has always perplexed me. And how does it apply to the proof?
I've done it this way for a problem solving class:
$$\sum_{k=1}^{p-1}\frac{1}{k}=\sum_{k=1}^{\frac{p-1}{2}}\frac{p}{k(p-k)}=p\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k(p-k)}=\frac{a}{b}$$.
AI: As $a$ runs through $U_p$, so does $a^{-1}$.
This means that the function $f:U_p\to\Bbb Z_p:a\mapsto a^{-1}$ is actually a bijection from $U_p$ onto $U_p$. Imagine calculating $f(a)$ one at a time for each $a\in U_p$ and listing the outputs as they appear: your list will be a listing of $U_p$, though in general in a different order from your input listing. |
H: Derivative of $n/x$
My assignment asks me to calculate the derivative of:
$$y = \frac{9}{x} + 6 \sec x$$
My first step, it seems, should be to break the equation up and find the derivatives of $\frac{9}{x}$ and $6 \sec x$, which would be $1$ and $6(\sec x\tan x)$, respectively.
This would leave with $ 1 + 6(\sec x \tan x) $ as a final answer. However, the answer given in my book is $-\frac{9}{x^2} + 6\sec x\tan x $.
Obviously, I'm incorrect in my assumption that the derivative of $\frac{9}{x} = 1$, however I don't understand why. Can it be said that the derivative of any $\frac{n}{x} = \frac{n}{x^2}$?
AI: Have you learned the power rule? It states that for all integers $k\neq 0$ (and actually real numbers), the derivative of $x^k$ is $kx^{k-1}$.
If we want to find the derivative of $\frac{n}{x}$, we can rewrite it as $nx^{-1}$, so its derivative must be $-nx^{-2}=-\frac{n}{x^2}$. |
H: Show that, if $\tan^2(x) = 2\tan(x) +1,$ then $\tan(2x) = -1$
(i) Show that, if $\tan^2(x) = 2\tan(x) + 1,$ then $\tan (2x) = -1$
AI: It's always the case that $\tan2x=2\tan x/(1-\tan^2x)$. If $\tan^2x =2\tan x+1$, then $2\tan x=\tan^2x-1$ and hence
$$\tan2x={2\tan x\over 1-\tan^2x}={\tan^2x-1\over1-\tan^2x}=-1$$ |
H: Question about Big-O notation
I'm learning Big-O notation in school and my friend and I have a hard time understanding some parts of it and we don't agree on some answers in the exercises.
There are two cases on which we don't agree, and they are particularly easy ones.
The following example contains both cases:
We need to find the Big-O of $f(n) = 6n^{10} + 3^{3n-6}$.
We know we have to separate the function in two distinct parts, namely:
$$f_1(n) = 6n^{10},$$
$$f_2(n) = 3^{3n-6}$$
and the maximum of the two is going to give the Big-O of the equation.
While, from what I've read in the textbook, $f_1$ should give a Big-O of $n^{11}$, my friend thinks it should be $n^{10}$.
As for $f_2$, I think the Big-O should be $3^{n}$ while he thinks it should be $3^{3n}$.
I'm looking for an explanation to know how I should be getting the correct results, no matter what they are. There are some theorems in the textbook for other types of functions (log, factorial, etc.), but I have no idea how to get the results for the two I mentioned earlier in this post.
AI: Both relations $f_1(n) = \mathrm O(n^{10})$ and $f_1(n) = \mathrm O(n^{11})$ are correct since Big-O is upper bound, not obligatory exact. However $\mathrm O(n^{10})$ gives you more tight bound.
Relation $f_2(n) = \mathrm O(3^{3n})$ is obviously correct, but $f_2 = \mathrm O(3^n)$ is not correct since for any $c$ there exist $N_c = 3 + \left\lfloor\frac{\log_3 c}2\right\rfloor + 1$ such that for all $n \ge N_c$ we have $3^{2n-6} > 3^{6 + (\log_3 c) - 6} = 3^{\log_3 c} = c$ and therefore $f_2(n) = 3^{3n - 6} > c 3^n$. |
H: Calculating future value and Present Value
I have been stuck on this one for hours ... not too great at math can someone help. Thanks.
Isaac borrowed $\$4000$ at $11.5\%$ compounded quarterly $5.5$ years ago.
One year ago he made a payment of $\$1500$. What amount will extinguish the loan today?
I've tried a bunch of different approaches, none were right.
From what I understand we should calculate the FV for $4.5$ years when $PV=\$4000$
then subtract $\$1500$ from answer and calculate FV for one more year.
But still no luck..
AI: First: The "value" of the loan after $4.5$ years would be
$$
FV(4.5) = 4,000\left(1 + \frac{0.115}{4}\right)^{4\cdot 4.5}.
$$
Now then after the $4.5$ years you would subtract $1,500$ from the debt and the add one years extra interest to what is left over. This will give you the future value after the $5.5$ years:
$$
(FV(4.5) - 1500)*(1 + \frac{0.115}{4})^{4\cdot 1}
$$
So this is equivalent to making a new loan of $FV(4.5)$ and then calculate what that load is "worth" after $1$ year with $11.5\%$ compounded quarterly. |
H: Can we turn $\mathbb{R}^n$ into a field by changing the multiplication?
Of course $\mathbb{R}$ is a field with usual addition and multiplication. When we move up a dimension into $\mathbb{R}^2$, however, there is not a clear way to multiply two vectors together to get something useful. In fact, if we define multiplication of two vectors component-wise (as is arguably the most natural way), we get something that isn't even an integral domain. However, if we implement the multiplication
$$ (a, b)(c, d) \mapsto (ac - bd, ad + bc), $$
then we obtain a copy of $\mathbb{C}$, which is again a field. Can we do this for higher dimensions? That is, is there some clever multiplicative structure on $\mathbb{R}^3$ that produces a field? $\mathbb{R}^n$?
AI: No, there is no field extension of $\mathbb R$ of degree $n$ other than for $n=2$, and we get something isomorphic to the complex numbers.
At least with hindsight, this is not so surprising, if not "obvious", because $\mathbb C$ is algebraically closed (from Liouville's theorem, a corollary of Cauchy's theorems).
Yes, in some ways this is unfortunate, since certain scenarios are precluded. An example of a way to circumvent these obstacles is Hamilton's creation of "quaternions", a four-dimensional $\mathbb R$-vectorspace, which unfortunately (and surprisingly, tittilatingly, in Hamilton's time) produces a non-commutative ... thing.
That is, $\mathbb R$ is "so close" to being algebraically closed that it admits only something isomorphic to $\mathbb C$ as literal field extension. And, perhaps even-more-disappointingly, only $\mathbb H$ as division-algebra extension. |
H: Finiding the integral [0,1] of an interesting function
I have been working in this problem for 4 days, and I just can't get anywhere, so here it is:
Be f the function with domain [0,1], defined in the following way:
for every x between 0 and 1, we have the decimal expansion, the infinite succession of their digits
[; x = 0,a_1 a_2 a_3 ....... ;]
We assume that this expansion does not have an infinite succession of 9, that is we use 0.15000000...... instead of 0.1499999999999999.....
Now we will define f as:
if the decimal expansion starts with:
[; x = 0, a_1 a_2 .... a_n 9... ;]
then
[; f(x) = 0, a_1 a_2 a_3.... a_n 9. ;]
if the decimal expansion of x does not have any 9, then:
[; f(x) = 0 ;]
i.e x = 1/4 then f(x) = 0, x = pi/10 then f(x)=0,314159
Using the definitions of lower and upper sums, find the integral from 0 to 1 of f. Or show that doesn't exist.
My work:
My first thought was that f is not integrable, my solution to the problem was to show that, it doesn't matter how we make the partition [0,1], there is always going to be a number on that subinterval on which f(x) = 0 (that being said a number on which the decimal expansion does not have a 9) and an x such that f(x) > 0 (that being said a number on which the decimal expansion does have a 9). So
The Upper Sums and the lower sums will never be equal but I do not know how to prove that. Help?
I tried also sandwich somehow, but I arrived at
[; 0< \int_0 ^1 (f(x))<8/9 ;]
but that doesn't tell me anything since f(x) approx. g(x)=x, then the integral should be less than 1/2
Also, my professor suggested that f is integrable and the hint was, that at some point we will have to use infinite geometric sums
AI: You might try this. Let $E_1 = [0.9,1]$. Then $f= 0.9$ on $E_1$. Then consider
$E_2 = [0.09, 0.1] \cup\cdots \cup [0.89. 0.9]$. Then $f$ is constant on each intervals. Inductively you can calculate the integral of each $f_1, f_2$ and $f_n$ is monotone increasing to $f$. Thus you can find the integral of $f$. |
H: probability question show that $P(A)>P(B)$
This question comes up several times in past exam papers so i would really like to work it out! Here is it:
Assume that $$0<P(C)<1.$$
How would I show that if
$P(A|C)> P(B|C)$ and $P(A|C^c)> P(B|C^c) $
Then $P(A)>P(B)$
My solution:
I have so far that the conditional probability of an event A given event B, denoted by
$P(A|B),$ is defined as
$$P(A|B) = P(A\cap B)/P(B)$$and similarly $$P(B|A) = P(A\cap B)/P(A)$$ with $$P(A)>0, P(B)>0$$
I know that i must need to use these but i dont really know how, i can see that if $P(A|C)> P(B|C)$ and $P(A|C^c)> P(B|C^c) $
Then $P(A)>P(B)$
but i dont know how to show this...
Any help would be much appreciated. Many thanks
AI: $P(A|C) > P(B|C) \implies P(A\cap C) = P(A|C)P(C) > P(B|C)P(C) = P(B\cap C)$.
$P(A|C^c) > P(B|C^c)\implies P(A\cap C^c)= P(A|C^c)P(C^c) > P(B|C^c)P(C^c) = P(B\cap C^c)$
So,
$P(A \cap C) > P(B \cap C)$ and $P(A \cap C^c) > P(B \cap C^c)$
$ \implies
P(A) = P(A \cap C) + P(A \cap C^c)>P(B \cap C) + P(B \cap C^c) = P(B)$ |
H: Find a positive integer $x$ less than $105$ satisfying the following simultaneous congruence equations.
$$x=2 mod 3$$
$$x=3 mod 5$$
$$x=4 mod 7$$
I have only learnt modulo for 2 weeks so far... really basic theorems.
My attempt using definitions of modulo
From Equation 1, $3a=x-2 \rightarrow 15a=5x-10$
From Equation 2, $5b=x-3 \rightarrow 15b=3x-9$
Adding them together, $15(a+b)=8x-19$ which implies, $8x=19mod15$ or $8x=4mod15$
and hence i am stuck....
please help, i can't seem to understand this at all.
AI: First, $x=3a+2=5b+3=7c+4$. From the second equality, you get that $3a+2=5b+3\Rightarrow 3a=5b+1$. If $b=3d$ or $b=3d+2$, you reach a contradiction, so $$b=3d+1\Rightarrow x=5b+3=5(3d+1)+3=15d+8.$$ From the last equation, $15d+8=7c+4\Rightarrow 15d+4=7c$. By contradiction you can check that $d=7e+3$, so $$x=15d+8=15(7e+3)+8=105e+53.$$ So, the number you ask for is $53$. |
H: Uniqueness Proof: Related to Division Algorithms
Regarding the statement: Let $a\in \mathbb {Z}$, $b\in \mathbb {Z}$. Then there exists integers q and r such that $a = qb+r$ where $0\le r \le b$.
Let $S$ $=$ $\lbrace a-qb: q\in \mathbb {Z}, a-qb \ge 0 \rbrace$
Why does S have to be non-empty?
AI: If $a>0$, let $q=0$: then $a-qb=a>0$, so $a\in S$. If $a<0$ and $b<0$, let $q$ be any integer greater than $\frac{a}b=\frac{|a|}{|b|}>0$; then $a-qb=q|b|-|a|>0$, and again we have a member of $S$. If $a<0$ and $b>0$, let $-q$ be any integer greater than $\frac{|a|}b=\frac{-a}b$; then $-qb>-a$, so $qb<a$, and $a-qb>0$, yet again giving us a member of $S$. The only remaining case is $a=0$; if $b>0$, let $q=-1$, and if $b<0$, let $q=1$. (The case $b=0$ should be excluded in the statement of the theorem.) |
H: Finding the vertical asymptote of a given function
Given any function, how can I find its vertical asymptote? I know that for rationals I can do this by letting the denominator equal to 0. But how about a function like:
$\ln(1-\ln(x))$? You can find the horizontal asymptote of any function by finding the limit of the function as it approaches positive and negative infinity. Is there a generalized algorithm like that for finding the vertical asymptote as well, even when the function is not a rational?
AI: You simply pick off values of $x$ that are not defined in an orderly manner. For example, in the case of $\ln(1-\ln(x))$, look to the inner nest. By a property of logarithms, $\ln(x)$ is undefined for $x \leq 0$. There is a vertical asymptote at $x=0$, and the function is not defined for $x \leq 0$. Take the next nest,
$$1-\ln(x) > 0 \Rightarrow \ln(x) <1 \Rightarrow e^{\ln(x)} < e^1 \Rightarrow x < e.$$
So now $x=e$ is another vertical asymptote, and $x<e$.
That is about it. Just collect exclusions for $x$ in an orderly manner. When an exclusion is a boundary as is the case here, it is generally going to be an asymptote. |
H: Prove that decomposition of second order tensors into symmetric and skew components is unique.
Title says it all.
$A=A_{sym}+A_{skew}$
$A_{sym}= \dfrac{1}{2}(A+A^T)$
$A_{skew}= \dfrac{1}{2}(A-A^T)$
Anybody can help?
AI: Assume $T=S_1+A_1=S_2+A_2$ are two symmetric skew-symmetric decompositions of rank two tensor. Then
$$0=T-T=(S_1+A_1)-(S_2+A_2)=(S_1-S_2)+(A_1-A_2)\\
0=0^T=(S_1-S_2)^T+(A_1-A_2)^T=(S_1-S_2)+(A_2-A_1)$$
Add up two equations produces
$$2(S_1-S_2)=0\Longrightarrow S_1=S_2$$
Minus equation 2 from 1 produces
$$2(A_1-A_2)=0\Longrightarrow A_1=A_2$$ |
H: How many multiples of 3 are between 10 and 100? (SAT math question)
In the figure above, circular region A represents all integers from 10 to 100, inclusive; circular region B represents all integers that are multiples of 3; and circular region C represents all squares of integers. How many numbers are represented by the shaded region?
a) 24
b) 25
c) 26
d) 27
e) 28
Here's my train of thought:
All we need to know is this set of numbers is from 10-100 (which I'm assuming does include 10 and 100?) and is multiples of 3. Instead of listing out every multiple of 3, I can have the largest multiple of 3 less than or equal to 100, which is 99, subtract the smallest multiple of 3 more than or equal to 10, which is 12.
99 - 12 = 87
Then to find the number of times 3 fits into 87, divide 87 by 3.
87 / 3 = 29
Agh! 29 is not one of the answer options provided!
Assuming what I've done so far is correct, now I'm thinking that for some reason 1 has to be subtracted from 29
29 - 1 = 28
to get 28, which is one of the options provided and also is the correct answer.
But why would I need to subtract 1 from 29? (Or is that not the right way to find the answer?)
AI: There are $30$ multiples of three between $10$ and $100$ (and it doesn't matter whether we include $10$ and $100$). You are correct that they range from $12$ through $99$, but you made a fencepost error by not adding one. So there are $30$ numbers in the shaded part plus the center of the diagram. The center has numbers that are between $10$ and $100$, squares, and multiples of $3$ (and hence $9$ as they are squares). These are $36$ and $81$. Subtracting those two gets you to $28$. |
H: Show convexity of the quadratic function
Given symmetric positive semidefinite matrix $A$, let
$$F(x) := x^TAx + b^Tx + c$$
Can someone show that $F$ is convex using the definition (without taking the gradient)?
AI: By definition of convex, for any $x,y\in\mathbb R$, we have
$$f(\frac{x+y}2)\leq\frac12(f(x)+f(y))$$
Thus it is sufficient to reduce and prove that
$$\frac12(x+y)^TA(x+y)\leq x^TAx+y^TAy\\
x^TAy+y^TAx\leq x^TAx+y^TAy$$
Namely
$$(x-y)^TA(x-y)\geq0$$
which is directly followed by positive semi-definite. |
H: Counterexample of certain non-primary ideals
Let $A$ be a commutative ring with unit. Let $I,J$ be primary ideals of $A$ such that $J$ is not contained in $I$ and $r(J)\subset r(I)$, $r(J)\neq r(I)$. Then $I\cap J$ is not necessarily primary.
I did not find counter examples in $\mathbb{Z}$. I am confused about polynomial rings. How to give an example?
AI: You could take $R = k[x,y]$ where $k$ is a field, $J = (x)$, and $I = (x^2,y)$. The ideal $J$ is primary since it is prime and $I$ is primary since $\sqrt{I} = (x,y)$ is maximal. Then one has $J \not\subseteq I$ and $J \subsetneq \sqrt{I}$. Finally, $J \cap I = (x^2,xy)$ shows that $x^2$ is part of a minimal generating set of $J \cap I$; hence it is not primary. Or one can use that $J \cap I$ is a primary decomposition of itself. |
H: How do I prove a "double limit"?
Prove $$\lim_{b \to \infty} \lim_{h \to 0} \frac{b^h - 1}{h} = \infty$$
I have never worked with double limits before so I have no idea how to approach the problem. Please don't use "$e$" in your solutions, since the above limit is part of the derivation of "$e$", so for all purposes "$e$" hasn't been discovered yet.
I know absolutely no Calculus rules except for the very basics (power, chain, quotient etc.). I also know the squeeze theorem and intermediate value theorem.
Thanks.
AI: Let $f_b$ be the function defined by $f_b(x) = b^x$ . Then: $$\lim_{b \rightarrow \infty} \lim_{h \rightarrow 0} \frac{b^h - 1}{h} = \lim_{b \rightarrow \infty} f_b'(0) = \lim_{b \rightarrow \infty} \log(b) = \infty$$ |
H: A First Order Definition of the Mod Function
Is there a good FOL definition of a $\bmod$ predicate in the language of Peano arithmetic? I tried $M(x,n,r) \equiv Ey(x=ny+r)$ but I don't like it very much.
AI: We want to say that $r$ is the remainder when $x$ is divided by $n$. I will assume that variables range over the non-negative integers.
So we want to say that there exists a $q$ such that $x=nq+r$ and $r\lt n$. To say $r\lt n$, we need to say that $\lnot (r=n)$ and there is a $t$ such that $t+r=n$. Putting things together, we get
$$\exists q\exists t((x=nq+r)\land \lnot(r=n)\land (t+r=n)).$$
Note that if we use this definition then we have decided that the predicate $M(x,n,r)$ will not hold if $n=0$. If we want to make another choice in the case $n=0$, suitable modification can be made. |
H: How to maximise functions of this shape $y=2\cdot3^{-x}$
How can I find the maximum of $2\cdot 3^{-x}$? I know its close to $1$ because I have seen its graph, but when I differentiate the function and set it equal to zero (to get a maximum) I get $-2\cdot 3^{-x}=0$. What does that mean? How can I solve for the $x$ that gives the maximum $y$?
AI: The derivative with respect to $x$ of $f(x)=2\cdot3^{-x}$ is
$$f\,'(x)=2\cdot 3^{-x}(-1)\ln 3=-2\cdot 3^{-x}\ln 3\;.$$
Since $\ln 3>0$ and $3^{-x}>0$ for all $x$, $f\,'(x)<0$ for all $x$, and indeed the function is decreasing everywhere and has no maximum. In fact $\lim\limits_{x\to-\infty}f(x)=\infty$. |
H: Finding a polynomial $k$ respects $e^{-\epsilon k}\leq \delta$
I am given $e^{-\epsilon k}$ and my goal is to find a polynomial $k$ (in $\epsilon$ and $\delta$) such that $e^{-\epsilon k}\leq \delta $ where $\epsilon,\delta,k>0$. The exercise shows that $k\geq$ $\frac{1}{\epsilon}$$ln(\frac{1}{\delta})$ is the solution. However, I do not know how the book derives it. How to go from $e^{-\epsilon k}$ to $k\geq$ $\frac{1}{\epsilon}$$ln(\frac{1}{\delta})$ ?
AI: Take the natural logarithm on both sides: $$\ln e^{-\epsilon k}= -\epsilon k \leq \ln \delta$$
We can then divide $-\epsilon$ over and using the laws of logarithms we have the following (since we are dividing by a negative number the sign of the inequality changes from $\leq$ to $\geq$): $$k \geq -\frac{1}{\epsilon}\ln\delta = \frac{1}{\epsilon}\ln(\delta^{-1})=\frac{1}{\epsilon}\ln\left(\frac{1}{\delta}\right)$$ |
H: What are the every possible sums of these numbers?
Let $S=\{\pm a,\pm(a+b)\}$. If we take the sum of arbitrary $2$ elements of $S$, including duplication, the every possible sums are $\{0,b,2a+b,2a,2(a+b),-b,-(2a+b),-2a,-2(a+b)\}$.
Now, if $S=\{\pm a,\pm(a+b),\pm(a+2b)\}$, and if we take the sum of arbitrary $m$ elements of $S$, including duplication, how can we express the every possible sums?
AI: The $a$'s can range from $-ma$ to $+ma$ by $2$'s. Given that the coefficient on $a$ is $n$, you must have had $\frac {m+n}2$ plus signs on $a$'s and $\frac {m-n}2$ minus signs. The $b$'s can range from $n-m$ to $m+n$ |
H: How to solve $ 2x-3-2x^{ -1/2 }= 0 $ for $x$?
Equation to solve
$$
2x-3-2x^{ -(1/2) }= 0
$$
The answer should be $2.1777$. However I'm not too sure how the steps in between are constructed. Anyone can guide me how do I solve x for this equation?
Progress
One way that I have tried solving this is
$$
2x-3-2x^{ -(1/2) }= 0
$$
$$
2x-2x^{ -(1/2) }= 3
$$
$$
2(x-x^{ -(1/2) })= 3
$$
$$
x-x^{ -(1/2) }= \dfrac32
$$
Not very sure how do I proceed on from here and whether it is the correct way of doing it?
AI: The equation is not a quadratic equation. If we let $x^{1/2} = u$, we then get that
$$2u^2 - 3 - \dfrac2u = 0 \implies u^3 - \dfrac32u - 1 = 0$$
This is a depressed cubic and solving such depressed cubic can be found here. The solution is given as follows:
Let $\alpha = \dfrac{\sqrt[3]{108-54 \sqrt2}}6$, $\beta = \dfrac{\sqrt[3]{2+\sqrt2}}{2^{2/3}}$ and $\omega = \dfrac{1+i \sqrt3}2$; then the three solutions are
$$\alpha + \beta; -\alpha \omega - \beta \bar{\omega}; -\alpha \bar{\omega} - \beta \omega$$ Of the three the only real solution is $\alpha + \beta$. Since, $x^{1/2} = u$, the only real solution for the initial equation is $(\alpha + \beta)^2$, which evaluates to $2.177650698804059954962643867932125475959166$ approximately. |
H: List the numbers in order
How would I list these numbers in order without using a calculator?
Thank you
List these numbers in increasing order: $2^{800}$, $3^{600}$, $5^{400}$, $6^{200}$
AI: Note that $$2^{800} = 16^{200}$$ $$3^{600} = 27^{200}$$ $$5^{400} = 25^{200}$$
so the order is: $6^{200}, 2^{800}, 5^{400}, 3^{600}$. |
H: If $A$ is null set, then $\int\limits_A f dm = 0 $
Define $ \int_E f dm = \sup Y(E, f) $
where $ Y(E,f) = \{ \int_E \phi : 0 \leq \phi \leq f \} $ $\phi$ is simple
Suppose $A$ is a null set. We show $Y(A, f) = \{ 0 \}$. Pick $x \in Y(A, f)$. So, we have $x = \int_A \phi dm $ for some simple $\phi \leq f$. Let $\phi = \sum_{i=1}^{n} a_i 1_{A_i} $.
$$ \therefore \int\limits_A \phi = \sum a_i m(A \cap A_i) \leq \sum a_i m(A) = 0 \implies \int\limits_A \phi = 0$$
Therefore, $Y(A,f) \subseteq \{ 0 \} $.
How can we show that $\{ 0 \} $ is in $Y(A,f) $ ? Or can we just assume this since the integral is never negative so it must be at least $0$, and hence the singleton $0$ must be there.
AI: Assuming that $f\geq 0$ (it seems that you are assuming this), it should be obvious that $0\in Y(E,f)$. Let $\phi\equiv 0$: then $\phi$ is simple, $0\leq\phi\leq f$, and $\int_E \phi=0$. |
H: Show that the quotient space $X^*$ is locally Hausdorff, but not Hausdorff.
Let $X$ be two disjoint copies of $\mathbb{R}$, that is say $X = (\{a\} \times \mathbb{R}) \cup (\{b\} \times \mathbb{R})$ for real numbers $a<b$ and consider X as a subspace of $\mathbb{R} \times \mathbb{R}$. Define an equivalence relation by $a \times t \sim b \times t$ for all $t \neq 0$. Show that the quotient space $X^{*}$ determined by this equivalence relation is locally Hausdorff, but not Hausdorff.
Definition of Locally Hausdorff: For any $x$ in $X$ there is a neighborhood $U$ of $x$ such that $U$, with the subspace topology, is Hausdorff.
AI: HINT: $X^*$ is almost homeomorphic to $\Bbb R$; there is exactly one pair of points in $X^*$ that cannot be separated by disjoint open sets, and knowing that should make the pair easy to find. |
H: biconnected components in a graph
i just started going through biconnected components can someone explain me this
Show that if G is a connected undirected graph, then no edge of G can be in two different
biconnected components
AI: HINT: Suppose that $G$ is connected, and $e=uv$ is an edge lying in the biconnected components $C_0$ and $C_1$, and let $C$ be the union of $C_0$ and $C_1$; show that $C$ is biconnected, contradicting the maximality of $C_0$ and $C_1$. (Recall that a biconnected component is a maximal biconnected subgraph.) It will be helpful to realize that if $H$ and $K$ are connected graphs that share a vertex, their union is also connected. |
H: Use residues to evaluate $\int_{0}^{\infty} \frac{dx}{x^2 + 1}$
Use residues to evaluate $\int_{0}^{\infty} \frac{dx}{x^2 + 1}$.
Okay so these are the integrals in complex analysis I am a little uncomfortable with. I purposely chose a simple problem out of a book so that I can save the slightly more difficult problems for when I actually understand them. The answer to this problem is $\frac{\pi}{2}$. Could someone please explain to me how to solve improper integrals in complex analysis using the problem I have provided as an example. I would appreciate it very much.
AI: The given function is even, so we can actually compute $$\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1}$$
and divide by $2$. Define $f(z) = \frac{1}{z^2 + 1}$; this is analytic with simple poles at $z = \pm i$.
Define a contour $\gamma_R$ to be the line segment between $-R$ and $R$, together with a semicircle $S_R$ around $0$ of radius $R$. Then
$$\int_{-R}^{R} \frac{dx}{x^2 + 1} = \int_{\gamma_R} f(z) dz - \int_{S_R} f(z) dz \,\,(*)$$
Assuming that $R > 1$, we can compute the first integral explicitly from the residue theorem: $f$ has a simple pole at $z = i$, with residue
$$\operatorname{Res}_{i} f(z) = \operatorname{Res}_{i} \frac{1}{(z + i)(z - i)} = \frac{1}{2i}$$
So the first integral is
$$2 \pi i \frac{1}{2i} = \pi$$
On the other hand, if $|z| = R$ is sufficiently large, then
$$|f(z)| \le \frac{1}{\frac{1}{2} R^2} = \frac{2}{R^2}$$
So the second integral can be estimated (arc length) (value of $f$), leading to
$$\left|\int_{S_R} f(z) dz\right| \le \pi R \frac{2}{R^2} = \frac{2\pi}{R}$$
This tends to $0$ as $R$ grows, and so taking a limit in $(*)$ leads to
$$\int_{-\infty}^{\infty} f(x) dx = \pi$$ |
H: Solving optimization problems using derivatives and critical points
I have a homework question which I have completed 2/3 of; however I am stuck on the last part of the question.
The question is:
A drug used to treat cancer is effective at low doses with an efficacy that increases with the quantity of the drug. However, at sufficiently high doses, the drug becomes lethal. For positive values of the constants k_1 and k_2, the fraction of patients surviving cancer with this drug treatment is given by
$$S(q)=\frac{q^2}{k_1^2+q^2}-\frac{q^2}{k_2^2+q^2}.$$
There are three parts to the question. I have answered the first two parts, but I don't know how to approach the last part, which in this case is question 3.
1) Find $S(0)$ and $\lim_{q\to\infty}S(q)$ and in each case explain what your findings mean in medical terms.
For part 1 I made $s(q)$ equal to $0$, which gave me
$$\frac 0 {k_1^2} - \frac 0 {k_2^2} = 0-0 = 0.$$
This indicates that when $q$ is equal to $0$, no patients have survived. I also used the limits for when $q$ is equal to $+\infty$; this gave me $(1-1)$, which is also equal to $0$. Therefore the conclusion here is that when $q$ is very high, hence infinity, the patients will not survive either.
2) What is the optimal daily drug quantity to administer in terms of $k_1$ and $k_2$?
For part two, I found the derivative of $s(q)$ and made is equal to $0$. So I found $s'(q)=0$.
This was a very long process which will take me forever to type here, however my final answer here was $q=\sqrt{k_2k_1}$, where $k_2$ and $k_1$ are constants. I am not sure if this is correct, but it looks pretty right to me.
3) Suppose Health Canada has only approved the use of the drug of up to 45 mg/day and suppose $k_1=25 \mathrm{mg}/\mathrm{day}$ is the same for all patients but $k_2$ varies from patient to patient. To determine a personalized treatment strategy it would be useful for physicians to have a plot of the optimal daily drug quantity as a function of $k_2$, call it $q∗(k_2)$. Sketch a plot of $q∗(k_2)$ and explain why you drew it that way. Hint: don't forget that $k_1<k_2$!
Tthis is the question that I need help approaching, any one care to help please? Am I meant to find the critical points..? Isn't that what I found in Part two? The graph it? If $k_2$ is meant to be greater than $k_1$ and $k_1$ is 25mg and overall it is 45mg, should $k_2$ be between 0 and 20mg?
Is this how my graph should be?
AI: Your calculation of the optimal $q$ is correct.
The rest is easy. As a function of $k_2$, we have $q(k_2)=\sqrt{k_1k_2}=5\sqrt{k_2}$, at least for $25\lt k_2\lt 45$. This is not difficult to plot: the curve $y=5\sqrt{x}$ has a familiar shape. Our curve is (part of) a rightward opening parabola. |
H: How is this set ascending?
In Royden's analysis, the proof for Lemma 10 in chapter 3 states the following:
For each $k$, the function $|f - f_k |$ is properly defined, since $f$ is real-valued, and it is measurable, so that the set $\{ x \in E : |f(x) - f_k(x)| < \eta\}$ is measurable. The intersection of a countable collection of measurable sets is measurable. Therefore $$E_n = \{ x \in E : |f(x) - f_k(x)| < \eta \text{ for all } k \geq n \}$$ is a measurable set. Then $\{E_n\}_{n=1}^\infty$ is an ascending collection of measurable sets.
It seems like such a set would be descending. Is this an error in the text, or am I missing something?
AI: When you increase $n$, you make it easier for a point $x$ to be in $E_n$, so the sets $E_n$ are non-decreasing. For example, in order for $x$ to be in $E_2$, $x$ must satisfy $|f(x)-f_k(x)|<\eta$ for $k=2,3,4,\ldots$. In order for $x$ to be in $E_3$, however, $x$ need only satisfy the inequality for $k=3,4,5,\ldots$; it no longer has to satisfy $|f(x)-f_2(x)|<\eta$. More generally, if $m<n$, and $x\in E_m$, then $|f(x)-f_k(x)|<\eta$ for all $k\ge m$ and therefore automatically for all $k\ge n>m$, so $x\in E_n$, and $E_m\subseteq E_n$. However, if there is a point $x\in E$ that satisfies $|f(x)-f_k(x)|<\eta$ for every $k\ge n$ but not for $k=m$, then $x\in E_n\setminus E_m$. |
H: Martingale: how do they simplify betting summation
$$
\sum_{i=1}^{n} B \cdot 2^{i-1}=B\left(2^{n}-1\right)
$$
How can I go from the summation to $B(2^n-1)$?
AI: It is a general fact that
$$\sum\limits_{i = 1}^{n} r^{i - 1} = \sum\limits_{i = 0}^{n - 1} r^{i} = \frac{1 - r^n}{1 - r}$$
Now select $r = 2$.
Proving this is a step in analyzing geometric series, and the above claim can be proven by induction or any number of other means. |
H: what does eventually mean in the following question ? thanks
Suppose $\phi < f $, $g_n \to f $ pointwise, $g_n = inf_{k \geq n} f_k $. MY books says that 'eventually' $g_n \geq \phi$. What do they mean by eventually? Also to show this, they do the following:
Let $A_k = \{ x : g_k(x) \geq \phi \} $.
I understand that $A_k \subseteq A_{k+1} $, but i dont understant why $\cup_{k}^{\infty} A_k = \mathbb{R} $. Can someone explain me this?
AI: "Eventually" = "When ignoring finitely many initial temrs"; for example the sequence $1,2,3,4,5,5,5,5,\ldots$ is "eventually constant.
Formally in your case:
There exists an $n_0$ such that $g_n\ge \phi$ for all $n\ge n_0$. It appears that this is meant pointwise, i.e. $n_0$ may depend on $x$.
Another handy formulations is: For almost all $n$, $g_n(x)\ge \phi$.
You can conclude $\bigcup A_k=\mathbb R$ because you know that for all $x$ there exixts a $k$ such that $g_k(x)\ge \phi$. |
H: problems on split exact sequence
Let $0 \to A \stackrel f\to B \stackrel g\to C \to 0$ be short exact sequence of modules ($f:A \to B$, $g:B \to C$). Suppose that there exists $\alpha: B \to A$ and $\beta: C \to B$ such that $g \beta = id_C$, $\alpha f = id_A$. How can I prove that $f \alpha + \beta g = id_B$?
AI: The claim is not correct (see below). However, the following is true: Assume that we have the short exact sequence, as well as some $\beta : C \to B$ with $g \beta = 1_C$. Then there is a unique $\alpha : B \to A$ with the desired properties:
The morphism $\beta g - 1_B$ has the property that $g(\beta g - 1_B) = g - g = 0$. Hence, by exactness, there is a unique $\alpha : B \to A$ such that $\beta g - 1_B = f \alpha$, i.e. $f \alpha + \beta g = 1_B$. It also follows that $f \alpha f = f$, hence $\alpha f = 1_A$ since $f$ is monic.
If $\alpha' f = 1_A$, this means that $\alpha' = \alpha + hg$ for some $h : C \to A$, and then $f \alpha' + \beta g = 1_B + f h g$ has no reason to coincide with $1_B$. |
H: what is the probality of taking first blue and last red ball when picking 6 balls?
There is 36 ballls.
12 are red R
12 are blue B
12 are yellow
What is the probability of taking first blue and last red ball when picking 6 balls? (Not returning them back).
Lets say A is any ball.
So the order would look like this:
BAAAAR
I so far I came to this:
$$\frac
{12 * 35 * 34 * 33 * 32 * X}
{36 * 35 * 34 * 33 * 32 * 31}
$$
now I am stuck of what to put instead of X? I dont know what is the probability of taking the last red ball.
Because it differs, depending on previuous 4 balls. How many are left? It can be 12 or 11 or 10 or 9 or 8 red balls left.
Edited:
Thanks for quick answers.
After writing post I thinked this could be correct way, which I calculated already:
Divide in the parts:
There is 0 red balls in the middle
$$\frac
{12 * 23 * 22 * 21 * 20 * 12}
{36 * 35 * 34 * 33 * 32 * 31}
$$
There is 1 red ball in the middle, lets say 2nd one:
$$\frac
{12 * 12 * 23 * 22 * 21 * 11}
{36 * 35 * 34 * 33 * 32 * 31}
$$
But in the middle there is 4 ways to put red ball, so multiply by 4.
There are 2 red balls in the middle:
$$\frac
{12 * 12 * 11 * 23 * 22 * 10}
{36 * 35 * 34 * 33 * 32 * 31}
$$
And 6 ways to put in the middle 4! / (2! * 2!) so multiply by 6
There are 3 red balls in the middle:
$$\frac
{12 * 12 * 11 * 10 * 23 * 9}
{36 * 35 * 34 * 33 * 32 * 31}
$$
and 4 ways to put them
There are 4 balls in the middle:
$$\frac
{12 * 12 * 11 * 10 * 9 * 8}
{36 * 35 * 34 * 33 * 32 * 31}
$$
And add all of them aproximatelly:
$$
0.02182163187 + 0.04800759013 + 0.0342911358 + 0.00935212794 + 0.00081322851 = \fbox{0.11428571425}
$$
So what do you think is this good way?
AI: Think of the balls as having distinct ID numbers. All sequences of $6$ balls (thinking of ID numbers, not colours) are equally likely.
The probability the first is blue is $\frac{12}{36}$. The probability the sixth is red, given the first was blue, is $\frac{12}{35}$. Multiply. |
H: Show that the vector field $X(x, y, z)=(xy-z^2, yz-x^2, x^2+z^2+xz-1)$ is tangent to the set $x^2 + y^2 + z^2 = 1$
I know I need to find functions $F(t)$, $G(t)$, and $H(t)$ such that $F(0)=x$, $G(0)=y$, and $H(0)=z$ and $F'(0)=xy-z^2$, $G'(0)=yz-x^2$, and $H'(0)=x^2+z^2+xz-1$. It's also necessary that $(F(t))^2 +(G(t))^2 + (H(t))^2 = 1$ for all $t$. Just not sure where to go from here...
AI: It sounds like you've been taught more from a pure differential geometry perspective, taking about curves on manifolds and such. I'll try to connect some of the disparate notions here into a coherent whole.
You've been given a set of position vectors ("points") that form a manifold. Let's not worry about what a manifold is. Let's just say it's a useful word that captures what is typically meant when talking about surfaces, volumes, and such, without necessarily choosing whether the object is 2d or 3d or whatever else.
At some position vector $p$, we can draw a curve $c(t)$ through the point, so that $c(0) = p$. If the curve is smooth enough and differentiable, we can take a derivative $c'(0)$. This is a vector, and a different kind of vector from a position vector. This is usually called a tangent vector.
The set of all possible tangent vectors at a point $p$ forms what's called the tangent space at $p$.
Your definition of what it means for a vector field to be tangent is then equivalent to the following:
A vector field $X$ is tangent to a manifold $A$ if at every point $p$, $X(p)$ lies entirely in the tangent space at $p$.
You should be able to see that this definition is equivalent to yours. If $X(p)$ does indeed lie in the tangent space, than you can draw some curve whose derivative is the tangent vector $X(p)$. We're not at all concerned with the behavior of the curve away from $p$, so finding these functions is overkill.
For your 2d manifold, the tangent space at any point $p$ is the plane tangent to the sphere at that point. To say that $X(p)$ lies entirely in the tangent space is to say that $X(p)$ has no component perpendicular to the tangent plane, or rather, that it has no component along the normal to the tangent plane. The normal vector could also be said to be $p$, which is exactly what MLT reasoned. MLT showed that the projection of $X(p)$ along the normal direction $p$ is zero, therefore $X(p)$ has no normal component and must lie entirely in the tangent plane at each point $p$. |
H: Definition of a group
What defines a group mathematically, please explain both in Mathematical language and in English if possible.
My current understanding:
Four things are required to define a group:
Closure - Any binary operation completed upon two elements of a group, must always equal a third element that is already contained within the group. So for all a,b element of G, g*h element of G.
Associativity - If completing a binary operation between three different elements of a group, the order is irrelevant.
Identity element - There must exist some element a in G such that a*b=b(so 1 in the real numbers for multiplication, or 0 for addition)
Inverse element - There must be inverse element such that $b*b^{-1}$ = e
AI: A group is a collection of objects, along with some defined binary operation, that meet the following four criterion:
The group exhibits closure under the operation. That is to say, when the operation is performed on any two elements in the group, the result is one of the elements in the group. For example, if the operation were addition, and I take two numbers from the collection, their sum is also a member of the group.
The group exhibits associativity under the group operation which we will call $*$. That is to say, given any three arbitrary elements in the group $a, b, c$, $(a*b)*c=a*(b*c)$.
There is a unique identity for every element in the group. That is to say, there exists one unique identity element $e$, such that for any arbitrary element $a$ in the group, $a*e=e*a=a$.
For any arbitrary element $a$ in the group, there exists a unique element $b$ in the group, such that $a*b=b*a=e$. That is to say that every element has one inverse, and when the operation is performed on the two elements, the result is the identity.
If these four properties hold, then we can say that this collection of elements forms a group.
The set of all integers $\mathbb{Z}=\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$ forms a group under the operation of simple addition. We can write this group in compact form as $<\mathbb{Z}, +>$. Is this statement true? Consider our four criterion for a group:
The group exhibits closure. If any two integers are added together, no matter which element they may be, the result will also be an integer.
Addition is associative. The order by which one chooses to add integers does not impact the result. for any integers $a, b, c$, $(a + b) + c = a + (b + c)$.
There is an identity element $e$, and it is the number $0$. Given any arbitrary element $a$ in the group, including the identity element, $a + 0 = a$.
Each element $a$ has its own unique inverse element $a^{-1}$, and $a + a^{-1} = e$. That element is the negation of $a$. For example, $2 + (-2) = 0$, or $(-3) + 3 = 0$. The identity is it's own inverse; $0 + 0 = 0$.
Since the four criterion are met, $<\mathbb{Z}, +>$ forms a group.
Welcome to groups. :)) |
H: $ \lim_{x\to 0}\frac{\tan x-\sin x}{\sin(x^3)}$
$$ \lim_{x\to 0}\frac{\tan x-\sin x}{\sin(x^3)} =[1]\lim_{x\to 0}\frac{\sin x/\cos x-\sin x}{x^3}\\ =[2]\lim_{x\to 0}\frac{1-\cos x}{x^2}\\ =\frac{1}{2} $$
My question is how [1]=[2]?
$\tan x-\sin x=\tan x(1-\cos x)=x(1-\cos x)$?
AI: We have $\frac{\sin x}{\cos x}-\sin x=\frac{\sin x}{\cos x}(1-\cos x)$.
Thus our function is equal to
$$\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x^2}.$$
The limit of $\frac{1}{\cos x}$ is $1$. So is the limit of $\frac{\sin x}{x}$.
You did not ask about the limit of $\frac{1-\cos x}{x^2}$. That can be obtained by multiplying top and bottom by $1+\cos x$, getting $\frac{1-\cos x}{x^2}=\frac{1}{1+\cos x}\left(\frac{\sin x}{x}\right)^2$.
Remark: The above answered your specific question. It did not address the question of finding
$$\lim_{x\to 0} \frac{\tan x-\sin x}{\sin(x^3)}.\tag{1}$$
For that, we rewrite (1) as
$$\lim_{x\to 0} \frac{\tan x-\sin x}{x^3}\frac{x^3}{\sin(x^3)}.$$
Since $\lim_{t\to 0}\frac{\sin t}{t}=1$, we have $\lim_{x\to 0}\frac{x^3}{\sin(x^3)}=1$, and we are at the question you asked. |
H: Changing the order of $\lim$ and $\inf$ for point-wise converging monotonic sequence of functions
Suppose $f_n: X \rightarrow \mathbb{R}$, where $X$ is some arbitrary subset of $\mathbb{R}^N$. Suppose that
$$ \forall n\geq0, \forall x \in X, \; f_n(x) \leq f_{n+1}(x) $$
Let $\{f_n\}$ be such that each $f_n(x)$ is continuous and $\forall x,\; \lim_{n\rightarrow \infty}f_n(x) = f(x)$ exists, but is discontinuous. Hence, $\{f_n\}$ is point-wise convergent but not uniformly-convergent. My question is:
Is there any such sequence for which
$$ \lim_{n\rightarrow \infty} \inf_{x\in X} f_n(x) = \inf_{x\in X} \lim_{n\rightarrow \infty} f_n(x) $$ (I know this is true for uniform convergence.)
If so, what is the best approach to prove it for that sequence?
Any solution or reference would be greatly appreciated. There have been similar questions before, like this one, however, I couldn't really relate to the answer. There's another one here.
AI: It is not true in general. Let $g : \mathbb R \to \mathbb R$ such that $g(x) = 0$ if $x\leq 0$ and $g(x)=1 $ for $x\geq 1$ and increasing. Let $f_n(x) = g(x+n)$. Then $f_n$ is an increasing sequence converging to $f=1$. So the left hand side of the limit is $0$, while the right hand side is $1$. |
H: Let $v$ and $w$ be eigen vectors of $T$ corresponding to two distinct eigen values $ \lambda _1$ and $ \lambda_ 2$ respectively
Problem:Let $v$ and $w$ be eigen vectors of $T$ corresponding to two distinct eigen values $ \lambda _1$ and $ \lambda_ 2$ respectively
Then which of the following is true ?
$1)$ For non zero scalars $\alpha_ 1$ , $\alpha_ 2$, the vector $\alpha_ 1v + \alpha_ 2 w$ is not eigen vector of $T$
$2)$ For all scalars $\alpha_ 1$ , $\alpha_ 2$, the vector $\alpha_ 1v + \alpha_ 2 w$ is not eigen vector of $T$
$3)$ $\alpha_ 1v + \alpha_ 2 w$ is eigen vector of $T$, if $\alpha_ 1 = \alpha_ 2$
$4)$ $\alpha_ 1v + \alpha_ 2 w$ is eigen vector of $T$, if $\alpha_ 1 = -\alpha_ 2$
Solution:
I know about eigen values and eigen functions
But I have no idea about this question
AI: $Tv=\lambda_{1}v$ and $Tw=\lambda_{2}w$ with $\lambda_1 \neq \lambda_2$. Since $T$ is (presumably)
linear, $$T\left(\alpha_{1}v+\alpha_{2}w\right)=\alpha_{1}Tv+\alpha_{2}Tw=\alpha_{1}\lambda_{1}v+\alpha_{2}\lambda_{2}w.$$
Since $\lambda_{1}$ and $\lambda_{2}$ are distinct, we cannot write
this as $\lambda\left(\alpha_{1}v+\alpha_{2}w\right)$. (1) is true.
Take $\alpha_{1}=0$. Then clearly $\alpha_{1}v+\alpha_{2}w$ is
an eigenvector of $T$. (2) is false.
(3) and (4) are false by (1). |
H: linearity of expectation in case of dependent events
I could understand the linearity of expectation in case of independent events, but why does it work in case of dependent events too. It seems counter - intuitive. In case of dependent events, each outcome influences subsequent outcomes, hence they cannot be just summed up to get expectation without taking care of this inter dependence.
Can anybody give an intuitive representation of why linearity works even in case of dependent events.
AI: By definition
\begin{align*}
\mathbb{E}\left[\alpha f\left(X\right)+g\left(Y\right)\right] & =\int_{\omega}\alpha f\left(X\left(\omega\right)\right)+g\left(Y\left(\omega\right)\right)d\mathbb{P}\left(\omega\right)\\
& =\alpha\int_{\omega}f\left(X\left(\omega\right)\right)d\mathbb{P}\left(\omega\right)+\int g\left(Y\left(\omega\right)\right)d\mathbb{P}\left(\omega\right)\\
& =\alpha\mathbb{E}\left[f\left(X\right)\right]+\mathbb{E}\left[g\left(X\right)\right].
\end{align*}
For a concrete example, consider a discrete random variable $X$ that
corresponds to rolling a fair 3-sided die.
\begin{align*}
\mathbb{E}\left[X+X^{2}\right] & =\frac{1}{3}\left(1+1^{2}\right)+\frac{1}{3}\left(2+2^{2}\right)+\frac{1}{3}\left(3+3^{2}\right)\\
& =\frac{1}{3}\left(1+2+3\right)+\frac{1}{3}\left(1^{2}+2^{2}+3^{2}\right)\\
& =\mathbb{E}\left[X\right]+\mathbb{E}\left[X^{2}\right].
\end{align*} |
H: The limit of $\sin(n^\alpha)$
(1) It is easy to prove that $\lim\limits_{n\to\infty}{\sin(n)}$ does not exist.
(2) I want to ask how to prove that $\lim\limits_{n\to\infty}{\sin(n^2)}$ does not exist.
(3) Furthermore, $\lim\limits_{n\to\infty}{\sin(n^k)}$ does not exist. ($k$ is a positive integer.)
(4) In addition, $\lim\limits_{n\to\infty}{\sin(n^{\alpha})}$ does not exist. ($\alpha$ is a positive real number.)
AI: Concepts I use here is "Weyl's Equidistribution Criterion"
Reference: http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/
(1) The limit does not exist simply because the sequence $\{e^{in}\}$ is equidistributed on the unit circle.
(2), (3) follows from Van der Corput's lemma:
We can bring down exponent $k$-case to exponent $1$-case.
Better way to put it is:
For any integer $k\geq 1$, the sequence $\{e^{in^{k}}\}$ is equidistributed on the unit circle.
(4) For $0<\alpha<1$, the sequence $\{e^{in^{\alpha}}\}$ is dense in the unit circle, so the limit of $\sin(n^{\alpha})$ does not exist.
For $\alpha>1$, and $\alpha=\frac{q}{p}\in\mathbb{Q}^{+}$, consider $\{\sin(m^p)^{\frac{q}{p}} \}$. This is a subsequence of $\{\sin (n^{\frac{q}{p}})\}$.
The subsequence $\{\sin(m^q)\}$ diverges by (3). Thus the original sequence diverges as well.
Remaining case is now $\alpha>1$ and irrational.
This case can be settled by the lemma:
(Lemma)
For a sequence of real numbers $\{x_n\}$, suppose that the set of all subsequential limits of $\{\exp(ix_n)\}$ is finite. Denote $D$ the difference sequence operator on the space of real sequences. Then for $\{y_n:=Dx_n=x_{n+1}-x_n\}$, we have also that the set of all subsequential limits of $\{\exp(iy_n)\}$ is finite.
If $x_n=n^{\alpha}$, and $\lfloor\alpha\rfloor=m$, then $z_n:=D^m x_n \sim cn^{\alpha-m}$ for some positive constant $c$. Thus, by the same reason for the case $0<\alpha<1$, we have $\{\exp(iz_n)\}$ is dense in the unit circle.
If $\{\sin(n^{\alpha})\}$ has a limit, then $\{\exp(ix_n)\}$ has a finite set of sequential limits. Then by lemma applied $m$-times, $\{\exp(iz_n)\}$ also has a finite set of sequential limits. This contradicts above.
Hence, $\{\sin(n^{\alpha})\}$ does not have a limit. |
H: Find dimension of even polynomials
Let $V$ be a the vector space over $\mathbb R$ of all polynomials with real coefficients. Let $W$ be the subset of all polynomials with only even powers in their expression.
So $p(X) \in W$ means $p(x)=\sum_{n=0}^k a_nx^{2n}$.
I showed that $W$ is a subspace of $V$, but I need to find out the dimension of $W$. Would I need to find a basis for $W$ first? How should I approach this?
Thank you.
AI: Your space $W$ is infinite dimensional. It is easy to see that every polynomial in $W$ can be expressed as a (finite) linear combination of the polynomials $1, x^2, x^4, x^6, x^8,\dots$.
For completeness, one should show that these polynomials form a linearly independent set, and are therefore a basis for $W$. This follows immediately from the definition of equality for polynomials. For an alternative argument, suppose that $P(x)=a_0+a_1x^2+a_2x^4+\cdots +a_nx^{2n}$ is identically $0$. We show that all the $a_i$ are $0$.
If the $a_i$ are not all $0$, we may assume without loss of generality that $a_n\ne 0$. From
$$0\equiv a_0+a_1x^2+a_2x^4+\cdots +a_nx^{2n},$$
by differentiating $2n$ times we get
$$0\equiv (a_n)(2n)!,$$
which is false. |
H: One-relator Groups and Subgroups.
I am currently working on "The Group Algebra of a torsion-free one-relator Group can be embedded in a Field." (Tekla and Jacques Lewin) In this Paper we find a corollary (the last one), which I am about to apply on something else. This corollary can only be aplied, if the group $G$ is a one-relator group. Since I am dealing with subgroups, I tried to figure out, when subgroups of one-relator groups are one-relator.
So let $G$ be a torsionfree one-relator group.
Is then every subgroup $H<G$ one-relator?
AI: Consider the group $G=\langle a,b \mid aba^{-1}=b^2 \rangle$. This 1-relator torsion-free group contains a rank 1 infinitely generated abelian subgroup (dyadic rationals). This subgroup is not 1-relator. As for finitely generated subgroups, I am not sure, but, I suspect that the answer is still negative.
Edit: Here is how to construct examples of finitely-generated subgroups which are not 1-relator. First, note that if $G$ is a 1-relator group, then $\operatorname{rank} H^2(G)\le 1$. In particular, the free product $A=\pi_1(S) \star \pi_1(S)$ (where $S$ is a closed oriented surface of genus $\ge 1$) is not a 1-relator group (as $\operatorname{rank} H^2(A) =2$). Now, many 1-relator groups $G$ contain such subgroups $A$ (with genus of $S$ at least 2). Once you have one such subgroup $A$, you can take its conjugate $gAg^{-1}$ via an element which does not normalize $A$ (assuming that such element exists). Then, I think, for large $n$ (at least if the group $G$ is torsion-free, hyperbolic and $A$ quasiconvex), the subgroup of $G$ generated by
$$
A, g^nA g^{-n}
$$
is isomorphic to the free product $A\star A$. Now, we just need a large supply of hyperbolic groups which contain quasiconvex surface subgroups $A$ and contain elements like $g$. I found such constructions in the paper
"One-ended subgroups of graphs of free groups with cyclic edge groups" by H.Wilton, http://arxiv.org/pdf/1102.2866v2.pdf.
There is further discussion of this in https://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces. |
H: Parametric surfaces - Parameterization of torus
A rotational surface area is created when a curve in the $xz$-plane, with parameterization $\def\i{\pmb{i}}\def\k{\pmb k}$ $r=x(t)\i + z(t)\k$ , $t \in [a,b]$, rotates around the $z$-axis. This surface is parameterized by;
$$t \mapsto \bigl( x(t)\cos\theta, x(t)\sin\theta, z(t)\bigr)^t $$
$t \in [a,b]$, $\theta \in [0,2\pi)$.
Use the above information to parameterize the torus that is created when a circle $(x-3)^2+z^2=4$ rotates around the $z$-axis.
The above is the first of a number of exercises on parametric surfaces which I will attempt to complete. I have not recieved any solved examples or specific guidance on the subject so I am hoping that someone here could show me how this is done.
I apologize for any spelling errors and the general lack of LaTeX.
Help is much appreciated!
AI: So we have to parametrize the given circle $(x-3)^2 + z^2 = 4$, in the form $\def\i{\pmb i}\def\k{\pmb k}$$$r(t) = x(t)\i + z(t) \k $$
on some $t$-intervall $[a,b]$, then we can apply the given information. Keeping in mind, the construction of sine and cosine, we have that $(\cos t, \sin t)$ describes a circle of radius 1 around $0$, we have a circle of radius $2 = \sqrt 4$ around $(3,0)$, now $(2\cos t, 2\sin t)$ is a circle of radius 2 around $0$, adding $(3,0)$ we are done and obtain
$$ r(t) = (3 + 2\cos t)\i + 2\sin t \;\k, \qquad t \in [0,2\pi] $$
for the given circle. Now apply the given information. |
H: Symmetric difference of sets and convergence in integration.
Let $(X,\mathcal{M},m)$ be a space of measure and $f_n,f \in L^1(m)$ such as $||f_n - f||_1 \rightarrow 0.$ Suppose that we also have $A_n,A \in \mathcal{M}$ and $m(A_n \triangle A) \rightarrow 0.$
I want to prove that $\displaystyle \int_{A_n}f_n dm \rightarrow \int_A f dm.$
AI: We have $\def\norm#1{\left\|#1\right\|_1}\def\abs#1{\left|#1\right|}$
\begin{align*}
\abs{\int_{A_n} f_n \, dm - \int_A f\, dm} &\le \abs{\int_{A_n} (f_n- f)\, dm} + \abs{\int_{A_n} f\, dm - \int_A f \, dm}\\
&\le \int_{A_n} \abs{f_n-f}\, dm + \int_{A_n \Delta A} \abs{f}\, dm\\
&\le \norm{f_n -f} + \norm{f \cdot \chi_{A_n \Delta A}}\\
&\to 0
\end{align*}
Where $\chi_{A_n \Delta A}$ denotes the charateristic function. For the second term on the second to last line, note that $\chi_{A_n \Delta A}f \to 0$ almost everywhere and $\abs f$ is a integrable bound, hence the bounded convergence theorem gives $\norm{f \chi_{A_n \Delta A}} \to 0$. |
H: Proving that representation is differentiable
Let $V$ be a real or complex finite-dimensional vector space, and let $\pi$ be a continuous representation of $(\mathbb{R}, +)$ on V with:
$$\pi(t + s) = \pi(t) \pi(s), t, s \in \mathbb{R} \: \: (1)$$
$$\pi(0) = I \: \:(2)$$
a) Prove that $\pi: \mathbb{R} \rightarrow L(V) is differentiable.
There is also a b) part, but I've solved that with assuming that $\pi$ is differentiable. From (1) it follows that it's enough to show differentiability in zero.
There is a pretty strange hint following the a) part... something about proving that there exists $\alpha > 0$ so that operator
$$B = \int_0^{\alpha} \pi(t) dt$$
is regular. I can't really understand what it means to take an integral of a linear operator, we haven't defined that anywhere. But it someone has an idea what that notation could mean, please explain it to me.
AI: It means $B: v \mapsto \int_0^\alpha \pi(t) v d t$ |
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