TITLE: On the boundedness of a certain sequence
QUESTION [1 upvotes]: Set
$\bullet\text{ } a_1 = 2$
$\bullet \text{ }a_{j+1} = \frac{a_{j}}{\log(a_{j})}$
Is $(a_{j})_{j=1}^{\infty}$ bounded or is it unbounded?
If $a_{1}$ were to be $e$ then it would be obviously bounded.
If the above sequence were to be bounded then it is clear that $\exists \delta >0$ so that $\min_{j \in \mathbb{N}}|a_{j}-1| \geq \delta$
Update it seems to be converging to $e$.

REPLY [1 votes]: It converges to $e$. Set $f(x) = \frac{x}{\log(x)}$ then it has a mimima at $x = e$ which is $e$. Thus $(a_{j})_{j=2}^{\infty} \geq e$.
We also have $a_{2} \geq a_{3} \geq...$ as in Brian's comment. Thus by M.C.T the above sequence converges to some $p \in [e, \frac{2}{\log(2)})$. If $p > e$ then note that if some d is sufficiently close to $p$ then $\frac{d}{\log(d)} < p$. Hence $p \leq e$ and therefore $p = e$.


TITLE: Show that dim$(U \cap V) \ge 2$ if U and V are 3-dimensional subspaces of $ K^4$
QUESTION [3 upvotes]: Show that dim$(U \cap V) \ge 2$ if U and V are 3-dimensional subspaces of $ K^4$
I'm absolutely stuck. I know that it is only small, easy proof, but can some show me how it is done?

REPLY [7 votes]: Hint:
$$\dim(U+V)=\dim U+\dim V-\dim U\cap V\;\ldots$$


TITLE: How could an element not in two sets be an element of their union with a third set?
QUESTION [0 upvotes]: In this problem, I don't understand why statement (f) is true.
Since it is possible that x is neither in A nor in B. And since it is not certain that x belongs to C, how is it possible to know that what statement (f) says is true? It possible that x belongs to none of these three sets and then the statement would be false.
The problem.
The solution provided by the book.

REPLY [0 votes]: It's just an ambiguity of wording. "For some nonempty set $C$ ..." means "There exists some nonempty set $C$ such that ..."
(It does not mean that the statement holds true for "just any" nonempty set $C$, as you seem to point out.)
To see it is true in this case, take $C=S$, or $C = \{x\}$.


TITLE: Closed form expression of Infinite Summation Squared
QUESTION [1 upvotes]: I am trying to simplify the following expression
$$
\left(\sum_{n=0}^{\infty} t^{2n}\right)^2
$$
I tried finding out a pattern for the coefficients by writing just the first three terms and squaring, i.e
$$
(1+t^2+t^4)^2=1+2t^2+3t^4+2t^6+t^8
$$
and I notice that the more and more terms I square, there seems to be a pattern,i.e for 3 terms, the coefficients are $ 1,2,3,2,1$ for 4 terms it is $1,2,3,4,3,2,1$ and so on. But could someone please tell me how I can get a closed form summation expression with this?

REPLY [2 votes]: Well, if you really insist on writing out $(\sum t^{2n})^2$ as a new summation $\sum a_n t^n$, rather than simply using the other answers for a closed form expression, it is going to really on the binomial theorem / collecting terms in the appropriate manner.
Looking at
 $$\left(\sum_{n=0}^\infty t^{2n} \right)^2 = (1 + t^2 + t^4 + \cdots)(1+t^2 + t^4 + \cdots)$$
we want to observe how many ways there are to produce terms of the appropriate exponent:

Constant terms: we can only pick $1$ from the first terms and $1$ from the second, i.e. $1\cdot 1$.
$t^{2k+1}$ terms: No possibilities--both expressions only include $t$ to even powers, so no multiplication will give odd exponents.
$t^2 terms$: we can pick $1\cdot t^2$ or $t^2 \cdot 1$, for a coefficient of 2
$t^4 terms$: we can pick $1 \cdot t^4$, $t^2\cdot t^2$, or $t^4 \cdot 1$. Thus the coefficient is 3.
$t^6 terms$: we can pick $1 \cdot t^6$, $t^2 \cdot t^4$, $t^4 \cdot t^2$, or $t^6 \cdot 1$. Coefficient is 4.
$t^{2n} terms$: we can pick $1 \cdot t^{2n}$, $t^2 \cdot t^{2n-2}$, $\dots$, $t^{2n} \cdot 1$. There are $n+1$ of these.

Thus the general formula would be
$$\left(\sum_{n=0}^\infty t^{2n}\right)^2 = \sum_{n=0}^\infty (n+1)t^{2n}$$
Note that if we were presented with $(\sum a_n t^{2n})^2$ instead, the answer would be much messier and would involve a double sum: $(\sum a_n t^{2n} )^2= \sum_n \sum_j a_j t^{2n}$


TITLE: Prove that $\frac{m}{n}+\sqrt{3}\frac{p}{q}$ is irrational
QUESTION [1 upvotes]: If $\frac{m}{n}$ and $\frac{p}{q}$ are rational. Prove that $\frac{m}{n}+\sqrt{3}\frac{p}{q}$ is irrational, given $p\neq 0$.

REPLY [1 votes]: HINTS:

Can you prove that $\sqrt{3}$ is irrational?
Can you prove that if $q \neq 0$ is a non-zero rational number and $x$ is an irrational number, then $qx$ is also irrational?
Can you prove that if $q$ is a rational number and $x$ is an irrational number, then $q+x$ is also irrational?

EDIT:
Well, actually, if I want to give you some further hints I'll give you too much information. But since you're still new to the subject I think it's not bad to see some methods first.
I'll only prove the 2nd one. I'll leave the first and third one to you to prove on your own. Suppose that $qx$ is not irrational. Therefore, $qx=p$ where $p \in \mathbb{Q}$. Since $q \neq 0$ I can divide by $q$ and get $\displaystyle x = \frac{p}{q}$. But since $p,q \in \mathbb{Q}$ and the division of two rational number is a third then $\displaystyle x = \frac{p}{q}$ which is contradiction because we assumed $x$ is irrational.
Now use the same trick for the third one. Also, I think by now you should know that $2$ or $3$ aren't any special numbers. You can prove that if $p$ is a prime number then $\sqrt{p}$ is irrational.That's a good exercise to do.


TITLE: Equation of motion of massless scalar field in F(L)RW spacetime
QUESTION [0 upvotes]: Consider the equation of motion of a scalar field $\phi(t,x^i)$, $$\nabla^\mu\nabla_\mu\phi=\frac{dV(\phi)}{d\phi}$$ where $V(\phi)$ is the potential. Specialise to a massless field ($V(\phi)=0$) in the FRW/FLRW metric (coordinates $\{t,r,\varphi,\theta\}$, with $\phi=\phi(t)$. I need to show that $\phi(t)-\phi(t_0)=k\int_{t_0}^tdt'a(t')^{-3}$ for some constant of integration $k$.
My attempt is as follows:
$$\nabla^\mu\nabla_\mu\phi=g^{\alpha\mu}(\partial_\alpha\nabla_\mu\phi-\Gamma^\lambda_{\alpha\mu}\nabla_\lambda\phi)=g^{tt}(\partial_t\partial_t\phi-\Gamma^\lambda_{tt}\partial_t\phi)+g^{ij}(\partial_i\partial_j\phi-\Gamma^\lambda_{ij}\partial_\lambda\phi)$$ where I used that $\nabla=\partial$ for scalars, and I carried out the implicit (Einstein) summation. Next, I use the following information

the only relevant non-zero Christoffel symbol is $\Gamma^t_{ij}=a\dot{a}\delta_{ij}$ (where $\dot{}$ denotes differentiation wrt $t$)
$\phi$ is a function of $t$ only, so $\partial_i\phi=0$
$g^{tt}=-1$ and $g^{ij}=\delta^{ij}/a^2$

to find
$$\nabla^\mu\nabla_\mu\phi=-\partial^2_t\phi-\frac{\delta^{ij}\delta_{ij}\dot{a}}{a}\partial_t\phi=-\partial^2_t\phi-\frac{3\dot{a}}{a}\partial_t\phi$$
Therefore, I am left with
$$-\partial^2_t\phi-\frac{3\dot{a}}{a}\partial_t\phi=0$$
No clue how to proceed. Any hints? Thanks.

REPLY [1 votes]: Set $\Phi= \dot \phi$ then your equation is
$$
\frac{\dot \Phi}{\Phi} =- 3 \frac{\dot a}{a}
$$
or
$$
\frac {d\ln \Phi}{dt}= -3 \frac{d\ln a}{dt}
$$
so $\ln[\Phi(t)a^3(t)]=constant$, or equivalently
$$
\Phi(t)= k a^{-3}(t).
$$
Therefore
$$
\phi(t)-\phi(0)= k\int^t_0 a^{-3}(t)dt.
$$
TaDa!


TITLE: What is the difference between a " line " and a " straight line "?
QUESTION [0 upvotes]: Is there actually a difference between a line and a straight line ?
Is figure 1 a line . ?

.
Should I take help from " Euclid "?
I believe according to " Euclid " the above figure is a valid line.

REPLY [0 votes]: According to Euclid, "A line is breadthless length" (Elements I, Definition 2, T. Heath translation). So your figure appears to be a line. Euclid says a straight line is "a line which lies evenly with the points on itself" (Def. 4). The exact meaning of this definition has been disputed, but it seems clear at least that, for Euclid, a line and a straight line are not the same thing.


TITLE: A linear operator that sends strongly convergent sequences to weakly convergent sequences
QUESTION [2 upvotes]: So I'm having trouble showing this. Its an old prelim problem. Suppose you have a linear operator $A$ on a Banach space which sends strongly convergent sequences to weakly convergent ones. Show that this operator is continuous. Now my first thought was to use closed graph theorem or hahn banach somehow, but I honestly have no clue what to do here.

REPLY [1 votes]: Recall, that
(1) linear operator is continious iff it is bounded.
(2) if $x_n \rightharpoonup x$ then sequence $\left\Vert x_n \right\Vert$ is bounded.
Suppose $A$ is not bounded and $\left\Vert Ax \right\Vert = \infty$. Then we can find sequence $x_n$ on a unit spere, such that $\left \Vert Ax_n \right\Vert \to \infty$. Now we construct another sequence $$y_n = \frac{x_n}{\sqrt{\left\Vert Ax_n\right\Vert}}.$$
We have now $y_n \to 0$ and $\left\Vert Ay_n\right\Vert \to \infty$ thus $Ay_n$ cannot converge weakly to anything.