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H: Intuition behind Doob's Optional Stopping theorem
I would like to ask about intuitively interpreting the results of Doob's Optiona Stopping Theorem applied to standard Brownian motion.
The theorem provides three conditions, under which a stopped process is a martingale. One of these conditions is that the stopping time $T_A$ (associated with an event "$A$" that results in stopping the process) is finite in expectation, i.e.: $\mathbb{E}\left[ T_A \right] < \infty$.
Case 1: Let's define the stopping time $T_A$ associated with $W_t$ as the first time the Brownian motion hits level "$a$". Since we have $\mathbb{E}[W_{T_A}]=a\neq\mathbb{E}[W_{T_A}|W_0]$, the stopped Brownian motion cannot be a martingale. Therefore it must be that none of the three conditions in Doob's theorem are true: therefore we must have that $\mathbb{E}[T_A]=\infty$.
Case 2: Let's define the stopping time $T_A$ associated with $W_t$ as the first time the Brownian motion hits either level "$a$" or "$-a$". It can be shown that in this case the probability $\mathbb{P}\left(T_a<t \right)\rightarrow1$ as $t\rightarrow\infty$. Therefore, we must have that $W_{T_A}$ is a martingale, so we must have that $\mathbb{E}\left[W_{T_A}|W_0\right]=0$.
How do these two cases fit together and how to interpret these intuitively?
Question Case 1: The first case tells us that the expected hitting time of a specific level is infinite: if we interpret expectation as the "average" over many paths, would it be correct to deduce that the result tells us that at least over some paths, Brownian motion tends to "diverge and never return"? I.e. it shoots off towards either positive infinity or negative infinity for at least some paths? Since even just one path that never hits the level "$a$" is enough for the expectation to diverge, this would intuitively explain why the stopping time is not integrable.
Question Case 2: Here, the stopping time is associated with $W_t$ breaking out of a "band" defined by the two boundaries $\{-a,a\}$: since this stopping time is finite, would it be correct to conclude that Brownian motion tends not to stay "confined to a band", not even for a single path out of infinite number of paths? I.e. can we conclude that Brownian motion always eventually breaks out of a range?
AI: Q1: No, that's not a correct interpretation. In fact, it is well known that Brownian motion is recurrent, which is to say that $T_A < \infty$ with probability 1. So actually there is no chance that the Brownian motion will diverge and never return. One can also show that with probability 1 we have $\limsup_{t \to \infty} W_t = +\infty$ and $\liminf_{t \to \infty} W_t = -\infty$, which says that Brownian motion makes wider and wider swings from positive to negative values, recrossing all the numbers in between every time, so that it eventually makes infinitely many widely separated visits to every number.
You can think of this as saying that Brownian motion is guaranteed to hit $a$, but takes on average an extremely long time to do so. Simple random walk has the same phenomenon: if you take a fair coin and flip it until the number of heads exceeds the number of tails (i.e. until the simple random walk hits +1), with probability 1 you will eventually finish, but the expected number of flips needed is infinite. Basically, there is a possibility that you start off with several tails in a row, which will tend to make it take an extremely long time before you get a corresponding run of heads.
Q2: Well, from your argument alone we can only conclude that there is a nonzero probability that the Brownian motion breaks out of the band $[-a,a]$. But it is in fact the case that this happens with probability 1. It follows from the facts mentioned above, since we know that the Brownian motion will hit the values $a+1$ and $-a-1$ with probability 1.
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H: The minimal number with a given number of divisors
My question comes here https://doi.org/10.1016/j.jnt.2005.04.004. Let $A(n)$ which assigns to each number $n$ the smallest number with exactly $n$ divisors. Is it true that $A(n)<A(n+1)$?
AI: No. The smallest number with exactly $5$ divisors is $2^4=16$. The smallest with $6$ divisors is $12$. The numbers are given in OEIS A005179 and you can see many decreases. It starts
$$1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144, 120, 65536, 180$$
The smallest number with $p$ divisors, for $p$ prime, is $2^{p-1}$ Often the next number will be smaller.
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H: Inequality in a Hilbert Space: $\sup_{||f||_{2}\leq 1}||fg||_{2}\leq C ||g||_{2}$
Let $f,g \in L^2(0,1)$ My question is the following: is there a constant $C>0$ such that
$$\sup_{||f||_{2}\leq 1}||fg||_{2}\leq C ||g||_{2},$$
all I know is that we have
$$\sup_{||f||_{2}\leq 1}<f.g>\leq C ||g||_{2},$$
using cauchy schwarz inequality.
AI: No, setting $f=g=\epsilon x^{-1/4}$ we have $\|f\|_2\ll 1$ for $\epsilon\ll 1$ but $$\|fg\|_{L^2} = \infty$$
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H: Closure of a topological space $Y$
Let $X$ be a topological space, and let $Y \subseteq X$. Prove that $\overline{Y} = \displaystyle\bigcap_{F \textrm{ is closed and }Y\subseteq F} F$.
$x \in \overline{Y}$ iff for all $U$ open set in $X$ such that $x \in U$ satisfy $Y \cap U \neq \emptyset$.
AI: Let $x \in \overline{Y} \Rightarrow \forall$ open set $U$ satisfy $U \cap Y \neq \emptyset$. Let $Z$ be a close set such that $Y \subseteq Z$. Suppose $x \notin Z$, then $x \in X \setminus Z$, that is a open set. Therefore $\exists U_0 = X \setminus Z$ a open set such that $x \in U_0$, then $U_0 \cap Y = \emptyset$. This is a contradiction, because $x \in \overline{Y}$.
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H: $T^2$ is measure preserving but $T$ is not measure preserving
I have had trouble with this exercise. Can anyone help?
Give an example of a probability space $(\Omega, \mathcal{F}, P)$ and a measurable mapping
$T : \Omega \rightarrow \Omega$ such that $T^2$ is measure preserving but $T$ is not measure preserving.
There is a hint that I should consider a space $\Omega$ containing only two points.
Best,
AI: Let $\Omega=\{a,b\}$ with $\mathbb P(\{a\})=1/3, \mathbb P(\{b\})=2/3$ and $T(a)=b$, $T(b)=a$. Then $T^2(a)=a$, $T^2(b)=b$ and this is measure-preserving transformation since it is identity.
But $T$ is not measure preserving: say, for $A=\{a\}$, $\mathbb P(TA)\neq \mathbb P(A)$.
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H: Can I square both sides of inequality for these functions?
I have two functions $f(x)$ and $g(x)$ for $x>0$. Both functions are monotonically increasing and $f(x)>5$ and $g(x)>0$ .
I know that $f(x)>\sqrt{g(x)}$.
Then, can I conclude that $f(x)^2>g(x)$ for $x>0$?
AI: Yes, because both $f$ and $\sqrt g$ are positive and $z\mapsto z^2$ is a strictly increasing function for positive $z$.
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H: If $T(p(t)) = p(t+1)$ then find its minimal polynomial where $T$ is a linear operator from $\Bbb{P_n} \rightarrow \Bbb{P_n}$
I tried substituting $p(t)$ with $p(t-1)$ and then taking the transformation to get some kind of annhilating polynomial but that just gave me trivial solutions. Also after spending an hour I think it can be done without knowing the annihilating polynomial itself.
Is it even possible to find a polynomial of $T$ in this case?
AI: If $p \ne 0$ then from $p(t+1) = \lambda p(t)$ by comparing leading coefficients we get that necessarily $\lambda = 1$. Therefore the only eigenvalue of $T$ is $1$ so the minimal polynomial must be of the form $(T-I)^k$. Indeed by Cayley-Hamilton certainly $(T-I)^{n+1}=0$ since $\dim \mathbb{P}_n=n+1$. Furthermore, for $1 \le k \le n$ and the polynomial $p(t) = t^k$ we have
$$((T-I)^kp)(t) = \sum_{j=0}^k {k \choose j} (T^jp)(t) = \sum_{j=0}^k {k \choose j}
p(t+j) = \sum_{j=0}^k {k \choose j} (t+j)^k$$
and for $t = 0$ it is
$$((T-I)^kp)(t) = \sum_{j=0}^k {k \choose j} j^k > 0$$
so clearly $(T-I)^k \ne 0$. Therefore the minimal polynomial has to be $(T-I)^{n+1}$.
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H: Elements of the ring of multipliers are integral
For $K$ a number field, $\{\alpha_1, \dots, \alpha_n\}$ a basis of $K/\mathbb{Q}$ and $M = \mathbb{Z}\alpha_1 + \dots + \mathbb{Z}\alpha_n$, the corresponding ring of multipliers is defined as
$$ \mathcal{O} = \{ \alpha \in K : \alpha M \subseteq M \}. $$
I want to prove that this is an order in $K$, which requires proving the inclusion $\mathcal{O} \subseteq \mathcal{O}_K$. I feel like I am missing something obvious here as I do not understand why this holds. I also have a hard time understanding what $\alpha$ are selected by the property $\alpha M \subseteq M$, and how $M$ and $\mathcal{O}$ are related in general, so any help about this is appreciated.
I found this related question, which gives the inclusion $d \mathcal{O}_K \subseteq \mathcal{O}$ for some $d$ in $\mathbb{Z}$ so the proof is complete after that.
This is also the only relevant result I found with the term "ring of multipliers" so could it be that this is not the proper terminology?
AI: By assumption, for all $i$, $\alpha\alpha_i=\displaystyle\sum_{j=1}^n a_{ij}\alpha_j,$ for some $a_{ij}\in\mathbb{Z}$.
Set $A=(a_{ij})$ and $v= (\alpha_1 \ \cdots \ \alpha_n)^t$, so that $Av=\alpha v$. In particular, $\det(\alpha I_n-A)=0$. Now $\det(X I_n-A)$ is a monic polynomial with integer coefficients for which $\alpha$ is a root.
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H: Prove that a Tower of Height $H$ can be built if $H*(H+1)/2 = R + G$
Let us define a Red-Green Tower:
Each level of the red-green tower should contain blocks of the same
color.
At every Increase in the level, number of blocks in that level is one less then previous level.
Prove that a Red-Green Tower of Height $H$ can be built if $H*(H+1)/2 = R + G$
R = total number of Red blocks in the tower.
G = total number of Green blocks in the tower.
Example:
H=4, R=4, G=6
H*(H+1)/2 = R + G
(Source)
AI: The total number of blocks of a $H$ tower is $N(H) = \sum_{k=1}^H n = \frac{1}{2}H(H+1)$. Now you are hence saying that for any $R$, $G$ such that $R+G=N(H)$ there is a red-green tower of $H$ levels.
You can procede by induction. If $H=1$ then $N(1)=1$ and you have a single block red or green constituting your tower.
Now you have to prove that if it's true for all $H$ then it is for $H+1$.
Fix hence a $H$. Now consider $R+G=N(H+1) = \frac{1}{2}H(H+1)$. Then we can distinguish three cases.
First case: $G=0$ or $R=0$. It is trivially true.
Second case: $G\leq H+1$ or $R\leq H+1$. Let's say that $R\leq H+1$. Then you fill the $R$-th level, which is made with $R$ blocks, with all the $R$ red blocks. You fill all the other levels with the green blocks and you are done.
Third case: $G>H+1$ and $R>H+1$. Then you can fill the last layer with $H+1$ green blocks. You are left with $N(H+1)-(H+1) = N(H)$ blocks and you have to fill the first $H$ layers. It is possible by induction, since you know that there exists a green-red tower of $H$ levels every time you have $N(H)$ blocks.
Note that as @Hagen has shown in his answer, it's easy to see that these three cases are exhaustive, i.e. you're always in one of them.
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H: Questions about counting the number of triples arranged in geometric progression
Problem: Three tickets are chosen from a set of $100$ tickets numbered $1,2,3,\ldots,100$. Find the number of choices such that the numbers on the three tickets are in geometric progression.
Solution: Let $k, n \in \mathbb Z_+$ s.t. $n \ge 2, \ kn^2 \le 100.$ Then the number of possibilities for $k$ is $\lfloor \frac{100}{n^2} \rfloor$. Hence total number of $k, kn, kn^2$ with integer common ratio is $\sum_{n = 2}^{10}\lfloor \frac{100}{n^2} \rfloor = 25 + 11 + 6 + 4 + 2 + 2 + 1 + 1 + 1 = 53.$ Now the number of $k, k\frac mn, k\frac{m^2}{n^2}$ where $n > 1, \ n^2 \mid k, \ m > n, \ \gcd(m, n) = 1$ is given by $f(n) = \sum\lfloor \frac{100}{m^2} \rfloor $. So the number of GPs with non-integer common ratio $\sum_{n= 2}^9f(n)$. Thus the total number of GP's in question is $53 + \sum_{n= 2}^9f(n) = 105.$
My questions:
When $n \in \mathbb Z_+$, we have the following
$k, 2k, 4k \iff 25, 50, 100$
$k, 3k, 9k \iff 11, 33, 99$
$k, 4k, 16k \iff 6, 24, 96$
$\ldots$
$k, 10k, 100k \iff 1, 10, 100$
That is, nine triples in geometric order. I am confused as to what exactly $\sum_{n = 2}^{10}\lfloor \frac{100}{n^2} \rfloor$ counts.
When the common ratio is rational, we have $km^2 \le 100n^2$. Why, then, is the number of $k$ is given by $\lfloor \frac{100}{m^2} \rfloor$ instead of, say, $\lfloor \frac{100n^2}{m^2} \rfloor$?
AI: Your first patterN, with a common factor of $2$, can start with any number from $1$ through $25$. $\lfloor \frac {100}{n^2} \rfloor$ is the number of progressions with common factor $n$. The third number is $n^2$ times the first, so the progression must start with a number small enough that the third term is no greater than $100$. That is all numbers from $1$ through $\lfloor \frac {100}{n^2} \rfloor$, where the brackets represent the floor function, so $\lfloor \frac {100}{9} \rfloor=11$
For the second, $k$ must be a multiple of $n^2$, so you should not have $n^2$ in the numerator.
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H: Remainder of $15^{81}$ divided by $13$ without using Fermat's Little theorem.
I was requested to find the congruence of $15^{81}\mod{13}$ without using Fermat's theorem (since that is covered in the chapter that follows this exercise). Of course I know that by property $15^{81} \equiv 2^{81} \pmod{13}$, but how could I find what is the congruence of $2^{81}$ without using Fermat? Needless it is to say that an exhaustive iterative method would be extremely long.
AI: $15\equiv2\bmod13$, so we have $$15^{81}\equiv2^{81}\bmod{13}$$$$\equiv512^9\equiv5^9\bmod{13}$$$$\equiv125^3\equiv8^3\bmod{13}$$$$\equiv512\equiv5\bmod13$$
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H: Large N limit of a particular sum
I'm working through a statistical physics book and one of the problems makes the claim that the quantity:
$$H =\frac{1}{N}\sum_{n=1}^N\frac{1}{\frac{a}{N} + \frac{b}{N^{5/3}}(n^{5/3}-(n-1)^{5/3})}$$
can be expressed in the large $N$ limit as the integral:
$$H = N\int_0^1 \frac{\text{d}u}{a+\frac{5}{3}b u^{2/3}}.$$
However, I am unable to arrive at that result. I've tried to use LRAM but I got a different result. Any tips will be appreciated.
AI: Observe that, for $n$ large,
$$
n^{5/3} - (n-1)^{5/3} = n^{5/3} \left[1- \left(1 - \frac{1}{n}\right)^{5/3}\right]
\sim n^{5/3} \frac{5}{3n} = \frac{5}{3} n^{2/3}
$$
Using the above approximation (even if, in my opinion, in this case it is not fully justifiable), you can write $H$ as
$$
H \sim N \cdot \sum_{n=1}^N \frac{1}{N} \cdot \frac{1}{a + \frac{5}{3} b \left(\frac{n}{N}\right)^{2/3}}\,.
$$
Finally, you can use the fact that, if $f$ is a continuous function, then
$$
\sum_{n=1}^N \frac{1}{N} f\left(\frac{n}{N}\right)
\to \int_0^1 f(u)\, du,
$$
as $N\to +\infty$.
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H: Partial Fraction Decomposition of $\frac{1}{x^2(x^2+25)}$
I have been reviewing some integration techniques and have been searching for tough integrals with solutions online. When I was going through the solution, however, I found a discrepancy between my solution and theirs and think what I did was correct instead.
I am trying to solve the indefinite integral: $\int\frac{dx}{x^2(x^2+25)}$. My first step was to break it into the fractions $$\frac{1}{x^2(x^2+25)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+25}$$ Then multiplying both sides by $x^2(x^2+25)$, we find our basic equation to be$$1=A*x(x^2+25)+B*(x^2+25)+(Cx+D)*x^2$$ Solving the system of linear equations, I found that $B=\frac{1}{25}$, $D=\frac{-1}{25}$, and $A=C=0$.
This is where I found the discepancy. The online solution has the basic equation as $$1=A*x(x^2+25)+B*(x^2+25)+(Cx+D)*x$$ so when they solve for coefficients they find that $B=\frac{1}{25}$, $C=\frac{-1}{25}$, and $A=D=0$.
Am I correct or are they? And if my answer is incorrect how does one of the $x$'s cancel out from the $(Cx+D)$ term? Thanks for any help!
AI: You are right. Here's an other proof:
If we put $X=x^2$ then
$$f(x)=\frac{1}{x^2(25+x^2)}$$
$$=\frac{1}{X(25+X)}$$
$$=\frac{B}{X}+\frac{D}{25+X}$$
$$Xf(x)=B+\frac{DX}{X+25}$$
with $ X=0$, it gives $ B=\frac{1}{25}$
$$(X+25)f(x)=\frac{B(X+25)}{X}+D$$
with $ X=-25$, we get $D=-\frac{1}{25}$
thus
$$f(x)=\frac{1}{25}\Bigl(\frac{1}{x^2}-\frac{1}{25+x^2}\Bigr)$$
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H: $ker(T)^{\bot} = \overline{im(T^*)}$ if $T$ is a linear operator between Hilbert spaces
Let $T$ be a linear operator.
For any underlying normed spaces it holds that
$$ker(T)^{\bot} \subset \overline{im(T^*)},$$
but if they are both Hilbert spaces we get
$$ker(T)^{\bot} = \overline{im(T^*)}.$$
Now my question is:
How to prove the inclusion from right to left?
AI: For a Hilbert space, and any vectors $x,y$, $$\langle T^*x,y\rangle = \langle x,Ty\rangle$$ Hence $x\in(im T)^\perp\iff x\in\ker T^*$, i.e., $(im T)^\perp=\ker T^*$. Taking a second perp gives $$(\ker T^*)^\perp=(im T)^{\perp\perp}=\overline{im T}$$ Apply this identity with $T^*$ instead of $T$ to get $$(\ker T)^\perp=\overline{im T^*}$$
Edit: This works in Hilbert spaces but not Banach spaces because $T^{**}=T$ is valid in the former but not the latter in general.
Edit 2: Direct proof. Let $x\in\overline{im(T^*)}$ and $y\in\ker T$, then there are vectors $z_n$ such that $T^*z_n\to x$, and $Ty=0$. So $$\langle y,x\rangle=\lim_{n\to\infty}\langle y,T^*z_n\rangle=\lim_{n\to\infty}\langle Ty,z_n\rangle=0$$
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H: Convergence in distribution - Gamma distribution/degenerate distribution
I got a gamma distribution which is defined as followed
$G_{n,\lambda}=\frac{\lambda e^{-\lambda x}(\lambda x)^{n-1}}{(n-1)!}$.
The paper I am currently reading says, that die $G_{n, n/t}$ converges in distribution to the degenerate distribution.
I don't know why this is true. Does someone has a hint or the answer for me ?
Thanks a lot !
AI: The Moment Generating Function of your rv is
$$MGF_{X}(k)=(1-\frac{kt}{n})^{-n}$$
Its limit for $n\rightarrow +\infty$ is
$$e^{kt}$$
That is exactly the MGF of a degenerate rv
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H: Understanding the double integral $\int_0^\infty\int_0^t f(x)g(t-x)dxdt$
I am dealing with the integral $\int_0^\infty\int_0^t f(x)g(t-x)dxdt$ and $g(x)=0$ if $x<0$.
I need to arrive to $\int_0^\infty f(x)dx\int_0^\infty g(t)dt$.
Using Fubini, I have:
$$\int_0^\infty\int_0^t f(x)g(t-x)dxdt=$$
$$\int_0^t f(x)\int_0^\infty g(t-x)dtdx=$$
$$\int_0^t f(x)\int_{-x}^\infty g(t)dtdx=$$
$$\int_0^t f(x)dx\int_{0}^\infty g(t)dt.$$
How can I proceed to get limit $\infty$ at first integral?
Many thanks!
AI: Well, you applied Fubini's theorem incorrectly. We have
$$\begin{align}
\int_0^\infty \int_0^t f(x) g(t-x)\,dx\,dt&=\int_0^\infty \int_x^\infty f(x) g(t-x)\,dt\,dx\\\\
&=\int_0^\infty f(x)\int_x^\infty g(t-x)\,dt\,dx\\\\
&=\int_0^\infty f(x)\int_0^\infty g(t)\,dt\,dx\\\\
&=\left(\int_0^\infty f(x)\,dx\right)\left(\int_0^\infty g(x)\,dx\right)
\end{align}$$
as was to be shown!
NOTE:
Alternatively, we could have exploited the fact that $g(x)=0$ for $x<0$. Then noting that $g(t-x)=0$ for $t<x$, we see that
$$\begin{align}
\int_0^\infty f(x)g(t-x)\,dx&=\int_0^t f(x) g(t-x)\,dx+\int_t^\infty f(x)\underbrace{g(t-x)}_{=0}\,dx\\\\
&=\int_0^t f(x) g(t-x)\,dx\tag1
\end{align}$$
Using $(1)$, we find that
$$\begin{align}
\int_0^\infty \int_0^t f(x) g(t-x)\,dx\,dt&=\int_0^\infty \int_0^\infty f(x) g(t-x)\,dx\,dt\\\\
&=\int_0^\infty f(x) \int_0^\infty g(t-x)\,dt\,dx\\\\
&=\int_0^\infty f(x) \int_{-x}^\infty g(t)\,dt\,dx\\\\
&=\int_0^\infty f(x) \int_0^\infty g(t)\,dt\,dx\\\\
&=\left(\int_0^\infty f(x)\,dx\right)\left(\int_0^\infty g(x)\,dx\right)
\end{align}$$
as expected!
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H: Integration with absolute value and a constant range in it
The question is
$\int_1^4 |a^2-x^2|dx=\cdots$ for $1<a<4$
Now that I am confuse since I don't have any idea on how to separate into several definite integrals. I mean, when $a=2$, then we can separate it into $\{1,2\}$ and $\{2,4\}$. However, when $a=3$, it will become $\{1,3\}$ and $\{3,4\}$, they're now different. How should I solve this? Thanks for any help!
AI: Separate in the two intervals $[1,a]$ and $[a, 4]$ as follows:
$$ \int_1^4 |a^2-x^2| dx = \int_1^a (a^2-x^2)dx + \int_a^4 (x^2-a^2)dx = \left[ a^2x-\frac{x^3}{3} \right]_1^a + \left[ \frac{x^3}{3} - a^2x \right]_a^4 $$
$$ = \frac{2}{3}a^3 - a^2 + \frac{1}{3} + \frac{64}{3}-4a^2+\frac{2}{3}a^3 $$
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H: Proof of the inclusion sets
I would know why if I have $A \subseteq B ,\:$ I obtain $P(B)\geq P(A)$
Thanks to everyone
AI: Hint:
Write $B= A\cup (B\smallsetminus A)$ and use the axioms of a probability measure.
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H: is there a function $\gamma(x)$ where when $a$ & $b$ and $a+1$ & $b+1$ are co-prime, $\gamma(\frac{a}{b})>\gamma(\frac{a+1}{b+1})$
is there a function $\gamma(x)$ where when $a$ & $b$ and $a+1$ & $b+1$ are co-prime, $\gamma(\frac{a}{b})>\gamma(\frac{a+1}{b+1})$
when you start with $\gamma(\frac{1}{2})$ you get an inequality for all n
$$\gamma(\frac{1}{2})>\gamma(\frac{2}{3})>\gamma(\frac{3}{4})>...>\gamma(\frac{n}{n+1})>\gamma(\frac{n+1}{n+2})>...$$
but if you start with $\frac{1}{3}$ you find no new information because $\frac{1+1}{3+1}$ isn't fully simplified so $3+1$ and $1+1$ isn't coprime so we don't know if $\gamma(\frac{1}{3})>$or$=$ or$<\gamma(\frac{1}{2})$
is there a function that follows this rule for all fractions $\frac{a}{b}$ and that's differentiable everywhere
And if there is a function $\gamma(x)$ then is it made up from elementary functions?
AI: Assuming that $a, b$ are positive integers:
$$
\frac{a}{b} < 1 \implies \frac{a}{b} < \frac{a+1}{b+1} < 1 \\
\frac{a}{b} > 1 \implies \frac{a}{b} > \frac{a+1}{b+1} > 1 \\
$$
(compare also How to prove that adding $n$ to the numerator and denominator will move the resultant fraction close to $1$?).
Therefore any function $\gamma$ which is strictly decreasing on $(0, 1)$ and strictly increasing on $(1, \infty)$ has the property that
$$
\gamma(\frac{a}{b})>\gamma(\frac{a+1}{b+1})
$$
for all $a \ne b$, and in particular for co-prime integers.
There are many such functions, a simple example is $\gamma(x) = x + \frac 1x$.
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H: every root of unity $x$ determines a degree one representation of $G$
I was told that if $G=\langle g \rangle$ is a cyclic group then every root of unity $x$ determines a degree one representation of $G$. But what is this representation? $\rho(x)=?$
AI: It is only $n$th roots of unity, where $g$ has order $n$. Then $\rho(g)=(\zeta)$ for $\zeta$ an $n$th root of unity is the representation.
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H: Calculate residues at all isolated singularities of $f(z)=\frac{z^2+4}{(z+2)(z^2+1)^2}$.
Calculate residues at all isolated singularities of $f(z)=\frac{z^2+4}{(z+2)(z^2+1)^2}$.
So I found the isolated singularities to be $z=-2$ and $z=i$.
Then I found the residue at $z=-2$.
$\operatorname{res}(f,i)=\lim_{z\to -2}$ $(z+2)\left(\frac{z^2+4}{(z+2)(z^2+1)^2}\right)$
$=\lim_{z\to -2} \frac{z^2+4}{(z^2+1)^2}=\frac{8}{25}$
but I don't know how to do the other one.
I tried this, since it's a double pole:
$\operatorname{res}(f,i)=\lim_{z\to i} \left((z-i)^2\frac{z^2+4}{(z+2)(z^2+1)^2}\right)'$, but I got stuck because this didn't allow me to cancel the singularity. What do I do?
AI: Let $g(z)=\dfrac{z^2+4}{(z+2)(z+i)^2}$. Then $g(i)=\dfrac{-6+3i}{20}$ and $g'(i)=-\dfrac{8+31i}{50}$. So, near $i$ you have$$g(z)=\frac{-6+3i}{20}-\dfrac{8+31i}{50}(z-i)+\cdots$$and therefore$$f(z)=\frac{g(z)}{(z-i)^2}=\frac{-6+3i}{20}(z-i)^{-2}-\dfrac{8+31i}{50}(z-i)^{-1}+\cdots$$So$$\operatorname{res}(i,f(z))=-\dfrac{8+31i}{50}$$and the computation of $\operatorname{res}(-i,f(z))$ is similar.
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H: Is every commutative ring isomorphic to a product of directly irreducible rings?
In the following, all rings are assumed to be commutative, with multiplicative identity.
A ring $R$ is said to be directly irreducible if it is not isomorphic to the direct product of two non trivial rings. An equivalent condition is that $R$ does not contain any idempotent element other than 0 and 1.
It is not hard to prove that any noetherian ring $R$ is isomorphic to a finite direct product of directly irreducible rings (if this wasn't the case, then you could "split" $R$ indefinitely and produce an infinite chain of ideals). Moreover, the factors are isomorphic up to reordering.
Is it true that every ring $R$ is isomorphic to a (possibly infinite) direct product of directly irreducible rings?
AI: No. For instance, let $R$ be the Boolean ring of subsets of $\mathbb{N}$ that are either finite or cofinite. Any quotient of $R$ is also a Boolean ring, and the only directly irreducible Boolean ring is $\mathbb{F}_2$. But $R$ is not a product of copies of $\mathbb{F}_2$, for instance because it is countably infinite.
More generally, in any ring $R$, the set of idempotent elements form a Boolean algebra $B$. If $R\cong \prod_{i\in I}R_i$ is a product of directly irreducible rings, then $B$ would be isomorphic to the power set algebra $\mathcal{P}(I)$. So, if $B$ is not a power set algebra, then $R$ cannot be a product of directly irreducible rings.
Note moreover that if $R\cong \prod R_i$ is a product of directly irreducible rings, then the projections $R\to R_i$ are exactly the quotient maps $R\to R/(1-e)$ where $e$ ranges over the atoms of the Boolean algebra $B$ (i.e., the minimal nonzero idempotents of $R$). So, a ring $R$ is isomorphic to a product of directly irreducible rings iff the canonical map $R\to\prod_{e}R/(1-e)$ is an isomorphism, where $e$ ranges over the atoms of $B$ (note that such a quotient $R/(1-e)$ always is directly irreducible).
Using this criterion, here is an example of a ring that is not a product of directly irreducible rings even though its Boolean algebra of idempotents is a power set algebra. Let $k$ be an infinite field, let $I$ be an infinite set, and let $R$ be the ring of functions $I\to k$ that take only finitely many values. Then the Boolean algebra of idempotents in $R$ is $\mathcal{P}(I)$, since the characteristic function of every subset of $I$ is in $R$. However, the quotient maps $R\to R/(1-e)$ for atoms $e$ are exactly the evaluation maps $R\to k$ at elements of $I$, so the canonical map $R\to\prod R/(1-e)$ is just the inclusion $R\to k^I$. Since $R$ is not all of $k^I$, it cannot be a product of directly irreducible rings.
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H: Solve for x, such that: $\sum_{n=0} \frac{x}{4^n} = \sum_{n=0} \frac{1}{x^n}$
The question is stated as the following to solve for x: $$x+\frac{x}{4}+\frac{x}{16}+\frac{x}{64}+...=1+\frac{1}{x}+\frac{1}{x^2}+...$$
I have so far determined their infinite series as shown in the title and have been treating both these series as geometric series. I am pretty confused in terms of next steps, thank you in advance!
AI: I will refer to the question, since the title has a couple of misleading indices I guess...
The LHS is equal to $x (1 + \frac{1}{4} + \frac{1}{16} + ...) = x \frac{1}{1 - \frac{1}{4}} = \frac{4x}{3}$. The RHS is equal to $\frac{1}{1 - \frac{1}{x}} = \frac{x}{x-1}$, but it converges only for $|x| > 1$. Now we solve
$$ \frac{4x}{3} = \frac{x}{x-1} \implies x = 0, \frac{7}{4}, $$
but $x = \frac{7}{4}$ is the only acceptable solution.
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H: Is the map $T \mapsto T^*T$ lower-semicontinuous in the weak operator topology?
Let $H$ be a separable Hilbert space and $\mathcal{L}(H)$ the set of bounded operators on $H$.
If $D \in \mathcal{L}(H)$ is a positive operator, $D \geq 0$, is true that the set
$$\{T \in \mathcal{L}(H)\mid T^*T \leq D\}$$ is closed in the weak operator topology (WOT) on $\mathcal{L}(H)$?
Note that the mapping $T \mapsto T^*T$ is not continuous in WOT, like the counterexample $H = \ell^2$, $T_n = S^n$ ($S$ - unilateral (right) shift) shows.
AI: Suppose $T_\alpha \to T$ in WOT and $T_\alpha^*T_\alpha≤D$. What we would like to have is that $T^*T^\mathstrut≤D$, ie
$$\langle x , (D-T^*T^{\mathstrut})x\rangle=\langle x,Dx\rangle - \|Tx\|^2\overset!≥0$$
for all $x$. Now we may note that $\|Tx\|=\sup_{\|y\|≤1}|\langle y, Tx\rangle|$, while
$$|\langle y, Tx\rangle |= \lim_\alpha |\langle y, T_\alpha x\rangle|≤\liminf_\alpha \|T_\alpha x\|$$
if $\|y\|≤1$, giving $\|Tx\|^2≤ \liminf_\alpha\|T_\alpha x\|^2$. Now using $\langle x ,(D-T^*_\alpha T_\alpha)x\rangle ≥0$ we find:
$$\langle x, Dx\rangle -\|Tx\|^2 ≥ \langle x, Dx\rangle -\liminf_\alpha \|T_\alpha x\|^2 =\limsup_\alpha \langle x, (D-T^*_\alpha T_\alpha)x\rangle$$
but for every $\alpha$ the term on the right is $≥0$, giving us the conclusion we wanted.
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H: Derivative of $h(x,t)=g\left(\frac{x}{t^2}\right)$
Can someone please explain me the method to get the partial derivative of a function like:
$h(x,t)=g\left(\frac{x}{t^2}\right)$ where $g$ is a differentiable function from $\mathbb{R}\to\mathbb{R}$.
I know that if we have a function $f(x(t,s),y(t,s))$ then
$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial t}$.
Also I can see that the given function $h$ can be written as a composition of two functions as follows:
For $g_1(x.t)=\frac{x}{t^2}$. So that, $$h(x,t)=g(g_1(x,t)).$$
But I really cannot see how to use them to partially differentiate $h$.
Appreciate your help.
AI: $$\frac{\partial h}{\partial x}=g'(g_1(x,t))\cdot \frac{\partial g_1}{\partial x}(x,t)=g'\left(\frac{x}{t^2}\right)\cdot \frac{1}{t^2}$$
and
$$\frac{\partial h}{\partial t}=g'(g_1(x,t))\cdot \frac{\partial g_1}{\partial t}(x,t)=g'\left(\frac{x}{t^2}\right)\cdot \frac{(-2x)}{t^3}$$
Edit: Note that when you wrote the formula for partial derivative of $f,$ you had $2$ dependent functions $x$ and $y,$ whereas your function is actually simpler in that you only have one dependent function $g_1.$ Drawing a tree diagram (for example, explained here) always helped me during undergrad while applying chain rule when dealing with partial derivatives.
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H: Orthonormal columns of block matrices expanded with Kronecker products
Let $W_{i,1},W_{i,2},W_{i,3} \in \mathbb{R}^{n \times n}$, $i \in {1,2}$ be such that
$$
\eqalign{
\Big[\matrix{W_{i,1}^T & W_{i,2}^T & W_{i,3}^T}\Big]
\left[\matrix{W_{i,1}\\W_{i,2}\\W_{i,3}}\right]
= I
}
$$
where $I$ is the identity matrix. Now, let
$$\eqalign{
W = \left[\matrix{
W_{2,1}W_{1,1} \\
W_{2,2}W_{1,1} \\
W_{2,3}W_{1,1} \\
W_{1,2} \\
W_{1,3}} \right]
}$$
Then, $W^{T}W = I$ but I don't see why. How can this idea be expanded towards the Kronecker products, i.e., $W_{i,j} \otimes W_{i,k}$ for $i \in \{1,2\}$ and $j,k \in \{1,2,3\}$?
AI: Let's start with showing $W^\top W = I$.
We can write the assumption about the $W_{i,j}$ as
$$
\begin{align}
\label{eq:condition}\tag{\(*\)}
W_{i,1}^\top W_{i,1} + W_{i,2}^\top W_{i,2} + W_{i,3}^\top W_{i,3}
= I,
\qquad i = 1,2.
\end{align}
$$
Now let's expand $W^\top W$ in a similar way.
$$
\begin{align*}
W^\top W
&= \Big[\begin{array}{ccccc}
(W_{2,1}W_{1,1})^\top &
(W_{2,2}W_{1,1})^\top &
(W_{2,3}W_{1,1})^\top &
W_{1,2}^\top &
W_{1,3}^\top
\end{array}\Big]
\left[\begin{array}{c}
W_{2,1}W_{1,1} \\
W_{2,2}W_{1,1} \\
W_{2,3}W_{1,1} \\
W_{1,2} \\
W_{1,3}
\end{array}\right]
\\
&= W_{1,1}^\top W_{2,1}^\top W_{2,1} W_{1,1}
+ W_{1,1}^\top W_{2,2}^\top W_{2,2} W_{1,1}
+ W_{1,1}^\top W_{2,3}^\top W_{2,3} W_{1,1}
+ W_{1,2}^\top W_{1,2}
+ W_{1,3}^\top W_{1,3}
\\
&= W_{1,1}^\top \left(W_{2,1}^\top W_{2,1}
+ W_{2,2}^\top W_{2,2}
+ W_{2,3}^\top W_{2,3} \right)W_{1,1}
+ W_{1,2}^\top W_{1,2}
+ W_{1,3}^\top W_{1,3}
\end{align*}
$$
Now recognize that the sum in the parentheses is \eqref{eq:condition} with $i = 2$.
Then we have a simplification:
$$
\begin{align*}
W^\top W
&= W_{1,1}^\top (I) W_{1,1}
+ W_{1,2}^\top W_{1,2}
+ W_{1,3}^\top W_{1,3}
\\
&= W_{1,1}^\top W_{1,1}
+ W_{1,2}^\top W_{1,2}
+ W_{1,3}^\top W_{1,3}
\\
&= I,
\end{align*}
$$
using \eqref{eq:condition} again but with $i=1$.
Not sure how this relates to the Kronecker product though, especially since $W$ has matrix-multiplied blocks instead of element-wise multiplications. Note though that $W$ has orthonormal columns, since $W^\top W = I$, but the $W_{i,j}$ don't necessarily, because $W_{i,j}^\top W_{i,j} = I$ for all $i,j$ would contradict \eqref{eq:condition}.
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H: What is the order and structure of Aut(SL$_2(p)$)?
Here $p$ is a prime. We know that for $Z(SL_2(p)) = \{ \pm I \}$ and $\lvert SL_2(p)\rvert= p^3-p$ so there are $\frac{p^3-p}{2}$ inner automorphisms. What is the outer automorphism group?
AI: It is $\mathrm{PGL}_2(p)$ for $p\geq 5$, which I assume you are thinking about. Any automorphism of $\mathrm{SL}_2(p)$ induces an automorphism on $\mathrm{PSL}_2(p)$. The automorphism group of $\mathrm{PSL}_2(p)$ is $\mathrm{PGL}_2(p)$. The only question is whether one can pull these back to automorphisms of $\mathrm{SL}_2(p)$. You can, and the group $\mathrm{SL}_2(p).2$ is in fact a subgroup of $\mathrm{SL}_2(p^2)$.
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H: How can I prove $A − A(A + B)^{−1}A = B − B(A + B)^{−1}B$ for matrices $A$ and $B$?
The matrix cookbook (page 16) offers this amazing result:
$$A − A(A + B)^{−1}A = B − B(A + B)^{−1}B$$
This seems to be too unbelievable to be true and I can't seem to prove it. Can anyone verify this equation/offer proof?
AI: \begin{align}
A - A(A+B)^{-1}A & = A(A+B)^{-1}(A+B) - A(A+B)^{-1}A \\ &= A(A+B)^{-1}(A+B - A)\\ &= A(A+B)^{-1}B \\ &= (A+B - B)(A+B)^{-1}B \\ &= (A+B)(A+B)^{-1}B - B(A+B)^{-1}B \\ &= B - B(A+B)^{-1}B
\end{align}
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H: Why is the solution to a non-homogenous linear ODE written in terms of a general fundamental solution and not a matrix exponential?
Why is the solution to a non-homogenous linear ODE written in terms of a general fundamental solution and not a matrix exponential? Generally, I see the solution to a non-homogenous linear ODE
$$
\dot{x} = Ax + b(t)\\
x(0) = x_0
$$
written as
$$
x(t) = \Phi(t)\Phi^{-1}(0)x_0 + \int_0^t \Phi(t)\Phi^{-1}(\tau)b(\tau)d\tau
$$
where
$$
\Phi^\prime(t)=A\Phi(t).
$$
At the same time, I thought $\Phi(t)=\mathbb{e}^{At}$, which means that $\Phi(0)=I$ and the above could at least be simplified to
$$
x(t) = \Phi(t)x_0 + \int_0^t \Phi(t)\Phi^{-1}(\tau)b(\tau)d\tau
$$
if not further to
$$
x(t) = \mathbb{e}^{At}x_0 + \mathbb{e}^{At}\int_0^t (\mathbb{e}^{A\tau})^{-1}b(\tau)d\tau.
$$
Why is it, then, that the non-homgenous solution is written as originally stated in terms of $\Phi$? Is there a change of basis or generalization that I'm not aware of, or is there a mistake in this simplication?
AI: There's nothing wrong with using the matrix exponential in this case, where $A$ does not depend on time. However, for non-autonomous equations $\dot{x} = A(t) x + b(t)$, you can't write $\Phi(t)$ as a matrix exponential (it's sometimes called a "time-ordered exponential").
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H: Ideal of whole numbers generates whole ring, can we find a linear combination of 1 with coefficients always whole numbers?
Let $R$ be a unitary ring. For $a,b \in \mathbb{Z}$ we can embed $a,b$ in $R$. Now consider the ideal $I:=(a,b)$ in $R$. If $(a,b )= R$ we find $c,d\in R$ such that $ac+bd = 1$. Can we always chose $c,d$ such that they come from the embeding of $\mathbb{Z}$ in $R$.
AI: Well, not really. Consider $R=\mathbb R$ and $I=(2,2)$. Then $2\cdot\frac12+2\cdot 0=1$, but you'll never be able to do it with integers.
It's because the abelian subgroup generated by $2$ can't ever escape the multiples of $2$. Being able to multiply by other elements of the ring to produce elements in the ideal expands it considerably.
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H: What does it mean for F(Lim X) to be 'naturally isomorphic' to Lim(F(X))
Let F be a left adjoint functor. Given any diagram X, what does it mean for F(Lim X) to be naturally isomorphic to Lim(F(X)). What are the functors here which are naturally isomorphic?
AI: The functors are $$\mathcal{C}^I\overset{F^I}{\longrightarrow}\mathcal{D}^I\overset{lim}{\longrightarrow}\mathcal{D}$$
and $$\mathcal{C}^I\overset{lim}{\longrightarrow}\mathcal{C}\overset{F}{\longrightarrow}\mathcal{D},$$
where $I$ is the domain of the functor $X$.
There is some possible abuse here, depending on whether or not $\mathcal{C},\mathcal{D}$ are actually (co)complete; an author could conceivably phrase this as if the limit functors exist when they don't really, as a kind of fiction to simplify the statement.
If they are not, then we can rephrase it, albeit in an uninteresting way. If $\bar{X}$ is the right Kan extension of $X$ along $!:I\to 1$, then we can say that $\overline{F\circ X}$ is naturally isomorphic to $F\circ\bar{X}$ as functors $1\to\mathcal{D}$. This is uninteresting because clearly we're just saying they're isomorphic as objects.
Personally, I don't think I've ever seen someone talk about natural isomorphism in the latter case, and I wouldn't recommend doing so.
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H: Distance between two closed subsets of $\mathbb{R}$
Suppose $A$ and $B$ are two closed subsets of $\mathbb{R}$, and assume that : $$\text{dist}(A,B)=\inf_{(a,b)\in A\times B}|a-b|=0.$$
Is it true that $A\cap B\neq\varnothing$ ?
I found a counter-example using arithmetics : $A=\mathbb{N}^\star$ and $B=\pi\mathbb{N}^\star$ are two disjoint closed subets of $\mathbb{R}$, but from rational approximations, I showed that $\text{dist}(A,B)=0$.
Now, if both $A$ and $B$ are compact, by considering the function $\Phi:A\times B\to\mathbb{R}$ defined by $\Phi(a,b)=|a-b|$, since it is continuous on the compact $A\times B$, it is bounded and there is a $(a_\ast,b_\ast)\in A\times B$ such that $\Phi(a_\ast,b_\ast)=\inf\Phi=0$, and this means that $|a_\ast-b_\ast|=0$, i.e. $a_\ast=b_\ast$, whence $A\cap B\neq\varnothing$.
My question is : What if we assume that only $A$ is compact ? ($B$ is still supposed to be closed.)
I considered the exact same $\Phi$ as before. Now, by the sequential characterization of $\inf\Phi$, there exist sequences $(a_n)_{n\in\mathbb{N}}\in A^\mathbb{N}$ and $(b_n)_{n\in\mathbb{N}}\in B^\mathbb{N}$ such that : $$\Phi(a_n,b_n)\underset{n\to+\infty}{\longrightarrow}\inf\Phi=0.$$
From (sequential) compactness of $A$, there exists $\psi:\mathbb{N}\nearrow\!\!\!\nearrow\mathbb{N}$ such that : $$a_{\psi(n)}\underset{n\to+\infty}{\longrightarrow}a_\ast\in A.$$
In particular, we still have : $$|b_{\psi(n)}-a_{\psi(n)}|\underset{n\to+\infty}{\longrightarrow}0.$$
Now I wrote :
$\begin{align}
|b_{\psi(n)}-b_{\psi(n+p)}|&\leqslant|b_{\psi(n)}-a_{\psi(n)}|\\
&+|a_{\psi(n)}-a_{\psi(n+p)}|\\
&+|a_{\psi(n+p)}-b_{\psi(n+p)}|.
\end{align}$
$(a_{\psi(n)})_{n\in\mathbb{N}}$ is Cauchy since it converges, which implies that $(b_{\psi(n)})$ is also Cauchy. From completeness, it therefore converges to some $b_\ast\in\mathbb{R}$, but since $B$ is closed, we have $b_\ast\in B$, thus the same conclusion as before applies : $a_\ast=b_\ast$ and $A\cap B\neq\varnothing$.
Does this seem correct to you ? Is there a simpler method to handle with this case ? Is there a way to do this without using completeness of $\mathbb{R}$ ?
AI: If you just assume that both sets are closed it does not follow that they intersect. Indeed, consider $A=\bigcup_{n=1}^\infty[2n,\text{ }2n+\frac{1}{2n}]$ and $B=\bigcup_{n=1}^\infty[2n+1,\text{ }2n+1+\frac{1}{2n}]$. As for the second question, note that the inequality $|b_{\psi(n)}-a_*|\leq |b_{\psi(n)}-a_{\psi(n)}|+|a_{\psi(n)}-a_*|$ already implies that $b_{\psi(n)}\rightarrow a_*$ so the completeness of $\mathbb{R}$ is not really needed.
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H: Understanding matrix norm and quadratic equations
Suppose I have two quadratic expressions
$$(a - b)^2$$ and $$(a-c)^2$$
where $a, b, c$ are real numbers, then I think the following holds:
$$(a - b)^2 < (a - c)^2$$
iff $|a - b| < |a - c|$.
Now suppose I have two expressions in matrix form:
$$((A - B)v)^T((A-B)v)$$
$$((A - C)v)^T((A-C)v)$$
where $A, B, C$ are $m\times n$ real matrices, and $v$ is a $n\times 1$ vector. In this case,
when does $$((A - B)v)^T((A-B)v) < ((A - C)v)^T((A-C)v)$$
hold? I was thinking that the above will hold if
$$\|A - B\| < \|A - C\|$$
where $||\cdot||$ is the Frobenius norm. However, I am not sure if this is correct. Is it correct to approach this by measuring the "similarity" between $A$ and $B$ using a matrix norm? If so, is the Frobenius norm the correct choice?
AI: The Frobenius norm of a matrix is the sum of all squared elements from the matrix. To obtain the Frobenius norm you choose $v$ to be a vector of ones.
But I think you want to allow $v$ be any real valued vector. If $A-B$ and $A-C$ are positive definite then you can create following norm $||M||_v=v^TM^TMv=(Mv)^TMv$ where $M$ is some $m\times n$ matrix and your idea holds.
EDIT: I am not allowed to comment yet, so I edit. Anyways, insert A-B and A-C where M is in the suggested definition of norm. Instead of M use $||A-B||_v<||A-C||_v$. Remember A-B and A-C need to be positive definite too.
https://en.wikipedia.org/wiki/Definite_symmetric_matrix
If they are not positive definite or if you cannot know if they are positive definite, then we need another norm.
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H: Calculating second derivatives with curves
I always thought that the following is true:
Let $M$ be a smooth manifold, $f \in C^2(M)$, and $X \in \Gamma(TM)$. Then
$$ X^2f(x) = \frac{d^2}{dt^2} (f \circ \gamma)(0) $$
for any curve $\gamma$ with $\gamma(0) = x, \gamma'(0) = X_x$.
Here, the LHS should be interpreted as $X_x(Xf)$ where $Xf$ is the function $x\mapsto X_xf$.
The LHS can be calculated in coordinates as
$$ X^2f = X(df(X)) = X(\frac{\partial f}{\partial x^i}\, X^i) = \frac{\partial^2 f}{\partial x^i\partial x^j} X^i X^j + \frac{\partial f}{\partial x^i}\frac{\partial X^i}{\partial x^j}X^j,$$
where Einstein convention is used. On the other hand, the RHS is
$$ \frac{d^2}{dt^2} (f \circ \gamma)(0)
= \frac{d}{dt}\Bigg|_0 df_{\gamma(t)}(\gamma'(t)) = \frac{d}{dt}\Bigg|_0 ( \frac{\partial f}{\partial x^i}\Bigg|_{\gamma(t)} \gamma'(t)^i) = \frac{\partial^2 f}{\partial x^i\partial x^j} X^iX^j + \frac{\partial f}{\partial x^i} \frac{d\gamma'(t)^i}{dt}\Bigg|_0. $$
So the claim is true if one can show that
$$ \left( \frac{d^2}{dt^2} (x^i\circ\gamma)(0) \right) = \frac{d\gamma'(t)^i}{dt}\Bigg|_0= \frac{\partial X^i}{\partial x^j}X^j. $$
However, I don't see how one can show this, since it seems that we need to have some local information about the tangent field $\gamma'(t)$ around zero, which we don't given the assumptions. What am I missing?
AI: You need to demand more on $\gamma$.
For example, let $M=\mathbb{R}$ and $X=\frac{\mathrm{d}}{\mathrm{d}x}$. The curve $\gamma(t)=t+t^2$ would have $(f\circ\gamma)''(0)=\frac{\mathrm{d}^2}{\mathrm{d}t^2}f(t+t^2)\vert_{t=0}=f''(0)+2f'(0)$ which is not $X^2f(0)=f''(0)$.
If $\gamma$ agrees with the integral curve of $X$ to second order, then indeed you have
$$
X^2f(x)=\frac{\mathrm{d}^2}{\mathrm{d}t^2}(f\circ\gamma)(0).
$$
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H: I don't understand Gödel's incompleteness theorem anymore
Here's the picture I have in my head of Model Theory:
a theory is an axiomatic system, so it allows proving some statements that apply to all models consistent with the theory
a model is a particular -- consistent! -- function that assigns every statement to its truth value, it is to be thought of as a "concrete" object, the kind of thing we actually usually think about. It's only when it comes to models that we have the law of the excluded middle.
My understanding of Gödel's first incompleteness theorem is that no theory that satisfies some finiteness condition can uniquely pin down a model.
So I am not really surprised by it. The idea of theories being incomplete -- of not completely pinning down a particular model -- is quite normal. The fact that no theory is complete seems analogous to how no Turing machine can compute every function.
But then I read this thread and there were two claims there in the answers which made no sense to me:
Self-referential statements as examples of unprovable statements -- Like "there is no number whose ASCII representation proves this statement".
A statement like this cannot be constructed in propositional logic. I'm guessing this has to do with the concept of a "language", but why would anyone use a language that permits self-reference?
Wouldn't that be completely defeat the purpose of using classical logic as the system for syntactic implications?
If we permit this as a valid sentence, wouldn't we also have to permit the liar paradox (and then the system would be inconsistent)?
Unprovable statements being "intuitively true/false" -- According to this answer, if we found that the Goldbach conjecture was unprovable, then in particular that means we cannot produce a counter-example, so we'd "intuitively" know that the conjecture is true.
How is this only intuitive? If there exist $\sf PA$-compatible models $M_1$, $M_2$ where Goldbach is true in $M_1$ but not $M_2$, then $\exists n, p, q$ such that $n= p+q$ in $M_1$ but not in $M_2$. But whether $n=p+q$ is decidable from $\sf PA$, so either "$\sf{PA}+\sf{Goldbach}$" or "$\sf{PA}+\lnot\sf{Goldbach}$" must be inconsistent, and Goldbach cannot be unprovable. Right?
In any case, I don't know what it means for the extension to be "intuitively correct". Do we know something about the consistency of each extension or do we not?
Further adding to my confusion, the answer claims that the irrationality of $e+\pi$ is not such a statement, that it can truly be unprovable. I don't see how this can be -- surely the same argument applies; if $e+\pi$'s rationality is unprovable, there does not exist $p/q$ that it equals, thus it is irrational. Right?
AI: This answer only addresses the second part of your question, but you asked many questions so hopefully it's okay.
First, there is in the comments a statement: "If Goldbach is unprovable in PA then it is necessarily true in all models." This is incorrect. If Goldbach were true in all models of PA then PA would prove Goldbach by Godel's Completeness Theorem (less popular, still important).
What is true is:
Lemma 1: Any $\Sigma_1$ statement true in $\mathbb{N}$ (the "standard model" of PA) is provable from PA.
These notes (see Lemma 3) have some explanation: http://journalpsyche.org/files/0xaa23.pdf
So the correct statement is:
Corollary 2: If PA does not decide Goldbach's conjecture then it is true in $\mathbb{N}$.
Proof: The negation of Goldbach's conjecture is $\Sigma_1$. So if PA does not prove the negation, then the negation of Goldbach is not true in $\mathbb{N}$ by Lemma 1.
Remember that $\mathbb{N}$ is a model so any statement is either true or false in it (in our logic). But PA is an incomplete theory (assuming it's consistent), so we don't get the same dichotomy for things it can prove.
Now, it could be the case that PA does prove Goldbach (so its true in all models of PA including $\mathbb{N}$). But if we are in the situation of Corollary 2 (PA does not prove Goldbach or its negation) then Goldbach is true in $\mathbb{N}$ but false in some other model of PA. (This would be good enough for the number theorists I imagine.) This is also where the problem in your reasoning is. It is NOT true that if Goldbach fails in some model $M$ of PA then there is a standard $n$ in $\mathbb{N}$ that is not the sum of two primes. Rather the witness to the failure of Goldbach is just some element that $M$ believes is a natural number. In some random model, this element need not be in the successor chain of $0$.
On the other hand, the rationality of $\pi+e$ is not known to be expressible by a $\Sigma_1$ statement. So we can't use Lemma 1 in the same way.
Edited later: I don't have much to say about the question on self-referential statements beyond what others have said. But I'll just say that one should be careful to distinguish propositional logic and predicate logic. This also goes for your "general picture of Model Theory". Part of the interesting thing with the incompleteness theorems is that they permit self-reference without being so obvious about it. In PA there is enough expressive power to code statements and formal proofs, and so the self-referential statements about proofs and so forth are fully rigorous and uncontroversial.
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H: Question concerning prime ideals of $\mathbb{C}[x,y]$
I know that $(0)$, $(x-a,y-b)$ for $(a,b)\in\mathbb{C}^2$ and $(f(x,y))$ for $f(x,y)$ irreducible in $\mathbb{C}[x,y]$ are all prime ideals in $\mathbb{C}[x,y]$.
What I'd like to understand is why they are the only prime ideals. In particular, I'd like to know why the following outline of a proof is valid (which comes from Vakil's algebraic geometry notes): Let $P$ be a prime ideal that is not principal. "Show you can find $f(x,y),g(x,y)\in P$ with no common factor. By considering the Euclidean algorithm in the Euclidean domain $\mathbb{C}(x)[y]$, we can find a nonzero $h(x)\in (f(x,y),g(x,y))\subset P$."
I have two questions.
Why can we find such $f(x,y),g(x,y)$ without a common factor? If we were to pick to distinct generators of $P$, I don't see why they must be coprime. Further, I don't think we can just factor out a common factor either.
Why does the Euclidean algorithm imply we can find such a nonzero $h(x)$? Might this not still be of the form $h(x,y)$ since the Euclidean algorithm only guarantees the existence of $q(x,y)$ and $h(x,y)$ with $f(x,y)=g(x,y)q(x,y)+h(x,y)$ with the $y$ degree of $h(x,y)$ less than that of $g(x,y)$? Does this somehow follow from $f(x,y)$ and $g(x,y)$ having no common factors?
AI: We know that $P$ can be generated by finitely many generators (because $\mathbb C[x, y]$ is Noetherian), so suppose that $f_1, \dots, f_n$ is a minimal set of generators for $P$. Since $P$ is not principal, $n \geq 2$.
Now recall that $\mathbb C[x, y]$ is a unique factorisation domain. So suppose that $h$ is a greatest common divisor for $f_1$ and $f_2$. Then $f_1 = hg_1$ and $f_2 = hg_2$, where $g_1, g_2$ have no non-trivial common factor.
Now $h$ can't be in $P$, otherwise $P$ would be generated by $h, f_3, \dots, f_n$, contradicting my assumption that the generators $f_1, f_2, \dots, f_n$ were minimal.
But $f_1 = hg_1 \in P$ and $P$ is prime, so either $h\in P$ or $g_1 \in P$. Since $h \notin P$, we have $g_1 \in P$. By a similar argument, $g_2 \in P$ too.
Thus we have constructed elements $g_1, g_2 \in P$ that have no non-trivial common factor.
For the second part, see this answer.
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H: Let $A\cup B$ be open, disconnected in $\Bbb{R}^2$ where $A,B$ are non-empty, disjoint. Are both $A,B$ open in $\Bbb{R}^2$?
I have tried it in the following manner-
Assume $A$ is not open. Then $\exists x\in A$ such that $x\notin A^\circ$ i.e. $\forall \epsilon>0, B(x,\epsilon)\not\subset A$ .
Now $x\in A\cup B$, open in $\Bbb{R}^2$. Hence $\exists r>0$ such that $B(x,r)\subset A\cup B$. But $B(x,r)\not\subset A$, so we must have $B(x,r)\cap B\neq\emptyset$. Again for any $0<\epsilon\le r, B(x,\epsilon)\subset A\cup B$ and $B(x,\epsilon)\not\subset A$, so we must have $B(x,\epsilon)\cap B\neq \emptyset$ for all $0<\epsilon\le r$. Hence $x\in \overline{B}$.
$\therefore x\in A\cap\overline{B}\implies A\cap \overline{B}\neq \emptyset$.
Now if both $A$ and $B$ are connected then $A\cup B$ will be connected (but here it is given that $A\cup B $ is disconnected). So we must have either $A$ or $B$ is disconnected.
Now from this stage I cannot proceed further to get a contradiction. Can anyone guide me to conclude? Thanks for help in advance.
AI: It's not true. For example, $A = ((0,1] \cup (3,4)) \times \mathbb R$,
$B = ((1,2) \cup [4,5)) \times \mathbb R$, so $A \cup B = ((0,2) \cup (3,5)) \times \mathbb R$ is open and disconnected, but neither $A$ nor $B$ is open.
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H: How do we apply the dominated convergence theorem to conclude the proposed claim?
Definition
Let $\{f_{\lambda}:\lambda\in\Lambda\}$ be a collection of functions in $L^{1}(\Omega,\mathcal{F},\mu)$. Then for each $\lambda\in\Lambda$, by the dominated convergence theorem and the integrability of $f_{\lambda}$,
\begin{align*}
a_{\lambda}(t) := \int_{\{|f_{\lambda}| > t\}}|f_{\lambda}|d\mu \rightarrow 0\,\,\text{as}\,\,t\rightarrow \infty
\end{align*}
The collection of functions $\{f_{\lambda}:\lambda\in\Lambda\}$ in $L^{1}(\Omega,\mathcal{F},\mu)$ is uniformly integrable if
\begin{align*}
\sup_{\lambda\in\Lambda}a_{\lambda}(t) \to 0\,\,\text{as}\,\,t\to\infty
\end{align*}
My difficulties
I am mainly concerned about the preliminary part of the definition. To be more precise, how do we apply the dominated convergence theorem to conclude the proposed claim?
Once someone has helped me to understand it, I would great appreciate if someone could explain me the intuition of such definition.
Any help is appreacited.
AI: Here $\lambda$ is fixed, so it is irrelevant. We are also concerned only with $|f|$, so we may assume $f\geq0$. The situation then is that you have $f\in L^1$, $f\geq0$. Let $f_t=f\,1_{\{f\geq t\}}$. Then $f_t\to0$ pointwise. As $f_t\leq f$, Dominated convergence applies and we have that
$$
\lim_{t\to\infty}\int_\Omega f_t=\int_{\Omega}\lim_{t\to\infty}f_t=0.
$$
"Uniformly Integrable" means that you can use the same $t$, for all $\lambda$, to get the integrals to be as small as you want. Or, equivalently, that you can approximate $\int_\Omega |f_\lambda|$ with $\int_{\{|f_\lambda|\leq t\}}|f_\lambda|$ with the same $t$ for all $\lambda$.
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H: Solve the initial value problem: $\frac{dy}{dx} = e^{x+y}$, given $y(0)=0$.
I have attempted the question several times so far, and I have always reached the same answer that differs from the solution, any advice would help greatly!
My attempt
$$\frac{dy}{e^y} = e^x dx$$
Taking the integral, I got $$-e^{-y} = e^x + C.$$
Solving for $y$, I got $$y=-\ln(C-e^x).$$
After subbing in $y(0)=0$, I got $$0=-\ln(C-1)$$ and solving for $C$, I got $$C=2.$$
Thus, I got $$y=-\ln(2-e^x).$$
However, the solutions have $y=-\ln(1-e^x)$ as the answer. Have I done something wrong? Thanks in advance!
AI: You are correct, and their solution is wrong. Their solution indeed satisfies the differential equation - however, ${y(0)}$ for their solution ${\neq 0}$. In fact, ${y(x=0)}$ doesn't even exist for their solution, since the function blows up to infinity
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H: If evey linear program can be transformed to an unconstrained problem, then the optimum is unbounded because the objective is linear?
Since optimization problems with linear equality constraints can be converted into an unconstrained problem this should apply for linear programs in standard form, right?
But doesn't this mean that the optimum is unbounded for every LP in standard form, since the objective is linear and the dual is infeasible?
AI: No, because the "unconstrained" LP is actually constrained. At least some variables are $\geq0$.
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H: Prove $\sum_{n=1}^\infty \frac{1}{a_n}$ is divergent if $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are both convergent
In my practice midterm there is a multiple choice question that I thought was relatively straight forward but the solutions gave an answer that was unexpected to me.
Question: If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are convergent series, which of the following is not necessarily true?
(A)$\sum_{n=1}^\infty a_nb_n$ is convergent
(B)$\sum_{n=1}^\infty (a_n+b_n)$ = ($\sum_{n=1}^\infty a_n$) + ($\sum_{n=1}^\infty b_n$)
(C)$\sum_{n=1}^\infty (a_n-b_n)$ = ($\sum_{n=1}^\infty a_n$) - ($\sum_{n=1}^\infty b_n$)
(D)$\sum_{n=1}^\infty ca_n$=$c\sum_{n=1}^\infty a_n$ for any constant c
(E)$\sum_{n=1}^\infty \frac{1}{a_n}$ is divergent (assuming $a_n\ne0$ for all n)
I understand why options B, C & D are true given the Algebraic Properties of Convergent Series and I thought that A is true as well. However, the solutions say that the correct answer is A.
Is there any proof that holds E to be true and under what situations would A be false in this scenario?
AI: If $\sum_\limits{n=1}^\infty a_n$ is convergent then there exists $\lim_\limits{n\to\infty} a_n=0$, hence $\lim_\limits{n\to\infty}\frac{1}{a_n}=\infty$ and $\sum_\limits{n=1}^\infty\frac{1}{a_n}$ cannot be convergent (assuming $a_n\ne0$ for any $n\in\mathbb{N}$).
So (E) is necessarily true.
But (A) is not necessarily true, indeed $\sum_\limits{n=1}^\infty a_n=\sum_\limits{n=1}^\infty(-1)^n\frac{1}{\sqrt[3]{n}}\;$ and $\;\sum_\limits{n=1}^\infty b_n=(-1)^n\frac{1}{\sqrt[6]{n}}\;$ are convergent, but $\sum_\limits{n=1}^\infty a_nb_n=\sum_\limits{n=1}^\infty\frac{1}{\sqrt{n}}$ is divergent.
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H: If $f(2n)=\frac1{f(n)+1}$ and $f(2n+1)=f(n)+1$ for all $n\in\Bbb N$, then find $n$ such that $f(n)=14/5$.
The set $\mathbb{N}$ is the set of nonnegative integers. Let $ f : \Bbb{N} \rightarrow \Bbb{Q}$ be defined such that
1.) $f(2n) = \dfrac{1}{f(n)+1}$ for all integers $n>0$, and
2.) $f(2n + 1 ) = f(n)+1$ for all $n\in\Bbb{N}$.
If $f(0)=0$, then determine the value of $n\in\mathbb{N}$ such that $f(n)=\dfrac{14}{5}$.
AI: Since $f(n)>0$ for $n>0$, $f(n)>1$ for odd $n$ and $f(n)<1$ for even $n$, you can see that if $f(n)=\frac{14}{5}$, $n$ must be odd, that is $n = 2n_1+1$, where $f(n_1) = \frac 95$. Proceeding in this manner, you get
$$
n_1 = 2n_2+1, n_2 = 2n_3, n_3 = 2n_4, n_4=2n_5+1, n_5=2n_6+1, n_6 = 2n_7+1, n_7=0
$$
going back, the answer is $n=115$.
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H: How does the process of simplifying imaginary numbers actually work?
Sorry in advance if this is a really stupid question
In class I've been told that $$\sqrt{-25} = 5j $$
Converting $\sqrt{-25} $ into $5j$ is straightforward for me, but I don't understand how it works
Doesn't the property of $\sqrt{xy} = \sqrt{x}\sqrt{y} $ only hold true for positive real number values of x and y, where $i^2$ is defined to be negative 1?
In the case of $$\sqrt{-25} = 5j $$ Would we treat j= $\sqrt{-1}$ as a positive real number in order to "break" the root using the elementary algebra associated with roots? I don't really understand how this works algebraically.
AI: There's no such thing as a stupid question!
As other people have pointed to, the square-root function isn't really well-defined for all complex numbers. Fix some complex number $z\neq 0$. We can try to solve the quadratic equation $w^2=z$ (that's what a square root seeks to do) and we will find two answers $\pm w$. Which one we call $+w$ and which one we call $-w$ is purely convention. The fact that these answers are negatives of each other comes from $(-w)^2=w^2$, which is really just a dressed-up version of $(-1)^2=1$. The fact that there are always two answers comes from the "Fundamental Theorem of Algebra."
If $z$ is a positive real number, then its square roots are of the form $\pm w$, where $w>0$. Of course, choosing $w>0$ was arbitrary, but it allows us to concretely write $\sqrt{z}$ as another way to denote this positive square-root $w$.
If $z$ is a negative real number, then you can set a similar convention and take its square-roots to be $\pm iw$, where $w>0$. With this, you can write $\sqrt{z}=iw$, although this notation is not always used. This is the same as defining $\sqrt{z}=i\sqrt{-z}$, using the existing convention for positive square roots.
As you've noted, even the very symbol $i$ is only defined up to a factor $\pm 1$, because $i^2=(-i)^2=-1$. This can be phrased by saying that complex conjugation $\overline{a+bi}=a-bi$ is an isomorphism of fields. This is probably getting beyond the scope of what you wanted to know (though I'm happy to answer further questions).
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H: If $\text{ord}_m(c)=n$, find $\text{ord}_m(c^2)$
I'm trying to solve the following problem:
if $\text{ord}_m(c)=n$, find $\text{ord}_m(c^2)$
Here's what I have so far (admittedly, not much):
We know that $c^n \equiv 1 \pmod{m}$, equivalently, $c^n = 1 + mp$, $p\in\mathbb{Z}$
We know $c^2$ = $c\cdot c$.
And, of course, $c^n\cdot c^n \equiv 1 \pmod{m}$
But what $p$ satisfies $(c^2)^p \equiv 1 \pmod{m}$?
AI: Note that $c^k\equiv 1$ (mod $m$) if and only if $n|k$. So in particular $(c^2)^k\equiv c^{2k}\equiv 1$(mod $m$) iff $n|2k$. So we are actually looking for the smallest natural number $k$ such that $n|2k$. If $n$ is even this smallest number is clearly $k=\frac{n}{2}$ while if $n$ is odd this is $k=n$.
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H: Kernels of commuting linear operators on infinite dimensional vector space
If $S$ and $T$ are commuting operators on an infinite dimensional vector space $V$, it is in general true that
$$\ker S + \ker T \subseteq \ker(ST),$$
but in general equality does not hold. A simple example is given by $S = T = \frac{d}{dx}$ on $C^\infty(\mathbb{R})$. I am looking for conditions on $S$ and $T$ that will give equality in the above equation, ie:
$$\ker S + \ker T = \ker (ST)$$
Writing $\ker T^\infty$ for $\cup_n \ker T^n$, I am currently trying to show that the conditions
$\mathrm{im} S = \mathrm{im} T = V$,
$\ker S^\infty \cap \ker T^\infty = \{ 0 \}$,
$\dim \ker S < \infty$ and $\dim \ker T < \infty$,
$ST = TS$
imply that $\ker S + \ker T = \ker(ST)$. I think the second condition can be weakened to $\ker S^2 \cap \ker T^2 = \{ 0 \}$, but I have this stronger condition for some operators I am interested in. Any help would be appreciated, thanks.
-edit- I am not confident that all these conditions are necessary.
AI: It suffices to assume that $ST=TS$, $\ker T$ is finite dimensional, and $\ker S\cap \ker T=0$. Given these assumptions, suppose $v\in \ker(ST)$. Let $A$ be the span of $\{v,Sv,S^2v,\dots\}$ and let $B$ be the span of $\{Sv,S^2v,\dots\}$. Since $TS^nv=S^{n-1}STv=0$ for any $n>0$, $B\subseteq \ker T$. Since $\ker T$ is finite-dimensional, $B$ must be finite dimensional, and hence so is $A$. Let $p$ be the minimal polynomial of $S$ on $A$. If $p$ has a nonzero constant term, then $v\in B$ and hence $v\in\ker T$. Thus we may assume the constant term of $p$ is $0$ and write $p(x)=xq(x)$ for some polynomial $q$.
Observe now that $q(S)v\in\ker S$, and is nonzero by the minimality of $p$. If the constant term of $q$ is zero, then $q(S)v\in B\subseteq \ker T$, but that is impossible since $\ker S\cap \ker T=0$. Thus $q$ has nonzero constant term, and we may multiply by a scalar to assume the constant term is $1$. But then $v-q(S)v\in B\subseteq \ker T$, and so $v=q(S)v+(v-q(S)v)\in\ker S+\ker T$.
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H: Linear programming with min of max function
I have to write the linear program which minimizes this function :
$$y = \max_j \sum_{i=1}^{n}c_{ij}x_{ij}$$
My book says that this is not a linear function but it can be trasformed into one using the minimizing program $\min y$ with the conditions :
$$ \sum_{i=1}^{n}c_{ij}x_{ij} \leq y \:\:, \:\:j = 1,...,m$$
(+ other conditions not related with $y$)
I really don't get why when these conditions are met then I should consider it a linear program, $y$ isn't neither a linear function nor a constant as far as I understand. Besides, I don't get neither how to calculate the maximum, can $y$ be traslated as :
$$\max (\sum_{i=1}^{n}c_{ij}x_{ij} \:\:, \:\:j = 1,...,m) $$
But, then I have a function with different variables, so how can I find a maximum, maybe considering the other restrictions ?
Maybe, I'm misunderstanding everything , I'm new to linear programming
AI: The $y$ in the linear program is being treated as a totally new variable that doesn't (directly) keep its old meaning as $\max_j \sum_i c_{ij} x_{ij}$. I suspect the reason for your confusion is that you're continuing to try to expand $y$ out to its old definition inside the linear program.
The point is that by adding a constraint that $\sum c_{ij} x_{ij} \leq y$ for each $j$, the linear program requires that the value assigned to the variable $y$ is at least $\max_j \sum_i c_{ij} x_{ij}$, so any optimal solution of the linear program, having made the variable $y$ as small as possible, must in fact make $y$ equal to $\max_j \sum_i c_{ij}x_{ij}$. (The constraints only force that $y$ is at least this max, but clearly there's no reason to have $y$ any larger than the max if there are no other constraints involving $y$, so an optimal solution makes it equal.)
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H: L'hopital rule fails with limits to infinity?
$$ \lim_{n \to \infty} \frac{1 +cn^2}{(2n+3 + 2 \sin n)^2} = ? $$
if I factor the $n^2$ out of denominator,
$$ \lim_{n \to \infty} \frac{ 1 + cn^2}{ n^2 ( 2 + 3n^{-1} + 2 \frac{ \sin n}{n} )^2}$$
And take limit directly, I get the answer as
$$ \frac{c}{4}$$
However, If I apply l'hopital rule, Iget
$$ \lim_{ n \to \infty} \frac{ 2cn}{2 (2n + 3 + 2 \sin n)( 2 + 2 \cos n)} $$
However this new limit gives a different value than original according to wolfram.. and neither am I able to compute it by hand, what am I missing?
Some people say of limit existing and not existing, but then suppose
$$ \lim_{x \to 0} \frac{1}{x} = \infty$$
Does this limit exist? how do you define a limit to be existing as in what is sufficent condition for it
AI: The rule of L'Hospital states that the limit of $\dfrac fg$ equals that of $\dfrac{f'}{g'}$ if the latter exists. You precisely found a case where this does not hold.
We can simplify the example as
$$\lim_{n\to\infty}\frac{n+\sin n}n=1$$
but
$$\lim_{n\to\infty}\frac{1+\cos n}1$$ is undefined.
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H: Finding the height of a Pyramid where the sides are given by an equation
Problem:
The vertex of a pyramid lies at the origin, and the base is perpendicular to the x-axis at $x = 4$. The cross sections of the pyramid perpendicular to the x-axis are squares whose diagonals run from the curve $y = -5x^2$ to the curve $y = 5x^2$. Find the volume of the pyramid.
Answer:
The first step is to graph the two curves. A graph can be found at the following URL:
https://www.wolframalpha.com/input/?i=plot+5x%5E2+%2C+-5x%5E2
Since the cross sections of the pyramid are perpendicular to the x-axis we will integrate with respect to x. Since each cross section is a square and $y$ goes from $-5x^2$ to $5x^2$, the the length of one side of the square will be $5x^2 - (-5x^2)$ or
$10x^2$. Let $V$ be the volume we are looking for.
\begin{align*}
du &= -dx \\
V &= \int_0^4 (10x^2)^2 \,\,\, dx = \int_0^4 100x^4 \,\,\, dx \\
V &= \frac{100x^5}{5} \Big|_0^4 = 20x^5 \Big|_0^4 \\
V &= 20(4^5) = 20(1024) = 20480
%
\end{align*}
The book's answer is:
$$ 10,240 $$
Where did I go wrong?
AI: Your method is correct but you lost a $1/2$ factor. The value of $10x^2$ is the diagonal, not the side of the square. So the integrand becomes $\frac 12 (10x^2)^2$.
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H: Sequence of Lebesgue integrable functions bounded in norm converges pointwise
I've come across a problem which states:
Given a sequence of integrable functions $\{f_k\}$ ($k≥1$) on $[0,1]$ with the property that $||f_k||_1 ≤ \frac{1}{2^k} $, then $f_k \rightarrow 0$ pointwise almost everywhere on $[0,1]$.
I'm not exactly sure how to proceed. I attempted to show this via contradiction, but couldn't see how to finish the argument.
AI: For a sequence of non negative measurable functions we can interchange infinite sum with integral, it follows from monotone convergence. So we have:
$\int_0^1(\sum_{k=1}^\infty |f_k|) dx=\sum_{k=1}^\infty\int_0^1 |f_k|dx=\sum_{k=1}^\infty ||f_k||_1<\infty$
So the integral of the non negative function $f:=\sum_{k=1}^\infty |f_k|$ is finite, and hence $f$ has to be finite almost everywhere. But at every point where the series is convergent we have $f_k\to 0$. So it happens almost everywhere.
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H: How should one understand the "indefinite integral" notation $\int f(x)\;dx$ in calculus?
In calculus, it is said that
$$
\int f(x)\; dx=F(x)\quad\text{means}\quad F'(x)=f(x)\tag{1}
$$
where $F$ is a differentiable function on some open integral $I$. But the mean value theorem implies that any differentiable function $G:I\to \mathbb{R}$ with the property $G'(x)=f(x)$ on $I$ can be determined only up to a constant. Since the object on the right of the first equality of (1) is not unique, we cannot use (1) as a definition for the symbol $\int f(x)\;dx$.
Formulas for antiderivatives are usually written in the form of $\displaystyle \int f(x)\;dx=F(x)+C$. For example,
$$
\int \cos x\;dx = \sin x+C\;\tag{2}
$$
where $C$ is some "arbitrary" constant.
One cannot define an object with an "arbitrary" constant. It is OK to think about (2) as a set identity:
$$
\int \cos x\; dx = \{g:\mathbb{R}\to\mathbb{R}\mid g(x)=\sin x+C,\; C\in\mathbb{R}\}. \tag{3}
$$
So sometimes, people say that $\int f(x)\;dx$ really means a family of functions. But interpreting it this way, one runs into trouble of writing something like
$$
\int (2x+\cos x) \; dx = \int 2x\;dx+\int \cos x\; dx = \{x^2+\sin x+C:C\in\mathbb{R}\}\;\tag{4}
$$
where one is basically doing the addition of two sets in the middle, which is not defined.
So how should one understand the "indefinite integral" notation $\int f(x)\;dx$? In particular, what kind of mathematical objects is that?
AI: Unless the equal sign "=" in the first identity of (1) is not considered [the same] as the equal sign in "3+5=8" ...
This is precisely what is done.
When you move on to studying measure theory and consider $L^p$ spaces, two functions are considered "equal" if they only differ on a "small" set of points (where "small" has a precise measure-theoretic definition). Mathematicians are not computers, and know how to use the context of a statement to understand what version of equals is being used.
In the world of computing anti-derivatives, "=" means "differ by a constant", or more generally, "differ only by a constant on each connected component of their domains".
You can get into problems when you forget which version of "=" is intended, and think "=" means more than it does. (There are a few math brain-teasers out there based on that.) I think of it as the same problem as if you went into the teacher's lounge and asked for "the calculus teacher", as you were expecting Professor Liang, who is 6'4" tall and you wanted help getting something off a high shelf, but you didn't realize that Professor Smith, who is 4'11", also teaches calculus, and that's who shows up. You thought that specifying "calculus teacher" carried with it Prof. Liang's height, but that's not the case.
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H: I need to help with a formula to find the length of a line.
I have a rectangle which I know the width and height of it.
I need to draw a line inside the rectangle and the information that I have include knowing the starting point of the line as well as the angle of the line.
My question is how can I calculate the length of the line given the informaiton that I have?
I don't know where the line will end up, I just have a starting point, the angle and the general width and height of the rectangle.
I tried to draw some sample lines to show you what I'm trying to do, I need to find the length of the red line so I can draw it in my App.
AI: Since the problem remains the same when the rectangle is rotated/flipped, we can assume for simplicity that the starting point is on the bottom border of the rectangle. (When rotating the problem, make sure to get the angle and width/height of the rotated problem correct.)
Let $w$ denote the width and $h$ the height of the rectangle. Let the starting point have distance $x$ from the left border of the rectangle. Let $\alpha$ be the angle between your line and the bottom line that you get by "rotating your line to the left."
Image:
We want to compute the point $\overline{EF}$. If $\operatorname{arcsec}$ denotes the inverse function of $x\mapsto\frac1{\cos(x)}$ and $$\alpha=\operatorname{arcsec}\left(\frac{\overline{CE}}{\overline{AE}}\right)=\operatorname{arcsec}\left(\frac{\sqrt{x^2+h^2}}{x}\right),$$
then your line goes straight to the top left corner and the length is $\sqrt{x^2+h^2}$. If $\alpha$ is less than that, then you will "bump" into the left border and we have $\overline{AF}=\tan(\alpha) x$. Hence the length of the line is $$\sqrt{(\tan(\alpha)^2+1) x^2}=\lvert\sec(\alpha)x\rvert.$$
Let $\beta=\pi-\alpha$. You get the same thing: If $$\beta=\operatorname{arcsec}\left(\frac{\overline{ED}}{\overline{BE}}\right)=\operatorname{arcsec}\left(\frac{\sqrt{(w-x)^2+h^2}}{(w-x)}\right),$$
then the line goes straight up to the top right corner. So the length is $\sqrt{(w-x)^2+h^2}$. If $\beta$ is less than that, then analogously to before, the length of the line is $$\lvert\sec(\beta)(w-x)\rvert.$$
Now, if $\alpha$ is big enough such that you don't bump into the left border and $\beta$ is big enough such that you don't bump into the right border, you will bump into the top border (as in my image). In this case, let $F^\top$ be the orthogonal projection of $F$ onto the bottom border. Then $\overline{F^\top E}=h\cot(\alpha)$ so that the length of the line is $$\sqrt{h^2\cot(\alpha)^2+h^2}=\lvert h\csc(\alpha)\rvert.$$
Example (30°): After flipping we see $h=10, w=6, \beta=\frac\pi6, x=2$. We compute $$\operatorname{arcsec}\left(\frac{\sqrt{(w-x)^2+h^2}}{(w-x)}\right)=\operatorname{arcsec}(\sqrt{29}/2)\approx1.19>\beta.$$
So we will bump into the right border with a line length of $$\lvert\sec(\beta)(w-x)\rvert=4\sec(\pi/6)=\frac{8}{\sqrt 3}.$$
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H: How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$?
How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\to 0}\dfrac{12^x\ln12-4^x\ln4}{9^x\ln9-3^x\ln3}&=\dfrac{\ln12-\ln4}{\ln9-\ln3}
\\ &=\dfrac{\ln(12/4)}{\ln(9/3)}
\\ &=\dfrac{\ln(3)}{\ln(3)}
\\ &=1
\end{align*}
My answer comes out to be $1$. Can I evaluate this limit without L'H$\hat{\mathrm{o}}$pital's rule? Thanks.
AI: A variation of other answers (that more closely parallels a common pattern when the numerator and denominator are polynomials) is "big part factoring". \begin{align*}
\lim_{x \rightarrow 0} \frac{12^x - 4^x}{9^x-3^x}
&= \lim_{x \rightarrow 0} \frac{12^x \left(1 - \left( \frac{4}{12} \right) ^x \right)}{9^x \left( 1-\left( \frac{3}{9} \right) ^x \right) } \\
&= \lim_{x \rightarrow 0} \frac{12^x \left(1 - \left( \frac{1}{3} \right) ^x \right)}{9^x \left( 1-\left( \frac{1}{3} \right) ^x \right) } \\
&= \lim_{x \rightarrow 0} \frac{12^x }{9^x} \\
&= \frac{1}{1} \\
&= 1 \text{.}
\end{align*}
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H: Show that $\mathfrak{m}_p$ is an ideal in $\mathcal{O}_V.$
I'm working through Algebraic Geometry: A Problem Solving Approach by Garrity et al, and I have found myself stuck on Exercise 4.13.1, which is the section Points and Local Rings.
Let $V = V(x^2 + y^2 - 1) \subset \mathbb{A}^2(k).$ Let $p = (1, 0)
\in V.$ Define $$\mathfrak{m}_p = \{f \in \mathcal{O}_V : f(p) =
0\}.$$
Show that $\mathfrak{m}_p$ is a maximal ideal in
$\mathcal{O}_V.$
We're told that $\mathcal{O}_V$ is a subring of the function field $\mathcal{K}_V = \{\frac{f}{g} : f, g \in \mathcal{O}(V), g \neq 0\},$ where $\mathcal{O}(V) = k[x, y]/I(V).$
It was easy enough to show that $\mathfrak{m}_p$ is an ideal, but I'm getting caught up on how to show that it is maximal.
I know that maximal ideals in $k[x, y]$ are all of the form $I = \langle x - a_1, y - a_2 \rangle,$ so my idea was to show that $\mathfrak{m}_p = \langle x - 1, y \rangle,$ and then I know that maximal ideals in $k[x, y]/I(V)$ correspond to maximal ideal in $k[x, y]$ containing $I(V).$ This feels close to something correct, but I think I just don't understand the different spaces well enough to put it all together. Also, $\mathfrak{m}_p$ is an ideal in $\mathcal{O}_V,$ while $\langle x - 1, y\rangle$ is an ideal in $k[x, y],$ so setting them equal to each other doesn't make any sense.
AI: Map $\mathcal{O}_V \to k$, $f \mapsto f(p)$. Show this is a ring homomorphism with kernel $\mathfrak{m}$.
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H: What is the difference between a semiconnected graph and a weakly connected graph?
These are the definitions I have found:
Semiconnected: if, and only if, for any pair of nodes, either one is reachable from the other, or they are mutually reachable.
Weakly connected: if, and only if, the graph is connected when the direction of the edge between nodes is ignored.
As far as I can tell, these definitions are identical.
AI: Semiconnected means that for every pair of vertices $(x,y)$, either there exists a path from $x$ to $y$ or a path from $y$ to $x$, with all steps of the path obeying the directionality of edges.
Weakly connected means that if you replace all the directed edges with undirected edges then the resulting undirected graph is connected.
Semiconnected implies weakly connected. The converse is not true, for example the graph $A \rightarrow B \leftarrow C$ is weakly connected but not semiconnected, because there is no path between $A$ and $C$ in either direction obeying the directionality of the edges. But if you forget about the directionality entirely then the graph becomes just $A \leftrightarrow B \leftrightarrow C$ which is definitely connected.
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H: A good description of the set of inner points of a polyhedron.
How can I get a good description of the inner points of a polyhedron? I am trying to calculate the volume of a polyhedron by change of variables, but I can't describe the set of points of the polyhedron properly (given its vertices).
I look for a description such as {(x,y,z):x+y+z<3;x<4y}.
AI: For a convex polyhedron this is not so hard: any convex polyhedron is the intersection of half-spaces defined by the faces (each face lies on exactly one plane, and the polyhedron lies entirely on one side of that plane). Thus, the set of points in a convex polyhedron can be specified by $F$ linear inequalities, where $F$ is the number of faces.
For a non-convex polyhedron, I think the easiest thing to do is to express the polyhedron as a union of convex polyhedra whose pairwise intersections all have measure $0$, then apply the same procedure as above. Then you do the change of variables to each convex piece and add up your answers.
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H: Inequality concerning functions quadratic decay of distribution $m(|f|>\lambda)\leq C\lambda^{-2}$
This was a problem in a prelim that I was not able to solve back in the day.
Let $m$ denote the Lebesgue measure on $\mathbb{R}^n$. Suppose $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a Lebesgue measurable function such that
$$
m\big[ |f|>\lambda\big]\leq C\lambda^{-2}
$$ for all $\lambda>0$. Show that there is a constant $C_1>0$ such that
$$
\int_E|f|\,dm \leq C_1\sqrt{m(E)}
$$
I've tried to use Holder's inequality but I don't get very far. Does any body have a better idea?
AI: I think I came across this problem in a test before. It is a nice one, but there is a slight trick to solve it.
It is enough to consider Lebesgue measurable set $E$ such that $0<m(E)<\infty$. Applying Fubini's theorem with the measure $m(\cdot\cap E)$ one gets that for any $t>0$
$$
\begin{align}
\int_E|f(x)|\,dx&=\int^\infty_0 m\big(E\cap\{|f|>\lambda\}\big)\,d\lambda\\
&=\int^t_0 m\big(E\cap\{|f|>\lambda\}\big) d\lambda + \int^\infty_t m\big(E\cap\{|f|>\lambda\}\big)\,d\lambda\\
&\leq t m(E) + C\int^\infty_t\frac{1}{\lambda^2}\,d\lambda\\
&= t m(E) + \frac{C}{t}=: \phi(t)
\end{align}
$$
The key now is to choose the best value for $t$, that the one that hopefully minimizes $\phi(t)= t m(E)+\frac{C}{t}$.
Solving for $\phi'(t)= m(E)-\frac{C}{t^2}=0$, one gets $t_E=\sqrt{\frac{C}{m(E)}}$. Furthermore, $\phi''(t)=\frac{2C}{t^3}>0$ on $(0,\infty)$ and $\lim_{t\rightarrow0}\phi(t)=\lim_{t\rightarrow\infty}\phi(t)=\infty$. Thus $\phi$ attains its minimum value at $t_E$ and $$\phi(t_E)=2\sqrt{C}\sqrt{m(E)}$$
Putting things together one gets
$$\int_E|f(x)|\,dx\leq 2\sqrt{C}\sqrt{m(E)}$$
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H: Character table of quaternions $Q_8$
I wanted to compute the character table of the quaternion group Q_8, and to this end I was reading this answer. For the degree 1 representation I used this question but for the degree 2 representation I don't understand how the hint from the first link works.
AI: The hint is to map your quaternions to matrices with complex entries of the form $$\left(\begin{array}{cc}z&w\\-\bar{w}&\bar{z}\end{array}\right).$$
This will give you a representation, on $\mathbb{C}^2$, and by taking the traces of the matrices you get the characters.
It is no surprise that the identity $e\in Q_8$ maps to the identity matrix.
So you are left with 3 quaternions, $i,j,k$ that you need to decide where to map to. Note the space of such matrices is 4 dimensional over the reals, parameterised by the real part of $z$, the imaginary part of $z$, the real part of $w$, and the imaginary part of $w$, from which you get a natural basis. You already sent $e$ to one of these basis vectors (corresponding to the real part of $z$). You could try mapping $i,j,k$ to the other three basis vectors.
Remember you need to check that the identities that hold between the elements of $Q_8$ also hold between the matrices you send them to.
You also need to decide to do with $-1\in Q_8$ but there is a really obvious candidate for where to send it. Also it is fixed once you decide where to send $i,j$ or $k$.
Also you may be using the notation $Q_8=\langle x,y|y^2=x^2,xyx=y\rangle$. In this case by $i$ I mean $x$, by $j$ I mean $y$, by $k$ I mean $xy$ and by $-1$ I mean $y^2$.
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H: Directly Calculating Birthday Paradox Probabilites
I am trying to calculate the probability of at least 2 people sharing a birthday in a group of 4 people. I understand that calculating it as 1-P(no shared birthdays) is simpler, but I would like to understand the counting method by doing it directly.
My attempt for $n=4$ is
P = P(2 people) + P(3 people) + P(4 people) = $\frac{1}{365}\binom{4}{2}+\frac{1}{365^2}\binom{4}{3}+\frac{1}{365^3}\binom{4}{4}=0.0164$...
but this does not match up with
P = $1-\frac{364}{365}\frac{363}{365}\frac{362}{365}=0.163...$
What am I doing wrong in the direct calculation?
AI: There are four sharing cases:
The probability two people share a birthday and the other two each have different birthdays is $\dfrac{{4 \choose 2}364\times 363}{365^3} \approx 0.01630349$
The probability three people share a birthday and the other one has a different birthday is $\dfrac{{4 \choose 3}364}{365^3} \approx 0.00002994$
The probability four people share a birthday is $\dfrac{{4 \choose 4}}{365^3} \approx 0.00000002$
The probability two people share a birthday and the other two share a different birthday is $\dfrac{{3 \choose 1}364}{365^3} \approx 0.00002246$
You might want to consider whether the last is like the first (no more than two people share any particular day) or the second (there are two days on which birthdays fall) or the third (all four share a birthday with somebody),
but in any case if you add these up, you get the same as you would have got with $1-\dfrac{364 \times363\times 362}{365^3} \approx 0.01635591$
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H: What is the ordinary differential equation for double exponential summation?
Given the following ordinary differential equation (ODE)
$$\frac{dy}{dx} = a y$$
its general solution is $y = c e^{a x}$, where $c$ is a constant. If we know
$$y = c_{1} e^{a_{1} x} + c_{2} e^{a_{2} x}$$
what is the corresponding ordinary differential equation for this solution?
AI: If $y''(x) + ay'(x) + by(x) = 0$, then it is well known that the general solution is:
$$c_1 e^{\alpha x} + c_2 e^{\beta x}$$
where $\alpha, \beta$ are the roots of the quadratic $x^2+ax + b = 0$.
This result can be obtained by letting $y = e^{rx}$, finding $y'(x)$ and $y''(x)$, and then substituting those values into the differential equation.
So the original problem has been transformed into: given the roots of a quadratic equation, how can I find the original quadratic? This is easy if you just expand $(x + \alpha)(x + \beta) = 0$, or use Vieta's formulas.
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H: What does the "$\bigwedge$" symbol mean in "$\bigwedge_{j=1,\ldots,M,j\neq i}\Delta_i(x)>\Delta_j(x)"$?
What does the "$\bigwedge$" symbol mean in the following?
I assume it means "and". Am I right?
AI: It is the conjunction of all predicates evaluated where the index is in the indicated domain.
EG: $\lower{1ex}{\left[\raise{1ex}{\bigwedge\limits_{j\in\{1,2,3\}} P_j}\right]} ~=~ P_1\wedge P_2\wedge P_3$
So we are taking the measure for the probability of the event where $\Delta_i(\chi)$ is greater than $\Delta_j(\chi)$ for all $j$ in the integer domain $\{1..M\}$ except when $j=i$.
$$\Pr\left\{\raise{1.25ex}{\mathop{\bigwedge\qquad\quad}_{j\in\{1..M\}: j\neq i}\hspace{-6ex}\bigl[\Delta_i(\chi)>\Delta_j(\chi)\bigr]}\right\} $$
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H: Proving a linear map is surjective
Suppose $V_1, \dots, V_m$ are vector spaces. Prove that
$\mathcal{L}(V_1 \times \dots \times V_m, W)$ is isomorphic to
$\mathcal{L}(V_1, W) \times \dots \times \mathcal{L}(V_m, W).$ (Note that $V_{i}$'s can be infinite-dimensional.)
I am having trouble showing that $\varphi$ defined below is surjective. For every $f \in \mathcal{L}(V_1 \times \dots \times V_m, W),$ I defined $f_{i}: V_{i} \to W$ by $$f_{i} (v_{i}) = f (0, \dots, v_{i}, \dots, 0).$$ Then, I defined $\varphi: \mathcal{L}(V_1 \times \dots \times V_m, W) \to \mathcal{L}(V_1, W) \times \dots \times \mathcal{L}(V_m, W)$ by $$\varphi (f) = (f_{1}, \dots, f_{m}).$$
Now, how would I show that $\varphi$ is surjective?
I know I have to show that for any $(g_{1}, \cdots, g_{m}) \in \mathcal{L}(V_1, W) \times \dots \times \mathcal{L}(V_m, W)$, there is a corresponding $g \in \mathcal{L}(V_1 \times \dots \times V_m, W)$ so that $\varphi (g) = (g_{1}, \dots, g_{m}).$
Can I simply define $g \in \mathcal{L}(V_1 \times \dots \times V_m, W)$ by
$$g (0, \dots, v_{i}, \dots, 0) = g_{i} (v_{i})? $$
I am not sure where to start.
AI: For $(g_1,\dots,g_m) \in \mathcal{L}(V_1,W) \times \cdots \times \mathcal{L}(V_m,W)$ define $g : V_1 \times \cdots \times V_m \to W$ by
$$g(v_1,\dots,v_m) = g_1(v_1) + \cdots + g_m(v_m).$$
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H: How to find the dominant eigenvalue of a Next generation matrix
I have been working with a compartmental model and I am now trying to compute the basic reproduction number. To do this I must find the spectral radius (dominant eigenvalue) of the following matrix.
$$K=\begin{bmatrix}\frac{\beta_{HH}( \pi_H/ \mu_H)}{(\alpha + \gamma + \mu_H)(\mu_H)}&\frac{\beta_{TH}(\pi_H/\mu_H)}{\mu_T(\mu_T * IT)}&\frac{\beta_{CH}(\pi_H/\mu_H)}{\mu_C(\mu_C * IC)}\\0&\frac{(\beta_{TTV} + \beta_{TTH})(\pi_T/\mu_T)}{\mu_T(\mu_T * IT)}&\frac{\beta_{CT}(\pi_T/\mu_T)}{\mu_C(\mu_C*IC)}\\0&\frac{\beta_{TC}(\pi_C/\mu_C)}{\mu_T(\mu_T * IT)}&0\end{bmatrix}$$
Next Generation Matrix K
I understand this is done by taking the determinant of $K - \lambda I$. I have found the determinant but cannot correctly compute the eigenvalues. I know when using a $2 \times 2$ Next Generation Matrix there is an easy formula to find the basic reproduction number but I am unsure for a $3 \times 3$. If anyone can give their insight that would be great. Thank you.
AI: The characteristic polynomial of $\pmatrix{a & b & c\cr 0 & d & e\cr 0 & f & 0\cr}$ is $(\lambda - a)(\lambda^2 - d \lambda - e f)$ so the eigenvalues are $a$ and $\dfrac{d \pm \sqrt{d^2+4ef}}{2}$. Assuming the parameters are all positive, the greatest eigenvalue is either $a$ or $(d + \sqrt{d^2+4ef})/2$, whichever is greater.
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H: Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
I tried to use Proof by Induction, but I'm stuck on the case for when $n=k+1.$
AI: Hint:
$5^{3^{k+1}}+1=(5^{3^k})^3+1=(5^{3^k}+1)(5^{2\cdot3^k}-5^{3^k}+1)$
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H: why $2\pi= c$ and $c=\pi ?$
Let $T:V\rightarrow V$ be the linear transformation defined as follows: If $f\in V, g=T(f)$means that $$g(x)=\int_{-\pi}^{\pi}\{1+\cos(x-t)\}f(t)~dt$$
Find all real $c \neq 0$ and all nonzero $f$ in $V$ such that $T(f)=cf$.
My attempt : I got the answer but i didn't understand the answer given below marked in red box
My thinking : $$f(x)=c_1+c_2\cos x+c_3\sin x $$
$$T(f)(x)= cf(x)$$
$$
T(f)(x)= c c_1 + c c_2 \cos x + c c_3 \sin x
$$
I m not getting how $\pi$ come in the given answer https://math.stackexchange.com/a/3290699/557708 ?
why $2\pi= c$ and $c=\pi ?$
AI: Now that you know the general form of $f(x) = c_1 + c_2\cos x + c_3\sin x$, you can find the coefficients in terms of the integral
$$\int_{-\pi}^{\pi}f(t)dt = 2\pi c_1$$
$$\int_{-\pi}^{\pi}\cos t(c_1 + c_2\cos t + c_3\sin t)dt = c_2\int_{-\pi}^{\pi}\cos^2 tdt = \pi c_2$$
$$\int_{-\pi}^{\pi}\sin t(c_1 + c_2\cos t + c_3\sin t)dt = c_3\int_{-\pi}^{\pi}\sin^2 tdt = \pi c_3$$
So we can write the following
$$T(c_1 + c_2\cos x + c_3\sin x) = 2\pi c_1 + \pi c_2 \cos x + \pi c_3 \sin x = cc_1 + cc_2\cos x + cc_3\sin x$$
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H: Lebesgue measure of boundary of an open set.
Let $g$ be a continuous function on $\mathbb{R}^n$ and $$O=\{x\in \mathbb{R}^n:g(x)\neq 0\}.$$
Is it true that Lebesgue measure of Boundary of $O$ always zero?
AI: In $\Bbb R$ let $C$ be a fat Cantor set. This is constructed
in a similar way to the usual Cantor set, but the removed open intervals shrink
quickly enough in length to ensure than $C$ has non-zero Lebesgue measure.
Let $O$ be the complement of $C$. The boundary of $O$ is the boundary of $C$
which is $C$ itself. The boundary of $O$ has non-zero measure.
If we define $g(x)=\text{distance}(x,C)$ then $g$ is continuous, and $O$ is the set of $x$ with $g(x)\ne0$.
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H: Bijection between $\mathrm{Hom}(D,-)$ functor exact sequnce and $F \leftrightarrow (g,f) $
The following is a paragraph from Dummit and Foote CH-10 (pp. 388).
Let $ 0 \rightarrow L \xrightarrow{\psi} M \xrightarrow{\varphi} N \rightarrow 0 $ be an exact sequence. Then,
\begin{equation}\label{key}
0 \rightarrow \mathrm{Hom}_{R}(D,L) \xrightarrow{\psi'} \mathrm{Hom}_{R}(D,M) \xrightarrow{\varphi'} \mathrm{Hom}_{R}(D,N) \rightarrow 0
\end{equation}
is exact if and only if there is a bijection $ F \leftrightarrow (g,f) $ between homomorphisms $ F : D \rightarrow M $ and
pairs of homomorphisms $ g : D \rightarrow L $ and $ f : D \rightarrow N $ given by $ F|_{\psi(L)} = \psi'(g) $ and $ f = \varphi'(F) $.
I don't understand what does $ F|_{\psi(L)}$ mean? How can we restrict $F$ to a subset of $M$? Even ignoring this, I could not find a way to prove $\Rightarrow$ result i.e. there is bijection.
AI: Say that two elements $f,g\in {\rm Hom}_R(D,M)$ are equivalent ($f\sim g$) if the image of $f-g$ lies in the image of $\psi$. Let $K$ denote a complete set of representatives of equivalence classes. Then we may bijectively identify: $${\rm Hom}_R(D,M)\cong K \times {\rm Hom}_R(D,{\rm im}(\psi))$$
Exactness of the Hom sequence is by definition equivalent to the following two maps being bijective:
(1) The map $\varphi'\colon K \to {\rm Hom}_R(D,N)$
(2) The map $\psi'\colon{\rm Hom}_R(D,L) \to {\rm Hom}_R(D,{\rm im{(\psi)}}) $.
Thus we may define a relation between ${\rm Hom}_R(D,M)$ and ${\rm Hom}_R(D,N)\times {\rm Hom}_R(D,L)$ which is bijective if and only if the Hom sequence is exact.
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H: What is the equation of the quadratic function whose vertex of the graph is on the $x$-axis and passes through the two points $(1,4)$ and $(2,8)$?
What is the equation of the quadratic function whose vertex of the graph is on the $x$-axis and passes through the two points $(1,4)$ and $(2,8)$?
Here is my attempt:
Use the vertex form of a parabola $f(x) = a(x – h)^2 + k$ where point $(h, k)$ is the vertex of the parabola.
Since the vertex is on the $x-axis$, the value of the ordinate $k$ of the vertex is $0$.
Hence, we will have $f(x) = a(x – h)^2$. I tried to substitute the values of $x$ and $y$ in each of the two given points on the parabola into the vertex form to solve for the values of $h$ and $a$.
Using point $(1, 4)$, $f(1) = a(1 – h)^2 = 4$
$(h^2 – 2h + 1)a = 4$ $(eq 1)$
Using point $(2, 8)$, $f(2) = a(2 – h)^2 = 8$
$(h^2 – 4h + 4)a = 8$ $(eq 2)$
After this, I tried to subtract $(eq 2)$ from $(eq 1)$.
$(eq 1) – (eq 2) = (h^2 – 2h + 1)a – (h^2 – 4h + 4)a = 4 – 8$
$(2h - 3)a = -4$
I am stuck after this solution. Any comments and suggestions will be much appreciated. Thank you!
AI: We divide both the equations
$$\frac{1}{2} = \frac{h^2-2h+1}{h^2-4h+4}$$
$$h^2-4h+4 = 2h^2 - 4h+2$$
$$\implies h^2 = 2 \implies h = \pm \sqrt2$$
For $h = \sqrt 2$
$$a = \frac{4}{3-2\sqrt2}$$
For $h = -\sqrt 2$
$$a = \frac{4}{3+2\sqrt2}$$
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H: Must a local homomorphism from a Noetherian local ring to an artinian local ring factor through a power of its maximal ideal
Let $f : B\rightarrow A$ be a local homomorphism of Noetherian local rings where $A$ is moreover Artinian. Must $f$ factor through $B/m_B^n$ for some $n$?
AI: Yes. Note that $m_A^n=0$ for some $n$, since $A$ is Artinian. Also $f(m_B)\subseteq m_A$ (local homomorphism) so $f(m_B^n)\subseteq m_A^n=0$.
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H: Show that $m(\{x\in[0,1]:\text{$x$ lies in infinitely many $E_j$}\})\geq\frac{1}{2}$ when $m(E_j)\geq\frac{1}{2}$
Question: Suppose $E_1, E_2,\ldots$ is a sequence of measurable subsets of $[0,1]$ with $m(E_j)\geq\dfrac{1}{2}$. Show that $m(\{x\in[0,1]:\text{$x$ lies in infinitely many $E_j$}\})\geq\dfrac{1}{2}$, where $m$ is the one dimensional Lebesgue measure.
My thoughts: I would imagine that one could use the Borel-Cantelli Lemma here. The issue is that for the Borel Cantelli Lemma I need $\sum_{j=1}^\infty m(E_j)<\infty$, but since $m(E_j)\geq\frac{1}{2}$, then I can't use it. Now, if the hypotheses of the Borel Cantelli Lemma were satisfied, then $m(\cap_{n=1}^\infty\cup_{n=k}^\infty E_j)=0$. Which, to my understanding, is the same things as saying that if $E=\{x\in[0,1]:\text{$x$ lies in infinitely many $E_j$}\}\implies m(E)=0$, where $E$ is the set we are dealing with in the question. So, I am wondering if there is a general way to sort of "shift" the Borel Cantelli Lemma. Or, is there another way to go about this problem? Any thoughts, ideas, answers, etc. are always greatly appreciated! Thank you.
AI: Let $F_j=[0,1]\setminus E_j$. Then $m (F_j ) \leq \frac 1 2$ and , by Fatou's Lemma, $\int \lim \inf I_{F_j} dm \leq \lim \inf m(F_j) \leq \frac 1 2$ Can you finish ?
[ Note that $\lim \inf I_{F_j}$ is the indicator function of $\lim \inf F_j$].
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H: Unspecified Constraint
In the following question-
Let a, b, c be positive real numbers. Prove that $$\sum_{cyc} {a^3\over a^3+b^3+abc} \ge 1.$$
In here, there is no constraint given.
But in the solution, the author assumes cyclic substitutions of $x={b\over a}$.
But that means xyz = 1
How can this happen if no constraint is given?
Thanks!
AI: Let $abc=k^3$, $a=k\frac{x}{y}$ and $b=k\frac{y}{z},$ where $x$, $y$ and $z$ are positives.
Thus, $$k\frac{x}{y}\cdot k\frac{y}{z}\cdot c=k^3,$$ which gives $$c=k\frac{z}{x}$$ and we need to prove that:
$$\sum_{cyc}\frac{\left(k\frac{x}{y}\right)^3}{\left(k\frac{x}{y}\right)^3+k^3+\left(k\frac{y}{z}\right)^3}\geq1$$ or
$$\sum_{cyc}\frac{\left(\frac{x}{y}\right)^3}{\left(\frac{x}{y}\right)^3+1+\left(\frac{y}{z}\right)^3}\geq1$$ which says that we can assume $k=1$ or $abc=1$.
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H: Suppose $z$ and $\omega$ are two complex numbers such that $|z|≤1$ and $|\omega|≤1$ and $|z+i\omega|=|z-i\omega|=2$. Find $|z|$ and $|\omega|$.
Suppose $z$ and $\omega$ are two complex numbers such that $|z|≤1$ and $|\omega|≤1$ and $|z+i\omega|=|z-i\omega|=2$. Find $|z|$ and $|\omega|$.
My attempt:
I squared the two given equations
$$|z+iω|^2=4 $$
Therefore $|z+iω||\overline z-i\overline ω|=4$. On simplifying,
$$z\overline z+iω\overline z-i\overline\omega z+ω\overline ω=4\tag{1}$$
Proceeding similarly for $|z-iω|=2$ we get another equation,call it $(2)$.
On adding $(1)$ and $(2)$ we get
$$z\overline z+ω\overline ω=4\;\text{or}\; |z|²+|ω|²=4 $$
which cannot simultaneously satisfy $|z|≤1$ and $|ω|≤1$ but the answer given is $|z|=|ω|=1$. What is the mistake in my method?
AI: Let's try this the old school way. Suppose $z = r_1 e^{i\theta}$ and $w = r_2 e^{i\psi}$ with $0 \le r_1, r_2 \le 1$ and $0 \le \psi \le \theta \le 2\pi$ without loss of generality. Then $$\begin{align}
|z+iw|^2 &= |(r_1 \cos \theta - r_2 \sin \psi) + i(r_1 \sin \theta + r_2 \cos \psi)|^2 \\
&= (r_1 \cos \theta - r_2 \sin \psi)^2 + (r_1 \sin \theta + r_2 \cos \psi)^2 \\
&= r_1^2 \cos^2 \theta + r_1 \sin^2 \theta + r_2^2 \sin^2 \psi + r_2^2 \cos^2 \psi - 2r_1 r_2 \cos \theta \sin \psi + 2r_1 r_2 \sin \theta \cos \psi \\
&= r_1^2 + r_2^2 + 2r_1 r_2 (\sin \theta \cos \psi - \cos \theta \sin \psi) \\
&= r_1^2 + r_2^2 + 2r_1 r_2 \sin (\theta - \psi).
\end{align}$$
Similarly,
$$|z-iw|^2 = r_1^2 + r_2^2 - 2r_1 r_2 \sin (\theta - \psi).$$
Thus $$4 = r_1^2 + r_2^2$$ which is impossible. Essentially, this is the same as your solution, just written explicitly with real-valued variables throughout.
The difference of the two equations yields $$0 = 4 r_1 r_2 \sin (\theta - \psi),$$ which of course is only possible if $|z| = 0$ or $|w| = 0$ or $\theta - \psi = k \pi$ for some integer $k$. It is trivial to dismiss the first two cases as these lead to immediate contradictions. The last case would imply that $z = -w$ or $z = w$; in either case, this implies $$|z \pm iw| = |z(1 \pm i)| = \sqrt{2}|z| \le \sqrt{2} \cdot 1 = \sqrt{2},$$ again showing impossibility.
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H: Where is the copy of $\mathbb{N}$ in the constructible hierarchy relative to a real closed field?
Let $X$ be a real closed field. Let us define a constructible hierarchy relative to $X$ is defined as follows. (This is slightly nonstandard terminology.). Let $L_0(X)=X$. For any ordinal $\beta$, let $L_{\beta+1}(X)=Def(L_{\beta+1}(X))$. For any limit ordinal $\gamma$, let $L_\gamma(X)=\cup_{\beta<\gamma}L_\beta$. And finally let $L(X) = \cup_\alpha L_\alpha$.
Now let $M=\{n1_X: n\in\mathbb{N}\}$. Then my question is, what is the smallest ordinal $\alpha$ such that $M$ is guaranteed to be an element of $L_{\alpha}(X)$?
Or is it consistent that $M\notin L(X)$? What if we were to add the axiom $V=L(X)$?
AI: To avoid notational clash, I'll use the notation $D_\alpha(X)$ to describe the hierarchy built on an RCF $X=(A;f_1,f_2)$ with underlying set $A$, addition function $f_1$, and multiplication function $f_2$, defined precisely as follows:
At the successor and limit steps we take definable powersets and unions respectively, as usual.
We start with $D_0(X)=A\cup A^2\cup\{f_1,f_2\}$.
Here are a couple quick comments to demonstrate that $D_0(X)$ really does have at least the "bare minimum" of expressive power we want for a set-theoretic implementation of an RCF:
We have $f_1, f_2\subseteq D_0(X)$ (and consequently $f_1,f_2$ are definable subsets of $D_0(X)$ since also $f_1,f_2\in D_0(X)$). This is because $A^2\subseteq D_0(X)$ and $f_1,f_2\subseteq A^2$.
We have that $A$ is a definable subset of $D_0(X)$ - e.g. as "The set of left coordinates of elements of $f_1$."
We can tell which of $f_1$ and $f_2$ is addition and which is multiplication, by asking which has an annihilator.
Now right away, we can make the following observation. As we go along the $D$-hierarchy, we "accidentally" wind up following the usual construction of $L$. In particular, we have $A^{<\omega}\subseteq D_\omega(X)$. This lets us implement the "natural" definition of $M$ in $D_{\omega+1}(X)$: "$M$ is the set of $m\in A$ such that there is some finite sequence of elements of $A$ whose first term is $1_X$, whose last term is $m$, and whose $(i+1)$th term is the $i$th term $+_X1_X$." This gives us the following:
$\alpha\le\omega+1.$
Can we do better? Well, at least for some presentations we can easily. Specifically, suppose that $$\mathcal{P}(A)\cap D_1(X)=Def(X),$$ where $Def(X)$ is the set of subsets of $A$ which are definable in the RCF $X$ in the model-theoretic sense. Then by o-minimality of RCF, we have that the following are equivalent for $U\in \mathcal{P}(A)\cap D_1(X)=Def(X)$:
$U$ is discrete, has $1_X$ as its least element, and for each $d\in U$ with $d\not=1_X$ we have $d-_X1_X\in U$.
$U=\{1\cdot 1_X, 2\cdot 1_X, ..., n\cdot 1_X\}$ for some $n\in\mathbb{N}_{\ge 1}$.
This gives us $M\in D_2(X)$: we have $m\in M$ iff there is some $U\in \mathcal{P}(A)\cap D_1(X)=Def(X)$ satisfying the above two bulletpoints with $m\in U$. Consequently, we have:
Restricted to "model-theoretically efficient" presentations of RCFs, that is, ones where $D_1(X)$ is "minimal," we have $\alpha=2$.
(It's easy to show $\alpha>1$.)
Moreover, we can get this unconditionally if $X$ is additionally Archimedean. This is because we can simply add the criterion that $U$ be bounded above and below; the only subsets of $A$ which are bounded above and below, contain $1_X$, and are closed under subtracting $1_X$ from ever non-$1_X$ element are the sets of the form $\{1\cdot 1_X, 2\cdot 1_X, ..., n\cdot 1_X\}$ for some $n\in\mathbb{N}$. That is:
If $X$ is Archimedean, then $\alpha=2$.
However, we run into a problem if $X$ is non-Archimedean and is presented in such a way that non-definable-in-$X$ subsets of $A$ show up in $D_1(X)$. In general I don't see a way to improve on the $\omega+1$ bound.
Conjecture: There is an RCF $X$ whose $\alpha$ is exactly $\omega+1$.
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H: Find the coefficient of $x^{24}$ in the binomial equation
Find the coefficient of $x^{24}$ in the equation ${\left( {1 - x} \right)^{ - 1}}.{\left( {1 - {x^2}} \right)^{ - 1}}.{\left( {1 - {x^3}} \right)^{ - 1}}$
My approach is as follow
The equation used is ${\left( {1 - x} \right)^{ - n}} = \sum\limits_{r = 0}^\infty {{}^{n + r - 1}{C_r}{x^r}} $
Then after expanding I get the following ${\left( {1 - x} \right)^{ - 3}}.{\left( {1 + x} \right)^{ - 1}}.{\left( {1 + x + {x^2}} \right)^{ - 1}}$
As ${\left( {1 - x} \right)^{ - 3}}$ is defined by the above formula I can expand it, but how I will do the expansion for other terms. Any other shortcut method.
AI: If you think about the product as $$\frac{1}{(1-x)(1-x^2)(1-x^3)} = (1+x+x^2+\cdots)(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots),$$ it becomes clear that the coefficient of $x^{24}$ is given by the number of ways to choose nonnegative integers $a,b,c$ such that $a + 2b + 3c = 24.$ Fortunately, these are not too difficult to enumerate since $24$ is small. The best way to proceed is to first choose $c$; i.e., for each $c \in \{0, 1, 2, \ldots, 8\}$, solve $$a + 2b = 24 - 3c = 3(8-c).$$ This in turn requires $$a = 3(8-c) - 2b \ge 0,$$ so $b \in \{0, 1, \ldots, 12 - \lceil \tfrac{3}{2}c \rceil\}$. So that means there are $$\sum_{c=0}^8 12 - \left\lceil \frac{3}{2} c \right\rceil + 1 = 13(9) - (0 + 2 + 3 + 5 + 6 + 8 + 9 + 11 + 12) = 61$$ such solutions, and this is the desired coefficient.
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H: Are these true for a martingale? $E\left[ \frac{X_{n+1}}{X_n} \right] = 1, E\left[ \frac{X_{n+2}}{X_n} \right] = 1 $
Let $(X_n)_{n \in \mathbb{Z}_+}$
be a martingale, $X_n(\omega) \neq 0$, and $X_{n+1}/X_n, X_{n+2}/X_n \in L^1 (n \in \mathbb{Z}_+)$
Do the following hold for $n \in \mathbb{Z}_+$?
$E\left[ \frac{X_{n+1}}{X_n} \right] = 1, E\left[ \frac{X_{n+2}}{X_n} \right] = 1 $
AI: $E(X_{n+m}|\mathcal F_m)=X_m$. This implies $E(\frac {X_{n+m}} {X_m}|\mathcal F_m)=1$ provided $\frac {X_{n+m}} {X_m}$ is integrable. Taking expectation on both sides we get $E(\frac {X_{n+m}} {X_m})=1$ .
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H: What does distance of a point from line being negative signify?
When we take distance from the line, we take
$$ d = \frac{ Ax_o + By_o + C}{ \sqrt{A^2 +B^2}}$$
usually with a modulus on top, now my question is if I evaluate this distance as negative what does it mean? Can I decide on which half-plane a point using this?
AI: Expanding on the answer of @Andrew Chin, the sign says in which of the two half-planes (in which the line divides the whole plane) lays the point. If it is from the same side of the vector $(A,B)$ then the result is positive, otherwise negative. Sometime called oriented distance.
For example, for the line of equation $x-2y+1=0$ we have ${\bf n}=(A,B)=(1,-2)$ and the following graphic representation
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H: One-step probability transition matrix
Start by rolling one die
If the outcome is even, roll two dice on the next turn
If odd, roll one die on the next turn
If two dice are rolled and sum is odd, roll one die next turn
If two dice are rolled and sum is even, roll two die next turn
Game ends, when a sum of 7 or 12 appears.
Write a one-step probability transition matrix for a Markov chain that can describe this situation.
I don't really get how to start on this question. I know the starting probability is 0.5 deciding even/odd.
AI: We can identify $3$ states:
About to roll one die.$\\[4pt]$
About to roll two dice.$\\[4pt]$
Terminal state.
For the above states, the transition matrix is the $3{\times}3$ matrix whose $ij$-th entry $p_{ij}$ is the probability to transition from state $i$ to state $j$ in one turn.
For example
\begin{align*}
p_{11}&=\frac{1}{2}\\[4pt]
p_{23}&=\frac{6}{36}+\frac{1}{36}=\frac{7}{36}\\[4pt]
\end{align*}
Can you work out the remaining $7$ transition probabilities?
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H: Is $X$ Hausdorff if its quotient space is Hausdorff?
Let $X$ be a topological space and $Y$ be a set.
Let $f:X\longrightarrow Y$ be a surjection and endow $Y$ with the quotient topology.
If $Y$ is Hausdorff, can I say that $X$ is also Hausdorff?
AI: If $X$ is any non-Hausdorff space, $Y=\{0\}$ and $f$ is the obvious map from $X$ into $Y$ then $Y$ is Hausdorff.
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H: Isomorphism between a power set $\{0, 1\}^n$ (regarded as a ring) and n digit binary number?
We can treat a power set of a set as ring, $A+B = A\cup B-A\cap B, A\times B = A\cap B$.
Can we use 'modified' modular binary additions and multiplications to represent addition and multiplication of a power set of finite set $X$.
I try this way: say $X=\{a_1, a_2, ..., a_n\}$, a subset containing $a_1, a_2$ will be represented by $1100...00$, other subsets will be represented similarly: if $a_i$ is in the subset, then $i$th digit will be $1$.
Then $\{a_2\}+\{a_2\}=\{\}$ while $010...0+010...0=100...0 _{n \ digits}\mod 1000...0 _{n+1 \ digits}$ which represents $\{a_1\}$.
It doesn't work.
But is there any way to make it work, i.e. is there a one one correspondence of a power set of n elements and an n-digit binary number which is also an isomorphism w.r.t. multiplication and addition?
AI: I don’t really understand, what you mean by modified modular binary additions and multiplications. For bitwise operation we get an isomorphism.
Let $X$ be a fixed set. We have the powerset ring $P=(\mathcal{P}(X),\Delta,\cap,\emptyset,X)$ as well as the ring of binary representations $B=\prod\limits_X \Bbb Z/2\Bbb Z$. We can define a map
$$\begin{array}{rcl}
P &\rightarrow &B\\
A & \mapsto & (\delta_{x,A})_x
\end{array}$$,
where
$$\delta_{x,A}=\left\{\begin{array}{ll}1&\text{if }x\in A\\0&\text{otherwise}\end{array}\right.$$
It is not hard to see that this map is a bijection of the underlying sets. It clearly maps $\emptyset$ to $(0)_x$ and $X$ to $(1)_x$. So we are left to show that it respects addition and multiplication. For $A,B\in P$ it suffices to note that for $x\in X$ the identities (in $\Bbb Z/2\Bbb Z$ of cause)
$$\delta_{x,A\Delta B} = \delta_{x,A} + \delta_{x,B}$$
and
$$\delta_{x,A\cap B} = \delta_{x,A}\delta_{x,B}$$
hold. So we have defined an isomorphism of rings $P \cong B$.
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H: Calculate new tile coordinates for 90 degree rotated square object
So I have a square (all sides even) with even AxA tiles (for example 3x3). Every tile has its x,y coordinates according its placement. After I rotate the object by 90 degrees around the center, the these world coordinates are no more valid (see image). So what I'm trying to figure out is a formula that takes 3 parameters: 1) old coordinates (x,y), 2) tile count (A) and 3) rotate direction.. and gives new nx,ny after 90 degree rotation. For example 1,1 after rotation becomes 1,3.. 3,2 become 2,1.. 2.2 stays 2.2 and so on. Any help appreciated.
AI: You need the rotation matrix centered at $(2,2)$
$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}2\\2\end{bmatrix} +\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta &\cos \theta\end{bmatrix} \begin{bmatrix}x-2\\y-2\end{bmatrix}$
EDIT
To clarify, this is in the sense that anti-clockwise rotations are positive, for clockwise rotations, just put the same angle, but negative.
EXAMPLE
$$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}2\\2\end{bmatrix} +\begin{bmatrix}\cos (-90) & -\sin (-90) \\ \sin (-90) &\cos (-90)\end{bmatrix} \begin{bmatrix}1-2\\1-2\end{bmatrix}$$
$$\implies \begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}2\\2\end{bmatrix} +\begin{bmatrix}0 & 1 \\ -1& 0\end{bmatrix} \begin{bmatrix}-1\\-1\end{bmatrix} = \begin{bmatrix}1 \\ 3\end{bmatrix}$$
Hence the point (1,1) becomes (1,3) which fits as the above diagram
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H: Capped Geometric Distribution
The Geometric Distribution is defined as the number of trials it takes for the first success to appear in a sequence of Bernoulli trials.
My question is what happens when the number of trials is capped at a number $N$? That is what would the distribution of $Y$ be if we define it as
$$Y = min(X, N) \quad X \sim Geom(p), \quad N \in \mathbb{N}$$
I found one article that references this type of problem (group of people playing a dice game, $Y$ determines the winner) but the main difference is that $Y$ gets reset to the support of $X$ and not set equal to $N$.
AI: The only difference in truncated and original geometric distribution is in "last value": $X$ can take values $1,2,\ldots$ and $Y$ can take values $1,2,\ldots, N$ only. And $Y=N$ iff $X\geq N$, so
$$
\mathbb P(Y=N)=\mathbb P(X\geq N) = \mathbb P(N-1 \text{ unsuccesses})=(1-p)^{N-1}.
$$
All the other values of $Y$ have the same probabilities as for $X$: for $k=1,2,\ldots,N-1$,
$$
\mathbb P(Y=k)=\mathbb P(X=k) = p(1-p)^{k-1}.
$$
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H: Example of Separable Product Space with cardinality greater than continuum?
In Willard, it's given that, for Hausdorff non-singleton spaces -
$\prod_{\alpha\in A}X_\alpha$ is separable iff $X_\alpha$ is separable $\forall\alpha\in A$ and $|A|\le\mathfrak{c}$
From reading the proof, I found that we could prove $\prod_{\alpha\in A}X_\alpha$ is separable $\implies$ $X_\alpha$ is separable without assuming $X_\alpha$ to be Hausdorff. Hausdorff-ness of $\prod_{\alpha\in A}X_\alpha$ was only used to show $|A|\le\mathfrak{c}$.
So, is there an example of a non-Hausdorff product space $\prod_{\alpha\in A}X_\alpha$ such that $\prod_{\alpha\in A}X_\alpha$ is separable $\implies$ $X_\alpha$ is separable, $X_\alpha$ is not a singleton, and $|A|>\mathfrak{c}$
EDIT:
Also, is there a non-Hausdorff $T_1$ product space $\prod_{\alpha\in A}X_\alpha$ which satisfies the above condition?
If not, then a non-Hausdorff $T_0$ product space?
AI: If you don't assume Hausdorff-ness, you can pretty much do whatever you want. You can take $X_\alpha$ to be all spaces with the trivial topology, and let $A$ be of as great a cardinality as you want - the product will have the trivial topology, and in particular will be separable.
By the way, for the theorem, you have to assume $X_\alpha$ are moreover not singletons (or rather the theorem says that only $\mathfrak{c}$ of them can have more than one point).
EDIT. Here's an example of a product of $T_1$ spaces. For any cardinality $\kappa$, the product of $\kappa$-many infinite countable spaces with the cofinite topology is separable. As bof pointed out in the comments, the set of constant functions (which is countable) is dense.
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H: find $f(n)$ that isn't little $o(n^2)$ and isn't $w(n)$
im having a problem with a question that tells me to find $f(n)$ so that
$f(n) \neq o(n^2)$ meaning also that $\lim_{\rightarrow\infty} \frac{f(n)}{n^2} \neq 0$
and $f(n) \neq w(n)$ meaning also that $\lim_{n\rightarrow\infty} \frac{f(n)}{n} \neq \infty$
I have tried solving each part and
the first equation tells me that if $f(n) = x^q$ then $2\leq q$
and the second tells me that $q\leq 1$.
is there even a solution to this problem?
AI: A function which oscillates back and forth between $x$ and $x^2$ will work.
Explicitly, consider the function $f$ given by $f(x)=x\,\sin^2(x)+x^2\,\cos^2(x)$.
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H: Relation between number of edges and the sum of number of vertices of degree $k$
Let $P$ be a polyhedron and let $G$ be its associated graph. Suppose $P$ has $V$
vertices, $E$ edges, and $F$ faces. For each $k$, let $V_k$ be the number of vertices of
degree $k$, and let $F_k$ be the number of faces of $P$ (or regions of $G$) of bound degree $k$. Since every edge of $P$
touches exactly two vertices and exactly two faces, we find that
$$\sum kV_k=2E= \sum kF_k$$
I don't understand why that relation is true.
AI: $V_k$ is the number of vertices that are connected to exactly $k$ edges. So $\sum kV_k$ is the sum, over all vertices, of the number of edges connected to each vertex. This is the same as the sum, over all edges, of the number of vertices connected to each edge. Since each edge is connected to exactly two vertices, this is equal to $2E$.
$F_k$ is the number of faces that have exactly $k$ edges. So $\sum kF_k$ is the sum, over all faces, of the number of edges each face has. This is the same as the sum, over all edges, of the number of faces that contain each edge. Since each edge is contained in exactly two faces, this is equal to $2E$.
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H: Point of intersection of pair of straight lines
We have our general equation is second degree in two variables:
$ax^2+2hxy+by^2+2gx+2fy+c=0$
Let's say this represents a pair of straight lines
Our professor told that we could find the point of intersection by partially differentiating it twice, once with x and once with y and solve those 2 equations.
$2ax+2hy+2g=0$
$2hx+2by+2f=0$
I don't really understand how partial derivative of the equation has any significance here. Why is this true? What do the partially differentiated equations represent?
AI: Suppose the two lines given to us are $L(x,y)=0$ and $J(x,y)=0$, where $L(x,y)=px+qy+r$ and $J(x,y)=sx+ty+z$. Then the equation of the pairs of these lines is
$$LJ=0. \tag{1}$$
Let us consider the line given by
$$L+\lambda J=0. \tag{2}$$
For different values of $\lambda$, we will get a line that passes through the intersection of the lines $L$ and $J$. The only point $(x,y)$ for which (2) will have a solution for every $\lambda$ is when $(x,y)$ is the intersection of the lines $L$ and $J$.
Coming back to equation (1), if we take the partial derivative with regards to $x$ and $y$, then we get
\begin{align*}
L_x\, J+J_x \, L & =0\\
L_y\, J+J_y \, L & =0.
\end{align*}
Note that $L_x,L_y,J_x,J_y$ are all constants (real numbers). Suppose say $J_x, J_y \neq 0$, then we can divide by $J_x$ and $J_y$ to get
\begin{align*}
L +\lambda_1 J & =0\\
L +\lambda_2 J & =0.
\end{align*}
Which is basically in the form of equation (2). So a common solution for this system will give us the solution for equation (2) that can work for all values of $\lambda$, and hence the intersection point.
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H: If $x$ is a fixed point of a continuous function $f$, there is an open neighborhood $N$ of $x$ with $f(N)\subseteq N$
Let $\Omega$ be an open subset of a topological space $E$ and $x\in\Omega$ be a fixed point of a continuous function $f:\Omega\to E$.
How can we show that there is an open neighborhood $N$ of $x$ with $f(N)\subseteq N$.
Clearly, by continuity, if $N_2$ is an open neighborhood of $f(x)=x$, there is an open neighborhood $N_1$ of $x$ with $f(N_1)\subseteq N_2$, but it's not immediately clear why we can take $N_1=N_2$.
AI: Consider $\Omega = (-1,1)$, $E = \mathbb{R}$ and $f(x) = 2 x$.
Clearly, $x=0$ is a fixed point of $f$ belonging to $\Omega$, but the condition $f(N) \subset N$ cannot be satisfied by any open set $N\subset\Omega$.
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H: How to verify this given homeomorphism in Munkres on stereographic projection
In munkres topology for theorem 59.3, he provided a homeomorphism between $S^n- p$ and $\mathbb{R}^n$,
where
$$f(x) = \frac{1}{1 - x_{n+1}} (x_1, \dots , x_n)$$
and
$$f^{-1}(y) = (t(y)y_1, \dots , t(y)y_n, 1- t(y))$$
given $t(y) = \frac{2}{1+ ||y||^2}$.
how do I check that they are actually the inverse of each other?
(I tried to plug in one to the other, but he algebra is just horrible.)
(This is on page 369 in munkres)
AI: I dispute the "horror" of the algebra. For $f^{-1}(f(x))$, let's write $a$ for $1/(1-x_{n+1})$. Then $y_i=ax_i$ and
$$\|y\|^2=a^2\|(x_1,\ldots,x_n)\|^2=a^2(1-x_{n+1}^2)=\frac{1+x_{n+1}}{1-x_{n+1}}$$
and
$$1+\|y\|^2=\frac2{1-x_{n+1}}$$
since $x$ is on the unit sphere. Then
$$t=t(y)=\frac2{1+\|y\|^2}=1-x_{n+1}.$$
So
$$f^{-1}(y)=((1-x_{n+1})y_1,\ldots,(1-x_{n+1})y_n,x_{n+1})=x$$
since $x_i=(1-x_{n+1})y_i$.
Doing $f(f^{-1}(y))$ is no more difficult.
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H: An area preserving diffeomorphism between a disk and an ellipse
This is a self-answered question. I post it here since (embarrassingly) it took me some time to realize that the solution is obvious.
Let $D \subseteq \mathbb R^2$ be the closed unit disk and let $E$ be an ellipse with the same area, i.e. with minor and major axes of lengths $a<b$ and $ab=1$.
$$
E=\{(x,y) \, | \, \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \}
$$
Question: Can we construct explicitly an area preserving diffeomorphism $f:D \to E$?
(i.e. $Jf=1$ identically on $D$).
AI: We have $A(E)=\pi ab$, so $A(E)=A(D)=\pi$ if and only if $ab=1$.
The linear map $f:(x,y) \to (\tilde x ,\tilde y)=(ax,by)$ is an area preserving diffeomorphism $D \to E$.
Clearly $Jf=ab=1$. We just need to make sure that $f$ maps $D$ onto $E$.
Indeed,
$$
(\tilde x ,\tilde y) \in E \iff (\frac{\tilde x}{a})^2 + (\frac{\tilde y}{b})^2 \le 1 \iff x^2+y^2 \le 1 \iff (x.y) \in D.
$$
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H: Find the area between $f(x) = x^2+3x+7 $ and $g(x) = xe^{x^3+4}$ for $x \in [3,5]$.
Calculate the area between the two functions, $f(x)$, $g(x)$, for $x \in [3,5]$.
$$f(x)=x^2+3x+7$$
$$g(x)=xe^{x^3+4}$$
To determine the area between the functions I used the formula $A= \int_a^b|f(x)-g(x)|dx$. Therefore, I have:
\begin{align}
A&=\int_3^5|x^2+3x+7-xe^{x^3+4}|dx \\
&= \int_3^5|x^2|dx+\int_3^5|3x|dx+\int_3^5|7|dx-\int_3^5|e^{x^3+4}|dx\\
&= \left(\frac{x^3}{3}+\frac{3x^2}{2}+7x\right)_3^5 - \int_3^5|xe^{x^3+4}|dx
\end{align}
Here I can use $u$-subsitution and the Incomplete Gamma Function to find the integral of $xe^{x^3+4}$. All in all I get:
\begin{align}
A&=\left(\frac{x^3}{3}+\frac{3x^2}{2}+7x-e^4\frac{1}{2}\left(-\frac{x^2Γ\left(\frac{1}{\frac{3}{2}},\:-\left(x^2\right)^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-\left(x^2\right)^{\frac{3}{2}}}}\right)\right)_3^5\\
&=-\frac{98}{3}-24-14+e^4\frac{1}{2}\left(-\frac{25Γ\left(\frac{1}{\frac{3}{2}},\:-25^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-25^{\frac{3}{2}}}}-\left(-\frac{9Γ\left(\frac{1}{\frac{3}{2}},\:-9^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-9^{\frac{3}{2}}}}\right)\right)
\end{align}
Can this be simplified further or is the given solution enough?
AI: $$\int{x\,e^{x^3+4}}\,dx=-\frac{e^4 x^2 \Gamma \left(\frac{2}{3},-x^3\right)}{3
\left(-x^3\right)^{2/3}}=-\frac{1}{3} e^4 x^2 E_{\frac{1}{3}}\left(-x^3\right)$$
$$\int_3 ^5 {x\,e^{x^3+4}}\,dx=\frac{e^4}{3} \left(9 E_{\frac{1}{3}}(-27)-25 E_{\frac{1}{3}}(-125)\right)$$ Since the arguments ar quite large, for an evaluation, you could use the expansion
$$E_{\frac{1}{3}}\left(-x^3\right)=-e^{x^3} \left(\frac{1}{x^3}+\frac{1}{3 x^6}+\frac{4}{9
x^9}+O\left(\frac{1}{x^{12}}\right)\right)$$
|
H: Representations of the formulas with the elements of the linear representations of $GF(2^2)$
The problem statement is as below.
Represent the following formulas with the elements which are of the linear representations of $GF(2^2)$ where the root $\alpha$ satisfies $x^2+x+1=0$ .
$(1)$ $\alpha^4$
$(2)$ $\alpha^2-(\alpha-1)$
$(3)$ $\frac{\alpha}{\alpha^4}$
I have the following formula but I don't know how should I use it.
$GF(2^2)=\{0,1,\alpha,\alpha+1\}$ $,(\alpha^2+\alpha+1)$
Can someone let me know the solutions or some website that explain about the solutions?
AI: I will do one example, that isn't on your list, and then that should show you what is meant. I choose $\alpha^2+1$.
Now the elements of the field are $0,1,\alpha,\alpha+1$, so this is equal to one of those four elements. We have the equation $\alpha^2+\alpha+1=0$, so (as we are over a field of characteristic $2$) $\alpha^2=\alpha+1$.
Then $\alpha^2+1=(\alpha+1)+1=\alpha$.
|
H: Finding $\frac{\cot\gamma}{\cot \alpha+\cot\beta}$, given $a^2+b^2=2019c^2$
This is a question that appeared in the $2018$ Southeast Asian Mathematical Olympiad:
In a triangle with sides $a,b,c$ opposite angles $\alpha,\beta,\gamma$, it is known that $$a^2+b^2=2019c^2$$ Find $$\frac{\cot\gamma}{\cot\alpha+\cot\beta} $$
Well, by the Sine Law we have $$\sin^2\alpha+\sin^2\beta=2019\sin^2\gamma$$ and by the Cosine Law, $$\cos\gamma=\frac{a^2+b^2-c^2}{2ab} = \frac{1009c^2}{ab}$$ I’m stuck here. I tried to convert everything in our target expression to sines and cosines, but that makes the expression more complicated. I guess we can use the fact that $\cot\gamma=-\cot(\alpha+\beta)$.
How can you tackle this question? (also, apparently there are no worked solutions online)
AI: Here is the complete solution from AoPS.
Continuing from Aqua's answer, we know that $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$. Furthermore, as you have found, $\cos C = \frac{1009c^2}{ab}$. Substituting these values in gives:
$$\frac{\cot\gamma}{\cot \alpha+\cot\beta} = \frac{\cos\gamma \sin \alpha \sin \beta}{\sin ^2\gamma} = \frac{\frac{1009c^2}{ab} \cdot \frac{a}{2R} \cdot \frac{b}{2R}}{\frac{c^2}{4R}} = \frac{\frac{1009c^2}{ab} \cdot ab}{c^2} = \frac{1009c^2}{c^2} = \boxed{1009}.$$
|
H: Properties of Lebesgue measure.
Let $A,B\subset \mathbb{R}$ such that $A$ is a set of positive Lebesgue measure and $B$ is a set of zero Lebesgue measure (hence $B^c$ is dense in $\mathbb{R}$). Is it true that
$$\overline{A\setminus B}=\overline{A}?$$
($\overline{A} $ denotes the closure of $A$ in $\mathbb{R}$)
AI: A counter-example is $B=$ Cantor set and $A = B\cup (1,2)$.
|
H: Compactness Interpretation?
We define a set $X$ as compact if: "for every open cover of $X$, there exists a finite subcover."
An Open Cover $C$ of $X$ is defined as the union of a collection of Open Sets:
$$C = \bigcup_{i \in I} A_i$$
and a finite subcover as the union over a finite subcollection:
$$F=\bigcup_{i \in F} F_i$$
If we define the collection of of open sets to contain only one element e.g. the reals $A=\{\Bbb R\}$. Then there exists a finite subcover of $\Bbb R$, since we have a finite subcover consisting of only one element i.e. $\Bbb R$. Thus, $\Bbb R$ is Compact (which is obviously incorrect.)
Take another example: consider any bounded set $E \subset \Bbb R$. Hence one could construct a superset of E given by: $F \supset E $. Let the collection of open sets consist of only element e.g. $A = \{F\}$. Hence there exists a finite subcover of $E$ consisting of only one element i.e. $F$. Hence $E$ is Compact (regardless of whether or not $E$ is closed or open, which is again incorrect.)
I am struggling to understand the concept of a finite subcover. We could always construct the Open Cover to be the union of a collection of a single set, hence a finite subcover will always exist. Is there a constraint on what a finite subcover should be, other than it should be a finite collection of sets? Do the sets in the collection have to be bounded for example, or is there something else? From what I've read, there is no imposed constraint on the finite subcover other than for it to consist of finitely many sets.
AI: It seems to me that you understood perfectly the meaning of “finite subcover”. The problem lies at “for every open cover”. Yes, $\{\Bbb R\}$ is an open cover of $\Bbb R$ which has a finite subcover. However, $\Bbb R$ is not compact since, for instance, the cover$$\bigl\{(n,n+2)\mid n\in\Bbb Z\bigr\}$$is another open cover of $\Bbb R$, and this cover has no finite subcover.
|
H: Do infinitely many points on earth have the same temperature as their antipodal?
Let $X=S^2$ be the unit sphere in $\mathbb{R}^3$ and $T:X\rightarrow \mathbb{R}$ be a continuous function.
My topology textbook claims that the set $A=\{x \in X\ |\ T(x)=T(-x)\}$ has an infinite number of elements.
The fact that $A$ is non empty is clear to me as a consequence of the intermediate-value theorem, since $$f:X\rightarrow \mathbb{R},\ x \mapsto T(x)-T(-x)$$
is continuous, X is connected and $f(X)$ contains a non-positive and a non-negative real number.
What's way less clear is how there can't be a finite number of points in $A$. My intuition is that there must be a (non-trivial) curve on the sphere that contains the antipodal of each of its points, but I really don't know how to show it, if that's even true.
AI: Take a point $p \in X \setminus A$ and call it northpole, call $-p$ southpole. Now we look at the longitudes, the circles (longitudes) through these poles on the sphere.
Take one longitude and call it $L$.
$f$ is defined on $L$ and it is nonzero on $p$. As $f(-p) = -f(p)$ we know that $f$ has both negative and positive points on $L$. As $L$ is connected and $f$ is continous this means that there is a point $z_L$ on $L$ where $f (z_L)=0$. So $ z_L \in L \cap A$. $p$ and $-p$ are not in $L \cap A$, though, so $z_L$ is not one of the poles.
As this is true for each of the infinite longitudes $L$ this means that there is a point in $A$ for every longitude. These points are distinct as the longitudes only meet in the poles (which are not in $A$). Therefore $A$ must have at least as many points as there are longitudes, so $A$ is inifinite.
qed
(Slightly away from your specific question, if you are working on something as similar to the Borsuk-Ulam theorem then you can't do yourself a bigger favour then watching this most excelent video from 3b1b about it: https://www.youtube.com/watch?v=yuVqxCSsE7c)
|
H: Problem with distribution of random variable which is a sum of function's values
I need to find distribution of the random variable $Y=\sum\limits_{i=1}^{n}f(U_i)$, where
$$
f(x_1,x_2) = \left\{ \begin{array}{ll}
1 & \textrm{when $x_1>x_2$}\\
0 & \textrm{otherwise}
\end{array} \right.
$$
and $U_1, U_2, \ldots, U_n$ are i.i.d. with the same distribution as $U$, $U=(X_1,X_2)$, $\gamma=P(X_1>X_2)$.
My question is: how to find distribution of $Y=\sum\limits_{i=1}^{n}f(U_i)$?
I have tried to do it this way (by checking the behaviour of distribution for different numbers):
$P(Y=1)=P(\sum\limits_{i=1}^{n}f(U_i)=1)=P(x_1>x_2)=\gamma,$
$P(Y=2)=P(\sum\limits_{i=1}^{n}f(U_i)=2)=?.$
I'm thinking about this problem for a long time and still I don't know how to proceed. Thanks in advance.
AI: You have $n$ independent trials, each with probability $\gamma$ to success and you count successes. Hence: $Y\sim Bin(n,\gamma)$ (https://en.wikipedia.org/wiki/Binomial_distribution).
So, for example:
$\Pr(Y=1)=n \gamma (1-\gamma)^{n-1}$ because you need one $U$ to be 1, and the rest zeros.
|
H: Help in finding $\int \frac{x+x\sin x+e^x \cos x}{e^x+x\cos x-e^{x} \sin x} dx$
I want to find
$$\int \frac{x+x\sin x+e^x \cos x}{e^x+x\cos x-e^{x} \sin x} dx.$$
But since algebraic, exponential and trigonometric functions are involved I am not able to solve it. Please help in finding it by hand.
AI: Note the law of equal peopoetions, we have: $$\frac{1+\sin x}{\cos x}=\frac{\cos x}{1-\sin x} \implies \frac{x(1+\sin x)}{x \cos x}=\frac{e^x \cos x}{e^x-e^x \sin x}=\frac{x+x\sin x+e^x \cos x}{x\cos x+ e^{x}-e^{x} \sin x}$$
So $$\int \frac{x+x\sin x+e^x \cos x}{e^x+x\cos x-e^{x} \sin x} dx= \int \frac{\cos x}{1-\sin x}dx=-\ln (1-\sin x)+C$$
|
H: Show $h(x) \in F[x]$
$Q)$ There are fields $\mathbb{Q} \subset F \subset K$ (Here the $\mathbb{Q}\subset $F and $\mathbb{Q}\subset $ K are Galois extension)
Say the $f(x) \in F[x]$ and $g(x) \in \mathbb{Q}[x]$ with $f(\alpha_1) = g(\alpha_1)$ for some $\alpha_1 \in K$. Plus $f$ has roots $\alpha_1, \alpha_2$ and $\alpha_3$ in $K$
(Here the $f$ is a irreducible over $F$ and the $g$ is a irreducible over $\mathbb Q$
)
There is a $\phi \in G(K/\mathbb{Q}) s.t. h(x) = (x-\phi(\alpha_1))(x-\phi(\alpha_2))(x-\phi(\alpha_3))$ with $h \neq f$
Show $h(x)$ is a irreducible in $F[x]$
I've already shown under hypothesis, $h(x) \in F[x]$ (I.e. $irr(\phi(\alpha_1), F) = h(x) $ by using $G(K/F) \lhd G(K/\mathbb{Q})$)
But the problem is I couldn't show $h(x) \in F[x]$ the hypothesis that I used.
So I checked the answer sheet but it said
$f(x) = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$ and $h = \phi(f)$
Hence $\alpha_1 + \alpha_2 + \alpha_3 \in F $, $\alpha_1\alpha_2 + \alpha_2\alpha_3+\alpha_1\alpha_3 \in F$ and $\alpha_1\alpha_2\alpha_3 \in F$,
So the $\phi(\alpha_1 + \alpha_2 + \alpha_3), \phi(\alpha_1\alpha_2 + \alpha_2\alpha_3+\alpha_1\alpha_3), \phi(\alpha_1\alpha_2\alpha_3) \in F$ (I.e. $h \in F[x]$).
Why does those results happening? I can't understand Why does the $\phi$'s image of those are element in $F$ at all. Any help always welcome. Thanks.
AI: I think what you're missing is that if $\mathbb Q \subset F \subset K$, with $F$ Galois over $\mathbb Q$ and $K$ Galois over $\mathbb Q$, and if $\phi \in {\rm Gal}(K / \mathbb Q)$, then for every $\alpha \in F$, $\phi(\alpha)$ is also in $F$.
Why is this true? Take an $\alpha \in F$, and consider its minimal polynomial $m(X)$ over $\mathbb Q$. Since $\alpha$ is a root of $m(X)$ and since $\phi$ fixes $\mathbb Q$, $\phi(\alpha)$ must also be a root of $m(X)$. But since $F$ is Galois over $\mathbb Q$, any irreducible polynomial in $\mathbb Q[X]$ with at least one root in $F$ splits completely in $F$. This applies to $m(X)$, which is irreducible and has at least one root in $F$ (namely $\alpha$). Hence $m(X)$ splits completely in $F$, i.e. all roots of $m(X)$ are contained in $F$, including $\phi(\alpha)$.
By the way, once you realise that $\phi$ maps $F$ into $F$, it's not too hard to convince yourself that restriction $\phi|_F$ of $\phi$ to $F$ is an automorphism of $F$ (which fixes $\mathbb Q$). Once you realise this, and once you realise that your $h$ is simply $\phi(f)$, it's not hard to see that $f$ being irreducible over $F$ implies that $h$ is irreducible over $F$.
|
H: What is the maximum value of the $4 \times 4$ determinant composed of 1-16?
If 1-9 is filled in the $3 \times 3$ determinant, and each number appears once,then the maximum value of the determinant is $412$.
For example, the following determinant can take the maximum value of $412$:
$$\left|
\begin{array}{ccc}
1 & 4 & 8 \\
7 & 2 & 6 \\
5 & 9 & 3 \\
\end{array}
\right|=412.$$
Question: if 1-16 is filled in the $4\times4$ determinant, and each number appears once, what is the maximum value of the determinant? Is it necessarily less than $16 \times 15 \times 14 \times 13= 43680$?
AI: The largest known value is
$$\left|
\begin{array}{cccc}
12 & 13 & 6 &2 \\
3 & 8 & 16 &7\\
14 & 1 & 9 &10 \\
5 & 11 &4 &15
\end{array}
\right|=40800.$$
See this paper and the OEIS sequence A085000 as a reference.
|
H: Characterization of Integral Domains
We can show that in an Integral Domain I if $x^2=1$ then $x=\pm 1$.
Is the converse true?
i.e if for any x in I with $x^2=1 \implies x=\pm1$ then I is an Integral Domain.
AI: No. Take $\mathbb{Z}/4$, for example, where it is well-known all squares are 0 or 1 (depending on how they reduce in $\mathbb{Z}/2$). So $x^2=1$ iff $x=\pm 1$ but this isn't an integral domain.
|
H: If $a_{1}=1$ and for n>1, $a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$ , then $a_{246}$ lies between two integers, what are they?
I tried to use the characteristic equation technique which basically led me to (n-2) roots.
what I did was put $a_n$ as $x^n$ and same for n-1. Taking $x^{(n-2)}$ led me to a higher degree polynomial. Then I resorted to python. I wrote a script so I could obtain the first n values for all integral inputs. The source code follows:
Python version 3
depth=input("enter depth : ")
depth=int(depth)
listis=[None]*depth
listis[0]=1
def f(n,listis):
if n == 1:
return 1
else:
listis[n-1]=listis[n-2]+1/listis[n-2]
return listis[n-2]+1/listis[n-2]
import time
for i in range(depth):
tt=time.time()
# print ("The value of f(",i+1,") is ",f(i+1,listis))
# print ("computed in ", time.time()-tt ," seconds")
print (f(i+1,listis))
#print (listis[depth-1])
You can check the following graph as well https://www.desmos.com/calculator/mtepe4pdsg
I found out that it increases always and the first decimal places showcases a pattern which is very much visible when the units digit is 9.
Pattern is as follows:
9.031846164717614
9.142565499586983
9.25194398859482
9.36002938137635
9.466866652847923
9.572498224166102
9.676964161540262
9.780302355555111
9.882548683290779
9.983737155217133
I have tried numpy polyfit as well and it doesnt seem to fit.
The coefficients for degree(1 upto degree 10 are as follows):
[0.11199654 3.9178784 ]
[-7.92924950e-04 1.92081958e-01 2.55642626e+00]
[ 1.24263633e-05 -2.67551898e-03 2.68517761e-01 1.89714183e+00]
[-2.60256961e-07 6.49982694e-05 -6.10038906e-03 3.46286630e-01
1.48684664e+00]
[ 6.40214942e-09 -1.87679969e-06 2.10497785e-04 -1.16532133e-02
4.28025960e-01 1.19246775e+00]
[-1.74582869e-10 5.93007587e-08 -7.96069525e-06 5.40080166e-04
-2.00822463e-02 5.15765103e-01 9.60398854e-01]
[ 5.10638718e-12 -1.97969074e-09 3.12182211e-07 -2.57784984e-05
1.20066428e-03 -3.23214527e-02 6.11265933e-01 7.64857997e-01]
[-1.56810140e-13 6.84576839e-11 -1.24466422e-08 1.22180547e-06
-7.02796874e-05 2.41588122e-03 -4.95017346e-02 7.16107540e-01
5.91920008e-01]
[ 4.98123182e-15 -2.42078000e-12 4.99472539e-10 -5.70348627e-08
3.94084176e-06 -1.69434350e-04 4.50932324e-03 -7.29491182e-02
8.31706919e-01 4.33414958e-01]
[-1.61948602e-16 8.67652759e-14 -2.00502065e-11 2.61608664e-09
-2.11920169e-07 1.10445757e-05 -3.71712108e-04 7.92828968e-03
-1.04170633e-01 9.59323448e-01 2.84241128e-01]
Here the last terms are constant terms and as we go on left, the power increases.
To summarize :
I was not able to approximate the solution without a computational aid. I would like a more intuitive way to approximate it.
#Without any computational aid
$a_{246}$=22.236857105192666
Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$
AI: $a_n$ is increasing and suppose that $k\lt a_n\lt k+1$ then $a_n+\frac 1{k+1} \lt a_{n+1} \lt a_n+\frac 1k$ so there is a fair estimate to be made that there are either $k$ or $k+1$ elements with integer part $k$. Refining this should get you pretty close to a good-enough estimate.
|
H: $L_2$ convergence maintains the sign
Let $X$ a finite dimensional space such that $X\subset H^1.$ Let a sequence of non-negative functions $f_n\in X,\,n\geq1$ and a function $f\in H^1$ such that
\begin{equation}
\|f_n - f\|_{L_2} \to0,\;\;\;n\to \infty.
\end{equation}
It is sufficient to conclude to the fact that $f\geq0?$
AI: Yes, $L^{2}$ convergence implies almost everywhere convergence for some subsequence. Hence $f \geq 0$ a.e.
|
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