7

As you noted, $H$ is always a closed unter multiplication, thus $H \subseteq G$ is a submonoid. The only way for $H$ to fail to be a subgroup is for it to fail to contain an inverse for one of its elements. Thus, to find a counterexample we have to ensure that no argument of the form $$gAg^{-1} \subseteq A \implies g^{-1}Ag \subseteq A$$ holds. As you ...


7

Since$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots,$$you know that$$\sin(x)+x=2x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$and that$$\sin(x)-x=-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots.$$Therefore both limits$$\lim_{x\to0}\frac{\sin(x)+x}x\text{ and }\lim_{x\to0}\frac{\sin(x)-x}{x^3}$$exist; they are equal to $2$ and to $-\frac16$ respectively. This explains why ...


5

Suppose you start with $\$1000$ and play $100$ games where you bet $\$1$ on each win. That's a fair game, so the expected value you end up with is $\$1000$. However, that doesn't mean you'll always end up with exactly $\$1000$. What if you lose all $100$ games? It's unlikely, but still possible, and you'd end up with $\$900$. With the same low probability ...


5

OK, let's talk through the thought process. When I see a difference of two fractions, I give them common denominators first, as per your second $=$. I can't help but factorise the new numerator's difference of two squares after that. Since there's a division by $\sin^2 x$, which $\to0$ as $x\to0$, I need to take out a $\left(\frac{\sin x}{x}\right)^{-2}$ ...


4

Once you know that $\dfrac{\sin x}{x} \to 1$, and similar, fairly simple limits, it becomes natural to try to make them appear by extracting them from more complicated expressions.


2

Identically distributed random variables are just random variables that have the same pdf. Here are some examples: Joe flips a coin 100 times and then writes down the number of heads that come up. Let's call this variable $X_{\textrm{Joe_Heads}}.$ Now Max does the same thing and writes down the number of heads he gets. Let's call this random variable $X_{\...


2

I understand the logic to be the same in the case of Bernoulli random variables as in the case of Binomial r.v.s because of independence. In the case of Bernoulli r.v.s, just as in the case of the binomial, the concept of variance is implicit in the notion of expected value itself. That is, the more the expected value per trial deviates from 1 or 0, the more ...


2

Let $\mathcal T$ be the sheaves of type T, for $X\in\mathsf C$ let be $\mathcal T\downarrow X$ be the category of arrows $T\to h_X$ for $T\in\mathcal T$ and let $\mathcal G$ be the Grothendieck construction of $X\mapsto \mathcal T\downarrow X$. Elements of $\mathcal G$ are of the form $r=\{r_{T,X}\overset{r_{f,X}}\to h_X\}_{X\in\mathsf C}$. Let $f:F\to G$ ...


2

Here's another proof (copied from brilliant.org) of the infinite series, but for arbitrary $r<1$. I wonder if it can be adapted for the finite case by thinking about a trapezoid like this instead of a triangle....


2

See this images: This are a graphic explaination of the sum of the geometric progression of ratio $\frac{1}{2}$.


2

There are some nice and succinct answers already. If you'd like even more intuition with as little math and higher level linear algebra concepts as possible, consider two arbitrary vectors $v$ and $w$. Simplest Answer Take the dot product of one vector with the projection of the other vector. $$ (P v) \cdot w $$ $$ v \cdot (P w) $$ In both dot products ...


1

If $V$ is a subspace of $\mathbb R^n$ and $V\neq\{0\}$, then there is some $v\in V$ such that $v\neq0$. But then the line $\{\lambda v\mid\lambda\in\mathbb R\}$ is a subset of $V$. So, no ball is a subspace of $\mathbb R^n$ and the same argument applies to spheres. Actually, the same argument applies to any bounded subset $\mathbb R^n$ other than $\{0\}$.


1

Ok, so I think the issue is your interpretation of the graph. Notice how the graph of $\{f(a_n)\}$ starts out at $n=1$ at a vertical value of $16$. This makes sense because $(3+\frac{1}{1})^2 = 16$. Then, notice how it decreases from $16$ and slowly flattens out to a value of $9$. This makes complete sense because \begin{align} \lim_{n\to \infty} \left(3+\...


1

I don't know if a vanishing cycle is always non zero in the case of an algebraic Lefschetz pencil, I recall something like that but I am not sure at all. If this is true, it will follow a posteriori from the cohomological study. At first glance, this is not obvious at all. Also, recall that vanishing cycles can be defined in a non algebraic context where ...


1

Draw a segment one unit long. Tick at the first third from the left. On the right side, tick at the first third from the left. On the right side, tick at the first third from the left. On the right side, tick at the first third from the left. … When you are done, you have the infinite sum for $a=\frac13,r=\frac23$. If you stop before infinity, the ...


1

The question is a little long and hard to follow (it would be better if it was posted as multiple different questions), but from what I can see this is what you want: Yes, when you have a multi-variate system you need to multiply the polynomial bases together as you have described. Yes, this can create a really large basis is in the end. This phenomenon is ...


1

@Neal has provided a superb explaination. There is also a Yotube video giving some detailed understanding: https://www.youtube.com/watch?v=BCWBT3OTzNk&list=PLpRLWqLFLVTCL15U6N3o35g4uhMSBVA2b&index=27


1

As per Mike Earnest argument in the comments, there are scenarios in which you need $\lceil n / 2 \rceil$ lines: If the given points all lie on the same circle, dividing that circle into $n$ arcs, then each line can only cross two arcs, so at least $n/2$ lines are necessary. (If any arc is uncrossed, then the two points at the end are not separated). Now ...


1

As a partial answer: we can always use about $\frac34n$ lines. (We need $\frac34n -1$ when $n$ is divisible by $4$, but slightly more or fewer in other cases.) This assumes we don't care whether the regions are finite or infinite; as mentioned in the comments, if you want all regions to be finite, we can just use $3$ more lines at the beginning to draw a ...


1

As I understood, (see here) a signed $n$-bit integer $a$ (that is a integer $a$ such that $-2^{n-1}\le a\le 2^{n-1}-1$) is presented in a memory in the unsigned form of its $n$-complement $a^*$, which is a binary representation of a unique integer $0\le a’\le 2^n-1$ such that $a=a’\pmod 2^n$. See also a bit below In two's complement notation, a non-...


1

You might try the sequence $$ 0, 1, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}, \frac{1}{5}, \ldots $$ enumerating all rationals in $[0,1]$, which has all sorts of interesting convergent subsequences.


1

Suppose we set n = 10. It is possible but unlikely that one player wins all 10 flips, scoring 10 points. There is a $(\frac 12)^{10}$ chance of this happening. There is a $10(\frac 12)^{10}$ that he scores 8 points. A $45(\frac 12)^{10}$ that he scores 6 points, a $120(\frac 12)^{10}$ that he scores 4 points, etc. And he can score negative points just ...


1

Your point is very interesting. I would say that both expected value standard deviation would make sense after some $n$ throws. In that game we would expect $Y \rightarrow 0$ because if the result is tails you subtract 1 from the score and over time you would expect to have roughly the same number of heads as you would tails. And the standard deviation ...


1

I don't see any contradiction. The standard deviation indicates what values we should expect, and the expected value, counter intuitive to its name, gives the average of the values.


1

This formula is a concise and expressive version of Koszul formula. This is just the matter of regrouping the terms. It shows that the Levi-Civita covariant derivative is given with a formula, which employs only the Lie derivative, the exterior derivative, and the given Riemannian metric. I find this formula very illuminating, because the Lie derivative ...


1

The integral on left-hand side (divided by $b-a$), which I denote by $\bar u$ in the visualization, can be interpreted as an average over the trajectory of $u$. I have tried to visualize it for two different trajectories in $\mathbb R^2$. In some sense, it is a weighted average over the points of the trajectory, since "segments" where the velocity is high ...


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