答:
There's a number of ways to approach this problem. Let me give you an idea of two ways to investigate this sort of thing. For this post, let $a(n)$ be the $n^{th}$ term of the sequence and $b(n)$ be the sum of the first $n$ terms of $a$. I doubt there's anything that might ordinarily be called a closed form, but you can certainly do better than just summing the terms.

Way 1: Look it up!

This is exactly the kind of question that should lead a mathematician to search on OEIS (the online encyclopedia of integer sequences) - if we type in the original sequence as a search query, we can find a page here regarding the sequence - described as "Highest power of $2$ dividing $n$." This is not super useful on its own - although, as we will see, the formal description of the sequence given in its title is useful - but there are a lot of comments beneath that might often be of use to this sort of question. In fact, the comments even list the sequence of partial sums of the original sequence - exactly what you are asking about (though you could surely also find this by calculating the first few partial sums and searching for them).

In the comments there, we find out a pair of recurrence relations for $b$:
$$b(2n)=2b(n)+n$$
$$b(2n+1)=2b(n)+n+1$$
which gives an extremely efficient way of computing it - for instance, the sum of the first $9$ terms is:
\begin{align*}b(9)&=b(2\cdot 4 + 1)\\&=2b(4)+4+1\\&=2b(2\cdot 2)+5\\&=2(2b(2)+2)+5\\&=4b(2\cdot 1)+9\\&=4(2b(1)+1)+9\\&=8b(1)+13\\&=21\end{align*}
where we note on the last step that $b(1)=1$. The trick here is that we can always rewrite whatever is inside $b$ as either $2n$ or $2n+1$, halving its size at each step - and you can check that this really does equal $1+2+1+4+1+2+1+8+1$

While you won't find justification of this on OEIS, knowing the answer makes finding justification much easier: note that $a(2n)=2a(n)$, so the sum of the terms of $a$ at positions $2,4,\ldots,2n$ is twice the sum of the terms at $1,2,\ldots,n$ (which is $b(n)$) - hence is $2b(n)$. The terms at the odd positions are all $1$ - and there are $n$ or $n+1$ of them below $2n$ or $2n+1$ respectively.

Way 2: Try to express it in terms of something simpler.

Another way to look at this is to note that it's "mostly" periodic, with a few exceptions - or, even more formally, that you can look at it as "mostly $1$, with some $2$'s added, and then some $4$'s added on top of $2$'s and so on...". More formally, let's consider an array where, on the first row, we write all ones. Then, we start a new row, every other entry of which is $1$. If we sum those rows, we'd get $1,2,1,2,1,2,\ldots$. To correct this, we add a third row, every fourth entry of which is $2$ - and if we add that on we get $1,2,1,4,1,2,1,4,\ldots$ - we then continue adding a $4$ every $8$ columns to fix it further - then an $8$ every $16$ columns, a $16$ every $32$ columns, and so on until we've generated enough of our sequence:
$$\begin{array}\\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 2 & 0 & 0 & 0 & 2\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 \end{array}$$
See how the sums of columns here are the entries of $a$. However, the sum of $n$ terms of any row is easy to figure out - clearly, the sum of $n$ terms of the first row is $n$. The second row has $\lfloor n/2\rfloor$ non-zero terms (where we use the floor function which rounds down its argument to the next integer) - each of which is $1$, so contributes $\lfloor n/2\rfloor$ in total. Then, the next row has $\lfloor n/4\rfloor$ non-zero terms, which are each $2$, so which contributes $2\lfloor n/4\rfloor$ to the sum. Continuing analogously, we will find that the sum of the first $n$ terms of the original sequence - which is the sum of the sums of the rows is:
$$n + \sum_{k=1}^{\infty}2^{k-1}\left\lfloor\frac{n}{2^k}\right\rfloor$$
where you only need to take this sum up to the last $k$ so that $2^k \leq n$, since the remaining terms after that are all zero.

Addendum: It might be worth noting that this function is pretty easy to describe on the large scale - the sum of the first $2^n-1$ terms is $n2^{n-1}$, which can be seen inductively by noting that the terms $1$ through $2^n-1$ are the same as those from $2^n+1$ to $2^{n+1}-1$ and the term at $2^n$ is just $2^n$ - and it actually looks more and more like a smooth curve the more you zoom out (though, note that the slope of the line that it looks like changes) - for instance, even the first 511 partial sums look pretty linear:

There's some small scale variation, but the function acts basically like $\frac{1}2 n\log_2(n)$ - here's that function overlaid on the prior plot:

Some more analysis can give that the deviations from this curve are actually quite small - their ratio tends to $1$ - but there's probably not a lot of hope for a much better representation since the deviations have some fractal nature. You could likely come up with some way to do this involving writing $n$ in binary and doing something weird to the bits - but that's largely just another way to encode the sorts of sums and recurrences that appear with this sort of sequence.

Title: Is there a closed form to this series?

Body:
Here is the series in question,

```<code>1 + 2 + 1 + 4 +
1 + 2 + 1 + 8 +
1 + 2 + 1 + 4 +
1 + 2 + 1 + 16 +
1 + 2 + 1 + 4 +
1 + 2 + 1 + 8 +
1 + 2 + 1 + 4 +
1 + 2 + 1 + 32 + ...
</code>```

so for example, given n = 4, the series sum is 1 + 2 + 1 + 4 = 8

I created this series inspired by the game 2048,
and I've been trying to find the closed-form with respect to n for a few hours to no avail.
Does there exist a closed-form? What's the name of this type of series?


问:
Divide by standard deviation and subtract mean or divide by max?
I am trying to do feature scaling to normalize features across different samples. So for example one sample has a feature in the range $[0,10]$ and another sample has the same feature to be in the range $[0,100]$.

The book that I am reading suggests to subtract the mean and then divide by the standard deviation. But why don't we just divide the different feature ranges with the maximum value? So we divide the fist sample features with $10$ and the next with $100$. Why do the former over the latter?

答:
The purpose of normalization is to place all features on the same scale.

Subtracting feature mean and dividing by feature standard deviation will lead to all features having mean 0 and standard deviation 1. This is known as "$z$-scaling".

If all your features range from 0 to some maximum value, then dividing by the feature max will lead to all features ranging from 0 to 1. This is often dubbed "max-scaling". (If a feature's values range from $a$ to $b$, you could achieve the same by subtracting $a$ and then dividing by $b-a$.)

Both of these strategies can be considered normalization, but they have different concepts of what normalization means. The former approach ($z$-scaling) ensures that each feature is centered at zero with the same spread; the latter (max-scaling) ensures that each feature is bounded between 0 and 1. But max-scaling doesn't work well when you have a feature with a skewed distribution, because you may end up bunching all your values to one end or the other, so you haven't really achieved a consistent centering of your features. Similarly if a feature has outliers at both ends, then max-scaling leads to compression of the bulk of your data into a tiny range, so you've failed to achieve consistent spread. Max-scaling is fine in the absence of skew or outliers.

On balance, if you want to apply a uniform normalization strategy to your features, the $z$-scaling approach produces more consistent results. (But it won't solve every problem with your data. If you are careful about cleaning up your data, you should first attempt to handle gross outliers before doing any normalization. And maybe transform your features to reduce skewness.)

ContentLicense: CC BY-SA 4.0


问:
How do I interpret this with logic notation? "not A doesn't mean not B"
$$
\neg A \rightarrow \neg B \equiv B \rightarrow A
$$

Is this correct?

答:
If not A, then not B. This is equivalent to if B, then A.

A truth table may be useful here to convince you that they are indeed equivalent:

A
B
¬A
¬B
¬A $\rightarrow$ ¬B
B $\rightarrow$ A

T
T
F
F
T
T

T
F
F
T
T
T

F
T
T
F
F
F

F
F
T
T
T
T


问:
product of limits proof
I'm hoping someone might have a hint as to how to prove/disprove the follow claim:

Claim: Let $f,g$ be real valued functions. If  $\lim_{x\rightarrow\infty}f(x)g(x)\text{ exists}\space\text{and}\space\lim_{x\rightarrow\infty}f(x)=0,$ then
$\lim_{x\rightarrow\infty}f(x)g(x)=0.$

Broader context: I'm trying to show that if
$\mathbb{P}\left(X\leqslant0\right)=1\space\text{and}\space\mathbb{E}\left[|X|^{n}\right]<\infty,$
then for $n>0, $
$\mathbb{E}\left[X^{n}\right]=-n\int_{-\infty}^{0}x^{n-1}F_{x}(x)dx.$

I get stuck on the following step:
$\mathbb{E}\left[X^{n}\right] =\lim_{L\rightarrow\infty}\left(\left[x^{n}F_{x}(x)\right]_{-L}^{0}-n\int_{-L}^{0}x^{n-1}F_{x}(x)dx\right)
 =\cdots.$
It's clear from the question that
$\lim_{L\rightarrow\infty}\left[x^{n}F_{x}(x)\right]_{-L}^{0}=0$
but I'm not clear on what makes
$\lim_{L\rightarrow\infty}\left[-(-L)^{n}F_{x}(-L)\right]$
evaluable.

答:
Your claim is not true. Consider $f(x) = \frac{1}{x}$ and $g(x) = x$. Then $\lim\limits_{x \to \infty} f(x) = 0$ but $\lim\limits_{x \to \infty} f(x) g(x) = 1$.

Now, let's solve your problem.

We know that $F_x(L)$ is defined to be $\mathbb{P}(x \leq L)$. We must show that $\lim\limits_{L \to \infty} (-L)^n \mathbb{P}(x \leq -L) = 0$.

We see that $\mathbb{E}(|X|^n$ given $X \leq -L) \mathbb{P}(x \leq -L) + \mathbb{E}(|X|^n$ given $X > -L) \mathbb{P}(X > -L) = \mathbb{E}[|X|^n]$. Then $\mathbb{E}(|X|^n$ given $X \leq -L) \mathbb{P}(x \leq -L) = \mathbb{E}[|X|^n] - \mathbb{E}(|X|^n$ given $X > -L) \mathbb{P}(X > -L)$.

Taking the limit as $L$ goes to infinity gives us $\lim\limits_{L \to \infty} \mathbb{E}(|X|^n$ given $X \leq -L) \mathbb{P}(x \leq -L) = 0$.

And we see that $L^n \leq \mathbb{E}[|X|^n$ given $X \leq -L]$. So therefore, $\lim\limits_{L \to \infty} L^n P(X \leq -L) = 0$.

Title: product of limits proof


问:
Define $f:(0,1) \to R$ by $f(x)=\frac{1}{\sqrt{x}}-\sqrt\frac{x+1}{x}$ Can one define $f(0)$ to make $f$ continuous at 0?
Define $f:(0,1) \to R$ by $f(x)=\frac{1}{\sqrt{x}}-\sqrt\frac{x+1}{x}$
Can one define $f(0)$ to make $f$ continuous at 0?

So from my understanding, $f$ is continuous if $Lim_{x\to \infty}f(x)=f(0)$ Is this the correct? If the limit exists at 0, does that mean that it is continuous at 0?

答:
$f(x)=\frac{1-\sqrt{x+1}}{\sqrt{x}}=\frac{1-(x+1)}{\sqrt{x}(1+\sqrt{x+1})}=\frac{-\sqrt{x}}{1+\sqrt{x+1}}$

Therefore $f(x)$ is continuous at $x=0$.


问:
Solving for inequality $-1 \leq 4x^2 \leq 1$
Bit rusty on my algebra, I'm trying to solve for the inequality of $-1 \leq 4x^2 \leq 1$.
If I divide everything by $4$ I get $-\frac{1}{4} \leq x^2 \leq \frac{1}{4}$, now you can't take the square root of a negative number so how come the solution is $-\frac{1}{2} \leq x \leq \frac{1}{2}$?

Would appreciate some help, thanks in advance!

答:
There is no need to take

$$-1≤4x^2≤1$$

Because, $4x^2≥0$, for all $x\in\mathbb R$. You have

$$0≤4x^2≤1$$

$$\implies 0≤ x^2≤\frac 14$$

$$\implies 0≤ |x|≤\frac 12$$

$$\implies -\frac 12≤x≤\frac 12.$$


答:
$ x^2 \geq 0 $ so your inequality becomes $ 0 \leq x^2 \leq \frac{1}{4}$

You can take the root of this equation, but remember that you have to take both the positive and negative root.

So you'll have:

$ 0 \leq x \leq \frac{1}{2}$ or $0 \geq x \geq -\frac{1}{2}$

Combine those and you have the result.

An alternate way to think about this is that

$-\frac{1}{4} \leq x^2 \leq \frac{1}{4}$ implies $|x^2| \leq \frac{1}{4}$, therefore $|x| \leq \frac{1}{2}$ which implies the solution listed.

Title: Solving for inequality $-1 \leq 4x^2 \leq 1$

Body:
Bit rusty on my algebra, I'm trying to solve for the inequality of $-1 \leq 4x^2 \leq 1$.
If I divide everything by $4$ I get $-\frac{1}{4} \leq x^2 \leq \frac{1}{4}$, now you can't take the square root of a negative number so how come the solution is $-\frac{1}{2} \leq x \leq \frac{1}{2}$?

Would appreciate some help, thanks in advance!


问:
What is the most common decision making math model?
I have been told that there are psychological math model to make decision. I want to know what is it and how it work.</p>

答:
What you are referring is probably 'prospect theory' which is developed by two psychologist Kahneman and Tversky. As one can expect from their background they developed a more descriptive theory of decision making than for example expected utility theory..
I can suggest the book "Prospect Theory: For Risk and Ambiguity" by Peter Wakker.

However, the most common model differs from this and it is certainly expected utility theory ( and its generalizations).

ContentLicense: CC BY-SA 3.0


问:
Doubt about a passage on page 24 of the Evans Partial Differential Equations book
Let $\Phi$ the fundamental solution of $\Delta v=0$, i.e.
$$
\Phi(x)=\left\{\begin{array}{ccc}-\frac{1}{2\pi}\log|x|& \mbox{if}& n=2\\ \frac{1}{n(n-1)\alpha(n)}\frac{1}{|x|^n-2}& \mbox{if}& n\geq 3\\ \end{array} \right.
$$
here $\alpha(n)$ denote the volume of unite ball in $\mathbb{R}^n$. In page 23 of Evans Partial Differential Equations book Theorem 1 ( Solving Poisson's equation ) we have for $f\in C_c^{2}(\mathbb{R}^n)$ and
$$
u(x)=\int_{\mathbb{R}^n}\Phi(x-y)f(y)\mathrm{d}y=\int_{\mathbb{R}^n}\Phi(y)f(x-y)\mathrm{d}y
$$
the following conclusions:
$$
\begin{array}{crl}
 \mathrm{(i)} &  u        &\in C^2(\mathbb{R}^n)
\\
\mathrm{(ii)}&  -\Delta u &= f
\end{array}
$$
In step 2 of the proof of this theorem, we have the following inequality
$$
\left|\int_{B(0,\epsilon)} \Phi(y)\Delta_x f(x-y)\mathrm{d} y\right|
\leq
C\cdot \|D^2f\|_{L^\infty(\mathbb{R}^n)}\int_{B(0,\epsilon)} |\Phi(y)|\mathrm{d} y
$$

Question. How can I prove that
  $$
\left|\int_{B(0,\epsilon)} \Phi(y)\Delta_x f(x-y)\mathrm{d} y\right|
\leq
C\cdot \|D^2f\|_{L^\infty(\mathbb{R}^n)}\int_{B(0,\epsilon)} |\Phi(y)|\mathrm{d} y\quad ?
$$

My attempt: We have
\begin{align}
\left|\int_{B(0,\epsilon)} \Phi(y)\Delta_x f(x-y)\mathrm{d} y\right|
\leq &
\int_{B(0,\epsilon)} |\Phi(y)\Delta_x f(x-y)|\mathrm{d} y
\\
&\leq
\mathop{\mathrm{sup \;ess}}|\Delta_x f(x-y)|\int_{B(0,\epsilon)} |\Phi(y)|\mathrm{d} y
\end{align}
This seems to indicate that I have to prove that
$$
\mathop{\mathrm{sup \;ess}}|\Delta_x f(x-y)|
\leq
\mathop{\mathrm{sup \;ess}}|D^2_x f(x-y)|_{\mathscr{L}(\mathbb{R}^n,\mathbb{R}^n)}
$$
where $|\;\cdot \;|_{\mathscr{L}(\mathbb{R}^n,\mathbb{R}^n)}$ is a matrix norm.

答: Evan's defines$$\vert D^2 f \vert  = \bigg (\sum_{i,j=1}^n (D_{ij}f)^2 \bigg)^{\frac 1 2} $$ and $$ \| D^2 f \|_{L^\infty(\mathbb{R}^n)} = \| \vert D^2 f \vert  \|_{L^\infty(\mathbb{R}^n} = \mathrm{ess}\sup_{x \in \mathbb{R}^n} \bigg (\sum_{i,j=1}^n (D_{ij}f)^2 \bigg)^{\frac 1 2}.$$ Then for all $x \in \mathbb{R}^n$, $D_{ij} f (x) \leqslant \vert D^2 f \vert \leqslant \| D^2 f \|_{L^\infty(\mathbb{R}^n)}$. Thus, by triangle inequality $$\vert \Delta f \vert \leqslant \sum_{i=1}^n \vert D_{ii}f \vert \leqslant n \| D^2 f \|_{L^\infty(\mathbb{R}^n)}.$$


问:
Probability of infection by staphylococcus aureus
Please forgive my innumeracy, but I have a question with which I am hoping someone might be able to help me.

Suppose the following be true. The chance of a prosthetic hip joint becoming infected by staphylococcus aureus is one per cent. The chance of a natural hip joint becoming infected by staphylococcus aureus is 0.1%. In other words (in case I am misusing the word 'chance'), one in one hundred people with prosthetic hip joints will become infected by staphylococcus aureus, whereas only one in one thousand people with natural hip joints will become so infected.

Now suppose that X has a prosthetic hip joint and that X's hip joint becomes infected by staphylococcus aureus.

Given only the information provided here, is it correct to say that X's hip joint probably would not have become infected but for the fact that X has a prosthetic hip joint (instead of a natural hip joint)? In other words (to make it clear what I mean by 'probably'), is it correct to say that there is a greater than 50 per cent chance that X's hip joint would not have become infected but for the fact that X has a prosthetic hip joint? Why or why not?

答:
Now suppose that X has a prosthetic hip joint and that X's hip joint becomes infected by staphylococcus aureus.

Given only the information provided here, is it correct to say that X's hip joint probably would not have become infected but for the fact that X has a prosthetic hip joint (instead of a natural hip joint)?

Let $A$ denote the event that hip joint becomes (or became) infected.

Let $B$ denote the event that hip joint prosthetic.

Let $C$ denote the event that hip joint natural, so $p(B) + p(C) = 1.$

Although the following analysis is not directly on point, it may be deemed pertinent.

That is, it may be pertinent to use Bayes Theorem, to calculate

$$p(B|A) = \frac{p(AB)}{p(A)}.\tag1$$

Note that this is not equivalent to the question that the OP is asking.  However,
re the OP's query, $p(B|A)$ may still be considered relevant.

The only data given in the problem is that:

$$\frac{p(AB)}{p(B)} = p(A|B)  = 0.01 \tag2$$ and

$$\frac{p(AC)}{1 - p(B)} = \frac{p(AC)}{p(C)} = p(A|C) = 0.001. \tag3$$

Since $B$ and $C$ are contradictory events, the roadblocks
to solving equation (1) above are

computing $p(AB)$

computing $p(A) = p(AB) + p(AC).$

Since $p(B)$ and (therefore) $p(C)$ are unknown, I see no way of

computing either $p(AB)$ or $p(AC)$.

However, it is reasonable to let $X = p(B)$, see where this leads,
and then make real world guesstimates of $p(B)$.  This will then
provide a real world estimate response to the query.

$p(AB) = 0.01 \times X.$

$p(AC) = 0.001 \times (1 - X).$

Therefore, $p(A) = 0.001 + 0.009X.$

Therefore, $p(B|A) = \frac{0.01X}{0.001 + 0.009X}.$

Taking a step back, it is clear that for small values of $X$, the
denominator in the above fraction will be $\approx 0.001$.

Anyway:

As an example, suppose that $X = 0.001.$

Then, $p(B|A) = \frac{.00001}{0.001009} \approx \frac{1}{100}$.

Suppose instead that $X = 0.01.$

Then, $p(B|A) = \frac{.0001}{0.00109} \approx \frac{1}{10}$.

Suppose instead that $X = 0.1.$ (An estimate that is clearly too high).

Then, $p(B|A) = \frac{.001}{0.0019} \approx \frac{10}{19}$.

Therefore, one concludes that $p(B|A)$ is probably not much above

$\frac{1}{10}$, and is perhaps significantly less.

Intuitively, the idea is that prosthetics only increased the chance of
infection by a factor of $(10)$, while the real world chance that
someone has a prosthetic is perhaps exponentially below $\frac{1}{10}.$

Now the fun starts.

The sample values examined for X suggest that
$p(B|A) \approx 10 \times p(B)$.

This strikes me as a reasonable guesstimate, since the chance of infection increases
by a factor of 10, given the prosthetic.

At this point, my only evaluation is that the whole issue is unclear.

In other circumstances, I might attempt to belabor this point, but the answer of jMdA
has already said everything that I might have thought of.  Lacking real world data on causation, but understanding the math as I do, my suspicion is that there is
an unknown causal link between prosthetics, and the increased chance of infection.

In fact, I will go further, and say that it is close to a virtual certainty that the
causal link exists, because how else could you explain that the chance of infection increases by a factor of 10, given the prosthetic?

Note that within the narrow confines of the OP's query, it is irrelevant whether the link is cellular, or because people with prosthetics might be more likely to have improper sleeping habits or dietary habits or exercise habits, or because they might be on some medication (e.g. antidepressants) that have weakened their immune system.  That is, even assuming the causal link, you can't say that the causation would evaporate if the amputee declined the use of the prosthetic.


问:
Determine the Probability of no bin being empty when 10 balls are randomly tossed with equal probability into 6 bins
Suppose we have 10 items that we will randomly place into 6 bins, each with equal probability. I want you to determine the probability that we will do this in such a way that no bin is empty. For the analytical solution, you might find it easiest to think of the problem in terms of six events Ai, i = 1, . . . , 6 where Ai represents the event that bin i is empty, and calculate P(AC1 ∩···∩AC6)using DeMorgan’s law.

Analytical= 1- P(bin one empty) + P(bin 2 empty) + P(bin 3 empty) + P(bin 4 empty) + P(bin 5 empty) + P(bin 6 empty) – P(1 and 2 empty) – P(1 and 3 empty) – P(1 and 4 empty) – P(1 and 5 empty) – P(1 and 6 empty) – P(2 and 3 empty) – P(2 and 4 empty) – P(2 and 5 empty) – P(2 and 6 empty) – P(3 and 4 empty) – P(3 and 5 empty) – P(3 and 6 empty) – P(4 and 5 empty) – P(4 and 6 empty) – P(5 and 6 empty) + P( 1 2 3 empty) + P(1 2 4 empty) + P (1 2 5) empty + P (1 2 6 empty) + P( 1 3 4 empty) + P(1 3 5 empty) + P( 1 3 6 empty) + P(1 4 5 empty) + P(1 4 6 empty) + P(1 5 6 empty) – P(1 2 3 4 empty) – P(1235 empty) – P(1236 empty) – P(2345 empty) – P(2346 empty) – P(3456 empty) + P(12345 empty) + P(23456 empty) + P(34561 empty) + P(45612 empty) – P(123456 empty)

1-(6P(1 specific empty bin) – 15P(2 specific empty bins) + 20P(3 specific empty bins) – 15P(4 specific empty bins) +6*P(5 specific empty bins)

This is what I have so far but I don't know how to determine the individual probabilities above.

答:
It can be written as,

$\displaystyle \small 1 - 6 \cdot \bigg(\frac{5}{6}\bigg)^{10} + 15 \cdot \bigg(\frac{4}{6}\bigg)^{10} - 20 \cdot \bigg(\frac{3}{6}\bigg)^{10} + 15 \cdot \bigg(\frac{2}{6}\bigg)^{10} - 6 \cdot \bigg(\frac{1}{6}\bigg)^{10}$

$\displaystyle \small \frac{5}{6}$ represents the probability that an item will land into one of the five specific bins out of six (i.e. a specific bin is empty). Similarly $\displaystyle \small \frac{4}{6}$ represents the probability that an item will land into one of the four specific bins and so on.


问:
Induced connections on embedded surfaces in 3 space
The unit normal to a two dimensional surface in 3 space maps the surface into the unit sphere. The induced vector bundle under this map, the Gauss map,is the surface's tangent bundle as can be seen from parallel translating its tangent planes to the origin.

The induced connection on the surface's tangent bundle - i.e. the pull back under the Gauss map of the standard connection on the sphere - defines a connection on the surface.

Here are some questions about the induced connection.

Given a second embedding of the surface, the Gauss map induces a second connection. This second Gauss map is homotopic to the first - since both classify the surface's tangent bundle. One can imagine deforming one surface into the other to obtain a homotopy and simultaneously a homotopy between the two connections. Does such a deformation always exist? If not can the homotopy be realized as a 1 parameter family of embeddings in a higher dimensional Euclidean space?

Is the induced connection just the Levi-Civita connection on the surface that it inherits from its embedding in 3 space?

If so this seems quite strange since the derivative of the Gauss map is not usually the induced map of vector bundles.

答:
The space of connections on a manifold is an affine space, literally you can take $(1-t)$ times your first connection plus $t$ times your second connection, this works at the level of Christoffel symbols, and also at the level of Ehresmann connections.   That answers your first question. </p>

<p>For your second question, yes there's a direct formula for the submanifold Levi-Cevita connection in terms of the Gauss map.  It's a little complicated but there is such a formula.  </p>

Body:
The space of connections on a manifold is an affine space, literally you can take $(1-t)$ times your first connection plus $t$ times your second connection, this works at the level of Christoffel symbols, and also at the level of Ehresmann connections.   That answers your first question.

For your second question, yes there's a direct formula for the submanifold Levi-Cevita connection in terms of the Gauss map.  It's a little complicated but there is such a formula.


问:
Doubts on proving there are only $2$ abelian groups of order $12$, up to isomorphism?
I have been tasked with the following exercise:

Prove there are only $2$ abelian groups of order $12$, up to isomorphism.

I have read some texts and watched some videos but got very confused. I want to understand what is the "algorithm" for doing it. I think it's the following:

Due to Lagrange's theorem, a group of order $12$ could have subgroups of orders $1, 2, 3, 4, 6, 12$.

Now, if one element of this group has order $12$, then it is isomorphic to $\Bbb{Z}_{12}$.

Question: What do we do now? I think we should do the following:

Suppose there is no element of order $12$, now we try to eliminate the remaining cases using Cauchy's theorem, Sylow theorems and certain considerations about groups. For example: If we have a subgroup of order $3$ and a subgroup of order $4$, then we must have an element of order $12$ and hence, this case must be discarded because it goes against what we assumed?

Is that it? I am asking because a lot of the material I read/saw until now do not state what are the "worthy" cases we should investigate and what cases are "unworthy" of investigation, such as the one I pointed out.

答: Hold on while I get out my sledgehammer: By the FTFAG, it's the product of cyclic groups.  Then the only possibilities are $C_2\times C_6$ and $C_{12}$.


问:
Why is $\mathbb{R}^2$ not a subset and /or a subspace of $\mathbb{R}^3$?
One thing this suggests--at least to me--is that the x-y plane and $\mathbb{R}^2$ are not necessarily equivalent.  For example, I could define the following: $X = \left\{ \begin{bmatrix} x\\y\\z\end{bmatrix} x,y \in \mathbb{R} \land z = 0\right\}$.  Am I wrong to think, one, that this is a subset of $\mathbb{R}^3$?  As I write this it occurs to me that while scalar multiplication is closed under the above rules, addition doesn't pass the smell test for a subspace... so, OK, it's certainly not a subspace.  I would welcome any insight readers of this query can provide.

答:
Assume $m<n.$  Then on the standard basis, $\mathbb{R}^m$ is, in fact a subspace of $\mathbb{R}^n.$

The vector space $\mathbb{R}^n$:
Begin by defining $\mathbb{R}^n$ (and similarly for $\mathbb{R}^m,$ etc.,) as the $n^{th}$ Cartesian product of $\mathbb{R}.$  Doing so requires an ability to distinguish each $\mathbb{R}$ appearing as a Cartesian factor from the others.  For example, in $\mathbb{R}^3$ we distinguish each by calling them the $x,y,z$ axes.  We  nominate each factor using a subscript natural number $i=1 \dots i=n$ so $\mathbb{R}^n=\mathbb{R}_1\times\dots\times\mathbb{R}_n,$ and assume the natural ordering on the indices.  These designations are arbitrary.  We could choose any unique subscript values in any ordering, just so long as we distinguish each Cartesian factor $\mathbb{R}$ from the others.

The standard basis $\left\{\mathfrak{\hat{e}}_i\right\}$ of $\mathbb{R}^n:$
From each $\mathbb{R}_i$ we identify its real number $1\in\mathbb{R}_i$ with the symbol $\mathfrak{\hat{e}}_i\in\mathbb{R}_i.$  The $\mathfrak{\hat{e}}_i$ are called the elements (vectors) of the standard basis of $\mathbb{R}^n.$  The set $\left\{\mathfrak{\hat{e}}_i\right\}$ is linearly independent, and spans $\mathbb{R}^n$.  That is, every element of $\mathbb{R}^n$ is given as a unique linear combination of the standard basis vectors.

$\mathbb{R}^m$ as a subspace of $\mathbb{R}^n$: We construct $\mathbb{R}^m$ and its standard basis $\left\{\mathfrak{\hat{f}}_j\right\}$ using the same method with $j=1 \dots j=m<n.$  The next natural step is to identify $\left\{\mathbb{R}_i=\mathbb{R}_{j=i\le{m}}\right\}$ which gives $\left\{\mathfrak{\hat{f}}_j=\mathfrak{\hat{e}}_{j=i\le{m}}\right\}.$

Therefore, assumming standard definitions and natural identities, $\mathbb{R}^m\subset\mathbb{R}^n.$

Of course, we could use different definitions, leading to different results.  For example, a typical definition is $\mathbb{R}^{n-m}=\mathbb{R}_{i=m+1}\times\dots\times\mathbb{R}_n.$
Other than an appeal to sanity, there is nothing to stop us from identifying $\mathbb{R}^{p}=\mathbb{R}^{n-m},$ where $p=n-m,$ and $\mathbb{R}^{p}$ is constructed as above.  Indeed $\mathbb{R}^{p}$ and $\mathbb{R}^{n-m}$ are isomorphic, so there is an equivalence relation between them.


问:
Under what conditions an subspace of a CDH space is CDH?
A space $X$ is countable dense homogeneous (CDH) if it is separable, and for every pair $(D,E)$ of countable dense subsets of $X$ there exists a homeomorphism $f:X\longrightarrow X$ such that $f(D) = E$.

Unless I'm wrong, being a CDH is not hereditary. I was working a while on it and I think that if $X$ is CDH and $ Y\subseteq X$ then at least it is necessary that $Y$ be dense in $X$ so that it "can" be CDH, although I think it is not enough.

Do you have an idea of what conditions I need?

答:
There are dense subsets of CDH-spaces that are not CDH. In fact, remember that $\mathbb{R}$ is CDH (back and forth). But $\mathbb{Q}$ is dense, non-discrete and not CDH. Because, we have the following

Theorem 1. Let $X$ be a non-discrete CDH space, then $\aleph_{1}\leq |X|$.

Proof.      Suppose otherwise, $X$ is countable. As $X$ is non-discrete, there is $x_{0}$ be a non-isolated point of
$X$ (that is, $\{x_{0}\}$ is not open). Note that $X$ and $X\setminus \{x_{0}\}$ are countable and dense in $X$, then there is $h\in\mathcal{H}(X)$ such that $X=h[X]=X\setminus \{x_{0}\}$, contradiction.


问:
Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$
I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$

Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$$

Where $H_n$ is the n-th harmonic number. I managed to get here after converting the integrand into a double summation, recognizing the digamma function hidden inside, and summing back to a harmonic number, but couldn't get past that.

Is there another way to solve the integral, or to solve the summation? I've tried breaking the sum down to its parts, but the best I could do was to find a few terms and was left with
$$\sum_{m=1}^{\infty} \frac{(-1)^m\psi(\frac{3m}{4})}{3m}$$
Where $\psi(x)$ is the digamma function, and still couldn't solve that.

答:
Maple gives a rather unpleasant expression involving complex dilogarithm functions:

$$-\frac{19 \pi^{2}}{144}+\frac{7 \ln \! \left(2\right)^{2}}{12}-\frac{\ln \! \left(2-\sqrt{3}\right)^{2}}{6}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+\left(-\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+\left(\frac{1}{2}-\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}+\frac{i}{2}+\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(1-\frac{i \sqrt{3}}{3}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}-\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(1+\frac{i \sqrt{3}}{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}+\frac{i}{2}-\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)}{3}
$$

I suspect this comes from writing the rational function $x^2/(1+x^3)$ in partial fractions and also writing $\ln(1-x^4) = \ln(1-x) + \ln(1+x) + \ln(1-ix) + \ln(1+ix)$.


问:
Integrating $\int x \arcsin x \,dx$ by parts
Solving the integral $$\int x \arcsin x\,dx$$

I know I have to solve integral using by parts

$$\int u\,dv = uv-  \int v\,du$$

$$\int x \arcsin x \,dx$$

Let $u = \arcsin x$ and $dv = x\,dx$

so $du = \dfrac1{\sqrt{1-x^2}} \, dx$

and $dv = x^2$.

but here is where I am stuck. because when I plug the variables into equation it becomes confusing on what to do next.

ps. New to using latex and still haven't perfected it so I apologize for formatting.

答:
Using OP's notation, we have $v=x^2/2$ and $\mathrm du=\mathrm dx/\sqrt{1-x^2}$, so that

$$
\int x\arcsin x\mathrm dx=\frac12x^2\arcsin x-\frac12\underbrace{\int{x^2\over\sqrt{1-x^2}}\mathrm dx}_I
$$

For the latter part, we have

$$
\begin{aligned}
I
&=\int{x^2\over\sqrt{1-x^2}}\mathrm dx \\
&=\int{x^2-1+1\over\sqrt{1-x^2}}\mathrm dx \\
&=-\int\sqrt{1-x^2}\mathrm dx+\int{\mathrm dx\over\sqrt{1-x^2}} \\
&=\arcsin x-\int\sqrt{1-x^2}\mathrm dx
\end{aligned}
$$

For the remaining integral, it follows from yet another IBP that

$$
\begin{aligned}
\int\sqrt{1-x^2}\mathrm dx
&=x\sqrt{1-x^2}+\int{x^2\over\sqrt{1-x^2}}\mathrm dx \\
&=x\sqrt{1-x^2}+I
\end{aligned}
$$

Plugging this back in, we have

$$
2I=\arcsin x-x\sqrt{1-x^2}+C
$$

Putting everything together, we get

$$
\int x\arcsin x\mathrm dx=\frac12x^2\arcsin x-\frac14\arcsin x+\frac14x\sqrt{1-x^2}+C
$$


问:
Is submodel of atomic model atomic
I think the submodel (elementary or arbitrary) of an atomic model may not be atomic. But I do not know an example.</p>

答:
Recall for readers that an atomic model is a structure such that each type realized in the structure is generated by a single formula.

Suppose $M\preccurlyeq N$, $N$ is atomic, and $m\in M$. By elementarity we have $tp^M(m)=tp^N(m)$, and by atomicity of $N$ we have that $tp^N(m)$ is generated by a single formula $\varphi$. Putting these together we get that $\varphi$ generates the type of $m$ in $M$ as well. So elementary submodels of atomic models are atomic.

If we look merely at substructures, however, everything breaks down immediately. It's a good exercise to show that every countable graph is an induced subgraph of an atomic countable graph. (HINT: think about gluing the starting graph to a copy of $\mathbb{N}$.)

OwnerDisplayName: anon

LastEditorDisplayName: anon

Title: Is submodel of atomic model atomic


答:
\begin{gather*}
I=\int x\arcsin x\,dx\\
\text{Let } x=\sin \theta \\
dx=\cos \theta \,d\theta \\
I=\int \sin \theta \cos \theta \,d\theta \cdot \theta =\frac{1}{2}\int \theta \sin 2\theta \,d\theta \\
\text{By integration by parts method,}\\
\int uv\,dx=u\int v\,dx-\int \frac{du}{dx}\left(\int v\,dx\right) dx\\
\text{Take } u=\theta \text{ and } v=\sin 2\theta .
\end{gather*}
Can you go ahead and take it from here?

Title: Integrating $\int x \arcsin x \,dx$ by parts

Body:
Solving the integral $$\int x \arcsin x\,dx$$

I know I have to solve integral using by parts

$$\int u\,dv = uv-  \int v\,du$$

$$\int x \arcsin x \,dx$$

Let $u = \arcsin x$ and $dv = x\,dx$

so $du = \dfrac1{\sqrt{1-x^2}} \, dx$

and $dv = x^2$.

but here is where I am stuck. because when I plug the variables into equation it becomes confusing on what to do next.

ps. New to using latex and still haven't perfected it so I apologize for formatting.


问:
Three dimensional graph to solutions of system of two differential equations
I've been reading Hirsch and Smale's Differential equations, dynamical systems and linear algebra.
The author mentions that it is possible to sketch the graph of a solution to a differential equation in two dimensions (say, $x' = f(t,x)$). This makes sense to me since we obtain a function $x(t)$ and we can simply graph this function.

However, the text then says that we can graph a solution to a system of two differential equations in 3 dimensions. I don't quite understand how this is possible. We will get as a solution to our system two functions, say $x(t)$ and $y(t)$. How can we plot these in three dimensions, since they both depend on the independent variable $t$? Does the author mean that we can plot the first solution in the $xt$ plane and the second in the $yt$ plane?

答: No. It means that you can graph the (parametric) curve $(t,x(t),y(t))$ in $txy$ space. If the equation is autonomous, you can project those curves onto the $XY$ plane (the so called phase plane) because they are invariant under $t$-translations.


问:
Show that operator is normal and determine its Singular Value Decomposition
could anybody please help me with the following task?

Consider the operator
  $$
Af(x):=\int\limits_{-\pi}^{\pi}\sin(x-y)f(y)\, dy, x\in [-\pi,\pi], f\in L_2(-\pi,\pi).
$$
  Show that the operator $A\in\mathcal{L}(L_2(-\pi,\pi))$ is normal. Determine the Singular Value Decomposition (SVD) of A.

In order to check if A is normal, I determined the adjoint operator with the result that
$$
A^* f(x)=\int\limits_{-\pi}^{\pi}\overline{\sin(x-y)} f(y)\, dy.
$$

Then I calculated $AA^*$ and $A^*A$. Here are my results:
$$
AA^* f(x)=\int\limits_{-\pi}^{\pi}\sin(x-y) \int\limits_{-\pi}^{\pi}\overline{\sin(y-z)} f(z)\, dz\, dy
$$
$$
A^*Af(x)=\int\limits_{-\pi}^{\pi}\overline{\sin(x-y)}\int\limits_{-\pi}^{\pi}\sin(y-z)f(z)\, dz\, dy
$$
And this is identical because $\sin(x)=\overline{\sin(x)}$.

Could you please write me in a comment if it is okay until now?

Thanks a lot.

Edit:

Concerning the SVD:

Is it right, that I have to determine the eigenvalues (resp. eigenfunctions) of $AA^*$?
Does one need the convolution theorem of the Fouriertransformation?
Explicitly:
To my opinion it is
$$
AA^*f(x)=(\sin\star A^*f)(x)
$$
and therefore
$$
AA^* f(x)=\lambda f(x)\Leftrightarrow \mathcal{F}(\sin\star A^*f)=(2\pi)^{1/2}\mathcal{F}(\sin)\cdot\mathcal{F}(A^*f)=\lambda\mathcal{F}(f),
$$
i.e.
$$
(2\pi)^{1/2}\mathcal{F}(\sin)\cdot\mathcal{F}(A^*f)=\lambda\mathcal{F}(f).
$$
Can one use that equation to determine now $\lambda$ resp. $f$?

Greetings

答:
Yes, Fourier transform is the way to go.

Recall the spectral theorem: a bounded operator on a Hilbert space is normal if and only it is unitarily equivalent to a multiplication operator. What you have here is a special case, with the implementing unitary being the Fourier transform $\mathcal{F}$.

As you noted, the convolution theorem says $\mathcal{F}$ maps convolution to multiplication. The Pontryagin dual in this case, $\mathbb{Z}$, is discrete. This means $A$ is diagonalizable, in particular normal. The $\mathcal{F}(h)$ for any $h \in L^1(\mathbb{T})$ is a sequence vanishing at infinity. So $A$ is compact and SVD makes sense.

In Fourier domain, your answer is very simple: $A$ is projection onto the basis element $\sin x$ (up to a scaling factor, possibly). So the eigenvalues of $A^*A$ is same as $A$: $1$ for the basis element projected on and $0$'s for every other element (again, you might need to adjust for a scaling factor that makes $\mathcal{F}$ unitary).


问:
Properties of prime ideals in ring extensions
Let $A \subseteq B$ be two commutative $\mathbb{C}$-algebras.
Let $q$ be an ideal of $B$ and denote $p=q \cap A$.

By [Lemma 3][1], if $A \subseteq B$ is an integral extension, then: $q$ is maximal in $B$ if and only if $p$ is maximal in $A$.

If $A \subseteq B$ is not necessarily integral, when is it possible to deduce the following half claim:
If $p$ is maximal (prime) in $A$, then $q$ is maximal (prime) in $B$.

I guess that this does not hold in general. Does one of the following additional conditions help:
(i) flatness of $A \subseteq B$; (ii) faithful flatness of $A \subseteq B$, see [this][2]; (iii) separability of $A \subseteq B$; (iv) $A$ and $B$ being UFD's.

Thank you very much!

Edit: One direction of the proof of Lemma 3 says:
"If $p$ is maximal then $A/p$ is a field and every element of $B/q$ is algebraic over $A/p$. Hence $B/q$ is a field" (so $q$ is maximal).

Therefore, if I am not wrong (but I may be wrong), the following claim is true:
If $A \subseteq B$ is algebraic, then: If $p$ is maximal in $A$, then $q$ is maximal in $B$.

Reason: If I am not worng, exactly the same proof holds; the critical line is:
and every element of $B/q$ is algebraic over $A/p$.

Am I right?

strong text
[1]: http://math.stanford.edu/~conrad/210BPage/handouts/math210b-going-up.pdf
[2]: Extensions and contractions of prime ideals under integral extensions

答:
For $(1), (2), (4)$, take $k \subseteq k[x]$ for any field.  This is a free (a fortiori faithfully flat) extension of UFDs in which the non-maximal ideal $0$ of $k[x]$ contracts to the maximal ideal $0$ of $k$.

I highly doubt that separability is even remotely related to the property in question, and will politely decline to think about it until you provide some motivation.

As for algebraic..... no, even if we restrict attention to $A, B$ domains this doesn't work out.  The key property of integral extensions in your "lemma 3" is that they are stable under quotients, but such is not the case for algebraic extensions.

For example $\mathbb{C}[x,xy] \subseteq \mathbb{C}[x,y]$ is an algebraic extension but the height $1$ (non-maximal) prime $(x)\mathbb{C}[x,y]$ contracts to the maximal ideal $(x,xy) \mathbb{C}[x,xy]$.


问:
Mobius transformation that maps $|z+i|=1$ to $|w|=4$
In Silverman's complex analysis textbook, he described how to find Mobius transformation that maps $|z+i|=1$ to $|w|=4$.

Let $T:\Bbb C_\infty\to\Bbb C_\infty$ be a Mobius transformation that we require. Then $Tz = \frac{az+b}{cz+d}$ for some $a,b,c,d\in\Bbb C$ such that $ad-bc \neq 0$. WLOG, let $T(-i) = \infty$. Then we can reduce the form to $Tz =\frac{az+b}{z+i}$. Since two poins $0,-2i$ in $|z+i| = 1$ should maps to points in $|w|=4$ and $T(0) = -ib$ and $T(-2i) = 2a+ib$, $|b| = 4$ and $\sqrt{4a^2+b^2} = 4$. Hence, we may choose $b = 4$ and $a = 0$. This gives $T(z) = \frac{4}{z+i}$. And then the book conclude that this is the desired map. I understand that $T$ maps circle to circle but circle that passing $4i$ and $-4i$ are infinitely many. How did he ensure this maps $|z+i| = 1$ to $|w| = 4$?

答: Note that if $|z+i|=1$, then $|T(z)| = \left|\frac 4{z+i}\right| = 4$.


问:
If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$
If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I put the value of $x= 0$ and the answer is coming out to be $13/6.$

But the actual answer is $2.$

Am I missing something ?

答: The derivative, $$\frac{2}{3}-\frac{3}{2 (x+1)^2}$$ is zero when $x=1/2$:


问:
Given a quadrilateral with 4 equal areas, prove that it is a parallelogram
I have the next quadrilateral with midpoints E F G H. The source of the problem is my class of geometry, I read the book but I don't find anything related to this.

I found in Google about Varignon's Theorem then I do the following construction:

My question is: it is correct to state that quadrilateral ABCD is a parallelogram? I think that is wrong because I don't use the areas of the statement.

答:
Let ABCD be a rhombus and EFGH be the rhombus formed by mid-points of sides.
Any projection of ABCD on an arbitrary plane has its area reduced by a factor $\cos \theta$ where this angle is between their plane normals.

The inner rhombus area is one-fourth the outer among all the possibilities.


答:
After completing the squares you can use the inequality between the arithmetic and geometric mean (AM-GM) in the form $a^2+b^2 \geq 2ab$:

Hence,

$$\frac{(2(x+1))^2+3^2}{6(x+1)}\stackrel{AM-GM}{\geq}\frac{2\cdot 2(x+1)\cdot 3}{6(x+1)}=2$$

Equality is reached for $2(x+1)=3 \Leftrightarrow x = \frac 12$.

Title: If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$

Body:
If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I put the value of $x= 0$ and the answer is coming out to be $13/6.$

But the actual answer is $2.$

Am I missing something ?


问:
Confusion about boundary point for polar integral
I'm having a bit of trouble with the following question.

Find the maximum height above the x-axis of the cardioid $r = 2(1 + \cos(\theta))$. What is the
area of the region that is inside of this cardioid and outside the circle $r = 6\cos(\theta) $?

I already found the maximum height. That wasn't difficult. However, I am not sure about the 2nd part.

The solutions does this and I get most of it. I got how the intersection points were found. That makes a lot of sense to me. I'm confused on how the $\theta = \frac{\pi}{2}$ and $\theta = \pi $ is found though. Where is that coming from and why is the integral split up at $\frac{\pi}{3}$ and again at $\frac{\pi}{2}$? Additionally, I notice the $\frac{5\pi}{3}$ isn't used anywhere.

I guess what I am confused about is the bounds for the integral. Why can't I just integrate from $\frac{\pi}{3}$ to $\frac{5\pi}{3}$?

Can someone clear up that part for me? I would greatly appreciate it.

答:
There are two things to understand -

i) The circle $r = 6 \cos \theta$ forms for $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

The circle is centered at $(3, 0)$ and has radius of $3$. Its equation is $(x-3)^2 + y^2 = 9$. When we represent this circle as $r = 6 \cos \theta$, we are basically taking distance of each point on the circle from the origin ($OA$ in the diagram). Geometrically, you can see $\angle OAB = 90^0, OB = 6$ and so $OA = r = 6 \cos \theta$. Now for $\theta = 0, OA = OB = 6$ and as we traverse counter-clockwise in the first quadrant, as $\theta$ increases $OA$ reduces and at $\theta = \frac{\pi}{2}, OA = 0 $ as  $A$ and $O$ coincide. The same logic in the fourth quadrant.

ii) We have to find area of cardioid outside of circle

I will only talk about the area above x-axis. We can double the area as the curves are symmetric about x-axis. Area of cardioid outside of the circle is only between $\frac{\pi}{3} \leq \theta \leq \pi$.

Now given the circle is only defined between $\frac{\pi}{3}$ and $\frac{\pi}{2}$, we need to divide the integral into two. The way the solution does is that it calculates the area of the cardioid between $\frac{\pi}{3}$ and $\pi$ and then subtracts the area of the circle ($\frac{\pi}{3} \leq \theta \leq \frac{\pi}{2}$) from it. Then it doubles it as the same area is formed below x-axis too.

You could also write the integral to find area as below,

$\displaystyle 2 \int_{\pi/3}^{\pi/2} \int_{6\cos\theta}^{2(1+\cos\theta)} r \ dr \ d\theta + 2 \int_{\pi/2}^{\pi} \int_0^{2(1+\cos\theta)} r \ dr \ d\theta$


问:
When to use Inclusion Exclusion principle?
I'm a bit confused as to when you should use the Incl/Excl principle. Take for example the following problem:

How many 13-card hands can be selected from the standard 52-card deck with exactly four spades or exactly four diamonds?

My university tutor solved this problem by using the principle. My theory would be that you need to use it in this case because a 13-card hand with exactly four spades can also contain four diamonds, and a hand with exactly four diamond can also contain four spades - so you end up counting too many times.

Is there a trick or a general rule for when you should use it?

Sub-question: My tutor wrote this for choosing exactly four spades, and I believe there might be a mistake:

$c(13,4) * c(9,39)$

$c(13,4)$ is the number of ways to choose four positions for the spades

$c(9,39)$ is the number of ways to choose the remaining 9 cards that are not spades

But isn't it missing another $c(13,4)$ to choose the four spades out of 13 spades?

答:
Don't really like to give general rules or tricks, since then looking for them can interfere with the analysis of a problem.  But the presence of or means we are trying to count the union of two sets $A$ and $B$.  And sometimes the best way to count a union is to count $A$, count $B$, count what they have in common, and do the obvious thing.  Your jntuitive explanation of the "why" in this case was clear and correct.

Your tutor's answer is right, but your description of it is not.  We are counting the number of hands, order is irrelevant.  The $c(13,4)$ in the answer comes not from the location of the spades.  It actually counts the number of ways of choosing $4$ spades from the $13$.  For every way of choosing the spades, there are $c(39,9)$ ways to choose the remaining cards (you kind of typed it backwards), so the number of $4$-spade hands is
$$C(13,4)C(39,9)$$


问:
Thom class: Why are the two definitions equivalent?
We know that the Thom class $\tau_W$ is defined on a disk bundle $W\rightarrow L$, where $L$ is a $p$-dimensional manifold and the rank of $W$ is $k$.

Let $[W]_0$ denote the fundamental class of the pair $(W,W\setminus L)$ or better, $(W,\partial W)$ as both are homotopy equivalent pairs.

Definition 1: $\tau_W$ is the Poincaré dual to the canonical zero section of $W$ w.r.t this class. That is, $\tau_W$ is the unique class satisfying $\tau_W\cap [W]_0=[L]$ where $L$ is the zero section of the disk bundle.

Definition 2: $\tau_W$ is the unique class that restricts to the positive generator of $H^k(D^k,\partial D^k)$

I tried arguing that if $W$ has a simplicial structure such that the fibre at some point is the $k^{th}$ front face and $L$ is the $p^{th}$ back face, then the cap product definition itself, straightaway gives the equivalence, if I start from Definition 2 and arrive at the former definition.

My question: Is there such a simplicial structure? I am thinking whether cell structures help.

OR

Is there a better way to prove the equivalence?

答:
A way of seeing this is to see $W$ as a vector bundle,  so that $H^\ast(W,W-L)$ is the same thing as $H^\ast_{c}(W)$, the compactly  supported cohomology. Now taking the cap-product with $[W_0]$ ressembles integrating along the fibre, and saying that $\tau_W$ is the unique class that restricts to 1 in $H^k_c(\mathbb{R}^k)$ amounts to say that the integration of $\tau_W$ along the fibers is the constant 1 function on $[L]$. Then juggling with Poincaré duality gives the answer.

For a more precise treatment, I recommend reading the chapter on the Thom isomorphism in Bott and Tu's excellent Differential Forms in Algebraic Topology.


问:
Prove the solution to this ODE is genuine
$$y'=\dfrac {x(1+e^y)}{e^y(1+x^2)}$$
I believe I have the correct solution
$$y=\ln ({C\sqrt {1+x^2}-1})$$
to the ODE shown but can't seem to prove the solution is genuine when substituting back into the original equation.
enter image description here

Body:
$$y=\ln ({C\sqrt {1+x^2}-1})$$
$$e^y={C\sqrt {1+x^2}-1}$$
Differentiate:
$$e^yy'=\dfrac {Cx}{\sqrt {1+x^2}}$$
Now the DE is:
$$e^yy'=\dfrac {x(1+e^y)}{(1+x^2)}$$
$$\dfrac {Cx}{\sqrt {1+x^2}}=\dfrac {x(1+e^y)}{(1+x^2)}$$
$$C=\dfrac {(1+e^y)}{\sqrt {1+x^2}}$$
$$C=\dfrac {C(\sqrt {1+x^2})}{\sqrt {1+x^2}}$$
$$C=C$$
Your solution is correct.


答:
If $y = \ln(C\sqrt{1+x^2} - 1)$, then $e^y = C\sqrt{1+x^2} - 1$. Now we verify whether this solution is correct:
$$\begin{align} \textrm{LHS} &= y'\\&= \frac{d}{dx} \ln(C\sqrt{1+x^2} - 1)\\&=\frac{1}{C\sqrt{1+x^2} - 1}\times\frac{1}{2\sqrt{1+x^2}}\times 2Cx\\ &= \frac{1}{e^y}\times \frac{Cx}{\sqrt{1+x^2}}\times \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\\&= \frac{x}{e^y} \times\frac{C\sqrt{1+x^2}}{1+x^2}\\&=\frac{x}{e^y}\times\frac{e^y+1}{1+x^2} \\&= \textrm{RHS}\end{align}$$
Thus the solution you found satisfies the ODE.

Title: Prove the solution to this ODE is genuine

Body:
$$y'=\dfrac {x(1+e^y)}{e^y(1+x^2)}$$
I believe I have the correct solution
$$y=\ln ({C\sqrt {1+x^2}-1})$$
to the ODE shown but can't seem to prove the solution is genuine when substituting back into the original equation.
enter image description here


问:
How differentiate the stereographic projection map
Let $f:\mathbb{S^2} \setminus {(0,0,1)}\to\mathbb{R^2} $ be the stereographic projection of a sphere onto a plane.

Find the derivative of $f$.

Note: You should get a linear map from 2-d space to 2-d space

=====================================================================
I'm new to the idea of differentiation of manifolds.

First I need two manifolds M and N and a smooth map between them

$M$ is the $\mathbb{S^2} \setminus {(0,0,1)} $ paramaterized as $\{(x,y,z)|x^2 + y^2 + z^2 = 1\}$

$N$ is the equatorial plane $\{z=0\}$

Now I need a smooth map between the two manifolds $f:M\to N $ $\space where \space(x,y,z)\mapsto (x,y)$
.

Now I need to associate each point $s\in M$ with a tangent space $TM_x$

This lets us define the derivative $df_x:TM_x\to TN_y$

According to my book (Topology from the Differentiable Viewpoint)
we choose a parametrization where $U$ is a subset of ${R^m}$ and M is a subset of ${R^k}$.

$g:U\to M$ of a neighborhood g(U) of x in M, with g(u) = x and the derivative is
$dg_u: {R^m} \to {R^k}$

We have our parametrization of $M$, $\{(x,y,z)|x^2 + y^2 + z^2 = 1\}$

$TM_x$ is the image of $dg_x \circ dg_u$

Now I am a little confused. I'd like a concrete example like this to work from where I can say,

This is my a manifold M, this is another manifold N, I have a smooth map between them, this is how I generate the tangent space, and this is how I calculate the derivative

答:
The stereographic projection map is

$f:\mathbb{S^2} \setminus {(0,0,1)}\to\mathbb{R^2} $

$(x,y,z) \mapsto (\frac{x}{1-z},\frac{y}{1-z})$

When we try and find our Jacobian, we know we want a $2 \times2$ matrix, but we have three variable?

When we make our Jacobian, we want local coordinates

$x^2+y^2+z^2=1 \leftarrow \rightarrow z=\sqrt{1-x^2 - y^2}$

so we can define our map as

$f:\mathbb{S^2} \setminus {(0,0,1)}\to\mathbb{R^2} $

$(x,y,z) \mapsto (\frac{x}{1-\sqrt{1-x^2-y^2}},\frac{y}{1-\sqrt{1-x^2-y^2}})$

which is in two variables

The Jacobian becomes

$$
    \begin{pmatrix}
    \frac{\partial}{\partial x} (\frac{x}{1-\sqrt{1-x^2-y^2}}) & \frac{\partial}{\partial y} (\frac{x}{1-\sqrt{1-x^2-y^2}})\\
    \frac{\partial}{\partial x}(\frac{y}{1-\sqrt{1-x^2-y^2}}) & \frac{\partial}{\partial y}(\frac{y}{1-\sqrt{1-x^2-y^2}})\\
    \end{pmatrix}
$$

Computing this, compute the determinant, and we are finished.


问:
Use mathematical induction to prove that for all natural numbers $4^n < 5^n$?
I'm stuck at the inductive step, which is $4^{k+1} < 5^{k+1}$ , where $k> 0$. Any help would be much appreciated on how to prove my inductive hypothesis.

Body: $$ 4^{k+1} = 4\cdot4^k < 4\cdot 5^k < 5\cdot5^k =5^{k+1}$$


问:
How many length-$n$ bitstrings containing $3$ consecutive $0$s and $4$ consecutive $1$s are there?

How many length-$n$ bitstrings containing $3$ consecutive $0$s and $4$ consecutive $1$s are there?

I thought that $a_n$ can be constructed in three ways:

The strings that contain both three consecutive $0s$ and four consecutive $1s$ and end up with a $0$. Hence, there are $a_{n-1}$ such  strings.

The strings that contain both three consecutive $0s$ and four consecutive $1s$ and end up with a $1$. Hence , there are $a_{n-1}$ such  strings.

The strings that end up with $\dots1111000$ or $\dots0001111$ . There are $2 \times 2^{n-7} = 2^{n-6}$ such strings.

As a result ,the number of length-$n$ strings that contain both three consecutive $0s$ and four consecutive $1s$ is equal to $$a_n = 2a_{n-1}+2^{n-6}$$ with $a_0, a_1,\dots, a_6 =0$ and $a_7 = 2$.  Is this correct? If not, can you write correct closed formula?

答:
Here's a recursive approach that might be useful.

Let $b_n$ be the number of length-$n$ bit strings that contain 3 consecutive 0s.  By conditioning on the prefix 1, 01, 001, or 000, we find that
$$b_n =
\begin{cases}
0 &\text{if $n < 3$} \\
b_{n-1} + b_{n-2} + b_{n-3} + 2^{n-3} &\text{if $n \ge 3$}
\end{cases}
$$
Note that $b_n$ is https://oeis.org/A050231.

Similarly, let $c_n$ be the number of length-$n$ bit strings that contain 4 consecutive 1s.  By conditioning on the prefix 0, 10, 110, 1110, or 1111, we find that
$$c_n =
\begin{cases}
0 &\text{if $n < 4$} \\
c_{n-1} + c_{n-2} + c_{n-3} + c_{n-4} + 2^{n-4} &\text{if $n \ge 4$}
\end{cases}
$$
Note that $c_n$ is https://oeis.org/A050232.

Now let $a^0_n,a^{00}_n,a^1_n,a^{11}_n,a^{111}_n$ be the number of length-$n$ bit strings that contain 3 consecutive 0s and 4 consecutive 1s and start with 0, 00, 1, 11, and 111, respectively.  Then conditioning on whether the next bit is 0 or 1 yields
\begin{align}
a_n &= a^0_n + a^1_n \\
a^0_n &= a^{00}_n + a^1_{n-1} \\
a^{00}_n &= c_{n-3} + a^1_{n-2} \\
a^1_n &= a^0_{n-1} + a^{11}_n \\
a^{11}_n &= a^0_{n-2} + a^{111}_n \\
a^{111}_n &= a^0_{n-3} + b_{n-4}
\end{align}


问:
How to integrate $\int \frac{8}{16-e^{4x}} \mathrm dx$ using trigonometric substitution?
How do I integrate this equation using trigonometric substitution?

$$\int \frac{8}{16-e^{4x}} \mathrm dx.$$

So far I figured out that $a = 4, u = e^{2x}$ then $e^{2x}=4\sin\theta \mathrm d\theta$ and $\sqrt{16-e^{4x}}$ and $2e^{2x}dx=4\sin\theta$. I don't know how to proceed from this because I don't know if $e^{2x}$ can be considered a constant and can be factored out of the integral or I have to use logarithmic functions.

Edit: I tried solving it, I had a final answer of
$$\frac{e^{2x}}{4\sqrt{16-e^{4x}}}+C.$$
Am I correct?

答:
In general，we have
$$\int \frac{1}{a-ce^{bx}}\mathrm{d}x＝\frac{bx-\log{(a-ce^{bx})}}{ab}$$
Proof: substitute $u＝bx$,then
$$I＝\int \frac{1}{a-ce^{bx}}\mathrm{d}x＝\frac{1}{b}\int \frac{1}{a-ce^u}\mathrm{d}u$$
substitute $t＝e^u$,
$$I＝\frac{1}{b}\int \frac{1}{t(a-ct)}\mathrm{d}t $$
The rest of the work is simple.


答:
$$
\int \frac{8}{16-e^{4x}} \, dx=\left| u=e^{2x} \atop du=2e^{2x}\,dx=2u\, dx  \right|
=\int\frac{8}{16-u^2} \cdot \frac{du}{2u}
$$
(we assume that $u\in(-4,4)$)
$$
=\left| u=4\sin t \atop du=4\cos t\,dt  \right|=
\int\frac{8}{16-16\sin^2t} \cdot \frac{4\cos t\,dt}{8\sin t}
=\frac14 \int \frac{\cos t\,dt}{\cos^2t \sin t}
$$
$$
=\frac12 \int \frac{dt}{2\sin t\cos t}=
\frac12 \int \frac{dt}{\sin 2t}=
\left| z= 2t \atop dz= 2\,dt  \right|=
$$
$$
=\frac14 \int \frac{dz}{\sin z}=
\frac14\ln\left| \tan\left( \frac{z}2 \right) \right|+C
=\frac14\ln\left| \tan t \right|+C
$$
$$
=\frac14\ln\left| \tan\left( \arcsin\frac{u}4 \right) \right|+C
=\frac14\ln\left| \tan\left( \arcsin\frac{e^{2x}}4  \right) \right|+C
$$
Since
$$
\tan^2(\arcsin x)=\frac1{\cot^2(\arcsin x)}=
\frac1{\frac1{\sin^2(\arcsin x)}-1}=
\frac1{\frac1{x^2}-1}=\frac{x^2}{1-x^2}
$$
and (considering the sign)
$$
\tan(\arcsin x)= \frac{x}{\sqrt{1-x^2}}
$$
the answer is
$$
\frac14\ln\left| \frac{e^{2x}/4}{1-\sqrt{\frac{e^{4x}}{16}}}  \right|+C
=\frac{x}2-\frac14\ln\left| \sqrt{16-e^{4x}} \right|+C
$$
$$
=\frac{x}2-\frac18\ln\left| 16-e^{4x} \right|+C.
$$

Title: How to integrate $\int \frac{8}{16-e^{4x}} \mathrm dx$ using trigonometric substitution?

Body:
How do I integrate this equation using trigonometric substitution?

$$\int \frac{8}{16-e^{4x}} \mathrm dx.$$

So far I figured out that $a = 4, u = e^{2x}$ then $e^{2x}=4\sin\theta \mathrm d\theta$ and $\sqrt{16-e^{4x}}$ and $2e^{2x}dx=4\sin\theta$. I don't know how to proceed from this because I don't know if $e^{2x}$ can be considered a constant and can be factored out of the integral or I have to use logarithmic functions.

Edit: I tried solving it, I had a final answer of
$$\frac{e^{2x}}{4\sqrt{16-e^{4x}}}+C.$$
Am I correct?


问:
Trefoil knot cannot be injectively projected to a plane?
I'm trying to do the following geometry qual problem from my university:

Give an example of a smooth embedding $f:S^1\rightarrow\mathbb{R}^3$, such that for each plane $P$ in $\mathbb{R}^3$, $\pi_P\circ f$ is not injective; here $\pi_P$ is the orthogonal projection of $\mathbb{R}^3$ onto $P$.

My guess is that the trefoil knot will work, but since I have zero knowledge of knot theory, can someone please let me know if there is a simple argument to see if (or why) this works? Alternatively, is there a simple example that I missed? Thank you!

Edit: Now I'm sort of confident that this is due to $R^3$-trefoil knot and $R^3-S^1$ have different fundamental groups. I'm still not sure about the details here. Can someone please help me finish the argument? Thank you!

答:
You can prove this in a variety of ways using some machinery from knot theory, but here is a more elementary approach.  Let's think about the question more geometrically.  To say that $\pi_P\circ f$ is not injective means that there are two distinct points $p$ and $q$ in the image of $f$ such that the vector $p-q$ is orthogonal to $P$.  So, we just need to arrange that the function $g:S^1\times S^1\to\mathbb{R}^3$ given by $g(x,y)=f(x)-f(y)$ hits every possible direction in $\mathbb{R}^3$.  Or, letting $U=\{(x,y)\in S^1\times S^1:x\neq y\}$, we can restrict to $U$ and normalize to get a smooth map $h:U\to S^2$ given by $h(x,y)=\frac{f(x)-f(y)}{\|f(x)-f(y)\|}$, and we just need this map $h$ to be surjective.

Now the basic idea is, $U$ is a smooth 2-dimensional manifold, and so is $S^2$, so it shouldn't be too hard to make $h$ surjective: generically, $h$ will be open, and we just need enough little open patches in its image to cover all of $S^2$.  In particular, let's think about the local behavior of $h$ near a point $(a,b)\in U$, where we may assume without loss of generality that $f(a)=(1,0,0)$ and $f(b)=(0,0,0)$.  Note that if you change the second coordinate of $f(a)$ a bit and the third coordinate of $f(b)$ a bit, then the vectors $f(a)-f(b)$ will trace out an entire rectangle tangent to the unit sphere at $(1,0,0)$.  When you project this onto the sphere, it will cover an entire open neighborhood $V$ of $(1,0,0)$.

So we've just discovered that if we pick our $f$ so that $f(a)=(1,0,0)$ and $f(b)=(0,0,0)$ for some $a$ and $b$ and that $f$ is moving along certain straight paths near $a$ and $b$, then the image of $h$ near $(a,b)$ will contain a nonempty open set $V\subseteq S^2$.  Now by compactness, $S^2$ is covered by finitely many rotations of $V$.  So to get $h$ to be surjective, we just need $f$ to contain certain finitely many rotations of our pair of straight paths.  By translating them, we can easily arrange that these rotated paths are all disjoint from each other, and then we have to just interpolate smoothly between them to get our loop $f$.


问:
For what values does the geothmetic meandian converge?
The geothmetic meandian, $G_{MDN}$ is defined in this XKCD as

$$F(x_1, x_2, ..., x_n) = \left(\frac{x_1 +x_2+\cdots+x_n}{n}, \sqrt[n]{x_1 x_2 \cdots x_n}, x_{\frac{n+1}{2}} \right)$$

$$G_{MDN}(x_1, x_2, \ldots, x_n) = F(F(F(\ldots F(x_1, x_2, \ldots, x_n)\ldots)))$$

The comic also (correctly) claims that $G_{MDN}(1, 1, 2, 3, 5) \approx 2.089$

There are two convergence questions I'm interested in:

For what values of $(x_1, x_2, \ldots, x_n)$ does $G_{MDN}$ converge to a single number?

For what values of $(x_1, x_2, \ldots, x_n)$ does $G_{MDN}$ converge, but not to a single number?

I've written up Python 3 code so that you can test out numbers yourself by changing the values at the bottom of the code. If the code never seems to stop running, then $G_{MDN}$ does not converge for your input. (In these situations, you can set verbose=True to see what's happening.)

```<code># assumes Python 3 because I assume that "/" always means float division
from typing import Iterable, Tuple, Union
from decimal import Decimal
import math
from functools import reduce  # Required in Python 3
import operator

# from https://stackoverflow.com/a/48648756
def prod(iterable):
    return reduce(operator.mul, iterable, 1)

def geothmetic_meandian(nums: Iterable[float], verbose=False) -> Tuple[bool, Tuple[float, float, float]]:
    def inner_geothmetic_meandian(nums: Iterable[float]) -> Tuple[float]:
        arithmetic_mean = sum(nums)/len(nums)
        geometric_mean = prod(nums)**(1 / len(nums))
        sorted_nums = sorted(list(nums))
        if len(nums) % 2 == 0:
            # even number of numbers
            higher_median_index = int(len(nums) / 2)
            lower_median_index = higher_median_index - 1
            median = (sorted_nums[higher_median_index] + sorted_nums[lower_median_index]) / 2
        else:
            # odd number of numbers
            median = sorted_nums[int((len(nums) - 1) / 2)]

        return (arithmetic_mean, geometric_mean, median)

    last_ans = None
    ans = inner_geothmetic_meandian(nums)
    converged = True
    while not (ans[0] == ans[1] == ans[2]):
        if ans == last_ans:
            converged = False
            break
        last_ans = ans
        ans = inner_geothmetic_meandian(ans)

        if verbose:
            print(ans)

    return converged, ans

if __name__ == "__main__":
    verbose = False
    values = (1, 1, 3, 2, 5)
    converged, results = geothmetic_meandian(values, verbose=verbose)
    if converged:
        print(f"The geothmetic meandian of {values} converged to: {results[0]}")
    else:
        print(f"The geothmetic meandian of {values} did not converge to a single value:\nArithmetic Mean: {results[0]}\nGeometric Mean:  {results[1]}\nMedian:          {results[2]}")
</code>```

I've tested this code to verify that $G_{MDN}(1, 1, 2, 3, 5) \approx 2.089$.

I have also not found any inputs that cause the program to not converge at all.

However, I have found that $G_{MDN}(1, 2, 3, 4, 5) = (2.8993696858822964, 2.899369685882296, 2.8993696858822964)$ which would mean that $(1, 2, 3, 4, 5)$ is in the second convergence class, where it converged, just not all to the same number. But I'm worried that this result is due to a rounding error in Python itself. (I quickly tried and failed to use the Decimal class due to the nth root operation.)

Thus, my hunch is that all inputs converge to a single number, but I have not been able to prove this yet. It looks like an epsilon-delta proof.

答:
Let
$$a_k=\min(F^{(k)}(x_1, x_2, ..., x_n))$$
and
$$b_k=\max(F^{(k)}(x_1, x_2, ..., x_n))$$
It follows from their definitions that the arithmetic and geometric means and the median of $F^{(k)}(x_1, x_2, ..., x_n)$ are all $\ge a_k$ and $\le b_k$, so $a_{k+1}\ge a_k$ and $b_{k+1}\le b_k$. We also know that $a_k$ and $b_k$ are bounded, so they must be convergent.

Let $\lim_{k\to\infty}a_k=a$ and $\lim_{k\to\infty}b_k=b$. Suppose for contradiction that $a\neq b$. Using the definition of limits, it’s fairly easy to show that you can pick $k$ such that the arithmetic and geometric means of $F^{(k)}(x_1, x_2, ..., x_n)$ are both greater than $a$ and less than $b$, which means that at least one of $a<a_{k+1}<b$ or $a<b_{k+1}<b$ is true, which contradicts $a_k$ is non-decreasing, $b_k$ is non-increasing, and their limits are $a$ and $b$. Thus $a=b$, and $G_{MDN}$ converges to a single number for all possible values.


问:
Deferred Annuity not working
A simple financial math problem:

Mack obtains $500\ 000$ repayable over $20$ years. If interest is compounded monthly at $9.25\%$ per annum, determine the monthly repayments if the repayment begins in $6$ months time.

I used the formula:

$$P_v = x[(1-(1+i)^{-n})/i]$$

but I'm not getting the right value.

What am I doing wrong?

答:
Because repayment begins in $6$ months, the effective principal is

$$P = 500,000\, \left ( 1 + \frac{i}{12} \right )^6$$

where $i=0.0925$ is the annual interest rate.  The monthly payment is then

$$m = \frac{P \, (i/12)}{1-\left [ 1 + (i/12) \right ]^{-240}}$$

Plugging in the numbers, I get a monthly payment of about $\$4795.25$.

Title: Deferred Annuity not working


问:
Finding the remainder of $2^{2021}$ divided by $2700$
Finding the remainder of $2^{2021}$ divided by $2700.$ I can't solve this because $2700= 3\cdot 9\cdot 100$ and we cannot use Little Fermat.

Body:
As $(2^{2021},2700)=4$

let us find $x=2^{2021-2}\pmod{\dfrac{2700}4}$

Now as $\phi(27)=18,2019\equiv3\pmod{18}, x\equiv2^3\pmod{27}$

and $\phi(25)=20,2019\equiv-1\pmod{20}, x\equiv2^{-1}\pmod{25}\equiv13$

We need $x=8+27a=13+25b\iff27(a+1)-25b=5+27=16(27-25)$

$$\iff27(a+1-16)=25(b-16), \dfrac{25(b-16)}{27}=a+1-16\text{ which is an integer}$$

So, $27$ must divide $b-16,b=16+27d$(say)

$x=13+25(16+27d)\equiv13+400\pmod{27\cdot25}$

Like How to find last two digits of $2^{2016}$,

multiply all sides by $2^2$ to find $$2^{2019+2}\equiv2^2\cdot413\pmod{2^2\cdot27\cdot25}$$


问:
How to determine the leading order asymptotic behaviour of this integral?
I tried to put it in the form f(t) exp(tx) to apply watson's lemma but I couldn't. Is there a trick?

$$\int_{0}^{1} e^{t}\left ( \frac{t}{1+t^{2}} \right )^{x} dt$$  as $ x \rightarrow {\infty} $

答:
$$I_x=\int_{0}^{1} e^{t}\left ( \frac{t}{1+t^{2}} \right )^{x} dt$$
$$\frac{t}{1+t^{2}}=\frac 1 2 \left(\frac 1 {t+i}+\frac 1 {t-i} \right)$$

$$I_x=\frac 1 {2^x}\int_{0}^{1} e^{t}\left(\frac 1 {t+i}+\frac 1 {t-i} \right)^x \,dt$$ So, for an integer value of $x$, you face linear combinations of exponential integral functions of conjugate complex arguments.

For example
$$I_3= \frac {1-e}4+\frac {A_3-B_3}{16}$$
$$A_3=(-1+3 i)\, e^{+i}\, \big[\text{Ei}(-i)-\text{Ei}(1-i)\big]$$
$$B_3=(+1+3 i) \,e^{-i} \, \big[\text{Ei}(+i)-\text{Ei}(1+i)\big]$$ Working the $I_x$, you could easily find interesting patterns.

As a result, for large values of $x$
$$I_x \sim \frac K{2^x}\qquad \text{with} \qquad K\sim \frac 1 3\color{red}{\text{    Wrong}}$$

Computing exactly for $x=100$, the value is $2.32\times 10^{-21}$ while the approximation gives $2.63\times 10^{-21}$.

Edit

After @Maxim's comments and @mathstackuser12's edit, I checked again my calculations and found some severe comptational errors.

I just performed a quick and dirty nonlinear regression for the model
$$I_x= a\, x^b\, 2^{-x}$$ for the range $5 \leq x \leq 1000$ (step=$5$) and obtained with $R^2=0.998378$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence
   Interval} \\
 a & +1.88546 & 0.01574 & \{+1.85441,+1.91650\} \\
 b & -0.41209 & 0.00175 & \{-0.41555,-0.40864\} \\
\end{array}$$


问:
Wirtinger derivative of composition of functions
So I have a very basic question : let $h : \mathbb{R} \rightarrow \mathbb{R}$ be a $C^1$ function, and let $g : \mathbb{C} \rightarrow \mathbb{R}$ be defined by $g(z)=h(z \overline{z})$. I want to compute $\frac{\partial g}{\partial \overline{z}}$ (which is defined because $g$ is defined on $\mathbb{C}$). I know the chain rule for $\overline{\partial}$. However, it involves terms like $\frac{\partial h}{\partial z}$ and $\frac{\partial h}{\partial \overline{z}}$, which confuses me because $h$ is only defined on $\mathbb{R}$. </p>

<p>Any help appreciated ! </p>

答:
You're absolutely right to be skeptical of the expressions $\partial h/\partial z$ and $\partial h/\partial\overline{z}$. Just like you say, since $h$ is only defined on $\mathbb{R}$, it isn't possible to take these derivatives. There are a couple of things you might do here in order to solve your problem.

(1) You can extend $h$ to a function $\tilde{h}$ on $\mathbb{C}$ in some way, and then define $g(z) = \tilde{h}(z\overline{z})$. This of course doesn't change the function $g$ in any way, but now you'll be able to take derivatives $\partial\tilde{h}/\partial z$ and $\partial\tilde{h}/\partial \overline{z}$. The most natural extension $\tilde{h}$ is the one given by $\tilde{h}(z) := h(\Re z)$.

(2) You can use the ordinary real chain rule as follows. Let $f(z) = z\overline{z}$, or, in terms of $x$ and $y$, this is $f(x,y) = x^2 + y^2$. Then $g = h\circ f$, so $$\frac{\partial g}{\partial x} = (h'\circ f)\frac{\partial f}{\partial x}\,\,\,\,\,\mbox{ and }\,\,\,\,\,\frac{\partial g}{\partial y} = (h'\circ f)\frac{\partial f}{\partial y}.$$ You can then use these expressions to compute $\partial g/\partial z$ and $\partial g/\partial\overline{z}$.


问:
Polar Form of a Matrix (functional calculus )
Let $T=
    \begin{pmatrix}
    1 & 1  \\
    0 & 0  \\
    \end{pmatrix}
$.

Find the polar form of $T= W|T|$, where $|T|= \sqrt{T^*T}$, and $W$ is unitary.

I was able to find

$|T|=
    \begin{pmatrix}
    \frac{1}{\sqrt{2}} &  \frac{1}{\sqrt{2}}  \\
    \frac{1}{\sqrt{2}} &  \frac{1}{\sqrt{2}}  \\
    \end{pmatrix}
$.

But I'm having trouble finding $W$, since $|T|$ here is not invertible.

Any help will be appreciated.

答:
The matrix
$$
W = \frac{1}{\sqrt{2}} \begin{pmatrix}
1 & 1
\\
1 & -1
\end{pmatrix}
$$
solves your equation.


问:
The radius of a circle, having maximum area, which touches the curve $y = 4 – x^2$ and the lines, $y = |x|$ is
The radius of a circle, having maximum area, which touches the curve $y = 4 – x^2$ and the lines, $y = |x|$ is

Note this is different from: this which is asking for minimum area.

If you click on the link and see the comments section of the answer the person said the area of the circle with maximum area touches the parabola at (0,4) and intersects the parabola at two other points and along with this touches the curve y = |x| at two distinct points.

My question is: isn't the circle with maximum area the one touching the parabola at (0,4) and touching the curve y=|x| at (0,0)? Can we call such a circle touching both y=|x| and the parabola?

Here is a diagram:

Here purple is the line y =|x|, green is the parabola, black is the circle with minimum radius (which I agree with) and red is the one they are claiming to be circle with the maximum radius which I'm not sure. Shouldn't it be the circle $x^2 + (y-2)^2 = 4$? That circle touches y=|x| at (0,0) and parabola at (0,4). The only difference is the derivative at the touching point isn't the same. What is right? Please clarify.

答:
The equation of the circle touching  $y=\pm x$ and center on $y$-axis be
$$x^2+(y-h)^2=h^2/2~~~(1)$$
Now if ith has to touch $y=4-x^2$, then the equation
$$4-y+(y-h)^2=h^2/2 \implies y^2-(2h+1)y+h^2/2+4=0$$ should have a unique root of $y$. Imposing $B^2=4AC$, we get
$$2h^2+4h-15=0 \implies h=\frac{-4\pm \sqrt{136}}{4}$$
Choosing positive value we have $h=\frac{\sqrt{136}-4}{4}$ in (1).

For the other possibility, (1) should give $y(0)=4$ and $h<4$.
This gives $(4-h)^2-h^2/2 \implies h(1\pm 1/\sqrt{2})=4$, we get
$$h=\frac{4 \sqrt{2}}{\sqrt{2}\pm1}~~~~(2)$$, in (1), choosing $+$ sign.

EDIT
So maximum radius from (2)  $$r=h/\sqrt{2}=\frac{4}{\sqrt{2}-1}.$$
as found by @Bright Star, below.


问:
How to solve the integral $\int \frac{y}{\cos^2(y)}~dy$.
I would like to compute the integral of

$$\int \frac{y}{\cos^2(y)}~dy.$$

Here is what I tried:

$$u=\cos(y),\quad du=-\sin(y)~dy,$$

$$dv=y~dy,\quad v=y^2/2,$$

$$uv-\int v~du=\frac{y^2}{2}\cos(y)+\int \frac{y^2}{2}\sin(y)~dy.$$

The expected answer is
$$y(\tan y)+\ln(\cos(y)).$$

How can I get to this? Thanks guys.

答:
Use integration by parts using
$$u=y \qquad du=dy$$
$$dv=\frac {dy}{\cos^2(y)}\implies v=\tan(y)$$ as a result
$$y\tan(y)-\int \tan(y) \,dy=y\tan(y)-\int \frac {\sin(y)}{\cos(y)} \,dy$$ which looks simple.


答:
Two curves $y=f(x)$ and $y= g(x)$ are said to be touching at $(x_0,y_0)$ if $y_0=f(x_0)=g(x_0)$ and $f'(x_0)=g'(x_0)$.

In your case:  Indeed the circle $x^2 + (y-2)^2 = 4$ and straight lines $y=|x|$ meet at $(0,0)$ but the function $y=|x|$ does not have a unique tangent at $(0,0)$.

Body:
The radius of a circle, having maximum area, which touches the curve $y = 4 – x^2$ and the lines, $y = |x|$ is

Note this is different from: this which is asking for minimum area.

If you click on the link and see the comments section of the answer the person said the area of the circle with maximum area touches the parabola at (0,4) and intersects the parabola at two other points and along with this touches the curve y = |x| at two distinct points.

My question is: isn't the circle with maximum area the one touching the parabola at (0,4) and touching the curve y=|x| at (0,0)? Can we call such a circle touching both y=|x| and the parabola?

Here is a diagram:

Here purple is the line y =|x|, green is the parabola, black is the circle with minimum radius (which I agree with) and red is the one they are claiming to be circle with the maximum radius which I'm not sure. Shouldn't it be the circle $x^2 + (y-2)^2 = 4$? That circle touches y=|x| at (0,0) and parabola at (0,4). The only difference is the derivative at the touching point isn't the same. What is right? Please clarify.


问:
Choosing Hymns -- Why Am I Wrong?
Let's say a choir director must select six hymns for a Sunday church service.  She has three hymn books, each containing 25 hymns, for a total of 75 different hymns.  The question asks for how many ways can she select the six hymns if she chooses at least one hymn from each book.

The correct answer given is as follows:

$$
3 \binom{25}{1}^2 \binom{25}{4} + 6 \binom{25}{1} \binom{25}{2} \binom{25}{3} + \binom{25}{2}^2
$$

I can see why that's correct.  But I want to understand why my following approach is wrong.

Specifically, I thought, let's first choose one hymn from each book.  There are 25 choices for each book, so there are $25^3$ ways to choose one hymn from each book.  Then, we just need to choose three of the hymns from the remaining 72 hymns--since we already have the three hymns from each book selected, it doesn't matter from where we get the remaining hymns.  So, in my mind, the answer should be:

$$
25^3 \times \binom{72}{3}
$$

That's clearly not the same answer as the given, and correct, answer, though.  I know I'm missing something basic, so any help would be greatly appreciated.  Thank you!

(For reference, this is Exercise 4(b) from the Supplementary Exercises for Ch. 1 in Grimaldi.)

答:
This is very common error in combinatorics. The problem is that your method does not uniquely specify the choice of hymns; certain selections are being counted multiple times.

For example, suppose the hymns are labeled $A_1$ to $A_{25}$ in the first book, and $B$ and $C$ for the other two books. Here is one way of choosing hymns according to your method:

Choose one hymn from each book: $A_1$ from the first, $B_1$ from the second, $C_1$ from the third.

Choose the remaining three hymns from any books: $A_2,A_3,B_2$

Here is another way of choosing hymns:

Choose one hymn from each book: $A_2$ from the first, $B_1$ from the second, $C_1$ from the third.

Choose the remaining three hymns from any books: $A_1,A_3,B_2$

Here is the problem; your method counts these two selections as different, when they both resulted in the same set of hymns.

Another way to state this: your method treats the hymns selected in the first stage as "special." So really, what you are counting is the number of ways to select six hymns, with at least one from each book, and to select one chosen hymn from each book to be "special." Since you are making extra choices, you will get a higher count for the number of ways.

Title: Choosing Hymns -- Why Am I Wrong?


问:
Determine the Probability of no bin being empty when 10 balls are randomly tossed with equal probability into 6 bins
Suppose we have 10 items that we will randomly place into 6 bins, each with equal probability. I want you to determine the probability that we will do this in such a way that no bin is empty. For the analytical solution, you might find it easiest to think of the problem in terms of six events Ai, i = 1, . . . , 6 where Ai represents the event that bin i is empty, and calculate P(AC1 ∩···∩AC6)using DeMorgan’s law.

Analytical= 1- P(bin one empty) + P(bin 2 empty) + P(bin 3 empty) + P(bin 4 empty) + P(bin 5 empty) + P(bin 6 empty) – P(1 and 2 empty) – P(1 and 3 empty) – P(1 and 4 empty) – P(1 and 5 empty) – P(1 and 6 empty) – P(2 and 3 empty) – P(2 and 4 empty) – P(2 and 5 empty) – P(2 and 6 empty) – P(3 and 4 empty) – P(3 and 5 empty) – P(3 and 6 empty) – P(4 and 5 empty) – P(4 and 6 empty) – P(5 and 6 empty) + P( 1 2 3 empty) + P(1 2 4 empty) + P (1 2 5) empty + P (1 2 6 empty) + P( 1 3 4 empty) + P(1 3 5 empty) + P( 1 3 6 empty) + P(1 4 5 empty) + P(1 4 6 empty) + P(1 5 6 empty) – P(1 2 3 4 empty) – P(1235 empty) – P(1236 empty) – P(2345 empty) – P(2346 empty) – P(3456 empty) + P(12345 empty) + P(23456 empty) + P(34561 empty) + P(45612 empty) – P(123456 empty)

1-(6P(1 specific empty bin) – 15P(2 specific empty bins) + 20P(3 specific empty bins) – 15P(4 specific empty bins) +6*P(5 specific empty bins)

This is what I have so far but I don't know how to determine the individual probabilities above.

答:
Another way is to use  Stirling numbers of the second kind which can be conceived  as putting n labelled objects (throws $1 \;thru\; 10$)
into k identical boxes with no box being left empty.
If we multiply $S2(10,6)$ by $6!$ to restore the identity of boxes, we immediately get the answer as $\dfrac{S2(10,6)\cdot 6!}{6^{10}} \approx 0.2718$

Addendum

@user2661923 has added a method for manually
working out S2(10,6). Here is another way  which is totally mechanical, without bothering about double counting. There are $5$ possible templates, viz $511111\;421111\;331111\;322111\;222211$ and the answer is

$10!\left[\dfrac 1 {5!\;5!} + \dfrac 1 {4!2!\;4!}+\dfrac 1 {3!3!\;2!4!}+\dfrac1 {3!2!2!\;2!3!}+\dfrac1 {2!2!2!2!\;4!2!}\right]$

To explain the third template say, to get the arrangements, you want $\binom{10}{3,3,1,1,1,1}\binom{6!}{2!4!}$ which simplifies to $\frac{10!6!}{3!3!1!1!1!1!2!4!}$ but we need to eliminate the $6!$ from the numerator and don't need to pad with $1!$


问:
Rotation of non-unit vector to another non-unit vector that share initial point.
So I have two non-unit vectors $f$ and $t $.  They are generated from 4 Cartesian positions $ f = [x_2, y_2, z_2 ] - [x_1, y_1, z_1 ] $ and $ t = [x_3, y_3, z_3 ] - [x_1, y_1, z_1 ] $

I know that the distance between the positions are equivalent, $r_{12} = r_{13}$.

But I need to rotate these vectors to be equal, where $\theta = arccos\left(\frac{f}{\vert \vert f \vert \vert} \cdot  \frac{t}{\vert \vert t \vert \vert}\right)= 0$ and $ f = t = [x_2, y_2, z_2 ] - [x_1, y_1, z_1 ]$

I need to generate a rotation matrix for this so that way the other Cartesian positions ([$x_4, y_4, z_4$], $\dots, [x_n, y_n, z_n]$) are rotated similarly.

I have tried to use the follow rotation matrix but it is not working since it breaks $r_{12} = r_{13}$ condition, meaning that  $ f \neq t \neq [x_2, y_2, z_2 ] - [x_1, y_1, z_1 ]$ and $r_{12} \neq r_{13}$.

Definitions

$ \begin{align}
u &= \frac{t}{\vert \vert t \vert \vert} \\
a &= cos\left(\frac{\theta}{2}\right)
&b &= u_{x} * sin\left(\frac{\theta}{2} \right) \\
c &= u_{y}* sin\left(\frac{\theta}{2}\right)
&d &= u_{z} *sin\left(\frac{\theta}{2}\right)
\end{align}$

Rotation Matrix
$\mathbf{B} = \begin{bmatrix}
 a^2 + b^2 - c^2 - d^2 & 2.0 (bc - ad) &  2.0(bd + ac) \\
2.0 (bc + ad) & a^2 + c^2 - b^2 - d^2 &   2.0(cd - ab) \\
2.0 (cd - ab) &   2.0(bd - ac) & a^2 + d^2 - b^2 - c^2  \\
\end{bmatrix}$

Then the rotation is $t' = \mathbf{B}t$

答:
Let $\vec{p}_i = (x_i, y_i, z_i)$, $\vec{v} = \vec{p}_2 - \vec{p}_1$ (corresponding to OP's $f$), and $\vec{u} = \vec{p}_4 - \vec{p}_3$ (corresponding to OP's $t$).

You need a rotation of angle $\theta$ around unit axis $\hat{a} = (a_x, a_y, a_z)$ to bring $\vec{u}$ parallel to $\vec{v}$, i.e.
$$\frac{\lVert \vec{u} \rVert}{\lVert \vec{v} \rVert} \vec{v} = \mathbf{R} \vec{u} \quad \iff \quad \vec{v} = \frac{\lVert \vec{v} \rVert}{\lVert \vec{u} \rVert } \mathbf{R} \vec{u}$$
where
$$\mathbf{R} = \left[ \begin{matrix}
c + a_x^2 u         &   a_x a_y u - a_z s   &   a_x a_z u + a_y s \\
a_x a_y u + a_z s   &   c + a_y^2 u         &   a_y a_z u - a_x s \\
a_x a_z u - a_y s   &   a_y a_z u + a_x s   &   c + a_z^2 u       \\
\end{matrix} \right] \tag{1}\label{G1}$$
and
$$\begin{aligned}
c &= \cos \theta \\
s &= \sin \theta \\
u &= 1 - \cos \theta = 1 - c \\
\end{aligned}$$

The unit rotation axis is perpendicular to both vectors:
$$\hat{a} = \frac{\vec{a}}{\lVert \vec{a} \rVert}, \quad \vec{a} = \vec{u} \times \vec{v}$$
and the angle fulfills
$$\begin{aligned}
\cos \theta &= \frac{\vec{u} \cdot \vec{v}}{\lVert \vec{u} \rVert \, \lVert \vec{v} \rVert} \\
\sin \theta &= \frac{\lVert \vec{u} \times \vec{v} \rVert}{\lVert \vec{u} \rVert \, \lVert \vec{v} \rVert} \\
\end{aligned}$$

In practice, you'll want to normalize the two vectors to unit length first,
$$\hat{u} = \displaystyle \frac{ \vec{p}_4 - \vec{p}_3 }{\lVert \vec{p}_4 - \vec{p}_3 \rVert} = \left[ \begin{matrix} u_x \\ u_y \\ u_z \end{matrix} \right] = \left[ \begin{matrix}
\displaystyle \frac{ x_4 - x_3 }{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2 + (z_4 - z_3)^2 }} \\
\displaystyle \frac{ y_4 - y_3 }{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2 + (z_4 - z_3)^2 }} \\
\displaystyle \frac{ z_4 - z_3 }{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2 + (z_4 - z_3)^2 }} \\
\end{matrix} \right] \tag{2a}\label{G2a}$$
$$\hat{v} = \displaystyle \frac{ \vec{p}_2 - \vec{p}_1 }{\lVert \vec{p}_2 - \vec{p}_1 \rVert} = \left[ \begin{matrix} v_x \\ v_y \\ v_z \end{matrix} \right] = \left[ \begin{matrix}
\displaystyle \frac{ x_2 - x_1 }{\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 }} \\
\displaystyle \frac{ y_2 - y_1 }{\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 }} \\
\displaystyle \frac{ z_2 - z_1 }{\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 }} \\
\end{matrix} \right] \tag{2b} \label{G2b}$$
Then,
$$\begin{aligned}
t_x &= u_y v_z - u_z v_y \\
t_y &= u_z v_x - u_x v_z \\
t_z &= u_x v_y - u_y v_x \\
t_n &= \sqrt{ t_x^2 + t_y^2 + t_z^2 } \\
c = \cos\theta & = u_x v_x + u_y v_y + u_z v_z \\
s = \sin\theta & = t_n = \sqrt{ 1 - c^2 } \\
a_x &= \displaystyle \frac{t_x}{t_n} \\
a_y &= \displaystyle \frac{t_y}{t_n} \\
a_z &= \displaystyle \frac{t_z}{t_n} \\
\end{aligned} \tag{3}\label{G3}$$
and with these (and $u = 1 - c$), use $\eqref{G1}$ to construct the rotation matrix $\mathbf{R}$.


答:
The answer should be the circle touching y=|x| and y=4-$x^2$ from above with r=  4(1+$\sqrt2$). Its radius is larger than 4($\sqrt2$-1), which is the radius of the other circle touching the parabola from underneath.

As shown in the graph, let r= radius of the circle centered at o'(0,x+r) , the circle touches y=x at A(x,x).

With  $\Delta$ OO'A, $(4+r)^2$=$r^2$+$(\sqrt2x)^2$

With $\Delta$ BO'A, $(4+r-x)^2+x^2$=$r^2$

Solving these two equations,

x($x^2-8x+8)=0, $r=$\frac{x^2-8}{4}$

Only solution x=4+2$\sqrt2$, r=4(1+$\sqrt2$) provides positive r.

Title: The radius of a circle, having maximum area, which touches the curve $y = 4 – x^2$ and the lines, $y = |x|$ is


问:
Differential equations question?
We have the equation $y=\ln(1+(y')^2)$.When I tried to solve this,I told my teacher that this is a Lagrange differential equation and he told me I was wrong. How can this be solved like a non-Lagrange equation? Should I put $p=y'$?</p>

答:
If you want the traditional solution:

We have $1+y'^2=e^y$, $y'=\sqrt{e^y-1}$, $dy/\sqrt{e^y-1}=dt$, $2\arctan{\sqrt{e^y-1}}=t+c$ hence $e^y-1=\tan^2((t+c)/2)$, $y=\ln(1+\tan^2((t+c)/2))$

Title: Differential equations question?


答:
The mistake is when you square $a+bi$, you should get $a^2-b^2+\color{red}2abi$ and not $a^2-b^2+abi$.

$$a^4+3a^2-4=0$$
$$(a^2+4)(a^2-1)=0$$
$$a=\pm 1, b = \pm 2$$
Hence,
$$x_{1,2}=1-2i\pm(1+2i)$$

which gives you the desired solution of $2$ and $-4i$.

Title: Complex Quadratic, what did I do wrong?

Body:
$$z^2-2(1-2i)z-8i=0$$
Here is my working:
$$\begin{align}
x_{1,2} &= \frac{2-4i \pm \sqrt{(4i-2)^2 + 4\cdot 8i}}{2} \\
&=\frac{2-4i \pm \sqrt{-16+4-16i + 32i}}{2} \\
&= \frac{2-4i \pm \sqrt{-12+16i}}{2} \\
&= \frac{2-4i \pm \sqrt{4(-3+4i)}}{2} \\
&= \frac{2-4i \pm 2\sqrt{(4i-3)}}{2} \\
&= 1-2i \pm \sqrt{4i-3}
\end{align}$$
Let $a+bi = \sqrt{4i-3}$

$$a^2-b^2+abi=4i-3$$

$$a^2-b^2=-3, ab=4$$

$$b=\frac4a$$

$$a^2-\left(\frac{4}{a}\right)^2=-3$$

$$a^4+3a^2-16=0$$

$$t^2+3t-16=0$$

$$t_{1,2}=\frac{-3\pm\sqrt{9+4\cdot 16}}{2}$$

Answers should be $z=2, z=-4i,$ I don't know what I am doing wrong, I can get it right when substituting $z=a+bi$ but I don't when using this method.


问:
Find a Quotient of Hausdorff spaces not is Hausdorff
Give a example of a quotient of a Hausdorff space that is not Hausdorff.

Attempt

Let $X=\mathbb{R}$, for $x,y\in X$ we define $x\sim y$ if $x=y$ or $|x|=|y|>1$.
I claim that the quotient $X\setminus \sim$ not is Hausdorff.

Let $\pi: X \longrightarrow  X\setminus \sim$ given by $\pi(x)=[x]$ and define the topology on $X \setminus \sim$ given by $U\subset X\setminus \sim$ is open if $\pi^{-1}([U])\subset X$ is open .

For $x\in X$ we have two possible cases.

Case I

$x\in \mathbb{Q}$, namely $x=p/q$ where $(p,q)=1,q\neq 0,p,q\in \mathbb{Z}$ WLOG $|x|>1$and then $[x]=\lbrace \frac{r}{s}\in \mathbb{Q} \mid ps=qr\rbrace$$\bigcup$ $ \lbrace -\frac{r}{s}\in \mathbb{Q} \mid ps=qr\rbrace$.

Other wise $|x|<1$ then $[x]=\lbrace \frac{r}{s}\in \mathbb{Q} \mid ps=qr\rbrace$

Case II

$x\in \mathbb{I}$ then $[x]=\lbrace x,-x\rbrace$.

We want take use the following:

If X is Hausdorff and $x_i\in X$ for $i\in 1,2,3, \cdots n$  then $\lbrace x_1,x_2,x_3, \cdots x_n\rbrace$ is closed

Consider $[\frac{1}{n}]\in X \setminus \sim $ for some $n\in \mathbb{N}$ fixed.

$\pi^{-1}(\lbrace[\frac{1}{n}]\rbrace)=\lbrace \frac{k}{kn}\mid k\in \mathbb{N} \rbrace=\lbrace \frac{1}{n}, \frac{1}{n}, \cdots \frac{1}{n} \rbrace =\lbrace \frac{1}{n}\rbrace\subset X$ wich is closed.

Hence $\lbrace[\frac{1}{n}]\rbrace$ is closed in $X\setminus \sim$.

Now consider $\bigcup_{n\in \mathbb{N}}\lbrace \left[\frac{1}{n}\right]\rbrace  \in X\setminus \sim $ and notice that
$$\pi^{-1}\left(\bigcup_{n\in \mathbb{N}}\lbrace \left[\frac{1}{n}\right] \rbrace \right)=\bigcup_{n\in \mathbb{N}}\pi^{-1}\left( \lbrace \left[\frac{1}{n}\right] \rbrace \right)=\lbrace\frac{1}{n}\mid n\in \mathbb{N} \rbrace \subset \mathbb{R} $$
Which is clearly not closed since $0$ is limit point of the set and is not contained in it.

Hence $X \setminus \sim$ cannot be Hausdorff.

I miss some?
Is possible get a intuitive representation of this quotient?

答:
The example works, but there is no reason to consider rational and irrational numbers separately, and your argument to show that $X/\!\sim$ is not Hausdorff is incorrect. I will first give a correct argument and then explain what is wrong with yours.

Let $p=\pi(-1)$ and $q=\pi(1)$; $|-1|=|1|=1\not>1$, so $p\ne q$. Suppose that $U$ and $V$ are disjoint open nbhds of $p$ and $q$ respectively. Then $\pi^{-1}[U]$ is an open nbhd of $-1$ in $X$, so there is a positive $\epsilon_p<1$ such that $(-1-\epsilon_p,-1+\epsilon_p)\subseteq\pi^{-1}[U]$. Similarly, there is a positive $\epsilon_q<1$ such that $(1-\epsilon_q,1+\epsilon_q)\subseteq\pi^{-1}[V]$. Let $\epsilon=\min\{\epsilon_p,\epsilon_q\}$, and let $x=1+\frac{\epsilon}2$. Then $|x|>1$, so $\pi(x)=\pi(-x)$, and therefore

$$\begin{align*}
\pi(x)&\in\pi[(1-\epsilon,1+\epsilon)]\cap\pi[(-1-\epsilon,-1+\epsilon)]\\
&\subseteq\pi\big[\pi^{-1}[U]\big]\cap\pi\big[\pi^{-1}[V]\big]\\
&=U\cap V\,,
\end{align*}$$

so that $U$ and $V$ are not disjoint. $U$ and $V$ were arbitrary nbhds of $p$ and $q$, so $p$ and $q$ do not have disjoint nbhds, and the quotient is not Hausdorff.

As for your argument, it is true that all finite subsets of a Hausdorff space are closed, so finding a finite set that is not closed would indeed show that the space is not Hausdorff, but the set $\bigcup_{n\in\Bbb Z^+}\left\{\left[\frac1n\right]\right\}$ is not finite, so showing that it is not closed tells you nothing. (You could even note that it’s also not closed in $\Bbb R$, which certainly is Hausdorff.) In fact a space is $T_1$ if and only if all of its finite subsets are closed, and $X/\!\sim$ is $T_1$.

There is a very easy way to picture the quotient. Bend the closed interval $[-1,1]$ around into a circle, but don’t close it: leave $-1$ and $1$ as separate points. Glue the rays $(1,\to)=\{x\in\Bbb R:x>1\}$ and $(\leftarrow,-1)=\{x\in\Bbb R:x<-1\}$ together, attaching $x$ to $-x$ for each $x>1$. You should end up with a picture something like this (but better!):

```<code>          _
         / \ 1
        |   :------------------------------
         \_/ -1
</code>```

Except for $1$ and $-1$, a point in the circular part has an open nbhd base of small open intervals. A point in the ray that is formed by gluing together $(1,\to)$ and $(\leftarrow,-1)$ also has an open nbhd base of small open intervals. A basic open nbhd of $1$ looks like an open interval that starts on the circle a little above $1$ and runs through $1$ into the ray. Similarly, a basic open nbhd of $-1$ looks like an open interval that starts on the circle a little below $-1$ and runs through $1$ into the ray. And such intervals necessarily intersect on the ray. (I apologize for the very shoddy sketch, but at the moment I do not have the ability to produce a proper diagram.)

Body:
The example works, but there is no reason to consider rational and irrational numbers separately, and your argument to show that $X/\!\sim$ is not Hausdorff is incorrect. I will first give a correct argument and then explain what is wrong with yours.

Let $p=\pi(-1)$ and $q=\pi(1)$; $|-1|=|1|=1\not>1$, so $p\ne q$. Suppose that $U$ and $V$ are disjoint open nbhds of $p$ and $q$ respectively. Then $\pi^{-1}[U]$ is an open nbhd of $-1$ in $X$, so there is a positive $\epsilon_p<1$ such that $(-1-\epsilon_p,-1+\epsilon_p)\subseteq\pi^{-1}[U]$. Similarly, there is a positive $\epsilon_q<1$ such that $(1-\epsilon_q,1+\epsilon_q)\subseteq\pi^{-1}[V]$. Let $\epsilon=\min\{\epsilon_p,\epsilon_q\}$, and let $x=1+\frac{\epsilon}2$. Then $|x|>1$, so $\pi(x)=\pi(-x)$, and therefore

$$\begin{align*}
\pi(x)&\in\pi[(1-\epsilon,1+\epsilon)]\cap\pi[(-1-\epsilon,-1+\epsilon)]\\
&\subseteq\pi\big[\pi^{-1}[U]\big]\cap\pi\big[\pi^{-1}[V]\big]\\
&=U\cap V\,,
\end{align*}$$

so that $U$ and $V$ are not disjoint. $U$ and $V$ were arbitrary nbhds of $p$ and $q$, so $p$ and $q$ do not have disjoint nbhds, and the quotient is not Hausdorff.

As for your argument, it is true that all finite subsets of a Hausdorff space are closed, so finding a finite set that is not closed would indeed show that the space is not Hausdorff, but the set $\bigcup_{n\in\Bbb Z^+}\left\{\left[\frac1n\right]\right\}$ is not finite, so showing that it is not closed tells you nothing. (You could even note that it’s also not closed in $\Bbb R$, which certainly is Hausdorff.) In fact a space is $T_1$ if and only if all of its finite subsets are closed, and $X/\!\sim$ is $T_1$.

There is a very easy way to picture the quotient. Bend the closed interval $[-1,1]$ around into a circle, but don’t close it: leave $-1$ and $1$ as separate points. Glue the rays $(1,\to)=\{x\in\Bbb R:x>1\}$ and $(\leftarrow,-1)=\{x\in\Bbb R:x<-1\}$ together, attaching $x$ to $-x$ for each $x>1$. You should end up with a picture something like this (but better!):

          _
         / \ 1
        |   :------------------------------
         \_/ -1

Except for $1$ and $-1$, a point in the circular part has an open nbhd base of small open intervals. A point in the ray that is formed by gluing together $(1,\to)$ and $(\leftarrow,-1)$ also has an open nbhd base of small open intervals. A basic open nbhd of $1$ looks like an open interval that starts on the circle a little above $1$ and runs through $1$ into the ray. Similarly, a basic open nbhd of $-1$ looks like an open interval that starts on the circle a little below $-1$ and runs through $1$ into the ray. And such intervals necessarily intersect on the ray. (I apologize for the very shoddy sketch, but at the moment I do not have the ability to produce a proper diagram.)


问:
What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$?
What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)

I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect each other in between x = 1 and 2 but I couldn't figure out a way to get to this conclusion on my own.

Is there any way by which I can determine where these functions intersect?

答: Notice that $x-1 < \lfloor x \rfloor \leq x$, so $x^3-x+1 > x^3- \lfloor x \rfloor \geq x^3-x$, and $x^3-x+1 > 3 \geq x^3-x$. The first part of the inequality shows that $x<2$, and the second shows that $x \geq 1$. This means that $\lfloor x \rfloor=1$, so $x^3-1=3$ and $x=\sqrt[3]4$.


问:
3 circles with radii 66, 77, and 88 externally tangent to each other, find the radius of the circle internally tangent to the other circles.

In the diagram, we see that there are 3 circles that are all externally tangent to each other and internally tangent to a much bigger circle. The radii of the 3 smaller circles are 66, 77, and 88. What is the radius of the bigger circle?

I actually solved this problem by drawing in the center of the bigger circle, drawing lines from that center to the 3 points of tangency on the large circle, then set up a system of equations and go for there (if you want me to be more specific just ask) but this is really messy. Is there a really nice intuitive way to go about this, or am I stuck doing the messy work?

答:
Hint:

Apply Descartes theorem:

$$(\frac 1{r_1}+\frac 1 {r_2}+\frac 1{r_3}+\frac 1 {r})^2=2(\frac 1{r_1^2}+\frac 1 {r_2^2}+ \frac 1{r_3^2}+\frac 1{r^2})$$

Where $r_1, r_2, r_3$ are radii of given circles and r is the radii of small(in the middle of circles) and big circle(the one you try to find it's radius).. That is you get a quadratic equation in term of r , solving it gives you the radius.


问:
If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$
If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I put the value of $x= 0$ and the answer is coming out to be $13/6.$

But the actual answer is $2.$

Am I missing something ?

Body:
Let $x+1=y^2$  as $x\ge0, y^2\ge1$

$$\dfrac{4x^2+8x+13}{1+x}=\dfrac{4(y^2-1)^2+8(y^2-1)+13}{y^2}=4y^2+\dfrac9{y^2}=\left(2y-\dfrac3y\right)^2+2\cdot2y\cdot\dfrac3y\ge12$$  the equality occurs if $2y-\dfrac3y=0\iff y^2=\dfrac32\iff x=?$


问:
Confusion about determinant properties for a matrix

Consider a $3 \times 3$ matrix
$$A=\begin{bmatrix}
a & b & c\\
d & e & f \\
g & h & i\\
\end{bmatrix}$$
Assume that we know $\det(A)=\frac{1}{2}$. Calculate $$\det\begin{bmatrix}
3a & 3b & 3c\\
d-a & e-b & f-c \\
g-2d & h-2e & i-2f\\
\end{bmatrix}$$

I attempted the following:

$$\det(A)=\begin{vmatrix}
3a & 3b & 3c\\
d-a & e-b & f-c \\
g-2d & h-2e & i-2f\\
\end{vmatrix}$$

$$\det(A)=\begin{vmatrix}
3a & 3b & 3c\\
d & e & f \\
g-2d & h-2e & i-2f\\
\end{vmatrix}-\begin{vmatrix}
3a & 3b & 3c\\
-a & -b & -c \\
g-2d & h-2e & i-2f\\
\end{vmatrix}$$

The second determinant is $0$ since the 1st and 2nd row are scalar multiples of each other. So now we have:

$$\det(A)=\begin{vmatrix}
3a & 3b & 3c\\
d & e & f \\
g-2d & h-2e & i-2f\\
\end{vmatrix}$$

$$\det(A)=\begin{vmatrix}
3a & 3b & 3c\\
d & e & f \\
g & h & i\\
\end{vmatrix}+\begin{vmatrix}
3a & 3b & 3c\\
d & e & f \\
-2d & -2e & -2f\\
\end{vmatrix}$$

The second determinant is $0$ since the 2nd and 3rd row are scalar multiples of each other. So now we have:

$$\det(A)=\begin{vmatrix}
3a & 3b & 3c\\
d & e & f \\
g-2d & h-2e & i-2f\\
\end{vmatrix}$$

$$\det(A)=\begin{vmatrix}
3a & 3b & 3c\\
d & e & f \\
g & h & i\\
\end{vmatrix}$$

Is this just $$\det(A)=3\begin{vmatrix}
a & b & c\\
d & e & f \\
g & h & i\\
\end{vmatrix}$$

$$\det(A)=3\left(\frac{1}{2}\right)?$$

Or is it $$\det(A)=3^3\left(\frac{1}{2}\right)?$$

I feel like it's the 1st one but I am not sure. Can someone clarify this last step?

答:
Result:
in an $ n×n$ matrix  $A$  , if a row  (column)  be multiplied by a scalar c then $det A$ is multiplied by c
So your first result is true.