问:
Generating function of total sales on a given day
Given the problem below:

The number of customers that shop at Elmo’s World of Food on a given day is geometrically distributed with parameter $p_1$. The number of items a customer buys at Elmo’s is geometrically distributed with parameter $p_2$. The distribution of the cost of each item is exponentially distributed with parameter $\lambda$. Assume that the amount customers buy and the amount of prices of items are independent. Let $X$ denote the total sales on a given day. Find the generating function of $X$.

In my attempt on solving this I let $N_c$ denote the number of customers that shop on a given fay, $N_i$ denote the number of items a customer buys, and $C$ cost of each item. Then we can write $X = \sum_{k=1}^{N_c} N_i\cdot C$. To find the generating function I used the MGF: $E[e^{tX}]$ and to compute it I conditioned on $N_c$ first and then on $C$. However, my work became very tedious and I'm stuck at simplifying the expression I got. Therefore, I was wondering if there is a different approach I could use in finding the generating function of $X$? This is just a practice problem as I study for my exam.

答:
Answer: The moment generating function for the total sales/profits is
$$ M_X(t) = \textbf{E}[e^{t X}] = \frac{\lambda p_1 p_2}{\lambda p_1 p_2 - t} \qquad \text{where} \qquad(\lambda p_1 p_2 > t).$$

Solution:
To clarify the question a bit: $X$ is the total profit (i.e., total costs of all products purchased) on a given day and since $X$ is effectively a continuous random variable and "generating functions" are typically defined as a way to encode the coefficients of an infinite sequence, I'll assume you want to calculate the moment generating function for $X$.

In order to compute the moment generating function for $X$, we need to compute the probability density function (PDF) for $X$.

By the prompt, we know that the probability that $k$ customers enter the store is $(1-p_1)^{k-1}p_1$. The probability that a single one of those customers purchases $\ell$ items is $(1-p_2)^{\ell-1} p_2$. And the probability density for the cost of a single item is $\lambda e^{-\lambda x}$.

In order to find the PDF for $X$, we need to compute the probabilities for sums of random variables. In particular, the probability that $k$ people purchase a total of $m$ items and that the total cost of these $m$ items is $X$. We note two things

The sum of $k$ geometric random variables with probability of success $p$ is a negative binomial random variable with parameters $k$ and $p$: $\binom{m-1}{k-1} p^k (1-p)^{m-k}$

The sum of $m$ exponential random variables with rate $\beta$ is a gamma $(m, \beta)$ random variable: $\beta^m x^{m-1} e^{-\beta x}/(m-1)!$

Thus the probability that $m$ items are bought by $k$ customers is $\binom{m-1}{k-1} p_2^k (1-p_2)^{m-k}$, and the PDF for the total costs (i.e., profits) from $m$ items is $\lambda^m X^{m-1} e^{-\lambda X}/(m-1)!$.

The PDF for $k$ customers visiting the store and purchasing a total of $m$ items which have a total cost of $X$ is then
\begin{align}p(k, m, X) &= (1-p_1)^{k-1} p_1 \times\binom{m-1}{k-1} p_2^k (1-p_2)^{m-k}\times\frac{\lambda^m }{(m-1)!}X^{m-1} e^{-\lambda X}\\[.5em]
& = \lambda p_1 p_2 \frac{\left(\lambda p_2(1-p_1) X\right)^{k-1}}{(k-1)!}\frac{\left(\lambda (1-p_2) X\right)^{m-k}}{(m-k)!} e^{-\lambda X}.
\end{align}
To find the PDF for the cost alone, we sum over all possible values of $m$ and $k$. We have
\begin{align}
\rho_{\text{cost}}(X) & = \sum_{k=1}^{\infty}\sum_{m=k}^{\infty}p(k, m, X)\\[.5em]
& = \lambda p_1 p_2 \exp\left( \lambda p_2(1-p_1)X\right) \exp\left( \lambda(1-p_2) X\right) \exp(-\lambda X)\\[.5em]
& = \lambda p_1 p_2 \exp(-\lambda p_1 p_2 X).
\end{align}
The moment generating function is then
$$ M_X(t) = \textbf{E}[e^{t X}] = \lambda p_1 p_2 \int^{\infty}_{0} dX\, e^{tX - \lambda p_1 p_2 X} = \frac{\lambda p_1 p_2}{\lambda p_1 p_2 - t} \qquad (\lambda p_1 p_2 > t).$$

Random Note: The problem's stated conditions/assumptions seem odd. I would expect the number of people in a store and the number of items each person purchases to follow something more like a poisson distribution and binomial distribution, respectively, rather than geometric distributions. Geometric distributions characterize discrete survival processes which doesn't seem to well model whether people enter a store or how many items they buy.


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: A nice example in Algebraic number theory is the solution of the $p=x^2+ny^2$ problem, described in details in Cox's book. Very much like Fermat's last theorem it has its roots in the 17th century (actually, it originated from Fermat...) and is completely solved only using some heavy machinery from number theory (although not as heavy as FLT). I'd say it is a good representative for the power and depth of the field that is still very much accessible.</p>

ContentLicense: CC BY-SA 3.0

Title: Landmarks of subjects of mathematics


问:
How can we make it easy to solve quadratics without the Quadratic Formula?
For example ,

$$5x^2 - 7x-2=0$$

Then, we only need to find is when
$$5x^2-7x=2$$
$$x(5x-7)=2$$
Since that when we get $2-2=0$.

Is there any way we can find them easily without solving quadratics using formula?

I am just getting very enthusiastic in this topic.

答:
Factoring the equation in the form of $(x-\alpha)(x-\beta)$ is the best way for quadratic equations if the coefficients are reasonably small (not always as in your case because the signs are different). But sometimes (if $\alpha, \beta$ turn out to have an imaginary component or the coefficients are large integers or fractions) the quadratic formula would have to be used. $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ In your case, the quadratic formula would have to be used (or completing the square as @Joe said).

If you want to calculate 'intuitively', there is another method. Note that if $\alpha$ and $\beta$ are the roots of a quadratic of the form $ax^2+bx+c$ ; $a \in \mathbb R - \{0\}$, then $$\alpha + \beta = \frac{-b}{a}$$ and $$\alpha \beta = \frac{c}{a}$$ This can work for equations like $x^2-50x+525$ where $\alpha + \beta = -(-50) = 15+35$ and $\alpha \beta = 525 = 15*35$.

Thus it comes to mind that the roots must be $15$ and $35$ (comparing $LHS$ and $RHS$ due to which the $(x-\alpha)(x-\beta)$ would be $(x-15)(x-35)$.

You need to practice such questions extensively to know which method is more efficient in which equation.

ContentLicense: CC BY-SA 4.0


问:
How to prove that following algorithm is correct?
Suppose I have some $n$ numbers which are powers of $2$, say $a_1,a_2,a_3.....a_n$ which are not necessarily all distinct. I have option to give them any sign. I have to find if I can make their sum after that equal to $num$.

I have following algorithm which I am sure will work, by lot of arguments and verification, but i am not able to prove it:

Sort the array and star.

Traverse from highest element to lowest element.

Let initialize another variable $temp = 0$

If $temp>num$ then subtract $a_i$ else add $a_i$ . If in end $temp=num$ then it's possible to assign signs such that we can make $num$ out the array else not possible.

See an example in comment section.

How can I prove it?

答:
So, you have a target value, $T$, that you want to reach by adding/subtracting some given numbers $a_1, \ldots, a_n$.  Each $a_i$ must appear in the sum,  but you are allowed to choose whether each $a_i$ is added or subtracted.  In other words, you want to choose signs $s_i\in \{ 1, -1\}$ so that
$$
\sum s_i a_i = T,
$$
or else determine that no such choice is possible.

If I understand correctly, your algorithm is: Start with the biggest $a_i$. (If there are no $a_i$, meaning that $n=0$, then there is nothing to do! So assume $n\geq 1$.)  If $T\geq 0$, then choose $s_i = 1$.  If $T< 0$, then choose $s_i = -1$.  (Note: if $T=0$ then it really doesn't matter what choice you make; if it works one way then it will work the other way by flipping all the remaining signs.)  Now update the target: the new target $T'$ is given by $T' = T - s_i a_i$.  Repeat until there are no more $a_i$ left.

There is another assumption which I haven't mentioned yet: you said that each $a_i$ is a power of $2$.  It's a good thing that we have that assumption: if we didn't, then this algorithm doesn't work.  Example: $a_1 = 6, a_2 = 6, a_3= 9$, and $T=3$.  The algorithm would say to choose $9$ to have a positive sign.  But now you need to make $-6$ out of the remaining $a_i$, and there is no way to add/subtract a pair of $6$'s and get a result of $-6$.  (Instead, you can do $3 = -9 + 6 + 6$,  a valid solution which the algorithm as stated would not find.)

OK, so we make the further assumption that every $a_i$ is a power of $2$: write $a_i = 2^{k_i}$ for each $i$.  We may as well assume that they are in order, so
$$
0 \leq k_1 \leq k_2 \ldots \leq k_n .
$$
We may assume WLOG that $T\geq 0$ (otherwise just flip all the signs).  This would mean that you your algorithm says to choose $s_n = 1$.

Now, to prove that your algorithm is optimal is kind of a tricky thing!  Optimality means that, IF it would work to make the OTHER choice, THEN it STILL works to make the choice you do.

OK, so: suppose that there is a valid choice of signs $(t_1, \ldots, t_n)$, with $t_n = -1$, such that $\sum t_i a_i = T$. We need to show that there is another valid choice $(s_1, \ldots, s_n)$ with $s_n = 1$.

That is, assume that
$$
t_1 a_1 + \ldots + t_{n-1} a_{n-1} = T + 2^{k_n},
$$
and show that there exist $s_1, \ldots , s_{n-1}$ such that
$$
s_1 a_1 + \ldots + s_{n-1} a_{n-1} = T - 2^{k_n}.
$$

Basically, we need to flip the signs on some of the $t_i$, and by doing so, decrease the sum by just the right amount.  This will certainly mean flipping some of the positive $t_i$ to negative.

Maybe we can get away with ONLY flipping from positive to negative!

So, for convenience let's restrict our attention to just the $a_i$ for which $t_i$ is positive; let these be $b_1, \ldots , b_m$.  Clearly there are some of these; indeed,
$$
b_1 + \ldots + b_m \geq T + 2^{k_n} > 0
$$
since counting only the positive terms (removing negative terms) can only increase the resulting sum.  (Thus $m\geq 1$.)  We still assume they are written in order; write $b_i = 2^{\alpha_i}$ where
$$
0 \leq \alpha_1 \leq \ldots \leq \alpha_m.
$$

Notice that subtracting these terms INSTEAD of adding them will reduce the sum by twice the sum of these terms.  We need to reduce the sum by $2\cdot 2^{k_n}$, so we need to show that there is a subset $C\subseteq \{ 1, \ldots, m\}$ such that
$$
\sum_{i\in C} b_i = 2^{ k_n}.
$$
Write $N=k_n$ for convenience.  Also, notice that $2^{\alpha_m} = b_m \leq a_{n-1} \leq a_n = 2^{N}$, so $\alpha_m \leq N$.

Now, if $b_m = 2^{N}$ then just flip that term and we're done.  So we can assume that $\alpha_m < N$.   That is, $\alpha_m \leq N  - 1$.

Here's the trick.  There must be some $i$ such that $b_i = b_{i+1}$.  For, if not, then
$$
b_1 + \ldots + b_m \leq 1 + 2 + 4 + \ldots + 2^{\alpha_m} \leq 2^N - 1 < 2^N,
$$
but we already said that $\sum b_i \geq 2^N + T \geq 2^N$.

So, there are two adjacent $b_i$ that share the same value.  Suppose that we REPLACE those two instances of $2^{\alpha_i}$ with a single instance of $2^{1 + \alpha_i}$.  The sum of the terms remains unchanged, and every term is still a power of $2$.

Now, continue combining duplicate terms until either there are no duplicates, or you make an instance of $2^N$.  Eventually you will make an instance of a $2^j$ for $j \geq N$, by the exact argument we already gave.  And the very first time you do so, the value of $j$ must be equal to $N$, because you have
$$
2^N \leq 2^j = 2^{j-1} + 2^{j-1}
$$
where $j-1 < N$.

So there is some collection of $b_i$ which sum to $2^N$. $\Box$

What we have shown is that choosing $s_n=1$ for $T\geq 0$ is not a mistake; it doesn't result in changing from "a solution exists" to "no solution exists".  (By contrast, in the $6,6,9$ example above, it was a mistake to add $9$.)  But if the algorithm never makes a mistake, then it is optimal.


问:
Counter example regarding bounded functions.

I feel that I completely missed the point in this question, as I didn't even use / don't know how to use the fact that $a,b$ are removable discontinuities..
Does that mean I need to construct a function $g$ that will cover the scenario regarding the discontinuity points?

Counter example:

Consider the function $f: R\ ➜ R$ , $f\left(x\right)\ =\ \frac{1}{x}$

Let $(a,b) = (o,1)$. Note that $f$ is continuous on $(0,1)$.

$\lim _{x\to 0^+}\left(\frac{1}{x}\right)=∞$

$\lim _{x\to 1^-}\left(\frac{1}{x}\right)=1$

Since the one sided limits tends to $∞$, $f$ is not bounded on
$(0,1)$.

答:
Let $c=\frac{a+b}2$.
By definition of "removable discontinuity", there exists a function $g\colon[a,b)\to \Bbb R$ such that $g|_{(a,b)}=f$ and $g$ is continiuous at $a$. it follows that $g$ is continuous, hence is continuous on the interval $[a,c]$, by compactness is bounded there, and $f$ is also bounded on $(a,c]$.

Similarly, the removable discontinuity at $b$ shows that $f$ is bounded on $[c,b)$, hence $f$ is bounded on all of $(a,b)$.


问:
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.

Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.

What I Tried: I only figured out that:- $$\rightarrow a = 2^\frac{2}{3} + 2^\frac{1}{3} + 2^\frac{0}{3}$$ Yet this does not help me anywhere. Perhaps I have to multiply something with $a$ only so that the expression becomes usable, what what to multiply?

Next, the expression we need to find is $\frac{3a^2 + 3a + 1}{a^3}$ , which I did not find any cool factorizations, and do not think anything will help here.

Can anyone help me?

答:
Note that
\begin{eqnarray*}
\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}= \frac{(a+1)^3-a^3}{a^3}.
\end{eqnarray*}
Note also that $a+1=\sqrt[3]{2}a$ so ...
\begin{eqnarray*}
\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}= \frac{2a^3-a^3}{a^3}=1.
\end{eqnarray*}


问:
Is this an exact differential or not?
I have the 1-form
$$dz=2xy\, dx+(x^{2}+2y)\, dy$$
And I want to integrate it from $(x_{1},y_{1})$, to $(x_{2},y_{2})$.

If I'm not drunk, checking mixed partials, I find that $dz$ is an exact differential. BUT, when I want to calculate explicitly the integral

$$\int_{\sigma}dz$$
where $\sigma$ is

i) $(x_{1},y_{1})\to(x_{2},y_{1})\to(x_{2},y_{2})$

ii) $(x_{1},y_{1})\to(x_{1},y_{2})\to(x_{2},y_{2})$

I find that the integral has a different value if I take the trajectory ii) instead of i). Why this happens?

Calculating the integral for i)

$\int dz=\int_{x_{1}}^{x_{2}}2xy_{1}\, dx+\int_{y_{1}}^{y_{2}}(x_{2}^{2}+2y)\, dy=y_{1}(x_{2}^{2}-x_{1}^{2})+x_{2}^{2}(y_{2}-y_{1})+y_{2}^{2}-y_{1}^{2}$

Calculating the integral for ii)

$\int dz=\int_{y_{1}}^{y_{2}}(x_{1}^{2}+2y)\, dy+\int_{x_{1}}^{x_{2}}2xy_{2}\, dx=x_{1}^{2}(y_{2}-y_{1})+y_{2}^{2}-y_{1}^{2}+y_{2}(x_{2}^{2}-x_{1}^{2})$

OwnerDisplayName: Anuar

Title: Is this an exact differential or not?

Body: $$\begin{array}{lr} & \color{Purple}{y_1x_2^2}-\color{Blue}{y_1x_1^2}+\color{Magenta}{x_2^2y_2}-\color{Purple}{x_2^2y_1}+\color{Green}{y_2^2-y_1^2} \\ & = \color{Teal}{x_1^2y_2}-\color{Blue}{x_1^2y_1}+\color{Green}{y_2^2-y_1^2}+\color{Magenta}{y_2x_2^2}-\color{Teal}{y_2x_1^2}  \end{array}$$


问:
Sketches of stable 6-pointed curves
I was wondering if I could ask about stable $n$-pointed curves (essentially intersecting copies of $\mathbb{P}^1$ with at least $3$ special points on each copy - these may be either a point of intersection or a marked point).

I was wondering whether someone could perhaps help me by sketching out the possible cases for stable $6$-pointed curves? I have found all the cases possible for $5$ marked points (see sketch below), but I lose track in terms of the types of intersections that may occur for $3$ components and beyond in the case of $6$ marked points.

I have for $n=6$:

1 component: analogous to $n=5$

2 components: $2$ potential cases (distributing the marks by $2,4$ or $3,3$).

3 components: where I find difficulty.

<img src="https://i.stack.imgur.com/sBrfK.jpg" alt="all possible cases for $n=5$" />

Would it be possible to construct the same sketch but for the case $n=6$?

Thanks you very much in advance!

答:
I think I've found the sketches for stable $6$-pointed curves. I was getting confused as I was forgetting the no closed circuits condition.

Thanks to @Tabes Bridges for their help!

Title: Sketches of stable 6-pointed curves


Body:
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.

What I Tried: I only figured out that:- $$\rightarrow a = 2^\frac{2}{3} + 2^\frac{1}{3} + 2^\frac{0}{3}$$ Yet this does not help me anywhere. Perhaps I have to multiply something with $a$ only so that the expression becomes usable, what what to multiply?

Next, the expression we need to find is $\frac{3a^2 + 3a + 1}{a^3}$ , which I did not find any cool factorizations, and do not think anything will help here.

Can anyone help me?

Body: Hint: As $$a=\frac{(\sqrt[3]{4}+\sqrt[3]{2}+1)(\sqrt[3]{2}-1)}{\sqrt[3]{2}-1}=\frac{1}{\sqrt[3]{2}-1} $$ $${\left(\frac{1}{a}+1\right)}^3=2$$ can you proceed ....


问:
Proving that the given piecewise function is not monotonic, but invertible
The piecewise function is $f(x)$=\begin{cases}
x, x \in \mathbb{Q} &\\
-x, x\in \mathbb{I}
\end{cases}
I have to prove that this function is not monotonic in any interval and that it's invertible.
I watched a lot of videos on youtube to understand this problem, and I thought if I can prove that this function is bijective, then it has to be invertible, but I don't know how I can prove that either.
And for monotonicity I tried proving some special cases, I don't think that's considered a formal proof.
Could you please help me?

Thanks in advance

答:
Note that $f(f(x))=x$, i.e., $f$ is its own inverse and so is invertible.

Assume $f$ is monotonic in some interval $(a,b)$. By density of rationals, there exist rational numbers $a<x_1<x_2<b$, for which $f(x_1)<f(x_2)$, i.e., $f$ is not weakly decreasing on $(a,b)$. Likewise, by densitiy of irrationals, there exist irrational numbers $a<y_1<y_2<b$, for which $f(y_1)>f(y_2)$, i.e., $f$ is not weakly increasing on $(a,b)$, either.


答:
The differential is exact, and you got the same answer. Just do some algebra and you will get
$$
-x_{1}^{2}y_1+x_{2}^{2}y_{2}+y_{2}^{2}-y_{1}^{2}
$$</p>

Title: Is this an exact differential or not?

Body:
I have the 1-form
$$dz=2xy\, dx+(x^{2}+2y)\, dy$$
And I want to integrate it from $(x_{1},y_{1})$, to $(x_{2},y_{2})$.

If I'm not drunk, checking mixed partials, I find that $dz$ is an exact differential. BUT, when I want to calculate explicitly the integral

$$\int_{\sigma}dz$$
where $\sigma$ is

i) $(x_{1},y_{1})\to(x_{2},y_{1})\to(x_{2},y_{2})$

ii) $(x_{1},y_{1})\to(x_{1},y_{2})\to(x_{2},y_{2})$

I find that the integral has a different value if I take the trajectory ii) instead of i). Why this happens?

Calculating the integral for i)

$\int dz=\int_{x_{1}}^{x_{2}}2xy_{1}\, dx+\int_{y_{1}}^{y_{2}}(x_{2}^{2}+2y)\, dy=y_{1}(x_{2}^{2}-x_{1}^{2})+x_{2}^{2}(y_{2}-y_{1})+y_{2}^{2}-y_{1}^{2}$

Calculating the integral for ii)

$\int dz=\int_{y_{1}}^{y_{2}}(x_{1}^{2}+2y)\, dy+\int_{x_{1}}^{x_{2}}2xy_{2}\, dx=x_{1}^{2}(y_{2}-y_{1})+y_{2}^{2}-y_{1}^{2}+y_{2}(x_{2}^{2}-x_{1}^{2})$


问:
Finding a subset cover in a special case where each element is covered by exactly two sets.
This question is taken from the problem here: bookcircle. I'll describe the important aspects here: There are $B$ boys and $G$ girls in a book club, as well as a set of books, lets call them $b_1, ..., b_n$. Each book has been read by exactly one boy and one girl. I have to find the smallest cardinality set of club members $S$ (can be a mix of boys and girls) such that every book has been read by someone in $S$.

Under these conditions, choosing $S$ to be all the boys or all the girls will cover all books, but does not guarantee that $S$ will be minimal. A greedy solution where at each step, you try to pick the person covering the most of the uncovered books, also does not seem to work (I can't see how to tiebreak between two sets correctly).

I have been stuck on this problem for super long, but without much progress. I have a strong hunch it should be reduced to a bipartite matching kind of problem, (or maybe even a flow problem) but I am having trouble finding what the left vertex set and right vertex sets should be.

Could anyone give a hint for this problem?

Some thoughts:

If it is really a matching problem, then I would need to find a property to be maximized, so that a max matching algorithm can find it. Intuitively we would want to maximize the amount of non-overlap between the members of $S$. But maximizing the non-overlapping does not translate to minimizing number of people chosen, so this can't be the way.

We could also try to maximize the number of people not in $S$, but this does not seem like it can be converted to maximum matching.

I try to make use of the fact that each book has been read by exactly 1 boy and 1 girl, which would suggest constructing a graph with vertices the boys and girls, and an edge between a boy and girl vertex if they have both read a book in common. But again, this graph does not seem to be very useful.

答:
I try to make use of the fact that each book has been read by exactly 1 boy and 1 girl, which would suggest constructing a graph with vertices the boys and girls, and an edge between a boy and girl vertex if they have both read a book in common. But again, this graph does not seem to be very useful.

On the contrary, this graph is very useful: a set $S$ of club members such that every book is read by at least one member corresponds to a vertex cover of this graph. Thus, your problem is equivalent to finding a minimum vertex cover. In general, this is NP-complete; luckily, your graph is bipartite, with classes made up by the boys and the girls, respectively. In this case, Kőnig's theorem asserts that the size of a minimum vertex cover is the same as the size of a maximum matching. Moreover, a minimum vertex cover can be produced efficiently from a maximum matching (see the "constructive proof" in the Wikipedia link above).


答:
Consider the given function as $f(x)$.

Consider a real number $y$, if it is rational then $f(y) = y$ and if it is irrational then $f(-y) = y$ so $f$ is onto. Suppose $f(x) = f(y)$ for some real $x$ and $y$, then we claim that $x=y$. Obviously we cannot have one is rational and other one is irrational, so we must have both of them is rational or both of them is irrational. Hence $f(x) = f(y)  \implies x = y $ or $-x = -y$ and we are done showing $f$ is bijective.

Title: Proving that the given piecewise function is not monotonic, but invertible

Body:
The piecewise function is $f(x)$=\begin{cases}
x, x \in \mathbb{Q} &\\
-x, x\in \mathbb{I}
\end{cases}
I have to prove that this function is not monotonic in any interval and that it's invertible.
I watched a lot of videos on youtube to understand this problem, and I thought if I can prove that this function is bijective, then it has to be invertible, but I don't know how I can prove that either.
And for monotonicity I tried proving some special cases, I don't think that's considered a formal proof.
Could you please help me?

Thanks in advance


问:
Finding the probability of a multi-state transition in a finite state machine
Finite State Machine Example (cannot embed yet)

I'm struggling to understand how to compute the probability of $q_{0} \rightarrow q_{3}$.

From my understanding, given that the transition from $q_{1}$ back to $q_{0}$ did not exist, the probability of reaching $q_{3}$ from $q_{0}$ would be $0.5 * 0.35$.

However, in the provided state diagram, we have a state transition from $q_{1} \rightarrow q_{0}$; in theory, this could occur an infinite amount of times before ever reaching $q_{3}$ - how do we incorporate this into our probability calculation?

答:
I'm not familiar with this exact kind of calculation, but I believe this procedure makes sense:

If we get to $q_3$, we have to pass through $q_1$ some nonzero number of times, let's call this number $n$ and the event of getting to $q_3$ passing $q_1 n$ times is $A_n$.

$n$ clearly can't take on more than one value for a single path, so we have disjoint spanning events. So, we can use the law of total probability to add up the probabilities for each value of $n$, which should give us a convergent infinite sum.

To begin with, if $n = 0$ it's pretty clear to see that the probability of getting to $q_3$ is simply $(0.5)(0.35) = 0.175$. Then, for any other value of $n$, starting from $q_0$ we have to run the loop once and then take the path with one fewer loop. This gives the recurrence relation $P(A_n) = 0.225P(A_{n-1})$. From here it should be clear that $P(A_n) = 0.175(0.225^n)$.

Now all we have to do is add up all of the probabilities: $P(A) = \Sigma_{n=0}^{\infty}0.175(0.225^n) = \frac{0.175}{1-0.225} = \frac{7}{31} = 0.2258$.

Hope this helps!


问:
Large N limit for $P(\mathbf{x} \cdot\mathbf{y}\cdot \sum\mathbf{x}\cdot \sum \mathbf{y}>0)$ for two binary vector $\mathbf{x}$ and $\mathbf{y}$
I have two binary vectors $\mathbf{x}$, and $\mathbf{y}$, each with $N$ elements; Each element of $\mathbf{x}$ and $\mathbf{y}$ belongs to {-1, 1}, and is drawn from uniform random distribution.

Now I would like to derive the large $N$ limit for $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})>0)$ and $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})<0)$, where $\mathbf{x} \cdot \mathbf{y}$ is the inner product between $\mathbf{x}$ and $\mathbf{y}$, and $\sum\mathbf{x}$, $\sum\mathbf{y}$ are the sum of all elements in $\mathbf{x}$ and $\mathbf{y}$, respectively.

I have derived the explicit form for $P(\mathbf{x} \cdot \mathbf{y} =k, \sum\mathbf{x}=x, \sum \mathbf{y}=y)$, which is
\begin{align}
P(\mathbf{x} \cdot \mathbf{y}=k, \sum\mathbf{x} = x, \sum \mathbf{y} = y) =\frac{{N \choose (N + x)/2}  \cdot{ (N+x)/2 \choose (N+x+y+k)/4 } \cdot {(N-x)/2 \choose (N+k-x-y)/4}}{2^{2N}}
\end{align}

From the above formula, I computed $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})>0)$ and $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})<0)$ from nested sum over all possible $x$, $y$ and $k$ values from $n=2$ to $n=400$, and the results suggest that $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})>0)$ and $P(\mathbf{x} \cdot \mathbf{y} \cdot (\sum\mathbf{x}) \cdot (\sum \mathbf{y})<0)$ are both converging to $0.5$, see the plot below:

However, I have hard time trying to derive the large $N$ limit in a formal way.

Another side question is that, is above $P(\mathbf{x} \cdot \mathbf{y}=k, \sum\mathbf{x} = x, \sum \mathbf{y} = y)$ an known probability distribution?

答:
Edited

Let $a^T=[1, 1, ..., 1]$, that is an all-one vector of length $N$. Then you can rewrite the random variable as:

$Z=(Y^T X)(X^T a)(a^T Y)$,

in which $T$ represent transpose, so $Y^TX$ is the inner product of $X$ and $Y$, $X^T a$ is the sum of the elements of $X$, and $a^T Y$ is the sum of the elements of $Y$. The reason I wrote it this way will be clear in a few moments!

You are interested in large $N$, so I suggest investigating the mean of $Z$ first, as follows.

$\mu = E\{Z\} = E\{Y^T X X^T a a^T Y\}$

Since $Y^T X X^T a a^T Y$ is a scalar, it equals to its trace. I will use the fact that $tr(AB)=tr(BA)$, also the fact that trace and expectation are both linear operators and interchangeable:

$\mu = E\{Y^T X X^T a a^T Y\} = tr(E\{Y^T X X^T a a^T Y\}) = E\{tr(Y^T X X^T a a^T Y)\} = E\{tr(Y Y^T X X^T a a^T)\} $

where I took $A=Y^T X X^T a a^T$ and $B=Y$ in $tr(AB)=tr(BA)$. So,

$\mu = E\{tr(Y Y^T X X^T a a^T)\} = tr(E\{Y Y^T X X^T a a^T\})
= tr(E\{Y Y^T\} E\{X X^T\} a a^T\})$.

Since $E\{Y Y^T\}=E\{X X^T\} = I_N$, where $I_N$ is the identity matrix of size $N$, and also $a a^T=1_N$ is an all-one matrix of size $N$:

$\mu = tr(I_N I_N 1_N) = N$.

The expected value of your random variable diverges with $N$. I suspect that this might change your question.


问:
Euler's Formula, from Needham's Visual Complex Analysis
I'm having trouble understanding this passage.

12 Geometry and Comples Arithmetic

Figure $[8]$

Since we now know that $|Z(t)|$ remains equal to $1$ throughout the motion, it follows that the particle's speed $|V(t)|$ also remains equal to $1$. Thus after time $t=\theta$ the partial will have travelled a distance $\theta$ round the unit circle, and so the angle of $Z(\theta)=e^{i\theta}$ will be $\theta$. This is the geometric statement of Euler's formula.

I don't understand the last paragraph.  Why is it that $|Z(t)|$ remains equal to 1 throughout the motion?

答:
I disagree with @MarkFantini's answer.

In this section ($\S$1.II.2 Moving Particle Argument), Needham is giving a heuristic argument in support of the fact that $e^{i \theta} = \cos \theta + i \sin \theta$. He has not defined $e^{i \theta}$ to be $\cos \theta + i \sin \theta$, and in fact he argues that doing so is "a low blow to Euler, reducing one of his greatest achievements to a mere tautology" ($\S$1.II.1 Introduction, page 10).

Instead, what Needham is arguing in $\S$1.II.2 is that if we suppose that it makes sense to talk about $e^{it}$, then a natural property we would want it to have is $\frac{d}{dt}e^{it} = i e^{it}$ (eq. (11) on page 11), analogous to the known property of the real exponential function, namely $\frac{d}{dx} e^{kx} = k e^{kx}$ for any real $k$.

Now, visualizing the function $t \mapsto e^{it}$ as describing the motion of a moving particle in the plane, the quantity $\frac{d}{dt} e^{it}$ is then the velocity of the particle. We know that multiplication by $i$ corresponds to a rotation of the plane by $\pi/2$ radians in the counter-clockwise direction. So, the position and velocity vectors are at right angles to each other at every point. From this, and the fact that at $t = 0$ the particle is at $e^0 = 1$, one can deduce that the particle is just moving around the unit circle.

Hence, $\lvert e^{it} \rvert = 1$ for all $t$.


答:
The successful paths are of the form $q_0 \rightarrow (q_1 \rightarrow q_0)^n \rightarrow q_1 \rightarrow q_3$ with $n \geqslant 0$. The corresponding probability is
$$
0.5 \times \Bigl(\sum_{n \geqslant 0} (0.45 \times 0.5)^n \Bigr) \times 0.35 = 0.5 \times \Bigl(\frac{1}{ 1 - (0.45 \times 0.5)} \Bigr) \times 0.35 = \frac{0.5  \times 0.35}{0.775} = 0.22580645161\cdots
$$

Title: Finding the probability of a multi-state transition in a finite state machine

Body:
Finite State Machine Example (cannot embed yet)

I'm struggling to understand how to compute the probability of $q_{0} \rightarrow q_{3}$.

From my understanding, given that the transition from $q_{1}$ back to $q_{0}$ did not exist, the probability of reaching $q_{3}$ from $q_{0}$ would be $0.5 * 0.35$.

However, in the provided state diagram, we have a state transition from $q_{1} \rightarrow q_{0}$; in theory, this could occur an infinite amount of times before ever reaching $q_{3}$ - how do we incorporate this into our probability calculation?


问:
Optimal bet given a repeated coin flip game
A brainteaser question obtained from here states:

Flip 10000 fair coins. You are offered a 1-1 bet that the sum is less than 5000. You can bet 1, 2, ..., 100
  dollars. How much will you bet. How much will you bet if someone tells you that the sum of the coins is
  less than 5100?

I believe that maximizing the expected value would give you 100 for the first situation because there are 5001 win cases and 5000 loss cases, and betting 1 (the minimum) in the second case because now the probability of losing is greater than 0.5.

However the question of "how much will you bet" is ambiguous to me. Is expected value the right metric for betting? Are there other reasonable metrics to consider before making a bet?

答:
Let's think about the problem with fewer coins, say 10. It is the same problem as if you had 1 fair coin with 10 flips. For 10 coins with 2 sides each there are $2^{10}$ possible outcomes. The number of ways to get exactly 4 heads from 10 is the binomial coefficient, 10C4 = 210. Dividing 210 by 1024, we get 20.5%. Now what's the probability that there are less than 5 heads? Sum up the binomial coefficients for $n < 5$ and divide by all potential outcomes, 210. That gives us 37.7%.

Now we can introduce the notion of expected value. You have a probability of 0.377 of the number of heads being less than 5, for 10 flips. That means there is a (1-0.377) probability of there being 5 or more heads. How much can you expect to make from betting $X$ dollars that there are less than 5 heads? That is simply the sum of the dollar amount associated with each of your probabilities. You've got X for 0.377, and you've got $-X$ for (1-0.377). The expected value is then $0.377X-X+0.377X=-0.246X$.

Whatever $X$ you have, you should expect to lose 24.6% of it!

Using Wolfram Alpha to solve the problem for 10000 coins and less than 5000 heads, you've got a probability of 0.496 of getting your money back. That means on average, you can expect to lose 0.8% of your money X.

For less than 5100 you get a probability of 0.977. That means your expected value is 0.954 and thus you should expect to get your money back plus 95.4% of it. An extra 100 heads of cushion makes this a slam dunk bet, despite intuition.

The correct number to bet depends on your risk tolerance and basically gives insight to your interpretation of risk-reward, given the probability of each outcome. The Kelly Criterion mentioned above could be a useful rule of thumb. Using that, we get 0.977*2-1=95.4% of capital.

Now moving on to the second question: how much will you bet if you are given the information that the sum of the coins is definitely less than 5100. That is a Bayesian conditional probability problem.

You're basically gaining a measurement, and that will affect your probabilities. The formula is given as:

$$
{\rm P}(A | B) = \frac{{\rm P}(B|A){\rm P}(A)}{{\rm P}(B)}.
$$

The formula is actually pretty easy to calculate for this situation. Event A is the probability of getting less than 5000 heads. Event B is the probability of getting less than 5100 heads. The bar means given. So the probaility of A given B is the probability of B given A times ${\rm P}(A)/{\rm P}(B)$. What is the probability of B given A? Well, if it's given that it is less than 5000, then it's certainly less than 5100! So ${\rm P}(B|A) = 1$. Plug in the two values we calculated above: ${\rm P}(A)=0.496$ and ${\rm P}(B)=0.977$. That means ${\rm P}(A|B)=0.508$. So that bit of extra knowledge gives you an edge in the game. If you are doing it a million times, then you can expect a positive return, no matter how much you put up.


问:
Not more than one perfect matching in a tree
I have to show that a tree $T = (V,E)$ has at most one perfect matching.

In order to show that, one has to assume that the tree $T$ has an even number of vertices, so that a perfect matching can exist. Due to that I think that one should work inductively by adding an even number of vertices (at least two) to the existing tree and showing that they "become" part of the already existing matching. The thing is the formality..

Thank you for your help

答: You are on the right track. Let $v$ be a leaf vertex with parent $u$. You should show that $T$ has a perfect matching if and only if $T - \{u,v\}$ has a perfect matching, and more precisely, the perfect matchings of these graphs are in a one-to-one correspondence. This will let you use induction on the number of vertices. Note that $T - \{u,v\}$ may be disconnected, but this is not a problem if you do the induction for forests instead of trees.


问:
Let $A,\,B$ be subsets of the set $X$. Prove one of de Morgan's Laws of Set Theory
Let $A,\,B$ be subsets of the set $X$. Prove one of de Morgan's Laws of set theory:$$X\setminus(A\cap B)=(X\setminus A)\cup(X\setminus B)$$

答:
Sketch of the proof. We want to prove that two sets are equal. Recall the definition of equality between sets.

Definition. Let $A$ and $B$ be any sets. We say that the sets $A$ and $B$ are equal, and we write $A = B,$ if $A \subseteq B$ and $B \subseteq A.$

Hence, we must prove two inclusions, by proving that for an arbitrary object $c$
\begin{align*}
c \in X\setminus(A\cap B) \implies c \in (X\setminus A)\cup(X\setminus B) && \text{and}\\
c \in (X\setminus A)\cup(X\setminus B) \implies c \in X\setminus(A\cap B).
\end{align*}

Proof. Let $x$ be an arbitrary object. Then,
\begin{align*}
x \in X \setminus (A \cap B) & \iff x \in X \wedge x \notin A \cap B & \text{(definition of set difference)}\\
& \iff x \in X \wedge \sim (x \in A \wedge x \in B) & \text{(definition of set intersection)}\\
& \iff x \in X \wedge (x \notin A \vee x \notin B) & \text{(De Morgan’s Law)}\\
& \iff (x \in X \wedge x \notin A) \vee (x \in X \wedge x \notin B) & \text{(distributive property of $\wedge$)}\\
& \iff x \in X \setminus A \vee x \in x \in X \setminus B & \text{(definition of set difference)}\\
& \iff x \in (X \setminus A) \cup (X \setminus B) & \text{(definition of set union)}
\end{align*}
Hence, we have prove that each of the sets $X \setminus (A \cap B)$ and $(X \setminus A) \cup (X \setminus B)$ is a subset of the other. Then, by definition of equality between sets we conclude that $X \setminus (A \cap B) = (X \setminus A) \cup (X \setminus B).$ $\square$

Remark. Note that in the above proof we only deal with equivalences and not just with implications, that’s because we are relying on the definitions to justify all of our steps plus the De Morgan’s Law which is itself an equivalence.


问:
whats is topology of $[0,1]$ in Urysohn Lemma?
Urysohn Lemma is

A topological space $(X, \mathcal{T})$ is normal if and only if for every
pair of disjoint nonempty closed subsets $C, D \subseteq X$ there is a continuous function $f : X \rightarrow [0, 1]$
such that $f(x) = 0$ for all $x \in C$ and $f(x) = 1$ for all $x \in D$.

are there some limits for the topology of $[0,1]$? if the topology of $[0,1]$ is trivial topology $\{\emptyset,X\}$,  i think such $f$ always exists?

so my question is that does Urysohn Lemma means for any topology of $[0,1]$ there is always such $f$ exists?

答: No, it is not valid for any topology. The topology used here is the standard topology, the one induced from the standard topology on $\Bbb R$.


问:
Find the value of $ A + B + C + D + E. $
KTOM Indonesia, October 2020

$\sin(5^{\circ})\sin(55^{\circ})\sin(65^{\circ})=\displaystyle\frac{A\sqrt{B}+C\sqrt{D}}{E}$

where $ B $ and $ D $ are two natural numbers that are not divisible by the square of any prime, $ A $ and $ C $ are integers, and $ E $ is a natural number. Find the value of $ A + B + C + D + E. $

答:
Continuing from what others said....
$$sin(5^{\circ})sin(55^{\circ})sin(65^{\circ})=\frac{sin(15^{\circ})}{4}$$

$$cos(30^{\circ})=1-2sin^{2}(15^{\circ})$$
$$2sin^{2}(15^{\circ})=1-cos(30^{\circ})$$
$$2sin^{2}(15^{\circ})=1-cos(30^{\circ})$$
$$2sin^{2}(15^{\circ})=1-\frac{\sqrt3}{2}$$
$$2sin^{2}(15^{\circ})=\frac{2-\sqrt3}{2}$$
$$sin^{2}(15^{\circ})=\frac{2-\sqrt3}{4}$$
$$sin(15^{\circ})=\frac{\sqrt{2-\sqrt3}}{2}=\frac{\sqrt{\biggl(\sqrt\frac{1}{2}-\sqrt\frac{3}{2}\biggr)^{2}}}{2}$$
$$=\frac{\biggr(\sqrt\frac{3}{2}-\sqrt\frac{1}{2}\biggl)}{2}\ \ (as\  sin(15^{\circ})\ is\ always\ positive)$$
$$\Rightarrow \frac{sin(15^{\circ})}{4}=\frac{\biggr(\sqrt\frac{3}{2}-\sqrt\frac{1}{2}\biggl)}{8}=\biggr(\frac{\sqrt3-1}{8\sqrt2}\biggl)=\biggr(\frac{\sqrt6-\sqrt2}{16}\biggl)$$
$$\Rightarrow\frac{A\sqrt{B}+C\sqrt{D}}{E}=\biggr(\frac{1\sqrt6-1\sqrt2}{16}\biggl)$$
1 and-1 are integers, 2 and 6 are natural numbers which is not divisible by square of any prime number and 16 is a natural number.

So,
$$A+B+C+D+E=1+6-1+2+16=24$$


问:
Weak topology of $X\times [0,1]$ with respect to a family of subspaces
This problem appeared to me at first when studying topology of CW-complexes. I also needed ths fact when trying to prove that some graphs are contractible, when seen as complexes. Basically, I need to prove that, if $X$ has the weak topology with respect to some subspaces $A_{i}$, and $\mathbb{I}$ is the unit interval $[0,1]$, then the product $X\times \mathbb{I}$ has too the weak topology with respect to the spaces $A_{i}\times [0,1]$.

What I mean in ''$X$ has the weak topology with respect to the subspaces $A_{i}$'' is that for every subset $B\subseteq X$, $B$ is open if and only if $B\cap A_{i}$ is an open set in $A_{i}$ for every $i$. (this definition appears in the book "Topology", by J. Dugundji, page 131).

What I wanted to prove is the following:

Let $X$ be a topological space and $\{A_{i}\}_{i\in I}$ a family of subspaces whose union is $X$. If $X$ has the weak topology with respect to $\{A_{i}\}_{i\in I}$, then $X \times \mathbb{I}$ has the weak topology with respect to $\{A_{i}\times \mathbb{I}\}_{i\in I}$.

At first, I tried to prove it by definition: the problem is that while I can prove that the product topology is contained in the weak topology (by means of basic open sets), the converse didn't seem as easy, since I don't have a clear basis for $X\times \mathbb{I}$ with the weak topology. I have tried now a quotient topology argument, which goes as follows:

Firstly, consider the mapping $p\colon\bigsqcup_{i\in I}A_{i}\to X$ (where the domain is the disjoint union of the $A_{i}$'s) generated by the inclusions $\iota_{i}\colon A_{i}\to X$. It is clear from the definitions that $X$ has the weak topology with respect to $\{A_{i}\}_{i\in I}$ if and only if $p$ is a quotient map.

Secondly, by repeating the argument in the case of $X\times \mathbb{I}$, we have that $X\times \mathbb{I}$ has the weak topology with respect to the family $\{A_{i}\times \mathbb{I}\}_{i\in I}$ if and only if the map $q\colon\bigsqcup_{i\in I}(A_{i}\times \mathbb{I})\to X\times \mathbb{I}$ induced by the inclusions $\iota_{i}\times \operatorname{Id}_{\mathbb{I}}\colon A_{i}\times \mathbb{I}\to X\times \mathbb{I}$ is a quotient map. That said, I expressed $q$ as the composition of $\alpha \colon\bigsqcup_{i\in I}(A_{i}\times \mathbb{I})\to \big(\bigsqcup_{i\in I}A_{i}\big)\times \mathbb{I}$, which is the ''identity map": for every $(a,t)\in A_{i}\times \mathbb{I}$, $\alpha(a,t)=(a,t)$, where I consider that $a$ lies precisely in $A_{i}$, and $p\times \operatorname{Id}_{\mathbb{I}}\colon \big(\bigsqcup_{i\in I}A_{i}\big)\times \mathbb{I}\to X\times \mathbb{I}$. On one hand, $\alpha$ is a homeomorphism, since it is open, continuous and bijective. On the other hand, Whitehead's Theorem asserts that since $p$ is a quotient map, and $\mathbb{I}$ is a locally compact space, then $p\times \operatorname{Id}_{\mathbb{I}}$ is a quotient map, hence $q=(p\times \operatorname{Id}_{\mathbb{I}})\circ \alpha$ is a quotient map as a composition of quotient maps.

Finally, since $q$ is a quotient map, it follows that $X\times \mathbb{I}$ has the weak topology with respect to $\{A_{i}\times \mathbb{I}\}_{i\in I}$, which would make the proof complete.

Is this argument correct? And would it still be applicable when we change $\mathbb{I}$ for, say, a locally compact space?

Edit: I'll add my attempt at proving that $\alpha$ is a homeomorphism for completeness' sake:

$\bullet$ $\alpha$ is clearly bijective by construction.

$\bullet$ $\alpha$ is continuous: given a basic open set $B\times C\subseteq \big(\bigsqcup_{i\in I}A_{i}\big)\times \mathbb{I}$, we have that $B\cap A_{i}$ is open in $A_{i}$ for all $i$, so that $B\times C$ is also open in $A{i}\times \mathbb{I}$ for every $i$. Therefore, $B\times C=\alpha^{-1}(B\times C)$ is open in $\bigsqcup_{i\in I}(A_{i}\times \mathbb{I})$.

$\bullet$ $\alpha$ is open: it is enough (since $\bigsqcup_{i\in I}A_{i}$ has the disjoint union topology) to prove that a set of the form $B\times C$, where $B\subseteq A_{i}$ (for some $i$) is open and $C\subseteq \mathbb{I}$ is open has an open image. But $\alpha(B\times C)=B\times C$ is open, since $B\cap A_{j}=\emptyset$ if $j\neq i$, and $B\cap A_{i}=B$, which are all open sets. Therefore, $\alpha(B\times C)$ is open as a product of open sets in their respective factors.

Thank you in advance!

答:
Is this argument correct? And would it still be applicable when we change $\mathbb{I}$ for, say, a locally compact space?

Yes,  the argument is essentially correct for any locally compact space $\mathbb{I}$. Good job! To make the argument rigorous, we need to replace $A_i$’s in the disjoints sums by its pairwise disjoint homeomorphic copies.


问:
A little help with elementary set intuition, specifically on intersection
Simple example, $A\cup 0=A$ is true because $x\in A$ or $x\in 0$ and that $0\subseteq A$. This I can understand (Actually, I hope I understood it)

But what I am having difficulty with is something like $A\cup(B\cap C)$. I know this implies $x \in A$ or $x\in (B\cap C)$. If $x\in A$, then $x\in(A\cup B)$ and $x\in (A\cup C)$ or alternitavely, $(A\cup B) \cap (A\cup C)$. What I am confused by is how can it be both? Isn't it that the OR in $x \in A$ or $x\in (B\cap C)$ implies that  $x$ is either in $A$ OR $x$ is in $(B\cap C)$ and not both? Obviously not, but I am having a bit of difficulty wrapping my head on why that is exactly.

答:
Sketch of the proof. We want to prove that two sets are equal. Recall the definition of equality between sets.

Definition. Let $A$ and $B$ be any sets. We say that the sets $A$ and $B$ are equal, and we write $A = B,$ if $A \subseteq B$ and $B \subseteq A.$

Hence, we must prove two inclusions, by proving that for an arbitrary object $c$
\begin{align*}
c \in A \cup (B \cap C) \implies c \in (A \cup B) \cap (A \cup C) && \text{and}\\
c \in (A \cup B) \cap (A \cup C) \implies c \in A \cup (B \cap C).
\end{align*}

Recall also the distributive property of the logical operation $\vee.$

Theorem. Let $p,q$ and $r$ be any propositions. Then
$$p \vee (q \wedge r) \iff (p \vee q) \wedge (p \vee r).$$

(The proof of this theorem is quite easy, you just have to construct the truth table for the proposition above stated and conclude that it is a tautology.)

Proof. Let $x$ be an arbitrary object. Then,
\begin{align*}
x \in A \cup (B \cap C) & \iff x \in A \vee x \in B \cap C & \text{(definition of set union)}\\
& \iff x \in A \vee (x \in B \wedge x \in C) & \text{(definition of set intersection)}\\
& \iff (x \in A \vee x \in B) \wedge (x \in A \vee x \in C) & \text{(distributive property of $\vee$)}\\
& \iff x \in A \cup B \wedge x \in A \cup C & \text{(definition of set union)}\\
& \iff x \in (A \cup B) \cap (A \cup C) & \text{(definition of set intersection)}
\end{align*}
Hence, we have prove that each of the sets $A \cup (B \cap C)$ and $(A \cup B) \cap (A \cup C)$ is a subset of the other. Then, by definition of equality between sets we conclude that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C).$ $\square$

Remark. Note that in the above proof we only deal with equivalences and not just with implications, that’s because we are relying on the definitions to justify all of our steps plus the distributive property of $\vee$ which is itself an equivalence.

In fact note that those two sets are the same. Always keep in mind that, in mathematics, we always use an inclusive or which means that saying $x \in A \cup B$ means $x \in A, x \in B$ or $x$ in both $A$ and $B,$ i.e., $x$ is in at least of of those sets $A$ or $B.$

Also, and I leave this as something for you to think about, what would happen if they are not the same? (I showed that they are the same, but answering this question is a really good way to test if you understood).


问:
Eigenvalues of a Hilbert-Schmidt operator
Let $\alpha>1/2$. What is the point spectrum of the operator $Tf(x)=\int_0^1(x^\alpha+y^\alpha)f(y)dy$ on $L^2([0,1],dx)$? Nobody ever showed me how to calculate the eigenvalues for this kind of operators.

答: Ok, I managed to solve this. It's just a matter of writing, for some $\lambda>0$ candidate and $f\neq0$ in $L^2$, $\lambda f(x)=x^\alpha\int_0^1 fdy+\int_0^1y^\alpha fdy$, so that $f(x)=Ax^\alpha+B$ for some $A,B\in\mathbb{C}$. Hence, $\int_0^1(Ax^\alpha+b)(x^\alpha+y^\alpha)dy=\lambda(Ax^\alpha+B)$. The left hand side gives a polynomial in $x$ that must be equal a.e. to the right hand side. Thus we have to solve a linear system $2\times 2$ and this is easy to do. Then, one has to verify that $A,B$ found satisfy the eigenvalue equation and for which $\lambda$ they do...


问:
Derivations on semisimple Lie algebra
First recall some definitions :

Let $B$ be a Killing form on Lie algebra $\mathfrak{g}$ over ${\bf R}$ such that
   $B(X,Y)\doteq Tr(ad_Xad_Y)$.

$\mathfrak{g}$ is semisimple if $B$ is non-degenerate.

Define $\partial \mathfrak{g} \doteq \{ D | D[X,Y]=[DX,Y] + [X,DY] \}$ Clearly it contains $ad(\mathfrak{g})\doteq \{ ad_X | X\in \mathfrak{g}\}$.

My question is about proof of $\partial \mathfrak{g}=ad(\mathfrak{g})$ :

I am reading Helgason's book. We can show that $ ad(\mathfrak{g}) = \mathfrak{g}$ easily.

I cannot proceed the proof since $B$ is defined on $\mathfrak{g}$ only.

How can I complete the proof ? Thank you in advance.

答:
Here's a proof from Humphrey's excellent book:

First of all, show that $I=ad(\mathfrak g)\subset\partial\mathfrak g$ is an ideal of the Lie algebra $\partial\mathfrak g$: if $x\in\mathfrak g$ and $\delta\in\partial\mathfrak g$, then $$[\delta,ad(x)]=\cdots\in I$$ This implies that the Killing form on $\mathfrak g$ coincides with that induced by the Killing form on $\partial\mathfrak g$. Since $\mathfrak g$'s Killing form is non degenerate by semisimplicity, we have
$$I\oplus I^{\perp}=\partial\mathfrak g$$
with respect to the Killing form on the derivations. Now consider $\delta\in I^{\perp}$. For any $x\in\mathfrak g$, since both $I$ and $I^{\perp}$ are ideals in $\partial\mathfrak g$, $$[\delta,ad(x)]\in I\cap I^{\perp}=\lbrace 0\rbrace$$ equals $ad(\delta(x))$ by the calculation above, so for all $x\in\mathfrak g,~\delta(x)\in\ker(ad)=\lbrace 0\rbrace$. That is, $\delta=0$ and $I^{\perp}=\lbrace 0\rbrace$, so that  $ad(\mathfrak g)=\partial\mathfrak g$.

Title: Derivations on semisimple Lie algebra


问:
Convergence/Divergence proof for Infinite products
Exercise 2.4.10 is about infinite products. I'd like someone to verify, if my convergence/divergence proof is technically correct and rigorous.

A close relative of the infinite series is the infinite product.
\begin{align*}
 \prod_{n=1}^{\infty} b_n = b_1 b_2 b_3 \cdots
\end{align*}

which is understood in terms of its sequence of partial products.
\begin{align*}
 p_m = \prod_{n=1}^{m} b_n = b_1 b_2 b_3 \cdots b_m
\end{align*}

Consider the special class of infinite products of the form
\begin{align*}
 \prod_{n=1}^{\infty}(1+a_n) = (1+a_1)(1+a_2)(1+a_3)\cdots
\end{align*}

where $a_n > 0$.

(a) Find an explicit formula for the sequence of the partial products in the case where $a_n = 1/n$ and decide whether the sequence converges. Write out the first few terms in the sequence of partial products in the case where $a_n = 1/n^2$ and make a conjecture about the convergence of this sequence.

(b) Show, in general, that the sequence of partial products converges if and only if $\sum_{n=1}^{\infty}a_n$ converges. (The inequality $1+x \le 3^x$ for positive $x$ will be useful in one direction.)

Proof.

(a) If $a_n = 1/n$, the partial product
\begin{align*}
p_m &:= \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{m}\right)\\
&= 1 + \sum_{m} \frac{1}{m} + \sum_{m < n} \frac{1}{m \cdot n} + \sum_{m < n < p} \frac{1}{m \cdot n \cdot p} + \ldots
\end{align*}

Since $\sum 1/m$ is the harmonic series, which is well-known to be divergent, this sequence of partial products is divergent.

Consider the case of $a_n = 1/n^2$. We are going to use the inequality $1 + x^2 \le e^{x^2}$. We have:
\begin{align*}
p_m &=  \left(1+\frac{1}{1^2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{3^2}\right)\cdots\left(1+\frac{1}{m^2}\right)\\
&\le e^{\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots + \frac{1}{m^2}}\\
&\le e^{\frac{1}{1\cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{m\cdot (m+1)}}\\
&\le e^{\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \ldots + \left(\frac{1}{m} - \frac{1}{m+1}\right)}\\
&= e^{1 - \frac{1}{m+1}}\\
&\le e
\end{align*}

Moreover, $p_1 = 2$, $p_2 = (2)(5/4)$, $p_3 = (2)(5/4)(10/9)$, $p_4 = (2)(5/4)(10/9)(17/16)$. So, $(p_m)$ is monotonic increasing and bounded by $e$. Consequently, by the Montone Convergence Theorem, $(p_m)$ is convergent.

(b) ($\longrightarrow$) direction.

The sequence of partial products,
\begin{align*}
p_n &= (1+a_1)(1+a_2)\cdots(1+a_n)\\
&\le e^{a_1}\cdot e^{a_2}\cdots e^{a_n}\\
&\le \exp \left({\sum_{k=1}^{n}a_k}\right)
\end{align*}

If the sequence of partial products is convergent and therefore bounded, the sequence of partial sums $\sum_{k=1}^{n} a_k$ must be bounded and convergent.

答:
For (a) what you have done is correct.

For (b) your argument is not valid. Note that $\sum \ln (1+a_n) <\infty$. This implies that $\ln (1+a_n) \to 0$ so $a_n \to 0$. Now, there exists $\delta >0$ such that  $\ln (1+x) \geq \frac 1 2 x$ for $0<x <\delta$ (because $\lim_{x \to 0}\frac {ln (1+x)} x=1$). Hence, $a_n <2 \ln (1+a_n)$ for $n$ sufficiently large. This proves that $\sum a_n <\infty$.

The converse part follows similarly by looking at $\ln (1+a_n)$ and noting that $\ln (1+x) \leq x$ for all $x >0$.


问:
Show groups of symmetries of a cube and a tetrahedron are not conjugate in isometry group.
I've shown that the symmetry group of a cube and a tetrahedron are both isomorphic to S4, but I am now trying to show that they are not conjugate when considered as subgroups of isometries of 3D space.

I cannot think of any kind of criteria that gives when two groups are not conjugate?

EDIT: I meant to say the rotations of a cube! So I've shown the full symmetry group of the tetrahedron is isomporphic to the rotations of a cube, however I need to show that they are not conjugate.

答:
I think I've answered it now. If we just think of all the group elements as matrices. Then we want to show that these two groups are not conjugate. So we try and find an element to "conjugate with" from the isometry group. So we get $A = C^{-1}BC$ where $A$ is an element of the rotations of a cube, and B an element from the rotations and reflections of the tetrahedron.

It then follows that $det(A) = det(B)$ and since for rotations $det(A) = 1$, and for reflections $det(B) = -1$, then we see that certain elements cannot possibly be conjugate.


问:
Why is $\sum_{k=2}^{\infty} \frac{k}{k^{2}-1}=\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$?
I am working through the problems on this page.  I am stuck on the fourth problem. In the hints they claim that $\displaystyle \sum_{k=2}^{\infty} \frac{k}{k^{2}-1}=\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$. But why does this equality hold?

We have:
$$\frac{1}{k-1}+\frac{1}{k+1} =\frac{k+1}{(k-1)(k+1)}+\frac{k-1}{(k+1)(k-1)}=\frac{(k+1)+(k-1)}{(k-1)(k+1)}=\frac{2k}{(k-1)(k+1)}=\frac{2k}{k^2-1}\neq \frac{k}{k^2-1} $$

And therefore shouldn't we have $\displaystyle \sum_{k=2}^{\infty} \frac{k}{k^{2}-1}\neq\sum_{k=2}^{\infty}\left(\frac{1}{k-1}+\frac{1}{k+1}\right)$?

答: This is likely just a typo, in any case it doesn't change the problem much at all because this is a dead end: you can't evaluate the sum this way because it diverges.</p>


问:
Terminology for a function following an exponential growth but being discrete for non-negative integer numbers
We currently have a discussion at work about the terminology for a function following this simple exponential law:

$$
y = C^x
$$

where x is following discrete steps
$$
x \in N _0 (0, 1, 2, ...)
$$

What would be the preferred terminology for this function? Is it still called an exponential function if it is discretized? Or would it be called an exponential series?

Sorry for this very basic question and thanks for any help in advance.

答:
Geometric sequence is probably the term you are looking for. You could say something like

Let $C$ be a fixed positive integer and let $u_n$ be the geometric sequence defined by $u_n = C^n$ for all $n \geqslant 0$.

Formally, a sequence of integers is a map from $\Bbb N$ to $\Bbb Z$, so this corresponds exactly to your need of having discrete steps. The sequence $n \to C^n$ is geometric, but the term geometric sequence covers the slightly larger class of sequences of the form $n \to ar^n$, where $a$ and $r$ are numbers.

EDIT (answer to the comment). The term sequence is sufficiently flexible to allow the case you describe. Just say: let $(u_n)_{n \geqslant 1}$ be the geometric sequence defined by $u_n = C^n$ for all $n \geqslant 1$. If $u_0$ is nevertheless defined, but if $u_0 \not= C^0$, you could still say that the restriction of the sequence to the positive integers is geometric or that the subsequence $(u_n)_{n > 0}$ is the geometric sequence $C^n$.


问:
How to integrate $\frac{\cos(x)}{e^x}$?
Surprisingly I haven't found such integral in the web, so I may ask this here bluntly.

A solution that Matlab suggest sounds like $\frac{{\mathrm{e}}^{-x}\,\left(\cos\left(x\right)-\sin\left(x\right)\right)}{2}$

答:
Just rewrite it as

\begin{equation*}
\int e^{-x} \cos x \ \mathrm{d}x
\end{equation*}
and integrate by parts.
That should lead you to your solution.

Title: How to integrate $\frac{\cos(x)}{e^x}$?


问:
Prove that $A\cap B\subseteq A$
I am unsure if I got it right, I proved it as follows:

$$A\cap B \longleftrightarrow x\in A \text{ and } x\in B \text{ is true }\forall x\in A \longleftrightarrow \subseteq A$$

Althought I kind of doubt the last part about it being true for all $x\in A$. Am I missing something?

I apologize in advance if this post is a duplicate.

答:
Sketch of the proof. We want to prove that one set is a subset of other set. Recall the definition of a subset.

Definition. Let $A$ and $B$ be any sets. We say that $A$ is a subset of $B$, and we write $A \subseteq B,$ if every element of $A$ is, also, an element of $B,$ i.e.,
$$(\forall x) x \in A \implies x \in B$$

Hence, to prove that $A \cap B \subseteq A$ we must prove the above implication, i.e., prove that for an arbitrary objet c we have that $c \in A \cap B \implies c \in A.$

Proof. Let $x$ be an arbitrary object. Suppose that $x \in A \cap B.$ By definition of set intersection, we have that $x \in A$ and $x \in B.$ In particular, we have that $x \in A.$ Hence, by definition of subset, we conclude that $A \cap B \subseteq A.$ $\square$

Remark. Also, (and maybe to get some intuitive sense), note that if you have the propositions $p$ and $q$ and you know that $p \wedge q$ is true, what can you say about each $p$ and $q?$

They have to be both true (by definition of conjunction). Suppose that (at least) one of them is false. Without loss of generality, suppose that $p$ is the one which if false. Hence, $p \wedge q$ would be false, which is an absurd because it contradict the fact that $p \wedge q$ is true. The absurd came from assuming that (at least) one of them is false. Hence, both $p$ and $q$ must be true.

From this discussion, observe that $x \in A cap B$ is logically equivalent to $x \in A \wedge x \in B.$ Hence, if this proposition is true, then both $x \in A$ and $x \in B$ are true. Therefore, $A \cap B \subseteq A$ and $A \cap B \subseteq B.$

Title: Prove that $A\cap B\subseteq A$


问:
Characterizing isolated singularities of $\frac{z}{e^{z} - z + 1}$
I intend to characterize the isolated singularities of $f(z) := \frac{z}{e^z - z + 1}$ which is defined on some open subset $\mathbb{C} \backslash f^{-1}({0}) \subset \mathbb{C}$. The possible singularities are only the zeros of $g(z) := e^z - z + 1$, so the approach should be to find those zeros, which is actually the difficult task.

At this point, what I know is the following:

Since no zeros of $g$ lie on $2\pi i \mathbb{Z}$, if we suppse that $z_{0} \in \mathbb{C}$ is a zero of $g$, then we will of course have $\lim\limits_{z \to z_{0}} |f(z)| = \infty$ and so $z_{0}$ will be a pole of $f$. Using then L'Hôpital's rule applied to $\lim\limits_{z \to z_{0}} (z-z_{0})^n f(z)$, we check that the pole will have to be of order $1$.

$g$ has zeros, which I found out by separating the function $g(z) = 0$ in two equations, concerning the real and imaginary parts of $g$ and then by ploting the resulting functions to see that they intersect.

My question is then: Can we see analytically that $g$ has zeros?

This is an exercise from a Reinhold Remmert's book, "Theory of Complex Functions" (line c) in exercise 1 on page 309) and I think I'm supposed to solve it analytically. Also, keep in mind that, at this point in the book, there are a lot of tools in complex analysis still not available to use, namely residue calculus and Laurent series.

Thank you in advance for all the help!

答:
Actually, the exercise is only about classifying those isolated singularities and, in the case of poles, to find their order. No reference is made to actually seeing them.

If $z_0$ is a zero of $g$, can it be a multiple zero? If it was, then it would be a zero of $g'$ too. But $g'(z_0)=e^{z_0}-1$. And $g(z_0)=0\iff e^{z_0}-z_0+1=0$. So, $e^{z_0}$ is equal to $1$ and also to $z_0-1$, which means that $z_0=2$. But $g(2)=e^2-1\ne0$. Therefore, $g$ has no multiple zeros and so each isolated singularity of $f$ is a simple pole.


问:
Let $x=\cos\theta$. Show that $\frac{d^{n-1}\sin^{2n-1}\theta}{dx^{n-1}}=\frac{(-1)^{n-1}}{n}\frac{(2n)!}{2^{n}n!}\sin n\theta$
I need some help with the following exercise which I don't know how to start with. Any tip for a start would be already great. IT's from the AAR Special Functions book.

Let $x=\cos\theta$. Show that:

$\frac{d^{n-1}\sin^{2n-1}\theta}{dx^{n-1}}=\frac{(-1)^{n-1}}{n}\frac{(2n)!}{2^{n}n!}\sin n\theta$

答:
Hint:
Use the chain rule for single variable differentiation, and then try to use mathematical induction over n.</p>


问:
$(,) = x^2$ How to determine if injective or surjective?
I just learned to determine wether $f(x)=x^2$ is injective or surjective but how can I solve this when there are two variables? I've been struggling to get my head around this.
I know that for a function to be injective the following condition must be met:
$()≠()$ implies $≠$, or the contrapositive.
Surjective:
$∀ ∈ , ∃ ∈  \text{ such that } f(x) = y$.

How could I scale these definitions to handle two variables?
$(,) = x^2$
mapping from $\Bbb R \times \Bbb R$ to $\Bbb R$

Edit: I want to thank you all for your time and help. It's greatly appreciated! I can clearly see through these problems now.

答:
Hints: If $f$ is injective, $f(x,y)=f(x',y') \implies x=x',y=y'$. What are $f(-1,0)$ and $f(1,0)$?

For surjective, is there a point $(x,y)\in\mathbb{R}^2$ such that $f(x,y)=-1$?


问:
Application of Urysohn's lemma
I am working on the following hw problem: If we have that $X$ is a compact Hausdorff space, with $\{U_\alpha\}_{\alpha\in A}$, then we can find a finite number of continous functions $f_1,...,f_k$, with $f_i:X\mapsto [0,1]$ such that $f_1(x)+...+f_k(x)=1$ for all $x$, and for each $i$ there exists a $\alpha_i$ such that $\overline{f_i^{-1}((0,1])}\subset U_{\alpha_i}$.

So this are my thoughts at the moment. Since $X$ is Hausdorff and compact it is normal, so we know that Urysohn's lemma applies. Meaning, if I have any two disjoint sets I can find a continous function that is $0$ is one of the sets and $1$ in the other one.

My intuition is telling me to use compactness and find $U_{\alpha_1},..., U_{\alpha_k}$ that cover $X$, and then by the last requirement of the problem we would need $f_i(x)=0$ if $x\notin U_{\alpha_i}$.

The problem is very clear if I can write my space $X$ as the disjoint union of finite closed sets cause then I could just apply Urysohns and get my functions very easily, but I do not see how to apply this lemma using the open cover given. Any hints would be greatly appreciated.

答:
Let us recall some important definitions fristly:

A family $\{f_s\}_{s\in S}$ of continuous functions from a space $X$ to the closed unit interval $I$ is called a partition of unity on the space $X$ if $\Sigma_{s\in S}f_s(x)=1$ for every $x\in X$.

A partition of unity $\{f_s\}_{s\in S}$ on a spaceis subordinated to a cover $\mathscr U$ of $X$ if the cover $\{f_s^{-1}((0,1])\}_{s\in S}$ of the space $X$ is a refinement of $\mathscr U$.

Theroem 1: For every $T_1$-space $X$ the following conditions are equivalent:

1) The space $X$ is paracompact;

2) Every open cover of the space $X$ has a partition of unity subordinated to it.

The proof of it can be seen the General Topology by Engelking Page 302.

Note that every compact Hausdorff space is paracompact.

Title: Application of Urysohn's lemma


问:
Find $a$ such that $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions

Find $a$ so that equation $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions.

My approach: Since that $$\log_{2}(x)=\frac{\ln(x)}{\ln(2)} \quad \text{and} \quad \log_{\sqrt{2}}(x)=\frac{\ln(x)}{\ln(\sqrt{2})}$$
then, we can re-write the equation as $$\left(\frac{\ln(x)}{\ln(2)}\right)^{2}-\left( \frac{\ln(x)}{\ln(\sqrt{2})}\right)^{2}=a-\sqrt{a+\frac{\ln(x)}{\ln(2)}}$$
Let $t:=\ln(x)$ so the equation above we can re-write in terms  of $t$ as $$\left(\frac{1}{\ln(2)}t\right)^{2}-\left( \frac{1}{\ln(\sqrt{2})}t\right)^{2}=a-\sqrt{a+\frac{1}{\ln(2)}t}$$

Then, the idea that I was thinking is squaring both sides of the equation, but there appears an equation of degree 4 that is difficult to solve. How could I continue from there? or maybe a simpler approach?

New approach: Using the point out of @Ross Millikan and since that $$\log_{\sqrt{2}}(x)=\frac{\log_{2}(x)}{\log_{2}(\sqrt{2})}=2\log_{2}(x)$$
and let $t:=\log_{2}(x)$ so the equation can be we-write in terms of $t$ as $$t^{2}-2t=a-\sqrt{a+t} \implies t^{2}-2t\color{blue}{+1}=a-\sqrt{a+t}\color{blue}{+1} \implies (t-1)^{2}=a+1-\sqrt{a+t}$$
or maybe $$ t^{2}-2t=a-\sqrt{a+t} \overset{y^{2}=a+t}{\implies} (y^{2}-a)^{2}-2(y^{2}-a)=a-y^{2}$$
solving a little the expression above we can see that $$y^{4}-2y^{2}a+a^{2}-2y^{2}+2a-a+y^{2}=0$$

how can I solve this?

答:
Consider the function $$f(x)=\left(\log_2 x\right)^2-\log_{\sqrt 2} x-a+\sqrt{a+\log_2 x},$$ defined for all $x\ge 2^{-a}.$ Then we find that $$f'(x)=\frac{1}{2x\log_e 2\cdot \sqrt{a+\log_2 x}}\left(4\log_2 x\sqrt{a+\log_2 x}-4\sqrt{a+\log_2 x}+1\right).$$ Letting $M=\log_2 x,$ we see that $f'(x)=0$ whenever $16(M+a)(M-1)^2=1,$ or in other words when $$g(M)=16M^3+16(a-2)M^2+16(1-2a)M-1=0,$$ which is to say, when $\log_2 x =m,$ for some real number $m$ a root of the cubic in $M.$ This happens either once or thrice.

Now for exactly two roots of the original equation we must have a unique solution to the cubic. We find that the discriminant of $g'(M)$ is given by $$\Delta=16^2(a-2)^2-3\times 16\times 16(1-2a)=16^2(a+1)^2>0$$ for all real values of $a.$ Thus to ensure a unique solution we must have that $$g\left(\frac{a-2+|a+1|}{3}\right)g\left(\frac{a-2-|a+1|}{3}\right)>0.$$ This gives us a preliminary condition on $a.$

Now for the values of $a$ satisfying the inequality above we have ensured that $f'(x)$ vanishes exactly once, namely whenever $x=2^m,$ where $m=m(a)$ is the unique solution of $g(M)=0.$ Finally, since for $x\to +\infty,$ we have that $f(x)>0$ and for $x=2^{-a}$ we have that $f(x)=a(a+1),$ it follows that if $a=-1$ or $a=0$ we already have one root, and if these values of $a$ satisfy the previous condition then we have found two values of $a.$ On the other hand if $-1<a<0$ then there cannot be two roots if $a$ also satisfies the previous condition.

This leaves us with the case when $a<-1$ or $a>0,$ so that $f(2^{-a})>0.$ Thus in this case there are also exactly two roots provided the previous condition is satisfied and also the minimum value $f(2^m)$ is negative. This gives us a further condition on $a,$ namely that $(m-1)^2-(1+a)+\sqrt{a+m}<0.$

In summary we must have that $a$ satisfies the following conditions:

(1) $g\left(\frac{a-2+|a+1|}{3}\right)g\left(\frac{a-2-|a+1|}{3}\right)>0,$

and exactly one of the following:

(2) $a=-1$ or $a=0,$

or

(3) $a<-1$ or $a>0$ and $(m-1)^2-(1+a)+\sqrt{a+m}<0.$

Given these, then $f(x)=0$ has exactly two roots.

PS. Note that the conditions are to be selected as (1) and (2) or (1) and (3).


问:
What's the most elementary way to solve this trigonometric problem?

A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.

This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.

Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).

Thanks in advance.

答:
We know the side $c=2$ and the altitude $h_c=1$, hence the area of the triangle is $S=\frac12ch_c=1$.

We also have Heron's formula for the area:

$$S=\sqrt{p(p-a)(p-b)(p-c)}$$

With $p=\dfrac{a+b+c}2=3$.

Hence

$$(p-a)(p-b)=\frac{S^2}{p(p-c)}=\frac13$$

$$p^2-(a+b)p+ab=p(p-a-b)+ab=\frac13$$

$$ab=\frac13+3=\frac{10}{3}$$

Now, we know $a+b$ and $ab$, hence:

$$(a-b)^2=(a+b)^2-4ab=16-\frac{40}{3}=\frac83$$

And since the larger side is $b$,

$$a-b=-2\sqrt{\frac23}$$

$$a=\frac12(a+b+a-b)=\frac12(4-2\sqrt{\frac23})=2-\sqrt{\frac23}$$

Now, with Pythagoras' theorem in the triangle BDC:

$$y=\sqrt{a^2-1}=\frac13\sqrt{33-12\sqrt6}=\frac13\sqrt{24-2\times3\times2\sqrt6+9}=\frac{2\sqrt6-3}3=\frac{2\sqrt6}3-1$$


问:
An interesting approach to an integral
I found this integral somewhere, wolfram alpha did spit out an answer but I am trying to solve it by myself.

The integral is

$\int\sqrt{2^x-1}\ln(2^x-1)dx$

My approach is noticing that the integrand resembles the derivative of some function $f(n)=(2^x-1)^n$

We can define $I(n)=\int(2^x-1)^ndx$

Thus $I'(n)=\int(2^x-1)^n \ln(2^x-1)$dx

And our integral is $I'(\frac12)$

I wasted enough time on this, and I'm asking if this approach will lead somewhere, or you could maybe guide me through it.

Thank you.

答:
We have the give integral as:
\begin{gather*}
\int \sqrt{\left( 2^{x} -1\right)} ln\left( 2^{x} -1\right) dx\\
let\ 2^{x} -1=t\\
2^{x} ln2\ dx=dt\\
dx=\frac{dt}{( t+1) ln2}\\
Substitute\ back:\\
\int t^{\frac{1}{2}} lnt\times \frac{dt}{( t+1) ln2}\\
\\
\end{gather*}
Can you attempt this integral using integration by parts?

Hint: Use ln(t) as the first part and

\begin{gather*}
\frac{t^{\frac{1}{2}}}{t+1}\\
\\
\end{gather*} as the second part.

Title: An interesting approach to an integral


问:
Prove $\prod_{i=1}^n \frac{3K_i+2}{2K_i+1}$ can never be a power of $2$
Suppose we have a finite set of numbers $K_i \in \mathbb N$.

Can we prove that the product $$\prod_{i=1}^n \frac{3K_i+2}{2K_i+1}$$

can never be equal a power of $2$?

As you can see, each term is between $\frac{3}{2}$ and $\frac{5}{3}$. The greater $K_i$, the closer it is to $\frac{3}{2}$. Ratio between exact power or $\frac{3}{2}$ and the next greater power of $2$ places the constraint on $K_i$ values: the lowest $K_i$ cannot exceed certain value. For example, for products up to $2000$ terms, the biggest lowest value of $K_i$ was found to be for $1636$ terms, and it is $291643$.

The value would be a power of $2$ if all factors of the numerator and the denominator, except of $2$s in the numerator, completely canceled each other. Note that the denominator is always odd. It appears that there's no set of $K_i$ to make such a product equal to a power of $2$.

But can it be proven?

答:
Its not true. For instance, define $\mathcal K:=\{5,6,8,24,27,32,41,47,69,92\}$. Then we have,
$$\prod_{i\in\mathcal K}\frac{3i+2}{2i+1}=2^6.$$


问:
Is $\lim\limits_{a^{-}}f = +\infty \Rightarrow \lim\limits_{a^{-}}f' = +\infty$ true? If so, proof?
Let $f:(-\infty,a) \to \mathbb{R}$ be a differentiable function verifying that $\lim\limits_{a^{-}}f = +\infty$. Since the function grows rapidly to infinity when getting close to $a$ in a "finite" space, I wondered if $\lim\limits_{a^{-}} f' = +\infty$. I didn't find any counterexample, so I think the implication is true. Now, for the proof, I thought of fixing an $\bar{a}$ arbitrarily close to $a$, and considering for all $x\in(\bar{a},a)$, the slope :
\begin{equation}
S:=S_{\bar{a}}(x)=\frac{f(x)-f(\bar{a})}{x-\bar{a}}
\end{equation}
When $x$ tends to $a$, $S$ goes to infinity and that's for all $\bar{a}$ (so if $\bar{a}$ tends to $a$, it'll be the same), which means that : \begin{equation}\lim\limits_{\bar{a}\to a} \lim\limits_{x \to a} S = +\infty \end{equation}
When $x$ approaches $\bar{a}$, S approaches $f'(a)$, and this being true for all $\bar{a}$, we can have :
\begin{equation}
\lim\limits_{\bar{a}\to a} \lim\limits_{x \to \bar{a}} S = \lim\limits_{\bar{a}\to a} f'(\bar{a})=\lim_{x\to a}f'(x)
\end{equation}
I feel that the two procedures are equivalent since we always get $x$ and $\bar{a}$ infinitesimally close, so both limits should be equal in some sense. So we get at the end that $f'(x) \to +\infty$.

Body: No. Consider $\sin(1/x^2)  - 1/x$, $a=0$.


答:
$$\int {e^{-x}} \cos x dx \\ = -e^{-x} \cos x -\int e^{-x} \sin x dx \\ =-e^{-x}\cos x +e^{-x}\sin x -\int e^{-x}\cos x dx $$

Move the integral to the other side and divide by $2$:

$$\int e^{-x}\cos x dx = \frac{e^{-x}}{2} (\sin x -\cos x) + \text{constant}$$

Title: How to integrate $\frac{\cos(x)}{e^x}$?

Body:
Surprisingly I haven't found such integral in the web, so I may ask this here bluntly.

A solution that Matlab suggest sounds like $\frac{{\mathrm{e}}^{-x}\,\left(\cos\left(x\right)-\sin\left(x\right)\right)}{2}$


问:
Covering space Hausdorff implies base space Hausdorff
There is an exercise problem in Hatcher's Algebraic Topology book asking to show that if $p:\tilde{X}\rightarrow X$ is a covering space with $p^{-1}(x)$ finite and nonempty for all $x\in X$, then $\tilde{X}$ is compact Hausdorff iff $X$ is compact Hausdorff.

I've managed to show each part of the statement except for $\tilde{X}$ Hausdorff $\Rightarrow X$ Hausdorff. The problem I encountered was that taking $x\neq y$ in $X$ and $\tilde{U},\tilde{V}$ disjoint open sets in $\tilde{X}$ with $p^{-1}(x)\subset\tilde{U}$ and $p^{-1}(y)\subset \tilde{V}$ (I'm using the assumption that $\tilde{X}$ is compact Hausdorff, thus normal), I could not use these to get disjoint open sets in $X$ separating $x,y$.

答:
Take $x \neq y$ in $X$ and choose $x' \in \tilde X$ mapping to $x$.  Choose a neighborhood $U'$ of $x'$ and neighborhoods $V_{y'}$ for each $y' \in p^{-1}(y)$ such that none of these neighborhoods intersect.  Let $U$ be the image of $U'$ in $X$.  Define $V$ by pushing the $V_{y'}$ to $X$ first and then intersecting.

Show that $U$ and $V$ are neighborhoods of $x$ and $y$ that don't intersect.

Edit:  Oops, forgot to say you'll want to choose the $V_{y'}$ so that no two intersect.  I've edited my answer above to reflect this.  You can do this by proving the following little lemma (if you don't already know it) and applying it to $\{x'\} \cup p^{-1}(x)$.

Lemma: Let $Y$ be Hausdorff and $\{v_1, \ldots, v_n\}$ a set of finitely many distinct points in $Y$.  Then there exist neighborhoods $V_i$ of $v_i$ such that no two of the $V_i$ intersect.


问:
Sum of the roots of an equation $\cos x=1-\frac{x[x]}{2}$

Find the sum of the roots of the following equation
$$\cos x=1-\frac{x[x]}{2}$$
I know the sum of the roots of the given equation is zero, I could prove it by using numerical method and confirmed it with wolfram alpha. I do not know how to do it algebraically.

答:
Hint: Use that $$-1\leq \cos x\leq 1\implies 0\leq x[x]\leq 4$$

 Let $k=[x]\in \mathbb{Z}$ then $x=k+r$ where $0\leq r<1$ then $$k^2\leq x[x] =k^2+kr <k^2+k$$

 So $k(k+1)>0$ which is true for all $k\in \mathbb{Z}\setminus\{-1,0\}$

 And $k^2\leq 4$ so $k\in\{-2,-1,0,1,2\}$. Thus $\boxed{k\in\{-2,1,2\}}$.

 - If $k=2$ we get $x=2+r$ so $0\leq 4+2r\leq 4$ so $r\leq 0$ so $r=0$ and thus $x=2$ which does not work since $\cos 2\ne -1$.

 - If $k=1$ we get $x=1+r$ so $$2\cos (r+1)=  1-r$$ This one needs to solved with some aproximation formula...

 - If $k=-2$ we get $x=-2+r$ so $$\cos (r-2)=  r-1$$ This one needs to be solved with some aproximation formula also...


问:
Convergence/Divergence proof for Infinite products
Exercise 2.4.10 is about infinite products. I'd like someone to verify, if my convergence/divergence proof is technically correct and rigorous.

A close relative of the infinite series is the infinite product.
\begin{align*}
 \prod_{n=1}^{\infty} b_n = b_1 b_2 b_3 \cdots
\end{align*}

which is understood in terms of its sequence of partial products.
\begin{align*}
 p_m = \prod_{n=1}^{m} b_n = b_1 b_2 b_3 \cdots b_m
\end{align*}

Consider the special class of infinite products of the form
\begin{align*}
 \prod_{n=1}^{\infty}(1+a_n) = (1+a_1)(1+a_2)(1+a_3)\cdots
\end{align*}

where $a_n > 0$.

(a) Find an explicit formula for the sequence of the partial products in the case where $a_n = 1/n$ and decide whether the sequence converges. Write out the first few terms in the sequence of partial products in the case where $a_n = 1/n^2$ and make a conjecture about the convergence of this sequence.

(b) Show, in general, that the sequence of partial products converges if and only if $\sum_{n=1}^{\infty}a_n$ converges. (The inequality $1+x \le 3^x$ for positive $x$ will be useful in one direction.)

Proof.

(a) If $a_n = 1/n$, the partial product
\begin{align*}
p_m &:= \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{m}\right)\\
&= 1 + \sum_{m} \frac{1}{m} + \sum_{m < n} \frac{1}{m \cdot n} + \sum_{m < n < p} \frac{1}{m \cdot n \cdot p} + \ldots
\end{align*}

Since $\sum 1/m$ is the harmonic series, which is well-known to be divergent, this sequence of partial products is divergent.

Consider the case of $a_n = 1/n^2$. We are going to use the inequality $1 + x^2 \le e^{x^2}$. We have:
\begin{align*}
p_m &=  \left(1+\frac{1}{1^2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{3^2}\right)\cdots\left(1+\frac{1}{m^2}\right)\\
&\le e^{\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots + \frac{1}{m^2}}\\
&\le e^{\frac{1}{1\cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{m\cdot (m+1)}}\\
&\le e^{\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \ldots + \left(\frac{1}{m} - \frac{1}{m+1}\right)}\\
&= e^{1 - \frac{1}{m+1}}\\
&\le e
\end{align*}

Moreover, $p_1 = 2$, $p_2 = (2)(5/4)$, $p_3 = (2)(5/4)(10/9)$, $p_4 = (2)(5/4)(10/9)(17/16)$. So, $(p_m)$ is monotonic increasing and bounded by $e$. Consequently, by the Montone Convergence Theorem, $(p_m)$ is convergent.

(b) ($\longrightarrow$) direction.

The sequence of partial products,
\begin{align*}
p_n &= (1+a_1)(1+a_2)\cdots(1+a_n)\\
&\le e^{a_1}\cdot e^{a_2}\cdots e^{a_n}\\
&\le \exp \left({\sum_{k=1}^{n}a_k}\right)
\end{align*}

If the sequence of partial products is convergent and therefore bounded, the sequence of partial sums $\sum_{k=1}^{n} a_k$ must be bounded and convergent.

答:
For (a) it says "explicit formula".  Yours is not.  Here it is:
$$
\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)
\cdots\left(1+\frac{1}{n-1}\right)\left(1+\frac{1}{n}\right)
\\
=\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{n}{n-1}\cdot\frac{n+1}{n}
= \frac{n+1}{1} = n+1
$$
Each numerator cancels the following denominator.


答:
The function g has infinite number of zeros.

Assume otherwise, and list all its zeros ( $z_{1}$, $z_{2}$, ...$z_{n}$) with multiplicity.

Set $$P(z) = \prod_{k=1}^{k=n}(z-z_{k})$$
(if g has no zeros at all set $P(z) = 1$)
and $$h(z) = \frac { g(z) } {P(z)}$$
Then h(z) is holomophic on the whole plain and has no zeros, hence it can be lifted.

In words, there is entire function, say f(z), such that h(z) = $e^{f(z)}$

Now note that $$|h(z)| <= e^{2|z|}$$
for |z| large enough and this implies that $$ |f(z)| <= 2|z| $$ for |z| large enough. In words f(z) must be a polynomial of degree at most 1, say $f(z) = az + b $.

Taking it all together, we showed that if g(z) has finite number of zeros then, for some $z_{1},z_{2}, ..., z_{n}, a, b$ we have
$$g(z) = e^{az+b}\prod_{k=1}^{k=n}(z-z_{k})$$ or $$e^{z} - z + 1 = e^{az+b}\prod_{k=1}^{k=n}(z-z_{k})$$
or
$$ P(z) = \prod_{k=1}^{k=n}(z-z_{k}) = \frac {e^{az+b}} {e^{z} - z + 1} $$
The last identity can not be true.

That is one can see that the coefficient a must be equal to 1 (otherwise a polynomial would grow as exponential at infinity) but if a = 1 then P(z) is bounded but not constant.

Title: Characterizing isolated singularities of $\frac{z}{e^{z} - z + 1}$

Body:
I intend to characterize the isolated singularities of $f(z) := \frac{z}{e^z - z + 1}$ which is defined on some open subset $\mathbb{C} \backslash f^{-1}({0}) \subset \mathbb{C}$. The possible singularities are only the zeros of $g(z) := e^z - z + 1$, so the approach should be to find those zeros, which is actually the difficult task.

At this point, what I know is the following:

Since no zeros of $g$ lie on $2\pi i \mathbb{Z}$, if we suppse that $z_{0} \in \mathbb{C}$ is a zero of $g$, then we will of course have $\lim\limits_{z \to z_{0}} |f(z)| = \infty$ and so $z_{0}$ will be a pole of $f$. Using then L'Hôpital's rule applied to $\lim\limits_{z \to z_{0}} (z-z_{0})^n f(z)$, we check that the pole will have to be of order $1$.

$g$ has zeros, which I found out by separating the function $g(z) = 0$ in two equations, concerning the real and imaginary parts of $g$ and then by ploting the resulting functions to see that they intersect.

My question is then: Can we see analytically that $g$ has zeros?

This is an exercise from a Reinhold Remmert's book, "Theory of Complex Functions" (line c) in exercise 1 on page 309) and I think I'm supposed to solve it analytically. Also, keep in mind that, at this point in the book, there are a lot of tools in complex analysis still not available to use, namely residue calculus and Laurent series.

Thank you in advance for all the help!


答:
If $g(z)$ had no zeroes, by Weierstrass factorization, we’d have to contend that there exists an entire function $h(z)$ such that $$e^{h(z)}=e^z-z+1\,,$$ or equivalently $$e^z(e^{h(z)-z}-1)=1-z\,.$$
It follows that $h(z)\ne z$ (that is, $h$ is not the identity function) and $h(1)-1=2\pi in_0$ for some integer $n_0$; furthermore, the entire non-constant function $h_1(z):=h(z)-z$ must never assume any value in the discrete set $\{2\pi in:n\in\mathbb{Z}\setminus\{n_0\}\}$ otherwise the left-hand side above will have at least two zeroes while the right-hand side has only one zero, which contradicts the Little Picard Theorem.


问:
The harmonic conjugate of $\Im e^{z^2}$?
It is obvious that $e^{z^2}$ is analytic, right? So the harmonic conjugate of $\Im e^{z^2}$ is $\Re e^{z^2}$, isnt' it?

However, the solutions manual I'm consulting gives the answer as $\Im (-ie^{z^2})$, which is not the same function, and I don't understand.

答:
Let $w = a + bi$. Note that $\text{Im } (iw) = \text{Im } (-b + ai) = a = \text{Re } (w)$.

So yes, it is the same function up to a minus sign, which I don't know where it comes from - perhaps it comes from the manual's specific definition of a harmonic conjugate.


答:
What you can conclude is that $\lim\sup_{x\to a^{-}}f'(x)=+\infty$. This is because: $\frac{f(x+h)-f(x)}{h}=f'(\xi)$ for some $\xi\in(x,x+h)$ (IVT), so letting $h\to (a-x)^{-}$ makes the left-hand side go to $+\infty$, so there will be points $\xi$ in each interval $(x,a)$ that produce $f'(\xi)$ arbitrarily large.

You cannot, however, conclude $\lim_{x\to a^{-}}f'(x)=+\infty$. One can found a counterexample by tweaking a well-known function, e.g. $f(x)=-\frac{1}{x}$ with $a=0$.

Take a sequence $x_n\to a^{-}$ monotonously, e.g. $x_n=-\frac{1}{n}$. Now make the sequence $y_n$ as follows:

For odd $n$, set $y_{2n-1}=f(x_{2n-1})$, i.e. in our example: $y_{2n-1}=2n-1$.

For even $n$, keep the same value, i.e. $y_{2n}=y_{2n-1}=f(a_{2n-1})$, i.e. in our example: $y_{2n}=2n-1$.

Now all it takes is to pass a smooth curve through the points $(x_n, y_n)$. This can be done in many ways - let your imagination go wild! You get a differentiable function $g(x)$ such that $g(x_n)=y_n$.

The function $g$ looks like smoothed staircase which climbs towards $+\infty$ at about the same rate as $f$: in fact, they coincide at points $x_{2n+1}$. However, $g$ has also got those intervals $(x_{2n-1},x_{2n})$ where it is (approximately) "flat".

This lets us conclude that, on each $(x_{2n-1}, x_{2n})$, there is a point $\xi\in(x_{2n-1}, x_{2n})$ such that $f'(\xi)=0$ (Rolle theorem). As $x_{2n-1}\to a^{-}$ as $n\to\infty$, we conclude that $\lim\inf_{x\to a^{-}}f'(x)\le 0$. However, because  $\lim\sup_{x\to a^{-}}f'(x)=+\infty$ (proven above), we conclude that $\lim_{x\to a^{-}}f'(x)$ does not exist.

Title: Is $\lim\limits_{a^{-}}f = +\infty \Rightarrow \lim\limits_{a^{-}}f' = +\infty$ true? If so, proof?

Body:
Let $f:(-\infty,a) \to \mathbb{R}$ be a differentiable function verifying that $\lim\limits_{a^{-}}f = +\infty$. Since the function grows rapidly to infinity when getting close to $a$ in a "finite" space, I wondered if $\lim\limits_{a^{-}} f' = +\infty$. I didn't find any counterexample, so I think the implication is true. Now, for the proof, I thought of fixing an $\bar{a}$ arbitrarily close to $a$, and considering for all $x\in(\bar{a},a)$, the slope :
\begin{equation}
S:=S_{\bar{a}}(x)=\frac{f(x)-f(\bar{a})}{x-\bar{a}}
\end{equation}
When $x$ tends to $a$, $S$ goes to infinity and that's for all $\bar{a}$ (so if $\bar{a}$ tends to $a$, it'll be the same), which means that : \begin{equation}\lim\limits_{\bar{a}\to a} \lim\limits_{x \to a} S = +\infty \end{equation}
When $x$ approaches $\bar{a}$, S approaches $f'(a)$, and this being true for all $\bar{a}$, we can have :
\begin{equation}
\lim\limits_{\bar{a}\to a} \lim\limits_{x \to \bar{a}} S = \lim\limits_{\bar{a}\to a} f'(\bar{a})=\lim_{x\to a}f'(x)
\end{equation}
I feel that the two procedures are equivalent since we always get $x$ and $\bar{a}$ infinitesimally close, so both limits should be equal in some sense. So we get at the end that $f'(x) \to +\infty$.


答:
I think the integrate-by-parts answers are best.  The alternative is to use complex numbers as rrv suggests:
$$
\frac{\cos x}{e^x} = \text{Re }e^{ix-x} = \text{Re }e^{(i-1)x}
$$
then
$$
\int e^{(i-1)x}dx = \frac{e^{(i-1)x}}{i-1} +C
\\
\int \frac{\cos x}{e^x}\;dx = \text{Re }\frac{e^{(i-1)x}}{i-1}+C
= \frac{e^{-x}}{2}(\sin(x)-\cos(x)) +C
$$

Title: How to integrate $\frac{\cos(x)}{e^x}$?


问:
Range of Real Inverse Hyperbolic Cosine -- can it be negative?
There are several results in Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968) concerning $\cosh^{-1}$ which are presented in the following format:

$$\cosh^{-1} x = \pm \left\{{\ln (2x) - \left({\dfrac 1 {2 \cdot 2x^2} + \dfrac {1 \cdot 3} {2 \cdot 4 \cdot 4x^4} + \dfrac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 \cdot 6x^6} }\right) }\right\}$$

$$[+ \text { if }\cosh^{-1} x > 0, x \ge 1, - \text { if }\cosh^{-1} x < 0, x \ge 1]$$

(item $20.40$)

and:

$$\int \cosh^{-1} \frac x a \, \mathrm d x = \begin{cases} x \cosh^{-1} (x/a) - \sqrt {x^2 - a^2} & : \cosh^{-1} (x/a) > 0 \\ x \cosh^{-1} (x/a) + \sqrt {x^2 - a^2} & : \cosh^{-1} (x/a) < 0 \end{cases}$$

(item $14.651$)

There are a number of such.

What confuses me is that I have been led to believe that the (real) inverse hyperbolic cosine is specifically defined as being the positive value of $y$ such that $\cosh y = x$.

So what is the motivation and conventional interpretation of "$f (\cosh^{-1} x)$ such that $\cosh^{-1} x < 0$"?

答:
Assuming $x, y$ real, and $x \ge 1$, since

$$\mathrm{cosh}(y) = \mathrm{cosh}(-y) =x $$

one has 2 possible choices for the definition of a single valued function $\mathrm{cosh}^{-1}(x)$, namely

$$\mathrm{cosh}^{-1}(x) =y$$

or

$$\mathrm{cosh}^{-1}(x) =-y$$

It seems like the author is just covering all the cases, since it's a reference text.


问:
Integration of $\cos x\cdot\cos 2x\cdot\cos 3x$
I study maths as a hobby and am trying to integrate
$\,\cos x\cdot \cos 2x\cdot \cos 3x\,$.

I am trying to split this up into a form in which I can integrate each component part. I have tried using the rule
$$2\cos A \cos B = \cos(A + B) + \cos (A - B)$$
but I can only get this far:
$$\begin{align}
\int\cos (x) \cos (2x)\cos (3x)\, dx
 & = \int\frac{1}{2}(2\cos (x)\cos (3x)\cos (2x), dx \\
 & = \int\frac{1}{2}( \cos (4x) + \cos (2x))\cos (2x), dx
\end{align}$$
but from here on I get stuck.

答:
Use the formula twice.

$\begin{align}\cos (x)\cos (2x) \cos(3x) &= \frac12(\cos (4x) + \cos (2x)) \cos (2x) \\&= \frac12\cos(4x)\cos(2x) + \frac12 \cos(2x)\cos(2x) \\&= \frac 14 (\cos(6x) + \cos(2x)) + \frac14(\cos(4x)+1)\end{align}$


问:
Change of measure s.t. stochastic process is in $L^p$
Assume a filtered probability space $(\Omega, \mathcal{G},(\mathcal{G}_t)_{t \in [0,1]}, \mathbb{P})$ and $(S_t)_{t \in [0,1]}$ an adapted stochastic process (no further path properties).

Fix $1 \leq p < \infty$. Is there an equivalent measure $\mathbb{P}' \sim \mathbb{P}$, s.t. $S_t$ is in $L^p(\Omega, \mathcal{G}, \mathbb{P}')$?

Backgound: I am reading a paper that poses this is true for a countable (but not uncountable) family of such processes, however without giving a proof. I am not even sure if that is true for the 1-dim case.

What I tried: A classical application of the Radon-Nikodym theorem. But, for a decent choice of density, I need $\mathbb{P}(\sup_t S_t < \infty) = 1$ as an assumption, what does not seem natural to me.

Thank you for taking your time.

答:
Consider simple case, when $\Omega = [0,1]$. In this case we may consider Dirac measure $\delta_0$ and get what you want. But new process will have nothing common with the initial one.

In general case also there's such a problem.


问:
Prove $\sum\limits_{n=1}^{\infty} \frac{1}{\int_0^1\ln^n(x)dx} = \frac{1}{e}-1$
Ok, that's really not an easy question to convey in text. What I'm asking is this:

$$\sum\limits_{n=1}^{\infty} \frac{1}{\int_0^1\ln^n (x)dx} = \frac{1}{e}-1$$

Numerically, I'm pretty sure it does (it's about -0.63212). But, I have no idea how I'd go about proving it.

Any insights?

答:
Hint: by the change of variable $\ln x=-u$,

$$\int_0^1\ln^n(x)dx=\int_{\infty}^0 -(-u)^n\exp(u)du=(-1)^n\int_0^\infty u^n\exp(-u)du=(-1)^n\Gamma(n+1)=(-1)^nn!$$


问:
Intuitive understanding of the derivatives of $\sin x$ and $\cos x$
One of the first things ever taught in a differential calculus class:

The derivative of $\sin x$ is $\cos x$.

The derivative of $\cos x$ is $-\sin x$.

This leads to a rather neat (and convenient?) chain of derivatives:

```
sin(x)
cos(x)
-sin(x)
-cos(x)
sin(x)
...
```

An analysis of the shape of their graphs confirms some points; for example, when $\sin x$ is at a maximum, $\cos x$ is zero and moving downwards; when $\cos x$  is at a maximum, $\sin x$ is zero and moving upwards.  But these "matching points" only work for multiples of $\pi/4$.

Let us move back towards the original definition(s) of sine and cosine:

At the most basic level, $\sin x$ is defined as -- for a right triangle with internal angle $x$ -- the length of the side opposite of the angle divided by the hypotenuse of the triangle.

To generalize this to the domain of all real numbers, $\sin x$ was then defined as the Y-coordinate of a point on the unit circle that is an angle $x$ from the positive X-axis.

The definition of $\cos x$ was then made the same way, but with adj/hyp and the X-coordinate, as we all know.

Is there anything about this basic definition that allows someone to look at these definitions, alone, and think, "Hey, the derivative of the sine function with respect to angle is the cosine function!"

That is, from the unit circle definition alone.  Or, even more amazingly, the right triangle definition alone.  Ignoring graphical analysis of their plot.

In essence, I am asking, essentially, "Intuitively why is the derivative of the sine the cosine?"

答: I don't think you can get an intuitive feel for the derivatives without looking at the plots personally. When you consider that a derivative is a rate of change, you need to be looking at a function that is varying, which implies you are looking at the plot/graph of the function. When you further consider that a derivative (by definition of it being a rate of change) is a gradient function, the intuitive answer is that cos is the gradient function of sin, and -sin is the gradient function of cos (and so on). So if you calculate the gradient of the sin curve at any point, the value you get will be the cosine value for that point.</p>

ContentLicense: CC BY-SA 2.5


问:
How to reduce lower one number, while another number increases goes up increments
This is such a basic question I'm sure, but I've been trying to find a robust solution to it for a while.

Lets say I have the numbers 0-100 in series.

As the number increases, I want another number say 1000, to decrease. so that when you got to 100, it was at 0

0 = 1000

50 = 500

100 = 0

What is this kind of series called?

答:
As you mention “series” in the question I was wondering if something as simple as the term “Arithmetic Progression” is what you're after.

Your first arithmetic progression being $$a_n=n$$ for n=0,1,...,100 and your second being
$$b_n=1000-10n=1000-10a_n.$$

Given your chosen name, I was wondering if your intended application has anything to do with optics? (I ask as my wife is an optician.)

Not really part of the answer but on a lighter note, here's a wonderful quote involving optics from the book “Littlewood's Miscellany”:

From time to time (since 1910) there were moves to get rid of the revolting optics and astronomy set in the Mathematical Tripos. It was discovered that over a period of years no wrangler attempted a question in either subject. An equivalent form of this proposition is that every attempt to do a question in optics or astronomy resulted in a failure to get a 1st.


问:
Show that $\mathcal{O}_k/(m)$ is finite.
Let $K$ be a number field and $\mathcal{O}_K$ be its ring of integers. It is easy to show that there is some nonzero rational integer $m$ in a prime ideal $\mathfrak{p}$. Now, I know that $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $n=[K:\mathbb{Q}]$.

I am now having trouble understanding why $\mathcal{O}_K/(m)$ is finite with size $m^n$. I can't really see why $(m)$ would be full rank here, and I think I'm missing something kind of obvious.

答:
Of course, by definition, $(m)=m \mathcal{O}_K$. So you want to show that $\mathcal{O}_K/m\mathcal{O}_K$ is finite. Now, the following is not that hard to prove:

"If $G$ is a free abelian group of rank $n$ and $m \in \mathbb{Z}$ then $G/mG$ is the direct sum of $n$ cyclic groups each of order $m$."

Applying this with $G=\mathcal{O}_K$ yields $$ \mathcal{O}_K /m\mathcal{O}_K \cong \underbrace{ (\mathbb{Z}/m\mathbb{Z}) \oplus \dots \oplus (\mathbb{Z}/m\mathbb{Z}) }_{\text{$n$ times}}  .$$ Therefore, $\vert \mathcal{O}_K /m\mathcal{O}_K \vert = m^n$.