问:
Limits and common sense
I'm stuck in understanding of limits.
It all makes sense, but at a certain point my answers which seem logical to me are not true.
Please can somebody explain why as a huge number gets divided by a big number plus the relatively small value $1$, the result gets closer and closer to zero:
$$
lim_{x\to \infty} \frac{\infty!}{(\infty + 1)!} =0.
$$
The next one is not equal zero as well.
$$
lim_{x\to \infty} \frac{\infty}{\infty-8}=1
$$
The relatively huge number gets divided by another relatively huge number minus 8, which is so little we could ignore it. Why is it approaching 1 then?
It can make sense from, I don't know, a visual point of view where
$$
lim_{x\to \infty} \frac{\infty}{\infty-8}=\frac{\infty}{\infty}=1
$$
but I could as well do that with first one and also get 1, which is wrong.</p>

答:
This answer is divided in two parts. In the first part I will share some of the theory behind infinity, and in part $2$ we will look at how to use this theory to solve limits.

Part 1 (Theory)

Before we start, I wanna make a comment on something you said. You said in a comment that you think we deal with same-size infinities all around. This is not true. Consider two limits
$$\lim _{x \to \infty} x = \infty$$

$$\lim _{x \to \infty} 2x = \infty$$

While we use the same symbol for the results of both of these limits, the infinity resulting from $2x$ is in some sense "bigger" or  (as I like to think about it) "stronger" than the infinity resulting from $x$. I think this is the root cause of your confusion, and it's the reason why in your question, one limit is equal to $0$, while another is equal to $1$. Anyway, here's the intro to the theory:

The symbol $\infty$ is not a number. Technically it's an extended real number. But no need to get into technicalities. Going by intuition is best.

The symbol $\infty$ behaves like a number in some cases, but not in others.

Here are some of the things you can always do with infinity. These should all be very clear intuitively. If they are not, let me know:

A) $\infty + a = a + \infty = \infty$ for any number $a$

B) $\infty + \infty = \infty$

C) $\frac {a}{\infty} = 0$ for any number $a$

D) $\frac {\infty}{a} = \pm \infty$, depending on the sign of $a$.

E) $\infty \cdot \infty = \infty$

F) $\infty \cdot a = a \cdot \infty = \pm \infty$, depending on the sign of $a$.

G) $\infty^2 = \infty, \infty^3 = \infty,$ $\infty^4 = \infty$, etc.

If you understand these intuitively, there should be no doubt that the above are true. I did not include everything here, but if you understand these intuitively, you should hopefully be able to come up with more of these by yourself (another one: $2^{\infty} = \infty$).

The following operations with $\infty$ are undefined (in the same way that, for example, $\frac 10$ is undefined).

$\infty - \infty$

$\frac {\infty}{\infty}$

Why is this? Consider for example $\infty + \infty$ compared to $\infty - \infty$. In the first expression, there is no doubt that the result should be $\infty$. If you add two huge numbers, you get a huge number. The huge numbers
are working together to make an even bigger number.

But for the second expression, there is doubt. If you subtract two really huge numbers, it totally depends on how big the two numbers are compared to each other. For example, $100000 - 1000$ is really big, but $1000 - 1000000$ is really small, and $100000 - 100000$ is zero. The two huge numbers are working against each other. The first number wants to pull the result up, but the second number wants to pull the result down. It all depends on which huge number is "stronger".

Here are examples with limits which demonstrate that with $\infty - \infty$, anything can happen, depending on how "strong" the specific infinities involved are. In their original form, all of these limits are of the type $\infty - \infty$ which is undefined. But after some easy manipulations, that are not of that type anymore, and we can compute them easily.

$\displaystyle\lim_{x \to \infty} (5x - x) = \lim_{x \to \infty} 4x = 4 \times \infty = \infty$

$\displaystyle\lim_{x \to \infty} [(x+4) - x] = \lim_{x \to \infty} 4 = 4$

$\displaystyle\lim_{x \to \infty} (x - x) = \lim_{x \to \infty} 0 = 0$

$\displaystyle\lim_{x \to \infty} [(x-4) - x] = \lim_{x \to \infty} -4 = -4$

$\displaystyle\lim_{x \to \infty} (x - 5x) = \lim_{x \to \infty} -4x = -4 \cdot \infty = - \infty$

These may seem like silly examples, but they illustrate the point.

Same story with $\frac {\infty}{\infty}.$ The huge number in the numerator wants to pull the result up (because for example $\frac {\infty}{5} = \infty$) but the huge number in the denominator wants to pull the result down towards zero (because for example $\frac {5}{\infty} = \infty$). Again, the result will depend on which huge number is "stronger".

Here are examples which illustrate that again, anything can happen, depending on the "strength" of the specific infinities involved.

$\displaystyle \lim_{x \to \infty} \frac {x^2}{x} = \lim_{x \to \infty} x = \infty$

$\displaystyle \lim_{x \to \infty} \frac {4x}{x} = \lim_{x \to \infty} 4 = 4$

$\displaystyle \lim_{x \to \infty} \frac {x}{x^2} = \lim_{x \to \infty} \frac 1x =  \frac {1}{ \infty} = 0.$

$\\$

Part 2 (Application)

So, how do you actually apply this stuff to solve a limit of the form $\lim_{x \to \infty} f(x)$? You start by plugging in $\infty$, and see what happens. There are two possibilities:

1) Once you plug in, you only have allowable operations to do. In this case, you do those operations, and get the answer. For example

\begin{align*}
 \lim_{x \to \infty} \dfrac {2x^3 + 3x - 4}{5} = \dfrac {2 \cdot \infty^3 + 3\cdot \infty - 4}{5} = \dfrac {\infty + \infty - 4}{5} = \frac{\infty}{5} = \infty
\end{align*}

2) Once you plug in, you get one of the forbidden operations, such as $\frac {\infty}{\infty}.$ If this happens, then you need to go back to the original and try to "fix" the issue. Either you do an algebraic manipulation, or (more advanced) use L'Hopital's Rule, or (even more advanced) Taylor series, etc.

Once you have done the manipulation (or whatever else) then you plug in $\infty$ again, and hope that you end up in case $1$. (notice that this is exactly what I did in the examples above when showing that $\infty - \infty$ or $\frac {\infty}{\infty}$ could be anything).If you end up in case $2$ again, then you repeat.

An example:

$$\lim_{x \to \infty} \frac {x^2 - 4}{x-2}$$

Plugging in gives us $\frac {\infty^2 - 4}{ \infty - 2} = \frac {\infty}{\infty }$ which is undefined, so we go back to the original and manipulate.

$$\lim_{x \to \infty} \frac {x^2 - 4}{x-2} = \lim_{x \to \infty} \frac {(x-2)(x+2)}{x-2} = \lim_{x \to \infty} x + 2 = \infty + 2 = \infty.$$

Now, this is what happens in your two examples in the question. Presumably, the first question started as

$$\lim_{n \to \infty} \dfrac {n!}{(n+1)!}$$
here we think of $n$ as a whole number, because factorials are only defined for whole numbers.

If you plug in $\infty$, you end up with $\frac{\infty}{\infty}$. So we go back and manipulate:

$$\lim_{n \to \infty} \dfrac {n!}{(n+1)!} = \lim_{n \to \infty} \dfrac {1 \cdot 2 \cdot 3 \cdots \cdot n}{1 \cdot 2 \cdot 3 \cdots \cdot n \cdot (n+1)!} = \lim_{n \to \infty} \frac {1}{n+1} = \frac {1}{\infty + 1} = \frac {1}{\infty} = 0.$$

In your second question. Presumably the problem started as

$$\lim_{x \to \infty} \frac {x}{x-8}$$

If you plug in you get $\frac {\infty}{\infty}$. So we go back and manipulate.

$$\lim_{x \to \infty} \frac {x}{x-8} = \lim_{x \to \infty} \frac {x}{x(1-\frac 8x)} = \lim_{x \to \infty} \frac {1}{1-\frac 8x} = \frac {1}{1 - \frac {8}{\infty}} = \frac {1}{1-0} = 1$$

These manipulations (especially for the second problem) might seem like they come out of nowhere. But if you study limits, you will see the same 3-4 types of manipulations over and over, and you will know how to use them. The ones I used are very common.

Title: Limits and common sense


问:
Ideal correspondence
I'm confusing the ideal correspondence theorem. Is the following  right?

Ideal correspondence: Let $f:A \to B$ be a ring homomorphism. Then there is a one-to-one order-preserving correspondence between ideals of $f(A)$ and ideals of $A$ which contain $\ker(f)$ by $\alpha \mapsto f(\alpha)$ (conversely $\beta \mapsto f^{-1}(\beta)$), and prime ideals correspond to prime ideals, and maximal ideals correspond to maximal ideals.

答: Yes.  You might make the direction of correspondence consistent by saying that ideals of $A$ which contain ker($f$) go to ideals of $f(A)$ under $\alpha$ going to $f(\alpha)$, but you clearly have the right idea.  In this mapping prime ideals that contain ker($f$) go to prime ideals of $f(A)$, and so also with maximal ideals containing ker($f$) going to maximal ideals of $f(A)$.  The statement could be a bit more concise if we require $f$ onto, so that $f(A)$ is replaced in the the statement by $B$.</p>

ContentLicense: CC BY-SA 3.0

Title: Ideal correspondence


问:
Finding how many times one number fits into the other
Is there a way to tell - for  example - how many $60$s are there in a number that cannot be divided by $60$, like $183$? I know that we should remove all numbers after the decimal point but how can we tell this mathematically?</p>

答:
It is still unclear whether you are looking for a trick, a formula, an algorithm, or a mathematical definition, so I will provide all four.

Trick: Probably the easiest way to do this by hand is do long division on $183 / 60$ but throw away the remainder (or remove all the numbers after the decimal point, if you prefer).

Formula: This is generally written $\left\lfloor \frac{183}{60} \right\rfloor$.  Here $\lfloor x \rfloor$ is called the "floor" of $x$ and means the greatest integer less than or equal to $x$.

Algorithm: This is called the Division Algorithm (as was mentioned by Sigur in the comments).  Basically, you subtract 60 from 183 until you get something on the interval [0, 60), and you count the number of times you subtracted 60 to get the answer.

Definition: Let $a$ and $b$ be integers.  The integer quotient of $a$ and $b$ is defined to be the largest integer $k$ such that $a > bk$.  (Note that this is well-defined if and only if $b > 0$).  In the case of 183 and 60, we want the largest integer $k$ such that $183 > 60k$.


问:
Proving that a function doesn't have a limit given $|a-b| = |f(a)-f(b)|$ for all $a,b\in \mathbb R$
For all $a,b\in \mathbb R$,

$|a-b| = |f(a)-f(b)|$.

Prove that $\;\lim\limits_{a\to \infty}|f(a)|=\infty$

What I was trying to do is show that if we assume that there is a limit, $\;\lim\limits_{a\to \infty}|f(a)|=L\;$ and therefore $\big||f(a)|-L\big|=\big||f(a)|-|L|\big| \leq |f(a)-L| < \epsilon$

Then we set $\epsilon$ to some arbitrary value and try to get a contradiction.

I also tried proving directly but couldn't get much out of it.

Is there a better approach for proving it?

答:
For any $\;a\in\mathbb{R}\;$ it results that

$\begin{align}
|f(a)|&\ge|f(a)-f(0)|-|f(0)|=|a-0|-|f(0)|=\\
&=|a|-|f(0)|\;,
\end{align}$

hence,

$\lim\limits_{a\to\pm\infty}|f(a)|\ge\lim\limits_{a\to\pm\infty}\big(|a|-|f(0)|\big)=+\infty\;,$

consequently,

$\lim\limits_{a\to\pm\infty}|f(a)|=+\infty\;.$

ContentLicense: CC BY-SA 4.0


问:
Status of The Triangle Book
I am interested in finding out about the current status of the planned book: The Triangle Book by John H. Conway and Steve Sigur. I understand that Steve Sigur died some time back. I got no reply from Prof. Conway; is there someone who knows the fate of this project. Is it abandoned or is it still under preparation? Here is some information about the planned book.</p>

答:
The Triangle Book is on hold (Conway died this year) but there are people trying to revive it trust me.

Colm Mulcahy

Title: Status of The Triangle Book


问:
f is continuous on [0,1]. f(0) = f(1). Show that f'(1 - c) = f'(c)
Let $f$ be continuous on $[0,1]$ and differentiable in $(0,1)$ s.t. $f(0) = f(1)$. Prove that $\exists c \ where \ 0 < c < 1 \ s.t. \ f'(1 - c) = f'(c)$.

Here is my attempt

Since $f(0) = f(1)$, Rolle's Theorem applies.

This means $\exists c \in (0,1) \ s.t. \ f'(c) = 0$

Since $c \in (0,1)$, we can see that $1 - c \in (0,1)$.

To show this is also $0$, examine the original function.

If $0 = 1 - c$, then we see $c = 1$. This will satisfy the inequality giving us $f(0) = f(1)$.

If $1 = 1 - c$, then $c = 0$. This will also satisfy the inequality giving us $f(1) = f(0)$. So whether we use $c$ or $1 - c$, Rolle's Theorem will be satisfied at both points.

I was told this is incorrect, and I can see why. Where I try to show that $1 - c$ works as well, I am only given information about the function (f(0) = f(1)). Not the derivative. So I can see why this doesn't work. I was told as a hint to examine the function $g(x) = f(x) + f(1 - x)$. However, I am not so sure how to apply this to the problem and proceed. Can anyone give me some guidance?

答: $g(x)=f(1-x)+f(x)$, $g(0)=g(1)$ implies there exists $c$ such that $g'(c)=0$.


问:
Discussion : Does Fourier transform of a distribution give classical Fourier transform of its associated function ? Why?
I just finished reading the distribution chapter of Brad Osgood's lecture notes on Fourier Transform : For someone more with an engineering background, he does an amazing job at giving intuition so you feel the mathematical background without entering too much in it (I'd love to understand this more deeply, but I mainly want to apply Fourier Transform without getting lost too much in maths !).

I understand that the framework of distributions allows us to compute Fourier transforms of "functions" we couldn't have before. It gives really nice and easy formulas to work with sines, cosines, dirac delta, Heaviside function, etc... Which we couldn't have with classical Fourier transform.

However, on an intuitive level, I do not understand (I) why having the Fourier transform of a distribution (so some functional whose associated function integrates nicely with a Schwartz function) somehow also gives us the classical Fourier transform of the associated function (how could it be if the classical Fourier transform does not exist like for sine, cosine, etc... ?) ? (II) What integration with some arbitrary class of rapidly decreasing function has to do with it ? Intuitively, it would seem unrelated to me ! How this ideal case of Fourier transform of distributions has to do with classical Fourier transform of functions ? Because above all : I want the frequency spectrum of the functions sine, cosine, heaviside function, etc... not the Fourier transform of some conceptual distribution whose associated function integrates nicely with Schwartz functions.

(III) Is there some theory to prove the equality between the Fourier transform of a distribution and the Fourier transform of its associated function ? So if we can't compute the classical Fourier transform of sine, we can still get it by the Fourier transform of its associated distribution ?

It seems the extract just below is also about this (but I'm not sure) : "If the classical Fourier transform of a function defines a distribution, then that distribution is the Fourier transform of the distribution that the function defines". (IV) But I don't know if it includes the following functions : sines, cosines, Heaviside unit step, dirac, etc... because the "classical Fourier transform" of the latter does not exist... So it can't define a distribution ?

In this extract from a lecture of Brad Osgood (47'13 to 47'33), he explains an equality that has to be understood as an equality of distributions, not of pointwise functions. (V) When and why can the line be blurred ?

答:
Every locally integrable function $f \in L^1_{\text{loc}}$ induces a distribution $T_f$ by integration against the function:
$$\langle T_f, \varphi \rangle := \int f(x)\,\varphi(x)\,dx.$$
Moreover, if $T_f = T_g$ then $f=g$ almost everywhere, so we can identify $T_f$ with $f \in L^1_{\text{loc}}$.

Not every distribution is given in this way, though. For example, the Dirac $\delta$ distribution is not. Those distributions that are given by a function constitute a subspace of the distributions. Because of the identification of $f$ with $T_f$ we can look at $L^1_{\text{loc}}$ as a subspace of the distributions.

In general, distributions are defined as linear functionals over $C^\infty_c,$ the space of infinitely differentiable functions with compact support. The space of such distributions is denoted by $\mathcal{D}'.$ Many operations on distributions are defined by moving the operation to the test function (with some modification of the operation). The definitions are usually based on how it works if $u$ is given by a function.
For example, if $f \in L^1_{\text{loc}}$ and $\varphi\in C^\infty_c$ then, by integration by parts, we have
$$
\int f'(x)\,\varphi(x)\,dx = - \int f(x)\,\varphi'(x)\,dx
$$
and therefore we define the derivative of a distribution $u$ by
$$\langle u', \varphi \rangle := \langle u, -\varphi' \rangle \quad \forall\varphi\in C^\infty_c.$$

The use of test functions with compact support (the ${}_c$ part of $C^\infty_c$) give the distributions a local behavior. We can say that $u=v$ on a region $\Omega$ if $\langle u, \varphi \rangle = \langle v, \varphi \rangle$ for all $\varphi \in C^\infty_c$ with support on $\Omega.$ For example, $\delta=0$ on $(0,\infty)$.

For the Fourier transform we have
$$\int \mathcal{F}f(\xi)\,\varphi(\xi)\,d\xi = \int f(\xi)\,\mathcal{F}\varphi(\xi)\,d\xi$$
so we want to define the Fourier transform of a distribution by
$$\langle \mathcal{F}u, \varphi \rangle = \langle u, \mathcal{F}\varphi \rangle.$$
There's a problem, however; it's not certain that $\mathcal{F}\varphi \in C^\infty_c$ so $\langle u, \mathcal{F}\varphi \rangle$ is not defined.

Therefore, when defining the Fourier transform of distributions one instead takes test functions from the Schwartz space $\mathcal{S},$ a space of rapidly decreasing functions (also the derivatives decrease rapidly). This space is closed under Fourier transform so the definition above makes sense. Unfortunately, since $\mathcal{S}\supset C^\infty_c$ the space of distributions over $\mathcal{S}$ is smaller than the space of distributions over $C^\infty_c.$ This space of distributions is denoted by $\mathcal{S}'$ and the distributions are called tempered.

The classical Fourier transform is not defined for $\sin, \cos, H, \delta$ since these are not in $L^1$ (or even $L^2$). But these functions define tempered distributions and as such they have a Fourier transform. For $\delta$ the Fourier transform is $T_{\mathbf{1}},$ where $\mathbf{1}$ is the function with value $1$ everywhere. And since we identify $T_f$ with $f$ one usually just says that $\mathcal{F}\delta=\mathbf{1}.$


问:
In what ways can the presidium of a meeting be elected?
The meeting of the Student Scientific Society was attended by $52$ students: with $13$ students each from $4$ faculties. In how many ways can four people be elected to the presidium of a meeting so that it includes students from $3$ faculties?

Attempt:  $4 \cdot 3 \cdot 13 \cdot 13 \cdot \dfrac{13 \cdot 12}{2!}$

答:
Your answer is correct.

For exactly three faculties to be represented in the presidium, two students must be elected from one of the four faculties and one student each must be elected from two of the remaining three faculties.

Choose which of the four faculties will have two representatives on the presidium, which two of the thirteen members of that faculty will be elected, which two of the remaining three faculties will have one representative each on the presidium, and which of the $13$ students in each of those faculties will be elected.  The number of ways the four members of the presidium can be elected so that the presidium includes students from exactly three of the four faculties is
$$\binom{4}{1}\binom{13}{2}\binom{3}{2}\binom{13}{1}\binom{13}{1}$$
as you found.


问:
What does this symbol means?
I saw this expression in an article, and i don't know what does the horizontal 【I】 means.
enter image description here</p>

答:
$\Xi$ is just a Greek letter called Xi; it's the capital version of $\xi$.

Presumably in this context it's just being used as a variable.

Title: What does this symbol means?


问:
Prove the addition form $cos(z_1+z_2) = cosz_1cosz_2 -sinz_1sinz_2$
Prove the addition form $cos(z_1+z_2) = cosz_1cosz_2 -sinz_1sinz_2$

I was trying this way:

$cos(z_1+z_2) =  \frac{e^{i(z_1+z_2)}+e^{-i(z_1+z_2)}}{2} $.

\begin{align}
cos(z_1+z_2) = \frac{e^{-(y_1+y_2)}[cos(x_1+x_2)+isin(x_1+x_2)]+e^{-(y_1+y_2)}[cos(-x_1-x_2)+isin(-x_1-x_2)]}{2}
\end{align}

how $cosx$ is even and $sinx$ odd, I rewrite like this:
\begin{align}
cos(z_1+z_2) = \frac{e^{-(y_1+y_2)}[cos(x_1+x_2)+isin(x_1+x_2)]+e^{-(y_1+y_2)}[cos(x_1+x_2)-isin(x_1+x_2)]}{2}
\end{align}

But it's getting to big and so complicate, there is another way??


问:
Algebraically determine the x-intercept of $y=2^{x+2}-3$
Algebraically determine the $x$-intercept of $y=2^{x+2}-3$

Any help would be much appreciated... looking at the question it is clear to me that there are no real intercepts for $x$ but there is for $y$. If I look on the graph I can point out that there is a X:INT of $x$=-0.415-

Body:
You want

$$2^{x+2}-3=0$$

so $$2^{x+2}=3$$ which means $$x+2=\log_2 3$$ and finally $$x=\log_23-2$$


问:
Quadratic equations with word problems
Find the two-digit number such that its ones digit is one bigger than its tens digit, and the product of the desired number and the sum of digits is $616$.
Answer is $56$.

答:
Say the tens digit is $A$, so the ones digit is $A+1$.

Then the number is $10A+(A+1)$, and the sum of digits is $A+(A+1)$.

You want $(10A+(A+1))(A+(A+1))=616$.

That's $(11A+1)(2A+1)=616$.

Equivalently, $22A^2+13A-615=0$.

Can you solve for $A$?

(There are two solutions, but only one is relevant.)

Title: Quadratic equations with word problems


问:
Possible value for $f(\frac{1}{2})$ for some function $f:\mathbb{D}\to\mathbb{D}$

Q.20.source) Let $\mathbb{D} = \{z \in \mathbb{C} : \left| z \right| < 1\}$ be the open unit disc and let $f : \mathbb{D} \to \mathbb{D}$ be a holomorphic function. Suppose that $f(0) = 0$ and $f^\prime(0) = 0$. Which of the followings are possible values of $f\left(\frac{1}{2}\right)$?

$1/4$

$-1/4$

$1/3$

$-1/3$

Here I know that 1, 2 are correct if we choose $f(z)=z^2$ and $f(z)=-z^2$. Please give some hint for options 3 and 4.

答:
The answers $3$ and $4$ are not correct.

Since $\;f(0)=f^\prime(0)=0\;,\;$ there exists a holomorphic function $\;g:\mathbb{D}\to\mathbb{C}\;$ such that

$f(z)=z^2g(z)\quad\forall\;z\in\mathbb{D}\;.$

If $\;|z|=r<1\;,\;$ then $\;|g(z)|=\dfrac{|f(z)|}{|z|^2}<\dfrac{1}{r^2}\;,$

and, by the maximum modulus principle, it follows that

$|g(z)|<\dfrac{1}{r^2}\quad$ for all $\;z\in\mathbb{C}\;$ such that $\;|z|\le r<1\;.$

As $\;r\to1\;,\;$ we get that

$|g(z)|\le1\;$ for all $\;z\in\mathbb{C}\;$ such that $\;|z|<1\;,$

hence,

$|f(z)|=|z|^2|g(z)|\le|z|^2\;$ for all $\;z\in\mathbb{C}\land|z|<1\;,$

consequently,

$\left|f\left(\frac12\right)\right|\le\frac14\;.$


问:
If $Ker(T)=\{0\}$, so does $AX=B$ have a solution?
I have to prove if it's true or not that, given $T$ a linear transformation on $\mathbb{R}^{n}$, such that
$$[T]_{\beta}=A$$
where $\beta$ is the canonical basis, then, if $Ker(T)=\{0\}$, the system
$$AX=B$$
has solutions and they are infinite.

I could prove that $AX=B$ has a solution iff $B\in Im(T)$. What I could think about was: since $Ker(T)=\{0\}$ then
$$\dim(Ker(T))=0$$
and therefore
$$\dim(Im(T))=n$$
but I really don't know if this is the right way and if so, how can it help me.

答: $\dim(\operatorname{im} T) = n$ means that $\operatorname{im} T = \mathbb R^n$, which in turn implies that $B \in \operatorname{im} T$.


答:
To find the intersection of $f(x):=2^{x+2}-3$ and $y=0$ you must solve the equation
$$
2^{x+2}-3 = 0
$$
which gives
$$
x= \log_23-2
$$
Comes handy to know:
$$
\log_a x = \frac{\log_b x}{\log_b a}
$$
You then have
$$
\log_23 = \frac{\ln 3}{\ln 2}
$$</p>

Title: Algebraically determine the x-intercept of $y=2^{x+2}-3$

Body:
Algebraically determine the $x$-intercept of $y=2^{x+2}-3$

Any help would be much appreciated... looking at the question it is clear to me that there are no real intercepts for $x$ but there is for $y$. If I look on the graph I can point out that there is a X:INT of $x$=-0.415-


问:
Upper bound on norm of Hermitian matrix
I encountered this statement in this paper about random walks on groups by H. Kesten (page 336, (1.2)). Although he gives a reference, I cannot find the proof there. The statement looks like a simple exercise in linear algebra (and it may very well be one), but I spent quite some time on it without finding a proof:

Let $A=(a_{ij})_{1\leq i,j\leq n}$ be a Hermitian matrix.

Then we have
$$||A||\leq \sup_{1\leq i\leq n} \sum_{j=1}^n |a_{ij}|,$$
where $||A||=\sup_{||x||=1}||Ax||$ is the operator norm with respect to the standard Euclidean norm on $\mathbb{C}^n$.

答:
Supposing one understands the operator 2 norm to be equivalent to the maximal singular value of $A$, this result follows from applying either (a) Gerschgorin Discs (since $A$ is Hermitian the modulus of each of its eigenvalues equals one of its singular values) or (b) the Schur Test.

proof of the Schur Test

for arbitrary $A \in \mathbb C^{m\times n}$

and $m$ dimensional $\mathbf x$ and $n$ dimensional $\mathbf y$, each constrained to have unit length (2 norm).

$R := \max_{i} \sum_{j=1}^n \vert a_{i,j}\vert$

$C := \max_{j} \sum_{i=1}^m \vert a_{i,j}\vert$

(max L1 norm amongst respective row and column vectors of $A$)

$$
\begin{align}
&\big \vert \mathbf x^*A\mathbf y\big \vert\\
& \leq \sum_{i=1}^m \sum_{j=1}^n \big \vert x_i a_{i,j} y_j\big \vert\\
&= \sum_{i=1}^m \sum_{j=1}^n \big(\vert x_i\vert \vert  a_{i,j}\vert^\frac{1}{2}\big) \big(\vert  a_{i,j}\vert^\frac{1}{2}\vert y_j\vert\big)\\
&\leq \Big(\sum_{i=1}^m \sum_{j=1}^n \big(\vert x_i\vert \vert  a_{i,j}\vert^\frac{1}{2}\big)^2\Big)^\frac{1}{2} \Big(\sum_{i=1}^m \sum_{j=1}^n \big(\vert  a_{i,j}\vert^\frac{1}{2}\vert y_j\vert\big)^2\Big)^\frac{1}{2}\\
&= \Big(\sum_{i=1}^m \vert x_i\vert^2 \big(\sum_{j=1}^n \vert  a_{i,j}\vert\big) \Big)^\frac{1}{2} \Big(\sum_{j=1}^n \vert y_j\vert^2 \big(\sum_{i=1}^m  \vert  a_{i,j}\vert\big)  \Big)^\frac{1}{2}\\
&\leq \Big(\sum_{i=1}^m \vert x_i\vert^2 \cdot R  \Big)^\frac{1}{2} \Big(\sum_{j=1}^n \vert y_j\vert^2 \cdot C  \Big)^\frac{1}{2}\\
&=  \Big(\sqrt{R\cdot C}\Big)  \Big(\sum_{i=1}^m \vert x_i\vert^2 \Big)^\frac{1}{2} \Big(\sum_{j=1}^n \vert y_j\vert^2   \Big)^\frac{1}{2}\\
&=  \sqrt{R\cdot C}  \\
\end{align}
$$where the inequalities are (1) Triangle Inequality, (2) Cauchy-Schwarz, (3) a point-wise bound (then summing over the bound)

remark:

i.)  for this problem, $R=C$ because $A$ is Hermitian so we recover the bound in the problem statement.

ii.) in general if $A=U\Sigma V^*$ with maximal singular value $\sigma_1$ then selecting $\mathbf x:=\mathbf u_1$ and $\mathbf y:= \mathbf v_1$ gives

$\sigma_1 =  \mathbf u_1^*A\mathbf v_1 = \big \vert \mathbf u_1^*A\mathbf v_1\big \vert\leq \sqrt{R\cdot C}$

Title: Upper bound on norm of Hermitian matrix


问:
What to do when the divergence results in z and you have to find the flux?
Suppose I want to find the flux of $F$ through the portion of the cylinder $z^2 + y^2 = 16$ from $x = 1$ and $x = 3$ and normal points outwards of the cylinder.

$F= (z^2 + y)i+(x^2 + z)j+5z^2k$

I solved for the divergence

$div \ F$ = $10z$

How do I go about solving this triple integral in cylindrical coordinates $dxdrdθ$?
I'm not quite sure how to convert the $z$ and make it in terms of $x$ or $r$. Do I just treat the $z$ as $r$?

答:
I am assuming you are calculating flux including flux through the part of the planes $x = 0, x = 1$ in the cylinder.

Given $z$ is an odd function and the cylinder is symmetric with respect to $XY$ plane, your integral will be zero.

You can use cylindrical coordinates as below -

$z = r \cos \theta, y = r \sin \theta, x$

$\displaystyle \iiint_S div F \, dV = 10 \int_0^{2\pi} \int_0^4 \int_1^3 \, r^2 \cos\theta \, dx \, dr \, d\theta = 0$


问:
$\int_{0}^{2}\int_{0}^{2}\sqrt{2-4z^2}\,dz\,dy$
How do you evaluate this integral? I have tried to start this, but I have not seen an integral like this in forever. Please help me out!

$$\int_{0}^{2}\int_{0}^{2}\sqrt{2-4z^2}\,dz\,dy$$

答:
$$I=\int_0^2\int_0^2\sqrt{2-4z^2}dzdy=\sqrt{2}\int_0^2\int_0^2\sqrt{1-(\sqrt{2}z)^2}dzdy=\sqrt{2}i\int_0^2\int_0^2\sqrt{(\sqrt{2}z)^2-1}dzdy$$
now we know that $\cosh^2x-\sinh^2x=1\Rightarrow \cosh^2x-1=\sinh^2x$. So we can make the substitution: $\sqrt{2}z=\cosh x\Rightarrow dz=\frac1{\sqrt{2}}\sinh xdx$ which transforms our integral into:
$$I=i\int_0^2\int_{\cosh^{-1}(0)}^{\cosh^{-1}(2\sqrt{2})}\sinh^2(x)dxdy=i\left[y\right]_{y=0}^2\left[\frac{\sinh(2x)-2x}{4}\right]_{x=\cosh^{-1}(0)}^{\cosh^{-1}(2\sqrt{2})}$$

Now the only hard part left is working out a value for this second part of the answer.


问:
multivariable limit of a piecewise function with a more general hypothesis
Find $\lim_{(x,y)\rightarrow(0,0)}f(x,y)$, where $$f(x,y)=\begin{cases}
      \dfrac{\ln(1+xy)}{x} & x\neq0 \\
      y & x=0
   \end{cases}$$

I have a hunch that this has something to do with $\lim_{(x,y)\rightarrow(0,0)}\dfrac{\ln(1+xy)}{x}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{\ln(1+xy)}{xy}y$ which would be equal to $0$ since $\lim_{t\rightarrow0}\frac{\ln(1+t)}{t}=1$. I shall hypothesize that $\lim_{(x,y)\rightarrow(0,a)}f(x,y)=a$ .

Is my hypothesis corect? if yes, how do I prove it? Thanks for the help!

答:
For $z \geqslant 0$ we have  $\displaystyle \frac{z}{1+z} \leqslant \ln(1+z)=\int_1^{1+z} \frac{dt}{t} \leqslant z$.

For $z < 0$ we have $\displaystyle -z = |z|\leqslant -\ln(1+z)=-\ln(1 - |z|)= \int_{1 -|z|}^{1} \frac{dt}{t} \leqslant \frac{|z|}{1-|z|} = \frac{-z}{1+z}.$

This implies that regardless of the sign  of $xy$ we have

$$\frac{xy}{1+xy} \leqslant \ln(1+xy) \leqslant xy$$

Thus, $f(x,y) = y$ for $x = 0$ and

$$\frac{y}{1+xy} \leqslant f(x,y) = \frac{\ln(1+xy)}{x} \leqslant y \quad \text{ for } x > 0,\\y \leqslant f(x,y) = \frac{\ln(1+xy)}{x} \leqslant \frac{y}{1+xy} \quad \text{ for } x < 0$$

By the squeeze theorem, it follows that $\lim_{(x,y) \to (0,a)} f(x,y) = a$ and the hypothesis is correct.


问:
Weakly convergent nets and compact operators
Let $ K \in B(H)$ be a compact operator on a Hilbert space. We call a map $x: I \to H$ a net, if $I$ is a (not necessary countable) directed set. One can view nets as a generalization of sequences.

Let $(v_i)_{i \in I}$ be a bounded net in $H$ that converges weakly to $0$.
Prove that the net $(Kv_i)_{i \in I}$ converges to zero in norm.

Conversely, if $K \in B(H)$ has the property that $\lim_{i \in I} || Kv_i || = 0 $, whenever $(v_i)_{i \in I}$ is a bounded net converging weakly to $0$, then $K$ is compact.

I know that for every functional $\omega \in H^* $ the sequence given by $\omega(v_i)$ must converge to $0$.
Moreover I know that a set X is compact if and only if any net in X has a convergent subnet.
But I have no idea how to use this in some way to show either that the image net converges or to proof the other direction.

答:
I dont see the reason to use nets since you are in a normed space and convergence is described nicely by sequences but since you wanna mess around with nets, i will write the first direction by using nets and the other one with sequences.

The direction "$\implies$" doesnt require the assumption of a Hilbert space, it works also for Banach spaces. The argument is the following:

Suppose that $K\in \mathcal{B}(H)$ is a compact operator and that there exists a  bounded net, say $(u_i)_{i\in I}$ that converges weakly to $0$ but $\lim_{i}Ku_i \nrightarrow 0$. Then, there is an $\epsilon>0$ and a subnet $(u_{i_\mu})_{\mu\in M}$ of $(u_i)_{i\in I}$ such that
$$\tag{1}||Ku_{i_\mu}||\geq \epsilon\  \ \text{for every}\  \mu \in M$$
Now, $(u_i)_{i\in I}$ is bounded, which means that there is a constant $C>0$ such that $||u_i||\leq C$ for every $i\in I$. This means that the net $(u_i)_{i\in I}$ belongs the open ball $B(0,2C)$ centered at $0$ of radius $2C$. Hence, the net $(Ku_{i_\mu})_{\mu \in M}$ belongs to $K\bigl(B(0,2C)\bigr)$. Now, by the compactness of $K$ we know that $\overline{K\bigl(B(0,2C)\bigr)}$ is compact. Hence, the net $(Ku_{i_\mu})_{\mu \in M}$ has a strongly convergent subnet, say $(Ku_{i_{\mu_\sigma}})_{\sigma \in \Sigma}$ to a point $y\in H$. Now, $(u_{i_\mu})_{\mu \in M}$ converges weakly to $0$, which means that $\langle u_{i_{\mu_\sigma}}, h \rangle \to 0\ $ for every $h\in H$. Using the adjoint operator $K^*$ of $K$ we see that
$$\lim_{\sigma \in \Sigma}|\langle Ku_{i_{\mu_\sigma}}, h \rangle|=\lim_{\sigma \in \Sigma}|\langle u_{i_{\mu_\sigma}}, K^*h\rangle|=0$$
Which means that $(Ku_{i_{\mu_\sigma}})_{\sigma \in \Sigma}$ converges weakly to $0$. But we  already know that $(Ku_{i_{\mu_\sigma}})_{\sigma \in \Sigma}$ converges strongly to $y$. Hence, $y=0$. But this means that $(Ku_{i_{\mu_\sigma}})_{\sigma \in \Sigma}$ can be arbitrary close to $0$ which contradicts $(1)$.

For the other direction "$\impliedby$" suppose that $K$ has the property that $\lim_{i\in I}||Ku_i||=0$ for every bounded net $(u_i)_{i\in I}$ which is weakly convergent to $0$. To show that $K$ is compact we only need to show that $K(B_H)$ is totally bounded (check this) where $B_H$ is the closed unit ball $\{h\in H:\, ||h||\leq 1\}$ of $H$. Supoose that $K(B_H)$ is not totally bounded. Then there exists an $\epsilon>0$ such that $K(B_H)$ cannot be covered by finitely many balls with centers from $K(B_H)$ and radius $\epsilon>0$. This means that for some $x_1 \in B_H$ the open ball $B(Kx_1,\epsilon)$ does not cover $K(B_H)$. Hence, there is some $x_2\in B_H$ such that $||Kx_1-Kx_2||\geq \epsilon$. Now, the open balls $B(Kx_1,\epsilon),\, B(Kx_2,\epsilon)$ cannot cover $K(B_H)$, hence there is some $x_3\in B_H$ such that $||Kx_1-Kx_3||,\, ||Kx_2-Kx_3||\geq \epsilon$. Continuing this way, recursively, we construct a sequence $x_n \in B_H$ such that
$$\tag{2} ||Kx_m-Kx_n||\geq \epsilon\ \text{ for every }\ n\neq m$$
Now, $x_n$ belongs to the closed unit ball $B_H$. Since, $H$ is a Hilbert space, it is also reflexive, this means that $B_H$ is weakly compact. Hence, $x_n$ has a weakly convergent subsequence, say $x_{k_n}$ converges weakly to $x$. Then, $z_{n}=x_{k_n}-x$ converges weakly to $0$. From our hypothesis, it follows that $\lim_{n\to \infty}||Kz_n||=0$. But, for every $n\neq m$
$$||Kz_n- Kz_m||=||Kx_{k_n}-Kx_{k_m}||$$
which contradicts $(2)$. Hence, $K(B_H)$ is totally bounded which implies that $\overline{K(B_H)}$ is compact.


问:
Finding as expressions for the $E[X^2_T]$ for a particular stochastic process.
Suppose that we have the following stochastic process and we want to find an expression for the expected value of this at time $t$ of the process at some future time $T$ of this process squared, $E[X^2_T]$.

$$dX_s=(\alpha X_s) ds + \sigma e^{-s} dW_s,\ X_t=x$$
In order to find an explicit solution for $E_{t,x}[X_T^2]$ I thought I should use Itô's lemma applied to the function $z(t,x)=x^2$.
$$dz(t,X_s)=\left( (\alpha X_s)(2X_s) + \frac{1}{2}\sigma^2e^{-2s}(2)\right)ds+\sigma e^{-s}(2X_s)dW_s$$
Now we integrate and simplify a bit, plugging in the boundary condition $X_t=x$.
$$z(T,X_T)= z(t,x) + \int_t^T \left( 2 \alpha X_s^2 + \sigma^2e^{-2s} \right)ds+ \int_t^T \sigma e^{-s}(2X_s)dW_s$$
We now take the expectation, which gives us the expression we are looking for, ie $E_{t,x}[z(T,X_T)] = E_{t,x}[X_T^2]$.
$$E_{t,x}[z(T,X_T)]= z(t,x) + E_{t,x}\left[ \int_t^T \left( 2 \alpha X_s^2 + \sigma^2e^{-2s} \right)ds \right]+ E_{t,x}\left[ \int_t^T \sigma e^{-s}(2X_s)dW_s \right]$$
As usual we take the expectation of the stochastic term to be zero.
$$E_{t,x}[z(T,X_T)]= z(t,x) + E_{t,x}\left[ \int_t^T \left( 2 \alpha X_s^2 + \sigma^2e^{-2s} \right)ds \right]$$

However, I am seeing that I have not really simplified anything since we sill have to take a expectation of $X^2_s$ in order to evaluate the integral. Am I missing something here or should I use a different technique besides Itô's lemma?

答:
Assume we have proven that $\mathbb{E}_{t,x}[X_T^2] = x^2 + \mathbb{E}_{t,x}[\int_t^T 2\alpha X_s^2 + \sigma^2e^{-2s} ds]$.

(Note: this can be easily done using the fact that both the coefficients of the original SDE are lipschitz in $x$ uniformly wrt $t$, and thus $\mathbb{E}[\sup\limits_{0\leq s\leq T} X_s^2]$ is bounded)

Now, our equation can be rewritten as $\mathbb{E}_{t,x}[X_T^2] = x^2 + 2\alpha\mathbb{E}_{t,x}[\int_t^T X_s^2 ds] + \frac{1}{2}\sigma^2(e^{-2t}-e^{-2T})$. This means, since everything is positive, that $$ \mathbb{E}_{t,x}[X_T^2] = x^2 + 2\alpha\int_t^T\mathbb{E}_{t,x}[ X_s^2 ] ds + \frac{1}{2}\sigma^2(e^{-2t}-e^{-2T}). $$

We now pose $f(T) = \mathbb{E}_{t,x}[X_T^2]$ for every $T\geq t$. We have $f(t)=x^2$ and $f'(T) = 2\alpha f(T) + \sigma^2e^{-2T}$ for every $T\geq t$.

Can you finish now?


问:
Existence of Phase Flow
Hi there I'm wondering if anyone can clear up my confusion:  What is the proof of the local existence of a phase flow for a differentiable vector field?</p>

答:
Assume $X^i:\mathbb R^n\to \mathbb R$ are $n$ coordinates of a vector field. We seek a set of functions $\phi^i(x_0,t)$ for small $t$ such that $\partial \phi^i\partial t =X^i(\phi(x_0,t))$.

By even just the Peano existence theorem (multidimensional version), we find such a $\phi$ exists by continuity of $X$.

Picard–Lindelöf in the multidimensional case along with continuous differentiability of $X$ lead to uniqueness of $\phi$.

Typical 'continuous dependence on initial conditions' results give continuity in $x_0$.

Title: Existence of Phase Flow


问:
Proving that a function doesn't have a limit given $|a-b| = |f(a)-f(b)|$ for all $a,b\in \mathbb R$
For all $a,b\in \mathbb R$,

$|a-b| = |f(a)-f(b)|$.

Prove that $\;\lim\limits_{a\to \infty}|f(a)|=\infty$

What I was trying to do is show that if we assume that there is a limit, $\;\lim\limits_{a\to \infty}|f(a)|=L\;$ and therefore $\big||f(a)|-L\big|=\big||f(a)|-|L|\big| \leq |f(a)-L| < \epsilon$

Then we set $\epsilon$ to some arbitrary value and try to get a contradiction.

I also tried proving directly but couldn't get much out of it.

Is there a better approach for proving it?

答:
Consider $b=0$.

As $a \to \infty,\ |f(a)| + |f(0)| \geq|f(a)-f(0)| = |a-0| = |a| \to \infty$.

Hence as $a \to \infty,\ |f(a)| \geq |a| - |f(0)| \to \infty.$


问:
solve $\lim _{x\to 2}\left(\frac{\sqrt{x^2-1}-\sqrt{2x-1}}{\sqrt{x+2}-\sqrt{x^2+2x-4}}\right)$ without l'Hopital
I came across this and I am struggling to solve it, without using L'Hopital:

$\lim _{x\to 2}\left(\frac{\sqrt{x^2-1}-\sqrt{2x-1}}{\sqrt{x+2}-\sqrt{x^2+2x-4}}\right)$

Any ideas?

答:
HINT

As @AndrewChin has mentioned, one has that
\begin{align*}
\sqrt{x^{2}-1} - \sqrt{2x-1} = \frac{(x^{2} - 1) - (2x-1)}{\sqrt{x^{2}-1}+\sqrt{2x-1}} = \frac{x^{2} - 2x}{\sqrt{x^{2}-1} + \sqrt{2x-1}}
\end{align*}
Similarly, we do also have that
\begin{align*}
\sqrt{x+2} - \sqrt{x^{2} + 2x - 4} = \frac{(x+2) - (x^{2} + 2x - 4)}{\sqrt{x+2} + \sqrt{x^{2} + 2x - 4}} = -\frac{x^{2} + x - 6}{\sqrt{x+2} + \sqrt{x^{2} + 2x - 4}}
\end{align*}

Now notice that $x^{2} + x - 6 = (x - 2)(x + 3)$.

Can you take it from here?


问:
Equivalent definition of supersingular elliptic curves
Using the definitions that an elliptic curve $E$ over a finite field $K$ of characteristics $p$ is ordinary if $E[p]\cong \mathbb{Z}/p\mathbb{Z}$ and supersingular if $E[p]\cong 0$, how can I show that $E$ is supersingular if and only if $\#E(K)\equiv 1 \mod p$?

答:
This is explained in the reference textbook Silverman AEC

From that $\phi_q(x,y)=(x^q,y^q)\in End(E)$ has a dual endomorphism $\phi_q^*$ such that $\phi_q^*\phi_q=q\in End(E)$ and $\phi_q^*+\phi_q=t\in \Bbb{Z}\subset End(E)$. Then $\phi_q$ is a root of $X^2-Xt+q$

With $a=\frac{t+\sqrt{t^2-4q}}{2},b=\frac{t-\sqrt{t^2-4q}}{2}$

and $$\#E(\Bbb{F}_{q^n})=\deg(\phi_q^n-1)= (a^n-1)(b^n-1)$$

If $X^2-Xt+q$ is irreducible then $\Bbb{Z}[\phi_q]$ is isomorphic to $\Bbb{Z}[a]\subset \Bbb{Q}(a)$, if one of $a,b$ is a unit of $\Bbb{Z}[a]/(p)$ then for some $n$ we'll have that $(a^n -1)(b^n-1)\in (p)$ so $p | \#E(\Bbb{F}_{q^n})$.

Thus $E$ is supersingular iff $a,b$ are not units which implies that $X^2-Xt+q = (X-0)(X-0) \bmod (p)$ ie. $\#E(\Bbb{F}_q)=q+1-t = 1\bmod p$.

If $X^2-Xt+q$ is reducible then $a,b$ are integers and $ab=q$, since $\deg(\phi)=\deg(\phi^*)=q$ then $a=b$, $q$ is a square and $X^2-Xt+q=X^2\pm 2q^{1/2} X+q$ so $\#E(\Bbb{F}_q)=q+1\pm q^{1/2} = 1\bmod p$.


问:
Solving for a differential equation Gompertz growth equation
What is the general solution of this differential equation?
$$ \frac{dy}{dt} = k \enspace  y \enspace \ln(\frac{a}{y})$$ where $a$ and $k$ are positive constants.

So far, my solution is:
$$ \frac{dy}{y \enspace \ln(\frac{a}{y})} = k \enspace dt$$
When I let $u=lny$,
$$ \int \frac{1}{y(\ln a-\ln y)} \,dy =\int k dt $$
$$ \int \frac{1}{\ln a-u} \,du= kt + C $$

How to continue this?

答:
$$\frac{dy}{dt}=ky\ln\left(\frac ay\right)$$
$$\frac{dy}{dt}=ky\left[\ln(a)-\ln(y)\right]$$
now make the substitution: $y=e^z$ so $\frac{dy}{dt}=e^z\frac{dz}{dt}=y\frac{dz}{dt}$ and so:
$$y\frac{dz}{dt}=ky\left[\ln(a)-z\right]$$
$$\frac{dz}{dt}=k\ln(a)-kz$$
if we let $\alpha=k\ln(a)$ you will see we have a simple equation:
$$z'=-kz+\alpha$$
$$\int\frac{dz}{kz-\alpha}=-\int dt$$
$$\frac1k\ln(kz-\alpha)=-t+C_1$$
$$\ln(kz-\alpha)=-kt+C_2$$
$$kz-\alpha=C_3e^{-kt}$$
$$z=C_4e^{-kt}+\ln(a)$$
$$y=a\exp\left(Ce^{-kt}\right)$$


问:
Universal property of field of fractions
Wikipedia states that

The field of fractions of R is characterised by the following universal property:

If $h : R \rightarrow F$ is an injective ring homomorphism from R into a field F, then there exists a unique ring homomorphism $g : \text{Frac}(R) \rightarrow F$ which extends $h$.

I don't understand how we can show this. I get that $g$ will be equal to $h \circ \iota$, where $\iota(x, 1) = x$ on all $(x, 1)$, but how can we uniquely that extend to the rest of $\text{Frac}(R)$?

答:
Consider $g : \operatorname{Frac}(R) \to F$ given by $g(a,b) = h(a)h(b)^{-1}$. Since $h(1) = 1$, $g(x,1) = h(x)$ for all $x \in R$, which means that $h = g \circ \iota$, where $\iota : R \to \operatorname{Frac}(R)$ is given by $\iota(x) = (x,1)$.

Added: Suppose that $g' : \operatorname{Frac}(R) \to F$ is another ring homomorphism such that $h = g' \circ \iota$. Then note that $g'(x,1) = h(x)$ for all $x \in R$, and then, for $b \in R \setminus \{0\}$ we have
$$1 = h(1) = g'(1,1) = g'(b,b) = g'((b,1)(1,b)) = g'(b,1)g'(1,b) = h(b)g'(1,b)$$ that is, $g'(1,b) = h(b)^{-1}$. Hence, for each $(a,b) \in \operatorname{Frac}(R)$ we have $$g'(a,b) = g'((a,1)(1,b)) = g'(a,1)g'(1,b) = h(a)h(b)^{-1} = g(a,b).$$


问:
Prove that for any Riemann integrable $f$ on the interval $[a, b]$, there is a $c$ such that $\int_{a}^{c} f = \int_{c}^{b} f$
This is one of the practice problems I am having a bit of trouble trying to figure out and prove.

Prove that for any Riemann integrable function $f$ on the interval $[a, b]$, there is a $c$ such that $\int_{a}^{c} f= \int_{c}^{b} f$.

I was thinking about like if $[a,b]$ is the entire interval, then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$. I am not sure how to proceed from here. Some help would be nice.

答: Let $g(x)=\int_x^bf-\int_a^xf$. Then, $g(a)=\int_a^bf$ and $g(b)=-\int_a^bf$. If $\int_a^bf=0$, you can use $c=a$. Else, you can use the Intermediate Value Theorem.


答: If $T$ is a linear transformation from $\mathbb{R}^n \rightarrow \mathbb{R}^n$ and $ker T = {0}$ this implies that $T$ is bijective. If $T$ is bijective that would imply that $Ax=b$ has only a single solution. Thus the statement that there is an infinite number of solutions is false.

Title: If $Ker(T)=\{0\}$, so does $AX=B$ have a solution?

Body:
I have to prove if it's true or not that, given $T$ a linear transformation on $\mathbb{R}^{n}$, such that
$$[T]_{\beta}=A$$
where $\beta$ is the canonical basis, then, if $Ker(T)=\{0\}$, the system
$$AX=B$$
has solutions and they are infinite.

I could prove that $AX=B$ has a solution iff $B\in Im(T)$. What I could think about was: since $Ker(T)=\{0\}$ then
$$\dim(Ker(T))=0$$
and therefore
$$\dim(Im(T))=n$$
but I really don't know if this is the right way and if so, how can it help me.


问:
How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
How can one prove the statement
$$\lim_{x\to 0}\frac{\sin x}x=1$$
without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.

This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.

答:
I am not sure if it counts as proof, but I have seen this done by a High Schooler.

In the given picture above,
$\displaystyle 2n \text{ EJ} = 2nR \sin\left( \frac{\pi}{n } \right ) = \text{ perimeter of polygon }$.

$\displaystyle \lim_{n\to \infty }2nR \sin\left( \frac{\pi}{n } \right ) = \lim_{n\to \infty } (\text{ perimeter of polygon }) = 2 \pi R \implies \lim_{n\to \infty}\frac{\sin\left( \frac{\pi}{n } \right )}{\left( \frac{\pi}{n } \right )} = 1$ and let $\frac{\pi}{n} = x$.


问:
Are there functions $f,g:\mathbb{R} \to \mathbb{R}$ such that they differentiate each other?
i.e.
$$f'(x) = (g \circ f)(x)$$
$$g'(x) = (f\circ g)(x)$$

I came up with this question a few years ago. A friend found the only example I know:

for $c\in\mathbb{R}$
$$f(x) = c$$
$$g(x) = cx - c^2 $$

After trying with some particular cases (with no success), I used the formula for the derivative of the inverse function and got that if $f$ and $g$ are bijective then,

$$f^{-1} = \int{}{dt \over g(t)}  $$
$$g^{-1} = \int {dt \over f(t)}$$

Assuming all conditions neccesary for this to be possible. I tried using this fact to construct the functions, again with no luck.

I would really appreciate any insight on how to tackle this problem.

答:
Some partial results, but too long for a comment.

The condition implies that $f,g$ are $C^\infty$.
In particular,
$$ f''=(g\circ f)'=f'\cdot g'\circ f=g\circ f \cdot f\circ g\circ f=(\operatorname{id}\cdot f)\circ g\circ f$$
so that every zero of $f'$ is also a zero of $f''$.
Let $h_0=f$ and recursively $$h_{n+1}=\begin{cases}g\circ h_n&n\text{ even}\\f\circ h_n&n\text{ odd}\end{cases}$$ So far, we have $f=h_0$, $f'=h_1$, $f''=h_1h_2$.
Note that
$$h_{n+1}'=\begin{cases}h_n'\cdot g'\circ h_n=h_n'\cdot f\circ g\circ h_n&n\text{ even}\\
h_n'\cdot f'\circ h_n=h_n'\cdot g\circ f\circ h_n&n\text { odd}\end{cases} $$
So at any rate,
$$ h_{n+1}'=h_n'h_{n+2}.$$
Thus by induction,
$$\tag1 h_n'=\prod_{k=1}^{n+1} h_k.$$

Let $$ H=\Bigl\{\,\prod_{k=1}^m h_k^{a_k}\Bigm| m\ge 1, a_1\ge a_2\ge a_m\ge1 \,\Bigr\}$$

If $\phi\in H$, then by the product rule and $(1)$, $\phi'$ is a finite sum of elements of $H$.
As $f'=h_1\in H$, we conclude that all $f^{(n)}$, $n\ge1$, are finite sums of elements of $ H$.
In particular, for all $x$ with $f'(x)=0$, we have $f^ {(n)}(x)=0$ for all $n\ge1$.
The same conclusion of course holds for $g$.

As a consequence of the identity theorem:

If $f$ is analytic, then $f$ is either constant (and so $g$ linear) or $f'$ has no zeroes (so in particular, $f$ is strictly monotonic). Same for $g$.

So if $f,g$ are analytic and neither is constant, then both are monotonic, hence all $h_n$ are strictly monotonic. If both $f,g$ are increasing, then all derivatives are increasing.


问:
Given this matrix $A$, prove analytically that $\det(A) = n^{n-1}+(n-1)^n$
A friend of mine asked me about the determinant of the following matrix.

$$A=\begin{pmatrix}
n   & 1   &  1 & \dots  & 1 & 1\\
n-1 & n   &  1 & \dots & 1 & 1\\
n-2 & n-1 &  n & \dots & 1 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
 2& 3 & 4 & \dots & n & 1\\
 1& 2 & 3 & \dots & n-1 & n
\end{pmatrix}$$

So I wrote the following Mathematica code:

```<code>function[n_Integer] :=
Det[Table[
  Which[i == j, n, i < j, 1, True, j - i + n], {i, 1, n}, {j, 1, n}]]
Table[function[i], {i, 2, 20}]
</code>```

which prints this result

$$3, 17, 145, 1649, 23401,...$$

and with the help of OEIS I found that the answer is

$$\det(A) = n^{n-1}+(n-1)^n$$

is there an analytical way of calculating this determinant?

答:
We exploit the fact that adding linear combinations of rows to another row does not change the value of the determinant to zero out all elements except the diagonal, subdiagonal and the top right corner and then we compute $\det A$ from Laplace expansion.

First, subtract $k$th row from $(k+1)$th row for $k = 1, \dots, n-1$ to obtain

$$B =\begin{pmatrix}
n   & 1   &  1 & \dots  & 1 & 1\\
-1 & n-1   &  0 & \dots & 0 & 0\\
-1 & -1 &  n-1 & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
-1 & -1 & -1 & \dots & n-1 & 0\\
-1 & -1 & -1 & \dots & -1 & n-1
\end{pmatrix}.$$

Next, add the bottom row to the top row to get

$$C=\begin{pmatrix}
n-1   & 0   &  0 & \dots  & 0 & n\\
-1 & n-1   &  0 & \dots & 0 & 0\\
-1 & -1 &  n-1 & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
-1 & -1 & -1 & \dots & n-1 & 0\\
-1 & -1 & -1 & \dots & -1 & n-1
\end{pmatrix}.$$

Subtract $k$th row from $(k+1)$th row for $k = 2, \dots, n-1$ to obtain

$$D=\begin{pmatrix}
n-1   & 0   &  0 & \dots  & 0 & n\\
-1 & n-1   &  0 & \dots & 0 & 0\\
0 & -n &  n-1 & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & n-1 & 0\\
0 & 0 & 0 & \dots & -n & n-1
\end{pmatrix}.$$

Finally, compute determinant of $D$ using Laplace expansion along the top row

$$
\det A = \det D = (-1)^{1+1} (n-1) \det D_{1,1} + (-1)^{n+1} n \det D_{1,n}
$$

where $D_{i,j}$ denotes the matrix obtained from $D$ by deleting the $i$th row and the $j$th column. Both $\det D_{1,1}$ and $\det D_{1,n}$ can now be easily computed by exploiting the fact that determinant of a triangular matrix is the product of the elements on its diagonal

$$
\det A = (n-1) (n-1)^{n-1} + (-1)^{n+1} n (-1) (-n)^{n-2} = (n-1)^n + n^{n-1}
$$

as claimed.


问:
Question about calculating norm of an operator
Let's define the following Hilbert space $\ell^2(\mathbb{Z})$ on the field $\mathbb{C}$:

$$l^2(\mathbb{Z}) = \{ x = (x_n)_{n=-\infty}^{\infty}: \sum_{n=-\infty}^{\infty}|x_n|^2 < \infty  \} $$
with the inner product $\langle x,y \rangle = \sum_{n = -\infty}^{\infty} x_n \overline y_n$

and let's define a linear operator $T$ on $\ell_2(\mathbb{Z})$ by
$$ T(...,x_{-1},x_0,x_1,...)_n = (...,\frac{x_{-2}+x_{0}}{2},\frac{x_{-1}+x_{1}}{2},\frac{x_{0}+x_{2}}{2},...)$$
(namely, $x_n \rightarrow \frac{x_{n-1} + x_{n+1}}{2}).$

I would like to compute the norm of this operator.

I started by noticing that we can write

$$T = \frac{1}{2} S^{+} + \frac{1}{2} S^- $$
where $S^+$ and $S^-$ are left and right shifts respectively. Since on this space shifting a sequence doesn't change the value of its norm (since we have infinite amount of elements for all integers) hence $\lVert S^+ (x) \rVert = \lVert x \rVert$ so
$$\lVert T \rVert = \frac{1}{2}\lVert S^+ + S^- \rVert \leq \frac{1}{2} [\lVert S^+ \rVert + \lVert S^- \rVert] = 1.$$

Now I'm unfortunately stuck with the other bound. Using the fact that space is Hilbert I computed
\begin{align}
\lVert Tx \rVert ^2 &= \frac{1}{4} \langle S^+ + S^-, S^+ + S^-\rangle = \frac{1}{4}\Big[\lVert S^+ \rVert^2 + \lVert S^- \rVert^2 +  \langle S^+,S^- \rangle + \langle S^-,S^+ \rangle \Big] \\
&=\frac{1}{4} \Big[2+ \langle S^+, S^- \rangle + \langle S^-, S^+ \rangle \Big].
\end{align}

So, now I'm looking for an element $x$ in $\ell^2(\mathbb{Z})$ such that $\lVert x \rVert = 1$ and

$$ \langle S^+ x, S^- x \rangle = 1, $$
that is
$$ \sum_{n = -\infty}^{\infty} x_n \overline{x_{n-2}} = 1 .$$

I was trying to do it with $x$ given by
$$x_n = \Big(\frac{1}{\sqrt{2}}\Big)^n$$
for $n>0$ and $0$ otherwise, so that $\lVert x \rVert = 1$ but then

\begin{align}
\sum_{n = -\infty}^{\infty} x_n \overline{x_{n-2}} &= \sum_{n=2}^{\infty} \Big[\Big(\frac{1}{\sqrt{2}}\Big)^{n-1}+\Big(\frac{1}{\sqrt{2}}\Big)^{n+1}\Big]^2 = \sum_{n=2}^{\infty} \Big(\frac{1}{2}\Big)^n = \frac{1}{2} \neq 1.
\end{align}

I also did some attempts with similar sequences (for example a sequence with $x_0 = 0$ and $x_{-n} = x_n = (\frac{1}{2\sqrt{2}})^n$) but unfortunately I wasn't able to reach 1 with any one them.

Could anyone share an idea which sequence could I consider, or if perhaps my initial estimate of $\lVert T \rVert \leq 1$ is too rough in the first place (or if I've made a mistake somewhere else)?

Thanks!

答:
For a fixed integer $N$, let $x=(x_n)_{n\in\mathbb Z}$ be the sequence such that $x_n=1$ if $\lvert n\rvert\leqslant N$ and $0$ otherwise. Then $\lVert x\rVert^2=2N+1$. Denoting by $y_n$ the $n$-th element of $Tx$, we get that
$y_n=0$ if $\lvert n\rvert\geqslant N+2$, $y_n=1$ if $\lvert n\rvert \leqslant N-1$  and $ y_{\pm N}=y_{\pm (N+1)}=1/2$ hence $\lVert y\rVert^2$ is of order $2N$. This shows that the norm of $T$ is $1$.


问:
Find scale, rotation, and translation that maps three points to three points in 3D
Given three source points $\mathbf{P}_1 = [x_1\ y_1\ z_1]^T,$
$\mathbf{P}_2 = [x_2\ y_2\ z_2]^T,$ $\mathbf{P}_3 = [x_3\ y_3\ z_3]^T,$ and

three destination points $\mathbf{P}'_1 = [x'_1\ y'_1\ z'_1]^T,$
$\mathbf{P}'_2 = [x'_2\ y'_2\ z'_2]^T,$ $\mathbf{P'}_3 = [x'_3\ y'_3\ z'_3]^T,$
how do we find the scale $S,$ rotation $R,$ and translation $T$ that transforms
the source points into the destination points as in the following equation?
$$
\left[\begin{array}{ccc}
s_x &     & \\
    & s_y & \\
    &     & s_z \\
\end{array}\right]
\left[\begin{array}{cccc}
r_{11} & r_{12} & r_{13} & t_x \\
r_{21} & r_{22} & r_{23} & t_y \\
r_{31} & r_{32} & r_{33} & t_z
\end{array}\right]
\left[\begin{array}{ccc}
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 \\
z_1 & z_2 & z_3 \\
1 & 1 & 1
\end{array}\right]
=
\left[\begin{array}{ccc}
x'_1 & x'_2 & x'_3 \\
y'_1 & y'_2 & y'_3 \\
z'_1 & z'_2 & z'_3
\end{array}\right]
$$
Since the $3 \times 3$ rotation $R$ is orthonormal and can be parameterized by three rotation
angles (e.g. yaw, pitch, roll) there are 9 equations and 9 unknowns. Ideally the scale would be uniform $(s_x = s_y = s_z)$ resulting in a similarity transformation, but then the system would be over constrained. As long as the source and destination points are not co-linear, this should have a unique solution.

One could solve for a general affine transformation $A$ and decompose this, but this would require 4 non-coplanar source and destination points and may include shears.

EDIT: A possible approach is to solve for the general affine case
$X A = B$
$$
\left[\begin{array}{cccc}
x_1 & y_1 & z_1 & 1\\
x_2 & y_2 & z_2 & 1\\
x_3 & y_3 & z_3 & 1
\end{array}\right]
\left[\begin{array}{ccc}
a_{11} & a_{21} & a_{31} \\
a_{12} & a_{22} & a_{32} \\
a_{13} & a_{23} & a_{33} \\
a_{14} & a_{24} & a_{34} \\
\end{array}\right]
=
\left[\begin{array}{ccc}
x'_1 & y'_1 & z'_1 \\
x'_2 & y'_2 & z'_2 \\
x'_3 & y'_3 & z'_3
\end{array}\right]
$$
which is under-constrained since there are 9 equations and 12 unknowns which means there are infinite solutions.
Therefore, following the recipe here, we will find the solution $A'$ that has the minimum norm $\|A\|.$
The normal equation for an underdetermined system is
$$
A^* = X^T (X X^T)^{-1} B
$$
from which we can get the optimal value via SVD.
The SVD for $X$ is
$$
\begin{eqnarray}
X &=& U \Sigma V^T \\
  &=&
\left[\begin{array}{ccc}
\cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot
\end{array}\right]
\left[\begin{array}{cccc}
  \sigma_1 & & & \cdot \\
     & \sigma_2 & & \cdot \\
     & & \sigma_3 & \cdot
\end{array}\right]
\left[\begin{array}{cccc}
\cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot
\end{array}\right]^T
\end{eqnarray}
$$
where $\sigma_1 \geq \sigma_2 \geq \sigma_3 \geq 0$ are the
singular values (if $\sigma_3 = 0$ then the source points are
co-linear, i.e., $X$'s rank is less than 3) and $U$ and $V$ are
orthonormal matrices.
Plugging this into our normal equation we have
$$
\begin{eqnarray}
A^* &=& V \Sigma^T (\Sigma \Sigma^T )^{-1} U^T B \\
  &=&
V
\left[\begin{array}{ccc}
\frac{1}{\sigma_1} & & \\
 & \frac{1}{\sigma_2} & & \\
 & & \frac{1}{\sigma_3} \\
 \cdot & \cdot & \cdot
\end{array}\right]
U^T B
\end{eqnarray}
$$
where the matrix above containing the reciprocals of the singular values
is the pseudo-inverse of $\Sigma.$
The hope is the solution $A^*$ which the minimal norm from the set of all solutions will be a similar transformation. Yet to be verified.

答:
The assumption is that this can be accomplished with a similarity transformation.
We use the following sequence of transformations to compose the desired transformation:

Translate $\mathbf{P}_0$ to the origin;

Orient $\triangle \mathbf{P}_0 \mathbf{P}_1 \mathbf{P}_2$
to align with $\triangle \mathbf{P}'_0 \mathbf{P}'_1 \mathbf{P}'_2$ via rotation $R;$

Uniform scale by
$s = \frac{\|\mathbf{P}'_1 - \mathbf{P}'_0\|}{\|\mathbf{P}_1 - \mathbf{P}_0\|}$;

Translate to $\mathbf{P'}_0$

$$
\begin{eqnarray}
S &=&
\left[\begin{array}{cccc}
1 & 0 & 0 & x_0' \\
0 & 1 & 0 & y_0' \\
0 & 0 & 1 & z_0' \\
0 & 0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cccc}
s & 0 & 0 & 0 \\
0 & s & 0 & 0 \\
0 & 0 & s & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cccc}
r_{11} & r_{12} & r_{13} & 0 \\
r_{21} & r_{32} & r_{23} & 0 \\
r_{31} & r_{32} & r_{33} & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cccc}
1 & 0 & 0 & -x_0 \\
0 & 1 & 0 & -y_0 \\
0 & 0 & 1 & -z_0 \\
0 & 0 & 0 & 1
\end{array}\right] \\
&=&
%\left[sI\ | \mathbf{P}'_0 \right] \left[R\ | -R \mathbf{P}_0 \right]
\left[\begin{array}{cccc}
s & 0 & 0 & x_0' \\
0 & s & 0 & y_0' \\
0 & 0 & s & z_0' \\
0 & 0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cccc}
r_{11} & r_{12} & r_{13} & -x''_0 \\
r_{21} & r_{32} & r_{23} & -y''_0 \\
r_{31} & r_{32} & r_{33} & -z''_0 \\
0 & 0 & 0 & 1
\end{array}\right]\ \ \ \ \mbox{(let $\mathbf{P}''_0 = R \mathbf{P}_0)$} \\
&=&
\left[\begin{array}{cccc}
s r_{11} & s r_{12} & s r_{13} & x'_0 - s x''_0 \\
s r_{21} & s r_{32} & s r_{23} & y'_0 - s y''_0 \\
s r_{31} & s r_{32} & s r_{33} & z'_0 - s z''_0 \\
0 & 0 & 0 & 1
\end{array}\right]
\end{eqnarray}
$$

To find the rotation $R$ we align the edge $\overline{\mathbf{P}_0 \mathbf{P}_1}$ with the edge $\overline{\mathbf{P}'_0 \mathbf{P}'_1}$ and align the plane of
$\triangle \mathbf{P}_0 \mathbf{P}_1 \mathbf{P}_2$  with the plane of
$\triangle \mathbf{P}'_0 \mathbf{P}'_1 \mathbf{P}'_2.$
We define the source orthonormal coordinate axes as
$$
\begin{eqnarray}
\mathbf{X} &=& \frac{\mathbf{P}_1 - \mathbf{P}_0}{\|\mathbf{P}_1 - \mathbf{P}_0 \|}\\
\mathbf{Z} &=& \frac{(\mathbf{P}_2 - \mathbf{P}_0) \times \mathbf{X}}{\| (\mathbf{P}_2 - \mathbf{P}_0) \times \mathbf{X}\|} \\
\mathbf{Y} &=& \mathbf{X} \times \mathbf{Z}
\end{eqnarray}
$$
and the target coordinate axes as
$$
\begin{eqnarray}
\mathbf{X}' &=& \frac{\mathbf{P}'_1 - \mathbf{P}'_0}{\|\mathbf{P}'_1 - \mathbf{P}'_0 \|}\\
\mathbf{Z}' &=& \frac{(\mathbf{P}'_2 - \mathbf{P}'_0) \times \mathbf{X}'}{\| (\mathbf{P}'_2 - \mathbf{P}'_0) \times \mathbf{X}'\|} \\
\mathbf{Y}' &=& \mathbf{X}' \times \mathbf{Z}'
\end{eqnarray}
$$
We are looking for the rotation $R$ that maps the source coordinate axes
into the target axes:
$$
\begin{eqnarray}
R \left[ \mathbf{X}\ \mathbf{Y}\ \mathbf{Z} \right]
&=&
\left[ \mathbf{X}'\ \mathbf{Y}'\ \mathbf{Z}' \right] \\
R &=& \left[ \mathbf{X}'\ \mathbf{Y}'\ \mathbf{Z}' \right]
\left[ \mathbf{X}\ \mathbf{Y}\ \mathbf{Z} \right]^T %\\
%   &=& \left[ \mathbf{X}'\ \mathbf{Y}'\ \mathbf{Z}' \right]
%    \left[ \begin{array}{c}
%        \mathbf{X}^T \\
%        \mathbf{Y}^T \\
%        \mathbf{Z}^T
%    \end{array}
%   \right]
\end{eqnarray}
$$
So we have a uniform scale $s,$ rotation $R,$ and translation
$\mathbf{P}'_0 - s R \mathbf{P}_0.$


问:
Schwartz functions after periodization has period 1
We define periodization of a function $f$ as
$$
P_1f(x) = \sum_{n\in N} f(x+n) = \lim_{N\rightarrow \infty} \sum_{|n|\leq N} f(x+n).
$$
To show that $P_1f$ has period 1, my idea was to do a change of variable so that $m=n+1$. However, this does not use any property of the Schwartz class and it seems applicable to any integer form, which looks wrong to me. Any help?

答:
The only property of a Schwartz function used here is that $$\tag{*}\left(1+(x+n)^2\right)\lvert f(x+n)\lvert\leqslant C,$$ which guarantees that $P_1f(x)$ is defined for every $x$ and makes all the computations with series valid.

Inequality $(*)$ comes from the fact that for each $k\geqslant 0$, there exists a constant $C_k$ such that $\sup_{t\in\mathbb R}\lvert t\rvert^k\lvert f(t)\rvert\leqslant C_k$; here we take $C:=C_0+C_2$.


问:
Is it correct to say that the eigenvalue decomposition of a matrix A is unique up to taking multiples of the eigenvectors?
In the notes I'm working through it mentions that the eigenvalue decomposition $A=Q\Lambda Q^T$ of a matrix A is unique apart from the signs of the eigenvectors (the columns of Q).

Considering the fact that if $q_i$ is an eigenvalue of A with corresponding eigenvalue $\lambda_i$ then so is any multiple of $u_i$(i.e. any vector parallel to $u_i$
.

Based off these two facts, I'd like to check that the following :deduction" is true:
The eigenvalue decomposition of a matrix is unique up to taking multiples of the eigenvectors.

答:
Yes, but I would say the correct statement is: "The eigendecomposition of a matrix is unique up to taking multiples of eigenvectors."
Eigenvalue decomposition is not so common.</p>


问:
convolution of sums
I have to find an infinite power series $B(x)$ in accordance to an already given infinite power series $A(x)$ so that $B(x)A(x) = 1 $

$A(x)$ is defined as follows:
$$1 + \sum_{n \ge 1 } a_n Z^n$$
when one considers the $0$. term as 1 then
$$a_0Z^0 + \sum_{n \ge 1 } a_n Z^n = \sum_{n \ge 0 } a_n Z^n$$

so far I tried to define $B(x)$ as
$$ B(x) :=  \sum_{n \ge 0 } b_n Z^n$$
with these limitations: $b_0 = 1 ; b_1= -1; b_i = 0$ for every $i > 1$

under the use of the convolution of sums
$$\sum_{n \ge 0}^{} (\sum_{k=0}^{n} a_k b_{n-k})$$
I could cancel many terms, but not the problem is that I also cancel the first and the second $1$ with $-1$

$$ a_0b_0 + (a_0b_1 + a_1b_0) + (a_0b_2 + a_1b_1 + a_2b_0) + ... = 1 + (-1 +a_1) + (0 + -a_1+a_2) + ...$$

as you can see every last term is being canceled with the term before the last one in the next brackets but the two first cancel each other as well.
How can one do it in this method so that it will fit nevertheless?

Thank you for your help

答:
We can conveniently use the convolution formula to calculate the coefficients $b_n, n\geq 0$. We have the following situation
\begin{align*}
B(x)A(x)&=\left(\sum_{k=0}^\infty a_k x^k\right)\left(\sum_{j=0}^\infty b_j x^j\right)\\
&=\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n a_kb_{n-k}}\right)x^n=1
\end{align*}

Using the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series we have
\begin{align*}
[x^0]B(x)A(x)&=a_0b_0=1\tag{1}\\
[x^n]B(x)A(x)&=\sum_{k=0}^na_kb_{n-k}=0\qquad n\geq 1\tag{2}
\end{align*}

From (1) we derive with $a_0=1$:
\begin{align*}
[x^0]B(x)A(x)&=a_0b_0=\color{blue}{b_0=1}
\end{align*}
From (2) we derive
\begin{align*}
[x^n]B(x)A(x)&=\sum_{k=0}^n a_kb_{n-k}=a_0b_n+\sum_{k=1}^{n} a_kb_{n-k}=0\tag{3}
\end{align*}
With $a_0=1$ we obtain from (3) for $n\geq 1$

\begin{align*}
\color{blue}{b_n}&\color{blue}{=-\sum_{k=1}^n a_kb_{n-k}}\\
\\
\color{blue}{b_1}&=-\sum_{k=1}^1 a_kb_{1-k}=-a_1b_0=\color{blue}{-a_1}\\
\color{blue}{b_2}&=-\sum_{k=1}^2 a_kb_{2-k}=-a_1b_1-a_2b_0\color{blue}{=a_1^2-a_2}\\
&\cdots
\end{align*}

Title: convolution of sums


答: Let $F(x)=\int_a^x f(x)dx-\frac{1}{2}\int_a^b f(x)dx$.  Since $F(b)=-F(a)$, and $F(x)$ is continuous, there is a point $x$ in the interval where $F(x)=0$, which is the desired point.

Title: Prove that for any Riemann integrable $f$ on the interval $[a, b]$, there is a $c$ such that $\int_{a}^{c} f = \int_{c}^{b} f$

Body:
This is one of the practice problems I am having a bit of trouble trying to figure out and prove.

Prove that for any Riemann integrable function $f$ on the interval $[a, b]$, there is a $c$ such that $\int_{a}^{c} f= \int_{c}^{b} f$.

I was thinking about like if $[a,b]$ is the entire interval, then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$. I am not sure how to proceed from here. Some help would be nice.


问:
How to solve recurrence relation $a_k = ba_{k−1} + cr^k$, assuming $b \neq r$
Solve for $a_k$ in terms of $a_0$ and the other parameters in the following recurrence relation:

$a_k = ba_{k−1} + cr^k$, assuming $b \neq r$.

答:
Welcome!!
Generating function is the answer. Take $f(x)=\sum a_kx^k$, multiply by $x^k$ each sides and sum over $k$..
$$
\sum a_kx^k=bx\sum a_{k-1}x^{k-1}+c\sum r^kx^k
$$
led to
$$
f(x)=bxf(x)+\frac{c}{1-rx}
$$
or
$$
f(x)=\frac{c}{(1-bx)(1-rx)}.
$$
Expand by partial fraction and you can find your $a_k$, or you can use the fact that
$$
f(x)=c\sum b^kx^k\sum r^kx^k
$$
can be written as
$$
f(x)=c\sum_{n\geq0}\sum_{0\leq k\leq n}b^kr^{n-k}x^k
$$
and so
$$
f(x)=\sum_nc\frac{b^{n+1}-r^{n+1}}{b-r}x^n.
$$


答:
$\det(A) = \begin{vmatrix} n & 1 & \ldots & \ldots & \ldots & 1 \\
                           n-1 & n & \ddots & &  & \vdots \\
                             n-2 &n-1 & n  &\ddots &  & \vdots \\
                            \vdots & \ddots & \ddots & \ddots & \ddots &\vdots \\
                               2 & \ldots & \ddots & n-1 & n & 1 \\
                               1 & 2 & \ldots & n-2 & n-1 & n
\end{vmatrix}$

For $i=n,n-1,\ldots,2$, we replace $L_i$ by $L_i-L_{i-1}$ and we obtain:
$\det(A) = \begin{vmatrix} n & 1 & \ldots & \ldots & \ldots & 1 \\
                           -1 & n-1 & 0 &\ldots  & \ldots & 0 \\
                             \vdots &\ddots  & \ddots &\ddots &  & \vdots \\
                            \vdots &  & \ddots & \ddots & \ddots &\vdots \\
                             \vdots &  & &\ddots & \ddots & 0 \\
                               -1 & \ldots & \ldots & \ldots & -1 & n-1
\end{vmatrix}$

For $i=1,2,\ldots,n-1$, we replace $C_i$ by $C_i-C_{i+1}$ and :

$\det(A) = \begin{vmatrix} n-1 & 0 & \ldots & \ldots & 0 & 1 \\
                           -n & n-1 & 0 &\ldots  & \ldots & 0 \\
                            0 &\ddots  & \ddots &\ddots &  & \vdots \\
                            \vdots & \ddots & \ddots & \ddots & \ddots &\vdots \\
                             \vdots &  &\ddots &\ddots & \ddots & 0 \\
                               0 & \ldots & \ldots & 0 & -n & n-1
\end{vmatrix}$

And, if we develop by the first row, we have:   $\det(A)=(n-1)^n+n^{n-1}$

Title: Given this matrix $A$, prove analytically that $\det(A) = n^{n-1}+(n-1)^n$

Body:
A friend of mine asked me about the determinant of the following matrix.

$$A=\begin{pmatrix}
n   & 1   &  1 & \dots  & 1 & 1\\
n-1 & n   &  1 & \dots & 1 & 1\\
n-2 & n-1 &  n & \dots & 1 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
 2& 3 & 4 & \dots & n & 1\\
 1& 2 & 3 & \dots & n-1 & n
\end{pmatrix}$$

So I wrote the following Mathematica code:

```<code>function[n_Integer] :=
Det[Table[
  Which[i == j, n, i < j, 1, True, j - i + n], {i, 1, n}, {j, 1, n}]]
Table[function[i], {i, 2, 20}]
</code>```

which prints this result

$$3, 17, 145, 1649, 23401,...$$

and with the help of OEIS I found that the answer is

$$\det(A) = n^{n-1}+(n-1)^n$$

is there an analytical way of calculating this determinant?


问:
How to calculate $\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}}$ when $x>1$?
Numerically, it looks that the limit is

$$\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} = \frac{1}{\sqrt{2}}\log(1 - x) + cte $$,

but I have not been able to demonstrate it analytically. Does anyone have a idea on how to deal with this limit?

答:
I have found how to handle with this problem. The key point is to use the following integral representation of $\frac{1}{\sqrt{x-y}}$:

$\frac{1}{\sqrt{x-y}} = \frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{da}{\sqrt{\pi}} e^{-a(x-y)}$.

Using Mathematica, we can carry out the integral over $y$ and $a$ (in this order), obtaining a complex result in terms of generalized hypergeometric functions

$\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} =
\frac{8 \sqrt{\frac{2}{\pi }} \left(2 x \Gamma \left(\frac{5}{4}\right) \Gamma \left(\frac{9}{4}\right) \, _3F_2\left(\frac{1}{4},\frac{3}{4},\frac{5}{4};\frac{1}{2},\frac{7}{4};\frac{1}{x^2}\right)+\Gamma \left(\frac{7}{4}\right)^2 \, _3F_2\left(\frac{3}{4},\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{9}{4};\frac{1}{x^2}\right)\right)}{15 x^{3/2}}.$

After that, the limit is easily obtained by Mathematica as

$\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} = \frac{1}{\sqrt{2}}\log(1-x) + \frac{75 \log (2)-4 \left(5 \, _3F_2\left(\frac{1}{4},1,\frac{3}{2};\frac{5}{4},\frac{7}{4};1\right)+\, _3F_2\left(\frac{3}{4},1,\frac{3}{2};\frac{7}{4},\frac{9}{4};1\right)\right)}{15 \sqrt{2}}$,

which is the expected result.

Thanks by your help.


问:
Why does $\chi_{(n,n+1)}$ converge pointwise to the zero function?
Let $g_n=\chi_{(n,n+1)}$ where $\chi$ is the characteristic function of the interval in the subscript. It is given as an example in 4.3 of Real Analysis by Royden and Fitzpatrick that $g_n$ converges pointwise to $g=0$. I don't really understand why though. Is it simply using the below logic? If so, can someone expound on why $\chi_{(\infty,\infty)}=0$. Thanks!

$$ \lim_{n\rightarrow\infty} g_n(x) = \lim_{n\rightarrow\infty} \chi_{(n,n+1)} = \chi_{(\infty,\infty)}=0  $$

答: Take $x\in\Bbb R$; you want to prove $\lim_{n\to\infty}\chi_{(n,n+1)}(x)=0$. But if $n>x$, then $\chi_{(n,n+1)}(x)=0$. So, the numbers of the sequence $\bigl(\chi_{(n,n+1)}(x)\bigr)_{n\in\Bbb N}$ are all $0$ if $n$ is large enough, and therefore $\lim_{n\to\infty}\chi_{(n,n+1)}(x)=0$.


问:
What is this animation about?
I saw this animation on a math telegram channel and I am curios to know what rule  of mathematic showed. We have a triangle at the end look like an equilateral triangle.

https://videoshome.com.au/Home/Play/5fcd58e63d68540841b3a998

答: It shows how to construct an equilateral triangle with a centre at a given point. This is different to the construction of an equilateral triangle given one side, which is a much simpler task.</p>

Title: What is this animation about?


问:
Automorphism of Riemann Sphere without Picard's Theorem
When I am trying to get familiar with the contributions that Gauss made to the quaternions, the following question comes to my mind:

Question: Is there a known proof that every 1-1 comformal mapping of Riemann Sphere to itself is a Mobius Transformation, without using Picard's Theorem about essential singularities?

答:
If $f$ is holomorphic $\Bbb{P^1(C)\to P^1(C)}$ then $f$ is meromorphic $ \Bbb{C\to C}$ and it has finitely many poles, there is a rational function such that $h=f-g$ has no pole.

$h$ is holomorphic $\Bbb{C\to C}$ and $\Bbb{P^1(C)\to P^1(C)}$, if $h(\infty)=\infty$ then  $1/h(1/z)$ is holomorphic and has a zero at of order $n>1$ at $0$ so $1/h(1/z)\sim c z^n$ as $z\to 0$,
$h(z)\sim c^{-1} z^n$ as $z\to \infty$, repeating with $h(z)-c^{-1} z^n$ we'll eventually obtain a polynomial such that $h-p$ is bounded.

So we obtained a polynomial such that $h-p$ is holomorphic $\Bbb{P^1(C)\to P^1(C)}$ and $h-p$ is $\Bbb{C}$-valued and bounded. Since $\Bbb{P^1(C)}$ is compact then $|(h-p)(s)|$ attains its maximum at some $s\in \Bbb{P^1(C)}$ which implies (maximum modulus principle) that $h-p$ is constant.

Whence $f$ was a rational function $u/v$ with $u,v$ coprime polynomials. $f$ is bijective iff $\deg(u)\le 1,\deg(v)\le 1,\deg(u)+\deg(v)\ne 0$ ie. $f$ is a Möbius transformation.


问:
How to solve the heat equation for functional boundary conditions and the step function as the initial condition?
We are trying to solve the one-dimensional heat equation:

$\frac{\partial u}{\partial t} = k \frac{\partial^{2}u }{\partial x^{2}}$.

The initial condition is:
$u(x,0)= u_{1} $ when $x < 1,$ and $0$ when $x \ge 1$.

The boundary conditions are as follows:

$u(0,t) = \frac{1}{t+a^{-1}}+a$

$u(L,t) = \frac{-1}{t+a^{-1}}+a$

$a$ here is a constant.

We are trying to figure out how to solve this equation. Should we go the Fourier series route to solve this(https://www.youtube.com/watch?v=ToIXSwZ1pJU) or should we do separation of variables?

答:
When solving the homogeneous problem on a finite domain with time dependent boundary conditions you want to get the equilibrium temperature. So if we have

$$\begin{cases} u_{t} = k u_{xx}  &  0 \leq x \leq L , t>0 \\ u(x,0) = f(x)  , & 0 \leq x \leq L \\ u(0,t) = A(t) , u(L,t) = B(t) & t > 0 \end{cases}$$

In this case we need a candidate solution for the equilibrium temperature so we create a function called the reference temperature, $r(x,t)$ and we want the function to satisfy.

$$ r(0,t) = A(t) \\ r(L,t) = B(t) $$

we can create one like so

$$ r(x,t) = A(t) + \frac{x}{L}\big[B(t) - A(t) \big]$$

Now, instead we have

$$ v(x,t) = u(x,t) - r(x,t) $$

and this solution $v(x,t)$ will solve the homogeneous problem.

Ok, now that it has been set up we'll let

$$ r(x,t) = A(t) + \frac{x}{L}\big[B(t) - A(t) \big]$$

and now we replace $A(t) = \frac{1}{t+a^{-1}} + a$ and $B(t) = \frac{-1}{t+a^{-1}} + a $

$$ r(x,t) = \frac{1}{t+a^{-1}} + a + \frac{x}{L} \big [\frac{-2}{t+a^{-1}} \big]$$

Now, instead we have this homogeneous problem you should be able to solve

$$\begin{cases} v_{t} = k v_{xx}  &  0 \leq x \leq L , t>0 \\ v(x,0) = f(x) - r(x)  , & 0 \leq x \leq L \\ v(0,t) = 0 , v(L,t) = 0 & t > 0 \end{cases}$$

The solution to this is

$$ v(x,t) = \sum_{i=1}^{\infty} a_{n} \sin(\frac{n \pi x}{L}) e^{-k t(\frac{n \pi}{L})^{2} }$$

you solve for the coefficients $a_{n}$ and we get

$$ a_{n} = \frac{2}{L} \int_{0}^{L} \big[ f(x) - r(x) \big] \sin(\frac{n \pi x}{L}) dx $$

where $f(x)$ is your function

$$ f(x) = \begin{cases}  u_{1}  &  x < 1 \\  0  & x \geq 1 \end{cases}$$

which looks like the complement of a Heaviside function multiplied by the constant $u_1$. Try that and see what happens.

This about the reference temperature function $r(x,t)$. We need a function such that it obeys the boundary conditions.

If $u(x,t) = v(x,t) + r(x,t)$, we have already said that $v(x,t)$ is the homogeneous solution. So $u(0,t) = v(0,t) + r(0,t)$ and $u(L,t) = v(L,t) + r(L,t)$. But $v(0,t) = 0$ and $v(L,t) = 0$.

$ x = 0$ we get

$$ r(0,t) = A(t) + \frac{0}{L}[B(t) - A(t)]  = A(t)$$

and at $x=L$ we have

$$ r(L,t) = A(t) + \frac{L}{L}[B(t) -A(t)] = B(t) + A(t) - A(t) = B(t) $$

which is what we wanted.


问:
Irreducibility of $x^n-x-1$ over $\mathbb Q$
I want to prove that

$p(x):=x^n-x-1 \in \mathbb Q[x]$ for $n\ge 2$ is irreducible.

My attempt.

GCD of coefficients is $1$, $\mathbb Q$ is the field of fractions of $\mathbb Z$, and $\mathbb Z$ is UFD. Hence, $p(x)$ is irreducible over $\mathbb Q$ iff it's irreducible over $\mathbb Z$ (Gauss's lemma).

Let $m\in \mathbb Z$ such that $\varphi(m)=n$ (Euler's totient). Make reduction of $p(x)$ by modulo $m$. Because of $\overline{x^n}=\overline{x^{\varphi(m)}}=\overline{1}$, we get $\overline{p(x)}=\overline{1-x-1}=\overline{-x}$, which is irreducible. Hence, $p(x)$ is irreducible.

Does this proof is correct?

UPDATE. Thanks to Calvin Lin. My mistake is: not for all $n$ we can find such $m$. OK, but as for the rest, does my proof is correct for such $n$, that $n=\varphi(m)$ for an integer $m$? And can it be some changed for all $n$, i.e. can we find such modulo that $\overline{p(x)}$ is irreducible for every $n\ge 2$?

答:
I doubt an easy proof of the irreducibility exists in general. If $n$ is a prime, then the polynomial is Artin-Schreier and handled easily.

Selmer gave a clever proof in the general case, working explicitly with the roots of the polynomial in $\mathbb{C}$. See E. S. Selmer, On the
irreducibility of certain trinomials, Math. Scand. 4 (1956), 287-302, available here.


问:
Convergence in measure of characteristic functions
I was having trouble starting this problem.  I would appreciate some help.  Thanks in advance.

Let $E_1, E_2, \ldots$ be measurable sets.  Suppose that the functions $f_j = 1_{E_j}$ converge in measure to a limit function $f$.  Show that $f$ is $a.e.$ equal to $1_E$ for some measurable set $E$.

答:
A similar approach but I think phrased much more simply (using an important concept which is often used): Pick a subsequence $E_{n}$ (I do not re-label it) s.t. $\chi_{E_{n}} \to f$ pointwise a.e. This implies that $f = \limsup_{n \to \infty} \chi_{E_{n}}$. Recall that for measurable sets $\{E_{n}\}$
$$
E:= \limsup_{n \to \infty} E_{n} := \bigcap_{N \in \mathbb{N}} \bigcup_{n \geq N} E_{n}
$$
It is straightforward to check that $\limsup_{n \to \infty} \chi_{E_{n}} = \chi_{E}$, so $f = \chi_{E}$ a.e.


问:
No. of 7 bit strings that are equidistant
I need the total number of 7-bit strings such that any pair disagree with each other on exactly 4 bits.</p>

答:
Let $X = \{0,1\}^7$ be the collection of $7$-bit strings.

For $x\in X$, let $|x|$ be the number of set bits in $x$.

For $x, y \in X$, let $x \oplus y$ be the string obtained by 'XOR' the bits.

It is easy to see $d(x,y) \stackrel{def}{=} | x \oplus y|$ is the distance between $x$ and $y$.

Let $x_1, \ldots x_n \in X$ be any $n$ bit strings with $d(x_i,x_j) = 4$.
Notice
$$d(x_i \oplus x_1, x_j \oplus x_1 ) = d(x_i,x_j) = 4$$
and $x_1 \oplus x_1 = 0 \stackrel{def}{=} (0000000)$, the bit string with all zero. In order to maximize $n$, we only need to consider the case $x_1 = 0$.
Under this assumption, $|x_i| = 4$ for all $i > 1$. ie. all the other $x_i$ has exactly 4 bits set. Permuting the position of bits if necessary, we can assume $x_2 = (0001111)$.

For other $x_i = (b_1b_2b_3c_1c_2c_3c_4)$. In order for $d(x_i,x_1) = d(x_i,x_2) = 4$, we need

two of $b_1,b_2,b_3$ are set.

two of $c_1,c_2,c_3,c_4$ are set.

The first constraint tell us there are 3 possible choices of $(b_1b_2b_3)$.
$$(b_1b_2b_3) = (011), (101)\;\text{or}\;(110)$$

For any choice of $(b_1b_2b_3)$, in order for
$d((b_1b_2b_3c_1c_2c_3c_4),(b_1b_2b_3c'_1c'_2c'_3c'_4)) = 4$,
we need $(c_1c_2c_3c_4)$ to be complement to $(c'_1c'_2c'_3c'_4)$.
There are at most two candidates of $x_i$ for any given $(b_1b_2b_3)$. As a result, $n$ is at most $2 + 3\cdot 2 = 8$.

$n = 8$ is actually possible. As an example,
$$(x_1,\ldots,x_8)^T = \left(\begin{smallmatrix}
(0000000)\\
(0001111)\\
(0110011)\\
(0111100)\\
(1010101)\\
(1011010)\\
(1100110)\\
(1101001)\\
\end{smallmatrix}\right)$$

This means $8$ is the largest size for collections of $7$ bit strings at mutual distance $4$.


问:
Resolving $16^x+4^{x+1}-3=0$ without completing the square?
I managed to solve this for $x$ by completing the square and using the properties of logarithms:

$$16^x+4^{x+1}-3=0$$

However, I was wondering if it's possible to resolve this without completing the square and simply using the properties of logarithms?

Thank you

答: You can note that $16^x=(4^2)^x=(4^x)^2$ and $4^{x+1}=4\cdot4^x$, so the equation becomes $$(4^x)^2+4\cdot4^x-3=0.$$ Let $t=4^x$. Then we have $t^2+4t-3=0$ where its solutions are $t_1=\sqrt{7}-2$, $t_2=-2-\sqrt{7}$. Since $t_2<0$ we don't have solution there; but for $t_1$ yes: $$4^x=\sqrt{7}-2\implies\boxed{x=\log_4(\sqrt{7}-2)}.$$


问:
Upper Triangular Matrix with zeros up
I have a linear algebra homework, but honestly I'vee been trying to solve this problem for some days without success:

Suppose T $\in$ $M_{5 \times 5}$ (K) is a triangulable matrix. Be $\beta = \{ \vec{v_1}, \vec{v_2}, \vec{v_3}, \vec{v_4}, \vec{v_5} \}$ a base of $K^5$ such that:

$$[T]_\beta = \begin{pmatrix}
4 & * & a & b & c \\
0 & 4 & d & e & f \\
0 & 0 & 5 & * & * \\
0 & 0 & 0 & 5 & * \\
0 & 0 & 0 & 0 & 5
\end{pmatrix}$$

And I must prove that exists a base $\beta'$ such that:

$$[T]_{\beta'} = \begin{pmatrix}
4 & * & 0 & 0 & 0 \\
0 & 4 & 0 & 0 & 0 \\
0 & 0 & 5 & * & * \\
0 & 0 & 0 & 5 & * \\
0 & 0 & 0 & 0 & 5
\end{pmatrix}$$

How can I prove it?

Thoughts: If a=b=c=d=e=f=0 means that there is nothing in the coordinates corresponding to the vectors related to 4, but I don't know how to prove that there is a basis with vectors that satisfy this.

答:
Hint:

Both matrices are Upper Triangular.

UT matrices are closed under addition and multiplication.

The matrix $R$ establishing the similarity $T'= R T R^{-1}$ will be UT as well.

You can partition $T, T', R$ in blocks anf for $R^{-1}$ you can apply Block Inversion

Title: Upper Triangular Matrix with zeros up


问:
f is differentiable with conditions. Show that f has a subsequence converging to a continuous function.
Let $f: \mathbb R \to \mathbb R$ be differentiable. Assume that $|f(x)| \leq 1, |f'(x)| \leq 1 \ \forall x \in \mathbb R$ and that $f(0) = 0$. Let $\{a_n\}$ be a sequence of non-zero real numbers. Prove that the sequence of functions $g_n(x) = \frac{1}{a_n} f(a_n x)$ has a subsequence converging to a continuous function.

I know that the subsequence exists because we have a bounded function. However, we are given a specific subsequence here. How do I show this is a subsequence that converges? Is there a good way to approach this? Or maybe showing that $g_n(x)$ is equicontinuous and then using Arzela-Ascoli? Any tips to do this one would be great!

答:
There are three cases.  The first is that there is a subsequence of $a_n$ that converges to a non-zero number.  The second is that there is a subsequence that converges to $-\infty$ or $\infty$.  The third is that there is a subsequence that converges to 0.

In each of the three cases, prove that the corresponding subsequence of functions converges to a continuous function.  (The three cases do have to be handled differently from each other.)

Here is the trick for the third.  By the definition of the derivative and the fact that $f(0) = 0$,

$\lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = f'(0)$

Which means for any $\epsilon > 0$ there exists $\delta > 0$ such that if $0 < |h| < \delta$ then $|\frac{f(h)}{h} - f'(0)| < \epsilon$.

Now let non-zero $x$ be fixed, $h = ax$ and suppose that there is a subsequence $a_{n_k}$ that converges to 0.  Then $\frac{f(ax)}{a} = \frac{f(ax)}{ax}x$.

And now for any $\epsilon > 0$ we can pick $\delta > 0$ such that if $|ax| < \delta$ then $|\frac{f(ax)}{ax} - f'(0)| < |\frac{\epsilon}{x}$ so that $|\frac{f(ax)}{a} - f'(0)x| < \epsilon$.  But since $a_{n_k}$ converges to $0$, for all sufficiently large $k$, $|a_{n_k}| < \delta$.  This lets us prove, directly from the definitions, that the subsequence converges pointwise to $f'(0) x$.

Note that this is actually a stronger result than you were asked for.  I only need $f$ bounded, continuous, and that $f'(0)$ is well-defined.


问:
Min and Max for $a\in R^n$
Let $a, x\in R^n$ with $a \ne 0$.

Show that

a.) $1/\|a^2\| =  \min\{\|x\|^2 :\langle a,x\rangle = 1\}$

b. ) $\|a\| = \max\{\langle a,x\rangle : \|x\|= 1\}$

I don't have any idea on how to start on this problem.

As $\langle a,x\rangle = 1$ means either $a$ and $x$ are pointing in the same direction and are unit. Or if they are both pointing in same direction, and if are not unit vectors then the length of one vector is reciprocal of other.

For $||x|| = 1$ and $ = ||a||. ||x_/||$, where $x_/$ is the component of $x$ parallel to $a$ . Not sure how to proceed for part b either.

What would be the solution?

答:
$\langle a,x\rangle=1$ geometrically means that $\|a\|\cdot\|x_\parallel\|=1$ where $x=x_\perp+x_\parallel$ with $x_\parallel\parallel a$ and $x_\perp\perp a$, and $x_\parallel$ points in the same direction as $a$ (and not the opposite).

The set of such points $x$ is the hyperplane orthogonal to $a$  which contains $b:=\dfrac a{\|a\|^2}$.

Prove that the minimum of exercise a) is taken at $x=b$, and the maximum of exercise b) is taken at $x=\dfrac a{\|a\|}$.

Title: Min and Max for $a\in R^n$


问:
Solving a Secret List (Inspired by 'Are you the One?')
Let's say there's a list of the numbers $1$ through $N$ in some sequence, unknown to us. This list is the "correct" order. You are able to arrange those numbers in any way, and after each step you are told how many of the items on your list are in the correct position, but not which ones. The goal is to guess the order of the secret list in as few steps as possible.
The 2 questions I've been thinking of are these:

Is there a strategy that guarantees you find the correct order in $M<N!$ steps?

Given a maximum number of guesses $G<M$, is there a strategy that maximizes your probability of finding the right order in $\leq G$ steps?

So far I haven't been able to make much headway. I though I could do something by keeping a fixed order and rotating it $N-2$ times, which would put every item in all but one spot once without double counting, and the $N-1$st could be deduced given the number of correct spaces would have to add up to $N$.

I don't think it gives useful information though. I imagine in order to find out which positions are correct you'd have to have that spot repeated at least once. Thoughts?

答:
We can find the correct listing in $O(N^2)$ moves. For any integer $i=1,\ldots,N$ let $b_i$ denote the integer that is supposed to be in the $i$-th position. Let $c$ denote the number of $i$ such that $a_i=b_i$.

Claim 1: Given 3 distinct integers $i_1,i_2,i_3$ we can determine in $O(1)$ queries precisely which if any of $i_1,i_2,i_3$ is $b_{i_1}$, which
if any of $i_1,i_2, i_3$ is $b_{i_2}$, and if any of $i_1, i_2, i_3$ is $b_{i_3}$.

Let us now present our algorithm. We call an integer $i=1,2,\ldots, N$ certain if $b_i$ has been found. We call an integer $i$ uncertain if it is not certain. Let us write the set of integers that are uncertain as $U$. Initially $U=\{1,2,\ldots, N\}$. We now describe our algorithm.

Let $i_1$ be any integer in $U$. We make $i_1$ certain in $O(N)$ moves as follows:

Finding $b_{i_1}$ in $O(N)$ moves.

For each $i \in U$ there may be an intermediate integer $a_i$ where $a_i$ may or may not be $i$.

$U' \leftarrow U \setminus \{i_1\}$.

WHILE $U'$ is nonempty and $i_1$ is still uncertain:

Let $i_2$ be another integer in $U'$, and remove from the set $U'$ the integer $i_2$.

Then check how $c$ changes when $a_{i_1}$ swaps positions with $a_{i_2}$.

If $c$ changes by 2 either way then either $a_{i_1}=b_{i_1}$ and $a_{i_2}=b_{i_2}$, or $a_{i_1}=b_{i_2}$ and $a_{i_2}=b_{i_1}$; either way $b_i$ was found for both $i=i_1$ and $i=i_2$ so remove both $i_1$ and $i_2$ from $U$.

If $c$ changes by 1 either way then $|U| \ge 3$ and one of $a_{i_1},a_{i_2}$ is $b_{i_1},b_{i_2}$. Let $i_3$ be a 3rd integer in $U$ and run Claim 1 to find
precisely which if any of $a_{i_1},a_{i_2},a_{i_3}$ is $b_{i_1}$, which
if any of $a_{i_1},a_{i_2},a_{i_3}$ is $b_{i_2}$, and if any of $a_{i_1},a_{i_2},a_{i_3}$ is $b_{i_3}$. For each $i \in \{i_1,i_2,i_3\}$
such that $b_i$ has not been found, reset $a_i$ to a remaining value
of what was before $a_{i_1},a_{i_2},a_{i_3}$.

Note that 2. take $O(|U|)$ moves. As there is an $i \in U$ such that $b_{i_1}=a_i$, the integer $i_1$ will become certain by the end of the steps 2--4 corresponding to $i$ being removed from $U'$ in step 1.
If $i=i_1$ then swapping $a_{i_1}$ with $a_{i_2}$ where $i_2$ is the next smallest integer in $U$, will change $c$ by -1 or -2; if $i \not = i_1$
then $c$ will change by 1 or 2 when $a_{i_1}$ is swapped with $a_i$.


问:
A function is smooth at a point and not smooth in any neighbourhood of it, exist or not?
Suppose that a function $f$ defined in an open set $U \subseteq \mathbb{R}^m$ is smooth at a point $p \in U$. Then we have that there exists an open set $U_n \subseteq U$ $($ say $U_{n+1} \subseteq U_{n} \subseteq U$ $)$ contains $p$ such that $f$ is $n$- derivable in $U_n$, that is, all $n$th-order partial derivatives of $f$ exist in $U_n$. Hence, the smooth domain of $f$ is $$V\overset {\text{def}}{=} \bigcap^{\infty}_{n=1}U_n ~.$$

My question:

Does there exist a function $f$ defined in an open set $U$ such that it is smooth at a point $p \in U$ and not smooth in any nonempty open subset of $U$ ? To put it another way, if we do not choose those $U_n$ too small artificially, then, is $V^{\circ}$ always an nonempty open set ?

答:
For continuous $f$ on $U:=(-1,1)$, let $I(f)$ be the function $x\mapsto \int_0^xf(t)\,\mathrm dt$.

Let $$f_n=I^n(x\mapsto \max\{|x|-\tfrac1n,0\}).$$
Note that this makes $|f_n|<\frac1{n!}$ on $U$. Then $f:=\sum_n f_n$ converges uniformly. We see that $f^{(n)}$ exists on $(\frac1n,\frac1n)$, but $f^{(n+1)}(\pm\frac1n)$ does not. So all derivatives of $f$ exist (and are zero) at $p=0$, but $f$ is not smooth in any neighbourhood of $p$. This is only halfway an answer to your question though (as $V$ does have a larger open subset).

So we repeat the above construction with a twist:
Let
$$ g_n=I^n(x\mapsto \tfrac1n\lfloor n|x|\rfloor )$$
Again, we have uniform convergence and $g^{(n)}$ exists on $(-\frac1n,\frac1n)$, but now $g^{(n+1)}(\tfrac kn)$ does not exist for $-n<k<n$, $k\ne 0$.  This time, we find that $V\cap \Bbb Q=\{0\}$, so $V$ has empty interior, as desired.


问:
Be $(u_n)$ and $(v_n)$ two sequences of real numbers converging to zero.

Be $(u_n)$ and $(v_n)$ two successions of real numbers converging to zero.  Let's suppose
that there is a number $M > 0$ such that: $\forall n \in \mathbb{N}, |v_{0}| + |v_1| + \ldots |v_n| \leq M$.

Show that: $\lim_{n -> \infty} u_{0}v_{n} + u_{1}v_{n-1}+ \ldots + u_{n}v_{0} = 0$

This is what I tried,
Prove:
We are gonna prove that $|u_{0}v_{n} + u_{1}v_{n-1}+ \ldots + u_{n}v_{0} - 0|< \epsilon $
so:
$|u_{0}v_{n} + u_{1}v_{n-1}+ \ldots + u_{n}v_{0}| \leq \sum_{k=0}^n|u_{k}v_{n-k}|$

but: $ \sum_{k=0}^n|u_{k}v_{n-k}| =  \sum_{k=0}^{n_{0}} |u_{k}||v_{n-k}| + \sum_{n_{0}+1}^n|u_{k}||v_{n-k}| $

then i can take:

$|v_{n}| + |v_{n-1}| + \ldots |v_{n-n_{0}}| + |v_{n-(n_{0}+1)}|+ \ldots + |v_{0}|\leq M$
therefore:

$|v_{n}| + |v_{n-1}| + \ldots |v_{n-n_{0}}| = \sum_{k=0}^{n_{0}}|v_{n-k}|\ \leq M $

on the other side, taking $n = n - (n_{0} +1)$:

$|v_{0}| + |v_1| + \ldots |v_{n -(n_{0}+1)}| = \sum_{k = n_{0}+1}^{n}|v_{n-k}| \leq M$

so, replacing:

$ \sum_{k=0}^n|u_{k}v_{n-k}| =  \sum_{k=0}^{n_{0}} |u_{k}||v_{n-k}| + \sum_{n_{0}+1}^n|u_{k}||v_{n-k}| \leq M( \sum_{k=0}^{n_{0}} |u_{k}| +  \sum_{n_{0}+1}^n|u_{k}|)$

and I'm stuck here; i thought, if the serie of $u_{k}$ were absl converged then i could put $\sum_{k=0}^{\infty}  |u_{k}|< \epsilon / 2M$. But i couldn't find the convergence.

Maybe this is not the way, can someone tell me a hint of what can i do to solve this problem pls.

答: Choose $N$ such that $|u_i| <\epsilon / M$ for $i \geq N$. $|\sum\limits_{i=0}^{n} u_iv_{n-i}|\leq |\sum\limits_{i=0}^{N} u_iv_{n-i}|+|\sum\limits_{i=N+1}^{n} u_iv_{n-i}|$ for $n >N$. Note that the second term does not exceed $\epsilon$ since $|u_i| <\epsilon / M$ for $n \geq N$. The forst term tends to $0$ as $ n \to \infty$. Hence $\lim \sup |\sum\limits_{i=0}^{n} u_iv_{n-i}| \leq 0+\epsilon =\epsilon$ whenever $n >N$.


问:
Multivariate gaussian integral with quadratic term outside exponent
I have searched around, but I have not found a solution to Gaussian integrals of the following form:
$$\int (x^\intercal R x)\exp\left[-\frac{1}{2} \left(x^\intercal Ax\right) + Bx\right]d^n x$$
Here, $B$ is an $n$-dimensional vector, $R$ and $A$ are $n\times n$ matrices, and we integrate over the space of $n$-dimensional vectors $x$ with real values.  Explicitly, $x^\intercal R x = \sum_{i,j=1}^n x_i R_{ij} x_j$ and $B x = \sum_{i=1}^n B_i x_i$.

Presumably, $A$ must be a positive-definite matrix to assure convergence.  Other than that, I think the elements of $R$, $A$, and $B$ can be arbitrary complex numbers.  However, I would be interested in evaluating this integral for restrictions, e.g., if $A$ is also real-valued (and therefore symmetric).

Wikipedia evaluates the special cases of $R=0$ and $n=1$.

(Note that the above integral is not addressed by similarly-worded question 187923, which uses "quadratic term" to refer to the term with $A$ in the exponent rather than $R$ outside the exponent.)

答:
Whoops, I missed the standard trick that is in fact listed on Wikipedia here.  For matrices with real elements,

$$\begin{align}
\int &(x^\intercal R x)\exp\left[-\frac{1}{2} \left(x^\intercal Ax\right) + Bx\right]d^n x \\ &= \sum_{ij} R_{ji} \partial_{B_i}\partial_{B_j}\int \exp\left[-\frac{1}{2} \left(x^\intercal Ax\right) + Bx\right]d^n x\\
&=\sum_{ij} R_{ji} \partial_{B_i}\partial_{B_j}\sqrt{\frac{(2\pi)^n}{|A|}}\exp\left[\frac{1}{2} \left(B A^{-1} B^\intercal\right)\right]\\
&=\sqrt{\frac{(2\pi)^n}{|A|}}\sum_{ij} R_{ji} \left[(A^{-1})_{ij} + (A^{-1}B^\intercal)_{i}(BA^{-1})_{j}\right]\exp\left[\frac{1}{2} \left(B A^{-1} B^\intercal\right)\right]\\
&=\sqrt{\frac{(2\pi)^n}{|A|}}\left[\mathrm{Tr}\left(R A^{-1}\right) + BA^{-1}RA^{-1}B^\intercal\right]\exp\left[\frac{1}{2} \left(B A^{-1} B^\intercal\right)\right]
\end{align}$$


问:
Proving that a function doesn't have a limit given $|a-b| = |f(a)-f(b)|$ for all $a,b\in \mathbb R$
For all $a,b\in \mathbb R$,

$|a-b| = |f(a)-f(b)|$.

Prove that $\;\lim\limits_{a\to \infty}|f(a)|=\infty$

What I was trying to do is show that if we assume that there is a limit, $\;\lim\limits_{a\to \infty}|f(a)|=L\;$ and therefore $\big||f(a)|-L\big|=\big||f(a)|-|L|\big| \leq |f(a)-L| < \epsilon$

Then we set $\epsilon$ to some arbitrary value and try to get a contradiction.

I also tried proving directly but couldn't get much out of it.

Is there a better approach for proving it?

答:
$$\forall a,b, \quad \frac{f(a)-f(b)}{a-b} = 1 \text{ or }-1.$$

Without loss of generality, take this difference quotient to be "1".  A similar argument will hold when this difference quotient is "-1".  Also, it should be clear that it is one or the other, not both.$^*$

$$f^\prime(x) = 1, \quad \text{everywhere}.$$

Take $f(0)= y_0$, some finite number.

Integrating,
$$f(x) = x+y_0$$

$$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} (x+ y_0) = \infty.$$

$^*$Here is a proof.  If, for $a,b,c$ all not equal, $f(a)-f(b) = a-b$ and $f(b)-f(c)=c-b$ then $f(a)-f(c)= a+c-2b$ so that $a+c-2b=a-c$ and $b=c$ or $a+c-2b=c-a$ and thus $a=b$.  Contradiction.  Thus $f(a)-f(b)=a-b$ everywhere or $f(a)-f(b)=b-a$ everywhere.