问:
Find the total surface area..
I'm unsure if I miss any pieces in the calculation for its surface area...

Question:

The diagram shows the uniform cross-section of a solid paper weight ABCDE, which is in the shape of a trapezium with a semi-circular hole ABC cut out of it. It is given that AC is parallel to DE, AC = $7$ cm, CD = $15$ cm, DE = $21$ cm, AE = $13$ cm, and the height of the trapezium is $12$ cm. The given height is $30$ cm.

The area for the cross section should be:

Area of trapezium - Area of semi-circle = $$0.5*(7+21)*(12) - 0.5*\pi*(3.5)^2 = 148.7  \ \text{cm}^2$$

Total Surface Area:

Area of side 15 + area of side 21 + area of side 13 + area of top and area of bottom + area of semi-cylinder =

$$(15*30) + (21*30) + (13*30) + (2*148.7) + \pi*(3.5)^2 + \pi*3.5*30 = 2135.2 \text{cm}^2$$

The answer for the surface area is $2430 \ \text{cm}^2$, but the closest I've got is $2135.2 \ \text{cm}^2$ so what did I miss?

答:
Think of the solid as having front and back, left and right, and top and bottom faces. The surface area is made up of the two cross-sectional areas (front and back), two flat rectangles (left and right), a half-cylindrical face (top), and a flat bottom face (bottom).

Thus the surface area is:

$$2(148.8) + 15 \cdot 30 + 13 \cdot 30 + \frac{1}{2} (2 \pi \cdot 3.5) \cdot 30+21\cdot30=2100 \ \text{cm}^2$$

to $3$ significant figures.

The $\pi \cdot (3.5)^2$ shouldn't be there as that region is just empty space, and everything else is correct.

Title: Find the total surface area..


问:
Proof for the fact that $(a,b)$ is not compact.
Here is the proof for the fact that $(a,b)$ is not compact.

Proof

$\bigg\{ \left(a+\dfrac{b-a}{2^n}, b \right) \Bigg| n \in \mathbb{N} \bigg\}$ is an open covering of $(a,b)$. Let $U_i:=\left(a+\dfrac{b-a}{2^i}, b\right)$.
Suppose we can cover $(a,b)$ with $U_1 \cup U_2 \cup \cdots \cup U_k $. Then, $a+\dfrac{b-a}{2^{k+1}} \in (a,b)$ and $a+\dfrac{b-a}{2^{k+1}} \notin U_1 \cup \cdots \cup U_k.$ This is contradictory.
Thus, $(a,b)$ is not compact.

I understood this proof, but why does this proof use
$\left( a+\dfrac{b-a}{2^n}, b\right)$ ?

I think $\left(a+\dfrac{1}{2^n},b \right)$ also works because $\lim_{n\to \infty} \left(a+\dfrac{1}{2^n} \right)=a.$

答:
The other answers seem to use a confusing definition of an interval. The set $(a,b)$ is $$(a,b) = \{x \in \Bbb R \mid a < x < b\}.$$

When $b \geq a$ it is not that this set is somehow `invalid'. It is just empty.

The set $(a+1/n, b)$ is always a subset of $(a, b)$. It's just that it is the empty subset when $a + 1/n \geq b$. The choice of $\frac{b-a}{2^n}$ instead of $\frac 1n$ is not useful, it just makes the typesetting more painful.

ContentLicense: CC BY-SA 4.0


问:
Number of digits in a number (i.e. log), based solely on the known number of digits in a different base?
I'm in a weird situation, with a cryptography system I'm working on.

I have very large binary integers, non-negative, thousands of bits long.

I cannot compute the actual value of the number (too big), but I DO know exactly how many bits are used.

(i.e. the leftmost non-zero bit)

So if I have an integer $x$ that is, let's say $10000$ bits long, I can safely assume:

$9999\leq\log_2(x)\lt10000$

Using that information alone, I need to determine the maximum possible number of digits required in a different base $b$. (which could be any base, e.g. $10$, $200$, $16$, $64$, $62$, $50$, etc.)

Normally, the simplest formula would be:

$\lceil\log_b(2^n-1)\rceil$ where $n=10000$

As an example with smaller numbers, let's say $n=12$, meaning my huge integer value $x$ is 12 bits long.

That means $11\leq\log_2(x)\lt12$

Using the previous formula $\lceil\log_b(2^n-1)\rceil$ where $n=12$ and $b=10$

results in the value $4$, which is correct, since a 12-bit integer is between $2048$ and $4095$ in base 10.

(4 digits long)

However, obviously this is impractical for end-user devices to compute in real-time, due to the CPU resources required when $n$ gets larger.

Is there a sneaky shortcut method to get (or even approximate) the result, without having to compute the huge value $2^n$?

答:
Clearly $\lceil\log_b2^n\rceil$ is a good upper bound. And $\log_bx=\frac{\log_2x}{\log_2b}$, so

$$\lceil\log_b2^n\rceil=\left\lceil\frac{\log_22^n}{\log_2b}\right\rceil=\left\lceil\frac{n}{\log_2b}\right\rceil$$

gives you a decent bound pretty easily.


问:
Relation between Cayley diagram and Abstract Group action.
So I have been reading a little about group theory and wanted to understand visually. I saw two different ways to visualize groups on many books and lectures.

$1$.Cayley graph:This is one way I found for visualizing groups . now , in a Cayley graph , there are some generators and relations .And by using that , I can make different kinds of groups.Here, they call each of the paths actions .This method is specially useful for visualizing things like Dihedral groups , orbits of a group (maybe homomorphism?) etc.

$2$.Group multiplication table : This one is another way to visualize the same group.It has a bunch of elements which combine with each other in the same manner as a multiplication table.This method is useful (for me at-least) to visualize things like cosets , lagranges theorem , normal subgroup , quotient group etc.

Now , one of the problems is that many books don't use both of the methods to describe the same group.So i get more biased for one method and can't understand anymore the concepts that were meant to be understood with the other method (or I am wrong again?)

In a abstract algebra textbook , I see two concepts :

$1$.Groups

$2$.Group Action

I saw the abstract definitions of groups and group actions (I will not state it here).Because Cayley diagram talk about actions ,I thought It could be related to group action somehow.And I already know that groups and group multiplication table fall hand in hand.

The problem is ,I know from the abstract definition that how is  group actions came from groups from defintions .But I can't really see how cayley diagram(with generators and relations) and the abstract group action is related to each other. And if they are related which one I should use for intuitive studying? Also books juggle around with the definitions of group and group action , so I want to show the intuitive way to relate them with each other?(I mean mathematicians didn't come up with these definition out of thin air , right?)

答:
Let $G $ be a group and $A $ a set. A (left) group action\begin{eqnarray} \star: G \times A &&\to A \\ (g,a) &&\mapsto g \star a \end{eqnarray} is a function from the cartesian product that is compatible with the groups structure ,i.e. $\forall a \in A, g,h \in G$
\begin{eqnarray} e\star a &&=a \\  g \star(h\star a) &&= (gh)\star a\end{eqnarray}
where e is the neutral element.

The prime example would be the generic group action, a group acting on its own underlying set via leftmultiplication with a group element, that is A=G and \begin{eqnarray} \circ:  G \times G &&\to G \\ (g,h)  &&\mapsto g\circ h= gh \end{eqnarray} where we did nothing but relabel the groups inner operation for clarity. So this is obviously a group action because per definition $G$'s operation is associative and $e$ acts trivially on every element.

So how does this lead us to graphs?

A group action on a set A gives us an associated graph $X=(A,E)$ via $E=\{(g \star a,a): g \in G\}$. in case of the generic operation on the group $G$ on itself we optain a graphic representation of the group operation.

So this graph is really just the same thing as the multiplication table, for each element you can see what comes out if you multiply any element from the left.

We may notice that a lot of the edges in this graph are redundant. So we may specify and put special attention to a certain subset $ S \subset G$ and only add edges that represent leftmultiplication with elements from $(S \cup S^{-1}) \backslash \{e\}$, where $S^{-1}=\{s^{-1}: s \in S\}$ (omitting the neutral element since all it does is produce loops in the graph).

Usually one looks at so-called generating sets $S$ (that is there is no proper subgroup of $G$ containing $S$) and thats what i refer to as the Cayley-Graph of $G$ by the generator $S$

Because Cayley diagram talk about actions ,I thought It could be related to group action somehow.

I interpret that as saying that the edges in the cayley graph represent the action of an element of the generating set. And although the "action" of S on G is not a group action in the forementioned sense (since $S$ is just a subset, not a group), it is however the restriction of the generic group action $\circ:  G \times G \to G$ to $(S \cup S^{-1}) \backslash \{e\} \times G$,

The problem is ,I know from the abstract definition that how is group actions came from groups from defintions .But I can't really see how cayley diagram(with generators and relations) and the abstract group action is related to each other. And if they are related which one I should use for intuitive studying?

There are quite a few theorems linking them. For example the cayley graph of a free group is a tree. What do you mean by relations in a Caleygraph?

Well use whatever works for you; for calculations tables seem more practical, for me graphs provide a better overview of the groups structure.


答:
So if you have $2^x=b^y$ and take logs to base $2$ you get $x=y\log_2 b$ where. $x$ is constrained and $b$ is known (or you can take logs to your favourite base $c$).

I find, with these $\log$ problems, that going back to the exponential form helps me.

Title: Number of digits in a number (i.e. log), based solely on the known number of digits in a different base?

Body:
I'm in a weird situation, with a cryptography system I'm working on.

I have very large binary integers, non-negative, thousands of bits long.

I cannot compute the actual value of the number (too big), but I DO know exactly how many bits are used.

(i.e. the leftmost non-zero bit)

So if I have an integer $x$ that is, let's say $10000$ bits long, I can safely assume:

$9999\leq\log_2(x)\lt10000$

Using that information alone, I need to determine the maximum possible number of digits required in a different base $b$. (which could be any base, e.g. $10$, $200$, $16$, $64$, $62$, $50$, etc.)

Normally, the simplest formula would be:

$\lceil\log_b(2^n-1)\rceil$ where $n=10000$

As an example with smaller numbers, let's say $n=12$, meaning my huge integer value $x$ is 12 bits long.

That means $11\leq\log_2(x)\lt12$

Using the previous formula $\lceil\log_b(2^n-1)\rceil$ where $n=12$ and $b=10$

results in the value $4$, which is correct, since a 12-bit integer is between $2048$ and $4095$ in base 10.

(4 digits long)

However, obviously this is impractical for end-user devices to compute in real-time, due to the CPU resources required when $n$ gets larger.

Is there a sneaky shortcut method to get (or even approximate) the result, without having to compute the huge value $2^n$?


问:
Can't understand homework assignment
Let $p\in\mathbb{N}$ be an odd prime. Prove that if $p \equiv 3 \pmod 4$ then $−1$ is not a square modulo $p$.

$\textbf{Hint}$ : recall that $\mathbb{Z}/p\mathbb{Z}$ is a ﬁeld, so that its multiplicative group is a cyclic group
of order $p − 1$. Prove that in any such group the order of $−1$ is $(p − 1)/2$, that the
order of any square is even and deduce that the equation $x
^2 = −1$ cannot be solved
if $p \equiv 3 \pmod 4$.

I can't understand the Hint. What does it mean "in any such group"? I mean, what is $-1$ in a generic cyclic group of order $p-1$? Suppose the group is the multiplicative group of $\mathbb{Z}/p\mathbb{Z}$. Why the order is claimed to be $(p-1)/2$?? Isn't the order of $-1$ always equal to $2$?

答: You're right, this seems to be a different usage of "order" than is standard. You should look at a fixed generator $\omega$ of the group; that is, $(\mathbb{Z}/p\mathbb{Z})^{\times} = \{\omega^n : n \in \mathbb{N}\}$, and call the order of $x$ the minimal $n$ such that $\omega^n = x$ - then the hint will work.</p>

Title: Can't understand homework assignment

ContentLicense: CC BY-SA 3.0


问:
Find points at which tangent intersects curve for variable x and y values
I am not sure if I titled this correctly. I am working through Calculus With Analytic Geometry by Simmons. Question $\# 17$ on page $87$ reads

Find the equation of the tangent to the curve $y = x^3$ at the point $(a, a^3)$. For what values of a does this tangent intersect the curve at another point?

I can calculate the equation of the tangent line to be $y=3a^2x-2a^3$ but I am not sure how to calculate the values of $a$ where the tangent intersects the curve at another point. Even a hint would be great, thank you.

答:
The values of $x$ where the tangent line and $y=x^3$ intersect are $x^3 = 3a^2x-2a^3$ by substituting $y=x^3$ in.

Now $x^3-3a^2x+2a^3 = 0$. However, you already know that the point $(a,a^3)$ lies on the tangent line and on the curve, so $x-a$ is a factor by the factor theorem.

Then use polynomial long division to get $(x^2+ax-2a^2)(x-a) = 0$, where you can now factor the quadratic.


问:
Find in closed form.$\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\sin^2(x)}\right)\cos(x)dx$
Find in closed form .

$\displaystyle\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\sin^2(x)}\right)\cos(x)dx$

My work

We put $t=\cos(x)$ we find $\displaystyle\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\sin^2(x)}\right)\cos(x)dx=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\beta(\frac{1}{2},\frac{2-n}{2})+\frac{3}{4}\eta(1)-\frac{\pi^2}{24}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2n^2+3n}$

But :

$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\beta(\frac{1}{2},\frac{2-n}{2})=?$ and $\sum_{n=1}^{\infty}\frac{1}{2n^2+3n}$

Where $\eta(s)$is eta function

答:
I did not check your calculations.

$$\frac{1}{2n^2+3n}=\frac{1}{3 n}-\frac{2}{3 (2 n+3)}$$

$$S_p=\sum_{n=1}^{p}\frac{1}{2n^2+3n}=\frac{1}{3}H_p-\frac{1}{3} \left(\psi \left(p+\frac{5}{2}\right)-\psi
  \left(\frac{5}{2}\right)\right)$$Using asymptotics
$$S_p=\left(\frac{8}{9}-\frac{2 \log (2)}{3}\right)-\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$
On the other side
$$T_p=\sum_{n=1}^{p}(-1)^n\frac{ B\left(\frac{1}{2},\frac{n-2}{2}\right)}{n}$$ is a nice monster in terms of hypergeometric function but its limit is simply
$$\frac{1}{2}-\frac{\pi ^2}{8}$$


问:
Dividing Two Infinities
I am Curious if the following is mathematically correct:

Let $a$ be the infinite set of all nonnegative integers $0,1,2,3...$.

Let $b$ be the infinite set of all nonnegative EVEN integers $0,2,4,6...$.

If I take the sum of $a$ and divide it by the sum of $b$ do I get 2?

Is this correct? Can one divide infinities like this?

If yes, does this mean that SUM $a$ > SUM $b$ (despite that both are infinite)?

(if this is not possible can one do some kind of equivalent division using mapping of sets?)

答:
I assume you don't know very much about number theory so I am well suited to answer you, also being an amateur.

You cannot sum up infinitely many natural numbers. You can take the sum of the first n numbers, S1(n) and divide one with the other, S2(n) and then divide to get S1(n)/S2(n) and the find a limit as n goes to eternity, which means it doesn't change very much as n goes bigger and bigger. In some cases you can do this, but the series to be summed have to consist of smaller and smaller terms, and only some of those will work. 1+1/2+1/4 +1/8 can be summed up but 1+1/2+1/3+1/4... cannot.

Title: Dividing Two Infinities


问:
Complex number can't be cross multiplied!
The question below confused me:

I think what confused me is that he asks to perform a x b i.e. cross product. However, Cross/vector product can be applied to VECTORS only and NOT to complex numbers where "j" is NOT the wye coordinate vector it's just the imaginary notation.
I mean if he asked
a x b
or a.b
they're both the same.
Am I right?

答:
The cross product only applies to vectors in three dimensions, so this is asking you to multiply normally.

However, you can think of complex numbers as vectors on the complex (Argand) plane. So you can take the dot product equivalent of two complex numbers.


问:
Given joint distribution, how do I find the conditional probabilities?
I have referred to quite a few questions being asked here but there aren't any solutions that point towards how to solve this (either that or I'm bad at searching).

Suppose we have 3 random variables, $X_1, X_2, X_3$, with joint pdf:

$$\pi(x_1,x_2,x_3) \propto x_1^{\alpha_1 -1}x_2^{\alpha_2 -1}x_3^{\alpha_3 -1}(1-x_1-2x_2-3x_3)^{\alpha_4-1}$$

where $0 < x_1 < 1$, $0 < x_2 < \frac 12$, $0< x_3 < \frac 13$, $0 < x_1+2x_2+3x_3 < 1$, $\alpha_i ,\forall i = 1, 2, 3, 4$ are known positive parameters.

What is the conditional distribution of $\pi(x_1|x_2,x_3)?$

I know that

$\pi(x_1,x_2,x_3) = \pi(x_1)\pi(x_2|x_1)\pi(x_3|x_1,x_2)$

$\pi(x_3|x_1,x_2) = \frac {\pi(x_1,x_2,x_3)} {\pi(x_1)\pi(x_2|x_1)}$  Eq(1)

$\pi(x_1|x_2,x_3) = \pi(x_1)\pi(x_2,x_3|x_1)$

Since $\pi(x_2,x_3|x_1) = \pi(x_3|x_1,x_2)$ (I found this online so I don't really know the validity of this)

$\pi(x_1|x_2,x_3) = \pi(x_1)\pi(x_3|x_1,x_2)$

Substituting Eq(1):

$\pi(x_1|x_2,x_3) = \pi(x_1)\frac {\pi(x_1,x_2,x_3)} {\pi(x_1)\pi(x_2|x_1)} = \frac {\pi(x_1,x_2,x_3)} {\pi(x_2|x_1)}$

Now, it leaves me to find $\pi(x_2|x_1)$ but I am not sure how to do it. I am not even sure if my above steps are correct.

Any help is really, really appreciated!

Body:
$\pi(x_1,x_2,x_3) = \pi(x_1|x_2,x_3)\pi(x_2,x_3) \implies \pi(x_1|x_2,x_3) = \frac{\pi(x_1,x_2,x_3)}{\pi(x_2,x_3)}$. To find $\pi(x_2,x_3)$, you need to marginalize the joint distribution over $x_1$ i.e. $\pi(x_2,x_3)=\int\limits_{x_1=0}^{x_1 = 1} \pi(x_1,x_2,x_3) dx_1$.
Use integration by parts maybe to get the integral. Just like one done here. May require more simplification but I hope the idea is clear.

$x_2^{\alpha_2}x_3^{\alpha_3-1}\int\limits_{0}^{1}x_1^{\alpha_1}(1-x_1-2x_2-3x_3)^{\alpha_4 -1} = x_2^{\alpha_2-1}x_3^{\alpha_3-1}\Big\{\big[-x_1^{\alpha_1-1}\frac{(1-x_1-2x_2-3x_3)^{\alpha_4}}{\alpha_4}\big]_{0}^{1} +\int\limits_{0}^{1} \frac{(1-x_1-2x_2-3x_3)^{\alpha_4}}{\alpha_4}dx_1\Big\} = x_2^{\alpha_2}x_3^{\alpha_3-1}\Big\{(-1)^{(\alpha_4+1)}\frac{(2x_2+3x_3)^{\alpha_4}}{\alpha_4}+\big[-\frac{(1-x_1-2x_2-3x_3)^{\alpha_4+1}}{\alpha_4(\alpha_4+1)}\big]_0^1\Big\} = x_2^{\alpha_2}x_3^{\alpha_3-1}\Big\{(-1)^{(\alpha_4+1)}\frac{(2x_2+3x_3)^{\alpha_4}}{\alpha_4}+\big[(-1)^{\alpha_4+2}\frac{(2x_2+3x_3)^{\alpha_4+1}}{\alpha_4(\alpha_4+1)}+\frac{(1-2x_2-3x_3)^{\alpha_4+1}}{\alpha_4(\alpha_4+1)}\big]\Big\} $


问:
What's the difference between a cyclic and periodic function?
I've seen the words "cyclic" and "periodic" used to describe characteristics of a given function. What do they mean? I can't seem to find a difference. Wikipedia says a periodic function is one that repeats values in a periodic interval.

Maybe I was mistaking about the phrase "cyclic" being used to describe functions. One place I do see the word used in math is cyclic group.

答:
The difference between a cyclic and periodic function is that one is the inverse of the other.

Take $f=\sin(x)$, which is periodic. An inverse $f^{-1}$ is normally done after restricting the domain of $f$ so that it's one-to-one and $f^{-1}$ isn't multi-valued. This does not interest us.

If you don't restrict the domain, $f^{-1}$ becomes multi-valued and we call it cyclic.

And if you invert a cyclic function you get a periodic function.


问:
How to calculate the p-value
I'm following a statistics course for the first time and I'm wondering how can I find the difference beteween two indepentent samples.  I read the explanation but still couldn't understand how it works and I'm still stuck at finding the p-value. Here is the data:

$$\bar x_1=5.9$$
$$\bar x_2=4.1$$
$$\sigma_1=4.1$$
$$\sigma_2=3.7$$
$$n_1=42$$
$$n_2=47$$
And I want to test:
$$H_0:\mu_1-\mu_2>0$$
$$H_a:\mu_1-\mu_2<0$$

Lets say at a 0.05 level.  So I find SE:
$$SE^2=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}$$
This gives SE = 0.83
So I calculate my t stat:
$$t=\frac{\bar x_1-\bar x_2}{SE}=2.16458$$

How can I calculate the p-value and what would it mean?

答:
The notation $\sigma_i$ rather than $s_i$ suggests you are treating the standard deviations as known rather than as being estimated based on the data. Furthermore, the way you find your standard error makes sense only if that is what you are doing. That is not realistic but is often done in early exercises. Therefore you should treat your test statistic $Z = (\overline x_1-\overline x_2)/\text{SE}$ as being normally distributed rather than as having a t-distribution.

Suppose your null hypothesis is that the means of the two populations are equal and the alternative hypothesis says only that they are not equal, without saying which direction the inequality goes. Then the p-value is the probability that $|Z|,$ the absolute value of $Z,$ exceeds the value actually observed, given that the null hypothesis is true. So, for example, suppose you have $|Z| = 2.3.$ then you need $\Pr(|Z|>2.3) = \Pr(Z>2.3) + \Pr(Z<-2.3) = 2\Pr(Z< -2.3).$ Normally you would find that probability by using either a table or a software package.

Title: How to calculate the p-value


问:
Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$
Can someone give me some hints on how to start solving $\sqrt{x+4}-\sqrt{x+1}=1$ for x?

Like I tried to factor it expand it, or even multiplying both sides by its conjugate but nothing comes up right.

答:
It's easy to guess that the answer is $x=0$.
However, one should prove that this is the only root. Here is how.

Because $\sqrt {x+4} > \sqrt {x+1} $, this function is strictly increasing.

Function $y =1$ is constant.

It is easy to proove that monotoniously increasing function and a constant can cross at no more than 1 point (either cross at 1 point or do not cross at all).

Since we have a point of $x = 0$, then there are no other roots for this equasion.

Title: Solve $\sqrt{x+4}-\sqrt{x+1}=1$ for $x$


问:
What is significance of the row space?
I became to study linear algebra a bit on my own. It is already  more than twenty years since my graduation. I ended up to my question due to practical problem that I have already solved. Having said that, my practical problem lead me to trying understand a concept of a column space, row space, null space and left null space.

I have developed a intuition or concept of a column space. In my thinking matrix columns are vectors in given space. Further on, in my thinking a column space is a space spanned by matrix columns. All that makes sense to me. I just rely on interpretation of a matrix as collection of vectors.
Also, somehow I have become to understand a null space as a special space that will lead to matrix equation of Ax=0. In my thinking, importance of a null space is that there can't be inverse for a such transformation.

Now, my difficulties arise in understanding a row space. It is useless to repeat definition of a row space as Col(A'). The definition doesn't really open this for me. I'd like to understand significance of a row space. In my thinking, columns of a matrix are vectors. This understanding doesn't really fit to a row or a row space. Rows are not vectors, hence a row space is not similar space as a vector space  spanned by vectors. Yes, technically it is so or may be, as I have understood. But I am still missing intuition or understanding of concept of a row space.
What is significance of a row space, how does it manifest itself?

And very similarly, what is importance of a left null space, how does it manifest itself?

You see, I am not a mathematician nor student of math. I am an engineer. And due to my personal way of learning, I am always trying to visualize or develop some kind of intuition before diving into the details and rigorous definitions.

Thanks in advance

Poincaré look-alike -

答:
Let's consider an $m \times n$ matrix $A$ of real numbers.

Although the row space is in general different than the column space, we have:
\begin{align} \text{A matrix $A \in \mathbb{R}^{m \times n}$ takes its row space onto its column space.} \tag{1}
\end{align}
Moreover, this is a one-to-one correspondence, so we can invert $A$ on the column space. The column space is an "isomorphic copy" of the row space.

Geometrically, if $A$ is viewed as a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$, then the subspace spanned by the row space (e.g. a plane) is mapped to a subspace of the same dimension (e.g. another plane).

So this tells us that row rank equals column rank. In any matrix, the number of linearly independent rows is the same as the number of linearly independent columns.

Since you also ask about the left null space $N(A^T)$, here is one way it might be used:

$$\text{Let $A \in \mathbb{R}^{m \times n}$. $Ax = b$ is solvable iff $b$ is orthogonal to $N(A^T)$.} \tag{2}
$$

This can be useful sometimes. For example, if we already know a basis of the left null space, then to check whether $b$ is a solution, it suffices to check whether its dot product with each basis vector of the left null space is zero. (An example involving Kirchoff's Voltage Law is given in section 3.1 of Gilbert Strang's Linear Algebra and its Applications.)

Proof of (1): The row space $R(A)$ is the orthogonal complement of the null space $N(A)$, so $\mathbb{R}^n = R(A) \oplus N(A)$; that is, every vector $x$ in $\mathbb{R}^n$ can be written uniquely as a sum $x = x_r + x_n$ where $x_r \in R(A)$ and $x_n \in N(A)$.

Given $y$ in the column space $C(A)$, say $y = Ax$ where $x = x_r + x_n$, we have $y = Ax = A(x_r + x_n) = Ax_r + Ax_n = Ax_r$, so every $y \in C(A)$ has a vector $x_r \in R(A)$ mapping to it.

This $x$ is also unique, because if $Ax' = y$ for another $x' = x_r' + x_n'$, then $0 = y - y = Ax - Ax' = Ax_r - Ax_r' = A(x_r - x_r')$. So $x_r - x_r'$ is in the null space of $A$. But $x_r - x_r'$ is also in the row space, and the row space and null space are orthogonal, so $x_r - x_r'$ is orthogonal to itself, hence it must be the zero vector. So $x_r = x_r'$. $\square$

Proof of (2): Note that "$Ax = b$ is solvable" just means that $b$ is in the column space $C(A)$, and "$b$ is orthogonal to $N(A^T)$" just means that $b \in N(A^T)^\perp$. But $C(A) = N(A^T)^\perp$, since the column space is the orthogonal complement of the left null space. $\square$

Title: What is significance of the row space?


答:
As others have indicated, half of the problem is performing the multiplication.  The other half is as follows:

Suppose that $z = x + iy$, and you are required to express this as

$re^{i\theta} = r[\cos \theta + i\sin\theta].$

Then $r = |z| = \sqrt{x^2 + y^2}~$ and, (assuming that $|z| \neq 0$) within a modulus of $(2\pi)$,

$\theta$ is completely determined by:

$\cos(\theta) = \frac{x}{|z|},~~\sin(\theta) = \frac{y}{|z|}.$

Title: Complex number can't be cross multiplied!

Body:
The question below confused me:

I think what confused me is that he asks to perform a x b i.e. cross product. However, Cross/vector product can be applied to VECTORS only and NOT to complex numbers where "j" is NOT the wye coordinate vector it's just the imaginary notation.
I mean if he asked
a x b
or a.b
they're both the same.
Am I right?


问:
Proved that angle subtended by a circle at centre is double at any other point on circle is wrong
There is a class 9 or class 10 theorem that "The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle." Now the proof which is taught to us is like-

Let there be a circle with centre O.

$$∠BOQ=∠OAQ+∠AQO       .....(1)
$$Also, in △ OAQ,
$$OA=OQ                  [Radii of a circle]
$$Therefore,
$$∠OAQ=∠OQA $$     [Angles opposite to equal sides are equal]
$$∠BOQ=2∠OAQ              .......(2)
$$Similarly, BOP=2∠OAP             ........(3)

Adding 2 & 3, we get,
$$∠BOP+∠BOQ=2(∠OAP+∠OAQ) $$
$$∠POQ=2∠PAQ               .......(4) $$

This I have understood. But let's say I altered the diagram like this.

In this O is the centre I tried using exterior property but failed. Mind you you might start proving it by same segment. Angle in same segment are equal. If we using this segment property then we can prove this but to prove that angle in same segment are equal we use angle subtended in circle is double at any point on circle and like using each other to prove each other seems a bit wrong. So is really the angle subtended at centre is wrong or I am missing to solve it here?

答:
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. This as described by you is a theoram of circle.

If you are not able to understand the diagram then let me explain it to you. I have joined OA and OQ. Let angle OAQ be 5 ( 5 is name not number), angle QAP be 4, OPA be 3, POQ be 6, OQA be 1 and AQP be 2.

OA = OB

So,$$ angle 4 + angle 5 = angle 3- eq 1 $$

Similarly, OA = OC

$$ angle 5 = angle 1 - eq 2 $$
Now,

$$ angle 3 + angle 4 = angle 1 + angle 6 $$
$$ angle 4 + angle 4 + angle 5 = angle 1 + angle 6 $$ ( replacing angle 3 as with eq 1)
$$ angle 4 + angle 4 = angle 6 $$ (cutting angle 5 and angle 6 from both sides as they are equal)

$$ 2 * angle 4 = angle 6 $$

So we have proved that angle subtended in centre is double the angle subtended at any property and thus I can claim that this theoram is correct


问:
p-adic numbers vs real numbers
Could anyone give a concrete example of a p-adic number that is not a "real number"? that is, do we create "new numbers" (non real numbers) by completing Q with a non Archimedean norm? If so, what are they?</p>

答:
But if you go into these things more deeply, you’ll see that the $p$-adic worlds (nonArchimedean systems) are entirely different from the real-complex world (the Archimedean system).

My favorite way of illustrating this is that within the complex numbers, any root of unity lies on the unit circle, as I hope you know. And they’re dense on the circle: every neighborhood of a root of unity contains infinitely many other roots of unity.

Entirely different for the $p$-adic universe: there, the roots of unity are a discrete set. Indeed, within a neighborhood of radius $1/3$ of a root $\zeta$ of unity, there are no other roots of unity.

Title: p-adic numbers vs real numbers


问:
$f_n(z)={z^n\over n}$, $z\in D$ open unit disk then
$f_n(z)={z^n\over n}$, $z\in D$ open unit disk then

1.$\sum f_n$ converges uniformly on $D$?

2.$f_n$ and $f'_n$ converges uniformly on $D$?

3.$\sum f'_n$ converges on $D$ pointwise?

4.$f_n''(z)$ does not converge unless $z=0$

clearly $1$ is false as $\sum {1\over n}$ is divergent, I know $f_n$ and $f'_n$ converges to $0$ on $D$ but dont know if the convergence is uniform or not, so not sure about $2$, $3$ is true as $|z|<1$  infinite geometric series $\sum z^{n-1}$ converges!, $4\Rightarrow$ $f_n''(z)=(n-1)z^{n-2}$ is convergent to $0$ on $D$ as $|z|<1$ so $4$ is false

答:
In what follows $\|\cdot\|_{C^0(D)}$ denotes the supremum norm $\|f\|_{C^0(D)} = \sup_{z\in D}|f(z)|$.

Your reasoning for (1) is right, perhaps you should mention that $\| \sum_{n=1}^N f_n\|_{C^0(D)} \ge \sum_{n=1}^N \frac 1n$, so as the partial sums are unbounded, we cannot have uniform convergence.

We have $\|f_n\|_{C^0(D)} = \frac 1n\to 0$, so $f_n \to 0$ uniformly. For $f_n'(z) = z^{n-1}$, note that $f_n'(z) \to 0$ pointwise, but $\|f_n'\|_{C^0(D)} = 1$, if $(f_n')$ were uniformly convergent, necessarily $f_n' \to 0$, contradicting $\|f_n'\|_{C^0(D)} \to 1$.

Yes, as you said, on $D$, we have $\sum_{n=1}^\infty z^{n-1} = \frac 1{1-z}$ pointwise (as fgp mentioned in his comment, not uniformly, but this isn't asked).

For (4) we have, as you say, $f_n''(z) = (n-1)z^{n-2} \to 0$ for $z\in D$.


问:
How does the modified Newton's method for multiple roots converge quadratically?
I am currently learning about the multiple root issue for Newton's method. My textbook hinted at tweaking the function using
$$h(x)=\frac{f(x)}{f′(x)}$$
and making the iteration function to actually be
$$g(x)= x - \frac{h(x)}{h'(x)}$$

My textbook claims that this method is also quadratically convergent but provided no proof and I am lost as to how it is quadratically convergent. I tried to do some more research but all I could find is this: https://mat.iitm.ac.in/home/sryedida/public_html/caimna/transcendental/iteration%20methods/accelerating%20the%20convergence/mrac.html

Can anybody show me how this quadratically convergent or at least point me in the direction to prove it myself?

答: Hint: That method is known as Schröder's Method and the iteration formula may be obtained by applying Newton's method to $f(x)/f'(x)$ instead of $f(x)$.


问:
Minimum values of multiple modulus function
Find the minimum value of the equation $|x-1|+|x-2|+|x+1|+|x+2|$.
I checked the values at x=1,x=2,x=-1 and x=-2 and getting the answer 6. Is there any other way to find the minimum values.

Body:
generalised Triangle inequality does it!

As $|x-2|=|2-x|$ and $|1-x|=|x-1|$ $$|1-x|+|2-x|+|x+1|+|x+2|\ge |1-x+2-x+x+1+x+2|=6$$


问:
The evaluation of $\int_{-\infty}^{\infty} \! e^{-2 |x - u| -3 |u|} \, \mathrm{d}u$
I am trying to solve this integral, however my attempts haven't worked out yet.

My general approach to solving equations involving modulus functions is to split them into cases, although I don't know if this will work here.

This is my attempt:

Let $x$ and $u$ both be positive.

Then we could have something like the following:

$\int_{-\infty}^{\infty} \! e^{-2x - 5u} \, \mathrm{d}u$

$e^{-2x} \int_{-\infty}^{\infty} \! e^{- 5u} \, \mathrm{d}u$

$e^{-2x} \biggl[-\frac{1}{5} e^{-5u} \biggr]_{-\infty}^{\infty}$

But this seems to diverge. Can anyone help me solve this integral? I apologise if I've made a silly error somewhere.

答:
Case I. $x \ge 0$

Then,
$$
\int_{-\infty}^\infty e^{-2|x-u|-3|u|} du
 = \int_{-\infty}^0 e^{-2(x-u)-3u} du
 + \int_0^x e^{-2(x-u)+3u} du
 + \int_x^\infty e^{2(x-u)+3u} du
$$
Can you integrate?

Case II. $x<0$

Can you write a similar decomposition over $(-\infty,x)\cup(x,0)\cup(0,\infty)$?


问:
Why $\boldsymbol{n} \approx \boldsymbol{n}^{+}$ contradicts the pigeon-hole theorem?
Can anyone draw a contradiction using the pigeon-hole theorem from $\boldsymbol{n} \approx \boldsymbol{n}^{+}$?

Pigeon-hole theorem is the following:

For all $\boldsymbol{n} \in \mathbb{N}$, $f: \boldsymbol{n} \to \boldsymbol{n}$ is onto if it is one-one.

$\boldsymbol{n}^{+} = \boldsymbol{n} \cup \left\{ \boldsymbol{n} \right\}$ is the successor of $\boldsymbol{n} \in \mathbb{N}$.

$\boldsymbol{n} \approx \boldsymbol{n}^{+}$ means there is a bijection between $\boldsymbol{n}$ and $\boldsymbol{n}^{+}$.

The formula $\boldsymbol{n} \approx \boldsymbol{n}^{+}$ may seem naively false, as $\boldsymbol{n}^{+}$ is of one more element than $\boldsymbol{n}$. But ironically, I find it difficult to prove that the formula is false.

I have used the approach provided in this post to prove a similar, but more general theorem:

If $\boldsymbol{m} < \boldsymbol{n}$, then $\boldsymbol{m} \prec \boldsymbol{n}$.

Assume $\boldsymbol{m} < \boldsymbol{n}$. Then $\boldsymbol{m} \subsetneq \boldsymbol{n}$. Define the identity map $f: \boldsymbol{m} \to \boldsymbol{n}$ as
\begin{equation*}
\left(\forall \boldsymbol{s} \in \boldsymbol{m}\right) \left(f\left(\boldsymbol{s}\right) = \boldsymbol{s}\right).
\end{equation*}
It is immediately clear that $f$ is one-one. Thus, $\boldsymbol{m} \preceq \boldsymbol{n}$.

Now assume that $\boldsymbol{m} \approx \boldsymbol{n}$. Then there exists a bijection $g: \boldsymbol{n} \to \boldsymbol{m}$. The restriction of $g$ to $\boldsymbol{m}$, $g\vert_{\boldsymbol{m}}: \boldsymbol{m} \to \boldsymbol{m}$ has to be one-one, thus it also has to be onto, according to pigeon-hole theorem. As $\boldsymbol{m} \subsetneq \boldsymbol{n}$, there exists $\boldsymbol{p} \in \mathbb{N}$, $\boldsymbol{p} \in \boldsymbol{n}$ and $\boldsymbol{p} \not\in \boldsymbol{m}$. As $g$ is onto, $g\left(\boldsymbol{p}\right) \in \boldsymbol{m}$. As $g\vert_{\boldsymbol{m}}$ is onto and $\boldsymbol{p} \not\in \boldsymbol{m}$, there has to exist $\boldsymbol{q} \in \boldsymbol{m}$ such that $\boldsymbol{q} \neq \boldsymbol{p}$ and $g\vert_{\boldsymbol{m}}\left(\boldsymbol{q}\right) = g\left(\boldsymbol{p}\right)$. Using the fact that $g\vert_{\boldsymbol{m}}\left(\boldsymbol{q}\right) = g\left(\boldsymbol{q}\right)$, we have $\boldsymbol{p} \neq \boldsymbol{q}$ and $g\left(\boldsymbol{p}\right) = g\left(\boldsymbol{q}\right)$, which contradicts the assumption that $g$ is one-one. Thus, there is no bijection between $\boldsymbol{m}$ and $\boldsymbol{n}$, and $\boldsymbol{m} \not\approx \boldsymbol{n}$. Combining $\boldsymbol{m} \preceq \boldsymbol{n}$ and $\boldsymbol{m} \not\approx \boldsymbol{n}$, we may conclude that $\boldsymbol{m} \prec \boldsymbol{n}$.

答:
Suppose $f:{\bf n^+}\rightarrow {\bf n}$ is a bijection. Think about the restriction $g$ of $f$ to ${\bf n}$. Since $f$ was a bijection, $g$ must be an injection: any restriction of an injection is again an injection. But by the pigeonhole principle, this means that $g$ is onto. Now, do you see why this contradicts the injectivity of $f$?

 Since $g$ is onto and has the same codomain as $f$, we must have $f({\bf n})\in ran(g)$, which is to say $f({\bf n})=g({\bf k})$ for some ${\bf k}\in dom(g)$. But the domain of $g$ is exactly ${\bf n}$ and so ${\bf k}\not={\bf n}$. Since $g$ is a restriction of $f$ we have $f({\bf k})=g({\bf k})$, and putting all this together we get $$f({\bf n})=f({\bf k})\quad\mbox{but}\quad {\bf n}\not={\bf k}.$$

Note that this is somehow "dual" to Yuz's argument in the comments, which replaces $f$ by its "codomain-only expansion" $$\overline{f}: {\bf n^+}\rightarrow{\bf n^+}: a\mapsto f(a).$$ Ultimately these are using the same idea, but they deploy it differently.


问:
Showing that the set of isomorphism class of simple left module is finite
Let $R$ be an Artinian Ring, then prove that the isomorphism classes of simple left $R$-modules is finite.

Morever show that the isomorphims class of simple left modules and the isomorphism classes of simple right modules has the same cardinality.

What am I thinking is to use descending chain condition, but I am unable to use it. Suppose that there are infinitely many isomorphism classes, then I am trying to use Artinian property of $R$, but it doesn't help

答:
If $S$ is a simple left $R$-module, it is cyclic (pick any non zero element $x \in S$, by simplicity $Rx = S$), hence $S \simeq R/\mathbf{Ann}_R(x)$. Since $S$ has no non-trivial submodules, we conclude that $\mathbf{Ann}_R(x)$ is maximal. Moreover, each quotient $R/ \mathfrak m$ is simple when $\mathfrak m$ is maximal.

This tells us that the classes of isomorphism of left (resp. right) simple modules are the same as the classes of isomorphism of quotients $R/\mathfrak m$ with $\mathfrak m$ a maximal left (resp. right) ideal. However, when $R$ is commutative the distinction between side ideals is irrelevant (and so the classes of isomorphism of right and left simple modules are in bijection) and we know there are finitely many maximal ideals, in particular, finitely many simple modules up to isomorphism.

Maybe this can be adapted to the non-commutative case, which I am not familiar with.


答:
Absolute values change behaviors when the sign changes. In your equation, the sign starts to change when $x= 1,2,-1,-2$ because some of the values in the absolutes become zero, so we examine what happens between these values.

Case $1$: $x \ge 2$. If $x \ge 2$, all the values in the absolute values are positive, and the absolute value of a positive number does nothing, so we remove all the absolute values and simplify to get $4x$. However, remember that $x \ge 2$ in this case, so the smallest $x$ we can use here is $x=2$, so the minimum for this case is $8$

Case $2$: $2 \ge x \ge 1$. If $2 \ge x \ge 1$, all values in $|x−1|+|x−2|+|x+1|+|x+2|$ are positive except for $|x-2|$. Taking the absolute value of a negative expression is equal to multiplying it by negative one. Example $|-1|$ = $1$ because $-1*-1$ = $1$. So, $|x-2| = -(x-2) = 2-x$. So, our equation becomes $(x−1)+(2-x)+(x+1)+(x+2)$ which simplifies down to $2x+4$. Since $2 \ge x \ge 1$, the smallest $x$ we can use is $1$, so our minimum value in this case is $6$.

Case $3$: $1 \ge x \ge -1$. If $1 \ge x \ge -1$, all values in $|x−1|+|x−2|+|x+1|+|x+2|$ are positive except for $|x-2|$ and $|x-1|$. So, our equation becomes $(1-x)+(2-x)+(x+1)+(x+2)$ which simplifies down to $6$. So, the minimum value is $6$ in this case.

Case $4$: $-1 \ge x \ge -2$. If $-1 \ge x \ge -2$, all values in $|x−1|+|x−2|+|x+1|+|x+2|$ are positive except for $|x-2|$ and $|x-1|$ and $|x+1|$. So, our equation becomes $(1-x)+(2-x)+(-x-1)+(x+2)$ which simplifies down to $-2x+4$. We are subtracting $2x$ from $4$, so we get the smallest value when we choose the biggest $x$ in $-1 \ge x \ge -2$, so $x=-1$, and thus the smallest value is $6$ in this case.

Case 5: $x \le -2$. If $x \le -2$, all values are negative in $|x−1|+|x−2|+|x+1|+|x+2|$, so our equation becomes $(1-x) + (2-x) +(-x-1) + (-x-2)$ which becomes $-4x$. Since we are subtracting $4x$, we make $x$ as big as possible to get smallest value. The biggest $x$ in  $x \le -2$ is $-2$, so our minimum value becomes $8$.

Thus, the smallest minimum in all of our cases is $6$. Took a lot longer, but at least you know for sure you didn't miss a smaller value.

Title: Minimum values of multiple modulus function

Body:
Find the minimum value of the equation $|x-1|+|x-2|+|x+1|+|x+2|$.
I checked the values at x=1,x=2,x=-1 and x=-2 and getting the answer 6. Is there any other way to find the minimum values.


答:
We can think of vectors as either rows or as columns, and can view matrix vector multiplication as acting with either: $x \mapsto Ax$ when $x$ is a column, $v \mapsto vA$ when $v$ is a row. So the column space and row space of a matrix have equal importance, they are essentially interchangeable depending on how we are working with the matrix and vectors.

In fact, you can find entire books which define vectors as rows and not columns, all of the definitions are adjusted slightly to account for this change and everything works as expected.

Sometimes we want a matrix to act on both rows and columns: for example take a bilinear form $B(u,v) = u^{T}Av$. Now consider the following question: is there a vector $u_{0} \neq 0$ such that $B(u_{0},v) = 0$ for all $v \in V$?

We will have $B(u_{0},v) = u_{0}^{T}Av = 0$ for all $v \in V$ precisely when $u_{0}^{T}A = 0$, in other words, when $u_{0}$ is in the left null space of $A$.

Note if you are interested in an advanced perspective on this: the formal connection between these two points of view (row vectors/column vectors) is provided by the concept of the dual space of a vector space; it states (abbreviated form) that if you are working with e.g. column vectors, then you can think of a one-row matrix as a linear function into the base field, and the collection of such functions also form a vector space (that in the finite-dimensional case is isomorphic to the original vector space).

Title: What is significance of the row space?

Body:
I became to study linear algebra a bit on my own. It is already  more than twenty years since my graduation. I ended up to my question due to practical problem that I have already solved. Having said that, my practical problem lead me to trying understand a concept of a column space, row space, null space and left null space.

I have developed a intuition or concept of a column space. In my thinking matrix columns are vectors in given space. Further on, in my thinking a column space is a space spanned by matrix columns. All that makes sense to me. I just rely on interpretation of a matrix as collection of vectors.
Also, somehow I have become to understand a null space as a special space that will lead to matrix equation of Ax=0. In my thinking, importance of a null space is that there can't be inverse for a such transformation.

Now, my difficulties arise in understanding a row space. It is useless to repeat definition of a row space as Col(A'). The definition doesn't really open this for me. I'd like to understand significance of a row space. In my thinking, columns of a matrix are vectors. This understanding doesn't really fit to a row or a row space. Rows are not vectors, hence a row space is not similar space as a vector space  spanned by vectors. Yes, technically it is so or may be, as I have understood. But I am still missing intuition or understanding of concept of a row space.
What is significance of a row space, how does it manifest itself?

And very similarly, what is importance of a left null space, how does it manifest itself?

You see, I am not a mathematician nor student of math. I am an engineer. And due to my personal way of learning, I am always trying to visualize or develop some kind of intuition before diving into the details and rigorous definitions.

Thanks in advance

Poincaré look-alike -


答:
The standard approach to the Inscribed Angle Theorem, as shown in most high school textbooks on geometry, is to start by proving it for the special case when one side of the angle is a diameter. They then treat the general case by drawing in an auxiliary diameter through the vertex of the angle, as shown below. Case 1 is from the source, and Case 2 is from the OP's drawing here.

Edit: At the original poster's request, I am providing an explicit proof for Case 2, rather than an argument sketch.

As in Case 1, the proof begins by drawing the diameter $AB$, showing that the measure of Angle $PAB$ is half the measure of Arc $PB$, and that the measure of angle $QAB$ is half the measure of Arc $QB$. As these proofs were given in the original post; as the arguments are identical in both cases; and as drawing in the radii would needlessly complicate the diagrams, I trust it is safe to omit full demonstrations of these preliminaries.

Once we've established that $m \angle PAB = \frac{1}{2} m \text{ Arc } PB$, $m \angle QAB = \frac{1}{2} m \text{ Arc } QB$, the proof of Case 2 looks a lot like Case 1, except we subtract instead of add: since $$m \angle PAQ = m \angle PAB - m \angle QAB$$ and $$m \text{ Arc } PQ = m \text{ Arc } PB - m \text{ Arc } QB,$$ we get

\begin{align*}
m \angle PAQ &= m \angle PAB - m \angle QAB \\
 &= \frac{1}{2} (m \text{ Arc } PB - m \text{ Arc } QB) \\
 &= \frac{1}{2} m \text{ Arc } PQ, \
\end{align*}
and our argument is complete.

Since you ask for a "reputable source," here is how Pearson's Common Core geometry textbook, widely used in US high schools, approaches the Inscribed Angle Theorem (pp. 780-781):

The pictures they draw for Exercises 26 and 27, in the same section of the textbook, are substantially equivalent to the ones I drew above for Cases 1 and 2 above (which the textbook refers to as "Case II" and "Case III" respectively). It seems noteworthy, per OP's request for a proof, that the student edition of the textbook does not explicitly fill in the proof of Cases 1 and 2 above, but assumes that a motivated and intelligent high school student would be able to come up with the details themselves, from seeing the demonstration when one side of the angle is a diameter.

Edit 2: Wanted to provide another source. Matthew Harvey (math professor at UVA) has a fascinating and readable free geometry textbook available on his university webpage. He gives an explicit proof of the Inscribed Angle Theorem on pages 12-13 of Chapter 16, "Circles". Harvey's proof is exactly the one given in my answer, and outlined in Pearson's geometry textbook. It covers the case which OP was confused about, where angle/arc subtraction is necessary.

Title: Proved that angle subtended by a circle at centre is double at any other point on circle is wrong

Body:
There is a class 9 or class 10 theorem that "The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle." Now the proof which is taught to us is like-

Let there be a circle with centre O.

$$∠BOQ=∠OAQ+∠AQO       .....(1)
$$Also, in △ OAQ,
$$OA=OQ                  [Radii of a circle]
$$Therefore,
$$∠OAQ=∠OQA $$     [Angles opposite to equal sides are equal]
$$∠BOQ=2∠OAQ              .......(2)
$$Similarly, BOP=2∠OAP             ........(3)

Adding 2 & 3, we get,
$$∠BOP+∠BOQ=2(∠OAP+∠OAQ) $$
$$∠POQ=2∠PAQ               .......(4) $$

This I have understood. But let's say I altered the diagram like this.

In this O is the centre I tried using exterior property but failed. Mind you you might start proving it by same segment. Angle in same segment are equal. If we using this segment property then we can prove this but to prove that angle in same segment are equal we use angle subtended in circle is double at any point on circle and like using each other to prove each other seems a bit wrong. So is really the angle subtended at centre is wrong or I am missing to solve it here?


问:
Calculate $\int^2_0\frac{\arctan x}{x^2-4x-1}\,dx$
Calculate
$\int^2_0\frac{\arctan x}{x^2-4x-1}\,dx$.

The only idea that I have is to substitute $t=x+1$, but I do not think it is a good one.

答:
As @Tito Eliatron commented, integrating by parts
$$I=\int\frac{\tan ^{-1}(x)}{x^2-4 x-1}dx$$
$$I=\tan ^{-1}(x)\frac{\log \left(-x+\sqrt{5}+2\right)-\log \left(x+\sqrt{5}-2\right)}{2 \sqrt{5}}+\frac J{2 \sqrt{5}}$$
$$J=\int\frac{\log \left(-x+\sqrt{5}+2\right)-\log \left(x+\sqrt{5}-2\right)}{x^2+1}dx$$
$$\frac 1{x^2+1}=\frac 1{(x+i)(x-i)}=\frac{i}{2 (x+i)}-\frac{i}{2 (x-i)}$$All of these make that we face now four integrals
$$I=\int \frac{\log (a x+b)}{x+c}dx=\text{Li}_2\left(\frac{a x+b}{b-a c}\right)+\log (a x+b) \log \left(1-\frac{a
   x+b}{b-a c}\right)$$ Integrated between the given bounds, this leads to a quite complicated expression which is hard to simplify because of the bunch of polylogarithms of complex arguments.

After a pretty tedious work
$$\int_0^2\frac{\tan ^{-1}(x)}{x^2-4 x-1}dx=-\frac{\tan ^{-1}(2) \sinh ^{-1}(2)}{2 \sqrt{5}}\sim -0.3573950303$$


问:
Integer valued polynomial implies integer coefficients

Let $f(x) = Ax^2 + Bx + C$ where $A, B, C$ are real numbers. If
$f(x)$ is an integer whenever $x$ is an integer, then

(a) $2 A$ and $A + B$ are integers, but $C$ is not an integer.

(b) $A + B$ and $C$ are integers, but $2 A$ is not an integer.

(c) $2 A$, $A + B$ and $C$ are all integers.

(d) None of these.

答:
HINT

Note $f(0)=C$, so $C$ is an integer, which would eliminate (a). What can you imply from $f(1)$ and $f(-1)$? Can you finish?


问:
Proving the recursive formula for "Virahanka Numbers"
So apparently Virahanka was an Indian mathematician that, in a way, discovered the Fibonacci sequence 500 years before Fibonacci. He was interested in finding the number of patterns of short syllables ($S$) and long syllables ($L$). We let $p$ be a pattern, that is, an ordered sequence of $S$'s and $L$'s, we assign a value of $1$ to $S$ and $2$ to $L$, and so the length of $p$ is: # of $S$'s + $2$(# of $L$'s). We let $P_n$ denote the set of patterns of length $n$, and let $v_n=|P_n|$

e.g. $P_0=\emptyset$, $P_1=${$S$}, $P_2=${$SS,L$}, $P_3=${$SSS,SL,LS$} and $v_0=1$, $v_1=1$, $v_2=2$, $v_3=3$

How can I prove that $v_n=v_{n-1}+v_{n-2}$

I honestly have no idea where to begin.

答:
More exactly, the elements of vn are obtained from elements of vn−1 by prefixing an S +  elements of vn−2 by prefixing an L.

For example, v4 = {LL, LSS, SSL, SLS, SSSS} = {LL, LSS} + {SSL, SLS, SSSS} where:

{LL, LSS}: L prefixed to each item of v2 = {L, SS}

{SSL, SLS, SSSS}: S prefixed to each item of v3 = {SL, LS, SSS}.

See code


问:
How can I evaluate $\int_0^\infty e^{\pm i \xi} d \xi $
I have encountered this problem while trying to evaluate
$$
\int_0^\infty e^{ix^2}dx
$$
by squaring it and using Fubini's theorem.

I just treat $i$ as a constant:
$$
\int_0^\infty e^{i \xi} d \xi = \frac{1}{i} \left[ e^{\pm i\infty} -1 \right]
$$
But what is '$i\infty$'?

答:
$$\int_0^p e^{+i x} dx=\sin (p)+i(1- \cos (p))$$
$$\int_0^p e^{-i x} dx=\sin (p)-i(1- \cos (p))$$

What happens when $p \to \infty$ ?


问:
A question on counting processes
Here is a question on counting processes. Could anyone try on it? :)

Let $\{X(t):t>0\}$ be a random process obtained by switching between the values $0$ and $1$ according to the event times in a counting process $\{N(t):t>0\}$. Let $P(X(0)=0)=p$ and

\begin{align}
P(N(t)=k)=\frac{1}{1+\mu t}\left(\frac{\mu t}{1+\mu t}\right)^k
\end{align}

for $k=0,1,\ldots$. Find the pmf of $X(t)$.

答: From the total law of probability we have for any $t>0$ that $$P(X(t)=0)=pP(X(t)=0|X(0)=0)+(1-p)P(X(t)=0|X(0)=1)$$ Again by the total law the probability $P(X(t)=0|X(0)=0)$ equals the sum $$\sum_{k=0}^{\infty}P(X(t)=0|X(0)=0,N(t)=k)P(N(t)=k|X(0)=0)$$ Note $P(X(t)=0|X(0)=0,N(t)=k)$ equals $0$ when $k$ is odd and $1$ when $k$ is even. With geometric series and the independence of $X(0)$ with $N(t)$ we have $$P(X(t)=0|X(0)=0)=\sum_{k=0}^{\infty}P(N(t)=2k|X(0)=0)=\frac{\mu t +1}{2\mu t +1}$$ A symmetric argument produces $$P(X(t)=0|X(0)=1)=\frac{1}{2\mu t +1}$$ Putting everything together, $$P(X(t)=0)=\frac{p \mu t}{2\mu t +1}$$ It's now easy to see that $X(t) \sim \text{Bernoulli}\Bigg(\frac{(2-p)\mu t +1}{2\mu t +1}\Bigg)$

Title: A question on counting processes


答:
As gt6989b notes, $f(0) = C$, and since $f(x)$ is an integer for any integer $x$, $C$ must be an integer as well.

Also, we have $f(1) = A + B + C$. This is equal to $B+A = f(1) - C$. However, note that $f(1)$ must be an integer, and we have already shown that $C$ is an integer. The difference of two integers is also an integer, thus $A+B$ is an integer as well.

We also have $f(2) = 4A + 2B + C$. Since $A+B$ is an integer, $2A+2B$ is an integer as well, and since $C$ is an integer, $2A+2B+C$ is also an integer. If we subtract this from $f(2)$, we get $f(2) - (2A+2B+C) = 2A$. Since $f(2)$ and $2A+2B+C$ are integers, their difference must also be an integer, thus $2A$ is an integer as well. So, we have $2A$, $A+B$, and $C$ as integers, thus the answer is C.

Title: Integer valued polynomial implies integer coefficients

Body:
Let $f(x) = Ax^2 + Bx + C$ where $A, B, C$ are real numbers. If
$f(x)$ is an integer whenever $x$ is an integer, then

(a) $2 A$ and $A + B$ are integers, but $C$ is not an integer.

(b) $A + B$ and $C$ are integers, but $2 A$ is not an integer.

(c) $2 A$, $A + B$ and $C$ are all integers.

(d) None of these.


问:
Integrating vertically giving different answer from intergrating horizontal

Find area in the region: $$ \frac{1}{x} <y < x^{\frac13}$$ , $1
<x<8$

So, I thought of integrating 'vertically' by:

$$ \int y dx = xy - \int x dy =16- \left[ \int_1^2 y^3 + \int_0^1 \frac{1}{y} dy\right] $$

Now, the problem is that $ \int_0^1 \frac{dy}{y}$ part, it seems to blow up. So, this method seems unfeasible, is there are any reason I could 'remove' the blow up and what exactly is the property of the function which causes this to occur?

Note the integral $ \int_0^1 \frac{1}{y} dy$ represents this following area:

Body:
Vertically:

$$\int_{1/8}^1 (8-1/y) \ dy + \int_{1}^2 (8-y^3) \ dy$$

Horizontally:

$$\int_{1}^8 (x^{1/3}-1/x) \ dx$$


问:
Does completeness of $L(E,F)$ implies completeness of $F$?
$\mathbf {The \ Problem \ is}:$ If $E$ and $F$ be two normed spaces with $E \neq \{0\}$  and $L(E,F)$, the set of all bounded linear maps from $E$ to $F$ is complete normed space under the operator norm .

Is $F$  complete?

$\mathbf {My \ approach}:$ Obviously, the converse is true . And I was thinking of $E = \mathbb R$ and $F = \mathbb{P}[0,1]$, set of polynomials on $[0,1]$, under supremum norm ; where $F$ is not complete, but I can't conclude about $L(E,F)$.

A small hint is warmly appreciated .

答: Let $(f_n)$ be a  Cauchy sequence in $F$. Let $x^{*}$ be any non-zero continuous linear functional on $E$. Define $T_n: E \to F$ by $T_n x=x^{*}(x) f_n$. Show that $\|T_n-T_m\|=\|x^{*} \| \|f_n-f_m\|$. So $(T_n)$ converges to some operator $T$. Now pick any $x$ such that $x^{*}(x)\neq 0$ and use convergence of  $x^{*}(x) f_n $ to see that $(f_n)$ converges.


问:
If $K$ is compact, then $m_*(K) = m^*(K)$
I was reading http://people.math.harvard.edu/~shlomo/212a/11.pdf

In page 25, it mentions that:

If $K$ is a comapct set, then $m_*(K) = m^*(K)$ since $K$ is a comapct set contained in itself.

I don't quite understand how the inner measure equal the outer measure here.

答:
I misread the definition of inner measure.

The inner measure is given as:
$$m_\ast(A)=\sup\lbrace m^*(K): K\subseteq A \text{ is compact}\rbrace.$$

Hence, when $K$ compact, by definition
$$m_*(K) = m^*(K)$$


问:
Smith Normal Form for PIDs

Could someone provide a good reference to look up the existence and uniqueness of Smith Normal Form (SNF) for a PID?

I have seen it done for Euclidean domains but not for a PIDs. I know the difference is not much, but I would like a good reference. Thanks.

答:
See Jacobson's Basic Algebra I, chapter 3.

See also the book Finitely Generated Abelian Groups and Similarity of Matrices over a Field by Christopher Norman.

Title: Smith Normal Form for PIDs


问:
Question: are quantum computers a type of Turing machine or something else?
I read that in quantum computations you can not examine the internal state of the computation while it is happening. This is something I thought is always possible with Turing machines. Does this mean that quantum computers are not Turing machines? Also as an extra question: If not then what languages do they recognize in the sense of the hierarchy of language grammars?</p>

答:
The Quantum Turing Machine (see, https://en.wikipedia.org/wiki/Quantum_Turing_machine) is indeed a universal model of Quantum Computing. However, the Quantum Circuit model is the more commonly used model.

We cannot ascertain the state of our quantum computation until we measure the state. However, the action of taking a measurement may impact the state. So when analyzing quantum algorithms, we are interested in the probabilities of ending up in certain states upon measurement. In the circuit model, the updated values percolate through the circuit as gates are applied. Measurement operators are treated as gates in the circuit model. For these reasons, the circuit model tends to be a more natural starting point than the Quantum TM.


问:
Sum of squares of natural numbers and geometric series
$$S_n = 0\cdot4^0 + 1^2\cdot4^1 + 2^2\cdot4^2 + 3^2\cdot4^3 + \ldots + n^2\cdot4^n$$

ie- $S_n =  \sum\limits_{i=0}^ni^2\cdot4^i$.

I need help with this problem. I integrated it using ILATE rule and found the leading term of this series of the order $O(n^2\cdot4^n)$. However, I want the exact sum of this series since I want the co-efficient of the leading term of this series.

I am working on a recurrence relation problem and after unfolding the recurrence relation, I got this series. However, I am not able to think of an approach to solve this series.

答:
Consider
$$S_n(x) =  \sum\limits_{i=0}^n i^2x^i= \sum\limits_{i=0}^n\big[ i(i-1)+i\big]x^i$$
$$S_n(x) =x^2 \sum\limits_{i=0}^n i(i-1)x^{i-2}+x\sum\limits_{i=0}^n ix^{i-1}$$
$$S_n(x) =x^2 \left(\sum\limits_{i=0}^n x^i \right)''+x\left(\sum\limits_{i=0}^n x^i \right)'$$
$$\sum\limits_{i=0}^n x^i=\frac{x^{n+1}-1}{x-1}$$  When done, let $x=4$.


问:
Isogonality in a complete quadrilateral

If $ABCD$ is a quadrilateral inscribed in circle $\omega$ with $E=AD\cap BC$ and $M,N$ midpoints of $AC$ and $BD$, prove that $EM$ and $EN$ are isogonal with respect to $AEB$. (Otherwise show that$\angle DEN=\angle MEC$).

I hunt for geometry properties as recreational activity and while drawing diagram I was noticed that this appears true. I attempted the use of Newton's Line through $MN$, to no avail. I am not that great at projecting geometry, please help! Thanks to all who can help.

答:
Observe that, $\triangle ACE\sim \triangle BDE$.

$\Rightarrow   \frac{CA}{CE}=\frac{BD}{DE}$

$\Rightarrow   \frac{CM}{CE}=\frac{ND}{DE}$      [Dividing both sides by 2]

$\Rightarrow   \triangle MCE\sim \triangle NDE$

$\Rightarrow   \boxed {\angle MEC=\angle DEN}$

Title: Isogonality in a complete quadrilateral


答:
Let $f(x)$ be the above expression. Since $f(x)$ is linear between integers, the minimum will occur at an integer. Since
$f(-x)= f(x),$ the minimum occurs at a nonnegative integer $n.$ Since
$f(0)=f(1)=6$ and $f(n) \ge 7$ for $n \ge 2,$ the minimum value is $6.$


问:
solving $\sqrt{3-\sqrt{3+x}}=x$.
Can we solve the following equation in $\mathbb{R}$ without expanding it into a fourth degree equation :

$$ \sqrt{3-\sqrt{3+x}} = x.$$

squaring both sides and squaring again is the only thing I could done, If you have any other idea just post hints.

答:
It's easy to guess that the answer is $x=1$.

However, one should prove that this is the only root. Here is how.

Function $ \sqrt {3- \sqrt {3+x} } $ is strictly decreasing.

Function $x$ is strictly increasing.

It is proven (ask if you need a proof), that a strictly increasing  and a strictly decreasing functions can cross at no more than 1 point.

Since we have a point of $x=1$, then there are no other roots for this equasion.

Title: solving $\sqrt{3-\sqrt{3+x}}=x$.


问:
Prove q-Vandermonde Identity using a lattice path counting argument
I have this problem, from Aigner's A Course in Enumeration:

Vandermonde's formula for the $q$-binomial coefficients is
      $${{n+m}\brack {p}}_{q} = \sum_{k=0}^p
q^{(m-k)(p-k)}{{m}\brack {k}}_{q}{{n}\brack {p-k}}_{q}$$
      Prove this formula by a lattice path counting argument.
      (Hint: Count lattice paths from $(0,0)$ to $(p,m+n-p)$, and consider the intersection between such paths with the line $x+y=n$.)

I've previously solved this for the standard Vandermonde identity (not the q-analog), but I'm having trouble generalizing this to the q-analog. I understand this proof, could I perhaps using a similar argument to prove this?

答:
Recall the $q=1$ version first. We know $\binom{n}{k}$ counts the lattice paths $(0,0)\to(k,n-k)$. Then

$$ \binom{m+n}{k}=\sum_{r+s=k}\binom{m}{r} \binom{n}{s}. $$

We can group the lattice paths $(0,0)\to(k,m+n-k)$ according to where they intersect the diagonal line through the points $(0,m)$ and $(m,0)$ - namely, which paths go through $(r,m-r)$ for various $0\le r\le m$.

Such paths can be split in two parts, those $(0,0)\to(r,m-r)$ and $(r,m-r)\to(k,m+n-k)$; the second-part paths can be shifted into paths $(0,0)\to(s,n-s)$ where $r+s=k$. Thus, the number of such paths is $\binom{m}{r}$ times $\binom{n}{s}$.

Now in the general case, the $q^b$ coefficient of $\left[\begin{smallmatrix} n \\ k\end{smallmatrix}\right]_q$ counts lattice paths $(0,0)\to(k,n-k)$ which enclose $b$ blocks. Again paths $(0,0)\to(k,m+n-k)$ split into first part $(0,0)\to(r,m-r)$ and a second part $(r,m-r)\to(k,m+n-k)$. The second path sits over a $(m-r)\times s$ rectangle. Therefore

$$ \left[\begin{matrix} m+n \\ k \end{matrix}\right]_q=\sum_{r+s=k} q^{(m-r)s} \left[\begin{matrix} m \\ r \end{matrix}\right]_q\left[\begin{matrix} n \\ s \end{matrix}\right]_q $$

by summing $q^{b(P)}=q^{(m-r)s}q^{b(P_1)}q^{b(P_2)}$ over all paths $P$ from $(0,0)\to(m+n-k)$ split into first part $P_1$ from $(0,0)\to (r,m-r)$ and second part $P_2$ from $(r,m-r)\to(k,m+n-k)$.


问:
Repeating digits in pi
Running an experiment to see how many times different integers of different lengths repeat themselves in the first million digits of pi

(searched number: how many times it appears)

When searching 1 digits:

1: 99758
2: 100026
3: 100230
4: 100230
5: 100359
6: 99548
7: 99800
8: 99985
9: 100106

When searching random 2 digits:

22: 9145 71: 10095 47: 10043 56: 10010 33:9125

When searching for random 3 digits:

742: 985 349: 1063 117: 1016 634: 988. 333:893

When searching for random 4 digits:

7562: 106 1974: 117 9255: 99 1213: 103 3333:94

When searching for random 5 digits:

12137: 11 32464: 5 67347: 11 87271: 7 33333:8

When searching for random 6 digits:

276582: 1 895732: 2 674215: 1 715627: 1 333333: 1

When searching for random 7 digits:

7689123: 0 4829544: 1 7928212: 1 5241928: 0 3333333: 1

When searching for random 8 digits:

83782749: 0 26372925: 0 53629572: 0 829471210: 0 33333333: 0

I am aware that it is just an heuristic but is it conjectured that there is a limit to how many digits can repeat themselves, and why does this "pretty" even distribution occurs?

答: This probably relates to the conjecture that $\pi$ is a normal number in base $10$, meaning that any $p$-digit number in the sequence of digits of $\pi$ has the same asymptotic density $1/p$ in base $10$. This, however, is still unproven, although widely believed.

Title: Repeating digits in pi


问:
Square root of a $2\times 2$ matrix with $Det(A)=Tr(A)=0$

Does there exist a matrix $A_{2\times 2}$ with real entries which has a square root and $Det(A)=Tr(A)=0$ ?

Let A be a $2 \times 2$ matrix with real entries, then a square root of the matrix is given by $$M = \frac{1}{t}(A+sI)$$ where $s^2 = det(A)$ and $t^2 = Tr(A)+2s$ (Square root can have complex entries). This follows, as $M^2 = \frac{1}{t^2}(A^2+2sA+s^2I) $ $= A + \frac{1}{t^2}(A^2-Tr(A)A+det(A)I) = A$. However, can we find a square root of $A$ if $t=0$ ? If $t=0$, that would imply both $Tr(A)-2\sqrt{Det(A)}$ and $Tr(A)+2\sqrt{Det(A)}$ are zero. Which implies $Det(A)=0$ and $Tr(A)=0$. I couldn't find such an example.

答:
Hint

The characteristic polynomial of $A$ is
$$
X^2-0X+0
$$
Show that if $B$ is a square root of $A$, then the characteristic polynomial of $B$ is $X^2$.
But then
$$
A=B^2=0
$$


问:
Is the Arabic "4" a value, numeral, or numerical expression?

What I understand:

A number is an abstract concept. We use numerals such as names (i.e. cuatro), symbols (i.e. Arabic "4"), and numerical expressions (i.e. "9 - 5") to represent numbers.

Where I get tripped up:

"Value" is defined as the number a numeral represents, where "4" is a value of "3 + 1". Doesn't this contradict the author's earlier definition of "4" being a symbol?

Is "4" a number or a numeral?

Bonus Question

"4" is then defined as a numerical expression.
Is a numerical expression a type of numeral, or are they synonymous to one another?

I value your insight!

答: The document is using the convention that "4" is a numeral, whereas $4$ [without quotations] represents a value. The problem with abstract ideas is that you need an abstract keyboard to type it out.


问:
Conditional Probability wording of question
My problem is in understanding the wording: Given a group of boys and girls- 2 students are selected at random one after another, without replacement. Do the statements: "probability that 1st student is male and 2nd student is female" mean the same as "probability of selecting a female student given that 1st is male." Would the implications be any different if the same selection was done with replacement?</p>

答:
No, in general $P(B \mid A)$ (probability of $B$ given $A$)
is not the same thing as $P(A \cap B)$ (probability of $A$ and $B$).

The answers to various probability questions will change, depending on whether you select students with or without replacement.
But you cannot treat $P(B \mid A)$ and $P(A \cap B)$ as if they were the same thing in either case.


问:
I don't understand how this concentration problem is modeled.
This is the problem:

Ignore the part about inverses. I got that. I am confused why C is modeled this way. I don't get where the 100 + n came from. When I look up concentration and mixture problems, I get problems that are not represented as rational functions, like this:

Can someone walk me through how this function was created from the information provided?

答:
The formula makes more sense if it's written like this.
$$C = \frac{0.2(100) + 0.4n}{1.0(100)+1.0n}$$


问:
If U=F(x,y,z), and z=f(x,y), then find the formula of $\frac{∂^2U}{∂x^2}$ in terms of derivatives of "F" and derivative of "z" repectively.
I have been attempting this question in many ways but I haven't yet reached any conclusion

As we are already given to find $\frac{∂^2U}{∂x^2}$

so I know that $\frac{∂U}{∂x}$=$\frac{∂F}{∂x}$
but how should I represent it as a derivative of z ?

答:
One thing that is to be noticed is that $z$, which is one of the "independent" variables on which the value of $U$ depends, it is itself a function of variables $x$ and $y$. Thus, if we are to write $\delta U/\delta x$ in terms of derivatives of $F$, we get
$$\frac{\delta U}{\delta x}=\frac{\delta F}{\delta x}+\frac{\delta F}{\delta z}\cdot\frac{\delta z}{\delta x},$$
which further gives,
$$\frac{\delta^2 U}{\delta x^2}=\frac{\delta^2 F}{\delta x^2}+\frac{\delta^2 F}{\delta x\ \delta z}\cdot\frac{\delta z}{\delta x}+\frac{\delta F}{\delta z}\cdot\frac{\delta^2 z}{\delta x^2}.$$
Thus, in standard notation, this can also be written as
$$U_{xx}=F_{xx}+F_{zx}\cdot z_x+F_z\cdot z_{xx}.$$

A simple way to go about finding partial derivatives of a dependent variable is to prepare a little chart to show variable relationships. So, in this case, we have that $U$ is a function of $x$, $y$ and $z$. So, we write $U$ at the extreme right of the diagram, and then write $x$, $y$ and $z$ on a (roughly) vertical line to the left of where we wrote $U$ earlier. We draw arrow-headed lines connecting each of $x$, $y$ and $z$ to $U$, showing that $U$ depends on these variables. Now, we have been given in the question that $z$ itself further depends upon $x$ and $y$. To account for that, we write $x$ and $y$ again to the left of $z$, and draw lines connecting $x$ and $y$ to $z$. This shows that $z$ depends upon $x$ and $y$. Now, the task of finding partial derivative gets reduced to finding all the "paths" in this diagram from $U$ back to $x$, and using the chain rule along each. We get two paths connecting $x$ to $U$: one directly, and one through $z$. The direct path gives us $\dfrac{\delta F}{\delta x}$ and the other one through $z$ gives us $\dfrac{\delta F}{\delta z}\cdot\dfrac{\delta z}{\delta x}$. We then simply add them up.


问:
Basis of a linear transformation
If $V$ and $W$ are vectorial spaces of finite dimension, we denote $\mathscr{L}(V,W)$ like the set of all linear maps of $V$  in $W$. Let a linear map $L\in \mathscr{L}(\mathbb{R}^{2},\mathbb{R}^{2})$ defined like $L(x,y)=(-x+y,2x+y)$ for each $(x,y)\in \mathbb{R}^{2}$. Find a basis $S=\{u_{1},u_{2}\}$  of $\mathbb{R}^{2}$ such that
$$[L]_{S} := M_{L}^{S \rightarrow S}= \begin{pmatrix} 0 & -1 \\ -3 & 0 \end{pmatrix}$$
Where $M_{L}^{S \rightarrow T}$ denote the matricial representation of the linear map $L\in \mathscr{L}(V,W)$ respective to the basis $S$ of $V$ and $T$ of $W$.

Let $u_{1}=(v,w)$,$u_{2}=(a,b)\in \mathbb{R}^{2}$. We have that the columns in the matricial representation are $L(u_{1})$ and $L(u_{2})$, so

$$L_{u_{1}}=(-v+w,2v+w)= (0,-3)^{T}$$
and
$$L_{u_{2}}=(-a+b,2a+b)=(-1,0)^{T}$$
Then, $v=w=-1$,$a=\frac{1}{3}$ and $b=-\frac{2}{3}$. Am I right?

答:
Notice that the linear map L can be represented by the matrix $A=\left(\begin{matrix}-1&1\\2&1\end{matrix}\right)$ such that $L(x)=Ax$. Using your representation of $u_1=\left(\begin{matrix}v\\w\end{matrix}\right), u_2=\left(\begin{matrix}a\\b\end{matrix}\right)$ to be the new basis, the following equations are gotten:

$\left(\begin{matrix}[Au_1]_\beta&[Au_2]_\beta\end{matrix}\right)= \left(\begin{matrix}0&-1\\-3&0\end{matrix}\right)$, where $\beta$ represents the new basis

This means

$\begin{cases}Au_1=0u_1+-3u_2\\ Au_2=-1u_1+0u_2\end{cases}$

Further elaborating,

$\begin{cases}w-v=-3a\\2v+w=-3b\\b-a=-v\\2a+b=-w\end{cases}$

$v=a-b, w=-2a-b$

Further, as mentioned in Cristoph's post, the basis vectors must be linearly independent. $u_1\ne cu_2\implies \frac v a\ne \frac w b \text{ or } bv-wa\ne0$

Picking $a=b=1,v=0,w=-3$, is a valid basis. Note that this basis is neither orthogonal nor are the vectors normalized. If you attempt to normalize the vectors, the transformation will not work due to the contraints in the equations above.

It is possible, however, to find an orthogonal basis. $\left(\begin{matrix}\frac 1 2 (13+5\sqrt {13})\\\frac 1 2 (13-\sqrt {13})\end{matrix}\right), \left(\begin{matrix}\sqrt {13}\\\frac 1 2 (-13-3\sqrt {13})\end{matrix}\right)$ is one such possible orthogonal basis. It is not possible to find an orthonormal basis. In fact, it is not possible to find a normalized basis i.e. where both vectors have euclidean norm 1.

Test transformation in standard basis and new basis:

$\left(\begin{matrix}-1&1\\2&1\end{matrix}\right)\left(\begin{matrix}1\\-2\end{matrix}\right)=\left(\begin{matrix}-3\\0\end{matrix}\right)$

$\left(\begin{matrix}0&-1\\-3&0\end{matrix}\right)\left(\begin{matrix}1\\1\end{matrix}\right)_\beta=\left(\begin{matrix}-1\\-3\end{matrix}\right)_\beta$

$\left(\begin{matrix}-3\\0\end{matrix}\right)=\left[\left(\begin{matrix}-1\\-3\end{matrix}\right)_\beta\right]_{\{e_1,e_2\}}$

Title: Basis of a linear transformation


问:
Power of adjacency matrix
What does ij element of the k-th power of adjacency matrix for a directed graph, i.e. A^k represent: number of paths with exact k length or number of paths with length k or less than k from i to j?</p>

答:
''number of paths with exact k length or number of paths with length k or less than k from i to j?''

Number of paths of length $k$ between nodes $i$ and $j$ in a graph given by its adjacency matrix.

Title: Power of adjacency matrix


问:
Prove by contradiction that there is an $i \in [n]$ such that $x_i \geq 2$ if $x_1,\ldots,x_n \in \mathbb{N} \cup \{0\}$
Prove this statement using a proof by contradiction:

Let $n$ be a natural number. If $x_1,\ldots,x_n \in \mathbb{N} \cup \{0\}$ and $\sum_{i=1}^{n}{x_i} = n+1$ then there is an $i \in [n]$ such that $x_i \geq 2$

I'm not sure how to approach this problem with the proof by contradiction. So far, I assume that there is no $i \in [n] : x_i \geq 2$. Then, $\sum_{i=1}^{1}{x_i} < 2$. Finally, because $\sum_{i=1}^{1}{x_i} = n + 1 = 2$, I can conclude that by contradiction this is false.

I'm not entirely sure if this logic is a valid proof by contradiction or what a better proof would be. Any help is appreciated

答:
If $x_i\leq 1$ for all $i$, then $x_1+\ldots+x_n\leq 1+\ldots +1 = n\cdot 1 = n$.

Contraposition:
If $x_1+\ldots+x_n> n$, there is an $i$ with $x_i\geq 2$.

$P\Rightarrow Q \Leftrightarrow \neg Q\Rightarrow\neg P$.


问:
How is the uniform boundedness principle compatible with this seemingly weak convergent sequence?
In showing that $(x_i\rightharpoonup x)\not\Rightarrow(x_i\to x)$ or similar noncorallaries, one frequently uses the counterexample
$$
  (u_i)_{i\in\mathbb{N}}\in \ell^2\colon \quad u_i = (\underbrace{0,\ldots,0}_{i-1},1,0,\ldots)
$$
This, if I can trust my professor, is a weak convergent sequence in $\ell^2$, because it is bounded (I apparently missed that bit, this was the cause of the confusion) and for each $\varphi_j=(0,\ldots,1,0,\ldots)$, there exists an $N\in\mathbb{N}$, namely $N=j$, such that for every $i>N$,
$$
  \langle u_i, \varphi_j \rangle_{\ell^2} = 0 + \ldots + 0\cdot1 + 0 + \ldots + 1\cdot 0 + 0\ldots = 0 < \varepsilon
$$
for all $\varepsilon>0$. Because the $\varphi_j$ form a complete orthonormal system in $\ell^2$, this is sufficient for $(u_i)_i$ to be weakly convergent.

The problem is now: consider the sequence
$$
  (v_i)_{i\in\mathbb{N}}\in \ell^2\colon \quad v_i = (\underbrace{0,\ldots,0}_{i-1},i,0,\ldots).
$$
You could show that this is weakly convergent in exactly the same way as I just did for $(u_i)_i$, but on the other hand, $(v_i)_i$ is obviously an unbounded sequence and according to the principle of uniform boundedness (which I don't quite understand) every weak convergent sequence is bounded.

So what's wrong here?

Well, as the answers pointed out that it was the requirement of $(u_i)$ being bounded which was necessary for using just the $\varphi_j$ to show weak convergence.

答:
The argument you attribute to your professor is incorrect.

Suppose we wish to show a sequence $(f_i)$ of vectors in $\ell^2$ (or any Hilbert space) is weakly convergent to some $f$.  It is not sufficient to show that $$\langle f_i, \phi \rangle \to \langle f , \phi \rangle \quad \text{ as } i \to \infty \quad (*)$$ for all $\phi$ in some complete orthonormal set.  Indeed, your counterexample $v_i$ shows that this is not sufficient.  Nor is it sufficient to show that (*) holds for all $\phi$ in some dense set.

However, if one knows a priori that the set $\{f_i\}$ is bounded in $\ell^2$ (i.e. there exists $M$ with $||f_i|| \le M$ for all $i$), then it is sufficient to show that (*) holds for all $\phi$ in some dense set, or even for all $\phi$ in some set whose linear span is dense.  Such as, for instance, a complete orthonormal set.  Proving this is a good exercise.

It can be shown, using an appropriate version of the uniform boundedness principle, that even without knowing $\{f_i\}$ is bounded, it is sufficient to verify that (*) holds for all $\phi$ in some nonmeager set.  However, nonmeager sets are not so easy to come by.  (In particular, the linear span of a countable set is always meager.)

So to give a better proof that the sequence $\{u_i\}$ converges weakly (to 0), one could first note that the sequence is bounded in $\ell^2$ norm, and then check that (*) holds for all the $\varphi_j$ you describe (and aren't the $u_i$ and $\varphi_j$ actually the same here?).  Actually, my preferred proof is to note that $u_i$ is itself an orthonormal set, and so for any $\phi \in \ell^2$ we have
$$\sum_i |\langle u_i , \phi \rangle|^2 \le ||\phi||^2 < \infty$$ by Bessel's inequality.  Since the sum converges we must have $\langle u_i , \phi \rangle \to 0$.


问:
Deeper question about Cayley's theorem
Cayley's theorem states that there exists an injective homomorphism $\phi:G \rightarrow S_n$ for any finite group with $|G|=n$.

Let $\pi: \phi(G) \rightarrow \mathbb{Z}_2$ be the sign map, sending even permutations to $0$ and odd permutations to $1$. If I compose the functions to get $\pi \circ \phi$, then the kernel of the homomorphism is still a normal subgroup of $G$. So, $\ker(\pi \circ \phi) \trianglelefteq G.$

Does this normal subgroup have any significance in group theory? I'm not sure if there would be a connection to a normal subgroup of elements of even order (I can think of a counterexample to this) or if there's even any remote connection with elements of even order for this group.

I also have speculated this means that simple groups must be isomorphic to a generating set of only even permutations. The idea of the proof would be to assume that the kernel of the composition of functions is equal to the identity. Thus, the nonidentity elements of $G$ must map to odd permutations. However, multiplying two odd permutations gives an even permutation. So, there must be a nontrivial element mapping to an even permutation.

If there's anything I should consider or if there are any mistakes in my argument, I'd love to hear it.

答:
Let $G$ be of order $n=2m$, $m$ odd, and $g\in G$ of order 2. then the left multiplication by $g$ is a product of $m$ transposition, so its signature is $(-1)^m=-1$.

In general if $g$ is of order $m$, and $k= {n\over m}$ the left multiplication is a product of $k$ cycles of length $m$, so its signature is $(-1)^k$ if $m$ is even, $1$ if $m$ is odd.

Title: Deeper question about Cayley's theorem


问:
Why contour integral is different from an approximate contour summation for complex functions with poles?
I am trying to calculate a contour integration of the form
$$
\frac{1}{2\pi i}\oint\limits_c f(z)\: dz
$$
numerically. I know the integral is the same as the sum over all the residues of the function on the poles within the contour $c$
$$
\frac{1}{2\pi i} \oint\limits_c f(z)\: dz = \sum_{z \in I} Res(f(z),z)
$$
$I$ is the set of poles within the contour $c$.

For simple poles, the residues can be calculated in a fairly simple manner like given in the answer, but for higher-order poles, this becomes very hard numerically as it involves higher-order
derivatives of $z$ (is there any other simple way without derivatives?) and calculating the contour integration directly rather seems like less of a daunting task to do. For example, say
$$
f(z) = 1/z
$$
The pole is at $z=0$ with residue $1$. So $\frac{1}{2\pi i} \oint\limits_c f(z)\: dz  = 1$.
I am choosing a dimond shaped contour centered at the origin like this,

And doing numerical integration in Mathematica

```<code>NIntegrate[1/(2 Pi I) 1/z, {z, 1, I, -1, -I , 1}]
</code>```

The result is $ 0.9999999999999979 - i\: 2.7755575615628914\times 10^{-17} \approx 1 $ as expected.

Till now everything is fine, the numerical result matches up to the analytical result up to machine precision.

Now, I am trying to do the same but using summation on the contour. I am approximating the contour integral as
$$
\frac{1}{2\pi i} \oint\limits_c f(z)\: dz \approx \frac{dz}{2 \pi i}\sum_{z \in c} f(z)
$$

I am parametrizing the curve as $p(t)$ as $t$ goes from $0 \to 1$ the contour is traversed once.
$$
p(t) =
\begin{cases}
1-4t+i\:4t &\mbox{ } 0 \leq t \leq \frac{1}{4} \\
1-4t+i\:(2-4t) &\mbox{ } \frac{1}{4} \leq t \leq \frac{1}{2} \\
4t-3+i\:(2-4t) &\mbox{ } \frac{1}{2} \leq t \leq \frac{3}{4} \\
4t-3+i\:(4t-4) &\mbox{ } \frac{3}{4} \leq t \leq 1
\end{cases}
$$

The summation is
$$
\frac{dz}{2 \pi i} \sum_{t=0}^{1} f(p(t)) =
\frac{dz}{2 \pi i} \sum_{t=0}^{1} \frac{1}{p(t)}
$$

```<code>path[t_] = Piecewise[{{{1 - 4*t, 4*t}, 0 <= t <= 1/4}, {{1 - 4*t, 2 - 4*t},
   1/4 <= t <= 1/2}, {{-3 + 4*t, 2 - 4*t}, 1/2 <= t <= 3/4},
     {{-3 + 4*t, -4 + 4*t}, 3/4 <= t <= 1}}, 0]

dz = 0.0001;
data = Sum[
  zi = #[[1]] + I #[[2]] &@path[t];
  dz*1/(2 Pi I) 1/zi
  , {t, 0, 1, dz}]
</code>```

the path function is the aforementioned $p(t)$

And the result in this case is $-6.009953532397927\times 10^{-18} - i \: 0.00001591549430919666 $. As I make dz even smaller the result doesn't improve much. This way off from the actual calculation. The numerical integration does much more than a vanilla sum, but as the spacing decreases, I would think the results would match.

NIntegrate in Mathematica by default uses some adaptive non-uniform sampling algorithm that samples the function at certain points and does a recursive bisection followed by a summation around that point until convergence up to certain decimal order is achieved.

Why are the results so much different?

答:
Let $\gamma: [0, 1] \to \Bbb C$ be a parametrization of the contour $c$. For a “sufficiently fine” partition $0 = t_0 < t_1 < \cdots < t_n = 1$ we have
$$
 \int_c f(z) \, dz = \int_0^1 f(\gamma(z)) \gamma'(z) \, dz \approx \sum_{j=1}^{n} f(\gamma(t_j)) \gamma'(t_j)(t_j - t_{j-1}) \\
\approx \sum_{j=1}^{n} f(z_j) (z_j - z_{j-1})
$$
where $z_j = \gamma(t_j)$. In particular, the “increments” $z_j - z_{j-1}$ are not constant along the curve.


答:
$\sqrt{x^2} = x$ if $x \geq 0$. Now try a negative number, like $x = -5$. Then
$$\sqrt{x^2} = \sqrt{(-5)^2} = \sqrt{25} = 5.$$
So when $x < 0$, it will become positive once we square; that is, $x^2 > 0$. Taking the square root of $x^2$ yields $-x > 0$.

In general, $\sqrt{x^2} = |x|$ for every real number $x$.

Title: why is $\sqrt{x^2} \ne x$?

Body:
If $\sqrt{3^2}= 3$, $\sqrt{2^2}\ne 2$.

why is $\sqrt{x^2}\ne x$?


答:
The short answer is that $x^{2}$ takes a negative number to a positive number, so $\sqrt{x^{2}}$ also takes a negative number to a positive number. Take as an example $-1$:

$$\sqrt{(-1)^{2}} = \sqrt{1} = 1$$

Title: why is $\sqrt{x^2} \ne x$?


问:
Is every closed subset of some ambient complete space complete?
https://math.stackexchange.com/questions/244661/showing-that-if-a-subset-of-a-complete-metric-space-is-closed-it-is-also-comple#:~:text=If%20A%20is%20a%20closed,Thus%20A%20is%20complete.

We know that if X is complete and its subset A is closed, then A is complete in a metric space. I assume this holds true in non-metric space because in the link above the proof didn't seem to use the fact that X is metric space. But I can't see the official theorem for this in any source without the mention of metric space. Is this really true?

答:
As to the linked question: The fact that you're even using the term complete (and Cauchy sequences) implies that we're working in a metric space and "metric complete" is the being notion used. The titel is a strong hint. Plus the fact that closedness can be decided by convergent sequences is already using the metric implicitly. So the proofs there do use (part of) the metric.

It's also possible to define completeness in any uniform space, which is a different kind of struture from a topology,though related, as any uniform structure induces a topology (like a metric does). For that notion of completeness the result (a closed subset of a complete structure is again complete ) is also true, and might be the general fact you've seen. A complete metric space has a complete uniform structure.

There also is a notion of being "topologically complete", which is also used sometimes, and is also preserved by closed subsets, and is a class of spaces in which Baire's theorem holds, and also includes locally compact Hausdorff spaces, as well as completely metrisable spaces.


问:
Question on convolution in measure theory to probability theory and the need for independence
In a classical measure theory context, with two measures $\mu_{1},\; \mu_{2}$ on $\mathcal{B}(\mathbb R)$ as the image of the product measure $\mu_{1}\otimes \mu_{2}$ under the map $+:\mathbb R\times \mathbb R \to \mathbb R$ is defined as a convolution. When switching over to probability theory, I notice that we also introduce the notion of independence, i.e. for independent random variables $X$ and $Y$ the convolution is the distribution of the sum $X+Y$.

Question: Why do we need the notion of independence for the random variables? Surely by the definition of the convolution in measure theory it immediately follows that we are evaluating the sum of the random variables? Or am I missing something obvious? I guess the notion of not knowing what the product measure looks like could be the pitfall. Thanks in advance!

答:
Remember that for any probability measure $\mu$ on $\mathcal B(\mathbb R)$ and measurable map $f:\mathbb R\to\mathbb R$, if $\mu$ is the distribution of a random variable $Z$, then the image of $\mu$ by $f$ is the distribution of $f(Z)$.

Here if $\mu_1\otimes\mu_2$ is the distribution of $Z$, then the convolution $\mu_1\star\mu_2$ is the distribution of $+(Z)$. Denoting by $X$ and $Y$ the marginals of $Z$, that is the random variables such that $Z=(X,Y)$, we have that if $(X,Y)\sim\mu_1\otimes\mu_2$, then $X+Y\sim\mu_1\star\mu_2$. But saying that $(X,Y)\sim\mu_1\otimes\mu_2$ is equivalent to say that $X$ and $Y$ are independent (it can even be taken as a definition of independence).

Indeed, you probably saw that $X$ and $Y$ are independent iff for any measurable Borel sets $A,B\in\mathcal B(\mathbb R)$, $\mathbb P(X\in A,Y\in B)=\mathbb P(X\in A)\mathbb P(Y\in B)$. Denoting by $\mu_1$ the distribution of $X$, $\mu_2$ the distribution of $Y$ and $\mu$ the distribution of $(X,Y)$, the latter equality can be rewritten as
$$
\forall A,B\in\mathcal B(\mathbb R),\quad\mu(A\times B)=\mu_1(A)\mu_2(B),
$$
which means that $\mu=\mu_1\otimes\mu_2$.


问:
Integrating vertically giving different answer from intergrating horizontal

Find area in the region: $$ \frac{1}{x} <y < x^{\frac13}$$ , $1
<x<8$

So, I thought of integrating 'vertically' by:

$$ \int y dx = xy - \int x dy =16- \left[ \int_1^2 y^3 + \int_0^1 \frac{1}{y} dy\right] $$

Now, the problem is that $ \int_0^1 \frac{dy}{y}$ part, it seems to blow up. So, this method seems unfeasible, is there are any reason I could 'remove' the blow up and what exactly is the property of the function which causes this to occur?

Note the integral $ \int_0^1 \frac{1}{y} dy$ represents this following area:

答:
Ok, I basically screwed up the figure,  the figure in the question goes from $0 $ to $8$ but we want from $1$ to $8$

The area of rectangle is given as $2 \cdot 8$

The blue area is given as:

$$ \int_1^2 y^3 dy $$

And the purple given as:

$$ \int_{\frac18}^1 \frac{1}{y} dy$$

Putting it together

$$ \text{required area} = 2 \cdot 16 -  \int_1^2 y^3 dy - \int_{\frac18}^1 \frac{1}{y} dy$$

And this gives the same as David P's answer


问:
$X$, an uncountable set, with the cofinite topology is not first-countable?
Is the following a valid proof that $X$, an uncountable set, with the cofinite topology is not first-countable?

By way of contradiction, suppose that $\mathscr{B}_x=\left\{B_n:n\in\mathbb{N}\right\}$ is a countable local base of open sets at the point $x$.

Since $X$ is $T_1$ and $\mathscr{B}_x$ is a local base at $x$,

$$\bigcap_{n\in\mathbb{N}} B_n = \{x\}.$$

By De Morgan,

$$X\setminus\bigcap_{n\in\mathbb{N}} B_n = X\setminus\{x\}$$

$$\bigcup_{n\in\mathbb{N}} X\setminus B_n = X\setminus\{x\}.$$

As $B_n$ is open the set $X\setminus B_n$ is finite, hence, it is countable. Notice, we have a countable union of countable sets, hence, a countable set. Therefore, it must be that $X\setminus\{x\}$ is countable, which implies $X$ is countable, a contradiction.

答:
Using $\bigcap_n B_n = \{x\}$ is not directly necessary, but is OK in your proof. So your solution is fine as soon as you can justify this intersection from $T_1$-ness (some earlier result in your text presumably).

You could also argue: $\bigcup_n X\setminus B_n$ is at most countable, so we can pick $y \in X$ that is not in this union and unqual to $x$, as $X\setminus \{x\}$ is uncountable. Then $U= X\setminus \{y\}$ is open (definition of cofinite topology, it has finite complement) contains $x$ and so must contain some $B_m$. But then $y$ was chosen to be not in $X\setminus B_m$, so $y \in B_m$ and we cannot have $B_m \subseteq X\setminus \{y\}$, contradiction.


问:
Solve recurrence equation
Could you Show me how to solve this equation:

$$x_n = \sqrt2x_{n-1} + \sqrt3$$

for $n \ge 1$ with $x_0 = 1$.

答: Hint: find a $b\in \mathbb{R}$ such that $x_n-b=\sqrt{2}(x_{n-1}-b)$, then you get a Geometric series.</p>

Title: Solve recurrence equation


问:
If $CC^*$ = $DD^*$ then C=DQ
If C and D be two m by l matrices and $CC^*$ = $DD^*$ then $C=DQ$ where $Q$ is a l by l unitary matrix .
If C and D  are  of order 1 then it is trivial  But for general case I was trying to use induction .
But I had faced slide problems . How can I do this for general case.

答:
from SVD we get C = $U\Sigma V^*$ and D = $U_1\Sigma_1 V_1^*$ and so $CC^*$ = $DD^*$
and $U\Sigma V^*$$V\Sigma^* U^*$ = $U_1\Sigma_1 V_1^*$$V_1\Sigma_1^* U_1^*$
and $\Sigma\Sigma^*$ = $Q_1\Sigma_1\Sigma_1^*Q_1^*$ where $Q_1$ is a unitary matrix.
But we have to show that C = DQ i.e., $\Sigma$ = $Q_2\Sigma_1 Y$ where $Q_2$ and Y are unitary matrix .
How can I do this.

Title: If $CC^*$ = $DD^*$ then C=DQ


问:
Differential equation of first order — omitting absolute value. Why is it omitted and when should I omit it?
I have the following simple exercise:
$$ \text{Solve: } y' = 2(x - xy) $$

The equation is the same as:

$$\frac{dy}{1-y} = 2x \ dx$$

Integrating both sides, we have:

$$-\text{ln}(|1-y|) + C_1 = x^2 + C_2 $$

So:

$$ |1-y| = e^{-x^2-C} $$

The expected result is:

$$y = e^{-x^2-C} + 1$$

Why is the absolute value omitted? It seems $|1-y|$ becomes $y-1$.

Thank you!

NOTE: I read a few posts here, about this, but I can't understand on this case, why the absolute value is omitted.

答:
Your integration is correct, $|1-y|=e^{-x^2-C} $ is indeed the (implicit) solution. However, note that due to the $|\cdot|$, we actually end up having two families of curves.

Note first that $e^{-x^2-C}= e^{-C}e^{-x^2}=Ke^{-x^2}$, where $K=e^{-C}$ is an arbitrary constant.

Another piece of information is that:

$$Ke^{-x^2}=|\pm(Ke^{-x^2})|$$

Use this to conclude that $|1-y|=|\pm(Ke^{-x^2})|$, so the solutions in $y$ are:

$$1-y=Ke^{-x^2} \implies y=-Ke^{-x^2}+1$$
$$1-y=-Ke^{-x^2} \implies y=Ke^{-x^2}+1$$

Since $K$ is an arbitrary constant, the $-$ sign can be absorbed. We can in fact say that $M=\pm K$ with $y=Me^{-x^2}+1$. We can then drop the absolute value sign after collecting constants.

Note: If an initial condition is specified, we would then have to worry about the constants and whether it is $+$ or $-$.


问:
Counting problem, number of moves
In a game, you are standing somewhere and a move consists of either moving forward any number of steps, not moving at all, or moving $1$ step backwards. If you complete $n$ moves, find the number of sequences that end with you in your starting position and which never go behind your starting position.

If the "never go behind the starting position" condition weren't there, the answer would just be $\binom{2n-1}{n-1}$. However I'm not on sure how to proceed from here.

答:
The answer is given by the Catalan numbers $C_n=\frac{1}{n+1}\binom{2n}{n}$.

A combinatorial reduction of the present problem to the interpretation of $C_n$ as the number of "balanced" strings containing $n$ pairs of parentheses is as follows. Given a sequence of $n$ moves, we transform each move (i.e. an integer) $m\geqslant-1$ into a string of $m+1$ opening parentheses, followed by a closing parenthesis: $$-1\mapsto\color{blue}{\texttt{)}},\quad 0\mapsto\color{blue}{\texttt{()}},\quad 1\mapsto\color{blue}{\texttt{(()}},\quad 2\mapsto\color{blue}{\texttt{((()}},\quad 3\mapsto\color{blue}{\texttt{(((()}},\quad\ldots$$ Clearly, this transformation is invertible (and one-to-one).

There's also an "assume nothing known" approach using generating functions. Let $c_{n,k}$ be the number of sequences of $n$ moves that take you $k$ steps forward (here $n,k\geqslant 0$). Then $$c_{0,k}=[k=0]=\begin{cases}1,&k=0\\0,&k\neq 0\end{cases};\qquad c_{n+1,k}=\sum_{j=0}^{k+1}c_{n,j}.$$

Now consider the "generating function" $g(x,y)=\sum_{n,k\geqslant 0}c_{n,k}x^n y^k$ as a formal power series (to stay ignorant of convergence issues; we could also consider it as an ordinary function, but then we would need to show its analyticity around $(0,0)$, for the "composition" argument below to be justified). Let $g(x):=g(x,0)$ (thus, $g(x)=\sum_{n\geqslant 0}c_{n,0}x^n$ is the "generating function" of the numbers we actually need). Then

\begin{align*}
\frac{g(x,y)-1}{x}&=\sum_{n,k\geqslant 0}c_{n+1,k}x^n y^k
\\&=\sum_{n,k\geqslant 0}x^n y^k\left(c_{n,0}+\sum_{j=0}^k c_{n,j+1}\right)
\\&=\sum_{n\geqslant 0}c_{n,0}x^n\sum_{k\geqslant 0}y^k+\sum_{n,j\geqslant 0}c_{n,j+1}x^n\sum_{k\geqslant j}y^k
\\&=\frac{1}{1-y}\left(\sum_{n\geqslant 0}c_{n,0}x^n+\sum_{n,j\geqslant 0}c_{n,j+1}x^n y^j\right)
\\&=\frac{g(x)}{1-y}+\frac{g(x,y)-g(x)}{y(1-y)},
\end{align*}
$$\text{i.e.}\quad\big(y(1-y)-x\big)g(x,y)=(1-y)\big(y-xg(x)\big).$$

Now - the "composition" argument! - we put $y=xg(x)$ here (this is a formal power series with no constant term, so all the compositions are well-defined). The RHS becomes identically zero, and since $g(x,y)$ is not, we must have $y(1-y)=x$.

Choosing the solution $y$ that gives $g(x)$ as a power series, we get $$g(x)=\frac{1-(1-4x)^{1/2}}{2x}.$$

It remains to use the binomial series for $(1-4x)^{1/2}$, or just recognize this generating function.

Title: Counting problem, number of moves


问:
How to evaluate $\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$
How can I approach:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$$
I tried the usual differentiation under the integral sign but it didn't work so well.

I also tried to rewrite it the following 2 ways:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\int _0^1\frac{\ln \left(\sqrt{x}+x\right)}{1+x^2}\:dx-\frac{1}{2}\underbrace{\int _0^1\frac{\ln \left(x\right)}{1+x^2}\:dx}_{-G}$$
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\underbrace{\int _0^1\frac{\ln \left(1-x\right)}{1+x^2}\:dx}_{\frac{\pi }{8}\ln \left(2\right)-G}-\int _0^1\frac{\ln \left(1-\sqrt{x}\right)}{1+x^2}\:dx$$
Yet I'm still stuck with those other integrals, any help is appreciated.

Managed to numerically find that:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\frac{\pi }{16}\ln \left(2\right)+\frac{\pi }{4}\ln \left(1+\sqrt{2}\right)-\frac{1}{2}G$$
But I still dont know how to approach it.

答:
Substitute $t=\sqrt x$ in $I= \displaystyle\int _0^1\frac{\ln (1+\sqrt{x})}{1+x^2}\:dx$

\begin{align}
I &
=2\int _0^1\frac{t \ln \left(1+t\right)}{1+t^4}dt
\overset{t\to\frac1t}= \int _0^\infty\frac{t \ln \left(1+t\right)}{1+t^4}dt - \int _1^\infty\overset{t^2\to t}{\frac{t \ln t}{1+t^4}}dt \\
&=  -\frac12\int _0^\infty{\ln \left(1+t\right)}\>d(\cot^{-1}t^2) -\frac14\int_1^\infty\frac{\ln t}{1+t^2}dt\\
&= \frac12 \int _0^\infty\frac{\cot^{-1}t^2}{1+t}dt - \frac14 G\tag1
\end{align}
Let $J(a) = \displaystyle \int _0^\infty\frac{\cot^{-1}(at^2)}{1+t}dt$ and differentiate under the integral
\begin{align}
J’(a) &= -\int _0^\infty\frac{t^2}{(1+t)(1+a^2t^4)}dt\\
&= \frac1{1+a^2} \int _0^\infty\left(-\frac1{1+t}+\frac{a^2t^3}{1+a^2t^4}-\frac t{1+ a^2t^4}+\frac{1-a^2t^2}{1+a^2t^4}\right)dt\\
&=\frac1{1+a^2} \left(\frac12\ln a -\frac\pi{4a}-\frac{\pi(a-1)}{2\sqrt{2a}}\right)
\end{align}
Then, with $J(\infty) =0$
\begin{align}
& \int _0^\infty\frac{\cot^{-1}t^2}{1+t}dt  = J(1)=- \int_1^\infty J’(a)da\\
=&-\frac12\int_1^\infty \frac{\ln a}{1+a^2}da +\frac\pi4 \int_1^\infty \frac{1}{a(1+a^2)}da
 +\frac\pi2\int_1^\infty \frac{a-1}{\sqrt {2a}(1+a^2)}da\\
=& -\frac12G +\frac\pi8\ln2 + \frac\pi2\ln(1+\sqrt2)
\end{align}
Plug into $(1)$ to obtain
$$I= \frac{\pi }{16}\ln 2+\frac{\pi }{4}\ln \left(1+\sqrt{2}\right)-\frac{1}{2}G$$


问:
Probability of dice roll
Rolling a fair $6$ sided die $2n$ times.

let $X$ be the sum of even sides from the first $n$ rolls.

let $T$ be the number of times the $i^{th}$ roll have the same side as the $n+i$ roll.

for example :

let $n = 3 $ and the rolls are :

$ r_1 = 4  ,r_2 = 3 , r_3 = 2,r_4 = 6 , r_5 = 3 , r_6 = 1 $.

then $ T = 1$ , and $X = 4+2 = 6$.

My question :

What is the distribution of $T$.

let $n=30$ what is the value of $E(X^2)=?$.

I think that $T$ is a binomial distributed random variable because it represents the number of successes from $(n+1,2n)$ after getting randomly values for the first $n$ rolls.

答:
Indeed $T$ is the number of success among $n$ trials where the probability of success is $1/6$, so it should follow the binomial distribution $\mathcal B(n,1/6)$.

More formally, let $X_i$ be the side of the $i^{th}$ roll. Then $T=\sum_{i=1}^n1_{\{X_i=X_{n+i}\}}$. The random variables $(X_i,X_{n+i})_{1\le i\le n}$ are independent, and for all $i\in\{1,\cdots,n\}$, $1_{\{X_i=X_{n+i}\}}$ equals $1$ with probability $1/6$ and $0$ with probability $5/6$ hence it follows the Bernoulli distribution $\mathcal B(1/6)$. Then $T$ is the sum of $n$ independent Bernoulli variables with parameter $1/6$, so $T$ is a Binomial variable with parameter $(n,1/6)$.

For the second question, you have $X=\sum_{i=1}^nX_i1_{\{X_i\in\{2,4,6\}\}}$. Let $Y_i=X_i1_{\{X_i\in\{2,4,6\}\}}$. Clearly the distribution of $Y_i$ is given by
$$
\mathbb P(Y_i=2)=\mathbb P(Y_i=4)=\mathbb P(Y_i=6)=\frac16\quad\textrm{and}\quad\mathbb P(Y_i=0)=\frac12,
$$
hence
$$
\mathbb E[Y_i]=2\quad\textrm{and}\quad\operatorname{var}(Y_i)=\frac{16}{3}\cdot
$$
Since the variance of the sum of independent random variables is equal to the sum of the variances, we have
$$
\operatorname{var}(X)=\sum_{i=1}^n\operatorname{var}(Y_i)=\mathbb E[X^2]-\mathbb E[X]^2,
$$
hence
$$
\mathbb E[X^2]=\sum_{i=1}^n\operatorname{var}(Y_i)+\left(\sum_{i=1}^n\mathbb E[Y_i]\right)^2=\frac{16}{3}n+4n^2.
$$

Title: Probability of dice roll