问:
MAT Q5 1996 Solution Verification (Combinatorics)
I have been doing some MAT past papers and I have got to papers where there is no official mark scheme. I checked on the student room but there seem to be no answers for the very old mat past papers. Here is the question I want to answer:

My solution:

a): There could be two values for each of the $4\times4=16$ places. Thus the total number of different arrangements is $2^{16}$

b): Picture the $4$ rows and let us consider the cases: " one $\times$ in each row", two $\times$ in each row, "three $\times$ in each row" and "four $\times$ in each row".

In the first case, we have $(^4C_1)^4$ different arrangements.

In the second case, we have $(^4C_2)^4$ different arrangements.

In the third case, we have $(^4C_3)^4$ different arrangements.

In the fourth case, we have $(^4C_4)^4$ different arrangements.

And thus the total number of different arrangements is $=1809$ (adding all of the above).

c): Consider the cases

$$\begin{pmatrix}
  \times & \text{either} & \text{either} & 0 &\\
   \text{either}& \times & 0 & \text{either} & \\
   \text{either} & 0 & \times & \text{either} &\\
   0& \text{either} & \text{either} & \times & \\
\end{pmatrix}
\text{and}
\begin{pmatrix}
  0 & \text{either} & \text{either} & \times &\\
   \text{either}& 0 & \times & \text{either} & \\
   \text{either} & \times & 0 & \text{either} &\\
   \times& \text{either} & \text{either} & 0 & \\
\end{pmatrix}
$$
In each of the cases, we have $8$ available positions giving $2^8$ possible arrangements. We need only now consider the case when

$$\begin{pmatrix}
  \times & \text{either} & \text{either} & \times &\\
   \text{either}& \times & \times & \text{either} & \\
   \text{either} & \times &  \times &\text{either} &\\
   \times& \text{either} & \text{either} & \times & \\
\end{pmatrix}$$

in which case we have yet again $2^8$ different arrangements. Thus the total number of different arrangements is $3\times 2^8=768$

Is this correct?

EDIT: I have realised I have interpreted the question way more generally than I should have. I thought that for each position I have a nought and across available. Anyways, I would like to ask if my generalization is correct.

答:
Part (a) is wrong, because you are constrained by having only 4 crosses.

Therefore, the answer is $\binom{16}{4}.$

Part (b) is also wrong, for much the same reason.

There are 4 choices where to put the cross in each row.

Therefore, the answer is $4^4.$

Part (c) is trickier.

Going from the top down, you will have to place a cross in each row.

In the 1st row, you have 4 columns to choose from.

Having made that choice, in the 2nd row, you have 3 remaining columns to choose from.

Final answer here is $4!$.

Addendum

Assume that crosses and noughts are unlimited.

For part (a), your answer of $2^{(16)}$ is correct.

Part (b) is complex.  I assumed that you intended that each row has at least one cross, but that you can have a different # of crosses from one row to the next.

I got

$$2^{(16)} - \binom{4}{1}2^{(12)} + \binom{4}{2}2^8 - \binom{4}{3}2^4 + \binom{4}{4}2^0.$$

The above formula employs the principle of Inclusion-Exclusion, as discussed at https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle.

The idea is to start will all possible possible placements, deduct possible placements of all noughts in the 1st row, with the other 12 rows unconstrained, multiplying this by 4, since the blanked out row could be any of the 4 rows.

Then you add back possible ways of blanking out two rows, since this have been over-counted by previous running total.  Then you continue adding and then deducting, in accordance with Inclusion Exclusion.

My answer actually agrees with Phicar's answer.  This can be verified by doing binomial expansion on

$$(2^4 - 1)^4.$$

Addendum-1

Part C.

I again used Inclusion-Exclusion, but in a much more convoluted fashion.  This answer is very complicated to explain.  Consider the following chart.

$$
\begin{array}{| l | l c  l  c l c l c l |}
  \hline
  T_0 & \binom{4}{0}\binom{4}{0}\times 2^{(16)}
 \\[8pt]
  \hline
  T_1 & \binom{4}{1}\binom{4}{0}\times 2^{(12)}
 &+& \binom{4}{0}\binom{4}{ 1}\times 2^{(12)}\\[8pt]
  \hline
  T_2 & \binom{4}{2}\binom{4}{0}\times 2^{(8)}
 &+& \binom{4}{1}\binom{4}{ 1}\times 2^{(9)} &+& \binom{4}{0}\binom{4}{ 2}\times 2^{(8)}\\[8pt]
  \hline
  T_3 & \binom{4}{3}\binom{4}{0}\times 2^{(4)}
 &+& \binom{4}{2}\binom{4}{1}\times 2^{(6)} &+& \binom{4}{1}\binom{4}{2}\times 2^{(6)} &+& \binom{4}{0}\binom{4}{ 3}\times 2^{(4)}\\[8pt]
  \hline
T_4 & \binom{4}{4}\binom{4}{0}\times 2^{(0)}
 &+& \binom{4}{3}\binom{4}{1}\times 2^{(3)} &+& \binom{4}{2}\binom{4}{2}\times 2^{(4)} &+& \binom{4}{1}\binom{4}{3}\times 2^{(3)} &+& \binom{4}{0}\binom{4}{4}\times 2^{(4)}\\[8pt]
  \hline
T_5 & \binom{4}{4}\binom{4}{1}\times 2^{(0)}
 &+& \binom{4}{3}\binom{4}{2}\times 2^{(2)} &+& \binom{4}{2}\binom{4}{3}\times 2^{(2)} &+& \binom{4}{1}\binom{4}{4}\times 2^{(0)}\\[8pt]
  \hline
T_6 & \binom{4}{4}\binom{4}{2}\times 2^{(0)}
 &+& \binom{4}{3}\binom{4}{3}\times 2^{(1)} &+& \binom{4}{2}\binom{4}{4}\times 2^{(0)}\\[8pt]
  \hline
T_7 & \binom{4}{4}\binom{4}{3}\times 2^{(0)}
 &+& \binom{4}{3}\binom{4}{4}\times 2^{(0)}\\[8pt]
  \hline
T_8 & \binom{4}{4}\binom{4}{4}\times 2^{(0)}\\[8pt]
  \hline
\end{array}
$$

Out of the $2^{(16)}$ possible configurations, you have to compute how many of them will have a cross in very row and also a cross in every column.

I construe the 4 rows and 4 columns as 8 "straight paths".  In the chart, you will see entries that look like this:

$$\binom{4}{a}\binom{4}{b} \times 2^c.
$$

In the first factor, $a$ corresponds to how many rows (from 0 through 4) inclusive are presumed to be filled with noughts.  Similarly, in the second factor, $b$ corresponds to how many columns (from 0 through 4) inclusive are presumed to be filled with noughts.   The $c$ exponent represents the # of unconstrained cells when $a$ rows and $b$ columns are presumed to be filled with noughts.

As an example, consider the following entry from row $T_4$:

$$\binom{4}{2}\binom{4}{2} \times 2^4.$$

If there are 2 rows and 2 columns each filled with noughts, there will automatically be 4 unconstrained cells, namely the 4 cells that are not in either of the two rows or two columns.

So the above expression indicates how many ways there are of choosing 2 rows to fill with noughts, simultaneously choosing 2 columns to fill with noughts, and allowing the remaining cells to be unconstrained.

In effect, the expression indicates that there are 36 distinct sets, each with 2 rows and 2 columns designated to be filled with noughts.  Further, in each of the 36 sets, there are $2^4 = 16$ elements, corresponding to the possible cross/nought possibilities for the remaining (unconstrained) cells.

For $k \in \{0,1,2,\cdots,8\},$ the row labeled $T_k$ identifies all of the sets where the number of rows to be filled with noughts + the number of columns to be filled with noughts $= k.$

In each row, the intent is that the variable $T_k =$ a sum of terms, where each term has 3 factors.

The actual answer to the problem is
$$T_0 - T_1 + T_2 - T_3 + T_4 - T_5 + T_6 - T_7 + T_8.$$

The fundamental idea in Inclusion-Exclusion, as I am using it, is that every unsatisfactory configuration (out of the $2^{(16)}$ possible configurations) will be subtracted, added back, subtracted, ... so that the net effect is that the unsatisfactory configuration is deducted once.

Consider the following example:

Rows $1, 2,$ and $3$ will be filled with noughts and columns $1,2$ will be filled with noughts.  The 2 remaining cells will be filled with crosses.  Of the $2^{(16)}$ possible configurations, this  example, represents one of the unsatisfactory ones.

In order for the algorithm to function correctly, the net effect must be that this particular example is deducted one time from the $2^{(16)}$ possible configurations,

Examining the representation of this example with respect to
$T_1, T_2, \cdots, T_8$:

In the first term in $T_1,$ the example is represented 3 times, since there are three rows to be filled with noughts,  Similarly, In the second term in $T_1$, the example is represented 2 times.  Therefore, the example is represented (3 + 2 = 5) times in $T_1$.

In the first term in $T_2$ this example is represented 3 times.,  This is because in this example, 3 rows are to be filled with noughts, and there are 3 different ways of selecting two out of the three rows.

Similarly, this example is represented 6 times in the second term in $T_2$.  This is because there are 6 different ways ($3 \times 2$) of choosing 1 row and 1 column.

Simlarly, this example is represented once with respect to the third term in $T_2$.

Therefore, this example is represented (3 + 6 + 1 = 10) times in $T_2.$

Using Similar analysis, this example is represented (1 + 6 + 3 + 0) = 10 times in row $T_3$.

Using Similar analysis, this example is represented (0 + 2 + 3 + 0 + 0) = 5 times in row $T_4$.

Using Similar analysis, this example is represented (0 + 1 + 0 + 0) = 1 time in row $T_5$.

This example is not represented in any of rows $T_6, T_7$, or $T_8$.  This is because those rows pertain to filling more than 5 "straight paths" with noughts, and this example represents filling only 3 rows + 2 columns with noughts.

Examining the representations for this example. the example is represented $(5, 10, 10, 5, 1)$ times in $T_1, T_2, T_3, T_4,$ and $T_5$ respectively.  Notice that these 5 terms represent all but the first term in the 5th row of Pascal's triangle.

Although it is unclear to me exactly why this is happening, it is related to the fact that this example specifically pertains to  (3 + 2 = 5) straight paths.  It is easy to demonstrate that when you add - subtract - add - subtract... across any row in Pascal's triangle,  where you start with the second term, the computation will equal 1.  This relates to the binomial expansion of $(1 + [-1])^k.$

The net effect of the algorithm's computation of $T_0 - T_1 + \cdots$ is that this specific unsatisfactory configuration is deducted one time from the $2^{(16)}$ total possible configurations.

The backbone of the Inclusion-Exclusion principle, as documented in the Wikipedia article is that each region will end up being added, then subtracted, then added, ... so that the net effect is that the region is counted once.


答:
No.

Hints:

For the first one you are allowing more than $4$ crosses. So you have to choose where are the crosses out of $16$ possibilities.

For the second one, I am really confused in how you are arguing. Where does the power of the combinatorial comes from? Notice that you have $4$ choices per row. Multiplication principle gives you that...

For the third one you can not have two crosses in the same column. So here are $4$ choices for the first row but $3$ for the second one...So multiplication principle again gives you...

Edit:
For the new problem, the first one is fine. The second one is not right. Notice that per row there are $2^4-1$ possibilities (this follows from your logic in problem 1). So you have $(2^4-1)^4.$

The third one is tricky cause you will have to do inclusion-exclusion. Try to do something like $$2^{16}-(4+4)*2^{16-4}+2\cdot \binom{4}{2}\cdot 2^{16-8}+4\cdot 4\cdot 2^{16-8+1}\cdots,$$
Essentially you are doing all of them and excluding when one row or one column has no crosses.

ContentLicense: CC BY-SA 4.0

Title: MAT Q5 1996 Solution Verification (Combinatorics)

Body:
I have been doing some MAT past papers and I have got to papers where there is no official mark scheme. I checked on the student room but there seem to be no answers for the very old mat past papers. Here is the question I want to answer:

My solution:

a): There could be two values for each of the $4\times4=16$ places. Thus the total number of different arrangements is $2^{16}$

b): Picture the $4$ rows and let us consider the cases: " one $\times$ in each row", two $\times$ in each row, "three $\times$ in each row" and "four $\times$ in each row".

In the first case, we have $(^4C_1)^4$ different arrangements.

In the second case, we have $(^4C_2)^4$ different arrangements.

In the third case, we have $(^4C_3)^4$ different arrangements.

In the fourth case, we have $(^4C_4)^4$ different arrangements.

And thus the total number of different arrangements is $=1809$ (adding all of the above).

c): Consider the cases

$$\begin{pmatrix}
  \times & \text{either} & \text{either} & 0 &\\
   \text{either}& \times & 0 & \text{either} & \\
   \text{either} & 0 & \times & \text{either} &\\
   0& \text{either} & \text{either} & \times & \\
\end{pmatrix}
\text{and}
\begin{pmatrix}
  0 & \text{either} & \text{either} & \times &\\
   \text{either}& 0 & \times & \text{either} & \\
   \text{either} & \times & 0 & \text{either} &\\
   \times& \text{either} & \text{either} & 0 & \\
\end{pmatrix}
$$
In each of the cases, we have $8$ available positions giving $2^8$ possible arrangements. We need only now consider the case when

$$\begin{pmatrix}
  \times & \text{either} & \text{either} & \times &\\
   \text{either}& \times & \times & \text{either} & \\
   \text{either} & \times &  \times &\text{either} &\\
   \times& \text{either} & \text{either} & \times & \\
\end{pmatrix}$$

in which case we have yet again $2^8$ different arrangements. Thus the total number of different arrangements is $3\times 2^8=768$

Is this correct?

EDIT: I have realised I have interpreted the question way more generally than I should have. I thought that for each position I have a nought and across available. Anyways, I would like to ask if my generalization is correct.


答:
I’m afraid that all of these are incorrect.

$2^{16}$ is the number of different ways to put either a nought or a cross in each of the $16$ positions without any restriction on the number of crosses. Here, however, we must have exactly $4$ crosses; there are $\binom{16}4$ ways to choose $4$ of the $16$ positions to get the $4$ crosses, and once we’ve done that, the other $12$ positions must be filled with noughts, so the correct answer to (a) is $\binom{16}4=1820$.

In (b) we must choose one position in each row. In any one row there are $4$ possible choices, so there are altogether $4^4=256$ ways to choose one position in each row to get the cross for that row.

In (c) there are $4$ ways to place a cross in the first row. Once that’s been done, there are only $3$ possible places for the cross in the second row, since it cannot lie in the same column as the cross in the first row. Similarly, once those two crosses have been placed, there are just $2$ possible positions for the cross in the third row, and after that there is just one possible position for the cross in the fourth row. Thus, there are $4\cdot3\cdot2\cdot1=4!=24$ possible arrangements of this type.


答:
Put the rows in line: you get a line of $16$ places divided in $4$ sectors, then

a) $\binom{16}{4}$ : you can put the four $\times$ in every place;

b) $4^4$ : four choices for each sector;

c) $4!$ : you can chose any permutation of $(1,2,3,4)$ and assign as place in first, 2nd , .. sector


问:
How to calculate the expected value of bivariate normal distribution?
Let $X= (X_1,X_2)$ be a random vector with bivariate normal distribution $X\sim N(\mu,\Sigma)$  such that $X_1$ and $X_2$ are positively correlated and we also have to: $P(X_1<1) = 0,84134 $, $P(X_2>6) = 0,02275 $, $Var[X_1] = 1 $ and $Var[X_2] = 2 $. How can I calculate the expected value  $E[X_1\mid X_2= 6]$ and the covariance $Cov(X_1,X_2)$?.

I have tried to do the integral according to the expected value and Conditional Variance, but I have not been able to arrive at something concrete. Is there a miraculous property that I don't know about and that works for me?

答:
We know $X_1$ and $X_2$ have variances $\sigma_1^2 = 1$ and $\sigma_2^2 = 2$ respectively.
$0.84134 = P(Z < 1)$ and $0.02275 = P(Z > 2)$ for a standard normal random variable $Z$.  That lets you figure out $\mu_1 = E[X_1]$ and $\mu_2 = E[X_2]$.  But you have given us no information about the covariance $\Sigma_{12}$ except that it is positive.


问:
Proving an identity with a combinatorial proof: $\binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{k-r}$

For any integers $n$, $k$, $r$ where $n\geq k\geq r \geq 0$, give a combinatorial proof of the following identity:
  \begin{align}
\binom{n}{k}\binom{k}{r}=\binom{n}{r}\binom{n-r}{k-r} .
\end{align}

The problem is that I can't come up with a good counting argument of what exactly the two sides are doing. The left hand side is quite mysterious, and the right hand side is apparently choosing $r$ items and then choosing $k-r$ items from the remaining, which should be equivalent to $\binom{n}{k}$ but somehow isn't. How should I approach this problem?

答:
Think of the problem of how to choose a sports team team consisting of $k$ people, and then to choose $r$ people from that team to play in a particular match (leaving $k-r$ people on the sidelines). How many ways are there to do this if you have $n$ total people from which to choose?

We can make these choices in two different ways. First of all, how many ways are there to pick the entire team first, and then pick the players for the match from the team? Secondly, how many ways are there to pick just the players for the match first, and then fill out the rest of the team from the people you didn't pick?

In more general terms, we you can think of this identity as saying that if we make a selection from a set of $n$ things so that $k$ of them have a property $1$ and $r$ of them have properties $1$ and $2$, it doesn't matter the order in which we assign the properties.

ContentLicense: CC BY-SA 3.0


问:
Find $a$ and $b$ such that probability is the biggest
Basketball player A hit the basket with probability $\frac{1}{2}$, and B with $\frac{1}{10}$. Player A can shoot $a$ times, and B $b$ times, $a+b=20$. Find $a$ and $b$ such that probability to make at least one point each is the biggest.

I just begging my course in probability and I have a problem with this. I tried to find max value of $(1-\frac{1}{2^{a}})(1-\frac{9^{20-a}}{10^{20-a}})$(complement of event that some of players doesn't make a point), and get solution a=1, b=19, but solution in book is a=5,b=15.

Sorry for my bad English, any help?

Body:
Denote the possible events of no hits/hits  by (no hits, hits). We get the following sequence for A:
$E_A=(0,1),(1,1),(2,1),....,(a-1,1)$ and similarly for b: $E_B=(0,1),(1,1),(2,1),....,(b-1,1)$. The probablities are:
$$
P(E_A)=(\frac{1}{2})^0(\frac{1}{2})^1+(\frac{1}{2})^1(\frac{1}{2})^1+(\frac{1}{2})^2(\frac{1}{2})^1...+(\frac{1}{2})^{a-1}(\frac{1}{2})^1\\=1-\frac{1}{2^a}\\
P(E_B)=(\frac{9}{10})^0(\frac{1}{10})^1+(\frac{9}{10})^1(\frac{1}{10})^1+(\frac{9}{10})^2(\frac{1}{10})^1...+(\frac{9}{10})^{b-1}(\frac{1}{10})^1\\=1-(\frac{9}{10})^b=1-(\frac{9}{10})^{20-a}
$$
Thus we have to maximize
$$P(a)=P(E_A)P(E_B)=(1-\frac{1}{2^a})(1-(\frac{9}{10})^{20-a})$$
Numerically I found that the maximum of $P(a)$ is at $a=4.764$. So the closest integer is:
$a=5$
and we have $b=15$


问:
Open in Product Topologies
Let $A=\{\langle a,b\rangle\in\mathbb{R}^2|b\leq a\}$

How would we show that $A$ is open in product topologies such as $\mathbb{R}_R\times\mathbb{R}_R$, $\mathbb{R}_U\times\mathbb{R}_U$, $\mathbb{R}_R\times\mathbb{R}_L$, or $\mathbb{R}_L\times\mathbb{R}_R$?

For reference $U,R$, and $L$ are the usual, right, and left topologies, respectively.

答:
The first step is to actually figure out what each of those topologies are. The easiest way to do this is via bases:

First, find a basis for each of the three "starting spaces," namely $\mathbb{R}_L$, $\mathbb{R}_R$, and $\mathbb{R}_U$.

Next, use the following fact (which you should prove as an exercise using the definition of the product topology): if I have bases $B_1,B_2$ for topological spaces $(X_1,\tau_1), (X_2,\tau_2)$ respectively, then $$\{U_1\times U_2: U_1\in B_1,U_2\in B_2\}$$ (= the set of "basic open rectangles") is a basis for $X_1\times X_2$ with the product topology coming from $\tau_1$ and $\tau_2$ (note that this is not the same as the Cartesian product of $\tau_1$ and $\tau_2$ as sets).

Once we've found bases $B_{R,R}, B_{R,L}, B_{L,R}, B_{U,U}$ for $\mathbb{R}_R\times\mathbb{R}_R$, $\mathbb{R}_R\times \mathbb{R}_L$, $\mathbb{R}_L\times\mathbb{R}_R$, and $\mathbb{R}_U\times\mathbb{R}_U$ respectively, we can use these to determine whether the set $A$ is open in each space in turn:

To tell whether $A$ is open in $\mathbb{R}_I\times\mathbb{R}_J$ (for $I,J\in \{R,L,U\}$), you need to figure out whether for each point $\langle u,v\rangle\in A$ there is some $W\subseteq\mathbb{R}^2$ with $\langle u,v\rangle\in W$ and $W$ is an element of the appropriate basis $B_{I,J}$.

It's a good idea to start trying to prove that it is open (in whichever topology you're looking at at the moment), and if that doesn't work use your failure to do so to help find a proof that it's not in fact open (again, in whichever topology).

Title: Open in Product Topologies


问:
Three-times differentiable function
Let $\mathbf a:\mathbb{R}\longrightarrow\mathbb{R^3}$ be a three-times differentiable vector valued function. If $\mathbf b(x)=\mathbf a(x)\cdot[\mathbf a'(x) \times\mathbf a''(x)]$   (eq $1$),

Show that:

$$\mathbf b'(x)=\mathbf a(x)\cdot\bigl[\mathbf a'(x) \times\mathbf a'''(x)\bigr]$$

What I have tried:

I think that to differentiate the RHS of the eq $1$ I use the product rule. So :

$\mathbf b'(x)=\mathbf a''(x)\cdot[\mathbf a(x)\times \mathbf a'(x)] -\mathbf a'(x)\cdot[\mathbf a(x)\times \mathbf a''(x)]$

答:
Hint:
Apply the rules for the dot product and the cross product:
\begin{align}
\mathbf b(x)'&=\mathbf a'(x)\cdot\bigl[\mathbf a'(x) \times\mathbf a''(x)\bigr]+\mathbf a(x)\cdot\bigl[\bigl(\mathbf a'(x) \times\mathbf a''(x)\bigr)'\bigr]\\
&=\mathbf a'(x)\cdot\bigl[\mathbf a'(x) \times\mathbf a''(x)\bigr]+\mathbf a(x)\cdot\bigl[\mathbf a''(x) \times\mathbf a''(x)+\mathbf a'(x) \times\mathbf a'''(x)  \bigr].
\end{align}

Title: Three-times differentiable function


问:
Connection between exponential function as repeated multiplication and $\lim_{n \to \infty} (1 + \frac{1}{n})^n$ (also known as compund interest)
I cannot find intuitive explanation for formula $e^x = \lim_{n \to \infty} (1 + \frac{1}{n} \cdot x)^n.$ I will write down my understanding of the exponential function so you can understand the question better.

Multiplication.
For hole numbers, multiplication is repeated addition (subtraction). Let's say we want to find $5 \cdot 3 = x.$ We know (or better, we defined this to be) $x = 3 + 3 + 3 + 3 + 3 = 15.$

For rational numbers it's simillar but it has two steps: $\frac{3}{2}7 = x.$ Problem on the left tells us to find number $y$ such that when I add it to itself gives me $7$ then I have to add that number to zero 3 times. So $y = 3.5.$ Now, I have to find $y + y + y.$ So it's kinda back and fort game, but it still can be desribed using repeated addition (subtraction).

Finally for irrational numbers multiplication can be approximated with error of $\epsilon \rightarrow 0$ using arithmetic in $\mathbb{Q}.$

Exponentiation.
Simillary as before we can do repeated multiplication. This arithmetically maybe isn't clear as before, but we can still do it (and for example doubling of something can easly be understood). Reasoning is simillar as before, just turn addition into multiplication. For example, number $7^{\frac{3}{2}} = x$ we can find as follows: find a number $y$ which multiplied by itself gives $7$ (so $y = 2.645...$). Then find $y \cdot y \cdot y.$ Thus $x = 18.520...$ For irrational exponent (base) we can  approximated it very well.

Now, we can investigate doubling of a money in hours, for example (suppose "money" can be any real number). In other words we are investigating function $f(x) = 2^x$ where $x$ is time passed in hours after 0th second.

Let's say we are starting from $2^0 = 1$ dollar. Right after 1 hour we have $2$ dollars, etc. Also 1 hour before starting moment we had $0.5$ dollars and so on. This starting from 1 (*) will be crucial.

We can observe that if more time passes doubling become larger, of course. Take intervals of one hour and see resulting change in amount of money, for example $ \frac{2^{0} - 2^{- 1}}{1 \ \text{hour}} <  \frac{2^1 - 2^0}{1 \ \text{hour}} <\frac{2^{10} - 2^9}{1 \ \text{hour}}.$

Of course, change in money over small time interval (dx) is proportional to amount of money we already have, $2^x$ for some fixed $x.$ After hypothetical calculation we get $\frac{d}{dx}2^x \approx 0.6931 \cdot 2^x$ and simillary for any positive base, $\frac{d}{dx}b^x = kb^x.$ We want to find base $a$ such that $k = 1$ (then obviously $\frac{d}{dx}a^x = 1 \cdot a^x = a^x$). After algebraic manipulation we can find that magical number and denote it by "e": $e:= a = \lim_{n \to \infty} (1 + \frac{1}{n})^n.$

I am interested in intuition why this is true. Simply: why we add to starting moment, 1 (*), this fraction $\frac{1}{n}$ and then multiply it by itself $n$ times for large $n$?

Simillary, why we add $x$ times this fraction $\frac{1}{n}$ and then again multilpy that $n$ times (I understand $e^x = e^{1 \cdot x}$ but I do not see it clearly.

Of course if you give good argument for $e^x = \lim_{n \to \infty}(1 + \frac{x}{n})^n$ then setting $x = 1$ we get basic formula for $e.$ Please, I am looking for intuition and logic, not just algebra. Thanks!

Also, is this limit well behaved, i. e. is inequality
$|e - \lim_{n \to \infty}(1 + (\frac{1}{n + 1})^{n + 1}| < |e - \lim_{n \to \infty} ((1 + \frac{1}{n})^n|$ true?

答:
$$e^m$$ is $e$ multiplied by itself $m$ times.

But $e$ is defined as

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n,$$ which turns out to exist and be the number $$1+1+\frac12+\frac1{3!}+\frac1{4!}+\cdots=2.718281828\cdots$$

Now, skipping some justifications,

$$e^m=\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^m=\lim_{n\to\infty}\left(1+\frac1n\right)^{nm}=\lim_{k\to\infty}\left(1+\frac mk\right)^k$$

by setting $nm=k$, so that $\dfrac1n=\dfrac mk$.


问:
Determining the Remainder
Is it possible to calculate the remainder of two given values with merely addition, subtraction, multiplication, and division? Is there an algorithm or formula if it is even possible?

For instance, say we have two values: 100 and 30. If we divide 100 by 30, we get 3.333.... But is there any way to reach 0.333... (without knowing from the get go how many wholes are included)?

To further illustrate what I mean, and sticking with the example above, one way to find the remainder would be the following formula: 100 / 30 / 10 * 30 = 0.333... BUT this obviously does not work for any two given numbers.

Some more examples (The values in the brackets are the values I am after):

100 / 40 = 2.5 (0.5)
450 / 50 = 9 (0)
11 / 4 = 2.75 (75)

Sorry, if this question is not clear. It is based on a programming challenge I have encountered. I want to use only CSS to calculate the remainder of a text's line height, given a specific vertical offset of the text. For instance, the height of the window might be 100, whereas the line height would be 30. But CSS has no function to easily determine the remainder, and I cannot use loops or recursion (i.e. 100 - 30 - 30 - 30 - 30 < 0), nor conditional (i.e. if N < 0, do...).

My math is awful. But I am wondering whether or not it is mathematically possible?

答:
I don't know whether it is possible to make the "remainder in integral division" in CSS using "thinking out of the box" somehow, but mathematically it cannot be done using just addition, subtraction, multiplication and division.

Namely, a function $f(x,y)$ made up of addition, subtraction, multiplication and division is a rational function (a quotient of two real polynomials in two variables $x$ and $y$).

Now suppose that $\text{remainder}(x,y)=\frac{P(x,y)}{Q(x,y)}$ where $P$ and $Q$ are polynomials. Fix $y=2$ and then we would have $\text{remainder}(x,2)=\frac{P(x,2)}{Q(x,2)}=\frac{p(x)}{q(x)}$ where $p(x)=P(x,2)$ and $q(x)=Q(x,2)$ - polynomials in one variable. Knowing that:

$$\frac{p(n)}{q(n)}=0$$

for each even $n$, we can conclude that $p=0$ (zero polynomial). However, this is then inconsistent with the other requirement, which is that:

$$\frac{p(n)}{q(n)}=1$$

for each odd $n$.

Note: I can see that CSS spec for $\text{calc}()$ says that, ultimately, when the result of CSS calculation is assigned to an attribute, it may be rounded if that attribute requires an integer. Rounding is similar to truncation ($\text{round}(x)=\text{ceil}(x+0.5)$) and truncation can be used for integral division ($\text{remainder}(x, y)=x-y\times\text{ceil}(x/y)$) so maybe this can be all cobbled together somehow - but I wouldn't know myself how to do that, as I am not a CSS expert...

Title: Determining the Remainder


问:
Find the direction of maximum decrease of the function $g$ in the point $(1,1)$
Let be $g(x, y)=f\left(x^{2}+y, 3 x y\right),$ where $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ is a differentiable function. Assume that the gradient of $f$ in (2,3) is the vector  $5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} .$ Find the direction of maximum decrease of the function $g$ in the point $(1,1)$

I don't know exactly what I have to do.
I know g :$\mathbb{R^2}\to\mathbb{R}$ because $f$ is a function $\mathbb{R^2}\to\mathbb{R}$
I and I know that the maximum decrease is the opposite to the gradient but how can I know what is $f$ any hint?

答: You want to find the gradient of $g$ at $(1,1)$. Take $u=x^2+y,v=3xy$, then$$\begin{align*}\begin{bmatrix}\frac{\partial g(x,y)}{\partial x}\\\frac{\partial g(x,y)}{\partial y}\end{bmatrix}_{x=1\\y=1}&=\begin{bmatrix}\frac{\partial f(u,v)}{\partial x}\\\frac{\partial f(u,v)}{\partial y}\end{bmatrix}_{u=2\\v=3}=\begin{bmatrix}\frac{\partial f(u,v)}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f(u,v)}{\partial v}\frac{\partial v}{\partial x}\\\frac{\partial f(u,v)}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f(u,v)}{\partial v}\frac{\partial v}{\partial y}\end{bmatrix}_{x=1,y=1\\u=2,v=3}\\&=\begin{bmatrix}2x\frac{\partial f(u,v)}{\partial u}+3y\frac{\partial f(u,v)}{\partial v}\\\frac{\partial f(u,v)}{\partial u}+3x\frac{\partial f(u,v)}{\partial v}\end{bmatrix}_{x=1,y=1\\u=2,v=3}\\&=\begin{bmatrix}2\frac{\partial f(u,v)}{\partial u}+3\frac{\partial f(u,v)}{\partial v}\\\frac{\partial f(u,v)}{\partial u}+3\frac{\partial f(u,v)}{\partial v}\end{bmatrix}_{u=2\\v=3}\end{align*}$$Now you are given $$\nabla f(u,v)=\begin{bmatrix}\frac{\partial f(u,v)}{\partial u}\\\frac{\partial f(u,v)}{\partial v}\end{bmatrix}_{u=2\\v=3}=[5,4]^T$$Can you take it from here?


答:
Doubling is a relative concept.

Starting with an amount $A_0$, after 1 hour you get a total amount $A_1=2 \cdot A_0$, which gives an absolute increase $A_1-A_0 = A_0$ and a relative increase $\frac {A_1-A0}{A_0} = 1 = 100 \%$, while the new amount relative to the old is $A_1/A_0 = 2 =200 \%$.

The contribution (or one of the factors) leading to Neper logarithm seems that was on question if given an interest $x$ per year, what was amount going to be if the interest was compound every year, very quarter of year, every month, .., every minute, .. , that is $(1+x/1)^1, (1+x/4)^4, (1+x/12)^{12}, \cdots$

Title: Connection between exponential function as repeated multiplication and $\lim_{n \to \infty} (1 + \frac{1}{n})^n$ (also known as compund interest)

Body:
I cannot find intuitive explanation for formula $e^x = \lim_{n \to \infty} (1 + \frac{1}{n} \cdot x)^n.$ I will write down my understanding of the exponential function so you can understand the question better.

Multiplication.
For hole numbers, multiplication is repeated addition (subtraction). Let's say we want to find $5 \cdot 3 = x.$ We know (or better, we defined this to be) $x = 3 + 3 + 3 + 3 + 3 = 15.$

For rational numbers it's simillar but it has two steps: $\frac{3}{2}7 = x.$ Problem on the left tells us to find number $y$ such that when I add it to itself gives me $7$ then I have to add that number to zero 3 times. So $y = 3.5.$ Now, I have to find $y + y + y.$ So it's kinda back and fort game, but it still can be desribed using repeated addition (subtraction).

Finally for irrational numbers multiplication can be approximated with error of $\epsilon \rightarrow 0$ using arithmetic in $\mathbb{Q}.$

Exponentiation.
Simillary as before we can do repeated multiplication. This arithmetically maybe isn't clear as before, but we can still do it (and for example doubling of something can easly be understood). Reasoning is simillar as before, just turn addition into multiplication. For example, number $7^{\frac{3}{2}} = x$ we can find as follows: find a number $y$ which multiplied by itself gives $7$ (so $y = 2.645...$). Then find $y \cdot y \cdot y.$ Thus $x = 18.520...$ For irrational exponent (base) we can  approximated it very well.

Now, we can investigate doubling of a money in hours, for example (suppose "money" can be any real number). In other words we are investigating function $f(x) = 2^x$ where $x$ is time passed in hours after 0th second.

Let's say we are starting from $2^0 = 1$ dollar. Right after 1 hour we have $2$ dollars, etc. Also 1 hour before starting moment we had $0.5$ dollars and so on. This starting from 1 (*) will be crucial.

We can observe that if more time passes doubling become larger, of course. Take intervals of one hour and see resulting change in amount of money, for example $ \frac{2^{0} - 2^{- 1}}{1 \ \text{hour}} <  \frac{2^1 - 2^0}{1 \ \text{hour}} <\frac{2^{10} - 2^9}{1 \ \text{hour}}.$

Of course, change in money over small time interval (dx) is proportional to amount of money we already have, $2^x$ for some fixed $x.$ After hypothetical calculation we get $\frac{d}{dx}2^x \approx 0.6931 \cdot 2^x$ and simillary for any positive base, $\frac{d}{dx}b^x = kb^x.$ We want to find base $a$ such that $k = 1$ (then obviously $\frac{d}{dx}a^x = 1 \cdot a^x = a^x$). After algebraic manipulation we can find that magical number and denote it by "e": $e:= a = \lim_{n \to \infty} (1 + \frac{1}{n})^n.$

I am interested in intuition why this is true. Simply: why we add to starting moment, 1 (*), this fraction $\frac{1}{n}$ and then multiply it by itself $n$ times for large $n$?

Simillary, why we add $x$ times this fraction $\frac{1}{n}$ and then again multilpy that $n$ times (I understand $e^x = e^{1 \cdot x}$ but I do not see it clearly.

Of course if you give good argument for $e^x = \lim_{n \to \infty}(1 + \frac{x}{n})^n$ then setting $x = 1$ we get basic formula for $e.$ Please, I am looking for intuition and logic, not just algebra. Thanks!

Also, is this limit well behaved, i. e. is inequality
$|e - \lim_{n \to \infty}(1 + (\frac{1}{n + 1})^{n + 1}| < |e - \lim_{n \to \infty} ((1 + \frac{1}{n})^n|$ true?


问:
Show $(\mathbb{R},0,+,-)$ has infinitely many automorphisms
Show $(\mathbb{R},0,+,-)$ has infinitely many automorphisms :

My idea is that if $\phi:(\mathbb{R},0,+,-) \rightarrow (\mathbb{R},0,+,-)$ is an automorphism. Then since here we don't have the order relation. For any $x_1, x_2, y_1, y_2 \in \mathbb{R}$, $(x_1,x_2 ) \rightarrow (y_1, y_2)$ will be an automorphism. In another words, any we can always map an open interval to another open interval to have automorphism.

Could someone let me know if my thinking is correct?

If not, could you let me know what went wrong, and how should I fix it?

Any hints or ideas will be appreciated!

Thanks!

答: Hint: $\mathbb R$ is an infinite-dimensional vector space over $\mathbb Q$.


问:
Expanding root polynomial form to general form
Let's suppose we have a set of n data points:
$$(x_0,y_0), (x_1,y_1),...,(x_{n-1},y_{n-1})$$

then Newton polynomial is given by:

$$P(x)=\displaystyle\sum_{k=0}^{n}\lambda_k\displaystyle\prod_{i=0}^{k}(x-x_i)$$

Suppose we already know every lambda. How to expand the Newton's form to the general form of:

$$P(x)=\displaystyle\sum_{k=0}^{n}a_kx^k$$

I would appreciate any help

答:
Through coefficient identification for the coefficients of $x^0, x^1, ... x^n$ in both written forms:

$$\displaystyle\sum_{k=0}^{n}\lambda_k\displaystyle\prod_{i=0}^{k}(x-x_i) = \displaystyle\sum_{k=0}^{n}a_kx^k$$

means

$\lambda_0(x-x_0) +\lambda_1 (x-x_0)(x-x_1)+\lambda_2 (x-x_0)(x-x_1)(x-x_2)+...+\lambda_n (x-x_0)(x-x_1)(x-x_2)...(x-x_n) = \displaystyle\sum_{k=0}^{n}a_kx^k$

The coefficient comparaison in the equation above leads to solving the following: I'll continue ...

$
\begin{pmatrix}1 &.&  .  &.& 0\\
1 &(x_1-x_0)&.&.&0\\
1&  (x_2-x_0)&   (x_2-x_0)(x_2-x_1)&.&0\\
.\\
.\\
.\\
1&  (x_k-x_0)&   .&.&\displaystyle\prod_{i=0}^{k-1}(x_k-x_i)\\
\end{pmatrix}.\begin{pmatrix}
\lambda_0\\
\lambda_1\\
.\\
.\\
.\\
\lambda_k\\
\end{pmatrix} = \begin{pmatrix}
a_0\\
a_1\\
.\\
.\\
.\\
a_k\\
\end{pmatrix}$

This can be solved iteratively.


问:
Transversal and intersection of two foliations
Let $\mathcal{F}_1$ and $\mathcal{F}_2$ be two foliations of a manifold. We say that $\mathcal{F}_1\pitchfork \mathcal{F}_2$ if $T_p L^{(1)}+T_pL^{(2)}=T_p M$ for any $p\in M$, where $L^{(1)}$ and $L^{(2)}$ are the leaves trough $p$.

Now if we have that $\mathcal{F_1}\pitchfork \mathcal{F}_2$, we define $\mathcal{F}_1\cap \mathcal{F}_2$ to be the foliation where the leaves are the connected components of $L^{(1)}\cap L^{(2)}$. Now we want to check that is a foliation with codimension the sum of the codimensions of $\mathcal{F}_1$ and $\mathcal{F}_2$.

I think the idea is to use the implicit function theorem but I am getting nowhere in concrete. That is take foliated charts for $\mathcal{F}_1$, $(x_1,...,x_k,y_1,...,y_{m-k})$, and foliated charts for $\mathcal{F}^2$, $(w_1,...,w_k',z_1,...,z_{m-k'})$ such that $p\cap L^{(1)}\cap L^{(2)}=\{p\in U: y_1(p)=ct,...,y_{m-k}(p)=ct,z_1(p)=ct,...,z_{m-k'}(p)=ct\}$ . Now from this I have tried and construct a new coordinate chart so that we get the result bu I got nowhere.

Any help or hints with this are aprecciated. Thanks in advance.

Attempt at solution :

First thing we note is that since $\mathcal{F}_1$ and $\mathcal{F}_2$ are foliations we will have that the leaves of this new foliation will cover $M$, are disjoint since we are taking the connected components, and will be path connected since we are taking the connected components and $M$ is locally path-connected. Now we need to find the foliated charts of $\mathcal{F}$ and check it's dimension. Let $p\in M$, note that since $L^{(1)}$ and $L^{(2)}$ are transversal we will have that $L^{(2)}\cap L^{(1)}$ is a submanifold of $L^{(1)}$.Using the local normal form we know that there exists open set of $p\in L^{(2)}, U,$ and coordinate chart $(V,\phi)=(V,x_1,...,x_k)$ for $L^{(1)}$ such that $U\cap L^{(1)}=\{p\in V : x^{k'}(p)=...=x^k(p)=0\}$, where $k'=dim L_1 -(dim L_1 + dim L_2 -dim M)= codim \mathcal{F}_2$. Now note that the way we gave the manifold structure to $L^{(1)}$ was that the topology was generated by the plaques $L^{(1)}\cap U'$ where $U'$  was a foliated chart for $p$, and the coordinate chart was the restriction of the foliated chart to the non-constant components. So we have the $(x_1,...,x_k)$ are associated with a chart of $M$, $(x^1,...,x^k,y^1,...,y^{d-k})$ such that the connected components of $L^{(1)}\cap U'$ are of the form $\{p\in U' : y^1(p)=ct,...,y^{d-k}(p)=ct\}$. Now we have $U$ is a plaque for $U''\cap L^{(2)}$. We also know that $p$ will be in a connected component of $L^{(1)}\cap L^{(2)}$, wich we now denote by $L$. If we consider the foliated chart to be $(W:=U'\cap U'',x^1,...,x^k,y^1,...,y^{d-k})$ we will have that the connected components of $W\cup L$ are of the form $\{p\in W: y^1(p)=ct,...,y^{d-k}(p)=ct,x^{k-k'}(p)=,...,x^k(p)=0\}$. To see the statement about the dimension we note that $codim \mathcal{F}_1+ dim L^{(1)}-(dim L^{(1)}+dim L^{(2)}-dim M)=codim\mathcal{F}_1 +codim \mathcal{F}_2 $.

答:
Here is, I think, an answer. Let $\mathcal{D}_1$ be the tangent ditribution of the foliation $\mathcal{F}_1$ and $\mathcal{D}_2$ those of $\mathcal{F}_2$. Theses distribution are totally integrable, that is $[\mathcal{D}_i,\mathcal{D}_i]\subset \mathcal{D}_i$. Let $\mathcal{D} = \mathcal{D}_1\cap \mathcal{D}_2$ be the intersection distribution. It has constant rank as at any $p\in M$, $\dim\mathcal{D}_p = \dim\left({\mathcal{D}_1}_p\cap{\mathcal{D}_2}_p\right)=\dim{\mathcal{D}_1}_p+\dim{\mathcal{D}_2}_p-\dim\left({\mathcal{D}_1}_p+{\mathcal{D}_2}_p\right)$, and then $\mathrm{codim}\mathcal{D}_p = \mathrm{codim}{\mathcal{D}_1}_p + \mathrm{codim}{\mathcal{D}_2}_p$ (recall that $\mathcal{F}_1\pitchfork \mathcal{F}_2$)

Now, let us show that $\mathcal{D}$ is integrable, that is $[\mathcal{D},\mathcal{D}]\subset \mathcal{D}$. Let $X,Y$ be vector fields in $\mathcal{D}$. As $\mathcal{D}\subset \mathcal{D}_1$, $[X,Y]$ is in a vector field in $\mathcal{D}_1$. Symmetrically, it is a vector field in $\mathcal{D}_2$. Then, it is a vector field in $\mathcal{D}_1 \cap \mathcal{D}_2 = \mathcal{D}$, and $[\mathcal{D},\mathcal{D}]\subset \mathcal{D}$. By Frobenius theorem, there exists a foliation $\mathcal{F}$ those tangent bundle is $\mathcal{D}$. The integral submanifolds are the intersections of the leaves of the previous foliations.

Edit Here is another strategy. Fist, show that the intersection of two transverse submanifolds is a submanifold. Let $N$ and $N'$ be two trasverse submanifold with $\mathrm{codim} N= p, \mathrm{codim} N' =q$. Let $x \in N\cap N'$ be fixed. Take a chart $\varphi : U \to \mathbb{R}^n$ centered at $x$. Then $\varphi (U\cap N)$ and $\varphi(U\cap N')$ are submanifolds of $\varphi(U)$ of codimension $p$ and $q$. There exist $f = (f_1,\ldots,f_p): \mathbb{R}^n \to \mathbb{R}^p$ and $g=(g_1,\ldots,g_q): \mathbb{R}^n\to \mathbb{R}^q$ smooth submersions such that $\varphi(U\cap N) = f^{-1}(\{0\})$ and $\varphi(U\cap N')= g^{-1}(\{0\})$. Let $h = (f_1,\ldots,f_p,g_1,\ldots,g_q) : \mathbb{R}^n \to \mathbb{R}^{p+q}$. The transversality condition assures that $h$ is a sumbersion, and thus, $\varphi(U\cap\left(N\cap N'\right)) = h^{-1}(\{0\})$ is a submanifold of $U\cap \mathbb{R}^n$ of codimension $p+q$, and $U\cap (N \cap N') = \varphi^{-1}(h^{-1}(\{0\}) = (h\circ\varphi)^{-1}(\{0\})$ is a smooth submanifold of codimension $p+q$.

Let $\mathcal{F}^1$ and $\mathcal{F}^2$ be two transverse smooth foliations on $M^n$. If $L^1$ and $L^2$ are leaves of $\mathcal{F}^1$ and $\mathcal{F}^2$, and if $L^1\cap L^2 \neq 0$, then the result above shows that $L^1\cap L^2$ is a smooth submanifold of $M$ with codimension $\mathrm{codim}L_1 + \mathrm{codim}L_2$. As the dimension of the leaves of a foliation is constant, if two transverse leaves intersect, their intersection is a submanifold of dimension $n - \mathrm{codim}\mathcal{F}^1 -\mathrm{codim}\mathcal{F}^2$.

It is staightforward to show that $\{L^1\cap L_2 ~|~ L_1 \in \mathcal{F}^1,L_2\in \mathcal{F}^2,L_1\cap L_2\neq 0\}$ is a smooth foliation.

Remark: this shows that the distribution $D_1\cap D_2$ associated is smooth.


问:
Exercise on Spectrum of Commutative Elements in a Banach Algebra

Given an element $A$ of a Banach algebra $V$ and $\epsilon > 0$, prove that if $0 \in \textrm{sp}(A)$, then there is $\delta > 0$ such that if $B \in V$ commutes with $A$ and $||A-B|| < \delta$, then there is a $\lambda$ in $\textrm{sp}(B)$ with $|\lambda | < \epsilon$.

My Attempt

My goal was to prove that by making $\delta$ small enough, we can guarantee $r(B)$ (spectral radius of $B$) to be less than $\epsilon$, and then use the fact that the spectrum is non-empty to prove the existence of such a $\lambda$.

Since $A$ and $B$ commutes, then $B-A$ and $A$ commutes, so we have:
$$r(B) = r(B-A+A) \leq r(B-A) + r(A) \leq ||B-A|| + r(A).$$
I am not sure how to proceed from here. I do not think this approach will work since I think the RHS of the above expression will not always be less than $\epsilon$ since $r(A)$ can be anything. However, I cannot think of any other approaches to take.

Could someone give a hint on how to approach this question?

答:
Lemma.  Given a unital Banach algebra $V$, and a
commutative subalgebra $W$,  there exists another commutative
subalgebra $W_1$, containing $W$, such that for any element $a$ in $W_1$, one has that $a$ is invertible relative to $W_1$
iff $a$ is invertible relative to $V$.   In particular, $$ \text{sp}_{V}(a)=\text{sp}_{W_1}(a),  $$
for every $a$ in $W_1$.

Proof. For every subset $S\subseteq V$, define the commutant of $S$ by
$$
  S'=\{a\in  V: as=sa: \text{ for all } s\in S\}.
  $$
It is
easy to see that

(1) $S'$ is always a unital subalgebra,

(2) $S$ is commutative iff $S\subseteq S'$,

(3) if $S\subseteq T$ then $S'\supseteq T'$.

(4) if $a\in S'$, and $a$ is invertible,  then $a^{-1}\in S'$.

Now, given $W$ as in the statement, we claim that  $W_1:= W''$ (that is, the commutant of the commutant of $W$) satisfies all of the required conditions.

First  observe that $W\subseteq W''$ by the following very trivial (if clumsy) reason:  every element of $W$ commutes with
everything that commutes with the elements of $W$.

Since $W$ is commutative, we deduce from (2) that  $W\subseteq W'$.  Using
(3) we get $W'\supseteq W''$, and   using (3) again we get $W''\subseteq W'''$. So the converse part of (2) implies that $W''$ is commutative.

Finally the last condition in the statement regarding invertible elements follows immediately from (4). QED

Back to the original question,  consider the commutative Banach algebra $W$ generated by $A$  and $B$,  and let $W_1$ be
as in the Lemma.  Then, for every element $a$ in $W_1$, we have that
$$
  \text{sp}_{V}(a)=\text{sp}_{W_1}(a) = \{\phi(a): \phi\in \text{Hom}(W_1, \mathbb C)\},
  $$
so the end result follows easily by the continuity of complex homomorphisms.


问:
Find number of normal subgroups in $F_3$ such that its factor is isomorphic to a given Abelian group
On the upcoming test I will be given a problem of type:

Find all normal subgroups $H$ in $F_n$ such that $F_n/H \cong G$.

Here $n$ is a small integer, likely 2 or 3, and $G$ is an Abelian group given as product of some cyclic groups.

Question: is there a more-or-less routine algorithm for such problems?

I do know an approach. I count two parameters: the number of epimorphisms $\#(F_n \twoheadrightarrow G)$ and the number of automorphisms $\#Aut(G)$. The answer then is $\frac{\#(F_n\twoheadrightarrow G)}{\#Aut(G)}$. This solution is based on the following facts: 1) every normal subgroup is a kernel of some homomorphism; 2) $F_n/Ker\,\phi \cong G$ if $\phi$ is an epimorphism; 3) $Ker\,\phi=Ker\,\psi$ iff $\psi = \alpha\circ\phi$ for some $\alpha \in Aut(G)$.

The problem is, very often I'm not sure how do I count one or both of these parameters. I could go through all homomorphisms and automorphisms manually, but the numbers in problem are usually too large.

Here are some cases to show how big numbers might be on the actual test. No need to solve them!

$n=3$, $G=\mathbb Z_{13}^{3}$

$n=3$, $G=\mathbb Z_{10}^3\times \mathbb Z$

$n=3$, $G=\mathbb Z_{70}\times \mathbb Z_{30} \times \mathbb Z_{15}$

$n=2$, $G=\mathbb Z_2 \times \mathbb Z_3 \times \mathbb Z_5$

$n=3$, $G=\mathbb Z_{55} \times \mathbb Z_{7} \times \mathbb Z_{77}$

$n=3$, $G=\mathbb Z_{75} \times \mathbb Z_{375} \times \mathbb Z_{125} \times \mathbb Z_{333}$

答:
To justify your approach you need one more fact:

4: the stabilizer of an epimorpism $F_n \to G$ with respect to the action of $\text{Aut}(G)$ is trivial.

All the groups $G$ are abelian, and to my mind the cleanest way to work with a finite abelian group is to factor it as the product of its $p$-torsion subgroups

$$A \cong \prod_p A_{(p)}$$

(equivalently, its Sylow $p$-subgroups). This is because of the following additional facts, which I'll leave as exercises:

5: $\varphi : F_n \to A$ is an epimorphism iff the projection to each $p$-part $F_n \to A_{(p)}$ is an epimorphism.

6: $\text{Aut}(A) \cong \prod_p \text{Aut}(A_{(p)})$.

This implies that the number $f_n(A) = \frac{|\text{Epi}(F_n, A)|}{|\text{Aut}(A)|}$ satisfies $f_n(A) = \prod_p f_n(A_{(p)})$ which reduces the problem in general to the problem for finite abelian $p$-groups (except for #2 where $A$ is infinite, but in #2 there are no epimorphisms, as you can see by tensoring with either $\mathbb{F}_2$ or $\mathbb{F}_5$).

If $A \cong C_p^k$ is elementary abelian (which occurs in #1, #3, #4, and #5) and $n < k$ then there are no epimorphisms $F_n \to A$. If $n \ge k$ every epimorphism $F_n \to C_p^k$ factors canonically through $C_p^n$ so it suffices to count epimorphisms $C_p^n \to C_p^k$. By taking vector space duals, this is equivalent to counting monomorphisms $C_p^k \to C_p^n$, and the standard argument involving picking one vector at a time gives that there are

$$(p^n - 1)(p^n - p) \dots (p^n - p^{k-1})$$

such epimorphisms, hence

$$f_n(C_p^k) = \frac{(p^n - 1)(p^n - p) \dots (p^n - p^{k-1})}{(p^k - 1) \dots (p^k - p^{k-1})} = {n \choose k}_p$$

which is a $q$-binomial coefficient for $q = p$. This occurs because we are counting subspaces of $C_p^n$ whose dimension is $n-k$ (so whose quotient is isomorphic to $C_p^k$), which gives an alternative argument for this count. This is a bit overkill for the small values of $n$ and $k$ occurring here but you asked for a routine algorithm and this will work no matter how big $n$ and $k$ get!

This handles every problem except #6 at the primes $p = 3, 5$, where we have $A_{(3)} \cong C_3^2 \times C_9$ and $A_{(5)} \cong C_{25} \times C_{125}^2$. I don't know how this goes off the top of my head but a modification of the above argument should work. For starters, every epimorphism from $F_n$ to one of these groups factors through $C_{p^k}^n$ where $p^k = 9, 125$ respectively. You may want to first work $\bmod p$ and then try to lift to $p^k$.

This may or may not have been the intended approach. An alternative is to argue that every epimorphism $F_n \to A$, for $A$ any abelian group, factors through the abelianization $\mathbb{Z}^n$, then try to count subgroups of $\mathbb{Z}^n$ with quotient isomorphic to $A$ using Smith normal form.


问:
How to integrate $e^{t \ln q} p e^{-t \ln q} dt$ for quaternion $q$ and $p$?
Trying to solve an inertial navigation problem in terms of quaternions,
I ended up with the following indefinite integral
$$
I = \int e^{t \ln q} \, p \, e^{-t \ln q} dt,
$$
where
$q \in \mathbb H$ is a unit quaternion,
$p \in \mathbb H$ is a pure imaginary quaternion,
and both $q$ and $p$ are constant with respect to
the integration variable $t \in \mathbb R$.

Assuming $q \neq 1$, I integrated by parts:
$$
I = \int
    \underbrace{\mathstrut{}e^{t \ln q} p}_u \,
    \underbrace{\mathstrut{}e^{-t \ln q}}_{v'} \, dt =
e^{t \ln q} p e^{-t \ln q} (-\ln q)^{-1} -
\int e^{t \ln q} (\ln q) \, p e^{-t\ln q} (-\ln q)^{-1} \, dt, \\
I = e^{t \ln q} p \, e^{-t \ln q} (-\ln q)^{-1} + (\ln q) \, I \,  (\ln q)^{-1},
$$
and got the equation
$$
(\ln q) \, I - I \, (\ln q) = e^{t \ln q} \, p \, e^{-t \ln q}.
$$

From this question I learned that
the general solution for the $ax + xb = c$ equation is
$$
x= \left( |b|^2 + 2b_0a + a^2 \right)^{-1} \left( ac +c b^* \right),
$$
but it did not work for my case because of zero quaternion inversion.
I also tried to expand this equation into quaternion components,
but came to a system of linear equations with a singular matrix.
This looks quite strange to me, since I'm pretty sure this integral should exist.

Could please someone point me to a mistake in my computations
and show how to calculate this antiderivative properly?
Might there be a well known solution for the problem?

答:
Note that conjugation preserves vectors, so the integrand and hence $\mathbf{I}$ are pure imaginary.

For pure quaternions $\mathbf{a}$ and $\mathbf{b}$,

$$ \mathbf{ab}=-\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\times\mathbf{b}. $$

Thus, writing $``\ln q "=\mathbf{v}$ we have

$$ (\ln q)\mathbf{I}-\mathbf{I}(\ln q)=2\mathbf{v}\times \mathbf{I}=\exp(t\mathbf{v})\mathbf{p}\exp(-t\mathbf{v}). $$

Note that $\mathbf{v}\times\mathbf{I}$ deletes the $\mathbf{v}$-component of $\mathbf{I}$, so this equation fails to "see" this component. However, this actually isn't a problem: conjugating by $\exp(t\mathbf{v})$ fixes the $\mathbf{v}$-component of $\mathbf{p}$, call it $\mathbf{p}_{\|}$, within the integrand for $\mathbf{I}$, so we may conclude the $\mathbf{v}$-component for $\mathbf{I}$ must be $\mathbf{I}_{\|}=t\mathbf{p}_{\|}$.

Applying $\mathbf{v}\times$ to $2\mathbf{v}\times\mathbf{I}_{\perp}=\exp(t\mathbf{v})\mathbf{p}\exp(-t\mathbf{v})$ (with $v=\|\mathbf{v}\|$) yields

$$ -2v^2\mathbf{I}_{\perp}=\mathbf{v}\times\big(e^{t\mathbf{v}}\mathbf{p}e^{-t\mathbf{v}}\big)  $$

from which we may conclude

$$ \mathbf{I}=t\big(\mathbf{p}\cdot\frac{\mathbf{v}}{\|\mathbf{v}\|}\big)\frac{\mathbf{v}}{\|\mathbf{v}\|}+\frac{1}{2}\big(e^{t\mathbf{v}}\mathbf{p}e^{-t\mathbf{v}}\big)\times\frac{\mathbf{v}}{\,\,\|\mathbf{v}\|^2}+\mathbf{C}. $$

Another route is to expand the exponentials $\exp(t\mathbf{v})$ using Euler's formula so you can multiply out and integrate directly. This works especially nice if, setting $\hat{\mathbf{v}}=\frac{\mathbf{v}}{\|\mathbf{v}\|}$ you write $\mathbf{p}$ with respect to an oriented orthonormal basis $\{\mathbf{u},\hat{\mathbf{v}},\mathbf{w}\}$.


问:
Weird $3^n$ in an identity to be combinatorially proved
Give a combinatorial proof of the following identity:

$$3^n=\sum_{i=0}^{n}\binom{n}{i}2^{n-i}$$

I can't see any counting argument that would yield $3^n$, and the right hand side is also pretty opaque. For some reason I really really suck at doing these proofs - I just started my first combinatorics course.

答: Hint: Color $n$ balls with 3 colors.

Body: Hint: Color $n$ balls with 3 colors.


问:
Bijection between the complex representations of a real Lie algebra and the complex representations of its complexification
Is the following obvious? I think it is, but wanted to make sure before an exam tomorrow!

"There is a bijection between the complex representations of a real Lie algebra and the complex representations of its complexification."

I think this is true (and very useful) because the complex representation of the complexification has the extra requirement of being complex linear, which fixes any extra freedom you'd get from complexifying.

Morally though, this seems weird - particularly as a complex Lie algebra may have non-isomorphic real forms. Does anyone have any comments on this which might make it seem more intuitive?

Many thanks in advance.

答:
It's 7 years too late for your exam but hopefully anyone else who is also wondering will appreciate an answer: yes, this is true (as long as you are careful to either say "bijection between isomorphism classes" or to talk about an equivalence of categories as Torsten says in the comments), and it follows directly from the universal property of complexification, namely that if $\mathfrak{g}$ is a real Lie algebra and $\mathfrak{h}$ is a complex Lie algebra then

$$\text{Hom}_{\mathbb{R}}(\mathfrak{g}, \mathfrak{h}) \cong \text{Hom}_{\mathbb{C}}(\mathfrak{g}_{\mathbb{C}}, \mathfrak{h}).$$

Then take $\mathfrak{h} = \mathfrak{gl}_n(\mathbb{C})$.

Morally though, this seems weird - particularly as a complex Lie algebra may have non-isomorphic real forms. Does anyone have any comments on this which might make it seem more intuitive?

Yes, and all of those real forms have the same complex representation theory, because they all have the same complex representation theory as their complexification. For example $\mathfrak{su}(2)$ and $\mathfrak{sl}_2(\mathbb{R})$ are non-isomorphic real forms of $\mathfrak{sl}_2(\mathbb{C})$ and they all have the same complex representation theory. Everything that the complex representation theory can detect about a Lie algebra factors through its complexification, again by the universal property of complexification.


问:
$C[0,1]$ is complete w.r.t. which norm(s)
$C[0,1]$ is complete w.r.t. which norm(s)

$\displaystyle\|f\|_\infty=\sup_{t\in[0,1]}|f(t)|$

$\displaystyle\|f\|_1=\int_0^1|f(t)| \, dt$

$\displaystyle\|f\|_\infty^{0,1}=\|f\|_\infty+|f(0)|+|f(1)|$

$\displaystyle\|f\|=\sqrt{\int_0^1|f(t)| \, dt}$

I know 1 is true. But I don't know how to check completeness of norm base on know complete norm. How can I solve it?

答:
Consider $$f_n(x) = \begin{cases} 0, & \text{if $x \leq 0 \leq \frac{1}{2}$} \\
 n\left(x - \frac{1}{2}\right), & \text{if $\frac{1}{2} \leq x \leq \frac{1}{2} + \frac{1}{n}$}\\
1, & \text{if $\frac{1}{2} + \frac{1}{n} \leq x \leq 1$}. \end{cases}$$

Then $f_n(x)$ converges in the $L^1$ norm to the function that is $0$ on $[0,0.5]$ and $1$ on $(0.5,1]$. For $|f_n(x) - f|$ is a non - increasing sequence converging pointwise to $0$ and so

$$\lim_{n \rightarrow \infty} \int | f_n - f| = \int \lim_{n\rightarrow\infty} |f_n - f| = 0.$$

by the monotonce convergence theorem. Your "norm" in 4) is not a norm because it does not satisfy $||a\textbf{x}|| = |a| ||\textbf{x}||$.


问:
Monty Hall problem generalized to $n$ doors
Generalize the Monty Hall problem where there are $n \geq 3$ doors, of
which Monty opens $m$ goat doors, with $1 \leq m \leq n$.
Original Monty Hall Problem: There are $3$ doors, behind one of which there is a car (which you want), and behind the other two of which there are goats (which you don’t want). Initially, all possibilities are equally likely for where the car is. You choose a door, which for concreteness we assume is Door $1$. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the car, will always open a goat door and offer the option of switching, and as above assume that if Monty Hall has a choice between opening Door $2$and Door $3$, he chooses Door $2$ and Door $3$ with equal probability .
Find the probability that the strategy of always switching succeeds, given that Monty opens Door $2$.

My approach:

Let $C_i$ be the event that car is behind the door $i$,
$O_i$ be the event that Monty opened door $i$ and $X_i$ be the event that intially I chose door $i$. Here $i=1,2,3,...,n$.

Let's start with case where I chose $X_1$. Then:

$P(O_{j_1, j_2, ..., j_m}|C_1, X_1) = {{n-1}\choose{m}}(\frac{1}{n-1})^m$, here $j \in$ {$m$ doors out of $n-1$, i.e., exclude Door$1$ }

$P(O_{k_1, k_2, ..., k_m}|C_t, X_1) = {{n-2}\choose{m}}(\frac{1}{n-2})^m$,
  here $k \in$ {$m$ doors out of $n-2$, i.e., exclude Door$1$ & Door$t$}, $t \in$ {$2,3, ..., n$}

Also, $P(C_r|X_s) = \frac{1}{n}$, here $r,s \in$ {$1,2,...,n$}

Probability of winning by switching is,

$$P(C_3 | O_{k_1, k_2, ..., k_m}, X_1) = \frac{P(O_{k_1, k_2, ..., k_m}|C_3, X_1).P(C_3|X_1)}{P(O_{m-doors}|X_1)}$$

$$= \frac{P(O_{k_1, k_2, ..., k_m}|C_3, X_1).P(C_3|X_1)}{P(O_{j_1, j_2, ..., j_m}|C_1, X_1).P(C_1|X_1) + \sum_{t=2}^n(P(O_{k_1, k_2, ..., k_m}|C_t, X_1).P(C_t|X_1))}$$

$$ = \frac{{{n-2}\choose{m}}(\frac{1}{n-2})^m.\frac{1}{n}}{(\frac{1}{n}).({{n-1}\choose{m}}(\frac{1}{n-1})^m + {{n-2}\choose{m}}(\frac{1}{n-2})^m.(n-1))}$$

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

However, the correct answer is $\frac{(n-1)}{(n-m-1).n}$. Any insights to what I have done wrong.

答:
While the original question has been answered, I'll post another solution to the actual problem at hand.

It will greatly simplify the calculations if we assume that the host momentarily identified a set $y > m$ doors (just in his mind) and opened $m$ doors from this $y$ set. Later we'll just make $y=n-1$.

In the beginning, the winning probability of group $y$ as a whole is simply $y \over n$. And, if $y$ wins, the winning probability of each door inside it is $1 \over y$. With $m$ non-winning doors now opened from $y$, the winning probability of each door in $y$ becomes $1 \over y-m$, provided $y$ wins. Hence the absolute winning probability of each door within $y$ is simply ${y \over n} \times {1 \over y-m}$, or ${1 \over n} \times {y \over y-m}$. Since ${y \over y-m} >1$, switching to any door within $\boldsymbol{y}$ is always advantageous.

Finally, on getting rid of $y$ by making it as large as $n-1$, i.e. it having every door except the player's original choice, we get the winning probability of switching which is $$ {1 \over n} \times {y \over y-m} = {1 \over n} \times {n-1 \over n-1-m} $$

Have a look at these two pages for comparing the real world results with the predictions from the equation - Monty Hall Paradox Test Tool and The Monty Hall Paradox Probability Equations.


问:
If $f$ is differentiable in a neighbourhood of $c$, is $f'(x)$ continuous at $x=c$?
I'm pretty sure the statement

"If $f$ is differentiable at $c$, is $f'(x)$ continuous at $x=c$"

is wrong and there are quite a lot of counterexamples to that, but what if we add a " differentiable on neighbourhood" condition?

Or, could the hypothesis be strengthened to:

"If $f$ is differentiable in a neighbourhood of $c$, is $f'(x)$ continuous at a neighbourhood of $c$?"

答: It is false. Take $f(x)=x^2\sin(1/x)$ for $x\neq 0$ and $0$ at $x=0$. This function is differentiable in a neighborhood of zero (in particular $f'(0)=0$) but $f'$ is discontinuous at zero since $\lim_{x\to} f'(x)$ does not exist. Derivatives cannot have jump discontinuities, but they can have essential discontinuities, like in the example.


问:
how can i solve the following integral
$$I = \int _{a}^{a+h} x^n \alpha  h^{-\alpha}(x-a)^{\alpha-1} dx$$

$$I = \alpha h^{-\alpha} \int_{a}^{a+h}x^n  (x-a)^{\alpha-1} dx$$

The result will be

$$\sum_{k=0}^{n} {n \choose k} h^k a^{n-k} \frac{\alpha}{\alpha+k}$$

答:
Make the substitution $x=a+y$, $dx=dy$.  The integral becomes

$$\alpha h^{-\alpha} \int_0^h dy \, (a+y)^n y^{\alpha-1}$$

You can expand the binomial, assuming $n \in \mathbb{N}$:

$$(a+y)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}y^k$$

You then reverse order of sum and integration and get

$$\alpha h^{-\alpha} \sum_{k=0}^n \binom{n}{k} a^{n-k} \int_0^h dy \, y^{k+\alpha-1} = \alpha h^{-\alpha} \sum_{k=0}^n \binom{n}{k} a^{n-k} \frac{h^{k + \alpha}}{k+\alpha}$$

The result follows.

Title: how can i solve the following integral


问:
How to write a polynomial function that has the roots $-2$ and $\sqrt7$?
I need to write a polynomial function with integer coefficients that has the roots $-2$ and $\sqrt7$. I'm able to do this correctly when I'm given roots like $-3+i$ & $-3-i$, in which I set the roots equal to zero and then multiply them by one another. However, when I try this with $-2$ and $\sqrt7$ and multiply $x+2$ by $x-\sqrt7$, I get $x^2+2x-\sqrt7 x-2\sqrt7$. I don't know where to go from here, and I don't think that this is the correct next step. What do I do next?

答:
Try multiplying by $x+\sqrt7,$ the conjugate of $x-\sqrt7$, as well.

$(x+\sqrt7)(x-\sqrt7)=x^2-7$ has integer coefficients, and its product with $x+2$ will too.


问:
Help proving a simple tautology $(p\land(p\to\lnot q))\to(\lnot q\lor p)$
Specifically this  $$(p\land(p\to\lnot q))\to(\lnot q\lor p)$$-

spent a lot of time going down the wrong path

答:
The statement is valid. To see this, we can use a proof tree, like so

This was generated here


问:
Defining a lambda expression to swap the order of lambda abstractions
I want to define a lambda term $\mathrm{swapQ}$ such that

$$\begin{align}
\mathrm{swapQ}\ &(\lambda P Q_1 Q_2. Q_1(\lambda x_1. Q_2 (\lambda x_2. Px_1x_2)))\\
\triangleright_\beta\
& (\lambda P Q_1 Q_2. Q_2 (\lambda x_2. Q_1(\lambda x_1. Px_1 x_2)))
\end{align}$$

So far I have come up with
$$\mathrm{swapQ^*} := \lambda S' P' Q_1' Q_2'. S' P' Q_2' Q_1' \\
\begin{align}
\mathrm{swapQ^*}\ &(\lambda P Q_1 Q_2. Q_1(\lambda x_1. Q_2 (\lambda x_2. Px_1x_2)))\\
\triangleright_\beta\
& (\lambda P Q_1 Q_2. Q_2 (\lambda x_1. Q_1(\lambda x_2. Px_1 x_2)))
\end{align}$$

but this only swaps the $Q_i$s, not the $\lambda x_i$s that should go with them.

After having fiddled around for a while, I can not seem to find a way to change order of the variable abstractions while leaving the variable occurrences inside the term intact (or vice versa for an equivalent expression). I probably need to abstract over the $(\lambda x_i. \ldots$) parts and reapply them in the same way I did with the $Q$'s, but I can not find the right way to "inject" this override application at the right place without taking the embedded applications with them.

Is what I want to do possible at all? I would appreciate hints for the right technique or perhaps pointers to the literature in case this has been done before.

Background:

In Montague grammar for formal natural language semantics, sentences can be formed by consecutively applying $n$ quantified expressions $Q_i$ (where quantified expressions are functions from sets of individuals to truth values) to an $n$-ary verb $P$ (where an $n$-ary verb is a function from $n$ individuals to truth values) to produce a statement. For instance,

$$
\begin{align}
& \mathrm{(sentence)(loves)((every)(woman))((a)(man))}\\
\triangleright_\beta
& (\mathrm{(every\ woman)}(\lambda x_1. (\mathrm{(a\ man)}(\lambda x_2. \mathrm{loves}x_1x_2))))\\
\triangleright_\beta
& \forall x (Wx \to \exists y (My \land Lxy))\\
& \text{"Every woman loves a man"}
\end{align}
$$
where
$$\begin{align}
\mathrm{sentence} = & \lambda P Q_1 Q_2. Q_1(\lambda x_1. Q_2 (\lambda x_2. Px_1x_2)))\\
\mathrm{every} = & \lambda P_1 P_2. \forall x (P_1x \to P_2x)\\
\mathrm{a} = & \lambda P_1 P_2. \exists x (P_1x \land P_2x)\\
\mathrm{loves} = & \lambda x y. Lxy\\
\mathrm{woman} = & \lambda x. Wx\\
\mathrm{man} = & \lambda x. Mx\\
\end{align}$$
and application is left-associative ($MNP = (MN)P$).

The purpose of the $\mathrm{swapQ}$ combinator is to swap the two quantified expressions to produce the inverted scope reading, such that

$$\begin{align}
& \mathrm{((swapQ)(sentence))(loves)((every)(woman))((a)(man))}\\
\triangleright_\beta
& (\mathrm{(a\ man)}(\lambda x_2. (\mathrm{(every\ woman)}(\lambda x_1. \mathrm{loves}x_1x_2))))\\
\triangleright_\beta & \exists y (My \land \forall (Wx \to Lxy))\\
& \text{"There is a man every woman loves"}
\end{align}$$
That is, the swapping combinator is applied to the sentence forming combinator such that the order in which the quantified expressions take scope over each other is swapped, while retaining their associated argument position in the verb (determined by which $x_i$ the quantifiers bind to) such that subject remains subject and object remains object.

The motivation is linguistic, but the problem is a purely lambda-technical one, so I figured it is best off at MathSE.

答:
Since $x_1, x_2$ are just dummy variables, one has
$$
(\lambda P Q_1 Q_2. Q_2 (\lambda x_2. Q_1(\lambda x_1. Px_1 x_2)))
=
(\lambda P Q_1 Q_2. Q_2 (\lambda x_1. Q_1(\lambda x_2. Px_2 x_1)))
$$

Thus you want
$$
\begin{align}
\mathrm{swapQ}\ &(\lambda P Q_1 Q_2. Q_1(\lambda x_1. Q_2 (\lambda x_2. Px_1x_2)))\\
\triangleright_\beta\
& (\lambda P Q_1 Q_2. Q_2 (\lambda x_1. Q_1(\lambda x_2. Px_2 x_1)))
\end{align}
$$
so $\mathrm{swapQ}$ should not only swap $Q_1$ and $Q_2$ but also the order of application on $P$:
$$
\mathrm{swapQ} := \lambda S' P' Q_1' Q_2'. S' (\lambda x y.P' y x) Q_2' Q_1' \\
$$


答:
$x = \sqrt 7 \implies$

$x^2 = 7\implies$

$x^2 - 7 =0$.  So $x^2 -7$ has $\sqrt 7$ as a root.

And $x=-20\implies x+2 = 0$ so $x+2$ has $-2$ as a root.

If $x^2-7=0$ and $x+2=0$ then $(x^2-7)(x+2)=0$ and $(x^2-7)(x+2)=x^3+2x^2-7x-14$ has $\sqrt 7$ and $-2$ as roots.

(It also has $-\sqrt 7$ as a root.)

=====

By the way.  There is no 2nd degree polynomial with two roots $-2$ and $\sqrt 7$.

A second degree polynomial has at most two roots and if the two roots are $r_1$ and $r_2$ the polynomial is $(x-r_1)(x-r_2)= x^2 -(r_1+r_2)x + r_1r_2$ and if $r_1$ or $r_2$ are irrational we have no reason to assume $r_1+r_2$ or $r_1r_2$ are integers.

Is if $r_1,r_2 = -2,\sqrt 7$ they won't be.  But to counter $\sqrt 7$ as a root we can add a third root of $-\sqrt 7$ to get a third degree polynomial:

$(x-\sqrt 7)(x+\sqrt 7)(x+2) = x^3+2x^2-7x-14$.

Title: How to write a polynomial function that has the roots $-2$ and $\sqrt7$?

Body: I need to write a polynomial function with integer coefficients that has the roots $-2$ and $\sqrt7$. I'm able to do this correctly when I'm given roots like $-3+i$ & $-3-i$, in which I set the roots equal to zero and then multiply them by one another. However, when I try this with $-2$ and $\sqrt7$ and multiply $x+2$ by $x-\sqrt7$, I get $x^2+2x-\sqrt7 x-2\sqrt7$. I don't know where to go from here, and I don't think that this is the correct next step. What do I do next?


问:
Does there exist a bijection that is not eventually equal to the identity, but has a finite number of "crosses"?
Consider a  bijection $f: \mathbb{N} \to \mathbb{N}$:

We define the (possible infinite) number of "crosses" or "intersections" by the following:

For every pair $i,j$ with $i<j, $ define:

$$
\alpha_{ij}=\begin{cases}
0&\ \text{if} \ f(j) > f(i)\\[8pt]
1&\ \text{if} \ f(j) < f(i)\\[8pt]
\end{cases}
$$

Then the number of (possibly infinite) intersections is $\ s(f) = \displaystyle \sum_{i=1}^\infty \left( \sum_{j=i+1}^\infty\alpha_{ij} \right)$.

(s depends only on the bijection $f$).

Also, define a bijection eventually equal to the identity to be one in which $ \exists N$ such that $f(n) = n \quad \forall n \geq N$.

My conjecture is that a bijection $f: \mathbb{N} \to \mathbb{N}$ is eventually equal to the identity $ \iff s$ is finite, but I don't know how to prove this formally.

Note that my conjecture would be false with bijections $f: \mathbb{Z} \to \mathbb{Z}$, for example:

The pigeonhole principle came to mind, but I'm not sure if it's useful or necessary for a proof.

答:
In my question, I should have drawn arrows to represent $f$, and not just straight lines without arrows. I intended $f$ to be arrows going from the top row to the bottom row.

$ \implies$ is easy, so we focus on $ \impliedby$: suppose s, the number of crosses, is finite. Then the number of integers in the top row involved in a crossing is finite (reasoning of this is left up to the reader). Because $f$ is a bijection, the number of integers in the bottom row involved in a crossing is the same as the number of integers in the top row involved in a crossing, and is therefore also finite.

Let $M_1 = \max$ { $n: \text{ the arrow } ( n, f(n) )$ is involved in a crossing } $= \max $ { $n:f(n) < f(m)$ for some $m < n$ }, and let $N_1 = \max$ { $f(n):\text{ the arrow } ( n, f(n) )$ is involved in a crossing } $= \max $ { $n:f^{-1}(n) < f^{-1}(m)$ for some $m < n$ }.

Note first that it is not necessarily the case that every $m<M_1$ is involved in a crossing. But $M_1$ and $N_1$ are involved in crossings by definition, so we use this fact in our proof of the Lemma and $(1)$ below.

$\bbox[lightgrey]{\text{Lemma: } f(M_1) < N_1\ \text{ and } f^{-1}(N_1)<M_1.\ }$ Proof of the first half of the Lemma: Suppose $f(M_1) > N_1.\ $ Then $f(M_1)$ is by definition involved in a crossing and is $> N_1$, contradicting the maximality of $N_1.$ If $f(M_1) = N_1,\ $ then since the arrow $(M_1, N_1)\ $ is involved in a crossing, either $M_1$ or $N_1$ will not be maximal, contradicting one of their definitions. The proof of the second half of the Lemma is very similar.

$\bbox[yellow]{(1): \quad m \leq M_1 \implies f(m) \leq N_1; \quad n \leq N_1 \implies f^{-1}(n) \leq M_1}.$ Proof of the first half of $(1)$: We must have $m<M_1 \implies f(m) < N_1+1\ $ else if such an $m'<M_1$ exists (such that $f(m') \geq N_1+1),\ $ then by the Lemma the arrow $(m', f(m'))\ $ crosses the arrow $(M_1, f(M_1)),\ $ and $f(m')>N_1,\ $ contradicting the maximality of $N_1$. The proof of the second half of $(1)$ is very similar.

$\bbox[yellow]{(2): \quad f(M_1+1) = N_1+1}.\  \text{Proof: by (1), }f(M_1+1) > N_1.\ $ Also,$ \ f(M_1+1) \not> N_1+1\ $ because then $N_1+1$ would be dead. This proves the result of $(2)$.

By the same reasoning as in $(2),\ f(M_1+2) = N_1+2,\quad f(M_1+3) = N_1+3,\ $ and so on. This shows that:

$\bbox[yellow]{(3): \quad \forall k \in \mathbb{N},\quad f(M_1+k) = N_1 + k}$, noting that I do not count $0$ as a natural number. Finally, going back to $(1)$, we have:

$\bbox[yellow]{(4): \quad M_1 = N_1 }$, else if $M_1 \neq N_1$ we would have a bijection between $N_1$ and $M_1 \neq N_1$ integers, which isn't possible.

Putting $(3)$ and $(4)$ together $ \implies$

$f(M_1 +k) = M_1 + k \quad \forall k \in \mathbb{N},\ $ i.e. $f$ is equal to the identity $\forall n \geq M_1+1.$


问:
Good undergrad level graph theory book?
I'm looking for a good undergraduate level graph theory book. I tried to read Bondy and Murty's, but I struggled with the first part about isomorphisms (I think, because I haven't gone through a Topology course yet). So,  before reading those more difficult books, I want one that is easier to get acquainted with the subject but that is not a dumbed-down-graphs-for-dummies type of text.

Any suggestions?

答: Introduction to Graph Theory, by Richard Trudeau (originally published under the title, Dots and Lines).</p>

Title: Good undergrad level graph theory book?


问:
Proof $(A-\{a\}) \sim (B - \{b\}) $
I must prove the following:

"let $A$ and $B$ two set, with $A \neq \emptyset$ and $B \neq \emptyset$ and $A \sim B$, and $a \in A$ and $b \in B$, then $(A-\{a\}) \sim (B - \{b\})$

The symbol $\sim$ is used for "equipotency"!!

Thanks in advance!

答:
Here is an example for finite $A,B$. The general case is similar.
<br>
Assume that $A=\{1,2,3,4,5,6,7\},B=\{p,q,r,s,t,u,v\}, \ a=2, \ b=s$ and the function $f$ which gives $A\sim B$ is given by
$$
\begin{pmatrix}
1&\rightarrow&q\\
\color{red}{2}&\rightarrow&\color{green}{r}\\
3&\rightarrow&p\\
4&\rightarrow&t\\
\color{green}{5}&\rightarrow&\color{red}{s}\\
6&\rightarrow&v\\
7&\rightarrow&u\\
\end{pmatrix}
$$
Then the function $g$ which gives $(A-\{a\})\sim (B-\{b\})$ is given by
$$
\begin{pmatrix}
1&\rightarrow&q\\
3&\rightarrow&p\\
4&\rightarrow&t\\
\color{green}{5}&\rightarrow&\color{green}{r}\\
6&\rightarrow&v\\
7&\rightarrow&u\\
\end{pmatrix}
$$ </p>

Title: Proof $(A-\{a\}) \sim (B - \{b\}) $

Body:
Here is an example for finite $A,B$. The general case is similar.

Assume that $A=\{1,2,3,4,5,6,7\},B=\{p,q,r,s,t,u,v\}, \ a=2, \ b=s$ and the function $f$ which gives $A\sim B$ is given by
$$
\begin{pmatrix}
1&\rightarrow&q\\
\color{red}{2}&\rightarrow&\color{green}{r}\\
3&\rightarrow&p\\
4&\rightarrow&t\\
\color{green}{5}&\rightarrow&\color{red}{s}\\
6&\rightarrow&v\\
7&\rightarrow&u\\
\end{pmatrix}
$$
Then the function $g$ which gives $(A-\{a\})\sim (B-\{b\})$ is given by
$$
\begin{pmatrix}
1&\rightarrow&q\\
3&\rightarrow&p\\
4&\rightarrow&t\\
\color{green}{5}&\rightarrow&\color{green}{r}\\
6&\rightarrow&v\\
7&\rightarrow&u\\
\end{pmatrix}
$$


问:
Is this trigonometric identity true? $\frac{\cos{x} - \cos{y}}{x-y}= \cos{x}$
I saw this example in a book:

$$\frac{\cos{x} - \cos{\frac{\pi}{2}}}{x - \frac{\pi}{2}} = \cos{x}$$

Is this a general rule?

$$\frac{\cos{x} - \cos{y}}{x-y}= \cos{x}$$

答: Of course not. With $y=0$ it says $\cos x - 1 = x \cos x$.


答:
($p \Rightarrow \sim q $ )
$ \wedge p$
by Modus Ponens

$\Rightarrow   \sim q $
and by adition $\sim q \vee p $

Title: Help proving a simple tautology $(p\land(p\to\lnot q))\to(\lnot q\lor p)$

Body:
Specifically this  $$(p\land(p\to\lnot q))\to(\lnot q\lor p)$$-

spent a lot of time going down the wrong path


问:
Every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without Lefschetz theorem.
I would like to know if there is a nice proof of the fact that every continuous map $f:\mathbb{C}P(2) \to \mathbb{C}P(2)$ has a fixed point, without use of the Lefschetz fixed point theorem.

答:
I don't understand the first paragraph of Connor's answer at all, but assuming it checks out, in general the total Chern class of $\mathbb{CP}^n$ is

$$(1 + \alpha)^{n+1} = \sum_{k=0}^n {n+1 \choose k} \alpha^k \in \mathbb{Z}[\alpha]/\alpha^{n+1}$$

(note that the $k = n+1$ term vanishes). If the (complex) tangent bundle of $\mathbb{CP}^n$ has a (complex) line subbundle then the total Chern class must factor as

$$(1 + c \alpha)(1 + c_1 \alpha + \dots + c_{n-1} \alpha^{n-1})$$

and since we never get a coefficient of $\alpha^{n+1}$ or higher, the problem of determining whether this is possible is equivalent to the problem of determining when $(1 + \alpha)^{n+1} - \alpha^{n+1}$ has a linear factor of the form $(1 + c \alpha)$, as an ordinary polynomial.

It will be easier to reverse the order of the coefficients: this is equivalent to determining when $\frac{(x + 1)^{n+1} - 1}{x}$ has a linear factor of the form $(x + c)$, where $c$ is an integer. This implies

$$(1-c)^{n+1} = 1$$

and if $n$ is even this gives $c = 0$, but $x + c = x$ is not a factor of the above polynomial because its constant coefficient is $n+1$ (this coefficient corresponds to the top Chern class and hence to the Euler characteristic $\chi(\mathbb{CP}^n) = n+1$ so we really are using that the Euler characteristic doesn't vanish). We conclude:

Claim: $\mathbb{CP}^{2m}$ has the fixed point property.

This is usually proven by Lefschetz (as far as I know anyway). If $n$ is odd then this gives either $c = 0$ or $c = 2$ so we don't get a contradiction, and I guess the existence of fixed-point-free maps in this case implies that the tangent bundle has a line subbundle with first Chern class $2$. I wonder if anyone knows a more explicit description of it.


问:
What is meant by 'stretching the neighbourhood'?
In the book I have it is written in an example -

To determine an expansion valid in the boundary layer, we need to stretch the neighbourhood of $x=1$. Thus, we let $\xi=(1-x)/\epsilon^v$.

What is meant by 'stretching the neighbourhood of $x=1$? I somewhat understand the concept of stretching the boundary layer, but what does the quoted part mean mathematically?

The problem is of 'boundary layer problems' of perturbation theory. The boundary layer is at the right end of $[0,1]$.

答:
I like to think of re-scaling in singular perturbation problems as "zooming in".

In this example, you have a small region near $x=1$ where your function changes really quickly. If you zoom in to the region near $x=1$ it will look like it's changing much more slowly. When solving equations, treating both the fast-changing and slow-changing parts together is difficult. But when you have zoomed in you can focus on just the fast-changing part of the solution and can forget about the rest of the solution. This gives you a simpler equation to try and solve. You can then solve for the slowly-changing part of the solution separately, and then finally combine both parts of solution.

Put another way, near $x=1$ your solution changes by $O(1)$ over a region of width $O(\epsilon)$. If you scale your independent variable by $\epsilon^{-1}$ then in the scaled coordinates your solution changes by $O(1)$ over a region of width $O(1)$. You can safely ignore terms that are very small relative to 1 (the $O(\epsilon)$ terms, for example).


答:
No, as a counter example, let $x=\pi$ and $y=\pi/2$.

$$\frac{\cos(\pi)-\cos(\pi/2)}{\pi-\pi/2}=\frac{-1-0}{\pi/2}=-2/\pi$$

whereas $\cos(\pi)=-1$

so in this case

$$\frac{\cos(x)-\cos(y)}{x-y}\neq \cos(x)$$

Also it is possible that in your book there might be other constraints because obviously you cannot have $x=y$. Maybe under certain conditions that statement follows, but in general that is not true.

Title: Is this trigonometric identity true? $\frac{\cos{x} - \cos{y}}{x-y}= \cos{x}$

Body:
I saw this example in a book:

$$\frac{\cos{x} - \cos{\frac{\pi}{2}}}{x - \frac{\pi}{2}} = \cos{x}$$

Is this a general rule?

$$\frac{\cos{x} - \cos{y}}{x-y}= \cos{x}$$


问:
How do I come up with a transformation matrix
What is the matrix that rotates by (pi/4) about an axis that is parallel to Y and passing through the point (9, 0, 0)?

Does anyone knows what this question means?

答:
Background:  Locations (positions) are represented by "augmented" vectors:  $(x,y,z, 1)^t$, and you operate on them with Affine transformation matrices to get the output $(X,Y,Z,1)^t$.

The matrix for the translation by $(\Delta X, \Delta Y, \Delta Z)^t$ is of the form:

$${\bf M}_{trans} = \left(
\begin{array}{cccc}
1 & 0 & 0 & \Delta X \\
0 & 1 & 0 & \Delta Y \\
0 & 0 & 1 & \Delta Z \\
0 & 0 & 0 & 1
\end{array}
\right)$$

(To check you understand this, compute ${\bf M}_{trans} (x, y, z, 1)^t$ by hand.  It takes just a few seconds to do that.)

Solution:  The Affine matrix for a rotation by $\theta$ about a three-dimensional vector ${\bf v}$ is given here.


问:
Comparing infinity with real numbers
This question is more about conventional notations in mathematics. Could we compare infinity with a real number and write something like $-\infty<0$? I have not seen this notation before but at the same I was not told not to use it. I tend to think of infinity as a concept rather than a number, but I was wondering such notations are allowed in mathematics.

The motivating reason behind this question is that I was trying to show the topology generated by all sets of the form $(-\infty,a)$ where $a \in \mathbb{R}$ does not contain the topology generated by the finite complement topology. For instance, $\mathbb{R}-\{0\}$ is in the finite complement topology but for the set $(0,\infty)$ it is impossible to have a basis of the form $(-\infty,a)$ because $-\infty<0$. If this notation is not allowed, how one would go about it?

答:
There are various ways to treat $\infty$ rigorously. The issue is that we have to be very careful how we do this (when we do, anyways). Here are some examples:

Positive or negative infinity as an extended real number - this is what you're thinking about, essentially. Expressions like "$(-\infty,0)$" and even "$(42, +\infty]$" are meaningful in this system, and - as long as we avoid square brackets at infinity! - amount to taking the usual interpretation and making the notation meaningful. The extended real numbers are quite convenient in certain situations, e.g. measure theory.

Infinity as a single point at infinity (so a "number circle," or "number sphere," or something similar).

Infinity as a measure of size of sets - that is, the transfinite cardinal numbers.

Infinity as a kind of generalized length of a process - that is, the transfinite ordinal numbers. Ordinals and cardinals each generalize the notion of natural number, not of real number. In a precise sense every cardinal is an ordinal, but not conversely.

Infinity as a "generalized real number," that is, an infinite element of a non-Archimedean ordered field (consider e.g. the surreal numbers). Here for example we can subtract infinities from each other in a well-behaved way - which we can't do in any of the other approaches above.

Those are the varieties of infinity which leap to mind, but I'm sure there are others. But we can't conflate these: e.g. the cardinal $\aleph_0$ and the extended real number $+\infty$ are simply unrelated objects.

Title: Comparing infinity with real numbers


问:
How to determine the angle from the origin for each of 2 planes when both are rotated about perpendicular axes
I'm sure this question has been asked before. In fact, there is likely a free online calculator out there that will solve it for me, except I'm not sure what terms to search for so I'm having a hard time finding them.  My math skills are fairly rusty, so please explain it to me like I'm in high school (or Jr. high... I'd be OK with that, too).

BACKGROUND:
I'm building a climbing wall in the corner of my basement. Note that this is the corner--we have 2 walls that meet at 90 degrees and I'm building onto both of them.  For the top part of the wall, I want to have the wall angle inward to make it more challenging.  The lower part of the wall is easy since it will go straight up from the floor--the plywood that goes up will sit one edge on the floor, one face against the wall, and then the edges of each of the 2 will meet up nicely with no modifications needed.  Since the top part is angling inward, though, I'll need to cut a triangle out where the edges of the boards meet. I want to know HOW to determine the angle to cut each sheet.  For simplicity, let's ignore the bottom section completely and assume the entire wall will be angled inward.

PROBLEM:
Assume I have 2 walls. One is in the X,Y plane and the other is in the Y,Z plane.  We will call the point where both walls and the floor meet the origin. I have a 2 sheets of plywood that are 4x4 squares. I want to mount the plywood such that:

the first starts on the X,Y plane, but is then rotated 20 degrees about the X axis such that the bottom is still on the X axis (On the line {0,0,0}, {0,4,0}), but the top comes in (positive Z value).

the second starts on the Y,Z plane, but is then rotated 30 degrees about the Z axis such that the bottom is on the line ({0,0,0}, {0,0,4}) and the top comes in (positive X value).
I need to know the angle to cut each sheet of plywood.

MUSINGS:
At first I thought I could just cut one at a 20 degree angle and the other at a 30 degree angle, but then I got to thinking: If I angled just one in at 30 degrees and left the other flat on the wall (0 degrees), then, yes--that one could be cut at 30 degrees and it would work.  However, if I angled the second all the way down to the floor (90 degrees) then it wouldn't need to be cut at all.  Since both are angling on each other, I think it is more complicated than simply cutting at the angle I want the other wall.

I know how to use trigonometry to determine how far away from the wall the top will be and stuff like that, but I don't know how to put the 2 together.  I've tried using GeoGebra, but I can't quite figure out how to get it to tell me the solution.  I did get a nice picture, though.  I include it for reference:

And since 3D is hard to show on 2D screens, here's another angle:

HOW do I determine the angles to cut?  Are you aware of any tools out there that would solve this for me or are there search terms I should use to look for one myself?

答:
I made a figure with GeoGebra, representing your two sheets (blue and red). As you can see, they intersect along a line forming an angle of $28.481°$ with the side of the red sheet, and an angle of $17.495°$ with the side of the blue sheet.

EDIT.

It is not so difficult to find an exact expression for those angles. Consider for instance the square on $yz$-plane, before rotation. This square then undergoes a 30° rotation, carrying it to the blue square. Its vertex $P=(0,4,0)$ is carried by the rotation to $P'=(2,2\sqrt3,0)$ (endpoint of a blue side), whose projection on $x$-axis is $H=(2,0,0)$. Point $P''$, which is the intersection of the side of the blue square parallel to $z$-axis with the red square, has then coordinates
$P''=(2,2\sqrt3,2\sqrt3\tan20°)$.

If $\theta=\angle P'OP''$, it follows
$$
\tan\theta={2\sqrt3\tan20°\over 4},
\quad\text{and}\quad
\theta=\arctan{\sqrt3\tan20°\over 2}\approx17.495.
$$
A similar reasoning leads to
$$
\tan\phi={2\cos20°\over 2\sqrt3},
$$
where $\phi$ is the other angle to be found.


问:
Proving $\frac { \Gamma(1-s) \Gamma(s) } {\Gamma(s/2) } = \frac{ 2^{s-1} \sqrt{\pi}} {\cos(\pi s/2) } $
I am trying a question in complex analysis and there solution was given but I am having trouble proving an identity involving Gamma Function.

Prove: $$\frac { \Gamma(1-s) \Gamma(s) } {\Gamma(s/2) } = \frac{ 2^{s-1} \sqrt{\pi}} {\cos(\pi s/2) } $$

I tried using Formula $\Gamma(s) =s\Gamma(s) $ and $\Gamma(s)\Gamma(1-s) = \frac{\pi}{ \sin(\pi s)}$, but in vain.

So, can you please tell how to prove them and which results you are using?

答:
A third very useful identity is the duplication formula $(18)$ :
$$\tag{1}\Gamma(2z)=(2\pi)^{-1/2}\,{2^{\,2z-1/2}}\;\Gamma(z)\,\Gamma\left(z+\frac 12\right)$$

\begin{align}
&\text{For $\,z=\dfrac s2\,$ this becomes}\\
\Gamma(s)&=(2\pi)^{-1/2}\,{2^{\,s-1/2}}\,\Gamma\left(\frac s2\right)\,\Gamma\left(\frac {s+1}2\right)\\
&\text{let's multiply by $\Gamma\left(\small{\frac {1-s}2}\right)$}\\
\tag{2}\Gamma(s)\;\Gamma\left(\frac {1-s}2\right)&=(2\pi)^{-1/2}\,{2^{\,s-1/2}}\,\Gamma\left(\frac s2\right)\,\Gamma\left(\frac {s+1}2\right)\Gamma\left(\frac {1-s}2\right)\\
&\text{and use the reflection formula :}\\
\tag{3}\Gamma\left(\frac {s+1}2\right)\Gamma\left(1-\frac {s+1}2\right) &= \frac{\pi}{ \sin\left(\pi \frac {s+1}2\right)}= \frac{\pi}{ \cos\left(\pi \frac s2\right)}\\
&\text{combining $(2)$ and $(3)$  gives :}
\end{align}
$$\tag{4}\boxed{\frac{\Gamma(s)\,\Gamma\left(\frac{1-s}2\right)}{\Gamma\left(\frac s2\right)}=\frac{2^{\,s-1}\,\sqrt{\pi}}{ \cos\left(\pi \frac s2\right)}}$$
There was merely a division by $2$ missing in one of the gamma functions!


问:
Show $(\mathbb{R},0,+,-)$ has infinitely many automorphisms
Show $(\mathbb{R},0,+,-)$ has infinitely many automorphisms :

My idea is that if $\phi:(\mathbb{R},0,+,-) \rightarrow (\mathbb{R},0,+,-)$ is an automorphism. Then since here we don't have the order relation. For any $x_1, x_2, y_1, y_2 \in \mathbb{R}$, $(x_1,x_2 ) \rightarrow (y_1, y_2)$ will be an automorphism. In another words, any we can always map an open interval to another open interval to have automorphism.

Could someone let me know if my thinking is correct?

If not, could you let me know what went wrong, and how should I fix it?

Any hints or ideas will be appreciated!

Thanks!

答:
The answer you propose in the comments is correct: For any non-zero $r\in \mathbb{R}$, the function $x\mapsto rx$ is an automorphism of $(\mathbb{R},0,+,-)$. I'll leave it to you to check the details.

On the other hand, what you've written doesn't make any sense to me. You say "$(x_1,x_2)\to (y_1,y_2)$ will be an automorphism". What function $\mathbb{R}\to \mathbb{R}$ do you have in mind here? It's not true that any open interval can be  mapped to any other open interval by an automorphism. For example, $(-1,1)$ can't be mapped to $(1,2)$, since the first interval contains $0$, the second does not contain $0$, and any automorphism maps $0$ to $0$.


问:
Three points $P,Q,R$ on a circle $O$. Draw circles with diameters $PQ,QR, RP$. Probability when $O$ lies inside the three circles drawn?
Sample three random points $P,Q,R$ on a unit circle $O$. Draw circles with diameters $PQ,QR,RP$. Find the probability when origin $O$ lies inside at least one of the three circles drawn?

There is a well know problem where we are simply looking for probability when $O$ is within $\triangle PQR$ and the probability is $\frac{1}{4}$. but is there a solution when we are looking for the probability when origin lies within circles with diameters $PQ,QR,RP$?

Probability that the center of a circle lies within three chosen points on a circle

What is the probability that the center of the circle is contained within the triangle?

答:
We will first look at probability of $O$ not falling in any of the $3$ circles.

As we are taking all $3$ segments as diameter of $3$ circles, their radii (say, $r$ for any one of them) has to be less than the perp distance from $O$ to the segment (say, $h$) for $O$ not to be in any of the circles.

Now, $r^2 + h^2 = 1^2$ given unit circle. As the radius has to be less than the distance from $O$ to the segment, $r \leq h \implies r \leq \frac{1}{\sqrt 2} (\leq 45^0$ angle).

So the segment should subtend an angle of $\leq \frac{\pi}{2}$ at $O$.

If we take point $P$ randomly on the unit circle, the probability of the next point $Q$  being within angle $\frac{\pi}{2}$ to it (on either side of it) will be $\frac{\pi}{2\pi} = \frac{1}{2}$.

Say points $P$ and $Q$ subtend an angle of $\theta (\leq \frac{\pi}{2})$ at $O$. Then point $R$ can be within $(\frac{\pi}{2} - \theta)$ angle on either side of $P$ or $Q$. But it can also fall between point $P$ and $Q$ hence the probability of point $R$ satisfying the condition such that point $O$ is not within the $3$ circles $= \displaystyle \frac{2(\frac{\pi}{2} - \theta) + \theta}{2\pi} = \frac{\pi - \theta}{2\pi}$ where $0 \lt \theta \leq \frac{\pi}{2}$.

The probability for point $R = \frac{1}{\pi/2} \int_0^{\pi/2} (\frac{1}{2} - \frac{\theta}{2\pi}) d\theta = \frac{2}{\pi} (\frac{\pi}{4} - \frac{\pi}{16}) = \frac{3}{8}$.

So probability that none of the circles have point $O = \frac{1}{2} \times \frac{3}{8} = \frac{3}{16}$.

So desired probability $ = 1 - \frac{3}{16} = \frac{13}{16}$


问:
Understanding the field of order $p^2$
So I understand that for any prime $p$, there exists a field of order $p^2$. What's confusing me, is that a field is also an abelian group, and the only abelian groups of order $p^2$ up to isomorphisms are $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p \times \mathbb{Z}_p$, neither of which are fields. Can anyone please help me wrap my head around how its possible for a field of order $p^2$ to both be an abelian group of order $p^2$, and simultaneously not be isomorphic to any abelian group of order $p^2$ that exists?

答:
Let $p$ be a prime. Then the field with $p^2$ elements has the underlying additive group $\mathbb{Z}_p\times \mathbb{Z}_p$.

You state that $\mathbb{Z}_p\times \mathbb{Z}_p$ is "not a field," and indeed it is not when you view it with the standard multiplication entrywise in each component. But the field with $p^2$ has a different multiplication structure which allows you to turn the additive group $\mathbb{Z}_p\times \mathbb{Z}_p$ into a commutative ring where every nonzero element is invertible with respect to this multiplication.

Title: Understanding the field of order $p^2$


问:
Solving probability problem with given information that doesn't add up
I hope you all are doing okay.

I have a quick question regarding a probability problem that I've been struggling to solve.

This is what we were given $$\Pr(A^c) = 2\Pr(A)\\ \Pr(A|B^c)=.2 \\ 2\Pr(B|A) = .2 $$

So I've started working on it.  First, trying to find $\Pr(A)$: $$1-\Pr(A) = 2\Pr(A) \\ 1 = 3\Pr(A) \\ \frac{1}{3} = \Pr(A)$$.

Then, trying to find $Pr(A\cap B)$: $$\Pr(B|A)=.1 \\ \frac{\Pr (A \cap B)}{\Pr(A)} = .1 \\ \Pr(A \cap B) = \frac{1}{30}$$

I only need $\Pr(B)$ now. Let's use infromation from $2\Pr(B|A)$: $$2\Pr(B|A) = .2 \\ 2[1-\Pr(B^c|A)] = .2 \\ 1-\Pr(B^c|A) = .1 \\ 1- \frac{\Pr(B^c \cap A)}{\Pr(A)} = .1 \\ 1 - \frac{\Pr(B^c \cap A}{\frac{1}{3}} = .1 \\ \Pr(B^c \cap A) = .3$$

Thus,  since $\Pr(A|B^c)=.2$, and $\Pr(A|B^c) = \frac{\Pr(A \cap B^c)}{\Pr(B^c)}$, we observe: $$ .2 = \frac{.3}{\Pr(B^c)} \\ \Pr(B^c) = \frac{.3}{.2} = 1.5$$

Which isn't right at all.  I can't really figure out what I'm doing wrong here and I would absolutely appreciate your help.

答:
Your calculations are correct. The problem is that the conditions are inconsistent, leading to a nonsensical result.

Another way to compute $P(B^c)$ is to rearrange the second two conditions:
$$
P(A\cap B^c) = .2 P(B^c)
$$
$$
P(B\cap A) = .1 P(A)
$$
Add the two equations, using $P(A) = P(A\cap B) + P(A\cap B^c)$:
$$
P(A) = .2 P(B^c) + .1 P(A)
$$
Given $P(A)=\frac13$, we solve for $P(B^c)$ and obtain $1.5$ again.


答:
You may indeed obtain absurd results while using valid reasoning but inconsistent facts.

These facts are inconsistent.

Let us take $\Pr(A^\complement)=2\Pr(A)$ and $2\Pr(B\mid A)=0.2$ as given, and derive $\Pr(A\mid B^\complement)$ to compare.

We have $\Pr(A)=1/3$ as you obtained, $\Pr(B\mid A)=1/10$ as is clear, and so $\Pr(A\cap B)=1/30$ by definition: $\Pr(A\cap B)=\Pr(A)\Pr(B\mid A)$

Likewise by definition, $\Pr (B\mid A)=\Pr(A\cap B)\div\Pr(B)$.

Therefore: $~\Pr(B)~{=\Pr(A\cap B)\div\Pr(B\mid A)\\=\tfrac 1{30}\div\tfrac 1{10}\\=\tfrac 13}\\\Pr(B^\complement)=\tfrac 23$

By Law of Total Probability: $\Pr(A\cap B^\complement)~{=\Pr(A)-\Pr(A\cap B)\\=\tfrac 13-\tfrac 1{30}\\=\tfrac{3}{10}}$

Therefore $\Pr(A\mid B^\complement)~{=\dfrac{3/10}{2/3}\\=1/20\\=0.05}$

Not $0.2$ as claimed.  Therefore the three facts, which you were provided, are inconsistent.

You will obtain different results for $\Pr(B^\complement)$ depending on which two of the three 'facts' you choose to derive it.

Title: Solving probability problem with given information that doesn't add up

Body:
I hope you all are doing okay.

I have a quick question regarding a probability problem that I've been struggling to solve.

This is what we were given $$\Pr(A^c) = 2\Pr(A)\\ \Pr(A|B^c)=.2 \\ 2\Pr(B|A) = .2 $$

So I've started working on it.  First, trying to find $\Pr(A)$: $$1-\Pr(A) = 2\Pr(A) \\ 1 = 3\Pr(A) \\ \frac{1}{3} = \Pr(A)$$.

Then, trying to find $Pr(A\cap B)$: $$\Pr(B|A)=.1 \\ \frac{\Pr (A \cap B)}{\Pr(A)} = .1 \\ \Pr(A \cap B) = \frac{1}{30}$$

I only need $\Pr(B)$ now. Let's use infromation from $2\Pr(B|A)$: $$2\Pr(B|A) = .2 \\ 2[1-\Pr(B^c|A)] = .2 \\ 1-\Pr(B^c|A) = .1 \\ 1- \frac{\Pr(B^c \cap A)}{\Pr(A)} = .1 \\ 1 - \frac{\Pr(B^c \cap A}{\frac{1}{3}} = .1 \\ \Pr(B^c \cap A) = .3$$

Thus,  since $\Pr(A|B^c)=.2$, and $\Pr(A|B^c) = \frac{\Pr(A \cap B^c)}{\Pr(B^c)}$, we observe: $$ .2 = \frac{.3}{\Pr(B^c)} \\ \Pr(B^c) = \frac{.3}{.2} = 1.5$$

Which isn't right at all.  I can't really figure out what I'm doing wrong here and I would absolutely appreciate your help.


问:
Prove that there exist arbitrarily long sequence of consecutive integers not of the form $x^k$ with $k > 1$. (Assuming that $x \in \mathbb Z$)
I'm not sure how to approach this question exactly, I've thought about assuming that the length of the sequence is finite and bounded above and somehow construct a contradiction.

Any hint on how to approach this question would be great, thanks.

答: Any number of the form $4n+2$ can't be a perfect power (because it's divisible by $2$ but not by $4$). Similarly for any number of the forms $9n+3$, $25n+5$, $49n+7$, and so on. So all you have to do is show that for any $m$ there exists $r$ such that $r\equiv2\bmod4$ and $r+1\equiv3\bmod9$ and and and $r+m-1\equiv p_m\bmod{p_m^2}$.


问:
find the estimate of the tangent line approximation
Suppose that $f(x)$ is a function with $f(130)=46$ and $f′(130)=1$. Estimate $f(125.5)$.

I looked at the problem multiple times, but I am still stumped!

答:
Use $f(x)\approx f(130)+(x-130)f'(130)$.

(The tangent line will have slope $f'(130)$ and contain the point $(130,f(130)$.)


问:
Weird $3^n$ in an identity to be combinatorially proved
Give a combinatorial proof of the following identity:

$$3^n=\sum_{i=0}^{n}\binom{n}{i}2^{n-i}$$

I can't see any counting argument that would yield $3^n$, and the right hand side is also pretty opaque. For some reason I really really suck at doing these proofs - I just started my first combinatorics course.

Body:
Hints:

$$(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$$


问:
Is there any discernible pattern for the number list $\frac{7}{3}, \frac{5}{4}, 1, \frac{8}{9}, ...$
I was looking at the notes of a calculus' student and one problem their teacher stated was to find a way to algebraically define the list of numbers $\frac{7}{3}, \frac{5}{4}, 1, \frac{8}{9}, ...$ into a sequence. Is there a "logical" answer for this? To me, one could add literally any other number into the list and find a 4-degree polynomial arbitrarily. This could be done with any finite list of 4 numbers however there are cases where 'logic' or 'common sense' is appealed (e.g. finding the next term of the sequence $1, 2, 3, 4, ...$).

I tried to analyze their differences in a spreadsheet and found nothing, as well as the ratio between the consecutive terms. Any ideas?

Edit: Here's an image of the notes (they're from a class in spanish).

It was part of a task where they were asked to find the $5311th$ term, this can only be doable by finding an algebraic expression.

答:
Hint: It might be easier to consider non-simplified fractions as terms of the sequence:

$\frac{7}{3},\frac{10}{8},\frac{13}{13},\frac{16}{18}$...