问:
Expected number of balls of each color, after drawing until 50 balls from an urn are removed, with each color having a different chance to be put back
Lets say we have an urn with 100 balls (20 red, 30 blue, 50, green). Whenever a ball is dawn there is a chance it will be kept and not put it back. This chance with which the ball is kept is different for each colour (red = 0.5, blue = 0.3, green = 0.2). So for example the first ball randomly drawn is blue and with a 0.3 chance it is kept otherwise it is put back into the urn and another random ball is drawn. This is repeated until 50 balls have been kept. What is the most likely distribution of coloured balls/expected number of balls of each colour of the 50 kept?

The application I need it for is because I am simulating animals eating a certain amount of total food (the 50 balls) made up of different plant types (different colours) that are available differently in the landscape (number of balls per colour and number of total balls) and that are eaten with a certain preference (probability to keep a ball).

I have calculated this by simulating this draw 10000 times in the programming language R and calculating the mean number of balls of each colour. Code is below in case you are interested but it is not needed to answer my question.
For 50 balls taken the most likely distribution is roughly:

Red: 14.3367 balls

Blue: 15.8905 balls

Green: 19.7728 balls

However I do not want to simulate it but just be able to calculate it without simulating this draw as it takes much longer especially if I have more than 3 colours (or plant types). I would very much appreciate your answer or if you can tell me where to find one or how this type of problem is called, as I was unsucessful in googling this problem. Thank you in advance!

```<code>#The find_prob is number of balls in the urn divided by total balls and the keep_prob is the chance the ball will be kept.
total_reds_pre<-20
total_blues_pre<-30
total_greens_pre<-50

total_balls<-total_reds_pre+total_blues_pre+total_greens_pre
red_find_prob<-total_reds_pre/total_balls
blue_find_prob<-red_find_prob + total_blues/total_balls

needed<-50

red_keep_prob<-0.5
blue_keep_prob<-0.3
green_keep_prob<-0.2

red_taken<-c()
blue_taken<-c()
green_taken<-c()

for (i in 1:10000) { # repeated 1000 times to get closer to the real number
  red<-0
  blue<-0
  green<-0

  total_reds<-total_reds_pre
  total_blues<-total_blues_pre
  total_greens<-total_greens_pre
  total_balls<-total_reds+total_blues+total_greens

while ((red+blue+green < needed)) {
  red_find_prob<-total_reds/total_balls
  blue_find_prob<-red_find_prob + total_blues/total_balls
  colour_drawn<-runif(1,min = 0, max = 1)
  keep_colour<-runif(1,min = 0, max = 1)
  if (colour_drawn < red_find_prob){
    if (keep_colour < red_keep_prob) {
      red = red + 1
      total_balls = total_balls - 1
      total_reds = total_reds -1
    }else {}
  }else if (colour_drawn < blue_find_prob) {
    if (keep_colour < blue_keep_prob) {
      blue = blue + 1
      total_balls = total_balls - 1
      total_blues = total_blues - 1
    }else {}
  }else {
    if (keep_colour < green_keep_prob) {
      green = green + 1
      total_balls = total_balls - 1
    }else {}
  }
}
red_taken[i]<-red
blue_taken[i]<-blue
green_taken[i]<-green
}
mean(red_taken)
mean(blue_taken)
mean(green_taken)
mean(red_taken)+mean(blue_taken)+mean(green_taken)
</code>```

Just to make sure it is not misunderstood: I do not need help with the code I need a mathematical formula to calculate the most likely distribution (of course if you find a griveous mistake in the code please point it out).

EDIT: I have finally found a similar question not sure if it exactly what I need, I'll have a closer look once I am back home: Drawing balls from a bin with a color-specific probabilistic discard step

答:
This is a more knowledgeable you from the future (wow!)
Here is a piece of code that is way faster (at least in R) because it uses vector calculation. It is a deterministic approach and just an approximation (although quite good)

The basis of how it works is that it just takes the find probability times the keep_probabilty. This number is taken times an "amount_removed". The closer you set that amount removed to 0, the more accurate it will be. Then that amount is removed and the find probabilities are updated.
The theory behind it is that normally you couldnt just take find probability times keep probability times total needed, because it does not take into account that once you remove something you lower the probability for that to be found. However, if the amount you remove goes close to 0 then it does take it into account the next time you remove a very small amount as you have updated the find probability.

Results in comparison to the approach in the question and saulspatz's answer:

Amount
keep_chance
question approach
saulspatz's answer
this answer

20
0.5
14.3367
14.325163888283411
14.32446

30
0.3
15.8905
15.908709486720031
15.90252

50
0.2
19.7728
19.76612662499662
19.77302

Code in R:

```<code>#Your input
balls_pre<-c(20,30,50)
needed<-50

#thinks for the algorithm
balls<-balls_pre
taken<-c(0,0,0)
counter_max_amount_removed<-0
max_amount_removed<-needed/100 #the lower you set max_amount_removed the more accurate but also longer it will take
counter<-0

while(needed > max_amount_removed) {
  #this is to go with smaller steps towards the end to get a lot closer to the total needed
  if (needed < 1.5 * max_amount_removed && counter_max_amount_removed < 50) {
    max_amount_removed<-max_amount_removed / 2
    counter_max_amount_removed <- counter_max_amount_removed + 1
  }
  balls<-balls-taken
  find_prob<-balls/sum(balls)
  preference<-c(0.5,0.3,0.2)
  taken<-max_amount_removed*(find_prob*preference) #this is the main step
  needed<-needed-sum(taken)
  counter<-counter + 1
}
total_taken<-balls_pre-balls
print(total_taken)
sum(total_taken)
print(counter) #shows how many iterations it took
<span class="math-container">```
```

Body:
This is a more knowledgeable you from the future (wow!)
Here is a piece of code that is way faster (at least in R) because it uses vector calculation. It is a deterministic approach and just an approximation (although quite good)

The basis of how it works is that it just takes the find probability times the keep_probabilty. This number is taken times an "amount_removed". The closer you set that amount removed to 0, the more accurate it will be. Then that amount is removed and the find probabilities are updated.
The theory behind it is that normally you couldnt just take find probability times keep probability times total needed, because it does not take into account that once you remove something you lower the probability for that to be found. However, if the amount you remove goes close to 0 then it does take it into account the next time you remove a very small amount as you have updated the find probability.

Results in comparison to the approach in the question and saulspatz's answer:

Amount
keep_chance
question approach
saulspatz's answer
this answer

20
0.5
14.3367
14.325163888283411
14.32446

30
0.3
15.8905
15.908709486720031
15.90252

50
0.2
19.7728
19.76612662499662
19.77302

Code in R:

#Your input
balls_pre<-c(20,30,50)
needed<-50

#thinks for the algorithm
balls<-balls_pre
taken<-c(0,0,0)
counter_max_amount_removed<-0
max_amount_removed<-needed/100 #the lower you set max_amount_removed the more accurate but also longer it will take
counter<-0

while(needed > max_amount_removed) {
  #this is to go with smaller steps towards the end to get a lot closer to the total needed
  if (needed < 1.5 * max_amount_removed && counter_max_amount_removed < 50) {
    max_amount_removed<-max_amount_removed / 2
    counter_max_amount_removed <- counter_max_amount_removed + 1
  }
  balls<-balls-taken
  find_prob<-balls/sum(balls)
  preference<-c(0.5,0.3,0.2)
  taken<-max_amount_removed*(find_prob*preference) #this is the main step
  needed<-needed-sum(taken)
  counter<-counter + 1
}
total_taken<-balls_pre-balls
print(total_taken)
sum(total_taken)
print(counter) #shows how many iterations it took
```


问:
Equivalency between a "mixed modular equation" of Gauss and a later theorem of Ramanujan.
In p. 476 of volume 3 of Gauss's collected works, appear several interesting identities on Jacobi theta functions which were used by the czech mathematician Karel Petr in an 1904 article "Bemerkung zur einer Gausschen Formel über die Thetafunktionen"  to derive relations giving the number of representations of a number $N$ by three quaternary quadratic forms: $x^2 + y^2 + 9z^2 + 9u^2$, $x^2+y^2+z^2+9u^2$, $x^2+9y^2+9z^2+9u^2$. The identities by Gauss are:

$$(\frac{3P^2-P^0\cdot P^0}{2})^2= p^4-4(\frac{pqr}{2})^{\frac{4}{3}}$$
$$(\frac{3Q^2-Q^0\cdot Q^0}{2})^2=q^4+4(\frac {pqr}{2})^{\frac{4}{3}}$$

where the relevant quantities are defined to be:

$$P(x^3,1)=P , P(x,1)=p, P^0=P(x^{\frac{1}{3}},1)$$
$$Q(x^3,1)=Q , Q(x,1)=q, Q^0 = Q(x^{\frac{1}{3}},1)$$
$$R(x^3,1)=R , R(x,1)= r$$

and the three functions $P(x,y),Q(x,y),R(x,y)$ are equivalent to Jacobi's theta functions $\vartheta_3,\vartheta_4,\vartheta_2$ and defined to be:

$$P(x,y)=1+x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})+x^9(y^3+\frac{1}{y^3})+...$$
$$Q(x,y)= 1-x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})-x^9(y^3+\frac{1}{y^3})+...$$
$$R(x,y)=x^{\frac{1}{4}}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})+x^{\frac{9}{4}}(y^{\frac{3}{2}}+y^{-\frac{3}{2}})+x^{\frac{25}{4}}(y^{\frac{5}{2}}+y^{-\frac{5}{2}})+...$$

To see the equivalency between Gauss's notation and Jacobi's theta functions, look at the post Interpretation of a certain general theorem used by Gauss in his work on theta functions.. User Paramanand Singh helped me understand the meaning of Gauss's identities and remarked that they are essentialy a "mixed modular equation" which connects theta functions of $x^{\frac{1}{3}},x,x^3$ (or, equivalently, $\tau,\tau^3,\tau^9$), and said that Ramanujan also gave this modular equation.

According to Paramanand Singh's comments, the following identity (from p. 142 of the book "pi and the AGM") of Ramanujan is equivalent to Gauss's:

$$\frac{\theta_3(q)}{\theta_3(q^9)} - 1 = (\frac{\theta_3^4(q^3)}{\theta_3^4(q^9)}-1)^{\frac{1}{3}}$$

However, he didn't provide proof of equivalency. Therefore, it's not certain this is the desired identity of Ramanujan, and it's not even certain that Ramanujan stated an equivalent identity at all.

Therefore, my questions are:

Can anyone familiar with Ramanujan's results on mixed modular equations say which of Ramanujan's theorems is equivalent to Gauss's identities? and where can i find the relevant fragment of Ramanujan's writings on the internet?

Can someone also give a proof of equivalency between Ramanujan's theorem and Gauss's identities? this doesn't have to be a proof of correctness of Gauss's identities (which can be quite complicated), just a proof of equivalency to Ramanujan's theorem.

答:
The Broweins' pi and the AGM identity

$$ \frac{\theta_3(q)}{\theta_3(q^9)} - 1 =
 \left(\frac{\theta_3^4(q^3)}{\theta_3^4(q^9)}-1\right)^{\frac{1}{3}} \tag{1} $$

appears in a slightly modified form in
Bruce Berndt, Ramanujan's Notebooks, Part III, page 218, equation (24.29)

$$  \frac{\phi(q^\frac13)}{\phi(q^3)} =
 1 + \left(\frac{\phi^4(q)}{\phi^4(q^3)} - 1\right)^\frac13. \tag{2}$$

The Gauss identity (with a typo fixed)

$$ \left(\frac{3P^2-P^0\cdot P^0}{2}\right)^2= p^4-4\left(\frac{p\,q\,r}2\right)^{\frac{4}{3}} \tag{3} $$

after translating the theta functions into the
Ramanujan notation

$$ q = x^{\frac13}, \quad \phi(q) := f(q,q), \quad f(-q) := f(-q,-q^2) \tag{4} $$

where the general Ramanujan theta function is

$$ f(a,b) := 1 +(a+b) +(a^2+b^2)(ab) +(a^3+b^3)(ab)^3 +\dots \tag{5} $$

and using the identity
$$ \frac{p\,q\,r}2 = \left(q^\frac14 f(-q^6)\right)^3 \tag{6} $$

becomes

$$ \left(\frac{3\phi(q^9)^2 - \phi(q)^2}2\right)^2 = \phi(q^3)^4
 - 4\,q\,f(-q^6)^4. \tag{7} $$

However, the identity in equation $(2)$ is very different than
this equation $(7)$. In the notation of my eta product identity
database, equation $(2)$ is $\texttt{q36_44_420}$ while equation
$(7)$ is not in the database but would be $\texttt{x36_52_520}$.
However, it is possible that several Ramanujan results combined
would imply the Gauss identity.

ContentLicense: CC BY-SA 4.0


问:
A D20 labelled {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,20} is rolled 8 times, is the probability of at least two consecutive 1s 25599203481/25600000000
This D20 has nineteen of its sides labelled as {1}, and the last one labelled as {20}.  If this die is rolled 8 times, there are 25,600,000,000 possible different outcomes.  Please confirm, prove or show that there are 25,599,203,481 out of all possible cases, which have at least two {1} rolled next to each other.

Two or more {20} do not satisfy this.  For example, the rolls {1,20,1,20,20,20,1,20} is a false instance, because there are no two consecutive rolls with {1} in this string.

The method that I used to get this answer was not brute forced, but I am wanting to compare my method with maybe a simpler method that may exist.

I am only an amateur mathematician, but also help people in doing the math of their board game designs, and while this math problem seems like a fringe question, I have constructed it as a fringe case of a project that I have been working on.

答:
Here is a way to proceed. I will replace the outcomes $1$ and $20$ by $1$ and $0$. We consider the following Markov chain with states $0,1,2$:

```<code>     1        1
[0] ---> [1] ---> [2]
 A        |
 |        |
  \______/

</code>```

which counts the consecutive ones.

The state $0$ means "no two consecutive ones so far, last result is empty or a $0$".

The state $1$ means "no two consecutive ones so far, last result is $1$".

The state $2$ means "two consecutive ones were already obtained". This state is absorbant, once we land here, we never quit.

Let $p=19/20$ be the probability to "read" an one, and $q=1-p=1/20$ the opposite probability.
The transition matrix for this chain is
$$
A =
\begin{bmatrix}
q & p & 0\\
q & 0 & p\\
0 & 0 & 1
\end{bmatrix}
\ .
$$
Then we can compute easily (with a computer of course) the value of $A^8$.

```<code>gp > p = 19/20; q = 1-p;
gp > A = [q, p, 0; q, 0, p; 0, 0, 1];
gp > A^8
%6 =
[10223/1280000000 592059/25600000000 25599203481/25600000000]
[31161/25600000000 173299/25600000000 1279989777/1280000000]
[0 0 1]
</code>```

So the probability to get from $0$ to $2$ in eight steps is that entry 25599203481/25600000000.

Body:
Here is a way to proceed. I will replace the outcomes $1$ and $20$ by $1$ and $0$. We consider the following Markov chain with states $0,1,2$:

     1        1
[0] ---> [1] ---> [2]
 A        |
 |        |
  \______/

which counts the consecutive ones.

The state $0$ means "no two consecutive ones so far, last result is empty or a $0$".

The state $1$ means "no two consecutive ones so far, last result is $1$".

The state $2$ means "two consecutive ones were already obtained". This state is absorbant, once we land here, we never quit.

Let $p=19/20$ be the probability to "read" an one, and $q=1-p=1/20$ the opposite probability.
The transition matrix for this chain is
$$
A =
\begin{bmatrix}
q & p & 0\\
q & 0 & p\\
0 & 0 & 1
\end{bmatrix}
\ .
$$
Then we can compute easily (with a computer of course) the value of $A^8$.

gp > p = 19/20; q = 1-p;
gp > A = [q, p, 0; q, 0, p; 0, 0, 1];
gp > A^8
%6 =
[10223/1280000000 592059/25600000000 25599203481/25600000000]
[31161/25600000000 173299/25600000000 1279989777/1280000000]
[0 0 1]

So the probability to get from $0$ to $2$ in eight steps is that entry 25599203481/25600000000.


问:
Does the left-hand limit of the unit tangent to a non-regular curve exist?
Let $ \gamma:[0,1] \to \mathbb{R}^3 $ be a  $ C^1 $-curve which is regular, that is $ \gamma'(t) \neq 0 $, except in $ t= 1 $ , that is $  \gamma'(1) = 0 $. Does the left-hand limit of the unit tangent to $ \gamma $ exist, that is does $ \lim_{t\nearrow 1} \frac{\gamma'(t)}{\vert \gamma'(t)  \vert} $ exist?

If $ C^1 $ is not sufficient, which type of differentiability would be sufficient?

答:
A possible example in $\mathbb R^2 = \mathbb C$.
Define $\gamma(t) = (1-t)^2\exp(i/(1-t))$, $t \in [0,1)$, with $\gamma(1) = 0$.

This is a spiral:

Then $\gamma'(1) = 0$ but $\displaystyle\lim_{t \to 1^-}\frac{\gamma'(t)}{|\gamma'(t)|}$ does not exist.

....

Variant example:
$$
\gamma_1(t) = \begin{cases}\exp\frac{i-1}{(1-t)},\quad &t<1
\\
 0, \quad &t\ge 1\end{cases}
$$
All properties given above, plus $\gamma_1$ is $C^\infty$ on $\mathbb R$.


问:
Determine if subspace - complex numbers
$$ \mathbb{F} := \mathbb{C} \\ V := \mathbb{C}_2[x] $$

$V$ is the set of all polynomials with degree less than or equal to $2$ with complex scalars.

$$ U := \left \{ p(x) \in \mathbb{C}_2[x] | p(x) = p(2x) \right \} $$

Given $ U \subseteq V$ I need to determine if $U$ is a linear subspace of $V$

My try:

$$ p(x) = ax^2 + bx + c \\ p(2x) = a(2x)^2 +2bx + c \\ p(x) = p(2x) \Rightarrow ax^2 + bx + c = 4ax^2 + 2bx + c \\ 3ax^2 + bx = 0 $$

Now, I know that I need to check $3$ things:

$\vec{0} \in U $

If $\vec{v_1} \in U$ and $\vec{v_2} \in U$ then $\vec{v_1} + \vec{v_2} \in U$

If $t \in \mathbb{C}$ then $t \cdot \vec{v_1} \in U$

But I only got the rule of thumbs that " $ 3ax^2 + bx = 0$ "

Of course the zero vector is in $U$ because we can substitude $x=0$.

But how do I check closure over addition? what is really the rule of the subspace? I would appreciate you help.. I am lost. Thank you!

答:
Your analysis shows that $a\,x^2+b\,x+c\in U$ if and only if $3a\,x^2+b\,x$ is the zero polynomial. This is the case if and only if $a=0$ and $b=0$. Hence $U$ is exactly the set of constant polynomials,
$$
U = \{\,0\,x^2+0\,x+ c\mid c\in\mathbb C\,\} = \mathbb C_0[x] \subseteq \mathbb C_2[x].
$$
This is of course a subspace of $\mathbb C_2[x]$.

Title: Determine if subspace - complex numbers


问:
Differential equation - solving for y.
$$ x\frac{dy}{dx} = y(y+1) $$

I'm unsure how to get rid of the two $y$'s I get further down.

I did $1/x\ dx = 1/ ( y^2+y )$.
After partial fractions
$$\ln x = \ln y - \ln (y - 1)$$

I have two $y$'s and I don't know how to get $y$ by itself. Help?

答:
\begin{align}
   x \frac{dy}{dx} &= y(y+1) \\
   \frac{dy}{y(y+1)} &= \frac{dx}{x} \\
   \frac{dy}{y} - \frac{dy}{y+1} &= \frac{dx}{x} \\
   &\phantom{n}\vdots
\end{align}

Title: Differential equation - solving for y.


答:
Simplify the notation to consider a coin with $P(H) = 19/20$ and $P(T) = 1/20$, and let $0 \leq n \leq 8$ be the number of $H$s in the sequence of $8$.  Clearly if $n = 0$ or $1$ there are no consecutive $H$s.  Likewise if $5 \leq n \leq 8$ you must have two successive $H$s.

Thus the only cases we must consider are $n = 2,3,4$.

For $n=2$, there are just $7$ cases:  $HHTTTTTT$, $THHTTTT$, etc.

For $n=4$, there are just two cases where there are no $TT$ sequences:  $HTHTHTHT$ and $THTHTHTH$.  Compute the total number with 4 Ts and 4 Hs, and then subtract $2$.

For $n=3$, place $HH$ in any of 6 paired slot, and for each consider the possible positions for the remaining $H$.  (Be careful to not overcount!)

Title: A D20 labelled {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,20} is rolled 8 times, is the probability of at least two consecutive 1s 25599203481/25600000000

Body:
This D20 has nineteen of its sides labelled as {1}, and the last one labelled as {20}.  If this die is rolled 8 times, there are 25,600,000,000 possible different outcomes.  Please confirm, prove or show that there are 25,599,203,481 out of all possible cases, which have at least two {1} rolled next to each other.

Two or more {20} do not satisfy this.  For example, the rolls {1,20,1,20,20,20,1,20} is a false instance, because there are no two consecutive rolls with {1} in this string.

The method that I used to get this answer was not brute forced, but I am wanting to compare my method with maybe a simpler method that may exist.

I am only an amateur mathematician, but also help people in doing the math of their board game designs, and while this math problem seems like a fringe question, I have constructed it as a fringe case of a project that I have been working on.


问:
Roots of $x^3+ax^2+bx+c=0$ are in arithmetic progression. Find $2a^3-9ab$
If the roots of the equation

$$x^3+ax^2+bx+c=0$$

are in an arithmetic, then what is the value of

$$2a^3-9ab$$

Please explain.

答:
Let $r$, $r\pm d$ be the three roots and, per the Vieta’s formulas

\begin{align}
&(r-d)+r+(r+d)= -a\\
&(r-d)r+r(r+d)+(r+d)(r-d)=b\\
&(r-d)r(r+d)=-c
\end{align}
or,
$$r=-\frac a3,\>\>\>\>\>d^2=3r^2-b,\>\>\>\>\>r^3-d^2r=-c
$$
Substitute the 1st and 2nd equations into the 3rd to obtain
$$2a^3-9ab=-27c$$


答:
This would be better expressed the other way around: the probability that a roll does not have two consecutive ones is $796519/(20^8)$, or in other words the number of ways of not getting two consecutive ones is $796519$.

The recurrence relation is pretty straightforward. Rather than have all these duplicate '1's kicking around, I'm going to use a regular d20 and look at the number of ways of not rolling two consecutive non-20 results. This is most simply looked at in terms of two sequences: $a_n$, the number of sequences without two consecutive non-20s where the last result is a 20, and $b_n$, the number of such sequences where the last result isn't a 20. Then we see that $a_{n+1} = a_n+b_n$, because putting a 20 on the end of a valid sequence gives a valid sequence regardless of whether it previously ended in 20 or not. Likewise, $b_{n+1}=19a_n$, because we can't put any non-20 result on a sequence that didn't end in 20, but we can put nineteen different results on a sequence that did.

Putting these together, we have that $a_{n+1}=a_n+19a_{n-1}$, and the result that you want for a sequence of length $n$ is $a_n+b_n=a_{n+1}$. This is a fibonacci-style recurrence relation, and it can easily be solved with all of the usual techniques for solving linear recurrence relations.


问:
What is the remainder when $P(x)$ is divided by $x^2-2x-35$?
Question:

When $P(x)$ is divided by $x-7$, the remainder is $11$.
When $P(x)$ is divided by $x+5$, the remainder is $-3$.

What is the remainder when $P(x)$ is divided by $x^2-2x-35$?

答:
Hint

The remainder of $P(x)$ divided by $x^2-2x-35=(x-7)(x+5)$ is of the form $ax+b.$ That is:

$$P(x)=(x^2-2x-35)Q(x)+ax+b.$$

Now, using that $P(7)=11$ and $P(-5)=-3$ we can get $a,b.$


问:
A basic birthday probability question
The problem: Suppose the members of a family of 3 were either born on Saturday or Sunday. What are the chances of this family having birthday parties on both Saturday and Sunday?

First approach (from the textbook):

With $x_{i}$ as the number of birthdays on the day $i$:

$n(S)=\binom{4}{3}=4$, as the number of the solutions of $x_{1}+x_{2}=3, x\geqslant0$.

$n(A)=\binom{2}{1}=2$, as the number of the solutions of $x_{1}+x_{2}=3, x\geqslant1$.

Thus $P(E)=\frac{n(A)}{n(S)}=\frac{1}{2}.$

Second approach (mine):

Since the events are independent, the chances of all 3 being born on Saturday would be $(\frac{1}{2})^3$, same with Sunday. So the chances of having birthdays on both days would be $1-2\times \frac{1}{8}=\frac{3}{4}$.

So which one of these approaches are correct? And what is the reason of the other giving the wrong result?

答:
Of course we assume that any given person is equally likely to be born on Saturday as Sunday, particularly given that no other explicit statement to the contrary is given.  This is the so-called "non-informative" prior, based on the "principle of indifference."  Also, on average there are as many Saturdays as Sundays within a year.  (As "meta-reasoning," I would point to the fact that the OP has a reputation of $31$, and so would likely have made this assumption but not voiced it explicitly.)

The second is correct.  Person $A$ is on one of the weekend days.  Call it Saturday (without loss of generality).  What is the probability that at either $B$ or $C$ (or both) are on Sunday?  Well, it is 1 minus the probability both are on Saturday, i.e., $1 - (1/2)^2 = 3/4$.

Another way:  What is the probability all three are on Saturday?  $(1/2)^3$. What is the probability all three are on Sunday?  $(1/2)^3$.  So what is the probability all three on one same day?  $1/8 + 1/8 =1/4$.  What is the probability they are not all on the same day?  $1 - 1/4 = 3/4$.

Title: A basic birthday probability question


问:
Textbook recommendation on group theory for graduate.
During studies I took a course on abstract algebra that involved elements of group, ring and field theories. Unfortunately the course was really basic and currently I'm searching for a book that maybe expand it a little bit more. Are there any good books that talk about these subjects focusing on more "advanced" topics? And by "advanced" I mean book for someone that already knows the basic definitions and theorems but want to explore that field of math a little bit more.

I'm primary interested in group theory - in particular group actions on a sets.

The basic idea is to learn something that would be useful in other parts on mathematics - for example in topology or geometry.

答:
It really depends on whether you want something a little more encyclopedic, or something with a different goal.

Here are some suggestions:

Encyclopedic. These are books that have a lot of stuff on a lot of topics, but generally do not delve very deep on any particular subject. They will have pointers to further work, though:

Joseph Rotman's An Introduction to the Theory of Groups; 4th Edition is from Springer. It starts pretty much from zero, but you should learn some new stuff from it about $G$-sets, and maybe even the Sylow Theorems. Once you hit Chapter 5 (of 12), it will probably be mostly new to you.

Lang's Algebra, revised third edition from Springer. The encyclopedia of Algebra. In my opinion, not a great book to learn from, but a great reference. It is not exclusively Groups, though, and quickly veers into Fields and other stuff.

Hungerford's Algebra is similar to Lang's.

Derek Robinson's A Course in the Theory of Groups, second edition, from Springer. Starts with the basics, but it hits Free Groups and presentations in Chapter 2 (of 15) and continues from there. It is a bit more focused on infinite groups rather than finite groups.

Kurosh's The Theory of Groups. There are two volumes, link is to volume 2, and I spotted a "free PDF" on a google search.

Somewhat more advanced, with a particular point of view

Peter Neumann, Gabrielle Stoy, and Edward Thompson's Groups and Geometry from Oxford Science Publication, is more focused on groups as measurements of symmetry, with a good chunk on actions and the connection between groups and geometry, as the title might imply.

I. Martin Isaacs, Character Theory for finite groups is the book to learn group characters from (it is related to representation theory). From Dover, it is fairly inexpensive.

I. Martin Isaacs, Finite Group Theory, is a kind of "A second course in group theory". It starts assuming you know basic group theory and goes into the Sylow Theorems (with a couple of interesting extensions you don't usually see). The goal of the book is to give a representation-free proof of Burnside's $p^{\alpha}q^{\beta}$ theorem. It is published by the AMS in their Graduate Studies in Mathematics series.

Magnus, Karrass, and Solitar's Combinatorial Group Theory is the classic book on combinatorial group theory.

Specific to group theory in topology

You mention applications of group theory to topology. There are a couple of books that deal with that directly.

From a more topological point of view, we have John Stillwell's Classical topology and combinatorial group theory.

For one of the classics, Max Dehn's Papers on Group Theory and Topology. Keep in mind that these are the papers he published. Influential, but possibly hard going to begin with.

There are a few other books that might contain interesting stuff depending on your interest. I'm afraid I don't have one that deals specifically with group actions. Some of the older books might contain interesting stuff that is not done much any more (Marshal Hall's classic book "The Theory of Groups" has a bunch of stuff on the Burnside problem and Jordan's theorem on multiply transitive groups). Some have specific topics in mind, such as Susan McKay's book on $p$-groups, or Hanna Neumann's classic book Varieties of groups, J-P Serre's Trees for groups acting on trees, etc., but those are the kind of books you should search for later, after you've decided you want to pursue this. If you want to work in abstract algebra, you'll probably want to get Lang or Hungerford at some point.

My recommendation if you are not wedded to looking at group actions is Isaacs' Finite Group Theory. If you definitely want to look at group actions, look at either Neumann, Stoy, and Thompson's book, or the relevant chapters in Rotman. But that's just my opinion, and I'm not a very big "geometry" guy, so someone else might be able to better advice you there.


答:
If you are looking for a book to read as a supplement, not as a textbook for preparing for graduate study, then I recommend Groups, Graphs, and Trees by John Meier.  It's lovely book with many interesting examples.  It does assume a first course in group theory, but it's not important that you remember any of the theorems---just that they are not ideas you are seeing for the first time.  I've used this book for reading courses and to as a source of lectures on specific topics, e.g. Thompson's group F, the lamplighter groups.  The emphasis is on group actions.  Professor Meier's so-called "Cayley's Better Theorem" ought to be part of any course on group theory with an emphasis on group actions.  This is the kind of book you read to learn more examples and because you are curious & hungry for beautiful ideas, not because you are trying to learn the formalities of a graduate level approach to algebra.

If you are looking more for a preparation for graduate studies, then the textbooks recommended in the comments are more suitable.

Rotman's book is very good for learning more group theory, but you have to read quite a few chapters before you see the connections to geometry and topology.  I learned a lot from this book, and I still use it as a reference from time to time.

A recent book that I like, but which may be a little terse, depending on your taste, is Introduction to Group Theory by Bogopolski.

Title: Textbook recommendation on group theory for graduate.

Body:
During studies I took a course on abstract algebra that involved elements of group, ring and field theories. Unfortunately the course was really basic and currently I'm searching for a book that maybe expand it a little bit more. Are there any good books that talk about these subjects focusing on more "advanced" topics? And by "advanced" I mean book for someone that already knows the basic definitions and theorems but want to explore that field of math a little bit more.

I'm primary interested in group theory - in particular group actions on a sets.

The basic idea is to learn something that would be useful in other parts on mathematics - for example in topology or geometry.


问:
Let $a_n = 1 . . . 1 $ with $3^n$ digits. Prove that $a_n$ is divisible by $3a_{n−1}$.
Let $a_n  = 1 . . . 1 $ with $3^n$ digits. Prove that $a_n$ is divisible by $3a_{n−1}$.
Is there any way to solve this question without mathematical induction?

Body:
$a_n=\dfrac{10^{3^n}-1}9$,

so $\dfrac{a_n}{a_{n-1}}=\dfrac{10^{3^n}-1}{10^{3^{n-1}}-1}=10^{2\cdot3^{n-1}}+10^{3^{n-1}}+1,$

which is $\equiv1+1+1\equiv0\pmod3.$


问:
Unit speed reparametrizations of proper patches
Suppose that $x:D\to M\subset\mathbb{R}^3$ is a proper patch, where $D$ is some open set in $\mathbb{R}^2$. The first fundamental form involves the real-valued functions,

$E=x_u.x_u$,

$F=x_u.x_v$,

$G=x_v.x_v$.

Now, of these three, $E$ and $G$ give the squared speeds of the $u$ and $v$ parameter curves of $x$. In some special cases (like surfaces of revolution), there exists a parametrization for which $E=1$. I am trying to understand what conditions can be imposed on $x$ such that both $E$ and $G$ equal 1.

A hand-wavy idea I had was to take the $u$ and $v$ parameter curves on $M$, and constructing $D$ by "unfurling" all the parameter curves to a plane. In that case, $u$ and $v$ become the arc-length parameters of the individual parameter curves, and consequently $E=G=1$. This may be possible at least for a subset of $M$, but I am not really sure if my intuition is right. Any help would be appreciated.

答:
$E=G=1$ and $F=0$ if and only if the surface is locally isometric to the plane, and this holds if and only if the surface is flat (Gaussian curvature $K=0$).

(By the way, geodesic normal coordinates always give you coordinates with $E=1$ and $F=0$.)

It's easy to see why the condition is necessary (e.g., the formula for $K$ in terms of $E$ and $G$), but sufficiency will involve some differential equations; my favorite proof uses differential forms.


问:
Normal distribution - calculating variance given mean and confidence interval
Consider a normally distributed random variable $X$, of which we know the mean $\mu$ but not the variance $\sigma^2$. If we know the $(1-\alpha)$ confidence interval of $X$. That is, we know the interval $X_\alpha =[x_{l,\alpha}, x_{u,\alpha}]$ such that $P(X \in X_\alpha) = 1-\alpha$. Note that the value of $\alpha$ is also known. How can we find the variance of $X$?

I know that I can write:

\begin{equation}
P(X \leq x_{u,\alpha}) = 1 - \frac{\alpha}{2} = \Phi \left( \frac{x_{u,\alpha} - \mu}{\sigma}\right) = \frac{1}{2}\left[1+erf\left( \frac{x_{u,\alpha} - \mu}{\sigma \sqrt{2}}\right) \right]
\end{equation}

But is there an efficient way to solve this for $\sigma$? Preferably using statistical packages in Python (e.g. scipy or numpy). Or perhaps a different approach? Thanks!

答:
Assume your confidence interval is $(x_{l,\alpha},x_{u,\alpha})$. We know the equation for a confidence interval is
$$\bar{X} - Z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \leq X \leq \bar{X} + Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}.$$
Just solving for both sides of the equation:
$$ \bar{X} - Z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = x_{l,\alpha} \\
   \bar{X} + Z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = x_{u,\alpha}, $$
you can find an equation for your variance.


问:
Why does the Gale-Ryser Theorem work?
"If a and b are partitions of a positive integer, then there
exists a binary matrix with row sums a and column sums b if and only if a is dominated by the conjugate of b."

This doesn't really seem very intuitive to me, and the proof I've found seems to be a bit too complicated for me to understand particularly well.

Does anyone have some intuition as to why this is the case?

Thank you

答:
It's easy to explain why the condition is necessary. The sum of the first $k$ rows is $a_1 + a_2 + \dots + a_k$. However, the $i^{\text{th}}$ column can only contain at most $\min\{b_i, k\}$ $1$'s in its first $k$ entries. So the sum of the first $k$ rows can be at most $\min\{b_1,k\} + \min\{b_2, k\} + \dots + \min\{b_n,k\}$. This gives us the condition $$\sum_{j=1}^k a_j \le \sum_{i=1}^n \min\{b_i, k\}$$ which is exactly the Gale-Ryser condition.

We can check this condition for any subset of the rows, but it's most constraining when we pick the $k$ largest rows for some $k$, which is why we check this condition with $\mathbf a$ sorted in decreasing order.

So the Gale-Ryser theorem is one of those "the obvious necessary condition is also sufficient" results, where by "obvious" we really only mean that proving it's necessary is much easier than proving the whole theorem.

Title: Why does the Gale-Ryser Theorem work?


问:
How would one show the evaluation map to $K(\alpha)$ is surjective directly?
Let $K$ be a field and $\alpha$ be an algebraic element. Consider the following ring homomorphism by which $\phi:K[X]\to K(\alpha)$ by which $\phi(X)=\alpha.$ I was wondering how would one show such $\phi$ must be a surjection directly? ($K(\alpha)$ denotes the smallest field to contain both $K$ and $\alpha$.)

I wanted to show this is a surjection thus the minimal polynomial of $\alpha$ must be irreducible. (I know other ways of proving this statement but I thought it was interesting trying out this way)

I know that we can use FIT and thus $\frac{K[X]}{\langle m_{\alpha}(X)\rangle}\cong K(\alpha)$; however, via this route we need to use $m_\alpha$, the minimal polynomial, is irreducible and thus I do not want to end up in a circular argument. Hence I was wondering if there exists other clever way of showing such map must be surjective.

Many thanks in advance!

答:
I think this is somewhat of an inefficient way of proving $m_\alpha(x)$ is irreducible (since you can just use the fact that $K[\alpha]$ is an integral domain). However, this is a proof using what you said.

This boils down to showing that $K[\alpha] = K(\alpha)$ since you already know that the image is $K[\alpha]$. Notice that $K[\alpha]$ naturally has the structure of a $K$-vector space and using the minimal polynomial you can see that this is finite dimensional. Hence, it is algebraic over $K[\alpha]$ and for $\beta \in K[\alpha]$ you can use that it has an inverse by using an annihilating polynomial. If $\sum_{i=0}^k \lambda_i \beta^i = 0$ for $\lambda_0,..., \lambda_k \in \mathbb{Q}$ we can write

$$\beta \left(\frac{1}{\lambda_0} \sum_{i=1}^k \lambda_i \beta^{i-1}\right) = 1$$

Note that we can assume $\lambda_0 \neq 0$ since we could just divide the polynomial by $x$ if $0$ was a root.

Hence, every element has an inverse and $K[\alpha] = K(\alpha)$ so the map is surjective.


问:
Laurent expansion of $f(z)= \frac{z}{(z-1)^2(z+2)}$
Compute the Laurent series of $f(z)= \frac{z}{(z-1)^2(z+2)}$ in a drilled disc of center 1. Specify the maximum radius of this disc.

I did the partial fraction decomposition of f:
$f(z) = \frac{2}{9} \frac{1}{z-1} + \frac{1}{3} \frac{1}{(z-1)^2} - \frac{2}{9} \frac{1}{z+2}$.
What shall I do next?


问:
Help solving recurrence relation, $a_n = 3a_{n-1} + 4a_{n-2} - 12a_{n-3}$
This is in my homework, and I am not sure how to go about this, I've read the book but I can't seem to grasp what to do. Help?

$$a_n = 3a_{n-1} + 4a_{n-2} - 12a_{n-3}$$

where $a_0 = 2$, $a_1 = -1$, $a_2 = 3$

答: Hint: the characteristic polynomial $x^3-3x^2-4x+12$ factors as $(x-3)(x-2)(x+2)$.</p>

ContentLicense: CC BY-SA 3.0


答:
Cases where the rolls do not have consecutive $1s$ need to be $796519$, easily computable on a pocket calculator as

$1 + \binom8 1\cdot19 + \binom 7 2\cdot{19}^2 + \binom6 3\cdot{19}^3 +\binom5 4\cdot{19}^4 = 796519$

The term $\binom 7 2\cdot{19}^2$, for instance is where there are $two\; 1's \;and \;six\; 20's$, giving $7$ spaces for the $1's$ in the interstices of the $20's$ including the ends.


问:
Composition of automorphisms can not reveal what the original automorphism is?
Let $f$ be an automorphism of a finite group $G$ generated by the elements $\{g_1, g_2, \cdots, g_i\}$. $f^m$ is the automorphism $f$ composed $m$ times. i.e. $f^m = f \circ f \cdots \circ f (m \,\,times)$.

Suppose for some $k \in \mathbb{Z}^{+}$, the values of $f^k(g_1), f^k(g_2), \cdots , f^k(g_i)$ are computed by one person (say $A$) and these values are transmitted to another person ($B$).

Since $B$ knows only $f^k(g_1), f^k(g_2), \cdots , f^k(g_i)$ values he can not find what the automorphism $f$ is, right?

Please help me to clarify regarding the above idea.

Thanks a lot.

答:
If you know the value of $k$ it might be possible to find out the automorphism. If $|G| = n$, since an automorphism fixes the identity, note that the automorphisms of $G$ are naturally a subgroup of $S_{n-1}$. Hence, it could happen that there is only one element in $S_{n-1}$ satisfying $g^k = \phi$ (where $\phi$ is the automorphism you've been given). For example, in $S_3$, for a given $3$-cycle it has a unique square root.

If this doesn't work, you can look at the group of automorphisms and do a similar analysis but that might be a bit more complicated.


问:
$\pi$ is just a number, or also the circumference of a sub-unit circle?
A unit circle defined in the Cartesian plane has a radius of $1$ and a diameter of $2$. So making a full round is $2 \pi$. Now, $\pi$ is the ratio of the circumference over the diameter, so if I have a circle with diameter $1$ (radius $0\mathord{,}5$), the circumference would be $\pi$ because its $c/d = c/1$ must equal $\pi$. Hence $c= \pi$. So $\pi$ is actually the circumference of a "subunit" circle? or I am missing something here?</p>

答:
I think you're confusing yourself by talking about two different circles, but you are correct that $\pi$ is the circumference of a circle of radius $0.5$.

For a circle of radius $1$, we have
$$r\text{ (radius)}=1\qquad d\text{ (diameter)}=2\qquad c\text{ (circumference)}=2\pi$$
The ratio of the circumference to the diameter is
$$\frac{c}{d}=\frac{2\pi}{2}=\pi.$$

For a circle of radius $0.5$, we have
$$r\text{ (radius)}=0.5\qquad d\text{ (diameter)}=1\qquad c\text{ (circumference)}=\pi$$
The ratio of the circumference to the diameter is
$$\frac{c}{d}=\frac{\pi}{1}=\pi.$$


答:
Here's how I'd go about solving this ODE.
\begin{align}
x\frac{dy}{dx} &= y(y+1)\tag*{[given]} \\
\int\frac{dy}{y(y+1)} &= \int\frac{dx}{x}\tag*{[separate variables & integrate]} \\
\int\left(\frac{1}{y}-\frac{1}{y+1}\right)\,dy &= \int\frac{1}{x}\,dx\tag*{[partial fractions on LHS]} \\
\ln{|y|}-\ln{|y+1|} &= \ln{|x|}+C_1,\ C_1\in\mathbb{R}\tag*{[perform integration]} \\
\ln{\left|\frac{y}{y+1}\right|} &= \ln{|x|}+C\tag*{[simplify LHS]} \\
\left|\frac{y}{y+1}\right| &= \exp\left(\ln{|x|}+C\right)\tag*{[exponentiate]} \\
&= C_2|x|,\ C_2\in\mathbb{R}\tag*{[simplify RHS]} \\
&= C_2x,\ x\geq 0.\tag*{[choose $x\geq 0$]}
\end{align}
It doesn't look like there's a convenient way to express $y$ as an explicit function of $x$, so that's as far as I'll go. I hope this helps.

Title: Differential equation - solving for y.

Body:
$$ x\frac{dy}{dx} = y(y+1) $$

I'm unsure how to get rid of the two $y$'s I get further down.

I did $1/x\ dx = 1/ ( y^2+y )$.
After partial fractions
$$\ln x = \ln y - \ln (y - 1)$$

I have two $y$'s and I don't know how to get $y$ by itself. Help?


问:
How to develop patience in mathematics?
How  to develop the patience in mathematics?

When  you read the books like Rudin, Conway, etc., it takes a lot of time for one problem; sometimes $10$ hours each.

I usually  devote one  problem for $30$ minutes. If I can't solve the problem within $30$ minutes, then I post the problem on this site for a solution.

Actually, I have no patience and can't wait for long hours for thinking about the problem.

How  can I improve myself and develop the patience?

答:
Always  remember the Alexander_Grothendieck quotes "Mathematics is not a competition sport.

Terence Tao statement taken from  https://terrytao.wordpress.com/

The objective in mathematics is not to obtain the highest ranking, the highest “score”, or the highest number of prizes and awards; instead, it is to increase understanding of mathematics (both for yourself, and for your colleagues and students), and to contribute to its development and applications. For these tasks, mathematics needs all the good people it can get.

Michael_Atiyah statement taken
from Advice to a YoungMathematician

''Be patient and persistent. When thinking about a
problem, perhaps the most useful device you can
employ is to bear the problem in mind all the time: it
worked for Newton, and it has worked for many a mortal as well. Give yourself time, especially when attacking
major problems; promise yourself that you will spend
a certain amount of time on a big problem without
expecting much, and after that take stock and decide
what to do next. Give your approach a chance to work,
but do not be so wrapped up in it that you miss other
ways of attacking the problem. Be mentally agile: as
Paul Erd˝os put it, keep your brain open''
You  have to be practice patience, because by practice  everything can be acheived .Practice make a man perfect

Abel prize winner lenart carleson statement taken from his interview

If you want to solve problems, as in my case, the most
important property is to be very, very stubborn.Stubbornness
is important; you don’t want to give up. But as I
said before, you have to know when to give up
also. If you want to succeed you have to be very
persistent. And I think it’s a drive not to be beaten
by stupid problems.

Title: How to develop patience in mathematics?


问:
Unit quarternion as rotation in $\mathbb{R}^3$
I'm pretty confused by the following statement on page 6 in the book The Seiberg-Witten Equations and Applications to the Topology of Smooth Four-Manifolds, which is meant to expain the fact that unit sphere $S^3$ in the quarternion algebra $\mathbb{H}$ is naturally associated with the group of rotation in $\mathbb{R}^3$:

I don't think the statemet is true at the very first place, $\alpha$ does not act trivially on $\mathbb{C}\alpha$, since by that we require for any complex number $z$, the equation
$\alpha (z \alpha) \alpha^{-1}=z \alpha$, i.e.
$$\alpha z \alpha^{-1}=z$$i.e. we need $\alpha$ commutes with any $z$. But this is not even true for $j \in S^3$ and $i \in \mathbb{C}$.

Can anyone tells me where I went wrong?

答:
I don't think it's referring to a $1$  dimensional $\mathbb C$ subspace, I think it's referring to a 2  dimensional $\mathbb R$ subspace copy of $\mathbb C$.

Apparently "the complex plane spanned by $\alpha$" means $P_1=\mathrm{span}_\mathbb{R}(\{1, \alpha\})$ and the other plane refers to the plane perpendicular to $\alpha$ in the pure quaternions.

Indeed, conjugation on $P_1$ by $\alpha$ leaves the basis fixed, so it leaves all elements fixed.


问:
How does $\frac{4{\pi}x +\cos(4{\pi}x)\sin(4{\pi}x)}{8{\pi}}+C$ become $\frac{\sin(8{\pi}x)+8{\pi}x}{16{\pi}}+C$?
I have an integral like

$$\frac{1}{4\pi}\int \cos^{2}(u)du$$

Where

$$u = 4{\pi}x$$

I get a result of $$\frac{4{\pi}x + \cos(4{\pi}x)\sin(4{\pi}x)}{8{\pi}} + C$$

But there is a further simplification for this that results in

$$\frac{\sin(8{\pi}x) + 8{\pi}x}{16{\pi}} + C$$

Can anyone go through each step to get from my result to the simplified result and list any concepts that are used, such as any trig identities, to find this simplification? Any help is appreciated, thank you!

答: They have simply used the identity $2\sin(x)\cos(x)=\sin(2x)$ and solved it further


问:
Sufficient condition for equality with order hypothesis
I have this very simple equation for arbitrary $a,b,c,d,e,f\in\mathbb{R}$ with $\mathbb{R}$ endowed with the usual order :

$$a+b+c=d+e+f$$

So, if this equality is verified, I know that

$$\left\lbrace a-d\right\rbrace =\left\lbrace e-b\right\rbrace +\left\lbrace f-c \right\rbrace\quad (*)$$

And consquently that

$$\text{if }e\geq b\text{ and }f\geq c\text{ then }a\geq d\quad (**)$$.

But all rearrangements of $(*)$ give a statement in the design of $(**)$ :

We can exchange $a$ in the left-hand side with $b$ or $c$ in the right-hand side, we get $3$ rearrangements. The others follow by exploiting the symmetries in $(*)$. Firstly, the braces can be switched rearranging the signs, multiplying by $3$ the number of rearrangements. Secondly, terms in the right-hand side can be permuted, multiplying by $2$ the number of rearrangements. Thirdly, sign can be reversed on both sides multiplying by $2$ the number of rearrangements. After all, there are $36$ rearrangements that generate $36$ statements in the design of $(**)$.

I denote by $\pi,\sigma,\ldots$ permutations in the set of permutations of $3$ distinct objects.

I can deduce that $(*)$ implies that for all $\pi,\pi^{\prime},\sigma,\sigma^{\prime}$ permutations of $3$ distinct objects

$$\text{if } \pi(e)\geq\sigma(b)\text{ and } \pi(f)\geq\sigma(c)\text{ then }\pi(a)\geq\sigma(d) \quad (\bullet)$$

and

$$\text{if } \pi^{\prime}(b)\geq\sigma^{\prime}(e)\text{ and } \pi^{\prime}(c)\geq\sigma^{\prime}(f)\text{ then }\pi^{\prime}(d)\geq\sigma^{\prime}(a) \quad (\bullet \bullet)$$

So my question is very simple in fact but difficult for me, if I suppose that $(\bullet)$ and $(\bullet\bullet)$ is true it is possible to show that

$$a+b+c=d+e+f$$

is true ?

if anyone knows the algebraic books that I should see ?

答:
Consider $(a,b,c,d,e,f)=(0,0,2,1,1,1)$. These numbers satisfy both sets of your order-based conditions (in most cases trivially, due to the premise of the implication being false), but $a+b+c\not= d+e+f$.

Very informally speaking, conditions based purely on comparison between some quantities are unlikely to be strong enough to say anything about their arithmetical properties (they might imply equality of the quantities involved though).


问:
Solve the differential equation: $\frac{dy}{dx}=1 + a\frac{y}{x}$
What are the steps to get to $y(x)$?

$$\frac{dy}{dx}=1 + a\frac{y}{x}$$

答: This is $y^\prime+Py=Q$ with $P=-\frac{a}{x},\,Q=1$. The integration factor $R=x^{-a}$, so $y=x^a\int x^{-a}dx$. If $\alpha=1$, the result is $x\ln|x|+Cx$, with $C$ locally constant but potentially dependent on the sign of $x$; otherwise, i's $\frac{x}{1-a}+Cx^a$, with $C$ globally constant.


问:
Find largest $N$ such that diophantine equation $13x+24y+31z=N$ has no solution for $x,y,z$ all positive integers
In a game of darts, the target is a bit special and has three regions which carry 13 points, 24 points and 31 points.
Question: What is the largest total number of points that is impossible to reach?

Here is my attempt, and where I get stuck:

In other words, the question is the diophantine equation $13x+24y+31z=N$ and asks the highest number $N$ such that the equation has no positive solutions $(x,y,z)$.
I managed to find the general solution of the equation:

$$x = -11(N+31k)+24m  \\  y=6(N+31k)-13m\\   z=-k$$

where $k$ and $m$ are any integer.

Demanding that $x$, $y$, and $z$ are all non-negative imposes conditions on $k$ and $m$ for all $N$:

$$k\leq 0 \\
\frac{143}{312}(N+31k) \leq m \leq \frac{144}{312}(N+31k)$$

Clearly, with $N$ large enough, there will always be an integer $m$ that satisfies the condition. But now which $N$ is the largest one for which there does not exist any $m,k$ integer that satisfy the conditions?

Note: the answer is $N=121$, as I confirmed with trying all possibilities... Very curious how to find this result! Thanks!

答:
When only two coprime numbers $a_1$, $a_2$ are given the largest number $N$ non representable in the form $N=ja_1+ka_2$ $(j, k\geq0)$ is given by
$$N=a_1a_2-a_1-a_2\ ,$$
as Stirling has proved. This is the simplest case of the so-called Frobenius problem. When $n=3$ three coprime integers $a_i$  are given there is no such simple formula. The following paper treats the $n=3$ problem. A linear diophantine problem involving the $a_i$ has to be solved during the computations.

https://people.math.ethz.ch/~blatter/geomfrob.pdf


问:
Verification of correct setup of the matrix form of the linear regression problem
The following is a question from Rice's Mathematical Statistics - Ch. 14 - Q-5

I wanted to know if the following approach would be the right set up of the problem:

I looked at this as a question of attempting to solve for the true of value of the points $p_{1}, p_{2}, p_{3}$.

To do this we have to first specify the linear model we are going to use. I specified it in the following way:

$$p_{i} = \beta_{0} + \beta_{1}y_{i,1} + \beta_{2}y_{i,2} + \beta_{3}y_{i,3} + e_{i}$$

Which if translated to matrix form would be:

$$\textbf{P} = \textbf{X} \textbf{$\mathbb{\beta}$}\ \text{where,} \\
\textbf{P} =
\begin{bmatrix}
p_{1} \\
p_{2} \\
p_{3} \\
\end{bmatrix}, \
\textbf{X} =
\begin{bmatrix}
1 & y_{1,1} & y_{1,2} & y_{1,3} \\
1 & y_{2,1} & y_{2,2} & y_{2,3} \\
1 & y_{3,1} & y_{3,2} & y_{3,3} \\
\end{bmatrix}, \
\textbf{$\mathbb{\beta}$} =
\begin{bmatrix}
\beta_{0} \\
\beta_{1} \\
\beta_{2} \\
\beta_{3} \\
\end{bmatrix} \\
\\
\text{where,} \
y_{i,j} =
\begin{cases}
|p_{i} - p_{j}| & \text{if}\ i\neq j \\
|p_{i} - 0| & \text{if}\ i = j \\
\end{cases}  $$

Which would put me in a position to be able to express the least squares expression:

$$S(\textbf{P}) = \sum_{i = 1}^{3} (p_{i} - \beta_{0} - \beta_{1}y_{i,1} - \beta_{2}y_{i,2} - \beta_{3}y_{i,3})^{2} \\
= ||\textbf{P} -\textbf{X} \textbf{$\mathbb{\beta}$}||^{2} \\
= ||\textbf{P} - \hat{\textbf{P}}||^{2}$$

So my concerns:

I looked at the question as regressing the $p_{i}$ on to the $y_{i}$, now the way they expressed things in the question it kind of feels as if they wanted me to do the opposite. I may be overthinking it here, but considering this is a first approach at linear regression I don't think they would deviate too much from how they use the $Y_{i}$ variable in the text.

I introduced an intercept term $\beta_{0}$, for me I was looking at it as perhaps the distance from the origin to the first point. That was my thinking behind it, but then I also have $\beta_{1}$ which is being estimated based off of distances that use the observed $p_{1}$ value. So would an intercept term be needed here?

How is the setup in terms of the overall idea?

答:
I think your setup is correct, but maybe it's not what you were asked to do. The question is about using the least-squares method to estimate (directly) the values of $p_i$ given the noisy measurements $Y_i$. If we wanted to state this question as a regression problem, we would be asked to regress $p_i$ given the values of $Y_i$.

The least-squares method is more general than regression and tries to find optimal parameters to some model given data. In our case:

We have latent values $p_1, p_2, p_3$ and we have information about $d_{12} = p_2 - p_1, ~ d_{13} = p_3 - p_1, ~ d_{23} = p_3 - p_2$.

We have noisy measurements of those quantities given by $Y_1, ..., Y_6$.

We now have to model our problem. The most common way to do it is to consider all measurements to have iid errors coming from a normal distribution, i.e. $Y_i = x_i + \varepsilon_i$ where $x_i$ is some data and $\varepsilon_i \sim N(0, \sigma^2)$ are iid samples from a gaussian, modeling noise (it can be justified by assuming all the measurements taken independently by the same person / instrument). In our model, we are taking measurements of $p_i$'s and $d_{ij}$'s with some noise, so we can state that each $Y_i$ comes from a gaussian centered in the true values of our quantities of interest.This gives us:
$$
\begin{aligned}
&Y_1 = p_1 + \varepsilon_1; ~~ Y_2 = p_2 + \varepsilon_2; ~~ Y_3 = p_3 + \varepsilon_3 \\
&Y_4 = d_{12} + \varepsilon_4; ~~ Y_5 = d_{13} + \varepsilon_5;~~ Y_6 = d_{23} + \varepsilon_6 \\
\end{aligned}
$$
Where $\varepsilon_i \sim N(0, \sigma^2)$ are iid random variables. Our problem looks like a linear regression given only by intercepts. When we deal with iid gaussians, the least-squares solution is equivalent to maximum likelihood estimation (MLE). To see it, let's write out our likelihood:

$$
P(Y_1,Y_2,Y_3,Y_4,Y_5,Y_6 | p_1,p_2,p_3) = \prod_{i=1}^{6} P(Y_i)
$$

We'd like to maximize that quantity by finding the optimal values of $p_i$'s, i.e. finding the most probable positions which explain our data. Observe that maximizing $\mathcal{L}(p_1, p_2, p_3) := P(Y_1,Y_2,Y_3,Y_4,Y_5,Y_6 | p_1,p_2,p_3)$ is the same as maximizing $\log \mathcal{L}(p_1, p_2, p_3)$, for what we have:

$$
\begin{aligned}
&\max_{p_1, p_2 , p_3} \log \mathcal{L}(p_1, p_2 , p_3) \\
& = \max_{p_1, p_2 , p_3}  -\frac{1}{2\sigma^2} \left ( \sum_{i = 1}^{6}
 (Y_i - p_i)^2 \right) \\
& = \min_{p_1, p_2 , p_3} \sum_{i = 1}^{6} (Y_i - p_i)^2
\end{aligned}
$$
Where $p_4 = d_{12}$, $p_5 = d_{13}$ and $p_6 = d_{23}$. The last expression is precisely the least squares solution. We were asked to find it's matrix form, for that let's define $T$ by:

$$
T = \left [
\begin{matrix}
1 & 0 & 0  \\
0 & 1 & 0  \\
0 & 0 & 1  \\
-1 & 1 & 0 \\
-1 & 0 & 1 \\
0 & -1 & 1 \\
\end{matrix}
\right ]
$$

Then, for $Y = [Y_1,Y_2,Y_3,Y_4,Y_5,Y_6]^T$ and $\mathbf{p} = [p_1, p_2, p_3]^{T}$ we finally have:

$$
\begin{aligned}
&\max_{p_1, p_2 , p_3} \log \mathcal{L}(p_1, p_2 , p_3) \\
& = \min_{p_1, p_2 , p_3} \sum_{i = 1}^{6} (Y_i - p_i)^2 \\
&= \| Y - T\mathbf{p} \|^2
\end{aligned}
$$
I hope it helps :)


问:
Equivalence between the structure theorems of finite abelian groups
I want to show that the primary decomposition and the invariant factor decomposition of finite abelian groups are equivalent. What I mean by this is that, given a finite abelian group $G$ is isomorphic to

$$\mathbb{Z}_{p_1^{k_1}}\oplus\dots\oplus\mathbb{Z}_{p_r^{k_r}},$$

where $r\in\mathbb{N}$ and every $p_i^{k_i}$ is the power of a prime (not necessarily different powers of primes), then $G$ is isomorphic to

$$\mathbb{Z}_{n_1}\oplus\ldots\oplus\mathbb{Z}_{n_u},$$

where $u\in\mathbb{N}$ and $n_i|n_{i+1}$ for every $i=1,\ldots,u-1$ and vice versa.

I read that this can be done using the Chinese remainder theorem, since this implies that $\mathbb{Z}_{ab}\cong\mathbb{Z}_a\bigoplus\mathbb{Z}_b$ iff $a$ and $b$ are coprime, but I can't seem to figger out how.

答:
Prove it by induction on the number of direct summands in the first decomposition. The result is trivially true if this is $0$.

Let $p_1, p_2, \ldots, p_s$ be the distinct primes dividing the group order (note that this is different from your notation). For each $p_i$, let $p_i^{k_i}$ be the highest power of $p_i$ that occurs as a summand ${\mathbb Z}_{p_i^{k_i}}$ in your first decomposition.  Let $m = p_1^{k_1} \cdots p_s^{k_s}$.

Then, letting $H$ be the sum of the remaining summands, we have $$G \cong H \oplus {\mathbb Z}_{p_1^{k_1}}  \oplus \cdots \oplus {\mathbb Z}_{p_s^{k_s}} \cong H \oplus {\mathbb Z}_m.$$

Now the result follows by applying induction to $H$, observing that the orders of all summands of $H$ divide $m$.


问:
Is it worth betting on this case?
Let's imagine a game which uses an unbiased coin.

Starting with $X$ dollars, your total increases $50\%$ every time you flip heads. But if the coin lands on tails, you lose $40\%$ of your total. You can play this game $N$ times. On each turn, you must bet the total amount you had on the last turn.

Is it worth betting on this case? How can we formalize this in order to have this answer in the general case?

Is it easier to formalize this for a specific value of $X$ and $N$? Let us assume, for example $X = 100$ dollars and $N = 100$ turns.

PS: This scenario appears in an article discussing some ideas of Ole Peters, a theoretical physicist who is claiming that Everything we know about modern economics is wrong.

答: Let $X_n$ represents the amont of money you have after the $n^{\text{th}}$ game played. Then $$X_n=\bigg(\frac{3}{2}\bigg)^{Q_n}\bigg(\frac{3}{5}\bigg)^{n-Q_n}X$$ where $Q_n \sim \text{Binomial}\bigg(n,\frac{1}{2}\bigg)$ and $X$ represents the initial amount of money that you had. The event of being "worth betting" corresponds to the event that $X_N>X$. Notice $$X_N>X \iff Q_N>\frac{N\ln(5/3)}{\ln(5/2)}$$ If $N$ is large we can say $Z=\frac{Q_N-N/2}{\sqrt{N}/2}$ is approximately $N(0,1)$ so $$P(X_N>X)\approx P\Bigg(Z>\sqrt{N}\bigg[\frac{2\ln(5/3)}{\ln(5/2)}-1\bigg]\Bigg)\approx 0$$

Title: Is it worth betting on this case?


答:
Just use trig identities.
$$\cos(x) \cdot \sin(x) = \frac{1}{2} \sin(2x)$$
so then you get:
$$\frac{\frac{\sin(8 \pi x)}{2} + 4 \pi x}{8\pi} + C$$
which simplifies to:
$$\frac{\sin(8\pi x) + 8\pi x}{16\pi} + C$$

Title: How does $\frac{4{\pi}x +\cos(4{\pi}x)\sin(4{\pi}x)}{8{\pi}}+C$ become $\frac{\sin(8{\pi}x)+8{\pi}x}{16{\pi}}+C$?

Body:
I have an integral like

$$\frac{1}{4\pi}\int \cos^{2}(u)du$$

Where

$$u = 4{\pi}x$$

I get a result of $$\frac{4{\pi}x + \cos(4{\pi}x)\sin(4{\pi}x)}{8{\pi}} + C$$

But there is a further simplification for this that results in

$$\frac{\sin(8{\pi}x) + 8{\pi}x}{16{\pi}} + C$$

Can anyone go through each step to get from my result to the simplified result and list any concepts that are used, such as any trig identities, to find this simplification? Any help is appreciated, thank you!


问:
Prove that $f: X \rightarrow Y$ is one-to-one, if and only if : $g: P(Y)\rightarrow P(X)$ , $g(Y)= \{ x\in X : f(x)\in Y \} $ is onto
My attempt (for first direction), I said that suppose that $f$ is one-to-one and $g$ is not onto,

then there exists $X'\subset X$, such that for all $Y'\subset Y$ : $g(Y')\ne X$

then if $f$ is one to one : $\forall x\in X, \exists! y\in Y : f(x)=y $

hence there must be a subset $Y'\subset Y$ that satisfies $g(Y')=X'$

which contradicts that $g$ is not onto.

but I really think I have made some mistakes since looking at it again I could have proved it in the same way without $f$ being one to one.
Any help or corrections or answers are appreciated.
Thanks in advance!

答:
The fact that $f$ is one-to-one does not mean that for each $x\in X$ there is a unique $y\in Y$ such that $f(x)=y$: that’s true simply because $f$ is a function. The fact that $f$ is one-to-one means that for each $y\in Y$ there is at most one $x\in X$ such that $f(x)=y$.

You’re right that your argument doesn’t work: it would apply even if $f$ were a constant function, and in that case it’s clear that $g$ is not onto.

Suppose that $f$ is one-to-one, and let $A\subseteq X$; we want to find an $S\in\wp(Y)$ such that $g(S)=A$. The natural candidate is $f[A]$, i.e., $\{f(x):x\in A\}$, so let $S=f[A]$. Certainly $g(S)=f^{-1}\big[f[A]\big]\supseteq A$, so we need only show that $g(S)\subseteq A$. Let $x\in g(S)$; then $f(x)\in S$. Let $y=f(x)$; $y\in S$, so there is an $a\in A$ such that $f(a)=y$. But then $f(x)=f(a)$, and $f$ is one-to-one, so $x=a\in A$. This shows that $g(S)\subseteq A$ and hence that $g(S)=A$, as desired, so $g$ is onto.

Now suppose that $f$ is not one-to-one. Then there are distinct $x_0,x_1\in X$ such that $f(x_0)=f(x_1)=y$, say. Let $A=\{x_0\}$, and suppose let $S\in\wp(Y)$ be such that $x_0\in g(S)$. By definition this means that $f(x_0)\in S$, so $y\in S$. But $f(x_1)=y$, too, so $x_1\in g(S)$ as well. Thus, $g(S)\supseteq\{x_0,x_1\}\supsetneqq A$, and we see that there is no $S\in\wp(Y)$ such that $g(S)=A$. That is, $g$ is not onto. This shows that if $g$ is onto, $f$ must be one-to-one and completes the proof.


问:
Proof of Contour Integral along Reversed Contour explanation
The following is provided as a proof on proofwiki.org, but I am not sure how the author went from line 2 to line 3 of their proof. If $u=\psi(t)$ then $\frac{du}{dt}=\psi'(t)$ implying that we can make the substitution $dt=\frac{1}{\psi'(t)}du$ into line 2, so I'm not sure where the minus sign in line 3 comes from. I see that $\psi'(t)=-1$, but surely this doesn't matter if we make that substitution for $dt$? I think I'm missing something obvious here, any help would be really appreciated.

答:
Expanding comment by @Diger.

In line 2 the integral in $t$ is taken from $t=a$ to $b$, while the integral in $u=\psi(t)$ is taken from $u=\psi(a)=b$ to $\psi(b)=a$. But $dt=-du$. So

\begin{eqnarray*}
I &=&\int_{t=a}^{b}f\left( \gamma \left( \psi \left( t\right) \right)
\right) \gamma ^{\prime }\left( \psi \left( t\right) \right) \,\psi ^{\prime
}\left( t\right) \,dt\quad\text{in line 2}, \\
&=&-\int_{t=a}^{b}f\left( \gamma \left( \psi \left( t\right) \right) \right)
\gamma ^{\prime }\left( \psi \left( t\right) \right) \,dt,\quad \psi \left(
t\right) =a+b-t,\psi ^{\prime }\left( t\right) =-1, \\
&=&\int_{u=\psi \left( a\right) =b}^{\psi \left( b\right) =a}f\left( \gamma
\left( u\right) \right) \gamma ^{\prime }\left( u\right) \,du,\qquad u=\psi
\left( t\right) ,du=\psi ^{\prime }\left( t\right) dt=-dt, \\
&=&-\int_{u=a}^{b}f\left( \gamma \left( u\right) \right) \gamma ^{\prime
}\left( u\right) \,du,\quad\text{in line 3}.
\end{eqnarray*}


答:
You can substitute $y=xv$ and $\frac{\mathrm dy}{\mathrm dx}=x\frac{\mathrm dv}{\mathrm dx}+v$ to get a separable equation:

$$x\frac{\mathrm dv}{\mathrm dx}+v=1+av\implies \frac{\mathrm dv}{1+(a-1)v}=\frac{\mathrm dx}x$$

Then consider the cases where $a=1$ and $a\neq1$. For instance, if $a=1$, then $v=\ln|x|+C$ and $y=x\ln|x|+Cx$.

Title: Solve the differential equation: $\frac{dy}{dx}=1 + a\frac{y}{x}$

Body:
What are the steps to get to $y(x)$?

$$\frac{dy}{dx}=1 + a\frac{y}{x}$$


答:
Thanks to @thedumbkid for providing the needed identity and @Han Xiong for the steps to implement it and get to the simplified form.

Needed Double Angle Identity:  $$2sin(x)cos(x) = sin(2x)$$
Steps to simplify: Follow @Han Xiong's answer

Thanks everyone :)


问:
Integral of product of first derivatives of Legendre polynomials
I have the integral,
$$I=\int_{-1}^1 dx P'_m(x) P'_n(x).$$
I tried using $$P_m'(x)=-(m+1)\frac{xP_m(x)-P_{m+1}(x)}{x^2-1},$$ but that yields,
$$I=\int_{-1}^1\bigg(\frac{x^2P_mP_n}{(x^2-1)^2}+\frac{P_{m+1}P_{n+1}}{(x^2-1)^2}-\frac{xP_mP_{n+1}}{(x^2-1)^2}-\frac{xP_{m+1}P_n}{(x^2-1)^2}\bigg),$$ which did not seem particularly convenient to me either. Alternatively, using integration by parts, along with $(1-x^2)P''_m-2xP'_m+m(m+1)P_m=0$
$$I=P'_mP_n\big\vert_{-1}^1-\int_{-1}^1 dx P_n P''_m=P'_mP_n\big\vert_{-1}^1-\int_{-1}^1\bigg(\frac{2xP_nP'_m}{1-x^2}-m(m+1)\frac{P_nP_m}{1-x^2}\bigg).$$
Stuck either way.
Lastly, I tried using the associated Legendre polynomials to cast the original integral as,
$$\int_{-1}^1 dx\frac{P_m^1P_n^1}{1-x^2},$$ which I could not find.

I guess the only way out might be to use the Rodrigues formula and churn through all the terms?

Addendum: A brute force evaluation on Mathematica, of both the last expression in terms of the associated Legendre functions, and the initial expression, seems to yield $$I=\int_{-1}^1 dx P'_m(x) P'_n(x)=\int_{-1}^1 dx\frac{P_m^1P_n^1}{1-x^2}=\begin{cases}&k(k+1),\quad k=\mathrm{min}(m,n);\quad m,n \text{ both odd or even}\\&0;\quad \text{otherwise}\end{cases}$$

答: After integration by parts, it's much easier to use orthogonality. If $m\leqslant n$ (say), then $$\int_{-1}^1 P_m'(x)P_n'(x)\,dx=P_m'(x)P_n(x)\Big|_{-1}^1-\int_{-1}^1 P_m''(x)P_n(x)\,dx;$$ the last integral is $\color{red}{0}$ because of orthogonality of $P_n$ to any polynomial of degree $<n$, and $$P_n(1)=(-1)^nP_n(-1)=1,\quad P_m'(1)=(-1)^{m-1}P_m'(-1)=m(m+1)/2.$$


问:
Proving Archimedes PROP VII-12: In an ellipse, the sum of the squares on conjugate diameters equals the sum of the squares on its axes
I was asked the following question:

"Prove analytically Proposition VII-12 that in any ellipse the sum of the squares on any two of its conjugate diameters is equal to the sum of the squares on it's two axes."

Above is my drawing of conjugate diameters. I know that the diameters are not perpendicular to each other - and this is clear from the picture. I would think I can use the pythagorean theorem in some way to prove this because what the proposition is essentially saying is that $HJ^2+ID^2=FG^2+AB^2$ - which has the look of the Pythagorean theorem.

Can anyone help me take the next step?

答:
One approach is to take the slope of $ED$ equal to $k$, making the point $D$ be $(x, kx)$, use calculus to compute the slope $K$ of $EH$ and then just compute the sum of squares. Namely, if the equation of the ellipse $ax^2+by^2=1$ then we have $x^2(a+bk^2)=1$ and $ED^2=\frac{1+k^2}{a+bk^2}$. The tangent at $D$ is given by implicit differentiation as $2axdx+2bydy=0$, so slope is $dy/dx=-a/(bk)$. Now we can compute $EH^2$ by the same formula just with the new slope instead of $k$. Then just algebraically check the sum is independent of $k$ and is indeed equal to what one would get when $k=0$ to which this computation is inaplicable.

Here is another approach (without derivatives): the ellipse with major axis of length $a$ and minor of length $b$ (NOT the same $a$ and $b$ as in the first approach) is a the image of the circle $x^2+y^2=1$ under the stretching map, and  so inherits a parameterization $(a\cos t, b \sin t)$ (note that $t$ is the angle along the circle, but NOT along the ellipse, so the conjugate diameters are not orthogonal). The stretching map takes tangents to tangents and since it is linear it preserves parallelism, so takes conjugate diameters on the circle to conjugate diameters on the ellipse. Thus to get, say, $H$ from $D$ one only needs to increase $t$ by $\pi/2$, so if $D$ has coordinates $(a\cos t, b \sin t)$ then $H$ has coordinates $(a\cos (t+\pi/2), b \sin (t+\pi/2))$. The sum of square lengths is evidently constant at $a^2+b^2$.

The linear algebra version is like this: let $A:\mathbb{R}^2\to \mathbb{R}^2$ be any linear map (in this problem, $A$ is any map taking the circle to the ellipse, say the stretching map) and $v_1$ $v_2$  an orthonormal basis for its domain. Then $|Av_1|^2+|Av_2|^2$ (which in this problem is the sum of squared lengths of the "conjugate axis")  is independent of the choice of the basis. Indeed, let $V$ be the rotation matrix with columns $v_1$ and $v_2$. Then

$$|Av_1|^2+|Av_2|^2=<Av_1, Av_1>+<Av_2, Av_2>=<v_1, A^TAv_1>+<v_2, A^TAv_2>=$$

$$v_1^T(A^TA)v_1+v_2^T(A^TA)v_2= $$

$$tr(V^TA^TAV)=tr(V^{-1}A^TAV)=tr(A^TA)$$

is constant.


问:
How can i calculate the minimum price based on a desired margin and fee structure?
I am trying to figure out how to get the breakeven price of a product where there is a fee charged based on the selling price of the item.

I know how to get the breakeven price if the fee is a fixed percentage:
(cost) x 100 / (100-(fee percentage + desired margin))

Example:

Cost $10

Desired Margin: %5

Fixed Fee: %15

$10 \times 100 / (100-(15 + 5)) = $12.5$

The issue I am having is that the fee is based on the final price of the product, for example if the product is under 15, the fee is 8%. If the products final price is over 15 the fee is 15%.

Is there a way I could calculate the final price of the product with a desired margin if the fee is not fixed?

答:
I don't know anything about the topic, but if the provided formula is correct and we need to find out the price,  using an uknown fee, then the equation should look like this:

$$p=\frac{100c}{100-f(p)-m}$$

Where $p$ is the price, $c$ the cost, $f(p)$ a variable fee, which is a function depending on price, and finally $m$ is the margin.

In your example we have:

$$f(p)=\begin{cases} 8, p<15 \\ 15, p \geq 15 \end{cases}$$

Solving the equation is a little bit tricky in this case, but possible.

The easiest way is simply to substitute whatever fee into the right hand side and see if the condition is fulfilled.

In this case setting $f=15$ gives $p=12.5<15$, while $f=8$ gives $p\approx 11.5$, which satisfies the condition, and so is the correct answer.


问:
Please help finishing the calculation to find the Entropy of Pareto distribution.
Let $X$ follow Pareto distribution with parameters $\alpha, a, h$. That is, $X\sim Pa(\alpha,a,h)$, where $\alpha>0$ is the shape parameter, $-\infty < a < \infty$ is the location parameter, and $h>0$ is the scale parameter.

Then the pdf of $X$ is $p(x) = \left(\alpha\cdot h^{\alpha}\right)/(x-a)^{\alpha+1}$; $x\geq a+h$.

Entropy, $H(X) = -\int_{a + h}^{\infty}p(x)\log p(x)\,dx$

$$H(X) = -\int_{a + h}^{\infty}\frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}}  \log\left(\frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}} \right)\,dx$$

$$= -\int_{a +h}^{\infty} \frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}} \left[\log(\alpha h^{\alpha}) -(\alpha+1)\log(x-a) \right]\,dx$$

$$= -\alpha h^{\alpha}\int_{a + h}^{\infty}\frac{1}{(x-a)^{\alpha+1}} \left[\log(\alpha) + \alpha \log(h) -(\alpha+1) \log(x-a) \right]\,dx$$

$$= -\alpha h^{\alpha}\int_{a + h}^{\infty} \frac{ \log(\alpha) + \alpha \log(h)}{(x-a)^{\alpha+1}}  dx + \alpha h^{\alpha}\int_{a + h}^{\infty} \frac{(\alpha+1) \log(x-a)}{(x-a)^{\alpha+1}}dx$$

$$= -\alpha h^{\alpha}(\log(\alpha) + \alpha \log(h))\int_{a + h}^{\infty} \frac{ 1}{(x-a)^{\alpha+1}}  dx +  \alpha h^{\alpha}(\alpha+1)\int_{a + h}^{\infty} \frac{ \log(x-a)}{(x-a)^{\alpha+1}}dx $$

Then how can I end up with the result

$$H(X) = -(1+2\alpha)\log(h) – \log(\alpha) + 1 + (1/\alpha)$$

答:
By using the following (which can be computed using standard integrals, integration by parts and this limit)

$$\int^\infty_{a+h}\frac{1}{(x-a)^{\alpha+1}}dx=\left.-\frac{1}{\alpha(x-a)^\alpha}\right|^\infty_{a+h}=\frac{1}{\alpha h^\alpha},$$

$$\int^\infty_{a+h}\frac{\log(x-a)}{(x-a)^{\alpha+1}}dx=\left.-\frac{\alpha\log(x-a)+1}{\alpha^2(x-a)^\alpha}\right|^\infty_{a+h}=\frac{\alpha\log(h)+1}{\alpha^2h^\alpha},$$

I get that $H(x)=-\log(\alpha)+\log(h)+1+\frac{1}{\alpha}$ which seems to match up with the entropy listed on wikipedia - are you sure that the end result you list above is correct? I could have mucked a minus sign somewhere - but I can't seem to find it.


问:
Question about Existential Instantiation
I was having some trouble understanding existential instantiation. My textbook (Rosen - Discrete Mathematics and its Applications) states this about existential instantiation:

Existential instantiation is the rule that allows us to conclude that there is an element c in
the domain for which P(c) is true if we know that ∃xP(x) is true. We cannot select an arbitrary
value of c here, but rather it must be a c for which P(c) is true. Usually we have no knowledge
of what c is, only that it exists. Because it exists, we may give it a name (c) and continue our
argument.

This makes sense to me for certain existential statements.

For example, consider the statement $\exists x\in \mathbb{Z}$ $(x + 1 = 2)$. There is only one integer that makes the propositional function ($x + 1 = 2$) true (namely, $1$). Therefore, it makes sense to me that a new symbol $c$ can be created to name "the one integer that makes $x + 1 = 2$ true".

However, consider the statement $\exists x\in \mathbb{Z}$ $(x * 0 = 0)$. There are many integers that make the propositional function ($x * 0 = 0$) true (ex. $1$, $2$, $3$).

In this case, when we create a new symbol $c$, is this symbol naming "one of the integers that makes $x * 0 = 0$ true"? I find this slightly ambiguous, so I was wondering if I was understanding the meaning of this symbol correctly.

Please clarify and thank you for your time.

答:
However, consider the statement $\exists x\in \mathbb{Z}$ $(x * 0 = 0)$. There are many integers that make the propositional function ($x * 0 = 0$) true (ex. $1$, $2$, $3$).

In this case, when we create a new symbol $c$, is this symbol naming "one of the integers that makes $x * 0 = 0$ true"?

Yes, that is exactly it.  So .. even though the use of $c$ suggests that we know exactly which object we are talking about, this is in fact not the case. We still only know that there is at least one object that satisfies the formula in question. But, in order to do our further reasoning, we need to be able to talk about 'one of those objects' and for that, this system uses an individual constant.... although you of course need to make sure that that very constant wasn't used elsewhere in the proof already to refer to some other object.

Please note that there are other formal proof systems that don;t use $c$ in this case, but keep the variable a variable, which has the advantage of the suggestion that you indeed do not know what specific object you are talking about ... but the drawback is that you now get lines in the proof that, when taken out of context of the rest of the proof, would have a free variable ... and that is indeed drawback enough for some people to use constants instead.

I have sometimes thought that maybe one way to deal with all this is to have a third set of ways to point to objects other than constants and variables: symbols that you would indeed use for this very existential instantiation, and which denote 'some object with some property, though we don;t know which one', i.e. not completely arbitrary (like a normal variable), but also not specific (like a constant). I have never sen formal systems do anything like that though.


答:
Okay, I think I might have come up with an answer for atleast one implication thanks to the examples at https://math.stackexchange.com/a/1874240/589 and https://math.stackexchange.com/a/2823996/589 .

Suppose a finite abelian group $G$ is indeed isomorphic to

$$\bigoplus_{i=1}^r\mathbb{Z}_{p_i^{k_i}}$$

One has to order the powers of the primes $p_i^{k_i}$ in increasing order. Then take the smallest one, say $p_a^{k_a}$ for some $a\in\{1,\ldots,r\}$ and all of its multiples in the list of powers of primes, $p_{a_1}^{k_{a_1}},\ldots,p_{a_\alpha}^{k_{a_\alpha}}$ with $\alpha\in\mathbb{N}$. This way $p_a^{k_a}|p_{a_1}^{k_{a_1}},\ldots,p_{a_\alpha}^{k_{a_\alpha}}$.

Now take the second smallest power of a prime from the list that is not a multiple of the previously taken smallest powers of primes, for now only $p_a^{k_a}$, say $p_b^{k_b}$ for some $b\in\{1,\ldots,r\}$ and again all of its multiples from the list, $p_{b_1}^{k_{b_1}},\ldots,p_{b_\beta}^{k_{b_\beta}}$ with $\beta\in\mathbb{N}$.

We keep repeating this procedure until all the powers of primes are ordered in this specific way. Now, place each collection in the row of a 'table' such that they are right aligned:

\begin{align*}
p_a^{k_a} \qquad &p_{a_1}^{k_{a_1}} \qquad p_{a_2}^{k_{a_2}} \ldots \space p_{a_\alpha}^{k_{a_\alpha}}\\
&p_b^{k_b} \qquad p_{b_1}^{k_{b_1}} \ldots \space p_{b_\beta}^{k_{b_\beta}}\\
&\ldots\ldots
\end{align*}
Now multiply each column and denote each product $P(a), P(a_1),\ldots,P(a_\alpha)$, referring to what column the product was made of. This way the gcd of all the terms in each of the products is 1 and $P(a)|P(a_1)|\ldots|P(a_\alpha)$. Now the Chinese remainder theorem come into play. This theorem implies that $\mathbb{Z}_{jk}\cong\mathbb{Z}_j\oplus\mathbb{Z}_k$ iff gcd$(j,k)=1$. Hence $\mathbb{Z}_{P(a_i)}\cong\mathbb{Z}_{p_{a_1}^{k_{a_1}}}\oplus\mathbb{Z}_{p_b^{k_b}}\oplus\ldots$ (a direct sum of cyclic groups, each with order equal to a term of the product $P(a_i)$) with $i=1,\ldots,\alpha$.

This means we finally have

$$\bigoplus_{i=1}^r\mathbb{Z}_{p_i^{k_i}}\cong\mathbb{Z}_{P(a)}\oplus\mathbb{Z}_{P(a_1)}\oplus\ldots\oplus\mathbb{Z}_{P(a_\alpha)},$$
where $P(a)|P(a_1)$ and $P(a_i)|P(a_{i+1})$ with $i=1,\ldots,\alpha-1$.

This is the best I could explain my method. I hope this is clear. If you have any comment feel free to leave it here.

Title: Equivalence between the structure theorems of finite abelian groups

Body:
I want to show that the primary decomposition and the invariant factor decomposition of finite abelian groups are equivalent. What I mean by this is that, given a finite abelian group $G$ is isomorphic to

$$\mathbb{Z}_{p_1^{k_1}}\oplus\dots\oplus\mathbb{Z}_{p_r^{k_r}},$$

where $r\in\mathbb{N}$ and every $p_i^{k_i}$ is the power of a prime (not necessarily different powers of primes), then $G$ is isomorphic to

$$\mathbb{Z}_{n_1}\oplus\ldots\oplus\mathbb{Z}_{n_u},$$

where $u\in\mathbb{N}$ and $n_i|n_{i+1}$ for every $i=1,\ldots,u-1$ and vice versa.

I read that this can be done using the Chinese remainder theorem, since this implies that $\mathbb{Z}_{ab}\cong\mathbb{Z}_a\bigoplus\mathbb{Z}_b$ iff $a$ and $b$ are coprime, but I can't seem to figger out how.


问:
System of first order differential equations problem
I have a system of differential equations which I need to solve and obtain $y(x)$ and $z(x)$. I tried elimination method and got to a point but I don't know what to do after here. Any help would be appreciated
Question:
$$y'(x)=\frac{x+y}{z}$$
$$z'(x)=\frac{x-y}{y}$$

My solution until the point I got stuck.
Any help would be appreciated

答:
$$y'(x)=\dfrac{x+y}{z} \\z'(x)=\dfrac{x-y}{y}$$
$$y'(x)z={x+y} \\z'(x)y={x-y}$$
Sum both differential equations:
$$y'z+z'y=2x$$
$$(yz)'=2x$$
$$\implies yz =x^2+C$$
The second DE is:
$$z'(x)y={x-y}$$
$$z'=\dfrac x y-1$$
$$z'=\dfrac {zx}{x^2+C} -1$$
That you can solve.

Edit1
$$z'=\dfrac {\color {red}{z}x}{x^2+C} -1$$
You can't integrate both sides the way you did because there is the $z$ function on RHS:
$$z'-\dfrac {zx}{x^2+C}= -1$$
$$\sqrt {x^2+C}z'-\dfrac {zx}{\sqrt {x^2+C}}= -\sqrt {x^2+C}$$
$$\left ( \dfrac  z {\sqrt {x^2+C}}\right)'=-\dfrac 1 {\sqrt {x^2+C}}$$
Now you can integrate both sides.
$$ \dfrac  z {\sqrt {x^2+C}}=-\int \dfrac {dx} {\sqrt {x^2+C}}+C_2$$
$$ z(x) ={\sqrt {x^2+C}}\left (C_2-\arctan \dfrac {x} {\sqrt {x^2+C}} \right)$$


问:
Relation Equivalence of $R \subseteq \mathbb Z\times \mathbb Z$

Consider the relation $R \subseteq \Bbb Z \times \Bbb Z$ given as
$$R = \{(x,y) \in \Bbb Z \times \Bbb Z \mid xy >0 \textrm{ or } x=y=0\}.$$
Prove that R is an equivalence and write down its equivalence classes.

Can anybody tell me how do I approach a task like this in general?

$[0] = \{0, 1, 4, 9\}$ for $xy>0$ would be my approach.

But then, what about $[1]$ for example?

I would like to see an example, let's say with $xy > 1$ or $xy>2$, as I am honestly a little bit lost (first time doing that kind of stuff) and my literature hasn't any kind of example for that, other than some theoretically stuff.

Thanks in advance everyone!

答:
To get you started: you are correct that $R$ must be symmetric, transitive, and reflexive. Prove these one at a time. Here's symmetry, for example:

Suppose that $(x,y) \in R$. Then (by definition of $R$), $xy>0$ or $x=y=0$. In the first case, since $xy=yx$ for all real numbers $x,y$, then $yx>0$. In the second case, $y=x=0$. Therefore $yx>0$ or $y=x=0$, so $(y,x) \in R$.

Now follow the same kind of reasoning for transitivity and reflexivity.

Once you've established that, figure out the classes. Take $x=0$, for example. For what $y$ do we have $(x,y)\in R$? That would require $xy>0$ or $x=y=0$. But $xy>0$ is impossible since $x=0$, so we must have $y=0$. In other words, $[0] = \{0\}$, which is to say that $0$ is the only element in its class. What happens if you do the same thing with $x=1$? What other number remain after this?

Finally, to give you an idea of why this might not work with other relations, let's change $xy>0$ to $xy>1$: let $S = \{(x,y) \in \Bbb Z \times \Bbb Z \mid xy >1 \textrm{ or } x=y=0\}.$

If this were an equivalence relation, it would be reflexive: that is, for every $x\in \Bbb Z$, we would have $x^2 > 1$ or $x=0$. This fails for $x=1$. (Exercise: is this relation transitive? symmetric?)


问:
Harder-Than-It-Seems: Finding inverse of $f(x)=x^3-x-12$?
This may seem simple, but I have had long days of frustration with finding the inverse of the function

$$f(x)=x^3-x-12.$$

I got this on some homework and it did not ask for the inverse. However I wanted to try anyways. So this is as far as  I got:

$$x=y^3-y-12$$
$$x+12=y^3-y$$
$$y(y^2-1)=x+12$$
$$y(y+1)(y-1)=x+12.$$

And that is absolutely as far as I go. In the end, I have no idea how this turns out to be the answer:

A step-by-step walk-through in finding the inverse is gratefully appreciated!

答:
The inverse function $y(x)$ satisfies $x=y^3-y-12$, or the cubic equation

$$y^3-y-(x+12)=0$$
Use Cardano’s formula to obtain

$$y(x)=\sqrt[3]{\frac {x+12}2+\sqrt{\frac{(x+12)^2}4-\frac1{27}}} +\sqrt[3]{\frac {x+12}2-\sqrt{\frac{(x+12)^2}4-\frac1{27}}}
$$

OwnerDisplayName: anon


问:
Finding an area of the portion of a plane?
I need help with a problem I got in class today any help would be appreciated!

Find the area of the portion of the portion of the plane $6x+4y+3z=12$ that passes through the first octant where $x, y$, and $z$ are all positive..

I graphed this plane and got all the vertices but I am not sure how my teacher wants us to approach this problem..

Do I calculate the line integral of each side of the triangle separately and add them together? because we are on the section of line integrals, flux, Green's theorem, etc..

答: I'm confident this isn't what your teacher has in mind, but once you have the three vertices, you can find the three distances between them, and use Heron's formula.</p>


问:
I have to solve the complex equation $2\operatorname{Im}(z)+1=\operatorname{Re}(z^2)$
$2\operatorname{Im}(z)+1=\operatorname{Re}(z^2)$ all complex solutions of the equation. So I started basically with stating $z=a+bi$ Therefore $\operatorname{Im}(z)=b$, $z^2=(a^2-b^2)+(2abi)$, and therefore $\operatorname{Re}(z^2)=a^2-b^2$. So I have that $2b+1=a^2-b^2 $ but I don't know what to do next. All solutions should be given in form of $z=a+bi$ Thanks for your help.

答: You are almost there! Note that\begin{align}2b+1=a^2-b^2&\iff b^2+2b+1=a^2\\&\iff(b+1)^2=a^2\\&\iff b+1=\pm a.\end{align}So, the solutions are the numbers of the form $a+(a-1)i$ ($a\in\Bbb R$), together with those of the form $a-(a+1)i$ (again, with $a\in\Bbb R$).


答: If you have the three vertices, you can calculate the length of the three sides and use Heron's formula</p>

Title: Finding an area of the portion of a plane?

Body:
I need help with a problem I got in class today any help would be appreciated!

Find the area of the portion of the portion of the plane $6x+4y+3z=12$ that passes through the first octant where $x, y$, and $z$ are all positive..

I graphed this plane and got all the vertices but I am not sure how my teacher wants us to approach this problem..

Do I calculate the line integral of each side of the triangle separately and add them together? because we are on the section of line integrals, flux, Green's theorem, etc..


问:
is this proof complete?
I am learning to write proofs in set theory but I am not sure if what I am doing is right...do you think I am on the right track? This is what I am trying to prove

Let $f$ be a function $f:A \to B$. Let $A_n$ be a sequence of subsets of $A$, and let $B_n$ be a sequence of subsets of $B$. Show that
$$f^{-1}(\bigcup_n B_n) = \bigcup_n f^{-1}(B_n).$$

This is what I have so far:

Proof first attempt

答:
Sorry, but no. The first error is in your first line: "We know that every subset of uncountable set is countable." This is not true; consider the uncountable set $\mathbb{R}$ with its uncountable subset, the irrational numbers.

The statement (4) is technically correct but it demonstrates a misunderstanding of what the question is asking. You write, "a function is invertible only if the function is bijective." However, when we write $f^{-1}(X)$, where $X$ is some set, we generally aren't referring to an explicit function $f^{-1}$. Instead, this is shorthand for the preimage or inverse image of that set. In other words, $f^{-1}(B_n)=\{a\in A\,|\,f(a)\in B_n\}$.

Most of the rest of the statements don't make sense to me as written; for example, (7) says that "$x \in$ union of functions," but you can't take a union of functions. (EDIT: In the comments, Brian M. Scott points out that you can take a union of functions in some circumstances. Still, the way the concept is applied in (7) doesn't make sense.)

I think that you're overcomplicating things. Go back to basics. Let $x \in f^{-1}(\bigcup\limits_n B_n)$. Then, that means that there exists some $b\in\bigcup\limits_n B_n$ such that $f(x)=b$. But $b\in\bigcup\limits_n B_n$ means that there exists some integer $k$ (I assume integers, but the problem should really specify) such that $b\in B_k$. But remember that $f(x)=b$, so $x\in f^{-1}(B_k)$ by definition. After a few more steps, you should be able to conclude that $f^{-1}(\bigcup\limits_n B_n)\subseteq \bigcup\limits_n f^{-1}(B_n)$.

Then, you have to go in reverse and prove that $\bigcup\limits_n f^{-1}(B_n)\subseteq  f^{-1}(\bigcup\limits_n B_n)$. Hopefully, you should be able to do this part yourself, and then your proof will be over.

Welcome to Math StackExchange.

Title: is this proof complete?


问:
Converting Polar equation to Cartesian by multiplying by r
I need to convert the polar equation $r = 3\sin(\theta)$ to cartesian form. Using the equations $x^2+y^2=r^2$, $x = r\cos(x)$ and $y = r\sin(x)$,

$$
r = 3\sin(\theta)\\ \implies
r^2 = 3r\sin(\theta)\\ \iff
x^2 + y^2 = 3x\\ \iff
(x-\frac 32)^2 +y^2 = (\frac 32)^2
$$
I understand that all $(r,\theta)$ that satisfy $r = 3\sin(\theta)$ also satisfy $r^2 = 3r\sin(\theta)$, so $r = 3\sin(\theta) \implies r^2 = 3r\sin(\theta)$.

The solutions manual says that $r = 3\sin(\theta)\iff r^2 = 3r\sin(\theta)$. From my understanding,

$ r^2 = 3r\sin(\theta) \implies r = 0$ or $r = 3\sin(\theta)$
How can I show that $r = 0$ or $r = 3\sin(\theta)$ and $r = 3\sin(\theta)$ are equivalent statements?

My thoughts:
For all values of $(r,\theta)$ that satisfy $r = 3\sin(\theta)$, $r = 3\sin(\theta)$ is true. This includes $(0,\pi),(3, \frac \pi2)$, etc. The ordered pair $(0,\pi)$ satisfies $r = 3\sin(\theta)$, but $r = 0$ itself does not.

I'm always unsure about multiplying both sides of an equation by a variable.
Any help is appreciated. Thank you

答:
When $r=0$ , $sin(\theta)=k\pi$

In either case, the converse will also hold and $r=0$ hence, you can multipy by $r$ since $r\neq0$ whenever $\theta \neq k\pi$


问:
Is it possible to evaluate $\lim_{n\to\infty}n(\sqrt[n]{4}-1)$ without applying L'Hospital's rule?
I can evaluate the limit with L'Hospital's rule:

$\lim_{n\to\infty}n(\sqrt[n]{4}-1)=\lim_{n\to\infty}\cfrac{(4^{\frac1n}-1)}{\dfrac1n}=\lim_{n\to\infty}\cfrac{\dfrac{-1}{n^2}\times 4^{\frac1n}\times\ln4}{\dfrac{-1}{n^2}}=\ln4$

But is there any way to do it without using L'Hospital's rule?

答:
Here's a trick to prove convergence of the continuous limit$$
\lim\limits_{x\rightarrow\infty} x \left(\sqrt[x]4 - 1\right)
$$
if you know also know how to integrate $2^x$.
Observe:\begin{eqnarray}
x \left(\sqrt[x]4 - 1\right) = x \left(\sqrt[x]2 - 1\right)\left(\sqrt[x]2 + 1\right) = 2x\left(\sqrt[2x]4 - 1\right)\frac{\left(\sqrt[x]2 + 1\right)}{2}
\end{eqnarray}
Thus we get $$
x\left(\sqrt[x]4 - 1\right) = \frac{\sqrt[x]2 + 1}2 2x\left(\sqrt[2x]4-1\right) \frac{\sqrt[x]2 + 1}2 \frac{\sqrt[2x]2 + 1}{2} 4x\left(\sqrt[4x]4-1\right) = \left(\prod_{k=0}^{n-1} \frac{\sqrt[2^kx]2+1}{2}\right) 2^n x \left(\sqrt[2^n x]4 -1\right)
$$
Note that the multiplicands are always bigger than $1$, so this implies that $2^n x \left(\sqrt[2^nx]4 - 1\right)$ is decreasing in $n$, so the limit exists for each $x$. It's also pretty clear that the function $x\to x\left(\sqrt[x]4-1\right)$ doesn't have any oscillations in the limit, hence we have that the continuous limit converges, so we just have to compute it for one value of $x$, say $x=1$. Thus we have $$
\lim\limits_{z\rightarrow\infty} z \left(\sqrt[z]4-1\right) = \lim\limits_{n\rightarrow\infty} 2^n \left(\sqrt[2^n]4-1\right) = \lim\limits_{n\rightarrow\infty} \frac{\sqrt[1]4-1}{\prod\limits_{k=0}^{n} \frac{\sqrt[2^k]2+1}{2}} = \frac2 {\prod\limits_{k=1}^\infty \frac{2^{1/2^k} + 1}{2}}
$$
Expanding the partial products, we can see that this product actually isn't hard to evaluate:\begin{eqnarray}
\prod_{k=1}^1 \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac12 + 1}2\\
\prod_{k=1}^2 \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac14 + 2^\frac12 + 2^\frac34 + 1}4\\
&\vdots&\\
\prod_{k=1}^N \frac{2^{\frac1{2^k}} + 1}{2} &=& \frac{2^\frac1{2^N} + 2^\frac{2}{2^N} + \cdots + 2^\frac{2^N-1}{2^N} + 2^\frac{2^N}{2^N}}{2^N}
\end{eqnarray}
The RHS is a left Riemann sum for $\int_0^1 2^t dt$ splitting the interval into $2^N$ subintervals. Thus we get $$
\prod_{k=1}^\infty \frac{2^{\frac1{2^k}} + 1}{2} = \int_0^1 2^t dt = \frac{2-1}{\ln 2} = \frac1{\ln 2}
$$
Hence we conclude $$
\lim\limits_{z\rightarrow\infty} z\left(\sqrt[z]4-1\right) = \frac{2}{1/\ln 2} = 2\ln 2 = \ln 4
$$


问:
When meet is strictly monotone? What is the name?
In a meet semilattice $(S, \wedge)$, the meet operation is monotonic. I.e. if $x\geq x'$ and $y\geq y'$, then $x\wedge y\geq x'\wedge y'$.

What if we assume that this is true for strict inequalities? Is there a name for such semilattices?

If $x> x'$ and $y> y'$, then $x\wedge y > x'\wedge y'$.

What is the name of such semilattices?

答:
I don't think there is any such name because that is not an interesting property.

Indeed, assume $x$ and $y$ are not related (that is, if the semilattice is not a chain);

it follows that $x \wedge y < x$ and $x \wedge y < y$.

If the operation would have such a property, then we would have
$$x \wedge y > (x\wedge y) \wedge (x \wedge y) = x\wedge y,$$
a contradiction.

So that property is only hold for semilattices which are chains, but in that case, it is trivial.


问:
Can a basis of $\mathbb R^n$ contain complex vectors?
I think that this should be false. But on the other hand I am thinking that if we multiply a complex vector with a complex number be can get a real vector. So just wanted to make sure. Any help will be appreciated.</p>

答: A complex vector $v$ is not a member of $\mathbb{R}^n$ and therefore, cannot be in a basis of $\mathbb{R^n}$.


问:
Integral of a derivative vs derivative of an integral?
I'm learning the fundamental theorem of calculus, and was confused on something. Is the original function $f(x)$ equal to the integral of the derivative, i.e. $\int f'(x)\ dx$, or equal to the derivative of the integral, i.e. $\frac{d}{dx} \int f(x)\ dx $?

Thanks.

答:
Short answer: derivative of an integral.

The first and second parts of the FTC respectively specify sufficient conditions, which we'll assume hereafter are correct, to obtain$$\tfrac{d}{dx}\int_a^xf(t)dt=f(x),\,\int_a^bF^\prime(t)=F(b)-F(a).$$So the first part recovers $f$ as the derivative of a definite integral $F(x):=\int_a^xf(t)dt$, while the second part recovers $F$ as a definite integral of the derivative $f:=F^\prime$

Note that if you differentiate an integral by the first part you get the original function, but if you integrate the derivative by the second part you get an arbitrary antiderivative, i.e. recover the original function up to an additive constant. (This "constant" is more generally a locally constant function, e.g. antiderivatives of $1/x$ can differ by different amounts either side of $x=0$.)


问:
Two definitions of $\limsup$
Here are two equivalent definitions of $\limsup_{n\rightarrow\infty} a_n$:

Let $u_n=\sup\{a_n, a_{n+1}, a_{n+2},\ldots\}$. Then
$$\limsup_{n\rightarrow\infty} a_n = \lim_{n\rightarrow\infty} u_n =
   \lim_{n\rightarrow\infty}\left(\sup\{a_n,a_{n+1},\ldots\}\right)$$

Let $E$ be the set of all subsequential limits of $\{a_n\}$. Then $$\limsup_{n\rightarrow\infty} a_n = \sup E$$

I'm curious as to which one people usually learn first, or which one people find more intuitive. Baby Rudin only has the second definition, but doesn't mention the first as far as I know.

As far as intuition, I think of the second definition as "the biggest limit point." I'm not sure how to think of the first definition though. I can see that $u_n \ge u_{n+1}$ for all $n$, but I'm not sure how to interpret the limit of those $u_n$ conceptually.

Are there any other ways to conceptualize $\limsup$?

答:
$\limsup_{n\to\infty} x_n $ and $\liminf_{n\to\infty} x_n $ naturally arise when trying to understand convergent subsequences of the sequence $ (x_n) $. Here's the idea ( We'll be looking at real sequences throughout ) :

Convergent sequences have nice properties ( e.g. boundedness, $ \lim_{n\to\infty} (xa_n + yb_n) = x(\lim_{n\to\infty} a_n) + y(\lim_{n\to\infty} b_n) $, etc ), and the notion is quite central since many other notions can be restated in terms of this ( e.g. "$ f : A (\subseteq \mathbb{R} ) \rightarrow \mathbb{R} $ is continuous at a point $ p \in A $" if and only if "for every sequence $ (a_n) $ in $ A $ converging to $ p $ we have $ f(a_n) $ converging to $ f(p) $" ). In general a sequence doesn't converge, but the next best thing we can have is a convergent subsequence (Recall a subsequence of $ x_1, x_2, x_3, \cdots $ is just a sequence $ x_{j_1}, x_{j_2}, x_{j_3}, \cdots $
with $ j_1 < j_2 < j_3 < \cdots $ ). When does a sequence $ (x_n) $ have a convergent subsequence ? And when it does, what can we say about $ \{ \text{ limits of convergent subsequences of } (x_n) \: \} $ ?

Let's first focus on the second question, i.e. on what happens if a sequence $ (x_n) $ does have a convergent subsequence $ (x_{n_k}) $. Assuming boundedness of $ (x_n) $ we have $ \beta_{n_k} \leq x_{n_k}, x_{n_{k+1}}, x_{n_{k+2}}, \cdots  \leq \alpha_{n_k} $ where $ \alpha_j := \sup \{ x_j, x_{j+1}, x_{j+2}, \cdots \} $ and $ \beta_j := \inf \{ x_j, x_{j+1}, x_{j+2}, \cdots \} $. Since $ (\alpha_j) $ is non-increasing & bounded below it has a limit $ \alpha = \inf \alpha_j $, and similarly $ (\beta_j) $ being non-decreasing & bounded above has a limit $ \beta = \sup \beta_j $. Now taking $ k \to \infty $ gives $ \beta \leq \lim_{k\to\infty} x_{n_k} \leq \alpha $.

To summarise : Let $ (x_n) $ be a bounded sequence with a convergent subsequence $ (x_{n_k}) $. Then $ \lim x_{n_k} \in [ \beta, \alpha ] $, where $ \alpha $ and $ \beta $ are the limits of $ \alpha_j := \sup \{ x_j, x_{j+1}, \cdots \} $ and $ \beta_j := \inf \{ x_j, x_{j+1}, \cdots \} $ respectively.

Now let's tackle the first question ( notice in the above discussion we only needed boundedness of $ (x_n) $ to talk about $ \alpha_j, \beta_j $ and their limits $ \alpha $, $ \beta $ ). If $ (x_n) $ is a  bounded sequence, because intuitively "there are terms of sequence sticking near $ \alpha_j $s, and the $ \alpha_j $s converge to $ \alpha $" and similarly for $ \beta $, we expect there are subsequences converging to $ \alpha $ & $ \beta $. It's actually true :
Let $ (x_n) $ be a bounded sequence. As usual $ \alpha_j := \sup \{ x_j, x_{j+1}, \cdots \} $, $ \beta_j := \inf \{ x_j, x_{j+1}, \cdots \} $, with respective limits $ \alpha = \inf \alpha_j $ and $ \beta = \sup \beta_j $. There exists an $ n_1 $ such that $ \alpha_1 - 1 < x_{n_1} \leq \alpha_1 $. Now there exists $ n_2 ( \geq n_1 + 1 > n_1 ) $ such that $ \alpha_{{n_1} + 1} - \frac{1}{2} < x_{n_2} \leq \alpha_{{n_1} + 1} $. And now there exists $ n_3 ( \geq n_2 + 1 > n_2 ) $ such that $ \alpha_{{n_2}+1} - \frac{1}{3} < x_{n_3} \leq \alpha_{{n_2}+1} $, and so on. Therefore we get $ n_1 < n_2 < \cdots $ such that $ \alpha_{{n_{j-1}}+1} - \frac{1}{j} < x_{n_{j}} \leq \alpha_{{n_{j-1}}+1} $ for all $ j \geq 1 $ ( we'll take $ n_0 = 0 $, to make sense of the $ j = 1 $ inequality ). So taking $ j \to \infty $, we finally get $ \lim_{j \to \infty} x_{n_j} = \alpha $. Similarly one can construct a subsequence of $ (x_n) $ converging to $ \beta $.

To summarise the entire discussion :
Let $ (x_n) $ be a bounded sequence. Then $ S := \{ \text{ limits of convergent subsequences of } (x_n) \: \} $ is non-empty, with $ \max(S) = \alpha $ and $ \min(S) = \beta $, where $ \alpha = \inf \alpha_j $ and $ \beta = \sup \beta_j $ are the limits of $ \alpha_j := \sup \{ x_j, x_{j+1}, \cdots \} $ and $ \beta_j := \inf \{ x_j, x_{j+1}, \cdots \} $ respectively.

Remark : Here the fact that $ S \neq \varnothing $, i.e. that every bounded sequence of reals has a convergent subsequence, is traditionally called Bolzano-Weierstrass theorem. It is central to Analysis, and can also be proved by a bisection argument.

Title: Two definitions of $\limsup$


问:
Deriving WKB approximation step query
I am working through the derivation of the WKB approximation and I can't seem to work out how one of the steps is done.  I've provided a picture from my lecture notes.

Screenshot for Question

How do they manage to obtain $Y_{1}=-\frac{1}{2}\log{Y_{0}}$ from the previous equation? I have tried working it out and haven't been able to reach the value provided.

答: They showed $Y_{1}' = - \frac{1}{2} \frac{Y_{0}''}{Y_{0}'}$.  Note that $\frac{Y_{0}''}{Y_{0}'} = \left( \log(Y_{0}') \right)'$.  Hence $Y_{1}' = -\frac{1}{2} \left(\log(Y_{0}')\right)'$ and the fundamental theorem of calculus gives $Y_{1} = - \frac{1}{2} \log(Y_{0}') + C$ for some $C \in \mathbb{R}$.  (It seems, in your attachment, they have the freedom to set $C = 0$.)

Title: Deriving WKB approximation step query