问:
Dividing by vector after factoring out
If I factor out a matrix from an equation and then am left with 0 on one side, can I effectively cancel out the vector by dividing both sides by it?

A trivial example being
$$
\begin{align}
AB\hat{x} - AC &= 0 \implies\\
A(B\hat{x} - C) &= 0 \implies\\
B\hat{x} - C &= 0
\end{align}
$$

while also obviously considering that $A=0$ is a possibility or has been assumed to not be $0$

答:
The correct analogy to the usual $xy=0$ implying $x=0$ or $y=0$ over the reals (or really any integral domain) uses invertibility of square matrices:

When $A$ is a (real or complex) square matrix and $x$ is a vector, the equation $Ax=0$ implies that $x=0$ or $A$ is not invertible (which allows more than just $A=0$).

Hence, the implication $$Ax=0\ \Longrightarrow\  x=0$$ is only true, when $A$ is an invertible matrix.

Title: Dividing by vector after factoring out


答:
Depending on the dimensions of the various things, which you do not define but I imagine $A,B,C$ to be square or rectangular matrices and $X$ to be a column vector, your equation does not make sense as it sums matrices of different sizes.

Even if the equation would make sense the answer would be no as you can simplify $A$ only if it is invertible (so you are actually multiplying by $A^{-1}$). In general with matrices you can have $AB=0$ with $A\neq 0$, $B\neq 0$.

ContentLicense: CC BY-SA 4.0

Title: Dividing by vector after factoring out

Body:
If I factor out a matrix from an equation and then am left with 0 on one side, can I effectively cancel out the vector by dividing both sides by it?

A trivial example being
$$
\begin{align}
AB\hat{x} - AC &= 0 \implies\\
A(B\hat{x} - C) &= 0 \implies\\
B\hat{x} - C &= 0
\end{align}
$$

while also obviously considering that $A=0$ is a possibility or has been assumed to not be $0$


问:
Analytic Solution to the Differential equation $(1-t^2)x'' + tx' - x = 0$.
I am looking to solve the following problem:

Consider the differential equation,

$$(1-t^2)x'' + tx' - x = 0$$

(a) Find a power series expansion about t = 0 for a general solution to the differential equation. Your answer should include a general formula for the coefficients.

(b) Solve the IVP $$(1-t^2)x'' + tx' - x = 0, x(0) = -1, x'(0) = 2$$

I have come to a dead end on this problem myself after my work and solution shown below;

$$x(t) =  \sum_{n=0}^\infty a_nt^n$$
$$x'(t) =  \sum_{n=1}^\infty a_nnt^{n-1}n$$
$$x''(t) =  \sum_{n=2}^\infty a_nn(n-1)t^{n-2}$$

Subsituting into the differential equation and simplifying gave:

$$\sum_{n=0}^\infty [a_{n+2}(n+2)(n+1) - a_nn(n-1)]t^n = 0$$

Since $t^n > 0$, we must have

$$a_{n+2}(n+2)(n+1) - a_nn(n-1) = 0 $$

This gives the recursive sequence:

$$a_{n+2} = \frac{a_nn(n-1)}{(n+2)(n+1)}$$

I am stuck here and am not sure how to proceed with solving the rest of the problem. I am looking for a solution to the initially stated problem so that I can compare my work to it and learn from it. As a note, I suspect my formula for the recursive sequence is incorrect since it terminates, although I am not sure.

Body:
$$
t^n: 0 = (n+2)(n+1)a_{n+2} - n(n-1)a_n + na_n - a_n = (n+1)(n+2)a_{n+2}-(n-1)^2a_n \Leftrightarrow
$$

$$
a_{n+2} = \frac{(n-1)^2}{(n+2)(n+1)}a_n:
$$

$$
\begin{aligned}
a_0 &= -1 \\
a_1 &= 2 \\
a_2 &= \frac{(-1)^2}{2\cdot 1}a_0 = -\frac{1^2}{1\cdot 2} \\
a_3 & = 0 \\
a_4 &= \frac{1^2}{4\cdot 3}a_2 = -\frac{1^2}{1\cdot 2\cdot 3\cdot 4} = -\frac{1^2}{4!} \\
a_5 &= \frac{2^2}{5\cdot 4}a_3 = 0 \\
a_6 &= \frac{3^2}{6 \cdot 5}a_4 = -\frac{1^2\cdot 3^2}{4!\cdot 5\cdot 6} = -\frac{1^2\cdot 3^2}{6!} = -\frac{\prod\limits_{j = 1}^{3-1}(2j-1)^2}{(2\cdot 3)!}\\
a_7 &= \frac{4^2}{7\cdot 6}a_5 = 0 \\
a_8 &= \frac{5^2}{8 \cdot 7}a_6 = -\frac{1^2\cdot 3^2 \cdot 5^2}{6!\cdot 7\cdot 8} = -\frac{1^2\cdot 3^2\cdot 5}{8!} = -\frac{\prod\limits_{j = 1}^{4-1}(2j-1)^2}{(2\cdot 4)!} \\
a_9 &= \frac{6^2}{9\cdot 8}a_7 = 0 \\
a_{10} &= \frac{7^2}{10 \cdot 9}a_8 = -\frac{1^2\cdot 3^2 \cdot 5^2 \cdot 7^2}{8!\cdot 9\cdot 10} = -\frac{1^2\cdot 3^2\cdot 5 \cdot 7^2}{10!} = -\frac{\prod\limits_{j = 1}^{5-1}(2j-1)^2}{(2\cdot 5)!} \\
&\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\\
a_{2k-1} &= 0, \quad k \geq 2 \\
a_{2k} &= -\frac{\prod\limits_{j = 1}^{k-1}(2j-1)^2}{(2k)!}, \quad k \geq 2
\end{aligned}
$$

$$
x(t) = -1 + 2t -\frac{1}{2}t^2 - \sum_{k = 2}^{+\infty}\frac{\prod\limits_{j = 1}^{k-1}(2j-1)^2}{(2k)!}t^{2k}
$$


问:
Trouble Understanding Opposite Category
I went through few posts on MSE but It was not sufficent. So I am studying opposite category and I cannot understand what does it mean by by reversing a map? I know that the objects are the same and thecomposition will make sense once I understand morphisms.

Suppose $\mathfrak{C}$ is a category and if $A,B$ are two objects in $\mathfrak{C}$, then for any morphism $f : A \to B$, we define a morphism from $f^{op}:B \to A$. But how is the map $f^{op}$ defined from $f$?

I cannot find a way in which $f^{op}$ is well defined. Can anyone help me?

答: I'm kinda new too to the concept of category, but I think you are just confused. Think about the category made by the couple $\big(X, \leq\big)$ where $\leq$ is a preorder. An arrow $x\rightarrow y$ means: $x\leq y$. Now, if you move to the opposite, or rather the dual, you might have something like $\big(X^{op},\geq \big)$, and now you can interpret the dual arrow as $x\rightarrow y \iff x\geq y$. But notice that this is just a particular case. In general, Categories of Lattices should provide enough example of how to interpret the dual, but in general it is just an abstract costruction not really bounded to the original idea. You should accept that in category we have no elements so you can't search the meaning of an arrow by the way it acts on the element of an object, simply because this last sentence makes no sense in this setting. Rather, the information you have on an arrow merely relies on the category you find it, and on its "direction".

Title: Trouble Understanding Opposite Category


问:
How can I find the surface area of a normal chicken egg?
This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis.  Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg?  Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

答:
Use any convenient method to determine the volume of the entire egg.

Remove contents of egg through a small hole without damaging eggshell. (Egg collectors have a number of techniques).

Fill eggshell with water to determine the volume of the inside of the egg.

The difference of these two measurements is the volume of the eggshell.

Break the egg shell and measure the thickness. $$\frac{\text{∆ Volume}}{\text{Thickness}} \approx \text{Area}$$

ContentLicense: CC BY-SA 3.0


答:
To specify a category, we need to specify the objects, the morphisms, the domain and codomain of each morphism, and the composition operation on morphisms. The objects and morphisms are just abstract sets (or classes); they do not have to have any internal structure themselves. (For example, there need not be any notion of "elements" of objects, and morphisms do not need to be functions or function-like entities.)

Given a category $\mathcal{C}$, the opposite category $\mathcal{C}^{op}$ is defined to have the same classes of objects and morphisms, but the domain and codomain of morphisms are reversed, as is the order of composition. So, by definition, given objects $X$ and $Y$ of $\mathcal{C}$, the set (or class) of morphisms $\operatorname{Hom}_{\mathcal{C}^{op}}(X, Y)$ is just $\operatorname{Hom}_{\mathcal{C}}(Y, X)$; that is, by definition, a morphism $X \to Y$ in $\mathcal{C}^{op}$ literally is the same thing as a morphism $Y \to X$ in $\mathcal{C}$.

Title: Trouble Understanding Opposite Category

Body:
I went through few posts on MSE but It was not sufficent. So I am studying opposite category and I cannot understand what does it mean by by reversing a map? I know that the objects are the same and thecomposition will make sense once I understand morphisms.

Suppose $\mathfrak{C}$ is a category and if $A,B$ are two objects in $\mathfrak{C}$, then for any morphism $f : A \to B$, we define a morphism from $f^{op}:B \to A$. But how is the map $f^{op}$ defined from $f$?

I cannot find a way in which $f^{op}$ is well defined. Can anyone help me?


问:
Probability of Type 1 Error when using $X_{min}$ as a test static
Let $X_1,X_2, ..., X_{15}$ be a random sample from the exponential distribution with $\lambda > 0 $. To test $H_0 : \lambda = 1/5$ versus $H_A : \lambda < 1/5$ use $X_{min}$ as a test statistic. If $X_{min} \geq 1$ reject the null hypothesis.

My problem is then to compute the probability of a type 1 error. I know that I have to calculate

$$\begin{align*}
  P(\text{Type 1 Error}) & = P(\text{Reject} \ H_0 \ | \ H_0 \ \text{True}) \\
  & = P(X_{min} \geq 1 \ | \ \lambda = 1/5) \\
  & =
\end{align*}
$$

However, I am not sure how to proceed now. As a hint in my book, I have to look an exercise where the PDF for $x_{min}$ is found. Do I have to find the pdf for $X_{min}$ now? I know that the PDF for $x_{min}$
is
$$
f_{min}(x) = n(1 - F(x))^{n-1}f(x)
$$
Do I have to use this? Furthermore, is there any way to calculate this probability with in r studio?

All help is appreciated.

TIA.

答:
You can say $P(X_{min} > 1) = (1-F(1))^{15}$ since you want all five samples to exceed $1$

If the density is $f(x)= \frac1\lambda e^{-x/\lambda}$ then $F(x)=1-e^{-x/\lambda}$, so $$P\left(X_{min} > 1 \mid \lambda = \tfrac15\right) = \left(e^{-1/5}\right)^{15}=e^{-3} \approx 0.049787$$ as you seem to have found by another approach


问:
Players $P_1,P_2...P_m$ of equal skill play a game consecutively in pairs

Players $P_1,P_2...P_m$ of equal skill play a game consecutively in pairs as $P_1P_2,P_2P_3....P_{m-1}P_m, P_mP_1$ and any player who wins two consecutive games (ie. k and (k+1)th game) wins the match. If the chance that the match is won at the rth game is $k$, then find relation between $k$ and $r$

I haven’t really understood the question or what it wants to do, so I don’t know how to begin. Can anyone explain the question and give a starter hint?

答:
Note:  what follows assumes that the match ends with the game between $P_m$ and $P_1$ even if no victor for the match is determined.  If, to the contrary, one is meant to replay the thing until a victor is set, then just use the infinite case below, setting $P_i=P_j$ if $i\equiv j\pmod m$. If you want the probability that a specified player wins, just sum the relevant probabilities.  If all you want is the probability that the match ends in a particular game, then the stated formula holds.

For simplicity, let us first consider the case where $m=\infty$ (Note that $P_1$ can never win this match).

A match can be considered a binary string, where a $0$ means that the player with the lower index won, and a $1$ means that the player with the higher index won.  Thus the match $0010$  means that the winners were $P_1,P_2,P_4, P_4$ so $P_4$ is the winner of the match.   in this way, we see that the match typically ends with the appearance of the first $10$ (player $P_1$ messes this up a little).

It is easy to count the strings with no $10$.  They must all be of the form $0^a1^b$  So there are $n+1$ such of length $n$.

The probability that a string of length $r$ has it's first $10$ ending in slot $r$ is then $$\frac {r-2+1}{2^{r-2}}\times \frac 14=\boxed {\frac {r-1}{2^r}}$$

Sanity Check:  Note that $$\sum_{r=2}^{\infty}\frac {r-1}{2^r}=1$$ so, in the case of infinite $m$, the match is eventually resolved with probability $1$.

The one exception to this analysis arises if $m$ is finite so $P_1$ might win.  In that case we need that there be no winner in the first $m-1$ games, but that $P_1$ must win the first and last.  Thus we have a word of the form $0W0$ where $W$ is a word of length $m-2$ with no $10$.  The probability of that is $$\frac 12\times \frac {m-2+1}{2^{m-2}}\times \frac 12=\frac {m-1}{2^m}$$

so the same formula holds in this case.

Sanity Check (for the exceptional case):  suppose $m=4$.  Then the paths in which $P_1$ wins are $$(P_1, P_2,P_3,P_1)\quad(P_1,P_2,P_4, P_1)\quad(P_1,P_3,P_4,P_1)$$ so the answer would be $\frac 3{16}$ in this case, matching the formula.

Remark:  there is no need for the sum of these probabilities to be $1$ since, in the finite case, the match could certainly be decided without a victor.


问:
Showing that a divergence free vector field is curl of a vector potential
(This must have been asked somewhere before but for some reason I could not find an answer)

In physics one often meets the criterion that a vector field has zero divergence, e.g.

$$\nabla\cdot\mathbf{v}= 0 \quad (1)$$

Based on this one often claim that the vector field can be written as the curl of another field,

$$\mathbf{v}=\nabla\times\mathbf{A} \quad (2)$$

since $\nabla\cdot (\nabla \times \mathbf{f})= 0$ for any (smooth) $\mathbf{f}$. But the implication here is only obvious from (2) -> (1) and not the other way around.

I tried the Helmholtz decomposition theorem where the vector field can be decomposed into curl-free and divergence-free components

$$\mathbf{v} = \nabla \phi + \nabla\times \mathbf{A}$$

Taking the divergence of this and combining with (1) gives

$$\nabla\cdot\mathbf{v} = \nabla \cdot\nabla \phi = 0$$

But I can't see how this would imply $\nabla \phi = 0$ which I guess would thus imply (2).

Am I missing something or completely on the wrong track?

答:
I am giving an answer which is differential form notation and Poincaré Lemma, so perhaps not very useful for OP. It is definitely possible to do this without using differential forms language so hopefully someone will provide another answer doing that.

It is a basic result in de Rham cohomology that closed differential forms, that is satisfying $d \alpha=0$ are locally exact, that is there is if $d \alpha = 0 $ at some point $p$ then there is a neighbourhood of $p$ where $\alpha = d \beta$. If the topology of the space is trivial (let us say contractible, $R^n$ is contractible) then $\alpha = d \beta $ is true globally.

An inner product gives a canonical isomorphism between a vector space $V$ and its dual, I will denote the map $V \to V^*$ by $\flat$ and its inverse by $\sharp$. In this language a vector being divergence-free translates to
\begin{equation}
d * v^\flat = 0
\end{equation}
hence we can, globally on $R^3$, write
\begin{equation}
* v^\flat = d A^\flat
\end{equation}
for some vector $A$. The map $*$ is the Hodge star, an isomorphism between $k$ forms and $(n-k)$-forms. In this setting it squares to the identity, so, also applying $^\sharp$, we get
\begin{equation}
v = (* d A^\flat)^\sharp
\end{equation}
and $(* d A^\flat)^\sharp = \mathrm{curl} A$ so we get the result we wanted.


问:
Posson and Binomic distribution
Exercise:
An ecologist wants to examine the deer population in two different areas, area 1 and area 2. He assumes the number of deer, X and Y, in areas 1 and 2, has Poisson distribution.

a.

The ecologist calculates that the expected number of deer is $2$ in area $1$ and $4$ in area $2$. Find $P(X=2)$ and $P(X>=3)$, and find an approximate expression for $P(X = Y)$. Specify the expectation you must use to calculate $P(x = y)$.

I have solved this exercise as written below:

Definition:

A random variable $X$ is said to have Poisson distribution with parameter $\lambda (>0)$ if $p(x)$ of $X$ is given by

\begin{align}\\
& p(x; \lambda) = \frac{e^{-\lambda}{\lambda ^x}}{x!}
\end{align}

For $\lambda_1 = 2$:

\begin{align}\\
& P(X=2)=\frac{e^{-2}{2^2}}{2!}=\frac{2}{e^2}=0,27067\\
& P(X=1)=\frac{e^{-2}{2^1}}{1!}=\frac{2}{e^2}=0,27067\\
& P(X=0)=\frac{e^{-2}{2^0}}{0!}=\frac{1}{e^2}=0,13533\\
\end{align}

For $P(X>=3)$:
\begin{align}\\
& P(X>=3) = 1-P(X<=2) \\
& P(X>=3) = 1-P(X=0)-P(X=1)-P(X=2) \\
& P(X>=3) = 1-0,27067-0,27067-0,13533 \\
& P(X>=3) = 0,32333
\end{align}

I assume this is correct, but please correct me if I got anything wrong.

b.

If a deer is in such an area, the ecologist assumes a probability of 0.4, that it will be observed. If there are actually 5 deer in the area, what is the probability that three of them will be observed? What is the expected number of deer observed?

I assume this is a Binominal distibution, because it satisfies the four axioms. There are two possible outcomes; observed or not observed. And $P(1)=0,4$. But I don't know how to continue. What am I doing wrong?

答: There are $\binom 53=10$ ways to choose the $3$ deer that will be observed.  The probability that they will be observed is $.4^3$ and the probability that the other $2$ will not be seen is $.6^2$  Therefore, the probability is $$10\left(\frac25\right)^3\left(\frac35\right)^2$$

Title: Posson and Binomic distribution


问:
can you solve this ? $ y'= \frac{x\sqrt{1+y^2}} { y\sqrt{1+x^2}} $
thanks, hard to write with keyboard
$$
y'= \frac{x\sqrt{1+y^2}} { y\sqrt{1+x^2}}
$$
i dont know what i start with

Body:
$$
\frac{ydy}{xdx}= \frac{\sqrt{1+y^2}}{\sqrt{1+x^2}}\leadsto \\
\frac{d(1+y^2)}{\sqrt{1+y^2}}= \frac{d(1+x^2)}{\sqrt{1+x^2}} \leadsto \\   \sqrt{1+y^2}=c+ \sqrt{1+x^2}.
$$


问:
Calculate $\int_0^{\frac{\pi}{2}} \frac{x \sin(x)}{\sqrt{\sin(x)+\sqrt{\cos (x)}}} dx $
I tried to calculate the following integral:
$$\int_0^{\frac{\pi}{2}} \frac{x \sin(x)}{\sqrt{\sin(x)+\sqrt{\cos(x)}}} dx $$

I used WolframAlpha and got $0.8$ but I don't know how

答:
I prefer to think that there is no antiderivative for such a monster.

We can have a decent approximation of the integral building the $[3,2]$ Padé approximant of the integrand. This would give
$$\frac{x \sin(x)}{\sqrt{\sin(x)+\sqrt{\cos (x)}}}\sim \frac {3 x^2 (3 x+2) } {2(2 x^2+6 x+3 ) }$$
Now, rewrite
$$\frac {3 x^2 (3 x+2) } {2(2 x^2+6 x+3 ) }=\frac {3 x^2 (3 x+2) } {4(x-a)(x-b) }$$ where
$$a=-\frac{3+\sqrt{3}}{2} \qquad \text{and} \qquad b=-\frac{3-\sqrt{3}}{2} $$ are the roots of the quadratic $2 x^2+6 x+3=0$.

Using partial fraction decomposition, we then have
$$\frac {3 x^2 (3 x+2) } {4(x-a)(x-b) }=\frac{3}{4} (3 a+3 b+2)+\frac{9}{4}x+\frac{9a^2(a+2)}{4 (a-b) (x-a)}-\frac{ 9 b^2( b+2)}{4 (a-b) (x-b)}$$ and we have very simple integrands.

Integrating, using the bounds and replacing $a$ and $b$ by their values, for the definite integral, we end with an explicit approximation which, numerically, is
$$\int_0^{\frac \pi 2}\frac {3 x^2 (3 x+2) } {2(2 x^2+6 x+3 ) }\,dx=0.842611\cdots$$ while the result of the numerical integration given by Wolfram Alpha is $0.833385$. So this simple work leads to an error of $1.1$%.


问:
$\left( {\begin{array}{ccc|c} 3 & 1 & 3 & 3 \\ 4 & 4a & 2a & 3 \\ 2 & 2 & 2a & 1 \end{array} } \right) a \in \mathbb Z_5$
$\left( {\begin{array}{ccc|c}
3 & 1 & 3  & 3 \\
4 & 4a & 2a & 3 \\
2 & 2 & 2a & 1
\end{array} } \right) a \in \mathbb Z_5$
I would need advice on how to proceed with this example. I need to find and those for which the equation has no solution. I know that in z5 I am limited to the set {0-4} and I try to perform Gaussian elimination
operations.

答: Compute the Gaussian elimination. Just take $a$ as it is any number. If this is to difficult, you can use 4 equality systems and check if there is a solution or not (set a=0 and solve then a=1 and solve, ...)


答: Hint: The determinant of the matrix is $8 (3 a^2 - 5 a + 3)$, which is zero iff $a \equiv \pm 2 \bmod 5$. So, for $a \equiv 0,1,4 \bmod 5$, the system always has a solution.

Body:
$\left( {\begin{array}{ccc|c}
3 & 1 & 3  & 3 \\
4 & 4a & 2a & 3 \\
2 & 2 & 2a & 1
\end{array} } \right) a \in \mathbb Z_5$
I would need advice on how to proceed with this example. I need to find and those for which the equation has no solution. I know that in z5 I am limited to the set {0-4} and I try to perform Gaussian elimination
operations.


问:
Borel set in an uncountably product space with product topology
Let $I=[0,1]$ be with its usual topology and $I^I$ be with product topology.

Let $1_A(x)$ be the indicator function of set $A$, i.e., $1_A(x)=1$ for $x\in A$ and $1_A(x)=0$ for $x\notin A$.

Let $f(x):=1_{\{x\}}$, so that $f$ defines a function from $I$ into $I^I$. Show that the range of $f$ is a Borel set in $I^I$.

My efforts:

For any $x\in I$, $U_x:=\{g\in I^I:g(x)\neq0\}$ is open. Thus $U_x^C:=\{g\in I^I:g(x)=0\}$ is closed and Borel. Similarly, $V_x^C:=\{g\in I^I:g(x)=1\}$ is closed and Borel.

Since arbitrary intersection of closed sets is closed, $V_y^C\cap\bigcap_{x\in I,x\neq y}U_x^C=\{1_{\{y\}}\}$ is closed and Borel for any $y\in I$.

Then I don't know how to continue.

答:
Let $X=I^I$. For $F\subseteq I$ finite and $h\in X$ let $B(F,h)=\{g\in X\mid g=h\text{ on }F\}$, which by definition of the product topology is a basic open set. Now note that the set of functions in $I^I$ that take the value $1$ at exactly one point is closed in $I^I$, being the complement of $$\bigcup_{\stackrel{\{a,b\}\in X^2}{a\neq b}}B(\{a,b\},1_{\{a,b\}}).$$

A similar argument establishes that the set of functions in $X$ that take values in $\{0,1\}$ is closed in $X$, hence the image of $f$, being the intersection of the two is also closed (and in particular Borel).

Note also that since convergence in $X$ is pointwise convergence of functions it is very easy to convince oneself that the image of $f$ is sequentially closed, and if you have familiarity with nets you can also show that it is closed in a more straightforward manner


问:
I need some help with this derivative problem
Suppose $f$ is  continuous at $x_0=2$ and
$\lim \frac{f(x)+x}{(x-2)}=3$ as $x\to2$.
Show that $f^{\prime}(2)=-1$.

答:
Multiply by $\displaystyle\lim_{x\to2}(x-2)$ to get
$$\lim_{x\to2}\Big(f(x)+x\Big)=0$$
$$\implies f(2)=-2$$
Can you use this in your original equation, and get your answer now?

NOTE: If you edit and provide context and/or your work, I will provide you the whole solution. This is the policy of MathSE.


问:
Determinant of $4\times4$ Matrix
I tried to solve for a $4 \times 4$ matrix, but I'm unsure if I did this properly, can anyone tell me if I did this correct? Or if there were any mistakes where at? Also, I know this is an inefficient method for finding the determinant, however I want to get practice with solving like so:

$$A=
\begin{bmatrix}
2 & 4 & 0 & 1 \\
0 & 8 & 0 & 2 \\
0 & 3 & 0 & 5 \\
1 & 2 & 1 & 1
\end{bmatrix}
$$

$$\begin{align*}
\det(A)&=2
\begin{vmatrix}
8 & 0 & 2 \\
3 & 0 & 5 \\
2 & 1 & 1
\end{vmatrix}
-4
\begin{vmatrix}
0 & 0 & 2 \\
0 & 0 & 5 \\
1 & 1 & 1
\end{vmatrix}
+0
-1
\begin{vmatrix}
0 & 8 & 0 \\
0 & 3 & 0 \\
1 & 2 & 1
\end{vmatrix}\\[0.3in]
&=2\left(8
\begin{vmatrix}
0 & 5 \\
1 & 1
\end{vmatrix}-0+2
\begin{vmatrix}
3 & 0\\
2 & 1
\end{vmatrix}\right)\\[0.1in]
&\quad{}-4\left(0-0+2
\begin{vmatrix}
0 & 0\\
1 & 1
\end{vmatrix}\right)\\[0.1in]
&\quad{}+0\\[0.1in]
&\quad{}-1\left(0-8
\begin{vmatrix}
0 & 0\\
1 & 1
\end{vmatrix}+0\right)\\[0.3in]
&=2(8(0-5)-0+2(3-0))\\[0.1in]
&\quad{}-4(0-0+2(0))\\[0.1in]
&\quad{}+0\\[0.1in]
&\quad{}-1(0-8(0)+0)\\[0.3in]
&= 2(8(-5)-0+2(3))\\[0.3in]
&=2(-45+6)\\[0.3in]
&=2(-39)\\[0.3in]
&=-78
\end{align*}$$

Sorry for the long post, I tried to make the readability easy for everyone.

答:
As pointed out by Amzoti in the comment: your calculations are all correct, save for the glitch at the end, when you erroneously computed $8(-5) = -45$, at the very end, which results in your final calculation being off by $2(-5) = -10$, giving you $-78$ instead of $-68$.

Correct that, and you're good to go!

But the "hard part" was all perfectly correct. (I double checked the rest of your work, too, and you did just fine.)  so if your point was to "practice" finding the determinant of a matrix by expanding along the first row, to get the process right, you did the key parts well.

ADDED: If you've learned how elementary row operations alter the determinant of the matrix on which you're operating, doing so can greatly simplify the computation of the determinant of a matrix! See, e.g. Java Man's comment: if you had performed the following elementary row operations $-2R_4 + R_1 \to R_1$, you could have expanded along the first column, greatly simplifying the process.

If you haven't yet learned how ERO's change the determinant, once you learn that, your work will be greatly reduced in the future! (For example, if you are able to use ERO's on some matrix to reduce it to an equivalent system/matrix containing a row, or a column, of all zeros, the determinant of the entire (original and reduced) matrix will be $0$.

The more practice you get, and the more "shortcuts" you learn, the less tedious computing the determinant of a matrix will become.

Title: Determinant of $4\times4$ Matrix


问:
Calculating the number of possible password combinations given a set of constraints
There are several password combination problems on this site but none is exactly like mine: An $8$-character password contains 3 numbers, at least one uppercase letter, at least one lowercase letter, and at least one special character. How many such password combinations are possible? (Note that, currently, there are $33$ publicly-known special characters in this world). Here's what I've been trying to do:

The three numbers will give us $10^3 = 1,000$ possible combinations. This now leaves us with $5$ characters that we have to account for. The sum of the pool of uppercase, lowercase and special characters is $26+26+33 = 85$. Thus, $85$ characters can fill $5$ slots in $85^5 = 4,437,053,125$ ways. However, this will not limit each uppercase, lowercase and special character to at least one, right? That is, my calculation also includes passwords like, say, 123abcde, right? I then tried $88^8 = 3,596,345,248,055,296$. The $88$ comes from the sum of $3$ numbers and the pool of $85$ characters we had earlier encountered. This also doesn't seem to work because, at the very least, the $3$ numbers are not specific.

How do I solve this?

Update: Here's what I tried next: In a table from one of the answers below, there are six major possibilities, so let's just work with one of them - a password containing $3$ numbers, $1$ uppercase, $1$ lowercase and $3$ special characters. Let's assume that the ordering is strictly that. That is, the first $3$ characters are numbers, the $4$th character is an uppercase, the $5$th character is lowercase, and the $6$th, $7$th and $8$th characters are special. Now, with this, how many such password combinations are possible? It's $220*26*26*33^3 = 5,344,550,640$. Yes, yes, yes, I know you're wondering where the $220$ comes from. These are all the triples where the first digit is less than or equal to the second digit AND the second digit is less than or equal to the third digit (I wrote a computer program to count them). We do this so that we can avoid duplicates. So far, this approach accounts for a password like, say, 123Aa+-* but ignores a password like 3Aa2+-*1. In order to account for it (and many others), we just permute. As such, our initial value of $5,344,550,640$ now gives us $5,344,550,640*8! = 215,492,281,804,800$ passwords. We do the same for the other $5$ major possibilities:

$3-1-2-2: 220*26*26^2*33^2*8!=169,781,797,785,600$
$3-1-3-1: 220*26*26^3*33*8!=133,767,477,043,200$
$3-2-1-2: 220*26^2*26*33^2*8!=169,781,797,785,600$
$3-2-2-1: 220*26^2*26^2*33*8!=133,767,477,043,200$
$3-3-1-1: 220*26^3*26*33*8!=133,767,477,043,200$

Adding it all up gives us: $956,358,308,505,600$

答:
So, first you noticed correctly that $10$ digits and $10^3$ ways to pick three digits in sequence (repetition allowed) and that there are $85$ non-digit characters.

I assume that your passwords must contain exactly three digits.

Let us begin by answering the question of how many passwords there are with exactly $3$ digits with the available characterset where we ignore the requirement of having at least one upper, lower, and special characters.  We will take care of those additional requirements later.

To do this, choose where in the password the digits go in $\binom{8}{3}$ ways and then what they were in $10^3$ ways.  For the remaining positions, choose what characters they were in $85^5$ ways for a total of

$$\binom{8}{3}10^385^5$$

Now... among these some of them were "invalid passwords" for one or more reasons because of not having any lower case characters or not having any upper case characters, etc...  Let us count how many of the passwords were bad.  Once having done so we will subtract it from our total to have the count of how many of them were in fact good.

Counting those who didn't have any lower-case characters, we again pick where the digits went, what digits they were, and then for the $85-26=59$ non-lowercase characters remaining choose which character each remaining position is.  There are $\binom{8}{3}10^359^5$ such passwords that were bad because they had no lowercase characters.

Similarly, we can count how many were bad because they had no upper-case characters as well as how many were bad because they had no special characters.

Now... you might think that we just subtract those three numbers and then we're done... BUT... this doesn't take into account that some passwords were bad for multiple reasons.  We must correctly apply inclusion-exclusion and after having subtracted these three numbers we must then add back what we subtracted too much of, namely the number of passwords that were bad for two reasons.  Note that no passwords exist that were bad for all three reasons simultaneously since otherwise we would have no available characters remaining to choose from, but in larger problems we might have needed to further correct our count and continued on to three, four, or more... alternating between adding and subtracting the counts.

Finish the problem by putting all of this together for a final answer.


问:
Why do the mapping from $A^m$ to $M_n$ is surjective?
I was unable to solve the problem 1.3.11(b) from Qing Liu's Algebraic Geometry and Arithmetic Curves.

Let $A$ be a commutative complete ring for the $I$-adic topology with unit, where $I$ is an ideal of $A$. Let $(M_n)_{n\geq 0}$ be $A$-modules such that $I^{n+1}M_n=0$
and that there exist a surjective homomorphisms $\pi_n:M_{n+1}\to M_n$ with
$\operatorname{Ker} (\pi_n)=I^{n+1}M_{n+1}$. Let $M=\varprojlim_n M_n$ and denote the (surjective) canonical homomorphism by $u_n:M\to M_n$.

Suppose that $M_0$ is generated over $A$ by a finitely number of elements $e_{0,1},\ldots, e_{0,m}$. Let $e_1,\ldots,e_m\in M$ be such that $u_0(e_i)=e_{0,i}$, and define $\phi_n:A^m\to M_n$ by $(a_1,\ldots,a_m)\mapsto \sum_ia_iu_n(e_i)$. How can I show that $\phi_n$ is surjective and $M$ is generated by the $e_i$? Do I need some kind of universal property?

答:
$M_n\require{AMScd}$ is finitely generated

We have an exact sequence
\begin{CD}
  0 @>>> IM_n @>{\hookrightarrow}>> M_n @>{\pi_{0n}}>> M_0 @>>> 0
\end{CD}
where $\pi_{0n} = \pi_0 \circ \cdots \circ \pi_{n-1}$ (see (a) of this exercise).
Let $m_1, m_2, \ldots \in M_n$, and $M_n$ is generated by $u_n(e_1), \ldots, u_n(e_m), m_1, m_2, \ldots$ over $A$:
\[ M_n = u_n(e_1)A + \cdots + u_n(e_m)A + \sum_{k\geq1} m_k A . \]

Let us consider $\pi_{0n}(m_1) \in M_0$. Since $M_0$ is generated by $u_0(e_1), \ldots, u_0(e_m)$, there exist $a_1, \ldots, a_m \in A$ such that
\begin{align}
  \pi_{0n}(m_1) &= a_1 u_0(e_1) + \cdots + a_m u_0(e_m) \\
  &= \pi_0( a_1 u_1(e_1) + \cdots + a_m u_1(e_m) ) \\
  &= \pi_0 \circ \cdots \circ \pi_{n-1} ( a_1 u_n(e_1) + \cdots + a_m u_n(e_m) ) \\
  &= \pi_{0n}(a_1 u_n(e_1) + \cdots + a_m u_n(e_m)).
\end{align}
Since $\ker \pi_{0n} = IM_n$, and $M_n$ is generated by $u_1, \ldots, u_m$ and $m_k~(k\geq1)$, there exits $i_1, \ldots, i_m \in I$ and $\tilde{i}_k~(k\geq1) \in I$ such that
\[ m_1 - (a_1 u_n(e_1) + \cdots + a_m u_n(e_m)) = i_1 u_n(e_1) + \cdots i_m u_n(e_m) + \sum_{k\geq1} \tilde{i}_k m_k. \]
Hence
\[  m_1 = (a_1 + i_1) u_n(e_1) + \cdots + (a_m + i_m) u_n(e_m) + \sum_{k\geq2} \tilde{i}_k m_k. \]
Since $A$ is complete respect to $I$-adic topology, $1 - \tilde{i}_1$ is invertible in $A$ (in fact, $1 - \tilde{i}_1 + \tilde{i}_1{}^2 - \cdots$ is its inverse). Hence we have
\[ m_1 = (1 - \tilde{i}_1)^{-1}\sum_{l=1}^m(a_l + i_l) u_n(e_l)+ \sum_{k\geq2} (1 - \tilde{i}_1)^{-1}\tilde{i}_k m_k. \]
Consequently, we can express $M_n$ as
\[ M_n = u_n(e_1)A + \cdots + u_n(e_m)A + \sum_{k\geq2} m_k A. \]

In the same manner, we can show that $M_n$ is generated by $u_n(e_1), \ldots, u_n(e_m)$:
\[ M_n = u_n(e_1)A + \cdots + u_n(e_m)A . \]
In other words,
\[ \phi_n \colon A^{\oplus m} \ni (a_1, \ldots, a_m) \mapsto \sum_{k=1}^m a_k u_n(e_k) \in M_n \]
is surjective.

$M$ is finitely generated

I just give a brief summary of proof. There would be much better ways.

Since $I^{n+1}M_n = 0$, we have a surjection
\[ \bar{\phi}_n \colon (A/I^{n+1})^{\oplus m} \ni ([a_1], \ldots ,[a_m]) \mapsto \sum_{k=1}^m a_k u_n(e_k) \in M_n . \]
Consider the commutative diagram
\begin{CD}
  0 @>>> \ker \bar{\phi}_{n+1} @>{\hookrightarrow}>> (A/I^{n+2})^{\oplus m} @>{\bar{\phi}_{n+1}}>> M_{n+1} @>>> 0 \\
  @. @VV{p_n'}V @VV{p_n}V @VV{\pi_n}V @. \\
  0 @>>> \ker \bar{\phi}_{n} @>{\hookrightarrow}>> (A/I^{n+1})^{\oplus m} @>{\bar{\phi}_{n}}>> M_{n} @>>> 0
\end{CD}
where $p_n \colon (A/I^{n+2})^{\oplus m} \to (A/I^{n+1})^{\oplus m}$ is given by
\[ p_n \colon (A/I^{n+2})^{\oplus m} \ni ([a_1], \ldots , [a_m]) \mapsto ([a_1], \ldots , [a_m]) \in (A/I^{n+1})^{\oplus m} . \]
Since $p_n'$ is surjective, by Lemma in the book, we have an exact sequence
\begin{CD}
  0 @>>> \varprojlim (\ker \bar{\phi}_{n}) @>>> \bigoplus_{k=1}^m \varprojlim (A/I^{n+1}) @>{\phi}>> \varprojlim M_n @>>> 0
\end{CD}
, which immediately follows ($A \simeq \varprojlim A/I^{n+1}$)
\begin{CD}
  A^{\oplus m} @>{\phi}>> M @>>> 0
\end{CD}
$\phi$ is given by
\[ \phi \colon A^{\oplus m} \ni (a_1, \ldots, a_m) \mapsto \sum_{k=1}^m a_k e_k \in M. \]


问:
Approximating $\frac{x-y}{w-z}$ in terms of $\frac{x}{w}$ and $\frac{y}{z}$
I have the expression $\frac{x-y}{w-z}$ and I would like to approximate it in two fractions, $x/w$ and $y/z$. Ideally a linear combination of $x/w$ and $y/z$ and their powers. If that is not possible, a non-linear expression will also do.

答: This is not possible.  We can multiply numerator and denominator of $\frac xw$ by $k$ without changing the value of the fraction.  As $k\to\infty$, $\frac{kx-y}{kw-z}\to\frac xw$, but as $k\to0$, $\frac{kx-y}{kw-z}\to\frac yz$.  That is, $\frac{kx-y}{kw-z}\to\frac xw$ can't be approximated as a fuction of $\frac wx$ and $\frac yz$.


答:
I doubt that you can. Suppose $y = z = 1$. Then you want to approximate
$$
A = \frac{x-1}{w-1}
$$
using $B = x/w$ and the number $1$. But $A$ can be any value between $1$ and $B$.

Title: Approximating $\frac{x-y}{w-z}$ in terms of $\frac{x}{w}$ and $\frac{y}{z}$

Body: I have the expression $\frac{x-y}{w-z}$ and I would like to approximate it in two fractions, $x/w$ and $y/z$. Ideally a linear combination of $x/w$ and $y/z$ and their powers. If that is not possible, a non-linear expression will also do.


问:
value of largest probability base on normal approximation
Let $X$ be the number of heads and $X \sim Binomial(87,0.5)$. Also, let $Y$ be the number of heads. $Y \sim Normal(np,npq)$ with $np =43.5, npq=21.75$, but I don't really know the way to find the number of heads which have the largest probability based on the Normal approximation in part d?

Suppose we flip a coin $87$ times. Assume that the coin is fair.

a) Compute the probability of obtaining $53$ heads.

b) Verify if the Rule of Five is satisfied? Compute the probability of obtaining $53$ heads based on Normal approximation. Is the answer close to part a)?

c) Compute the probability of obtaining at least $53$ heads based on the Normal approximation.

d) Find the number of heads which have the largest probability based on the Normal approximation

答: Hint : a normal distribution attains its maximum at the mean $\mu$ and the normal distribution is simmetric to the mean, increasing when $x < \mu$  and decreasing when $x > \mu$. What can you do, then, to find the largest probability?


问:
Confusion about definition of faithful group action.
Here are some excerpts from my lecture note:

An action from a group $G$ to a set $X$ is a homomorphism $\alpha: G \to \text{Sym}(X)$, where $\text{Sym}(X)$ is the group of bijections $X \to X$.

Equivalently, an action $\alpha$ of $G$ on $X$ can be viewed as a map $G \times X \to X$, which we usually write as $(g,x) \mapsto g \cdot x := \alpha_g(x) \in X$.

An action $\alpha$ is called faithful if it is injective.

The action of left translation of a group $G$ on itself is defined as $L_g(h) = gh$, for some fixed $g \in G$. This action is faithful.

Am I correct to assume that the statement that $L_g$ is faithful is vacuously true, since we are restricting the domain of the action to just one element $g$? Or am I missing something?

答:
It's $L$ that is faithful, not $L_g$.

That each $L_g$ is a bijection is indeed obvious. To show $L$ is faithful you need to show that $L_g \ne L_h$ when $g \ne h$.

That is not hard.


问:
Math Olympiad 2020 (BdMO-Secondary-National)
Let $x_1, x_2,\ldots$ be real numbers so that for all $n > 0,$
$$x_{n+3} = x_{n+2} - 2x_{n+1} + x_n. $$
Suppose $x_1 = x_3 = 1$ and you're given that $x_{98} = x_{99}$.

Find the sum $x_1 + x_2 + \ldots + x_{100}$.

答:
See this image for solution

Could not write this in MathJax, sorry for that


Title: Math Olympiad 2020 (BdMO-Secondary-National)

Body:
Let $x_1, x_2,\ldots$ be real numbers so that for all $n > 0,$
$$x_{n+3} = x_{n+2} - 2x_{n+1} + x_n. $$
Suppose $x_1 = x_3 = 1$ and you're given that $x_{98} = x_{99}$.

Find the sum $x_1 + x_2 + \ldots + x_{100}$.

Body:
Given that

$$
x_{n+3} = x_{n+2} - 2x_{n+1} + x_n,
$$

one can rewrite this as

$$
x_{n+3} - x_{n+2} =  - 2x_{n+1} + x_n \Leftrightarrow
$$

$$
\begin{aligned}
x_{4} - x_{3} &=  - 2x_{2} + x_1 \\
x_{5} - x_{4} &=  - 2x_{3} + x_2 \\
x_{6} - x_{5} &=  - 2x_{4} + x_3 \\
\ldots \ldots \ldots &\ldots \ldots \ldots \ldots \ldots \\
x_{98} - x_{97}  &=  - 2x_{96} + x_{95} \\
x_{99} - x_{98}  &=  - 2x_{97} + x_{96} \\
x_{100} - x_{99} &=  - 2x_{98} + x_{97}
\end{aligned}
$$

Now, let us add all these equalities:

$$
x_4-x_3 + x_5-x_4 + x_6-x_5 + \ldots x_{98}-x_{97} + x_{99}-x_{98} + x_{100}-x_{99} = -2\sum_{n = 2}^{98}x_n + \sum_{n = 1}^{97}x_n.
$$

In the left-hand side, many terms are canceled out:

$$
\begin{aligned}
&-x_3 + x_{100} = -2\sum_{n = 2}^{97}x_n - 2x_{98} + x_1 + \sum_{n = 2}^{97}x_n \Leftrightarrow \\
&-x_3 + x_{100} = -\sum_{n = 2}^{97}x_n - 2x_{98} + x_1 \Leftrightarrow \\
&-x_3 + \sum_{n = 2}^{97}x_n + 2x_{98} + x_{100} = x_1 \Leftrightarrow \\
&x_1 + \sum_{n = 2}^{97}x_n + 2x_{98} + x_{100} = 2x_1 + x_3 \Leftrightarrow  \\
&\left|
\begin{array}{l}
\text{since }x_{98} = x_{99}, \text{ then }2x_{98} = x_{98} + x_{99} \\
\text{since }x_{1} = x_{3} = 1, \text{ then }2x_{1} + x_{3} = 3
\end{array}
\right| \Leftrightarrow  \\
&\sum_{n = 1}^{100}x_n = 3.
\end{aligned}
$$


问:
which parameters always make this rational equation evenly divisible?
Hi guys I have the following equation:

$$x = \dfrac{a + b \times c - b}{c}$$

This is what I know about each variable:
$$a \ge 64$$
$$b \ge 0$$
$$8 \le c \le a$$

My questions is there a concise way for me to pick a, b, and c so that x will always be a positive integer >= 8.

For example if $a = 64$; $b= 0$; $c = 8$ (the smallest possible values for each)

$x = (64 + 0 \times 8 - 0)/8$

$x = 8$ which is a positive integer $\ge 8$

However, if let's say $a = 200$; $b = 2$; $c = 10$

$x = (200 + 2 \times 10 - 2)/10$

$x = \dfrac{218}{10}$ which is $21.8$ and is not a positive integer greater than or equal to $ 8$

How can I always know that $x$ will be a positive integer?? Thanks!

答:
If
$$x=(a+b*c-b)/c$$
You can split it into
$x=\frac{a-b}{c}+b$

Then, the only divisibility condition is
$$\frac{a-b}{c}=k$$
$$a-b=ck$$
$$a=b+ck$$
So, you can pick any $b$,$c$ and $k$ integer such that x is an integer.

Replacing a you get
$k+b \ge 8$

Finally, you can pick any integers $b$ and $k$ such that $k+b \ge 8$ and any integer $c$ that follow your initial inequalities, then operate for $a$.


问:
Calculate profit or defecit of a hazard game
Hi i have a hazard game that costs $\$5$ for entry. When i pay for the game and spin a wheel i have:

$40\%$ chance to get $\$2$

$30\%$ to get $\$4$

$20\%$ to get $\$15$

$9,9\%$ to get $\$30$

$0,1\%$ to get $\$200$

I am looking for general formula how to calculate if this game be profitable or not? And about how much?
For example final answer could be game will be profitable and after millions of spins I get back $\$5.13$ average per spin.

Its for my computer game, I am a programmer that is dumb in math.

答:
All you need to do is multiply each amount by the probability of getting that amount, and add them up. So you get $0.4\times 2+0.3\times 4+\cdots$.

You should find that the game is very profitable.


问:
$\left[A, e^{\lambda B}\right]=\lambda e^{\lambda B}[A, B]$ proof?
In a derivation, my professor used the following property:
$$\left[A, e^{\lambda B}\right]=\lambda e^{\lambda B}[A, B]$$
Where $A$ and $B$ are operators that commute with their commutator $[B,[A,B]]=[A,[A,B]]=0$. How can this be proven?

答:
It is easy to prove that, in general, the adjoint action is but
$$
e^{-\lambda B} A e^{-\lambda B}= A -\lambda [B,A]+ {\lambda^2\over 2!} [B,[B,A]]+ ...
$$
But, in your case, the series truncates after the leading two terms.

It is then evident that
$$
[A,e^{\lambda B}] =  e^{\lambda B} ( e^{-\lambda B} A e^{\lambda B} -A)=- e^{\lambda B} \lambda [B,A].
$$


问:
Showing Lebesgue integral over set of measure zero is in fact zero zero.

Question: Let $E\subset\mathbb{R}$ be a Lebesgue measurable set for which $\operatorname{m}(E)=0$, where $\operatorname{m}$ denotes the Lebesgue measure. I'm trying to show that this implies
$$\int_{E}e^{-x^2/2}dx=0.\tag{1}$$

(Attempted) Solution: Let $f:=e^{-x^2/2}$. Since $0 < f \leq 1$ over $\mathbb{R}$, and thus over any $E\subset\mathbb{R}$, it follows

$$0\leq\int_{E}e^{-x^2/2}dx\leq\int_{E}1dx = 1\cdot\operatorname{m}(E) =0.$$

That is,

$$0\leq\int_{E}e^{-x^2/2}dx\leq0.$$

Therefore, indeed we have that $(1)$ holds.

My question is, is this solution correct? If not, can it be repaired? And if not, any help would be appreciated!

答: This works. Your argument shows that if $f$ is bounded on a set $E$ of Lebesgue measure zero, then $\int_E f = 0$. In fact the stronger claim that $\int_E f = 0$ without any hypothesis on $f$ holds! Indeed the integral is, by definition, the supremum over $g$ with $g$ simple and $g \leq f$ of $\int_E g$. But any such $g$ has integral zero.


问:
Order of elements in the symmetric group
What are the elements of order $2$ of the symmetric group $S_n$?

I've tried listing some of the sets for small $n$ to get a better understanding but I'm really not sure:

$S_1$ has only one element and that's order $1$.
$S_2$ has elements $\begin{pmatrix}1&2\\1&2\end{pmatrix}$ and $\begin{pmatrix}1&2\\2&1\end{pmatrix}$ ; the first of these is order $1$ and the second is order $2$.

I don't know how to proceed though.

答: Hint: any permutation is a a product of disjoint cycles. If $\sigma_1, \sigma_2, \ldots, \sigma_k$ are disjoint cycles, the order of $\sigma= \sigma_1\sigma_2\cdots\sigma_k$ is the least common multiple of the orders of the $\sigma_i$. So what can you say about the orders of the $\sigma_i$ if the order of $\sigma$ is $2$?


问:
How can I generate a number of matrices with specified attributes?
For my researches, I need to have a list of different matrices (about 400 matrices) with some specified attributes such as determinant, trace, the number of zero eigenvalues, etc. as a dataset.

How can I generated these matrices and save them as an Excel file?

Thank in advance.

答:
Obviously this will depend on exactly what you want the matrices to look like, but you might recall the characteristic polynomial $\chi_A(t)$ has

the eigenvalues as roots

the determinant as a constant term (up to sign)

the trace as the second highest term (up to sign)

Moreover, given a characteristic polynomial of interest, we can build a matrix which has that as its characteristic polynomial by looking at the companion matrix of a polynomial $p$.

So, what does this mean for your project? Well, you know what attributes you want. You should find a way to generate polynomials with these attributes. Then by looking at the companion matrices for the polynomials you get in this way, you can generate matrices with exactly the attributes you want.

You can do much of this efficiently in sage, and then export it as a csv file to open in excel. If you want to do it directly in excel, I'm afraid this isn't the right place to ask.

I hope this helps ^_^


Title: I need some help with this derivative problem

Body:
Suppose $f$ is  continuous at $x_0=2$ and
$\lim \frac{f(x)+x}{(x-2)}=3$ as $x\to2$.
Show that $f^{\prime}(2)=-1$.

Body:
This is my “answer”.
Firstly I found $f(2)=-2$ then I supposed that $k(x)= \frac{f(x)+x}{x-2}$ and then
$$f’(2)= \lim_{x\to2}\frac{f(x)-f(2)}{x-2}$$
$$ = \lim_{x\to2} \frac{(x-2)k(x) -x -f(2)}{x-2}$$
$$=\lim_{x\to2}\frac{(x-2)(k(x)-1)}{x-2} = \lim_{x\to2} (k(x)-1)$$ which equals $k(2)-1=3-1=2$ and not $-1$.


答:
The elements of $S_n$ of order $2$ are those permutations $\varphi$ such that $\varphi(\varphi(x)) = x$ for all $x$.

One way for this to happen is to have $\varphi(x) = x$, however if we have $\varphi(x) = y$ with $y\neq x$ then we must have $\varphi(y)=x$.

What this means is that our function $\varphi$ will split the numbers $\{1,2,\dots,n\}$ into two different numbers. We first have the set $F$ of numbers that are fixed by $\varphi$. We also have the set of numbers $M$ that are moved by $\varphi$, and these numbers get split into pairs of numbers, such that for each pair $\{a,b\}$ we have $\varphi(a)=b$ and $\varphi(b)=a$.

The problem is that this description is a bit confusing haha, and especially when looking at your way of representing permutations it can be a bit confusing.

However if you look at the "disjoint cycle representation" you just need all the cycles to have size $2$.

Also, permutations of order $2$ are called involutions, so perhaps you can find more stuff by searching that.

Title: Order of elements in the symmetric group

Body:
What are the elements of order $2$ of the symmetric group $S_n$?

I've tried listing some of the sets for small $n$ to get a better understanding but I'm really not sure:

$S_1$ has only one element and that's order $1$.
$S_2$ has elements $\begin{pmatrix}1&2\\1&2\end{pmatrix}$ and $\begin{pmatrix}1&2\\2&1\end{pmatrix}$ ; the first of these is order $1$ and the second is order $2$.

I don't know how to proceed though.


问:
Why/How was Hardy impressed with Ramanujan’s work prior to seeing any proofs of their validity?
Hardy’s famous correspondence with Ramanujan highlights an interesting aspect of the mathematics discipline; the appreciation of a mathematical expression prior to knowing its validity.

Why/how was Hardy impressed with Ramanujan’s early work on continued fractions if he didn’t even know they were true?

There is the proclamation by Hardy:

Must be true, because, if they were not true, no one would have the
imagination to invent them

— G. H. Hardy, 1913

...which suggests there is some aesthetic to the expression that is noticed prior to any proof. What is this aesthetic? Is it the apparent symmetry? Some immediate familiarity to other known expressions?

Or is it simply plugging in numbers into the RHS and being able to immediately validate that it “must/should” be true?

答:
See Robert Kanigel's book, The Man Who Knew Infinity, in particular the chapter "I Beg to Introduce Myself...". There is a decent amount of detail there about Hardy and Littlewood's process of investigation. Here are some excerpts:

The first wasn't new to Hardy, who recognized it as going back to a mathematician named Bauer. The second seemed a little different... Hardy and others would show how these series were derived from a class of functions called hypergeometric series...

... they struck [Hardy] as "much more intriguing, and it soon became obivous that Ramanujan must possess much more general theorems and was keeping a great deal up his sleeve." Some theorems in Ramanujan's letter, of course, did look comfortably familiar... Hardy  has proved theorems like it, had even offered a similar one as a mathematical question in the Eudcation Times fourteen years before. Some of Ramanujan's formulas actually went back to the days of Laplace and Jacobi a century before.

[Hardy and Littlewood] began to reach a judgement... "There is always more in one of Ramanujans formulae than meets the eye... In some the interest lies very deep... but there is not one which is nor curious and entertaining."

So we can chalk up the decision to not classify Ramanujan as one of the many cranks that used to send letters as part familiarity with some of the results, part entertainment value of the results, and partially that there seemed to be more general results, leading to curiosity. You can find more details in the book I mentioned.


问:
Related to Problem 2.12-e of Convex Optimization book (by Boyd)
I have a question from Boyd book -

The set of points closer to one set than another, i.e.,
$\{x|\text{dist}(x,S) ≤ \text{dist}(x,T)\}$,
where $S,T⊆R^n$, and $\text{dist(x,S)} = \inf\{|x−z|_2|z\in S\}$.

In the solution manual of the convex optimization book, we have the solution for problem 2.12 -e.

The solution is:
In general, this set is not convex, as the following example in R shows. With $S =\{-1,1\}, T=\{0\}$.

$\{x|\text{dist}(x,S) ≤ \text{dist}(x,T)\} = \{x|x \le -0.5,x\ge 0.5\}$
which clearly is not convex.

Can someone explain the solution?

答:
$S$ consists two points on the real line, $-1$ and $1$. $T$ is formed by just one point, $0$. If you consider point $x=3$, what is its distance to $S$? The distance between $x=3$ and $1$ is $2=|x-1|$ and the distance between $3$ and $-1$ is $4=|3-(-1)|$. You are suposed to take the smallest, so $dist(3,S)=2=\inf\{|x-1|,|x-(-1)|\}$.

If you think of $x=0.5$, you see that it is halfway between $0$ and $1$. Any $x>0.5$ will, therefore, be closer to $1$ than to $0$ and, hence, closer to $S$ than to $T$. Likewise for $x<-0.5$. The limiting cases $x=\pm 0.5$ need to be included as well, because the set must contain all points such that the distance to $T$ is at least as large as the distance to $S$.

The resulting set is not convex, because it has a gap. In order for a set to be convex you need to be able to choose any two points of it, say $x_1$ and $x_2$, and all points that lie on the straight line segment that connect them need to be elements of the set as well. Let $A=\{x|x \le -0.5,x\ge 0.5\}$. If we choose $x_1=-1$ and $x_2=1$ we see that $y=\alpha x_1+(1-\alpha)x_2$ only is an element of $A$ for some $\alpha\in\left[0,1\right]$. For instance, $y=\frac{1}{2} x_1+\left(1-\frac{1}{2}\right)x_2=0\not\in A$.


问:
Nonzero nilpotents of $\mathbb{Z}_{210}$?
An exercise asks me to write two nilpotents in $\mathbb{Z}_{210}$. I honestly doubt that there is any nonzero nilpotent in $\mathbb{Z}_{210}$, since $210=2\cdot 3 \cdot 5\cdot 7$ and by CRT there should be no nonzero nilpotent element. Am I missing anything?

答:
You're quite right.

In general, the nilpotent elements of $\mathbb{Z}_{n}$ correspond to the multiples of $rad(n)$. In particular, the only nilpotent element is $0$ when $n$ is square- free.


答: If $a$ is nilpotent that means $a^k$ is a multiple of $3\cdot2\cdot5\cdot7$ for some $k$ which means each of those primes must divide $a$ which means $a$ is zero.

Title: Nonzero nilpotents of $\mathbb{Z}_{210}$?

Body: An exercise asks me to write two nilpotents in $\mathbb{Z}_{210}$. I honestly doubt that there is any nonzero nilpotent in $\mathbb{Z}_{210}$, since $210=2\cdot 3 \cdot 5\cdot 7$ and by CRT there should be no nonzero nilpotent element. Am I missing anything?


问:
Atomic measure on generators
In measure theory, I've sometimes seen arguments of the sort of the following, if I'm interpreting it correctly.

Let $(X,\Sigma)$ be a measurable space where the $\sigma$-algebra $\Sigma$ is countably generated. Let $\mu$ be a probability measure on $\Sigma$, and suppose that for a set of generators $\{S_i\}_{i\in\mathbb{N}}$ of $\Sigma$ we have that for all $i\in\mathbb{N}$, $\mu(S_i)\in\{0,1\}$.
Then for all measurable sets $A\in\Sigma$, $\mu(A)\in\{0,1\}$ as well.

First of all, is this correct? If so:

How can we prove it?

Is the "countably generated" assumption necessary?

答:
Let $\mathcal{A}$ be the algebra generated by $\{S_i\}$: its easy to see that its sets have measure $\{0,1\}$.
Since $\mu$ is a probability measure, we know that its extension from $\mathcal{A}$ to $\Sigma$ is: $\mu(A)=\inf\{\sum A_i:A\subseteq \cup A_i; A_i\in \mathcal{A}\}$. Being the infimum of integers, it must be an integer ($\le 1$ as it is a probability) and the result follows.

Title: Atomic measure on generators


问:
Determinant of $4\times4$ Matrix
I tried to solve for a $4 \times 4$ matrix, but I'm unsure if I did this properly, can anyone tell me if I did this correct? Or if there were any mistakes where at? Also, I know this is an inefficient method for finding the determinant, however I want to get practice with solving like so:

$$A=
\begin{bmatrix}
2 & 4 & 0 & 1 \\
0 & 8 & 0 & 2 \\
0 & 3 & 0 & 5 \\
1 & 2 & 1 & 1
\end{bmatrix}
$$

$$\begin{align*}
\det(A)&=2
\begin{vmatrix}
8 & 0 & 2 \\
3 & 0 & 5 \\
2 & 1 & 1
\end{vmatrix}
-4
\begin{vmatrix}
0 & 0 & 2 \\
0 & 0 & 5 \\
1 & 1 & 1
\end{vmatrix}
+0
-1
\begin{vmatrix}
0 & 8 & 0 \\
0 & 3 & 0 \\
1 & 2 & 1
\end{vmatrix}\\[0.3in]
&=2\left(8
\begin{vmatrix}
0 & 5 \\
1 & 1
\end{vmatrix}-0+2
\begin{vmatrix}
3 & 0\\
2 & 1
\end{vmatrix}\right)\\[0.1in]
&\quad{}-4\left(0-0+2
\begin{vmatrix}
0 & 0\\
1 & 1
\end{vmatrix}\right)\\[0.1in]
&\quad{}+0\\[0.1in]
&\quad{}-1\left(0-8
\begin{vmatrix}
0 & 0\\
1 & 1
\end{vmatrix}+0\right)\\[0.3in]
&=2(8(0-5)-0+2(3-0))\\[0.1in]
&\quad{}-4(0-0+2(0))\\[0.1in]
&\quad{}+0\\[0.1in]
&\quad{}-1(0-8(0)+0)\\[0.3in]
&= 2(8(-5)-0+2(3))\\[0.3in]
&=2(-45+6)\\[0.3in]
&=2(-39)\\[0.3in]
&=-78
\end{align*}$$

Sorry for the long post, I tried to make the readability easy for everyone.

答:
I know you said that you know that this is not the most efficient way of finding the determinant, but may I point out a quick way? In fact, you use the same technique, but apply it to the third column.

$$\det(A) = (-1)^{3+4}
\left| \begin{array}{lll}
2 & 4 & 1 \\
0 & 8 & 2 \\
0 & 3 & 5
\end{array}\right| = -2 \left| \begin{array}{lll}
8 & 2 \\
3 & 5
\end{array}\right| = -68
 $$

Title: Determinant of $4\times4$ Matrix


答:
a)

$$\mathbb{P}[X=53]=\binom{87}{53}\frac{1}{2^{87}}\approx 1.079\%$$

b)

$$\mathbb{P}[52.5<Y<53.5]=\Phi\left(\frac{53.5-43.5}{\sqrt{21.75}}\right)-\Phi\left(\frac{52.5-43.5}{\sqrt{21.75}}\right)\approx 1.081\%$$

c)

$$\mathbb{P}[Y\geq 53]=1-\Phi\left(\frac{52.5-43.5}{\sqrt{21.75}}\right)\approx 2.682\%$$

d) as already noted, normal distribution is symmetric around its mean thus 43 or 44 have the same largest probability

Title: value of largest probability base on normal approximation

Body:
Let $X$ be the number of heads and $X \sim Binomial(87,0.5)$. Also, let $Y$ be the number of heads. $Y \sim Normal(np,npq)$ with $np =43.5, npq=21.75$, but I don't really know the way to find the number of heads which have the largest probability based on the Normal approximation in part d?

Suppose we flip a coin $87$ times. Assume that the coin is fair.

a) Compute the probability of obtaining $53$ heads.

b) Verify if the Rule of Five is satisfied? Compute the probability of obtaining $53$ heads based on Normal approximation. Is the answer close to part a)?

c) Compute the probability of obtaining at least $53$ heads based on the Normal approximation.

d) Find the number of heads which have the largest probability based on the Normal approximation


问:
Why can't I substitute $u=ie^{-it}$ on integral $\int_{-\frac{\pi}{2}}^\frac{\pi}{2} (e^{-2it}+1)^x\ln(e^{2it}+1)dt$?
I wanted to use substitution $u=ie^{-it}$ to simplify following integral: $$\int_{-\frac{\pi}{2}}^\frac{\pi}{2} (e^{-2it}+1)^x\ln(e^{2it}+1)dt$$
After substituting I got following integral:
$$i\int_{-1}^1 \frac{(1-u^2)^x\ln(1-\frac{1}{u^2})}{u}du$$
This is the integral of the odd function over symmetric interval and therefore it is equal to $0$. However this is not proper result (For example when $x=1$ the integral should be equal to $\pi$).

What did I do wrong?

答: In the first integral, when $t$ is from $-\frac{\pi}{2}$ via $0$ to $\frac{\pi}{2}$, $ie^{-2it}$ is from $-1$ via $i$ to $1$. However $u$ is from $-1$ via $0$ to $1$ in the second integral. They go through different paths. They coincide only under certain conditions.


问:
System of Pythagorean Quadratics
I have a system of quadratics, obtained from three mechanical links, fixed at one end and free at the other. The intersection point of the three free ends is required.

$AC=\sqrt{(A_x-C_x)^2+(A_y-C_y)^2+(A_z-C_z)^2}$

$BC=\sqrt{(B_x-C_x)^2+(B_y-C_y)^2+(B_z-C_z)^2}$

$FC=\sqrt{(F_x-C_x)^2+(F_y-C_y)^2+(F_z-C_z)^2}$

Where C_x,y,z are the unknowns. I am halfway through solving by substitution and it is really messy. Does anyone know of a slightly more elegant way of solving these? Any techniques I should research. I want to avoid doing it numerically if possible.
Many Thanks
James

答:
So basically what you have are three circles and you want to find the point where they intersect. I am going to show you how to use linear algebra to solve your problem with an example.

Suppose that your three circles are: $C_1=(-3, 50), r_1=41$, $C_2=(11,-2), r_2=13$, and $C_3=(13,34), r_3=25$. Let us write the equations of the three circles:

\begin{align*}
x^2 + 6x + 9 + y^2 - 100y + 2500 &= 1681\\
x^2 - 22x + 121 + y^2 + 4y + 4 &= 169\\
x^2 - 26x + 169 + y^2 - 68y + 1156 &= 625
\end{align*}

After re arranging them we have:

\begin{align*}
x^2 + y^2 &= -6x + 100y - 828\\
x^2 + y^2 &= 22x - 4y + 44\\
x^2 + y^2 &= 26x + 68y - 700
\end{align*}

And finally this can be transformed in the following linear system:

\begin{align*}
-6x + 100y - 828 &= 22x - 4y + 44\\
-6x + 100y - 828 &= 26x + 68y - 700
\end{align*}

which reduces to:

\begin{align*}
-28x + 104y &= 872\\
-32x + 32y &= 128
\end{align*}

whose solution is: $(6,10)$. The tale of the story is the following, The system of linear equations is solely formed from the data: centers and radii.

Title: System of Pythagorean Quadratics


问:
Numbering a octahedron

In the following figure, on the left, it is represented in an o.n.
Oxyz, a regular octahedron [ABCDEF], whose vertices belong to the
coordinated axes. Assume that the [ABC] face of the octahedron is
numbered with the number 1, as shown in the figure on the right.
It is intended to number the octahedron's remaining faces with
numbers from 2 to 8 (a different number in each face).

How many different ways
can the remaining seven faces be numbered, so that, after the
octahedron have all the faces numbered, at least three of the
concurrent faces at vertex A are numbered with odd numbers?

I've tried to solve this problem with the following reasoning:

$${}^{3}\textrm{C}_{2}*(^{3}\textrm{P}_{2})*5! = 2160$$

Where:

${}^{3}\textrm{C}_{2}$ is the number of ways we can select the 2 faces that will have odd numbers from the 3 faces that are concurrent at vertex A;

$_{}^{3}\textrm{P}_{2}$ are the possibilities of choosing 2 of the 3 odd numbers available (3,5,7) to number the 2 faces that need to be concurrent at vertex A;

5! that is the numbers of ways we can then number the 5 left faces with the 5 numbers that are remain.

However, I can't understand why my answer is wrong. The correct answer to this problem is 1872, given with the reasoning of:

$$3!*4! + 4*3*(_{}^{3}\textrm{P}_{2})*4!=1872$$ In which:

$3!*4!$ is the numbers of ways in which all of the 4 faces that are concurrent at vertex A have odd numbers;

$4*3*(_{}^{3}\textrm{P}_{2})*4!$ is the numbers of ways in which 3 of the 4 faces that are concurrent at vertex A have odd numbers:

$4$ are the four even numbers available (2,4,6,8)

$3$ are the numbers of faces that can have the even numbers at the top of the octahedron;

$_{}^{3}\textrm{P}_{2}$ are the possibilities of choosing 2 of the 3 odd numbers available (3,5,7) to number the 2 faces that need to
be concurrent at vertex A;

$4!$ that is the number of ways we can then number the 4 left faces with the 4 numbers that remain.

Even though I could understand the solution to this problem, I still don't know why mine is wrong. I am assuming I am counting extra scenarios (since $2160>1872$), but which one's?

答: Your formula counts the cases where all four faces are labeled with odds numbers $3$ times each, once for each pair of odd numbers $>1$.  So, you need to subtract twice the number of such labelings.  The offical solutions shows that there are $144$ such labelings, and indeed $$2160-288=1872$$

Title: Numbering a octahedron


问:
Find $L^{-1}{\frac{{{e^{-\sqrt{p}}}}}{p}}$
Find $$L^{-1}{\frac{{{e^{-\sqrt{p}}}}}{p}}$$.
where $L^{-1}$ denotes the laplace inverse .

I tried expanding the exponential but got stuck in the middle.

答:
All done in Mathematica:

$$\mathcal{L}_p^{-1}\left[\frac{\exp \left(-\sqrt{p}\right)}{p}\right](t)=\\\mathcal{L}_p^{-1}\left[\frac{\sum
   _{j=0}^{\infty } \frac{\left(-\sqrt{p}\right)^j}{j!}}{p}\right](t)=\\\sum _{j=0}^{\infty }
   \mathcal{L}_p^{-1}\left[\frac{(-1)^j p^{-1+\frac{j}{2}}}{j!}\right](t)=\\\sum _{j=0}^{\infty } -\frac{2 (-1)^j
   t^{-\frac{j}{2}}}{j j! \Gamma \left(-\frac{j}{2}\right)}=\sum _{j=0}^{\infty } \frac{(-1)^j t^{-\frac{j}{2}}}{\Gamma \left(1-\frac{j}{2}\right) \Gamma (1+j)}=\\1-\text{erf}\left(\frac{1}{2
   \sqrt{t}}\right)=\text{erfc}\left(\frac{1}{2 \sqrt{t}}\right)$$


问:
Doubt in the Proof on Dehn-Nielsen-Baer theorem
I'm reading the proof of Dehn-Nielsen-Baer theorem given in the book 'A primer on mapping class groups' by Margalit and Farb. I'm having trouble understanding the following line from the proof of lemma 8.5.
"Assume for the purposes of contradiction that  $\Phi(\gamma)$ and
$\Phi(\delta)$ are linked at infinity. It follows that the polygonal paths $\{\Phi(\alpha_i)\}$ and $\{\Phi(\beta_i)\}$ have to cross."
Since $\Phi(\gamma)$ and
$\Phi(\delta)$ are linked at infinity thier axes intersect. But what is the exact relation between the axes of $\Phi(\gamma)$ and
$\Phi(\delta)$ and the polygonal paths that forces them to intersect?

答:
To review the setting in more detail, we have two embeddings of $[0,1]$ in the closed disc $D^2$ with images $A$ and $B$:

$A$ is the closure of $\Phi(\gamma) = \text{interior}(A) = A \cap \text{interior}(D^2)$ with endpoints $\partial A = \{a_0,a_1\} = A \cap S^1$;

$B$ is the closure of $\Phi(\delta) = \text{interior}(B) = B \cap \text{interior}(D^2)$ with endpoints $\partial B = \{b_0,b_1\} = B \cap S^1$.

To say that "$\Phi(\gamma)$ and $\Phi(\delta)$ are linked at infinity" means that the two sets $\partial A,\partial B \subset S^1$ are linked in $S^1$, which in turn means that there exists a parameterization $[0,2\pi] \mapsto S^1$ in the usual fashion (continuous, surjective, and injective except that $0,2\pi$ are mapped to the same point of $S^1$), and there exist parameters $0 < s_0 < t_0 < s_1 < t_1 < 2\pi$, such that
$$s_0 \mapsto a_0, \quad t_0 \mapsto b_0, \quad s_1 \mapsto a_1, \quad t_1 \mapsto b_1
$$
Another way to say this without parameterizations is that each of the two components of the set $S^1 - \{a_1,a_2\}$ contains a point of $\{b_1,b_2\}$.

That answers your question about what the exact relation is.

The way that you then get the desired conclusion is by applying a theorem of topology:

For any two embeddings of $[0,1]$ in $D^2$ denoted $A,B \subset D^2$, if $A \cap S^1 = \partial A$ and $B \cap S^1 = \partial B$, and if $\partial A$ and $\partial B$ are linked in $S^1$ (as defined above), then $A \cap B \ne \emptyset$.

This can be proved by doubling $D^2$ along $S^1$ to obtain a 2-sphere, in which there is an embedded circle obtained by doubling $A$ along $\partial A$, and then applying the Jordan Curve Theorem.


问:
Matrix exponential - inequality
Hi so I'm supposed to solve an inequality about Matrix exponentials. I'm pretty sure it follows directly from some theorem but I just can't seem to find it in my script.

Let $A = (a_{ij})_{1\leq i,j\leq n}$ be a square matrix with real coefficients. We set $\lVert A \rVert$ $=$ max$_{1\leq i,j\leq n} \lvert a_{ij} \rvert$. Show that for every integer $k \geq 1$, the inequality $\lVert A^k \rVert \leq n^{k-1} \lVert A \rVert^k$ holds.

Now I've seen some different inequalities and proofs, however I don't really know how to approach this. In general, I think it would make sense to prove it via induction, although it seems to me that there is a much simpler proof. I've tried doing that, however I'm not sure how to work with $\lVert A^k \rVert $and$ \lVert A \rVert^k$ as for the first the exponent is inside the norm and for the second it isn't. How can I go about proving such an inequality?

Title: Matrix exponential - inequality

Body:
Lemma If $A$ and $B$ are $n\times n$ then $||AB||\le||A||\,||B||$.

Proof: Say $A=(a_{jk})$, $B=(b_{jk})$ and $AB=(c_{jk})$. Various definitions show that $$|a_{jk}|\le||A||,$$ $$|b_{jk}|\le||B||,$$and $$c_{jk}=\sum_{l=1}^na_{jl}b_{lk}.$$ So $$|c_{jk}|\le\sum_{l=1}^n|a_{jl}|\,|b_{lk}|\le\sum_{l=1}^n||A||\,||B||=n||A||\,||B||.$$

This takes care of the induction step in the inequality you ask about. Assume $||A^k||\le n^{k-1}||A||^k$. Then $$||A^{k+1}||=||AA^k||\le n||A||\,||A^k||\le n||A||n^{k-1}||A||^k= n^{(k+1)-1}||A||^{k+1}.$$


问:
computing the limit and stating the basic limit used $\lim_{x \to 2} \frac{1}{x^2}$
I'm trying to compute the following limit whilst using the basic limit theorems, although I'm slightly rusty and would appreciate you honest feedback.

$\lim_{x \to 2} \frac{1}{x^2}$

Given that when $|x-2| < \delta$

Then $\frac{1}{x^2}=\frac{1}{4}$

This is my reasoning thus, far, and from this I proceeded with:

$\lim_{x \to 2} \frac{1}{x^2} = \frac{1}{4}$

$|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$

Though i'm not sure where to go next.As for the basic limit theorems, I tried approaching this with:

$\lim_{x \to p} \frac{f(x)}{g(x)}=\frac{A}{B}$

答:
You seem to be incorporating some version of an $\epsilon$-$\delta$ argument, which is unnecessary if you're using basic limit theorems. For the record though, your statement
"Given that when $|x-2|<\delta$ then $\frac{1}{x^2}=\frac{1}{4}$" is a misrepresentation of $\epsilon$-$\delta$ arguments. It would be $|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$, as you have later down. With regards to that portion, though
"I proceeded with: $\lim_{x\to 1}\frac{1}{x^2}=\frac{1}{4}$, $|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$" is a bit of a reversal of the chain of reasoning you would present in an $\epsilon$-$\delta$ argument.

Anyhow, I just wanted to point those things out, but since this is not supposed to be an $\epsilon$-$\delta$ argument, I'll leave it there. As for the question at hand, Correa is correct in his comment.

Let $f(x)=1$ and let $g(x)=x$. You can use the property you stated, namely $\lim_{x\to p}\frac{f(x)}{g(x)}=\frac{f(p)}{g(p)}$ (provided each limit exists separately and $g(p)\neq 0$), but you must first establish that $\lim_{x\to 2}x=2$. You can then use $\lim_{x\to p}f(x)g(x)=\lim_{x\to p}f(x)\lim_{x\to p}g(x)$ to get the final answer.


问:
Converting $9e^{(3+i\pi/6)}$ into cartesian form
For a question of $$9e^{(3+i\pi/6)}$$ I thought the answer would be $$9((\cos(3+\pi/6)+i \sin(3+\pi/6))$$ which would give out a "a + bi" form of $9\cos(3+\pi/6)+9 \sin(3+\pi/6)i$, but I quiz showed me an error. Is that not the correct answer? Why not, since my other similar answers were fine? Thanks if you could point out my mistake. *Hope that the formatting works

Body:
$9((\cos(3+\pi/6)+i \sin(3+\pi/6))=9^{i(3+ \pi/6)} \ne 9e^{3+i\pi/6}.$

$9e^{3+i\pi/6}= 9e^3 e^{i \pi/6}.$

$ \cos( \pi /6)=\sqrt{3}/2$ and $ \sin ( \pi/6)=1/2.$


问:
Solve Non-lineaer Differential Equation (Step-by-Step)
Being able to solve Linear and Non-liner differentional equations is an important part of mathematics. I am trying to work out the exact solution for this non-linear differential equation and require some help from the community (Please could you provide the full step-by-step-solution), I want to undertstand the mathematics and be able to apply this do other questions.

$$\dfrac {du}{dx}+ u^2 = 0  \tag 1 $$

Boundary conditions $0<x<1; u(0)=1$

What is the exact solution?

答:
Note that you can rewrite it as:
$$-\frac{1}{u^2}\frac{du}{dx} = 1,$$
and you might recognize on the left-hand side the derivative with respect to $x$ of $x\mapsto \frac{1}{u(x)}$. Indeed, by the chain rule, denoting $f(x)=\frac 1x$:
$$\frac{d}{dx}\bigl(f(u(x))\bigr) = \frac{df}{dx}(u(x))\cdot\frac{du}{dx}(x) = -\frac{1}{u^2}\frac{du}{dx}.$$

Thus integrating over $x$ on both sides gives, for the left hand side:
$$\int_0^x -\frac{1}{u(s)^2}\frac{du}{dx}(s)ds = \int_0^x \frac{d}{dx}\bigl( f(u(s))\bigr) ds = f(u(x)) - f(u(0))$$
and the right hand side:
$$\int_0^x 1ds = x.$$
Equating both and using the condition $u(0)=1\implies f(u(0))=1$:
$$f(u(x)) - 1 = \frac{1}{u(x)} -1 = x,$$
$$\iff \frac{1}{u(x)} = x+1 \iff u(x) = \frac{1}{x+1}.$$
The last equality is possible since $0\leq x \leq 1$.


问:
$|A|=\lim_{t\to\infty} |A\cap (-t,t)|,\ A\subset\mathbb{R}, t>0$
I have proved the following statement, but I am not yet sure that everything is correct so I would appreciate comments and corrections, thanks.

$|A|=\lim_{t\to\infty} |A\cap (-t,t)|,\ A\subset\mathbb{R}, t>0$

NOTE: $|\cdot|$ refers to outer measure, i.e. for $A\subset\mathbb{R},\ |A|:=\inf\{\sum_{k=1}^{\infty}l(I_k): I_1,I_2,\dots\text{ are open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$; the length of an open interval $I\subset\mathbb{R}$ is defined as

$l(I):=\begin{cases}
b-a & \text{if }I=(a,b),\ a,b\in\mathbb{R}, a<b; \\
0   & \text{if }I=\emptyset; \\
\infty & \text{if } I=(-\infty, a)\text{ or } I=(a,\infty);\\
\infty & \text{ if }I=(-\infty,\infty)
\end{cases} $

My proof:
We study the two possible cases: $|A|=\infty$ and $|A|<\infty$.

CASE $|A|=\infty$| Let $I_1,I_2,\dots$ be a sequence of open intervals such that $A\subset\bigcup_{k=1}^{\infty}I_k$. Then $\sum_{k=1}^{\infty}l(I_k)=\infty$ (since $|A|\leq\sum_{k=1}^{\infty}l(I_k)$) so if we take $M>0$ arbitrary there must be $K\geq 1$ such that $\sum_{k=1}^{K}l(I_k)>M$. We note that $I_k\cap(-t,t)$ is an open interval and $$A\cap(-t,t)\subset\bigcup_{k=1}^{\infty}(I_k\cap(-t,t))$$ so $$|A\cap(-t,t)|:=\inf\{\sum_{k=1}^{\infty}l(U_k):U_1,U_2,\dots\text{ open intervals such that }A\cap(-t,t)\subset\bigcup_{k=1}^{\infty}U_k\}=\inf\{\sum_{k=1}^{\infty}l(I_k\cap(-t,t)):I_1,I_2,\dots \text{ open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$$ Now, if one the the $I_k$ (say, $I_{k*}$) is of the form $(-\infty,a),(a,\infty)$ ($a\in\mathbb{R}$) or $(-\infty,+\infty)$ we have that, in the first case $$I_k\cap(-t,t)=(-t,a)\Rightarrow l(I_k\cap(-t,t))=a-(-t)=a+t>M$$ for $t>\max\{M-a,a\}$, in the second case $$I_k\cap(-t,t)=(a,t)\Rightarrow l(I_k\cap(-t,t))=t-a>M$$ for $t>M+a$ and in the third case $$I_k\cap(-t,t)=(-t,t)\Rightarrow l(I_k\cap(-t,t))=2t>M$$ for $t>\frac{M}{2}$ thus in this case it's enough to take $t>\max\{M-a,M+a,\frac{M}{2}\}$ to have $\sum_{k=1}^{\infty}l(I_k\cap(-t,t))\geq l(I_{k*})>M$. If the $I_k$ are all finite ($I_k=(a_k,b_k)$) it's enough to take $t>\max\{|a_1|,|b_1|,|a_2|,|b_2|,\dots, |a_K|,|b_K|\}$ to have $\sum_{k=1}^{\infty}l(I_k\cap(-t,t))\geq\sum_{k=1}^{K}l(I_k\cap(-t,t))>M$. Thus we can conclude that if $t>\max\{|a_1|,|b_1|,\dots, |a_K|,|b_K|, M-a,M+a,\frac{M}{2}\}$ we have $\sum_{k=1}^{\infty}l(I_k\cap(-t,t))>M$ and taking the $\inf$ we get $|A\cap (-t,t)|>M$. Since $M>0$ was chosen arbitrarily we can conclude that for all $M>0$ there exists $t\geq 0$ such that $|A\cap (-t,t)|>M$ i.e. $\lim_{t\to\infty} |A\cap (-t,t)|=\infty=|A|$, as desired.

CASE $|A|<\infty|$ since $|A|=|A\cap (-t,t)|+|A\cap\mathbb{R}-(-t,t)|$ it is enough to show tha $\lim_{t\to\infty}|A\cap\mathbb{R}-(-t,t)|=\lim_{t\to\infty}|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|=0$.

So, let $I_1,I_2,\dots$ be a sequence of open intervals ($I_n =(a_n, b_n)$ whose union contains $A$ (note that there cannot be an infinite one like $I_{n*}=(-\infty,a)$, $(a,\infty)$ or $(-\infty,\infty)$ since in that case $\sum_{n=1}^{\infty}l(I_n)\geq l(I_{n*})=\infty$).

For each $n\geq 1$ let $J_n:=I_n\cap (-\infty,-t]$ and $L_n:=I_n\cap [t,\infty)$. Since $A\cap (-\infty,-t]\cup A\cap [t,+\infty)\subset \bigcup_{n=1}^{\infty} (J_n\cup L_n)$ we have $$|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|\leq |\bigcup_{n=1}^{\infty}(J_n\cup L_n)|\leq\sum_{n=1}^{\infty}(l(J_n)+l(L_n))=\sum_{n=1}^{\infty}l(J_n)+\sum_{n=1}^{\infty}l(L_n)$$
Now, let $\varepsilon >0$: there must be some $N\geq 1$ such that $|\sum_{n=1}^{\infty}l(J_n)-\sum_{n=1}^{N}l(J_n)|=|\sum_{n=N+1}^{\infty}l(J_n)|<\varepsilon/2$ and similarly there must be $N'\geq 1$ such that $|\sum_{n=1}^{\infty}l(L_n)-\sum_{n=1}^{N'}l(L_n)|=|\sum_{n=N'+1}^{\infty}l(L_n)|<\varepsilon/2$ so if $t>\max\{|a_{1}|,\dots,|a_{N}|,|b_{1}|,\dots, |b_{N'}|\}$ we have that  $\sum_{n=1}^{\infty}l(J_n)+\sum_{n=1}^{\infty}l(L_n)<\varepsilon\Rightarrow |A\cap (-\infty,-t]\cup A\cap [t,+\infty)|<\varepsilon$. Thus, $\lim_{t\to\infty}|A\cap (-\infty,-t]\cup A\cap [t,+\infty)|=0$, as desired.

答:
To me it seems that your argument works out if $A$ is bounded.
To avoid infinite sums, you may then take a finite covering of $A$ with open intervals of radius $\epsilon>0$ (since the closure of $A$ is compact).
In other words, you can take the sequence $(a_n,b_n)$ such that $b_n-a_n=\epsilon>0$ for any $\epsilon>0$ and finitely many such intervals suffice to cover $A$.

I'm not convinced that your argument works directly for unbounded sets.
I think the partial sums are not guaranteed to be bounded by $\epsilon/2$.

Here is a similar approach which seems to work generally.
The idea is the same, but I choose the intervals $(a_n,b_n)$ carefully, to benefit from the result you provided.
Since you know:
$$|A|=|A\cap (-t,t)|+|A\setminus (-t,t)|$$
It also holds that for $t'>t$:
$$
|A\setminus (-t,t)|=|A\setminus (-t,t)\cap (-t',t')|+|A\setminus (-t,t)\setminus (-t',t')|
$$
$$
=|A\cap((-t',-t]\cup[t,t'))|+|A\setminus (-t',t')|
$$
So, we may write:
$$
|A|=\sum_{n=0}^\infty|A\cap ((-n-1,-n]\cup[n,n+1))|
$$
To see that the sum indeed converges to $|A|$, note that if $|A|=+\infty$, the sum must diverge and if $|A|<+\infty$, then the sum converges monotonely to $|A|$.
Then, you have for $t>0$:
$$
\sum_{n=0}^{\lfloor t\rfloor}|A\cap ((-n-1,-n]\cup[n,n+1))|\leq |A\cap (-t,t)|
$$
$$
\leq \sum_{n=0}^{\lceil t\rceil}|A\cap ((-n-1,-n]\cup[n,n+1))|
$$
As $t\rightarrow\infty$, both the left and right side converge monotonely to $|A|$, hence so does $ |A\cap (-t,t)|$.


问:
Probability inequality for the case $E(X^2)=1$
I am trying to solve the following exercise from Kai Kai Chung's A Course in Probability Theory:

If $E(X^2)=1$ and $E(|X|)\ge a>0$ then $P[|X|\ge \lambda a]\ge (1-\lambda)^2 a^2$ for every $0\le \lambda \le 1$.

I know that the second condition implies that $a\le 1$ by Jensen's inequality and hence the proposition holds for trivial cases $\lambda \in \{0,1\}$. I guess I could integrate some indicator functions but I have no clue. Could someone give me a hint?

答:
Welcome to MSE!

Hint:

It's often useful to split up a random variable into the places where it is "good" and "bad". Then we can control those regions separately to get whatever inequality we're interested in.

For this, we might try looking at the decomposition

$$
X =
X \cdot \mathbf{1}_{\{X \leq \lambda a\}} + X \cdot \mathbf{1}_{\{X > \lambda a\}}
$$

The expectation of the first term is at most $\lambda a$, more or less by definition.

The expectation of the second term is at most $\sqrt{E[X^2]} \cdot \sqrt{\Pr[X > \lambda a]}$ (do you see why?)

Do you see how to combine these facts in order to get the claim? If you get stuck, you might want to google the "Paley–Zygmund inequality".

I hope this helps ^_^


问:
which parameters always make this rational equation evenly divisible?
Hi guys I have the following equation:

$$x = \dfrac{a + b \times c - b}{c}$$

This is what I know about each variable:
$$a \ge 64$$
$$b \ge 0$$
$$8 \le c \le a$$

My questions is there a concise way for me to pick a, b, and c so that x will always be a positive integer >= 8.

For example if $a = 64$; $b= 0$; $c = 8$ (the smallest possible values for each)

$x = (64 + 0 \times 8 - 0)/8$

$x = 8$ which is a positive integer $\ge 8$

However, if let's say $a = 200$; $b = 2$; $c = 10$

$x = (200 + 2 \times 10 - 2)/10$

$x = \dfrac{218}{10}$ which is $21.8$ and is not a positive integer greater than or equal to $ 8$

How can I always know that $x$ will be a positive integer?? Thanks!

答:
As $$x = \frac{a+ b\cdot c -b}{c} = \frac{a-b}{c} + \frac{b\cdot c}{c} = \frac{a-b}{c} + b,$$ we can say that $x$ will be an integer whenever $a-b$ is divisible by $c$. That is, we need an integer $n$ such that $$a-b=cn \leftrightarrow a = cn + b.$$ Finally, as we want $x \geq 8$ this means we need $$\frac{a-b}{c} + b \geq 8.$$ By solving for $a$, we have $a \geq b -bc +8c$.

Combining our expressions involving $a$ results in $$cn+b \geq b-bc+8c$$ $$n+b\geq 8.$$ Now you may choose any $b \geq 0$ and any $n$ satisfying $n+b \geq 8$ and you will have $x \geq 8$.


答:
You can write
$$
E(|X|1(|X|\geq \lambda a)) \leq \left(E(X^2) P(|X|\geq \lambda a)\right)^{1/2} = P^{1/2}(|X|\geq \lambda a)
$$
The first inequality is Cauchy-Schwarz. Then
$$
E(|X|1(|X|\geq \lambda a)) =E(|X|) - E(|X|1(|X|< \lambda a))\geq a-\lambda a.
$$

Combining the last two displays and using $\lambda\in (0,1)$
$$
P(|X|\geq \lambda a)\geq (1-\lambda)^2a^2.
$$

Title: Probability inequality for the case $E(X^2)=1$

Body:
I am trying to solve the following exercise from Kai Kai Chung's A Course in Probability Theory:

If $E(X^2)=1$ and $E(|X|)\ge a>0$ then $P[|X|\ge \lambda a]\ge (1-\lambda)^2 a^2$ for every $0\le \lambda \le 1$.

I know that the second condition implies that $a\le 1$ by Jensen's inequality and hence the proposition holds for trivial cases $\lambda \in \{0,1\}$. I guess I could integrate some indicator functions but I have no clue. Could someone give me a hint?


问:
Does every bounded, divergent sequence contain only convergent subsequences with at least two different limits?
Claim: Every bounded and divergent sequence $(a_n)$ contains only convergent subsequences.

A bounded, divergent sequence $(a_n)$ implies that there is at least one convergent subsequence (by Bolzano Weierstrass) $(a_{nk}) \rightarrow l$. Now, because $(a_n)$ diverges, there exists some subsequence $(a_{jk})$ which isn't in the  $\epsilon -$ neighborhood of $l$. This means that either $(a_{jk})$ is divergent or converges to some other point (as it cannot be unbounded). If it converges to some other point $(a_{jk}) \rightarrow k$ we have shown that there exist two limit points and we are done. If, however, $(a_{jk})$ is divergent itself, I can apply Bolzano Weierstrass again and basically have a recursive loop until I am only left with only convergent subsequences that all converge to a different limit.

Is this valid? I'm not really sure if this is sufficient or even correct...

答: As the comments already mentioned, the claim is incorrect (The sequence itself is a subsequence for example). The flaw in your reasoning is in your recursive loop. You implicitly assume this loop will end in finite steps. This is by no means clear, since we can have infinitely many different subsequences</p>


问:
For what $n$, can we find $n$ positive integers such that $\frac{1}{a_1} + \frac{1}{2 a_2} + \dots + \frac{1}{n a_n} = 1$
For what $n$, can we always find $n$ positive integers such that $\frac{1}{a_1} + \frac{1}{2 a_2} + \dots + \frac{1}{n a_n} = 1$

I have tried a few numbers, it seems like it works for any $n$.

When $n=1$ we have $a_1 = 1$. When $n=2$ we have $a_1 = 2, a_2 = 1$

But I'm curious how to prove that this exists for all $n$?

答:
Notice $a_i = i+1$ for $1\leq i < n$ and $a_n=1$ works (proof by induction is straightforward).

The way I came up with it is by noticing $\frac{1}{a} - \frac{1}{b}$ is of the form $\frac{1}{k}$ when $b-a$ divides $ab$ and I wanted the $b$ to be the newer $n$ that we were incorporating, and then I noticed that if $a=n-1$ then it works, and I got lucky I guess.

So the idea was to do an inductive construction, so if you have that $(a_1,\dots,a_n)$ works for $n$ and you can substract $\frac{1}{n}$ from one of the previous fractions so that it is still of the form $\frac{1}{k}$ then you can grow the construction.


问:
Calculating the number of possible password combinations given a set of constraints
There are several password combination problems on this site but none is exactly like mine: An $8$-character password contains 3 numbers, at least one uppercase letter, at least one lowercase letter, and at least one special character. How many such password combinations are possible? (Note that, currently, there are $33$ publicly-known special characters in this world). Here's what I've been trying to do:

The three numbers will give us $10^3 = 1,000$ possible combinations. This now leaves us with $5$ characters that we have to account for. The sum of the pool of uppercase, lowercase and special characters is $26+26+33 = 85$. Thus, $85$ characters can fill $5$ slots in $85^5 = 4,437,053,125$ ways. However, this will not limit each uppercase, lowercase and special character to at least one, right? That is, my calculation also includes passwords like, say, 123abcde, right? I then tried $88^8 = 3,596,345,248,055,296$. The $88$ comes from the sum of $3$ numbers and the pool of $85$ characters we had earlier encountered. This also doesn't seem to work because, at the very least, the $3$ numbers are not specific.

How do I solve this?

Update: Here's what I tried next: In a table from one of the answers below, there are six major possibilities, so let's just work with one of them - a password containing $3$ numbers, $1$ uppercase, $1$ lowercase and $3$ special characters. Let's assume that the ordering is strictly that. That is, the first $3$ characters are numbers, the $4$th character is an uppercase, the $5$th character is lowercase, and the $6$th, $7$th and $8$th characters are special. Now, with this, how many such password combinations are possible? It's $220*26*26*33^3 = 5,344,550,640$. Yes, yes, yes, I know you're wondering where the $220$ comes from. These are all the triples where the first digit is less than or equal to the second digit AND the second digit is less than or equal to the third digit (I wrote a computer program to count them). We do this so that we can avoid duplicates. So far, this approach accounts for a password like, say, 123Aa+-* but ignores a password like 3Aa2+-*1. In order to account for it (and many others), we just permute. As such, our initial value of $5,344,550,640$ now gives us $5,344,550,640*8! = 215,492,281,804,800$ passwords. We do the same for the other $5$ major possibilities:

$3-1-2-2: 220*26*26^2*33^2*8!=169,781,797,785,600$
$3-1-3-1: 220*26*26^3*33*8!=133,767,477,043,200$
$3-2-1-2: 220*26^2*26*33^2*8!=169,781,797,785,600$
$3-2-2-1: 220*26^2*26^2*33*8!=133,767,477,043,200$
$3-3-1-1: 220*26^3*26*33*8!=133,767,477,043,200$

Adding it all up gives us: $956,358,308,505,600$

答:
Since there is only 8 characters in password, we could evaluate this directly.

First, you were right, since there is exactly 3 digits, there are $10^3$ ways to pick the digits. Then we have to choose the position of those digits: $8\choose 3$

For the remaining 5 characters, we need at least 1 uppercase, 1 lowercase, and 1 special and two others to be determine.  This gives us 6 possible scenarii of Uppercase - Lowercase - Special :
$$1-1-3, 1-2-2, 1-3-1, 2-1-2, 2-2-1, 3-1-1$$
For the first scenario, the password has 3 digits, 1 uppercase, 1 lowercase and 3 special characters. We start by choosing the digits' positions, then we choose the digits (with replacement)
$${8\choose3}10^3$$
From the five remaining positions, we choose one for the uppercase, then choose the uppercase character
$${8\choose3}10^3{5\choose1}26$$
From the four remaining positions, we choose one for the lowercase, then choose the lowercase character
$${8\choose3}10^3{5\choose1}26{4\choose1}26$$
Finally, the three remaining positions are filled with special characters.
$${8\choose3}10^3{5\choose1}26{4\choose1}26{3\choose3}33^3$$

We do the same for each of the 6 possible scenarii.
$$
\begin{array}{c|c}
\text{Digits-Upper-Lower-Special} & \text{possibilities} \\
\hline
3-1-1-3 & {8\choose3}10^3{5\choose1}26{4\choose1}26{3\choose3}33^3 \\
3-1-2-2 & {8\choose3}10^3{5\choose1}26{4\choose2}26^2{2\choose2}33^2 \\
3-1-3-1 & {8\choose3}10^3{5\choose1}26{4\choose3}26^3{1\choose1}33 \\
3-2-1-2 & {8\choose3}10^3{5\choose2}26^2{3\choose1}26{2\choose2}33^2 \\
3-2-2-1 & {8\choose3}10^3{5\choose2}26^2{3\choose2}26^2{1\choose1}33 \\
3-3-1-1 & {8\choose3}10^3{5\choose3}26^3{2\choose1}26{1\choose1}33
\end{array}
$$

The numbers of possible passwords with these restrictions is $150\ 634\ 323\ 840\ 000$


答: No. Consider the bounded sequence $1,2,3,1,2,3,1,2,3,\cdots $. One of its proper subsequences is $1,2,1,2,1,2,\cdots$ and it does not converge.

Title: Does every bounded, divergent sequence contain only convergent subsequences with at least two different limits?

Body:
Claim: Every bounded and divergent sequence $(a_n)$ contains only convergent subsequences.

A bounded, divergent sequence $(a_n)$ implies that there is at least one convergent subsequence (by Bolzano Weierstrass) $(a_{nk}) \rightarrow l$. Now, because $(a_n)$ diverges, there exists some subsequence $(a_{jk})$ which isn't in the  $\epsilon -$ neighborhood of $l$. This means that either $(a_{jk})$ is divergent or converges to some other point (as it cannot be unbounded). If it converges to some other point $(a_{jk}) \rightarrow k$ we have shown that there exist two limit points and we are done. If, however, $(a_{jk})$ is divergent itself, I can apply Bolzano Weierstrass again and basically have a recursive loop until I am only left with only convergent subsequences that all converge to a different limit.

Is this valid? I'm not really sure if this is sufficient or even correct...


问:
Square Root Time Complexity
I'm deeply sorry for this, I don't know if it's the appropriate place to post this but I just can't figure out what's the time complexity of this algorithm:

```<code>int fun(int n)
{
    int j = 10;
    while( j < n)
        j+= sqrt(j);
    return j;
}
</code>```

I know that at the $k^{th}$ iteration we have $f(k)=f(k-1)+\sqrt{f(k-1)}$, with $f(0)=10$, I tried solving this function, but I couldn't find any way to do so.

答:
You can see that the second derivative of the function is a constant value, so you can approximate it with a quadratic function.

The values for $a$ and $b$ in the quadratic function take on values of around $0.00001$, so $$f(n) \approx 0.00001n^2+0.00001n+10$$

Title: Square Root Time Complexity


问:
Groups acting on polynomials
I am a little confused about actions right now.

Say I have a group $G$ that acts on the roots of a polynomial, say $(x - x_1)(x - x_2)(x - x_3)$.  We know that this group permutes the roots of a polynomial, that is you can get to any of the roots by having an element of the group act on another root.  Does that mean the group fixes the polynomial?

I feel like it should, but I'm not sure how this actually works.  Supposedly we would have to write $g \cdot [(x - x_1)(x - x_2)(x - x_3)] = (x - x_1)(x - x_2)(x - x_3)$.  But is this actually possible?  Because $G$ acts on the set of the roots, not the polynomial.  I'm not sure if we can say a polynomial is fixed by certain things acting on the roots.  When they say the polynomial is fixed, does that mean the coefficients are fixed?

Also, can you say something like $g \cdot (x_1x_2 + x_3)$ = $(g \cdot x_1)(g \cdot x_2) + g \cdot x_3$?

Thanks!

答:
Let's back up a little; you seem to be confused about more fundamental things right now.

Fix a field $F$. If $f(x) \in F[x]$ is a monic polynomial, we can talk about the splitting field of $f$, which is a field extension $K/F$ obtained by adjoining all of the roots $r_1, ... r_n$ of $f$. For various reasons it is natural to consider a certain group $G = \text{Aut}(K/F)$, the group of all invertible field homomorphisms $g : K \to K$ which fix $F$. This means that

$g(a + b) = g(a)+g(b)$

$g(ab) = g(a) g(b)$

$\forall a \in F : g(a) = a$.

This should hopefully answer your last question. The properties above imply that $g(f(r)) = f(g(r))$ for any $r \in K$, from which it follows that $g(f(r)) = 0$ if and only if $f(g(r)) = 0$. Since $g$ is an automorphism, this implies that $r$ is a root of $f$ if and only if $g(r)$ is a root of $f$, hence that $G$ permutes the roots of $f$.

Now, the elements of $G$ act by automorphisms $K \to K$, but we can extend this action to an action by automorphisms $K[x] \to K[x]$ by declaring that $x$ is sent to $x$ by every element of $G$ and extending by linearity and multiplicativity. So one way to make sense of the statement that elements of $G$ fix the polynomial $f(x)$ is that they fix the element $f(x) \in F[x] \subset K[x]$. There are (at least) two ways to see this.

First, elements of $G$ fix elements of $F$ by definition, and since $f(x)$ is a sum of elements of $F$ and powers of $x$ (which $G$ also fixes by definition), the result is clear.

Second, note that in $K[x]$ we can write

$$f(x) = (x - r_1)(x - r_2)...(x - r_n)$$

so it follows by the homomorphism property that

$$g(f(x)) = (x - g(r_1))(x - g(r_2))...(x - g(r_n))$$

and since $g$ permutes the roots of $F$ it follows that the expression on the RHS is just a reordering of the factors of $f(x)$.

Title: Groups acting on polynomials


问:
normal distribution finding b
For part b, I wrote the answer like this:

Pr(− < ( − )/ < ) = 0.90

Pr(− < Z < ) = 0.90

b=（1.64＋1.65)/2 =1.645

Am I correct doing like this? I'm not sure.
Do I need to do the inversion of centralise by making it times  and plus ?

Question: If random variable  follows Normal distribution

(a)If we know () = 75 and () = 100 please find Pr(X < 60) and
Pr(70 < X < 100)

(b)  Find  so that Pr(− < ( − )/ < ) = 0.90

答:
b)

the request is the same as

$$\mathbb{P}[Z<b]=0.95$$

Reading the result on Z-table you get

$$b\approx 1.64$$

this because you have to calculate the two quantiles excluding 5% on the left tail and another 5% on the right one.

Title: normal distribution finding b


问:
Vectors and Planes with parametric equations
Any help would be appreciated!

Find a parametric equation of a line $L$ that is obtained as the intersection of the planes $P$ and $Q$ with scalar equations $$P: x+2y+3z=4$$ and $$Q: x-3y+z=1$$

Related to this is the question:

Give a parametric equation of a line $M$ contained in $Q$ and orthogonal to $L$.

Note: This is from a Calculus course so any advice involving linear algebra may not be of much help!

答:
Even though this is a question from a Calculus course, it boils down to linear algebra, since the whole problem is a system of linear equations. You need to solve the system

$\begin{align*}
x+2y+3z &= 4 \\
x-3y+z &= 1
\end{align*}$

When you solve this, which you can do directly using substitution, you find that

$\begin{align*}
x &= -\frac{11}{5}z + \frac{14}{5} \\
y &= -\frac{2}{5}z + \frac{3}{5}
\end{align*}$

This is a parametrization of the line you are looking for. If you prefer the vector equation, you could write it as

$$\mathbf{r}(t) = \frac{1}{5}(-11t+14, -2t + 3, t).$$


答:
Let first rewrite the equation: $$T(n^2)=n^2T(n)+n^4$$

First step is to homogenize the formula.

Let set $T(n)=n^2U(n)$, after substitution we get $$U(n^2)=U(n)+1$$

Then we linearize it.

Let set $n=2^p$ and $V(p)=U(n)$, after substitution we get $$V(2p)=V(p)+1$$

Let set $p=2^q$ and $W(q)=V(p)$, after substitution we get $$W(q+1)=W(q)+1$$

Solve the formula (notice it is a telescoping sequence)

$$W(q)=W(0)+q$$

Last step is to go back up the chain of substitutions

$W(0)=V(2^0)=V(1)=U(2^1)=U(2)=\dfrac{T(2)}{2^2}=\dfrac 14$

$q=\log_2(\log_2(n))$

$T(n)=n^2U(n)=n^2V(p)=n^2W(q)=n^2(\frac 14+q)$

$$T(n)=\frac 14n^2 + n^2\log_2(\log_2(n))$$

Title: Solve the recurrence $T(n) = nT(\sqrt{n}) + n^2, T(2) = 1$

Body:
$$T(n) = nT(\sqrt{n}) + n^2, T(2) = 1$$

The master theorem does not apply here, only recursion tree. Assume $n = 2^{2^k}$.

The solution for self-check is $T(n = 2^{2^k}) = k \cdot n^2 + \frac{n^2}{4}$.

Edited:

My approach:

$T(2^{2^k}) = 2^{2^k}T(2^{2^{k-1}}) + 2^{2^{k + 1}}$

$= 2^{2^k + 2^{k - 1}}T(2^{2^{k-2}}) + 2^{2^{k+1}} + 2^{2^{k+1}}$

...

$=2^{\sum_{i = 1}^{k} 2^i} + k \cdot 2^{2^{k+1}}$

$=2^{\sum_{i = 0}^{k} 2^i - 1}  + k \cdot 2^{2^{k+1}}$

$=2^{2^{k+1} - 2} + k \cdot n^2$

$=\frac{n^2}{4} + k \cdot n^2$


答:
The definition for limit of a function at a point is as follows:

If for all $\epsilon>0$, there exists $\delta>0$ such that $0<|x-x_0|<\delta$ implies $|f(x)-L|<\epsilon$, then the limit of $f(x)$ as $x$ approaches $x_0$ exists and is equal to $L$.

Here is a correct version of the $\epsilon,\delta$ argument using the definition:

Let $\epsilon>0$ be given and define $\delta=\min\left\{\epsilon,\frac{1}{2}\right\}$. Then $|x-2|<\delta$ implies $\frac{3}{2}<x<\frac{5}{2}$. However, we also have that

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|=\frac{|4-x^2|}{4x^2}=\frac{|2-x||2+x|}{4x^2}=\frac{|x-2|(x+2)}{4x^2}$$

and therefore

$$<\frac{\delta(5/2+2)}{4\cdot (3/2)^2}=\frac{1}{2}\delta$$

Now, if $\delta=\epsilon$ then we are done as

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|<\frac{1}{2}\delta=\frac{1}{2}\epsilon<\epsilon$$

If not, then $\delta=\frac{1}{2}<\epsilon$ and therefore

$$\left|\frac{1}{x^2}-\frac{1}{4}\right|<\frac{1}{2}\delta=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}<\frac{1}{2}<\epsilon$$

and we are done.

Title: computing the limit and stating the basic limit used $\lim_{x \to 2} \frac{1}{x^2}$

Body:
I'm trying to compute the following limit whilst using the basic limit theorems, although I'm slightly rusty and would appreciate you honest feedback.

$\lim_{x \to 2} \frac{1}{x^2}$

Given that when $|x-2| < \delta$

Then $\frac{1}{x^2}=\frac{1}{4}$

This is my reasoning thus, far, and from this I proceeded with:

$\lim_{x \to 2} \frac{1}{x^2} = \frac{1}{4}$

$|\frac{1}{x^2}-\frac{1}{4}|<\epsilon$

Though i'm not sure where to go next.As for the basic limit theorems, I tried approaching this with:

$\lim_{x \to p} \frac{f(x)}{g(x)}=\frac{A}{B}$


答:
Normal to the plane P is $\vec p=\vec i+ 2\vec j+ 3\vec k$
Normal to plane Q is $\vec q=\vec i-3 \vec j+\vec k$
The direction vector for the line of intersection is $\vec L=\vec p \times \vec q= 11 \vec i+2 \vec j-5 \vec k.$
For a common point in two plane, let $z=t$
$$x+2y=4-3t~ \& x-3y=1-t \implies y=(3-2t)/5, x=(14-11t)/5$$ Let $t=-1$, the common point is $(5,1,-1)$, the Eq. og line is
$$\frac{x-5}{11}=\frac{y-1}{2}=\frac{z+1}{-5}=s.$$
The parametric Eq. of line is
$$x=11s+5, y=2s+1, z=-5s-1, s \in \Re$$

EDIT  $\vec M. \vec L=0$ and also $\vec M. \vec p=0$
$$\implies \vec M= \vec L \times \vec p=2(8\vec i-19 \vec j + 10\vec k).$$ One can again use the point $(5,1,-1)$ to write the equation of M as
$$\frac{x-5}{8}=\frac{y-1}{-19}=\frac{z+1}{10}=u, u \in \Re $$

Title: Vectors and Planes with parametric equations

Body:
Any help would be appreciated!

Find a parametric equation of a line $L$ that is obtained as the intersection of the planes $P$ and $Q$ with scalar equations $$P: x+2y+3z=4$$ and $$Q: x-3y+z=1$$

Related to this is the question:

Give a parametric equation of a line $M$ contained in $Q$ and orthogonal to $L$.

Note: This is from a Calculus course so any advice involving linear algebra may not be of much help!