问:
Pure Bayesian regression approach
Based on my understading estimating the Maximum a posteriori (MAP) for the regression coefficients stil heavily relies on a 'frequentist approach'. This because I'm simply assming a Gaussian prior $p(\textbf{w}|\alpha) = \mathcal{N}(\textbf{w}|\textbf{0},\alpha^{-1}I)$ and then using Bayes theorem which states that

$$p(\textbf{w}| \textbf{X,t}) \propto p(\textbf{t,X}|\textbf{w})p(\textbf{w}|\alpha)$$

Where on the right hand side I have the prior and likelihood function.

Note: On many books I find the notation $p(\textbf{t}|\textbf{X,w})$ for the likelihood but that sounds a bit confusing to me because the idea is that I'm considering a function of parameters $\textbf{w}$ that is conditional on a dataset $\mathcal{D}$ so I also find the notation $p(\mathcal{D}|\textbf{w}$) that I like much more because for me $\mathcal{D} = \{\textbf{X,t}\}$ but then when it comes to consider inputs and targets the notation becomes the one I initially wrote on top and this confuses me a bit sometimes.. why this? END NOTE

Estimating the MAP still relies on frequentist approach because I'm still going to do Maximum Likelihood this time on likelihood multiplied by prior and then maximizing so I still find a 'punctual' value for the coefficients.

For a 'Pure Bayesian' treatment, I consider the parameters $\textbf{w}$ as random variables and I want to find the predictive distribution directly from the dataset. So basically I'm reading that:

\begin{align}
p(t|\textbf{x};\textbf{X,t}) =& \int p(t,\textbf{w}|\textbf{x};\textbf{X,t}) \, d\textbf{w}\\
=& \int p(t|\textbf{w,x};\textbf{X,t})p(\textbf{w}|\textbf{x};\textbf{X,t}) \,d\textbf{w}\\
=& \int p(t|\textbf{w,x})p(\textbf{w}|\textbf{X,t})d\textbf{w}.
\end{align}

I would like to understand this last chain of equalities: first of all, what is the quantity $p(t|\textbf{x};\textbf{X,t})$ supposed to represent? It is not a likelihood function because it does not depend on parameters $\textbf{w}$. Also why are we computing these transformations inside the probability distributions?

答:
I don't have enough reputation to comment on your post to request clarification as there exist some quantities you haven't defined. However, I am inferring from the notational conventions you have adopted that you are operating in the area of statistics/machine learning. On the queries you have outlined, it's best perhaps to clarify the likelihood function in this setting, MAP estimation and then the predictive distribution, highlighting aspects of your query as I proceed.

Context.

I am going to follow the notational conventions of the book Pattern Recognition and Machine Learning by Christopher Bishop. It should be remarked that the notation used in this book explicitly supports Bayesian interpretations of statistics, rather than frequentist interpretations.

Consider a training dataset $\{\mathbf{x}_n, t_n \}$ for $n = 1, ... ,N$ consisting of $N$ IID input vectors $\mathbf{x}_i$ and $N$ IID scalar target variables $t_n$. We stack the $N$ input vectors into a matrix $\mathbf{X}$, and the scalar target variables into a vector $\mathbf{t}$. Denote the parameter vector $\mathbf{w}$ which consists of
regression coefficients.

We assume that we can model the target variable $t_n$ with the deterministic function $y(\mathbf{x}_n, \mathbf{w})$ with additive Gaussian noise (i.e. $\epsilon \sim \mathcal{N}(0, \beta^{-1})$ where the precision/inverse variance is $\beta$):

$$t_n = y(\mathbf{x}_n; \mathbf{w}) + \epsilon = \mathbf{w}^T\mathbf{x}_n + \epsilon$$

Due to the assumption of additive Gaussian noise, the likelihood of a single $t_n$ will be $p(t_n | \mathbf{x}, \mathbf{w}, \beta) = \mathcal{N}(t_n | y(\mathbf{x}_i, \mathbf{w}), \mathbf{\beta}^{-1})$.

Emphasising that the target variables are multiple observations of a single target variable, rather than a single observation of a multivariate target vector, encoded in the independence of the scalar target variables $t_n$, we have the following likelihood function on $\mathbf{t}$:

$$p(\mathbf{t} | \mathbf{X}, \mathbf{w}, \beta) = \prod^N_{n=1} \mathcal{N}(t_n | \mathbf{w}^T \mathbf{x}_n, \beta^{-1})$$

Now we define a Gaussian prior on the parameter vector $\mathbf{w}$, so we have $p(\mathbf{w} | \alpha) = \mathcal{N}(\mathbf{w} | \mathbf{0}, \alpha^{-1}I)$.

The likelihood function.

Now on the query concerning whether the likelihood should be $p(\mathcal{D} | \mathbf{w}) = p(\mathbf{t, X} | \mathbf{w})$ or whether it should be $p(\mathbf{t} | \mathbf{X}, \mathbf{w})$.

I am fairly certain that is the latter rather than the former, through consideration of the context of the linear regression in statistics/supervised learning in machine learning. That is, we are interested in modelling the distribution of the target variables $\mathbf{t}$ using the information contained in the regressors $\mathbf{X}$ (i.e. conditioning on these variables), and on a suitable value of the parameter $\mathbf{w}$. As we have assumed that the $\mathbf{t}$ are a function of the $\mathbf{X}$ that is linear in the parameter $\mathbf{w}$, and additive Gaussian noise, the likelihood function $p(\mathbf{t} | \mathbf{X}, \mathbf{w})$ arises naturally from this assumption. So to clarify, the likelihood $p(\mathbf{t} | \mathbf{X}, \mathbf{w})$ is both a function of the input matrix $\mathbf{X}$ and of the parameter $\mathbf{w}$, and the latter remains to be estimated according to a procedure with desirable statistical properties.

Now I think the confusion possibly arises from the fact that you using the notation $\mathcal{D} = \{\mathbf{t}, \mathbf{X} \}$ to refer to the "dataset", which is an acceptable form of notation, but it runs the risk of masking the distinction, crucial to the linear regression/supervised learning context, between the parts of the dataset $\mathcal{D}$ you are seeking model i.e. $\mathbf{t}$, and the information you conditioning on, the input variables/regressors, $\mathbf{X}$.

Hence your specification of the posterior distribution in this setting should read:

$$p(\mathbf{w} | \mathbf{t}, \mathbf{X}, \alpha, \beta) \propto p(\mathbf{t} | \mathbf{X}, \mathbf{w}, \beta) p(\mathbf{w} | \alpha)$$

And in Bishop you will find he drops the $\alpha$ and $\beta$ for notational brevity, and also the input matrix $\mathbf{X}$. The latter is because in supervised learning/discriminative modelling in machine learning we are not interested in modelling the distribution of the input variables $\mathbf{X}$. Yielding:

$$p(\mathbf{w} | \mathbf{t}) \propto p(\mathbf{t} | \mathbf{w}) p( \mathbf{w})$$

MAP estimation.

You are correct concerning MAP estimation being frequentist. Even though we take the Bayesian route and assume that the parameter $\mathbf{w}$ is a random variable, and we compute a posterior distribution on $\mathbf{w}$, the MAP estimator is a point estimator. That is we select a value of $\mathbf{w}$ so as to maximise the posterior $p(\mathbf{w} | \mathbf{t}, \mathbf{X}, \alpha, \beta)$:

$$\mathbf{w}_{MAP} = \underset{\mathbf{w}}{\text{argmax}} \space p(\mathbf{w} | \mathbf{t}, \mathbf{X}, \alpha, \beta) = \underset{\mathbf{w}}{\text{argmax}} \space p(\mathbf{t} | \mathbf{X}, \mathbf{w}, \beta) p(\mathbf{w} | \alpha)$$

And yes you are correct concerning the distinction between MAP estimation and maximum likelihood estimation. In the latter case you maximise the likelihood $p(\mathbf{t} | \mathbf{X}, \mathbf{w}, \beta)$. However, in doing MAP estimation, you are estimating the parameter $\mathbf{w}$ under a different model where you have added $L_2$/Tikhonov regularisation, known as ridge regression. See references at the bottom for further info.

Predictive distribution.

In order to understand what is going on here, you need to make a distinction between the notation for the test-set, and the training set. So modifying your notation to denote a new input vector belonging to the test set as $\mathbf{x}_{0}$ we are interested in making a prediction of the target variable, which we denote $t_0$ (this is unknown).

Now via a marginalisation argument where we integrate out the parameter $\mathbf{w}$, the predictive distribution is:

$$\begin{align}
p(t_0 | \mathbf{x}_0, \mathbf{t}, \mathbf{X}, \alpha, \beta) &= \int p(t_0, \mathbf{w} | \mathbf{t}, \mathbf{x}_0, \mathbf{X}, \alpha, \beta) d\mathbf{w} \\
&= \int p(t_0 | \mathbf{w}, \mathbf{t}, \mathbf{x}_0, \mathbf{X}, \beta) p(\mathbf{w} | \mathbf{x}_0, \mathbf{t}, \mathbf{X}, \alpha, \beta) d\mathbf{w} \\
&= \int \underbrace{p(t_0 | \mathbf{w}, \mathbf{x}_0, \beta)}_{\text{likelihood}} \underbrace{p(\mathbf{w} | \mathbf{t}, \mathbf{X}, \alpha, \beta)}_{\text{posterior}} d\mathbf{w}
\end{align}$$

Similar to above and in Bishop, you can drop all the $\alpha$ and $\beta$ for notational brevity. In going from the 1st to the 2nd equality we just probability rules. In going from the 2nd to the 3rd equality we use conditional independence:

$$p(t_0 | \mathbf{w}, \mathbf{t}, \mathbf{x}_0, \mathbf{X}, \beta) = p(t_0 | \mathbf{w}, \mathbf{x}_0, \beta)$$

$$p(\mathbf{w} | \mathbf{x}_0, \mathbf{t}, \mathbf{X}, \alpha, \beta) = p(\mathbf{w} | \mathbf{t}, \mathbf{X}, \alpha, \beta)$$

If you aren't comfortable with why we can invoke conditional independence just say in the comments and I will reply, as this post is getting too long for my liking. Notice that the two terms on the right above are just the likelihood and posterior we computed previously.

Concerning your query on what the predictive distribution $p(t_0 | \mathbf{x}_0, \mathbf{t}, \mathbf{X}, \alpha, \beta)$ is supposed to represent.

It is just the product of the likelihood and posterior, with the parameter $\mathbf{w}$ integrated out. Intuitively, you are evaluating the likelihood of a value $t_0$ given $\mathbf{x}_0$ for a particular $\mathbf{w}$. Then weighting that likelihood by your current belief about the parameter $\mathbf{w}$ given data $\mathcal{D} = \{\mathbf{t}, \mathbf{X} \}$. Then integrating over all possible values $\mathbf{w}$ can take.

This not a fully Bayesian approach, but more Bayesian than say attempting to predict $t_0$ using the MAP point estimator $\mathbf{w}_{\text{MAP}}$ and computing $t_0 \approx \mathbf{w}_{\text{MAP}}^T\mathbf{x}_0$.

References

For appropriate references on Bayesian linear regression, look at Chapter 3 of "Pattern Recognition and Machine Learning" by Christopher Bishop (from which I have used notation, although some of the arguments I've made above aren't fully explained). And there is a good table on how Bayesian one can get on p175 of "Machine Learning: A Probabilistic Perspective" by Kevin Murphy. There is also very clear exposition from John Paisley which is excellent for the link between MAP estimation, ridge regression, and Bayesian linear regression here - see lecture 5 in particular . It's free to enroll so if you sign up you can download all the course materials for free.

Title: Pure Bayesian regression approach


问:
Proof that $x^2$ is not uniformly continuous on $\mathbb R$
This is my attempt. My instructor said it was incorrect. I would like to know exactly where my mistake is.

Suppose $x^2$ is uniformly continuous on $\mathbb R$.

So, $\forall \epsilon > 0 \ \exists \delta > 0 \ \forall x,y \in \mathbb R \ s.t. \ if \ |x - y| < \delta \ \text{then} \ |f(x) - f(y)| < \epsilon$

In this case, $|x - y| < \delta \ \text{and} \ |x^2 - y^2| < \epsilon$

$|(x - y)(x + y)| < \epsilon$

$|(x - y)| |(x + y)| < \epsilon$

$\delta|x + y| < \epsilon$

Notice, $|x + y|$ is unbounded on the real line. So $x$ and $y$ can be made arbitrarily large. So there will not always be a $\delta$ that will make all values of $x$ and $y$ less than an arbitrarily small $\epsilon$.

Therefore, $x^2$ is not uniformly continuous on $\mathbb R$.

答:
Your intuition as to why it's not uniformly continuous is correct, but your proof doesn't make any sense. Start with the definition of not uniformly continuous, which is what you're trying to prove.

There exists $\epsilon_0>0$ such that for all $\delta>0$ there exists $x,y$ with $|x-y|<\delta$, but $|f(x)-f(y)|\geq \epsilon_0$.

In this case, just choose $\epsilon_0=1$. Then for every $\delta$, try to come up with $x,y$ such that $|x-y|<\delta$, but $|x-y||x+y|\geq 1$. Let $x$ be arbitrary and pick $y=x+\delta/2$. Then $|x-y|=\delta/2<\delta$ is satisfied and
$$
|x-y||x+y|=\frac{\delta}{2}\left(2x+\frac{\delta}{2}\right).
$$
If $x$ is sufficiently large....

ContentLicense: CC BY-SA 4.0


问:
Fun with Powerball history. What probability theories apply?
I wanted to play around with probability theories and see if I find any statistical inferences from the Powerball lottery history going back to 11/1/1997.  What theories could I apply besides simple linear and multiple regression analysis?

I have broken the numbers down based on Wikipedia description of the game with a simple console program.

答:
Studying the statistical properties of the Powerball draws is likely not going to turn up anything interesting. I wouldn't play Powerball, but I would bet on the fact that the numbers are uniformly drawn.

However, there might be something interesting to learn about how people play Powerball. If one knew about how people tend to select numbers, it would be possible to take advantage of this by playing combinations that are less likely to end in a shared prize. The ideal data-set for this would be the actual combinations selected by people.

However, short of this, it would be possible to extract some information about patterns-of-play using the actual draws from the data set you have provided along with any data you can get on number of matches, prizes paid, etc. (tickets sold or at least jackpot size would be useful here as well).

ContentLicense: CC BY-SA 3.0


问:
Doubts in question and solution about Jordan measurable sets and sets that has measure zero. Would you help me?
QUESTION a) Let $f:A\rightarrow \mathbb{R}$ be a integrable function defined in a block $A\subset\mathbb{R}^m$. If $A_0\subset\mathbb{R}^m$ is a block such that $A\subset A_0$, then the function $\overline{f}:A_0\rightarrow \mathbb{R}$ given by

\begin{equation*}
    \overline{f}(x) = \begin{cases}
              f(x) & \text{if } x \in A,\\
               0 & \text{if } x \in A_0\backslash A.
          \end{cases}
\end{equation*}
is integrable (in $A_0$) and $$\int_{A_0}\overline{f}(x) dx= \int_{A} f(x) dx$$

QUESTION b) Let $A=\displaystyle\Pi_{i=1}^{n}[a_i, b_i]$ be a block in $\mathbb{R}^m$, such that $A \subset A_0$, then $$\int_{A_0}\chi_{{}_{A}} dx = \Pi_{i=1}^{m} (b_i-a_i)$$

MY ATTEMPTY:

a) Because $A\subset A_0$ let's divide in two cases:

Supposing  that $A_0\cap A=A$ by definition of $f$ we have $$\int_{A_0} \overline{f}(x) dx = \int_{A} f(x) dx.$$

Now, in the set $A_0\backslash A$ we have $\overline{f}(x) = f(x) =0$. Considering $A_0\backslash A \subset \displaystyle\bigcup_{i=1}^{\infty} C_i$, where $C_i's$ are open blocks, by the Lindelöf's theorem has a countable subcover, let's define it as $A_0\backslash A\subset \displaystyle\bigcup_{i=1}^{k} A_i =D$, however all countable set has zero measure (because is a set of points which has zero measure). Then we can write $\{x \in A_0\backslash A; f(x)=0\}= D$. Thus $$\int_{A_0}\overline{f}(x) dx = \int_{A} f(x) dx.$$

b) Supposing $A\subset A_0$, let's divide in two cases:

if $\overline{A_0}=\partial A$, so because $\partial A$ is defined as an edge of the block $A=\Pi_{i=1}^{n}[a_i, b_i]\subset \mathbb{R}^m$, then by definition has zero measure. Thus in this case, $$\int_{\overline{A_0}} \chi_{{}_{A}}(x) dx = \int_A \chi_{{}_{A}}(x) dx= \Pi_{i=1}^{m}(b_i-a_i).$$

if $A$ is a proper subset of $A_0$, then we have by definition of characteristic function that $\chi_{{}_{A_{0}\backslash A}}(x) = 0$. Therefore, by the question one, we have $$\int_{A_0} \chi{{}_{A}} dx = \int_{A} \chi{{}_{A}} dx = \Pi_{i=1}^{m}(b_i-a_i).$$

MY DOUBTS: Are my answers right? Would you correct for me or help me to improve it, please?

答: Your proof is fine, but it might be simpler just to observe that when you take upper and lower sums for the Jordan measure of a set, it's the same as taking upper and lower Darboux sums for a Riemann integral, so your integrals are simply Riemann integrals over cubes. And since it's given that $f$ is integrable, and $\chi_{A}$ certainly is, then $g:=\overline f\cdot \chi_{A}$ is also integrable, and this gives $a).$ For $b).\ $ we go the other way, that is, integrals of the constant function $f=1$ over cubes give Jordan content, so we have $\int_{A_0} \chi{{}_{A}} dx=\int_{A} dx=\operatorname{vol} A=\Pi_{i=1}^{m}(b_i-a_i).$


问:
How to solve the differential equation $yy' =-x$?
I just started learning about Differential Equations and there's this equation that I had problems with:

$$yy' =-x$$

I know that it's a separable ODE and that I can integrate both sides. The right side is then going to be $-\frac{x^2}{2}+c$ but I'm not sure how to go about the left side. I know that the solution is $\frac{y^2}{2} = -\frac{x^2}{2}+c$ but why is the left side equal to that? Usually $\frac{y^2}{2}$ is the anti-derivative of $y$ and not $yy'$. What am I missing here ?

Thanks.

答:
hint

You can easily check that The antiderivative of

$\sin(x)$ is not $\frac{\sin^2(x)}{2}$

But, you can be sure that the antiderivative of

$\sin(x)\cos(x) $ is $\frac{\sin^2(x)}{2} $.

do not forget to add a constant $ C$.

$$\int x.1.dx =\frac{x^2}{2}$$
and by the same
$$\int y(x).y'(x)dx=\frac{y^2(x)}{2}$$


问:
which of the following statements are true in ring theory
which of the following statements are true in ring theory?

(a) Let $R$ is a commutative ring with unity and $I$ be an ideal. Then $R/I$ is an integral domain.

  (b) If $R$ is a commutative ring with unity the units of $R[x]$ are the units in $R$.

(a) I think it is not true as $R/I$ is an integral domain iff I is an prime ideal but I have no counter example.

(b) I have no idea.

can I get some help.Thanks for your cooperation.

答:
(a) Which ideals are prime in $\mathbb Z$?

(b) Hint: $(1-aX)(1+aX)=1-a^2X^2$.

What happens if $R$ is a ring and $a \in R$ is such that $a^2=0$? Can you think of any such ring?


问:
Trying to understand a double summation formula
I am trying to understand a double summation mentioned in this paper: https://doi.org/10.1016/0024-3795(95)00696-6. How can we prove that this is true?

$$\sum_{l = 1}^n\;\; \sum_{k = l}^{n+l} = \sum_{k \,=\, 1}^n \;\; \sum_{l \,=\, 1}^{k} \,+\, \sum_{k \,=\, n+1}^{2n}\;\; \sum_{l \,=\, k-n}^{n}$$

答:
One way  to handle these changes of variables in a more formal way is to re-write the summation in terms of indicator variables on unrestricted sums, so that interchanging sums is as simple as switching the order. Then we need to manipulate the inequalities defining the indicators until we get the standard form of a sum.

Define $1_P$ to be $1$ when the proposition $P$ is true, otherwise $0$. If we're given variables $a_{l,k}$ with $1 \leq l,k \leq 2n$, we extend $a_{l,k}$ to be $0$ for $l,k$ outside of this range, and use $\sum_l$ to mean $\sum_{l=-\infty}^\infty$.    Then we have

\begin{align*}
\sum_{l = 1}^n\;\; \sum_{k = l}^{n+l} a_{l,k} &= \sum_l  \sum_{k}  1_{1 \leq l \leq n, \, l \leq k \leq n+l}  a_{l,k}\\
&= \sum_k  \sum_{l}  1_{1 \leq l \leq n, \, l \leq k \leq n+l}  a_{l,k}\\
&= \sum_{k=1}^{2n}  \sum_{l}  1_{1 \leq l \leq n, \, l \leq k, k \leq n+l}  a_{l,k}\\
&= \sum_{k=1}^{2n}  \sum_{l}  1_{1 \leq l \leq n, \, l \leq k, k-n \leq l  }  a_{l,k}\\
&= \sum_{k=1}^{2n}  \sum_{l}  1_{1 \leq l \leq n, \, k-n \leq l \leq k}  a_{l,k}\\
&= \sum_{k=1}^{2n}  \sum_{l=\max(k-n,1)}^{\min(k,n)} a_{l,k}\\
&= \sum_{k=1}^{n}  \sum_{l=\max(k-n,1)}^{\min(k,n)} a_{l,k} + \sum_{k=n+1}^{2n}  \sum_{l=\max(k-n,1)}^{\min(k,n)} a_{l,k} \\
&= \sum_{k=1}^{n}  \sum_{l=1}^{k} a_{l,k} + \sum_{k=n+1}^{2n}  \sum_{l=k-n}^{n} a_{l,k}. \\
\end{align*}


问:
Is this space normed or not?
This particular question was part of my analysis assignment I was unable to prove / disprove it.

Let $V$ denote the vector space of all sequences $a=(a_1 , a_2,...,a_n,...)$ of real numbers such that $\sum_{n\geq 1} 2^n |a_n|$ converges. Define ||.||: $V \to \mathbb{R}$ by $||a|| =\sum_{n\geq 1}2^n |a_n|$. Then is $V$ complete normed space or not.

There were four questions to be answered regarding $V$ . I have done 3 but unable to do this. I know about completely normed space but I was unable to do this . The  idea is to show that every sequence of sequence converges to a limit and that limit is to be constructed or guessed but I am unable to do it

Kindly guide on how this should be done

答:
Define a measure space as $(\mathbb N,\mu),$ where for any subset $E\subset \mathbb N,$

$$\mu(E)=\sum_{n\in E} 2^n.$$

Then $L^1(\mathbb N,\mu)$ is precisely the set $V$ you describe, with

$$\|a\|=\int_{\mathbb N} |a|\,d\mu =\|a\|_{L^1(\mathbb N,\mu)}.$$

Since every $L^1$ space is a complete normed linear space, we are done.

Of course, if you haven't studied measure theory, the above is not too useful for this problem. But it's still worth knowing.

Title: Is this space normed or not?


答:
It's true that:
$$\dfrac {dy^2}{\color {red}{dy}}=2y$$
But when you differentiate $y^2/2$ with respect to the variable $x$ you get:
$$\dfrac 12\dfrac {dy^2}{\color {red}{dx}}=\dfrac 12\dfrac {dy^2}{dy}\dfrac{dy}{dx}=y\dfrac {dy}{dx}=yy'$$
And you have also:
$$ \int yy'dx=\int y\dfrac {dy}{dx}dx=\int ydy= \dfrac {y^2}2+C$$

Title: How to solve the differential equation $yy' =-x$?

Body:
I just started learning about Differential Equations and there's this equation that I had problems with:

$$yy' =-x$$

I know that it's a separable ODE and that I can integrate both sides. The right side is then going to be $-\frac{x^2}{2}+c$ but I'm not sure how to go about the left side. I know that the solution is $\frac{y^2}{2} = -\frac{x^2}{2}+c$ but why is the left side equal to that? Usually $\frac{y^2}{2}$ is the anti-derivative of $y$ and not $yy'$. What am I missing here ?

Thanks.


答:
$$y y' = -x$$
$$y \frac{dy}{dx} = -x$$
$$y dy = -x dx$$
This is what it means to separate it. Now only terms involving $y$ are on the left, and terms involving $x$ are on the right. So integrate...
$$y^2/2 + C_1 = - x^2/2 + C_2$$
$$y^2 = -x^2 + C$$
$$y = \pm \sqrt{C-x^2}$$

As another user pointed out, you can also just recognize that $$(y^2)' = 2y y'$$


答:
(a): Think of a ring $R$ that is not an integral domain, and let $I$ be the zero ideal, i.e. $I=\{0\}$.

(b): Remember, given a ring $S$, we say that an element $y\in S$ is a unit when there is some $z\in S$ such that $yz=1_S$ (the symbol $1_S$ denotes the multiplicative identity element of $S$).

The polynomial ring $R[x]$ consists of polynomials whose coefficients are elements of $R$:
$$a_0+a_1x+a_2x^2+\cdots+a_nx^n,$$
where $n$ can be any integer greater than $0$, and each of the $a_i$'s come from $R$. They are added and multiplied like any other polynomials you have worked with before.

Clearly, if $a\in R$ is a unit (so that there is some $b\in R$ such that $ab=1_R$), then the polynomial
$$a+0x+0x^2+\cdots$$
is a unit in $R[x]$, because
$$\begin{align*}
(a+0x+0x^2+\cdots)(b+0x+0x^2+\cdots)&=ab+(a0+0b)x+(a0+0+0b)x^2+\cdots\\
&=ab+0x+0x^2+\cdots\\
&=1_R+0x+0x^2+\cdots
\end{align*}$$
and $1_R+0x+0x^2+\cdots$ is the mutliplicative identity of $R[x]$. (Usually, we don't bother writing all of these $0x^i$ terms, and just think of elements of $R$ as also being elements of $R[x]$. These elements of $R[x]$ are called constant polynomials.)

The question is whether there can be units of $R[x]$, other than the units of $R$ themselves. Thus, we're looking for non-constant polynomials (i.e., polynomials with some $x$'s in them) that can be multiplied by other polynomials to produce $1$. For simplicity, you might first take a look at linear polynomials, i.e. polynomials with an $x$ term, but no $x^2,x^3,\ldots$ terms. Of course, there's no guarantee that if there are non-constant polynomials that are units, that any of them will be linear; but this is a good place to start.

So, let's say we have our ring $R$, and elements $a,b\in R$ (with $b\neq 0$, so that we're looking at a non-constant polynomial) such that the polynomial $a+bx\in R[x]$ is a unit. In fact, let's make an even bigger simplifying step: assume that $a+bx$ is its own inverse, or in other words, that
$$(a+bx)^2=a^2+2abx+b^2x^2=1.$$
(Again, we don't know that even if there are any polynomials that are units, and are linear, that they would do this; but again, it's a first thing to look at. Remember, mathematicians are lazy.)

This can only happen if $2ab=0$ and $b^2=0$ in $R$. Can you think of a ring $R$ and $a,b\in R$ satisfying these relations?

Body:
which of the following statements are true in ring theory?

(a) Let $R$ is a commutative ring with unity and $I$ be an ideal. Then $R/I$ is an integral domain.

  (b) If $R$ is a commutative ring with unity the units of $R[x]$ are the units in $R$.

(a) I think it is not true as $R/I$ is an integral domain iff I is an prime ideal but I have no counter example.

(b) I have no idea.

can I get some help.Thanks for your cooperation.


问:
Find the canonical Jordan form of $T\left(p(x))=p(x)^{\prime}+2 p(x)\right.$
In $P_{2}(R),$ let be $T: P_{2}(R) \rightarrow P_{2}(R),$ the operator given by  $T\left(p(x)=p(x)^{\prime}+2 p(x)\right.$

Find the canonical Jordan form  $J$ for $T$. and find a basis of vectors  $P_{2}(R)$ where the associated matrix to  $T$ is the matrix $J$.

First I find the matrix associated to T it gives me

$M=\begin{bmatrix} 2 & 1 & 0 \\0 & 2 & 2 \\ 0 & 0 & 2\end{bmatrix}$ in the canonical basis ${1,x,x^2}$

Then its eigenvalues are 2 with multiplicity 3,

and then the Jordan form is
$J=\begin{bmatrix} 2 & 1 & 0 \\0 & 2 & 1 \\ 0 &0  & 2\end{bmatrix}$

and I find the $Ker(M-2I)$  and it gives me $gen={t(1,0,0)}$

and now im stuck finding the other vector to from a basis

Body:
Find a nonzero vector $w \in \Bbb{C}^3 \setminus \ker(M-2I)^2$. Now $$\{u,v,w\}=\{(M-2I)^2w, (M-2I)w, w\}$$
is a basis for $\Bbb{C}^3$ and we have
$$Mu = M(M-2I)^2w = (M-2I)^3w + 2(M-2I)^2w = 2u$$
$$Mv = M(M-2I)w = (M-2I)^2w + 2(M-2I)w = u+2v$$
$$Mw = (M-2I)w+2w =v+2w$$

so it is a Jordan basis for $M$. Picking $w = (0,0,1)$ we get the basis
$$\{u,v,w\} = \{(2,0,0),(0,2,0),(0,0,1)\} = \{2,2x,x^2\}.$$


答:
You have to take some care when you are dealing with multiple variables which may be dependent on each other. Here $y'=\frac {dy}{dx}$ is a derivative with respect to $x$ and you are taking the antiderivative with respect to the variable $x$ and not the variable $y$.

You need to take the same antiderivative on both sides of your equation. There are a number of ways of saying this, and a number of ways of using different notations for the same thing, so it is easy to get confused. But with any equation "do the same to both sides" is the basic rule (and even then there are some operations - eg taking square roots - which require care). Here the key is identifying what is "the same thing" that you are doing.

If you differentiate $y^2$ with respect to $x$ you will find that you get $2yy'$, while if you differentiate with respect to $y$ you get $2y$. Likewise taking the antiderivative is different in the two cases.


问:
Is there any way to bound the integral with mean value theorem?
For small $s$, I try to find some coefficient $M=M(s)$ where for any smooth real function $f$

$$\int_0^1 (f(t+s)-f(t))dt=\int_0^1 s\cdot f'(\eta(t))dt\leq M\int_0^1 f'(t)dt$$

. In here, $\eta(t)$ is the real number satisfying

$$\frac{f(t+s)-f(t)}{s}=f'(\eta(t))$$
. I try to make some coefficient $M$, but it is hard to find. Could anyone help me?

答:
It's clear there can be no such inequality, because the first integral involves values of $f(x)$ for $x\notin(0,1)$. And it's clear the upper bound should be $M\int_0^1|f'(x)|\,dx$, since $M\int_0^1f'(x)\,dx=M(f(1)-f(0))$.

Perhaps (for $s>0$) we can get $\int_0^{1+s}(f(s+t)-f(t))\,dt\le M\int_0^1|f'(t)|\,dt$.

Using FTC seems more likely to work than MVT. $$\begin{align}\int_0^1(f(t+s)-f(t))\,dt&=\int_0^1\int_t^{t+s}f'(x)\,dx\,dt.
\end{align}$$

If you apply Fubini here you get three terms, due to the shape of the region. In each term the inner integral is trivial, of the form $\int_a^b f'(x)\,dt=(b-a)f'(x)$; put it together and you get $$\left|\int_0^1(f(t+s)-f(t))\,dt\right|\le s\int_0^{1+s}|f'(x)|\,dx.$$


问:
Show that the Fourier series for $f$ doesn't converge uniformly on $[-\pi,\pi]$.
Let $\quad f(x)=\begin{cases}-1,&-\pi<x<0\\{}\\1,&0<x<\pi\end{cases}$

Show that the Fourier series for $f$ doesn't converge uniformly on $[-\pi,\pi]$.

My approach:

The Fourier series for $f$ is: $S_n(f)(x)=\frac{1}{\pi}\left[\int_{0}^{\pi}D_n(t-x)dt-\int_{-\pi}^{0}D_n(t-x)dt\right]$. Where $D_n$ is the Dirichlet kernel defined like $\quad D_n(t)=\frac{sin(n+1/2)t}{2sin(t/2)}, t\in[-\pi,0)\cup(0,\pi]$

I tried to suppose that the series converges and get to a contradiction without success. Any suggestions would be great!

答: Every uniformly convergent Fourier series has a continuous sum, but $f$ is not continuous.


问:
Can a discrete random variable be manipulated just like a algebraic variable?
Can the discrete random variable X in  P(-2X>10Y) be manipulated into P(X<-5Y)

And furthermore can this be done to any other operations on X

答:
Indeed, you can do this. This is clear when you go back to the definition. If $X$ and $Y$ are defined on some probability space $\Omega$, that is,
$$X,Y:\Omega\to\mathbb R,$$ then
$$\mathbb P\{-2X>10Y\}:=\mathbb P\{\omega\in\Omega\mid-2X(\omega)>10Y(\omega)\}.$$

Clearly, conditions in a set can be transformed into equivalent conditions, so that we obtain:
$$\mathbb P\{-2X>10Y\}=\mathbb P\{\omega\in\Omega\mid X(\omega)<-5Y(\omega)\}.$$

As before,
$$\mathbb P\{X<-5Y\}:=\mathbb P\{\omega\in\Omega\mid X(\omega)<-5Y(\omega)\},$$
so that finally
$$\mathbb P\{-2X>10Y\}=\mathbb P\{\omega\in\Omega\mid X(\omega)<-5Y(\omega)\}.$$

This can of course also be done with other kinds of operations.


问:
A complex number problem of equation ${z}^4 + {z}^3 + {2z}^2 + {z} + 1 = 0$
If $z$ be a complex number satisfying ${z}^4 + {z}^3 + {2z}^2 + {z} + 1 = 0$, then find the value of $|\bar{z}|$

How to you proceed with this problem?

Replacing $z$ with $a + {i}b$ doesn't seem to work, as you just get a huge biquadratic equation. I do recognize that $|\bar{z}| = |z|$, but not sure how to manipulate this fact to simplify the given biquadratic equation

Body:
Note
$${z}^4 + {z}^3 + {2z}^2 + {z} + 1 =( z^4 + z^3 + z^2 )+(z^2 + z+ 1 ) \\= (z^2+1)(z^2 + z+ 1 ) =\frac{  (z^2+1)(z^3-1 ) }{z-1}=0
$$
which leads to $z^2=-1$, or $z^3=1$. As a result, $|\bar z|=1$.


问:
Comparing asymptotics to alternate series test
I am a bit confused when applying asymptotics:

Let's consider $\sum\limits_{k=1}^\infty(-1)^k\frac{1}{k}$. We know that it is convergent (alternate series test). However, if I apply asymptotics I get the following:
$$
(-1)^k\frac{1}{k}= O\left(\frac{1}{k}\right)\implies \sum\limits_{k=1}^\infty(-1)^k\frac{1}{k}=\sum\limits_{k=1}^\infty O\left(\frac{1}{k}\right).
$$
The resulting series is no longer convergent although it has the same asymptotic behavior?! How does this fit together?

答:
Let's say we have two series $$\sum_{k=1}^{\infty}a_{k}\quad \text{and}\quad \sum_{k=1}^{\infty}b_{k}$$
and we know that $$a_{k} = \mathcal{O}(b_{k})\quad \text{as} \quad k\to\infty.$$
Remember that this means that there is some constant $M>0$ and some index $n$ such that $$|a_{k}|\leq M|b_{k}| \quad\text{for all } k\geq n.$$

Now, this is useful for determining convergence of $\sum_{k=1}^{\infty}a_{k}$, but only in the case that we already know that $\sum_{k=1}^{\infty}b_{k}$ converges absolutely.  Indeed, if $\sum_{k=1}^{\infty}b_{k}$ converges then we can say that
$$\sum_{k=1}^{\infty}|a_{k}| = \sum_{k=1}^{n-1}|a_{k}| + \sum_{k=n}^{\infty}|a_{k}| \leq \sum_{k=1}^{n-1}|a_{k}| + M\sum_{k=n}^{\infty}|b_{k}| $$ which converges because $\sum_{k=n}^{\infty}|b_{k}|$ is the tail of a convergent series.  However, if $\sum_{k=1}^{\infty}|b_{k}|$ diverges, then all we've done is bound $\sum_{k=1}^{\infty}a_{k}$ above by a divergent series, which we know from the comparison test doesn't tell us anything about the convergence of $\sum_{k=1}^{\infty}a_{k}$.


问:
For each of the following relations, determine whether it is reflexive, symmetric or transitive
a) $X$ is the set of real numbers, $n$ is a natural number
 $$R =  \{(x, y) \mid x, y \in X, x^n = y^n\}.$$

b) $X$ is the set of people in the world
$$R = \{ (x, y) \mid x, y \in X, x\text{ and }y\text{ share a parent}\}$$

I believe both relations are reflexive, symmetric and transitive, aren't they?

I mean, for a) you can have $x^n = x^n \rightarrow (x,x)$, you can have $(x,y)$ and $(y,x) \rightarrow$ symmetric and you can have transitivity by $\leq.$

And for b) it's pretty much the same. As it doesn't say $x$ mustn't be equal to $y$, then you can have $(x,x)$ which is true. You can also have $(y,x) \rightarrow$ symmetry and you can also get transitivity for "if $(x,y)$ is true and $(y,z)$ is true, then $(x,z)$ is true," right?

答:
a) $X$ is the set of real numbers, $n$ is a natural number

$R$ =  {($x$, $y$) | $x, y \in X, x^n = y^n$ }

You need to be careful, as was pointed out, with your phrasing of "can have" which implies "there exists", and your invocation of the $\leq$ relation to address problem (a).

First, reflexivity, symmetry, and transitivity of a relation requires that the properties are true for all elements of the set in question.

And secondly, to establish the transitivity of the relation $R$ in (a), you must demonstrate that the given defined relation is transitive (as was pointed out above, the $\leq$ relation is not the relation for which you need to establish transitivity.)

So, if you are going to argue that the relation given in (a) is transitive, you need to show that for all $x, y, z \in N,$ if $x^2 = y^2$ and $y^2 = z^2$, then it must follow that $x^2 = z^2$.  To prove transitivity, you assume that the "if" conditions hold: Let $x, y, z$ be arbitrary elements of the set of natural numbers. Assume $x^2 = y^2$ and $y^2 = z^2$.  Let $X = x^2, Y = y^2, Z = z^2$.  Then our assumption can be expressed as $X = Y$ and $Y = Z$. Since equality is an equivalence relation, it is transitive, and hence $X = Z$, or $x^2 = z^2.$  Since $x, y, z$ were chosen arbitrarily, it follows that for all $x, y, z \in N,$ we have that $R$ is indeed transitive.


问:
If $R$ is a commutative ring with unity and $R$ has only one maximal ideal then show that the equation $x^2=x$ has only two solutions
If $R$ is a commutative ring with unity and $R$ has only one maximal ideal then show that the equation $x^2=x$ has only two solutions.

I know that $0$ and $1$ are the solutions, but I can't proceed from there.

答: Let $M$ be the maximal ideal of $R.$ Then $R= \{ units \} \cup M.$ Now suppose $x(x-1)=0.$ If $x$ is a unit then multiplying by the inverse gives $x=1.$ If $x-1$ is a unit then multiplying by its inverse gives $x=0.$ Thus if $x\neq 0,1$ is to solve $x(x-1)=0$ then $x$ and $x-1$ are both not units, and hence both in $M.$ But then $ 1= x - (x-1)\in M$ which is nonsense. </p>


问:
When is the system of first order differential equations asymptotically stable?

Given the system $$\mathbf{x}'=A\mathbf{x} $$
where $A=\begin{bmatrix}a &0 & 4 \\ -1 & -1 & 0 \\ -2-a & 0 & -3 \end{bmatrix}$

In what interval of $a$ is the system asymptotically stable, and for what value of $a$ is the system stable/unstable if such a case even exists?

Attempt

For a system to be asymptotically stable the real part of all eigenvalues must be negative. $$\Re(\lambda)<0, \: \: \: \text{for all} \: \: \lambda $$

Since we are dealing with a $3\times3$ matrix, I used maple to find the eigenvalues from the system matrix. I also tried to use Maple to solve the inequality case for the eigenvalues, but I don't think I'm getting correct results.

The results I get from Maple state that the system is asymptotically stable when $$-8 < a \leq5-4\sqrt{3} $$
I spoke with my peers, and they said that this interval of $a$ is wrong. I posted my Maple document below. Can anyone see what is going wrong?

答:
Below I use Maple 2020.1 (the current version).

```<code>restart;

A := <a,-1,-2-a|0,-1,0|4,0,-3>:

elist := convert(LinearAlgebra:-Eigenvalues(A),list):

map(lprint, elist):
   -1
   1/2*a-3/2+1/2*(a^2-10*a-23)^(1/2)
   1/2*a-3/2-1/2*(a^2-10*a-23)^(1/2)

W := map(L->evalc(Re(L)<0),elist):

map(lprint, W):
   -1 < 0
   1/2*a+1/4*abs(a^2-10*a-23)^(1/2)*(1+signum(a^2-10*a-23)) < 3/2
   1/2*a-1/4*abs(a^2-10*a-23)^(1/2)*(1+signum(a^2-10*a-23)) < 3/2

# solving for all three conditions together
solve(W,a);

        {-8 < a, a < 3}

plot(map(Re,elist),a=-9..4,thickness=1,
     view=-2.75..1/4,size=[500,200]);
</code>```

```<code># for fun (only), solve each separately

solve(W[1],a); # true for all `a`

                a

solve(W[2],a);

      RealRange(Open(-8), Open(3))

solve(W[3],a);

      RealRange(-infinity, Open(3))
</code>```

Body:
Below I use Maple 2020.1 (the current version).

restart;

A := <a,-1,-2-a|0,-1,0|4,0,-3>:

elist := convert(LinearAlgebra:-Eigenvalues(A),list):

map(lprint, elist):
   -1
   1/2*a-3/2+1/2*(a^2-10*a-23)^(1/2)
   1/2*a-3/2-1/2*(a^2-10*a-23)^(1/2)

W := map(L->evalc(Re(L)<0),elist):

map(lprint, W):
   -1 < 0
   1/2*a+1/4*abs(a^2-10*a-23)^(1/2)*(1+signum(a^2-10*a-23)) < 3/2
   1/2*a-1/4*abs(a^2-10*a-23)^(1/2)*(1+signum(a^2-10*a-23)) < 3/2

# solving for all three conditions together
solve(W,a);

        {-8 < a, a < 3}

plot(map(Re,elist),a=-9..4,thickness=1,
     view=-2.75..1/4,size=[500,200]);

# for fun (only), solve each separately

solve(W[1],a); # true for all `a`

                a

solve(W[2],a);

      RealRange(Open(-8), Open(3))

solve(W[3],a);

      RealRange(-infinity, Open(3))


问:
Integral using Gamma and Beta function
the integral I have found this integral on past exams papers for a course in Mathematical Methods of Physics. I really cant find any material on the textbook which could be helpfull for the solution. Do you have any idea how to solve this or can you suggest any material I could use? I post the question in the picture bellow. I have to express the solution using Beta and Gamma functions. Any help would be valuable. Thanks in advance!

Thanks for the edit, its the 1st time I post here, sorry if my post was wrong

$$\int_{\mathbb{R}^n}(a+|x|)^m\exp\left(-\lambda\left[\sum_{i=1}^nx_i^2\right]^k\right)d^nx$$
$$x=\mathbb{R}^n,k>0,m\in\mathbb{N},\Re(\lambda)>0$$

答:
I think one thing you can do is use hyperspherical coordinates i.e:
$$r^2=\sum_{i=1}^nx_i^2$$
making:
$$\exp\left(-\lambda\left[\sum_{i=1}^nx_i^2\right]^k\right)=\exp\left(-\lambda r^{2k}\right)$$
$$\left(a+|x|\right)^m=(a+r)^m$$
and now we have have to use the Jacobian to convert from cartesian to hyperspherical:
$$d^nx=r^{n-1}dr\prod_{i=1}^{n-1}\sin^{i-1}(\varphi_i)d\varphi_i$$
now in terms of our domain for each of these, we have:
$$r\in[0,\infty),\,\varphi_{1}\in[0,2\pi),\varphi_{i\ne1}\in[0,\pi]$$
so now if we reconstruct our integral:
$$I_n=\int_{r=0}^\infty\int_{\varphi_1=0}^{2\pi}\int_{\varphi_{i\ne1}\in[0,\pi]^{n-2}}r^{n-1}(a+r)^m\exp\left(-\lambda r^{2k}\right)dr\prod_{i=1}^{n-1}\sin^{i-1}(\varphi_i)d\varphi_i$$
now we can split this up as the function is separable, and we can write it as:
$$I=\left(\int_{r=0}^\infty r^{n-1}(a+r)^m\exp(-\lambda r^{2k})dr\right)\left(\int_{\varphi_1=0}^{2\pi}d\varphi_1\right)\prod_{i=2}^{n-1}\left(\int_{\varphi_i=0}^\pi\sin^{i-1}(\varphi_i)d\varphi_i\right)$$
Now the rest should be fairly easy :)

Note that:
$$\int_0^\pi\sin^{i-1}(\alpha)d\alpha=B\left(\frac12,\frac i2\right)$$
so your integral can be simplified to:
$$I=2\pi\prod_{i=2}^{n-1}B\left(\frac12,\frac n2\right)\int_0^\infty r^{n-1}(a+r)^m\exp(-\lambda r^{2k})dr$$

I think this integral could get quite messy due to the power in the exponential, so the next step I would take is use the fact that:
$$\exp(-\lambda r^{2k})=\sum_{j=0}^\infty(-1)^j\frac{\lambda^j r^{2jk}}{j!}$$
then the integral we have left could be represented in terms of the incomplete beta function

Title: Integral using Gamma and Beta function


问:
Existence of a generic filter $G$ such that $p\in G$ and Axiom of choice
When we prove that "If $M$ is countable, $(P,\le)$ is a partial order and $p\in P$, then there is a $G$ that is $P$-generic over $M$ such that $p\in G$", the axiom of (countable) choice should be used?

If we let $M$ be a transitive set, then we can prove this without using the axiom of countable choice.

But I don't know whether the axiom should be used to prove the general case.

答:
No. The fact that $M$ is countable is already enough, at least assuming that $M$ is a model of [enough] set theory.

The reason here is that it is enough to find a generic filter on the partial order $P\cap M$, which is a countable set. So we can enumerate it, $\{p_n\mid n<\omega\}$ and since $M$ is countable, $\{D\in M\mid D\subseteq P\text{ is dense}\}$ is countable, and we can enumerate it as $\{D_n\mid n<\omega\}$. And now the usual thing works: recursively construct $G$ by taking the least possible condition in $D_n$.

If we do not assume that $M$ is countable, then the axiom of choice might come into play, but there's more finesse to this, and it's best to set this aside for now.


问:
Composing with an epimorphism preserves image in an abelian category
Let
$$
A \overset{\pi}{\twoheadrightarrow}B \overset{f}{\to} C
$$
be maps in an abelian category $\mathcal{A}$, where $\pi$ is an epimorphism. I would like to show that $\operatorname{im}(f\circ \pi) \cong \operatorname{im} f$.

I can prove the result using the Freyd-Mitchell embedding theorem, but I would like a more elementary proof, using the axioms of abelian categories.

My proof using F-M Embedding

It is clearly true for $R$-modules. The embedding theorem tells us that there is a fully faithful, exact functor $F:\mathcal{A} \to R\mathrm{Mod}$ for some ring $R$. Then $F(f\circ \pi) = F(f)\circ F(\pi)$ and by exactness $F(\pi)$ is an epimorphism, so $\operatorname{im} F(f\circ \pi) = \operatorname{im} F(f)$.

Lemma: For any morphism $\alpha:X\to Y$ in $\mathcal{A}$, we have $F(\operatorname{im} \alpha) = \operatorname{im} F(\alpha)$.

Proof: By definition of $\operatorname{im}$ there are morphisms $X \overset{e}{\twoheadrightarrow} \operatorname{im} \alpha \overset{m}{\rightarrowtail}Y$ such that $m$ is mono and $m\circ e = \alpha$. Also, since abelian categories have all equalisers, $e$ is an epimorphism. This is taken by $F$ to a diagram
$$
F(X) \overset{F(e)}{\twoheadrightarrow} F(\operatorname{im}\alpha) \overset{F(m)}{\rightarrowtail} F(Y),
$$
where $F(m)$ is mono and $F(e)$ is epi since $F$ is exact, and $F(m)\circ F(e) = F(m\circ e) = F(\alpha)$. Since $F(m)$ is mono, up to isomorphism we may assume that $F(\operatorname{im} \alpha)$ is a submodule of $F(Y)$ and $F(m):F(\operatorname{im} \alpha) \hookrightarrow F(Y)$ is the inclusion.

Let $y \in \operatorname{im} F(\alpha)$. Then $y = F(\alpha)(x)$ for some $x \in F(X)$. But then $F(e)(x) = F(\alpha)(x) \in F(\operatorname{im}\alpha)$. So $\operatorname{im} F(\alpha) \subseteq F(\operatorname{im}\alpha)$.

Conversely let $y \in F(\operatorname{im}\alpha)$. Then since $F(e)$ is onto, $y = F(e)(x)$ for some $x \in F(X)$, which means that $y = F(\alpha)(x) \in \operatorname{im} F(\alpha)$. So $\operatorname{im} F(\alpha) = F(\operatorname{im}\alpha)$, as required.
$$\tag*{$\blacksquare$}$$

So finally we have that $F(\operatorname{im} (f\circ\pi)) = F(\operatorname{im} f)$, and it follows by full faithfulness of $F$ that $\operatorname{im}(f\circ\pi) = \operatorname{im} f$.

My Question

I find this proof to be pretty inelegant, especially as it uses the sledgehammer of F-M Embedding. Can anybody give me a simpler proof, especially one that just relies on the axioms of abelian categories? Thank you.

答:
The image of $f:B\to C$ is $\ker \newcommand\coker{\operatorname{coker}}\coker f$ (regarded as a subobject of $B$). So we want to show that $\ker \coker f = \ker \coker (f\pi)$ as subobjects of $B$. It suffices therefore to show that $\coker f \cong \coker (f\pi)$.

Now the cokernel of a morphism is the coequalizer of the morphism and $0$, so it suffices to show that in general, coequalizers are not affected by precomposition with epimorphisms.

Suppose we have a diagram $A\twoheadrightarrow B \rightrightarrows C$, where the epimorphism is $\pi$, and the parallel arrows are $f,g$. We want to show that $q:C\to Q$ is a coequalizer of $f$ and $g$ if and only if it is a coequalizer of $f\pi$ and $g\pi$.
This follows from the fact that $hf = hg$ if and only if $hf\pi = hg\pi$, since $\pi$ is an epimorphism. This tells us that we have a natural (in $D$) bijection
$$\{ h:C\to D \mid hf\pi = hg\pi \}\leftrightarrow  \{ h : C\to D \mid hf = hg\},$$
and $q:C\to Q$ is a coequalizer of $f$ and $g$ if and only if $q^*$ induces a bijection between maps $Q\to D$ and the set on the right, and $q$ is a coequalizer of $f\pi$ and $g\pi$ if and only if $q^*$ induces a bijection between maps $Q\to D$ and the set on the left, so this bijection establishes the equivalence.


问:
How do you account for the number of students in a classroom when people go in and leave?
The problem is as follows:

In two classrooms there are the same number of people. For every five
people who leave the first, three leave the second to enter the first
and one more retires home. When there are $50$ people in the first
room, in the second there are $20$ people. How many people were
initially in each room?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{100 people}\\
2.&\textrm{90 people}\\
3.&\textrm{85 people}\\
4.&\textrm{80 people}\\
5.&\textrm{75 people}\\
\end{array}$

I'm confused exactly how to assess this problem?. The part which makes me confused is how to understand the fact that for every five people who leave the first and three enter the first. Does it mean that in total there are those five people plus one who retires home from the second classroom?.

I don't know how to set the equations in this particular problem. Can someone help me with this using a step-by-step approach so I can better understand?.

From the looks of this problem it seems to be that this problem is to be solved using a system of two linear equations, but as mentioned, I don't know how to translate those words into an equation.

答:
The answer seems to be 80, but I agree with the commenter lulu that the question is ambiguously phrased.

For every 5 people leaving room A, (as I read it) three are moving from room B to room A, and one is leaving room B to go home. So Room A loses 5 people but gains 3, for a net change of -2, while room B loses 4 people.

We are given the condition that at some point in time room A has 50 people in it, room B has 20. We are also told that initially there are the same number of people in each room. So
$$
20 + 4x = 50 + 2x,
$$
where $x$ counts the number of times at which students are leaving the room.
Solving this equation gives $x = 15$, so each room had 80 occupants initially.


问:
How do you visualize polynomials rings and their quotient?
When working in the framework of group theory I could easily visualize the group by imagining its elements and I could see the group "collapse" when forming the quotient group by a normal subgroup by simply imagining the elements of the subgroup all being collapsed onto the identity, same thing for the other cosets.

But for polynomial rings I have a very hard thing visualizing this and it has lead to a poor understanding of ideals and quotient rings since I cannot rely on my visualize intuition anymore. I cannot make sense of $\mathbb{R}[x]$, its ideal $x^2+1$ and its quotient ring by that ideal.

I was wondering if you had any help in this regard because I am struggling here.

答: It's the same. Literally. Ideals of a ring are normal subgroups of their additive group, and the quotient of the ring by the ideal is just the quotient of the additive group by this normal subgroup. And then the multiplicative structure can be added on top. At their core, rings are groups with additional structure, and their quotients reflect this.</p>


问:
Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm
I am a little confused about taking derivatives w.r.t. the norms.

$L_0$-norm: $L_0$ means number of non-zero elements  in a vector.  Say, I am interested in an $x_i$.

$$\displaystyle\min_{i}(y_i-x_i)^2+c\|x_i \|_{0}$$
The answer depends on $x_i=0$ or not?

My work: take the norm of $x_i$, which is a constant, then, derivative, so it's 0.

$L_1$-norm: Manhattan distance.  What should I do?
$$\displaystyle\min_{i}(y_i-x_i)^2+c\|x_i \|_{1}$$

$L_2$-norm:Euclidean distance.  What should I do?

$$\displaystyle\min_{i}(y_i-x_i)^2+c\|x_i \|_{2}$$

答:
I'm going to assume that you're talking about partial derivatives and gradients. All of the norm functions that you stated are non-differentiable somewhere:

[$L_0$] This is zero (as you pointed out), but only in places where it isn't interesting. Where it is interesting it's not differentiable (it has jump discontinuities).

[$L_1$] This norm is not differentiable with respect to a coordinate where that coordinate is zero. Elsewhere, the partial derivatives are just constants, $\pm 1$ depending on the quadrant.

[$L_2$] Usually people use the 2-norm squared so that it's differentiable even at zero. The gradient of $\|x\|_2^2$ is $2x$, but without the square it's $x/\|x\|$ (i.e. it just points away from zero). The problem is that it's not differentiable at zero.

If you're trying to do gradient descent with these formulations, the only one that will really work is the squared $L_2$ norm, because that's the only one that's differentiable everywhere.


问:
Let X be a topological space. a) If X is Hausdorff and it has an countable base, so X is normal?
Let X be a topological space.

a) If X is  Hausdorff  and it has an countable base, so X is normal?

b) If X is regular, and if each point of X has a fundamental countable neighborhood system and is separable, then is X normal?

Being normal means that X is $T_1$ and $T_4$.

We say that:

X is $T_4$ if for any A, B $ \subset $ X closed there are U, V $ \subset $ X open disjoint such that A $ \subset $ U and B $ \subset $ V.

X is regular if X is $ T_3 $ and $ T_1 $.

X is $ T_3 $ iff for each x $ \in $ X and each F $ \subset $ X closed such that x $ \notin $ F there are U, V $ \subset $ X open disjoint such that x $ \in $ U and F $ \subset $ V.

If X is $T_3$ is $T_2$ and $T_1$.

If X is $T_2$ is also $T_1$.

if X is Separable it contains a subset A where for any point x in X, any neighborhood of x contains at least one point from A and A is countable. (A space is called separable if it contains a countable, dense subset).

a) and b) are false, right?

In both cases, there are $T_2$ spaces that we want to prove are normal.

Being $T_2$ means that U, V $ \subset $ X open disjoint such that A $ \subset $ U and B $ \subset $ V.But I don't see how to find a closed for any A, B, even with the countable base (first case), as in the case of having a fundamental system of countable neighborhood and being separable (second case).

答:
No to both: the Sorgenfrey plane is a counterexample to the second.

The first has the $\Bbb R_K$ as a counterexample.

Both are treated in Munkres' text book.


问:
Showing $\int_{0}^{\infty} \frac{\sin(x^{3})}{x}\,dx = \frac{\pi}{6}$
$$\int_{0}^{\infty} \frac{\sin(x^{3})}{x}\,dx = \frac{\pi}{6}$$
How to derive the above equality?

Looking for non complex analysis methods.

Body:
Hint:

$$\int\frac{\sin(x^3)}{x}dx=\int\frac{\sin(t^{3\alpha})}{t^\alpha}d(t^\alpha).$$

$$\int\frac{\sin(x^3)}{x}dx=\int\frac{\sin(t^{3\alpha})}{t^\alpha}d(t^\alpha)=\alpha\int\frac{\sin(t^{3\alpha})}tdt.$$


问:
Polynomial with geometric progression
If the roots of an equation with real coefficients $ax^3+bx^2+cx+d=0, a\ne0$ form a geometric progression, prove $ac^{3}=db^{3}$

I have no idea how to get started, so if it can a few hints?

Title: Polynomial with geometric progression

Body:
Start here:

$$(k_1 - x) (k_2 k_1 - x) (k_2^2 k_1 - x) = a x^3 + b x^2 + c x + d = 0$$


问:
Limit of weighted average
Let $x_1, x_2, \dots, x_n$ be a finite set of real numbers and $w(x, \theta)$ a weight function such that

$$w(x, \theta) = e^{x\theta}.$$

Denote $a(\theta)$ the weighted average of the $x_i$'s with weights given by $w$, i.e.,
$$a(\theta) = \frac{\sum_{i = 1}^n x_i e^{x_i\theta}}{\sum_{i = 1}^n e^{x_i\theta}}.$$

Intuitively, I think that $\lim_{\theta \rightarrow \infty} = \max_i x_i$ and $\lim_{\theta \rightarrow -\infty} = \min_i x_i$ because when $\theta$ is too large, then the relative weight of the maximum value will make all others negligible. I don't know what tools I can use to prove these limits. Is there any trick I can use here?

答:
Really interesting question hope I can help! This might seem very long and wordy but I have just explained things in a lot of detail, its not that horrible don't worry!
Before we begin to answer the question we are going to need to prove a tiny lemma!

$\textbf{Lemma:} $

Let $a_{max}$ = max {$a_1,a_2,a_3,..,a_k$} then $\lim_{n \to +\infty} \frac{a_{max}^n}{a_1^n+a_2^n+a_3^n+...+a_k^n} =1$

We will use the following simple law of limits (*) : $\lim_{n \to +\infty} \frac{1}{s_n} = \frac{1}{\lim_{n \to +\infty} s_n }$

Proof of Lemma:

$\lim_{n \to +\infty}\frac{a_1^n+a_2^n+a_3^n+...+a_k^n}{a_{max}^n} = \lim_{n \to +\infty}((\frac{a_1}{a_{max}})^n + (\frac{a_2}{a_{max}})^n  + ... + (\frac{a_{max}}{a_{max}})^n + ...+ (\frac{a_k}{a_{max}})^n )$

(All we have done is split the sum into its $k$ pieces. And there is one $a_i$ such that $a_{i} = a_{max}$, from definition of $a_{max}. $ Yes this $i$ could be $1,2$ or $k$ but this doesn't matter, the expansion is just for a visual aid. )

$\lim_{n \to +\infty}((\frac{a_1}{a_{max}})^n + (\frac{a_2}{a_{max}})^n  + ... + (\frac{a_{max}}{a_{max}})^n + ...+ (\frac{a_k}{a_{max}})^n)  = \lim_{n \to +\infty}(r_1^n + r_2^n + ... + 1^n + ... + r_k^n)$

Where $r_i := \frac{a_i}{a_{max}}$

Notice that for all i we have $r_i < 1 $

Hence:  $\lim_{n \to +\infty} r_i^n = 0 $ $\forall$ $i $

And then $\lim_{n \to +\infty}(r_1^n + r_2^n + ... + 1^n + ... + r_k^n) =$ 0 + 0 + ... + 1 + ... + 0 = $1$

Hence:

$\lim_{n \to +\infty}\frac{a_1^n+a_2^n+a_3^n+...+a_k^n}{a_{max}^n} = 1 $ and then from (*) law we finally have:  $\lim_{n \to +\infty} \frac{a_{max}^n}{a_1^n+a_2^n+a_3^n+...+a_k^n} =\frac{1}{1} = 1$

Don't worry! That was the hard part! The rest is easy dont worry! Now onto the last stretch:

Firstly let us define $a_i := e^{x_i} $

$ \lim_{\theta \to +\infty} a(\theta)  = \lim_{\theta \to +\infty} \frac{\sum_{i = 1}^n x_i a_i^{\theta}}{\sum_{i = 1}^n a_i^{\theta}} = \lim_{\theta \to +\infty} (\frac{x_{1}a_1^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + \frac{x_{2}a_2^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + ... + \frac{x_{max}a_{max}^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + ... + \frac{x_{n}a_n^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}})$

This is the same trick as before, writing a sum as all of its components but separately listing the maximum. We must now try to evaluate this limit for each component. We will split these into 2 cases to do seperately. Those with $ a_j \not =  a_{max} $ and then $a_{max} $

$\textbf{Case 1} $

To evaluate $\lim_{\theta \to +\infty} (\frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}) $ for $ a_j \not =  a_{max} :$

Notice that $ \sum_{i = 1}^n a_i^{\theta} $ (the denominator of our components) is greater than $ a_{max}^{\theta} $.

Hence $ \frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}  < \frac{x_{j}a_j^{\theta}}{a_{max}^{\theta}}   $ as it has a greater denominator

Hence: $\lim_{\theta \to +\infty} \frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}  < \lim_{\theta \to +\infty} \frac{x_{j}a_j^{\theta}}{a_{max}^{\theta}} $

Now, like before, define $r_j := \frac{a_j}{a_{max}} $ < 1

And hence: $\lim_{\theta \to +\infty} \frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}  < \lim_{\theta \to +\infty} \frac{x_{j}a_j^{\theta}}{a_{max}^{\theta}} =  \lim_{\theta \to +\infty} = a_j \cdot r_j^{\theta} = 0   $

So $\lim_{\theta \to +\infty} (\frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}) $ for $ a_j \not =  a_{max}  = 0 $

$\textbf{Case 2} $

$\lim_{\theta \to +\infty} (\frac{x_{max}a_{max}^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}) = \lim_{\theta \to +\infty} x_{max} \cdot \frac{a_{max}^{\theta}}{\sum_{i = 1}^n a_i^{\theta}} = x_{max} \cdot 1 $ (From the Lemma we proved)

Hence:

$ \lim_{\theta \to +\infty} a(\theta)  = \lim_{\theta \to +\infty} \frac{\sum_{i = 1}^n x_i a_i^{\theta}}{\sum_{i = 1}^n a_i^{\theta}} = \lim_{\theta \to +\infty} (\frac{x_{1}a_1^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + \frac{x_{2}a_2^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + ... + \frac{x_{max}a_{max}^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + ... + \frac{x_{n}a_n^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}) = 0 + 0 + ... + x_{max} + ... + 0 = x_{max} $

Finally for  $ \lim_{\theta \to -\infty} a(\theta) $:

Notice $ \lim_{\theta \to -\infty} a(\theta)  =  \lim_{\theta \to +\infty} a(-\theta) = \lim_{\theta \to +\infty} \frac{\sum_{i = 1}^n x_i e^{-x_i\theta}}{\sum_{i = 1}^n e^{-x_i\theta}} = -\lim_{\theta \to +\infty} \frac{\sum_{i = 1}^n -x_i e^{-x_i\theta}}{\sum_{i = 1}^n e^{-x_i\theta}}$ =

$-$Max{$-x_1,-x_2,...,-x_n$} = $- - $Min{$x_1,x_2,...,x_n$} = Min{$x_1,x_2,...,x_n$}

Your intuition was correct!

Hope this helped.

Oskar

Title: Limit of weighted average


问:
Let $a_{n+1}=\sqrt[3]{\frac{1+a_{n}}{2}}$ show that $0 \leq a_{n} \leq 1$
Let $0 \leq a_{1} \leq 1$ and $a_{n+1}=\sqrt[3]{\frac{1+a_{n}}{2}}$ for all $n \geq 1$.

Show that:

(a) $0 \leq a_{n} \leq 1$ for all n.

(b) $ a_{n+1} \geq a_{n}$ for all n.

I don't know if the following is correct.

My attempt.

By using induction:

(a) Show that $0 \leq a_{n} \leq 1$ for all n.

We have $0 \leq a_{1} \leq 1$ and we want to show that $0 \leq a_{2} \leq 1$ for all n.

$0 \leq a_{1} \leq 1 \Rightarrow \frac{1}{2} \leq \frac{a_{1}+1}{2} \leq 1 \Rightarrow  \sqrt[3]{\frac{1}{2}} \leq  \sqrt[3]{\frac{a_{1}+1}{2}}  \leq \sqrt[3]1 \Rightarrow 0.8 \leq a_{2} \leq 1 \Rightarrow 0 \leq a_{2} \leq 1 $

Now we suppose that the relation is true for n.

We want to show that it is also true for n+1.

We have:

$0 \leq a_{n} \leq 1 \Rightarrow \frac{1}{2} \leq \frac{a_{n}+1}{2} \leq 1 \Rightarrow  \sqrt[3]{\frac{1}{2}} \leq  \sqrt[3]{\frac{a_{n}+1}{2}}  \leq \sqrt[3]1 \Rightarrow 0.8 \leq a_{n+1} \leq 1 \Rightarrow 0 \leq a_{n+1} \leq 1 $

(b) Show that $ a_{n+1} \geq a_{n}$ for all n.

$ a_{n+1} \geq a_{n} \Rightarrow \sqrt[3]{\frac{1+a_{n}}{2}} \geq a_{n} \Rightarrow 1+a_{n} \geq 2a_{n}^3$

This is true because we know that $0 \leq a_{n} \leq 1$ and that $ a_{n} \geq a_{n}^3$.

答:
For $(b)$ it doesnt look like you've shown the base case nor the implication that if $a_{n+1} \geq a_{n}$, then $a_{n+2} \geq a_{n+1}$. All you did was reach a true statement...

I see what you have done now, but it is less obvious that you are meaning to say that since these steps are reversible, the inequality is satisfied.

I would still write out that you have reversible steps, or even start your implications starting at $1+a_{n} \geq 2{{a_n}^3}$ and showing this is true since by the inequalities you have listed we have $1+ a_{n} \geq a_{n}+a_{n}=2a_{n} \geq 2a^3_{n}$. It looks like your way might be a bit better because it follows directly from proving the base case in my proof.

I did it a different way:

Suppose $a_{n+1} \geq a_{n}$ is true for some $n$ (it is clear from the previous work this works for $n=1$).

Then $a_{n+2} = \sqrt[3]{\frac{1+a_{n+1}}{2}} \geq? \sqrt[3]{\frac{1+a_{n}}{2}} = a_{n+1}$ which is true by the inductive hypothesis.

In fact, the base case for $(b)$ might not be so clear without using your inequality anyways.


问:
Does the limit $\lim_{(x,y)\to(0,0)} (x^2+4y^2)^{x^2y^2}$ exist?
I have to study the existence of$$\lim_{(x,y)\to(0,0)} (x^2+4y^2)^{x^2y^2}$$
I've first considered the limit $$\lim_{x\to 0} f(x,mx) = \lim_{x\to 0} \ (x^2+4m^2x^2)^{m^2x^4} = e^{\lim_{x\to 0} m^2x^4\log(x^2+4m^2x^2)}$$

Using L'Hospital Rule, I get $$\lim_{x\to0}\ m^2x^4\log(x^2+4m^2x^2)=\lim_{x\to0}-\frac{m^2x^4}{2}=0$$
So $$\lim_{x\to 0} f(x,mx) = e^0 = 1$$
Now how can I compute the limit differently?

Body:
hint

We know that
$$\lim_{X\to0^+}X\ln(X)=0$$

thus
$$\lim_{(x,y)\to(0,0)}(x^2+4y^2)\ln(x^2+4y^2)=0$$

On the other hand
$$0\le \frac{x^2y^2}{x^2+4y^2}\le y^2$$

So,
$$\lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+4y^2}=0$$

You will find that the limit is $ e^0=1$.


问:
covering of a connected genus?
Let $\Sigma_n$ be a connected sum of $n$ tori for $n \in \mathbb{N}$. So basically, $\Sigma_0$ is the sphere, $\Sigma_1$ is the torus and then it's basically connected tori and the whole thing has $n$ holes. (I hope you get the picture.) I want to know for which values of $n$ does the sphere $S^2$ cover $\Sigma_n$.

I think the $\Sigma_n$ can be written as $S^2\times S^2\times \cdots \times S^2$ with $n$ many $S^2$'s, but does it mean that $S^2$ can cover the $\Sigma_n$ for all values of $n$?

答:
Firstly, $\Sigma_n$ cannot be written as a product $S^2 \times \cdots \times S^2$.

Secondly, using the fact that the Euler characteristic of a covering is the product of the Euler characteristic of the base space and the degree of the cover, we see that $S^2$ can only cover $\Sigma_0$ (i.e., itself) since $\chi(\Sigma_n) = 2 - 2n$.

Title: covering of a connected genus?


答:
If $X$ is an $n$-fold cover of $Y$ then (under mild hypotheses which are satisfied here, say, for finite CW complexes) we have $\chi(X) = n \chi(Y)$ where $\chi$ is the Euler characteristic. The Euler characteristic of $\Sigma_g$ is $2 - 2g$, so it follows that if $\Sigma_g$ covers $\Sigma_h$ (any such covering map is necessarily finite, because $\Sigma_g$ is compact) then $1 - h$ must divide $1 - g$.

Setting $g = 0$ gives that $1 - h$ must divide $1$ (and the quotient must be positive, since it must be the degree of the cover) which gives $h = 0$. In other words, the only closed orientable surface that $S^2$ covers is itself.

An alternate argument is thinking about universal covers. $S^2 = \Sigma_0$ is simply connected, $T^2 \cong \Sigma_1$ has universal cover $\mathbb{R}^2$, and for $g \ge 2$, $\Sigma_g$ has universal cover the upper half plane $\mathbb{H} \cong \mathbb{R}^2$ (e.g. by the uniformization theorem although one can also write down a covering map explicitly). In particular the universal cover of $\Sigma_g$ is noncompact for $g \ge 1$ (this is is equivalent to the fundamental group being infinite) and so cannot be $S^2$.

Title: covering of a connected genus?

Body:
Let $\Sigma_n$ be a connected sum of $n$ tori for $n \in \mathbb{N}$. So basically, $\Sigma_0$ is the sphere, $\Sigma_1$ is the torus and then it's basically connected tori and the whole thing has $n$ holes. (I hope you get the picture.) I want to know for which values of $n$ does the sphere $S^2$ cover $\Sigma_n$.

I think the $\Sigma_n$ can be written as $S^2\times S^2\times \cdots \times S^2$ with $n$ many $S^2$'s, but does it mean that $S^2$ can cover the $\Sigma_n$ for all values of $n$?


问:
Improper Integral with product of series
I am trying to solve $$ \int_0^\infty \left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6 }+\cdots\right)\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2\cdot 6^2 }+\cdots\right)\text dx$$
but failed, even the left hand side can be expressed as $x\cdot e^{-x^2 /2}$, I can still not deal with the second parenthesis. Is there anyway to evaluate this integral when integrand involved series which is a sum of squares ?

答:
Rather than knowing the series on the right outside of a general term, try distributing what you do know and make another series that you can integrate with.

I know it seems vague, but that's how I got the answer and I figured you want to solve this on your own.

If you want me to explain what I did, I can.


问:
How does a single $d$ in Leibniz notation work? Eg, $y\,\mathrm{d}x-x\,\mathrm{d}y\equiv-x^2\,\mathrm{d}\left(\frac{y}{x}\right)$
Recently a lecturer used this notation, which I assume is a sort of twisted form of Leibniz notation:

$$y\,\mathrm{d}x - x\,\mathrm{d}y \equiv -x^2\,\mathrm{d}\left(\frac{y}{x}\right)$$

The logic here was that this could be used as:

$$\begin{align}
-x^2\,\mathrm{d}\left(\frac{y}{x}\right) &\equiv -x^2\,\left(\frac{\mathrm{d}y}{x} -\frac{y}{x^2}\,\mathrm{d}x\right)\\
&\equiv y\mathrm{d}x - x\mathrm{d}y
\end{align}
$$

Why is this legal?

I can see some kind of differentiation going on with the second term in the above equivalence, producing the $\frac{1}{x^2}$, but having the single $\mathrm{d}$ seems like a really weird abuse of notation, and I don't quite follow why it splits the single $\frac{y}{x}$ fraction into two parts.

答:
In the context for this discussion you have two variables $x$ and $y$ that are somehow related. Perhaps you have $y$ as a function of $x$, or perhaps both depend on some other parameter $t$. You are interesting in knowing how small changes in $x$ and $y$ change the    quotient $y/x$, written as the product
$y \times (1/x)$. So what you are looking for is $d(y/x)$. If $y$ depends explicitly on $x$ you can think of this as calculating $d/dx$. If the dependence is just implicit it's easier to work with the differentials.

The actual algebra uses  the product rule and the  rule for differentiating $1/x$. You could do it directly with the quotient rule.

To see more intuitively what is going on, simplify the expression
$$
-x^2\left( \frac{y+dy}{x+dx} -  \frac{y}{x} \right)
$$
and remember that $dx$ and $dy$ are small.


问:
Does L'Hopital's rule imply that $\lim_{x\to a}\frac{f'(x)}{g'(x)} = \lim_{x\to a}\frac{f(x)}{g(x)}$ always?
My question is summarised as follows:

Is $$\lim_{x\to a}\frac{f'(x)}{g'(x)} $$
always equal to $$\lim_{x\to a}\frac{f(x)}{g(x)} $$
or can we only deduce values of limits using differentiation in the forward direction, ie we can't deduce that the value of $\lim_{x\to a}\frac{f'(x)}{g'(x)} $ is the same as $\lim_{x\to a}\frac{f(x)}{g(x)} $?

Logically I cannot see why this shouldn't be true (ie I can't see why we can't also integrate the numerator and denominator). HOWEVER, the following result seems to throw some doubt on this:

Consider the identity
$$f(x)^2\frac{d}{dx}\frac{g(x)}{f(x)}+g(x)^2\frac{d}{dx}\frac{f(x)}{g(x)}\equiv0$$
$$\implies\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv-\frac{f(x)^2}{g(x)^2}$$
On applying L'Hopital's rule we obtain
$$\lim_{x\to a}\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}\implies\lim_{x\to a}\frac{\frac{f(x)}{g(x)}}{\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
$$\implies\lim_{x\to a}\frac{f(x)^2}{g(x)^2}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
which is obviously false.

Any help explaining this apparent proof would be appreciated.

答:
The standard theorem says that both limits are equal if

$f$ and $g$ are differentiable on an interval containing $a$, with the possible exception of $a$ itself

$\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or $\lim_{x \to a} |f(x)| = \infty$ and $\lim_{x \to a} |g(x)| = \infty$

$\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists.

You can use it to find the limit of $f(x)/g(x)$ from the limit of $f'(x)/g'(x)$ or to find the limit of $f'(x)/g'(x)$ from the limit of $f(x)/g(x)$, but you do need to know that the limit of $f'(x)/g'(x)$ exists.

EDIT: Your "apparent proof" is bogus because it violates assumption (2).


答:
I believe it is being used as shorthand for:
$$d\left(\frac yx\right)=dx\frac{d\left(\frac yx\right)}{dx}=dx\left(\frac{dy}{dx}x-\frac{y}{x^2}\right)=xdy-\frac{y}{x^2}dx$$

Title: How does a single $d$ in Leibniz notation work? Eg, $y\,\mathrm{d}x-x\,\mathrm{d}y\equiv-x^2\,\mathrm{d}\left(\frac{y}{x}\right)$

Body:
Recently a lecturer used this notation, which I assume is a sort of twisted form of Leibniz notation:

$$y\,\mathrm{d}x - x\,\mathrm{d}y \equiv -x^2\,\mathrm{d}\left(\frac{y}{x}\right)$$

The logic here was that this could be used as:

$$\begin{align}
-x^2\,\mathrm{d}\left(\frac{y}{x}\right) &\equiv -x^2\,\left(\frac{\mathrm{d}y}{x} -\frac{y}{x^2}\,\mathrm{d}x\right)\\
&\equiv y\mathrm{d}x - x\mathrm{d}y
\end{align}
$$

Why is this legal?

I can see some kind of differentiation going on with the second term in the above equivalence, producing the $\frac{1}{x^2}$, but having the single $\mathrm{d}$ seems like a really weird abuse of notation, and I don't quite follow why it splits the single $\frac{y}{x}$ fraction into two parts.


问:
Prove that every subgroup of an infinite cyclic group is characteristic
Prove that every subgroup of an infinite cyclic group is characteristic.

I know that every infinite cyclic group is isomorphic to $\Bbb Z$, and any automorphism on $\Bbb Z$ is of the form $\alpha(n) = n$ or $\alpha(n) = -n$.  That means that if $f$ is an isomorphism from $\Bbb Z$ to some other group $G$, the isomorphism is determined by $f(1)$.  But from here I can't figure out how to show that it's characteristic.

答: For each $n \in \mathbb N^*$, there is exactly one subgroup of $\mathbb Z$ of index $n$, namely $n\mathbb Z$. Therefore, it must be fixed by every automorphism.


问:
Linear Difference Equation Solution

$a[n + 2] - ea[n + 1] = 3f[n]; f[n] = e^n, a[-1] = \frac{e^2 - 3}{e^3}$

Wolfram Alpha gives the solution $a[n] = 3e^{n - 2}n + e^n$, but also says that there is no solution.  The solution it gives does not seem to satisfy the difference equation.

My solution is $a[n] = e^{n - 1} - 3e^{n - 3} + 3ne^{n - 4}$, which seems to satisfy the difference equation.  My solution does not even have the same number of terms as Wolfram's solution, and the two are not identical when graphed.  Is Wolfram just bad at these, or am I doing something wrong?

答:
$$a_{n+2} - ea_{n+1}=3e^n \implies \frac{a_{n+2}}{e^{n+2}} - \frac{a_{n+1}}{e^{n+1}} = 3e^{-2}$$

So $\frac{a_n}{e^n}$ is an arithmetic sequence,
$$
\frac{a_n}{e^n}=(n+1)3e^{-2}+\frac{a_{-1}}{e^{-1}} \implies a_n=e^n+3ne^{n-2}.
$$

Your solution gives $a_{n+2} - ea_{n+1}=3e^{n-2}$, a little off.

Wolfram Alpha is good but not always reliable.

Title: Linear Difference Equation Solution


答:
The answer to your first question is No.

Take $$f(x)=x^2\sin(\frac 1x)$$
and
$$g(x)=x$$

then

$$\lim_{x\to 0}\frac{f(x)}{g(x)}=0$$

But

$$\lim_{x\to0}\frac{f'(x)}{g'(x)}$$
Does not exist.

Title: Does L'Hopital's rule imply that $\lim_{x\to a}\frac{f'(x)}{g'(x)} = \lim_{x\to a}\frac{f(x)}{g(x)}$ always?

Body:
My question is summarised as follows:

Is $$\lim_{x\to a}\frac{f'(x)}{g'(x)} $$
always equal to $$\lim_{x\to a}\frac{f(x)}{g(x)} $$
or can we only deduce values of limits using differentiation in the forward direction, ie we can't deduce that the value of $\lim_{x\to a}\frac{f'(x)}{g'(x)} $ is the same as $\lim_{x\to a}\frac{f(x)}{g(x)} $?

Logically I cannot see why this shouldn't be true (ie I can't see why we can't also integrate the numerator and denominator). HOWEVER, the following result seems to throw some doubt on this:

Consider the identity
$$f(x)^2\frac{d}{dx}\frac{g(x)}{f(x)}+g(x)^2\frac{d}{dx}\frac{f(x)}{g(x)}\equiv0$$
$$\implies\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv-\frac{f(x)^2}{g(x)^2}$$
On applying L'Hopital's rule we obtain
$$\lim_{x\to a}\frac{\frac{d}{dx}\frac{f(x)}{g(x)}}{\frac{d}{dx}\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}\implies\lim_{x\to a}\frac{\frac{f(x)}{g(x)}}{\frac{g(x)}{f(x)}}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
$$\implies\lim_{x\to a}\frac{f(x)^2}{g(x)^2}\equiv\lim_{x\to a}-\frac{f(x)^2}{g(x)^2}$$
which is obviously false.

Any help explaining this apparent proof would be appreciated.


答:
Such arguments can always be rephrased to avoid treating $dx$ and $dy$, etc, as individual "infinitesimal" quantities. (On the other hand, "infinitesimal intuition" is a powerful and intuitive way to derive calculus formulas, so I can see why physicists are drawn to it.)

I'll assume that $y$ is a function of $x$. Let $h(x) = y(x)/x$. Then
$$
h'(x) = \frac{x y'(x) - y(x)}{x^2}
$$
so
$$
\tag{1} -x^2 h'(x) = y(x) - x y'(x).
$$
That's probably how I would write it, because the meaning is perfectly clear.

We could also write (1) using Leibniz notation:
$$
-x^2 \frac{dh}{dx} = y - x \frac{dy}{dx}.
$$
If we then "multiply through by $dx$", we obtain
$$
- x^2 dh = y dx - x dy
$$
which is what your lecturer wrote.

I can imagine that some people think the version using infinitesimal notation is more beautiful or more intuitive.


问:
Verify that $M_{X}(t) \ge e^{t\mu}$ using Jensen's identity
Suppose that a real valued random variable with a probability density of $$f(x) = \begin{cases}
\frac{1}{8}(x+1) &:-1 <x<5\\
0 &: \text{else}
\end{cases}$$.

I need to verify that $M_{X}(t) \ge e^{t\mu}$, where $M_{X}(t)$ is the moment generating function of X, $t=\frac{1}{2}$ and $\mu$ is the mean of the probability density function, using Jensen's inequality.

From what I understand, I need to show that $E[w(X)] \ge w(\mu)$.

My question is what would I use for the value of $w(X)$? would I use $M_{X}(t)$ or should I use the probability density function?
Thanks for any help.

答:
For each $t$, the function $x\mapsto e^{tx}$ is convex. Hence, Jensen's inequality shows that
$E[e^{tX}]\geq e^{tE[X]}=e^{t\mu}.$

The problem is pretty straightforward in that the left is
$$
\int_{-1}^5 e^{tx}f_X(x)\,dx
$$
and the right is just $e^{t\mu}$ where
$$
\mu=E[X]=\int_{-1}^5 xf_X(x)\,dx.
$$


问:
If $R$ is a commutative ring with unity and $R$ has only one maximal ideal then show that the equation $x^2=x$ has only two solutions
If $R$ is a commutative ring with unity and $R$ has only one maximal ideal then show that the equation $x^2=x$ has only two solutions.

I know that $0$ and $1$ are the solutions, but I can't proceed from there.

答:
Suppose that $r \in R$ satisfies $r^2=r$.  Since $(r)+(1-r)=R$, one of the principal ideals $(r)$ or $(1-r)$ is not contained in the maximal ideal $M$.  But then $(r)=R$ or $(1-r) = R$, i.e. $r$ or $1-r$ is a unit.

If $x$ is a unit satisfying $x^2=x$, then $x=1$ (dividing by $x$).  Thus $r=1$ or $1-r=1$.


问:
Suppose that $f(z) = u(x,y) + iv(x,y)$ is a holomorphic function and that $2u(x,y) + v(x,y) = 5$, for all $z=x+iy$ in $C$, show that $f$ is constant.
Any hints on where to start? I'm stuck.

I understand that is suffices to show that $f'(z) = 0$

In order to do that I must find what $u$ and $v$ in terms of $x$ and $y$ first, can you point me in the direction of how to do so?

答:
$2v_y+v_x=0$ and $-2v_x+v_y=0$ by Cauchy-Riemann equation so
$$
2v_y=-v_x=-\frac{v_y}{2}
$$

thus $5v_y=0$, so $v_y=0$ so again by Cauchy-Riemann we have $u_x=v_y=0$, by first equation we have $v_x=0$ so $-u_y=v_x=0$ and $u,v$ are constant's functions, in particular $f$.


问:
Exercise 6.21 Isaacs's Character theory of finite groups

Let $1= H_0\lhd H_1\lhd\cdots\lhd H_n=G$. Assume that $H_i/H_{i-1}$ is nonabelian. Show that there exists $\chi\in Irr(G)$ with $\chi(1)\ge 2^n$.

Hint Use Corollary 6.17(Gallagher): Let $N\lhd G$ and let $\chi\in Irr(G)$ be such that $\chi|_N=\theta\in Irr(N)$. Then the characters $\beta\chi$ for $\beta\in Irr(G/N)$ are irreducible, distinct for distinct $\beta$ and are all of the irreducible constituents of $\theta^G$.

My attempt:

We can argue by induction on the length $r$ of the subnormal series. The base case is clear. Suppose it is true for all $r<n$, then for $r=n$, we notice that $H_n/H_{n-1}$ is non abelian implies there exists an irreducible character $\beta\in Irr(H_n/H_{n-1})$ such that $\beta(1)\ge 2$. By the induction hypothesis, there exists a $\psi\in Irr(H_{n-1})$ with $\psi(1)\ge 2^{n-1}$. To use Corollary 6.17, it suffices to show that $\psi$ is extendable to $H_n$.

Then I got stuck... Any hint?

答:
I figured it out. With the same notations in the question we have the following:

Consider $\psi^G$. Note that $\psi^{G}(1)=|G:H_{n-1}|\psi(1)\ge 2^{n}$.

Prove by contradiction. Suppose any irreducible character of $G$ has degree less than $2^{n}$.

Let $\gamma$ be an irreducible constituent of $\psi^G$, that is $[\psi^G,\gamma]_{G}\ge 1$, by the Frobenius reciprocity, we have $$[\psi,\gamma|_{H_{n-1}}\ ]_{H_{n-1}}=[\psi^G,\gamma]_{G}\ge 1. $$

Clifford's theorem gives us $\gamma|_{H_{n-1}}=e\sum_{i=1}^t\psi^{(i)}$ where $e=[\psi^G,\gamma]_G$ and $\psi^{(i)}$ denotes the conjugate of $\psi$.

So $\gamma(1)=et\psi(1)\ge et2^{n-1}$. But $\gamma(1)<2^n$. So $e=t=1$. Thus $\gamma|_{H_{n-1}}=\psi$. And we know that $\psi$ is extendible to $G$. Hence $\gamma\beta(1)\ge 2^{n}$ and $\gamma\beta\in Irr(G)$ by Corollary 6.17 which is a contradiction.


问:
What is the " base step" in this proof by induction regarding an elementary fact of FOL's syntax? ( "Classical logic" in SEP)
Source : https://plato.stanford.edu/entries/logic-classical/#BuilBloc

Here is the proof ( where the " rules (2)-(7) ) " refered to are the formation rules of  formulas) :

 Let us temporarily use the term “unary marker” for the negation symbol (¬) or a quantifier followed by a variable (e.g., ∀x,∃z).

Lemma 2. Each formula consists of a string of zero or more unary markers followed by either an atomic formula or a formula produced using a binary connective, via one of clauses (3)–(5).

Proof: We proceed by induction on the complexity of the formula or, in
other words, on the number of formation rules that are applied.

The Lemma clearly holds for atomic formulas.

Let n be a natural number, and suppose that the Lemma holds for any formula constructed >from n or fewer instances of clauses (2)–(7). Let θ be a formula constructed
from n+1 instances. The Lemma holds if the last clause used to
construct θ was either (3), (4), or (5). If the last clause used to
construct θ was (2), then θ is ¬ψ. Since ψ was constructed with n
instances of the rule, the Lemma holds for ψ (by the induction
hypothesis), and so it holds for θ. Similar reasoning shows the Lemma
to hold for θ if the last clause was (6) or (7). By clause (8), this
exhausts the cases, and so the Lemma holds for θ, by induction.

My question is : does the base case consist in  the assertion that the property holds for atomic formulas?

I can see that the induction hypothesis is that the property holds when n instances of rules (2)-(7) are applied.

I see that after that, the induction step is proved : if the property holds for $n$ applications, it also holds for $n+1$ applications.

But I cannot detect the base step Does it consist in the assertion that the property is true of atomic formulas?

If it is the case , what relation has the base step to do with number $n$?

Does the atomic formulas case represent $n=0$ ?

答:
Yes, the base case is the case for atomic formulas, and this corresponds to "$n=0$:" when $n=0$ we're limited to formulas constructed without using clauses $(2)$ through $(7)$ at all, that is, only those formulas constructed via clause $(1)$. And all clause $(1)$ does is give us the atomic formulas:

All atomic formulas of $\mathcal{L}1K=$ are formulas of $\mathcal{L}1K=$.


问:
Prove $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$ for $c = \min(a, b, c)$ and $ab + bc + ca = 2$

Problem 1: Let $a, b, c \ge 0$ with $c = \min(a, b, c)$ and $ab + bc + ca = 2$. Prove that
$\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$.

Background: This problem was proposed by csav10@AoPS
(https://artofproblemsolving.com/community/q1h2321937p18543869)

which is a variant of the following:

Problem 2: Let $a, b, c \ge 0$ with $ab + bc + ca = 2$. Prove that $\sqrt{a+ab} + \sqrt{b+bc} + \sqrt{c+ca} \ge 3$.

(Proposed by KaiRain@AoPS. See https://artofproblemsolving.com/community/c6h1905412.

Also see: https://artofproblemsolving.com/community/q1h2308247p18537783.)

For Problem 2, yleo@AoPS and csav10@AoPS gave nice solutions, respectively.
Also, I solved Problem 2 by the Buffalo Way (BW) which is a computer solution.

For Problem 1, I have not yet solved it by the Buffalo Way (BW).

By the way, Equality case: $(a, b, c) = (2, 1, 0)$.

Any comments and solutions are welcome and appreciated.

Add the Buffalo Way solution (outline) for Problem 2

Fact 1: If $b \ge a \ge c \ge 0$, then
$$\sqrt{a + ab} + \sqrt{b + bc} + \sqrt{c + ca} \ge \sqrt{b + ba} + \sqrt{a + ac} + \sqrt{c + cb}.$$

From Fact 1, we assume that $a \ge b \ge c$.

After homogenization, we need to prove that
$$\sqrt{\frac{a \sqrt{\frac{ab+bc+ca}{2}} + ab}{\frac{ab+bc+ca}{2}}} +
\sqrt{\frac{b \sqrt{\frac{ab+bc+ca}{2}} + bc}{\frac{ab+bc+ca}{2}}} +
\sqrt{\frac{c \sqrt{\frac{ab+bc+ca}{2}} + ca}{\frac{ab+bc+ca}{2}}} \ge 3.$$
Denote
$$X = \frac{1}{4}\cdot\frac{a \sqrt{\frac{ab+bc+ca}{2}} + ab}{\frac{ab+bc+ca}{2}},
Y = \frac{b \sqrt{\frac{ab+bc+ca}{2}} + bc}{\frac{ab+bc+ca}{2}},
Z = \frac{c \sqrt{\frac{ab+bc+ca}{2}} + ca}{\frac{ab+bc+ca}{2}}.$$
We need to prove that $2\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge 3$.

By using $\sqrt{u} \ge \frac{2u}{1+u}$ for $u\ge 0$, it suffices to prove that
$$\frac{4X}{1+X} + \frac{2Y}{1+Y} + \frac{2Z}{1+Z} \ge 3.$$
After some manipulation, it suffices to prove $A\sqrt{\frac{ab+bc+ca}{2}} + B \ge 0$ where $A, B$ are some polynomials and $A \ge 0$.

It suffices to prove that $A^2\cdot \frac{ab+bc+ca}{2} \ge B^2$.

BW works.

答:
Partial answer

Let $f = \sqrt{a + ab} + \sqrt{b} + \sqrt{c} - 3$ and $g = ab + bc + ca -2$, so we are to show $f\ge 0$ under the condition $g =0$.

Start by noting that if $a=b=c=c_0$, we have  $c_0 = \sqrt{2/3} \simeq 0.8165$ and $f = 0.0251$ so the condition is very tight.

Nevertheless, in this partial answer the inequality can be proved for small $c$, and for $c$ in the (larger) vicinity of $c_0$. Indications are given to extend these calculations.

Treat $c$ as a parameter. First, note that for $a\ge b$, we observe for the exchange of $a\leftrightarrow b$ that  $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} - 3 \le \sqrt{b + ab} + \sqrt{a} + \sqrt{c} - 3$, hence we need to consider the case $a\ge b$ only.

Since $c$ is required to be the smallest variable, we consider $a\ge b \ge c$.

Case 1: small $c$.

We show that the equality condition $(a,b,c) = (2,1,0)$ is actually a minimum of $f$. A convenient way to do so is variational calculus, where $c$ is arbitrary but fixed. This gives
$
0 = dg = (b+c)da + (a+c)db$
and
$ 2 \sqrt{a + ab}\sqrt{b} \; df = \sqrt{b} (1+b) \; da + (a \sqrt b + \sqrt{a + ab}) \; db$

For the RHS to become zero we insert the $dg=0$-equation and get
$  d \tilde f = [\sqrt{b} (1+b) (a+c) -   (a \sqrt b + \sqrt{a + ab}) (b+c)] \, da  $.

For $c=0$, $  d \tilde f =  0$ gives
$   \sqrt{a}=   \sqrt{1 + b} \sqrt{b}$. Now inserting, from $g=0$, the condition $a = 2/b$, gives $2 = b^2 + b^3$ which has the only positive solution $b=1$, from which follows again $a = 2$.  This establishes that $(a,b,c) = (2,1,0)$ is actually a minimum of $f$.

Now in the vicinity of that minimum, i.e. for small $c > 0$, the $f$-equation itself guarantees the inequality. The reason is that $  d \tilde f = 0$ tells us that the value of $f$  will shift away in linear dependence with $c$, as $a$ and $b$ vary to obey $  d \tilde f = 0$.  In contrast, in $f$ the additional term $\sqrt c$ increases $f$, where this term has an infinite growth rate at $c=0$ (illustrated also in  the second figure below).  Hence for small  $c > 0$, $f$ will always move in positive direction with $c$.

This behavior can be utilized with the following sketch: use $d \tilde f = 0$ and $g=0$ to calculate the (small) change of $a$ and $b$ once $c$ moves away from $0$. With this change, calculate the new $f$ where the first terms change linearly and the last term goes with $\sqrt c$.  Then, equate the "worst" change in the first terms with  $\sqrt c$ to get an estimate for the highest $c$  where $f \ge 0$ holds.

Case 2: large $c$.

The largest $c$ that can be attained is  $c_0 = \sqrt{2/3}$, as $c$ must be the smallest variable.  In this case,  $a=b=c=c_0$, $f = 0.0251$, and variations of $(a,b,c)$ under the condition $g=0$ change $f$ linearly with $c$. Hence, in some environment $f$ is guaranteed to stay positive.

Indeed, here we can estimate $f$. Method 1: As $a$ is the largest of the variables, the smallest $a$, for given $c$, that can be attained, is observed when $a=b$. Under $g=0$, this gives $a = \sqrt{2 + c^2} - c$. The smallest $b$ is $b=c$. For $a=b=c=c_0$ both estimates are tight, so for some range of  $c < c_0$, we can ensure  $f\ge 0 $ to hold. We have
$f \ge  \hat f = \sqrt{(\sqrt{2 + c^2} - c)} \, \sqrt{1 + c} + 2\sqrt{c}  - 3$ and this holds until $\hat f = 0$ which is attained for $c_{min}  \simeq 0.793$.

Method 2: Inserting $a$ from $g =0$ gives $f = \sqrt{2-b c} \, \frac{\sqrt{1 + b}}{\sqrt{c + b}} + \sqrt{b} + \sqrt{c} - 3$. Now  $\frac{\sqrt{1 + b}}{\sqrt{c + b}} \ge \frac{\sqrt{1 + b_{max}}}{\sqrt{c_{max} + b_{max}}}$ and, by AM-GM, $\sqrt{b} + \sqrt{c} \ge 2 \sqrt[4]{b c}$. Letting $bc = x$ this gives   $$f \ge \hat f =  \sqrt{2-x} \, \frac{\sqrt{1 + b_{max}}}{\sqrt{c_{max} + b_{max}}} + 2 \sqrt[4]{x} - 3$$ which is a function of $x$ only, with $b_{max} = \max\{\sqrt{2 + c^2} - c\} = \sqrt{2}$ and $c_{max} = c_0$. Again, all estimates are tight for $a=b=c=c_0$, which entails $x = \frac23$ and $f = 0.0251$.  So $\hat f = 0$ can be used to calculate the smallest $x$ where the estimate holds, which is given for $\hat x \simeq 0.632$. Since $c = \hat x/b$, this makes it feasible to reduce $c$ while increasing $b$, however this is only possible until $b$ reaches its highest value. For given $c$, this value is  $\sqrt{2 + c^2} - c$ which gives $c(\sqrt{2 + c^2} - c) = 0.632$ or $c_{min} =  0.737$, which is better than the bound with method 1.

Again, this can be improved, if better values for $b_{max}$ and $c_{max}$  are found. For example, with the result just obtained, one can iterate the method and reduce $c$ once again, where now it is established that $c_{max} = 0.737$ can be used. Also, the estimate $b_{max} = \max\{\sqrt{2 + c^2} - c\}$ needs the smallest value of $c$ which goes into consideration, where $c=0$ was the most conservative one. Here, once, from the extension of case 1, a highest value of $c$  is known for which the inequality holds, this value can be used.

Case 3: "small $c$" $ < c < 0.737$.

To be done. Indications are given above.

Here is a figure which shows what to expect, and consequently what to do:

In green is the location of the minimum, i.e. the solution of $df = 0$ under $g=0$. In red is the range for $b$, i.e. $c \le b \le \sqrt{2 + c^2} \, - c$. For  $c=0$, the minimum is to be found at $b_{min}=1$, as was calculated above. As $c$ increases, $b_{min}$ gets smaller. For $c \simeq 0.345$, the location of $b_{min}$ reaches the lower limit  of the range for $b$. Hence for the remaining $ 0.345 < c < \sqrt{2/3}$, the smallest $f$ is to be found at $b_{min} = c$. This makes it easy to show the inequality in this interval of $c$: we have
$f \ge f_{min}^* = \sqrt{2- c^2} \, \frac{\sqrt{1 + c}}{\sqrt{2 c}} + 2\sqrt{c} - 3$ and this is clearly positive, see the following plot (created with values from numeric evaluations). In that plot, $f_{min}$ is plotted over the first $c$-interval $\mbox{[0 $\,$ 0.345]}$ where the value of $b$ is taken as the free minimum, from the solution of $df = 0$ under $g=0$ (blue section). The green circle is there to mark the transition to the second $c$-interval [$0.345 \,$ $\sqrt{2/3}$] where $f_{min}^*$ is plotted (see above) which is  taken at $b = c$ (orange section). Note the rise of $f_{min}$ with infinite rate at $c=0$ acording to the $\sqrt{c}$-term in $f$, as discussed above under case 1: small $c$.

Hence the agenda for analytic work is:

establish for $c < 0.345$ that $f$ has a free minimum w.r.t. $b$, and that at that minimum, $f_{min}(c) > f_{min}(c=0) =0$.

establish for $c > 0.345$ that $f(b,c) \ge f(c,c)$, and then  $f(c,c) > 0$ can be shown easily (see above), which gives the desired result.


问:
Showing $\int_0^1\frac{\sin^{-1}\left(\frac{2x}{1+x^2}\right)}{1+x^2}dx=\frac{\pi^2}{16}$
Could someone please explain how to arrive at this answer?

$$\int_0^1\frac{\sin^{-1}\left(\frac{2x}{1+x^2}\right)}{1+x^2}dx=\frac{\pi^2}{16}$$

I'm having trouble to approach this competition problem. I'm not sure if a clever $u$-substitution($u=1+x^2$?!), series approach (maybe the Basel problem?! because $\pi^2$!!), or some other form would be a suitable approach.

答:
Substitute $x=\tan\theta, dx=\sec^2\theta d\theta$ to get

$$\int_0^\frac\pi4\dfrac{\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)\sec^2\theta d\theta}{1+\tan^2\theta}=\int_0^\frac\pi4\sin^{-1}\frac{2\tan\theta}{\sec^2\theta}d\theta=$$
$$\int_0^\frac\pi4\sin^{-1}(2\sin\theta\cos\theta) d\theta=\int_0^\frac\pi42\theta d\theta$$


答:
For (a) you don't need to prove the statement for $a_2$.

You can simplify the argument by proving first that $a_n\ge0$. This is true for $a_1$; suppose it is true for $a_n$; then $a_{n+1}=\sqrt[3]{(a_n+1)/2}\ge0$.

Now the other inequality, which you did well: if $a_n\le1$, then $(a_n+1)/2\le1$ and therefore $a_{n+1}=\sqrt[3]{(a_n+1)/2}\le1$.

For (b) the arrows are in the wrong direction! You need to prove that $a_{n+1}\ge a_n$ and you can't take this as an assumption. However
$$
a_{n+1}\ge a_n
\quad\text{if and only if}\quad
a_{n+1}^3\ge a_n^3
\quad\text{if and only if}\quad
\frac{a_{n}+1}{2}\ge a_n^3
$$
so we are reduced to proving that $2a_n^3\le a_n+1$.

Your argument is fine: $a_n\le1$ implies $a_n^3\le a_n$ and $a_n^3\le 1$.

Title: Let $a_{n+1}=\sqrt[3]{\frac{1+a_{n}}{2}}$ show that $0 \leq a_{n} \leq 1$

Body:
Let $0 \leq a_{1} \leq 1$ and $a_{n+1}=\sqrt[3]{\frac{1+a_{n}}{2}}$ for all $n \geq 1$.

Show that:

(a) $0 \leq a_{n} \leq 1$ for all n.

(b) $ a_{n+1} \geq a_{n}$ for all n.

I don't know if the following is correct.

My attempt.

By using induction:

(a) Show that $0 \leq a_{n} \leq 1$ for all n.

We have $0 \leq a_{1} \leq 1$ and we want to show that $0 \leq a_{2} \leq 1$ for all n.

$0 \leq a_{1} \leq 1 \Rightarrow \frac{1}{2} \leq \frac{a_{1}+1}{2} \leq 1 \Rightarrow  \sqrt[3]{\frac{1}{2}} \leq  \sqrt[3]{\frac{a_{1}+1}{2}}  \leq \sqrt[3]1 \Rightarrow 0.8 \leq a_{2} \leq 1 \Rightarrow 0 \leq a_{2} \leq 1 $

Now we suppose that the relation is true for n.

We want to show that it is also true for n+1.

We have:

$0 \leq a_{n} \leq 1 \Rightarrow \frac{1}{2} \leq \frac{a_{n}+1}{2} \leq 1 \Rightarrow  \sqrt[3]{\frac{1}{2}} \leq  \sqrt[3]{\frac{a_{n}+1}{2}}  \leq \sqrt[3]1 \Rightarrow 0.8 \leq a_{n+1} \leq 1 \Rightarrow 0 \leq a_{n+1} \leq 1 $

(b) Show that $ a_{n+1} \geq a_{n}$ for all n.

$ a_{n+1} \geq a_{n} \Rightarrow \sqrt[3]{\frac{1+a_{n}}{2}} \geq a_{n} \Rightarrow 1+a_{n} \geq 2a_{n}^3$

This is true because we know that $0 \leq a_{n} \leq 1$ and that $ a_{n} \geq a_{n}^3$.


问:
Simple Triangle-like Inequality for a general norm
I came up with this inequality as I was thinking about norms. I don't know if it's true though.

Given three points $x,y,z$ in a normed space $(X,\|\cdot\|)$, is it true that if

$$
\|x-y\|\le\|y\|
$$

and

$$
\|x-z\|\le\|z\|
$$

then

$$
\|x-(y+z)\|\le\|y+z\| ?
$$

In other words, for a given vector $x$ in a normed space $X$, is the set $K:=\{y:\|x-y\|\le\|y\|\}$ additive?

First, if it were true, it almost looks like we could change the $y$ and $z$ around so that, e.g. the 1st and 3rd inequalities imply the 2nd. As it is, given any arrangement $x+y=z$ (none of which is $0$), we can tell which is which by simply adding.

I thought proving $X=\mathbb{R}$ would easily turn into $X=l_p$, but that seems naive.

Inner Product Space

The first two inequalities reduce to

$$
\langle x,x\rangle\le 2\langle y,x\rangle
$$

and

$$
\langle x,x\rangle\le 2\langle z,x\rangle
$$

Adding these together gets

$$
\|x\|^2\le \langle y+z,x\rangle\le 2\langle y+z,x\rangle
$$

Working the same algebra backwards then gets the third inequality.

Other norms?

Note that if $y\in K$, then $cy\in K$ for all $c\ge 1$. So there is a curve/surface in $X$, symmetric about $0$ (AKA odd), that divides $X$ into two parts, one part being $K$. It looks like if the unit ball $B$ of $X$ is oblong and $x$ is not perpendicular to an axis of symmetry of B, the curve/surface will not be linear and hence not convex (symmetric about $0$ and convex implies linear).

So, the factor of $2$ cannot be dismissed for general $X$ as it was for inner product spaces (?). I feel like that argument says the 2 cannot be dismissed for some inner product spaces as well. Mistake?

Thanks!

Body: The inequality can fail for $X=\Bbb R^2$ endowed with $\|\cdot\|_\infty$ norm. Indeed, let $x=(1,0)$, $y=(-1,2)$, and $z=(-1,-2)$. Then $\|x-y\|=\|y\|=\|x-z\|=\|z\|=2$, but $\|x-(y+z)\|=3>2=\|y+z\|$.


问:
Matching in biregular graph
I have to solve this exercise:

Let G = (S, T; E) a biregular graph. Prove that in this graph there exist a matching that either saturates all the vertices in
S or saturates all the vertices in T.

From what I know, I have to use Hall's Theorem, but I have no clue how to do so.

Any hint is appreciated. Thank you.

答:
Suppose WLOG that $|S| \leq |T|$. Let the degree of each vertex in $S$ be $d_S$ and the degree of each vertex in $T$ be $d_T$.

Take any subset $W \subseteq S$. Let $N(W)$ denote the set of neighbors of verties in $W$. Let $G'$ be the induced bipartite subgraph with vertex sets $W$ and $N(W)$. Then $|E(G')|$, the number of edges in $G'$, satisfies
$$ |E(G')| = d_S \cdot |W| .$$
Since we have drawn $|E(G')|$ edges which are incident to $N(W)$ vertices, there must be a vertex in $N(W)$ with degree $\geq \frac{|E(G')|}{|N(W)|}$. So
$$ d_T \geq \frac{d_S \cdot |W|}{|N(W)|} .$$

But now notice that since $|S| \cdot d_T = |E(G)| = |T| \cdot d_S$, we have
$$ d_S \geq d_T. $$

Combining these ineqalities, we get
$$ d_S \geq \frac{d_S \cdot |W|}{|N(W)| } ,$$

which simplifies to $|W| \leq |N(W)|$. By Hall's theorem, we are done.

Title: Matching in biregular graph


问:
What is the area of the part of the plane $2x + 2y - z = 10$ that lies in the cylinder $x^2 + y^2 = 4$?

Calculate the area of the part of the plane $2x + 2y - z = 10$ that lies in the cylinder $x^2 + y^2 = 4$.

What can I use to solve this? Can I use the Green's theorem?

答: Think about projecting onto the $xy$-plane. What is the relation between the two areas? Hint: The angle between the two planes is relevant.


问:
Nonconstructive proof of the irrationality of a power
I'm given some rational $x$ and irrational $y$, I want to show nonconstructively that that $\exists x \exists y(x^y)$ is irrational.

I'm given the example $4^{\sqrt 2}$. It's just a one-liner. $4$ is rational, $\sqrt 2$ is irrational, and so $4^\sqrt{2}$ is irrational.

I intuitively understand this, but is there a slightly more detailed explanation I can give to show that $4^\sqrt 2$ is irrational?

答:
Your "one-liner" is wrong.  $4$ is rational, $\log_4(9)$ is irrational, but $4^{\log_4(9)} = 9$ is rational.

EDIT:

What is true (by the Gelfond-Schneider theorem) is that $4$ is an algebraic number other than $0$ or $1$, $\sqrt{2}$ is an irrational algebraic number, and therefore $4^{\sqrt{2}}$ is irrational (in fact transcendental).

But I think you were looking for something like this.  Consider $x = 2$, $y = 1/\sqrt{2}$.  If $2^{1/\sqrt{2}}$ is a rational number $r$, then
$$ r^{1/\sqrt{2}} = (2^{1/\sqrt{2}})^{1/\sqrt{2}} = 2^{1/2}$$
which we know is irrational.  Therefore there is at least one example with $y = 1/\sqrt{2}$ irrational and $x$ rational, either $x = 2$ or (if $2^y$ happens to be rational) $x=2^y$.


问:
Estimate of $\prod_{d\mid n}(d+1)$.
I would like to ask if there is any cool estimate of $\prod_{d\mid n}(d+1)$.

I know that $\prod_{d\mid n}d=n^{\tau(n)/2}$ ($\tau(n)$ is the number of divisors of $n$) so we have the trivial estimate $\prod_{d\mid n}(d+1)>n^{\tau(n)/2}$.

The question is, is there any hope for something better? The answer seems to be yes, since for big $\tau(n)$ the difference $\prod_{d\mid n}(d+1)-\prod_{d\mid n}d$ seems to be large, but I was unable to find something precise.

答:
If you're looking for asymptotic formulae on the average $\sum_{n<N}\prod_{d|n}(1+d)$, then I do not think that you will find any. Let $F(n)=\prod_{d|n}(1+d)$ and let us assume that we have indeed found representation in the form

$$\frac{1}{N}\sum_{n<N}F(n)=M(N)+R(N)$$

where $R(N)$ is some "small" error term and $M(N)$ is a main term. The weakest possible equation of this form would be $R(N)=o\left(M(N)\right)$, i.e

$$\frac{1}{N}\sum_{n<N}F(n)\sim M(N)$$

This would imply that

\begin{align*}
\left|\frac{1}{N}\sum_{n<N}\log(F(n))-\log(M(N))\right|&=\left|\frac{1}{N}\sum_{n<N}\log\left(\frac{F(n)}{M(n)}\right)\right|\\
&=\left|\frac{1}{N}\sum_{n<N}\log\left(1+\frac{F(n)-M(n)}{M(n)}\right)\right|\\
&\sim \left|\frac{1}{N}\sum_{n<N}\frac{F(n)-M(n)}{M(n)}\right|\\
&=\frac{R(n)}{M(n)}\\
&=o(1)
\end{align*}

The sums $\sum_{d|n}\log(1+d)$, however, do not go to zero and are not perfectly behaved and so it would be unreasonable to expect a formula in the form

$$\frac{1}{N}\sum_{n<N}\sum_{d|n}\log(1+d)=M(N)+o(1)$$

we can, however, examine

\begin{align*}
\frac{1}{N}\sum_{n<N}\sum_{d|n}\log(1+d)&=\frac{1}{N}\sum_{n<N}\sum_{d|n}\log\left(1+\frac{1}{d}\right)+\frac{1}{N}\sum_{n<N}\sum_{d|n}\log\left(d\right)
\end{align*}

We get now that

\begin{align*}
\frac{1}{N}\sum_{n<N}\sum_{d|n}\log\left(1+\frac{1}{d}\right)&=\frac{1}{N}\sum_{n<N}\sum_{d|n}\frac{1}{d}+O\left(\frac{1}{N}\sum_{n<N}\sum_{d|n}\frac{1}{d^2}\right)\\
&=\frac{1}{N}\sum_{d<N}^{ }\sum_{n<\frac{N}{d}}^{ }\frac{1}{d}+O\left(\frac{1}{N}\sum_{d<N}^{ }\sum_{n<\frac{N}{d}}^{ }\frac{1}{d^{2}}\right)\\
&=\sum_{d<N}^{ }\frac{1}{d^{2}}+O\left(\frac{\log\left(N\right)}{N}\right)+O\left(\sum_{d<N}^{ }\frac{1}{d^{3}}\right)\\
&=O\left(1\right)
\end{align*}

and so

$$\frac{1}{N}\sum_{n<N}\sum_{d|n}\log(1+d)=\frac{1}{N}\sum_{n<N}\log(n)\frac{\tau(n)}{2}+O(1)$$

you can now estimate this RHS with whatever classical results you want on the distribution of $\tau(n)$. The best results have a worse than $O(1)$ error (obviously) and so you'll get a claim of equal strength to what is state of the art.

Title: Estimate of $\prod_{d\mid n}(d+1)$.


问:
Polynomial with geometric progression
If the roots of an equation with real coefficients $ax^3+bx^2+cx+d=0, a\ne0$ form a geometric progression, prove $ac^{3}=db^{3}$

I have no idea how to get started, so if it can a few hints?

答:
Say, $x_1,x_2,x_3$ are the roots, then by Vieta we have $$x_2^3= x_1x_2x_3 = -{d\over a}$$ and $$x_2(x_1+x_2+x_3) ={c\over a}$$
and
$$x_1+x_2+x_3 =-{b\over a}$$ So from the alst two equation we get $x_2=-{c\over b}$ so putting this one in first equation we get $$-{b^3\over c^3} = -{d\over a}$$ and thus a conclusion.

Title: Polynomial with geometric progression