问:
Is ${s_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{{3^{i - 1}}}}} = \dfrac{3}{2}\left( {1 - \dfrac{1}{{{3^n}}}} \right)$?
Here in example 4 this result is shown.

${s_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{{3^{i - 1}}}}}  = \dfrac{3}{2}\left( {1 - \dfrac{1}{{{3^n}}}} \right)$

Now without looking at its solution I was trying like this:

${s_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{{3^{i - 1}}}}}=3\sum\limits_{i = 0}^n {\dfrac{1}{{{3^{i }}}}}=3\bigg(\dfrac{1-(\frac{1}{3})^{n+1}}{1-\frac{1}{3}}\bigg)=\mathop {\lim }\limits_{n \to \infty }\dfrac{3}{2}\cdot3(1-(\frac{1}{3})^{n+1})=\dfrac{9}{4}$

What am I doing wrong why I am not getting answer $\dfrac{3}{2}$

答:
$S_n=\sum_{i=1}^n \frac{1}{3^{i-1}}=1+\frac 13+...+\frac{1}{3^{n-1}}=\frac{1-\frac{1}{3^n}}{1-\frac13}=\frac 32(1-\frac{1}{3^n})$
Thus $\lim_{n \to\infty}S_n=\frac 32$.
If you are considering the infinite geometric series with $|r|<1$ then sum=$\frac{a}{1-r}$ where $a$ is the first term of the series.  Here $a=1,r=\frac 13$. So sum is $\frac{1}{1-\frac 13}=\frac 32$


问:
What is the explicit smooth structure of an embedded submanifold?
Here is a quotation from Chapter 5 in "Introduction to smooth manifolds" by J. Lee.

Suppose $M$ is a smooth manifold with or without boundary. An embedded submanifold of $M$ is a subset $S \subseteq M$ that is a manifold (without boundary) in the subspace topology, endowed with a smooth structure with resect to which the inclusion map $S\to M$ is a smooth embedding.

My questions are the following:

What is the topology on $S$?

What is the smooth structure on $S$?

In my thought, $\{i^{-1}(U\cap S)\mid U \in \mathcal{O}(M)\}$ is the topology on $S$ with the topology $\mathcal{O}(M)$ on $M$.

For the second question, I think $\mathcal{A}_{S}:=\{(i^{-1}(U\cap S), \phi\circ i)\mid (U,\phi)\in \mathcal{A}_{M}\}$ is the smooth structure on $S$ with the smooth structure $\mathcal{A}_{M}$ on $M$, but it seems to be odd for me if $\mathrm{dim}(S)<\mathrm{dim}(M)$ because the dimension of the image under $\phi\circ i$ is $\mathrm{dim}(M)$, which is absurd.

Is there anything wrong with the above discussion?

答: A priori, the smooth structure on $ S $ can have no relationship to that of $ M $. However, as mentioned and proven later in the book, for any embedded submanifold $ S\subseteq M $, its smooth structure must be given by the slice charts of $ S $. More specifically, any chart $ (U,\phi=(x^1,\ldots,x^n)) $ of $ M $ such that $ \phi(S\cap U) $ is given by $ x^{k+1}=\cdots=x^n=0 $, where $ k=\dim S $, gives rise to the chart $ (U\cap S,(x^1,\ldots,x^k)) $ of $ S $.

ContentLicense: CC BY-SA 4.0


问:
How to evaluate the integral $\int e^{x^3}dx $
How to evaluate the integral
$$\int e^{x^3}dx \quad ?$$

I've tried to set $t=x^3$, but it seems to be a blind alley; I don't know what to do with $\int\frac{e^t}{3\sqrt[3]{t^2}}dt$.

ContentLicense: CC BY-SA 3.0

Body:
$$\int e^{x^3}dx=\int \sum_{n=0}^{\infty }\frac{x^{3n}}{n!}dx$$

$$\int \sum_{n=0}^{\infty }\frac{x^{3n}}{n!}dx=\sum_{n=0}^{\infty }\frac{x^{3n+1}}{(3n+1)(n!)}+c$$

$$\frac{1}{3n+1}=\frac{(\frac{1}{3})^{(n)}}{(\frac{4}{3})^{(n)}}$$

$$\sum_{n=0}^{\infty }\frac{x^{3n+1}}{(3n+1)(n!)}+c=x\sum_{n=0}^{\infty }\frac{(\frac{1}{3})^{(n)}(x^3)^n}{(\frac{4}{3})^{(n)}(n!)}+c$$

$$x\sum_{n=0}^{\infty }\frac{(\frac{1}{3})^{(n)}(x^3)^n}{(\frac{4}{3})^{(n)}(n!)}+c=\ x\ 1F1(\frac{1}{3};\frac{4}{3};x^3)+c$$

so $$\int e^{x^3}dx=\ x\ {}_1F_1(\frac{1}{3};\frac{4}{3};x^3)+c$$

where ${}_1F_1$ is Hypergeometric Function of the First Kind


问:
detail in proof of the mean value theorem for integrals
I am trying to prove the following result

If $f$ is a continuous function on a closed interval $[a,b]$, then
there exists a $c \in (a,b)$ such that,

$$f(c)=\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x$$

I have worked out the following proof

Because $f$ is continuous on the interval, it reaches a minimum $m$, and a maximum $M$ so that we have
$$\begin{aligned} m&\leq f(x)\leq M \\ m(b-a)\leq  & \int_a^bf(x)\;\mathrm{d}x\leq M(b-a) \\  m\leq &\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x\leq M\end{aligned}$$
and the IVT guarantees the exxistence of $c$ as above.

How do I prove that $c \neq a\ \& \ c\neq b$?

Thanks for the help !

答:
Here is another approach. Assume that $$f(x) > K=\frac{1}{b-a}\int_a^b f(x) \, dx$$ for all $x\in(a, b) $. This means that $g(x) =f(x) - K>0$ for all $x\in(a, b) $. Hence $$\int_{a} ^{b} g(x) \, dx>0$$ ie $$\int_{a} ^ {b} f(x) \, dx>K(b-a) $$ which is absurd.

Similarly one can show that $f(x) <K$ for all $x\in(a, b) $ can't hold. Thus by intermediate value theorem we must have $f(c) =K$ for some $c\in(a, b) $.


问:
Meaning of complex trigonometric ratios
I have encountered about finding trigonometric ratios of complex arguments (a+ib) those can be computed via series expansion and Euler formula. But my question is what is the definition of these trigonometric ratios as i have learnt these ratios where defined via triangles and for real arguments. It's also said that cosine and sine both have periods 2nπ for both complex and real arguments. What's the justification for all these as these where defined and found for real arguments only,how do we extend this to complex numbers. Also i want to know the meaning of these ratios as a real trigonometric ratios have meaningful meaning. Any help please. Thank you</p>

答:
There are relations between trigonometric and hperbolic functions. For example
$$\sin(ix)=i \, \sinh(x)$$
$$\cos(ix)=\cosh(x)$$
$$\tan(ix)=i \, \tanh(x)$$

Suing the classical expansions, then
$$\sin(a+ib)=\sin (a) \cosh (b)+i \cos (a) \sinh (b)$$
$$\cos(a+ib)=\cos (a) \cosh (b)-i \sin (a) \sinh (b)$$

Title: Meaning of complex trigonometric ratios


问:
Prove that a wff built up only with $\lnot$ and $\leftrightarrow$ is a tautology iff $\lnot$ and each statement letter occur an even number of times.
First of all, I know how to prove this theorem below:

A statement form that contains $\leftrightarrow$ as its only connectives is a tautology iff  each statement letter occurs an even number of times.

To prove the theorem, I first derived this corollary:

A statement built up using only $\leftrightarrow$ is true iff there are an even number of false terms.

But now, I have to prove the theorem below:

Proving that a statement form that contains $\lnot$ and $\leftrightarrow$ as its only connectives is a tautology iff $\lnot$ and each statement letter occur an even number of times.

The bold texts are the only differences comparing it to the previous question that I could solve. I tried modifying the corollary to help me prove the theorem but it didn't help me, or modifying the corollary made it impossible to prove. Is there anything else I could do?

答: I believe your goal follows from the corollary and the fact that $\lnot\phi$ is equivalent to $\phi\iff\text{false}$.


问:
Trig values of $45^\circ+\alpha$
If $\sin\alpha=\dfrac{4}{5}$ and $\alpha\in(45^\circ;90^\circ)$ find the values of the other trig functions of $\alpha$ and the values of the trig functions of $45^\circ+\alpha$.

Using the indentity $\sin^2\alpha+\cos^2\alpha=1$ I got that $\cos\alpha=\dfrac35,\tan\alpha=\dfrac43$ and $\cot\alpha=\dfrac34.$ How can I find the values of the trig functions of $45^\circ+\alpha?$

答:
Hint:

$\sin(A+B) = \sin A \cos B+ \cos A \sin B$

$\cos(A+B) = \cos A\cos B-\sin A \sin B $

To prove using unit circle:

Equate $PQ_1 = RT$ because both the angles are equal

https://medium.com/@nubtrek/easy-with-sin-a-b-proof-vuja-de-50c15fca3344

Title: Trig values of $45^\circ+\alpha$


问:
Midpoint of a line segment passing through two points.

Let $C_1$ be a circle defined with $X(-2,7)$ and $Y(2,-5)$ as the endpoints of the diameter of the circle.
Let $C_2$ be a circle defined with $Y(2,-5)$ and $Z(4,-11)$ as the endpoints of the diameter of the circle.

Let $A = C_1 \cup C_2 - Y$

Let $B$ be the set of all midpoints of the set of all straight line segments joining two distinct points in $A$ and passing through the point $Y$. Find $B$.

My attempt:

Since $XYZ$ are collinear $C_1$ and $C_2$ meet at a single point: $Y$. I can easily find the set of all midpoints of the line segments joining a point (say $T$) in $C_1$ and $Y$. Let $M(x,y)$ be the midpoint of $TY$ and $N(0,1)$ be the center of $C_1$. Notice that $MN$ is perpendicular to to $MY$. Therefore we can build the equation $\dfrac{y-1}{x-0}*\dfrac{y+5}{x-2}=-1$ i.e product of their slopes equal -1. Similarly, the set of all midpoints of the line segments joining a point (say $U$) in $C_2$ and $Y$ can be found.
The issue now is, I don't know how to use this information to compute the required locus $B$. I don't even know if this information is useful.

答:
Notice that transformation $A\mapsto B$ from bigger circle which has radius $2\sqrt{10}$: $$XY  = \sqrt{4^2 +12^2} = 4\sqrt{10}$$ and center at $O(0,1)$ to smaller circle with radius $\sqrt{10}$ and center at $O'(3,-8)$ is homothety with center at $Y$ and dilatation factor $-1/2$. So, for all $A$ we have $${BY\over YA} = {1\over 2}\implies {AB\over YA} ={3\over 2}$$

Then $${YM \over YA} = 1- {AM\over YA}  = 1-{AB\over 2YA}  = {1\over 4} $$

So $M$ is actually a picture of $A$ under homothety at $Y$ with dilatation factor $1/4$, so $M$ descibes a circle with radius $r=\sqrt{10}/2$ and center at $({1\over 2},-{7\over 2})$.


问:
Why is there a negative sample space in this probability question?
So there are multiple ways of solving the following problem but when I try the table method this happens:

Given $A$ and $B$ are events of a sample space $S$ and $P(A\cap B) = \frac{2}{5}$ and $P(A\cap B') = \frac{3}{7}$ then what is $P(B'\mid A)$?

My table looks like this

A
B

A
15
14
29

B
14
???
6

29
6
35

$B$ would have to have a negative value which is impossible. Any help?

答:
You have been promised that $\mathsf P(A\cap B)=14/35, \mathsf P(A\cap B')=15/35$, which entails that $\mathsf P(A)=29/35$ and hence $\mathsf P(A')=6/35$.

However we have no way to evaluate $\mathsf P(B)$.   All we can say is that $14\leqslant\mathsf P(B)\leqslant 20$.

We might indicate that $\mathsf P(B\cap A')=x/35$ for some unknown $x: 0\leqslant x\leqslant 6$.

Hence the table shall be:

$$\begin{array}{|c|c:c|c|}\hline &A&A'\\[1ex]\hline B&14&x&14+x\\[1ex]\hdashline B'&15&6-x&21-x\\[1ex]\hline & 29& 6&35\\\hline\end{array}$$

Now, you can use this table to evaluate $\mathsf P(B'\mid A)$....


问:
Convex conjugacy and differentiating under the "sup" operator
I am working through example 3.10 (page 71) of a book on variational inference in statistics and the authors discuss taking the gradient of the following function with respect to $\mu$:
$$
A^*(\mu) = \sup_{\theta \in \mathbb{R}} \{\, \theta \cdot \mu - \log(1 + \exp(\theta))\,\}.
$$

They state that the stationary condition $\nabla_\mu A^*(\mu) = 0$ is attained at the location $\mu = \exp(\theta)/(1+\exp(\theta))$.

I'm wondering how to derive this since $\nabla A^*(\mu)$ seems to require pushing the gradient operator through the "sup" operator which has no closed form and which may not even be differentiable!

Any guidance or trick for how this stationary condition is derived is very much appreciated

答:
I think you might be confusing something here. The authors construct the conjugate function of $A(\theta)$, noted as $A^*(\mu)$. In order to do so, they solve the maximization problem $(3.47)$, which is done by setting $\nabla_{\theta}A=0$. They then analyze the solution of $\nabla_{\theta}A=0$ for different values of of $\mu$, which they denote as $\theta(\mu)$.

They arrive to the conclusion that $A^*(\mu)=\mu\log\mu+(1-\mu)\log(1-\mu)$ over the domain $\mu\in[0,1]$. When $\mu<0$ or $\mu>1$ the original conjugate maximization problem is unbounded, therefore $\mu$ is restricted to that line interval.


答:
Let $A(\theta)=\log(1+\exp(\theta))$. Then, $A$ is a proper, continuous and convex function.

In this case, it holds $(\partial A)^{-1} = \partial A^*$, where $\partial A$ stands for the subdifferential of $A$
(see Corollary 16.30 in "Convex Analysis and Monotone Operator Theory in Hilbert Spaces, 2nd edition").
By using this formula, we obtain $\nabla A^*(\mu)= (\nabla A)^{-1}(\mu)$.

Title: Convex conjugacy and differentiating under the "sup" operator

Body:
I am working through example 3.10 (page 71) of a book on variational inference in statistics and the authors discuss taking the gradient of the following function with respect to $\mu$:
$$
A^*(\mu) = \sup_{\theta \in \mathbb{R}} \{\, \theta \cdot \mu - \log(1 + \exp(\theta))\,\}.
$$

They state that the stationary condition $\nabla_\mu A^*(\mu) = 0$ is attained at the location $\mu = \exp(\theta)/(1+\exp(\theta))$.

I'm wondering how to derive this since $\nabla A^*(\mu)$ seems to require pushing the gradient operator through the "sup" operator which has no closed form and which may not even be differentiable!

Any guidance or trick for how this stationary condition is derived is very much appreciated


问:
Is this a correct translation from english into symbolic logic?
"You can fool some of the people all of the time, and you can fool all of the people some of the time, but you can’t fool all of the people all of the time." (Abraham Lincoln)

Let

$P$ be "fooling some of the people all of the time",

$Q$ be "fooling all people some of the time",

$R$ be "fooling all of the people all of the time".

$(P \lor Q) \rightarrow \neg R$

Is this a correct translation in propositional logic?

答:
No, a correct formalization of Lincoln's sentence in propositional logic is the following:

$$(P \lor Q) \land \lnot R$$

Indeed, from a logical point of view, "but" as the same meaning as "and".
Note that I translated the "and" between the first two propositions by an "or", because in this context the two propositions express an alternative.

By the way, propositional logic is not the best logic to formalize this kind of sentences. First-order logic and modal logic can express a more faithful formalization of Lincoln's sentence.


问:
Finding the maximizer
Let be $ A\in \mathbb{R}^{2,2} $ a symmetrical and positive definite matrix with different eigenvalues $ \lambda_1>\lambda_2>0 $. Further we have the rayleigh quotient $$ \max_{x\neq 0}\frac{\overline{x}^T\cdot A\cdot x}{\overline{x}^T\cdot x}=\max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$.
For how many points $ x\in \mathbb{R}^2 $ the maximum $$ \max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$ is reached?

My idea:
Let $$ A:=\begin{pmatrix}a&b\\b&c\end{pmatrix},\quad x:=\begin{pmatrix}x_1\\x_2\end{pmatrix} ,\quad \|x\|_2^2=1=x_1^2+x_2^2$$
and define
$$ F(x)=x^T\cdot A\cdot x=(x_1, x_2)\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}x_1\\x_2\end{pmatrix}\\=(x_1, \sqrt{1-x_1^2})\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}x_1\\\sqrt{1-x_1^2}\end{pmatrix}=:F(x_1) $$.

Derivative:
$$ \nabla F(x_1)=\left(1,\frac{-x_1}{\sqrt{1-x_1^2}}\right)\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}1\\\frac{-x_1}{\sqrt{1-x_1^2}}\end{pmatrix}\stackrel{!}{=}0 $$.

From here I get stuck and I don't see how can I use the information about $ A $, the eigenvalues and the rayleigh qoutient.

答:
Note that the set $\|x\|=1$ is compact, so a $\max$ exists.

Solve $\max_{ \|x\|^2 =1 } {1 \over 2}x^T A x$ using Lagrange multipliers to get
$Ax = \lambda x$, so we see that the Lagrange multiplier must be an eigenvalue.

Hence $\max_{ \|x\|^2 =1 } x^T A x = \lambda_\max$ and in the given example,
since the eigenvalues are distinct we see that $x$ must be an eigenvector and hence the two solutions are $\pm x$.

Title: Finding the maximizer


问:
Indefinite integral of $\int \frac 1 x \operatorname{arsech} \frac x a \, \mathrm d x$
Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.668$ gives the indefinite integral of the area hyperbolic secant (that is, the "inverse" hyperbolic secant) as:

$$\int \frac 1 x \operatorname{arsech} \frac x a \, \mathrm d x = -\frac 1 2 \ln (a/x) \ln (4 a/x) - \frac {(x/a)^2} {2 \cdot 2 \cdot 2} - \frac {1 \cdot 3 (x/a)^4} {2 \cdot 4 \cdot 4 \cdot 4} - \cdots$$

This can be expressed as:

$$\int \frac 1 x \operatorname{arsech}  \frac x a \, \mathrm d x = -\frac 1 2 \ln \left({\frac a x}\right) \ln \left({\frac {4 a} x}\right) - \sum_{n \mathop \ge 0} \frac {(2 n)!} {2^{2 n} (n!)^2 (2 n)^2} \left({\frac x a}\right)^{2 n} + C$$

Getting nearly there is easy enough.

I take as my starting point the power series expansion of the area hyperbolic cosine of $\dfrac x a$:

$$\operatorname{arcosh} x = \ln \frac {2 x} a - \left({\sum_{n \mathop = 1}^\infty \frac {(2 n)!} {2^{2 n} (n!)^2 (2 n)} \left({\frac a x}\right)^{2 n} }\right)$$

(This is derived from the result in Spiegel, item $20.40$.)

From it I get the same for the area hyperbolic secant $\operatorname{arsech}$ of $\dfrac x a$, as it's just the $\operatorname{arcosh}$ of $\dfrac a x$:

$$\operatorname{arsech} x = \ln \frac {2 a} x - \left({\sum_{n \mathop = 1}^\infty \frac {(2 n)!} {2^{2 n} (n!)^2 (2 n)} \left({\frac x a}\right)^{2 n} }\right)$$

and integrate term by term (justified by Fubini's theorem, I believe) to get me eventually to:

$$\int \frac 1 x \operatorname{arsech} \frac x a \, \mathrm d x = -\frac 1 2 \ln^2 \left({\dfrac x {2 a} }\right) + \sum_{n \mathop = 0}^\infty \frac {(2 n)!} {2^{2 n} (n!)^2 (2 n)^2} \left({\frac x a}\right)^{2 n} + C$$

During the course of the above I inverted the reciprocal in the logarithm of the integrand to get it into a standard form for integration $\displaystyle \int \dfrac {\ln (c x)} x \mathrm d x = \dfrac {\ln^2 {c x} } 2$ which I think I got right.

The above is consistent with the result quoted in Schaum for the indefinite integral for the area hyperbolic cosine, where the logarithm term was left as a square.

So there are $2$ questions outstanding:

How do you actually get from $\ln^2 \left({\dfrac x {2 a} }\right)$ to $\ln \left({\dfrac a x}\right) \ln \left({\dfrac {4 a} x}\right)$? I get that they will differ by a constant which can be subsumed into a constant of integration, but manipulation of $\ln^2 \left({\dfrac x {2 a} }\right)$ gets me only as far as $\left({\ln \left({\dfrac a x}\right)}\right)^2 + 2 \ln 2 \ln \dfrac a x + \left({\ln 2}\right)^2$ and at this point I can't see how to proceed. I can't reduce the $2 \ln 2 \ln \dfrac a x$ and get it to go the way I want it to.

How did Spiegel ever get to that $\ln \left({\dfrac a x}\right) \ln \left({\dfrac {4 a} x}\right)$ term in the first place? His quoted result for the indefinite integral for the area hyperbolic cosine has that term in the $\ln^2$ form, the same as what I got for the area hyperbolic secant. That is, what direction could he have taken with his integration so as to land upon a result in that form? I can't see why he would deliberately manipulate it into that form, as he is happy enough to leave the indefinite integral for the area hyperbolic cosine in the $\ln^2$ form.

答:
Since you are calculating indefinite integral, the inconsistency is obvious, the difference between two results must be a constant!

The difference:

$$\ln \left(\frac{a}{x}\right) \ln \left(\frac{4 a}{x}\right)-\ln ^2\left(\frac{x}{2 a}\right)=
\ln \left(\frac{a}{x}\right) \left[\ln \left(\frac{a}{x}\right)+2 \log (2)\right]-\left[\ln \left(\frac{a}{x}\right)+\ln (2)\right]^2
=-\ln ^2(2)$$

The reason: the real domain of ${\rm arcsech(x)}$ is $[0,1]$, thus when you change the operations of expansion and integral, you must be careful, i.e.
$$
\int\to\int_a^x
$$
In other words, you should have
$$
\int_a^x \frac{\ln (2 a)-\ln (z)}{z} \, dz=-\frac{1}{2} \ln \left(\frac{a}{x}\right) \ln \left(\frac{4 a}{x}\right)
$$


问:
Prove that the optimal value is $f(x)=\sum_{i=1}^{k}x_{[i]}$ where $x_{[i]}$ is the ith largest values in the vector $x$

Given the problem $\max x^Ty$

s.t $e^Ty=k$

$0\leq y\leq \mathbb{e}$

Prove that the optimal value is $f(x)=\sum_{i=1}^{k}x_{[i]}$ where $x_{[i]}$ is the ith largest values in the vector $x$

I've tried at first with lagrangian but it didn't work(if someone can prove it like this I would like to see).
So I tried the following assume that $y^*$ is optimal vector of the given problem where $y_i^*,y_j^*<1$(wlog only 2 indices) such that $y_i^*+y_j^*=1$ and $x_j\geq x_i$
Looking at the optimal value only at the two coordiants $i,j$ we get $x_iy_i^*+x_jy_j^*\leq(y_i^*+y_j^*)x_j=x_j$ and therefore the optimal value is $f(x)$.

I'm not sure if the way I've done it is right

Note-this excerise is given from duality section

答:
Since the constraint set is compact we know there is a maximiser.
We must have that the dimension $n \ge k$ otherwise there are no solutions. I am assuming that $k$ is a natural number.

We can assume that $x_1 \ge x_2 \ge  \cdots $.

Suppose $y$ is an optimal solution.

If $y_1 <1$ then $\sum_{i>1} y_k = k-y_1 > k-1 \ge 1$ and so we can
reduce $y_n, y_{n-1},...$ and add the combined reduction to $y_1$ so that $y_1 = 1$
without increasing the cost. (There must be enough non zero elements in $y_2,...$ so that we can do that and the cost does not decrease because of the ordering of the $x_i$.)

We can see that $y'=(y_2,...,y_n)$ must solve the problem
$\max_{e^T y'=k', \ 0 \le y' \le e} (x')^T y'$, where $x' = (x_2,...)$ and $k'=k-1$.

We can repeat as long as $k' >0$.

Hence we can take a maximiser to be $(1,....,1,0,...,0)$ where there
are $k$ ones and hence the optimal value is
$x_1+\cdots + x_k$.


问:
How to solve the coupled differential equations with 3 variables?
I have the following set of differential equations,
\begin{aligned}
& x' - \alpha(yz' - zy') =  -\beta(yP_z -zP_y) \\
& y' - \alpha(zx' - xz') =  -\beta(zP_x -xP_z) \\
& z' - \alpha(xy' - yx') =  -\beta(xP_y -yP_x)
\end{aligned}
I know how to implement runga kutta 4th order or Euler or Heun's method...what I can't figure out is how to update the $x'$ or $y'$ or $z'$ term while I am solving it. Any help or direction towards possible resources would be appreciated.

答:
On the left side you have a linear system in the first derivatives,
$$
\begin{bmatrix}
1&αz&-αy\\
-αz&1&αx\\
αy&-αx&1\\
\end{bmatrix}
\begin{bmatrix}
x'\\y'\\z'
\end{bmatrix}
=
\begin{bmatrix}
 -β(yP_z -zP_y)\\ -β(zP_x -xP_z) \\-β(xP_y -yP_x)
\end{bmatrix}
$$
In the evaluation of the ODE function, that is, the right side of the explicit first order system, you just have to construct these matrix and vector and call a solver for linear systems.

Then proceed as usual for explicit first-order systems.


问:
Solutions of the equation $x^x=\frac{1}{256}$
Find the solutions of the equation
$$x^x=\frac{1}{256}$$
I know that the function $f(x)=x^x$ is defined for $x>0$, so the solutions, if they exists, must be $>0$; as far as I know, this is because to maintain the formal properties of power, we must impose that $x>0$ or it is easy to get contradictions like $-1=(-1)^{\frac{2}{2}}=[(-1)^2]^{\frac{1}{2}}=1^{\frac{1}{2}}=1$.

So, since the function $x^x$ has an absolute minimum at $x=\frac{1}{e}$ and it is $\left(\frac{1}{e}\right)^{\frac{1}{e}}>\frac{1}{256}$, it follows that there isn't an $x>0$ such that $x^x=\frac{1}{256}$ and it follows that the equation hasn't real solutions.

However, Wolfram|Alpha says that the equation has the integer solution $x=-4$ and it is indeed a solution, as one can check by substitution. However, for what I said before, the function $x^x$ doesn't exist for $x<0$ and so it can't be evaluated for that value.

Why is this happening? Has it something to do with complex numbers? Or finding a solution of that equation is different to consider the function $x^x$ involved? Thank you.

Body: Wolfram seems to take as $$x^x = \frac{1}{4^4} = \frac{1}{(-4)^4} = (-4)^{-4}$$


问:
Method of moments estimator and Delta Method
I’ve been a couple of days trying to figure out a satisfactory solution to this exercise but I’m stuck. By how the exercise is phrased I think it wants me to use the moments method and I also think it is a Delta Method exercise.

The statement

I am given the following discrete distribution with $\theta>0$

$$p(x) = \left(\frac{\theta}{1+\theta}\right) ^{2-x}\left(\frac{1}{1+\theta}\right)^{x-1} \hspace{1cm} x=1,2$$

I need to calculate an estimator of $\theta$ (call it $T_n$) using a sample $x_1,x_2,...,x_n$, deduce its distribution and calculate a confidence interval for $\theta$.

What I did

If $X_1,...,X_n$ are n copies of $X$, the sample is $x_1=X_1(\omega),..., x_n=X_n(\omega)$

Using the method of moments we can relate the sample mean to the expectation

$$\overline X_n = E[X] = 1+\frac{1}{1+\theta}=\mu$$

and define the estimator of $\theta$

$$T_n=\frac{1}{\overline X_n-1}-1$$

supposing n is big enough so that $\overline X_n$ is not 1.

I also calculated the variance of X: $Var(X)=\frac{\theta}{(1+\theta)^2}=\sigma^2$

By the Central Limit Theorem

$$\sqrt{n}\,(\overline X_n-\mu) \rightarrow N(0,\sigma^2)$$

We can apply the Delta Method with the function $g(t)=\frac{1}{t-1}-1$ to get

$$\sqrt{n}\,(g(\overline X_n)-g(\mu)) \rightarrow N(0,\sigma^2g’(\mu)^2)$$

i.e.

$$\sqrt{n}\,(T_n-\theta) \rightarrow N(0,\theta(1+\theta)^2)$$

My Problem

Now I need to determine the distribution of $T_n$ but I don’t know the value of $\theta.$ I could approximate $\theta$ by $T_n$ and use Slutsky’s theorem to conclude that

$$\sqrt{n}\,\frac{(T_n-\theta)}{\sqrt{T_n}(T_n+1)} \rightarrow N(0,1)$$

But again I don’t know how to obtain the distribution of $T_n$.

答:
Your pmf can be rewritten in the following way

$$P(X=x)=\frac{1}{\theta+1}\theta^{2-x}$$

$X=1,2$ that is the following rv

$$ X =
\begin{cases}
\frac{\theta}{\theta+1},  & \text{if $X=1$} \\
\frac{1}{\theta+1},  & \text{if $X=2$}
\end{cases}$$

Now we can transform it into

$$Y= X -1=
\begin{cases}
\frac{\theta}{\theta+1},  & \text{if $Y=0$} \\
\frac{1}{\theta+1},  & \text{if $Y=1$}
\end{cases}$$

That is

$$Y=
\begin{cases}
1-p,  & \text{if $Y=0$} \\
p,  & \text{if $Y=1$}
\end{cases}$$

Concluding... $Y$ is bernulli $B(p)$.

A CSS (complete and sufficient statistic) for $p$ is

$$\Sigma_i Y_i=(\Sigma_i X_i) -n$$

this is a nice estimator for $p$. Now you can calculate the confidence interval for $p$ as an approximate interval using CLT (if $n$ is greater enough) or also an exact confidence interval using, i.e. the Statistical Method and a binomial table (or any calculator).

here I calculated a similar exact CI with Excel (both for Poisson and Bernulli distribution, thus I think it can be  useful for you).

Confidence interval for $\theta$

$\frac{1}{\theta+1}=p$ is a monotonic function, thus given the CI for $p$ you can easily derive also the CI for $\theta$.

Distribution of $\hat{\theta}$

$\Sigma_i  Y_i$ is binomial distributed, thus also the distribution of $\hat{\theta}$ is binomial...only with a modified support.

Note that using this method of estimating $\theta$, you can derive the same estimator you found with MoM.


答:
Yes, this does have to do with complex numbers. As you correctly observed; blindly taking powers like $(-1)^{\frac12}$ will lead to seeming contradictions.

Thus in complex-analysis, we make sure to be very specific in what we mean with $(-1)^{\frac12}$. This is done by defining the complex log, where we need to pick a branch. This way we can consider $x^{x}$ on a suitable space, determined by the choice of the branch. We can then vary the branch, to check all of $\mathbb{C}$.

Hope this was enough to show you that the contradictions can be avoided and the $x<0$ cases can be considered.

Title: Solutions of the equation $x^x=\frac{1}{256}$

Body:
Find the solutions of the equation
$$x^x=\frac{1}{256}$$
I know that the function $f(x)=x^x$ is defined for $x>0$, so the solutions, if they exists, must be $>0$; as far as I know, this is because to maintain the formal properties of power, we must impose that $x>0$ or it is easy to get contradictions like $-1=(-1)^{\frac{2}{2}}=[(-1)^2]^{\frac{1}{2}}=1^{\frac{1}{2}}=1$.

So, since the function $x^x$ has an absolute minimum at $x=\frac{1}{e}$ and it is $\left(\frac{1}{e}\right)^{\frac{1}{e}}>\frac{1}{256}$, it follows that there isn't an $x>0$ such that $x^x=\frac{1}{256}$ and it follows that the equation hasn't real solutions.

However, Wolfram|Alpha says that the equation has the integer solution $x=-4$ and it is indeed a solution, as one can check by substitution. However, for what I said before, the function $x^x$ doesn't exist for $x<0$ and so it can't be evaluated for that value.

Why is this happening? Has it something to do with complex numbers? Or finding a solution of that equation is different to consider the function $x^x$ involved? Thank you.


问:
Why do we represent each column by x and y values in column picture of a system of linear equations
I started watching Gilbert Strang's Linear Algebra lectures. In the first lecture, he has the following system of linear equations

$$\begin{cases}
 2x - y = 0 \\
 -x + 2y = 3
\end{cases}
$$

When he talks about column picture of the equations, he writes the equations as the following linear combination of columns

$X\begin{bmatrix} 2 \\ -1\end{bmatrix}$ + $Y\begin{bmatrix} -1 \\ 2\end{bmatrix}$ = $\begin{bmatrix} 0 \\ 3\end{bmatrix}$

And then, he shows how it looks geometrically as follows

One thing that i don't understand from the above representation is, how can we consider two coefficients of X (2 and -1) and plot them as if they are X and Y coordinates of a vector?

答:
We didn't represent X,Y tables of data in X,Y plots before 1600.

It was a sort of discovery, and you can test it, predict, and so on.

It is useful.

In the same way, you can look at those coefficients as being vectors, and it's very insightful. Because from their shape and values, you can derive properties of solutions etc.

It is exactly the same with matrices.


问:
when a bounded set in $W^{2,p}(\Omega)$ has a strong limit point in $C(\overline{\Omega})$?
I just have a question in functional analysis. The question says:

If $\Omega \subset \mathbb{R}^3$ is bounded, and suppose that $H$ a bounded set in the Sobolev's space $W^{2,p}(\Omega)$. Then what values of $p \geq 1$ makes $H$ have a strong  strong limit point in $C(\overline{\Omega})$?

Thank you!

答:
I will assume $\Omega$ has sufficiently regular boundary (Lipschitz is sufficient).

There is a lot of unnecessary information in the question which makes it seems complicated, but this appears to boil down to whether the embedding
$$ W^{2,p}(\Omega) \hookrightarrow C(\overline\Omega)$$
holds. By the general Sobolev inequality this is the case for $p > \frac32;$ you can see this follows from the Gagliardo-Nirenberg and Morrey inequalities noting that if $p \in (\frac32,3)$ we have
$$ W^{2,p}(\Omega) \hookrightarrow W^{1,\frac{3p}{3-p}}(\Omega) \hookrightarrow C^{0,2-\frac3p}(\overline\Omega) \hookrightarrow C(\overline\Omega),$$
and if $p > 3$ we have
$$ W^{2,p}(\Omega) \hookrightarrow C^{1,1-\frac3p}(\overline\Omega) \hookrightarrow C(\overline\Omega).$$
(If $p=3$ you can note that $W^{2,3}(\Omega) \hookrightarrow W^{2,2}(\Omega) \hookrightarrow C(\overline\Omega)$ for instance.)

Given this, for $p>\frac32$ if $H \subset W^{2,p}(\Omega)$ we have any limit point $u \in \overline{H} \subset W^{2,p}(\Omega)$ evidently lies in $C(\overline\Omega)$ by the above embedding. Therefore $p>\frac32$ is a sufficient condition.

If $p \leq \frac32,$ it well known that this embedding fails, and you can find $u \in W^{2,p}_0(\Omega) \setminus C(\overline\Omega).$ Then take a sequence  $(u_n)$ of $C^{\infty}_c(\Omega)$ functions such that $u_n \to u \in W^{2,p}_0(\Omega),$ and consider $H = \{u_n\} \subset C(\overline\Omega).$


问:
how many integers $n$ have the property that there exist positive integers $a,b,c$ such that $a^n+b^ n=c^n$?
I want to know how to solve this question. I found this question in an online test but was not able to find the answer. Later after the test, when I reviewed the test, it showed the answer as $4$ but I want to know how the answer ends up in $4$.

I tried a lot but wasn’t able to find the answer.

Keep in mind that I’m a beginner and
$a,b$ and $c$ belong to $Z^+$ and $n$ belongs to $Z$

Thanks in Advance

答:
Fermart last theorem says that if $a,b,c,n\in\mathbb{Z}$ with $n\ge3$, then the equation $a^n+b^n=c^n$ means that at least one of $a,b,c$ are zero.

So, $n<3$. If $n\le-3$, $-n\ge3$, we have $$\begin{split}a^n+b^n&=c^n\\\bigg(\frac1a\bigg)^{-n}+\bigg(\frac1b\bigg)^{-n}&=\bigg(\frac1c\bigg)^{-n}\\(abc)^{-n}\Bigg(\bigg(\frac1a\bigg)^{-n}+\bigg(\frac1b\bigg)^{-n}\Bigg)&=(abc)^{-n}\bigg(\frac1c\bigg)^{-n}\\(bc)^{-n}+(ac)^{-n}&=(ca)^{-n}\end{split}$$

which has no solutions from Fermart Last Theorem.

So it remains to check the case when $n=-2,-1,0,1,2$. The casework is left for an exercise.


问:
How to determine if the process is martingale?
I have that $\{X_t; t=1,2... \}$ is an i.i.d random variables such that $P(X_t=1)=p=1/4$ and $P(X_t=-1)=1-p=3/4$. Let $S_0=0$ and $S_t=\sum_{i=1}^tX_i$ for $t=1,2...$

Let finally $Y_t=a^{S_t}$ for $t=0,1,2...$ where $a$ is a constant (not zero).

The question is, for which values of $a$ is $Y_t; t=0,1,2..$ a martingale with respect to $X_t; t=0,1,2... ? \quad$

Could somebody help me with this? I have tried with the definition of a martingale $E[a^{\sum_ {i=1}^{t+1}X_i}|X_t, X_{t-1}...]$ but I am stuck here

答: $E[a^{\sum_ {i=1}^{t+1}X_i}|X_t, X_{t-1}...]=a^{\sum_ {i=1}^{t+1}X_i}Ea^{X_t}$. So we get a martinagle if and only if $Ea^{X_t}=1$ for all $t$. Can you compute $Ea^{X_t}$ and solve $Ea^{X_t}=1$ for $a$?


问:
Second-largest singular value of a row-stochastic matrix
I am working on this problem. Consider a non-singular row-stochastic matrix $A$; we know that its 2-norm, $\|A\|_2$, is strictly greater than one if and only if it is not doubly stochastic.

Let $\sigma_1\geq\sigma_2\geq\dots\geq \sigma_n\geq0$ be the singular values and assume that the matrix $A$ is not doubly stochastic, so that $\|A\|_2=\sigma_1>1$ and assume that $\sigma_2\lneq\sigma_1$.

My question is about the second-largest singular value. Is it true that $\sigma_2\leq1$? Are there any conditions for this?
Thank you.

答:
$\sigma_2(A)$ is not always $\le1$. An obvious counterexample is a direct sum $A=X\oplus Y$ of two stochastic matrices $X$ and $Y$ such that $\sigma_1(X)>\sigma_1(Y)>1$. By perturbing such an $A$, one can also obtain a counterexample in which $A$ is positive.

I am not aware of any useful condition that guarantees that $\sigma_2(A)\le1$.


问:
Why does this formula have the summation sign?
I'm doing a programming tutorial and an algorithm to sum each number up to N can also be done by the below formula. E.g. if n = 10, 1+2..+10 = 55.

Forgive me for the stupid question :) but why does the formula use the sigma notation? Could we not use this formula without the summation and get the same result?

$\sum_{i=1}^n$$i= \frac{(n)(n+1)}{2}$

答: Note that $$1+2+\dots + n = \sum_{i=1}^n i = \frac{n(n+1)}{2}$$ So, both notations are identical.


问:
Prove $\sqrt{a_{1}a_{n}}\leq\sqrt[n]{a_{1}a_{2}\cdot...\cdot a_{n}}$
Let {$a_{n}$} be an arithmetic sequence with positive terms.
Prove that for any $n \in \mathbb{N}$ $$\sqrt{a_{1}a_{n}}\leq\sqrt[n]{a_{1}a_{2}\cdot...\cdot a_{n}}$$
When proving that using induction, induction step would be if it is true for some n,
we show that $$\sqrt{a_{1}a_{n}a_{n+1}}\leq\sqrt[n+1]{a_{1}a_{2}\cdot...\cdot a_{n}a_{n+1}}$$ or
$$\sqrt{a_{1}a_{n+1}}\leq\sqrt[n+1]{a_{1}a_{2}\cdot...\cdot a_{n}a_{n+1}}$$ and why?
Also solutions for this problem are appreciated.

答:
First show that
\begin{align}
a_1 a_n \le a_k a_{n-k+1}
\end{align}
for any $k=1,\ldots,n$. By using the assumption that $\{a_n\}$ is an arithmetic sequence with positive terms, it can be easily shown. (Equality holds when $k=1$, $k=n$, or $d=0$, where $d$ is the common difference.)

Then multiplying the above inequalities for $k=1,\ldots,n$, we obtain
\begin{align}
  (a_1 a_n)^n\le (a_1a_2\ldots a_n)^2.
\end{align}
(Note that all terms of $\{a_n\}$ appear twice. For example, $a_2$ appears when $k=2$ and $k=n-1$.)
This completes the proof.


问:
Not able to find abcissa of absolute convergence of this general dirichlet series
I am reading about general dirichlet series from my class notes and I also referred to wikipedia here:

https://en.m.wikipedia.org/wiki/General_Dirichlet_series

In the 2nd section : Abcissa of convergence I am unable to follow with this: If $\sigma_{c} \leq 0$ then $\sum a_n$ converges. But how in this case $\sigma_{c} $ is given by $\sigma_{c} = lim n\to \infty sup \frac{ log |a_{n+1} + a_{n+2}+... | } {\lambda_{n}} $, which is clearly different from the case in which $\sigma_{c} \geq 0$.

答:
In this answer, we are going to prove

Proposition: Suppose the series $\sum_n a_ne^{-\lambda_ns}$ converges for some $s=s_0$ with $\sigma_0=\Re(s_0)<0$, then

$$
\limsup_{N\to\infty}{\log|a_N+a_{N+1}+\cdots|\over\lambda_N}\le\sigma_0
$$

Proof. For convenience, let's define

$$
A(x)=\sum_{n\le x}a_n
$$

and

$$
S(x)=\sum_{n\le x}a_ne^{-\lambda_n s_0}
$$

Then by the assumptions, we know that there exists $K>0$ such that $|S(x)|\le K$ holds for all positive $x$. Thus, if we let $N>M$ be two positive integers, then

$$
A(N)-A(M)=\sum_{M<n\le N}a_n=\sum_{M<n\le N}[S(n)-S(n-1)]e^{\lambda_n s_0}
$$

Now, it follows from summation by parts that

$$
A(N)-A(M)=S(N)e^{\lambda_Ns_0}-S(M)e^{\lambda_Ms_0}+\sum_{M\le n<N}S(n)[e^{\lambda_{n+1}s_0}-e^{\lambda_ns_0}]
$$

Now, by triangle's inequality, we have

$$
\begin{aligned}
|A(N)-A(M)|
&\le2Ke^{\lambda_M\sigma_0}+K\sum_{M\le n<N}|e^{\lambda_{n+1}s_0}-e^{\lambda_ns_0}| \cr
&\le2Ke^{\lambda_M\sigma_0}+K|s_0|\int_{\lambda_M}^{\lambda_N}e^{\sigma_0u}\mathrm du \cr
&\le K\left(2+\left|\frac{s_0}{\sigma_0}\right|\right)e^{\lambda_M\sigma_0}
\end{aligned}
$$

Since the resulting upper bound does not depend on $N$, we have that for each fixed $s_0$, there exists $B=B(s_0)>0$ such that

$$
|a_N+a_{N+1}+a_{N+2}+\cdots|\le Be^{\lambda_N\sigma_0}
$$

Taking logarithms on both side concludes the proof.

It can be easily shown that the limit supremum is exactly the abscissa of convergence. Hope this can address your concern!


问:
Integrate: $\int_0^{\pi} \log ( 1 - 2 r \cos \theta + r^2)d\theta$
If $r \in \Bbb R$ how to integrate $\displaystyle \int_0^{\pi} \log ( 1  - 2 r \cos \theta + r^2)d\theta$?

I need some hints. Special case, if $r = 1$ then I know the above integral is zero.

Here is my working
\begin{align*}
\int_0^{\pi}\log (1 - 2 r \cos \theta + r^2)d\theta &= \int_0^\pi\log ((1 - re^{i \theta})(1 -re^{-i\theta} )) d\theta\\
 &= \int_0^\pi \log(1 - r e^{i\theta})d\theta + \int_0^\pi\log(1 - re^{-i\theta})d\theta\\
 &= \int_0^{2\pi} \log( 1 - re^{i\theta})d\theta \\
 &= 0
\end{align*}

答:
Two hints:

The integrand is an even function of $\theta$ $\Rightarrow$ the integral can be written as $\frac12\int_0^{2\pi}$.

$1-2r\cos\theta+r^2=(1-re^{i\theta})(1-re^{-i\theta})$.


问:
Prove formula of matrix norm $\|A\|$
Let $\|\cdot \|:\mathbb{R}^n\rightarrow \mathbb{R}$ be the norm defined by $$\|x\|=\max_{1\leq i\leq n}|ix_i|, \ x\in \mathbb{R}^n$$ Show that the corresponding matrix norm to vector norm $\|\cdot \|$ can be computed by $$\|A\|=\max_{1\leq i\leq n}\sum_{j=1}^n\frac{i}{j}|a_{i,j}|, \ A\in \mathbb{R}^{n\times n}$$

We have that $$\|A\|=\sup_{x\neq 0}\frac{\|Ax\|}{\|x\|}$$ So we have to write the numerator and the denominator using the given formula, right?

It is :\begin{align*}\frac{\|Ax\|}{\|x\|}&=\frac{\displaystyle{\max_{1\leq i\leq n}\left |i\sum_{i,j}a_{i,j}x_i\right |}}{\|x\|}\leq \frac{\displaystyle{\max_{1\leq i\leq n}i\sum_{i,j}|a_{i,j}|\cdot |x_i|}}{\|x\|} \leq \frac{\displaystyle{\left (\max_{1\leq i\leq n}\sum_{i,j}|a_{i,j}|\right )\cdot \left (\max_{1\leq i\leq n}i |x_i|\right )}}{\|x\|}\\ & =\frac{\displaystyle{\left (\max_{1\leq i\leq n}\sum_{i,j}|a_{i,j}|\right )\cdot \|x\|}}{\|x\|}=\max_{1\leq i\leq n}\sum_{i,j}|a_{i,j}|\end{align*}  How do weget the missing term $\frac{i}{j}$ ?

答:
On one hand, we have
\begin{aligned}
\frac{\|Ax\|}{\|x\|}
&=\frac{\max_i\left|i(Ax)_i\right|}{\|x\|}\\
&=\frac{\max_i\left|i\sum_j a_{ij}x_j\right|}{\|x\|}\\
&=\frac{\max_i\left|\frac{i}{j}\sum_j a_{ij}jx_j\right|}{\|x\|}\\
&\le\frac{\max_i\frac{i}{j}\sum_j |a_{ij}|\|x\|}{\|x\|}\\
&=\max_i\frac{i}{j}\sum_j |a_{ij}|.
\end{aligned}
On the other hand, tie occurs in the inequality above when $|x_1|=|2x_2|=\cdots=|nx_n|$ and the sign of each $x_j$ is chosen such that $a_{i_0j}x_j\ge0$ for every $j$ on row $i_0$, where $i_0=\arg\max_i\frac{i}{j}\sum_j |a_{ij}|$. Therefore $\sup_{x\ne0}\frac{\|Ax\|}{\|x\|}=\max_i\frac{i}{j}\sum_j|a_{ij}|$.

If you are familiar with the induced maximum norm $\|B\|_\infty=\sup_{x\ne0}\frac{\|Bx\|_\infty}{\|x\|_\infty}$ and know that it is identical to $\max_i\sum_j|b_{ij}|$, here is a cleaner solution. Let $D=\operatorname{diag}(1,2,\ldots,n)$. Then $\|x\|=\|Dx\|_\infty$ and hence
$$
\|A\|
=\sup_{\|x\|=1}\|Ax\|
=\sup_{\|Dx\|_\infty=1}\|DAx\|_\infty
=\sup_{\|y\|_\infty=1}\|DAD^{-1}y\|_\infty
=\|DAD^{-1}\|_\infty.
$$
It is now evident that the $\frac{i}{j}$ factor comes from $DAD^{-1}$ because $(DAD^{-1})_{ij}=\frac{i}{j}a_{ij}$.

Title: Prove formula of matrix norm $\|A\|$


问:
Change of variables for step function before differentiating
I have some confusion regarding the Heaviside step function and how and when variables can be changed. The issue comes from considering the function $H(e^x - e^{x_0})$. Taking the derivative with respect to $x$ gives $\frac{d}{dx} H(e^{x} - e^{x_0}) = e^x \, \delta(e^x - e^{x_0})$. However if you take the derivative of the function $H(x - x_0)$ you get $\frac{d}{dx} H(x - x_0) = \delta(x - x_0)$. The two step functions appear to be the same, but their derivatives aren't equivalent since using the composition rule for the Dirac delta function gives $ e^x \, \delta (e^x - e^{x_0}) = \frac{e^x}{e^{x_0}} \delta(x - x_0)$. Clearly something is going wrong here, but where?

答:
If $g$ is smooth, we have (if chain rule is applicable for distributions)
$$
\frac{d}{dx} H(g(x))
= g'(x) \delta(g(x))
= g'(x) \sum_{x' \text{ s.t. }g(x')=0} \frac{1}{|g'(x')|} \delta(x-x')
.
$$

For $g(x)=e^{x}-e^{x_0}$ we then have $g'(x) = e^{x}$ and $g(x')=0$ for $x'=x_0.$ Thus,
$$
g'(x) \sum_{x' \text{ s.t. }g(x')=0} \frac{1}{|g'(x')|} \delta(x-x')
= e^{x} \frac{1}{e^{x_0}} \delta(x-x_0)
= e^{x_0} \frac{1}{e^{x_0}} \delta(x-x_0)
= \delta(x-x_0)
.
$$

Thus,
$$
\frac{d}{dx} H(e^{x}-e^{x_0}) = \frac{d}{dx} H(x-x_0)
.
$$


问:
Find all the integer solutions of $x^2-4y^2=1$

Find all the integer solutions of the equation: $x^2-4y^2=1$.

I know I can't solve it like a PELL equation because d is a square in this case.

Would appreciate your help:)

答:
Rewrite is as$$(x-2y)(x+2y)=1$$
Then, there will be two cases
$$\begin{cases} x-2y = 1 \\ x+2y = 1 \end{cases} \qquad \text{or}\qquad \begin{cases}x-2y = -1 \\ x+2y = -1 \end{cases}$$


答:
We have $1=x^{2}-4y^{2}=x^{2}-(2y)^{2}=(x-2y)(x+2y)$, and since we are working over the integers, both of these must be integers which implies that both are either $1$ or $-1$.

Hence, we must have $x-2y=x+2y \implies -2y=2y \implies -2 = 2$ for $y \neq 0$, and so there are no solutions where $y \neq 0$.

If $y = 0$, then $x^{2} = 1$ $\implies x= 1$ and $x = -1$ are the only solutions.

Therefore, the solutions to $x^{2}-4y^{2}=1$ are $(1,0)$, and $(-1,0)$.

Title: Find all the integer solutions of $x^2-4y^2=1$

Body:
Find all the integer solutions of the equation: $x^2-4y^2=1$.

I know I can't solve it like a PELL equation because d is a square in this case.

Would appreciate your help:)


问:
How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital?
$\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$

$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:

$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$

 I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?

Body:
fast solution,as another idea$$\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}=\\
\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+0n^2+n+1}-\sqrt{n^2-n+2}}=\\
\lim_{n \to \infty}\frac{1}{\sqrt[3]{(n+0)^3-0+0n^2+n+1}-\sqrt{(n-\frac12)^2-\frac14+2}}=\\
\lim_{n \to \infty}\frac{1}{n-(n-\frac12)}=\\2$$

REMARK:
$$n^3+an^2+bn+c=\\(n+\frac a3)^3-(3n^2.\frac a3+3n(\frac  a3)^2+(\frac a3)^3)+bn+c\\(n\to \infty) \implies n^3+an^2+bn+c\sim (n+\frac a3)^3 $$ so $$n^3+n+1=(n+0)^3-0^3+n+1$$ also for $n^2+an+b=(n+\frac a2)^2-(\frac a2)^2+c$


问:
Every normal subgroupoid is the kernel of a groupoid morphism
Sorry for the block of text from Kirill Mackenzie's chapter on groupoids.

I have proven that $\natural,\natural_0$ is a groupoid morphism and that this quotient construction satisfies the groupoid axioms. I am having trouble showing that the kernel is $\Phi$.

Suppose $\xi\in \ker(\natural)$. Then $\exists x\in B$ such that $[x]=\natural_0(x)\in B/\Phi$ and $\natural(\xi)=[\xi]=[\tilde{x}]$. Now
$[\xi]=[\tilde{x}]\Leftrightarrow \zeta,\zeta'\in \Phi$ such that $\xi=\zeta\tilde{x}\zeta'=\zeta\zeta'$, where the last equality comes from the fact that if $\zeta\tilde{x}\zeta'$ is defined then $\alpha(\zeta)=\beta(\tilde{x})=x$ so $\zeta\tilde{x}=\zeta\widetilde{\alpha(\zeta)}=\zeta$. Now $\xi=\zeta\zeta'\in \Phi$ since $\Phi$ is a subgroupoid. I think this shows $\ker(\natural)\subseteq \Phi$. I am having a trouble showing the reverse inclusion.

The only insights I can see that might help are:

Let $\xi\in \Phi$ with $\alpha(\xi)=x$,$\beta(\xi)=y$. $\xi = \xi\xi^{-1}\xi$, so $[\xi]=[\xi^{-1}]$ and hence $\natural(\xi)=\natural(\xi^{-1})$.

$\bar{\alpha}(\natural(\xi))=\natural_0(\alpha(\xi))=[x]$

$\bar{\alpha}(\natural(\xi^{-1}))=\natural_0(\alpha(\xi^{-1}))=[y]$

Hence $[x]=[y]$ and so $x\sim y$ in B. This seems like it might be useful but I am completely stuck here and not sure how to show that $\natural(\xi)=[\tilde{x}]$ for some $x\in B$.

答:
I think I got it but it's very unsatisfying and kind of fishy:

Let $\xi\in \Phi$. Then $\widetilde{\alpha(\xi)} = \xi^{-1}\xi\widetilde{\alpha(\xi)}$, and since $\xi^{-1},\widetilde{\alpha(\xi)}\in \Phi$, so $[\widetilde{\alpha(\xi)}]=[\xi]=\natural(\xi)$. Hence $\xi\in \ker(\natural)$.

$\therefore \Phi=\ker(\natural)$.


问:
Reduce the Cauchy-Euler equation to a smaller expression?
Given the Cauchy-Euler second order differential equation:
$$ \left( x^2\frac{d^2}{d x^2} + x\frac{d}{d x} -1\right)y(x)=0 $$
is it possible to find a function $u(x)$ , which satisfies the following simple form:?
$$ \frac{d^2 y(u)}{d u^2}=0 $$
***notice that $t=\ln(x)$ yields:
$$ \frac{d^2 y(t)}{d t^2}-1=0 $$

答:
we suppose that there exits a $u(x)$, if so; using chain rule we have:
$$ \frac{d^2y}{du^2} = \left(\frac{dx}{du}\right)^2 \frac{d^2y}{dx^2} + \frac{dx}{du} \frac{d\left(\frac{dx}{du}\right)}{dx} \frac {dy}{dx}$$
now setting this equal to the main equation:
$$ x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} -y $$
there is an extra term in the main equation ($-y$) which cannot be found.
however if this extra term didnot exist, we'd have:
$$ \left(\frac{dx}{du}\right)^2 \frac{d^2y}{dx^2} = x^2\frac{d^2y}{dx^2} $$
$$ \frac{dx}{du} \frac{d\left(\frac{dx}{du}\right)}{dx} \frac {dy}{dx} = x\frac{dy}{dx} $$
these two equations are dependant, i.e. a solution to one of them is a solution to the other. therefore solving the first one suggests:
$$ u = \ln(x) $$
this is valid for the second one, too.


问:
Show that $b_nb_k\mid b_{n+k}$
Let ${a_n}$ be sequence such that $\forall n,k\in\mathbb{N}, a_n\mid a_{n+k}-a_k$ and, let $b_n=\Pi^{n}_{i=1} a_i$. Then prove that $\forall n, k\in\mathbb{N}, b_nb_k\mid b_{n+k}$.

$a_i=ci, c\in\mathbb{N}$ satisfies the condition:
I'll prove $m!k!\mid (m+k)!$. $p\in\mathbb{P}$, let $\upsilon_p(a)=k\iff p^k\mid\mid a$. $\forall p\in\mathbb{P}$, $\upsilon_p(\frac{(m+k)!}{m!k!})=\sum_{i=1}^{\infty}([\frac{m+k}{p^i}]-[\frac{m}{p^i}]-[\frac{k}{p^i}])$ and since $[x+y]\ge [x]+[y]$, $[\frac{m+k}{p^i}]\ge [\frac{m}{p^i}]+[\frac{k}{p^i}]$for all $i\implies \forall p\in\mathbb{P}, \upsilon_p(m!k!)\le\upsilon_p((m+k)!)\implies m!k!\mid (m+k)!$. And I want to prove that there is no more solutions. Can anyone help me?

答:
First, we show inductively that
$$
a_1a_2\cdots a_n \,|\, a_{l+1}a_{l+2}\cdots a_{l+n}, \quad \text{for all $\,l\ge 0$ and $n\ge 1$.}
$$
For $n=1$ is clear, since $\,a_1\,|\,a_{l+1}-a_1,\,$ and hence $\,a_1\,|\,a_{l+1}$. Assume it is true for $\,n=k$. We need to show that
$$
a_1a_2\cdots a_na_{n+1} \,|\, a_{l+1}a_{l+2}\cdots a_{l+n}a_{l+n+1}=f(l)
$$
Meanwhile, for every $j\ge 0$,
$$
f(j+1)-f(j)=(a_{j+n+2}-a_{j+1})a_{j+2}\cdots a_{j+n+1}.
$$
By the induction hypothesis, $\,a_1\cdots a_n \,|\, a_{j+2}\cdots a_{j+n+1},\,$ and is given that $\,a_{n+1} \,|\,a_{j+n+2}-a_{j+1}$. Hence
$$
a_1\cdots a_{n+1}\,|\, f(j+1)-f(j), \quad \text{for all $j\ge 0$}
$$
and since
$$
a_1\cdots a_{n+1}| f(0)=a_1\cdots a_{n+1},
$$
then
$$
a_1\cdots a_{n+1}\,|\, f(0)+\big(f(1)-f(0)\big)+\cdots+\big(f(l)-f(l-1)\big)=
f(l), \quad \text{for all $l\in\mathbb N$}
$$

Consequently,
$$
b_kb_n=(a_1\cdots a_k)(a_{1}\cdots a_{n})\,|\,  (a_1\cdots a_k)(a_{k+1}\cdot a_{k+n})=b_{k+n}.
$$

Title: Show that $b_nb_k\mid b_{n+k}$


问:
Calculus Differentiable Functions, Limits
Regarding this question:

Let $f$ be a function that is defined on a neighborhood of $0$.

Suppose that $\lim_{x \to 0} \frac{f(x)}{|x|} = 1$. Then $f$ is not differentiable at $0$.

--

So I think this is true, and I want to separate it to 3 options $f(0)=0$, $f(0)<0$, $f(0)>0$:

If $f(0)=0$, then $\lim_{x \to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0} \frac{f(x)}{x}$

From here, I want to separate it to $0^+$ and $0^-$, so:

$\lim_{x \to 0^+} \frac{f(x)}{x}$ and from the fact that $\lim_{x \to 0} \frac{f(x)}{|x|} = 1$, I can say that $\lim_{x \to 0^+} \frac{f(x)}{x} = 1$.

$\lim_{x \to 0^-} \frac{f(x)}{-x}$ and from the fact that $\lim_{x \to 0} \frac{f(x)}{|x|} = 1$, I can say that $\lim_{x \to 0^-} \frac{f(x)}{x} = -1$.

Is that legit? I'm not sure if I can say that.

Thanks a lot!

答:
You proved correctly that $\lim_{x\to0^+}\frac{f(x)}x=1$ and $\lim_{x\to0^-}\frac{f(x)}x=-1$; so $\lim_{x\to0}\frac{f(x)}x=$ doesn't exist.

Note that $\lim_{x\to0}\frac{f(x)}{|x|}=1\implies\lim_{x\to0}f(x)=0$ (since $\lim_{x\to0}|x|=0$). so, there are now two possibilities: either $f(0)\ne0$ or $f(0)=0$. In the first case, $f$ isn't even continuous at $0$, and therefore it cannot be differentiable there. Otherwise, what was done above shows that the limit $\lim_{x\to0}\frac{f(x)-f(0)}x$ doesn't exist, and so, again, $f$ cannot be differentiable at $0$.


问:
How easily to check: $f$ is a closed map?
Assume that $f:\mathbb R\to\mathbb R$ is a function where $\mathbb R$ is the real number and the usual topology is defined on $\mathbb R$.

I have two questions:
   1. Let $C$ be closed. Then what is the proper form of $C$? For example, without the loss of generality, open set is replaced by the open interval.
   2. To check that $f$ is a closed map, we need some information: $f(C)$ is closed. How to easily check this?

答:
There are several ways to check that a map is closed.

The formula $\overline{f(A)}\subseteq f(\overline A)$ is a description of closed maps. However, since the proof that this formula is equivalent to closedness of $f$ is very easy, I think you could as well show directly that your $f$ is closed.

Here are some hints:

A characterization of closedness of $f$ is: For each $y\in Y$ and every open $U\supseteq f^{-1}(y)$ there is an open $V\ni y$ with $f^{-1}(V)\subseteq U$.

Closedness is local in the codomain. This means that for each point $y\in Y$ there is a neighborhood $V$ such that the restricted $f:f^{-1}(V)\to V$ is closed. Equivalently, $Y$ is covered by the interiors of subsets $(V_i)_{i\in I}$, where $f_i:f^{-1}(V_i)\to V_i$ is closed for each $i\in I$. As the restriction $f:S\to f(S)$ of a closed map to a saturated set $S$ is again closed, this condition is also necessary.

The second point can be applied if $f$ is a proper map (preimages of compact sets are compact) and the codomain is locally compact Hausdorff or, more generally, each point in $Y$ has a compact Hausdorff neighborhood. In that case it is easy to prove that $f$ is a closed map. For example, a continuous map from $\Bbb R$ to $\Bbb R$ is closed if the preimage of a bounded set is bounded, so $f(x)=x^2$ is closed.

Yet another fact: If $f:X\to Y$ is proper and $Y$ is Fréchet-Urysohn and UCC, then $f$ is closed. UCC (unique convergent clustering) means that every convergent sequence has a unique cluster point.


问:
Finding the derivative of $ \cos(\arcsin x)$
I study maths as a hobby. I am trying to find the derivative of $ \cos(\arcsin x)$

This is how I have been proceeding:

Let u = $\arcsin x$

Then $\sin u = x$

Differentiating:
\begin{align}
\cos u \frac{du}{dx} &= 1 \implies \frac{du}{dx} = \frac{1}{\cos u} \\[4pt]
\cos^2 u + \sin^2 u &= 1 \\[4pt]
\cos u &= \sqrt {1 - \sin^2 u} = \sqrt {1 - x^2}
\end{align}
But that is as far a I get.

The text book says the answer is $\frac{x}{\sqrt {1 - x^2}}$  but I cannot see how this is arrived at.

答: With your notation, $x = \sin u$, we have $$\frac{d}{dx}\left[\cos (\sin^{-1} x)\right] = -\sin (\sin^{-1} x) \cdot \frac{d}{dx} \left[ \sin^{-1} x \right] = -x \frac{d}{dx} \left[ \sin^{-1} x \right].$$  Since $$\frac{dx}{du} = \cos u,$$ we have $$\frac{du}{dx} = \frac{1}{\cos u} = \frac{1}{\sqrt{1 - \sin^2 u}} = \frac{1}{\sqrt{1-x^2}}.$$  Therefore, $$\frac{d}{dx}\left[\cos(\sin^{-1} x)\right] = - \frac{x}{\sqrt{1-x^2}}.$$


答:
Let $y=\cos(u)=\cos(\arcsin x)$. You have correctly worked out that $du/dx$ is
$$
\frac{1}{\sqrt{1-x^2}}
$$
but you need to multiply this result by $dy/du$. The textbook's answer actually contains a sign error.

This result can be obtained more directly if you write the chain rule in Lagrange notation. If $y=f(g(x))$, then
$$
\frac{dy}{dx}=f'(g(x)) \cdot g'(x) \, .
$$
Here, $f=\cos$ and $g=\arcsin$. Hence, $dy/dx$ equals
$$
-\sin(\arcsin(x)) \cdot\frac{1}{\sqrt{1-x^2}} \, ,
$$
which can be simplified further.

Title: Finding the derivative of $ \cos(\arcsin x)$

Body:
I study maths as a hobby. I am trying to find the derivative of $ \cos(\arcsin x)$

This is how I have been proceeding:

Let u = $\arcsin x$

Then $\sin u = x$

Differentiating:
\begin{align}
\cos u \frac{du}{dx} &= 1 \implies \frac{du}{dx} = \frac{1}{\cos u} \\[4pt]
\cos^2 u + \sin^2 u &= 1 \\[4pt]
\cos u &= \sqrt {1 - \sin^2 u} = \sqrt {1 - x^2}
\end{align}
But that is as far a I get.

The text book says the answer is $\frac{x}{\sqrt {1 - x^2}}$  but I cannot see how this is arrived at.


答:
Since the the condition is met, you can use Gauss-Seidel's recursion. The formulae you stated are right. I checked the iterations using Mathematica and found out that the convergence is a bit slow. I think the reason for that is the coefficients of the $x_i$ in the $i$th equation are not big enough in absolute value than the other in the same equation.

I hope that answers your question.

Title: Did I write the steps of Gauss-Seidel's method correctly?

Body:
I want to solve this system of equations with Gauss-Seidel's method:

$${ \begin{cases}
2x_1&+x_2&&&&=3 \\
-x_1&+2x_2&+x_3&&&=2 \\
&-x_2&+2x_3&+x_4&&=2 \\
&&-x_3&+2x_4+x_5&&=2 \\
&&&-x_4+2x_5&&=1
\end{cases} }$$
I wrote equations as:

$$x_1=\frac12(3-x_2)$$

$$x_2=\frac12(2+x_1-x_3)$$

$$x_3=\frac12(2+x_2-x_4)$$

$$x_4=\frac12(2+x_3-x_5)$$

$$x_5=\frac12(1+x_4)$$

For this method if I want to use algorithm to solve the system is the following steps right ?

$$x_1^{(k+1)}=\frac12(3-x_2^{(k)})$$

$$x_2^{(k+1)}=\frac12(2+x_1^{(k+1)}-x_3^{(k)})$$

$$x_3^{(k+1)}=\frac12(2+x_2^{(k+1)}-x_4^{(k)})$$

$$x_4^{(k+1)}=\frac12(2+x_3^{(k+1)}-x_5^{(k)})$$

$$x_5^{(k+1)}=\frac12(1+x_4^{(k+1)})$$


答:
Hint: Note that when $r<1$ this is the real part of
$$\frac12 \int_{|z|=r} \log(1-z)^2\,\frac{dz}{iz}$$
for the usual branch of $\log$.
When $r>1$, you're going to need to make a branch cut.</p>

Title: Integrate: $\int_0^{\pi} \log ( 1 - 2 r \cos \theta + r^2)d\theta$

Body:
If $r \in \Bbb R$ how to integrate $\displaystyle \int_0^{\pi} \log ( 1  - 2 r \cos \theta + r^2)d\theta$?

I need some hints. Special case, if $r = 1$ then I know the above integral is zero.

Here is my working
\begin{align*}
\int_0^{\pi}\log (1 - 2 r \cos \theta + r^2)d\theta &= \int_0^\pi\log ((1 - re^{i \theta})(1 -re^{-i\theta} )) d\theta\\
 &= \int_0^\pi \log(1 - r e^{i\theta})d\theta + \int_0^\pi\log(1 - re^{-i\theta})d\theta\\
 &= \int_0^{2\pi} \log( 1 - re^{i\theta})d\theta \\
 &= 0
\end{align*}


问:
Why does the discriminant in the Quadratic Formula reveal the number of real solutions?
Why does the discriminant in the quadratic formula reveal the number of real solutions to a quadratic equation?

That is, we have one real solution if
$$b^2 -4ac = 0,$$
we have two real solutions if
$$b^2 -4ac > 0,$$
and we have no real solutions if
$$b^2 -4ac < 0.$$

答:
Maybe the computation is even more obvious when you avoid fractions.

Multiply the equation $ax^2+bx+c=0$ by $4a$ (note that $a\neq0$) and observe
$$ax^2+bx+c=0 \quad \Leftrightarrow \quad 4a^2x^2+4abx+4ac=0 \quad \Leftrightarrow \quad (2ax+b)^2=b^2-4ac.$$
Looking at the last equation, it is obvious that the number of solutions depends on the sign of $b^2-4ac$. In my opinion, it is also easy to see that if $b^2-4ac>0$, we may proceed further to get the well-known solutions:
$$2ax+b=\pm\sqrt{b^2-4ac}\quad \Leftrightarrow \quad x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$


问:
Odd perfect numbers and egyptian fraction conjecture
I have reached the conclusion that the non-existence of odd perfect numbers is related to the following

Conjecture Let it be $R=\{d_1,d_2,...,d_n\}$ the set of distinct proper divisors less than $\sqrt{O}$ of some odd positive integer $O$, such that $d_1<d_2<...<d_n$, and such that for some divisor $d_k\in R$ holds that $d_k\leq d_n<3d_k$. Then, it exist no solution to the egyptian fraction $$\frac{2}{d_k}+\frac{1}{d_1}+\frac{1}{d_2}+...=1$$

The process to reach the conjecture is the following (for you to check it is correct):

Let $P$ be some odd perfect number.

Let $R=\left\{ d_{1},d_{2},...,d_{n}\right\}$  be the set of proper divisors of $P$ less than $\sqrt{P}$ excluding $1$, and $T=\left\{ \frac{P}{d_{1}},\frac{P}{d_{2}},...\frac{P}{d_{n}}\right\}$  be the set of proper divisors of $P$ greater than $\sqrt{P}$ excluding $P$.

As $P$ is a perfect number,

$$1+d_{1}+d_{2}+...+d_{n}+\frac{P}{d_{n}}+...+\frac{P}{d_{2}}+\frac{P}{d_{1}}=P (1)$$

Operating,

$$1+d_{1}+d_{2}+...+d_{n}=P-\frac{P}{d_{1}}-\frac{P}{d_{2}}-...-\frac{P}{d_{n}}$$

$$1+d_{1}+d_{2}+...+d_{n}=P\left(1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}\right)$$

$$\frac{1+d_{1}+d_{2}+...+d_{n}}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}=P$$

As $1+d_{1}+d_{2}+...+d_{n}$ is an integer, it follows that $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ must be integer. Therefore, $P$ is a perfect number only if both $1+d_{1}+d_{2}+...+d_{n}$ and $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ are proper divisors of $P$.

It is trivial to show that one of the two expression is less than $\sqrt{P}$, and therefore belongs to $R$, and the other is greater than $\sqrt{P}$ and belongs to $T$.

It is clear that $1+d_{1}+d_{2}+...+d_{n}$ is greater than the greatest element of $R$, so subsequently we can state that

$$\left(\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}\right)=d_{k}\in R (2)$$

$$1+d_{1}+d_{2}+...+d_{n}=\frac{P}{d_{k}}\in T (3)$$

The proof showing that $d_k\leq d_n<3d_k$ is a bit long, so I skip it for the shake of briefness. And operating with $(2)$, the conjecture follows. I have found (thanks to some help provided here in MathStackExchange) examples of odd positive integers for which $(3)$ alone holds, but not any example of an odd positive integer for which $(2)$ holds with the restrictions exposed.

Any help on how to attack this conjecture would be welcomed. Otherwise, if someone is able to show any counterexample to the conjecture (that is, provide some odd positive integer for which $(1)$ holds), it would be more than welcomed to save efforts proving it. I have not seen this conjecture elsewhere, so it is possible that it is mistaken.

Thanks in advance!

Edit

I post the proof showing that $d_k\leq d_n<3d_k$, as it is fundamental for the conjecture and for the solution I have provided for verification.

From (3), and operating,

$$1+\sum_{j=1}^{n}d_{j}=\frac{P}{d_{k}}$$

$$P=d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$

As by definition we have that $d_{n}<\sqrt{P}$, just substituting we can set that

$$d_{n}<\sqrt{d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)}$$

$$d_{n}^{2}<d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$

$$\frac{d_{n}}{d_{k}}<\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}} (4)$$

Other hand, taking (1) and dividing by $\sqrt{P}$, we get that

$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{P}{d_{n}\sqrt{P}}+...+\frac{P}{d_{2}\sqrt{P}}+\frac{P}{d_{1}\sqrt{P}}=\sqrt{P}$$

$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{\sqrt{P}}{d_{n}}+...+\frac{\sqrt{P}}{d_{2}}+\frac{\sqrt{P}}{d_{1}}=\sqrt{P}$$

As $\forall d_{j}\,\frac{\sqrt{P}}{d_{j}}>1$, and $1+\sum_{j=1}^{n}d_{j}>\sqrt{P}$, then we get inmediately that $n<\sqrt{P}-1$; otherwise, the sum of all the terms would be greater than $\sqrt{P}$ and $P$ would not be a perfect number. Thus, we can affirm that

$$n\leq\sqrt{P}-2 (5)$$

As the minimum gap between consecutive elements of $R$ is $2$, the maximum sum of elements of $R$ with the minimum gap between them is

$$d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$

Therefore, we can establish that

$$1+\sum_{j=1}^{n}d_{j}\leq d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$

$$1+\sum_{j=1}^{n}d_{j}\leq nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)$$

Subsequently, we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\frac{nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq n-\frac{\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$

Substituting $n$ by the inequality obtained in (5), we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-2-1\right)^{2}+\left(\sqrt{P}-2-1\right)\right)}{d_{n}}$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)\right)}{d_{n}} (6)$$

Operating with the numerator of the third term of the right handside of (6), we get that

$$\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)=$$

$$P+9-6\sqrt{P}+\sqrt{P}-3=$$

$$P-5\sqrt{P}+6$$

As by definition $d_{n}<\sqrt{P}$, we can affirm that

$$\frac{P-5\sqrt{P}+6}{d_{n}}>\sqrt{P}-5+\frac{6}{\sqrt{P}}$$

Therefore, substituting at (6), we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<\sqrt{P}-2-\left(\sqrt{P}-5+\frac{6}{\sqrt{P}}\right)$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3-\frac{6}{\sqrt{P}}$$

Subsequently,

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3$$

$$1+\sum_{j=1}^{n}d_{j}<3d_{n}$$

Finally, substituting in (4), we get the desired result

$$\frac{d_{n}}{d_{k}}<3$$

答:
(1) You are trying to prove that if $P$ is an odd perfect number, then both $1+d_{1}+d_{2}+...+d_{n}$ and $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ are proper divisors of $P$. As Arnie Bebita-Dris pointed out, you have an error in the following part :

$$\frac{1+d_{1}+d_{2}+...+d_{n}}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}=P$$As $1+d_{1}+d_{2}+...+d_{n}$ is an integer, it follows that $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ must be integer.

Take $(a,b)=(5,\frac 57)$ for which $P=\frac ab$ and $a\ (\lt P)$ are integers, and $\frac 1b$ is not an integer.

(2) You are trying to prove that if $P$ is an odd perfect number, then $d_n\lt 3d_k$. Your proof starts with $1+d_{1}+d_{2}+...+d_{n}=\frac{P}{d_{k}}$ which you have not proved. This means that you have not proved that if $P$ is an odd perfect number, then $d_n\lt 3d_k$.

(3) In your proof for the conjecture, I think that you have correctly proved that $p_k$ is a composite number.

However, you have an error in the following part :

$$d_{j}=\frac{d_{k}\left(1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)\right)}{d_{k}-2}$$As $d_{k}$ and $d_{k}-2$ are odd integers, it follows that $\gcd\left(d_{k},d_{k}-2\right)=1$. As each $d_{j}$ is some odd positive integer, then necessarily for each $d_{j}$ we have one of the following two options: either $d_{k}=3$, or $d_{k}-2$ divides $1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)$.

You are implicitly assuming that $1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)$ is an integer, which you have not proved.

(4) I'm going to prove the following claims :

Claim 1 : If the conjecture is true, then $n$ is even larger than $4$.

Claim 2 : $(d_1,d_2,d_3)$ is either

$(p,p^2,p^3)$ where $p$ is a prime number

$(p,p^2,q)$ where $p,q$ are prime numbers

$(p,q,p^2)$ where $p,q$ are prime numbers

$(p,q,r)$ where $p,q,r$ are prime numbers

Claim 1 : If the conjecture is true, then $n$ is even larger than $4$.

Proof :

Seeing
$$\sum_{\substack{j=1\\j\neq k}}^{n}\left(\prod_{\substack{s\in S\\s\neq j}}d_{s}\right)+2\prod_{\substack{s\in S\\s\neq k}}d_{s}=\prod_{s\in S}d_{s}$$ in $\text{mod $2$}$ where $S=\left\{ 1,2,...,n\right\}$, we have $(n-1)+0\equiv 1\pmod 2$, so $n$ has to be even.

If $n=2$, then $$\frac{2}{x}+\frac 1y=1\implies (x-2)(y-1)=2\implies (x-2,y-1)=(1,2),(2,1)$$
implying $(x,y)=(3,3),(4,2)$ which don't satisfy our conditions.

If $n=4$, then, according to this answer, there is no such $O$. $\quad\blacksquare$

Claim 2 : $(d_1,d_2,d_3)$ is either

$(p,p^2,p^3)$ where $p$ is a prime number

$(p,p^2,q)$ where $p,q$ are prime numbers

$(p,q,p^2)$ where $p,q$ are prime numbers

$(p,q,r)$ where $p,q,r$ are prime numbers

Proof :

Suppose that $d_1$ is a composite number. Then, there is a prime number $p$ satisfying $p\lt d_1$ and $p\mid O$. This contradicts that $d_1$ is the smallest divisor of $O$ larger than $1$. So, $d_1$ is a prime number.

Next, suppose that $d_2$ has two distinct prime factors $p\lt q$. Then, $p,q$ are divisors of $O$ satisfying $1\lt p\lt q\lt d_2$. This contradicts that $d_2$ is the second smallest divisors of $O$ larger than $1$. It follows that $d_2$ is of the form $q^{k}$ where $q$ is a prime number. If $q=d_1$, then $k=2$. If $q\not=d_1$, then $k=1$.

If $d_1=p,d_2=p^2$, then $d_3$ is either a prime number or $p^3$.

If $d_1=p,d_2=q$, then $d_3$ is of the form $s^k$ where $s$ is a prime number. If $s=p$, then $k=2$, If $s\not= p$, then $k=1$. $\quad\blacksquare$


问:
Help with sum of a series of fractions
I was expected to have this series $$S = \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... + \frac{1}{2^{k}} $$
summed up to 1. However, I couldn't do that using this calculation:
$$ S = \frac{1}{2} + S - \frac{1}{2} $$ where $$ \frac{1}{2} + S = \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... + \frac{1}{2^{k}} $$ was a geometric series.
Thus, $$ S = \frac{1}{2} + S - \frac{1}{2} = a(\frac{1 - r^{k}}{1 -r}) - \frac{1}{2} $$
where $ a = \frac{1}{2} $, and $ r = \frac{1}{2} $.

Substituting the values of $a$ and $r$ into the expression above, I got $$ S = \frac{2^k - 1}{2^k} - \frac{1}{2} $$
Then taking limit on the right hand side $$ \lim_{k \to \infty} (\frac{2^k - 1}{2^k} - \frac{1}{2}) $$
I got $$ S = \frac{1}{2} $$
What did I do wrong?

EDIT: The reason I wanted to have $S$ summed to 1 because of this requirement (highlighted in yellow):

Title: Help with sum of a series of fractions

Body:
$$\sum_{k=0}^{n} r^k=\frac{r^{n+1}-1}{r-1}$$
Here
$$S = \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}}... + \frac{1}{2^{k}} $$
$$S=\frac{1}{2^2}\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{k-2}}\right)$$
$$S=\frac{1}{4}\frac{2^{-(k-1)}-1}{2^{-1}-1}=\frac{1}{2}(1-2^{1-k})$$
So if $k\to \infty$, $S\to \frac{1}{2}$


Title: How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital?

Body:
$\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$

$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:

$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$

 I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?

Body:
binomial theorem for rational exponents:

(1+n)^(1/3) = 1 + n/3 + ...
(1+n)^(1/2) = 1 + n/2 + ...

s1= (n^3 + n - 1)^(1/3) = [n^3 (1+ 1/n^2 + ...)]^(1/3) = n (1+ 1/(3 n^2) ... )
s2= (n^2 -n  + 2)^(1/2) = [n^2 (1- 1/n^2 + ...)]^(1/2) = n (1 -1/(2n) ... )
s1-s2 = 1/(3n) + 1/2 ...

lim  1/(s1-s2) = 2

n->infinity


答:
Alternative approach:

The range of ArcSin is $[-\pi/2, \pi/2]$, so the $\cos[$ArcSin$(x)]$ will be $\geq 0.$

Further, if $\theta = $ArcSin$(x)$, then $\cos(\theta)$ [which by the above statement will be non-negative] will be $\sqrt{1 - x^2}.$

Therefore, the question immediately reduces to computing

$$\frac{d}{dx} \sqrt{1 - x^2}.$$


问:
What's the value of two digits from a certain number composed by six digits when it comes from a prime?
The problem is as follows:

The number $\overline{ab0ab0}$ is the result of the product between
two consecutive prime numbers. Assuming the $0$ in the number
represents zero. Find the value of $\overline{ab}$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&38\\
2.&85\\
3.&51\\
4.&34\\
4.&17\\
\end{array}$

In my attempt to solve this problem I was only able to spot this by decomposing the six number digit as follows using base-10.

$\overline{ab0ab0}=10^5a+10^4b+10^3 \times 0+10^2a+10b+0=100100a+10010b$

Then I could note this can be rearranged as follows:

$10010(10a+b)$

Upon performing a canonical decomposition in the number I could spot that:

$10010= 2 \times 5 \times 7 \times 11 \times 13$

Had this question asked all those I would have landed in the answer. But that's where I'm stuck. Can someone help me from here?.

Am I heading in the right direction with this approach?. It would help me a lot if someone could help me from where I had came stuck.

答:
The problem makes some sense if we drop the word "two" and ask for just "a product of consecutive prime numbers".  In such a case, having derived

$10010=2×5×7×11×13$

we observe that $3$ is missing.  We multiply that in to get $30030$, which does not qualify because that would require the initial digit $a$ to be zero.  To remove that blockage, try introducing the next prime factor, $17$, to get

$2×3×5×7×11×13×17=\color{blue}{51}0\color{blue}{51}0$.


问:
Infinite many roots of an equation
Let $k,A>0$. How may roots the equation $f(x)=\frac{1}{k} (1+\frac{A}{2}x)\sin(kx)=c,\ x\in \mathbb{R}, \ c\in \mathbb{R}$ has? I beliave the roots are infinite and one tric I thought is by using Bolzano theoreme I may find sequences $(x_n),(y_n)$ s.t $\ x_n,y_n\to \infty$ and $f(x_n)f(y_n)<0$. But I cannot compose it. Any help please?

答:
$f\left(\displaystyle\frac {\dfrac {\pi} 2+2n\pi} k\right) \to \infty$ as $n \to \infty$ and $f\left(\dfrac {\dfrac {3\pi} 2+2n\pi} k\right) \to -\infty$ as $ n \to \infty$

Hence, $f$ takes the value $c$ between $\dfrac {\dfrac {\pi} 2+2n\pi} k$ and $\dfrac {\dfrac {3\pi} 2+2n\pi} k$  for all sufficiently large values of $n$.

Title: Infinite many roots of an equation


问:
Determine the Automorphisms of the extension field $\mathbb{F}_{19683}$ of $\mathbb{F}_{27}$
I am trying to find $\text{Aut}_{\mathbb{F}_{27}}(\mathbb{F}_{19683})$. I have a theorem that looks promising but can not make it work.

For a field $\mathbb{K}$ with $\text{char}(\mathbb{K})=p$ prime we have the Frobeniusmorphism $\text{Fr}_p:\mathbb{K} \rightarrow \mathbb{K}, \ \alpha \mapsto \alpha^p$. Then $(\text{Fr}_p)^n$ is also a morphism. The theorem I am trying to use states that

$\text{Aut}_{\mathbb{F}_{p}}(\mathbb{F}_{p^n}) = \langle Fr_p \rangle$.

Obviously my problem is that 27 is not prime. Rewriting the problem $\text{Aut}_{\mathbb{F}_{3^3}}(\mathbb{F}_{3^{3^3}})$ and the fact that the theorem was mention just above this exercise make me think that one can make use of it to solve the problem.

Any help appreciated.

答:
Note that for general field extensions $E/F/K$ we have a restriction homomorphism $$\psi:\operatorname{Aut}_K(E)\to\operatorname{Aut}_K(F)$$
with kernel $\operatorname{Aut}_F(E)$. Applying this to the situation $E=\mathbb{F}_{19683},F=\mathbb{F}_{27},K=\mathbb{F}_{3}$ we see that the group $\operatorname{Aut}_F(E)$ is exactly that subgroup of $\langle Fr_3\rangle=\operatorname{Aut}_K(E)$ which fixes $\mathbb{F}_{27}$. Since $\langle Fr_3\rangle$ is cyclic it shouldn't be too hard to find a generator for that.


问:
Closed form for $n$th derivative of exponential of $f$
What is the closed form for:

$$\frac{\partial^n}{\partial x^n}\exp(f(x))=\exp(f(x))\cdot[????]$$

答:
So we know that if $f,g$ are infinitely differentiable, we can compute
$$
\frac{d}{dx} \left( e^{f(x)} g(x) \right) = \left( f'(x) e^{f(x)} \right) g(x) + e^{f(x)} g'(x) =  e^{f(x)} \left( f'(x) g(x) + g'(x) \right).
$$
So since $e^{f(x)} = e^{f(x)} \left( 1 \right)$ and that we know that all derivatives of $e^{f(x)}$ are going to be of the form $e^{f(x)}g(x)$ for some functions $g$, we can apply the above formula and assume that
$$
\frac {d^n}{dx^n} e^{f(x)} = e^{f(x)} g_n(x), \qquad g_0(x) = 1.
$$
We can use the formula found above to find a recurrence relation for $g_n(x)$ :
$$
g_n(x) = f'(x) g_{n-1}(x) + g_{n-1}'(x), \qquad g_0(x)
$$
The formula from Wolfram Alpha provided by Integral looks wrong because there should be no $f(x)^j$ in there ; it's impossible to make the function $f$ appear in the factor because all derivatives of $e^{f(x)}$ have the derivatives of $f$ appear in the factor (i.e. the function $g_n$), but never the function $f$ itself.

If you wanted this for yourself to compute higher derivatives of $e^{f(x)}$ I suggest you just apply the recurrence relation to get the first few. For a general formula I must admit I don't know. Would require more thinking.


问:
proof any finite sequence of real numbers is bounded
For any sequence of real numbers over a finite domain such as,
$ \langle a_n \rangle = a, a+1, \ldots, b $

Can someone show me (in formal terms) how it is I can go about defining a bijection between the range $ \left\{ f(n) : a ≤ n ≤ b \right\} $ and subset of natural numbers so that I can go on to use the fact that any finite set of real numbers is bounded to assert the existence of a maximal and minimal elements of the range.

Or if this is not the way to formally show that maximal and minimal elements of the range exist (obviously they do), can someone show me the way.

Thank you!!

答:
As you already said yourself, the statement is obviously true -- and this should usually count as a perfectly fine answer. It is not the point of formal proofs to be joylessly pedantic about things which are indeed clear to us!

One situation where it is useful to ask for a formal proof is this: For some reason you want to develop a formal theory for the mathematics that you are interested in. For this you usually start by expressing some basic facts (axioms) and some modes of reasoning that you are allowed to use (such as induction). Then in order to show that your formal system is strong enough, you can try to formally prove simple true facts (such as boundedness of finite sequences) using only the formal system that you developed. If this is possible, it shows you that you have chosen your axioms and inference rules wisely.

Long story short, to ask for a formal proof of a statement makes only sense if you tell us which are the axioms and inferences that you are allowed to use. If on the other hand you are interested in the truth of a statement, then any convincing argument is allowed, incuding appeal to intuition (which would be appropriate here).


问:
when a bounded set in $W^{2,p}(\Omega)$ has a strong limit point in $C(\overline{\Omega})$?
I just have a question in functional analysis. The question says:

If $\Omega \subset \mathbb{R}^3$ is bounded, and suppose that $H$ a bounded set in the Sobolev's space $W^{2,p}(\Omega)$. Then what values of $p \geq 1$ makes $H$ have a strong  strong limit point in $C(\overline{\Omega})$?

Thank you!

答:
On the negative side, for sets with bad boundary, it can fail: Let $\Omega = \mathbb B(0,1)\setminus([-1,1]\times \{0 \})\subset \mathbb R^2$ be the open unit disk with a horizontal cut. Then $\overline\Omega=\overline{\mathbb B(0,1)}$ but $f:\Omega\to\mathbb R$, defined by $$f(x,y)=\begin{cases} 1 &y>0\\ 0& y<0\end{cases}$$
belongs to every $W^{k,p}(\Omega)$, and not to $C(\overline \Omega)$. Its also easy to make an example with a connected set e.g. $B(0,1)\setminus([0,1]\times \{0 \})$, but this was easier to type.

In the above cases the boundary is not even locally a graph, but IIRC you can see a different behavior with $p$ if the boundary is only locally $\alpha$-Hölder: I believe one should check when $1/\sqrt{x^2+y^2}$ belongs to $W^{k,p}(\Omega)$ where now $\Omega = \mathbb B(0,1) \setminus \{ (x,y) : x\ge0,\ |y|^\alpha \le x\}$.

Also worth mentioning is that, even if $\Omega$ is as smooth as you like, $C(\overline\Omega)$ is not even a subset of $W^{k,p}(\Omega)$, $k\ge 1$, even if $k$ or $p$ is very large. So in particular, it is not dense in $W^{2,p}$ as claimed in the comments. The reason already appears in dimension 1 in the form of e.g. the devil's staircase. It's continuous, and  it's a.e. derivative (and therefore the only candidate for the weak derivative) is zero. But its not constant; so it has no weak derivative. (In fact, the derivative is just a measure.)


问:
Why don't these two normal vectors have a 0 or a 180 degree angle between them when graphed?
There's a plane that is given with the equation $2x-3y+z+2=0$ and there's a line equation $\frac{x+3}{-1}=\frac{y}{3}=\frac{z-1}{P}$. I need to find the missing number $P$, when the line is parallel to the plane.
As number $P$ is a coordinate of the normal of the line and we also have the normal of the plane, I assumed we could use the dot product here and write $n_1 \cdot n_2 = 0$ as the normals need to be parallel in order for that to happen. I solved that and I got $P=11$, but when I graph the normals they don't seem to be parallel at all. Is there something wrong? Please help me. Thanks in advance.

答:
$v=(-1,3,P)$ is the direction vector of the line and not the normal (which would be a plane).

And indeed, the solution is to find $P$ for which $n\cdot v=0$ and $P=11$ seems correct.

To verify, show that the line and the plane don't have any common points.


问:
How do we check that a polynomial is irreducible in $\mathbb{C}[x,y]$
we know that  $\mathbb{C}$ is an algebraically closed field  so every polynomial $f$ in $\mathbb{C}[x]$ can be represented as

$f=(x-a_{1})^{\alpha_{1}} ....(x-a_{n})^{\alpha_{n}} $

and the only irreducible polynomials are the constant polynomials and of polynomials of the form  $g=x-a $ ,$a \in \mathbb{C} $

Let $f \in \mathbb{C}[x,y]$ be a polynomial

When do we say that $f$ is irreducible in $\mathbb{C}[x,y]$? Do the irreducible polynomials  have an exact form like in the case of $\mathbb{C}[x]$? if not how we can check whether $f$ is irreducible or not ?

For example  is this polynomial irreducible in $\mathbb{C}[x,y]$ or not

$h(x, y)=x^{p-1}+x^{p-2}+....+x+1-y$

答:
The definition of irreducible polynomials $\mathbb{C}[x,y]$ is the same as in general commutative rings: A polynomial $f$ is irreducible if

it is not a unit or $0$, i.e. constant in the case $\mathbb{C}[x,y]$

if there is a factorization $f=ab$ then $a$ or $b$ has to be a unit.

I don't know if there is some kind of exact form for irreducible polynomials over $\mathbb{C}[x,y]$, the generalization of the fact that the irreducibles in $\mathbb{C}[x]$ are exactly those of the form $x-a$ is the Nullstellensatz concerning maximal ideals, but not single polynomials.

For your example: $h(x,y)=x^{p-1}+\dots+x+1-y$. This is irreducible since it has degree 1 in $y$.


问:
Proving that this map is neither open nor closed

Let X={(x,y)$\in \mathbb{R}^2  :x \geq$ 0 or y=0}, and let T be the subspace topology on X induced by induced by usual topology on $\mathbb{R}^2$ and let U be the usual topology on $\mathbb{R}$. Define f:$X \to \mathbb{R}$ by f((x,y)=x for all (x,y) $\in X$. Prove that it is neither open nor closed.

I am really confused on how this problem should be solved . I tried to make some counterexamples but couldn't sucessfully disprove it.

This question was about proving f to be a quotient map and I proved that by definition but I am struck in this.

Can you please spare some time for it?

答:
$\{(x,y): 1<y<2, 0 \leq x <1\}$ is an open set in $X$ whose image is not open. $\{(x,\frac  1 x)
: x>0\}$ is a closed set in $X$ whose image is not closed.


问:
Exterior powers of tensor products
I am trying to understand Cauchy formula for exterior powers of tensor products.

Let $k$ be a characteristic 0 field. Let $E$ and $F$ be two vector spaces over $k$. Then we have $$ \wedge^d(E\otimes F) \cong \bigoplus_{|\lambda|=d} L_\lambda E \otimes L_{\lambda'}F, \ \text{where $\lambda'$ is dual of $\lambda$}.$$ Here $L_{\lambda}$ is the Schur functor and $L_\lambda E$ is the Schur module. I am following the notations of Weyman's book( Cohomology of vector bundles and syzygies).

I am curious in the map of this natural isomorphism. There is a proof given in the Weyman's book but I am unable to understand it. If someone can explain the maps even in $d = 2 \ or \ 3$ case that will work for me. Any ideas\hints are welcome.

答:
This fact can be deduced from Schur-Weyl duality.

Consider the vector space $(E \otimes F)^{\otimes d}$, which carries commuting actions of the three groups $GL(E)$, $GL(F)$, and $S_d$. We can rearrange things by writing $(E \otimes F)^{\otimes d} = E^{\otimes d} \otimes F^{\otimes d}$, so now $GL(E)$ acts on the first $d$ factors, $GL(F)$ on the last $d$ factors, and this time the symmetric group acts along the diagonal: $\sigma$ acts by permuting the first $d$ and last $d$ factors simultaneously. It is helpful to think of this $S_d$ action as being diagonal: $S_d \hookrightarrow S_d \times S_d$.

By Schur-Weyl duality, considering the pairs $GL(E) \times S_d$ and $GL(F) \times S_d$ indepdently, we get
$$ \begin{aligned}
E^{\otimes d} \otimes F^{\otimes d}
&\cong \left( \bigoplus_{|\lambda| = d} L_\lambda E \otimes S^\lambda \right) \otimes \left( \bigoplus_{|\mu| = d} L_\mu F \otimes S^\mu \right) \\
& = \bigoplus_{|\lambda| = |\mu| = d} L_{\lambda} E \otimes L_{\mu} F \otimes S^\lambda \otimes S^\mu. \end{aligned}$$

Now take the sign-isotypic component of the $S_d$ action on both sides. On the left, we get the sign-isotypic component of $(E \otimes F)^{\otimes d}$, which is isomorphic to the exterior power $\bigwedge^d(E \otimes F)$ because we are in characteristic zero. On the right, the tensor product $S^\lambda \otimes S^\mu$ of Specht modules contains the sign irrep once if $\mu = \lambda'$, and does not contain the sign irrep otherwise. (This is hom-tensor-dual to the fact that tensoring with the sign representation sends $S^\lambda$ to $S^{\lambda'}$). We get
$$ \bigwedge^d(E \otimes F) \cong \bigoplus_{|\lambda| = d} L_\lambda E \otimes L_{\lambda'} F.$$

This is essentially the argument given in 4.1 of Howe's "Perspectives on invariant theory", and he calls this theorem "skew $(GL_n, GL_m)$-duality" (others sometimes call it "skew Howe duality").

As for what the isomorphism looks like, Howe also offers some explanation. Fix bases $e_1, \ldots, e_n$ and $f_1, \ldots, f_m$ of $E$ and $F$ respectively, so that $E \otimes F$ has a basis $\{v_{ij} = e_i \otimes f_j\}$. The highest-weight vector corresponding to some partition $\lambda$ on the right can be written down by imagining the $v_{ij}$ in a rectangular matrix, overlaying the Young diagram of $\lambda$, and wedging everything together. For instance, for the partition $(4, 2, 1)$ would have highest weight vector
$$ v_{(4, 2, 1)} := (v_{11} \wedge v_{12} \wedge v_{13} \wedge v_{14}) \wedge (v_{21} \wedge v_{22}) \wedge (v_{31}),$$
which we can see has $GL(E)$-weight $(4, 2, 1)$ and $GL(F)$-weight $(3, 2, 1, 1)$.

Title: Exterior powers of tensor products


问:
Determining an Interval such that f(x) contracts
Question: Let $a,b \in \mathbb{R}$ with $a>b>0$. Find an Interval D $\subset \mathbb{R_+}$ such that $f: \mathbb{R_+}\to \mathbb{R}$ with $f(x)= \sqrt {a+\frac bx}$ contracts for $f_{|D}$. First off I computed the Derivative: $$f'(x)= \frac {-b}{2\sqrt{a+ \frac bx}\ *x^2}$$ Next step I tried to find where $\vert f'(x)\vert=1 \ $. Since $\lim \limits_{x \to \infty} \uparrow 0$, I specifically tried to find the case where $f'(x)=-1$. This way I can use the mean value theorem to determine from where on the function contracts. Sadly this is where I get stuck. No matter how I rewrite or simplify the function, I cannot seem to solve it for $x$. Is there any merit to this approach, or am I overlooking something here? If yes I'd be very grateful for a nudge in the right direction.

答:
We know that

If $f'$ is bounded by $1$ on $D$ then $f$ is contractive on $D$.

Thus, we may choose any set $D$ such that $f'$ is bounded by $1$ on $D$.

In your setting, $f'$ is continuous and smooth on $A=(-\infty, -\frac{b}{a}) \cup( 0,+\infty)$ with limits:

$\lim_{x\rightarrow -\infty} f'(x)=\lim_{x\rightarrow +\infty} f'(x)= 0$

$\lim_{x\rightarrow (-\frac{b}{a})^-} f'(x)=\lim_{x\rightarrow 0^+} f'(x)=+\infty $

Thus, we may choose, for example, $D=[x_0,+\infty)$ where $x_0$ is large enough.

Edit. An explicit formular for $D$ is
\begin{align}
D=\left[\sqrt[3]{\frac{b}{4}}, +\infty\right)
\end{align}

Proof.
Observe that
\begin{align}
\frac{1}{|f'(x)|} =2\sqrt{ \frac{a}{b}+ \frac{1}{x} } \frac{x^2}{b^{1/2}}>
2\frac{x^{3/2}}{b^{1/2}}\geq 1
\end{align}
whenever
\begin{align}
x\geq \sqrt[3]{\frac{b}{4}}.
\end{align}


问:
Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$?
I know there must be something unmathematical in the following but I don't know where it is:

\begin{align}
\sqrt{-1} &= i \\\\\
\frac1{\sqrt{-1}} &= \frac1i \\\\
\frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\\\
\sqrt{\frac1{-1}} &= \frac1i \\\\
\sqrt{\frac{-1}1} &= \frac1i \\\\
\sqrt{-1} &= \frac1i \\\\
i &= \frac1i \\\\
i^2 &= 1 \\\\
-1 &= 1 \quad !!?
\end{align}

答: Guaranteeing that all $a$ and $b$ is larger than $0$, $\sqrt{a}\sqrt{b} = \sqrt{ab}$. This also can be used when only one is larger than zero. However, if those are all negetive, it comes out to be $-\sqrt{ab}$. This can be proven by figuring out the square roots of each two negative numbers. For instance, think about the numbers $-2$ and $-3$. You'll see that $\sqrt{-2}\sqrt{-3} = i\sqrt 2 i \sqrt 2 = i^2\sqrt 6 = -\sqrt 6.$ The fact that $\frac{\sqrt a}{\sqrt b}=-\sqrt{\frac{a}{b}}$ when a is positive and b is negative can be proven in the same way.


问:
union of $[-r,r] ,r\in \mathbb{R}$
I have some basis questions:

I know if  $r \in \mathbb{R}$,We cay say $\cup_{r \in \mathbb{R}}(-r,r)=(-\infty,\infty)$.

Even if $r \in \mathbb{Q}$, $\cup_{r \in \mathbb{Q}}(-r,r)=(-\infty,\infty)$ because $\mathbb{Q}$ is dense in $\mathbb{R}$.

But What is $\cup_{r \in \mathbb{R}}[-r,r]$?

I know $(-r,r) \subsetneq [-r,r] \implies\cup_{r \in \mathbb{R}}(-r,r) \subsetneq \cup_{r \in \mathbb{R}}[-r,r]\implies (-\infty,\infty)\subsetneq \cup_{r \in \mathbb{R}}[-r,r]$ where I am wrong?

Thanks in advance!

答: You can have $A_n$ and $B_n$ are subets of $X$ and $A_n$ is strictly included in $B_n$ and $\cup_n A_n=\cup_nB_n$. You just point a particular example of that situation.

Title: union of $[-r,r] ,r\in \mathbb{R}$


问:
Derivation step in paper on LDA
I am wondering how do the authors of this paper perform the following step (in the paper it is on page 29):

$$\dfrac{\partial L}{\partial \gamma_i}=\Psi^\prime(\gamma_i)\left(\alpha_i + \sum_{n=1}^N\phi_{ni}-\gamma_i\right)-\Psi^\prime\left(\sum_{j=1}^k\gamma_j\right)\sum_{j=1}^k\left(\alpha_j+\sum_{n=1}^N\phi_{nj}-\gamma_j\right)=0$$

Maximized for:

$$\gamma_i=\alpha_i+\sum_{n=1}^N\phi_{ni}$$

Does anyone know the calculations for this step?
($\Psi'$ is the derivative of the digamma function, everything else is a variable)

答: Just plugging the $\gamma_i$ and $\gamma_j$ into the second and fourth brackets turns those brackets into 0. And thus the expression into 0.

Title: Derivation step in paper on LDA


问:
Question about Theorem 3.54 Rudin.

Let $\sum a_n$ be a conditionally convergent series. Suppose :
$$
-\infty \leq \alpha\leq \beta \leq +\infty$$
Then there exist a rearrangement $\sum a'_n$ with partial sums $\{s'_n\}$ such that

$$
\lim\limits_{n\to\infty} \inf s'_n=\alpha \quad \quad \lim\limits_{n\to\infty}\sup s'_n=\beta$$
Let $$p_n = \frac{|a_n| + a_n}{2}, \ q_n = \frac{|a_n| - a_n}{2} \ (n = 1, 2, 3, \ldots). $$
Then $p_n - q_n = a_n$, $p_n + q_n = |a_n|$, $p_n \geq 0$, $q_n \geq 0$. The series $\sum p_n$, $\sum q_n$ must both diverge.

For if both were convergent, then $$\sum \left( p_n + q_n \right) = \sum |a_n|$$ would converge, contrary to hypothesis. Since $$ \sum_{n=1}^N a_n = \sum_{n=1}^N \left( p_n - q_n \right) = \sum_{n=1}^N p_n - \sum_{n=1}^N q_n,$$ divergence of $\sum p_n$ and convergence of $\sum q_n$ (or vice versa) implies divergence of $\sum a_n$, again contrary to hypothesis.

Now let $P_1, P_2, P_3, \ldots$ denote the non-negative terms of $\sum a_n$, in the order in which they occur, and let $Q_1, Q_2, Q_3, \ldots$ be the absolute values of the negative terms of $\sum a_n$, also in their original order.

The series $\sum P_n$, $\sum Q_n$ differ from $\sum p_n$, $\sum q_n$ only by zero terms, and are therefore divergent.

We shall construct sequences $\{m_n \}$, $\{k_n\}$, such that the series $$ P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} + P_{m_1 + 1} + \cdots + P_{m_2} - Q_{k_1 + 1} - \cdots - Q_{k_2} + \cdots \tag{25}, $$ which clearly is a rearrangement of $\sum a_n$, satisfies (24).

Choose real-valued sequences $\{ \alpha_n \}$, $\{ \beta_n \}$ such that $\alpha_n \rightarrow \alpha$, $\beta_n \rightarrow \beta$, $\alpha_n < \beta_n$, $\beta_1 > 0$.

Let $m_1$, $k_1$ be the smallest integers such that $$P_1 + \cdots + P_{m_1} > \beta_1,$$ $$P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} < \alpha_1;$$  let $m_2$, $k_2$ be the smallest integers such that $$P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} + P_{m_1 + 1} + \cdots + P_{m_2} > \beta_2,$$ $$P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} + P_{m_1 + 1} + \cdots + P_{m_2} - Q_{k_1 + 1} - \cdots - Q_{k_2} < \alpha_2;$$ and continue in this way. This is possible since $\sum P_n$, $\sum Q_n$ diverge.

If $x_n$, $y_n$ denote the partial sums of (25) whose last terms are $P_{m_n}$, $-Q_{k_n}$, then $$ | x_n - \beta_n | \leq P_{m_n}, \ \ \ |y_n - \alpha_n | \leq Q_{k_n}. $$ Since $P_n \rightarrow 0$, $Q_n \rightarrow 0$ as $n \rightarrow \infty$, we see that $x_n \rightarrow \beta$, $y_n \rightarrow \alpha$.

Finally, it is clear that no number less than $\alpha$ or greater than $\beta$ can be a subsequential limit of the partial sums of (25).

Why is $| x_n - \beta_n | \leq P_{m_n}$? Why $P_n\to 0$? What if $\sum a_n$ converges absolutely?

I don't know how this inequality is true but makes sense because it says something like "By definition $m_n$ is the smallest integer such that $x_n > \beta_n$" which might be key to it. I think $P_n\to 0$ is because $a_n\to 0$ but it is still not clear since there might be exception. $\sum a_n$ shouldn't converge absolutely for this theorem makes sense but I don't know why.

答:
By definition, $x_n>\beta_n$. If $x_n-\beta_n>P_{m_n}$, then $x_n-P_{m_n}>\beta_n$, contradicting the choice of $m_n$.

The sequence $\{P_n\}$ is a subsequence of $\{a_n\}$; as the latter goes to zero, so does the former.

If the series converges absolutely then any rearrengement converges to the same limit. So in that case the result can only be possible when $\alpha=\beta=\sum_na_n$.

To see what happens with $\sum_n|a_n|<\infty$, let $\sigma:\mathbb N\to\mathbb N$ be a bijection. Let $L=\sum_na_n$. Fix $\varepsilon>0$. By  hypothesis, there exists $n_0$ such that $\sum_{n>n_0}|a_n|<\varepsilon$. Let $F=\sigma^{-1}(\{1,\ldots,n_0\})$ and put $m=\max\{k:\ k\in F\}$. Then
$$
\Bigl|L-\sum_{k=1}^ma_{\sigma(k)}\Bigr|=\Bigl|\sum_{k>m}a_{\sigma(k)}\Bigr|
\leq \sum_{k>m}|a_{\sigma(k)}|\leq\sum_{k>n_0}|a_k|<\varepsilon,
$$
where the second to last inequality is justified by the fact that if $k>m$, then $k\not\in F$ and so $\sigma(k)\not\in\{1,\ldots,n_0\}$, that is $\sigma(k)>n_0$.

Title: Question about Theorem 3.54 Rudin.


问:
Midpoint of a line segment passing through two points.

Let $C_1$ be a circle defined with $X(-2,7)$ and $Y(2,-5)$ as the endpoints of the diameter of the circle.
Let $C_2$ be a circle defined with $Y(2,-5)$ and $Z(4,-11)$ as the endpoints of the diameter of the circle.

Let $A = C_1 \cup C_2 - Y$

Let $B$ be the set of all midpoints of the set of all straight line segments joining two distinct points in $A$ and passing through the point $Y$. Find $B$.

My attempt:

Since $XYZ$ are collinear $C_1$ and $C_2$ meet at a single point: $Y$. I can easily find the set of all midpoints of the line segments joining a point (say $T$) in $C_1$ and $Y$. Let $M(x,y)$ be the midpoint of $TY$ and $N(0,1)$ be the center of $C_1$. Notice that $MN$ is perpendicular to to $MY$. Therefore we can build the equation $\dfrac{y-1}{x-0}*\dfrac{y+5}{x-2}=-1$ i.e product of their slopes equal -1. Similarly, the set of all midpoints of the line segments joining a point (say $U$) in $C_2$ and $Y$ can be found.
The issue now is, I don't know how to use this information to compute the required locus $B$. I don't even know if this information is useful.

答:
Locus of points dividing segments joining points of $S_1$ with points of $S_2$ in ratio $\lambda : (1-\lambda)$ is given by $$\lambda S_1 + (1-\lambda) S_2=0$$

$(2,-5)$ should lie on this locus, which it clearly does. For midpoints, put $\lambda = \frac{1}{2}$. The coefficient of $x^2$ and $y^2$ will turn out to be same; this locus is a circle.

Given $$S_1 : (x-2)(x+2) + (y+5)(y-7)=0$$
$$S_2 : (x-2)(x-4) + (y+5)(y+11)=0$$

Required locus is $$S : (x-2)(x-1) + (y+5)(y+2)=0$$

with center at $$\Big(\dfrac{2+1}{2},\dfrac{-5-2}{2}\Big)=\Big(\dfrac{3}{2},\dfrac{-7}{2}\Big)$$


问:
Help with sum of a series of fractions
I was expected to have this series $$S = \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... + \frac{1}{2^{k}} $$
summed up to 1. However, I couldn't do that using this calculation:
$$ S = \frac{1}{2} + S - \frac{1}{2} $$ where $$ \frac{1}{2} + S = \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... + \frac{1}{2^{k}} $$ was a geometric series.
Thus, $$ S = \frac{1}{2} + S - \frac{1}{2} = a(\frac{1 - r^{k}}{1 -r}) - \frac{1}{2} $$
where $ a = \frac{1}{2} $, and $ r = \frac{1}{2} $.

Substituting the values of $a$ and $r$ into the expression above, I got $$ S = \frac{2^k - 1}{2^k} - \frac{1}{2} $$
Then taking limit on the right hand side $$ \lim_{k \to \infty} (\frac{2^k - 1}{2^k} - \frac{1}{2}) $$
I got $$ S = \frac{1}{2} $$
What did I do wrong?

EDIT: The reason I wanted to have $S$ summed to 1 because of this requirement (highlighted in yellow):

答: Note that there are two cases for each case containing $k$ letters (e.g. for 2 letters: aa,bb; for 3 letters: acc, bcc). So the probability of all the cases actually sum to $$2\cdot \frac{1}{2^2}+2\cdot \frac{1}{2^3}+\cdots =\frac{1}{2}+\frac{1}{2^2}+\cdots =\frac{1/2}{1-1/2}=1$$

Title: Help with sum of a series of fractions


答:
For real constants $a_1, a_2,$ The number of times $a_1x\sin(x)$ passes through any real number $c,$ which is $ \infty,$ is not affected by adding $a_2\sin(x)$ to it.

This is because the “amplitude”/oscillations* of $x\sin(x)$ increases without bound as $x \to \infty$ ( and $x \to -\infty$), so $a_1x\sin(x)$ passes through every real number an infinite number of times, whereas $a_2\sin(x)$ is bounded with amplitude $a_2$.

And replacing $x$ with $kx$ does nothing to the number of roots either, because it just scales it horizontally.

By amplitude/oscillations* I mean the distance from the "centre" of your function, which is $0$.

Title: Infinite many roots of an equation

Body: Let $k,A>0$. How may roots the equation $f(x)=\frac{1}{k} (1+\frac{A}{2}x)\sin(kx)=c,\ x\in \mathbb{R}, \ c\in \mathbb{R}$ has? I beliave the roots are infinite and one tric I thought is by using Bolzano theoreme I may find sequences $(x_n),(y_n)$ s.t $\ x_n,y_n\to \infty$ and $f(x_n)f(y_n)<0$. But I cannot compose it. Any help please?


问:
Suppose $x$ and $y$ are two real numbers such that $x + y = 6$ and $x^2 + y^2 = 43$. Find the value of $x^3 + y^3$.
I tried quadratic and substituting and cubing the equations but it doesn't work and I'm pretty sure i did not make any careless mistake.</p>

答:
The following $2$ facts are very useful generally:

$$x^2+y^2=(x+y)^2-2xy$$
$$x^3+y^3=(x+y)^3-3xy(x+y)$$
Using the first fact, we can easily find the value of $xy$ (using $x^2+y^2=43,~~~x+y=6$), and we already know the value of $x+y$. Hence we can easily find the value of $x^3+y^3$.

If you need any more help please don't hesitate to ask :)


问:
Is it true that $\mathbb{R}^m \cap \mathbb{R}^n = \emptyset$ where $m \neq n$
Is it true that $\mathbb{R}^m \cap \mathbb{R}^n = \emptyset$ where $m \neq n$ ?

I would say that it is true because the elements in each set are different entities. This means that one set cannot be a subset of the other. An example would be $\mathbb{R} \cap \mathbb{R}^2$. You cannot say that $1 \in \mathbb{R}^2$ for example.

答:
Well, these sets are disjoint.

To see this, define the ordered pair
$$(x,y) = \{\{x\},\{x,y\}\}$$
and extend this definition to $n$-tuples
$$(x_1,\ldots,x_n) = ((x_1,\ldots,x_{n-1}),x_n).$$
So $n$-tuples are sets and we have clear notion of equality of two sets:
$$A= B :\Longleftrightarrow \forall x[x\in A\Leftrightarrow x\in B].$$
Equality of two $n$-tuples is then given by
$$(x_1,\ldots,x_n)=(y_1,\ldots,y_n)\Leftrightarrow \forall i[x_i=y_i].$$
Taking this into accout you see that the above sets are disjoint if $m\ne n$.


问:
Reduce the Cauchy-Euler equation to a smaller expression?
Given the Cauchy-Euler second order differential equation:
$$ \left( x^2\frac{d^2}{d x^2} + x\frac{d}{d x} -1\right)y(x)=0 $$
is it possible to find a function $u(x)$ , which satisfies the following simple form:?
$$ \frac{d^2 y(u)}{d u^2}=0 $$
***notice that $t=\ln(x)$ yields:
$$ \frac{d^2 y(t)}{d t^2}-1=0 $$

答:
Starting backwards from the solution $y=c_1x+c_2x^{-1}$, if you set $xy(x)=f(x^2)$, then you get for the derivatives
$$
xy'+y=2xf'(x^2)\\
xy''+2y'=4x^2f''(x^2)+2f'(x^2)
$$
so that
$$
0=x^2y''+xy'-y=4x^3f''(x^2).
$$
This gives the expected equation $f''=0$, but not exactly the way you were looking for.