问:
Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $

If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $  find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$

Here's what I have tried so far,

I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\frac{c}{a} $

So I can express $b=-a(\alpha+\beta)$

$c=a.\alpha\beta$

Once I substitute for b and c in the equation I can get, $$\alpha\beta x^2+(\alpha+\beta)(\alpha\beta+1)x+(\alpha\beta+1)^2=0$$

I want to know whether there is any different approach other than this method?

Any hint is higly valued. thank you!

答:
Solve the two equations to find

$$ay^2+by+c=0\implies y=\dfrac{-b\pm\sqrt{b^2-4ca}}{2a}\ \ \ \ (1)$$

$$cax^2-b(c+a)x+(c+a)^2=0\implies x=(c+a)\cdot\left(\dfrac{b\pm\sqrt{b^2-4ca}}{2ca}\right)\
 \ \ \  (2)$$

Considering the opposite signs of the roots,

$$\dfrac yx=-\dfrac c{c+a}$$

Considering the same signs of the roots,

$$\dfrac yx=-\dfrac c{c+a}\cdot\dfrac{b+m\sqrt{b^2-4ca}}{b-m\sqrt{b^2-4ca}}\text{ where }m =\pm1$$

So, the ratios are not the same!


问:
In how many ways can you select $4$ different numbers from the numbers $1,2,..,15$
In how many ways can you select $4$ different numbers from the numbers $1,2,..,15$ so that no pair of numbers among the selected ones are subsequent numbers.

I turned this into a minor problem. I assumed that we have $5$ numbers and we want to choose two different numbers so that no pair of numbers among the selected ones is subsequent. I came to the conclusion that it will be $5$ choose $2$ and then minus all subsequent pairs that I calculated in my head. But I wonder if there is a better method that can be applied to $15$ numbers.

Also, how would you make this problem more difficult.

答:
I remember writing a generalization to this a long time ago. I'll paste it here. Overall it's pretty intuitive to come up.

Generalization

The number of ways to create a subset of $k$ elements denoted by $e_1<e_2<..<e_k$ from $\{ 1,2,...,n \}$ such that the members of the subsets differs at least $y$ is $\binom{k+n-y(k-1)-1}{k}$

Proof:

To construct a unique subset we need:

A variable ($a$) that states the distance between $1$ and the first element of the subset.

$k-1$ variables ($d_1,d_2,...,d_{k-1})$ that states the distance between $e_1$ and $e_2$, $e_2$ and $e_3$, and so on.

A variable ($b$) that states the distance between $e_k$ and $n$.

And the sum of all the variables above will always be equal to $n-1$. So, we have:
$$a+d_1+d_2+...+d_{k-1}+b=n-1 $$
To apply stars and bars for non-negative integers we need to remove the restriction of $d_i \ge y$, we do this by adding $y$ to each of $d_i$:
$$a+d_1+d_2+...+d_{k-1} + y(k-1)+b=n-1$$
$$a+d_1+d_2+...+d_{k-1}+b=n-1-y(k-1)$$
By stars and bars, the amount of non-negative integer solutions the equation is $$\binom{k+n-y(k-1)-1}{k}$$

For the question here, $n=15,k=4,y=2$. So the answer is
$$\binom{4+15-2(4-1)-1}{4}=\binom{12}{4}$$

ContentLicense: CC BY-SA 4.0


问:
$ \mathcal{F}_n = \sigma(\{0\},\{1\},\{2\}, \dots , \{n\})$ showing $\cup_{n\geq0} \mathcal{F}_n $ is not a sigma algebra, do not understand.
Isn't this countable union equal to the sigma algebra generated by the singletons of the natural numbers, which is the class of subsets of the natural numbers where the set is either or countable or its complement is countable. And we know that this class IS a sigma algebra.

So isn't saying this isn't a sigma algebra a contradiction.

答:
$$\mathcal F =\bigcup_{n\ge 0} \mathcal F_n$$ isn’t closed under countable union.

Namely all the $\{n\}$ belong to $\mathcal F$. However $A=\{0,1, 2, \dots\} \notin \mathcal F$ as $A$ belongs to none of the $\mathcal F_n$.

$$\mathcal F \neq \sigma(\{0\}, \{1\}, \dots )$$


问:
Finding a basis for polynomial vector spaces?
Find a basis of the following vector space.

The vector space over $\Bbb C$ of polynomials $p(z)$ of degree at most $4$ with coefficients in $\Bbb C$ satisfying $p(z) = p(−z)$ for all $z ∈ \Bbb C$ and $p(−2) = 0$.

How do you do this?

答:
Any fourth degree polynomial can be written $p(z)= az^4+ bz^3+ cz^2+ dz+ e$.  The requirement that $p(z)= p(-z)$ means that $az^4+ bz^3+ cz^2+ dz+ e= az^4- bz^3+ cz^2- dz+ e$ so that $bz^3+ dz= -bz^3- dz$ and then $bz^2+ dz= 0$ for all $z$ so that $b= d= 0$.  That is, since this is an "even" function only the even powers of $z$ have non-zero coefficient.  The condition that $p(-2)= 0$ then gives $16a+ 4c+ e= 0$.  We can write that as $e= -16a-  4c$.

The polynomial can be written as $az^4+ cz^2-16a- 4c= a(z^4- 16)+ c(z^2- 4)$


问:
Show $ (2 - b^2c^2) (1 + b^2) \ge (b^2 + c^2)(b + c - 3)^2 $ in some range.
Show $ F = (2 - b^2c^2) (1 + b^2) - (b^2 + c^2)(b + c - 3)^2 \ge 0$ in the following two ranges of $(b,c)$:

$b \in [0 \quad 1]$ and $0 \le c\le b$,

$b \in [1 \quad 1.2]$ and $0 \le c\le 3-2b$.

Equality occurs only at $(b,c)= (1,0)$.

This can be solved by treating $F$ as a polynomial in one of the variables (the other a parameter), and then using Budan's theorem. This has been performed here, but this is a very lengthy business. Does someone have a more direct method? Thanks in advance!

答:
Not an answer just a remark for the first case:

If we substitute $b=\frac{z}{z+y}$ and $c=\frac{y}{x+y}$ we get this .Remains to plays properly with the three variables or perhaps use Buffalo's way I don't know .

Hope you find something good !


问:
Generating function: Find a closed form of $\sum_{k=0}^n (-3)^k(k+1)$
Find the closed form of $\sum_{k=0}^n (-3)^k(k+1)$.

So the generating function would be: $$A(x)=1-6x+18x^2-108x^3...$$ So what I did notice is that its closed form is perhaps some variation of $1\over {1+x}$ but I didn't manage to find a general formula.

Thanks for any help in advance!

ContentLicense: CC BY-SA 3.0

Body:
Let $f(x) = \sum_{k=0}^n x^{k+1}
= \frac{x-x^{n+2}}{1-x}
$.
Then $f'(x) = \sum_{k=0}^n (k+1)x^k$,
so $f'(-3) = \sum_{k=0}^n (k+1)(-3)^k$.

So

$\begin{align}
f'(x) &= \frac{(x-x^{n+2})'(1-x) - (1-x)'(x-x^{n+2})}{(1-x)^2}\\
&= \frac{(1-x)(1-(n+2)x^{n+1})+x-x^{n+2}}{(1-x)^2}\\
&= \frac{1-(n+2)x^{n+1}-x+(n+2)x^{n+2}+x-x^{n+2}}{(1-x)^2}\\
&= \frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(1-x)^2}\\
\end{align}
$

Putting $x = -3$,

$\begin{align}
f'(-3)
&= \frac{1-(n+2)(-3)^{n+1}+(n+1)(-3)^{n+2}}{4^2}\\
&= \frac{1-(-3)^{n+1}((n+2)+(n+1)3)}{4^2}\\
&= \frac{1-(-3)^{n+1}(4n+5)}{16}\\
\end{align}
$

As a check (since I did this off the top of my head):

If $n=0$,
$f'(-3) = \frac{1-(-3)^{1}(5)}{16}
=\frac{1+15}{16}
=1
$
and $1*(-3)^0 = 1$.

If $n=1$,
$f'(-3) = \frac{1-(-3)^{2}(4+5)}{16}
=\frac{1-9*9}{16}
=-5
$
and $1+2*(-3) = -5$.

If $n=2$,
$f'(-3) = \frac{1-(-3)^{3}(8+5)}{16}
=\frac{1+27*13}{16}
=22
$
and $-5+3*(-3)^2 = -5+27 = 22$.


问:
Local martingale $\Leftrightarrow$ every component is a local martingale
I want to prove the following corollary:

A $\mathbb K^d$-valued process $M$ is a local martingale if and only if every component process is a $\mathbb K$-valued local martingale.

($\mathbb K$ means $\mathbb R$ or $\mathbb C)$

I know that the statement holds true for martingales.
I shall use the following exercise to prove the corollary:

Does anyone have an idea how to apply this exercise here?

答:
For the "$\Rightarrow$"-direction you don't use this corollary. You just take the localizing sequence, then you know the stopped process is a martingale and use that you know that every component has to be also a martingale.

But for the "$\Leftarrow$"-direction you have to build a new localizing sequence of the localizing sequences of the components. This is where the corollary comes into play. You take for each $n$ the minimum of the localizing stoppingtimes of all components. Now the corollary tells you that this new stopping time resp. sequence is a localizing sequence for each component. Then you use again that the statement holds for martingales and you are done.


答:
You have to impose that for a generic polynomial $\alpha x^4+\beta x^3+\gamma x^2+\delta x+\varepsilon \in\mathbb C_{\le 4}[x]$ the conditions:
$$P(-2)=0\implies16\alpha-8\beta+4\gamma-2\delta+\varepsilon=0 \text{ and }$$
$$\forall x\in\mathbb C:P(x)=P(-x)\iff2\beta x^3+2\delta x=0\iff\beta x^3+\delta x=0.$$
From the first condition we get $\varepsilon=2\delta-4\gamma+8\beta-16\alpha=-16\alpha-4\gamma$, since $\beta$ and $\delta$ are $0$.

Then we have two free parameters $\alpha$ and $\gamma$ and a basis of this space is
$$\mathcal B=\{x^4-16,x^2-4\}$$

Title: Finding a basis for polynomial vector spaces?

Body:
Find a basis of the following vector space.

The vector space over $\Bbb C$ of polynomials $p(z)$ of degree at most $4$ with coefficients in $\Bbb C$ satisfying $p(z) = p(−z)$ for all $z ∈ \Bbb C$ and $p(−2) = 0$.

How do you do this?


问:
Are all critical points either inflection points, local minimum or local maximum?
Question:

Let $f$ be a differentiable function. If the point c is a critical number, then either  it is a local maximum, or  local minimum, or an inflection point. $T/F$ ?

My opinion:

If c is a critical point then f'(c)=0 or undefined. So it may local maximum and local minimum.

If  f '(c)=∞ then c is inflection point at the same time and if f '(c)=0 it may inflection point again.

But i can't find instance disproves this thesis.

答:
What about

$$f(x)=\begin{cases}
x^2\sin\left(\frac{1}{x}\right) & x \neq 0\\
0 & x=0
\end{cases}$$

at $x=0$?

 $f$ is differentiable on $\mathbb R$ but $0$ is not a minimum, not a maximum and not an inflexion point.


问:
How many ways can I partition a string of length n into k parts? k can be >n.
How many ways can I partition a string of length n into k substrings? k can be >n, and hence some parts can be empty also. For example, if the string is 'abc' and k = 4, then the possible partitions are (, seperates the parts)</p>
```<code>abc,,,,
ab,c,,,
ab,,c,,
ab,,,c,
a,bc,,,
a,b,c,,
a,b,,c,
a,,bc,,
a,,b,c,
a,,,bc,
,abc,,,
,ab,c,,
,ab,,c,
,a,bc,,
,a,b,c,
,a,,bc,
,,abc,,
,,ab,c,
,,a,bc,
,,,abc,
</code>```

答:
As far as I understood, you want to get an ordered list of substrings with empty ones. If the length of your string is $m$, you have to place $n-1$ delimiters in
$m+n-1$ possible positions. So the answer is
$$
C_{m+n-1}^{n-1} = \bigl(\substack{m+n-1\\ n-1}\bigr) = \frac{(m+n-1)!}{(n-1)!\,m!}
$$
In your example $m=3, n=4$ it is $$ C_6^3 = \bigl(\substack{6\\3}\bigr) = 20 $$


问:
an orthonormal basis of $L^{2}([0,2\pi]\times [0,2\pi])$
define $(n,m) \in \mathbb{Z}^{2}$ the vectors $e_{n,m} \in L^{2}([0,2\pi] \times [0,2\pi])$ with
\begin{align}
e_{n,m}=\frac{1}{2\pi}e^{inx}e^{imy}
\end{align}

Prove that $(e_{n,m})_{(n,m) \in \mathbb{Z}^{2}}$ is an orthonormal basis of $L^{2}([0,2\pi]\times [0,2\pi])$.

I tried to prove this by calculation $\langle e_{n,m},e_{n,m}\rangle=1$ and $\langle e_{n,m},e_{i,k}\rangle =0$ for $n\neq i, m\neq j$ I got a really long expression and I probably made a mistake, but I think this can be to done an easier way. Has someone an idea?

EDIt: to prove this by calculation $\langle e_{n,m},e_{n,m}\rangle=1$ and $\langle e_{n,m},e_{j,k}\rangle =0$ for $n\neq j, m\neq j$ you see an answer in the answers.

Now I wonder, is this enough to talk about an orthonormal basis? No, it's only enough to talk about orthonormal. But Now I need to prove that those vectors are a basis, in other words that a function $f \in L^{2}([0,2\pi] \times [0,2\pi])$ can be written as a lineair combination of the vectors. How do you start with this if  you don't know the function $f$?

答:
Hint: use this
$$ \left<f,g\right>=\int_0^{2\pi}\int_0^{2\pi}\bar{f}(x,y)g(x,y)dxdy. $$
Using
$$ \int_0^{2\pi}e^{i(l-j)x}dx=2\pi \delta_{jl} $$
you have
$$ \left<e_{j,k},e_{l,m}\right>=\frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi}e^{i(l-j)x}e^{i(m-k)y}dxdy=\delta_{jl}\delta_{km}. $$


问:
Inverse Laplace transform after derivative of transform.
I've been using the following theorem in my intro ODE course:

If $F(s) = \mathscr{L}\left\{ f(t) \right\}$ and $n\in\mathbb{N}$, then $\mathscr{L}\left\{t^n f(t) \right\} = (-1)^n\frac{d^n}{ds^n}F(s).$

What is the inverse of this transform? For instance, say I wanted to take $$ \mathscr{L}^{-1}\left\{ \frac{s}{(s^2+16)^2} \right\}.$$ I can see that the result would be $\frac{1}{8}t\sin(4t)$. But, how would one go about manually working this out (and particularly in general)?

Body:
Observe that $$\int \frac s{(s^2+16)^2}ds=-\frac1{s^2+16}+C\iff \frac{d \left(\frac1{s^2+16}\right)}{ds} =- \frac s{(s^2+16)^2}$$

and from this, $$\mathscr{L}^{-1}\left\{ \frac1{(s^2+16)} \right\}=\frac{\sin4t}4$$

So, $$\mathscr{L} \left\{ t^1\cdot\frac{ \sin 4t}4 \right\}=(-1)^1\frac{d^1}{ds^1}\left(\frac1{s^2+16}\right)=+\frac{2s}{(s^2+16)^2} $$


问:
Area of square inside a triangle
Considering the attached image, please compute the area of the square.

$\\$To elaborate suppose we have a triangle $\Delta ABC$, that its angle $\angle ABC$ is equal to $45^o$. On side $\overline{AB}$ we have a point $E$ that divide the side to two segments, $\overline{BE}=4$ and $\overline{EA}=6$.

Find the area of the square $\square CDEF$, such that its $\overline{FC}$ side reside on $\overline{AC}$ side of the triangle $\Delta ABC$, its one vertex is on E point, and the alternate vertex of E is on vertex C.

答:
Let the square have side $L$ and consider triangles $AEF$ and $BCE$.

Simply using the sum of angles in a triangle we see that $\angle AEF =\angle BCE=X$ say.
Then
$$6\sin X=\sqrt{36-L^2}, \sqrt2L\sin X=4\sin 45.$$
Eliminating $\sin X$ gives a quadratic for the area $L^2$,  $$L^2(36-L^2)=12^2,$$
with solutions $L^2=18\pm 6\sqrt 5.$

Title: Area of square inside a triangle


问:
$f$ is Derivative in $x_0=0$ and for all $n \in \mathbb{N}\rightarrow f(\frac{1}{n})=a$ for ($a\in \mathbb{R}$), calculate $f'(0)$.

Let $f$ be differentiable at $x_0=0$ and for all $\mathbb{N}\ni n\longmapsto f\left(\frac{1}{n}\right)=a$ for some $a\in \mathbb{R}$. Then calculate $f'(0)$.

Hi there, I'm stuck in this question and I have a feeling I'm missing something, Perhaps using some theorem idk.

By typing "Derivative" I mean that the $\lim_{x\to0 }\frac{f(x)-f(0)}{x}$ is finite and exists.

I'm not sure how to use the idea of

"for all $n\in \mathbb{N}\rightarrow f(\frac{1}{n})=a$"

Thank you in advance.

答:
If the derivative exists at zero, i.e., the limit
$$\lim_{x\to0}\frac{f(x)-f(0)}{x}$$
exists, then it equals
$$\lim_{n\to\infty}\frac{f\big(\frac1n\big)-f(0)}{\frac1n} = \lim_{n\to\infty}\frac{a-f(0)}{\frac 1n}.$$
Now $f$ is continuous at $0$ (since it is differentiable there), so $f(0) = f(\lim\frac1n) = \lim f(\frac1n) = \lim a = a$, so the derivative is equal to
$$f'(0)=\lim_{n\to\infty}\frac{a-a}{\frac 1n}=0.$$


问:
How can I show that this set is bounded?
How can I show that this set is bounded?

Montrez que l'ensemble
$$A = \left\{\frac{\sin(xy)}{x^2+y^2}; (x,y)\in\mathbb R^2, (x,y)\ne (0,0)\right\}$$
est borné.

(Source)

this is my try answer

FIRST: I tried to put an R that belongs to A so R=sin(xy)/x²+y²

SECOND: I turnd the equality into inequality and I put the absolute value like that: |R|<=|sin(xy)|/x²+y² and after that I stucked because i didn't find the way how to solve this inequality and my target is to find borders for sin(xy)/x²+y²,Once I find them i can say that's bounded.

答:
$\newcommand{\R}{\mathbb{R}}$
If you know that

$|\sin(a)| \leq |a|$ for all $a \in \R$.

$2|xy| \leq x^2 + y^2$ for all $x,y\in\R$ because $(|x|-|y|)^2 \geq 0$

$$
\frac{|\sin(xy)|}{x^2 + y^2}
\leq
\frac{|xy|}{x^2 + y^2}
\leq
\frac{1}{2}
$$
for all $(x,y) \in \R^2\setminus \{(0,0)\}$.


问:
Find integral with derivate from function values and other integral
It is known that $f(x)$ has a continuous derivative with $f(6)=36$, $f(9)=243$ and $\int\limits^{9}_{6}{\frac{f(x)}{x^3}dx}=4$. Find the value of the integral $\int\limits^{9}_{6}{\frac{f'(x)}{x^2}dx}$.

The answer to the question is $10$ and I am really out of ideas on how to solve it. I have a feeling it is done with the FTC but I can't seem to get any results. Any tips?

Body:
Integration by parts:
$$4=\int_6^9\frac{f(x)}{x^3}\,dx=-\frac12\left[\frac{f(x)}{x^2}\right]_6^9+\frac12\int_6^9\frac{f'(x)}{x^2}\,dx$$
$$4=-\frac12(243/81-36/36)+\frac12\int_6^9\frac{f'(x)}{x^2}\,dx=-1+\frac12\int_6^9\frac{f'(x)}{x^2}\,dx$$
$$\int_6^9\frac{f'(x)}{x^2}\,dx=2(4+1)=10$$


问:
prove or disprove Lipchitz function
Prove or disprove that $f(x)= -7x-3$ is Lipchitz on $\mathbb{R}$

My answer is:

$|f(x)-f(y)|=|-7x-3-(-7y-3)|$

$=|-7x+7y|=|-7(x-y)|=7|x-y|$

If the answer is equal not less than or equal can we say it is Lipchitz?
And also they asked if it is Lipchitz in $\mathbb{R}$ so is this an answer on $\mathbb{R}$?

答:
Your proof is fine. For your query in comment:

The definition of Lipschitz function is:

If $A\subseteq\mathbb{R}$ and $f:A→\mathbb{R}$ is function. Then, $f$ is said to be Lipschitz on $A$ if there exists a constant $K>0$ such that,

$|f(x)-f(y)|≤K|x-y|$ for all $x,y\in A$

See that, in definition $A$ is subset of $\mathbb{R}$. So $A$ may be equals to $\mathbb{R}$ itself.

Title: prove or disprove Lipchitz function


问:
F(x) is a function differentiable and monotonic strictly increasing on the OPEN interval (a,b).

$f(x)$ is a  differentiable and  strictly increasing function on the open interval $(a,b)$.

Prove or disprove that $f'(c)>0\; \forall c\;\in (a,b)$.

I was trying to apply Lagrange's derivative theorem, but the condition that the function must be continuous on the closed interval $[a,b]$ is not given. Im assuming the answer should be false, but I can't prove it's false or give a counterexample. Can we find a way to apply Lagrange's theorem? Then for sure, it's a true affirmative.

答:
$f:x\to x^3$ on $(-1,1)$ is a counter-example for your statement, as $f'(0)=3\times(0)^2=0$  .

The true implication is that if $f$ is differentiable on an open interval $(a,b)$ and strictly increasing, then $f'\geq0$ and there exists no open interval $(x,y)\subset(a,b)$ such that $\forall{t}\in(x,y) : f'(t)=0$, which means that $f'$ is never constant at zero on an open interval of $(a,b)$, since if it was the case, we'd get that $f$ is constant on that interval  which would contradict the strict monotony.

In fact, suppose by contradiction that there exists $(x,y)\subset(a,b)$ on which $f$ is always zero, then let $t,h\in(x,y)$ such that $t\neq h$. By the mean value theorem, there exists $c\in(x,y)$ such that :
\begin{equation}
f'(c)=\frac{f(t)-f(h)}{t-h}
\end{equation}
Since $f$ is always zero on $(x,y)$, so in particular $f'(c)=0$, hence $f(t)=f(h)$ which contradicts the strict monotony of $f$.


问:
Can I get a light hint for this binomial proof?

Consider the geometric series

$$S=1+(1+x)+(1+x)^2+\dots+(1+x)^n$$

(a) Show that
$$S=\frac{(1+x)^{n+1}-1}{x}$$

(b) Hence show that

$$S={n+1\choose1}+{n+1\choose2}x+\dots+{n+1\choose r+1}x^r+\dots+{n+1\choose n+1}x^n$$

(c) Hence prove that

$${n\choose r}+{n-1\choose r}+\dots+{r\choose r}={n+1\choose r+1}$$

I have successfully solved parts a) and b). However, I am struggling with c). I can tell it's the hockey stick identity but the thing that's getting me is the hence. How can I use the previous parts to prove it? I just need a little push. Perhaps it has to be a specific binomials relationship. I tried using the definition of symmetry but that didn't work.

答:
Another way to do:
Note
$$S=1+(1+x)+(1+x)^2+\dots+(1+x)^n. \tag1$$
Multiplying (1) by $1+x$ gives
$$(1+x)S=(1+x)+(1+x)^2+(1+x)^3+\dots+(1+x)^{n+1}.\tag2$$
Now using (1) to subtract (2) will give the answer.


问:
The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place
Say I was trying to find the derivative of $x^2$ using differentiation from first principles. The usual argument would go something like this:

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

Throughout this argument, I assumed that
$$
\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
$$
was actually a meaningful object—that the limit actually existed. I don't really understand what justifies this assumption. To me, sometimes the assumption that an object is well-defined can lead you to draw incorrect conclusions. For example, assuming that $\log(0)$ makes any sense, we can conclude that
$$
\log(0)=\log(0)+\log(0) \implies \log(0)=0 \, .
$$
So the assumption that $\log(0)$ represented anything meaningful led us to incorrectly conclude that it was equal to $0$. Often, to prove that a limit exists, we manipulate it until we can write it in a familiar form. This can be seen in the proofs of the chain rule and product rule. But it often seems that that manipulation can only be justified if we know the limit exists in the first place! So what is really going on here?

For another example, the chain rule is often stated as:

Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $(f \circ g)$ is differentiable at $x$, and
$$
(f \circ g)'(x) = f'(g(x))g'(x)
$$

If the proof that $(f \circ g)$ is differentiable at $x$ simply amounts to computing the derivative using the limit definition, then again I feel unsatisfied. Doesn't this computation again make the assumption that $(f \circ g)'(x)$ makes sense in the first place?

答:
Proposition: Let $c \in \mathbb{R}$. Suppose $f$ and $g$ are defined and equal to each other on some punctured open ball $(c - \delta) \cup (c + \delta)$ of $c$, where $\delta > 0$. Then $\lim_{x \to c} f(x)$ exists if and only if $\lim_{x \to c} g(x)$ exists. And if either limit exists, so does the other, and they are both equal.

Sketch of proof: Observe that the definition of limit at a point $c$ concerns itself only with points close to $c$ but not equal to $c$. So whatever the value of $f$ or $g$ at $c$, or for that matter whether or not they are defined there, does not matter. Since $f$ and $g$ are equal at points close to $c$ but not equal to $c$, our limit statement about either function at $c$ must therefore also hold for the other. $\square$

This justifies the various limit calculations that we often do, such as the one you showed. In fact, let us go through your example step by step.

If $f(x)=x^2$, then \begin{align} f'(x) &= \lim_{h \to
0}\frac{(x+h)^2-x^2}{h} \\ &= \lim_{h \to 0}\frac{2hx+h^2}{h} \\
&= \lim_{h \to 0} 2x+h \end{align} As $h$ approaches $0$, $2x+h$
approaches $2x$, so $f'(x)=2x$.

What do these sequences of calculations really mean or imply? Well, in the final step/equality, we computed
$\displaystyle \lim_{h \to 0} 2x + h$, which we agree exists and equals to $2x$. Since the function $\displaystyle \frac{2hx + h^2}{h}$ equals $2x + h$ in some punctured neighborhood of $0$, we can now use the proposition to conclude that $\displaystyle \lim_{h \to 0} \frac{2hx + h^2}{h}$ equals $\displaystyle \lim_{h \to 0} 2x + h$, which equals $2x$. So going from line (3) to line (2) is justified. Next, the function $\displaystyle \frac{(x+h)^2 - x^2}{h}$ equals $\displaystyle \frac{2hx + h^2}{h}$ in some punctured neighborhood of $0$, so again we can use the proposition to justify going from line (2) to line (1).

So we have sort of reasoned backwards, but practically speaking this is not necessary in ordinary limit calculations. Our reasoning also "works" even when the limit does not exist. If at the end we arrive at a limit that exists, then necessarily we can work backwards and guarantee that the initial first limit exists; and if at the end we arrive at a limit that does not exist, then necessarily the initial first limit cannot exist, otherwise we could go down the series of equivalences guaranteed by the proposition to guarantee that the final limit exists.

So in all cases things "work out fine". The important thing to note is simply that we have certain logical equivalences at each step: the limit exists at some step if and only if it exists at any earlier or later step.


问:
How to solve this equation $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$?
$(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$

$Domain: x\in (-\infty;-2]\cup(1;+\infty)$

$x = 2\longrightarrow$ Done

How to prove $x = \dfrac{-1-3\sqrt{17}}{2}$ is the last solution? With raise both side by power of two?

Any better way such as factoring, grouping,...?

Please help!!

答:
Hint:

Note that $$(x-1)\sqrt{\frac{x+2}{x-1}} = \sqrt{(x-1)(x+2)}$$ when $x>1$ and
$$\begin{align}(x-1)\sqrt{\frac{x+2}{x-1}} &= -(1-x)\sqrt{\frac{x+2}{x-1}} \\[1mm]&= -\sqrt{(1-x)^2\frac{x+2}{x-1}}\\[1mm]&=-\sqrt{(x-1)(x+2)}\end{align}$$ when $x \le -2$.

Substitute $$t = \sqrt{(x-1)(x+2)}$$ Then, how will the equation look like?


答:
Assuming $x\neq 1$ you can rewrite your equation as

$$(x-1)(x+2) + 4\sqrt{(x-1)(x+2)} = 12$$

Then call $z = \sqrt{(x+1)(x-2)}$ and solve the quadratic associated equation, and then you pullback to $x$.

Title: How to solve this equation $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$?

Body:
$(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$

$Domain: x\in (-\infty;-2]\cup(1;+\infty)$

$x = 2\longrightarrow$ Done

How to prove $x = \dfrac{-1-3\sqrt{17}}{2}$ is the last solution? With raise both side by power of two?

Any better way such as factoring, grouping,...?

Please help!!


问:
On the definition of the direct sum of not necessarily abelian groups
Consider the following definition for the direct sum of groups:

Definition. Let $(A_i)_{i\in I}$ be a family of groups. Then we define the direct sum of the family $(A_i)_{i\in I}$ as the subgroup of the direct product $\displaystyle \prod_{i\in I}A_i$ that contains all the elements $(a_i)$ such that there are only finitely many $i$ such that $a_i$ is not the identity element of $A_i$.

Does this definition make sense for families of groups that are not necessarily abelian (I know that this definition commonly applies to families of abelian groups)? My understanding is the following: this is going to be a subgroup of the direct product even if the groups are not abelian, but it will not retain the properties it has if all the groups are abelian. If we want to retain the properties of the direct sum of abelian groups, then we have to use the free product, which is a different construction from the one in my definition. So, am I right?

答:
Yes, you are right. There is a notion of direct sums of (infinitely many) groups, see e.g. Wikipedia, though in my eyes it could be an ambiguous term, as one usually uses speaks about coproducts when people talk about direct sums.

You are also right that this definition does not satisfy the universal property of coproducts. As you noted, the coproduct in the category of groups $\mathbf{Grp}$ is given by the free product. Since coproducts (of the same components) are canonically isomorphic, surely your notion is not the coproduct in $\mathbf{Grp}$.

That is, the forgetful functor $\mathbf{Ab} \to \mathbf{Grp}$ is not cocontinuous (i.e. does not preserve small colimits) which in particular shows that it does not admit a right adjoint.


问:
If $a_n=100a_{n−1}+134$ , find least value of $n$ for which $a_n$ is divisible by $99$

Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $\mathrm{n}$ for which $\mathrm{a}_{\mathrm{n}}$ is divisible by $99 ?$

So, In this post,the present answer there get to the result that $a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$

But after that what i did is to multiply both sides by $9$ in the congruence

$35n - 11 \equiv 0 \pmod{99}$

$315n \equiv 0 \pmod{99}$

$18n \equiv 0 \pmod{99}$

dividing by $18$ ,we get
$n \equiv 0 \pmod{11}$ but I did not get the other part of the congruence which was arrived in that answer by this process ?

thankyou

答:
From $a_n= 24+(n - 1)35 \equiv 0 \pmod{99} $ you can see that $n-1$ is a multiple of $3$. Let it be $3m$, then dividing throughout by 3, $$0\equiv8 +35k \equiv 8+2k \pmod{33} $$ and therefore $$k \equiv -4 \pmod{33}$$

Then take $k=29$ and $n=88$.

N.B.
Another way to look at your 'multiplying by 9' idea is that you can, of course, do this but you must multiply the base by the factor as well! In general this is unlikely to be a useful procedure. In your case you would change an equation mod $99$ into one mod $891$.


问:
limits of sequences exponential and factorial: $a_n=e^{5\cos((\pi/6)^n)}$ and $a_n=\frac{n!}{n^n}$
Compute the limit $\lim\limits_{n\to\infty}a_n$ for the following sequences:

(a) $a_n=e^{5\cos((\pi/6)^n)}$

(b) $a_n=\frac{n!}{n^n}$

For part (a) do I just take the limit of the exponent part and then the answer would be $e$ raised to whatever the limit is?

And would the limit be $1$ or $-1$? because $\cos$ goes between those two.

For part $b$ it is in the form of infinity over infinity but how do you take the derivative of $n!$? Will it ever break out of infinity over infinity?

答:
(a)
$$
\lim_{n \to \infty} e^{5 \cos((\frac{\pi}6))^n} = e^{\lim_{n \to \infty} 5 \cos((\frac{\pi}6))^n} = e^{5 \cos(\lim_{n \to \infty}(\frac{\pi}6)^n)} = e^{5 \cdot 1} = e^5
$$

(b)
$$
 \frac{n!}{n^n} = \frac{1\cdot 2 \cdots n}{n\cdot n \cdot n} \leq \frac{1}{n}
$$
From here
$$
\lim_{n \to \infty}  \frac{n!}{n^n} = 0
$$
since
$$
\lim_{n \to \infty} \frac{1}{n} = 0
$$


问:
Show that the chromatic number of a Series Parallel graph is at most 3
Although it is "well known" that a (simple) series parallel graph has chromatic number at most 3, I cannot seem to find a proof of this statement anywhere.

Does anyone know how to prove the statement?

It may be related to the fact that series parallel graphs are K4 minor free.

答:
Every series-parallel graph (SPG) may be coloured with at most $3$ colours such that their source and sink are coloured differently (a P-colouring). This is clearly true for the base case, a single edge.

Suppose we have two P-coloured SPGs; the first's source and sink are coloured with colours $a$ and $b$, the second's $c$ and $d$ respectively. By permuting colours

we can make $b=c$ and $a\ne d$, whereby we can form a P-coloured series composition of the two graphs

we can make $a=c$ and $b=d$, whereby we can form a P-coloured parallel composition of the two graphs

Since all SPGs are formed by series and parallel compositions from the single edge, this shows by induction that all SPGs have a P-colouring, i.e. a $3$-colouring.


问:
Prove or disprove: function is uniformly continuous
Let $f$ be a function on $\mathbb R$. If $$|f(x)-f(y)|<4|x-y|$$ for all real numbers $x$ and $y$, then $f$ is uniformly continuous on $\mathbb R$.

My answer is as follows:

The definition of Lipschitz function is:
If $A⊆R$ and $f:A→R$, then $f$ is said to be Lipschitz on $A$ if there exists a constant $K>0$ such that
$$|f(x)−f(y)|≤K|x−y||f(x)−f(y)|≤K|x−y|,$$

for all $x,y∈A.$

So $f$ is a Lipchitz function and every Lipchitz function is uniformly continuous function. Is it true?

答:
Given $\epsilon > 0$, we want to find $\delta > 0$ such that
$$
|f(x) - f(y)| < \epsilon \quad  \text{whenever} \quad |x-y| < \delta
$$
and $\delta$ is independent of $x$ and $y$. By Lipschitzness, if $|f(x) - f(y)| < L|x-y|$, we can define $\delta = \epsilon/L$, so the uniform continuity criterion will hold, as $L$ is independent of $x$ and $y$.


问:
What is an extension of an inner product space?
Chapter 5 of Alan Macdonald's "Linear and Geometric Algebra" begins by saying,

The geometric algebra $\mathbb{G}^n $ is an extension of the inner product space $\mathbb{R}^n.$

What is an extension of an inner product space?

答:
According to dictionary.com, "extension" can mean "an instance of enlarging the scope of something". And according to Wikipedia, an "inner product space" is "a vector space with an additional structure called an inner product" and while this is alluded to early on, later it clearly states "the real n-space ${\displaystyle \mathbb {R} ^{n}}$ with the dot product is an inner product space".

Putting this together, an "extension of the inner product space $\mathbb R^n$" means taking the vectors and inner product (standardly the dot product) of $\mathbb R^n$, and somehow enlarging the scope. That's a little vague, but the default interpretation would be that, at minimum: $\mathbb G^n$ is a superset of $\mathbb R^n$, and you can still take the inner (dot) product of the $\mathbb R^n$ vectors inside of $\mathbb G^n$. (The previous chapter on $\mathbb G^3$ and the later text in this chapter clarifies what $\mathbb G^n$ is more precisely than this one-sentence summary.)


问:
Non-parallel vectors confusion
I've got a section in my textbook about non-parallel vectors, it says:

For two non-parallel vectors a and b, if $\lambda a + \mu b = \alpha a + \beta b$
then $\lambda  = \alpha $ and $\mu  = \beta $

Okay I get that you can equate coefficients and solve for mu and lambda, but how are the two sides of the equation equal in the first place? How can you just equate two different vectors to each other like that? I'm just confused and i'm not entirely sure what about. I've tried googling but not much turns up.. I'd love it if someone could explain in basic terms what this equation is telling me.. (that non parallel vectors are equal?) as I've only just been introduced to this topic recently.. Thank you.

答:
I believe what the book is showing is that if $a$ and $b$ are not parallel vectors (read: linearly independent), and say $w$ is a third vector such that $w=\alpha a+\beta b$ for some scalars $\alpha$ and $\beta$, then that expression of $w$ is unique.

If you thought that you could be clever and write $w=\lambda a+\mu b$, because of the fact that $a$ and $b$ are not parallel it turns out that $\alpha = \lambda$, and $\beta = \mu$.

Title: Non-parallel vectors confusion


问:
Solving limit without L'hopital $\lim_{x\to0} \left(\frac{\frac{xe^x}{e^x-1}-1}{x}\right)$
I'm trying to solve this limit using transformations & common limits but I've been unable to:
$$\lim_{x\to0}\left( \frac{\frac{xe^x}{e^x-1}-1}{x}\right)$$

This limit simplifies to $1/2$ using L'Hôpital's rule but I had to use it 3 times in order to get the result. (I'm trying to prove that the function is differentiable at $x=0$.)

So if anyone is able to solve it using transformations only without L'hopital I would be grateful.

Thanks for your time and stay safe...

答: You can do it with Taylor polynomials if you take $y:=1-e^{-x}$, so the limit is$$\lim_{y\to0}\left(\frac1y+\frac{1}{\ln(1-y)}\right)=\lim_{y\to0}\frac1y\left(1-\frac{1}{1-y/2+o(y)}\right)=-\frac12.$$


问:
What's meant by radius vector of a curve?

Find the length of the minimum radius vector of the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$.

Let $r$ be the radius vector. So, $x^2+y^2=r^2$. Taking out the value of $x^2$ from the given curve and putting it here, we get value of $r$ in terms of $y$. Taking derivative and equating it to zero, we get the value of $y$ and $x$. Thus, $r$ comes out to be $a+b$.

But what's meant by radius vector of a curve? We are calculating its minima. So, can it have other values as well?

答:
I'm not sure @B.Goddard's comment is right. Here's what I took it to mean:

Just as a point on a curve has a tangent to the curve that is a line agreeing with the curve on $y$ and $y^\prime$ there, you can also choose a circle meeting the curve at a given point so that the circle and curve agree on $y,\,y^\prime,\,y^{\prime\prime}$. The term radius of curvature determines to this circle's radius, so that's the quantity you're asked to minimize here. Presumably, the radius vector is from said circle's centre to the point it touches the curve.

If $\tfrac{a^2}{x^2}+\tfrac{b^2}{y^2}=1$ then$$\begin{align}y^\prime&=-\tfrac{a^2y^3}{b^2x^3}\\\implies y^{\prime\prime}&=-\tfrac{3a^2y^2}{b^2x^3}y^\prime+\tfrac{3a^2y^3}{b^2x^4}\\&=\tfrac{3a^2y^3(a^2y^2+b^2x^2)}{b^4x^6}\\\implies r&=\tfrac{(1+y^{\prime2})^{3/2}}{y^{\prime\prime}}\\&=\tfrac{(a^4y^6+b^4x^6)^{3/2}}{3a^2b^2x^3y^3(a^2y^2+b^2x^2)}.\end{align}$$The parameterization $x=a\sec t,\,y=b\csc t$ gives$$r=\tfrac{(a^2\sec^6t+b^2\csc^6t)^{3/2}}{3a^3b^3\sec t\csc t},$$which shouldn't be hard to minimize.


答:
Write the recursion $\bmod 99$
$$a_n = a_{n-1} + 35\bmod 99;\;a_1=24$$
we get the general solution:
$$a_n=35 n-11$$
and $35n-11\equiv 0\bmod 99$ if $$35n\equiv 11\bmod 99\tag{1}$$

Consider the inverse of $35\bmod 99$, that is $17$.

$17\cdot 35\equiv 1 \bmod 99$

Multiply both sides of $(1)$ by $17$

$$n\equiv 187\bmod 99\to n\equiv 88\bmod 99$$

Title: If $a_n=100a_{n−1}+134$ , find least value of $n$ for which $a_n$ is divisible by $99$

Body:
Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $\mathrm{n}$ for which $\mathrm{a}_{\mathrm{n}}$ is divisible by $99 ?$

So, In this post,the present answer there get to the result that $a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$

But after that what i did is to multiply both sides by $9$ in the congruence

$35n - 11 \equiv 0 \pmod{99}$

$315n \equiv 0 \pmod{99}$

$18n \equiv 0 \pmod{99}$

dividing by $18$ ,we get
$n \equiv 0 \pmod{11}$ but I did not get the other part of the congruence which was arrived in that answer by this process ?

thankyou


问:
convergence of series question

how do you determine if a series converges or diverges? Do you just look at their behavior?

答:
Hint:

For (a), calculate the limit of general term $\lim\limits_{n\to\infty}{\dfrac{n^3-1}{\sqrt{n^6+n^2}}}$ and use the necessary condition for convergence.
For (b) also use the necessary condition.

Title: convergence of series question


问:
Find value $c$ if you have integral result
Find $c$ if $$\int_1^{e^3}{\frac{16+\ln{x^c}}{x}\ln xdx}=1881$$

I am trying to do this with substitution and integration of parts but without any luck. I divided this into two integrals and I get $8\ln{x}$ for the first one but I cant seem to solve the second one. Any help?

答:
You can easily rewrite your integral as

$$\int_1^{e^3} \left(\dfrac{16}{x}\ln(x) + \dfrac{c\ln^2(x)}{x}\right)\ \text{d}x = 1881$$

Where I have used $\ln(x^c) = c\ln(x)$

Now, hints:

$$\int \dfrac{\ln(x)}{x}\ \text{d}x = \dfrac{\ln^2(x)}{2}$$

$$\int \dfrac{\ln^2(x)}{x}\ \text{d}x = \dfrac{\ln^3(x)}{3}$$

Now you can proceed...


问:
In the graph, why 'z' is not considered?
https://www.geogebra.org/m/dv86k7uy

The graph shows error when you visit the above site.
But if you click the "undefined" text twice and click somewhere else, the graph gets plotted.
Question is: why the graph has no '$z$' points plotted?

Explanation of the question:
The graph $1=xyz$ should have $z$ that looks like this: https://www.geogebra.org/3d/n5m2jjx7 (example $xz = 1$) but in the graph $1 = xyz$ the $'z'$ points are not considered?

答:
Implicit 3D plot are not yet implemented in GeoGebra.

You must enter $z=\frac{1}{xy}$.


Title: Find value $c$ if you have integral result

Body:
Find $c$ if $$\int_1^{e^3}{\frac{16+\ln{x^c}}{x}\ln xdx}=1881$$

I am trying to do this with substitution and integration of parts but without any luck. I divided this into two integrals and I get $8\ln{x}$ for the first one but I cant seem to solve the second one. Any help?

Body:
Hint:

$\ln(x^c)=c\ln(x)$ and $\dfrac{d }{dx}(\ln^3(x))=\dfrac{3\ln^2(x)}{x}$


问:
limit of series exponential

Compute the limit of the series $$\sum\limits_{n=4}^\infty 3\frac{2^{n+1}}{5^{n-2}}$$

How do you approach these types of problems?

I'm thinking that this one is in indeterminate form, is that correct?

答:
HINT:

So, the $r\ge 4$ the term  $$t_r=3\cdot \frac{2^{r+1}}{5^{r-2}}=3\cdot \frac 2{5^{-2}}\cdot \left(\frac25\right)^r=150\cdot \left(\frac25\right)^r$$

Clearly, this is an  infinite geometric series with the common ratio is $\frac25$ and the first term $=150\cdot \left(\frac25\right)^4$

Title: limit of series exponential


答:
You need to solve $35n \equiv 11 \pmod{99}$. This can be done via a variation of the (Extended) Euclidean Algorithm.

$$
\begin{array}{ll}
99n \equiv 0 \pmod{99} & (1)\\
35n \equiv 11 \pmod{99} & (2)\\
-6n \equiv -33 \pmod{99} & (3) = (1)-3\times(2)\\
-n \equiv -187 \equiv -88 \pmod{99} & (4) = (2) + 6\times(3)
\end{array}
$$
Therefore $n\equiv 88 \pmod{99}$.


问:
Stuck in finding the area of a convex quad
Problem:

A convex quadrilateral has sides $10$, $6$, $8$ and $2$ (in that order). If the angle between the diagonals is $π/4$, then find the area of the quadrilateral.

I tried applying the Law of Cosines for the angles in between the diagonals, but arrived at a weird relation.

Any hints?

PS: I don't know much Latex, so I had to use radians instead of degrees.

答:
If distances from the intersection point of the diagonals to the vertices are $a, b, c, d$, here is the relations between them -

$a^2+b^2 - 2 ab \cos 135^0 = 64$

Please note the area of $\triangle AOB$ is $A_1 = \frac{1}{2} ab \sin 135^0 = - \frac{1}{2} ab \cos 135^0$

So, $a^2+b^2 + 4A_1 = 64$ ...(i)

Similarly, $c^2+d^2 + 4A_3 = 100$ ...(ii)

You also get,

$b^2+d^2 - 4A_2 = 36$ ...(iii)

$c^2+a^2 - 4A_4 = 4$ ...(iv)

Subtract (iii) from (i), (iv) from (ii) and then add. What do you get?


问:
discontinuous functions
Prove or disprove.
If a function f is discontinuous at a point c then |f| is discontinuous at c.
I tried to graph and think about many discontinuous functions like: 1/x,√x   and sin 1/x in all the cases their absolute is also discontinuous. but how can we prove that without graph?</p>

答: If you define $f\colon\Bbb R\longrightarrow\Bbb R$ by$$f(x)=\begin{cases}1&\text{ if }x\in\Bbb Q\\-1&\text{ otherwise,}\end{cases}$$then $f$ is discontinuous at every point of $\Bbb R$, but $|f|$ is continuous everywhere.

Title: discontinuous functions


答:
Note

$$\lim_{x\to0}\left( \frac{\frac{xe^x}{e^x-1}-1}{x}\right)
 =\lim_{x\to0}\left(\frac1{1-e^{-x}}-\frac1x\right)
= \lim_{x\to0}\left(\frac1x +\frac12 +O(x)-\frac1x\right)=\frac12
$$

Title: Solving limit without L'hopital $\lim_{x\to0} \left(\frac{\frac{xe^x}{e^x-1}-1}{x}\right)$

Body:
I'm trying to solve this limit using transformations & common limits but I've been unable to:
$$\lim_{x\to0}\left( \frac{\frac{xe^x}{e^x-1}-1}{x}\right)$$

This limit simplifies to $1/2$ using L'Hôpital's rule but I had to use it 3 times in order to get the result. (I'm trying to prove that the function is differentiable at $x=0$.)

So if anyone is able to solve it using transformations only without L'hopital I would be grateful.

Thanks for your time and stay safe...


Title: Please check my solution

Body:
$x=5^{5^{5^{5^5}}}$

$x^{\frac{1}{5}}=5^{5^{5^5}}$

$x=5^{x^{\frac{1}{5}}}$

$x^5 = 5^x$

$x=5$

which is obviously false

Body:
From the first step take the $\frac{1}{5}$th power on both sides:
$$x=5^{5^{5^{5^{5}}}}\longrightarrow x^\frac{1}{5}=\left(5^{5^{5^{5^{5}}}}\right)^\frac{1}{5}=5^{(5^{5^{5^{5}}})\times \frac{1}{5}}\ne 5^{5^{5^{5}}}$$


答: Hint:-$\frac{16\ln(x)}{x}dx+\frac{c(\ln(x))^2}{x}dx$ put $\ln(x)=u \implies \frac{dx}{x}=du$ so we have integral as $8\ln^2(x)+c\frac{(ln(x))^3}{3}|_1^{e^3}$ . Can you take it from here?


问:
Showing that $p$ is a Fermat prime if and only if every quadratic non-residue mod $p$ is also a primitive root mod $p$
I want to show that $p$ is a Fermat prime $\iff$ every quadratic non-residue of $p$ is also a primitive root mod $p$

These are some facts that I know:

$F_n = 2^{2^n} + 1$

Every prime divisor $p$ of $F_n$ is of the form $2^{k + 1}k + 1$

$\mathrm{ord}_{F_n}(2) = \mathrm{ord}_{p}(2) = 2^{n + 1}$

This is what I wrote for the ($\Rightarrow$) direction and I want to make sure it is correct:

$p = F_n = 2^{2^n} + 1$ is prime, $\phi(p) = p - 1 = 2^{2^n}$

let $a$ be a quadratic non-reside mod $p$ so $x^2 \equiv_p a$ has no solutions. Suppose $ord_p(a) = h$. Since $h \ | \phi(p) = 2^{2^n}$ then $h = 2^k$ for some $k \leq 2^{n}$
Then $(x^2)^h \equiv_p x^{2h} \equiv_p a^h \equiv_p 1$ which is the same as $x^{2^{k + 1}} \equiv_p 1$ has no solutions therefore $2^{k + 1}\nmid 2^{2^n}$ We also know $2^k \mid 2^{2^n}$ so from these two we can conclude that $2^k = 2^{2^n}$ so the order of $a$ is $\phi(p)$ hence it is a primitive root mod $p$.

I don't know what to do for the other direction. Hints would be appreciated.

答:
We will use the fact that the set of quadratic non-residues mod $p$ is the same as the set of primitive roots mod $p$. Which means $\phi(\phi(p)) = \phi(p - 1) = \frac{p - 1}{2}$  Hence $2 \mid p - 1$ so we can write $p = 2^km$ where $m$ is odd.  Since $2^k$ and $m$ are relatively prime and the Euler Phi function is multiplicative we have $\phi(p - 1) = \phi(2^k)\phi(m) = 2^{k - 1}\phi(m) = \frac{p - 1}{2} = 2^{k - 1}m$ thus $m = \phi(m)$ which means $m = 1$.

So now we have $p - 1 = 2^k \rightarrow p = 2^k + 1$. Now suppose $k = 2^st$ where $t$ is odd. Now we can write $p = (2^{2^s})^t + 1$. If $t > 1$ then $p$ could be factorised since $t$ is odd therefore $t = 1$ so now we have $p = 2^{2^s} + 1 = F_s$.


答:
On the second line, to remove one $5$, take $\log_5$:

$$\log_5 x = 5^{5^{5^5}}$$

The general misconception here is that

$$a^{b^c} = a^{(b^c)} \ne (a^b)^c$$

So both the second and fourth lines were done wrongly.

Title: Please check my solution


问:
A general term for complex exponentiation, i.e. $(a+ib)^{c+id}$
So I've tried to make out a general term for complex exponentiation, that is
$$(a+ib)^{(c+id)}$$I could not find any solution to this calculation online, so if anyone could verify my solution or point out any mistake, it would be very helpful.

Steps are as follows:

Let
$$y = (a+ib)^{c+id}$$

Now, using

$$a^{b} = e^{b\ln{a}}$$

$y = e^{(c+id)(\log{(a+ib)})}$      ......a

Also, using
$$\log{(x+iy)} = \frac{1}{2}\log{(x^{2}+y^{2})}+ i\tan^{-1}({\frac{y}{x}})$$

$\implies(c+id)(\log{(a+ib)}) = (\frac{c}{2}\log{(a^{2}+b^{2})}-d\tan^{-1}{(\frac{b}{a})})+i(c\tan^{-1}{(\frac{b}{a})}+\frac{d}{2}\log{(a^{2}+b^{2})})$

$\implies y = e^{(\frac{c}{2}\log{(a^{2}+b^{2})}-d\tan^{-1}{(\frac{b}{a})})+i(c\tan^{-1}{(\frac{b}{a})}+\frac{d}{2}\log{(a^{2}+b^{2})})}$

Using
$$e^{ix} = \cos{(x)}+i\sin{(x)}$$

$\implies y = e^{(\frac{c}{2}\log{(a^{2}+b^{2})}-d\tan^{-1}{(\frac{b}{a})})}(\cos{(c\tan^{-1}{(\frac{b}{a})}+\frac{d}{2}\log{(a^{2}+b^{2})}})+i\sin{(c\tan^{-1}{(\frac{b}{a})}+\frac{d}{2}\log{(a^{2}+b^{2})}))}$

Any confirmation or pointed out mistake would be very helpful. Thanks in advance...

答:
If you want to be strict and avoid confusion because of the arctangent, write it is as
$$\left(a^2+b^2\right)^{c/2} e^{-d \theta}\Bigg[\cos \left(\frac{d}{2}  \log \left(a^2+b^2\right)+c \theta\right)+i \sin \left(\frac{d}{2}  \log \left(a^2+b^2\right)+c \theta\right) \Bigg]$$ where $\theta=\arg (a+i b)$.


问:
Rudin's RCA Q3.4
I'm trying to solve the following question from Rudin's Real & Complex Analysis. (Chapter 3, question 4) :

Suppose $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, and
  $$\varphi(p) ~=~ \int_X |f|^p \; d\mu ~=~ \|f\|_p^p,~~~~~~~~~~(0 < p < \infty).$$
  Let $E := \big\{ p :~ \varphi(p) < \infty\big\}$. Assume $\|f\|_\infty > 0$.

(a) If $r < p < s$, $r \in E$, and $s \in E$, prove that $p \in E$.

(b) Prove that $\log(\varphi)$ is convex in the interior of $E$ and that $\varphi$ is continuous on $E$.

(c) By (a), $E$ is connected. Is $E$ necessarily open ? Closed ? Can $E$ consist of a single point ? Can $E$ be any connected subset of
  $(0,\infty)$ ?

(d) If $r < p < s$, prove that $\|f\|_p \leq \max\big( \|f\|_r, \|f\|_s\big)$. Show that this implies the inclusion

  $$ \mathcal{L}_r(\mu) \cap \mathcal{L}_s(\mu) ~\subseteq~\mathcal{L}_p(\mu).$$

(e) Assume that $\|f\|_r < \infty$ for some $r < \infty$ and prove that $$ \|f\|_p \xrightarrow[p \rightarrow \infty]{}\|f\|_\infty.$$

I got a solution to (a), (d) and (e). While typing my question, MSE suggested me to look at $a\mapsto \log\left(\lVert f\lVert_{1/a}\right)$ is a convex map
which seems to be related to (b). However I'm clueless about (c). Where should I start from ?

Edit

I believe the idea from $a\mapsto \log\left(\lVert f\lVert_{1/a}\right)$ is a convex map can be applied to get a proof that
$p \mapsto \log\|f\|_{\frac{1}{p}}$ is continuous on the interior of $E$. But this is not quite what we are asked to demonstrate... I'm puzzled.

Second Edit

Based on zhw's answer, it appears that $E$ is not necessarily open, nor necessarily closed and that it can be a singleton. The question whether or not $E$ can be any connected subset of $(0,\infty)$ remains. But I think I can come up with a proof that it can.

答:
I am going to add something for $b)$ because I have been working on this problem too and want anyone else that searches for this problem to have easy access to. The first fact that we need is for any function $f\in L^\alpha$ and for any $\beta > 0$ that
$$||f||_\alpha^{\beta}= \Big(\int_X |f|^{\alpha}d\mu\Big)^{\beta/\alpha}=\Big(\int_X (|f|^{\beta})^{\alpha/\beta}d\mu\Big)^{\beta/\alpha} = || |f|^\beta||_{\alpha/\beta}$$
Thus $|f|^\beta\in L^{\alpha/\beta}$ if $f\in L^\alpha$. With this fact in mind, let $p,q\in int(E)$ and $t\in (0,1)$. Thus $f\in L^p$ and $f\in L^q$. Using the previous fact, we have that
\begin{align*}
||f||_p^{(1-t)p} &= || (|f|^{(1-t)p})||_{p/[(1-t)p]}= || (|f|^{(1-t)p})||_{1/(1-t)}\\
||f||_q^{tq} &= || (|f|^{tq})||_{q/tq}= || (|f|^{tq})||_{1/t}\\
\end{align*}
So we have that $|f|^{(1-t)p}\in L^{1/(1-t)}$ and $|f|^{tq}\in L^{1/t}$. Then as $$\frac{1}{1/(1-t)}+\frac{1}{1/t}=1-t+t = 1 $$
We can use Holder's inequality to see that $|f|^{(1-t)p+tq}\in L^1$ and that $$||(|f|^{(1-t)p+tq})||_1 \leq || (|f|^{(1-t)p})||_{1/(1-t)} || (|f|^{tq})||_{1/t}=||f||_p^{(1-t)p}||f||_q^{tq} = \varphi(p)^{1-t}\varphi(q)^t$$
At the same time we have that
$$\varphi((1-t)p+tq)=||f||_{(1-t)p+qt}^{(1-t)p+qt} = || (|f|)^{(1-t)p+qt}||_1$$
So if we take the $\log$ on both sides and use properties of the $\log$ then
$$
\begin{align*}\log(\varphi((1-t)p+tq)) &\leq \log(\varphi(p)^{1-t}\varphi(q)^t)\\
&=\log(\varphi(p)^{1-t}) + \log(\varphi(q)^t) \\
&= [1-t]\log(\varphi(p)) + t \log(\varphi(q))
\end{align*}$$
Since $p,q \in int(E)$ and $t\in (0,1)$ were arbitrary, we may conclude that $\log \varphi$ is a convex function on the interior of $E$. Now, $\log\varphi$ being convex implies that $\varphi$ is convex and hence is also continuous on $int(E)$.

Title: Rudin's RCA Q3.4


问:
Does there exist a universal formula of first-order logic that is satisfiable only by structures with infinite domains?
Following up this question - Here:

Does there exist a universal sentence $A:=\forall x_1 \dots \forall x_n \varphi(x_1 ,\dots ,x_n )$ (where $\varphi$ does not contain quantifiers) that is satisfiable but not in finite models (i.e. only in infinite models)?

答:
Yes. For example, let $\varphi$ be the sentence $$\forall x\forall y \forall z\, ((f(x) = f(y)\rightarrow x = y)\land \lnot (f(z) = c))$$
where $f$ is a unary function symbol and $c$ is a constant symbol. Any model of $\varphi$ is a set $X$ equipped with a function $f\colon X\to X$ which is injective but not surjective. Thus $X$ is infinite.

A set $\mathcal{S}$ of sentences has the finite model property if every satisfiable sentence $\varphi\in \mathcal{S}$ has a finite model. The above example shows that the set of universal $L$-sentences in an arbitrary language $L$ does not have the finite model property. However, it is clear that in a relational language, the set of universal sentences does have the finite model property: If a universal sentence $\varphi$ has a model $M\models \varphi$, any finite subset $A\subseteq M$ is the domain of a substructure of $M$ (since we are in a relational language), and hence $A\models \varphi$.

A classic paper by C. J. Ash classifies all first-order languages $L$ with the property that the set of universal $L$-sentences has the finite model property. In first-order logic with equality, these are exactly the languages which fall into the following cases:

No function symbols (only relation symbols and constant symbols).

No function or relation symbols of arity $\geq 2$, no constant symbols, and at most one unary function symbol.

In first-order logic without equality, the classification is a bit simpler:

No function symbols (only relation symbols and constant symbols).

No function or relation symbols of arity $\geq 2$.

Ash also gives many explicit examples of satisfiable universal sentences with only infinite models to show that the results are the best possible (e.g., in a language with a binary function symbol, a language with two unary function symbols, etc.).


问:
To prove $AE$ bisects $\angle BAC$
In the figure below, given that $DE\:$ bisects $\angle BDF$ and $FE\:$ bisects $\angle DFC$. Prove that $AE\:$ bisects $\angle BAC$.

My try:
I dropped perpendiculars from $E$ to $DB$ and $FC$. But could not figure out.

答:
$E$ is an excentre of $\triangle ADF$ -- it is the centre of a circle tangent to all three sides, as
$$\operatorname{dist}(E,\overline{AC})=\operatorname{dist}(E,\overline{DF})=\operatorname{dist}(E,\overline{AB}).$$

So automatically we get that $AE$ bisects $\angle DAF$.

Title: To prove $AE$ bisects $\angle BAC$


问:
$u_t+(u(1-u))_x=a(1-2u)$, method of characteristics for traffic flow equation with riemann initial data
We consider the non conserving equation
$$u_t+(f(u))_x=af'(u)$$
where $a$ is a constant and $f(u)=u(1-u)$.

I am trying to solve this equation by method of characteristics with the initial condition
$$u(x,0)=\begin{cases}
u_l & x\leq0 \\
u_r & x>0 \\
\end{cases}
$$
By method of characteristics, I have $\displaystyle \frac{dt}{1}=\frac{dx}{1-2u}=\frac{du}{a(1-2u)}$, this means that the characteristics equation is $$\displaystyle \frac{dx}{dt}=1-2u$$ along with $\displaystyle \frac{du}{dx}=a, \displaystyle \frac{du}{dt}=a (1-2u).$

Solving these equations, I reached upto $u(x,t)=ax+ g(t)$ where $g$ is some function of $t$ alone. I don't know how to proceed further.

I was able to solve this when we had the equation $$u_t+(f(u))_x=0$$ as there $u$ was constant along the line of characteristics.
Thanks in advance for any help.

答:
Note that the initial data $u(x,0)$ consists of a jump discontinuity from $u_l$ to $u_r$, thus this initial-value problem is a Riemann problem. The popular Lighthill-Whitham-Richards (LWR) traffic flow model is recovered when $a=0$, and the corresponding Riemann solution is described in this post. Let us tackle the case of arbitrary $a$, e.g. by following a similar approach to this post. Setting $v = 1 - 2u$ provides the PDE
$$
v_t + vv_x = -2av
$$
for which the method of characteristics yields $v = c_1e^{-2at}$, $\frac{v-c_1}{2a} = -x+c_2$ and
$$
v = f\!\left(x - v\,\frac{e^{2at}-1}{2a}\right) e^{-2at} \, ,
$$
which is equivalent to the solution found in the answer by @Dmoreno. However, for discontinuous initial data, the method of characteristics is not sufficient (it is only valid where $u$ is smooth). Thus, we use appropriate methods for solving this problem in the weak sense, see related post. Here, we find the shock-wave solution
$$
v(x,t) = \left\lbrace
\begin{aligned}
&v_le^{-2at} &&\text{if}\quad x< x_s(t) \\
&v_re^{-2at} &&\text{if}\quad x> x_s(t)
\end{aligned}\right. ,\qquad
x_s(t) = \frac{v_l+v_r}{2}\frac{1-e^{-2at}}{2a} .
$$
if $v_l > v_r$, and the rarefaction wave solution $$
v(x,t) = \left\lbrace
\begin{aligned}
&v_le^{-2at} &&\text{if}\quad x< v_l (e^{-2at} - 1) \\
& \frac{x e^{-2at}}{e^{-2at} - 1} && \text{if}\quad v_l (e^{-2at} - 1)\leq x\leq v_r (e^{-2at} - 1) \\
&v_re^{-2at} &&\text{if}\quad x> v_r (e^{-2at} - 1)
\end{aligned}\right.
$$
if $v_l < v_r$. One could check that the same solution $u = \frac{1-v}2$ is obtained by tackling the initial PDE problem directly (without changing variables).


问:
limits of sequences exponential and factorial: $a_n=e^{5\cos((\pi/6)^n)}$ and $a_n=\frac{n!}{n^n}$
Compute the limit $\lim\limits_{n\to\infty}a_n$ for the following sequences:

(a) $a_n=e^{5\cos((\pi/6)^n)}$

(b) $a_n=\frac{n!}{n^n}$

For part (a) do I just take the limit of the exponent part and then the answer would be $e$ raised to whatever the limit is?

And would the limit be $1$ or $-1$? because $\cos$ goes between those two.

For part $b$ it is in the form of infinity over infinity but how do you take the derivative of $n!$? Will it ever break out of infinity over infinity?

答:
Note that
$$\lim_{n\to\infty}\left(\frac{\pi}{6}\right)^n=0.$$
Since $\cos x$ is continuous, it follows that
$$\lim_{n\to\infty}\cos\left(\left(\frac{\pi}{6}\right)^n\right)=\cos(0)=1.$$
Since the function $e^x$ is continuous, it follows that the required limit is $e^5$.

For the second problem, please see the answer by UrošSlovenija.


答:
Drop altitudes from E to AB, DF and AC, with respective feet X, Y and Z.

The right triangles DXE and DYE are congruent; so are FYE and FZE, Thus, EX = EY = EZ and EXA and EZA are congruent, hence the angle bisector AE.

Title: To prove $AE$ bisects $\angle BAC$

Body:
In the figure below, given that $DE\:$ bisects $\angle BDF$ and $FE\:$ bisects $\angle DFC$. Prove that $AE\:$ bisects $\angle BAC$.

My try:
I dropped perpendiculars from $E$ to $DB$ and $FC$. But could not figure out.


答:
Same answer as the others, but with different words:

The circle with center at $E$ that has $AB$ and $DF$ as tangents (which exists because $DE$ is a bisector) is necessarily the same as the circle that has center at $E$ with $AC$ and $DF$ as tangents.

Thus there is a circle centered at $E$ with both $AB$ and $AC$ as tangents, so $AE$ bisects the angle between those lines.

Title: To prove $AE$ bisects $\angle BAC$


问:
Analytical approach to find area of parallelogram
I know that to find the area of parallelogram, we have to either find the cross product of its adjacent sides or half the cross product of its diagonals. However I've encountered a question to find the area of a parallelogram through an analytical method. What exactly is analytical method? Simple algebra ? If then too, I'm stuck cause the question provides me the coordinates of the parallelogram and the simple method to find the area of a parallelogram is the multiplication of its base to its height. Again I'm stuck at finding the height

I then tried to cut it into two triangles but I can't find the area as the coordinates possess three dimensions and all areas of triangle formula I've learnt is of two coordinates .

What then is the solution ?

答:
Using an analytical approach means you derive the formula for the area of the parallelogram, and use that formula to find the exact result.

Let's say you know the location of vertex $\vec{c}$, and two vertices adjacent to it, $\vec{a}$ and $\vec{b}$.  You know
$$\vec{a} = \left[ \begin{matrix} a_x \\ a_y \\ a_z \end{matrix} \right ], \quad
\vec{b} = \left[ \begin{matrix} b_x \\ b_y \\ b_z \end{matrix} \right ], \quad
\vec{c} = \left[ \begin{matrix} c_x \\ c_y \\ c_z \end{matrix} \right ]$$

To find an analytical solution to the area of the parallelogram, you need to find the function
$$A(a_x, a_y, a_z, b_x, b_y, b_z, c_x, c_y, c_z)$$
If you are not given three vertices of the parallelogram, but say edge lengths and angles, you need to find the area of the parallelogram in terms of those variables you are given.

For example, the two edges meeting at vertex $\vec{c}$ can be described using vectors $\vec{u}$ and $\vec{v}$,
$$\vec{u} = \left[ \begin{matrix} u_x \\ u_y \\ u_z \end{matrix} \right] = \left[ \begin{matrix} a_x - c_x \\ a_y - c_y \\ a_z - c_z \end{matrix} \right], \quad \vec{v} = \left[ \begin{matrix} v_x \\ v_y \\ v_z \end{matrix} \right] = \left[ \begin{matrix} b_x - c_x \\ b_y - c_y \\ b_z - c_z \end{matrix} \right]$$
and we know that
$$\begin{aligned}
A &= \left\lVert \vec{u} \times \vec{v} \right\rVert \\
~ &= \sqrt{ (\vec{u} \times \vec{v}) \cdot (\vec{u} \times \vec{v}) } \\
~ &= \sqrt{ (u_y v_z - u_z v_y)^2 + (u_z v_x - u_x v_z)^2 + (u_x v_y - u_y v_x)^2 } \\
\end{aligned}$$
You can simplify this to find the desired function $A$.

Another example:
The part of $\vec{v}$ perpendicular to $\vec{u}$ is $\vec{v}_\perp$,
$$\vec{v}_\perp = \vec{v} - \vec{u} \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}}$$
and is the "height" of the parallelogram, so the area is
$$\begin{aligned}
A &= \left\lVert \vec{u} \right\rVert \left\lVert \vec{v}_\perp \right\rVert \\
~ &= \sqrt{ \vec{u} \cdot \vec{u} } \sqrt{ \vec{v}_\perp \cdot \vec{v}_\perp } \\
~ &= \sqrt{ u_x^2 + u_y^2 + u_z^2 } \sqrt{ \bigr( \vec{v} - \vec{u} \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}} \bigr) \cdot \bigr( \vec{v} - \vec{u} \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}} \bigr) } \\
~ &= \sqrt{ u_x^2 + u_y^2 + u_z^2 } \sqrt{ \vec{v} \cdot \vec{v} - 2 \vec{u} \cdot \vec{v} \frac{\vec{u} \cdot \vec{v}}{\vec{u} \cdot \vec{u}} + \frac{ (\vec{u} \cdot \vec{v})^2}{\vec{u} \cdot \vec{u}} } \\
\end{aligned}$$
This, too, simplifies to the same desired parallelogram area function.


问:
Non-parallel vectors confusion
I've got a section in my textbook about non-parallel vectors, it says:

For two non-parallel vectors a and b, if $\lambda a + \mu b = \alpha a + \beta b$
then $\lambda  = \alpha $ and $\mu  = \beta $

Okay I get that you can equate coefficients and solve for mu and lambda, but how are the two sides of the equation equal in the first place? How can you just equate two different vectors to each other like that? I'm just confused and i'm not entirely sure what about. I've tried googling but not much turns up.. I'd love it if someone could explain in basic terms what this equation is telling me.. (that non parallel vectors are equal?) as I've only just been introduced to this topic recently.. Thank you.

答:
If you are in $\mathbb{R}^n$ of in any vector space then you can always compare two vectors. They are just two elements of a set. Equality perfectly make sense. Perhaps the following description can help you.<br>
 If you rewrite this equation, you will get $$(\lambda -\alpha)a=(\beta -\mu)b$$ which is same as $$a=(\beta -\mu)/(\lambda -\alpha)b.$$ That is a is a scalar multiple of b. Therefore if they are not parallel (if x=cy for two vectors x and y and scalar c  then x and y are parallel) then the denominator should be 0 hence you get the result. </p>

Title: Non-parallel vectors confusion

Body:
If you are in $\mathbb{R}^n$ of in any vector space then you can always compare two vectors. They are just two elements of a set. Equality perfectly make sense. Perhaps the following description can help you.

 If you rewrite this equation, you will get $$(\lambda -\alpha)a=(\beta -\mu)b$$ which is same as $$a=(\beta -\mu)/(\lambda -\alpha)b.$$ That is a is a scalar multiple of b. Therefore if they are not parallel (if x=cy for two vectors x and y and scalar c  then x and y are parallel) then the denominator should be 0 hence you get the result.


问:
If $M/N_1, M/N_2$ are Noetherian then $M/(N_1∩N_2)$ is Noetherian.

Problem:  Let $R$ be a comutative ring and $M$ be  an $R$-module. Let
$N_1$ and $N_2$ be submodules of $M$.   If $M/N_1, M/N_2$ are
Noetherian then $M/(N_1∩N_2)$ is noetherian.

Solution: Let's take $M/N_1, M/N_2$ be Noetherian and $N:=N_1+N_2$.

Let $S_1⊆S_2⊆⋯⊆M/N$ be a chain of submodules.  Let us define

$A_i:=\{m+N_1:m+N∈S_i\}$ for each $i$. Then $A_i$ is a submodule of $M/N_1$.
If $m+N_1∈A_i$ then $m+N ∈ S_i ⊆ S_{i+1} \implies m+N_1 ∈ A_{i+1}$.

Now, $A_1 ⊆ A_2 ⊆...⊆ M/N_1$ is a chain of submodules. Next, $M/N_1$ is Noetherian, so  for each $a$ with $a ≤ i ≤ k $, we have
$A_i = A_k.$ So for each  $a$ with $a≤i≤k$ we have $$S_i =  \{ m+N: m+N_1 ∈ A_i \} = \{ m+N : m+N_1 ∈ A_k \} = S_k.$$

Now $A_j:=\{ m+N_2 : m+N ∈ S_j \}$ for each $j$, and $A_j$ is a submodule of $M/N_2$.
If $m+N_2 ∈ A_j$ then $m+N ∈ S_j ⊆ S_{j+1} \implies m+N_2 ∈ A_{j+1}$.

$A_1 ⊆ A_2 ⊆...⊆ M/N_2$ is a chain of submodules.
$M/N_2$ is Noetherian so for each $a$ with $a ≤ j ≤ h$ we have $A_j = A_h$. Hence,  for each $a$ with $a ≤j ≤ h$ we get
$$S_j = \{ m+N : m+N_2 ∈ A_j \} = \{ m+N : m+N_2 ∈ A_h \} = S_h.$$

Hence $M/N := M/(N_1+N_2)$ is Noetherian. $N_1 ∩ N_2 ⊆ N_1+N_2$, so $M/(N_1∩N_2)$ is Noetherian.

Is everything right in my solution? Please help me.

答:
I suggest another way to approach this problem, because I don't like too much index juggling. Note also that we don't need that $R$ is a commutative ring.

Define the $R$-linear map
$$\psi: M\to M/N_1 \oplus M/N_2: m\mapsto (m + N_1, m + N_2).$$
Then $\ker \psi = N_1 \cap N_2$ and by the first isomorphism theorem
$$M/(N_1 \cap N_2) \cong \operatorname{Im}(\psi). \quad (*)$$

By assumption, $M/N_1$ and $M/N_2$ are Noetherian, so their direct sum $M/N_1 \oplus M/N_2$ is also Noetherian. Hence, $\operatorname{Im}(\psi)$ is Noetherian, as a submodule of a Noetherian module. In view of the isomorphism $(*)$, it follows that $M/(N_1 \cap N_2)$ is Noetherian.


问:
Proves with $K\subset L^{2}(\mathbb{R})$
Define a subset $K \subset L^{2}(\mathbb{R})$ as

$K=\{f:\mathbb{R}\to \mathbb{C}|f \in L^{2}(\mathbb{R})$ and $f(x)\in \mathbb{R}$ for almost every $x\in \mathbb{R} \}$

if $f\in L^{2}(\mathbb{R})$, prove that $f \in K$ if and only if $\langle f,g\rangle \in \mathbb{R}$ for every $g \in K$.

My prove:

First we assume that $f \in K$. Take $g \in K$random. Then \begin{align}
\langle f,g\rangle =\int_{\mathbb{R}}f(x)\overline{g(x)}d\lambda(x)
\end{align}
Because $g\in K$ is $g(x) \in \mathbb{R}$ for almost every $x \in \mathbb{R}$.
This means that $\lambda(\{x \in \mathbb{R}|g(x) \notin \mathbb{R})\}=0$. Because this is $0$ we can say that \begin{align}
\langle f,g\rangle &=\int_{\mathbb{R}}f(x)\overline{g(x)}d\lambda(x)\\
&=\int_{\mathbb{R}}f(x)g(x)d\lambda(x)\\
&=\int_{\mathbb{R}}\overline{f(x)g(x)}d\lambda(x)\\
&=\overline{\int_{\mathbb{R}}f(x)g(x)d\lambda(x)}\\
&=\overline{<f,g>}
\end{align}
This means $\langle f,g\rangle \in \mathbb{R}$

The other way is analogous.

Prove that $K$ is a closed and convexed subset of $L^{2}(\mathbb{R})$

This is a difficult part for me. I already tried to prove it's closed and convex but I got stuck.

My attempt:

Proving $K$ is closed means proving $K^{c}$ is open. So we define $K^{c}=\{f:\mathbb{R}\to \mathbb{C}|f\in L^{2}(\mathbb{R})$ and $f(x) \notin \mathbb{R}$ for almost every $ x\in \mathbb{R}\}$.
Take $h \in K^{c}$. Now we look for an $\epsilon >0$. take $g\in L^{2}(\mathbb{R})$ random. Then $g$ is measurable and $g=0$ almost everywhere but $<g,g> \neq 0$.
\begin{align}
\lVert h-g\rVert &=\langle h-g,h-g\rangle\\
&=\int_{\mathbb{R}}(h(x)-g(x))\overline{(h(x)-g(x))} d\lambda(x)\\
&=...
\end{align}

I got stuck form here so if someone could help me to finish it?

if $f \in L^{2}(\mathbb{R})$, witch element of $K$ lies the closet with $f$? Wht is the distance from $f$ to that element

I think you need to find the orthonormal projection on $K$. But I can't find the point that we're looking for.

答:
For the "closed"-part: For $g\in L^2(\Bbb R)$ define the bounded linear functional $F_g:L^2(\Bbb R)\to \mathbb{C}, F_g(f)=\langle f,g\rangle$. By the first part we have the equality $$K=\bigcap_{g\in K}F_g^{-1}(\Bbb R)$$
Why is this closed?

For the closest element: If you take any complex number $z$ the closest real number to that is $\operatorname{Re}z$, this suggests to take $\tilde{f}=\operatorname{Re}f\in K$ for $f\in L^2(\Bbb R)$. Now you just need to prove that for all $g\in K$ we have $$\left\lVert g-f\right\rVert\geq \left\lVert \tilde{f}-f\right\rVert$$

Edit: (for the last part) If we have any complex number $z$ and any real number $x$ we have $$\left\vert x-z\right\vert\geq\left\vert \operatorname{Re}z-z\right\vert$$
(Maybe draw a picture)

So if we take any $g\in K$ we get $$\left\vert g(x)-f(x)\right\vert\geq \left\vert (\operatorname{Re}f(x))-f(x)\right\vert$$
for almost every $x\in \Bbb R$. Now square and integrate.

Edit 2:
$$\left\lVert g-f\right\rVert^2=\int_{\Bbb R}\left\vert g(x)-f(x)\right\vert^2dx\geq\int_{\Bbb R}\left\vert (\operatorname{Re}f(x))-f(x)\right\vert^2dx=\left\lVert (\operatorname{Re}f)-f\right\rVert^2$$


问:
Proof of "a sequence of uniformly convergent bounded functions converges to a bounded function"
Let $A \ne \emptyset$ a subset of $\mathbb{R}$, let $f_n:A\to\mathbb{R}$ be a sequence of functions and let $f_n \to f$ uniformly in $A$. My book proves the theorem "a sequence of uniformly convergent bounded functions converges to a bounded function" using the uniformly convergence definition given by the Cauchy convergence, meaning $f_n \to f$ uniformly in $A \iff \forall \varepsilon>0 \ \exists N_{\varepsilon} \in \mathbb{N} \ \text{s.t}. \forall x \in A \ \text{if} \ n,m>N_{\varepsilon}\ \text{then} \ |f_n(x)-f_m(x)| < \varepsilon $.

I've tried to prove it in another way, but I'm not sure if this is correct. Here is my attempt: by hypothesis $f_n$ converges to $f$ uniformly in $A$, meaning that for all $\varepsilon>0$ exists $N_{\varepsilon} \in \mathbb{N}$ such that for all $x \in A$, $n>N_{\varepsilon} \implies |f_n(x)-f(x)|<\varepsilon$. This mean that $-\varepsilon+f_n(x) < f(x) < f_n(x)+\varepsilon$. By hypothesis $f_n$ is bounded in $A$, meaning that there exists $M \in \mathbb{R}$ such that $-M \leq f_n(x) \leq M$ for all $n\in\mathbb{N}$. So it is $-\varepsilon-M \leq -\varepsilon+f_n(x) < f(x) < f_n(x)+\varepsilon<M+\varepsilon$ and this implies that $-\varepsilon-M < f(x) < \varepsilon+M$, and since $\varepsilon>0$ is arbitrary it follows that $-M \leq f(x) \leq M$, so $f$ is bounded.

I suspect that this is wrong because I don't see where I used the uniform convergence, indeed I believe that when I estimate using $-M \leq f_n(x) \leq M$ it isn't correct because the boundedness of $f_n$ is referred to "for all $n\in\mathbb{N}$" and I don't have information if it is valid "for all $x\in A$" too, so to do that I should have something like a "uniform boundedness", meaning I should have that there exists $M \in \mathbb{R}$ such that for all $n \in \mathbb{N}$ and for all $x \in A$ it is $-M \leq f_n(x) \leq M$. It seems like that in my proof the constant $M$ depends on $x$ and this is not ok, but I'm not sure if it is the problem. If it is wrong, can someone explain me why? Thanks.

答:
The main issue in what you did is when you say "by hypothesis $f_n$ is bounded in $A$, meaning that there exists $M \in \mathbb{R}$ such that $-M \leq f_n(x) \leq M$ for all $n\in\mathbb{N}$."

The problem is that $M$ depends on $n$ (not on $x$). To get rid of the issue, you can proceed as follows.

As $\{f_n\}$ converges uniformly to $f$ on $A$, it exists $N \in \mathbb N$ such that $\vert f_n(x) - f(x) \vert \le 1$ for all $x \in A$. Therefore

$$\vert f(x) \vert \le 1 + \vert f_N(x) \vert$$ for $x \in A$. As $f_N$ is supposed to be bounded (we speak of only one map here), it exists $M \gt 0$ such that $\vert f_N(x) \vert \lt M$ for all $x \in A$ and

$$\vert f(x) \vert \le 1 + \vert f_N(x) \vert \le 1 + M$$ for all $x \in A$. Proving as desired that $f$ is bounded on $A$.


问:
Revised GRE Math Probability section
The table shows the distribution of a group of $40$ college students by gender and class
$$
\begin{array}{c|lcr}
 & \text{Sophomores} & \text{Juniors} & \text{Seniors} \\
\hline
\text{Males} & 6 & 10 & 2 \\
\text{Females} & 10 & 9 & 3 \\
\end{array}
$$

If one student is randomly selected from this group, find the probability that the student chosen is (i) a female or a sophomore?

答:
This question uses the inclusion exclusion principle:

$\Pr(F \cup S) = \Pr(F) + \Pr(S) - \Pr(F \cap S)\tag{1}$

$\cup$ is the math notation for "union" which is equivalent to "OR"

$\cap$ is the math notation for "intersection" which is equivalent to "AND"

Pr(F) = Probability of Females

Pr(S) = Probability of Sophomores

$\Pr(F \cap S)$ = Probability of Females and Sophomores

There are 22 females, 16 sophomores, 10 females who are sophomores, and 40 students total.  So plug and chug baby!

$\Pr(F \cup S) = \cfrac{22+16-10}{40}=0.7$

You can think of the equation (1) as a venn diagram.  How many elements do you have in all the circles?  Its the one you have in A + B minus the elements in A + B!.

Title: Revised GRE Math Probability section


问:
Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$

Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.

I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.

This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?

答:
EDIT: As pointed out, this answer contains a flawed application of Cauchy-Schwarz. In particular, the L.H.S. of the inequality can be zero, while the R.H.S. is positive, an absurdity. I tried to remedy this flaw but ultimately failed. Hence, I have come up with a new approach that proceeds by a "smoothing principle" argument. Any feedback on the new proof will be much appreciated.

Note: In your post, you claimed the following:

$x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}.$

I do not see how the last equality follows. If it were true, then we would have $a_k=\dfrac{1}{a_1a_2...a_n}=1$, which is not necessarily the case. But anyways, it is possible to solve the problem using AM-GM and Cauchy-Schwarz.

Firstly, the case where $x_1=x_2=...=x_n=1$ is trivial. We only consider the case where $\exists \ i \in \{1,2,..,n\}$ such that $x_i \neq 1$. Note that:

$$\dfrac{1}{n-1+x_1} + \dfrac{1}{n-1+x_2} + ... + \dfrac{1}{n-1+x_n} \leq 1$$
\begin{align}
 & \iff \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_1}\right) + \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_2}\right) + ... + \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_n}\right) \geq 0 \\
& \iff \dfrac{x_1-1}{n(n-1+x_1)} + \dfrac{x_2-1}{n(n-1+x_2)} + ... + \dfrac{x_n-1}{n(n-1+x_n)} \geq 0 \\
& \iff \dfrac{x_1-1}{n-1+x_1} + \dfrac{x_2-1}{n-1+x_2} + ... + \dfrac{x_n-1}{n-1+x_n} \geq 0 \ (\because \dfrac{1}{n} > 0 ) \\
\end{align}
By Cauchy-Schwarz,
$$\left(\dfrac{x_1-1}{n-1+x_1} + \dfrac{x_2-1}{n-1+x_2} + ... + \dfrac{x_n-1}{n-1+x_n} \right) \left[(x_1-1)(n-1+x_1) + ...  + (x_n-1)(n-1+x_n) \right]$$
$$\geq \left[(x_1-1)+(x_2-1) + ... + (x_n-1) \right]^2$$
The R.H.S. of the preceding inequality is clearly non-negative. It thus suffices to prove that:
\begin{align}
& (x_1-1)(n-1+x_1) + (x_2-1)(n-1+x_2) + ... + (x_n-1)(n-1+x_n) > 0 \\
& \iff n(x_1-1)+(x_1-1)^2 + n(x_2-1)+(x_2-1)^2 + ... +  \ n(x_n-1)+(x_n-1)^2 > 0 \\
& \iff (x_1-1)^2 + ... + (x_n-1)^2 + n\left[(x_1-1) + ... + (x_n-1) \right] > 0. \\
\end{align}

Since $\exists \ i \in \{1,2,..,n\}$ such that $x_i \neq 1$, it is obvious that $(x_1-1)^2 + ... + (x_n-1)^2 > 0$. Now, we claim that $n\left[(x_1-1) + ... + (x_n-1) \right] \geq 0 $, which concludes our proof. To see why this is true, note that:

\begin{align}
n\left[(x_1-1) + ... + (x_n-1) \right] & = n(x_1 + ... + x_n -n) \\
& \geq n(n \sqrt[^n]{x_1...x_n} - n) \ (\text{By applying AM-GM inequality})\\
& = n(n-n) \ (\because x_1...x_n=1) \\
&=0.
\end{align}


问:
Prove upper bound from gaussian quadrature
Let $I = \int_{0}^1 ln(1+sin(x))dx$, and P be the polynomial with degree less or equal than 5 using the minimum square method to approximate the function $ln(1+sin(x))$ on [0,1].
Show that $$2\underset{x\in[0,1]}{sup}|ln(1+sin(x))-P(x)|$$ is an upper bound of the maximum absolute error of I using gaussian quadrature with 3 points

答:
I've tried to several things like:
By the mean value theorem of integrals, there exists an $\alpha \in [0,1]$ such that $ln(1+sin(\alpha)) = I $, so if a is the gaussian quadratue with 3 points, $|I-a| = |I-P(x) + P(x) - a| \leq |I-P(x)| + |P(x) - a|$, if I could show that $|P(x) - a| \leq |I-P(x)|$ I'd have the problem solved, but can't progress from here...

Also, I tried to rewrite $I = \displaystyle \sum_0^5 1/(n+1)a_n $, but couldn't progress from here as well (this is the first equation to approximate f(x) using the minimum square method).


问:
Density Function of Uniform Distribution?
I had this problem from my last question:

Given that:
$X \sim \text{Uni}(a,b)$.

I know that the density function is:

$$f_X(x) = \begin{cases} \dfrac{1}{b-a} &x \in [a,b] \\ 0 & \text{otherwise}. \end{cases}$$

Then How can this help me to find the density function of : $-X$?

Note: what is the right way to write the density function of : $-X$? will it be $f_{-X}(-x)$ or $f_{X}(-x)$ or $f_{-X}(x)$?

答:
Just consider the cumulative distribution function:
$$F_X(x)=\int_a^x f_X(x) \quad dx=P(X<x)
$$ Then if $Y=-X$, by definition we have:
$$
F_Y(y)=P(Y\leq y)=P(-X\leq y)=P(y\geq -X)=1-F_X(-x)
$$

So, again by definition: $f_Y(y)=F_Y'(y)=(-F_X(-x)
)'
=
f_X(-x)
$.

There you go. And the right way to write the density function of $-X$

would be $f_{-X}(-x)$, or more commonly:
$f_Y(y)$.


问:
How many ways to arrange 5 different dogs, 1 cat and 1 rat such that the rat is always left to the cat (not necessarily near).

How many ways to arrange 5 different dogs, 1 cat and 1 rat such that the rat is always left to the cat (not necessarily near).

I started out by arranging the 5 different dogs which is $5!$, and from here basically I got stuck, I tried to do it by cases, like if I arranged the 5 dogs, I have 6 places to put the cat in,starting from the right, the rat also can be put in 6 places (can be near the cat from the left), and moving one to the left each case, so I got: $6*6+5*5+4*4+3*3+2*2+1*1=91$ and this multiplied by the number of ways to arrange the dogs gave me the answer: $91*5!=10920$, now this seems clearly wrong and I messed up since $7!=5040$, which is the number of ways to arrange all seven, I am trying to understand my mistakes and how to deal with this question.

Thanks in advance to any help.

答:
Hint:

Suppose the rat is the leftmost animal. Than it is always left to the cat and you have $6!$ ways to permute the other animals.

Suppose the rat is in the 2nd place from the left. Again there are $6!$ permutations of the animals but if the cat is at the leftmost side your condition is not satisfied. So yuo have to subtract to $6!$ the number of ways in which the cat is at the first place. Theese are exactly $5!$ (you have 7 places, 2 fixed)

And so on. The total number is the sum of all the cases you have to analize.


问:
GCSE: The straight lines $mx+3y-7=0$ and $2x-ny+8=0$ are parallel. Find $m \cdot n$.
looking please for any help.

The straight lines $mx+3y-7=0$ and $2x-ny+8=0$ are parallel. Find $m \cdot n$.

My attempt was the following:
I replaced $m \to 2$ so from $mx+3y-7=0$;
$2x+3y-7=0$ because  the angular coefficient is the same if two straight lines are parallel, but I don’t know how to find the $n$?!

All help/solutions appreciated.

Thanks.

答:
HINT: Two lines $ax+by+c=0$ and $ex+fy+g=0$ are parallel if
$$\frac{a}{b}=\frac{e}{f}.$$


问:
For any $S \subseteq \mathbf{R}$, if $\partial S$ denotes the boundary of $S$, prove that $\partial(\partial S) \subseteq \partial S$.
I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott. I learnt the trick of constructing a $\delta$-neighborhood of a random point $y \in (x - \epsilon, x+ \epsilon)$ to always reside in this $\epsilon$-neighborhood of $x$.

I tried to write a proof for the below assertion, and would like to ask if my proof checks out.

For any $S \subseteq \mathbf{R}$, if $\partial S$ denotes the boundary of $S$, prove that $\partial(\partial S) \subseteq \partial S$.

My Attempt.

Let $x$ be an element of $\partial(\partial S)$.

By definition of the boundary of a set $S$, for every $\epsilon > 0$, the open interval $(x - \epsilon, x + \epsilon)$ contains at least one element of $\partial S$ and at least one element of $(\partial S)^C$.

We are interested to prove that, $\forall \epsilon>0$, the open interval $(x - \epsilon, x + \epsilon)$ contains at least one element of $S$ and at least one element of $S^C$.

Pick an arbitrary element $y \in (x - \epsilon, x + \epsilon)$, such that $y \in \partial S$. Then, for every $\delta > 0$, the open interval $(y - \delta, y + \delta)$ contains atleast one element of $S$ and atleast one element of $S^C$. If we choose,
\begin{align*}
\delta = \delta_0 = \min \left\{\frac{y - (x-\epsilon)}{2},\frac{(x + \epsilon) - y}{2}\right\}
\end{align*}

Then, $(y - \delta_0,y + \delta_0) \subseteq (x - \epsilon, x + \epsilon)$.

So, we found a small $\delta_0$-neighborhood, $(y - \delta_0,y + \delta_0)$ inside of $(x - \epsilon, x + \epsilon)$, which contains at least one element of $S$ and at least one element of $S^C$, for all $\epsilon>0$.

Consequently, we have shown that, $\forall \epsilon > 0$, $(x - \epsilon, x + \epsilon)$ contains at least one element of $S$ and at least one element of $S^C$.

So, $x \in \partial S$.

Hence, $\partial(\partial S) \subseteq \partial S$. This closes the proof.

答:
What you did looks good to me.

Below another proof, that you can use depending on what you already proved.

If you know that:

The frontier $\partial S$ of a subset $S$ is closed.

$ \partial A \subseteq B$ whenever $ A \subseteq B$ and $B$ is closed.

Then is it almost immediate as $\partial S \subseteq \partial S$ and $\partial S$ is closed. Therefore $\partial(\partial S) \subseteq \partial S$.


问:
$C^{-1/2}$ in hyperbolic distance equation (conformal representation)
Penrose introduces hyperbolic geometry in the second chapter of his book "The Road to Reality". While discussing the conformal representation he introduces a formula to find the hyperbolic distance between two points A and B, which is

$$\log{\frac{QA\cdot PB}{QB\cdot PA}},$$

where Q and P are the points in which the hyperbolic straight line connecting A and B intersect the bounding circle at right angles. Further, he mentions that you can multiply the above expression by
$C^{-1/2}$ if you want. My question is, what does he means by want? Are there some cases in which it is necessary to include this constant? Or is it just some type of normalization?

I understand that $C$ is a measure of curvature and depends on $R$, the radius of the hyperbolic plane which is an imaginary number. What is the formula that relates this two values? And how can the radius of the hyperbolic plane be an imaginary number?

答:
Side note: the conformal representation of hyperbolic geometry is most commonly reffered to as the Poincaré disk model.

My question is, what does he means by want? Are there some cases in which it is necessary to include this constant? Or is it just some type of normalization?

The constant $C^{-1/2}$ will always be equal to $1$ in the two dimensional representation of the the Poincaré disk model (which is what I was working with in the original question) since the curvature of the hyperbolic disk  is always equal to $-1$ (source). Whenever the curvature of the hyperbolic disk is not equal to $-1$, the constant $C^{-1/2}$ needs to be introduced into the equation. However, Penrose points out that we can always scale things such that $C=1$.

What is the formula that relates this two values? And how can the radius of the hyperbolic plane be an imaginary number?

The formula that relates $C$ and $R$ is obtained from Albert Girard's equation for the area of a spherical triangle. Which is the following for a sphere of radius $R$

$$\Delta = R^2(\alpha + \beta + \gamma - \pi)$$

What Lambert noticed is "that certain formulae of hyperbolic geometry can be obtained by replacing, in certain formulae of spherical geometry, distances by the same distances multiplied by the imaginary number $\sqrt{−1}$, and by keeping angles untouched" (source). So, replacing $R$ with the imaginary radius of $\sqrt{-1}R$, we obtain the following formula

$$\Delta = -R^2(\alpha + \beta + \gamma - \pi) = R^2(\pi - \alpha - \beta - \gamma),$$

which is the same as the formula for the area of a triangle with angles $\alpha, \beta, \gamma$ in the hyperbolic plane of constant curvature $-\frac{1}{R^2}$. Therefore, we obtain the formula

$$C = -\frac{1}{R^2},$$

so that $C\Delta = \pi - \alpha - \beta - \gamma$, which is Lambert's formula for the area of a hyperbolic triangle.

From this derivation of the formula for $C$, we can see that the fact that $R$ is an imaginary number is a necessary condition.

Further reading: Hyperbolic geometry in the work of Johann Heinrich Lambert


问:
How to find $1+\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}$
Let $a_{i}>0,i=1,2,\cdots,n$,

If
$$1+\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}\ge k\;\left(\dfrac{1}{\sqrt{1+a^2_{1}}}+\dfrac{1}{\sqrt{1+a^2_{1}+a^2_{2}}}+\cdots+\dfrac{1}{\sqrt{1+a^2_{1}+\cdots+a^2_{n}}}\right)$$

then find the maximum value for $k$

My idea was to use

$$\left(\sum_{i=1}^{n}\dfrac{1}{\sqrt{1+a^2_{1}+a^2_{2}+\cdots+a^2_{i}}}\right)^2\le\left[\sum_{i=1}^{n}\dfrac{a_{i}}{1+a^2_{1}+a^2_{2}+\cdots+a^2_{i}}\right]\left(\sum_{i=1}^{n}\dfrac{1}{a_{i}}\right)$$
and to use the approaches discussed here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=119163&sid=cd139cff36102a55e4f91a7c8dcd4f21#p119163

Alas! I failed to figure out how to do it.

Body:
Substituting $a_i=\sqrt{2}^{i-1}$ gives

\begin{align}
1+\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}=1+\sum_{i=1}^{n}{(\frac{1}{\sqrt{2}})^{i-1}}=1+\sum_{i=1}^{n}{\frac{1}{a_i}} & \geq k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}}\right)\\
& =k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{2^{j-1}}}}}\right) \\
& =k\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{2^i}}}\right) \\
& =k\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}
\end{align}

Thus
\begin{align}
k \leq \frac{1+\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}{\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}=\sqrt{2}+\frac{1}{\frac{1}{\sqrt{2}}\frac{1-\frac{1}{\sqrt{2}^n}}{1-\frac{1}{\sqrt{2}}}}=\sqrt{2}+\frac{\sqrt{2}-1}{1-\frac{1}{\sqrt{2}^n}}
\end{align}

Since this must hold for all $n$, we have $k \leq 2\sqrt{2}-1$. We now show that $k=2\sqrt{2}-1$ works, so that $2 \sqrt{2}-1$ is the maximum value of $k$. In other words, we will show that

$$1+\sum_{i=1}^{n}{\frac{1}{a_i}} \geq (2\sqrt{2}-1)\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}}\right)$$

By Cauchy Schwarz inequality we have

$$\sqrt{2}^i\sqrt{1+\sum_{j=1}^{i}{a_j^2}}=\sqrt{1+\sum_{j=1}^{i}{a_j^2}}\sqrt{1+\sum_{j=1}^{i}{2^{j-1}}} \geq 1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}$$

By weighted AM-HM inequality we have

$$\frac{1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}}{2^i}=\frac{1+\sum_{j=1}^{i}{2^{j-1}\frac{a_j}{\sqrt{2}^{j-1}}}}{1+\sum_{j=1}^{i}{2^{j-1}}} \geq \frac{1+\sum_{j=1}^{i}{2^{j-1}}}{1+\sum_{j=1}^{i}{2^{j-1}\frac{\sqrt{2}^{j-1}}{a_j}}}=\frac{2^i}{1+\sum_{j=1}^{i}{\frac{(2\sqrt{2})^{j-1}}{a_j}}}$$

Thus

\begin{align}
\sum_{i=1}^{n}{\frac{1}{\sqrt{1+\sum_{j=1}^{i}{a_j^2}}}} \leq \sum_{i=1}^{n}{\frac{\sqrt{2}^i}{1+\sum_{j=1}^{i}{\sqrt{2}^{j-1}a_j}}} & \leq \sum_{i=1}^{n}{\frac{1+\sum_{j=1}^{i}{\frac{(2\sqrt{2})^{j-1}}{a_j}}}{(2\sqrt{2})^i}} \\
&=\sum_{i=1}^{n}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{(2\sqrt{2})^{j-1}}{a_j}\sum_{i=j}^{n}{\frac{1}{(2\sqrt{2})^i}}} \\
&=\sum_{i=1}^{n}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{1}{a_j}\sum_{i=1}^{n-j+1}{\frac{1}{(2\sqrt{2})^i}}} \\
& \leq \sum_{i=1}^{\infty}{\frac{1}{(2\sqrt{2})^i}}+\sum_{j=1}^{n}{\frac{1}{a_j}\sum_{i=1}^{\infty}{\frac{1}{(2\sqrt{2})^i}}} \\
&=\frac{1}{2\sqrt{2}-1}(1+\sum_{i=1}^{n}{\frac{1}{a_i}})
\end{align}


问:
Variance of sample variance?
What is the variance of the sample variance? In other words I am looking for $\mathrm{Var}(S^2)$.

I have started by expanding out $\mathrm{Var}(S^2)$ into $E(S^4) - [E(S^2)]^2$

I know that $[E(S^2)]^2$ is $\sigma$ to the power of 4. And that is as far as I got.

答:
I will use this as an example of this theorem(from Seber, G.A. and Lee, A.J. (2012))

Let $X_1, X_2, ... , X_n$ be independent rvs with means $(\theta_1, \theta_2, ... ,\theta_n)$,common $\mu_2,\mu_3,\mu_4$. If A is any n x n symmetric matrix and $a$ is a column vector of the diagonal elements of A, then

$$var[X'AX]=(\mu_4-3\mu^2_2)a'a+2\mu^2_2tr(A^2)+4\mu_2\theta'A^2\theta+4\mu_3\theta'Aa
$$
denote $1_n$ as n-dim column vector that all elements are 1, notice that for sample variance
$$S^2=\frac{1}{n-1}X'AX, where A=I_n-\frac{1}{n}1_n1_n'
$$
and we have$A^2=A$, $a=(1-\frac{1}{n})1_n$

since $X_i$ in our case are iid, let's say their mean is $\mu$, then $\theta=\mu1_n$

so the third and fourth term is $0$,since
$$
A^2\theta=A\theta=\mu(1_n-\frac{1}{n}1_n(1_n'1_n))=0\\
Aa=(1-\frac{1}{n})(1_n-\frac{1}{n}1_n(1_n'1_n))=0
$$
then$$
var[S^2]=\frac{1}{(n-1)^2}[(\mu_4-3\mu_2^2)(1-\frac{1}{n})^2n+2\mu_2^2(n-1)]=\frac{\mu_4}{n}-\frac{n-3}{n(n-1)}\mu_2^2
$$

Title: Variance of sample variance?


问:
Misunderstanding on Taylor series expansion
Sometimes, I try to find the series expansion of some random functions but I am usually wrong, especially when using already known formulas such as $$\frac{1}{1-X} = 1 + X + X^2 + X^3 + ... \quad (1)$$
Now imagine I want to find the series expansion of $\frac{1}{2+x^2}$, here is what I do :

$$\frac{1}{2+x^2}=\frac{1}{1-(-x^2-1)}=1+(-x^2-1)+(-x^2-1)^3+... \quad (2)$$

But the result is not the same as the results in the internet that finds : $$\frac{1}{2+x^2}= \frac{1}{2}- \frac{1}{4}x^{2}+\frac{1}{8}x^{4}- \frac{1}{16}x^{6}$$.

My guess is that the $X$ in (1) must verify $\lvert X \rvert < 1$ and when I take $X = (-x^2-1)$ in (2) then this condition is not verified. I also think that to find the right answer, computers do $$\frac{1}{2+x^2} = \frac{1}{2}\frac{1}{1+(\frac{x^2}{2})}$$

答:
The expansion $\frac{1}{1-x} = 1 + x + x^2 + x^3 + ... $ only works if $|x|<1$ as you pointed out yourself. Because $2+x^2>1$ the substitution you make is guaranteed to give you a divergent sum.

Now, by factoring out $\frac{1}{2}$ you solve that problem, since now it works whenever $|x|<\sqrt{2}$. So the substitution is fine (that is, it results in a convergent series).


问:
Evaluate $ \int \frac{3x+1}{x^2+4x+4}dx $
$$
\int \frac{3x+1}{x^2+4x+4}dx
$$

How can I solve this integral? I can't find any way that lead to $$\frac{f^\prime(x)}{f(x)}$$

I tried with WolframAlpha but the explanation wasn't very intuitive this time!

答: The denominator is $(x+2)^2$, so the integrand is $\frac{3(x+2)-5}{(x+2)^2}=\frac{3}{x+2}-\frac{5}{(x+2)^2}$, which I leave you to integrate.


问:
Evaluate sum of the series $\sum { 1\over n!(n^4+n^2+1)}$

Evaluate  sum of the series:  $\displaystyle \sum_{n=0}^\infty {1\over n!(n^4+n^2+1)}$

Comparison test $\left(\text {with } \sum {1\over n^2}\right)$ confirms the convergence of the series. But what's the sum of the series.
Using comparison tests I managed to show that ${19\over 14}\le$ sum $\le{e+121\over 91}$.

The series seems to converge to $\frac e2$. But how to show that ?

答:
Let $ n\in\mathbb{N} $, notice that :\begin{aligned} \frac{1}{n^{4}+n^{2}+1}=\frac{1}{\left(n^{2}-n+1\right)\left(n^{2}+n+1\right)}&=\frac{1}{2}\left(\frac{n+1}{n^{2}+n+1}-\frac{n-1}{n^{2}-n+1}\right)\\ &=\frac{1}{2}\left(\frac{n+1}{n^{2}+n+1}-\frac{n^{2}}{n^{2}-n+1}+1\right) \end{aligned}

Thus : \begin{aligned} \small\sum_{n=0}^{+\infty}{\frac{1}{n!\left(n^{4}+n^{2}+1\right)}}&\small=\frac{1}{2}\sum_{n=0}^{+\infty}{\frac{1}{n!}\left(\frac{n+1}{n^{2}+n+1}-\frac{n^{2}}{n^{2}-n+1}+1\right)}\\ &\small=\frac{1}{2}\sum_{n=0}^{+\infty}{\left(\frac{\left(n+1\right)^{2}}{\left(n+1\right)!\left(\left(n+1\right)^{2}-\left(n+1\right)+1\right)}-\frac{n^{2}}{n!\left(n^{2}-n+1\right)}+\frac{1}{n!}\right)}\\&\small=\frac{1}{2}\sum_{n=0}^{+\infty}{\left(a_{n+1}-a_{n}\right)}+\frac{1}{2}\sum_{n=0}^{+\infty}{\frac{1}{n!}}\\ &\small=\frac{1}{2}\left(0-0\right)+\frac{\mathrm{e}}{2}\\&\small=\frac{\mathrm{e}}{2} \end{aligned}

Were $ \left(a_{n}\right)_{n\in\mathbb{N}} $ is the sequence defined by $ \left(\forall n\in\mathbb{N}\right),\ a_{n}=\frac{n^{2}}{n!\left(n^{2}-n+1\right)} $.

In the third line, we were able to split the sum apart, because $ \sum\limits_{n\geq 0}{\left(a_{n+1}-a_{n}\right)} $ and $ \sum\limits_{n\geq 0}{\frac{1}{n!}} $ both converges, which means $ \sum\limits_{n\geq 0}{\left(a_{n+1}-a_{n}+\frac{1}{n!}\right)} $ converges, and $ \sum\limits_{n=0}^{+\infty}{\left(a_{n+1}-a_{n}+\frac{1}{n!}\right)}=\sum\limits_{n=0}^{+\infty}{\left(a_{n+1}-a_{n}\right)}+\sum\limits_{n=0}^{+\infty}{\frac{1}{n!}} $.


问:
Open Ball in Supremum Space of Continuous Functions on Interval
I'm having trouble with an exercise in W.A. Sutherland's "Introduction to Metric and Topological Spaces" 1st Edition (Oxford Science Publications, 1975), Chapter $2$, Exercises $2.6: 25$.

Let $\mathscr C [a, b]$ be the set of all continuous real functions $f: [a, b] \to \mathbb R$, where $[a, b]$ denotes the closed real interval from $a$ to $b$: $[a, b] = \{x \in \mathbb R: a \le x \le b\}$.

Let $d: \mathscr C^2 \to \mathbb R$ be the supremum metric on $\mathscr C [a, b]$ defined as:

$\displaystyle \forall f, g \in \mathscr C: d (f, g) := \sup_{x \mathop \in [a, b]} |f(x) - g (x)|$

where $\sup$ denotes the supremum function.

Let $f, g \in \mathscr C [a, b]$ be such that $\forall x \in [a, b]: f(x) < g(x)$

Consider the set $S$, defined as:

$S = \{h \in \mathscr C [a, b]: \forall x \in [a, b]: f(x) < h(x) < g(x)\}$

Is $S$ an open ball of $(\mathscr C [a, b], d)$?

The answer given is that $S$ is an open ball of $(\mathscr C [a, b], d)$ if and only if $g(x) - f(x)$ is constant.

Proving the necessary condition is straightforward: if $g(x) - f(x)$ is constant, then $S$ is the open ball whose center is $\rho$ defined such that $\rho(x) = f(x) + \dfrac {g(x) - f(x)} 2$. Proving this is an open ball is easy.

Going the other way is a little more difficult.

This is what I've done so far.

Let it be assumed that $S$ is an open ball.

Then there exists $\phi \in \mathscr C [a, b]$ and $\epsilon \in \mathbb R_{>0}$ such that $B_\epsilon(\phi) = S$.

That is:

$$\forall \rho \in S: \sup_{x \mathop \in [a, b]} | \rho (x) - \phi (x) | < \epsilon$$

Aiming for a contradiction, suppose it is not the case $f$ and $g$ are such that $\forall x \in [a, b]: g(x) - f(x) = c$ for some constant $c \in \mathbb R$.

Then
$$\exists \xi, \zeta \in [a, b]: g (\xi) - f (\xi) > g (\zeta) - f (\zeta)$$

Then:
$$\sup_{x \mathop \in [a, b]} |g (x) - f (x)| \ge g (\xi) - f (\xi) > g (\zeta) - f (\zeta)$$

As $f$ and $g$ are continuous:

$$\exists (p, q) \subset [a, b]: \forall x \in (p, q): |g (x) - f (x)| > g (\zeta) - f (\zeta)$$

Let $h \in S$ such that:

$$\exists r, s \in (p, q): |h(r) - h(s)| > g (\zeta) - f (\zeta)$$

That is, in the interval $(p, q)$, the oscillation on $h$ is greater than the minimum distance of $g - f$.

The intention is to prove that $h$ cannot be in the open ball that is $S$, but I've lost the train of my thinking and I can't see my way through to the end.

Can someone guide me out of the woods?

答: Suppose that $g-f$ is not constant. Take $h\in S$ and $r>0$ such that $B_r(h)\subset S$; I will prove that $B_r(h)\varsubsetneq S$. Take $N\in\Bbb N$ such that $\frac1N<r$. For each $n\geqslant N$, let $g_n=h+r-\frac1n$. Then $g_n\in B_r(h)$ and therefore $g_n\in S$, which implies that $(\forall x\in[a,b]):g_n(x)<g(x)$. But then, for each $x\in[a,b]$,\begin{align}h(x)+r&=\lim_{n\to\infty}h(x)+r-\frac1n\\&=\lim_{n\to\infty}g_n(x)\\&\leqslant g(x).\end{align}By a similar argument,$$(\forall x\in[a,b]):h(x)-r\geqslant f(x)$$and therefore$$(\forall x\in[a,b]):g(x)-f(x)\geqslant2r.$$But $g-f$ cannot be the constant function $2r$ and so there is some $x_0\in[a,b]$ such that $g(x_0)-f(x_0)>2r$. So, you must have $g(x_0)>h(x_0)+r$ or $f(x_0)<h(x_0)-r$ (or both). If$$g(x_0)>h(x_0)+r,\tag1$$then the function $g-h$ is continuous, it is always greater than or equal to $r$, and $(1)$ holds. Take $k<g(x_0)-h(x_0)-r$ and such that $g-k>f$. Then $g-k\in S$. But $g-k\notin B_r(h)$, since $g(x_0)-k-h(x_0)>r$.


答:
The bottom factorises into $(x + 2)^2$ and so you can substitute $z = x + 2$.

It then becomes an integral in $(a z + b) / z^2$ which should be trivial.

Title: Evaluate $ \int \frac{3x+1}{x^2+4x+4}dx $

Body:
$$
\int \frac{3x+1}{x^2+4x+4}dx
$$

How can I solve this integral? I can't find any way that lead to $$\frac{f^\prime(x)}{f(x)}$$

I tried with WolframAlpha but the explanation wasn't very intuitive this time!


答:
Choose two of the seven positions for the cat and the rat.  Since the rat must be to the left of the cat, there is only one way to arrange the cat and the rat in these positions.  The five dogs can be arranged in the remaining five positions in $5!$ ways.  Hence, there are
$$\binom{7}{2}5!$$
ways to arrange the seven animals so that the rat is somewhere to the left of the cat.

Body:
How many ways to arrange 5 different dogs, 1 cat and 1 rat such that the rat is always left to the cat (not necessarily near).

I started out by arranging the 5 different dogs which is $5!$, and from here basically I got stuck, I tried to do it by cases, like if I arranged the 5 dogs, I have 6 places to put the cat in,starting from the right, the rat also can be put in 6 places (can be near the cat from the left), and moving one to the left each case, so I got: $6*6+5*5+4*4+3*3+2*2+1*1=91$ and this multiplied by the number of ways to arrange the dogs gave me the answer: $91*5!=10920$, now this seems clearly wrong and I messed up since $7!=5040$, which is the number of ways to arrange all seven, I am trying to understand my mistakes and how to deal with this question.

Thanks in advance to any help.